/
Text
Equations
of Applied Mathematics
SECOND EDITION
ERICH ZAUDERER
Wiley-Interscience Srries in Pure and Applied Mathematics
PURE AND APPLIED MATHEMATICS
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A complete list of the titles in this series appears at the end of this volume
PARTIAL DIFFERENTIAL
EQUATIONS OF APPLIED
MATHEMATICS
Second Edition
ERICH ZAUDERER
Polytechnic University
New York
A Wiley-Interscience Publication
JOHN WILEY & SONS, INC
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Library of Congress Cataloging in Publication Data:
Zauderer, Erich.
Partial differential equations of applied mathematics / Erich
Zauderer.—2nd ed.
p. cm.
(Pure and applied mathematics, ISSN 0079-8185)
"A Wiley-Interscience publication."
Bibliography: p.
Includes index.
I. Differential equations. I, Title.
II. Series: Pure and applied mathematics (John Wiley & Sons)
QA337.Z38 1988
515.3*53—del 9
ISBN 0-471-61298-7
ISBN 0-471-31516-8 (Paperback)
Printed in the United States of America
10 98765432
To my Mother
and the memory of my Father,
PREFACE
The study of partial differential equations of applied mathematics involves
the formulation of problems that lead to partial differential equations, the
classification and characterization of equations and problems of different
types, and the examination of exact and approximate methods for the
solution of these problems. Each of these aspects is considered in this book.
The first chapter is concerned with the formulation of problems that give
rise to equations representative of the three basic types (parabolic, hy-
hyperbolic, and elliptic) to be considered in this book. These equations are all
obtained as limits of difference equations that serve as models for discrete
random walk problems. These problems are of interest in the theory of
Brownian motion, and this relationship is examined. Although some
elementary concepts from probability theory are used in this chapter, the
remaining chapters proceed independently of this one.
Chapter 2 deals with first order partial differential equations and presents
the method of characteristics for the solution of initial value problems for
these equations. Problems that arise or can be interpreted in a wave
propagation context are emphasized. First order equation also play an
important role in the methods presented in Chapters 9 and 10.
In Chapter 3 partial differential equations are classified into different
types and simplified "canonical" forms are obtained for second order linear
equations in two independent variables. The concept of characteristics is
introduced for higher order equations and systems of equations, and its
significance for equations of different types is examined. In addition, the
question of what types of auxiliary conditions are to be placed on solutions
of partial differential equations so that the resulting problems are reasonably
formulated is considered. Further, some physical concepts such as energy
conservation and dispersion which serve to distinguish equations of different
types are discussed. Finally, the concept of adjoint differential operators is
presented.
Chapter 4 presents the method of separation of variables for the solution
of problems given in bounded spatial regions. This leads to a discussion of
eigenvalue problems for partial differential equations and the one-dimen-
vii
viii iCE
sional version thereof known as the Sturm-Liouville problem. Eigenfunc-
tion expansions, in general, and Fourier series, in particular, are considered
and applied to the solution of homogeneous and inhomogeneous problems
for linear partial differential equations of second order. It is also shown that
eigenfunction expansions can be used for the solution of nonlinear problems
by considering a nonlinear heat conduction problem.
In Chapter 5 the Fourier, Fourier sine, Fourier cosine, Hankel, and
Laplace transforms are introduced and used to solve various problems for
partial differential equations given over unbounded regions in space or time.
As the solutions of these problems are generally obtained in an integral
form that is not easy to evaluate, approximation methods for the evaluation
of Fourier and Laplace integrals are presented.
Not all problems encountered in applied mathematics lead to equations
with smooth coefficients or have solutions that have as many derivatives as
required by the order of the partial differential equations. Consequently,
Chapter 6 discusses methods whereby the concept of solution is weakened
by replacing the differential equations by integral relations that reduce the
number of derivatives required of solutions. Also, methods are presented
for dealing with problems given over composite media that can result in
singular coefficients. Finally, the method of energy integrals is discussed and
shown to yield information regarding the uniqueness and dependence on the
data of solutions of partial differential equations.
Green's functions, which are discussed in Chapter 7, depend on the
theory of generalized functions for their definition and construction. There-
Therefore, a brief but self-contained discussion of generalized functions is pre-
presented in this chapter. Various methods for determining Green's functions
are considered and it is shown how initial and boundary value problems for
partial differential equations can be solved in terms of these functions.
Chapter 8 contains a number of topics. It begins with a variational
characterization of the eigenvalue problems considered in Chapter 4, and
this is used to verify and prove some of the properties of eigenvalues and
eigenfunctions stated in Chapter 4. Furthermore, the Rayleigh-Ritz method
which is based on the variational approach is presented and it yields an
approximate determination of eigenvalues and eigenfunctions in cases where
exact results are unavailable. The classical Riemann method for solving
initial value problems for second order hyperbolic equations is briefly
discussed, as are maximum and minimum principles for equations of elliptic
and parabolic types. Finally, a number of partial differential equations of
mathematical physics are studied, among which the equations of fluid
dynamics and Maxwell's equations of electromagnetic theory are discussed
at length.
Chapters 9 and 10 conclude the book. They deal with perturbation and
asymptotic methods for solving both linear and nonlinear partial differential
equations. In recent years these methods have become an important tool for
the applied mathematician in simplifying and solving complicated problems
PREFACE ix
for linear and nonlinear equations. Regular and singular perturbation
methods and boundary layer theory are discussed in Chapter 9. Linear and
nonlinear wave propagation problems associated with the reduced wave
equation that contains a large parameter are examined in Chapter 10. These
include the scattering and diffraction of waves from various obstacles and
the problem of beam propagation in linear and nonlinear optics. It is also
shown in Chapter 10 how singularities that can arise for solutions of
hyperbolic equations can be analyzed without having to solve the full
problem given for these equations. Finally, an asymptotic simplification
procedure is presented that permits the replacement of linear and nonlinear
equations and systems by simpler equations that retain certain essential
features of the solutions of the original equations.
The Bibliography contains a list of references as well as additional
reading. The entries are arranged according to the chapters of the book and
they provide a collection of texts and papers that discuss some or all of the
material covered in each chapter, possibly at a more elementary or ad-
advanced level than that of the text.
The material in this book has been developed and expanded from a set of
lectures I have given over a number of years in a course on partial
differential equations. It is intended for advanced undergraduate and begin-
beginning graduate students in applied mathematics, the sciences, and engineer-
engineering. The student is assumed to have completed a standard calculus sequence
including elementary ordinary differential equations, and to be familiar with
some elementary concepts from advanced calculus, vector analysis, and
matrix theory. (For instance, the concept of uniform convergence, the
divergence theorem, and the determination of eigenvalues and eigenvectors
of a matrix are assumed to be familiar to the student.) Although a number
of equations and problems considered are physically motivated, a knowl-
knowledge of the physics involved is not essential for the understanding of the
mathematical aspects of the solution of these problems.
In writing this book I have not assumed that the student has been
previously exposed to the theory of partial differential equations at some
elementary level and that this book represents the next step. Thus I have
included such standard solution techniques as the separation of variables
and eigenfunction expansions together with the more advanced methods
described earlier. However, in contrast to the more elementary presenta-
presentations of this subject, this book does not dwell at great length on the method
of separation of variables, the theory of Fourier series or integrals, the
Laplace transform, or the theory of Bessel or Legendre functions. Rather,
the standard results and methods are presented briefly but from a more
general and advanced point of view. Consequently, it has been possible to
present a variety of approaches and methods for solving problems for linear
and nonlinear equations and systems without having the length of the book
become excessive.
There is more than enough material in the book to be covered in a
x PREFACE
year-long course. For a shorter course it is possible to use the first part of
Chapter 3 and Chapters 4 and 5 as a core, and to select additional material
from the other chapters as time permits.
The book contains many examples. Very often new approaches or
methods are brought out in the form of an example. Thus the examples
should be accorded the same attention as the remainder of the text. There
are more than 800 exercises in the book. They are placed at the end of each
chapter and are listed according to the sections in the chapter. With a few
exceptions, no substantially new theories or concepts are introduced in the
exercises. For the most part, the exercises are based on material developed
in the text, and the student should attempt to solve as many of them as
possible to test his or her mastery of the subject.
In preparing the second edition of this text, a strong effort was made to
clarify the presentation where necessary. This entailed the inclusion of
additional comments and observations throughout the text. In some cases,
such as in Sections 6.7 and 10.2, a large part of the section was entirely
rewritten. New material was added throughout the text. We cite examples
from each chapter. In Chapter 1 a derivation of the Fokker-Planck and the
backward Kolmogorov equations is given. In Chapter 2 a discussion of
shock waves is presented. Normal forms for quasilinear hyperbolic systems
are given in Chapter 3. Nonself-adjoint boundary value problems are
discussed in Chapter 4. Chapter 5 contains an expanded discussion of
Laplace transforms. Weak solutions of Galerkin form are considered in
Chapter 6. Generalized power functions are introduced in Chapter 7. An
expanded discussion of maximum principles is given in Chapter 8. Chapter 9
in the first edition has been split into two chapters in the second edition.
Gaussian beam propagation is discussed in Chapter 10. *
A substantial number of new exercises, as well as a number of new
examples have been included. In addition, solutions to selected exercises are
provided at the end of the book. The Bibliography has also been expanded
to aid the reader of this text.
I would like to thank Maria Taylor, Mathematics editor at Wiley-
Interscience for her support of this project. I acknowledge my gratitude to
my wife, Naomi, for her assistance and understanding during the many
hours that were spent on this book. Also, I thank my children Joshua, Chani
and David, and my granddaughter Elana for their patience while I was
involved with this work.
Erich Zauderer
New York
April 1989
CONTENTS
1* Random Walks and Partial Differential Equations 1
1.1. The Diffusion Equation and Brownian Motion, 2
1.2. The Telegrapher's Equation and Diffusion, 15
1.3. Laplace's Equation and Green's Function, 27
Exercises for Chapter 1, 34
2. First Order Partial Differential Equations 46
2.1. Introduction, 46
2.2. Linear First Order Partial Differential Equations, 48
2.3. Quasilinear First Order Partial Differential Equations, 63
2.4. Nonlinear First Order Partial Differential Equations, 81
Appendix: Envelopes of Curves and Surfaces, 93
Exercises for Chapter 2, 96
3. Classification of Equations and Characteristics 109
3.1. Linear Second Order Partial Differential Equations, 109
3.2. Characteristic Curves, 117
3.3. Classification of Equations in General and Their
Characteristics, 121
3.4. Formulation of Initial and Boundary Value Problems, 136
3.5. Stability Theory, Energy Conservation, and Dispersion, 139
3.6. Adjoint Differential Operators, 146
Exercises for Chapter 3, 150
4. Initial and Boundary Value Problems in Bounded Regions 163
4.1. Introduction, 163
4.2. Separation of Variables, 168
4.3. The Sturm-Liouville Problem and Fourier Series, 178
4.4. Series Solutions of Initial and Boundary Value Problems, 193
4.5. Inhomogeneous Equations: Duhamel's Principle, 201
xi
CONTENTS
4.6. Eigenfunction Expansions: Finite Fourier Transforms, 207
4.7. Nonlinear Stability Theory: Eigenfunction Expansions, 218
Exercises for Chapter 4, 225
Integral Transforms 244
5.1. Introduction, 244
5.2. One-Dimensional Fourier Transforms, 246
5.3. Fourier Sine and Cosine Transforms, 260
5.4. Higher Dimensional Fourier Transforms, 271
5.5. Hankel Transforms, 280
5.6. Laplace Transforms, 287
5.7. Asymptotic Approximation Methods for Fourier Integrals, 295
Exercises for Chapter 5, 308
Integral Relations 324
6.1. Integral Relations, 325
6.2. Composite Media: Discontinuous Coefficients, 329
6.3. Solutions with Discontinuous First Derivatives, 338
6.4. Weak Solutions, 342
6.5. The Integral Wave Equation, 349
6.6. Concentrated Source or Force Terms, 363
6.7., Point Sources and Fundamental Solutions, 370
6.8. Energy Integrals, 390
Exercises for Chapter 6, 396
Green's Functions 412
7.1. Integral Theorems and Green's Functions, 412
7.2. Generalized Functions, 426
7.3. Green's Functions for Bounded Regions:
Eigenfunction Expansions, 446
7.4. Green's Functions for Unbounded Regions, 467
7.5. The Method of Images, 484
Exercises for Chapter 7, 497
Variational and Other Methods 516
8.1. Variational Properties of Eigenvalues and Eigenfunctions, 516
8.2. The Rayleigh-Ritz Method, 536
8.3. Riemann's Method, 549
8.4. Maximum and Minimum Principles, 556
8.5. Solution Methods for Higher Order Equations and Systems, 563
Exercises for Chapter 8, 596
CONTENTS xiii
9. Perturbation Methods 616
9.1. Introduction, 616
9.2. Regular Perturbation Methods, 619
9.3. Singular Perturbations and Boundary Layer Theory, 647
Exercises for Chapter 9, 688
10. Asymptotic Methods 700
10.1. Equations with a Large Parameter, 700
10.2. The Propagation of Discontinuities and Singularities for
Hyperbolic Equations, 778
10.3. Asymptotic Simplification of Equations, 808
Exercises for Chapter 10, 832
Bibliography 840
Solutions to Selected Exercises 845
Index 883
PARTIAL DIFFERENTIAL
EQUATIONS OF APPLIED
MATHEMATICS
Second Edition
1
RANDOM WALKS AND
PARTIAL DIFFERENTIAL
EQUATIONS
It is traditional to begin a course on partial differential equations of applied
mathematics with derivations of the basic types of equations to be studied
based on physical principles. Conventionally, the problem of the vibrating
string or the process of heat conduction is considered and the corresponding
wave or heat equation is derived. (Such derivations and relationships will be
discussed later in the text.) We have chosen instead to use some elementary
random walk problems as a means for deriving and introducing prototypes
of the equations to be studied in the text: A) the diffusion equation, B) the
wave equation, or more specifically, the telegrapher's equation, and C)
Laplace's equation. More general forms of each of these types of equations
are also derived. The equations describing the random walks are difference
equations whose continuum limits are the aforementioned partial differen-
differential equations. Only elementary and basic concepts from probability theory
will be required for our discussion of the random walk problems and these
concepts will not be used in the sequel.
Our discussion of the limiting processes will be somewhat heuristic and
formal but they can be rigorously justified. We shall not be discussing
numerical methods for solving partial differential equations in this text, but
the discrete formulation of the differential equations based on the random
walk problems yields a useful insight into the differences in the numerical
approaches required for each of the differential equations considered.
Furthermore, some basic properties of each of the three types of equations
to be derived are elementary consequences of the discrete random walk
formulations. These properties will be verified directly for each of the
limiting partial differential equations.
2 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
1.1. THE DIFFUSION EQUATION AND BROWNIAN MOTION
We begin by considering the unrestricted one-dimensional random walk
problem. A particle starts at the origin of the дс-axis and executes random
steps or jumps each of length 8 to the right or to the left. Let xt be a random
variable that assumes the value 8 if the particle moves to the right at the /th
step and the value -8 if it moves to the left. We assume that each step is
independent of the others, so that the xt are (identically distributed)
independent random variables.
Let the probability that the particle moves to the right or left equal p or
q, respectively. Since these probabilities are identical for each step, we have
Probfo = 8) s />(*. = 8) =p and P(xt = -8) = q. The particle must move
either right or left so that p + q = 1. The position of the particle at the nth
step is given by
A.1) Хя = хг+х2 + --- + хн.
Using the binomial distribution, the probability that the particle is located at
a fixed point after a given number of steps can be determined explicitly.
Since we are more interested in the continuum limit of the random walk
problem as the step length 8—>0 and the number of steps n —»«>, we shall
not consider the exact solution of the above problem. (See, however,
Exercise 1.1.11.) We do, however, determine the mathematical expectation
and the variance of the random variable Xn. They shall be related below to
certain physically significant constants when the above random walk model
is connected with the theory of Brownian motion of a particle. The
expectation of Xn yields the expected or mean location of the particle at the
nth step whereas the variance of Xn is a measure of how much the actual
location of the particle varies around the expected location.
Example 1.1. Expectation and Variance: The Discrete Case. For a
discrete valued random variable x that assumes the values am with probabili-
probability pm, the expected value or the mathematical expectation is defined as
A.2) E(x) = 2amPm.
m
Thus for the random variable xt we have
A.3) £(*,) = (+8)P(xt = 8) + (-8)P(Xi = -8) = (p-
Since E(x) is a linear function of x we have
A.4) E(Xn) = дB x) = 2 E{xt) = (P- q)8n
THE DIFFUSION EQUATION AND BROWNIAN MOTION 3
in view of A.3). In the physical literature the mathematical expectation is
expressed as
A.5) E(x) = (xO
and this notation will be used below. Thus A.4) yields
A.6) (Xn)=(p-q)8n.
Note that if p — q = \, so that the particle is equally likely to move to the
right or to the left, we have (Xn) = 0. Then the expected location of the
particle after n steps is the origin (i.e., the starting point). If p>q, the
expected location is to the right of the origin, whereas the converse is true if
p<q.
The variance of a random variable x is defined in terms of the expectation
as
A.7) V(x)=((x-(x)J) = {x2)-{x)\
where the last equation is an easy consequence of the properties of the
expectation. It can also be shown that the variance of a sum of independent
random variables equals the sum of the variances of the random variables.
Thus since Xn = EJL1 xt and the xt are independent random variables,
A.8) V(Xn) = v(t x) = 2 V(xt) = 2 [<*?> - (x,J].
\ = 1 ' i = \ i = l
But
A.9) (х2)^^8JР(х, = 8) + (-8JР(х^-8) = 82(Р + д) = 82
since p + q = 1. Recalling A.3) we easily obtain
A.10) V(Xn) = t[82-(p~ qJ82] = 4pq82n
on using the identity 1 = (p + qJ.
The foregoing results will now be applied to yield a mathematical
description of one-dimensional Brownian motion. This motion refers to the
ceaseless, irregular, and (apparently) random motion of small particles
immersed in a liquid or gas. A given particle is assumed to undergo
one-dimensional motion due to random collisions with smaller particles in
the fluid or gas. In a given unit of time many collisions occur. Each collision
is assumed to be independent of the others and to impart a displacement of
4 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
the particle of length 8 to the right or to the left. The apparent randomness
of the motion is characterized by assuming that each collision independently
moves the particle to the right or to the left with probability p or q,
respectively. Clearly, the random walk problem described can be used to
simulate the observed motion of the particle.
Experimentally, it is found for a particle immersed in a fluid or gas
(undergoing one-dimensional motion) that the average or mean displace-
displacement of the particle per unit time equals c, whereas the variance of the
observed displacement around the average equals D > 0. We assume that
there are r collisions per unit time (note that r is not known a priori but is
assumed to be large). Then for our random walk model of the observed
motion we must have (approximately) after r steps
A.11) (p-q)8r~c,
A.12) 4pq82r~D ,
in view of the expressions A.6) and A.10) for the mean and the variance.
Since the motion of the particle appears to be continuous, we must
examine the limit as the step length S-»0 and the number of steps r-»°°.
This must occur in such a way that с and D [as given in A.11) and A.12)]
remain fixed in the limit. Now if p Ф q and p — q does not tend to zero as
S—»0 and r—»oo, we have from A.11)
which implies
A.14) 4pq82r->
However, 4pq8 2r must tend to D Ф 0 in the limit as 8 -~» 0 and r-> oo? in view
of A.12). Therefore, we must have p — #—>0 in the limit. Combined with
p + q -1 this implies that both p and q tend to \ in the limit. If p - q = \ in
the discrete model, we have с = 0. But if p - q Ф 0 so that с Ф 0, the particle
exhibits a drift to the right or to the left depending on the sign of c.
Additionally, if D - 0 there is no variation around the average displacement
с and the motion of the particle must appear to be deterministic and not
random or irregular.
These results can be realized if we set
A.15) p=J(l
with the constant b (independent of 8) to be determined and chosen such
THE DIFFUSION EQUATION AND BROWNIAN MOTION 5
that 0 < p, q<l. Then, p + q = 1 and p — q = b89 and the foregoing limits
imply that as 5—»0 and r—»°o, we have 52r—»D, />^5, <?—»i, and
b = c/Z>. More generally, if we set
A.16) p=i(* + M), fl = *(«-M)
with 0< a < 1, we find that p + q = a. Thus, if a < 1, there is a nonzero
probability l-p-# = l-a that the particle rests at each step, that is, it
takes a step of zero length. Since the additional value that the random
variable хй can assume is zero, we see that the expectation and the variance
of the random variable are again given by A.3) and A.9), except that
p + q = a in A.9). As a result, A.10) becomes V(Xn) = (a - b2d2)82n.
Finally, A.11) and A.12) (suitably modified) yield a82r^>D and b82r->c,
so that b = aclD. It is seen that a and b cannot be determined uniquely in
terms of the observed quantities с and D. The more precise details regarding
the probability that the particle rests or moves at each step cannot be
determined from its observed motion.
For r steps to occur in unit time, each individual step of length 8 must
occur in 1/r = r units of time whereas n steps occur in nlr — nr time units.
To describe the motion of the particle that starts at the point x = 0 at the
time t = 0, we use the random walk model to obtain the probability that the
particle is at the position x at the time /. That is, we have approximately
after n steps
A.17) Xn = x and nr = t.
If x >0, for example, we must have
A.18) Xn = k8 = x and nr = t
where к equals the excess of the number of steps taken to the right over
those taken to the left. A similar expression with к <0 is valid if x <0.
We define
A.19) y(jc, f) = P(Xn = x) at the time t = nr
to be the probability that at the (approximate) time t, the particle is located
(approximately) at the point я\ If p + q = 1, we could determine an explicit
expression for v(x, t) on using the binomial distribution. However, since we
are interested in a continuum limit of the random walk problem as 8 and т
tend to zero, we construct a difference equation satisfied by v(x, t). Again,
we do not solve the difference equation but show that in the continuum limit
it tends to a partial differential equation that serves as a model for the
Brownian motion of a particle. (See, however, Exercise 1.1.11.)
The probability distribution v(x, t) satisfies the difference equation
6 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
A.20) v(x, t + t) = pv(x - 8, t) + qv(x + 5, f) .
This states that f/ie probability that the particle is at x at the time t + т equals
fue probability that it was at the point x — 8 at the time t multiplied by the
probability p that it moved to the right in the following step plus the
probability that the particle was at the point x + 8 at the time t multiplied by
the probability q that it moved to the left in the following step. The
plausibility of the above equation, where p + q = l, is apparent.
Expanding in a Taylor series with remainder we have
\v(x, t + t) - v(x, t) + rvt(x, t) + 0(r2)
<L21>
[ v(x ± 5, i) = v(x, i) ± 8vx(x, i) + \8vxx{x, t) + 0E3)
where O(yk) means that lim^o у~кО{ук) is finite. Substituting A.21) into
A.20) and using p + q = 1 we readily obtain
A.22) vt(x, t) = [(q-p) *]vx(x, 0+1 (v)u«(^ 0 + O(t) + o(^
As 5->0 and r-»0, A.22) tends to the limiting partial differential equation
A.23) ^£ = _C^+1D^
v ' dt dx 2 dx
on using A.11) and A.12) with r — l/r and noting that p and q tend to \ in
the limit. In view of A.18) which requires x and t to remain fixed in the
limit, we must also have |Jfc| —> oo and n^oo such that k8->x and nr^>t.
Further, v(x, t) must now be interpreted as a probability density associated
with the continuous random variable x at the time t, rather than as a
probability distribution in the discrete random walk model. This means that
the probability that x lies in a given interval [a, b] is given by
rb
A.24) P(a<jc<b)= v(x, i)
Ja
dx at the time t.
If p and q are defined as in A.16), the difference equation A.20) must be
replaced by
A.25) v(x, t + r) - A - p - q)v(x, t) + pv(x - 5, i) + qv(x + 5, t)
since there is a nonzero probability that the particle was at the point x at the
time t and did not move from there. However, if we expand v as in A.21)
and note the results following A.16), we again obtain the equation A.23) in
the limit. This is consistent with our earlier observation that in the con-
THE DIFFUSION EQUATION AND BROWNIAN MOTION 7
tinuum limit the results are insensitive to the detailed structure of the
discrete random walk model.
The equation A.23) is known as a diffusion equation and D is called the
diffusion coefficient. We examine the significance of the coefficients с and D
in A.23) in Example 1.3, with calculations based directly on A.23) rather
than on the results obtained from the random walk model.
To completely determine the solution v(x, t) of A.23) we need to specify
initial conditions for the density function. Since the particle was initially
(i.e., at the time t = 0) assumed to be located at the point x = 0, it has unit
probability of being at x = 0 at the time t = 0 and zero probability of being
elsewhere at that time. This serves as initial data for the difference equation
A.20). In terms of the density function v(x, t) that satisfies A.23), we still
must have v(x, 0) = 0 for x ^ 0, and the density must be concentrated at
x = 0. Since the total probability at t = 0 satisfies
A.26) P(
we see that v(x, 0) behaves like the Dirac delta function; that is,
A.27) v(x,0) = 8(x)
[the delta function 8(x) is not to be confused with the step length 8
introduced in the random walk problem.]
Example 1.2. The Dirac Delta Function. The Dirac delta function is
discussed more fully in Section 7.2. For the purposes of our present
discussion, we may characterize it as the limit of a sequence of discontinuous
functions 8€(x) defined as
— I
0, Ы>е.
A.28) 8e(x) = <
Each 8€(x) has unit area under the curve and in the limit as €—»0, Se(jt)-»O
for all jt#O. However, on formally interchanging the limit process with
integration, we obtain
/•00 />0O
A.29) J ^8(x)dx = hm]_m8e(x)dx = l.
It also follows for continuous functions f(x), on using the mean value
theorem for integrals before passing to the limit, that
A-30) J' J(x)8(x)dx=f@)
8 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
so that in effect f(x)8(x) = f@)8(x). Further, 8(x - £) vanishes for all x * £,
and if 8(x) is replaced by 8(x - f) in A.30), we obtain/(f) instead of ДО)
so that f(x)8(x - g) =f(t)8(x - g).
The combined equations A.23) and A.27) constitute an initial value
problem for the partial differential equation A.23). It is derived later and it
can be shown directly by substitution that the function
is a solution of the initial value problem A.23) and A.27). For fixed f, the
probability density function A.31) is the density function of a normal or
Gaussian distribution with mean (x) = ct and variance {{x - (x)f) = Dt.
In view of the exponential decay of v(x, t), we find that the density is
concentrated around the curve x — ct = 09 so that the particle appears to
move with the (drift) velocity dxldt = c. This also follows from the equation
(x)It— c. The variance of the particle location around the path x = ct [in
the (jc, f)-plane] increases linearly in t and is given by Dt.
Although the foregoing results for the mean and the variance are well
known for the normal distribution, they may be verified directly by consider-
considering the limit of the discrete random walk problem. In Example 1.3 these
results for the mean and the variance are obtained directly from the partial
differential equation A.23).
The random walk model considered above yields A.6) and A.10) as the
mean and the variance. As n-»o°, S-»0, and 1/r—>0, we obtain from A.6)
A.32) <jc> = Um(^) = lim(p - q)8n = lim(/> - q)8r -
- q)8r]nr = ct
on using A.11) and A.18). Similarly, A.10) yields
A.33) <(*- (x)f) =lim((Zrt - (Xn)J)=\im4pq82n
= lim 4pq82rnr = Dt.
(In carrying out the limit in A.33) we have assumed that /? + <? = l.
However, both A.32) and A.33) remain valid if p and q are defined as in
A.16).) Using A.32) and A.33) and applying the central limit theorem of
probability theory to Xn (which is a sum of n independent random variables)
as n-*«, we can conclude directly that the limiting distribution for x is
Gaussian with mean ct and variance Dt.
In the following example we show how the mean and variance of the
random variable x with density v(x, i) can be obtained directly from
THE DIFFUSION EQUATION AND BROWNIAN MOTION 9
differential equation A.23) and the initial conditions A.27) without actually
solving for v(x\ t).
Example 1.3. Expectation and Variance: The Continuous Case. If a: is a
continuous random variable, the kth moment of x is defined as
A.34)
where v(x91) is the probability density function at the time t. We assume
that v(x, t) and its x-derivatives vanish sufficiently rapidly at infinity, so that
all terms evaluated at infinity in the following integrations vanish. [In fact,
A.31) shows that v(x, t) vanishes exponentially at infinity.]
First we show (jc°) = 1 for all t so that v(x, t) is, in fact, a probability
density for all f>0. Integrating in A.23) we have
A.35) \_Jt \_\\_go
since v and vx are assumed to vanish at infinity. Thus (x°) is constant in
time. But A.26) shows that (jc°) = 1 at t = 0 so that (jc°) = 1 for all time
t>0.
Next, we consider the expected value {x) of the continuous random
variable jc. We multiply A.23) by x and integrate from -oo to +o°. This gives
A.36) j
= -c I xvx dx+ - D I xvxx dx- c\ vdx- c(x°) =
on integrating by parts, using A.35) and (jc°> = 1. At the time f = 0 since
v(x, 0) = 8(x), the property A.30) of the delta function implies that (x) = 0
at f = 0. Therefore, we obtain an initial value problem for (jc),
037) i <„>-«. м
with the solution
A.38) <*> = <*,
which agrees with the result A.32).
The variance of x is given by
A.39) v(x) = (x2)-(*
= 0
10 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
The second moment (x2) clearly vanishes at r = 0 in view of A.27) and
A.30). Multiplying A.23) by x2 and integrating, we have
A.40) £ A dx = | (x>) = -c £ *Ч Л + | D
= 2c £ то Лс + Z) £ уЛс = 2c(jc) + D{x°)
on integrating by parts. Using A.38) and (jc°) = 1 we obtain the initial value
problem for (x )
A.41) jt(x2)=2c2t+D, <*2>|,=0 = <>,
which yields
A.42) <jt2) = cV + D*.
Thus the variance is given by
A.43) V{x) = (x2) -(xJ = c2t2 + Dt- c2t2 = Dt,
which agrees with A.33).
The foregoing results, that is, the random walk problem and the limiting
initial value problem, provide a valid description of the motion of a
Brownian particle in an unbounded region. If the region is bounded on one
or both sides, certain boundary conditions must be added. We now consider
two random walk problems in which the particle is restricted to move in the
region x < /, and at x -1 > 0 there is either an absorbing or reflecting
boundary.
At an absorbing boundary, once the particle reaches the boundary at
x = I it is absorbed and can not longer move into the region x < I. With
v(x, t) as the probability that the particle is (approximatley) at the point x at
the time t we have at x — /,
A.44) v(l,t + r) = pv(l-8,t),
since the particle cannot reach x -1 from the right. We assume that
p + q = 1. Using the Taylor series A.21) gives
A.45) A -p)v(l, t) = O(t) + O(8) .
Since p-> \ as 8 and r tend to zero, we have in the limit
THE DIFFUSION EQUATION AND BROWNIAN MOTION 11
A.46) i>(U) = 0,
as the boundary condition for the density function at an absorbing
boundary.
For the case of a reflecting boundary, the boundary is assumed to have
the properties of an elastic barrier. When the particle reaches the barrier at
the time t9 there is a probability p that the particle moves to the right
beyond x = / and returns to x = / at the time t + т. (Of course, it can also
move to the left with probability q when it reaches the barrier.) Thus the
boundary condition for the probability v(x, t) is
A.47) v(lt t + T) = pv(l - S,
Again using A.21) we have
A.48) v(l, t) + тоД t) + 0(r2) = 2/w(/, t) - p8vx(l91) + O(82) .
On multiplying across by 8/r, noting that 2p — l=p — q and going to the
limit in A.48), we obtain
A.49) lim( - )(p - q)v - limf — )pvx - lim Svt +
lim O(St) + lim o(^-) = cv - \Dvx = 0
on using A.11). Therefore, the boundary condition at the reflecting boun-
boundary point x = / is
A.50) cv(l9t)-\Dvx(l,t) = Q.
Within the interval -o°<jc</, the density function v(x, t) satisfies the
diffusion equation A.23) with the initial condition A.27) at t = 0. At x = /,
v(x, t) satisfies either the boundary condition A.46) or A.50). The solutions
of these initial and boundary value problems are considered in Chapter 5. If
we study the motion of a Brownian particle in a finite interval, say
-l<x<l9 each end point can represent an absorbing or a reflecting
boundary. The boundary conditions at x = -/, in the case of an absorbing or
relecting boundary are given as in A.46) and A.50), respectively, except
that / is replaced by -/. In the interior of the interval, v(x, t) satisfies A.23)
with the initial condition A.27). Such problems are considered in Chapter 4.
In a further generalization of the foregoing results, we assume that the
probabilities p and q [defined as in A.16)] are functions of the position of
the particle. Then, the probabilities associated with the point x are given as
A.51) P(x) = \[a{x) + b(x)8] , q(x) = \[a{x) - b(x)8].
12 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
We assume that 0 < a{x) < 1, and that b(x) is chosen such that 0 < p, q ^ 1-
The probability v(x, t) that the particle is at the point x at the time f, satisfies
a difference equation of the form A.23) where p and q are now given by
A.51), so that
A.52) v(x, t + r) = [1 - p(x) - q(x)]v(x, t) + p(x - 8)v(x - S, t)
On expanding all the terms in Taylor series around x and t as in A.21), and
assuming that as S->0 and т->0, а(хN2/т-> D(x) and b(x)82/r—> c(x), we
obtain, in the limit
A.53) *, = -(с«0, + *(Л>)*,.
This is known as the Fokker-Planck equation in the physical literature, and
as the forward Kolmogorov equation in the mathematical literature. It
represents a generalization of the diffusion equation and describes the
motion of a Brownian particle in an inhomogeneous medium, on the basis of
our model. The initial condition for the probability density function v(x, t) is
again A.27). However, the functions c(x) and D(x) can no longer be
interpreted as in A.38) and A.43), since the calculations in Example 1.3 are
not valid for the Fokker-Planck equation with variable с and £>. Neverthe-
Nevertheless, the coefficients c(x) and D(x) are known as the drift and diffusion
coefficients, respectively. At an absorbing boundary we find as before that
the density function must vanish. The boundary condition at a reflecting
boundary point is considered in the exercises.
Until now we have expressed the probability that the Brownian particle is
at the point x at the time t in terms of its location at the preceding time
period. This analysis gave rise to the difference, diffusion and Fokker-
Planck equations determined above. Now, however, we examine the prob-
probability that the particle is at the point x at the time t in terms of its location
in the following time period. With p and q given as in A.51), we find that
the particle can only have reached the points x - 5, x and x + 8 at the time
/+т. Thus, v satisfies the difference equation
A.54) v(x, 0 = [1" P(x) - q(x)]v(x, t + r) 4- p(x)v{x + 5, t + r)
+ q(x)v(x-69t+T).
In contrast to A.52) all the probabilities p and q are evaluated at x since
they are all related to the motion of the particle from the point x at the time
t. To determine the probability that the particle is at some point x on the
real axis, we must know the probabilities for all points x at some later time.
Thus, if v@,0) = l and v(x, 0) = 0 at any other point, A.54) gives the
probability that the particle is at some point x at an earlier time
THE DIFFUSION EQUATION AND BROWNIAN MOTION 13
Consequently, A.54) is a backward difference equation. Proceeding as in
the discussion following A.52), we obtain in the limit as 5—»0 and r—»0 in
A.54),
A.55) vt = -cvx~\Dvxx.
This is known as the backward Kolmogorov equation. If the condition
A.27) is assigned for the density function v(x91) at the time t = 0, it
represents an end condition for A.55), as the equation is to be solved for
r<0, as we have indicated. While we have assumed that с and D are
functions of x in A.55), even if they are taken to be constant, the forward
and backward Kolmogorov equation still differ in the sign of the second
derivative term. In fact, A.55) is the adjoint equation of A.53), as follows
from our discussion of adjoint operators in Section 3.6.
The forward Kolmogorov equation determines the probability density at
each point x in terms of the position of the particle at an earlier time. The
backward Kolmogorov equation determines the probability density at each
point x in terms of the position of the particle at a later time. Even if we
reverse the direction of time in the backward equation, that is, we replace t
by — f, the distinction in how the solution is to be interpreted remains.
However, we do not pursue these matters any further. The form of the
boundary conditions for the backward equation will be considered in the
exercises. Apart from its intrinsic interest, the backward equation also arises
in the process of determining the Green's function for the forward equation,
as will be seen in Chapter 7.
We conclude this section by noting a number of consequences for the
solutions of initial and boundary value problems for the diffusion equation
A.23) and its generalizations that follow from the difference equation
formulation of the associated random walk problems. To begin with, we
observe that the difference equations A.20), A.44), A.47), and A.52)
imply that the solutions of the initial and boundary value problems evolve in
time. That is, the solution of any point at a given time t depends only on
initial and boundary data given at earlier times. This characterizes the
causality property that only past events influence future events. For the
backward Kolmogorov equation, however, there is backward causality. That
is, events in the future influence the past.
Secondly, we consider the initial value problem or the initial and boun-
boundary value problem with the boundary condition A.46). The difference
equation A.20) for v(x, f + т) represents it as a weighted average of и
evaluated at two points at an earlier time t. (We have a weighted average
since p>0, q>0, and p + q = l). Since v vanishes initially everywhere
except at one point and v vanishes on the boundary, we conclude that
0^ v < 1 everywhere and the maximum value of v (i.e., v = 1), as well as
the minimum value of v (i.e., v — 0), is attained either on the boundary or
on the initial line. This maximum (and minimum) principle carries over to
14 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
the diffusion equation for which it is shown to be valid in Chapter 8. The
effect of the diffusion process is to distribute the densities in the interior in a
fairly uniform fashion so that the maximum and minimum values occur on
the initial or boundary line. The difference equation A.52) that yields the
Fokker-Planck equation does not express v(x, t + r) as a weighted average
of v evaluated at the earlier time t unless p and q are constants. Otherwise,
since the p and q are all evaluated at different points, the coefficients of v
need not add up to one. Consequently, we do not expect the general
Fokker-Planck equation to satisfy a maximum principle. However, in the
difference equation A.54), v(x, i) is given as a weighted average of v
evaluated at the time t + r9 since p and q are both evaluated at the same
point. As a result there is a maximum principle for the backward Kol-
mogorov equation, as will be shown in Chapter 8.
Finally, we note that the requirement that S2/r tends to a finite nonzero
limit as 8 and т tend to zero implies that 5/т-»°° in the same limit. This
means that the speed of the particle in Brownian motion (which is given by
the limit of Sfr) is infinite. This fact is also implied by the solution A.31) of
the initial value problem A.23) and A.27). It shows that v{x, t) is instanta-
instantaneously nonzero for all x when t > 0 even though v(xy t) vanishes for all
x Ф 0 at t = 0. Thus there is a nonzero probability, however small it may be,
that the particle is located in the neighborhood of any point as soon as t
increases from zero.
The foregoing limitation of the theory of Brownian motion based on
solution of the diffusion equation was already noticed by Einstein who was
the first to derive a diffusion equation to describe Brownian motion. He
recognized that the diffusion equation yields a valid model only as t gets
large. Since (with c-0) A.23) also represents the equation of heat conduc-
conduction, with v equal to the temperature and D equal to a thermal diffusion
coefficient, the same difficulty with regard to the interpretation of the
physical processes involved occurs for the heat equation.
A different analysis of the theory of Brownian motion was given by
Ornstein and Uhlenbeck in an effort to overcome this shortcoming. It is
based on the Langevin equation for the velocity u(t) = x'(t) of the particle in
Brownian motion. It has the form
A.56) u>(t) = -fu(t) + F(t).
This is just Newton's law of motion for the particle. The term mu(i) is the
momentum of the particle, and we have divided through by m in the
equation. The positive constant / is a frictional resistance coefficient, while
the term F(t) represents the effects of the random collisions with the other
particles in the fluid or gas. They showed that A.56) implies finite particle
velocities, in particular for small times, and yields a description of the
motion of the particle for large times based on the diffusion equation A.23).
THE TELEGRAPHER'S EQUATION AND DIFFUSION 15
Equation A.56) and its generalizations that describe Brownian motion in
inhomogeneous media, are all examples of stochastic differential equations,
in that they contain random terms. The theory and application of such
equations has been the subject of much research in recent years.
In the following section we show, on the basis of a modified random walk
model, how to construct a limiting differential equation that yields a finite
speed for the particle undergoing Brownian motion for small t and reduces
to the preceding results as *->«>.
1.2. THE TELEGRAPHER'S EQUATION AND DIFFUSION
The random walk problem considered in the previous section was shown to
imply an infinite speed for a particle in Brownian motion in the continuum
limit modeled by the diffusion equation A.23). It may be reasonably argued
that this is a consequence of the basic assumption that each step in the
random walk is independent of the previous steps. Thus in the limit as the
step length, as well as the time lapse between steps, tends to zero, the
probability that the particle moves right or left tends to \. The particle is,
therefore, equally likely to move to the right or to the left regardless of the
direction in which it was moving previously. The limiting path is, con-
consequently, totally irregular and the particle cannot be said to have a fixed
finite velocity. The same is true for each of the models considered.
An assumption in the random walk problem that might be expected to
yield a smoother path of motion for the particle in the limit (at least in the
initial stages) is that there exists a positive correlation between two adjacent
steps. This correlation is expected to increase to a maximum value of unity
as the step length and the time between steps tend to zero. The correlation
implies a tendency for the particle to continue moving in a given direction
once it begins to move in that direction. If, initially, probabilities for motion
to the right or left are established, the particle will maintain its tendency to
move in a fixed direction for a certain time at a finite speed. After a while,
the inherent randomness of the process reduces the motion to that obtained
in the previous section. The foregoing assumption was introduced into the
random walk problem by R. Furth in a study of Brownian motion. He
showed that it implies a finite velocity for the particle at small times and
yields the results of Section 1.1 for large times.
The correlated random walk was independently considered by G. I.
Taylor in a discussion of diffusion processes. It was reexamined by S.
Goldstein who formulated a difference equation characterizing the random
walk and constructed its limiting partial differential equation. We shall use
the results of Taylor and Goldstein in our discussion but retain the notation
of Section 1.1. The correlated random walk is described in the following
example.
16 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
Example 1.4. The Correlated Random Walk. The particle is assumed to
start at x = 0 and move to the right or to the left with probability equal to \
initially. Subsequently, the particle has a constant probability of persistence
in or reversal of direction at each step. At the ith step the random variable
xt assumes the values +5 or —5 and is readily seen to do so with probability
equal to \. However, the steps are no longer assumed to be independent, as
was the case in the preceding section. With Xn again giving the position of
the particle after n steps [as in A.1)] we now have for the mean value of Xn,
A.57)
i=l
since the equal likelihood of a step to the right or left implies (xt) = 0.
For the variance of Xn we have
A.58) V(Xn) = <(*„ - <*„»2> = (X2n)
on using A.57), so that the variance equals the second moment of Xn. Also,
we easily obtain
A.59) V{Xi) = {{Xi-{Xi)f) = {x]) = b\
The second moment {X2n) is given as
=((i xY) - (i x?+2 i *,*,
*=1
and we evaluate it in terms of the correlation coefficient between xt and jc;.
The correlation coefficient between two random variables xt and xi+k is
defined as
and it is seen to vanish if the xt are independent random variables. We shall
assume that partial correlations between two nonadjacent random variables
xt and jc; (i.e., \j - i\ > 1) equal zero. Since the xt are identically distributed
random variables, the correlation coefficient between any two adjacent
random variables is equal, so that
A.62) P(xi9xi+1)*p.
THE TELEGRAPHER'S EQUATION AND DIFFUSION 17
Further, the correlation between xt and x^k (with k>l) occurs only
through the intermediate random variables xi+19 xi+29. . . , xi+k_x since
partial correlations are assumed to vanish. Thus
A.63) p(*,.,x,.+Jk) = p* .
Since V(xt) = S2 for all i and <jc,.> =0, we obtain from A.61)
A.64) <**■+*> = *V.
Introducing A.64) into A.60) gives
A.65) (X2n) = 82[n + 2{n - l)p + 2(n - 2)p2 + • -
This series is easily summed in terms of the finite geometric series, and we
have
пыл /r2*-*2L+ 2np
A.66) <^M 1Л + Г=^
Proceeding as in Section 1.1, we set n = tlr and consider the limit as
5-»0, т->0, and w—>o° with f fixed and Xn-*x. Equation A.66) takes the
form
We assume that lim 8/r = y, the finite velocity of the particle. For A.67) to
have a finite nonzero limit as т—>0 we must have
A-68) lim_L^ = _L
т-»о 1 — p 2A
where A is a nonzero positive constant. [The factor 2 in A.68) is introduced
for convenience.] This is consistent with the fact that as r—>0 the correlation
coefficient p must tend to unity. It then follows that the limit of p'/r as т-»0
is
A.69) limp"T=<T2A',
and A.67) tends to
A.70) <^) =
18 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
An expression of the form A.70) for the variance of x was obtained by
Ornstein and Uhlenbeck and Furth in their (improved) theories of Brow-
nian motion. It was also assumed in their theories (as we have done) that the
mean displacement (x) of the particle equals zero. For large values of kt,
A.70) reduces to
A.71) <*2>
which agrees with the result A.33) for the mean square displacement of the
particle in Brownian motion (in the case where (x) = 0) if we set
A.72) D=2X'
where D is the diffusion coefficient. For small values of \t, on expanding
e~2Xt in a Taylor series we obtain
A.73) {x2)~yzt\
This shows that in the initial stages of the motion
A.74) Щ^- «у,
which means that the motion of the particle is essentially uniform with speed
y. Ornstein and Uhlenbeck using Langevin's equation obtained a density
function for the random variable x that gives the displacement of the particle
in Brownian motion. It reduces to the density function A.31) for the normal
distribution as *-»<», has zero mean, and its variance is given by A.70).
They also constructed a differential equation that their density function
satisfies but were not able to show that their equation reduces to the
diffusion equation A.23) (with c = 0) as f—»oo.
Following the approach of S. Goldstein, we now show how to construct a
difference equation and its limiting partial differential equation that charac-
characterize the correlated random walk described.
Let a(x, t) be the probability that a particle is at the point x at the time /
and arrived there from the left, whereas /3(xyt) is the probability that a
particle is at x at the time t and arrived there from the right. Thus a(x, t)
and /3(x,t) characterize right and left moving particles, respectively. Also let
p be the probability that the particle persists in its direction after completing
a step, whereas q is the probability that it reverses its direction after
completing a step. The probabilities p and q are assumed not to vary from
step to step, and we have p + q = 1. Thus if a particle arrives at x from the
left, p is the probability that it continues to the right in the next step,
THE TELEGRAPHER'S EQUATION AND DIFFUSION 19
whereas q is the probability that it reverses itself and goes to the left in the
next step. (Note that p and q were defined differently in the previous
section.)
With steps of length 8 occurring in time intervals of length т, we
immediately obtain the following coupled system of difference equations for
a(x, /) and /3(x, *),
A.75) a(x, t + r) = pa(x - 5, 0 + qf$(x - 5, 0
A.76) /3(x, t + t) - pp(x + 8, t) + qa(x + S, 0
on using the preceding definitions of a, /3, p, and <y.
The assumption introduced above that as т~-»0 the correlation coefficient
p-*l [see A.68)] implies that as т—>0, the probability p of persistence in
direction should tend to unity, whereas the probability q of reversal should
tend to zero. This means that we should have for small r,
A.77) /? = 1-At+O(t2)
A.78) ? = At+O(t2),
where A is the rate of reversal of direction. Now, it can be shown that the
correlation coefficient p is related to p and q as
A.79) P=P~q.
Thus since p + q = l, A.77) and A.78) yield
A.80) p = p-q = l-2q = l-2\T+O(T2).
Recalling A.68), we have
so that A = A.
It was observed by Kac that in view of the equations A.77) and A.78)
and the independence of the probabilities p and q for each step, the
probability of reversal of direction in a given time span is determined by
what is known as a Poisson process. This fact will be exploited later when an
elementary property of the Poisson process is used.
Introducing Taylor expansions for a and /3 in A.75) and A.76), using
A.77) and A.78), and taking the limit as 5 and т tend to zero, with 8/т-*у,
we easily obtain the following coupled system of partial differential equa-
equations:
20 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
In the limit, a(*, i) and /3(x,t) are to be interpreted as probability density
functions for right and left moving particles, respectively.
Since at time t = 0 the particle is located at x — 0 and is equally likely to
move to the right or to the left, the probabilities a and j3 vanish for x Ф 0
and equal \ at x = 0. In terms of the density functions a(x, i) and j3(x, i) this
yields the initial conditions
A.84) a(x,0) = p(x,0)=$8(x),
where 8(x) is the Dirac delta function. The reasoning that results in the
delta function initial data A.84) is similar to that given in Section 1.1.
The system A.82) and A.83) together with A.84) constitute an initial
value problem for a(x, t) and /3(jc, t) that can be solved directly. However,
to compare our results with those of the previous section, we introduce the
probability density function v(x, t) associated with the point x and the time t
without regard to the direction in which the particle is moving. This is given
by
A.85) ф,0 = «(*,') +0OM),
since the particle must have arrived at x either from the right or from the
left. Adding A.82) to A.83) gives
A.86) jt(<* + P) + y£(a-p) = 0,
Subtracting A.83) from A.82) gives
A.87) ~(«-0) + У^(« + /3) = - 2A(« -13).
Differentiating A.86) with respect to t and A.87) with respect to j^after
multiplying by y, and subtracting the second equation from the first yields
the following equation for v = a + /3:
V * ' dt2 ' dx2 dt
This partial differential equation is a special case of what is called the
telegrapher's equation, an equation which governs the propagation of signals
THE TELEGRAPHER'S EQUATION AND DIFFUSION
21
on telegraph lines. It can also be characterized as a wave equation with a
damping effect due to the term Ikdvldt. That is, if A = 0, A.88) reduces to
the (one-dimensional) wave equation. A full discussion of the solution of
A.88) is found in later chapters; however, the damping effect will be
considered below.
In contrast to the diffusion equation A.23), which contains only one time
derivative, the telegrapher's equation A.88) has two time derivatives and,
therefore, requires two initial conditions. From A.84) and A.85) we obtain
A.89) v(x,0) = 8(x)9
whereas A.86) implies, since a-K = 0atf = 0,
A.90)
> 0
Bt
In the following example, we derive the moments of the continuous
random variable x whose density function v(x, t) satisfies A.88)-A.90). The
procedure follows that of Example 1.3 considered in the preceding section
and yields the expectation and variance of x without having to determine the
explicit form of v(x, t).
Example 1.5. Expectation and Variance: The Continuous Case. To
obtain the first three moments of the random variable дс, A.88) is multiplied
by 1, jc, and x2, respectively, and the result is integrated from — °° and oo with
respect to x. Integrating by parts, assuming v(x, t) vanishes sufficiently
rapidly at infinity so that no contributions from the limits at infinity occur,
and using the initial conditions A.89) and A.90), we easily obtain the
following equations.
A.91) ^
A.92) 4
A-93) |5
1> -u
-
dt
f=0
= 0.
dt
<*>
1=0
= 0, з:
r-0
= 0.
f-0
= o,
!<*'>
t=0
-0.
The moments (xk) are defined as in A.34).
The solution of A.91) is (x°) = 1, which shows that v(x, t) is indeed a
probability density function for all f>0. The solution of A.92) is clearly
(*)=0, whereas that of A.93) is identical with A.70). These results
confirm that v(x, t) characterizes a random variable whose expectation and
variance agree with those obtained above in the continuum limit of the
correlated random walk.
22 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
From A.71) and A.72) we see that as f->oo, the mean square displace-
displacement (jc2> of the particle tends to the form given in Section 1.1 for a particle
undergoing Brownian motion. We may expect, therefore, that in some sense
the telegrapher's equation A.88) tends to the diffusion equation A.23) (with
с = 0) as t-> oo. To establish a means of assessing the magnitude of the terms
in A.88) for large f, we introduce the change of scale
A.94) f=|,
where 0< € < 1. Even for moderate values of a since € is small, t takes on
large values. We have dldt = e(dldo) and d2ldt2 = €2(S2/Sa2) so that A.88)
becomes
If we assume that v does not vary rapidly with respect to a, we may discard
the term e2(d2vld<x2) compared to 2\e(dv/da) since e<\ implies that
e2 < в. Thus A.88) can be approximated for large t by the equation
dl) l У2\ д2\)
(L96) Tt=\k)J7>
Ol \Z*A/ OX
where a was replaced by / in A.95) after discarding the second time
derivative term. This equation is to be compared with the diffusion equation
A.23) where we must put c = 0. It has already been shown in A.71) and
A.72) that with D - y2/A, the mean square displacements obtained from the
random walk models in this and the previous section agree for large t.
Consequently, if the parameters are identified in this manner, A.96) is
identical with A.23). The validity of the diffusion equation is limited to
large values of t in its use as a model for Brownian motion as we have seen.
However, A.88) may be taken to represent a valid model for Brownian
motion for all t > 0. For small values of t it yields the required finite particle
velocities, and for large t it reduces to the diffusion equation.
To examine the behavior of the solution of the initial value problem
A.88)-A.90) for small t, we note that the speed of the particle is y. Thus
since the particle is at x = 0 at the time t - 0 and it can travel to the right or
the left, it can never reach the set of points x for which |*| > yt. The density
function v(x, t) should, therefore, vanish for |jc| > yt and this will be verified
when the full solution of A.88)-A.90) is given in a later chapter. The
location of the particle on one of the lines x = ±yt can only be the result of
the particle never having reversed its direction from the time t = 0 on, when
it started on one of the paths x = ±yt in the (jc, t)-plane. The probability of
reversal of direction is characterized by a Poisson process as described by
A.77) and A.78) with rate of reversal A. It is well known for this process
THE TELEGRAPHER'S EQUATION AND DIFFUSION 23
with rate A, that the probability of nonreversal (i.e., persistence) in direction
is given by e~xt at time t. As the particle is equally likely to be on the line
x = yt or x = — yt, we conclude that the probability that the particle is on
either line is
A.97) v(x>t)\x_±yi=\e-kt.
For small values of /, e~Kt ** 1 so that most of the probability is concen-
concentrated on the lines x — ±yt. Therefore, we have essentially deterministic
motion with speed у along the lines x = ±yt. As t increases, e~kt decays
rapidly and the density v(x, t) begins to become more concentrated in the
interior region |jc| < yt. Eventually, the motion appears to become totally
random and can be described by the random walk model and the limiting
diffusion equation given in the previous section.
In a further analysis of the effect of the randomness assumptions that led
to the partial differential equations A.23) and A.88), it is of interest to
consider the reduced equations obtained from A.23) and A.88) when D and
A, respectively, are equated to zero. Since D is a measure of the variance
around the mean particle path, we expect that the reduced equation with
D = 0 should yield deterministic motion with the mean speed с Also, A is
the rate of direction reversal of the particle, so that if A = 0, the particle
should move with speed у either along the path x = yt or x = — yt.
Putting D = 0 in A.23) gives
(O8) *♦,£-..
With the initial condition A.27), the formal solution of A.98) is
A.99) v(x9t) = 8(x-ct)9
where the Dirac delta function 8(x - ci) vanishes for x-cf^O. If we
formally differentiate A.99), we find that it satisfies A.98), and A.99)
equals 8(x) at t - 0. The density function v(x, i) is concentrated on the path
x - ct = 0 so that the particle moves (deterministically) along that path.
Equivalently, the particle moves with the fixed velocity c. The case where
D^O in A.23) (i.e., the diffusion equation) may, therefore, be character-
characterized as representing a random motion around a deterministic path for the
particle given by x = ct in the (x, f)-plane. The equation A.98) is called a
wave equation representing unidirectional wave motion with velocity c, for
reasons that will become apparent when such equations are discussed in
Chapter 2.
Putting A = 0 in A.88) yields the wave equation
24 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
With the initial conditions A.89) and A.90), the formal solution of A.100)
is
A.101) v(x, t) = |S(x - yt) + \S(x + yt) .
This result may be obtained from A.98) and A.99) by noting that with
A = 0, the system A.82) and A.83) reduces to two unidirectional wave
equations
— + — =0
Ж
Ж
with initial data a(x, 0) = /3(x, 0) = \S(x). Recalling that v = a + /3 and the
result A.99) immediately yields A.101). The definition of the Dirac delta
function implies, by way of A.101), that the particle is restricted to move
along the deterministic path x = yt or x = — yt with speed y. The factor \
before each delta function is a consequence of the random choice of
direction at the time / = 0 for the particle (i.e., either to the right or to the
left). Once the particle chooses a direction it must continue to move in that
direction, since A —0 implies that the probability of reversal of direction is
zero. If A 7*0, we have seen in the foregoing that there is some initial
directionality to the particle motion, but this rapidly disappears and the
motion becomes completely random. The wave equation A.100) permits
two directions of motion (to the right and to the left) with speed y.
Another important limiting form of the telegrapher's equation A.88)
occurs if we divide by 2 A and let A-*°° and y^oo but require that y2/A
remain fixed and nonzero. Then, A.88) reduces to A.96). As the rate of
reversal A and the particle speed у both tend to infinity, but with y2l\ — D
as in A.72), the Brownian motion model of this section effectively reduces
to that of the preceding section. However, it is not correct to state that both
models are completely equivalent under the assumptions of large reversal
rates and speeds. For, as we have shown in A.71), the mean square
displacement of the particle predicted by both models agrees only if kt is
large. Thus, even for very large A, if Г is sufficiently small, the results of both
models disagree. Further, if we replace A.88) by A.96), we lose a second
order time derivative term, and A.96) cannot accommodate both initial
conditions A.89) and A.90). The neglect of the term vtt because it is
multiplied by a small parameter represents a singular perturbation of the
equation. Such problems are studied in Section 9.3.
As was done for the Brownian motion model considered in Section 1.1,
we now introduce persistence and reversal probabilities that depend on the
position of the particle. In a further generalization, we assume that these
probabilities depend on the direction of motion of the particle and admit the
THE TELEGRAPHER'S EQUATION AND DIFFUSION 25
possibility that the particle rests at each step. In the present correlated
random walk model, we define the probabilities a(x, i) and P(x,t) as
before. However, p+ and q+ represent the probabilities of persistence and
reversal in direction if the particle is moving to the right, and p~ and q
represent the same probabilities for leftward motion. These probabilities are
given as
A.103) p±{x) - a(x) - X±(x)t , q±{x) = A±(jc)t .
The probability that the particle rests at each step is given byl— p± — <7±==
1 - a, and is taken to be independent of the direction in which the particle is
moving. As т tends to zero, the reversal probabilities both tend to zero. This
characterizes the correlated nature of the random walk. We require as
before that as 5 and т tend to zero, the ratio S/т—>y, the finite particle
velocity.
The coupled system of difference equations satisfied by a(x, i) and /3(x, t)
is easily found to be
A.104) a(x, t+ r) = [1 - p+(x) - q\x)\a{x, t) + p\x - 8)a(x - 5, t)
A.105) /3(jc, t + t) = [1 -p'(x) - q-(x)]P(x, t)+p~(x + 8)f3(x + 8, t)
On expanding the terms in Taylor series around x and t, and letting 8 and т
tend to zero, we obtain
A.106) at + y(<ra)x + \+a - Л~K = О
A.107) pt - y{<rp)x - X+a + X~p = 0 .
To obtain a single equation for the probability density function v(x, t),
defined as in A.85), we assume that the sum of the reversal probabilities A +
and A~ is a constant. More precisely, we set
A.108) \±(x) = \ + V\ilt(x)9
where A is a constant. On manipulating equations A.106) and A.107), as
was done for A.82) and A.83), we obtain
A.109) vtt - y2[<r(<rv)x]x + 2VXy(^v)x + 2Xvg = 0 .
This equation is the analogue of the telegrapher's equation A.88) for the
case of Brownian motion in an inhomogeneous medium. The initial data for
v(x, i) at f = 0 are again given by A.89) and A.90).
26 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
If a = 1 and ф = 0, A.109) reduces to A.88). The difference A" - A+ in
the reversal rates gives rise to the drift term (that is, the first order
jc-derivative term) in A.109). This term arises even if a = 1, in which case
there is a zero rest probability. To relate A.109) to the Fokker-Planck
equation A.53), which was shown to characterize Brownian motion in an
inhomogeneous medium on the basis of an uncorrelated random walk
model, we divide by 2A in A.109) and let A and у tend to infinity, while
requiring that y2/A have a nonzero limit. This yields
A.110) vt = ~(cv)x + \[VD{VDv)x]x ,
where VD(x) = lim(y/V\)<r(x) and c(x) = lim(y/V\)*p(x)<r(x). We observe
that A.110) does not have the same form as A.53), unless the diffusion
coefficient D is a constant.
The difference in the form of the diffusion equations A.53) and A.110) is
well known in the theory of stochastic differential equations. The Fokker-
Planck equation A.53) results on solving generalized Langevin equations by
using the I to stochastic calculus. If the Stratonovich stochastic calculus is
used, an equation of the form A.110) results. Further, it has been shown
that on solving certain generalized Langevin equations in which the random
terms contain correlation effects, and then letting the correlation effects
tend to zero, a diffusion equation of the type A.110) results. This conclusion
is consistent with our derivation of A.110). It is also possible to construct
backward difference equations, telegrapher's equations and their generaliza-
generalizations, in the manner of the results in Section 1.1. These matters will be
considered in the exercises.
Boundary value problems based on the random walk model formulated in
this section are considered in the exercises. A full discussion of initial and
initial boundary value problems for the telegrapher's and related equations
will be given in later chapters. We conclude by noting that the difference
equation formulation of the random walk problems in this section again
indicates, as was the case in the preceding section, that the solution at the
time t depends only on data for the problem given at earlier times. The fact
that the particle speed у is finite implies, as was indicated, that the density
function v(x, t) with the data A.89) and A.90) concentrated at the origin
vanishes when |jc| > yt. This is true for the solutions of the telegrapher's and
wave equations. For the equation A.109) with the data A.89) and A.90),
the region outside of which v(x, f) = 0 has a more complicated form. It will
be determined later in the text. The existence of a finite (maximal) speed of
propagation of disturbances is a fundamental property of hyperbolic partial
differential equations of which the wave and telegrapher's equations are
prototypes. This property is not shared by diffusion equations (which are
equations of parabolic type), as shown in the preceding section. (The
classification of equations into various types will be given in Chapter 3.)
LAPLACE'S EQUATION AND GREEN'S FUNCTION 27
1.3. LAPLACE'S EQUATION AND GREEN'S FUNCTION
In the random walk problems of the preceding sections we were concerned
with finding the probability that a particle located initially at the point x = 0
and moving in a random manner is located at a point x at a later time t.
Consequently, we were dealing with time dependent problems. In this
section we consider three time independent random walk problems in plane
regions. The one- and three-dimensional versions of these problems will be
considered in the exercises.
Each of the problems we study involves a random walk in a bounded
plane region with an absorbing boundary. The first problem examines the
probability that a particle starting at some point in the region reaches a
certain point on the boundary and is absorbed before it reaches and is
absorbed by the remaining portion of the boundary. The second problem is
essentially concerned with the probability that the particle reaches a fixed
interior point before it is absorbed by the boundary. The third problem
involves the determination of the expected or mean time it takes for a
particle starting at some point in the interior until it is absorbed at the
boundary. This is known as a mean first passage time problem. The number
of steps required for the particle to reach the fixed boundary or interior
point is not relevant. That is, the time it takes for the particle to reach that
point is not important so that these problems are time independent. For the
first passage time problem, the possible times until absorption for each
interior point are averaged out, so that there is no explicit time dependence.
The first two problems may be thought to represent stationary or steady-
state versions of appropriate modifications of the problems considered in
Sections 1.1 and 1.2. In fact, the time independent form of the diffusion and
telegrapher's equations (in two space dimensions) is Laplace's equation
which we now derive. We remark that none of these problems is meaningful
if there is a reflecting boundary. For in that case, the particle does not cease
its motion on reaching the boundary. However, we can consider a mixed
boundary condition, in which part of the boundary is absorbing and the rest
of the boundary is reflecting.
Let A represent the bounded region under consideration and dA its
boundary curve. We enclose A and its boundary by a rectangle with sides
* = a, x = b, y~c, and у = d where a < b and с < d. With 8 as the step
length in the random walk, we assume the intervals [a, b] and [c, d] can be
subdivided into the set of points xk = a + 8k and yt = с + 81, respectively,
with 0<£<n,0:</</n, xn = by and ym = d. Each point that lies within A
(i.e., does not lie on dA) is called an interior point. Each interior point
(xk> У1) has four neighboring points in four perpendicular directions. If one
of the neighboring points lies on dA or is exterior to Л, we call it a
boundary point. The points in the rectangle that are neither interior nor
boundary points will not be considered.
28 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
In the first random walk problem we ask for the probability that a particle
starting at an interior point of A reaches the fixed boundary point (xf, y;)
before it reaches any other boundary point. Let v(x, y) be the probability
that the particle starts at the interior point (дс, у) and reaches the boundary
point (jci? y;). We assume that the particle is equally likely to move to any of
its four neighbouring points from the interior point (x, y). Thus the prob-
probability that it moves to any of its four neighbors equals 1/4. Since the
probability that the particle reaches the boundary point (x,, yf) from the
point (x, y) can be expressed in terms of the probability that it moves to any
of its four neighboring points and reaches (x(, ys) from one of these points,
we obtain the difference equation
A.111)
v(x, y) = \[v(x + 5, y) + v(x - 5, y) + i/(x, у + 5) + v(x, у - 8)].
If (x, y) is a boundary point, we have
A.112) Ф,У) =
since when the particle is at a boundary point (х, у) Ф (xi9 y;) it is absorbed
and cannot reach the point (xi9 y;). If one of the neighboring points of (x, y)
is a boundary point, A.112) is to be used in A.111).
In the limit as the step length 5-»0, the number of points in the
subdivisions of [a, b] and [c, d] tend to infinity, and the boundary points
defined actually lie on the boundary В A, Using Taylor's formula gives
A.113) v(x ±8,y) = v(x, y) ± 8vx(x, y) + h82vxx(x, y) + O(83)
A.114) v(x, у ±8) = v(x, y) ± 8vy(x, y) + \82vyy{x, y) + O(83).
Inserting A.113) and A.114) into A.111), dividing by 82, and letting 5->0
yields
which is known as Laplace's equation. The function v(x, y) is now inter-
interpreted as a probability density.
We further assume that arc length s is defined on the boundary ЭА and
that as 5->0, the point (xi9 yf) tends to the (boundary) point (i, y) on dA.
The boundary conditions A.112) are easily found to take the form
A.116) v(x9 y) - 0 , (x, y) e dA , (x, y) * (£, У)
LAPLACE'S EQUATION AND GREEN'S FUNCTION 29
A.117)
For example, if (x, y) is a point in an open interval on the jc-axis that
comprises a portion of the boundary dAy we can set
A.118) v(x,y) = 8(x-£)
in that interval [where S(x - x) is the Dirac delta function] and v(x9 y) = 0
elsewhere on SA.
The second random walk problem we consider asks for the probability
that a particle starting at an interior point (x, y) in the region A reaches a
fixed interior point (£,17) before it reaches a boundary point and is
absorbed. The region A is subdivided as in the first problem and interior and
boundary points are defined as before with the step length again equal to 5.
Since the problem is time independent, and the particle does not stop its
motion once it reaches (£, 17) for the first time, it is possible for the particle
to pass through the point (f, 17) more than once before it reaches and is
absorbed at the boundary. Consequently, if the particle begins its motion at
(£, 17), it has unit probability of reaching (£,17) since it is there already.
However, it can also move to one of its four neighboring points and reach
(£, 17) from there if the neighbor is not a boundary point. Therefore, if we
introduce a function w(x, y) that characterizes the prospects of a particle
reaching (£, 17) from the starting point (x, y), we cannot consider w(x, y) to
be a probability distribution since it may assume values exceeding unity. In
particular, w(£, 17) > 1, as we have seen.
The preceding random walk problem (as well as the first one of this
section) was considered by Courant, Friedrichs, and Lewy in an early paper
on difference methods for the equations of applied mathematics. They
introduced the probabilities that a particle starting at (x, y) reaches (£, 17) in
0, 1, 2,... , n,. . . steps and defined a function of (*, y) that equals the sum
of all these probabilities. This function gives the expected number of steps
that it takes for a particle starting at (x, y) to reach (£,17) before it is
absorbed at the boundary. We shall take the function w(x, y) introduced
above to be defined in this manner. This function satisfies a difference
equation as we now show.
If the point (x, v) is a boundary point we must have
A.119) w(x, y) = 0 , (x, y) a boundary point,
since the particle is absorbed at a boundary point and cannot reach (£,17)
from there. If (jt, y) is an interior point not equal to (£,17), we obtain the
difference equation
30 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
A.120)
w(*, У) = *И* -8,y)+ w(x + 5, y) + w(x, у - 8) + w(jc, у + 8)]
on expressing the expectation w(x, y) of reaching (£, 17) from (x, y) in terms
of the expectation of reaching (£,17) from each of the four neighboring
points (jc ± 5, y) and (jc, у ± 8), on condition that the particle goes from
each of these points to the point (x, y) in the next step. Since the particle is
equally likely to move to each of its four neighboring points, the conditional
probability equals \. If (*, y) = (£, 17), we have
A.121)
J )
since the particle has unit probability of reaching (£,17) considering that it is
there to begin with and it retains the possibility of reaching (£,17) from each
of its four neighboring points as before.
For small 5, A.120) and A.121) take the form
2 2 O(82),
a2w a2w y J
<1122> 17 + 1?
4
on using A.113) and A.114). This shows that w(x, y) satisfies Laplace's
equation at interior points (x, у)¥>(£, 17) as S->0. At the point (£,17), the
right side of A.122) blows up as 5->0. Now as 5—>0, w(x, y) is to be
understood as a density function, so that the integral of w(x, y) over some
small neighborhood of (jc, y) characterizes the property in which we are
interested. If we consider a square with center at (£,17) and with side
proportional to the step length 5, the area of the square multiplied by 4/52
tends to a finite nonzero limit as 5->0. Since the right side of A.122)
vanishes for (x, у)^(£> 17) and its integral over the aforementioned square
has a finite nonzero limit as 5 —> 0, we conclude that it must be proportional
in the limit to the two-dimensional Dirac delta function.
The two-dimensional Dirac delta function with singular point (£,17) is
defined as (see Section 7.2):
A.123) 8(
A.124) I jR 8(x - £)8(у - 7,) dx dy = 1,
where R is any open region containing the point (£, 17). As in Example 1.2,
it follows that for continuous /(*, y), we have the further property
LAPLACE'S EQUATION AND GREEN'S FUNCTION 31
A.125) j jRflx,y)8(x-()8(y-7i)dxdy=f{€,ri).
As 5->0, the boundary points of the discrete problem tend to points on
dA and A.119) now states that w(x, y) vanishes on ЭА. To analyze the
properties of the solution of this boundary value problem, it is convenient to
replace w(x, y) by the so-called Green's function K(x, у; £, tj) for this
problem. The Green's function for Laplace's equation with homogeneous
boundary conditions is defined to be a solution of
A.126) ^ + ^ = -S(x- {ЩУ -V) , (*, У)е A ,
ax ay
which satisfies the boundary condition
A.127) K(x, y;t,v) = 0, (*, у)ЕЭА.
As was shown above, the density function w(x, y) differs from Green's
function only by a constant factor. Thus both w and К are solutions of
(special) inhomogeneous forms of Laplace's equation. (Green's functions
are a useful tool for solving boundary value problems for partial differential
equations and are discussed in Chapter 7.)
The two random walk problems considered so far in this section are not
completely unrelated, since one problem asks for the probability that an
interior point is reached and the other seeks the probability that a boundary
point is reached. In the following example we show that the two problems
are indeed connected, by establishing a relationship between the density
function v(x, y) for the first random walk problem and the Green's function
K(X9 У\ £, v) f°r the second random walk problem.
Example 1.6. Green's Theorem. We formally apply Green's second
theorem to the functions v(x, y) and K(x, y; £, 17). Integrating over the
region A and its boundary dA we have
where V2 = d2ldx2 + d2ldy2 is the Laplacian operator, dldn is a derivative in
the direction of the exterior normal to the boundary dA, and s is arc length
on dA. We have V2v = 0 and V2^ - -8(x - £)8(y -17), in view of A.115)
and A.126). Further, A.116) and A.117) imply that on the boundary
v(x, y)dK(x, y\ ij,r))/dn equals v(x, у)дК(х, у; £,г)Iдп. Combined with
the fact that К vanishes on dA we obtain from A.128)
32 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
0.Ш) «^.
where A.117) and A.125) were used. This shows that if the Green's
function К can be determined, the boundary value problem A.115)-A.117)
can be solved since (£, tj) is an arbitrary interior point in A.
Equation A.129) has the following interpretation. The probability density
v( £, г]) characterizes the probability that particles starting in a neighborhood
of (£, 17) reach the boundary point (x, y). The normal derivative
ЭК(х, у; £, 7])ldn is a measure of the flux of particle density associated with
particles originating near the boundary point (x, y) and reaching the interior
point (£,17). Therefore, —dKldn is a measure of the flux in the reverse
direction and this should essentially equal u(£, 17).
The third random walk problem we consider determines the mean or
expected time it takes for a particle starting at an interior point (jc, y) in the
region A until it is absorbed at the boundary. This yields what is known as
the mean first passage time for each point and we denote it by u{x, y).
Assuming the particle takes steps of length 8 at intervals of time т, and is
equally likely to move to each of its four neighboring points from the point
(x, y) we obtain the difference equation
A.130)
m(jc, y) - т 4- \[u(x ~8,y) + u(x + 8, у) + и(х, у - 5) + u(xy у + 5)].
The expected time until absorption u(x, y) is expressed in terms of the
expected time until absorption for each of the four neighboring points,
multiplied by the probability \ that the particle moves to each of these
points. We must also add the time т it takes for the particle to reach one of
the neighboring points in a single step. If (x, y) is a boundary point,
u(x, y) = 0 since the time until absorption at any boundary point is zero. In
the limit as S->0 and r-»0, while S2/2t = D, we obtain
A.131) \D(uxx + uyy)=-l
on using A.113) and A.114). The inhomogeneous form of Laplace's equa-
equation, of which A.131) is a special case, is known as PoissorCs equatioA. The
boundary condition for и is
A.132) w(*,y) = 0, (x,y)EdA.
Each of the three random walk problems discussed in this section can be
generalized to permit position-dependent jump probabilities as was done in
Sections 1.1 and 1.2. To do so, we assume that the probabilities that a
LAPLACE'S EQUATION AND GREEN'S FUNCTION 33
particle located at the point (jc, y) moves to the right or to the left are given
by Pi and gl9 respectively, while those for upward or downward motion are
given by p2 and q2i respectively. These probabilities are defined as
A.133) Pi(x, y) = \[ай(х, у) + Ьй(х, y)S], i = 1,2 ,
A.134) qAx, У) = Цай(х, у) - Ьй(х, y)8], i = 1,2 .
The at and bt are chosen such that 0 <px + qx + p2 + q2 — 1*
For the first random walk problem, the difference equation A.111) is
replaced by
A.135)
v(x9 y) = [1 - рг - qt ~ p2 ~ 92M*> >0 + Р1Ф + 5> У)
The coefficients of the v terms are all evaluated at (jc, y), and the coefficient
of v(x, y) on the right of equation A.135) represents the probability that the
particle rests at the point (jc, y). On expanding all the terms in A.135) in
Taylor series, as in A.113) and A.114), we easily obtain
A.136) аг(х9 y)vxx + a2(x, y)vyy + 2Ъг(х, y)vx + 2b2(x, y)vy = 0
on dividing by S2 and letting 5 tend to zero. The boundary condition is again
A.116) and A.117).
For the second random walk problem, the difference equation A.120) is
replaced by an equation of the form A.135), with v replaced by w.
However, in view of our definition of w(x, y), the probabilities p( and qt
must all be evaluated at the same points as the expectations w that they
multiply. The same is true for the modified form of A.121). As a result,
when all the terms are expanded for small 8, and we let 5 tend to zero, we
obtain for the Green's function K(x, y; £, 17) associated with this problem,
the equation
A.137) (axK)xx + (a2K)yy - 2(ЬгК)х - 2(b2K)y = ~8(x - f )8(y -1,).
Again, К must vanish on the boundary. The differential operator acting on
К in A.137) is the adjoint of the operator acting on v in A.136), as follows
from our discussion in Section 3.6. This observation yields a relationship
between v and К similar to given in Example 1.6, and based on a
generalization of Green's theorem. This will be considered in the exercises.
Similarly, the generalization of the third random walk problem considered
in this section will be examined in the exercises.
We conclude this section with a number of comments regarding the
34 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
properties of boundary value problems for Laplace's equation and the
related equations derived in this section, that are suggested by the difference
equation treatment considered above.
The difference equation A.111) together with the boundary condition
A.112), when applied to all interior points in the subdivision of A, yields a
simultaneous system of equations for the functions v(xk, yt) where (xk, yt)
is a typical interior point. All the boundary data play a role in the solution
for v at the interior points. The same is true for the other equations
considered in this section. This contrasts with the situation encountered in
the preceding sections. Though the solution was required for all f >0, the
solution for fixed t did not depend on data for later values of t.
Further, the difference equation A.111) characterizes the value of v(x, y)
at the center of a square as the mean or average of its values at the four
vertices. With (jc, y) as its center, the square has the points (x ± S, y) and
(jc, у ± 8) as its vertices. This mean value property carries over to functions
v(x, y) satisfying Laplace's equation. It is shown later than any solution
v(x, y) of A.115) equals the average of its values on a circle with center at
(x, y) as long as the circle is interior to the region A.
The above mean value property for the difference equation implies that
the maximum and minimum values of v(x, y) must be assumed at boundary
points of A. This follows since v(x, y) is the average of its neighboring
values and, therefore, cannot be greater than or less than those values. This
maximum and minimum principle will be shown to be valid for certain
boundary value problems for Laplace's equation. In fact, it follows from
A.111) that if the maximum or minimum of v(x, y) is attained at an interior
point of Ay v(x, y) must be constant throughout Л. For if (x, y) is an
interior maximum or minimum point, v(x, y) cannot be represented as the
average of its four neighboring values if any one of them is, respectively,
smaller or larger than the value of v at (*, y). This property carries over to
solutions of Laplace's equation, as will be shown, and is known as the strong
maximum and minimum principle.
The representation A.135) for v(x, y) implies that it also satisfies a
maximum and minimum principle. Consequently, we expect such a principle
to be valid for solutions of A.136), and this will be demonstrated. Further
questions relating to the positivity of solutions of Poisson's equation A.131)
and the positivity of the Green's function determined from A.126) and
A.127), will also be examined later in the text.
EXERCISES FOR CHAPTER 1
Section 1.1
1.1.1. Verify, by direct substitution, that the function v(x, t) given in A.31)
is a solution of the diffusion equation A.23).
EXERCISES FOR CHAPTER 1 35
1.1.2. Show that the function v(x, t) given in A.31) satisfies the following
conditions when t >0,
(а) /Гв v(x, t)dx = \
Conclude thereby that v(x, t) is a probability density function with mean ct
and variance Dt.
Hint:
e~~x dx = Vtt .
1.1.3. Again considering the function v(x, t) of A.31), show that the limit
as t | 0 of u(jt, t) is zero when x Ф 0. Conclude from this and the result of
Exercise 1.1.2(a) that lim,i0 u(jc, i) = 8(x), so that the initial conditon A.27)
is satisfied.
1Л.4. Show that if />0 the functions
v±(x,t) =
satisfy the diffusion equation A.23) (with с = 0), the initial condition A.27),
and, respectively, satisfy the boundary conditions A.46) and A.50) with
с = 0. (The construction of these solutions may be based on the method of
images which is discussed in Section 7.5).
1.1.5. Derive the diffusion equation in two dimensions
by constructing an appropriate difference equation in the manner leading to
A.20), but assuming the particle is equally likely to move to the points
(x ± 8, y) and (x, у ±8) from the point (jc, y). Use the Taylor series and
define D = limT^0F2/2r) to obtain the diffusion equation from the differ-
6-»0
ence equation.
1-1.6. Derive the Fokker-Planck equation in two dimensions
*>t = -(cxv)x - (c2v)y + И(ад„ + (D2v)yy],
by assuming the probabilities for motion to the right or to the left and up or
down are given as in A.133) and A.134), and constructing an appropriate
difference equation. With D given as in Exercise 1.1.5, and the coefficients
in the Fokker-Planck equation defined analogously to those in A.53),
obtain the differential equation from the difference equation as 5 and r tend
to zero.
36 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
1.1.7. With the probabilities defined as in Exercise 1.1.6, derive a backward
difference equation and the corresponding backward Kolmogorov equation
in two dimensions.
1.1.8. Derive the diffusion equation in three dimensions
with £> = 1ш1т-И)E2/Зт), by using assumptions similar to those given in
Exercise 1.1.5.
1.1.9. Generalize the results of Exercise 1.1.6 and derive a difference
equation and a Fokker-Planck equation for the three-dimensional case.
1.1.10* With the probabilities given as in Exercise 1.1.9, derive a backward
difference equation and a backward Kolmogorov equation for the three-
dimensional case.
1.1.11. With x = k8 and t = пт, where к = 0, ±1, ±2,. . . , and n = 0, 1,
2,. . ., v(x, t) = v(k8, пт) gives the probability that the particle is at the
point k8 at the time пт.
(a) Express the difference equation A.20) in terms of the function
v(k8, пт).
(b) Given the initial conditions v@,0) = 1 and v(k8, 0) = 0 for
fc 7*0, solve the difference equation recursively for all к with
/z = l, 2, 3, and 4.
(c) Show that for each of the foregoing values of n we have
(d) Show that the solution of the initial value problem of part (b) is
v(k8, пт) = <
0, otherwise.
Using this solution, verify the results of part (b).
1.1.12. With a = 1, D = constant and c(x) = -a>x in A.51) and A.53), the
Fokker-Planck equation describes Brownian motion with the particle sub-
subjected to an elastic "restoring" force. As |jc| increases, the probability that
the particle moves further away from the origin decreases.
(a) Show that this "restoring" property follows from the definition
of p(x) and q(x).
(b) Using the methods of Example 1.3 and retaining the initial
condition A.27), determine the first three moments of the
random variable x for this case.
(c) Obtain the expectation and the variance using the results of part
(b) and compare them with the expectation and variance given
in the text for the Brownian motion of a "free" particle.
EXERCISES FOR CHAPTER 1 37
1.1.13. For the random walk problem modeled by A.52), using the assump-
assumptions made in the text, show that at a reflecting boundary, A.47) is replaced
by
- 5, t) + p(l)v(l, t)
Show that in the limit as 8 and т tend to zero this yields
i(Dv)x-cu=0
as the boundary condition for the Fokker-Planck equation at a reflecting
boundary point x = /.
1.1.14. (a) Since the particle must remain fixed at an absorbing boundary
point once it reaches that point, show that the boundary condi-
condition for the backward Kolmogorov equation A.55) at an absorb-
absorbing boundary point is v = 0.
(b) Show that at a reflecting boundary point, the boundary condi-
condition associated with the backward difference equation A.54) is
v(l, t) = pv(l, t + t) + [1 - p ~ q]v(l, t + t) + qv(l - 5, t + r) ,
with p and q both evaluated at x = L Verify that as 5 and т tend
to zero, this tends to
This is the boundary condition for the backward Kolmogorov
equation A-55) at a reflecting boundary.
1.1.15. If x = -/ and x = / are reflecting boundary points, by integrating the
Fokker-Planck equation A.53) from -/ to +/, show that the integral of
v(x, t) over that interval remains constant. Thus, the probability that the
particle remains in the interval is fixed for all time since no absorption takes
place. This probability is initially equal to one, and it remains at one for all
time.
1.1.16. (a) Show that at an absorbing boundary in the two- or three-
dimensional case, we must have i>=0, both for the Fokker-
Planck equation and the backward Kolmogorov equation,
(b) If the particle is equally likely to move in all directions in the
two- and three-dimensional cases, show that the normal deriva-
derivative of v at the boundary (that is, the derivative of v in the
direction of the exterior unit normal at the boundary of the
region) must vanish at each reflecting boundary point. This
boundary condition applies for both the Fokker-Planck and the
backward Kolmogorov equations.
38 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
1.1.17. Show that the function
with constant с and D, is a solution of the diffusion equation A.23) and the
time reversed backward Kolmogorov equation
vt = cvy+ \Dvyy.
(This equation results from A.55) if we replace t by -t and x by y.) For
both equations the initial condition is v(x, 0; у) = д(х - у).
Section 1.2
1.2.1. Verify that A.65) sums to A.66).
1.2.2. Show that the limit in A.69) is correct.
1.2.3. Obtain the results of Example 1.5 for the indicated moments by
solving the given initial value problems for the ordinary differential equa-
equations.
1.2.4. Consider the solution A.31) of the initial value problem A.27) for
the diffusion equation A.23). Using the approach of Exercise 1.1.3, show
that the limit of this solution as Z>—»0 is the delta function (L99).
1.2.5. (a) Let v = Ve~kt in the telegrapher's equation A.88) and show that
V satisfies the equation
(b) Put z = \Jy2t2-x2 and show that V(x91) = W(z) satisfies the
ordinary differential equation
[Note that the region z ^ 0 is where the solution v(x, t) was
shown to be nonzero in the text.]
(c) Let W(z) = A/Vz)w(z) and show that w{z) satisfies the equation
w"(z) + [1 /4z2 - (Л V)]w(z) = 0 .
(d) Observe that for \x\ < yt, as уt-*■<*> we have
±.,
EXERCISES FOR CHAPTER 1 39
and, consequently, z—>°° as y/—»«>. Thus we can approximate
the equation for w(z) when yt is large, by
Using A.72), show that as yf-»oo we have approximately
v(x9t) «s (a/y/yi)e~(x /2Dt) where a is an arbitrary constant. This
has the form of the solution A.31) of the diffusion equation with
c = 0.
1.2.6. Put x = k8 and * = лт, as in Exercise 1.1.11 and set a(x, t) =
a(k8, пт) and j8(x, /) = 0(fc5, лт). Show that the difference equations A.75)
and A.76) take the form
(a[k8, (n + 1)t] =pa[(k - 1M, пт)] + qp[(k - 1M, пт]
t ДО, (n + 1)т] =pj8[(* + 1M, пт] + ?а[(* + 1M, nr],
with the initial conditions a@,0) = /3@,0) = \, and a(A:5,0) = j8(A:5,0) = 0,
for A:^0.
1.2.7. (a) Solve the difference equations of Exercise 1.2.6 recursively for all
к and for n = 1, 2, 3, and 4.
(b) Show that £*=_«, a(A:5, от) = Е^_то )8(А:5, пт) = J for the above
values of n.
1.2.8. (a) Solve the difference equations of Exercise 1.2.6 for a(±n8, пт)
and C(±п8, пт). Show that a(n89 пт) = P(-n8, пт) = \pn~l and
а(-п5, пт) = )8(п5, azt) = 0 for n > 1.
(b) With x = ±n8 and * = пт so that x = ±E/т)*, show that as n-► oo?
5~»0, t-»0, and 5/т-^у, we have a(y^, 0~^ 5e"Ar»
P(-yt,t)-> \e~kt- These results are in agreement with A.97)
since i; = а + )8.
1.2.9. With v{k8, пт) = а(к8, пт) + j8(A:5, пт) show that v satisfies the differ-
difference equation
v[k89 (n + 1)t] - pv[(k - 1M, пт] + pv[(k + 1M, пт]
with the initial conditions u@,0) = l, y(A:5,0) = 0, A:^0, and v(-8, т) =
u(s> т) = §, u(A;5, т) = 0, A:^ ± 1. [We observe that with p = q = \, the
correlation coefficient p-p-q vanishes and the difference equation re-
reduces to A.20). If p =p - q t*0, the probability v(k8, пт) depends not only
on the location of the particle at the previous step but the previous two
steps.]
1.2.10. (a) Solve the difference equation of Exercise 1.2.9 for v(k8, пт) with
n = 2, 3 and 4, and for all /с.
40 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
(b) Solve the equation for v(±n8, лт), for л>2.
(c) Show that v(k8, nr) = 0 for \k\>n.
1.2.11. Let A+ and A~ be constants and cr = 1 in A.103), and A.106) and
A.107). Putting i; = a + 0, show that v(x91) satisfies the telegrapher's
equation
1.2.12. (a) Apply the method of Example 1.5 to show that as f-»<», the
expectation (x) and variance V(x) of the continuous random
variable characterized by the equation in Exercise 1.2.11 are
given as
(b) Compare the results of part (a) with those for the Brownian
motion model of Section 1.1. Explain why the coefficient of t in
the expression for (x) may be characterized as a "drift" vel-
velocity.
(c) Show that V(x) « у V for small t.
1.2.13. Generate the results of Exercise 1.2.12 for the expectation and
variance by proceeding as follows. Assuming x and t are large and of the
same order of magnitude (say, x = xle and t = tie where 0 < e <l 1), we can
neglect the second derivative terms in.the equation for v(x, t) given in
Exercise 1.2.11, and obtain in a first approximation,
Using this equation to obtain an expression for vlt, show that in the next
approximation we obtain the following diffusion equation for v(x, t):
L Л ~г А
Demonstrate that this equation implies the results of Exercise 1.2.12(a).
1.2.14. With the persistence and reversal probabilities p± and q± defined as
in A.103), obtain the difference equations
a(jc, 0 = [1 - p+ - q+]a(x91 + т) + p+a(x + 8, t + т) + q+p(x - 5, t + T)
P(x, t) - [1 -p" - tf~]j3(x, Г + t) + ^~a(x + 8, t + т) +р"Ж^ ~ 8>* + r) ,
EXERCISES FOR CHAPTER 1 41
where the p± and q~ are all evaluated at the point x. These equations
determine the probabilities that right and left moving particles are located at
the point x at the time /, in terms of their possible locations at the time t + т.
1.2.15. As 8 and r tend to zero, show that the system of difference
equations derived in Exercise 1.2.14 yields the system of differential equa-
equations
at + y<rax - X+a + A+j8 = 0
Pt - yapx + k~a - \~p = 0 .
This system is the analogue of the backward Kolmogorov equation for the
model of Brownian motion considered in this section.
1.2.16. Put v = a + p and let A+ and A" be defined as in A.108). Show that
the system of Exercise 1.2,15 yields
vtt - y2a{avx)x - 2л/1уафих - 2kvt = 0 .
This is the backward form of the equation A.109). It is to be solved for t < 0
in terms of the data A.89) and A.90) given at f = 0. It is also the adjoint
equation of A.109). (See Section 3.6.) We note that if a = 1 and ф = 0, this
is known as the backward telegrapher's equation.
1.2.17. Carry out the limit described in the discussion immediately preced-
preceding equation A.110) for the equation of Exercise 1.2.16 and obtain
vt = -cvx-\VD(VDvx)x.
This is the backward form of the equation A.110), and is again its adjoint.
1.2.18. Put a = 1 in A.103) and ф = -х in A.108). Show that the limiting
diffusion equation A.110) has the form of the equation given in Exercise
1.1.12. Explain why the behavior of the reversal rates A+ and A" for x <0
and дс>0, suggests that this problem is similar to that of Exercise 1.1.12
where the particle is subjected to an elastic restoring force.
1.2.19. Consider the motion of a particle in the semi-infinite interval x < I,
and let x = / be an absorbing boundary.
(a) Show that for the discrete problem p(l, t) = 0, since the particle
cannot reach that point from the right. For a particle that moves
to the right we have
(b) In the limit as 8 and т tend to zero, conclude that a(/, t) =
P(l, r) = 0 are the required conditions at an absorbing boundary.
1.2.20. Let the particle move in the interval x < I and assume that x = / is a
reflecting boundary.
42 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
(a) Show that for the discrete problem we have
№, t + r) = p\l)a(l, t) + A - p{l) - <Г(/)H(/, г)
a(l,t + r) =p+(/- в)а(/- 5, 0 + *"(/- fiH(/ " 5, r)
(b) In the limit as 5 and r tend to zero, obtain the equations
a(/, 0 = jB(/, 0 , a, + y(<ra)x = (A" - A + )a
as the boundary conditions at a reflecting boundary.
(c) Show that if A+ and A" are defined as in A.108) and A and у
tend to infinity, the boundary conditions of part (b) assume a
form that differs from that given in Exercise 1.1.13 for the
Fokker-Planck equation in the case of a reflecting boundary.
Section 1.3
1.3.1. (a) Show that in the one-dimensional case, if the particle is equally
likely to move to the right or to the left, A.111) is replaced by
(b) Show that in the three-dimensional case, when the particle is
equally likely to move to any of its six neighboring points, we
obtain
v(x9 y, z) = %[v(x + 8, y9 z) + v(x - Й, y, z) + v(x, y~8,z)
+ v(xf у + 5, z) + v(x, y, z - 8) + v(x, y, z + 8)].
1.3.2. Show that the limiting differential equations for Exercise 1.3.l(a) and
(b) are
(a) и„«0;
(b) w« + iiyy + i>«=0,
respectively.
1.3.3. Consider the interval O^x^l. Let v(x) be the probability that the
particle reaches the boundary point x = 1 and is absorbed before it reaches
and is absorbed at the boundary x = 0.
(a) Show that the appropriate boundary conditions in this case are
u@) = 0 and u(l) = l.
(b) Show that v(x) = jc, which satisfies the boundary conditions of
part (a), is a solution of both the difference equation of Exercise
1.3.1 (a) and the (one-dimensional) differential equation of Exer-
Exercise 1.3.2(a).
EXERCISES FOR CHAPTER 1 43
1.3.4. We define the Green's function K(x; £) in the one-dimensional case,
for the interval 0< x < 1, to be the solution of the boundary value problem
dx
_
(a) By integrating the equation for K(x; f) over a small neighbor-
neighborhood of x = £ show that
дх
= -1
(i.e., the jump in dKldx at x = f is -1). Assuming that K(x; f)
is continuous at * = f, define A^x, £) as #(x; £) = ^(x; £),
0 < Jt < f, and K{x\ f) = ^(^5 f), f ^ x < 1. Show that we then
obtain the equations
and the supplementary conditions
(b) Show that the solution of part (a) is
Кг(х; f) = A - ()x , K2(x; {) = A - x)f .
1.3.5. (a) Apply a one-dimensional form of Green's theorem A.128) to
show that with v(x) and K(x; £) defined as in the two preceding
exercises we have
(b) Demonstrate that the result of part (a) is correct by using the
explicit forms for the functions v(x) and K(x; f).
1.3.6. Solve the difference equation A.111) in the square 0<Jt<3 and
0 :s у < 3 at the points with coordinates x = 1, 2 and у = 1, 2, assuming 5 = 1
and (xif^) = @,1) in A.112).
1.3.7. Show that v(x, y) = 1 is a solution of the boundary value problem for
Laplace's equation in the square 0^jc<1, O^y^l, with the boundary
conditions v(x, 0) = 1 and dvfdn = 0 on the other three sides of the square.
44 RANDOM WALKS AND PARTIAL DIFFERENTIAL EQUATIONS
Noting that dvfdn^Q corresponds to a reflecting boundary condition,
interpret the solution.
1.3.8. (a) Show that in the one-dimensional case, if the particle is equally
likely to move to the right or to the left, the mean first passage
time u(x) satisfies the difference equation
u(x) = т + \[u(x - 8) + u(x + 8)].
(b) Show that in the three-dimensional case, if the particle is equally
likely to move to any of its six neighboring points, the mean first
passage time u(x, y, z) satisfies the equation
u(x, y, z) - т + \[u(x + 8, yf z) + u(x - 5, yf z) + w(x, у + 5, z)
+ u(x, у - 5, z) + u(x, y9 z + 5) + n(x, y9 z - 8)].
1.3.9. Show that if lim 82/r = D in part (a) and lim 52/Зт = Z> in part (b) of
Exercise 1.3.8, the limiting differential equations for the one- and three-
dimensional difference equations are, respectively,
(a) iZ>n"(*) = -l,
(b) *Я[и„ + и„+ и„] = -1.
1.3.10. Show that in the one-dimensional case, the equations A.136) and
A.137) take the form
L*[K\ - (aK)xx - 2{bK)x = -8(* - *).
1.3.11. Demonstrate that with the operators L and L* defined as in
Exercise 1.3.10, we have
vL*[K] - KL[v] = [v(aK)x - aJCvx - 2bKv]x .
1.3.12. Given the interval 0 < * <= 1, with y@) = 0, v(l) = 1, and K(Q; f) =
ХA;{)-0, integrate the expression given in Exercise 1.3.11 over the
interval. Use the equations in Exercise 1.3.10 to conclude that
This is the analogue of the result A.129) and Exercise 1.3.5. A similar result
is valid in the two- and three-dimensional cases.
1.3.13. Write down the equations that correspond to A.136) and A.137) in
the three-dimensional case.
EXERCISES FOR CHAPTER 1 45
1.3.14. Show that the mean first passage time u{x) for an inhomogeneous
one-dimensional medium, with p and q given as in A.51), satisfies the
equation
where с and D are defined as in A.53).
1.3.15. Consider the interval 0<x<l, and the mean first passage time
problem of Exercise 1.3.9(a).
(a) Find the solution w(x), if w@) = w(l) = 0.
(b) Determine u(x), if w@) = u'(l) = 0. In this case, the boundary
at x - 0 is absorbing and that at x = 1 is reflecting.
(c) If w'@) = м'A) = 0, in which case both boundary points are
reflecting, show that the problem has no solution.
1.3.16. Use Green's theorem A.128) with v(x, y) = l, to show that the
Green's function К that satisfies A.126) and whose normal derivative
vanishes on the boundary of A, does not exist. That is, in the case of a
reflecting boundary, the Green's function problem has no solution.
1.3.17. Show that u(x, y) = (x + /y)n, where i = V—1, is a solution of Lap-
Laplace's equation A.115) for all integers w>0.
1.3.18. Consider v = (x + iyf and write it in the form v(x, y) = f(x, y) +
/g(jc, y), where / and g are the real and imaginary parts of the function
v(x, y). Show that/(jc, y) and g(x, y) have no relative maxima or minima in
the (jc, y)-plane and that both satisfy Laplace's equation.
1.3.19. Show that v(x, y, z) - (x + iy cos в + iz sin 6)n is a solution of Lap-
Laplace's equation in three dimensions (see Exercise 1.3.2) for all integers n >0
and for О<0<2тг.
1.3.20. Write v = (x + iy cos в + iz sin вJ in the form v = F(x, y, z) +
iG(x> У> z), where F and G are real valued functions. Show that F and G are
solutions of Laplace's equation and that they have no relative maxima and
minima in (x, y, z)-space.
2
FIRST ORDER PARTIAL
DIFFERENTIAL
EQUATIONS
2.1. INTRODUCTION
This chapter initiates our study of methods for solving partial differential
equations. We begin with single first order equations, that is, equations in
which the highest derivative of the dependent variable is of first order. We
restrict our discussion to problems involving two independent variables and
deal almost exclusively with the method of characteristics for solving these
problems. The presentation is thereby simplified, but as shown in the
exercises, this method is easily extended to handle equations with more
independent variables. Other solution methods for first order equations
exist, but apart from the method of the complete integral, they are not
discussed here.
First order equations can occur directly as models for physical processes
or can arise as approximations to higher order equations or systems of
equations. Thus the unidirectional wave equation A.98), which is studied in
Example 2.2, is an approximation to the diffusion equation of Section 1.1, in
case the diffusion coefficient is negligibly small. The quasilinear wave
equation B.65), or more generally B.104), serves as an approximation to
Euler's equations of fluid dynamics (see Example 8.9) or represents^ simple
model for traffic flow. The nonlinear eiconal equation B.147) plays an
important role in the theory of geometrical optics and represents an
approximation to wave optics as shown in Section 10.1. In fact, linear,
quasilinear, and nonlinear first order equations play significant roles in the
approximation methods for higher order equations which are discussed in
Chapters 9 and 10.
The formulation of mathematical models can lead naturally to systems of
first order equations, as in the case of the correlated random walk of Section
46
INTRODUCTION 47
1.2 and the system A.82), A.83) or as in Euler's hydrodynamic equations of
Example 8.9. Alternatively, as shown in Example 2.1, systems can be
constructed through the reduction of higher order equations. Conversely,
linear systems of equations with constant coefficients can always be reduced
to single higher order equations, as was shown in Section 1.2 for the special
system A.82), A.83) and as can be demonstrated in general. The methods
of this section cannot be used for the solution of first order systems except in
special cases, such as in the derivation of d'Alembert's solution of the wave
equation in Example 2.4. Although we do not present any general methods
for solving first order systems in this text, specific problems involving linear
and nonlinear systems will be studied.
Example 2.1. Reduction of Higher Order Equations to Systems. The
wave equation A.100) can be written in the factored form
B.1) ^ - Г2 д~Л = (*? - У**> = V, + ydM ~ УЫ V = 0,
at dX
where the differential operators Sx9 St, and Bt ± ydx are defined in an obvious
way. If we set (St ~ ydx)v = u, B.1) implies that (<?, + удх)и = О and we
obtain the system
r dv dy
Though this system is coupled, it is possible to solve the second equation
independently of the first. The solution u(x, t) can then be introduced into
the first equation and v(x, t) can be determined. A solution of the system
B.2) is found in Example 2.4 using the method of characteristics.
The damped wave equation A.88) after a transformation of the depen-
dependent variable of the form w = ektv becomes
B-3) —г - У2 —т ~ Л2н> = 0 .
дг дх
This factorization method can be applied to yield
f dw dw
B.4) p7 " Г ** " U
du du. _ 2
In this case both first order equations must be solved simultaneously, and
48 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
the methods developed in this chapter do not lead to a straightforward
solution of B.4). (A direct factorization of equation A.88) is presented in
the exercises.)
If we formally apply the factorization technique to Laplace's equation
A.115), we obtain
B.5) ^ + ^ = {dl + <?> = (dx + idy)(dx - idy)v = 0,
oX ay
where i = V^T. Proceeding as for the wave equation leads to a first order
coupled system with real and imaginary coefficients. Since we generally
require real solutions of Laplace's equations, this reduction procedure is
unsatisfactory. A system of equations more customarily associated with
Laplace's equation can be obtained by putting dvldx = и and dvldy = w.
Assuming the mixed partial derivatives of v are equal, we easily obtain on
using B.5),
B.6)
dw _ ди
дх ду
dw __ ди
ду " дх
These are the Cauchy-Riemann equations which are well known in the
theory of complex variables. Clearly, they must be solved simultaneously.
For the diffusion equation A.23), no simple factorization is possible.
However, if we put и — dvldx, we obtain the system
<2-7) \Z lau
Again, this system must be solved simultaneously, and no obvious benefit
appears from writing the diffusion equation in system form.
2.2. LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
The most general first order linear partial differential equation has the form
B.8) a{x, О Ц + Kx, t) j-t = c(x, t)v + d{x, t) ,
where a, b, c, and d are given functions of x and t. Unless stated otherwise,
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 49
these functions are assumed to be continuously differentiable. At each point
(x, t) where the vector [a, b] is defined and nonzero, the left side of B.8) is
(essentially) a directional derivative of v(x, t) in the direction of [a, b]. The
equations
dx , ч dt
B.9) -^ = <*> 0 > ^ = 6(*> 0 >
determine a family of curves x = x{s), t=t{s) whose tangent vector
\x'(s)i t'(s)] coincides with the direction of the vector [a, b] at each point
where [a, b] is defined and nonzero. Therefore, the derivative of v(x, t)
along these curves becomes
dv _ dv[x(s), t(s)] dv_ dx dv_ A
BЛ0) ds~ ds dx ds dt ds
on using the chain rule and B.8) and B.9).
The family of curves x = x(s), t~ t(s), v = v(s) determined by the solu-
solution of the system of ordinary differential equations B.9), B.10) are called
the characteristic curves of the partial differential equation B.8). Since the
equations B.9) can be solved independently of B.10), the curves in the
(x, f)-plane determined from B.8) are occasionally also referred to as
characteristic curves or characteristic base curves. The approach we develop
to solve B.8) by using the solutions of B.9), B.10) is called the method of
characteristics. It is based on the geometric interpretation of the partial
differential equation B.8).
The existence and uniqueness theory for ordinary differential equations,
assuming certain smoothness conditions on the functions a, by c, and d>
guarantees that exactly one solution curve (x(s), t(s), v(s)) of B.9), B.10)
(i.e., a characteristic curve) passes through a given point (jc0, f0, v0) in
(x, t, v)-space. As a rule, we are not interested in determining a general
solution of the partial differential equation B.8) but rather a specific
solution v = v(xy t) [i.e., a surface in (jc, t, v)-space] that passes through or
contains a given curve С This problem is known as the initial value problem
for B.8).
The method of characteristics for solving the initial value problem for
B.8) proceeds as follows. We assume the initial curve С is given parametri-
cally as
B.11) x = x(t), *=*(t), v = v(t)
for a given range of values of the parameter т. The curve may be of finite or
infinite extent and is required to have a continuous tangent vector at each
50
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
point. Every value of т fixes a point on С through which a unique
characteristic curve passes. The family of characteristic curves determined
by the points of С may be parametrized as
B.12)
x = x(s, т) , t = t(s, t) , v = v(s, t) ,
with 5 = 0 corresponding to the initial curve С That is, we have x@, т) =
х(т), t@, t) = <t), and u@, r) = v(j).
The equations B.12), in general, yield a parametric representation of a
surface in (jc, t, v) -space that contains the initial curve С Assuming the
equations x = x{s,t) and t=t(s, т) can be inverted to give s and т as
(smooth) functions of x and f, and this is the case if the Jacobian A(s, т) =
xstT - tsxT ^0 at C, these functions can be introduced into the equation
v = v(s, t). The resulting function v = V(x, i) satisfies B.8) in a neighbor-
neighborhood of the curve С in view of B.10), the initial condition B.11) {i.e.,
V[x(t), /(t)] = u(t)}, and is the unique solution of the given initial value
problem. The smoothness requirements placed on the functions a, b, c, and
d in equation B.10) imply that V(x, t) must be continuously differentiable
near the curve С Figure 2.1 shows how the integral surface is constructed in
terms of the initial and characteristic curves.
If the foregoing method does not lead to a solution, the initial value
problem may not have a solution at all or it may have infinitely many
solutions. The latter situation arises if the initial curve С is itself a
characteristic curve, in what is known as a characteristic initial value
Figure 2.1. Construction of the integral surface.
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 51
problem. It may also be that the integral surface is not differentiate or is
discontinuous along some curve.
In the following we illustrate the use of the method of characteristics in
solving initial value problems for linear first order equations and show how
problems with nonsmooth or nonunique solutions can arise and how to deal
with them. The role of characteristics in relation to the nonexistence or
nonuniqueness of solutions of initial value problems is re-examined from a
different point of view in Section 3.2. Finally, we observe that although the
method of characteristics reduces the problem of solving first order partial
differential equations to that of solving a system of ordinary differential
equations, the solution of this (generally) nonlinear system is most often not
an elementary task.
Example 2.2. Unidirectional Wave Motion. It was stated in Chapter 1
that A.98), that is,
B.13, *♦,£-,.
represents unidirectional wave motion. (The coefficient с is assumed to be
constant.) We examine the initial value problem for B.13) with the initial
condition at t = 0
B.14) v(x,0) = F(x),
where F(x) is a given function.
To apply the method of characteristics, we parametrize the initial curve С
as follows,
B.15)
The characteristic
B.16)
х = т,
equations
[i
t = 0,
•e., B.9),
dt
ds~
i
B-
v = F(r) .
10)] become
ds
Solving B.16) subject to B.15) with 5 = 0 corresponding to the initial curve,
gives
BЛ7) x(s, r) - cs + r , t(s, t) = 5 , v(s9 t) = F(r) .
using the first two equations to solve for s and r as functions of x and t
yields
B.18)
T = X — Ct .
52 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Substituting this result in the equation for v in B.17), we obtain
B.19) v(s9 t) = F[t(x, t)] = F[x - a].
Clearly, if F(x) is differentiable, the solution v(x, t) = F{x - ct) satisfies
B.13) as well as the initial condition B.14). The formal solution v(x, t) =
8(x-ct) of A.98) given in Chapter 1 follows from B.19) since u(*,0) =
8(x) in that problem.
If we consider the initial function v(x9 0) = F(x) to represent a waveform,
the solution v(x, t) = F(x - ct) shows that a point x for which x - ct =
constant, will always occupy the same position on the wave form. If с > 0,
that point x moves to the right with the speed dxldt = c. Since x is a typical
point, we see that the entire initial wave form F(x) moves to the right
without changing its shape with speed с (if с < 0, the direction of motion is
reversed). Consequently, B.13) characterizes unidirectional wave motion
with velocity c.
The characteristic base curves for B.13) are given by x — ct = r in view of
B.18). For each value of the parameter т, B.17) shows that v is constant on
the characteristic base curves. This implies that the initial data are trans-
transmitted along the characteristic base curves, in the sense that whatever the
value of v is at t = 0 at some point jc, it is retained at all points x that lie on
the characteristic curve through that initial point. As a result, the initial data
are transmitted at the characteristic velocity dxldt = с (to the right or to the
left) as t increases from zero.
The expression v-F(x-ct) obtained in B.19), represents a general
solution of B.13) since F(z) is an arbitrary function. It can be used, as an
alternative to the method of characteristics, to solve initial value problems
for B.13). For example, if we specify that v = f(t) on the curve x = h(t), we
obtain v[h(t), t]=f(t) = F[h(t) - ct]. With z = h{t) - ct, there exists a dif-
differentiable inverse function t~g{z) if zf(t) = h'(t) - c#0, that is, if the
initial base curve x = h(t) is not tangent to a characteristic base curve at any
point. Then the unique solution is v(x, t) — f[g(x — ct)].
If h'(t) — с along the entire curve, we must have x — ct = constant, so that
the initial base curve is a characteristic. If we denote the constant by a, we
have v =/(/) = F(a) on the initial curve, so that the problem has no splution
unless f{t) = A = constant. If this is the case, this constitutes a characteristic
initial value problem for B.13), since it follows from B.16) that the initial
curve in (jc, t, v)-space is characteristic. The solution to this problem is not
unique, for any differentiable function F(z) with F(a) = A yields a solution
v = F(x — ct) of the problem.
If h'(t) = с at only a finite number of points, a unique continuous inverse
function t - g(z)y as defined above, can be found. However, g(z) will not be
differentiable at these points. As a result, the solution v =f[g(* - ct)] will
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 53
be continuous but not differentiable along the characteristic base curves
through these points.
Example 2.3. Unidirectional Wave Motion: Specific Initial Data. Con-
Consider the unidirectional wave equation B.13) with the initial value и = f(t)
on the parabola x = h(i) = t\ Since h'{t) = 2t=c when t = c/2, we see that
the parabola is tangent to the characteristic base curve x - ct = -c2/4 at the
point (x, t) = (c2/4, c/2). Using the general solution v = F(x - ct) we obtain
F(t2 - ct) = f(t) on applying the initial condition. With z = t2 - ct, the
inverse function t = g(z) is defined as t = g_(z) = \c - (z + c2/4I/2 for
f < c/2 and * - g+(z) =\c + (z + c2/4I/2 for f > c/2. This function is con-
continuous for z ^ - c2/4 but it is not differentiable at z = -c2/4.
Thus, on the characteristic base curves that intersect x = t2 with f < c/2
we have the solution у = v_(x, t) =f[g-(x - ct)], while those curves that
intersect the parabola with t> c/2 yield v = v+(x, t)=f[g+(x - ct)]. How-
However, in the region x~- ct^ — c2/4, where the solution to the initial value
problem is defined, each characteristic intersects the parabola twice except
at the point (x, t) = (c2/4, c/2) where the characteristic is tangent to the
parabola. Because the initial value u=f(t) is assigned arbitrarily and
v_(x, t) and v+(x, t) are constant on each characteristic, we see that the
solution is double valued for x — ct > -c2/4. Since any neighborhood of the
point of tangency on the parabola contains points where the solution is
double valued, we cannot restrict the domain of definition of the integral
surface to obtain a nonsingular solution. Consequently the initial value
problem has no solution.
If we reformulate the problem and specify that v = f(t) on x = t2 for
^^ c/2, the unique solution is v =f[g-(x — ct)]. This solution is continuous
for x - ct> — c2/4 but it is not differentiable along the line x — ct— — c2/4.
As such it must be interpreted as a solution in a generalized sense, in a
manner to be defined below. However, if we require that t^to< c/2, the
solution has the same functional form as before but it is now differentiable
wherever it is defined. Similarly, we can consider an initial value problem on
x = t2 with t>c/2 and obtain a unique solution.
With regard to the initial value problem for the general first order
equation B.8), we remark that if the characteristic base curves, determined
from B.9), intersect the initial base curve more than once, the problem
generally has no solution. For the solution v(x,t) is completely determined
along a characteristic base curve by its value at a point on the initial base
curve. Thus, if the projection of this characteristic on the (лг, t)-plane (that
is» the characteristic base curve) intersects the initial base curve more than
°nce, the solution will in general be multivalued. Each intersection point
carries an initial value for v and thereby determines a characteristic curve
that differs from those characteristics determined at other intersection
54 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
points, except in the unlikely case that the data at all these points are
compatible.
Example 2.4. d'Alembert's Solution of the Wave Equation. In this
example we derive the solution of the initial value problem for the wave
equation B.1) which was first obtained by d'Alembert. We encountered the
wave equation in Section 1.2 as a special case of the telegrapher's equation
and it describes many physical phenomena as we shall see, but it is often
first introduced as an equation describing the transverse displacement of a
tightly stretched string (it is discussed from this point of view in Example
4.9). The constant y2 then equals the tension Г divided by the density p of
the string and v(x, t) is the vertical displacement of a point x on the string at
the time t. Appropriate initial conditions for the wave equation are
B.20) v(x,0)=f(x),
For the case of a vibrating string, these represent its initial displacement and
velocity. It is then assumed that the string is sufficiently long that distur-
disturbances arising at the ends of the string do not affect its motion within the
time span in which it is observed.
We solve the initial value problem B.1) and B.20) by using the system
B.2). The initial value of v(x, t) in the system is given as in B.20), but the
initial condition for u(x, t) is obtained from the first equation of B.2) as
B.21) u(x,0) = g(x)-yf'(x).
Using B.19) we immediately obtain
B.22) u(x, t) = g(x - yt) - yf\x - yt)
since F(x) — g(x) — yf'(x) in this case. Substituting into the first equation in
B.2) gives
B.23) ~-t - у £ = g(x - yt) - yf\x - yt) ,
with the initial condition v{xy 0) =/(*)•
From B.9) and B.10) we obtain the characteristic equations for B.23) as
<**> f-r. I"- Z-'-rr.
with the initial curve v(x9 0) = f(x) parametrized as
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 55
B.25) x = r9 f = 0, v=ftr).
The solutions of B.24) and B.25) yield the system of curves
B.26) x = -у5 + r , t = s, v = |o (g - у/')*г + Дт).
Now, both £ and /' are functions of л; — yf, and the first two equations in
B.26) imply that x — yt = —2ys 4- т. Thus v(s, т) can be expressed as
B.27) v(s, t) - |o [g(-2ycr + r) - y/'(-2ya + r)] da + Дт).
The change of variables A = -2ya + т in B.27) yields
B.28) v(s, т) = - щ I [g( A) - y/'( A)] dA + Д
T) + I
Again the first two equations in B.26) imply that т = x + yt and — 2ys +
r = x- yt, so that in terms of the variables x and t, the solution B.28) is
B.29) v = V(*, 0=5 [/(* + уО +Л* ~ У0] + Щ
This is d*Alember?s solution of the initial value problem for the wave
equation. It can be verified by direct substitution that it satisfies B.1) and
the initial conditions B.20).
Let G(z) = Jz g(A) dky that is, G is the antiderivative of g. Then B.29)
can be expressed as
B.30) „(X| о - [ \ f{x + yt) + ^ G(*
Since у > 0, the first bracketed term represents a wave traveling to the left
with speed у and the second bracketed term, a wave traveling to the right
with speed y. Each wave travels without change of shape. However, the
presence of these wave forms in the solution of the wave equation is not
always apparent because of the interference caused by the interaction of
these traveling waves when they are superposed or summed as in B.30).
56 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
The family of characteristics B.12) yields a smooth solution v = V(x, t) of
the initial value problem B.8) and B.11) only if the data B.11) and the
functions a, b9 c, and d in B.8) are smooth and the Jacobian A(s, т) is not
zero. If the initial value и = v(t) in B.11) is singular at т = т0, this
singularity is transmitted along the characteristic B.12) that passes through
that point on the initial curve B.11). Then, the integral surface is singular
along that characteristic and, depending on the singularity of и(т), it may
not haye first derivatives or it may be discontinuous there.
For example, if the first derivative of F(x) in B.14) has a jump discon-
discontinuity at x = 0, the solution v = F(x — ct) of B.13) has discontinuous x- and
^-derivatives along the characteristic (base) curve x = ct. If F(x) has a jump
discontinuity at x = 0, then v = F{x — ct) has a jump discontinuity along
x = ct. In each case, v - F(x - ct) is not a solution of B.13) in the
conventional sense since it is not differentiable along x = ct. We must treat it
as a generalized solution and now show how this is to be done in the general
case.
Let v(x, t) be a continuously differentiable solution of B.8) except along
the arbitrary curve x = x(s), t = t(s) where it is continuous but its first
derivatives have (finite) jump discontinuities. We evaluate B.8) at two
points (x, t) on either side of the curve and consider the difference of these
two equations. In the limit as these points approach a common point on the
curve we obtain
B.31) a[vx] + b[v,] = 0,
where the brackets represent the jumps in the first derivatives across the
curve. It has been assumed that all other terms in B.8) are continuous
across the curve. A continuous solution of B.8) that is continuously
differentiable except along a curve across which the jumps in the first
derivatives satisfy B.31), is said to be a generalized solution of B.8).
In fact, the curve x- x(s), t= t(s) across which the jumps in the first
derivatives satisfy B.31) cannot be arbitrary but must be a characteristic
base curve. Since [v] = 0 by assumption, if we differentiate this equation
along the curve we find that
B.32) *J + ^)Ws0.
Combined with B.31) this yields a simultaneous homogeneous system for
the jumps in vx and vr By assumption, they are not both zero so that we can
effectively conclude that x'(s) = a and t'(s) = b. In view of B.9), the curve
must be a characteristic and B.31) states that d[u]fds = 0 along the charac-
characteristic. It is shown in the exercises that the jumps in vx and vt satisfy
ordinary differential equations of first order along the characteristics. From
these equations we conclude that unless the derivatives undergo jumps in
the data for the problem, the solution cannot have any jumps in the
derivatives.
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 57
If v(x, t) is a continuously differentiate solution of B.8) except along a
curve x = x(s), t = t(s) where v has a jump discontinuity, we must generalize
the concept of a solution of B.8) even further. To do so we write B.8) in
conservation form
B.33) (<*v)x + (bv)t = (c + ax + bt)v + d ,
and integrate B.33) over an arbitrary closed and bounded region R in the
(x f)-plane with a piecewise smooth boundary curve 5. On applying
Green's theorem in the plane we obtain
B.34) I avdt-budx = J I [(c + ax + bt)v + d]dxdt,
with integration over S taken in the positive direction. This integral relation
is equivalent to the differential equation B.8) in any region where u(x, t) is
continuously differentiable, as follows from Green's theorem and the arbit-
arbitrariness of the region R (see Exercise 8.1.9). However, the integral relation
remains valid even if v(x, t) has a jump discontinuity across some curve.
If we apply the integral relation B.34) to a region R that surrounds an
arbitrary portion of the curve x — x(s), t = t(s) across which u(x, t) has a
jump discontinuity (see Figure 6.3), and let the boundary curve S collapse
onto the discontinuity curve, we obtain
B.35) (аф)-Ьх'(з))[и] = 0у
since the limiting line integrals are taken in opposite directions and v(x, t)
has different limits on both sides of the curve. While the limit only implies
that the integral of B.35) over a portion of the discontinuity curve vanishes,
the arbitrariness of that portion implies that the integrand must itself be zero
(see Exercise 8.1.9).
Since [u] t^O in B.35) by assumption, we conclude that at\s) - bx\s) =
0. But the vector [t'(s)9 -xf(s)] is normal to the curve x = x(s), t= t(s), so
we again conclude that the discontinuity curve must be a characteristic base
curve. Thus, if u(jc, t) has (together with its first derivatives) a jump
discontinuity across a characteristic base curve B.9) but is a continuously
differentiable solution of B.8) elsewhere, it is a generalized solution of
B.8). For example, v - x ~ ct for x < ct and v - 1 for x > ct is a generalized
solution of B.13), while v = x - ct for x < 0 and v = 1 for x > 0 is not a
generalized solution since x = ens a characteristic base curve of B.13) while
* = 0 is not.
If we apply the discussion that precedes equation B.31) to the present
case, where both v and its first derivatives have jump discontinuities across
the characteristic base curve B.9), we obtain
B-36) «[v,] + b[vt] = c[v],
58 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
since v also has a jump discontinuity. As has been shown above, B.36) is
equivalent to the ordinary differential equation d[v]/ds = c[v], along the
discontinuity curve. This equation not only determines the variation of the
jump along that curve, but also shows that there can be no discontinuity in a
generalized solution of B.8) unless it arises in the data for the equation.
The case when the functions a, b, c, and d have jump discontinuities also
requires the introduction of generalized solutions, and these are considered
in the exercises.
Example 2.5. Equations with Singular Points. In our discussion of equa-
equation B.8) it has been assumed that the coefficients a and b do not both
vanish at some point (x, t). If they are both zero at x - x0, t = t0, (x0, /0) is a
critical point for the (autonomous) system B.9) x'(s) = a(x, t), t'(s) = b(x, t)
from which the characteristic base curves are determined. Either two or
more solution curves of this system intersect at the critical point or they
spiral around or encircle that point. In each case difficulties arise in
formulation initial value problems for B.8) in the neighborhood of the
critical point or singularities occur at that point in solutions with initial data
given outside its neighborhood. Consequently, we refer to that point as a
singular point for the equation B.8).
The partial differential equation
/л „ч dv dv
B.37) x_ + ,_eCB>
where с is a constant, has a singular point at (д;, i) = @,0). As the initial
condition for B.37) we take
B.38) ф,1) =/(*),
and examine the effect of the singular point on the solution of the initial
value problem.
The initial curve С can be parametrized as
B.39) x-t, f = l, у=/(т),
and the characteristic equations are
B.40) — = jc , -r = t, — = cv.
ds ds ds
The solutions of B.40) satisfying the initial conditions B.39) at s = 0 are
B.41) x(s, t) = res , t(s, r) = es, v(s9 r) = f(r)ecs .
Solving for s and r in terms of x and t, gives
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 59
B.42) s = log t;
т = -
t '
which is valid for f>0. Inserting this into the equation for v, we obtain as
the solution of the initial value problem B.37), B.38)
B.43) v(x,t)=
The critical point (jc, t) = @,0) for the system x'(s) = *, tr(s) = t is called
a proper node. The characteristic base curves are the pencil of straight lines
x = \t (—°° < A < °°), together with t = 0, all of which intersect at the origin.
Thus, if с = 0 in B.37) so that B.43) reduces to v=f(x/t), the solution is
not defined when t = 0 and x Ф 0, and is infinitely multivalued at the origin
when x = 0, for arbitrary f(x). Iif(x) = 1, the solution B.43) is not defined
on the Jt-axis if c<0; is continuous but not differentiable there if с = 1/3;
but is a valid solution for all x and t if с = 1.
The partial differential equation
B.44) tvx -xvt = 0
also has a singular point at (jc, i) = @,0). As the initial condition we take
B.45) K*,0) =/(*), *>0.
The characteristic equations for B.44) are
B.46) x\s) = t, t\s) - -x , v'(s) = 0.
If we parametrize the initial curve С as
B.47) х = т, t = 0, w-/(t),
we obtain the solutions of B.46) that satisfy B.47) at 5 = 0 in the form
B.48) x(sy т) = т cos 5 , t(s, т) = -т sin s , v(s, т) = f(r) .
On solving for т in terms of x and f, we find that r = (jc2 + f2I/2, so that
B.49) »(*,') =/[(*2 + '2I/2b
The solution v(x, i) is completely determined for x2 + t2 > 0 in terms of the
data given on the positive x-axis. It is not differentiable at the origin.
The critical point @,0) for the system x'(s) = t, t'(s) = -x is called a
center. The characteristic base curves are concentric circles whose center is
60 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
at the origin. They are completely determined by specifying a point that
they pass through on the positive Jt-axis. The solution v(x, t) is constant on
each circle.
If the Jacobian Д($, т) = xstT — tsxT associated with the first two equations
in B.12) vanishes on the initial (base) curve jc = х(т), t= t(r), the tangent
vector to the curve has the direction of the vector [a, b] at each point.
Therefore, it must be a characteristic base curve. We change the variable on
the initial base curve from r to 5, with 5 chosen so that the components x'(s)
and t'(s) of the tangent vector satisfy B.9). Then, we have a characteristic
initial value problem if the initial value v, expressed in terms of the
parameter 5, satisfies equation B.10). If this equation is not satisfied, the
initial value problem either has no solution or a nonsmooth solution.
As an example we consider the equation B.37) with the initial condition
v = f(i) on x = / for t > 0. On representing the initial curve parametrically as
х = т, t — т and f=/(r), we find that the solution of the characteristic
equations is x = те\ t = res and v =f{r)ec\ The Jacobian A(s, r) vanishes
not only at s — 0 but identically. (As shown in the exercises this is always the
case if the initial base curve is characteristic and the coefficients a and b in
B.8) are continuously differentiable.) To determine what condition f(t)
must satisfy for this to be a characteristic initial value problem, we set
r = t(s) and find that the first two characteristic equations in B.40) are
satisfied if r = es so that x = t = es. Then v = f(es) and the third equation in
B.40) requires that esf\es) = cf(es). The general solution of this equation is
f(es) = fiecs, where C is an arbitrary constant. In terms of x and t this
requires that on x = t we must have v = /@ = ptc to obtain a characteristic
initial value problem. If v = Д0 does not have the required functional form,
the problem has no solution.
If а Ф 0 in equation B.8), we can divide through by a in B.8) and replace
the characteristic equations B.9) and B.10) by
dv с d dt b
B.50) _ = _(; + _ on =_
dx a a dx a
Similarly, if b ^0 in B.8) we obtain the characteristic equations
/« ел ч dv с d dx a
B.51) ~Г = Г^+Г on т = г •
v } dt b b dt b
Given the initial value problem v =f(x) on t = h(x) for equation B.8), if
t = h(x) is a solution of t'(x) = ft/a, we have a characteristic initial value
problem if v=f(x) satisfies the equation f'(x) = (c/a)f(x) + dla. If the
initial values are v=f(t) on x = g(t), and x = g(t) satisfies the equation
jc'(O = alb, we must have/'@ = (clb)f(t) + dlb for this to be a characteris-
characteristic initial value problem. Both characteristic initial value problems have
LINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 61
nonunique solutions and this is most easily shown by constructing a general
solution of B.8) for each case.
Let <p(x, t) = A represent a family of characteristic base curves for equa-
equation B.8). In terms of equation B.50) we assume these curves can be
expressed as t = h(x, A) and for each A we solve the linear ordinary
differential equation for v. If we express the general solution as v — F(x, A),
the general solution of B.8) is v = F[x, <p(x, t)]. In connection with B.51)
we express the characteristics as x = g(t, A) and for each A we solve for v.
With the general solution given as v = G(t, A), the general solution of B.8)
is и = G[t, (p(x, t)]. (We note that (p(x, i) satisfies the equation a<px + b<pt =
0.) Each of the foregoing general solutions of the ordinary differential
equations contains an arbitrary constant that can vary with A. Consequently,
the corresponding general solutions of B.8) depend on an arbitrary func-
function. The arbitrary functions are constant on each characteristic base curve.
Therefore, a characteristic initial value problem has infinitely many solutions
since all that is required of the arbitrary functions is that their value at
(p(x, t) = k0, the initial characteristic (along which they are constant), be
chosen so that the initial condition is satisfied.
Example 2.6. Characteristic Initial Value Problems. The partial dif-
differential equation B.37) can be expressed in either of the forms B.50) or
B.51), say if x and t are both positive. We obtain either
B.52) £ = C-v on
' dx x
or
dt
dx~
dx
dt
_ t
X
X
t
dv c
—- = - v on
at t
The family of characteristic base curves is <p(x, t) = xlt = A. This yields
* = h(x, A) = xlk orjc = g(t, A) = Af as the explicit representations of these
curves. Along t - x/X, we find that v - F(x, A) = j3( A)xc, while on x = kt we
obtain v = G(t, k) = p(k)tc. Thus, the general solution of B.37) has the
form v(x, i) = p(x/t)xc or v(x, t) - p(x/t)tc and they both contain an arbit-
arbitrary function p(x/t).
If we specify that v = 5xc on the line t = x/3, we have a characteristic
initial value problem. The (nonunique) solution is v(x, t) = P(x/t)xc, with
/КЗ) = 5 but /3(x/t) otherwise arbitrary. However, if v = sin x on t - x/3 the
initial value problem has no solution, since v does not satisfy the ordinary
differential equation in B.52).
An interesting characteristic initial value problem occurs for the equation
B.54) i/x
62 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
with f>0. The characteristic equations can be given in the form B.50) as
do dt r-
B.55) -r = о on -r- = V7.
We are interested in studying the characteristic base curves that originate on
the x-axis. Thus we must solve the equation t\x) = \/Ф0 with t(\) = 0. This
equation is nonlinear and the solution of the initial value problem for each
value of the parameter A is not unique. Two solutions are t = 0 and
t= \{x~ AJ, and there are infinitely many others.
We see that t = \ (x - AJ is a family of characteristics for B.54), and each
curve in this family is tangent to the x-axis at its point of intersection. Thus,
the line t = 0 is an envelope of characteristics (see the Appendix) in addition
to being a characteristic, as we have shown. In terms of the discussion that
precedes this example, the characteristic family is given as <p(x, t) = x -
2Vt = A, and we easily find from B.55) that the general solution of B.54) is
B.56) v(x,t) = p(x-2Vi)ex
for arbitrary j3(A). On the characteristic f = 0, B.55) yields v = fiex.
Noting the form that v must have on the characteristic t = 0, we expect
that the initial value problem for B.54) with v(x, 0) = f(x) does not have a
solution unless f(x) = fiex with constant /3. The solution is then given as
v(x, t) = fie*. We may however use the general solution B.56) to obtain, for
arbitrary f(x), the solution v(x9t)=f(x-2Vi) expBV7). This function,
while it is continuous at t — 0, is not differentiable there, so it must be
understood as a generalized solution. If f(x) = jSe*, this solution agrees with
that obtained above and is differentiable at t = 0. In either case the solution
is uniquely determined.
The foregoing results are at variance with our general conclusions
concerning characteristic initial value problems. If v{x, O) = /(jc) has the
required exponential form, the initial value problem is characteristic and yet
the solution was found to be unique. If f(x) does not have the required
form, the initial value problem should have no solution and yet we found a
generalized solution. The reason for these inconsistencies can be traced to
the fact that the coefficient Vt of v in B.54) is not differentiable at t = 0.
The line t = 0 is a singular solution of t'(x) = V*00 anc* it cannot be obtained
from the family of characteristics determined above for any specific choice
of A. Therefore, the general solution of B.54) given above does not yield
nonuniqueness here. Because each of these characteristics intersects the line
t — 0, the general solution can be used to solve an initial value problem given
on the x-axis. Each characteristic intersects the x-axis only once, so the
solution is unique. The intersection is, however, tangential, so the solution
is not differentiable at t = 0.
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 63
2.3. QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL
EQUATIONS
A first order partial differential equation of the form
B.57) a(x9 Uu)j^ + b(x, '> w) ^ = c^''' u^'
is said to be quasilinear since it is linear in the derivative terms but may
contain nonlinear expressions of the form uduldt or u2. Such equations
occur in a variety of nonlinear wave propagation problems and in other
contexts but we will emphasize the wave propagation aspect in the exam-
examples. Proceeding as in Section 2.2, we interpret B.57) geometrically in order
to construct a solution.
We assume that a solution и = u(x, i) of B.57) can be found and examine
the properties of the solution as implied by B.57). The solution и = u(x, t)
is called an integral surface, and we express it in implicit form as F(x, t, u) =
m(jc, t) - и = 0. The gradient vector VF= [ди/дх, duldt, -1] is normal to
the integral surface F(x, t, u) = 0. By transposing the term с in B.57) we can
express the resulting equation as a scalar or dot product,
/^^«ч du . du _ , , \ ди
<2-58> aTx + bTt-c={a>b>cn
The vanishing of the dot product of the vector [a, b, c] and the gradient
vector VF implies that these vectors are orthogonal. Accordingly, the vector
[a, by c] lies in the tangent plane of the integral surface и = и(х, i) at each
point in the (jc, ty w)-space where VF is defined and nonzero.
At each point (jc, t, w), the vector [a, b, c] determines a direction that is
called the characteristic direction. As a result, the vector [a, ft, c] determines
a characteristic direction field in (jc, t, w)-space, and we can construct a
family of curves that have the characteristic direction at each point. If the
parametric form of these curves is x — x(s), t = f(s), and и = u(s), we must
have
B.59) | = e(x, i9uh | = b{%9 tfUh *L = ф,,, u) 9
since [dx/ds,dt/ds,du/ds] is the tangent vector along the curves. The
characteristic equations B.59) differ from those in the linear case, since the
equations for x and t are not, in general, uncoupled from the equation for w.
The solutions of B.59) are called the characteristic curves of the quasilinear
equation B.57).
Assuming a, ft, and с are sufficiently smooth and do not all vanish at the
same point, the theory of ordinary differential equations guarantees that a
unique characteristic curve passes through each point (jc0, t0, м0). The initial
64
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
value problem for B.57) requires that u(x, t) be specified on a given curve in
(x, t)-space. This determines a curve С in (x, ty u)-space referred to as the
initial curve. To solve this initial value problem we pass a characteristic
curve through each point of the initial curve С If these curves generate a
surface, this integral surface is the solution of the initial value problem. We
now state and outline the proof of a theorem that gives conditions under
which a unique solution of the initial value problem for B.57) can be
obtained.
Let a, b9 and с in B.57) have continuous partial derivatives in all three
variables. Suppose the initial curve С given parametrically as х = х(т),
t = /(r), and и = и(т) has a continuous tangent vector and that
B.60) Д(т) - fT ф(т), t(r), и(т)] - fr b[x(t), t(r), и(т)] Ф 0 ,
on С Then, there exists one and only one solution и = u(x, i), defined in
some neighborhood of the initial curve C, that satisfies B.57) and the initial
condition u[x(t), t(r)] = и(т).
The proof of this theorem proceeds along the following lines. The
characteristic system B.59) with initial conditions at 5 — 0 given as x =
jc(t), t = t(r), and и = м(т) has a unique solution of the form
B.61) x = x(s, r) , t = t(s, т), и - u(s, t) ,
with continuous derivatives in s and r, and with
B.62) *@,т) = х(т), *@,т) = Г(т), и@,т) = и(т).
This follows from the existence and uniqueness theory for ordinary differen-
differential equations. The Jacobian of the transformation x = x(s, r), t = t(s, т) at
,y = 0 is
B.63)
dx
ds
dt
Ts
dx
дт
dt
dt dx L
dr dr
s=0
= Л(т)
and it is nonzero in view of the assumption B.60). By the continuity
assumption, the Jacobian does not vanish in a neighborhood of the initial
curve. Therefore, the implicit function theorem guarantees that we can solve
for 5 and r as functions of x and t near the initial curve.
Then
B.64)
u(j, t) = u[s(x, t), t(x, t)] = U(x, t) ,
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 65
jS a solution of B.57). This is readily seen on substituting u[s(x, t), r(x, t)]
into B.57), using the chain rule and the characteristic equations B.59). The
uniqueness of the solution follows since any two integral surfaces that
contain the same initial curve must coincide along all the characteristic
curves passing through the initial curve. This is a consequence of the
uniqueness theorem for the initial value problem for B.59). This completes
our proof.
The condition B.60) essentially means that the initial curve С is non-
characteristic or is not the envelope of characteristic curves. If the initial
curve is characteristic, so that Д(т) = О and in addition it is a solution of the
characteristic equations B.59), solutions exist but they are not unique. This
is called a characteristic initial value problem. If the initial curve is the
envelope of characteristic curves (see the Appendix), so that Д(т) = О along
the curve but the curve does not satisfy B.59), the integral surface may not
be differentiable along the initial curve. If the initial curve is noncharacteris-
tic but Д(т) = 0 at a discrete set of points, special problems that originate at
these points occur with the solution. Examples of each of these types are
considered in the exercises.
We will now consider a single example of a quasilinear equation, the
simplest case of unidirectional nonlinear wave motion. The initial value
problem for this equation is discussed in great detail and various features of
nonlinear wave motion, which distinguish it from linear wave motion, are
emphasized.
Example 2.7. Unidirectional Nonlinear Wave Motion. The simplest
quasilinear equation characterizing one-directional nonlinear wave motion is
B.65, u + .£...
In this example we consider the initial value problem for B.65) with
B.66) w(jc,0) =/(*).
where f(x) is a given smooth function. First a general discussion of the
solution of B.65), B.66) is given and then three specific choices for f(x) are
considered.
To solve the initial value problem we parametrize the initial curve as
<2-67) jc = t, t = 0, м=/(т).
Applying the condition B.60) we obtain
B.68) Д(т) = -1,
66 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
so that Д(т) is nonzero along the entire initial curve. The characteristic
equations are
dx dt ч du
1
with initial conditions at 5 = 0 given by B.67). Denoting the solutions by
x(s, r), t(s, r), and u(s, т), we find that dulds = 0 implies u(s, т) is constant
along the characteristic curves. Therefore,
B.70) w(*,t) = w@,t)=/(t),
on using B.67). Inserting the expression for и in the equation dxlds = wwe
can immediately solve for jc(s, т) and t(s, т) to obtain
B.71) jc(j, т) = т + 5/(t) , ;(s, t) = 5 .
If the conditions of the theorem are satisfied, this system can be inverted
near s = 0 to give s = t and т = т(х, i). Then B.70) yields the solution
B.72) u=f[T{x9t)] = U(x9t).
Since s = / and r = x — j/(t) = jc — to in view of B.70), the solution of B.65)
can also be given in implicit form as
B.73) n=/[x-fti].
We consider the solution и = u(xy t) of B.65) to represent a wave. The
wave form at the time t is given by the curve и = m(jc, f) in the (jc, u)-plane
with t as a parameter and и =f(x) as the initial wave form. The (numerical)
value of m(jc, i) gives the height of the wave at the point x and the time t. To
determine the motion of the wave, we must find the velocity dxldt of each
point on the wave. The first two characteristic equations in B.69) imply that
B.74) f = и .
The greater the amplitude \u(x, t)\ of the wave, the greater the speed of the
corresponding point x on the wave. [This contrasts with the situation for
linear wave motion determined by equation B.13) where dxldt^c-
constant, so that each point and, consequently, the entire wave form moves
with a single speed.] If m(jc, f)>0, the point x moves to the right, if
u(x, t) = 0, it remains fixed, whereas if u(x, t) < 0, it moves to the left. Thus
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 67
if u(x, t) takes on both positive and negative values, individual points
maintain a fixed direction of motion but different portions of the wave form
can move either to the right or to the left. Therefore, the wave motion need
not be totally unidirectional as is the case for linear wave motion. Assuming
и > 0, points x where u(x, t) has larger values and the wave is higher move
more rapidly to the right than points x where u(x, i) has smaller values and
the wave is lower. If, initially, there are higher portions of the wave form
located to the left or the rear of lower portions, the higher points may
eventually catch up with and pass the lower points, at which time the wave is
said to break. At the time t of breaking, the function w(jc, /) becomes
multivalued and is no longer a valid solution of B.65). Similar difficulties
can occur if m(jc, /) is not restricted to be positive. The geometric aspects of
the breaking process are demonstrated in Example 2.8.
In many physical processes described by the quasilinear wave equation
B.65), the function u(x, t) represents a physical quantity such as density,
which is intrinsically expected to be single valued. Thus when the wave
u(x, t) breaks and becomes multivalued, the equation B.65) describing the
physical process is no longer an acceptable model for the physical process.
In general, this means that certain higher derivative terms that were
neglected in the derivation of the quasilinear equation become significant
and must be retained. It will be shown in Section 10.3 for a model equation,
known as Burgers' equation, how the addition of a second derivative term to
B.65) can yield solutions valid for all time. A similar situation has already
been observed in our discussion of the diffusion equation A.23) in Chapter
1. With delta function initial data, the solution of the diffusion equation
given in A.31) is a smooth function for all x and for t>0. However, if the
diffusion coefficient D is equated to zero and the first order unidirectional
wave equation A.98) results, the solution A.99) remains sharply singular for
all /> 0. This indicates that the inclusion of higher derivative terms tends to
smooth out the solution.
An alternative method for dealing with the breaking phenomenon is to
introduce a discontinuous solution of B.65) known as a shock wave that
extends the validity of the solution beyond the breaking time. Such solutions
of B.65) must be interpreted in a generalized sense, as was done for
discontinuous solutions of the linear equation B.8). This approach has the
advantage that a complete analysis of the breaking wave solution can be
carried out on the basis of the first order equation B.65) without having to
analyze a higher order nonlinear equation. However, as will be seen, the
correct determination of the (shock wave) discontinuity solution is not as
straightforward as it was in the linear case. In general, an appeal to the
Physical origin of the problem is needed to make the correct determination.
In any case, even if the wave breaks at the time /, the solution B.73) of
the initial value problem remains valid until that time. Thus it is important
to determine the time when the wave u(x, t) first begins to break. In
addition, to determine the shock wave it is necessary to know not only the
68 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
time but also the point(s) x at which the wave breaks. We consider two
methods for doing so.
First, we use implicit differentiation to determine the slope of the wave
form и = u(x, t)~f(x - tu) at the time f. The slope duidx is readily found
to be
B75)
y }
dx l + tf'(x-tu)'
From B.71) we see that the characteristic base curves for B.65) are the
family of straight lines x - tf(r) = т, where r is a parameter. When the
denominator on the right side of B.75) [i.e., 1 + tf\x - tu) = 1 + tff(r)] first
vanishes, the slope duldx becomes infinite and the wave begins to break.
Since the denominator vanishes when
B-76) t — jfa,
the breaking time is determined from the value of т at which t has its
smallest non-negative value.
The second method for determining the breaking time considers the
characteristic curves B.70), B.71) of the equation B.65). We have shown
that и — /(r) on the characteristic base curves
B.77) x-tf(r) = r,
where r is a parameter. If for two or more values of r the straight lines
B.77) intersect, и will in general be multivalued at the intersection point,
for u(x, i) must equal /(r) at that point, and /(r) may have different values
on each of the lines B.77) that intersect at the point. To find the possible
intersection points we must determine the envelope of the family of straight
lines B.77). As shown in the Appendix, if F(x, t, r) = 0 is a one-parameter
family of curves, the envelope of this family is obtained by eliminating the
parameter r from the system F(x, f, r) = 0 and Ft(jc, t, r) = 0. In our case,
F = x - tf(r) - r = 0 and dFldr = -tf'(r) -1 = 0. The equation dFldr = 0
shows that the times at which the characteristic curves touch the envelope
are given by t= -1//'(т), so that the initial breaking time is in agreement
with that determined from the first method.
Once the value of r that yields the breaking time t is determined, the
point x at which the wave breaks is found from B.77). We remark that there
may be more than one value of the parameter r that yields the same
breaking time t. In general, each of the values of r yields a different
breaking point x. It can also happen that different parts of the wave break at
different times. While the classical (differentiable) solution is not valid after
the first breaking time, the later breaking times play a role in the construc-
construction of shock waves.
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 69
We note that if /'(т) >0 for all r, the right side of B.76) must always be
negative, so that the wave never breaks. Geometrically, this means that the
initial wave form и ~f(x) is a monotonically increasing function of x. Since
the larger the values of м, the faster the corresponding points on the wave
move, no point x on the wave can overtake any other point and no breaking
occurs. However, if f'{r) <0 for all r, the initial wave form is a monotoni-
monotonically decreasing function. Thus points in the rear portion of the wave move
faster than points in the front portion and eventually overtake them. When
this happens, the wave breaks. If /'(т) takes on both positive and negative
values, wave breaking generally occurs.
The foregoing results concerning the breaking of the wave u=f(x- tu)
can be interpreted from a different point of view. From B.76) and B.77) we
conclude that the breaking time is determined from the point on the curve
x = т - Дт)lf\r), t = -1 /ff(r)9 u = /(T) where t has its smallest value. As is
easily verified, this curve has the property that Д(т) [as defined in B.60)]
vanishes on it. Thus if this curve were chosen as an initial curve for B.65),
the theorem stated for the general case would not guarantee that this initial
value problem has a unique (smooth) solution. In fact, B.75) implies that
the solution would not be differentiable along the initial curve. We infer
from the above that even when the intial data guarantee a unique solution
for the given quasilinear equation, this solution breaks down along a curve
where the condition B.60) (i.e., Д(т)^0) is violated.
The breaking curve determined is not a characteristic curve for B.65)
since и is not constant on it. For the sake of completeness, we show that if
the initial curve for B.65) is characteristic, the initial value problem has
infinitely many solutions. Let w = с = constant on the line x — ct. [This curve
is a characteristic for B.65) since with t = s, x = cs the characteristic
equations B.69) are satisfied.] Then, u=f[x-tu] is a solution of the
characteristic initial value problem as long as ДО) = с but / is an otherwise
arbitrary but smooth function.
We continue our discussion of B.65) in the following example and select
three specific initial wave forms and examine the resulting wave motions.
Example 2.8. Unidirectional Wave Motion: Specific Initial Data. In
this example three special choices of initial values for B.65) (i.e., duldt +
uduldx = 0) are given and the corresponding solutions analyzed.
1. The initial value is
B-78) w(XH) =/(*) = -*.
Since f'(x) = -1, we expect (in view of the foregoing discussion) that the
wave will break and the solution u(x, i) will become multivalued. In fact
(since Дт) = -r), B.76) gives t = 1 as the breaking time.
Now, B.73) gives the implicit form of the solution as
70
B.79) u
from which it follows that
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
B.80)
u(x, t) =
We easily verify that B.80) satisfies B.65) and the initial condition B.78),
as well as the fact that it blows up at the time t = 1.
The motion of the wave w(jc, t) = x/(t — 1) is indicated in Figure 2.2. As t
increases, the wave form w(jc, 0) = —x executes a clockwise rotation around
the origin in the (jt, u)-plane. Since и = 0 at x = 0, that point (i.e., x = 0) is
stationary. Also, |w| increases linearly with |jc| and points x further away
from the origin have a linearly increasing velocity, yielding the effect
indicated in the figure. At t = 1 the wave form u(x91) coincides with the
w-axis and becomes infinitely multivalued.
Figure 2.2. The wave motion.
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
71
The breakdown of the solution and its multivaluedness at t = 1 may also
be determined by considering the characteristic base curves, which for this
initial value problem are given by
B.81) x«(l-0r,
with r as a parameter. When t = 1, these characteristic lines all intersect at
the point (jc, i) = @,1). Since each of the lines carries a different value of т,
and m(jc, t) is constant on each of these lines, it must become infinitely
multivalued at t= 1. This is indicated in Figure 2.3.
If the initial condition B.78) is replaced by/(x) = jc, we have Дт) = т and
/'(t) = 1. Then, according to B.76), the wave never breaks. This fact is
borne out by the solution u(x, i) = xl(t +1) of this problem, which is
defined for all f>0.
More generally, if u(x, 0) = f(x) = a + 0jc, we determine from B.73) that
the solution is m(jc, t) = (/3jc + a)/(pt+ 1). Thus if 0 <0, the wave breaks
when t = -l/<3, but if 0 > 0 the wave never breaks. The characteristic lines
are given as jc = at + (fit + 1)т and at the breaking time Г = -1 /0, we have
jt = -a/0. At this point w(jc, f) is stationary and the wave form rotates
around this point. At the breaking time the wave form u(x, t) is vertical and
is infinitely multivalued.
2. The initial condition is
B.82) m(jc,0)-/(jc) = 1-jc2.
Using B.73) gives the implicit form of the solution as
Figure 2.3. The characteristic curves.
72 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
B.83) и =f(x -tu) = l-(x-tuJ.
Solving this quadratic equation for и gives two branches
,2.84, „(,,„. f-IiVT
To satisfy the initial condition B.82) we must choose the minus sign in
B.84) and obtain
, ч , ч x l-Vl
B.85) ф, r)=7 —^
Differentiating B.85) with respect to x9 we find that the slope duldx\
becomes infinite when the radical in B.85) vanishes; that is, when
B.86) x = i+h'
As long as x < t + 11 At, the radical in B.85) is real valued and the solution
m(jc, t) is well defined. An easy calculation shows that jc< f+ II At for all
x < 1 since t > 0. At x = 1, the radical vanishes when t = \ and as x increases
towards +0°, the time t at which it vanishes decreases to zero. Thus if the
initial condition is given over the infinite interval — °o<jt<<», the wave
begins to break immediately at x — +oo.
The characteristic base curves for this problem are [see B.77)]
B.87) x-t{\ -т2) = т,
where r is a parameter. The solution m(jc, t) is constant along the charac-
characteristics. The envelope of this family of characteristic curves is readily found
to be
B.88) *-,-1=0,
which is identical with B.86). Since two neighboring curves of the charac-
characteristic family intersect on the envelope corresponding to different ,values of
т, the solution m(jc, i) becomes double valued at the envelope. In fact, the
radical term in the solution B.84) vanishes on the curve B.88) in the
(jc, f)-plane, so that m(jc, t) splits into the two branches given in B.84) along
the envelope B.88). An indication of how the wave m(jc, t) propagates is
given in Figure 2.4.
If the initial data B.82) are restricted to the interval -&<x<a where
a > 1, the foregoing discussion implies that the resulting wave will not break
immediately. However, some care must be exercised in determining the
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
73
Figure 2.4. The wave motion
breaking time. On applying the formula B.76) to this problem, we obtain
t = 1 /2т, so that the wave breaks when t = 112a. Yet the solution B.85) with
x<a and t>0 predicts the earlier breaking time t= \(a-^Ja2 - 1). This
discrepancy may be resolved by observing that for this problem, B.85) is
valid only in the region jc< a + A - a2)t, t>0. That is, the region is
bounded by the characteristic base curve x = a + A - a2)t that passes
through the point (jc, i) = (a, 0). In that region the breaking time is easily
found to be t = 112a. The point at which the wave breaks is jc = {all) +
(
In general, w(jc, 0) must be specified over the entire jc-axis while in the
foregoing it is given only for x^a. The simplest approach is to set
u(x, 0) = 1 — a2 for x > a. The resulting initial value /(jc) is continuous for all
x, but it is not differentiable at jc = a. The solution m(jc, i) is given by B.85)
to the left of the characteristic jc = a + A - a2)t while m(jc, i) = 1 - a2 to the
right of the characteristic. The solution is continuous across the characteris-
characteristic but the first derivatives of и have jump discontinuities there. [General-
[Generalized solutions of B.65) with jumps in the first derivatives will be considered
below.] The breaking time for this wave is again t - 112a. This result is valid
for all a and not just for a > 1. However, the wave does not break unless a is
positive.
3. The initial condition is
B-89) w(jc,0)-/(jc)
The implicit form of the solution is given as
74 ™*ST ORDER PARTIAL DIFFERENTIAL EQUATIONS
B.90) и = f(x - tu) = sin {x - tu) ,
and the characteristic curves are
B.91)
Using B.76) we have
B.92)
x — f sin т — т = 0.
t=-
1
COST
so that the breaking time occurs when t — 1, since when
B.93)
т = Bn + 1)тг , л = 0, ±1, ±2 ,...
we have cosr= -1. At all other values of r, t as given by B.92) either
exceeds unity or is negative. When t — 1 the wave breaks simultaneously at
the points x = Bл + 1)тг.
A qualitative picture of the wave motion is indicated in Figure 2.5 where
the interval 0<х^2тг is considered. The critical nature of the point
(jc, и) = (тг, 0) is apparent since the wave form rotates in a clockwise motion
near that point.
In a generalization of the foregoing we replace B.89) by u(x, 0) = f(x) =
a + b sin jc, a > b > 0. Then /(jc) is positive for all jc and each point on the
initial wave form moves to the right. From B.76) we find that t = -1 /6cos т,
and t has a minimum when т is an odd multiple of тг, as in B.93). The
breaking time is t=\lb, and the wave breaks simultaneously at the points
Figure 2.5. The wave motion.
QUASILINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 75
If the initial wave form/(jc) for equation B.65) is a smooth function, the
solution u(x, t) is smooth until the time when the wave breaks, if indeed it
breaks at all. However, if f(x) is continuously differentiable for x # jc0, but is
rnerely continuous at x = x0 and has a jump in the first derivative there, it
follows from the results of Example 2.7 that u=f(x0) on the line x =
x + tf(x0). The first derivatives of и have jumps across this line and they
satisfy the equation
B.94) [ut] + u[ux] = [и,] +/(*0)[mJ = 0 .
This equation is derived exactly as was B.31) for the linear case. The
solution u(x, i) which is continuous but has a jump in the first derivatives
across a characteristic line is called a generalized solution. It is shown in the
exercises that the jumps in the first derivatives satisfy ordinary differential
equations along the characteristic base curve.
For example, if f(x) = x for x < 1 and f(x) = 1 for x > 1, the solution is
u(x, t) = xl(t + 1) for x < t + 1 and u(x, t) = 1 for x > t + 1. Then, и = 1 on
the line x = t + 1, while [i*J = 1 /(f + 1) and [wj = -1 /(f + 1) across the line
x = 1 + t, so that equation B.94) is satisfied.
If/(x) has a jump discontinuity at jc = jc0, the concept of solution must be
generalized even further. To do so, we express B.65) in conservation form
B.95) (w), + Gw2),=0.
This is also referred to as a conservation law. We integrate B.95) over an
arbitrary closed and bounded region R in the (jc, t)-plane with a piecewise
smooth boundary curve 5. Then, Green's theorem in the plane yields
B.96) f udfc-
js
with integration taken in the positive direction. When u(x, t) is continuously
differentiable, this integral relation is equivalent to the differential equation
B.65).
If a solution m(jc, t) of B.65) has a jump discontinuity across the curve
x-x(s), t-t(s), we let the boundary curve 5 collapse onto an arbitrary
portion of [x(s), t(s)] and conclude as in the linear case [see equation B.35)]
that
B-97) jc'E)[m]-|^)[m2] = 0.
the discontinuity curve be represented as jc = x@ (i.e., let t = s) and
iu] = м2 — ul9 where u2 and ut are the limits of и as the curve is approached
from the right and the left, respectively. Since [u2] = u22-u\, B.97) reduces
to
76 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
B.98) *'@ =!<«i +и2)-
If the discontinuity in и originates at the point jc0 when t = t0, we have the
initial condition x(t0) = x0 for the equation B.98).
Let x'(t) = U(t). Then, the discontinuity in the solution moves with the
velocity U. This velocity is the the average of the two (limiting) velocities ux
and u2 on either side of the discontinuity curve. Since u(x, i) is constant on
each characteristic, the discontinuity curve is intersected by the characteris-
characteristics x ~ tux + r2 and x = tu2 + т2, with тг < т2. These lines can intersect only
if ux > u2. Since U is the average of иг and w2, we must have
B.99) u1>U>u2.
In the absence of the discontinuity curve the solution would be multivalued
and this curve prevents that from happening.
The propagating jump or discontinuity in the solution is referred to as a
shock wave and the velocity U of the jump is called the shock velocity.
When a wave breaks at x = x0 at the time f0, the solution w(jc, t) becomes
multivalued and loses its validity. By introducing a shock wave for which the
shock curve is determined from B.98) with jc(*0) = jc0, the solution can be
extended beyond the breaking time. In the following examples we show this
can be done for a number of problems related to those considered in
Example 2.8.
Example 2.9. Compression and Expansion Waves. In this example we
consider waves of the form studied in part 1 of Example 2.8. However, in
this and the examples that follow, the initial wave form w(jc, 0) = /(jc) is
assumed to be bounded for all jc. In all cases the initial values are given for
the equation B.65).
If /(jc) = 1 for x ^ 0, f(x) = 1 - xla for 0 < x < a, and /(jc) = 0 for x > a,
we obtain the solution w(jc, 0—1 for x < /, u(x, t) = (a — x) I {a — t) for
t<x<a9 and m(jc, t) = 0 for x ^ a. As t increases the distance between the
two constant wave heights и — 1 and м = 0 decreases until the time t = a is
reached and the wave breaks. The wave form appears to undergo compres-
compression with increasing time, so it is called a compression wave.
At the time t = a, all the characteristics that issue from the interval
0 < x < a intersect at the point x - a. Since /(jc) assumes all the yalues of x
between zero and one in that interval, the solution m(jc, t) which is constant
on each characteristic assumes all the values between zero and one when
t = a. That is, it is infinitely multivalued. Technically, however, u(x,a)
merely has a jump discontinuity at x = a. It equals one for x < a and it
vanishes for x> a. Therefore, the solution can be extended beyond t = a by
introducing a shock wave.
The wave u(x91) breaks at jc = a when t = a and ux = 1 while u2 = 0. The
condition B.99) is met and B.98) becomes x'(t) = \ with x(a) = a. The
QUASIUNEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 77
solution is x(t)= \(a + t), for t>a. Consequently, we obtain the shock
wave for t ^ a
u(x9t) = l, x<l(a + t),
BЛ00) и(х,0-0, *>*(« +О-
The continuous (generalized) solution given above remains valid until t = a.
At that time it is replaced by the shock wave B.100). The shock speed is
U = \ and the shock wave form is a unit step that moves to the right with
speed \.
If f(x) = 0 for x < 0, f{x) = xla for 0 < x < a and f(x) = 1 for x > a9 we
obtain the solution u(x91) = 0 for x < 0, u(x, f) = x/(a + f) for 0 < x < a + t
and u(jc, t) = \ for x ^ a + t. For this problem as / increases, the distance
between the two constant wave heights и = 0 and и = 1 increases. Therefore,
this is called an expansion wave. This wave does not break and the solution
is valid for all time.
It is of interest to consider the limits of the two foregoing problems as a
tends to zero. In both cases this means that, in the limit, the initial function
f(x) has a jump discontinuity at x = 0. In the first case, f(x) — 1 for x < 0 and
f(x) = 0 for x > 0. On solving this initial value problem by the method of
characteristics we find that the characteristics intersect at x = 0 as soon as t
increases from zero. Thus the solution is double valued and the wave breaks
immediately. In the second case, f(x) — 0 for x < 0 and f(x) = 1 for x > 0.
Again solving by the method of characteristics, we find that u(x, t) = 0 for
x < 0 and u(x, t) = l for x > t. In this problem the characteristics do not
intersect so that the wave does not break. However, since none of the
characteristics pass through points (jc, t) in the sector 0 < x < t, the solution
remains undetermined there.
In the first problem the wave breaks immediately at x = 0 when / = 0.
Since ux = 1 and u2 = 0, the shock wave is immediately found to be
u(x, t) = l for x<\t and m(jc, r) = 0 for x>\t. This also follows from
B.100) on taking the limit as a tends to zero.
For the second problem we obtain a continuous solution valid for all time
on letting a tend to zero in the original version. This yields m(jc, t) = 0 for
* =£ 0, u(x, t) = xltioxQ<x<t and m(jc, 0 = Иогх> t.
We remark that it is possible to construct a shock wave for the second
problem defined as m(jc, t) = 0 for x < \t and m(jc, i) = 1 for x > \u However,
this solution must be rejected because the shock wave construction fails to
satisfy the condition B.99) since the wave speed on the left of the shock
(i.e., Mj=0) is smaller than that on the right (i.e., u2 = 1). Thus, the
absence of condition B.99) could lead to nonunique solutions of problems
with discontinuous data. It is also shown in Section 10.3 that shock waves
that fail to satisfy the condition B.99) cannot be limits of traveling wave
solutions of Burgers' equation.
78 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Example 2.10. Triangular Waves. If the initial condition of part 2 of
Example 2.8 is replaced by f(x) = l-x2 for |jc| == l and f(x) = 0 for |x| > 1,
the resulting wave breaks at the point x = 1 when t = \. The jump in the
solution is zero at that time and it increases gradually. As a result, the shock
wave for this problem has a variable speed, in contrast to the situation
encountered in Example 2.9. Rather than study this problem we consider a
simpler qualitatively similar problem.
Let u(x, 0) = f(x) = 0 for |*| > 1 and f(x) = 1 for |*| < 1. The compressive
part of this wave at x = 1 breaks immediately and requires the introduction
of a shock wave with shock speed U - \. The expansive part of the wave at
x — —1 can be dealt with by introducing a continuous wave, as was done in
the preceding example.
Since the shock originates at x = 1, the equation of the shock front is
x = \t + 1. The jump at x = -1 yields w(x, t) = (x + 1)// for -1< x < t - 1
(see Exercise 2.3.18). The full solution to the problem is then easily found
to be
B.101)
w = 0,
This solution is valid only until the time t = A when the expansion wave
и = (x + 1) It overtakes the shock wave. This occurs at the point x = 3.
Within the sector — l<x<t — l, the characteristics are the lines x + 1 =
Af that issue from the point (x, i) = (—1,0) and on which и = {x + l)/t = A.
The characteristics that issue from the x-axis for x > 3 are all vertical lines
on which м - 0. These two sets of characteristics begin to intersect at x = 3
when t = 4 and the solution becomes double valued at that time. A new
shock wave is required and the shock front is determined from x'(t) =
J[(* + l)/f] with *D) = 3. The solution is x{i) = 2Vt-\ and the shock
speed is U = (\ltI12.
Thus, the solution for t > 4 is
B.102)
и = 0 , л; > 2 V? - 1.
The solution has the form of a triangular wave (see Figure 2.6). As t tends
to infinity, the jump in the solution tends to zero. The wave form for the
QUASIUNEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
79
х =
-1
и .
х + 1
t
х**2\
п-
w = 0
1
X
Figure 2.6. A triangular wave.
problem originally formulated in this example assumes a similar triangular
shape as t gets large.
Example 2.11. N Waves. In view of the periodicity of the sinusoidal wave
forms considered in part 3 of Example 2.8, it is sufficient to study the
problem over a single period. However, to simplify matters we consider an
initial wave form that is constant outside a finite interval but has a
quasisinusoidal shape within the interval.
Let и(х,0) - f(x) = 2 for |jc|a=2, fix) = -x for -2<x< -1, f(x) = x + 2
for |jc| < 1, and f{x) = 4-jcforl<jc<2. The wave form has two compres-
sive parts whose slope is -1 and an expansive part with unit slope. Using
the results of the first part of Example 2.8 the solution is easily constructed.
The wave и(дг, t) breaks at the time t = 1 and this occurs at the two points
x = 0 and x = 4.
When f > 1 the solution is given in terms of the characteristics as follows.
For x<2*-2 we have w = 2, for f-K*<3f + l, и = (я: + 2)/A +1),
while и = 2 for x > It + 2. The characteristics intersect at x = 0 and x = 4
when t = 1 and two shock waves are required with shock fronts originating at
these points. The equations for the shock speed B.98) are identical for both
shocks and are given as x\t) -1 + (x + 2)/B + 2/). With jr(l) = O the
solution is л:@ = 2*-B* + 2I/2, and for jcA)=4 the solution is *(/) =
(; + 2I/2.
Thus, the solution for t > 1 is
B.103)
и —
дг + 2
shown in Figure 2.7, the wave form has the shape of an inverted N.
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Figure 2.7. An iV wave.
Therefore it is called an N wave. As t tends to infinity the magnitudes of the
two jumps decrease on the order of 1/V7.
If f(x) = a + b sin x with a > b > 0, the solution takes the form of a
periodic set of N waves after the breaking time. While the foregoing N
waves all travel to the right with increasing f, if f(x) = sin x the shock waves
are all stationary because of the symmetry of the data with respect to the
x-axis. However, they still assume the general form of N waves after the
breaking time.
The equation
B.104) w, + c(w)wx=0,
where c(u) is an arbitrary function of и, is the quasilinear equivalent of the
linear equation B.13) and is more representative of the type of first order
quasilinear wave equations studied in applications. If we multiply across by
c'(u) in B.104), we obtain ct + ccx = 0 and this is exactly B.65) with и
replaced by c. Thus, we do not expect the behavior of the solutions of
B.104) to be qualitatively much different from those of B.65) as long as the
solutions remain smooth. However, if the solution becomes multivalued and
shock waves are required the transformed version of B.104) does not
determine the correct shock fronts unless c(u) = с + си. То see t$iis, let the
conservation law associated with B.104) be given as ut + (C(u))x = 0, with
C'(u) = c(u). Then the shock speed U for this equation is U = [C(w)]/[u].
For the transformed equation we have U= \(сг + сг)9 where c2 and cx are
the values of c(u) on the right and left sides, respectively, of the shock.
These shock velocities are unequal, in general, unless c(u) = с + си with
constant с and c, as is easily verified directly. Nevertheless, the simpler
transformed version of B.104) remains valid until the breaking time and can
be used to determine that time.
NONLINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 81
2 4. NONLINEAR FIRST ORDER PARTIAL DIFFERENTIAL
EQUATIONS
In this section the method of characteristics is extended to deal with
nonlinear first order partial differential equations in two independent vari-
variables. Although solutions of linear and quasilinear equations can be ob-
obtained as special cases of the general method developed below, these
equations are of sufficient special interest to merit the separate discussion
accorded them. Following a general discussion of the method of characteris-
characteristics, we consider several examples among which the classical eiconal equa-
equation of geometrical optics is analyzed in some detail
In its most general form the first order partial differential equation can be
written as
B.105) F(x,t, w, ux,ut) = 0,
where ux = duldx and ut = ди/dt. Let p-ux and q — ut and consider an
integral surface и = u(x, t) satisfying B.105). Its normal vector has the form
[ux, un -1] = [p, q, -1], and B.105) requires that at the point (*, t, u) the
components p and q of the normal vector satisfy the equation
B.106) F(x,t,u,p,q) = 0.
Since an integral surface и = u(x, t) of B.105) is not known a priori, we
think of B.106) as characterizing a collection of admissible solutions or
integral surfaces u = u(x, f). That is, their normal vectors at the points
(x, t9 u) must satisfy B.106). Each normal vector determines a tangent plane
to the surface, and B.106) is seen to generate a one-parameter family of
tangent planes (to possible integral surfaces) at each point in (x, t, u)-space.
We require that F\ + F2q¥>0.
For example, if B.105) has the form uxut -1 = 0, then F = pq - 1 = 0
and q = \lp. As p ranges through all real values, q~l/p determines a
one-parameter family of normal vectors [p, q, -1] = [p, lip, -1] at each
point (x, t, и). Similarly, if F = xu + tu) - (sin x)ux + 1 = 0 then at (jc, t, u)
we have F(x, t, u, p, q) = Xu + tq - (sin x)p + 1 = 0 which yields a set of
values q — q(p). Note that q cannot always be specified as a single valued
function of p. However, we shall always assume that one branch of the
possible set of solutions q = q{p) has been chosen.
In general, the (tangent) planes determined by p and q envelop a cone
known as the Monge cone. Then, if и = m(jc, t) is a solution of B.105), it
must be tangent to a Monge cone at each point (л;, t, u) on the surface. The
Hitersection of the Monge cones with the surface determines a field of
directions on the surface known as the characteristic directions, as shown in
Rgure 2.8.
To find the characteristic directions we proceed as follows. The planes
82
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Characteristic
directions
Figure 2.8. The Monge cones.
determined by the set of normal directions [p, q, -1] at a point (x0, r0, uQ)
satisfy the equations
B.107) H[x91, u,p,q)=zu- uo~p(x - x0) - q(p)(t - t0) = 0 .
It is assumed in B.107) that F[x0, t0, w0, p, q] = 0 is solved for q as a
function of p and one of the set of possible solutions expressed as q — q(p)
is selected. The envelope of the planes B.107) (where p is a parameter) is
determined by eliminating p from B.107) and the equation
B.108)
дН
dp
dq
If we solve for p =p(x, t) from B.108) and substitute the result in B.107)
we obtain the equation of the Monge cone.
For example, if F = pq - 1 = 0, then q-l/p and dq/dp = -lip2. From
B.108) we obtain q =l/p2 = (x- xo)/(t- f0). Substituting in B.107) gives
the equation of the Monge cone as (w - u0J = 4(jc - xo)(t - t0).
Continuing our discussion, we observe that F[x0, t09 м0, р, q] - 0 implies
Since q'(p) = -Fp/Fq, substitution in B.108) yields
B.11U)
x~xo __ t~to
-
Using B.110) in B.107) gives
NONLINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 83
u-u0__x-x0 . Fp ш __pFp + qFq
Combining B.110) and B.111) yields
u-u0 x - x0 t - t0
0 0
B.И2) PF+qFp Fp F
Recalling that (jc, t, u) are the running variables in the tangent planes
B.107) and that the denominators in B.112) are constants evaluated at
(*o> 'o> Mo) on *e integral surface, we see that a fixed direction on each
tangent plane (for particular values of p and q) is determined by B.112).
The direction is determined by the vector [Fp9 Fq, pFp + qFq]. As p and q
range through all their values, these directions, known as the characteristic
directions, determine the family of lines B.112) that generate the Monge
cone at (x0, t0, w0).
We now construct equations for curves x — x(s)91 = t(s), and и = u(s) that
have a characteristic direction at each point, (i.e., the direction of the vector
[F , Fq, pFp + qFq]). The ordinary differential equations for these curves
are given as
BИЗ) f = Fp ,
B-114) | = F,,
B.115) f-/*, + 9V
Since p and q can vary from point to point, the curves B.113)-B.115)
must be chosen such that they lie on a single surface и = u(x, i). To achieve
this result, we assume that a surface и = u(x, t) is given so thatp and q (i.e.,
ux and ut) are known. Then we determine the values that p and q must have
along curves x = x(s), t = t(s)y и = u[x(s), t(s)] that lie on that surface. Since
we must also have p =p[x(s), t(s)] and q = q[x(s), t(s)] on that surface, we
obtain
B.116) ± = PxX'(
B.117) d-l = qxxX
on using B.113), B.114).
84 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Further, if и = u(x, i) is a solution of B.105), we have F[x, t, u, p, q] =
0, which implies
B.118) % = Fx + FuP + FpPx + Fqqx=0,
B.119) ^ = F, + Fuq + Fp/7, + F, *, = 0 .
But ux = p and u, = £ imply that мх, - pt = мгд. = ^. Substituting pt = </x into
B.118), B.119) and inserting the result in B.116), B.117) yields
BЛ20) ± = -Рх-РыР9
B.121) g = _F/_F^.
The five equations B.113)-B.115) and B.120), B.121) now constitute a
completely self-contained system of equations for the functions x(s), t(s),
u(s), p(s), and #(s). [The integral surface и - u(x, t) no longer needs to be
given a priori in order to specify the values of p and q.] They are known as
the characteristic equations for the differential equation B.105).
The initial value problem for B.105) requires that the integral surface
и — u(x, i) contain a curve C, the initial curve, which we give parametrically
as
B.122) * = *(t), ; = '(t), и = и(т).
In terms of the functions x(s), t(s), and u(s), these are taken to correspond
to initial values given at 5 = 0, as was done in the previous sections.
However, in contrast to the situation for linear and quasilinear equations,
we must now also require initial values for p(s) and </(s) if we hope to obtain
a unique solution of the characteristic equations and, consequently, of
the initial value problem for B.105). The initial values p(r) and q(r) on the
curve С cannot be arbitrary, since p and q must be components of the
normal vector to the integral surface и — u{x, t).
If we Jet p = р(т) and q = q{r) be the initial values of p and q on the
curve C, these values must be determined from the equation
B.123) F[*(t), f(r), u(r), p{r), q(r)] = 0 ,
and the so-called strip condition
BЛ24)
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
85
These equations follow from the fact that p{r) and q(r) must be components
of the normal vectors to the surface и = и(х91) evaluated along the initial
curve C. The data р(т) and ^(т) are determined by solving B.123) and
B.124) simultaneously. As indicated previously, there may be more than
one set of solutions for р(т) and q(r). However, in each problem we select a
specific set of initial values jc(t), г(т), m(t), р(т), and q{r). Further, the data
n(r) and q{r) may be complex valued, in which case the solution u(x9 i) will
also be complex valued.
Given the characteristic equations and initial conditions for x, t, u, p9 and
л, together with the requirement (see the Appendix)
B.125)
on the initial curve C, we can obtain a unique solution of the initial value
problem for B.105) in a neighborhood of the initial curve С We note that
the assignment of x, t and и together with p and q on the curve С means that
we are specifying an initial strip, that is, a space curve х(т), t(r), м(т)
together with a family of tangent planes at each point having a normal
vector with components [р(т), q{r), -1], as shown in Figure 2.9. This
explains why B.124) is called the strip condition.
The solutions x(s, т), t(s, т), and m(s, t) of the characteristic equations for
fixed т determine a space curve. The functions p(s, т) and q(s9 r) determine
a tangent plane with normal vector [p, q, — 1] at each point of the space
curve. The combination of a curve and its tangent planes is called a
characteristic strip. The integral surface и = u{x> i) is constructed by piecing
together these characteristic strips to form a smooth surface. Its equation is
found by solving for s and т as functions of x and t from x = x(s, т) and
f = t(s, t) and inserting the results in и = u(s9 т).
In the following examples we demonstrate the use of the method of
characteristics for solving nonlinear equations by applying it to several
specific initial value problems.
Figure 2.9. The initial strip.
86 FIRST ORDER PARTIAL DIFFERENTIAL EQUATION!
Example 2.12. A Nonlinear Equation. We consider the equation
B.126) F(x91, м, ux, ut) = uxut -1 = 0,
with the initial condition
B.127) u(jc,O) = jc.
In parametric form the initial conditions are given as
B.128) * = r, f = 0, w = r.
The equations B.123) and B.124) for determining/?^) and ^(т) are
and, therefore,
B.130) Р(т) = 1, *(t)-1.
Also, on the initial curve we have x'(r)Fp - t'(r)Fq = q = 1 so that B.125) is
satisfied.
The characteristic equations are found to be
B.131) £-,, ^=p, %2Pq, ^0, ^
v ' ds ^ ds ^ ds ™ ds ds
The initial curve B.128) corresponds to s — 0. Since p and q are constant on
the characteristics we have
B.132) p(s, r) = p@, т) = 1 > q(s, r) = q@9 r) = 1 .
Solving for jc, t, and u and using B.128) and B.132) gives
B.133) Jt($, r) = 5 + r , f($, r) = 5 , mE, t) = 2j + t .
We invert the first two equations to determine s and r as functions of x and f
and obtain
B.134) s = f, r = jc-f.
Inserting this result into the third equation in B.133) gives
B.135) u(x,t) = x + t,
which clearly satisfies B.126) and the initial condition B.127).
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 87
Example 2.13. A Nonlinear Wave Equation. We consider the equation
B.136) F(x, t, и, ux, ut) = ut + u\ = 0 ,
with the initial condition
B.137) u(x,Q) = ax,
where a is a constant. The equation B.136) is of the general form ut +
c(u,ux)ux=0, of which two special cases с = constant and c = u were
considered in the previous sections. We might expect that B.136) has some
features in common with the linear and quasilinear wave equations consi-
considered earlier and refer to B.136) as a wave equation. The nonlinearity of
the x-derivative term in B.136) implies, as we demonstrate, that the
solution и = u(x, i) is not constant along the characteristic curves. However,
the velocity of specific points x on the wave is related to the nature of the
wave form at those points as was the case for the quasilinear equation
B.65).
To solve this initial value problem we parametrize the initial curve as
B.138) x = r9 t = 0, м = дт.
From B.123) and B.124) we obtain
B.139) <7(т)+р(тJ = 0, a=p{r),
which yields as the initial data for p and q,
BЛ40) p{r) = a9 q(r) = -a\
Since x\r)Fq - t'(r)Fp = 1, condition B.125) is also satisfied.
The characteristic equations for this problem are found to be
dx dt du д 7
„MI) *=2p' =■'■ *-q+2p'
ds U' ds "•
Using the data B.139), B.140), we obtain
B-142) x(s, t) = las + т , t(s, т) = s , u(s,t) = a2s + ат
B.143) p(s,r) = a, q(s,r) = -a2.
88
Solving for
gives
FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
t, r), t = t(*, t) from the first two equations in B.142)
B.144) s = t, r = x-2at,
and the solution of B.136), B.137) becomes
B.145) w(x, г) = a[x - at].
The solution u(x, i) = a[x ~ at] represents a plane wave moving with velocb
ty a. If a > 0, it moves to the right, whereas if a < 0, it moves to the left as
shown in Figure 2.10.
Although the solution u — a[x — at] behaves like the solution of a linear
unidirectional wave equation with velocity a, the wave u = a[x — at] is not
constant on the characteristic base curves, x — 2at = r. Points on these
curves travel with velocity 2a, whereas points on the wave и = a[x - at]
have velocity a.
In fact, if we replace the parameter s by t in the equations B.141) we
obtain
B.146) f=
!-•■
Since p = constant and p = ux, we conclude that points x move with a
velocity equal to twice the slope of the wave form и = и(х, t). However, the
fact that и Ф constant on the characteristics implies that this velocity is not
that of points on the wave moving parallel to the jc-axis. This is consistent
with the foregoing results. It does indicate that if the initial wave form
Figure 2.10. The wave motion.
NONLINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 89
w(x, 0) =/(*) has variable slope, points x where/(x) has positive slope move
t0 the right, whereas points where f(x) has negative slope move to the left,
дко, the greater the magnitude of the slope, the greater the velocity. This
process continues throughout the wave motion and can lead to the break-
breakdown of the solution even for smooth initial wave forms. This difficulty did
not arise for the problem B.136), B.137) since the initial slope is constant.
In the exercises another initial value problem for B.136) is considered
where the problem of the breaking of the wave и — м(лг, t) does occur.
Example 2.14. The Eiconal Equation. The eiconal equation of geometrical
optics in two space dimensions has the form
B.147) ux + u] = n2 ,
and we shall assume that n = constant. We present a derivation of the
eiconal equation in the exercises that exhibits its connection with wave
propagation problems. In this example we replace the variable t of B.105)
by the variable у so that q = uy and у takes the place of t in B.105)-B.125).
As the initial condition for B.147) we take
B.148) w(jc, y)\xs=y = u(y, y) - ay ,
where a = constant. Parametrically, this can be expressed as
B.149) x = r, у = т, м = ат.
The initial values р(т) and q(r) on the curve B.149) are determined from
B.123) and B.124) as
B.150) />2 + ?2-n2 = 0,
B*151) a=p(r) + q(T).
Equation B.150) states that the vector [p/n, qln] is a unit vector. Let
0 = constant with 0 < в < 2тг, and set
B*152) p{r) = n cos в , ^f(r) = n sin в ,
so that B.150) is satisfied and then B.151) implies
B*!53) cos в + sin в = л/2 sin( в +£) = -.
V 4/ n
[We select one of the solutions of B.153). If, for example, aln = 1, we can
have в = 0 or в = тг/2.] The condition B.125) requires that
90 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
B.154) 2(sin 0 - cos 0) * 0 .
Since this expression vanishes only at 0 = тг/4 or 0 = 5тг/4, we must exclude
those values of 0 and the related values of a; that is, a — ±wV2.
The characteristic equations are
B.155)
dx dy du л
^-=2/7, ТГ=2<Ь —=2p
Thus p and # are constant on the characteristics so that
B.156) p(s, t) - p@, т) = л cos 0 , q(s9 r) = ^@, r) = nsinO ,
and we obtain the straight line characteristics:
B.157) x(s, r) = 2/w cos 0 + r , y(s, т) = 2и.у sin в + r
B.158) w(s, r) = 2w2.y + (sin в + cos 0)«r ,
where B.153) was used. Solving for s and r as functions of x and у from
B.157) yields
B 159)
2w(sin 0 - cos 0) ' sin 0 - cos 0
Inserting this into B.158) gives the plane wave solution
B.160) u(x, y) = n(x cos 0 + у sin 0).
It can be verified directly that B.160) satisfies the eiconal equation and the
initial condition B.148), on using B.153). A more general plane wave
solution of the eiconal equation can immediately be constructed from
B.160) in the form
B.161) m(x, y) = u0 4- n[(x - x0) cos 0 + (y - y0) sin 0],
where Jt0, y0, and u0 are constants.
A second initial value problem for the eiconal equation, where the values
are given at a point rather than on a curve, leads to a singular solution of
B.147) which is called a cylindrical wave. Let the initial point be (л:, y9 u) =
(jc0, y0, u0) so that the initial values are
B.162) x(r) = x0 , y(r) = y0 , m(t) = "o •
NONLINEAR FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 91
The "strip condition" B.124) is satisfied automatically for all/? and q. Thus
p(r) and q(r) are determined from the single equation
B.163) F[x09 yQ, u0, р(т), q(r)] = p(rf + q{rf - n2 = 0 .
Let
B.164) p(r) = n cos т , q(r) = л sin т ,
and B.163) is satisfied. Proceeding as in the preceding initial value problem
we again find that p(s, r) and q(s, r) are constant on the characteristics and
B.165) x(s, t) = 2ns cos т + дс0 , }>(.*, r) = 2ns sin т + y0
B.166) ф,т) = 2п25 + м0.
Squaring and summing the equations for x and у in B.165) and solving for s
readily yields, if we take the positive square root,
B.167) u(x, y) = u0 + wV^-XoJ-^-^J
as a singular solution of the eiconal equation. It clearly satisfies B.147)
everywhere except at the initial point (x0, y0) where the derivatives ux and
uy are singular. In fact, B.167) determines a cone with vertex at
(*o> Уо> wo)> which is identical with the Monge cone through that point.
In the context of geometrical optics, if и = w(jc, y) is a solution of the
eiconal equation, the level curves m(jc, y) = constant represent wave fronts,
whereas the characteristic base curves x = x(s, т) and у = y(s, r), for fixed т
represent light rays. Thus the level curves of the solutions B.160) or B.161)
are straight lines. In terms of the physical three-dimensional problem for
which we are considering two-dimensional cross sections, the straight lines
correspond to planes. Hence they are called plane waves. Similarly, the level
curves of the solution B.167) are circles that in the three-dimensional
problem correspond to cylinders and result, therefore, in cylindrical waves.
The tangent vectors to the characteristic base curves (i.e., the light rays) are
found from B.155) to be [dx/ds, dy/ds] = 2[p, q]. These curves are straight
lines since p and q are constant for fixed r. Further, the normal vectors to
the level curves u(x, y) = constant are Vw = [ux, uy] = [p, q]. This shows
that the light rays are normal to the wave fronts.
The intensity of the light in geometrical optics is characterized by the
convergence or divergence of the light rays. As the rays converge the
intensity increases, whereas the intensity decreases as the rays diverge. The
singular point (x0, y0) in the above initial value problem is known as a focal
Point or a focus, since all the rays are seen to converge at that point or
diverge from it. It represents a point of high intensity of light. Another
92 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
region of high intensity of light is given by curves that are envelopes of the
characteristic base curves or the light rays. Such curves are known as caustic
curves. Although the solutions of the eiconal equations break down at foci
or caustic curves, other methods not based solely on the eiconal equation
lead to valid results in such regions. They are discussed in Chapter 10.
To complete our discussion of the eiconal equation we show how the
singular solution B.167) may be used to solve a general initial value
problem for the eiconal equation. The method we use is related to the
complete integral method for solving first order partial differential equations,
A complete integral of the equation B.105) is a solution of the form
B.168) М[х,^и,а,Ь]=0,
where a and b are arbitrary constants. In our problem the solution B.167),
in fact, contains three arbitrary constants, xQ, yQ, and м0. We now indicate
how B.167) can be used to solve the initial value problem B.149). For
simplicity, we assume that the constants a and n are chosen such that a = n.
This implies that we either have 0 = 0 or в = тг/2 in B.160), so that
m(jc, y) = nx or u(x, у) = ny.
We set хо — т, y0 = r, and ио = пт in B.167) so that (jc0, y0, u0)**
(тут9пт) represents the initial curve B.149). This yields the family ol
solutions
B.169) и(х,у) = пт + пУ(х-т)
Expressing the solution in implicit form as
B.170) #(jc, у, м, т) = и - пт- пУДх^
and differentiating with respect to r gives
B.171) ^ = -„ + „
where B.170) and B.171) are the equations that determine the envelope of
the family of solutions B.169). We easily conclude from B.171) that
B.172) (
so that either r = x or т = у.
With r = x in B.170) we have
B.173) H = и - nx - пУ/(у-хJ = и-пх-п\у-х\=0,
and with т = у
APPENDIX: ENVELOPES OF CURVES AND SURFACES 93
BД74) H=u-ny- ri\J(y-xf = и - ny - n\y - x\ = 0 .
Two continuously differentiable solutions can be obtained from the above in
the form
B.175) w(x,y) = iw,
B.176) Ф> У) = пу,
which correspond to the values 0 = 0 and в = тг/2 in B.160), respectively.
Both solutions clearly satisfy the initial condition и = nx = ny on the line
х-У-
The fact that there are two solutions to the initial value problem B.147),
B.148) merits discussion. It signifies that merely specifying и on a curve
without also specifying p and q (i.e., an initial strip) does not yield a unique
solution, in general. The eiconal equation, F = p + q2 - n2 - 0 is a quad-
quadratic expression for which q = ± Vw2 - p2, so that for every choice of p there
are generally two values of q. This means that there are, in general, two
differentiable solutions passing through a given initial curve. The import-
importance of the existence of these two solutions in applications of the eiconal
equation is indicated in the exercises, and is emphasized in our discussion of
the eiconal equation in Chapter 10.
APPENDIX: ENVELOPES OF CURVES AND SURFACES
Let F(jc, f, r) = 0 be a one-parameter family of curves in the (дс, t)-plane
with parameter r. The envelope of the family of curves is determined from
the equations
dF
F(x,t,r) = 0, —(x,t,T) = O.
ОТ
Solving for т = t(x, i) from the second equation and substituting in the first
equation gives the envelope
It is generally assumed that each curve in the family intersects the envelope
tengentially at least once. Otherwise the resulting curve is not referred to as
an envelope. If F is a function of more than two variables and depends on a
Parameter r, the foregoing procedure for constructing the envelope also
applies.
As an example, let a curve be defined as t = f(x). If [т, /(т)] is a fixed
Point on that curve, the tangent line to the curve at that point has the
equation
94 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
F(x, t, r) = t-f(r) - (x - T)f'(T) = 0 .
As the parameter r varies, we have a one-parameter family of tangent lines
to the curve. Now,
implies if /"(т) ^ 0 for all r, that r = Jt. Inserting this in the equation
F(x, f, r) = 0 yields the envelope
Thus the given curve t — f(x) is the envelope of its family of tangent lines.
The family of curves is tangent to the envelope at, at least, one point.
If we consider the family of curves t — |x| + r, and express them as
we find that FT = -2(t - r) = 0. This yields т = t and we conclude that x = 0.
However, the f-axis is not an envelope of this family of broken lines, since
none of them is tangent to the f-axis. It does represent the locus of singular
(nondifferentiable) points of this family. As another example we consider
the pencil of straight lines through the origin F = t — rx — 0. Combined with
the equation FT — 0 we obtain x = t = 0. While the origin is not an envelope
for this family of lines, it is the point where they all intersect.
As an example of a three-dimensional problem we consider the family of
spheres
F(x, y, z, r) = (jc - rJ + у2 + z2 - a2 - 0 ,
whose radii equal a and whose centers are the points (r, 0,0), where т is a
parameter. Since
2(*тH,
we find that т = jc. The envelope of the family of spheres is, therefore, given
by
This is the equation of a circular cylinder of radius a, whose axis coincides
with the x-axis.
If the family of curves is given in parametric form
: ENVELOPES OF CURVES AND SURFACES 95
x = x(s, т) , t = t(s, t) ,
we let s be the running parameter along a single curve and т be the
parameter specifying a member of the family. Geometrically, the curves of
the family are tangent to the envelope at their points of intersection, so we
may characterize т as a parameter along the envelope curve (each value of т
specifies a point of intersection on that curve). Now, [дх/ds, dt/ds] is a
tangent vector of a member of the family for each fixed т, whereas
[StISr, ~dxldr\ is a normal vector of the envelope curve when s is evaluated
at the envelope. Since the tangent vector of a member of the family of
curves is orthogonal to the normal vector of the envelope at the point of
intersection, we have for their dot product
This equation describes the envelope of the family.
The envelope equation is to be compared with the conditions B.60) and
B.125). Noting B.63), B.113), and B.114) we see that both of the
conditions B.60) and B.125) are equivalent to
In terms of the initial value problem for the equations B.57) and B.105) the
condition Д@, т) Ф 0 requires that the initial curve С be neither a charac-
characteristic curve nor an envelope of characteristic curves. Otherwise the
solution may not exist or, if it does exist, need not be unique. Additionally,
even if the initial curve satisfies the condition B.60) or B.125), when the
family of characteristics forms an envelope, the solution breaks down, as is
seen for the quasilinear equation B.65) and is shown for the eiconal
equation in Chapter 10.
For example, the circle x2 + t2 = a2 in parametric form is
x = a cos т , t = a sin т ,
and the tangent lines are
x(s, т) = a cos т — as sin т , t(s, т) = a sin т + as cos т ,
as is readily seen. Here 5 is the running parameter along a tangent line and т
characterizes a particular tangent line. We have
A(s, t) = xstr - t$xT = a2s ,
s° that Д^, т) = 0 implies that s = 0 is the envelope. Then, x@, r) = a cos т,
*@, t) = a sin т yields the circle x2 + t2 = a2, as was expected.
96 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
EXERCISES FOR CHAPTER 2
Section 2.1
2.1.1. Show that if we set и = vt and w = vx in the wave equation B.1), we
obtain the system
и* = wt '■ wr = У V, .
Note that this system is coupled, and in contrast to B.2) neither of the
equations can be solved independently of the other. Thus the manner in
which an equation of higher order is represented as a system can play a
significant role.
2.1.2. Express the telegraphers equation A.88) in the form
(<?, + ydx)(dt - ydx + 2A)u - 2kyvx = 0 ,
and obtain the coupled system
vt - yvx = и - 2\v , ut + уux - 2X^17^ = 0.
2.1.3. Consider the equation vtt — у 2vxx + avx + bvt + cv = 0 with constant
coefficients. Use factorization to obtain the system nt + Au^ + Cu = 0 where
=[bc o1]-
Show that if a = yb and с = 0, the differential equation for и can be solved
independently of that for v.
2.1.4. Show that the functions и and w in the Cauchy-Riemann equations
B.6) both satisfy Laplace's equation.
Section 2.2
2.2.1. Solve the initial value problem for the damped unidirectional wave
equation,
vt + cvx + \v = 0 , u(x, 0) = F(x) ,
where A > 0 and F(x) is given.
2.2.2. (a) Solve the initial value problem for the inhomogeneous equation
where f(x, t) and F(x) are specified functions,
(b) Solve this problem when f(x, t) = xt and F(x) = sin x.
EXERCISES FOR CHAPTER 2 97
2.2.3. Discuss the solution of the wave equation B.1) in the following cases:
(a) v(x, 0) = /(*) - x; vt(x9 0) = g(x) = 0.
(b) f{x) = 0;g(x) = x.
(c) f(x) = sin x; g(x) = -y cos x.
(d) f(x) = sin jc; g(x) = у cos *.
Based on the results obtained, observe that solutions of the wave equation
may or may not have the form of traveling waves.
2.2.4. Consider the inhomogeneous wave equation
(a) Apply the method of Example 2.1 to reduce it to the system
vt-yvx = ut щ + уux = F .
(b) Solve the initial value problem for the inhomogeneous wave
equation with the initial data v{x, 0) = /(jc), u,(jc, 0) = g(jc).
2.2.5. Solve the "signaling" problem for B.13) in the region jt>0 with the
boundary condition v@, t) = G(t) for -«<t< +«.
2.2.6. Show that the initial value problem
vt + vx = 09 v = x on x2 + t2 = 1
has no solution. However, if the initial data are given only over the
semicircle that lies in the half-plane x + f ^0, the solution exists but is not
differentiable along the characteristic base curves that issue from the two
end points of the semicircle.
2.2.7. Solve the initial value problem
2.2.8. (a) Show that the initial and boundary value problem for B.13) with
с > 0 can be solved in the quarter-plane x > 0, t > 0 with the data
v(x, 0) = F(x) and u@, t) = G(t), if F and G are arbitrary smooth
functions that satisfy the conditions G@) = F@) and G'@) =
-cF'@).
(b) Show that for с <0, the problem can be solved only if the data F
and G satisfy a compatibility condition. In this case, the charac-
characteristic base curves intersect both the x- and the f-axes.
*-2.9. If Vx has a jump across the characteristic base curve x = x(s)91 = t(s),
but the generalized solution v of B.8) is continuous there, show that
±
98 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
If vx has a jump at a point on the initial curve, this determines the variation
of [vx] along the characteristic through that point. Once [vx] is determined
[vt] can be obtained from B.31).
2.2.10. Let f(x) in B.38) be given as f(x) = 0 for jc<0 and f(x) = x for
x>0. Obtain the generalized solution of this problem and use it to
determine the jump in vx. Show that this jump satisfies the equation found
in Exercise 2.2.9.
2.2.11. If the function d(x, t) in B.8) has a jump discontinuity across a
(noncharacteristic) curve but the coefficients a, fe, and с are smooth func-
functions, obtain an equation for the jumps in vx and vt across that curve.
(Assume that v is continuous there.) Show that if d(x, i) is discontinuous
across a characteristic base curve, the solution v must also have a jump
along that curve.
2.2.12. Solve the initial value problem of Exercise 2.2.2 if (a) f(x, t) = H(x)
and (b) /(*, t) = H(x - ct). In both cases set F(x) = 0. Here H(z) is the
Heaviside function defined as H(z) = 0 for z < 0 and H(z) = 1 for z > 0,
Verify the conclusions of the preceding exercise.
2.2.13. Let/(x) = H(x) in B.38), where H(x) is the Heaviside function (see
Exercise 2.2.12). Find the generalized solution of this initial value problem
for B.37) and verify that B.36) is satisfied for this equation. Also, show that
the jump in v satisfies the ordinary differential equation given in the
discussion in the text that follows equation B.36).
2.2.14. Let the coefficients a and b in B.8) have jump discontinuities acros$
a curve x = x(s), t ~ t(s). Assume that v also has a jump across that curve
and use the integral relation B.34) to show that that the values of a, b, and
v on opposite sides of the curve (with unit normal vector n) are related by
(The subscripts denote values on opposite sides of the curve.)
2.2.15. Consider the initial value problem
vt + c(x)vx = 09 c = cl9 x<0, c = c29 x>0, ф,0)=/(*)
where сг and c2 are positive constants. Use the jump condition obtained in
the preceding exercise to solve this problem.
Hint: First solve for v in the region *<0 and f>0. Then use the jump
condition as well as the initial condition to solve for v in x > 0 and / > 0.
Note that v has a jump across x = c2t unless Д0) = 0.
2.2.16. Show that the initial value problem
(t-x)vx-(t + x)vt = Q, v(x,Q)=f(x)> x>0>
has no solution if f(x) is an arbitrary function.
EXERCISES FOR CHAPTER 2 99
Hint: The singular point at the origin gives rise to characteristics that spiral
around the origin.
2 2.17. Show that the equation a(x)vx + b(t)vt = 0, has the general solution
v(x, t) = F[A(x) - B(t)], where A'(x) = l/a(x) and B'(t) = 1/&(')•
2 2 18. Show that the equation a(t)vx + b(x)vt = 0 has the general solution
v(x, 0 = F[B(x) - A(t)], where B'(x) = b(x) and A'(t) = a(t).
2.2.19. Given the initial value problem
tvx + xvt = cv , у(х, дс) ^/W , Jc > 0
where с is a constant, determine a condition on the function f(x) so that this
becomes a characteristic initial value problem. Supposing that f(x) satisfies
that condition, obtain a (nonunique) solution of the problem.
2.2.20. Show that the initial value problem
ut + ux = x , u(x9 x) — 1
has no solution. Observe that the initial curve t — x is a characteristic base
curve and explain why this is not a characteristic initial value problem.
2.2.21. (a) Show that the problem
ut + cux = F(x, t), w (*,-*)= f(x)
is a characteristic initial value problem if f'(x) =
(l/c№,(l/c)*].
(b) Verify that
is a solution of the characteristic initial value problem of part (a)
if g@) =/@) but g(z) is an otherwise arbitrary function,
(c) Let F(x, t) = cos(* + t) in part (a). Determine the appropriate
choice of f(x) and obtain the general solution of the problem.
2-2.22. Show that the Jacobian Д(,у, т) satisfies the equation
Thus, if ДE, T) vanishes at s = 0 and ax and bt are continuous, A(s, т)
vanishes identically.
Hint: Use equations B.9).
2.2.23. Determine Д(.у, т) for equation B.54) with initial data on t = 0. Note
«lat the Jacobian vanishes at s = 0 but not for s > 0. Show that Д(.у, т)
100 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
satisfies the equation of the preceding exercise and explain why the result of
that exercise is not contradicted.
2.2.24. Obtain the general solution of the equation
Show that t = 0 is a characteristic for this equation, but the initial value
problem with v(x,0) - /(x) has a unique but nondifferentiable solution (at
t = 0) for arbitrary /(x). Show that the characteristics are all tangent to the
x-axis.
2.2.25. Generalize the method of characteristics to problems in three di-
dimensions. Given the linear equation
avx + bvy + cvz = dv + e ,
where a, b9 c, d, e, and v are functions of (x, y, z), obtain the characteristic
equations
dx dy dz dv
ds ds ds ds
The initial value problem for this case specifies an initial hypersurface given
parametrically in (x, y, z, y)-space, as
х = х(А,т), У-у(А,т), z = z(A,r), u = i;(A, t),
with A and т as the parameters. The family of characteristic curves is now
given as
x = xO, А, т) , у = yE, A, r) , z = z(s, А, т) , v = у(.у, A, r) ,
with j as the running parameter along a curve and (А, т) as a two parameter
family specifying the individual curves. If the equations for (x, y9 z) can be
inverted to yield {s, A, r) as smooth functions of (x, y, z), the function
v = u(x, y, z) obtained thereby is a solution of the initial value problem.
Discuss situations in which the method of characteristics may not give a
solution or yields a nonunique solution.
2.2.26. Using the method of characteristics solve the initial value problem
vx + vy + v2 = 0 , y(x, y, 0) = /(x, y) ,
where Дх, у) is given.
(a) Parametrize the initial data and let x - A, y = r, z = 0, and
v =/(A, r) and set up the characteristic equations.
EXERCISES FOR CHAPTER 2 101
(b) Show that the solution of these equations is
(c) Conclude that the solution of the initial value problem is
v(x, yyz) = f(x-z, y-z).
2.2.27. Extend the method of characteristics to linear equations in n-
dimensions. Given the equation
2 e,(xlf ...,*„) — = b(x\> • • * >xn)v + c(jfi» • • • > *J >
where у = v(xx,.. . , xn), obtain the characteristic system
dxt ... du
-^ = «,. (i-l,...,n), ^-ta + c.
The initial values for this problem are given as xt = xt(rl9. . . , т„_х) and
i? = u(t1s ... , тп_!) at j = 0, with r19. . . , Tn-1 as parameters. The charac-
characteristic curves are xt = *,.($, Tlf. .., rn^x) and v = u(£, т1?. . ., rn_x) where
i = 1,2,..., л. If 5, т19. . . , Tn_t can be solved for in terms of xl9.. . , xn,
then v = v(xl9. .., xn) is a solution of the initial value problem.
2.2.28. Use the method of characteristics to solve the initial value problem
n
2 v =0, v(x1,x2,...,xn.l,O)=f(x1,x2,...,xn^.1),
1 = 1
where / is given.
(a) Show that the solution of the characteristic equations can be
given as
xn=s, »=Лп Vi)«
(b) Conclude that the solution of the problem is
!>(*!,..., Xn) - f(xx - Xn, X2 - Xn, . . . , X||_1 - xj .
Action 2.3
^•3-l. Using implicit differentiation, verify that и = /(jc - m) is a solution of
lfte wave equation B.65).
102 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
2.3.2. Consider the damped quasilinear wave equation
ut + uux + cu — 0 ,
where с is a positive constant.
(a) Using the method of characteristics, construct a solution of the
initial value problem with u(x, 0) =f(x), in implicit form. Discuss
the wave motion and the effect of the damping.
(b) Determine the breaking time of the solution by finding the
envelope of the characteristic curves and by using implicit dif-
differentiation. With т as the parameter on the initial line show that
unless f'(r)< —c, no breaking occurs. Contrast this result with
that for the undamped case discussed in the text.
2.3.3. Let f(x) ~ ax in Exercise 2.3.2. Obtain an explicit representation of
the solution and show directly that the wave does not break if a > - с If
a<~c the wave breaks at the time t = (lie)log[a/(a + c)].
2.3.4. Consider the one-dimensional form of Euler's equations for isen-
tropic flow (8.315), (8.322) and (8.325) and assume that the pressure p is a
constant. The equations reduce to
pt + pux + upx = 0 , ut + uux = 0 .
Let u(x, 0) = f(x) and p(x, 0) = g(x). By first solving the equation for и and
then the equation for p, obtain the implicit solution
u = f(x - ut), p = g(x - ut)l[l + tf'(x - ut)}.
2.3.5. Obtain explicit expressions for the solution given in the preceding
exercise if (a) f(x) = x, g(x) - 1 (b) f(x) - x, g(x) = x.
2.3.6. Solve the initial value problem for the equation
ut + cux + u2 = 0 , u(x, 0) = x ,
where с is a constant.
2.3.7. Obtain the solution of the initial value problem
where с is a function of u, in the implicit form и =f[x - tc(u)]. Discuss the
solution in the cases where c'(u)>0 and cr(u)<0.
EXERCISES
FOR CHAPTER 2 103
2 3.8. Using the implicit form of the solution obtained in Exercise 2.3.7,
determine the breaking time of the wave.
2.3.9. Solve the initial value problem
u, + u2ux = 0 , u(x, 0) = x .
Determine the breaking time of the solution and compare it with that
obtained through the result of Exercise 2.3.8.
2.З.Ю. Solve the initial value problem
ut + uux = x , u(x, 0) = f(x) ,
using the method of characteristics.
(a) Using the parameters s and т as defined in the text, show that
the solution can be expressed as x(s9 r) = £[/(T) + T\e$ ~
\[f{r) - r]e"\ t = 5, u(s, t) = \[f{r) + r]es + \[f(r) - т]е~\
(b) Obtain the solution in the form и = u(x, t) when f(x) - 1 and
when f{x) = x.
2.3.11. Consider the initial value problem
ut + uux = 0 , u(x, 0) = м0 + eF(x) ,
where ио = constant, 0<e<^l, and F(x) is uniformly bounded for all x.
With € = 0, и = w0 represents a constant solution of the wave equation. The
initial condition represents a small perturbation around the constant u0, so
we look for the solution in the form of a perturbation series around the
constant state и = u0.
(a) Let
и = u0 + eu^x, t) + €2u2(x, t)-\
and substitute this series formally into the equation for и.
Collecting like powers of e and equating their coefficients to
zero, obtain a recursive system of equations for the functions ut
(i > 1). [Determine only the equations for ux and w2.] Show that
the appropriate initial conditions for the щ are их{х^)~
F(x),M,(x,0) = (U> 2.
(b) Show that the solutions for иг and u2 are
мх(х, 0) - F(x - uot), м2(х, 0 = -tF'(x - uot)F{x - uot) .
(c) Since u = uo + eu1-\- e2u2 + • • •, show that when -etF' «1, the
term e2u2 is of the same order of magnitude as еиг and the
terms in the series do not get smaller as was assumed. Compare
104 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
the time t at which the series breaks down with the breaking
time given in the text, as applied to the initial value problem,
(d) Obtain the foregoing results for иг and u2 by inserting the series
for и directly into the implicit form of the solution given in
B.73), expanding the function /in a series in € and comparing
like powers of e.
2.3.12. Show that the initial condition и = t2/2 on x = t3f6 for the equation
w, + uux — t, yields a characteristic initial value problem. Obtain the implicit
solution x = ut - /3/3 + F(u - t2i2) where F@) = 0 but F(z) is otherwise
arbitrary.
2.3.13. Given the initial condition и = 1 — Alt2 on x = t + 1I At for equation
B.65), express the initial curve in parametric form by setting t = r. Show
that Д(т) = О but the initial curve is not characteristic. Obtain two continu-
continuous solutions for this problem neither of which is differentiable on the initial
curve.
Hint: See Example 2.8, Part 2.
2.3.14. Let x - |t2, t = т and и = т be the initial curve for equation B.65).
Verify that Д(т) = О but that this is not a characteristic initial value problem.
Obtain two solutions for this problem.
2.3.15. Solve the equation of Exercise 2.3.12 with the initial curve given as
in the preceding exercise. Show that the initial curve is not characteristic but
that Д(т) = 0 there. Obtain an implicit solution of this problem.
2.3.16. Solve equation B.65) with the initial value w@, t)= —t. Express the
initial curve as x = 0, t = т, and и = — т and show that Д(т) = О only when
т = 0. Use the method of characteristics or B.73) to obtain the solution
Observe that u(x, t) is not differentiable on t2 + Ax = 0 and is not continuous
at t = 0, x > 0.
2.3.17. Given the initial values for B.65), u(x,0) =/(*), *<a and
u(x, 0) = g(*), *>a, with f(a)<g(a), obtain the continuous solution (in
implicit form) w(*, t) =f(x - tu), x<tf(a) + a, u(x,t) = (x~a)/t, tf(a) +
a<x< tg(a) + a, and u(x, i) = g(x - fw), я:> tg(a) + a.
Hint: Consider the characteristics that issue from the point (x, t) = (a, 0).
2.3.18. Let /(*) = A and g(*) = В (A < B) in Exercise 2.3.17. Obtain an
explicit expression for the solution u(x, i) for all x.
2.3.19. If the initial condition for B.65) is u(x9 0) = A, x < a and u(x, 0) -
B, x>a9 with A > B, show that the wave breaks immediately. Determine
the form of the shock wave and show that it is stationary if В = - A.
2.3.20. Let u(x, 0) = A, x < - A, u(x, 0) = -x, -A<x<A and u(x, 0) =
- A, x > A, with A > 0, be the initial values for B.65). Solve for u(x, t) and
EXERCISES FOR CHAPTER 2 105
determine the breaking time. Construct the appropriate shock wave and
show that it is stationary.
2.3.21. By introducing a shock wave when necessary, obtain a solution of
the initial value problem for B.65) with u(x, 0) = 1, x < -1, u(xy 0) = -*,
-1 < x <0, and u(xy 0) = 0, x >0, that is valid for all time.
2.3.22. Let u(x, 0) = 2, л: < 0, и(х, 0) = 1, 0 < x < 1 and u(x, 0) = 0, x> 1 be
the initial values for B.65). Show that the shock that originates at jt = O
overtakes the shock from x = l at the time t=\ at the point * = 3/2.
Introduce a new shock wave and extend the validity of the solution for all
time.
2.3.23. Using graphical means or otherwise show that if x = тг and t > 1 in
B.91) there is one root тг for equation B.91) that lies in the interval
0 < т < тг. Verify that r0 = тг and т2 = 2тг - тг are also roots of this equation.
Since и is constant on each characteristic, show that w(tt, t) is triple valued
for t s: 1. Show that w is equal in magnitude but opposite in sign on the two
characteristics that intersect the line x = тг from the left and from the right.
Conclude that the corresponding shock wave is stationary.
2.3.24. Multiply B.65) by и and obtain the conservation law (u2/2)t +
{иъ1Ъ)х = 0. From the corresponding integral relation determine the equa-
equation for the shock front that takes the place of B.98). Show that if
u(x9 0) = 1, x < 0 and u(x, 0) - 0, x > 0, the shock speed U = 2/3. On the
basis of the conservation law B.95), however, the shock speed [/=1/2.
Thus, even though both conservation laws are equivalent to B.65) if the
solutions are smooth, ^iey give rise to different shock waves. The correct
form of the conservation law must be known to guarantee the correct choice
for the discontinuity solutions.
2.3.25. Verify that if c(u) in B.104) is a linear function of w, the shock
velocity has the form given in the text.
2.3.26. Differentiate B.65) with respect to x and show that on a characteris-
characteristic curve along which x'(t) = w, ux satisfies the equation
Let ux = A at t — 0 and show that the solution of this ordinary differential
equation becomes singular when t=-l/A. Consequently, any point on the
initial wave form where the slope A is negative is associated with a breaking
time t=-l/A. This result agrees with that given in equation B.76).
2.3.27. Generalize the method of characteristics to deal with the three-
dimensional quasilinear equation
aux + buy + cuz = d ,
where a, by c, and d are functions of x, yy z, and и. Show that the
characteristic equations are
106 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Л - n dy - к dz du j
~fc~a> *"*' ^=C' ^=rf;
discuss appropriate initial conditions.
2.3.28. Solve the initial value problem
ux + uy + uz = m2 , w(*, y, 0) = x + у .
Section 2.4
2.4.1. Solve the initial value problem
u\ut -1 = 0, w(jc, 0) = x .
2.4.2. Use the method of characteristics to solve
2.4.3. Obtain two solutions of B.136) if u(xyx) = 1. Verify that B.125) is
satisfied for both solutions.
2.4.4. Obtain two complex valued solutions of B.126) if u(x, x) - 1.
2.4.5. Solve B.136) if м@, t) = L
2.4.6. Solve the initial value problem
u, + u\ + и = 0 , m(jc, 0) = x ,
using the method of characteristics.
2.4.7. Consider the wave equation in two dimensions
+
where с may be a function of (дс, у).
(a) Let ф, y, t) = V(^, у)е"'ш/ (i = V^I) and show that V(x, y)
satisfies the reduced wave equation
(The constant a> is the "angular" frequency of the solution.) Since
с represents the speed of wave propagation (as we shall see later
in the text), we introduce a constant reference speed c0, say, the
speed of light, and define the index of refraction n as n(x, y) -
co/c(x, y) and the wave number к as к = <o/c0. Then the reduced
wave equation takes the form
eXErCISES FOR CHAPTER 2 107
(b) Assume that к > 1 and look for a solution of the reduced wave
equation in the form V= Aeikuy where A and и are functions of
(x, y). Show that on substituting this form in the reduced wave
equation, collecting like powers of k, and equating the coeffici-
coefficients of k2 and к to zero, we obtain the eiconal equation B.147)
for u(x, y) and the following equation for A(x, y),
2Axux + 2Ayuy + (uxx + uyy)A = 0 .
This is known as a transport equation for A (which represents an
amplitude term).
(c) Show that if n, A = constant and и is the plane wave solution
B.160), V(xy y) is an exact solution of the reduced wave equa-
equation. Otherwise, the function V= Aelku represents what is known
as the geometrical optics approximation to the solution of the
reduced wave equation. (A farther discussion of this approxima-
approximation is given in Section 10.1.)
2.4.8. Let v = A{i)eu in equation A.23) and solve the resulting equation for
A and и by first taking и to be a solution of
ut + cux- \Du2x=Q.
Let u~ ax + bt and obtain the family of solutions и = rx + (§2>r2 — cr)t.
Determine the envelope of this family and thereby obtain the singular
solution и = -(l/2Dt)(x - cif. If u(x, t) is the singular solution, solve for
A(t) and show that the resulting solution for v(x, t) is of the form A.31).
2.4.9. Solve
w, = c2u2x ,
2.4.10. Let и = Ух2 + у2. Solve the transport equation for A(x, y) given in
Exercise 2.4.7, subject to the condition A(x, y) = 1 on the circle x2 + y2 = 1.
Show that A(x, y) is a singular at the origin @,0), which represents a focal
Point for u(x, y).
2-4Л1. Consider the eiconal equation
"* + му-л U, у), п(л:, У)-[„2> х>0,
n2 > nx and both equal to a constant.
(a) Given the boundary condition w@, y) = nxy cos в where в =
constant, solve for all possible u(x, y) in the regions x < 0 and
x > 0. (Hint: They will be plane wave solutions.) These solutions
«* FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
play a role in the problem of a plane interface between two
media with differing indices of refraction.
(b) Obtain expressions for the directions of the rays found in part
(a), as given by Vu and evaluated on the interface x = 0. These
formulas represent what are known a SnelVs laws of reflection
and refraction.
2A.12. The characteristic equation for the wave equation A.100) is given as
(see Section 3.2). Use the method of characteristics to obtain the plane wave
solutions и = x ± yt and the singular solution и = constant on the pair of
straight lines x2 = у V.
2.4.13. Extend the results of the text for nonlinear equations to n dimen-
dimensions. With the equation given as
F(xl9 x29...,xn9u, uXi,. .. , ux) = 0 ,
obtain the characteristic equations with pt = ux,
where i = 1,2,. .. , n. Also, obtain conditions equivalent to B.123)
B.124).
2.4.14. Obtain plane wave solutions and singular solutions equivalent tcft
B.161) and B.167), respectively, for the eiconal equation in three dimen*
sions,
u2 + u2y + u2z ~ n2, n = constant.
2.4.15. Solve the initial value problem
щ + u2x = 0 , m(jc, 0) = -ж2 .
Show that the solution breaks down when t — \.
2.4.16. Differentiate B.136) with respect to x and let v = ux. Show that v
satisfies an equation of the form B.104) and interpret the results of Example
2.13 accordingly.
2.4.17. Implicitly differentiate F(x, t, u, p, q) with respect to s, and use
B.113)-B.115) and B.120), B.121) to conclude that Fis constant along the
characteristics.
2.4.18. Use implicit differentiation and the equations of Exercise 2.4.13 to
show that F(jcx, . . . , pn) is constant along the characteristics.
3
CLASSIFICATION OF
EQUATIONS AND
CHARACTERISTICS
The telegrapher's and wave equations, the diffusion equation, and Laplace's
equation, derived in Chapter 1, are prototypes of the three basic types of
partial differential equations encountered most often in applications. They
are equations of hyperbolic, parabolic, and elliptic type, respectively. The
classification of equations and systems of equations, in general, into these
three (or possibly other) types is considered in this chapter.
Furthermore, formulations of initial and boundary value problems appro-
appropriate to equations of different types are given. Additionally, characteristics
that were shown to play an important role in the theory of first order partial
differential equations are reintroduced from a somewhat different point of
view. Their significance in the theory of hyperbolic partial differential
equations is demonstrated. Further, certain basic concepts relating to second
order linear equations with constant coefficients are introduced and dis-
discussed. Finally the important concept of adjoint differential operators is
presented.
ЗЛ. LINEAR SECOND ORDER PARTIAL
DIFFERENTIAL EQUATIONS
We begin by considering the general second order linear partial differential
equation in two independent variables
(ЗЛ) ^и« + 2Buxy + Cuyy + Dux + Euy + Fu~G,
the terms A, B,.. . , G are given real valued functions of x and y.
cases of C.1) were derived in Chapter 1 (with у replaced by t in
109
И0 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
some instances). It will now be shown that the principal part of C.1), that is,
the terms containing second derivatives of w, can be transformed by a
change of independent variables into a form similar to that of the wave,
diffusion, and Laplace's equation considered in Chapter 1. This indicates
that these equations are, in fact, prototypes of second order linear equations
of the general form C.1). The transformed versions of these equations are
referred to as canonical forms.
Without loss of generality, it may be assumed that A(x, y)^0 in som#
region R and we divide by Л in C.1). Then, with dx = dldx and dy = d/dyt
we express the principal part of the differential operator in C.1) as follows^
C.2) a\
dx dy
where a>+ and w" are defined as [on comparing both sides of C.2)]
+ IB + _ С
C.3) o> + ft> = - — , <y<w=-.
Solving the system C.3) for <o+ and ш~ gives
C.4) <y"(x, y) =,
so that «у* are the roots of the quadratic equation
Aw2 + 2B<o + С = 0.
In obtaining C.2) we have followed the procedure used in Section 2.1,
where the differential operators for the wave and Laplace's equation were
factored. Since the coefficients A, B, and С may depend on x and y, we
have the additional term involving dy in C.2). Instead of using C.2) to
express C.1) as a system of two first order equations (as was done in Section
2.1), we use C.2) to simplify the form of the given equation C.1).
Since we are generally interested in obtaining real valued solutions of
C.1), the possibility of using the factorization C.2) to simplify C.1)
depends on whether ^(x, y) are real or complex valued functions of x and
y. This is determined by the sign of the discriminant B2 - AC in C.4). We
shall assume that B2 - AC is either of one sign or vanishes identically
throughout the given region /?. On that basis we classify C.1) as belonging
to one of the following three types in the region R:
C.5) B2 - AC>0, hyperbolic type ,
LINEAR SECOND ORDER PARTIAL DIFFERENTIAL EQUATIONS 111
C.6) B2 - AC = 0, parabolic type ,
B2 - AC < 0, elliptic type .
Applying these criteria to the equations of Chapter 1, we find that the
wave (and telegrapher's), diffusion, and Laplace's equation are of
hyperbolic, parabolic, and elliptic type, respectively. We now determine
canonical forms for C.1) of each of the types C.5)-C.7).
3.1.a. Equations of Hyperbolic Type
When B2 - AC>0 in the region R, the functions <o±(x, y) are real valued
and distinct in that region. We may then express the operators dx - o>±^ as
directional derivatives. Along the family of curves
C.8) % = ~m±{*'y)*
we have, for any differentiable function v = v(x, y),
on using C.8).
The (one-parameter) families of curves determined by the solutions of
C.8) are called the characteristic curves of C.1). They form two indepen-
independent families of curves in the (x, y) — plane since dyldx — — (o± with <o+ Ф
<o~ implies that they intersect nontangentially. We express these curves as
where f = constant corresponds to the curves у'+ ш+ = 0, whereas 77 =
constant corresponds to the curves yf + w~ = 0. For example, if the coeffici-
coefficients А, Б and С in C.1) are constants, so are ш+ and a>~. Then we can set
I = у + о) +x and 4 = y + <o~x.
We introduce £ and 77 as new coordinates in C.1), and this is referred to
as the characteristic coordinate system. On the characteristic curves £ =
constant and 77 = constant,
C.11) 0 = £ + /(*)£, = £-*>4>
C'12> O-iJ + /(x),^i,]-«-i,y.
h follows from our assumptions that £у ^ 0 and 77^ ^ 0. Since w + and a> ~ are
solutions of the quadratic equation Aa>2 + 2B<o + С = 0 as shown above, we
conclude that <p = £(*, y) and 9 = v(x, y) both satisfy the characteristic
equation for C.1)
112 CLASSIFICATION OF EQUATIONS AND CHARACTERISTIC^
C.13) A<p2x + 2B<px<py + C<p2y = Q.
Also, if и = м(£, 7]) = м[£(х, у), Ф, y)]y we have
so that
C.14)
in view of C.11). Similarly, C.12) implies that
C.15) Tx-w~Ty=^- ш~ду> = K&
On using C.11) and C.12) to express £ and r)x in terms of ^ and 17^
obtain in view of C.14) and C.15),
C.16) dx - a>+dy = -yy(a>+ - <o~)dv
(ЗЛ7) дх-а>-ду = £у(а>+-а>-)д€,
which are both nonzero operators since ш+ Ф a>~. Thus
C.18) (dx - »4)(*. ~ ) = ["^(^+ " »~Kl[f,(® + " »"H1
Substituting this result in C.1), as modified in C.2), and expressing
everything in terms of £ and tj, we obtain (on dividing by the nonzero
coefficient of м^)
C.19) u^ + аи; + buv + cu = d,
where a, b, c, and d are specified functions of f and 17. This is one of the
canonical forms for C.1) when it is of hyperbolic type.
While we have used the factorization procedure to determine the princi-
principal part of the canonical form C.19), it is generally simpler to use a more
direct approach to obtain C.19). Once the characteristic coordinates (З.Ю)
UNEAR SECOND ORDER PARTIAL DIFFERENTIAL EQUATIONS 113
have been found, we can transform C.1) directly into the form C.19), using
the chain rule for differentiation. The principal part of the equation in
(P т?)-coordinates is given as in C.53). It is shown in the exercises that the
coefficient of u€v in C.53) equals D/A)(AC- B2)£yvy and this is nonzero
in R-
The transformation
C.20) £ = a + j3, ri = a-p
in C.19) yields an alternate canonical form for the hyperbolic case,
C.21) "«a * «w + Sua + bup + tu = d,
with specified functions a, by c, and d of a and /3. НА, В and С in C.1) are
constants, the variables a and /3 are given as a = у - (B/A)x and fi =
A/A)(B2- ACI/2x. We see that in either the form C.19) or C.21) the
principal parts of the canonical forms have constant coefficients. The
telegrapher's and wave equations of Section 1.2 are of the form C.21).
3.1.b. Equations of Parabolic Type
When B2-AC = 0 in the region R, we have <у+ = й>~ = о> and C.2)
becomes
with ш = —BIA in view of C.4). Since there is only one value of o>, we
obtain only one family of characteristic curves defined by
Let
C-24) f = «*,y), Ч = Ч(*,У),
where £ = constant is the family of characteristic curves determined from
\ 23) and 17 = constant is an arbitrarily chosen independent family of
^rves, such that the Jacobian determinant of the transformation C.24)
^'e> Sx% - (yVx) ^s nonzero in the region R.
Again <p = g(x, y) is a solution of the characteristic equation C.13). For
example, if А, В and С in C.1) are constants, so is <o and we can set
ъ~~У + а>х and 7} = (oy + cx where сФы2 but is an otherwise arbitrary
°nstant. Proceeding as for the hyperbolic case, introducing the coordinates
114 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
C.24) and using the chain rule to transform C.1), we obtain (see C.53) and
the exercises)
C.25) um + au€ + bur) + cu = d,
where a, b, c, and d are specified functions of f and tj. Again, the
coefficients in the principal part of C.25) are constant. The diffusion
equation of Section 1.1 has the form C.25)
З.1.С. Equations of Elliptic Type
When B2 - AC<0, co+ and o>~ as defined in C.4) are complex valued in
the region Л. Hence the characteristics defined by C.8) are complex curves.
Assuming the terms A,B9...9G in C.1) can be defined for complex
arguments, the change of variables C.10) may still be applied to bring C.1)
into the form C.19). Again the coefficient of u^ is given as D/A)(AC-
В2)^ут]у and this is nonzero. However, the variables £ and 17 are complex,
so that the usefulness of the canonical form C.19) is questionable. For
example, if А, В and С in C.1) are constants, we obtain £ = у - (B/A)x +
i(l IA)(AC - B2)V2x and 77 - у - (B/A)x - i(l /A)(AC - B2)ll2x where i =
The further transformation
C.26) i - a + ф, >q = a-ip
in C.19) introduces the real valued variables a and /3. (This follows since
o>* and o>~ are complex conjugates in the elliptic case, when the terms A, fi,
and С are real as assumed.) If A, B, and С are constants, a = у - {BIA)x
and j8 = A/A)(AC - B2)i/2x. We then easily obtain the canonical form
C.27) uaa + ufifi + aua + bup + cu = d
for the elliptic case where a, b, c, and d are specified real valued functions of
a and /3. [The canonical form C.27) can also be obtained by another
method (which we do not consider) that does not require the introduction of
complex variables.] Laplace's equation of Section 1.3 has the form C.27).
If the terms A, B,.. . , F in C.1) are identically constant, the above
canonical forms for the hyperbolic, parabolic, and elliptic cases are valid
over the entire domain of definition of the inhomogeneous term G, and they
all contain constant coefficients. In that case a further simplification of the
equation can be achieved, as shown in the exercises.
Example 3.1. An Equation of Mixed Type. The equation
C.28) uxx+yuyy = 0,
SECOND ORDER PARTIAL DIFFERENTIAL EQUATIONS 115
. said to be of mixed type, since it is hyperbolic for у < 0, parabolic for
1 = 0, and elliptic for у > 0. This follows immediately from the fact that
V JAC = —y for this equation. We now obtain canonical forms for C.28)
in all three regions of the (x, y) - plane given above.
The parabolic region is the simplest to deal with since at y = 0 we
.mmecliately obtain the canonical form
C.29) и** = °-
In this case <o = 0, so that the characteristic curve determined from dyldx =
0 is у = 0- That *s> *e x-sxis *s *© characteristic curve, and is represents a
curve across which a transition from hyperbolic to elliptic type takes place.
For this reason, the solution of initial and boundary value problems for
C.28) in a neighborhood of the jc-axis is rather difficult since C.28) is of
three different types in such a neighborhood.
In the hyperbolic region у < 0 we have
C.30) <o±(x,y) =
and the characteristic equations
C.31) /(x)=-«±
yield the two characteristic families
C.32) £ = x + ly/^y = constant,
C.33) 7}-x- ly/^y = constant.
From C.2) we obtain for the differential operator in C.28)
C.34) a\ + yd* = {dx - УГгуду){дх + V=jW,) - Уy ,
and from C.18)
Since ду = €,3с + ъэч = -A/V=y)[*« - ^] and ^-77 = 4x^7, we obtain
ae canonical form for C.28) in the hyperbolic region as
C.36)
116
CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
The characteristic curves for the hyperbolic case are the two branches o:
the parabolas y = -\(x- cJ, where с is a constant, as shown in Figure 3.1
The branches with the positive slopes give the curves f = constant, wherea,
7} = constant gives the branches with the negative slopes. Note that botl
branches are tangent to the x-axis, which is the single characteristic curve ir
the parabolic region. In fact, the jc-axis is the envelope of the characteristic
curves for the hyperbolic region у < 0.
In the elliptic region у >0, we have ш±(х9 у) = ±V=y = ±iy/y, so tha
the characteristic curves are complex. The equations C.32) and C.33
remain valid but take the form
C.37)
Using C.26) gives
C.38) a = x, p=2Vy.
With a and /3 as new variables, we have
C.39) uxx = uaa, uy = ^=up, uyy = -up
so that the canonical form is
C.40) Mj
2y
3/2
We conclude our discussion of this example with the observation that the
canonical forms C.36) and C.40) in the hyperbolic and elliptic cases have
coefficients that are singular when £ = rj and /3 = 0, respectively. Both
singular regions correspond to the лс-axis, across which the equation C.28)
Figure 3.1- The characteristic curves.
CHARACTERISTIC CURVES 117
undergoes a transition from hyperbolic to parabolic to elliptic type. The lack
of validity of the hyperbolic and elliptic canonical forms at the до axis is
signaled by the singularity of the coefficients there. Although the study of
equations of mixed type is of interest in a number of applications, we shall
deal (almost) exclusively with equations of a single type in the region under
consideration.
3.2. CHARACTERISTIC CURVES
To introduce and motivate our discussion of characteristic curves for second
order partial differential equations, we begin by reconsidering linear first
order equations and their characteristics. However, we now discuss them
from a somewhat different point of view than that of Section 2.2.
The characteristic (base) curves of the linear first order equation
C.41) a(x, y)ux + b(x, y)uy = c(x, y)u + d(x, y)
can be obtained (assuming a ^ 0) as solutions of
C-42)
dx a
We suppose that у = h(x) is a solution of C.42) and look for a solution и of
C.41) with the initial value u[x9 h(x)] =/(*). Using the chain rule and
C.41), C.42), we conclude that a duldx = af'(x) = cf + d on the charac-
characteristic base curve у — h(x). Unless/(jc) satisfies this compatibility condition,
the initial value problem has no solution. If /(jc) does satisfy the compatibili-
compatibility equation, the curve y = h(x), u[x, h(x)]= f(x) in (jc, у, и)-space is a
characteristic curve for C.41) and there are infinitely many solutions or
integral surfaces и = u(jt, y) of that equation that contain the characteristic
curve.
These results were obtained in Section 2.2 (where у is replaced by t). [See
the discussion that follows equation B.51).] We are assuming that a and b in
C.41) are not both zero. If b Ф 0 but a can vanish we can express C.41) in a
form equivalent to B.51) and adapt the discussion of that equation to obtain
compatibility conditions in the present case. In this section, however, we
deal with the case where a is nonzero.
It was also shown in Section 2.2 that if we ask for a solution of C.41) that
ls continuous along the curve у = h(x) but has discontinuous derivatives
across it, then this curve must be a characteristic base curve [that is, it
satisfies C.42)]. We demonstrate this again, using a different method. This
method can be easily generalized to deal with higher order differential
equations.
118 CLASSIFICATION OF EQUATIONS AND CHARACTERISTIC»
Let у = h(x) be the curve of discontinuity and represent it implicitly as
<p(x, у) = у - h(x) = 0. We consider the family of curves <p(x, y) = constant
and the family of orthogonal trajectories ф(х, у) = constant [the orthogonal
trajectories are solutions of the equation у1 = -1/Л'(*)]. Introducing the
(£,77) -coordinate system defined by
C.43) €=Ф,У), Ч = *(х,у),
we obtain for и = w(£, 77) = w[£(x, y), i7(*> у)],
C.44) и, = и^, + h4ij, = ftnf + 0яич ,
C.45) иу = и^у + нч1|у = <ру^ + фуи^ .
Thus
C.46) аих + Ьм, = (а^ + Ь?у)"{ + (л& + Ьф^и^ = cu + d .
Now let w^jc, >^) and w2(x, >^) represent solutions of C.46) defined in the
regions £=£0 and £^0, respectively, with ux = u2 along the curve ^ =
у - h(x) = 0. With w = w2 for £ < 0 and и = u2 for ^ > 0, the solution и and
the derivative uv are continuous on the curve £ = 0. However, w^ represents
a normal derivative across the curve £ = 0 and wf has a jump discontinuity
there by assumption (note that uv is an interior derivative along the curve
£ = constant). Since ux and u2 are solutions of C.41) for £ ^0 we have
C.47) [(a<px + bb)u, + (afc + Ьфу)ищ -си- d]\lZ^ - 0 ,
where we consider the difference of the bracketed expression across the
curve £ = 0 along the orthogonal curves 77 = constant. In the limit as £->0
we obtain from C.47)
C.48) (*ft + 4)M«-o = O
where [w^]||=0 is the jump in u€ across the curve £ = 0. The other terms
vanish since they are continuous at £ = 0. By assumption [wf]|^=0^0, so
that C.48) implies
C.49) a<px + bf,\im0 = ~ah\x) + b = 0.
since £ = 0 corresponds to the curve <p(x9 y) = y- h(x) = 0. As a result,
h'(x) - y'(x) = bla so that у = h(x) is a solution of the characteristic equa-
equation C.42) as was to be shown. That is, discontinuities in derivatives of
continuous solutions и = u(x, y) of C.41) can occur only across characteris-
characteristic (base) curves. This is so only if a, b, c, and d are continuous (see
Exercise 3.2.7).
CHARACTERISTIC CURVES 119
We now demonstrate that if initial data u(x, y) are prescribed for the
equation C.41) along a characteristic (base) curve у = h(x), it is impossible
to solve for ux and uy uniquely along that curve, if a solution exists at all.
Thus if there is a solution, no unique tangent plane can be defined along the
initial curve and it is not possible to specify a unique integral surface
containing the initial curve that satisfies the initial value problem. (This
result has been obtained previously, based on the method of characteristics,
but we now obtain it in a different way, one that can be easily generalized to
deal with systems of equations.)
Given the initial value u[x, h(x)] = f(x) on the curve у = h(x) that
satisfies C.42), we have
dulx- h(x)] ,,, ч .,, v
C5O) dx = Ux + H {X)U> ^fix) '
Combining C.50) with C.41) evaluated on у = h(x) yields the simultaneous
system for ux and uy,
К+ *'(*Ч =/'(*)
\aux + buy = cf(x) + d.
The determinant of coefficients of this system vanishes; that is,
C.52) D =
II h'(x)
\a b
-ahf(x) + b =
since y'(x) = h\x) = bla in view of C.42). Consequently, either ux and uy
cannot be determined at all on у — h(x) or, at best, they can be solved for
nonuniquely if f(x) satisfies a consistency condition. This condition is readily
found to be af'{x) ~ cf(x) + dy which is identical with the compatibility
condition given above. Conversely, the above discussion shows that a curve
у = h(x) on which и is prescribed and for which ux and uy either cannot be
determined or can only be determined nonuniquely, must be characteristic
since we must then have D = 0 in C.52). The determinant D vanishes only if
У = h(x) is a solution of C.42).
To determine the characteristic curves for the second order equation
C.1), we apply the first of the two techniques developed for first order
equations. (The second approach is presented in the exercises.) We shall
specify the possible curves у = h(x) across which the second derivatives of a
solution и = u(x, y) of C.1) can have discontinuities. The solution u(x, y)
and its first derivatives are assumed to be continuous across these curves.
Let <p{x, y) = y- h(x) = 0 and define the orthogonal family of curves
f = constant and т/ = constant as in C.43). In terms of the variables £ and 17,
C.1) takes the form
120 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
C.53) (Acp2x + 2B<px<p 2
x<py
2[Асрхфх + В(срхфу + ^^) + C^^]Mfn
(Аф2х + 2tf «^ + Сф2у)и^ + • • • = G ,
where the dots represent first derivative and undifferentiated terms in «.
Since и and w^ are assumed to be continuous across £ = v - A(jc) = 0, we
conclude that м^, wf4, and иЧ1| are also continuous since <?/<?т/ is an interior
derivative operator along £ = 0.
Again we consider solutions иг(ху у) and w2(jc, у) of C.1) defined in the
regions £ <0 and £ ^0, respectively. Evaluating C.53) in each of the above
regions, taking the difference of these equations, and going to the limit as
£—»0, we obtain
C.54) (A<pl+2B<px<py + C<p2y)[u((]\(=0 = 0 ,
where [w^]|^=0 *s the jump in the second derivative of w across the curv&
f = 0. All other terms vanish in view of their assumed continuity across that
curve. Since [w^5]|^=0 is taken to be nonzero, we must have
C.55) Aq>2x + 2Bq>x?, + Cq>2y = 0.
This is the characteristic equation for C.1) [see C.13)].
Assuming (as was done in Section 3.1) that the coefficient A in C.1) i
nonzero, we can factor C.55) as
C.56) A(<px - (o+<py)(<px - «>y) = 0,
with (o+ and w~ defined as in C.4). If <o+ and a>~ are real valued, we set
that a curve £ = <p(x, y) = 0 or, equivalently, у = h(x) across which u^ isl
discontinuous must be one of the characteristic curves C.8). This follows?
since <p = у - h(x) implies that <px - (o±<py = ~h'(x) - <o±. Thus one of the
two equations y' = h' - -ш± must be satisfied in view of C.56).
Using similar arguments, it can be shown that discontinuities in higher
derivatives of a solution must also occur only across characteristic curves.
Example 3.2. The Wave, Diffusion, and Laplace's Equations. In this
example we discuss the characteristic equations C.55) for the wave, diffu-
diffusion, and Laplace's equations derived in Chapter 1,
For the wave equation A.100), the characteristic equation has the form
C.57) <p) - yVx = (<pt + y<px)W, - m) = 0.
Equating each factor to zero separately gives as the characteristic curves
C.58) <p(x, t) = x±yt = constant,
CLASSIFICATION OF EQUATIONS IN GENERAL 121
which are straight lines in the (ж, t)-plane with slopes dtfdx = ±l/y. Since
the telegrapher's equation A.88) has the same principal part as the wave
equation A.100), it has the same characteristic curves C.58).
For the diffusion equation A.23) we have the characteristic equation
C.59)
which implies that <p = <p(t) = constant, so that the straight lines t = constant
are the characteristics.
Laplace's equation A.115) yields the characteristic equation
C.60)
For real valued <p(x, y) we must have <px = <py = 0 so that no real solutions
other than <p s constant exist. Thus Laplace's equation has no real charac-
characteristic curves.
The results of the above example are typical for equations of hyperbolic,
parabolic, and elliptic type. Elliptic equations have no real characteristic
curves so that solutions cannot have discontinuous derivatives. Consequent-
Consequently, solutions и = u(x, y) are extremely smooth functions and this is consis-
consistent with the nature of elliptic equations that describe equilibrium processes
where everything has already smoothed itself out.
Parabolic equations of the type we generally consider, have the lines
t — constant as characteristics, and discontinuities in derivatives must occur
across these lines. Since we are concerned with the evolution of solutions as
t (i.e., the time variable) increases, if these discontinuities occur initially
when t = 0, they cannot be spread into the region t > 0 and again solutions
are smooth functions.
It is in the theory of equations of hyperbolic type that characteristics play
the most significant role. The preceding example shows that for the wave
and telegrapher's equations the characteristics extend from the (initial) line
' = 0 into the region />0. Thus in their role as curves across which
discontinuities or singularities in the solution occur, they act as carriers of
singular initial data for the solution and the effects of these singularities are
felt for all time. The importance of characteristics in the solution of various
problems for hyperbolic equations is demonstrated throughout the text.
3-3. CLASSIFICATION OF EQUATIONS IN GENERAL
AND THEIR CHARACTERISTICS
^e begin by considering the second order linear partial differential equation
in n variables
122 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
n n -2 n .
C.61) 2j 2. ^7-TT + Z Ь/Т7+си + ^-0,
i = l j=l CtXiOXj i = l ОХг
where u, aij9 bi9 c, and d are functions of Jtj,. . . , xn. This equation cannot,
in general, be reduced to a simple canonical form over a full region, as was
done in the case of two independent variables in Section 3,1. For the
purpose of classifying C.61) into different types, we shall generalize
the factorization procedure of Section 3.1. It will then be seen that if the
coefficients a^ are identically constant or if we restrict ourselves to a single
point in (xx,... ,xn)-space, it is possible to bring the principal part of
C.61) (i.e., the highest derivative terms) into a canonical form.
With dXi = S/Sxi9 i=l,...,/i, and dx = [dXi,. . . , <?XJ, as a row|
(gradient) vector (which is the transpose of the column vector dx), it is
easily seen that we may express the principal part of the differential
operator in C.61) as
C.62) 2?1Шг»^ + '"»
where the dots stand for first derivative terms and the n by n matrix A has
the coefficients atj as its elements. Note that if the atj are constants, the firsi
derivative terms in C.62) are absent. Since we assume that mixed partia
derivatives of и are equal (i.e., uXiX = и*.*.), C.61) may be arranged so thai
its coefficients have the property that at- — ajt and we assume that this haa
been done. Thus the matrix A is symmetric and it is assumed to be real
valued in the present discussion.
It is well known from matrix theory that a real valued symmetric matrix
A has only real eigenvalues X19 A2,... , Лл (counted with their multiplicities)
and that there exists a corresponding orthonormal set of n eigenvectors
r1,. . . , rn. Forming the matrix R with the eigenvectors r, as its n columns,
we find that R is an orthogonal matrix with the property
о
C.63) RTAR = D =
О А.
That is, using R we can diagonalize the matrix A.
We now introduce the directional derivative operators
C.64) ^rfex, / = l,...,/z,
and form the vector operator
C.65) d, = RTdx .
CLASSIFICATION OF EQUATIONS IN GENERAL 123
[The operators dt may be compared with the operators df and dv in C.16),
/3 17).] Since R is an orthogonal matrix so that RT = R~\ (i.e., its transpose
equals its inverse), we have from C.65) dx = Rde. Introducing this expres-
expression into C.62) gives
C 66) BTxAdx = (Rae)TA(Rdt) = d]{RTAR)di + ■■■
where the dots represent first derivative operators (since the elements in the
preceding matrices need not be constant) and where C.63) has been used.
Recalling the classification method for equations in two independent
variables given in Section 3.1, we find that in terms of the result C.66), the
elliptic case corresponds to the situation when Ax and A2 have the same sign.
The hyperbolic case requires that Хг and A2 have opposite signs whereas the
parabolic case occurs when At or A2 equals zero. For the w-dimensional case
in C.61), we base our classification of the equation at a point P in
(xl9 ..., xn)-space on the result obtained in C.66). Thus it is characterized
by the properties of the eigenvalues of A.
With Al9 A2,. .. , An as the eigenvalues of A, we introduce the following
classification for the equation C.61):
C.67)
A,>0, all/,
or elliptic type .
А(<0, all/,
Г One of the A,<0or A,>0, , f
1 All other A, have opposite sign , hyperbolic type .
C.69) One or more of the A; = 0, parabolic type .
For the two- and three-dimensional cases the above classification exhausts
all possibilities. However, if n > 4 in C.61), it may happen that two or more
of the A, have one sign, whereas two or more of the remaining A(. have the
opposite sign. Such equations are said to be of ultrahyperbolic type. Since
they do not occur often in applications they are not studied in this text.
К C.61) is of fixed type at every point in a region, the equation is said to
be of that type in the region. If the type changes within a region, the
equation is said to be of mixed type. When the coefficients atj of the
principal part of C.61) are identically constant, the equation is clearly of
one type everywhere in (xl9. .. , Jtj-space.
As basic examples of second order partial differential equations we have
C.70) v2w = uxx + uyy + uzz = 0 , elliptic type .
C.71) utt - V2u = utt - uxx - uyy - uzz = 0 , hyperbolic type .
124 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
C.72) ut - V2m = ut - uxx - uyy ~uzz=Q, parabolic type .
These are the higher dimensional Laplace's, wave, and diffusion equations.
The operator V = дг1дхг + дг1дуг 4- d2ldz2 is the Laplacian operator in
three dimensions. A further example is given by Schrodingefs equation,
where i = V-T, h is Planck's constant, m is the mass and V is a given
potential function. This equation is clearly of parabolic type.
Once we have exhibited how second order equations in n variables are to
be classified, it is appropriate to ask whether there exist transformations of
the independent variables (similar to those considered in Section 3.1) that
bring C.61) into a canonical form, say, of the type given in C.66) but where
the principal part has constant coefficients. It is shown in the exercises that
this is not possible, in general, if n > 3 in C.61). However, if the coefficients
ац of the principal part are identically constant, simple canonical forms do
exist. In fact (noting C.64)), the linear transformation
C.73) & = rfx, i = l,...,n,
where xr = [^i» - - - ,*„]> transforms the principal part of C.61) into the
form
C.74) SJ,"»1^/",?,^**'
where the Л, are the (constant) eigenvalues of A. Clearly, an elementary
further transformation of the variables £,...,£, can reduce the principal
part to a form in which the coefficients are either 0, +1, or -1.
To determine the characteristics for C.61) we look for surfaces
<p{xX)..., jcn) = constant across which second derivatives of solutions of
C.61) can have discontinuities. This can be done by introducing n (indepen-
(independent) families of surfaces <рA)(хг,... , xn) - constant (with <pA) = <p) and
the corresponding coordinate system ц - <p{l\x1,. .. , xn), i = 1,. . . , w.
Transforming from the xt coordinates to the ijf. coordinates in C.61) we
easily obtain, with т)г~ <p.
C.75) [S 2 а„ърх]и„ + - • • « 0 ,
where the dots represent second derivative terms in the variables ту2, ...,!?„
as well as first derivative terms and undifferentiated terms in all variables.
Now u^ is a second derivative of и across the surface <p = constant. Thus if
<p = constant satisfies the equation
CLASSIFICATION OF EQUATIONS IN GENERAL 125
n n
C.76) ХЕ
у cannot be uniquely determined in terms of u^ and w, as well as interior
derivatives of w, given on the surface <p = constant. Consequently, the only
surfaces across which и may have discontinuities in its second derivatives are
solutions of C.76). These surfaces are the characteristic surfaces for C.61),
and C.76) is the characteristic equation.
For example, Laplace's equation C.70) has the characteristic equation
C.77) 9* + ?* + *«=0>
and this equation has no real solutions. Thus Laplace's equation has no real
characteristics. This indicates that its solutions are smooth functions. The
three-dimensional wave equation C.71) yields the characteristic equation
C.78) <Р*-<Р2Х-<Р2У-<Р]=О-
Among the important solutions of C.78) are the plane waves,
C.79) <p = <ot — кхх — к2у - k3z = constant,
where <o2 = к\ + к\ + к\у and a singular solution, the characteristic cone
C.80) <p = (t - tQf - (* - x0J -(y- y0J - (z - z0J = 0 ,
where PQ = (x0, y0, z0, t0) is a given point. The three-dimensional diffusion
equation C.72) has the characteristic equation
C-81) <р2х + <р2у + <р]=0,
which yields the characteristic surfaces
C-82) (p(x, y, z,t) = t- constant.
Example 3.3. Classification of an Equation of Mixed Type. We consider
the equation
C.83) uXiXi + 2A + сх2)иХ2Хз = uXiXi + A + сх2)иХ2Хз + A + схг)иХъХг = 0,
where с is a constant and apply the above classification procedure. From
C.62) we have
C.84) a\x + A + схг)дхдхъ + A + сх2)дХгдХ2 = дтхЛдх - сдхъ,
where
126
CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
C.85)
A =
10 0
0 0 A + cx2)
0 A + cx2) 0
The eigenvalues and the corresponding orthonormalized eigenvectors of
are
C.86)
r,= 0 ,
A3 = -A + cx2), r3 -
0
J_
V2
-Л
lV2J
The orthogonal matrix R is
C.87)
10 0 "
0 Vf V!
0 V2 VlJ
r, =
0
1
V2
1
V2
and it may be verified that R- RT — R l and that
0
C.88)
0
RTAR = \0 (l + cx2) 0
0 0 -(l + c*2)j
Further,
C.89)
so that, since /? is a constant matrix,
CLASSIFICATION OF EQUATIONS IN GENERAL
C.90)
1 0
0 A + cjc2)
0 0
0
0
127
where we have used C.86) and C.89).
In view of C.68), C.69), we conclude that C.83) is parabolic when
x2 = -l/c (if c^O) and is hyperbolic in the half-spaces x2>-\lc and
x2<-l/c. If c = 0, all the eigenvalues are constant with kx - A2 = 1 and
A3 = -1, so that C.83) is hyperbolic everywhere. Applying the transforma-
transformation C.73), we have
and C.83) takes on the canonical form
C.92) иы + ий € - ийй = 0 .
This has the form of the wave equation in two dimensions with £3 as the time
variable.
We continue our discussion by considering the classification of first order
systems of linear partial differential equations in two independent variables.
The equations may be written as
C.93)
Эх
ду
where Л, В, and С are n by n matrix functions of x and у and u(jc, y) and
d(*> y) are /z-component vectors. At least one of the matrices Л and В is
assumed to be nonsingular, that is det Л Ф 0, det В Ф 0 or both are not zero.
Our classification procedure is based on the properties of the characteristic
curves that can occur for C.93).
128 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
Now C.93) represents a natural generalization of the linear first order
equation C.41) with the scalar dependent variable replaced by a vector
variable. Consequently, if A is nonsingular, it is appropriate to formulate an
initial value problem for C.93) that assigns the value u = f(x) on the curve
у = h(x). Proceeding as in Section 3.2, the characteristic curves of C.93) are
defined to be those curves on which nx and uy cannot be uniquely specified
(if indeed they can be determined at all) in terms of the initial data given.
The alternative procedure presented in Section 3.2 whereby the characteris-
characteristics were determined as the curves across which continuous solutions of
C.93) can have discontinuous first derivatives can also be applied here. It is
considered in the exercises.
On the initial curve у = h(x) we have u[jc, h(x)] = f(jc). Thus
C.94)
Solving for ux in terms of uy and substituting in C.93), gives
C.95) [B - h'(x)A]uy = Cf(x) - АГ(х) + d ,
where A, By C, and d are evaluated on the curve y~h(x). Now C.951
represents a system of equations for u^ [on the curve у - h(x)], and it has m
unique solution only if the determinant of the coefficient matrix of ny iu
nonzero, that is,
C.96) det[B - h'{x)A] = \B - h'{x)A\ ^0 .
A curve у = h(x) for which this determinant vanishes identically is called a
characteristic curve for C.93). If the initial conditions u[jc, h(x)] — f(x) are
such that the system C.95) has a (nonunique) solution, we have a charac*
teristic initial value problem. The singularity of the matrix В - h'A gives risd
to a compatibility condition on the right side of C.95) for a solution to exist»
which, in turn, implies a restriction on the data f(jc). Even if f(jc) does not
satisfy this condition, the curve у = h(x) is called a characteristic curve.
To determine the full set of (possible) characteristic curves, we express
the characteristic determinant C.96) in the form
C.97) \B[x, y(x)) - y'(x)A[x, y(x)]\ = 0,
with y = y(x) as the characteristic curve. This determinant yields an nth
degree algebraic equation for y'(x) and this is the characteristic equation for
C.93).
If all the roots y\x) of C.97) are real and distinct in some region, the
system C.93) is said to be totally hyperbolic in that region. Denoting these
CLASSIFICATION OF EQUATIONS IN GENERAL 129
roots by the functions <аг{ху y) (i = l, ...,w) we obtain n families of
characteristic curves as solutions of the equations
C.98) У'(х) = a>,(x, у), i = 1,2,. . . , n .
If all the roots o>,(jc, y) of C.97) are complex valued, the system C.93) is
said to be of elliptic type. Then there are no real characteristic curves as
carriers of possible discontinuities in derivatives of solutions and the solu-
solutions are expected to be smooth functions. The case where some roots are
real and other are complex will not be considered. When all the roots are
real but one or more is a multiple root, additional conditions which we now
consider must be given to determine if C.93) is of hyperbolic or parabolic
type.
If the matrix A in C.93) is singular, C.97) does not yield an nth degree
algebraic equation for y'(x), as the lines jc = constant are characteristics in
that case and they cannot be expressed in the form у = у(х). Since A is
singular, В must be nonsingular according to our assumptions. Then we
consider the initial value problem u = f(y) on the curve x = g(y). Proceed-
Proceeding as before and solving for u^ we find that if x = x( y) is a solution of the
characteristic equation
C.99) \A(x(y)9 y)-x'(y)B(x(y), y)|=0,
the solution for u^ is not unique, if indeed it exists at all. This is an nth
degree algebraic equation for x'(y) and each root determines a characteris-
characteristic curve. The classification of the system C.93) based on the roots x'(y) of
C.99) parallels that given above based on the roots y'(x) of C.97) and will
not be repeated here.
If we set <p = у - h(x) = 0 or <p = x - g(y) = 0 both characteristic equa-
equations C.97) and C.99) can be written as
(ЗЛ00) \<pxA + <pyB\=0.
This is often referred to as the characteristic equation for C.93).
To complete the classification process we now assume that the matrix В
ш C.93) is nonsingular throughout the region under consideration so that it
has an inverse. Multiplying through by the inverse matrix B~l in C.93), we
obtain an equation of the form
V
that is, we have effectively put В = /, the identity matrix. [With у replaced
ЬУ t, C.101) has the form of many time dependent systems that occur in
applications.] Then, with В = I in C.99) and k.(i = 1,. . . , л) as the eigen-
eigenvalues of A counted with their multiplicities, we have x'(y) = Л,.. If all the
130 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
eigenvalues are real and there exist n linearly independent eigenvectors r.
(i = 1,. . . , n) for the matrix A, the system C.101) is of hyperbolic type[
The multiplicity of the eigenvalues plays no role, but it is true that when all
the eigenvalues are distinct there exist n linearly independent eigenvectors.
In particular, if Л is a (real) symmetric matrix, there are always n linearly
independent eigenvectors. (We note that if there are multiple eigenvalues,
there are fewer than n independent families of characteristic curves.)
However, if there are multiple (real) eigenvalues and fewer than n linearly
independent eigenvectors for A, the system C.101) is of parabolic type.
When C.101) is of hyperbolic type it can be transformed into the
following canonical form. We form the matrix R whose column vectors
rx,. . . , rn are eigenvectors of A. Then let
C.102) u = i?v
in C.101) and since R is nonsingular, we have for v,
C.103) v, + (R~1AR)yx = Cv + d ,
where R~lAR is a diagonal, not necessarily constant, matrix whose diagonal
elements are the eigenvalues A, of A. (This follows since R~*AR is a
similarity transformation which diagonalizes the matrix Л.) The matrix
С = RXCR - R l Ry - R~lARx, where Rx and Ry are derivatives of the
matrix /?, and d = R~ld. Since each eigenvalue A, determines a characteristic
direction x'(y), that is, x'(y) = \i9 the system C.103) can be written in
component form as
C.104) $=|1^Л + 8" »"-1.---.»•
Each vt is differentiated along a characteristic curve determined from
x'(y) = A(. As a result, this system is said to be in characteristic form.
Example 3.4. First Order Systems in Two Variables. In Section 2.1 the
wave, diffusion, and Laplace's equations were reduced to first order sys-
systems. We now reconsider their classification in system form based on the
above discussion.
The system B.2) for the wave equation may be written as
C.105) u, + Anx = Си ,
where
-[:]■ -[v :]■ c-Bii-
CLASSIFICATION OF EQUATIONS IN GENERAL
putting у = t in C.97) we have
131
C.106) \I-t'(x)A\
1 + yt'(x)
0
0
1 - уф)
l-y2[t'(x)]2 = 0,
as the characteristic equation because В = / in this case. Since t'(x) = ±1 /у,
we obtain, as expected, the characteristic curves x ± y* = constant. The
roots of the characteristic equation are <ог = 1/y and a>2 = -1/y, and these
are real and distinct. Thus the system C.105) is totally hyperbolic.
The Cauchy-Riemann equations B.6) have the form
C.107)
where
Thus C.97) becomes
C.108) \B-y'<
= 0,
? -si-
u =
1 -y'(x)
Since y'(x) — ±i (i = V—7), there are no real characteristics and the system
C.107) is elliptic.
The system B.7) representing the diffusion equation has the form
C.109)
where
Anx + Bnt = Cu
H *-bj]. чтя- -га-
. 0
We multiply across in C.109) by
obtain
C.110)
where
2^
L 0 lj
u^ + Bu, = Си ,
132
CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS?
Lo о J l i o_
This has the basic form of C.101) and has the characteristic equation
_ v л _ !
C.111) \B-t'(x)l\= D
0 -t'(x
Thus t'(x) = 0 is a double root and the characteristic curves are t = constant.
Since the matrix В has only one linearly independent eigenvector г =
(as is easily shown), the system C.110) is of parabolic type.
The system
C.112) u, + ux = Cu,
where С = I n and u = I is equivalent to the single equation
C.113) ц„ + 2uxt + ua - и = 0.
\ / XX XI II
Comparing C.112) with C.101), we set t = у and A = I. The identity matrix
/ has the double real eigenvalues Ax = A2 = 1 and the linearly independent
eigenvectors rl = i and r2 =j. Thus C.112) is a hyperbolic system but it is
not totally hyperbolic. Yet according to the classification procedure of
Section 3.1 [with y = Mn C.1)], we have A = B = C=1 C.113) so that
B2 - AC = 0 and C.113) is of parabolic type. This example shows that the
definition of hyperbolicity given above for systems of equations is somewhat
more general than given for single equations.
Turning to the consideration of a single linear partial differential equation
of order m in n independent variables, we have
(З.П4)
= 0,
1-1 '2=1
where we write only the principal part of the differential equation. The dots
represent lower derivative and undifferentiated terms in u(xl9... , xn).
Adapting the procedure given above for the second order equation C.61), it
is readily shown that the characteristic surfaces <p(xl9 ...,*„) = constant are
solutions of the characteristic equation
C.115) !Г\|Л -Л,- ^=°-
These are surfaces in n-spaoe across which mth order derivatives of solutions
CLASSIFICATION OF EQUATIONS IN GENERAL 133
of C.114) can have discontinuities. If there are no real surfaces that satisfy
C.115), the equation C.114) is of elliptic type. Classification into other
types is somewhat more complicated and is not considered. However,
specific equations of higher order are discussed in the text.
As an example we consider the biharmonic equation in two variables,
C.116) V2V2m - uxxxx + 2uxxyy + uyyyy = 0 .
xxyy + uyyyy
Its characteristic equation is
C.117) <P*X + 2<p2x<p2y + <p\ = (<p2x 4- <p2yf = 0 .
Clearly, it has no real solutions so that C.116) is elliptic.
Next, we consider a system of first order equations in n variables
n „
2 Л'
C.118)
7
i
where A1 and В are A: by A: matrix functions ofxl9...9xn and u and с are
fc-component column vector functions of x19...9xn. The characteristic
surfaces for C.118) are surfaces <p(x19..., xn) = constant on which the
derivatives ux(i = 1,. . ., n) cannot be specified uniquely (if at all) for given
initial data u = f on those surfaces. It can be shown that such surfaces must
be solutions of the characteristic equations
= 0.
Again, if no real surfaces satisfying C.119) exist, C.118) is of elliptic type.
We shall have occasion to characterize systems as being of hyperbolic or
parabolic type in specific examples and exercises but do not classify them in
general. We also do not discuss the classification of systems of higher order
equations. Technically, they can always be reduced to first order systems.
Finally, it is necessary to discuss nonlinear equations and systems. If the
principal parts of these equations are linear and the nonlinearities are
confined to the lower order terms, the classification of these equations
proceeds as above. For example, the equation
C-120) V2M-e"
ls of elliptic type, whereas the system
C.121)
where у = constant is of hyperbolic type.
134 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
As to equations where nonlinearities occur in the principal parts, we deal
only with quasilinear equations. That is, the principal part is linear in the
highest derivative terms, but it may have coefficients that contain lower
derivative or undifferentiated terms. The classification proceeds as for the
linear case, but it depends on the specific solution under consideration. The
characteristics also depend on the specific solution. (This fact was already
observed in our discussion of first order quasilinear equations in Chapter 2.)
As an example we consider the system
[vt-c'(u)ux=0,
where c(u) is a given differentiable function. (This system occurs in a
number of branches of continuum mechanics in one form or another.)
Identifying у with t, C.122) has the form C.101) with
= [
-c\u)
The eigenvalues of the matrix A are Аг - \Jc\u) and A2 = -y/cr(u). If the
solution и is such that c'(u)>0, we find that C.122) is hyperbolic. If
c'(w) = 0, the system is clearly parabolic, whereas it is elliptic if c'(u) < 0. In
the hyperbolic case, the characteristic curves are determined from the
equations dxldt = ±yjc'(u). They are obviously dependent on the particular
solution и under consideration.
Putting v = wt and u — wx in C.122), we obtain the second order
quasilinear equation
C.123) wlt-c'(wx)wxx=0.
Considering a specific solution w in C.123) and recalling the results of
Section 3.1 we have
C.124) B2-AC = c'(wx).
Thus C.120) is hyperbolic if cr(wx) > 0, parabolic if c\wx) = 0, and elliptic
if c'(wx)<0. This is consistent with the above results for the system C.122)
since u — wx.
More generally, we consider the quasilinear system
C.125) Aux + 5u, = c,
where u and с are n-component vectors and the nby n matrices A and B, as
well as c, are functions of x, t, and u. We assume that the matrix В is
nonsingular and consider the eigenvalue problem
CLASSIFICATION OF EQUATIONS IN GENERAL 135
C.126) ATs = \BTs
where Ar and BT are the transposes of the matrices A and B. Since BT is
nonsingular this represents a generalized eigenvalue problem for AT. (It can
ajso be interpreted as a generalized eigenvalue problem for A with respect
t0 By in terms of left (row) eigenvectors sr rather than right (column)
eigenvectors. That is, we consider the eigenvalue problem sTA = AsrB.) If all
the eigenvalues are real and there exist n linearly independent eigenvectors
the system C.125) is of hyperbolic type. If there are multiple (real)
eigenvalues and fewer than n linearly independent eigenvectors, the system
C.125) is of parabolic type. If all the eigenvalues are complex, it is of elliptic
type. [We remark that this classification also applies to the linear system
C.93) if В is a nonsingular matrix.]
If the system C.125) is hyperbolic it is possible to reduce it to a simpler
form known as the characteristic normal form. To do so we multiply C.125)
on the left by the transpose of each of the n independent eigenvectors
determined above. This yields
C.127)
[dui Зи
\-£ + A, -^
Now the characteristic curves for this system are the solutions of x\i) — \(.
Thus, along each characteristic and for each eigenvector, we have
C.128) iy^ = «i on ^ = A,., / = l,...,n.
This is the normal form for the system. In each of the n equations the
components of u are differentiated in a single direction, the characteristic
direction. While this normal form is also valid for the linear system C.93), it
is not as strong a simplification of the linear system as was achieved in the
canonical form C.104). In the canonical form only a single unknown
function is differentiated in each equation.
Example 3.5. The Characteristic Normal Form. We construct the charac-
characteristic normal form for the quasilinear system C.122) in the hyperbolic
case. This system has the matrix form C.125) with A and u defined as in the
Paragraph that follows equation C.122). The matrix B = I and the vector
c = 0. The eigenvalues for the problem C.126) are the same as those for the
eigenvalue problem for A. They were found to be Aj = V^T") and A2 =
~~Vc'(M), in the preceding discussion of C.122). [We assume that c'(u) is
Positive.] Two generalized eigenvectors are easily determined to be sf =
[V?(jg, -1] and s2r = [V?00. 1].
136 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
To obtain the normal form we multiply the equation Aux + u, = 0 by the
eigenvectors sf and sj. This yields
C.129) У/Щ$[Щ ± У/с'(и)их] + [vt ± V^(u)vx] = 0 ,
where the upper signs correspond to Лг and the lower signs to A2. Along the
characteristics x'(t) = ±\Jc'(u) these equations can be written as
. du do dx
C.130) V?(^-t-=o on ^
The equations C.129) and C.130) represent the characteristic normal form
for the given system.
For this problem, however, a further simplification is possible. Equation
C.130) can be written as
C.131) jt[j \/F(u)du + v\=Q on ^ = ±
This shows that the bracketed terms are constant on the corresponding
characteristics. They are called Riemann invariants. They occur in a number
of fluid mechanics problems and play an important role in their solution.
3.4. FORMULATION OF INITIAL AND BOUNDARY VALUE
PROBLEMS
As we have seen in our discussion of the model equations derived in
Chapter 1 and the first order partial differential equations of Chapter 2, we
are not merely interested in finding arbitrary functions that satisfy the given
differential equations. Rather, we ask for specific solutions that satisfy
certain auxiliary conditions associated with the given problem.
The diffusion and telegrapher's or wave equations which are of parabolic
and hyperbolic types, respectively, both contain a time dependence. We
have found it appropriate in our discussion in Chapter 1, to prescribe values
of the solutions of these equations at an initial time t = 0. For the diffusion
equation, initial values for the density function v(x, t) were assigned,
whereas for the telegrapher's and the wave equation, both v(x, t) and
dv(x, i) idt were prescribed at t = 0. In case the values of x are unrestricted,
this constitutes an initial value problem for each of these equations.
When the values of x are restricted to lie in a bounded or semi-infinite
interval, v(x, t), dv{x, t)ldx, or possibly a linear combination of both must
be prescribed on the boundaries for all t > 0, as was seen in our discussion of
the diffusion equation in Chapter 1. Similar conditions are appropriate for
the telegrapher's or wave equation when x is similarly restricted. The
FORMULATION OF INITIAL AND BOUNDARY VALUE PROBLEMS 137
cOmbined prescription of v(x, t) and/or its derivatives on the initial line t - 0
and on the boundary line(s), constitutes an initial and boundary value
problem for each of these equations for v(x, t).
Laplace's equation was shown in Chapter 1 to characterize an equilibrium
or steady-state situation where time dependence plays no role. The un-
jcnown function v(x9 y) was specified on the boundary of the region under
consideration. This constitutes a boundary value problem for Laplace's
equation.
As we have seen, the equations discussed in Chapter 1 are representative
of second order equations of parabolic, hyperbolic, and elliptic type. As a
general rule (in terms of the equations we shall consider), parabolic and
hyperbolic equations are characteristic of problems that contain a time
dependence. Initial value or initial and boundary value problems are
appropriate for such equations depending on whether spatial boundaries
occur for such problems. Elliptic equations represent equilibrium or steady-
state situations in regions with boundaries (in which time dependent effects
play no role), and boundary value problems are appropriate for such
equations.
The number of initial and /or boundary conditions that should be assigned
for a given differential equation depends on several factors. It may be stated
that if the initial line or plane t - 0 is noncharacteristic and the equation
contains к time derivatives, then the function and its first к - 1 time
derivatives must be prescribed at t = 0. This is valid if we deal with a vector
or scalar function. The dependence of the number of boundary conditions
on the order of the differential equations is more complicated and requires
separate discussions for different classes of equations.
More generally, if the data for the problem are not given at t = 0 or on
some spatial boundary for all time, it becomes necessary to examine if they
have the character of initial or boundary data. This determination again
depends on the given partial differential equation. When data are given in a
region that has the character of an initial curve or surface, the problem is
known as a Cauchy problem. Also, for certain equations, boundary data are
referred to as Dirichlet or Neumann data, and these are discussed at the
appropriate place in the text. Further, we consider data given on charac-
characteristic curves and surfaces and it is necessary to determine under what
circumstances such problems can be solved. (We have already encountered
certain problems of this type in Chapter 2 and in the preceding sections of
this chapter.)
The problem of deciding what form of initial and/or boundary data are
appropriate for given partial differential equations is fairly complicated. A
set of guidelines was proposed by Hadamard, who listed three requirements
that must be met when formulating an initial and /or boundary value
Problem. A problem for which the differential equation and the data lead to
a solution satisfying these requirements is said to be well posed or correctly
Posed. If it does not meet these requirements, it is incorrectly posed.
138 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
Hadamard's conditions for a well posed problem are:
1. The solution must exist.
2. The solution must be uniquely determined.
3. The solution must depend continuously on the initial and I or boundary
data.
The first two conditions require that the equation plus the data for the
problem must be such that one and only one solution exists. The third
condition states that a slight variation of the data for the problem should
cause the solution to vary only slightly. Thus since data are generally
obtained experimentally and may be subject to numerical approximations,
we require that the solution be stable under small variations in initial and/or
boundary values. That is, we cannot permit wild variations to occur in the
solution if the data are altered slightly. These conditions represent reason-
reasonable requirements for a problem arising in a physical context.
Thus for any given differential equation defined over a certain region,
one must check whether the data assigned for the problem meet the
Hadamard criteria. It was seen in Chapter 1 that the equations we derived
were naturally associated with a set of initial and/or boundary conditions.
These conditions are, in fact, appropriate for the differential equations that
were considered and the problems given in Chapter 1 are well posed. It may
well be argued that any problem arising in a physical context comes with
built-in data, so that there is no need to decide which data are relevant.
However, since any partial differential equation representing a physical
process is a mathematical model obtained, in general, under various sim-
simplifying assumptions, it is not a priori obvious that the formulation of the
mathematical problem is reasonable or well posed.
The problems we consider in this text are sufficiently standard that their
appropriate formulations are well understood. Nevertheless, we comment
occasionally on questions of uniqueness or continuous dependence on data.
It should be noted that certain problems representing physical processes
are, in fact, incorrectly posed in the sense of Hadamard, and this does not
appear to be due to a weakness in the mathematical model. Such problems
have been studied and methods for dealing with such problems are continu-
continuing to be developed. We do not consider such problems in this text.
We now present two examples of incorrectly posed problems and discuss
further questions relating to this matter in Section 3.5.
Example 3.6. Incorrectly Posed Problems. Boundary value problems are,
as a rule, not well posed for hyperbolic and parabolic equations. This
follows because these are, in general, equations whose solutions evolve in
time and their behavior at later times is predicted by their previous states.
STABILITY THEORY, ENERGY CONSERVATION, AND DISPERSION
139
У
<
it
о
и{х, 0)=f,(x)
S
ГМ
tl
"а
X
Figure 3.2. A boundary value problem.
Thus a boundary value problem that arbitrarily prescribes the solution at
two or more separate times is not reasonable.
As a simple example, we consider the hyperbolic equation ux = 0 in the
unit square 0<x<land0<)><l with boundary values assigned on the
sides of the square. We show that this problem has no solution if the data
are prescribed arbitrarily. Since uxy = 0 implies that ux(x, y) = constant, we
have ux(x,0) = ux(x, 1). Taking note of the boundary conditions given in
Figure 3.2, we have ux(x, 0)=f{(x) and ux(x, 1) -/;(*). Thus unless fx{x)
and/3(x) are prescribed such that/{(x) =/£(*), the boundary value problem
cannot be solved. Therefore, it is incorrectly posed.
For Laplace's equation, the Cauchy problem is, in general, not well
posed. This is shown by the following example credited to Hadamard. We
consider* the equation uxx + uyy = 0 in the region у > 0 with the Cauchy data
Ф,0) = 0 and Uy(xy0) = (sinnx)In. The solution is easily obtained as
w(*> У) = [sinh(n>0 $in(nx)]/n2. Now, as n^oo, Uy(xy 0)-»0 so that for large
n the Cauchy data м(*,0) and uy(x,0) can be made arbitrarily small in
magnitude. However, the solution m(jc, >>) oscillates with an amplitude that
grows exponentially like eny as n^> oo. Thus arbitrarily small data can lead to
arbitanly large solutions and the solution is unstable. This violates the third
condition of Hadamard requiring continuous dependence of the solution on
the data.
^STABILITY THEORY, ENERGY CONSERVATION,
AND DISPERSION
Ь this section we consider certain general properties of partial differential
equations that distinguish equations of different types beyond or apart from
140 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
the classification process given in the previous sections. We recall that
classification depends only on the form of the principal part of the equation.
Here we examine the role played by the lower order terms in the equation in
determining the behavior of the solutions.
For simplicity we restrict our discussion to second order linear partial
differential equations with constant coefficients. The general ideas carry
over to higher order linear equations and systems of equations with constant
coefficients. Also certain aspects can be generalized to apply to equations
with variable coefficients and to nonlinear equations as indicated later in the
text.
Given the linear second order homogeneous equation in two variables
C.132) Auxx + 2Buxt + Cutt + Dux + Eut + Fu = 0 ,
where A, B,..., F are real constants and и = w(jc, t), we look for exponen-
exponential solutions of the form
C.133) u(x, t) = a(k) exp[ikx + \(k)t],
where a(k) is a constant and i — V—T. The parameter к is assumed to be
real and A (A:) must be chosen such that C.133) satisfies C.132). The
solutions m(jc, t) for each к are called normal modes of the equation C.132).
Since the coefficients in C.132) are assumed to be real, the real and
imaginary parts of the normal modes C.133) are also solutions of C.132).
(If a is a complex valued quantity and is written as a = ax + /a2, where ax
and a2 are real valued, then ax - Re[a] and a2 = Im[a] are the real and
imaginary parts of a, respectively.)
We assume that t represents time and discuss the behavior of the normal
modes in the region *>0. As shown in later chapters, the solution of an
initial value problem for C.132) with initial data at t = 0 can be represented
as a superposition of normal modes either by the use of Fourier series or
integrals. Thus the behavior of the solution of the initial value problem is
largely determined by that of the normal modes.
Inserting C.133) into C.132) yields the quadratic equation
C.134) СЛ2 + BiBk + £)A + (- Ak2 + iDk + F) = 0.
Solving for A = A(fc), we let \{k) represent either of the two possible
solutions of C.134). Then the normal mode can be written as
C.135) u(x, t) = a(k) exp[i(kx + Im \(k)t)] exp[Re \(k)t].
The magnitude of u(x, t) is given as
C.136) \u(x9 01 = |л(*)| exp[Re X(k)t].
STABILITY THEORY, ENERGY CONSERVATION, AND DISPERSION 141
Assuming \a(k)\ is bounded for all k, we see that the growth of \u(x, t)\ as t
increases from zero is determined by the expression Re \(k).
We consider two possibilities. Either Re A(A;) is bounded above for all
real к or it is unbounded. We define the constant ft to be the least upper
bound (denoted as lub) of Re A(A:) as к ranges through all its values; that is,
C.137) n = /Mf>ReA(A:), -oo<A;<oo.
If Re X(k) is not bounded above, we set ft = +00 and this is the case we now
study. The case when ft < +00 is considered later.
Assuming ft = +00 we put a(k) = 1 /X2(k) in the normal mode C.133) and
consider the initial value problem for the normal mode with data at t = 0.
We have, on evaluating C.133) and its derivative at t = 0,
Since Re \{k) is unbounded, there exist values of к for which \u(x, 0)| =
l/|A(ifc)|2 and 111,(x, 0)| = l/|A(i)| are arbitrarily small and yet \u(x9 t)\ [as
given in C.136)] is arbitrarily large for any t >0. This follows since l/|A(fc)|
decays algebraically whereas exp[Re X(k)t] grows exponentially. Thus
Hadamard's stability criterion for the initial value problem for C.132) is
violated if ft=+o°, and the initial value problem is not well posed. If
ft< +0°, it can be shown that the initial value problem for C.132) is well
posed.
Before considering the case where ft < +°°, we consider an example in
which the foregoing discussion is applied to specific equations of elliptic,
parabolic, and hyperbolic types.
Example 3.7. Well Posedness for Elliptic, Parabolic, and Hyperbolic
Equations. Given the elliptic equation,
(ЗЛ39) uxx + utt + pu = 0,
with p = constant, the A(fc) in the normal mode solution C.133) has the
form
Selecting the positive root, we see that A(jfc) is real for k2>p and that
Re A(A:)-> +00 as \k\ -» +00. Consequently, ft = +00 and the Cauchy problem
f°r C.139) with data at t = 0 is not well posed.
The parabolic equation
C.141) Р«« + и, = 0,
142 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
where p = constant, has the form of the diffusion or heat equation if p <0
and is known as the backward diffusion or heat equation if p > 0. One form
can be obtained from the other if the time direction is reversed; that is, if r is
replaced by -f. Normal mode solutions for C.141) yield
C.142) k{k) = pk2.
Now if p < 0, we have £1 = 0, whereas if p > 0, we have ft = +«> so that the
Cauchy problem is not well posed if p > 0. The appropriate data (in view of
our discussion) would be u(x, 0) = (l/pk2)elkx with a(k) = 1/A(fc), since only
one time derivative occurs in C.141). The normal mode solution has the
form
C.143) u(x, i) - [-p] exp[*fcc + pk2t],
and it is apparent that for p >0 and к—»°°, |м(дс, 0)| is small and |w(x, i)\ is
large.
The hyperbolic equation
C.144) utt-uxx + pu = 0,
where p = constant, has normal mode solutions C.133) with K{k) given as
C.145) A(*) = ±iV*2 + p.
For sufficiently large |*|, if p < 0, we have Re \(k) = 0. With p > 0, Re X(k)
vanishes for all k. Thus fl < +°° for C.134) and the Cauchy problem is well
posed.
In all the foregoing cases we have | A(A:)|->°° as |/:|-»a>, so that if data of
the form C.138) are chosen, |m(jc, 0)| and |u,(x,0)| can be made arbitrarily
small by choosing |&|>1. But if ft <+<*>, \u(x, t)\ is also small since
exp[Re X(k)t] does not grow unboundedly with к for fixed values of t.
The constant п defined in C.137) is known as the stability index for the
differential equation. We have already seen that when ft = +°o, solutions of
the initial value problem for C.132) are unstable. That is, solutions whose
magnitude is initially arbitrarily small can grow arbitrarily large even at
finite times. In the following discussion we assume that П< +°о and ask
whether solutions of initially bounded magnitude can grow unboundedly as
t-> +oo. Equations for which this can happen are said to be unstable.
Otherwise they are said to be stable. Thus even if the Cauchy problem is
well posed, the equation may be unstable.
Physically, instability indicates that, even in the absence of external
THEORY, ENERGY CONSERVATION, AND DISPERSION 143
effects due to forcing terms in C.132), internal mechanisms generate a
growth in the solution as time increases. If the Cauchy problem is not well
posed (i.e., ft=+°°), the validity of the mathematical model must be
reexamined. If the Cauchy problem is well posed, instability plays an
important role if the equation was derived by a linearization procedure
under which it was assumed that solutions remain small in magnitude for all
time. Examples of stability analyses where the present results play a role are
given later in the text.
Given the normal mode solution C.133) and its magnitude C.136), we
note the following results. If ft<0, then |w(x, r)|—>0 as f—>°° for all k. If
(i>0 there are normal modes with к near ft, for which \u(x, f)|->oo as
/-»oo. If ft = 0, there may be values of к for which \u(x, t)\ is bounded but
does not tend to zero as f-»°°, although \u(x, t)\-»0 for the remaining
values of A:. [We remark that since ft is a least upper bound it may happen
that Re \(k) never equals zero even if ft = 0.]
Noting the preceding discussion, we introduce the following classification
for C.132). If ft <0, the equation is said to be strictly stable. If ft>0, the
equation is unstable. If ft = 0, the equation is said to be neutrally stable, but
it may be unstable in this case.
For example, the parabolic equation
C.146) ut - c2uxx + aux + bu = 0
has \(k) = -k2c2 ~b- ika so that ft = -b. It is stable if b >0, unstable if
b< 0 and neutrally stable if b = 0.
A simple example indicating the problems in the neutrally stable case is
given by the equation
C.147) uxx + 2uxt + utt = 0,
for which \(k) = -ik so that Re A(A:) = 0 = ft. The equation has solutions
C.148) U(x, t) = aeik{x~t} + bteik{x'° ,
where a and b are arbitrary constants. In particular, the solution of the
Cauchy problem m(jc, 0) = 0 and ut(x, 0) = 1 is given by u(x, t) = t. Thus
even though the data are uniformly bounded in magnitude, the solution
grows unboundedly as f-*°°. It is generally true that for a neutrally stable
case, if there is instability, the growth of the solution will be algebraic rather
*an exponential. [We remark that the second term on the right of C.148) is
n°t a normal mode solution of C.147) as defined in C.133).]
It must again be emphasized, that for the data at t = 0 of the normal
*ode solution C.133), we have |и(х,0)| = |л| and |и,(*,0)| = |лЛ|, which
*** uniformly bounded for all x and fixed k. However, |ы(дг, r)l =
la|exp[Re \{k)t] and if ft<°°, \u(x, i)\ can tend to infinity only as r-»oo.
144 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
This is in contrast to the case where П=+оо and \u(x, t)\ can become
arbitrarily large for fixed t.
If Re \{k) = 0 for all k, C.132) is said to be an equation of conservative
type. Then we have
C.149) |и(л,01Н«1Чи(*,0)|
in view of C.136), so that the amplitude |w| of the normal mode solution is
constant in time. Since |w|2 is generally a measure of the energy of the
normal mode solution, we find that the energy is conserved in time and say»
that the equation C.132) is conservative.
In the conservative case, \{k) may be expressed as \(k) = — i(o(k), where
<o(k) is real valued for all k. The normal mode solution then take the forrn
C.150) u(x, t) = a(k) exp{i[fcr - o>(k)t]} .
Inserting C.150) into C.132) yields
C.151) <*> = o>(k) ,
which is known as the dispersion relation for the differential equation. If
<o(k) is real valued for all real к as we have assumed and, additionally,
<o"(k)^0 (i.e., a) is not a linear function of k), the equation C.132) is said
to be of dispersive type. [We note that regardless of the assumptions made
above on <o(k), C.151) is known as the dispersion relation for C.132).]
Recalling the discussion in Example 2.2, we find that C.150) for fixed к
has the form of a wave that travels with velocity dxldt — <o(k)/k. The term
C.152) e = kx-<o(k)t
in C.150) is called the phase of the normal mode and dxldt - <o(k)/k is the
phase velocity. If <o"(k)^0, we assume a>@)~0 and find that <o(k) = ck
where с = constant. Thus the phase velocity dxldt = cklk = c, has a constan
value с for all normal modes. Since the general solution, as we have
indicated, can be constructed as a superposition of normal modes it will alsc
represent a wave traveling with velocity с [In general, <o(k) has two values
so that the general solution is a sum of two waves.]
However, if to"(fc)|^0, the phase velocity dxldt = <o(k)fk will be differen
for different values of к so that different normal modes have differen
velocities. The general solution obtained as a superposition of the normal
modes yields a wave that disperses since its components or modes all travel at
different velocities. We shall see in Section 5.7 that the relevant concept for
dispersive wave motion is not the phase velocity o)(k)/k, but the group
velocity defined as da)(k)ldk.
Finally, if the stability index П<0 and Re X(k) is negative for all except
STABILITY THEORY, ENERGY CONSERVATION, AND DISPERSION 145
a finite number of values of k, the equation C.132) is said to be of
dissipative type. In that case C.136) shows that \u(xy t)\ decays to zero as
for all but a finite number of values of к and the energy \u(x, t)\2 is
t
dissipated as f-^°°. When ft<0 all solutions tend to zero as
However, if п = 0, those modes for which Re \(k) = 0 do not decay as *-» oo
and they are expected to constitute the major contribution to the solution
for large values of /. This fact is demonstrated when dissipative equations
are studied in Section 5.7.
We now consider two examples of equations of dissipative and dispersive
type. The telegrapher's equation derived in Chapter 1 is an equation of
dissipative type. The Klein-Gordon equation which occurs in relativistic
physics is an equation of dispersive type. Both these equations are common-
commonly used as model equations of dissipative and dispersive type.
Example 3.8. The Telegrapher's and Klein-Gordon Equations. The
telegrapher's equation A.88)
C.153) utt-y2uxx + 2\ut = 0,
[where A has been replaced by A to avoid confusion with A = A(A:)], yields
for the normal mode solutions C.133)
C.154) A(Jfc) = - A ± У\2-у2к2 .
Since A>0 by assumption, the real part of \(k) is easily seen to be negative
for all к Ф 0. At к = 0 either A@) = 0 or A@) = -2A so that п = 0 for
C.153). Consequently, the telegrapher's equation is neutrally stable and is
of dissipative type. Note that when A = 0 C.153) reduces to the wave
equation and X(k) = ±iyk in that case. Then Re k(k) = 0 for all к so that
the wave equation is also neutrally stable and is of conservative type. For
this reason when A>0, C.153) is often called the damped wave equation.
All normal mode solutions of C.153) are damped as f->°° except for the
solution corresponding to A@) = 0. It is shown in Section 5.7 that the main
contribution to the solution of the Cauchy problem for C.153) as
comes from the normal modes with к ** 0.
The Klein-Gordon equation has the form
where у and с are constants. For its normal modes C.133) we have
(ЗЛ56) A(jfc) = ±i\jy2k2 + c2 - ±ш(к),
so that the dispersion relation is
146 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
C.157) a> = a>(k) - Уу2к2 + с2 .
Since a>(k) is real valued for all к and <o"(k)^Q9 we conclude that, in
addition to being neutrally stable, the Klein-Gordon equation is of con-
servative and dispersive type. Again, it may be noted that if с = 0 in C.155),
it reduces to the wave equation for which a>(k) — yk. Since a)"(k) = 0 in this
case, the wave equation is not of dispersive type.
To conclude our discussion of stability theory, it must be emphasized that
the foregoing definitions are relevant, in general, only for the initial value
problem for C.132). When initial and boundary value problems are consi-
considered, the parameter к may be restricted to a discrete set of values. The
solution of the problem is then given as a superposition of a discrete set of
normal modes that correspond to the aforementioned values of k. Stability
is then defined in terms of the growth or decay of these normal modes. If
one or more of the normal modes is unbounded as f—»oo? the problem is
unstable. If all the normal modes decay as t—»o°, the problem is stable.
Examples of stability analyses for initial and boundary value problems are
given later in the text.
3.6. ADJOINT DIFFERENTIAL OPERATORS
The concept of adjoint differential operators has been encountered in
connection with each of the continuum limits of random walk problems
considered in Chapter 1. In the study of time dependent random walk
problems it was noted that the forward and backward differential equations
are adjoints of one another. For time independent problems, it was indi-
indicated that Laplace's equation and its generalizations, and the equations for
the corresponding Green's functions involve adjoint differential operators.
In each case the equations and their adjoints were derived independently of
one another. However, a key relationship between them involves the use of
Green's theorem and its generalizations. This was demonstrated in Example
1.6 for Laplace's equation and the corresponding Green's function.
The basic result that is used in obtaining Green's theorem and its
generalizations is the divergence theorem. Thus, if L is Laplace's operator
L = V2, we have
C.158) wL(u) - uL(w) = wV2u - wV2w =V- [wVu - iNw],
where the term on the right is a divergence expression. On integrating
C.158) over a bounded region, the divergence theorem expresses the
integral of the left side of C.158) in the interior of the region in terms of the
integral of the normal component of the vector on the right in C.158) over
the boundary.
ADJOINT DIFFERENTIAL OPERATORS 147
However, if the operator L is given as L = dldt-V2, we find that to
obtain a divergence expression in the manner of C.158) it is necessary to
introduce a new operator L* = -dldt-V2 in terms of which
C.159) wL{u) - uL*(w) = V- [-wVu + uVw, wu],
where V- [V, dldt] is the gradient operator in space-time. [The divergence
theorem in space-time can be applied to equation C.159).] The operator L*
is called the adjoint operator of L.
As shown above, if L is Laplace's operator the expression wL(u)-
uL*(w) is in divergence form with L* = L = V2. Since the adjoint operator is
identical with Laplace's operator, the Laplacian is a self-adjoint operator.
Equations in which these differential operators appear are called self-adjoint
or nonself-adjoint equations according as the operators are self-adjoint or
not self-adjoint.
More precisely, if L* is an operator for which wL(u)- uL*(w) is a
divergence expression, it is the formal adjoint operator of L, and if L* = L,
the operator L is formally self-adjoint. (The same is true for the related
differential equations.) As will be seen in our discussion of eigenvalue
problems and Green's functions later in the text, the boundary (or initial
and boundary) conditions associated with the given equation and its formal
adjoint also play a role in determining whether or not the problem is
self-adjoint. Thus it is possible for an operator to be formally self-adjoint
but for the problem associated with this operator and its adjoint to be
nonself-adjoint. But if the operator is not self-adjoint, the problem cannot
be self-adjoint. In this section, however, we deal only with the adjointness
of the operators themselves, and we now present a formula for the adjoint
of the most general linear second order differential operator.
The general linear second order differential operator L can be written as
C.160) L[u] =
I==1
where the coefficients a(j, b(, and с may be functions of xt, x2,.. . , xn. The
adjoint operator L* is given as
It can be verified directly that
C-162) wL[u] -
where
148 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
<з.ш, г,-
The expression on the right of C.162) is a divergence expression, and the
w-dimensional form of the divergence theorem yields
C.164) f f {wL[u]-uL*[w]}dv= f P-nds,
J JG JdG
where P is a vector with n components Px,..., Pn and n is the exterior unit
normal vector to the boundary dG. As has already been indicated, the
possibility of using the divergence theorem as in C.164) is a key reason for
the importance of the adjoint operator L*.
To see how the adjoint operator L* is constructed we consider a typical
term. We have
<n uo a%u д \ ЗиЛ д Г диЛ d(anw) ди
C.165) wau —г= и>вп ~ waw
v } n dx\ n dxx Ldx^ dxx L u dxx\ dxx dxx
This yields
C.166) wau —2 - и \— = -— axiw и — .
V } n dx\ dx\ dxx L n dxx dxx J
By carrying out all the differentiations in C.161) and comparing L* wit$
L, it is easy to determine that if
C.167) 6,= S-r~, / = l,...,n,
/-1 dxi
the operator L is self-adjoint. In the self-adjoint case the operator L takei
the form
C.168) L[u] = ±±j
i=i /=i «
We observe that if the operator L has constant coefficients, it cannot hava
any first derivative terms if it is to be self-adjoint. The foregoing results cafl
be specialized, by putting n = 1, to apply to second order ordinary different
tial equations as well.
Example 3.9. Adjoint Differential Operators. In this example we con-
consider three differential operators of elliptic, hyperbolic and parabolic types
ApjOlNT DIFFERENTIAL OPERATORS 149
^d determine their adjoints. Equations involving these operators are
studied throughout the remainder of this text.
We begin with the elliptic operator L defined as
C.169) Lm = -V- (pVu) + qu ,
where p>0. Since
,~ Yjq\ wLu — uLw = V • [— pwVu + puVw],
we see that the operator L is self-adjoint.
For the hyperbolic operator L defined as
C.171) Lm = putt + Lu ,
with Lu given as in C.169), we have
C.172) wLu — uLw = V* [-pwVu + puVw, pwut — pwwj ,
where V is the space-time gradient operator. (The positive function p is
assumed to be independent of t.) Thus, the operator L is self-adjoint.
For the parabolic operator L defined as
C.173) Lm-PM, + Lm,
where L is defined as in C.169), we have
C.174) wLu - uL*w=V-[-pwVu + puVw, pwu],
where V is again the space-time gradient operator and the adjoint operator
L* is given as
C.175) L*^ = -pm, + Lm.
Jhus, in the parabolic case, the operator L is not self-adjoint.
Adjoints can also be defined for systems of equations. We consider only
linear systems in two variables. Let the matrix operator M be defined as
<3л76) M[u] - Aux + Buy + Си ,
where Д, B, and С are n by n matrix functions of x and y. We define the
adjoint operator M* as
(ЗЛ77) M*M = -(^w), - (BTw)y + Crw ,
150 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
where the superscript T denotes the transpose of the matrix or the vector.
With this choice of M * we obtain
C.178) wrMu - urM*w = V- [wrAu,
where V is the two-dimensional gradient vector. That is, we obtain a
divergence expression and this motivates the definition of the adjoint
operator. If A = —AT, B=—BT (that is, A and В are antisymmetric),
A^ + BTy - 0, and С = Cr, the operator M is self-adjoint. Some examples of
systems of equations and their adjoints are considered in the exercises.
If the equations or systems are quasilinear it is not possible to use the
foregoing results to define adjoint operators because the dependence of the
coefficients on the unknown function(s) invalidates the results, as is easily
seen. Nevertheless, it is useful to obtain results similar to those given above
for linear problems, for the purpose of defining weak or generalized
solutions of quasilinear equations. This can be accomplished if the equation
is given in conservation form. We consider only a single equation in two
variables.
Let Lu be defined as
C.179) Lu=V-u = (u1)x + (u2)t
where ux -f(x, t, u) and u2 = g(x, t, u) with и as the unknown function.
Then, it is easily verified that
C.180) wLu + u-Vw = (wul)x + (wu2)t = V- (wu)
We do not explicitly define an expression that serves as the adjoint of Lm,
but note that C.180) does express the left side of the equation in divergence
form. Thus, if we integrate C.180) over a bounded region R in (x, f)-space
and w vanishes on the boundary, we have
C.181) J J wLudxdt=-j ju-Vwdxdt.
R R
On this basis we can define weak solutions of Lu = 0 which are not
differentiable everywhere, as merely piecewise continuous solutions foi
which the integral on the right in C.181) vanishes for each w. Weak
solutions of linear equations are discussed from this point of view in Section
6.4.
EXERCISES FOR CHAPTER 3
Section 3.1
3.1.1. Show that the coefficient of u^ in C.53) equals D/A)(AC -
in the hyperbolic case, if f and rj are defined as in Section 3.1.a.
EXERCISES FOR CHAPTER 3 151
3.1.2. Show that if <p = f and ^ = 77, as defined in C.24), are used as
coordinates in transforming C.1) in the parabolic case, the coefficient of uVJ)
in C-53) is the only one that is nonzero, and the equation C.25) results.
3.1.3. Show that if the transformation C.10) is introduced into C.1) in the
elliptic case, the coefficient of u€v in C.53) equals D/A)(AC- B2)^yy]y.
3.1.4. Show that the equation
uxx + 4uxy + 3uyy + 3ux - uy + 2м = О
is of hyperbolic type. Determine its characteristic curves and bring it into
the canonical form C.19). Introduce the further transformation
m(£, 17) = exp(a£ + j3t)M£, 17) ,
and choose the constants a and 0 to eliminate the first derivative terms in
the resulting canonical form.
3.1.5. Show that the equation
uxx + 2uxy + uyy + 5ux + Ъиу + и = 0
is of parabolic type and bring it into the canonical form C.25). Using the
transformation of the dependent variable given in Exercise 3.1.4, show that
the terms involving uv and и in the canonical form can be eliminated.
3.1.6. Show that the equation
uxx - 6uxy + I2uyy + Aux - и = sin(xy)
is of elliptic type and obtain its canonical form C.27). Proceeding as in the
above exercises, show that the first derivative terms in the canonical form
can be eliminated.
3.1.7. Determine the regions where TricomVs equation
Uxx + ™yy = 0 '
|s of elliptic, parabolic, and hyperbolic types. Obtain its characteristics and
*ts canonical form in the hyperbolic region.
3.1.8. Show that the equation
has the simple canonical form uiv - 0 in the region where it is of hyperbolic
type. Use this result to show that it has the general solution
w = /(* + ^V^J) + g(x ~ 2V=y) ,
where / and g are arbitrary functions, in the hyperbolic region.
152 CLASSIFICATION OF EQUATIONS AND CHARACTERISTIC^
3.1.9. Classify the following equations into hyperbolic, elliptic or parabolic
type. ^ 2
(a) 5uxx - 3uyy + (sin x)ux + ey uy + и = 0.
(b) uyy - 10ux + 4мл + (cosh x)w = 0.
(c) 10^ + uyy -ux + [log(l + x2)]u = 0.
3.1.10. Classify the following equations into hyperbolic, elliptic, or
parabolic type.
(a) exyuxx + (cosh x)uyy + ux - и = 0.
(b) [log(l + x2 + y2)\uxx - [2 + cos *]uyy = 0.
(c) uyy + [1 + *2K - и, + w = 0.
3.1.11. Show that if В = 0 in C.1), the equation is of fixed type if AC < 0,
AC-09 or АО0 everywhere in the region where C.1) is defined.
Section 3.2
3.2.1. If (p{x, y) = constant is a family of characteristics for C.41), show
that C.46) reduces to
Assuming и is continuous across £ = <p(x, y) = 0 while u^ has a jump
discontinuity there, demonstrate that the jump [u^] at £ — 0 satisfies the
ordinary differential equation
by differentiating C.46) with respect to £. This result determines the
variation of the jump [wf] along the characteristic £ = 0.
3.2.2. Consider the equation ux + uy = 0 and the continuous solution и =
ai(y ~x)fory<x and и = a2(y - x) for у > x. Let £ = у - x and show that
for a^aj, [u€] = a2- ax across | = 0 and that this jump satisfies the
condition determined in Exercise 3.2.1 for the variation of the jump.
3.2.3. Consider the equation
ux + uy - cu - 0 ,
where с = constant. Show that the appropriate (£,17) coordinates in C.43)
are
where £ = у - x = constant are the characteristics. Obtain the equation
2мп - cu = 0,
and the equation for the jump [u€],
EXERCISES FOR CHAPTER 3 153
Verify that both u(x, у) = рх(у -л:)ехр[^с(^ + *)], У^х and u(x, y) =
a(y-x) exp[§c(v + x)]9 у > x are solutions of the given equation for у Ф x
and that the jump in u$ across f = у - x = 0 satisfies the equation for the
jump (Pi and Pi are constants).
3.2.4. If ?>(*, jO - constant is a family of characteristics for C.1) show that
C.53) reduces to the form
20«*4 + У«™ + «ил + puf + Am + cr-0.
If м^ has a jump across £ = <р(лг, у) = 0 and the terms in the above equation
are continuous across £ = 0, conclude on differentiating the equation with
respect to £ that the jump [u€4] satisfies
assuming smooth coefficients.
3.2.5. Given the hyperbolic equation
show that u(x, y) = pr(y -xf\ y<x and u(x, y) = P2(y - xJ, y>x are
solutions with discontinuous second derivatives across the characteristic
у = x if the constants ft and /32 are unequal. If £ = v - x, show that [wff ]
satisfies the appropriate form of the equation for the jump across £ = 0 given
in the preceding exercise.
3.2.6. Let ube a continuous solution of the diffusion equation A.23) and
assume that vt has a jump across a characteristic t = constant. Show that the
jump [vt] satisfies the equation [vt] ~ 0, so that no jump can exist under the
given conditions.
3.2.7. Let the inhomogeneous term d(x, y) in C.41) have a jump across the
curve £ = <p(xy y) = 0 and assume that the solution u{x, y) is continuous
there. Proceeding as in the text, show that the equation for the jump in u€
across | = 0 is
(a<Px + b<py)[u4] = [d]
Thus, if £ = (p(x, v) = 0 is a characteristic curve we must assume that и itself
**as a jump across £ = 0 to avoid a contradiction.
3«2.8. Show that the results obtained in Exercise 2.2.11 are consistent with
those that follow from the preceding exercise.
3«2.9. By differentiating C.53) as often as necessary with respect to £, show
that jumps in дпи1д(;п, п^Ъ must also occur across characteristics, assuming
the appropriate lower order derivatives are continuous.
154 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
3.2.10. If и and the normal derivative duidn are specified on a curve
у = h(x), show how to specify ux and uy on that curve. Then, assuming
ux[x, h(x)] = a(x) and uy[x, h(x)] = p(x), where a and /3 are known func-
functions, obtain the equations
Auxx + 2Buxy + Cuyy = -Da - E@ - Ff + G ,
for the specification of uxx, uxyJ and uyy along the curve у = Л(дг). {We
assume u[x, h(x)] = /(*).} Show that if
the second derivatives of и cannot be specified uniquely on у = h(x), if they
can be determined at all. If we set <p =y - h(x), show that the above
equation becomes the characteristic equation C.55).
3.2.11. Consider the hyperbolic equation
Apply the method of the preceding exercise and show that uxx, uxyy and uyy
can be determined nonuniquely on a characteristic if the compatibility
condition arhf - fi* = 0 is satisfied. Construct a set of initial conditions for и
and duldn on the characteristic у - x = 0 such that the compatibility condi-
conditions are satisfied and show that this problem for the given equation has
infinitely many solutions.
3.2.12. Show that the compatibility condition that guarantees that the
problem in Exercise 3.2.10 has a (nonunique) solution along the characteris-
characteristic у = h(x) is
Ah'p' - 2Bf3' -ЛаЧС- Ff- Ep - Da = 0 .
Section 3.3
3.3.1. Show that the equation
- 2иХ2Хз + 3u,rt + 5uX2 - и,з + Юн = 0
is of elliptic type by determining that the matrix A [see C.62)] has the
eigenvalues A, = 1, A2 = 3, and A3 = 4. Determine a transformation C.73)
that yields the equation
1Ом =
EXERCISES FOR CHAPTER 3 155
Construct transformations of the independent and dependent variables that
replace the coefficients of the second derivative terms by unity and eliminate
the first derivative terms.
3.3.2. Show that with и = и(хи jc2, x3), the equation
is of parabolic type. Use C.73) to bring it into canonical form.
3.3.3. Classify the following equations into elliptic, parabolic, or hyperbolic
type. 2
(a) uxx + 2uyz + (cos x)uz - ey и = cosh z.
(b) uxx + 2uxy + uyy + 2u2Z - A + xy)u = 0.
(c) 7uxx - Wuxy - 22uy2 + 7nw - \6uxz - 5uzz - 0.
(d) ezuxy - uxx = log[*2 + y2 + z2 + 1].
3.3.4. Show that all linear second order equations of elliptic type with
constant coefficients can be brought into the form
3.3.5. Show that all linear second order equations of hyperbolic type (in
n +1 variables) with constant coefficients can be transformed into
2 uXiXi - uXqXq +cu = F(x0, xl9...,xH)
3.3.6. Determine the regions where
is of hyperbolic, elliptic, or parabolic type.
3.3.7. Demonstrate that the equations
(a) V-(pVu) + qu = F,
(b) put -
(c) putt-
are of elliptic, parabolic, and hyperbolic types, respectively, in a region R
where p and/? are positive. [Assume that и = u(;t, y, /) or и = m(jc, y, z, f) in
the hyperbolic and parabolic cases, and u = u(x, y) or и = u(x, y, z) in the
elliptic case.]
^•3,8. Considering that if we disregard the order of differentiation, the
function w(fx,.. . , ^и) has exactly \n(n -1) mixed partial derivatives of
°*<ler n, show that if n > 3 there are, in general, an insufficient number of
equations of transformation leading from C.62) to C.66) to eliminate the
156
CLASSIFICATION OF EQUATIONS AND CHARACTER!».
mixed partial derivative terms in ft,. . . , £n throughout a given region
n = 3, the mixed partial derivative terms can be eliminated but the coc
ents of the u^ terms cannot, in general, be made to equal plus Oi iu
one, as was done in the two-dimensional case.
3.3.9. Use the transformation C.91) to bring the equation C.83) of Exa
pie 3.3 into the form
Attempt a transformation of this equation to bring it into the form C.9Г
3.3.10. Show that the system uy + Aux = 0, where
is of parabolic type.
3.3.11. Show that the system Aux + Bny = Cu 4- d, where
л-[_\_°\- -[-?-!]■
is of elliptic type.
3.3.12. Determine the characteristic curves of the (totally) hyperbolic s
tern u^ + Aux = Cu, where
Г-2 1 5]
= 0 3 7 .
L i -3 -loJ
Reduce this system to the canonical form C.103).
3.3.13. Show that uy + Aux = 0 where
is a hyperbolic system and reduce it to the (diagonal) canonical form
C.103).
3.3.14. Use the canonical form obtained in Exercise 3.3.13 to show that the
general solution of the equation u^, + Aux = 0 given in that exercise is
u(x, y) =
f3(x~2y)
FOR CHAPTER 3 157
Use this result to solve the initial value problem for uy + Aux - 0 with the
initial condition
3.3.15. Show that the system uy + Aux =0, where
is totally hyperbolic. Determine its characteristic curves and reduce it to the
canonical form C.103).
3.3.16. If the constant-coefficient system uy + Aux = 0 is hyperbolic (with u
as an w-component vector) show how to obtain a general solution by using
the canonical form C.103).
3.3.17. Use the method given in the text that leads up to equation C.48) to
show that if u is a continuous solution of C.93) whose first derivatives have
a jump across the curve £ = (р{х, y) = 0, the jump in the derivative across the
curve satisfies the equation
Since the jump is nonzero by assumption, conclude that <p(x, y) = 0 satisfies
C.100) and must be a characteristic curve.
3.3.18. Let у = h(x) be a characteristic curve for the totally hyperbolic
system C.93). Then, the matrix В - h'A is singular. Multiply on the left in
C.95) by a left nullvector of the singular matrix and determine a com-
compatibility condition on the initial value u = f(jc) for the problem to be a
characteristic initial value problem.
3.3.19. Consider the totally hyperbolic system uy + Aux = 0, with
■Г1
L2
Show that у = — x and у = A /Ъ)х are characteristic curves for this system. If
we set u = f(x) on each of these curves, determine conditions on/(jc) so that
Jn each case we have a characteristic initial value problem. For appropriately
chosen f(x) determine the (nonunique) solution in each case.
3-3.20. Reduce Euler's equations of one-dimensional isentropic fluid flow
(see Example 8.9),
Pt + UPX + PUX = 0 ,
", + (c2/p)Px + uux = 0 ,
158 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
to characteristic normal form. Also, determine the Riemann invariants for
this system.
3.3.21. Reduce the shallow water equations (see Example 10.15)
ht + uhx + hux = 0 ,
where h is the height of the water above a horizontal bottom, w is its velocity
and g is the gravitational constant, to characteristic normal form. Also,
determine the Riemann invariants for this system.
3.3.22. Reduce Euler's equations of one-dimensional adiabatic flow (see
Exercise 8.5.19)
Pt + uPx + pux = 0,
pt + upx + c2pux = 0 ,
to characteristic normal form.
3.3.23. Show that the equation
= 0
is of hyperbolic type and determine its characteristic curves. (It is of
hyperbolic type if the characteristic curves are real and distinct.)
Section 3.4
3.4.1. Obtain the solution of the Cauchy problem for Laplace's equation
given in Example 3.6 by expanding u(x, y) as a power series
, ч v 1 **«(*, 0) *
Determine the derivatives <?*u(jc?0) by differentiating Laplace's equation
uyy = -uxx along the initial line у = 0 and using the Cauchy data given there.
3.4.2. Construct the solution of the Cauchy problem for the heat equation
ut = c2uxx with the initial condition w(jc, 0) =/(*), as a power series
df
assuming that f(x) is sufficiently differentiable. [Determine the derivatives
dntu(xy 0) from the heat equation and the initial condition.]
EXERCISES FOR CHAPTER 3 159
3,4.3. Use the method of Exercise 3.4.2 to show that the solution of
is given as u(xy t) = cos(jc) exp(-c2f).
3.4.4. Obtain a solution of the telegrapher's equation
where у and A are positive constants, in the form of a power series in t9 if
the Cauchy data are given as u(x, 0) = cos x, m,(jc, 0) = 0.
3.4.5. Show that the solution of the initial value problem for the wave
equation B.1) is continuously dependent on the initial data B.20), by
showing that over the finite time interval 0 < t ^ T9 d'Alembert's solution
B.29) is uniformly small in magnitude if the same is true for the initial
values f(x) and g(x).
3.4.6. Consider the wave equation utt = uxx in the square 0 < jc < тг, 0<t
< тг, with и = 0 on the boundary. Show that u(x, t) — sin x sin t is a solution
of the equation that vanishes on the boundary. Explain why the boundary
value problem for the wave equation in the given square with и = / on the
boundary is not well posed.
3.4.7. Use the solution u = (l/ri)exp(pn2t)sin(nx) of the backward heat
equation C.141) with p>0, to show that the initial and boundary value
problem for that equation in the interval 0<x< it with t>0, where и is
specified on x = 0, x = тг, and t = 0, is not well posed.
3.4.8. Show that the exterior boundary value problem for Laplace's
equation
where p2 = x2 + y2 4- z2 and u(x, y, z) is specified on the sphere p = a as
MlP=a = A (A = constant) has (at least) two solutions,
Aa
u- Ay u- — .
P
(The solution can be shown to be unique if we require that и—»0 as p-*«.)
3-4.9. Apply the divergence theorem to the expression V-V« to show that
the boundary value problem for V2w = 0 in the region A with duldn = / on
the boundary ЭА of A has no solution unless
JL
160 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
Also, observe that if a solution does exist for this problem, it is not uni<~
since the solution и = a (a = constant) can be added to the given result.
Section 3.5
3.5.1. Show that the stability index ft for the hyperbolic equation
is given as ft = c. (The constant с is assumed to be positive.) Thus although
the Cauchy problem for this equation is well posed, the equation is unstable
3.5.2. Consider the hyperbolic equation
By examining the solutions Л = Л(А:) of C.134) as specialized to the abov
equation show that the equation is unstable if c2<a2.
Hint: Look at small values of k.
3.5.3. Demonstrate that the following equations are all of dispersive type b
considering normal mode solutions of the form C.150). Also, determine th<
relevant dispersion relations.
(a) utt-y2uxx-a2uxxtt = 0.
(b) нй + Дш = 0.
(c) ut + aux + puxxx = 0.
3.5.4. Show that the diffusion equation A.23) is of dissipative type.
3.5.5. Consider the hyperbolic equation
Obtain the relationship Л = \(k) for the normal mode solutions C.133) oi
the above equation. Using the known criterion that the polynomial P(X) =
A3 + ax A2 + a2k + a3 has roots with negative real parts if a3 > 0, at > 0, anc1
axfl2>fl3, conclude that the given hyperbolic equation is neutrally stabk
and is of dissipative type if у > с2.
3.5.6. Given the system of equations
Anx + But + Cu = 0 ,
where A, B, and С are constant n by n matrices and u is an n-component
vector, normal mode solutions are given as
u(x, i) = a(Jfc) exp[ikx + X(k)t],
where a(Ar) is a constant vector. Show that u is a solution of the given
equation if we have
EXERCISES FOR CHAPTER 3 > ^ *■* 161
Explain how concepts of well-posedness and stability may be introduced for
the given system on the basis of the above result.
3.5.7. Apply the method of Exercise 3.5.6 to the system A.82), A.83) and
to the systems given in Example 3.4. [replacing у by t in the Cauchy-
Riemann system C.107)]. Discuss stability and well-posedness questions for
these systems and compare the results with those obtained for the scalar
equivalents of these systems.
3.5.8. With к = [kl9 k2, ,k3] and x = [x, v, z]9 obtain the appropriate equa-
equation A = A(k) for the normal mode solutions
и(х) = a(k) exp[ik • x + A(k)f]
of the following equations:
(a) «,,-y2V2M + AM,=0,
(b) utt-y2V2u + c2u = 0,
(c) M,r + V2M = 0,
(d) ut-y2V2u = 0,
where A>0, all the coefficients are constants, and V2 is the Laplacian
operator in three dimensions. Discuss well-posedness and stability for these
equations in the manner of Section 3.5.
Section 3.6
3.6.1. Verify the result C.162).
3.6.2. Determine the adjoint operators for the following operators:
(a) Lu = exuxx + x2uxy + yuy - 10m;
(b) Lu = a(x, y)ux + b(x9 y)uy + ф, у)и;
(c) Lu = ut-c2[uxx + uyy].
3.6.3. Show that the backward Kolmogorov equation A.55) is the adjoint of
the Fokker-Planck equation A.53).
3.6.4. Show that the differential operator in A.137) is the adjoint of the
operator in A.136).
3.6.5. Show that the equation derived in Exercise 1.2.16 is the adjoint of
equation A.109).
3.6.6. Show that the system derived in Exercise 1.2.15 is the adjoint of the
system A.106), A.107).
3.6.7. Determine the adjoints of the systems in Example 3.4.
3-6.8. Verify that C.180) is valid for the equation B.95).
3»6.9. Obtain an adjoint operator L* for the second order linear ordinary
differential operator L given as
Ly = ay" + by' + cy .
162 CLASSIFICATION OF EQUATIONS AND CHARACTERISTICS
3.6.10. Obtain an adjoint operator L* for the third order operator L given
as
Lu = ut-ux + y2uxxx ,
such that wLu — uL*w is a divergence expression.
3.6.11. Show that the operator
Lu = utt + uxxxx
is self-adjoint by expressing wLu - uLw in divergence form.
3.6.12. Demonstrate that
Lu = un + V2V2«
is self-adjoint by expressing wLu - uLw in divergence form.
4
INITIAL AND BOUNDARY
VALUE PROBLEMS IN
BOUNDED REGIONS
4.1. INTRODUCTION
This chapter deals with initial and boundary value problems for elliptic,
parabolic, and hyperbolic equations given over bounded spatial regions. The
basic method for solving these problems is the technique of separation of
variables. This method requires that eigenvalue problems for differential
equations be studied and expresses solutions as series of eigenfunctions.
Only scalar problems in one, two, or three space dimensions will be treated
in this chapter. The one-dimensional eigenvalue problem, known as the
Sturm-Liouville problem, will be studied in detail. Higher dimensional
eigenvalue problems will be considered in Chapter 8.
The separation of variables method is applicable to the study of linear
homogeneous equations with homogeneous boundary conditions. Tech-
Techniques for solving inhomogeneous problems, such as Duhamers Principle
and eigenfunction expansions, will also be presented in this chapter. Fur-
Further, it will be demonstrated by means of a detailed example in Section 4.7
how eigenfunction expansions may be used in the solution of nonlinear
problems.
To achieve a unity of presentation, we consider only second order
equations of each of the three basic types given in a special form. This form,
however, is representative of a large class of partial differential equations of
Physical interest. All equations with constant coefficients derived in Chapter
1 are either of this form or can be reduced to this form. To indicate how
these equations which may have variable coefficients can arise in applica-
applications, we now present a brief derivation of a parabolic equation of the form
to be considered. This derivation in contrast to those given in Chapter 1, is
^ore representative of the approaches generally used in deriving partial
differential equations as models for physical processes.
163
164 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
Throughout this chapter, G will represent a bounded interval in one
space dimension or a bounded region in two or three dimensions and dG
will represent its boundary. (In one dimension dG is just the end points of
the interval G.)
In the following derivation we restrict our discussion to three space
dimensions and let G be an arbitrary closed region within G, with dG as the
boundary surface. Let ds be an element of surface area of dG and dv be a
volume element. The function u(x, i) where x is a point in space and t is
time is assumed to represent a scalar physical quantity such as temperature
or the concentration of a substance, for example.
With n given as the exterior unit normal vector on the boundary dG,
p(x)Vu - n ds represents the flux or rate of flow of the quantity ы(х, /)
through ds at the time t. The positive function p(x) is given and assumed to
be time independent. The operator V is the gradient operator in three-
dimensional space. The time rate of change of u(x, t) in an element dv at the
time t is given by p(x)(du/dt) dv, where the positive quantity p(x) is given
and time independent. Additional effects occurring in the element dv at the
time t are given as H(x, t) dv. The function H(x, t) is assumed to be of the
form H(x, i) = — q(x)u(x1, i) + F(x, t), where for notational convenience in
our discussion we set F(x, t) = p(x)F(x, t). Again, q(x) is a given non-
negative time independent function and qu represents (internal) effects due
to changes proportional to u(x, t), whereas F(x, t) represents external
influences on the medium under consideration (See Section 8.2 for the
consequences of permitting q to be negative.)
The basic physical balance or conservation law for the arbitrary region G
requires that
<4л>
This states that the time rate of change of u(x, i) in the region G equals the
flux of m(jc, i) through the boundary of G and the changes due to internal
effects proportional to u(x, t), as well as external effects acting throughout G
(i.e., the effects must balance each other out).
Applying the divergence theorem to the surface integral in D.1) yields
D.2) f pVu n ds = I I V- (pVu) dv .
JdG J JG
Inserting D.2) into D.1) and combining terms gives
D.3) f jo[p%-
INTRODUCTION 165
Assuming a continuous integrand, the arbitrariness of the region G implies
the vanishing of the integrand in D.3) (see Exercise 8.1.9). Consequently,
we obtain the inhomogeneous parabolic equation
D.4) p^-V-O>Vn) + *n = pFf
which is valid at an arbitrary point in the region G. A similar argument in
two space dimensions leads to the same equation with V interpreted as a
two-dimensional gradient operator. In one space dimension, the resulting
equation has the form
ди д I du
In the context of heat conduction, u(x, t) is the temperature and p(x) is
the thermal conductivity of the medium. The relationship between the heat
flux J and the temperature u, when expressed in the form J= —pVu, is
called Fourier's law. (The minus sign indicates that heat flows from warmer
to cooler regions.) The term p(x) is the product of the specific heat с and the
density p(x) of the medium. In the case of one-dimensional heat conduction
in a rod, if there is lateral heat transfer between the rod and the surrounding
medium, the function q(x) results from Newton's law of cooling which states
that the heat flow through the (lateral) boundary is proportional to the
difference between the internal temperature и of the rod and the external
temperature.
If D.4) or D.5) describes the diffusion of a substance, u(x, t) equals the
concentration of that substance. The relationship between the flow of the
substance and its concentration и has the same form as Fourier's law and is
known as Fick's law in diffusion theory. The quantity p(x) is called the
diffusion coefficient or the diffusivity. In the simplest models of diffusion
p(x) ~ 1 and q — 0. (We remark that in the case of heat conduction if p and p
are constants, the ratio pip is called the thermal diffusivity.) The present
derivations of the heat and diffusion equations are based on macroscopic
considerations where observable phenomena are taken into account. In
Chapter 1 we derived the diffusion equation by considering the motion of a
particle or a group of particles. From these microscopic considerations we
derived equations that involve observable quantities such as particle den-
densities, drifts and variances.
In view of our continual use of the above and related equations, we
define the differential operator L in two or three dimensions as
in one space dimension as
166 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
It is assumed that p and p are positive functions and that q is a non-negative
function. Then the parabolic equations D.4) and D.5) take the form
du
D.8) p — + Lu = pF.
If the problem leading to D.8) is stationary or time independent, that is,
F and all conditions given on the problem depend on x only, we may look
for a solution in the form и = и(х) and D.8) reduces to
D.9) Lu = pF.
This is an equation of elliptic type.
Further, it can be shown that a large class of hyperbolic equations of
interest in applications can be expressed as
D.10) P?-j + Lu = pF.
at
For example, in one space dimension the equation governing the longi-
longitudinal vibration of a rod has the form D.10). We assume that the rod has a
constant cross-sectional area A and let u(xy t) be the displacement of the
section of the rod at the point x as a function of time. By Hooke's law the
tension force acting on the section at the point x is given as p(x)Aux(x, t),
where p(x) is Young's modulus. A segment of the rod extending from x to
x + Ax, of length Длс, has the momentum pAAxun where p(x) is the
density. Then, Newton's law of motion applied to this segment of the rod
yields p(x)A Axutt = Ap(x + Ax)ux(x + Ax, t) - Ap(x)ux(x, t). On dividing
by A Ax and letting Ax tend to zero we obtain the one-dimensional form of
D.10) with q = F = 0.
As has already been remarked in Example 2.4 and will be demonstrated
in Example 4.9, the equation of the vibrating string has the form D.10) (in
one dimension) with p—T, the constant tension of the string. Similarly, the
transverse vibration of a tightly stretched thin membrane with density
p(x, y) is governed by the two-dimensional form of D.10) where p equals T,
the constant tension of the membrane. In both cases и is the vertical
displacement of a point on the string or on the membrane. If a restoring
force proportional to the displacement acts at each point of the string or
membrane, in addition to other external forces, the full equation D.10)
applies.
The telegrapher's equation of Chapter 1 is not of the form D.10) since it
contains a duldt term. However, a simple change of the dependent variable
INTRODUCTION " 167
reduces it to the above form. In fact, any equation of the form D.10) that
contains an additional term pduldt with p independent of t, can be trans-
transformed into the form D.10) by a change of variable if pip = constant.
In the following sections we shall consider initial and /or boundary value
problems for the equations D.8)-D.10). Appropriate initial values for the
parabolic and hyperbolic equations D.8) and D.10) will be introduced as
required. However, the boundary values on dG for all three equations
D.8)-D-Ю) are assumed to be of a standard form common in applications.
In the case of two or three space dimensions we consider boundary
conditions of the form
D.11)
did
dn
dG
where a(x), P(x), and B(x, t) are given functions evaluated on the boundary
region dG. [Note that in the parabolic and hyperbolic cases, the region G
and its boundary dG are extended parallel to themselves in (jc, f)-space. In
the elliptic case В = B(x) is independent of /.] The expression duldn
denotes the exterior normal derivative on dG. The boundary condition
D.11) relates the values of и on dG and the flux of и through dG. We
require that a(x) >0, p(x) >0, and a + p >0 on dG. If A) a #0, p = 0,
B) a = 0, p Ф 0, C) a Ф 0, 0 ^ 0, D.11) are known as boundary conditions
of the first, second, and third kind, respectively. Boundary conditions of the
first and second kind are often referred to as Dirichlet and Neumann
conditions. In addition, we consider boundary conditions of the mixed kind.
This means that a mixture of boundary conditions of the first, second, and
third kind is specified for и on the boundary. To make this precise, we
assume that dG is subdivided into three disjoint subsets Sl9 52, and S3. On
Sl9 S29 and 53, и satisfies a boundary condition of the first, second and third
kind, respectively. Each of these sets may be an empty set. If at least two of
these sets are not empty, the boundary condition is of the mixed kind. If
only one of the three is not empty, the boundary condition is of the first,
second or third kind according as the set is S19 S2 or S3. To simplify the
presentation of formulas and results in the remainder of the text, we use this
notation throughout when dealing with integrals over the boundary and
boundary conditions.
For the one-dimensional case, G is assumed to represent the interval
®<x<l and dG comprises the points x = 0 and x = l. The boundary
conditions D.11) in the hyperbolic and parabolic cases take the form
D.12)
where the conditions on (a, p) given above carry over in a natural way to
the pairs (a1? px) and (a2, K2) both of which are constant. The minus sign in
168 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
the condition at r = 0 in D.12) follows since the exterior normal derivative
at that point is Ыдп = -duldx.
When F and i (or g1 and g2) in D.8)-D.12) are nonzero, the equations
and boundary conditions are of inhomogeneous type. The separation of
variables method introduced in the following section applies only to the
homogeneous esses when F and В are both zero. However, when this
method is appliable to boundary value problems for the elliptic equation
D.9), we must hive F = 0 while В = 0 on only part of the boundary, as will
be seen.
4.2. SEPARATION OF VARIABLES
In this section we present the method of separation of variables and apply it
to the solution d initial and/or boundary value problems for homogeneous
versions of the eolations and boundary conditions introduced in the preced-
preceding section.
In the hyperblic case we consider the homogeneous equation for и ^
u(x, i)y
D.13) p(x) »^ + Lu = 0 , xE G, t>0 ,
where the bounded region G, the coefficient p(x), and the operator L ar©
defined as in Section 4.1. The homogeneous boundary conditions are, in two
or three dimensions,
D.14)
' v ' дп dG
and in one dimension
= 0, t>0,
On the boundaiy dG the coefficient in D.14), D.15) must satisfy the
conditions given in Section 4.1. The initial conditions for D.13) are
D.16) M(t,0)=/W, u,(x,O) = g(x),
In the parabolic case, the equation for u(x, t) is given as
D.17) p(x)^ + Lu = 0, xeG,t>0,
at
sEPARATION OF VARIABLES *«>
with the boundary conditions D.14) or D.15) and the initial condition
D.18) и(*,0) =/(*), ^G.
In the elliptic case, we introduce a function и = u(x, y), where x is a point
. tne region G (which is now one- or two-dimensional) and у is a scalar
variable given over the interval 0 < у• < /. The equation for u(x, y) has the
form, with p, p and # given as functions of x only,
D.19)
The boundary conditions in x are of the form D.14) or D.15). These are
combined with the inhomogeneous boundary conditions at у = 0 and у = I,
D.20)
u(x,Q) = f{x) , и(х, I) = g{x),
G .
Example 4.1. Equations with Constant Coefhcients. We assume that x
is a one-dimensional variable and that p(x) = p(x) ~ 1 and #(jc) = 0 in D.13),
D.17), and D.19), and formulate the appropriate problems for each of these
equations. The regions wherein the solutions are to be determined and the
initial and/or boundary conditions for these problems are presented in
Figures 4.1-4.3.
О 1
II
"**
сГ
а1*
О
а
сГ
@,0)
у turn
uix,
щ
Ш
Щ
Ш
Ж
щ
0)
Ш
^ ■
В
Я
г{лг, 0) =
Ш-
Ш \
о
II
-^
^t
f
»
g(x) (/, 0) х
Figure 4.1. Hyperbolic case.
170
INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
@,0)
u(x, 0) =* fix) {I, 0)
Figure 4.2. Parabolic case.
(ID
@,0)
Figure 4.3. Elliptic case.
(/, 0)
It should be noted that for the hyperbolic and parabolic cases we are
considering initial and boundary value problems, whereas in the elliptic case
we have *k strict boundary value problem. The equation D.19) for the
elliptic problem is written in the indicated form to permit a unified presenta-
presentation of the separation of variables technique for all three cases.
The separation of variables method asks for a solution of D.13) and
D.17) in the form
SEPARATION OF VARIABLES 171
D.21) u(x, t) = M(x)N(t) ,
and of D.19) in the form
D.22) u(x,y) = M(x)N(y),
^ith the function M(x) required to satisfy the boundary conditions D.14) or
D.15). Substituting D.21) and D.22) into the appropriate equations and
dividing through by pMN yields
D.23)
D.24)
and
D.25)
N"(t)
N(t)
N'(t)
N(t)
N"(y)
LM(x)
p(x)M(x) '
LM(x)
p(x)M(x) '
LM(x)
hyperbolic case .
parabolic case .
elliptic case.
Since the left and right sides of these equations depend on different
variables, they cannot be nonconstant functions of their respective variables.
Thus each side of the equations must equal a constant. We denote this
(separation) constant by -A. As a result, we obtain the following equations
for M and N,
D.26) LM(x) - Xp(x)M(x),
and
(N"(t) + XN(t) = 0 , hyperbolic case
N'(t) + \N(t) = 0 , parabolic case
N"( y) - kN(y) = 0 , elliptic case
In addition to being a solution of D.26), M(x) is required to satisfy the
boundary conditions D.14) or D.15). [Consequently, u(x, t) = M(x)N(t) or
u(x, y) = M(x)N(y) will also satisfy these boundary conditions.] Since both
the equation and the boundary conditions for M(x) are homogeneous,
Щх) = О is a solution of the problem. This solution must be rejected since it
*s of no value in solving the given problem for u. Thus we require nonzero
solutions of the boundary value problem for M(x), and such solutions exist
°nly for certain values of the parameter A in D.26).
The problem of determining the nonzero M(x) is known as an eigenvalue
Problem. The values of A for which nonzero M(x) exist are known as
eigenvalues. The corresponding solutions of the eigenvalue problem for
172 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
M(x) are called eigenfunctions. In the one-dimensional case, the eigenvalue
problem is known as the Sturm-Liouville problem in the mathematical
literature and this problem is discussed in the following section.
The differential operator L [or more precisely, (l/p)L], which occurs in
D.26), when applied to functions that satisfy the boundary conditions D.14)
or D.15) has the property of being a self-adjoint as well as a positive
operator. These properties are defined and discussed in the following
example, after which important consequences of these properties for the
eigenvalue problem associated with the operator L are examined. Although
a large part of our discussion is presented in notation appropriate for the
two- and three-dimensional cases, the general results are valid for the
one-dimensional case as well.
Example 4.2. Self-Adjoint and Positive Operators. We consider the
(smooth) functions и and w defined over the two- or three-dimensional
region G and satisfying the boundary condition D.14) on BG. It will be
shown that for these functions
D.28) /1 [WLu - uLW] dv = fta P[u % - w %\ ds = 0,
where dldn is the exterior normal derivative. While the operator L ha|
already been shown to be formally self-adjoint in Example 3.9, we demon-
demonstrate this fact here for the sake of completeness and go on to show that the
boundary value problem is self-adjoint.
To prove D.28) we note that
D.29) ivV- (рЩ = V- (pwVu) - pVw -Vw .
Interchanging w and и in D.29) and subtracting one expression from the
other gives
D.30) wV- (pVu) - wV- (pVw) = V- (pwVu) -V-(puVw) .
In view of D.6) (i.e., the expression for L), we obtain
D.31)
I I [wLu - uLw] dv = -I I V- [pwVu - puVw] dv
dw З
on applying the divergence theorem. Both и and w satisfy the boundary
condition D.14) so that
SEPARATION OF VARIABLES 173
D.32)
H dn
dw
«" + /* —
dG
о
dG
= 0.
We may think of D32) as a simultaneous system of homogeneous equations
for a and f$ at each point of dG. Since a + /3>0on<?Gby assumption, the
above system must have a nonzero solution. This can occur only if the
determinant of the coefficients of the system (i.e., udwldn- wduldri)
vanishes on dG. Consequently, the surface integral in D.31) vanishes as was
to be shown. (We remark that the result is valid as long as a and j8 are not
both zero.)
Given two functions / and g defined and integrable over the region G, the
inner product of these two functions with weight p(x) > 0 is defined to be
D.33)
These functions are assumed to be real valued and need not satisfy the
boundary conditions D.14). Since (/, g) = (g, /), the inner product is
symmetric. (If the functions are complex valued, a different inner product,
introduced in the next section, is defined.) If (/, g)- 0, the functions/and g
are said to be orthogonal. In terms of the inner product D.33) we define a
norm of the function f(x) as
<4-34)
The norm ||/|| which is non-negative for real valued/is a measure of the
magnitude of/ If/is continuous in G, then ||/|| =0 if and only if/=0.
Most often one defines the inner product D.33) with p(x) = 1, but for our
discussion we require the more general definition given above.
In terms of the inner product D.33), the equation D.28) can be ex-
Pressed in the form (и>, (l/p)Lw) = ((l/p)Lw, и). An operator L with the
Property that (w, Lu) = (Lw, u) is said to be self-adjoint. If р{х)Ф 1, the
operator (l/p)L satisfies the self-adjointness condition in terms of the inner
Product D.33), whereas if p(x) = 1, the operator L is self-adjoint. We
remark that the self-adjointness of (l/p)L or L is determined not only by
Ле definition of L but also by the boundary conditions D.14) that и and w
^ust satisfy. (The discussion of adjoint differential operators given in
Section 3.6 involved only the concept of formal adjointness.)
Next, we consider a function и that satisfies the boundary condition
D-l4). We have
174 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIONS
D.35) J £ [uLu] dv=j ja u[-V- (pVM) + qu] dv
= -/ /G [V- (риЩ - p(VMJ] dv + jjG qu2 dv
where D.29) and the divergence theorem have been used. On the subset $
of <?G, we have from D.14)
whereas on St and 52, и and duldn vanish, respectively. Consequently
D.35) can be expressed as
D.37) \ jo [uLu] dv = jjQ [p(Vuf + qu2] dv + ^ p(j)u2 ds .
The assumptions on py q, a, and K given previously imply that the right sidi
of D.37) is non-negative if и is real valued.
In terms of the inner product D.33), D.37) implies that if и is real
valued,
D.38) J Jg [uLu] ^ = (-
A self-adjoint operator L with the property that (Lw, u) >0 is said to be
positive operator. Thus if p^l, we see that (l/p)L is a positive operator
whereas if p = 1, the operator L is itself positive. Again the positivity о
(l/p)L or L is based not only on its definition but also on the boundar
conditions D.14) that и must satisfy. If we relax the requirement that q h
D.4) and a and fi in D.11) must be non-negative, the operator (l/p)L H
still self-adjoint as long as a and /3 are not both zero simultaneously.
However, for (l/p)L to be a positive operator, these conditions cannot b&
relaxed.
We now apply the results of the foregoing example to the eigenvalue
problem D.26). Let Mk{x) and МДх) be eigenfunctions corresponding to
the distinct eigenvalues кк and A;, respectively. Since Mk(x) and Mj(x) both
satisfy the boundary condition D.14), the result D.28) implies
D.39) j jG (MkLMj - M,LMk) dv=0.
EPARATION OF VARIABLES 175
gut LMj = A,-pAfy and LMk = ХкрМк9 so that D.39) yields
D.40) \\c (bjpMkMj - ккрМ,Мк) dv = (Ay - Хк)(Мк9 Mf) = 0 .
gy assumption, A;# Ал, so that D.40) implies
D.41) (M,,M,) = 0.
Consequently, eigenfunctions corresponding to different eigenvalues must
be orthogonal in terms of the inner product D.33).
Using D.41) and the fact that the coefficients in the differential operator
I are assumed to be real valued, it is easy to show that the eigenvalues for
our problem must be real and that the corresponding eigenfunctions may be
chosen to be real valued. (This is proven for the one-dimensional eigenvalue
problem in Section 4.3 and the method of proof carries over immediately to
the higher dimensional problem.)
Further, let Mk(x) be a real valued eigenfunction that corresponds to the
real eigenvalue Afc. Since LMk = XkpMk9 we obtain from the positivity
property D.38)
D.42)
Because Mk(x) is real valued and nonzero by assumption, we conclude that
the eigenvalue A^ >: 0.
Thus the self-adjointness property of the operator (l/p)L implies that the
eigenvalues are real and that eigenfunctions corresponding to different
eigenvalues are orthogonal. The positivity property implies that the eigen-
eigenvalues are non-negative. It can be shown that there are a countably infinite
number of eigenvalues \к (k = 1, 2,. . .) whose only limit point is at infinity
(ieM Ал—»oo as k—><*>). We denote the corresponding set of (real valued)
eigenfunctions by Mk(x). Further properties of eigenvalues and eigenfunc-
eigenfunctions are presented later, and the eigenvalue problem is reexamined from
the variational point of view in Section 8.1.
For each eigenvalue Xk we obtain an equation for Nk in D.27). Assuming
for simplicity that A^ > 0 for all k, we have for Nk
D.43)
Nk(t) = ak c°s(VV) + bk sin(VV) , hyperbolic case
Nk{t) ~ ak exp(—Akt), parabolic case
Nk(y) = ak exp(VTky) + bk exp(-yfYky) , elliptic case
ал and bk are arbitrary constants that must be determined.
Forming the product MkNk, we obtain
176 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.44) uk
which for each value of к satisfies the appropriate equation D.13), D.17)
D.19) together with the boundary condition D.14) or D.15). ' ;>Cr
To satisfy the additional conditions on u(x, t) or u(x, y) given above w
consider the formal superposition of the solutions uk and construct the ser" ^
oo oo
D.45) и = 2 uk = 2 MkNk .
The constants ak and bk in the terms Nk are chosen to satisfy the additional
conditions placed on и that have not as yet been accounted for by our choice
of the Mk{x).
In the hyperbolic case the initial conditions D.16) must be satisfied and
we obtain formally
и(х,0)= 2 Mk(x)Nk@)=2 akMk(x)=f(x)
D.46)
«/*, 0) = 2 Mk(x)N'k@) = 2 VX~kbkMk(x) -;
The expressions D.46) represent what are referred to as eigenfunction
expansions off(x) and g(x). The validity of this representation is character-
characterized by the concept of completeness of the set of eigenfunctions Mk(x). If
the set is complete, under certain conditions on the functions f(x) and g(x)
the expansions D.46) with appropriately chosen ak and bk, will converge to
these functions. The more stringent the conditions, the better the converg-
convergence. Since the Mk(x) are eigenfunctions for a self-adjoint operator as we
have shown, even under very mild conditions on f(x) and g(x) the series of
eigenfunctions will converge in some sense to these functions. This question
is discussed more fully for the one-dimensional case in Section 4.3 and for
the higher dimensional problem in Chapter 8.
For the hyperbolic case considered above and the further discussion ffl
this section, the Mk(x) are assumed to be an orthogonal set, that is,
(Mki My) = 0 for к Ф]. As shown above, this assumption is valid if each
Mk(x) corresponds to a different eigenvalue. If the eigenvalue is multiple
and a finite number of linearly independent eigenfunctions correspond to a
single eigenvalue, the Gram-Schmidt orthogonalization process, which wiU
be discussed later in the text (see Exercise 8.2.1), can be used to orthogonal-
ize the finite set of eigenfunctions. (For the eigenvalue problem under
consideration, there cannot exist more than a finite number of linearly
independent eigenfunctions for each eigenvalue.)
To determine the ak in D.46) we multiply the series by p(x)Mj(x) afl<*
integrate over G. This gives
PARATI OF VARIABLES 177
D.47) / « v »
(/(*), M,(x)) = ( 2; akMk(x), Mfr)) = £ ak(Mk, Mf) = a,{M/t M,)
using D-41) and assuming summation and integration can be inter-
interchanged in the series. Thus
_ (/(*),
D.48) fl* (Mk(x),Mk(x))'
and, by the same argument,
(g(x),Mk(x))
D.49) ** ~
The formal solution of the initial and boundary value problem for the
hyperbolic equation D.13) is then given as
00
D.50) u(x, t) = S [ak cos(VV) + bk $т(^ТкО]Мк(х).
For the parabolic case, the initial condition D.18) must be satisfied. This
gives
00 00
D.51) u(x, 0) = X Mk(x)Nk@) = 2 akMk(x) =/(*) .
kir±k\
Thus ak is specified as in D.48) and the formal solution of the initial and
boundary value problem for the parabolic equation D.17) is
00
D52> Ф, t) = 2 ak cxp[-Xkt]Mk(x).
k=\
In the elliptic case, the boundary conditions D.20) must be satisfied. This
elds the eigenfunction expansions
D.53)
Ar = l
Ф, Ь = X [ak exp(VTkt) + bk схр(-\ПГк!)]Мк(х) = g(x) .
Pplying the technique used in the hyperbolic case yields
178 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED R:
D.54) Г *""<"*(*). "*<*))
Unique solutions for the ak and bk can be determined from the foregoin
system. Then the formal solution of the boundary value problem for th
elliptic equation D.19) is
D.55) и(х, у) = £ [ak txp(VTky) + bk exp(- Vhy)]Mk(x) .
The above solutions are often termed classical solutions of the given
equations if the formal operations carried out are valid and the series
expansions of и can be differentiated term by term as often as required by
the equations and the initial and boundary data. However, even if term by
term differentiability is not valid, the sum и may still be characterized as a
generalized solution if certain conditions specified later are met.
We have shown that the separation of variables method reduces each of
the initial and boundary value problems given in this section to the study of
an eigenvalue problem. Unless the eigenvalues A^. and the eigenfunctions
Mk(x) can be determined, the formal series solutions obtained remain
unspecified. For many of the problems considered in this text the eigen-
eigenvalues and eigenfunctions can be determined exactly, as we demonstrate in
this and later chapters. When exact results are not available, approximation
methods such as the Rayleigh-Ritz method discussed in Section 8.2 yield
effective approximations to the leading and most significant eigenvalues and
eigenfunctions.
4.3. THE STURM-LIOUVILLE PROBLEM AND FOURIER SERIES
The one-dimensional version of the eigenvalue problem introduced in the
preceding section
D.56) L[v(x)] = -fx [p(x) ^] + q(x)v(x) = Xp(x)v(x) ,
where 0<jc</ and v(x) satisfies the boundary conditions D.15), that is,
is known as the Sturm-Liouville problem. We require that p(x) > 0, p
0, and ?(x)>0, and that p{x), p(x)> q{x), and p\x) be continuous in tn
€TUrM-LIOUVILLE PROBLEM AND FOURIER SERIES 179
ed interval 0 < x < /. Also, we must have a, ^ О, Д ^ 0, and a, + Д > 0
c ,- = 1, 2. Then we have a regular Sturm-Liouville problem. If one or
fOf e of the conditions on the coefficients in D.56) are relaxed, say, if p(x)
0Ot(x) or both are permitted to vanish at either or both of the endpoints
°Г-О and x - /, we have singular Sturm-Liouville problem. Both cases are
X'iinterest in applications and examples of each type are considered. We
*агк that in discussions of Sturm-Liouville problems in the literature it is
Ге erally not required that q(x) and the coefficients in the boundary
Editions D.57) be non-negative.
Before discussing and deriving some of the important properties of the
eenvalues and eigenfunctions of the Sturm-Liouville problem, we intro-
introduce several relevant definitions and concepts. Some of these have already
been given in the preceding section but are repeated here in their one-
dimensional form. Unless otherwise specified, the functions considered in
the following discussion will be real valued.
The inner product of two functions <p(x) and ф(х) (bounded and inte-
integrable over the interval 0 < x < /) is defined as
D.58) (<P, Ф) = Jo р(х)<р(х)ф(х) dx ,
with the weight function p(x) > 0 in 0 < x < L [Most often one uses a unit
weight function p(x) = 1; however, for our purposes the above definition is
more appropriate.] The inner product is clearly symmetric; that is, (<p, ф) =
{ф, <р). The norm of a function ^>(jc) defined in terms of the inner product
D,58) (or induced by it) is
D.59)
The norm, which is clearly non-negative, is a measure of the magnitude of
the function <p(x) over the interval 0 < x < /. If <p(x) is continuous in that
interval, then || <p \\ = 0 if and only if <p(x) - 0. Any function <p(x) with a finite
norm (i.e., ||<p||<oo) is said to be square integrable over the interval
^<* < /. If (p(x) is such that ||<p|| = 1, then <p(x) is said to be normalized to
^ty. Any square integrable function can be normalized by defining a new
[All**00 ^*) = ^WIMI for which it is immediately seen that \\<p\\ = 1.
I square integrable functions will be required to be integrable as well (see
Zeroises 4.3.1 and 4.3.2.)].
functions <p and ф for which
to be orthogonal over the interval 0<л:</. A set of functions
fc = l, 2,... , for which
180 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGION
D.61) Op*,<P;-) = 0, кФ)
is said to be an orthogonal set. If, in addition, \\<pk\\ = 1 for all <pk, the set'
said to be orthonormaL Any orthogonal set can be orthonormalized j/
normalizing each of the <pk in the manner shown above. У
If the functions <p, ф or <pk can assume complex values, we replace th
inner product D.58) by the (weighted) Hermitian inner product
D.62) (*>, ф) = JQ р(х)ф)ф(х) dx ,
where the bar denotes complex conjugation. Since we now have (<p, ф\,
(ф, <p), the inner product has Hermitian symmetry. The induced norm I
|2
D.63) IMHVOp, <e) = VJ0 p(x)\<p(x)\2 dx,
so that
We now return to the consideration of real valued functions. Given the
orthonormal set of square integrable functions {<pk(x)}, к = 1, 2,.. ., and
the square integrable function <p(x) over the interval 0<x<l, the set of
numbers (<p, <pk) are called the Fourier coefficients of <p(x). The formal series
00
D.64) ?(x)=2(?,ft)ftW
is called the Fourier series of <p(x). Even though we have equated <p(x) to its
Fourier series in D.64), we have yet to specify in what sense the Fourier
series converges to the function <p(x), if it converges at all. Further, although
we use the term "Fourier series" to denote any expansion of a function in
terms of a set of orthonormal (or more generally orthogonal) functions, in
our later discussions, as well as in the literature, this term is often used to
denote an expansion in a series of trigonometric functions.
To discuss the convergence properties of the series D.64) we note that
for the finite sum E*=1 (<p, <рк)<рк(х), we have
D.65)
IN i|2 / N N \
<p(x) - E Op, <pfc)<p*(*)| = \9 - 2 Op, <pk)<pk, <p ~ ]£ (<p, <Pk)<Pk)
(Р,р) 2(р,рк)(рРк) Ы\(Р
Jt=i k=i k=i
where we have used elementary properties of the inner product,
stUrM-LIOUVILLE PROBLEM AND FOURIER SERIES 181
rthonormality of the set {<pk(x)}, and the non-negativity of the norm. Since
? II <oo and D,65) is valid for all TV, we obtain in the limit as N^>°°
D.66) i(^ft)
h'ch is known as BesseVs inequality. This shows that the sum of squares of
^he Fourier coefficients of any square integrable function <p(x) converges,
Lid implies that (<p, <pk) tends to zero as fc-><*>.
A sequence of (square integrable) functions {^N(x)}, N = 1, 2,. . . , is
aid to converge to a function <p(*) in the mean if
D.67)
This type of convergence, which differs from and does not imply pointwise
convergence, is called mean square convergence. If the partial sums of the
series D.64) are denoted by
N
D.68) Фх(х) = 2 (<p, <Pk)<Pk(x) >
we see that if ParsevaVs equality
D.69) Zj (<p, <pk) = ||^>||
is satisfied, D.65) implies that
D.70) Urn ||„(Х) - ^W||> = lim {||И|2 " S (<P, <Pkf} - 0.
Thus the Fourier series D.64) converges in the mean square sense to the
function <p(x) in the interval 0< x< /, if Parseval's equality is satisfied.
A set of square integrable functions {<pk(x)} is said to be complete (with
respect to mean square convergence) if, for any square integrable function
П*), its Fourier series D.64) converges to it in the mean. It follows from
D.70) that if Parseval's equation D.69) is satisfied for all square integrable
functions <p(x), the set of functions {д>к(х)}9 к = 1, 2,.. ., is complete.
Although few restrictions need to be placed on a function <p(x) in order to
achieve mean square convergence for its Fourier series, in applications of
°urier series to the solution of boundary value problems, stronger forms of
^nvergence such as uniform convergence are generally required.
Returning to our consideration of the Sturm-Liouville problem, we now
st the basic properties of the eigenvalues and eigenfunctions and derive the
SlniPlest of these properties.
182 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGION
1. Eigenfunctions corresponding to different eigenvalues are orthog0n
Let Л; and Ay be two distinct eigenvalues and vt(x) and vfa) two correspond
ing eigenfunctions for the problem D.56), D.57). Then
D.71) |o [ViLvj - VjLvJ dx = jQ [v}(pv\y - v^pv))'} dx
as is easily verified on using the boundary conditions D.57) for u. and v
But we also have *
D.72) |o [ViLv, - VjLv;] dx = (A, - Ay) jQ pvivi dx = (A, - A .)(<;,, vf) .
Since А,-т^Ау, by assumption, we conclude that
D.73) (<;,.(*), uy(x)) = 0,
which implies orthogonality. [This result should be compared with D.39)-
D.41) and follows from the self-adjointness of the operator (l/p)L.]
2. The eigenvalues are real and non-negative and the eigenfunctions may
be chosen to be real valued. Suppose an eigenvalue A, is complex valued.
Then A, = A; represents a second distinct eigenvalue. Because the coeffici-
coefficients in the operator L are real valued, eigenfunctions corresponding to A,
and A; = A, are v^x) and v^x) = vt{x) where vfa) is obtained by complex
conjugation in the equation for vt(x). Then since А, Ф A;, D.73) shows that
D.74) (Vi(x)9 и/*)) = (v}<x), ил(х)) -1 p(x)\v§(x)\2 <& = 0 •
But y.(x)^0 so that the integral in D.76) must be nonzero, and we have a
contradiction. Consequently, we must have A; = A, and A, is real valued.
Since the coefficients and the eigenvalues in the equation D.56) are now
real valued, the real and imaginary parts of any solution (i.e., eigenfunction)
must also satisfy D.56). Therefore, the eigenfunctions can be chosen to be
real valued and we assume them to be such.
The non-negativity of the eigenvalues can be shown as follows. Let v(x)
be an eigenfunction. Then
D.75) (и, i Lv) = - £ , £ (pi,') dx + { qv2 dx
= -pw'lo+ I pv'1 dx + Jo qv2 dx>0,
since, say, at x = / we have
E sTURM-LIOUVILLE PROBLEM AND FOURIER
D.76)
SERIES 183
0,
. a similar result valid at x = 0 (the last two integrals in D.75) are clearly
non-negative). But we also have
D.77) \v> ~ Lv) = (u> A") = A("' w) = AIIUH2 '
«ith llull>0, by assumption. Combining D.75) and D.77) gives
D.78) A= —[klP—~ '
which proves that the eigenvalue A is non-negative. (Note that the positivity
of the operator (l/p)L has been used to prove this result. This may be
compared with the discussion in the preceding section.)
The above discussion may be used to show that if q(x) > 0 in the interval
0 < x ^ /, then A = 0 cannot be an eigenvalue of the Sturm-Liouville prob-
problem. If q(x) = 0, then A = 0 is an eigenvalue if and only if аг = a2 — 0. This is
shown in the exercises.
3. Each eigenvalue is simple. Since D.56) is a second order equation,
there can be at most two linearly independent eigenfunctions for each
eigenvalue A. However, for the Sturm-Liouville problem, as shown in the
exercises, there is only one linearly independent eigenfunction for each
eigenvalue. That is, each eigenvalue is simple.
4. There is a countable infinity of eigenvalues having a limit point at
infinity. The set of eigenvalues can be arranged as follows: 0<A1<A2<
*з < * * * with Afc-»oo as fc-»°°. The set of eigenvalues is called the spectrum
of the operator L, and we see that the spectrum is discrete, non-negative,
and has a limit point at infinity.
5. The set of eigenfunctions {vk(x)}, k — 1, 2,... forms a complete
Orthonormal set of square integrable functions on the interval 0<x<L As a
fesult, if the function v(x) is square integrable over the interval 0 < x < /,
he Fourier series or, equivalently, the eigenfunction expansion
00
D79) v(x)=2(v,vk)vk(x)
k=l
converges to v(x) in the mean. It can be shown, however, that if v(x) is
°ntinuous and, in addition, has a piecewise continuous first derivative in
s.*s', and v(x) satisfies the boundary conditions D.57), the Fourier
ries D.79) converges absolutely and uniformly to v(x) in the given
184 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGION
interval. If v(x) has a jump discontinuity at some interior point *0 in tL
interval, the Fourier series converges to the value (U2)v{x -л.?
A /2)v(x0 + ) at that point, where v(x0 -) and v(x0 + ) represent onesided
limits of u(x) as x approaches x0 from the left and right, respectively.
Proofs of properties 4 and 5 are not given here. However, the discussio
in Section 8.1 which deals with higher dimensional eigenvalue problems саП
be adapted to yield a proof of these properties for Sturm-Liouville problem
The eigenvalues and eigenfunctions of the Sturm-Liouville problem can
be formally obtained as follows. Let V(x; A) and W(x; A) be solutions of the
initial value problems
D.80)
D.81) W@; A) - 0 , W'@; A) = 1 ,
respectively, for the equation D.56). Then, the function
D.82) u(x; A) = frV(x; A) + atW(x; A)
is a solution of D.56) that satisfies the boundary condition D.57) at x = 0.
To satisfy the condition at x - / we must have
D.83) a2/W> A) + a2axW(l; A) + ftfcf'(/; A) + ft^W'(h A) = 0 ,
which results from the substitution of D.82) into D.57). The eigenvalues of
the Sturm-Liouville problem A = Xk (fc = 1,2,.. .) are determined as the
roots of the equation D.83). The corresponding eigenfunctions are given as
D.84) vk(x) = v(x; Xk) = ^V(x; Xk) + axW(x\ Afc) , к = 1,2,. . . .
In the following examples we consider a number of regular and singular
Sturm-Liouville problems that lead to trigonometric, Bessel, and Legendre
eigenfunctions.
Example 4.3. Trigonometric Eigenfunctions. We set p(x) = p(x) = 1 and
<?(*:) = 0 in D.56) and retain the general boundary conditions D.57). The
resulting equation for the eigenfunction v(x) is
D.85) -v"(x) = Av(x).
The solutions of the initial value problems D.80), D.81) for D.85) are
found to be
D.86) V(x; A) = cos(VAx) , W(x; A) =
STURM-LIOUVILLE PROBLEM AND FOURIER SERIES 185
we have assumed A>0. The case where A = 0 is considered in
4.3.10. Then
г- ч s
4 g7) v(x; A) = ft cos(V Ax) + ax
d the eigenvalues are determined from the transcendental equation
D.88) VAfoft + ft«2) cos(VA/) + (o^ - A/32fr) sin(VA/) = 0 .
Some properties of the solutions of this equation are considered in Exercise
43,9. Once the eigenvalues Лл (fc= 1,2,. ..) are determined from D.88),
we obtain the eigenfunctions
. sin(VTLt)
D.89) vk(x) = ft cos(\A» + аг v^_* , Л = 1,2,....
Example 4.4. Fourier Sine Series. In this example we again consider the
eigenvalue equation D.85), but we simplify the boundary conditions D.57)
and assume that
D.90) ai = a2 = l5 ft = ft = 0.
Then A = 0 is not an eigenvalue (see Exercise 4.3.10) and the eigenvalue
equation D.88) reduces to
D.91) sin(VA/) = 0,
and this yields the eigenvalues
D-92)
1 ne normalized eigenfunctions are given as
D-93> **(*)
as is easily verified.
O the basis of the exact determination of the eigenvalues and eigenfunc-
for the foregoing Sturm-Liouville problem, the first four properties for
general eigenvalue problem listed above can be verified directly. The
ansi of a function in a series of eigenfunctions D.93) as given in D.79)
as a Fourier sine series.
186 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
Example 4.5. Fourier Cosine Series. We again consider D.85) but wjtu
boundary conditions D.57) for which
D.94) ai = a2 = 0, ft = 02 = l.
The eigenvalue equation for A > 0 becomes
D.95) sin(VA/) = 0
and since A = 0 is an eigenvalue (see Exercise 4.3.10) this yields the
eigenvalues
D.96) A* = lT/ ' * = <U.2,.-..
The corresponding normalized eigenfunctions are found to be
к = 1,2,
Again, the first four properties for the general Sturm-Liouville problem can
be verified. The series expansion of a function in terms of the eigenfunctions
D.97) in the form of D.79) is known as a Fourier cosine series.
Example 4.6. Complete Fourier Series. In this example we again consider
the equation D.85) but extend the interval of definition of v(x) to -/<*£
/. In addition, the boundary conditions D.57) are replaced by what are
known as periodic boundary conditions. Specifically, we study the eigen-
eigenvalue problem
D.98) -v"{x) = Au(x), -/ < x < /,
with the (periodic) boundary conditions
The eigenvalue problem D.98), D.99) is not of the Sturm-Liouville type-
Nevertheless, most of the properties listed for the Sturm-Liouville problem
remain valid in this case as we shall see.
Using the general solution
D.100) v(x; A) = a cos(Vbr) + b si
sTtfRM-LlOUVILLE PROBLEM AND FOURIER SERIES 187
f D 98) (where we have tacitly assumed A > 0), we easily conclude that the
eigenvalues are given as
D.Ю1) Л*:
applying D.99). By considering the solution of D.98), D.99) for A <0 it
0П be shown that v(x) = 0 is the only possible result so that there are no
^eative eigenvalues. However, Ao = 0 is a eigenvalue for this problem.
Since for each A* except Ao = 0, соз(л/Ткх) and sin(\A^) are indepen-
independent eigenfunctions, we conclude that each Xk>0 is a double eigenvalue.
Thus property 3 given for the Sturm-Liouville problem is not valid in this
case.
The full set of normalized eigenfunctions in terms of the (new) inner
product (defined over -/ < x < I)
D.102)
and the induced norm
D.103)
is given as
D.104) *о(*) = ^=, **(*) = ^cos(^x), к = 1,2,...,
D.105) U(tW=i,sin(^^, * = 1,2,....
For &>1, vk(x) and vk(x) are linearly independent normalized eigenfunc-
eigenfunctions that correspond to the eigenvalues Xk. It can be verified directly that
D106) (<5*Л) = К,о,) = 0, к* I,
D'107) (»*,»,) = 0, *,/ = 0,l,2,....
the set of eigenfunctions {vk(x), v^x)}, к = 0, 1, 2, ...,/ = 1, 2,... is
j*n orthonormal set with respect to the inner product D.102). Further, it can
e shown that the eigenfunctions form a complete set with respect to square
ttrtegrable functions v(x) over the interval -/ < x < I Such functions have a
mPlete Fourier series or, more simply, a Fourier series expansion
D*108) u(x) = (ll> eN + f [(v, vk)vk(x) + (v, vk)vk(x)]
at c°nverges to v(x) in the mean.
188 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
If v(x) is continuous and, in addition, has a piecewise continuous
derivative in the interval -/<*</ and v(x) satisfies the boundary cond
tions D.99), the series D.108) converges absolutely and uniformly to V(J\
It may be noted that all essential properties of the Sturm-Liouville eige
values and eigenfunctions carry over to the present case.
The trigonometric Fourier series considered in the Examples 4,4-4.6 ar
of sufficient interest to be studied without relating them to any eigenvalu
problem. Their basic common feature is that they all involve expansions i
the functions {cos[(irk/l)x]}9 k = 0, 1, 2,... and {sin[(<irk/l)x]}y k~\
2,.... These functions are all periodic of period 2/, the cosine functions are
all even functions of jc, and the sine functions are odd functions of jc. [\ye
recall that a function is periodic of period P if <p(x + P) = <p(x) for all x. It
is an even function if <p{-x) = (p{x) for all x and it is odd if <p{-x) = -<p(*)
for all jc.] Thus although the functions v{x) that are expanded in these
trigonometric series are defined over the interval 0 ^ x < / or -/ < x s /, the
Fourier series themselves are defined for all jc.
Given that the trigonometric (Fourier) series converge, they may be
considered to provide an extension of the definition of the given function
v(x) from its given interval to the entire real line. Thus in the case of the
(complete) Fourier series, the given function v(x) is extended as the function
V(x) as follows
D.109) \V(x + 2l) = V(x) -oo<*<oo.
In the case of the Fourier sine series, v{x), which is defined over the
interval 0s^</, is extended to Vo(x) which is given as
D.110) \VJx) = -v(-x)9 -/<jc<0
Thus Vo(x) first extends u(jc) as an odd function into the interval -/ < x <0
and then as an (odd) periodic function of period 2/ over the entire x-axis.
The odd extension is relevant here since all the sine functions are odd.
For the Fourier cosine series we extend u(jc), which is defined over the
interval 0^ jc< /, to the entire jc-axis via the function Ve(x) as follows,
D.111)
Ve(x) = v(-x), -Kx<0
Ve(x + 21) = V,(x), -°o<;c<oo.
Thus Ve(x) is an even periodic extension of period 2/ of the function v(x)-
The significance of the above discussion lies in the fact that the Fourief
;TURM-LIOUVILLE PROBLEM AND FOURIER SERIES
189
series, cosine series, and the (complete) Fourier series obtained in
[aples 4.4-4.6 not only represent the given function v(x) over the
iate interval (i.e., either 0 < x < / or -/ < jc < /) but also the func-
• ns K,(*)i ve(x), and V(x), respectively, over the entire real line. Con-
tl0 uently, if V(x) is an odd periodic function, the (complete) Fourier series
SC\t reduce to the form of a sine series. Similarly, if V(x) is periodic and
n the (complete) Fourier series must reduce to a cosine series. Further-
eore' the graphs of the extended functions V(x), Vo(x), and Ve{x) provide
some insight into the pointwise convergence properties of the appropriate
Fourier series.
д8 an example, we consider the function v(x) = x over the interval
q< jc ^ /. The odd and even periodic extensions of v(x) are shown in Figures
4 4 and 4.5. Now, if v(x) - x in -/ < x < /, the periodic extension of v(x) to
y(x) with V(x + 21) = V{x) is identical to the function Vo(x) graphed in
Figure 4.4. Thus in either case the extended function Vo(x) = V(x) has jump
discontinuities at the points x = ±Bn + 1)/, (n = 0,1,2,. . .). Because of
these discontinuities, the pointwise convergence of the Fourier series in
these cases is slow away from these points. At these points the Fourier series
converges to zero. Note that in the case of the sine series v(x) — x does not
satisfy the appropriate boundary condition D.57) at x - /, whereas in the
case of the complete Fourier series the function v(x) = jc is not periodic over
the interval -/ < jc < / since v(~l) Ф v(l). For the even periodic extension,
the function Ve(x) is continuous for all jc, but is not differentiable at the
Figure 4.4. Odd extension of v(x) = jc.
190
INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
Figure 4.5. Even extension of v(x) = x.
points x = nl (n = 0, ±1, ±2,. . .). Pointwise convergence is more rapid in
this case than for the odd or periodic extension considered previously.
However, it is still fairly slow because the curve is not smooth. In fact, with
v(x) = jc we have v'@) = v'(l) = 1, so that the boundary condition D.57)
appropriate for the cosine eigenfunctions is not satisfied.
If the boundary conditions for the relevant eigenvalue problem are
satisfied by the given function v(x), the smoother the function v(x) is, the
better the convergence. This is evidenced by the smoothness of the relevant
extended functions Vo(x), Ve(x), and V(x). Generally speaking, the greater
the number of continuous derivatives the extended functions have, the more
rapid the pointwise convergence of the Fourier series. This matter is further
examined in the exercises.
Example 4.7. Bessel Function Expansions. We consider a singular
Sturm-Liouville problem that arises from the eigenvalue problem for Ba-
Basel's equation of order n, where n is an integer or zero. The equation we
consider is
D.112) Цфс)], _
on the interval 0 < x < L [On replacing x by VXjc, D.112) becomes Bessel's
equation of order п.] Since p(x) = p(x) = x and q{x) = n2lx, we see that
these functions are positive for 0 < x < I but p and p vanish at x = 0 and q{x)
is singular there. Nevertheless, x = 0 is a regular singular point for Bessel s
equation and bounded solutions of D.112) do exist. To obtain these
solutions we impose the boundary conditions
D.113)
[ v(x) bounded at jc = 0
cTURM-LIOUVILLE PROBLEM AND FOURIER SERIES 191
the condition at x = / can be replaced by a boundary condition of the
nd or third kind. The eigenvalues and eigenfunctions for these problems
S!p°considered in the exercises. In general, the boundedness condition
13) at *e sinSular point is supplemented by the condition
p(x)v'(x) = 0. This guarantees that not only are the eigenfunctions
*° i but that eigenfunctions corresponding to different eigenvalues are
' 1. To see this, we recall from D.71), D.72) that with А, Ф Ay,
D.114) (A< " xj)(Pt> y>)
For the right side to vanish so that orthogonality results, we must not only
uave у and Vj bounded at x = 0 and x = l but also lim p(x)v\(x) -
lim pv'-(x) = 0 as x-»0 and jc—»/ if a singularity exists at x = 0 and x = /.
For'our eigenvalue problem D.112), D.113), the singular solution of
Bessel's equation is rejected in view of D.113) and the eigenfunctions must
be determined from the Bessel functions of order n, that is, Jn(\fkx). Since
j s.1 + O(x2) and Jn(V\x) = O(xn) as x->0, we find that not only are
these functions bounded at jc = O but also xJ'0(y/~kx) = O(x2) and
xJ'n(yf\x) = O(xn) as x—> 0. Since p(x) — jc, in this case, the added condition
lim^o P(x)v'(x) = 0 is satisfied.
The eigenvalues are determined from the boundary condition v(l) = 0.
We denote the positive zeros of the Bessel function Jn(x) by akn, fc = l,
2,..., so that Jn(otkn) = 0. [There are infinitely many real positive zeros of
the Bessel functions Jn{x) with a limit point at infinity and they are
tabulated.] Then v(l) = 0 implies that /„(VA7) = 0 and the eigenvalues for
each n are given by
D.115) Аь. = (^)\ к = 1,2,....
Using the properties of the Bessel function, it is shown in the exercises that
the square of the norm of the eigenfunctions vk(x) = Jn{\f\^nx) is
h view of the orthogonality property demonstrated above, the set
D.117) „ W
кЛ)
ls an onhonormal set of eigenfunctions for the Sturm-Liouville problem
^ ^2), D.113) for each n. These eigenfunctions form a complete set, and
bny smooth function v(x) defined over the interval 0 < x < / that satisfies the
°undary condition v(l) = 0 can be expanded in a convergent series
192 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.118) v(x) = E (v(x), vk,n(x))vk,n(x) .
Example 4.8. Legendre Polynomial Expansions. The eigenvalue probi
associated with the self-adjoint form of the Legendre equation is Ш
D.119) L[v(x)) = - ^ [A - x2) fx ] = Ai; ,
where v(x) is defined over the interval - К x < 1 and satisfies the boundary
conditions
D.120) v(x) bounded at x = +1.
Since p(x) = 1 - x2 is positive for - К x < 1 and vanishes at x = ± 1, we
have a singular Sturm-Liouville problem in D.119), D.120). The points
x = -1 and x = 1 are regular singular points for Legendre's equation, and
bounded solutions for D.119) exist.
The Frobenius theory of power series solutions for D.119) shows that the
only bounded solutions of the Legendre equation D.119) over the interval
-1 < x ^ 1 are polynomial solutions and these occur only if A assumes the
eigenvalues
D.121) kk = к(к + 1) , Jfc = 0,1,2,
The eigenfunctions are usually normalized to yield the Legendre polyno-
polynomials. These are polynomials of degree к and are defined as
D.122)
with the normalization chosen such that Pk(l) = l for all к. Since
limx_±1(l-jc2)P;(jc) = 0, the supplementary condition that p(x)v'(x) be
finite at the singular points is satisfied [see the discussion leading to D.114)].
Consequently, the Legendre polynomials form an orthogonal feet over the
interval -1<jc<1. Further, it can be shown that
D.123) \\P^x)\\2 = jlpUx)dx=j~[, к = 0,1,2,
Thus the orthonormal set of eigenfunctions is given as
/2fc + 1
D.124) vk(x) = V-^— P*W ' * = 0,1,2,. . . .
s SOLUTIONS OF INITIAL AND BOUNDARY VALUE PROBLEMS 193
set {vk(*)} is complete and any smooth function v(x) defined over the
larval -1 - x - 1 can be exPand^d in a convergent series of eigenfunctions
D.125)
Problems in which the singular Sturm-Liouville problems involving the
sel and Legendre equations play a role are presented in Exercises 4.2.10
and 4.2.П.
4.4. SERIES SOLUTIONS OF INITIAL AND BOUNDARY VALUE
PROBLEMS
In this section the method of separation of variables is applied to specific
initial and boundary value problems for partial differential equations. Three
examples dealing with equations of hyperbolic, parabolic, and elliptic types
are presented, and the solution of each of the problems is discussed in some
detail.
Example 4.9. Vibrations of a Fixed String. Let u(x, i) represent the
transverse displacement of a tightly stretched string of length /. Under
various simplifying assumptions it is found that u(x, t) satisfies the wave
equation
D.126) utt - c2uxx = 0, 0<Jt</,/>0,
с = y/T/p, where T is the tension and p is the density of the string, both
of which are assumed to be constant.
Now и = u(x, i) is the graph or shape of the string at the time t and
ux(x, t) is the slope at a point x on the string. We assume that \ux\ < 1 and
ftis implies that the curvature of the string approximately equals uxx(x, t).
The mass per unit length of the string equals the density p of the string.
ГЬеп Newton's law of motion states that putt at a point x on the string
equals the transverse force acting on it. All forces acting on the string are
neglected compared to the force due to the string's tension. This force is
Proportional to the curvature uxx with a constant of proportionality equal to
e tension Г. The greater the curvature, the greater this (tension) force
nose effect is to straighten the string out. As a result, u(x, t) satisfies the
wave equation D.126).
th enck °* *e str*nS are assumed to be fixed for all time and this yields
ne boundary conditions
DЛ2?)
м@,0 = 0, w(/, 0 = 0, t > 0.
194 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGIom
The given displacements and velocities at t — 0 yield the initial conditio
D.128) u(x, 0) = f{x) , ug(x9 0) = g(x), 0 < x < I.
To solve the initial and boundary value problem D.126)-D.128) w
apply the separation of variables method of Section 4.2 and set u(xy л =
M(x)N(t). Then M(x) satisfies the following eigenvalue problem,
D.129) -c2M\x) = AM(jc) , M@) = M(l) = 0 .
An equivalent Sturm-Liouville problem was considered in Example 4.4 and
yielded the following results. The eigenvalues for D.129) are
D.130) A* = l
and the eigenfunctions are
D.131) Mk(x) = ^sin(?fx), к = 1,2,3,
From D.43) we obtain
D.132) Nk(t) - ak cos^ t) + bk sin^ ; j , * = 1,2, 3,. ..,
and the corresponding solutions uk(x, t), which are known as normal modes,
D.133)
ч Г (тткс \ . . I тгкс \1 /2 . Гтг/с 1
мл(х, 0 = \ак cos^-^- rj + Ьк sm^— /JJyу sin[— *,J ,
Each of the uk satisfies the wave equation D.126) and the boundary
conditions D.127).
To satisfy the initial conditions D.128) we form the sum D.45), set
00
D.134) ф,0=Еи^0,
and apply the initial conditions to u(x, t). This gives
D.135) u{x, 0) = £ uk(x, 0) = £ etV7 Ц— *j = /W ,
IES SOLUTIONS OF INITIAL AND BOUNDARY VALUE PROBLEMS 195
,0) А /тгкс\, [2 . (ттк
are Fourier sine series with normalized eigenfunctions Mk =
ffiijsin[(irk/l)x] and D.48), D.49) imply that
D.137)
/2 (' I vk \
Uk = (fix), Mk(x)) = Vy Jo f(x) Sin\ ТГ' * = 1,2,.. .,
and
D.138)
bk = (-^ gW) Mkix)) = ^ { e(x) sin(f x) Л , * - 1,2,....
This completes the formal determination of the solution.
To assure that the formal series solution D.134) represents an actual
solution of the wave equation and satisfies the initial conditions we must
place conditions on the data f(x) and g(jc). The function f(x) is required to
have two continuous derivatives and a third piecewise continuous derivative
and, in addition, we must have /@)=/(/) = 0 and /"@) = /"(/) = 0. The
function g(x) must be continuously differentiable and have a piecewise
continuous second derivative as well as satisfying the conditions g@) =
g(/) = 0. Under these conditions the series D.134) can be differentiated
term by term twice and the resulting series converge uniformly. Thus w(jc, t)
is a solution of the wave equation since each uk(x, t) satisfies the equation.
The conditions on f(x) and g(x) guarantee that the sine series as well as the
series of its derivatives converges, since it is required that f(x) and g(x)
satisfy the boundary conditions for the appropriate Sturm-Liouville prob-
problems to assure the uniform convergence of the Fourier series as was
indicated in the preceding section (see Exercises 4.3.11 and 4.3.12).
We now give a brief interpretation of the solution D.134). Each of the
normal modes uk(x, t) [i.e., D.133)] can be expressed as
"*(*, 0 - ak cos[^ (t + 8,)]sin(f x) , * « 1,2,
D.140)
°r each solution uk(x, t), a fixed point x0 executes harmonic vibrations with
an4>Htude
196 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.141) Ak = с
Solutions of the wave equation are called waves; we refer to uk(x, t) a
standing wave, since each point x on the string oscillates in place as t varie *
The points xm = milk (m = 1,2,. . ., к - 1) at which sin[GrA://)jc ] ^0
remain fixed in the vibration process and are called the nodes of the standin
wave uk. The points xn = B/z + l)l/2k (n = 0,1,2,. . . , к - 1) at which
sin[Grfc//)jtn] = ±1, are those where the standing wave has maximum
amplitudes and these are called antinodes. We also note that
D.142) <ок = ^=л/Х~к, к= 1,2,3,...,
equals the frequency of vibration of the kth harmonic uk(x, i) (the standing
wave uk is also known as the fcth harmonic).
The energy of the vibrating string at the time t is given by the integral
which represents a sum of the kinetic energy of motion and the potential
energy due to the tension. The total energy of the string is a sum of the
energies distributed among the harmonics uk(x, i). It is readily seen that the
energy of the fcth harmonic uk{x, i) is
D.Ш,
where m = pi is the mass of the string. We observe that Ek(i) is independent
of t and thus the total energy
00 00
D.145) E(t) = 2 Ek{t) - у 2 a>l(a2k + b2k)
is also constant in time. That is, E{t) = £@), so that the initial energy is
conserved as was to be expected, since no dissipative effects are assumed to
be present. This is consistent with our characterization of the w^ve equation
in Example 3.8 as being of conservative type.
We have seen in Example 2.4 that the solution of the initial value
problem for the wave equation can be written as the sum of two propagating
or traveling waves; that is, it has the form
D.146) m(jc, 0 = F(x - ct) + G(x + ct) ,
where F and G are arbitrary functions. The normal modes uk(x, t) can also
be expressed as the sum of propagating waves as
rIES SOLUTIONS OF INITIAL AND BOUNDARY VALUE PROBLEMS 197
D.147) uk(x, t)=^= [ak sin[^ (x + ct)] + ak sin[^ (* - ct)]
+ bk cos[^ (x - ct)] - bk cos[^ (x + ct)]} .
Although there are four propagating waves in D.147), through interference
h v combine to form a standing wave. In this representation, the solution
u(x, t) has the form
u(x, t) « 2 uk(x, 0=2 л?! йк V 7 sin[ "T (* + СГ)J
D.148)
1 V [2 . [тгк
+ 2 ^ WL
where we have assumed for simplicity that g(x) = 0, so that bk = 0 for all A:.
Apart from the factor 5, these series are identical in form to the Fourier
sine series D.135), if the argument x is replaced by x ± ct. We note that/(x)
is not defined outside the interval [0, /]. Proceeding as in Section 4.3, we
define the function F(x) as
D.149)
F(x)=f(x), 0<x<l
F(x) = -Д-х), -I < x <0
F(x + 21) = F(jc) , -00 < x < 00 ,
so that F(;t) is an odd periodic extension of f(x) of period 2/ over the entire
x-axis. The Fourier series expansion of F(x) is identical to the sine series
D.135) for/(jc), since F(x) is an odd periodic function. Now F(x) is defined
for all x so that the Fourier series for the periodic function F(x ± ct) can be
constructed. Its form is precisely that of D.135) with x replaced by x ± ct.
Consequently, the solution D.148) can be expressed as
D-150) u(x, t) = \F(x + ct) + \F(x - ct) , 0<x < /, t >0 .
The conditions under which D.150) represents a solution of the initial and
boundary value problem for the wave equation [with g(x) = 0] can be shown
0 be slightly weaker than those given for the standing wave representation
of the solution.
he propagating wave representation D.150) of the solution of the wave
Rtion can be interpreted in terms of waves being continually reflected off
e ends of the string. The interference of these waves with one another
^Plains why only standing wave motion seems to occur when the vibration
^string is observed over a length of time.
The Fourier series representation of F(x) retains the discontinuities of
vvjmd does not smooth them out as was indicated previously. Thus unless
198 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
F(x) is a smooth funtion, the solution u(x, t) may have discontinuitie
discontinuous derivatives along the curves jc ± ct = constant, in view °Г
D.150). This shows that the discontinuities in solutions or derivative °*
solutions of the wave equation can occur only along the character* °*
curves, consistent with the results of Section 3.2. It is shown later in the ^ °
how the concept of solution may be generalized to include functions thatT
not have the required number of derivatives. °
Example 4.10. Heat Conduction in a Finite Rod. Let m(jc, t) represent
the temperature in a homogeneous, laterally insulated rod of length / at th
point jc @ < jc < /) and at the time t. Since no heat can escape through the
sides of the rod, the problem is effectively one-dimensional and in the
absence of heat sources m(jc, i) satisfies the heat equation
D.151) ut = c2uxx, 0<x<l,t>0,
where c2 is the coefficient of heat conduction. As indicated in Section 4.1, c2
equals the thermal conductivity divided by the product of the specific heat
and the density of the medium, all of which are taken to be constant in this
problem. The ends of the rod at x = 0 and jc = / are assumed to be kept at
zero temperature so that
D.152) w@, t) = m(/, 0 = 0,
whereas initially there is a given temperature distribution
D.153) m(jc,O)-/(jc), 0<jc</.
We apply the separation of variables technique to solve this problem and
proceed as in Example 4.9. With uk = Mk(x)Nk(t), the eigenvalues Л* and
the eigenfunctions Mk are given as in D.130), D.131) and Nk is found to be
[see D.43)]
The solution u(x, i) becomes
DЛ54)
Л54) - - F5 - t*k\
u(x, t) = 2 «*<*, 0 = 2 Mk(x)Nk(t) = Vf 2 -,*-<*"*sin(T x) ■
Initially, we have
D.155) u(x, 0) = 2 aW 7 sin( -y- xJ = f(x)
l%$ SOLUTIONS OF INITIAL AND BOUNDARY VALUE PROBLEMS 199
that the ak are defined as in D.137). Thus the formal solution of
*л i5lH4-153) is comPletely specified.
To examine the validity of the expansion D.154) we note for t>0 the
_s in the series decay exponentially. If we assume that the initial
perature/(*) is bounded over the interval [0, /], the Fourier coefficients
t are bounded. As a result, the series for u(x, t) with t>0 can be
Iffferentiated term by term as often as required. Since each term uk(x, t)
t'sfies the heat equation, so does u(x, i) for t>0 and we also have
S^0 t) = w(/, О ~ 0* To assure that the initial condition is satisfied, we must
h ve the Fourier series D.155) converge uniformly. To achieve this we
uire tkat дх) ье continuous and piecewise continuously differentiable in
0<д:< / and that /@) = /(/) = 0. The uniform convergence of the Fourier
sine series D.155) implies that the series D.154) for u(x, t) converges
uniformly for t ^ 0 and that the solution is continuous for 0 ^ x ^ / and / > 0.
In the vibrating string problem of the previous example (which involved a
hyperbolic equation) the solution as given in D.148) essentially has the form
of a Fourier series. This has the effect that any discontinuities in the data are
preserved and transmitted along the characteristic lines x ± ct = constant in
(x, r)-space. For the heat equation, however, even if the initial temperature
f(x) is discontinuous but bounded, the solution u(x, t) is continuously
differentiable as often as is required. Now the more derivatives a function
has, the smoother it is, so that heat conduction is seen to be a smoothing
process. Even if the heat is initially concentrated near one point on the
interval 0 < x < /, it is distributed instantaneously according to our solution
in an even and smooth fashion throughout the interval. Further, as f-к»,
the temperature u(x, t)^>0 since there are no heat sources in this problem
and the rod is continually being cooled at its end points. /
We have already indicated in our discussion of the diffusion equation in
Chapter 1 that disturbances move at infinite speeds for the heat equation,
and this fact is borne out by the properties of the solution obtained. It may
also be noted that the characteristics of the heat equation are the lines
' = constant. Since characteristics are carriers of discontinuities or rapid
variations of the solution, there is no mechanism whereby these effects can
be transmitted into the region t > 0 from the initial line t = 0 which is itself a
characteristic curve.
Example 4.11. Laplace's Equation in a Rectangle. We consider the
boundary value problem for Laplace's equation
D156) «« + «„-0
m the rectangle 0 < jc < / and 0 < у < I, with the boundary conditions
D.157> (
200
INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
The function u(x, y) may represent the steady state (or equilibrium)
placement of a stretched membrane whose rectangular boundaries are гъЛ
according to D.157). [Proceeding in a similar fashion, we can solve anotrT*
problem for D.156) with nonzero boundary conditions onjc = 0 and x^
and zero boundary conditions on у = 0 and у = /. On adding the solutions !
both problems we can thereby solve Laplace's equation with arbitra
boundary conditions on a rectangle.] ^
Applying the separation of variables method of Section 4.2 and proceed
ing as in Example 4.9, we obtain the eigenfunctions Mk(x) as in D.131) and
the eigenvalues кк = {тгк11J. With uk(x, y) = Mk(x)Nk(y), the Nk(y) are
given as [see D.43)]
D.158)
-?f y],
The functions Nk(y) can be expressed as linear combinations of hyperbolic
sine functions and we write uk(x, y) as
D.159) uk(x,y) = {aksmh[?f
bk siiih[^- (y -
y sin(~ x)
with the ak and bk as yet arbitrary. A formal solution of Laplace's equation
obtained by superposition is
D.160) «(xj)=E«fc(jfj).
The boundary conditions at у = 0 and y = l imply that
D.161) w(x, 0) = 2 bk sinhj - ^- / k/7 sin( 2- x) = /(*) ,
D.162) m(jc, l)= 2j ak sinh| -y /1 \ у sin\"T"x) = *W »
The Fourier coefficients of these sine series are determined to give ak and h
as [see D.137), D.138)]
(f(x),Mk(x)) t_, „
D.163)
«* =
), Mk(x))
inhl —r- Ij
■
sinh
OMOGENEGUS EQUATIONS: DUHAMEL'S PRINCIPLE 201
• competes the formal determination of the solution of the boundary
valTUfewPe assume that the integrals Д \f\ dx and Д \g\ dx are bounded by the
^ber m, and note that sinh x = (ex - e~x)/2 < A /2)ex for x > 0, we easily
find that
Jirk _
k sinh[— (y
D.164)
, У27/техр[-тг*у//]
1-ехр[-2тг///]
with a similar bound valid for \ak sinh[GrA://)y]|. Thus the terms in the
eries u(x, y) = £*=i ик(х> У) аге bounded by exponentially decaying terms
for large к in the open interval 0 < у < /. (The exponential bound for the bk
and ak terms breaks down at у = 0 and у = /, respectively.) Consequently,
the series can be differentiated term by term as often as desired in the
interval 0<y < / and m(jc, y) satisfies Laplace's equation and the boundary
conditions at x = 0 and x = /, since each of the uk does so. To assure that
u(x, y) is continuous up to у = 0 and у = / and assumes the boundary values
there, we must assure that the Fourier series D.161), D.162) are uniformly
convergent in 0 < x < /. This is the case if /@) = /(/) = g@) = g(l) = 0, if
f(x) and g(x) are continuous and/'(jc) and g (x) are piecewise continuous in
that interval.
For w(jc, y) to satisfy Laplace's equation inside the rectangle we merely
require that f(x) and g(x) be bounded or that jl0 \f(x)\ dx and J^ \g(x)\ dx
exist, (i.e., / and g may even be singular). Then u(x, y) is defined by the
series to be an infinitely differentiable function within the rectangle. How-
However, the boundary values at у = 0 and у = / need not be assumed continu-
continuously at all points unless the additional conditions on / and g given
previously are met. Again, this contrasts with the results obtained for the
wave equation, where discontinuities are seen to spread into interior regions
from the boundary. Since Laplace's equation has no real characteristics,
boundary data discontinuities must be confined to the boundary and the
interior solution is smooth. Physically, since Laplace's equation character-
"^ steady-state or equilibrium situations, we may expect that effects due to
in the data have smoothed themselves out.
nis concludes our discussion of the separation of variables method for
mogeneous problems. Further examples involving other eigenvalue prob-
ms are considered in the exercises and in Section 4.6.
4'5- INHOMOGENEOUS EQUATIONS: DUHAMEL'S PRINCIPLE
on of variables was applied in the foregoing to i
boundary value problems for homogeneous partial
So]e niethod of separation of variables was applied in the foregoing to obtain
utions of initial and boundary value problems for homogeneous partial
202 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
differential equations. In this section we construct solutions of inhon
neous equations by using a technique known as DuhameVs principle, Wh'
effectively relates the problem to one involving a homogeneous equatj1Cil
The method is valid for Cauchy (i.e., initial value) problems and initial °П
boundary value problems for hyperbolic and parabolic equations. ^
We consider equations of the form
\p(x)utt + L[u] = g(x, t) , hyperbolic case
[p(x)ut 4- L[u] = g(x, t) , parabolic case
with L[u] defined as in D.6) or D.7), p(x)>0 and g(x, t) a given forcing
term (note that x may be vector or scalar variable). If we consider the
Cauchy problem for D.165), we assume w(x, i) satisfies homogeneous initial
conditions at t - 0. For the initial and boundary value problem for D.165) in
a bounded region G, we again assume homogeneous initial conditions for
u(x, t) at * = 0 in addition to the homogeneous boundary conditions D.14)
or D.15). Thus
f u(x, 0) = ut{x, 0) = 0 , hyperbolic case
D.166) | ф^ o) = 0 , parabolic case .
DuhameFs principle proceeds as follows. Consider a homogeneous ver-
version of D.165), that is,
p{x)vtt + L[v] — 0 , t > r
p(x)vt + L[v] == 0, t>r
for the function v(x, t), which is assumed to satisfy the same boundary
conditions (if any are given) as u(x, t). Let v{x, t) satisfy the following initial
conditions given at t = т where т ^ 0:
DЛ68)
v(x, r) = 0, v£x, t) = / I , hyperbolic case
я(х г) PW
v(x, r) = /v , parabolic case ,
P\X)
with p and g given as above.
It is assumed that the above problem for i; can be solved either by
separation of variables for the initial and boundary value problem or by
other means for the Cauchy problem. Since the solution depends on the
parameter т (i.e., the initial time), we write it as v = v(x,t;r). Then
Duhamel's principle states that the solution u(x, t) of the given inhomoge-
neous problem is
D.169) w(jc, t) = Jo y(x, t\ r) dr
OGENEOUS EQUATIONS: DUHAMEL'S PRINCIPLE 203
Ovation for the method is obtained by noting that the effect of the
. term g(x, t) can be characterized as resulting from a superposition of
ulses at times t = т over the time span 0 ^ т ^ t].
P verify that u(x, t) as given in D.169) is a solution of the problem, we
note that
ut(x, t) = v(x, t; t) + Jo vt(x, t; r) dr ,
utt(x, t)= — [v(x, t\ t)] + vt(x, t; t) + Jo vtt(x, t; r) dr ,
In view of D.168) we have in the parabolic case
D.170) V(X>t;t)=^M~ '
and in the hyperbolic case
D.171) v(x, t;t) = O, vt(x, t; t) =
Therefore
D.172) put + L[u] = g(x, t) + Jo [pvt + L(u)] dr = g(x91),
and
pUtt + L[u] = g(jc, 0 + JT [р1;в + L(v)] dr = g(jc, r).
u(x, 0) = 0 and ut(x, 0) = y(x, 0; 0) = 0 so that u(x, t) as defined by
D.169) satisfies all the conditions of the problem.
Next we consider two examples where we apply DuhameFs principle to a
auchy problem for the inhomogeneous wave equation and an initial and
°undary value problem for the inhomogeneous heat or diffusion equation.
MPLE 4.12. The Inhomogeneous Wave Equation. We consider the
ave equation with a given forcing term g(x, t),
Utt - fuxx = g(x, 0, -
ncl the homogeneous initial conditions
INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
w(x,0) = w,(*,0) = 0 , -
204
D.175)
[Note since we can add to и(дс, /) a solution of the homogeneous wav
equation with arbitrary data, we are effectively able to solve the gene ^
initial value problem for D.174)].
Applying Duhamel's principle, we consider the function v(x, t\ r) th
satisfies the equation
D.176) vtt-c2vtt = 09 *>r,
and the initial conditions at t — т,
D.177) v(x, t; t) - 0, vt(x, r; r) - g(x, r) .
From d'Alembert's solution (see Example 2.4) we easily obtain
D.178) viX,V,r)=j-XZl^r)da
and from D.169), u{x, i) takes the form
D.179) m(jc, 0 = 2c Jo J^c(,_t) S(^ r) da dr .
If we replace the initial data D.175) by the arbitrary data
4
* • *o - "o
<xo,to)
/\
X = X + Ct X
Figure 4.6. Domain of dependence.
H0M0GENEOUS EQUATIONS: DUHAMEL'S PRINCIPLE
205
Figure 4.7. Domain of influence.
D.180)
u(x,0) =
u,(x,0)=G(x),
the solution of the initial value problem D.174) and D.180) is immediately
obtained as
D.181) u(x, 0=| [F(x + ct) + F(x - ct)] + ^ j^ G(a) da
+ ;r~ I I £(°">
2C JO }х'ф-т) 5V '
At the arbitrary point (jc0, t0) with t0 >0, the solution m(jc0, ^o) depends
only on values x and Г within the characteristic triangle pictured in Figure
4.6. An inspection of the argumens of F(x0 ± ct0) and the domains of
integration for G(<r) and g{a, r) in D.181) shows this to be the case. As a
result, the characteristic triangle is called the domain of dependence of the
solution at the point jc0 and at the time f0. For a similar reason, the sector
pictured in Figure 4.7 that is bounded by the characteristic lines issuing from
fte point (xo,0) is called the domain or region of influence of the initial
P°int (xo,O). All points (x, t) that lie in this sector have the point (*0,0)
within their domain of dependence. In particular, if g(x, t) = 0 and F(x) and
U) are concentrated at the point jc0, say, they have Dirac delta function
i we see that the solution u(x, t) vanishes identically outside the
^ of influence of the point *o. Within the sector, m(jc, 0 may or may
be zero. The two characteristic lines x± ct = xo± ct0 represent wave
onts for the solution, since и = 0 ahead of the wavefronts (which move to
e right and to the left) and и Ф 0, in general, at points (x, i) that lie behind
J^a fronts, that is, within the sector.
206 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
Example 4.13. The Inhomogeneous Heat Equation. The heat equation i
a finite interval 0 < x < /, with a heat source, has the form П
D.182) щ - c2uxx = g(xy t), 0<x<l,t>0.
We assume a homogeneous initial condition u(x7 0) = 0 and the boundary
conditions w@, i) = w(/, t) = 0 at x = 0 and x-L
Applying DuhamePs principle and using the solution obtained in D.154ч
for the homogeneous heat equation with t replaced by t — r, we obtain
D.183) v{x, t; т) = д/у 2 ак(т) exp[- {^f)\t - r)]dn(j *) ,
with the ak(r) determined from the initial condition
D.184) v(x, r; r) - g(x, r)=J-| ak{r)sin(^-
Note that the Fourier coefficients ak are functions of the parameter r [i,e.r
for each value of r we obtain a Fourier sine series of g(x, r)]. Then the
solution of the initial and boundary value problem for D.182) is
DД85)
m(x, t) = J v(x,t;r)dr
where summation and integration have been interchanged. This is a valid
procedure since the series converges uniformly, say, if we assume g(x, t) is a
bounded function. If u(x, 0) - f(x) and is nonzero, we may add the solutioa
D.154) to D.185) to obtain the solution of the new problem.
To consider a specific problem we suppose that
D.186) g(*,0 = sin(
Then it can be seen by inspection that we have ak = 0, к > 1, and ax =
so that the solution is
D.187) "(*>') = {/„'exp[-(y) (t-T)]dr}sm(™)
[Note that we again have u(x, 0) = 0.]
As f—»oo we see that
ENf UNCTION EXPANSIONS: FINITE FOURIER TRANSFORMS 207
D.188) *^(^)Цт)'Ф)'
The linking function v(x) in D.188) is a solution of the steady-state problem
f ^he heat equation with g = sinGrjt/Z) which is given as
D.189) А^Т)' «*<>)-i>(/)-0.
This problem results when we assume w(x, f) = u(x) (i.e., и is independent
of t) and the initial condition is dropped.
More generally, if the inhomogeneous term g(x, t) is independent of
time, that is, g{xy t) = G(x), with G(x) arbitrary, we obtain from D.185)
D.190) «(*, о - Vf S {(ж) 1 -exp[~(iT£) Wt *)} '
where we have used the fact that ak is independent of т to carry out the т
integration. As f-*<» in D.190) we have
D.191) Jim ф, O=Vf S (^) •* sin(^x) - v(x).
It can be verified directly that v(x) satisfies the steady-state problem for the
heat equation
D.192) ~c2vxx = G(x), v@) = u(/) = 0 .
Now even if u(x, 0) =f(x) Ф 0, we have already seen in Example 4.10 that
the effect of the nonzero initial heat distribution dies out as f—»<», so that
when the heat source term g(x, i) = G{x) is independent of time, the
temperature distribution (when the ends of the rod are fixed at zero
temperature) is well approximated by the solution of D.192) for large t.
This completes our discussion of Duhamel's principle. The method of
te Fourier transforms and other techniques for handling inhomogeneous
problems are discussed in the following section.
JJ- EIGENFUNCTION EXPANSIONS: FINITE FOURIER
TRANSFORMS
he method of separation of variables presented in Section 4.2 led to the
^presentation of the solution of the given problem in a series of eigenfunc-
lons of the differential operator L defined in D.6), D.7). To generate the
208 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED Ш(Ц0
appropriate eigenvalue problem it was necessary that both the equation а н
the boundary conditions be homogeneous. For the self-adjoint operato
(l/p)L under consideration, the eigenfunctions have the property of Co ^
pleteness, which means that under certain conditions a function f(x) can h*
expanded in a (convergent) series of eigenfunctions. The orthogonality ^
the eigenfunctions renders the determination of the coefficients in the set-
to be quite simple. es
In this section we reverse the above approach and begin directly h
expressing the solution of the problem as a series of eigenfunctions. Th
coefficients in the series are arbitrary and must be determined from the
differential equation and the data for the problem. Rather than substituting
the series directly into the equation, we convert the partial differential
equation into a hierarchy of ordinary differential equations for the determi-
determination of the unknown coefficients, termed Fourier coefficients. Thereby
the equation is transformed into a collection of equations for these Fourier
coefficients, and this technique is often called the method of finite Fourier
transforms. More specifically, if the method involves the use of Fourier sine,
Fourier cosine, or Bessel function expansions of the form considered in
Section 4.3, it is known as the method of finite sine, cosine, or Hankel
transforms, respectively. The reason for this becomes clearer when we
discuss (infinite) Fourier transforms in Chapter 5. This method applies to
inhomogeneous equations with inhomogeneous initial and boundary data.
Technically, therefore, this method enables us to solve the most general
problems formulated in Section 4.1.
We consider equations of the form
D.193) pKu + Lu = pF,
where L is defined as in D.6) or D.7) and p(x)>0 and F are j>iven
functions. The differential operator К can be given as К = dldt, К = д Idi,
or К=-д2/ду2, so that D.193) has the form of D.8), D.10) or D.19),
respectively. Or К = 0 and D.193) reduces to D.9). In fact, L may have the
form associated with any of the eigenvalue problems considered in Section
4.3. All that is required of К is that it does not depend on any of the variables
that occur explicitly in the eigenvalue problem associated with L. Thus
equations of parabolic, hyperbolic, or elliptic type can be treated by this
method. The boundary conditions for the problem are of inhomogeneous type
as given in D.11) or D.12) and initial conditions are assigned in the hyperbolic
and parabolic cases as in D.16) or D.18). Conditions of the form D.20) for the
elliptic case can also be given.
The solution и of each of the foregoing problems is expanded in a serie
of eigenfunctions Mk(x) determined from the eigenvalue problem
D.194) LMk = kkpMk , k = 1, 2,. . .
mrv I EXPANSIONS: FINITE FOURIER TRANSFORMS 209
. homogeneous boundary conditions of the form D.14) or D.15). It is
wl ffled that the Mk form a complete orthonormal set for the purposes of
^discussion. Thus (M,, M;) = 0 for к Ф] and (Mk, Mk) = 1 for all к. We
Expand и as
D-195) u = ^NkMk(x),
here the Fourier coefficients Nk (which may depend on t or on y) are to be
W ecified. In terms of the inner product D.33), the Nk are formally given as
D.196) Nk = («, Mk) ,
since the Mk are orthonormalized, but the solution и is not as yet known. To
specify the Nk we do not substitute the series D.195) directly into the
equation D.193) since the series may not be differentiable term by term.
Instead, we transform D.193) into a system of equations for the Fourier
coefficients Nk. For each k, Nk is a finite Fourier transform of u. Once the
full set of the Nk is known, the function и can be reproduced from the series
D.195). In this sense D.195) inverts the transforms Nk and the series
representation of и is called the inverse (finite) Fourier transform.
To proceed, we multiply D.193) by Mk(x) and integrate over the
underlying region G to obtain (on using notation appropriate for the two- or
three-dimensional case)
D.197) J £ pMkKu dv = KJjG puMk dv = K(u, Mk)
= - J jGMkLudv + J jGpFMk dv ,
where we have pulled the differential operator К out of the integral. Now
from D.31) we have
D.198) -j j^ MkLu dv - - J |c uLMk dv
^ [ ( ди <?МЛ
+ p\Mk- и —-*■ I ds .
JdG^\ к дП дП /
)hen, using D.194) and the boundary condition satisfied by и and Mk in
И.198) we readily obtain
210 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
We denote the boundary integral terms in D.199) by Bk. It will be assumed
in our discussion that zero is not an eigenvalue for any of the problems w
consider. e
Combining D.199) and D.197), we obtain the hierarchy of ordina
differential equations for Nk, ^
D.200)
where Fk is the Fourier coefficient of F [i.e., Fk = (F9 Mk)]. In the elliptic
case that corresponds to D.9) and for which К = 0 in D.193), the equations
D.200) are algebraic and since kk>0 for all к by assumption, the Nk are
uniquely determined. For the elliptic case corresponding to D.19), D.200) is
a second order equation and the Nk must satisfy the boundary conditions
obtained by setting и = / and и = g in D.196) at у = 0 and у = /, respective-
respectively, in view of D.20).
For the parabolic case, K= dldt and we have for Nk(t),
D.201) ^^ + \kNk(t) = Fk(t) + Bk{t) ,
with the initial condition
D.202) W*@) = (/,MJ,
which is obtained from the eigenfunction expansion of u(x,0)=f(x). The
solution of this problem is
D.203)
Nk(t) = (/, Mk) exp[-\kt] + Jo [Fk(r) + ВД]ехр[-Л*(* - r)] dr.
In the hyperbolic case, with K= d2ldt2 we obtain the initial conditions
from
u(x,0)=f(x)=tNk@)Mk(x)
D.204)
so that at t = 0 we have for Nk{t),
D.205) N,@) - (/, Mk), N'k@) = (*, Mk).
The solution of the equation
FUNCTION EXPANSIONS: FINITE FOURIER
TRANSFORMS 211
D.206) ^*VUf)-W+W
satisfying D.205) is
D.207) Nk(t) = (/, Mk) cos(VV) + y= (g, Mk) si
+ ^L J[ [F,(r) + Вл(т)]вш[лД;(Г - т)] dr .
The foregoing results for the parabolic and hyperbolic cases reduce to those
obtained in Section 4.2 by means of separation of variables when Fk and Bk
are equated to zero.
We note that the solutions и obtained as a series of eigenfunctions
cannot, in general, be expected to be classical solutions that satisfy the
differential equation D.193) and assume the initial and boundary values
pointwise. Thus we observe that each of the eigenfunctions Mk(x) satisfies a
homogeneous boundary condition, whereas the solution и given as a sum of
these eigenfunctions may satisfy an inhomogeneous boundary condition.
Nevertheless, these boundary conditions are certainly accounted for in the
equations D.200) for the Nk. Thus convergence up to the boundary may
have to be interpreted in a generalized sense, perhaps using mean square
convergence. Additionally, although there may be pointwise convergence in
interior regions, the lack of convergence up to the boundary generally slows
the rate of convergence for the series everywhere. Consequently, we present
a method (following Example 4.15) that often enables us to circumvent the
foregoing difficulty by converting the problem to one with homogeneous
boundary conditions.
Example 4.14. Hyperbolic Equations: Resonance. We consider the
inhomogeneous hyperbolic equation
D-208) p(x)utt + Lu = p(x)Mt(x) sin Ш ,
where M.(x) is one of the eigenfunctions determined from D.194) and
•ц *}= ^i(x) sin <ot is a periodic forcing term in t with frequency of
^bration o) = constant. The initial conditions as well as the boundary con-
mons are assumed to be homogeneous; that is, f(x) = g(x) = B(x, t) = 0.
Now
D.209) ркф = (Fj Mk) = sin
8ik = о for i\ф к and 8U = 1 for all L Since (/, Mk) = (g,Mk)
к in this problem, we have from D.207)
212 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED Rt
D.210) Nt(t) = гт= /0 sin(a>T) sinhA^ " т)] dr
and Nk(t) = 0 for кФи Thus the series D.195) reduces to the single te
D.211) u(x, t) = ^Лу [ 77= si
which is, in fact, the solution of the problem. Clearly, D.211) is valid only if
<o Ф yXt. If a> = \Д], the above solution is not valid, but we may obtain the
solution in that case by going to the limit as о>-*\/л) in D.211). Using
l'Hospital's rule we obtain
D.212) u(x, t) =
To inteфret these results, we observe that the numbers a>k = \/Тк (к =
1,2,...) represent the natural frequencies of vibration of the solution of the
homogeneous initial and boundary value problem for D.208) (i.e., with
F = 0) in view of D.43). Thus when w Ф (ot (one of the natural frequencies),
the solution D.211) oscillates with the imposed external frequency со, as well
as the natural frequency &>. = VV The external energy fed into the system is
distributed between these two frequencies. However, if <o — <o( = V^*> we
find that the amplitude of the oscillatory solution increases unboundedly
with t as t—»°°. The effect is known as resonance. It results because the
entire external energy is concentrated within a single natural frequency of
vibration and the continuous input builds up its amplitude. _
Example 4.15. Poisson's Equation in a Circle. The inhomogeneous Lap-
Laplace's equation
D.213) V2m = -F,
is often referred to as Poisson's equation. We consider the boundary value
problem for D.213) in a circle of radius R with center at the origin and an
inhomogeneous boundary condition on the circle. Expressing the problem i
polar coordinates, we obtain for u(r, в) the equation
within the circle (i.e., for r<R) and the boundary condition
EXPANSIONS: FINITE FOURIER TRANSFORMS 213
D.215) «(*••)-Л*)-
view of our discussion following equation D.193), this problem can be
d in terms of the eigenfunctions Mk determined from the problem
*°щ _ _д Uk for r<R with Mk = 0 on г = R. Since this higher dimension-
*envalue problem is not considered until Chapter 8, we present an
^ ative approach to this problem that involves an expansion in terms of
dimensional eigenfunctions.
The solution of D.214)-D.215) is expected to be single valued at any
int within the circle, so we must have
„ЙЬ a similar result for F(r, 0) and f@). That is, w(r, 0), F(r, 6) and f@)
are periodic of period 2ir in 0. This suggests that the eigenvalue problem
associated with Lu = -д2и1дв2 with periodic boundary conditions of period
2тг is appropriate here. Accordingly, we write D.214) in the form
D.217) Ku + Lu = - r2urr - rur + Lu = r2F ,
and expand the solution in a series of eigenfunctions for the operator L. The
foregoing eigenvalue problem was discussed in Example 4.6, where with
/= 7Г we obtained the eigenfunctions {cos кв}, к - 0, 1, 2,. . . and {sin kO},
* = 1, 2,..., and the eigenvalues \k- kz, Jfc = 0, 1, 2,.... Using the
orthonormal set of eigenfunctions as defined in D.104), D.105) we obtain
the eigenfunction expansion
D.218) «(r, S)^~= ao(r) + 2x J* {efc(r) cos *0 + bfc(r) sin M}
which corresponds to D.195). Even though this eigenvalue problem is not of
Murm-Liouville type, the finite Fourier transform method can be applied.
To begin, we note that the Fourier coefficients ak and bk are given in
terms of was
<4-2W) °»
D.220)
°*(Г) = Vм' ^ «* кв) - ^ / "(r- 0) cos(fce> d0 - fc = 1,2,
D.221)
= 1,2,
r) = («. ^ sin кв) = ^ I_ u(r, в)
214 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGm
The equations for the Fourier coefficients ak and bk are obtained f
D.214) on multiplying across by the eigenfunctions and integrating frOm ^
to 7Г, as ~~n
D.222) ^ + 1*Й)-^-д.И, „0,1,2,
D.223) &р + 1»М-*ш-,м, t.ll2
where Ak and Bk are the Fourier coefficients of F(r, в). Since r = 0 is a
singular point in the equations for ak{r) and bk(r) and we require the
solution u(r, в) to be bounded at r — 0, we obtain the boundary condition
D.224) ak(r), bk(r) bounded at r - 0 , к = 0,1,2,
At г = /?, D.218) must represent the Fourier series expansion of f@). Thus,
the boundary condition D.215) implies
D.225) ао(Л)=/о, а*(Л)=Д, bk(R) = fk, *al,
where /0, Д and Д are the Fourier coefficients of /@).
Noting that D.222), D.223) are both inhomogeneous forms of Euler's
equation
D.226) C&r) + \ C'k{r) - ^ Ck(f) = 0 , A: = 0,1, 2,...
for which a fundamental set of solutions is
D.227) Co(r) =
we obtain on using variation of parameters, the following solutions of the
boundary value problems D.222)-D.225):
D.228)
with the bk(f) identical in form to the ak(r) (k > 1) except that Ak(t)
are replaced by Bk{t) and/fc, respectively.
215
JUNCTION EXPANSIONS: FINITE FOURIER TRANSFORMS
Thus the Fourier coefficients in the series D.218) for u(r, 0) are com-
jv specified. Under suitable conditions on F(r, 0), say
^tf« f2(r,0)rdrd0<™ with/@) continuous and piecewise smooth, the
° • r series D.218) can be shown to converge uniformly. Then u(r, 0) is
An interesting special case for the parabolic and hyperbolic problems
. cussed arises when the functions F(jc, t) and B(x, t) are independent of t.
!Sce these terms cause the problems to be inhomogeneous, we say that
* homogeneities are stationary (i.e., time independent). The effect of this
ssumption is that the terms Fk and Bh are time independent constants. As a
result, the integrals in D.203) and D.207) can be evaluated, and we obtain
for Nk(t),
D.230) Nk(t) = Г (/, Mk) ~y(Fk + Bk)\e-Kkt + у [Fk + Bk],
D.231) Nk(t) = [(/, Mk) - f (Fk + 5,Icos(Va;0
L Ak J
+ дт ^ Mk) sin<Vv)+T- [Fk+вк].
V "-к к
in the parabolic and hyperbolic cases, respectively.
In both cases we can introduce the decomposition of Nk(i) in the form
D.232) Nk(t) = Nk(t) + Nk ,
where Nk = (l/\k)[Fk + G^] is independent of /. Correspondingly, the solu-
solution u(x, t) can be formally expressed as
D-233) m(x, 0=S Nk(t)Mk(x) + 2 NkMk(x)
k=l k=\
К we set F(x, f) = F(jc) and B(x, t) = B(x) since F and 5 are independent of
< and define
k=l
us that V^ is the solution of the stationary version of D.193);
he boundary condition (in two or three dimensions)
216 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.236) Ф)ф) + №)
dG
[On solving D.235), D.236) by the finite transform method we oht •
D.234).] ' tain
Then if we put
D.237) u(x,t) = w(x,t) + v(x),
with v(x) defined as above, w(x91) satisfies the homogeneous equation
D.238) pKw
with the boundary condition
D.239) aw
SG=0>
where K=dldt and K=d2ldt2 in the parabolic and hyperbolic cases,
respectively. The respective initial conditions are
D.240) w(x,0)=f(x)-v(x),
D.241) w(x,0) = f(x)-v(x), w,(x,0) = g(x).
Once the possibility of the decomposition D.237) has been recognized, it
may be introduced directly when solving initial and boundary value prob-
problems with stationary inhomogeneities. Since the equation for w(x, t) and the
boundary conditions are homogeneous, the rate of convergence of the
formal series solution of the problem for w(jc, i) is accelerated. Additionally,
if the problem involves only one space dimension, we obtain an ordinary
differential equation for v(x). If the solution for v(x) can be obtained
without the use of eigenfunction expansions, we expect that the solution of
the given problem и = w + и takes a form better suited for numerical
evaluation.
In a similar vein, recognizing that the nonhomogeneity of the boundary
conditions weakens the convergence rate of the eigenfunction expansion
D.195), the following method is occasionally useful. Given the boundary
condition D.11) or D.12), we seek a function V(x, t) such that
D.242)
or
D.243)
*,-**•'>
EXPANSIONS: FINITE FOURIER TRANSFORMS 217
function V(x, t) must be differentiable as often as required in the given
Inferential equations. If such a function can be constructed we set
D244) u(x,t) = W(x,t) + V(x,t),
d find that W(x, t) satisfies the equation
D.245) PKW + LW= pF-pKV- LV ,
with initial conditions of the form D.240) or D.241) (where v is replaced by
V and a term -Vt is added to the second equation in D.241)), depending on
whether the equation is parabolic or hyperbolic. However, the boundary
conditions for W are homogeneous since, for example,
dW
—
dn
dG
= 5-5-0.
Duhamel's principle or the finite transform method can now be applied to
solve for W(x, t).
Example 4.16. The Inhomogeneous Heat Equation. We consider the fully
inhomogeneous initial and boundary value problem for the one-dimensional
heat equation in a finite interval. We have
D.246) ut - c2uxx = g(x, t), 0 < x < /, t > 0
D.247) ф;,0) =/(*), 0<x<l
D.248)
where the functions g(x9 r), f(x), gx(t), and g2(i) are prescribed.
Using linear interpolation we readily construct the function
satisfies the boundary conditions D.248). Then with и = W + V, we
nnd that W satisfies the equation
D.250) Wt-c2Wxx = g{xt)_l
Or 0 < x < I and t > 0 with the initial condition
D-251) щХ9 0) = дх) _ 1 [%@) + (/ _
lI1e homogeneous boundary conditions
218 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.252) W@, t) =
It may be noted that if g(x, t)^0 and the functions gt(t) and g2(A
constants independent of time so that that the inhomogeneities are stati ^
ary, we have °n~
D.253) V(x, t) - V{x) = у [xg2 + (/ - x)gl].
The function V(x) is a solution of the stationary form of the heat equatio
uxx = 0 with the boundary conditions w@, t) = gx and u(/, t) = g2. However
this does not mean that there is an equivalence between the two decomposi'
tion methods given above in all cases in which the given equation is
homogeneous and the boundary conditions are stationary. For example
another possible choice for V(x, t) in the above problem is
D.254) V(x9 0 = 5^7 {(«nh x)g2(t) + [sinh(/ - x)]gl(t)} ,
and this does not reduce to a solution of the stationary case if g1 and g2 are
constant.
4.7. NONLINEAR STABILITY THEORY: EIGENFUNCTION
EXPANSIONS
In this section we consider a nonlinear heat conduction equation in one
dimension and study the stability of the equilibrium or zero temperature
distribution. That is, we wish to determine if small perturbations at the
initial time around the zero temperature distribution decay to zero as /
increases or grow, perhaps, into a new stationary solution with increasing t.
We are concerned with a problem in a finite interval with the ends of the
(insulated) rod being kept at zero temperature for all time. However, to
compare results of the present stability analysis with that carried out in
Section 3.5, we briefly consider the stability problem for a rod of infinite
extent.
Let u(x, t) represent the temperature distribution of an insulated rod and
assume that there is a nonlinear heat source of strength -ЛмA - и ). Then
the (nonlinear) heat equation to be studied has the form
D.255) ut - uxx = AmA - u2) ,
where A is a parameter that depends on the properties of the rod. Since we
are interested in the development in time of small initial perturbations
jfljNEAR STABILITY THEORY: EIGENFUNCTION EXPANSIONS 219
und the equilibrium solution uo(x, t)=0 of D.255), we introduce the
Sal condition
a.256) u(x> °) = €h(x)'
here Л(х) is uniformly bounded and 0 < € <§ 1.
W We are interested in two problems. The basic problem is that in which the
od is of finite extent (it is assumed to have length тг), and the boundary
conditions at the end points x = 0 and x = тг are
D.257) и@,/) = 0, и(тг,0 = 0.
In this case the equations D.255), D.256) are valid in the interval 0 < x < тг.
The other problem deals with a rod of infinite extent so that D.255), D.256)
are given over the interval -oo<jc<oo. Since the data for either problem
are small (of order of magnitude e), we look for a solution of each problem
in the form
D.258) u(x, t) = ew(x, t),
and trace its evolution in time. If this perturbation around the solution u0 = 0
remains small as t increases, the zero solution is stable. However, if w(x, i)
grows without bound as time increases, the equilibrium solution is unstable.
Since we are unable to solve either of the above nonlinear problems exactly,
we use approximate methods for studying the stability of the equilibrium
solutions of the above problems.
Inserting D.258) into D.255) and dividing by e gives
D.259)
We first carry out a linear stability analysis. Assuming the solution и - ei
stable so that it does not grow without bound and noting that e2 <tl, we
linearize the equation D.259) by dropping the term proportional to w\ The
resulting linear equation is
= ew is
we
*,-*„ = Aw,
and we perform a stability analysis of D.260) with the initial and boundary
conditions carried over from the nonlinear problem. If the solutions of the
Jnear problem are stable, our assumption leading to the neglect of the
nonlinear term in D.259) is valid for all time. We then say that the nonlinear
Problem is stable and is well approximated by the linearized version of the
Problem.
However, if the linearized problem exhibits instability, the solution
v*> 0 grows in time so that, eventually, the term Ле2и>3 may attain an order
220 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
of magnitude equal to that of the term w. Consequently, the nonlinear
in D.259) cannot be neglected and the full nonlinear equation must be
Although the linear stability analysis predicts that the equilibrium solu^
u0 = 0 is unstable and that the perturbation grows without bound, a 1СП
linear analysis may show that, in fact, the perturbation grows only
h th ilibi lti Thi i d
y y , , p grows only umv
reaches another equilibrium solution. This is demonstrated for the iJ ^
and boundary value problem for D.255). ltla*
We begin by considering a linear stability analysis of the Cauchy probl
for D.255). Thus we study the normal mode solutions 6m
D.261) н<х, 0 = a(k) exp[ikx + X(k)t]
of the linearized equation D.260). Inserting D.261) into D.260) gives
D.262) A(Jt) = Л - к2 .
Since A is a real constant, the stability index ft defined as in Section 3.5 is
given by ft- A. Thus D.260) is strictly stable if A<0 and is unstable if
A > 0. For A > 0, the linear stability analysis predicts unbounded growth for
the perturbation ew and indicates that the equilibrium solution мо = 0 is
unstable.
To see that the perturbation ew need not necessarily grow without bound
but may reach another (stable) equilibrium solution, we proceed as follows.
We first note that ux = 1 is a solution of D.255). To check for stability of this
(additional) equilibrium solution we set
D.263) u(x, 0 = 1 + ev(x, t),
with 0< € < 1 and analyze the linearized equation for v(x, t); that is,
D.264) vt-vxx =
For the normal mode solutions D.261) (with w replaced by v) we now
obtain
D.265) A(A:) = -2A-A:2
so that the stability index П = -2A. Thus for A > 0, in which case the linear
stability analysis showed the zero (equilibrium) solution u0 = 0 to be un-
unstable, we now have stability for the equilibrium solution ux = 1. This
suggests that perturbations of zero solution do not grow unboundedly bu
grow only until they reach the stable equilibrium ux = 1. Although we do no
prove this is the case in general, we exhibit an explicit solution of a Cauchy
problem for D.255) where this is exactly what happens.
Let h(x) = \ in D.256) and look for a solution of D.255) that is
STABILITY THEORY: EIGENFUNCTION EXPANSIONS 221
«dent of x; that is, и = U(t). Then U(t) satisfies the ordinary differen-
266) U'(t) = \U(t)[l-U(tJ),
with the initial condition
D.267) "<°> —
equation D.266) can be solved by separation of variables and the
Station of D.266), D.267) is
For small values of f, we may expand this solution in powers of б and find
that и ~ eekt = ew(x, t), where w(x, t) satisfies D.260) and w(x9 0) = 1. As
,_*oo, w(x, O-^00 if A > 0, but the solution D.268) does not tend to infinity
but approaches the equilibrium solution ut = 1 as is easily seen. If A < 0, the
term e2kt in the denominator of D.268) tends to zero as *-►*>. Thus since
e2kt - 1« 0 for small t and e2 < 1, we conclude that the solution w(jc, f) given
in D.268) is well approximated by ew(x, t) = eekt for all time. That is, the
linearization procedure is valid for all time if the linearized problem is
stable.
Turning now to the consideration of the initial and boundary value
problem D.255)-D.257) in the interval 0<д:<тг, we begin by applying
linear stability theory. We look for normal mode solutions of the linearized
equation D.260) that satisfy the homogeneous boundary conditions D.257).
Although D.261) yields solutions of D.260) for all real k, to find solutions
that vanish at x = 0 and x = тг we must take linear combinations of the
normal modes and restrict the values of k. We obtain
D-269) Wn(x91) = hn exp[( A - n2)t] sinnx, n = 1,2,... ,
where hn is a constant and we have set к = n since w(x, t) vanishes at x = 0
and x = тг for these values of k. [The solutions D.269) are precisely what
results if separation of variables is applied to D.260) with the boundary condi-
conditions D.257)].
Hie solution of the initial and boundary value problem for the equation
I 260) with the conditions D.256), D.257) is a linear combination of the
| °rmal mode) solutions D.269). Thus the stability properties are de-
ttmmed from the discrete set of solutions wn(x, t) given in D.269), rather
an from the continuous set D.261) with -« < к < oo, which is relevant for
he Cauchy problem. We see that if A < 1, the wn(x, t) in D.269) all decay to
222 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
zero as *->», whereas if A > 1, at least one of the wn(x, t) grows e
tially as f-юо. Consequently, the linear stability analysis predicts th
equilibrium solution wo = O is unstable to small perturbations if д^
whereas if A < 1, it is stable and all perturbations eventually die out *'
increases. Again, if A>1, the linear analysis that predicts instability к '
comes invalid after a finite time since the nonlinear term in D.259) beco ^
equally important with the linear term in view of the growth of ц>(х ?
Therefore, a nonlinear analysis of the growth of the perturbation te
u = ew must be carried out.
The values A = 0 and A = 1 given for the Cauchy and the initial and
boundary value problems for the linearized equation D.260) determine the
threshold of instability (in the linear theory) for the given problems. These
values of A are denoted by Ac, the critical value of A. That is, if A > л , we
have instability and if A < Ac, we have stability. It should be noted that the
critical value is determined not only from the given (linearized) equation but
also by the boundary conditions if any are given. As shown above, the
critical values Ac differ in the case where no boundary conditions are
assigned and in the case where boundary conditions are given.
The initial and boundary value problem D.255)-D.257) cannot be solved
exactly. There are several approximate methods for analyzing this nonlinear
problem, but we consider only one approach which is based on eigenfuno
tion expansions or equivalently finite Fourier transforms.
The operator L = ~d2ldx2 in D.255), together with the boundary condi-
conditions D.257), is associated with the set of eigenfunctions
-f-
' IT
D.270) MJx) = V - sin kx , к = 1,2,.. .
[i.e., LMk = XkMk with Xk = к2 and Мк@) = Мк(тг) =0]. The normalized
set {Mk(x)} is complete and we represent the function w(x, t) (recall that
и = ew) in the form
where
D.272) Nk(t) = (w,Mk).
Using the initial condition D.256), we have
D.273) Nk@) = (h,Mk).
[The inner product (/, g) = J^ fg dx in this problem.]
We obtain an equation for Nk(t) by taking a finite sine transform l
ftfEAR STABILITY THEORY: EIGENFUNCTION EXPANSIONS 223
coy
y that is, we multiply by Mk(x) and integrate between 0 and тт. This
gives
274) N'k(t) + k2Nk(t) = ANk(t)-Ae2jo w3Mk(x) dx , к = 1,2,....
The integral term yields
D.275) /о" "X Л " Г [2 Щ)М,(х)\ Мк(х) dx
*,/■*-1
where the coefficients aj^ are obtained by cubing the series and integrating
term by term. Thus we obtain the infinite system of coupled equations for
Nt(t),
D.276) N&t) + (k2 ^^
/,/,/ = 1
with the initial conditions D.273).
Neglecting the terms of order e2 in the equations D.276) is equivalent to
linearizing the problem and has the effect of uncoupling the equations. Then
the terms Nk(t)Mk(x) have the form D.269) where к is replaced by n. We
again observe that for A < Ac == 1 all the Nk(t) in the linearized case tend to
zero as f-нюо, whereas for A>AC = 1, at least one of the Nk(t) grows
exponentially in t.
We wish to examine the behavior of the solution if A is slightly larger than
the critical value Ac; that is, A «1 but A > 1. Then к2 - A > 0 for к > 2 but
l-A<0. It is of interest to study the behavior of the solution in the
neighborhood of the critical value where according to the linear theory a
transition from stability to instability takes place.
Since for к > 2, к2 - A > 0, we take as a first approximation
*a* is, we neglect the terms of order e2 in D.274) with k>2. A similar
^PProximation in the equation for Nx(t) leads to an exponentially growing
Я*щ of the form D.277) with k = l. Inserting these expressions for Nk(t)
| "-1,2,. ..) into the series on the right side of D.276) shows that all the
^ in the sum decay exponentially except for the terms af^Nl(t).
e linearized form of the equation for N^t) implies it is exponentially
f T^n8' whereas all other Nk(t) decay exponentially, so we retain the term
€ amNl(t) in the equation for Nx(t) and obtain
224 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
D.278) N[(t) + A - X)Nx(t) + \€2a[\\Nl(t) = 0 .
For the coefficient a\\\ we have from D.275)
D.279) a[\\ = [ [ЛЗД]4 dx = 4j [ sin4 x dx = ^ .
Thus N^t) satisfies
D.280) N[(t) + A - А)ВД + |£ €2N\(t) = 0.
To solve D.280) we multiply across by N^t) and obtain
D.2Я,
which is a Riccati equation for AfJ. The solution of D.281) satisfying
ЛГДО) = (h, MJ is found to be
D-282) "■"■p.^iiJ^t-HK^-flf
For small values of t this expression reduces to
D.283) Nx(t) ~ (A, Mx)e{k~l)t,
which is the solution of the linearized equation, as was expected. However,
as t—»oo? since A — 1 >0 we have
so that N&) tends to a finite (stationary) value, rather than growing without
bound as predicted by the linear theory.
We remark that the above results are self-consistent since we now find
that the terms a{£[Nl(t) in the equations D.276) for к > 2 are uniformly
bounded for all t and it is correct to approximate Nk(t) as in D 277) since
the right side of D.276) is of order e2.
Thus we obtain from D.271)
„(,,,) = «^,0 = «i|«.<0M,w-|^I(^)[bi]'"sin,
as *-*oo in view of D.277) and D.284). We assume that (A, Ma)^° and
FOR CHAPTER 4 225
285) shows that the solution at large time depends only on the sign (i.e.,
(h M\)l\(h> M*)\) of the leadinS term of the Fourier series of M(*' °) and
t °n its magnitude. The large time behavior of the solution is independent
Пг, so that the solution u(x, t) with AsAc does not grow unboundedly as
° edicted by the linear theory. Instead, it approaches a steady state whose
^proximate description, (correct to order e) is given in D.285). We note
h t D.285) is an approximate solution of the stationary form of D.255)
(i e., with w, = 0) when Л « kc = 1.
The foregoing approximation method may be extended to deal with
nonlinear heat sources of the form A/(u), where /@) = 0 and Дм) has a
Taylor expansion around и = 0. Also, hyperbolic and elliptic equations with
similar types of nonlinearities can be treated by this method. Some examples
are presented in the exercises.
EXERCISES FOR CHAPTER 4
Section 4.1
4.1.1. Derive the heat or diffusion equation D.5) from a "balance law"
appropriate to the one-dimensional case.
4.1.2. Show that if p = cp, where с is a constant, the equation
can be brought into the general form D.10) if we set u = exp[at]v and
choose a appropriately.
4.1.3. Carry out the transformation of Exercise 4.1.2 for the telegrapher's
equation
where Л is a positive constant.
Section 4.2
4-2.1. Verify the result D.28) in the two-dimensional case if L = -V2,
«Г 1-~~ Г ' W " ^ ~ rfy where fl = *2 + У2> tf the ^gion G is the interior of
e Clrcle r = 1. Note that both и and w vanish on the boundary r = 1.
•2-2. Let u = (i-rJ and vv = cos7rr, so that duldn and dwldn both
anish on the circle r = 1. With L - -V2 and G given as the region r < 1,
enry that the integral over the region G in D.28) vanishes.
.2.3. With R2 - x2 + y2 + z2, let if = exp[l - R] and w - (R - 1J? so that
. /dn + w and (?w/^n + w both vanish on the sphere R = 1. If G is the
tenor of the unit sphere (i.e., R < 1) and L = - V2 + 10, show that D.28)
valid for the above functions.
226 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
-'4)
4.2.4. If the region G is the unit square O<jt<l,O<y<l, and u(x, y\ ^
xy{\ - x)(l - y) so that u(x, y) vanishes on dG, show that D.38) is satisfy
II Li == V .
4.2.5. Show that A = 0 is an eigenvalue and M(x) = с = constant is
eigenfunction for the eigenvalue problem D.26) if and only if q = о in tj*n
operator L and a = 0 in the boundary condition. e
Hint: Use D.37).
4.2.6. Show that if A = 0 occurs as an eigenvalue when the elliptic proble
D.19) is solved by separation of variables, with the boundary conditio
D.20) replaced by uy(x90) = f(x) and uy(x91) = g(x), the problem has no
solution unless / and g satisfy compatibility conditions. (Use the results of
Exercise 4.2.5 to show this.) Conclude that the solution is not unique if the
compatibility conditions are met and verify that these results are consistent
with those obtained in Exercise 3.4.9.
4.2.7. Show that МДх, у) = 1 and Mk(xf y) = cos тгх cos iry are both eigen-
functions for the problem
with the boundary condition dMIdn =0 on the unit square. Determine the
corresponding eigenvalues and verify directly that the orthogonality condi-
condition (Mp Mk) = 0 is satisfied.
4.2.8. Show that M = 1 is an eigenfunction for the problem
in the region G with dM/dn = 0 on dG. Determine the corresponding
eigenvalue and explain why the occurrence of a negative eigenvalue does
not contradict D.42).
4.2.9. Assuming the eigenfunctions Mk{x) are uniformly bounded, interpret
the fact that m(jc, i) as given in D.52) vanishes as *-><*> if the Kk are all
positive, in terms of the physical significance of the heat conduction problem
and its boundary conditions. Consider also the special case referred to in
Exercise 4.2.5 which yields A = 0 as an eigenvalue.
4.2.10. Consider the two-dimensional eigenvalue problem in polar coordi-
coordinates r and 0,
in the region r<l with the boundary condition (a) M(/, в)-^, (b)
M,(/, 0) - 0, or (c) Mr(/, в) + hM(l, 0) = 0(h> 0). By expressing the Lapla-
cian in polar coordinates and assuming that M is independent of 0, show tha
M(r) satisfies Bessel's equation of order zero. In the general case, with
M = M(r, 6), set M = Mx(r)M2(B) and use separation of variables to show
227
e%eKcises for chapter 4
if M(ry в + 2тг) = M(r, 0), the function Mx(r) must satisfy a Bessel
in of integral order. For each of these eigenvalue problems we
that the solutions be bounded at r = 0 (see Example 4.7 for a further
discussion).
2 11. Express Laplace's equation V и = 0 in spherical coordinates (r, 0, <p)
•th rsO, О^0^2тг, and 0<<р<тг (see Example 8.3). Assume u =
*/> <p) (i.e., w is independent of 0) and set и = N(r)M(<p). Use separation
of variables to show that M(<p) satisfies the equation
- — [sin <pM f(<p)] = A sin <pM(cp),
with A equal to the separation constant. Let x = cos <p and conclude that
$(jt) = M(<p) satisfies the Legendre equation. If we require that M(x) be
bounded at x = -1 and x = +1, we obtain the eigenvalue problem discussed
in Example 4.8.
4.2.12. Given any two functions / and g for which the inner product and
norm are defined as in D.33) and D.34), show that they satisfy the
Cauchy-Schwarz inequality
l(/.*)l* И/И 1Ы1-
Hint: Expand out the non-negative integral S fo P(f+ aSJ du> where a is a
constant, as a quadratic expression in a and show that the discriminant must
be negative or zero.
4.2.13. Suppose that q(x) in the operator L [defined in D.6)] can have
negative values, but that q > q, with q as a negative constant. Also, let
ppc) > p, with p as a positive constant. Put a == qfp, and define the operator
Lu = Lu- apu. For functions и that satisfy D.14), show that (l/p)L is a
positive operator. Since the eigenvalue problem for L must yield only
non-negative eigenvalues, conclude from the fact that the eigenvalues have
no finite limit points, that the eigenvalue problem for L, with q{x) given as
above, can have only a finite number of negative eigenvalues.
4.2.14. Consider the operator L defined as Lu = Lu + b(x) • Vm where Lu
has the form D.6) and b(x) is a given vector function. Determine the formal
a4Oint L* of the operator L. Also, determine the boundary conditions that
w must satisfy so that the integral of wLu-uL*w over the region G
vanishes, if и satisfies a homogeneous boundary condition of the (a) first
kln<i, (b) second kind or (c) third kind on dG.
•z-i5. Consider the eigenvalue problem LM = ApM, for each of the three
. °undary conditions determined for L in the preceding exercise. Proceed as
n the text and use the conclusion of the preceding exercise to show that
^enfuctions for the eigenvalue problem for L and the adjoint eigenvalue
for L* are orthogonal if they correspond to different eigenvalues,
two sets of eigenfunctions are referred to as a biorthogonal set.
228 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED ]
Section 4.3
4.3.1. Show that (with p(x) = 1), the function <p(x) = 1/Vx is integrable
an improper integral) over the interval 0<x<l, but that it is not son
integrable. ч are
4.3.2. Show that (with p(x) = 1), the function <p(x)9 defined to equal 1 ^
x is a rational number and to equal -1 when x is irrational, is not integral
over any finite interval but is square integrable. e
4.3*3. Determine that the set of square integrable functions фк(х)-
sinGrfclogje/log2), fc&l, is orthogonal over the interval 1<jc<2 with
respect to the inner product with weight function p(x) = llx. Obtain the
norms of these functions and construct an orthonormal set.
4.3.4. Show that the set of complex valued functions фк(х) = ехр[йЫ
k~0y ±1, ±2,..., is orthogonal with respect to the Hermitian inner
product D.62) over the interval - тт < x < 7r, if the weight function p(x) = l.
Determine the norms of the фк(х) and construct an orthonormal set.
4.3.5. Obtain the Fourier coefficients of the function <p(x) = x with respect
to the orthonormal set of functions obtained in Exercise 4.3.3.
4.3.6. Adapt the argument given following D.32) to show that the boundary
terms in D.71) vanish.
4.3.7. Conclude from D.77) that if A = 0 is an eigenvalue of the (regular)
Sturm-Liouville problem, the corresponding eigenfunction v(x) must satisfy
[pvt2 + qv2] dx~pvvf =0,
and we must have qv2(x) = p[v'(x)]2 = 0, since the boundary contributions
are non-negative. Noting that v(x) cannot vanish identically, show that
q(x) = 0 and u(x) = constant. This must, in turn, imply that ax = аг^Ь,
since the boundary conditions could not otherwise be satisfied.
4.3.8. Suppose there are two linearly independent eigenfunctions v(x) and
w(x) that correspond to the eigenvalue A. Since they both satisfy the same
equation and boundary conditions, show that we must have
aiv@) ~ fav'@) = 0 , axw@) - Pxw'@) = 0 .
Noting that ax and ft cannot both vanish, show that the determinant of the
above system, which equals the Wronskian of v and w evaluated at zero,
must vanish. Thereby, conclude that v(x) and w(x) are linearly dependent,
so that each eigenvalue must be simple.
4.3.9. Consider the transcendental equation D.88) for the determination of
the eigenvalues in Example 4.3. Let A = p2 and show that the equation can
be written as
FOR CHAPTER 4 229
P
x = tan(p/) and x = pfoft + a2)81)/(p2j81j82 - a^). Graph both func-
s in the (p, x)-plane and show thereby that there are an infinite number
^eigenvalues kk and that Лл-»оо as £-><*>. Show that for large k,
t2
j*3 10. Obtain the equation that corresponds to D.88) if we put A = 0 in
' 054 show that zero is an eigenvalue only if at = a2 = 0 in the boundary
conditions D.57) and find the corresponding eigenfunction.
i 3 11. Use D-56) and integration by parts to show that if v(x) satisfies the
same boundary conditions as the eigenfunctions vk(x) and is sufficiently
smooth, the Fourier coefficients of v have the form
It can be shown that Xk^c2k2 as fc—><*> for any (regular) Sturm-Liouville
problem. (The constant с may be different for each problem.) Conclude
thereby that if \vk(x)\ and \(vk,(l/p)Lv)\ are uniformly bounded, the
Fourier series D.79) converges uniformly and absolutely.
Hint: Use the Weierstrass M-test.
4.3.12. Verify the results of Exercise 4.3.11 for the case of the Fourier sine
series assuming v(x) has two continuous derivatives and v@) = v(l) = 0.
4.3.13. Determine the eigenvalues and eigenfunctions of the problem
Show that the properties 1-4 in the text are valid for this problem and
normalize the set of eigenfunctions.
4.3.14. Determine the eigenvalues (approximately) and the eigenfunctions
of the problem
v" + Xv - 0 , v@) = 0 , v'(/) + fiv(l) = 0 ,
where 0>O. Show that the properties 1-4 in the text are satisfied and
normalize the set of eigenfunctions.
•15. Obtain the eigenvalues and eigenfunctions of the proble
•lem
[A + xJv']' 4- Аи = 0 , v@) = v(l) = 0 .
1ГУ the properties 1-4 in the text and normalize the eigenfunctions.
nt: Let t = 1 + x to transform the given equation into an Euler (homoge-
(homogenous) equation.
230 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGln
4.3.16. Obtain the Fourier sine and cosine series of the following fUncti
given over the interval 0 < x < L °ns
(a) v(x) = x.
(b) ф) = *(*-/).
(c) ф) = х2B
()K) (^)
Construct the even and odd extensions of each of the above functions
Compare the rates of convergence for each of the Fourier series based o^
the smoothness properties of the even and odd extensions.
4.3.17. Obtain the complete Fourier series expansion of the function v(x) =
sin2(irx/l)(x -1)\ defined in the interval -l<x<L Discuss the converge
ence properties of the series.
4.3.18. With А = АЛв, i; = /jVX~r] and f = VX~t in D.112), multiply
across by tJ'n and integrate from 0 to VA^/ to obtain
"Jo Jt^'^^^i
and conclude that
f - n2){
(a) In the case where v(t) — 0 determine the normalization constant
D.116) by using the formula /; = -Jn+l + (n/x)Jn.
(b) Determine the eigenvalues A^ and the normalization constant
for the eigenfunctions if the boundary condition is v'(l) = 0-
(c) Determine the eigenvalues kkn and the normalization constant
for the eigenfunctions if the boundary condition is i/@ +
hu(l) = 0, with positive Л. [In parts (b) and (c) the eigenvalues
are given in terms of zeros of equations that involve Bessel
functions and/or their derivatives.]
(d) Show that zero is an eigenvalue only if n = 0 and the boundary
condition is ^'(O^0- Determine the normalized eigenfunction
in that case.
4.3.19. Given the Taylor expansion
show that the first three (positive) zeros of J0(x) are (approximately)
alo = 2.41, a20 = 5.52, and азо = 8.65. [Note that the series for Jo(*) iS
alternating.]
FOR CHAPTER 4 231
i 20. Use the asymptotic formula
cos(VAx- § n - f) ,
hich is valid as \/Лл:-»оо, to show that the large eigenvalues for Example
4 7 are approximately given as kkn «(irkflJ.
13 21 Expand the function v(x) = 1 in a series of eigenfunctions ул 0 as
len in D.117), D.118).
Hint: Use the formulas in Exercise 4.3.22.
4 3.22. Expand v(x) = /2 - x2 in a series of eigenfunctions vk0(x) as given in
D 117) D.118). Evaluate the Fourier coefficients by using integration by
parts and the formulas [xnjn(x)]f = xnJn_x(x) and /_„(*) = (-lR(x).
4.3.23. Obtain the first five Legendre polynomials P0(x),. . ., P4(x) and
show that they are mutually orthogonal.
4.3.24. Use the derivative definition of the Legendre polynomials Pk(x)
given in D.122) (this is known as Rodrigues' formula) to obtain the
normalization constants D.123).
Hint: Integrate by parts as often as necessary.
4.3.25. Expand the following functions in a series of Legendre polynomials.
(a) v(x) = x.
(b) u(x) = 5 - Ax 4- IOjc3.
(c) v(x) = x\
4.3.26. Determine the eigenvalues and eigenfunctions for the problem
/ + Ay = 0, 0<x</, with y'@) + y@) = 0 and ;y@ = 0. Note that the
boundary condition at x = 0 does not have the required form D.15) and
show that there is one negative eigenvalue. (Show by graphical means how
the eigenvalues are distributed.)
4.3.27. Determine the eigenvalues and eigenfunctions for the eigenvalue
problem
/ + 2;у' + A + Л);у-0, 0<х<тг, /@) = /(тг) = 0.
(Note that this is not a Sturm-Liouville problem since the equation is not
^lf-adjoint. It can be reduced to self-adjoint form via a transformation of
tfte dependent variable.)
•3-28. Show that the eigenvalue problem
*"-2/ + (i + A)>, = 0 , 0< jc < 7Г , /@) = 2^@) , y'(ir) = 2y(ir)
^ the adjoint problem for that given in Exercise 4.3.27, in the sense of
^rc* 42
p g ,
a ^ 4.2.14. Show that the eigenvalues for both problems are the same
^at *e eigenfunctions for the original and the adjoint problem are
hogonal if they correspond to different eigenvalues.
232 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
Section 4.4
4.4.1. Determine the motion of a plucked string. That is, find a soluti
u{x, t) of the wave equation D.126) with w@, i) = u(l, t) = 0 and the miti*
conditions
h
-x, 0<x<a
I l-a '
4.4.2. Solve the initial and boundary value problem for the wave equation
D.126) with w@, t) = u(/, t) = 0 and the initial conditions
/W = 0 , g(x) = S(x - a) , 0<a<l,
where 8(x) is the Dirac delta function. This yields the motion of a string due
to a point impulse at x = a administered at the time / = 0.
4.4.3. Solve D.126)-D.128) if
4.4.4. Solve D.126)-D.128) if
4.4.5. Express the solution of Exercise 4.4.1 in the form D.150) and use this
result to describe the displacement u(x, i) of the string at various times t > 0.
4.4.6. Use separation of variables to obtain the general form of the (series)
solution of the problem
utt - c2uxx = 0, 0<Jt</,f>0,
with the initial conditions
u(x,0)=f(x), «,(*,()) = £(*),
and the boundary conditions
4.4.7. Obtain the solution of the problem in Exercise 4.4.6 if f{x) = 0 an
g(x) = 1. Interpret the result.
4.4.8. Solve the problem of Exercise 4.4.6 if
f(x) = x\x-l)\
EXEKCISES FOR CHAPTER 4 233
4 9, Using separation of variables solve the following initial and boundary
value problem
utt - c\t = 0,
и(х,0)=Дх),
4 4 10. Solve the problem in Exercise 4.4.9 if
f(x) = x{x-l)\ g(x) = 0.
4.4.11. Apply the method of separation of variables to solve the wave
equation D.126) with the initial data D.128) and the boundary data
и@, 0 = 0, ux(l, t) + /Зи(/, /) = 0, P >0 .
4.4.12. Obtain the solution of the problem in Exercise 4.4.11 if the initial
data are
4.4.13. Multiply the wave equation D.126) by ut and use the identity
uxxut = (utux)x - \{u2x)t to obtain
iW + <4i-«'i[Mj-e.
Integrate over the interval 0 < x < I and show that if either и or ux vanishes
atjr = O and x = l, the energy integral D.143) is a constant (recall that
T~cp). Obtain an appropriate energy integral if ux@, t)-13^@, t) = 0
and «,(/, t) + j82m(/, 0 = 0 with ft, j82 > 0.
4-4.14. Verify the result D.145).
•4.15. Use separation of variables to solve the following problem for the
telegrapher's equation
0, 0<x<l,t>0,
whereA>0, with
°w that
the solution v(x, t) tends to zero as f-»oo.
234 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGlo
4.4.16. Solve the initial and boundary value problem for the Klein-Gord
equation °n
un ~ У 2uxx + c2u = 0 , 0 < x < /, t > 0 ,
Use separation of variables.
4.4.17. Solve the heat equation D.151) with the boundary data D.152) and
the initial data
u(x,0)=f(x) = x(x-l).
4.4.18. Solve the problem D.151)-D.153) if
where 8(x) is the Dirac delta function.
4.4.19. Apply separation of variables to solve,
u,-c2uxx = 0, 0<x<l,t>0,
u(x,0)=f(x),
The boundary conditions imply that no heat escapes through the ends of the
rod. Show that as t—»°° we have
1 f
lim u(x, t) = 7 I f(x) dx .
Г-><ю / JO
This represents the average of the initial temperature distribution.
4.4.20. Solve the problem in Exercise 4.4.19 if
(a) /(*) = *.
(b) f(x) = sin2(irx/l).
4.4.21. By integrating the heat equation D.151) over the interval 0 < x < U
show that if ux@91) = ux(l, t) = 0, we have /p u(x, t)dx = constant. Explain
why this is consistent with the result in Exercise 4.4.19 regarding the limit of
u(x, i) as /->oo.
4.4.22. Use separation of variables to solve the following problem for the
heat equation
и, - c2uxx = 0, 0<*</,*>0,
и(*,0)=Л*),
м@, 0=0, ux(l, t) + /3u(/, t) - 0 , p >0 .
X£RCISES FOR CHAPTER 4 235
^e boundary condition of the third kind at x = / results from Newton's law
f cooling if there is convective heat exchange between the rod and a
edium adjacent to the rod at x = / that is kept at zero temperature.
4 23. Solve the problem in Exercise 4.4.22 if
a 4 24. И the lateral portion of a rod undergoes convective heat exchange
with a medium kept at zero temperature, the temperature u(x, t) in the rod
satisfies the equation
ut - c2uxx + a2u = 0 , 0 < x < /, t > 0 .
Assume that w@, f) = м(/, f) = 0, и(х, 0) = /(x), and that a = constant. Find
the temperature u(x, i) by the method of separation of variables. Obtain the
limit as t->°° of the temperature u(x, t).
4.4.25. Use separation of variables to solve Laplace's equation V2w = 0 in
the rectangle 0<x</, 0<y<l, with the boundary conditions of the second
kind
uy{x> 0) - f{x), uyix, t) = gix) , 0 < x < I.
Determine conditions on fix) and gix) for a solution to exist. Discuss the
rate of convergence of the solution.
4.4.26. Solve the following boundary value problem for Laplace's equation
«хх + и^^О, 0<x</,0<>></,
uyix90)=fix), и,(х, t) = gix), 0<x</ .
4-4.27. Solve Laplace's equation V2m = 0 in the rectangle 0<x</, 0<y<l
Wlth the boundary conditions
40, у) - 0, их(/, у) + Kм(/, у) = 0 f /3 > 0 ,
•4-28. Let Дх) = xix - I) and gix) = 0. Solve the following boundary value
Problems
(a) Example 4.11.
(b) Exercise 4.4.25.
(c) Exercise 4.4.26.
236 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED REGta
4.4.29. Consider Laplace's equation V2u = 0 in the unbounded region n
*</, y>0. Let ^
u@,y) = u(hy) = 0, y>0,
as well as requiring that u(x, y) be (uniformly) bounded as y^oo f0
0<x</. Show that the solution of this problem can be obtained bv
separation of variables. Find the limit of u(x, y) as y^>°°.
4.4.30. Apply separation of variables to solve the boundary value problem
for the elliptic equation
uxx + uyy ~ c2" = ° > 0 < x < /, 0 < у < I
in the given rectangle, with the boundary conditions
4.4.31. Use separation of variables to show that the reduced wave equation
with the (homogeneous) boundary condition u(x, y) = 0 on x = 0, x = /,
у = 0, у = /, can have nonzero solutions for certain values of k. Determine
these values of к — kn. (They correspond to the eigenvalues for Laplace's
equation in a rectangle.) By applying Green's theorem [see D.31)] conclude
that if к - kn, the Dirichlet problem for the reduced wave equation has no
solution unless the boundary values satisfy a compatibility or orthogonality
condition. [Let w = wn be a solution of V2wn + k2nwn - 0, wn = 0 on the
boundary of the rectangle, in the application of D.31)]. Show that if a
solution does exist for к = кп, it is not unique.
4.4.32. Solve the initial and boundary value problem for the two-dimension-
two-dimensional wave equation in a disk,
ив-с2[«„+ «„]=<>, x2 + y2<l2, t>0,
with the initial conditions
u(x, y, 0) = f{x, y), u,(x, y, 0) = g(x, y)
and the boundary condition
u(x,y,t) = 0, x2 + y2 = l2, t>0.
FOR CHAPTER 4 237
Use separation of variables and the results of Exercise 4.2.10 and Example
4 7 (The solution of this problem describes the vibration of a circular
ernbrane with a fixed edge if c2 = Tip, where T= tension and p = density.)
4 4 33. Solve the initial and boundary value problem for the heat equation
in a disk,
ut-c2[uxx + uyy] = 0, x2 + y2<l\t>0,
with the initial condition
u(x, y,0)=f(x, v),
and the boundary condition
Use separation of variables and see Exercise 4.4.32.
4.4.34. Express the Laplacian in cylindrical coordinates (r, 0, z) and solve
the Dirichlet problem for V2w = 0 in the finite cylinder 0< r < #, O<0<
2тг, 0 < z < I with u(R, 0, z) = f(z) and w(r, 0,0) = и(г, 0, /) = 0, by looking
for a solution in the form и — w(r, z).
Hint: The solution must be bounded at r — 0 and involves the modified
Bessel function IQ(x).
4.4.35. Solve the problem of the preceding exercise if the boundary condi-
conditions are replaced by u(r, 0,0) = 0, w(r, 0, /) = g(r) and u(R, 0, z) = 0, by
looking for a solution in the form и = u(r, z).
Hint: Use Example 4.7.
4.4.36. Express V2w = 0 in spherical coordinates (r, 0, <p) with r >: 0, 0 < 0 <
2тг, and 0 < <p < 7Г. Consider the boundary value problem for V2w = 0 within
the sphere r < I for the function и = w(r, 0, 9) with the boundary condition
Oe., the boundary values are independent of в). Look for a solution in the
torm и ~ U(ry <p) and use separation of variables to construct an eigenfunc-
ion expansion of the solution in terms of Legendre polynomials (see
bcercise 4.2.11 and Examples 4.8 and 8.3).
Section 4.5
•s-l. Solve the initial value problem
Utt - C2UXX = g(x, t), -oo<X<oo, f>0
with «(*, 0) = ut(x, 0) = 0 if g(x, t) has the following form
238 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED
(a) g(x, t) = (sin x)e~\
(b) g(x, t) = u
(c) g(x, 0 = S(x) cos L
(d) g(x, 0 = S(x - a)S(f - ft) , a, ft = constant, ft > 0.
In (c) and (d), 5(z) is the Dirac delta function.
4.5.2. Use Duhamel's principle to solve,
, 0 > 0<* < /, f >0 ,
w@, 0 = m(/, 0 = 0-
4.5.3. Solve the problem in Exercise 4.5.2 if g(x, t) assumes the following
forms
(a) g(x,t) = x{x-l). ^
(b) g(x, i) = sin(u>*) sin(a>ct) , a> # — .
(c) g(x, 0 = sin(y x) sin(^ tj .
4.5.4. Apply Duhamel's principle to obtain the solution of the following
problem
и — с it = q(tc t\
Ut C Uxx 5 Vя' */ >
4.5.5. Consider the special case in which g = g(x) (i.e., it is independent of
0 in Exercise 4.5.4 and examine the limit of the solution u(x91) as f->°°.
What condition must g(x) satisfy so that this limit exists?
4.5.6. Let g(x, t) = cosGrxll) in Exercise 4.5.4 and obtain the solution of
the initial and boundary value problem in that case.
4.5.7. Use Duhamel's principle to solve the telegrapher's equation
«i» ~ c\x + 2*ut = g(x) , 0 <x </, />0
where A > 0, if ut(x, 0) = ut(x, 0) = 0 and w@, 0 = u(l, t) = 0. Find the limit
of the solution as f->°° and show that it satisfies an appropriate steady-state
problem. Verify this result if g(x) = sinGrx//).
4.5.8. Use Duhamel's principle to obtain the solution of the inhomogeneous
Klein-Gordon equation
utt - uxx + c2u = g(x,t), 0<x<l,t>09
with m@, 0 = Ф, 0) = и(/, 0 = »t(x, 0) = 0. Use the results of Exercise
4.4.16.
239
rCISES FOR CHAPTER 4
Section 4.6
z i Obtain the coefficients Nk in the eigenfunction expansion D.195) for
h one-dimensional hyperbolic and parabolic problems.
, 2 Let the operator К in D.193) be given as К = d2ldt2 + IKdldt, where
s a positive constant. Given the same initial and boundary conditions as
the hyperbolic problem discussed in the text, determine an expression
Д the Fourier coefficient Nk analogous to equation D.207) in the text.
4 6 3. К was assumec* *n ^е text that the eigenvalues Л* are all positive.
Discuss the necessary modifications if Ao = 0 is an eigenvalue and the term
N M0(x) occurs in the expansion D.195).
4.6.4. Use the finite sine transform to solve the problem
utt - c2uxx = xe~*, 0<л; < /, f >0 ,
w@, 0 = sin t, и(/, 0 = 1,
4.6.5. Apply the finite cosine transform to solve the problem
utt - c2uxx = 0, 0 < д: < /, Г > 0 ,
4.6.6. Use the finite sine transform to solve the problem
utt - c2uxx = F(a:) sin a>r, 0<x < l,t>0 ,
where со ^ Trfcc// and the initial and boundary data are
u(x, 0) = u,(*, 0) = m@, 0 = и(/, 0 = 0.
Consider the limit of the solution u(x,t) as а>-*тгкс/1 and obtain a
^sonance effect.
4-6.7. Consider the two-dimensional Laplace's equation V2w = 0 in the disk
r<R with the boundary condition u(Ry0)=f(e). Solve this problem by
using finite Fourier transforms.
•6.8. Use the finite sine transform to solve the Dirichlet problem for
^place's equation in a rectangle
uxx + uyy = 0 , 0 < x < /, 0 < у < I,
и(х,0)=Дх), u(x,t) = g(x),
240 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED Reqj
4.6.9. Apply the finite cosine transform to solve the Neumann problem f
Poisson's equation in a rectangle Or
uxX + uyy = -I\x, y), 0<x<l9Q<y<l,
hx(/,j0 =
и,(*. I) =
Determine conditions on F(x, y) and the boundary values of м(лг, у) so that
a solution exists.
4.6.10. Solve the Neumann problem for Laplace's equation V2u = 0 in the
disk r < R if
. з
= sin3 в .
dr
4.6.11. Consider the equation
Ku - V2m = F(r, t) , 0 < r < Л, 0 < f, 0 < в < 2тг ,
where К is either a first or second order partial derivative operator in t with
constant coefficients. If u(R, в, t) = B(t) and the initial data are functions of
r only, we may set и = м(г, f). Show that the equation can then be written in
the form
rKu + Lu = rF ,
where L is the Bessel operator (with n = 0) defined in Example 4.7. Show
how the results in the text and in Example 4.7 can be used to develop a
finite transform method for solving the given problem. This transform which
involves Bessel functions is called the finite Hankel transform.
Hint: See Section 5.5.
4.6.12. Use the finite sine transform to solve the following problem for the
heat equation
u, - c2uxx = e~', 0 < x < /, t > 0 ,
w(*,0) = 0,
m@, t) = a, u(l, 0 = J3 ,
where a and f$ are constants. Obtain the behavior of the solution as t t
large.
4.6.13. Use the finite cosine transform to solve the following problem
er
CISES FOR CHAPTER 4 241
ut ~ c\x = 0 , 0 < jc < /, f > 0 ,
4 6Л4- Consider the problem
«r - c\x = ° > 0<л:</, f >0,
h(jc,O) = O,
м@,0 = 0, и(/,0 = 1.
(a) Solve the problem by using D.249) to eliminate the inhomoge-
neous boundary terms.
(b) Solve the problem by using the finite sine transform.
(c) Compare the rates of convergence of the series solutions ob-
obtained in parts (a) and (b).
4.6.15. Given the following problem
ut - c2uxx = 0,
и(х,0) = 0,
Determine conditions on с and / such that there is a solution of the
homogeneous heat equation in the form u(x, i) — v(x)e~l that satisfies the
above boundary conditions. Use this solution to solve the given initial and
boundary value problem.
4.6.16. Obtain a solution of the heat equation
ut - c2uxx = 0, 0<л;</, f >0 ,
with the boundary conditions
*n the form of a polynomial in x and t. Show how this solution can be used
to solve the initial and boundary value problem for the heat equation with
the above boundary data and u(x, 0) = /(*), 0 < x < I. Discuss the behavior
of the solution of the latter problem for large f.
4-6-17. Solve the problem
utt - c2uxx = 0, 0< jc< /, f >0 ,
и(*,0) = и,(*,0) = 0,
и@, t) = t9 и(/, 0 = 1,
242 INITIAL AND BOUNDARY VALUE PROBLEMS IN BOUNDED ReGi
by introducing a change of the dependent variable that renders the b0
dary conditions homogeneous. Ulb
Section 4.7
4.7.1. Show that D.284) is a solution of the equation D.280).
4.7.2. Obtain the solution D.282) of the equation D.280).
4.7.3. Consider the nonlinear heat equation
ut~ uxx = Aw(l - w) ,
Carry out a linear stability analysis around the two solutions uo(x, t) = Q an^
ux(x,t) = l.
4.7A. Solve the Cauchy problem for
Ut-Uxx = ku(\ -U) , -оо<д:<оо) f>0,
with the initial condition
u(x9 0) = e , 0 < e ^ 1
by looking for a solution independent of x. Expand the solution for small e
and compare the result with that obtained by solving a linearized version of
the problem. Obtain the limit of the solution of the nonlinear problem as
4.7.5. Apply the method of eigenfunction expansions to solve (approximate-
(approximately) the nonlinear heat equation.
Щ - uxx = AmA - и) , 0<ж < я-, *>0
if
u(x9 0) = eh(x), 0 < e <Z 1,
u@,t) = u(ir,i) = 0.
Determine the critical value Ac of A and solve the problem if A is slightly
larger than Ac.
4.7.6. Consider the nonlinear elliptic equation
"« + uyy = -^i1 ~ ) ' 0<x<ir,y>09
with the boundary conditions
и@, у) - м(тг, у) = 0 , и(х, 0) - еА(х) .
FOR CHAPTER 4 243
uiring that u(x, y)^>0 as v->oo, apply a linear stability analysis to
n a critical value Ac for A. Expand the solution in the form
obtain a
u(x, y) =
k=l
Nk(y)(y[l sin kx)
a obtain an (approximate) nonlinear equation for Nx(y) and linear
Й ations for the Nk(y) with fc^2. Discuss the behavior of the nonlinear
nation for Nx(y) as far as possible without necessarily solving it and
consider the behavior of the solution u(x, y) as y->oo.
477. Consider the nonlinear hyperbolic equation
(a) Show that if the equation is linearized for small и and A > 0, there
exist exponentially growing solutions.
(b) Set и = 1 + ev and obtain a linearized equation for v in the form
of the Klein-Gordon equation if A > 0, and show that it is neutrally
stable.
4.7.8. Consider the initial and boundary value problem for the nonlinear
hyperbolic equation
Utt - Uxx = *U( 1 - U2) , 0 < X < 7Г, t > 0 ,
Use a linear stability analysis around the solution wo = 0 to determine a
critical value Ac such that exponentially growing normal mode solutions exist
rf A>AC. Using eigenfunction expansions in the manner presented in the
text, attempt a discussion of this nonlinear initial and boundary value
Problem if A is slightly greater than Ac.
5
INTEGRAL TRANSFORMS
5.1. INTRODUCTION
The preceding chapter dealt with initial and boundary value problems for
partial differential equations given over bounded spatial regions. The
method of separation of variables and the closely related finite Fourier
transform method were used to obtain solutions of these problems. The
present chapter deals for the most part with partial differential equations
defined over unbounded spatial regions. The tools we shall use for solving
Cauchy and initial and boundary value problems are integral transforms.
Specifically, we shall consider the Fourier transform, the Fourier sine and
cosine transforms, the Hankel transform, and the Laplace transform.
Instead of proceeding directly to a discussion of each transform to be
considered, we begin by showing how the separation of variables method
motivates some of the results and suggests which transforms are appropriate
for given problems.
We again consider the second order equations D.13), D.17), and D.19),
of hyperbolic, parabolic, and elliptic types, respectively. Applying separa-
separation of variables we set u(x, t) = M(x)N(t) (with t replaced by у in the
elliptic case) and obtain as before
E.1) LM(x) = \2p(x)M(x),
where we have replaced the separation constant A by A2 for convenience in
our further discussion. For N we have the equations
(N"(t) + A2N(t) = 0, hyperbolic case
N'(t) + X2N(t) = 0, parabolic case
N"(y)-X2N(y) = 0, elliptic case .
Since we are dealing with unbounded spatial regions, we shall generally
require that the solution M(x) of E.1) be bounded or vanish at infinity. И
the region is a semi-infinite interval or a half-plane, one of the boundary
244
245
ditions given in Chapter 4 will be applied on the finite portion of the
conndary. AH boundary conditions are assumed to be of homogeneous type,
boUthat д#(*) = 0 satisfies E.1) and the boundary conditions. To obtain
*° zero M(x), restrictions must be placed on the parameter A2. The values
°f °A for which nonzero M(x) can be found are the eigenvalues and the
ponding щх) are the eigenfunctions for the eigenvalue problem.
C°The fundamental distinction between the eigenvalue problem for E.1)
an unbounded region and that for a bounded region considered in
fhapter 4 is that, in general (and this is the case for the problems considered
here) the spectrum (i.e., the set of eigenvalues) is continuous in the
unbounded case, whereas in the bounded case the spectrum is discrete as we
have seen in Chapter 4. Consequently, if \k (£ = 1,2,. . .) is the set of
eigenvalues in the discrete case and uk = MhNk is the corresponding set of
solutions of the given differential equation, the general solution is obtained
by superposition of the uk as
00 00
E.3) «=E"rX MkNk .
In the case of a continuous spectrum, the eigenvalues Л range over the set
D, which may be the interval — oo<a<ooj 0^A<<», or some other un-
uncountable set. Let w(A) = M(k)N(k) be the separated solution, correspond-
corresponding to A, of the given partial differential equation. We then obtain a general
solution by the formal superposition
E.4) и = jD u( A) dk = J^ M( k)N( A) dk .
Since the spectrum is continuous and an orthogonality property for the
eigenfunctions M(A) does not occur in a simple and natural way (as was the
case for the discrete spectrum), the specification of the N(k) in E.4) in
terms of initial and boundary data cannot be carried out in a general way as
was done in Chapter 4.
Therefore, we shall concentrate on a number of specific eigenvalue
problems for E.1) in this chapter which lead to the consideration of Fourier,
Wankel, and Laplace transforms. (In this respect our discussion parallels
that given in Section 4.3 for the Sturm-Liouville problem.) Only the most
asic properties of each of the transforms will be presented and their use
*"• be demonstrated in a number of examples. Once it is determined that a
Particular transform is relevant for the solution of a given problem, the
^° ution is obtained by transforming the given equation and solving for the
^ansform function. (This approach is identical to that used in the finite
°urier transform method of Chapter 4.) Since a simple and useful expres-
1On *or the solution does not often result once its transform is known, we
P^sent approximate methods for evaluating Laplace and Fourier transforms
and integrals in the concluding sections of this chapter.
246 INTEGRAL TRANSFORM
5.2. ONE-DIMENSIONAL FOURIER TRANSFORMS
We consider the one-dimensional form of E.1) with p = p = 1 and
given over the infinite interval -oo<x<oo; that is, '
E.5) M"(x) + k2M(x) = 0, -oo < x < oo ,
with the auxiliary condition that M(x) remain bounded as |x|-*oo.
general solution of E.5) in complex form is
E.6) M(x; A) = a{k)eiXx + j3(k)e~iXx ,
where a and /3 are constants. The only restriction placed on the eigenvalue
parameter A by the boundedness condition is that A be real valued. Thus the
eigenvalues A lie in the interval
E.7) -оо<л<оо,
so that the spectrum is continuous.
Corresponding to the eigenfunction expansions given in Chapter 4, we
would like to be able to represent an arbitrary function f(x) (under suitable
conditions) in terms of the eigenfunctions M(x; A) in the form
iXx + /B(k)e'iXx] dk = ] y(k)e'ikx
E.8) /(*) = j_№ [a(k)eiXx + /B(k)e'iXx] dk = ]_ y(k)e'ikx dk 9
where y(A) = /3(A) + a(-A) [the integral in y(A) is obtained from the first
integral via a simple change of variables]. Unfortunately, since the eigen-
functions for this eigenvalue problem are not orthogonal (in an elementary
sense), it is not immediately apparent how to determine y(A) in terms of
f(x) in E.8). [An "orthogonality" property for the eigenfunctions M(x; A) is
defined when Dirac delta functions are discussed in Chapter 7.]
To see how y(A) in E.8) is determined, we may examine the expansion
D.108) for the complete Fourier series of the function v(x) given in
Example 4.6. Expressing the trigonometric functions in complex form and
letting /—>oo, it can be shown (see the exercises) that a plausible result is the
Fourier integral formula
E.9) /(*) = ± /_ /_ e-ix(x^f(t) dt dk ,
where we have used/(x) in place of v(x). If we define the function F(k) in
terms of the inner integral in E.9) to be
E.10) F(A) = ^/_>**/(*)<**
247
^-DIMENSIONAL FOURIER TRANSFORMS
here we have changed the variable of integration from * to л;), we obtain
The coefficients in E.10), E.11) have been chosen to be llVba for the
purpose of symmetry.
The function F(A) in E.10) is called the Fourier transform of f(x) and
,c y\\ js ^e inversion formula that gives the function f(x) in terms of its
transform F(A). The function/(*) is referred to as the inverse transform of
Я А). ВУ comparing E.11) with E.8) we conclude that y(A) =
л/Ж)
Once the Fourier integral formula has been obtained, sufficient condi-
conditions for its validity can be verified directly. Thus if f(x) is piecewise
continuously differentiable on each finite interval and $ZJ\f(x)\ dx<™, it
can be shown that the integral in E.9) converges pointwise to the function
f(x) at points of continuity of f(x). At a point x0 where f(x) has a jump
discontinuity, the integral converges to \f(xo-)+ 5/(хо+). If we weaken
the conditions on f(x) and merely require that it be square integrable over
-oo < jc < oo, it can be shown that the integral converges to f(x) in the mean
square sense.
In applications of Fourier transforms it is often necessary to find the
inverse transform of #(A) = F( A)G( A) where F(A) and G( A) are transforms
of the known functions/(jc) and g(x), respectively. The inverse transform of
Я(А) is given as
E.12)
2тг J-oo J_«
we have assumed the interchange of orders of integration is valid.
result is known as the convolution integral theorem for Fourier trans-
transforms.
Putting G(A) = f(A), the complex conjugate of F{A) in E.12), it is easily
°wn that we obtain the Parseval equation for Fourier transforms
EЛЗ)
Ть*
n*s result should be compared with the Parseval equation D.69) for
Curler series.
248 INTEGRAL
An additional result that relates the Fourier transform of derivatives
functions to the transform of the functions themselves is important f
applications of transforms to differential equations. For the transform °*
f'(x) we have Ot
EЛ4) vfc f.*'A™ * e Ш f
dx = "/
where F(A) is the Fourier transform of/(jc), assuming that/(jc) is a smooth
function that vanishes at infinity. More generally, if f(n\x) is the nth
derivative of /(*), and /(jc) and its first n -1 derivatives are smooth
functions that vanish at infinity, we have
E.15) yL J_ eikYn\x) dx = (-/A)"F(A),
The formulas E.14), E.15) are obtained on integrating by parts.
If f{x) is absolutely integrable, the Riemann-Lebesgue lemma asserts that
its Fourier transform F(A) tends to zero as |A| tends to infinity. (See the
exercises for a proof.) The rate at which F(A) tends to zero depends on the
smoothness of/(jc) and on how rapidly/(jc) decays as |jc| tends to infinity.
Further useful properties of Fourier transforms and transforms of specific
functions are given in the exercises and may be found in tables of Fourier
transforms. We now consider several examples whose solutions involve
Fourier transforms. In each example one of the independent variables has
the infinite interval (—°o, +°°) as its domain of definition.
Example 5.1. An Ordinary Differential Equation. A simple example
that exhibits the essential features of the Fourier transform method for
solving differential equations is given by the following boundary value
problem for the ordinary differential equation
E.16) y\x) - k2y(x) = -/(jc), -oo < x <oo ,
where к is constant and /(jc) is specified. We require that
E.17) y(x),/(*)->0as|*|->°o.
Assuming F(A) is the Fourier transform of f(x) we set
E.18) Y(A) = ^= \_m eikxy{x) dx ,
so that Y(A) is the Fourier transform of the solution y(x).
To solve this boundary value problem, we (Fourier) transform the
oNE>I)lMENSIONAL FOURIER TRANSFORMS 24V
ation Eл6) by multiplying across by (l/V2ir)eiAx and integrating both
C'des with respect to x from -да to +oo. By our assumption on y(x) and y'(x)
at infinity we have, on using E.15),
E 19) -JL |_ eikxy»{x) dx = (-*AJY(A) = -Л2У( А) .
The transformed equation is, therefore, algebraic and is given as
E.20) -(Л2 +
so that
E.21)
To find y(x) we must invert Y( A). This can be done using the convolution
theorem E.12) if the inverse transform of G(A) is known. Using a table of
transforms or by direct verification based on complex integration theory it
can be shown that
E22) vrzLe G^dx=
with /:>0. [Note that once the inverse transform of G(A) is given, as in
E.22), the Fourier transform of that function is easy to evaluate and thereby
shown to equal G(A).] Using E.22) in the convolution theorem we obtain
the solution of E.16), E.17) in the form
E.23) y(x) = -j= £ e-y( A) dk = _^ £ e-F(A)G(A) dX
ft is of interest to apply the solution formula E.23) for two special
cjcnces of f(x). First we set f(x) = 1 and then we set fix) - S(x - f), where
\* f) is the Dirac delta function. With f(x) = 1 we have
E-24)
—
k2
250 INTEGRAL
Putting f(x) = d(x-g) and recalling the properties of the delta
given in Example 1.2 gives
E.25) y(x) = -^ £ е"*1*""^ - () dt = i е-*1*-* .
However, each of these "solutions" has a shortcoming. The functio
y(x) = 1/Л2 satisfies E.16) but fails to vanish at infinity, as required by th
boundary condition E.17). The function y(x) = (l/2k)e~klx^1 satisfies
E.16) at all x Ф g and vanishes at infinity, but is not differentiable at x = (
We do not expect the above "solutions" to satisfy all the conditions of the
boundary value problem E.16), E.17) since in both cases the Fourier
transforms of the functions/(jc) are not defined in a conventional sense. The
function f(x) = 1 has a Fourier transform equal to л/2тгд(\). This can be
seen by putting F(A) = л/2тгв(Л) in the inversion formula E.11). Similarly,
the Fourier transform of f(x) = 8(x - £) is F(X) = (\lVtrr)eiH. In either
case we are required to deal with generalized functions and without extend-
extending the theory of Fourier transforms to include such functions and an
appropriate formulation of the corresponding boundary value problems, it is
not clear that the solution formula can be applied in such cases. Neverthe-
Nevertheless, the solution y{x) = 1 Ik2 in the case where /(jc) = 1 is certainly plausible
(even if it does not vanish at infinity), and the solution y(x) = (l/2fc)e~*|jc~**
when f(x) - S(x - £) is known as Green's function for the boundary value
problem E.16), E.17). (Green's functions play an important role in the
theory of boundary value problems for both ordinary and partial differential
equations.) Although a direct approach can be developed to deal with each
of the above cases which may or may not depend on Fourier transform
theory, we now show how each of the foregoing solutions can be obtained as
limits of sequences of solutions of the boundary value problem E.16),
E.17), each of which satisfies the conditions of the problem. As such, the
two solutions obtained previously are called generalized solutions.
We define the sequence of functions
\x\>N,
where N> 0 and consider two cases. First, we set a = 1 for all N and find
that lim^^ fN(x) = 1 for all x. This corresponds to the case f(x) = 1 given
above. Second, we set a(N) = 112N and obtain in accordance with our
discussion in Example 1.2, Ит^_0 fN(x) = 8(x). This corresponds to the
case f(x) = S(x) where we have put £ = 0 for simplicity.
Clearly, each of the fN(x) has a Fourier transform, and the problem
E.16), E.17) may be solved in the manner given above with / = /*♦ The
solution уы(х) corresponding to fN(x) is found to be
piMENSIONAL FOURIER TRANSFORMS 251
cosh kx]a(N) , ,
к2 ' ' '
[e~*M sinh kN]a(N)
k2
\x\>N.
For each yN(x) we have lini|xbee >>„(*) = 0 so that E.17) is satisfied.
First we put a(N) = 1 in E.27). Then as N-><*>, the pointwise limit of
(x) is clearly yN(x)-* Ilk2 for all finite jc. Next we set a(N) = 1/2ЛГ and
the limit as ЛГ->0 is found to be, on using l'Hospital's rule, yN(x)~*
A/2Л)е"*'х|. (This limit equals the Green's function with £ = 0.) Since the
limits are nonuniform over the infinite interval, the smoothness properties
and the vanishing at infinity are not preserved in both cases, yet we have
obtained these solutions as limits of strict solutions of the boundary value
problem. It may be stated as a general principle that solutions of a given
problem obtained by formal and plausible means can be characterized as
limits obtained in an appropriate fashion of a sequence of strict solutions of
the problem. We shall have further occasion to construct solutions of
problems by formal means, but we will not always indicate how they may be
obtained as a limit of strict solutions. It should be remarked that in Chapter
4, the eigenfunction expansions did not always constitute strict solutions of
the given problem and had to be interpreted as generalized solutions of the
given problem even though this was not always explicitly indicated.
If we replace E.16) by
E.28) /'(*) + k2y(x) = -Дх), -a><x <oo,
we can no longer require that y(x) and y'(x) tend to zero as |x|-*oo, and
apply the Fourier tranform as was done for E.16). Indeed, if /(jc) = 0 for
M > N, the solution of E.28) for |x| > N is a linear combination of sin kx
and cos kx. Thus it is merely bounded at infinity but does not vanish there
unless we require it to be identically zero. Further if we formally apply the
Fourier transform to E.28) and neglect contributions from infinity, we
obtain
E.29)
and the inverse transform y(x) does not exist for arbitrary/(jc) in view of the
angularity of Y(A) at A = ±*.
To proceed with the solution of E.28), we adapt the solution formula
v5-23) to fit the present problem. [This parallels the approach used in the
^° special cases considered for E.16).] We observe that if we replace к by
252 INTEGRAL TRANSFORM
ik or by -ik in E.16), the equation takes the form E.28). Thus, if w
replace к by ±ik in the solution formula E.23) we must obtain a solution erf
E.28) if f{x) is such that the resulting integral converges. This gives
E.30) y±(x) = ±(//2*) j_ e±ik{x~t{f(t) dt,
where y±(x) correspond to the choice + ik, respectively. Both y+(x) and
У-(х) аге solutions of E.28) and are bounded at infinity. The solution
cannot be required to vanish at infinity to obtain a unique result, as was the
case for E.16). Furthermore, asking that the solution be bounded does not
yield uniqueness as shown by the two solutions given in E.30). If E.28)
arises in the context of wave propagation as a reduced (inhomogeneous)
wave equation (see Example 6.13) the radiation condition at infinity
picks out the solution y+(x) in E.30) and guarantees a unique solution. If
the source function f(x) vanishes outside a finite interval, the radiation
condition E.31) signifies that a wave traveling away from the source towards
infinity is generated. The solution y_(x) corresponds to a wave that travels
inwards from infinity. These matters are elaborated on in Example 6.13 and
Section 10.1.
Example 5.2. The Cauchy Problem for the Heat Equation. We consider
the heat (or diffusion) equation
E.32) ut-c2uxx=0, -oo<x<oo,*>0,
where c2 is a constant, with the initial condition
E.33) u(jt,O)=/(jt), -oo<jc<oo,
and solve this problem using a Fourier transform in the x-variable. We
assume f(x) has a Fourier transform and that и and ux vanish at infinity so
that E.15) is applicable. It will be found, however, that the solution formula
obtained is valid under weaker conditions than are necessary for the Fourier
transform method to be applicable.
Denoting the Fourier transform of u{x, t) by
E.34) I/(A, 0 = y= j_ eixxu(x, t) dx ,
we multiply through by A/V2ttVAjc in E.32) and integrate with respect to X
^ „B1 FOURIER TRANSFORMS 253
^ -oo to +°°. Using E.15) we obtain an ordinary differential equation
from
iotU(X,t)>
E.35)
~ dt
Fourier transforming the initial condition E.33) yields
E36) U( A, 0) = F( A) = ^ f_m e^Kx) dx .
The solution of the initial value problem E.35), E.36) is
E 37) U( A, t) = F( A)e~(cAJ' .
To find m(jc, 0 we must invert the Fourier transform C/(A, /). We have
E.38)
where we have used E.36) and interchanged the order of integration. To
evaluate the inner integral we first transform it as
E.39) Г e-iA(*-s)-A2c2'dA= Г e'
'ixix-s)-k2c2t
dX
cos[A(x - s)] dX .
The last integral in E.39) can be evaluated explicitly. Let the integral I(a)
be defined as
ш view of the exponential decay of the integrand we can obtain dllda by
differentiating under the integral sign. We have
_ о
Now /@) is given as
254 INTEGRAL
E.42)
4
С t
The solution of the initial value problem E.41), E.42) is easily found to h
E.43) I(a)\a=x_s = 2 [ e~M cos[A(jc - s)] dX
ПГ Г (jc-sJ]
Introducing E.43) into E.38) yields the solution of the Cauchy problem
E.32), E.33) as
E.44) и(л:,0 = -7=[ exp[-^-^l/(s)<fc.
V4ttc t •'0 L 4c f J
The term
E.45) *-М-4т4*|Й
VAirct L 4c/ -I
is known as the fundamental solution of the heat or diffusion equation. With
the formal substitution f(s) = 8(s — f) in E.44) we obtain the fundamental
solution E.45). Thus E.45) is a solution of the heat equation corresponding
to a point source of heat at the initial time t — 0 located at the point x = £.
The solution E.44) can be thought of as a superposition of solutions due to
point sources distributed along the jc-axis with density /(jc).
We have already encountered a solution of the form E.45) in Section 1.1.
In fact, given the diffusion equation for u(jc, t),
E.46) vt + cvx = l Dvxx ,
considered in A.23), if we set
E.47) jc = jc-cf, i=t,
and w(jc, t) = y(jc, t), then и satisfies E.32) with c2 replaced by \D and
jc, f-> jc, I Further, w(jc, 0) = w(jc, 0) = u(jc, 0). Thus with the initial condi-
condition A.27) for y(jc, t) [i.e., u(jc, 0) = 5(jc)], we obtain the solution A.31) lot
y(jc, t) on comparing with the fundamental solution E.45).
Either from the solution E.44) with /(jc) = 0 outside a finite interval ot
from the fundamental solution E.45), one sees that m(jc, i) is instantaneous-
instantaneously greater than zero at each jc, for any time t > 0. [If w(jc, t) represents
temperature, it is generally assumed that u(x, 0) =/(jc) >0.] Thus heat
DIMENSIONAL FOURIER TRANSFORMS
255
at infinite speed according to the heat equation E.32). A similar
was made and discussed in connection with the diffusion
equation of Section 1.1.
For the nonhomogeneous heat equation
E.48) ut ~ c\x = F(x9t),
vrith homogeneous initial data
E 49) u(x, 0) = 0 ,
we easily obtain, on using Duhamel's principle,
-oo < x < oo, f > 0 ,
E.50) u(x, t) - jo ]_ G(x -s,t- r)F(s, t) ds dr ,
where G(x - s, t - r) is defined as in E.45). The result E.50) can also be
obtained by the use of Fourier transforms. The fundamental solution E.45)
is thus seen to play an important role in the solution of the Cauchy problem
for the heat equation. As we shall see, it also plays a role in the solution of
problems over semi-infinite intervals and even problems over a finite
interval.
The function G(x - £, t) is also of interest for the reason that as f-^>0,
G(x - £, t)~^> 8(x - £). Thus we have in G(x - f, i) a smooth function
which in the limit tends to the Dirac delta function. This is in contrast to the
functions 8€(x) defined in Example 1.2, which tend to 8(x) as с->0 but have
jump discontinuities at x = ±€ and are not smooth functions.
Example 5.3. The Cauchy Problem for the Wave Equation. The Cauchy
problem for the wave equation asks for a function u(x, t) that satisfies
E-5l) utt - c2uxx = 0, -оо<л:<оо, f>0,
and has the initial values
<5-52) m(jc, 0) = /(*), -oo < x < oo ,
Fourier transformed problem is
E.54)
dt
t>0 ^
f/(A,0) =
256 INTEGRAL TRANSF()RM
E-56) ^-Z = G(A);
where t/, F, and G are Fourier transforms of u, f, and g, respectively ь •
assumed that the operations necessary to obtain E.54)-E.56) [e g the *S
of E.15)] are valid. ' Use
The solution of the initial value problem E.54)-E.56) for U(\ л •
easily found to be
E.57) U(k, t) - [J F( A) + ^ G(\)]eixct
Inverting this transform gives the solution
E.58)
\\-^= Г e'a(x-c°F(\)d\ + ^= Г
2 1у/2тг 1-°° V2lr J
cwbrr J-00 fA
cV2^ J— tA
The integrals involving G(A) in E.58) are to be interpreted in the Cauchy
principal value sense in the neighborhood of A = 0. If we express the
function g(x) as
j ~ax
E.59) g(x) - ^= j_ e~axG(\) dX ,
the indefinite integral of g(x) is given as
with the integral in G(A) inteфreted in the Cauchy principal value sense.
Then E.58) can be expressed as
Dl]vtt;NSlONAL FOURIER TRANSFORMS
0.61) w(*' t)=2 [f(* " Ct) + /(* + Ct)]
1 (x~ct 1 fx
257
using E-11) an(* E.60). The expression E.61) is just d'Alembert's
° lution which was obtained in Example 2.4 by another method.
Example 5.4. Laplace's Equation in a Half-Plane. We begin by consid-
considering the first boundary value problem, that is, Dirichlefs problem, for
Laplace's equation in a half-plane. We require that m(jc, y) satisfy
E.62) мяя + иуу = 0,
with the boundary condition
E.63) и(х,0) = Дх),
and the additional condition
E.64) u(x, y) -» 0 as |jc| -> », у —> oo .
To solve E.62)-E.64) we introduce the Fourier transform in x of the
function u(x, y) defined as
E.65) f/( А, У) = y= £ e^uix, y) dx .
Then the Fourier transformed problem E.62)-E.64) becomes
E.66) ^_A2[/ = 0> 0<y
By
*»th the conditions
E-67) U(X, 0) = F(A), t/(A, >>)^
ere F(A) is the transform of f(x) and f/(A, ^)»0 as y^oo i
equence of the condition E.64). The solution of E.66)-E.67) is
E'68)
as
as y-^oo is a con-
258 INTEGRAL TRANSP0RMs
where the |A| occurs, since we require that i/-»0 as у-юо and A ran
from -oo to +00. Inverting the transform Щ\, y) gives 8es
E.69) M(jc,y) = -^£\
" Ъг J— J-
dk
on using the inverse transform for F(A). The inner integral can be evaluated
explicitly as
E.70)
J-oo J-oo JO
A=0 eX
+ —:
A=0 (JC - Л" +
Thus we have
It may be verified directly that for y>0, E.71) satisifies Laplace's
equation. However, a careful examination of the integral is required to show
that the boundary condition E.63) is satisfied. Although the coefficient у of
the integral vanishes at у = 0, the integral is singular at the point t = x when
у - 0. The combination of these two effects leads to the satisfaction of the
boundary condition. We do not carry out this verification.
In the exercises, we apply the solution formula E.71) to the case where
f(x) is either constant or piecewise constant, and may not be Fourier
transformable iff(x) does not vanish at infinity. In all cases the solutions are
bounded at infinity and it can be shown that the boundedness requirement
implies that the solutions are unique.
If we consider the second boundary value problem for E.62), that is, the
Neumann problem, the boundary condition E.63) is replaced by
E.72)
and we require that u(x, y) be bounded in the upper half-plane. Solving this
problem directly leads to certain difficulties with the convergence of the
relevant Fourier transforms that we prefer to avoid. Instead, we solve this
problem by using a general principle known as Stokes' rule. For our problem
it takes the following form.
Let m(jc, y) satisfy
- PUVIENSIONAL FOURIER TRANSFORMS 259
E73) "„ + "„=0, -oo<*<oo, y>0
j the boundary condition
0.74) ">•(*> °) = *(*) ■
put
E.75) Ф. Л
Then
E.76) V2,=V4
E.77) »(*,0) = w
Thus if we can solve the Dirichlet problem E.76), E.77) for v(x, y), the
solution of the Neumann problem E.73), E.74) is given as
E.78) u(x,y) = j v(x,s)ds,
and is determined up to an arbitrary constant. Using the solution E.71) of
the Dirichlet problem we have
verified that E.79) satisfies E.73), E.74).
We have considered the application of Fourier transforms to some basic
f . Sln*ple problems. In each case it was possible to obtain the solution in a
^lrly straightforward manner by inverting relevant Fourier transforms.
quations of parabolic, hyperbolic, and elliptic types were chosen to show
e aPplicability of the Fourier transform technique for equations of each of
e three basic types. Further discussion of the solutions obtained in the
Samples is given later in the text and in the exercises.
*n general applications of transform methods it is not expected that the
INTEGRAL
inverse transforms can be simplified as far as was done in the abo
examples. However, approximation methods are available that yield result
in various regions of interest in the domain of the dependent variabl S
Certain methods of this type are presented in Section 5.7.
5.3. FOURIER SINE AND COSINE TRANSFORMS
On considering the one-dimensional form of E.1) over the semi-infinite
interval 0 < x < <» with p = p = 1 and q = 0, we obtain the equation
E.80) M"(x) + \2M(x) = 0, 0 < x < oo.
Two sets of boundary conditions are considered for the function u(xy t) [or
u(x, y)] which gives rise to the eigenvalue problem under consideration
when the variables are separated as in Section 5.1. First, we assume that и
vanishes at x — 0 and is bounded as x—>&. This implies the (homogeneous)
boundary conditions
E.81) M@) = 0, M(x) bounded as лг^оо ,
for the solution M{x) of E.80). Second, we assume that ux vanishes at x = 0
and that и is bounded as x-^oo. This yields the (homogeneous) boundary
conditions
E.82) M'@) = 0, M(x) bounded as *-» oo,
for E.80).
For the eigenvalue problem E.80) and E.81) we immediately obtain the
eigenfunctions
E.83) M(x;\) = sin\x,
with the continuous spectrum
E.84) 0<A<oo.
Negative values of A need not be considered since sin(-Ax) = -sin(Ax).
For the eigenvalue problem E.80) and E.82) we obtain the eigenfunc-
eigenfunctions
E.85) M(x; A) = cosAjc,
with the continuous spectrum
iiRlER SINE AND COSINE TRANSFORMS 261
E.86) 0<A<°°.
ain negative values of A are not included since cos(-Ax) = cos(Ax).
Let the function f(x) defined for x> 0 satisfy conditions equivalent to
hose given in Section 5.2 for Fourier transformable functions. Then corre-
onding to the eigenvalue problem E.80) and E.81) we have the repre-
representation
E.87) Fs(A) = V| / sin(ХхШ*) dx
0
with Fs(\) defined as the Fourier sine transform of f(x). The inversion
formula giving f(x) in terms of its transform is
E.88) fix) = V| /0 sin<A*) W) dk ■
The formula E.88) yields an odd extension of f(x) to the entire x-axis
[i.e., /(-*) = -f(x)]. By considering the Fourier transform E.10) and its
inversion formula E.11) as applied to the odd extension of /(*), we obtain
the formulas E.87), E.88), as is shown in the exercises. Thus E.88)
converges to f(x) at points of continuity of f(x) and to its mean value at
points of jump discontinuity.
Corresponding to the eigenvalue problem E.80) and E.82) we have the
representation
E.89) Fc(A) = V| [ cos(\x)f(x) dx ,
FC(X) defined as the Fourier cosine transform of /(*). The inversion
formula for the cosine transform is
dk .
formulas result from the Fourier integral formulas E.10), E.11)
when applied to an even function /(*) [i.e., /(-jc) = /(jc)], as is shown in the
e*ercises. It is seen that E.90) yields an even extension of f{x) to the entire
<°us and again, the convergence properties for E.90) are equivalent to those
Or the Fourier transform.
m applications of the sine and cosine transforms to differential equations,
ls again necessary to express the transform of derivatives of /(*) in terms
1 the transform of/(jc). For the sine transform we have
262
INTEGRAL TRANSFORms
E.91) ^/||о sm(kx)f\x)dx
E.92) V~
-n~Av^/(*)cos(A*)
Thus if we assume that f(x) and /'(*) vanish at infinity, we obtain on using
the defintions E.88) and E.89),
E.93)
E.94)
-Jo sin(Ax)f'(x)dx=-\Fc(\),
= Ад/| /@) - A2F,(A) .
We note that the Fourier sine transform of a first derivative of a function is
given in terms of the Fourier cosine transform of the function itself. The
Fourier sine transform of a second derivative is given in terms of the sine
transform of the function. There is an additional boundary term aV27tt/@)
that vanishes if /@) = 0; that is, if f(x) satisfies the boundary condition of
the relevant eigenvalue problem. The formulas E.93) and E.94) show that
the use of the sine tranform is not expected to be effective unless the
differential equation contains only derivatives of even orders in the trans-
transformed independent variable and if only one boundary condition is assigned
for each two derivatives of the transformed function.
A similar result is true for the Fourier cosine transform. We easily find on
using E.87) and E.89) that
E.95)
E.96)
'^Jo f'(x)cos(Xx)dx = -yJz
AFf(A),
f- I f"(x) cos(Xx)dx = -^^ /'@) - A2Fe(A) ,
if we assume that f(x) and/'(*) vanish at infinity. Again, if /'@) = 0 so that
f(x) satisfies the boundary condition at x = 0 of the associated eigenvalue
SINE AND COSINE TRANSFORMS 263
oblem, the cosine transform of f"(x) is given strictly in terms of the cosine
рГ sfOrm of the function/(*) as shown in E.96).
The formulas E.94) and E.96) for the Fourier sine and cosine transforms
f the second derivative of a function require that the function and its
derivative, respectively, be specified at x = 0 if these transforms are to be
eful in solving differential equations. Thus, the sine or cosine transforms
appropriate for problems over semi-infinite intervals in a spatial variable
• which the function or its derivative are prescribed on the boundary. The
Fourier sine and cosine transforms are generally not useful when applied
with respect to a time variable in connection with initial conditions at t = 0.
For example, the heat equation has only one time derivative un so that its
sine transform is given in terms of its cosine transform. The wave equation
contains a second time derivative utt, but since both и and ut are specified at
t = 0 neither the sine nor the cosine transform can account for the full initial
data. As shown in Section 5.6, the Laplace transform is the proper choice
when the time variable is transformed.
We now consider several applications that involve the use of Fourier
cosine and sine transforms. It should be noted that some problems that can
be solved by use of Fourier transforms in one of the variables, such as
Example 5.4, can also be solved by the use of cosine or sine transforms in
the other variable(s). Cosine or sine transforms can be used when the
transformed spatial variable is restricted to a semi-infinite interval, and the
choice of the cosine or sine transform is dictated by the boundary condi-
conditions, as indicated. However, in applications to partial differential equa-
equations, since more than one variable occurs, it can happen that one variable is
unrestricted whereas the other variable is bounded on one side. Thus either
the Fourier transform or the cosine or sine transform may be used to solve
the given problem.
Example 5.5. The Heat Equation in a Semi-Infinite Interval. We con-
consider the heat (or diffusion) equation
w, - c2uxx = 0, 0<*<oo,f>0,
*nere c2 is a constant, over the semi-infinite interval 0<x<Q°, with the
filial condition
E*98) u(x,0)=f(x)9 0<*<oo,
and either the boundary condition of the first kind
r the boundary condition of the second kind
264 INTEGRAL
E.100) ux(O,t) = h(t), t>0.
For the first boundary value problem [i.e., where u@, t) = g(t)] we ары
the Fourier sine transform in x, since the term м@, t) corresponds to th
/@) term that occurs in E.94). For the second boundary value proble *
[i.e., where ux@, t) = h(t)] we use the Fourier cosine transform, in view of
the relation between ux@, i) and/'@) in E.96).
Applying the sine transform to E.97), we multiply through in E.97) bv
V2/tt sin( Ajc) and integrate from 0 to °o to obtain
E.101) ^^ + (XcfUs{X, t) = aV|
on using E.94) and E.99). The function C/5(A, t) is the sine transform of
w(x, t)\ that is,
E.102) £/,( A, t) = л/ - sin( A*)m(jc, /) dx .
УТ JO
From the sine transform of the initial condition E.98) we have
E.103) C/,(A,0) = F,(A).
The solution of the initial value problem E.101) and E.103) is
E.104) f/5(A, 0 = F,(A)<TAV' + Ад/| £ е-А2с2('-^(т) dr .
The inverse transform of C/5(A, t) [see E.88)] yields the solution
E.105) ф( t) = y]- f/ДА, 0 sin(A^) dA
" 7Г JO
/2 Г . _A2C2,
= v— I F.(A)sin(Ax)e dk
v7rJ0
- Г Г
7Г JO JO
AND COSINE TRANSFORMS 26S
= - f f e"A2A{cos[A(x - s)] - cos[Л(лг + s)]} f(s) dX ds
7Г JO JO
= L [°(x - s>') - G(x + 5' ')]/(*) *
*G(s,*-t) , , .
-2* Г
Jo
where we have used the notation G(x, t) for the fundamental solution of the
heat equation defined in E.45). The foregoing result was obtained by
interchanging the order of integration in both integrals and evaluating the
inner integrals following the approach presented in E.39)-E.43). We have
also used the identity 2 sin(Ax) sin(As) = cos[A(jc - s)] - cos[A(* + s)].
Using Duhamel's principle and E.105) it is easy to see that the solution
of the inhomogeneous heat equation
E.106) vt - c2vxx = F(x, t), x>0,t>0
with data v(x, 0) = f(x) and v@, t) = g(t) is
E.107)
Г Г
v(x, t) — u(xy t) + I I [C?(jt — s, i — r) — G(x + s, t — t)\F(s9 t) ds dr
Jo Jo
where u(x, t) is given by E.105).
In the case of a uniform initial temperature, u(x, 0) = f(x) ~ u0 -
constant, we readily obtain the following result on changing the variable of
integration,
E.108)
[ edr[
~x/2cVt Jx/2cVl
2UO r'2cVi _гг
= j e
*here ФB) is the (tabulated) error function integral defined
E.109) $B)serf(z)
as
266 INTEGRAL ANSFO
Th
and normalized such that Ф(оо) = 1 since Jq e~r2 dr = Vn/2. Thus
lti f h iiil d bd bl E9)
solution of the initial and boundary value problem E.97)-E.99)
u(x, 0) = u0 and m@, t) = 0 is given as
E.110) u(x,t) = uo<l
We note that u(x9 0) = м0Ф(оо) = u0, м@, t) = моФ@) = 0 and u(x, f)-^0 as
f-><». Even though f(x) = u0 does not have a sine transform, the solution
E.110) satisfies all conditions of the problem. Again it can be shown that
E.110) is the limit of a sequence of problems with Fourier transformable
initial data but we do not demonstrate this.
We observe that if the initial temperature f(x) is not constant but
1/001 < M < oo for all jc > 0 (as we indeed assume), we have
E.111) ||o
J [G(x-s,t)-G(x
since G(x ~ s, 0 ^ G(x + s, t) for x, s ^ 0 and Ф(г)-> 0 as z -► 0. [We have
used the extended mean value theorem for integrals applied to the finite
interval [0, N) and then let Af-» » to obtain the inequality in E.111).] Thus
as f->°°, the solution w(jc, t) [i.e., E.105)] of the initial and boundary value
problem E.97)-E.99) tends to the steady state
E.112) ф, t) ~ -2c21 dG{X;lxT) g(r) dr9
in the sense that the effect of the initial temperature distribution u(x, 0) =
f(x) is dissipated (the term steady state does not signify time independence
for this problem).
In fact, if we consider the initial value problem for the heat equation
E.97) with 0<x<oo and t>t0, and specify that u(x,to)=f(x) and
w@, t) = g(t), the solution is
E.113) u(x, 0 = I [G(x -s,t-t0)- G(x + s,t- to)]?(s) ds
as
follows from E.105). We assume that \f(x)\ < M and that g(t) is defined
RIER SINE AND COSINE TRANSFORMS 267
all t. The estimate E.111) shows that the absolute value of the first
fof ral is bounded by МФ(х/2слД=Т0). Then as tQ->-*>, if Ф»0 is
1П unded for all time, we obtain in the limit
„■.<> «<*,o=-2
Th's is the solution of the steady-state problem for E.97) over the interval
ft<JC<oo) where no initial condition is prescribed and the boundary condi-
condition is given for alW> -« as w@, О - g(t). If
Г 9(t) t > 0
E.115) «@,0-«')«{£'' t<Q
E.114) reduces to E.112).
If the initial temperature u{x, 0) = f(x) - 0 and the boundary temperature
wm f) = g(t) = ux = constant, it is readily verified (as shown in the exercises)
that the solution E.105) can be expressed as
E.116) u(x, t) = Ml[l - Ф
The bracketed term in E.116) is often referred to as the complementary
error junction and it tends to unity as /—»«>. Thus from E.111) we conclude
that the temperature u(x, t)-+0 as f-»oo5 if the boundary temperature
g@ = 0 and the initial temperature u(x, 0) =f(x) is uniformly bounded. If
the boundary temperature g(t) = ux = constant, and the initial temperature
is uniformly bounded, E.116) shows that the temperature w(x, i) tends to
the constant state ux as *—►<».
Finally, we remark that the function
which occurs in the last integral in the solution E.105), is readily verified to
be a solution of the homogeneous heat equation E.97) that satisfies the
conditions
^5tl18) lim w(x,0 = 0,
EЛ19) Urn tKx.0-0,
+0+
owever, as we approach the origin (jc, i) = @,0) along the curve x = 2cVt
We have
268 INTEGRAL
E.120) hm -2c2 —^-^ - lim 575 exp - ~r~
t-+o+ dx x*=2cVt r-»-o+ L 2y/lfct V 4c 7
2cV7
lim —-^->oo.
Thus и>(;е, г) is unbounded at the origin. Although w(x, t) is a solution of th
heat equation that satisfies homogeneous initial and boundary conditions in
view of E.118), E.119), it is unbounded at the origin. To be certain that
solutions of the initial and boundary value problem E.97)-E.99) are
uniquely determined, we must require the solution u(x, t) to be bounded
Otherwise, we could add the function w(x, t) multiplied by an arbitrary
constant to any solution, without altering the initial and boundary data for
the problem. In fact, w(x, t) is a solution of the heat equation for -<»<*<
oo and t > 0 and its limit as t tends to zero along any line x = constant is zero.
However, it is not bounded at the origin in view of E.120). Consequently,
unless we require that the solution of the initial value problem for the heat
equation is bounded for all t ^0, the solution is not unique.
Using the Fourier cosine transform and proceeding as above, we readily
obtain the solution of the second boundary value problem E.97), E.98),
and E.100) in the form
E.121)
Ф> 0 = I [G(x - 5, 0 + G(x + 5, t)]f(s) ds
~2c2 jo G(x,t-T)h(r)dT
+ | i [G(x -s,t-r)+G(x + s,t- r)]F(s9 T)dsdr,
where the last term occurs if the inhomogeneous problem for the heat
equation [i.e., E.106)] is considered.
With constant initial temperature u(x, 0) = /(*) = w0 = constant and h «
F = 0 in E.121) we have, after some manipulation of the integrals,
E.122) w(x, 0 = |o [G(x - s, i) + G(x + s, OK
This is to be expected since du@,t)ldx determines the amount of heat
passing through the boundary x = 0 and we have <?w@, i)ldx = 0. Als°>
F(x, t) = 0 so that there are no heat sources. Thus no heat escapes througo
SINE AND COSINE TRANSFORMS 269
boundary x = 0 and no heat is generated. Consequently, the temperature
llains fixed at u(x,t) = u0.
It should also be noted that the function
123) w(x, t) = -2c2G(x, 0
hat occurs in the second integral in E.121) has the property that
E.124) Mm w(x,t) = 0, x>0,
E.125) !
as is readily seen. Also, w(x, i) is a solution of the heat equation E.97).
However, w(x, t) is unbounded at (x, t) = @,0) if we approach the origin
along the curve given in E.120), so we must require the solution of the
second boundary value problem to be bounded in order to obtain a unique
solution.
The solution of the third boundary value problem for the heat equation in
the semi-infinite interval with the boundary condition
E.126) Wjc@, 0 - hu@, t) - r{t),
where h>0 is a constant and r(t) is given, is considered in the exercises.
Some of the above results are rederived in Chapter 7 using Green's
function methods and have been expressed in terms of the fundamental
solution G(x, t) for the purposes of comparison with later results.
Example 5.6. Laplace's Equation in a Strip and a Quarter-Plane. We
consider Laplace's equation in a semi-infinite strip
the boundary data
E.128) M@, y) = 0, u(xy }>)->0 as *->«>, uniformly in у ,
E129> w(*,a) = 0, u(x,0)=f(x).
Since u(x, y) is given on the boundary x = 0, the Fourier sine transform is
aPPropnate for this problem. Let the sine transform of u(x, y) be defined as
Eлз0) ц(А, У)
J^d multiply through in E.127) by V2/irsin(Ax) and integrate from 0 to *>.
Usil*g E.94) and noting that и@, у) = 0, we obtain
270 INTEGRAL TRANSFORM
E.13.) £!^_дЧ(д>у),0.
With
E.132) t/,(A,O) = F,(A),
where FS(A) is the sine transform of f(x) and
E.133) £/ДЛ,а) = 0,
in view of E.129), we obtain for t/,(A, y),
E 134}
E.134)
The inverse sine transform yields
E.135) u(x,y) =
= 1 Г Г Л*)sin(A,)sin(Ax) Sinh[Ah(ra,7] Л<*A .
тг Jo Jo yv ' v ' v ' smh[aA]
Note that if aA > 1, we have
E.136) sinhjAta-^
v ' sinh[aA]
Thus the Л-integral in E.135) converges exponentially. The ^-integral also
converges well if /(дг) —> 0 as x—»°° as is, in fact, required by the boundary
condition E.128). We remark that if there are inhomogeneous boundary
conditions at y = a and x-0, the resulting problem can be solved in a
similar fashion. Also, if we replace the Dirichlet boundary condition in
E.128) with the Neumann condition мл@, y) = g(y) at x = 0 the problem
can be solved with the use of the Fourier cosine transform.
In the limit as a -> oo in the above problem we obtain the quarter-plane
problem for Laplace's equation, that is,
E.137) uxx + uyy=0, x>0,y>0
with the boundary conditions
E.138) w@, y) = 0, u(x, ^-^Oasjc^», uniformly in у ,
E.139) и(*,0) =/(*), *>0.
HIGHER
DIMENSIONAL FOURIER TRANSFORMS 271
This problem may be solved by a direct application of the sine transform
by g°in8 to the 'im^ as a*~>0° in the solution E.135). Using the latter
approach and noting E.136), we obtain as the solution of E.137)-E.139),
E.140) Ф> У) = ~ Jo Jo /(*) sin( Aj) sin( Ах)*ГЛу <fc dA .
Interchanging the order of integration, we have for the inner integral,
f00
E.141) 2jo sm(\s)sin(\x)e~Ay dA
r°°
= I e~A>{cos[ A(x - s)] - cos[A(x + я)]} d\
'о
У
on adapting the result in equation E.70). Then the solution u(x, y) takes
the form
E..42) ^.j
The formula E.142) may be compared with that given in E.71) for the
solution of the half-plane problem for Laplace's equation. If we let f(t) in
E.71) be defined as an odd function of t [i.e., /(-/) = -/(*)], the integral
E.71) can easily be transformed into E.142). This is equivalent to extending
the above quarter-plane problem to a half-plane problem by extending the
solution u(x9y) to the full infinite interval as an odd function [i.e.,
u(~~x> У) = ~u(x, y)] with м@, у) = 0. Then if f(x) is also extended as an
odd function, we would expect the solutions E.71) and E.142) to agree, as
they indeed do. A further connection between half- and quarter-plane
problems is considered when we discuss Green's functions in Chapter 7.
5-4. HIGHER DIMENSIONAL FOURIER TRANSFORMS
Higher dimensional Fourier transforms may be characterized (as in the
Preceding sections) in terms of higher dimensional eigenvalue problems for
J^ 1) with p = p = l and q = 0 over the entire space. Alternatively, they may
e °btained by a repeated application of one-dimensional Fourier trans-
Orms in each of the variables. The conditions of validity for the transforms
<*e then readily carried over from the one-dimensional case. Instead of
lscussing the properties of the higher dimensional transforms, we merely
272 INTEGRAL
define the transforms and the inversion formulas and cite the
properties as they are required in the following examples to be
Let x = [*!,...,*„] and X = [Al5 . . . , Aj be n-component vecto?
Under hypotheses on /(x) analogous to those given in the one-dimensio S
case we have for the Fourier transform F(X) of /(x) the representation ^
EЛ43) F(x)=(vfer Г- • • • С е'Ху(х) dx'
and the inversion formula
E.144) /(x) =
The integrals E.143), E.144) are both n-dimensional, dx and dk are
«-dimensional volume elements, and X-x = E"=1 A,*, is the scalar or dot
product of the vectors \ and x. The formulas E.14), E.15) relating the
transforms of derivatives to the transforms of given functions are valid in the
higher dimensional case as well. Here they must be applied to transforms of
partial derivatives of /(x) and we require that / and its partial derivatives
vanish at infinity.
It is also possible to consider transforms equivalent to the sine and cosine
transforms given above if the problem is given over a semi-infinite space.
We remark that although Fourier transforms lead to a formally simple
approach to the solution of initial and boundary value problems for partial
differential equations, the evaluation and simplification of the resulting
integral representation of the solution is generally not a simple task. This
was already observed in the examples we have considered. We restrict our
discussion only to two and three-dimensional Fourier transforms in the
examples we consider.
Example 5.7. The Cauchy Problem for the Three- and Two-Dimensional
Wave Equations: Hadamard's Method of Descent. The Cauchy or initial
value problem for the wave equation in three dimensions is formulated as
E.145) utt - c2[uxx + uyy + uzz] = 0, *>0, -«<*, y, z <« ,
where c2 is a constant, with the initial conditions
E.146) n(x,y,z,0) = 0
E.147) ug(x,y,z,0)=f(x,y,z).
We consider only the case where u(x, y, z, 0) = 0, since according to Stokes
rule (see Example 5.4), the solution of the wave equation E.145) with data
HIGHER
DIMENSIONAL FOURIER TRANSFORMS 273
v z 0)=ftx9y,z)9 and ut(x, y, z, 0) = 0 is given as dufdt where
/ v z, t) satisfies E.145)-E.147). This is easily verified. Then the solu-
• ' of the general initial value problem for E.145) is just the sum of these
solutions.
To solve E.145)-E.147) by the Fourier transform method, we multiply
145) by 1/(V2ttK exp[/( Хгх + X2y + A3z)] and integrate with respect to x,
nd z from -°° to oo. Let C/(X, t) denote the Fourier transform of
У\ z> ^). Using the analogues of the formulas E.14), E.15), we obtain
die transformed equation
E.148) "
on assuming that и and its first partial derivatives vanish at infinity. The
initial conditions for U(\, t) are
E.149)
where F(X) is the Fourier transform of f(x, y, z).
The solution of E.148), E.149) is
E.150)
where |X| = уЛ? + k\ + A3. We determine «(*, >^, z, r) by inverting the
Fourier transform and have
00
A151) „(,, ,, ,,0-^ ///
where X = [A,, A2, A3], x = [*, y, z] and dX = dX1d\2dX3.
By writing sin[|X|rt] = (l/2i)(e'|x|cl - e-'|x|c/), we can express u(x, y, z, t)
as
E.152)
X {exp[i|X|(ct - ^p)] - op[-i|X|(e» + ^p)]} dk
^e terms exp[±i|X|(rt + X-x/|X|)] in E.152) represent plane wave solu-
lons of the wave equation E.145). That is, the solutions remain constant on
Planes X • x = constant that move parallel to themselves at the speed с Thus
274 INTEGRAL ANSFo^
E.152) represents a superposition of plane wave solutions traveling i
possible directions. (It should also be noted that these plane waves repre ^
normal mode solutions of the wave equation.) Sem
The integral E.152) can be simplified by expressing F(X) in te
f(x9 y, z). It is then possible to integrate out four of the resulting int
and one method for doing so that involves the use of the Dirac dl
function is given in Example 7.9. Instead of carrying out this process *
merely quote the final result and indicate in the exercises an alternativ ^
simpler method for deriving the final expression.
In terms of spherical coordinates with their origin at the point (*, y z\
defined as
E.153)
£ - x + r sin <p cos 0, j] = у + г sin <p sin 0, £ = z + r cos <p ,
the solution is
E.154) и(х, у, z, 0 =
t Г f2"
T~ f(x + ct s*n 9 cos 99 y + ct sin <p sin 0, z + ct cos <p) sin <p dO d<p >
щтт Jo Jo
Given a sphere of radius a with center at (x, y9 z), we define the average
value of the function /(£, 17, () over the sphere to be given as
E.155)
s
where Air a2 is the area of the sphere S and ds is the area element on the
sphere. In the spherical coordinates E.153) the area element is given as
ds^a2 sin <p dB d<p. Then it is easy to show that the solution E.154) can be
written as
E.156) u(x,y,z,t) = tMet[f],
where Mct[f] is the average value of / over the sphere of radius ct with
center at the point (x, y, z).
Recalling Stokes' rule and using superposition of the two solutions we^
obtain as the solution of the wave equation E.145) with initial data
E.157) u(x, y, z, 0) - g(x, y, z), ut(x, y, z, 0)=f(x, y, z),
the following result,
E.158) и(х, у, z, г) = tMct[f] + Jf {*Mct[g]} .
DIMENSIONAL FOURIER TRANSFORMS 275
expression E.158) is often called Kirchhoff's formula for the solution
^the wave equation.
Using Duhamel's principle (see Section 4.5), we readily obtain the
tion of the inhomogeneous wave equation
159) utt - c2[uxx + uyy + uxx] = G(x, y, z, t) ,
initial data E.157), as
E.160) u(x, y, z91) = tMct[f] + j( {tMct[g}}
+ /0 (* - t)Mc(,.t)[G(x, y, z, t)] dr .
The integral term on the right in E.160) can be expressed as a triple integral
over the interior of the sphere r = ct with center at (x, y, z) and radius ct,
where r = V(€ "~xJ + i7!~~ З'J + ( £ ~ zJ- ^n rectangular coordinates, the
integral takes a form referred to as a retarded potential
E.161)
In the retarded potential the contribution to the solution at the point
(*, y9 z) and at time t come only from the points in the interior and on a
sphere of radius ct with center at (xy y,z). Since the speed of wave
propagation is c, the function G, evaluated at the retarded or earlier time
t - rlc, is integrated over a sphere of radius r, as the effect of these points
reaches the center of the sphere at the time t.
Before discussing the solution E.158), we obtain from it by means of
Hadamard's method of descent the solution of the initial value problem for
the wave equation in two dimensions. In this method we look for a solution
of the wave equation E.145) independent of z with initial data
f=0
So l"at u(x> y91) satisfies the wave equation in two dimensions
Eл63) и„ - c2[uxx + uyy] = 0, t > 0, -oo < jc, у < x .
th ^h6 introduce /(*' У) and 8(x, y) into the solution E.158) and because
^ e data are independent of z we may take the center of the sphere г = ct to
ljtt_(x, yy Q). Introducing spherical coordinates on the sphere r =
fS i ^ + ^ " y^2 + £2 = Ct With the surface element ds, we can write
w-158) as
276 INTEGRAL
E.164) „(,,,,,),-Л,-
The sphere r = ct has its center at (x, y, 0) and the (f, T7)-plane cuts it in the
circle У(€~~хJ + (r]-yJ = ct. Since the integrands in E.164) are indepen-
independent of £ we can integrate out the £ dependence by effectively projecting the
upper and lower hemispheres onto the circle in the (£, 17)-plane. Using the
relation
E.165) ds = sec у d£ dv ,
between the surface element ds on the sphere and the area element dg dr) in
the plane, where
E.166) secy=
and noting that we get the same contribution from each hemisphere, we
obtain
E.167)
where r = }/({-xf + D - yf. The integration in E.167) is carried out
over the interior of the circle of radius ct centered at (x, y). For the
nonhomogeneous problem
E.168) м„ - c2[Uxx + uyy] = G(x, y, t),
with homogeneous initial data, Duhamel's principle yields the solution
E.169) «..,.
where г = V(£*) + (*/H- , - nm
This technique is called the method of descent since we descend iron*
three to two space dimensions to construct a solution for the lower dimen-
dimensional problem in terms of the solution of the higher dimensional one. ItlS
DIMENSIONAL FOURIER TRANSFORMS 277
possible to descend from two to one dimension and construct d'Alem-
t's solution from the solution E.167).
berWe nOw discuss the solution E.158) and E.167) of the three- and
-dimensional wave equations, respectively. The basic distinction between
^two solutions is that E.158) represents integration over the surface of a
* here whereas E.167) represents integration over the interior of a circle.
This points out a sharp difference between wave propagation in three and
two dimensions.
We suppose that the initial data / and g are concentrated in a neighbor-
h od of a point Po in the two- or the three-dimensional case. That is, we
assume that they vanish outside an arbitrarily small neighborhood of Po. In
the three-dimensional case, E.158) states that the solution u(x, y, z, i) of
the wave equation, at the time / and at the point (x, y, z), depends on the
data / and g only at points intersected by the sphere of radius ct with center
at (*, У, z)- К this sphere does not intersect the given neighborhood of Po,
the solution u(x, y, z, t) = 0. As t increases, the sphere must eventually
intersect a neighborhood of Po (no matter where (x, y, z) is located). The
first time a disturbance is felt at (x, y, z) represents the time t it takes the
wave front, originating at the point of the neighborhood of Po nearest to
(x, y, z), to reach the point (x, y, z). Since the distance from the nearest
point in the neighborhood of Po to (x, y, z) is ct, the speed of the
propagation of the wave front is c. Because the data / and g are concen-
concentrated near Fo, as t increases the sphere will ultimately no longer intersect
the given neighborhood of Po and the solution u(x, y, z, t) will again be
equal to zero.
The preceding discussion shows that sharp signals or disturbances can
occur in three-dimensional wave motion. Not only does it take a finite time
for disturbances to travel from one point to another, but after the disturb-
disturbance has passed, the solution returns to zero and its effect is no longer felt.
This phenomenon is known as Huygens' principle, Huygens used this
principle to construct solutions of wave propagation problems. By consider-
lng a disturbance surface (or a sharp wave front) at the time t0, one
constructs a sphere of radius ct around each point on the surface. Then the
envelope of these spheres gives the location of the disturbance surface at the
toe t0 + л Although this construction only gives the front of the disturbance
ln 8eneral, in the case of concentrated sources (i.e., initial data concentrated
at a point Po) the construction gives the full disturbance.
. F°r the two-dimensional case, if the initial data / and g are concentrated
to a neighborhood of the point Po, the solution E.167) shows that u(x, y, i)
*ш be zero until the circle with center at (x, y) and radius ct intersects the
neighborhood of Po. Once the circle begins to intersect the neighborhood of
0) the solution at the point (x, y) will not vanish as t increases. This results
ecause the integration in E.167) is carried out over the interior of the
Clrcle, not just its boundary. Thus although sharp wave fronts do exist in
w°-dimensional wave propagation, since it takes a finite time for the effect
278 INTEGRAL TRANSFORM
of the data near Po to reach the point (x, y), the effects of the disturb
linger and do not disappear sharply as in the three-dimensional case. Inste
they diffuse slowly to zero at t->°o. As a result, Huygens' principle in tK
above form is not valid in the two-dimensional case. However, Huyge *
construction as given can be used to give the location of the wave
The solutions given for the two and three-dimensional wave q
can be used to determine domains of dependence and influence as d
earlier in the one-dimensional case, but this is not carried out here. Also
seen in the exercises, Hadamard's method can be applied to other
tions.
Example 5.8. The Modified Helmholtz Equation. We consider the
(inhomogeneous) elliptic equation
E.170) uxx + uyy - k2u = f(x, y)y -°°<x,y<oo,
where the constant к > 0, /(jc, y) is specified, and w(jc, y) satisfies the
additional condition
E.171) u(x,y)-+0as\xl\y\-+°°.
We assume that /(jc, y) has a Fourier transform. The equation E.170) is
sometimes referred to as the modified Helmholtz equation, in contrast to the
case where - k2 is replaced by 4- k2 which is called the Helmholtz equation.
The equation E.170) can arise as a stationary version of two-dimensional
diffusion equations.
To solve E.170), E.171) we apply the two-dimensional Fourier trans-
transform to E.170) and easily obtain
EЛ72, „,*)--jj
where X = [Als A2], and t/(X) and F(X) are the Fourier transforms of u(x, y)
and /(jc, y), respectively. [We have assumed in obtaining E.172) that not
only и but also its first partial derivatives vanish at infinity.]
Inverting the transform gives the solution
Since F(k) has the form
E.174) F(X) =
we obtain, on inserting E.174) into E.173),
GHER DIMENSIONAL FOURIER TRANSFORMS 279
«.175) u(x,
00
"£?! fff exp{-i[AI(Jc -i) + hb " n)]}
nterchanging the order of integration. Introducing polar coordinates in
Jhe two inner integrals in E.175), that is,
/5Д76) Ai = P cos <P> A2 = P sin <P у
and also expressing x - f and у - т/ in polar form
we readily obtain
E.178) Ax(* - £) + X2(y -17) = pr cos((jp - B) .
Then transforming to polar coordinates gives
E.179) Г Г е«Р{--[А,(Гтл,(,-,I}
Using a well-known integral representation of the Bessel function of order
zero J0(z),
E-180) /O(Z) = J- Г е" «"<*-•) rf<p,
Ztt Jo
we obtain, since /0(z) = /0(-z),
Г " «"<*-•) rf = constant,
A further result from the theory of Bessel functions yields
where #0 is the modified Bessel function of the second kind, some of whose
Properties are given in Chapter 6. When E.182) is introduced into E.175)
we obtain
280 INTEGRAL
E.183) u(x, y) = -^ j_ J^ K0(kr)f({, v) d£ dV ,
where r = V(x - £J + (y - tjJ.
With /(*, y) = S(x - xo)8(y - y0), the two-dimensional Dirac delta func.
tion, E.183) reduces to
E.184) u(x, y)=^ K0(kr), r = V(x-xoJ + (y-yof .
This is known as the free-space Green's function for the modified Helmholtz
equation. As shown in the exercises, E.170) becomes a modified Bessel
equation when expressed in a polar coordinate form independent of the
angular variable. We return to this problem in later chapters.
If we replace k2 by -k2 in E.170) we obtain the (inhomogeneous)
Helmholtz equation. On attempting to solve by using the Fourier transform,
we find that the transform of the solution is given by E.172) with k2
replaced by -k2. Thus, the transform is singular and the inversion formula
for the determination of u(x, y) cannot be applied. [The situation is
analogous to that encountered for E.28).] Rather than consider the general
case, we obtain the free-space Green's function for the Helmholtz equation.
To do so, we replace к by ~ik in E.184). Using a known relationship
between the function Ko of imaginary argument and the Hankel function of
the first kind Я(о1} yields
E.185) u(x9y) = (i/4)H{*\kr).
This solution satisfies the radiation condition A0.15) (where we put n = 1)
in view of the behavior of the Hankel function for large argument given in
F.206). Replacing к by ik in E.184) leads to a Hankel function of the
second kind, and it does not satisfy the radiation condition.
5.5, HANKEL TRANSFORMS
We have seen in Chapter 4 and the exercises for that chapter that certain
problems lead naturally to the representation of the solutions in terms ot
Bessel functions. In bounded regions we are led to the Bessel function
expansions considered in Chapter 4. For unbounded regions, in carrying out
separation of variables in two or three space dimensions in cases where the
Laplacian operator is expressed in polar or cylindrical coordinates, we are
often led to consider the following eigenvalue problem for the equation
involving the radial variable
n2
E.186) L[v(r)] = -(rv'(r))' + — v(r) = A2ru(r), 0< r <»,'
TRANSFORMS 281
h re Л2 is the eigenvalue parameter and the equation is written in the
tf-adjoint form D.56). (This is Bessel's equation of order n in the variable
Se \ Here n can be any positive integer or zero. We require that the
* enfunctions v(r) are bounded for all r. This implies that the eigenfunc-
tions are
E.187) v(r) = Jn(\r)
here Jn(z)-is the Bessel function of order n and the eigenvalues A can be
non-negative real number (i.e., 0<A<°°) so that the spectrum is
continuous. Since Jn(~z) = (-lOn(z), we need not consider negative val-
ues of A.
A function /(r) defined for r > 0 and satisfying the conditions given in
Section 5.3 has the representation in terms of the eigenfunctions E.187),
E.188) ^
where the Hankel transform F(\) is given by
E.189) ^A) = /o rJn(\r)f(r)dr, n>0.
[We note that an alternative form of these formulas is given in some texts,
with the terms A and r in E.188), E.189) both replaced by VXr, thereby
giving the formulas a more symmetric form.] The formulas E.188), E.189)
can be obtained from the two-dimensional Fourier transform formulas, and
this derivation is considered in the exercises.
For each value of the integer n>0, the integral E.189) determines a
Hankel transform of order n. As has been demonstrated, the usefulness the
Fourier, sine and cosine transforms in solving boundary value problems for
differential equations rests on the fact that transforms of derivatives of
functions can be related to the transforms of the functions themselves. The
corresponding property for the Hankel transform is contained in the follow-
following equation. (The bracketed term on the left arises directly or after some
intermediate calculations when the Laplacian is expressed in polar or
cylindrical coordinates.)
In applying the Hankel transform of order n to a given equation we are
le<* to consider an expression of the form
E.190)
282
INTEGRAL TRANSFORM
This result is obtained on integrating by parts and is closely related
D.114). In general, the bracketed term on the right is required to vani h°
This occurs if / and fr are bounded at the origin and Vrf and Vr/r vanish
infinity. Note that for n^O the Bessel function Jn(kr) is bounded at ze^
and vanishes like 1/Vr at infinity. In certain problems, we permit /to h°
singular at r = 0 but in such a fashion that the limr^0 rfrJn and limr^0 rfj' \?
finite.
Although comprehensive tables of Hankel transforms are available w
list several results for use in the examples. The equation
E.191) |o e~XzJ0(\r)d\= r-L—2, z,r>0
shows that the Hankel transform (n = 0) of/(r, z) = 1 /Vr2 + z2 is F( Л, z)
(\lk)e~kz. Using the transform formula E.189) we obtain
E.192)
The result E.191) can be verified by using the integral representation of the
Bessel function /0( kr) and interchanging orders of integration (see Exercise
5.6.10). A second result that can be obtained by using the series expansion
of the Bessel function /„(Ar) and integrating term by term is
E.193) JT A7n(Ar)e-V dX = 1 (i)" exp(-£) .
We cite additional results as they are required and now consider several
examples in which Hankel transforms are used to solve problems for
equations of elliptic, hyperbolic, and parabolic types.
Example 5.9. Laplace's Equation: An Axial Source. We consider Lap-
Laplace's equation in three dimensions with a source concentrated on the z-axis.
Introducing cylindrical coordinates (г, в, z) and noting the axial symmetry
we ask for a solution of
2m +
E.194) V2m - urr + - ur + uZ2 - 0, r > 0, -oo < z <
with the conditions on the z-axis given as
E.195)
lim r2u{r, z) = 0, lim 2тгг ^ (r, z) = -/(z), -co < z < »
TRANSFORMS 283
that/(z) is a measure of the strength of the source. [Note that in E.194)
s° 0.dependence of the Laplacian is dropped because of the assumed axial
*e metry. A discussion of concentrated source problems is given in Section
6 Apply"* the zer°-°rder Hankel transform to E.194) [in view of E.190)]
multiply across by r/0(Ar) and integrate from 0 to ». We use /0(Ar) in
^u- case because the term (n2/r2)u does not occur in equation E.194).
Defining the Hankel transform f/(A, z) of w(r, z) as
E.196) ^'^"Jo rJo(br)u(r, z) dr
and using E.190) we obtain
E.197) 0 = JT r/0(kr)[urr + \ ur + uZ2] dr = - A2t/( A, z) + ^^ Z)
Now 70@) = 1, Jt0=-Jl and /х(Аг)«Аг/2 as r-»0 as is well known.
Assuming that the contributions from the limit at infinity vanish in E.197)
[u(r, z) is assumed to be of that form] and noting that ruJ&kr) » ~{kr2l2)u
as r-»0, we obtain on using E.195),
E.198) Uxz - A2f/= - ^- /B), ~oo < z <oo .
Requiring that (/(A,z)-»0 as |z|-»<x>, we have precisely the problem
considered in Example 5.1 at the beginning of this chapter. The solution is
given as
E199)
on comparing with E.23). Inverting the transform U(X, z) gives
E.200)
u(r, z) = |o A/0(Ar)U(A, z)d\=-^ | [ e-*-%(\r)f(s) d\ ds
4ir J- yr2 + B _ sJ
On mterchanging the order of integration and using E.191).
The term {XIA-afti2 + (z - sJ]" is the source function or free space
reen's function for Laplace's equation, with the source point at (x, y, z) =
284 INTEGRAL TRANSFOftM
@,0,5), (see Example 6.13). Thus, the solution E.200) represents
superposition of point source functions over the z-axis with density ft \
Example 5.10. The Wave Equation in Two Dimensions. We consider th
initial value problem for the two-dimensional wave equation over the em* *
space with radially symmetric initial data. We introduce polar coordinat^
(г, 0) and look for a solution w(r, t) independent of 0. Thus we ^
E.201) utt -c2(urr+|Wr)=0, r > 0, * > 0 ,
with initial data
E.202) M(r,0)=/(r), n,(r,0) = g(r).
Applying the (zero-order) Hankel transform to E.201), E.202) gives
E.203) Utt + c2\2U = 0, t>0
E.204) £/( A, 0) = F( A), tf,(A, 0) = G( A)
where U, Fy and G are the (zero-order) Hankel transforms of w, /, and g,
respectively, and E.190) with n = 0 has been used. [It is assumed that и is
such that the boundary terms in EЛ90) vanish.] Solving E.203), E.204)
gives
E.205) Щк, t) = F(A) cos(AcO + —^ sin(AcO .
Inverting the transform yields
E.206)
f°° 1 f°°
u(r, t) = AF( A) cos( kct)J0{kr) dk + - G( A) sin( Ac/)/0(A?") dA •
jo с jo
For specific choices of f(r) and g(r) we can occasionally evaluate the
integrals in closed form. Thus if
E.207) f(r) = > 2 ^, g(r) = 0, A, a = constant,
we have on using E.192)
E.208) F( A) = 4^ е~Ав' а > ° '
A
NKEL TRANSFORMS 285
^209)
This gives
M<r' ') = Aa L e~Xa со<*сОМ*г) d\ .
210) M<r'
The integral may be evaluated by expressing it as
where Re[ ] = the real part of the bracketed quantity. Using E.191) we
have
E.212)
"^""U. + W' + r']
By requiring the complex square root in E.212) to be positive when f = 0
(i.e., when it is real valued), the initial conditions E.207) are satisfied by
this solution. [On equating the bracketed term in E.212) to the complex
number a + ib, squaring both terms and solving the resulting algebraic
equations, the real value of E.212) may be determined, but we do not carry
this out].
Example 5.11. The Diffusion Equation with Axially Symmetric Data.
We consider the three-dimensional diffusion equation in cylindrical coordi-
coordinates where the concentration u(x, y,z,i) has an initial distribution that
depends only on r = V*2 + y2- We introduce cylindrical coordinates (r, 0, z)
and look for a solution w(r, i) of the equation
E*213) ut = D\urr + - urj , r>0,f>0
where D>0 is the diffusion constant, with the initial condition
E214) M(r,0)=/(r), r>0.
Applying the (zero-order) Hankel transform to E.213) and E.214) we
obtain on using the notation of Example 5.10,
E.215) ^ + DA2t/ = 0, t>0
at
E-2l6) f/(A,0) = F(A).
Thus
286 INTEGRAL
E.217) 2
and
E.218) u(r,t) = jo \Jo(\r)U(k,t)d\ = jo \e-°*2'J0(\r)F(\) d\
-Г Г
Jo Jo
AsJ0(\r)J0(\s)e~DK2tf(s) d\ ds ,
where we have interchanged the order of integration in the last integral. \y<
cite a known result to evaluate the inner integral. That is,
(S.219)
where /0(z) is the modified zero-order Bessel function discussed in Chaptei
6. We note the property of /0(z) that /0@) = 1, so that for s = 0 in E.219)
[since /0@) = 1] we find that E.219) reduces to E.193) with n = 0 and
D = 1. Inserting E.219) into E.218) gives
E.220) u(r,,) = ± jT exp[ - ^
It is of interest to show how this solution reduces to the fundamental
solution for the two-dimensional heat or diffusion equation if we allow /(r)
to represent a concentrated source at r = 0. We require that with f(r)^0
E.221) lim 2ir J[ f(r)r dr = 1,
where we assume the source is concentrated in a circle of radius e and let the
radius tend to zero but keep the source strength fixed at unity. It may be
assumed that f(r) vanishes for r > e. Using the generalized mean value
theorem for integrals we have [since f(z) ^ 0]
E.222) u(r, г) = ± «p[- £±p]j,( A) /; f(s)s
where s = £(€)-»0 as €—>0. Then as e-*0 we obtain
E.223) M(r, 0
on using the result E.221) and /0@) = 1. This is the fundamental solution of
the diffusion equation in the two-dimensional case. It should be compare
with the one-dimensional fundamental solution G(x, t) defined in E.45)-
дрьАСЕ TRANSFORMS 287
If the initial value f{r) is concentrated on a circle r = a, we require that
E.224)
that the concentrated source has unit strength. To determine the solution
^ this case, we assume that /(r)>0 and vanishes outside the interval
f1 - e a + e) thereby reducing the integral E.220) to one with finite limits,
p ceeding as in the point source problem we obtain the solution (on letting
До),
, ч 1 / r2 + a2\J ar
E.225) «(г, t) (A
In the limit as a tends to zero, E.225) reduces to E.223).
5.6. LAPLACE TRANSFORMS
The solution of initial and boundary value problems for hyperbolic and
parabolic equations by the use of the transform methods presented previous-
previously in this chapter involves the transformation of the spatial variable(s) in the
problems, as we have seen. The problems are thus reduced to the solution
of initial value problems for the transformed dependent variables. The
Laplace transform, however, solves these problems by acting on the time
variable in the hyperbolic and parabolic equations and yields boundary
value problems for the transformed dependent variables. (In its application
to Cauchy problems, the boundary conditions are given at infinity.) Even
though the nature of their applications is rather different, the Laplace
transform and its inversion formula can be derived from the corresponding
formulas E.10) and E.11) for the Fourier transform and its inverse. This is
the approach that we use. As the Laplace transform commonly acts on the
time variable, we formulate its properties in terms of the variable t.
Given a function j{t) defined for / > 0 and integrable over any finite
interval (say, it is piecewise continuous), we extend its definition to /<0 by
putting f(t) = о for t < 0. [It is not required that /@) = 0.] We assume that
Д0 is of exponential order as /->». That is, there exist real constants c>0,
*> and M such that |/(/)| < Mekt for r>c.Letabe an arbitrary constant
^th a > k. Then, the Fourier transform F(A) of e~atf(t) is given as
E26)
s«ice f{t) vanishes for negative t. (The notation for the transform has been
cnanged slightly.) Let A = a - i\ in E.226) and we have
288
E.227)
INTEGRAL TRANSFORM
-I e-"f{t)dt
where F(A) = Bтт) F[i(\- a)]. From the expression E.11) for the inverse
Fourier transform we obtain
E.228)
"КО-
I
Transforming again to the variable A = a — г'Л, we find that A = +00 corre-
corresponds to A= a ± /oo and that dk = -(l/i) dk. Therefore, the integration
path along the real A-axis in E.228) is transformed into a path along the line
L in the complex A-plane parallel to the imaginary axis. With A = kx + ;л2, l
is given as A = a + /A2, — °° < A2 < °° and has the direction shown in Figure
5.1. In terms of A the inversion formula E.228) takes the form
E.229)
since the term e at cancels out.
The formula E.227) determines the Laplace transform F(A) of f(t). The
complex inversion formula for the Laplace transform is given by E.229).
For a function f(t) that satisfies the conditions given above, the formulas
E.227) and E.229) are valid for all A such that Re(A) = Ax = a > k.
Given the transform F(A), the function/(;) can often be determined from
E.229) by using the residue theory of complex variables. Otherwise,
approximation methods are available to deal with E.229). As we do not
Figure 5.1. The line L.
[ e~ktfn\t) dt = A" I e^f(t) dt - \n'lf@) /«"""(О)
TRANSFORMS 289
resupp°se a knowledge of complex integration theory, the formula E.229)
does not play a (significant) role in our discussion. However, Laplace
sforms are extensively tabulated, and we can often invert Laplace
transforms by reducing F(\) to a form for which the inverse transform is
known. In this respect the linearity of the Laplace transform and the
onvolution theorem (see Exercise 5.6.3) play an important role. Thus if
°p( л) can be expressed as a finite (or even an infinite) sum each of whose
terms has a known inverse transform, we can invert F(A) given that the
nverse of the sum is the sum of the inverses. It is not always a straightfor-
straightforward matter to carry out this procedure, as we shall see.
The usefulness of the Laplace transform in solving problems that involve
initial conditions is brought out by the following result which relates the
transform of derivatives to the transform of the given function. If f(n\t) =
dnf{t)ldt\ we have
E.230)
on using integration by parts. The transform of the nth derivative is given in
terms of the transform of the function and n initial values of the function
and its derivatives at t = 0. By contrast, the sine or cosine transform of an
even order nth derivative of a function involves only nil values of the
function and its derivatives at x = 0. Additional«general properties of
Laplace transforms are given in the exercises and are found in tables of
transforms. We now consider two examples in which the Laplace transform
is used to solve initial and boundary value problems for the heat and the
wave equation.
Example 5.12. The Heat Equation in a Finite Interval. We again con-
consider the heat equation in a finite interval 0 < x < /,
E-23l) u, - c2uxx = 0, 0<x<U>0
with the homogeneous boundary conditions
и@,0-0, m(/,0 = 0, t>0,
and the initial condition
E>233) "(*,<))=/(*).
*s problem was solved by separation of variables in example 4.10.
owever, we now obtain the solution in a different form and compare both
results.
problem E.231)-E.233) is solved by applying the Laplace transform
290 INTEGRAL
in the t-variable. We multiply in E.231) by e~kt and integrate with respect
t from 0 to oo. Defining the transform of u{x, i) as °
E.234) U(x, A) - Jo е~А'и(х, t) dt
and using E.230) we have
E.235) A£/(jc, A) - f(x) - с2 1Л^Л = 0, 0< x < I.
ox
From the transform of the boundary conditions E.232) we obtain
E.236) (/@, A) =[/(/, A) = 0.
The solution of the boundary value problem E.235), E.236) is found to be
E.237)
^ (/- J)]/M <fc) .
This result may be obtained by using the Green's function or variation of
parameters method for ordinary differential equations.
The inversion of the transform U(x, A) is not straightforward, even if the
inversion formula E.229) is used, unless the transform is simplified first. In
view of the discussion preceding E.229) we may assume that A (or | A|) is not
small. Consequently, we have the following result
E.238)
1 2 2exp(-VX//c)
sinh(VI//c) ~ exp(VA//c) - exp(-Vll/c) " 1 - exp(-2VX//c)
-VX2k[\
с /
= 2exp( —— I Z expl
\ с / ^=0 \
on using the series 1/A - x) = E£=o xk which is valid for |jc| < 1, since it may
be assumed that A is large enough so that exp[-2VX//c]<^l. Further, the
hyperbolic functions in the integrals E.237) can also be expressed in terms
of exponentials. We combine the exponentials and assume that in the
inversion process, summation and integration, as well as orders of integra-
integration, can be reversed (this can be shown to be valid for this problem)-
Further, we find from a table of Laplace transforms that
aPLACE
TRANSFORMS 291
ing E-239)' *e solution «(*> 0 can then be written as (after a tedious but
iehtforward manipulation of the resulting series expansions),
E.240)
u(x, t) = f E № - s - 2kl, t)-G(x + s- 2kl, t)}f{s) ds ,
JO £=_„>
where G(x - s, t) is the fundamental solution of the heat equation defined in
E.45).
The solution E.240) has a form quite different from th;;t given in terms
of Fourier sine series in Example 4.10. By the uniqueness theorem for the
initial and boundary value problem for the heat equation proven in Chapter
8, both solutions must be identical. The equivalence of both sums is
demonstrated in the exercises at the end of Chapter 7 based on the Poisson
summation formula.
We remark that both solution forms are useful in evaluating the solution
u(x, i) for different ranges of values of f. Because of the rapid exponential
decay of the terms in the Fourier series D.154), the series converges very
rapidly for large values of Л However, the series of fundamental solutions
converges most rapidly for small values of t since all the terms in the series
except G(x- s, /), which has delta function behavior in the interval 0<x
< I, vanish at t = 0. Thus for the purposes of computation each of the series
is useful.
Example 5.13. The Wave Equation in a Finite Interval. We consider an
initial and boundary value problem for the wave equation in a finite interval.
We have
E241) utt - c2uxx = 0, 0<x</,*>0,
with the homogeneous initial conditions
E242) и(*,0) = 0, u,(jc,0) = 0,
and the boundary conditions
E*243> "@,0 = 0, «(/.O
his problem can be solved by the method of finite sine transforms given in
action 4.6, but we now obtain the solution in a different form by using the
LaPlace transform to solve the problem.
292
INTEGRAL TRANSFORM
Let U(x, A), defined as in E.234), denote the Laplace transform of
solution u(x, t). We transform E.241) and use E.230) and the
conditions E.242) to obtain
E.244) K2U - c2Uxx = 0, 0 < x < I,
E.245)
where G( A) is the Laplace transform of g(t). The solution of the boundarv
value problem for U(x, A) is
E.246) Щх, A) - G(A) sinh[(A/c)x]/sinh[(A/c)/].
To invert this transform we proceed as in Example 5.12 and expand
U(x, A) for large A, by expressing the hyperbolic functions in terms of
exponentials. This yields
E.247)
Щх, A) = X G(A){exp[£ (x - / - 2/A:)] - exp[-£ (x + I + 2flfc)]}
The inverse transform of each term in this series can be found from the
result of Exercise 5.6.4, and the solution to the given problem is
E.248)
u(x, t) = f IhU - - [A + 2k)l - x]}g\t - - [A + 2k)l - x]}
- H[t - -c [A + 2k)l + x]}g{t - \ [A + 2k)l + x]}}
where H(z) is the Heaviside function.
The form of this solution differs from that obtained by means of the finite
sine transform. That solution involves trigonometric functions and effective-
effectively represents u(x91) as a sum of standing waves. The solution form E.248) is
given in terms of traveling waves. The occurrence of the Heaviside functions
signifies that at given times additional right and left traveling waves or wave
fronts are generated as a result of reflection from the boundary. (A direct
method for constructing this type of solution is given in Example 6.7)
Clearly, this form of the solution is only useful for small t. For large t, the
standing wave representation is preferable. ____^
As we have seen in the foregoing examples, it is not always a simp'®
matter to invert a Laplace transform resulting from the solution of a partial
differential equation. Consequently, it is of interest to examine methods that
pLACE TRANSFORMS 293
eld аррг°х*та1е expressions for inverse Laplace transforms. The Abelian
\d Tauberian asymptotic theories for Laplace transforms that relate the
\lues of the transform F( A) as Л-» oo and A-> 0 to the values of the function
l(t) as f-»0 and *-»», respectively, are useful in that regard. Given the
resence of the exponential term in the formula E.227) for F(A) and
Pssuming f{t) is reasonably behaved, we expect that the main contributions
f the integral for F(A) come from small values of A*. Thus if A is large, t
must be small and vice versa.
A useful Abelian result states that if as f->0 we have
E.249)
then for the transform F(A) we have, as A
E.250) F( A)« 2 otn ^r,
=0
The converse is also true, so that E.250) implies E.249). The result E.250)
can be obtained by formally substituting E.249) in the integral F(A) an<J
integrating term by term. The series E.249) and E.250) need not converge,
but may have only asymptotic validity. This means that for small t and large
A, f(t) and F(A) are well approximated by the leading terms in the given
series even if the series diverge. (See Section 5.7 and Chapter 9 for a more
complete discussion of the meaning of asymptotic equalities and series.) In
the theory of asymptotic expansions of integrals the relation between F(A)
and /@ given by E.250) and E.249) is a special case of a general result
known as Watson's lemma. The converse, which is more useful for our
purposes, is attributed to Heaviside.
A simple application of the foregoing Abelian result can be given in
connection with the initial value problems formulated in Exercises 3.4.2 and
3.4.4 for the heat and telegrapher's equations. If these problems are solved
РУ means of the Laplace transform and the transform function is expanded
jn inverse powers of A as in E.250), the power series expansions of the
inverse transforms coincide with the results obtained in these exercises. The
exPansions obtained in the two examples in this section are also Abelian
results that are valid and useful for small t, since they were obtained by
expanding the transform for large A. However, the solutions are not in the
onn of power series in t, since the large A expansion does not have the form
small values of A, a useful Tauberian result which requires a
dge °f comPlex variables theory for its application is as follows. Let
be analytic for Re(A)> k. Assume further that F(A) has isolated real
ncl simple poles in the finite A-plane located at the set of points {Art} with
294 INTEGRAL
Aa > A2> A3> • • •. With f(t) assumed to be real, suppose that F(\) u
(real) residues {pn} at the poles {An}. Then we have as
E.251)
which is valid as f-> oo. This result is a special case of a general technique f
inverting Laplace transforms known as the Heaviside expansion theorem To
apply this theorem we suppose that the transform function F(A) can be
written as F(A)= G(A)/#(A), where G(A) is an entire function of A, and
H(\) has simple zeros at the points A = Art. Thus, F( A) has only simple poles
at A = Art. (They need not be real valued.) Then, on using residue theory to
evaluate the inversion integral E.229), we obtain the series
E.252)
The residue of F(A) at А„ is given by G(AJ/#'(AJ. [The series E.252)
may have only asymptotic validity.] If the poles Art satisfy the additional
conditions given above, we obtain a Tauberian result in that the terms in the
series E.252) grow or decay exponentially, so that ДО is well approximated
by the leading terms in the series for large t. However, if the Art are all pure
imaginary numbers for example, each of the exponentials in E.252) has unit
magnitude so that no simplification results for large values of t.
To obtain a Tauberian result for Example 5.12 we note that the transform
function U(x, A) is apparently singular at the zeros of sinh(V'A//c). These
are located at the points Art = -(ttc//Jw2 (n = 0,1,2,. . .). It is easily seen,
however, that Ao = 0 is not a singular point of U(x, A). Since the zeros {Л„}
(n > 1) are all simple, U(x, A) has simple poles at these points. The Art
satisfy the condition \l > A2 > • • • and it can be verified that the residues
{($n(x)} of U(x, A) at {An} are all real so that the conditions for the
aforementioned Tauberian theorem are met. The Tauberian result u(x, 0 я*
E"=1 pn(x) ехр[АпГ] can be shown to be identical with the Fourier sine series
D.154). Again, that series is convergent and not merely asymptotic and is
useful for large t.
In Example 5.13 the poles of the transform function E.246) are de-
determined from the zeros of sinh[(A/c)/] and the poles of G(A). The
hyperbolic function has simple zeros at the points \n = iircn/l, n~
±1, ±2,... (Ao = 0 does not contribute a pole). If we assume that the poles
of G(A) are all simple and do not coincide with the Art, the expansion form
E.252) (appropriately modified) is valid for this problem. However, if the
poles of G(A) have a nonpositive real part, it does not yield a usefu
Tauberian result. The pure imaginary zeros of the hyperbolic sine function
contribute the most significant terms for large t, and none of these terms
decays in magnitude as / tends to infinity.
APPROXIMATION METHODS FOR FOURIER INTEGRALS 295
7 ASYMPTOTIC APPROXIMATION METHODS FOR FOURIER
JjTEGRALS
this section we discuss two useful approximation methods for evaluating
rals that result from the application of Fourier transform techniques.
We give only a brief discussion of these methods and apply them to two
roblems relating to dispersive and dissipative wave motion (these concepts
^ere defined in Section 3.5). We refer to the literature for additional details
carding the validity and applicability of these and other asymptotic
methods for evaluating integrals. We say that the function f(x) is asymptotic
to the function g(x) as jc-> a and denote the asymptotic equality by
Я*) «*(*), x^a
if the ratio f(x)/g(x) tends to unity as x-» a.
We first consider the method of stationary phase which we apply to a
solution of the Cauchy problem for the Klein-Gordon equation representa-
representative of dispersive wave motion. Then we consider a method developed by
Sirovich (which is related to the given Tauberian theory for Laplace
transforms) and apply it to a solution of the Cauchy problem for the
telegrapher's equation representative of dissipative wave motion. (Some
properties of both these equations were given in Example 3.8.) Neither of
these methods requires complex variables theory for its application.
It was seen in our discussion of transform methods in this chapter that
one cannot, in general, expect to evaluate integrals resulting from transform
methods in closed form. Consequently, the methods discussed are essential
m that they yield simple and useful representations of the solutions.
Example 5.14. Dispersive Wave Motion: The Method of Stationary
Phase. The method of stationary phase is an asymptotic approximation
method for evaluating integrals of the form
wnere A: is a large real parameter, f(t) is a real or complex valued function
and the real valued function cp(t) is called the phase term. [The case of finite
endpoints in the integral E.253) is considered below.] Since к is large and
*K0 real valued, the integrand in E.253) oscillates rapidly and cancellation
^curs over most of the domain of integration. However, near the points t
t- ere 4>'{t) = 0, that is, the stationary points of the phase, no such cancella-
*°n occurs because of the diminished oscillation of the integrand. Thus the
aJ°r contribution to the integral comes from the neighborhood of these
Points.
We assume that <p(t) and f(i) are smooth functions and proceed as
296 INTEGRAL
follows. Let t0 be a stationary point and assume that <p"(*o) ^ ° and/00) *0
Expand <p(t) and f(t) in a Taylor series around t = t0 and obtain °Uin '
<pVo) = o] e
E.254) <p(t) = <p(t0) + W\tQ){t - t0J + • • •, №=f(t0) + • -..
Only the terms given in E.254) are retained in the (leading order) stationary
phase approximation. Let e>0be a small number. Then the method of
stationary phase asserts that the main contribution to the integral I(k) fOr
large k, if there are no other stationary points, is given as
E.255) I(k)~[°_\xp[ikcp(to)+ | <p"(to)(t-toJ]f(t0)dt
= exp[ikcp(to)]f(to) £_' exp[y <p"(to)(t - tof] dt
]/(/„) J_и ехр[ у V"('o)('" ^o)
where we have replaced the finite limits to± e by ±o° since the main
contribution to the infinite integral (see the discussion below) comes from
the neighborhood of t0. The last integral in E.255) can be evaluated and we
obtain the asymptotic result
E 256^ I(k)^e
E.256) /(*) е
The formula E.256) represents the dominant contribution to the integral
I(k) for к > 1 if t = t0 is the only stationary point. If there is more than one
stationary point, there is a contribution of the form E.256) from each such
point and we sum all the contributions. If <p"{t0) - 0 so that there is a higher
order stationary point, the above method fails and a different discussion
which requires the expansion of <j>{i) up to cubic or higher powers is
necessary to yield the contribution from that stationary point.
If the phase <p{t) has no stationary points, so that <p'@^° for a11 '' we
determine the behavior of E.253) by integrating by parts. To do so, we
replace the infinite limits in the integral by the finite limits a and b. After
integrating by parts in the modified integral, we let a and b tend to minus
and plus infinity, respectively. On integrating by parts once, we obtain
ikla dt [(р'^Г^41
297
gVjylpTOTIC APPROXIMATION METHODS FOR FOURIER INTEGRALS
tf the endpoint contributions vanish as a and b tend to infinity, we conclude
hat I(k) decays at least like I/A: as A;-*oo. [Even if the endpoint contribu-
l. s do not vanish in the limit, we can transform E.253) into a Fourier
Utegral and conclude from the Riemann-Lebesgue lemma that the integral
fishes as A;—>°°, but perhaps at a slower rate than I/A:.] If n integrations
L parts can be carried out and at each step the endpoint contributions are
его, the integral I(k) decays at least like l/kn as A:-»<». Clearly, if an
unlimited number of integrations by parts can be carried out, I(k) is in effect
exponentially small for large k.
If one or both of the endpoints in E.253) are finite (we denote them by a
and b, as above), we must include the contributions from these endpoints in
our asymptotic approximation of /(/:). [We assume that the endpoints are
not stationary points of <p(t).] The contributions from the endpoints are
obtained from the integration by parts formula E.257). Thus, if f(t) does
not vanish at the endpoints, their contribution is of order I/A: as &—><». If
<p(t) has no stationary points, the endpoint contributions are dominant.
Otherwise the stationary point contribution, given in E.256), which is of the
order of 1/Vfc is dominant. [If an endpoint is also a stationary point, E.256)
multiplied by \ gives the asymptotic value of I(k) at that point.] If the
function/(/) in E.253) is piecewise smooth, we break up the integral into a
sum of integrals in each of which f(t) is smooth. We do not discuss the case
of a higher order stationary point.
We now apply the method of stationary phase to the solution of the
Cauchy problem for the Klein-Gordon equation
E.258) utt - y2uxx + c2u = 0, t > 0, -<» < x < *>,
with initial data w(x,0) and n,(x,0) specified at t = 0. Let t/(A, t) be the
Fourier transform of u(x, t), that is,
^e Fourier transformed equation E.258) is given as
E.260)
df
Producing the Fourier transforms of the initial data, we easily determine
at ЩЛ, t) can be written as
U = U+(\, t) + U_(\, t)
+ F_(A)
298
INTEGRAL TRANSFORM
The terms F±(k) can be specified in terms of the transforms of the injt-
data for u(x, t). Since we are only interested in a general characterization ^
dispersive wave motion, we are not concerned with the specific form of th *
functions F±(A). We do require, however, that F±(A) be smooth function
This is the case if the initial data vanish sufficiently rapidly at infinity i%
example. ' Or
Inverting the Fourier transform U(k, t) gives
E.262) u(x, t) = ^= j m F+(A) ехр{г[о>+(А);- Ах]} dk
+ ^ /_ *-(A) exp{*>_( A); - Ax]} dk
where
E.263) o>±(A) = ±V?2A2 + c2 - ±ЦA) .
The equation w = <o( A) is the dispersion relation for the Klein-Gordon
equation and the terms exp{i[±a)(k)t- kx]} are the normal mode solu-
solutions. (In comparing with the results in Example 3.8, the parameter к used
there is replaced here by -A.) The solution E.262) is thus a superposition of
normal mode solutions with the phase terms
E.264) <p±(x, t, A) = ±u)(k)t - kx .
The phase velocity, that is, the velocity dxldt for which <p±(x, t, A) remains
constant, is given as
We see that for each real value of A we have \dxldt\ > y, so that the phase
speed of the normal modes exceeds the characteristic or wave front speed у
of the Klein-Gordon equation. The characteristic speed represents the
maximum speed of propagation of disturbances for hyperbolic equations.
This property has already been demonstrated for the wave equation and is
shown to be valid for more general hyperbolic equations such as E.258)
later in the text. Thus the physical significance of the phase velocity must be
examined more closely.
Noting that the general solution E.262) of the Klein-Gordon equation is
a superposition of the normal modes with phase terms E.264), each о
which has a different phase speed, it appears that the appropriate object to
consider is not a single normal mode but a group of normal modes or a wav*
packet. A wave packet is obtained by superposing a collection of norma
modes with wave numbers A ranging over some interval. Since each mode
FOURIER INTEGRALS 299
мрТОТ1С APPROXIMATION METHODS FOR
a different phase velocity, the group of normal modes disperses as it
haS agates and fails to retain its shape. We now show by using the method
Pf stationary phase, that although individual modes in a group travel at
Afferent velocities, the wave packet as a whole retains its group character
d travels with a velocity appropriately known as the group velocity. (It can
аП hown that an energy associated with the wave packet travels with group
locity, but we do not demonstrate this.) We apply the stationary phase
^ethod'to each of the integrals in the solution E.262), choosing x and t,
which are parameters in the integrals, to be large. A single stationary point
i is found for each choice of x and t. Since the method determines the major
contribution to each integral as coming from values of A near the stationary
noint, we thereby obtain a wave packet whose properties we examine.
It Is shown in Exercise 5.7.3, that the F_(A) integral in E.262) is the
complex conjugate of the F+(A) integral. Thus it suffices to consider only
the F+(A) integral, and the solution w(x, t) equals twice the real part of that
integral. Let
E.266) /(x, 0 = ^= j_ F+ ( А) ехр[/(У?2А2 + Л - Ax)] dk ,
and assume that both x and t are large. If we were to set x = at where a is
constant, then either x or t would play the role of the large parameter к in
the integral E.253) so that the stationary phase method could be applied to
E.266) and we do so without bringing the integral into the exact form
E.253).
The stationary points of the phase <p+(x, ty A) are determined from
E.267)
This implies that
E.268) * = ,
and the stationary value A is given as
300
INTEGRAL
The expression (o'(X) = у2\Г\/у2\2 + с2 is known as the growp velocity
As is seen from E.268), there is a common stationary value A for all (x \
that satisfy the equation x/t= o/(A) [that is, different values of x and t$
not lead to different stationary values A if the x and t are related b
x = <o'(A)t]. It follows from the stationary phase method that the solutio
retains a fixed character or form at a point x moving with velocity <o'(\) *
the wave packet with wave numbers A near the stationary value is observed
We note that the group speed \a>'(\)\ is less than у (the characteristic speed)
for all values of A. In fact, there are (real) stationary values A only for x and /
such that y2t2-x2>0 as is seen from E.269). If x and t are such that
yV-jc2<0, the stationary phase result is I(x, t)**0 since there are no
stationary points and we assume that an unlimited number of integrations by
part are possible.
Now the lines x = ± yt represent the boundaries of the domain of
influence of the point @,0) for the Klein-Gordon equation (this has already
been shown for the wave equation and is shown to be valid for the
Klein-Gordon equation in Chapter 7). Thus in the context of the stationary
phase result we have I(x, t)**0 and, equivalently, the solution u(x, t)^h
outside the domain of influence corresponding to a concentrated or point
initial source at x - 0. This suggests that for large x and /, the solution
м(лг, t) (under the foregoing assumptions on the initial data) appears to
correspond to the solution of a problem with initial data concentrated at the
origin. The group velocity lines x-o)\k)t lie within the domain of in-
influence, as shown in Figure 5.2. (This contrasts with the phase velocity lines
all of which lie outside the domain of influence.) As the values x and /
approach the characteristic lines jc= ±yt, E.269) shows that the stationary
value A tends to infinity. Then the stationary phase method, as presented,
Figure 5.2. The group line.
APPROXIMATION METHODS FOR FOURIER INTEGRALS 301
and a different discussion, which may be found in the literature, is
ssary to discuss the solution near the characteristic lines. We observe
t as (AH00 E.268) implies that |<w'(A)|->% so that the group speed
thads to the characteristic speed.
To complete the determination of the asymptotic approximation of I(x, i)
e note that *>"( A) = угс\у2A2 + c2)'3'2 > 0 and <p'l(A) = a>"(\)t > 0 so that
*"(A)/k+(A)i = 1# Consequently, we obtain from E.256)
1 Г Г 7г
E.270) /(*> 0 ~ г ff(A4ii/2 f+(A) exp|i[oi(A)r - A* + -
where E.269) was used to express A as a function of xlt and H+ -
F [о*"]1 П- This asymptotic formula determines the amplitude and the phase
of a wave packet moving with the group velocity dxldt — o>'( A) (-» < A <
oo) and is valid for |jc| < yt. The amplitude of the wave packet decays like
l/v7as/->°°. Asjc-» ±ytor |A|->», we see that o>"(A)-»0, so that E.270)
becomes invalid near the characteristic lines as was indicated. In the region
|jc| > yt9 that is, outside the characteristic sector indicated in Figure 5.2, we
have I(x, t)« 0 if the conditions given above are met, since the integral has
no stationary points.
We conclude our discussion by observing that the initial value problem
for the Klein-Gordon can be solved exactly in terms of Bessel functions
(this is demonstrated in Chapter 7). However, the foregoing asymptotic
characterization of dispersive wave propagation and the significance of the
group velocity and wave packets are not apparent from the exact solution
without further analysis. In fact, in a discussion of asymptotic methods for
partial differential equations in Chapter 10 we shall use the general form of
E.270) to construct approximate solutions of the Klein-Gordon equation
directly. Finally, we emphasize that E.270) is only valid for large times t as
our construction shows.
Example 5.15. Dissipative Wave Motion. We consider the Cauchy prob-
em for the dissipative wave equation (or the telegrapher's equation),
E*271) utt - uxx 4- ut = 0, *>0, -oo<jc<oo?
**h smooth initial data u(x, 0) and ut(x, 0) at t = 0. Using an asymptotic
Pproximation method for evaluating the integrals that result when the
auchy problem for E.271) is solved by means of Fourier transforms, we
etermine the behavior of u(x, t) as t tends to infinity. An exact solution of
e Cauchy problem can be given in terms of modified Bessel functions as
h°wn in Chapter 7. However, the large time behavior of the solution
302
INTEGRAL TRANSFORMS
obtained by our approach is not apparent from the Bessel function repre,
sentation unless further analyses are carried out. Because of the dissipative
character of the equation, the method of stationary phase is not suitable for
this problem and another approach is used.
We define the Fourier transform t/(A, i) of u(x, t) as in E.259) and
obtain the transformed equation of E.271) as
E.272)
d2U
эи
t>0,
This initial values £/(A, 0) and dU(\, O)/dt of t/(A, i) are given in terms of
the transforms of m(jc,0) and ut(x, 0). We are not concerned with the
specific form of the data since we want to characterize the solution in
general terms. Solving E.272) we have
E.273)
t/(A, 0 =
and the inverse transform yields the solution as
E.274)
^
where F+(A) and F_(A) are specified in terms of the initial data. We assume
that
E.275)
dk < M < oo .
To discuss the behavior ofjhejolutionfpxlarge values of t, we first notice
that for A2 > \ we have Vl-4A2 = A/4A2 - 1, since 4A2 - 1 > 0. Thus
E.276)
exp[(_ 1 ± 1 VT^P).] = e" exp[±| DA2 -1I'2/], A2 > \
and, consequently, we obtain
SVMPTOTIC APPROXIMATION METHODS FOR FOURIER INTEGRALS
E-277)
303
Me
~tl2
•n view of E.275). Therefore, the contributions to both integrals in E.274)
for values of A such that A2> J, are exponentially small for large t. For
jB<|we have
E.278)
E.279)
exp[(- \
as is easily seen. Thus the exponential in E.278) is maximal at A- ±5,
whereas that in E.279) has its maximum at A = 0. Combining the result
E.278) with E.277) we obtain an estimate for the F_(A) integral in E.274)
over the full interval of integration. We have
E.280)
V2tt
Г
Therefore, the entire F_(A) integral is bounded by {MlVl7r)e tl2 and we
have shown it to decay exponentially uniformly in x as f—>oo. The F+(A)
integral does not exhibit uniform decay as f->°°, and we expect that the
major contribution to the solution u(x,t) as f—»<» must come from
this integral. Furthermore, since the exponential E.279) that occurs
in this integral decays for all А Ф 0, we expect that the major contribution to
this integral for large t must come from the neighborhood of A = 0.
To evaluate the first integral in E.274) asymptotically, we introduce a
method developed by Sirovich for approximately evaluating Fourier integ-
integrals of the type considered in this example and which generally arise in
Problems of dissipative type. This method differs from the method of
stationary phase in that the integrals considered decay exponentially as the
Levant parameter gets large. Thus the major contribution comes from the
JJe*ghborhood of the point in the interval of integration where the integrand
.as its maximum absolute value. In the stationary phase method, the
Jntegral oscillates rapidly as the relevant parameter gets large and the main
^°ntributions come from the points where the oscillation is minimal.
irovich's method is closely related to approaches used in approximating
,aplace transform integrals, since for those integrals we also have exponen-
IaI decay as the transform parameter gets large.
304
INTEGRAL TRANSFORM
We now present Sirovich's method and then apply it to the F+(A) i
in E.274). We consider the integral
E.281) I(x, t) = |_да F( A) exp[-g( A); - iXx] dk
for which the following conditions are satisfied:
*2)
(i) J_jF(A)|</A<M<oo,
(ii) max|F(A)|<M,
(m) Re[g(A)]>0,
(iv) g( A) = 0 if and only if A = 0,
(v) g(A) = iaX + )8A2 + O(| A|3), a, K real, K >0, A-*0,
(vi) #(A) continuous in A.
Then as f—»<» we have
E.283) I(x, t) = J_e ехр[-)8А2Г - iakt - i\x\F{\) d\ + o\~\ >
where б is a small positive constant. The order term O[ • • • ] is understood to
mean that if F(x) = O[G(x)], then |F(jc)/G(jc)|-> A for some constant Л as
x-+ a. Thus O[\ltl~B] means that as *-><» the error term in E.283) decays
like A/tl~s. To simplify the integral in E.283) we introduce a further
approximation and expand F(A) in a Taylor series around A = 0 retaining
only the leading term F@). We then obtain
E.284) I(x, t)~\ exp[-j8A2* - iakt - bUr]F@) d\ .
J-00
This integral can be evaluated exactly [see E.39)-E.43)] and we have
E.285) /(*, 0
For a discussion of the error involved in replacing F(A) by F@) in E.283)
and for proofs, further developments, and refinements of this method, *e
refer to Sirovich's work. .
We are now in a position to apply the preceding method to the F+{*>
integral in E.274). To do so we identify F+(A)/V2tt with F(A) and the term
\ - £Vl-4A2 with g(X) in E.281). We assume that F+(A) satisfies the first
APPROXIMATION METHODS FOR FOURIER INTEGRALS
305
0 conditions in E.282) so that, in addition to being absolutely integrable,
с (A) must be uniformly bounded. Further, we have
E.286)
E.287)
E.288)
8( A) = \ ~ IV1-4A2 = 0 only at A = 0 ,
\ ~ ^Vl-4A2^ - *[1 A2 + O(|A|3)]
A2 + O(|A|3), A-»0
on using the binomial expansion. Also, g(A) is continuous for all A. Thus all
the conditions on g(A) are met, and we have a =0 and j8 = 1 in condition
(v) in E.282). Then E.285) yields
E.289)
which is valid as /^-»oo.
It may be remarked that to apply Sirovich's method to the integral in
E.281), we simply expand g(X) in a Taylor series around A = 0 retaining
terms up to quadratic order only. This yields the exponent in E.283). The
conditions given on g and F insure that over the full range of integration the
maximum contribution for large t comes from the neighborhood of A = 0.
Combining the results E.280) and E.289) we obtain
Recalling the form of the fundamental solution of the heat equation [see
E45) with £ = 0 and с = 1], that is,
we find that
E.292) M(x, t)«V2^F+@)G(x, i), *-► oo .
№ce the exponential in E.290) essentially differs from zero only in the
306
INTEGRAL
Figure 5.3. The large time behavior.
parabolic region x2/4t = 0A), the large time behavior of the solution u(x, t)
is as shown in Figure 5.3.
The solution E.292) has the form of a constant multiple of the fundamen-
fundamental solution of the heat or diffusion equation with a source point at the
origin. (Note that at large times the solution appears to represent the effect
of a source at the origin just as it was the case for the large time solution of
the Klein-Gordon equation.) The diffusion effect E.292) occurs in the wake
of the wave fronts or characteristics x = ±t after a long time, and yields the
major contribution to the solution u{x, /) since the dissipative effect has
damped out everything else. Although E.292) also dies out as f-к», it
decays only algebraically like Cxn for x2<4t rather than exponentially as
occurs for all other values of x.
For the more general dissipative wave equation or telegrapher's equation,
E.293)
ut- aux = 0, t>0, -
where a and с are positive constants and appropriate initial data u(x, 0) and
ut(x, 0) are assigned, the Fourier transform method readily yields
E.294)
Ф, 0- y= £ F+(A)exp[(-i + iVl«4c2A2-4/Aa)/-/Ax] dX
+^£F-(A)exp[D4
with F±{k) specified in terms of the initial data. The coefficient of t in the
exponential of the first integral has the following expansion around Л ~ ®-
vMPTOTIC APPROXIMATION METHODS FOR FOURIER INTEGRALS 307
E.295)
^l + i\Tl - 4c2A2 - 4/Aa - - \ + \[l - Щс2кг + 4/Aa)
= _ I + I _ /Aa - c2A2 + a2A2 + O(A3)
Unless a2<c2, the coefficient of A2 is positive, and the exponential grows as
t increases if A is small. Thus for c2<a2 the Cauchy problem for E.293) is
nstable since there are normal modes for small A that grow unboundedly as
t increases.
We assume, therefore, that a <c in E.293). It can then be shown that
the second integral in E.294) decays exponentially as *->«> and that
Sirovich's method can be applied to the first integral. We obtain
, ч F+(°) Г (X+Cttf
As was shown in Example 5.2, the expression E.296) is—apart from a
multiplicative constant—the fundamental solution of the diffusion equation
E.46) if we set с = -a and D/2 = c2 - a2 in that equation. In constrast to
the result obtained for E.271), the diffusion effect is now concentrated
along the line x - -at rather than along the /-axis. To observe the diffusion
as t-*oo9 We must let x-> -» such that x 4- at remains small. Since a < c,
the line of diffusion x + at = 0 lies within the characteristic lines x = ±ct. As
the diffusion effect may be thought to be traveling along the line x = -at
with velocity -a, we may consider E.293) to possess two speeds of wave
propagation. For small times t, disturbances travel at the characteristic
speed с to the right and/or to the left. At large times t, the dissipative effects
have damped out most of the initial disturbances and what remains in the
wake of the wave fronts is a diffusion effect characterized by E.296) that
propagates with speed a to the left.
The relationship between dissipative wave propagation and diffusion has
discussed in the models presented in Chapter 1. The foregoing
asymptotic results yield a further demonstration of the close relationship
between both effects. We have further occasion to touch upon this question
j^eU the text.
To conclude our discussion of asymptotic methods, we remark that even
, en exact solutions of problems are available, asymptotic approximations
3^1d useful and often significant insights into the behavior of solutions in
rent regions. In addition, by suggesting general forms for solutions in
308 INTEGRAL TRANFORMs
different regions, it is often possible to construct approximate solutions of
equations (say, when they have variable coefficients) for which no exact
solution is available by assuming that these solutions have the same gener I
asymptotic form. This idea is exploited in Chapter 10 when we discuss direct
asymptotic methods for partial differential equations.
EXERCISES FOR CHAPTER 5
Section 5.2
5.2.1 Show that the complete Fourier series of Example 4.6 may be
expressed in complex form as
with the Fourier coefficients ck given as
1 f'
l
This result may be obtained directly by using the orthogonality property of
the functions ехр[/(тгА://)х] over the interval -l<x<l, in the manner of
Exercise 4.3.4.
5.2.2. Show that the results in Exercise 5.2.1 may be expressed as
fix) = ± ^^ f'_, АО exp[-i T (* "
Assume that the A-axis (-<»<X<a>) is subdivided into intervals [Afc_1? Aj,
with kk = irkll, whose length is AAft = ttIL Rewrite the expression as
and show that as /-»oo this (formal) Riemann sum tends to the Fourier
integral formula E.9) if the limit exists.
5.2.3. Prove that if f(x) is absolutely integrable its Fourier transform E.10)
tends to zero as |л|->». (This is the Riemann-Lebesgue lemma.)
Hint: Let x = у + тг/А in E.10) and show that F( A) can be expressed in the
form E.10) with/(x) replaced by -Дх + тг/А). Take the average of the two
expressions for F(A) and let |Aj—>°°.
5.2.4. Consider the Fourier transforms obtained in Exercises 5.2.7, 5.2.9»
and 5.2.10, and determine how the rates at which these transforms tend to
zero as |A|->oo are related to the smoothness properties of the inverse
transforms f(x).
FOR CHAPTER 5 309
2#5# Show that the Fourier transform of xf(x) is -/F'(A).
'2 6. Obtain the Fourier transform У(Л) of the Airy equation y"{x)-
y(x) = °> with У^°) = 1/BъУ/2- Show that the inverse transform Ai(x) has
fhe integral representation given in Exercise 5.7.2.
5 2 7- show that the Fourier transform of the function fN(x) defined in
E.26) is given as
2" /жгч sin \N
5 2.8. Let a(N) = 1/2N in Exercise 5.2.7 and show that the limit of FN(\)
a§ #->0 is l/V^zr. Conclude that the Fourier transform of 8(x) should be
lh/2ir.
5.2.9. Obtain the Fourier transform of f(x) = exp[-fc|x|], A:>0 and verify
the result in E.22).
5.2.10. Adapt the results of Example 5.2 to obtain the Fourier transform of
Дх) = ехр(-сУ).
5.2.11. Use the formula E.14) and the result of Exercise 5.2.10 to obtain
the Fourier transform of the error integral
5.2.12. Obtain the (closed form) solution of the problem E.16), E.17) if
f(x) is defined as
5.2.13. Obtain the result E.50) by using
(a) Duhamel's principle.
(b) Fourier transforms.
5'2.14. Show that the fundamental solution G(x - f, i) of the heat equation
^ defined in E.45) has the property that as f-^0,
BmG(xf,0 0,
A1so demonstrate that for t>0 we have
p°nclude from these results that G(x-£, t) tends to the Dirac delta
faction 8(*-£) as f-*0.
310 INTEGRAL
5.2.15. Assuming и and its derivatives vanish as |x|-*°°, integrate the heat
equation E.32) over the x-axis and obtain the result
I u(x, t)dx=\ f{x) dx ,
where u(x, 0) = /(*). Using Exercise 5.2.14, verify the result for the solution
given in E.44).
5.2.16. Let the initial temperature f(x) in E.44) be given as
\x>ay
where мо = constant. Express E.44) in terms of the error integral Ф(х)
defined in Exercise 5.2.11.
5.2.17. Use Fourier transforms to solve
utt - c2uxx = F(x, t), -oo < x < oo, t > 0 ,
5.2.18. Solve the initial value problem for the hyperbolic equation
with the initial data
M(*,0) =/(*), м,(*,0) = 0.
5.2.19. Use Fourier transforms to solve
u,, + uxxxx = 0, -oo<x<oo,t>0,
М(ж,0) =/(*), и,(х,0) = 0
Hint: The inverse transform of cos(A2f) is
1 (x2 ir\
t"*\Tt At'
Wt
5.2.20. Let ДО = 1 in E.71) and evaluate the integral in terms of tan1 to
show that the solution becomes u(x, y) = 1. Note that u(x, y) does not tend
to zero as y-»oo in this case and that /(jc) = 1 does not have a Fourier
transform.
5.2.21. Noting the result in Exercise 5.2.20 and the fact that
lim K(x -t,y) = O, x Ф t,
y*0
FOR CHAPTER 5 311
-t,y) = (y/7r)/[(x-tJ + yz], show that K(x -1, y) tends to
where K(x-t,
J5fr~0 as>"~*0'
52.22. Let f(t) = a, t<0 and f(t) = b, t>0 in E.71) and evaluate the
'l in terms of the inverse tangent function.
teg
2 23. Use Fourier transforms to solve the boundary value problem
uxx + uyy - c2u = 0, -oo < x < oo, у > 0 ,
with the boundary conditions
uy(x, 0) = -/(*), u(x, y) bounded as у-» oo .
Hint: The inverse transform of exp(-yVA2 + c2)/Va2 + c2 is
(VThr)K0[c\fx2 + y2] where J^o is the zero-order modified Bessel function
of the second kind.
5.2.24. Solve Laplace's equation in a strip using Fourier transforms
uxx + uyy = °> -oo < x < oo? 0 < у < L,
u(x,0) = e~lx\ u(x,L) = O,
u(x, y) = 0 as |x|->oo.
Section 5.3
5.3.1. Show that if f(x) is an odd function of x [i.e., f{-x) = -/(jc)], the
formulas E.10), E.11) for the Fourier transform yield the formulas E.87),
E.88) for the Fourier sine transform.
5.3.2. Show that if f(x) is an even function of x [i.e., f(~x)=f(x)], the
formulas E.10), E.11) yield the formulas E.89), E.90) for the Fourier
cosine transform.
5.3.3. Determine the Fourier sine and cosine transforms of the following
functions:
(a) /(*)-*"**, ■
l0, x>a.
(c)
(d) Дх) = ехсо&х.
•3.4. By specializing the convolution theorem for the Fourier transform to
even and odd functions f(x) and g(x) as needed, obtain the following
convolution theorems for sine and cosine transforms.
(a)
312 INTEGRAL TRANSFORM
(b) |o cos(A*)Fc(A)Gc(A)^A
= \l Л')[«<1*-'I)+ *(* + *)]*.
5.3.5. Use the Fourier sine transform, Exercise 5.3.3(a) and the convoluti
theorem to solve °n
5.3.6. Use the Fourier cosine transform, Exercise 5.3.3(a) and the convolu-
convolution theorem to solve,
f(x)-k2y(x) = -f(x), x>09
5.3.7. Use the Fourier sine transform to solve the boundary value problem
for the ordinary differential equation
у"(x) - k2y(x) = e~\ 0 < x < oo, к Ф 1,
y@) = 1, y(x)->0 as x->oo .
5.3.8. Use the Fourier cosine transform to solve the problem in Exercise
5.3.7 if we replace the condition y@) = 1 by the boundary condition
y'@) = l.
5.3.9. Use Duhamel's principle to obtain the result E.107).
5.3.10. Show that the solution E.105) takes the form E.116) if /(*) = 0 and
g(t) ~ ux = constant.
5.3.11. Suppose that u(x, t) satisfies the heat equation E.97), the initial
condition E.98), and the boundary condition E.126). Let
u(x, 0 = ux(x, i) - hu(x, t) .
Show that v(x, t) is also a solution of the heat equation; that is,
vt - c2vxx = 0, 0<x<oo,
with the initial condition
v(x,0)=f'(x)-hf(x)>
and the boundary condition
FOR CHAPTER 5 313
XERCISES
v(O,t) = r(t), t>0.
a suming that u(x, t) is bounded as *-*°°, show that the solution of the
Ven problem for и is given in terms of v(x, t) as
u(x, t) = ehx £ e~hsv(s, t) ds .
5 3 12. Use the Fourier sine transform to solve the following initial and
boundary value problem for the wave equation
utt-c2uxx = Q, 0<x«*>,t>0,
5.3.13. Solve the following problem for the wave equation using the Fourier
cosine transform
5.3.14. Solve the Dirichlet and Neumann problems for Laplace's equation
in a half-plane given in Example 5.4 by the use of Fourier sine and cosine
transforms, respectively.
5.3.15. Solve the boundary value problem E.137)-E.139) for Laplace's
equation using the Fourier sine transform.
5.3.16* Use the Fourier sine transform to solve the problem
uxx + Uyy=Q> 0<x<°o,0<y<a ,
u@, y) = 1, u(x, >>)->0 as x->oo 9
u(x, a) = 0, w(x, 0) = e~x .
•3'17. Apply the Fourier cosine transform to solve the problem
uxx + uyy =0, 0<jc<o°,0< )>
«,@,у) = 0, у>0,
1' 0<д:<1
314 INTEGRAL
5.3.18. Use Stokes' rule in the manner indicated in Example 5.4 to Co
struct a solution of the problem n*
uxx + uyy = 0, 0 < x < oo, 0 < у < oo 9
from the solution of the problem E.137)-E.139) given in E.142).
Section 5.4
5.4.1. Express the three-dimensional wave equation in spherical (spatial)
coordinates and show that if we look for a solution и — u(r, i) where r is the
radial variable, the function v(r, i) = rw(r, i) is a solution of the one-
dimensional wave equation
vtt-c2vrr = 0.
Obtain the general solution
и = - F(r - ct) + - G(r + ct) ,
where F and G are arbitrary functions. These solutions represent spherical
(propagating) waves.
5.4.2. Consider the function S€(x) defined in A.28) and the integral
where r2 = (x~ fJ + (у -т/J + (z - fJ. Since 8€(r-ct) vanishes when
\r — ct\ > e9 the integral extends only over a finite region. Conclude, in view
of the fact that the integral represents a superposition of spherical waves,
that I(x, y, z, t\ e) is a solution of the three-dimensional wave equation.
Introduce the spherical coordinates defined in E.153) and show that the
integral takes the form
/ = -— I I I f(x + r sin <p cos в, у + r sin <p sin в, z + r cos <p)
x 8€(r - ci)r sin <p dr dO d<p .
Carry out the limit as €-^0, use the property A.29) of S€ and the fact that
lime^0 S€(r- ct) vanishes when r^ct, and conclude that I(x9y,z,t\e)
tends to E.154) as e->0.
5.4.3. Verify by direct differentiation that E.154) is a solution of E.145)-
E.147).
FOR CHAPTER 5 315
с 4 4. Solve the Cauchy problem for the wave equation
utt - c\uxx + uyy + uZ2) = 0, -°o<x, y, z <oo, t>0
tf the initial data are
u(x,y,z,0) = 05 ut(x,y,z,0) = [0y Г>Д)
where r2 = x +y + z .
5 4 5. Solve the following problem for the wave equation using the spherical
wave solutions obtained in Exercise 5.4.1.
utt - c2V2u = 0, -oo < x, y, z < oo, t > 0 ,
M(x,;y,z,0) = l, ut(x,y,z,0) = r2,
where r2 = x2 + y2 + z2.
5.4.6. Obtain the solution of the Cauchy problem for the two-dimensional
wave equation
Determine the value of и@,0, i).
5.4.7. Solve the following Cauchy problem.
u(x,y,0) = x9 u£x,y,Q) = 0.
5.4.8. Apply the method of descent to solve the Cauchy problem for the
Klein-Gordon equation:
vtt - c2vxx + a2v = 0, -oo < x < oo, * > 0 ,
K*,0) = 0, vt(x,0)=f(x).
^ u(x> У, t) = cos(ay/c)v(x91) and show that u(x, y, t) satisfies the wave
equation in two dimensions. Then descend from two dimensions to one.
•*A Use the method of descent to solve the Cauchy problem for the
Allowing hyperbolic equation
vtt - c2vxx - a2u = 0, -oo < x < oo, f > 0 ,
i;(x,0) = 0f vt(x,O)=f(x).
316 INTEGRAL
Let u(x, y, t) = exp(ay/c)v(x, t), show that u(x, yt i) satisfies the tw
dimensional wave equation, and descend to one dimension. °*
5.4.10. Express the Laplacian in polar coordinates and show that E.184^ •
a solution of the homogeneous form of E.170) when r>0.
5.4.11. Use the two-dimensional Fourier transform to solve the follow'
Cauchy problem for the heat equation. *
5.4.12. Use the three-dimensional Fourier transform to solve the Cauchy
problem for the heat equation (as formulated in Exercise 5.4.11) in the case
of three dimensions.
5.4.13. Consider the following Cauchy problem
Ы — С (Ы *"H it H~ It I = JriJC V Zl€ —oo ^ X V Z ^ oo / ^> 0
u(x, y9 z, 0) = ut{x, у у z9 0) = 0 ,
where a> = constant. Assume that F(x, y9 z) vanishes outside the bounded
region V. Use the retarded potential E.161) to show that for large t the
solution of this problem has the form
w(x, y, z, i) = * I I I F(£, 17, L) — af dr\ d£, f—>oo
where к = o>/c. Noting that и has the form w(x, y9 z, /) = v(x, y, z)e ш\
show that v satisfies the inhomogeneous reduced wave equation
V2v + k2v = - — F(x, y, z), -oo < x, y, z < oo .
с
The function v = A/4ят)е'*г is the free space Green's function for the
reduced wave equation (see Section 6.7).
Section 5*5
5.5.1. Consider the Fourier transform formulas E.143), E.144) in the case
w = 2. Let
xx = r cos в, x2 = r sin в, Ax = A sin <p, A2 = A cos <p ,
and put Дхих2) = е~ш/(г). Show that the transform F(A,, A2) [i-e"
E.143)] takes the form
gC!SES FOR CHAPTER 5 317
F( At, A2) = ^ Jo J ^ exp[*Ar sin@ + ?
prom the integral representation
1 f"
ЛЛ Лг) = ;r" exp(iAr sin 0 - w
2тг J-ir
for the Bessel function Jn(kr) of integral order, conclude that
e-*"F{bi,*2) = l rJn(kr)f(r)dr=F(k).
Insert ein<pF{k) and e~inef(r) into the inversion formula E.144) to obtain
1 f00 Г
ftr\ = —- ехр[т(в + q>) - r'Ar sin@ + <p)]F( A)A rfip dA .
Use the integral representation of /„(Ar) given previously to derive E.188).
5.5.2. Verify the result given in E.190).
5.5.3. Integrate by parts in E.191) to obtain the result
Г e'kzJA\r) dk = i (l -
Jo lV ^ г V
5.5.4. Evaluate the integral E.200) if /(z) = 1.
5.5.5. Use the zero-order Hankel transform to solve the Neumann problem
for Laplace's equation
wrr+ - мг + uzz = 0, r>0, z>0, M->0asz->°°,
uz(r, 0) j, r < 6, nx(r, 0) = 0, r > € .
Show that the transform is U{kJz)^{ll7rek2)Jl{\e)e'k\ Let e->0 and
show that the solution w(r, z)-^(l/27r)(r2 + z2)~1/2 in the limit.
5*5.6. Use the Hankel transform to solve the problem
urr + - ur + uzz- k2u = 0, r > 0, -oo < z < oo ,
the boundary conditions E.195) given on the z-azis.
^'S-7. Use the Hankel transform to construct a (formal) solution of the
following problem
INTEGRAL
TRANSFORM
7 иг) + *(r91),
where D > 0 and u(r, 0) = /(r).
5.5.8. Solve the Cauchy problem for the damped wave equation
utt + a\ - c2(urr + - ur) = 0, r >0,*>0 ,
with the initial conditions
w(r, 0) = f(r), w,(r, 0) = g(r), r > 0
by using the Hankel transform.
5.5.9. Consider the Cauchy problem for the wave equation
un-c \urr+-ur + - ивв)=0, t>0, г>О,О<0<2тг,
with the initial conditions
u(r, 6,0) - /(r) cos(«0), ut(r, 0,0) - g(r) cos(n$),
where n is a positive integer. Let и = u(r, t) cos(nO), and solve the resulting
Cauchy problem for v(r, t) by means of the wth order Hankel transform.
5.5.10. Consider the diffusion equation in three dimensions. Introduce
spherical coordinates and if the initial value is given as и = /(r), where r is
the radial variable, show that w(r, t) satisfies the equation
r>0,f>0,
with D as the diffusion constant. This problem cannot be solved by means of
the Hankel transform. However, let v(r, t) = гм(г, t) and show that v
satisfies the one-dimensional diffusion equation with u@, t) = 0 and v(r, 0)
= rf(r). Use the results of Example 5.5 to solve this problem and obtain the
solution of the given problem for u(r, i). As in Example 5.11, let f{r)
represent a concentrated source at r = 0, and, in analogy with E.221),
suppose that
Determine the limit of the solution u(r, t) as e->0. This limit is the
fundamental solution of the diffusion equation in the three-dimensional
case.
FOR CHAPTER 5 319
Section 5.6
с 6 1 Determine the Laplace transforms of the following functions:
'"(a) /@ 1
(b)
(c) f(t) = cos <ot.
(d) f(t) = sin cot.
(e) f(t) = (l/n\)t
(f) Л0 = *('-*)
5 6 2. Show that if F(A) is the Laplace transform of /(f), then the Laplace
transform of e@ is ^( А + a).
5.6.3. Let/@ andg@ have Laplace transforms F(A) and G(A), respective-
respectively. Show that the Laplace transform of $'о f(s)g(t - s) ds is given as
f(A)G(A). (This is the convolution theorem for the Laplace transform.)
5.6.4. Show that the Laplace transform of H{t- a)f(t- a), a>0, where
H{t) is the Heaviside function, is F(k)e~ak.
5.6.5. Show that the Laplace transform of $q f(s) ds is given as F(A)/A.
5.6.6. Let/(f; a) have the Laplace transform F(A; a), where a is a parame-
parameter. Show that
(a) Sfida has the transform dFlda.
(b) ja f(t; s) ds has the transform Ja F( A; s) ds.
5.6.7. Use E.239) and Exercise 5.6.6 to show that the Laplace transform of
)]e-a2f4t is e'aV~\
5.6.8. Use E.239) and Exercise 5.6.6 to show that the Laplace transform of
f(/V7[l^e~aVI]/A
5.6.9. Use E.239) and Exercise 5.6.6 to show that the Laplace transform of
erfc(a/2V7)is(l/A)e"aVI.
5.6.10. Determine the Laplace transform of the zero-order Bessel function
Hint: Use the integral representation given in Exercise 5.7.1.
5.6.11. Solve, using the Laplace transform
ut - c2uxx = 0, 0< jc<oo, f >0 ,
•6-12. Solve, using the Laplace transform
ut-c2uxx = 0y x>0,t>0,
и@,0 = *@, и(*,0) = 0.
•*-13. Solve, using the Laplace transform
320 INTEGRAL TRANSFORM
u,-c2uxx = 0, x>0,t>0,
ux@,t) = h(t), u(x,0) = 0.
5.6.14. Use the Laplace transform to solve the Cauchy problem
u,-c2uxx=0, -<»<jc<oo, r>0,
u(x,0) = S(x-a).
5.6.15. Use the Laplace transform to solve the following problem
ut - c\x = °> 0<л;</, f >0,
u@9t)^e'\ ux(l,t) = 09 t>0,
м(л;,0) = 0, 0<лг</.
5.6.16. Solve, using the Laplace transform
utt - c2uxx = 0, 0<л; <oo, f >0 ,
u(x, 0) = и Ax, 0) = 0, и@, 0 - fit).
5.6.17. Solve the problem of Exercise 5.6.16 if the Dirichlet boundary
condition is replaced by the Neumann condition ux@, t) = -g(t).
5.6.18. Use the Laplace transform to solve the Cauchy problem
utt - c2uxx = 0, -oo < x < oo, t > 0,
5.6.19. Solve the Cauchy problem for the telegrapher's equation using the
Laplace transform
utt - y2uxx + 2Xut = 0, -oo < x < oo, t > 0,
Hint: Eliminate the ut term by a transformation of the dependent variable.
Use the fact that the Laplace transform of /0(aV?2 - b2)H{t - a) equals
(A2 - a2)/2 exp(-bVA2-a2), where /0 is the modified Bessel function of
zero order and H(z) is the Heaviside function.
5.6.20. Solve the initial and boundary value problem for the telegrapher's
equation of Exercise 5.6.19 with x > 0 andt > 0, if u(x, 0) = и Ax, 0) - 0 and
«*(<U) ="/(')•
XERCISES FOR CHAPTER 5 321
5 6.21. Verify the relationship E.249), E.250) by formally integrating term
by term in the formula E.227).
5 6.22. Use the Laplace transform to obtain the exact solution of the
following problem
u(x, 0) = x + sinf — xj .
Use the Abelian result E.249), E.250) to discuss the solution of this
problem for small t on the basis of the behavior of the Laplace transform.
Compare your results with those obtained from the exact solution.
5.6.23. Suppose the Laplace transform F(A) is a rational function, that is,
F(A)= G(k)IH(k) where G and H are polynomials. If #(A) has only
simple zeros, show that the partial fraction expansion of F(A) yields a
(finite) Heaviside expansion E.252) for the inverse transform if G(A) is a
polynomial of lower degree than #(A).
5.6.24. Invert the transform E.246) by use of the convolution theorem and
the Heaviside expansion of the ratio of hyperbolic functions. Show that the
resulting series solution agrees with that obtained by using finite sine
transforms to solve the problem in Example 5.13.
5.6.25. Let g(i) = e~' in E.243) and invert the resulting transform E.246)
by a direct application of the Heaviside expansion E.252).
5.6.26. Demonstrate that the Laplace transform of the solution of the
following problem satisfies the conditions required for the validity of the
Tauberian result given in the text.
ut - c2uxx =0, 0 < jc < /, *>0,
и@,*) = и(/,0 = 1, 0<t,
u(x, 0) - 0 .
Use the Tauberian result E.251) to obtain an expansion of the solution
u{x, t) for large t and show that it is, in fact, an exact solution of the given
Problem.
5-6.27. Solve, using the Laplace transform
utt - c2uxx = e~\ 0<x<oo, t>0,
u(x, 0) = ut(x, 0) = 1, 0 < x < oo ,
m@, 0 = sin t, t>0.
322 INTEGRAL TRANSFORMS
Section 5.7
5.7.1. The Bessel function of integral order Jn(x) has the integral repre-
representation
1 Г
Jn(x) = — I exp(£c sin в - inB) dB .
Apply the method of stationary phase to obtain the asymptotic formula
T
valid as x—»<*>.
Hint: There are two stationary points.
5.7.2. The Airy function Ai(x) has the integral representation
Let x < 0 and т = V^xt and use the method of stationary phase to show that
the Airy function has the asymptotic expansion
as x~~> —oo.
Hint: There are two stationary points.
5.7.3. Express the functions F+(A) and F_(A) in E.262) in terms of the
transforms of the initial data u(x, 0) = f(x) and ut(x,0) = g(x). Show that
F+(~A) = F_(A) (the complex conjugate) and that the solution u(x, t) in
E.262) is real. Obtain the asymptotic form of u{x, t) by considering the
contributions from the F+(X) and F_(X) integrals.
5.7.4. Show that
is a solution of the Klein-Gordon equation E.258). Let x = at in the Bessel
function with \a\ < y, and expand Jo asymptotically for large x and t. Show
that the result agrees with that obtained in Exercise 5.7.3 for an appropriate
choice of F+ and F_.
5.7.5. Use the Fourier transform to obtain the solution of the Cauchy
problem for the linearized Korteweg-deVries equation (see Example 10.15)
ut + a2ux + b2uxxx = 0, -
with the initial condition
EXERCISES FOR CHAPTER 5 323
Apply the method of stationary phase to discuss the solution u(x, i) for large
X and t.
5 7.6. Use the Fourier transform to solve the initial value problem for the
dissipative wave equation
with the initial data
u{x, 0) = /(*), ut(x9 0) = a2f"(x), ~«><x<«>.
Apply Sirovich's method to obtain an asymptotic expression for the solution
as t gets large.
5.7.7. Apply the Fourier transform to solve the boundary value problem for
the elliptic equation
uxx + utt - ut=f(x, t), -oo<*, t<oo ,
assuming u(x, i) and f(x, t) are suitably behaved at infinity. Apply Sirovich's
method to discuss the asymptotic behavior of the solution u{x, t) for large t.
5.7.8. Verify that the solution E.294) of the initial value problem for
E.293) has the behavior given in E.296) for large t.
6
INTEGRAL RELATIONS
The problems considered in the preceding chapters dealt mostly with partial
differential equations whose coefficients, inhomogeneous terms, and initial
and/or boundary data were smooth functions. Consequently, the solutions
were expected to be smooth functions as well.
It is often the case, however, that the medium for which the differential
equation models some physical property is heterogeneous and some of its
characteristics, such as density or conductivity, change discontinuously
across some region. For example, this situation arises when we consider the
longitudinal vibration of a composite rod composed of two rods of different
constant densities joined at some point. Then the coefficients in the equa-
equation for the vibrating rod may be singular at the point where the rods are
attached.
In addition, it is of interest to consider problems where the inhomoge-
inhomogeneous term in the equation, which may represent a forcing term or a source
(or sink), is concentrated over some lower dimensional region such as a
curve or a point. Also, the data for the problem may have discontinuities or
singularities. As a result, the solutions of these problems are no longer
expected to be smooth functions in general. Thus it becomes necessary to
attach a meaning to "solutions" of differential equations that are not
differentiable as often as required by the equation. In some cases these
solutions are not even continuous everywhere. (Some of these matters were
considered in our discussion of first order equations in Chapter 2.)
This chapter deals with the foregoing questions by showing how the given
differential equations can be replaced by equivalent integral relations. These
relations will be derived directly from the partial differential equations and
may involve derivatives of the unknown function. In any case, fewer
derivatives than are required in the solution of the partial differential
equation are needed for the solution of the equivalent integral relation.
Thereby, we generalize or weaken for some problems the concept ot
solution of a differential equation and the results obtained are called
generalized or weak solutions.
The wave equation will be discussed in some detail and an equivalent
324
JP4TEGRAL RELATIONS 325
ntegral wave equation will be derived. We have seen in Chapter 3 that
hyperbolic equations, in contrast to elliptic and parabolic equations, do not
smooth out discontinuities or singularities in the data. Since the wave equation
*s a prototype equation of hyperbolic type, we concentrate our discussion on
it A further discussion of singular solutions is found in Section 10.2.
Finally, we introduce the concept of energy integrals for partial differen-
differential equations of each of the three types. We show how they can be used to
prove uniqueness and continuous dependence of the solutions on the data.
Our discussion will be restricted to second order partial differential
equations of the form considered in Chapter 4. However, the ideas and
methods introduced can be carried over to other equations.
6.1. INTEGRAL RELATIONS
In Section 4.1 it was shown how to derive the parabolic equation D.4) by
applying a balance law over a certain region. By assuming all terms
occurring in the balance law are continuous (or more generally smooth), it is
possible to derive the parabolic equation D.4). If the smoothness require-
requirement is not met throughout the region, for example, there is some subregion
across which the properties of the medium undergo a sharp discontinuous
change, the arguments leading to D.4) may fail. In that case, the balance
law can be used to derive matching conditions for the solutions of the partial
differential equation D.4) that remain valid on both sides of the region of
discontinuity. These conditions show how to connect solutions across the
region of discontinuity. A similar approach may be applied if the inhomoge-
neous term of the equation is singular in some region.
Generally speaking, if the partial differential equation is derived from
some physical principle applied over an arbitrary region as in Section 4.1, it
is possible to take into account the foregoing special situations and incorpo-
incorporate the appropriate results into the derivation. We have derived various
equations in Chapter 1 from a different point of view, one from which it is
not obvious how to take into account discontinuities or other singularities in
our derivation. Therefore, we adopt the following approach. We assume the
Partial differential equation is given and convert it into an equivalent
integral relation (thereby, we reverse the commonly used method of deriva-
ll<>n of the differential equation from the integral relation). On using this
aPproach in Chapter 2 to construct generalized solutions of quasilinear
equations, it was found that different integral relations (each of which is
equivalent to the given differential equation when the solutions are smooth)
Save rise to different generalized solutions. This nonuniqueness in the
^termination of generalized solutions does not occur for linear equations,
^nich we deal with exclusively in this chapter. A detailed discussion is given
Or the hyperbolic equation D.10) and the appropriate results are given
326 INTEGRAL RELATION
without derivation for the parabolic and elliptic equations D.8) and D m
respectively. No other equations are discussed. ' *'
We consider the hyperbolic equation D.10) in two or three dimension
that is, S;
F.1) P^j-V'(pVu) = pF-qu,
at
where p, p, and q are functions of л; only and have the properties given in
Chapter 4. Let Л be a closed and bounded region in (x, 0-space (л; is a two-
or three-dimensional variable) with dR as its boundary. We integrate F.1)
over R to obtain
F.2)
Let the gradient operator in (x, f)-space be defined as
so that F.2) may be written as [since p = p(x)]
F.4) jjj- [pVw, -pwj dv = ffR [qu - pF] dv ,
where [pVu, -put] is a three- or four-component vector. Applying the
divergence theorem to the first integral in F.4) gives
F.5) J^ [pVu, -put]-nds = j jR (qu - pF) dv ,
where n is the exterior unit normal vector to dR. Equation F.5) is the
equivalent integral relation (or, more precisely, the integro-differential
equation) we are seeking.
We now choose a special form for the region R that commonly occurs
when the partial differential equations are derived using a balance law or a
similar physical principle. Fixing f, we choose a closed and bounded region
Rx in x-space, and extend it parallel to itself between t = t0 and t = tx ifl
(jc, 0-space to form the region R shown in Figure 6.1. Let dRx be the
boundary of Rx and let dRx be the lateral boundary of R. Also, denote the
upper and lower caps of R by dRx and dR0, respectively (note that
dR0 ^ Rx). The exterior unit normal vector n on these surfaces assumes the
following form. With n^ equal to the exterior unit normal vector to dRx, we
find that on dRx,n = [nx90]9 on dRx,n = [0,1], whereas on dRo>n^
[0, -1]. Then F.5) takes the form
RELATIONS
327
Figure 6.1- The region R.
С С С f f
JSRq JBR^ JdRx J JR
The integration over dR0 and dRx in F.6) is effectively taken over the
region Rx. On the lateral surface dRx we have ds = dsx dt, where dsx is the
surface differential over dRx, and in the region R we have dv — dxdt (here
dx is an area or volume element in two or three dimensions, respectively.)
The equation F.6) can, therefore, be written as
F.7) f [p(x)ut(x, tx) - p(x)ut(x, t0)] dx
jkx
= ff p-~-dsxdt-fl t (qu-pF)dxdt,
Jfg J#RX &ftx ° X
where duldnx is the exterior normal derivative of и on the surface dRx. In
one space dimension with Rx equal to the interval (xo>*i)> F-7) has the
вдгщ
i, 0 -
Г'1
- f' / ' [«(*)«(*, 0 - pWF(jc 0] dx dt
328
INTEGRAL RELATl0Nfs
To consider a concrete example, we assume that F.1) describes th
longitudinal vibration of a rod and that the problem is one-dimensional
Then the integral relation F.8) characterizes the change in momentum of
segment of a rod (x0, x^) in the time interval (f0, tx). The displacement of
the rod at the time t is given by u(x, t), the density is p(x), and the tension is
T(x, t)=p(x)ux(x, t)~as required by Hooke's law—where p(x) equals
Young's modulus at the point x. The momentum per unit length equal
p(x)ut(x, i) and we put q(x) = 0 in F.8). Also p(x)F(x, t) represents the
external force density. Thus F.8) equates the change in momentum to the
forces acting on the segment (x0, хг) of the rod. In the limit as x0—> xx and
to->t19 we obtain the partial differential equation F.1) with q = 0 and V
replaced by dldx. For the limiting procedure to be valid, u(x, t) must have
continuous second derivatives and p, p\ and F must be nonsingular. This
enables the mean value theorems for derivatives and integrals to be applied
to F.8). At points where the limit process is not valid, F.8) does yield
matching conditions as will be shown.
For completeness we present the appropriate integral relations for the
elliptic and parabolic equations of Section 4.1. Their derivations are similar
to those given in the hyperbolic case and are not presented. For the
parabolic equation D.4) we have
F.9) \aR [pS/u, -pu]-nds = jjR (qu - pF) dv
and, in a form equivalent to that in F.7),
F.10) f [р(х)и(х,^)-Р(х)и(х,10)]<Ь
JRX
Jt0 JaRxF дПх х
For the elliptic case, given the equation
F.11) -V-(/7Vm) + ?u = p
we have the equivalent integral relation
F.12)
The regions R and Rx and their boundaries are defined as in the foregoing
hyperbolic problem for the parabolic case, whereas R = Rx in the elliptic
case. We do not write down the one-dimensional forms of F.9)-F.12).
In each of the above integral relations in two- or three-dimensional
jt-space, if we choose Rx to be a rectangular region (i.e., xo<x<*v
oMPOSlTE MEDIA: DISCONTINUOUS COEFFICIENTS 329
<y<y1, zo<z<zxin three dimensions), it is an easy matter to retrieve
^he partial differential equations from the integral relations in the limit as Rx
brinks to a point and f0—> tx. However, the function и and the coefficients,
Ss well as F, must be sufficiently smooth for this limiting process to be valid.
This implies the equivalence of the differential and integral representations
f the given equations in regions where everything is smooth. In the
following sections we show how the integral relations are to be used in cases
where the foregoing limit process breaks down in some region, so that the
given differential equation is not valid there.
6.2. COMPOSITE MEDIA: DISCONTINUOUS COEFFICIENTS
We consider any one of the three integral relations F.7), F.10), or F.12)
and assume that a Cauchy, an initial and boundary value or a strict
boundary value problem, whichever is appropriate for the equation, is given
over a region G in jc-space. (The region G may be bounded or unbounded.)
Let G be divided into two subregions Gx and G2 with SQ as the boundary
region separating Gt and G2. We assume that the properties of the given
medium vary discontinuously across SQ (for all time t if the problem has a
time dependence), so that one or more of the functions p(x), p(x), or j?(x)
is discontinuous across SQ (they are permitted to have, at most, jump
discontinuities). For example, if two strings of different densities are at-
attached at a point xQ and we consider the equation for the vibration of the
composite string, the density p(x) will be discontinuous at the point л:0.
At points x or (x, i) that do not lie in the region of discontinuity and
where the coefficients and the solution are assumed to be smooth functions,
we may go to the limit in the integral relations and derive the appropriate
partial differential equation. Thus at points x or (x, t) such that x is interior
to Gx or G2, the partial differential equation is valid. Let us denote the
solution in Gx by ux and the solution in G2 by u2. It is meaningful to require
that the solution of the problem be continuous across 50. Thus if w is a
temperature distribution or represents the displacement of a string, we
expect it to be continuous in the interior of the region G. Therefore, the first
matching condition is
To obtain a second matching condition, we apply the appropriate integral
Elation over a region Rx that contains a portion of So in its interior, as
Pictured in Figure 6.2. The region Rx is extended parallel to itself from t0 to
i Ш the hyperbolic and parabolic cases (as was done above). We now
consider the integral relations F.7), F.10), and F.12) [with R = RX in
^•!2)]. Let the region Rx collapse onto So. Since и (as well as ut) is
^ntinuous across So and p, p, q, and, possibly, F have, at most, jump
330
INTEGRAL RELATION
Figure 6.2. The composite region.
discontinuities across So, the only contribution that results in the limit comes
from the normal derivative terms that may have different values on both
sides of So. Because this result is valid over any portion of 50, we conclude
on the basis of the duBois-Reymond lemma (see Exercise 8.1.9) that the
second matching condition is
F.14)
Pi
dux
~~dn
du2
where pt(x) and p2(x) represent the limiting values of p(x) as x approaches a
point in 50 from the subregions Gx and G2, repectively. Both normal
derivatives (in So) in F.14) are taken in the same direction. We remark that
F.13) and F.14) state that both и and pduldn are continuous across So, and
these conditions are valid for all time if и is time dependent. If p js
continuous across 50 but one or more of the other terms in the equation is
discontinuous, then и and its normal derivative are continuous across 5. in
that case, apart from the change in the matching condition F.14), where
Pt=p2, the problem is solved in the same way.
In the one-dimensional case, if 50 corresponds to the point x0, and (j} an
G2 correspond to values of x less than and greater than x09 respectively,
F.13), F.14) are replaced by
диг
F.15)
F.16) />,(*„) 7;
For the sake of concreteness we formulate an initial and boundary value
problem for a hyperbolic equation in a composite medium with So as W
COMP°SITE MEDIA: DISCONTINUOUS COEFFICIENTS
Figure 6.3. The region G.
331
region of discontinuity, (see Figure 6.3). We have, with G equal to the
union of Gl9 So, and G2,
= PlFx , л;e G1? f >0
P2M -^T "V- (p2VM2) + ?2м2 = p2F2 , x e G2>
F.17)
The initial conditions are
F.18) M(*,0) =/(*), u,(x>0) = g(x), M
The boundary conditions given on the boundary 5 of the region G are
FЛ9)
—
dn
д: G <?G2 .
and the matching conditions are
F.2i) £i!i = £«2
e subscripts 1 and 2 indicate functions evaluated in the regions G, and
°2, respectively. '
n case we consider the Cauchy problem for the preceding hyperbolic
connv the region G is Of infinite exte«t so we drop the boundary
the к F19) and ГеЫп aU °ther e<luations-In either case, we can solve
Problem by considering F.17) in the regions Gt and G2 separately. We
332
INTEGRAL RELATION
apply the appropriate initial and boundary conditions for each subregi0
with the additional conditions ux = ux and u2 = u2 on So where ux and й а °
as yet unknown. Once the solutions ux and w2 are determined, the functio *
ux and w2 are then specified by applying the matching conditions F Ш
F.21). " J'
For the parabolic case we replace дгик1вгг, i = 1, 2, by dut/dt, / = 1,2 i
F.17) and drop the second initial condition ut(x, 0) = g(x) in F.18). In the
elliptic case, the time derivatives and the initial conditions in F.17) and
F.18) are dropped. However, the other conditions remain for both cases
Example 6.1. Vibration of an Infinite Composite Rod. We consider the
longitudinal vibration of a composite rod of infinite extent. The rod is made
up of two homogeneous rods joined at x = 0, each of which has a constant
density p and a constant Young's modulus p (see the foregoing discussion).
The region G is the infinite interval — <» < x < «> with Gx equal to — oo < x < 0
and G2 equal to 0 < x < «> and the discontinuity region So is the point x = 0.
We set q and F equal to zero in F.17) and replace V by dldx. Also, pu ^
and pl9 p2 are constants that are given for the problem.
The appropriate wave equations in the regions Gx and G2 are
^2 Ci .2 U> *-A,Z,
where с] ~рг1рг, / = 1,2. Thus there are different speeds of wave propaga-
propagation сг and c2 in the two regions Gx and G2, respectively.
We assume that a wave f(t - х/сг) is approaching the junction point x = 0
as / tends to zero (from negative values) and we want to determine the
resulting reflected and transmitted waves. To do so, we formulate this as a
Cauchy problem with the initial data
F.23)
F.24) u2(*f0)*0, ^(*,0) = 0, x>0,(xSG2).
Note that a solution of F.22) with i = 1 and the initial data F.23) is just
"i(*, t)=f(t-xlcx)—that is, a wave traveling to the right with speed cv
For small t, this is assumed to be the only disturbance in the region Gv As t
increases, the wave reaches the junction point д: = 0 and gives rise to a
reflected and transmitted wave. This is the case if we assume that/(-*) **
for x>xx where ;Ci<0. In addition to the conditions F.23), F.24), the
solutions Mj and u2 must satisfy the matching conditions F.15), F.16) where
xo = O.
The general solutions of the wave equations F.22) are
E MEDIA: DISCONTINUOUS COEFFICIENTS 333
F.25) ««<*. 0 -/i(» - f) + g,(t + i) , * -1.2.
From F.23), F.24), we have
F.27) /i^M^Mlf). *<0'
F.28)
F.29)
On differentiating F.26) and F.28) we easily conclude that gx(z) = 0 for
z<0, ft(*)e° for z>0,/xB)=/(z) for z>0, and/2(z) = 0 for z<0. The
matching conditions
F.30) /,@ + gl@=/a(o + g2(t), t>o,
]t>0
yield a simultaneous system of equations for f2(t) and g^t) with f >0.
Solving these equations and introducing the results into F.25) yields the
solution as
F-32) Ui{Xft)=f(t-L\ + Rf(t+±.\ x<0,t>0,
F-33) u2(x,t) = Tf(t--), x>0,t>0,
\ C2'
where the reflection coefficient R and the transmission coefficient T are
given as
F-34)
R =
P2P2
^ Near / = 0 we have ux = /(Г - */c?) and w2 = 0, where f(t - х/сг) repre-
ents a wave traveling to the right (i.e., towards the juncture point x = 0).
334
INTEGRAL RELATION
Figure 6.4. Vibration of a composite rod.
This wave is called the incident wave. When the incident wave reaches the
interface * = 0, two additional waves arise: the reflected wave Rf{t + xlcx)
which travels to the left and the transmitted wave Tf(t - xlc2) which travels
to the right. Since, apart from a constant factor, each wave has the same
waveform /(*), we call R and T the reflection and transmission coefficients.
These coefficients are a measure of the amplitudes of the waveforms. We
observe that 1 + R — T so that there is an equal distribution of amplitudes
(in a sense) in both the regions x < 0 and x > 0. If p1pl = p2p2 we have R = 0
and T— 1, so that the incident wave is transmitted without undergoing
reflection. The solution is described in Figure 6.4.
We have indicated that to solve initial and boundary value problems in
composite media, the problems should first be solved in the two subregions
Gx and G2 with (as yet) unspecified boundary data on the discontinuity
region 50. The solutions are then fully determined by applying the matching
conditions. This procedure was used in the foregoing example. There is an
alternative approach that we now discuss for problems in bounded spatial
regions. In the absence of a discontinuity region, such problems were solved
in Chapter 4 with the use of eigenfunction expansions. The eigenfunctions
were determined from the eigenvalue problem associated with the given
(homogeneous) equation and boundary conditions. We now demonstrate
that it is also possible to construct eigenvalue problems and solve the
problems considered for a composite medium by means of eigenfunction
expansions.
Again considering the same region G with So as a discontinuity region for
the coefficients, we introduce the following eigenvalue problem for the
function v(x)
,oMPOSlTE MEDIA: DISCONTINUOUS COEFFICIENTS 335
l ъЬЛ l*\v\ = ~V*(pVy) + qv = Xpv , ;t6G,.
s i£\ OLD + В —
,, <ун\ v and p — are continuous across So .
6.3/J ^ дп и
fhe coefficients p, q, and p may have jump discontinuities across So. All the
-i ificant properties of eigenvalues and eigenfunctions given in Chapter 4
remain valid for this eigenvalue problem.
We demonstrate the orthogonality of eigenfunctions corresponding to
different eigenvalues in the one-dimensional case. With the operator L
defined as in D.56) and vt and v}- given as eigenfunctions corresponding to
different eigenvalues A,- and A; (А, Ф A;), we have from D.71)
f1 Г0 d
F.38) Jo {vtLv} - VjLvt) dx = Jo — (pvjv\ - pvtv)) dx
{' —
JXq dX
where x0 is the point of discontinuity and 0 < x0 < L The contribution to
each integral vanishes at the boundary points x = 0 and x = / as before. At
the point x = x0, the functions pvtvj andpy;y- both occur with opposite signs
and their contribution vanishes since vi9 Vj9 pv'i9 and pv'j are assumed to be
continuous at x = xQ. The inner product for this eigenvalue problem is
defined as
pvtVj dx = pvp; dx + I pvp* dx ,
0 JO Jx0
and the norm is
using D.72) and F.38) we conclude that vt(x) and Vj(x) are orthogon-
«• The other properties listed and proven in Section 4.3 for the Sturm-
Liouville problem can be proven for the present problem as well.
In the following example we solve the problem of heat conduction in an
Jflsulated composite, finite rod with the ends kept at zero temperature, in
erms °f the eigenvalue problem that results on solving this problem by
^aration of variables.
6.2. Heat Conduction in a Finite Composite Rod. We consider
aeat conduction in an insulated piecewise homogeneous rod of length /, with
336
INTEGRAL RELATloNs
the ends kept at zero temperature for all time. With the rod situated
in th
d in th
interval 0 < x < / and the junction of the two homogeneous portions occu °
ring at the point x = jc0, we have the following initial and boundary valu
problem for the temperature u(x, t) with u = ux{x, t) for x<jc0 and u-
u2(x, t) for x > x0
F.41)
д2и.
xo<x<l,t>0.
The coefficients рх,рг,рх, and/?2 are all constant. (See Section 4.1 for their
physical significance.) The boundary conditions are
F.42) м^О, t) - О, u2(l, 0 = 0,
The initial condition is
F.43) u(x,0)=f(x), 0<x<l.
The matching conditions at x = x0 are
'ui(xo,t) = u2(xQ,t)9 t>0
f>0.
F.44)
дх
= au2(x0,t)
иг дх
Using separation of variables we are led to the following eigenvalue
problem for the function v(x) with v = vt(x) for x< x0 and v = v2(x) for
x>x0
+ Ap,u,(x) = 0, 0 < x < x0
( }
The homogeneous boundary conditions are
F.46) ^@) =0, i>2(/
and the matching conditions at x = x0 are
xo<x<l.
F.47)
dVl(x0) _ dv2(x0)
—^T-P-J—
To determine the eigenfunctions we solve F.45) for vx(x) and v2(x)
applying the boundary conditions F.46) to obtain with arbitrary constants/1
and B,
337
^ <* - ')] •
F.49) <
A sin
OMP°SITE MEDIA: DISCONTINUOUS COEFFICIENTS
F.48) v^^Asin^-^x), v2(x) = Bsin
Applying the matching conditions gives
x0 — /)J = 0
[/Ap2 1
v — (x0 - /) = 0 .
Рг ^
We require a nonzero solution of this homogeneous system for A and В so
that the determinant of the coefficients must vanish. This can be written as
/vx \ rvx/7 л .
F.50) qpi cot \—xb) + c2p2 cot — (/ - x0) = 0 ,
V \ Cx / L C2 J
where cx = VPi^Pi an^ сг = ^PjPi- This equation determines the eigen-
eigenvalues kk (к = 1,2,. . .). We do not solve this equation (which can be done
graphically and numerically) but note that the Afc are real and countably
infinite and positive according to the general theory. The eigenfunctions
vik\x) corresponding to the kk are
F.51)
0<Jt<;tn
шЫТк{1-х)/с2]
[sin[VA;(/»xo)/cr2] '
For the norm of v(lc\x) we have
xn<x<l
F-52) ||0
= Г л(»1к)WJ л + f ft^WJ
" *o)
(/ - x0) /c2) *
orthogonality of the set of eigenfunctions vk(x) may be verified directly,
an<*, using F.52), the set may be orthonormalized. The completeness
Property of the eigenfunctions asserts that a function <p(x) under suitable
^nditions can be expanded in a series of eigenfunctions.
Applying the foregoing results to the initial and boundary value problem
* the heat equation F.41)-F.44), we obtain on separating variables or
lng finite Fourier transforms, the series representation for the solution
> 0 in the form
338 INTEGRAL RELATloNs
oo
F.53) u(x9 0) = 2 ake~kktv{k\x) ,
where the Л* and i/*}(;c) are the eigenvalues and eigenfunctions of th
problem F.45)-F.47). The ak are determined as the Fourier coefficients f
the initial temperature distribution f(x)\ that is,
00
and they are given as
l Г [X{
F.55) tt*= ||p(*)W||2[ftJ0
for к = 1,2 ... and ||у(/с)(д:)||2 given in F.52).
The series F.53) represents the formal solution of the problem. The
nonhomogeneous version of the foregoing heat conduction problem can
be solved using the eigenfunctions vik)(x) and the finite Fourier transform.
Similarly, related problems for hyperbolic and elliptic equations also can be
solved in terms of these eigenfunctions.
6.3. SOLUTIONS WITH DISCONTINUOUS HRST DERIVATIVES
It has been shown in Section 3.2 that discontinuities in second derivatives
for second order partial differential equations must occur across characteris-
characteristic curves or surfaces. A large class of second order partial differential
equations were shown in Section 6.1 to have equivalent integral relations in
the case where the coefficients and the unknown functions were smooth.
Since the integral relations contain, at most, first derivatives of the unknown
functions, they can be used to attach a meaning to solutions of differential
equations that are continuous and have only piecewise continuous first
derivatives.
We consider the integral form F.5) of the hyperbolic equation F.1) <n
three dimensions. Assume that a solution u(x, t) of F.5) is continuous
across a surface <p(x, t) = 0 but has jump discontinuities in its first derivatives
across that surface. [Here x represents (jc, y, z).] Denote the surface by So
and assume it divides the given region R into subregions Rx and R2- ^e
solution u(x, t) is assumed to be smooth in R1 and R2 and the functions p, P>
q, and F in F.1) and F.5) are assumed to be smooth throughout R. A uni
normal vector n to 50 can be given in terms of the gradient vector V<p a
F.56)
tUT!ONS WITH DISCONTINUOUS FIRST DERIVATIVES 339
We apply ^е integral relation F.5) over Яг and R2 and allow these
gions to collapse onto 50. With the exception of w, and Vw, all functions in
'he integral relation F.5) including n(jc, t) are continuous across 50. There-
Therefore, we easily obtain
F.57)
where [V] denotes the jump in V across So. Since the same expression can
be obtained over any arbitrary subregion of 50, we conclude on the basis of
the duBois-Reymond lemma (see Exercise 8.1.9) that the integrand itself
must vanish and obtain
F.58) Р[Щ -Vq> - p[ut]<pt = 0 .
Now u(x, t) is continuous across 50 and is continuously differentiable in
the interior of 50-that is, directional derivatives of u(x, i) in tangential
directions on 50 are continuous. Thus we can obtain additional equations for
the jumps in ut and Vm across So (note that there are four derivatives to be
determined in the three-dimensional case so that four equations are
needed). We observe that the vectors пг = [<p,,0,0, -<px], n2 =
[0, <pt9 0, — <py], and n3 = [0,0, <pt, —<pz] are linearly independent and are all
orthogonal to the normal vector n. Thus the scalar product of the (space-
time) gradient vector Vw into any of the vectors, nl9 n2, and n3 yields an
interior derivative in 50. Denoting u(x91) by ut(x91) in Rt and by u2(x91) in
R2, we differentiate иг and u2 in each of the directions nl9 n2, and n3. The
resulting interior derivatives of и are continuous across So so that the
difference of the results for ut and u2 must vanish. This yields the following
three additional equations
F.59) [Vu.ni] = KM-KK = 0,
for the jumps in Vm and ut across So.
The four equations F.58)-F.61) are a homogeneous linear system for
Jumps [«J, [Uy]9 [uz]9 and [ut] across So. Since we assume that one or
ore of these jumps is nonzero, the determinant of the coefficients of this
ystem must vanish to guarantee a nonzero solution. Evaluating the de-
ermjnant gives the equation
F-62) <Р<[Р<Р<-Р(У<РJ] = О.
Equating the bracketed term in F.62) to zero yields the result
340 INTEGRAL
F.63) P
On comparing with C.76) we note that F.63) is the characteristic equatio
for the hyperbolic equation F.1). Thus <p(x, t) = 0 must be a characteri^
surface and discontinuities in first derivatives can occur only across
teristic surfaces. It is shown in the exercises that <p, = 0, which is
consequence of F.62), does not lead to a surface of discontinuity, \у
conclude that a continuous, piecewise continuously differentiable solution of
F.1) or, more precisely, of F.5), satisfies F.1) in regions where it is
smooth and the jumps in the first derivatives must satisfy F.58) on all
characteristic discontinuity surfaces.
Equation F.63) can be expressed (apart from a nonzero factor) as a dot
product of the normal vector n [i.e., F.56)] to the characteristic surface
<p-0 and the vector t = [pV<p,-p<pf]. Since n-t = 0, the two vectors are
orthogonal and t is tangent to the characteristic surface. But F.58) can be
written as [Vw]*t = 0, and this states that the (interior) derivative of [u] in
the direction of t is zero on the characteristic, consistent with the fact that и
is continuous there. Consequently, the condition F.58) is automatically
satisfied if the solution и is continuous across the characteristic but has
jumps in the first derivatives.
For the parabolic case F.9), where ut does not appear in the integral
relation, the foregoing analysis yields F.58) with the ut term absent. The
further calculations are the same as before, and instead of F.62) we obtain
F.64) P<P*(V<PJ = O,
which characterizes the surface So across which ut and Vw can have jump
discontinuities. Again <pt = 0 does not yield a surface of discontinuity (see
the exercises). This leaves (V<pJ = 0, which is the characteristic equation for
the parabolic equation D.4), and we conclude that <p - <p(t), so that the
surfaces t = constant are the characteristics. As a result, Vm is continuous
across the characteristics, since it represents interior differentiation on the
surfaces t = constant. Thus it appears that only ut can have a jump across the
characteristics. However, by considering the jump in the parabolic equation
D.4) across t = constant (in the manner of Section 3.2), we obtain p[ut] = 0,
since м, its interior derivatives, and all other terms in the equation are
assumed to be continuous across t = constant. Consequently, ut must also be
continuous across t — constant. 2
For the elliptic equation F.12) we obtain the result F.64) with the tpt
term absent. Since the surface So is given as <p(x, y9z) = 0 in this case (i.e.»
there is no r-dependence), we again conclude that there are no real
characteristics or discontinuity surfaces for the elliptic equation F.11).
Example 6.3. The Cauchy Problem for the Wave Equation. We consider
a Cauchy problem for the one-dimensional wave equation with discontinu-
discontinuous initial data. Let u(x, t) satisfy
oLl;TlONS WITH DISCONTINUOUS FIRST DERIVATIVES
F.65) w« " C4, = 0 , -oo<*<°o, t>0
ith the initial conditions
F.66) w(x,0) = 0, -
341
F.67)
O,
Since u(x, t) is initially continuous and ut(x, 0) has a jump discontinuity, we
expect the solution to have discontinuous first derivatives on the characteris-
characteristics that issue from the point (x, i) = @,0).
A formal application of d'Alembert's solution B.29) yields
F.68)
with w/jt,0) given as in F.67). We divide the half-plane t>0 into three
sectors as shown in Figure 6.5. The sectors I, II, and III are separated by the
characteristic curves x= ±ct that issue from the initial discontinuity point
@,0) of the velocity ut(x, t). The integration in F.68) is easily carried out
and yields
F.69) u(x, t) =
0;
с/'
JC
JC
JC
- ct > 0 :
-сГ<0<л
+ ct < 0 ;
с + ct :
I
II
III
u(X> 0) = и, (x,0) = О «Ос, 0) « 0, ut(x, 0) = 1
Figure 6.5. The solution of the Cauchy problem.
342 INTEGRAL RELATloNs
q () rior of
sector I, II, and III. Also, the solution u(x, t) is continuous across
characteristics ±t I t I h д/д 0 d
Clearly, u(x, t) satisfies the wave equation F.65) in the interior of ea к
ctor I, II, and III. Also, the solution u(x, t) is continuous across th
characteristics x = ±ct. In sector I we have ди/дх = 0 and duldt^\ |
sector II, duldx^lllc and du/dt= 1/2, whereas in sector III, duldx^
dld 0 Th h drii f ( ) j T
, d
duldt = 0. Thus the derivatives of u(x, t) have jump discontinuities acrc
the characteristics x = ±ct. Therefore, u(x, t) is not a strict solution
F.65) since it lacks the required number of derivatives.
To show that F.69) satisfies the integral form F.5) of the wave equation
we show that ut and ux satisfy the one-dimensional form of the jum
condition F.58) across the characteristics x = ±ct. We have already see
that u(x, t) is a solution of the wave equation F.65) away from these
characteristics. For the characteristic x = ct, we have <p(x, t) = x - ct = 0 so
that <px = 1 and <pt = -c. Also, p = 1 and p = c2 so that F.58) becomes
(,70,
On the characteristic x - -ct, we have <p = x + ct = 0 with <px = 1 and <p, = c.
Thus F.58) becomes
F.71) с \иЛ - c[ut] - с —- —-~ \ ~~ ч — Г
v y L xJ l (J l дх дх J I <?/ ^/ J
We have demonstrated that u(x, t) is a twice differentiable solution of the
wave equation F.65) away from the characteristics and that it satisfies the
jump condition F.58) across the characteristics x = ±ct. Thus F.69) satis-
satisfies the integral form F.5) of the wave equation.
6.4. WEAK SOLUTIONS
The results of the preceding section have extended the concept of the
solution of second order partial differential equations to the case of discon-
discontinuous first derivatives. Had we introduced initial displacements u(x,0)
with jump discontinuities in Example 6.3, a formal application of d'Alem-
bert's solution would have shown that the solution has discontinuities across
characteristics for the wave equation. Such "solutions" cannot be discussed
on the basis of the foregoing methods where the continuity of solutions was
assumed. To deal with such problems, we weaken the concept of solution
even further and obtain a new integral expression for each of the hyperbolic,
SOLUTIONS 343
rabolic, and elliptic equations considered. Again, this expression is equi-
f^lent to the given differential equation when the solutions are smooth.
However, it remains valid even if the solution has jump discontinuities.
VVe discuss the hyperbolic equation F.1) in detail and then state the
suits for the parabolic and elliptic equations of Section 4.1. Given the
r ion #s we consider the smooth function v(x, t) that is assumed to vanish
"identically near the boundary dR of R if R is of finite extent. If R is of
nfinite extent, we assume v(x, t) vanishes outside some bounded region as
well as near any finite boundary of R. We define the operator L as
F.72) L[u\ = putt - V- (рЩ + qu ,
where we are again considering two- or three-dimensional spatial regions.
Then we readily obtain, since the operator L is formally self-adjoint (see
Section 3.6),
F.73) vL[u]-uL[v]
= p[vutt - uvtt] - [uV* (pVu) - uV-(pVv)]
= — [pvut - puvt] - V- [pvVu-puVv],
since p = p(jc). Using the space-time gradient vector V= [V, dldt] we have
F.74) vL[u] - uL[v] = -V- [pvVu - puVv, ~pvut + puvt].
Integrating over the region R we obtain
F.75) J £ {v(L[u] - pF) - (uL[v] - vpF)} dV
= - I I V- [pvVu - puVv, -pvut + puvt] dV
= - I [pvVu - puVv, ~pvut -I- puvt] -nds = 0
JdR
where n is the exterior unit normal to dR. The last integral in F.75) results
pn using the divergence theorem, and it equals zero since v vanishes
identically near dR. If R is unbounded, the result is still zero since v(x, t)
vanishes for sufficiently large x and t. The divergence theorem is then
aPpHed to a region bounded by dR and by portions of circles or spheres
whose radius is subsequently allowed to tend to infinity.
We conclude from F.75) that
v^[u] ' pF) dV=l L (u£t[v] "
vpF) dV *
1
344 INTEGRAL RELATloNs
If u(x, t) is a smooth solution of F.1), the integral on the left in Fj(>\
vanishes and we conclude that for any admissible v(x, t) we have
F.77) J |д (uL[v] - vpF) dV= 0 .
Conversely, if u(x, t) satisfies the integral expression F.77) for all admiss-
admissible v(x, t) and u(x, t) is twice continuously differentiable, we obtain from
the equation F.76)
F.78) J Jr v(L[u] - pF) dV= 0 .
Using the fundamental lemma of the calculus of variations (see Exercise
8.1.8), we conclude from the arbitrariness of v(x, t) that L[u] - pF = 0 so
that u(x, t) satisfies F.1).
If u(x, t) is a twice continuously differentiable solution of the hyperbolic
equation F.1), we shall say that u(x, t) is a classical solution of the
differential equation. If u(x, t) is a solution of F.77) for all admissible
v(x, i) but does not have the required number of derivatives to be a classical
solution of the differential equation F.1), we say that u(x, t) is a weak or
generalized solution of the differential equation. Note that the solutions of
the integral relation F.5) may also be termed weak solutions if they are not
twice differentiable.
Weak solutions of initial value problems for F.1) are defined as follows.
We proceed as before but assume that a portion of dR, say dR0, lies on the
initial region t = 0, where u(x, 0) =/(*) and ut(x,0) = g(x) are specified.
The function v(x, t) need not vanish on dR0 but it must vanish on the
remainder of dR. As a result, the integral over dR in F.75) is nonzero when
taken over the subset SR0 of dR. Thus, u(x, t) is a weak solution of the
aforementioned initial value problem for F.1) if
F.79) | jR (uL[v] - vpF) dV +1^ (pfu, -pgv)ds = 0,
for all admissible v(x, t).
It can again be shown that if u(x, t) is a classical solution of F.1) in two
subregions Rx and R2 of R and has a jump discontinuity across the region 50
separating R1 and R2, then 5? must be a characteristic. This is done by
applying F.76) to the two regions Rx and R2 and considering the nonzero
contributions-the integrals over the region So. We do not carry out the
discussion which is similar to that given in Section 6.4.
We now define weak or generalized solutions for the parabolic ano
elliptic equations of Section 4.1. We say that u(x, t) is a classical solution of
the parabolic equation
SOLUTIONS 345
F.80) £["] == ри, - V- (pVw) + qu=pF
*f w(x, 0 has a continuous time derivative and two continuous x-derivatives.
\Vith the region R and the smooth function v(x, i) as previously defined, we
easily conclude by adapting the foregoing procedure for classical solutions
u(x, t), that
f fR[u{-pvt-
The operator L* in the curly bracket is the formal adjoint of L (see Section
3 6). If u(x91) satisfies F.81) for all admissible v(x, t) and is not a classical
solution of F.80), we say that it is a weak or generalized solution of the
differential equation F.80). Weak solutions of initial value problems for
F.80) are considered in the exercises.
Similarly, if u(x) is a twice continuously differentiable solution of the
elliptic equation
F.82) L[u] = -V- (pVu) + qu = pF,
и is said to be a classical solution of F.82) and it is easily shown to satisfy
the integral relation
F.83) | jR [m{V- (pVv) - qv} + vpF] dV= 0 ,
for all admissible v(x, t) since L is a formally self-adjoint operator. If u(x)
satisfies F.83) but is not a classical solution, then it is said to be a weak or
generalized solution of F.82). In view of the smoothness properties of
solutions of elliptic equations, there is generally no distinction between
weak and classical solutions in the elliptic case.
We now consider some examples. In the first example we consider a weak
solution of an initial value problem for the wave equation F.65). In the
second example we examine the Fourier series solutions obtained in Chapter
4 and show how they may be interpreted as weak solutions.
Example 6.4. Weak Solutions of the Wave Equation. The Cauchy prob-
problem for the wave equation
uu - c2uxx = 0 , -oo<x <oo, t>0 ,
wi*h the initial data
the formal d'Alembert's solution
346
INTEGRAL RELATION
F.86)
A, jc-cf>0
§, x~ct<0<x
0, jc
Since u(x, t) is discontinuous across the characteristics x~±ct, we mu
interpret the solution in a generalized sense.
We consider the rectangular region R displayed in Figure 6.6 and assume
that the smooth function v(x, i) vanishes near the boundary SR of R except
for the part of SR that coincides with the лг-axis. The region R is divided into
sectors I, II, and III, and we now show that the solution F.86) satisfies
F.79) for all admissible v(x, i) so that it is a weak solution of the Cauchy
problem for F.84). We have
F.87)
u[vtt - c2vxx] dV= \ I l[vtt - c\x) dx dt
Jln0[vtt-c\x]dxdt.
For the integral over the region A, where A represents I or II we have
(xv T)
(x2,T)
lx2,0)
Figure 6.6. The region R.
SOLUTIONS 347
- I u, dx + c2yx Л ,
JdA
on using the divergence theorem and the fact that nds = [dt,-dx\. The
integration over dA is taken in the positive direction.
Since v(x, i) vanishes on dR if t > 0, and onjc + c^ = 0we have vtdx +
we obtain
r@,0) r@,0)
F.89) jj[\]ddl djj
-\c\ do
J@,0)
Since T is arbitrary, we have shown that м(дс, f) satisfies F.79) (appropri-
(appropriately modified for this example) for all x and for t > 0, and it is therefore a
weak solution of the Cauchy problem F.84), F.85).
Example 6.5. Fourier Series and Weak Solutions. The series solutions
obtained in Chapter 4 by the method of separation of variables or by finite
Fourier transforms were shown to require substantial conditions on the data
for the given problems to render them classical solutions. This was especially
the case for the hyperbolic equations that were considered. It was indicated
ш Chapter 4 how the series may be understood as generalized solutions in
case they are not classical solutions, and we now use the preceding results to
show in a more precise manner how the series solutions are to be inter-
interpreted as weak or generalized solutions.
Our discussion is restricted to the hyperbolic equation D.10) with the
Homogeneous boundary conditions D.14) and the initial conditions D.16).
Proceeding as in Section 4.6 we expand u(x, t) in a series of eigenfunctions.
348
INTEGRAL RELATION
with Nk(t) defined as in D.207) except that Bk = 0 for all к since th
boundary conditions are homogeneous. We assume that each of the eige C
functions Mk(x) is twice continuously differentiable and so are the N (a
The partial sum }t
F.91) un(x9t) = ^Nk(t)Mk(x)
is easily seen to be a classical solution of the equation
F.92) Р^Г~*
where Fn(x, i) is the nth partial sum of the Fourier series of F(xy t). The
initial conditions for the un(x, t) are
n n
F.93) un(x,0) = 2 Nk@)Mk(x) - 2 (/, Mk)Mk(x),
*=1
F.94) ""^' } = g Л^(О)А#к(дс) = ^2 (g, M,)M,W ,
and the boundary conditions are
F.95)
= 0.
We consider the time interval 0 < t < Г and the region G on which the initial
data are assigned together with its boundary dG, Let region G and its
boundary <?G be extended parallel to themselves from f = 0 to t = T and
denote the resulting region by Л. Let v(x, i) be a smooth function that
vanishes near SR. Then since each un is a classical solution of F.92), we
have from F.77)
F.96)
We now assume that asn^oowe have un(x, 0)-»/(x), dun{x,0)/dt-*g(x)>
Vun(xy 0)^V/ and Fn->F where / and g are the initial data and pF is the
inhomogeneous term of the hyperbolic equation. The convergence ot
un(x9 0) to f(x) is assumed to be uniform, whereas all other cases represent
mean square convergence [see D.67)]. It can then be shown that the partial
sums un(x, t) converge to a function u(xy t) in the region R in the mean
square sense.
If we proceed to the limit as n—>*> in F.96) we conclude that
INTEGRAL WAVE EQUATION 349
F.97)
since
lim I/1 [(u - un)Lv - p(F - Fn)v] dV
li
where L is the operator in the curly brackets in F.97) and the Cauchy-
Schwarz inequality (see Exercise 4.2.12) has been used. Thus u(x, t) is a
weak solution of D.10). On using F.79) instead of F.77) to define a weak
solution, the foregoing analysis shows that u(xy t) is a weak solution of the
given initial and boundary value problem.
We have not taken into account the boundary conditions assigned for the
problems considered in this section because v(x, t) was chosen to vanish on
the boundary. An alternative weak formulation of initial and boundary
value problems for more general forms of (second order) equations than
those considered in this section is given in the exercises. Boundary terms do
play a role in the resulting weak solutions. This formulation is used in the
Galerkin method for obtaining approximate solutions of the given initial and
boundary value problems (see Section 8.2).
6.5. THE INTEGRAL WAVE EQUATION
The integral relation F.5) that corresponds to the one-dimensional wave
equation F.84) leads to a number of interesting and useful expressions that
can be applied to obtain the solution of various initial and boundary value
problems for the wave equation as we now demonstrate.
If u(x, i) is a solution of the one-dimensional wave equation F.84), then
F.5) takes the form of a line integral
F.98) f [c2ux,-ut]-nds= t ut
JdR1 XJ fl J3R '
Slnce n ds = [*; - dx]. When the wave equation F.84) has a nonhomoge-
«eous term F, this term must be integrated over the region R. We shall call
F.98) the integral wave equation, and we now choose various regions R and
Integrate F.98) over their boundaries.
First we let R be a region in (x, t)-space bounded on four sides by
characteristic lines x ± ct = constant as pictured in Figure 6.7. Thus dR is a
350
INTEGRAL
С =* (x - k, t + A)
В = (x + ch, t - k/c)
Figure 6.7. The characteristic quadrilateral.
characteristic quadrilateral. Now on the lines x ± ct — constant we have
dx = +c dt and ut dx + c2ux dt=+c du. The integral F.98) takes the form
r rB rC rD rA
F.99) utdx + c2ux dt= cdu- с du + cdu- cdu
JdR ' X JA JB JC JD
= 2c[u(B) + u(D) - u(A) - u(C)] = 0 ,
with w(C), for example, given as u(C) = u(x-k,t+ A), where к and h are,
given positive constants. We, therefore, obtain the difference equation
F.100) u(x -k,
= u\x + ch,t J + u\x - ch, t + -) .
Using Taylor's expansion it can be shown that any twice continuously
differentiable solution of the difference equation must satisfy the wave
equation in view of the arbitrariness of h and к (see Exercises 6.5.1 and
6.5.2).
. iNTEGRAL WAVE EQUATION
351
(x,0
Figure 6.8. The characteristic triangle.
As our next choice for the region R, we choose a characteristic triangle
with base on the ж-axis and the two other sides equal to characteristic line
segments as shown in Figure 6.8. Then BR is the boundary of the charac-
characteristic triangle, and using the foregoing results we obtain
F.101) j^
J4*>0 Hx~ct,0) r(x+ct,O)
cdu+l cdw+ ut(x,0)dx
= -2cu(x, t) + cu(x + ct, 0)
rx + ct
+ cu(x - ct, 0) + J ^ u,(x, 0)dx = 0.
Given the Cauchy problem for the wave equation F.84) with initial data
(бЛ02> и(*,0) =/(*), u,(x,0) = g(x),
we obtain from F.101),
U(x, i)=\ [f(x - ct) + f{x + ct)] + *- Г" g(x) dx ,
** Z*C Jx~ct
is just d'Alembert's solution. For the inhomogeneous wave equation
a term F on the right of F.84), we must add the integral of F over the
cnaracteristic triangle R to the solution F.103).
A third choice for the region of integration R in F.98) is the region
P ctured m Figure 6.9 with the boundary dR, three of whose sides are
aCtenSticS and whose f°Urth Side is an interval on the дг-axis. We apply
over the boundary of the region R to obtain
352
INTEGRAL RE
(Ott-x/c}
<x,t)
(ct -x, 0) эй (* + <*. 0)
Figure 6.9. The quadrilateral.
J r<*,0 r(o,i-*/c)
ut dx + c2ux dt = -I с du + I c dw
r(ct~x,0) r(x+ct,0)
cdu+\ ut(x, 0) dx
-см(сГ-х,О)
+ f w,(jc,O)dx = O.
With the initial conditions u(xy0)=f(x),ut(x,0) = g(x) and the boundary
condition w@, f) = Л@» we obtain
F.105)
u(x, t) = \ [f{x + <*) - Act - x)]
4- —
Differentiating F.105) with respect to x and evaluating the result at x
gives
F.106) M0, 0 =/'() £ () А'()
= 0
\ u,ict, 0) - - u,@, 0 .
INTEGRAL WAVE EQUATION
353
Using these results, any initial and boundary value problem for the wave
uation over a finite, semi-infinite, or infinite interval can be solved and
domains of dependence for the solutions can be established as we now show.
Example 6.6. The Wave Equation in a Semi-Infinite Interval. We con-
confer the initial and boundary value problem for the inhomogeneous wave
uation jn a semi-infinite interval; that is,
F.Ю7) *n ~ c\* = F(x,t), 0 < x < oo, f > 0 ,
with initial conditions
F.108) u(x, 0) = /(*) , u£x, 0) = g(x) , 0 < x < oo ,
and the boundary condition
F.109) aw@, t) - fiux(O, t) = B(t) , t > 0
where a ^ 0, £ ^ 0, and a + /3 > 0 (with a and f$ both equal to constants)
and F, /, g, and В are given functions.
To solve this problem we break up the first quadrant in the (jc, t)-plane
into two regions. In region I, x > ct and in region II, x < ct. These regions
are separated by the characteristic curve x- ct which issues from the origin
as shown in Figure 6.10. For a point (jc, i) in region I, we integrate over a
characteristic triangle and obtain d'Alembert's solution F.103) plus an
integral of F(jc, i)\ that is,
Figure 6.10. The first quadrant.
354 INTEGRAL
F.110) u(x, 0=| [f(x - ct) + f(x + ct)]
fx + ct 1 Ct Гдс + с(/-т)
with the last integral taken over the characteristic triangle. If the point (x
is in region II, we integrate over the quadrilateral pictured in Figure 6 lo
three of whose sides are characteristics, and obtain from F.105)
F.111) u(x, t) = \ [f(x + ct) - f(ct - *)] + ^ JH*g(s) *
where h(t) = w@, t) which is as yet unspecified if /3 Ф 0 in F.109) and F(x, t)
is integrated over the interior of the quadrilateral. If F~ 0, we have
F.112) ux@, t) = f\ct) + - g(ct) - - h'(t).
с с
Substituting F.112) into the boundary condition F.109) gives
F.113) au(Q9t)-pux@9t)
= ah(t) - p[f'(ct) + \ g(ct) - \ h'(t)] = B(t),
which is an ordinary differential equation for w@, t) = h(t). The initial
condition is A@) = lim,_0 u@, t) = м@,0), assuming w(jc, t) is continuous at
@,0). In that case м@,0) = lim^0 w(jc, 0) = lim^0 f(x)=f@).
Now if /3 = 0 and a = \ in F.109), then u@,t) = h(t) = B(t) so that
F.111) is already completely specified. However, if 5@)^/@), the solu-
solution F.110), F.111) is discontinuous across the characteristic x = ct. As the
characteristic is approached from points in region I, the limit of u(x, t) is
/@) + constant and the limit of u(x, t) as x = ct is approached from region II
is B@) + constant, where the constant is identical for both regions. We
remark that even if 5@) = Д0), the derivatives ux and ut may be discontinu-
discontinuous across the characteristic x = ct, say, if 5'(°) = limr-o Mr(°> 0 ^ 8^ ~~
Нт,_0 ut(x, 0). In these cases, the solution of the initial and boundary value
problem must be interpreted in the generalized sense. If 0 =0 in F.109),
w@,f) is unspecified and we may put м@,0)=/@) to make the solution
continuous across the characteristic line x = ct. The solution of F.113) wi
when £ ?*0. If F 9*0, it contributes an additional term to h{t).
INTEGRAL WAVE EQUATION 355
In the special case where /=g = F = Owe obtain the following solution of
the initial and boundary value problem FЛ07)-F.109)
F.114) fO, x>ct
• а ф 0. If a - 0 and /3 = 1, we have
F.115) "(*> 0 =
0, x> ct
rt-xfc
с I B(r) dr , 0 < x < ct,
whereas if /3 = 0 and a = 1, we obtain
@, jc>cr
*(,_£) o<X<ct
In each case, the boundary condition F.109) gives rise to a wave of the form
h(t - xlc) that travels to the right with speed c. For this reason the foregoing
problem is often referred to as a signaling problem.
The domain of dependence of a point (x, t) in the first quadrant is given
by the characteristic triangle in region I and by the quadrilateral in region II.
This follows from the solutions F.110), F.111).
Example 6.7. The Wave Equation in a Finite Interval. The initial and
boundary value problem for the wave equation in a finite interval can be
solved by separation of variables and by the use of finite transform methods
in terms of standing waves as was shown in Chapter 4. We now use the
foregoing results to construct a solution that yields a useful small time
description in terms of propagating waves (see Example 5.13).
Given the interval 0 < x < I we consider the problem
utt - c2uxx = F(x9t), 0<x<l,t>0,
the initial data
F.118) ("(*,<>)=/(*)
an<* the boundary data
F.119) {<*u@,t)-pux@9
*here a >0, /3 >0, and a + /3 >0.
356
INTEGRAL RELATIONS
To solve the problem we break up the strip 0<Jt=£/, f>0 into
collection of regions Rl9 R2> R3,. . . , as shown in Figure 6.11. The region
Rk (Л = 1,2,...) are bounded by portions of the initial line f = o th
boundary lines x = 0, /> and portions of characteristic lines x ± ct = constant
The problem can be solved successively by starting with the solutions in R
Rly and /?3. In Rl9 u(x> i) is given by d'Alembert's solution F.110). in r
and R3 the solution is obtained as in Example 6.6 from F.111) or a slight
modification thereof. In R4 we find w(jc, i) by using a characteristic quadrila-
quadrilateral and F.100) which gives u(x, t) in terms of its (known) values in
RlyR2, and Л3. Similarly, it is possible to construct quadrilaterals and to use
F.100) or F.111) to determine the solution in all the regions Rk with к >4
on proceeding step by step from R5 to R6 to R7, and so on. The values of
u(x> t) on the characteristic lines that separate the regions Rk depend on the
Figure 6.11. The regions Rk.
357
ТЦЕ INTEGRAL WAVE EQUATION
mpatibility of the data at @,0) and (/, 0) which determines whether the
Slution is continuous.
The hatched strips in Figure 6.11 indicate how the solution u(x, i)
ropagates, if the initial or boundary data are concentrated in a small x or a
P ц t interval. Two cases are depicted. In one case we assume that
s,x o)=/W vanishes outside the interval (xo>*i) and that the other data
(x), F(x> ')' Bi(f)> and вг@ а11 vanish. This results in two waves with
harply defined wave fronts if f(x) is discontinuous at xQ and xl9 traveling to
the right and to the left with speed c. As they hit the boundaries x = 0 and
s= / they are reflected and move in the opposite direction, as shown, again
with sharp wave fronts. This process continues indefinitely.
In the second case, we assume Bx{i) = 0 everywhere outside the interval
(t ,t) and all the other data vanish. This yields a wave traveling to the right
with speed с until it reaches x = /. It is then reflected and reverses direction
until it is again reflected from the boundary x = 0 and so on. In both cases if
fl#0 in F.119) the waves do not have sharp trailing edges.
For the general case, it may be possible to detect at the initial stages of
the wave motion (if the data are concentrated in small intervals or regions),
propagating waves resulting from the effect of the data. However, as time
increases, the interference from all waves makes it impossible to distinguish
individual waves and the representation of the solution in terms of standing
waves given in Chapter 4 becomes more useful.
Example 6.8. Moving Boundaries. We consider an initial and boundary
value problem for the wave equation in a region initially given as x ^ 0. As t
increases from zero, the boundary point x = 0 begins to move according to
the equation x = h(t), with Л@) = 0. Thus, u(x9t) satisfies the wave
equation
F.120) utt - c2uxx = 0, h(t) <x«»9t>0,
with the initial conditions
F121) u(x, 0) = f{x) , ut(x, 0) = g(x), 0 < x < oo .
On the boundary we have
FЛ22) u(x, 0LA(/) = u(h(t)91) = B{t) , t > 0.
If the speed of the moving boundary \dxldt\ - |ft'(f)| exceeds the speed с
PosWHVe proPagation for F-120), the problem F.120)-F.122) is not well
(W TWs iS Seen by examinin8 FiEure 6.12, in the case where A'@>0.
C°nCentrate Our discussion on this case) The solution u(xy t) at a point
0 on the boundary curve is uniquely determined (by means of
358
INTEGRAL RELATIONs
Figure 6.12. The case when h'(t) > с
d'Alembert's solution) by the initial data/(*) and g(x). Thus, unless the data
u(h(t), t) = B(t) are such that the values of и equal those resulting from the
initial data, a solution does not exist. Consequently, the problem is not well
posed in this case. [If h'(t)<-c9 boundary condition F.122) does not
determine the solution u(x, t) uniquely, as will be shown in the exercises.
Consequently, the problem is again not well posed.]
When the boundary speed \dxldt\ = \h'(t)\ < c, the speed of wave propa-
propagation, the problem F.120)-F.122) is well posed and the solution can be
obtained in the manner indicated in Figure 6.13 for the case where h'{t) > 0.
[The result, however, is also valid for ft'(f) <0.] We consider the regions Rx
and R2 separated by the characteristic line x — ct. At a point (x, t) in Rv
d'Alembert's solution is valid and it expresses u(x, t) in terms of the initial
data. At a point (x, t) in R2, we use the quadrilateral indicated in Figure
6.13 to obtain the result,
F.123)
u(x, t) = u(a, r) + \ [f(x + а) -Л*1 + <*i)l + Yc Lcr 8(S) * '
The point (<r, r) is where the characteristic x - ct - constant through (*>
intersects the curve x = h(t). It is determined from the equations
F.124)
o* — ct = x — ct,
h(r) - a = 0 .
INTEGRAL WAVE EQUATION
359
Figure 6.13. The case when 0 < Л'@ < с.
It follows that x = Л(т) + c{t - т), so that r is a function of x - cf. Thus,
м(сг, т) = #(т) is a right traveling wave. Once (о*, т) is determined, we may
express u(x, t) in R2 as
F.125) ф, 0 = Д(т) + i [/(x + <*) - Дет + ст)] + —
As (ху О-* @,0), we also have (ст, т)-> @,0) so that with
FЛ26) lim u(h(t), t) = lim B(t) = B(O) ,
we find that unless limx^0 u(x, 0) = Итл_0 /(ж) == ДО) - #@), the solution
!s discontinuous across the characteristic x = ct and must be interpreted in a
generalized sense.
As a concrete example we consider the curve
F.127) y - u(t\ — с t \c I < г
ад^ the boundary condition
Fl28) u(h{t), t) = B{t) = M(cof, t) = ,4 cos wf
*here Л an(j w are constants. We assume that f(x) = g(x) = 0. It is then
ea!% seen that
360
F.129)
u(x,
0-'
\c-c0
0;
INTEGRAL
ct < x.
This result has the following interesting interpretation. The wave u(x t) =
A cos [(co)/(c - co))(t - x/c)], which results from the effect of the boundary
motion-which may be thought of as a moving energy source—has
frequency of oscillation о)г = [сI(с - co)]a> which exceeds the frequency of
oscillation a> of the source term (i.e., A cosa>f) if co>O. The converse is
true if c0 < 0. In both cases the wave u(x, t) moves to the right. Thus, if the
source move in the direction of the resulting wave motion the frequency o>
increases, while if it moves in the opposite direction the frequency de-
decreases. This phenomenon is known as the Doppler effect.
Generally speaking, a curve x = h(t) in the (x, f)-plane for which
|h\t)\ < с where с is the characteristic speed of the wave equation utt = c2uxx
is called a time-like curve. If |Л'@1 >с it is called a space-like curve. Any
space-like curve can be used as an initial curve for the wave equation on
which и and ди/дп, the normal derivative on the curve, must be specified.
Any time-like curve can play the role of a boundary curve on which one
condition on u(x, i) can be assigned, as is the case for the time-axis x = 0. We
now demonstrate how the foregoing results can be used to solve an initial and
boundary value problem with initial data given on a space-like curve and
boundary data prescribed on a time-like curve.
Let u{x, t) be a solution of the wave equation F.120) and suppose that
x = hx{i) is a space-like curve and x = h2(t) is a time-like curve that bound
the region R depicted in Figure 6.14. Let u(x, i) and du{x, t)ldn, the normal
derivative, be specified on x = hx(t) as
F.130) u(hx(t), t)
and let u(x, i) be prescribed onx = h2{t) as
F.131) u(h2(t), t) = B(t) .
It follows from F.130) that
F.132)
[-1,^KOJ .Vu = -u +h[(t)u, =
Vl + h[(tJ Vl + *i@2
INTEGRAL WAVE EQUATION
361
Figure 6Л4. Time-like and space-like curves.
It is possible to determine ux and Mfonjc = hx{t) in terms of f(t) and g(t)
from F.132). In the region Rx of Figure 6.14 the solution at an arbitrary
point (*, t) can be specified in terms of f(t) and g(t) by applying the formula
F.98) to the characteristic triangle shown in the figure. In the region R2 of
Figure 6.14 the solution at a point (x, t) can be specified in terms of f(t),
g(t), and B(t) by using F.98) and integrating over the quadrilateral shown
in the figure. We observe from the figure that of the two characteristics
issuing from the point (x, t), only one intersects the time-like curve while
both intersect the space-like curve. This signifies that only one wave is
generated by the data on a time-like curve while two waves are generated by
data on a space-like curve.
Another problem that is of great interest for the wave equation is the
characteristic initial value problem. In this case u(x, t) is specified on the two
characteristics issuing from the point (xo,O) and we look for a solution of
the wave equation in the interior of the sector bounded by the characteris-
ics. (The boundaries move with the speed of wave propagation.) We
insider the equation
utt - c2uxx = F(x, t), x0- ct<x<x0
the characteristic data
'6.134) JwfoOJx-cf-xo
B2(t)
t>0,
362
INTEGRAL
and the compatibility condition Bx@) = £2@). Using the difference en
tion F.100) we obtain the solution u{x, i) as a*
F.135)
where the integration is carried out over the characteristic quadrilateral
depicted in Figure 6.15.
It may be noted that in contrast to the results for first order equations
where characteristic initial data resulted in nonunique solutions, the solution
F.135) is unique. This can be attributed to the fact that the data are given
on two intersecting characteristics and are compatible at the point of
intersection. If u(x, t) is specified on only one characteristic line, the
solution is not unique.
The characteristic initial value problem plays a role in Riemann's method
of solution of the Cauchy problem for the second order hyperbolic equation
in two variables discussed in Section 8.3. The equation is assumed to be in
the canonical form for hyperbolic equations (apart from the fact that с Ф1),
F.136)
utt - c2uxx = aux + but + du + F(x, t),
U0,0)
Figure 6.15. The characteristic initial value problem.
363
CONCENTRATED SOURCE OR FORCE TERMS
here й» Ь, and d таУ be functions of x and t. We assume that the
Characteristic initial data for u(x9 i) are given as in F.134) and are compat-
compatible By treating the right side of F.136) as an inhomogeneous term, F.135)
1 elds the solution in the form of an integral equation
F.137)
+ h i7of(cr' t) da dT+h J7<>(aM<r+bUr+du) d<T dr
= Я(х, 0 + T[u,ux9ut],
where R represents the terms that involve B19 B2, and F9 and Г represents
the integral terms in ux, u,, and w.
The integral equation F.137) can be solved by iteration. We select an
arbitrary initial approximation u@)(x, i) [we can set u@) = 0 in this case] and
insert м@) into the right side of F.137). In the next approximation we have
F.138) u(l\x, t) = R(x, t) + T[u@\ Mf, M/<°>].
Iterating this process gives
F.139) u(n\x, t) = R(x, t) + T[uin'l\ u{;~l\ u?~l)].
It can be shown if the data and the coefficients in F.136) are smooth, that
the sequence of functions u{n\x, t) converges to a unique solution of the
characteristic initial value problem; that is, u{n)(x, t)-* u(x, i) as n->o°. We
also have du{n)ldx^> duldx and du{n)ldt--> duldt, which implies that u(x9 i)
satisfies F.137). The details of the proof are not presented here.
6-6. CONCENTRATED SOURCE OR FORCE TERMS
It has been generally assumed in the preceding sections and chapters that
the inhomogeneous term pF which occurs in the equations of Section 4.1,
and which represents a source or force term, is fairly smoothly distributed
throughout the region where the problem is to be considered. Often it
nappens, however, that the source or force term is effectively concentrated
near some lower dimensional region, such as a curve for a problem in
^o-dimensional space-time. In such cases it is convenient to idealize the
^homogeneous term as having infinite density in the region of concentration
and vanishing outside that region. Thus pF must be singular in the region
where it is concentrated, and the given differential equation must be
364 INTEGRAL
interpreted in terms of the integral relations given in Section 6.1. We assu
that although pF is singular, the integral of pF over the region of \t
concentration is finite, and we now show how to characterize the effect f
the inhomogeneous term on the solution.
We consider the hyperbolic equation F.1) and its integral relation F s\
If the equation is defined over (w +1)-dimensional space-time, th
inhomogeneous term pF is assumed to be concentrated over an
sional region 50 and to have the property that
F.140)
where ds is a surface or line element. The result F.140) is a consequence of
the fact that F(x, t) = 0 for (x, t) not on 50, but F(x, t) is concentrated on 50
such that the integral over R collapses to an integral over 50. The function
pF0 represents the surface distribution of the force or source term over 50.
[pF is also known as a single layer distribution.] The solution u(x, t) of F.5)
is assumed to be continuous across 50. We do not consider double layer
distributions of F(x, t) over 50. In that case, the solution u(x, t) is discon-
discontinuous across 50 (see exercises 7.3.23 and 7.3.24).
To determine the variation of и across 50, we let the region R collapse
onto 50 so that in the limit dR coincides with 50 and, in fact, covers it twice
as shown in Figure 6.16. The limiting normal vectors n of dR [in the integral
relation F.5)] have opposite direction at common points on So so that they
differ only in sign. This implies that
F.141) |5o [(pVu, -put])SQ.nds = -jSQ PF0 ds ,
with the term involving qu vanishing in the limit since p, p, and q are
assumed to be smooth functions. The bracket [.. ,]Sq in the integral repre-
represents the jump in the quantity across 50. Since F.141) is also valid if the
Figure 6.16. The region R.
CONCENTRATED SOURCE OR FORCE TERMS MS
•ntegrals are taken over an arbitrary portion of So, we conclude from the
ju3ois-Reymond lemma (see Exercise 8.1.9) that the jump condition
F.142) [№, -P",]L0*n = -pF0
is valid at any point (jc, t) on So. The continuity of u(xy t) across So yields a
second jump condition
F.143) Mso = O.
Let <p(x, t) = 0 represent the surface 50. Applying interior differentiation
to F.143) in 50 as in F.59)-F.61), yields an additional three equations (in
the three-dimensional case) for the jumps in the first derivatives across 50.
In combination with F.142), we have a system of equations from which the
jumps in the first derivatives of и can be determined. However, if 50 is a
characteristic surface, the system has no solution for nonzero Fo. In that
case, the homogeneous version of this system [that is, F.58)-F.61)] has a
nontrivial solution. The difficulty arises because u(x, t) itself must be
discontinuous across 50, if 50 is characteristic. Since the (source) surface So is
moving at a characteristic speed, it cannot generate a disturbance that
travels ahead of it, as that would require a speed greater than the (max-
(maximum) characteristic speed. It does, however, leave a disturbance in its wake
and, as a result, u(x, i) has a jump across 50.
In the special case that 50 is a cylindrical surface with generators parallel
to the f-axis, we conclude that [ut]So = 0 since [u]Sq = 0. Then, we can
express F.142) in terms of the jump in the normal derivative of и across 50
as
ldnh0 p
]
p and p evaluated on 50.
We only consider problems for F.1) in which (each) 50 divides the region
^ over which the problem is defined into two disjoint regions. Away from
^o> u(x,t) satisfies a homogeneous version of F.1) since F{x,i) vanishes
there. To solve the given problem, we can solve for и in each of the two
regions, using the given data for the problem and matching these solutions
jjc|oss 50 with the aid of F.142)-F.143). This approach is used in Example
•10. A more direct approach to the solution of such problems is presented in
Example 6.9.
In the case of one space-dimension if pF is concentrated at the point
^^o for all time in F.1), we have instead of F.142), F.143),
1 - P(
U p(x0)
366 INTEGRAL
With appropriate modifications the foregoing matching conditions are valid
for the elliptic and parabolic equations of Section 4.2.
Example 6.9. Vibration of a Loaded String. We consider the vibration of
a finite string with fixed endpoints and with concentrated masses m. (/ =
1,2,. . . , n) placed at points xt (i = 1,. . . , n) on the string, with 0<x <Tl
As the string vibrates, the masses mt exert concentrated forces Ft(t) at the
points jc(, and these are determined from Newton's second law of motion to
be
F.146) F£t) = -т,ий(х„ t), i = 1,2,.. . , n , t>0 ,
where u(x, t) is the displacement of the string. Since the acceleration of the
point mass m, is just the acceleration of the point xt on the string and this is
given by utt(xi91), we obtain the formulas in F.146).
The problem to be solved for an arbitrary, not necessarily homogeneous,
string requires that u(x, t) satisfy
F.147) p(x)utt = y yp(x) y-J , 0 < x < I, t > 0, x Ф xt,
(generally, p(x) represents the constant tension Г), with the conditions
F.148) м@, r) = 0, m(/, t) = 0, t>0,
at the boundary, and the initial conditions
F.149) u(x, 0) = f{x) , i#,(jc, 0) = g(x) , 0 < x < I.
The jump conditions at the points x = x( are
F.150)
= mIM,,(jcI,0, / = 1,2, ...,n.
where F.146) was used and pF is replaced by F in this case.
The initial and boundary value problem F.147)-F.150) can be solved as
follows. Let
F.151) u(x,t) = M(x)N(t),
and we obtain, on separating variables,
F.152) N"(t) + XN(t) = 0 , t > 0 ,
CONCENTRATED SOURCE OR FORCE TERMS ^7
F.153) (р{х)М'{х)У + Xp{x)M{x) = 0 , 0 < x < l9x * xt,
^th the boundary conditions
F.154) M@) = 0, M(/) = 0,
and the jump conditions
f[Af(x)L( = 0 =
FЛ55) [т,М(;ОЛГ@ -^,)[M'W]^@ '
Using F.152) we can replace the second matching condition in F.155) by
F.156) р(х,)[М'(х)]Х{ + Лт,Л#(х,) - 0 , i - 1,2,. . . , n .
The equations F.153)-F.156) constitute an eigenvalue problem for
M(x), with the interesting aspect that A, the eigenvalue parameter, enters
not only in the differential equation F.153) but also in the jump conditions.
It can be shown that the eigenvalues and eigenfunctions for this problem
have essentially the same properties as those occurring in the Sturm-
Liouville problem (see the exercises). In the exercises a vibrating string
problem with a load (or mass) placed at one end of the string is considered.
Once the eigenfunctions are determined, the solution of the initial and
boundary value problem is found in the usual manner by using eigenfunction
expansions.
Example 6.10. Moving Concentrated Forces or Sources. Let us consider
the one-dimensional versions of the hyperbolic or parabolic equations of
Section 4.1 and assume the force or source term moves and is concentrated
along the curve jc = h(t) which plays the role of So in the foregoing
discussion. In the hyperbolic case we use the one-dimensional form of the
matching or jump conditions F.142) and F.143) onjc = h(t). Since u(x9 t) is
assumed to be continuous across x = h(t), a relationship between the jumps
1MJ and [ut] across the curve can be established by differentiating [u] along
x~h(t). Also, an expression relating the jumps [ux] and [ut] is determined
from F.142). This yields
F.157)
hio p-ph'(tf
where p and p are evaluated onx = h(i). In the parabolic case it is easily
sfen that the jump condition corresponding to F.142) does not involve a
me derivative of u. The matching conditions in both the hyperbolic and
Parabolic cases can be given in terms of a known function fo(t) (expressed in
ter of F0(r)) as
368
INTEGRAL RELATIONS
F.158)
An interesting problem of this type is the Cauchy problem for the wave
equation with a concentrated source term moving along the line x = с t with
|co| < c, where с is the speed of wave propagation. We assume that the
source term is oscillatory and obtain a result similar to that for the problem
considered in Example 6.8. The problem is formulated as follows. The
function u(x, i) satisfies the wave equation
F.159) utt-c2uxx = 09 -*><x<cot, c0t<x<™9 co>0,
where
F16°) "-Wx.0,
The matching conditions are
Cot < X < oo
F.161)
where A and <o are constants. The initial conditions are
F.162)
In the regions Rx and R2 of Figure 6.17 we have и1 = и2 = 0 since the
domains of dependence of points in these regions contain only initial data
that are all zero. In Rt we look for a solution in the form ut(x> t) — g(x + ci)
and in R2 u{x, t) has in the form u2(x, t)-f{x - ct) since the moving source
generates waves moving to the right and to the left. The curve along which
the source moves is time-like and characteristics issue from either side of the
curve. (A more systematic approach based on the integral wave equation of
Section 6.5 can also be used.) Applying the matching conditions F.161) we
readily obtain
F.163) <
c°'< x < ct
CONCENTRATED SOURCE OR FORCE TERMS
369
Figure 6.17. The moving source problem.
We again observe the Doppler effect in F.163). The wave иг(х, t) that
moves in a direction opposite to that of the source if co>O has the
frequency щ — [cl{c + co)]a> which is smaller than the frequency a> of
the source term. The wave u2(x, t) that travels in the same direction as the
source has the frequency <o2 = [c/(c - cQ)]a> which exceeds the frequency <o
of the source term.
If the source speed c0 exceeds the characteristic speed c, the foregoing
results are not valid. The problem is formulated as in F.159)-F.162),
except that we consider an arbitrary source term/0(f) (that is, it need not be
oscillatory) and assume that c0 > c. Thus, the line x = cot lies between the
characteristic line x- ct and the positive лс-axis, and is not situated as shown
Ь Figure 6.17. Because the initial data F.162) are zero, we find that
Wj(x, t) = 0 in the sector between the negative jc-axis and the line x = —ct.
Similarly, u2(x, t) = 0 in the sector between the positive x-axis and the line
* = cot. Then, the matching conditions F.161) determine the (Cauchy) data
°n the space-like line x = cot to be
F.164)
ux(cotf
дЫл
-^ (cot, i) = fo(t) .
0 solve for иг(х, i) in the sector bounded by x-cot and x = ct, we
^nstruct a characteristic triangle whose base is a segment of the line x = cot.
Using the integral wave equation F.98) gives
370
INTEGRAL RELATION
Now u(x, t) is continuous across the characteristics x = ±ct since th
initial data are continuous at the origin. Consequently, ux at x = cr equ Г
F.165) evaluated on that line, and иг on x--ct equals zero since S
vanishes to the left of x = —ct. To complete the determination of u, we mu t
solve the characteristic initial value problem in the sector bounded by th
lines x = ±ct, with the foregoing data. From F.135) we obtain
c2 _ c2 f(x + ct)/{co+c)
F.166) «iOM) = ^^J0 fo(s)ds9 -<****<*.
Thereby, the solution is completely determined.
In the problem considered previously where the source speed is less than
the characteristic speed, the moving source point gives rise to two waves,
one of which precedes the source. In the present problem, the source point
again gives rise to two waves, one moving to the right and the other to the
left. However, no wave precedes the source point, since the wave speed is
less than the source speed.
If the source speed c0 = c, the characteristic speed, the solution u(x, t)
must be discontinuous across the path of the source, as was noted previous-
previously. In that case the source gives rise to only one wave traveling in the wake
of the source. Ahead of the source the solution is zero if the initial data are
zero. Thus, if the solution were continuous on the source path, this wave
would have zero height on the source path. Since the wave is constant on
each characteristic that intersects the source path, it must have zero height
everywhere and this is not possible.
6.7. POINT SOURCES AND FUNDAMENTAL SOLUTIONS
In this section we assume that the inhomogeneous term pF in the differen-
differential equations of Section 4.1 that represents a source or force term is
effectively concentrated at a point Fo. We consider two cases. To begin, we
deal with higher dimensional hyperbolic and parabolic equations and assume
that pF is concentrated at the point Po = x? for all time t == 0. The one^
dimensional version of this problem was considered in Section 6.6. Next w^
assume that the point of concentration in the hyperbolic or parabolic case i
*о = (*о>'о) and in the elliPtic case' Po = *o- For the time depe^. ^
problems this constitutes an instantaneous point source. [We recall tna
and x0 represent a one-, two-, three-dimensional spatial variable.]
In both cases pF can be expressed in terms of the Dirac 8-function brie у
discussed in Chapter 1. Since the theory of the б-function and otn
generalized functions is not presented until Chapter 7, we discuss
POINT
SOURCES AND FUNDAMENTAL SOLUTIONS 371
going problems, as far as possible, on the basis of the integral relations
i Section 6.1. However, the theory of Green's functions developed in
rhapter 7 is closely related to the point source problem.
\Ve start with the integral relations F.7) and F.10) for the hyperbolic
d parabolic cases, respectively. We assume that pF is concentrated at the
**oint Po - *o in the re8ion Rx for all r > 0 and that it has the property
F.167)
where ДО is a (smooth) function that characterizes the source strength for
f>0 and vanishes for t<0. In the limit as Rx shrinks down to the point Po,
we obtain from F.7) and F.10)
Although F.168) is, in fact, integrated over the interval to<t<tl in F.7)
and F.10), our result is valid since the interval of integration is arbitrary.
The integrals in F.7) and F.10) that involve и or ut vanish in the limit as
Rx-+P0 since these terms are not more singular than x-derivatives in и and
the relevant integrals are higher dimensional than the integral in F.168).
The result F.168) is valid for an arbitrary region Rx and its boundary
BRX. It takes on a simpler form if Rx is chosen to be a specific region. In the
two-dimensional case let dRx be the circle of radius r with center at
^о = (*о>.Уо) anc* introduce polar coordinates with the pole at Po. Then
F.168) becomes
F.169)
IT" ^x =lun f p^ rdB = lim (ivpr ^) = -fit) ,
дПх * r-*0jo r дГ r->0 V И дГ/ /W
3RX
where the integral is taken over the circle of radius r. In the three-
dimensional case let dRx be a sphere of radius r with center at Po =
)*o>)>o,zo) and introduce spherical coordinates with the pole at Po. Then
F168) easily yields
*(*o, y0, z0) JdRx r dnx * r-
over the sphere of radius r. In F.169) and F.170) we have
fact that ^"/<?лх = duldr on dRx and assumed that the mean value
rJ г°г integrals can be applied in both cases.
\ P°int source Probiem can now be formulated as follows. The function
' 0 satisfies the homogeneous form of the relevant hyperbolic or
372 INTEGRAL *ELATl0Ns
parabolic equation for x¥^x0 and f>0, since pF vanishes away from tk
source point Po. At PQj u(x, i) must satisfy the condition F.169) or F.170^
Initially, at * = 0 it is required that u(x91)9 as well as ut(x,t) ш tk
hyperbolic case, vanish. If the problem is given over a bounded spatH
region, u(x, t) must satisfy homogeneous boundary conditions.
The foregoing problem requires a different approach from that used i
solving the concentrated source problem of the preceding section or, for that
matter, all the problems of the previous sections. In the earlier problems it
was possible to divide the region R over which the problems were formu-
formulated into two (or more) subregions separated by the curves or surfaces on
which the sources or singularities were concentrated. Consequently, these
curves or surfaces could be treated as additional boundaries for the prob-
problems, on which boundary conditions (i.e., matching or jump conditions)
could be assigned. For the preceding point source problem and the one
treated in the following, such a subdivision is not a priori possible because
the source is concentrated either on a line or at a point and, as such, cannot
represent a boundary for the region, given the space-time or space dimen-
dimension of the problems we are considering. (However, if the problem has
certain symmetries, such as axial symmetry, it may be possible to reduce the
number of independent variables and treat the source points as boundary
points as was done in Example 5.9.)
The nature of the singularity of the solution m(jc, t) of this problem at the
source point may be determined by treating F.169) and F.170) as approxi-
approximate equations near the point Fo. Integrating F.169) yields
F.171) u(x9 y, t) ~ ~y}v v log[(x - x0J + (y - уоJ]ш
гтгр(х9 у0)
in the two-dimensional case. In three dimensions F.170) gives
F.172)
(x, y, z)**(xo,y09z0).
In both cases only the most singular part of the solutions has been retained.
Example 6.11. The Point Source Problem for the Wave Equation. Let и
satisfy the two- or three-dimensional wave equation
F.173) uu - c2V2u = 0 , P * pQ91 > 0
everywhere except at the source point Po. At t = 0 we assume that и - ut -
and at Fo, the condition F.169) or F.170) is satisfied.
INT SOURCES AND FUNDAMENTAL SOLUTIONS 373
To solve this problem we let г be the (spatial) distance from the source
nt p0 and look for a solution of F.173) in the form и = u(ry t). [The
P°proximate results F.171) and F.172) suggest that the solution depends
lv on these two variables locally, even in the case of variable coefficients.]
ТЪеп F.173) becomes
F.И4) и„ = с2[игг+?-у^иг], r>0,*>0,n = 2,3
in the case of two or three (space) dimensions.
We begin with the three-dimensional case since it is easier to handle. Let
n = 3 in F.174) and put v = ru. Then we easily verify that v satisfies the
one-dimensional wave equation vtt - c2vrr = 0 which has the general solu-
solution v(r, t) = F[t - (r/c)] + G[t + (rlc)l Thus u(r, t) = A lr)F[t - (r/c)] +
(\lr)G[t +(r/c)]. Since м, must vanish at t = 0 and и satisfies the source
point condition F.170) where f(t) = O for t<0, we readily conclude that
G[t + (r/c)] = 0 and the solution is
F.175)
Because f(i) vanishes for negative t, u(r, t) vanishes at t = 0 when r > 0.
Near r = 0, F.175) behaves like F.172) where we must set p = c2.
The solution F.175) represents a propagating spherical wave that starts to
spread out from the source point Po at the time t = 0 with speed c. The
spherical wave fronts are given as r = ct because и = 0 when r> ct since f(t)
vanishes when its argument is negative.
In the two-dimensional case F.174) cannot be simplified as nicely as was
done for n = 3. However, the point source problem for n = 2 can be solved
in terms of Hankel transforms in which case the condition F.169) at the
source can effectively be treated as a boundary condition at r = 0 (see
Examples 5.9 and 5.10).
A simpler method for solving this problem uses the three-dimensional
result F.175) and proceeds as follows. We think of the point source at
*o == (xQ, y0) as representing a line source for the three-dimensional prob-
problem located on the line x = x0, у = y09 z = z in space. By summing (i.e.,
integrating) over all source points (л:0, yQ9 z0) along this line, we obtain the
solution to the two-dimensional problem. Thus we integrate F.175) with
respect to z0 from -oo to +» and obtain
F176) u(x,y9t)-
db
374 INTEGRAL RELATl0Ns
where z = z - z0, f2 = (x-xof + (y -y0J, r = t- (r/c) and we have used
the fact that f{t) = 0 for t<0. It can be shown that F.176) is a solution of
the two-dimensional wave equation and satisfies the condition F.169)
the exercises). Also, F.176) vanishes at t = 0 since ДО vanishes for i
ega
values of its argument. In fact, F.176) vanishes when f>ct so that tl^
circles г = ct are circular (or cylindrical) wave fronts for the two-dimension 1
point source problem.
For the sake of completeness, we include the solution of the point source
problem for the one-dimensional wave equation. This problem can be
solved by the methods of the preceding section. If the source acts at the
point x0, we require that и and ut vanish at * = 0 and that the jump
conditions F.145) are satisfied, with p = 1, p = c2 and F0(t) = f(t). To solve
we set и = F[t + A /c)(x - *0)] for x < jc0 and и = G[t - A /c)(x - x0)] for
jc>jc0. Using the initial and jump conditions, we easily obtain the solution
F.177) "^'^Jo f(s)dS
We recall that f(t) = 0 for t < 0, so that w(jc, 0 = 0 for |jc — л:0| > ct.
Example 6.12. The Point Source Problem for the Heat Equation. Let м
satisfy the two- or three-dimensional equation of heat conduction
F.178) u, - c2V2m = 0 , РФ Ро, t>0
everywhere except at the source point Fo. At t = 0 we assume that и = 0 and
at Po, и must satisfy the condition F.169) or F.170).
To solve this problem we look for a solution in the form и = u(r, t),
where r is the distance from the source point Po. Then F.178) becomes
F.179) ut = c2[urr + ^-^ ur] , r>0, f >0, n = 2,3
in two or three dimensions.
For the three-dimensional case, with n = 3 in F.179), we put v = ru and
find that v satisfies the one-dimensional heat equation vt = c2vrr. At t = 0 we
have u(r, 0) = 0. We assume that v and vr are bounded at r = 0, so that
F.170) yields the condition u@, 0 = A/4тгс2)/@. The resulting initial
value problem for v(r, t) for r>0 and t>0 was solved in Example 5.5 [see
E.105)]. In terms of that result for v(r, t) we obtain the solution
F.180) M(r, 0 = {[4irc\t - r)]/2 exp[- ^,т)
In the two-dimensional case, we set n = 2 in F.179). At t = 0 we have
POINT
SOURCES AND FUNDAMENTAL SOLUTIONS 375
^ 0 and at r = 0 the condition F.169) must be satisfied. We use the zero
^rder Hankel transform to solve this problem and proceed as in Example
rii We easily determine that the Hankel transform U(Xyt) satisfies the
equation Ut + (сЛJС/ = A/2тг)/@. Also, J7(A,0) = 0. Solving this equa-
t'on, inverting the transform, interchanging the order of integration in the
resulting double integral, and using E.193) yields the solution
F.181) u(r, 0 = I [4«2(*- т)Г ехр[- 4с2(^.т)]/(т) dr .
For the one-dimensional case, with the source located at x = x0, we cite
the result of Exercise 6.6.8, which yields the solution
F.182) w(x, i) = Jo G(x -xQ9t- r)/(r) dr ,
where G(x, t) is the fundamental solution of the heat equation defined in
equation E.45). In fact, in each of the three problems considered in this
example, the solution of the point source problem is given as a product of
the fundamental solution of the heat equation with the density function/(r),
integrated from 0 to t. (Fundamental solutions of the heat equation are
discussed later in this section.)
Next we consider point source problems for elliptic equations and instanta-
instantaneous point source problems for parabolic and hyperbolic equations. In the
elliptic case the source point is Po = *o, whereas in the hyperbolic and
parabolic cases it is Po = (xQ,t0). The source is assumed to have unit
strength, and the concentration of pF at the point Po implies that
F.183) j jRpFdv =
for any region R in space or space-time that contains Po, whereas pF
vanishes when РФ Po. Thus pF is identical with the Dirac 3-function with
singular point at Po.
Since pF = 0 for РФ Po, the solution u{x) or w(jt, i) of the point source
Problem satisfies the homogeneous form of the relevant differential equation
or the equivalent integral relation away from Po. The behavior of и at Po
*nust be determined from the integral relation since и is not expected to be
smooth at Po.
Proceeding formally, we consider the integral relation F.5) for the
hyperbolic case and assume the region R contains the source point Po. Then
as R shrinks down to Po we obtain the formal limit
lim I [pVu, -
376 INTEGRAL RELATlO\s
on using F.183) and assuming the other terms in the integral relation vanish
in the limit. Similarly, the integral relation F.9) for the parabolic cas
yields, as /?->P0,
F.185) lim J [pVw, -pu]-nds = -1.
In the elliptic case the integral relation F.12) gives
F.186) lim f p^ds = -l.
SR^Pq JdR dtl
In obtaining F.185) and F.186) we have used F.183) and assumed that all
other terms in the integral relations F.9) and F.12) vanish in the limit as
Now for the hyperbolic problem the singularity of u(x, t) does not remain
isolated at the source point Po but is transmitted along the characteristics
that contain that point. Thus u(x, i) may not possess the derivatives near Po
required to make the integral relation F.5) and its limit F.184) valid. In
that case we must replace the foregoing integral relation by that obtained in
Section 6.4 for which the solutions u(x, i) were not required to be differenti-
able. The relevant integral relation is F.77) and in terms of that expression
it is easy to see that F.184) should be replaced by
F.187) Jim f f u[pvu -V-(pVv) + qv] dV= v(P0) .
oti—*rQ J JK
This condition must be satisfied for every smooth function v(x, t) that
vanishes near the boundary of R. More precisely, we require that м be a
weak solution of the given hyperbolic equation, in the sense that it satisfies
F.77), and this takes the form of F.187) with the limit process removed.
In the hyperbolic problem, the family of characteristic curves or surfaces
containing a given point /^"(^o'o) envelops a (singular) characteristic
known as the characteristic conoid which has a vertex at Po. This conoid
reduces to the two characteristic curves that pass through the point Po in the
case of one space dimension. For the wave equation the conoid coincides
with the characteristic cone C.80). The solution of the point source problem
is thus expected to be singular along the entire characteristic conoid and it
appears that F.187) rather than F.184) is the relevant condition at Po for
the hyperbolic case.
In fact, if we formally set f(t) = 8(t -10) [where 5@ is the Dirac
8 -function and t0 > 0] in the point source problem for the wave equation
considered in Example 6.11, the problem reduces to that for an instanta-
instantaneous point source at (x0, t0). It has the formal solution w*
[1/Dтгс r))8[t- t0 - (r/c)] as follows from F.175). This shows the solution
to be singular not just at Po but along the entire forward characteristic cone
pOlNT SOURCES AND FUNDAMENTAL SOLUTIONS 377
r= c(t~ 'о)- We remafk that in the continuous source problem treated in
the beginning of this section, we assumed that /(*) is smooth for t > 0 so that
nce the wavefront r~ct passes, the solution is smooth for r>0 and the
ntegral relation can be used for t>0 without encountering the foregoing
difficulty.
In the parabolic problem, the characteristic curves or surfaces that
contain the point Po = (x0, t0) are the lines or planes t = t0, as follows from
the results of Chapter 3. However, as has been shown previously in this
chapter, the singularity in u(x, t) is confined to the point Po, and does not
spread throughout the characteristic t = t0. Consequently, it is not necessary
to replace the condition F.185) by a condition based on the weak formula-
formulation of the problem given in F.81). While this simplifies the parabolic case,
in comparison with the hyperbolic case, the condition F.185) is not always
easy to apply.
The source point condition F.186) for the elliptic case poses the fewest
problems. Since elliptic equations have no real characteristics, the singulari-
singularity of u(x) must be confined to the source point F0 = jc0, and и and its
derivatives are expected to be smooth on dR in F.186). Effectively, the
elliptic point source problem can be treated in the same way as the
continuous point source problem for hyperbolic and parabolic equations
discussed at the beginning of this section, as shown in the following.
As a result of the foregoing observations, we discuss the elliptic point
source problem on the basis of the condition F.186), while the point source
problems for the hyperbolic and parabolic cases are dealt with in a different
way. We remark that it is generally easier to deal with the latter problems by
expressing pF as a Dirac 8-function and using the theory of generalized
functions to solve the problem. This approach is indicated in Chapter 7.
Nevertheless, a substantial body of results for the point source and related
problems was achieved before the theory of generalized functions was
developed.
Solutions of the foregoing point source problems are called fundamental
solutions of the given differential equation. As such they are merely
required to satisfy the homogeneous form of the equation away from the
source point Po and the appropriate condition F.184), F.185), F.186), or
F.187) at Po. Fundamental solutions are, therefore, not uniquely de-
determined since any smooth solution of the homogeneous differential equa-
equation can be added to them. Here we are mostly concerned with fundamental
solutions that are required to satisfy additional conditions that serve to
specify them uniquely.
For the hyperbolic and parabolic problems we require that the fundamen-
fundamental solution u(x, t) satisfies a causality condition. This states that the solution
4*, t) must vanish identically for t<t0 when P0 = (xQ,t0) is the source
Point. Thus we are considering the effect of the source point Po on the
solution u(x, t) in the region t>t0. This is referred to as the causal
fundamental solution. The causality assumption translates into the condition
378 INTEGRAL
that и and ut vanish at t = t0 for all P^ Po. That is, we effectively have a
initial value problem for the given equations with zero data away from p
and the data at Po determined from the source point conditions. Since t = t°
is a characteristic for the parabolic problem, we can arbitrarily set u{x, t) = л
for t < t0 since singularities in the solution are confined to the point P p0
the hyperbolic problem (as we have already observed in a number of special
cases), the domain of influence of a point Po = (x0, t0) is the forward
characteristic conoid—that is, that part of it for which t> t0—with vertex at
Po. Then with u(x, i) = 0 for t< t0, the causal fundamental solution vanishes
identically in the exterior of the forward characteristic conoid. It is shown in
Chapter 7 how the causal fundamental solutions for hyperbolic and
parabolic problems can be determined as solutions of homogeneous equa-
equations for t> t0 with appropriate (singular) initial data at t = f0.
In the elliptic case we require that the fundamental solution have a
specified behavior at infinity. Then the fundamental solution is known as the
free space Green's function. In this section we only consider fundamental
solutions in unbounded regions. If the point source problem is given for a
bounded region, the fundamental solution must satisfy homogeneous condi-
conditions on the boundary. Solutions of such problems are generally referred to
as Green's functions or influence functions. Green's functions for equations
of all three types are considered in Chapter 7. The Green's functions that
are determined in that chapter for hyperbolic and parabolic equations in
unbounded regions differ from the causal fundamental solutions obtained in
this section.
Since elliptic problems are easiest to deal with, we begin our discussion of
fundamental solutions with the elliptic case. In the one-dimensional case,
the source point condition F.186) is easily seen to reduce to a jump
condition for ux at the point x0. In two and three dimensions the condition
F.186) is essentially of the form F.168) if we set/@ = 1. Consequently, the
simplified forms of F.168) given in F.169) and F.170) are valid for
the present case as well if we set f(t) = 1 in these equations. With n equal to
the number of dimensions, we conclude that the source point condition
F.186) can be expressed as
F.188)
Г-1
L dx J*
/1 = 1
Эи\
— ) = -l, л = 2
dr)
, n — э ,
дг
—
or
The bracket in the case where n = 1 represents the jump in ux at x = x0, that
is, ux(x0+) - ux(x0-). For n = 2 we have r2 = {x - x0J + (y- yof and for
n = 3, r2 = (x - xof + (y- yof + (z- z0J.
POINT SOURCES AND FUNDAMENTAL SOLUTIONS 379
Away from the source point Po, the fundamental solution и satisfies the
homogeneous equation
F.189) -V.(/>VW) + ?W = 0, P*PO,
where V is replaced by dldx in the one-dimensional case. We note that for
the one-dimensional problem, the source condition F.188) can be treated as
a matching condition for two solutions of F.189) in the intervals л; < д:0 and
у>л: This is not the case, however, for the higher dimensional problems.
It is also seen that the conditions F.188) are insufficient to determine
solutions of F.189) uniquely.
Fundamental solutions can be found by looking for singular solutions of
F.189) that have the appropriate behavior F.188) at the source point. This
approach is used when we discuss fundamental solutions of equations with
constant coefficients in Example 6.13. Even if these solutions do not satisfy
the auxiliary conditions placed on the problem, they are important because
we can add smooth solutions of F.189) to the problem to account for the
additional conditions. (This technique will be used in our discussion of
Green's functions in Chapter 7.)
To examine the form of the singularity at the source point Po, we may
consider F.188) to represent approximate equations near Po, just as was
done in obtaining F.171) and F.172). On solving these (approximate)
equations we obtain
F.190)
\x - xo[, n = 1
2p(x0)
1
0, Уо)
1
Q, y09 zo)r *
log r , n = 2
/2 = 3,
which is valid for points P near Po and where r is defined as in F.188) for
n==2 and n = 3. The one-dimensional form of u(x) in F.190) was chosen to
roake the solution symmetric with respect to x0. In all cases only the most
singular terms of the solution were retained. We note that the singularity of
u is localized at the source point PQ in all cases, and that the strength of the
Singularity increases as the dimension n increases.
In the following example we construct fundamental solutions of the
elliptic equation F.189) in the case of constant coefficients.
Example 6.13. Elliptic Equations with Constant Coefhcients. We con-
consider the elliptic equation
380
INTEGRAL
where p>0 and q are constants. Guided by F.190) we look for a
damental solution of F.197) that depends on the distance r from the so
point Po. In one dimension F.191) is an ordinary differential equation ан
г = |* - xo|. In two and three dimensions r is the radial variable in polar a ri
spherical coordinates, respectively, with the pole at Po.
Let и = п(г) in F.191) and we obtain
F.192) p[urr + ^i ur] - qu = 0 , г>0, л = 1,2,3 ,
with n equal to the number of dimensions in the problem. Note that for
n = 2 and n = 3 we are assuming the fundamental solution is independent of
the angular variables.
To begin, we set q = 0 in F.191) which reduces it to Laplace's equation
when n — 2 and n = 3. For each n we require a singular solution of F.192),
and a set of such solutions is easily found to be given by
F.193)
и = u(r) =
cr = c\x~xo\ ,
с log г = с logV(* - *oJ + (У ~
1-1/2
with the constant с to be determined in each case. On using the conditions
F.188) at Po or, more simply, by comparing F.193) with F.190) we find
that
F.194)
с =
2irp '
Атгр '
Other solutions of F.192) exist that also satisfy the conditions F.188) at Po.
However, the fundamental solutions F.193), F.194) are in the standard
form associated with F.191) when q = 0, particularly when n = 2 or n = 3, in
which case it reduces to Laplace's equation.
If q 7^0 in F.191), F.192) and n = 1, we obtain
F.195) и = п(г) =
p 7
p exp(~ V| r) '
exp( J y-J rj + c2 exp( - i y-| rj ,
q < 0,
lNT
sources and fundamental solutions
the constants ct and c2 to be determined. This is done so that we obtain
Ге free space Green's functions for these equations by specifying the
behavior of the fundamental solutions at infinity as well as at the source
P°For q>0 we require that F.195) tend to zero at infinity—that is, as
, Xq^oo—and that it satisfy F.188) at x0. Since r = |*-*ol in F.195)
we conclude that
F.196) u(x)= \ (^r1/2exp(-Vf l*-*ol) > Я>0
is the (uniquely determined) free space Green's function for this case.
With q <0 and n = 1, F.191) has the form of the Helmholtz or reduced
wave equation in one dimension. Given the wave equation vtt = pvxx, if we
ю1 v(x9t)-u(x)exp(-iV=q0 with q<0, we find that u(x) satisfies the
one-dimensional form of F.191). The free space Green's function for the
Helmholtz equation requires that the fundamental solution represent a wave
traveling away from the source point. (This requirement is related to the
Sommerfeld radiation condition discussed in Chapters 7 and 9.) In combina-
combination with the source point condition F.188) it implies that F.195) has the
form
F.197) u(x) = \ {-pq)~xl2 exp(iдР| \x - xo|) , q < 0 .
When multiplied by exp^/V^/), F.197) is a plane wave traveling away
from the source point x = x0 with speed Vp.
In both of the foregoing problems, the presence of exponentially growing
solutions at infinity when q > 0 and the existence of waves traveling from
infinity to the source point when q <0 is physically unreasonable. A physical
interpretation of the case q > 0 is given below.
With n = 2 and <?<0 F.192) has the form of Bessel's equation of zero
order. We require a singular solution of this equation and
F-198) и = «(r) = c0Y0[ V^| V(x - xQf + (y- yof] , q<0,
where c0 is a constant and Y0(z) is the Neumann function of zero order is
singular at r-0. With n = 2 and q>0 in F.192) we have the modified
vessel equation of zero order and a singular solution is
F.199) и = u(r) = сгК0
cx is a constant and K0(z) is the modified zero-order Bessel function of
second kind. For z « 0 these functions have the following behavior
382
INTEGRAL RELATION
ye(*)~-logz,
F.200)
Comparing F.198), F.199) with F.190) and using F.200), we conclude
tnat
F.201)
с, = ■
_1_
1
Г1 2тгр ■
With n = 3 and <? ^0 in F.191) we easily obtain the solutions
F.202)
w(x, y, z) =
Ы-VB
exp'V":
where r2 = (x -x0J + (y ->>oJ + (z - zof. As
these fundamental
solutions have the required behavior F .190).
The foregoing two- and three-dimensional fundamental solutions were
specified somewhat arbitrarily without regard to their behavior at infinity.
To obtain the free space Green's functions for these problems we require
that the fundamental solutions have a prescribed behavior at infinity. If
<jr>0, F.191) represents a stationary form of the diffusion equation vt +
qv = pV2v, that is obtained on setting v(x, t) s u(x). We require that v(x91)
and, consequently, u(x) tend to zero as the distance r from the source point
Po tends to infinity. If q <0, F.191) is again the reduced wave equation or
Helmholtz's equation. If v(x, t) is a solution of the wave equation vtt =
pV2v, we require that v(x,t)-u(x)exp(~-iVzrqt) represent a diverging
wave traveling away from the source point PQ as r—»«>.
To apply the conditions at infinity we must know the behavior of the
Bessel functions Yo and KQ for large values of the argument. We have
F.203)
We conclude that for q>0 the fundamental solutions F.199) and F.202)
both decay exponentially as r—> « and are, therefore, the free space Green s
functions in two and three dimensions for this case.
In the three-dimensional case with q<0, the fundamental solution
F.202) is easily seen to represent a diverging spherical wave traveling away
POINT SOURCES AND FUNDAMENTAL SOLUTIONS 383
f om the source point Po with speed Vp- Thus it is the free space Green's
function for this problem. However, since sin z = (l/2i)[eiz - e "], we find
that the fundamental solution F.198) contains both diverging and converg-
converging cylindrical waves for large r in view of F.203). Thus F.198) is not the
free space Green's function for this problem.
The correct form for the free space Green's function in two dimensions is
given by
F.204) ы(х,у)=^Я(
where H(q\z) is the zero-order Hankel function of the first kind. H(q\z) is a
singular solution of the Bessel equation of zero order with the following
behavior
F.205) H%\z) « -J log z , z * 0 ,
^) exp[/(z- ~
F.206)
Consequently, F.204) exhibits the correct behavior F.190) at the source
point and represents a diverging cylindrical wave as the distance r from the
source point tends to infinity.
We now turn to the consideration of fundamental solutions for hyperbolic
and parabolic equations. Because of the aforementioned complications with
the interpretations of these solutions and the determination of their be-
behavior at the source point, we concentrate mostly on one-dimensional
problems for equations with constant coefficients. Some results for higher
dimensional problems are also presented.
The solution of the instantaneous point source problem for hyperbolic
and parabolic equations can be determined from the solution of the continu-
continuous source problem discussed previously, by specializing the result to the
case where source acts instantaneously at the time t = t0. We show how this
can be done for the wave and heat equations, for which the continuous point
source problem was solved in Examples 6.11 and 6.12. In each of these
Problems we put f(t) = 8(t - r0), the Dirac delta function, with t0 > 0. Since
the delta function vanishes for f#f0, the continuous source problem is
ormally converted into an instantaneous source problem. The solution of
the instantaneous source problem is obtained by replacing f(t) by the delta
function in each of the solution formulas obtained in these two examples.
For the wave equation, we obtain from F.175) in the three-dimensional
case
384 INTEGRAL RELATl0Ns
F.207) u(r,t) = -A- s(t-t„--)•
4тгс r \ u c/
In two dimensions, F.176) yields (see A.30))
F.208) и(;,,) 1 *р«-'>-;1
In the one-dimensional case, F.177) yields
F.209) tt(x, 0 = yc H№ ~ *o) "I*' *oll •
In F.208) and F.209), #(z) is the Heaviside function. Its presence signifies
that the solutions vanish outside the forward characteristic cone or sector
with vertex at Fo and f0. The causal fundamental solution F.207) vanishes
everywhere except on the forward characteristic cone. We observe that each
of these solutions is singular not only at the vertex of the characteristic cone,
but along the entire forward characteristic cone. This shows why they must
be interpreted in the weak sense, as indicated previously. (These solutions
are rederived and discussed in Section 7.4.)
For the heat equation, we find from Example 6.12 that the causal
fundamental solution has the form
F.210) u(r, t) = G(r, t - to)H(t - *0) ,
where r is the (spatial) distance from the source point PQ, H(z) is the
Heaviside function and G(r, t - t0) is the fundamental solution of the heat
equation in one, two, or three dimensions. The Heaviside function is
required in F.210) to make u(ryt) = 0 for t<to. However, as shown
previously, since G tends to zero at an exponential rate as f—> t0 + , u(r, t) is
a smooth function across t = f0, for all points P^P0. Therefore, the integral
condition F.185) can be used to determine the behavior of the solution at
the singular point (r, t) - @, f0). Its use will be demonstrated below for the
one-dimensional problem.
The construction of the foregoing fundamental solutions for the wave and
heat equations by specializing the results obtained for the continuous point
source to the case of an instantaneous point source may be considered
somewhat unsatisfactory, because such an approach was not necessary for
the elliptic point source problem whose solution was obtained in a more
direct manner. Because fundamental solutions and the related Greens
functions are expected to play a role in the solution of more general
problems as shown in Chapter 7, we would like to determine the fundamen-
fundamental solutions independently of having to solve more general problems.
Although such direct determinations of fundamental solutions are given in
POINT SOURCES AND FUNDAMENTAL SOLUTIONS 385
Chapter 7 by using the theory of generalized functions and transform
methods, we now present two alternative methods for obtaining fundamen-
fundamental solutions for hyperbolic and parabolic equations with constant coef-
coefficients.
Example 6.14. Hyperbolic Equations with Constant Coefhcients. We
consider the hyperbolic equation
F.211) putt-pV2u + qu = 0,
with constant coefficients and present a method for determining fundamen-
fundamental solutions of F.211) similar to that given for the elliptic equations of
Example 6.13. Let Po be the source point in one, two or three (spatial)
dimensions and /0 the time when the source acts. In space-time the instanta-
instantaneous source point is Po = (Fo, t0). With r as the distance from Po (in one
dimension r=\x — xQ\), we define the hyperbolic distance у as
F.212) y =
where c2 =p/p. In a hyperbolic (nonEuclidian) geometry appropriate for the
hyperbolic equation F.211), у represents distance from the point (Fo, t0).
The distance function у is positive when r2 < c2(t -10J and is imaginary
when r2 > c\t - t0J. It vanishes when r2 = c2(t - t0J, and this is the equa-
equation of the characteristic cone for F.211) with vertex at (Po, /0). [In one
dimension this yields the two characteristic lines through the point (x0, t0).]
Since the fundamental solutions for F.211) are expected to vanish outside
the characteristic cone, as has been shown for the wave equation, this
distance function is appropriate for our problem.
Proceeding as in Example 6.13, we look for a singular solution of F.211)
in the form и = w(y). This gives
F.213) p[Uyy + - uy] + qu = 0 , n = 1,2,3 .
This equation has the general form of F.192). To avoid having to use the
complex valued variable у when considering the fundamental solution
outside the characteristic cone, it is often convenient to introduce the
squared hyperbolic distance a = y2. We note that a >0 within the cone and
ff<0 outside the cone. With и = й(о) in F.213) we obtain
F.214)
Hrst we set q = 0 and n = 1 in F.214) to obtain (тйаа. + йа = 0. [With
n * 1 and q = 0, F.213) and F.214) have the same form.] The general
386 INTEGRAL
solution of this equation is й = a + blog <r, where a and b are arbitrary
constants. While the solution й = blog a is singular on the characterise
cone, it differs from the fundamental solution F.209) obtained previously
The second (independent) solution w = a is not singular at all, but it *
constant within the (forward) characteristic cone as is the case for F.209) и
we formally put й = 0 outside the characteristic cone, the solution has a
jump discontinuity across the cone and is no longer a classical solution of the
equation for й. It must be interpreted as a generalized solution.
To show that such a solution is possible, we require some elementary
concepts from the theory of generalized functions, whose properties are
fully discussed in Section 7.2. If H(x) is the Heaviside function, then
H'{x) — 0 for all x ¥" 0, but the derivative is not defined at x = 0. However if
we assume that H'(x) can be defined at x = 0 as well (this will be done'in
Section 7.2), we have J~mH'(x)dx = l9 since #(<*>) = 1 and #(-») = o.
Noting the results of Example 1.2 we put Hl(x)-8(x), the Dirac delta
function. Additionally, since f(x)8(x) =f@)8(x) (see Example 1.2), we find
that x8(x) = 0. Formally differentiating this equation gives x8'(x) + 8(x) = 0.
Consequently, й = aH{a) is a generalized solution of айаа. + йа = 0. It
vanishes outside the characteristic cone where a < 0, and й = a within the
cone where a > 0.
For и to be a fundamental solution of F.211) we must specify the value
of the constant a. If и is to be a causal fundamental solution, it must vanish
for t < tQ as well. We cannot obtain a causal fundamental solution strictly in
terms of the variable <r, as it can distinguish only between the interior or the
exterior of the full characteristic cone, and not the forward and backward
parts of the cone. Therefore we set и = aH{&) for t > t0 and и = 0 for t < r0.
In effect, we define the causal fundamental solution of F.211) (with
<jr = 0) in the one-dimensional case to be и = a for \x - лго| < c(t- t0) and
и = 0 for |jc - xo\ > c{t - *0). It follows from the discussion in Example 6.4
that this is a weak solution of the differential equation for (jc, /) ¥^ (x0, t0).
We now determine the constant a so that the source point condition FЛ87)
is satisfied.
Let R in F.187) be a rectangular region, with the point (x0, t0) in its
interior, whose boundary is parallel to the coordinate axes. The function
v(x, t) in F.187) is assumed to vanish on and near dR. Noting the definition
of м, the integral F.187) reduces to an integral over the region Л, whose
boundary comprises a portion of the rectangle and the two characteristics
\x - xo\ = c(t - f0). Thus,
F.215) | \r u[pvn - pvxx] dV=jjA u[pvtt - pvxx] dV
= ajjA [Pv« " PvxX] dV=-ajdApvtdx + pvx dt
= -aVpp dv + aVpp I dv = 2aVppv(P0) = v(P0) ,
JPq JP%
POINT
SOURCES AND FUNDAMENTAL SOLUTIONS
387
where Ро = (хо,*о); Л and P2 are the points where the characteristics
intersect the rectangle, and the techniques used in Example 6.4 have been
applied. We find that a = l/2Vpp = l/2cp. Consequently the causal fun-
fundamental solution of the one-dimensional version of F.211) with q = 0 is
F.216)
u(x,
I*-
0,
This result coincides with F.209) if p = 1 and c2 =p.
jf n = i and ? > 0, F.213) has the form of Bessel's equation of zero
order. The nonsingular solution is и = aJQ[(q/pI/2y], where a is an arbitrary
constant and /0(z) is the zero-order Bessel function. To obtain a causal
fundamental solution for F.211) in this case, we put a = 1 llcp and set и = 0
outside the forward characteristic cone. This gives
F.217)
u(x, i)
1
с
О,
- *ol
If n = 1 and <jf <0 in F.213), we have a modified Bessel equation of zero
order. With /0(z) as the modified Bessel function of zero order, the causal
fundamental solution of F.211) for this case is
F.218)
Ф, 0 =
0,
\x-xo\<c(t-to)
\x-xo\>c(t-to)
The verification that F.217) and F.218) are weak solutions of F.211) is
somewhat more complicated than that given above for the case when q = 0,
and will not be given here. We note that when #=0, both F.217) and
F.218) reduce to F.216). The derivation of causal fundamental solutions of
F.211) in higher dimensions by the foregoing method will not be given
here. It requires a greater familiarity with the properties of generalized
functions than was needed for the one-dimensional case. These solutions
iM be considered in Section 7.4.
The causal fundamental solution for the heat equation in one, two or
three dimensions was given in F.210). By looking for a solution of the heat
equation that depends on a similarity variable (as defined in the exercises),
the heat equation is reduced to an ordinary differential equation. The
388 INTEGRAL
fundamental solution can then be constructed in terms of the solutions
the ordinary differential equation. This approach parallels that given i
Examples 6.13 and 6.14 for elliptic and hyperbolic equations. It will h
considered in the exercises. In the following example we show how th
causal fundamental solution for the one-dimensional heat equation can b
obtained as a limit of the causal fundamental solution for the telegrapher'
equation. This result is of interest in connection with the Brownian motion
problems discussed in Sections 1.1 and 1.2.
Example 6.15. The Heat Equation in One Dimension. The (hyperbolic)
telegrapher's equation
F.219) euu-c\x + ut = 0,
where e > 0 formally reduces to the heat equation if we set e = 0. We might
expect that as €—»0, the causal fundamental solution of F.219) correspond-
corresponding to a source point at Po = (x0, t0) reduces to the causal fundamental
solution of the heat equation. Indeed, the causality condition for both
problems requires that u(x, t) — 0 for t<t0. Furthermore, the causal fun-
fundamental solution of F.219) vanishes outside the forward characteristic
sector |jc — jco| < c(t — t0) where с = c/Ve. As e —> 0, the boundary
Ve\x — xo\ —c(t—t0) of this sector tends to the line t — tQ, which is a
characteristic line for the heat equation. In the limit, therefore, the fun-
fundamental solution of F.219) is expected to be nonzero in the region t> tQy a
result appropriate for the heat equation.
Let
F.220) m(jc, 0 = exp[- ^ (t - tQ)\u(x, i)
in F.219) and we obtain
F.221) evtt-c2vxx-j-€V = 0.
At the source point Po = (jc0, f0), F.220) shows that w(*0, t0) = v(x0, t0) so
that the fundamental solution of F.221) should yield the fundamental
solution of F.219) by means of F.220). With p = e, p = c2, and q = -1 /4c,
we obtain the causal fundamental solution of F.219) from that of F.211) in
the form
F.222)
u(x, t) =
0, I* ~ *ol > ^ (' ~ *o)
SOURCES AND FUNDAMENTAL SOLUTIONS 389
where F.220) was used in connection with the result F.218). As z~»°o, the
" Bessel function /0(z) has the asymptotic behavior
F.223) /oB)
Also, for t > t0 and 0 < € < 1 we have
F.224)
which follows from the binomial expansion. Thus
F.225)
As 6->0 in F.222), the argument of /0 tends to infinity. Using F.223)
and F.225), we easily obtain in the limit as e->0,
F.226) u(x,t)
which is identical to the causal fundamental solution of the heat equation
given in F.210).
We now demonstrate that the integral condition F.185) is satisfied for the
causal fundamental solution F.226) of the heat equation ut = c2uxx. For
simplicity we assume that the source point (jc0, to) = @,0). Let Л be a
rectangular region whose sides are parallel to the coordinate axes with
~~a < x < a and - T < t < T. Then F.185) yields
F.227) f [c2u -M]. n ds = f c\ dt+udx
JdR1 x9 J JdR x
= сг Г их{а, 0 dt - [ u(x, T)dx-c2 Г их(~а, t) dt
u(x, t) = 0 for t < 0. In carrying out the limit as <?Д~> @,0), we first let
0 and then a-*0. In the integral over the line r=Iwe set xllcVT = a
then let T-»0. Thus
390 INTEGRAL
F-228) H
ficsff
The other two integrals in F.227) vanish as Г~»0, since G(x> t) and its
x-derivative vanish as f-»0. Finally, as a—►() (it no longer appears explicitly
in the result), we find that F.185) is satisfied. The foregoing analysis can be
used to determine the arbitrary constant that results when the causal
fundamental solution is obtained by the similarity transformation method. It
is easily extended to the two- and three-dimensional cases.
6.8. ENERGY INTEGRALS
Until now we have been concerned with the construction of solutions of
partial differential equations but not with the question of uniqueness and
continuous dependence on the data. As was stated in Section 3.4, existence,
uniqueness, and continuous dependence on data are required if an initial
and/or boundary value problem is to be well posed. Using the concept of
energy integrals, we now briefly discuss the uniqueness and continuous
dependence on the data of the solutions of the second order differential
equations we have considered in the last three chapters. (See Section 4.1 for
a description of the boundary conditions.)
We begin with the hyperbolic equation
F.229) putt - V- (pVu) + qu = pF, x E G, t>0 ,
where G is a region in x-space with the boundary dGy p(x) and p(x) are
positive, and q(x) is non-negative in G. Initial and boundary data are given
as in D.16) and D.11). The coefficients and the data for F.229) are
assumed to be such that classical solutions can be constructed. We show that
these solutions are unique and depend continuously on the data.
To construct an energy integral for F.229) we multiply across by ut and
integrate over G. Since
F.230) w,V- (pVu) - V- (putVu) -pVu -Vw,,
we obtain
F.231)
/ \G u\pu« -V- (PVu) + qu] dv = fj jG [\ pu) + \ Р(Щ2 + \ qu2} dv
INTEGRALS 391
where we have used the divergence theorem and the fact that the region G
and the coefficients in F.229) are time-independent.
The energy of the system in the absence of external effects is
F.232) E(t) =\\\G [pu] + p(Vuf + qu2] dv+
where the surface integral term comes from substituting for duldn in F.231)
on using the boundary condition. Note that based on the assumptions on p,
and q, as well as a and /3, all of which are non-negative, each of the
integrals in F.232) is non-negative.
Thus on using F.232) in F.231) we have
' dn Js2
l tds
The terms on the right in F.233) represent changes in the energy of the
system due to internal and boundary forces.
To prove that the initial and boundary value problem for F.229) has a
unique solution, we show that on assuming that two solutions exist, we
conclude that they must be identical. Let ux{x,t) and u2(x, t) be two
solutions of the initial and boundary value problem for F.229). Since the
problem is linear, the difference u{xy t) = иг — и2 must be a solution of the
homogeneous form of F.229), that is, with F-0 and with zero initial and
boundary data. Since F= # = 0, we find from F.233) that E'(t) = 0. Thus
E(t) = constant and energy is conserved, and
F.234) £@ = £@) = \ j ja [pu](x, 0) + p(Vu(x, 0)J + qu\x, 0)] dv
since the initial data u(x, 0) and ut(x, 0) vanish. Note that E(i) = constant
implies that E(t) = E@). Then for any f >0 we have
F.235) E(t) = \ \ \ [Pu2 + piyuf + qu2] dv + \ f p -u2 ds = 0.
Z J JG £ J53 p
Since p, p are positive and q is non-negative in G, the integrand in F.235) is
nonnegative and we conclude that
м, = 0 and Vm = 0,
Slnce a and p are positive on S3, so that the boundary integral is non-
negative as well. The vanishing of all the first derivatives of u(xy t) implies
that u(x, t) = constant. But u(x, 0) = 0 so that u(x, t) = 0 and we have
392 INTEGRAL RELATIONS
ux(x, t) = u2(x, t). Thus we have shown that the (classical) solutions of the
initial and boundary value problem are unique. In obtaining F.236) we have
admitted the possibility that q = 0. If q >0 in G, we conclude directly from
F.235) that w(jc, 0 = 0.
If the coefficient q in F.229) is permitted to be negative, this result fails
to prove uniqueness since we can no longer conclude from the vanishing
of E{t) in F.235) that the integrands and, consequently, ut and Vu vanish. To
obtain uniqueness for this case we employ the following device. Let
F.237) u(jM) = e'4M),
where r is a positive constant to be specified. Inserting F.237) into F.229)
yields
F.238) pvtt + 2rpvt -V-(pVv) + (r2p + q)v = e~rtpF .
We assume that p and q are uniformly bounded in the region G with | q\ < M
where M is some constant. Then we choose r such that r2p > M (recall that
p>0), and this implies that r2p + q, the coefficient of и in F.238), is
positive.
We multiply across by vt in F.238) and integrate over G. Proceeding as
before, we easily obtain
F.239)
I £ e'rtBvt ds + f f e~rtpFvt dV
Js2vs3 jg ' J Jg r '
I
Js2vs3 jg
Let E(t) represent the term in curly brackets in F.239). Assuming there are
two solutions ux and u2 of the initial and boundary value problem for
F.229), we consider the difference u = u1-u2. This leads to a homoge-
homogeneous initial and boundary value problem for the solution of the homoge-
homogeneous form of F.238). Then since F= В = 0 in F.39) we have
F.240)
inasmuch as p and r are positive. From F.240) we determine that E(t) is
nonincreasing in time. But E@) = 0 since v = vt = 0 at the time t = 0, and
E(t) >0 for all t since p, p, and r2p + q are positive in G and alp is positive
in 53. Since E(t) cannot increase from its value Д0) = 0 as t increases, we
ENERGY INTEGRALS 393
conclude that E{t) = 0. Again the non-negativity of the integrals implies that
v = 0, so that vx = v2 and, consequently, щ = и2 and uniqueness is proven.
The energy integral method can also be used to prove continuous
dependence of the solution of the initial and boundary value problem for
F.229) on the data. For simplicity we assume that F = В = 0; that is, the
equation and the boundary data are homogeneous (to consider the case with
nonzero F and В we would need additional estimates of the given integrals
in F and B). Let u(xy 0) = f(x) and ut(x, 0) = g(x). We assume there are two
sets of initial data fx{x), f2(x) and g^x), g2(x) with corresponding solutions
u (x, t) and w2(x, i). The data are assumed to be close to each other in the
sense that
F.241) ЦЛ-ЛН^х, Ш~Ш<Ч> lift" fell <«э>
where the norm || • • • || represents the absolute value of the maximum
difference between the indicated functions over the region G and the
constants el9 e2, and e3 are small. We consider the case where q>0 and
/3 = 0 in the boundary condition. Thus the energy E(t) is conserved.
Let u(x, t) = мДх, t) - м2(х, i). The energy integral for и yields
F.242) E(t) = E@)
since, for example, ut(x9 0) = gx(x) - g2(x). Using F.241) we can bound
E(t) in terms of ег, е2, and €3. We write this as
F.243) £@^M(€l,€2,€3),
where M(el9 e2, €3)^0 as el9 e2, and €3^0. Thus we can make E(t)
arbitrarily small, say, less than €, if the €- (/ = 1,2, 3) are small. This yields
F.244)
positivity of p, p, and q implies that each term in the integral is small,
so that ux and its first derivatives are close to u2 and its first derivatives.
Thus if the data /, V/, and g undergo small perturbations, the solutions are
^У perturbed slightly. This signifies continuous dependence on the data.
We remark that if F and В are not zero, slight variations in their values lead
f° variations of the corresponding solutions that increase in time. However,
ln a fixed time interval 0 < t < T the difference between the original and
Perturbed solutions can be kept uniformly small.
394 INTEGRAL RELATIONS
If the domain G is of infinite extent and we are considering the Cauchy
problem for F.229) with initial data in G, we must assume that the data and
the solution u(x, t) vanish for sufficiently large |*|. Then uniqueness can be
proven by the foregoing energy integral approach. The surface integrals are
absent and we can conclude that zero data implies that и = 0. In the same
fashion, continuous dependence on the data can be shown. In addition, if q
is a semi-infinite region, say, a half-space, the preceding arguments also
carry through.
We have seen that in absence of interior or boundary force terms F and
B, the energy E(t)—that is, F.232)—is conserved if ^>0. However, for
F.238), which is essentially a dissipative hyperbolic equation with 2rpvt
representing the dissipative effect, the energy E(t) decreases in time as seen
from F.240). That is, the kinetic and potential energy of the system is
dissipated as t increases and is converted into another energy form such as
heat energy. This is consistent with our results in Chapter 1 relating the
telegrapher's equation to the diffusion or heat equation.
We conclude our discussion of the hyperbolic case by noting that more
precise results could have been obtained for F.229) that would have given
domains of dependence for both the Cauchy and the initial and boundary
value problems for F.229). For the sake of simplicity we did not present the
more complicated analysis required to obtain these sharper results.
Turning now to the parabolic equation
F.245) put -V- (pVu) + qu = pF, xEG,t>0,
with the initial condition u(x,0)=/(x) and the boundary data D.11), we
again assume that p, p are positive, q is non-negative in G, and u(x9 i) is a
classical solution of the initial and boundary value problem.
To obtain an energy integral for F.245) we multiply by и and integrate
over G. Since
F.246) mV- (pVu) = V- (риЩ -р(Щ2
we obtain on using the divergence theorem
F.247) j-( [\ \ \g pu2 dv) + / /g [р(Щ2 + qu2] dv
= \ pu 4^ ds + f f p Fи dv .
}dG И дП i JGr
We have
F.248)
f pu^ds = -l %pu2ds
JdG * dn Js3 ft r
f uBds+l B
We define the energy term for F.245) as
ENERGY INTEGRALS 395
F.249) E(t)=±fjapu2dv.
Assuming there are two solutions иг and u2 for the initial and boundary
value problem for F.245), the difference и = ux - u2 satisfies the homoge-
homogeneous form of F.245) with homogeneous initial and boundary data. Thus
we obtain from F.247), F.248),
'(t) = - //G
E'(t) = - //G [Р(Щ2 + Яи2] dv -js^
F.250)
because p, q9 p, ot, and /3 are non-negative. Since £(f)>0, E@) =
\jjGpu(x, 0J dv = 0, and E(t) is nonincreasing in view of F.250), we
conclude that E(t) = 0 for all t. Thus u(x, t) = 0 and ux = u2 so that
uniqueness is established. Continuous dependence on the data follows by an
argument similar to that in the hyperbolic case. If the region G is of infinite
extent and u(x, t) vanishes as |х|-»«>, we can also prove uniqueness and
continuous dependence on the data.
Again if q <0, this method fails but if q is uniformly bounded in G, the
transformation F.237) reduces F.245) to a form for which this approach
proves uniqueness.
The equation F.250) shows that the energy E(i) is not conserved and this
is the case even if и = 0 on the boundary and q = 0. There is, however,
another conserved quantity if q = 0 and duldn vanishes on the boundary
(i.e., there is no flux of и through the boundary). With F= q = 0 in F.245)
we obtain on integrating the equation over G,
(«•25D /1 iP", -V- (pVu)] dv = Jt
Thus the quantity J JG pu dv is conserved and if, initially, u(x, 0) = f(x), we
have
F.252) | jG pM(jc,,) du = | ^ pfl,) dv .
The integrals in F.252) represent the total amount of heat, say, in the
region G. Even if G is of infinite extent and u(x, t) vanishes as |jc|-*°°, we
still obtain F.252).
For the elliptic equation
F.253) -V- (pVw) + qu - PF , x e G ,
396 INTEGRAL RELATl0Ns
with the boundary condition D.11), we multiply across by u(x) and integrate
over G. This gives the energy integral, on using the divergence theorem am*
F.246), d
F.254) j \g [p(Vuf + qu2] dv = j^ pu j~ds + j £
The energy term is
F.255) E = \\g [p(VuJ + 9M2] dv
Proceeding as before, we easily prove uniqueness of classical solutions if
а Ф 0 in the boundary condition аи + рды/дп\ю - В. However, if a and q
are zero, the solution is unique only up to an additive constant since
u = constant is a solution of V-(/?Vm) = 0 with the boundary condition
fiduldn\aG = 0. Furthermore, uniqueness can only be demonstrated if q >0
in F.253). If q <0, the energy integral method (in the present form) cannot
be used to prove uniqueness. In fact, as shown in Chapter 8, the solution
need not be unique in that case.
The uniqueness theorem can also be extended to exterior boundary value
problems for F.253) where G is a semi-infinite region or a region of infinite
extent outside a closed and bounded region with the conditions D.11)
assigned on the finite boundaries. The solution is required to satisfy
appropriate conditions at infinity. (See the exercises for a discussion of
Laplace's and Poisson's equations.) Also, the continuous dependence on the
data for solutions of the boundary value problem for F.253) can be proven
by the energy integral method.
The energy integral method can be applied to systems of equations. It can
also be used to prove existence of solutions, not only uniqueness and
continuous dependence on data, but we do not consider this here.
To conclude our discussion, we note that for parabolic and elliptic
equations there exist maximum principles whereby uniqueness and continu-
continuous dependence on data can be proven for equations of a more general form
than those considered in this section. The appropriate maximum principles
are considered in Chapter 8.
EXERCISES FOR CHAPTER 6
Section 6.1
6.1.1. Derive the one-dimensional form of F.1) from the integral relation
F.8) by using mean value theorems for derivatives and integrals, assuming
the functions are sufficiently smooth.
6.1.2. Obtain the integral relation F.9) from the parabolic equation D.4).
EXERCISES FOR CHAPTER 6 397
6.1.3- Obtain the integral relation F.10) from the parabolic equation D.4).
6.1.4- Specialize F.10) to the case of one space dimension and derive the
(one-dimensional) parabolic equation D.5) from it using appropriate mean
value theorems.
6.1.5- Derive the integral relation F.12) from the elliptic equation F.11).
6.1.6. Consider a two-dimensional form of F.12) and let R be the rectangle
x < x < xx and y0 < у < yx. Assuming the functions are smooth, derive
F.11) from F.12) by using appropriate mean value theorems.
Section 6.2
6.2.1- Formulate, in the manner of F.17)-F.21), an initial and boundary
value problem for the parabolic equations D.4) and F.9) if So is a
discontinuity region.
6.2.2. Formulate, in the manner of F.17)-F.21), a boundary value prob-
problem for the elliptic equations F.11), F.12) if SQ is a discontinuity region.
6.2.3. (a) Show that if p2~0 in Example 6.1, then Л«1 and Г«2 in
F.34). Demonstrate that the (approximate) solution to this prob-
problem (i.e., with p2^0) can be obtained by solving for ux{x, i) in
jc<0 with the boundary condition диг@, t)ldx = 0, and then
obtaining u2(x, t) in ;c>0, by using the boundary condition
M2@, t) = ut(O,t).
(b) Show that for /?2-»oo we have /?—> —1 and Г—>0 so that there is
no transmitted wave. Introduce the assumption thatp2>px into
the matching condition for Example 6.1, set up (approximate)
boundary values for the two regions x < 0 and x > 0, and de-
determine that their solutions are consistent with the results ob-
obtained from F.32), F.33) for large p2.
6.2.4. Consider the problem of heat conduction in an infinite composite rod
composed of two homogeneous rods connected at x — 0. Put и = щ(х, t) for
x < 0 and и = u2{x, i) for x > 0 and assume they satisfy the equations
Px
Рг
~lt
du2
dt
~Pi
Рг
£4
дх2
S2u2
Лх2
= 0,
п
-oo<X<0,t>09
0<x<°o,t>0.
^et ut(x, 0) = A and м2(х, 0) = В where A and В are constants and apply
the matching conditions F.44) at x0 = 0. Determine the temperature u(x, t)
for t > 0.
Hint: Let
9 0 = «x + ft*[- 2^д] ' U2(x> 0 = «
398 INTEGRAL RELATIONS
where a( and Д are constants and c2 -pjp., for / = 1,2. The function Ф(г\
is the error integral defined in E.109).
6.2.5. Reconsider the problem of Exercise 6.2.4 and replace the constant
initial data by the arbitrary initial conditions иг(х, 0)=/г(х) and u2(x, 0) =
/2(x). Solve the respective problems for ut(x, t) and u2(x, t) by assuming
that ё
is a known function. The solution of the initial and boundary value problems
for ux and u2 may then be obtained from the result E.121). Use the
matching condition ux@, t) = w2@, i) to show that g(t) is the solution of the
Abel integral equation
where G(t) is a known function. The solution of this equation is
/4 1 d Г G(t) j
5W it dt Jo Vt^r
6.2.6. Show that the solution of the problem in Exercise 6.2.4 may be
obtained by applying the Laplace transform in the time variable.
6.2.7. Adapt the discussion of Example 4.2 to show that the operator
(l/p)L associated with the eigenvalue problem F.35)-F.37) is a positive
operator. Conclude that the eigenvalues for the problem are non-negative.
6.2.8. Discuss the eigenvalue problem F.45)-F.47) if the boundary condi-
conditions F.46) are replaced by v[@) = vf2(l) = 0. Obtain an equation for the
determination of the eigenvalues and obtain an orthonormal set of eigen-
functions. Show that A = 0 is an eigenvalue and v(x) = 1 is an eigenfunction
for this problem.
6.2.9. Expand the function f(x) = 1 in a series of the eigenfunctions v(k)(x)
given in F.51) and use this result to obtain a solution of the problem
F.41)-F.44) if w(x,0) = l.
6.2.10. Consider the initial and boundary value problem for the hyperbolic
equations
д2и д2и
р^ 2~ p ^ 2~ == U , 0 ^ X "^ Xq у t ^ U
д2и2 д2и2 л f
-^г-/>2-^г = 0> xo<x<l,t>0
in a composite medium. Let
u(x, 0) = f(x) , w,(x, 0) = g(x) , 0 < x < I,
FOR CHAPTER б 3"
where и = ux for x < x0 and и = u2 for x > x0. Assuming ux and u2 satisfy the
boundary conditions F.42) and the matching conditions F.44), apply
geparation of variables and show how the solution can be expressed in terms
of the eigenfunctions obtained in Example 6.2.
6.2.11- Develop a finite Fourier transform approach for the eigenfunctions
obtained in Example 6.2 for the purpose of solving the problem F.41)-
F.44) if the equation and the boundary conditions are inhomogeneous.
6.2.12. Show how Laplace transforms can be used to solve the problem
F.41)-F.44).
6.2.13. Let иг(х, у) and u2(x, y) satisfy the Helmholtz equations
V2ux + k\ux = 0, -o
V2W2 + к\иг = 0 , -oo<j;<oo;x>0,
with kx and k2 as constants and let ux and ux be related across x = 0 by the
matching conditions
<?"i@> y) duJO, y)
Represent иг and u2 as
ui(x> У) - exp[ifcj(x cos в + у sin 0)]
+ R expfifc^jc cos Й + у sin Й)]
мг(^; У) = Г ехр[Л2(х cos </> + у sin </>)],
where -тг/2< ^, <р < тг/2, and тт/2<6 <Зтг/2. The angle 0 is assumed to
be specified. Determine the constants 0, <p9 R, and T such that the
Helmholtz equations and the matching conditions are satisfied. This prob-
problem characterizes the scattering of a plane wave at an interface. A general
discussion of scattering problems is given in Chapter 10. The restrictions on
the angles 6 and <p are required to guarantee a unique solution to this
Problem and to correspond to physically motivated radiation conditions. The
relationships between the angle of incidence в and the angles of reflection
and refraction 0 and <p are known as SnelVs laws.
Section 6.3
6'3.1. Show that for the one-dimensional wave equation F.65) the jump
conditions F.58), F.59) reAice to the single condition
across the characteristics <p = x =p ct = constant.
400 INTEGRAL RELATlO\s
6.3.2. Determine that the function
„^ л-1Я*-<*), x~ct<0
with /@) = 0 and /'@) Ф 0 satisfies the jump condition across the charac-
characteristic <p = x - ct = 0 for the wave equation given in Exercise 6.3.1.
6.3.3. Show that
~'Л*~0> x-r<0
x-t>0,
where / is a smooth function with /@) = 0 and /'@) Ф 0 is a solution of the
hyperbolic equation
with discontinuous first derivatives across x - t = 0. Derive the jump condi-
conditions for this equation and verify that the jump condition across the
characteristic jc - t - 0 is satisfied for u(x, t).
6.3*4. Determine that the function
n ip , <p <0
where <p = x cos 0 + у sin 0 — ct with 0 = constant is a solution of the two-
dimensional wave equation
whose derivatives are discontinuous across the characteristic <p = 0. Show
that the jump conditions at <p = 0 are satisfied.
6.3.5. Show that
, ч -sinh(r-cf), r-ct<0
u(x, y,z,t) = \r
0, r-ct>0,
where r2 = x2 + y2 + z2 satisfies the three-dimensional wave equation
and that its first derivatives satisfy the jump conditions across the charac-
characteristic r — ct = 0.
6.3.6. Show that if we have <pt = 0 in F.62), we conclude from F.59)-
F.61) that [ut] =0 across the surface <p(x, y, z) = constant, and find from
EXERCISES
FOR CHAPTER б 401
F 58) ^а1 *е norma^ derivative of и must be continuous across <p -
onstant. Therefore, both и and its first derivatives must be continuous
across <p = constant.
6 3.7. Obtain F.64) from the integral relation F.9) in the parabolic case.
6 3.8. Show that <pt = 0 in F.64) does not yield a surface of discontinuity for
the' first derivatives of solutions of the corresponding parabolic equation.
6.3-9- Show that the solution of the initial and boundary value problem for
the heat equation,
- c2uxx = 0, 0<л;<оо?/>0
w(jc, 0) = 0 , x > 0
"(°'V-M ^0<t
has the form
'0, 0<x<°o?o<f<f0
u(x, t) =
4sb:]-
where erfc[z] is the complementary error integral defined in E.116). Verify
that even though и@, t) has a discontinuity at t= t09 u(x, t), ux(x, t), and
ut(x, t) have a zero jump across t = t0 for x > 0. Note that t = t0 is a
characteristic for the heat equation.
Section 6.4
6.4.1. Show that
0, x-ct<0
1, jc - ct > 0
is a weak solution of the two-dimensional wave equation
Hint: Use F.75) and apply it to regions on both sides of the characteristic
surface <p = x - ct = 0. Show that the boundary integral over the surface
9 = 0, which results from the integration over the region where <p > 0,
vanishes by showing that the terms in v can be expressed as a directional
derivative.
6-4.2. Verify that
. f cos(* + ct) , x + ct<0
M(*> ^ 1 sin(;t + cf) , x + ct > 0
ls a weak solution of the one-dimensional wave equation.
402 INTEGRAL RELATIONS
6.4.3. Obtain the integral F.81) that characterizes weak solutions for the
parabolic equation F.80).
6.4.4. Obtain an integral relation [similar to F.79)] that characterizes weak
solutions of initial value problems for the parabolic equation F.80).
6.4.5. Using the convergence properties of Fourier sine series discussed in
Chapter 4, set up an initial and boundary value problem for the one-
dimensional homogeneous wave equation in the finite interval 0 < jc < / with
m@, t) — m(/, t) = 0 such that its separation of variables result is a weak
solution and not a classical solution. That is, choose the initial data/(;t) and
g(x) such that the conditions given in Example 6.5 are met, but the series
solution is not twice differentiable term by term.
6.4.6. Solve the initial and boundary value problem for the heat equation
given in Exercise 6.3.9, with the initial condition u(x, 0) = 1 and the
boundary condition м@, 0 = 1 for 0 < f < f0 and w@, t) = 0 for t > t0. Show
that the solution is continuous across t= t0.
6.4.7. Show, by using separation of variables, that a formal solution of the
steady state problem for the heat equation
Щ ~ c\x = ° >
with the boundary condition
sin (ot, t>0
0, t<0,
where o> = constant, is given as
Г Г5Г 1 [ fw" 1
exp — v—т jc sin -V7i x + tot\ , t>0
L v 2c J L v 2c -I
0, t<0
if we require that u(x, t)-*Q as x-»«>. Determine that u(x, i) is not a weak
solution of the heat equation. Show that E.114) with g(t) = м@, 0 (as
defined in this exercise) is a weak solution of the heat equation.
6.4.8. Consider the (nonself-adjoint) elliptic equation
— V • (pVw) + qu + b • Vm = pF
in the bounded region G, with one of the (four) boundary conditions given
in Section 4.1 assigned on dG. Multiply across by a function v(x) and
integrate the equation over the region /?. Use Green's theorem to obtain the
integral relation
И [pVu-Vv + quv + vb-Vu] dV= I I pvF dV + I pV -^ ds
J Jg J JG JdG on
FOR CHAPTER 6 403
Determine the form of the integral relation for each of the four boundary
conditions given in Section 4.1. If u(x) is specified on a part of the
boundary, put v(x) = 0 there. For each case this yields an integral relation of
the Galerkin form associated with the boundary value problem for the
elliptic equation. The function u(x) is said to be a weak solution of the given
boundary value problem, in the Galerkin sense, if it satisfies the correspond-
corresponding integral relation for all admissible v(x). Only in the case of Dirichlet data
on all or part of the boundary must u(x) be chosen to satisfy the boundary
condition and v(x) be equated to zero there. Otherwise, и and v are
unrestricted. One of the forms of the Galerkin method for obtaining
approximate solutions of boundary value problems is based on the foregoing
results. An integral relation of Galerkin form can also be associated with
initial and boundary value problems for hyperbolic and parabolic equations.
Section 6.5
6.5.1. Use Taylor's series to show that a solution of the difference equation
F,100) must be a solution of the wave equation F.84) in view of the
arbitrariness of h and fc.
6.5.2. Show that the general solution of the wave equation F.84), that is,
u(x, t) = F(x - ct) + G(x + ci) ,
is also a solution of the difference equation F.100).
6.5.3. Obtain the solution u(x, t) for each of the formulas F.114)-F.116) if
B(t) = sin tat. Verify that in each case the solution obtained satisfies the
boundary condition.
6.5.4. Let f(x) = 1 and g = j8 = Bx = B2 = F = 0 in the problem F.117)-
FЛ19). Use the method of Example 6.7 to obtain the solution u(x, i) of this
problem in the regions Rl9R29...,R7 indicated in Figure 6.11.
6.5.5. Show that the problem
u(x,0) = x, ut(x,0) = 0,
uBt,t) = l, t>0
has no solution.
<>*5.6. Solve the initial and (moving) boundary value problem F.120)-
F.122) if f(x) = g(x) = 0, h(t) = -cor, 0 < c0 < c, and B(t) = A cos <oL Ob-
Obtain the Doppler effect and show that the solution oscillates at a frequency
below the input frequency o>.
6-5.7. Solve the problem F.120)-F.122) if Ддг) = 1, g(x) = 0, h(t) = -cot,
co>cJ and B(t) = 1. Show that the problem has a nonunique solution.
404 INTEGRAL RELATIONS
6.5.8. Solve the characteristic initial value problem F.133), F.134) if
F(x, t) = cos <ot, Bx(t) = e~\ and B2(t) = 1.
6.5.9. Let x = hl(t) = 2t be a space-like curve on which the data F.130)
where f(t) = sin t and g{t) = 1 are given. Solve the Cauchy problem for the
wave equation utt = uxx in the region 2/ - x > 0 with the given Cauchy data
on x = 2t.
6.5.10. Solve the Goursat problem for the wave equation м„ = c2u^ in the
region 0 < x < ct, t > 0 with the data
"@, 0 = g(t) , и(с*, 0 = ДО > ' > 0 ,
assuming that ДО) = g@). (A problem in which the data are given on a
characteristic curve and on an intersecting time-like curve is called a
Goursat problem.)
6.5.11. Solve the modified Goursat problem for the wave equation utt =
c2uxx in the region cf<x<°°, t>0 with the data
assuming that /@) = g@). [Here the data are given on a characteristic line
and the (space-like) Jt-axis.j
6.5.12. Solve the characteristic initial value problem for the hyperbolic
equation
M = 0, -1 < x < t, t > 0 ,
with the data
by using the iteration procedure given in F.137)-F.139).
Hint: The solution is w(x, t) = e~x.
6.5.13. Show that the initial and (moving) boundary value problem for the
heat equation
ut - c2uxx = 0, cot < x < oo, / > 0
w(jc,O) =/(*), 0<*<oo
can be solved by introducing the moving coordinate system
a = x - cot, r = t.
xERCISES FOR CHAPTER 6 405
In the new variables a and т we obtain a problem for й(а9 г) in the region
a > 0, т > 0 with data given at a = 0 and r = 0. Introduce a change of the
dependent variable that yields an initial and boundary value problem for a
heat equation with data given at a = 0 and r = 0.
Section 6.6
6.6.1. Adapt the method presented in D.71), D.72) to determine that the
orthogonality condition for the eigenvalue problem F.153)-F.156) is given
as
I Mf{x)Mk{x)p(x) dx + X /п,МДж,)Мл(ж,) = 0 ,
where M; and Мл correspond to the eigenvalues A; and A*, with A^A^.
Conclude that the appropriate inner product associated with this eigenvalue
problem is
f' v
(<P, Ф) = L <р(х)Ф(х)р(х) dx+Z т^(х1)ф(х^ .
The (induced) norm is given as
rl n
' = (<P, <P) = Jo <P2P dx + 2 ™,[<K*,)]2 •
Show that the eigenvalues are real and positive. Determine a formula for
the Fourier coefficients ck in an expansion of a function f(x) in a series of the
eigenfunctions Mk(x).
6.6.2. Let p(x)=p(x) = 1, / = 7Г, n = 1 and xx = 1 in the problem F.153)-
F.156). Determine an equation for the eigenvalues and obtain the corre-
corresponding eigenfunctions.
6.6.3. Expand f(x) = 1 in a series of eigenfunctions determined in Exercise
6.6.2.
6.6.4. Assume that the load in the vibrating string problem of Example 6.9
is placed at the end of the string at x = /. The boundary condition u(l, t) = 0
is then replaced by
mxutt{U i) = ~p(l)ux(l, t) .
Show why this follows from F.150). Use separation of variables to obtain
the eigenvalue problem for this case that replaces the one given in Example
6.9.
6.6.5. Let p = p = 1 in the problem of Exercise 6.6.4. Show that the
eigenvalues for the resulting eigenvalue problem are determined from the
equation
406 INTEGRAL RELATIONS
cot[/VA] =
Demonstrate graphically that there are infinitely many eigenvalues Л = л
Show that the eigenfunctions Mk(x) are given as **
and that they are orthogonal with respect to the inner product of Exercise
6.6.1 where n - 1 and xx = /. Calculate the norm of these eigenfunctions,
6.6.6. Obtain the limit of the solutions F.163) as c0—>0 and discuss the
resulting problem and its solution in that case.
6.6.7. Let fQ(t) = A cos <ot in F.164) and obtain the solution of the given
problem in that case. Discuss the Doppler effect.
6.6.8. Use the matching conditions F.158) to solve the (heat) equation
put - puxx = 0, -oo<x<oo^>0,jc^O,
where p and p are positive constants, if u(x, 0) = 0 and h(t) = 0 in F.158).
Use the Laplace transform.
6.6.9. Solve the problem of Exercise 6.6.8 by means of the Fourier
transform.
Hint: Use the jump conditions F.158) to obtain an expression for the
transform of the second derivative of и in terms of the transform of u.
6.6.10. Derive the jump condition F.157) for the hyperbolic case and show
that the corresponding condition for the parabolic case is obtained by
replacing the denominator in F.157) by p.
6.6.11. Consider the heat equation
-oo<jc<oo?f>0,
with u(x, 0) = 0 and a source moving along the line x = h(t) = cot, across
which the jump conditions [u] = 0 and [ux] = -(Vl + c20/c2)FQ(t) are satis-
satisfied. Use the transformation of Exercise 6.5.13 to change this problem into
one with a stationary source point and solve the given problem.
6.6.12. Consider the following moving boundary (and source) problem
и@, t) = u09 u[h(t), t] = 0, ux[h(t), t] = -ah'(t) ,
where ft@) = 0, and the determination of h(t) is part of the problem. (The
constants a>0 and uo>0 are given.) This is an example of a one-phase
Stefan problem, in which the region дг>0 is filled with ice, taken to be at
EXERCISES FOR CHAPTER 6 407
zero temperature, and the boundary x = 0 is kept at a positive temperature
u As t increases from zero, the ice begins to melt and the curve я: = h(t)
separates the water and the ice. In effect we are solving a heat conduction
problem in a composite region composed of water and ice and the condi-
conditions along the separation curve я: = h{i) are jump conditions determined
from the properties of heat flow across such a boundary. However, since the
temperature u(x, t) of the ice remains at zero while in its frozen state, we
may treat the separation curve as boundary curve for the liquid state, and
treat the jump conditions as boundary conditions. Nevertheless, this prob-
problem is complicated by the fact that x = h(t) is not known a priori.
To solve this problem, assume that и has the form of u2 given in Exercise
6.2.4. and substitute it into the three boundary conditions. Conclude that
h(t) must be a constant multiple of Vr, and show how this constant can be
determined numerically.
Section 6.7
6.7.1. Use F.188) to determine that the constants с in F.193) are given as
in F.194).
6.7.2. Derive the result F.177).
6.7.3. Derive the result F.181).
6.7.4. Show that the results F.207)-F.209) follow from the corresponding
results for a continuous point source.
6.7.5. Derive the equations F.213) and F.214).
6.7.6. Show that if u(x, t) is a solution of the heat equation ut = c2uxx, so is
the function u(ax, a2i) where a is a constant. Put a = 1 /Vt and show that the
heat equation has a similarity solution of the form и = F{xlVi) where F{z)
satisfies the equation
and that F'(z) = a exp(-z2/4c2). Noting that ux is a solution of the heat
equation if u(x, i) is a solution, conclude that
(with a ~ 1/V47TC2 this is the fundamental solution of the heat equation).
<j-7.7. Let r2 = x2 + y2 and u(r, t) be a solution of the heat equation in two
dimensions. Show that u(ar, a2t) is also a solution and construct a similarity
solution of the form и - F(rfVt). Noting that w,(r, t) is also a solution of the
beat equation if w(r, t) is a solution, conclude that
408
INTEGRAL RELATIONS
satisfies the heat equation ut = с V и (with 5=1 /4тгс2 this is the fundameib
tal solution of the heat equation).
6.7.8. Let r2 = x2 + y2 + z2 and show that the heat equation ut = c2V2u j
three dimensions has a solution of the form и = F(r/Vf) /r, where ra = ^ =
F(r/Vt) is a similarity solution of the one-dimensional heat equation v =
2 '"""
c2vrr. Show that
satisfies the heat equation. With a = A/V4ttc2K this is the fundamental
solution of the heat equation in three dimensions.
6.7.9. Obtain the solution of the (continuous) point source problem for the
two-dimensional heat equation discussed in the text, by constructing a
superposition of appropriately modified fundamental solutions as given in
Exercise 6.7.7.
6.7.10. Proceeding as in Exercise 6.7.9. and using Exercise 6.7.8, solve the
(continuous) point source problem for the three-dimensional heat equation.
Section 6.8
6.8.1. Use the energy integral method to prove uniqueness for the solution
of the Cauchy problem for the hyperbolic equation F.229).
6.8.2. Construct the appropriate energy integral for the one-dimensional
form
putt-(pux)x + qu = pF
of the hyperbolic equation F.229) and prove uniqueness and continuous
dependence on the data for the initial and boundary value problem for this
equation. (In proving continuous dependence, set F and the boundary terms
equal to zero.)
6.8.3. Prove that the solution of the initial and boundary value problem for
the parabolic equation F.245) is continuously dependent on the data
assuming that F and the boundary terms are zero.
6.8.4. Determine conditions at infinity that would guarantee uniqueness for
the exterior Dirichlet problem for Laplace's and Poisson's equations on the
basis of the energy integral method.
Hint: Apply the energy integral method to the region bounded by the curve
or surface where the data are assigned and by a circle or sphere that
completely contains the boundary on which that data are given. Allow the
radius of the circle or sphere to tend to infinity and obtain conditions on
the solution that make the integral over the circle or sphere tend to zero as
the radius tends to infinity.
6.8.5. Show that u(x, y) = log(V*2 + y2la) and и = 0 are both solutions of
u + u =o for x2 + y2 > a2 with и = 0 on x2 + y2 = a2. Use the conditions
XERCISES FOR CHAPTER 6 409
at infinity determined in Exercise 6.8.4 to eliminate one of these solutions
and to obtain uniqueness for the foregoing exterior Dirichlet problem.
6 8.6. Show that u(x, yy z) = a/y/x2 + у2 + z2 and и = 1 are both solutions
of the exterior Dirichlet problem for uxx + uyy + u2Z = 0 in the region
^2 + y2 + z2 > a2 with the boundary condition и = 1 on x2 + y2 + z2 = a2.
Use the conditions at infinity obtained in Exercise 6.8.4 to eliminate one of
the above solutions so that this problem has a unique solution.
6.8.7. Develop an energy integral approach for the first order linear
equation
by multiplying across by w, using the identity auux -{\au2)x - \axu2 and
integrating with respect to x from -» to -». Assuming all coefficients have
the necessary behavior at ±oo and using the change of variable F.237), show
how the energy integral can be used to prove uniqueness for the Cauchy
problem for the given equation.
6.8.8. Consider the first order hyperbolic system with constant coefficients
u, + Anx + Bu = 0
where Л is a real symmetric matrix. Multiply across by the vector ur and
integrate from -«to +°° with respect to x. Assuming that u(x, /) vanishes
as \x\ -* oo, show that the energy integral E(t) = /* „ |u|2 dx < 0 if В is a
non-negative matrix with the property that urSu s> 0 for all vectors u. If В is
not non-negative, the transformation u = ert\ with r > 0 yields a system for v
in which the coefficient В of v can be made non-negative for an appropriate
choice of /\ Use the energy integral E(t) to prove uniqueness for the Cauchy
problem for the given system.
6.8.9. Use the method of Exercise 6.8.8 to show that the Cauchy problem
for the system B.2), which is equivalent to the wave equation, has a unique
solution.
6.8.10. Apply the method of Exercise 6.8.8 to prove uniqueness for the
solution of the Cauchy problem for the system A.82), A.83), which is
equivalent to the telegrapher's equation.
6.8Д1. Use an energy integral to show that the initial and boundary value
problem for the equation of a vibrating rod (see Section 8.5)
uu + ««« =0, 0<х</, f >0 ,
w|th u(xy 0) and ut(x9 0) specified and u(x, t) and ux(x, t) given at x = 0 and
x- l> has a unique solution,
Hint: Use the following identity
utuxxxx - (utuxxx)x - (utxuxx)x + \(u2xx)t.
410 INTEGRAL RELATIONS
6.8.12. Show that the boundary value problem for the biharmonic equation
(see Section 8.5)
V2V2m = 0, w = u(x, y)
in a bounded region R9 with w and duldn specified on dR has a unique
solution by the use of an appropriate energy integral.
Hint:
MV2V2w = wV- (VV2w) = V- (wVV2w)
Deduce that if и = duldn = 0 on dR, we must have V2w = 0, and conclude
from the uniqueness theorem for Laplace's equation that и = 0.
6.8.13. Combine the methods of Exercises 6.8.11 and 6.8Л2 to develop an
energy integral that yields uniqueness for the initial and boundary value
problem for the equation of a vibrating plate (see Section 8.5)
utt + c2V2V2m = 0 , и = w(x, y, t)
in a bounded region R where u(x, y, 0) and ut{xy y> 0) are prescribed and и
together with duldn is specified on dR.
7
GREEN'S FUNCTIONS
The method of Green's functions is an important technique for solving
boundary value, initial and boundary value, and Cauchy problems for
partial differential equations. It is most commonly identified with the
solution of boundary value problems for Laplace's equation and a Green's
function has already been introduced in that context in Chapter 1. In
Example 1.6 it was shown by way of an application of Green's second
theorem that the Dirichlet or first boundary value problem for Laplace's
equation can be solved if the Green's function for that problem can be
determined. It was also seen in Section 1.3 and in our study of point source
problems in Section 6.7, that the Green's function is often worthwhile
determining in its own right rather than as a tool to be used only for solving
another problem.
In this chapter we begin by constructing generalizations of Green's
second theorem that are appropriate for the second order differential
equations introduced in Chapter 4. These integral theorems (which are
special cases of the general result C.164) given in Section 3.6) are then used
to show how boundary value, initial and boundary value, and Cauchy
problems can be solved in terms of appropriately defined Green's functions
for each of these problems. Even though the construction of Green's
functions requires that a problem similar to the original (given) problem
must be solved, it is often easier to solve the Green's function problem in a
number of important cases as we shall see. In this regard the fundamental
solutions, of which Green's functions are a special case, considered in
Section 6.7 play an important role. Since the determination and use of
Green's functions require the use of generalized functions such as the Dirac
delta function, a brief discussion of the theory of generalized functions will
be given in this chapter. Most of the chapter, however, is devoted to the
construction and use of Green's functions for problems involving equations
°f elliptic, hyperbolic, and parabolic types.
411
412 GREEN'S FUNCTIONS
7.1. INTEGRAL THEOREMS AND GREEN'S FUNCTIONS
In this section we construct integral theorems appropriate for the elliptic
hyperbolic, and parabolic equations introduced in Section 4.1. Each of these
theorems follows from an application of the divergence theorem and
represents a generalization of Green's second theorem. These theorems
form the basis for the construction of the Green's functions we consider in
this chapter. Technically, the theorems are valid only if the functions
occurring in the integrals are sufficiently smooth and, as we have seen in
Section 6.7, this is generally not the case for Green's functions. Neverthe-
Nevertheless, we shall assume these theorems are formally valid in all cases and rely
on the theory of generalized functions presented in Section 7.2 to form a
basis for their validity, even though this is not demonstrated. We begin our
discussion with problems in two or three space dimensions and present the
one-dimensional results at the end of this section. Even though these
integral theorems are special cases of the general result C.164), we include
some details of their derivation.
We start with the elliptic equation
G.1) Lu = -V- (pVu) + qu = pF ,
in two or three dimensions given over a bounded region G with the
boundary conditions
G.2)
dG
The conditions on the coefficients in G.1) and G.2) given in Section 4.1 are
assumed to remain in effect. Introducing a function w(x) whose properties
are to be specified and proceeding as in Example 4.2, we obtain
G.3) J | (wLu - uLw) dv = - J J^ V- (pwVu - puVw) dv
= - p(wVu - uVw) • n ds
JdG
f / dW du\
= p\ и — w — I as
JdGy\ dn dnl
on applying the divergence theorem with n equal to the exterior unit normal
on dG. Equation G.3) is the basic integral theorem from which the Green's
function method proceeds in the elliptic case.
The function w(x) is now determined such that G.3) expresses и at an
arbitrary point £ in the region G in terms of w and known functions in G.1)
and G.2). Let w(x) be a solution of
THEOREMS AND GREEN'S FUNCTIONS 413
G.4) Lw = 8(x-U),
where S(x - f) is a two- or three-dimensional Dirac delta function. The
substitution property of the delta function then yields
G5) J Jg uLw dv = J Jg u8(x - £) dv = u( £).
In view of G.1) we also have
G.6) J jG wLu dv = J JG PwF dv •
It now remains to choose boundary conditions for w(x) on dG so that the
boundary integral in G.3) involves only w{x) and known functions. This can
be accomplished by requiring w(x) to satisfy the homogeneous version of the
boundary condition G.2); that is,
dw
G.7) aw + P~^
= 0.
dG
If x E 5j on dG, we have
/„ Лч dw dU \ I dW du\
G.8) и w -— = — I аи — aw — I
I dw I ^QSu\ 1 „dw
a dn \ ^ dn) a dn
in view of G.2). If x G 52 U S3 on <?G, we have
(n n4 (9w du 1 л
The function w(x) is called the Green's function for the boundary value
problem G.1), G.2). To indicate its dependence on the point £, we denote
the Green's function by
as in Section 1.3. In terms of the Green's function K(x; £) the foregoing
results imply that G.3) takes the form
pKFdu-f 2-B^ds
r Jsl a dn
f B^ds+l
Jsl a dn Js2us3 ft
The Green's function K(x; £) thus satisfies the equation
414 GREEN'S FUNCTIONS
GЛ2) — V* (pVK) + qK — S(x — £) , jc, £ E G ,
and the boundary condition
Ж
G.13)
dG
with the derivatives taken in the x-variable. It follows from G.12) and our
discussion in Section 6.7 that the Green's function is a fundamental solution
of G.1). This fact will be exploited in the construction of certain Green's
functions. It must be noted that not all problems G.12), G.13) have
solutions. In certain cases that are considered later, a generalized or
modified Green's function must be constructed that satisfies an equation that
differs from G.12) or boundary conditions that differ from G.13). However,
once the Green's function has been determined, the formulas G.10) and
G.11) or slightly modified ones in the generalized case yield the solution
u(x) of the boundary value problem G.1), G.2) at any point in G. By
introducing appropriate assumptions on the behavior of the solutions at
infinity, the Green's function technique can also be applied to problems over
unbounded regions. We construct Green's functions for specific elliptic
equations of the form G.1) over bounded and unbounded regions in this
chapter.
Turning now to a consideration of hyperbolic problems, we examine the
initial and boundary value problem for the hyperbolic equation
G.14) pw,, + LM = pF, jc£G,/>0,
where the operator L is defined as in G.1) and G is a bounded region in two
or three-dimensional space. The initial conditions for u(x, i) are
G.15) u(*,0) =/(*), «,(*,<>) = ;
The boundary conditions for u(x, t) are given on dG for all f > 0 in the form
G.16) аи + p d-
0П dG
as in Section 4.1.
The integral theorem appropriate for the problem is given over the
bounded (cylindrical) region R in (x, f)-space obtained by extending the
region G parallel to itself from t = 0 to t = T, as shown in Figure 7.1 (T >0
is an arbitrary number). The lateral boundary of R is denoted by dRx and
the two caps of the cylinder, which are portions of the planes t = 0 and
t= T, are denoted by dR0 and dRT, respectively. We remark that SR0 is
identical to the region G and the initial conditions for u(x, i) are assigned on
THEOREMS AND GREEN'S FUNCTIONS
415
Figure 7.1. The region R.
it. The boundary conditions for u(x, t) are assigned on SRX. The exterior
unit normal n to dR has the form n = [nx, 0] on dRx, where nx is the exterior
unit normal to dG. On SR0, n has the form n = [0, -1], and on dRT, it has
the form n = [0,1].
It follows from the results at the beginning of Section 6.4 that
G17) j jR [w(puu + Lu) - u(pwtl + Lw)] dv
= I I V* [—pwVu + pmVvv, pwut — puwt] dv
= I [—pwVu + puVw, pwut — puwt] • n ds
— I (—pwVu + puVw) • nx ds + I (pwut — puwt) dx
JdRx JdRj-
- \ (pwut- puwt)dx
where V= [V, dldt] is the gradient operator in space-time, and we have used
the divergence theorem as well as the foregoing results concerning the
exterior unit normal to the boundary dR. The integral relation G.17) forms
the basis for the Green's function method for solving the initial and
boundary value problem G.14)-G.16).
416 GREEN'S FUNCTIONS
We now show how w(x91) is determined so that the solution u(xy t) Of
G.14)-G.16) can be specified at an arbitrary point (£, т) in the region R
from G.17). First we require that w(x, t) be a solution of
G.18) pwtt + Lw = 8(x - €)8(t-r) , £ G G,O<t< Г.
The product of the Dirac delta functions in G.18) has the effect
G.19) J J r u(pwtt + Lw) dv = J J r u8{x - g)8(t -r)dv = w(£, т) .
In addition, we obtain from G.14)
G.20) J J w(putt + Lu) dv = I I pwF dv ,
so that this term is known once w(x, i) is specified.
Since
•~ ~*ч Г /« «ч if / <?м dw\
G.21) pi — w\u + u\w) *nvo5= pi — w Km — I ds ,
JdRx ry x jdRx r \ dn dn/
we see that if we require, as in the elliptic case, that
G.22)
we obtain
G.23)
JtRxy\ dn dn) J£t a dn Js2u$3
f
where Sl9 52, and 53 are the portions of SRX that correspond to 5X, 52, and
53 on <?G, respectively.
To complete the determination of w(x, r) we expect that initial conditions
must be assigned to it for some value of t. If w and wt are specified at t = 0,
the integral over <?i?0 in G.17) is completely determined since и and ut are
given at t = 0. However, м and w, at f = T (i.e., on <?ЛГ) are not known. И
we specify w and wT at t - Г, it must be done in such a way that the
unknown values of и and ut play no role in the integral over dRT. The only
possible choice is to set
G.24) w(x, T) = 0, w,(x, Г) = 0,
so that the entire integral over <?tfr vanishes.
INTEGRAL THEOREMS AND GREEN'S FUNCTIONS 417
The equation G.18) together with the boundary condition G.22) and the
conditions G.24) at t = T constitutes a backward initial and boundary value
problem for the function w(x, t). It differs from the types of problems
considered previously for hyperbolic equations (see, however, Section 1.2)
where initial conditions were assigned at t = 0 and the problem was solved
for t > 0. Here we assign end conditions at t = T and solve the problem for
t< T. The problem for w(x, t) is well posed because if t is replaced by -/ in
the hyperbolic equation in G.18) (thereby reversing the direction of time),
the equation is not altered. We refer to problems with initial conditions or
end conditions as initial value problems. The function w(x, i) determined
from G.18), G.22), and G.24) is called the Green7s function for the initial
and boundary value problem G.14)-G.16) for u(x, i). It is denoted as
w(x, t) = K(x, t\ £, r).
Once the initial and boundary value problem for К is solved, the values
of К and Kt at / = 0 are known. Then the foregoing results yield the solution
и at an (arbitrary) point (£, r) as
G.25) u(€,T)
a dn JS2uS3 /J
Thus, the Green's function K(x, t; g, r) satisfies the equation
G.26) pKtt + LK=8(x-€)8(t-r), x, £ E G,t, r< T, r>0 ,
with the end conditions
G.27) K(x, T; £, r) = 0 , Kt(x, Г; (, r) - 0,
and the boundary condition
G.28)
дп
BR,
= 0, t<T.
It is shown in the exercises that K(x, t\ £, т) - K(£, -т; х, -t). Therefore,
as a function of £ and т, К satisfies the same differential equation but with
time running forwards instead of backwards. In these variables it represents
the causal fundamental solution for the given hyperbolic operator and the
boundary condition G.28).
The parabolic equation
the initial condition
418 GREEN'S FUNCTIONS
G.30) м(*,0) = /(*),
and the boundary condition
G.31)
t>0,
can be treated in the same way as the hyperbolic problem G.14)-G.16)
The operator L, the regions R and G, and their boundaries are defined as it
the foregoing hyperbolic problem.
We introduce the function w(x, i) and consider the integral relation
G.32) j jK [w(pu, + Lu) - u(-pw, + Lw)] do
= I I V- [—pwVu + puVw, pwu] dv
[I du dw\ ( [
= I ~pw — + pu —-1 л$+ puwdx-l puwdx.
JdRx \ r dn * dnl JaRT ^ idRQ *
Again, V= [V, dldt] is the gradient operator in space-time and the region R
and its boundaries are as shown in Figure 7.1. The result G.32) is a
consequence of the divergence theorem, but it differs from the preceding
integral theorems for the elliptic and hyperbolic problems in the following
respect. The operator p(dldi) + L that occurs in the parabolic equation
G.29) is not self-adjoint as was the case in the preceding problems. Its
adjoint operator is given as —p(dldi) + L. With this choice for the adjoint
operator we find that w(put + Lu) — u(—pwt + Lw) is a divergence expres-
expression as is shown in G,32) (see Example 3.9).
If we require w(x, t) to be a solution of
G.33) -pwt + Lw = 8(x - f )8(f - t) , f e G,0<r< T ,
with the end condition
G.34) w(jc,r) = 0,
and the boundary condition
G.35) aw + p
we immediately obtain from G.32)
G.36)
= 0,
jjjTEGRAL THEOREMS AND GREEN'S FUNCTIONS 419
where we have set w{x, t) = K(x, t; £, r). K(x, t; £, т) is the Green's func-
function for the initial and boundary value problem G.29)-G.31).
Thus the Green's function K(x, f; £, т) satisfies the equation
G.37) -рК, + ЬК=8(х-$)8A-т), х,^С,()т<Г)т>0,
with the end condition
G.38) K(x,T; £,т)=0,
and the boundary condition
dK
G.39)
= 0, t<T.
The equation G.37) satisfied by the Green's function К is a backward
parabolic equation that results on reversing the direction of time in the
(forward) parabolic equation G.29). However, since the problem for the
Green's function is to be solved backwards in time, the initial and boundary
value problem G.37)-G.39) for К is well posed. (That is, we must
determine К for t < T with an end condition given at t - Г.) Once К has
been determined, all the terms on the right side of G.36) are known and the
solution w(jc, t) of the initial and boundary value problem G.29)-G.31) is
completely specified.
It is shown in the exercises that K(x, t; f, т) = К(£, —т; х, -t). Thus, as
a function of £ and т, К satisfies a forward parabolic equation [of the form
G.29)] with time now running forwards instead of backwards. In these
variables it represents the causal fundamental solution for the given
parabolic operator and the boundary condition G.39).
Each of the Green's functions К defined above is a fundamental solution
of the equation G.12), G.26) or G.37) in the elliptic, hyperbolic or
parabolic case, respectively. Each of these equations is the adjoint of the
given equation for u. Since the elliptic and hyperbolic equations are
self-adjoint, the Green's function is also a fundamental solution of the given
equation. In the parabolic case, since the given equation is not self-adjoint,
the Green's function is only a fundamental solution of the adjoint equation.
We have shown, however, that as a function of f and т, К is a fundamental
solution of the given parabolic equation.
If we consider an instantaneous source point problem for the hyperbolic
and parabolic cases, with the source acting at the time t = r and located at
the point x = £, and require that the solution и satisfy the homogeneous
boundary condition аи + pduldn = 0 on dG for t>r and that w = 0 for
t<T, the solution is called a causal fundamental solution for the boundary
value problem. (In Section 6.7, causal fundamental solutions were defined
°ver unbounded regions.) We have already indicated how to obtain these
solutions in terms of the Green's functions K.
420 GREEN'S FUNCTIONS
The Green's function method can also be used to solve Cauchy problems
for hyperbolic and parabolic equations. In the hyperbolic case we assume
that u(xy t) is a solution of G.14) with initial data G.15) and in the parabolic
case u(jc, i) is a solution of G.29) that satisfies the initial condition G.30)
Both problems are given over the entire two- or three-dimensional space
The integral theorems G.17) and G.32) can be used for these problems if
we assume that the (spatial) boundary dRx tends to infinity and the solution
и and the Green's function К are such that the contributions from these
integrals vanish in the limit.
The Green's function К for the hyperbolic case is taken to be the solution
of the backward Cauchy problem G.26) and G.27) and the solution of the
Cauchy problem G.14), G.15) is given as
G.40) ii(&t) = |o j №pKF dxdt + j_JpKg- PKtf)t=odx ,
as is easily seen from the (modified) integral relation G.17). In the parabolic
case the Green's function К is chosen to satisfy the backward Cauchy
problem G.37)-G.38) and it then follows from the (modified) integral
theorem G.32) that the solution of the Cauchy problem G.29), G.30) takes
the form
G.41) u({,t)
The foregoing results are easily modified to yield Green's functions and
solution formulas for initial and boundary value problems for hyperbolic and
parabolic equations given over semi-infinite spatial regions.
There is an alternative approach to the construction of Green's functions
that applies in the hyperbolic and parabolic cases. Instead of having
K(x, t\ £, r) satisfy the inhomogeneous equations G.26) and G.37), we
require that they be solutions of the homogeneous equations
G.42) pKtt + LK = O, x,€SG,t<T,
G.43) -pK, + LK = O, x,£EG,t<T,
in the hyperbolic and parabolic cases, respectively. The homogeneous initial
conditions G.27) and G.38) are replaced by
G.44) K(x,t;£,t) = 0, K,(x, r; f, т) = - *(* ~ *> , ^G, ,
G.45) K(x, t; f, т) =
THEOREMS AND GREEN'S FUNCTIONS 421
respectively. The boundary conditions G.28) and G.39) for K(x, t\ f, r) are
retained.
In this formulation we obtain
G.46) \aR [pKut - puKt] dx = jaR uS(x - £) <fc = и(& т) ,
G.47) I puKdx=\
JdRT
for the hyperbolic and parabolic cases, respectively, when G.44) and G.45)
are used. The solutions w(£, т) then have the form G.25) and G.36) in the
hyperbolic and parabolic cases as is easily seen. The only difference is that
the domain of integration in the original formulation of the Green's function
problem extends from 0 to Г while in the present formulation it extends
from 0 to т < Г. However, since the equation and the data for К are all
homogeneous for r<t<T, the Green's function К vanishes identically in
that interval. Consequently, the domains of integration are, in effect,
identical for both formulations. The relation between these two approaches
is connected with Duhamel's principle (see Section 4.5) which relates
inhomogeneous equations with homogeneous initial conditions to homoge-
homogeneous equations with inhomogeneous initial conditions.
The preceding results are valid in two or three space dimensions. The
one-dimensional case for the elliptic equation G.1) leads to the Green's
function for a boundary value problem for an ordinary differential equation.
The Green's function K(x; f) satisfies the equation
G.48) LK=-4~(p 4^) + 9K = 8(x ~ О
ox \ dX I
in the interval 0 < x < I with the boundary conditions
G.49)
a2K(l,O + p2
dx
where a1; ot2, /3X, /32 satisfy the conditions given in Chapter 4.
The one-dimensional versions of the hyperbolic and parabolic equations
G.14) and G.29) lead to the consideration of the region R given as 0 < x < I
and 0<f<T. The boundary dR is made up of the portion SRx that
comprises the lines x = 0 and x = I with 0 < t < T and dR0 and dRT which
Represent the lines t = 0 and t = T, respectively, with 0 < x < L The region R
is depicted in Figure 7.2.
For the hyperbolic case we have
422
GREEN'S FUNCTIONS
<0,Г)
Figure 7.2. The region R,
0 < x < I,
G.50) Ри«-{
with the initial conditions
G.51) u(x, 0) - /(*) , и,(х> 0) =
and the boundary conditions
G'52)
With the operator L defined as in D.7) we have
G.53) | jR [К(ри„ + Lu) - и(рК„ + LK)] dx dt
where V= [д/дх, dldt] is the space-time gradient operator and we have used
Green's theorem in the plane, which is just the planar version of the
divergence theorem.
In a similar fashion for the one-dimensional parabolic case we consider
the equation
lNTEGRAL THEOREMS AND GREEN'S FUNCTIONS 423
G.54) pul-j^(p^) + qu = PF, 0<x<l,t>0,
with the initial condition
G.55) и(дс,0) =/(*),
and the boundary conditions G.52). Proceeding as in the higher dimensional
case we obtain
G.56) /1 [K{put + Lu) - u(-pK, + LK)] dx dt
= J £ V- [~pKux - puKx, pKu] dx dt
= -jQ [puK),.o dx + jo [-pKux + puKx]x=l dt
/0[puK],.T dx-jo [-PKux + puKx]x=0 dt.
With (£, t) as an interior point in the region R, we choose K(x, t; (■, т) to
be a solution of
G.57) pKtt + LK = 8(x - O4t ~r), t<T ,
G.58) -pK, + LK = S(x-OS(t-T), t<T,
with 0 < x, I- < I, and 0 < t, т < T in the hyperbolic and parabolic problems,
respectively. In addition, we have the end conditions
G.59) K(x, T;£,r) = 0, K,(x, T; f, т) = О
in the former case and
G6°) K(x, 7U,r) = 0
in the latter case. In both cases K(x, t\ f, т) is required to satisfy the
homogeneous version of G.52). If we have ^ = )82 = 0 in the boundary
conditions, we obtain
G.61) |оГ [-pKUx + puKx]x=l dt = \l (Cfil) g2(t)Kx(h t; f, r) dt,
G.62) JoT [-pKux + PuKx]x=0 dt = \l (£ffi>) gl(t)Kx(o, n f, r) dt.
424
GREEN'S FUNCTIONS
Using these results, we easily obtain the following expression for the
solution и at the arbitrary point (£, т) in the hyperbolic case
G.63)
"(*' T) = / L pFK dxdt + [ P(x)[g(x)K(x, 0; €, t) - f(x)Kx(x, 0; f, T)] dx
when /3j = /32 = 0 in the boundary conditions. For the parabolic case we have
G.64) u(Z,T) = ffRpKFdxdt + jQ p(x)f(x)K(x,0;{,r)dx
with /8г = K2 = 0 in the boundary conditions. If j3x and j32 are not zero or if
there are mixed boundary conditions, a somewhat different expression for
the solution is readily obtained. Also, if the problem is given over a
semi-infinite interval or we are dealing with the Cauchy problem over the
infinite interval, appropriate expressions for the solution are easily found in
a manner similar to that previously used in the higher dimensional prob-
problems. Further, K(x, t; f, r) can be characterized in an alternative manner as
was done in G.42)-G.45) for the higher dimensional case.
The initial (or end) data for the Green's functions К in the hyperbolic
and parabolic cases are assigned either at t — T or at t — т. For the given
problem for u(x, t), however, the data are given at f = 0. This results
because the Green's function is the solution of an adjoint problem in which
time runs backwards rather than forwards. The form of the boundary
conditions is the same for К and и in the elliptic, hyperbolic, and parabolic
problems considered. This occurs because the elliptic operator L that occurs
in all three equations is self-adjoint, and it determines the choice of the
boundary condition for the (adjoint) Green's function problem in each case.
To see what happens if the elliptic (spatial) operator is not self-adjoint, we
now consider a boundary value problem for a nonself-adjoint elliptic
equation and determine the corresponding Green's function problem.
Consider the elliptic equation
THEOREMS AND GREEN'S FUNCTIONS 42S
G.65) Lu = Lu + b • Vw = pF
in a bounded region G, where L is the operator defined in G.1). With
the operator L is not self-adjoint. The boundary conditions for u(x) are
G.66)
dG
as in the problem G.1), G.2). The operator L*, defined as L*u-
Ьи - V- (иЬ), is the adjoint of L as is easily checked, and we have
G.67) I I (wLu - uL*W) dv = I I V- [puVw - pwVu + wuh] dv
= Ipw pw —- + wwb • n \ds
JdG Iй дП Г дП J
on using the divergence theorem.
Proceeding as in the problem G.1), G.2), we put w = K(x; £), the
Green's function, and require that К be a solution of
G.68) L*K = 8(x-£), x,{EG.
Then, К must be specified on dG so that u(g) is completely determined in
terms of the boundary values for и and K. lfxE:S1ondG (see Section 4.1)
we have
G.69) pu d-f -pK ^ + Kuhn = £ В d-f + к(иЬ-п-р ^
^ дп * дп а дп \ у Эп
а дп
if we set К = 0. If x G 52 U 53 on dG, we have
G.70)
if we set /><?#/<?* + (pa/p)K+ ATb-n = 0. Thus, the (adjoint) boundary
conditions for К are
G.71)
426 GREEN'S FUNCTIONS
In terms of the Green's function K(x; f) that satisfies the adjoint
equation G.68) and the adjoint boundary conditions G.71), the solution
u(x) of the problem G.65), G.66) is given by the formula G.11). We
observe that only in the case of Dirichlet boundary conditions for u(x) does
the Green's function К satisfy the (homogeneous) Dirichlet condition К = о
on the boundary. For boundary conditions of the second and third kind, к
satisfies modified boundary conditions as given in G.71) (unless b-n = 0 on
the boundary). The form of the adjoint boundary conditions in the one-
dimensional case are considered in the exercises. If we replace the operator
L in G.14) and G.29) by the operator L, but leave the initial and boundary
conditions for these equations unchanged, we find that the Green's functions
К for these problems are defined in terms of the adjoints of the modified
equations. They have the same end conditions as before, but on the (lateral)
boundary dRx they satisfy G.71).
Because of the greater difficulty involved in determining Green's func-
functions for the foregoing nonself-adjoint problems, we restrict our discussion.
to problems with self-adjoint elliptic operators. However, before proceeding
to construct Green's functions for various problems, we present the theory
of generalized functions in the following section since these functions play
an important role in Green's function theory.
7.2. GENERALIZED FUNCTIONS
Throughout the text and especially in connection with fundamental solutions
and Green's functions, we have used the Dirac delta function in various
calculations. Though the delta function is neither integrable nor differenti-
able in the conventional sense, we have integrated and differentiated this
function. In this section, the Dirac delta and other functions are character-
characterized as generalized functions and it is shown how the formal operations
carried out on these functions are to be interpreted. Only the elementary
and basic ideas of the theory of generalized functions are presented. Our
discussion is restricted mostly to the one-dimensional case, as we shall see
that the results for higher dimensional problems can often be characterized
in terms of the one-dimensional case. The approach we use was developed
by Laurent Schwartz and defines generalized functions in terms of what are
known as linear functionals.
We begin by considering the collection of test functions <p(x) that are
assumed to be infinitely differentiable (i.e., they have derivatives of all
orders) and vanish identically outside a bounded region. (Such functions are
said to have compact support and in the exercises it is shown that such
functions exist.)
The generalized function f(x) is defined with respect to the set of test
functions <p(x) as the linear functional (/, <p). For each <p(x) the functional
has a uniquely defined numerical value. Since/(x) is, therefore, technically a
GENERALIZED FUNCTIONS 427
function of a function, assuming a (numerical) value for each <p(x), we call
(/, <p) a functiona^ Strictly speaking, the generalized function f(x) is not a
function of x (i.e., it does not assume values for each x) but rather of each
test function <p(x). The <p(x) are called test functions since/is determined by
specifying its effect on the test function <p(x). However, for the cases we
consider, generalized functions can be characterized as ordinary functions of
x for almost all values of x.
The functional (/, <p) is assumed to be linear; that is, if <p(x) and ф(х) are
two test functions and a and b are constants,
G.72) (/, aq> + Ьф) = a(f, <p) + b(f, ф).
In addition, (/, <p) is required to be continuous in the following sense. If
{<p.(x)} is a sequence of test functions, all of which vanish outside a
common region and {<pk} converges (say, uniformly) to a function <p(x) as
к ->°°, then we have
G.73) lim (/, <pk) = (/, <p) .
The collection of generalized functions/(jc) is a linear (vector) space; that
is, they can be added together and multiplied by constants. Given a
sequence of generalized functions {fk(x)} for which
G.74) lim (fk, (p) = (/, <p) ,
we say thdXfk{x) converges to the (generalized) function f(x) as /:->oo. This
type of convergence is called weak convergence and has been considered
earlier in the text in a different context.
A concrete general representation of the linear functional (/, <p) is given
in the case where f(x) is an (ordinary) integrable function over any finite
region. We then associate the functional (/, <p) with an integral given as
= j
= jf(x)<p(x)dx,
where the integration is carried out over the region where <p(x) is nonzero.
The linearity of the functional follows from the linearity of the integral and
the continuity of the functional is not hard to show. In view of the
representation G.75) the reason for the use of the inner product notation
(/> (p) for the functional becomes clear.
and g(x) are two integrable functions such that
for all <p(x), we must have f(x) = g(x) for almost all x. This follows since
428 GREEN'S FUNCTIONS
G.77) (/, <P) - (g,<p) =/(/-£)? Л = 0.
The vanishing of the integral for all test functions <p(x) implies (say, by using
the fundamental lemma of the calculus of variations) that / = g for almost all
x. This shows that the functional representation G.75) of f(x) essentially
specifies f(x) uniquely if f(x) is integrable. Thus with each integrable
function a unique generalized function (/, <p) can be identified. Such
generalized functions are often called regular generalized functions.
Generalized functions that are not ordinary integrable functions are often
called singular generalized functions. The basic example is given by the
Dirac delta function 8(x) which is defined as
G.78) (8, <p) = j 8(х)ф) dx = <p@) .
Although we have formally carried over the integral representation of (/, <p)
given in G.75) to this case, the delta function is not an integrable function.
We recall that 8(x) was defined to be zero for all x Ф 0. Thus the improper
integral in G.78)—since 8(x) is assumed to be singular at zero—must vanish
for all x. Consequently, since <p@) need not vanish for all <p(x), we find that
8(x) is a singular generalized function. It may be noted that the integral
representation G.75) for singular generalized functions is not strictly valid
but it is generally used.
The generalized function f{x) can be identified with (the values of) an
ordinary function g(x) in a region G if we have
G-79) (f,<p) = (g,<p)
for all test functions <p(x) that vanish outside G. We then say that f(x) = g(x)
in the region G. In particular, if g(x) = 0 in G, we say that f(x) vanishes in
the region G. As an example, we now show that 8(x) vanishes for all x ^0.
We consider all test functions <p(x) that vanish in a neighborhood of x = 0 so
that, in particular, <p@) = 0. Then using the definition G.78) of 8(x) we find
that E, <p) = <p@) = 0 = @, ф). Thus 8(x) = 0 for all x Ф 0.
In Example 1.1 we characterized the delta function as a limit of the
sequence of integrable functions fk(x),
{
0, \x\>{
We have
GENERALIZED FUNCTIONS
429
G.80)
on using the mean value theorem for integrals with -1/k-^x^l/k. Since
q>(x) is continuous at x = 0 we obtain
G.81)
lim (fk,
<p(x) = <p@) = E,
Recalling our definition of the convergence of sequences of generalized
functions, we conclude that the functions fk(x) converge to the delta
function 8(x) as /c-»oo.
The properties of generalized functions are usually derived from results
valid for the functional (/, <p) if f(x) is a regular generalized function. These
results are then defined as properties of generalized functions. Thus in the
one-dimensional case we have
G.82)
«[ f(cx)<p(x)dx =
Joo
c>0
c<0,
and
G.83)
=\ f(x)<p(x - a) dx ,
J ОС
where с and a are constants. Therefore, we define for the generalized
function f(x)
G.84)
and
G-85) (f(x + а), ф)) = {f{x), cp(x - a)) .
As a consequence of these definitions we have for the delta function 8(x),
G-86) («(„), ф
-щ (d(x), <p(±)) = ± 9(fi) = (ij S(x),
430 GREEN'S FUNCTIONS
so that 8(cx) = (l/\c\)S(x) formally. In particular, if с = -1, we obtain
8(-x) = 8(x) so that 8(x) may be said to be an even function. Similarly, We
have
G.87) (8(x - y\ <p(x)) = (S(x), <p(x + y)) = <p(y) ,
which is known as the substitution property of the delta function and is
generally written as
G.88)
If f(x) and g(x) are regular functions, we have
G.89)
(g(x)f(x), ф)) = | (8Ш(х))Ф) dx = j f(x)(g(x)<p(x)) dx
= (f(x),g(x)<p(x)).
Thus if g(x) is infinitely differentiable, g(x)q>(x) is also a test function and
we therefore define the product of an infinitely differentiable function g(x)
and the generalized function f(x) as
G.90)
An example of this result is given by
G.91) g(x
which follows from
G.92) (g(x)8(x - y), <p(x)) = (8(x - y), g(x)<p(x)) = g(y)<p{y)
in view of G.87). A special case of the above is
G.93) x8(x) = 0-8(x) = 0.
If f{x) and f'(x) are regular generalized functions, we obtain the result
G.94) (/'(*)>*(*))-[ f'(x)<P(x)dx=t <pdf
J —00 J ~*t*)
FUNCTIONS 431
since <p(x) vanishes as |*|->oo. lff(x) has n derivatives, repeated integration
by parts yields
G.95) (ЗЙ*.*))- (-.."(Л* *&). «I-
We take G.95) to be the definition of the wth derivative of the generalized
function /00. Since the test functions <p(x) are assumed to be infinitely
differentiable, we conclude that any generalized function/(jc) has derivatives
of all orders. In higher dimensions a similar result is valid for partial
derivatives of/(x).
As an example of the preceding we consider the function
G.96) /WU
Ibis function is continuous at x = 0 but is not differentiable there. To obtain
the generalized derivative we have
G.97) (/'(*), Ф)) = -(/(*), *'(*)) = -J_. f(xW(x) dx
= -| x<p'(x)dx
Jo Jo Jo
Now the functional representing the Heaviside function H(x) defined as
x<0
G.98) Щх)
is given as
G.99) (Щх), <p(x)) = | ^ Щх)ф) dx = |o «>(*) Л .
Thus G.97) shows that
G-!00) /'W = «W,
*fth /W defined as in G.96).
For the derivative of the Heaviside function we have
G.101)
(#'(*), <p(*)) = -(Я(лс), <p'(x)) = -Jo V'(*) Л = <p@) = E, <p) ,
so that
432 GREEN'S FUNCTloNs
G.102) H'(x) = S(x).
Further, we have
G.103) («'(*), Ф)) = -(«(x), <p'{x)) = -<p'{0)
and
G.104) (8"(x), <p(x)) = (S(x), <p"(x)) = <p"@) .
An additional result that has already been used in Section 6.7 is the
following
G.105)
(xS'(x), <p(x)) = (8', x<p) = -(8, (xcp)') = -E, x<p' + <p)
This may be formally expressed as the equation
G.106) x8'(x) + 8(x) = 0.
Example 7.1. Generalized Power Functions. The functions xm, where m
is a negative integer, and xaH(x), where a is any real number and H(x) is the
Heaviside function, are either ordinary power functions or the zero function
for jct^O. However, either the functions themselves or their derivatives
(from some order on) are singular at x = 0. We introduce generalized
functions associated with these functions that retain the basic properties of
(nonsingular) power functions.
First we define the generalized functions
G.107) fn(x) = xnH(x) , n = 0,1,2,....
If n = 0 this is just the Heaviside function. For x > 0, it is identical with xn. К
n >0 we obtain from G.94)
G.108)
xn <p(x) dx = (nfn_u <p) ,
so that/; = nfn_x. Also, xfn(x) =fn+1(x). We see that fn(x) retains the basic
properties of the power function xn (at x^Q) but it has generalized
derivatives of all orders. The kth derivative of fn(x) is given by
G£NERALIZED FUNCTIONS 433
G.Ю9)
in view of G.95).
Next we define the generalized functions
G.110) /.(*) = xaH(x), a > -1, а Ф integer.
Since a> -1» the integral representation of (fa(x), <p(x)) converges. Thus,
f(x) has generalized derivatives of all orders. If a<-l, the foregoing
integral representation of fa(x) diverges. In that case we define the general-
generalized functions
G.111)
where к can be chosen as the smallest integer such that a + к > -1, and a is
not an integer. In terms of /„(*; k), we define fa(x) as
G.И2) /„(*) = ^[/a(*;*)L
with the kth generalized derivative of fa(x; k) understood to mean
G.113)
^1)k
., 9) -
on using G.95). (The integral converges since a + к > -1.) We note that for
e>-l, we would set k = 0 in G.111) and fa(x;0)=fa(x), so that the
definitions G.110) and G.112) agree in that case.
If x>0, both G.110) and G.112) yield fa(x) = xa. It is shown in the
exercises that f'a(x) = afa^(x) and that xfa(x)=fa+l(x). The generalized
definition of fa(x) can be used to assign a meaning to divergent improper
integrals with algebraic singularities. This matter and its connection with
Hadamard's method of finite parts for integrals is considered in the exer-
exercises.
The product of the function 1/x", where n is a positive integer, with a test
function <p(x) is not integrable at x = 0, in general. To deal with this
Problem, we first let n = 1 and define the generalized function gx{x) = 1 Ix as
dx
G-U4) (gl(x), ф)) = (I, <p(x)) = P.V. |_№ i <p(x)
434 GREEN'S FUNCTION^
where P.V. signifies that the principal value of the integral is to be taken. ln
terms of gx(x) we define the generalized functions gn(x) as
G.115) gn(x) = t^Ll £-L [gi{x)] t w = 2j 3,....
This means that
G.116) (gn(x), ф)) = (n_lv P-V.
on using G.95). The relation between gn(x) and l/xn, and some additional
properties of these functions are considered in the exercises.
Finally we consider the function (l/xn)H(x)9 where n is a positive
integer. The singularity at x = 0 is not integrable. It is not possible to define
a unique generalized function associated with (l/xn)H(x) that retains the
basic properties of the power function l/xn, as shown in the exercises. We
define the generalized functions hn(x) as
G.117) hn(x) = ("У ±-n [(logx + c)H(x)], n = l,2,...,
I ft J. I • CtvV
where the arbitrary constant с can be different for each n. (If x>0,
Ля(х) = 1/х".) We have
G.118) (Л„(х), *(x)) = ^^y /o (log x + c)<p(n\x) dx .
Some additional properties of hn(x) are considered in the exercises.
Example 7.2. The Derivative of a Discontinuous Function. Let the
function f(x) be continuously differentiable everywhere except at x = «
where it has a jump discontinuity. Consequently, f(x) does not have an
ordinary derivative at x - a, but the generalized function associated with
f(x) is differentiable everywhere and we now obtain its first derivative.
Using G.94) we have
>—(/, <P') = " J_ №*'(*) dx -1 f(x)<p'(x) dx
= ~f(x)<p(x)\l№ + J^ f'(x)(p(x) dx -f{x)<p(x)\Z
+ f f'(x)<p(x)dx
Jet
geNErauzed functions 435
where [/(*)L«« is the jump in f(x) at x = a and the derivative in the last
integral is defined at all x except x = a. Since f'(x) is assumed to have a
finite limit as x approaches a from the left and from the right, the value of
f'(x) at x = ct plays no role in the integral. We can express [f(x)]x=a<p(a) in
the form {[f(x)]x_a8(x-a)9 <p(x)} and the last integral in G.119) can be
written as (/'(*), <p(x)). Thus the generalized derivative of f(x) can be
expressed as
G.120) /' = [Л*)],-*(* " «) + /'MU •
It may be noted that G.102) is a special case of this result.
An application of G.120) may be made to the Green's function K(x; %)
determined from the ordinary differential equation G.48). We have
G.121)
£ [p(x) д-^Щ = "8(* " i) + Я(х)К(х; i).
Using G.120) we conclude that qK is the ordinary derivative of pKx for
x Ф ij, and -1 is the jump otpKx at x = £. Since p(x) is continuous at x = £,
the jump in dK(x; Qldx at д: = £ is
G-122)
a result that agrees with F.188).
The expression G.120) for the generalized derivative of f(x), suggests
that we can express f(x) as a sum of a continuous function and a piecewise
constant function with a jump at x = a. In fact, we can write
G.123) /(*) = /(*) + [f(x)UaH(x - a).
Alternatively, we can set
G.124) Kx) = f{x) - [f(x)]x=aH(a - x) .
The functions f(x) and f(x) are given as the difference of f(x) and the jump
terms on the right of G.123) and G.124), respectively. Each function is
continuous and has^a jump in the first derivative at *=a. Thus, the
derivative functions f'(x) and f'{x) are regular generalized functions and the
need for a generalized derivative of f(x) is restricted to the terms that
involve the Heaviside functions.
If f(x) has derivatives of all orders for x < a and for x > a and/(x) and its
derivatives have (at most) jumps at x = a, we can express f(x) as the sum of
an infinitely differentiable function and a series of terms that accounts for
the jumps at x = a. To do so we introduce the generalized functions
43« GREEN'S FUNCTIONS
G.125) Я.(х) = i /„(,) = 1 х"Щх) , я -0,1. • ■ ■ ,
given in terms of G.107). Then, we express /(*) as
G.126) f(x) =/(*) + £ [/(n)ML=atfn(* - a) ,
«=o
where we have used the form G.123). The function f(x) equals the
difference between f(x) and the sum onjhe right side of G.126). Noting that
H'n(x) = Hrt_1(x), we easily verify that/(x) is infinitely differentiable. If f(x)
has a finite (or even an infinite) number of jumps, the foregoing representa-
representations can be carried out at each singular point.
As an example we consider the function
G.127) Ax) = x, x<a, f(x) = x\ x>a.
We easily obtain
G.128) 2
f(x)=f(x) + 2 [f(n\x)]x=aHn(x - a)
л = 0
= x + (a2 - a)H(x - a) + Ba - 1)^(jc - a) + 2H2(x - a) .
In determining Green's functions for differential equations it is often
necessary to take transforms of the delta function. Therefore, we now
discuss how Fourier transforms of generalized functions are to be defined.
Since we are mostly concerned with transforms of the delta function and
possibly its derivatives, we begin by considering generalized functions that
vanish outside a bounded region.
For the one-dimensional problem, let the generalized function/(jt) vanish
outside the interval [-Я, R] (i.e., it has compact support). It is possible to
construct test functions ф(х) that equal unity in [-Я, R] and vanish outside
an interval containing [-Я, /?]. Assuming for the moment that f(x) is a
regular function, the Fourier transform F(A) of f(x) is given as
G.129) FIX) = ^= £ ГЮ dx = ^== f_R e"*f(x) dx
Since A/\г2тг)€1Ххф(х) is again a test function, we define the Fourier
transform of the generalized function f(x) to be
GENERALIZED FUNCTIONS 437
G.130)
For example, the Fourier transform of the delta function 8(x) is given as
G.131)
since ф@) я1 by assumption. This result is formally equivalent to the
following
G.132)
All the properties of Fourier transforms given in Section 5.2 are valid for
this case. Higher dimensional Fourier transforms are defined in a similar
way.
To define Fourier transforms for generalized functions without compact
support we proceed as follows. If f(x) and g(x) are regular functions whose
Fourier transforms are F(A) and G(A), respectively, we obtain from E.12)
on setting x = 0,
G.133) ]_ F(A)G( A) d\ = j_ f(x)g(x) dx ,
since G(A) is the Fourier transform of g(-x).
Noting the formulas E.10), E.11) relating the transform F(A) and its
inverse transform /(*), we see that g(x) may be considered to be the
transform of G(A). Then if G(A) is assumed to be a test function <p(\) and
we denote its transform by Ф(*), G.133) may be written as
G.134) I F( A)<p( A) d\=l /(х)Ф(х) dx .
J— 00 J — 00
Clearly, both sides of the equation are linear functionals of the form G.75)
and we can write G.134) as
GЛ35> (F,<?) = (/,<t>).
is taken as the definition of the Fourier transform F(A) of the
generalized function /(*).
Unfortunately, even though every test function q> has a Fourier trans-
1*** the transform is not itself a test function unless <p vanishes identically,
the right side of G.135) does not define a generalized function in
8eneral since Ф is not necessarily a test function, so that G.135) does not
^el in a meaningful definition of the Fourier transform for all generalized
ions. To obtain a useful definition of the Fourier transform on the basis
438 GREEN'S FUNCTIONS
of G.135) we must introduce a new class of test functions whose properties
are preserved under Fourier transformation and, correspondingly, a new
class of generalized functions.
Accordingly, we define the class of test functions of rapid decay that are
required to be infinitely differentiable and to vanish, together with all their
derivatives, more rapidly than any negative power of |x| as |jc| tends to
infinity. The function exp(-jc2) belongs to this class of test functions but not
to the previously defined collection of test functions since it does not have
compact support. However, every test function with compact support be-
belongs to this class of test functions. It is not hard to show that the Fourier
transform of the present class of test functions is again a test function of the
same class.
If <p(x) is a test function of rapid decay, the linear functional (/, <p)
determines a generalized function of slow growth. The basic definitions and
properties given for the previously defined generalized functions carry over
to this class of generalized functions. Thus they are infinitely differentiable.
(We do not discuss convergence of test functions and weak convergence of
generalized functions for this case.) Any regular function f(x) that is
integrable over any finite interval and does not grow more rapidly than any
power of |x| as |jc| tends to infinity, determines a (regular) generalized
function of slow growth by means of the formula G.75). Since our main
interest lies in the definition of the Fourier transform for the class of
generalized functions of slow growth, we do not discuss all of their prop-
properties. We conclude with the observation that if f(x) is a generalized function
of slow growth and <p(x) is a test function of rapid decay, then the linear
functional on the right side of G.135) determines a generalized function of
slow growth that we define to be the Fourier transform F(A) of f(x).
As an example, we consider the Fourier transform of the (generalized)
function of slow growth f(x) - 1. The conventional Fourier transform is not
defined for this function and it does not have compact support so that the
definition G.130) cannot be used. Using G.135) with F(A) equal to the
Fourier transform of f(x) = 1, we have
G.136) (F,<p) = (l,<&)
= V2^[^- Г Г
where we have used the Fourier integral formula E.9) after interchanging
the order of integration. Thus
G.137) (F, <p) =
GENERALIZED FUNCTIONS 439
s0 that the Fourier transform of/(ж) = 1 equals V2tt8(x). This result can be
obtained formally from the inversion formula for the Fourier transform
obtaine y the inversio
E.П). where the use of G.131) gives
G.138)
Multiplying across by л/2тг and applying complex conjugation on both sides
yields the required relationship.
In connection with the application of Fourier series methods for the
solution of partial differential equations, we consider a further property of
generalized functions that assigns a significance to such series even if they do
not converge everywhere. Let £~=1 gn(x) be an infinite series of functions
uniformly convergent in any bounded region and let the sequence of partial
sums of the series be defined as
к
G.139) fk(x) - 2 gn(x), к = 1,2,....
Then if f(x) is the limit of the sequence {fk(x)}, we have
G.140)
и,т (/*(*)> <P(*)) = Jim fk(x)<p(x) dx
= J lim Шф) dx = j f(x)<p(x) dx = (/, <p) ,
since the uniform convergence of the sequence fk(x) permits the interchange
of the limit process and integration. Note that the vanishing of the test
function <p(x) outside a bounded region implies that the integration is
carried out only over that region so that the convergence is uniform.
The result G.140) shows that the sequence {fk(x)} and, consequently,
the given series converges to the function f(x) in the sense of weak
convergence. Thus f(x) and the fk(x) can be interpreted as generalized
functions. If each term gn(x) in the series has m derivatives, so does each
term in the sequence fk(x). Using G.95) we have
Since the right side of G.141) converges as к tends to infinity (for the term
d <p/dxm vanishes outside a bounded interval), we conclude that the se-
sequence of derivatives {dmfkldxm} {k = 1,2,...) converges (weakly) as
**~*00 to the mth derivative of f(x). In fact, we have
440 GREEN'S FUNCTIONS
G.,42) fe (Ztp., «#) - (-1)- Jin, (/„ g!)
As an example of the use of these results we consider the Fourier sine
series
00
G.143) F(x)= 2 ап$тпх,
where we assume that \an\ < M < °o for all w. By formally differentiating the
series
00
G.144) G(x) = - 2 % sin nx
twice term by term, we obtain the series G.143). Since the terms in the
series G(x) are majorized by M/n2, the series is uniformly convergent. Thus
even though the series for F(x) may not converge pointwise everywhere, we
conclude on the basis of the foregoing results that it converges in the
generalized sense to the generalized function F(x) which equals the second
(generalized) derivative of G(x).
Example 7.3. Generalized Solutions of the Wave Equation. It was
shown in Example 6.6 that the solution of the wave equation
G.145) utt - c2uxx = 0 ,
in the semi-infinite interval x > 0, with homogeneous initial data
G.146) w(jt,0) = 0,
and the boundary value
G.147) "@,0
is given in terms of the Heaviside function as
G.148) M
Since H(x) is infinitely differentiable in the generalized sense, we have
G.149) «« =
GENERALIZED FUNCTIONS
441
G.150) «~-7f7
Thus G.148) is formally a solution of the wave equation if we admit
generalized functions as solutions. This interpretation of a generalized
solution of the wave equation is closely related to the concept of weak
solutions introduced in Section 6.4.
We next consider the wave equation G.145) in the finite interval 0 < x < I
with the boundary conditions
G.151) и@,0
and the initial conditions
G.152) m(x,0) = 1,
Using the results of Example 4.9 we obtain the (formal) Fourier sine series
solution
u(x, 0 = Vу 2 ak cos(^ tj sin(^y- xj .
G.153)
The Fourier coefficients ak are given as
G.154)
ттк
\2\/2l
0, fe = 2,4,....
Thus G.153) can be written as
/i 1 к\ , ч 4 v' 1 / тткс \ . I irk \
G.155) U(x, t)=-Z -г cos( —r- t) sin! -y- x) ,
ъ k-i к \ I i \ I I
with the summation taken only over the odd numbers k = l, 3, 5..., as
indicated by the prime.
In view of D.148) we can write G.155) as
G.156)
■*•" - \
where each term in the series represents a propagating or progressive wave.
As was noted in Section 4.3, the Fourier sine series G.153) evaluated at
' = 0 represents the odd periodic extension of the function f(x) = 1 defined in
442
GREEN'S FUNCTIONS
the given interval 0 < x < L The graph of the extended function is shown in
Figure 7.3.
Let F(x) denote the extended function of /(ж). It is seen that F(x) has
jumps of magnitude 2 at the points x = nl, n = 0, ± 1, ±2,. ... Then G.156)
shows that u(x, t) has the form
G.157)
u(x, t) = \F(x + a) + \F(x - ct) .
The Fourier series does not converge pointwise to F(x) atx = nl,n = Q, ±i9
±2,..., and F(x) is certainly not differentiable there. Consequently,
G.157) cannot be interpreted as a classical solution of the initial and
boundary value problem for the wave equation.
However, each of the Fourier sine series in G.156) converges in the
generalized sense to the functions \F(x ± ct) in view of our discussion. In
addition, they can be differentiated twice to show that G.157) is a general-
generalized solution of the wave equation. For example, we have for F(x + ct),
A °° 1 Г lr 1
G.158) F(x + ct) = - 2 ' -j- sin[ ^- (x + <*)J ,
and on using Example 7.2 we obtain,
G.159)
dF(x + ct) _ £, 4 ^ ,
-2/
2/
Figure 7.3. The odd periodic extension of f(x) « 1.
GENERALIZED FUNCTIONS 443
since F(x + ct) = ±1 at x + ct ^ nl and has jumps of magnitude 2 at x + c* =
rt/. A second derivative yields
G.160)
Since 5'(*)~0 ^ог *^0> ^e series of derivatives of delta functions in
G.160) vanishes for x + ct^nl. Therefore, the sine series must also vanish
for those values of л: and t. However, at x + ct = nl each term in the sine
series is zero. Thus it would appear that the sine series is identically zero
and this is not so. [We have shown in G.106) that х8г(х)т^0 but equals
-8(x).] This observation demonstrates that infinite series that are valid in a
weak or generalized sense cannot always be dealt with in the same way as
ordinary series. Again, this interpretation of the solution is closely related to
that given in Example 6.5 where such Fourier series were characterized as
weak solutions.
We have defined the product of an ordinary function and a generalized
function in G.90). However, there does not seem to exist a useful definition
of the ordinary product of two generalized functions. Thus we cannot define
the product 8(x)8(x) or H(x)H(x) in a consistent manner (see Exercise
7.2.19). Yet 8(x)8(x - a), with a ^0 is well defined since the singular points
of both delta functions do not coincide. Thus at x = 0, 8(x - a)|xse0 ~
8(-a) = 0 and at x = a, 8(x)\x=a = 8(a) = 0. Consequently, we are effective-
effectively considering the product of an ordinary function and the delta function and
we have 8(x)8(x - a) = 0.
In a similar fashion it is possible to consider the product of two
generalized functions if they involve different variables. (This is sometimes
called the direct product.) For example, the two-dimensional delta function
8(x, y) can be written as
Using the definition of the delta function G.78) which is valid in any number
of dimensions, we have
G.162)
00
J 18(x, у)ф, y)dxdy = J_ J_e 8(х)8(у)ф, у) dx dy
— 00
= J_ 8(У)<Р(°> У) dy =
444 GREEN'S FUNCTIONS
so that the two expressions in G.161) are seen to be equivalent. Similarly
we obtain in three dimensions
G.163) 8(x, y, z) = 8{x)8{y)8{z).
By changing variables in the integral in GЛ62) it is possible to obtain delta
function representations in polar or other coordinate systems. This can also
be done in the three-dimensional case.
In addition, it is possible to define generalized composite functions such
as 5[#(л;)] where g{x) is assumed to be an ordinary function. The interpreta-
interpretation of these functions is based on the integral representation G.75) and a
change of variables, in the manner used in G.82), G.83).
For example, if g(x) is a smooth monotonic function that vanishes at
x - x0 and is such that g'(x) > 0 for all x, we have formally
G.164) (8[g(x)l <p(x)) = j_ 8[g(x)]<p(x) dx
where we have used the transformation
G.165) <r = g(x)
with a = 0 corresponding to x = x0. Thus
G.166) l
] = e(xx0).
If g\x) < 0 for all x, we find that the change of variables G.165) reverses the
order of integration in G.164). Then for both g'(x) > 0 or g'(x) < 0 for all x
with g(x0) = 0 (and x = x0 as the only zero) we obtain
G.167) ^
If g(x) has more than one zero and gr(x) does not vanish at each of the
zeros, we have, on applying the preceding argument in the neighborhood of
each zero (which we assume to be isolated),
G.168)
where we sum over all zeros.
FUNCTIONS 445
Example 7.4. Some Properties of the Delta Function. We show that
G.169) S(ax + by)8(cx + dy) = |^ГЩ *(*Ж JO *
Using a formal approach, we would consider the two-dimensional integral
representation of the functional (8(ax + by)S(cx + dy), <p(x9 y)) and change
the variables to obtain G.169). However, we use a more straightforward
approach that treats the problem as one involving the product of two delta
functions with different arguments.
We first assume that a = 0 but b and с are not zero. Since
8(ax + by)\a=0 = 8(ЬУ) = 0 for у ^0, on using G.91) we find that
G.170) 8(by)8(cx + dy) = 8(by)8(cx) .
Then G.86) implies that
G.171)
and this yields G.169) in the case a -0. If a ^0, we have
G.172)
8(ax + by)8(cx + dy) = 8(ax + b)S[(
where we have used G.91) and G.86) several times.
In a similar fashion we may consider the generalized function S[g(*)] in
the case where g(x) = 0 only at x = xQ and g'(x) Ф 0 for all x. We have [since
'()O], on using Taylor's theorem with remainder,
S[g(x)] = s[g(x0) + g'(xo)(x ~xo)+\ gT(€)(x - xof]
Since 5[g(x)] is singular only at x = x0 we may write it as
G-174) S[g(x)] = в[*'(*о)(* - *o)l = J^fi S(x - x0) ,
ich is in agreement with G.167).
446 GREEN'S FUNCTIONS
Both results obtained show that taking account of the localized (point)
singularity of the delta function can lead quickly to simplified expressions in
problems that involve the Dirac delta function. The basic property we have
used is that the delta function vanishes everywhere except where its
argument equals zero. However, this approach must be used with caution
when dealing with derivatives of delta functions.
We conclude our discussion of generalized functions by noting that
generalized functions are often referred to as distributions. This is because
generalized functions may be thought of as densities of masses, charges, or
other physical quantities. For example, the delta function 8(x) has infinite
density at x = 0 but
G.175) E(x),l)=f 8(x)dx = l,
J—co
so that the total amount of mass, say, is unity. Similarly, the derivative of
the delta function 8 '(x) may be thought of as the density of a dipole located
at x = 0. The total charge is given as
G.176) (в'(х),1)=
that is, the total charge vanishes as expected. However, the moment of the
dipole around x = 0 is given as
G.177)
|_ xS '(*) dx = (8 '(*), x) = - («(*), ^) = - (8(*), 1) = -1.
Thus -d'(x) is a dipole distribution located at x = 0 with a unit moment.
Further properties of one-dimensional or higher dimensional generalized
functions are considered as needed in our discussion.
7.3. GREEN'S FUNCTIONS FOR BOUNDED REGIONS:
EIGENFUNCTION EXPANSIONS
A general procedure for determining Green's functions for problems given
over bounded (spatial) regions is the method of finite Fourier transforms
presented in Section 4.6. For the equations G.12), G.26), and G.37) given
in Section 7.1, the Green's function is expanded in a series of eigenfunctions
of the elliptic operator L defined in G.1). The coefficients in the series ot
eigenfunctions are specified in the manner shown in Section 4.6. Although
this procedure is identical to that given in Section 4.6, the solutions of the
given problems expressed in terms of the Green's functions have a some-
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 447
what different form than that given earlier. However, the uniqueness
theorems guarantee that there can be only one solution for each of these
problems. We do not demonstrate in the general case the equivalence of the
various solution forms.
We begin by constructing the Green's function K(x; f) for the elliptic
problem. As shown in Section 7.1, the function K(x; £) satisfies the
equation
Gel78) LK^-V- (pVK) + qK=8(x-£),
and the boundary condition
G.179)
■o,
where derivatives are taken with respect to the variable x. Let Mk(x) be the
orthonormalized set of eigenfunctions of the operator L; that is,
G.180) LMk = kkpMk , k = 1,2,. . . ,
where the \k are the eigenvalues of L and p(x) is a given weight function.
The boundary condition for the Mk is G.179) with К replaced by Mk.
We express K(x; f) as a series of eigenfunctions
CO
G.181) K(x;()=4ZNk(()Mk(x),
k = l
as in D.195) with the (Fourier) coefficients Nk((j) to be determined.
Proceeding as in Section 4.6, we multiply G.178) by Mk(x), (k = 1,2,...)
and integrate over the region G. (Note that the operator К introduced in
Section 4.6 is unrelated to the Green's function К of the present chapter.)
Using the results D.197)-D.200) and noting that both K(x; £) and Mk(x)
satisfy homogeneous boundary conditions of the form G.179), we obtain
GЛ82) \kNk(t) = jjG8(x-t)Mk(x)dx = MkW, it = 1,2,...,
since feG. Then if all the kk > 0, we have
this yields the bilinear expansion of the Green's function K(x; g),
G.184) Afr;l)-j
448
GREEN'S FUNCTIONS
We notice that K(x; £) = K(£; x), so that the Green's function for the
elliptic problem is symmetric. This result can be proven directly for the
Green's function without the use of the bilinear expansion (see Exercise
7.1.1). The symmetry is a consequence of the fact that the operator A/pU
taken together with the boundary conditions is self-adjoint. It implies that
the interchange of the source point g and the observation point x does not
alter the solution.
It has already been shown in Section 4.2 that the eigenvalues Xk for the
eigenvalue problem G.180) with the boundary condition G.179) [with К
replaced by Mk(x)] are non-negative. It was also demonstrated in Exercise
4.2.5 that Ло = 0 is an eigenvalue for this problem if and only if q = 0 in the
operator L and a - 0 in the boundary condition. The corresponding eigen-
function M0(x) is clearly a constant in that case. Consequently, unless zero is
an eigenvalue, the bilinear expansion G.184) yields a formal solution of the
boundary value problem G.178), G.179) for the Green's function K(x; £).
To verify that G.184) is a (generalized) solution of G.178), G.179), we
apply the operator L to the series G.184) term by term. We have
G.Ш) LK= t
£ р{х)
*1 лк Jfc=l
in view of G.180). The eigenfunction expansion of 8(x - f) is given as
G.186)
S(x - £) = S (8(jc - a Mk(x))Mk(x) = 2 p@MkU)Mk(x) ,
)t = l jfc=l
since the Fourier coefficients are
G.187) (8(x - f), Mk{x)) = \\g p(x)8(x - f )*#*(*) dx -
From G.186) we obtain, on using G.91),
<7188» *Sif-^-I.-•'«-•«-яо
so that LK= 8(x - £). Since each eigenfunction Mk(x) satisfies the boun-
boundary condition G.179), the bilinear series in the generalized sense also
satisfies the boundary condition G.179). We have, therefore, shown that if
K(x; £) is interpreted as a generalized function so that term by term
differentiation in the above series is possible, its representation as a bilinear
series satisfies the problem G.178)-G.179).
For certain elliptic problems it is possible to construct the Green's
function in terms of eigenfunctions for lower dimensional problems. Pro-
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 449
ceeding as in Section 4.2 we consider the function и = u(x, y) where x is a
point in the (bounded) region G and у is a scalar variable defined over the
interval 0<y < I In place of G.1) we now consider the equation
G.189) p(x)uyy -Lu = -p(x)F(x, y), xEG,0<y<J
and the boundary conditions G.2) and D.20) if G is a two-dimensional
region. If G is one-dimensional and is given as 0 < x < /, w(x, y) satisfies the
boundary conditions G.52) where t is replaced by y.
The Green's function for the problem satisfies the equation
d2K
G 190) p —T " LK= — 8(x~ €)8(y —17) » x, £ E G,0<y, 7/</
v dy
and the boundary condition G.179)—if G is two-dimensional—as well as a
homogeneous form of the boundary condition D.20). The differential
operator in G.189) is self-adjoint. The Green's function is expressed in the
form К = K(x, y; £, 77) and the derivatives in G.190) are taken with respect
to the variables x and y. If the region G is one-dimensional, К satisfies
G.52) with и replaced by К and gx = g2 = 0.
To determine K(x, y; f, 7/) we consider the eigenfunctions of the
operator L in G.190)—that is, the set Mk{x) (k = 1,2. ..)—and construct
the eigenfunction expansion
G.191) K(x, y; f, ч) = 2 A
The Л^(у) are the Fourier coefficients of К given as
G.192) Nk{y) = (K, Mk) = \\g pK(x, y; f, 4)Jlffc(x) dx , к > 1,
assuming G is two-dimensional. We use the finite Fourier transform proce-
procedure of Section 4.6 and multiply G.190) by Mk(x) and integrate over the
region G. Again using the results D.197)-D.200), we obtain
G.193) N%y)-\kNk(y)=-Mk(€)8(y-4), * = 1,2,....
Since K(x, 0; f, v) = K(x, I; (, n) = 0, we must have
as the boundary conditions for Л^(у). То determine the Nk(y) we must
essentially obtain a Green's function for an ordinary differential equation in
a finite interval. Although this problem may also be solved using eigenfunc-
450 GREEN'S FUNCTIONS
tion expansions, we solve it instead in a more concrete manner in the
following example.
Example 7.5. Green's Function for an Ordinary Differential
Equation. Using the notation given in Example 7.1, we consider the
Green's function K(x\ £) defined over the interval 0<jt</ and satisfying
the equation
G.195) д *(*;f) -c2K(x;t) = -8(x-t),
dx
with the homogeneous boundary conditions
G.196) *@;£) = 0, *(/;£) = 0.
From our discussion in Example 7.2 we find that K(x; £) satisfies the
homogeneous equation
G.197)
and the jump conditions at x = £,
G.198) K(x; £) continuous at x = £ ,
where the square bracket represents the jump in the first derivative of
K{x\ £) across x = £.
To solve this problem we denote К by Kx for x < f and K2 for x> £. Both
Кг and ЛГ2 satisfy the homogeneous equation G.197) and Kx vanishes at
x - 0, whereas K2 vanishes at x = I. We easily conclude that
where ax and a2 are as yet unspecified constants (note that с is assumed to be
a constant). The continuity condition G.198) implies that
G.201) ax sinh(c£) = a2 sinh[c(/ - £)],
and the jump condition G.196) requires that
G.202) -a2c cosh[c(/ - £)] - axc cosh(cf) = -1.
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 451
solution of the system G.201), G.202) for ax and a2 is readily found to
be
sinh[c(/-g)]
G-203) fli - csinh[c/] '
sinh(c£)
G-204) fl2=
Inserting these expressions into G.200) we obtain the Green's function
K(x; S) as
_ 1 Г sinh[c(/ - f)] sinh[cx], 0 < x < f
G.205) *(*;«- csinh[c/] { sinh[c£] sinh[c(/ - *)], f < x < /.
Notice that this Green's function is symmetric since Jf(jr, £)
In the limit as c—>0 we obtain the Green's function K{x\ £) satisfying the
equation
G.206)
2
and the boundary conditions G.196). It is given as
A201) &ЛЛ
since sinhx^jt as x—>0.
We continue our discussion of G.190) and assume the eigenvalues A* of
G.180) are all positive. Then on using the results of the preceding example,
we find that the solution of G.193)-G.194) is given as
G.208)
N ( ) = M*U) f sinhfVA# - v)] sinh[\JTky], 0 < у < ij
sin[ \А*Л (sinh[\/A^ 1?] sinh[VA^(/ -у)], -q<y<l.
Inserting G.208) into G.191) completes the formal solution of the problem
G.190) for the Green's function K(x, y\ £, 77) for the given special region.
To see that the series G.191) represents a generalized solution of G.190),
^ find as in G.185) that
G.209) LK = p(x) J: XkNk(y)Mk(x) .
Also, on using G.193) we have
452 GREEN'S FUNCTIONS
G.210) p(x) ^-~ = p(x) 2 N%y)Mk{x) = p(x) £ \kNk(y)Mk(x)
oy *=1 ki
Then
G.211)
in view of G.188).
The preceding technique can be generalized to deal with elliptic equa-
equations of the form
G.212) p(x)Lu -Lu= -p(*№, y) ,
where и = u(x, y) and
G.213) Lu = -j
so that L is a self-adjoint operator. The (x, y)-region may be defined as in
the foregoing, and the boundary conditions in the y-variable may be of the
more general form G.52) where t is replaced by x, x is replaced by y9 and /
by/.
The Green's function К (jc, y\ f, 7/) satisfies the equation
G.214) PLK-LK=-8(x - t)8(y-V) ,
and К satisfies a homogeneous version of the boundary conditions for
m(jc, y). Expanding К as in G.191), we conclude that Nk(y) satisfies
G.215) LNk{y) - \kNk(y) = -Мк(()8(У ' ч), к = 1,2,.. . ,
and Nk{y) satisfies appropriate boundary conditions at у = 0 and у — I. Thus
we again obtain a Green's function problem for a self-adjoint ordinary
differential equation. A simple case of the Green's function problem was
considered in Example 7.5, and further cases are studied in the exercises.
We have further occasion to consider the technique of reducing the Green's
function problem to a lower dimensional one later in this chapter.
In the following example we construct the Green's function appropriate
for the first boundary value problem for Laplace's equation in a rectangle,
using each of the given methods.
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 453
Example 7.6. Laplace's Equation: Green's Function for a Rect-
Rectangle. We consider the rectangle G, defined as 0 < x < I and 0 < у < I, and
construct the Green's function K(x, y;$,v) that satisfies the equation
G.216) V2K=^ + ^ = -8(x
v c?X ay
with 0< f <' and 0 < tj < / and the Dirichlet boundary condition
G.217) K(x, y; £, i») - 0 , (jc, y)EdG.
Applying the first method presented, we must solve the following eigen-
eigenvalue problem in the rectangle G
G.218) -V2M(x, у) = ЛЛ#(*, у) , (x,y)EG,
with the boundary condition
G.219) M(x,y) = 0, (x,y)SdG.
The eigenvalues and eigenfunctions are determined by using separation of
variables.
Let M(x, y) = F(x)G(y) and insert this expression into G.218). We have
G.220)
Mxx + Myy + AM = F"(x)G(y) + F(x)G"(y) + \F(x)G(y) = 0 .
Dividing by F(x)G(y) and separating variables gives
where k2 is the separation constant. The equations for F and G are
G-222) F"(x) + ( A - k2)F(x) = 0
G-223) G"(y)
The boundary condition G.219) implies that
G-224)
G-225)
Consequently, we are led to consider one-dimensional eigenvalue problems
for F(x) and G(^) of a type studied in Section 4.3.
454 GREEN'S FUNCTIONS
For the eigenvalue problem G.223) and G.225) we obtain as the
eigenvalues
G.226) k2 =
and the eigenfunctions
G.227) Gm{y) = sin[ ™ y] , w = 1,2,....
For each of these eigenvalues, we have an eigenvalue problem G.222) and
G.224) for F(jc). As in Section 4.3, we find the eigenvalues
G.228) a-*2
and the eigenfunctions
G.229) Fn(x) = sin[ ™ x] , n = 1, 2,3,.. . .
Combining the results obtained, we conclude that the eigenvalues A for
the problem G.218), G.219) are given as
G.230) Anm = (^) + (^)\ n,m = 1,2,3,...,
and the corresponding eigenfunctions are
G.231) Mnm = sin( ™ xj sin(^ уJ , л, m - 1,2, 3,....
We denote the eigenfunctions as Mnm = FnGm since they are not as yet
normalized. It may be noted that the eigenvalues knm are positive, infinite in
number, and tend to infinity as n and m tend to infinity.
The appropriate inner product for the rectangular region G is given as
G.232) ^^
If we consider the two pairs (n, m) and (у, к) with (n,m)^ (j, k), we have
FUNCTIONS FOR BOUNDED REGIONS 455
G.233)
= 0
since {sin[(irn/l)x]} and {sin[(irm/!)y]} are orthogonal sets and m Ф к
and/or я 7*/. This shows that the set of eigenfunctions {Mnm(x, y)} is an
orthogonal set. To orthonormalize the set we proceed as follows. The
square of the norm of Mnm is given as
G.234) \\Mnm f = (Mrtm, Mnm) = /o J[ sin2(^ x) sin2(^ y) dx dy
pn using the results of Section 4.3. Thus an orthonormal set of eigenfunc-
eigenfunctions is given to be
G.235) Mn
nm{x, у) = щ sin( ™ x) sin(^ y) , n, m = 1,2,....
It should be noted that different sets of values of («, m) do not necessari-
necessarily yield distinct eigenvalues Anm. For example, if 1 = 1, we see that A12 =
(тг//J + Bir/lJ = A21. Nevertheless, the set of eigenvalues Anm can be
arranged in a sequence that corresponds to the positive integers k = 1, 2,
3,... with an equivalent arrangement for the eigenfunctions Mnm. It is then
possible to speak of the set of eigenvalues kk and the eigenfunctions
^t(*> У), with к = 1, 2, 3,.. ., and even if Xk is a multiple eigenvalue, the
corresponding eigenfunctions are orthogonal, as we have shown. This
notation was used in our earlier discussions of multidimensional eigenvalue
problems. Here we retain the double subscript notation for the eigenvalues
and eigenfunctions.
Given a function f(x, y) defined in the rectangle 0 < x < I and 0 < у < I
^id satisfying certain smoothness conditions we have the expansion
<7B6> л">
with the Fourier coefficients cnm given as
456 GREEN'S FUNCTloiys
G.237) cnm = (f, Mnm) = ~ \oj'Q f(x, y) sin(^ jc) sin(^ y) dx dy .
The expansion G.236) is known as a double Fourier sine series.
Having determined the eigenfunctions and eigenvalues for the problem
G.218), G.219) we are now in a position to construct the Green's function
K(x, y\ €,v)- In view of G.184), we obtain
G.238) K(x,y;t,>n)
= i Y Y sm[(<jrn/l)x] sin[Grn//)£] sin[Grm//)y] si
it ^
it „=i ^i (тгп//J + (ттт/tJ
as the eigenfunction expansion of the Green's function for the rectangle G.
The alternative approach for the construction of the preceding Green's
function has better convergence properties than those of the series G.238),
as we now demonstrate. The equation G.216) for К is written as
G.239) v'Ar-0@H
with the operator L given as L = -(d2ldx2). The appropriate eigenvalue
problem is
G.240) LM{x) = -M"(x) = \M(x) , 0 < x < I,
with the boundary conditions
G.241) M@) = M(/) = 0.
This represents a Sturm-Liouville problem which was solved in Example
4.4. The eigenvalues are
^-j , * = 1,2,...,
and the orthonormal set of eigenfunctions is
G.243) Mk{x) - Уу sin(^ jc) , к = 1,2,. ...
With Nk(y) defined as in G.208) the Green's function is given by the
eigenfunction expansion
00
G.244) K(x, y; & v) = 2 Nk(y)Mk(x) .
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 457
роГ у different from 17, the hyperbolic functions that occur in Nk(y) can be
approximated by exponentials as was done in Example 4.11, and the series
can be shown to converge fairly rapidly if |y - 171 is not small.
Both expressions G.238) and G.244) can be used in solving the boundary
value problem of the first kind for Laplace's equation in a rectangle. We
have already seen in Example 4.11 how a direct separation of variables
approach can be used to solve that problem with quite satisfactory results.
The eigenfunction expansion method for constructing Green's functions
for the foregoing elliptic equations fails if Ло = О is an eigenvalue of the
associated eigenvalue problem. As indicated, this occurs for the Green's
function problem G.178), G.179) if and only if q(x) = 0 in G.178) and a = 0
in the boundary condition G.179).
Let Ao = 0 be an eigenvalue of the operator L [see G.180)] and M0(x) be
the eigenfunction corresponding to Ao. Then G.182) implies that AoiVo = 0 =
Afo(£), and this is not possible since Mo cannot vanish identically. Clearly
then, the Green's function cannot be constructed in the given manner if
Ao = 0 is an eigenvalue. There are two methods whereby a modified Green's
function can be constructed in the special case that Ao = 0 is an eigenvalue.
For each of the two modified Green's functions it is possible to solve the
corresponding boundary value problem for u(x) in a manner similar to that
presented for the (ordinary) Green's function K(x\ £) in Section 7.1.
We consider the equation
G.245) Lu = -V- (pVu) = pF , x E G ,
with the Neumann boundary condition
G.246) в —
дп
dG
The eigenvalue problem that corresponds to G.245), G.246) is given as
G.247)
дМ
дп
dG
= 0,
and Ao = 0 is an eigenvalue. Thus it follows from D.200) that
G-248) A0W0 = 0 = J \c PFM0 dx + /^ | M0B ds .
Since the eigenfunction M0(x) must be a constant (see Exercise 4.2.5), we
conclude that
458 GREEN'S FUNCTIONS
Unless F and В are specified such that G.249) is satisfied, the boundary
value problem G.245), G.246) has no solution. We assume that G.249) is
satisfied and proceed with the construction of the modified Green's function
In the first method the modified Green's function, which we denote bv
K(x\ £), is expanded in a series of eigenf unctions of the operator L
00
G.250) K(x;t)=I,Nk(£)Mk(x),
k = l
where the Mk(x), к — 1,2,. .. correspond to the positive eigenvalues of L.
This expansion differs from that given in G.181) since the eigenf unction
M0(x) is absent from the series, even though Ao = 0 is an eigenvalue for this
problem. We have removed the term N0(i;)M0(x) from the expansion so as
to avoid obtaining the contradictory result АоЛ/о = 0 = М0(£) that was
derived in the foregoing.
Since the complete set of eigenfunctions is {Мк(х)}9 к = О, 1, 2,. . . , the
eigenfunction expansion of 8(x - |) is
G.251)
00 00
8(x - t) = £ (S(x - £), Мк(х))Мк(х) = p(t) 2 Мк@Мк(х) .
O fc 0
Jt=O
As shown, this series can be expressed as
00
G.252) S(x - t) = p(x) £ МкШМк(х) .
Further, we have
00 00
G.253) Lk = 2 Nk(i)LMk(x) = p(x) 2 *Mt)Mk(x) .
Comparing G.251) with G.253) shows that if we set
G.254) л
the modified Green's function satisfies the equation
00
G.255) LK = 2 р(х)Мк(£)Мк(х) = д(х - £) - Р{х)М0Ц)М0(х) .
k = \
Noting the foregoing discussion, we define the modified Green's function
K(x\ f) to be a solution of the equation
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 459
G 256) Lk = - V- (pVK) = 8(x -£)- p(x)
the Neumann boundary condition
dn
SG
= 0.
A.151)
The bilinear series of eigenfunctions for K(x; f) has the form
^ --V MkWMk(x)
G.258) K{x
к
Given the boundary value problem G.245), G.246), we now construct a
solution formula for u(x) in terms of the modified Green's function K(x\ £).
Following the procedure given in Section 7.1, we set w = К in G.3) and
note that G.5) gives
G.259) | \c uLK dv = \\G u[8(x - f) - р(х)М0Ц)М0(х)] dv
p(x)u(x)M0(x) dv .
The last integral in G.259) is the Fourier coefficient of u(x) with respect to
the eigenfunction M0(x), and it can be expressed as (м, Mo). Then we obtain
from G.11) the solution formula
G.260) M(O = /\cpKFdv + J^| KB ds + (w, M0)M0(t).
The solution exists only if the compatibility condition G.249) is satisfied,
and even then it is determined only up to an arbitrary constant multiple of
the constant eigenfunction M0(x) as seen from the last term in G.260).
Consequently, as has already been noted in Section 6.8, the solution to the
boundary value problem G.245), G.246) is not unique. Any constant can be
added to it.
Similarly, the modified Green's function K(x; |) is determined only up to
an arbitrary constant. By expressing K(x; fj) in the form G.250) and G.258)
we have, in effect, equated the arbitrary constant to zero. As a result, the
modified Green's function K(x; £) is symmetric.
The modified Green's function K(x; £) differs from the ordinary Green's
function K(x\ f) inthat LK does not equal 8(x- £), but is given as in
G.256). However, K(x; £) does satisfy the homogeneous boundary condi-
condition G.257). There is an alternative construction of a modified Green's
function, which we denote by K(x; £), associated with the boundary value
Problem G.245), G.246), for which we have
460 GREEN'S FUNCTIONS
G.261) LK=-V(pVK) = 8(x~i), x,iGG.
An application of the divergence theorem shows that
G.262)
Consequently, if we set LK = 8{x - £) we cannot have дК/Эп = 0 on дG, as
was the case for K{x\ £). Instead, we must set
G.263)
dK
дп
dG
ъ,
where b(x) is any function for which
G.264) f pbds = -l,
JdG
in view of G.262). (Note that b may be chosen to equal a constant.) The
solution formula for u(x) that satisfies G.245), G.246) is thenpbtained from
the results given at the beginning of Section 7.1. With w = К we obtain in
place of G.9)
and the solution is obtained in the form
G.266) и{{) = \ \q pKFdv +1^ ^BKds- jac pub ds ,
on using G.11). The last integral in G.266) is an arbitrary constant.
The expansion of £(*; £) in a series of eigenfunctions is not as
straightforward as that for K{x\ f) since К does not satisfy a homogeneous
boundary condition. Nevertheless, it can be carried out using the finite
Fourier transform techniques of Section 4.6. We do not present this
expansion here. It may be noted that К and К may be related to each other
by way of the procedure developed in Chapter 4, whereby inhomogeneous
boundary conditions can be transformed to homogeneous conditions. This is
considered in the exercises.
Example 7.7. Laplace's Equation: The Modified Green's Function in a
Rectangle. Given the rectangular region G defined as 0 < x < I and 0 <
y<t, we construct the modified Green's function К that satisfies the
equation [(see G.256)]:
GREEN'S FUNCTIONS FOR BOUNDED REGIONS 461
G-267)
ox ay
with (£,v)^G and the Neumann condition
dk
G.268) -^ (x, y;€,v) = 0> (x> y)e#G,
where dkldn is an exterior normal derivative.
To solve for K, we must determine the eigenvalues and eigenfunctions of
the problem
G.269) -V2M(x, у) = Ш(х, у) , (x,y)eG,
where M(x, y) satisfies the boundary condition
, y)
G.270)
dn
dG
= 0.
Using separation of variables, we set M(x, y) = F(x)G(y) in G.269) and
obtain
G.271)
Mxx + Myy + KM = F'(x)G(>0 + Д*)О"(у) + A^x)G(y) = 0.
Dividing by F(x)G(y) and separating variables gives
G.272)
with Л as the separation constant. Then F(x) satisfies the boundary value
problem
G.273) F"(x) + (A - k2)F(x) = 0 , 0 < x < I,
with
G-274) F'@) = 0, F'(l) = 0.
Also, G(y) satisfies
G-275)
with
462 GREEN'S FUNCTIONS
G.276) G'@)=0, G'(/) = 0.
Each of these eigenvalue problems for F and G was studied in Section 4 3
The eigenvalues for G.275), G.276) are
G.277) k2
and the eigenfunctions are
G.278) Gm(y) = cos[™ y] , m-0,1,2,....
For F(x) we obtain the eigenvalues
G.279) k-
and the eigenfunctions
G.280) Fn(x) = cos[™ x] , n =0,1,2
Then the eigenvalues for G.269), G.270) are
G-281) Anm
and the eigenfunctions are
G.282) Mnm(x, y) = cos(^ x) cos(^y y) , n, m = 0, ,1,2,....
The inner product for the region G is given as in G.232), and it is easily
shown that {Mnm} is an orthogonal set using the results of Section 4.3. Also,
the square of the norm of Mnm is
G.283) \\Mnm\\2 = (Mnm,Mnm)=j , *,m = 1,2,3....
For Moo we have
G.284)
and for Mn0 and MOm we obtain
GREEN'S FUNCTIONS FOR BOUNDED REGIONS
463
G.285)
Consequently, the orthonormal set of eigenfunctions is given by
G.286)
1
ш
m = n = 0
m = 0, n = l,2,...
n = 0, m = l, 2,...
/4 /тгл \ ( тгт \
yj-f cos^— xj cos^-y- уj , n, m =* 1,2,
We note that Aoo = 0 is an eigenvalue for which the eigenfunction M^, is a
constant.
Having determined the eigenfunctions Mnm(x, y), we expand the mod-
modified Green's function in a series of these functions as in G.258). The
eigenfunction М00(дс, у) must be excluded from the series. We have
G.287)
, y, t, V) - Jt
cos[(irm/t)y]
щ
ff "i (тгл//J
4 у у cos[Grn//)g] cos[Grn//)x] cos[Grm//)i7] cos[(irm/l)y]
It »-i ^1 (тгп//J + (тгт/?J
The alternative procedure presented above for the construction of
Green's functions in terms of eigenfunctions for lower dimensional prob-
problems, can also be used to determine modified Green's functions. Its applica-
application to the problem of Example 7.7 is considered in the exercises. We do not
construct Green's functions for nonself-adjoint problems because of difficul-
difficulties that arise in connection with eigenfunction expansions for nonself-
adjoint eigenvalue problems.
The Green's function for the hyperbolic problem considered in Section
7-l is expressed as K(x, t; £, т) and satisfies the equation
G.288)
S2K
2
at
464 GREEN'S FUNCTIONS
where x and £ are points in the bounded region G and t and r are less than
Г. In addition, K(x, t; £, r) satisfies the initial conditions or, equivalents
the end conditions
G.289) K(x, T; f, r) = **&>£ *• T> = 0 ,
and the boundary condition
G.290) aK + p —
an
= 0, t<T.
If the region G is one-dimensional (i.e., 0< jc< /), the boundary condition
G.290) is replaced by a one-dimensional form as in G.49).
Proceeding as in Section 4.6, we expand K(x, t; £, r) in a series of
eigenfunctions,
G.291) K(x91; {, т) - 2 Nk(t)Mk(x),
where Mk(x) are the eigenf unctions of the operator L, satisfying a boundary
condition of the form G.290). To determine the Nk(t), G.288) is multiplied
by Mk(x) and integrated over G. This yields the equations
G.292) Nl(t) + XkNk(t) - Mk(£)8(t - т) , Г, т < ГД = 1,2,. . . .
The initial conditions for Nk(t) are
G.293)
On the basis of Example 7.2 we conclude that Л^(/) is continuous at t = т
and N'k(t) has a jump discontinuity
G.294)
The conditions at t = Г imply that
G.295) ^@ = 0, r<t<T.
The continuity of Л^(г) and the jump condition on N'k(t) att = r imply that
G.296) Nk(r) = 0 , N'k(r) = -
These serve as initial conditions for Nk(t) in the interval t<T. Since
5(/ - T) = о for t < r, JV*(/) satisfies the equation
GREEN'S FUNCTIONS FOR BOUNDED REGIONS
G.297) N%t) + \kNk(t) = 0 , t < r .
The solution of G.296), G.297) is easily found to be
G.298) Nk(t) - —= sm[VTk(r - t)]MkW , t < т .
Combining G.295) and G.298) we can write Nk(t) in the form
G.299) Nk(t) = ^= sin[V^(T - t)]Mk(f)H(r - 0 ,
where H(x) is the Heaviside function G.98). The Green's function
K(x, t; g, t) thus has the form
G-300)
дс) Я(т - /) .
J
Notice that К is symmetric in x and £ but not in t and r. As noted in the
discussion following G.28), the function S(x, t; £, r) = K(x, —t\ f, -r) is
the causal fundamental solution for the given problem.
It may be verified directly that G.300) is a generalized solution of
G.288). Let G(jc, t) represent the bracketed term in G.300) and write К as
G.301) K(x91; g, t) = G(x, t)H(r - 0 .
Then
G.302) ^ = ^я(т-0-
since G(x, t)8(r -t)= G(x, t)S(t - t) and G(x, т) = О. Further,
G.303) £^«^h(t-0-— «(t-0-
Now
G.304)
501 r - i;(T-o]Mfc(OM,(*)L = -i m,
view of G.188). Also,
466 GREEN'S FUNCTION)
G.305)
LK = LGH(r - t) = [J£ ^ MVKfr - t)]Mk
- Г)
= — о —5~ Нут ~~ ч .
Combining these results we have
G.306) p~
Recall that й(т - f) = й(* - r). It has been assumed that Ao = 0 is not an
eigenvalue. If Ao = 0 is an eigenvalue, an additional term (т — t)M0( f )Afo(jc)
must be added to the series in G.300), as shown in the exercises.
The Green's function K(x, t\ £, т) for the parabolic problem of Section
7.1 satisfies the equation
Ж
G.307) -p — + LK = 8(x-t)S(t-T), x,{eG,t,r<T.
at
the initial condition
G.308) *(х,Г;&т) = 0,
and the boundary condition
G.309)
0, t<T,
if G is a two- or three-dimensional region.
As was done for the hyperbolic problem, we expand K(x91\ £, т) in a
series of eigenfunctions
00
G.310) K(x, t; {, т) = Д Nk(t)Mk(x).
Multiplying G.307) by Mk(x) and integrating over the region G we obtain
G.311) -N'k(t) + kkNk(t) = Mk(ki)8(t-r), к = 1,2
Since 8(t - т) = 5(t - /) we can write G.311) as
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 467
G.312) | [е~А*Ч@] = -Мк(£)е-к*Ь(т - t)
where G.91) was used. Integrating G.312) and using the initial condition
N (Г) = 0, we obtain
G.313) **(') = e^~7)Mk( i)H(r - 0 .
[We recall that dH(x)/dx = 8(x).] Thus
G.314) K(x91; (, r) = [ £ е^'г)МкШМк(х)]н(т - t) .
[As noted in the discussion following G.39), the function S(x, t; g, r) =
д^ _f. £ —T) is the causal fundamental solution for the given problem.] It
is not difficult to show directly (see the exercises) that the series G.314) is a
generalized solution of G.307). If Ao = 0 is an eigenvalue, the series G.314)
must be modified as shown in the exercises. Some specific examples of
eigenfunction expansions of Green's functions for hyperbolic and parabolic
problems are also considered in the exercises.
We conclude this section with the observation that the Green's functions
G.300) and G.314) are solutions of the homogeneous versions of G.288)
and G.307), respectively, for t<r. At f = r the Green's function G.300)
vanishes, and its time derivative equals -S(x - £)/p. The Green's function
G.314) equals 8(x - ij)lp at t = r. Also, the boundary condition G.290) or
G.309) is satisfied by both Green's functions for t < r. These results follow
easily from our discussion. Consequently, we find that for t < т the Green's
functions determined by each of the methods given in Section 7.1 for the
hyperbolic and parabolic cases are identical.
7.4. GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS
The transforms introduced in Chapters 4 and 5 (i.e., finite and infinite
transforms) can be used to determine Green's functions for problems given
over unbounded spatial regions. Rather than give a general discussion for
equations of different types, we consider several examples in which specific
Green's functions are obtained.
Example 7.8. The Heat Equation in an Unbounded Region. We begin by
^nsidering the heat equation in one dimension over the infinite interval
~"°°<*<oo. The medium is assumed to be homogeneous. In view of our
*scussion in Section 7.1, the Green's function K(x, t; £, r) is a solution of
the equation
4«8 GREEN'S FUNCTIONS
G.315)
2 (т), *,т<7\-оо<д;<оо,
and satisfies the initial condition at t — T
G.316) K(x, Г;£,т) = 0.
The initial value problem for К will be solved by using the one-
dimensional Fourier transform in jc. Let the Fourier transform of К be
denoted by k(X, t; £, r). Then
/ iAx
G.317) fc(A, t\ €, т) = y= /_ eiAxK(x, t; f, r)
To obtain an equation for k, we multiply G.315) by (l/V27r)eM* and
integrate from — «> to +<*>. Using E.15) we obtain the equation
G.318) —-(cAJfc = ^==e'Afe(/-T), t<T.
Also, /c satisfies the condition
G.319)
The equation G.318) has the form of equation G.ЗП) with Nk replaced by
k, Afc equal to (cA)z, and Mk given as A/V2n)e^e. The initial condition at
t = Г is identical for both functions Л/* and k. Therefore, the solution of
G.318), G.319) is obtained from G.313) as
[(AJ(
G.320) k( A, t; (, t) = ^= exp[(cAJ(/ - r) + iAf ]Я(т - 0 .
The inversion formula for the Fourier transform then gives
G-321)
K(x, t; €, t) = ^= J_e «""**( A, f; f, r) dA
= ^- Я(т - 0 f exp[-/A(* - ^) - c2A2(t - 0] dk .
This integral was evaluated in Example 5.2 [see E.39)-E.43)]. Using the
results of that example gives the Green's function К as
'S FUNCTIONS FOR UNBOUNDED REGIONS 469
VVe note that К is symmetric in x and g but not in t and т. In fact, there is a
reversal in time in relation to the fundamental solution G(x - f, f - т)
defined in E.45). This results because К satisfies the backward heat
equation, whereas G satisfies the forward heat equation
By using the two-dimensional Fourier transform it is easy to show that
the two-dimensional Green's function is
Similarly, the three-dimensional Fourier transform yields
G.324)
K(x, y, z, t\ £, 77, С, г)
3 L 4c2(t-0
as the three-dimensional Green's function. We will refer to G.322)-G.324)
as the free space Green's functions for the heat equation. In each case we
obtain the causal fundamental solution for the heat equation by replacing t
by —t and т by — r. The results agree with F.210).
Given the initial value problem for u(xy t),
G.325) ut - c2uxx = F(x, t), -oo<jc<oo?f>0,
with the initial condition u(x, 0) -f{x), the Green's function К may be used
to obtain the solution w. With -oo < g < oo and т > 0, the solution formula is
G.326) u(£fT)=f f KFdxdt+l K(x9O;€9T)f(x)dx9
JO J-oo J-°°
as follows from the expression G.41) in Section 7.1, with К given in G.322).
Now
G.327)
Jo J-«
^, 0
4c\r-t)
since Я(т - /) = 0 for t > т. Also,
470
G.328)
GREEN'S FUNCTIONS
, 0;
> exp[- ^] A
since H(r) = 1 for r>0. It is immediately seen that the solution G.326)
agrees with that given in Example 5.2.
Example 7.9. Green's Function for the Wave Equation. In this example
we construct the Green's function K(xy y9 z, t; f, 17, f, т) appropriate for the
solution of the Cauchy problem for the wave equation in three dimensions.
Using this function, the two-dimensional Green's function is obtained. In
addition, the Green's function for the one-dimensional problem is discussed.
In three (space) dimensions, the Green's function К satisfies the equation
G.329)
2
Э2К
with -oo<jc, y9 z, f, 17, f <°°, and t, r< T. The conditions at f = T are
G.330)
= 0.
To solve, we use the three-dimensional Fourier transform (see Section 5.4)
and define
G.331)
00
*(A,, A2, A3, t; f,,, {, r) =
2)K dxdydz.
То obtain an equation for k, multiply 1/(V2ttK exp[/(X,x + X^ + X3z)] into
G.329) and integrate with respect to jc, y9 and z from -oo to + «>. Using the
properties of the Fourier transform (see Example 5.7) gives
G.332)
d2k
and the initial conditions
G.333) k
/-r~ dt
0.
FUNCTIONS FOR UNBOUNDED REGIONS 471
The problem G.332), G.333) is equivalent to that for Nk(t) given in
G.292), G.293). Using the result G.298) gives the solution
G.334) к = ^-}3 sin[c|A|(r - 01 exp[i(A,£ + А2т, + Л3£]#(т - t) ,
where |A| = VAi + A2 + Аз- Inverting the Fourier transform yields
G.335)
Я(т-0 Г f f sin[c|Al(T-Q]
X exp{-i[A,(x - f) + A2(y - т,) + A3(z - f)]} dk, dk2 d\3.
This integral, which is to be interpreted in a generalized sense, can be
evaluated by transforming to spherical coordinates. Let |A|, в, and <p be
spherical coordinates with |A|^0, O^0s2tt, and 0<<p^ir. The polar
axis (i.e., <p = 0) is chosen to coincide with the (half) line connecting the
origin (i.e., Ax = A2 = A3 = 0) to the fixed point (x - g, у - tj, z - f). Since
(x - £, у — tj, z — f) is on the polar axis, we have
G.336) A,(x - f) + A2(y ~v) + A3B " С) = IA|rcos <p
where r = V(* — £J + (y — vJ + (z ~ fJ- The volume element is given as
G.337) d\, dk2 d\3 = | A|2 sin <p d\ A| dO d<p ,
and G.335) takes the form
G.338)
К=^гФ~ I [I {sinMA|(T-0]e""A|rcoi*|A|sin9}d9^rf|A|
C\&Tf\ Jv J" J"
The two inner integrals are easily evaluated, and their contribution is found
to be Dir/|A|r)sin[|A|r]. Then G.338) is evaluated as follows:
G.339)
" ') - rl> - cos{|A|[c(r - 0 + r]}) d\X\
JJ I (exp{'lA|[c(r " V rl> exp{i|A|[c(t - Г) + r]}) </| A|
- 0 " r] - 8[c(r -t) + r]} ,
472 GREEN'S
where G.138) has been used. Now 8[с(т - t) + r] = 0 since for т - t > 0 and
r>0, с(т~0 + г^0. F°r *e same reason, 8[c(r-t)-r] vanishes fOr
т — t < 0 so that there is no need to include the Heaviside function in
G.339). Thus К is given as
G.340) K(x9 y, z, t; (9 V, l, т) =
V(« (y) ()
To obtain the Green's function in two (space) dimensions, we integrate
the Green's function G.340) with respect to I from -»to +». The resulting
Green's function K- K(xy v, t\ g, 77, r), is independent of z and £ and
satisfies the equation
G.341) ^ - 4^ + 0] = «(» - «Жу - v)8(t - r)
for -oo<jc, I, y, 77<°o, and tJ т< Т. In effect, we are considering a line
source in three-dimensional space and because of the homogeneity of the
medium (there are constant coefficients in the equation) this corresponds to
a point source in two dimensions. We remark that if we formally integrate
G.329) with respect to ( and assume К is independent of z, G.341) results.
On integrating over (f the two-dimensional Green's function becomes
G.342) K(X, y, t; (,,, r) -
Introducing the change of variables s = z- ( gives
G.343) g-J-Г «WrQ^
Г
4irc J-* г 2тгс Jo
where r = \J(x - £J + (у -rjJ + s2 and the second integral from 0 to °°
results since the integrand is an even function of s. A further change of
variables
G.344) r2 - p2 + s2 ,
with p2 = (jc - ^J + (у - уJ yields
G.345)
and G.343) becomes
G.346) K=—1
473
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS
If c(r - 0< P» *e integral vanishes, since the argument of the delta
function is negative in that case. If c(r - t) > p, the substitution property of
the delta function yields К = A /2тгс)A ЫЛт ~ {f " P*)- with the use of
the Heaviside function both results can be combined into a single expression
G.347)
, y, t\ f, 7), T) ••
where p2 = (x~ tJ+ (y-r,J-
In the one-dimensional problem the Green's function K(x, t; g, r) satis-
satisfies the equation
with — °° < x, £ < +o°, and т < 7\ Again the conditions at t = T are
G.349)
dK
dt
t=T
= 0.
It is a simple matter (see the exercises) to adapt the general solution of the
Cauchy problem to show that К is given as
G.350)
\x-Z\>c(r-t).
In terms of Heaviside functions К can be expressed as
G.351) K(x, t;Z,T) = -l-H[x-ct-(Z- ст)]Н[£ + ст-(х + ct)],
and we now show that К is a solution of G.348), G.349).
Putting t = T in G.351) gives
G.352)
с, T; Lt) = ^- H[x -cT-{£- ct)]H[£ + cr - (x + cT)] = 0
because if the argument of the first Heaviside function [that is, x - g -
C(T~ t)] is positive, we have x - §> c(T~ т)>0 since Tis greater than т.
Consequently, (£ - x) - c(T- т) = -(x - £) - c(T - т) <О, and the second
Heaviside function vanishes. A similar result holds if the argument of the
^cond Heaviside function is positive.
Further, dKldt is given as
474 GREEN'S FUNCTIONS
дК. 1
G.353) — = - - 8[x - ct - Ц - ct)]HU + or - (x + <*)]
and /C, vanishes at f - T for the same reason that К vanishes there.
Finally, we easily obtain
G.354) Ktt - c2Kxx =2c8[x-ct-U- ст)]8[{ + cr - (x + с*)]
- 2cS [x - cr - ( f - ct)]5 [2c(t - r)]
where the properties of the delta function given in the preceding section
have been used.
A common feature of the Green's functions G.340), G.347), and G.351)
is that they are symmetric in x and £ where jc is the observation point and f
is the source point in one, two, or three dimensions. We shall refer to them
as the free space Green's functions for the wave equation. If we replace t by
-t and т by -г, we obtain the causal fundamental solution for each of the
problems. Let the fundamental solutions be denoted by 5. In the three-
dimensional case we have
G.355) 5=
In two dimensions
whereas in one dimension
G.357) 5 = ^ H[c(t - r) + (x - {)]H[c(t - т) - (* - ()].
Each of these functions satisfies the same equation as the Green's function
K, except that for each of them we have
G.358)
dS
dt
= 0.
Even though they do not all have the same form as the causal fundamental
solutions given in F.207)-F.209), they can be shown to be identical on
using the properties of the delta and Heaviside functions.
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 475
The fundamental solutions yield a vivid distinction between the nature of
wave propagation in two and three dimensions as characterized by solutions
of the wave equation. (These differences have already been discussed in
Example 5.7.) Given the source point Ц, v, £) and the time r, the forward
characteristic cone for the wave equation is
G.359) c(t - r) - У(х-£J + (у-чУ + (г ~ £ f , * * т .
For the two-dimensional problem with the source point (£, 77) and the time т
we have
G.360) c(t-T) = ^(x-{J + (y-vJ, **t.
Noting the behavior of the delta function, G.355) indicates that the
disturbance due to a point source at (£, 77, £) acting at the time т, is
concentrated on a sphere of radius c(t — r) with center at (£,1?, £) at the
later time t and vanishes elsewhere, in the three-dimensional case. For the
two-dimensional problem, the disturbance resulting from a point source at
(£,17) acting at the time r, is distributed throughout a circle of radius
c(t-r) with center at (£,17) at the later time t. Consequently, both
disturbances travel with speed с in all directions. Hpwever, in the three-
dimensional case there is a sharp wave front that leaves no wake and the
disturbance is only felt instantaneously at any point. In the two-dimensional
problem the solution does not return to zero as soon as the wave front
passes but there is a wake that decays like l/c(t-r) as t increases. The
sharp signals that occur in three-dimensional wave propagation and are
lacking in the two-dimensional problem characterize what is known as
Huygens9 principle.
Example 7.10. Green's Functions for the Klein-Gordon and the Mod-
№ied Telegrapher's Equations. In connection with the Cauchy problem
for the one-dimensional Klein-Gordon equation E.258), we are led to
consider the Green's function K(x, t; £, r) that satisfies the equation
G.361) £!£ _ y»££ + лс e 8(X - mt - T),
with -00 < x^ £ < 00 and t, т < Г. The conditions at t = T are
G.362) К
ЭК
dt
= 0.
On replacing t by —t and т by — т in the causal fundamental solution
F.217) and adjusting the parameters, we conclude that the Green's function
^ should be given as
476 GREEN'S FUNCTIONS
G.363)
°. \x-€\>y(r~t),
where /0 is the Bessel function of order zero. Noting G.351) we can express
К as
G.364) K(x, t; £, т) =
хЯ[лг-уГ-(£-уг)]Я[£ + уг-(х + уО].
Since /0@) = 1 we see that G.364) reduces to the Green's function for the
one-dimensional wave equation if we set с = 0.
Inasmuch as we have not derived the Green's function К directly by
Fourier transforms or some other means, we now show that it satisfies
G.361), G.362). We set
G.365) K=J0K,
where К is the Green's function for the wave equation, and satisfies G.348)
with с replaced by y. Then
G.366)
дгК 2 дгК
~У ~TT
дх* I at дх*
In view of the results of Example 6.14 [see F.211)-F.213)] we conclude
that the first bracketed term in G.366) vanishes. Also, we have
G-367)
2 «„ „л l cJi[(c/y)y/y\t-TJ-(x-tf}
ax 1
dt dt y
x {[-y(t -t) + (x- £)]8[x -yt-Ц- yr)W[U + ут ~
+ [~y(t- r) ~ (x - t)]H[x -yt-U- yr)]«[f + ут-(х + yt)]} ■
The expression G.367) vanishes since
G.368) [-y(t -
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 477
G.369) [~y(t -r)-(x- £)]8[£ + yr - (x + yt)]
on using x8(x) = 0. Finally, we have
G.370)
- т)/0@) = 8(x - f )8(Г - t)
since /0@) = 1 anc* ^ satisfies G.348). Thus A" is a solution of the equation
G.361). The conditions G.362) are also satisfied because К and its time
derivative vanish at t = Г.
It may be noted that with с = it (where i = V^T) in the Klein-Gordon
equation G.361) we obtain the modified telegrapher's equation
G.371) 5 " Y2 T? " c2K=8(x - f )в(Г- т).
ot aX
This equation results when the first time derivative term is eliminated from
the telegrapher's equation (see the exercises). Since JQ(ix) = /0(jc), the mod-
modified Bessel function of zero order, we obtain for К in place of G.364) (see
also F.218)),
G.372) K(x, t; {, r) = ^ /0[| \Jy\t- rJ - (x - ^
Again, G.364) and G.372) are referred to as free space Green's functions.
Given the initial value problem for the Klein-Gordon equation,
G.373) utt - у 2uxx + с2м = F(x, t),
with the initial conditions
G.374) и(х,0)=Дх), u,(x,0) = g(x
the solution at an arbitrary point (£, r) is given in terms of G.364) as
G.375)
where т< Г and G.40) has been appropriately modified.
478 GREEN'S FUNCTIONS
In the double integral in G.375) we have К = 0 for t > т, so that the limjt
in the t integral extends only up to т. Also, from G.363) we conclude that К
vanishes unless \x - £| < у(т - i) and this is equivalent to
G.376)
Therefore, we obtain
G.377)
ГГ.FKdxdt=Ту ГГГ-ГFix'f)
Further, we have
G.378)
so that
G.379)
g(x)K(xfQ;£9'.,™ 2y is-
since the product of the Heaviside functions vanishes outside the interval
(f-уг, £ + yr). Finally,
G.380)
c,0;&t)
-14 у
- a - rim* + yr -
In view of the substitution property of the delta function, the last two terms
in G.380) reduce to - \8[x - Ц - yr)] - i&[x - (f + yr)] since /0@) = 1
and #Byr) = 1. Therefore, we obtain
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 479
G.381) }_J(x) — (x,0-<:,r)dx
2
Combining these results and noting that -J'0(x) = Jt(x), the Bessel function
of order one, gives the solution u(x, t) of the initial value problem G.373),
G.374) as
u{x, t) =
*■.
This solution formula reduces to that for the Cauchy problem for the
inhomogeneous wave equation if we set с = 0. Also, if we set с = ic in
G.382) and note that J0(iz) = /0(z) and J^iz) = il^z), we obtain as the
solution of the modified telegrapher's equation
G.383) utt - y\x - c2u = F(x, t),
with the initial conditions G.374),
G.384)
Я
Ту Jo i-^
Ту Jo i^ F^' T4y Vr^r)(,|)] « A-.
480 GREEN'S FUNCTIONS
Both solutions G.382) and G.384) exhibit the domains of dependence and
influence for the Klein-Gordon and modified telegrapher's equation. (The
telegrapher's equation is discussed in the exercises.) It has already been
indicated that these domains, which characterize the maximum speed at
which disturbances or signals travel, are determined by the principal parts of
the given equations (i.e., the second derivative terms) and do not depend on
the lower order terms. Our results show that these equations and the wave
equation have identical domains of dependence and influence.
Before considering Green's functions for elliptic equations, we examine
the relationship between the Green's functions for hyperbolic and parabolic
equations defined in terms of the inhomogeneous equations G.26) and
G.37), and those defined in terms of the homogeneous equations G.42) and
G.43). The relationship between these two determinations of Green's
functions for the case of bounded regions was discussed at the end of
Section 7.3.
Let the Green's functions К be defined in terms of the (backwards) initial
value problems G.42)-G.45) for the hyperbolic and parabolic cases with the
initial data given at t=r and the problem defined over the entire space.
Then if we set K' = КН(т — t) with т<Т, it is easy to verify from the
properties of К and by direct differentiation that K* is indeed the solution of
the Green's function problem given in terms of the inhomogeneous equa-
equations G.27) and G.37) for the hyperbolic and parabolic cases, respectively.
Thus for t < т both of these formulations lead to the same Green's functions.
In a similar fashion, it can be shown that the causal fundamental solutions
for hyperbolic and parabolic equations can be characterized in terms of
initial value problems. If the instantaneous source point is at (jc, t) = (f, r),
the initial conditions for the homogeneous form of the hyperbolic equation
G.14) are и = 0 and ut = S(x - £)lp at t - т. For the homogeneous form of
the parabolic equation G.29) with the same source point, the initial
condition is и = 8(x ~ £)lp at t = r. Then the causal fundamental solutions
for these problems, which we denote by 5, can be expressed in terms of
solutions и of the aforementioned initial value problems in the form
S = uH(t - r). The verification of this result is almost identical to that for
the Green's function problems discussed in the preceding paragraph.
Consequently, instantaneous point source problems and the related caus-
causal fundamental solutions need not necessarily be characterized strictly in
terms of inhomogeneous differential equations as has been done in Section
6.7 and Example 7.9, for instance. We can determine the solutions of these
problems in terms of homogeneous equations with appropriate Dirac delta
function initial conditions. This approach is used in Section 10.2 when we
discuss methods for analyzing the propagation of singularities for hyperbolic
equations.
Since the free space Green's functions for (self-adjoint) elliptic equations
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 48l
with constant coefficients have already been fully discussed in Example 6.13,
we restrict our analysis of Green's functions for elliptic problems in un-
unbounded regions to a single problem involving the reduced wave equation.
Example 7.11. Green's Function for the Reduced Wave Equation. In
this problem we construct a Green's function for a boundary value problem
for the reduced wave equation which is of interest in the theory of ocean
acoustics.
The ocean which we assume to be homogeneous, is taken to be of infinite
extent in the x and у directions and is assumed to have a constant depth
ft>0. Thus — °°<jc, y<°°, and — h<z<0. A point source is located at
х=у = 0 and z = £ with -Л<£<0. Then the Green's function K =
K(x, y, z; £) satisfies the equation
G.385)
V2K + k2K=^ + 0 + ^ + k2K=-8(x)8(y)8(z-O,
aX ay oZ
and the boundary conditions
G.386) *(*,у,0;£) = 0,
G.387) "^7 (*'^'~Л; £)==0*
The real part of Ке~ш\ where <o is the frequency and k = We, represents
the time harmonic acoustic pressure and satisfies a wave equation with
constant wave speed с in the absence of any source terms (see Exercise
2.4.7). Thus the equation for К is known as the reduced wave equation (It is
also called the Helmholtz equation.). The plane z = 0 represents the surface
of the ocean, and the plane z = —h is the (rigid) bottom. The conditions
G.386) and G.387) are appropriate for the sound pressure at the surface
and at the bottom, respectively. The homogeneity of the ocean implies that
k in G.385) is a constant. Ibis problem for К does not have a unique
solution unless a radiation condition is imposed. This is done in our
discussion.
To solve for the Green's function К we use the method of eigenfunction
expansions. We consider an eigenvalue problem for the function M{z) with
-Л < z < 0 associated with the operator L = - d2ldz2 ~ k2. Thus
G.388) -M"(z) - k2M(z) = AM(z), -h < z < 0 ,
where Л is the eigenvalue parameter. The boundary conditions are
(?.389) M@) = 0, M'(-h) = 0.
482 GREEN'S FUNCTIONS
On comparing G.388), G.389) with the Sturm-Liouville problem of Section
4.3 we find that, apart from the fact that the coefficient -k2 of M{z) is
negative, all conditions are the same. All properties of the eigenvalues and
eigenfunctions given in that section are valid for G.388), G.389) except that
a finite number of the eigenvalues are negative as shown.
The general solution of G.388) is
G.390) M(z) = c, sin[zVfc2 + A] + c2 cos[zVfc2 + A].
The condition M@) = 0 implies that c2~0. The condition M'(-/i) =
implies that
G.391)
so that A is specified in terms of the zeros of the cosine function. The
eigenvalues Xn are given as
G.392) A|l
and the eigenfunctions are
G.393) Mn(z) - яп[(л + \)(j)z] , n = 0,1,2,... .
Note that some of the Xn may be negative.
The inner product appropriate for this eigenvalue problem is
G.394) (9,t)
and (Mn, Mm) = 0 for m Ф n as can easily be shown. In addition,
G.395) (Ma, Mn) = \_h sin2[(n + \)(j-
Thus the normalized set of eigenfunctions are
G.396) Mn(z) = yfl sin[(n + |)(f )z] , n = 0,1,2,.. . .
In terms of the Mn(z) we construct the eigenfunction expansion of the
Green's function К
GREEN'S FUNCTIONS FOR UNBOUNDED REGIONS 483
G.397) K(x, y, z; £) = 2 *„(*> y)Mn{z) .
v o
n=o
The coefficients ЛГЛ are determined by multiplying G.385) by Mn(z) and
integrating from -A to 0. Using the procedures of Section 4.6 we obtain
G.398) —f + —г
v дХ ду
for n = 0, 1, 2,....
To obtain a unique solution of G.398) we must specify the behavior of Nn
as x and у tend to infinity. Disregarding the multiplicative constant Mn(g),
we are essentially interested in constructing a free space Green's function
for G.398). If k2 > [n + A/2)]2(тг/ЛJ, G.398) is an inhomogeneous Helm-
holtz or reduced wave equation, whereas if k2 < [n + A /2)]2(тг/АJ, we have
a modified Helmholtz equation.
With r = yx2 + y2, a function <p(x, y) is said to satisfy the Sommerfeld
radiation condition at infinity if
[dw 1
-~ - iaxp \= 0 ,
dr J
and <p satisfies the equation
G.400) <pxx + <pyy H- u) V = 0
for large values of r. With a>2n - -An and Ал defined in G.392), we write
G.398) as
With w2>0 we use the results of Example 6.13 to conclude, in view of
F.204), that
G*402) Nn(x, у) = (^Мп(ОН%\<опг) , n - 0,1,. . . ,
where H(ol\z) is the (zero-order) Hankel function of the first kind. With the
use of F.206) it is easy to see that G.402) satisfies the radiation condition
G.399) if o>2>0. If w2<0 (that is <on is imaginary), the appropriate
solution of G.401) must be given in terms of the modified Bessel function
Ko(z) as in F.199). However, since H$\iz) = B/iri)K0(z) for z>0 we
conclude that G.402) is a suitable solution for all values of n. Since ЛГ„(л;, у)
decays exponentially at infinity, in view of F.203), when o>2<0, the
radiation condition is certainly satisfied.
484 GREEN'S FUNCTIONS
Collecting our results, we obtain the Green's function К in the form
G.403, «(,. ,. z; О - Ш 2 *[(- ♦ i)(f )t] Ц(. + |)(f),]
where w2 = A:2 - [n + A/2)]2(тг/ЛJ. As n increases, the argument of the
Hankel function eventually becomes imaginary and the terms in the series
decay exponentially for large л.
7.5. THE METHOD OF IMAGES
Given any of the partial differential equations of Section 7.1, the Green's
function К can be expressed in the form
G.404) K=KF + KG,
where KF is the free space Green's function. We recall that the free space
Green's function satisfies the same differential equation as the Green's
function K. In addition, KF satisfies (backward) causality conditions in the
hyperbolic and parabolic cases, and appropriate conditions at infinity in the
elliptic case. Consequently, KG satisfies a homogeneous differential equation
with homogeneous end conditions at t — T if these are relevant. The
boundary conditions for KG are no longer homogeneous, however. For
example, if К is required to vanish on dG, then KG = — KF on dG. Although
it may be possible to use eigenfunction expansions or transform methods to
determine KG, we do not use these approaches since К itself can just as
easily be determined in the same fashion. (However, the results for KG may
be better suited for numerical evaluation than those for K, since KG is not
singular within G.) Instead we shall use the method of images to specify KG
in case the equations of Section 7.1 have constant coefficients and the
boundary dG has a special form as described in the following.
We assume that the boundaries for the given problem are made up of
(portions of) lines or planes, or (portions of) circles or spheres. Given the
singular point of the delta function in the equation for K, we consider all
possible image points obtained by reflection through lines and planes and
inversion through circles and spheres. (The inversion process is defined
later.) If none of the resulting image points lies in the interior of the region
in which the problem is specified and certain additional conditions are met,
the Green's function К can be specified in a simple manner. We do not
describe the most general regions and equations for which the method of
images works. Instead we consider a number of examples and exercises that
exhibit the basic features of the method.
ТЦЕ METHOD OF IMAGES 485
To apply the method of images, it is necessary to know the free space
Green's functions for the given equations and most of the relevant ones
have already been determined.
Example 7.12. Laplace's Equation in a Half-Space. We consider Lap-
Laplace's equation in the half-space z>0. The Green's function K =
K(x, y, z\ £,17, £) satisfies the equation
G.405) V2^=0 + 0 + 0 = -8(x
in the region G defined as the half-space z > 0. On the boundary dG [that is,
the (x, y)-plane z = 0] we have
n IT
G.406) aK+fi-
aK+fidG
z=0
We assume that a and /3 are constants and consider three cases. For the
Dirichlet problem (i.e., the first boundary value problem) we set a — 1 and
j8 = 0; for the Neumann problem (i.e., the second boundary value problem)
we put a = 0 and f$ = — 1; for the third boundary value problem we set
a = h and j3 = 1 with h > 0. The Green's function К is required to vanish at
infinity.
To begin, we recall that the free space Green's function for Laplace's
equation is
G.407) KF ш -L [(, - tf + {y- rtf + (z - ОГЩ
as was shown in Example 6.13. It follows from our discussion in that
example that V2KF = -8(x - £)8(y - -q)8{z - {). The Green's function К is
then expressed as
Since we must have
G.409) V2K = V2KF + V2XG = -8(x - £)8{y - v)8(z -
we conclude that
so that Kc is an harmonic function (i.e., a solution of the homogeneous
^place's equation). In view of G.406), the boundary condition for KG is
486
GREEN'S FUNCTIONS
G.411)
z=0
The point (£, 17, £) is the source (or singular) point for the Green's
functions К and KF. Now KF is given in terms of the distance from the
observation point (jc, yf z) to the source point (£, 17, £). As shown in Figure
7.4, if we introduce the image (source) point (f, 17, ~C)—that is, the
reflection of (£,17, £) in the plane z-0—then as the observation point
(jt, y, z) tends to a boundary point (x, y,0), its distance from (f, rj, f)
equals its distance from (£, 17, -f).
Consequently, if we introduce the function
we have in effect a free space Green's function corresponding to the image
source point (£, 17, -f). At the boundary z = 0, KG = KF and
G.413)
i-3/2
z=0
4тг
2*0
dz
z=0
Furthermore,
G.414)
-5(x - ()8{y -
te j', 2)
Figure 7.4. The source point (^, 17, £) and its image point.
METHOD OF IMAGES 4»7
and the right hand side vanishes in the half-space z>0 since 8(z + £) = 0
there. Therefore, if we set KG = -KG for the Dirichlet problem and
К - KG for the Neumann problem, the Green's function К = KF + KG
satisfies G.405) in the half-space z >0 and the boundary conditions K\z=0 =
0 and <?Ar/<?z|2=0 = 0 which are appropriate for the Dirichlet and Neumann
problems, respectively.
Xhe Green's function for the third boundary value problem cannot be
obtained solely in terms of an image source point at (f, tj, -£)• Instead we
must introduce an entire line of image sources on the line x = £ and у = tj
with г extending from z = -ftoz = -«> and a source density function to be
determined. Let
G.415) KG = -^ [(* - {J + (у - Vf + (z + О2}
1 ГС P(sh
4тг J-« [(x-ex> + (v-v
2ГШ
where p(s) is the source density. When h = 0 in the boundary condition
<?X7<?z -А£|г=0 = 0, we expect G.415) to reduce to the form appropriate
for the Neumann problem so that p = 0 when h = 0. For this reason we have
added the free space Green's function corresponding to an image source at
(£,i), -f) to the integral term in G.415).
It is assumed that p(s) decays sufficiently rapidly at infinity that the
integral in G.415) converges and that differentiation under the integral sign
is permitted. We then find that V2KG = 0 for z>0, since all the singular
points in G.415) occur in the lower half-space z <0. Applying the boundary
condition at z = 0 gives, in view of G.411),
G.416)
1 f{ ч ,? r
41Г J-« <?5
1-1/2
488
GREEN'S FUNCTIONS
The operator didz has the same effect as -dlds at z = 0, and the use of
dlds in the integral term enables us to integrate by parts. We have
G.417) | _ p(s) - [(x - О2 + (У- vf + г2]'112 с
>\2 | /„ \2 , c.2-1-1/2
Combining results gives
t 1 p [P4»-ftpfr)]*2i/2_0
Therefore, the boundary condition is satisfied if we set
G.419) p'(s) - hp(s) = 0 , s < - С ,
G.420) />(-£) = -2A.
The solution of the initial value problem G.419), G.420) is
G.421) p(s) = -2heH(s+n .
We note that p(s) vanishes for h = 0 and that it decays exponentially as
s—* —oo.
The Green's function for the third boundary value problem thus has the
form
G.422)
J_ A A 1} 1/2
K(x, y, z\ f, 77, £) = — [(x - £) + (y - 77) + (z - f) ]
With h = 0 this reduces to Green's function for the Neumann problem. The
Green's function for the Dirichlet problem is
METHOD OF IMAGES 489
G.423) K(x, y, z; f, tj, f) = j" [(* - £J + (у - *?J + (z -
i-l/2
On using the free space Green's function for Laplace's equation in two
dimensions, it is easy to obtain the Green's function for the half-plane
problem in two dimensions. Furthermore, on using the formulas given in
Section 7.1, one can readily obtain the solutions of boundary value problems
in the half-space or the half-plane for Laplace's equation, as shown in the
exercises.
Example 7.13. Hyperbolic Equations in a Semi-inhnite Interval. The
free space Green's functions corresponding to the one-dimensional hy-
hyperbolic equation G.50) with constant coefficients were obtained in our
discussion in Examples 7.9 and 7.10.
For the Klein-Gordon equation, the free space Green's function KF
satisfies G.361) and is given as
G.424) KF(x, V, i,-i) = Yy J<[ ~y
X H[x -yt-U- yr)]H[g + ут-(х + yt)]
in view of G.364). For the modified telegrapher's equation G.371) we have
G.425) KF(x, t; f, r) = ^ /0[^ y/7\t-rf-(x- £
X H[x- yt-(£- yr)\HU + Ут-(х + yt)].
The free space Green's function for the one-dimensional wave equation
G.348) is
G.426) KF(x, t;£,T)=±H[x-ct-(t- ct)]H[ £ + cr - (x + ct)].
It should be noted that for the wave equation KF can also be expressed as
G.427) KF(x, t; (, r) - ^ H(r - t)H[c\t - rJ - (x - tf].
It is readily seen that G.427) is consistent with G.350) and that the product
°f the Heaviside functions given in G.424), G.425), can also be expressed in
the form G.427) with appropriately modified constants. In connection with
490 GREEN'S FUNCTIONS
the application of the method of images, it is apparent from the form of
I'S
in G.427) that an image source can be introduced at x = -f if the Green'f
function in the interval jc>0 is to be obtained.
We now show by direct substitution that G.427) is a solution of G,348)
h
We have
& J
G.428) & = - yc «<т - t)H[c\t - rJ -(x
+ c(t - т)Я(т - t)8[c2(t - tJ - (x - tf] ■
The first term on the right of G.428) vanishes since 5(т - t)H[-(x - f J] =
8(t - t)H[c\t - tJ - (x - £J] an<l H(x) vanishes for negative values of the
argument. Then
G.429)
д K
dt2 dx2
= -c(t - tM(t - t)8[c\t - tJ - (x - O2]
+ 2cH(r - t){[c\t - tJ - (* - tJ]8'[c2(t - tJ - (x - if]
+ 8[c2(t-TJ-(x-tJ]}. .
The second term on the right of G.429) vanishes in view of G.106). Using
G.168) gives
G.430) 8[c2(t-rJ-(x-tJ]
1
2ф-т|
{8[c(t -t)-(x-
In the region t < т, which is of greatest interest, we have \t-r\- -(t- т).
Thus
G.431) - c(t - tM(t - t)8[c\t - tJ - (x - £J]
= 8(t - т)8(х -
тНЕ METHOD OF IMAGES 491
since «(-*) = 8(z) and 8(T - /)«[c(f - т) ± (x - £)] = 8(r - t)S[±(x - {)]
For each of the hyperbolic equations considered we now obtain Green's
functions for the semi-infinite interval x>0. In the case of Dirichlet
boundary conditions at x = 0 (i.e., К = 0 at x = 0), the Green's function is
given as
G.432) K(x, t; f, r) = KF(x, t; £, т) - KF(x, t\ -f, т) ,
with £F e4ual to the appropriate free space Green's function G.424),
G.425), or G.426). If a Neumann boundary condition is given at x = 0 (i.e.,
\=0 = 0), the Green's function is
G.433) K(x, t; ^ t) = KF(x, t; f, т) + iCF(x, r; -f, т) .
Finally, if a boundary condition of the third kind is given (i.e., dKldx-
hK\x=0 = 0 with Л>0), the Green's function is
G.434) K(x, t; f, т) = KF(x, t\ f, r) + /STf(jc, r; -f, т)
-2Л eA(j+oii:F(jM; 5, т) Л .
J — 00
By comparison with the methods used in the preceding example or by direct
verification it can be determined that G.432)-G.434) give the required
Green's functions for each of the boundary value problems considered.
Example 7.14. The Heat Equation in a Finite Interval. In this example
we construct the Green's function K(x, t\ f, т) for the equation of heat
conduction in a finite interval 0 < x < I. Thus К satisfies the equation
G.435) ££ + c»L*«-e(x
It is assumed that К vanishes at the endpoints so that
G.436) K@,t;(,T) = K(l,t;€,T) = 0.
In addition, we have the end condition
The free space Green's function KF for the one-dimensional heat equation
was found in Example 7.8 to be [see G.322)],
492 GREEN'S FUNCTIONS
-2/-* -2/+* -f * 2/-? 2/+$
—' о 1 о 1 о 1 о ' о 1 i ,
-3/ -21 -I 0 /2/ зГТ
Figure 7.5. The source point and the image sources.
We express К in the form G.404), (i.e., К = KF + KG) and use the method
of images to specify KG. The source point x = £ must have an image with
respect to x = 0 and x = /, and each of the image sources, in turn, must also
have images with respect to x = 0 and x = /. Consequently, we are led to
consider an infinite sequence of source points at the points fn = ± £ ± 2nl
with n = 0,1,2,3,.... Some of these points are shown in Figure 7.5.
The Green's function К can then be written as an infinite series
G.439)
K(X9 t\ ^> T) ~
„-—I *1 4c2(t-0 (t0
The term with n = 0 and a positive coefficient corresponds to KF. Clearly,
К — 0 at t = Г since r<T. It can be shown that the series can be differen-
differentiated term by term. Inasmuch as each of the functions in the series except
that term corresponding to KF has it source point £„ outside the interval
0<x < /, we see that G.439) satisfies G.435). Also, it is not difficult to see
that at x = 0 and x - I there corresponds to each term in the series with a
positive coefficient an identical term with a negative coefficient. For
example, Figure 7.5 shows that x = f and x = -f are images with respect
to x = 0, and x - f and x = 21 - f are images with respect to x = /. The
terms in the series G.439) that correspond to the points x = £ and
x = -f are exp[-(x - £J/4с2(т - t)] - exp[-(x + f J/4с2(т - f)L and this
difference vanishes at jr = O. Also, for x = f and x = 2/-f we have
ехр[-(х-^J/4с2(т-0]-ехр[-(х + ^-2/J/4с2(т-0], and this differ-
difference vanishes when x = /. In fact, if we replace n by -n in the second series
it becomes identical to the first series at x = 0 and their difference equals
zero. At x = /, we replace n by -n + / in the second series and find that the
two series are equal.
The solution of the initial and boundary value problem for the heat
equation in a finite interval when the formula G.41) of Section 7.1 is used
has a form similar to that given in Example 5.12. As was shown there, this
result is expected to be useful for small values of t, whereas that given by the
finite Fourier transform method is more useful for t large. ^
Example 7.15. Green's Function for Laplace's Equation in a
Sphere. As shown in the preceding examples, the method of images can be
METHOD OF IMAGES 493
applied to equations with constant coefficients of all three types if there are
linear or planar boundaries. For Laplace's equation it is possible to extend
the image method to problems that involve circular or spherical boundaries
as demonstrated in this example.
We now construct the Green's function for the Dirichlet problem for
Laplace's equation in the interior of a sphere using inversion with respect to
the sphere. The Green's function K^ K(x, y, z; £, 17, £) satisfies the
equation
G.440) V2K= —2 "* ' 2" = ~&{x — £)8(y — t/M(z — £)
^ ' дх ду dz
and the boundary condition К = 0 on the sphere of radius a with center at
the origin. Let the observation point P be denoted by P = (x, y, z). The
source point is Po = (f, 17, £) and the origin of coordinates (i.e., the center
of the sphere) is О = @,0,0).
As shown in Figure 7.6, we introduce the image source point P'o =
(£f>V> (')• The point Pq lies on the radial line extending from the origin О
through the source point Po. Its distance from the origin equals p1? and the
distance of Po from the origin is p0. These distances are related by the
equation
G.441) P0Px = a2
where a is the radius of the sphere. The foregoing process of obtaining P'o
from Po is called inversion with respect to the sphere. Note that if the radius
a = 1, we have px = l/p0, so that the distances p0 and px are inverse to one
another. The observation point P is assumed to lie on the sphere in Figure
7.6.
The triangles &OPP0 and AOPPq in the figure are similar since they have
a common angle^<POP0 and proportional sides OP0/OP= OPIOP^ This
follows since OP = a (the radius of the sphere), OF0 = p0, (Щ = pl9 and
Figure 7.6. Inversion with respect to the sphere.
494 GREEN'S FUNCTIONS
(OP0)(OP'0) = PePi = (OPJ = a2 in view of G.441). The similarity of the
triangles implies that all three sides are proportional and we have
G.442) A-±-Ue
To complete the solution of the problem we set К = KF + KG where К is
the free space Green's function G.407) and KG is a constant multiple of the
free space Green's function with source point at Р^ = (^\г}', £'). That is
G.443) К = ± {[(x - if + (y - r,J + (z - ОТ
with the constant с to be specified. Since Pr0 lies outside the sphere, the
second term in G.443) is a solution of (the homogeneous) Laplace's
equation within the sphere. Thus G.443) is a solution of G.440). On the
sphere—that is, when P=(x, y, z) lies on the sphere—we have, in the
notation of Figure 7.6,
G.444)
[ +
4тг L r0 rx J 4тгг0
in view of G.442). Thus if с = -a/p0, the Green's function К vanishes on
the sphere. Therefore, we obtain the Green's function К as
4тгЬ0 porj
G.445) [
Let G represent the interior of the sphere x2 + y2 + z2 = a2 with dG
equal to the sphere itself. The Dirichlet boundary value problem for
Poisson's equation
G.446) V2m = uxx + uyy + uzz~-F(x, y,z)9 (x,y,z)EG,
with the boundary condition
G.447) u\dG = B(x, y, z) ,
has the solution
G.448) ^V
with (£, 77, C) equal to an arbitrary point within the sphere, as follows from
G.11).
METHOD OF IMAGES 495
To determine G.448) more explicitly it is necessary to evaluate the
exterior normal derivative dKldn on the surface of the sphere. Now
dK 1 Г в ( 1 \ a (? /
G.449) -^ - ^ [— (-) ~ - ^ ^-
4тг !_<?r0 \r0/ <?w Po drx \rj dn
4тг L Го <?и p0 r\ В
We recall that r2 = (x - f J + (y - туJ + (z - £J and that r\ = (x - f'J +
/ у _ ^'J + B — £ 'J, with (x, y, z) given as a point on the sphere in the
present discussion. On introducing coordinate systems with the origins at
(£>Ч> О an^ (£'>T?'> ^')» we immediately conclude that
G.450) ^ = cos 60 ,
G.451) -^ = cos ^ ,
respectively. The angle в0 is the angle between the exterior unit normal
vector to the sphere at the point P = (x, y, z) and the vector extending from
the source point Po = (f, 17, £) to P. Similarly, вг is the angle between the
normal vector at P and the vector from Pq = (f', 17', С') to P. Referring to
Figure 7.6 and using the law of cosines we obtain
G.453)
Using G.442) to replace px and r, by />0 and r0 in G.453) gives
G-454) cos в = a2 + (a2/P°)ro-(fl4/Po) = Po + ro-fl2
1 2a(a/p)r 2p0r0
Combining these results we conclude that
G455^ дЛ
2аг0 i + ?^lp;A 2p0r0
the solution formula G.448) reduces to
496 GREEN'S FUNCTIONS
G.456)
By transforming to spherical coordinates with center at the origin, the
second integral in G.456) can be expressed as, with у = <POP0,
G.457) JLff [f^Wf rrtf-Pl)B ****»_
Atra J Jsg L rl J 4тг Jo Jo [a2 - 2ap0 cos у + p2]3'2 *
This expression is known as Poisson's integral for the sphere and can be
obtained by using separation of variables for Laplace's equation in the
sphere. It represents the solution of the Dirichlet problem for Laplace's
equation, that is, G.446), G.447) with F = 0.
If we put F = 0 in G.456) and evaluate и at the origin, we obtain
G.458) «°>Wtbl
since Po = 0 and r0 = a in that case. Now Air a2 equals the surface area of the
sphere and В is the value of и on the surface of the sphere. Thus и at the
center of the sphere equals the average of its values on the surface of the
sphere. This mean value property is valid for harmonic functions (i.e.,
solutions of Laplace's equation) in both two and three dimensions. It also
arises in discrete formulations of Laplace's equation as was seen in Section
1.3. This property can be derived in a more direct manner using Green's
theorem as shown in the exercises.
The preceding method of inversion used to obtain Green's function for
the Dirichlet problem does not work for the case of Neumann's problem or
the third boundary value problem for the sphere. Similarly, it cannot be
applied to other elliptic equations or, for that matter, to hyperbolic or
parabolic equations. For example, we consider the reduced wave equation in
three dimensions
G.459) V2m + k2u = uxx + uyy + u2Z + k2u - 0
in a sphere and try to use inversion to find the Green's function for the first
boundary value problem. As shown in Example 6.13, the free space Green's
function for this problem is
G.460) KF
1
4тгг
ikr
EXERCISES FOR CHAPTER 7 497
with r2 = (x - U? + (у - vf + О - £J- This function satisfies the (three-
dimensional) radiation condition of Sommerfeld
G.461) Um
The Green's function К is sought in the form
G.462) *-
r0 and rt defined as in the preceding example and the constant с to be
determined. Clearly, с must have the same value given in the foregoing, that
is, с = -#/Po- However, on the sphere rx = (a/po)ro so that the exponentials
have different arguments and no cancellation occurs. It is not possible to
replace eikri by elk(p°/a)ri in which case both exponentials would be equal on
the sphere, since (l/rt) exp[ik(po/a)rl] is not a solution of the reduced wave
equation.
There are other techniques for obtaining Green's functions that we have
not discussed. For example, conformal mapping methods are extremely
useful in obtaining Green's functions for Laplace's equation in two dimen-
dimensions. However, they require a knowledge of the theory of complex
variables for their application. In addition, it is often possible to use known
Green's functions to convert a differential equation to an integral equation
and thereby construct other Green's functions. Again, it is rarely a simple
matter to solve the integral equations that result, and most often this must
be done approximately. Finally, there exist perturbation and asymptotic
methods for finding certain Green's functions. An indication of how to use
such methods is given in Chapters 9 and 10 which are devoted to approxima-
approximation methods.
EXERCISES FOR CHAPTER 7
Section 7.1
7-l.l. Show that for the elliptic case the Green's function K(x; f) de-
determined from G.12), G.13) is symmetric [that is, K(x; £) = K(£, x)].
Hint: Let и = K(x\ |) and w = K(x; £) in G.3).
J'1.2. Show that the Green's function for the hyperbolic problem G.26)-
(J.28) satisfies the equation K(x, t; £, т) = К(£9 -т; jc, -f).
№nt: Let и = K(x, -t, £, -т) and w - K(x, t; f, r) in G.17).
J-1.3. Show that the Green's function for the parabolic problem G.37)-
£p9) satisfies the equation K(x, t; £, т) = К(£, -т; х, -t).
Hint: Let и - K(x, -t; £, -т) and w = K(x, t; f, r) in G.32).
498 GREEN'S FUNCTIONS
7.1.4. Let F = 0 in G.11) and let В have delta function behavior with the
singular point at x ~ x on the boundary so that G.11) reduces to
and
Use these results to show that not only can the solution of G.1), G.2) with
5 = 0 and F Ф 0 be expressed as the superposition of the solutions of point
source or singularity problems, but the same can be done for G.1), G.2) if
F = 0 and В Ф 0. In the latter case the point sources lie on the boundary.
7.1.5. Use the expression G.25) to characterize the solution of the initial
and boundary value problem for the hyperbolic equation G.14) as a
superposition of solutions of point source problems, in the manner of
Exercise 7.1.4.
7.1.6. Use the expression G.36) to characterize the solution of the initial
and boundary value problem for the parabolic equation G.29) as a super-
superposition of solutions of point source problems.
7.1.7. Obtain an expression for the solution ы(£, т) of the hyperbolic
equation G.14) based on the Green's function obtained from G.42) and
G.44).
7.1.8. Using the Green's function determined from G.43) and G.45) obtain
an expression for the solution м(£, т) of the parabolic equation G.29).
7.1.9. Verify that w(£, т) is given as in G.63) if ft = ft = 0 and find the
appropriate form for w(f, т) if ax = a2 = 0.
7.1.10. Verify the expression for иЦ, т) in G.64) if ft = ft = 0 and obtain
the correct form for u( f, т) if ax - a2 - 0.
7.1.11. Consider the initial and boundary value problem for the (damped)
hyperbolic equation
putt + 2kput + Lu = pF,
where A is a positive constant and L is given as in G.1), with the initial and
boundary conditions G.15), G.16). Determine the appropriate Green's
function problem and obtain a solution formula for и that corresponds to
G.25).
7.1.12. Let
Mu = utt + c2uxxxx .
Express wMu - uMw in divergence form and use this result in the manner
EXERCISES FOR CHAPTER 7 499
done in the text for the hyperbolic equation G.14) to determine how a
Green's function K(x, t; £, т) should be constructed for the initial and
boundary value problem for Mu = F. Here w and w, are specified at t = 0
and, say, и and ux are given on the boundary of the interval 0 < x < L
7.1.13- Generalize the result of Exercise 7.1.12 to the equation
utt + c2V2V2m = F .
Hint: Use (8.308).
7.1.14. Consider the hyperbolic system
A/u = u, 4- Aux + Su = с , Л, В constant,
where A is a (real) symmetric matrix and В and с are real valued. With M*
as the adjoint of M and BT equal to the transpose of В show that
- uTM*w = wr[u, + Aux + Яи] - ur[-w, - Av/X + Srw]
has the form of a divergence expression. Use the formula to construct a
Green's matrix (or tensor) К that is a solution of
where / is the identity matrix, and use К to obtain a formula for the solution
of the Cauchy problem for the given equation for u with u(x, 0) = f(jc).
7.1.15. Determine the equation for the Green's function and the adjoint
boundary conditions for the one-dimensional form of G.65), G.66).
7.1.16. Let k(x\€) be a solution of ЬК = 8{х-£) [see G.65)] with a
homogeneous form of the boundary condition G.66) given on dG. Show
that K(x; £) = K(€; x), where К is the Green's function determined from
G.68) and G.71).
Section 7.2
7.2,1. Let the sequence of functions fk(x9 y9 z) be defined as
f VL I
where r2 = x2 + y2 + z\ Let <p(x, y, z) be a test function. Show that
00
lim (fk, <p) = lim J J J fk<p dx dy dz = <p@,0,0),
— 00
500 GREEN'S FUNCTIONS
so that the sequence fk converges to the delta function 8(x, y, z) in three
dimensions.
7.2.2. Show that the functions
О, |,|
where x is a one-, two- or three-dimensional variable, are test functions.
That is, they are infinitely differentiable and vanish outside the region
7.2.3. (a) Show that the generalized derivative of |jc| is sgn jc, the signum
function, defined as sgn x = -1 for x < 0 and sgn x = 1 for x > 0.
(b) Show that the generalized derivative of sgn x equals 28{x).
7.2.4. Show that xn8{m){x) = 0 if m < n and if m > n we have xn8{m\x) =
(-l)n[m\/(m-n)\]8im-n\x).
7.2.5. If fa(x) is the power function defined in G.113), show that/X*) =
afa-\(X) and Xfa(X) =/a + lW-
7.2.6. Show that xgt(x) = 1 and that xgn(x) = £„_!(*) for n > 2, where ^п(л:)
is the generalized power function G.114), G.115).
7.2.7. Consider the product xhn(x), with hn(x) defined as in G.117).
Express this product as (xhn(x), <p(x)) = (An, x<p). Use G.118) with n=2
and <p replaced by x(p to conclude that xh2{x) ¥>hx(x) unless we choose two
different constants in the definition G.117).
7.2.8. Show that h'n(x) = -nhH+l(x), for hn(x) defined by G.117).
7.2.9. Determine the generalized derivative of the following functions:
(b) /(jc) = x28(x).
(c) f(x) = ex8 \x + 3) - H(x) cos2 jc.
(e) f(x) = e*{
7.2.10. Obtain the formal Fourier sine and cosine series of the delta
function 8(x — jc0),
7rkx
and
(тгкхЛ (ъкх^
1 2 v /■"kxo\ (ттк
Six - *0) = у + у 2 cos^-pj cos(
EXEgCISES FOR CHAPTER 7 501
where 0 < x, x0 < /. Show that these series are convergent in the generalized
sense.
Hint: Consider test functions <p(x) that vanish outside a closed subinterval of
@, /)• show that the sine and cosine series converge weakly to <p(x0).
7 2.11. Express 8(x2 - a2) as a sum of delta functions in the manner of
G.168).
7.2.12. Express S[sinjc] as a series of delta functions by using G.168).
7.2.13. Show that the sequence of functions
tends to 8(x) as €->0.
7.2.14. Show that
lim = 8(x)
in the sense of weak convergence.
7.2.15. Let f(x) be a positive continuous function in [a, b] that vanishes
outside [a, b] and has unit area under the curve. Show that
lim -/l-l = i
«-►o+ e \6/
in the weak sense.
7.2.16. Show that 8'(-x) = -8'(x).
7.2.17. Let f(x) be defined as 8(x) in the interval - тг < x < к and extend it
as a periodic function of period 2 тт. Show that the extended function f(x)
can be expressed as
00
/(*)= X 8[х-2ктг]
and that it has the generalized (complex) Fourier series
f(x) = — 2) exp[*7cx].
7.2.18. Express the principal value integral in G.114) as
ф) - »(-x) ^
show that the generalized derivative of ^(д:) = log |jc| is given by the
generalized function g^x) = l/x.
502 GREEN'S FUNCTIONS
7.2.19. Nothing that x8(x) - 0, show that
= @, <p) - (x8(x) l><p) = («(*), <p) = «p@)
0
where ga(x) = \lx is the generalized function defined in equation G.114)
Conclude from this that if the product of the generalized functions x8{x) and
\lx were defined, we would have the result 0 = 8{x).
7.2.20. Show that the function F(x) in Example 7.3 can be expressed as
0 00
F(x) = 1+ | 2(- l)n+1H(nl - x) + E 2(- l)nH(x - nl) .
7.2.21. Given the divergent integral
[xMx)dx, -2<a<-l,
where q>@) Ф 0, Hadamard defines its finite part to be
(We do not consider his definition of the finite part for the case where
a<-2 and nonintegral.) Use the generalized function fa(x) defined in
G.112) to obtain the result
dx = - ^ JT jT V(x) dx + £L «c).
Integrate by parts once in Hadamard's finite part formula to show that the
two expressions for the value of the divergent integral are equivalent.
7.2.22. Determine the finite part of the integral
7.2.23. Given a (piecewise smooth) surface 5, consider the generalized
function F(x, y, z)S(S) that vanishes at all points P not on the surface 5 and
has the property that
The function F8(S) is called a single layer distribution and represents a
generalization of the one-dimensional Dirac delta function. Show that if
f(x, y, z) is a piecewise smooth function with a jump discontinuity across the
surface S, then
FOR CHAPTER 7 503
ax
P0S
where n is a unit normal to the surface S and [/]5 is the jump in / across S,
given as the difference between the value of/on the side of 5 that n points
to and the side it points away from. Also, obtain the corresponding formula
for the other two partial derivatives of /.
Hint: Use the definition of the generalized (partial) derivative of/and the
divergence theorem.
7.2.24. With the single layer distribution F8(S) defined as in Exercise
7.2.23, the double layer distribution is defined as -d/dn[F8(S)]9 where dldn
is the normal derivative on S and we have
This represents a generalization of -S'(*)- Show that
a2f _ a2f
ах2 dx2
P0S dX
and
Single and double layer distributions can be used to describe some of the
results in Chapter 6 relating to concentrated source terms in the language of
the theory of generalized functions.
7.2.25. Verify the following results.
If
(a) T~ 1 exp[«Ajc] exp[-/A*] dx = 5(A - A).
Lit J —°°
(c) - f
V 7 7Г Jo
- |ft cos Ajc cos Ajc dx = S( A - A) , A, A >0.
sin Ajc sin Ajc dx - 8( A - A) , A, A ^ 0.
Jo
Show that these results can be interpreted as orthogonality condi-
conditions for the functions arising in the Fourier transform, the Fourier
cosine transform, and the Fourier sine transform. In part (a) the
Hermitian inner product must be used.
Section 7.3
7.3.1. Obtain the Green's function G.207) for the problem G.206) and
G.196) directly—that is, not as a limit of the function G.205). Use this
Green's function to solve the problem
504 GREEN'S FUNCTIONS
u"(x) = f(x), 0<x<l
и@) = u(l) = 0.
7.3.2. Solve the boundary value problem for the ordinary differential
equation
u"(x) - c2u{x) = -/(*) , 0<*</
using the Green's function G.205).
7.3.3. Determine the Green's function K(x; ij) that satisfies the equation
and the boundary conditions
7.3.4. Consider the problem of Exercise 7.3.3 with the boundary condition
at x = 0 replaced by
0
Sx
and determine the Green's function K(x; £). Show that K(x; ^) has no limit
as c—»0 and explain why this limit fails to exist.
7.3.5. Obtain the Green's function K(x; £) that satisfies the following
problem
aX
7.3.6. Find the Green's function £(x; £) determined from the following
problem
£) = 0, K(x; £) bounded at x = O.
7.3.7. Solve for the Green's function K(x; £)
505
gXERCISES FOR CHAPTER 7
7.3.8. Obtain a one-dimensional version of the formula G.11), for the
solution of the boundary value problem for u(x)
0<x <
where ax, o^, ft, ft are non-negative, ^ + ft > 0, a2 + ft > 0, p > 0, q > 0,
p>0, and F is a given function.
Hint: The solution formula is given in terms of the Green's function K(x; £)
determined from G.48), G.49).
7.3.9. Use the results of Exercises 7.3.3 and 7.3.8 to solve the following
problem
u"{x) - c2u(x) = ex , 0<jc</
w@) = 3, w'(/) = 10.
7.3.10. Obtain an eigenfunction expansion for the Green's function K(x; f)
determined from G.206) and G.196) and show that the result obtained is
the Fourier sine series representation of the function K(x; £) given in
G.207).
7.3.11. Show that the following Green's function problem has no solution.
дх
дх
Hint: Integrate from x - 0 to x = /.
7.3.12. Obtain the modified Green's function K(x; £) that satisfies the
problem
дх
дх
j)
the form
if* s
f
506 GREEN'S FUNCTIONS
(Note that К is determined up to an arbitrary constant.) Show how K(x; g)
may be used to solve the boundary value problem for u(x)
~(pu')'=f(x), 0<x<l
assuming this problem has a solution. Integrate the equation for u(x) from 0
to / to determine a condition for the solution to exist.
7.3.13. Construct a modified Green's function K(x; f) that satisfies the
problem
дх А' dx * •
Choose the constants A and Л such that the problem for К has a solution.
Then show how К can be used to solve the boundary value problem for u{x)
given in Exercise 7.3.12.
7.3.14. Verify that G.299) is the solution of G.292), G.293).
7.3.15. Verify that G.313) is the solution of G.311) with Nk(T) = 0.
7.3.16. Show that if the Green's function K(x; £) determined from G.178),
G.179) has the eigenfunction expansion G.184), then the Green's function
K(x; £) determined from the problem
with the boundary condition G.179) has the expansion
k=\ лк~ А
(Note that this Green's function is not defined in the preceding form if
A-A,.)
7.3.17. Use the result of Exercise 7.3.16 to obtain the Green's function
K(xy у; £,7)) for the elliptic equation
V2K + KK = -8(x - S)8(y - n)
in the rectangle 0<x</, 0<y<l with the Dirichlet boundary condition
G.217) from the result G.238).
7.3.18. Construct the Green's function K(x, y; £, t)) for Laplace's equation
in the rectangle 0<x</, 0<y<t, where К satisfies G.216) and the
boundary conditions
EXERCISES FOR CHAPTER 7 507
= 0> 0<X<l,
)=0, 0<y<t.
oX
Use the method presented in the discussion following G.189).
7.3.19. Obtain the Green's function К for Laplace's equation in a disk
qU r :s R with the Dirichlet boundary condition К = 0 on r = Л. Express the
Laplacian in polar coordinates and obtain the equation for K(r, в; р,(р),
дг2 г дг г2 дв2 r
where 0<r, p<R and 0^0, <p ^ 2тг, and the right side of the equation is
the two-dimensional delta function in polar coordinates. Construct an
eigenfunction expansion of К based on the eigenvalue problem for the
operator -д21дв with periodic boundary conditions; that is, — M"@) =
ХМ(в) (-тг < 0 < тг) with М(-*тг) = М(тт) and M'(-if) - M'(ir). Obtain an
equation of the general form G.215) in the radial variable and solve it
subject to the conditions that the solution is bounded at r = 0 and vanishes
at r = R.
7.3.20. Solve the problem of Exercise 7.3.19 if (the disk is replaced by an
annular region Rx<r<R2 and К vanishes on the boundary.
7.3.21. Obtain the Green's function in the disk 0<r<7? for the reduced
wave equation with a Dirichlet boundary condition. That is, the Green's
function К satisfies the equation
V2K + k2K=-8 , r<R
and the boundary condition К = 0 on r = R (here 8 is the two-dimensional
delta function). Proceed as in Exercise 7.3.19, the only difference being that
the equation in the radial variable now has the form of Bessel's equation.
Two independent solutions of Bessel's equation of order n are Jn(z) and
^„(z), the Bessel and Neumann functions, respectively.
7.3.22. Consider the modified Green's functions К and К for the Neumann
problem for Laplace's equation in the disk 0^ r < Л. Construct a function
f(r) that is smooth in the disk and satisfies the condition f'(R) = -U2irR.
^se Лг) to transform the problem for X to a new problem with a
homogeneous boundary condition. Discuss the relationship between this
new problem and that for the function K.
7.3.23. Obtain the form of the eigenfunction expansion G.300) for
^(*> t; £, T) if Ao = 0 is an eigenvalue.
7.3.24. Obtain the form of the expansion G.314) if Ao = 0 is an eigenvalue.
7.3.25. Determine the Green's function for^ the wave equation utt - uxx -
uyy = 0 in the rectangle 0<x<l, 0<y<[ with the boundary condition
508 GREEN'S FUNCTIONS
м = 0оп the rectangle. That is, find the solution of G.288)-G.290) with
p - p = 1 and q = 0 in G.288) and a = 1, p = 0 in the boundary condition
G.290).
7.3.26. Construct the Green's function for the one-dimensional wave equa-
equation utt - uxx = 0 in the interval 0 < x < / with the following boundary
conditions:
(a) и@,0 = и(Л 0 = 0, '>0;
(b) m@,0 = mx(U) = 0, *>0;
(c) их@,0 = и,(/,0 = 0, f>0.
7.3.27. Use the Green's functions determined in Exercise 7.3.26 to write
down the solution of the wave equation utt = uxx in the interval 0 < x < /,
with the initial data w(x, 0) = sin x, w,(x,0) = l, and the boundary condi-
conditions:
(a) ii@,0-e"', «(/,0 = 0, t>0;
(b) и@,0 = 0, ux(l,t)
(c) iix@,0 = sinf, 11,G
7.3.28. Verify that G.314) is a generalized solution of G.307).
7.3.29. Construct the Green's function for the one-dimensional heat equa-
equation ut — uxx in the interval 0 < x < I with the following boundary conditions:
(a) ii@,0 = и(/, 0 = 0, t>0;
(b) и@,0 = mx(U) = 0,
(c) мя@, 0 = мх(/, 0 = 0,
7.3.30. Use the Green's functions obtained in Exercise 7.3.29 to solve the
initial and boundary value problems for the heat equation ut = uxx in the
interval 0 < x < I with u(x, 0) = 1, and the boundary data:
(a) и@,0 = е-', м(/,0 = 0, *>0;
(b) m@, 0 = sin t, «,(/, 0 = 0, t > 0;
(c) nx(o,o = i, «,(/,0 = 0, ;>o.
7.3.31. Determine the Green's function for Laplace's equation in the cube
0<л:</, 0<у</, 0<z</ with Dirichlet conditions on the boundary. Use
the eigenfunctions for the rectangle obtained in Example 7.6.
Section 7.4
7.4.1. Use the two-dimensional Fourier transform to obtain the Green's
function G.323).
7.4.2. Use the three-dimensional Fourier transform to obtain the Green's
function G.324).
7.4.3. Apply the Fourier sine transform to obtain the Green's function for
the one-dimensional heat equation ut = с ихх in the semi-infinite interval
0 < x < oo with u(x, t) specified at x = 0 and t = 0.
7.4.4. Apply the Fourier cosine transform to obtain the Green's function for
ut = c2uxx in 0 < x < oo, t > 0 with u(x, 0) and ux@, t) specified.
FOR CHAPTER 7 509
7.4.5- Use the general solution of the Cauchy problem for the wave
equation in one dimension to obtain the Green's function К as given in
G.350).
7.4.6. Apply the formula G.40) to obtain the solution of the Cauchy
problem for the wave equation utt = c2V2w in three dimensions if the initial
data are u(x, y, z, 0) = 0 and w,(x, y, z, 0) =/. The solution has the form
given in E.156).
74.7. Obtain the result E.167) for the Cauchy problem for the two-
dimensional wave equation utt = c2V2u if u(x, y, 0) = 0 and ut(x, y, 0) = /,
using the Green's function G.347) and the formula G.40).
7.4.8. Use the Green's functions for the wave equation in one, two, and
three dimensions obtained in Example 7.9 to determine domains of depen-
dependence for the solutions of the Cauchy problem for the inhomogeneous wave
equation in each of the three cases.
7.4.9. The Green's function for the Klein-Gordon equation utt - y2V2u +
c2u = 0 in three dimensions is given as
«*->-*■
where r2 = (x- f J + (y - rjf + (z - £ J, as can be shown by using the
Fourier transform. Let r2 = у2(т - tf - r2 in the Klein-Gordon equation
and obtain the equation
where й = u{f) [see F.213)]. Show that this equation has the nonsingular
solution
where Jx is the Bessel function of order one. Argue that the preceding
expression for the Green's function К is reasonable in view of the fact that it
reduces to the Green's function for the wave equation when с = 0 and has
the same singularity on the characteristic cone.
7'4.10. Apply the discussion given in Exercise 7.4.9 to obtain the Green's
function К for the hyperbolic equation utt - y2V2u -c2u = Q in three dimen-
dimensions.
Hint: It can be obtained by replacing с by ic in the Green's function for
Exercise 7.4.9.
510 GREEN'S FUNCTIONS
7.4.11. Obtain the Green's function K(x, y, t\ f, 17, т) for the two-dimen-
two-dimensional Klein-Gordon equation utt — y2V2u + c2u — 0 in the form
1 cos[(c/y)Vy2(r - if - p2]
where p2 = (jc - f J + (у -17J, by following the procedure given in Exercise
7.4.9.
7.4.12. Obtain the Green's function K(x, y, t; £, 17, r) for the two-dimen-
two-dimensional hyperbolic equation utt — y2V2u — c2u = 0 in the form
co»h[(c;y)Vy'(T-<)'-p']
where p2 = (x - f J + (у -17J, by following the method presented in Exer-
Exercise 7.4.9.
7.4.13. Obtain the (free space) Green's function for the telegrapher's
equation A.88).
Hint: Eliminate the first derivative term in the equation for the Green's
function.
7.4.14. Obtain the Green's function appropriate for the Cauchy problem for
Schrodinger's equation
л Эй h2 д2и л
dt 2m dx
where ft is Planck's constant, m is the mass, and i = V—1.
Hint: Use the Fourier transform.
7.4.15. Show that the solution K(x\ £) of the problem
K(x; f)-^-0 as
is given as
(see Examples 5.1 and 6.13).
7.4.16. Obtain the solution K(x; £,) of the problem
lim
FOR CHAPTER 7 511
(that is, K{x\ i) satisfies the radiation condition at infinity) in the form
' 2k
(see Example 6.13).
7.4.17. Solve the following problems for K(x, £)
ax
(a) K@; £) = 0, K(x; ^)-*0 as x
7.4.18. Obtain the Green's function for the following problem
7.4.19. Determine the Green's function K(x, y; ij, 17) that satisfies the fol-
following problem
d2K , d2K .. ,.e/ . @<x, £<l
17 + lJFm~t(x'my~vh l
К = 0 at л: = 0 and x = /,
^C is bounded as |_y| —> oo .
Hint: Use the eigenfunctions for the interval 0 < x < I with zero boundary
conditions.
7.4.20. Show that the following problem has no solution.
ж
—- = 0 at x = 0 and x = /,
if^O as Ы->«>.
Hint: Use the cosine eigenfunctions for 0 < x < I.
7.4.21. Obtain the Green's function K(x, y;^v)
^ = 0 onx = 0, x =
K-+0 as b|->°°.
Hint: Use the eigenfunctions for the interval 0 < x < L
512 GREEN'S FUNCTIONS
7.4.22. Set up a two-dimensional version of the problem in Example
7.11—that is, drop the у dependence in the problem—and solve for the
Green's function K{x, z; £).
7.4.23. Solve for the Green's function K(x, y, z; f, т/, f)
*к - гк- -8(x - my „жг о, {Л5z>
A:-0 onjc-0, x = /
Я"—»0 as |y|—»°°, \z\ —>oo .
Hint: Use the free space Green's function found in Example 6.13.
Section 7.5
7.5.1. Use the Green's functions obtained in Example 7.12 and appropriate
forms of the solution formulas obtained in Section 7.1 to construct the
solution of Laplace's equation V2u = 0 in the half-space z > 0 if the following
boundary conditions are given:
(a) u(x, y,0);=/(x, y),
me
in the half-plane у >0 for the Dirichlet problem, in the form
7.5.2. Use the method of images to obtain the Green's function for V2m = 0
K(x, y;£,i7) = - — log? ; 0<у,т;<оо, -oo< X,
Ztt p
where p2- (x- £J + (y - r/J and p2 = (jc - £J + (у + т;J. Use this
Green's function to obtain the solution E.71) of the boundary value
problem E.62)-E.64).
7.5.3. Apply the method of images to obtain the Green's function for
V2m = 0 in the half-plane у > 0 for the Neumann problem, in the form
K(x9y;(,ri) = -—log pp, 0<у,ч<оо,-оо<х,£<оо,
where p and p are defined as in ^xercise 7.5.2. Use this Green's function to
obtain the solution E.79) of the problem E.62), E.72).
7.5.4. Obtain the Green's function that corresponds to G.422) in the
two-dimensional case.
7.5.5. Use the method of images to obtain the Green's function for Lap-
Laplace's equation in a composite medium in three dimensions. That is,
determine the function K(x, v, z; f, v> О that satisfies the equation
FOR CHAPTER 7 513
pV2K =-8(x-g)8(y- V)8(z -£),
with -°°<*, f, y9 17, z<oo (z#0), the condition
^-►Oas |z|-h>oo,
as well as the jump conditions at z = 0,
A^ continuous at z = 0 ,
dK
P T~ continuous at z = 0 ,
aZ
where p = A for z<0, p=p2 for z >0, and p2 and p2 are constants.
Hint: Let K= Кг for z < 0 and K= K2 for z > 0. Since the source point lies
in z > 0 use the free space Green's function and its image for z > 0 and the
free space Green's function with a source at (£, 17, £) in z <0. Then apply
the matching conditions.
7.5.6. Use the procedure of Example 7.14 to construct a Green's function
for Laplace's equation V2w = 0 in the region — °o<x<<x>, — co<y<<x>9
0< z < I for the case of Dirichlet boundary conditions.
7.5-7. Apply the method of images to construct the Green's function for the
Helmholtz equation V2w + k2u = 0 in the half-space z > 0, in the case where
K(x, y, z\ g, 17, £) satisfies one of the following boundary conditions:
(a) AT = 0atz = 0;
(b) —-0 at z = 0;
OZ
йК
(c) ^-hK = 0atz = 0,
dz
with -oo<jc, y, f, 77<°o and where К satisfies the radiation condition at
infinity in all cases.
Hint: Use Example 6.13.
7.5.8. Construct the two-dimensional forms of the Green's functions ob-
obtained in Exercise 7.5.7 for the Helmholtz equation V2m + k2u = 0 in the
half-plane у >0.
7.5.9. Construct the Green's function for V2u - c2u - 0 in the half-space
z > 0 if the Green's function K(x, y9 z; £, 17, £) satisfies the following boun-
boundary conditions
(a) K = 0atz = 0,
(b)~0atz-0,
for all -oo<JC) y<co and K->0 as z-^00.
Hint: Use Example 6.13 and the method of images.
7*5.10. Carry out the solution of the two-dimensional version of Exercise
7-5.9 in the half-plane у > 0.
514 GREEN'S FUNCTIONS
7.5.11. Use the Green's function G.432) for the wave equation in the
semi-infinite interval 0<x<», to solve the following initial and boundary
value problem.
utt - c2uxx = 0,
u(x, 0) = u£x, 0) = 0,
w(<U) = gW, *>0.
7.5.12. Solve the following problem by use of the Green's function G.433)
u(x, 0) = ut{x, 0) = 0 , x > 0 ,
7.5.13. Verify that G.433) is indeed the appropriate Green's function for
the modified telegrapher's equation G.371) in the case of Neumann boun-
boundary conditions.
7.5.14. Noting the result of Exercise 7.2.17,
ik*
2тг £ 8[х-2ктг]= 2 e
k= — » k— — oo
multiply both sides by a Fourier transformable function <p(x) and conclude
upon integrating that
Jfe=-o
<р[2ктг]= 2 F[k]
where F[k] is the Fourier transform [see E.10)] of <p{x). This result is
known as Poisson's summation formula.
7.5.15. Use the Poisson summation formula (Exercise 7.5.14) to show the
equivalence of the Green's function expression G.439) for the heat equation
and that obtained by the method of eigenfunction expansions.
7.5.16. Use the method of Example 7.14 to construct the Green's function
К for the heat equation in a finite interval if
(a) dKldx = 0 at x = 0 and x = /,
(b) K = 0 at x = 0 and dKldx = 0 at x = /.
7.5.17. Apply the method of images to construct a Green's function for the
heat equation in the semi-infinite interval 0 < x < oo with Dirichlet boundary
conditions and use it to solve the following problem.
ut-c2uxx=0, 0<x<oo,r>0,
u(x,0)=f(x), 0<Ar<oo,
"@, t) = g(t) , 0<r<oo.
FOR CHAPTER 7 515
7 5.18. Use the inversion process of Example 7.15 to obtain the Green's
function for the Dirichlet problem for Laplace's equation in a disk x2 + у <
a -
7.5.19. Verify the result G.457).
7.5.20. Obtain Poisson's integral for the circle [see G.457)] by using the
Green's function obtained in Exercise 7.5.18.
7.5.21. Use the method of images to determine the appropriate Green's
function for Laplace's equation V2w = 0 in the quadrant 0<x<oo90<y<<»
with the following boundary conditions for K(x, y; £,??):
(a) К = 0 on x'= 0, у >0 and у = 0, x >0;
(b) К = 0 on x = 0, у > 0; dKldy = 0 on у = 0, x > 0;
(c) dKldx = 0 on x = 0, у >0, <?AT/<?)> = 0 on у = 0, jc >0.
7.5.22. Apply the method of images of construct the Green's function for
Laplace's equation within the hemisphere x2 + y2 + z2 < a2, z ^ 0, with
Dirichlet boundary conditions.
Hint: Use inversion in the sphere and reflection in the plane z-0.
7.5.23. Obtain the Green's function for the Dirichlet problem for Laplace's
equation within the quarter-circle x2 + y2 < a2, x > 0, у > 0.
Hint: Reflect in the x- and y-axis as in the quarter-plane problem and then
invert each source and image source in the circle.
7.5.24. Let и be a solution of Laplace's equation V2w = 0 in the region G
and let R be the interior of a circle or a sphere centered at Po and
completely contained within G.
(a) Use Green's theorem or the divergence theorem to show that
— d -0
dn " *
(b) Let К be the free space Green's function for Laplace's equation
with singular point at Po. Apply Green's theorem in the region
R to the harmonic function и and the Green's function K.
Deduce the mean value theorems
and
in three and two dimensions, respectively, where a is the radius
of the sphere and the circle with center at Po.
8
VARIATIONAL AND
OTHER METHODS
In this chapter we present a number of methods and results that either apply
to problems studied previously or to a new class of problems. We begin with
a presentation of a variational characterization of the eigenvalue problems
introduced in Chapter 4, and show how the variational approach enables us
to prove some of the properties of eigenvalues and eigenfunctions given in
that chapter. An important consequence of this approach is the Rayleigh-
Ritz method that yields approximate determinations of eigenvalues and
eigenfunctions and this method is considered next.
We continue with a discussion of the classical method of Riemann for
integrating linear hyperbolic equations of second order. This is followed by
a presentation of maximum and minimum principles for the diffusion and
Laplace's equations, and their consequences for uniqueness and continuous
dependence on data for problems for these equations. In addition, we
consider a positivity property for the telegrapher's equation.
We conclude with a discussion of some basic equations of mathematical
physics. These are the equations of fourth order governing the vibration of
rods and plates and the equations of fluid dynamics, Maxwell's equations of
electromagnetic theory, and the equations of elasticity theory, which are
systems of equations. Techniques for simplifying and solving these equations
will be considered.
8.1. VARIATIONAL PROPERTIES OF EIGENVALUES AND
EIGENFUNCTIONS
It has been demonstrated in the preceding chapters that eigenvalue prob-
problems play an important role in various methods for solving problems for
partial differential equations. Up to now we have determined the eigen-
eigenvalues and eigenfunctions by using separation of variables for partial
516
VARIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 517
differential equations and general solutions of ordinary differential equa-
equations for the Sturm-Liouville problem. In this section we show how the
eigenvalues and eigenfunctions can be specified by means of a variational
principle. Thereby, it is possible to prove some of the properties of the
eigenvalues and eigenfunctions given at the beginning of Chapter 4. Further-
Furthermore, the variational approach leads to a useful approximate method for
determining the first few eigenvalues and eigenfunctions. This method,
known as the Rayleigh-Ritz method, is discussed in Section 8.2.
To motivate the use of the variational principle, we begin by considering
the (self-adjoint) eigenvalue problem
(8.1) LM = -V- (pVM) + qM= kpM
for the function M(x) defined in the bounded region G. As in Sections 4.1
and 4.2, we assume that p(x) > 0 and p(x) > 0 in G, whereas q(x) > 0 in G.
The eigenfunction M(x) is assumed to satisfy the boundary condition on <?G,
(8.2)
dG
= 0,
where a(jc)>0 and /3(x)>0, whereas a + )8>0 on <?G. The boundary
conditions may be of the first, second, third or mixed kind, as described in
Section 4.1. (We use notation appropriate for the two- or three-dimensional
eigenvalue problem and occasionally indicate the appropriate forms for the
one-dimensional Sturm-Liouville problem.)
As was discussed in Section 4.2, we assume that there are a countable
infinity of eigenvalues kk{k = 1,2,...) that are real valued and non-nega-
non-negative. They are numbered according to increasing values as
(83) 0< kx < A2 < A3 < A4 < • • • < \k < • • • .
The associated eigenfunctions Mk(x) are assumed to form a complete
orthonormal set. That is,
where 5^ = 1 for k = j and 8kJ = 0fotk*j9 with*, j -1,2,3,.... Further-
Furthermore any function u(x) that is sufficiently smooth and satisfies the boundary
condition (8.2)—where M is replaced by u—can be expanded in a series of
eigenfunctions
The Fourier coefficients Nk are given as
51» VARIATIONAL AND OTHER METHODS
(8.6) Nk = (u(x),Mk(x)) = j jPuMkdv.
G
We assume that u(x) is normalized to unity so that
(8.7) \\u\\2 = (u,u) = ljpu2dx = l.
G
To proceed, we consider the energy integral
(8.8) E(u) = jj [uLu] dv = jj [p(Vuf + qu2] dv
G G
- pu — ds
JdG * дП
where D.35) has been used. Now u(x) satisfies the boundary condition
(8-9) au+fi%
BG
= 0.
Noting the formulation of the boundary conditions in section 4.1, the energy
integral becomes
(8.10) E(u) = j j [p(Vuf + qu2] dv + £ (|) pu2 ds .
G
We refer to E(u) as an energy integral on the basis of our discussion in
Section 6.8.
On inserting (8.5) into (8.8) we have
(8.11) E(u) = f f Г 2 NkMkLu\ <fc = X Nk\ \ [MkLu] dv
G G
G
= ]S Nkjj [мк(х) I NjLMjix)] dv
G
= 2Nk f f f i \М{р(х)Мк(хЩ(х)}] dv
G
= 2 aJ2 Щ(Мк, Mj)} = i xkN2k ,
since LM;- - klpMj and (Л#л, М7) = Skr Also,
G
dv
VAKIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 519
(8.12) 1 = (и, и) = ( 2 NkMk(x), 2 *W(
V 4 = 1 / = 1
= 1 / = 1
р(х)Мк(х)
^ JJ [р()к()
G
since (Mk9 Afy) = 5Л/. We have assumed that the interchanges .of summation,
differentiation, and integration used in (8.11), (8.12) are valid.
Noting (8.3) and (8.12), we obtain from (8.11)
2
(8.13) B(«)=ZAX
Thus for all admissible functions u(x) that satisfy (8.9) and (8.7) we have
£(u)>At. If u(x) satisfies the additional constraints
(8.14) (u, Mk) =
the first n - 1 coefficients in the series (8.5)—that is, Nl9 N2,. .., Nn_1—all
vanish. Then (8.11) yields
k=n k=n
on using (8.3) and (8.12). In addition, for all n,
(8.16) E[Mn] = \\MnLMn dv = kn\\pM2H dv = \n,
G G
since the Mn(x) are normalized by assumption.
In view of these results, the energy integral E(u) is greater than or equal
10 the smallest eigenvalue Ax associated with the eigenvalue problem (8.1),
(8-2). The minimum value Ax is assumed if и = Mx(x)y the first eigenfunc-
tion. If u{pc) satisfies the constraints (8.14), we have E(u)>\n and the
nunimum is assumed if и = Mn(x). Consequently, we may characterize the
eigenvalues and eigenfunctions Аи and Mn(x) in terms of the following
minimum problem. The nth eigenvalue of the problem (8.1), (8.2) is the
minimum value of the energy integral E(u), where the admissible functions
"(*) satisfy the conditions (8.7), (8.9), and (8.14). Ifn = l, the conditions
(8Д4) are absent. Further, among all admissible functions u(x), the minimum
v<due of E(u) is assumed if и is one of the eigenfunctions of (8.1), (8.2) and
we minimum value is equal to the corresponding eigenvalue.
520 VARIATIONAL AND OTHER METHODS
In a slight variation of this procedure, we drop the restriction that the
admissible functions u(x) are normalized. If w(x) is a function that satisfies
the boundary condition (8.9) and the constraints (8.14) (where и is replaced
by w), we can set
(8.17) ф) **>
Then u(x) satisfies all the conditions stated in the minimum problem
presented. We have
(8.18) E(u) = Ei-r^rr) = -^ E(w) = -^Щ- .
\\\w\\/ \\w\\2 (w,w)
With E(w) defined as in (8.10), the ratio E(w)/(w, w) is known as the
Rayleigh quotient. The minimum values of the Rayleigh quotient still
determine the eigenvalues, and these minima are assumed by the eigenfunc-
eigenfunctions. However, the resulting eigenf unctions are no longer normalized but
are determined up to an arbitrary multiplicative constant.
For the purpose of comparing the properties of eigenvalues arising in
different eigenvalue problems, our formulation of the variational problem is
unsatisfactory because it characterizes the nth eigenvalue for a given prob-
problem in terms of the first n — 1 eigenfunctions of the problem in view
of (8.14). That is, we must first determine the eigenfunctions
Mx(x)... , Mn_l(x) in order to specify the eigenvalue \n. It is of interest to
obtain a variational formulation that permits the determination of any
eigenvalue directly and independently of first having to solve any other
problem. Such an approach was developed by Courant and is known as the
maximum-minimum principle.
Given the eigenvalue problem (8.1), (8.2) we consider a collection of
(sufficiently smooth) functions {mk(x)}> к = 1,2,3,. .., defined in the
region G. These functions need not satisfy any specific conditions in G or on
the boundary dG. To determine the nth eigenvalue Аи, we introduce the
function w{x) that satisfies the boundary condition (8.9) and the л -1
constraints
(8.19) (w, mk) = j J pwmk dv = 0, к = 1,2,. . . , n - 1.
G
By varying over all admissible functions vv(jc), we minimize the Rayleigh
quotient E(w)/(w, w). For each set of functions m^x),. . . , mn-i(*) we
obtain a minimum for the Rayleigh quotient. We then vary these minima
over all possible sets of functions тг(х),. . . , тп_г(х). The maximum of
these minima is equal to the nth eigenvalue An. The maximum-minimum
principle can be stated as
VARIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 521
min -, —r J — К •
To prove this result, we note that on selecting w(x) as
n
(8.21) ф) = Еда
where the Mk(x) are the eigenfunctions for (8.1), (8.2), we may specify the
constants ck so that the n -1 constraints (8.19) are satisfied. In fact, we
have
n
(8.22) (w, mh) = S Cj(Mj, mk) = 0 , к = 1,. .. , w - 1.
This is an underdetermined system of homogeneous linear equations for the
л constants cJ9 and it has a solution with at least one of the c; remaining
arbitrary. Therefore, we can satisfy the additional condition
(8.23) (w, iv)=Sc^l
/=1
by choosing the undetermined constant(s) appropriately. Note that w(x)
satisfies the boundary condition (8.9) since each of the Mj(x) does so.
Therefore, w(x) is an admissible function for the variational problem.
We have, for the w(x) given above,
in view of (8.11) and (8.23), since
(8-25)
Therefore, for each set of functions mx{x),. . ., mn_x{x) we have found an
admissible w(x) for which the Rayleigh quotient is less than or equal to An.
Consequently, the maxima of all the minima of the Rayleigh quotient
cannot exceed An. However, if w(x) = Mn(x), the nth eigenfunction for the
Problem (8.1), (8.2), we have
(8.26)
°n using (8.16), so that the maximum-minimum principle (8.20) has been
verified.
522 VARIATIONAL AND OTHER METHODS
Example 8.1. The Eigenvalue Problem for a Square. In this example we
reconsider the eigenvalue problem for the Lapladan operator, that is,
(8.27) V2M + AM = Mxx + Myy + AM = 0
in the square 0 < x < тг and 0 < у < тг, with the Dirichlet boundary condition
(8.28) M(x,y)\dG~0.
The region G is the aforementioned square, and dG is given by the sides of
the square. The eigenvalues and eigenfunctions for the problem (8.27)
(8.28) were determined in Example 7.6. The first eigenvalue An was found
to be
(8.29) An=2,
and the associated normalization eigenfunction Mn{x, y) is
2
(8.30) Mn(x, y) = — sin x sin у .
тг
The appropriate energy integral for this problem is
(8.31) E(u) = [ [[u2x + u2y)dxdy,
and an admissible function u(x, y) must vanish on the boundary of the
square. Also, u(x, y) must be normalized so that
(8.32) (u9u)
Now if m(jc, y) = Mn(x, y), we already know that
л с™ c*
(8.33) (Mn,Mn) j sin2 x sin2 у dx dy = 1,
7Г JO JO
as can be verified directly. Also,
4
(8.34) E(Mn) = — I I [cos2 л: sin2 у + sin2 л: cos2 y]dxdy = 2,
so that E(Mn) = An, as was to be expected in view of (8.16).
As was shown, if u(x, y) is any admissible function other than Mn(x, y)>
we must have E(u) > An = 2. To see this in a particular case we consider the
function
523
vaKIational properties of eigenvalues and eigenfunctions
(8.35) и(х, у) = ^ ху(тг - x)(ir - у) .
^e coefficient 30/тг5 has been selected so that (8.32) is satisfied, and
u(x, У) clearly vanishes on the boundary of the square. We readily find that
(8.36)
E(u) = P [ [ [х2(тг - xJGT - 2yf + у\тг - у)\тг - 2xf] dx dy
= 2.03.
Thus E(w) = 2.03>£(Mn) = 2 as follows from (8.15). Furthermore, the
value 2.03 is a remarkably good approximation to the exact eigenvalue
д = 2 given that u{x, y) was chosen quite arbitrarily.
Although the given function u(x, y) yields an excellent approximation to
the eigenvalue An, it does not appear to be a good approximation to the
eigenfunction Mxx{xy y). In fact, if we measure the difference between
m(jc, y) and Mu(x, y) in the mean square norm, we obtain
(8.37) \\MU - u\\2 = [ [ [Mn - uf dxdy = 0.73 ,
so that j|Mn - u\\ = 0.85. The norm is not very small. This indicates that the
variational approach may be expected to yield a much better approximation
to the eigenvalues than to the eigenfunctions. The reason for this is
demonstrated below.
We have shown that subject to appropriate constraints the minima of the
Rayleigh quotient (8.18) are the eigenvalues of (8.1), (8.2) and the minima
are assumed when the minimizing functions are eigenfunctions. Reversing
this process, we now formulate a variational problem for the Rayleigh
quotient and show directly that the eigenvalues and eigenfunctions for (8.1),
(8.2) result from the minimization problem. The type of variational problem
considered depends on the boundary conditions for the eigenvalue problem
(8.1), (8.2). It has already been seen in (8.10) that the form of energy
integral E(u) varies with the boundary conditions (8.9).
In the case of Dirichlet boundary conditions (i.e., M-Q on dG) the
aPpropriate Rayleigh quotient is
Ь addition, in the case of Neumann boundary conditions (i.e., SM/dn = Q
°n dG) the Rayleigh quotient again has the form (8.38). However, for
524 VARIATIONAL AND OTHER METHOD^
boundary conditions of the third kind (i.e., dMIdn + hM = 0 on dG with
A = a//3) we obtain, in view of (8.10),
(839) Ш=jk2{(/lpiVwJ+qw2]dv+Lhpw2*}■
It is assumed that A > 0. In the case of boundary conditions of the mixed
kind and if S3 is not an empty set, the Rayleigh quotient has the form (8.39)
with the surface integral taken over 53.
The variational problem is given as follows. The function w that yields a
minimum for the variational problem
(8.40) -г—p = minimum ,
II HI
over all functions w that vanish on dG and satisfy the constraints
(8.41) (iv, Mt) = 0v, M2) = • ■ • = (w, Mn_t) = 0,
is—upon normalization-the nth eigenfunction Mn of the problem (8.1), (8.2)
with a = 1 and f$ =0 in (8.2). The nth successive minimum value of the
Rayleigh quotient (8.40) is the eigenvalue Xn that corresponds to Mn. The
eigenvalues X- are ordered as 0^ \t ^ A2^ • • • ^ Аи. For the eigenvalue
problem (8.1), (8.2) with boundary conditions of the second and third kinds,
there are no restrictions placed on the boundary values of admissible func-
functions w(x). In the case of mixed boundary conditions, we require that w = 0
on St.
To demonstrate this result we assume that a sufficiently smooth minimiz-
minimizing function for the variational problem exists. (This is difficult to prove, in
general.) Let the minimizing function be denoted by wn(x), and the admiss-
admissible functions w(x) be represented as
(8.42) w = wn + eW,
where e is a constant and W satisfies the same admissibility conditions as
w(x).
In terms of (8.42) the variational problem becomes
E{w) E{wn + eW)
(8-43) # =
As a function of the parameter e, the Rayleigh quotient in (8.43) assumes a
minimum value when € = 0, that is, when w = wn. Consequently, € = 0 must
be a critical or stationary point of the function £:(>v)/||>v|| , so that its
derivative with respect to e must vanish at e = 0. We have
VARIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 525
д
K + eW]f + q[wn + e
€ = 0
J €=0
2
IKII
Vwn VW + qwnW-
-П72 f hpwnWds = 0.
wm\\2 J*o
If we are dealing with the Dirichlet or Neumann problem, the surface
integral is absent in (8.44). Using D.29) and the divergence theorem gives
(8.45)
J J pVwn VWdv - -J J WV-(pVwn) dv + J J V-(pWVwn) dv
Introducing (8.45) into (8.44) yields
(8.46) \\w{-V-(pVwn
qWn -
dv
For the eigenvalue problem with Dirichlet boundary conditions, the
surface integral in (8.46) is zero since W=0on the boundary. From the
arbitrariness of W in the region G (see Exercises 8.1.7 and 8.1.8 for some
additional details regarding this point) we conclude that the bracketed term
ui the volume integral must vanish. Therefore, wn satisfies the equation
526 VARIATIONAL AND OTHER METHODS
(8.47) -V. (pVwJ + qwn = [^J]pW- ,
and wn = 0 on dG. This implies that
(*лял E[Wn]-x
(8.48) |к„|ГЛ-
and wn(x)/\\wn(x)\\ = Mn(x), in view of (8.1), (8.2).
In the case of Neumann conditions for the eigenvalue problem, we must
set A = 0 in the surface integral in (8.44). The arbitrariness of W again
implies that the integrals over G and dG in (8.46) must vanish separately.
Thus wn(x) again is a solution of (8.47) but the vanishing of the surface
integral leads to the natural boundary condition
(8.49)
dw
dn
dG
= 0.
We conclude that wn/||wn(jc)|| = Afn, the nth eigenfunction for the problem
(8.1) with a =0 and /3 = 1 in (8.2), and that (8.48) is again valid.
For the third boundary value problem (i.e., with ft^O), the arbitrariness
of W implies that both integrals in (8.46) vanish separately. The natural
boundary condition for wn becomes
dw
(8.50) -^ + /nv,
and wn satisfies (8.47). Consequently, и>л/||и>л(л;)|| - Mn, the nth eigenfunc-
eigenfunction for (8.1), with a/p = h in (8.2), and (8.48) yields the eigenvalue.
In the case of mixed boundary conditions we conclude as before that wn
satisfies (8.47). On Sx where Dirichlet conditions are given we have wn =0.
On 52, wn satisfies the natural boundary condition (8.49) (where dG is
replaced by S2). On 53, wn satisfies the natural boundary condition (8.50)
(where dG is replaced by S3).
We observe from (8.42) that ||w - wj| = 0(e) while for the approximate
eigenvalue A = £(н>)/||н>||2 we have A - An = 0(e2), since the O(e) term in
the expansion of A in powers of e vanishes in view of (8.44). This accounts
for the fact that we consistently obtain better approximations for the
eigenvalues than for the eigenfunctions when using the variational method.
This concludes our direct demonstration that the eigenvalues and eigen-
eigenfunctions for (8.1), (8.2) can be determined from a variational problem.
Since the Rayleigh quotient is non-negative, so is its minimum and we must
have A, > 0. The ordering of the eigenvalues is considered in Exercise 8.1.6.
The preceding variational problem requires that the first n - 1 eigenfunc-
VARIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 527
tions must be known if we are to determine the nth eigenvalue and
eigenfunction. However, Couranfs maximum-minimum principle permits
the determination of each eigenvalue independently of the other (lesser)
eigenvalues. Restricting ourselves to the variational eigenvalue problem with
pirichlet boundary conditions, we now show that the eigenvalues Ля-»о° as
п_»оо. Although the variational problem shows (see Exercise 8.1.6) that the
eigenvalues can be ordered as in (8.3), it may happen that the eigenvalues
have a finite limit point and that infinitely many (independent) eigenfunc-
eigenfunctions can be associated with a single eigenvalue. However, Ля-»о° implies
that each eigenvalue has finite multiplicity; that is, there are only a finite
number of linearly independent eigenfunctions associated with each distinct
eigenvalue.
The use of the maximum-minimum principle permits the comparison of
variational problems with different sets of coefficients in the Rayleigh
quotients and problems defined over different regions G. We assume the
coefficients p, p, and q that occur in the Rayleigh quotient have maximum
and minimum values in the region G together with its boundary dG and still
require that p > 0, p > 0, and q >: 0 in G. Let the constants pM, pM, and qM
represent the maximum values and pm, pm, and qm represent the minimum
values of the functions p, p, and q, respectively. (We assume that pm and pm
are not zero.) Further, the admissible functions w(x) will be required to
vanish on dG and to satisfy the constraints (8.19).
On examining the Rayleigh quotient appropriate for the Dirichlet
problem
(8.51)
jj[p(VWf
I I pw2 dv
it is clear that if we replace the functions p, p, and q by the constants pM, pm
and Qm> we obtain a new Rayleigh quotient, denoted by #M0v)/||vv||^ for
which
(8.52) E№ ВД
NIL " Ml2
for each admissible function w. Now the inner product that occurs in the
constraints (8.19) for the new problem with constant coefficients can be
expressed as
(8.53) (vv, mk) = / / Pmwmk dv = j j pw{mk ^) dv .
528 VARIATIONAL AND OTHER METH()DS
Thus the constraints for the new problem must be given in terms of the s
of functions mk = (pmlp)mk if we want the new problem to have the san?
set of constraints. But in the maximum-minimum principle the set f
functions mk(x) ranges over all sufficiently smooth functions and, therefor
so dees the set thk(x). Consequently, we may conclude that max-min of
Ем(и>)/||и>||^ ls greater than or equal to the max-min of E(w)/||vv||2.
In a similar fashion, if we replace the functions /?, p, and q in (8.51) b
the constants pm, pM, and qm, we obtain a modified Rayleigh quotient
Em(w)f\\w\\2M for which
(8.54)
The foregoing procedure may be applied to the constraints (8.19), and we
conclude that the max-min of £т(и>)/||и>||^ does not exceed the max-min
of E(w)/||w||2 Since for the set of admissible functions w, vanishing on dG
and satisfying the constraints (8.19), each of the aforementioned max-min
yields the nth eigenvalue for the given variational problem, we conclude that
/о <<л \ (OT) <=- \ «^ \ t^)
where A^m), An, and A^M) are the nth eigenvalues associated with the
variational problem for the Rayleigh quotients £m(w)/||w>||^, £(w)/||w||2,
and Ям(и>)/|И|2 , respectively.
Next we consider changes in the region G. Suppose we consider a
subregion G contained in G (a portion of the boundary dG may coincide
with dG). Let £(w)/||w||2 be the Rayleigh quotient for the variational
problem in G, and suppose we require that all admissible functions w(x) in
addition to vanishing on dG and satisfying the constraints (8.19), must
vanish on dG and in that part of G exterior to G. Since the additional
requirement on the functions w(x) effectively reduces the number of admiss-
admissible functions for the variational problem the maximum-minimum for the
new problem cannot be smaller than the max-min An attained when the
added restriction is absent. However, the new set of admissible functions is
precisely that which is appropriate for the variational problem in the
subregion G. The max-min for the variational problem in G is An the nth
eigenvalue for that problem. In view of the preceding observation, we
obtain
(8.56) АЛ<А„.
To complete our discussion we consider two rectangular regions RM and
Rm. (In three dimensions these are rectangular boxes.) The region Rm
contains the region G within it, whereas the region Rm is completely
contained within G. First we consider the max-min variational problem for
VAg!ATlONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 529
the Rayleigh quotient EM(w)/\\w\\2m, with the region G replaced by Rm but
the problem otherwise unchanged. In view of (8.56) we conclude that
(8.57) A^} < A<M) ,
where AJ;M) and A^A/) are the wth eigenvalues for the problem associated with
E (wO/lMlL over t^e regi°ns G and Rm, respectively. Next, we consider
the max-min variational problem for £m(w)/||n>||^ with the region G
replaced by RMJ but the problem otherwise unchanged- Using (8.56) gives
(8.58) ^i - A«w> >
where A* and A^m) are the nth eigenvalues for the problem associated with
^(^VIMIm over the regions RM and G, respectively.
Combining (8.55), (8.57), and (8.58) yields the set of inequalities
(8.59) A^m) < Aj,m) < А„ < A<M) :
The eigenvalues A*m) and A**° can be determined exactly be separation of
variables since the equations have constant coefficients and the regions are
rectangular. This has already been carried out for a special two-dimensional
problem in Example 7.6, and the general case is considered in Example 8.2
for two and three dimensions. Since the eigenvalues A*m) and A^M) tend to
infinity as rt ^ oo (as shown in Example 8.2) so do the eigenvalues Ал.
In the following example we consider some specific problems that illus-
illustrate the results obtained in our discussion. There are many further prop-
properties of eigenvalues that may be obtained by arguments similar to those
given. For example, it can be shown that the eigenvalues Art for the
Neumann, the third boundary value problem, and the mixed problem also
tend to infinity as w->«> but we do not prove this here. We use the property
that An -» oo as n -> <» to prove completeness in the mean square sense of the
eigenfunctions for the Dirichlet problem in our discussion following Exam-
Example 8.2.
Example 8.2. Dirichlet Eigenvalue Problems for Equations with Con-
Constant Coefhcients. The eigenvalue problem
XpM
in the region G with the Dirichlet boundary condition
(8.61) M(x)|,o = 0,
where py qy and p are assumed to be constants, is associated with the
^national problem
530 VARIATIONAL AND OTHER
\\[p(VwJ + qw2]dv
(8.62) -rr^l = ~ = minimum ,
Hi f Г 2
J J p>v2 dv
G
where all admissible functions w(x) must vanish on BG and satisfy арргоргь
ate additional constraints.
Now (8.60) can be written as
(8.63) V2M
that is, A = (Ap — q)/p. Similarly, the variational problem (8.62) can be
expressed as
dv
г г
2dv
— = minimum.
P
Both (8.63) and (8.64) imply that if we can solve a simplified version of
(8.60)-(8.61) or (8.62) with p = p = l^and q =0 and we denote the eigen-
eigenvalues of that simplified problem by Art, then the eigenvalues Xn of (8.60),
(8.61) or (8.62) are given as
(8.65)
The relationship (8.65) between Art and kn demonstrates the property
discussed in the general case of variable coefficients. If the coefficients/? and
q are increased, the eigenvalues increase, whereas the reverse is true if p
and q are decreased. Also if p is decreased, the eigenvalues are increased,
whereas if p is increased, the eigenvalues decrease.
In the two-dimensional case, if we specialize G to be a rectangle as given
in Example 7.6 with 0 < x < I and 0 < у < /, we obtain the eigenvalues for
(8.63) and (8.61) as
(8.66) Xnm = (^) + (iff, n, m = 1,2,3,.. .,
as shown in that example. Similarly, on considering the rectangular box
0<jc</, 0<y</, and 0<z<f in three dimensions and separating vari-
variables, we readily obtain the eigenvalues for (8.63) and (8.61) as
VARIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 531
(8.67) Kmk 7
In both cases the eigenvalues can be expressed as a sequence {Art} and we
have A*-*00 as n-*oo, since \nm and \nmk both tend to infinity as their
subscripts tend to infinity. Consequently, in view of (8.65), An-»°° as n-»a>
since p and p are positive. Furthermore, the rectangular regions RM and Rm
introduced in our earlier discussion may be assumed to have sides parallel to
the coordinate lines or planes in the two- or three-dimensional cases,
respectively. Since the eigenvalue equation in the case of constant coeffici-
coefficients is invariant under the translation of axes, the eigenvalues are given as in
(8.66) or (8.67). Therefore, we conclude that the eigenvalues \(nM) and A^m)
tend to infinity asn-»« and (8.59) implies that An—>« as n—►«.
It is seen from (8.66) and (8.67) that as the dimensions of the rectangular
region G are changed, the eigenvalues change in a specific manner. As the
lengths of the sides are decreased, the eigenvalues Xnm and knmk increase,
whereas the reverse is true if the lengths of the sides are increased. This
result is consistent with our conclusion relating the eigenvalues of region G
to those of a subregion G.
To conclude this example we discuss the Dirichlet eigenvalue problem for
the unit circle. We first obtain the eigenvalues and eigenfunctions by means
of separation of variables and then use the inequalities (8.59) to estimate the
first eigenvalue for the circle in terms of the eigenvalues for inscribed and
circumscribed squares.
Expressing the Laplacian operator in polar coordinates, we consider the
eigenvalue problem
(8.68) V2M + KM = Mrr + J Mr + \ MBe + AM = 0
in the unit disk 0 < г < 1, with the Dirichlet boundary condition for M{r, в),
The solutions M are required to be single valued in the disk so that we have
<8*70) M(r, в + 2тг) = M(r, 0) .
To solve for Af(r, в) we use separation of variables and set
foserting (8.71) in (8.68) gives
F(r)
532 VARIATIONAL AND OTHER METHODS
where m2 is the separation constant. The equation for GF) is
(8.73) G"@) + 2
The condition (8.70) requires GF) to be periodic of period 2тг, and this
implies that m = n, an integer. The equation for Fn(r) (i.e., for each n) is
(8.74) F'n{r) + \ F'n(r) + (a - ^)fn(r) = 0 ,
which is Bessel's equation of order n. This equation was discussed in
Example 4.7. The solutions Fn that are finite at r = 0 are the Bessel functions
and are given as
(8.75) Fn
The boundary condition (8.69) requires that
(8.76)
so that the eigenvalues A are the squares of the zeros of the Bessel functions.
These were denoted by akn(k = 1,2,. . .) in Example 4.7 so that we have
(8.77) А,„ = К„J, * = 1,2,..., n = 0,l,2,....
Since with m = n (8.73) yields the trigonometric functions cos nO and
sin nO, the eigenfunctions Mkn are obtained from the functions
(8.78) Jo(*kOr), Jn(<xknr) cos пв, Jn(aknr) sin пв.
Using D.116) it is not difficult to normalize this orthogonal set of eigenfunc-
eigenfunctions. At present we are mainly concerned with the eigenvalues Xkn, and the
first eigenvalue is given in terms of the first zero of the Bessel function J0(x),
that is,
(8.79) A10 = (a10J~B.40J~5.76.
To obtain an upper and lower bound for the first eigenvalue A10 for the
unit circle, we consider inscribed and circumscribed squares of sides V2 and
2, respectively. They are pictured in Figure 8.1. The leading eigenvalue An
for the circumscribed square is—on using (8.66)—
(8.80) Au=
The first eigenvalue for the inscribed square is
properties of eigenvalues and eigenfunctions
533
f
\
Ю,
(V2/2, 0I
0)
(■Jin, 0)
J
A.0) *
(8.81)
Thus we have
(8.82)
Figure 8.1. Eigenvalues for a circle.
<9.87.
) ^ ^11 9
a result consistent with (8.59). Clearly, (8.82) yields a very poor method for
approximating the eigenvalue A10, since the squares considered do not
approximate the circular region very well. However, this method can be
used to provide bounds for all the zeros of the Bessel functions. In Section
8.2 we use the Rayleigh-Ritz method to approximate the first eigenvalue Alo
for the circle.
We are now prepared to prove the completeness of the orthonormalized
set of eigenfunctions {Mk(x)} that satisfy (8.1) in G and the Dirichlet
conditions Mk(x) = 0 on <?G. Let u(x) be a smooth function defined in G
that vanishes on dG\ that is, u(x) satisfies the admissibility conditions for the
variational problem. We expand u(x) in a series of eigenfunctions as in (8.5)
with the Fourier coefficients given in (8.6). It is not required that u(x) be
normalized as in (8.7). Let the remainder Rn be defined as
(8.83) r (x\ = u(x\ - У N м {x\
n we show that
(8-84) ton И*,(*)|| = Urn \u(x) - 2 NkMk(x)\ = 0 ,
534 VARIATIONAL AND OTHER METHODS
so that the series (8.5) converges to u(x) in the mean square sense. Thereby
the completeness of the eigenfunctions {Mk(x)} is demonstrated. The norni
\\Rn\\ is defined as
Since u(x) and the eigenfunctions {Mk(x)} are admissible for the varia-
tional problem, so is the remainder term Rn(x). We have, in addition,
(8.85) (Д„, Mj) = (u - 2 NkMki M,) - (и, Ms) - 21 Nk(Mk, M,)
in view of (8.4) and (8.6). Consequently, Rn(x) satisfies the constraints for
the minimum problem that determines the (n + l)st eigenvalue An+1, and
this yields for the Rayleigh quotient (8.51)
(8-86) ^ ^ An+1,
E(Rn)
Ы2
since the minimum is assumed for the eigenfunction Mn+l(x). We rewrite
(8.86) as
(8.87) ||ЯИ||2:
We already know that Ля+1—>°o as n-»oo so that we need only show that
E(Rn) is bounded as n—»<» to conclude that ||i?J|—»0 as n—*°°.
Now, as is easily shown,
(8.88) E(u) = e\Rn + 2 NkMk] = E(Rn) + fi[ 2 A^MJ
+ 2 i iV, f f [pVRn-VMk + qRHMk]du.
Using (8.45), the integral term in (8.88) can be expressed as
(8.89)
VM, + qRnMk] dv=j JRn[-V-(p™k) + qMk] dv
RnMkdv = Xk(Rn,Mk) = 0,
G
yAgIATIONAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS 535
here we have used the fact that Rn = 0 on BG, the orthogonality properties
(8.85), as well as the fact that Mk(x) satisfies (8.1) with A = AA. Further-
Furthermore, we obtain from (8.11),
(8.90) e\2 NkMk{x)} = f f t N,M,(x)l[ 2 NkMk(x)\ dv
n n
= 2 2 кккщмк, м,) = ]£ aX -
Combining these results gives
(8.91) ВД.) = E(u) - ^2 AX * ВД >
since £(m) is finite by assumption and the eigenvalues Xk are non-negative.
We conclude from (8.91) that E{Rn) is bounded for all n. Therefore, (8.87)
shows that ||i?J|-»0 as n->», and the mean square convergence of the
series (8.5) to u(x) and the completeness of the Mk(x) is demonstrated. As
was indicated earlier, if we can show that the eigenvalues Art for the second,
third, and mixed boundary value problems tend to infinity, we can use the
given argument to show that the corresponding set of eigenfunctions is
complete.
For the one-dimensional Sturm-Liouville eigenvalue problem discussed in
Section 4.3, the appropriate Rayleigh quotient is
(8.92)
Jopw dx
when the admissible functions w{x) satisfy the Dirichlet conditions
(8.93) и<0) = и</) = 0.
The appropriate Rayleigh quotients for the second, and third, and mixed
boundary value problems are presented in the exercises. All the basic results
given for the higher dimensional eigenvalue problem carry over to the
Sturm-Liouville problem. Thus the completeness of the eigenfunctions (i.e.,
the sine functions) of Example 4.4, and the convergence of the Fourier sine
series in the mean square sense can be demonstrated.
We note that we can drop our assumption that the function u(x)
expanded in a series of eigenfunctions must vanish on the boundary SG.
This follows since any admissible function u(x) that does not vanish on BG
can be approximated arbitrarily closely by a function that does vanish on BG
in the mean square norm. In this norm the values of functions at specific
Points or lower dimensional regions do not play a significant role, in general.
536 VARIATIONAL AND OTHER METHODS
Solutions of the boundary value problems for the self-adjoint еЩрцс
equations considered in Section 4.1 can also be determined by means of °
variational principle. Indeed, Dirichlefs principle is a classical variation
principle that determines the solution of the Dirichlet problem for Laplace'
equation to be the minimum of an energy integral. Dirichlet's principle and
more general variational principles are considered in the exercises. We have
chosen to discuss variational principles for eigenvalue problems in this
section because the results are useful in solving problems for equations of all
three types as has been indicated.
8.2. THE RAYLEIGH-RITZ METHOD
The formulation of the eigenvalue problem in variational form has led to
some interesting and useful conclusions concerning the general properties of
eigenvalues and eigenfunctions as we have seen in the preceding section.
For the most part, however, we have had to rely on separation of variables
or exact solutions in the one-dimensional case to determine the eigenvalues
and eigenfunctions for a given problem. The Rayleigh-Ritz method pre-
presented in this section is an elegant approximation method for the determina-
determination of the first few eigenvalues and eigenfunctions. It is, in fact, most
effective in approximating the lowest eigenvalue and this is often the most
important one in applications. As we shall see, the method yields an
algebraic problem for the determination of the eigenvalues and eigenfunc-
eigenfunctions.
To begin, we must select n functions <px{x)y.. ., <pn(x) that are admissible
for the variational problem under consideration. That is, they must be
sufficiently smooth and vanish on dG in the case of Dirichlet boundary
conditions. For the case of Neumann or boundary conditions of the third
kind, they need not satisfy any conditions on dG. These functions are
chosen to be linearly independent and are selected, if possible, to be good
approximations to the eigenfunctions for the problem.
We form the linear combination
(Q QA\ ьи(г} = ^? r en (r\
V / v*/ — < * krk\ )
with as yet undetermined coefficients ck and insert this sum into the
Rayleigh quotient £"(iv) /1| w||2. This gives for the Rayleigh quotient (8.39),
(8.95)
2 c,ck\ I I [pV<Pj ■ V% + q<Pj<Pk] dv + I hp<pj<pk ds\
E(w) = M-' Uo Js*
2j c.ck I | p<pj<pk
dv
RAYLEIGH-R1TZ METHOD 537
*ye recall that if there are only Dirichlet or Neumann conditions on dG, the
set S3 is empty and the surface integral over 53 is absent in (8.95). If the
boundary conditions are strictly of the third kind, the set 53 coincides with
$Qt Otherwise, S3 is a subset of dG.
Let the n by n matrices A and В be defined by their respective elements
/g 96) ajk = j I [P^Py *^Р* + W/^Jfe] ^У
(8.97) */*==/J i
We define the n-component column vector с to have the components
с ,. • • > cn anc* cT to be ^ts transpose—that is, [cl9 c2,.. ., сл]. Then the
equation (8.95) can be expressed as
as is easily verified. We remark that both c'kc and crfic are quadratic
forms. Further, the (real valued) matrices A and В are symmetric as follows
from (8.96), (8.97). Both matrices are also positive definite if <y(jt)>0,
whereas В is always positive definite, as shown in the exercises. That is, for
q(x) > 0, c^c and cTBc are positive scalars for any real vector с Ф 0 and
they vanish only if с = 0. If q{x) = 0, the matrix A is, at least positive
semi-definite. That is, it can vanish for с Ф 0 if we take the <pk to be constants
and there is a Neumann boundary condition on all of <?G.
The Rayleigh quotient £(и>)/||и>||2 is now a function of the vector с and
we wish to minimize this expression. If the functions <pl9... , <pn are part of a
complete set of functions, it is possible to represent any admissible function
w(x) in an infinite series of those functions, and the minimum problem can
technically be replaced by a problem where the coefficients ck in the
expansion are chosen to achieve a minimum. We replace the infinite sum by
a finite sum and expect this to yield approximate eigenvalues and eigenfunc-
tions. We do not prove the validity of the method.
To proceed with the method we assume that c = a is the vector that
minimizes the expression (8.98) and represent the arbitrary vector с in the
form
(8.99) c = a+ 2 €лел,
where the vectors e19. . . , е„ are the standard basis vectors for n-component
Actors with ek having unity as its fcth component and zero as its remaining
f°niponents. The constants ek characterize the variation around the minimiz-
lng vector a. Inserting (8.99) into (8.98) gives
538 VARIATIONAL AND OTHER METHODS
ад _ [°+2уф[» + ,|, «л]
<8100) ЯГ
[n -iT
To determine the equation satisfied by cr we differentiate (8.100) with
respect to ek (k = 1,. .., n) and then set all the ek and the derivatives equal
to zero since the minimum occurs when all the ek = 0. This gives
(8.101)
9 гад]
<*JlkH2Jr
= о.
Since A and 5 are symmetric matrices and q^e* and &TBek are scalars we
have
(8.102) (<rTAek
and a similar result with A replaced by B. Thus (8.101) reduces to
(8.103) e[[Aa - (^)i»ir] =0, к = 1,..., n .
Since the bracketed vector in (8.103) is orthogonal to the n basis vectors
e1?.. . ,en, we conclude that it must be the zero vector. As a result we
obtain
(8.104) Atr =
where we have set
(8.105) A =
Now if the functions <px. . . , <pn are an orthonormal set, we find from
(8.97) that bjk = 8jk, the Kronecker delta (i.e., 8jk = 0 for }Ф к and 8Jk = 1
for j=k). Then В is a unit matrix, and (8.104) reduces to the standard
eigenvalue problem for the matrix A with cr as the eigenvector and A as the
eigenvalue. If В is not a unit matrix, we also refer to cr as an eigenvector of
A with respect to the positive definite matrix В and to A as the correspond-
corresponding eigenvalue.
The characteristic equation for the determination of the eigenvalues A is
(8.106)
RAYLEIGH-RITZ METHOD 539
^kis is an nth degree algebraic equation for A, and the symmetry of A and В
implies that all the eigenvalues are real. Since the matrices A and В are
positive definite if q(x)>0, we also find that the eigenvalues must be
positive. (If q = 0, they are, at least, non-negative.) Therefore, they can be
ordered as
(8.107) 0<Aa<A2<--^An.
For each eigenvalue kk there is at least one independent eigenvector a(/c).
AH together there are n linearly independent eigenvectors {а(л)}, к =
I . , л. It is easy to show that eigenvectors corresponding to different
eigenvalues are orthogonal with respect to the matrix B. In fact, we have
from (8.104),
Т[
(8.108) 0 = <т(к)Т[А -
where the symmetry of A and В was used. Thus if Хк Ф A;, we have
(8.109) Л(Л
If В is the identity matrix, (8.109) yields the scalar or dot product of the
two vectors a(fc) and a(;) and shows that they are orthogonal. If В is not a
unit matrix, we say that the vectors are orthogonal with respect to B. Using
a Gram-Schmidt process (see Exercise 8.2.1) it is possible to orthonormal-
ize the set of eigenvectors with respect to В and obtain
(8.110)
я
where 8kj is the Kronecker delta.
Each eigenvector <ra) determines an approximate eigenfunctions МДх),
(8Л11) %)=ic[\D / = 1,...,л,
where the c[J) are the components of the vector a0). If is of interest to
observe that the inner product (Af., Mk) of the set of approximate eigen-
ftmctions (M.(x\\ is
" **»4fc illV X111J
functions {Щх)} is
(8.112)
540 VARIATIONAL AND OTHER METHODS
Thus approximate eigenfunctions corresponding to different approximat
eigenvalues are orthogonal. Furthermore, if the <x(/) are orthonormaliZeci
with respect to B, so are the Mj(x) with respect to the above inner product
and its (induced) mean square norm.
By using the Courant maximum-minimum principle of Section 8.1 w
may compare the approximate eigenvalues kk with the exact eigenvalues Л
In replacing the given variational problem with the Rayleigh-Ritz formula-
formulation we note that we are placing an additional constraint on the problem bv
requiring all admissible functions to be linear combinations of the function
<Рг(х),..., <pn{x). This means that every eigenvalue obtained by means of
the Rayleigh-Ritz process cannot be smaller than the exact eigenvalue and
we obtain
(8.113) ЛЛ<АЛ, Л = 1 л.
In Example 8.1 we chose a single admissible function м(лс, у) for the
purpose of approximating the lowest eigenvalue and eigenfunction for the
Dirichlet eigenvalue problem in the rectangle. As the preceding discussion
shows, this is equivalent to using the Rayleigh-Ritz method in terms of a
single function <рг(х, у) that vanishes on the boundary of the given region.
Example 8.3. The Eigenvalue Problem for a Sphere. We consider a
sphere of unit radius with center at the origin and the eigenvalue problem
within the sphere
(8.114) V2M(jc, y, z) + AM(jc, y, z) = 0 ,
with the Dirichlet boundary condition
(8.115) M(x,y,z)\dG = 0,
where dG is the surface of the sphere. Introducing spherical coordinates
(г, 0, <p) with 0<r<°°, О<0^2тг, and 0<^<тг, we have
(8.116)
r sin <p
4^
sin <p dB
where M = M(r, 0, <p). The boundary condition (8.115) becomes
(8.117) M(l,0,<p) = O.
Using separation of variables, we set
RAYLEIGH-RITZ METHOD 541
(8.118) M(r,e,<p) = F(r)Y(e,<p)
and insert (8.118) into (8.116) to obtain after some simplification
(r2F'(r))' , 1 Г 1 д I . dY\
(8-П9) '-j+f- + Ar2 = - 7(ё-7) [^ - (sm , -j
sin2^ ^2J
with fc2 equal to the separation constant. The equation for Y@, <p) is
^y=0
^ ид ^^ + ^ + ^
sm ^ d<p L ^ J sin2 <p дв
The conditions on Y@, ^>) are that it be periodic in в [i.e., Y(9 + 2тг, ^р) =
YF, q>)] and that it be bounded at the poles of the sphere (i.e, when <p = 0
and <p = тг) as the coefficients in (8.120) are singular there. These conditions
2
determine the values of k2 to be
n = 0,1,2,. . . ,
and yield the solutions YnF9 <p) known as the spherical harmonics. These are
specified by means of a further separation of variables in (8.120) to be
products of Legendre functions and sine and cosine functions (they are
discussed in the exercises).
The equation for F(r) then becomes
with the conditions
(8Л22)
and that F(r) is bounded at the origin. The substitution
(8.123) Р{г) = Щ
reduces (8.121) to the Bessel equation of order n + \,
(8-124) fXr) + \ f'(r) + [ A - (n+r12/2J]/(r) = 0.
boundedness condition of F(r) at r = 0 implies that
542 VAMATIONAL AND OTHER METHOu$
(8Л25) /W=/n+1/2(VAr)
and the boundary condition (8.122) requires that
(8.126) /„+i/2<VA) = 0.
If amn(m = 1,2,. . .) are the roots of the Bessel function /n+1/2(a), we
obtain the eigenvalues
(8.127) Amn = (amn)\ n = 0,1,2,... , m = 1,2,....
The roots of the Bessel functions are all real and for each n are listed in
increasing order. It is well known that
(8.128) /1/200=^™*
and, in fact, all the functions Jn+U2(x) can be expressed in terms of powers
of x and the sine and cosine functions.
We have shown that the eigenvalues for the sphere are given by (8.127)
and the associated eigenfunctions are
(8.129) Mmn = ±=
(The Mmn are not normalized.) The lowest eigenvalue is given by A10 = oj0,
and al0 is the smallest positive zero of the Bessel function /1/2(*)- In view of
(8.128) we have
(8.130) А10 = а?0 = тг2.
To approximate the eigenvalue A10 using the Rayleigh-Ritz method, we
introduce the function
(8.131) ft(r,ef*) = l-r,
which appears to be the simplest function that vanishes at r = 1 and is
bounded at r = 0. Then
(8.132) vv = c1<Pl
is an admissible function for the variational problem associated with (8.114),
(8.115). We have
^E RAVbEIGH-RITZ METHOD 543
(V<px Jr2 sin <p dr dO dip
о Jo Jo l
(8.I33) Hwll2 Г f2n f1
K " " c2 <p\r2 sin <pdrd6 d<p
1 Jo Jo Jo *
/0 '2 dr
= 10.
Since the exact eigenvalue is тг2 = 9.87, the approximate eigenvalue A10 = 10
given in (8.133) yields excellent agreement. We do not discuss how well the
first eigenfunction is approximated nor approximations for higher eigen-
eigenvalues and eigenfunctions.
The previous determination of the lowest eigenvalue for the sphere that
was carried out using separation of variables and the Rayleigh-Ritz proce-
procedure is of interest in the following problem. Let u(x, y, z,t) be the
concentration of some diffusing substance and let the strength of the sources
in the substance be proportional to the concentration. According to our
discussion in Section 4.1, the equation for и is given as
(8.134) ut = c2V2u + yu,
where c2 and у are positive constants, both of which are characteristic of the
substance. Note that —y which corresponds to q in D.4) is now negative
rather than non-negative. An initial and boundary value problem of interest
for (8.134) occurs if we consider the diffusion process within a bounded
region G, with the initial condition
u(x,y,zy0)=f(x,y9z)
and the boundary condition
(8Л36) «<*,**,OU = o.
Applying the separation of variables method to the problem (8.134)-
(8.136) we set
и = N(t)M(x, у, z)
insert (8.137) into (8.134). This leads to
544
VARIATIONAL AND OTHER METHODS
where A is the separation constant. Consequently, we must consider th
eigenvalue problem
(8.139) V2M + AM = 0, (jc, yjz)SG,
(8.140) A#(x,y,z)|,o = 0,
for the function M. The eigenvalues kk{k = 1,2,.. .) for this problem are
positive and can be arranged as
(8.141) 0 < kx s= A2 < • • • < Ад, < . . . ,
as has been shown previously. The associated eigenfunctions are denoted by
Mk(x9 y,z). For each Xh we obtain a solution Nk{i) of the separated
equation for N in the form
(8.142) Nk(t) = ak exp[(y - Afcc2)r], * = 1,2,. . . ,
with arbitrary constants ak. Finally, the solution of (8.134)-(8.136) is given
as
(8.143) и(х, у, z, t) = £ akMk(x, у, z) exp[(y - A,c2)*]
with the ak given as the Fourier coefficients in the eigenfunction expansion
of the initial concentration Ддс, у, z).
An important question regarding the solution (8.143) is whether the
concentration u(x, y, z, t) grows without bound as the time t increases, in
which case the problem (8.134)-(8.136) is unstable. Since the concentration
и at any time Г is a result of reactions that occur within the substance as was
indicated, we say that when the concentration и increases exponentially, a
chain reaction is taking place because of the rapid increase in u. The solution
(8.143) shows that exponential growth in t occurs if у - кксг > 0. In view of
(8.141) we see that the lowest eigenvalue Ax is the most significant and if
(8.144) ^>A15
с
a chain reaction takes place in the substance. Now the eigenvalue varies as
the region R is changed, and the parameters c2 and у can presumably be
adjusted for given substances. Therefore, if we fix two of the three con-
constants, say, у and c2, we may define a critical value of the third constant Ar
If Ax > ylc\ no chain reaction takes place, whereas if Ax < y/c2, a chain
f HE RAYLEIGH-RITZ METHOD 545
reaction does occur. Thus the critical value of X19 which we denote by Af \ is
given as
For a specific problem it is important to determine the first eigenvalue
(i.e., the lowest) to see if it is greater or less than the critical value. We have
found that for the sphere of radius one, the lowest eigenvalue equals тг2
The Rayleigh-Ritz method yielded a slightly higher value. It is clearly
important to be able to approximate the lowest eigenvalue as accurately as
possible. (See also the waveguide problem in Exercises 8.5.31 and 8.5.32.)
Example 8.4. A Sturm-Liouville Problem. The eigenvalue problem
(8.145) M"{x) + AM(jc) = 0, 0 < x < тг ,
with the mixed boundary conditions
(8.146) M'@) = 0, М(тг) = 0,
has the eigenvalues
(8.147) A, = (*+±J, * = 0,l,2,...
and the (unnormalized) eigenfunctions
(8.148) Mk(x) = cos[(* + k)x]y к = 0,1,2,. . . ,
as is easily shown. We shall use the Rayleigh-Ritz method to approximate
the lowest two eigenvalues.
The appropriate Rayleigh quotient for this problem is
(8.149)
г
Jo
о w2dx
admissible functions w(x) are required to vanish at x — тг—that is,
w(tt) = 0. At jc = O there is a natural boundary condition so that no
restrictions need be placed on w(x) at that point.
To apply the Rayleigh-Ritz method we select two functions ^(д:) and
fz(x) that satisfy the admissibility conditions; that is, ^(тг) = <^(тг) = 0. As
ш (8.94) we set
546
VARIATIONAL AND OTHER METHODS
c2<p2(x),
with ^ and <p2 chosen as
(8.151) <pl(x)
V-*3.
We note that ^(ir) = <p2(?r) = 0 and that, in addition, <p[@) = p^()
Although admissible functions need not have vanishing derivatives at x = о
we expect to get improved results with our choices for <px and <p2 since they
are expected to more closely approximate the eigenfunctions (8.148) whose
derivatives do vanish at x = 0.
Evaluating the integrals (8.96), (8.97), with p = p = 1 and q = 0 and V<p
replaced by <p\ we obtain the matrices
(8.152)
w.
u7
The two roots of the characteristic equation
(8.153) det(A-AB) =
are
(8.154)
t = 0.25,
We have retained only two decimal places, and ix is, in fact, slightly larger
than 0.25. Comparing the approximate eigenvalues At and A2 with the exact
eigenvalues Aa =0.25 and A2 = 2.25 [see (8.147)], we observe that there is
excellent agreement between Ax and \г but that A2 is not as close to A2. This
seems to be the norm for the Rayleigh-Ritz procedure in that the higher
eigenvalues are more poorly approximated than the lower ones. We do
observe that for both eigenvalues we have \г < ix and A2 < A2 as required by
(8.113).
Given the approximate eigenvalues A, and A2 we can determine approxi-
approximate eigenfunctions by specifying two sets of coefficients cx and c2 in
(8.150). We do not carry out this calculation. _
Example 8.5. A Sturm-Liouville Problem with a Variable Coefhcient.
We consider the equation
(8.155) M"(x) + ( A - ex)M(x) = 0, 0 < x < tr ,
with the boundary conditions
RAYLEIGH-RITZ METHOD 547
(8.156)
^e assume that 0 < € < 1 so that the term q(x) = ex is uniformly small in the
interval 0<Jt=£7r and is non-negative. With 6=0, (8.155) reduces to an
exactly solvable problem. In any case, (8.155) can be solved exactly in terms
of Airy functions, which are solutions of the equation
(8.157) y"(x)-xy(x) = 0
and are tabulated. However, we do not use the exact solutions of (8.155)
but use approximate techniques to determine the lowest eigenvalue for the
Sturm-Liouville problem (8.155), (8.156).
First we apply the Rayleigh-Ritz method. In view of (8.92) the Rayleigh
quotient appropriate for (8.155), (8.156) is
(8.158)
И 1 I '* ^
dx
The admissible functions w(x) are required to vanish at x — 0 and jc = тт.
Since € is small and for e - 0 the lowest eigenvalue for (8.155), (8.156) is
Ax = 1 and the corresponding eigenfunction is Mr(x) = y/Jlhrjsin x, we pick
<Pi(x) in the Rayleigh-Ritz approximation (8.94) to be
(8.159) ft(x) = sinx.
Clearly, ^(jc) is admissible for the variational problem.
With w = cx<px{x) we have
(8.ш) за.^""""-х'».,+=
sin xdx
Jo
Now in the interval 0 < x < ir9 the function q(x) = ex satisfies the
inequalities
4m-0 and qM = eir representing the minimum and maximum value of
4\x)y respectively. Then (8.55) implies that
(8.162) A^m) ^ A
iW)
548 VARIATIONAL AND OTHER METHODS
where Ajm) = 1 and \[M) = 1 + етг, as is easily seen on replacing A - ex by \
and by A - етг to determine A{m) and АAл/), respectively. The Rayleigh-Rjt2
procedure implies that
(8.163) A^A^l + y ,
and we see that kx <\^*\ The smaller e is, the better the approximation to
X1 is expected to be.
An alternative procedure that may be applied directly to the problem
(8.155), (8.156) is a perturbation method that we now present. The
solutions M and the eigenvalues A are functions of e; that is, Af = M(x\ e)
and A = A(e). Consequently, we expand both M and A in a power series in e
in the form
(8.164) M(x;e) = M@)(x)
(8.165) А(е) = А@) + €АA) + • ■ •.
We insert (8.164), (8.165) into (8.155), (8.156) to obtain
(8.166) M" + (A - €x)M = [d Mz + A@)M@)J
2 *" ^ M + (A — jc)M + • • • = 0
ax J
(8.167) M(o)@) + eMA)@) + • • • = М(о)(тг) + еМA)(тг) + • ■ • = 0.
Equating coefficients of like powers of e to zero in (8.166), (8.167) leads
to the following problems. For M{0\x) we have
,2д,@)
(8.168) =-£j- + A@)M@) = 0, M@)@) = Mw(ir) = 0 .
The leading eigenvalue and normalized eigenfunction for this problem are
(8.169) А(!0) = 1, Mf\x) = y^ sin x .
The equation for MA)(jc) is
(8Лщ «J*_
with the boundary conditions
METHOD 549
(8.171)
With A@) = Al0) = l the first eigenvalue for (8.168), the inhomogeneous
equation (8.170) has no solution unless the inhomogeneous term is ortho-
orthogonal to the eigenfunction М^(х). This follows on multiplying (8.170) by
д/@)(jc) and integrating from 0 to тт. We have
(8.172)
f м[0) i^r+A(i°)MA)]* e Г *A)[^-+х™м™]л=°
on integrating by parts and using (8.168) and (8.171). This implies that
(8.173)
from which we conclude that
(8.174) AA> = A<*> = f
f .
Having determined A^ we can specify M[l)(x) but this is not carried out.
Inserting (8.169) and (8.174) in (8.165) gives
(8.175) Ai(€)«1+^.
The approximation (8.175) to the lowest eigenvalue is identical to that given
by the Rayleigh-Ritz method. This appears to be due to our choice of <px(x)
in the Rayleigh-Ritz method to equal the leading eigenfunction for (8.155)
with e = 0. However, the perturbation method readily yields approximations
to the higher eigenvalues, whereas this is not so easily accomplished by way
of the Rayleigh-Ritz method as we have seen (see also Section 9.2.).
The Rayleigh-Ritz method can also be used to obtain approximate
solutions of the variational form of (self-adjoint) elliptic boundary value
problems. For nonself-adjoint elliptic boundary value problems, the Galer-
kin method can be applied to the Galerkin form of these problems (see
Exercise 6.4.8) to yield approximate solutions. These matters are considered
ш the exercises.
8-3. RIEMANN'S METHOD
Riemann's method is a classical technique for solving the Cauchy problem
for hyperbolic linear equations in two variables. Although this method does
550 VARIATIONAL AND OTHER METHODS
not yield closed form solutions except in a few cases, it does provide usefui
information about domains of dependence and influence for solutions i
general. Now it was shown in Section 3.1.a that second order hyperboli
equations in two variables can be brought into one of the two canonical
forms C.19) or C.21), and the Riemann method is generally applied to on
of these forms. We do not consider this method in its full generality but we
also do not use either of the two canonical forms for hyperbolic equations in
order to show how to deal with curvilinear characteristics.
We consider the equation
(8.176)
utt - у\x)uxx + a(x)ux + /3(x)ut + c(x)u = F(x,t), -« < x < *, t > 0,
with coefficients that depend only on x. The initial conditions given at t = 0
are
(8.177) u(x, 0) = f(x), ut(x, 0) - g(x),
with -o
With M defined as
(8.178) M[u] = utt - y\x)uxx + a(x)ux + fi(x)ut + c(x)u ,
the adjoint operator M* is given as
(8.179) M*[w] = wtt - (y2w)xx - (aw)x - /3wt + cw ,
in view of C.161). Introducing a bounded but as yet arbitrary region G in
(x, f)-space with boundary <?G, we obtain on using C.162)-C.163)
(8.180) Jj {wM[u]-uM*[w]}dxdt
G
I [-y2wux + u(y2w)x + auw] dt + [uwt - wut - puw] dx ,
jdG
with integration over dG carried out in the positive direction. We assume
that the two-dimensional divergence theorem or Green's theorem in the
plane is valid for the region G.
The characteristic curves for (8.176) are solutions of the equations [see
C.8)]
so that they are given by
METHOD
551
(8.182)
■• constant.
Let (£, t) be a point in the (jc, r)-plane with T>O:The two characteristic
curves (8.182) that pass through that point are given as
(8.183)
Г ds
Je y(s)
If the curves are extended backwards in t until they intersect the я-axis, we
obtain a characteristic triangle whose base is a segment of the дс-axis and
whose other sides are characteristic curves. The interior of this triangle is
denoted as the region G and the triangle itself by <?G, as shown in Figure
8.2. The boundary dG is divided into three parts <?G0, dG+J and <?G_, and
we apply (8.180) to this region with the positive direction of integration
displayed in the figure.
On the line segment dG0 the line integral in (8.180) becomes
(8.184)
1g0
" Wu< -
=}A
)Wt(x, 0)
since df = 0, u(x,0) =/(*), and ut(x,0) = g(jc). On the characteristic seg-
segments dG± we have, since we assume that y
Figure 8.2. The characteristic triangle.
552
VARIATIONAL AND OTHER METHODS
(8.185) *=+- dx
and for any function v(x, t) defined and differentiable along the charac-
characteristics,
do dt _ 1
(8.186) ta=V*+&iV* = V*+yVt-
Thus
(8.187)
I [-y2wux + u(y2w)x + auw] dt + [uwt - wut ~ /3uw] dx
JdG±
= I + ywux + - u(y2w)x ^ ~ auw + WH;f"" wut ~ Puw dx
yuw\pA + ум 2-^ + -^- w+ — aw+ - /3w \
' 1Л* JaG± ' V dx у у у J
\dx
у
Collecting the results, we obtain
(8.188)
= {-y/ivL + yfw\AJ
[f(x)wt - (g(x) + P(x)f(x))w] dx
Ja+ ' L dx у у1 у
dx у
f J {wM[m] - wM
To obtain an expression for the solution u(x, t) at the point P = (£, т) we
require that w{x, t) satisfy the following conditions. First we set
METHOD 553
(8.189) M*[w] = 0
n the characteristic triangle G. To eliminate the integrals over the charac-
ltics dG+ and <?G_ we set
zf + ^
dx y{x) у
on these characteristics. Finally, to obtain an expression for u(x, t} at P in-
independent of the value of w(xy t), we set
(8.191) 2y(x)w(x, t)\P = 2y(i)w(i, т) = 1.
The equations (8.190), (8.191) represent an initial value problem for w on
the characteristic curves. The solution is easily found to be
with w± evaluated on the characteristics dG±. The equation (8.189) for w
and the conditions (8.191), (8.192) comprise a characteristic initial value
problem for w(x, t). (The problem is to be solved backwards rather than
forwards in time.) Such a problem is well posed and has been discussed at
the end of Section 6.5 where an iteration method for solving the characteris-
characteristic initial value problem for the canonical form of second order hyperbolic
equations was presented. We assume a solution can be found for the present
problem as well and obtain explicit solutions for two simple cases in
Example 8.6.
The solution w is called the Riemann function. Since it depends on the
point P = (f, t), we express it as
Then we have from (8.188), since M[u] = F,
(8Л94) u{(,r)
[f(x)Rt - (g(x) + p(x)f(x))R] dx .
The solution formula (8.194) shows that the domain of dependence of the
solution u(x, t) of (8.176), (8.177) is the interior of the backward charac-
characteristic triangle with vertex at the point (jc, i). Reversing the argument shows
№at the domain of influence of a point (x, 0) is the interior of the forward
554 VARIATIONAL AND OTHER METHODS
characteristic sector formed by the two characteristics issuing from the
()
(,)
To establish a connection between the Riemann function, the Green'
function, and the fundamental solution we set
(8.195)
in (8.176) and (8.194) and assume the initial data/(x) and g(x) vanish. Then
(8.194) yields, with (f, r) replaced by (x t),
(8.196) u(x, 0 = J JG Д(х, h x, t)8(x - #M(Г- f) dx dt
(|,r)eG
О,
This shows that the Riemann function Л(£, r; x, t) represents the effect of a
point source located at the point (f, r). As shown in Figure 8.3, if the point
(x, i) lies in the forward characteristic sector with vertex at (£, r), the
solution is nonzero. Otherwise, u(x, t) vanishes since disturbances cannot
move faster than the characteristic speed x'(t) = y(x). In fact, the Riemann
function /?(£, r;x, t) is equal to the causal fundamental solution for the
operator Mu [see (8.178)] in the forward characteristic sector where that
solution is nonzero.
Similarly, the Riemann function R(x, t\ £, r) (now expressed in terms of
the original variables) agrees with the free space Green's function for the
operator Mu in the backward characteristic sector that issues from (£, т)
Figure 8.3. Forward and backward characteristic sectors.
METHOD 555
(see Figure 8.2). The causal fundamental solution and the free space
Green's function both vanish outside the forward and backward characteris-
characteristic sectors, respectively. Thus, each of these functions is completely de-
determined once the Riemann function together with its domain of definition
is known. Until the advent of the theory of generalized functions, the
importance of whose role in the construction of Green's functions and
fundamental solutions was demonstrated in the preceding chapters, most
efforts to solve Cauchy problems for hyperbolic equations were based on
generalizations of Riemann's method.
In the following example we obtain explicit representations for the
Riemann function in the case where (8.176) has constant coefficients.
Example 8.6. Riemann's Function for Equations with Constant Coef-
Coefficients. With у-constant and a-/3 = c = 0 in (8.176) we obtain the
inhomogeneous wave equation
(8.197) un - y2uxx = F(x, t) .
In this case the operator M as given in (8.178) is self-adjoint. The Riemann
function equals l/2y on the characteristics and, in fact, we immediately
conclude that
(8.198) /?(*,*; f,r)=-^
in the entire characteristic triangle. The causal fundamental solution for
this problem has the form F.216) which agrees with (8.196) and it is easily
shown that (8.194) reduces to d'Alembert's solution of the initial value
problem for (8.197).
К у = constant, a = 0 = 0, с = constant, and с < 0, we obtain for (8.176),
(8199) utt-y2uxx + cu = F(x,t).
The operator M in (8.178) is self-adjoint and we have R = Illy on the
characteristic curves. By looking for a solution of the equation for R that
depends on the hyperbolic distance defined in F.212) with с replaced by у
we find as in Example 6.14 that the Riemann function R is given as
(8-200) R(X, t; t, r) = £ /в[^
sJnce the modified Bessel function I0(z) has the property that /0@) = 1. The
causal fundamental solution (8.196) agrees with F.218) and (8.194) agrees
^th G.384), as shown in the exercises.
556 VARIATIONAL AND OTHER METHODS
8.4. MAXIMUM AND MINIMUM PRINCIPLES
In Chapter 1 the diffusion, telegrapher's, and Laplace's equations wer
derived as limiting forms of certain random walk problems. It was indicated
that some of the properties of the discrete random walk models carry over
to the limiting partial differential equations. For example, a mean value
property for the discrete form of Laplace's equation was shown in Example
7.15 to be valid for solutions of Laplace's equation. In this section we show
that the maximum and minimum properties valid for discrete forms of the
diffusion and Laplace's equation carry over to their limiting forms. These
properties can be used to prove uniqueness and continuous dependence on
data of the solutions of these equations. In addition, we will show for the
telegrapher's equation that if the initial data for the Cauchy problem are
positive, then the solution must be positive. This is consistent with the
interpretation of the solution in Section 1.2 as a probability density function
that must be non-negative.
We begin by discussing the diffusion or heat equation
(8.201) ut-c\x=0
in the closed region 0 < x < / and 0 < t < T. We assume that c2 is a constant.
A more general equation of parabolic type with constant coefficients in the
form
(8.202) vt - c\x = avx + bv
can be brought into the form (8.201) by way of the transformation
v = exp(ax + J3t)u
with a = -a/2c2 and 0 = b - a2/4c2. The maximum and minimum principle
for (8.201) is given as follows. Let u(x,t) be a solution of (8.201) in
0<*</ and 0<f< T which is continuous in the closed region 0 < x ^ / and
0 < t < Г. Then the maximum and minimum values of u(x, i) are assumed
on the initial line t = 0 or at points on the boundary x = 0 and x = /.
In our proof of this principle we make use of a corresponding principle
for the inhomogeneous diffusion equation
(8.203) ut-c2uxx = F(x,t)
given over the closed region 0 < x ^ / and 0 < t < Г. Let u(x, t) be a solution
of (8.203) that is continuous in the closed region 0<x^ I and 0^ f — ^
Then if F(x, t) is negative in that region, u(x, t) attains its maximum values
on t = 0, jc = 0 or x = / and not in the interior of the region or at t = Т. И
F(x, i) is positive in the region, u(x, t) attains its minimum values on t = 0,
x = 0 or x = /and not in the interior of the region or at t = T.
AND MINIMUM PRINCIPLES 557
The proof of this principle for the inhomogeneous equation (8.203) is
d on showing that if a maximum or minimum occurs at an (interior)
int (*o> 'o) with 0< *o < I and 0< t0 < Г, we are led to a contradiction.
V/e consider the case with F < 0 first. Since u(x, t) is continuous in a closed
d bounded region it must assume its maximum there. At the assumed
interior maximum point (x0, tQ) we must have
d2u(x0,Q
(8.204)
дх2
dt
as is known from calculus. If t0 < T, then ut = 0 at (x0, t0) since both ux and
и vanish at an interior maximum point. However, we may have t0 = T in
which case (jc0, t0) = (jc0, T) is on the boundary of the closed region and we
can merely assert that ut ^ 0 at the point since u(x, t) may be increasing at
that point.
If we insert the inequalities (8.204) into (8.203) we find that the left side
of the equation is non-negative while the right side is strictly negative.
Consequently, the maximum must be assumed on the initial line or on the
boundary. If F>0, we assume there is an interior minimum point (x0, t0)
and obtain the inequalities (8.204) with the signs reversed. Again this leads
to a contradiction, so that the minimum must be assumed on the initial line
or on the boundary.
The inequalities (8.204) at a maximum or their reversed form at a
minimum do not yield a contradiction when they are inserted into (8.201)
since uxx and ut may both vanish at (x0, f0). To obtain a proof of the
maximum principle for the homogeneous diffusion equation (8.201) based
on the foregoing argument, we introduce the auxiliary function
(8-205) vv(x, 0 = u(x, t) + ex2 ,
where e >0 is a constant and и satisfies (8.201). Now, w(x, t) is continuous
ш 0 < x < / and 0 < / < Г so that it has a maximum at some point (xt, tt) in
the region. If we assume that 0 < хг < I and 0 < tx < T, we again conclude
that
(8*206) wt(xu tx)>0, *„(*!, tx) <0 .
But
(8.207) Wf - c2wxx = ut- c2uxx - 2c2e = -2c2e < 0 ,
Slr*ce и satisfies (8.201). Inserting (8.206) into (8.207) now leads to a
contradiction since the left side is non-negative and the right side is strictly
558 VARIATIONAL AND OTHER METHODS
negative. Therefore, w(x91) assumes its maximum on the initial line or on
the boundary since w satisfies (8.203) with F < 0.
Let M be the maximum value of u{xy t) on t = 0, x = 0, and x = / (i.e. 0
the initial and boundary lines). Then
(8.208) w(x, t) = w(jc, t) + €jc2 < M + el2
for 0< jc ^ / and O^t^Tsince w has its maximum on t - 0, x = 0, or jc = /
Consequently,
(8.209) м = w - ел:2 < w < Л/ + e/2 .
Since e is arbitrary we can let e—>0 in (8.209) and conclude that
(8.210) m(jc, Г) < M
throughout the closed region and the proof is completed.
To obtain a minimum principle, we consider the function —u where
m(jc, f) is a solution of (8.201). Clearly, -u is also a solution of (8.201) and
the maximum values of — и correspond to the minimum values of u. Since
— и satisfies the maximum principle, we conclude that и assumes its min-
minimum values on the initial line or on the boundary lines. In particular, this
implies that if the initial and boundary data for the problem are non-
negative, then the solution must be non-negative. This result is consistent
with the interpretation of the solution as a probability density or as a
temperature.
It can also be shown for the Cauchy problem for (8.201) over the initial
line t-0 with -o°<jt<oo that the solution u(x, t) is non-negative if the
initial values m(jc, 0) = f(x) are non-negative. This follows directly from the
representation of the solution given in Example 5.2. We have
(8.2H) u(x> о = -^L= £ «р[-^]Л.) ds .
Since the exponential term and the algebraic coefficient in (8.211) are
positive, we conclude that w(jt, t) is non-negative as long as the initial data
fix) are non-negative.
On examining the proofs of the maximum and minimum principles for
(8.201) and (8.203) we observe the following basic difference between the
results. For the inhomogeneous equation (8.203) it was shown that the
maximum or minimum values must be attained either on the initial line or
the boundary lines and that they cannot be assumed in the interior. This
result is known as a strong maximum or minimum principle. For the
homogeneous equation (8.201) we determined that the maximum and
AND MINIMUM PRINCIPLES 559
minimum values are attained on the initial line or on the boundary lines.
However, the possibility that the solution can also assume these maximum
nd minimum values at interior points was not excluded. (This would be the
^ase, for example, if the solution is identically constant.) This result is
referred to as a weak maximum and minimum principle. It can be proven
(but we do not carry this out) that a strong maximum and minimum
principle is valid for (8.201). It states that if the solution does assume the
maximum and minimum values in the interior, then it must be identically
constant.
We now show that if the solution u(x, t) of (8.201) has a maximum value
at a point on the boundary x = 0 or x = /, then the normal derivative duldn
is positive at that point unless m(jc, i) is identically constant. This means that
M >0 at jc = / and ux <0 at x - 0. At a minimum point the normal
derivative is negative. (We note that at x = /, say, we must have ux >: 0 at a
maximum point since the point lies on the boundary, but the theorem states
that ux is strictly positive there.)
A separate discussion is required at x = 0 and at x = L We consider the
boundary line x = / and assume that (jc, i) = (/, t0) is a maximum point for
the solution u(x, t) of (8.201) with u(l, t0) = Л/. Let 0< a < t0 < j3 < T and
0< a < /, and construct the rectangle R bounded by x = a, x = I, t= a, and
t = j8. We assume that u(l, t) < M for t Ф t0 on the side of R that coincides
with the boundary and we also have u(x, i) < M on the three other sides of
the rectangle because of the strong maximum principle. Given the auxiliary
function
(8.212) v(x, t) = m(jc, t) + e(x2 - 4/jc + 3/2)
where e > 0, we have v(l, t) - w(/, t) and we choose e to be sufficiently small
so that v(x, t) < M on the sides x = a and t = a of the rectangle. (This is
possible because и is strictly less than M on these sides.) Further,
(8.213) vt - c\x = ut- c2uxx - 2c2e = -2c2e < 0 .
Thus, the maximum principle for the inhomogeneous diffusion equation
applied to the rectangle R implies that v(x, i) has its maximum at the point
С» 'о)- As a result, ux(l, t0) > 0 and this means that ux{l, t0) > 2/e > 0, as was
to be shown. (This result is valid even if the maximum point is not isolated.)
" u(x> t) is a constant the proof is clearly not valid. The proof that ux < 0 at
a maximum point on jc = 0 is left as an exercise.
The maximum and minimum principles may be used to prove uniqueness
and continuous dependence on the data for solutions of initial and boundary
value problems for the diffusion equation. They can also be extended to
higher space dimensions by using similar arguments and to equations with
variable coefficients. These matters are considered in the exercises.
Before discussing the maximum and minimum principle for Laplace's
560 VARIATIONAL AND OTHER METHODS
equation, we discuss a corresponding maximum and minimum principle f
Poisson's equation Or
(8.214) V2u(x, y) = -F(x, y)
in two dimensions. We show that if we consider a bounded region G and t
boundary dG, then the maximum values of a solution u(x, y) of (8.214) a
attained on dG if F(x, y)<0 in G and the minimum values of u(x7 y) ar
assumed on dG if F(x, y) > 0 in G.
To prove this result we note that since u(x, y) is continuous in a closed
and bounded region by assumption, it must assume its maximum in G or in
dG. Let us suppose the maximum is assumed at a point (jc0, y0) in G and
consider the case where F(x, y)<0 in G. Then at the interior maximum
point (jc0, y0) we must have
(8.215) uxx(x0, y0) ^ 0, uyy(x0, y0) < 0 .
But since F<0, (8.214) yields at (jt0, y0),
(8.216) uxx + uyy>0,
which is a contradiction. Consequently, the maximum of u(x, y) must occur
on dG.
To show that the minimum of m(jc, y) is attained on dG if F(jc, y) >0 in
G we replace и by —u in the preceding argument. This is equivalent to
replacing F by ~F in (8.214). Since F>0, we obtain -F<0 and conclude
that — и assumes its maximum on dG. Therefore, и assumes its minimum on
dG and the minimum principle is proven.
Now the Green's function K(x, y; ij, rj) for Laplace's equation with
Dirichlet boundary conditions satisfies [see A.126), A.127)]
(8.217) Kxx + Kyy = -8(x - £)8(y -17), (x, y), (^)£G,
and the boundary condition K = 0 on dG. We would like to conclude that
К > 0 in G in view of its probabilistic interpretation in Section 1.3. Since the
delta function is not an ordinary function, it is not obvious that the
preceding theorem can be applied. However, we may obtain the delta
function in two dimensions as a limit of sequences of ordinary non-negative
functions in the manner of Example 1.2 and Exercise 7.2.1 as is easily
shown. For each of these functions we obtain a non-negative solution of
(8.217) where the delta function is replaced by a member of the sequence of
functions. This follows since К and each of its approximations vanishes on
dG and the right side of (8.217) is nonpositive. Technically, the minimum
principle proved in the foregoing requires the right side of (8.214) to be
strictly negative in G, and the members of the (delta) sequence are such that
AND MINIMUM PRINCIPLES 561
^f in (8.214) can only be said to be nonpositive. However, the minimum
rincipte is valid in that case also, as easily follows from the proof of that
rincipte given for Laplace's equation. In the limit as the sequence tends to
[he delta function the sequence of approximate Green's functions tends to
the Green's function К and this, in turn, must also be non-negative.
Xhe maximum principle for the (homogeneous) Laplace's equation is
proven by following the procedure used for the heat equation. We consider
a solution u(x, у) of Laplace's equation
(8.218) **xx + uyy=°
in the region G with boundary dG. Let M be the maximum value of u(x, y)
ondG and suppose that the (bounded) region G can be enclosed in a square
of side 2/ with center at the origin. Then, the function
(8.219) M>, У) = u(x, y) + ex2 ,
with e > 0, satisfies the equation
(8.220) V2w = V2[u + ex2] = 2e > 0 .
Based on our considerations for Poisson's equation with F<0, we conclude
that w(x, y) attains its maximum on dG. Then for (jc, у) Е G, we have
(8.221) w(jc, y)
since /2>:jc2 for all x on dG by our assumption. Finally,
(8.222) и - w - ex2 < w < M + el2 .
Since e is arbitrary, we conclude that
(8.223) И(х, у)<М
°n letting €—>0, so that the maximum is assumed on the boundary. By
considering -u in place of и we conclude, as was done earlier, that the
ntinimum is also assumed on the boundary. As we have not excluded the
possibility that u(x, y) can assume the maximum and minimum values in the
region G, our result is a weak maximum and minimum principle. It can also
be shown for Laplace's equation that unless m(jc, y) is identically constant
the normal derivative duldn on dG is positive at a point where и assumes its
Maximum and negative at a point where и assumes its minimum.
These results can be used to prove uniqueness and continuous depen-
dependence on data for boundary-value problems for Laplace's and Poisson's
equations. They can be extended to three dimensions and to certain
562 VARIATIONAL AND OTHER METHOD
problems with variable coefficients of the type considered in this text. This *
done in the exercises. Is
The minimum principle for Laplace's equation again shows that no
negative boundary data implies non-negative solutions as required by 0
random walk formulation in Section 1.3. We have already derived the me Г
value property for solutions of Laplace's equation (see Exercise 7.5.21)
This property can be used to prove the maximum and minimum principle
for solutions for Laplace's equation. It can be used to show that wheneve
the maximum or minimum is attained in the interior of the region (as well as
on the boundary), the solution и must be identically constant. This is the
strong maximum and minimum principle for Laplace's equation. It js
discussed in the exercises.
To conclude this section, we consider the telegrapher's equation A.88) of
Section 1.2, that is,
(8.224) vtt-y2vxx + 2\vt = Q
and express it in the form of a system
(8.225) at + yax = -ka + \p
(8.226) j3r-y& = Aa-A/3,
as in A.82), A.83), with v = a + j8. Since a and C are interpreted as
probability densities, we require that a and j8 be non-negative. In particular,
if the initial data
(8.227) a(jt,0) =/(*), p(x,0) = g(x), -oo<jc<oo
are non-negative, we must have a(x, t)>Q and p(x, t)>Ofort> 0. In terms
of v(x, t) we should have v(x, i) >: 0 if v(x, 0) > 0 when solving a Cauchy
problem for u. We will show that if the data f(x) and g(x) in (8.227) are
strictly positive, the solutions a(x, i) and £(*, i) must also be positive. The
case of non-negative data is considered in the exercises.
To demonstrate these results we consider the characteristic curves for
(8.224)-(8.226), which are given as
(8.228) x ± yt = constant.
On the curves x ± yt - constant we have for any function w(x, t)
dw dx
(8.229) -£ = wx -jt +wt
This yields for (8.225)
= constant •
4 METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 563
da
(8.230) «i + У** = ~Jt = "Aa + ЛK, * - 7< = constant,
and for (8.226)
dp _
(в.^1/ ™ x dt
These equations can be written as
(8.232) jt [ex'a] = A<?A'j3, * - yt = constant
(8.233) "г [extf$] = kexta, x + yt = constant.
Now if a(*,0)>0 and /3(jc,0)>0, (8.232) and (8.233) signify that
(d/dt)[exta]\t=0>0 and (d/dt)[eXtp]\t=0>0 since A>0. Initially, therefore,
exta and eA'j3 must be increasing functions of time. In view of the fact that
a(x, t) and fi(x, t) are initially positive, it is therefore not possible that they
become negative at any later time t since (8.232), (8.233) are valid for all
time. Consequently, v(x, t) = a(x> t) + /3(jt, t) is also positive.
8.5. SOLUTION METHODS FOR HIGHER ORDER EQUATIONS AND
SYSTEMS
With the exception of Chapter 2 which deals with first order equations, we
have mostly studied second order equations in this text. In this section we
show by means of a number of detailed examples how to solve problems
that involve higher order equations and systems. In a number of cases the
techniques used previously for second order equations can be applied
whereas in other cases special methods need to be introduced. We begin by
considering the higher order equations that govern the vibrations of rods
^d plates. Then we discuss the equations of fluid dynamics, Maxwell's
e4Uations of electromagnetic theory, and the equations of elasticity theory.
Both linear and nonlinear problems are considered.
Example 8.7. The Vibration of Rods. The lateral vibration of a thin
homogeneous rod is governed by the fourth order equation
(8.234) u + c2u =0
7 tt *"* **XXXX
where w(x, t) is the displacement or deflection of the rod at the time f. The
c°nstant c2 depends on the physical properties of the rod. We do not give
fte derivation of (8.234) but note that in contrast to the wave equation
564 VARIATIONAL AND OTHER METHODS
which contains the term uxx proportional to the curvature, we have th
fourth derivative term uxxxx. That is, because of the extreme rigidity of th
rod, the internal force on a portion of the rod is proportional to a fourth
rather than a second derivative of the displacement u. Invoking the result
of Section 3.3, we find that (8.234) is an equation of parabolic type and th
lines t = constant are the characteristics.
The Cauchy problem with data w(jt, 0) and ut(x, 0) assigned on the *-axis
is well posed. To see this we will follow the procedure of Section 3.5 and
look for normal mode solutions
(8.235) u(x, t) = a(k) cxp[ikx + \(k)t].
This yields
(8.236) A2(fc) + c2Jfc4 = 0
and
(8.237) A(A:) = ±/cA;2
for all real k. The stability index О = 0, so that the Cauchy problem is well
posed. Furthermore, with
(8.238) <o(k) = ck2
we have
(8.239) k(k) = ±ia>(k)
so that (8.234) is conservative as well as being of dispersive type. The
dispersion relation is ш{к) - ck2 and the group velocity d<o(k) Idk ~ 2ck
exceeds the phase velocity <o(k)/k = ck in magnitude. It is worth noting that
even though (8.234) is not of hyperbolic type it is, nevertheless, an equation
of dispersive type.
The Cauchy problem for (8.234) with initial data
(8.240) u(x, 0) -/(*), Mf(jr, 0) - g(jc), -oo < x < oo
may be solved by Fourier transform methods. Let the Fourier transform of
u(x91) be
(8.241) U( A, 0 - yj= /.. *"*«<*, t) dx .
Proceeding as in Section 5.2, we obtain the Fourier transformed problem
UXION METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 565
(8.242) ^r-^+cVtfO
(8.243) ЩХ.0У-
(8.244) d;
here Fand G are the Fourier transforms of/and g, respectively. It has the
solution
, 0 = [\ F(A) + ^
(8.245)
Inverting the transform yields
| ~iXx
(8.246) u(x, t) = y= |_w e~iXxU( A, /)
Я+(А) ехр[/(б>( А); - A*)] dk
Я"(Л) СХр[/("w(A)r" KX)] dk >
where ЦА) = cA2 and Я±(А) are the coefficients of е±1схЬ in (8.245).
Rather than evaluate (8.246) for specific choices of Ддс) and g{x) we use the
method of stationary phase to analyze the behavior of the solution u(x, t) as t
gets large. This method was presented in Example 5.14. The integrals in
(8.246) may be written as
(8.247) I±(x91) = ~= |_ HJ, A) exp[i>±(*, fi A)] <*A
with
(8-248) <p± = ±л>( A)? - kx = ±cA2^ - Ajc .
We assume that x and t are large as was done for the solution E.266) of the
Klein-Gordon equation in Example 5.14. Then the stationary points are
given as
(8.249) ^g- :
so that
566 VARIATIONAL AND OTHER METHODS
(8.250) A-A± = ±—.
Thus, say, if *>0, we have A+ = */2cf and A_ = -xllct as positive and
negative stationary points for the integrals /+ and /_, respectively. Also
(8.251) **l*'* **)
and [d2<p±(\±)]/d\2 = ±2ct so that
(8 252^ {
(8252) K
Furthermore,
(8.253)
so that on using an appropriate modification of E.256) we have
(8.254) u(x, t) = /+(x, 0 + /_(*, 0 - ^^ exp[- |£ + / j
If the initial data/(jc) and g(x) are real valued, it is easy to see, on using the
definition of the Fourier transform, that Я+(А) and Я_(-А) are complex
conjugates. Since A+ = -A_, we conclude that (8.254) is real valued.
Thus near the (group) lines xllct = constant, the solution of the initial
value problem for (8.234) for large t is approximated by (8.254) and the
solution decays like 1/V7 as f-»oo.
If the rod is of finite extent, say, of length /, we must introduce boundary
conditions at x = 0 and x = /. The boundary conditions
(8.255) и@,/) = ия@,0 = 0,
(8.256) м(/,0 = "х(/,0 = 0
imply that the rod is clamped at both edges so that и and the slope ux both
vanish. The initial and boundary value problem for m(jc, t) in the interval
0< x < I with и satisfying (8.234), the initial conditions (8.240) in the finite
interval, and the boundary conditions (8.255), (8.256) can be solved by
separation of variables.
We set
f METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 567
u(x, t) = F(x)g(t)
•n (8.234) and obtain
G\t) = F"\x) __ x2^
with -A2 as the separation constant. This gives
(8.259) G"@ + ( AcJG@ = 0,
(8.260) F"'(x)-A2F(*) = 0.
The conditions (8.255), (8.256) imply that
(8.261) F@) = F'@) = F(l) = Ff(l) = 0 ,
so that (8.260), (8.261) is an eigenvalue problem for F(x). We do not discuss
general properties of higher order eigenvalue problems but solve the
preceding problem directly.
The general solution of (8.260) is (for A>0)
(8.262) F(x) = a cos(VAjc) + b cosh(Vb:) + с sin(Vbr) + d sinh(VAx).
The conditions F@) = F'@) = 0 imply that b = -a and d = -c. Thus
(8.263) F(x) = a[cos(v"Ajc) - cosh(VAjc)] + c[sin(Vbr) - sinh(VrAjc)].
At x = / we obtain
(8.264) a[cos(VXl) - cosh(V^V/)] + c[sin(VA/) - sinh(VA/)] = 0 ,
(8.265) a[-sin(VA/) - sinh(VA/)] + c[cos(Vl/) - cosh(Vl/)] = 0 .
For a and с to be nonzero, the determinant of the coefficients of the system
(8.264), (8.265) must vanish. That is,
(8.266) ^(VA/) - cosh(VT/) sin(VA/) - sinh(VI/)
-sin(VA/) - sinh(VT/) cos(Va/) - cosh(VI/)
= cos2(VA/) + sin2(VA/) + cosh2(VA/) - sinh2(VA/)
- 2 cos(VA/) cosh(VAZ) = 2[1 - cos(VA/) cosh(VA/)] = 0 .
■*"e eigenvalue equation (8.266) can be written as
(8-267) cosh(VAZ) - sec(VAZ).
568
VARIATIONAL AND OTHER METHODS
Figure 8.4. The graphs of cosh x and sec x.
As the graph in Figure 8.4 shows, the curves for cosh(VT/) and sec(VA/)
have infinitely many intersections. [Zero is not an eigenvalue since F(x)
vanishes for A = 0.] Clearly, the eigenvalues tend to infinity. Since
cosh(VAV) grows exponentially with VX, the large eigenvalues are given
approximately by the zeros of cos(VAZ). Let Ak(k = 1,2,...) denote the
eigenvalues. Then a set of eigenfunctions is given by
- cosh(\A>)]
(8.268) Fk(x) = [si
The Fk(x) are not normalized. By direct verification it can be shown that
(8.269)
if \k Ф Xr Furthermore, it can be shown that the set {Fk(x)} is complete and
any smooth function in 0 < x < I can be expanded in a series of the {Fk(x)}-
On solving for Gk(t) from (8.259) we have
(8.270)
Gk{t) = ak cos( kkct)
with ak and j3* as arbitrary constants. The formal solution of (8.234) is
obtained by superposition as
(8.271) u(x, 0=2) Fk(x)Gk(t) .
Formally applying the initial conditions to the series (8.271) leads to the
lUTION METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 569
cification of ajfc and pk jn terms of the (Fourier) coefficients of the
functions f(x) and g(x) when expanded in a series of eigenfunctions Fk(x).
We consider other types of boundary conditions for (8.234) and prop-
rties of the corresponding eigenvalue problems for F(x) in the exercises.
Example 8.8. The Vibration of Plates. The vibration of a thin plate is
governed by the equation
where и = u(x, y, t) is the transverse displacement of the point (x, y) on the
plate at the time t, and V2 is the two-dimensional Laplacian. The constant c2
depends on the physical properties of the plate. The wave equation that
governs the vibration of a membrane is given as utt - c2V2u = 0 and contains
only the Laplacian operator, whereas (8.272) contains the iterated Laplacian
because of the greater rigidity of the plate. If we assume that the plate
occupies the region G in the (*, y)-plane and is clamped at the boundary or
edge dGj we have the boundary conditions
(8.273)
= 0 ^
' дП
= 0.
dG
Also, m(jc, y, t) must satisfy the initial conditions
(8.274) u(x, y, 0) = f(x, y)9 utix9 y, 0) = g(x9 y) .
Because (8.272) is of fourth order in the spatial variables, we require two
boundary conditions. However, since there are only two time derivatives in
(8.272) we have two initial conditions. The problem (8.272)-(8.274) may be
shown to be well posed.
We attempt to solve this initial and boundary value problem by using
separation of variables. Let
This yields, on inserting (8.275) into (8.272),
(8.276) jg40, = .V2V2F(x,y) =
c2G(t) F(x,y)
where -A is the separation constant. The equation for G(t) is
(8-277)
F(x, y) must satisfy
570 VARIATIONAL AND OTHER METHODS
(8.278) V2V2F-AF = 0
for (x, y)GG with the boundary conditions
(8.279) F(x, у)
dG дП
dG
= 0.
That is, we obtain an eigenvalue problem for F(jc, y).
To make any further progress with this problem we must be able to
determine the eigenvalues A and the eigenfunctions Ffor specific regions. It
turns out, however, that (8.278) is only separable in polar coordinates so
that more complicated or, possibly, approximation methods are needed if
the problem does not involve a circular clamped plate. Indeed, in rectangu-
rectangular coordinates (8.278) takes the form
(8.280) Fxxxx + 2Fxxyy + Fyw - AF = 0 .
With
(8.281) F(x,y) = A(x)B(y)
we have
(8.282) А"" В + 2Л"Я" + ВтА - KAB - 0
and it is not possible to separate the variables x and у in (8.282).
For the case of a circular plate it is possible to obtain the eigenfunctions
F(x, y) by proceeding as follows. Let A = k4 for simplicity of notation and
factor (8.278) as
(8.283) (V2V2 - k4)F = (V2 + A:2)(V2 - k2)F = 0 .
This factorization can be carried out irrespective of the choice of coordi-
coordinates, but only for the polar coordinate problem considered does it lead to a
solution of the eigenvalue problem (8.278), (8.279). We express F in polar
coordinates as F= F(r, 0), and assume the region G is the interior of the
circle r < R with r = R as the boundary dG. Then F(r, 0) must be bounded
at r = 0 and must be periodic of period 2тг in 6—that is, F(r, 0 + 2тг)-
/r(r? 0)_since F must be single valued in G. Thus if F(r, 0) satisfies either of
the equations
(8.284) (V2 ± k2)F = Frr + \ Fr + « Fee ± *2F = 0,
it is a solution of (8.278) since the order of the operators in (8.283) can be
interchanged.
oLljTION METHODS FOR HIGHER ORDER EQUATIONS
The equations (8.284) can be solved by separation of variables. Let
(8.285) F(r,0) = V(r)W(e),
and (8.284) can be expressed as
(8.286) V\r) + i V\r) ± k2V{r) - ^ y(r) = 0 ,
(8.287) - W'W + «2WW = 0,
after separating the variables with n2 equal to the separation constant.
The periodicity of F(r, 0) implies that W($ + 2тг) = W@) so that n must
be a (positive) integer or zero (i.e., n = 0,1,2,...). Consequently, we have
W (в) = cos w0 and WnF) = sin w0 as the appropriate solutions of (8.287).
For each value of n, (8.286) is a Bessel equation of order n if the plus sign
(i.e., +k2) is chosen. When the minus sign (i.e., -k2) is selected in (8.286),
we have the modified Bessel equation of order n.
The product functions Fn(r, 0) = Vn(r)Wn@) must be bounded at r = 0, so
that the appropriate solution of (8.286) with the term +k2 is Vn = Jn(kr),
the Bessel function of wth order. For the case -k1 we obtain Vn - In(kr),
the modified Bessel function of nth order. Consequently, we find that for
each integer n we have the solutions
(8.288) Fn(r, 6) = [anJn(kr) + yjn(kr)] cos пв
of (8.283) where an, /3n, yn, and 8n are constants. On applying the boundary
conditions (8.279) to each of the Frt, we determine the eigenvalues A = к .
For the circular region G, the normal derivative dFldn becomes dFldr on
r = R. Since the boundary conditions must be valid for all 0, we see that the
coefficients of cos пв and sin пв must vanish separately once the boundary
conditions are applied. The coefficient of cos пв yields the system
A second system with an and yn replaced by /3n and 8n, respectively, results
from the coefficient of sin пв. Since the coefficients an and yn must not
vanish, we conclude that the determinant of the coefficients in (8.289)
vanishes and this yields the transcendental equation
(8.290) Jn(kR)Vn{kR) - In(kR)J'n(kR) = 0
for the determination of the eigenvalues. (The second system yields the
572 VARIATIONAL AND OTHER METHODS
same result.) The appropriate values of к must be specified approximate!
and we do not consider this matter here. Once the full set of eigenvalues h *
been found we may solve (8.277) for the function G(t). Then on summinS
over all the product solutions (8.275) we may specify the two sets Of
arbitrary constants in the solution of (8.277) so as to satisfy the initial
conditions (8.274).
Since the general problem is quite complicated, we consider a simplified
case where the initial data (8.274) are functions of the radial variable r only
that is,
(8.291) u(r, в, 0) = /(r), ut(r, 0,0) = g(r) .
Then the eigenfunctions F may be chosen to be independent of 0, and we
obtain instead of (8.288)
(8.292) F(r) = Oo/0(*r) + yolo(kr) .
The equations (8.289), (8.290) remain valid with л = 0. Let km(m =
1,2, 3,. . .) represent the (real) roots of (8.290) with n = 0. The (unnormal-
ized) eigenfunctions Fm(r) then have the form
(8.293) Fm(r) - I0(kmR)J0(kmr) - J0(kmR)I0(kmr) .
It is not hard to show that the {Fm(r)} are an orthogonal set over the
interval 0 < r < R with weight function r. With Am = kAm the solutions of
(8.277) are
(8.294) Gm(t) = am cos(k2mct) + bm sin(*»
with arbitrary constants am and bm. The formal series и = Е^=1 Fm(r)Gm(t)
satisfies the equation (8.272) in the disk 0 < r < R, the boundary conditions
(8.273) on r = Л, and the initial conditions (8.291) once the constants am
and bm in (8.294) are specified in terms of f(r) and g(r). Using the
orthogonality properties of the eigenfunctions {Fm(r)} this is easy to carry
out. The eigenfunctions for the eigenvalue problem (8.278), (8.279) are
known to form a complete set, so that the expansions discussed are valid if
the initial data satisfy certain smoothness conditions.
It is of considerable interest to study the stationary version of the
vibrating plate equation (8.272). We consider displacements и that are
independent of time so that и = u(x, y), and find that u(x, v) satisfies the
biharmonic equation
(8.295) V2V2m = 0.
Clearly, any harmonic function и (i.e., a solution of Laplace's equation
SOLUTION METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 573
V2w = 0) Win also satisfy (8.295). This fact can occasionally be used to
dvantage in solving boundary value problems for the biharmonic equation.
Using direct verification it may be shown that the function
(8,296) « = (r2 - R2)v + w
is a solution of the biharmonic equation if v and w are both harmonic
functions, r2 = x2 + y2, and R2 is a constant. Given the boundary value
problem for (8.295) in the disk 0< r < R with the boundary conditions
(8.297)
аи
дГ
we may solve this problem by choosing v and w in (8.296) appropriately.
The boundary condition u(R, 6) =f implies that
(8.298) u(R,e) = w(R,6)=f.
Thus w must be a solution of Laplace's equation in the disk 0 < r < R which
satisfies the Dirichlet condition w =/on the boundary r = R. This problem
has been solved previously and completely specifies w. The condition
duldr = g on r = R implies
(8.299)
dr
dr
Since w has already been specified, dw{R, B)ldr is a known function and
(8.299) yields v(R, 0) = (l/2R)g- (l/2R)dw(R, B)ldr as the Dirichlet
boundary condition for the harmonic function v. Again, и is thereby
uniquely specified and we have obtained a solution of the boundary a
solution of the boundary value problem (8.295) and (8.297) in terms of the
solution of two boundary value problems for Laplace's equation in a circle.
We complete our discussion of the vibrating plate by briefly considering
the fundamental solution and Green's function methods for the biharmonic
equation.
We proceed as in Section 6.7 and consider the inhomogeneous bihar-
monic equation
(8-300) V2V2u = -F(x,y).
With Po equal to the point (x0, ;y0), we integrate over a region R with
boundary dR that contains Po in its interior and obtain
(8.301) J jR
574 VARIATIONAL AND OTHER METHOD
where we have applied the divergence theorem and chosen F(x, y)
represent a point source or, equivalently, a Dirac delta function with
singularity at Po so that the integral of F equals unity. Introducing poi
coordinates with the pole located at Po and taking the limit in (8.301) Г
<?/?-> Po, we obtain s
rde
x=xo + r cos в
y=yo+r sine
if we let R represent a disk of radius r centered at Po. Continuing as in
Section 6.7 we assume that и is a function r only and we have
(8.303) *
As a result we obtain the approximate equation
(8.304) 2 '[If (,£")].-!.
v y <?r Lr dr \ dr)\
Integrating out this equation and retaining only the most singular terms
yields the fundamental solution
(8.305) w = -^-r2logr
where r2 = (x - x0J + (y - y0J. It can be verified that (8.305) is a solution
of the biharmonic equation everywhere except at r = 0. In fact, apart from
the constant factor, the solution (8.305) may be derived by looking for a
solution of (8.295) that depends only on r.
On the basis of the identity
(8.306) mV2V2u - yV2V2w =V-p
where
(8.307) p = wVV2y 2 (V2)V (V2)V
we obtain the Green's theorem
(8.308) 11 [mV2V2u - uV2V2m] dx dy
on using the divergence theorem.
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 575
Green's function К for the biharmonic equation with clamped plate
boundary conditions is defined to be a solution of
(8.309) V2V2K=-8
where the two-dimensional Dirac delta function 8 is singular at the point
<£ n) in G and К satisfies the boundary conditions
Ж
(8.3Ю) К dG = —
dG
-0.
No if "(*> У) satisfies the inhomogeneous equation (8.300) in G and
assumes given boundary conditions и and duldn on dG, the Green's
theorem (8.308) (where we set v = K) yields the result
(8.311) «U, v) = / / KF dx dy + \bg [(V*) £ - и д-^р] ds
G
where all the terms on the right of (8.311) are known.
Using special techniques it is possible to obtain the Green's function К
for the case of circular clamped plate. We do not derive or exhibit this
Green's function but note that in view of (8.305) К can be expressed as
(8.312) K(x, y, (, v) = - -^ r1 log r + V
where У is a regular solution of the homogeneous biharmonie^equation.
Applying the boundary conditions (8.310) to К yields a set of boundary
conditions for V and dVldn on dG. Since the boundary value problem for
the biharmonic equation with clamped plate boundary conditions was solved
previously, it is possible to determine V and, thereby, К for the circular
clamped plate.
Example 8.9. The Equations of Fluid Dynamics. We begin by deriving
the Euler equations of motion for a nonviscous or inviscid fluid. Let
и - n(x, y9 2, t) be the velocity vector of a fluid particle situated at the point
(*» У, z) at the time t. Also let p = p(x, y, z, t) and p = p(jt, y, z, t) repre-
represent the pressure and the density, respectively, of the fluid. We assume that
no external forces are acting on the fluid.
We consider a volume of the fluid occupying a region R with boundary
. Я. Assuming there are no sources or sinks for the fluid in R, the rate of
^crease of the fluid in R in unit time must be balanced by the flux or inflow
01 fluid through the boundary dR. Analytically, this may be expressed as
576 VARIATIONAL AND OTHER METHODS
where n is the exterior unit normal vector and the minus sign occurs
we are concerned with the inflow. (We note that p do is an element of m^
and that pwnds represents the flow of mass through the boundary element
ds.) Applying the divergence theorem to the integral over dR and bringin
the time derivative inside the integral over R in (8.313)—since R is fixed
time—we obtain
(8.314)
Since (8.314) must be valid for an arbitrary region R, we conclude that
integrand vanishes identically and obtain
(8.315) p, + V-(pu) = 0,
which is known as the equation of continuity.
Next we consider the law of conservation of momentum for the fluid.
Since we are neglecting frictional forces due to the effects of viscosity and all
external forces, the only force acting on the volume of fluid in the region R
is due to the pressure along the boundary dR. The total pressure is given as
(8.316) p=-f
JdG
which is effectively a resultant of all the internally directed normal pressures
at the boundary. According to Newton's second law of motion, the pressure
force equals the mass times the acceleration of the fluid element or particle.
The acceleration of a particle is given by du/dt, but since the particle is
moving along a path x = x(t), у = y(t), z = z(t), we obtain from the chain
rule
/o *>*^ du <?u <?u dx du dy du dz , _4
Here we have used the fact that
(8-318) u = [x'(t),y'(t),z'(t)]
and
(8.319) m.V±± ±
mVx+y + x.
The derivative du/dt in (8.317) is often referred to as a material derivative.
The total momentum of the fluid particles in R must equal the resultant
pressure forces and we obtain
[ METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 577
(8.320)
R
ince p dv is an element of mass. Using the gradient theorem, which is a
Simple consequence of the divergence theorem, the integral over dR can be
Converted into an integral over R and (8,320) becomes
(8.321)
The arbitrariness of R implies the vanishing of the integrand and we obtain
the equation of conservation of momentum
(8.322) u, + (u • V)u + [-)Vp = 0 .
Further we assume that we are dealing with adiabatic flow. This means
that the entropy sy a thermodynamic quantity, remains constant for each
fluid particle as it moves along the particle path (jc(j), y(i), z(t)). (A
thermodynamic relation between the entropy and other physical properties
of the fluid is given below in our discussion of the Navier-Stokes equations.)
Thus the material derivative of the entropy dsldt must equal zero. On using
the chain rule and (8.318), (8.319) we obtain
(8.323) s, + u-Vs = 0.
The equations (8.315), (8.322), and (8.323) are incomplete as they stand
since we have only five equations and six unknowns (i.e. p, p, s, and the
three components of u). To complete the system we need an equation of
state which we give as
(8.324) P=f(P,s),
where/(p, s) is a function that is determined from thermodynamic conside-
considerations and that differs for different fluids.
The equations (8.315), (8.322), (8.323), and (8.324) constitute Euler's
Equations of fluid dynamics in the case of inviscid, adiabatic fluid flow. We
note that if the entropy s is initially uniform, say s-s0, then dsldt -0
unplies that it remains uniform for all time, since each particle carries the
same constant value s = s0 for the entropy. (The flow is assumed to remain
smooth.) A fluid flow with uniform entropy is said to be isentropic. In that
case we can drop the equation (8.323) and replace the equation of state
(8.324) by
578 VARIATIONAL AND OTHER METHODS
(8.325) P=/(p),
since s = s0 for all time. The reduced set of equations (8.315), (8.322) a d
(8.325) are Eulefs equations for inviscid, isentropic flow. Each of the*
systems is a quasilinear first order system and it is not an easy matter
solve these systems in general. Consequently, certain simplifying assum °
tions are introduced to deal with specific types of fluid flow.
In dealing with the flow of liquids rather than gases we assume the flow is
isentropic and that the density p is identically constant; that is,
(8.326) p = constant.
This means that we are dealing with an incompressible fluid and (8.326)
replaces the equation of state (8.325). In view of (8.326), the equation of
continuity (8.315) reduces to
(8.327) V-u = 0.
We further assume that the fluid flow is irrotational which means that the
velocity field u has the property that its curl vanishes; that is,
(8.328) Vxu = 0.
On the basis of Stokes' theorem from vector analysis we conclude that if
(8.328) is satisfied everywhere, the integral of the velocity around any closed
curve vanishes; that is,
(8.329) Ф и • dx = 0 .
The integral in (8.329) defines the circulation of the fluid around the closed
curve and irrotationality means that no (nonzero) circulation exists. Since
the Euler equations describe the fluid motion as it develops in time, we
cannot necessarily assume that if the flow is irrotational at one time it may
not become rotational at a later time. It can be shown, however, that the
fluid remains rotation free for all time (based on Euler's equations) if it was
so at a given instant of time. Rather than consider this question we examine
steady fluid flow for which the pressure p and the velocity u are independent
of time.
Now it is easy to show that if (8.329) is satisfied for all closed curves, the
velocity vector u can be derived from a potential function <p and can be
expressed as
(8.330) u(x, y, z) = V<p(x, y, z) .
Inserting (8.330) into (8.327) gives
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 579
(8.331) V-u = V-V<p=VV = 0,
that <p satisfies Laplace's equation. Once Laplace's equation for the
velocity potential <p is solved, we may determine the pressure p from (8.322)
where we set u, = 0. We see that for steady, incompressible, and irrotational
aow, the study of the fluid motion is reduced to the solution of Laplace's
eauation, subject, of course, to appropriate boundary conditions.
A different analysis of Euler's equations for isentropic flow that results in
a study of the (linear) wave equation is a linearization procedure that leads
to the theory of acoustics. We perturb the solutions u, p> and p around a
(constant) equilibrium state u = 0, p =p0, and p = p0 [with p0 =/(p0)] and
write
(8.332) \p=Po + €Pi+"
[p = Po + €P\ + " '
where 0 < e <€ 1 with e = constant and the dots refer to terms that are of
higher order in e. Inserting (8.332) into Euler's equations and retaining only
terms of order e gives
(8.333) ^
(8.334) ^Ь
as the linearized versions of the Euler equations.
From (8.334) we find that
(8.336) Ul(x, y, z, 0 = Ul(x, y, z, 0) - - V Г Pl dt.
Po Jo
K, initially, u, = 0 or if ux can be written as
0-e., it can be derived from a potential), we obtain
Ul(jc, y9 z, t) = -v[^r + 1 £ Pl Л] ^ _v*
<p equals the bracketed term in (8.338). Inserting (8.338) into (8.334)
yields
580 VARIATIONAL AND OTHER METHODS
(8.339)
and we may choose
(8.340) pt
Then (8.335) implies that
(8.34D
Inserting (8.338) and (8.341) into (8.333) yields
(8.342) 9,,V
so that q> satisfies the wave equation if/'(po)>O. In fact, for a polytropic
gas we have
(8.343) f(p) = Ap\
where A and у are positive constants with у > 1 and given as у = cplcv with
cp and cv equal to the specific heats of the gas at constant pressure and
constant volume, respectively. We set
(8.344) c2=/'(p),
so that c(po) = V/'(Po) *n (8.342) represents the propagation speed for
disturbances. We note that not only <p but p19 px, and each component of nx
also satisfy the wave equation (8.342).
For the one-dimensional case we replace u by и and obtain the one-
dimensional wave equation for <p(x, t) with the general solution
(8.345) <p(x, t) = F(x - cot) + G(x + cQt) ,
with c0 = c(p0). The function <p must satisfy appropriate initial and boundary
conditions and then w1? рг, and pt can be expressed in terms of <p by means
of (8.338), (8.340), and (8.341). In particular, if the solution <p results in
unidirectional wave motion, that is,
(8.346) <p(x, t) = F(x - cot) ,
and only right moving waves occur, we find that
oLl[TION METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS
581
(8.347)
Ul = -F'(x-cot)
Pi = -c0p0F'(x - cot)
so
that both p, and p, can be expressed in terms of Mt in the form
(8.348)
Pi =
This result suggests that we look for special solutions of the full nonlinear
system of Euler's equations of isentropic flow for the one-dimensional case
in the form
(8.349)
P=p(u),
where u(x, t) is the velocity. Using (8.344), the one-dimensional form of
Euler's equations are
(8.350)
(8.351)
p, + pux
u, + uux
Px=0.
Since p has been eliminated from (8.351) we need only use p = p(u).
Inserting this into (8.350) and (8.351) gives
(8.352)
(8-353)
where p' = dpldu. Equating the right sides of these equations yields
(8.354) (cp'f = p2
so that
(8.355) p' = ±P-.
с
Inserting (8.355) in (8.352) gives, with c(u) = c[p(u)],
(8-356) ut + [u±c(u)]ux=0,
a first order quasilinear (wave) equation studied in Chapter 2. Once u(x, t) is
582 VARIATIONAL AND OTHER METHODS
determined from (8.356), we may use the solution of (8.355) to specify
Rather than deal with the general case, we consider a poly tropic gas fo
which (8.343) is valid and choose the plus sign in (8.355), (8.356). The
(8.357) с* = ±
and (8.355) takes the form
(8.358)
This ordinary differential equation is separable. Assuming that at и = 0 we
have p = p0 and c(p)\u=0 = c(p0) = c0, the solution in terms of c(u) is
(8.359)
from which p can be determined. Then (8.356) takes the form (with the plus
sign chosen)
(8.360) и,
If the initial waveform of u(x, t) at t = 0 is given as u(x, 0) = h(x), the
result of Exercise 2.3.7 implies that
(8.361)
is the implicit form of the solution of (8.359). This represents a wave
traveling to the right for positive и and, according to the results of Section
2.3, each point x on the waveform u(x, t) travels with speed
(8.362) % = Co
In fact, for the general case of (8.356) we have
(8.363) ^ = и ± с .
We see from (8.362) and (8.363) that ±c represent an addition to the actual
velocity и of the fluid at which disturbances propagate in the fluid. There-
Therefore, с = V/'(p) is called the (local) speed of sound of the fluid or gas. For
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 583
the linearized equations of acoustics where a perturbation around a zero
velocity was carried out, c0 = V/(Po) represents the (constant) speed of
xvave propagation or speed of sound for the gas. By choosing the minus sign
in (8.355), (8.356), we may construct solutions that represent waves travel-
travelog to the left. If we assume that the velocity и is small and approximate c(u)
by с(ио) = сО9 our results reduce to those for the equations of linear
acoustics.
In general, solutions of (8.356) in the form
(8.364) u = h[x-(u± c(u))t]
are known as simple waves and represent unidirectional wave motion. An
alternative method for deriving simple waves is based on the use of the
Riemann invariants for the one-dimensional Euler equations either in the
adiabatic or isentropic case (see Example 3.5 and the exercises for this
section). Simple waves play an important role in a number of problems in
fluid dynamics, a particular case of which is the piston problem. This
concerns the (one-dimensional) flow that results when a piston is pushed
into or withdrawn from a long tube filled with gas at rest in a uniform state.
(That is, the velocity и = 0 and the density, pressure, and entropy are all
constant.) As a result, the flow can be assumed to be isentropic and can be
described in terms of a simple wave, under certain circumstances to be
considered in the exercises.
However, as was found for solutions of quasilinear wave equations in
Section 2.3 the simple wave solutions can break down after a finite time
because they become multivalued or singular. In that case the solution can
be continued beyond the breaking time by introducing shock waves as was
done in Section 2.3. A set of shock conditions for the one-dimensional case
is considered in the exercises.
Alternatively, we may conclude that effects such as viscosity and depen-
dependence on temperature variations that were neglected in our derivation of the
fluid flow equations must be included. These effects introduce higher
derivative terms into the system of fluid dynamics equations and, as we have
seen on a number of occasions, the inclusion of higher derivative terms may
have a smoothing effect on the solutions so that no breakdown occurs at all.
Rather than discuss the general case, we restrict ourselves to one
dimension. Without presenting a derivation we write down the Navier-
Stokes equations for viscous fluid flow in one dimension. We choose as the
unknowns the velocity u, the density p, the pressure p, and the tempera-
temperature Г. We have
(8-365) p, + pux + upx = 0 ,
584 VARIATIONAL AND OTHER METHOfts
(8.367) pcv[Tt + uTx] + pux = kTxx + | n(uxf ,
(8.368) p = RPT.
In these equations /i is the (constant) coefficient of viscosity, к is th
coefficient of heat conduction, cp and cv are the specific heats at constant
pressure and volume, respectively, and R = cp - cv is the gas constant. We
have assumed there are no external forces or heat sources. (We note that in
the literature the Navier-Stokes equations are often written in different
forms.)
The first two equations (8.365), (8.366) are the equations of continuity
and conservation of momentum. The new term in (8.366) represents forces
due to viscous effects and vanishes when /i = 0. The equation (8.367)
characterizes the conservation of energy. If we neglect conductivity and
viscosity (i.e., we put к = /i = 0), we obtain
(8.369)
From (8.365) we can replace ux by —(l/p)(p, + upx) and obtain
(8.370) p[e, + uex] - (^)[Pl + upx] = 0,
where the interned energy e is given as
(8.371) e = coT.
Using the material derivative gives
(8.372)
dt \p2/ dt dt F dt \p
Then the thermodynamic relation
(8.373) Tds = de+p
where s is the entropy reduces (8.372) to
(8.374) Tf=0-
We see that the entropy for individual fluid particles remains constant so
that this represents adiabatic flow. If at the initial time we have uniform
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS
585
ntropy, it remains so for all time and this corresponds to the isentropic flow
considered.
The final equation (8.368) is an equation of state for a perfect gas that
takes the place of (8.324). We do not discuss the relationship between these
equations, say, if /i = к ~ 0. In any case, the introduction of viscosity and
heat conduction increases the order of the system of equations from first to
second order by means of the addition of derivatives in x. It is shown in the
exercises that for isentropic flow the Euler equations are a hyperbolic system
whose characteristic curves for a given solution are (8.363) with с replaced
ky c = c(p). The equations of adiabatic flow are also hyperbolic as shown in
Exercise 3.3.22. The full Navier-Stokes equations are of parabolic type but
in view of their complexity not much appears to be gained from this
information. We do not discuss the solution of these equations here but in
Section 10.3 we present an approximation method that enables us to draw
some conclusions about the effect of additional higher derivative terms in
the Navier-Stokes equations on certain solutions of the fluid flow equations.
We conclude our discussion of fluid flow by examining the Euler equa-
equations in the case of steady (i.e., time independent) two-dimensional isen-
isentropic flow. With м(х, у) and v(x, y) as the velocity components, p(x, v) as
the density, and c2 =/'(p), we have from (8.315) and (8.322),
(8.375) puux + pvuy + c2px = 0 ,
(8-376) puvx + pWy + c2py = 0 ,
(8.377) pux + pvy + upx + vpy = 0.
The system (8.375)-(8.377) can be written in matrix form as
(8.378) Ayvx + Bmv = 0
where the matrices A and В and the vector w are given as
(8.379)
pu
0
. P
0
pu
0
г
0
и
, в =
pv 0 0
0 pv с2
0 р v
■-Н-
This first order quasilinear system may be classified according to the method
of Section 3.3 for a fixed solution w. The characteristic curves у = у(х)
determined from [see C.97)]
are
(8.380)
and this gives
p(v-y'u) 0
0 p{v-y'u)
-РУ' P
-c2y'
(v - y'u)_
= 0,
586 VARIATIONAL AND OTHER METHOD
(8.381) [v - uyf][(v - uy'f - A + уf2)c2] = 0 .
The first factor of (8.381) yields the characteristic curves
(8.382) у = ^ = - ,
v ' J dx и
and these represent the streamlines of the flow, that is, the curves tangent to
the velocity vectors whose components are и and u.
The second factor in (8.381) can be written as
(8.383) (c2 - u2)y'2 + Buv)yr + c2 - v2 = 0 .
The roots yf of this quadratic equation may be characterized by means of
the discriminant, which is
(8.384) D = 4c2[w2 + v2 - c2],
as is easy to show. If u2 + v2 > c2, both roots y' are real and distinct and
neither is equal to the third root (8.382). Thus there are three real and
distinct characteristics, and the system is (totally) hyperbolic. Since с is the
sound speed and Vw2 + v2 is the speed of the fluid, this case corresponds to
supersonic flow. If u2 + v2<c2, the corresponding roots yf are complex.
This case corresponds to subsonic flow. (According to our classification
scheme, since not all the y* are complex this is not an elliptic system.) If
u2 + v2 = c2 the root is real and double and the situation is complicated in
that case. The flow may be termed sonic flow since the fluid speed and
sound speed coincide.
The distinction between subsonic and supersonic flow can be brought out
in another, simpler manner by linearizing the system (8.375)-(8.377). We
set
(8.385)
v = evx +
p = p0 + е
where u0 and p0 are constant and 0< e < 1. The dots correspond to higher
order terms in €. In (8.385) we are perturbing around a constant state in
which there is a uniform flow with velocity u0 >0 parallel to the x-axis.
Inserting (8.385) in (8.375)»(8.377) and retaining only the terms of order
e yields
(8.386) ^М°5 + С«1г = 0'
(8-387) AHo_l + cgJi»o,
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 587
(8.388) "о^Г + 'о^ + Ио^О.
where c0 = c(p0). From (8.388) we have
(8.389) Po jr = -"o aJ
дих
in
M° dx u0 dx '
view of (8.386). Since д\х1дхду = д\/дудх,
д\ c20 д\
Po дудх щ ay1 '
^2-
a2vv г Co ] <?Vi
which follow from (8.387) and (8.389) and imply that
(8.392) (l-M2)^ + ^=0,
where the Mach number M is defined as
(8.393) M=^,
co
that is, it is the ratio of the underlying constant velocity to the uniform
sound speed co = c(po).
If the Mach number exceeds unity (i.e., M>1), (8.392) is clearly of
hyperbolic type. This corresponds to supersonic flow since u0 > c0 in the
unperturbed state. If M < 1, (8.392) is of elliptic type and we have subsonic
flow since M0<c0. If M = 1, the equation is parabolic and we have sonic
flow since u0 = c0. We note that the characteristics corresponding to the
streamlines for the nonlinear system [i.e., (8.382)] play no role in the
linearized problem. In fact, if we insert (8.385) into (8.382), we obtain to
leading order у' = еиг/и0 and from (8.383) we have у '2 = M2 - 1. Thus with
€ = 0 the characteristic equations are y' = 0 and y' = ±VM2 - 1. The solu-
solutions of y' = 0 (i.e., у = constant) are the streamlines of the steady uniform
fl^___M_=M0, and u = 0. The characteristics of (8.392) (i.e., y±
Y^_ - 1 x = constant) are solutions of the equation y' = ±VM2 - 1.
Example 8.10. Maxwell's Equations of Electromagnetic Theory. The
electric and magnetic field vectors E and H, respectively, which are studied in
588 VARIATIONAL AND OTHER METHODS
electromagnetic theory, are solutions of the fundamental set of equatio
known as Maxwell's equations. Each of these equations is based on ^
experimentally observed property of the electric and magnetic fields. We /
not present a derivation of these equations but write out the equation in щк°
units and discuss some of their properties. s
Maxwell's equations are given as follows:
(8.394) VxE=~,
at
(8.395) VXH=4?+J,
at
(8.396) VB = 0,
(8.397) VD = p.
In addition there are the constitutive relations that express B, D, and J in
terms of E and H. These equations are
(8.398) D = €E ,
(8.399) В = fMH ,
(8.400) J=<rE.
The vectors D, B, and J are the electric displacement, the magnetic
induction, and the conduction current density, respectively. Further, p is the
density of electric charges, e is the dielectric constant, and jx is the magnetic
permeability.
We now introduce some simplifying assumptions that render this system
more tractable. In problems of electrostatics we set all time derivatives equal
to zero. From (8.394) we then find that
(8.401) VxE = 0
so that E is an irrotational field. As shown in the preceding example, this
implies that E can be derived from a potential so that
(8.402) E=-V<p,
with <p equal to the electrostatic potential. Further, if € is a constant in
(8.398), we have from (8.397)
(8.403) V-E=^ = -V-V<p,
which shows that
METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 589
(8.404) VV = -f
<j (p is a solution of Poisson's equation. If the charge density p vanishes, <p
antisfies Laplace's equation. If the medium is nonconducting, so that a = 0
Snd J = °> and' in addition> M is a constant, the same result is obtained for
the magnetic field H. However, since there are no magnetic charges, the
magnetostatic potential satisfies Laplace's equation.
If the medium is homogeneous (i.e., €, fx, and a are constant) and we
also have p = 0, the following simplification of Maxwell's equations results.
We apply the curl operator to (8.394) and obtain
(8.405) VxVxE=-Vx ^^- = -/i^(VxH).
Using the vector identity
(8.406) V x V x E = V(V- E) - V2E
and (8.395) for H gives
(8.407) V(V-E)-'
(^) r (fia)
But V-D - €V-E = 0, in view of (8.397) so that
(8.408) V2E-(^)^^
with an identical equation satisfied by H. Thus each Cartesian component of
E and H satisfies a telegrapher's equation of the form
(8-409) utt-c2V2u + Xut = 0,
where с2 = A/б^) and Л=о7б. If the medium is nonconducting so that
a==0, each component satisfies the wave equation. In either case distur-
disturbances are propagated with the speed с We note that if p ^ 0, we obtain an
^homogeneous form of (8.408).
It is also possible to introduce scalar and vector potential functions to
feat the time dependent Maxwell's equations. We assume the medium is
homogeneous as well as nonconducting so that a = 0 and J = 0.
The equation (8.396), that is,
(8.410) V-B = 0,
p that В is a solenoidal field and it is shown in vector analysis that
there exists a vector potential A such that
590
(8.411)
Then (8.394) yields
(8.412)
and we obtain
(8.413)
VxE=-
Vx
B = Vx
~ dt
VARIATIONAL AND OTHER METHODS
A.
-Vx(-),
tI-o,
so that E + A, is an irrotational vector. Consequently, it can be derived from
a potential <p and we have
(8.414) E+^
dt
From the identity V x Vi/r = 0, we see that if A is replaced by
(8.415) A = A0 + V<A,
(8.411) will still be satisfied by Ao. Further, in terms of <p we have from
(8.414)
(8.416) E + ^ = -Щ - V<p = -V
so that a new scalar potential <p0 can be defined as
(8.417) <Р = <Ро~Ф{.
To remove this ambiguity we impose the Lorentz condition,
(8.418) va=^J
With this choice A and <p both satisfy wave equations as we now demon-
demonstrate.
From (8.414), (8.397), and (8.418) we have
(8.419)
so that
6
tUXlON METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 591
(8.420) с2у2*-<р<< = -?^'
which is an inhomogeneous wave equation. To obtain an equation for A we
use (8.395) in the form
(8.421) V x (fiH) = V x В = (e/t) —
and insert the expression (8.411) to obtain
(8.422) VXB = VXVXA = V(V-A)-V2A
where (8.406) and (8.414) were used. Rearranging terms and using the
Lorentz condition (8.418) yields
(8.423) (eM) ф- -V2A= -v|Va + (щ) ^] =0
so that A satisfies the homogeneous wave equation. Once A and <p are
specified, the electric and magnetic fields are given as
(8.424) E = _^_V^,
at
(8.425) H--VXA.
We have shown the relationship between Maxwell's equations and Lap-
Laplace's and Poisson's equations in the time independent case and the wave
and telegrapher's equations in the time varying case. Many problems for
Maxwell's equations can be reduced to a study of these simpler equations
subject to appropriate initial and boundary conditions. As a simple example
we consider the initial value problem for Maxwell's equations in a vacuum.
We have € = ц = constant, <r = p = 0, and J = 0 in (8.394)-(8.400). The
equations can then be written as
(8.426) VxE=-,f,
at
(8.427) VxH = € —,
dt
592 VARIATIONAL AND OTHER METHOD
(8.428) V-E=V-H = 0.
These equations are valid throughout space and initially we have
(8.429) E(jc, 0) = F(x), H(x, 0) = G(x) .
The procedure that led to (8.408) can be applied to (8.426), (8.427) to yield
the equations
(8.430) V2E
(8.431) У2Н
Each Cartesian component of E and H satisfies a scalar wave equation.
However, we have only one initial condition for E and H in (8.429) and we
must also prescribe E, and H, initially to obtain a unique solution. From
(8.426) and (8.427) we find that
(8.432)
The initial value problem for the scalar wave equation has been solved
previously in the text in one, two, and three space dimensions. Thus E(x, /)
and H(x, i) can be determined. To complete the verification of the solution
we must show that (8.428) is satisfied for *>0 provided it was true at the
time t = 0. Thus we must assume that the initial fields F and G are such that
(8.433) V-F-V-G = 0.
Since V-V x A = 0 for any vector A, we have
(8.434) 0 = V-VxE=-jtbV-H, = -/i^(V-H),
(8.435) 0 = V-VxH=eV-E, = e |-(V-E)
at
on using (8.426), (8.427). Thus V- E and V- H are constant in time and since
they vanish at t-0, they do so for all time.
For the sake of concreteness we consider a specific problem in one
dimension. Let
(8.436) E(x, 0) = ¥(x) = (sin x)j, H(x, 0) = G(jc) = (cos x)k .
Then (8.433) is satisfied and
^lON METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 593
V x F = (cos *)k, V x G = (sin x)j
so that
(8.438) E,(*, 0) - \ (sin *)j, Ht(x, 0) = - - (cos x)k .
Each component of E and H satisfies the one-dimensional wave equation
(8.439) Их*-(ч0и« = 0.
y-component of E has the initial conditions
(8.440) w(x,0) = sin*, ut(x, 0) = - sin x ,
whereas the z component of H has the initial conditions
(8.441) u(xy 0) = cos x, ut(x, 0) = - - cos x .
All other components have zero initial conditions and, therefore, vanish for
all time. Using d'Alembert's solution of the wave equation we easily obtain
(with c2 = 1 /e/it) the standing wave representations
(8.442) Е(лг, t) = (sin x)(cos ct + — sin ctjj ,
(8.443) H(jc, 0 = (cos x)(cos ct-~- sin ctjk .
Clearly, Е(лг, t) and Щх, t) satisfy the initial conditions (8.436). Further,
(8.444) V x E = (cos x)(cos ct + — sin ctjk
(8.445) -ц —- = (cos jc)(ixc sin cr + cos ct)k ,
<7f
so that (8.426) is satisfied since /гс = Vm^ = 1/ce. It can be easily verified
that (8.427), (8.428) are also satisfied. Finally, it may be noted that although
the fields E and H propagate along the x-direction, the nonzero field
^Qinponents lie in the transverse у and z-directions.
Example 8.11. The Equations of Elasticity Theory. Let u(jc, y, z, t) be
the vector representing the displacement of a point (x, y, z) in an elastic
Medium at the time /. Assuming the medium is isotropic, the equation of
Motion satisfied by u, which is known as Naviefs equation, is
594 VARIATIONAL AND OTHER
(8.446) p —5- = (A + 2^)V(V-u) -/LtVxVxu
at
METHODS
Here p is the density of the medium, A and ft are the Lame constants wh' ь
depend on the properties of the medium, and F represents the exter
forces. Using the vector identity (8.406), we can put (8.446) into th
equivalent form e
(8.447) p-5- = (A +
al
We assume in our discussion that p, A, and fi are constant.
We begin our analysis of the equations of elasticity by assuming that
u = u(jc, t) so that the problem is one-dimensional. Let the components of ц
be w, v9 and w (i.e., u = ui + vj + и>к). The divergence V*u has the form
(8.448) Vu= — ,
since u is independent of у and z. Thus (8.447) becomes
(8.449) p ~ = (A + д) ^4 ! + /* ^~T + F •
With F = Fti + FJ + F3k, we can uncouple the three equations (8.449) fon
the Cartesian components of u and obtain
(8.450)
F2
Thus each component of u satisfies a wave equation in one dimension.
However, the speed of wave propagation for the component и equals
[(A + 2fi)/p]1/2, whereas the corresponding speed for the components v and
w is given as (/u/pI/2.
These results show that in the one-dimensional case there are two modes
of wave propagation that can occur for elastic waves. Since the waves
associated with the component и have the property that the displacement is
in the direction of wave propagation, they are called longitudinal waves and
travel with the speed [( A + 2fx)/p]in. The second wave motion is associated
with the components и and w and thus corresponds to displacements in a
plane perpendicular to the direction of wave propagation. Therefore, they
are called transverse waves and they travel with the speed [fxlp]112. Since A,
/jl, and p are all positive, we see that the longitudinal waves have greater
speed than the transverse waves. (The longitudinal waves are often called
compression waves and the transverse waves are called shear waves.)
It may be noted that if
uTl0N METHODS FOR HIGHER ORDER EQUATIONS AND SYSTEMS 595
(8.452) u = u(*> 0 = <x> 0* + K*> 01 + ^(x, 04 >
= u\ + V ,
have (since и and V are functions of x and * only)
(8.453) Vx(m) = 0, V-V-0.
Also, as is easily verified,
V-o = V-(ui), Vxu=VxV.
This suggests that we try to decompose the displacement vector u in the
general case into two parts, one involving V*u and the other V x u. (It is
shown in vector analysis that such a decomposition is always possible for a
vector field. That is, it can be decomposed into a sum of two vectors one of
which has zero divergence whereas the other has zero curl.)
To achieve this decomposition we multiply (8.446) with V- and then with
V x ; that is, we take the divergence and curl of these equations. We have
(8.454) p — (V-u) = (A + 2/i)V2(V-u) »/iV-(VxVxu)
dt
since the divergence of a curl vanishes. Also,
(8.455) P^(Vxu) = (A + 2/,)Vx [V(V- u)]
= /nV2(V x u) + V x F
because V x V<p = 0 and V x [V x V x u] = -V2(V x u) + V(V-V x u) =
-V2(Vxu), asV-Vxu = 0.
We have shown that in the general case, the elasticity equations can be
reduced to the study of two wave equations for V*u and V x u. The term
v#u characterizes the compression and expansion of the body and its waves
Javel with the speed [(A + 2^)/p]1/2. The term Vxu characterizes the
^stortion of the body and its waves travel with the speed [ti/p]in. In
sealing with initial value problems for elastic wave propagation with no
boundaries present, this decomposition can be used to great advantage since
^ know how to solve the initial value problem for the wave equation.
However, if there are boundaries, the boundary conditions couple the
ggiponents of u and the problems are more complicated.
596 VARIATIONAL AND OTHER METHODS
We have treated a number of examples of physical problems that give rise
to differential equations that are not of the general form studied previou 1
in the text. It has been shown in some cases that the problems can h
reduced to a consideration of equations studied previously. However u^
have only touched upon some of the methods used for these problems and
each of the theories of fluid dynamics, electromagnetics, and elasticity has
rich body of mathematical methods associated with it.
In Chapters 9 and 10 we consider approximate methods for solving partial
differential equations and these techniques are useful not only for the
equations studied in the main body of this text but also for the types of
problems considered in this section.
EXERCISES FOR CHAPTER 8
Section 8.1
8.1.1. Given the eigenvalue problem (8.27), (8.28) of Example 8.1, de-
determine all the eigenvalues and the orthonormalized eigenfunctions (see
Example 7.6). Show that the eigenvalues can be ordered as in (8.3). (Note
that there are multiple eigenvalues for this problem.) Also show that the
energy integral (8.31) satisfies the equation (8.16) for each eigenfunction
and its corresponding eigenvalue.
8.1.2. Consider the eigenvalue problem of Example 8.1 with the Dirichlet
boundary condition (8.28) replaced by the Neumann condition dMldn\dG =
0 (see Example 7.7). Show that the eigenvalues can be ordered as in (8.3)
and that the appropriate energy integral [that is, (8.31)] satisfies (8.16).
8.1.3. Obtain the appropriate form for the energy integral (8.10) for the
Sturm-Liouville eigenvalue problem
-{ри'У + qu = Apw, 0< x < I,
ыЩ - AlM@) - 0, u'(t) + h2u(l) = 0
where hx and h2 are positive constants.
8.1.4. Show that the energy interal (8.10) for the eigenvalue problem
u\x) + ku{x) = 0,
u@) = 0, м'(/) + Ц/) = 0, А>0,
has the form
FOR CHAPTER 8 597
T t u (x) = sin(VV0 be the (unnormalized) eigenfunctions for this prob-
problem with the Art determined from V^ cosO/A^/) + h sin(VXO = 0. Demon-
Demonstrate that
ex-
ex£Ю Ап(ип,м„).
Hint: Use the eigenvalue equation to simplify the energy integral
pression.
8.1.5. Consider the function
u(x, y) = cxy sin jc sin у .
Determine the constant с such that the norm (8.32) of this function is unity.
Observe that u(x, y) vanishes on the boundary of the square 0 < x < тг,
0<y < 7Г and evaluate the energy integral (8.31) for the function. Compare
your result with the eigenvalue An = 2 for the square and evaluate the norm
\мп - м|| as in Example 8.1.
8.1.6. Show that the eigenvalues obtained from the variational principle
(8.40) with the constraints (8.41) can indeed be ordered as O^Aj^A^
• ••<ABs---, by noting that to determine An an additional constraint is
added to those already given for the preceding eigenvalues so that the class
of admissible functions is reduced.
8.1.7. It is required in the text that the function W defined in (8.42) satisfy
the constraints (8.41). Show that if we set
n-l
where W is arbitrary (apart from having to vanish on dG if W vanishes
there), we may choose the constants a; = -(ТУ, Af;.) so that W satisfies the
constraints (8.41). Replace W by the preceding expression in the discussion
following (8.44) and show that (8.46) results with W replaced by W.
8.1.8. Show that if
jjfgdv = \
f°r a continuous function f in G and arbitrary functions g, then we must
have / = 0. This result is often called the fundamental lemma of the calculus
°f variations.
Assume /V0 at some point in G, so that it is nonzero in the
neighborhood of that point because it is continuous, and then choose g
aPpropriately.
598 VARIATIONAL AND OTHER METHODS
8.1.9. Show that if
я.
for arbitrary regions G within which / is continuous, we must have
Hint: Proceed as in Exercise 8.1.8. This result is known as the du B
Reymond lemma and is closely related to the lemma of Exercise 8.1.8
8.1.10. The eigenvalues for the problem
(xw')'+-w = 0, Kx<2y
(see Exercise 4.3.3), are given as
With p = x, q = 0, p = 1 Ixj determine the constants pm, pM, pm9 and pM for
the interval \<x<2, where the subscripts m and M correspond to the
minima and maxima of these functions. Obtain the eigenvalues for the
appropriate constant coefficient problems obtained from this problem and
verify the result (8.55) for the eigenvalues of all three problems.
8.1.11. Obtain upper and lower bounds for the eigenvalues for the following
problem.
-V2M + xyM = AM, 0<jc<7t,
M@, у) = М(тг, y) = M(x, 0) = M(x, ir) = 0.
8.1.12. Use the method of Example 8.2 to estimate the lowest eigenvalue
for the following problem.
x2 v2
0, — + 73^1, <*<b,
a b
8.1.13. Obtain an upper and lower bound for the lowest eigenvalue of the
following problem in a sphere.
V2M + AM = 0, x2 + y2 + z2 < 1,
M(x,y,z) = 0, x2 + y2 + z2 = l.
eXERCISES FOR CHAPTER 8 599
tj the method of Example 8.2. The exact eigenvalue is found in Example
g 3 to equal rr .
"j j4. Use the energy integral (8.10) to show that the leading eigenvalue Ax
for the problem (8.1), (8.2) increases as alfi increases. (Using the Courant
ax-fflin principle, it can be shown that all the eigenvalues kn have this
property.)
« 1 15. Use separation of variables in polar coordinate to determine the
eigenvalues and eigenfunctions for the following problem.
M(r, 0) = 0 on the boundary .
Hint: Proceed as in Example 8.2. The eigenvalues are the squares of zeros
of Bessel functions.
8.1.16. Use the results of Exercise 8.1.15 and the method of Example 8.2 to
obtain an upper and lower bound for the leading eigenvalue for the
following problem.
yy 0, (x, y)EG,
M(x, y) = 0on dG,
where the G is the interior of the triangle bounded by the lines у = 0, у == jc,
and x = 1.
8.1.17. Show that the (exact) eigenvalues and eigenfunctions for the prob-
problem of Exercise 8.1.16 are given as
Km = ir\n2 + rn2), n < m, n = 1,2,.. ., m = 2,...
Mnm(x> У) = sin(?rmx) sin(Trny) - sin(irnx) sm(jrmy)
Verify that the bounds on the leading eigenvalue obtained in Exercise 8.1.16
are correct.
8.1.18. Show that the energy integral E(w) [as denned in (8.38)] for
unctions w that equal /on dG is minimized by the solution и of the
°irichlet problem
t: Let w = и + W where и is the (assumed) solution of the problem and
is an arbitrary function that vanishes on G. Show that E(w) = E(u) +
) > E(u), if W Ф 0. If p = 1 and q = 0, this is a classical result known as
ls principle.
600 VARIATIONAL AND OTHER
8.1.19. Adapt the procedure given in the text to show that the solutio
the problem °*
£(w) - I I 2wF do = minimum ,
G
where E{w) as defined in (8.38) or (8.39) is a solution of the problem
with homogeneous boundary conditions of the first, second, third or mixed
kind. In the case of Dirichlet boundary conditions the admissible functions
for the variational problem must vanish on dG for the Dirichlet problem or
on the subset St of dG in the mixed problem. The boundary conditions of
the second and third kind are natural and no restriction must be placed on
the admissible functions in that case. In the latter cases, the boundary
condition that the solution и satisfies depends on the choice of E{w).
Hint: Let w - и + eW, where и is the (assumed) solution of the minimum
problem.
8.1.20. Show that the solution of the problem
E(w) - I I 2wF dv- j 2wf ds = mini
G
yields the solutions of the boundary value problems
minimum
with the boundary conditions
Эй диги s а ы*
-=/ or ^ + hu=f, xESG,
depending on whether E(w) has the form (8.38) or (8.39), respectively. The
boundary conditions for the minimum problem are natural. To obtain the
solution of the inhomogeneous equation with the Dirichlet boundary condi-
condition u=f on dG, we must proceed as in Exercise 8.1.19 except that all
admissible functions must equal / on dG Adapt the foregoing results to
obtain the appropriate minimum problem in the case of mixed boundary
conditions.
Hint: Let w = и + eW where и is the (assumed) solution of the problem.
Section 8.2
8.2.1. The Gram-Schmidt Process. Let (<jp, ф) represent a (weighted) inner
product for either vectors or scalars. Assume the set of functions <Pk
FOR CHAPTER 8 601
/^ s* 1,2,. . •) is linearly independent. Show that if ||<р||2 = (<p, Ф), the set of
[unctions фк formed as follows,
Ш
ft = ft ~
фъ
and so on, is an orthonormal set.
8.2.2. Let (<p, ф) = Jo ФЖ*) dx. Let <рг(х) = 1, ^(x) = x, <p3(x) = л2, and
<рА{х) = хъ. Construct an orthonormal set фг, ф2, ф3, and ф4 using the
Gram-Schmidt process.
8.2.3. Given the three (independent) row vectors
*f- [1,2,-1], ^2r« [0,1,3], «if-[-1,1,0],
form an orthonormal set of vectors фг, ^, and ^ from the preceding set
using the Gram-Schmidt method and the dot product of the vectors as an
inner product.
8.2.4. Conclude from the fact that E(w) with <?>0 and ||и>|| are both
positive if w vanishes on dG and is not identically zero, that the matrices A
and В defined by (8.96), (8.97) must be positive definite in view of (8.98).
8.2.5. Separate variables in equation (8.120) for the spherical harmonics
with k2 = n(n + 1), and let Yn@, <p) = H(<p)G@) to obtain the equations
4- [sin <p ^1 + \n(n
d<p I d<p J L v
sin <p d<p I d<p J L v ' sin ^
where д2 is the separation constant. Conclude, since GF) is periodic of
Period 2тг, that ц = w = integer so that the G@) are trigonometric functions.
Let t = cos <p in the equation for #(</>) and obtain the associated Legendre
equation
| [n(n +1) - y^5 ]я@ = о
JJ the interval -1</<1 where H(t) = H(cos <p) = H(<p). The solutions
Щ<р) must be bounded at <p = 0 and <p = тт. Consequently, H(t) must be
b°unded at *=±1. If m = 0, the bounded solutions are the Legendre
Polynomials Pn(cos <p) discussed previously. If m ^0, bounded solutions are
602 VARIATIONAL AND OTHER METHOD
obtained only if m < n and are given as the associated Legendre functio
P"(cos <p). The functions P™(t) can be defined in terms of the Legend^
polynomials Pn(t) as re
8.2.6. Differentiate the Legendre equation
m times and show that the function H(t) = A - t2)mf2(dm/dtm)[Pn(t)] is a
solution of the associated Legendre equation of Exercise 8.2.5. Determine
that P™(t) = 0 if m > n from its definition in terms of Pn(t)9 and show that
P™(t) is bounded at f = ±1.
8.2.7. Determine the coefficients ct and c2 corresponding to each eigenvalue
Аг and A2 for the problem of Example 8.4 and show that the resulting
functions are orthogonal.
8.2.8. Use the Rayleigh-Ritz method to approximate the leading eigen-
eigenvalue for the following problem
V2M + AM = 0, x2 + y2 < 1,
M 0 2 + 2 l
Use the approximating function ipt = cosGrr/2), where r2 = jc2 + y2. Com-
Compare the result with that given in Example 8.2.
8.2.9. Construct a function <px(x, y) that vanishes on the triangle given in
Exercise 8.1.16 and use the Rayleigh-Ritz method to approximate the
leading eigenvalue for the problem as given in Exercise 8.1.17.
8.2.10» Approximate the lowest eigenvalue for the problem in Example 8.4
by using only the function ^(jc) as defined in (8.151). Compare the
approximate result with that obtained in Example 8.4.
8.2.11. Using the appropriate Rayleigh quotient, approximate the first
eigenvalue for the following problem
M@) = 0, M'(l)
Let <px(x) = 3x-2x\
8.2.12. Solve the problem of Example 8.5 for the leading eigenvalue by
letting w = c1ip1(x) + c2ip2(jc) where <рг(х) is given by (8.159) and (p2(x) =
xcos(jx).
FOR CHAPTER 8 603
Я2.13- Explain why if the approximation functions <pk(x) in (8.94) are
''sen t0 be elements of a complete set {<pk}, the more terms one takes in
the series (8.94), the better the approximation to the leading eigenvalue and
eigenfunction.
8 2.14- Use the Rayleigh-Ritz method to approximate the leading eigen-
eigenvalue for the problem in Exercise 8.1.10. Let <рг(х) = (х- l)(x - 2).
g 2.15. Show how the Rayleigh-Ritz method can be applied to the varia-
ti'onal problems given in Exercise 8.1.19.
Hint: Express w(x) as in (8.94) with w required to vanish on dG for the
nirichlet problem or on Sx for the mixed problem, but arbitrary on dG or
on a subset thereof where boundary conditions of the second or third kind
are assigned.
8.2.16. Use the Rayleigh-Ritz method (see Exercise 8.2.15) to approxi-
approximately solve the following Dirichlet problem for Poisson's equation within
the square.
V2M = -Sin Jt, 0 < X < 7Г, 0 < у < 7Г ,
w(jc, y) - 0 on the boundary .
Use a one term Rayleigh-Ritz approximation.
8.2.17. Show how the Rayleigh-Ritz method can be applied to the varia-
tional problems of Exercise 8.1.20.
Hint: See Exercise 8.2.15 and for Dirichlet conditions, let w - <p0 + E ск<рк
where <po=f on dG or Sx and the <pk vanish on dG or St. If boundary
conditions of the second or third kind are given on all or part of dG, put
(p0 = 0 and let the <pk be unrestricted there.
8.2.18. Solve the Dirichlet problem,
x2
using the Rayleigh-Ritz method [see Exercise 8.2.17 and let <po = Jt2 and
<Pi = cos(tt/72) as in Exercise 8.2.8].
8*2.19. Consider the nonself-adjoint elliptic equation and the integral rela-
relation of Galerkin form given in Exercise 6.4.8. Depending on the boundary
conditions assigned for и on G, expand и in the same form as the function w
m Exercise 8.2.17. Then choose n independent functions v19 v2,. . . , vn that
are required to vanish on whatever part of dG that и itself is specified on,
but are otherwise arbitrary. Insert the expansion of и into the Galerkin
integral relation and successively set v equal to vl9 v2,. . . , vn. This yields n
equations for the coefficients сг, c2,.. . , cn. This technique for obtaining an
approximate solution to the given boundary value problem is known as the
Galerkin method.
604 VARIATIONAL AND OTHER
(a) Determine the system of equations for the ck.
(b) Show that the system obtained in part (a) is identical to the syste
determined from the Rayleigh-Ritz method if vk = <pk, k = 1,. . . y n /J11
Exercise 8.2.17) and the vector b = 0, so that the original equation *
self-adjoint.
8.2.20. Use the Galerkin method to solve the Dirichlet problem
V2u - ux + uy = -1, x2 + y2 < 1,
where u satisfies the same boundary condition and is expanded exactly as in
Exercise 8.2.18, and v = 1 - r2.
Section 8.3
8.3.1. Determine the conditions under which equation (8.176) is self-adjoint
and show that the Riemann function is symmetric in that case. That is,
R() R(£)
8.3.2. Show that (8.194) reduces to d'Alembert's solution of the wave
equation if R is given by (8.198), у = constant and /3 = 0.
8.3.3. Verify that (8.194) and G.384) are identical if R is given by (8.200)
and c = -c.
8.3.4. Obtain the Riemann function for (8.199) with c>0 (i.e., the Klein-
Gordon equation). Show that the solution formula for this problem agrees
with G.382).
Section 8.4
8.4.1. Use the maximum principle to show that the solution of the initial
and boundary value problem for the one-dimensional heat equation is
unique. That is, show that
ut - c2uxx = F(jc, 0, 0<x < l,t>0 ,
м(дг,0) =/(*), 0<jc</,
has a unique continuous solution.
Hint: Assume there are two solutions щ and u2 and consider the function
и = ux — u2.
8.4.2. Show that if ux(x, t) and m2(jc, /) are solutions of the heat equation
ut = c2uxx whose initial and boundary data at x = 0, t = 0, and x = / satisfy
the inequality
\u1(x,t)-u2(x,t)\<e, x = 0, f = 0, jc = /,
then
proves
FOR CHAPTER 8 605
we must have \ux - u2\ < € for all (jc, t) in 0< x < l,t >0. This result
that solutions of the first boundary value problem for the heat
tion depend continuously on the data,
jf Consider the solutions vl = ~€, w = мх - w2 and u2 = £ of the heat
nuation and use the maximum principle to conclude that vx < и < и2 by
examining these functions two at time.
g 4 3. Prove that the Cauchy problem for the heat equation
м(х,0) =/(*), -oo<JC<oo,
\u{x,t)\<M, -oo<jc<oo,r>0
has a unique solution. Assume the problem has two solutions mx(jc, t) and
u (jc, i) and consider the difference й = мг — w2. Consider the interval
-l<x^l and the function u(jc, f) = DM/I2)(c2t + jc2/2) which is a solution
of the (homogeneous) heat equation. Show that \u(x, t)\ < v(x, t) on the
boundary л: = — /, t = 0, jc = / and conclude from the maximum principle in
the bounded interval |jc| ^ / that |й| < v in |дг| < /, and / >: 0. By letting /-> oo9
conclude that m(jc, t) = 0.
8.4.4. Consider the initial and boundary value problem given in Exercise
8.4.1. Let R be the region 0<л:</ and 0<*<Г, and introduce the
auxiliary function
w(x,t) = u(x9t)+l ~max|F|,
с R
as well as the function w(x, t) which has the form of w(xy t) except that и is
replaced by -u. Use the maximum principle for the inhomogeneous heat
equation to show that
/2
\u(x, 0|=£шах|и| + \ ~2 max|F|,
2
2
where S represents the part of dR that coincides with / = 0, л: = 0, and x = I
Use this inequality to prove continuous dependence on the data (including
the inhomogeneous term in the equation).
8.4,5. Show that the initial and boundary value problem of Exercise 8.4.1
with Neumann instead of Dirichlet boundary conditions at jc = 0 and x = I,
has a unique solution.
Hint: Use the fact that ux Ф 0 at a maximum or minimum point on jc = 0 and
* - / to conclude that the homogeneous version of the problem has its
extreme values at t = 0.
8-4.6, Show that the problem of Exercise 8.4.1 with boundary conditions of
the third kind or mixed boundary conditions has a unique solution.
606 VARIATIONAL AND OTHER
Hint: Consider what happens at a maximum or minimum point on th
boundary and use the hint in Exercise 8.4.5.
8.4.7. Determine a maximum and minimum principle for the heat equati
u,-c2V2M=0
in a bounded region in two or three dimensions.
Hint: Proceed as in the one-dimensional case and replace w in (8.205) b
w=u + er2 where r2 = x2 + y2 and r2 = x2 + y2 + z2 in two and three di-
dimensions, respectively. Note that the bounded region may be assumed to be
contained in a circle or sphere of sufficiently large radius.
8.4.8. Obtain a maximum principle for the parabolic equation
where p > 0 and p > 0 are bounded in any bounded region. Similarly, obtain
a maximum principle for the equation in Exercise 1.1.17.
Hint: Let w - и + € exp[ox] and then choose a such that pwt - V- (pVw) <
0. Conclude that w must have its maximum at t — 0 or at the boundary of the
given region for f>0. Proceed as in the one-dimensional case in the text.
8.4.9. Use the maximum principle of Exercise 8.4.8 to prove that the first
boundary and initial value problem for put - V- (pVu) - pF has a unique
solution.
8.4.10. Consider the elliptic equation
= -F, p>0
in two or three dimensions in a bounded region G with the boundary dG.
Show that if F<0 in G, the solution и assumes its maximum on dG and if
F>0 in G, the solution и assumes its minimum on dG.
Hint: The first partial derivatives of и vanish at an interior maximum point.
8.4.11. Obtain a maximum principle for the elliptic equation
with p > 0, in the two- and three-dimensional cases. Similarly, obtain a
maximum principle for equation A.136).
Hint: Proceed as in Exercise 8.4.8 and use Exercise 8.4.10
8.4.12. Use the maximum principle to prove continuous dependence on the
data for the Dirichlet problem for the elliptic equation
= 0, p>0,
in a bounded region.
Hint: Adapt the method of Exercise 8.4.2.
*UK
4 13. Use the fact that the normal derivative duldn must be nonzero at a
im or minimum point on the boundary dG, to prove that the second,
i i f
um p y p
^ and mixed boundary value problems for the elliptic equation of
T? ercise 8.4Л0 have a unique solution. In the case of the second boundary
lue problem the solution is unique up to an arbitrary constant,
я 4Д4. Consider the elliptic equation
in a bounded region G where p > 0 and q > 0. Show that и < 0 at an interior
maximum point and that и ^ 0 at an interior minimum point. Conclude from
this that the Dirichlet problem for the equation with u-fondG must have
a unique solution.
8.4.15. Let u(x, y9 z) satisfy V2m = 0 in G, and let Po and the sphere So
centered at Po with radius a lie completely within G. Then the mean value
theorem for harmonic functions states (see Example 7,15).
with the integration taken over the sphere So. Assuming и is continuous up
to dG so that it has a maximum in the closed region, use the mean value
theorem to show that the maximum is achieved on the boundary dG.
Hint: Assume that the maximum point Mo is interior to G. Apply the mean
value principle to all spheres centered at Mo extending up to the boundary
*?G. Since w(M0)>u(P) for each P in G or dG by assumption, conclude
that if u(P) < u(M0) we would be led to the contradiction that u(M0) <
w(Af0). This argument can be extended to conclude that if и does attain its
maximum in G, it must be a constant function.
8.4.16, Show that the solution u(x, t) of the Cauchy problem for the wave
equation
ut - с uxx = 0, -oo < x < ooy t > 0
u(x, 0) = /(*), ut(x, 0) = g(x), -oo < x < oo
is non-negative if f(x) > 0 and g(x) > 0.
Hint: Use d'Alembert's solution.
8-4.17. Use the solution formula G.384) for the Cauchy problem for the
Modified telegrapher's equation G.383) and the properties of the modified
Bessel functions /0(z) and Ix{z) to show that if the data /(*), g(jt), and
"(, 0 are non-negative, then so is u(x, t). Show how this relates to the
l for the telegrapher's equation given in the text.
608 VARIATIONAL AND OTHER METHODS
Section 8.5
8.5.1. Show that if f(x) = 8(x) (the Dirac delta function) and g(x) = о i
(8.240), the asymptotic solution (8.254) of (8.234) takes the form
1 \ X1 7Г
u(x, t) = Га cos —
v } yjA>nct I Act 4
and this is the exact solution of the initial value problem.
8.5.2. Use the result of Exercise 8.5.1 to show that the solution of (8.234)
with the data u(x, 0) = f(x), u,(jc, 0) = 0 is
8.5.3. A rod is said to be simply supported at я: = 0 if н@, i) = uxx@, t) = 0.
Use the Fourier sine transform to solve the following problem for the
vibration of a semi-infinite rod.
u,, + с V^ = 0, 0<*<оо,*>0,
u(x9 0) = f(x), ut{x, 0) = g(x), 0 < x < oo ,
u@,t) = uxx@,t) = 0, t>0.
8.5.4. Apply the Fourier cosine transform to solve the following problem
и„ + с2и«„ = 0, *>(U>0,
Ф, 0) = f(x), ut(x, 0) = g(x), x>0,
8.5.5. The vibration of a finite rod simply supported at both ends is
determined by the solution of the following problem
utt + c2uxxxx = 0, 0<x<l,t>0,
u(x,0)=f(x),ut(x,0) = g(x), 0<*</,
u@, t) = nw@, 0 = m(/, /) - и„(/, 0 = 0, t > 0 .
Obtain the solution by using separation of variables.
8.5.6. A rod is free or unsupported at a point if uxx and uxxx vanish there.
Use separation of variables to solve the following problem for the vibration
of a rod clamped at one end and free at the other end.
FOR CHAPTER 8 609
utt + c2uxxxx = 0, 0<x<I, t>0
u(x, 0) = /(*), ut(x, 0) = g(x), 0<x<l,
м(о, о = ux(o9 0 - «„(/, 0 = um{i9 0 = 0, * > о.
8 5.7. Consider the eigenvalue problem
v""(x) - A2y(x) = 0, 0<л;</
with the following boundary conditions:
(a) v@) = v'@) = v(l) = v'(l) = 0;
(b) i;@) = ^@) = KO = ^'@^0;
(c) y'@)=yw@) = i;4/) = ^'@==0;
(d) u@) = v '@) = y"(/) - *>'"(/) = 0 .
Let kn and Am be distinct eigenvalues for each of these problems and vn{x)
and vm(x) the corresponding eigenfunctions. Use the method of Section 4.3
to show that the eigenfunctions are orthogonal with respect to the following
inner product
vn(x)vm(x)dx = O, пФт.
8*5.8. Construct the Green's function K{x\ £) for the following problems
where K(x; f) satisfies the boundary conditions (a), (b), (c) or (d) given in
Exercise 8.5.7.
Hint: К satisfies a homogeneous equation at x Ф f and the jump in дъК1дхъ
a* x = £ is -1. In case (c) a modified Green's function must be constructed.
8-5.9. Show how the finite sine transform method can be used to solve the
Problem in Exercise 8.5.5 if the equation and the boundary data are
inhomogeneous.
8-5.10. A simply supported plate satisfies the boundary conditions и = 0 and
S uldn2 = 0 at the edge. (Here д2и1дп2 is the second exterior normal
derivative.) Use separation of variables to solve the problem of the vibrating
simply supported rectangular plate. That is, solve the following problem for
610 VARIATIONAL AND OTHER METHODS
utt + c2V2V2m = 0, Q<x<l,Q<y<t,t>0,
u(x, y9 0) - f(x, y), ut(x, y9 0) - g(x, y) ,
и@, у) = и„@, у) - и(/, у) = Wjcjc('> у) = 0, 0<у<19
и(х, 0) = иуу(х, 0) = и(*, /) = иуу(х, 1) = 0, 0< х < I.
Hint: Obtain (8.282) and set An = sin[Grn//)x] to satisfy the conditions at
x = 0 and x = L Conclude that Bm = sin[(irm/t)y]9 and obtain the solution in
terms of a double Fourier sine series.
8.5.11. Use the Fourier transform to solve the Cauchy problem for the plate
equation
utt + c2V2V2m = 0, -oo < x, у < oo, t > 0 ,
u(x, y, 0) = f(x, y), ut(x, y, 0) = g(x, y) ,
with appropriate conditions at infinity.
8.5.12. Use the Green's theorem (8.308) to show that eigenfunctions corre-
corresponding to different eigenvalues of the problem (8.278), (8.279) are
orthogonal.
8.5.13. Verify that (8.296) satisfies the biharmonic equation.
8.5.14. Solve the boundary value problem
by using a Fourier series in the angular variable 0.
8.5.15. Solve the problem of Exercise 8.5.14 using the method presented in
the text if the boundary conditions are
8.5.16. Show that (8.305) satisfies the biharmonic equation if
8.5.17. Solve the following problem by looking for a solution of the form
и = u(r):
V2V2w = F0, x2 + y2<R2
where Fo is a constant.
eXERcises for chapter 8
a 5.I8. Obtain the solution of the problem
by looking for a solution in the form
и = - — r2 log r + v(r)
where r2 = x2 + /.
8.5.19* Take the total or material derivative of the equation of state (8.324)
with respect to t and obtain dpldt = c2dpldt on using (8.323) and defining
2=
Use this result to replace (8.323) by the equation
Pi + (" • V)p " c2[p, + (u • V)p] - 0 .
Show that the one-dimensional version of the modified form of Euler's
equations for adiabatic flow is given as in Exercise 3.3.22.
8.5.20. Use the equation
_ ds de d
that follows from (8.373) and (8.324), together with the equations (8.315)
and (8.322) to replace the (entropy) equation (8.323) by the equation
-
This the equation of conservation of energy,
8.5.21. Use the result of Exercise 8.5.20 to express the one-dimensional
form of Euler's equations for adiabatic flow in conservation form (see
Section 2.3),
(\pu2 + pe)t + [{\pu2 + pe)u + pu]x - 0 .
is from this system of conservation laws that the Rankine-Hugoniot shock
612 VARIATIONAL AND OTHER METHODS
conditions are derived, in the manner of Section 2.3. We do not write the
down. ш
8.5.22. Multiply (8.350) by c(p), add (8.351) to it, and then subtract
(8.351) from it to obtain ct
c[pt + (и ± c)Px) ± p[ut + (и ± c)ux] = 0 .
Conclude from these equations that
dp du dx
cTt+pTt=Q on Tt=u + C>
dp du dx
cTt-pTt=0 on -&'*-*>
where dldt = dldt + {dxldi)dldx. Since с - c(p), show that these equations
can be written as
г
с(р) dx
-*—*■ dp + u = constant on — = и + с ,
р at
Р с(р) , dx
dp- u = constant on — = и - с .
р at
They are the Riemann invariants,
8.5.23. Show that if we put one of the Riemann invariants in Exercise
8.5.22 identically equal to a constant for all x and t (that is, it is not just
constant on the characteristics), then we must have p = p(w). Conclude that
the other Riemann invariant has the form of a simple wave.
8.5.24. In the case of a polytropic gas [see (8.343)], show that the Riemann
invariants of Exercise 8.5.23 take the form
2 dx
c±u — constant on — = u±c .
8.5.25. Consider the problem of the withdrawal of a piston from a poly-
polytropic gas initially at rest. The problem is one-dimensional and at the time
t = 0 the gas is located at x > 0 with velocity и = 0, a constant propagation
speed c = c0 and constant entropy s = s0. The path of the withdrawing piston
is jt = g(f), with £@) = 0 and g'@<0- Use the Riemann invariants of
Exercise 8.5.24 and conclude that
с- и-
since each of the characteristics associated with this invariant issues from the
positive jc-axis. Show that и - 0 and с = c0 in the sector cot < x < °°. In the
region between tte piston curve and the line x = cot, show that
XERCISES FOR CHAPTER 8
613
и = constant on -77 = c0 H
dx
dt
that is, the solution is a simple wave there, Since the fluid velocity on the
iston equals the piston velocity, we have the boundary condition
и = g'{i) on x = g(t).
Obtain the equation of the characteristic x'(t) - c0 + [(у + 1)/2]м in the
parametric form
The solution is given as
y-1
c0 + —— g'(r),
with т determined as a function of x and t from the equation for the
characteristic. We do not discuss the formation of shocks.
8.5.26. Consider the Hertz vector тг and express the scalar and vector
potentials <p and A for Maxwell's equations as
_ 1 <?тт
where ep = l/c2. Show that the Lorentz condition (8.418) is thereby satis-
satisfied. Express E and H in terms of it, and if p = a = 0 show that it satisfies
the equation
8.5.27. Determine the form taken by Maxwell's equations if p = <r = 0, e, ц
are constant, and E and H are given as
where e and h depend only on the spatial variables and a> is a constant.
8-5.28. Verify that V-F = V-G = 0 for the data (8.436) and show that the
s°lution (8.442), (8.443) also satisfies V-E = V-H = 0.
**-5.29. Formulate Exercise 8.5.26 for the Hertz vector if the problem
depends only on x and t. Assume the Hertz vector ir(*, i) is given initially as
•jt(jc, 0) = (cos x)\ + j, тг,(*, 0) = k, -oo< x < oo .
614 VARIATIONAL AND OTHER METHODS
Determine <n(xt), E(x,t), and H(x,t) for t>Q by solving the Cauchy
problem for it(x, t). y
8.5.30. Let the Hertz vector it be given as тг = ae~lti>t where a depends only
on the spatial variables (see Exercise 8.5.26). Assuming the conditions given
in Exercise 8.5.27 apply, show that
V2ot + k2a = 0
where k~ <o/c and
Vxa.
8.5.31. To study the possibility that electromagnetic waves can propagate in
a waveguide, we consider Maxwell's equations in a cylindrical region whose
walls are perfectly conducting. The interior of the guide is such that e and д
are constant and a = 0. In addition, there are no sources so that p = 0 and E
and H are assumed to have harmonic time dependence as in Exercise 8.5.27.
The tangential component of E vanishes on the waveguide wall. Assume the
cylinder has its generators parallel to the 2-axis and show that if we set the
Hertz vector it equal to
the tangential component of E vanishes if
и = 0 on the cylinder wall
(use Exercise 8.5.30). Also, conclude that и satisfies the reduced wave
equation
V2w + A;2w = 0, Л = л>/с.
Let u(x, y, z) = M(x, y)v(z) and determine M(x, y) from the eigenvalue
problem
V2M + AM = 0, (x9 y)£G ,
where G is a perpendicular cross-section of the cylinder. Let Xn and
Mn(x, y) be the eigenvalues and eigenfunctions for this problem with
x"^ A2 < • • *. Then vn(z) must satisfy
xEgClSES FOR CHAPTER 8
615
Considering only waves traveling in the positive z-direction, show that the
Hertz vectors тги(*, у, z, t) must take the form
wn = anMn(x, у) exp[;(<rwz -
where ^„ = V^ ~ К anc* ал *s a constant. Determine that no traveling
waves exist in the waveguide if k2 < A2 and that only a finite number of such
waves exist if A^ < k2 < Ал. (With k2 < Ая, the field ттл decays as z->oo.)
8.5.32. Since к = cole, show that if a>2<\lc2 for a particular waveguide
where Aj is determined as in Exercise 8.5.31, no traveling waves can exist in
the waveguide. Thus a> = VA^c is termed a cutoff frequency. Determine the
cutoff frequencies for the waveguides with the following cross-sections:
(a) 0<
(b)
8.5.33. Solve the initial value problem for the one-dimensional Navier
equation (8.447) with F = 0 and the initial data
u(jc, 0) = (cos x)\ + j + (sin д:)к ,
8.5.34. Show that if the equation (8.454) is solved for V* u we may obtain u
from (8.447) by treating the term involving V- u as an additional inhomoge-
neous term. Indicate how we may express the solution of the initial value
problem for (8.447) with u(x, y, z9 0) and ut(x, y, z, 0) specified, using the
general solution formula for the wave equation.
9
PERTURBATION
METHODS
9.1. INTRODUCTION
Chapters 9 and 10 deal with a collection of methods that yield approximate
solutions for a large class of initial and boundary value problems for partial
differential equations. Generally speaking, these methods are used when a
small parameter (or a large parameter) occurs in the given equation or data
for the problem. Then the (assumed) solution is expanded in a series of
powers (or inverse powers) of the parameter and this expansion is inserted
into the equation and data for the problem. By equating like powers of the
parameter, a collection of problems results whose solution is expected to be
simpler than that of the given problem.
If the series expansion of the solution converges, or is expected to
converge, the aforementioned technique is often referred to as a perturba-
perturbation method. If the series is divergent but asymptotic, so that the first few
terms yield a good approximation when extreme values of the parameter are
considered, the above technique is called an asymptotic method. The
terminology is, however, not uniform in the literature and what are termed
perturbation series may, in fact, be asymptotic and vice versa.
In general, it is difficult to specify all the terms in a perturbation or
asymptotic series for problems involving partial differential equations. Thus
only the first few terms in the series are determined and the distinction
between a convergent or asymptotic series often becomes irrelevant. Al-
Although for many problems precise results are available regarding the
convergent or asymptotic nature of the result, we shall concentrate on the
formal aspects of constructing the series solutions and will generally only
obtain the first few terms of the series.
In this chapter we discuss regular and singular perturbation methods. In
Chapter 10 we discuss three methods that are not generally referred to as
perturbation methods. Therefore we collect them under the heading of
asymptotic methotk.
616
PRODUCTION 617
The appropriate expansion forms for the solution of a given problem are
no means obvious in all cases. Consequently, it is often useful to study
solutions of particular problems and expand these solutions in a series
p p p
giving the relevant small or large parameter. This not only suggests
Ш propriate expansion forms for more general problems but may also yield
eful information for specifying undertermined quantities that occur in the
U neral case. For example, in our study of the Klein-Gordon equation in
Example 5.14 as a model for dispersive wave motion, it was shown that the
olution for large x and / has a specific asymptotic form. By inserting that
form directly into the equation it can be shown that the results of that
example can be reproduced apart from an unspecified constant. In addition,
related dispersive equations can be solved by using similar expansion forms.
Although no large or small parameter occurred explicitly in the Klein-
Gordon equation of Example 5.14, by rescaling the x and t variables such
that large values are emphasized, say, x = kx, i=kt with k>l, a large
parameter к can be introduced into the equation.
Perturbation techniques can also be used to replace given equations by
simpler ones whose solutions contain many of the features of the solutions
of the original problem. This is especially important for nonlinear equations
where perturbation methods are used to linearize the problem, as has been
done for Euler's equation of fluid dynamics in Example 8.9, for instance.
Even if the linearization procedure breaks down in certain regions, it may
still be possible to replace the given equation or system by a simpler
nonlinear equation. The equation of simple wave motion that was obtained
as an approximation to Euler's hydrodynamic equations is representative of
this approach. This idea will be developed further in this chapter as well as
in Chapter 10.
In our study of hyperbolic equations we have seen that discontinuities or
singularities in the solutions, interpreted in the weak sense, occur across
characteristics. In fact, any rapid variation in the data for these equations
must be carried along the characteristics. Near the characteristics, the
solutions may be described by means of series expansions which may contain
a small parameter as we will show. These results are closely related to the
asymptotic expansions that will be given for the Helmholtz or reduced wave
equation and they are both discussed in Chapter 10.
It will be seen that more than one type of expansion may be necessary to
completely describe the perturbation or asymptotic solution of a given
Problem. For example, an expansion may break down in some region or
n^y be insufficient to satisfy the data for the problem. These difficulties
signify that the given expansion is not uniformly valid over the entire region
°f interest. Techniques such as the boundary layer method or the method of
Multiple scales need to be used to remedy this problem. They are introduced
щ this chapter but are used in both chapters.
Even though asymptotic equalities, asymptotic expansions, and order of
Magnitude symbols have often been defined and used in the preceding
618 PERTURBATION METHOD$
chapters, we now redefine these concepts in view of their particular rel
vance for the material in this and the following chapter.
The order symbol О is defined as
if |F(x)/G(*)|-> A as л:-> a, where A is some constant (here a may be ±00)
If (9.1) is valid, we say that F is of order G near x = a.
The function/(л:), which depends on the parameter k, has the asymptotic
(power) series
valid as &-»<», if for each N we have
(9.3) ЛдО-21/.(*)*"+ o[*-*L *-«.
The result is assumed to be uniformly valid for all x in some region. (Note
that with e = 1 Ik, we have an asymptotic power series in €, valid as
through positive values.)
The order symbol о is defined as
(9.4) F(*) =
if F(jc)/G(x)->0 as *-> a. Thus we can write (9.3) as
(9.5) Л*)=2
We shall emphasize the use of the (large) О order symbol in our discussion.
The basic feature of an asymptotic series is that the remainder is of lower
order than the last term retained. The full series may converge and
convergence is not necessary for the result to be useful.
Often more general forms of asymptotic expansions are required, such as
series in fractional powers of k, and these are also defined as having
remainders of lower order than the last term retained.
As stated previously, we do not prove the the formal series solutions we
obtain are either convergent or asymptotic. Occasionally, therefore, strict
equality rather than asymptotic equality signs are used even though the full
series may diverge. Since only a few terms in the series are found, this
distinction is not always significant. It is a general property of asymptotic
series that finding additional terms need not improve the approximation
PERTURBATION METHODS 619
•nCe the series is generally divergent. As the parameter к is (9.2) increases,
the approximation provided by the series gets better. Thus we are effectively
suflring in our discussion that the parameter is very large (or very small if
a = l/&) since we obtain only the leading terms. However, it is often
♦he case that the results are very good even for moderate values of the
arameter. Finally, we note all formal operations that we carry out on the
svmptotic series, such as term by term differentiation and integration, are
assumed to be valid.
9.2. REGULAR PERTURBATION METHODS
We consider a linear or nonlinear differential equation
(9.6) L(u,e) = 0
that depends (smoothly) on the small positive parameter e and a problem
for (9.6) given over a bounded or unbounded spatial region G. If (9.6) is of
elliptic type, appropriate boundary conditions are assigned on dG or at
infinity. If (9.6) is of hyperbolic or parabolic type, in addition to the
boundary conditions assigned on dG or at infinity for all t>0, initial data
are given in G at the time t = 0. The boundary or initial data may depend on
c, but the boundary dG is for the present assumed to be specified indepen-
independently of e.
The reduced or unperturbed problem associated with the problem for
(9.6) is obtained on formally setting e = 0 in (9.6) and its data. That is, we
consider the equation
(9-7) L(v,0) = 0
with the reduced data obtained from the data for the given problem for
(9.6). If the reduced problem has a unique solution, then the given problem
is called a regular perturbation problem. If this is not the case, we have a
singular perturbation problem. Problems of the latter type are studied in
Section 9.3.
Generally speaking, if the reduced equation is of different type or order
|han the given equation, we have a singular perturbation problem. It may
haPpen, however, that the reduced problem can be solved even if the order
Or type of the given equation is changed.
For example, the signaling problem for the hyperbolic equation
eutt-c2uxx + ut = 09 x>09-oo<t<oo
the boundary condition
62° PERTURBATION
(9.9) M@,0=/(
reduces to the parabolic problem
(9.10) c2vxx - vt = 0,
with the boundary condition
(9.11) y@,0
Both the given and reduced problem can be solved in this case. However
the signaling problem should be interpreted as a large time limit of an initial
and boundary value problem with zero initial data and in this case the initial
value problem for (9.8) has an excess of initial conditions in relation to the
reduced problem (9.10).
Similarly, the hyperbolic equation
(9.12) e(utt - c2uxx) + ux=0, x>0, -»< t<oo
with the boundary condition (9.9) reduces to
(9.13) vx = 0,
with the condition (9.11). The (unique) solution of the reduced problem for
v(x, t) is v=f(t). Since (9.13) is a first order equation and (9.12) is of
second order, the comments relating to the prescription of initial values
apply to (9.12) and (9.13).
In connection with boundary conditions that depend on the parameter c,
it may happen that the conditions for the reduced problem render it
unsolvable. For example, we consider the boundary value problem for the
biharmonic equation
(9.14) V2V2m = 0, xEG
with the boundary conditions
(9.15) n=/, e^ + w = g, xEdG.
an
Since (9.14) does not depend explicitly on e, the reduced equation for и is
also the biharmonic equation, but the reduced boundary conditions require
that и =f and u = g on <?G. Thus unless /= g, the reduced problem for и
has no solution. Further, even if /= g, the solution of the reduced problem
is not unique since a boundary condition is lost. Thus (9.14), (9.15) is, *n
fact, a singular perturbation problem.
perturbation methods 621
proceeding with our discussion of the regular perturbation method, we
nand the solution и of (9.6) in the perturbation series
(9-16) - .?,"•'"
The difference between и and u0 (i.e., и - u0), is referred to as a perturba-
perturbation on the solution м0 of the reduced or unperturbed problem. Inserting this
expansion into (9.6) gives
(9.17)
=0
We assume that L(u, e) can be expanded in a power series in и and e. As a
result (9.17) can be expressed in the form of a series
00
(9.18) Дм,б)=2 Ln(MB,un_i,...u1,Mo)en = 0,
n = 0
where the Ln represent differential operators which may be linear or
nonlinear, and which act on the functions u0, ul9.. ., un. The series (9.16)
is also inserted into the given initial and/or boundary conditions for the
problem.
To solve the given problem by means of the perturbation method, we put
the coefficients of e" in (9.18) equal to zero and obtain
(9.19) Ln(un, un_u ... , wo) = O , п -0,1,....
Similarly, we equate coefficients of like powers of e in the initial and/or
boundary data. This yields the system of equations (9.19) with appropriate
data that we solve recursively.
That is, we first solve the reduced equation
with the relevant data. Once u0 is specified, the equation for ui9
its data, is solved and then the equations for w2, w3,.. . with their data
аде solved successively.
If the given equation (9.6) is linear, the equations Ln(un9. . . , u0) = 0 are
generally nonhomogeneous versions of the equation L0(u0) = 0. However,
^o(w0) = 0 may itself be a nonhomogeneous equation. Even if the given
Problem is nonlinear but the reduced problem is linear, all the equations
622 PERTURBATION METHODS
(9.19) are homogeneous or nonhomogeneous equations of the same
For example, if we consider the equation
(9.22) L(u, e) = ut + uux - eu = 0 ,
we find that
(9.23)
whereas
(9.24) M"i,"o)=:77
However, the problem
(9.25) L(u, e) = u, + euux -uxx = 0
yields
(9.26) i,(Bi)_^._£4_o,
,9.27) ,,(„„„„) = £_^ +„„£.„.
For (9.22) the reduced equation is nonlinear and (9.24) for ux is linear,
whereas for (9.25) the reduced equation as well as all the higher order
equations for the un are linear and of the same form.
We now consider a number of examples of the perturbation method.
Each of the examples emphasizes a different aspect of the method. In the
first example we discuss a problem for which a closed form exact solution is
easily obtained and compare the results of perturbation theory with the
exact result.
Example 9.1. Helmholtz's Equation with a Small Parameter. We con-
consider the two-dimensional Helmholtz equation
(9.28) uxx + uyy + €2u = 0
in the unit disk x2 + y2 < 1 with the Dirichlet boundary condition
(9.29) u(x, y) = 1 , x2 + y2 = 1.
PERTURBATION METHODS <>23
Tjje parameter e2 is assumed to be small, such that the solution of (9.28),
(9.29) is unique.
To solve, we introduce the perturbation series
(9.30) "(*> У) =
2
where the expansion is in powers of e2 We insert (9.30) into (9.28) and
(9.29) to obtain
(9 31) V2u + e2u = V2[ S И„б
(9.32) u(x, y) = uo(x, y) + Z, un(x, y)e2n = 1, x2 + v2 = 1.
n=i
We have interchanged summation and differentiation in (9.31) and have
collected like powers of e2. On equating like powers of e2 to zero in (9.31)
and to unity or zero in (9.32) we obtain
(9.33) V2m0 - 0,
(9.34) V2un = -*„_!, „al,
and the boundary conditions
The equations for the un can be solved recursively starting with (9.33)
and using the boundary conditions (9.35). The perturbation method has
replaced the Helmholtz equation (9.28) by the system of Laplace and
Poisson equations (9.33), (9.34). Introducing polar coordinates r and в, the
Helmholtz as well as the Laplace and Poisson equations can be solved by
looking for solutions independent of в since the problem has no angular
dependence.
With и = и(г), (9.28) becomes
°n expressing the Laplacian operator in polar coordinate form and dropping
the ^-derivative. Now (9.36) is just Bessel's equation of zero order. The
s°lution of (9.36) that is bounded at r = 0 and that satisfies the boundary
^di mA) = 1 is clearly
624 PERTURBATION METHODS
(9.37) "= Л^Р
To solve for the un we again assume that un = un(r) and obtain for и (A
(9.38) ^-^ + i^=0, M0 = latr=l
The bounded solution of (9.38) is
(9.39) ио = 1.
The equation for ux{r) is
and a simple integration yields the bounded solution
(9.41)
Thus to
(9.42)
Since /0(z)
(9.43)
we have
(Q AA\
leading
has the
orders the
и — 1 н
expansion
1-r2
Mi 4 •
solution is
€2A-Г2) 4
4
) = l-Zj + O(z>),
/0(e)
The series obtained from (9.44) converges for sufficiently small e, and it
agrees to leading orders with the perturbation result (9.42). Note that e
must be smaller than the first zero of J0(z) which occurs at z «2.4. Thus the
perturbation solution yields a good approximation to the exact solution
throughout the unit circle, even if we retain only the first few terms, as long
as e is small.
PERTURBATION METHODS 625
Although the perturbation method yielded an approximate solution valid
over the entire unit circle in the preceding example, the perturbation result
. tke following example is not uniformly valid over the entire region of
* terest. We consider heat conduction with a slow radiation loss and show
that the perturbation solution is not valid for all time. The main reason for
the distinction between the two cases is that the region of interest is
bounded in the preceding example and unbounded in the following ex-
example-
Example 9.2. Heat Conduction with Slow Radiation. We consider the
Cauchy problem for the parabolic equation
(9.45) ut + eu = uxx,
with initial data
(9.46) m(jc, 0) = /(*), -oo < x < oo .
The change of variables
(9.47) u(x, t) = e'€tv(x, t)
yields
(9.48) v( = vxx
and v{x, 0) -f(x), so that v satisfies the heat equation. The equation (9.45)
describes heat conduction in a rod in which there is heat loss due to
radiation on the surface. The radiative effect gives rise to the term eu in
(9.45) and (9.47) shows that this yields a slow heat loss since 0< € <^ 1.
The solution of (9.45), (9.46) is immediately given in terms of the
solution of the Cauchy problem for the heat equation (9.48) studied
previously. However, we want to apply the perturbation method to solve
(9.45), (9.46) in order to study the effect of the unboundedness of the (jc, t)
domain for the problem on the perturbation result.
Let
and insert (9.49) into (9.45), (9.46). On equating like powers of e we have
(9.50) iH
Эх2
626 PERTURBATION METHODS
(9.51) ^-^ = -Мя1) „:>!.
The initial conditions are
(9.52) ио(*,О) =/(*),
(9.53) и„(*,0) = 0, я>1.
Putting
(9.54) uo(x, t) = v(x,t),
where v(x, t) is the solution of (9.48) with the initial condition v(x, 0) = /(*)
[see E.44)], we easily conclude that
(9.55) un(x, t) = ^- v(x, t) ,
on using mathematical induction. We verify that
л-1
since v(x, i) satisfies (9.48).
The full perturbation solution is
00 I- / v к *n 00 , v fi
(9.57) u(x, /)=2 ^T" w(jc, 0 «" = 2 ^Г" »(*. 0
n=0L «! J я=0 «!
and this is identical with the exact solution (9.47). However, the purpose of
the perturbation approach is to approximate the exact solution by determin-
determining and retaining only the first few terms in the perturbation series. If we
retain only the first two terms in the series (9.57) and write
(9.58) u(x91) = v(x, t) - etv(x91) + O(e2),
we conclude on comparing with the exact solution that for e/<^l, (9.58)
yields a good approximation.
But if et=O(l) or, equivalent^, r=O(l/e), we find that the first
perturbation ещ = -etv(x, t) is of the same order of magnitude in e as the
PERTURBATION METHODS 627
leading term i*0. In fact, (9.57) shows that every term enun in the series is of
the same order in e as the leading term u0 if t = O(l/e). Consequently, no
matter how small e is, there is a time t at which all terms in the perturbation
eries are of the same order in e and cannot be neglected on the basis that
tjjey constitute small corrections for small e. Thus even though the leading
terms of the perturbation series yield a good approximation for et< 1, the
result is not uniformly valid for all time. Terms of the form etv(x9t) are
called secular terms and the difficulty caused by the occurrence of such terms
in a perturbation series is known as secular behavior.
There are several methods for remedying the difficulties caused by
secular behavior in a perturbation series. One approach involves the sum-
summation of all the secular terms in the series and thereby assigning equal
importance to all of them. This process is called renormalization and can be
carried out in one form or another for a number of problems. In our
example each term in the perturbation series is a secular term, so that the
summation of the series yields the exact solution (9.47).
Another approach is known as the method of multiple scales and takes
note of the fact that the perturbation solution appears to exhibit two time
scales for our problem. TTiere is a slow time scale et and a (comparatively)
rapid time scale t, as shown by (9.57). Therefore, we look for a solution of
the form
(9.59) u(x, t) = и(х, t9 r), r = et
that depends on jc, t9 and r - et. That is, we treat the fast and slow time
scales as independent variables.
Since
dt ~ dt * !fr ~di ~ ~dt * € ~дт '
we have, instead of (9.45),
ut + €(иг + и) = йхх .
Expanding и in a perturbation series
Yields the recursive system
628 PERTURBATION METHOD*
Since uQ = f{x) at t - 0, we find that
(9.65) йо(х, t9 t) = c(t)v(x, i) ,
where v(x, t) is defined as before and с(т) is an arbitrary function of т that
need only satisfy the condition c@) = l, since f = 0 implies that r = €f
vanishes. The equation for ux is
(966) '-Ж ~ 17 -
with the initial condition ux = 0 at f - 0. Since the terms с'(т) and с(т) are
constants, as far the operator on the left side of (9.66) is concerned, we
obtain the solution
(9.67) йх(х, и т) = ~t[c'(r) + c(r)}v(x, t) + d(r)v(x, /),
where d(r) vanishes at т = 0 and u(jt, 0 satisfies (9.48) but are both otherwise
arbitrary. Consequently, we still obtain a secular term that grows with t in the
solution (9.67). However, we are now in a position to remove the secularity by
specifying с(т) such that
(9.68) c'(t) + c(t) = 0, c@) = 1
which yields
(9.69) c(r) = e~T=e~e'.
We also set d(r) ~ 0, since otherwise a further secular term would arise at
the next level of approximation. Consequently, йх = О and, as a result,
un = 0 for all n > 1. The series (9.62) thus terminates with the first term and
yields the exact solution (9.47).
In contrast to the preceding examples where exactly solvable problems
were solved by the perturbation method, we now consider a nonlinear
problem that cannot be solved explicitly. The perturbation method reduces
it to a class of solvable linear problems.
Example 9.3. A Nonlinear Klein-Gordon Equation. The hyperbolic
equation
(9.70) wtt - y2wxx + c2w - aw3 = 0
PERTURBATION METHODS 629
duces to the Klein-Gordon equation when the parameter tr is equated to
x& and arises in a number of physical contexts. The coefficients in (9.70)
%tt assumed to be constants. We consider the initial value problem for
/o 70) over the infinite line -oo < % < <*> with the data
w(x, 0) = e cos kx , w£x, 0) = 0 ,
where 0< e < 1 and A: is a specified constant.
Since the initial data (9.71) are uniformly small in magnitude, we look for
a solution of (9.70), (9.71) in the form
(9.72) w = €U
and apply the perturbation method to the problem for u(x> t). We have
(9.73) utt - у 2uxx + c2u - € W = 0
with the data
(9.74) u(x, 0) = cos kx , m,(jc, 0) = 0.
Expanding w(jc, f) as
00
(9.75) u(x,t)=Zun(x,t)en
and inserting (9.75) into (9.73), (9.74) yields as the leading order equations
(9.76) М«.)-^-У^ + Л.-0, —0,1,
(9.77) L2(u2, Ul,u0)=^-y^ + c% ~
01 OX
2 2
(9.78) L3(«3, иг, и1г и0) = —^ - у2 —% + с2м3 - 3au20ut = 0.
f/f ОХ
initial conditions are
(9.80)
„.0.
The solution of the initial value problem for uo(x, t) is
630 PERTURBATION METHODS
(9.81) uo(jt, t) = cos <ot cos kx
where
(9.82) oJ = )V + c2,
which is the dispersion relation for the linear Klein-Gordon equation. Th
equation for ux is uncoupled from that for u0 and has homogeneous data
Thus
(9.83) ut(x,t) = 0.
Since
(9.84) cos3 kx = | cos kx + \ cos 3fcc ,
the equation for u2(x, t) may be written as
•л «,-v ^2 2 ^2«2 2 Зог з о- з
(9.85) —2— У —Г + с w2 = — cos со* cos kx + -r cos cof cos 3fcc.
(?^ ^дс 4 4
On setting
(9.86) u2(x, t) = Fx{t) cos kx + F2(/) cos 3fcc
and inserting (9.86) into (9.85), we obtain the ordinary differential equa-
equations
(9.87) Fl(t) + (y V + с2Ж(г) - ^ cos3 tot 9
(9.88) F;@ + (9y2k2 + c2)F2@ = j cos3 o>r.
The initial conditions for u2(x, t) require that Fl9 F[9 F2, and F2 all vanish at
f = 0. On using the identity (9.84), the equations for Fx and F2 are easily
solved by the method of undetermined coefficients.
Using these results, the leading terms of the perturbation series solution
of (9.73), (9.74) are found to be
(9.89)
u(x, t) = cos a>t cos kx
+ e | ;-r— t sin a>t + —-—2 (cos <*>t — cos Ъш) cos kx
L 32u> 128o> J
gULar perturbation methods
631
2Г 3a
+ € [128<y2fe2 (COS (Ot - COS
128c2
(cos A/ - cos 3wt) cos 3fcc + 0(e3),
-I
where A2 - 9y2k2 + c2. The secular term
/n Q(\\ U(X. t) = ^rr— t Sin (Ot COS i
2)
in the series shows that when t = O(e 2), this term is of the same order of
magnitude as the leading term. As a result, the perturbation series becomes
invalid at those values of t. In constructing the perturbation series we are
assuming that terms containing higher powers of e are much smaller than
those with lower powers. We have already indicated in the preceding
example that the conventional perturbation expansion is generally not
uniformly valid over unbounded regions.
The result (9.89) is valid and useful only for times t such that t < 0(e~2).
To extend the validity of the result we note that
(9.91) cos <ot+ -^ /sin <ot =
as is easily seen by expanding the right side of (9.91) for small €. Using
(9.91) to replace the first two terms on the right in (9.89) yields the
improved result
(9.92) u(x, t) = cos[(o> - j£)t] cos kx + €2[ • - • ] + 0(e3),
in which the secular term is no longer present. There may be other secular
terms in the series but they occur for higher order terms in € and the validity
of the perturbation solution is presumably extended in time when written in
the form (9.92).
The method used to remove the secular term is essentially the method of
renormalization discussed in Example 9.2. Secular terms in the perturbation
series are removed by summing a part of that series. In Example 9.2 a full
Perturbation series was summed, whereas in the present example, since only
the leading terms of the perturbation series were determined, the renormali-
grtion method could only be carried out in an approximate manner.
In the following example we consider a linear equation with nonconstant,
but slowly varying, coefficients. As we see, the perturbation method appro-
appropriately modified yields a simple approximate solution.
632 perturbation
Example 9.4. The Wave Equation With a Slowly Varying Wave Speed
The hyperbolic equation
(9.93) un-c2(ex)uxx = 0,
where c2(ex) is assumed to have the expansion
/О ол\ г2(ех\ = с2 + У r \erl"
with constant cn, reduces to the wave equation with wave speed c0 if e = 0
Since dc2ldx = O(e) and e is assumed to be small, we may approximate c2 by
cl and state that the function c(ex) represents the slowly varying wave speed
for (9.96). We note that (9.93) is not expressed in self-adjoint form and that
the (variable) speed of disturbances as determined by the domain of
influence for concentrated initial data for (9.93) (see Section 8.3) is given by
c. We apply the perturbation method to an initial value problem for (9.93)
to determine the effect of the slow variation of c2(ex) around c\.
Let the initial conditions for (9.93) be
(9.95) u(x, 0) = /(*), ut(x, 0) = 0 , -oo < x < oo .
We expand u(x, t) as
(9.96) u(x,t)=2 un(x,t)en
rt=0
and insert (9.96) into (9.93) and (9.95). Equating like powers of e yields as
the leading order equations
where (9.94) has been used. The initial conditions for u0 and ux are
(9.99) «„(*,(>) =/(*), ^g^=0,
(9.100) ut(x90)= 1Уд/ =0.
We first discuss this problem in general terms, noting the difficulties that
• PERTURBATION METHODS 633
rise and indicating how they can be resolved. Then we consider a specific
\oice for c(€x) and obtain an explicit result for that case.
° The solution of (9.97) and (9.99) is
(9.101) «o(*> 0 = \Ax - cot) + \f(x + cot) .
Then the equation for ux becomes
д2и г дгих 1 1
(9.102) -^r " co -J^T = 2 xcJ'b ~ coi) + ^ xej"(x + co0 .
Using the method of undertermined coefficients, we easily solve (9.102)
with the data (9.100) and obtain
(9.103) ux(x9 0 = 1ГГ if(x - co0 ~ Ax + cot)]
We note that when t = O(e~112) or xt = О(е~х), the term €WX is of the same
order in e as the leading term м0, assuming / and its derivatives are
uniformly bounded. Since both the x and the t intervals relevant to this
problem are unbounded, the perturbation series would appear to be invalid,
in general, since values of x for which x = O[{et)~l] must occur if -» < x <
°°, for any f >0. However, if/vanishes outside a bounded interval, so does
«i(x, t) and large values of x do not play a role as long as t is small. In that
case, the perturbation theory solution may be valid for a substantial length
of time.
The general difficulty with the preceding perturbation series may be
traced to the fact that the solutions (9.101) and (9.103) are expressed in
terms of the characteristics
(9-104) ф± (х, t) = x±cot = constant
of the reduced equation—that is, (9.93) with € = 0—rather than the exact
characteristics of (9.93) that are given as solutions of
(9.105) 4
dx c(ex) '
°iven a point (x^0) on the дс-axis, the difference between the exact and
reduced characteristics issuing from that point becomes appreciable for large
634 PERTURBATION METH()DS
values of x and t. Consequently, we may expect that the perturbatio
solution becomes invalid for x and t large. n
To resolve the difficulties with the secular terms that arise in th
perturbation result (9.101) and (9.103), we introduce the independent
families of exact characteristic curves <p(x, t) = constant and ф(х л ^
constant, determined from (9.105), which reduce to the curves x - с r =
constant and x + cot = constant, respectively, when we set e = 0 in (9.105}
Noting the results (9.101) and (9.103) we look for a solution of (9.93) in the
form
(9.106) U(X, t) « g(€X9 €t)f(<p) + h(€X,
Inserting (9.106) into (9.93) yields
(9.107) utt-c2uxx = [<p*-cV
[ф2 - с2ф2Ж(Ф) + [M<PtgT ~ c2<px
{<ptt - c2<pxx)g]n<p) + [2в(«АЛ " c
(ф„ - с2фххЩПф) + е2[8тт - с2^
where a = ex and т = et. Since <p = constant and ф = constant are charac-
characteristics, they satisfy the characteristic equations for (9.93) and we have
(9.108) ^-cV' = ^-cVx = 0.
Also, since <p is a solution of
(9.109) «>, + <%-0,
we find that
(9.110) <ptt = -c<pxt = -aptx - c2<pxx + €cc>x
with a similar result valid for ф. This gives
(9.111) 2e[<ptgT - €2<pxg<r] + [<ptt - c2<pxx]g - -2ec<px[gT + cga-^ 8\
and a similar expression for A, where с and c' are replaced by -c and -c'.
The coefficients of f"(<p) and /"(ф) in (9.107) vanish in view of (9.108)^
The leading terms of order e are obtained by equating the coefficients of
f'(<p) and/'(*A) to zero. It follows from (9.111) and the corresponding result
for h that
gULAr perturbation methods
(9.112) gr + cg
The initial conditions (9.95) imply that
(9.114) Ф, 0)« g(ex, O)f((p) + h(ex, 0)/(ф) = /(*) ,
(9.115) u,(x, 0) ~ g(ex, Щ<р,П<р) + h(ex, 0)ф,/'(ф) + O(e) = 0 .
We conclude that the initial values for the characteristics <p(x, t) and ф(х, t)
are
(9.116) <р(х,0) = ф(х,0) = х.
Since <pt = -c<px and ^, = сфх, we find that at t = 0,
(9.117) «>,(*. 0)--*,(x,0)=-c.
Inserting (9.116), (9.117) into (9.114), (9.115) gives
so that
Using the methods of Chapter 2, the initial value problems for <p, ф, g, and A
are readily solved. Rather than present the general solution, we restrict our
attention to a specific problem that is exactly solvable.
Consider (9.93) in the interval 0<jc<<» with
(9Л20) c(ex) = co + €*,
and «@, t) = 0 for t > 0 and the data (9.95) for x > 0. Comparing (9.94) with
(9.120) we conclude that cx = 2c0, c2 = 1, and cn = 0 for n > 2. The charac-
characteristics <p and i/r corresponding to this choice of c(ex) and the data (9.116)
are
(9-121) 9(*,0 = «-g' + ^(e-*-l)f
(9.122) #*,*)-«в'+а(е"-1).
636 PERTURBATION
We discuss the solution in the region where <p(x, i) > 0. The equatio
(9.112), (9.113) for g and h with the data (9.119) can be solved by lookin
for solutions that are independent of tr since c' = 1 in this case. We obta'^
(9.123) g = | exp(^) , A = \ exp(- Ц) .
Then the approximate solution (9.106) is
(9.124) ф, t) ~ \ exp(y)/[«-' + ^ («-' -1)]
On expanding (9.124) in powers of e it is easily shown that the first two
terms in the expansion agree with (9.101) and (9.103) in the region <p >0.
Although the secular terms found in the conventional perturbation expan-
expansion are absent in (9.124), we do observe a slow exponential growth in r,
say, if f(x) is uniformly bounded. Though it would appear that we have used
the method of multiple scales to obtain (9.124), this problem is somewhat
more complicated since we not only introduced ex and et as variables but
also the characteristics <p and ф.
We have seen that the separation of variables and transform methods for
the solution of initial and/or boundary value problems for partial differen-
differential equations are useful only for problems given over special regions such as
squares, circles, spheres, or half-spaces. If the boundary of the given region
varies only slightly from a boundary for which the foregoing methods can
successfully be applied, it is possible to use perturbation theory to effect an
approximate solution to the problem. We now present a brief discussion of
boundary perturbation methods for a two-dimensional problem and consider
an example.
Let the given problem be specified in a two-dimensional region G with
dG as its boundary. We assume that dG can be expressed in parametric
form as
(9.125) x - g(s) + €g(s) , у = ft(s) + €h(s),
where 0 < e <^ 1, .s is a parameter, and the functions g, g, A, and ft are given.
We could also consider a more general e dependence such as x = g(s, e) and
у = h(s, e), where g and h can be expanded in powers of e, but we do not do
so for the sake of simplicity. Assuming that the function u(x, y) is specified
on the boundary curve dG, we have
(9.126) ч(х9 y)U = <8 + «& h + eh) =f(s) ,
PERTURBATION METHODS 637
here f(s) is prescribed. Further, u(xy y) is assumed to satisfy a linear
econd order partial differential equation with no explicit e dependence
(9.127) Lu = 09 {x9y)e6.
Using the perturbation method we expand u(x, y) as
00
(9.128) u = n2 unen
and insert (9.128) into (9.126), (9.127). Since L is a linear operator we find
that
(9.129) 4=0» *^0.
In (9.126) we first expand и in powers of e as
(9.130) u(g + eg, h + eh) = u(g, h) + e[gux(g, h) + huy(g, A)] + O(e2)
and then obtain from (9.128)
(9.131) uo(g,h)=f,
(9.132) иЛe, h) = —s —-—■ h —™—■— .
v } lV5) } 6 dx dy
Thus the perturbation method reduces the given problem to a collection
of problems given over the unperturbed region G with the boundary curve
dG whose equations are x = g(s) and y = h(s). It is assumed that the
boundary value problems for the functions un(x, y) in the region G can be
solved. It is a straightforward matter to extend these results to three-
dimensional problems and to cases with other types of boundary conditions.
We now consider an example.
Example 9.5. Boundary Perturbations for Laplace's Equation. The so-
solution of the Dirichlet problem for Laplace's equation in a rectangle was
presented in Example 4.11 where the method of separation of variables was
used. We now consider a small perturbation of the rectangular region
0<*</ and 0<y<L. Let the region G be given as the interior of the
trapezoid defined as 0<x<l and ex<y<L, so that the boundary line
У = 0 is replaced by у = ex where 0 < e < 1 (it is assumed that el<L).
The Dirichlet problem is given as
(9l33> «tt + Bw=0, 0<*</, €x<y<L
i the boundary conditions
«8 PERTURBATION METHODS
(9.134) „@,^ = 0, 0<y<L,
(9.135) u(x,L) = 0, 0<x<l,
(9.136) и(/,у)-0, el<y< L,
(9.137) u(x,€x)=f(x), 0<x<l,
where f(x) is a prescribed function.
To solve (9.133)-(9.137) we first expand u(x, ex) as
(9.138) u(x, ex) = u(x, 0) + exuy(x, 0) + O(e2) = /(x).
Then the expansion (9.128) of u(x, y) yields
(9.139) uo(x, 0) + e[Ml(*, 0) + x
Each of the un(x> y) satisfies Laplace's equation. For мо(х, у) we have the
boundary conditions
(9.140) ио(х,у) = 0, x = 0, y = L, jc = /,
(9.141) ио(х,О)=Дх),
in view of (9.139), with
(9.142) V2w0 = 0,
The function иг(х9 у) also vanishes on the three sides of the rectangle (i.e.,
x = 0, у = L and x = /), whereas on the fourth side
(9.143) Ui^0) = -X^0)^ 0<x<l,
and satisfies Laplace's equation
(9.144) V2Ml=0, 0<jc</,0<y<L.
Using the results of Example 4.11, we find the solution of the problem for
uo(x, y) to be
(9.145) «,(*, У) = Vy 2 ** sinh [т (y "
where bk is defined in D.161) with / replaced by L. To determine the
PERTURBATION METHODS 639
boundary condition (9.143) for щ(х, у) we differentiate (9.145) term by
rm with respect to у and evaluate it at у = 0 to obtain
(9.146)
"i(*> 0) = " V7 2^ хЦ-j cosh [—J sm [—J .
дат the results of Example 4.11 can be used to specify иг(х, у) by putting
fix) equal to the right side of (9.146) and setting g(x) = 0 in that example.
For the sake of concreteness we set
(9.147) /to = sin —
in (9.137). Then uo(xy >>) has the form
(9.148) uofe У) = sinh (^L/Z) m Ы
and the boundary condition (9.146) for ux(xy y) is
(9.149) ttl(x, 0) - — coth [—J sm [—J .
We note that ut(x, 0) vanishes at x = 0 and x = /, and is a smooth function in
the interval 0 < x < L Therefore, its Fourier sine series converges uniformly
in the given interval and, as a result, the solution wa(x, y) obtained by the
methods of Example 4.11 is continuous and, therefore, bounded in the given
region. Thus €UX represents a small perturbation around u0 for small e.
Similarly, if u^x, 0) in (9.146) is a smooth function in 0 < x < /, we conclude
that €iil is a small perturbation around u0 since иг@,0) = ux{l, 0) = 0.
Eigenvalue problems can also be treated by perturbation methods. A
special case has already been considered in Example 8.5. We use the
notation of Chapters 4 and 8 in our discussion and consider the problem
(9.150) LM^ -V' (pVM) + qM = (A + er)pM
ш the bounded region G with the boundary condition
дМ
дп
dG
= 0.
parameter e is small (i.e., 0<e<^l) and r is a given function. It is
assumed that the reduced eigenvalue problem with 6 = 0 is solvable and that
the eigenvalues are simple so that there is one linearly independent eigen-
fanctio for each eigenvalue.
640
PERTURBATION METHODS
To determine the eigenvalues and eigenfunctions of (9.150), (9.151) w
expand both M and A in powers of e and set
oc
(9.152) M{x; e) = 2 Min)(x)e" ,
я=0
(9.153) A(e)=S A(/V.
n=0
Inserting (9.152), (9.153) into (9.150), (9.151) gives
(9.154) LM@) = A@)PM@),
(9.155) LMA) = A@)pMA) + rpM@) + AA)pM@),
(9.156) LMB) = A@)pMB) + rpMA) + AA)pMA) + AB)pM@)
as the equations for the first three M(n). The boundary conditions for each of
the M(n) are
(9.157)
,=0, n>0.
дП BG
Let a£0) = kk and Mk0)(x) = Mfc(x) (A: = 1,2,.. .) represent the (simple)
eigenvalues and orthonormalized eigenfunctions of the unperturbed prob-
problem (9.154) and (9.157) where n = 0 (these are assumed to be known). The
equation (9.155) is an inhomogeneous version of (9.154), with M( (x)
required to satisfy the homogeneous boundary condition (9.157). This
problem has no solution unless the inhomogeneous term satisfies a com-
compatibility condition, as we now show. We put M(n) and А(л) equal to M(kn) and
\[n) in (9.154)-(9.157). Then (9.155) takes the form
(9.158) LMhl) - ккрМ^ = (г +
since A£0) ^ Xk and м£0) = Мк.
Applying the results of Example 4.2, we multiply (9.158) by Mf and
integrate over G. Since both Mj and Af[l) satisfy the boundary condition
(9.157) we obtain, in view of D.28),
(9.159) j \g ЩЬМ[1) - A,pMX}] dv
: PERTURBATION METHODS
= (\J-\k)(M[1\MJ)
= \ jG rpM1Mk dv + А^ЧЛ/;, Мк)
where the inner product (/, g) is defined as
, if / = k, so that Ay = kk, the left side of (9.159) must vanish and this
implies that \k1] must be specified as
(9.160) A<1} =
since (Mk,Mk) = l. For j^k, the Fourier coefficients (М(к\М;) of the
function Mkl) are given as
(rMk,MA
(9.161) {Mkl\Mf)= } , j*k
Aj Ak
where we have used the fact that (Mj9 Mk) = 0 for / Ф k.
The eigenfunctions Mj(x) are assumed to form a complete set and
Mkl){x) can be expanded in the series
(9.162) M^\x) = X (Mj^, M^M^x), к = 1,2,....
Each of the Fourier coefficients is specified as in (9.161) except for
(Мь,Мк) which remains undetermined. If we normalize each of the
eigenfunctions Мк(х; е) we have
(9.163) 1 = (Мк(х; е), Мк(х; е))
= (Мк°\Мк0))
since the eigenfunctions M[o) = Mk are already normalized. On equating like
Powers of 6 we conclude that
Consequently, the series (9.162) for Mkl\x) is completely specified.
For each unperturbed eigenvalue kk and eigenfunction Mk(x) we have
determined the first perturbation €\k1} and eM^jc). Higher approximations
642 PERTURBATION METHODS
can be obtained by applying the finite Fourier transform method to th
equations for M^}(x) and proceeding as in the foregoing. A modification of
this procedure, which we do not present, is needed to deal with the case of
multiple eigenvalues. We note that the Sturm-Liouville problem has bee
shown to have simple eigenvalues and these results can easily be expressed
in a one-dimensional form. In higher dimensional problems multiple eieen
values often occur. However, the perturbation method can be applied for
every simple eigenvalue in a set of eigenvalues that may include multiple
eigenvalues. The following example considers such a case.
Example 9.6. Eigenvalue Perturbations in a Square. We consider the
square of side тг, with the interior region G defined as 0 < x < тг and
0 < у < 7гу and consider the perturbed eigenvalue problem
(9.165) -(Mxx + Myy) + exyM = AM , (x, y)EG
and the boundary condition
(9.166) M(x,y;c)\,G=0,
where 0 < 6 < 1. The reduced eigenvalue problem with € = 0 was considered
in Example 7.6. There it was found that the eigenvalues \nm are given as
(9.167) \nm = n2 + m2 , n, m = 1,2,...
and the normalized eigenfunctions Mnm(x, y) are
2
(9.168) Mnm(x> У) = ~~ sin nx sin my , n, m = 1,2,
7Г
We see that An is a simple eigenvalue with only one linearly independent
eigenfunction Mu. However, A12 equals A21 and to this eigenvalue there
correspond two independent eigenfunctions M12 and M21. There are infinite-
infinitely many other multiple eigenvalues.
As has been indicated previously, it is often of greatest interest to
determine the lowest eigenvalue for a given problem. Since the lowest
unperturbed eigenvalue An is simple, we may apply our perturbation proce-
procedure and determine the corresponding perturbed eigenvalue as well as its
eigenfunction. The situation is complicated somewhat because of the double
subscript notation for the eigenvalues and the eigenfunctions of the reduced
eigenvalue problem. Therefore, we shall only obtain the first perturbation ot
the eigenvalue An but not the perturbation of the eigenfunction Mn(x, У)-
In the notation of the previous discussion we have p — \ and
(9.169) r(x, y) = -xy.
Then with
PERTURBATION METHODS «43
(9.170) Au(€) = An + ek$ + O(e2)
representing the lowest (perturbed) eigenvalue, we have from (9.160)
(9.171) A">e^?Jo Jo xy&fxwfybdy
/2 Г . 2 Л2 к2
= - /sin2 tdt) =~r
\тг Jo I 4
to that
(9.172) Au(e) = 2 + (ir2/4)e + 0(«2).
The nonlinear Klein-Gordon equation
(9.173) и>„ - yV^ + c2w - о-и»3 = О
discussed in Example 9.3 was found to have an approximate solution of the
form
(9.174) w(x, t) = e cos kx cos[(u> - ^^)t] + O(e3)
as follows from (9.92) and (9.72). This can be rewritten as
(9.175) w(x, f)=| cos[fcc - (« - |^
(We recall that к is a constant and a>2 = у 2Jfc2 + c2.) Now each of the terms in
(9.175) represents a traveling wave of the form
(9-176) w(jc, 0 = a cos[kx + a>(k)t]
where the amplitude term a = e/2 is small and the phase term в = Ь: +
w(^)^, with
has the phase velocity
(9.178) * - -
644
PERTURBATION METHOD»
If the amplitude a = ell were assumed to be infinitesimal, we would neglect
the term involving a2 in (9.178) and the phase velocity would reduce to tha
for the linear Klein-Gordon equation as given in Example 3.8. However
for small but finite amplitude traveling waves where a2 is not neglected w'
find that the phase speed or, correspondingly, the speed of the wave (9Л7б\
depends not only on the wave number к but also on the amplitude a.
The dependence of the speed of the wave on its amplitude has already
been observed in our discussion of nonlinear unidirectional wave motion in
Chapter 2. The dependence of the speed of normal mode solutions of linear
hyperbolic equations on the wave number к has been characterized as
dispersive wave motion in Section 3.5. The preceding results show that the
nonlinear Klein-Gordon equation has (approximate) traveling wave solu-
solutions that exhibit the combined effects of having the wave speed depend on
both the wave number and the amplitude.
The theory of nonlinear dispersive wave motion has recently undergone
much study, especially by Whitham. One aspect that has been investigated is
whether periodic finite amplitude traveling waves of the form (9.176) or some
more general form can be constructed for these equations. Since the
amplitude of these waves is assumed to be small, it can serve as a
perturbation parameter. It is assumed that when a small amplitude solution
is inserted in the given equation and the equation is linearized, the result is a
dispersive wave equation. The effect of the nonlinear terms is to introduce
higher harmonics as in (9.89) and a dependence of the dispersion relation
ox = a>(k) on the amplitude as seen in (9.177).
We do not attempt to characterize the general form of nonlinear disper-
dispersive wave equations for which periodic traveling wave solutions can be
found. Rather, we discuss a specific equation in the following example and
consider further examples in the exercises. We refer to the literature for
general results. One of the earliest applications of the perturbation method
for determining periodic traveling waves for nonlinear equations was given
by Stokes in his study of water waves. In the following example we construct
approximate traveling wave solutions of the nonlinear Korteweg-deVries
equation which plays an important role in the theory of water waves (see
Example 10.15). Even though exact periodic traveling wave solutions can be
found for this equation, we do not do so but apply the perturbation method
instead. The main feature in our approach is that not only the solution but
the dispersion function is expanded in powers of the small parameter. The
need for this is suggested by (9.177).
Example 9.7. Periodic Traveling Wave Solutions of the Korteweg-
deVries Equation. The Korteweg-deVries equation is given as
(9.179) ut + (c + u)ux + $uxxx = 0 ,
where с and /3 are given constants. In its linearized form
PERTURBATION METHODS
(9.180) ut + cux + puxxx
the dispersion relation is (see Section 3.5)
a>(k) = ck-
so that (9.180) is of dispersive type. If we drop the third derivative term in
(9.179), we have
(9.182) и, + (с + и)их = 0,
which is a quasilinear first order wave equation whose wave speed depends
on the amplitude.
The linearized equation (9.180) has traveling wave solutions
(9.183) u(x, t) = a cos[fcc - cot],
where a = constant and ox is given by (9.181). The quasilinear equation
(9.182) has the (implicit) solutions
(9.184) и = a cos[kx - k(c + u)t].
The wave speed for (9.183) is dxldt - oxlk = c- /3k2 and the wave speed for
(9.184) is, formally, dxldt = с + м. In both cases the wave speed is perturbed
around the constant c, in the linear case by a function of the wave number
and in the nonlinear case by the amplitude. Note that if a is small, then |u| is
small in view of (9.184). We may also characterize k{c + u) as a nonlinear
frequency term and then (9.183), (9.184) show that the frequency of the
traveling wave solution for the full problem (9.179) should depend on the
wave number к and on the amplitude a.
Thus we look for a perturbation solution of (9.179) in the form
where a is a small positive constant and
Л expanded as
<9'187) & = i Wn(k)a" .
л = 0
We attempt to determine the unF) to be periodic functions, so that (9.185)
646 PERTURBATION METHo&S
is a periodic traveling wave solution of (9.179). To accomplish this it i
necessary to specify the terms in series expansions of the frequency A
appropriately. We note that we have included a dependence on к and a f0
the frequency w in (9.187).
Inserting (9.185) and (9.187) into (9.179) yields the equations
(9.188) (ca0 - ck)u[ - f$k3urr; = 0 ,
(9.189) (a>0 - ck)u2 - pk3u' = kuxu[ - <oxu[ ,
(9.190) (<o0 - ck)ur3 - $къи% = kuxu2 4- ku2u[ - <oxu2 - w2u[
on equating like powers of a.
As the solution of (9.188) for ux we take
(9.191) 1^@) = cos 0
and obtain
(9.192)
This yields a periodic traveling wave solution corresponding to the linearized
equation (9.180). Neglecting higher powers of a in (9.185)-(9.187) yields
the result (9.183). With this choice for ul9 the right side of (9.189) becomes
(9.193) kuxu\ - <axu[ = -k sin в cos в + шх sin в
к
= - - sin 20 + й>! sin в .
The term o>x sin в yields a secular term in the expression for u2 proportional
to в sin 0. Since we seek periodic solutions, we must remove this term and
this is the case if we set a>x = 0. Then a solution of (9.189) is
(9.194) M2@) ^
Continuing with (9.190) we obtain for the right side (since a>x =0)
(9.195) kuxu'2 + ku2u[ - a>2u[ = [- ^~ + o>2] sin в - щ sin 3$ .
Again the term involving sin в must be removed to avoid obtaining a secular
term. Thus we set *
nGULAB PERTURBATIONS AND BOUNDARY LAYER THEORY 647
(9.196)
and the solution w3@) is
(9.197) (*)
a2 a3
20 +
The full solution to order a is
a2 a
(9.198) u@) = acosO+ щ^ cos 20 + ~Щ1р cos 3$ + 0(a )
with 0 defined as in (9.186) and <b given as
(9.199) Л - &{k, a) = ck~ f3k3
The frequency w and, consequently, the wave speed are seen to depend
both on the wave number к and the amplitude a.
We conclude this example by noting that on inserting the expression
м = m@)—with в given as in (9.186)—into the Korteweg-deVries equation,
it is possible to find an exact periodic traveling wave solution in terms of
elliptic functions. If that solution is expanded for small amplitudes, the
preceding results are reproduced. We do not carry out this discussion, which
is given in the literature.
9.3. SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY
In singular perturbation theory we,are concerned with the study of partial
differential equations that contain a small parameter that multiplies one or
more of the highest derivative terms in the equations. Thus when that
parameter is equated to zero either the order or the type (or both the order
and type) of the given equation is changed. Generally, this means that a
regular perturbation series solution proves inadequate to handle the initial
and/or boundary data for the given problem. It thus becomes necessary to
introduce boundary or initial layers where the solution of the given problem
undergoes a rapid transition from a form that satisfies all the data given for
the problem to a form represented by the perturbation series. The determi-
determination of the boundary layers and the approximate forms of the given
equations in those regions forms the subject of boundary layer theory. The
Procedure whereby solutions valid in the boundary layers are identified with
the perturbation series solution valid in the so-called outer region, is often
CaUed the matching process. The general process of determining perturba-
648 PERTURBATION METHODS
tion and boundary layer expansions and matching these results is sometimes
referred to as the method of matched asymptotic expansions. By combining
the perturbation and boundary layer solutions, a fairly good approximate
description of the solution of the given problem can often be found for
problems where the exact solution is difficult or impossible to determine or
where the solution is not easy to interpret or evaluate.
It should be noted that singular perturbation theory is sometimes taken to
encompass any problem where regular perturbation theory is inadequate for
any reason. This may not involve the presence of a small parameter
multiplying the highest derivative, but may be due to the presence of secular
terms that result in the nonuniformity of the solution over an infinite region
or the occurrence of a small parameter in the data for the problem
(examples of these types have been given in Section 9.2). Nevertheless, we
shall restrict our discussion in this section to the type of problem discussed
in the preceding paragraph.
As it is somewhat complicated to present a general theory that en-
encompasses all types of singular perturbation and boundary layer problems,
we begin by considering two simple examples for a first order equation.
Then we shall consider second and higher order equations from a general
point of view and in a number of examples.
Example 9.8. Singular Perturbation of a First Order Equation. The
initial value problem for the equation
(9.200) e(ut + ux) + u = sint9
with
(9.201) w(jc,0) =/(*),
where 0< e < 1 and f{x) is a prescribed smooth function, has the solution
(9.202) u(x, t) = j^p (sin t - e cos t) + [ f(x - t) + j^-pJ exp(-^) >
as is easily verified. (The solution may be obtained by the method of
characteristics given in Chapter 2.)
If we attempt to solve (9.200), (9.201) by using a conventional perturba-
perturbation series, we set
00
(9.203) u(x, 0=2 un(x, t)e" .
n = 0
Inserting (9.203) into (9.200) and equating like powers of e yields the system
(9.204) u0 = sin t
PERTURBATIONS AND BOUNDARY LAYER THEORY 649
(9.205) u» =
We easily conclude that
(9.206) u2n = (-1)" sin t, м2п+1 = (-l)rt+1 cos t, л >0.
Thus the terms in the perturbation series (9.203) are uniquely specified
without regard to the initial value (9.201) that m(jc, t) must satisfy. Further-
Furthermore, the perturbation series
(9.207) u(x, t) = [ S (-1Гб2л1 sin t - e\ 2 (-l)
COS
certainly does not satisfy the initial condition (9.201). The reason for this is
that the reduced problem for (9.200)—where e is set equal to zero—is not
even a differential equation and, therefore, cannot absorb arbitrary initial
values. This reduced problem characterizes the form of all equations arising
from the conventional perturbation approach, as shown by (9.204), (9.205),
and none of them is a differential equation.
On comparing the perturbation solution (9.207) and the exact solution
(9.202) we see that difference between both solutions is a term significant
only in the region 0<f<O(e). That is, for t>O(e)9 the exponential
exp(-f/e) is small and can be neglected. Then the solution of (9.200),
(9.201) is well approximated by the perturbation result (9.207). However,
within a layer of width O(e) near the jc-axis, the exponential term in (9.202)
is significant and cannot be neglected. As we can see, it is this term in
combination with the perturbation result (9.207) that enables the solution to
satisfy the initial condition.
The existence of an initial layer of width O(e) where the conventional
perturbation series is not valid may be inferred from the fact that for t *= e
we have sin t ** sin e « e and cos t« cos e «1. Thus the first two terms in the
perturbation series behave like
(9.208)
sinf-ecosf«e-e =
which implies that the two terms are of the same order in e. Therefore, the
Perturbation series is not well ordered in the region where t = O(e) and it is
not expected to be a valid representation of the solution. Of course, we
know that (9.207) is not valid near t = 0 since it fails to satisfy the initial
condition. However, the foregoing argument determines an approximate
description of the size of the region where the conventional perturbation
Method is invalid. This type of argument is similar to that given in Section
9-2 in connection with secular terms.
650 PERTURBATION METH0d
The significant conclusion that we have reached is that the perturb
result need not be discarded completely because of its failure to /
initial condition. It need only be replaced by a different or
approximation in an initial layer of width O(e) near the x axis. To stud
equation in the initial layer we introduce the stretching transformation
(9.209) Г=€гт
where the positive constant r is to be specified. Since e is small, the
variable is large even for small or moderate values of t. Thus the region ne ?
the jc-axis is stretched out. The equation (9.209) indicates that we wish to
study (9.200) in a region where t = O(er). Now we already have shown that
the choice r = 1 is appropriate for our problem, but we wish to show this
directly from the equation (9.200) by using boundary layer arguments.
We have
I
(9.210) €l~% + ейх + и = sin er = err - ^- + 0(e5r)
on using (9.209) in (9.200) with й(х, т) = и(лг, err). Now if r = 1, there is a
balance between the terms ur and и in (9.210), and they represent the
leading terms for small e in that equation. We need to retain the йт term so
that we have a differential equation in the initial layer that can absorb an
initial condition. We would also like to retain the term и that occurs in the
reduced equation for (9.200). Thereby, a smooth transition from the initial
layer to the outer region, where the perturbation series (9.207) is valid, is
expected to result. With г > 1, the term ur is the leading term. Although this
choice of r leads to equations for which the initial condition can be
accounted, the initial layer seems to be too thin since the equations do not
retain any part of the reduced equation. For r < 1, й is the leading term and
nothing has been accomplished.
Putting r = 1 in (9.210) we obtain
eV s
(9.211) йт + и + ейх = sin er = er - — + O(e ).
The initial condition at т = 0 is
(9.212) w(jc,0) =/(*).
We solve (9.211) by the perturbation method and set
(9.213) й(х,т) = ^йп(х,т)еп.
Inserting (9.213) into (9.211) and equating like powers of e yields
uLAR PERTURBATIONS AND BOUNDARY LAYER THEORY 651
(9.214) дт ° '
дйх д _ SUq
(9.215) <?т wi -т дх 9
(9.216) 7Г + й2 = ""^Г
the leading equations. The initial conditions for these ordinary differen-
differencial equations are
(9.217) йо(х, 0) = f(x) , йн(х9 0) = 0 , h > 1.
It is often the case that the boundary layer equations are ordinary differen-
differential equations.
The solutions of the initial value problems for the first two boundary
layer terms йп(х, г) are
(9.218) йо(х,т)=/(х)е~т,
(9.219) йг{х, т) = т - 1 + [1 - rf'(x)]e-* ,
Thus
(9.220) m(jc, t) * f(x)e~T + e{r - 1 + [1 - rff(x)]e'T} + O(e2).
In general, we now have to apply the matching process to specify
unknown quantities that occur in the outer perturbation expansion. That is,
we assume that the perturbation series (9.207) and the boundary layer
(perturbation) series (9.213) have a common region of validity. Then we
express both series in terms of a single set of variables and identify
corresponding terms. Since the (outer) perturbation series in the present
Problem is completely specified, we carry out the matching to show how the
solution undergoes a transition from its boundary layer form to the form
valid in the outer region where t> O(e). We have
u(x, 0-2 un(x, бт)€п « 2 й„(х, т)еп
m a common region of validity of both series that is assumed to exist. Thus
^ l222) u0 + eux = sin(eT) - e cos(€t) « er - e 4- O(e3)
and
К + *«i =f(x)e~T + €lr - 4 + Ф - rff(x)]e~r^er - e ,
652 PERTURBATION METHoDs
where for the purpose of matching we have assumed that т is large so th
the exponential e~T can be neglected, but that ет is small so that sin(er) at
cos(er) can be approximated by the leading terms in their power se ^
expansions. This procedure can be formalized by assuming, for examT8
that т= O(e'U2) so that т is large for small e but ет= O(e1/2) which '
small. Then the region where the matching is carried out correspond 1S
t = €T = O(e1/2). The matching process can be carried out systematically b°
using the method of intermediate limits. However we do not require tt/
method, as we consider only leading terms in our analysis, and do not
discuss this technique.
On comparing (9.222) and (9.223) we see that they both agree to the
order of € retained. Consequently, it follows from our perturbation and
boundary layer results that the approximate solution u{x, i) of (9.200)-
(9.202) is given as
(9.224) u(x9 0 ~ t - e + [f(x) - tf\x) + e]e-'l€ , 0 < f < O(e)
and
(9.225) u(x, t) « j sin l 2 cos * ** sin * ~ € cos t +
for t > O(e). In the initial layer of width O(e) the solution undergoes a rapid
transition from the form (9.224) which satisfies the initial condition to the
form (9.225) which is the (outer) perturbation solution. The solution (9.224)
in the initial layer corresponds to the expansion of the exact solution (9.202)
for small t as is easily verified. The fact that/and its derivatives are given as
functions of x rather than x - /—corresponding to the characteristics x-t =
constant of (9.200)—as is the case in the exact solution (9.202), is not that
important. Because of the rapid exponential decay of the term that involves
/, this difference in the argument of that function is not significant.
To conclude this problem we note we note that if the term и is replaced
by -u in (9.200), we may still use the regular perturbation method to
construct a series solution of the form (9.203). However, the leading order
boundary layer equation will now be
(9.226)
as is easily seen on proceeding as in the foregoing. The solution м0 =/(у
now grows exponentially as т increases so that we do not obtain an initia
layer within which a rapid transition of the solution to the form of the outer
perturbation expansion takes place. The reason for the breakdown of the
perturbation method and the boundary layer theory is that the problem with
the term и replaced by -u is unstable and has exponentially growing
GUtAR PERTURBATIONS AND BOUNDARY LAYER THEORY 653
tions. This is easily shown by finding the exact solution of the initial
50 problem or by the use of a stability analysis.
The general first order linear equation
(9 227) €(aux + but) + cu = d, -oo<x<oo, t>0,
here 0<€<l and л, Ь, с, and d are specified functions of x and t, with the
initial condition
(9.228) Ф> 0) = /W» -°° <x < °° >
yields a singular perturbation problem that may require the introduction of
additional internal boundary layers apart from the initial layer for its
solution. Thus if с vanishes on a curve in the region t > 0 where a and b do
not vanish, the conventional perturbation series may become singular there.
However, the solution of the full problem (9.227), (9.228) is not expected to
be singular along that curve. Consequently, a modified result that can be
obtained by boundary layer methods must be introduced. In the following
example a problem of this type is considered.
Example 9.9. An Internal Boundary Layer. The first order equation
(9.229) е(м, + м,) + (*-1Jм = 1, -оо<д;<оо, t>0
with the initial condition
(9.230) m(jc,O) = O, -oo<jc<oo
has the exact solution
as is easily verified. Because of the jc-independent initial condition, we may
consider (9.229), (9.230) to be an ordinary differential equation problem. It
*s> nevertheless, of interest to study the phenomena that occur for the
Perturbation solution of this problem, as they occur for more general
Problems as well.
We begin by considering the conventional perturbation series solution
(9.232) u= 2 unen
of (9.229), (9.230). We easily obtain as the leading terms
«4 PERTURBATION METH0Ds
(9.233) M"M° + eMl" (t-lJ + (/-IM *
Not only does (9.233) not satisfy the initial condition (9.230) but it also *
singular at t = l, whereas the given equation (9.229) has no singular'^
there. ly
To deal with the initial condition we introduce the stretching tra
formation
(9.234) t=er,
which yields for (9.229)
(9.235) йт + ейх + й - 2етй + € Vw = 1,
where m(jc, t) = u(x, er). To obtain (9.234) we can use the approach given in
Example 9.8. With
00
(9.236) ~~
n=0
we have
(9.237)
as the leading equations. The initial data for the un are
(9.239) йп(д:,О) = О, n>0.
The solutions are
(9.240) щ = 1 - e~T,
(9.241) йг = 2(r - 1) + B - т2)е~т ,
so that
(9.242) й * 1 - e-T + e[2(r - 1) + B - r2)e'T].
It is readily verified that (9.233) and (9.242) match one another if (9.233) is
expanded for small t and (9.242) is expanded for large r.
At / approaches unity, the (outer) perturbation series (9.233) breaKS
PERTURBATIONS AND BOUNDARY LAYER THEORY 655
To determine the values of t where (9.233) first begins to break
we note that when
(9.243) ,-l = O(el/3),
he perturbation series (9.233) becomes disordered in that the second term
• of the same order in e as the leading term. This suggests that the width of
the internal boundary layer region near t = 1 is given as \t — 1| - 0(e1/3).
Further, we find that in the boundary layer
(9.244) w>
Therefore, to study the solution of (9.229), (9.230) near f = lwe set
(9.245) Г-1 = €1/3т, u = €~2/3v.
This gives
(9.246) vT + T2v + €Xi\ = l.
The appropriate stretching exponents for € in (9.245) could have been
determined by the use of an argument similar to that given in the preceding
problem, whereby a balancing of significant terms in the resulting equation
(9.246) is carried out. A new feature for this problem is that not only an
independent, but also the dependent variable was stretched. This is usually
required for nonlinear problems and often arises when dealing with
inhomogeneous linear equations.
To specify the solution v(x, r) of (9.246) we require that it match the
outer solution (9.233) as we approach the region O(e)<t<l - O(e1/3)
where the perturbation series is valid. That is, we require that as
t-> -oos v(x, t) tends to the form (9.233) expressed as a function of т.
The boundary layer expansion for v(x, r) is given in powers of e1/3, in
view of the form of (9.246), and we have
(9.247) v = 2 vnen/3 .
/t=0
leading equation is
whose general solution can be written as
Vo{x, T) = exp[- £]£ ^p[~]da + b exp[- £] ,
656
PERTURBATION METHODS
where the arbitrary constant b is as yet unspecified. To match (9.249)
(9.233) we integrate by parts to obtain
(9.250) jl exp[^] * = Д, exp[^] + 2£ £ exp[^] *,
with т assumed to be negative. A further integration by parts yields
(,25!) I e*p£] * - £ ♦ A
Thus
(9.252) Vo(x, т) = -2 + -5 + 10exp[- ^-1 Г Л ехр[^
where we have put 6 = 0, since the solution would otherwise grow exponen-
exponentially as r-> -oo, and that is ruled out by the form of (9.233). If we now
express, (9.233) in terms of т, we have
(9.253) " = е-2/3[Л + Л1-
Since u = e~2/3v—in view of (9.245)—we find that the internal boundary
layer expression (9.252) matches the outer solution (9.233).
To complete our discussion of the solution we need to examine what
happens in the region above the internal boundary lavjer, that is, when
t> 1 + O(e1/3). It is expected that the outer solution (9.233) should be valid
there. However, for that to be the case, the outer solution must match the
internal boundary layer solution as т^> +оо. To show this we write vo(x, r)
as
(9.254) v0 = expl - — l| J expj — I dor + J expl — J daj
where both a and r are positive and a < r—note that we have put b = 0 in
(9.249). Since 0 < a < т, the first integral in (9.254) contributes an exponen-
exponentially small term to the overall result. Integrating by parts in the second
integral as was done previously, readily yields an expression that matches
the outer solution (9.233) as т->оо, since the contributions from the lower
limit result in exponentially small terms.
We have thus shown that a satisfactory description of the solution oi
(9.347), (9.348) can be obtained by combining perturbation and boundary
layer methods. Although an exact solution is available for this problem, the
approximate results are generally easier to evaluate.
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 657
we discuss the singular perturbation of a second order linear
hyperbolic equation that yields an initial layer as well as secular terms.
Example 9.10. The Singular Perturbation of a Hyperbolic Equation.
The second order hyperbolic equation
(9.255) <utt - c2uxx) + ut + aux=0
^th constant coefficients and 0< e < 1, has the property that with e = 0 it
reduces to an equation of first order. To examine some of the problems that
can result from this fact we consider the Cauchy problem for (9.255) with
initial data оц the jc-axis given as
(9.256) u(x, 0) = f(x), ut(x, 0) = g(x) .
We attempt to solve (9.255), (9.256) by the use of a perturbation series
00
(9.257) и(*,0=2 un(x,t)e\
Inserting (9.257) into (9.255) and equating like powers of e yields the
recursive system
(9.258)
дип дип
dt дх дГ
with the initial conditions
(9.259)
Since the equations for Mn(jc, t) are each of first order, only one initial
condition can be assigned for each un at t = 0. Therefore, the initial value
Problems for the un cannot be solved in general. The singular nature of the
Perturbation is thereby brought into evidence. As a result, we cannot expect
h perfurbation seri^s to be valid near the лт-axis. Nevertheless, we expect
the series (9.257) to provide an approximate description of the solution
tt(*, 0 away from the jc-axis.
Using the methods H Chapter 2, we easily obtain as the general solutions
of the first few equations (9.258) for the un(x, t)9
658 PERTURBATION METHOD
(9.261) Ul(x9t) = t(c2 - a2)F"(x ~ at) + G(x - at),
(9.262) u2(x, t) = \t\c2 - a2)F»»(x - at) + 2at(c2 - a2)F"'(x - at)
+ t(c2 - a2)G\x - at) + H(x - at) ,
where F, G, and H are arbitrary functions.
Assuming that F(*), G(x), and H(x) are uniformly bounded functions
together with their derivatives for all x, we see from (9.260)-(9.262) that
secular terms with coefficients et and (etJ arise in the perturbation expan-
expansion (9.257). Thus the series (9.257) is not expected to be valid for
€t= O(l) or when t= O(l/e).
We have demonstrated that the perturbation series (9.257) is not only
invalid near the initial line but that it also breaks down after a sufficiently
long time. Additionally, we observe from (9.260)-(9.262) that the terms
un(x, t) in the perturbation series are waves that travel to the right or left
with speed \a\ (depending on the sign of a). Now we have already shown
that the maximum speed at which disturbances for the hyperbolic equation
(9.255) can travel is the characteristic speed. This speed equals с and if
\a\> с it can happen that disturbances as described by our perturbation
approximation can travel at a speed exceeding the characteristic speed.
Since this is theoretically not possible, it would appear that we have to reject
the perturbation series completely for this reason if \a\ > c.
An application of the stability analysis of Section 3.5 shows that (9.255) is
unstable if |a|>c. If we insert the normal mode solution C.133) into
(9.255) we obtain for A(Jfc),
(9.263) eA2 + A 4- ec2k + iak = 0 .
Noting that 0 < e <^ 1 and assuming moderate values of k, we obtain two
approximate solutions for A,
(9.264) Aj « - - , A2 * -iak - e(c2 - a2)k2 .
€
Thus if \a\ > c, the real part of A2 is positive and the problem is unstable. We
shall, therefore, assume that \a\<c in (9.255) so that disturbances as-
associated with the equations (9.258) travel at slower speeds than the charac-
characteristic speed с It can be shown that the real parts of Ax and A2 are negative
for all к 7*0 if \a\ < c, so that (9.255) is an equation of dissipative type.
To specify the unknown functions in the perturbation series we mus
relate it to the initial data and this is done by the use of boundary layer
theory. Since the perturbation series is expected to be valid from some time
t > 0 on, we assume there exists an initial layer near the jt-axis where trie
solution of (9.255), (9.256) undergoes a rapid transition from a form tna
satisfies the initial data to the perturbation series form.
PERTURBATIONS AND BOUNDARY LAYER THEORY 659
To determine the appropriate form of (9.255) in the boundary or initial
t yer we introduce the stretching transformation
t
(9.265) T " 7 '
here the positive constant г is to be specified. We insert (9.265) into
(9.255) to obtain
(9.266) ** 4r - есгйхх + e ~ruT + aux = 0,
where we have set
(9.267) й{х, т) - u(x, err).
The initial data for и at r = 0 are
(9.268) u(x, 0) = Дх), йт(ж, 0) = 6rg(x).
To specify the constant r in (9.265) we argue that the most significant
terms in (9.265) are those with the lowest power (possibly negative) in e. On
solving the equation (9.266) by a perturbation method as is our intention,
the most significant terms in (9.266) determine the basic form of the
equations (in either homogeneous or inhomogeneous form) that must be
satisfied by the terms in the perturbation series. Since each term is required
to satisfy two initial conditions at r = 0, we require that мтт be retained as a
leading term. Clearly, this requires that 1 - 2r < -r, as is seen on comparing
terms in (9.266). If we choose r> 1, only йтт occurs as the leading term.
However, with r= 1, wrT and wT are both of the same order in e, and this
choice also yields a balance between the term йтт that is significant in the
boundary layer region, and the term йт that occurs (as ut) in the reduced
problem in the outer region where the perturbation series (9.257) is valid.
Therefore, we set r = 1. Multiplying through by e in (9.266) yields
йт + €айх - €2с2йхх = 0
w*th the data
<9-270) й(х, 0) = f(x) , йг(х, 0) = eg(x).
To solve (9.269)-(9.270), we introduce the boundary layer expansion
mto (9-269), (9.270) and obtain the recursive system of equations
660 PERTURBATION METHODS
(9.272) ^-^ + ^2=0,
' дтг дт
{921 А) —£ + —- = -а —2-1 + с2 ^ , п>2
V } дт2 дт Эх Эх2 '
on equating like powers of e. The initial data are
(9.275) йо(х, 0) - /(*) , йп(х9 0) = 0, п>1,
(9.276)
o. #>. e. na2.
<7T aT *7T
The boundary layer equations are all ordinary differential equations with
initial data at r - 0 and are easily solved. We find that for n = 1 and 2,
(9.277) mo(jc,t) =/(*),
(9.278) ЙД*, r) = (g(x) + ef (*))(! - e~T) - атПх).
~t€
The significant feature in йх is the presence of the exponential e~T — e
which decays rapidly as т or t increases and thus does not play a role when
we consider the (outer) perturbation expansion (9.257). To specify the
unknown functions in (9.257) we need to match (9.271) and (9.257), each of
which are different representations of the unique solution u(x, t). Athough
(9.271) is assumed to be valid in the boundary layer region near the x-axis
whose width is of order e, and the perturbation series is valid in some region
away from the л;-axis, it is assumed that they have a common region of
validity.
In terms of the perturbation and boundary layer expansions for u(x, t) we
have
00 00
(9.279) u{x, t)« 2 и„(х, er)en ~ 2 йп(х, т)еп , t = er .
n=0
We assume that er is sufficiently small such that the terms un(x, ет) can be
well approximated by the leading terms in their series expansions in t = *r-
However, r must be large enough that the exponentials e~T in the йп(х, *)
terms can be neglected. This is the case if t = ет = О(е1/2). Using (9.260),
(9.261) gives
PERTURBATIONS AND BOUNDARY LAYER THEORY 661
(9.280) "o + €Mj = F(x) - earF'(x) + eG(x) + O(e2) ,
and neglecting the e~T terms gives
(9.281) Щ + €ЙХ = /(*) - eer/'(*) + «*(*) + eaf'{x) + O(e2) .
Comparing like powers of € in (9.279) yields
(9.282) F(x) = /(*) , G(jc) = g(x) + e/'(x)
on using (9.280), (9.281). Consequently, the perturbation series (9.257) has
the form
(9.283) u(x91) - f{x - at) + e[t(c2 - a2)f"{x - at)
+ g(x- at) + afXx- at)] + O(e2) .
Further terms in the perturbation series can be determined by carrying this
matching procedure out to higher orders.
The perturbation series (9.283) breaks down when t= 0A/e). A leading
order approximation to the solution valid for t ^ 0A le) may be obtained as
follows. On inserting (9.257) into (9.255) we obtain
(9.284) ^ + fl^ + J^l + «^l + ^-c2^
' dt dx L dt dx dt dx
From (9.284) we see that
(9.285)
so that
(9.286)
dt
d2u0
dt2
~a~dx~
2 *X
dx2
0F)
if we assume that the solutions un(x, t) are smooth. Inserting (9.286) into
(9.284) yields, to the same level of approximation,
Electing like powers of €, we obtain equations for w0 and ux whose
s°htions are (9.260) and (9.261). Clearly, the term (a2 - c2)d2u0/dx2 is the
Ol*e that gives rise to the secular term in (9.261). We may avoid this
662 PERTURBATION METHODS
secularity by regrouping the terms in (9.287) and writing the equation a
The first bracketed term on the left is equated to zero so that и (х л
satisfies a diffusion equation. The initial data for uo(x, t) may be taken to be
(9.283) for u(x, t) evaluated at some time t = t0 < 0A/e). It may be noted
that according to results given in Sections 1.2 and 5.7 we expect the solution
of (9.255) to approximately satisfy a diffusion equation for large t. This
approach is related to the parabolic equation method discussed below.
We continue our consideration of singular perturbation problems for
second order equations by studying the linear elliptic equation
(9.289)
e(Auxx + 2Buxy + Cuyy + Dux + Euy + Fu) + aux + buy + cu = g9
where 0< e < 1 and u(x, y) is specified on the boundary of the given region.
We assume that a2 + b2>0 so that the reduced equation for (9.289),
(9.290) aux + buy + cu = g,
obtained on setting e = 0 in (9.289) is of first order.
The difficulties that can arise out of an attempt to solve (9.289) by using a
regular perturbation series (9.16) result from the fact tLat each term in
the series satisfies a first order equation of the form (9.290). Consequently,
the data for the given equation (9.289) may overdetermine the solutions of
the equations for the terms in the perturbation series. Also, singularities in
the data for (9.289) must be carried along the characteristics of the reduced
equation (9.290) and, therefore, they occur in the terms of the regular
perturbation series. The elliptic equation (9.289) has no real characteristic
and its solutions must, therefore, be smooth functions. Further, it may
happen that characteristic initial value problems can occur for the terms in
the perturbation series or that these terms may become singular in the
interior of the given region for the boundary value problem for (9.289).
These and other difficulties can be remedied by the introduction of appro-
appropriate boundary layers near portions of the boundary or in the interior of
the given region.
The construction of a perturbation solution for the boundary value
problem for (9.289) over an arbitrary region is not a simple matter, m
general. To appreciate some of the problems that can arise in solving
(9.289), we first consider a simple example and then proceed to a more
general discussion.
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 663
Example 9.11. A Singular Perturbation Problem for an Elliptic
Equation. We consider the elliptic equation
(9.291) e(uxx + uyy) + ux+ buy = 0,
oVer the semi-infinite region 0<jc<*> and 0<y<L, with 0<е<^1 and
h = constant. The boundary conditions are
(9.292) u(x, 0) - fix), w@, y) = g(y), u(x, L) = h(x) .
We assume that / and A vanish as jc-*°° and require that u(x, y)—>0 as
for 0<y^L.
As a first step in solving (9.291), (9.292) by the perturbation method, we
expand и as
(9.293) u(x, y) = 2 un(x, y)en
0
n = 0
and insert (9.293) into (9.291), (9.292). Then, on equating like powers of e,
we obtain for w0,
(9.294) ^
V } дх
and for the un,
(9.295) —
The boundary conditions for u0 are identical to those for и [i.e., (9.292)],
whereas un(x, y) vanishes on all three boundary lines if n > 1.
The general solution of (9.294) is
where F(z) is an arbitrary function. For ux(x9 y) we obtain
<9-297) ux(x, y) = Fx(y - bx) - (^-^-) yF"(y - fex),
where Fx(z) is arbitrary. Alternatively, the solution can be written as
where Gx(z) is arbitrary.
664 PERTURBATION METHOD
The lines у - bx = constant are the characteristic lines for (9.294)
(9.295). The solution M0 = F(y- bx) is constant on the characteristic \[пП^
If b^O, the characteristics intersect two of the boundary lines within n!
given region for the problem. If b = 0, the boundaries у = 0 and у = f
coincide with characteristic lines. In either case it is not possible to sat' ft
the full set of boundary conditions on mo(jc, y) since the boundary values?
g, and h are assigned arbitrarily. If b >0, for example, we may set
(9.299) uo = g(ybx),x<;
and satisfy the conditions at у = 0 and x = 0, or we may set
(9.300)
and satisfy the condition at у — L. We cannot determine from the perturba-
perturbation series (9.293) and the resulting equations for its terms, which of these
boundary conditions should be assigned. In any case, since the full set of
boundary conditions for u0 and the un cannot be applied, we must introduce
boundary layers to deal with the lost boundary conditions.
Since each boundary line is a possible candidate for a boundary layer, we
introduce stretching transformations that emphasize neighborhoods of each
of the boundary lines. If b #0, we set
(9.301) у = €V
near у — 0 and
(9.302) у - L - €V
near у = L. With v = v(x, 17) we then have the boundary layer equation
(9.303) »чч + Ч + €^ + Л« = 0»
which replaces (9.291). With v expanded in powers of e and v0 as its leading
term, we find that
and
(9.305) vo(x, 7]) = a(x) + р(х)е~Ь1) ,
where a and p are arbitrary functions. Now if b > 0, the exponential term
PERTURBATIONS AND BOUNDARY LAYER THEORY 665
/Q 305) decreases as 17 increases. Therefore, the boundary layer for this case
ust lie near У = 0- И b<0, the exponential decreases as 17 decreases
^d--in view of (9.302)—the boundary layer should be located near у = L.
тп either case the boundary layer width is 0(e).
However, if b = 0 and we retain the stretching transformations (9.301),
(9.302), we find that vQ is given as
(9.306) vo(x, 17) = <*(x) + p(x)t),
so that no boundary layer effect occurs at either у = 0 or у = L, since there
is no exponential decay. In this case we easily conclude that the appropriate
stretchings are
(9.307) y = c1/2ri
and
(9.308) y-L = €U2riy
and the boundary layer equation becomes
(9.309) wn4 + i>, +€*„=<>•
Then vo(x, 17) satisfies the parabolic equation
(9.310) ^ + ^=0.
' dt]2 dx
The boundary layer width is 0(e1/2), and there are boundary layers near
У = 0 and у = L as shown in the following.
Near x = 0 we set
(9-311) * = €£
u* (9.291) and obtain the boundary layer equation
+ €2и>^ = 0 ,
with w= w(£, y). If и> expanded in powers of e, we easily find that the
fading term vv0 is given as
(9'313) "O(£,y)
So that a boundary layer can be located at x = 0 because of the exponentially
decaying term in (9.313).
666
PERTURBATION METHOD»
For the case b Ф 0, we now assume that b > 0 and complete the perturba-
perturbation solution of the problem. Since we have shown that there can be
boundary layers near x = 0 and у = 0 if b > 0, we specify u0 = F{ у - bx) to
satisfy the boundary condition at у = L. This gives the outer solution
(9.314)
The boundary layer solution vo(x> 17)—that is, (9.305)—near у = 0 is re-
required to satisfy the boundary condition
(9.315) vo{x,O)=f(x)
and the matching condition
(9.316) lim vo(x, 17) = lim uo(x, у)
This states that for large 17 the boundary layer solution vo(x, 17) must agree
with the outer solution uo(x, y) evaluated for small y. We readily find that
vo(x, 17) is given as
(9.317) vo(x, V) = h(x + |) + [fix) - *(* + \)]e~brt.
For the boundary layer near x = 0 we have the boundary condition
(9.318) H>o@,y) = g(>0
and the matching condition
(9.319) \\mwo{^y) = \imuo{x,y)
with w0 defined in (9.313) and w0 given in (9.314). Solving for wQ gives
(9.320) wo{{9 y) = k(^) + [g(y) - h(^)]e~t.
The matching procedure given in (9.316) and (9.319) is effectively
equivalent to that carried out in the preceding examples. Since we are only
matching the leading terms in each case it is convenient to carry out the
matching by the use of a limit process. However in the hyperbolic problem
of Example 9.10 where two terms in the perturbation and boundary layer
expansion were used, the matching was carried out by expressing both series
in terms of a common variable. This approach could have been used here as
well.
By inspection of the outer and boundary layer solutions we observe that
the solution of (9.291), (9.292) in the case where b > 0 can be expressed as
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 667
(9.321) u(x, y) - h(x - £^k) + [/(*) - h(x + \)] exp[- ^
Th terms containing the exponentials are significant only within their
respective boundary layers. Apart from exponentially small terms, u(x, y)
satisfies the boundary conditions at у = 0, x = 0, and у = L and vanishes as
x-*00 since/and h vanish there. We do not consider effects due to possible
incompatibilities of the boundary values at the points (jc, y) — @,0) and
/^ y) = (o, L). The uniform expression (9.321) is often termed a composite
expansion and can be constructed in a systematic fashion. We do not pursue
this matter here.
If ft <0 in (9.291), the form of the solution is similar to that in (9.321)
except that the boundary layer is shifted to у = L. If b = 0, the leading term
in the (outer) perturbation expansion is
(9.322) uo(x,y) = F(y).
Since F(y) is constant on у =0 and у = L, it cannot satisfy the boundary
conditions there. We could choose F(y) = g(y) and thereby satisfy the
boundary condition at x = 0. However, u(x, y) must vanish as jc-*<*>, and
the solution
(9-323) uo(x,y) = g(y)
remains constant as x varies and does not satisfy the condition at infinity
unless g(y)-0. Since g(y) does not vanish, in general, and is, in fact,
specified arbitrarily, we must put F(y) = 0 and obtain as the outer solution
(9324) М~м0 = 0,
for otherwise и would not vanish as jt-»<». Consequently, we must use not
only the boundary layers at у = 0 and у = L but also the boundary layer at
The boundary layer at x = 0 is the easiest to consider. Using (9.318),
(9.319) we conclude that
(9.325) *o({,y)
since w0 = 0. Near у = 0 and y = Lwe have the parabolic boundary layer
equation for vb « vo(x, ?/),
(9.326) ^° + ^0
dri2 dx J
668 PERTURBATION
as given in (9.310). In the boundary layer near у = 0 the boundary condition
is
(9.327) vQ(x,0)=f(x),
whereas near у = L the boundary condition is
(9.328) vo(x,O) = h(x).
As jc-»oo we require that vo(xy 17) -►(). Also, as 17-» 00 in the boundary layer
near у = 0 and 77-» -°° in the boundary layer near у = L, we must have
u0—>0, because v0 must match the outer solution u0 that vanishes.
With x thought of as the time variable, (9.326) is a backward heat
equation. Thus we cannot assign values for vo(x, 17) at * = 0 and solve
(9.326) for jr>0, for such a problem would not be well posed. Instead, we
must place conditions on v0 at some jc = jco>0 and solve for д:<д:о. The
only conditions given are at infinity, as we have shown. We may think of the
problem as being given with data at x = x0 and then let Jto-»°°- The solution
is then given for all x < «. The result is equivalent to a steady-state problem
for the ordinary heat equation where the effects of the initial temperature
distribution have died out. We can think of that problem as being given for
all t > -00. The solution to the steady-state problem for the heat equation
and to our problem for (9.326) is unique if we assume the solution is
uniformly bounded.
By adapting the results of Example 5.5, we conclude that the solution of
(9.326), (9.327) with уо(°°^) = 0 is
(9.329) ,„(,,,)
For the problem (9.326) and (9.328) the solution is
(9.330) „e(x,,) = ^ f exp[- ^L] ^75 da .
We note that as Ы-»°° both solutions decay exponentially, and they both
vanish as x—»°°.
A composite expression for the solution may be given as
(9.331)
(- -) + -7=== f expf- -r-^- r] f^\3n d<r
\ €j V4eir J» L 4c(<7 - x) J (<r - x)
expr uw=7)J (^p d° ■
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 669
We again do not consider the behavior of u(x, y) at @,0) and @, L). The
solution is exponentially small when x>O(e), y>O(Ve) and |y-L|>
We remark that near x = 0 and in the case where b Ф 0, we were led to
consider ordinary differential equations in the boundary layer regions, as
had been the case in the problems considered previously. However, with
^ = 0 the boundary layer equations valid near у = 0 and у = L turned out to
be parabolic partial differential equations. This can be attributed to the fact
that these boundary lines coincided with the characteristics of the reduced
equation for (9.291) with 6 = 0. The occurrence of parabolic boundary layer
equations when portions of the boundary coincide with the characteristics of
the reduced equation is generally the case, as shown later.
To study the singular perturbation problem for the elliptic equation
(9.289), it is convenient to transform the coordinates so as to simplify the
form of the reduced equation
(9.332) aux + buy + cu = g.
We assume that the characteristic curves for (9.332) do not intersect in the
region given for (9.289). Then the characteristics and their orthogonal
trajectories may be introduced as a new set of coordinates in (9.289). The
transformed equation can be written as
(9.333) e(Auxx + 2Buxy + Cuyy + Dux + Euy + Fu)- ux + cu = g ,
where we have retained the original notation for the coefficients and the
variables for the sake of convenience. The type of an equation is invariant
under nonsingular transformations so that (9.333) is again of elliptic type.
We now consider the Dirichlet problem for (9.333) in a bounded region
G, and note that our discussion applies to the transformed region obtained
in going from (9.289) to (9.333). To begin, we assume that G is a convex
region as shown in Figure 9.1.
The points P and R in the diagram are points where the tangent line to
the (smooth) boundary curve is horizontal. To the left of those points the
curve is given as x = hx(y), and u(hx(y), y) =fx(y) is the boundary condi-
condition. To right of P and R we have u(h2(y), y)=f2(y). The characteristics
for the reduced equation for (9.333) are the lines у = constant. Thus the
characteristics passing through G intersect the boundary curve twice, and
they are tangent to the boundary curve at the points P and R.
The conventional perturbation series
(9-334) и = 2o une" ,
*hen inserted into (9.333) yields for u0,
670
PERTURBATION METHODS
x=hjy)
Figure. 9.1. The convex region G.
(9.335)
with similar equations satisfied by the wn. Now u0 must satisfy the boundary
conditions
(9.336)
and this is not possible, in general, since u0 is already specified uniquely
either by its values on x - Нг(у) or on jc = h2(y). Consequently, a boundary
layer must be introduced near x = hx(y) or x = h2(y), and we do not know
in advance where to place the boundary layer.
We proceed as in Example 9.11 and stretch the neighborhoods of each of
the boundary curves x = hx{y) and x = h2{y) to determine where the
boundary layer is to be placed. The stretching transformations are given as
(9.337) *-A,.(y) = €£, i = l,2
(9.338) y = ri.
We have anticipated that the boundary layer width is O(e) as is easily
shown. The boundary layer equation obtained from (9.333) is
(9.339)
(A-2h\B
O(e)
where v = v(£, ri) and A, By and С are evaluated at f = 0. Since (9.333) is
elliptic, A and С are both positive or both negative and the quadratic form
PERTURBATIONS AND BOUNDARY LAYER THEORY 671
(9.340) A - 2KB + Л2С ^ Q( A)
is nonzero, so that the coefficient of vH in (9.339) where A = h'n is either
strictly negative or strictly positive.
To solve the boundary layer equation, we expand v in powers of e with i;0
as its leading term. Then v0 satisfies the equation
where h\ = Н'((г)). The general solution of (9.341) is
(9.342) i>o«, i?) = «(t?)
Now £ increases and decreases with x in view of (9.337). Thus if Q > 0, the
boundary layer must be placed near * = A2(v), whereas if Q<0, the
boundary layer must be near х = кг(у). For with these choices, the ex-
exponential in (9.342) decreases as we move away from the boundary into the
region G. Since Q(X) is either greater than zero or less than zero for all A,
we conclude from (9.340) that if A and С are positive in and on the
boundary of G, then Q >0, whereas if A and С are negative, then Q<0.
Consequently, the boundary layer is at jc = h2(y) if A and С are positive,
and it is at x = АД v) if A and С are negative.
Assuming A and С are positive, we now know that a boundary layer (if it
is necessary) must lie near x = h2(y). Thus the (outer) equation (9.335)
must be solved subject to the boundary condition given at x~ hx(y)—that
is, uo{hx{ v), y)-f1(y)- The solution is easily found to be
(9.343) uo(x, у) = Л(у) ехр[£у) ф, у) ds]
For the boundary layer function i>0(£, 77) we require that
(9344) *о@,ч)=/2(ч)
since v = 7/, whereas the matching condition gives
(9 345^ г (& \— v
(9.342) yields
PERTURBATION METHODS
672
(9.346)
»о(&ч)-«о
Combining (9.343) and (9.346) we may express u(x, v) in a composite
form as
|ехрЫ
(9.347) u(x, y) « uo(x9 y) + [f2(y) - uo(h2(y)9
where Q = Q(h'2(y)). Since hf2(y) becomes infinite at the points P and R on
the boundary curve, these results are not valid near P and R. (The behavior
of the solution near those points is given below.) A similar result for the
solution obtains if A and С are negative.
A different approach is needed if the region G has a portion of its
boundary that coincides with a line v = constant. A situation of this type is
pictured in Figure 9.2. The point P separates the curves x = hl(y) and
x = h2(y). The transition from these curves to the line у = v0 at the points R
and T is assumed to be smooth. In the region у > y0 within G and away from
the point P, the result (9.347) again yields an approximate solution if A and
С are positive. The line у = y0 is a characteristic for the reduced equation
sf2(y>
Figure 9.2. A characteristic boundary curve.
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 673
(9,335) and the equation for w0 cannot be solved there subject to arbitrary
data «o =/W- A boundary layer must be introduced near у = y0.
We set
(9.348) УУо^^
to emphasize the neighborhood of у = y0 and insert (9.348) into (9.333).
This gives, with w = w(x, rj),
(9.349) *l~2rCwm -wx + cw + O(€lr) = g ,
where C, c, and g are evaluated at у = y0. To achieve a balance between the
reduced equation and the term wvv, we must set r = |. On expanding w in
powers of e1/2 we obtain for w0,
(9.350) C(x, v0) -^ - -^ + c(x, yo)wo = g(x, y0) ,
which is a parabolic equation. At 17 = 0 we have the boundary condition
(9.351) и>0(*,0) = /(*),
whereas the matching condition requires that
(9.352) lim wo(x, tj) = lim uo(x, y)
so that the boundary layer and the outer solution (9.343) agree. An
additional condition should be placed on wo(x917) regarding its behavior
near x = hx(y) or x = h2(y). Thinking of x as a time-like variable in (9.350),
we see that (9.350) corresponds to a forward parabolic equation. Thus it is
appropriate to specify the values of wo(jc, 17) at x — h^y) since we are
considering a region where x > hx(y). However, near x = h^y) away from
the point R, the outer solution w0 is valid, so that we may consider the
(initial) condition near x = ht( v) to be equivalent to the condition (9.352).
That is, wo(x, ту) should agree with uo(x> y) away from the line v = v0.
The solution of (9.350)-(9.352) is not straightforward if the coefficients С
and с are not constant, and special techniques or possibly numerical
methods must be used to solve for wo(jt,i7). If С and с are constants and
8 * 0, the solution can easily be obtained on following the procedure given
m the preceding example, but we do not present it here.
We conclude our discussion of (9.333) by examining the behavior of the
solution of the Dirichlet problem for (9.333) near points on the boundary
^here the tangent line is horizontal. As shown in Figure 9.3, we are
uiterested in points where the curve is either concave up or concave down or
674
PERTURBATION METHODS
Figure 9.3. The nonconvex region G.
has a point of inflection. This contrasts with the situation considered
previously where an entire portion of the boundary was horizontal. At the
point P in the figure, the curve is concave down, at R it is concave up,
whereas at S the curve has a point of inflection and a zero slope.
If Dirichlet boundary conditions are given for (9.333) and A and С are
both positive, we find, as was shown earlier, that a boundary layer must be
introduced along x — h2(y). The outer equation (9.335) can be solved with
the data given on x = hx(y) and x = Л3(>>). We can even solve for uo(x, y)
with data given at the point of inflection. However, the derivatives of hu h2,
and h3 are singular at one or more of the points P, R, and 5. Thus the
boundary layer solution breaks down at P and R and the second term in the
ordinary perturbation series is singular at 5, since it contains derivatives of
hi or Л3. The horizontal line that extends from 5 in the figure indicates that
the outer solution cannot be determined on that line ar.d an internal
boundary layer must be introduced along that line.
Let (x0, y0) represent the coordinates of the boundary point P, R, or 5.
We assume the boundary curve can be expressed as у = F(x) near (x0, y0).
Since the slope vanishes at (jc0, y0) we have
(9-353) у - y0 = %F"(xo)(x - x0J + • • •
near P or R, and
<9*354)
у - y0 =
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 675
near S with y0 = F(x0). We assume that F"(*o) and F'"(x0) are not zero in
these equations.
To emphasize the neighborhood of the point (x0, yo)9 we introduce a
(double) coordinate stretching
\y*3JJj л ло c ь > У Уо с ч
with r and s to be determined. Inserting (9.355) into (9,333) and with
v = v(£, 7}), we obtain
(9.356) €l'2rAvu + 2el-r'sBv^ + e'^Cv^ - e~rvu + • • • = 0 ,
where A, B, and С are evaluated at (x0, y0) and only the leading terms in e
have been retained.
Now both r and s must be positive, otherwise the neighborhood of
(x0, y0) is not emphasized. Also, the term vu must be retained in (9.356) to
maintain a balance between terms from the reduced equation and higher
derivative terms. Finally, we require that r and s be such that an approxima-
approximation to the boundary curve у = F(x) of the form (9.353) or (9.354) be
retained near (xQ9 y0). Thus if (9.355) is inserted in (9.353), we have
(9.357) 6^=^2rF'(xo)f2 + --.,
and from (9.354) we obtain
(9.358) €'4=i€3rFw(x0)f3 + ---.
Consequently, for points of the type P or R we must put
(9.359) 2r = s ,
whereas for points of type 5,
(9.360) 3r = s
in order to balance the leading terms in (9.357) and (9.358). It is then easily
seen that the only way to balance v€ with a second derivative term in (9.356)
is to set
(9.361)
-25--r.
Thus for points of type P or R we have r = \ and s = \, whereas for
points of type S, г = \ and s=\. In either case, if v is expanded in
appropriate powers of c, we find that the leading terms satisfy the equations
676 PERTURBATION METH0Ds
(9.362)
д\
(9.363) C^f-f-^-^
G7} &Ь иХ
where all coefficients, as well as g, are evaluated at (x0, y0). The last term
on the right of (9.363) is absent if the point is of type 5.
To leading order in e, the boundary curve has the form tj = \F"{x)£2 near
points of type P and /?, whereas it has the form 17 = \Ftn{xQ)^ near points
of type 5. To facilitate the study of the boundary layer equations (9.362)
(9.363), we introduce the new variables
(9.364) a =
where a1 = F"(x0) and a2=\Fm(x0). In these coordinates, a=0 corre-
corresponds to the boundary curve. With wk = vk(a(g, tj), /3(£, tj)) we obtain, in
place of (9.362) and (9.363),
c£ + ,,. £-£-.. .-..2.
(9.366) ^ ^^ИС
да2
where к = 1 corresponds to a point of type P or R, к = 2 corresponds to a
point of type 5, and the last two terms on the right of (9.366) are absent if
к =2.
The solutions of (9.365), (9.366) must be matched to the outer solution
and must satisfy the boundary conditions. For definiteness we now restrict
our discussion to a point of type R and assume that the coefficient С in
(9.333) is positive. In that case the matching with the outer solution takes
place in the region x<x0, and the outer solution u0 has the form (9.343).
Also, near a point (x0, y0) of type R the boundary curve is given as (9.353)
withF'(;to)>O.
We assume that the boundary condition near (x09 y0) is given as
(9.367) u(x,F(x))=f(x).
For x < jc0, the boundary curve is expressed as x = ht(y), and the boundary
condition is given as
PERTURBATIONS AND BOUNDARY LAYER THEORY 677
(9.368) "(M>0> зО^ЛСзО-
Consequently, f(x) and f^y) are related by
(9.369) /1(У)=/(Л1(У)).
since у = F(x) is expressed as in (9.353) we have
. . . T
(9.370) x =
with ax = ^"(*о)» where (9.364) with к = 1 has been used and the minus sign
follows from the fact that x - x0 < 0. Also,
(9.371) x = x0 + (x - x0) = x0 + €1/3j8 .
Expressing the outer solution uo(x, y) as given in (9.343), in terms of the
boundary layer variables we obtain—to leading terms—
(9.372)
"о = /(*o) + €l"fr(xo)yj~ + P2 + (c/(*o) - <
where с and g are evaluated at (jc0, y0). We note that j8 <0 in the matching
region so that V/iF = |/3| = -j3. On the boundary we have
(9.373) и =f(x) = Д*о) + €1/3/'(^o)i8 + O(e2/3) .
Since
(9.374)
we find that on the boundary we must set
<9-375) wo@, /3)=/(j
and
We remark that a = 0 is only an approximate leading order expression for
the boundary curve у = F(x). Since и>0(а, /3) is shown to be identically
instant, additional effects that might influence the boundary condition for
^i do not play a role. The matching conditions require that
678 PERTURBATION METHODS
wo(a, fi)-*f{x0) ,
(9.378) Wl(.a, P)^-f'(xo)y]^+{i2 + (c/(*0) - g)(fi + yj~
as /3 —> — oo with a > 0.
The solution of the problem for wo(a, 0) is immediately found to be
(9379) (a,/3 W(jc0) ,
where, in fact, f(x0) = m(jc0, y0). Since w0 is a constant we may set
(9.380) Wl(a9 /3) = (cf(x0) - g)fi 4- W(a, fi)
in (9.366) (with к = 1) and conclude that W(a, /3) satisfies the homogeneous
equation
(9.381)
The boundary and matching condition for W must be determined from
(9.376) and (9.378) with the use of (9.380). The solution of (9.381) can be
expressed in terms of products of exponentials and Airy functions (see the
exercises). The process of satisfying the boundary and matching conditions
for that solution is not straightforward and is not given here.
Although we do not discuss the boundary layer equations near points of
type P or 5, we note that in the case of a point 5 there is an additional
(internal) boundary layer of parabolic type. The solution of the boundary
layer equation near 5 must be used to provide initial data for that problem.
If (9.289) is of hyperbolic or parabolic type, perturbation and boundary
layer methods may again be used in the manner given to solve initial and
boundary value problems. A particular hyperbolic problem was considered
in Example 9.10 where the special problem arising because of an excess of
initial conditions for the reduced (perturbation) equations was examined. A
general discussion of these problems akin to that given for the elliptic
problem, is not given here.
A common problem for the elliptic, hyperbolic and parabolic cases is
where to place the boundary layer(s) if the data for the given problem result
in an overdetermined problem for the reduced equation. This question can
be answered, as has been done previously, by introducing boundary layer
coordinates along the appropriate parts of the boundary. Those portions of
the boundary along which the boundary layer equations have solutions that
decay exponentially away from the boundary are the regions where boun-
boundary layers can be located. An alternative method for determining the
location of the boundary layer for elliptic and hyperbolic equations is now
presented.
To begin, (9.289) is replaced by
PERTURBATIONS AND BOUNDARY LAYER THEORY 679
(9.382) e(Auxx + 2Buxy + Cuyy) + aux + buy=09
which may be of elliptic or hyperbolic type. The additional terms in (9.289)
have been neglected for the sake of simplicity since they do not play a
significant role in our discussion. Now (9.382) states that
(9.383) <*"x + buy = O(e) .
Assuming that a, b, u, and their derivatives are smooth functions, we may
express uxx, uxy, and uyy either in terms of uxx and ux or in terms of uyy and
и correct to O(e). For example, if a ^0, we have
(9.384) ux = - - u
y
so that
(9.385)
(9.386) uxy = - ~a
Inserting (9.385), (9.386) into (9.382) yields
(9.387)
Similarly, if b ¥=■ 0, we obtain
(9.388) €[CAJ-^ + a]uxx
aux
The parabolic equations (9.387), (9.388) may be solved by using a
^nventional perturbation series. The two leading terms in the series satisfy
equations identical to those for the leading terms in the regular perturbation
solution of (9.382) as is easily shown. Therefore, we expect the solutions of
(9-387) or (9.388) to coincide, approximately, with the outer solution of
(9.382) in regions where that solution is valid. However, as we shall see, the
^lutions of (9.387) and (9.388) remain valid in certain regions where the
°uter solution breaks down. Also, information about the location of boun-
680 PERTURBATION METHODS
dary layers for (9.382) can be determined from the parabolic equations
(9.387), (9.388).
For example, if the data for the given equation (9.382) are not smooth
the discontinuities or singularities are propagated into the interior along the
characteristics of the reduced equation (9.383). If the given equation is
elliptic, the solution must be smooth in the interior and if it is hyperbolic
the singularities must occur across the characteristics of the given equation'
These characteristics generally do not coincide with those of the reduced
equation. The parabolic equations (9.387), (9.388) generally smooth out the
singularities in the data, so that its solutions are smooth across the charac-
characteristics of the reduced equation.
To determine the location of boundary layers for the elliptic or hy-
hyperbolic equation (9.382), we consider initial value problems for (9.387) and
(9.388) along the boundary of the region. Those portions of the boundary
for which the initial value problems for the parabolic equations (9.387) or
(9.388) are not well posed, determine where the boundary layers must be
located. To see how this works, we consider a simpler form of the elliptic
equation (9.289) or (9.382)
(9.389) <Auxx + 2Buxy + Cuyy) - ux - 0 ,
and the convex region G considered previously. Since a - -1 and b = 0 in
relation to (9.382), the parabolic equation (9.387) takes the form
(9.390) eCuyy-ux = 0,
where we have dropped the O(e2) terms. Referring to the diagram of the
convex region G given in Figure 9.1, we see that if 00, the initial value
problem for (9.390)—with x treated as the time variable—is well posed for
data given on x = hl(y). But if C<0, the data must be assigned on
x = h2(y) for the initial value problem to be well posed. Thus the boundary
layer must be placed along x = h2(y) if С >0 and along x = hx(y) if С <0.
This result is consistent with that given earlier based on a direct boundary
layer construction. It is not difficult to show that both, the present and the
earlier method for determining the location of boundary layers lead to the
same result. However, this is not demonstrated here.
Given the hyperbolic equation with constant coefficients
(9.391) e(utt - c\x) + ut-aux=0,
a related parabolic equation is
(9.392) e(a2 - c2)uxx + ut-aux=09
since ut - aux + O(e). The parabolic equation (9.392) cannot absorb arbit-
arbitrary initial data u(x, 0) =/(*) and ut(x, 0) = g(x). Thus a boundary layer is
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 681
generally required at f = 0. However, if a2 > c\ we find that even if
appropriate initial values are determined for (9.392), the problem is not well
posed in the region t > 0 since the coefficients of uxx and ut are both positive.
We considered a similar problem in Example 9.10 where it was shown that if
\a\>c, the initial value problem for (9.391) is unstable. We assume,
therefore, that |a|<c and note that the parabolic equation approach
indicates whether the given problem is stable or unstable.
If we consider an initial and boundary value problem for (9.391) in the
interval 0 < x < I and with t > 0, we find that a boundary layer is required
either along x = 0 or x = I if the conventional perturbation approach is used.
Now once initial data for the parabolic equation (9.392) have been de-
determined, it is possible to solve (9.392) with the given boundary conditions
on x = 0 and x = /. That is, a boundary layer need not be introduced at all
on either boundary line. Nevertheless, it is usually simpler to solve the given
problem with the use of boundary layers.
Since we are concerned with the solution near the lines x = 0 and x = /,
we replace (9.392) with the parabolic equation
(9.393)
where we have used ux = A la)ut + O(e). If a > 0, the coefficients of utt and
ux have the same sign since c>|a|, whereas if a<0, they have opposite
signs. Thus if (initial) data are given at x = 0 and (9.393) is to be solved for
x > 0, the problem is well posed only if a < 0. Consequently, a boundary
layer must occur near x - 0 in the case a > 0. Similarly, if data are assigned
at x = I and (9.393) is solved for x < I, the problem is well posed only if
a > 0, so that a boundary layer near x = / occurs when a < 0. We note that
if a = 0, the lines x = 0 and x — l are characteristics of the reduced equation
м, = 0 and (9.392) is, in fact, the parabolic boundary layer equation. Again,
these results concerning the location of the boundary layers are identical
with those obtained by means of a boundary layer analysis.
In the following example we apply the parabolic equation method to
specific equations of elliptic and hyperbolic types to demonstrate how
problems can be solved by this method.
Example 9.12. The Parabolic Equation Method. We begin by re-examin-
Щ the problem of Example 9.11. That is, we consider the elliptic equation
within the region 0 < x < » and 0 < у < L, with b - constant. The boundary
conditions are
u(x9 0) = fix) , u@, y) = g(y), U(x, L) = h(x) ,
u(x, y) is required to vanish as x->oo in the interval 0<y <L.
682 PERTURBATION METHODS
Since ux = -bux + O(e) and uy = -A lb)ux + O(e) if b Ф 0, the parabolic
equations for (9.394) are
(9.395) e(l + Ьг)иуу + ux + buy=Q,
(9.396)
From (9.395) we see that a problem with initial data given at x = 0 and
м(х, у) to be determined for x>0 is not well posed. Also, (9.396) shows
that if b > 0, a problem with initial data at у = 0 to be solved in the region
у > 0 is not well posed. If b < 0 and initial data are given at у = L and
(9.396) is to be solved for у < L, the resulting problem is not well posed. If
6 = 0, (9.396) is not valid and in this case there are parabolic boundary
layers at у = 0 and у = L.
We conclude from this discussion that a boundary layer should occur at
x = 0 in all cases. If b > 0, a boundary layer occurs at у = 0, whereas if fe < 0,
a boundary layer occurs at у — L. These results are consistent with those
obtained in Example 9.11 by a different method.
We now suppose that b < 0 so that the boundary layer must be located at
y = L. It is then possible to solve the parabolic equation (9.396) with the
data given along the lines у = 0 and x = 0. Even though x = 0 was shown to
be the site of a boundary layer, the initial and boundary value problem for
(9.396) is well posed. Thus we consider the equation
(9.397) e\2uxx + ux- c2uy = 0 ,
where we have put A2 = 1 + (libI and b = -c2. The initial and boundary
conditions are
(9.398) и(х,0)=Л*Ь Ф,У) = 8(У)-
Though the given problem (9.291), (9.292) was formulated in the region
0 < x < oo and 0 < у < L so that g(y) in (9.398) is unspecified for у > L, we
may solve (9.397), (9.398) in the first quadrant 0 < x < °° and 0 < у < °° The
boundary function g(y) may be extended arbitrarily over the interval y>L.
Because of the causality property for (9.397), the solution u(x, y) at any
value of у such that 0 < у < L depends only on data given for smaller
(non-negative) values of y. Thus the values assigned to g(y) for у > L are
irrelevant in the region 0 < у ^ L.
To solve (9.397) we set
(9.399) u(x, y) = exp(- ^ ~ ^yI**, У)
and find that v(x, y) satisfies
PERTURBATIONS AND BOUNDARY LAYER THEORY 683
(9.400) Vy
with the data
(9.401) v(x9 0) =/(*) exp(^) , v@9 y) = g(y) ехр
Using the results of Example 5.5 [see E.105)], we find that u(x9 y) can be
expressed as
c2xsJ
, ч « Г Г Г (y + c2x-sJ
(9.402) ф9 y) - 4— I |exP[- ^V(y,^
where G(x9 у) is given as
which is the fundamental solution of (9.400).
In the first integral in (9.402) we have у - s > 0 and x > 0. Thus when
x > 0 we see that the contribution from the entire integral is exponentially
small if € <^ 1 since the exponential is uniformly small. This term yields the
boundary layer behavior near x = 0. From the expression for G(x9 y) in
(9.403) we see that as e\2y/c2^>09 G(x9 y)-> S(x) in view of the Dirac delta
function behavior of the fundamental solution discussed previously. Thus as
e->0, we find that G[(y/c2) + x ± s9 y]-> S[(y/c2) + x ± s]. In the region
where у >0, x > 0, and s > 0 the delta function S[(y/c2) + x + s] vanishes.
Even though esUX grows exponentially as €—»0, its product with the term
G[(y/c2) + x + 5, y] may be shown to vanish as e-*0. Thus if e <^ 1, we find
that (9.402) reduces to
(9.404) U{x9 y)~
to leading order in €. The expression (9.404) represents the leading term in
J*fc outer solution of (9.394) that satisfies the boundary condition at у = 0.
Гпе term involving g{y) in (9.402) is significant only in a boundary layer
near x = 0. To satisfy the boundary condition at у = L, a boundary layer
must be introduced as was done in Example 9.11. The solution can then be
exPressed in a composite form in the manner of (9.321). It may be noted
684 PERTURBATION METHODS
that the solution form (9.402) smooths out discontinuities or singularities i
the data/(jc) and g(y).
Next we consider a signaling problem for the hyperbolic equation
(9.405) e{un - c2uxx) + ut-aux = 09 x,t>0,
with the homogeneous initial data
(9.406) w(x,0) = 0, м,(д;,0) = 0, д:>0
and the boundary condition
(9.407) u@9t) = g(t), t>0.
It is assumed that \a\ < c, both of which are constant, so that the problem is
stable.
The relevant parabolic approximations for (9.405) are
(9.408) ut - aux - e(c2 - a2)uxx = 0 ,
(9.409) ((^)
Since c2 > a2, the initial and boundary value problem for (9.408) is well
posed. However, in the case of (9.409) if initial data are given on x = 0, the
problem is well posed for a<0 and is not well posed for a>0, as is
immediately seen. This is consistent with the fact that the characteristics of
the reduced equation ut - aux = 0 are the lines x + at = constant. For a <0
these characteristics do not simultaneously intersect the initial line t = 0 and
the boundary line x - 0, so that the given initial and boundary data (9.406),
(9.407) can be assigned to the perturbation series solution of (9.405) without
encountering any difficulty. However, if a>0, the characteristics x + at-
constant intersect both the initial and boundary lines and a boundary layer
at x = 0 is needed.
Since the initial data are homogeneous there is no need for an initial layer
near t = 0 in this problem. We may use и = 0 as the initial condition when
solving (9.405) either by the conventional perturbation and boundary layer
approach or by the use of the parabolic equation (9.408) as we now
demonstrate. We solve (9.408) which we rewrite as
(9.410) ut - aux -ey2uxx=09 x,t>09
with y2 = c2 - a2. The initial condition is
(9.411) u(x,0) = 0, jc>0
and the boundary condition is
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 685
(9.412) и@,0 = *<0> t>0-
To solve (9.410)-(9.412) we proceed exactly as for the initial and boundary
value problem for (9.397) treated earlier. Accounting for the slight change
in notation we find that
^ t a « Г I \x + a(t-T)f\ g(r)
Now if a > 0, the argument of the exponential in (9.413) vanishes only at
x = 0, in the region x >0 and *>0. Since 0< € <€ 1, we find that м(х, 0 is
exponentially small for x > 0 and f > 0, and it is only significant in a small
(boundary) layer near x = 0. Thus the outer solution is given as
(9.414) и(х,0~0, *>(U>0
apart from a layer of thickness O(e) near x = 0.
If л < 0, the solution (9.413) is significantly different from zero for values
of x and t in the first quadrant for which
(9.415) * = -а(/-т).
Since f - т > 0 and f, т s: 0, (9.415) corresponds to the characteristics of the
reduced equation ut - aux = 0 that occupy the sector between the line
x + at = 0 and the f-axis. Now if а = 0 in (9.413), we know that as jc—>0 the
expression (9.413) tends to g(f) with the basic contribution from the integral
coming from the value r = t. Similarly, we expect that as e—»0 in (9.413)
the main contribution to the integral comes from the value т = (x/a) + t and
that w(x, *) tends to g[(x/a) + /]. We note that (9.413) has been written in
terms of e(t — т) and €jc, to indicate that small values of x and t — т are
equivalent to considering small values of €. Further, in the sector between
x + at = 0 and the jc-axis, the exponential term in (9.413) and, consequently,
the integral is small. Therefore, we conclude that for small € and a < 0 we
have
Ы ) x + at<0
. is result is consistent with that obtained from the conventional perturba-
юп method as is easily shown. However, the expression (9.413) remains
and useful for large values of x and ty as well as in the case where the
faary function g(t) is not smooth, or if g@) #0.
К the initial data u(x, 0) and ut(xy 0) for (9.405) are not homogeneous, an
mal layer of width O(e) is required. The value of the solution at the edge
ot this initial layer may then be used as an initial condition for the parabolic
e4uation (9.408).
686 PERTURBATION METHODS
We conclude our discussion of singular perturbation methods by consider-
considering exterior boundary value problems for the Helmholtz and modified
Helmholtz equations. Rather than give a general discussion, we examine a
specific example that brings out the basic distinction between the two
problems.
Example 9.13. The Singular Perturbation of the Helmholtz and Mod-
Modified Helmholtz Equations. We begin by considering the modified
Helmholtz equation in the region exterior to the unit disk
(9.417) e\uxx + uJ-m = 05 r > 1,
with the boundary conditions
(9.418) и\гш1 = 1, u^O as r->oo,
where r2 = x2 + y2. The exact solution of this problem is
(9.419) -Crt-**'"
where Ko is the (zero-order) modified Bessel function of the second kind.
If we introduce the regular perturbation expansion
00
(9.420) u= E un(x,y)e2n ,
into (9.417) we find that each term in the expansion equals zero. Thus the
outer solution of this singular perturbation problem is u(x, y) = 0. This
solution satisfies the boundary condition at infinity but not on the finite
boundary r = 1. To deal with that boundary condition we require a boun-
boundary layer at r = 1 and introduce the stretching transformation
(9.421) r - 1 = €<r
into (9.417) expressed in polar coordinates with и - м(г), since the problem
has no angular dependence. With v{&) = мA + ecr) we obtain
(9.422) y^ - v + eva + O(e2) = 0
as the boundary layer equation. We expand v(a) as v = v0 + evx + O(e )>
and the boundary condition at r = 1 requires that i>0@) = 1 while the
remaining terms in the expansion vanish at a = 0. The vanishing of the outer
solution implies that each term in the expansion of v must vanish as а-*ф-
We easily find that
SINGULAR PERTURBATIONS AND BOUNDARY LAYER THEORY 687
2)
(9.423) v(a) = e~a- \eaea + O(e2) .
In terms of the radial variable r we have
(9.424) u(r) ~ [1 - \{r - 1)] exp[- \{r- 1)] .
This result agrees with the asymptotic approximation of the exact solution
(9.419) near r = l, obtained by using the formula F.203), with 1/e-x».
Outside the boundary layer, the exact solution is exponentially small for
small e, so that its magnitude is smaller than any power of e. Therefore the
regular perturbation expansion (9.420) of the solution must be zero, as is
indeed the case.
The situation is quite different if we consider a similar exterior boundary
value problem for the Helmholtz or reduced wave equation,
(9.425) €2(uxx + uyy) + и = 0 , r > 1,
with u = l at r=l. As r-><», we require that и satisfy the radiation
condition G.399), where o> is replaced by 1/e. The exact solution of this
problem is
(9.426) и(х, у) ■■
#<>/€)
where Hq1} is the zero-order Hankel function of the first kind.
The regular perturbation expansion (9.420) for this problem again yields
u(x, y)-0 as the outer solution. (We observe that и = 0 satisfies the
radiation condition.) To deal with the boundary condition at r=l, we
introduce the stretching transformation (9.421) and find that the leading
term v0 in the boundary layer expansion satisfies the equation
(9-427) ^ + ,o = O.
At o- = 0 we have v0 = 1. As cr-> <x> the boundary layer term must match the
outer solution w = 0, so we require that uo->0 as <г->оо. However, the
boundary layer equation has only trigonometric (oscillatory) solutions and
the only solution that tends to zero at infinity is the zero solution. That
solution does not satisfy the boundary condition at a = 0.
Consequently, no boundary layer exists for the Helmholtz problem and
the methods of this section are inappropriate for this problem. Indeed, the
asymptotic approximation of the exact solution [see F.206)] for small e is
(9-428) м«
688 PERTURBATION METHODS
Thus, the outer solution и = 0 has no asymptotic validity, since the solution
of the given problem does not decay exponentially for large r and small
but merely oscillates rapidly. A different approach is required for dealing
with this and related problems and this is given in Section 10.1.
EXERCISES FOR CHAPTER 9
Section 9.2
9.2.1. Use the perturbation method to solve the following boundary value
problem.
y2 < 1,
Obtain the first two terms in the expansion.
9.2.2. Obtain the exact solution of the following boundary value problem
for the Helmholtz equation in the unit sphere.
V2u + е2и = 0 , x2 + y2 + z2 < 1,
Also, solve the problem using perturbation theory and compare the exact
with the approximate solution up to terms of O(eA).
9.2.3. Use the perturbation method to solve
[1 + e(x2 + y2)]uxx + uyy = 1, x2 + /<1,
9.2.4. Apply the perturbation method to obtain an approximate solution of
the following problem.
9.2.5. Let u(x, y) satisfy
Uxx + A + ey) Uyy = 0 , 0 < X < 77, 0 < у < 7Г ,
and the boundary conditions
EXERCISES FOR CHAPTER 9 689
M@, y) = u(<7T, y) = u(x, 7Г) = 0 ,
Use the perturbation method to obtain an approximate solution.
9.2.6. Consider the exterior boundary value problem for the modified
Helmholtz equation
with the boundary conditions
u\r=1 = e~e , ы—>0as r-»oo 9
where r2 = x2 + y2 + z2. Use a perturbation expansion to solve the problem
and show that it is not possible to make each term in the series vanish as
r—>o°. Determine that the series exhibits secular behavior and use the
method of multiple scales to overcome this difficulty and thereby obtain the
exact solution of the problem.
9.2.7. Let K(x, yy z; £, т/, f) represent the free space Green's function for
the modified Helmholtz equation
V2K- €2K= -S(x - t)8(y-v)8(z - C) ,
with -oo < x9 y9 z, £, г), £ <oo. Assuming e is small, apply the perturbation
method to obtain К in terms of the Green's function for Laplace's equation.
Using the results of Example 6.13, discuss the behavior of the exact and the
approximate Green's functions at infinity. Explain the differences in their
behavior at infinity as a result of the occurrence of secular terms in the
perturbation series.
9.2.8. Use the perturbation method to solve Laplace's equation V2m = 0 in
the disk r2 = x2 + y2 < 1 with the boundary conditions:
(a) €^ + ll = i, r = l,
(b) ^ + «i = lf r«l.
Show that the problem of part (b) cannot be solved by using the perturba-
tioiytnethod and obtain the exact solution of the problem to show why the
^gular perturbation theory fails.
9-2.9. Consider the boundary value problem of the third kind for the elliptic
equation
690 PERTURBATION METHODS
in a region G with the boundary condition
BG
with a >0 and /3 >0. Assuming either a or /3 are uniformly small on <?G
construct a perturbation method whereby the solution of the given third
boundary value problem can be approximated by solutions of Dirichlet or
Neumann problems. Comment on the possibility that the data for the
reduced Neumann problem are incompatible and explain how to resolve
that difficulty [see Exercise 9.2.8(b)].
9.2.10. Use the perturbation method to solve the Cauchy problem for the
parabolic equation
ut + eux = uxx , -oo < x < oo, t > 0 .
Determine that the perturbation series exhibits secular behavior and use the
method of multiple scales to deal with this problem. Also, convert the given
problem to an initial value problem for the heat equation by introducing the
change of variable т = t, a = x- et. Obtain the solution of the transformed
problem and compare it with the perturbation result.
9.2.11. Solve the Cauchy problem for the weakly damped wave equation
и« + €и,-с2и„ = 0, -«x*<oo,f>0
with the initial data
u(x,0)=f(x), n,(*,0) = 0
using perturbation theory. Show that secular terms arise in the perturbation
expansion and use the method of multiple scales to eliminate them.
9.2.12. Construct a solution of the Cauchy problem (9.73), (9.74) in the
form
71=0
e2nvn(t)cos[Bn
Note that this should lead to the expansion (9.89) but show that vo(t) can be
specified so that the O(e2) secular term is eliminated. Indicate how i>i@
must be chosen to eliminate secular behavior at the O(e ) level.
Hint: Expand vo(t) and v^t) in a series in powers of e2.
9.2.13. Show that if c(ex) in the wave equation (9.93) is given as in (9.120),
the equation can be transformed into an equation with constant coefficients
in the region x > 0. Let a = A le) log[(c0 + ex)/c0] and т = t in (9.93) and
obtain the equation
EXERCISES FOR CHAPTER 9 691
jf the data for the wave equation (9.93) are w@, t)=09 u(x, 0) =/(*), and
Uf(x, 0) = 0, so that the problem is given in the region x > 0, t > 0, determine
the appropriate region and data for the transformed problem for v(a, r).
piscuss the approximate solution of the initial and boundary value problem
for v(<r, t) in the region a - т>0 and compare with the result (9.124).
Hint: Do not expand the transformed initial data in powers of e, on using a
perturbation series to solve for u(<r, т). Show that secular terms arise in the
perturbation series and apply the method of multiple scales in a judicious
manner to reproduce the result in the text.
9.2.14. Consider the initial and boundary value problem for the parabolic
equation.
ut - (c0 + exfuxx = 0 , x>09t>0 ,
where c0 > 0 and 0 < e < 1, with the data
Solve by using the perturbation method and show that secular terms occur in
the perturbation series. Attempt to eliminate this difficulty by using the
method of multiple scales. Use the transformation given in Exercise 9.2.13
to transform the problem into one with constant coefficients that can be
solved exactly,
9.2.15. Using the boundary perturbation method, solve the Neumann prob-
problem for Laplace's equation
uxx + uyy = 0 , 0 < x < /, ex < у < L
with the data
, y) = du(l9 y) = du{x, L)
дх дх ду
ди{х, ех)
16. Solve the following Dirichlet problem using the boundary perturba-
approach.
uxx + uyy = 0 9 e sin у < x < /, у > 0 ,
м(е sin у, у) - иA, у) = 0 , у > 0 ,
u{x9 y) bounded as у -» oo .
692 PERTURBATION METHODS
9.2.17. Apply the boundary perturbation method to solve the following
problem.
utt - c2uxx = 0,
u(x,0) = u,(x,0) = 0, x>0,
u(et, t) = Л cos cot, f>0.
Show that secular terms result in the perturbation series. Compare the
perturbation result with the exact solution given in F.129) when c0 = 6.
9.2.18. Solve the Dirichlet problem for Laplace's equation V2u = 0 in an
ellipse x ~ A + ae) cos а; у = A + be) sin a (i.e., a slightly perturbed unit
circle) using the boundary perturbation method. The boundary condition on
the ellipse is u(x, у) = х .
9.2.19. Use the eigenvalue perturbation method to determine up to O(e)
terms the leading eigenvalue and eigenfunction for the following problem.
M"(x) + A( 1 - € sin2 x)M = 0, 0<х<7г,
9.2.20. Approximate the leading eigenvalue for the following problem.
-V2Af + exyM =
9.2.21. Use the results of Example 8.3 to approximate the leading eigen-
eigenvalue for the following problem in a sphere r - x2 + y2 + z2 = 1.
M(x,y,z) = 0, r = l.
9.2.22. Develop a perturbation method to obtain the eigenvalues for the
following Sturm-Liouville problems.
dx \ dx I
with the boundary conditions
(a) eM'@)-M@) =
(b) M'@) - eM@) = 0 , M{1) = 0 .
9.2.23. Apply the method of Exercise 9.2.22 to obtain the leading eigen-
eigenvalue for each of the following problems.
EXERCISES FOR CHAPTER 9 693
M"(x) + \M(x) = 0 , 0<x<ir,
with the boundary conditions
(a) еМЩ-М@) = 0, М(тг) = О.
(b) M'@) - €M@) = 0 , М(тг) = 0.
9.2.24. Reproduce the results (9.175)-(9.178) by looking for a solution of
the nonlinear Klein-Gordon equation (9.173) in the form
л=1 л = 1
where 6 = kx- a>t, в = кх- Ш and
00 00
Л = 2 <оя(к)еп , E=2 wa{k)en
л=0 п = 0
Proceed as in Example 9.7 and choose the <on and £>n to eliminate secular
terms in the expansion of w.
9.2.25. Obtain traveling wave solutions of the nonlinear hyperbolic
equation
wtt — y2wxx — c2w — aw3 = 0
by using the expansion forms given in Exercise 9.2.24. (Note that the
linearized version of this equation does not have a real dispersion relation
for all values of k. Consider the problem only for those к that yield traveling
waves in the linearized case.)
9.2.26. The nonlinear system
occurs in the theory of nonlinear optics. Show that if we consider the linear
case and set cr = O, and look for a solution Е = Е(в), Р = Р(в) where
0 = kx — <ot, we obtain the dispersion relation
К — <j
For'4he (weakly) nonlinear case look for a solution of the given system in
the form
E = a cos в + ax cos 30 + • • • ,
P = b cos в + Ъх cos 36 + • - • ,
694 PERTURBATION METHODS
where the constants a, b, atJ and bt are small, but ax and bx are smaller in
magnitude than a and b, and в — kx — <ot. Insert these expansions into the
nonlinear equations and equate the coefficients of like trigonometric terms
From the coefficients of the cos в terms obtain an expression giving b in
terms of a and an equation for a = y2k2 - <o2. Expand a in powers of a2 and
show that
cV
]Л^Ш £_^ Зас W 2
-2 _ 2/ 2 i\ л 2/ 2 -«\4
Section 9.3
9.3.1. Obtain the exact and the regular perturbation solution of the initial
value problem
e{ut + ux)+ M = si
u{x, 0) = sin x , -oo < x < oo
(note that the solution of the reduced equation satisfies the initial condition),
Construct a boundary layer solution, if it is needed, to obtain an approxi-
approximate solution valid for all t > 0.
9.3.2. Construct a perturbation and exact solution for the following problem
and compare both results.
e(ut -ux)-u = 0, x > 0, -oo < t < oo ,
9.3.3. Give a complete discussion of the perturbation solution of the
following problem.
e(ur+ ux) + u2 = 0,
u(x, 0) = sin x , -oo < x < oo .
9.3.4. Obtain a perturbation and boundary layer solution of the following
problem.
(t x) t>0,
u(x, 0) = sin x , -oo < д: < oo .
Hint: The initial layer has width O(V?).
9.3.5. Derive the exact solution of the problem in Example 9.9 and
compare it with the perturbation result for t ** 1 and t» 0.
9.3.6. Let /(*) = sinjt and g(x) = 0 in the problem of Example 9.10 and
obtain the boundary layer solution near t = 0 and the outer solution (9.283)
EXERCISES FOR CHAPTER 9 695
At t = T > 0 let (9.283) serve as initial data for the parabolic equation for u0
in (9,288) and let ux{xy t) = 0 in that equation. Determine uo(x, t) for t> T
as a solution of the parabolic equation and characterize the range of values
of t in which each of the approximate solutions obtained is valid.
9.3.7. Discuss the initial value problem
vt + cux = \Dvxx , -» < x
(considered in Section 1.1) from the point of view of a singular perturbation
problem if we assume the diffusion coefficient D is small. Show that even
though the reduced equation can absorb the initial condition, a boundary
layer must be introduced near x = ct since the solutions of the regular
perturbation equations are singular there.
9.3.8. Consider the singular perturbation problem for the ordinary differen-
differential equation
ey"(x) + a(x)y'(x) + b(x)y(x) = 0, 0 < x < 1 ,
with the boundary conditions
y@) = a , y(l) - p .
Show that it is not possible, in general, to solve this problem using regular
perturbation methods.
(a) Assume that a(jt)>0 in 0<jc<1 and obtain an approximate
solution of the problem by introducing a boundary layer at x = 0.
(b) Assume that a(x)<OinO<jt<l and solve the problem approxi-
approximately by introducing a boundary layer at x - 1.
9.3.9. Let f{x) = x2e~x, /*(*) = 0, and g(y) = sinGry/L) in the problem of
Example 9.11. Give a full discussion of the perturbation and boundary layer
solutions of the given problem in the following cases,
(a) 6 = 1.
(b) 6 = 0.
(c) b«-l.
9.3.10. Use perturbation and boundary layer methods to obtain an approxi-
approximate solution of the following problem.
€(uxx + uyy) + ux + uy = 0, jc >0, у >0 ,
u(x, >>)y
••3.11. Obtain an approximate solution of the following singular perturba-
perturbation problem.
696
PERTURBATION METHODS
u@, у) = м(тг, y) = u(jc, 0) - u(*, тг) = 0 .
9.3.12. Given the Dirichlet problem,
ea(uxx + uyy) - ux + и = 0 , jt2 + y2 < 1,
where a is a constant, determine the location of the boundary layer and
obtain an approximate solution, away from the points (jc, y) = @, -1) and
(x,y) = @,l), if:
(a) a = -l;
(b) fl = l.
9.3.13. Obtain a perturbation and boundary layer solution of the following
problem.
e(uxx + uyy) - wx = 0 , *2 + y2< 1, у >0 f
u(x, 0) = 1; ф, у) = у f x2 + y2 = 1, у > 0 .
Discuss the solution away from the points (—1,0), A,0), and @,1).
9.3.14. Consider the singular perturbation problem for the elliptic equation
with the Dirichlet boundary conditions
(a) Show that the point @, -1) on the circle is a point of type R and
obtain the approximate boundary layer equation (9.365) at that
point.
(b) Show that the point @,1) is of type P and obtain a boundary
layer equation of the form (9.365) at that point.
9.3.15. Show that with the transformation
W(a, P) = exp(aap + \bp*)v{a9 p)
and appropriately chosen constants a and b, (9.381) can be brought into the
form
EXERCISES FOR CHAPTER 9 697
vatx + Xav + <0Vp =0,
vatx
where A and <o are constants. Separate variables in the equation for v(a, /3)
and show that v can be expressed as a product of Airy functions and
exponentials.
9.3.16. Determine the appropriate matching conditions for the problem in
Exercise 9.3.14 near the point @, -1) on the circle if the boundary
conditions are u(x, y) = x2 on x2 + y2 = 1. That is, specialize the discussion
following (9.367) to this problem and proceed up to (9.381).
9.3.17. Use perturbation and boundary layer methods to obtain an approxi-
approximate solution of the following parabolic problem.
e uxx + ux - ut = 0 , 0 < x < /, t > 0 ,
with the initial and boundary conditions
u(x, 0) = /(*), и@, 0 = g(t) , u(l, t) = h{t).
9.3.18. Obtain an approximate solution of the singular perturbation
problem,
euxx + aux - ut = 0 , x > 0, t > 0,
with the initial and boundary conditions
u(*,0) =/(*), u@,t) = g(t).
Consider the cases
(a) a = l,
(b) e = -l.
9.3.19. Use the parabolic equation method to determine the location of the
boundary layers for the elliptic equation
if the Dirichlet conditions are assigned on the boundaries of the following
regions.
(a) 0<JC<77,
(b) 0<*<7r, 0<y<oo.
(c) х2 + у2<1.
9-3.20. Complete the solution of the problem for (9.394) given in Example
9-12 by obtaining a boundary layer solution near у = L. Combine the result
698 PERTURBATION METHODS
of this exercise and of the example in the text to yield a composite
approximate solution of the problem in the manner of (9.321).
9.3.21. Solve the following initial and boundary value problem using the
parabolic equation method.
e(utt - c2uxx) + щ - aux = 0 , 0<x < /, t>0 ,
«@,0-1, «(/, 0 = 0, f>0.
Assume that \a\ < с and consider these cases.
(a) a<0.
(b) a = 0.
(c) a>0.
9.3,22. Consider the problem
with u(x, y)-+0 as x-* oo. Apply the parabolic equation approach and obtain
the equation
Solve the problem using the parabolic equation method and the perturba-
perturbation and boundary layer method and compare results.
9.3.23. Use perturbation and boundary layer methods to solve
€И„ + 1*„ + И,=0, 0<ДС<7Г,0<у<7Г,
w(x, 0) = и@, у) = w(x, тг) = 0 ,
9.3.24. Apply boundary layer and perturbation methods to solve the folio*
ing initial value problem.
€{um - a2uxxt) + utt - c2uxx = 0,
u(x9 0) = f(x) , ut(x, 0) = g(x) , и„(х, 0) = k(x) ,
with a > c.
Hint: Proceed as in Example 9.10.
EXERCISES FOR CHAPTER 9 699
9.3.25. Apply perturbation and boundary layer methods to the following
problem.
"@, 0 = ux@, t) = и(/, /) = ux(l, t) = 0, t > 0 .
Determine the outer solution and show that there are parabolic boundary
layers at x = 0 and x — L
9.3.26. Determine the location of the boundary layers and the correspond-
corresponding boundary layer equations for the following singular perturbation
problem.
Hint: (jc, y) = @,0) is a point of type 5.
9.3.27. Obtain the exact and approximate solution of the modified Hel-
mholtz equation in the region exterior to the unit sphere
2r?2
with the boundary conditions (9.418) where r2 = x2 + y2 + z2. Proceed as in
Example 9.13.
10
ASYMPTOTIC METHODS
This concluding chapter continues our presentation of methods that yield
approximate solutions of initial and/or boundary value problems for partial
differential equations. We refer to them as asymptotic methods, for reasons
indicated in Section 9.1, even though they do not all fit precisely into that
category.
We begin with a consideration of equations with a large parameter and
concentrate mostly on the reduced wave equation or the Helmholtz equa-
equation. Linear and nonlinear problems are considered and the solution tech-
techniques are also applied to the (hyperbolic) Klein-Gordon equation. A
common feature in all the problems is that the solutions contain rapidly
oscillating terms.
Next we consider methods for describing the propagation of singularities
of hyperbolic equations. They permit the solutions to be analyzed near
singular regions, such as wave fronts, without having to solve the full
problems. Even though the methods and results appear to have a somewhat
different character than those considered in the remainder of this chapter
and in Chapter 9, they are related to the techniques of Section 10.1 as will
be seen.
Finally, we present a method for simplifying linear and nonlinear equa-
equations and systems of equations by approximating them, in a systematic way,
by simpler partial differential equations. The simpler, more easily solvable
equations generally model one or more of the basic physical features
associated with the full equations or systems. The study of simplified models
of given problems is fundamental in applied mathematics.
10.1. EQUATIONS WITH A LARGE PARAMETER
In the study of boundary value problems for the reduced wave equation
A0.1) V2w + A;Vw = 0,
700
EQUATIONS WITH A LARGE PARAMETER 701
a case of greatest interest in the context of wave propagation occurs when
the (constant) parameter к is large. The given function n in A0.1) is then
known as the index of refraction. The connection between A0.1) and the
Wave equation is given in Exercise 2.4.7.
Assuming к is large, we apply a perturbation procedure to A0.1) and
expand и in inverse powers of к as
"'
A0.2) и = X и,*
Inserting A0.2) into A0.1) and equating like powers of fc, we immediately
conclude that u0 and all further terms in the expansion must vanish. Putting
e - I/A:, we recognize that A0.1) is, in fact, a singular perturbation problem
and asymptotic expansions of A0.1) must have a form different from that
given in A0.2) (see Example 9.13). To determine appropriate expansion
forms we consider some exact solutions of A0.1) in the case where
n = constant.
For the two-dimensional problem, the simplest solutions of A0.1) are the
plane wave solutions (see Example 2.14)
A0.3) u(xy y) = exp[*A;n(* cos в + у sin в)],
where 0 and n are constants. With r2 = (x - £J + (y -17J another (simple)
solution of A0.1) in two dimensions is
(Ю.4) ф,у) = МЬи>),
where /0 is the zero-order Bessel function. For large values of fe, assuming nr
is not small, the asymptotic expansion of the Bessel function (see Exercise
5.7.1) yields
A0.5)
(* \ / \
iknr - jj + (lirknry1'2 exp^-iknr + jj .
Each of the exponential terms in A0.5) represents a cylindrical wave
solution of A0.1).
In the three-dimensional case with r2 = (x - £J + (y - туJ + (z - ()\ a
spherical wave solution of A0.1) is
A0.6) w(jc, y, z) = - eiknr,
38 is easily verified.
702 ASYMPTOTIC METHODS
The exact and asymptotic solutions (Ю.З)-(Ю.б) of A0.1) indicate that
when we expand solutions of A0.1) asymptotically for large k, they have the
form of a rapidly varying exponential term multiplied by an amplitude term
On considering further terms in the asymptotic expansion of Jo we find that
the amplitude terms in A0.5) are given as series in inverse powers of к. ц
the exponentials in A0.5) are also expanded in powers of k, we find that the
solution u(Xy y) contains both negative and positive powers of к in its
expansion. Consequently, the series A0.2) does not lead to a useful result
Noting these results we look for asymptotic solutions of A0.1) in the
form
A0.7) м = veik<p
where q> is the phase term and и is the amplitude term. Inserting A0.7) into
A0.1) gives
A0.8) V2m + k2n2u = {k2[n2 - (V<pf]v + ik[2V<p • Vv + vV2<p]
Since к is assumed to be large, we equate the coefficient of the highest
power of к in A0.8) to zero and obtain
A0.9) (V<pJ = n2.
The eiconal equation A0.9) was studied in Example 2.14 for the case of two
dimensions. The phase term <p must be specified from A0.9) subject to
appropriate conditions that result from the data given for the reduced wave
equation A0.1).
Once the phase <p is specified, the amplitude v must be determined from
A0.10) ik[2Vq> Vv + vV2<p] + V2v = 0 ,
in view of A0.8), A0.9). Dividing by ik in A0.10) and letting fc-юо reduces
the order of A0.10) from second to first order. Since к is assumed to be
large, we conclude that the problem of determining v from A0.10) is a
singular perturbation problem. Therefore, difficulties of the type encoun-
encountered in Section 9.3 are likely to occur in the present problem.
Nevertheless, we look for an asymptotic solution of A0.10) in the form
2
Inserting A0.11) into A0.10) and equating like powers of к yields the
recursive system of equations
EQUATIONS WITH A LARGE PARAMETER 703
A0.12) 2V<p-Vv0 + v0V2q> = 0,
A0.13) 2V<p >VVj + vjV2<p=-V2vj-i, J^1-
These first order partial differential equations for the u; are known as
transport equations since they describe the variation of the amplitude terms
v. along the rays or characteristics determined from the eiconal equation, as
shown later. The (initial) conditions for the u; must be determined from the
data for the reduced wave equation A0.1).
Much of the terminology associated with the large к solutions of the
reduced wave equation A0.1) is drawn from the theory of optics. The
propagation of light waves, when described in terms of A0.1), corresponds
to the case where к is large. The asymptotic solution A0.7) characterizes
geometrical optics since it is given in terms of the rays determined from the
eiconal equation. Often geometrical optics is taken to mean the results
determined strictly from the eiconal equation without regard to the trans-
transport equations A0.12), A0.13). Solutions based directly on A0.1), or some
approximation thereof that involves second order differential equations,
yield results that characterize wave optics. Wave optics is required when
geometrical optics fails to describe the solution of A0.1) correctly, either
because the solution breaks down in the geometrical optics description or
because it fails to account for certain effects due to the diffraction of light.
However, a geometrical theory of diffraction has been developed by J. B.
Keller that shows how to take into account diffraction effects within the
context of a geometrical description of the solution similar to that given by
geometrical optics. These asymptotic analyses of wave propagation are
examined in some detail in this section.
In the following example we consider Green's functions for the two-
dimensional reduced wave equation. The exact results obtained yield useful
information about the asymptotic solutions of A0.1) and serve to motivate
our discussion.
Example 10.1. Green's Function for the Reduced Wave Equation. The
free space Green's function K(x, y; £,17) for the reduced wave equation
A0.1) in two dimensions satisfies the equation
A0.14) V2K + k2n2K = -S(x -
and the radiation condition at infinity
A0.15)
with г2 = (* - £ f + (у - ri)\ and n assumed to be constant. In Example
"•13, this Green's function was found to be
704 ASYMPTOTIC METHODS
where H$\z) is the zero-order Hankel function of the first kind.
In view of F.206), the leading term in the asymptotic expansion of
with knr > 1 is
A0.17) K(x, y; Z, V) - - (^) exp{iknr - l±) .
Since with г Ф 0, К is a solution of the (homogeneous) reduced wave
equation A0.1), we conclude that A0.17) is the leading term in an
asymptotic solution of A0.1) in the form A0.7) and A0.11). To see this
directly, we now construct an asymptotic solution of A0.1) that depends
only on r.
Let
00
A0.18) К = eik*{r) 2 i;,(r)(flfc)-'
with r > 0. The eiconal equation for <p takes the form
A0.19) 2 2 2
The solution of A0.19) for which the radiation condition A0.15) can be
satisfied by the expansion A0.18) is clearly given as
A0.20) <p(r) - nr .
The transport equation for vo(r) is
A0.21) 2V<p -Vv0 + i/0V2«p = 2«p'i/J+ (<p" + \ <p')v0 .
- nv0 ~ 0,
and its solution is
A0.22)
where c0 is an arbitrary constant that may depend on /:. With <p and Vq given
by A0.20) and A0.22), respectively, we find that voeik<p agrees with the
asymptotic form of К given in A0.17). However, the constant c0 in A0.22)
cannot be specified by the preceding (direct) asymptotic method since it
depends on the behavior of the Green's function К at the source point r = 0.
EQUATIONS WITH A LARGE PARAMETER 705
jsfow both the Green's function A0.16) and its asymptotic representation
A0.17) are singular at r = 0. But the Green's function A0.16) has a
logarithmic singularity at r = 0, whereas its asymptotic form A0.17) is
algebraically singular at r = 0. Since the derivation of A0.17) from A0.16)
required that r>0, we cannot expect A0.17) nor the equivalent result
obtained from the direct asymptotic method to be valid at r = 0.
Thus although the asymptotic expansion A0.18) generates the correct
form for the Green's function К with r > 0, it cannot be directly related to
the behavior of the Green's function at r - 0, so that the arbitrary constants
that occur in A0.18) can be specified. An indirect approach such as the
boundary layer method must be used to construct a modified expansion for
К valid near r = 0. By matching this expansion with A0.18), the arbitrary
constants in the latter expansion can be specified. The boundary layer
method in this case yields the exact solution A0.16) as is easily shown.
However, in the case of a variable index of refraction the boundary layer
method yields useful results since the exact solution is not, in general,
available.
The free space Green's function A0.16) is now used to construct a
Green's function in the half-plane x > 0 with the boundary condition
A0.23) *@,у;&ч) = 0.
The source point (f, т/) lies in the right half-plane so that £ >0. The exact
solution of this problem can be obtained by means of the method of images
to be
A0.24) K(x, y; (,,) = l- H«\knr) - | Н^(кпг) ,
where f1 = (jc + f J + {у - туJ. This Green's function also satisfies the radia-
radiation condition at infinity.
If we expand A0.24) asymptotically for r>0 in the right half-plane, we
have
1/2
л ittI
Introducing the oscillatory term e~lmt where a> is a constant, we form the
function
A^.26) v{x9y90 = K(x9y;{9fi)e"lmt.
We easily find that v satisfies the two-dimensional wave equation when
(*> У) *(€, v) (see Exercise 2.4.7). We set
706 ASYMPTOTIC METHODS
A0.27) Ке~ш = ^ Н^\ ^
- w7<Tiu"
where м7 and us are the incident and scattered waves, respectively.
This terminology is employed because with the introduction of the
asymptotic expression A0.25) for К into A0.27), the term vt = ut exp(-/a>f)
represents a cylindrical wave traveling away from the source point. That
part of the wave that hits the boundary x = 0 is moving toward the
boundary. Thus it is characterized as a wave incident upon the boundary.
The cylindrical wave us = us exp(-icat) is a wave traveling away from the
(fictitious) reflected source point and that part of the wave the intersects the
boundary x = 0 moves away from it towards the half-plane x > 0. Since its
existence is due to the presence of the boundary, we refer to it as a scattered
wave. (Cylindrical waves are discussed in Example 2.14.)
Noting these results, we formulate the asymptotic Green's function
problem as follows. We seek a function К given as
A0.28) K~Ul + us,
where w7 and us are asymptotic solutions of the reduced wave equation
A0.1). The incident wave w7 is completely specified—it is the free space
Green's function A0.16), A0.17)—and the scattered wave us is to be
determined. The boundary condition A0.23) for К implies that
A0.29) us(P,y)=-ug@9y).
The scattered wave us is required to be an outgoing wave at the boundary
x - 0. That is, use~ltat must represent wave motion away from x = 0 into the
interior region x > 0. The outgoing condition takes the place of the radiation
condition in the asymptotic formulation of this problem.
We now demonstrate that the solution of the foregoing Green's function
problem, carried out to leading terms only, agrees with A0.25). The
asymptotic form of м7 is given in A0.17), and us is expressed as
A0.30) us~Ve&*.
In view of A0.29) we must have
A0.31) ~ *
(Ю.32) m y) = - J Ш
that is, we equate both the phase and the amplitude terms in A0.29).
EQUATIONS WITH A LARGE PARAMETER 707
There are two possible solutions of the eiconal equation satisfied by the
(scattered) phase term <p with the initial condition A0.31). They are
(Ю.зз) <p±(x, y) = ny/(x±tf + (y-vr •
We reject the minus sign in A0.33) since it does not yield an outgoing
cylindrical wave for us. In fact, the phase will then be identical to that of the
incoming wave w7. Selecting the plus sign in A0.33) and defining f2 =
(x + £J + (у ~ vJ> we specify the phase as <p = nr, and readily conclude
that
i / 2 >1/2
A0.34) V(x9 y)=~-
on using A0.21), A0.22) and the boundary condition A0.32). The
asymptotic result A0.28) agrees with A0.25) which was obtained from the
exact solution.
We now continue our discussion of the eiconal equation A0.9) and the
transport equations A0.12), A0.13). These first order partial differential
equations are solved using the method of characteristics as developed in
Chapter 2.
Our present discussion is restricted to the three-dimensional problem, but
the results are presented later for the two-dimensional case as well. With
P~ Фху 4 = <Py> and r = <pZJ we set
A0.35) F = p2 + q2 + r2-n29
so that F = 0 corresponds to the eiconal equation. Using the results of
Exercise 2.4.13, we find that the characteristic equations for x, y, z, <p, p, q>
and r are
dx __ dy _ dz _
A0.37) ^r = 2(p2 + q2+r2) = 2n2,
A0.38) ^ - ± (n2) *£ = ± {n^ dL = A (и2}
ds dx v n ds dy ds dz v
where s is a parameter along the characteristics and the fact that F - 0 along
the characteristic curves was used to obtain A0.37).
The characteristic (base) curves x = x(s) = [x(s), y(s), z(s)] are called
r<*ys, and the surfaces of constant phase <p = constant are called wave fronts.
Since the vector [p, q, r] = V<p is normal to the wave front <p = constant, the
equations A0.36) that can be written as
708 ASYMPTOTIC METHODS
A0.39) ^
show that the rays are orthogonal to the wave fronts.
The full first order system of characteristic equations A0.36)-A0.38)
must be solved simultaneously, particularly if the index of refraction л is a
function of x. However, it is possible to obtain a separate second order
system of equations for the rays x = x(s). On differentiating A0.39) with
respect to s and using A0.38) we obtain
A0.40) ^
This vector equation together with
determines the rays x(s) and the variation of the parameter s along the rays.
These equations for x(s) are known as the ray equations, and they show that
the rays are specified in terms of the index of refraction n. In particular, if
the index n is a constant, we see that \(s) is linear in s so that the rays are
straight lines.
Example 10.2. Rays in a Strath^ed Medium. We consider wave propaga-
propagation in two dimensions and assume that n — n{x) so that we have a stratified
medium. The ray equations A0.40), A0.41) take the form
From the equation for y(s) we conclude that y'(s) = 2a, where a is an
arbitrary constant. We multiply the equation for x(s) by x'(s) and obtain
d2x d fl
from which we conclude that [jt%s)]2 = 4w2 + constant. Using the^ third
equation in A0.42) and /(s) = 2a, we obtain x'(s) = ±2(n2 - a ) . But
jc'(^) = xf(y)y\s) = 2ax'(y), so that the ray equations reduce to
If we specify a point (jc0, )>o) that the ray must pass through and the slope of
EQUATIONS WITH A LARGE PARAMETER 7W
the ray at that point, the constant a (as well as the sign) are determined and
the resulting ray equation has a unique solution.
In the context of geometrical optics, it is of interest to determine the
structure of the rays in a stratified medium without determining the full
field. We consider two cases.
First we assume that all rays issue from the y-axis with the same (positive)
slope. Thus, we set
7Г
A0.45) y@) = y0, y'@) = tan 0, 0 < в < -
We use the given initial slope y'(Q) to determine the arbitrary constant a in
A0.44), and find that a = w@) sin 0. As a result, the ray equation becomes
(since /@)>0)
dy
and the (formal) solution is
A0.47) y(X)
f2
~ n\0) sin2 в
We only consider the rays in the right half-plane x > 0.
To proceed we must specify the behavior of the index of refraction n(x).
The simplest case occurs if n(x) is monotonically increasing or nondecreas-
ing. Then the denominator in A0.47) is never zero and the solution is valid
for all x > 0. As yQ ranges from minus to plus infinity, the rays cover the
entire right half-plane.
If n(x) decreases monotonically to zero as x^°°, the slope y'(x) increases
monotonically until x reaches the value x = xc, at which point n(xc) =
n@)sinB and the slope y'(xc) is infinite. Each of the rays is continuously
refracted as x increases and at the point x = xci each ray has a vertical
tangent line and begins to reverse its direction. Since none of the rays
penetrate into the region x>xcJ it is called the refraction shadow region.
While the slope y\x) of each ray is infinite at x = xc, the integral A0.47)
converges at x = xc and y(xc) serves as the initial value for the (turned) ray
with negative slope. The equation for the new set of rays is
A0.48)
y(x\ = _ Г и@) sin в ds + fXc n@) sin в ds
к Уn\s)-n\0) sin2 в J° VOO-(O)sin20
We note that each ray has a negative slope for x < xc, and an infinite slope
710 ASYMPTOTIC METHODS
If n(x) decreases monotonically to a value nx>0, we see that if n >
w@) sin 0, the slope of the rays never becomes infinite and there is Xno
shadow region. The rays cover the entire right half-plane. However, if We
determine the angle 6X from the equation nx = n@) sin^), we find that for
в > 6X a shadow region will exist. Thus for decreasing w(jc) if the initial ray
slope is sufficiently large, a refraction shadow will occur.
Next we consider an initial value problem for A0.44) in which all the rays
issue from the origin in the (jc, y)-plane. The initial values are
A0.49) y@) = 0, y'(O) = tan0, О<0<тг.
We assume that w(jc) is a smooth function defined as follows. For jc<0
n(x) = w@) and for jc > xx > 0, w(jc) = n(xx) = nx. Thus, w(jc) is constant for
jc <0 and x > xx. As jc increases from 0 to xx, n{x) decreases monotonically
from w@) to nt.
The rays whose initial direction angle satisfies тг/2< в < тг, are straight
lines whose equation is
A0.50) у = jc tan 0, | < 0 < 7Г .
The ray that corresponds to 0 = тг/2 is the line jc = 0 with у > 0. In the first
quadrant the rays are
A0.51) *ш£у/ягТ-Уо**пЧ- °<в<5-
Now the rays with an initial direction angle 0 for which n@)smd<nXi
never achieve a vertical slope as jc increases from zero. As these rays pass
the line jc — xx their slope becomes constant and they have the equation
A0.52)
The rays for which sin0>w1/n(O) have an infinite lope at the point
хв<хг determined from w(jce) = n@) sin 0. At that point the rays turn
backwards. Their equation is given by A0.48) where we replace xc by xe and
put y0 = 0. There is a limiting diffracted ray whose angle вс is determined
from sin0c = nx/n@). All rays with 0>0c are refracted backwards while
those with в < 6C have a finite slope for all positive jc. The refraction shadow
is the region above the extended limiting ray with negative slope. Since this
ray turns around at jc = jcx where its slope is infinite, the shadow region lies
to the left of the line jc = xt (see Figure 10.1). We remark that each of the
turned rays has a constant slope on the part of the ray that lies in the second
quadrant.
EQUATIONS WITH A LARGE PARAMETER
у 4 shadow
711
Figure 10.1. Rays in a stratified medium.
Now A0.41) shows that s is not generally equal to an arc length
parameter on the rays. In fact, with da2 = dx2 + dy2 + dz2 so that a is an arc
length parameter, we have from A0.41)
A0.53) d<j = 2nds.
In terms of a the ray equations A0.40) become
whereas A0.41) is replaced by
A0.55)
as is easily shown.
Using the parameter a in place of s and noting that d/ds = 2nd/da, we
have for the phase <p,
A0.56) ^- = n,
instead of A0.37). Integrating along a ray x(cr) from aQ to a we have
where <p and n are functions of x(o-) and <p0 is a specified constant that may
vary from ray to ray. Thus once the rays are known the phase <p can be
determined.
712 ASYMPTOTIC METHODS
The (standard) initial value problem for <p requires that we specify <p on a
given initial surface. Let the initial surface be expressed parametrically as
x = R(a, p) and let ^ = <po(a, p) on that surface. To use the procedure
given for determining q> we must find the rays that pass through the initial
surface and then integrate along the rays as indicated in A0.57). However
since the ray equations are of second order, an initial point, as well as an
initial direction must be assigned for each ray on the initial surface x =
R(a, /3). Now A0.39) shows that the direction of a ray x (i.e., dxlds or
dx/da) is given in terms of p, qy and r. That is, to solve for <p uniquely, we
must not only specify x and <p initally, but also p, q, and r. This is consistent
with the fact that the full set of characteristic equations A0.36)-A0.38) for x,
<p, />, q, and r is what we are effectively solving.
To determine p, q, and r initially, and thereby dx/da, we follow the
general procedure given in Section 2.4 and determine conditions equivalent
to the strip conditions found in that section. From the initial condition
<p[R(a, p)] = <po(a, p) we obtain
A0.58) £*-V,.£ = i£.£-»£.£
v 7 da da 2 ds da da da
and, similarly,
d<p0 „ <?R dx dR
where A0.39) and dlds = 2nd I da have been used. An additional condition,
which is equivalent to setting F = 0 in A0.35) and which is identical with
A0.55) is
This yields three equations for the three components of the vector dx/da
evaluated on the initial surface. We assume that a = a0 at the point where
each ray intersects the surface x = R(a,/3) so that dx/da evaluated at
a = a0 is determined from A0.58)-A0.60).
To simplify the determination of the possible solutions for dx/da at
a = a0 we assume that a and p measure arc length in an orthogonal
coordinate system on the initial surface x = R(a, K). Then dx/da, dR/d<*>
and dR/dp are all unit vectors and, in addition, dR/da and dR/dP are
mutually orthogonal. Let В and a> denote the angles between the vector
dx(ao)/da and the vectors dR/da and dR/dp, respectively. Then A0.58)-
A0.60) yield
00.61) -.-i* —i$-
EQUATIONS WITH A LARGE PARAMETER 713
The preceding equations generally determine exactly two directions for
each ray on the initial surface. To see this, we choose an arbitrary point P
on that surface and introduce a (local) Cartesian coordinate system with its
origin at that point. Let the unit vectors i, j, and к of the coordinate system
correspond to the vectors dR/da, SR/др, and a unit normal vector to the
initial surface, respectively. Then our definition of 0 and a) implies that
A0.62) cos0=^"i> cosa> = STj
at the point P on the initial surface. The angles 0 and <o are uniquely
specified from A0.61). If the angle у is defined as
A0.63) cos у = -^ • k,
we conclude from A0.60) that
A0.64) cos2 0 + cos2 <o + cos2 y = l.
Now by definition we have 0 < 0, o>, у ^ тт. If 0 and o> are such that
cos у Ф 0 in A0.64), we see that the quadratic expression A0.64) yields two
possible values of cos y, one positive and one negative, with identical
absolute values. If we choose the value of y, say у = y0, for which cos y0 > 0,
then 0 < y0 < тг/2 and the vector dx(ao)/dcr points toward the same side of
the initial surface as the normal vector in view of A0.63). If we set
У = Yi - it - y0, then cos yx = -cos yQ < 0 and тг/2 <yx<ir. In this case
dx((T0)/dcr and the normal vector have opposite directions. The vectors
d\(ao)/da that correspond to y0 and y19 respectively, and the normal vector
to the initial surface are all coplanar at each point P.
If the initial data are such that cos у = 0 at a point on the initial surface, it
follows from A0.63) that dx(cro)/da lies in the tangent plane of the surface
at that point. If the rays are tangent to the initial surface at every point,
there are two cases that are referred to as characteristic initial value
problems for <p. In one case the rays are completely contained within the
initial surface. Since the rays do not leave the initial surface x = R(a, £), the
equation A0.56) that describes the variation of q> along the rays is, the fact,
a condition on the variation of the initial value <p0 along the rays on the
initial surface. If the conditions on the data are met, we find that our result
does not determine the phase function q> outside the initial surface.
A more interesting case occurs if the rays are merely tangent to the initial
surface but do not lie within it. Then the initial surface is an envelope or a
caustic surface of the system of rays. Again the initial value <p0 must be such
that A0.56) is satisfied when a = tr0, that is, at the point of tangency of the
rays. One ray issues from each point on the initial surface and <p is
714 ASYMPTOTIC METHODS
determined from A0.57) along each ray. A more complicated situation that
occurs often in asymptotic problems is when the rays are tangent to the
initial surface only along a curve or some other subregion but are not
tangent elsewhere on the surface. Although it is possible to determine the
rays and the phase function <p for many of these problems, the fact that the
rays intersect at the initial surface leads to difficulties with the full
asymptotic solution.
In the noncharacteristic case (i.e., when cos у Ф 0 in the preceding) there
are two possible ray directions determined at each point on the initial
surface. To obtain a unique solution for the phase q> we must choose one of
the two directions determined from A0.61)-A0.64). That is, we select the
set of rays issuing either on one side of the initial surface or on the other
side. Thereby, x(cr) and dx/da are uniquely specified at each point of the
initial surface. Consequently, the ray equations determine x(cr) uniquely.
Finally, the phase <p(x, y9 z) is found from A0.57) where integration takes
place along the ray that passes through the point (jc, y, z).
The transport equations A0.12), A0.13) can also be solved by integrating
along the rays. We have from A0.39)
A0.65) 2V* •*,, = § • Vb, = In £ -Voj = In ^ ,
where dvj/da is directional derivative along the ray x(cr). Thus the transport
equations reduce to ordinary differential equations along the rays and are
given as
A0.66) 2w^ + u0VV = 0,
dv} ~ ~
A0.67) In ^ + VjV2<p = -V2u._i> /* 1 >
where all functions are evaluated on the ray x(cr).
To solve these equations, we express V2<p in terms of its variation along a
ray. To do so, we consider a region R bounded laterally by a tube of rays
and capped by two segments of wave fronts <p = constant which we denote
by So and 5j as shown in Figure 10.2. The lateral boundary is denoted by 5.
Let N be the exterior unit normal on the boundary of R. Applying the
divergence theorem in R gives
A0.68) f fv2<pdV=f fv-V<pdv=js V^-Nda
1
The surface integral over 5 vanishes since V<p lies in the direction of the rays
EQUATIONS WITH A LARGE PARAMETER 715
5
Figure 10.2. A ray tube.
and N is orthogonal to that direction on S. Assuming the parameter a
increases as we move from So to 5a, we observe that V<p has the direction of
N on Sp whereas it has the direction of — N on So, since So and 5a are
portions of phase surfaces. Also, since \Vq>\ = n, in view of A0.9), we have
A0.69) I I V2<p dV= nda-\ nda.
J J JSi JSq
R
To determine a simple and useful expression for the element of surface
area da on a wave front, we introduce a reference wave front <p = 0 and a
coordinate system (£, 17) on that wave front. We then determine all the rays
orthogonal to that wave front. Each surface <p = constant is parallel to the
wave front <p = 0 and is orthogonal to the rays. With a equal to arc length on
the rays, we may introduce a (ray) coordinate system (£,17, a) in space.
Further, we can replace a by the wave front coordinate <p and, in view of
A0.56), we have
A0.70) dip = nda
along the rays. In the (£, 17, ф) coordinate system, the surfaces <p = constant
are orthogonal to the rays which are given as £ — constant and 17 = constant.
Consequently, if x = r(£, 17, <p) are the equations of transformation from
Cartesian coordinates to (£, 17, <p) coordinates, we find that the area element
^ on a wave front is given as
wher^ /=|rf xrj, as is well known from vector analysis. The volume
dement dV in (£, 17, <p) coordinates has the form
(Ю.72) dV= dad<r=-Jd€ di) d<p ,
716 ASYMPTOTIC METHODS
since a is arc length and the distance between two infinitesimally close wave
fronts q> = constant and <p + d<p = constant is given by da = (l/n) d<p, [n
view of A0.70).
We now assume that the surface elements So and Sl considered corre-
correspond to the wave fronts <p = constant and ф + d<p= constant, respectively
Then as the region R (i.e., the ray tube) shrinks down on a particular ray
we have in (£,17, <p) coordinates
A0.73) J J V2<p
and
A0.74) |S nda- jS nda~[nJ\^dv - nJ\v] d£ dv ,
so that
In the limit as d<p—>0 we obtain
•чЛ„^ч «2 w d , T4 1 d , T4
A0.76) V> = - -j- (nJ) = - — (w/)
on using A0.70).
We are now ready to solve the transport equations A0.66)-A0.67).
Inserting A0.76) into A0.66) yields
A0.77) In — + (nJ)v0 - 0 .
This can be written as
A0.78)
Integrating along the ray x(o-) from a0 to a with vo\ao = V0(<r0), we obtain
where the dependence of the solution on variables other than a has been
suppressed. Similarly, the transport equations A0.67) can be written as
EQUATIONS WITH A LARGE PARAMETER 717
Again integrating along a ray with uyL0 - Vj((r0) we have
/as 1.
Although the derivation was carried out for the three-dimensional case,
the results are valid in two dimensions as well. We need only define J
appropriately. On the reference wave front <p - 0 we now introduce the
single coordinate £, and determine all the rays orthogonal to that wave
front. In the resulting (£, <p) coordinate system we set x = r( £, <p) and find
that the linear element ds on a. wave front is given as
A0.82) ds
where /= |rf |. The area element dA then has the form
A0.83) dA = ds da = - J d% dip .
since A0.56) is valid in two dimensions as well. In deriving A0.75) the
integration is carried out over a plane region bounded by two wave fronts
and two rays. The same expression for V2q> as was found in three dimensions
results, and the expressions for the phase function <p and the amplitude
terms V; are as given.
In both cases the function / represents an expansion or contraction factor
for the rays. It is proportional to an area or linear element on the wave
fronts that increases in magnitude as the rays diverge and decreases in
magnitude as the rays converge. Noting the form of v0 in A0.79) we find
that, to leading order in k, the amplitude of the solution increases as the
rays converge and decreases as the rays diverge. This result is consistent
with a basic principle of geometrical optics according to which the field is
stronger the greater the concentration of the rays.
If the rays intersect at same point or curve, the function / vanishes and
the amplitude terms v- all become infinite. The solution of the reduced wave
equation of which we have found the asymptotic expansion, either remains
finit^in that region or has a different type of singularity, as we have seen in
Example 10.1. Thus our asymptotic result is not valid in regions where the
fays intersect and modified (asymptotic) results must be found in such
regions.
It is worthwhile to observe that the dependence of v0 on the divergence
718 ASYMPTOTIC METHODS
or convergence of the rays may be obtained by means of an extremely
interesting alternative method. If we multiply the equation A0.12) by vQ9 we
have
A0.84) vo[2V<p • Vv0 + v0V2<p] = V- (v20V<p) = 0 .
Integrating this divergence expression over the ray tube R as defined and
using the divergence theorem gives
A0.85) 0 - f \v-(vlV<p) dV= f v2Qn da - f v20n da ,
J J JSi JSq
R
as is easily seen on applying the results that lead from A0.68) to A0.69). In
an approximate sense, A0.85) may be thought to represent a conservation of
energy along the ray tubes of the solution of the reduced wave equation. If
the ray tube shrinks down on a ray x(o-), we have
A0.86) vlndal^vlnda^
where o*ls and <r0 correspond to two points on the ray. In view of the relation
A0.71) between da and /, we find that v0 at some arbitrary point on a ray is
given in terms of Vq]^ = V0(<t0) precisely as in A0.79).
In the case where the index of refraction n ~ constant, it is possible to
find some simple explicit forms for the function / and thereby for the
amplitude term v0. With n = constant, the rays are straight lines and we may
assume without loss of generality that n = 1. Then, as shown by A0.56),
both a and <p measure arc length on the rays.
If the rays are parallel lines, the area element da remains constant on all
wave fronts so that / = constant as well. This implies that v0 — constant on
the rays. If v0 retains the same constant value on all rays, the vf may be
chosen to vanish and the result is a plane wave solution of the reduced wave
equation. (The wave fronts are clearly planes for this case.) If v0 assumes
different values on different rays, we obtain,what is called a general plane
wave. The wave fronts <p = constant are again planes but the amplitude term
y0, although it is constant on each ray, may vary in magnitude from point to
point on the wave fronts. A similar result applies in the two-dimensional
problem.
For the case of spherical wave fronts we introduce spherical coordinates
A0.87) x = rsin в cos a>, у = rsin 0sin <», z = rcos0,
where r > 0, 0 < в < «•, and 0 < o> ^ 2тг. With n = 1, the phase term can be
written as
A0.88) <p = r-rl9
EQUATIONS WITH A LARGE PARAMETER 719
where r1 is a constant and the reference wave front <p = 0 is given as the
sphere r= rx. The ray equation A0.54) becomes
A0.89) 0=0,
and has the solution
A0.90) x = a + bo-,
where a and b are constant vectors and |b| = 1 in view of A0.55). Replacing
the variable a by r and introducing the initial data
A0.91) x(rj) = a + Ъгг = [rx sin в cos <o> rx sin в sin o>, rx cos 0],
A0.92) —-j-^- = b = [sin в cos o), sin 0 sin <y, cos 0],
we find that
A0.93) x(r) = br = [r sin 0 cos o>, r sin в sin &>, r cos 0].
The data A0.91), A0.92) imply that the rays x(r) pass through and are
orthogonal to the sphere r = rl9 and A0.93) shows the rays to be radial lines
issuing from the origin. The radial parameter r clearly measures arc length
on the rays and the spheres <p = constant are orthogonal to the rays.
To determine the function / for this case, we identify the coordinates £
and 17 introduced previously on the wave front <p - 0 with the angular
variables в and o>. Since n = 1, both <p and r measure arc length on the rays.
The transformation from Cartesian to ray coordinates is given as
A0.94) x = r(r, $y <o) — [r sin в cos a), r sin в sin <o, r cos в]
as seen from A0.87). Then da is an element of area on a sphere and we
have
(Ю.95) da = r2 sin в d$ cho = / d$ da> ,
where / = \re x r J - r2 sin ft Thus
°n each ray. If we assume that V0(r0) is independent of 0 and a> so that it has
the same value on each ray, we easily show that the uy(/ > 1) may be chosen
720 ASYMPTOTIC METHODS
to vanish and the solution of the reduced wave equation is the spherical
wave
A0.97) и = (constant) - eikr.
For this case, the asymptotic approach yields an exact solution of the
reduced wave equation valid up to the singular point r = 0. (The corre-
corresponding cylindrical wave solution found in Example 10.1 is, however, not
valid up to the singular point.) The point r = 0 represents a focus for the
rays since they all intersect there. Even though the (asymptotic) solution
may have a valid form up to r « 0, it does blow up there since / vanishes at
that point.
More generally, the term V0(rQ) in A0.96) may vary from ray to ray; that
is, it can have a nonconstant dependence on в and <o. In that case the wave
fronts are still spheres, and u0 decays like IIr along a given ray, but a more
complicated expression results for the amplitude distribution of the field that
involves all the vr
Next we consider a two-dimensional example in which the rays form a
(smooth) envelope. If x = R(f) is the equation of this envelope or caustic
curve where £ is an arc length parameter, the rays are just the tangent lines
of this curve. With a equal to arc length along the rays, an orthogonal
coordinate system in £ and a is given by the equations of transformation
A0.98) x = r(f,G) = R(f) + (^-f)R'(O.
Since £ is arc length, we have |R'(f )| = 1, R' • R" = 0, and |R"| = 1/p where
is the radius of curvature of the caustic curve. Then
imply that rf та = 0, so that the (£, or) coordinate system is orthogonal.
The lines £ = constant are the rays, whereas the curves a = constant are the
(orthogonal) wave fronts. This may be verified directly by noting that with
A0.99) ^ = |rf| = l^il, К = \та\ = \
we have
(lo.ioo) (v9f = ±9] + ±*l = ^--г <p\ + *l.
Thus <p ™ a is a solution of the eiconal equation (V^J = 1. The function /
for this system of rays is given as
EQUATIONS WITH A LARGE PARAMETER 721
A0.101)
and the solution v0 has the form
A0.102) Vo = VoitrJ^\ = V0(<r0)
1/2
The amplitude term v0 is singular when a = f, and this occurs at points
on the caustic x = R(£) as follows from A0.98). The rays intersect at the
caustic and, therefore, / = 0 on that curve. Although each ray intersects the
caustic at <r = f, the set of points a > £ and a < £ represent two groups of
rays, each of which terminates at the caustic. The asymptotic solutions given
for <p and v0, combined with the results that can be obtained for all the vj9
are valid for each group of rays away from the caustic curve. On the caustic
curve the solution of the reduced wave equation of which this is the
asymptotic expansion remains bounded. A boundary layer approach pre-
presented later yields a finite asymptotic result at the caustic.
A typical boundary value problem for the reduced wave equation A0.1),
which is to be solved asymptotically for large k, is formulated in the manner
indicated in Example 10.1. An incident wave ut completely specified as an
asymptotic series of the form A0.7) and A0.11) approaches a boundary
region and is scattered by it. The resulting scattered wave us is again taken to
be an asymptotic series of the type A0.7) and A0.11) and must be
determined. The total field и = uf + us, which is an asymptotic solution of
A0.1), satisfies a homogeneous boundary condition of the first, second,
third, or mixed kind on the boundary region. We mostly assume that и = 0
on the boundary in our discusssion. The problem is to be solved in an
exterior unbounded region and we require that us satisfy a radiation
condition at infinity. In our asymptotic version of the problem the radiation
condition is replaced by an outgoing condition at the boundary. If <p is the
phase term of the asymptotic expansion of us, then us is said to be outgoing
at the boundary if the exterior normal derivative of <p (i.e., d<pldn =V<p *N)
is positive for every exterior unit normal N to the boundary. The outgoing
condition indicates that us travels away from the boundary and is, in fact,
radiating towards infinity.
The specification of the incident wave un the boundary condition, and
the outgoing condition uniquely specify us and thereby the asymptotic
solution u, provided the phase and the amplitude terms in the asymptotic
expansion of us can be determined. We assume that иг is given as
(Ю.103) «/ = *'** 2
=o
722 ASYMPTOTIC METHODS
where the phase ф and the amplitude terms w;. are known. The scattered
wave is expanded as
A0.104) «5в«Л*2у
/-о
with cp and the vf to be determined. If и = иг + us = 0 on the boundary, we
find that
A0.105) eik* E ,
on the boundary. This leads to the conditions
A0.106) <p = ф ,
A0.107) у/ = ~н;;» /-0
on the boundary.
As we have seen, the (initial) condition A0.106) for <p is by itself
insufficient to specify the phase uniquely. In determining the rays associated
with the phase <p we must also specify a direction for these rays on the
(boundary) region where <p is given. We have found that there are, in
general, two possible ray directions at each point where <p is specified.
However, V<p has the direction of the rays as shown by A0.39). It has been
demonstrated that if the rays are not tangent to the surface where <p is
specified, their two possible directions lie on opposite sides of the surface.
Thus the outgoing condition on the scattered wave us uniquely specifies a
ray direction at each point as is easily seen.
Once the rays and the phase are determined, the amplitude terms vj can
be obtained by integrating along the rays as in A0.66), A0.67) and using the
data A0.107) as initial conditions.
As an example we now consider the problem of the reflection of a
cylindrical wave by a parabolic cylinder. The line source of the (incident)
cylindrical wave lies on the z-axis and the generators of the parabolic
cylinder are parallel to the z-axis. Thus the problem may be treated as
two-dimensional and we look for a solution in the form и = и(х, y)>
Example 10.3. Reflection of a Cylindrical Wave by a Parabola. We
consider the parabola
1 2 a
Х=2аУ 'I*
where a > 0 is a given constant. A cylindrical wave is generated by a source
EQUATIONS WITH A LARGE PARAMETER 723
located at the origin, which is also the focus of the parabola. Thus the total
field и satisfies the equation (we put n = 1)
A0.108) V2« + k2u = -8(x)8(y) ,
where w = w(jc, y) and d(x)8(y) is the two-dimensional Dirac delta function.
We assume that и vanishes on the parabola; that is,
A0.109) u(x9 y) = 0, x=Yay2~2*
and и satisfies a radiation condition at infinity. An outwardly radiating wave
generated by the source is
A0.110) щ = ± H$\kr)
as we have shown in Example 10.1. We are concerned with the asymptotic
problem for large k. Thus we replace A0.110) by its asymptotic expansion.
The leading terms are given as
A0.111) щ « l- ^/2ЫГг(\ + -~У exp(i*r - j) ,
as follows from the asymptotic expansion of the Hankel function for large
arguments.
The wave ut is a cylindrical wave incident upon the parabola. The
asymptotic problem then looks for a solution
A0.112) и^щ + и^
where the incident wave w7 is given in A0.111) and the scattered wave us
expanded as in A0.104) is to be determined. The boundary condition
A0.109) requires that
A0.113) us = - uIy x = — y2 - - ,
whereas the radiation condition implies that us must be an outgoing wave at
the parabola.
Qn comparing with the notation given in our general discussion, we find
that
A0.114) ф = г, wo = -7=» H;=:^L^)
724 ASYMPTOTIC METHODS
with the other terms in the expansion of ut not having been specified in
A0.111). The constant с is given as с = (М4)л/2Ысе~'"/4. The appropriate
data for the scattered or reflected wave us on the parabola are
and <p "outgoing" on the parabola. The first two amplitude terms v0 and и
satisfy *
on the parabola in view of A0.114).
To solve for the phase <p of the scattered wave us we first represent the
parabola in parametric form as
A0.117) у = т, JC=^(r2~a2)> -о°<т<оо
(we note that r is not an arc length parameter here). With a as arc length
along the rays we must solve the characteristic equations for jc(ct, т), у (а, т),
<р((т, r), /?(cr, t), and q(a, r). The initial conditions are given for a — 0. Thus
A0.118) y@, т) = г, *@, т) = Ya (т* " ^ '
as follows from A0.117). That is, cr = 0 corresponds to the parabola
A0.117). The condition A0.115) for q> yields
A0.119) <р@,т)=^(т2 + я2)
on expressing A0.117) in terms of the variable r.
The initial conditions for p and q are determined by proceeding as in
Section 2.4. The eiconal equation requires that
A0.120) />2@,t) + <72@,t) = 1.
The strip condition states that on the parabola,
(Ю.Ш) 2-'£+'£ -О.
and this yields
A0.122) ^^
EQUATIONS WITH A LARGE PARAMETER 725
on using A0.118), A0.119). On solving A0.120) and A0.122) for p@, r)
and #@, r), we obtain two sets of solutions. We have
A0.123) p@,r) = l, <?@,t) = 0,
and
22
A0.124) p@, r) = -4—T > 9@, t) =
One of these sets of initial conditions for p and q must be rejected on the
basis of the outgoing condition satisfied by w5. This states that if N is the unit
normal pointing towards the inside of the parabola, we must have V<p • N > 0
on the parabola. We easily find that the unit normal N is
A0.125)
Since V^> = [p, q] we see that on the parabola
A0.126) V<p-N = [p, fl]-N= 7Гф==>0
if A0.123) is used, whereas
A0.127) V<p.N = [p,q].N=7T=^==<0
if A0.124) is used. Therefore, we conclude that A0.123) are the appropriate
initial conditions for the problem.
Since n = 1 and d<r = 2ds9 the two-dimensional version of the characteris-
characteristic equations A0.36)-A0.38) for this problem is
A0.128) ^=P, £ = q, £«1, ^=0, ^=0.
dcr r d<r da da da
The last two equations in A0.128) state that p and q are constant along the
characteristics. Thus they are equal to their initial values A0.123) for all a
and we have
Inserting these results into the equations for x and у in A0.128) and using
(Ю.118) gives
A0.130) x(<r, t) = a- + ^-a О ~ fl2) '
726 ASYMPTOTIC METHODS
A0.131) у(<7,т) = т.
Finally we obtain for <p, in view of A0.128) and A0.119) the result
A0.132) <р(<т,т) = <7+^(т2 + а2).
If we invert the system A0.130), A0.131) and express a and T as
functions of x and y, we easily obtain
A0.133) <т = х-±(у2-а2), r = y.
Inserting A0.133) into A0.132) yields
A0.134) <p(x,y) = x + a.
Thus the phase of the scattered wave us is that of a plane wave. However,
since the amplitude terms of us are not constant, us is, in fact, a general
plane wave.
To solve for the amplitude terms v0 and vx we note first that V2<p = 0 in
view of A0.134). Further, we find that r = (l/2a)(r2 + a2) on the parabola.
Thus
A0.135) 2V<p • Vv0 + V2<pv0 -2^=0,
so that
A0.136) vo(a,T) = vo@
on using A0.116). In terms of x and у we have
(Ю.137) ^y) =
since т - y. The equation for u2 is
A0.138) 2^ = -V4,
where the right side of A0.138) is a function of r only, in view of A0.137).
That is, we operate on vQ with the Laplacian in (jc, y)-coordinates and then
use у = т to convert back to (cr, r) variables (we do not carry out this
calculation). On using A0.116) we obtain
EQUATIONS WITH A LARGE PARAMETER 727
A0.139) vM, t) - - \ (V2i;0)o- - | (^27)
In x, у variables A0.139) has the form
A0.140) »,(*, У) = У,(у)х + V2(y),
with given V, and V2.
We have shown that us is given as
V2(y))} exp[iA:(x + a)],
and this has the form of a general plane wave. We have carried out the
calculation for us to two terms to show that the asymptotic result A0.141) is
not uniformly valid. As x increases to the point where x = O(k), it follows
from A0.141) that the term (l/ik)vl is of the same order of magnitude as
v0. Therefore, the asymptotic expansion of us becomes disordered, in that
our initial assumption that (l/ik)v1 is of lower order in к than v0 is not valid.
The asymptotic expansion we have obtained for us is, consequently, useful
only if jc<O(A:).
The growth of the amplitude term vx, as well as the growth of the further
terms in the expansion of ws, corresponds to the secular behavior en-
encountered and discussed in a number of problems in the preceding chapter.
To analyze this question more closely we insert
A0.142) us = v exp[ik(x + a)]
into the reduced wave equation A0.1) with и = 1, and find that
(Ю.143)
Inserting the expansion и = E°°=o vl{ik)'i into A0.143) and solving for the vj
shows that uy = O{xJ) as Jt->o°, as we have seen for v1.
The secularity problems encountered may be removed by using a stretch-
Щ transformation similar to that used earlier in boundary layer theory.
Since our asymptotic results must be modified only for x > O(k), we set
(Ю.144) x = k€
in A0.143) to emphasize the region where x is large. This yields
728 ASYMPTOTIC METHODS
(Ю.145) 2»* + ^ + р»«=0-
To leading order in к we obtain a parabolic Schrodinger equation
A0.146) 2^ + 6^ = 0.
We must find a solution of A0.146) that matches the asymptotic result given
above for v, to leading order terms as f-»0. Effectively, we are considering
an initial value problem for A0.146) in the region f >0 with v specified at
some positive value of f. We remark that if the parabolic equation method
of Example 9.12 is applied to A0.143), we are led to consider the same
equation A0.145).
A general solution of A0.146) is
A0.147) d(i, y) = £ {^= exp[^^]}/(s) ds .
The bracketed term in the integrand in A0.147) is, apart from a constant
factor, a fundamental solution of A0.145). It may be verified by direct
substitution into A0.146) that it satisfies the equation for £>0. The
function f(s) in A0.147) is arbitrary and must be specified by matching
A0.147) with the asymptotic results given previously.
To carry out the matching procedure, we rewrite A0.147) in terms of
x — kg and у to obtain
A0.148) v{x, y)
Now if x< O{k)y we see that the exponential in A0.148) is rapidly oscillat-
oscillating since к is large by assumption. We may, therefore, evaluate A0.148) by
the method of stationary phase, as presented in Example 5.14. The phase
term <p(s) = (y- sfllx in A0.148) has a stationary point where <p'(s0) = 0,
that is, at s0 - y. Then the leading order result as given in E.256) is
A0.149) v ~ V2tt exp(^J f(y) .
This result, which is valid for small g = x/k, must be equated to the leading
amplitude term v0 given in A0.137). Thus
A0.150)
and v(x, y)—that is, A0.148)—is completely specified.
We might expect that it should be possible to generate the amplitude
EQUATIONS WITH A LARGE PARAMETER 729
term vx from the asymptotic expansion of the integral A0.148). To see how
this occurs we expand the function /E) in A0.148) around the stationary
point so = у and obtain
A0.151) f(s) = f(y) + E - y)f'(y) + h(s - yff'Xy) + ■ ■ ■ •
Inserting A0.151) into A0.148), we find on integrating term by term
that the leading term f(y) yields the result A0.149). The second term
(s-y)f'(y) yields an exact differential and is asymptotic to zero if we
neglect contributions from infinity as we do. For the third term we have, on
integrating by parts,
(Ю.152) \ V£ Ay) £ (у ~ *
where we have neglected contributions from infinity and noted that f(y) =
(l/V2^r)i;oexp[-/7r/4]. On comparing A0.152) with the expression for
(l/ik)vl obtained from A0.140), we find that the asymptotic expansion of
the integral A0.148) agrees with the terms v0 + (l/ik)uv provided we
replace/E) in A0.148) by/E) + {l!ik)F(s) and specify F{s) as to obtain full
agreement between the O(\lk) term obtained from the integral and the
term (\lik)vx.
We have shown, formally, how to construct a solution of A0.146) that
matches the first two terms v0 and vx in the asymptotic expansion of the
amplitude of м5. However, the integral representation of the solution of
A0.146) does not exhibit secular behavior for large x, as was the case for the
(conventional) asymptotic result given previously.
Examples 10.1 and 10.3 have shown that the geometrical optics results, in
which the scattered and incident fields are expanded in the form A0.7) and
A0.11), are not always adequate. In regions where the rays intersect, or in
the far field as in the preceding example, the geometrical optics solution is
not valid and a modified expansion or result is needed. In the following
example we show how to construct a valid asymptotic result near a two-
dimensional caustic where the geometrical optics result fails. The boundary
layer method will be used.
Example 10.4. The Asymptotic Expansion at a Caustic. We consider a
two-dimensional problem for the reduced wave equation A0.1) with n = 1
730 ASYMPTOTIC METHODS
and an asymptotic solution of the form A0.7) and A0.11). The rays for this
solution are assumed to have a smooth envelope. The envelope is a caustic
curve for this problem and the asymptotic solution breaks down on the
caustic, as we have seen. We retain the notation introduced in our discussion
of this problem [see A0.98)-A0.102)].
The equation of the caustic is assumed to be x = R(£) where £ is arc
length on the curve. The rays for this problem are the tangent lines to the
caustic and are given as
A0.153) x = r(f,<r) = R«) + (<r-f)R'(f),
with a equal to arc length on the rays, as was shown following equation
A0.98). The phase of the wave is given as <p = a and, in view of A0.102),
the leading term in the asymptotic result is
A0.154) „„v0(ob)[^r?
1/2
Jka
From A0.153) we see that when a— £, the rays intersect the caustic and
A0.154) shows that the asymptotic result is infinite on the caustic. We
assume that cr, as well as сг0, in A0.154) is less than £ and that V0((t0) is a
given function. Thus A0.154) represents a wave approaching the caustic
x = R(£) whose amplitude becomes infinite there.
Although the asymptotic field is singular at the caustic, the actual field is
finite there. After the incoming rays pass the caustic they become outgoing
rays, and the resulting outgoing field is well behaved, and we must de-
determine its geometrical optics representation. In addition, although no rays
penetrate to the concave side of the caustic in Figure 10.3 so that the
geometrical optics field must vanish in that region, the actual field is, in fact,
nonzero there. We wish to^ study the field at and near the caustic and to
determine the transition undergone by the geometrical optics field on its
passage through the caustic region, that is, how its amplitude and phase
terms change.
Figure 10.3. The caustic curve.
EQUATIONS WITH A LARGE PARAMETER 731
To study the field near the caustic x = R(£) we express the reduced wave
equation in (£, &) coordinates as defined by A0.153). The coordinate
system is orthogonal and we have from A0.99)
A0.155) h LlZy л =i,
* P
where p = p(£) is the radius of curvature of the caustic. We are presently
only considering values of cr for which cr < £, since we would otherwise have
a double covering of the region above the caustic curve. That is, the full
tangent lines cover the region twice, so we restrict ourselves only to
half-lines.
The reduced wave equation takes the form
A0.156) V u + ku = T^rJo
in £, cr coordinates. Let
A0.157) и = weikcr
and insert A0.157) into A0.156). We are retaining the phase term of the
geometrical optics solution A0.154) in the expression A0.157) since it is the
amplitude term of A0.154) that blows up at the caustic. We obtain for w,
A0.158) J2^--!- 7f
L da- £-(T J {; — <тд(т
p д Г p ^1_0
£-a dZlZ-a dO
To emphasize the neighborhood of the caustic we set
(Ю.159) t-* = lc-% f = A,
where r>0 is to be determined. Since к is assumed to be large, the
stretching transformation A0.159) implies that small values of g - a are to
be considered. Now dld£ = dldk + k'dldi) and д1дсг = -кгд1дт), so that
A0^158) becomes
df\ V J 77 dr] L ' ^77
732 ASYMPTOTIC METHODS
In assessing the significance of the terms in A0.160), we note that the
coefficient of k1+r should be retained, since in the absence of the stretching
transformation it is this term that yields the geometrical optics amplitude
results in A0.154). In addition, it is the absence of second derivative terms
in the transport equations that invalidates the geometrical optics result at
the caustic. Therefore, we must retain at least one second derivative term in
A0.160). In effect, we have a singular perturbation problem in A0.158) and
it is the neglect of the higher derivative terms that leads to difficulties in the
geometrical optics expansion. The highest power of к that occurs in the
second derivative terms in A0.160) comes from kA\ Thus we choose r so
that /c1+r and kAr are equally significant. The yields
A0.161) l + r = 4r
so that r = \.
We set r = \ in A0.160) and expand w in a series in powers of k~in. With
w0 as the leading term in the series, we find that vv0 satisfies the boundary
layer equation
This ordinary differential equation can be simplified by setting
A0.163) n>o
where
A0.164)
and W(z) satisfies the equation
A0.165) rB) + 2f(z) = 0.
With z replaced by -z, A0.165) becomes the Airy equation which has
two linearly independent solutions Ai(z) and Bi(z), some of whose prop-
properties we now present. The Airy function Ai(z) decays exponentially as
z—»+а>? whereas Bi(z) grows exponentially as z increases. Both functions
are oscillatory for z < 0. The asymptotic behavior of the Airy function Ai(z)
for complex valued z expressed as z = \z\el$, with \z\ > 0 and -it ^ в ^ ^ *s
given as
EQUATIONS WITH A LARGE PARAMETER 733
A0.166) ^(z)-^-(z)-i
and as z-* -oo we have
A0.167) ^(z)«^(-z)-1
From the functional relations
A0.168) Ai(z) = -(oAi(a>z) - a>2Ai(a>2z) ,
A0.169) Bi(z) = iw>l/(wz) - fa>2Ai(a>2z) ,
where a> = ехр(-2тп/3), we can determine the asymptotic behavior of the
solutions Bi(z), Ai(<oz), and Ai(a>2z) of the Airy equation in terms of the
results A0.166), A0.167) for Ai(z).
For the equation A0.165) for W(z), a general solution is given as
A0.170) W(z) = ct(€)Ai(-z) + c2(t)Bi(-z) ,
where the functions q and c2 must be determined by matching the boundary
layer result woelk<r with the geometrical optics result A0.154). It is easily
seen from A0.166) that the function Ai{—<oz) has the asymptotic behavior
A0.171)
as z->oo, and A0.168), A0.169) show that
A0.172) AHr»z) = ^ [Bi(-2)
Putting Cj = -ic2 in A0.170), we express W(z) as
(Ю.173) WB)
where c(|) is to be specified. As we now demonstrate, this choice for W(z)
is appropriate for matching the boundary layer and geometrical optics
solution.
hk the (|, a) variables the boundary layer solution u0 = woeika takes the
form
A0.174) u0^,a)-
734 ASYMPTOTIC METHODS
To match A0.174) with the geometrical optics result A0.154) we expand the
Airy function asymptotically for f - cr>0 and к large. In view of A0.171)
we obtain
Comparing with A0.154) shows that c(£) must be chosen as
A0.176) cU) = *"VW25 V1/3V0K)(| - <r0) exp(|) ,
and thereby the boundary layer term щ given in A0.174) is completely
specified.
This result shows that the field is not infinite at the caustic as predicted by
the geometrical optics result. Instead, we see from A0.174) and A0.176)
that the field has a finite value at the caustic, that is, when £ = a. However,
the amplitude of the geometrical optics term is 0A) in к away from the
caustic, whereas at and near the caustic, the boundary layer term has an
amplitude that is 0(fc1/6). Since к is assumed to be large, we do find that
there is a growth in the amplitude of the field at the caustic, but it does not
become infinite there.
The preceding results are not completely satisfactory for the following
reasons. First, they predict that the field below the caustic (i.e., in the
region not penetrated by the rays) grows without bound as we show. The
field below the caustic is expected to be weaker than the geometrical optics
field. Thus it should decay rather than grow below the caustic. Second, it is
not apparent from our results what role is played by the outgoing geometri-
geometrical optics field that results after the incident wave passes through the caustic
and how to determine that field. We now proceed to modify these results so
as to remove these difficulties.
To begin, we show that the boundary layer solution A0.174) grows
without bound below the caustic. To do so, an approximate expression for
the distance along the normal lines to the caustic valid in the boundary layer
region is found. The wave fronts associated with the system of rays A0.153)
are the curves a = constant. Since the wave fronts are orthogonal to the rays
and the rays are tangent to the caustic, we see that the wave fronts are
orthogonal to the caustic. Now distance along the wave front a = constant is
given as ds = (|£ - <r\lp) d£. The boundary layer region is determined by
£ = a + O(k~1/3) in view of A0.159). On integrating the previous expres-
expression for ds on the wave front a = constant in the boundary layer region we
obtain
A0.177)
EQUATIONS WITH A LARGE PARAMETER 735
where A0.159) and A0.164) are used. Within the bounday layer, the
variable s yields a measure of distance along the normal line to the caustic,
positive values of s correspond to points above the caustic (i.e., in the ray
region). Negative values of s correspond to points below the caustic.
Using A0.177) we express the boundary layer solution A0.174) in terms
of the variables £ and s. To leading order we have
A0.178) Mo - [
where we have set a = £ + O(k'l/3) in the boundary layer. If A0.178) is
evaluated for negative values of s in the region below the caustic, we find
that k2/3s becomes large and negative. Then the asymptotic result A0.166)
for the Airy function in A0.178) shows that \uo\ grows exponentially as
k2/3s-*-oo.
This difficulty may be resolved by considering the behavior of not only
the incident field, but also of the outgoing field near the caustic. The rays for
both the incident and outgoing fields are given by A0.153). On a ray of the
incident field we have a < g, whereas on a ray of the outgoing field we have
a > f. At the caustic both rays meet and <r = £.
The leading term of the geometrical optics form of the incident field и is
given in A0.154). We denote the outgoing field by й. Since the ray structure
is the same for both fields, the leading term in the geometrical optics
representation of й must have the form
A0.179)
v-l/2 ika
The difference between A0.179) and A0.154) lies in the fact that <r> £ in
A0.179), whereas <г<€ in A0.154). Further, the term %Ц) is as yet
unspecified. Both A0.179) and A0.154) are singular on the caustic (i.e.,
when o- = £).
The total field at a point (jc, y) above the caustic is a sum of the incident
and outgoing fields and is given as и + м. We must note that even though the
incident and outgoing fields are expressed in terms of the variables £ and <r,
at each point (x, y) these fields are evaluated at two different sets of values
of £ and o\ The determination of these values is indicated in Figure 10.4.
To determine V0(f) in A0.179) and to resolve the difficulty concerning
the behavior of the field below the caustic, we must consider the behavior of
the total field и + й near the caustic. Now the boundary layer form of the
incident field и was already given in A0.175), A0.176). To study the
boundary layer behavior of й we may carry over the preceding boundary
layer results directly. The only difference is that the magnification element
h€ has the form h^ = (<r - £)/p if a > £ and is not given as in A0.155).
However, A0.156) shows that the form of the reduced wave equation in
the (£, a) variables is unchanged if £ - a is replaced by a - £. Thus every
736 ASYMPTOTIC METHODS
(x,y)
Figure 10.4. The incoming and outgoing rays.
result obtained up to A0.165) remains valid for the outgoing field. There-
Therefore, if we define the leading order boundary layer term to be w0, we find
that
A0.180) *o
where 17,2, and W are given as in A0.159), A0.164), and A0.165),
respectively.
Noting the asymptotic result A0.166), we find that
A0.181)
f
as z-* 00. If we then set
A0.182) W(z) =
in A0.180), we obtain for u0 = wQeik<T,
A0.183)
й0Ц, a) = e(t;)Ai[(^p)V\t - aJ] exp{ik[* + i^
To specify c(f) we expand the Airy function in A0.183) asymptotically for
<r - \ > 0 and к large. Since £ - a occurs in the Airy function in squared
form we must use the asymptotic form A0.181) valid for large positive z.
Thereby, the matching between the boundary layer result й0 and the
geometrical optics result w is carried out.
The variable z that occurs in A0.180)-A0.182) is defined in A0.164) and
we have
EQUATIONS WITH A LARGE PARAMETER 737
A0.184) fz^JL^^-JL^-^,
since £-<r<0 so that |f - cr\ = -(f- <*■)• This result shows whY Jt is
appropriate to choose the Airy function Ai[-a>2z] for this case (i.e., the
outgoing field), for the asymptotic form of мо(£, a) is then found to be
A0.185) uo(£, a)
Comparing A0.185) with the geometrical optics result A0.179) yields
A0.186) c(£) = kl'6V*25l6p-ll3V0W exp( - g) .
Although A0.186) expresses c(£) in terms of V0(g), we still do not know
what form Vo has. We expect that V0(i) must be expressed in terms of the
(given) incident field. In order to establish this relationship we examine the
total field м0 + й0 (to leading order) in the boundary layer region. To do so
it is not convenient to use the (£, cr) variables since they have different
values for u0 and й0 as we have indicated. Instead, we use the variables £
and 5, where s measures distance along the normal from the caustic in the
boundary layer region. To the level of approximation used, it follows from
A0.177) and the discussion preceding it, that the incident and outgoing
fields are both evaluated at the same point (f, s). Thus (to leading order)
the total boundary layer field u0 + u0 is given as,
A0.187) «o + Bo-*1
We require that u0 + w0 decay in the region below that caustic, that is, when
s<0 in A0.187). In view of the asymptotic result A0.166), this implies that
the sum of the Airy functions in the bracketed term of A0.187) add up to a
muhiple of Ai[-Btp)lf3k?3s]9 for Ai(-z) decays exponentially as z^> -«,
whereas Ai{-a>z), Ai(-a>2z), and Bi(-z) all grow exponentially as z-> -».
Using A0.168) with z replaced by -z and noting that Ai[-a>z] =
Ai[exp(i<rr/3)z] and Ai[-w>2z] = Л/[ехр(-/тг/3)г], we easily conclude that
*() must be chosen such that
738 ASYMPTOTIC METHODS
A0.188) V(t) exp( - g) = a>V0(v0)(i - <г0) exp(g) .
Then A0.187) becomes
A0.189) u0 + u0- *1/вл/5пГ"/423 V1/sVo(<»o)(* - <roI/2
and the field decays below the caustic when k2/3s gets large and negative
Requiring that the field must decay below the caustic has enabled us to
determine Vo(() as given in A0.188). Inserting the result for V0(f) into
A0.179) yields the outgoing geometrical optics field
A0.190) U = V0(a-0)f i^l exp(*far - Щ) ,
with a > £. To leading order, the incident and outgoing geometrical optics
field have an amplitude term |<r-£|~1/2 which is a consequence of the
convergence and divergence of the rays as the wave approaches and leaves
the caustic. However, on comparing the phase terms in A0.154) and
A0.190), we find that the incident wave undergoes a phase shift by an
amount of тг/2 as it passes through the caustic and becomes the outgoing
wave.
It has been noted in the preceding example that according to geometrical
optics, the field in the region below the caustic must be zero since no rays
penetrate into that region. However, the boundary layer result shows that
there is a nonzero field below the caustic that decays exponentially. In the
theory of optics when there are nonzero fields in regions where geometrical
optics predicts that the field must be zero, we attribute these effects to the
process of diffraction of light. In the following two examples we consider
scattering problems in which diffraction effects play a significant role.
Example 10.5. Scattering by a Half-Plane. We assume that the plane
wave m7(jc, y, z) = elkz is incident upon the half-plane z = 0, у ^ 0. The
resulting field и = м7 + us where us is the scattered wave satisfies the reduced
wave equation A0.1) with a unit index of refraction. The boundary condition
is that the total field u(x,y,z) vanishes at z = 0, y<0 (i.e., on the
half-plane). Thus
A0.191)
A0.192)
u(x,
V2u
y,0)
+ k2u
= 0,
= 0,
y<0.
Following the general procedure as given, we expand us as
EQUATIONS WITH A LARGE PARAMETER 739
A0.193) и,-*1" 2 »,(*)"'.
On the boundary surface z = 0, у < 0, the initial condition for <p(x, y, z) is
A0.194) (p(xyy,0) = 0,
and the initial conditions for the vf are
A0.195) vo(x, y,0) = -l,
A0.196) Vj(x9 y9 0) = 0, у < 0, / > 1.
These conditions follow from the fact that the incident phase term ф = z
vanishes at z = 0, whereas the leading order amplitude term w0 = 1 is the
only nonzero term in the (asymptotic) expansion of the incident field ur We
remark that this problem could be treated as being strictly two-dimensional
in terms of the y, z variables, but we prefer to deal with it in three-
dimensional form.
To determine the phase term <p(x> у, z) of us, we require that <p satisfy
the outgoing condition at the boundary z=0, in addition to A0.194).
Clearly, the two solutions of the eiconal equation (V^>J = 1 that vanish at
z = 0 are <p+ = z and <p_ = —z. By requiring that us be an outgoing wave at
the boundary, we determine the appropriate choice for <p. To apply this
condition we must distinguish between the two sides of the half-plane z = 0,
у ^ 0. Accordingly, we express the scattered wave us in the form us =
v±elkip±. On the side of the half-plane facing the half-space z<0 the
appropriate choice for <p is <p = <p_ = -z, since the wave us = v_e~lkz travels
away from the half-plane into the half-space z < 0. On the opposite side of
the half-plane facing the region z > 0, we must set <p = <p+ == z, since
us = v+e travels towards the half-space z>0 away from the half-plane
z = 0.
Since the scattered field in the region z<0, y<0 is a plane wave, we
immediately conclude from the initial conditions A0.195), A0.196) for the
amplitude terms that vo(x9 yy z) = uo(jc, y,0) = -1 and vf{xy y, z) =
vj(x> У у 0) = 0 for all jf a 1. Consequently, the scattered wave in this region is
given as
(Ю.197) Us(x, у, z) = -e~ik\ z < 0, у < 0 .
In the region z > 0, у s 0, ms is also a plane wave and A0.195), A0.196)
imply that vo(x, y,z)=-\ and Vj(x, y,z) = 0 for у а 1. Thus the scattered
wave in this region is given as
A0.198) us{x, y,z) = - eik\
740
ASYMPTOTIC METHODS
There is no boundary surface in the half-space у > 0, and the scattered
waves constructed previously do not penetrate into the region since the rays
are orthogonal to the half-plane z = 0, у < 0. Thus the total field и = и + u
predicted by the geometrical optics approach is s
A0.199)
u(x, y, z) ■•
eikz;
eikz -
-oo<z<
e~ik\ z<0,
z>0,
», y>0
y<0
y<0
As shown in Figure 10.5, the plane у = 0 is a surface of discontinuity for
the geometrical optics solution A0.199). Even though each of the repre-
representations of и given in A0.199) is an exact solution of the reduced wave
equation A0.191) in the interior of the three indicated regions, the full
expression for и is not a regular solution of A0.191). The half-plane у = 0,
z > 0 is called a shadow boundary since it separates the (shadow) region
z > 0, у < 0 where no (incident) rays penetrate, from the region z > 0, у > 0
where the incident field acts. In a similar fashion we refer to the half-plane
У = 0, z < 0 as a reflection boundary. It separates the region where the
incident and reflected field are observed from the region where there is only
the incident field.
In order to obtain a smooth transition for the asymptotic solution of the
given scattering problem across the discontinuity boundaries, we make use
of the boundary layer theory. We consider the boundaries at у - 0 with z < 0
and z > 0 separately.
ЩЛ<Щх?иReflection boundary : _
Figure 10.5. The geometrical optics solution.
EQUATIONS WITH A LARGE PARAMETER 741
At the shadow boundary y = 0, z > 0, the field on one side is и = 0
whereas the field on the other side is и = eikz. Thus we set
A0.200) и = weikz
in A0.191) and obtain
A0.201) 2ikwz + wxx + wyy + wzz = 0.
To emphasize the neighborhood of у - 0, we introduce the stretching
transformation
A0.202) у = k~rt],
with the exponent r>0 to be determined. In the jc, 17, 2 variables, A0.201)
becomes
A0.203) 2ikwz + fc2^ + w^ + wzz = 0 .
To ensure that the original leading term 2ikwz remains as significant as a
second derivative term in A0.203), we must set r = \. Then
A0.204) kBiwz + wvv) + ^ + wzz = 0 ,
and the equation for the leading order boundary layer term w0 is
This parabolic Schrodinger equation has already been encountered in Exam-
Example 10.3 [see A0.146)]. We require that for large values of 7/ = kll2y the
solution w0 of A0.205) match with the geometrical optics result A0.199).
We rewrite A0.205) in the original у and z variables and obtain
A0.206) ^ ^
dz dy2
Noting the results A0.147), A0.148) of Example 10.3, we consider a general
solution of A0.206) in the form
where f(s) is to be specified by means of the matching process. Since к is
b, we evaluate A0.207) asymptotically using the method of stationary
742 ASYMPTOTIC METHODS
phase as in Example 10.3. There is exactly one stationary point at s = у and
in view of A0.149) we have the asymptotic result
A0.208) wo
so that
A0.209) woeikz ~ V2^/( y) exp(/Jfcz + i J) .
The expression A0.209) must be matched with the geometrical optics
field A0.199) in the regions z>0, y>0 and z>0, y<0. Since the
geometrical optics field vanishes in the (shadow) region z > 0, у < 0 we must
set f(y) = 0 for у < 0. For z >0, у > 0, the geometrical optics field equals м7
so that /(y) = A/V2tt) ехр(-Ьг/4) for у >0. Thus
A0.210)
У<0,
and the boundary layer result w0 becomes
With the change of variables
A0.212) "
A0.211) takes the form
A0.213) ".-•/а
where the Fresnel integral F(i) is defined as
A0.214)
The field near the shadow boundary у = 0, z > 0 is thus given as
A0.215) и « tt"i/2f( - yy]—J cxpyikz - ™) ,
to leading order.
Since F(-oo) = Vtt ехр(/тг/4) and F(+oo) = 0s we find that wo-*0 as
z -^ о with у < 0, whereas w0-* 1 as z -> 0 with у > 0. Thus wo(z, y) as given
EQUATIONS WITH A LARGE PARAMETER 743
in A0.213) is, in fact, a solution of the parabolic equation A0.206) with the
discontinuous initial data w@y у) = 0 for у <0 and w@, y) = l for у >0.
Had we applied the parabolic equation method of Example 9.12 to this
problem, we would have been led to consider A0.206) with the aformen-
tioned data. Further, F@) = \y/H exp(*V/4), so that the amplitude term w0
equals \ on the shadow boundary у = 0, z > 0. It equals the average of the
amplitude terms of the field in the shadow and illuminated regions, respec-
respectively.
The matching procedure carried out only considered the leading term in
the asymptotic expansion of the boundary layer term w0. On evaluating the
representation A0.211) of w0 by the method of stationary phase, we find
that there is only one stationary point located at s = y. Since the domain of
integration is 0<s<°o, there is no stationary point in the region y<0,
z > 0. In the region у ^ 0, z > 0, the stationary point coincides with the end
point s = 0 of the integration interval if the value of у equals zero.
Otherwise, the stationary point s = у and the end point 5 = 0 are distinct.
We now apply the stationary phase results of Example 5.14 to the integral
A0.211) for w0. If у < 0, there is no stationary point in the given interval of
integration. Thus the main asymptotic contribution to A0.211) comes from
the endpoint 5 — 0. In view of E.257) we have
If у >0, the integral A0.211) has a stationary point contribution from the
point s = у y as well as an endpoint contribution from 5 = 0. The stationary
point result was found to be w0 « 1, whereas the endpoint result is identical
to A0.216). Thus
We note that both A0.216) and A0.217) are singular at у = 0 as well as at
Since wQ is valid only in the region near the shadow boundary z>0,
У = 0, we conclude from A0.216), A0.217) that the total field u(x, y, z) has
the form
A0.718)
744
ASYMPTOTIC METHODS
near the edge of the shadow boundary. On comparing A0.218) with the
geometrical optics result A0.199), we observe that there is no counterpart to
the O(k~in) term of A0.218) in A0.199). In fact, the geometrical optics
result was expressed as an (asymptotic) series in integral powers of k~\ Thus
terms of order k~112 could not arise in that series. These additional terms
are due to diffraction effects. In the preceding example, the field due to
diffraction below the caustic was found to decay exponentially. In this
problem, the diffraction field is smaller than the geometrical optics field by a
multiplicative factor of O(k~l/2). Since it is not exponentially small, we
must account for it in our discussion of the scattering problem.
An asymptotic method closely related to the geometrical optics approach
and accounting for the effects of diffraction is J. B. Keller's geometrical
theory of diffraction. In the context of the present problem this theory
introduces a set of diffracted rays that emanate from the edge of the
boundary surface (i.e., the half-plane z = 0, y^O). The incident rays, in
addition to giving rise to the reflected rays in the region z<0, y<0,
generate a set of diffracted rays. Each incident ray that hits the edge yields a
family of edge diffracted rays that emanate from the edge as shown in Figure
10.6. These rays correspond to those of a cylindrical wave with a focal point
at the edge. If we set
the resulting diffracted field uD-~which supplements the geometrical optics
field—is shown in the geometrical theory of diffraction to have the form
Sl
Diffracted rays|||l|
Incident
ray
Figure 10.6. Edge diffracted rays.
Edge
EQUATIONS WITH A LARGE PARAMETER 745
A0.219) uD(x, y,z)~-j= u,(x90,0)eikr.
Here D is known as a diffraction coefficient, and щ(ху 0,0) is the value of
the incident field at the point of diffraction (x, 0,0) on the edge of the
half-plane.
The diffraction coefficient D must be determined either by solving a
canonical problem or by using boundary layer theory near the edge. In
general, a canonical problem is one in which the local geometry and other
properties are the same as those for the given problem near the point of
diffraction. Using either of these methods leads one to consider the problem
of the scattering of a plane wave by a half-plane with the same boundary
condition as for our original given problem. That is, the canonical problem
is identical to the given problem in the present case. (In general, of course,
the canonical problem is much simpler than the given one.) The exact
solution of our problem was found by Sommerfeld. On expanding hisjresult
asymptotically for large kr it is found that the diffraction coefficient D must
be chosen as
where 0 is the angle that the particular diffracted ray—when projected onto
the y, z-plane—makes with the negative z-axis. Also, Uj(x, 0,0) = 1 for all
values of x.
The terms of order к'1П in A0.218) must be identified with the diffracted
field uD in the region near the shadow boundary. To effect this comparison
more easily we introduce the cylindrical coordinate system
A0.221) x = x, y = rcos0, z = rsin0.
The angles 0 of A0.220) and 0 of A0.221) are related by
A0.222) * = e+f •
For small у with z > 0, we have
A0.223)
Usiag A0.221) in A0.223) we easily conclude that we must have
A0.224) sin 0 ~ 1, cos 0 «0
if A0.223) is valid. Also,
746 ASYMPTOTIC METHODS
A0.225) 1 f sec \ в + esc \ в] = ""&*?) + *>&/2)
2 L 2 2 J 2sin@/2)cos@/2)
_ cos(fl/2) + sin(fl/2)
sine
Noting A0.221)-A0.224) we have
1/2
since cos 0 = sin 0 and sin 0 «1. Further, A0.224) implies that в « тг/2 so
that 0«^ Consequently, cos@/2)«O and sin@/2)«l and, in view of
A0.225), D has the approximate form
A0.227)
Inserting A0.227) into A0.219) we see that A0.226) and the diffracted field
uD agree in the neighborhood of the shadow boundary since w7(jc, 0,0) = 1.
The field near the reflection boundary is considered in the exercises.
Example 10.6. Scattering by a Circular Cylinder. A plane wave u} = el x
is incident upon a circular cylinder of radius a, whose axis coincides with the
z-axis. The total field и = w7 4- w5where us is the scattered wave satisfies the
reduced wave equation A0.1)—with n = 1—and vanishes on the surface of
the cylinder. This problem can clearly be formulated as a two-dimensional
one.
Thus we consider the function и = u(x, y) that satisfies the reduced wave
equation
A0.228) uxx + uyy + k2u = 0
in the exterior of the circle x2 + y2 = a2. On the circle we have
A0.229) u(x,y) = 0, x2 + y2 = a2. ♦
With и = м7 + us where w7 = etkx, we require that the scattered field us
satisfies the radiation condition at infinity.
EQUATIONS WITH A LARGE PARAMETER 747
Assuming к is large, we determine an asymptotic solution for us in the
manner described. The scattered field us is expanded as
A0.230) «,(*, У) = e*9 2 vjii
On the circle the phase term (p(x, y) satisfies the (initial) condition
A0.231) ф,у) = х, x2 + y2 = a2
as well as an outgoing condition. The amplitude terms vf(x, y) satisfy the
initial conditions
A0.232) vo(x, y) = -1, x2 + / = a ,
A0.233) u,(x,,y) = 0, x2 + y2 = a2, />1.
To solve for <p we represent the circle in parametric form as
A0.234) x - a cos т, у - a sin r, 0 < т < 2тг .
We note that r is not an arc length parameter unless a - 1. With a equal to
arc length along the rays we must solve the characteristic equations for
x((r, r), y(<r9 т), <р((т, r), p(<T, r), and q(<r, r) in order to determine <p(x, y).
The initial conditions for x and y, which are given at <r = 0, are
A0.235) д:@, т) = a cos r, y@, r) = a sin r .
From A0.231) we have
A0.236) <p@, r) = x@, r) = a cos r .
The initial conditions for /? and ^r must be found from the eiconal
equation and strip condition as in Section 2.4. We have from the eiconal
equation
(Ю.237) />2@,t) + <?2@,t) = 1,
and the strip condition
A0.238) -a sin r = -p@, т)а sin r + #@, r)a cos r .
On solving A0.237), A0.238) we easily conclude that there are two possible
sets of solutions. They are
748
A0.239)
and
A0.240)
P@,
P(
r) =
:o,r)-i,
-cos 2t,
ASYI
9@,t)-0
q@9 t) = -sin 2t .
ASYMPTOTIC METHODS
The appropriate solutions must be determined from the outgoing condition.
This states that if N is the exterior unit normal to the circle, then V<p -N =
[p, q] • N > 0. The normal N for x2 + y2 = a2 is N = [cos т, sin т]. Thus
A0.241) V<p • N = [p, q] • N = cos т
if A0.239) is used, and
A0.242) V<p • N - [p, ?] • N = -cos т
if A0.240) is used. Therefore, since cost>0 for 0<т<тг/2 and Зтг/2<
т < 2тг and cos т < 0 for тг/2 < т < Зтг/2, we see that V<p • N > 0 on the left
semicircle (i.e., тг/2<т<Зтт/2) if A0.240) is chosen, whereas V<p-N>0
on the right semicircle if A0.239) is selected. At the points @, a) and
@, —a) where т = тг/2 and т = Зтг/2, respectively, the rays of the scattered
wave are tangent to the circle.
It has been shown that on the right semicircle (i.e., x2 + y2 = a2 with
x>0) we must set p@, т) = 1 and #@, т) = 0. Since the characteristic
equations are
A0.243)
dx dy d(p dp dq
we see that p(a, r) = 1 and q(a9 r) = 0. Then A0.235), A0.236) imply that
A0.244) jc(<7, т) = o- + a cos т, у (о-, т) = a sin т ,
A0.245) <р(а, т) = о- + a cos т = jc.
Thus the rays of the scattered wave that issue from the right semicircle are
identical to those of the incident wave. The phase term <p equals x, the phase
of the incident wave. Further, since V2<p = 0, we conclude that
A0.246) vo(x9 y) = -1, v,(x, y)~0, j^l
from the transport equations A0.66), A0.67) and the initial conditions
A0.232), A0.233). Therefore, the scatteredv|iel^/M5 is given as
A0.247) us(x, y) = -eik\ x2 + y2 > a\ x > 0, \y\ < a ,
EQUATIONS WITH A LARGE PARAMETER 749
in the region directly behind the circle where no incident rays penetrate.
The total field и = w7 + us in that region is
A0.248) u(x9 y) = eikx - ёкх = 0, x2 + / ^ a\x > 0, |y\ < a .
The geometrical optics approach thus predicts a zero total field or, equival-
ently, a shadow in the region behind the circle not illuminated by the
incident rays.
On the left semicircle (i.e., x2 + y2 = a2 with x<0) we must have
p@, T) = -cos2t and q@, r) = -sin 2т, as has been shown. Since p and q
are constant along the characteristics they retain their initial values along
these curves. Consequently, the equations for x and у are
A0.249) ~a\r~p = ~cos 2r' ~a\r = q = ~sin 2r '
with ^/2^t^3it/2. Noting the initial conditions A0.235), we find that
A0.250) x(<r, r)- a cost - a cos 2т, >>(сг, т) = a sin т - tr sin 2т .
Also, from d<p I da = 1 and A0.236) we conclude that
A0.251) <p(a, t) - o- + a cosт, f~T-T"*
It is not possible to express a and т as functions of x and у in simple form.
Therefore, we leave our result for the phase in parametric form.
The rays of the scattered or reflected field us are determined from
A0.250), and their direction is given by A0.249). The direction of the
incident rays is that of the unit vector i. The exterior unit normal vector to
the circle is N = [cos т, sin т]. It is a simple matter to show that the angle of
incidence ax of the incident ray equals the angle of reflection aR of the
reflected ray. These angles are defined in Figure 10.7. In fact, we have
a/=a* = k-T| with 1г/2<т<Зтг/2. When t=it/2 and т = 3*г/2, the
reflected rays coincide with the incident rays. That is, the incident rays are
tangent to the circle at the points @, a) and @, -a) and their extension
beyond these points into the region x > 0 coincides with the reflected rays.
We note that the equality of the angles of incidence and reflection is a
general principle of geometrical optics. The region covered by the reflected
and the incident rays is given as x + y2 > a2, with -oo<y<ooforx<0 and
\y\ ^ a for x > 0. This region is referred to as the illuminated region.
$e now proceed to determine the leading order amplitude term u0 of the
reflected wave us. To do so we express the transport equation for v0, which
is given as
A0.252) 2§ + l,ovV = 0>
750
ASYMPTOTIC METHODS
Figure 10.7. Angles of incidence and reflection.
in the ray coordinates a and т determined from A0.250). The Laplacian V2
must be expressed in the ray coordinate system so that V2<p can be
evaluated. It is readily found that
Inserting A0.253) into A0.252) and using the initial condition vo(O, r) = -1,
we obtain
7Г 3ir
<<
,„лл^ч , ч / (all) cos т
A0.254) Uo(or,T) = -^J-_, 2
as is easily verified. Thus the scattered field is given as
A0.255)
/ {all) cos т
"До-, т) - V (а/2) cos т -
2<T<T'
to leading order.
The curve
A0.256)
a = ^ c°s т, — — T
Зтг
in the envelope of the reflected ray system A0.250,, as is easily verified by
evaluating the Jacobian of the transformation A0.?50) or, equivalently, by
using the results in the Appendix of Chapter 2. When тг/2< т<Зтг/2,
EQUATIONS WITH A LARGE PARAMETER 751
cost<0 so that a is negative in A0.256). Therefore, the envelope lies
within the circle x2 + y2 = a2 for those values of r, since a > 0 corresponds
to points on or outside the circle. However, if т = тг/2 or т = Зтг/2, we have
cos тг/2 = cos Зтг/2 = 0, so that a = 0 there. Consequently, the envelope or
caustic curve A0.256) intersects the circle at the points @, a) and @, -a). It
is for this reason that us is not well defined at т = тг/2 and т = Зтг/2.
The asymptotic (geometrical optics) result we have obtained is the
following. In the illuminated region the solution is expressed in the ray
coordinates a and r as
A0.257) u(a, r) ~ exp[*'A:(a cos т - о- cos 2т)]
-v
(а/2) cost
exp[ik(cr + a cos t)J ,
(a/2) cos т - a
and this is valid for тг/2<т<Зтг/2 and o->0. In the shadow region we
have
A0.258) m(jc, H«O, x2 + />a2, jc>0, |y|<a.
The lines y = ±a, jc^O are the shadow boundaries that separate the
illuminated region from the shadow region. The asymptotic solution is not
continuous across these lines.
The field near the shadow boundaries may be studied by the use of the
boundary layer theory. The parabolic equation A0.205) obtained in the
preceding problem may be derived. However, the solution must be matched
not only with the incident and reflected waves in the illuminated region and
the zero field in the shadow region. The field in the shadow boundary must
also be matched with the field near the points of diffraction @, a) and
@, -a). The field near these points may be analyzed by deriving an
equation of the form (9.381) by the use of boundary layer methods. The
problem is rather complicated and has undergone much investigation. We
note that the geometrical theory of diffraction introduces surface diffracted
rays which yield an additional nonzero but exponentially small field in the
shadow and illuminated regions. However, we do not pursue these matters
here any further.
We have seen in Example 10.3 that the geometrical optics expansion
associated with a general plane wave is not valid over large distances
because of the occurrence of secular effects. In the following example we
consider the propagation of a narrow beam of light, as might be generated
by a laser. The analysis of this problem by means of a geometrical optics
expansion gives rise to difficulties similar to those encountered in Example
Ю.З and they are resolved in a similar manner.
752 ASYMPTOTIC METHODS
Example 10.7. The Propagation of a Gaussian Beam. We consider the
propagation of a (light) beam in the two-dimensional case. The beam
propagates in the direction of the positive x-axis and its amplitude is
concentrated near the x-axis. The function и = u(x, y) that represents the
beam is a solution of the reduced wave equation A0.1) where we set n = l
We express и in the form A0.7) and A0.11) assume that at x = 0, the phase
term <p and the amplitude term v are given as
A0.259) *@, y) = 0, v@, y) = A exp( - Q ,
where A and a are positive constants. The solution is determined in the
half-plane x>0, so that A0.259) represents a boundary condition for the
beam field u. The solution is required to be outgoing at x = 0.
We do not consider how the beam is formed in the half-plane x < 0, but
assume that it has the phase and amplitude given in A0.259) when it reaches
the y-axis. Since the amplitude is in the form of a Gaussian distribution (see
Section 1.1) we refer to it as a Gaussian beam. The beam half-width is a
measure of the distance from the x-axis within which the amplitude of the
beam is concentrated and it equals V2a on the y-axis.
Now <p(x, y) is a solution of the eiconal equation A0.9) with n = 1. The
initial condition A0.259) together with the outgoing condition yields the
solution <p(x9 y) = jc. Thus, и = uetkx and v satisfies
A0.260) 2ikvx + vxx + vyy = 0.
On expanding v(x, y) as in A0.11), we find that v0 satisfies the boundary
condition A0.259), while the remaining Vj all vanish at x = 0. On solving for
v0 and vt we obtain
A0.261) 0 - v0 + 1 Vl = A exp( - 01 - ^ g
Since the amplitude term is concentrated near у = 0 we determine that
the asymptotic expansion of v(xy y) begins to be disordered when
A0.262) x~2ka.
At this value of x the second term in the expansion of v has the same
magnitude as the leading term, when у = 0. The geometrical optics descrip-
description A0.261) of the beam is not valid as x reaches this value and diffraction
effects begin to play a role. Consequently, the distance x = 2ka is referred to
as a diffraction length.
The geometrical optics representation of the beam is a general plane
wave. To study the solution in the secular ^6gion where x ^ O(ka) we
EQUATIONS WITH A LARGE PARAMETER 753
proceed as in Example 10.3. The results leading from A0.144) to A0.149)
apply to the present problem as well and the expression in A0.149) is
equated to the Gaussian amplitude term in A0.259). This yields
00-263) ^
The integral can be evaluated explicitly by completing the square in the
exponential term and introducing a complex valued change of variables that
reduces the integral to the form given in Exercise 1.1.2. (The transformation
may be carried out formally and the resulting expression can then be shown
to satisfy the conditions of the problem.) The Gaussian beam representation
is thus given as
A0.264) ^ [
If we express A0.264) in the form и = velk<p with <p = x, we find that it
satisfies A0.259) at x = 0. Further, v(x, y) is a solution of the paraxial wave
equation
A0.265) 2ikux + v,y = 0,
which is, in fact, the parabolic equation approximation to A0.260). The
term "paraxial" signifies that it describes the field и — velkx near the axis of
propagation (in this case the jc-axis).
An interesting feature of the beam representation A0.264) is that it
shows how the general plane wave representation of the beam obtained
above loses its validity for large x. The beam spreads out and assumes the
form of a (general) cylindrical wave. This effect is a result of diffraction. To
see this we introduce polar coordinates x = r cos 0, у = r sin 0, and assume
that x t> ka and that y/x < 1. Then the radial variable r can be approximated
as r**x + y2/2x and A0.264) takes the form
u(x, y)~A —^ exp^ - - к a tan 6) txp^ikr- —j
This has the form of a cylindrical wave with source point at the origin. Each
value of в determines a ray and we see that as |0| increases, the amplitude
term decreases so that we have a cylindrical beam.
Our discussion has so far been limited to the linear reduced wave
equation A0.1) and its asymptotic solutions. However, in the theory of
nonlinear optics one is led to consider a nonlinear version of A0.1). If a
strong laser beam propagates in a medium, it may happen as a result of
754 ASYMPTOTIC METHODS
various effects that the index of refraction is altered as the beam traverses
the medium. Assuming the medium to be otherwise homogeneous, the
index of refraction n is found to become a function of the intensity of the
field. If и represents the (complex) field, we have n = л(|и|2) and the field
satisfies the nonlinear reduced wave equation
A0.266) V2m + AV(|m|2)m = 0
in two or three space dimensions. We again assume that the wave number к
is large and look for asymptotic solutions of A0.266). It is immediately
possible to construct plane wave solutions of A0.266). For example,
A0.267) и = A exp[ikn(A2)x]
where the amplitude term Л is a real constant satisfies A0.266). We note
that in contrast to the linear problem studied in the foregoing, the phase in
A0.267) is a function of the amplitude.
Proceeding as in the linear case, since there are solutions of A0.266) in
the form A0.267) we look for a solution of the nonlinear reduced wave
equation in the form
A0.268) и = Aeik*
where the amplitude term A and the phase term ц> are now assumed to be
real valued (we no longer require that A be a constant). Inserting A0.268)
into A0.266) and equating real and imaginary parts to zero gives
A0.269) [(V<pJ - n\A2)]A -pV24 = 0,
A0.270) 2V<p VA + AV2q> = 0 .
In the geometrical optics approximation we let £-»<» in A0.269) and obtain
A0.271) (V<pJ = n\A2) ,
A0.272) V*B
where A0.84) has been used in A0.270). These equations are referred to as
the equations of nonlinear geometrical optics. In contrast to the situation in
linear optics, A0.271), A0.272) are coupled and must be solved simulta-
simultaneously. Nevertheless, the eiconal equation A0.271) determines phase sur-
surfaces (p = constant and rays orthogonal to these surfaces. Also, the transport
equation A0.272) determines the amplitude in terms of the divergence and
convergence of the rays, as was the case in the linear problem. However, we
cannot determine the rays without knowing tteamplitude and vice versa.
EQUATIONS WITH A LARGE PARAMETER
755
We do not consider how the system A0.271), A0.272) may be solved in
general. Instead, we examine how a beam propagates in the nonlinear
medium on the basis of the geometrical optics equations. A beam is
characterized as follows. Let <p = constant represent a particular phase
surface or wave front. The amplitude A is assumed to have a maximum at a
point P on the wave front, and it decreases sharply in value as we move
away from P in all directions on the wave front. Thus the significant values
of the field are concentrated near P on the wave front <p = constant, and
these values are propagated along the rays in the form of a beam. We now
assume that the index of refraction n(A2) decreases as the amplitude A
decreases. Thus since the amplitude falls off as we move away from P on
<p = constant, so does n(A2). On considering the two neighboring wave
fronts <p = constant and <p + d<p = constant, the distance between them
measured on the (orthogonal) rays is given as
in view of A0.70). Since d<p is fixed, we find that as A2 and, consequently,
n(A2) decrease (on the wave front <p — constant) as we move away from the
point P, the distance da between the two wave fronts increases. That is,
(p = constant and <p + d<p = constant are nearest to each other at P and
become increasingly separated away from P. The rays, however, always
remain orthogonal to the wave fronts. On applying this argument to a
succession of wave fronts, we obtain the situation pictured in Figure 10.8.
The wave fronts become increasingly concave and the (orthogonal) rays
begin to converge. If the field is symmetric with respect to the point P, the
rays may eventually focus at some point. In any case, the rays begin to
converge and this property of nonlinear optics is known as the self-focusing
effect. Even if the wave front <p - constant is plane or initially convex, the
Figure 10.8. The self-focusing effect.
756 ASYMPTOTIC METHODS
dependence of the index of refraction n on the amplitude will bend the rays
and force them to converge, possibly even to a focus. A medium for which
the index of refraction has this property is called a focusing medium.
However, as the rays converge, the amplitude of the beam begins to
increase in view of the relation between the amplitude and the ray structure.
If the amplitude A becomes extremely large or possibly singular, the
geometrical optics approximation that predicts the self-focusing effect no
longer remains valid and diffraction effects must be included. That is, we
must retain second derivative terms in A0.269). In the linear problems
discussed earlier the inclusion of diffraction effects was shown to remove the
singularities predicted by the geometrical optics approach and to lead to a
spreading of the rays in beam propagation. Here, because we have a
nonlinear problem, it may happen that a beam that undergoes focusing in
the geometrical optics approximation may actually focus and become singu-
singular even if diffraction effects are included.
To see that it is possible to obtain beams that do not undergo self-
focusing we construct an exact solution of A0.266) or, equivalently, of
A0,269), A0.270), which represents a beam. We consider the two-dimen-
two-dimensional case and assume that л(|м|2) is given as
A0.273) n\\u\2) = n2o + nl\u\2>
that is, we have a quadratic nonlinearity. The constant term n0 is the linear
index of refraction that occurs when nonlinear effects can be neglected. The
constant nl is positive, so that n(A2) decreases as A2 decreases. To solve
A0.269), A0.270) we set
A0.274) ф, y) = Ax, A(xy y) = A(y)
where A is a constant. Then A0.270) is identically satisfied and A0.269)
becomes
A0.275) A\y) - k\k2 - n20)A(y) + k2nxA\y) = 0.
It is easily verified that a solution of A0.275) that vanishes together with its
derivatives as \y\—>°° is
A0.276) A(y) = 2(Л "o) sech[*(A2 - n20)v2y]
L ft j J
if we assume that | A| > n0 > 0. The amplitude term A = \u\ decays exponen-
exponentially as Ы->°° and is significantly different from zero only in the interval
Ы ^ O[k~\k2 - ПоГ1/21* Thus the solution
A0.277) ф, y) = [2{X^nl)Y sech[*(A2 -
EQUATIONS WITH A LARGE PARAMETER 757
represents a beam that travels without change of amplitude along the x-axis.
We have shown in Example 10.7 that a plane wave solution in the form of a
beam of the linear reduced wave equation is described by a parabolic
equation in the far field. Although the geometric optics result in that
problem predicted that the plane wave travels without change of amplitude
to leading order, the diffraction effects as characterized by the parabolic
equation, diffuse and spread the field out at large distances. In the nonlinear
problem considered here, the diffraction effects again tend to spread out the
beam, whereas the nonlinear effects tend to focus the beam. Since these
effects appear to be balanced in A0.277) and the beam travels without
spreading, the solution A0.277) is referred to as a self-trapped beam.
It should be noted that if the index n(A2) is such that it increases as A
decreases, this discussion would show that the rays diverge as a result of the
nonlinearity. An initially thin beam would begin to spread out as it moved
through the nonlinear medium. Therefore, we refer to such media as
defocusing media. For example, if л2(|м|2) = л2, + wJm|2 with nx<0, the
index n has the defocusing property. Self-trapped solutions of the form
A0.277) are not possible for such a medium since the amplitude A is real by
assumption, and if nx <0, the amplitude term A0.276) is purely imaginary.
In fact, since both diffraction and nonlinear effects tend to spread out the
beam, we would not expect that a self-trapped beam would occur in a
defocusing medium.
To determine whether the self-trapped beam A0.277) can be expressed in
some simpler form, we rewrite the hyperbolic secant in terms of exponen-
exponentials and easily conclude that
[9f a2 — п2\Л112
A 0) exp[ik\x -k(k2- nlfl2\y\\
if |y|>0. Since the amplitude term is exponentially small, we could not
generate A0.287) by looking for an asymptotic solution of A0.266) in the
form и — Aelk<p where A is expanded in inverse powers of k. However, since
A is arbitrary (apart from the requirement that A2 > n\) we may consider
solutions of the form A0.277) in which k2 ~n20 +c2/k2, where с is a
constant. Then A0.277) becomes
A0.279) u(x9 y)~
This suggests that if we ask for solutions of A0.266) in the form |w| =
O(k~r) with r > 0, an asymptotic approach similar to that used for the linear
Problem can be applied to the nonlinear problem. We note that with the
above assumption on |м| = Л, we are essentially considering a weakly
nonlinear problem since к > 1 by assumption.
758 ASYMPTOTIC METHODS
With \u\ = O(k~r), r>0, we assume that л(|м|2) can be expanded as
A0.280) п2(|и|2) = n20 + nM2 + пМ* + ■■■
where the coefficients лД/ = 1,2,...) are all constants and n, > 0. As we
have indicated, with л, positive the index of refraction A0.280) corresponds
to a focusing medium if \u\ is small. We define the amplitude \u\ as
A0.281) |«| = А=рв, r>0
and with и = Ae'k<f as in A0.268) we obtain for ф and a,
A0.282) {[(V?J - n20]a - p ща3 - ^ n2as -■■■}
A0.283) 2V<p • Va + aV2<p - 0 .
To proceed with the solution of A0.282), A0.283) we must specify the
exponent r and then expand <p and a in inverse powers of k. The most
meaningful and interesting choices for r are r = \ and r = 1. With r=^we
find that <p and a should be expanded in powers of k~l, whereas if r = 1, the
expansion of <p and a should contain powers of k~2. Now if r = 1, we see
from A0.282) that the cubic term a3 and the term V2a are of the same order
in k. Thus at the second level of approximation the (leading) nonlinear term
and the diffraction term V2a are balanced. A particular case of this occurs in
the result A0.279) where A = O(\ik) and we have a self-trapped wave.
However, if r - \, the nonlinear effect due to a3 enters at the second level
of approximation. The diffraction effect due to I2a does not occur until the
third level of approximation as we shall see. We consider the case where
r = \. The case where r = 1 plays a role in the following example.
With r = \ in A0.282), we expand <p and a as
00
A0.284) <p = 2 <Pjk~',
00
A0.285) a - 2 «,*"' >
and insert these expansions in A0.282), A0.283). Collecting like powers of
\lk and equating their coefficients to zero gives
A0.286)
EQUATIONS WITH A LARGE PARAMETER 759
A0.287) 2V<p0 - Va0 + a0V2<p0 = 0 ,
A0.288)
A0.289) 2V<p0 - Vax + a
A0.290)
2V<p0 - V^ = n2a* + Зпгаоаг + — V2a0 - -^
and
A0.291)
2V<p0 la2 + e2V2ft = -2V<p2 • Va0 - a0V2^ -
as the leading equations. We note that the nonlinear term nxa20 enters in the
equation for <рг, whereas the diffraction term V2a0 enters in the equation for
<p2. The preceding equations and all further equations for the <pj and af are to
be solved recursively. The equations A0.286) and A0.287) for <p0 and a0 are
just the eiconal and transport equations of linear geometrical optics. Each of
these equations reduces to an ordinary differential equation along the rays
associated with the phase term <p0.
Rather than discuss the solution of these equations for a general class of
boundary value problems, we consider a specific two-dimensional problem
in the following example. This example exhibits some of the basic features
of beam propagation in a nonlinear medium.
Example 10.8. The Propagation of a Beam in a Nonlinear Medium. The
x, y-plane is assumed to be divided into a linear and nonlinear region. The
interface between the two regions is the line x = 0. A Gaussian beam is
incident on the nonlinear region x > 0 from the linear region x < 0. We do
not consider the field in the linear region but only in the nonlinear region.
At the interface x = 0 the field w(jc, y) equals the incident field w7 and this
is assumed to be
A0.292) u@, у) = и,@, У)=
where E and a are real positive constants (see the discussion in Example
Ю.7). We express the nonlinear field w(jc, y) in the form
A0.293) и = ~ aeik<p
~
where a(x, y) and <p(x, y) are real valued and are to be expanded in inverse
powers of к as in A0.284) and A0.285). The field и in the region jc>0
760 ASYMPTOTIC METHODS
satisfies the nonlinear reduced wave equation A0.266) with n expanded as in
A0.280). The terms <pf and a. in the expansion of <p and a satisfy A0.286)-
A0.291) for У = 0, 1, and 2. The initial values for the <p- and af are found
from A0.292) to be
A0.294) >
A0.295) eo(O, y) = E exp( - ^), «ДО, у) = 0, у > 1.
We also require that A0.293) be outgoing from the interface x = 0 toward
jc > 0, so that %(*, y) must satisfy the outgoing condition at x = 0.
Now %(*, y) satisfies the eiconal equation A0.286), the initial condition
%@, y) = 0, and the outgoing condition at x = 0. We immediately conclude
that
A0.296) <po(
The equation A0.287) for ao(x, y) then becomes
A0.297) 2no^=°-
In view of A0.295) we find that
A0.298) ао(х9у)
To leading order we have
A0.299) u{x, y)~~ exp( - ^
This represents a beam propagating in the jc-direction. The beam half-width
is given as a since for \y\ > a the amplitude term of и is exponentially small
(see Example 10.7). At this level of approximation neither the nonlinear nor
the diffraction effects have been accounted for and we consider those effects
next.
The equation A0.288) for <рг(х, у) has the form
A0.300) 2n0 ^ = nxa\ = n,E2 exp( - %) .
ox \ ot f
Noting A0.294) gives
♦
A0.301) ft (*, y) = 5 ^ E2 exp( - ^4)^ .
Thus the phase term <p is given as
EQUATIONS WITH A LARGE PARAMETER 761
A0.302) ф, y)~<Po+l<Pl = nox+±'^E2 exp( - ?£)x .
The rays, which are orthogonal to the wave fronts, have the direction of V<p,
and this is
(Ю.303)
At y = 0, V<p has the direction of the vector i. For y^O, we see from
A0.303) that for fixed у as л; increases, the j component of V<p becomes
increasingly negative if у > 0 and increasingly positive if у < 0. At x = 0, V<p
has no j component. This indicates the presence of a self-focusing effect, in
that the rays start bending towards the jc-axis as x increases, even though
they were originally parallel at x = 0. We note that if nx <0, the reverse is
true and defocusing takes place. We also observe the dependence of the
phase on the amplitude in A0.302). This represents a strictly nonlinear
effect.
The equation A0.289) for ax becomes
A0.304) 2n0 —± =
ox
ox \ noa noa / \ a
so that
ъх2
A0.305) „««.„.(J^—J^) „,(-.£).
since ay@9 y) = 0. This can be written as
and we obtain
A0.307)
The presence of terms of order x2 in A0.306) and, consequently, in A0.307)
shows that ax{x, y) undergoes secular growth asjc-^w. Since the exponen-
exponential term in A0.306) is small for large \y\ and the term of order y2 is small
for small |у|, we conclude that the first term in the parenthesis in A0.306) is
762 ASYMPTOTIC METHODS
the most significant for large jc. With x fixed, the term that corresponds to it
in A0.307) attains its maximum at у = 0. Thus it can be said that when
A0.308) Й^
we have the first value of x at which a0 and (l/k)al are essentially of the
same order of magnitude and the asymptotic series for a(jc, y) begins to
become disordered.
We denote the value of x determined from A0.308) by xf and we see that
Apart from a (dimensionless) constant factor, this determines the value of x
at which the initially plane wave is expected to focus and we refer to it as a
focal length. We note, however, that if лх<0 we would obtain a similar
expression for x with y/JT[ replaced by V|wx|, at which the secular behavior
becomes significant. Yet with nx <0 we have a defocusing medium and no
focus is expected to appear. Nevertheless, we associate the region where
secular growth becomes significant with the onset of the self-focusing region
for the nonlinear medium. We now demonstrate that such a relationship
does exist.
Since our geometrical optics result for <p and a breaks down when
jc = O(Vk), as follows from A0.309), we study that region using a boundary
layer approach. It is easily seen from A0.290) that <рг(х, у) contains a term
of the order jc3. Thus (l/A:)^ + A Ik2)<p2 are of the same order of magnitude
when jc = O(Vk) since <px = 0(jc). However, the leading term <pQ = nox is
never of the same order of magnitude as the other terms in the expansion of
<p. Thus secular behavior occurs in the expansions of both <p and a, but % is
not involved in it. As a result we shall modify only the series expansion of <p
beginning with the <px term. Since A/к)<рг = O(l/Vk) in the secular region
jc = O(V)t), we set
A0.310) f
in A0.282), A0.283), where r = §. Then
A0.311)
A0.312) 2n0ax + -щ Щ • Vfl + ^= а!гц> = 0 .
Since x = O(Vk) in the secular region we set
EQUATIONS WITH A LARGE PARAMETER 763
(Ю.313) x = Vki
in A0.311), A0.312). Then dldx = (llVk)dld£ so that to leading order
A0.314) i {2поф; + ф2у - П1а2}а + o(p) = 0,
A0.315) ^= {2noae + 2фуау + афуу) + o( |) = 0,
where & = &({■, y) and ф = <p(f, зО- То solve, we expand d and ф in powers
of 1/Vfc, and obtain to leading order for a0 and ^0,
(Ш16) 2no^
,10.317, ^«**4f..
Although these equations are coupled as is the case for the nonlinear
geometrical optics equations A0.271), A0.272), they are, nevertheless,
easier to solve.
A number of solutions of A0.316), A0.317) have been obtained by
Akhmanov et al. and other authors. We shall consider one of these solutions
and match it with our geometrical optics results. To carry this out we require
that the field be considered only near the center of the beam (i.e., near
y = 0). Thus
A0.318) аЦх, y) = E2 exp( - ^) ~ E2(l - ^
A0.319) л(х> у) - i Ь E2 exP( - Щ
| J E2x -
An exact solution of A0.316), A0.317) can be found that has the form
A0.320)
(Ю.321)
whe*c % )8, and / are specified on substituting A0.320), A0.321) into
A0.316), A0.317). We match ф0 and a\ with ^ and a\ by noting that
x = Vlf, so that for small f we have moderate or large values of x. On
comparing the equations for the two sets of variables, we find y(f) and
£(f) must tend to zero as £->0, whereas /(f) must tend to unity as f-»0.
Thus we set
764 ASYMPTOTIC METHODS
A0.322) y@) = )8@) = 0, /@) = l.
Before inserting A0.320), A0.321) into A0.316), A0.317) we multiply
A0.317) by d0 to simplify the calculations. We obtain
A0.323)
from A0.316) and
A0.324)
/U) «2/3U) fXi)
a2f*(€)
from the modified form of A0.317).
Now if we set
in A0.323), the two terms in that equation that do not contain a factor y2
cancel out. We are left with
A0.326)
Similarly, if we set
A0.327)
/U)
the two terms in A0.324) tha} do not contain a factor y2 drop out. It then
turns out that with this choice of /3(£) the remaining terms in A0.324)
cancel out, so that this equation is satisfied regardless of the form of /(£)•
Therefore, /(£) must be specified from A0.326). Inserting A0.327) into
A0.326) yields
A first integral of A0.328) may be obtained by multiplying through by
/'(£). This gives
EQUATIONS WITH A LARGE PARAMETER 765
where we have used A0.327) to conclude that /'@) = 0 since 0@) = 0 and
ДО) = 1. Now A0.329) is exactly solvable by means of a special transforma-
transformation that we do not exhibit. With /@) = lwe obtain in implicit form
Then y(f) and £(£) can be specified from A0.325) and A0.327) but we do
not carry this out.
Since /(f) appears in the denominator in the expression A0.321) for &l,
we see that when/(£) = 0, the amplitude blows up. This corresponds to a
focal point for the field. To determine the location of this focal point we set
= 0 in A0.330) and obtain as the (first) focal length
since sin тг = 0. There are, in fact, infinitely many focal points predicted by
A0.330) but the others are not of interest to us.
We want to match the preceding boundary layer results with the previous-
previously given geometrical optics results. This requires that we consider y(f), /3($,
and /( f) for small £. Since /@) = 1, /'@) - 0, and /"@) = -2nxElln%a \ in
view of A0.328) we have
(.0.332,
As a result we find that
(.0.333)
(Ш34,
Inserting these expressions into A0.320) yields
A0.335)
, y) - <P0(x, y)+± фо[-^ х>
766 ASYMPTOTIC METHODS
which agrees with the geometrical optics result A0.302) when that is
evaluated near the axis of the beam as in A0.319).
For ao(£, y) we have
■«(•♦2S).
for small y. This agrees with the expression A0.307) for a(x, y) near the axis
of the beam. Furthermore, if we determine the focal point for a0 to the level
of approximation used in A0.336), we find that the focal length is
an0Vk
A0.337) xf
since /(£) vanishes at this point if it is expressed as in A0.332). Comparing
A0.337) with A0.331) we find that they differ by a factor of u72V2* 1.1.
The difference is not substantial. The result given in A0.309) on the basis of
the geometrical optics approach differs from A0.337) by a factor of V2. We
see from A0.336) that this difference in results is due to the factor \ which
occurs when the binomial expansion is used in A0.336). Nevertheless, all
expressions for the focal length contain the same dependence on the
parameters a, n0, nl9 E, and k.
Even though the boundary layer equations A0.316), A0.317) predict that
focusing occurs, as we have seen, it may yet be that focusing is prevented if
diffraction effects, which have so far been neglected, are taken into account.
That is, we must consider equations for the field that include second
derivative terms in the amplitude. The self-trapped solution A0.279) shows
that plane wave solutions (in an approximate form) do exist that do not
undergo self-focusing. To analyze this problem we now obtain the terms <^
and a2 in the expansion of the geometrical optics solution A0.293).
The phase term <p2 satisfies A0.290). Using the expressions obtained
above for <p0, <px, a0, and аг and the initial condition <p2@, y) = 0, we easily
obtain
A0.338)
30a2 8n30 nla4 t F\ a
We then insert A0.338) into A0.291) for a2 and integrate, retaining only the
most significant terms near у = 0 and for x large. This gives
EQUATIONS WITH A LARGE PARAMETER 7<S7
A0.339) в2(х,
a
In the absence of nonlinear effects (i.e., nx = n2 = 0) the first term on the
right in A0.339) is the only contribution to аг. It represents a diffraction
effect. Since ax = 0 if ^ = 0, we have
A0.340) «(*, у) « «o + p? «2 « (l - 1Г7-4)^ exp( - ^) .
Thus for a linear medium, secular effects occur when
A0.341) xd~knQa2
where xd is the diffraction length [it agrees with A0.262) if we put a2 = 2a].
We have shown in Example 10.7 that when x = O(k), the Gaussian beam
field must be described by means of a Schrodinger equation. However, since
diffraction effects occur when x = O(k) and self-focusing occurs when
x = O(Vk) in view of A0.309), we might expect that diffraction cannot, ill
general, influence and counteract self-focusing.
With пг and n2 not equal to zero, we obtain near у = 0,
A0.342) a(x, )+ +
2п\а2к п\а\2 п\а2к>
Now secular effects occur when (l/k)a1 =»a0 in the nonlinear problem and
when (l/k2)a2» a0 in the (strictly) linear problem. The combined effects in
A0.342) that result in a0 - (l/k)^ - (l/k2)a2 yield
Noting A0.309) and A0.341), we have
A0.344) i-i-3
and we assume that n2>0. Arguing as above, we could state that A0.344)
determines a first value of x at which self-focusing becomes significant if we
768 ASYMPTOTIC METHODS
include diffraction and higher order nonlinear effects. We have established a
relationship between the secularity and self-focusing regions previously
Thus A0.344) shows that diffraction need not prevent self-focusing but it
does delay its onset. The higher order nonlinear effect enhances the
self-focusing property and brings the focal region closer to the origin. This
result is not unexpected in view of our general discussion of the self-focusing
effect. If к is large and the other terms in A0.343) are of moderate size, the
diffraction and higher order nonlinear effects do not appear to be sig-
significant.
An interesting situation arises if xf = xd so that the focal and diffraction
lengths are equal. This implies that
A0.345) E2 =
пха2к
so that E is small for large k. With this choice for E, the last term in
A0.342) is of order k~4E so that
A0.346) a«ao + -a1 + -Ta2«E + O(k~4E) .
К /с
Also, in A0.343) we may eliminate the n2 term since it is of order /Г4
compared to the other two bracketed terms which are of order k~\
Consequently, A0.344) becomes
A0.347) —~0,
and the focal point is moved off to infinity at this level of approximation.
Furthermore, we have
A0.348) ?>«%+|<р1
for у = 0 as is easily seen. Thus with E given as in A0.345) we obtain
1
A0.349) «(*>>0 =
2 \1/2
using the given approximation. We have replaced a~E by a~
E exp[-y2/a] in A0.346) since the preceding was valid near у = 0 and toe
field, which has the form of a beam, is exponentially small away from у - u-
We have shown that if E is given by A0.345), the .esulting field is a beam
that travels without change of shape to a high level of approximation tor
large Jt. Since this is effectively a self-trapped beam for which an (exact;
EQUATIONS WITH A LARGE PARAMETER 769
balance between the nonlinear self-focusing effect that the defocusing
diffraction effect exists, it is of interest to compare A0.349) with the
approximate form A0.279) of the exact self-trapped solution found earlier.
The parameter с that occurs in A0.279) may be interpreted as an inverse
beam half-width for that solution, since when cy s: 1, sech(cy) is small. Thus
if c = l/«, the amplitude term in A0.279) evaluated at the center of the
beam is equal to E as defined in A0.345). We recall that a is the half-width
of the beam A0.349). Even though the beams A0.279) and A0.349) have
somewhat different phase and amplitude representations, they have a
common beam half-width a and have the same amplitude E at the center of
the beam. In addition, they are both (approximately) general plane waves
and are self-trapped.
We now present a boundary layer analysis of the field in the region
x = O(k) where diffraction effects become important. The phase in that
region should equal nox plus lower order correction terms, and the am-
amplitude should be bounded or decreasing as к increases. From A0.301) and
A0.305) we see that A/к)<рг = 0A) and A/к)аг = O(k) when x = O(k).
However, <p0 = O(k) and a0 = O(l) when x = O(k). Since we cannot have
(l/k)a1 be of higher order than a0, it must be assumed that a0 —
Eexp(-y2/a2) = O{k"s) for some positive s and this means that E =
0{k'5). Then (l/*)ft = 0(*'2') and (Uk)ax = O{kx'u)9 so that these
correction terms are of reduced orders in fe.
To proceed we set
A0.350) ^
in A0.282), A0.283), since when x = O(k) we have seen that the correction
term to <p0 = nox is one order of к lower than <p0 itself. With r - \ in
A0.282), A0.283) we have
A0.351)
{f w + p
2noax + | V£ • Ve + \ al2y = 0
with
A0.353) x = kit
We set
A0.354) «=1«, £
770 ASYMPTOTIC METHODS
where s > 0 and is to be determined. This implies that the amplitude a is
O(k~s) and the correction term to <^, = nox is O{k~2s) as was required
previously. We obtain
A0.355)
where we have neglected lower order terms. We must choose s = \ so that
the significant terms in A0.355), A0.356) include the leading terms in
A0.314), A0.315) valid in the focal region. However, we do have an
additional second derivative term in the amplitude to account for dif-
diffraction.
The leading terms in the expansions of ф and a, which we denote by <p0
and a0, satisfy
A0.358, 2^+2= = +„^_0.
Introducing the function
A0.359) v( £, y) = ao( £, y) exp[/%( £, y)],
we easily find that
A0.360) 2m0^ + vyy + Wl|y|2y = 0 .
This nonlinear Schrodinger equation reduces (as it should) to the linear
Schrodinger equation A0.146) or A0.265) when ^=0. Both equations
describe the far field in the region x^O(k) where diffraction effects are
important. In terms of u(x, y) our result is now given as
A0.361) u(x, y)~^ v(j x, y)eikn^
where v satisfies the equation
A0.362) 2iknovx + vyy + nx\v\2v = 0 .
EQUATIONS WITH A LARGE PARAMETER 771
The nonlinear Schrodinger equation A0.360) is one of a class of non-
nonlinear equations for which an exact solution of the initial value problem can
be found. The inverse scattering method and related methods that yield
these solutions are too complicated to be considered here. A self-trapped
solution of A0.360) may be found by setting
A0.363) & = *& «о^о(Й
in A0.357), A0.358). Then A0.358) is identically satisfied and A0.357)
reduces to the ordinary differential equation
A0.364) 2nokao(y) - nxal{y) - a&y) -0 .
This has the form of A0.275), and a solution that vanishes as Ы->°° is
A0.365) ao(y) = [^] sech[Bn0AI/2y]
in view of A0.276).
The field u(x, y) is given as
A0.366) u(x,
on using A0.359) and A0.361). In we set A = c2/2n0, this expression is
identical to A0.279). However, we are interested in obtaining a solution of
A0.357), A0.358) that matches A0.349). To do so we set
A0.367) £0 = 0; flo
where Ё is a constant that is to be determined. Inserting A0.367) into
A0.357), A0.358) yields
-20-$•*(-#—.*■«»(-$
from A0.357), since A0.358) is satisfied identically. Near the center of the
beam where у «0, we may^approximate exp[~2y2/a2] by 1 - 2y2/a2. Doing
so in A0.368) shows that Ё must be given as
Thus the field u(x, y) near у = 0 is
772 ASYMPTOTIC METHODS
A0.369)
The expression A0.369) for и agrees with A0.349) which was obtained
from the geometrical optics solution under the asumption that the self-
focusing and diffraction lengths are equal. The field A0.369) represents a
self-trapped beam since the amplitude is independent of x and decays
exponentially in |y|. If we define Bn0A)/2 = a to be the beam half-width
of A0.366), we obtain for A0.366)
A0.370) ф, у) - (^pI'2 sech(^)
The fact that A0.369) and A0.370) are both self-trapped beams of
half-width a and \u(x, 0)|2 = 2/n1a2k2 ~ E\, suggests that Eo plays a critical
role in determining whether beams are self-focused, self-trapped, or effec-
effectively propagate as in a linear medium. If a beam u(x, y) has |ii(jc, 0)| > E09
the focusing effect is stronger than the defocusing diffraction effects and
self-focusing occurs. If \u{x, 0)| — Eo, self-trapping occurs and if \u(x, 0)| <
Eo, the nonlinear effects contribute small corrections to linear wave propa-
propagation. The concept of a critical intensity El associated with beam propaga-
propagation has been discussed in the literature and is somewhat consistent with
these results, but is difficult to verify in general.
We conclude our discussion of asymptotic methods for equations with a
large parameter by considering an initial value problem for the Klein-
Gordon equation. It is shown how the methods presented to deal with
elliptic equations can be carried over to hyperbolic equations. Rather than
give a general discussion, we concentrate on a specific problem for the
Klein-Gordon equation. As we have indicated in Examples 3.8 and 5.14,
this equation is representative of dispersive wave motion. It is for dispersive
equations that our method is applicable and most useful.
Example 10.9. An Asymptotic Solution of the Klein-Gordon Equation.
With м = u(<r, t) we begin by considering the Klein-Gordon equation
A0.371) мтт - у 2йа(Г + с2й = 0^~- -ос < о- < oo, т > 0 ,
with the initial conditions
A0.372) й(а, 0) = 0, ит(<г, 0) = 8(а) ,
where S(a) is the Dirac delta function and у and с are specified constants.
To obtain a large parameter in the equation we set
EQUATIONS WITH A LARGE PARAMETER 773
A0.372) (r = kx, T = kt
where к > 1 and this gives
A0.374) u,,-y2uxx + k2c2u = 0,
where и = u(x, i) = u[kx, kt]. The initial conditions for и are
A0.375) u{x, 0) = 0, u,(x, 0) = 8(x)
since й>, 0) = | u,(x,0) = S[kx] = \ S(x).
The solutions й and и of these problems yield the causal fundamental
solutions for the appropriate Klein-Gordon equations as follows from the
results of Chapter 7. The transformation A0.373) indicates that we are to
study these fundamental solutions at large times and large distances from
the origin. In Example 5.14 the method of stationary phase was applied to
the Klein-Gordon equation to examine its solutions for large |jc| and t. The
asymptotic result given in that example suggests that we look for a solution
of A0.374), A0.375) in the form
A0.376) u(x,t) = a(x,t)eik<p{xt)
f
/=o
We note that the assumed expansion has the same form as that considered
for the reduced wave equation.
On inserting A0.376) into A0.374) and equating like powers of Ilk we
obtain
A0.377) tf-y'vl-c'-O,
A0,78) .«.„.j^^.^.,
i
2 '
dt2 +У дх
The equation A0.377) for the phase term <p(x, t) is known as the dispersion
equation. If we put
A0.380) <» = -<?„ \ = <px
in A0.377) we obtain
774 ASYMPTOTIC METHODS
A0.381) a>2 = y2A2 + c2,
and this leads to the dispersion relation w = w(A) as defined in C.157). The
equations A0.378), A0.379) are the transport equations for the amplitude
terms a;. We shall solve the dispersion equation using the method of
characteristics and solve the leading order transport equation for aQ.
To begin, we solve A0.377) for <pt and obtain
A0.382) ft = ±Vy>I + ?.
Each of these equations must be dealt with separately. We study the case
with the minus sign in some detail and then state the result for the other
case. Thus we consider
A0.383) <pt + V?V* + c2 = 0 .
With p — <px and q = <ptwt have
A0.384) F(p, q) = q + У/у У + с2 = 0 .
Noting the results of Section 2.4 we obtain the characteristic equations for x,
t9 <p, p, and q as
A0.385) f-F,-^-^-,, £-Ff-l.
, 2 2 2 _ 2
A0.386) ~r- = pFB + oFfl = -^L£_ + ? = _ =
ds F p H * q q
A0.387) ^ = 0, ^r=0.
ds ds
Here 5 is a parameter along the characteristic curves. The curves x = x(s),
у = y(j) are called rays, and F = 0 along the rays.
Since the initial data for u(x, t) are concentrated at the point x = 0, the
data for the system A0.385)-A0.387) are concentrated at the origin (x, 0 -
@,0). Therefore, we set
A0.388) x@) = 0, *@) = 0, 9@) = <p0 ,
where <p0 is a constant that has to be specified. Also,
A0.389) q@) + УуУ@) + с2 = 0 .
Since the initial curve reduces to a point in this case, there is no strip
condition that p and q must satisfy. Now A0.387) shows that p(s) and q(s)
EQUATIONS WITH A LARGE PARAMETER 775
are constant along the characteristics. Thus A0.389) determines q in terms
of the parameter p where — <» < p < °°-
With the given data we easily find that
A0.390)
A0.391)
The parameter p identifies the specific characteristic curves and rays. Since
s -1, we can replace A0.390) by
y2p
A0.392) x = , J[ t, -oo < p < oo,
and these are the rays for this problem. Also,
(Ю.393) <p= ~\
Setting G(p) = у VVrV + c1 we find that G'(p)>0 for a11 P? so that
is an increasing function. As p->-», G(p)->-y, and as /?->°°,
*y- Thus
A0.394)
Since dxldt— G(p) is the ray velocity on a ray p = constant, we see that
|&/A| < y. Now у is the maximum speed of disturbances for the Klein-
Gordon equation as was seen in Chapter 7. Also, given the dispersion
relation <o = <o(A) = Уу2Х2 + с2 for the Klein-Gordon equation, the quanti-
quantity co'(A) = y2A/Vy2A2 + c2 is known as the group velocity. Identifying p
with A, we find that G(p) = <o'(p) is the group velocity. Thus the rays
A0.392) may be termed group lines, and they are located within the
characteristic sector with vertex at the origin and bounded by the charac-
characteristics x = ±yt of the Klein-Gordon equation, as shown in Figure 10.9.
The ray equation A0.392) may be used to express p as a function of x and
t. If the result is inserted in A0.393), we obtain
A0.395) Ф, t) = - ~ VyV-*2 + <Po •
This shows explicitly that <p is not real valued outside the characteristic
sector and that <p = <p0 on the lines x = ±yt.
776
ASYMPTOTIC METHODS
Figure 10.9. The characteristic sector and a ray.
Turning to the solution of the transport equations we show that they
reduce to ordinary differential equations along the rays. Since <рх=р and
<pt = q, we have
t
A0.396) 2— -^-2у2 — — =2q\—L- JUL- -1 =:
dt dt Эх дх I dt Jv ■
in view of A0.385). Also, we easily show that
A0.397)
We solve only the equation A0.378) for ao(x, i), and this becomes
A0.398)
along a ray p = constant. Thus the general solution of A0.398) is
A0.399) *o = ^.
Since p depends on x and t in the special form x/t, as follows from
A0.392), we find that the general solution of A0.378) is, in fact,
A0.400) aQ(x, t) =
V *
where / is an arbitrary function, as may be verified directly.
EQUATIONS WITH A LARGE PARAMETER 777
To complete the solution of the problem we must add the asymptotic
result that corresponds to choosing the plus sign in A0.382). Proceeding as
above, we easily show that
A0.401) ф91)=- V?2'2-*2 + ft >
A0.402)
in that case, with <pQ and/as yet undetermined. The asymptotic solution of
A0.374), A0.375) is the sum of both the results obtained and to leading
order we have
A0.403) «(*. () - ± /() [*( C
If we are to specify the unknown functions and constants in A0.403), we
must relate u(x, i) to the initial data A0.375). The form of the data has been
used only to determine that all rays issue from the origin (x, t) = @,0), but
no further information about u(x, 0) and ut(x, 0) has been used. However,
since the amplitude terms in A0.403) are singular at t = 0, u{x, i) is not valid
near the initial line so that no direct matching with the initial data can be
carried out. This is consistent with the fact that our results are expected to
be valid at large times and at large distances from x = 0.
The use of boundary layer methods near the initial line leads to an
interesting result. If we simply put t = (l/k)r in A0.374) to emphasized the
neighborhood of the initial line, it is easily found that the boundary layer
result
A0.404) v(x, r) = - 8(x) sin(cr)
с
is inadequate for the purpose of matching with A0.403). Since A0.403) is
not valid for \x\ > yt, we see that not only t but also x must be stretched to
emphasize the origin (x, t) = @,0). The stretching clearly has the form
x = (l/k)a and t = A/к)т and is identical with A0.373). Thus the boundary
layer problem is just the original problem A0.371), A0.372), which must
now be solved exactly. Although this result appears to raise doubts about
the usefulness of our asymptotic approach, it must be emphasized that the
asymptotic method also works if the equation has variable coefficients. In
that case, the boundary layer problem would reduce to the study of an
equation with constant coefficients so that a simplification is achieved.
778 ASYMPTOTIC METHODS
The exact solution of A0.374), A0.375) in the region |*| < yt is found
from G.382) to be
A0.405) u(x, t)=Y
Using the large argument asymptotic form for J0(z), that is,
A0.406) J0(z) ~ yf^ sin(z + j), z-*oo ,
we find that
A0.407)
1 -1/2 ii i ~м± (ike r^r? 5 тг\
M(*> 0 те ^Г B777/rc) \y4 - x ) expl — Vy ' — д: + f — I
- — B7rykc)~1/2(y2t2 - x2)~1/4 expf - — *\Jy2t2-x2 -/7],
Ll \ у 4 /
Comparing A0.403) and A0.407) shows that we must set
A0.408) /(у) -/(у) = (8тгу*сГ1/2(у2- Т-)
A0.409) ft=-ft5S5^L.
It is seen from A0.407) that u(x, t) when expanded asymptotically is not
only singular at (jc, t) = @, 0) but also on the characteristics x~±yt. Thus
A0.403) is valid only in the interior of the characteristic sector. Near
x=±yty which determine the wave fronts for this problem, the form
A0.405) must be used.
10.2. THE PROPAGATION OF DISCONTINUITIES AND
SINGULARITIES FOR HYPERBOLIC EQUATIONS
Discontinuities and singularities in the data for (linear) hyperbolic equations
are propagated along characteristics, as we/have demonstrated on a number
of occasions. The behavior of solutions in the neighborhood of characteris-
characteristics across which discontinuities or other rapid variations of the solution
occur can be studied separately. Although the results obtained do not, in
general, yield a full description of the solution, they do describe it in a
region of interest such as at a wave front. Other approximate methods of the
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 779
type given in Sections 10.1 and 5.7 may be used to complete the description
of the solution in certain other regions of interest.
To begin, we consider the hyperbolic equation
A0.410) utt - V- (pVu) + 2Ли, + qu = 0 .
This has the general form of the hyperbolic equations considered in Chapter
4 except that we have put p = 1 in D.10), as well as F = 0, and added a
damping term 2Лм, where Л is a positive function of the space variables only.
The coefficients p and q are also functions of the space variables only and
satisfy the conditions given in Section 4.1.
We study the form of the solutions of A0.410) near a wave front surface
<p(x, t) = 0, where x equals jc, (jc, y), or (jc, y, z) in the case of one, two, or
three dimensions. [In the one-dimensional case <p(jt, i) = 0 is a curve.] It is
assumed that the solution и vanishes on one side of the surface and is a
nonzero function that satisfies A0.410) on the other side of the surface. The
solution may be represented as
A0.411) w(x, 0 = ф, t)H[<p),
where H[<p] is the Heaviside function that vanishes when <p <0 and equals
unity when q> ^ 0. This requires that у be a solution of A0.410) in the region
<p > 0 and that the value of v on <p = 0 equals the jump in the solution и
across the wave front.
To determine what conditions must be placed on <p and v we insert
A0.411) into A0.410). This gives, on using H'[<p] - 8[<p],
A0.412) 8'[q>]{tf ~p(V<pJ}v + 8[?]{2ftu, - 2pV<p ■ Vv + 2\<ptv
+ H[<p]{vtt -V-(pVv) + 2\vt + qv}=0,
with V= д/дх in the one-dimensional case. Since и is a solution of A0.410),
the coefficient of H[<p] in A0.412) must vanish in the region where
H[<p] 9^0. The delta function 8[<p] and its derivative 8'[<p] both vanish when
<p ^ 0. Since at <p = 0, 8 f[<p] is more singular than 8[<p], we require that their
coefficients vanish separately so that A0.412) is satisfied. This implies that
9 = 0 must be a solution of the characteristic equation for A0.410)
A0.413) «p,2-/>(V?02 = O.
This result is not unexpected since we have already shown that all discon-
discontinuities must occur across characteristics.
The vanishing of the coefficient of 8[<p] in A0.412) implies that on the
characteristic surface <p(x, t) = 0 the function i; must satisfy
780 ASYMPTOTIC METHODS
A0.414) 2<ptvt - 2pV<p Vv + 2\<ptv + [<ptt - V- (pV<p)]u = 0 .
The appropriate solution <p(x, t) = 0 of A0.413) may be determined by
the method of characteristics. The relevant equations are given in Exercise
2.4.13. The initial conditions for determining the requisite characteristic
function (p must be found from the data for the hyperbolic equation
A0.410). Then A0.414) can be solved as we demonstrate. Again the
conditions on v in A0.414) come from the data for A0.410).
Before doing so, we use the same approach on the parabolic equation
D.17), where we put p = 1. Thus, we insert A0.411) into D.17) and obtain
A0.415) -8'[<p]p(V<pJv + 8[<p]{<ptv - 2pV<p Vu
-V- (pV<p)u} + H[<p]{vt - V- (pVv) + qv) = 0 .
Arguing as before, we conclude that (V<pf ~ 0 so that <p = <p{i). Any jumps
that occur in the solution must occur across the characteristics t = constant.
Since <p is independent of x, the coefficient of 8(<p) reduces to <ptv and this
must vanish on <p = 0. But <p = <p(t) - t - с ~ 0 on the characteristic t = с so
that <pt7*0. Therefore, v — 0 on the characteristic and the solution of the
parabolic equation cannot have a jump across the characteristic. This is
consistent with our results in Chapter 6. We have again determined that
jumps in the solutions of homogeneous hyperbolic and parabolic equations
with smooth coefficients can occur only across characteristics. Since elliptic
equations do not have real characteristics, the foregoing analysis applied to
the elliptic equation D.9) with F = 0 shows that its solutions cannot have
any jumps. In the parabolic case, even though there are real characteristics,
no jumps can occur across them. Only in the hyperbolic case can there be
nonzero jumps and we now resume our discusion of their properties.
We now apply the method of characteristics to A0.413). Let a = <pn
P = V<p, and we have
A0.416) F(x, a, p) = a2 - p(x)|p|2 = 0 .
The characteristic curves for A0.413) are called bicharacteristics to dis-
distinguish them from the characteristics of the hyperbolic equation A0.410).
With x(s), t(s), <p(s), a(s)9 and p(s) representing the bicharacteristics of
A0.410) we have the equations (see Exercise 2.4.13)
A0.417) f =
A0.418) g =
A0.419) f = 0, f = |P!2Vp,
where d<pids — 0 since F = 0 along the bicharacteristics.
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 781
From A0.418) we see that <p = constant on the bicharacteristics. Thus if
initially (p = 0, we conclude that <p(x, t) = 0 is the resulting characteristic
surface. Also, A0.419) shows that a = constant along the bicharacteristics.
Then A0.417) indicates that the parameter s can be replaced by the time
parameter t along the bicharacteristics, with dt = 2a ds. Thus A0.417) is
replaced by
A0.420) ^ = ~"P'
and a similar transformation can be carried out for A0.418) and A0.419).
The bicharacteristics can be used to reduce A0.414) to an ordinary
differential equation. Given the (bicharacteristic) curve x = x(t), t = t on the
characteristic <p(x, i) — 0, we have
A0.421) 2<ptvt - 2pV<p • Vu + 2X<ptv = 2au, - 2/?p • Vu + 2\av
= 2a[vt + xf@ • Vv + \u] = 2a [ ^ + Ay]
in view of A0.420), with i; = u(x@, 0- Tbus A0.414) reduces to
A0.422) ± ±
along the bicharacteristics. We recall that a = <pt = constant on the bicharac-
bicharacteristics.
If p = constant and <p = 0 is the plane wave front
A0.423) <p
where 7 and с are constants, with |y| = 1, A0.422) becomes
A0.424) ^ + Aw=0.
at
Since A>0, this implies that v—when evaluated on the characteristic-
decays (exponentially) with increasing time. As a result, the presence of the
damping term 2Aw, in A0.410) has the effect of smoothing out the discon-
discontinuity A0.411).
In the two-dimensional case, if p = constant and <p = Vpt - r = 0 is a
cylindrical wave front, A0.422) takes the form
A0.425) f
782 ASYMPTOTIC METHODS
Since we can replace r by y/pt on the wave front and a = <pt — y/~p on the
bicharacteristics, we obtain the solution (for constant A)
A0.426) v = Vte~Xt>
where с is a constant that varies from one bicharacteristic to the next. Again
v decays exponentially as a result of the damping effect and also algebraical-
algebraically because of the spreading of the bicharacteristics. Also, v is singular at
t = 0 because all the bicharacteristics intersect there.
In three dimensions, if p = constant and we consider the spherical wave
front <p = Vpt - r = 0, we find that (for constant A)
A0.427) v=- e~kt,
where the constant с may have different values on different bicharacteristics.
We again observe the combined exponential and algebraic decay of v and its
singularity at t = 0.
Although this result only describes the variation of the solution within the
region of discontinuity, it is possible to extend it to obtain an expression
valid near the discontinuity region as well. To do so we consider the
characteristic surfaces <p = constant, which are parallel to the characteristic
surface <p = 0. For small values of <p it is possible to expand v(x, t) in a
Taylor series of the form
A0.428) vM = 2v*L
/-o J-
where the vf are evaluated on the characteristic <p = 0. Effectively, we are
introducing <p as a new coordinate in addition to a set of coordinates on the
characteristic surface <p = 0. We insert A0.428) into A0.411) and obtain
A0.429) u(x,0
The generalized functions
A0.430) Щх) - [^]ВД, у = 0, lf 2.. .,
where H(x) is the Heaviside function^ have the property that
A0.431) H'jW^fy^C*), />1
(see G.125) and Example 7.1). Thus we write A0.429) as
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 783
00
A0.432) M(x, 0=2 VjHji
J=o
Inserting A0.432) into A0.410) and using A0.431), we again find that 8'[<p]
is the most singular term and this implies that <p must satisfy the characteris-
characteristic equation A0.413). The remaining terms are
A0.433)
8[<p]{2 £t%- 2pV<p • V,o + 2A f( v0 + [<ptt -V-
7 = 1
+ [<Pu - Ч>
+ 2A-^i + ^_1] = 0.
Equating the coefficients of 8[<p] and Hj[<p] to zero yields the system of
transport equations
dip dvn дф
A0.434) 2 -f —2 - 2pV9 • Vu0 + 2A ^7 u0 + [д, - V •
A0.435) 2^d-£-
In view of A0.421), each of the transport equations can be expressed as an
ordinary differential equation along the bicharacteristics.
In solving these equations in terms of the discontinuous data given for the
problem, we do not merely determine the Vj along the bicharacteristics that
issue from the discontinuity region. Instead, the data are represented in the
form A0.432) at t = 0 or on the boundary region and the us are determined
in terms of the coefficients in these series. (The equations for the Vj can be
expressed as ordinary differential equations along the bicharacteristics of the
surfaces <p = constant.) Thereby the vj are specified over the entire region
where the solution is valid. The jumps in и and its normal derivatives at
<P = 0 can be determined from this more general result.
This problem can be simplified somewhat if the characteristic surface
<p = 0 is expressed in the form
784 ASYMPTOTIC METHODS
A0.436) <p(x,0 = '-<W*) = 0.
Then ф(х) = constant represent the position of the wave fronts at different
times t. [It is sometimes convenient to write <p as <p = ф(х) — t.] The
functions u; in A0.428) that are evaluated on the characteristic may be taken
to be functions of x. Thus A0.432) becomes
A0.437) m(x, 0=2 vXx)H,[t - ф(х)].
The characteristic equation A0.413) becomes
A0.438) (V(«2 = j ,
and this has the form of the eiconal equation. The transport equations
A0.434) and A0.435) take the form
A0.439) 2pVtj,-Vv0 + 2Au0 + V- (pVty)vQ = 0 ,
A0.440)
[If we set <p = ф - t, A0.439) is unchanged but the signs on the right side of
A0.440) are reversed.]
Apart from the term 2Au; in A0.439) and A0.440), these equations are
similar to the transport equations obtained in Section 10.1. If A = 0 in
A0.439) and we multiply through by v0 we can write the equation as
A0.441) V-(pV*uJ) = 0.
By integrating along the rays determined from A0.438) as was done in the
preceding section, the variation of v0 along the rays can be found. With
A > 0, the medium is dissipative and u0 depends not only on the converg-
convergence or divergence of the rays but undergoes exponential decay along each
ray.
The expressions A0.432) or A0.437) are to be interpreted as weak
solutions of A0.410) since they involve generalized functions. To see how
initial conditions are to be assigned for the bicharacteristic or ray equations
and for the transport equations for the u;, we consider the one-dimensional
telegrapher's equation in Examples 10.10 and 10.11. The results that can be
obtained for various problems involving that equation suggest how the
general problem ought to be treated^ A method for the discussion of the
propagation of singularities other than jump discontinuities for A0.410) is
also given.
the propagation of discontinuities and singularities 785
Example 10.10. An Initial and Boundary Value Problem for the Tele-
Telegrapher's Equation. We begin by considering an initial and boundary
value problem for the telegrapher's equation
A0.442) utt - y2uxx + 2Лм, = 0, x > -a - ct, t > 0 ,
where y, A, a, and с are positive constants with с < у. The initial data are
A0.443) u(x, 0) = H(x), ut(x, 0) = y8(x), x>~a
and on the moving boundary x = -a - ctf u(x, i) satisfies
A0.444) u(-a-ct, t) = 09 t>0.
We require that с < у so that the boundary line is time-like (see Section 6.5)
and, therefore, only one condition need be assigned on it.
In the absence of the damping term 2\ut, (i.e., if A = 0), the solution of
A0.442)-A0.444) is
A0.445) *(*, г) = H(x + 70 - h[j^Z (x - yt) - ^
as is easily verified. The second term in A0.445) is the wave reflected from
the boundary line so that the singularity in the initial data gives rise to an
incident and reflected wave.
With A>0 in A0.442), we consider a solution u(x,t) expanded in the
form A0.432) and given as
A0.446) u(x9 0 = 2 v,(x9 ОЩф, 0] .
/-о
Inserting A0.446) into A0.442) and equating coefficients of the Щ<р], 8[<р],
and 8 '[<p] to zero yields the characteristic equation
(Ю.447) «\-yV = 0,
and the transport equations
A0.448) 2<p, ^a - 2y V, ^2 + 2\<p,v0 + [<ptl - y2<pxx]v0 = 0 ,
A0.449) 2<p, -jf ~ 2y\x -£
786 ASYMPTOTIC METHODS
Since it takes a finite time for the disturbance resulting from the initial
data to reach the boundary, we first use A0.446) only to account for the
initial conditions. Afterwards, we deal with the boundary conditions. In
addition, we note that A0.446) is a formal solution of A0.442) if A0.447)^
A0.449) are satisfied. That is, we do not require that <p(x, t) = 0 but we can
consider any characteristic <p(x, i) = constant. Thus even though the results
are expected to be useful and meaningful with Vj(x, i) evaluated on the
characteristic <p = 0, we apply the initial data A0.445) directly to A0.446) so
that we are effectively dealing with the whole family of characteristic lines
<p(x, t) = constant.
At / = 0 we have
A0.450) u(x, 0) = H(x) = 2 v,(x, 0Щ[<р(х9 0)],
A0.451)
n,(*, o) = -yew - 2 [v,(x, o)g>t(x, ощ^ф, о)] + Щг- Ц[Ф> о)]}.
On comparing terms in these equations, we find that cp(x, 0) should be
chosen as
A0.452) <p(x,0) = x.
Since A0.447) has the solutions <p = x ± yt - constant, we conclude that
there are two possible solutions of A0.447) and A0.452),
A0.453) $(x, t) = x- yt, ф(х9 t) = x + yt.
Since there is no basis for selecting one of these characteristics above the
other as they both result in waves that are outgoing from the point x, we
must use both of them in our result.
Thus A0.446) must be replaced by
A0.454) u(x, 0 = 2 v,(x, Щ[$\ + 2 Цх, Щ[ф].
Consequently, we are led to consider two solutions valid near the charac-
characteristic lines <£(*, 0 = 0 and ф(х, t) = 0, each of which issues from the
singular data point (jc, t) = @,0).
If A0.454) is inserted into A0.442), we find that vj and uf each satisfy the
transport equations A0.448), A0.449) with <p replaced by <p and ф, respec-
respectively. We determine only leading terms v0, v0, vl9 and vx. Inserting
A0.454) into the initial data—as in A0.450), A0.451)—we obtain
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 787
A0.455)
H[x] = [vo(x, 0) + vo(x, 0)]H0(x) + [v^x, 0) + vx(x,
A0.456)
yS[x] = [~yvo(x, 0) + yvo(x, 0)]H_t(x)
Since H0(x) = H(x) and Н_г(х) = Н'(х) = 8(х), equating coefficients in
A0.455), A0.456) gives
A0.457) vo(x, 0) + i30(jt, 0) = 1, -vo(x, 0) + vo(x, 0) = 1,
and this yields
A0.458) vo(x, 0) = 0, vo(x, 0) = 1.
Clearly, the solution of the transport equation for vo(x, t) is vo(x, t) = 0. The
equation for i30(jc, t) is
A0.459) 2y —- - 2y2 —- + 2Ayu0 = 0 ,
with the initial condition A0.458). Using the method of characteristics, or
otherwise, we obtain
A0.460) бо(*,/) = е-А'.
Then the equations for the data иДя:, 0) and ux(jc, 0) are found from
A0.455), A0.456) to be
A0.461) vt(x, 0) + vx(x, 0) = 0, -vx(x9 0) + йх(х, 0) = » ,
so that
A0.462) ^(^0) = - — , vt(x,0)= — .
Since vo(x9 i) = 0, the transport equation for vx(x, t) is
A0.463) -2y ~~- - ly1 -p- - 2kyvx = 0 .
For vx(x9 i) we have
788 ASYMPTOTIC METHODS
A0.464) 2? ^ - 2y2 ^ - 2А-УУ1 = Л Vх'.
Using A0.462) we easily find that
A0.465) щ(х,1) = -±е-к1, v^t)
We have, therefore, found for small t,
A0.466) u(x, t) = e-x'H(x + yt) + (^- + ^)e"A'(jt + yt)H(x +
This shows that the jump in u(x, t) across £ = * + yf = 0 equals e~k\
whereas there is no jump in w(jc, i) across <p = jc - yr = 0.
We now introduce the boundary condition A0.444) into our discussion.
The solution found in A0.466) is evaluated on the boundary line and this
gives
A0.467) и(-в - c$9 i) = e~ktH[-a + (y - c)t]
Now on x = —a - ct,
A0.468) ЯДЙ = ЯД-л - (y + c)r] = 0, />0
since ЯДдг) vanishes for x < 0. Also, on x = -a - ct,
A0.469) ЯДft = ЯД-a + (у - c)t] = 0, Г < ^ .
Therefore, the terms ЯД£] that represent /igAf traveling waves or distur-
disturbances do not play a role in the boundary condition since these waves do not
intersect the boundary line. The terms Щф] represent left traveling waves
that do not reach the boundary until the time f = a/(y-c)>0. Thus
A0.466) remains valid until that time.
If / > a/(y - c), the solution A0.466) must be modified. We express it as
00 OO
A0.470) u{x, t) = U{x, 0+2 Щ[§\ + 2
У-0 "—' >-0
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 789
where U(x, t) is to be specified and the two series are as previously given in
terms of the initial data. The boundary condition A0.444) implies that
00
A0.471) (/(-a - ct, t) = -2 Ц~а - ct, Щ[-а + (у - c)t],
j = 0
in view of A0.468). Since u(x, t) is given by A0.466) in the region
t<a/(y- c), we conclude that U(x> i) vanishes identically in the region.
The boundary value A0.471) for U(x, i) suggests that we expand U(x, t)
as
A0.472) U(x, 0 = 2 v,(x9 Щ[<р].
/"-О
Since U must satisfy A0.442), <p must be a solution of the characteristic
equation A0.447) and the Vj must satisfy the transport equations A0.448),
A0.449). From A0.471) we determine that
A0.473) <p(-a - ct91) = -a + (y - c)t,
A0.474) i;.(-a - cf, 0 s -6,(-« - ct, t) ,
with the 0j having been specified in terms of the initial data.
Of the two solutions of the characteristic equation A0.447) that satisfy
A0.473), the appropriate one is
A0.475) ф, t) = £1 (x - yt) - 22
The second solution contains characteristic lines that intersect the дг-axis in
the region x > -a. Since U(x, 0) = 0 in that region, the Vj(x, t) correspond-
corresponding to that solution must be chosen to vanish.
With <p(x, t) given by A0.475), vo{x, t) satisfies
(,0.476) ^ + т^ + д„0 = „.
Since vo(x, t) = e~A/, we easily conclude from A0.474) that
(Ю.477) Уо(л^)=_е-Л/.
The equation for vx(x, t) is
A0.478) ^ + У-^ + АУ1 = !
and the boundary condition is
790 ASYMPTOTIC METHODS
A0.479) Ui(
We readily find that
A0.480) ,,(x,«)
Combining these results we find the leading terms of our solution to be
A0.481)
u(x, t) = е-Жх + yt) - е~л'я[^ (» - yt) - ^]
ф+yt)~ Ty e""<x ~
a2c
We now bring A0.481) into a form where each coefficient of Ho and Hx is
evaluated on the appropriate characteristic. To do so we express vx(x, t) [see
A0.480)] as
^ C /,- *\-~Ar
r(c - y)
where ^ is defined in A0.475). Then A0.481) can be written as
A0.483) u(x, t) = e-uH(x + yt) - е~х'н[^ (x - yt) - -^
X2ac A \2t
A
J 2
(
r(c-yJ 2a 2y\c-y
ХНЕ PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 791
In contrast to A0.445), which gives the full (weak) solution of A0.442)-
A0.444) in the case where Л = 0, the expression A0.483) is only valid near
the three characteristics that occur in the expansion of u(x, i). We note that
A0.483) reduces to A0.445) when A = 0, but with Л > 0 there are additional
waves of the form Щ(х - yt) that have no counterpart in A0.445). Each
term in A0.483) is multiplied by an exponential decay factor e~Ar. The jumps
in и(х, 0 across x_+ yt = 0 and [(c - у)/(с + y)](x - yt) - [2ay/(c + у)] = 0
are e xt and -e л', respectively. The jump across the third characteristic
curve x — yt = 0 is seen to be zero. In contrast to the case Л = 0 where the
jumps are +1 or —1, we see that the jumps decay exponentially in the
damped case where A>0.
If the characteristics $ = 0, ф = 0, and <p = 0 are each expressed in a form
t= ф(х), the points at which ф(х) = constant are the wave fronts for the
problem A0.442)-A0.444). The aforementioned jumps are, in fact, the
jumps in the solution across these wave fronts at different times t. The
expansion A0.483) not only determines the jump in u(x, i) across the wave
fronts but also describes the wake behind these fronts. As indicated in
Example 5.15, for instance, what remains in the wake of the wave fronts at
large times is a diffusion effect. This conclusion cannot be reached, how-
however, by considering only the leading terms in the given expansion.
Adopting a slightly different point of view we express the solution of
A0.442H10.444) as
A0.484)
«(*, 0 - е-"Н(х + yt) - e"Ar»[f^ (x - yt) - J2Z-] + v(x91)
where we have retained the two leading terms in A0.483). If we insert
A0.484) into A0.442), we obtain
A0.485)
vtl - y\x + 2\vt = Л VA7/(* + yt) - \2е-х'н\^-1 (X - yt) - i^-1 .
L С i* у С "т" у J
AW = 0 A0.443) implies
A0.486) v(x, 0) = 0, vt(x9 0) = ЛВД, x > -a ,
and on the boundary x = -a- ct,
A0.487) i;(-a-cr,0 = 0,
The data for v(x, t) are less singular than those for u(x, t). Thus v(x, t) is
expected to be a smoother function than u(x, t) since it is the solution of a
problem with smoother data. If additional terms from the series A0.483) are
792 ASYMPTOTIC METHODS
included in A0.484), the resulting function v(xy i) becomes even smoother
Thus this method may be used to take into account singularities in the
solution and to reduce the problem to one that has smooth data and a
smooth solution.
In the preceding problem the data A0.443) were expressed directly in
terms of H(x) and 8(x), If the data are given in terms of a discontinuous
function F(x) defined as
A0.488) **>
and fix) can be defined or extended as a smooth function for all x, we can
write
A0.489) Fix) = fix) + H(x)[g(x) - /(*)].
If g(x) - f(x) can be expanded in a Taylor series around x = 0, we can
express A0,489) in the form
00
A0.490) Fix) = f{x) + 2 {gu\0) -/О)@)}ЯДлг)
/-о
/«о
where [FU)(x)]x=t0 represents the jump in the /th derivative of F(x) at x = 0.
More generally, if F(x) is a smooth function for x ^ a and for x ^ a, with at
most jumps in its derivatives at x = a, we can represent F(x) as in G.126)
and obtain
QO
A0.491) F{x) = F{x) + 2 [Г°\х)]х-аН,(х - a) .
/-o
At each point where F(x) has a jump it can be expressed in this manner.
If Fix) is given as in A0.490) or A0.491), its generalized derivatives can
be obtained by differentiating the series term by term. Apart from addition
of a finite number of terms that involve the delta function and its deriva-
derivatives, the derivatives of Fix) are still represented in terms of a series of the
functions Hjix) or Hjix - a).
Example 10.11. Initial Value Problems for the Telegrapher's Equation.
To begin, we consider the initial value problem for the telegrapher's
equation
A0.492) utt - y2uxx + 2Xut = 0, -o
with the initial data
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 793
r-i, *<-i
A0.493) U(x, 0) = bin x, -\<x<\, ut(x, 0) = 0 .
The function u(x, 0) can be written as
A0.494) u(x, 0) = -1 + A + sin x)H(x + 1) + A - sin x)H(x - 1) ,
as is easily checked.
To account for the jumps in the data at x = 1 and x = — 1 we introduce
the expansions
A0.495) £/(*, 0=2 vj(x, Щ[<р{х, 01 + 2 Wj(x, (Щ[ф(х, t)].
Inserting A0.495) into A0.492), we conclude that both <p and ф must satisfy
the characteristic equation A0.447) and the y; and wj satisfy the (appropri-
(appropriate) transport equations A0.448), A0.449). At f = 0,
A0.496) U(x, 0) = 2 v,(x, 0Щ[ф, 0)] + 2 ",(*, 0Щ[ф(х, 0)],
and this can account only for the singular terms in the data A0.494). This
implies that we should set
A0.497) <p(xy 0) = x + 1, ф(х, 0) - x - 1.
Again we are led to consider both solutions of the characteristic equation
A0.447) that satisfy A0.497) and they are
A0.498) $(x, t) = x + l-yt, 4>(x,t)
A0.499) ф(х, 0 = * - 1 - У, «£(*, 0 = х - 1 + yt.
For each characteristic line <p = ф = ф = ф = 0 we introduce a series of
the form A0.446), and thus we replace A0.495) by
A0.500)
00 да
U(x, 0=2 Vj(x, Щ{х +1 - yt) + 2 v,(x, t)HXx + 1 + yt)
+ 2 *>j(x, t)H,(x - 1 - yt) + X й-Дх, {Щ(х - 1 + yt) .
Considering only the leading terms we have
794 ASYMPTOTIC METHODS
A0.501) U(x, 0) = [vo(x, 0) + vo(x, 0)]H(x + 1)
= A + sin x)H(x + 1) + A - sin x)H(x - 1),
A0.502) U,(x, 0) = [-yvo(x, 0) + yvo(x, 0)]S(x + 1)
+ [~ywo(x, 0) + ywo(x, 0)]S(x - 1) + • • • = 0 ,
since we are only concerned with the singular part of the data. Equating
coefficients of like terms in A0.501), A0.502) and solving the resulting
equations yields
A0.503) vo(x9 0) = vo(x, 0) = |A + sin x) ,
A0.504) wo(x, 0) = wo(x, 0) = |A - sin x) .
The transport equation for v09 tf0, w0, and w0 are easily solved, and we
obtain, on using A0.503), A0.504),
A0.505) U(x9t) = |[1 + sin(;c - yt)]e'kiH{x + 1 - yt)
+ |[1 - sin(A: + yt)]e~ktH{x - 1 + yt) + • • •.
The coefficients of the Heaviside functions in A0.505) are not evaluated
on the relevant characteristics x + \±yt = 0otx-\±yt = 0. Although our
original discussion of these expansion forms required that the transport
coefficients be evaluated on the characteristics, we showed in the preceding
that these expansions are formal solutions even if the coefficients are not
evaluated on the characteristics. However, each of the coefficients in
A0.505) can be expanded in powers of the relevant characteristics. For
instance,
A0.506) iT0(jc, 0 = \e~Kt[l + sin(* - yt)]
= \e~kt[{\ - sin 1) - |(cos l)(x + 1 - yt) + • • •]
and similar expansions can be carried out for v0, wQ, and w0. If this is done,
we obtain to leading terms
A0.507) U(x, t) = \{1 - sin l)e~xt[H(x + 1 - yt) + Щх + 1 + yt)
+ H(x-1- yt^ + H(x - 1 + yt)] + ' "
and the jump across each characteristic is |A — sin l)e~k\
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 795
Alternatively, we can expand 1 ± sin x in A0.494) in a Taylor series
around я: = +1, and A0.494) takes the form A0.491) where F(x) = -1 and
there are two series involving #Д* + 1) and Щ(х -1). We proceed as
before, the only difference being that the initial value of U(x, t) is now given
in terms of the foregoing series in H}(x ± 1). Then, the leading terms in
U(x, t) are given as in A0.507).
The solution u(x, t) of the given problem A0.492), A0.493) may be
represented as
A0.508) u(x, t) = v(xy t) + U(x, i),
with U(x, t) as given earlier. In view of A0.493), A0.494), the initial data
for v(xy i) are
A0.509) v(x, 0) = -1 + (sin x + sin l)H(x + 1)
- (sin x - sin l)H(x - 1) + • • •,
A0.510) vt(x, 0) = A(sin 1 -
In addition, U(x, t) gives rise to an inhomogeneous term that occurs in the
equation for v(x, t). Although u(xy 0) is discontinuous at x = -1 and x - 1,
v(x, 0) is continuous for all x. Consequently, v(x, i) is a smoother function
than u(x, t). We do not solve for у(лг, i).
Next we show how to treat problems in which the data do not lead to
jump discontinuities but result in stronger or, possibly, weaker singularities
in the solution. For instance, the initial value problem for the telegrapher's
equation in Section 1.2 has m(jc, 0) - 8{x) so that the solution is expected to
have a delta function singularity. Also, the problem considered for у(лг, i) in
A0.508) leads to a solution with discontinuous derivatives but that is
continuous everywhere.
These problems can be dealt with by observing that A0.446) can be
replaced by the expansion
00
A0.511) u(x, 0 = 2 v,.(x, OSjMx, г)]
where the (generalized) functions Sj[x] are related to each other by
(Ю.512) SJM = SHW,
but are otherwise arbitrary. If we set S0[x] = H[x]9 the Heaviside function,
A0.512) yields 5Длг] = Щ[х] and A0.511) reduces to A0.446). If A0.511) is
inserted into the telegrapher's equation A0.442), A0.512) is used, and
coefficients of the Sj[<p] are equated to zero, we conclude that <p must satisfy
the characteristic equation A0.447) and the v, satisfy the transport equations
A0.448), A0.449).
796 ASYMPTOTIC METHODS
For a given problem we may use the general series A0.511) and then
decide on the basis of the data what form So and the 5; should take. As an
example, we consider the problem of Section L2 in which u(x, t) satisfies
A0.513) utt-y2uxx
with the data
A0.514) u(x, 0) = 8(x), ut(x, 0) = 0 .
Expanding u(x, t) as in A0.511) we find that
A0.515) u(x, 0) = 2 vj(x, ОЩф, 0)] = 8(x).
Since the 5; are increasingly less singular for increasing /, we find that
A0.516) S0[x) = 8(x), ф,0)=х9
and consequently
A0.517) St[x]^H[x), S;M = #/_,[*], /al,
with the Hj defined as in A0.430). Again, A0.516) implies that we must
consider the two characteristics
A0.518) <p(jt, t) = x - yt = 0, ф(х, t) = x + yt = 0.
It is easily shown on using the transport equations for v0 and v0 that
A0.519) u(x91) - \e~kt8[x + yt] + \e~kt8[x -?*] + •••,
to leading terms. It is of interest to note that A0.519) confirms the result
given in A.97).
As another example we consider the initial value problem that yields the
causal fundamental solution for the telegrapher's equation as follows from
Chapter 7. The function u(x, i) satisfies A0.513) and the data are
A0.520) «(*, 0) = 0, u,(*, 0) = 8{x) .
Expanding u(x, t) as in A0.511) we have
00
A0.521) u(x, 0) = 2 vf(x, 0)Sj[<p(x, 0)] = 0,
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 797
A0.522)
u,(x, 0) = vo(x, 0)<р,(х, 0)S_l[<p(x, О)]
Xx, 0) + ^^-}si[<p(x, 0)] = S(x) .
Since S_x[x] - SqM is the most singular term in A0.522), we conclude
that
A0.523) S&x] = S(x), <p(x, 0) = x,
so that
A0.524) S0[x] = #[*].
Consequently, the series A0.511) for this problem is identical with A0.446)
since Sj[x] = Hj[x].
Adapting the result F.222) it is easy to show that the causal fundamental
solution of the telegrapher's equation is
A0.525) u(xy t) = j- e~A70[- y/yY^^Hiyt-x)H(yt + x) ,
where /0(z) is the modified Bessel function of zero order. On using the
Taylor series for I0(z) and considering A0.525) near each characteristic
x - yt = 0 and x + yt = 0 for t > 0, it can be shown that the expansion based
on the use of A0.511) agrees completely with the expression obtained from
the exact solution A0.525). To accomplish this, however, the terms in the
expansion must be evaluated on the appropriate characteristics x — yt = 0 or
x + yt — 0. This is considered in the exercises.
An important conclusion that can be drawn from the preceding examples
is that we need not consider only solutions of A0.410) having the expansion
form A0.432). Instead, we can expand и in the more general form
(Ю.526) и = 2 v,S,[<p],
where the Sj[x] satisfy the relations
A0-527) 5;[jc] = 5/_1[jc].
This represents a formal solution of A0.410) if <p satisfies the characteristic
equation
798 ASYMPTOTIC METHODS
A0.528) <p] - p(V<pf = 0
and the vt satisfy the transport equations
A0.529) 2<pt -£ - 2pV<p - Vt,;. + 2\<plVj + [<plt - V• (pV<p)]Vi
д
where it is assumed that v^t=0y so that v0 satisfies a homogeneous
equation.
The expansion A0.526) no longer need be restricted to describing the
propagation of singularities. For example, if we set
A0.530) S0[q>] = еы\ S,[9] = ^7 e"» ,
where to is a large (real) parameter, A0.526) may be regarded as an
asymptotic solution of A0.410). It would occur if the problem contains
rapidly oscillating data. For instance, if initially we have
A0.531) m(x, 0) = f(x)ei<O8ix\ м,(х, 0) = 0
where a> is large, we look for a solution of A0.410) in the form A0.526) with
the St given by A0.530). From the initial data A0.531) we conclude that
<p(x, 0) = g(x). There are two solutions <p(x, t) and £(x, t) of A0.528) that
satisfy this initial condition. For each of those functions we construct a series
A0.526) with coefficients vj and tf;, respectively. On inserting these series
into the data A0.531) we easily determine initial conditions for vj and vp
each of which satisfies the transport equations A0.529).
With A = q = 0 and p = y2 = constant in A0.410) the choice
A0.532) <Р = Ф(х)~1, 1>у = |>,(х)
with the Sj defined as in A0.530) leads to the geometrical optics expansion
for the reduced wave equation as is immediately seen. In fact, the series
A0.533) ^=2„(Х) * ^-«»>
is a solution of the reduced wave equation as followi on setting
A0.534) М(х,0 = ^1л>'Ф)в^(х).
With a> = к A0.534) reduces to the form considered in Section 10.1. In view
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 799
of the fact that solutions of A0.410) undergo rapid variations across
(characteristic) singularity regions, it is not surprising that solutions of the
type A0.526) and A0.530) with rapidly oscillating terms are similar in form
to solutions that describe the propagation of singularities.
In the present section, however, we are only interested in discussing the
propagation of singularities for solutions of A0.410). The preceding exam-
examples have dealt with one-dimensional problems and we now consider a
problem in two dimensions.
Example 10.12. The Two-dimensional Wave Equation. Let u(x, v, t)
satisfy the wave equation
A0.535) и„-У2(и„ + и„) = 0,
where у is a constant. The initial data for и are
{222
% Xx2tyy2iaa2> »,(*.* 0) = 0.
With p2 = x2 + y2, A0.536) can be written as
A0.537) u(x9 y,0)- a + (j3 - a)H[p - a], и,(х, j>,0) = 0,
where H[x] is the Heaviside function and a and {$ are constants. We may
expand и as
00
A0.538) u(x, y, t) = a + 2 vfr, y, t)S,[q>(x9 y, t)].
At t = 0 we have
A0.539)
u(x, y, 0) = a + 2 Vj(x9 y, 0)S.[<p(x, y, 0)] = a + (K - а)Я[р - fl].
This suggests that we set
A0.540) S0[<p] - H[q>], S.[g>] = ВД
with the Щ defined as in A0.430) and
A0.541) <p(*,J>,0) = p-e.
Since <p must satisfy the characteristic equation <p2 - -/(<p2x + ^) = 0, we
obtain two possible solutions
800 ASYMPTOTIC METHODS
A0.542) ф(х, y,t) = p-a-yt, ф(х, y,t) = p-a + yt.
To account for <j> and ф we replace A0.538) by
00 00
A0.543) u(x, y,/) = a + 2 Щ[$\ + 2 Щ[Ф] -
Inserting A0.543) into the initial conditions A0.537), we easily conclude
that
A0.544) vo(x, y,0) = vo(x, y,0)=i(fi ~ a) .
The transport equations for v0 and v0 are
and the solutions of A0.544)-A0.546) are
A0.547) vo(x, y, t) = UP - «
p J
These expressions may be expanded in powers of <p and ф. Then
A0.548) ^[jV^
1 Г 11/2
A0.549) Ы
Thus the leading terms in the expansion of u(x, y, i) are
1 Г Л112
A0.550) м(х, у, t) = а + - (j3 - а)[ ^ j Я[р - а - у-
11/2
The coefficients of the Heaviside functions in A0.550) could have been
determined by expressing A0.545), A0.546) as ordinary differential equa-
equations along the bicharacteristics of <p == 0 and ф = 0, as was done in A0.425).
Alternatively, since the coefficients are evaluated on the characteristics, we
can write A0.550) in the equivalent form
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 801
A0.551) u(x, y,t) = a + ±(p- flOJj]'Пн[(^-у) - t]
where the coefficients are determined from a ± yt = p.
The characteristic surface 4> = p-a-yt = 0 represents cylindrical wave
fronts that travel outward from the circle p- a with speed dpldt = y. The
jump in и across the wave fronts decays like t'112 for increasing t. The
characteristic surface <p-p-a + yf = 0 yields cylindrical wave fronts that
travel inward from p = a with speed y. At the time t = aly the wave fronts
collapse on the point p = 0, and A0.550) is singular there since the
coefficient of H[p - a + yt] blows up. The singularity results because the
origin is a focal point for the bicharacteristics. We do not consider how
A0.550) should be modified to deal with this singularity.
The usefulness of the expansion A0.526) in solving problems for the
hyperbolic equation A0.410) depends on the possibility of representing the
data for A0.410) in the form A0.526) and the ability to determine the
coefficients in the expansion of the solution in terms of those in the data. It
may happen, however, that the bicharacteristics of <p = 0 intersect in the
initial or boundary region. This occurs, for example, if <p = 0 is a characteris-
characteristic cone or conoid with vertex at t = 0, in the two- or three-dimensional case.
As a result, the leading coefficient v0, at least, in A0.526) must be singular
where the bicharacteristics intersect and it is not possible to determine its
value directly in terms of the data.
As an example, we consider the causal fundamental solution for the
Klein-Gordon equation
A0.552) utt - y2[uxx + uyy + uzx] + c2u = 0, -oo<x, v, z
It follows from our discussion in Chapter 7 that it can be determined as a
solution of A0.552) with the initial data
A0.553) u(x, y, z, 0) - 0, ut(x, y, z, 0) - 8(x)S(y)8(z) .
It is given as (see Exercise 7.4.9),
A0.554) „(*, y, z, t) = — S[yt - r] - 4JVy2f2_r2 H[yt - r],
for t > 0, where r2 = x2 + y2 + z2 and Jt is the Bessel function of order one.
Using the Taylor series for Jx{z) we have
802 ASYMPTOTIC METHODS
A0.555)
1 S[yt-r]
4тгуг
с у (-l)'(
Since [yt + r]J = [2r + yt ~ r]J and can be expanded in powers of yt - r, we
see that A0.555) has the form
00
A0.556) u(x, y, z, t) = 2 wy(*. У, г, O-Si[y»" ^ ,
where
A0.557) S0[x] = 8[x), ад-ЯДх], /al,
with ЯДх] defined as in A0.430). The transport coefficients Vj are evaluated
on the characteristic surface <p — yt - r = 0. The leading terms are
<10-558> ^
We now attempt to reproduce these results by using the direct expansion
method and set [with the Sj given as in A0.557)]
oo
A0.559) и = 2 wA
j0
j=0
and insert this into A0.552). Then <p must satisfy the characteristic equation
A0.560) tf-y2(V<pJ = 0,
and the i>; satisfy the transport equations
A0.561) 2<p, -£■ - 2y2V<p -Vvj + [<ptt - r2VV]U/
at
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 803
where 1^=0. Since the data for this problem are concentrated at
(x, у9 z) = (°> °> °)> we must choose <p = 0 to be the forward characteristic
cone
A0.562)
The equation
A0.563)
and this has a
A0.564)
where Vo may
A0.565)
for v0 becomes
~!t~
solution
be an arbitrary
and this has the solution
/1П ^А£Л
<p~yt r-0.
У dr rV°~
Oo=7>
function of yt — r.
dvx у с2
dr r l 2y
V c2V
For vx we have
vo
with Vj as an arbitrary function of yf - r (yx and V2 are constant on <p — 0).
If we set VJ, = 1 /4тгу and Vj = 0, the first two terms in A0.559) agree with
those in A0.556). The singularity of v0 at r = 0 results from the fact that the
bicharacteristics of the forward characteristic cone A0.562) all intersect
there. Nevertheless, we have shown that the causal fundamental solution
can be expanded in the form A0.526). However, for this approach to be
useful, we must determine the coefficients in A0.559) and the 5;- in terms of
the initial data A0.553) and not by comparing terms with the expansion of
the exact solution as we have done. While some progress in this direction
can be made by expressing the three-dimensional delta function in a
spherical coordinate form, this technique is not easily carried over to more
complicated problems. Therefore we now present an alternative method for
constructing causal fundamental solutions of hyperbolic equations.
We represent the causal fundamental solution A0.554) of the Klein-
Gordon equation in terms of the square of the hyperbolic distance и defined
as
A0.567) a = ,2,I_
У
(see Example 6.14). It has the form, for f >0,
804 ASYMPTOTIC METHODS
A0.568) u(x, j-,z,0-
as is easily verified on using the properties of the delta function given in
Section 7.2.
On expanding Jx in a Taylor series, as in A0.555), we find that и takes
the form
A0.569)
Then, if we set S0(a) = 8(cr) and Sj(a) = НИ1(о), /s> 1, we have Sfo) =
Sj_1(a) and и can be written as
(Ю.570) «=£«да. -/
While this series has the form of A0.526) and the Sf satisfy A0.527), it
differs from A0.526) in that the surfaces a = constant are not characteristics
of the Klein-Gordon equation unless the constant equals zero. In fact,
A0.571) <r*-y2(VaJ = 4cr
and only a = 0, which is the characteristic cone, satisfies the characteristic
equation A0.560). We further observe that with S_t(a) = 8'(a) and
S^2(cr) = 8"(a) we have
A0.572) aSjia) - JSJ+1(a)9 j = -2, -1,0,...,
as is easily shown.
We now show how to determine the coefficients yy in A0.570) directly
from the Klein-Gordon equation and the initial conditions. To do so, we
insert A0.570) into A0.552) and make use of A0.572) to obtain
A0.573)
after some elementary calculations, with y_t =0. On equating the coeffici-
coefficients of the Sj to zero, the expression in the braces in A8*573) yields the
transport equations for the vr They differ from ths transport equations
obtained previously because a = constant is not a characteristic.
As we only need to integrate these equations over the forward charac-
characteristic cone t= r/y, and r-V= r{d!dr), we have
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 805
A0.574) t± + rA = rj. on ,.r/y.
We require that each Vj be bounded at the vertex of the cone where r = 0,
and this implies that v0 and all the vf must be constants. By induction, we
find that
(-l)'c2'
A0.575) vi = K 2J v0, / = 1,2
With v0 = 1/2тгу3, the result agrees with A0.570).
The determination of uQ may be based on comparing the solution with the
causal fundamental solution of the wave equation near the singular point at
the vertex of the characteristic cone where the most singular part of both
fundamental solutions must agree. While we have specified the form of the
5- in the expansion A0.570) that was inserted into the Klein-Gordon
equation, this need not have been done. As will be seen in the following
example, the form of the 5; can also be determined from the conditions
given for the problem.
Example 10.13. Causal Fundamental Solutions of Hyperbolic
Equations. We consider the hyperbolic equation
A0.576) L(u) = utt - V- (pVu) + 2Aw, + qu = 0 ,
in two or three dimensions, where p —p{r)—with r as the distance from the
origin in two or three dimensions—and A is a constant. With the data и = 0
and ut - 8 at t = 0 (the delta function is singular at r = 0), the solution of
A0.576) for t> 0 over the entire space R2 or R3 yields the causal fundamen-
fundamental solution for A0.576). We construct an expansion of the fundamental
solution in terms of the square of the hyperbolic distance associated with
A0.576) and determine the leading term in the two- and three-dimensional
cases. This method can also be used for the one-dimensional problem. In
that case, however, the solution can be determined completely as an
expansion in terms of the relevant characteristics, as we have shown.
The hyperbolic distance element s for A0.576) is given as
A0.577) ds2 = dt2--dr\
P
so that the square of the hyperbolic distance a measured from r = 0 and
' = 0 is
A0.578) 1г-|*
We easily verify that a satisfies
806 ASYMPTOTIC METHODS
A0.579) <r2-p(V<rJ=4a.
The fundamental solution of A0.576) is expanded in the series
00
A0.580) и = и(г, 0 = 2 v,(r9 t)Sj(*), S'f = SH1 ,
with the Vj and the 5y. to be determined. On inserting A0.580) into A0.576)
we obtain
A0.581) 2 {[a2 -
- V- (pVir))^.! + L(i;.M;] - 0 .
Now or = 0 represents a characteristic conoid for A0.576) with vertex at
(r, t) = @,0). We determine the form of the S, in A0.580) by requiring that
the solutions of the (transport) equations for the v- be nonsingular at the
vertex of the characteristic conoid. Only the forward characteristic conoid
t= Jo ds/y/p(s) concerns us and we have for v0,
A0.582)
2*t ^f - 2p(r)ar ^ + 2Xatv0 + [<rtt - V
^n A
in the two-dimensional case, with differentiation taken over the forward
characteristic conoid.
On collecting the coefficients of v0 in the braces in A0.582) we determine
that they behave like 3/2r + 0A) near r = 0. Consequently, if we equate
A0.582) to zero to obtain a transport equation for v0, the solution is singular
at r = 0. However, in view of A0.579) the leading term in the expansion
A0.581) is given as 4aS_2v0. Thus if we equate this term to -eS^o» we
find that
A0.583) 4<rS_2 + 65_a = 4aS% + 65^ = 4[aS^ + | sj - 0 ,
THE PROPAGATION OF DISCONTINUITIES AND SINGULARITIES 807
and on equating the resulting coefficient of 5_x in A0.581) to zero, we now
have
A0.584)
and the singularity in the coefficient of v0 is eliminated.
Noting the properties of the function fa{x) defined in Example 7.1, we
determine from A0.583) that
A0.585) S0(a) --
The solution of A0.584) is
A0.586) vo = co\—tL= f -A=] e"Ar on t=\-^=,
where c0 is an arbitrary constant. With p = constant and q = A = 0, A0.580)
must reduce to the fundamental solution of the two-dimensional wave
equation. On comparing our result with F.208) we find that co =
1/2ttVp@). Then, the causal fundamental solution is approximately given
as
A0.587)
i
1-1/4
r Jo y/p@)p(s) J е Г Tf
where the argument <r of the Heaviside function has been replaced by
1 ~~ Jo 1 Np(s) ds, which is valid for t > 0. We do not obtain any further
terms in the expansion A0.580).
The only difference that occurs in the case of three dimensions is that the
Laplacian operator in spherical coordinates has the dldr term multiplied by
2 instead of by 1 as is the case for cylindrical coordinates. This has the effect
that we must set
A0.588) 4o-5_2 + 85.t = 4[<rS% + 2S£ = 0 ,
808 ASYMPTOTIC METHODS
in order to eliminate the singularity in the equation for v0 at the vertex of
the characteristic conoid. The solution of this equation for So is
A0.589) S0{o) = 8{&).
On solving for v0 and choosing the arbitrary constant by comparing the
result with G.355) we obtain
(Ю.590) u(r, t)- ^ [p@)p(r)]"l'VA'a[vmt - [ тЩ ds],
which is valid for f>0. Again we do not obtain further terms in the
expansion of u(r, t).
We have shown in our general discussion and in the examples how the
behavior of solutions of hyperbolic equations near singularity regions can be
determined without having to find the full solution. Further examples are
given in the exercises.
10.3. ASYMPTOTIC SIMPLIFICATION OF EQUATIONS
In the preceding sections of this chapter and in Chapter 9 we discussed
techniques for obtaining approximate solutions of initial and boundary value
problems for partial differential equations. In this, the concluding section of
the text, we consider methods for replacing the equations themselves by
simpler, more easily solvable, equations. Although the use of perturbation
and asymptotic methods also results in the solution of simplified equations,
the approach we use differs from that employed in the aforementioned
methods. We show how to construct the simplified equations but do not, in
general, indicate how they may be used to solve specific initial and boundary
value problems. As shown, general information about the behavior of the
solutions of the given full equations can be inferred from their simplified
forms.
To fix ideas we discuss the dissipative wave equation E.293), that is, with
constant a and c,
A0.591) utt - c2uxx + и, - aux = 0 .
Assuming that the point (jc0, t0) is a discontinuity or singular point for the
data for A0.591), we study its effect on the solutions of A0.591) by setting
x - x0 = €<r and t - /0 = €т. This yields
A0.592) urT - с2и^ + €[ur - aua] = 0 .
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 809
Although we have, in effect, introduced a boundary layer stretching to
obtain A0.592) we do not proceed as in boundary layer theory by expanding
и in powers of e. Instead, we introduce an approximate factorization of
A0.592). If €=0, then A0.592) can be factored exactly as was done in
B.1), but if e #0 we cannot factor A0.592) in a simple fashion, yet since e
is small it can be factored approximately. Letting da = dldv and dT = dldr in
A0.592) we have
A0.593) K2 - c2d2a + edr - eada]u
= [dr - cda + €a+ O(e2)][dT + cda + eb + 0(e2)]w = 0 ,
where a and b are to be specified. (We may assume that a and b are
constants.) Multiplying out the two first order operators in A0.593) and
comparing the results with the second order operator in A0.593), we find
that
c— a , c + a
A0.594) « = -2Г, * = ^Г-
Since the order of the operators in A0.593) can be interchanged, we see that
to order e2, A0.592) is equivalent to the two first order equations
(Ю.595)
Neglecting the O{e2) terms in A0.595), we obtain the solutions
Г /с~ау
/(о- + ст)ехр -[-ТГГ.
g(a - ст) exp[ - у-^~)ет\ ,
A0.596)
where / and g are arbitrary functions. If |a|<c, these solutions decay
exponentially as /=er-»<*>, and it is easy to show that the initial value
problem for equation A0.591) is unstable if \a\ > c. Since discontinuities in
the solutions are carried along the characteristics a ± cr = constant, we see
that A0.596) is useful in describing the propagation of discontinuities in the
solutions of A0.591). In fact, the neglected O(e2) terms in A0.595) are
antiderivative or integral operators that have the effect of smoothing out
singularities in the solutions. The expressions for и in A0.596) have the
general form of the solutions given in the preceding section.
To study the solutions of A0.591) at large times and in the far field we set
x ~ ale and / = т/е and obtain
810 ASYMPTOTIC METHODS
(Ю.597) e[urr - c2uatr\ + мг-ам,=0.
Then using A0.597),
A0.598) urr = auar-e[urTr-c2ua<rr]
= ^N + O(e) = aV + O(e),
and A0.597) can be replaced by
A0.599) uT - aua + e(a2 - c2)uva = O(e2) .
If \a\ < c, so that the initial value problem for A0.591) is stable, we see that
on neglecting the O(e2) terms in A0.599) we have a diffusion equation. In
the original variables this equation has the form
A0.600) ut - aux + e(a2 - c2)uxx = 0
and this shows that for large x and t the dissipative wave equation yields a
wave motion along the line x + at — 0 modified by a diffusion effect repre-
represented by the second derivative term. That is, if e ~ 0 in A0.600), we have
undirectional wave motion along x + at = 0. The O(e2) terms neglected in
A0.599) are higher derivatives of м. Since the diffusion effect smooths out
the solutions, these terms can be disregarded if they are multiplied by higher
powers of e. Our conclusions are consistent with those obtained in Example
5.15 and the discussion in Section 1.2.
There are two types of wave motion associated with A0.591). The
principal part of A0.591)—that is, un — c2uxx—yields waves moving to the
right or left with speed c. The reduced part ut - aux gives rise to a traveling
wave moving with speed |a|. We have shown that wave motions with both
speeds play a role in the solution of A0.591) in different regions. In A0.595)
we obtained approximate equations corresponding to waves moving with
speed с and in A0.599), A0.600) we obtained an equation characterizing
wave motion with the speed \a\. In this section we show how to construct
such simplified approximating equations that retain certain aspects of the
given equations. Since our method is most easily presented in terms of
matrix theory, we restrict our discussion to systems of equations.
With v - ux and w = ut, the dissipative equation A0.591) can be written
as a first order system
A0.601) а, + Ли, + Bu = 0 ,
where
Then A0.592) is equivalent to
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 811
A0.603) uT + Aua + eBvk = 0
and A0.597) corresponds to
A0.604) e[uT + Aua) + Bu = 0
We introduce decompositions of A0.603) and A0.604) that correspond to
the reduced equations obtained on setting € = 0 in each of these systems of
equations. To carry out these decompositions we require a result from
matrix theory.
Let M be an n by n matrix and C(A) = |M-A/| its characteristic
polynomial We factor C(A) as C(A) = Q(A)C2(A) where the factors CX(A)
and C2(A) contain no common eigenvalues as roots. Introducing the partial
fraction decomposition
A0.605) C(A) - C(A) + Cz(A) ,
we have
A0.606) 1 = c,(A)C2(A) + c2(A)C,(A).
Then the matrices
A0.607) P1 = Cj(M)C2(M), P2 =
are projection matrices that project vectors into the eigenspaces spanned by
the eigenvectors corresponding to the factors CX(A) and C2(A), respectively.
They have the following additional properties
A0.608) Pi + P^I, PtP2^P2Px^09 Р1М
We do not require any additional facts regarding these matrices for our
discussion beyond what was stated. We assume in our discussion that the
matrix M has n linearly independent eigenvectors.
As an example, we construct projection operators for the matrices A and
В defined in A0.602). The characteristic polynomial for A is C(A) =
\A - A/| = A2 - c2 = (A + c)(A - c). With CX(A) = A + с and C2(A) = A - c,
we have
-c) + (l/2c)(A
A2-c2
812 ASYMPTOTIC METHODS
Thus Cl(A) = -c2(A) = l/2c, cx(A)C2(A) = (-l/2c)(A - c), and c2(A)C1(A)
= (l/2c)(A + c). Finally, the projection operators are given as
A0.610) ^—^-^[с l]f
A0.6И) ^=^И + с/] =
The matrix A has two eigenvalues Xx = -c and A2 = с The corresponding
eigenvectors are
A0.612) г, =
and they are linearly independent. We have P1r1 = rls Pxr2 =0, Р2тх =0,
and P2r2 = r2. Since any two-component vector can be expressed as a linear
combination of тг and r2, this shows that P1 and P2 project vectors into the
required eigenspaces (these are one-dimensional in this case). Also, it is
easily seen that Px and P2 have the properties given in A0.608).
For the matrix В in A0.602) we have C(A) = \B - A/j = A(A - 1). With
Q(A) = A and C2(A) = A-lwe have
so that
The eigenvectors r2 and r2 corresponding to the eigenvalues Хг = 0 and
A2 = 1, respectively, of 5 are
A0.615) Г1 = [^J, Вт, = 0, r2 = [JJ, 5r2 =r2 .
Again, rx and r2 are linearly independent, and Px and P2 have all the
properties required of projection operators.
We now use these projection operators to construct asymptotic decompo-
decompositions of A0.603) and A0.604). Beginning with A0.603) we set
A0.616) u(<r, r) - V(<r, т)тг + eW(a, r)r2
where V and W are scalar functions that may depend on e, and r2 and r2 are
the eigenvectors of A given in A0.612). Inserting A0.616) into A0.603)
gives
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 813
A0.617) [VT - cVJr, + e{(WT + cWJr2 + v[c ° e
Applying the projection operator Px to A0.617)—that is, multiplying across
on the left by Pi—gives
(Ю.618)
since P,r2 = 0. Applying P2 to A0.617) yields
A0.619) e{wT + cWv - {^f)v}r2 + ^2(£jf)Wr2 = 0 .
Retaining terms up to order e in A0.618) as was done in our discussion of
the scalar example, we find that
A0.620)
and this agrees with A0.595). The leading term of A0.619) shows that W
can be expressed in terms of V apart from an arbitrary function of a — cr. If
this arbitrary function is disregarded, we can replace W in A0.618) by a
function of V given in integral form. Higher approximations may be
obtained by expanding W in a power series in e and solving A0.619) for the
coefficients as functions of V, but we do not pursue this matter here. We do
note that if r: and r2 are interchanged in A0.616), we obtain the second
form of the equations for и given in A0.595) as the equation satisfied by V.
Next we consider the system A0.604) and again represent u as in
A0.616), with r2 and r2 as defined in A0.615). Inserting A0.616) into
A0.604) yields
A0.621) €{утГ1 + V9[ Iасг] + Wr2} + e2{WTr2 + Wa[ ™J]} = 0 ,
since Bvx = 0. Applying the projection operator Pl9 as given in A0.614), to
A0.621) gives
A0.622) {VrJ^
as is easily seen. Applying P2 = В to A0.621) results in
A0.623) {(a2 - c2)V; + W}r2 + e{Wr + aWJr2 = 0 .
We solve A0.623) by expanding W as
814 ASYMPTOTIC METHODS
A0.624) W= S W,€J .
Then Wo has the form
(Ю.625) W0(<7,r) = (c2-a2)V;.
The expansion A0.624) must also be inserted into A0.622), and, retaining
only terms up to order e, we have
A0.626) VT - a Va - e(c2 - a2)VafT = 0 .
This result has the form of A0.599). Clearly, the Wj in A0.624) are all
expressed in terms of increasingly higher order derivatives of V, as is seen
from the form of A0.623). Thus in neglecting O(e2) terms to obtain
A0.626) we are disregarding presumably smoother higher derivative terms
in V, in agreement with the discussion given for the scalar version of this
example.
For A0.604), we do not consider the effect of interchanging rt and r2 in
the assumed solution form A0.616) because on setting e = 0 in A0.604) we
obtain Bn = 0, and this has the solution u = Vrx where V is arbitrary. It is
this expression that was perturbed around in A0.616). Similarly, with 6=0
in A0.603) we have uT + Anff - 0, and this has the solutions u = V(a + cr)rl
and u = V(a - cr)r2 with arbitrary V and VI Therefore, we considered
perturbations corresponding to each of the eigenvectors rx and r2 in our
discussion of A0.603).
The system obtained on setting e = 0 in A0.603) or A0.604) is known as
the reduced system. We constructed asymptotic simplifications of the full
systems corresponding to solutions for the reduced system. The solutions of
the reduced system are related to the properties of a relevant coefficient
matrix. Based on a decomposition of that matrix according to its eigenvec-
eigenvectors (even though this was not done explicitly), a decomposition of the full
system was carried out. We now show how this technique may be applied to
a general class of systems of equations.
We consider systems of equations of the form
A0.627) u, + Anx + eN[u] - 0
where u is an «-component vector, A is a constant n by n matrix, and N is a
linear or nonlinear differential operator. The matrix A is assumed to have n
linearly independent eigenvectors. With C(A) = \A ~ A/| as the characteris-
characteristic polynomial of A, we assume it can be factored as C(A) — Q(A)C2(A)
where Q(A) and C2(A) have no common eigenvalues as roots. Further, it is
assumed that 0 < m < n independent eigenvectors are associated with the
eigenvalues of Cx(k) and the remaining n - m independent eigenvectors are
associated with C2(A). The eigenspaces spanned by these two sets of
eigenvectors have the property that any w-component vector can be ex-
expressed as a sum of two vectors, each of which is one of the eigenspaces.
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 815
Proceeding as before, but replacing M by A in the equations, we can
construct the projection operators Рг and P2 associated with the eigenspaces
corresponding to Ct(A) and C2(A), respectively. If v lies in the eigenspace
related to Ct(X) and w lies in the eigenspace related to C2(A), we have
p v = v, PjW = 0, P2w = w, and P2\ = 0 as easily follows from the properties
of Px and P2. We recall that the projection operators were defined in
A0.607).
With v and w lying in the two eigenspaces defined previously, we look for
a solution of A0.627) in the form
A0.628) u = v + «гw .
In doing so it is assumed that we are interested in perturbing u around a
solution of the reduced system u, + Aux = 0 which lies in the eigenspace
associated with v.
Inserting A0.628) into A0.627) yields
A0.629) v, + A\x + c{w, + Av/X + N[\ + ew]} = 0 .
We assume that N[\ + ew] can be expanded in powers of e and that
N[\ + ew] = No[\] + O(e). Multiplying A0.629) by the projection operator
Px gives
A0.630) v, + A\x + €PxN[y + ew] = 0 .
To obtain A0.630) we have used the facts that PXA = APX and that Px is a
constant matrix. Thus, for example, PXA\X = АРг\х = A{dldx)[Pxy\ = A\x
and Pxvit = {dldi)[Pxvi] = 0. Similarly, we obtain on multiplying A0.629) by
A0.631) e{w, + Avfx + P2N[\ + ew]} = 0 .
Next, we expand w as
A0.632) w=2w,€'"
; = 0
and insert this series into A0.630) and A0.631). Using the expansion of
ЛГ[у + ew] in powers of e, we equate like powers of e in A0.631). This yields
A0.(833) *£ + л*£ + РМу1]„0
for w0, and similar equations for the remaining w-. The equations for w are
to be solved only in terms of v and any solutions of the homogeneous
versions of these equations (which do not depend on v) are to be discarded.
The resulting expressions for the wy. are then inserted into A0.630). This
yields the simplified system for v since v belongs to a lower dimensional
816 ASYMPTOTIC METHODS
space than u. Thus A0.630) effectively contains fewer equations in fewer
unknowns than A0.627).
The equation A0.630) for v may have a very complicated form. How-
However, we do not expand v in powers of e since this—in combination with
A0.632)—would yield the conventional perturbation solution of A0.627)
That solution may contain secular behavior that we are trying to avoid
Instead, we truncate A0.630) at some order of e. This new equation might
then be used as a basis for a perturbation result. For example, retaining only
O(e) terms in A0.630) gives
A0.634) v, + A\x + eP.N^y] = 0.
This equation corresponds to A0.620) in the example considered earlier.
Combined with A0.633), this yields a leading order approximation to a
solution of A0.627). The leading order perturbation result can always be
retrieved by expanding v in a series in powers of e. Thus A0.630) represents
a more general result and it can be constructed for each eigenspace
associated with the matrix A.
For example, if A = A2 is a simple eigenvalue of A and тг is the
corresponding eigenvector, we may consider Сг(А) = A - A2. The eigenspace
for Q(A) is then one-dimensional and is spanned by тг. Then we can set
\~Угг, where V is a scalar function, in A0.628). Since Px projects vectors
into the eigenspace spanned by rl9 we have PjiV[v + ew] = a(V, w)rl where
a is a scalar function of V and w. Thus A0.630) takes the form
A0.635) [Vt + XXVX + ea(V, w)]r, = 0 ,
and this is scalar equation for V once the w. are specified in terms of V.
In the two examples considered we show how equations of the form
A0.627) arise in fluid dynamics and the theory of water waves. The
aforementioned theory is then used to obtain simplified forms for these
equations. We note that the simplification method presented can also be
applied if the matrix A in A0.627) is a slowly varying function of x and t.
Also, a related simplification method can be used even if the given equation
is not in the form A0.627) as we have shown for the dissipative wave
equation. However, we restrict our discussion only to the preceding constant
coefficient case. An extension of this method to higher space dimensions is
not obvious since the reduced system would generally contain more than
one coefficient matrix to be dealt with in such manner. If one of the spatial
variables can be distinguished from the others for some reason, this proce-
procedure can be carried out but this is not considered here.
Example 10.14. The Navier-Stokes and Burgers' Equations. The
Navier-Stokes equations that describe viscous fluid flow in one dimension
were given in Example 8.9 as [see (8.365)-(8.368)]
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 817
pux + upx = Q,
A0.636)
A0.637)
A0.638)
A0.639)
P,
Pu,
PCV[T,-
p =
Here p is the density, и is the velocity, p is the pressure, and T is the
temperature. The parameter fi is the viscosity coefficient, к is the coefficient
of heat conduction, cp and cv are the specific heats at constant pressure and
volume, respectively, and R - cp - cv is the gas constant.
To study these equations we perturb p and T around a constant density
and temperature and и around a zero velocity. Thus we set
A0.640) u = eu, p = po + ep, T=T0 + €f,
where e >0 is a small parameter. Also, we consider the solution in the far
field and at large times and let
A0.641) *=-, *=-.
Inserting A0.640), A0.641) into A0.636)-A0.639) and using A0.639) to
express p as a function of p and T, we readily obtain
A0.642) pT + рои„ + е[рйа + йра] = 0 ,
A0.643) *,+ {&)*. * Rt, * .[(£)*, ♦ (£Ц + Ш.
(Ш.644,
In these equations we have only retained terms up to order e. If we set
у = cplcv, the ratio of the specific heats, we find that with cp = 1 /(y - 1), we
have cv = l/y(y - 1), R = 1/y, and Rlcv = у - 1. By rescaling the variables
and parameters in A0.642)-A0.644), it is possible to eliminate the constants
p0 and TV Thus we may set p0 = To = 1. Also, for simplicity of notation we
drop the carets on the variables p, w, and t and replace a and т by л: and U
As a result, the system A0.642)-A0.644) can be written in the form
818
A0.645)
where
A0.646)
ASYMPTOTIC METHODS
u, + Aux + e[Bux + Cu, + Duxx] = 0 ,
и
T
У
0
1
У
0
0 (у-
1
0
у-1
р
и
1)(ГЧ
0'
1
У
0_
■р)
0"
р
У
и
D =
0
0
0
с=
_
0
0
0
4м
3
0
0
р
0
к
0"
0
р
0
0
A-у)у
A0.647) В
This system is of the form A0.627), and we note that the operator
multiplying e is nonlinear. The linear system u, + Anx = 0 obtained on
setting e = 0 in A0.645) yields the equations of linear acoustics.
The characteristic polynomial for A is C(A) = \A - A/| = A(l - A)(l + A)
so that the eigenvalues of A are Ao = 0, Ax = 1, and A2 = -1. The corre-
corresponding linearly independent eigenvectors are
A0.648) г0 =
Using these results we easily find that the reduced system
A0.649) ut + Aux = 0
has the solutions
" 1 ■
1
у-1.
> 2 ™~
' 1 '
-1
.У-1.
A0.650) u = a(jc)r0, u = p(x -
u = 8(x + 0^
where a, /3, and 8 are arbitrary functions. The general solution of A0.649) is
a linear combination of these three solutions and it shows that two waves
traveling to the right and to the left with unit speed occur in the solution of
the acoustic equations.
Using the (conventional) perturbation method to solve A0.645) we set
A0.651)
u = un
Then u0 satisfies the reduced equation A0.649). If we choose u0
P(x — /)r1? we readily find that Uj satisfies
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS
819
A0.652) —1
where
A0.653)
Ё =
1
у-
У
0
1
1
1
1
Ку-1)
0
1
У
1
Г0 0 0]
C= 0 1 0
Lo о iJ
Either by expressing ux and all the vectors in A0.652) as linear combinations
of the eigenvectors A0.648) or by using the projection operators associated
with the eigenspaces for A, the system A0.652) can be solved. It is then
seen that secular terms result in the expression for ux and the expansion
A0.651) becomes invalid for large t, specifically when t = 0A/e). We do not
carry out the solution of A0.652). Although we have shown for several
problems in this chapter how secular effects can be removed, we do not
apply those methods here. Instead, we show how the asymptotic simplifica-
simplification method presented enables us to avoid any problems with secular terms.
The given perturbation result suggests that as we follow the wave
u0 = K(jc - f)ri that satisfies the reduced system for a long time, the
nonlinear and higher derivative terms in A0.645) become significant. To
determine their effect, we look for a solution of A0.645) in the form
A0.654) u = j3rx + e[ar0 + 5r2].
This corresponds to A0.628) with v = pr1 and w = ar0 + Sr2. The functions
P(x, i), a(x, t), and 8(x, t) are to be specified, but pr1 lies in the eigenspace
spanned by rx whereas ar0 4- 8r2 is a typical vector in the eigenspace
spanned by r0 and r2. To obtain the projection operators for these eigen-
eigenspaces we set С(А) = |Л-А/| = С1(А)С2(А) where Cj(A) = l-A and
Q(^) = A(l + A). The eigenvalue of A corresponding to Q(A) is Ax = 1, and
the remaining eigenvalues Ao = 0 and A2 = -1 correspond to the factor
C2(A). Since
A0.655)
1/2 l + A/2
A(l-A)(l + A) 1-A A(l + A) '
we find that the projection operator Px which projects vectors into the
eigenspace spanned by rx is
A0.656)
1
У
У
The projection operator P2 corresponding to the eigenspace of r0 and r2 is
820 ASYMPTOTIC METHODS
A0.657) P2 = i
x = ±
2y-l -y -1
-1 у -1
i-y y(i-y) i + y
Recalling the definition of u given in A0.646), we see that A0.654)
implies
A0.658) P = j3 + OF), м = 0 + 0(е), Г = (у-1H + О(€).
If these expressions for p, м, and Г are inserted into the matrices В and С as
given in A0.647), we obtain
A0.659) В = pB + O(e), С = j8C + O(e)
with В and С given in A0.653). Thus on inserting A0.654) in A0.645) we
have
A0.660) [ft + /Bjr, + e{[8, - 8x]r2 + a,r0
+ /S&fir, + /ЭДСг, + fi^Dr,} = O(e2) .
On multiplying A0.660) with the projection operator P, we obtain
A0.661)
since Р,г0 = PYr2 = 0. Now A0.661) implies that /3, = -& + 0(e), thus we
can replace A0.661) by
A0.662)
Neglecting the O(e2) terms in A0.662) we obtain Burgers' equation for right
traveling waves. It contains both a nonlinear term fiCx and a second
derivative term fixx. The combination of these terms in the equation leads to
some interesting solutions, as we show when we discuss Burgers' equation.
Next we multiply A0.660) by P2 and this yields
A0.663) [8, - Sx)r2 + a,r0 + №хР2Вгг
+ №tP2Crx + PxxP2Dtx - O(e) ,
since P2rx = 0. This equation is to be solved by expanding a and S as
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 821
oo ос
A0.664) a = 2 a*cJ9 5=2 ft*' .
>=o o
>=o
It is easiest to solve A0.663) by construcing projection operators Po and P2
that project vectors into the eigenspaces spanned by r0 and r2, respectively.
Using this method and the fact that Px + Po + P2 = I, we easily establish that
A0.665)
y-1 0 -1
0 0 0
1-у 0 1.
2у
1 -у 1
-1 у -1
у-1 уA-у) у-1.
Then Poro = r0) Por2 = 0, P2r2 = r2, and P2r0 = 0. Since /3, = -ft + O(e),
we may express A0.663) as
A0.666) [8, - 5jr2 + a,r0 + fifix[P2& - P2C]r, + P^D^ = O(e) .
Inserting A0.664) into A0.666) and multiplying by Po yields for a0,
A0.667) ^ + A - y)fifix + k(y - 1)%X = 0.
Multiplying A0.666) by P2, we obtain for 50,
We do not consider the equations for a- and 5; with /^1.
If we set
A0.669) u = Sr2 + e{ar0 + jSrJ ,
it may be shown by following this procedure that 8(x, t) satisfies a Burgers'
equation for left traveling waves. The derivation is carried out in the
exercises. However, if we set
A0.670) u=aro + eiPr, + Sr2} ,
we find that a satisfies a heat conduction equation to the previous level of
approximation as we now show.
With u given by A0.670) we have
A0.671) p = a + O(e), и = O(e), T = (y - \)a + O(e) ,
and the matrices В and С become
822
ASYMPTOTIC METHODS
A0.672)
B = a
0
У
0
1
0
у(у-
1)
0
У
0
aB
and C = aC+ O(e) and £ defined as in A0.653). Thus inserting A0 6701
into A0.645) yields " '
A0.673) a,r0 + e{(/3, + px)Tl + (8, - Sx)r2
+ aaxBr0 + aa,tr0 + axxDr0} = O
Multiplying A0.673) by the projection operator Po gives
A0.674) a, + e{ ^ «a, -k(y- 1)«J = O(e2) .
Since at = O(e), we may replace A0.674) by
A0.675) a,-ek{y-l)axx = O{e2).
Since у > 1, this has the form of the heat conduction equation if the O(e2)
terms are neglected. On using the projection operators P1 and P2 in A0.673)
we may express fi and 8 in terms of a, but this is not carried out here.
Truncating A0.662) at the O(e2) level, we find that /3(x, t) satisfies the
Burgers' equation
A0.676)
The leading terms a0 and 50 in the expansions of a and 8 are expressed in
terms of p by Wans of A0.667) and A0.668). The perturbed density,
velocity, and temperature as defined in A0.640) are thus given as
A0.677) p = 0 + ea0 + eS0, й = fi - e80 ,
In order to compare our present results with those obtained in Example
8.9, we discuss the preceding approximations as they relate to the velocity
и = ей [see A0.640)]. We have
A0.678) w=eu = e/3-€2S0,
and to leading order we set и = e/3. Multiplying A0.676) by e gives
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 823
A0.679) ut
The diffusion term uxx in A0.679) disappears if e = 0 or if /л = к = 0.
Then A0.679) takes the form of the quasilinear one-dimensional wave
equation (8.360) except that c0 -1 in our case. The equation (8.360) was
obtained on the basis of Euler's equations of motion where viscosity and
conductivity are neglected (i.e., we have /t = fc = 0). It was also assumed
that the gas is polytropic; that is, (8.343) is valid. Consequently, A0.679) is
seen to represent a modification of the quasilinear equation (8.360) whose
solutions were called simple waves in Example 8.9. Since both (8.360) and
A0.679) can be solved exactly, we can compare their solutions for identical
initial values and determine the role of the diffusion term in smoothing out
the solutions.
With
A0.680)
A0.679) becomes
A0.681) и, + [l + {Ц^и}^ - *vuxx = 0.
Putting e or v equal to zero in A0.681), the solution of the initial value
problem for the resulting first order equation with the initial condition
A0.682) u(x,0) = h(x)
is given implicitly as
A0.683) и = h[x - (l + Z±l M^J e
It was observed by Cole and Hopf that if we set
A0.684) 1
v
in A0.681), then v satisfies the heat conduction equation
A0.685) v, = evvxx ,
as is readily verified. The initial condition A0.682) for u(x, 0) takes the form
A0.686) v(x, 0) = exp{ - ^- JT [l + 1±1 h(s)] ds) - H(x) ,
824 ASYMPTOTIC METHODS
and the solution of the initial value problem A0.685), A0.686) is given as
A0.687) *,, 0 = ^L- £ HW exp[ - !
The solution u(x, t) of A0.681) and A0.682) can then be obtained from
A0.684) as
A0.688)
[2/(y + 1)] f [(x - {Щ cxp(-gl2ev) d*
»(*> о
J — o
where
A0.689) g = \l [l + 2±!
We study A0.688) for the special initial condition
(Ю.690) «(*,o)=*(*)
where 0<Ax<A2 and Л15Л2 are constants. Then g can be expressed in
terms of the functions gt defined as
A0.691) ft = e<f + fci£
where at = 1 + [(у + 1)/2]Л/(/ = 1,2), in the form
(Ю.692Т ,
In terms of the fundamental solution G(x, t) of the heat equation
A0.693) G(x,t)
we can express A0.688) as
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 825
A0.694)
1 + ±— u(x, t)
ехр[А(х - \t)lev\ f_m [(x - *)/*]Ga(f) <g + jQ [(x -
ехр[А(л: - kt)/ev] /_ G2(f) </£ + J"q G,
where A = (аг + a2)l2, k = (a1-a2)/2 and G,(£) = <?(* - af - £, €*f),
(/ = 1,2), as is easily verified.
Now as ev—>0 we have
A0.695) lim G,( f) - lim G(x - a(Y - £, €W) = 5(x - a(Y - f) ,
ev—>0 £v—>0
as follows from the well-known property of the fundamental solution
G(x, t). Also, since ax<a2, the exponentials in A0.694) tend to infinity as
ev -> 0 if x - kt < 0, and to zero if x - kt > 0. Noting these results we readily
linU4hat
A0.696) lim u(x, t) = ^
h2, x
since
A0.697)
It is expected that A0.696) is a solution of the reduced equation for
A0.681), that is
A0.698) M,
with the initial condition A0.690). However, the implicit solution A0.683)
of A0.698) shows that и is a constant on the lines x - {1 + [(y + l)/2]u}t =
£ = constant. And at t = 0, и = Л2 if x <0, and u = /Tj if x>0, so that
Since at < a2, these regions in the (x, t)-plane are given as shown in Figure
10.10. In the sector between x = axt and x = a2t the solution A0.699) is
826
ASYMPTOTIC METHODS
Figure 10.10. The multivalued region and the shock line.
multivalued and since this multivaluedness occurs as soon as t exceeds zero,
this solution is not valid for any time f>0.
However, the result A0.696) yields a (discontinuous) solution of A0.698)
that ^satisfies the initial condition A0.690) and has a jump across the line
x — kt that lies between x = axt and x = a2t as picturedjn Figure 10.10. The
jump in u(x9 i) travels to the right at the speed A and this does not
correspond to either of the two characteristic speeds ax and a2 for this
problem. In fact, the traveling wave solution A0.696) of the quasilinear
equation A0.698) is a shock wave and it has been obtained as the limit of
the smooth solution A0.694) of Burgers' equation as ev-»0.
In Section 2.3 we constructed shock waves for first order quasilinear
equations directly without having to solve a higher order equation and
evaluating a limit, as we have done here. [The direct construction of the
shock wave A0.696) is considered in the exercises.] We recognize, however,
that the shock wave is an idealized and simplified form of the solution of the
given problem for Burgers' equation in which ev is small but nonzero.
Therefore, the solution of Burgers' equation for small but nonzero ev is
valuable in that it yields a solution of the problem that varies rapidly near
the shock front but is nevertheless smooth. This more detailed description of
the solution near the shock gives what is called the shock structure.
An alternative approach to determining the shock wave solution A0.696)
of A0.698), which is also useful for more general (quasilinear) equations
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 827
that do not have an exact solution as in the case of Burgers' equations, is to
look for a traveling wave solution
A0.700) w(x, г) = v(x - Xt)
in A0.681) with the function v and the constant A to be specified. We also
require that
A0.701) в(ж-Л0->{£; * I ££"<»,
with hx and ft2 defined as in A0.690).
Inserting A0.700) into A0.681) gives
A0.702) A - A)w'
Integrating once in A0.702) and noting that u'-»0 as \x - Ar| —> +o°, in view
of A0.701), we have
A0.703) A - k)v
where v0 is a constant of integration. As x —
A0.704) A_A)Ai
and as x - kt~+ — <x>9
A0.705) A - A)A2
Solving A0.704), A0.705) for A and v0 yields
(Ю.706) x = i (l±iy
as in A0.697) and
A0.707) V
On inserting these values into A0.703), we find that the equation can be
written as
828 ASYMPTOTIC METHODS
(ю.708) y'
Using partial fractions we easily find that
A0.709)
u(x, t) = v(x - kt) = hx +
+ exp{[y + l)!Aev](h2 - ht)(x - A/)}
and this satisfies A0.701). Furthermore, as ev-*Q, A0.709) tends to
A0.696) since h2> hx, as is readily checked. Thus for small but nonzero ei>,
this solution which is easier to evaluate than A0.694) yields the shock
structure and describes how the initial discontinuity A0.690) propagates as a
smooth step whose transition from the value h2 to the value hx moves at the
shock speed A.
We have assumed that 0< hx < h2 in the initial condition A0.690). This
gives rise to a contraction wave which breaks immediately, as a solution of
A0.698), and requires the introduction of a shock wave. However, if
0<h2<hx in A0.690) we have an expansion wave for which continuous
solutions can be found as shown in Section 2.3. Now if we look for a
traveling wave solution of Burgers' equation with h2 < hx in A0.701), we see
from A0.708) that there is no traveling wave solution that increases
smoothly from h2 at — o° to hx at +o°, since for h2<v <hx we have vr <0.
Thus, i; must be a decreasing rather than an increasing function and this is
not possible. As a result, we find that expansion shocks do not result as
limits of traveling wave solutions of Burgers' equation.
More general initial data for A0.681) and the behavior of the correspond-
corresponding solution A0.688) as ei>—»0 may be discussed by evaluating the integrals
in A0.688) by the method of steepest descents or the saddle point method.
We do not pursue this matter, however.
Example 10.15. The Boussinesq and Korteweg-deVries Equations. It
can be shown that the theory of water waves for the case of shallow water
and waves of small amplitude can be (approximately) described by the
Boussinesq equations
A0.710) vt + wx + e[vwx + wvx] = 0 ,
A0.711) wt + vx + e[wwx - \wxxt] = 0 .
If we set gh(x, t) = l + ev(x, t) and w(jt, /) = ew(x, t), where g is the gravita-
gravitational constant, then h represents the height of the water above a horizontal
bottom and и is the velocity of the water. If we drop the third derivative
term in A0.711) and express the resulting first order system in terms of h
and u, we obtain the equations of Exercise 3.3.21. They are known as the
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS 829
equations of shallow water theory. The small parameter e in A0.710),
A0.711) signifies that we are considering small wave amplitudes and shallow
water, or, equivalently, long waves.
With
A0.712)
» 11 n
o J'
the system A0.710), A0.711) can be written as
A0.713) u, + Aux + e[Bux + Cuxxt] = 0 .
The system A0.713) is of the form A0.627). If we put e = 0 in A0.713), we
obtain the linear reduced system ur + Aux = 0, which is equivalent to the
one-dimensional wave equation with unit wave speed. In fact, the charac-
characteristic polynomial for A is C(A) = \A - A| = (A - 1)(A + 1). With Ax = 1
and A2 - — 1 as the eigenvalues of Л, the corresponding eigenvectors are
A0.714) ri
With u = art we have u, + Aux = (a, + ax)vx = 0, and when u = /3r2, we
have u, + Aux = (ft - Px)r2 = °- Thus a " a(x ~ 0 and P = P(x + 0-
Since Л has only two (orthogonal) eigenvectors we do not require the full
machinery developed in our general discussion. To study the effect of the
nonlinear and higher derivative terms on the right traveling wave obtained
from the reduced equation, we set
A0.715) u = атг + epr2 ,
where a(x, t) and p(x, i) are to specified. We insert A0.715) into A0.713)
and obtain
A0.716) [a, + ojr, + e{[ft - &]r2 + ааЦ] - | axx[ J]} = O(e2),
since v = a + O(e) and w - a + O(e). Since rx and r2 are orthogonal, we
need not construct projection matrices corresponding to their eigenspaces.
Instead, we simply take dot products with rx and then with r2 in A0.716).
Dotting with r2 in A0.716) gives
A0.717) <*, + «*+ \eaax - \eaxxt = O(e2) .
Dotting with r2 gives
830 ASYMPTOTIC METHODS
A0.718) pt - ft + \aax + \axxt - O(e) .
We solve A0.718) by expanding f in a series in powers of e with В =
Po + O(e). Thus
Since a, = -ax + O(e) in view of A0.717), we may replace that equation by
A0.720) a, + ax + § eaa, + | e^ = 0,
on dropping the O(e2) terms. This is the Korteweg-deVries equation. As we
have indicated in Example 9.7, this is a nonlinear dispersive equation. Thus
it differs from the Burgers' equation which is of dissipative type. It turns out
that the Korteweg-deVries equation is exactly solvable; however, the
method for doing so is substantially more complicated than that given in the
preceding example for Burgers' equation and it is not presented here.
We note that if A0.720) is solved by expanding a in powers of e (i.e., by
using the perturbation method), secular terms result in that expansion. Thus
we leave A0.720) as it stands, and in terms of its solution we obtain for v
and w,
A0.721) v = a + €jS0, w = a - eft,
in view of A0.715) and A0.719) where Д, is solved for in terms of a. If we
interchange атг and /3r2 in A0.715) we again find on proceeding as earlier,
that /3 satisfies a Korteweg-deVries equation for left traveling waves as
shown in the exercises.
We have already discussed approximate periodic solutions of the
Korteweg-deVries equation in Example 9.7. We now obtain a special
traveling wave solution of A0.720) that represents what is known as a
solitary wave. This wave consists of a single hump of constant shape that
moves at constant speed.
Let
A0.722) a(x, t) = af(x - Xt) ,
where the constants a and Л and the function / are to be specified. We
assume that f(x - A/) vanishes together with its derivatives as \x - A|°
Inserting A0.722) into A0.720) gives
A0.723) A - A)/' + \ eaff' + | /'" = 0 .
Integrating once in A0.723) gives
ASYMPTOTIC SIMPLIFICATION OF EQUATIONS
831
A0.724)
with no constant of integration since / and /" vanish at infinity. Next we
multiply A0.724) by /' and integrate once more to obtain
A0.725)
Again, there is no constant of integration. Then A0.725) can be written as
The solution of A0.726) that vanishes at infinity is readily verified to be
_ „ i1/2
A0.727) f(x-kt) =
If we set
A0.728)
the solution a = af(x -At) can be written as
A0.729) a(xy t) = a sech2{(^ uf^x - (l + у )t]} ,
with the solitary wave speed given as
A0.730)
A - 1 + у .
Figure 10.11. The initial waveform.
832 ASYMPTOTIC METHODS
At the time t = 0,a(x,t) has the form given in Figure 10.11. The
maximum value of a(x, 0) occurs at x = 0 and equals a. The waveform
a(x, 0) is symmetric with respect to x = 0. This form travels to the right
without change in shape at the speed A = 1 + ей/2. Thus as a increases, so
does the wave speed, and higher solitary waves move more rapidly than
lower ones. This leads to some interesting problems regarding the interac-
interaction of two or more solitary waves moving at different speeds. Also, by
modifying the foregoing discussion it is possible to obtain exact periodic
traveling wave solutions of the Korteweg-deVries equation. We do not
discuss these questions here.
The Burgers', Korteweg-deVries, and nonlinear Schrodinger equations
that were discussed in this and the preceding sections occur as the relevant
approximate equations for a number of significant problems of physical
interest. Each of these equations can be solved exactly, and the study of the
solution of these and related nonlinear partial differential equations has
been the subject of intense investigation in recent years.
EXERCISES FOR CHAPTER 10
Section 10.1
10.1Л. Let и/ = A/4ягг)е1'Лпг be the free space Green's function for the
reduced wave equation A0.1) (n = constant) with source point in z > 0. Use
the method of images to construct the Green's function for the half-space
z > 0, in the form К = w7 + us if К = 0 on z = 0. Note that both ut and us
are of the form A0.7) and show that the phase and amplitude terms of щ
and us are equal on z = 0. Also show that us satisfies the outgoing condition
at z=0.
10.1.2. For the two-dimensional problem considered in Example 10.1,
determine the Green's function К if the Dirichlet condition A0.23) is
replaced by the Neumann condition dKldx = 0 on x = 0.
Hint: Use the method of images. Expand the result asymptotically and
show how the leading amplitude term and the phase term of the scattered
wave us can be specified by applying the boundary conditjdn to th$
asymptotic expansion of the solution.
10.1.3. Apply the method of images to construct the Green's function for
A0.1) (n = constant) in the quarter-plane jc>0, у >0 if the Green's func-
function is required to vanish on the boundary.
Hint: If (f, 77) is the source point for the free space Green's function in the
first quadrant, there are image sources at (-£,17), (-£, -17), and (£, -17).
Expand the Green's function asymptotically and show how the leading terms
may be obtained by expressing the incident and reflected waves in the
asymptotic form A0.7).
EXERCISES FOR CHAPTER 10 833
Hint: Consider that the incident rays may undergo a double reflection from
x = 0 and y = 0.
10.1.4. Solve the problem of Exercise 10.1.3 if the Green's function satisfies
the following boundary conditions: dKldx = 0 at x = 0 and К = 0 at у = 0.
10.1.5. Consider the equation
which is the one-dimensional version of A0.1). Obtain an asymptotic
solution of this equation by expanding u(x) as in A0.7) and A0.11).
Determine and solve the equations for the phase and amplitude terms. (This
technique in one dimension is known as the WKB method and there is a
large body of results available for this method.)
10.1.6. Determine the rays and the phase function <p if cp is specified to
equal 1 on the circular cylinder x2 + y2 — a2 and the index of refraction
n = 1. Find both sets of rays and phase functions.
10.1.7. Determine the rays and the phase function <p in the two-dimensional
case if we set <p = в (the polar angle) on the unit circle x2 + y2 = 1. Also
solve the leading order transport equation A0.66) for v0 in this case. Let
/i = L
10.1.8. Show that for a cylindrical wave with phase <p — nr = ri\jx2 + y2 and
a constant refractive index n, the amplitude term v0 decays like 1/Vr, by
using the relation A0.86).
10.1.9. Consider the asymptotic scattering problem for u — Uj + us where w7
and us are given in A0.103), A0.104). Determine the appropriate condi-
conditions on the phase term <p and the amplitude terms Vj if the following
boundary conditions are given on the scattering surface.
« £-•■
(b) — + /ш = 0.
10.1.10. Obtain the geometrical optics solution of the problem in Example
10.3 in the field u(x, y) satisfies the Neumann condition duidn =0 on the
parabola.
10.1.11. Assume the plane wave ut = elkx is incident on the parabola of
Example 10.3 from the left and that the total field u = uI + us vanishes on
the parabola. Determine the asymptotic expansion of the scattered wave м5.
10.1.12. Assume that the plane wave м7 = elky is incident on the parabola of
Example 10.3 and that the total field и = м7 + us vanishes on the parabola.
Determine the geometrical optics expansion of the scattered field ms. Obtain
only the leading amplitude terms for us and leave it in parametric form.
834 ASYMPTOTIC METHODS
Show that us is singular at the vertex of the parabola and that the field is
discontinuous across the shadow boundary x = -a/2,y>0.
10.1.13. Rework the problem discussed in Example 10.4 under the assump-
assumption that the caustic curve is the circle x2 + y2 = a2 and that V0(a0) = i
Obtain expressions for the incident and outgoing geometrical optics fields
and for the field in the boundary layer in terms of variables associated with
the circular caustic. Determine the asymptotic value of the field on the
caustic x2 + y2 = a2.
10.1.14. Consider the problem of Example 10.5. Introduce a boundary layer
at the reflection boundary у = 0, z < 0 and proceed as in the text to obtain a
boundary layer solution that yields a smooth transition across the reflection
boundary.
10.1.15. Construct the geometrical optics solution for the problem of Exam-
Example 10.5 if the boundary condition A0.192) is replaced by the Neumann
condition ди(х, у, 0) Idz — 0 for у < 0.
10.1.16. In two dimensions, consider the problem of diffraction by a slit.
Let the incident plane wave uf(x, y) = elkx approach the y-axis which has an
opening or slit for \y\ < 1. For x = 0 and \y\ > 1, the boundary condition for
the total field и = ut + us is given as м@, у) = 0, |y| > 1. Using appropriate
outgoing conditions, determine the geometrical optics field and show that
there are two shadow and reflection boundaries. Show how boundary layer
methods can be used to obtain smooth transitions across these shadow and
reflection boundaries. Compare these results with those that might be
obtained by applying the geometrical theory of diffraction at each edge—
that is, at @, -1) and @,1).
10.1.17. Use the parabolic equation
2ikvx + vyy = 0, x> 0
with the data
to obtain an approximate solution to the problem of Exercise 10.1.16 in the
half-plane x > 0. Compare the solution obtained by the parabolic equation
method with that obtained in Exercise 10.1.16 using geometric^optic>and
boundary layer methods.
10.1.18. Obtain the geometrical optics solution of the problem of the
scattering and diffraction of a plane wave by a circular aperture. That is,
consider the plane z = 0 with a circular hole given as x2 + y2 < a2, and a
plane wave u^e*2 approaching the x,y-plane. Let the total field
w(jc, y, z) = uI + us vanish on z = 0, x2 + y2 > a. Determine the geometrical
optics field us and show that the total field и is discontinuous across the
cylinder x2 + y2 = a2.
EXERCISES FOR CHAPTER 10 835
10.1.19. Apply the parabolic equation method for the problem of Exercise
10.1.18 in the region jc>0. Obtain the equation
2ikvz + vxx + vyy = 0, z > 0 ,
with the boundary condition
, m fl,
Noting that the fundamental solution for this parabolic equation is given as
where r2 = (x - f J + (y - r}J, obtain the solution of the Cauchy problem
for the parabolic equation as an integral over x2 + y2 < a2. Introduce polar
coordinates and show that the double integral can be expressed as a single
integral involving the Bessel function /0.
10.1.20. Solve the problem of Example 10.6 if the boundary condition
A0.229) is replaced by the Neumann condition duldn = 0 on x2 + y2 = a2.
10.1.21. Write A0.264) in the form и = veikx and show that у is a solution of
the paraxial wave equation A0.265) and satisfies the boundary condition
A0.259).
10.1.22. If v(x, z) and v(y, z) are solutions of the paraxial wave equations
2ikvz + uXJC=0 and 2ikvz + vyy = 0, respectively, show that v(x, y, z) =
v(x, z)v(y, z) is a solution of the paraxial wave equation in three dimen-
dimensions
10.1.23. Use the results of Example 10.7 and Exercise 10.1.22 to obtain the
Gaussian beam
( л *( ~ika
z + 2(z-ika)
in three dimensions.
10.1.24. Verify that A0.276) is a solution of A0.275).
10.1.25. Discuss the problem of Example 10.8 in the three-dimensional
case. Introduce cylindrical coordinates (r, 0, z) and assume that the initial
conditions at z - 0 are
836 ASYMPTOTIC METHODS
Proceed as in the two-dimensional case and assume that the phase and
amplitude are independent of 0. Determine an approximate focal distance z
as in A0.309) and a focal length zf as in A0.331), based on the solution of
the appropriate nonlinear geometrical optics equations obtained by proceed-
proceeding as in the two-dimensional problem. Note that the equation for / that
occurs in three dimensions and corresponds to A0.329) is easier to solve
that its two-dimensional analogue.
10.1.26. Obtain the exact solution of the initial and boundary value problem
for the Klein-Gordon equation
u(x, 0) - 0, ut(x, 0) = 8(x), x>-l,
u(-l,t) = 0, t>0
by applying the method of images. Assume that A; is a large parameter. Use
the asymptotic forms and results obtained in Example 10.9 to obtain an
asymptotic solution to this problem. Assume that the boundary gives rise to
an expansion of the form A0.376) and determine its phase and amplitude
terms in the manner employed in solving scattering problems for the
reduced wave equation. Compare the asymptotic results with those obtained
from the exact solution when it is expanded asymptotically.
Section 10.2
10.2.1. Use expansions of the form A0.432) to account for the singularity in
the data for the following problem.
u(x, 0) = H(x), ut(x, 0) = yS(x), ~oo < x <oo .
10.2.2. Solve the problem of Exercise 10.2.1 if the equation is replaced by
utt-y2uxx-c2u = Q, -oo<jc<oo, t>0,
and the data remain the same.
10.2.3. Expand the solution A0.525) near the characteristic x - yt*= 0 and
show that the first two terms agree with those obtained by applying the
expansion method directly.
10.2.4. Consider the solution и of the problem A0.513) and A0.520), as
given in A0.525). Show that ut + 2ku is the solution of A0.513), A0.514).
Determine the solution and verify the result A0.519).
10.2.5. Use an expansion of the form A0.526) to discuss the approximate
solution of the initial value problem
EXERCISES FOR CHAPTER 10 837
2 =:0> -оо<л:<оо, t>0,
, xil u,(x,0) = 0.
Hint: Use the generalized functions fa(x) of Example 7.1.
10.2*6. Solve the wave equation utt=V2u in three dimensions with the
Cauchy data
и(х, у, z,0) = 0, u£x, у, z,O) = j^' r>a
if r2 = x2 + y2 + z2 and a, C, and a are constants. Use an expansion of the
form A0.526).
10.2.7. Replace с by ic in the fundamental solution A0.568) of the Klein-
Gordon equation to obtain the solution of the following problem and discuss
the solution and its expansion in terms of the hyperbolic distance, in the
manner in which the Klein-Gordon equation was analyzed in the text.
c, y,
u(x, y, z, 0) = 0, u£x, y, z, 0) = 8(x)8(y)8(z).
10.2.8. Let p(r) = 72A + a2r2) in A0.576). Determine the explicit forms of
и as given in A0.587) and A0.590) in the two- and three-dimensional cases,
respectively.
10.2.9. Given the hyperbolic system
uf + Aux + fiu = 0 ,
consider an expansion
with the S, given as in A0.527). Insert the expansion for u into the equation
for u and equate coefficients of like terms 5;. Show that <p must be a
characteristic of the hyperbolic equation and that the v; are given in terms of
eigenvectors of the matrix from which the characteristics are determined.
10.2.10. Express the telegrapher's equation A0.442) as a system of equa-
equations by setting ut = v and ux = n>. Use the initial data A0.520) to obtain
n(x, 0) = [v(x, 0), w(x, 0)] = [8(x), 0] = 8(x)i. Solve the Cauchy problem for
this system with the preceding data using the method of Exercise 10.2.9.
Compare the results with those obtained by solving the telegrapher's
equation directly by the expansion method given in the text.
838 ASYMPTOTIC METHODS
Section 10.3
10.3.1. Consider the equation
e{uttt - a2uxxt) + utt - c2uxx = 0,
where a>c. Let vl = utn v2 = uxn and v3 = uxx and express the equation for
и as the system
where u has the components v19 u29 and v3. Construct a decomposition of
this system based on two linearly independent null eigenvectors of the
matrix B—that is, the eigenvectors that correspond to the eigenvalue Л = 0.
Hint: Each of these equations is a parabolic equation.
10.3.2. Express the equation
as a system by setting v = ut and w = ux and obtain
u, + Aux + eBnxx = 0 ,
where u has the components и and w. Obtain a decomposition of the system
based on the eigenvalues and eigenvectors of the matrix A.
10.3.3. Obtain a Burgers' equation for left traveling waves by using A0.669)
and proceeding as in the text.
10.3.4. Determine the shock wave A0.696) by using the method presented
in Section 2.3.
10.3.5. Obtain a Korteweg-deVries equation for left traveling waves by
setting
u = ar2 + е/3гх
in the system A0.713) of Example 10.15.
10.3.6. Show that A0.709) is the solution of A0.702) with Л and v0 given as
in A0.706), A0.707) and verify that it tends to A0.696) as €v-*0,
10.3.7. The nonlinear hyperbolic system
represents a model for Boltzmann's equation in the kinetic theory of gases.
Let
EXERCISES FOR CHAPTER 10 839
V=1 + €V, W=l+ew, t=ery x = ea,
and obtain the system
Си + €[u, + Aux + Bn] = 0
where u has the components и and w, and the matrices A, B, and С are
defined appropriately. Show that the matrix С has a null vector rr = [1,1].
Carry out a decomposition of the system corresponding to this vector and
obtain a Burgers' equation as the approximating equation.
10.3.8. The nonlinear dispersive (wave) equation
models the vibration of a nonlinear string. Let v = ut and w = ux, and
express the equation in the form of a nonlinear system
where ur = [v, w] and the matrices A, B, and С are appropriately defined.
Show that the matrix A has eigenvalues A = +1 and A - -1 with orthogonal
eigenvectors. Construct a decomposition of the system corresponding to the
eigenvalue A = 1 and obtain a Korteweg-deVries equation as an approxi-
approximation.
BIBLIOGRAPHY
CHAPTER 1
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Gordon and Breach, New York, 1970.
Berg, H. C, Random Walks in Biology, Princeton University Press, Princeton, 1983.
Courant, R., K. O. Friedrichs, and H. Lewy, Uber die partiellen differenzengleichungen der
mathematischen physik, Math, Ann., 100, 1928, 32-45. (English transl. in IBM J. Res.
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Doyle, P. G., and J. L. Snell, Random Walks and Electric Networks, Cams Math. Monogr.,
No. 22, Mathematical Association of America, Washington, 1984.
Einstein, A., Investigations on the Theory of the Brownian Movement, R. Furth, Ed., Dover,
New York, 1952.
Feller, W., An Introduction to Probability Theory and its Applications, Vol. 1, 3rd ed., Wiley,
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Furth, R., Die Brownsche bewegung ..., Z. Phys., 2, 1920, 244-256.
Gardiner, C. W, Handbook of Stochastic Methods, Springer, Berlin, 1983.
Goldstein, S., On diffusion by discontinuous movements and the telegraph equation, Q. J.
Mech. AppL Math., 4, 1951, 129-156.
Horsthemke, W, and R. Lefever, Noise-Induced Transitions, Springer, Berlin, 1984.
Kac, M., A stochastic model related to the telegrapher's equation, Rocky Mt. J. Math., 4,1974,
497-509.
Lin, С С, and L. A. Segel, Mathematics Applied to Deterministic Problems in the Applied
Sciences, Macmillan, New York, 1974.
Okubo, A., Diffusion and Ecological Problems: Mathematical Models, Springer, Berlin, 1980.
Risken, H., The Fokker-Planck Equation, Springer, Berlin, 1984. ^ ___^
Schuss, Z., Theory and Applications of Stochastic Differential Equations, Wiley, New York,
1980.
Taylor, G. L, Diffusion by continuous movements, Proc. London Math. Soc, 20, 1920,
196-212.
Uhlenbeck, G. E., and L. S. Ornstein, On the theory of Brownian motion, Phys. Rev., 36,
1930, 823-841.
Van Kampen, N. G., Stochastic Processes in Physics and Chemistry, North-Holland, Am-
Amsterdam, 1981.
Wax, N.. Ed., Selected Papers on Noise and Stochastic Processes, Dover, New York, 1954.
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BIBLIOGRAPHY 841
CHAPTER 2
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Courant, R., and К. О. Friedrichs, Supersonic Flow and Shock Waves, Springer, New York,
1976.
Garabedian, P., Partial Differential Equations, 2nd ed., Chelsea, New York, 1986.
Rhee, H., R. Aris, and N. R. Amundson, First Order Partial Differential Equations, Prentice-
Hall, Englewood Cliffs, N.J., 1986.
Smoller, J., Shock Waves and Reaction-Diffusion Equations, Springer, New York, 1983.
Whitham, G. В., Linear and Nonlinear Waves, Wiley-Interscience, New York, 1974.
CHAPTER 3
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Jeffrey, A., Quasilinear Hyperbolic Systems and Waves, Pitman, Boston, 1976.
CHAPTER 4
Abramowitz, M., and I. Stegun, Handbook of Mathematical Functions, Dover, New York,
1964.
Andrews, L. C, Elementary Partial Differential Equations, Academic, Orlando, 1986.
Berg, P. W., and J. L. McGregor, Elementary Partial Differential Equations, Holden-Day, San
Francisco, 1966.
Birkhoff, G., and G. С Rota, Ordinary Differential Equations, 3rd ed., Wiley, New York,
1978.
Churchill, R. V, and J. W. Brown, Fourier Series and Boundary Value Problems, 4th ed.,
McGraw-Hill, New York, 1987.
Courant, R., and D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New
York, 1953.
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Duff, G. F. D., and D. Naylor, Differential Equations of Applied Mathematics, Wiley, New
York, 1966.
Haberman, R., Elementary Applied Partial Differential Equations, 2nd ed., Prentice-Hall,
Englewood Cliffs, 1987.
Hanna, J. R., Fourier Series and Integrals of Boundary Value Problems, Wiley-Interscience,
New York, 1982.
Iooss, G., and D. D. Joseph, Elementary Stability and Bifurcation Theory, Springer, New York,
1980.
Joseph, D. D., Stability of Fluid Motions, 2 Vols., Springer, Berlin, 1976.
Lin, С. С, and L. A. Segel, Mathematics Applied to Deterministic Problems in the Natural
Sciences, Macmillan, New York, 1974.
Pinsky, M., Introduction to Partial Differential Equations with Applications, McGraw-Hill,
New York, 1984.
Stakgold, L, Green's Functions and Boundary Value Problems, Wiley-Interscience, New York,
1979.
842 BIBLIOGRAPHY
Taylor, A. E., and W. R. Mann, Advanced Calculus, 3rd ed., Wiley, New York, 1983.
Tikhonov, A. N., and A. A. Samarskii, Equations of Mathematical Physics, Macmillan TsW
York, 1963. ' W
Tolstov, G. P., Fourier Series, Dover, New York, 1976.
Vladimirov, V. S., Equations of Mathematical Physics, Dekker, New York, 1971.
Weinberger, H. F., A First Course in Partial Differential Equations, Blaisdell, New York, 1965
Zachmanoglou, E. C, and D. W Thoe, Introduction to Partial Differential Equations with
Applications, Dover, New York, 1987.
CHAPTER 5
Bleistein, N., and R. Handelsman, Asymptotic Expansions of Integrals, Dover, New York,
1986.
Churchill, R. V., Operational Mathematics, 3rd ed,, McGraw-Hill, New York, 1972.
Copson, E. Т., Asymptotic Expansions, Cambridge University Press, Cambridge, 1965.
Davies, В., Integral Transforms and their Applications, Springer, New York, 1984.
Duff, G. F. D., and D. Naylor, Differential Equations of Applied Mathematics, Wiley, New
York, 1966.
Dym, H., and H. P. McKean Jr., Fourier Series and Integrals, Academic, New York, 1972.
Erdelyi, A. et al., Tables of Integral Transforms, McGraw-Hill, New York, 1953.
Olver, F. W. J., Asymptotics and Special Functions, Academic, New York, 1974.
Pearson, С. Е., Ed., Handbook of Applied Mathematics, Van Nostrand, New York, 1974.
Sirovich, L., Techniques of Asymptotic Analysis, Springer, New York, 1971.
Sirovich, L., Asymptotic evaluation of multidimensional integrals, /. Math. Phys., 11, 1970,
1365-1374.
Sneddon, I. N., Fourier Transforms, McGraw-Hill, New York, 1951.
Sneddon, I. N., The Use of Integral Transforms, McGraw-Hill, New York, 1972.
Wolf, К. В., Integral Transforms in Science and Engineering, Plenum, New York, 1979.
Weinberger, H. F., A First Course in Partial Differential Equations, Blaisdell, New York, 1965.
CHAPTER 6
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Garabedian, P., Partial Differential Equations, 2nd ed., Chelsea, New York, 1986.
Hadamard, J., Lectures on Cauchy's Problem, Dover, New York, 1952.
Logan, J. D., Applied Mathematics, Wiley-Interscience, New York, 1987.
Tayler, А. В., Mathematical Models in Applied Mechanics, Oxford University Press, Oxford^
1985.
Tikhonov, A. N., and A. A. Samarskii, Equations of Mathematical Physics, Macmillan, New
York, 1963.
Vladimirov, V S., Equations of Mathematical Physics, Dekker, New York, 1971.
Webster, A. G., Partial Differential Equations of Mathematical Physics, Dover, New York,
1955.
BIBLIOGRAPHY 843
CHAPTER 7
Carrier, G. F., and С. Е. Pearson, Partial Differential Equations, Academic, New York, 1976.
Courant, R., and D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New
York, 1953.
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Duff, G. F. D., and D. Naylor, Differential Equations of Applied Mathematics, Wiley, New
York, 1966.
Greenberg, M. D., Applications of Green's Functions in Science and Engineering, Prentice-
Hall, Englewood Cliffs, 1971.
Jones, D. S., The Theory of Generalised Functions, 2nd ed., Cambridge University Press,
Cambridge, 1984.
Kanwal, R. P., Generalized Functions'. Theory and Technique, Academic, New York, 1983.
Keller, J. В., and J. Papadakis, Eds., Wave Propagation and Underwater Acoustics, Springer,
New York, 1977.
Lighthill, M. J., Introduction to Fourier Analysis and Generalised Functions, Cambridge
University Press, Cambridge, 1960.
Morse, P M., and H. Feshbach, Methods of Theoretical Physics, 2 Vols., McGraw-Hill, New
York, 1953.
Roach, G. F., Green's Functions, 2nd ed., Cambridge University Press, Cambridge, 1982.
Stakgold, I., Green's Functions and Boundary Value Problems, Wiley-Interscience, New York,
1979.
Vladimirov, V. S., Equations of Mathematical Physics, Dekker, New York, 1971.
Weinberger, H. F., A First Course in Partial Differential Equations, Blaisdell, New York, 1965.
CHAPTER 8
Ames, W. F., Numerical Methods for Partial Differential Equations, 2nd ed., Academic, New
York, 1977.
Courant, R., and D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New
York, 1953.
Courant, R., Methods of Mathematical Physics, Vol. 2, Wiley-Interscience, New York, 1962.
Duff, G. F. D., and D. Naylor, Differential Equations of Applied Mathematics, Wiley, New
York, 1966.
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Garabedian, P., Partial Differential Equations, 2nd ed., Chelsea, New York, 1986.
Ockendon, H., and A. B. Tayler, Inviscid Fluid Flows, Springer, New York, 1983.
Protter, M. H., and H. F.Weinberger, Maximum Principles in Differential Equations,
Springer, New York, 1984.
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York, 1963.
Weinberger, H. F., A First Course in Partial Differential Equations, Blaisdell, New York, 1965.
Widder, D. V., The Heat Equation, Academic, New York, 1975.
844 BIBLIOGRAPHY
CHAPTER 9
Bender, C, and S. Orszag, Advanced Mathematical Methods for Engineers, McGraw-НШ
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York, 1981.
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Whitham, G. В., Linear and Nonlinear Waves, Wiley-Interscience, New York, 1974.
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Keller, J. В., The geometrical theory of diffraction, /. Opt. Soc. Am., 12, 1962, 116-132.
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York, 1981.
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Math., 35, 1978, 575-596.
SOLUTIONS TO SELECTED EXERCISES
CHAPTER 1
SECTION 1.1
1.1.5. v(x, y,t + r)= \[v(x - 8, y, t) + v(x + 8, y, t)
+ v(x, y-8,t) + v(x, у+ 8, t)].
1.1.7. v, = -Clvx - c2vy - lDlVxx - \D2vyy.
1.1.11. (a) v[kS, (n + 1)t] = pv[(k - 1)8, m] + qv[(k + 1)S, nr].
00
(c) 2 v[k8,nr] = (p + q)" = l, и = 0,1,2,....
1.1.12. (b) <*°> = 1, <^> = 0 , (x2) = (D/2a>)(l - e"") .
(c) E(x)=0, V(x) = (D/2w)(l-e-2a").
SECTION 1.2
00 00
1.2.7. (b) 2 а[к8,пт]= 2 ДО, /it] = \(P + Я)"~Х = 5 .
fc=— 00 Л=-00
1.2.10. (b) u(nS, пт) = и(-и5, m)=\pn~l .
SECTION 1.3
1.3.6. юA,1) = 0.292, vB,1) = o(l, 2) = 0.083, vB,2) = 0.040.
1.3.15. (a) M(*) = (!
(b) и(х) = (±)
845
CHAPTER 2
SECTION 2.2
2.2.1. v = F(x - ct) e'kt
2.2.2. (a) v = }q /(cs + x - ct, s) ds + F(x - с*).
(b) v = \xt2 - -ct3 + sin(x - ct).
2.2.3. (a) v = x;
(b) У = ^;
(c) i; = sin(x-yO;
(d) i/ = sin(x + yt).
2.2.4. (b) i; = — Jo J^^e> F(cr, .) </<r & + - /(x - yt) + -
2.2.5. v =
2.2.6. y= 1(^-0" \^l2-(x-
2.2.7. y = -l
2.2.8. (a) v = G\t-~),0<x^ct, v = F{x-ci),x> ct.
(b) The data are compatible if F(x) = g(—J in which case
2.2.10. и = xtc~\ x > 0, and v = 0, x < 0. Thus [i>J = te~\ and we use the
fact that d/ds = frf/dt on л: = О.
846
SOLUTIONS TO SELECTED EXERCISES 847
2.2.12. (a) v = - xH{x) --{x- ct)H(x - ct).
с с
2.2.13. For f si, v=0, x<0, and v = tc, x>0.
2.2.15. v=f(x-Clt), x<0, v=^ f\-(x-c2t)\,0<x<c2t,
C2 L C2 J
2.2.19. We must have /(*) = (constant)xc.
2.2.21. (c) /(*) = (j^) sin[(±±-^)*] + constant,
and и =g(x- ct) + ^y^J[sin(;t + t) - sin(x - ct)],
with g@) equal to the constant given above but g(z) otherwise
arbitrary.
2.2.24. v =
2.2.26. (a) x'(s) = 1, /C0 = l, z'E) = l, u'(*)
At s = 0, jc = А, у = т, z = 0, and v = /(А, т).
2.2.28. *;(*) = 1, i = 1,2,... , n , v'(s) = 0.
At s = 0, jc, = T;, i = 1,2 и - 1, xn = 0, and
SECTION 2.3
2.3.2. u = e~ctf[x - i (e" - l) и].
-. , - axe~ct
2.3.3. "-1
2.3.5. (a)M = ^TT7y,p = ^T-y.
1 1
2.3.9. и^~2( + 2t V* + 4**- T*16 wave breaks at t = 0.
2.3.10. If f{x) = 1, w = x tanh r + sech L If Дх) = x, then и = x.
2.3.14. u = t
848 SOLUTIONS TO SELECTED EXERCISES
2.3.18. u = A,x<At + a, u = (x- a) It, At+ a<x<Bt+ a,
u = B, x> Bt+a.
2.3.19. u = A, x< %(A + B)t + a , и = В, х> \(А + B)t + a.
If A + В = 0, the shock is stationary.
2.3.20. For t < 1, w = A, x<At- A, u= ( * , At- A^x^A- At,
u = -A,x>A-At.
The wave breaks at Г = 1 and then w = А, л: < 0, and и = - A, x > 0.
2.3.21. и = 1, дг<Г-1, м = л:/(Г-1), Г-Кх<0, w=0, jc^O.
The wave breaks at t = 1 and then и = 1, л: < ^(r - 1) and м = 0,
2.3.28. и=^±*--2*_
1 - z(* + у - 2z)
SECTION 2.4
2.4.1. w = * + f.
2.4.2. м=|г2
2.4.3. и = 1 and w = л: - Г + 1.
2.4.4. m = ±i(jc-0 + 1-
2.4.5. u = ±ix + t.
2.4.6. w = (л: - l)e-/ + e~2'.
2.4.9. u = ±-ix-t.
с
2.4.10. A(*, >) = [л:2 + у2]~1/4.
2.4.11. Let u — ux for x < 0 and u = u2 for jc > 0.
Then, ux = «x[jc sin в + у cos в] or ux = nx[—x sin в + у cos 0], and
м2 = «2[* cos Ф + У sin *A] or U2 = w2[-;c cos Ф + У s^n *A] where
I —
2.4.14. Plane waves:
u = u0 + /z[(jc - дг0) cos a + (y - y0) cos i8 + (z " zo) co| ?]
with constant w0, дг0, y0, z0> and cos2 a + cos2 )8 + cos2 у = 1.
Singular solutions (spherical waves):
и = м0 + nV(* " ^oJ + (У ~ УоJ + (^ - z0J.
2.4.15. u.
2.4.16. With и = м^, we have y, + 2uy^ = 0.
CHAPTER 3
SECTION 3.1
3.1.4. With £ = у - x and т/ - у - Здг, и( £, 17) satisfies
и^ + wf + - ич - - w = 0. Then, w = exp(- - | - 17JU yields
v€v-3v=0.
3.1.5. The characteristics are £ = у - x = constant. Let 17 - у + дг.
Then, 11^-^+211, +Jm = 0.
Let w - exp(- § | -17I;
and obtain vvv — |u^ = 0.
3.1.6. With a = у + 3* and /3 = V3*, w(a, /3) satisfies
M«« + "ээ + 4w« + ^Д H " 3 u = 3 sin[ V5 ^(a " ^
Then, the transformation и = exp(^-2a - 7^= j8 jy yields
3.1.7. Hyperbolic for x < 0, parabolic at x — 0, and elliptic for x > 0.
With x<0 and £ = >>-- (-*K/2 and rj = у + - (-*K'2,
1 г i ^
we have ut H г Г и* -м 1 = 0
3.1.8. In the hyperbolic region у < 0, £ = jc - ly/^y and 17 = x 4- l-^^y.
3.1.9. (a) Hyperbolic.
(b) Parabolic.
(c) Elliptic.
3.1.10. (a) Elliptic.
(b) Hyperbolic except at the origin where it is parabolic.
(c) Parabolic.
849
850
SECTION 3.2
3.2.5. [ufi] = 2(ft - /3,) so that [ик]„ = О.
SOLUTIONS TO SELECTED EXERCISES
SECTION 3.3
3.3.1. The transformation £ = —j= (xx +'.
+ x3), £2 = ~/f (*i ~
i3 = —j= (xj - x2 + x3) yields the given equation.
/9 1 3 \
Then, и = ехру- -т= Pl - -щ р2 + y— p3jv gives
with pj = ilt P2 = y5 f2> А»з = 2 ^з-
3.3.3. (a) Hyperbolic.
(b) Parabolic.
(c) Hyperbolic.
(d) Parabolic.
3.3.6. Hyperbolic if \x\ > 1, parabolic if |jc| = 1 and elliptic if
3.3.11. \<pxA + <pyB\ = -(h'(x)f - 1 = 0, so that h' = ±i.
3.3.12. <px = у - x, <p2 = у - \x, <p3 = y + x. Let u = R\ with
Г1 О П
R= 0 1 2 andv
Lo о -2]
1 0
0 2
0]
0 \vx =
j
L0 0 -l
l -6
L-J \
| < 1.
1
3.3.13. vv +
У
0
0
.0
0
2
0
0
0
2.
vv = 0 if R =
0 0 1.
3.3.14. u = \\ sin x + I sin(* - 2>>), 1 - \ sin jc + 5 sin(* - 2y), exp(x -
3.3.15. Characteristics: £ = у + x ~ constant, ту = у ~ |лг - constant. Let
u = R\ with
3.3.20. See Exercise 8.5.22.
3.3.21. Riemann invariants: и ± I'sfgh- constant on— = w ±
at
SOLUTIONS TO SELECTED EXERCISES 851
3.3.22. [pt + (и + c)px] + cp[ut + (и + c)wj = 0 ,
[Pt + (и - <?)pj ~ cp[w, + (и - c)wj = 0 ,
3.3.23. Characteristic equation: <p3t - c2q>l<pt = 0. The real and distinct
characteristics are x — constant, x + ct = constant, and x — c* =
constant.
SECTION 3.4
* In Jpnft \
3.4.2. u= ^ г , т»„ *
3.4.4. If A = у, и = A + A/)e"Ar cos x.
SECTION 3.5
3.5.1. A(fc) = ±Vc2 - k\
3.5.2. One of the roots \(k) has the form
А(Л) = (a2 - c2)A:2 + iak + О(Л3), and if c2 < a\ Re[A(fc)] >0.
(b) o>(
(с) Ш(
3.5.5. A3 + A
3.5.8. (a) A2
(b) A2
(c) A2
(d) A-
SECTION 3.6
к)=±у1?а2к2IП
к) = рк3 - ак.
2 + у2к2Х. + с2к2 = 0.
+ АА + Г2|к|2 = 0.
-|l|2=o+C
fy2|k|2 = 0.
3.6.2. (a) L*(w) = (exw)xx + (x2w)xy
(b) L*(w) = -(aw)x - (bw)y + cw.
(c) L*(w) = -w, - c2(wxx + wyy).
3.6.10. L*(w) = -w, + wx - yzwxxx.
3.6.11. wLu - uLw = — (wu, - uwt)
+ ~dX (WU*** ~ W*U*
3.6.12. wLu - uLw = — (wu, - uw,)
V- (wVV2m - mVV2w + (V2vc)Vm -
CHAPTER 4
SECTION 4.1
4.1.2. a = -\c.
4.1.3. With и = e~ktv, we obtain vtt - y2vxx - \2v = 0.
SECTION 4.2
4.2.6. The compatibility condition is I I fp dv ~ I I gp dv.
G G
4.2.8. The operator is not positive in this case.
4.2.14. (a) If и = 0 then w = 0.
(b) If duldn = 0 then pdwldn + (b • n)w = 0.
(c) If duldn + hu = 0 then pdwldn + pftw + (b * n)w = 0.
SECTION 4.3
4.3.5. x = Zj ~,—772—7j—^TT s^n ( —!__/■»— ) ) 1 <x <2 .
4.3.10. Eigenfunction: vo(x) = 1.
/ \2/ 1\2
4.3.13. Eigenvalues: A^^fy) (A:—-),
Normalized eigenfunctions: vk(x) - ( т ) cos (A: - J) у jcJ , A: s 1.
4.3.14. Eigenvalues \k > 0 are determined from the equation
VA^ cos(Va^/) + /3 sin(VA^0 = 0.
Normalized eigenfunctions:
/ч Г/ 1
852
1/2
SOLUTIONS TO SELECTED EXERCISES 853
4.3.15. Eigenvalues: At = j + [ iog(^ + /)J , * > 1.
Normalized eigenfunctions:
4.3.16. (a) x = -
7Г
д:=- + — Е p[(-l)*-l]coe -j-x
(d) co.(^*)=
/5тг \ /5тг \
cosl — x) = cosl — jc I .
4.3.23. P0 = l, P,=x, Р2=\(Ъхг-\),
Ръ = \ E*3 - Зх) , P4 = | C5*4 - Ш2 + 3) .
4.3.25. (с) хл = ^ [8P4 + 20P2 + 7P0].
4.3.26. With A = — к2, the negative eigenvalue is found from the equation
tanhjfc/=l/jfc.
4.3.27. There is one negative eigenvalue A0 = -l and the corresponding
eigenfunction is yo(x) = 1.
SECTION 4.4
2hl2 ^ 1 (irkc \ . (irk \ . (irk \
АЛЛ. и = — —г 2j ti cosl —r- t) sinl —r- a) sinl -7- x) .
a(l-a)it k=i к V / / \ I / \ I I
я „^ 2 $ 1 . (irkc \ . (irk \ . (ттк \
4-4-2- u =«hism\—')sm\-a)si4tx)■
4.4.3. »4i
or k=\
854 SOLUTIONS TO SELECTED EXERCISES
алс 1/ 2Ш2 v 1 . (irk \ . \ттк
2W
2
4.4.6. M=
where MQ -
4.4.7. и = t.
4.4.9. и=2{(/,Л4
— xj
where Mk = у у sin[(A; - - j j
4A12. и = 2^
where the Ал and the Мл are defined as in Exercise 4.3.14.
4.4.15. Nk(t) satisfies N£ + 2\Nfk + AjX == 0, with Ал = (тгА://J.
To solve, we must consider if Л < л/Х~ку, A = V^y or A >
4.4.16. „ =
where Mfc(x) = B//I/2 sinf^ л:] .
ЛЛ1О 2 v Г /irikc\2l . («к \ . (irk \
4.4.18. и = у 2 «4>HtJ 'J Sln(— ai SmlT x) ■
4.4.19. и = )
4.4.20.
SOLUTIONS TO SELECTED EXERCISES 855
The Xk are defined in Exercise 4.3.14.
- - 7 2 [Г /
sin(f x) *] exp[-(»° + (^)>] *,(* x).
4.4.25. Compatibility condition: I f(x) dx = I g(*) dx.
4.4.27. w = 2 К cosh(VV>0 + 6л sinhCV^)]^*) with ak = (f,Mk)
and bk = ~{fj Mk) coth[\/A^/]* The Ал and Mk are defined in Exer-
Exercise 4.3.14.
4.4.28. (b) The problem has no solution since the data are not compatible.
4.4.29. u = ]£ (/, Mk) exp[- ^ y]il#fc(x) , Mk(x) = д/| sin(^ x) .
4.4.32. и = 2 2) [^ c
where Mkn = [akn cos(n6) + bkn sm(n0)]vkn(r),
with the АЛп and vkn defined as in Example 4.7.
The determination of the arbitrary constants in terms of the data
involves a combination of Fourier series and Bessel function
expansions.
1/2
4.4.36. w(r, <р) = 2 (—^—) akrkPk(cos q>) ,
with вл = -f ^x), V"^— p*wj *
SECTION 4.5
4.5.1. (а) и = 2 sin дг [sin(c/) - с cos(ct) + с e~f]
(d) « = - 2^' l*-fll<cl'-*l'
856 SOLUTIONS TO SELECTED EXERCISES
.5.2. « - >/f 2 [£ bk(r) sin[^ (, - r)] *] sin(^ x) ,
where ftt(r) = -^j- Jq g(x, т) sm^— x) dx .
4.5.3. (c) u .
4
.5.4. B - Vf 2 {( *4(r) exp[-(^H - r)] dr} cos(f
If If'
+ vi Jo 6°^dr'where 6°^ = vi Jog(jc'т) л» 6*(т
4.5.5. A steady-state limit exists if I g(x) dx = O.
4.5.7. With g = sin( -j x) and A<— , u->{ —) sin( у х)
as
SECTION 4.6
4.6.3. The eigenfunction Л/0(дг) = constant.
Parabolic case: N0(t) = (/, Mo) + jQ [F0(r) + fio(r)] Jr.
Hyperbolic case: N0(t) = (/, Mo) + (g, M0)t +
irlcc
4.6.6. If <o = —г— for some k, the corresponding iVfc has the form
4.6.9. The compatibility condition is given as
3 1 з
4.6.10. и = - г sin в - ——5 г sm 30 + constant.
SOLUTIONS TO SELECTED EXERCISES 857
4.6.12. lira u(x, () = ЛЁт[«- (-!)*£] sin(^ x) .
+ у [1 - e~'\, if икс ФI for к > 1.
4.6.15. If / Ф тгпс for n > 1, и(л:) = -ЛД • To solve the full problem,
set и = w + v e~\ In the exceptional cases use the finite sine
transform.
4.6.16. The polynomial is v(x, t) - I —j) x3 - ( —2 ) jc + ( 7) xt.
SECTION 4.7
4.7.3. The perturbation around u0 is stable if A < 0 and unstable if Л > 0.
The reverse is true for ux.
4.7.4.
4.7.5.
4.7.6.
4.7.7.
4.7.8.
u(x
к-
к-
(а)
А =
,0
= 1.
= 1.
м =
= 1.
-!/(') = -
If А а \с,,
: ew yields
+
н-
w
ее"
ГЗ',
-1)
гг(А-
8Л
деде
. К А-
I •
1 si]
Aw = 0.
CHAPTER 5
SECTION 5.2
5.2.9. F(A) =
5.2.10. F(A) =
5.2.11. F(A) =
5.2.12. y(x)-
5.2.16. M = i
2
-^ e~kM sinh(b),
C
a .
5.2.18. и = у - jo Re[e"lA'F(A)] cosh[Va2 - (c\Jt] dk
+ лД f Re[e~lAxF(A)] cos[V(cAJ - a2t] dk
where Re(z) is the real part of z.
5.2.23. м=-
858
ds
SOLUTIONS TO SELECTED EXERCISES
859
тг J-~ A + A2)si
sinh[AL]
dk.
SECTION 5.3
5.3.3. (a) F,(A)=ViH;
V4 + 4
5.3.5.
5.3.6.
5.3.7.
5.3.8.
,(,)
JL
y(x) = ±
dt.
dt.
y(x) =
\-кг
кг-2
l-к2
5.3.12. м = '
5.3.13. и
{-!)■
W
0,
0<*<- .
с
-с
Jo
0,
X
t>-
с
с
c^i^ 2 Г ,ь Jl-cosh(Ay) /cosh(Aa)-l\ . u/ J , л
5.3.16. и = - I sin( Ax) Г1-^ + ( , ■ w, v ) sinh( Ay) Uy + й,
ir Jo 7L A \ Asinh(Aa) / J ^
5.3.17.
5.3.18.
ih(Aa)
with й given by E.135) where f(s) = e.
2 Г 1
м = - - Jo - sin(A) cos(Ajc) e~Xy dk .
SECTION 5.4
5.4.4. (i) u = t,
a- r
0<r<a.
SOLUTIONS TO SELECTED EXERCISES
2-' — <« —
(iii) w = 0, *>-—-, 0<r<a.
с
(iv) м = 0, 0<*< —-, r>a.
с
(vi) м = 0, f>^—-, r>a.
с
5.4.5. м = 1 + r2t + czt3 .
5.4.6. и = (x2 + y2)t + | cV .
5.4.7. и = x.
A fX + Ct Г T
5.4.8. v = Yc l-c, КЩ с V(<*J-(*-*JJ ds .
-J ГХ + Ct Г
SECTION 5.5
5.5.4. The integral diverges, but if we disregard the singular contribution
to the integral from the limit at infinity we obtain и - -log r.
5.5.6. M = ^-f f exp[-VA2 + k2\z - s\)J0(Ar)f(s) d\ ds .
Ц-7Г Jt°° ./0
\
with uH given as in E.220).
SECTION 5.6
5.6.1. (b)
SOLUTIONS TO SELECTED EXERCISES
1
861
(e)
~~«A
t-(xfy)
5.6.10. F(A) =
5.6.20. u = yH\t~-
where H(z) is the Heaviside function.
5.6.22. и = x + expi -(-— J t\sin(^-y xj + / 2 erfcl
5.6.26. м = 1 + -
1)/-
2cVt
x) .
0,
5.6.27. u = e~
SECTION 5.7
5.7.3. и « ^~ F
where Re[z] is the real part of z.
V? \2тгс/
0<t<-
c
5.7.6. м
4a
exp[- ^^1} .
l 4a t JJ
" 2^ £ f
as f—>oo,
CHAPTER 6
SECTION 6.2
6.2.4. ux =
PlC2
. x<e.
p2ctВ
P2C1 PlC2
6.2.8. Eigenvalue equation:
ф[
= о with h >0.
Eigenfunctions:
, 0 < х < х0 ,
< Г < /
6.2.9. With \k, v(k\ and \\vm\\ defined as in Example 6.2,
where
6.2.10. With kk and v(k) defined as in Example 6.2,
where /(jc) = 2 <*kv(k\x) and g(jc) = 2 VX~kpkv(k\x)
862
SOLUTIONS TO SELECTED EXERCISES
6.2.13. в=тг-в, <p = sin г1 sin в ,
863
^ cos в
кх cos 0 + Vк\~к\ sin2 0
and/?=T-l.
SECTION 6.4
6.4.5. For example, let u(x, 0) = м@, t) = u(l, 0 = 0 and u,(x, 0) = дс.
1,
6.4.6. м = •{ _„ = , t>f
SECTION 6.5
6.5.3. (a) au@, t)-pux(Q, t) = sm(<ot);
u(x, t)= —
(b) -Wjc(o, о - sin(oiO; Ф, 0^
(c) m@, t) = sin(co/); m(jc, /) = sin <oU - - J
Each solution is valid for 0 < x < ct.
6.5.4.
6.5.6. м = 0, 0<с/<л:; м== Л cos Ш [t- -) , -
6.5.7. и = 1 + F(x - ct) -
(x +
where F(z) = 0 for z > 0 but is otherwise arbitrary
.5.8. u =
со2
864
SOLUTIONS TO SELECTED EXERCISES
6.5.9. M = ^sin U
6.5.11. и-Дх-сг) +J*±£?l-J£_2].
SECTION 6.6
6.6.2. Eigenvalues: ml\/Yk sin[V~Aj si
- 1)] =
- x)]
Eigenfunctions: Mk{x) =
with the МДх) chosen such that Mk{\) = 1.
00
6.6.3. /(jc) = 1 = 2 ckMk(x)y
where cfc - —-
-1)] l
К >о
6.6.4.
6.6.5.
6.6.6. и
2 sin2[A/Aj sin2[VA^(^ - 1)]
(jc)M(jt) = O, 0<x^
'(/) - kmxM(l) = 0, M@) = 0 .
2 J
+
2 '
0,
|x| > ct,
-ct<x<0,
-■£-Aain\-(x-ct)], 0<x<ct.
2o> L с J
^]«0*.
SOLUTIONS TO SELECTED EXERCISES 865
SECTION 6.8
6.8.4. In two dimensions, u(x, y) must be bounded at infinity. In three
dimensions, u(x, y9 z) must vanish at infinity.
6.8.7. E(t) = \\ u2dx and E'(t) = - (b - \ax)u dx.lfb- \ax => 0,
J-00 J-00
uniqueness follows, otherwise F.237) must be used.
6.8.8. E'(t) = - I ur5utic. If urJ5u>0 for all u, uniqueness follows.
Otherwise, let u = erty.
6.8.9. Putu = e1/2'v.
6.8.10. Here ur5u = A(a - pJ > 0, with ur = [a, fi].
1 f'
6.8.11. E(t) = - J [и? + «L] dxiiu and w^ vanish at x = 0, /.
6.8.12. £ = I (V2mJ jy if и and — vanish on dR.
6.8.13. E{t) = ^ ||д [м2 + (V2wJ] dV if u and |^ vanish on *?/?.
CHAPTER 7
SECTION 7.1.
7.1.4. u( £) = - f ^ £fe_£2 ад л if « ^ 0 on <?G.
ve/ JaG а{х) дп
{х)
i ^)ад ^if =° °n <?g-
7.1.9. u( f, t) = | j pFK dx dt + |o [pg/C - pfK,],.o dx
R
+Г f- ^ ь *+Г it gi^Uo *•if ai=
7.1.10. u( f, t) =
7.1.12. M[K] = 8(x-£)8(t-r), 0<x<l,t<T,
with К = Kt = 0 at t = T and К = Kx =0 at x = 0, I.
7.1.14. u( i, t) = f f *r(jt, *; f, т)с <£с Л + f ^r(x, 0; & r)f ^x
J~oo JO J—oo
SECTION 7.2
7.2.9. (a) /'(*) = S(*-1)sin 1 + #(jc-1)cos*.
(b) /'(*) = 0.
(c) /'(*) = S'(x + 3)e~3 + 8"(x + 3)e~3 - 8(x)
+ 2H(x) сов(дг) sin(x).
866
SOLUTIONS TO SELECTED EXERCISES
867
(d) f'(x)=-H(x-l).
2л
2a
(e) /'(*)
7.2.11. S(x2 - a2) = ^ 5(x - a) + ^ 5(* + a) ,
7.2.12. 5(sinx) = £
7.2.22. 2Vc-2/Vc.
SECTION 7.3
7.3.1. u(x) =
7.3.2. u(x) -
ds + у
csinh[c/]
Г si
Jo
ds
^h
cosh[c(/ -
7.3.3.
с cosh[c/] '
^] cosh[c(/ - x)]
с cosh[c/]
7.3.5. K(x; {) =
с sin[c/j
?)sin[c(/-
a>0.
с sin[cl] '
7.3.6. /C(jc; £) = —log ^, 0 < дс < ^ , ^(д:; £)
7.3.8. For example, if /3, = /32 = 0 and 0 < £ < /,
u(£) = Jo p(x)F(x)K(x; t)dx + p@) ^. ^@; f) -
ccosh(c/) {"Iе* coshM/ ~ *I sinhtCJCl Л
- J e* sinh[c|] cosh[c(/ - л:)] йдс
+ 3 cosh[c(/ - {)] + 10 sinh[c£]} ,
868 SOLUTIONS TO SELECTED EXERCISES
7.3.12. ii(f) = |o K(x; £)f(x) dx+p(l)bk(h f) -p@)o£@; f) + i £ „(jc)
The last term is an arbitrary constant. For a solution to exist we
must have ap@) - bp(l) = I f(x) dx.
7.3.13. A (nonunique) Green's function K(x; f) is
where Bp(l) - Ap@) = -1.
r2
7.3.22. K(x, y,£,v) = K(x, y,i,v)~ J—gi + constant.
7.3.23. There is an additional term which is given as
N0(t)M0(x) = (t - t)M0(i)M0(x)H(r - t).
7.3.24. There is an additional term N0(t)M0(x) = М0(£)М0(х)Н(т - t).
7.3.26. (a) K(x, t; £, т) =
»АI * s14t (т - '}J sm[— ч si4t *J) я(т -') •
7.3.27. (а) и(х, /) = - 2,Д ^ sm[— дг] sin[— rj Jq sin[— sj dsj
2 v / • \vk 1 [wife 1 f . /ч . (ттк \
+ |Z |sin[— *J cos[— fj JQ sin(s) sin^— s)
7.3.29. (а) ^(л:, /; $, т) =
~'
7.3.30. (а) иЦ, т) = |о К(х, 0; f, т) djc + Jo е~'Кх@, V, £, т) dt
with AT given as in Exercise 7.3.29(a).
SOLUTIONS TO SELECTED EXERCISES
SECTION 7.4
7.4.10. K(x, y, z, t; £, v, £, т) =
1
869
4iryr
7.4.13. K(x, t; €, r) = Yy
S[y(r-t)-r]
^ \J y\t - rf - {x -
H[y(r-t)-r\
7.4.14. K(x, t; £, т) =
7.4.17. (a)
- yt- (f -
V'2
Згтг
\ sinh(ta) e~ki,
- sinh(A:£) e
-кх
7.4.18.
j-eiki
eiki sin(kx),
7.4.19. а:(*. у; €, п) = - Sx ^ sinLT ^J smLT ^J expL" T|у " ч|J
7.4.22. JC(x. г; О = ^ ^ - sm[(n + -J T\ sm[{n + - j T Je ,
(f)
7.4.23. К(х, y,z;i,r,,() = —t 2, sm[— £j sin[— x\
x
where IC0 is the modified Bessel function of zero order.
SECTION 7.5
i3/2
870
7.5.5.
SOLUTIONS TO SELECTED EXERCISES
2ir(Pi+p2)
\-ll2
where r2 = (x - f J + (y- r,J + (z - tf,
and r2 = (x-tJ + (y-vJ + (z + O2-
7.5.7.
7.5.8. (a) K(xyy;t,<n)=-H\
where p2 = (x - £J + (y -tjJ, and p2 = (x - £J + (у + т/J.
7.5.9. (a) K(x, y, z; ^ 17, £) = -— eT - -т—z e~c\ with r and r defined as
. „ . «,-« ^7rr ™r
in Exercise 7.5.7.
7.5.10. (b) K(x, y; i;, 7/) = — X0(cp) + т— #0(cp), with p and p defined
as in Exercise 7.5.8.
7.5.12.
-rjr
7.5.18. jR:(x, y; ^, 7/) = т— log ^^ , where p0, r0, and rx are the two-
Ztt l пт^ j
dimensional analogues of the terms defined in Example 7.15.
7.5.20.
7.5.21. (с) К{х9у\£9ц) = -тг-
7.5.23.
SOLUTIONS TO SELECTED EXERCISES 871
i , i>o V(*+r)+(y-v:
_j_i гро
2ir °gL a
where pi = ^2 + ту2 and (£\ 77') is the inverse point, with respect to
x2 + y2 = a2, of the point (£, 17).
CHAPTER 8
SECTION 8.1
г/
8.1.3. E(u) = Jo [р(и'У + qu2] dx + hlP@)u2@) + h2p{l)u\l) .
8.1.5. c= 312^ , £(м) =
з , E(u) з .
2тг — Зтг 2тг — Зтг
8.1.10. рт = 1, рм =2, рж = ^ and pM = 1.
Thus, k[M) = BttwJ and A^m) = (тгпJ.
8.1.12. Use inscribed and circumscribed circles of radius a and b, respectively.
Then, --j- < Aj < —г- ,
where a10 is the first positive zero of /0(jc).
8.1.13. Use inscribed and circumscribed cubes with centers at the origin and
sides 2/V3 and 2, respectively. Then, Зтг2/4< Л2 = тг2 <9тг2/4
8.1.15. Eigenvalues: АЛи = -^г .
а
Unnormalized eigenfunctions:
{) ()
Мкп{г, в) = Jml9{*f) sin(^) , *, n = 1,2,...,
where the akn are the positive zeros of Jvnlip(x).
SECTION 8.2
8.2.2. £, = 1, ft =
20 /3 3 2 3 9
+X ~X~
8.2.8. A, = -=r
1 4
872
SOLUTIONS TO SELECTED EXERCISES 873
8.2.9. Let ft(jc, y) = (x - l)y(y ~ x). Then, At = q
8.2.10. L =
8.2.11. ^=25/6.
8.2.12. Ax = 1 + ^ €7Г + °
8.2.14. A^ 22.124.
8.2.18. w(x, y) = x2 + @.801) cos(y).
SECTION 8.3
8.3.4. ^,/;?,t)=^
SECTION 8.5
8.5.5. w(x, 0 = V7 2 (a* cos[(l~) C
where ал and йл are found from
f(x) =y-j Zj ak sm^— xj and g(x) = у у Z ^—J 6fc sinV— x
8.5.8. For the problem of Exercise 8.5.7(c), the Green's function does not
exist since Ao = 0 is an eigenvalue. A modified Green's function
K(x; |) that satisfies Kxxxx = -8(x - f) + 1//, with Kx\x=0 =
Kxxx\x=0 = Kx\x=l = Kxxx\x=l = 0 can be found.
8.5.10. u(x, y, t)=-i= S {anm oos(VA^cr) + bnm
7ГП \ . ( 7ГГП
where
Grn\2 (ттт\2
and where anm and Ьпш are found from
2c
Vlt =iUnm smw / sin\~ y)
. (тгп \ . /тгт \
874 SOLUTIONS TO SELECTED EXERCISES
8.5.15 u(r, в) = (r2 - R2)[-^ rsind-^r3 sinЗв] + L
8.5.17. u(r) = ^ F0(R4 + r4) - ^ F0/?2r2.
8.5.18. м(г) = ^- (log Л - log r)r2 +z^z(r2- R2) .
8ж
8.5.29. тт(х, t) = [\ cos(jc + rt) + \ cos(jc - cr)]i + j + tk, E(x, t) = 0 and
8.5.32. (a) o, = c[(f
(b) w = —— , where a10 is the first positive zero of /0(jc).
8.5.33. u(x, t) =
\ cos(x - cfy + ^ cos(x + cfi) + ^- [(x i- c,rJ " (x - c,02]}i
(l - ^- exp[~(x + c,0] + ^- exp[-(x - crr
if + - sin(x - cti) + - sin(x + ctt)|k ,
where c7 = and <r = — .
P P
CHAPTER 9
SECTION 9.2
9.2.1. u = \(r2-\) + <
9.2.3. и = !(г2-1) + £еA-г4) + 0(е2), r2 = x2 + y2.
9.2.4. u = l+\e2(r2-
9.2.5. u(x, y) = | | " " Л1гГГ "" «ИМ
[тг2 - ir/k] siah]ky] [y2 - у Ik] cosh[A:(y -
[sinh(A:7j-)]2
] cosh[A:(y - эг)] 1
inh(*7r) J
9.2.7. Exact result, X = -— exp(-er).
4тгг rj г r i
Perturbation result, К = -— + e2 r— + c0 + O(e4),
4тгг L отг J
where c0 is an arbitrary constant and
Г2 = (jC - £J + (y — 7}) + (Z — ^J.
9.2.8. (b) и — - u_x = - , is the exact solution.
9.2.9. (b) If q = 0 and a = ea, set и = - m_j + X *kuk.
9.2.11. Perturbation result,
m(jc, 0 = - И - - €t)[f(x - cf) + /(jc + ct)] + ^- I /E) ds
Multiple scales result, with r = et,
U = u(X,t,T)
= exp(- ^j[- Дл: - rt) + - f(x + с
875
876 SOLUTIONS TO SELECTED EXERCISES
u. _,. „ - [а±2]и 4 £] j
where G(x, t) is the fundamental solution of vt — vxx.
9.2.16. и = м0 + eii! + O(€2)
^ / 7гу\ . /тгх\
Then, м0 = exp^- — J sm^—J
and Mi satisfies V2m2 = 0, 0 < л: < /, у > 0,
with ux bounded as у-»00, м^х, 0) = щA, у) = О
and мДО, у) = - у sin(y) exp(- у у).
9.2.17. u(x, t) = A a
0 < x < ct, and u(x, t) = 0, x > ct.
9.2.18. «(x, y) = i(l + x2 - y2) + e[i(a + b)+ \{a - b)(x2 - y2)] + O(e2) .
9.2.20. An(e) = 2+^ + O(e2).
9.2.21. Aw(e) = 7Г2 + e + O(e2) .
9.2.23. (a) A,(e) = 1 - — + O(e2) .
(b) A1(e)=i + ^ + O(e2).
SECTION 9.3
9.3.1. Exact solution, u(x, t) =
f " 2 )(sin X ~ € COS X) + f 2 ) eXp( jl^2 Sin(^ ~ 0 + € COS(X
Regular perturbation solution, m(jc, ^) — ( 2 )[s*n x ~ € cos xl*
Boundary layer expansion, й(х, г) =
sin x — e cos x - €2 sin x + e~T[e cos x + €2A + r) sin x] + O(e ) ,
where r = - .
e
SOLUTIONS TO SELECTED EXERCISES 877
_ € sin(x - t)
9.3.3. Exact solution, u(x, t) - € + t sin/x _ a •
sinx
Boundary layer solution, й(х, т) = ——:— + O(e), where r = tie.
1 T T Sin X
Both solutions break down at t = e.
9.3.4. Regular perturbation solution, м = - + 2 2n+i ~^ €л.
In the boundary layer set a — t/Ve and м = u/Ve and let
у - 2 en/V Then u0 = exp(- — ) exp(T) dA, and
„=0/^.2 4 \ I / JO \ Z /
vx =exp^- — jsinx.
9.3.7. With D = c, the regular perturbation expansion is given as
v(xy t) = S(x - ct) - $€t8"(x - ci) + O(€*).
A boundary layer analysis yields the exact solution of the full problem.
9.3.8. (a) Outer expansion, y(x)« y0 = C exp - I -j~ ds .
Boundary layer expansion,
y~yo = a + {f}cxp[-[ %£ ds]-a}{l-exp[-a@)<r]}
where a = xle.
9.3.9. For each case we give a composite result.
(a) и « x2e'x exp(- ^J + sin[^ yj exp(- J) .
(b) и = sin[y y] exp(- -) + Dney1/2y Г а\и - x)~v2
x exp - -r~~ - cr Ida .
rL 4e(a - x) J
(c) и «(x + уJ exp[-(x + y)] - (x + LJ exp[-(x + L)]
9.3.10. Outer solution, и « м0 = |jc - y| exp[-|x- y\]. There is an
internal boundary layer at у = x.
9.3.11. Composite result,
+ЯФ[- ^
878 SOLUTIONS TO SELECTED EXERCISES
9.3.13. Composite result, u**y + erfc , J .
where a = V - \ f, p = £, x - ell% у + 1 = е2/Ч
9.3.19. (a) Parabolic boundary layers at у =0 and у = тт.
(b) Parabolic boundary layer at у = 0 and ordinary boundary layer
at jc = тт.
(c) Ordinary boundary layer at x = У1 — у2.
9.3.21. In all cases the solution is given as
9.3.23. Composite result,
2 [ Г /() i(fa) &1 [А:2(
и(х, у) ~ - 2 [ Г /E) sin(fa) &1 ехр[А:2(* -
2 °°
/(*) sin(fc5) di ехр[-£2тг] sin ку ехр( )
CHAPTER 10
SECTION 10.1
10.1Л. us = - —- exp(iknr), where f2 = (x - |J + (y - ?jJ + (z + £J .
10.1.4. /(Г(д:, у; f, 77) = j if o^t^vC^ " ^J + (У + vJ]
\
00 rx
10.1.5. m(x) = 2 v£x)(ik)~J ехр[/А:<р(д;)] where <р(л;) = ±
with arbitrary cy, and y_! =0.
10.1.6. Rays, л: = (a ± a) cos a, у = (a ± a) sin a, z = ^3.
Phase functions, <jp+ = 1 — a + r, <p_ — 1 + a — r,
where r is the radial variable in cylindrical coordinates.
10.1.9. (a) <p - ф on the boundary.
дФ д<р дф
If -21 ф о on the boundary, -~ = ± -21 there.
on the boundary.
879
880 SOLUTIONS^) SELECTED EXERCISES
10.
.1.10. us - I( AI/2( 2^_)Ш exp[ik{x + fl) _ iE] .
10.1.14.
near у = 0, 2 < 0,
where F(z) is the Fresnel integral A0.214).
1 Г iVl rd-rtfr/a»I"
10.1.17. ф, y) - ^= ехр[йд; - T J J_A+>)(fc/2jtI/2 exp(«2) Л .
10.1.19. и(л, у, z) =
к
SECTION 10.2
c2t
10.2.1. и(х, t) = Я(д; + yt) - — (д: + yf)&(* +
1 A2f
10.2.3. и(х, t) = - ^- е~л'Я(;с - у*) + —j е~х\х - yt)H(x - yt) +
10.2.6. ф, у, z,t) = at+j(p- e)[£7^](»P" в " 0Я(г - а + 0
10.2.10.
SOLUTIONS TO SELECTED EXERCISES 881
SECTION 10.3
10.3.1. The parabolic equations are vt - cvx - ~€(a2 ~ c2)vxx = O(*2)
and wt + cwx - - e(a2 - c2)wxx = O(e2) .
10.3.3. The approximate Burgers' equation is given as
3 1
10.3.5. The Korteweg-deVries equation is at - ax - -z eaax - - ea,^ = 0 .
- - €«жл. = 0.
10.3.7. The Burgers' equation is at — - eaa^ - -
10.3.8. The Korteweg-deVries equation is at + ax — - eaax + - eaxxx = 0 .
INDEX
Abel's integral equation, 398
Acoustics, 579, 818
Adjoint equation, 13, 41, 161, 425
Adjoint operator, 33, 146, 147, 148, 149, 161,
162, 550
formal, 147, 227
Admissible function, 519, 524
Airy function, 322, 547, 678, 697, 732, 733,
735, 736, 737
asymptotic form, 322, 733, 736, 737
Airy's equation, 547, 732
Akhmanov, S.A., 763
Amplitude, 643, 644, 645, 702, 717, 754, 828
Angle of incidence, 750
Angle of reflection, 750
Antinode, 196
Approximate factorization, 809
Associated Legendre equation, 601
Associated Legendre function, 602
Asymptotic decomposition, 812
Asymptotic equality, 295, 618
Asymptotic expansion, 293, 618, 702, 721
Asymptotic method, 616, 702, 772
Asymptotic simplification, 808
Axial source, 282
Backward difference equation, 12, 40
Backward initial and boundary value problem:
heat equation, 469, 668
hyperbolic equation, 417
parabolic equation, 159, 419
Backward initial value problem:
heat equation, 142
hyperbolic equation, 420
parabolic equation, 420
wave equation, 473
Balance law, 164, 225, 325
Beam half-width, 752, 760
Bessel function, 191, 230, 231, 279, 281, 322,
387, 532, 542, 571, 624, 701, 778, 801,
835
asymptotic form, 231, 322, 701, 778
eigenfunction expansion, 190
zeros, 191, 532, 542
BesseFs equation, 190, 280, 381, 532, 541,
571,623
BesseFs inequality, 181
Bicharacteristics, 780
Biharmonic equation, 133, 572, 620
Bilinear expansion, 447, 459
Biorthogonal set, 227
Boltzmann's equation, 838
Boundary condition:
Dirichlet, 167
first kind, 167
mixed kind, 167
natural, 526
Neumann, 167
second kind, 167
third kind, 167
Boundary layer, 647, 664, 670, 682
equation, 651, 654, 659, 664, 665, 670, 678,
686, 687, 732
theory, 617, 647, 650, 658, 663, 705, 721,
740, 765, 777, 809
Boundary perturbations, 636
Boundary value problem, 137
Boussinesq equations, 828
Breaking time, 68
Breaking of waves, 67, 828
Brownian motion, 3, 15, 22, 24, 25
absorbing boundary, 10, 28, 41
backward Kolmogorov equation, 13, 38, 41
backward telegrapher*s equation, 41
deterministic motion, 23, 24
diffusion coefficient, 7
883
884
INDEX
Brownian motion (Continued)
diffusion equation, 7 , 12, 22, 35, 36, 40
drift velocity, 4, 12
elastic restoring force, 36
finite particle velocity, 15, 25, 26
Fokker-Planck equation, 12, 26, 35
forward Kolmogorov equation, 12
infinite particle velocity, 14, 24
inhomogeneous medium, 12, 25
Langevin's equation, 14, 18, 26
Laplace's equation, 28
mean displacement, 2, 4, 18
mean first passage time, 32, 44
mean square displacement, 18, 22
random motion, 23
random walk:
correlated, 15
independent steps, 2
position dependent, 11, 25, 33
time independent, 27, 30, 42
rate of reversal, 19, 22, 41
reflecting boundary, 11, 37, 42
telegrapher's equation, 20, 25
unidirectional wave equation, 23, 24
Burgers' equation, 67, 820, 822, 826, 838, 839
Canonical form, 110
elliptic equation, 114
hyperbolic equation, 112, 113, 362
hyperbolic system, 130
parabolic equation, 114
Cauchy principal value, 256, 433
Cauchy problem, 136. See also Initial value
problem
elliptic equation, 139, 141
hyperbolic equation, 139, 142
parabolic equation, 142
Cauchy-Riemann equations, 48, 131
Cauchy-Schwarz inequality, 227, 349
Causal fundamental solution, 377. See also Free
space Green's function
hyperbolic equation, 384, 385, 417, 419, 465,
474, 475, 480, 555, 773, 796, 801, 805
parabolic equation, 384, 388, 419, 467, 480
Causality, 13, 26, 377
Caustic:
curve, 92, 720, 729
surface, 713
Chain reaction, 544
Characteristic:
base curves, 49, 117, 707
cone, 125, 376, 385
conoid, 376, 378, 806
coordinate system, 111
curves, 49, 63, 117, 128, 550, 582, 633,
664, 669, 672, 774, 778
direction, 63, 81, 83
equation, 49, 63, 84, 111, 120, 125, 128,
129, 132, 340, 774, 779, 785, 797
initial value problem, 50, 52, 60, 61, 65, 69,
128, 361, 370, 662, 713
polynomial, 811
strip, 85
surface, 125, 340, 344, 782
velocity, 52, 370. See also Speed of wave
propagation
Characteristic form, 130
Characteristic normal form, 135
Classical solution, 178, 211, 390
Classification of equations:
mth order, n variables, 132
second order:
n variables, 122
two variables, 110
Classification of systems, first order:
n variables, 132
two variables, 127, 134
Cole, J. D., 823
Compact support, 426, 436
Compatibility condition, 117, 154, 457
Complete integral, 92
Completeness criterion, 176, 181
Composite expansion, 668, 672
Composite medium, 329
Composite rod:
heat conduction, 336
longitudinal vibration, 332
Compression wave, 76, 828
Concentrated source, 363, 372
Concentration, 165, 543
Conservation form, 57, 75, 105
Conservation law, 75
Conservative type, equation of, 144, 390
Consistency condition, 119
Constraints, 519, 520
Continuous dependence on data, 138, 393, 395,
559, 561
Convolution theorem:
cosine transform, 312
Fourier transform, 247
Laplace transform, 319
sine transform, 312
Correlation coefficient, 16
Courant, R., 29, 520
Courant's maximum-minimum principle, 520,
527, 540
Critical intensity, 772
Critical value, 222, 544
INDEX
885
Cutoff frequency, 615
Cylindrical wave, 90, 91, 374, 383, 701, 722,
744, 753, 781, 801
d'Alembert's solution, 51, 341, 345, 351, 358,
555
Defocusing medium, 757, 762
Difference equation:
correlated random walk, 19
mean first passage time, 32, 44
position dependent random walk, 12, 33
random walk, 6
time independent random walk, 28, 30, 42
wave equation, 350
Diffracted ray, 744
edge, 744
surface, 751
Diffraction, 703, 744, 760
Diffraction coefficient, 745
Diffraction length, 752, 767
Diffusion effect, 306, 307, 791
Diffusion equation, 7, 12, 22, 35, 36, 40, 48,
121, 131, 285, 543, 662. See also Heat
equation
Dirac delta function, 7, 30, 255, 311, 370, 375,
429, 445
Dirichlet data, 137, 167, 527
Dirichlet's principle, 536, 599
Discontinuities, 56, 57, 342, 364, 434, 778,
792, 809
Discontinuous derivatives, 56, 75, 117, 120,
338, 342, 354
Dispersion relation, 144, 630, 645, 774
Dispersive type, equation:
of linear, 144, 564
of nonlinear, 645, 830
Dispersive wave motion, 144, 295, 644
Dissipative effect, 307
Dissipative type, equation:
of linear, 145, 301, 394, 658, 808, 810
of nonlinear, 830
Dissipative wave motion, 145, 301
Distribution, 446
Divergence form, 146, 150
Divergence theorem, 146, 164, 172
Domain of dependence, 205, 480, 553
Domain of influence, 205, 480, 553
Doppler effect, 360, 369
Double Fourier sine series, 456
Double layer distribution, 364, 503
duBois-Reymond lemma, 330, 339, 365, 598
Duhamel's principle, 201, 421
Eiconal equation, 89, 702, 704, 707, 754, 784
Eigenfunction expansion, 176, 183, 192, 193,
207, 209, 347, 446, 458, 464, 466, 481,
517
Eigenfunctions, 171
completeness, 176, 337, 517, 533
orthogonality, 175, 182, 337, 540
orthonormality, 183, 517
real valued, 175, 182
Eigenspace, 811
Eigenvalue perturbations, 548, 568, 640,
642
Eigenvalue problem, 171, 178, 184, 185, 186,
190, 192, 194, 208, 213, 334, 367, 447,
453, 456, 461, 517, 522, 524, 529, 540,
567, 570, 614, 639, 642
adjoint, 227, 231
Eigenvalues, 171
continuous spectrum, 245
countability, 175, 183, 337, 517, 529
discrete spectrum, 175, 183, 245
limit point, 175, 183, 337, 529
multiple, 176
non-negativity, 175, 182, 337
real valued, 175, 337
simple, 183
Einstein, A., 14
Elasticity theory, 593
compression wave, 594
isotropic medium, 593
Lame constants, 594
longitudinal wave, 594
Navier's equation, 593
shear wave, 594
transverse wave, 594
wave equation, 594
Electromagnetic theory, 587
constitutive relations, 588
dielectric constant, 588
electric field vector, 587
electrostatic potential, 588
Hertz vector, 613
irrotational field, 588
Laplace's equation, 589
Lorentz condition, 590
magnetic field vector, 587
magnetic permeability, 588
magnetostatic potential, 589
Poisson's equation, 589
scalar potential, 589
solenoidal field, 589
telegrapher's equation, 589
vector potential, 589
wave equation, 589
Electrostatics, 588
886
INDEX
Elliptic equation, 114, 123, 141, 166, 169, 328,
345, 379, 395, 402, 412, 447, 452, 457,
587, 662, 663, 669, 679, 681
Energy conservation, 139, 144, 196, 391, 395,
584, 718
Energy dissipation, 145, 394, 395
Energy integral, 390, 518
biharmonic equation, 410
dissipative hyperbolic equation, 392
elliptic equation, 395
hyperbolic equation, 390
hyperbolic system, 409
parabolic equation, 394
vibrating plate, 410
vibrating rod, 409
vibrating string, 196, 233
Envelope, 93, 713
Error function, 265
complementary, 267
Euler's equations of fluid dynamics, 157, 158,
575, 577, 578, 579, 581, 585
equation of conservation of energy, 611
equation of conservation of momentum, 577
equation of continuity, 576
equation of state, 577, 578
Even function, 188
Expansion wave, 77, 828
Expectation, mathematical:
continuous case, 9
discrete case, 2, 16, 21
Extension of functions, 188
even, 188
odd, 188
periodic, 188, 197, 441
Fick's law, 165
Finite Fourier transform, 208, 214
inverse, 209
Fluid dynamics, 575
circulation, 578
density, 575
entropy, 577, 584
Euler's equations, 577, 578
gas constant, 584
heat conduction coefficient, 584
internal energy, 584
Laplace's equation, 579
Mach number, 587
material derivative, 576
Navier-Stokes equations, 583, 816
polytropic gas, 580, 582, 823
potential function, 578
pressure, 575
quasilinear wave equation, 581, 823
specific heat, 584
speed of sound, 583
streamline, 586
velocity, 575
viscosity coefficient, 584
wave equation, 580
Fluid flow:
adiabatic, 577, 584
compressible, 577
incompressible, 578
inviscid, 575, 577
irrotational, 578
isentropic, 577, 585
sonic, 586, 587
steady, 578, 585
subsonic, 586, 587
supersonic, 586, 587
viscous, 583, 816
Focal length, 762, 766
Focal point, 91, 744, 765, 801
Focus, 720. See also Focal point
Focusing medium, 756
Fourier coefficients, 180, 209, 338, 518
Fourier cosine series, 186
Fourier cosine transform, 261, 268
of derivatives, 262
finite, 208
inversion formula, 261
Fourier integral formula, 246
Fourier series, 180, 186, 308, 347
Fourier sine series, 185, 440
Fourier sine transform, 261, 264, 269
of derivatives, 262
finite, 208
inversion formula, 261
Fourier*s law, 165
Fourier transform:
generalized, 436, 437
higher dimensional, 271
of derivatives, 272
inversion formula, 272
one dimensional, 246
convolution theorem, 247, 249
of derivatives, 248
inversion formula, 247
Free space Green's function, 378, 381. See also
Causal fundamental solution
heat equation, 469, 491
Helmholtz equation, 381
Klein-Gordon equation, 477, 489
Laplace's equation, 380, 485
modified Helmholtz equation, 280, 383
INDEX
887
modified telegrapher's equation, 477, 489
ordinary differential equation, 380
reduced wave equation, 382, 496, 703
wave equation, 474, 489
Fresnel integral, 742
Friedrichs, K.O., 29
Fundamental lemma of calculus of variations,
344, 597
Fundamental solution, 370, 377
adjoint equation, 419
biharmonic equation, 574
elliptic equation, 380, 414, 419
heat equation, 254, 286, 375, 384, 407, 824
hyperbolic equation, 385, 419, 554
parabolic equation, 419, 835
wave equation, 384
Furth, R., 15, 18
Galerkin method, 349, 403, 549, 603
Gas dynamics, see Fluid dynamics
Gaussian beam, 752, 759, 835
Gaussian distribution, 8
mean, 8
variance, 8
Generalized derivative, 431, 434
Generalized functions, 250, 386, 426
composite, 444
product of, 443
regular, 428
singular, 428
of slow growth, 438
Generalized power functions, 432
Generalized solution, 56, 57, 75, 178, 211, 250,
344, 346, 354, 359, 440
Geometrical optics, 89, 107, 703, 732, 738,
740, 749, 765, 798
nonlinear, 754
Geometrical theory of diffraction, 703, 744, 751
Goldstein, S., 15, 18
Goursat problem, 404
modified, 404
Gram-Schmidt process, 176, 539, 600
Green's function, 33, 43, 250, 378, 412, 446
biharmonic equation, 575
elliptic equation, 413, 447, 452
heat equation, 467, 491
Helmholtz equation, 513
hyperbolic equation, 417, 420, 422, 463, 510,
554
Klein-Gordon equation, 475, 491, 509
Laplace's equation, 31, 48, 199, 453, 485,
492, 560
modified, 457, 460
Laplace's equation, 460
modified Helmholtz equation, 280
modified telegrapher's equation, 491
nonself-adjoint elliptic equation, 425
ordinary differential equation, 250, 421, 435,
450
parabolic equation, 419, 420, 423, 466
reduced wave equation, 481, 705
Schrodinger's equation, 510
wave equation, 470, 491
Green's theorem, 31, 412, 550, 574
Group velocity, 144, 299
Hadamard, J., 138, 139, 141
Hadamard criteria, 138
Hadamard's method of descent, 275, 315
Hankel function, 280, 383, 483, 687, 704, 723
Hankel transform, 281, 282, 373, 375
of derivatives, 281
finite, 208, 240
inversion formula, 281
Harmonic function, 485
Harmonic vibrations, 195
Heat conduction, 14, 198, 625
infinite speed, 14, 199, 255
Heat equation, 14, 198, 206, 217, 252, 263,
289, 336, 384, 388, 407, 467, 556, 821,
823
nonlinear, 218
steady state, 207, 266, 668
Heaviside function, 431
Helmholtz's equation, 278, 381, 483, 622, 686.
See also Reduced wave equation
modified, 278, 483, 686
Hooke's law, 166, 328
Hopf, E.,823
Huygen's principle, 277, 475
Hydrodynamics, see Fluid dynamics
Hyperbolic distance, 385, 803, 805
Hyperbolic equation, 111, 123, 142, 166, 168,
202, 211, 326, 331, 343, 347, 362, 385,
390, 414, 550, 587, 619, 632, 657, 679,
680, 779
Hyperbolic system:
canonical form, 130
characteristic form, 130
characteristic normal form, 135
Illuminated region, 749
Incident wave, 334, 706, 721
Incorrectly posed problem, 137, 138, 682
Index of refraction, 106, 701, 708, 754
Initial and boundary value problem, 137
888
INDEX
Initial layer, 650. See also Boundary layer
Initial strip, 85
Initial value problem, 8, 20, 49, 64, 136. See
also Cauchy problem
Inner product, 173, 179, 335
Hermitian, 180
Integral relation, 75, 325, 364
Galerkin form, 403
elliptic equation, 328
hyperbolic equation, 326
parabolic equation, 328
Integral surface, 50, 63
Integral theorem, 412, 499
elliptic equation, 412
hyperbolic equation, 415
parabolic equation, 418
Integral wave equation, 349
Interface, 108, 759
Internal boundary layer, 653, 656
Inversion, 493
Iteration method, 363, 553
Jacobian, 50, 56, 64
Jump, 56, 75, 340
Jump conditions, 98, 339, 365, 367. See also
Matching conditions
Kac, M, 19
Keller, J.B., 703, 744
Kirchhoff s formula, 275
Klein^Gordon equation:
linear, 145, 297, 477, 617, 772, 801
nonlinear, 628, 643
Korteweg-deVries equation:
linear, 322
nonlinear, 644, 830, 838, 839
Kronecker delta function, 538
Laplace's equation, 28, 48, 121, 199, 212, 257,
269, 282, 380, 485, 573, 579, 589, 623,
637
Laplace transform, 288
Abelian asymptotic theory, 293
convolution theorem, 319
of derivatives, 289
Heaviside expansion theorem, 294, 321
inversion theorem, 288
Tauberian asymptotic theory, 293
Laser beam, 751, 753. See also Optical beam
Legendre function, 541, 601, 602
Legendre polynomial, 192, 231, 237, 601
eigenfunction expansion, 193
Legendre's equation, 192
Lewy, H., 29
Linear functional, 426
Linearization, 219, 579
Lower order terms, 140
Matching conditions, 325, 329, 365. See also
Jump conditions
Matching process, 647, 651, 654, 666, 672,
728, 743
Matrix:
characteristic equation, 538, 546
diagonal, 122, 130
eigenvalues, 122, 123, 130, 135, 538
eigenvectors, 122, 130, 135, 538
orthogonal, 122
positive definite, 537
symmetric, 122, 537
Maximum principle, 13, 31, 556, 560, 561
strong, 558, 562
weak, 559, 561
Maxwell's equations, 588
in vacuum, 591
Mean square convergence, 181, 348, 534
Mean value theorem, 34, 496, 515
Method of canonical problems, 745
Method of characteristics, 49, 100, 101, 105,
108, 707, 780
Method of finite parts, 433, 502
Method of images, 484, 705
Method of matched asymptotic expansions, 648
Method of multiple scales, 617, 627, 636
Method of stationary phase, 295, 565, 728, 742
Minimum principle, 13, 31, 556, 560, 562
strong, 558, 562
weak, 559, 561
Minimum problem, 519, 524, 600
Dirichlet boundary conditions, 524
mixed boundary conditions, 524
natural boundary conditions, 524
Modified Bessel equation, 381
Modified Bessel function, 237, 279, 286, 301,
320, 381, 387, 389, 483, 555, 686, 797
Moment of random variable, 9
Monge cone, 81
Moving boundary, 387, 785
Moving source, 367
Natural boundary condition, 526
Navier's equation, 594
Navier-Stokes equations, 583, 817
Neumann data, 137, 165
Neumann function, 381
Newton's law of cooling, 165, 235
Newton's law of motion, 14, 166, 193, 366
Node, 196
INDEX
889
Norm, 173, 179
Normal distribution, see Gaussian distribution
Normal mode, 140, 194, 564
Ocean acoustics, 481
Odd function, 188
Optical beam, 755, 757, 759. See also Laser
beam
Optics, 703
nonlinear, 693, 753
Order symbol:
O, 618
o, 618
Ornstein, L.S., 14, 18
Orthogonal, 173, 179, 503
set, 176, 180
Orthonormal set, 180
Outer region, 650, 656, 667
Outgoing condition, 706, 721
Outgoing wave, 706
Parabolic boundary layer, 667, 673. See also
Boundary layer
equation, 667, 673
Parabolic equation, 113, 124, 141, 165, 168,
202, 328, 345, 390, 417, 587, 620, 625,
679, 680, 682, 780
Parabolic equation method, 681, 834
Paraxial wave equation, 753
Parseval's equation:
Fourier series, 181
Fourier transform, 247
Partial differential equation:
linear first order, 48
linear mth order, 132
linear second order, 109, 122, 140
nonlinear first order, 81
nonlinear higher order, 133
quasilinear first order, 63
quasilinear higher order, 134
Periodic:
boundary conditions, 186, 213
function, 188
Perturbation method, 103, 219, 548, 616, 625,
628, 632, 701, 818
Perturbation problem:
regular, 619, 621, 622, 625, 628, 632, 637,
642,644
singular, 24, 619, 647, 657, 662, 663, 686,
701, 732
Perturbation series, 103, 621, 648 653, 663,
669, 686
Phase, 295, 643, 702
Phase shift, 738
Phase velocity, 144, 298
Piston problem, 583, 612
Plane wave, 90, 91, 273, 701, 718, 746, 781
general, 718, 726, 752
Plucked string, 232
Point impulse, 232
Point source, 284, 370, 375
continuous, 365, 368, 371, 374, 383
instantaneous, 325, 383
Poisson process, 19, 22
Poisson's equation, 32, 212, 494, 560, 589, 623
Poisson's integral, 496, 515
Poisson summation formula, 291, 514
Positive operator, 174
Principal part, 110, 140, 810
Probability density, 6
Projection matrix, 811
Projection operation, 811, 819, 821
Propagating wave, 196, 373, 441. See also
Traveling wave
Radiation condition of Sommerfeld, 252, 381,
483, 497, 687, 703, 706, 721
Random walk:
absorbing boundary, 10, 28, 29, 41
correlated, 16, 19
independent steps, 2
position dependent, 11, 25, 33
reflecting boundary, 11, 42
time independent, 27, 37, 42
unrestricted, 2
Rapidly oscillating data, 798
Ray, 707, 708, 717, 722, 755, 774, 784
equation, 708
incoming, 736
outgoing, 736
tube, 714
Rayleigh quotient, 520, 524, 536
Dirichlet boundary conditions, 523, 527
mixed boundary conditions, 524
Neumann boundary conditions, 523
Sturm-Liouville problem, 535, 545
third kind boundary conditions, 524
Rayleigh-Ritz method, 536, 545
approximate eigenfunction, 539
approximate eigenvalue, 538
Reduced equation, 662, 672, 825
system, 811, 814, 818, 829
Reduced part, 810
Reduced problem, 619, 649, 662, 814
Reduced wave equation, 106, 382, 481, 496,
617, 700, 703, 721, 731, 738, 746, 798.
See also Helmholtz's equation
nonlinear, 754
890
INDEX
Reduced wave equation (Continued)
nonuniqueness of solutions, 236
Reflected wave, 334, 357
Reflection:
boundary, 740, 834
region, 740
Refraction shadow, 709
Renormalization, 627, 631
Resonance, 212, 239
Retarded potential, 275
Riccati equation, 224
Riemann function, 553, 555
Riemann invariants, 136, 158, 612
Riemann-Lebesgue lemma, 248, 297, 308
Riemann's method, 362, 549
Rodrigues' formula, 231
Scattered wave, 706, 722, 738, 746
Scattering problem, 721, 722, 738, 746
Schrodinger's equation, 124, 510, 728, 741
nonlinear, 770
Schwartz, L., 426
Secular behavior, 627, 662, 727, 761
Secular terms, 627, 631, 633, 646, 658, 819,
830
Self-adjoint operator, 147, 149, 172, 424
formal, 147
Self-focusing, 755
Self-trapping, 757, 769
Separation of variables, 168, 193, 244, 566, 569
Shadow:
boundary, 740, 751
region, 740, 749
Shallow water, 158, 829
Shock conditions, 76
Rankine-Hugoniot, 611
Shock speed, 828
Shock structure, 826
Shock velocity, 76
Shock wave, 67, 76, 583, 826
N wave, 79
stationary wave, 104, 105
triangular wave, 78
Signaling problem, 97, 355, 619, 684
Similarity solution, 407
Simple wave, 583, 823
Single layer distribution, 364, 502
Singularities, propagation of, 778, 798
Singular point, 58
Singular solution, 90, 372, 379
Sirovich, L.,295, 303
Sirovich's method, 304
Snell's laws, 108, 399
Solitary wave, 830
Sommerfeld, A., 745
Sound wave, 580, 582
Space-like curve, 360, 369
Spectrum, see Eigenvalues
Speed of wave propagation, 26, 277, 298, 358,
370, 580, 582, 644
Spherical harmonics, 541, 601
Spherical wave, 314, 373, 382, 701, 718, 782
Square integrable function, 179, 183
Stability index, 142, 564
Stability theory:
boundary value problem, 219, 221, 544
Cauchy problem, 143, 160, 220, 222, 653,
658
linear, 143, 219, 221, 658, 681
nonlinear, 218, 222
Standing wave, 196, 292
Stationary inhomogeneities, 215
Stationary point, 295
Stefan problem, 406
Stochastic calculus:
Ito, 26
Stratonovich, 26
Stochastic differential equation, 15, 26
Stokes, G.G., 644
Stokes' rule, 258, 272, 274
Stratified medium, 708
Stretching transformation, 650, 654, 659, 664,
670,675, 686, 727, 731,741
Strip condition, 84
Sturm-Liouville problem, 172, 178, 535, 545,
546
regular, 178,
singular, 178, 190, 192
Substitution property, 430
Superposition of solutions, 176, 245
Symmetry of Green's function, 448, 451, 459,
465, 497
Systems of equations:
linear first order, 20, 127, 579, 810
nonlinear second order, 583, 817
nonlinear third order, 828
quasilinear first order, 134, 585
Taylor, GX, 15
Telegrapher's equation, 20, 25, 145, 225, 301,
388, 562, 589, 785, 792, 797. See also
Wave equation, damped
modified, 477, 479
Test function, 426
of rapid decay, 438
Threshold of instability, 222
Time-like curve, 360, 368, 785
Totally hyperbolic, 128
INDEX
891
Transmitted wave, 334
Transport equation, 703, 704, 714, 754, 783,
785, 798
Traveling wave, 196, 381, 643, 645, 785, 820,
821, 826, 830. See also Propagating wave
periodic, 644, 832
Tricomi's equation, 151
Trigonometric eigenfimctions, 184
Type of equation:
elliptic, 111, 114, 123
hyperbolic, 110, 111, 123
mixed, 114, 123
parabolic, 111, 113, 123
ultrahyperbolic, 123
Type of system:
elliptic, 129, 135
hyperbolic, 130, 135
parabolic, 130, 135
totally hyperbolic, 128
Uhlenbeck, G.E., 14, 18
Uniform convergence, 183, 195, 229, 348
Uniqueness of solutions, 64, 393, 395, 396,
410, 559, 561
Variance:
continuous case, 9, 21
discrete case, 3, 16
Variational principle, 517
Variational problem, 524, 529. See also
Minimum problem
Vibrating beam, see Vibrating rod
Vibrating membrane, 166, 237
steady state, 200
Vibrating plate, 569
circular, 570
clamped, 569
simply supported, 609
Vibrating rod:
clamped, 566
composite, 332
free, 608
laterally, 563
longitudinally, 328
simply supported, 608
Vibrating string, 54, 166, 193
loaded, 366
nonlinear, 839
Water waves, 828
Watson's lemma, 293
Wave breaking, 67, 828
Wave equation, 23, 47, 120, 130, 193, 203,
255, 272, 284, 291, 341, 345, 349, 353,
355,.361, 368, 372, 384, 440, 470, 555,
580, 589, 632, 705, 799, 829. See also
Telegrapher's equation
damped, 21, 47, 145, 498, 808
nonlinear, 87
unidirectional:
linear, 23, 24, 51
nonlinear, 65, 102, 581, 645, 825
Wave form, 52, 66
Wave front, 277, 373, 377, 707, 714, 718, 779,
784
Waveguide, 614
Wave motion, unidirectional:
linear, 23, 51, 53
nonlinear, 65, 69, 583
Wave number, 106, 644
Wave optics, 703
Wave packet, 298
Weak convergence, 348, 427, 439
Weak solution, 342, 347, 376. See also
Generalized solution
Well posed problem, 137, 141, 142, 160
Whitham, G.B., 644
WKB method, 833
Young's modulus, 166, 328, 332
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