Methods Based on the Wiener-Hopf Technique for the Solution of Partial Differential Equations

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OTHER TITLES IN THE SERIES ON PURE AND APPLIED MATHEMATICS

Vol. 1. Introduction to Algebraic Topology by A. H. WALLACE

Vol. 2. Circles

by D. PEDOE

Vol. 3. Analytical Conics

by B. SPAIN

Vol. 4. Integral Equations

by S. MIKHLIN

Vol. 5. Problems in Euclidean Space: Application of Convexity

by H. G. EGGLESTON

Vol. 6. Homology Theory on Algebraic Varieties by A. H. WALLACE

METHODS BASED ON THE

WIENER-HOPF TECHNIQUE for the solution of partial differential equations by B. NOBLE

Senior Lecturer in Mathematics The Royal College of Science and Technology Glasgow

PERGAMON

PRESS

LONDON · NEW YORK , PARIS , LOS ANGELES 1958

PERGAMON PRESS LTD. 4 & 5 Fitzroy Square, London W.l PERGAMON PRESS, INC. 122 East 55th Street, New York 22, N.Y. P.O. Box 47715, Los Angeles, Calif. PERGAMON PRESS S .A.R.L. 24 Rue des Ecoles, Paris Ve

Copyright

©

1958

B. NOBLE

Library of Congress Card No. 58-12676

Printed In Northern Ireland at The Universities Press. Belfast

CONTENTS Preface Home basic notation and results from Chapter I

PAGE

vii x

I. COMPLEX VARIABLE AND FOURIER TRANSFORMS Introduction 1 1.2 Complex variable theory 5 J.3 Analytic functions defined by integrals 11 21 I .4 The Fourier integral 27 1.1) The wave equation 31 l.a Contour integrals of a certain type 36 1.7 The Wiener-Hopf procedure Miscellaneous examples and results I 38 1.1

2.1

2. 2 2.3 2.4

2.1i 2. 6 2.7 2.H 2.9

:1.1

:1.2 :I.:J

:1.4 :1.[; :I.a

II. BASIC PROCEDURES : HALF-PLANE PROBLEMS Introduction Jones's method A dual integral equation method Integral equation formulations Solution of the integral equations Discussion of the solution Comparison of methods Boundary conditions specified by general functions Radiation-type boundary conditions Miscellaneous examples and results II III. FURTHER WAVE PROBLEMS Introduction A plane wave incident on two semi-infinite parallel planes :Eadiation from two parallel semi-infinite plates l�adiation from a cylindrical pipe Semi-infinite strips parallel to the walls of a duct A strip across a duct Miscellaneous examples and results III

48 52 58 61 67 72 76 77 83 86

98 1 00 1 05 110 1 18 122 125

lV. EXTENSIONS AND LIMITATIONS OF THE METHOD ·1.1

4·.2

4·,3 4.4 4.5 4. (}

Introduction The Hilbert problem General considerations Simultaneous Wiener-Hopf equations Approximate factorization Laplace's equation in polar co-ordinates Miscellaneous examples and results IV v

141 141 147 1 53 160 1 64

167

VI

5.1 5.2

5.3 5.4

5.5 5.6

UONTEN'rs V. SOMJ<: APPHOXIMATE ME'l'HODFl

Introduction Some problems which cannot be solved exactly General theory of a special equation Diffraction by a thick semi-infinite strip General theory of another special equation Diffraction by strips and slits o f finite width Miscellaneous examples and results V

178 180 184 187 196 203 207

VI. THE GENERAL SOLUTION OF THE BASIC

WIENER-HOPF PROBLEM

6.1 6.2

Introduction The exact solution of certain dual integral equations Miscellaneous examples and results VI

Bibliography Index

220 222 228 237 243

PREFA CE The methods described in this book solve certain boundary-value problems of practical importance involving partial differential equations. A typical problem requires solution of the steady-state wave equation in free space when semi-infinite boundaries are present. Examples are given from electromagnetic theory, acoustics, hydrodynamics, elasticity and potential theory. The twin aims of this book are: to take the student from ordinary dogree studies into the research field covered by the Wiener-Hopf technique, and to provide the research worker with a reasonably comprehensive summary of what can and what cannot be done at the moment by the technique. The reader's attention is drawn particularly to the various methods for approximate solution of problems. One of the remarkable features is the range of apparently unrelated topics covered by ramifications of the technique. It is hoped that some of the comments in the text and in examples may suggest suitable lines for further research. The Wiener-Hopf technique was invented about 1931 to solve an integral equation of a special type. During the war it was noted by J. Schwinger (and independently by E. T. Copson) that problems involving diffraction by semi-infinite planes could be formulated in terms of integral equations which could be solved by the Wiener Hopf technique. The solution depends on the use of Fourier integrals to obtain a complex variable equation which is solved by analytic continuation. Most of this book is based on a different but equivalent approach due to D. S. Jones. Fourier transforms are applied. directly to the partial differential equation and the complex variable equation is obtained directly without the necessity for formulation of an integral equation. From this point of view the Wiener-Hopf technique provides a significant and natural extension of the range of problems that can be solved by the use of Fourier, Laplace and Mellin transforms. I started this book with the intention of running the integral equation and Jones's method alongside each other, but as the writing progressed it seemed pointless and confusing to elaborate two equivalent methods. Jones's method seems simpler to me and I have included only sufficient details of the integral equation method to enable the reader to follow the literature. The material in this book should be accessible to anyone who is familiar with the Laplace transform, its complex inversion formula, and integration in the complex plane. The first chapter is intended to supplement the usual undergraduate course in complex variable vii

Vlll

PREFACE

theory, and to familiarize the reader with the use of the Fourier transform in the complex plane. As this book has been written for workers whose interests are primarily in applications of the theory rather than in the theory itself, the standard of rigour may not satisfy the pure mathematician though it should suffico for practical purposes. It is important to emphasize that from the point of view adopted in this book the essence of the Wiener-Hopf technique is that it can be used to obtain numerical values for physical quantities. For various reasons I have decided to omit numerical tables although as far as possible results are given in a form suitable for numerical computation and references are given to the few existing sets of tabulated values. Practically no discussion is given of the physical implications of results. For electromagnetic theory this gap should be filled by the volume in this series by D. S. Jones. In the examples treated in the text I have carried the analysis far enough in each case to obtain at least one result of physical signifi cance in simple form. This is partly to encourage the beginner who might think that complicated formulae can be interpreted only with the aid of an electronic computer. The stimulus to write this book came originally from a course of post-graduate lectures suggested by Prof. D. C. Pack. Among other things I am grateful to Prof. Pack for organizing ideal working conditions in his department at the Royal College of Science and Technology, Glasgow. I am also grateful to Prof. I. N. Sneddon for asking me to write this volume for his series, and for helpfllI sugges tions in connexion with shortening a manuscript that was much longer than originally envisaged. It is perhaps worth mentioning that I have had to omit a chapter dealing with applications of Bessel function dual integral equations to disk problems. A referee for the Proc. Camb. Phil. Soc. (whose name cannot now be traced) suggested to me that these problems are best tackled by the Wiener Hopf method. Having written this book I am not entirely convinced that the suggestion is correct, but in a sense the referee ' s comment provoked the book! I am indebted to Prof. D. S. Jones for various references and correspondence and to Dr. W. E. Williams who carefully checked Chapter V. My thanks are due to the printers for accurate work on a difficult manuscript. B. N. 31. 12.57.

SOME

BASIC NOTATION AND RESULTS FROM CHAPTER 1

The following may prove useful for reference. A time factor exp (-iwt) is used throughout this book.

oc =

(f

Y

(oc2 - k2 ) 1/2

(k - OC ) 1 / 2

Y

=

= =

+ iT

:

k =

i(oc - k) 1/ 2

-ik if oc

=

=

>

0, k2

>

-i(P - OC2 ) 1/2 .

:

°

(k 1

k l + ik2'

( - k - oc) 1/ 2 :

y!":::i

=

0). [(1.14)]

-i(oc + k) 1/ 2 .

[(1.12)]

loci if oc is real and large.

- 00

where

- 00

We write, for instance,

where any of these forms may be used according to convenience provided that there is no risk of confusion. If Ic/>(x )\ If \ c/> (x) \

<

<

A exp (T_X) as X� +<Xl, B exp (T+X) as X� -<Xl,

+(oc) is regular in T > _(oc) is regular in T <

�

}

T_ T+ [ §1.3]

If c/> (x) "" x'1 as X� +0, then + (oc) "" oc - '1- 1 as oc�<Xl n T > T_ If c/> (x) "" x'1 as X� -0, then _(oc) "" oc -'1 - 1 as oc� <Xl III T < T+

}

[Cf. (1. 74)] (j(x) ....., g(x) as x --+ a means that j(x) = g(x) + h(x) where hjg --+ 0 as x --+ a. The number a may be infinite. Occasionally as in the last paragraph we use j ....., g to mean j = Cg + h for some constant C, if the value of C is not important. ) x

CH APTER I

COMPLEX VARIABLE AND FOURIER TRANSFORMS

1.1

Introduction

One of the remarkable features of the mathematical description of natural phenomena by means of partial differential equations is the comparative ease with which solutions can be obtained for certain geometrical shapes, such as circles and infinite strips, by the method of separation of variables: in contrast, considerable difficulty is usually encountered in finding solutions for shapes not covered by the method of separation of variables. The Wiener-Hopf technique provides a significant extension of the range of problem that can be solved by Fourier, Laplace and Mellin integrals. To illustrate these remarks and to remind the reader of the real variable Fourier integral consider three problems connected with the steady-state wave equation (1.1)

Suppose we wish to find a solution of this equation in the semi infinite region -00 < x < 00, y � 0, such that rfo represents an outgoing wave at infinity in each of three separate cases (i)

rfo = f(x)

on y=O,

- oo < x < oo ;

(ii)

orfoloy = g(x)

on y=O,

-oo<x < oo ;

(iii)

rfo = f(x) orfoloy = g(x)

on y = O, on y=O,

° < x < oo , - 00 < x < 0.

}

(1.2)

Separation-of-variables solutions exist for (1.1) in the form rfo=

X(x) Y(y) with· X(x )

=

e±i=

Y(y) = e ±Y1I,

y=

( ex2

_

k2 ) 1 / 2 ,

where ex is a parameter. Together with the fact that the range of x is infinite this suggests use of the Fourier integral in - 00 < x < 00, and in fact we show that the first two problems can be solved exactly by Fourier integrals: the third leads to equations which can be solved by the Wiener-Hopf technique. 1

2

T H E W I E N E R - H O P F T E C HN IC� UE

Although it will appear that we must use the Fourior integral in the complex plane, consider in this section the oniinn,ry form 00

' f (oc, y) = (27T)1/2 cp(x,y)e'rxa; dx ' cp(x,y) = )1/2 f <1>( oc,y)e -,rxa; doc , 1

- 00

00

1 (27T

(1.3)

- 00

where oc is real. We use the method of Fourier transforms (1. N. Sneddon [1] , C. J. Tranter [1] , or E. C. Titchmarsh [1] Chapter X, "formal solutions" ) . Multiply both sides of (1.1) by (27T)-1 / 2 exp (iocx) and integrate throughout with respect to x from 00 to 00:

-

(1.4) - co

Integrate the following expression by parts :

OCP e''C<JX A - �oc. [..I..'l'e!'C<xJA - oc2 f A..I..'l'e'.C<x dx. fA02CP -ax2 e',rxa; dx = [aX -A -A -A -A

Let A --+ 00 and assume that contributions from the bracketed terms at upper and lower limits tend to zero. (This is connected with the condition that cp represents an outgoing wave at infinity and is investigated further in §1.5.) Equation (1.4) then becomes

d dy2

2<1> - 1'2<1> = 0,

1'2 = oc2 - k2). Define oc k , where in this section we assume that k is real. A difficulty arises since we need to define for oc k and it is not clear, for example,= whether to take the upper or lower sign in the formula ±i(k2 - OC2)1/2, l oc' l k. This question is examined in by using analytic continuation arguments developed in §1 .2. Assuming I'

where

(1.5)

(

>

<

I'

§1.5,

<

that the definition of I' has been settled, the solution of (1.5) which must be used is (1.6)

since it will appear that I' = + (oc for oc < so that this solution is bounded for all oc as y tends to plus infinity whereas the solution in exp (+yy) increases exponentially as y tends to

2 - k2)1/2

-k

O O M P L E X VA R I A B L E A N D

3

FO URIER TRA N S FORMS

A(oc)

infinity for l OCi > The function is an arbitrary function dotermined from the boundary condition on 0. Now consider problems (i)-liii) in turn.

k.

y=

(i) Application of the boundary condition on

( <
= A(oc) = (2:)1/2 A(oc)

co

y

= ° to (1.6) gives

f f(�)eiCt. � d�.

- co

Substitute this value for in (1.6) and use the Fourier inversion formula (1.3). This gives the solution

cp

= 2� f

co

co

f

e - iCt. x -yy f(�)e iCt. � d� doc.

- 00

- 00

(ii) In an exactly similar way the second problem gives co

(�:) Y�O = -yA(oc) = (2:)1/2 f g(�)eiCt. o d �, cp =

-

2

- co

co

1

co

(1.7 ) - co

- co

(iii) In this case there are three methods of procedure which are basically identical but deserve separate mention. A. In (1.2) extend

f (x) to denote the (unknown) value of

cp on y = 0, x < 0, and g (x) to acpjay on y = 0, x > 0. Define

denote the (unknown) value

«P+(oc,y) = (27T)1 l/2 «P+(oc,O), «P_(oc,O)

of

co

f cp(x,y)e'Ct.X. dx,

o

- co

cp(x,O) =y,

Then are the corresponding integrals of !(x) . Use a dash to denote differentiation with respect to so that

«P'+(oc,O) = {a«p+(oc,y)jay}y�O = (2�1/2

co

f g(x)eiCt.X dx, o

4

T H E W IE N E R -H O P F T E C H N IQUE

with a corresponding definition for <1>'_( oc,O). conditions now yield

Tho boundary

Eliminate A (oc) from these equations :

The functions <1> '_, <1>+ are known but there are two unknown functions, <1>' + and <1>_. It will appear that if oc is taken as a complex variable in the original Fourier transform (1.3), a process involving analytic continuation and Liouville ' s theorem can be used to determine the unknown functions in (1.8 ). This process, which is described in §1.7, is the "Wiener-Hopf technique" . B . Next consider an integral equation formulation of the problem. In (1.7 ) interchange orders of integration, let y tend to zero, introduce boundary condition 11.2) for x> 0, split the range of integration in � into ( - 00 ,0), (0,00 ) and rearrange: 00

fK(x - �)g(�) d� = f(x) f g(�)K(x - �) d�, 0

-

- 00

o

where

K(x - �)

=

-

(x > 0),

00

;7T fy-1eic«; - x) doc, - 00

and the quantities on the right· hand side of the equation are known. This is an integral equation for the unknown function g(�), � > 0. The important feature from the present point of view is that the kernel K(x �) is a function of (x �). Such integral equations can be solved by the Wiener-Hopf technique. In the literature the usual procedure is to obtain this type of equation by a Green' s function technique, and then to reduce the integral equation to (1.8 ) by Fourier transforms. A more detailed discussion is given in §§2.4, 2.5. -

-

O. Finally consider the formulation in terms of dual integral equations. From (1.6),

cp = (2:)1/2

00

f A(oc)e -ic<x -rv doc.

- 00

C O M P L E X VA R I A BL E A N D

FO URIER TRANSFORMS

5

The boundary conditions give 00

1 f A(oc)e -"'"". doc

(27T) 1/ 2

- 00

=

f(x) ,

(x > O),

( 1 .9a)

=

g(x),

(x
( 1 .9b)

00

-

: f yA (oc)e - ic<x doc

(2 ) 1 / 2

- 00

These are dual integral equations for the unknown function A(oc). It will be shown in §2.2 that these equations can be solved directly by a procedure depending on the essential step in the Wiener-Hopf technique. This completes our introductory discussion. In order to solve problem (iii) above by any of the methods A, B, C, it is necessary to consider complex oc. This requires a discussion of certain topics in complex variable and Fourier transform theory which are examined in § §1 .2-1 .5. Various questions which have been left unanswered in the above treatment are solved in §1 .5 with the help of § §1 .2, 1 .3, 1 .4. The equation which replaces (1 .8) when complex oc is considered is stated in §1 .7, where a summary of the Wiener-Hopf method for solution of the equation is given. The reader who is prepared to refer back to this chapter can proceed directly to Chapter II where a problem equivalent to (iii) above is solved in detail. 1.2

Complex variable theory

We start with a brief summary of complex variable theory required for succeeding developments. Our references will be the standard texts of E. T. Copson [1], and E. C. Titchmarsh [2] . Greek letters will be used to denote complex variables, e.g. , = � + i'l}, oc = Cf + iT. When the complex variable is associated with the Fourier transform we invariably use oc = Cf + iT. Latin letters a, b, k, etc., will be used for constants. It will be clear from the context whether these are to be regarded as real or complex. We recall the following definitions and results. If to each point , in a certain region R there correspond one or more complex numbers, denoted by X, then we write X = f(' ) and say that X is a function of the complex variable , . If the function has a uniquely defined value at each point of the region R it is said to be single-valued in R. The crucial property possessed by useful

6

T H E W I E N E R - H O P F T E CHN I Q Ul�

functions is that at most points of R they are differentiable, i.e. = lim Im '

d�O

f(' + (J ) - fm (J

exists and is independent of the direction along which the complex number (J tends to zero. The function X = f(' ) is said to be analytic at the point ' when it is single-valued and differentiable at this point. The functionf is said to be regular in a region R if it is analytic at every point of R. The phrase "fm is an analytic function in a region R" means that the function is analytic at every point of a region except for a certain number of exceptional points : this will be defined more precisely later in connexion with analytic continuation. Points at which the function is not analytic are called singularities. The singularities of a function are very important since they characterize the function. The next idea required is that of a line integral. The central result concerning line integrals is Oauchy ' s theorem of which we quote the following form (E. T. Copson [1], p. 61 ) : If fm is an analytic function, continuous within and on the simple closed rectifiable curve 0, and ifI'' ( ) exists at each point within 0, then

J fm d ,

a

=

o.

From this can be deduced Oauchy ' s integral formula: If fm obeys the same conditions as for Cauchy ' S theorem and if at is any point within 0, then 1 fm

f(at ) = 2 . 7r�

J-y

a

",-at

d,.

Some familiarity is assumed with the application of these theorems to evaluation of contour integrals by residues and shifting of contours in the complex plane, particularly when branch points are present. An analytic function which is regular in every finite region of the '-plane is called an integral function, e.g. a polynomial in , is an integral function:· also exp , is an integral function. Liouville ' s theorem states that iffm is an integral function such that Ifm l � M for all " M being a constant, then f(' ) is a constant. It is easy to extend this result to the following : If f' ( ) is an integral function such that Ifm l � M I W as 1'1-- 00 where M,p are constants, then fm is a polynomial of degree less than or equal to [P] where [p] is the integral part of p. Tayl or's the orem states that iff(' ) is an analytic function regular in the neighbourhood I' - al < R of the point , = a, it can be

C O M P L E X VA R I A B L E A N D

F O U R I E R T R A N SF O R MS

7

expressed in this neighbourhood as a convergent power series of the form

� (' _ a) r

fW = f( a ) + L ar

r�1

r!

'

A zero of an analytic function fW is a value of , such thatfW = o. It can be deduced from Taylor ' s theorem that the zeros of an analytic function are isolated points, i.e. iff(') is regular in a region including , = a then there is a region I' - al < p, (p > 0), inside which f ( ,) has no zeros except possibly' = IX itself. If a singularity is isolated it is possible to specify an annulus in which the function is analytic and to deduce a Laurent expansion. If this expansion is of the form 00

fW= L aA' - a)r, T= -n

(n > 0),

then the function is said to have a pole of order n at the point a. The remainder of this section is devoted to analytic continuation, which is important for later developments. Since we shall need analytic continuation mainly in connexion with the Fourier integral we use the complex variable IX = (] + iT. It often happens that a representation of a function of a complex variable is valid only for restricted IX. Thus the series f(lX) = 1 + IX + 1X2 + . . . converges only for I IX I

<

1 . However for I IX I < 1 we have f(lX) = (1

-

IX) -I.

The extension of the definition of f(lX ) by identifying f(lX ) with ( 1 - IX) -I for I IXI > 1 is called analytic continuation. It is possible to carry out analytic continuation systematically by means of power series but we do not go into details. We assume merely that the functions f(lX) with which we deal are defined in such a way that if we start at any point IX = a in the complex plane and draw a continuous curve to another point say IX = b in such a way that no singularities of the function lie on the curve, then the values of f(lX ) vary continuously along the curve and can be determined from the definition of f(IX ) . The expression "the analytic function f(IX)" can now be defined as the totality of all values of f(lX ) which can be obtained by analytic continuation as just described, starting at a given point IX = a. The natural question which arises is whether a function which is continued along two different curves from IX = a to IX = b will have

8

T H E W I E N E R- H O P F T E C H N IQUE

the same final value for the two ways. This question i s partly answered by the following theorem (E. C. TitchmafHh [2], p. 145): If we continue an analytic function f( oc) along two differ'ont routes from a to b and obtain two different values of f (b) then f( oc) must have a singularity between the two routes. Of courso the converse is not true, that if there is a singularity between the two routes we necessarily obtain two different values off ( b) : it needs to be a special type of singularity, namely a branch-point to produce a difference in value. If the values of a function found by analytic continuation are unique, independent of the paths of continuation, then the function is called single-valued. Otherwise the function is called many-valued. A branch point a is a singular point such that there exists no neigh bourhood l oc - al < e in which f(oc) is single-valued. By inserting certain lines in the complex plane and stating that paths of analytic continuation must not cross these lines it is possible to specify a branch of a many-valued function which in itself is single-valued. Such lines are called branch-lines or cuts. Branch points always occur in pairs, and branch lines join branch points. A simple example of a many valued function is X = oc1 /2• One branch point is oc = O. The point at infinity is also a branch point as can be seen by setting oc = ,-1 . A branch line in this case is any line joining these two points. For instance draw the branch line along the negative real axis, -00 < a ::::;;; O. In order to specify the branches consider the behaviour of the function at only one point say oc = p, where p is a positive real quantity. Adopt the convention that p1/ 2 refers to the positive square root of p. Then the two branches of X OC1/2 correspond to analytic continuation from the two values X ±pl/2 at oc = p. If we choose the branch 2 specified by X = p1 / then on the upper side of the negative real axis, oc = r exp (in ), X ir1 /2 , and on the lower side, oc = r exp (-in ), X = -irl / 2 . Suppose next that we cut the oc- plane by a straight line from the origin to infinity in the upper half-plane and define two functions X = OC1 / 2 , and 1fJ = ( _ OC) 1 / 2 , where branches are chosen so that if p is a positive real number, X +p1/ 2 when oc p and 1fJ = +p1 / 2 when oc = -p o Then by analytic continuation X = _ ip1 / 2 when oc = -p and 1fJ = + ip1/2 when oc = +p. Thus on any segment of the real axis, =

=

=

=

=

(1.10) There is an important theorem which states that if two functions are equal along any line of finite length in the complex plane they are equal in any region into which they can both be continued along a

C O M P L E X VA R I A B L E A N D

F O U R I E R T R A N SF O R MS

9

common line, starting from any point at which they have the same value. Thus ( 1 . 10) holds at any point in the cut plane, with the branches as previously defined. We must remember that the second equation in (1 . 10) is not an ordinary algebraic equation, e.g. if we change the sign of at on both sides we obtain ( _ at) 1 / 2 = -i(at) 1 / 2 which seems to contradict the original equation in (1 .10). The /Cut

-�7-k

/{/

+k":"'�L�

--------70�--a

,,'/

'Cut

FIG. 1.1.

reason for the apparent contradiction is of course that if we change the sign of at we also change the position of the cut. Now define X2 = (at + k) 1/ 2 X l = (at - k) 1 / 2 , X4 = ( - k - at) 1/ 2 Xa = (k - at) 1/ 2 ,

}

where the at-plane is cut as in Fig. 1 . 1 . Define the branches ofXl andX2 so that both functions tend to + a1 / 2 as at----+ +00 along the positive real axis, and define the branches of Xa , X4 so that both functions tend to ( _ a) 1 / 2 as at ----+ -00 along the negative real axis. Then we readily prove by the method of the last two paragraphs that Xl = -ikl/ 2 , at = O X 1 ----+ -i(-a) 1 / 2 , at = a----+ -00; l 2 / X2 = k , at = O X2 ----+ + i( _ a) 1 / 2 , at = a----+ -00; at = a----+ +00; Xa ----+ +ia1 / 2 , Xa = kl/ 2 , at = O l 2 1/2 / X4 ----+ -ia , X4 = -ik , at= O at = a----+ +00. We see that with the specified branches, Xa = iXl> X4 = -iX2 ' everywhere in the cut plane, i.e. ( -k - at) 1 / 2 = -i(at + k)1/ 2 . ( 1 . 12) (k - at) 1 / 2 = i(at - k) 1 / 2 We noted that it was not permissible to change the sign of at on both sides of the equation ( 1 .10). In defining the Xi care has been taken to choose the branches so that it is permissible to change the Hign of at in any equation involving theXi. Thus in (1 .12), changing

10

T H E W I E N E R- H O P F T E C H N I Q U E

the sign of ot merely changes one equation into the other. On the other hand we cannot change the sign of k in these equations, and if we let k -+ 0 in ( 1 .12) we obtain the apparently contradictory ( ot) I/ 2 = - i(otJ1/ 2 . results : _ ( ot)I/ 2 = i(ot)I/ 2 , _ Later we shall meet expressions like (k cos 0 - k)I/ 2 where o < 0 < 7T. This could be obtained by writing ot = k cos 0 in Xl or ot = -k cos 0 in X4. With the branches specified, the reader will readily prove that these are identical, and that in fact (k cos 0 - k)I/ 2 = -i(k - k cos 0)1 / 2 = -i(2k)I/ 2 sin t o . ( 1 . 13a) Similarly ( - k cos 0 - k)I/ 2 = -i(k + k cos 0) 1 / 2 = -i(2k) I / 2 cost0. ( 1 .13b) The most important example of all from our point of view is the function y ( ot y-+ a if ot a-+ + 00 . We can write, in notation 1 ( .11), y = (ot - k)I/ 2 (ot + k)I/ 2 =X X2 =XaX4 and the results obtained above I give y = -ik when ot = 0 and y-+ l a l when ot = a -+ - 00 . Similarly we can consider K= (k2 - o(2 ) 1/ 2 where we choose the branch such that K = k if ot = o. Then we can write K =X2Xa = -XIX4 and K--+- +ia if ot = a-+ + oo ; K-+ i l a l if ot= a-+ - oo . Equations ( 1 .12) now give the important results ( 1 . 14) y = -iK K= iy, =

=

everywhere in the plane cut as in Fig 1 . 1 , the branches being defined as above. As a special case suppose that k is real. Then on the real axis in the ot-plane y= + ( a2 k2 ) 1/ 2 K = +i(a2 - P)I/ 2 , (a > k) ; (2 ) 1 / 2 = -i(k2 ( -k< a< k) ; ( 1 . 15) = + (P (2 ) 1/ 2 , 1 2 2 2 2 2 P / / k) = + (a = +i(a - )I , (a < -k). _

_

_

_

If the angle A in Fig. 1 . 1 tends to zero, the branch lines lie on the real axis. Results ( 1 . 15) still hold providing that a is taken on the upper side of the branch cut a < -k, and on the lower side of the branch cut a> +k. When reading literature on the Wiener-Hopf technique it is essential to start off by checking the conventions used. Some authors take k = k l - ik 2 (k l>k 2 > 0), which means that branch ' cuts have to be chosen in a different way. Others use Laplace transforms instead of Fourier transforms which means that ot is replaced by (is) and instead of y defined above, it is necessary to consider the function S ( 2 + k2 ) 1 / 2 . Some examples to clarify the situation are given at the end of the chapter.

C O M P L E X VA R I A B L E A N D

1.3

11

FOURIER TRANS FORMS

Analytic functions defined by inte grals

We shall often meet functions defined by integrals of the type

G(IX)

=

J g(IX,' ) d"

a

( 1 . 1 6)

where g(IX,' ) is a function of the complex variables IX and " and 0 is a contour in the complex'-plane. The variable IX will be assumed to lie inside a region R, i.e. the boundary of R, if any, is excluded. The contour 0 is assumed to be smooth, i.e. it is possible to specify position on the contour by means of a parameter t such that , = �(t) + i'f}(t), to � t � tv and �'(t), 'f}'(t) exist and are continuous. Before stating conditions for G(IX) to be regular we make some general remarks. Any line integral like ( 1 . 16) can be reduced to real integrals and we shall assume that these are Riemann integrals. It would be convenient in some ways to assume knowledge of Lebesgue integration. Theorems would then be expressed more concisely and elegantly. On the other hand the Riemann integral requires conditions which correspond more directly to the pro perties we intuitively associate with the functions occurring in physics. Also most books on complex variable and Laplace trans form are written in terms of Riemann integrals. Where conditions of validity are given below, they are usually sufficient but not necessary. We now state conditions under which G( IX ) is regular (cf. E. C. Titchmarsh [2], pp. 99· , 100) : THEOREM A. Let g( IX,' ) = f (')h( IX ,') satisfy the conditions (i) h (IX ,' ) is a continuous function of the complex variables IX

and , where IX lies inside a region R and , lies on a contour o. (ii) h(IX,' ) is a regular function of IX in R for every ' on o. (iii ) f( ') has only a finite number of finite discontinuities on 0 and a finite number of maxima and minima on any finite part of o. (iv) f m is bounded except at a finite number of points. If '0 is such a point, so thai g( IX, ' ) ---+ 00 as ,---+'0' then

exists where the notation (0 - c5) denotes the contour 0 apart from a small length c5 surrounding '0' and lim (c5 ---+ 0) denotes th e limit as this excluded length tends to zero. The limit must be approach ed uniformly when IX lies in any closed domain R' within R. (v) If 0 goes to infinity then any bounded part of 0 must be smooth and conditions (i) a,,!-d (ii) must be satisfi ed for any bounded part

12

T H E W I E N E R-H O P F T E C H N I Q U E

of o. The infinite integral defining G(at) must be uniformly con vergent when at lies in any closed domain R' within R. Then G(at) defined by ( 1 . 16) is a regular function of at in R.

As a special case' may be real, say' = $, and the contour may consist of the portion a � $ � b of the real axis. Then ( 1 . 1 6 ) is an ordinary real integral :

G(at)

b

=

f g(at,$) d$.

a

We shall generally prove uniform convergence by using the com parison test in the following form. Suppose that g( at,') satisfies the conditions of the above theorem, and I g(at") 1 � M(t) for any at in R where position on the contour 0 is specified by a parameter t,

,

=

b

f M(tM' (t) + i'f)'(t) 1 dt converges,

W) + i'f)(t), a � t � b, and

a

then

f g(at,' ) d , is uniformly and absolutely convergent in R' .

a

We can now deduce some important results. In the following 0'+,

0'_, T+, T_ are real constants.

( 1 ) If

+(at)

F

co

f f(x)eirxx d

=

x,

o

0' + i T , f(x) satisfies conditions (iii), (iv) above, and A exp ( T_X) as x ---+ + 00 , then F(at) is regular in the upper half-plane T > T_. Similarly if f ( x ) satisfies (iii) , (iv) and If ( x) 1 < B exp ( T+X ) as x ---+ -00 then

where at

If(x) 1

=

<

F_(at)

=

o

f f(x)eirxx d

x

- co

is regular in the lower half-plane T < T+. These statements follow immediately from the theorem since exp (iOtX) obviously satisfies (i), (ii) and the restrictions as x ---+ ±oo ensure uniform con vergence. (2) If 1

Fp( S ) = f f(p)ps-l dp , o

co

FN (s) = f f(p)ps-l dp , 1

s

=

0'

+ iT ,

where f ( p ) satisfies conditions (iii ) , (iv) and If ( p ) 1 < A p-a_ as p ---+ O , If( p ) 1 < Bp-a+ as p---+ + 00 , then Fp(s) is regular in a

C O M P L E X VA RI A BL E A N D

F O U R I E R T R A N S F O R MS

13

right-hand half-plane a> a_, and FN (s) is regular in a left-hand half-plane a < a+ . (3) If

f

00

f

oo + ic

fm f ( � + ic) d = F(oc) d�, , oc , (� a) + i(c T) - oo + ic - 00 where oc = a + iT, , = � + ic, c is a given constant, f( � + ic) regarded as a function of � satisfies (iii ) , (iv) and I f ( � + ic) 1 < al� l -le, k > 0, for I �I > M, say, then F(oc) is a regular function of oc in T > c and it is also a (different) regular function of oc in T < c. Again the result follows from the theorem since under the stated conditions ( ' - OC)-1 satisfies (i) , (ii) . Also if we consider the region R' of (v) such that c + e � T � K, a � a � b, we =

_

_

_

have

�

00

f If(u(u2++ae2+)1/2ic) 1 du,

- 00

on introducing u = � - a. Divide the range of u into (-oo,A), (A,B), (B,oo) where the finite number of discon tinuities that f is allowed from (iii ) , (iv) lie in (A,B), and A < - (M + b ), B > (M - a) . Under these conditions If(u + a + ic) 1 < alu + al -le, k > 0, in (-oo,A) and (B,oo) and the integral is absolutely convergent, independent of the position of oc in R'. Next consider THEOREM B. Let f( oc) be an analytic function of oc = a + iT, regular in the strip T_ < T < T+ , such that If ( a + iT) 1 < al al-p, p > 0,

for lal � 00, the inequality holding uniformly for all T in the strip T_ + e � T � T+ - e, e > O. Then for T_ < c < T < d < T+, f(oc)

=

f+(oc) + f_(oc),

( 1 .17 )

where f+(oc) is regular for all T > T_, and f_(oc) is regular for all T < T+ .

14

THE

W I E N E R - H O P F T E C H NI Q U E

The statements regarding regularity are proved as in (3) above. To prove ( 1 . 17) apply Cauchy ' s integral theorem to the rectangle with vertices ±a + ie, ±a + id. From our assumption as regards the behaviour of f( oc) as 1 0'1 � 00 in the strip, the integrals on ±a tend to zero as a � 00 and we are left with the required 0' equation. As an example consider =

f( oc )

=

(oc

_

1 k cos O)( oc + k) 1/ 2 '

k

=

k 1 + ik 2 ' (1 .18)

where it is assumed that - 1T < 0< 1T . f( oc ) has a simple pole at oc = k cos 0 and a branch point at oc = -k. Cut the plane by a line running from -k to infinity in the lower half-plane, choose the point oc so that -k 2 < T < k2 cos 0, and complete the contours for both f+ ( oc ) and f_( oc ) by semi-circles in the upper half-plane. The integrand for f+(oc) has simple poles at , = oc and , k cos 0 giving =

f+( oc )

=

(ci

_

{I

I

}

1 k cos 0) (oc + k) 1/ 2 (k + k cos 0) 1/ 2 '

regular in T> - k 2 • The integral for f_(oc) has only a simple pole at , = k cos 0 giving

1 f- (oc) (oc - k cos O)(k + k cos 0) 1/2 '

regular in T < k cos 0.

These results are obviously correct and of course they could have been guessed without using the theorem. (See ex. 1 .7 for a generali zation.) Next consider the decomposition of a function K(oc) in the form of a product, K(oc) = K+ (oc)K'-(oc) , where K+ and K_ are regular and non-zero in upper and lower half-planes T> T_ , T < T+ respec tively, where T_ < T+ . Sometimes this decomposition can be guessed, e.g. if K(oc) = (oc2 .- k2 ) 1/ 2 , k = k 1 + ik 2 ' (k 1 ,k2 > 0), with the oc-plane cut as in Fig. 1 . 1 we can write K+ (oc) = (oc + k)1/ 2 regular and non-zero in T> -k 2 ' and K_(oc) = (oc - k) 1/2 regular and non-zero in T < k2 • Obviously any decomposition of this type is not unique since we can multiply K+ by any integral non-zero function providing that we divide K_ by the same factor. This remark will be important later. Another kind of example where the answer can be guessed is given in ex. 1 . 1 1 . If K(oc) is an integral function which can b e expressed as an infinite

C O M P L E X VA R I A B L E A N D

F O U R I E R T R A N S F O R MS

15

product (see ex. 1 .9 below) then the decomposition is immediate. The important case from our point of view occurs when K(oc.) is an even function of oc.. Then the roots occur in pairs, say oc. = ±oc... , and K'(O) = O. From ex. 1 .9 we can write co

K(oc.) = K(O)IT {I - (oc./oc.n)2}. .. �l This can be decomposed in the form

K±(oc.)

=

co

{K(O)}1/2e'fx(Ot>IT {I ± (oc./oc.n)}e'f(Ot/P,,>, .. �l

( 1 . 19)

where upper and lower signs go together and the terms have been arranged so that all the zeros of K+(oc.) lie in the lower half-plane and vice-versa. Hence K+(oc.) is regular and non-zero in the upper half plane (Im oc.)> - (1m oc.l). The function X(oc.) is arbitrary and can be chosen to ensure that K+, K_ have suitable behaviour as oc.---+ 00 in appropriate half-planes. The infinite product will in general have exponential behaviour as oc.---+ 00 whereas the functions which need to be decomposed later have algebraic behaviour at infinity due to additional terms multiplying the infinite product. These facilitate the choice of x( oc.) and it is convenient to postpone further discussion until we require the decomposition of concrete examples in §3. 2 (see exs. 3.3-3.6). It is emphasized that the correct choice of X(oc.) is crucial for the successful application of the Wiener-Hopf techmque. When the function has a branch point the infinite product method will break down. We now give a theorem which proves the existence of the decomposition K(oc.) K+(oc.)K_(oc.) for a general class of K(oc.). The theorem also provides a practical method for performing the decomposition in complicated cases. THEOREM C. If In K(oc.) satisfies the conditions of theorem B, which =

implies in particular that K(oc.) is regular and non-zero in a strip T_ < T < T+, -00 < a < 00, and K(oc.)---+ +1 as a---+ ±oo in the strip, then we can write K(oc.) = K+(oc.) K_(oc.) where K+(oc.), K_(oc.) are regular, bounded, and non-zero in T> T_, T < T+, resp ective ly.

(The conditions of the theorem are more restrictive than necessary but cover the applications needed in this book. A more general theorem is stated in ex. 1 .12 so as to bring to the reader ' s notice certain points which must be borne in mind in more complicated examples-in particular the more general theorem shows how to deal with zeros of K(oc.) in the strip, and with cases where K(oc.)---+ exp (ip,), exp (i'JI) as a---+ +00, -00 respectively, or IK(oc.) I '" lalp as Ial---+ 00, in the strip.) I

16

T H E W I E N E R- H O P F T E C H N I Q U E

To prove theorem C, apply theorem B to f (oc) = In K(oc). In K(oc)

=

1 27T �·

ic+ co

J

In KW

�- co

1 -r-- d, - 2 7T�· ", - oc

id+ co

J

U- co

In KW

-r--

", - oc

d,

( 1 .20)

where c, d are any numbers such that T_ < c < T < d < T+. The integral for f+(oc) is convergent for all oc such that T> c. Hence f+(oc) is bounded and regular in T > T_ since we can choose c as near as we please to L. Similarly f_(oc) is bounded and regular in T < T+. If we set then In K+(oc) + In K_(oc) = In K(oc),

K+(oc) K_(oc) = K(oc). From the properties off+(oc), it is seen that K+(oc) is regular, bounded, and non-zero in T> L. Similarly K_(oc) is regular, bounded, and non-zero in T < T+. Therefore the theorem is proved since K+(oc), K_(oc) have been constructed which satisfy the necessary conditions. i.e.

The presence of the logarithms in ( 1 .20) often makes the integrations difficult. Sometimes simpler integrals are obtained as follows. We have In K(ex) = In K+ (ex) + In K_ (ex). Differentiate with respect to ex:

K'(ex) K(ex)

=

K'+ (ex) K + (ex)

+

K'-(ex) . K _ (ex)

( 1 . 22)

If the expression on the left-hand side is decomposed by theorem B we can set

and an integration with respect to ex will give In K+ (ex), In K_ (ex). An example is given in ex. 2. 10. It is clear that in the practical application of theorems B and C the chief difficulty is the evaluation of the contour integrals ( 1 . 1 7 ) , ( 1 .20). When the integrals have simple poles and no branch points the situation is comparatively easy since the integrals can be evaluated by residues, but in this case the reduction or separation can usually be carried out more directly by simpler methods, e.g. in the case of theorem B we pointed out that ( 1 . 1 8 ) could be decomposed by inspection, and in the case of theorem C the reader will readily verify that evaluation by residues is equivalent to the infinite product method given above. We now derive two results which are useful when evaluating integrals of type (1.17), ( 1 . 20). Suppose that j(ex) is an even function of ex,

C O M P L E X VA R I A B L E A N D

regular in -k2 < 'T for -k2 < d < k2 1-(-tX) = -

< k2'

1

satisfying the conditions of theorem B. Then

id+ 00

27J1,.

-

f

17

F O U R I E R T R AN S F O R M S

fm

--

�+

tX

-id+ 00

1

d� =

27T�.

-

�- oo

f

-�- oo

f( -�) --d � = j+(tX), � tX -

( 1 . 2 3)

i.e.

Instead of evaluating the integrals in ( 1 . 20) directly it is usually convenient to introduce a quantity g(tX) in the following way. Suppose that K(tX) is even, regular in -k2 < 'T < k2' and satisfies the conditions of theorem C. We have Set Then

f+(tX) - 1-(tX) = g(tX),

say.

f+(tX) = V(tX) + tg(tX)

1-(tX) = V(tX) - tg(tX).

( 1 . 24) (1.25)

Since 1-(tX) = f+(-tX) we have g(tX) = -g(-tX). Introducing K(tX) into

( 1 .25), we obtain :

In K_(tX) = tIn K(tX) - !g(tX),

K_(tX) = {K(tX)}l/2e-tU(ex).

( 1 .26)

It is sometimes easier to evaluate the integral for g(tX) rather than the integrals for f+(tX), f_(tX) separately. If tX is real, say tX = �, and the contour in ( 1 .20) can be taken as the real axis (c ->- - 0, d ->- + 0) then on using the result in ex. 1 .24 we find

with

g(�) =

!.. p 7Ti

00

f

- 00

f(x) d�, x-�

( 1 . 2 7b )

where P denotes a Cauchy principal value. The most important branch points in our applications appear when integrands are functions of y = (tX2 - k2)1/2. To illustrate the procedure suppose that f(tX) in ( 1 .17) is a function only of ya = (tX2 - k2)1/2a, sayf(tX) = F(ya), where the constant a has been added for convenience. Consider the integral for f_(tX) and examine three methods of procedure in turn. We have ( 1.28)

where 'T < d and -k2 < d < k2• Assume that the contour can be deformed into an upper half-plane if necessary and that in any such deformation no poles are crossed. (Contributions from poles can be

18

T H E W I E N E R-H O P F T E CH N I Q U E

taken into account in the usual way. ) Similarly contributions due to poles at CI. = ±k will be ignored. The three procedures to be examined correspond to different ways of choosing branch cuts, as shown in Fig. 1.2. It is assumed that CI., k are real and that -k < CI. < k. A convenient method of finding the positions of indentations is to imagine the diagrams for the general case 1m CI. < 0, 1m k > 0, and then let 1m CI. ->- - 0, 1m k ->- + 0. If CI. lies outside the range (O,k) the indenta tions and contributions from indentations must be altered accordingly. (a) First of all consider simply the limiting case of ( 1 .28) when k2 ->- 0, the complex plane is cut from + k to + 00 , -k to - 00 , along the positive and negative real axes respectively, and the contour becomes the real axis, passing above the branch point at , = -k, below the branch point at, = + k and above the indentation at, = CI.. The values of arg yare shown in Fig. 1 . 2(a). For clarity set CI. = C;, a real variable. As in the argument leading to ( 1 .27),

( ) = t F{(c;2 - k2 ) 1/2a} - ig(c; ), L c; ( 1.29 )

where

k

g(c; ) = � P m

+

f F{(x2 - k2)1/2a} �

x - c;

-k -k

�

m

+

f F{(x2 - k2)1I2a} � x -c;

- 00

+

�

7T�

co

. f F{(x2 - k2)1/2a} � x -c;

k

Change x to ( -x) in the second integral and combine with the third. Split the first integral into integrals over ( -k,O), (O,k) and combine into one integral over the range (O,k ) . Change variable to u defined by (x2 - k2 )a2 = u2 , (k2 - x2 )a2 = u2 in the integrals over (k, 00 ) , (O,k) respectively. This gives

2c;a f { co

+

.

7Ti

o

F(u) u du . 2 2 + 2 112 . c; 2 2 2 2 + u) a (k - )} ( k a u

( 1 . 30)

(b) In Fig. 1 . 2(b) the contour is deformed along two sides of a branch cut which consists of the real axis of, from + k to + 00. If, = x + iy, then on the lower side of the cut , ( 2 - k2 ) 1/2 = + (x2 - k 2 ) 1/2 , and on the upper side , ( 2 - k2)1/2 = _ x ( 2 - k2 ) 1I 2 . Hence forc; < k,

f-(c; ) = - �

�

co

f [F{(x2

k

-

k2 ) 1/2a} - F{ _ (x2 - k2 ) 1/2a}] � . ( 1 . 3 1 ) x -c;

C O M P L E X VA R I A B L E A N D

19

FOURIER TRAN SFORMS

As in ( 1 . 23) i+(;) = L( - ; ) and as in ( 1 . 24), ( 1 .25) it is convenient to consider g(;) =i+(;) - L(;) . If we change variable to u defined by (x2 - k2 )a2 = u2 , we find ( 1 . 3 2) '1'

erg')l=O

erg/,=- t'IT

+t-------- 0' orgy=O (0) •

ergy=

-k

-

'IT

+k

o

.

2�

2

?I =1 (1(-+ y )

erg?,= t11 --_"k--Hr.E�---

0'

erg'Y=O

org1=- t 11

(b)

(c)

FIG. 1.2. - - - Branch cut.

--

Contour.

(c) Finally consider the contour in Fig. 1 .2(c). On combining the integrals along the two sides of the branch cut from 0 to k in x and 0 to OCJ in y,

- �

21T

+

00

f [F{i(k2

o

p � 21Tt

+

y2 ) 1/2a} - F{ -i(k2 + y2 ) 1/2a}] � �y - ;

f [F{i(k2 - x2)1/2a} k

o

-

+

. F{ -i(k2 - x2 ) 1/2a}] � X - ;

20

THE

W I E N E R - H O P F T E C H N I Q UE

If -k < � < 0 the first term on the right does not occur and the integral is not a principal value. On following a procedure similar to that used in (b) we obtain

g(�) = � � [F {i(k2 - �2 ) 1/2a} -F{ -i(k 2 �2 _ ) 1/2a}] 2 1 �1 ct)

f [F{i(k2 y2) 1/2a} - F{ -i(k2 y2) 1/2a}] � �a f {F2(iu)2 - F2 ( -iu)}2 2 2u du 2 1/2 · ( 1. 33) -� w

+

+

�+r

o

ka

wi

P

o

{a (k

-

�)

_

u } (k a

_

U)

The form in (a) is the simplest to obtain : in (b) and (c) care has to be exercised regarding contributions from poles. The form in (b) may be convenient if {F(u) - F ( -u)}, (u real), is simple : also no principal value occurs if I � I < k. The form in (c) may be convenient if {F(iu) - F ( -iu)} is simple : also no principal value occurs if I �I > k. Instructive examples of the use of integrals of the above type to perform factorizations are given, for instance, in H. Levine and J. Schwinger [1], L. Vajnshtejn [1], [3], and W. Chester [1]. In the general case k = kl + ik 2 ' k2 > 0 , there are various forms of contour which reduce to the forms used above when k 2 O. Thus we can take the contour from +k to infinity as (i) a straight line from k = kl + ik 2 to 00 + ik 2 parallel to the x-axis, (ii) a continuation of the straight line joining the origin to k, (iii) the part of the hyperbola through k defined by �1] = klk2 ' (� = � + i1]), on which I' = (�2 - k2 ) 1/2 is real. As an example of case (b) above consider --+

K(ex) =

e-yd

In K(ex) =

-I'd = _ (ex2 - k2 ) 1/2d.

( 1.34)

We cannot apply theorem B to In K(ex) as it stands, since it increases as ex --+ 00 but we can write

The factor B to (ex2 -

A standard integral is

f

1

1,

(k2 - ex2 ) is an integral function and we can apply theorem k2 )-1/2 = r l . Then ( . 32 ) with F (u) = u-l , a = gives

du = (u2 + p ) (u2 + q) I/2 pl/2 (q

1

_

{

}

u(q - p) 1/2 arc tan , (q> pl · I l/2 p) /2 p (u2 + q ) I/2

O O M P L E X VA R I A BL E A N D

Hence for

0

<:

�

<:

F O URIER TRAN S F O RMS

21

k.

2 � g(�) = 1+(�) - L (�) = i(k2 _ �2 ) l/2 arc tan (k2 _ �2 ) 1/2 ' 1T' But 1+(� ) + L (�) = (�2 - k2 ) -1/2 and i(k2 _ �2 ) 1/2 = _ (�2 _ k2 ) 1/2 . Hence 1+( � ) = 7T-l (�2 - k2 )-1/2[!7T - arc tan {�(k2 _ �2 )-1/2}] = 7T-l (�2 _ k2 ) -1/2 arc cos (�/k)

where the branch of the inverse cosine is chosen so that when � = 0, arc cos 0 = In. By analytic continuation we can now define 1+ (rx) by these expressions with � replaced by rx, provided that branches are defined to agree with the above values when rx = �, � real, 0 <: � <: k. In this case arc cos

(rx/k) = +i In {(rx + y)/k},

By analytic continuation

-i In (2rx/k) + 0(lrxl -2 ) , (rx/k) 1 (rx2 - k2 ) /2 = - rx + 0 ( 1 rxl -2 ),

arc cos

a s a ->- 00 in an upper half-plane. Returning t o ( 1 .34) w e have therefore the decomposition

K_ ( a ) = exp { - T_ (rx)}; K+(rx) = exp { - T+( a )} ; T (rx) = 7T-1d(rx2 - k2 ) 1/2 arc cos (a/k) ; T_(rx) = T+ ( - rx) ; ( 1 .35) + T+(rx) = (id a/7T) In (2 a/k) + 0(l a l -1 ), as rx ->- 00 in an upper half-plane. An alternative method for obtaining these results is given in ex. 1 . 25. 1 .4

The Fourier integral

At the outset is is necessary to decide whether to use Fourier or Laplace transforms. Both have been used in the literature in connexion with the Wiener-Hopf technique and in the complex plane the two transforms are completely equivalent as will be clear later in this section. In this book we use Fourier transforms for no special reason apart from the need to standardize on one or the other. The Fourier integral can be stated in the form 1

F(rJ.) = 21T 1/2 ( )

00

f f(x)e'lX. X dx

- 00

f(x) =

1

(21T) 1/2

f F(rJ.)e - '.1XX drJ.,

- 00

( 1 .36)

22

THE

W I E N E R-H O P F T E C H N I Q U E

where Ot is usually taken to be real. The Laplace integral is usually stated in this form

� f

co

�(8) =

f

c + i co

f(x) e - 8X dx

f(x) = 2 i

o

�(8) e 8X d8,

(c >

co ) ,

c - i oo

( 1 .37) where the second integral gives automatically that f(x) = 0 for x < o. The Laplace integral corresponding to ( 1 .36) is the two-sided Laplace integral 00

�(8) =

f f(x)e - 8X dx

- 00

� f �(8)e8X d8, c + i oo

: f(x) = 2 i

(c1

c - i oo

< c < c

2) ·

(1 .38)

Corresponding to (1 .37) we have the Fourier integral 1

F(Ot) = 27T) 1/2 (

co

f f(x)e'lXX. dx o

1

f(x) = 27T) 1/2 (

co

f F(Ot)e - "X. X dOt,

- co

(1 .39) where the second integral gives automatically that f(x) = 0 for x < o. This form of the Fourier integral is not often used in the literature. It is sometimes stated that it is preferable to use the Laplace integral instead of the Fourier integral since the Laplace integral can deal with more general situations. The example that is usually quoted is f(x) = 1 , (x > 0) : f(x) = 0, (x < 0). The Laplace trans form of this function, from ( 1 .37) , is S-1 whereas the first integral in ( 1 .39) does not converge for this f(x), if Ot is real. This is true if we are dealing with real Ot and real 8, but if Ot and 8 are both complex we have merely the equivalent statements : if f(x) = 1 , (x > 0) : 0, (x < 0), then the Laplace transform of f(x) is 8-1 provided that Re 8 > 0; the Fourier transform of f(x) is iOt-1 provided that 1m Ot > O. This complete equivalence of Fourier and Laplace trans forms will be empliasized later in this section. One of the advantages of using the Laplace transform would be that we are accustomed to considering the parameter 8 as complex: thus the limits on the integral and the statement c > Co in (1 .37 ) automatically imply that 8 is complex. We shall start by assuming ( 1 .36 ) for real Ot and show how the formulae are modified when Ot is allowed to take complex values. It is not necessary from our point of view to state the exact con ditions which must be satisfied by f(x). We content ourselves with

C O M P L E X VA R I A B L E A N D

FOURIER

23

TRAN S F ORMS

stating that integrals will be understood in the Riemann sense, and that f(x) may have a finite number of infinite discontinuities and need not be absolutely integrable in the infinite range, e.g. we have Fourier transforms of the following types :

{f (x) F(oc) { f (x)

} O, (x < O) ; }

=

( i x l < a) (a2 - X2 )-1/2; = (i7T)1/2 Jo(aoc) . xv - I , (0 < y < 1 , x > 0) F(oc ) = ( 27T ) -1/2 r( y ) e t i 1Tv oc-v .

0, ( I x l > a) ;

=

(It is worth noting that the original work of N. Wiener and E . Hopf is rigorous and uses Lebesgue integration. There is a temptation to use the Plancherel theory as starting point because of the elegance and generality of the results. It was felt that any such attempt at rigour would be out of character with the remainder of this book.) Now suppose that oc is a complex variable, oc = a + i T . We state and prove the following : THEOREM A. Let f(x) be a function of the real variable x such that If(x) 1 � A exp (T_X) as x -- + 00 and I f(x) 1 � B exp (T+X) as x __ - 00 , with T_ < T+ : suppose that for 80me given TO ' T_ < TO < T+,

Fourier ' s theorem for real variable8 a8 given by (1 .36) applies to th e function f(x) exp ( - TOX). The n if we define F(oc)

=

1 ( 27T) 1/2

co

f f(x)e!"X. dx,

at =

- co

F(oc) i8 an analytic function of oc, regular in f(x)

=

1 ( 27T) 1/2

i'T + co

f F(oc)e- t". X doc,

i'T - co

a

+ iT,

T_ < T < T+,

( 1 . 40 )

and

for any T, T_ < T < T+. ( 1 .41 )

The statement regarding the analytic nature of F( oc) follows from §1 .3, theorem B, example (1 ) . Substitute ( 1 . 40 ) in the right-hand side of ( 1 .41 ), which becomes

1 27T

00

oo +i'T

f(�) ei"" d�e - iet.x doc f f - CO + iT L eTX f e - i<1X f {J ( �)e -T"}ei<1< d� da - co

= =

co

- 00

eTX {J ( x) e -TX }

co

- 00

=

f(x).

24

THE

W I E N E R-H O P F T E C H N I Q U E

In the second line we changed the variable by setting IX - iT a, dlX = da. The third line is obtained from the Fourier integral (1 .36) , since if a single TO exists as postulated in the theorem, Fourier's integral applies to f(x) exp ( - n) for any T in T_ < T < T+ because of the assumed behaviour of f(x) as x � ±oo. As an example, if f(x) = 1, (x > 0), f(x) = eX, (x < 0), then T_ = 0, T+ = 1 , and F(IX) should be regular in 0 < T < 1 . In fact =

{ ( 2:) 1i2 f

co

=

F(IX)

o

eirJ-X

f e(l+ i<X)X dX} 1 � ilX } ,

dx +

- co

1

: {- i1X

(2 ) 1/2

=

0

+

( 1 .42)

where the first integral can be evaluated only for T > 0, and the second only for T < 1 . F(IX) has singularities on the lines T = 0 and T = 1 as we should expect. The converse of theorem A is given in THEOREM B. Supp08e that F(IX), IX = a + iT, i8 regular in the 8trip T_ < T < T+, and i F(IX) i � 0 uniformly a8 i ai � 00 in the 8trip T_ + S � T � T+ - S where S i8 an arbitrary p08itive number. If we

define a function f(x) by f(x)

1 ( 21T ) 1/2

=

i'T + co

f

.

F(IX) e -trJ-X dlX

i-r - 00

(1 .43)

for a given T, T_ < T < T+ , and any given real x, then

f f (x) et<x. x dx. co

F(IX)

=

( 21T1) 1/2

( 1 .44)

- co

A l80 if(x) i < exp (T_ + (J)x a8 x � + 00 , and if(x) i < exp (T+ - (J)x as x � -00 wh ere (J i8 an arbitrarily small p08itiv e number. The function f(x) defined by (1 .43) can be regarded as the 80lution of the integral equation (1 .44) . T o prove this theorem choose c , d such that T_ < c < T < d < T+ where T is the imaginary part of IX in (1 .43). Substitute (1 .43) in the right-hand side of (1 .44) : I =

Lf

o

- co

eirJ-X dx

=

id+ co

f

id - co

00

Lf - co

e

f

i'T + co

ei<xx dx

iT - CO

xF({3) d{3 +

- if3

e - if3xF({3) d{3 co

Lfe

irJ-X

0

ic + co

dx

f

ic - oo

e - if3x F({3) d{3.

C O M P L E X VA R I A B L E A N D

FOURIER TRANSFORMS

25

The splitting of the integral and the use of different though equivalent T-values has been arranged so that when orders of integration are interchanged the inner integrals are convergent since 1m (IX. - {3) < 0, x < 0 in the first integral and 1m (IX. - {3) > 0, x > 0 in the second. On interchanging orders of integration and evaluating the inner integrals, we find 1

id+ 00

: f 1 F({3) d{3 , = 2 . f {3 =-

2 ;

" u

id - oo

-

n�

a

-

F({3) 1X. d{3 {3 -

�

+ 2n�

ic + 00

f

� - oo

F({3) d{3, {3 - 1X.

IX.

where 0 is a closed contour consisting of the limit of the rectangle joining the points (±A,c), (±A,d) as A ----+ 00, the rectangle being traversed in a positive direction. The integrals along the vertical sides of the rectangle vanish in the limit because of the assumption that F(IX.) tends to zero uniformly as l a l ----+ 00 in the strip. Cauchy' s in tegral theorem then gives 1 = F(IX.) as required. If If(x) 1 < exp (px) as x ----+ + 00 and f(x) < exp ( qx) as x ----+ -00 then from ( 1 .44) F(IX.) would be regular in p < T < q . But we are given that F(IX.) is regular in T_ < T < T+. Hence by reductio ad absurdum we can take p = T_ + 15, q = T+ - 15, where 15 is an arbitrarily small positive number. A special case of the above theorem, namely when either T_ or T+ is infinite, is sufficiently important to warrant statement as a separate theorem : THEOREM C. Suppose that F(IX.) , IX. = a + iT, is regular in T > T_ and F( IX.) ----+ 0 as r ----+ 00 where r is defined by IX. - iT_ - is = r exp (ifJ) for arbitrary s > 0 and the limit is approached uniformly for o � fJ � n . If we define f(x) as in (1 .43) for any given T > T_

and any given real x then

F(IX.) =

1

( 2 n ) 1/ 2

00

f e''lXXf(x) dx,

(1 .45)

o

and the function f(x) defined by ( 1 .43) can be regarded as a solution of the integral equation ( 1 .45). Also f(x) 0 for x < 0, and If(x) 1 < exp (T_ + b)x as x ----+ + 00 , where 15 is an arbitrarily small positive number. .

The proof is left to the reader. If follows exactly the same lines as before except that the contour for the application of Cauchy ' s theorem is now completed by an infinite semi-circle in the upper half plane. A similar situation arises when F(IX.) is regular in some lower

26

THE

W I E N E R-H O P F T E CH N I Q U E

half-plane T < T+ Then under appropriate conditions f(x) defined by ( 1 .43) satisfies an integral equation similar to ( 1 .45) except that the limits (0, 00 ) are replaced by ( - 00 ,0) . Next consider the relation of theorems A and B to the two-sided Laplace transform. In the integrals ( 1 .40), ( 1 .41) make the change of variable iat = -s where s a ' iT' , another complex variable. Since at = a iT, then a ' + T, T' - a . Define §'(s) (21T) -1/ 2 F(at) and find, for some constant value of a ' which we denote by b, •

+

§'(s) =

=

=

J

+

=

-

co

f(x)e - 8X dx

f(x)

=

- co

� 2m

J

b + i co

§'(s)e8X ds. ( 1 .46)

b - i co

This is the two-sided Laplace transform. To make clearer the relation between at and s the paths of integration and strips of regularity in the two complex planes are plotted in Fig. 1 .3.

0. - plane

FIG. 1 . 3 .

Similarly the form given in theorem 0, with the same change of variable i at = -s, reduces to the standard Laplace transform

J f(x)e - 8X dx co

§'(s)

=

f(x) =

o

� 2m

J

b + i co

§'(s)e8X ds.

( 1 .47 )

b - i co

As already mentioned some of the literature on the Wiener-Hopf technique has been written in terms of the Laplace transform and in order to translate the notation of this book into terms of the Laplace integral and vice-versa it is important that the reader should under stand the complete equivalence of Fourier and Laplace integrals in the complex plane. As an example, the Laplace integral correspond ing to ( 1 .42 ) gives 1 __ . §'(s) = 1 _·s s The statement that ( 1 .42 ) represents a function which is regular in

{� +

}

C O M P L E X VA R I A BL E A N D

FO URIER TRAN SFORMS

27

o < T < 1 is equivalent to stating that the corresponding Laplace result is regular in 0 < a' < 1 . The reader will find a great deal of material on the subject matter of this section in books on the Laplace transform, e.g. the standard texts of Doetsch, Van der Pol and Bremmer, Widder. An elementary treatment of some topics is given in R. V. Churchill, Modern opera tional mathematics in engineering, McGraw-Hill (1944), Chapters V, VI. Finally consider the relation of the above results to the Mellin transform. Consider the substitution

p=

eX

dx = d p l p

f(x) - g( p ) .

If we use this in ( 1 .40), ( 1 .41 ) we find that reciprocal formulae are : i'T + co

co

1 f

F(IX) = 2 1 / 2 ( 7T )

o

.

g(p)ptrx - l d p

g(p)

=

dIX. 1 -f F(IX) p -(1trx. .48)

(27T) 1 / 2 iT

co

Similarly in (1 .46) make the substitution p = exp ( -x) . This gives 00

ff(s)

=

f g( p ) pS-l dp

o

T + i oo

� f ff(s) p-S ds.

g( p ) = 2 i

( 1 .49)

-r - i oo

These two forms are completely equivalent. The second is the well known Mellin transform and we use it in this book. Perhaps it would be slightly more consistent to use the form ( 1 .48) since we have standardized on the Fourier transform instead of the Laplace transform. We shall have to talk of right and left half-planes when using the Mellin transform instead of upper and lower half-planes for the Fourier transform. However the reader will see that this causes no confusion in applications. There is of course a theorem for Mellin transforms corresponding to theorem C above (ex. 1 . 14) . I n the applications of the Mellin transform made later it would be possible to apply the substitution p = exp (-x) to the original partial differential equation, and then apply the Fourier transform to the new equation, so that theoretically we need never use the Mellin transform at all. But in practice it is convenient to use the Mellin transform directly. 1 .5

The wave equation

In this section we wish to clarify several points concerning the wave equation. For concreteness consider the two-dimensional ('quation

(1 .50)

28

THE

W I E N E R -H O P F T E C H N I Q U E

For reasons which will become apparent presently it is desirable to treat this equation as the limiting case of the equation with damping. ( 1 .51)

B

where is a damping factor greater than zero. We consider two types of problem : (i) Steady· state problems in which 'IfJ(x,y,t) = cp(x,y) exp (±iwt) ; (ii) Transient problems in which it is necessary to apply a Fourier transform in time. A point which has not been mentioned previously arises in connexion with the Fourier transform, namely, that we can choose one of two forms : 'IfJ

1 = (21T ) 1/

co

2 f 'I"e 'F ,.wt dw, - co

( 1 .52)

where upper and lower signs go together, and it is assumed that w is real. In previous sections we have tacitly assumed the upper sign. On applying each of these to ( 1 .51) in the usual way, we find

02'1"2 + Q2'Y2 + {W2 ± iBW} 'Y = 0, c2 ox oy

( 1 .53)

where the upper and lower signs correspond to those in ( 1 .52). Consider next a steady-state problem with 'IfJ proportional to exp (±iwt) . On substituting in ( 1 . 5 1 ) we see that for exp ( + wt ) we obtain an equation corresponding to the lower sign in ( 1 .52) and ( 1 .53), and for exp (-iwt) the equation corresponds to the upper sign in ( 1 .52) and ( 1 .53). There is no nece ssity to use the same sign in the exponential for Fourier transforms in space and time, or to choose a steady-state time factor which gives an equation corresponding to a particular sign of the Fourier transform used for space variables, but for convenience we standardize on the upper sign in ( 1 .52) for all Fourier transforms. This has been used in previous sections, e.g. equation ( 1 .3). Corresponding to this we use the steady-state time factor exp ( X iwt) . This means that when we deal with the steady state equation in the form

i

( 1 .54a)

29 then k2 , when damping is present, has a positive imaginary part (cf. (1 .53), k2 = ( w2 + isw)/c2 ) . We write C O M P L E X VA R I A B L E A N D

F O U R I E R TRANS F O R M S

( 1 .54b) Next consider application of a Fourier transform in a space variable to ( 1 .54a). Assume that (1m k) = k 2 > O. For definiteness consider problem (i) of §1 . 1 : find a solution of (1.54a) in - 00 < x < 00 , y :;:, 0 such that cp = f(x) on y = 0 , - 00 < x < 00 . To solve this problem apply a Fourier transform in x. Define 1 <1>( lX ,y) = (27T ) 1/ 2

00

f cp rx,y)e'rxx. dx.

- 00

From ( 1 .54a), as in § l . l ,

d2<1>(IX,y)/dy2 - (1X2 - k2 )(IX,y)

=

0;

( 1 .55a) (1 .55b)

As in § 1 . 1 the following question arises : In this solution there are branch points at IX = ±k. How should the cuts be arranged in the IX-plane so that ( 1 .55b) represents a solution of ( 1 . 55a) which can be inverted to give cp(x,y) ? We have already implicitly assumed: (a) For any given y, (IX,y) exists in a certain strip c < T < d, - 00 < a < 00 of the IX-plane. We make the further assumption : (b) <1>( lX,y) is bounded as y ---+ + 00 for all IX in the strip c < T < d. (In our applications (IX,y) in fact tends to zero exponentially as y ---+ 00 but the weaker assumption just given is easy to verify when conditions on cp are specified and it is sufficient for our purpose.) If we cut the IX-plane by straight lines from IX ±k to infinity, both of the straight lines going to infinity in the lower half-plane, then it will be necessary to invert ( 1 .55b) for a value of T, say TO ' such that TO > k2. But in this case it can be proved by analytic continuation, as in §1 .2, that if we choose the branch of A such that A---+ l or l as a---+ + 00 , IX a + iTO ' then A---+ - I IXI as a---+ - 00 . Thus i n (1 .55b) it will not be possible to choose A and B such that is bounded as y ---+ 00 for all IX on the line IX a + iTO ' A similar argument holds if we cut the plane by two lines both of which go to infinity in the upper half-plane or if we cut the plane by a straight line joining + k to -k (cf. ex. 1 .4 ) . The only remaining possibility is to cut the IX-plane by a line from + k to infinity in the upper half plane and - k to infinity in the lower half-plane as in Fig. 1 .1 . By analytic continuation we have that A ---+ + I IX I as a---+ ± oo so that =

=

=

30

T H E W I E N E R- H O P F T E C H N I Q U E

( 1 .55b) represents a solution satisfying condition (b) if we take = O. In fact we can identify A with y defined at the end of §1 .2

B(ex) and

cI> (ex,y)

=

A ( ex) e-w .

(This argument will not be repeated in detail again. ) To determine A (ex) we set y = 0 and use the boundary condition on y = O. This gives cI> (ex, y ) =

00

: fJ(u)eilXu du e - W•

(2 ) 1/2

- co

rp(x,y) is obtained by inversion. It is clear that a third condition is needed: (c) The strips c < T < d and -k2 < T < k2 overlap. In any particular case it is necessary to verify that rp is such that conditions (a)-(c) are satisfied (cf. §2.2 ).

,

,

,

� ,

,

,

,lCut

,// cut

+ k/ ====="=_� k oF=======+a , ,

,

, , ' Cut ,

,

,

,

,

I

,

"

,

,'

/' - k

+k

/

,

,

,/

ICut FIG. 1 0 4.

-

- - Branch cut.

--

Contour.

Finally consider what happens when k 2 --+ o. It will be remem· bered that k 2 arises from the damping factor in (1 .51 ) and k 2 0 means that the damping is zero. When k 2 > 0 the contours in the ex·plane are shown in Fig. 1 .4a. When k 2 = 0, the contour is along the real axis but by performing the limiting process 'k 2 --+ 0, we see that the contour has to be in dented as in Fig. 1 .4b, i.e. ahove at -k, and below at + k. If we had used exp ( + iwt) as the steady-state time ff1Ctor, or used the lower sign in the Fourier transform (1.52) then k wou ,d have a negative imaginary part and by drawing diagrams similar to Fig. 1 .4, it is clear that when k 2 --+ 0 in this case, the contour along the real axis has to be indented below at ex = -k and above at ex = + k. The case of zero damping factor requires modifications in our conditions (a)-(c) above. Physically when k 2 > 0, rp behaves =

C O M P L E X VA R I A B L E A N D

F O U R I E R T R A N S F O RM S

31

exponentially at infinity because of the finite damping-outgoing waves must decrease exponentially as we go to infinity. If no damping is present this is no longer true. Instead we need to assume some condition such as Sommerfeld ' s "radiation condition at infinity" . In the two dimensional case this states : as r --+ ct) , if c/> represents an outgoing wave at infinity, and if the time factor is taken as exp (-iwt) then rl/2 {(ac/>/ar) - ikc/>} --+ 0; if the time factor is exp ( + iwt) then rl/2 {(ac/>/ar) + ikc/>} --+ O. These conditions must be satisfied uniformly with respect to B. In this monograph we shall always work with finite k2 and obtain the case of zero damping by letting k 2 tend to zero. The Sommerfeld radiation condition will not be used directly although it always provides a useful check on the final solutions. 1 .6 Contour integrals of a certain type Consider a type of integral which occurs frequently later,

f

ia +

I

=

co

ia - 00

f(oc) e - itXX - Ylvl doc,

( 1 .56)

where y = (oc2 - k2 ) 1/2 as defined at the end of § 1 . 2 , and - 1m k < a < 1m k. f(oc) is an analytic function such that the integral converges. Further necessary restrictions will be clear from the discussion below. First consider a useful shift of contour in the oc-plane which enables us to evaluate two special integrals explicitly. Suppose that x = r cos (), I YI = r sin () where we assume without loss of generality that 0 .;;;; () .;;;; rr. Consider the contour oc

=

-k cos (() + it ) ,

(-

00

00 ) .

( 1 . 57)

The reader will easily prove that this represents one half of a hyperbola passing through oc = -k cos () with symmetry about the line j oining -k to + k . A typical case is shown in Fig. 1 . 5. An easy way of visualizing the picture is to note that if k2 is small then a

�

-kl cos () cosh t

T �

kl sin () sinh t.

For 0 < () < rr/2, ( 1 .57) represents the half of the hyperbola in the left-hand half-plane, and for rr/2 < () < rr, the half in the right-hand half-plane. If t goes from - 00 to 00, the direction in which the contour is traversed is shown by arrows in the figure. If the line j oining -k to +k makes an angle A with the a-axis then the asymptotes make angles (A + ()) and (A - ()) with the a-axis. We have

y

=

-i(k2 - o(2 ) 1/2

=

-ik sin (() + it),

(0 .;;;; () .;;;;

)

rr ,

( 1.58)

where the sign o f the last term i s settled b y analytic continuation since when () = rr/2, t = 0, i.e. oc = 0, then y = - ik. Therefore -iocx

- ylyl

=

ikr{cos () cos (() + it ) + sin () sin ( () + it ) }

=

ik r cosh t.

32

T H E W I E N E R- H O P F T E C H N I Q U E

Now suppose that the contour in ( 1 .56) can be deformed on to the appropriate half of the hyperbola. In the usual way connect the two contours by arcs at infinity. For 0 < B < 7T/2 these are shown by DE and A BO in Fig. 1 . 5. For 7T/2 < B < 7T we need to use the right-hand half of the hyperbola and the arcs are then obtained by extending DE and contracting ABO. Assume that the behaviour of the integrand at infinity is such that the contributions from the arcs at infinity are zero (see ex. 1 . 6) . Assume also that no singularities of j(rx) fall inside the

c _ - -- �;6 Cut

I I I

--

- - 8 - '"

,./

"

/

I

I

/

FIG. 1 . 5 .

closed contour. (As shown in example (2) below it is easy to deal with poles inside the contour.) Finally therefore, since d rx = ik sin ( B + it) dt, ( 1 .56) becomes <Xl

f j{ -k cos ( B + it)}eikr

I = - ik

cosh

t sin ( B + it) dt,

( 1.59)

- <Xl

where r2 = x2 + y2. Consider two examples

f

ia + <Xl

I =

(1)

ia - 00

y-1e

-

ilXX

-

v l ll l d rx,

( 1 . 60)

No poles are enclosed on shifting the contour. Equation ( 1 . 59) reduces to a well known representation of a Bessel function (G. N. Watson r l], p. 1 80, ( 10) ) : I

=

<Xl

f eikr

- <Xl

cosh t

dt = wiH�l) (kr) = wi {J o (kr) + i Yo(kr)}. ( 1 . 6 1 )

C O M P L E X VA R I A B L E A N D

33

F O URIER TRA N S F OR M S

k _rriH�2)(kr) . ia+ (k e - k)1I2 f - k)1( -iI2 (ex --ky iy/ le) d ( 1 . 62) (2) iae - k)1/2 (k e - k)1/2 -i(2k)1/21T, -k2te_iP/2 k( 21 3 e,) =k e. (1 .57) () 1T, 1T - e () - 1 e 1T - e () 1T, ( 1.63 ) __ {J + 21Ti { -ikr (() + e)}, J () 1T - e 1T, J 2i te f (ikr(() + �t)t) + t(() e+ it) dt ( 1.64) 1.22 ( 1.63), ( 1 . 64) t (() - e)} = 21T1 I2ei1T/4[ _ e -ikr (O - El) F{ (2kr)1/2 (O+El)F ikr {(2kr)1/2 t(() + e )}] , ( 1 . 65) +e (), e, () .;;; 1T, e 1T, F(v) = f eiu' duo ( 1 . 66)

Note that i f i s taken t o have a negative imaginary part then the integral represents I

00

=

exp

cos

exx

(ex

co

ex,

cos

where is a constant angle, 0 < e < < a < cos and (ex is defined as the branch which tends to as ex tends to zero. Similarly cos = sin as in . I a . The integrand has a pole at ex cos When the contour is shifted by the pole is inside the contour when < cos < - cos () for any (), 0 < < i.e. 0 < < . We find exp cos 0 < < < I

where

00

=

sin

o <

exp

- 00

From the results in ex.

for all values of

.

<

sin

.

cos

it can be shown that

cos

I

cosh

cos

<

0

cos

cos

cos

0

.;;;

<

00

<

give

+

where

v

Thus the integral has been evaluated in terms of the complex Fresnel function which has been extensively tabulated. Next consider a change of variable by means of which ( 1.56) can be considered from a more general point of view. For simplicity suppose that is real and positive. The transformation ex = cos {J where {J = ft ex = a + transforms the ex-plane cut along the real axis from 00 to and from to + 00 into the region 0 .;;; ft 00 < < 00 in the {J-plane, where we have chosen one of the infinitely many strips in the {J-plane into which the ex-plane is mapped. The situation is depicted in Fig. 1 . 6 where corresponding points bear the same letter. The contour EBAD along the real axis lying above the cut and below the cut transforms into the path E' B'A ' D'. The half-hyperbola defined by ex = cos ( () + 00 < < 00 , with () a constant such that 0 < () < (cf. equation ( 1 . 57) above) transforms into the line ft = () = constant, in the {J-plane (Q' P' R'). W e have y = = sin {J and therefore the change of variable ex = cos {J in the integral ( 1. 56) gives

k iT, + iv - - v -k

-k

+k

( - oo, -k) - t

.;;; 1T,

(k,oo)

-k (ex2 - k2J1/2 -ik f -k {J)eikr ({J- O) I

=

k

a

f(

cos

cos

-k1T

sin {J d{J.

it),

( 1.67)

34

T H E W I E N E R- H O P F T E C H N I Q U E

where the contour 0 must be indented appropriately if there are indentations in the contour in the oc-plane. As an example, ( 1.60) becomes I

,,

,

,

F

" ,

,

,

,

"

\

J eikr

i

C

r

cos

(f3 - 0)

dfJ

=

7riH�l)(kr). 11

G

/"? " \

=

I

----

I

I

.

\ S'

\ \ \

\

-k

E'

. , , ,

\p 0

,

:

,

A

c

D:

+k

, , ,

:

;/ R/______

\ \ \

I!- - _ /

.-

10 a ' :

c'

'

'

:

,

' , :p' 0' . A 0H�----' ', B'f'---� +o:-, JL ,

(]'

:, I , , ,

:'

I

( 1 .68)

,

,

I I I

'y' \ \

,

,�"

F ' R':L _ H '

u- = u + ir

oc

FIG. 1 . 6.

=

-

k cos fJ.

\

0"-� --

f3 = fL + ill

We now apply the method of steepest descent to ( 1. 67) . The idea of this method for approximate evaluation of the contour integral

J

=

J G(fJ) exp {zg(fJ)} dfJ,

( 1. 69)

C

is that we should deform the contour 0 so that it passes through a point, say �, such that the major part of the integral is given by the integration over the part of the deformed contour near �, with G(fJ) slowly varying around fJ = �. We write approximately

J

�

am

J exp {zg(fJ)} dfJ,

( 1 .70)

D

where D is the new contour into which 0 has been deformed : then try to integrate the new contour integral exactly or approximately. It turns out that the criterion which fixes � is dg(fJ) fdfJ O. For (1.67) this gives sin (fJ - 0) = 0, i.e. fJ O. If the real part of g(fJ) i cos (fJ - 0) is plotted in the neighbourhood of fJ = 0 it will be found that there is a "path of steepest descent" . On this path Re g(fJ) is a maximum at 0 = fJ and decreases rapidly on either side of the maximum. A suitable path is shown by the dashed line E'S' P'T' D' in Fig. 1.6 provided that the tangent t o S' P'T' a t P ' makes an angle o f 1350 with the It-axis. Assume then that the contour 0 in ( 1 .67), namely E' B'A' D' in Fig. 1.6 can be deformed into the contour E'S'P' T'D' which we =

=

=

C O M P L E X VA R I A B L E A N D

F O URIER TRAN S F O RM S

35

denote b y D. Contributions from poles can b e included i n the usual way and are ignored here. Then ( 1 . 70) gives I

�

f

kf( -k cos 0) sin 0 eikr cos (f3 - 0) df3 . D

The contour D in this integral can now be deformed back into the original contour a in which case the integral becomes identical with ( 1 . 68) i.e . . I � 1Tkf( -k cos 0) sin OH�l) ( kr ) .

This approximation is valid only when r i s large since i t i s only then that a pronounced saddle point exists. Thus we can introduce the asymptotic expansion of the Hankel function to obtain the final result that as r -l" 00 (2k1T) 1/2e -t;irrf( -k cos 0) sin Or - l /2eikr . I (1.71) �

( A general reference on asymptotic expansions is A . Erdelyi [1]. I t is often convenient to obtain a complete asymptotic expansion for integrals of the above type, in the form of a series, by using the method of steepest descent in conjunction with Watson's lemma (G. N. Watson [ 1 ] , p . 235) cf. ex. 1 . 23 (i) . ) As an example apply ( 1 . 7 1 ) t o ( 1 . 62). Then a s r -.. 00 , for 0 < 0 < 1T - 0 , I

�

sin �2 O sin .12 0 -2(27r/k) 1/ 2e - iirrr - l/2eikr

cos 0 + cos 0

.

( 1 .72)

However it is clear that this result is valid only when (cos 0 + cos 0) is appreciably different from zero. From the point of view of the method of steepest descent the reason for the breakdown of the asymptotic formula ( 1 . 7 1 ) is clear, ' for when (cos 0 + cos 0) is nearly zero, the pole at CI. = k cos 0 in the original integral ( 1 . 62) is very near the saddle point CI. = -k cos O. We mention two methods for dealing with this situation. Suppose that we can write ( 1 . 69) as

J

=

f H(f3)h(f3) exp {zg(f3)} df3,

e

where H(f3) varies slowly near the saddle point f3 = C , and h( f3 ) has a simple pole at f3 = f30 which is the cause of the difficulty. Suppose that

Lr

=

f (f3 - c)rh(f3) exp {zg(f3)} df3

e

can be evaluated exactly. The two methods of procedure are (a) Set

J =

f {H (f3) - H(f3o)}h(f3) exp {zg(f3)} df3 + H( f3o)Lo'

e

36

THE

If {H(fJ)

WIE N E R-H O P F T E CHN I Q U E

- H(fJo )}h(fJ) varies slowly near the saddle point we can write

J = {H( C ) - H(fJ o )}h(�)

I exp {zg(fJ)} dfJ + H (fJo) Lo,

( l .73)

a

where the first integral can now b e evaluated b y steepest descents. (b) Suppose that H(fJ) can be expanded in a Taylor series about fJ = � . Then

J

=

H(�)Lo + H'(�)Ll + . . . .

Under certain circumstances this gives an asymptotic series of functions of z instead of the usual asymptotic series in inverse powers of z. The first method is the one usually used in connexion with the Wiener-Hopf technique (see also F. Oberhettinger, J. Math. PhY8. 34 ( 1 955), 245-255) . An application is given in §5.4. The second method is discussed, for instance, by P. C. Clemmow, Quart. J. Mech. and Applied Math. 3 ( 1 950) , 24 1-256. We conclude this section by quoting certain well-known results concerning asymptotic relations between functions and their Fourier and Laplace transforms. These will be used frequently later. Define �(8)

=

co

f f(x)e - SX dx.

o

Then if, for - 1

< 'YJ <

0,

f(x) ,-,., Axfl, we have

(x � + O ) (x � 00 ) ( IX ->- oo ) ( IX ->- + 0 ) (8 ->- OO ) (8

--->-

0)

} }, }

,

( l . 74)

,

where upper and lower limiting processes go together. IX tends to zero or infinity along pa�hs in the upper half-plane, (1m IX ) > O. 8 tends to zero or infinity along paths in the right half-plane, (Re 8) > O. (Cf. theorems in A. Erdelyi [1], Chap. II, G. N. Watson [ 1], pp. 230, 235, and the Abelian theorems given in most books on the Laplace transform e.g. Doetsch, Van der Pol and Bremmer, Widder. ) 1.7

The Wiener-Hopf procedure

The practical details of applying Fourier transforms in the examples considered later tend to obscure the essential simplicity of the complex variable procedure, which is therefore summarized in

C O M P L E X VA R I A B L E A N D

FOURIER TRAN SFORMS

37

this section. The typical problem obtained by applying Fourier transforms to partial differential equations is the following. Find unknown functions
(1 .75)

where this equation holds in a strip T_ < T < T+ , -00 < (J < 00 of the complex Ot-plane, T_, 'Y_(Ot) is regular in T < T+ , and certain information which will be specified later is available regarding the behaviour of these functions as Ot tends to infinity in appropriate half-planes. The functions A (Ot), B(Ot) , O(Ot) are given functions of Ot, regular in the strip. For simplicity we assume that A, B are also non-zero in the strip. The fundamental step in the Wiener-Hopf procedure for solution of this equation is to find K+ (Ot) regular and non-zero in T > T_, IC(Ot) regular and non-zero in T < T+ , such that

( 1 .76) Sometimes K+ , K_ can be found by inspection but in any case, for the A , B which occur in our applications, they can always be found with the help of theorem C of §1.3. (The precise details will become clear when specific examples are considered later. ) Use (1 .76) to rearrange ( 1 .75) as K+ (Ot)
(1 .77)

Decompose K_(Ot)O(Ot)/B(Ot) in the form

(1 .78) where O+ (Ot) is regular in T > T_, O_(Ot) is regular in T < T+ . In the general case this can be done by using theorem B of §1 .3. With the help of (1 .78) rearrange ( 1 .77) so as to define a function J(Ot) by J(Ot)

=

K+ (Ot)
( 1 .79)

So far this equation defines J(Ot) only in the strip T_ < T < T+ . But the second part of the equation is defined and is regular in T > T_, and the third part is defined and is regular in T < T+ . Hence by analytic continuation we can define J(Ot) over the whole Ot-plane and J(Ot) is regular in the whole Ot-plane. Now suppose that it can be shown that

( 1 .80)

38

T H E W I E N E R - H O P F T E C H NI Q U E

Then by the extended form of Liouville ' s theorem J(oc) is a poly nomial P(oc) of degree less than or equal to the integral part of min (p, q), i.e. K+(oc)
(1 .81 ) K_(oc)'I"_(oc) + O_(oc) = -P(oc). These equations determine
which must be determined otherwise. The reader who desires a concrete example can work through the solution of equation (2.24) in the next chapter, as given in equations (2.28), (2.30). The crucial step is the finding of K+ {oc), K_(oc) to satisfy ( 1 .76). The methods for solution o f partial differential equations described in this book do not all involve an equation of form (1 .75) explicitly. But if we say that a method is "based on the Wiener-Hopf technique" we imply that at some stage of the solution a decomposition of form (1 . 76) is involved. Miscellaneous Examples and Results I

Unless otherwise stated, IX = a + iT, I' = (1X2 - k2 ) 1/2, k

=

k1 + ik2' (kl ' k2 > 0), as in § 1 . 2 . (1X2 - k2 ) 1/2, K1 = (k2 _ 1X2 ) 1/2 , where k = k1 - ik2' (k1 ,k2 > 0),

1 .1 If 1'1 = and the IX-plane is cut by lines going to infinity in the upper and lower half-planes for IX = -:k, + k, respectively, show that if we choose branches such that 1'1 = +ik, K 1 = k, when IX = 0, then 1'1 = + iK 1 everywhere in the cut plane (cf. ( 1 . 14) and Fig. 1 . 1 ) . Show also that if the branches of the four functions ( ±IX ±k) l/2 are defined as in connexion with ( l o U ) then the equations corresponding to ( 1 . 12) are

(k - 1X) l/2

=

-i(1X - k) 1/2

1 .2 Define a complex variable s = a + iT, and let k = k 1 - ik , 2 (kl 'k2 > 0). Cut the s-plane by straight lines from ik to infinity in the right half-plane (a ?- 0) and from -ik to infinity in the left half-plane (a < 0). Let p/2 be the square root which tends to + 14/2 when k 2 tends

to zero. Define

C1 , C2 = + kl/2 when s o. C2 (k - is) 1/2, l /2 C4 = ( -k + is) , C a, C4 = ik1/2 when 8 = 9· i C2, and if we Show that everywhere in the cut plane C a = i C1 , C4 define the branch of (S2 + k2 ) 1/2 so that ( S2 + k2 ) 1/2 = k if s = 0, then

C1 = (k + is) 1/2 C a = ( -k - is) 1/2

=

=

=

( S2 + k2 ) 1/2

=

C1 C2

=

- C a C4 = -iC1 C4 = -i C2 C a .

C O M P L E X VA R I A B L E A N D

F O U R I E R TRANS F O R M S

39

s = i7, and k i s real, then ( 1 71 < k), (k 2 - 72 )1 /2, -i(72 - k2 ) 1/2, ( 1 71 > k).

Show that i f s lies o n the imaginary axis,

( S2 + k2 ) 1/2 (s2 + P ) 1/2 1 .3 Prove (i) I' always has a positive real part when ex lies in the strip -k2 < 7 < k2 : (ii) if (k2 - a2 ) 1/2 = p + iq then Iql :> k 2 , Ip i <: k1· 1 .4 Show that if 1' 2 (ex2 - k 2 ) 1/2 where the ex-plane is cut by a line j oining + k to - k and 1'2 ->- + a as ex a ->- + 00 , then 1'2 ->- - I al as ex = a ->- - 00 . 1 . 5 Choose the branch o f f(ex) = In (ex - c), c a + ib, such that f(ex) ->- In a if a ->- + 00, 7 = O. Cut the ex-plane by the line ex - c r exp (iO), 0 <: r < 00. Show that as a ->- - 00 , 7 0, (i) if 0 < 0 < 1T then f(ex) ->- -i1T + In l al : (ii) if - rr < 0 < 0, then f(ex) irr + In l al . =

=

=

=

=

=

=

->-

1 .6 In connexion with the discussion of integral ( 1 .56) at the beginning of § 1 . 6 suppose that any point on the arc A BO in Fig. 1 . 5 i s given b y ex = R exp (i0), 0 :> 0 > - rr + 0 + A., 0 < 0 < rr. By definition I' ->- ex as ex ->- 00 in this sector so that as R ->- 00

-icxx - I' I YI

->-

=

- R ei 0 (ir cos 0 + r sin 0) Rr sin (0 - 0) - iRr cos ( 0 - 0) .

But the restriction on 0 j ust given ensures that the real part o f this expres sion is negative. Hence on A BO the exponential factor in ( 1 .56) goes to zero exponentially as the radius of the arc tends to infinity. Similarly on DE, but we must remember that if R ->- 00 in A. < 0 < rr + A. then I' R exp i ( 0 - rr ) . ->-

1 . 7 Letf(ex) be an analytic function regular in the strip 7_ < 7 < 7+, the only singularities of f(ex) in the upper half-plane 7 > 7_ being a finite number of poles al ' . . . , an ' Show by inspection that

where f_ (ex) is defined by the series and is regular in 7 < 7+ , and then f+ (ex) is defined by {j(ex) - f_ (ex) } and is regular in 7 > L . Similarly if the only singularity off(ex) in the upper half-plane is a double pole at ex = a, show that

f (ex) -

=

b

-ex - a

c + ( ex _ a ) 2 '

similar decomposition can be applied to meromorphic functions (ox. 1 . 8 ) .

A

1.8

Expansion of meromorphic functions in partial fractions (E. C.

'.l'itchmarsh [2], p. 1 10).

A

function is meromorphic in

a

region if it is

40

T H E W I E N E R-H O P F T E C H N I Q U E

regular in the region except for a finite number of poles. Let j(ex) be a function whose only singularities except possibly at infinity are poles. For simplicity suppose that all the poles are simple. Let them be exl ' ex2, , where 0 < l ex1 1 < l ex2 1 <: . . . , and let the residues at the poles be �, a2, , respectively. Suppose that there exists an increasing sequence of numbers Rm such that Rm --+ 00 as m --+ 00 and such that the circles Om with equations l exl = Rm pass through no pole of j(ex) for any m. Suppose that f(ex) is bounded on Om for all m. Then •

•

•

•

•

•

j(ex) = f(O) +

� an ( ex - exn + .!..exn) 1

_

_

n=l

for all

oc

except the poles.

ex

cosec

-

1

-

oc

As =

an example

L:oo, (-I)n {

n = - co

where the dash means that the term

n

=

I

I}

, ex - n7T + n7T

___

_

0 is omitted.

1 .9 The infinite product theorem (E. C. Titchmarsh [2], p. U3) . If f(ex) is an integral function of ex with simple zeros at exl ' ex2 , , then it can be shown that J'(ex)/ j(ex) is a meromorphic function of ex which •

•

•

can be expanded in partial fractions as in ex. 1 . 8 . On integrating this expansion the following infinite product representation of j(ex) is found

j(ex)

=

j(O) exp {exj' (O)/j(O) }

fl ( :J e<X/a... 1

-

In this form the exponential factors are necessary to ensure convergence, since exn an + b as n --+ 00 in the examples considered below. If j(ex) is an even function of ex, the roots occur in pairs, ± exn , and f'(O) = 0, so that we can write �

j(ex) = j(O)

II'

n = - co

{I - .:::.CXn..} e<X/<X" = j(O) IT { I

_

n=l

where ex_n = - exn and the dash denotes that the. term n As examples of this type we have :

}

2 .:::... , n I'X = 0 is omitted.

cos exa, exna = (n - t)7T ; exna = nrr (=)-1 sin exa, 1 l 2 exna (k a2 - n27T2 ) 1/2; (ya t sinh ya = (Ka)- sin Ka, cos �, exna = {k2a2 - (n - t) 27T2}1/2. cosh ya =

=

1 .1 0 The gamma junction. We frequently require certain well-known properties of the gamma function which it is convenient to state here for reference. r(ex) is an analytic function, regular except when

41

C O M P L E X VA R I A B L E A N D F O U R I E R T R A N S F O R M S

ex

C(

= =

0 , - 1 , - 2, . . . . At these points there are simple poles, the residue at - n being ( - I ) n /(n!). 00

r(ex)

=

I e-tf" - l dt

r(ex + 1 )

o

7Tex/sinh ex

=

r( l - iex) r ( I + iex) { r(exn-l

where 0

r(ex)

=

r(ex)/ r(2ex)

{I i {.:n

n=l In

7T/cosh 7Tex = r(t + iex) r(t - iex) .

00

exeCa II { I + (ex/n) }e -a/n, n=l

0·5772 . . . , the Euler constant. Stirling's formula states that e -aexa -t (27T)1/2{ l + ( I 2ex)-1 + . . . } as ex -+ 00 , l arg exl < 7T. r(ex + b)/ r(ex)

IT

exr(ex).

=

�

n= l

=

2eCa

=

�

__

}

ex + __ e - a/an an + b I_ _ n +z =

r'( I )

}

00 ,

l arg exl

ex __ } e -a/(n - !) , { IT I + (n - 1 n=l

� /22 1 - 2a/r(ex + !)

=

-

exb as ex -+

=

2)

�

2 ! - 2aea+ tex -a , as ex -+

(

e -ca/a r � + I

=

0 + 'P'(z + 1 )

-0

r'(i)

7T.

<

=

a

00.

)/ r (�

+

'F(z)

r'(z)/ r(z).

a

=

)

�+ 1 .

a

_ �/2(0 + 2 In 2).

all these expressions 0 i s Euler's constant 0·5772 . . . .

1 .1 1 The product decomposition of certain functions can be written down directly in terms of gamma functions e.g.

K(C( ) = K+ (ex) K_(ex) = ( 7Tex) coth 7Tex, K +(ex) = 7T1/2r(I - iex)/ r( t - iex)

lC,(ex) is regular and non-zero in (1m ex) ex -+ 00 in an upper half-plane.

>

- ! , and I K+(ex) 1

�

I ex 11 /2

as

1 .12 The following is a more general form of theorem 0 of § 1 . 3, based on the original theorem of N. Wiener and E . Hopf [ 1 ] . (See also K C. Titchmarsh [1], p. 339. ) Let K(ex) be an analytic function, regular in the strip 'T _ < 'T < 'T + such that K (ex) -+ exp (ift) as a -+ + 00 , and 00 in the strip (ft,V real) ; 1 1 - IK(ex)l 1 < K(ex) exp (iv) as a -+ alai -v, p > 0, as l a l -+ 00 in the strip, and the inequality holds uni formly in any interior strip (c,d), where 'T _ < c < 'T < d < 'T +. Then in any such interior strip K(ex) has only a finite number of zeros. If j,hose are exl ' ex2, , exn we can write -

-+

•

•

•

(a)

42

THE

WI ENER-H O P F T E C H N I Q U E

where K+(ex), K_(ex) are regular and free from zeros in 'T :> respectively, and in their respective half-planes of regularity

01 I ex l - 1n - A 

ex � 00 ,

'T <

c,

'T

<:

d

C;

d,

= ( 21T ) 1 ( ,u - v ) . The proof is left to the reader with the hint that one considers the function -

where L <: 'T _, t+ :> 'T +. The factor on the bottom has been inserted to ensure that F(ex) has no zeros in the strip and the powers in n ensure that I F ( ex ) 1 � 1 as l al � 00 in the strip. If the ex-plane is cut from it+ to i 00 and iL to -i 00 and A. is chosen as (21T) 1 ( ,u - v ) then the branches can be arranged so that F(ex) � 1 = exp (iO) as a � ± 00 in the strip (cf. ex. 1 . 5 ) . I n theorem 0 o f § 1 . 3 and above, i t i s assumed that I K(ex) 1 � 1 as a � ± 00 in the strip. If I K(ex) 1 l al p as a � ± 00 in the strip the simplest procedure is to apply the theorems to -

�

where t_ <: 'T _, t+ :> 'T +, and 1] is a suitable constant. If K is even it is convenien� to choose 1] = !p and then I K+(ex) 1 l exl tp as oc � 00 in 'T > 'T _. It is also possible to apply the decomposition theorems to such K(ex) directly but the integrals which occur must be understood . in the sense �

ia + A

lim

A...,. 00

(Of.

H.

f

ia - A

In K(O

� - ex

d� .

Levine and J. Schwinger [1], §V and Appendix B . )

1.13

The generalized Fourier integral. The following theorem over

comes the restriction 'T _ [1], p. 4). Assume that

F +(ex)

=

1

<

'T

+ of theorem A, § I A (E. O. Titchmarsh

00

�

(21T) /2

o

f j(x)e · irxx dx

F _(ex) =

o

: f j(x)eirxx dx,

(2 ) 1/2

- 00

exist, the former for 'T > 'T _, the latter for 'T < 'T + where now the only restriction is that 'T _, 'T + are finite. Then for a > 'T _, b < 'T +, oo + �

j(x) -

1

(21T)l /2

f

- oo + �

oo + �

F + (ex)e -·rxx dex

.

1

+ -(21T)1/2

f

- oo + �

.

F (ex) e -'rxx d ex . -

C O M P L E X VAR I A B L E A N D

FOURIER TRANSFORMS

43

The first integral equalsf(x) for x > 0 and is zero for x < 0 : the second is zero for x > 0 and equals f(x) for x < O. This theorem can be regarded as a superposition of two cases of theorem C, § 1 .4. 1 .1 4 Prove the Mellin transform analogue of theorem C, § 1 . 4 : Suppose that F(s), s = G + iT, i s regular i n G > G _ and s- l F(s) --+- 0 as r --+- 00 where r is defined by s G_ Ii = r exp (iB) , for arbitrary Ii > 0, the limit being approached uniformly in tn- .;;; B .;;; tw . Then the solution of the integral equation -

-

-

1

F(s)

=

f f(p)pB - l dp

o

a + i oo

f( p) =

is

� f

F(s)p - B ds,

2

(a)

a - i oo

Similarly if F(s) is regular at infinity, the solution of

in

F(s) is (a) with 1.15

G

<

G

G+

<

and satisfies suitable conditions

00

=

f f(p)p8 - 1 dp

1

G+.

If

d21ll (y)

�

-

,.,21ll (y) = f(y),

then (i) the solution in ( is!

-

Ill (y) =

00 < y < 00) such that

-

� 2,.,

00

- 00

00

-

III

--+-

� ff(1]) {e - l'l'1 -1I1

2

--+-

f f(1])e - l'llI - '11 d1] :

(ii) solutions in (0 .;;; y < 00 ) such that

Ill (y) =

III

0 as y --+- +

0 as y --+-

00

±

00

are :

± e - l'l'I +!II } d1] ,

o

where the upper sign gives a solution with dlll /dy the lower sign gives III = 0 on y O. =

=

0 on y = 0, and

1 . 1 6 For convenience we occasionally use the Dirac delta-function, l5(x -- g). The fundamental property of this function is

j

tI

f(x) 13 (x _ g) dx =

{

f(g) 0

if

if a,

a

b

g

b,

or

a,

b < g.

44

T H E WIE N E R-H O P F T E C H N I Q U E

As an example we find a solution of

r/>",,,, + r/>w + k2r/> = - 41T!5 (x - xo ) <5 (y - Yo ) in free space, such that r/> represents an outgoing wave at infinity. The expression on the right-hand side represents a line source at (xo ,Yo ) ' A transform in x gives d2(J)/dy2 - y2(J) = - 2(27T)1/2ei"':o <5(y - Yo )' From ex. 1 . 1 5 or from first principles the solution such that (J) ....... 0 as y ....... ± <Xl is (J) = (27T)I/2y-1ei"':o - 1'11I - llol . Inversion and ( 1 . 6 1 ) gives

r/> (x,y) = 7TiH�I) (kR), 1 .1 7 Show that i f i n ex. 1 . 1 6 w e assume that imaginary part then the solution is

r/> (x, y ) 1 .1 8

=

k has a negative

_7TiH�2) (kR) = -?Ti{Jo (kR) - i Yo (kR) }.

Consider, in free space,

tp",,,, + tpw - c- 2tp tt - eC-� t = -4rr8 (X,y,t), where the expression on the right represents a source function. Apply a triple Fourier transform in x, y, t (all limits from - <Xl to + <Xl ), 'Y =

1

__

(27T)3/2

In the usual way we find --

fff tpei(...: +.BII + rut) dx dy dt.

(r:x.,(1,w)e - i(","' + fJlI + rut) dr:x. d(1 dw. . (12 - c- 2 (w2 + �ew) If we wish to evaluate the integral in r:x., the poles are given by r:x. = ±C-1 {(w2 - (12c2 ) + iew}l/2 . In the limit as e tends to zero, if w2 > (12c2 so that these roots lie on the tp

=

47T (27T) 3/2

fff r:x.2S

+

real axis, the contour must be indented as in Fig. l Ab. But if we wish to evaluate the integral in w instead of the integral in r:x., the integrand has poles when

w 2 + iew - c2 (r:x. 2 + P2 ) = 0, i.e. w = -fie ± {C2 (r:x. 2 + (12 ) _ le2}I/2 . For e > 0 both of these lie in the lower half-plane. If e ....... 0 the contour

in the w-plane is along the real axis indented by semi-circles passing

above both the points w = ±c(r:x.2 + (12 ) 1/2 .

1 .1 9 Show that if in ( 1 .56) the integrand contains exp (ir:x.x) instead of exp ( -ir:x.x) then the appropriate equations corresponding to ( 1 . 57 ) , ( 1 . 58) are y = -ik sin (0 + it). Cl. = k cos (0 + it)

C O M P L E X VAR I A B L E A N D

45

F O U R I E R T R A N SF O R M S

1 .20 Show that if k = kl - ik ' k > 0, in ( l . 56) then the equations 2 2 corresponding to ( l . 57), ( l . 58 ) are + ik sin (0 + it) . IX = k cos (0 + it) y I n Fig. l . 5 the cuts will lie in the second and fourth quadrants and the arcs at infinity will be drawn differently. Show that the equation corresponding to ( l . 7 1 ) is I (2krr)1/2e Hrrf (k cos 0) sin Or-1 /2eikr . =

�

The asymptotic behaviours of the Hankel functions as r ->-

1 .21

Ho( l )(kr)

�

/2 (�) 1 ei(kr _ ! rr) rrkr

H0( 2) (kr)

�

Hence H�l ) (kr) exp ( -iwt), H�2) (kr) exp ( + iwt) waves at infinity.

are

2 (�) 1/ e _ i(kr _ trr) . rrkr

represent

outgoing

In order to evaluate ( l . 64) note that

1 .22

Sin !(0 + it) Sin !0 cos (0 + it) + cos 0 so that where H(A)

{

<XI

00

I

1

= -

4

- 00

J

1

= "4 =

1 cos t(it + 0 + 0) cos Wt

- 2iH(0 - 0) + 2iH(0

exp (ikr cosh t) dt . cos ! (�t + A)

Make the substitution 7

=

To evaluate the integral write {eikr cos AH(A) }

=

I

=

0

00

I

o

0 - 0

J

(a)

COS !A cosh it exp (ikr cosh t) dt . cos A + cosh t

sinh !t, cosh t

H(A) = e - ikr cos A cos 1A 2

!

00

+ 0),

�

=

1

+ 272 . This gives

exp { 2ikr(72 + cos2 !A) } . d7 72 + COs2 !A

e3irr/4 cos tA(rrk/2r)1/2 exp (2ikr cos2 !A) ,

where the integral that occurs o n differentiation has been evaluated by the well-known result 00

I eiaT'd7

Integration gives where G(A) F(v)

=

{

=

F{(2kr)1/2 cos tA } ,

f eiu' duo

II

! (rrja ) 1/2eirr/4•

o

- F{ _ (2kr)1/2 cos tA},

00

=

if if

cos tA

cos tA

> 0,

e.g.

< 0,

e.g.

- 1T

rr

< <

A < 1T.

A < 21T.

46

THE W I E N ER-H O P F T E C H N I Q U E

The constants of integration have been arranged so that G r --+ w . Note that F (v) + F( -v) = -r?-/2ei rr/4 .

--+

0 as ( b)

Also 0 .;; () .;; 7T, 0 < 0 < 7T, so that -7T < () - El < 7T in all cases. For (() + El) there are two cases depending on whether (() + 0) is greater or less than 7T. If (() + 0) > 7T, on using (b), it can be shown that

H ((} + 0)

=

_ 7Te - ikr cos (8 H » +

+ -r?-/2e - i rr/4e - ikr cos

(8 H » F{ (2 kr )1/ 2 cos i((} + 0) } .

Equation ( 1 .65) can be obtained from these results on using ( 1 .63) and (a) above. From the standard result F(v) i(2v)-1 exp (iv2 ) as v --+ w we can show that H( J.) !(7T/2kr)1/2 (cos !J.j-1ei(kr + H, as r --+ w . �

�

Then the result ( 1 .72) can be deduced from the exact evaluation ( 1 . 65 ) . 1 .23 Two other methods for deducing the asymptotic behaviour of ( 1 .56) may be useful . We assume that g ( x) Ax2 1' - 1 as x o. (i) In conjunction with the method of steepest descent we can use the result that under certain conditions �

00

I g(x)e -PX' dx

o

�

!A I' C u )p - l' ,

--+

as p

--+

w.

This can be applied to ( 1 .67) if we make the substitution cos (fJ - ()) = 1 + i'T2 . (ii) In conjunction with the method of stationary phase we can use the lemma that under certain conditions 00

I g(x)eiqx' dx

o

�

!A I' (.u)q - l'ei l'rr/2.

This can be applied to ( 1 .59) with the substitution sinh it = 'T. In each case we can verify that the formula ( 1 . 7 1 ) is obtained for the asymptotic expansion of ( 1 . 5 6 ) . (Cf. A. Erdelyi [1], Chap. II, G. N. Watson [ 1 ] , pp. 230, 235, and the Abelian theorems at the end of §1.6). 1 .24

Show that

f

C

f a -a I; da, b

f( rx) drx = ± 7Tif(l;) + rx - I; .

p

f(

)

a

where a, b, I; are real, a < I; < b, and the contour 0 goes from a to b along the real axis and avoids the pole at rx = I; by an indentation below the real axis for the plus sign and above the real axis for the

C O M P L E X VA R I A B L E A N D

FOURIER TRANSFORMS

47

minus sign. The integral on the right-hand side is a Cauchy principal value : P

Jb a -

J(a)

a

c;

da

=

lim

.--> 0

( Jb

<+ .

f(a)

a - c;

da +

"J-'

a

j

f( a) d . a a - c;

The above is a special case of the Plemelj formulae. 1 .25 An ingenious suggestion has been made by W. S. Ament [ 1 ] for factorization of a function K(ex) in the form K+ (ex)K_(ex) . We have seen in § 1 . 3 that if the only singularities of K(ex) are poles the factorization is comparatively easy, but if branch points are present we need to use a general method which may involve complicated integrations. If, how ever, we can invent a transformation of the ex-plane, say ex = X Ul ) such that the transformed function has no branch points in the fJ-plane, then it may be possible to decompose the function in the fJ-plane by elementary methods. Consider an example treated in § 1 .3, K(ex) exp (ya). This was reduced to the decomposition of g(ex) (ex2 - k 2 )-1/2 into the sum of two functions regular in upper and lower half-planes. Assume k real and use the transformation ex -k cos fJ considered in § 1 . 6 in con nexion with ( 1 . 6 7 ) . Then g(ex) = + i(k sin fJ)-l . The transformed plane is shown in Fig. 1 . 6 where the contour is indented below at ex = k, i.e. fJ = 7T, and above at ex = -k, i.e. fJ O. There are now only simple poles at these points and the factorization can be carried out by inspection (cf. ex. 1 . 7 ) : fJ 1 7T - fJ = + sin fJ 7T sin fJ 7T sin fJ • =

=

=

=

If fJ = f-! + iv , the second function on the right-hand side has no singularities in 0 <: f-! <: 7T, v ;;;. 0 above and on the contour. The first function has no singularities in 0 <: f-! <: 7T, v <: 0, below and on the contour. Transform back into the ex-plane.

'1'-1

=

( 7T')' )-1 arc cos ( - exJk) + ( 7T')' )-1 arc cos ( exJk ) ,

which agrees with the result at the end of § 1 . 3 . In some cases it may b e useful t o consider poles i n the whole o f the fJ-plane instead of only the strip into which the ex-plane transforms . W. S. Ament [ 1 ] has indicated in a general way how the method can be applied to K(ex) K(ex)

=

=

{y-1 + a(b2 - ex2 )-1} , (a, b are constants) ; {c(ex2 - k2 ) 1 /2 + (ex2 - l2 ) 1/2}, (c, k, 1 are constants) .

This last case requires a transformation involving elliptic functions.

CHAPT E R II

BAS I C PRO CE D URES : HALF - PLANE PROBLEMS

2.1

Introduction

As already stated at the end of §1 .7, when we say that a method of solution of a partial differential equation is "based on the Wiener Hopf technique" we mean that at some stage of the solution a function K (ex) , ex = (] + iT, is given which is regular and non-zero in a strip T_ < T < T+ , -00 < (] < 00 , of the complex ex-plane. The method of solution requires that this function be decomposed in the form (2.1) where K+ is regular and non-zero in T > T K_ is regular and non-zero in T < T+ . In order to illustrate some basic procedures for using (2 . 1) to solve problems involving partial differential equations, this chapter begins with an extended discussion of the Sommerfeld half-plane diffraction problem. The solution will be obtained by three methods, all depending on the Wiener-Hopf technique. In §2.2 the problem is solved by a straightforward method due to D. S. Jones [3] . In §2.3 the problem is solved by a dual integral equation approach. In §2.4 the problem is formulated as an integral equation which is solved in §2.5. We give pride of place to Jones's methDd because of its simplicity. It provides a routine procedure for problems which can be solved exactly by the Wiener-Hopf technique. An appropriate transform is applied directly to the partial differential equation. No integral equation is formulated. The method will be used extensively in this book for both . exact and approximate solutions of various problems. A dual integral equation method which will also be used later is described in §2.3. It will be particularly useful in connexion with problems involving general boundary conditions, and also in certain approximate methods of solution. Although it depends on an apparently special method of procedure, it will be found that this is no limitation in this book. The method does not seem to have been used previously. _,

48

BASIC PRO C E D UR E S :

HALF -PLAN E PROBLEMS

49

The integral equation method of §§2.4, 2.5 has been used exten sively in the literature. The Wiener-Hopf technique was originally invented to solve an integral equation of the type 00

f f(�)K(x - �) d� = g(x) ,

o

(0 <

I:

< (0) ,

(2.2)

where K and g are given, j is to be found. The prollems encountered in this chapter and the next can be formulated in terms of integral equations of this type and it was in fact for this reason that Schwinger and Copson noted that these problems could "be solved by the Wiener-Hopf technique. Much of the literature iE written in terms of the integral equation approach, the integra� equation being formulated by a Green's function method. We shaII give a detailed description of the formulation and solution of the Sommerfeld half-plane diffraction problem from this point of view which should enable the reader to follow the method as used in the literature. Our treatment is also designed to illustrate the cOllnexion between transform and Gree�'s function methods. Howe'rer, we shall not make much use of the Green's function-integral equation method in this book since Jones's method is equivalent and TIore mechanical. This book could have been written in severa! different ways. We could have regarded the above Green's function.integral equation method as fundamental ; or we could have made much more use of a different dual integral equation approach associated with work of Vajnshtejn, Karp and Clemmow (§4.3) ; or we eould have based the whole discussion on the "Hilbert problem" instead of the Wiener-Hopf technique, which has certain advantages (§4.2) . However the point we wish to emphasize here is 1hat the principal method used in this book is Jones's method descrited in §2.2. Once the reader has mastered this section he should be in a position to read most of the book, providing that the relevant parts of Chapter I have been understood. Now consider the half-plane problem which we are to study in detail. Steady-state waves with time factor exp (-iwt) exist in two-dimensional (x,y) space. There is a rigid boundary along the negative x-axis and waves

CP i

=

exp ( -ikx cos 0

-

iky sin 0) ,

0 < 0 < 1T ,

(2.3)

are incident on the screen (Fig . 2 . 1 ) . Denote f'le total velocity potential at any point by CPt and define cP by the equation (2.4)

50

T H E W I E N E R-H O P F T E C H N I Q U E

Then 4> satisfies

( 2 . 5) where k has a positive imaginary part (§1 .5) . The following conditions apply : =

° on y = 0, - 00 < x � 0, so that o4>/oy = ik sin 0 exp ( - ikx cos 0) , y = 0,

(i) o4>t/oy

x � 0. (2.6) (ii) o4> t/oy and therefore o4>/oy are continuous on y = 0, - 00 < x < 00 . (iii) 4> t and therefore 4> are continuous on y = 0, ° < x < 00 . -00 <

Incident wave

I I

(2)

,

I

I I I I I

I

FIG. 2 . 1 .

In addition to these conditions we need certain assumptions concerning the behaviour of 4> at infinity and near the edge of the screen at the origin. When considering 4> we need to subtract out the incident wave by definition (2.4) . If x = r cos e, y = r sin e then as r tends to infinity the (x,y) plane can be divided into three regions as shown in Fig. 2 . 1 : ( 1 ) Region ( 1 ) in which 4> consists of a diffracted wave and a reflected wave. (2) Region (2) in which 4> consists of a diffracted wave minus an incident wave. . (3) Region (3) in which 4> consists of only a diffracted wave. As r tends to infinity, since the diffracted wave can be regarded as produced by line sources in the screen, the diffracted waves behave, for any fixed e ( - 7T < e < 7T ) , as (ex. 1 .21 ) : lim O lH&l ) (kr) roo.; 0 2r-l/2e + iklTe -k.T ,

r� co

0 1 > 0 2 constants).

B ASI C PRO C E D URE S :

HAL F -PLANE

P R O B L E MS

51

In region ( 1 ) the reflected wave is given by exp ( - ikx cos 0 + iky sin 0) . From these statements we deduce : (iv) For any fixed y, y ;;::: 0 or y :::::;; 0 (a) 1 4> 1

<

03 exp (k 2X cos 0

-

k 2 1 y l sin 0)

for - 00 < x < - I y l cot 0 . 2 2 ( b ) 1 4> 1 < 04 exp { - k 2 (X + y )1/ 2 } for - I y l cot 0 < x < 00 . Finally near the edge of the screen at the origin we assume on y = 0, as (v) (o4>t/oy) ---+ 05X-1/ 2 x ---+ + 0 as x ---+ + O on y = O, 4>t ---+ 06 on y = +0, as x ---+ -0 4> t ---+ 07 on y = -0. x ---+ -0 as 4>t ---+ Os (For convenience we assume more information than is strictly necessary. cf. §2 .6.) In these conditions, the expression x � ·-O, y = + 0, for example, means that x tends to zero through negative values of x on the upper side of the screen. The 0i are constants which need not be known explicitly. It can be considered that the "edge conditions" (v) are given by the physics of the problem. They play an important role in the solution since they are concerned with the uniqueness of the result. A discussion in general terms is given in §2.6. Many methods are available for solution of the problem we have just stated. One virtue of the Wiener-Hopf technique is that the method used to solve this simple problem generalizes immediately to deal with problems involving more complicated conditions on the half-planes, and more complicated equations. (The method also generalizes to deal with parallel half-planes : this is discussed in Chapter III. ) .

Connected with the problem which has just been stated we have the following variants and generalizations. (a) The "absorbent" half-plane with boundary conqition
0 and
X

}

-

52

THE WIEN ER-H OPF T E CHNI Q U E

Another type of generalization occurs when the incident plane wave (2.3) is replaced by a source distribution. For example our original problem would become : Find a solution of the equation - 00

<

x,

y <

00 ,

(2.8)

where o/ oy = ° o n y = 0 , - 00 < x ..;; 0 , and represents an outgoing wave at infinity. (See ex. 2.2.) We shall show that the Sommerfeld problem stated in detail at the beginning of this discussion is equivalent to the following : Find a solution of the two-dimensional steady- state wave equation in y ;> 0, - 00 < x < 00 such that o/ oy

= ik sin 0 exp ( -ikx cos 0),

y

= °

y

= 0, = 0,

and represents an outgoing wave at infinity. Connected with this formulation there are two generalizations : Find a solution of (2.5) in y ;> 0, - 00 < x < 00 under the following boundary conditions. A . No incident wave, but o/ oy

= j(x), = g(x),

y = 0, y

°

- 00

= 0,

x < OO ' < x < 0. <

}

This was considered in § 1 . 1 and will be solved in §2.8. B. Incident wave (2.3) with p( o/ oy) - q r( o/ oy) - 8

=

0,

= 0,

y y

=

0,

= 0,

°

- 00

< <

x x

<

<

OO ,

0.

(2.9)

}

(2.10)

(See ex. 2 . 1 2 (A). cf. §2. 9 . ) Another variant occurs when the equation i s different from the two dimensional steady-state wave equation. Some equations similar to (2.5) are mentioned in ex. 2 . 12(C) (D) (E ) . Diffraction of an electromagnetic wave in three dimensions by a half-plane is considered in ex. 2. 14. Diffraction of a wave by a half-plane in an elastic medium and a viscous medium are mentioned in exs. 2 . 1 5, 2 . 1 6 respectively. All the problems considered in this chapter involve only the half planes - 00 < x < 0, ° < x < 00 , for y = O. Generalizations involving more complicated geometry are considered in Chapter III. 2.2

Jones's method

In this section the Sommerfeld half-plane diffraction problem formulated in §2. 1 is solved by means of a straightforward method due to D. S . Jones [3] . (Our conventions differ from those of Jones since he uses the Laplace transform, and assumes that k has a negative imaginary part. Also in [3] he solves the problem CP t = ° on x > 0, y = 0. But these are details : see exs. 2 . 1 , 2.3.)

B A S I C PRO C E D UR E S :

HAL F - P LAN E

PROBLEMS

53

We introduce the Fourier transforms

oc

- co

«1>_(oc,y) =

1

= (]

+ iT,

o

( 27T) 1/2

f cpe'.Ct.x dx.

- co

(2 . 1 1 )

From condition (iv) in §2 . 1 , for a given y, / cp/ < D 1 exp ( - k 2X) as < D 2 exp (k 2 cos 0x) as x � - 00 where D1, D 2

x � + 00 and / cp/

T

FIG. 2.2.

are constants. Therefore as in deduction ( 1 ) from theorem A, §1.3, «1>+ is analytic for T > -k 2' «1>_ is analytic for T < k 2 cos 0, and «1> is analytic in the strip k 2 < T < k 2 cos 0 (cf. Fig. 2.2) . Also -

co

: J / cpe - TX / dx

/ «1>(oc,y) / � (2 ) 1 /2

- co

=

/ «1> 1 / + / «1> 2 / '

where, from conditions (iv) (a) , (b), §2 . 1 , we can set -

/ «1> 1 / < K 1 / «1> 2 / < K 2

Iv l cot 0

f

- co co

f

- co

exp {( k 2X cos 0 - k 2 / y / sin 0 )

-

T1X} dx,

exp { - k 2 (X2 + y2 ) 1/2 - T 2x} dx .

It is easily proved that / «1> 1 / is bounded as /y/ � 00 if T 1 < k 2 cos 0 and / «1>2/ is bounded if - k 2 < T2 < k 2 . Hence «1>(oc,y) is bounded as /y/ � 00 if T lies in the strip -k 2 < T < k 2 cos 0 .

54

T H E W I E N E R-H O P F T E C H N I Q U E

If now we apply a Fourier transform in x to (2.5) we find, as in §1.5 by the argument following equation ( 1 .55), that 1 d 2 B I > A 2 , B 2 are functions of IX only. There are two forms of

=

=

A 1 (IX)

=

Hence

-

B 2 (IX)

{

,

=

A(IX) ,

say.

(y � 0), (2.14) (y � 0). For brevity we shall sometimes write
A (IX)e - w , -A ( IX)eW ,

=

(2.15)

- co

where in the usual way (lim, y ----+ + 0) means the limit as y tends to zero approached from positive values of y, etc. Similarly for

=

=

(2.16)

We also write mo l

'l' _ (IX,y) From (ii) in §2 . 1 ,
=

=

=

1 (27T) 1 /2

o

f ocpoy

- 00

. d . e"XX x (2. 1 7)

B A S I C PR O CE D U RE S :

H A L F - P L A N E P R O B L E MS

55

On applying these definitions to (2. 14) we find

<1>+(0) + <1>_( +0) = A (oc), <1>+(0) + <1>_( -0) = -A(oc), <1> � (0) + <1> � (0) = - yA ( oc ) .

(2. 1 8a) (2.18b) (2 . 1 8c)

The procedure at this point is of general application and it is important to understand clearly the logic behind it. We wish to deal with equations which contain only functions whose regions of regularity are known. Hence eliminate A (oc) from the above three equations. (2 . 1 8a) and (2 . 1 8b) are obviously analogous. Add these two equations to find (2.19) Next subtract (2.18b) from (2. 1 8a) and eliminate A(oc) between the resulting equation and (2.18c). Then

<1>� (0) + <1> '- (0) = - t y {<1>_ (+ O) - <1>_( -0)}. In this equation <1>� (0) is known. In fact from (i), §2. 1 , <1> '- (0)

=

1 (2 7T ) 1 /2

o

' J e''XX( ik sin 0e- ' x cos . ) dx Ok

e

- co

=

(2 .20)

k sin 0 k cos 0)'

(2 7T ) 1 /2 ( OC

_

(2.2 1 )

For simplicity write

where D_ is short for D_(oc), 3_ for 3_(oc), and the letters D and 3 are used merely to remind us that these are derived from the differences and sum of two functions. Both D_ and 3_ are regular in T < k2 cos 0. Equations ( 2 . 1 9 ) and (2 .20) now become

<1>+(0) k sin 0 <1> ,+ (0) + (2 7T 1 /2 OC k cos 0) ) ( _

=

=

-3_,

(2 .23)

-yD_ .

(2.24)

In these two equations there are four unknown functions <1>+, <1> � , < T < k2 cos 0.

3_, D_ . Each equation holds in the strip -k2

Both o f these equations are in the standard Wiener-Hopf form (cf. § 1 . 7 , ( 1 .75) ) : ( 2 . 25) where R, S, T are given, upper half-plane, T > T_ ,

iD+ , 'Y_ iD_ in a

are unknown, iD+ is regular in an lower half-plane T < T+, and R. S

56

T H E W I E N E R-H O P F T E C H N I Q U E

are regular in the strip T_ < T < T+ . The object of the elimination of A (CI() from ( 2 . 1 8 ) was to obtain equations of this type and it is easy to see that the above procedure is the only correct one. If for example we eliminate (CI() from A (2. 18b) and (2. 1 8c), ( 2 . 26) In this equation 11>'+ (0) and 11>+(0) are both unknown, and both regular in the upper half-plane T > - k2 • Therefore (2. 26) is not of form (2.25). From a slightly different point of view, in (2. 1 8c) 11>� (0) is known and 11>+ (0 ) is unknown. If 11>+(0) is also unknown it is useless to try to obtain an equation involving both 11>+ and 11>'+ as this will certainly be of type ( 2 . 26). Since 11>� (0) is known we shall need at least one equation involving 11>+ (0). If in obtaining such an equation we wish to use other equations which involve 11>+(0), we must first of all eliminate 11>+(0) between these other equations as we did above.

Now return to the solution of the problem. First deal with (2.24) . We have y = (oc2 - k2 ) 1 / 2 = (oc - k) I / 2 (OC + k) I / 2 , where the branch of each factor is defined so that (oc ± k ) 1/ 2 --+ OC1 / 2 as (J --+ + 00 in the strip -k 2 < 'T < k 2 (§1.2). The factor (oc + k) I/2 is regular and non-zero in 'T > -k 2 . Divide (2.24) by (oc + k) I/2 :

k sin 0 <1>+ (0) + (oc + k) I /2 ( 27T ) 1/ 2 (OC + k) I/ 2 (OC - k cos 0)

=

- ( oc

_

k) I / 2 D- . (2.27)

The first term on the left-hand side is regular in 'T > -k 2 : the term on the right-hand side is regular in 'T < k2, cos 0. The remaining term is regular only in the strip -k 2 < 'T < k 2 cos 0 , but we can write it in the following way (cf. ( 1 . 1 8) onwards)

k sin 0 2 1 / + 27T ( ) (OC k ) I/ 2 (OC - k cos 0) k sin 0

{ I

I

= ( 27T) 1 /2 (OC - k cos 0) (oc + k) I / 2 - (k + k cos 0) 1/ 2 k sin 0

}

+

(2.28) where H+ is regular in 'T > -k 2 ' H_ is regular in Insert in (2.27) and rearrange to find

J(oc) = (oc + k)-1 / 2<1> + (0) + H+(oc)

=

'T

<

k 2 cos 0.

-(oc - k) I / 2 D_ - H_(oc). (2.29)

BASIC PR O C E D U R E S :

HALF -PLANE PROBLEMS

57

In this form the equation defines a function J(IX) which is regular in T > -k 2 ' and also regular in T < k 2 cos 0 i.e. in the whole of the lX ·plane, since these two half-planes overlap. Providing that J(IX) has algebraic behaviour as IX tends to infinity we can use the extended form of Liouville's theorem (§1 .2) to determine the exact form of J(IX). Thus we proceed to examine the behaviour of the functions in (2.29) as IX tends to infinity. From (v) in §2.1 and the Abelian theorem at the end of § 1 .6, 1 <1>_( +0) 1 < S I IXI -1 as IX -+ Cl) in T < k 2 cos 0 , 1 <1>� (0) 1 < c 2 1 1X 1 -1/ 2 as IX -+ Cl) In T > -k 2 . Also from definition (2.28), I H_(IX) I < Ca l lXl -1 as IX -+ Cl) I H+ (IX) I < c4 1 1X 1 -1 as IX -+ Cl) Hence from (2.29) IJ(IX) I < c5 1 1X 1 -1/2 as IX -+ Cl) as IX -+ Cl) < c6 1 1X 1 -1

in T < k 2 cos 0, in T > -k 2 .

in T < k 2 cos 0 , In T > -k 2 . i.e. J(IX) is regular in the whole of the IX-plane and tends to zero as IX tends to infinity in any direction. Hence from Liouvill e 's theorem J(IX) must be identically zero, i.e. (2.30a) <1>� (0) = - (IX + k) I/2H + (IX), 2 D_ = -(IX - k) -1/ H_(IX) . (2.30b) To complete the solution note first of all that from (2.18c) , (2.2 1 ) , (2 .28) , (2.30a), 1 k sin 0 . (2.31) IX) A ( = - (2 7T ) 1 / 2 (k + k cos 0) 1 / 2 (1X k) I / 2 (1X _ k cos 0) _

Use (2. 14) and the Fourier inversion formula to find oo + ia e - ilXX 'f YY 1 k cos 0 ) 1/2 �--��--�--�� cp = =F (k 27T (IX - k) I / 2 (1X - k cos 0) dlX , - oo + ia (2.32a) oo + ia (IX + k) I/2e - ilXX 'f YY ocp 1 - = - (k - k cos 0) 1/2 (2.32b) -'----'-o-----:=-- dlX, (IX - k cos 0) oy 27T - oo +ia where -k 2 < a < k 2 cos 0 and the upper sign refers to y ;?: 0, the lower to y ::S; 0. This solution is discussed in §2.6.

J

J

58

THE

W I E N E R- H O P F T E C H N I Q U E

We finish our treatment by Jones's method by noting that the remaining equation, (2.23), can be solved in exactly the same way as (2 .24). The two sides of the equation together define a function which is regular in the whole Ot- plane and from the behaviour of the two sides as Ot tends to infinity this function must be identically zero. Thus $+(0) = 0. Also S_ = ° or from definition (2.22), $_( + 0) = -$_(-0). The result $+(0) ° means that 4> = ° or 4>t = exp ( -ik cos 8) on y = 0, x > 0. Alternatively we see immediately from (2.32a) that 4> = ° on y = 0, x > ° by completing the contour in the lower half-plane. Hence $+ (0) = ° and from (2. 1 8a, b) we then have $_( + 0) -$_ 1 -0). Thus (2.23) is, strictly speaking, superfluous and the complete solution can be obtained from (2.24) . =

=

A

2.3

dual integral equation method

In this section we consider another method of procedure. Apply Fourier transforms to the partial differential equation as in §2 .2 and obtain (2.14) . Invert by the Fourier inversion formula and find A.. �t

= e - ikX COS 8 - ikY Sin 8 ±

I

oo + ia

___

��

- 00

f

+ ia

A(Ot)e 'F yy - iax dOt ,

(2.33)

where -k 2 < a < k 2 cos 8 and the upper sign applies for y ;?: 0, the lower for y � 0. Since 4>t is continuous on y = 0, x > ° we have

:

1/ (2 ) 2

oo +ia

f

- oo + ia

A(Ot) e - iax dOt = 0,

(x >

0).

(2.34)

(Note that the deduction 4>t = exp ( -ikx cos 8) on y 0, x > ° is immediate. In Jones's method this was established either from the final solution or from an analytic continuation argument. Cf. the end of §2.2.) Since o4>t!oy ° on y = 0, x < 0, =

=

1 / (2 7T ) l 2

oo + ia yA (Ot)e - irxx dOt = -ik sin 8e - ikx sin 8 ' - 00 +ia

f

(x

< 0).

(2.35)

Equations (2.34) and (2.35) are dual integral equations from which the unknown function A (Ot) can be found by the following procedure. Replace x by (x + �), x > 0, � > 0, in (2.34), multiply throughout by some function JV 1 (�) to be determined and integrate with respect to � from ° to 00. Assume that orders ofintegration can be interchanged.

B A S I C PR O C E D U R E S :

We find

:

( 2 ) 1 /2 >

00 + ia

HALF -PLA N E

co

J J %l (�)e-i<xo d� A (IX)e - i<x< dlX

- 00 + ia

=

0,

PROBLEMS

x>

0.

59

(2 .36)

0

Similarly in equation (2.35) replace x by (x - �), x < 0, � > 0, multiply through by some function % 2 (�) to be determined, and integrate with respect to � from 0 to infinity. Interchange orders of integration. This gives co

oo + ia

:

( 2 ) 1/2 =

J J %2 (�)ei<xo 0

- oo + ia

-ik sin 0e - ikxcos E>

d�

yA ( IX) e - iax dlX

00

Jo % 2(�)eikO COS E> d�,

(x < 0).

(2 .37)

= N+ (IX).

(2.38)

Introduce functions N+ (IX), N_(IX) defined by

00

J o

% l ( � ) e - i<X< d�

=

_

N (IX)

00

Jo %2 (�)ei<xo

d�

The functions N_ and N + are regular in some lower and upper half planes respectively (Example ( 1 ) following theorem A of §1.3). If we could choose % 1 and % 2 so that

N + (1X)(1X2 - k2 ) 1 /2 = N(IX), say, (2.39) then the left-hand sides of (2.36) and ( 2.37 ) would become identical. N_(IX)

=

N + (lX)y

=

In fact we should have

:

oo + ia

(2 ) 1/2

J i N(IX)A(IX)e - iaX dlX

- 00 + a

(

0

= -ik sin 0e - ikx cos E>

00

f

,

%2 (�) eikx cOS E> d �,

(x > (x <

0),

0).

Then A ( IX ) could be found directly by the Fourier inversion formula. From (2.39) the obvious choice for N + , N_ is N + (IX) (IX + k) -1/ 2 , N_(IX) = (IX - k) 1 /2 , where we have used the crucial Wiener-Hopf decomposition of y (1X2 - k2 ) 1 /2 into functions regular in upper and lower half-planes. The choice of N is not suitable since from its definition (2.38), N_ must tend to zero as IX tends to infinity in a lower half-plane. However we shall see that it is suitable to choose =

=

_

60

T H E W I E N E R- H O P F T E C H N I Q U E

N_ = ( ex - k) -1/ 2 . The reason for this will be clear later (see also the treatment of the general case in §6.2). Theoretically, having chosen N + , N_ we could regard (2.38) as integral equations for .;V 1 > .;V 2 ' which can be solved generally by theorem C, §1 .4. However in this case it is sufficient to use the elementary results of ex. 2.4 and we find ,

.;V 1 (�)

=

:

n-1 /2 �-1 /2eil;k Hrr/4

.;V 2 (�)

=

n-l /2 �-1 /2eil;k - i 7r/ 4 .

(2.40)

Equations (2.36), (2.37) become

oo + ia

1

( 2n ) 1/2

:

( 2 ) 1/2

-

(ex - k ) -1/ 2 f a oo +i

A ( ex ) e - icxx dex

oo +ia

(ex - k ) 1/2 f - oo +ia

A ( ex ) e - icxx

= _n-1 /2ei7r/4 k sin 0e - ikx cos 0

= 0'

(x > 0),

dex

(2.41a )

(2.41b)

00

f �-1/2eil;(k+ k cos 0 ) d�,

(x < 0).

o

The integral on the right can be evaluated (ex. 2.4) . If we multiply (2.41a) by exp (ikx) and differentiate with respect to x, the left-hand sides of the resulting equation and (2.41b) are identical. These steps give

:

(2 ) 1 /2

oo +ia

f

- 00

(ex a +i -

{

-

k )1 /2A ( ex ) e iCXX dex -

0 -

ik sin 0 ( k +

k cos

The Fourier inversion formula then yields

, (x > 0), 0) -1 /2 e - ikxsin 0 , (x < 0). o

ik sin 0 ei(<x - kcos0 ) x dx, A ( ex ) = 1 /2 + cos 0) 1/2 (ex k) 1/2 ) 2 k k ( ( - n - 00 k sin 0 . ( 2 n ) 1/2 ( k + k cos 0)1/2 ( ex - k ) 1/2 ( ex - k cos 0 ) _

f

This is identical with (2.3 1 ) found by Jones's method. One of the pleasing features of this method is that analytic continuation is not involved. The crucial step of decomposing a function in the form K(ex) = K+(ex) K_( ex ) is still present. It is this step which gives the clue to the correct N+ , N_ and hence .;V 1 > .;V 2 ' In the above case .;V 1 > .;V 2 can be found explicitly and the solution

B A S IC P R O C E D UR E S :

HALF-PLAN E PROBLEMS

61

can b e carried through formally with only nominal reference t o the Wiener-Hopf technique. We shall see later that this use of explicit .;V 1 ' .;V 2 and consequent avoidance of the machinery of the Wiener Hopf technique is possible in several important cases (Chapter VI) . The real utility o f the method will appear in connexion with specific applications later. It may seem to the reader that the method of this section does not involve so much analysis of detail as Jones's method. This is partly illusory since we have assumed various results that were proved in §2.2 in connexion with Jones's method, e.g. the analysis leading to (2.33). The edge conditions (v) of §2 . 1 do not occur explicitly in this section but they are assumed implicitly in that we assume that certain orders ofintegration can be interchanged and certain integrals are convergent. Some further comments are given in §2.6 below. 2 .4

Integral equation formulations

In this section we consider the formulation of the Sommerfeld half-plane diffraction problem of §2. 1 in terms of integral equations. This is carried out in more detail than is strictly necessary in order to illustrate some points of general interest. We shall compare Green's function and transform methods of approach and clarify the reasons for the occurrence of divergent integrals in some cases. We first of all consider the Green's function approach and then show that the same results can be obtained by transform methods. A Green's function G(x,y : xo 'yo ) for the two-dimensional steady state wave equation (2.5) is defined as a solution representing the potential at a point (x,y) caused by a line source of unit strength at the point (xo 'Yo ) in a region of any shape with given boundary conditions . Thus G can be found by solving the equation (2.42)

For brevity represent the point (x,y) by r and (xo 'Yo ) by ro . The basic equation used to establish integral equations by a Green's function method is that if cp is a solution of cp",,,, + CP 1/1/ + Pcp = 0 in a region R, then o cp ( r0 ) o G( r l r0 ) 1 S r (2.43) cp ( ro ) = 47T G( r l ro) o n o no cp( o) d o' o

I{

SO

}

where the point r 0 i.e. (xo 'Yo ) lies on the boundary So of the region R, and the integral is taken over this boundary. o lano refers to differen tiation along the outward normal at the point (xo 'Yo ) on So. The region R must be contained in the region D in which the Green's function is defined. R and D are often taken to be the same. The

62

THE WIEN ER-HO P F T E CHNIQ U E

boundary conditions on any finite part of the boundary of D are usually taken as homogeneous conditions of the form ao G/o no + bG = 0 on So. If D extends to infinity the appropriate condition is that G represents an outgoing wave at infinity. For further details the reader is referred to P. M. Morse and H. Feshbach [1], Chapter 7. Now consider formulation of integral equations for the Sommerfeld half-plane problem of §2. 1 : (la) Greens ' function method : unknown function o 4>t/oy h (x) on y 0, x > o. Consider separately the regions y :?: 0, y � o. In y :?: 0 apply (2.43) to the region bounded by the x-axis and a semi circle at infinity. Then =

=

- 00

- 00

where there is an additional minus sign since % no = -% Yo ' and the contribution from a semi-circle at infinity is zero if G represents outgoing waves at infinity. We wish to formulate an integral equation in terms of o 4>t! oy. (2.44) involves the two unknown functions o 4>/oyo and 4>. The term involving 4> will disappear if we choose a G such that o G/oyo 0 on Yo o. A suitable choice for G is obviously (cf. ex. 1 . 1 6 and (2.42)) (2.45) G( x , y : xo ' yo) = 7ri {Hb1 ) (kR) + Hb1 ) (kR')}, where R = {(x - XO ) 2 + ( y - YO ) 2 }1/2 : R' {( x - XO ) 2 + ( y + YO ) 2 }1/2 . =

=

=

We cannot apply (2.44) to 4>t since this contains a wave incident from infinity. But define 4> by 4>t ( x , y ) = e - ikxcos 8 - ikysin8 + e - ikxcos8 +ikysin8 + 4> , (2.46) where we have removed both the incident and reflected waves. The object of this is to make o 4>/oy = o 4>t/oy on y = O. Then (2.44), (2.45) and (2.46) give for y :?: 0 4>t ( x , y) = e - ikxcos 8 - ikl/sin8 + +

where

co

e - ikxcos 8 +ikysin8 - !i J Hbl) (kRo)h(�) d�, o

(2.47)

BASIC PRO C E D URE S :

HAL F - PLAN E

PRO BLEMS

63

Green's theorem can be applied directly to CPt in y � o. In this case, on y 0, % no + % Yo, and we find for y � 0, =

=

CPt( x,y)

co

=

ti J H&l)(kRo )h( � ) d � .

(2 .48)

o

But CPt is continuous on y = 0, x > 0 and on letting y � 0 in (2.47), (2.48) and equating the results on y 0, x > 0 we obtain the integral equation =

i

co

J H&l)(k /x - �/ )h( �) d�

o

=

2e -ikx cos 8 .

( 2 .49)

( lb) Green ' s function method : unknown function 2 e (x) = CPt(x, +O) - CP t(x , -O) for x < o. Adopt the procedure in (la) but now eliminate o cp/oYo instead of cP from (2.44) by choosing G so that G 0 on Yo 0 i.e. choose (cf. 2.45) =

=

G(x,y : xo 'Yo )

=

1Tj {H&l)(k R ) - H&l)(kR')}.

Write (cf. 2.46)

CP t(x,y)

e - ikx cos 8 - iky sin 8 - e - ikx cos 8 +ikY Bin 8

=

+

(2 .50)

cP, (y � O), (2.51 )

where we have removed the incident wave and inserted an extra term so that CPt = cP on y = o. Apply Green's theorem (2.44) to cP in y � 0 and to CP t in y � O. This gives

CPt

=

+

+

e - ikx cos 8 - iky sin 8 - e - ikx cos 8 +iky sin 8 o

L f (�:) - co

CPt = -

0

f (xo ' + O) dxo +

co

L f (�:) f (xo ) dxo ' (y 0

0

(2.52a)

co

o

� 0).

L f (�:) f (xo , - O) dxo - L f (�:) / (xo ) dxo , (y � 0) , 0

- co

0

(2.52b)

where oG/oyo is to be taken on Yo = 0, and we have denoted the cP on y = 0, x > 0 by ( x), and on y ±O, x < 0 by value of CPt f f (x,±O) respectively. We now use the fact that o CP t/oy 0 on y 0, x < 0 to obtain an integral equation. Differentiate the above equations with respect to y and let y tend to +0 and -0 respectively, x < o. We have =

=

=

oG yo-> + o oYo lim

-

=

oG oyo

lim- 0 -

yo->

=

0 {H&l)(kR)} . lIo->o oyo

2 1Ti lim

-

=

64

T H E W I E NE R - H O P F T E C H N I Q U E

Eliminate terms involving f(x) by adding (2 .52) and introduce

e (x) defined by Then lim i

Y,Yo""'O

f(x, +O) - f(x, - O)

o

l(kR )}e (xo) dxo f� oyoYo {Hb1

- 00

=

=

2 e (x) .

(2.53)

2 ik sin 0e - ikx cos 8 ,

x < O.

(2.54) We cannot go to the limit explicitly since the differentiations under the integral sign give a kernel such that the integral would be divergent. We have

02 02 Hbl ) (kR) = -lim 3"2 Hb1 ) (kRo) y Yo y"",o u y Y,Yo""'O u u lim

�

=

(:;2 + p) Hb1) (k l x - xo J ) ,

where R� = (x - XO ) 2 + y2 . Hence (2.54) can also be written

02 -lim i 2 oy y O

.....

or

i

o

f Hb1)[k{(x

- 00

�

=

o

�) 2 + y2F/2] e (�) d� 2 ik sin 0e - ikx cos 8 ,

(:;2 + k2 ) f Hb1) (k l x - � J )e(�) d� - 00

=

2ik sin 0e - ikx cOB 8 ,

(x < 0) ; (2.55)

(x < 0). (2.56)

We shall see in §2.5 that either of these forms can be solved by the Wiener-Hopf technique. If we interchange orders of differentiation and integration in (2.56) the resulting integral is divergent. The divergency obviously comes from the region x === � where

H� ) (k l x - � J ) === O ln I x - �I : 0 2jox2 {Hb1 ) (k l x - � I ) } === -O(x - �)-2 . Thus the divergent integral is of the type o

f (x - �)-2F(�) d�,

- 00

( - 00

< x < O).

This is exactly the kind of integral involved in Hadamard's theory of the finite part of an infinite integral invented in connexion with hyperbolic equations. However it is not necessary to invoke any such theory here since it is possible to apply Fourier transforms

B A S I C P R O C E D UR E S :

HAL F - PLANE P R O B L E M S

65

directly to integral equations of type (2.55), (2.56), as we see in the next section. In ( Ia), (I b) different but equivalent integral equations have been obtained, one from a Green's function such that G = 0 on finite boundaries, the other such that o G/ on = 0 on boundaries. There are an infinite number of intermediate integral equations which can be obtained by using a radiation condition poG/o n + qG = O. It will appear that although the Green's function is to some extent arbitrary and the integral equations may seem different, all integral equations lead to exactly the same Wiener-Hopf problem . The function K(oc) which has to be decomposed in the form K + (oc) K_(oc) is the same in all cases. We next show that results equivalent to the above can be obtained by transform methods. (2a) and (2b) correspond to ( la) and ( lb) above. ( 2a ) . Transform method : unknown function otl oy = h(x) on y = 0, x > o. As in ( la) introduce defined by (2.46). A Fourier transform can be applied to and we find in the usual way
oI oy where

=

y

0,

h(x)

=

0,

x <

oI oy =

0

is unknown. Hence

A (Ol)

=

-

1

-

h(x), y

=

0,

x >

0,

00

__

(21T)1/2y

f h(;)eirxe d; .

o

Apply the Fourier inversion formula and find t

=

e - ikx cos 0 - iky sin 0 + e - ikx cos 0 +iky sin 0 00 00 +i b 1 irxx w e h(;) eirx< d; dOl, y- - 2 1T 0 - 00 +ib

I f

f

( y ;;;. 0).

(2.57)

An analysis of the behaviour of t as x ->- ± 00 shows that this is valid for k cos 0 0 and by letting y ->- 0 in (2.57) (2.58), and subtracting the two equations find the following integral equation for h(;) :

oo +ib

� f

- oo + ib

00

f

y-1e - irxx h(;) eirxo d; dOl 0

=

2e -ikx cos 0 ,

(x > 0) .

(2.59)

66

THE

W I E N E R- H O P F T E O H N I Q U E

In this equation it is permissible to interchange orders of integration. If we use the result ( 1 . 60), ( 1 . 6 1 ) namely

: J

co + ib

y-1 eia( " - x) d rJ. = iH�l) (kI X - � I ) ,

- co + ib

then (2.59) reduces to (2.49) found by a Green's function method. ( 2b ) . Transform method : un known function CPt(x, + O) - CP t (x, -0) on y = 0, x < o . As in ( 2b) introduce cP defined by ( 2 . 5 1 ) . Fourier trans forms can be applied to cP and we find A (rJ.) exp ( -yy), =

o

A ( rJ. )

=

: J

( 2 ) 1 /2

: J

co

f(�, + O)ei<x" d� +

/ ( 2 )1 2

- co

f( � )ei<x" d�,

0

where cP = f (x, + O ) on y = + 0, x < 0, cP = f (x) on y = 0, x > 0 defined in connexion with ( 2. 52). Hence CP t = e -ikx cos e - iky sin e -e - ikx cos e

L J

co + �

+

e - iexx - yy

- co + �

0

{J

+

iky sin e

+

J

f(�, + O)ei<x< d� +

0

J

co + ib

e - iexx - )y

{J

0

J

in

y

<:

O. This gives

f(�, _ O)ei<xo d� +

0

(2.61)

}

co

- co

- co + ;b

(y ;;, 0).

f(�)ei<xo d� drJ.,

A Fourier transform can b e applied directly t o CP t 1 CP t = 27T

}

co

- co

as

(y

f(�)ei<xo d� drJ.,

<

0) .

(2. 62)

Apply the condition ocpt/ oy = 0 on y = 0, x < 0 to (2.6 1 ) , ( 2.62), and eliminate f(�), � > 0 by adding the resulting equations. We find, using (2.53),

: J

co + ib

- co + ib

ye - iexx

0

J

- co

e(�)ei<X" d� drJ.

�

_

2ik sin

El e - ikx co s

e

,

(x

<

0).

(2.63)

This is an integral equation for e(�). It is important to note that we cannot now interchange orders of integration since the inner integral would then diverge (cf. (2.59 ) ) . This integral equation can be solved

:B A S I C P R O C E D U R E S :

HALF - PLAN E

PROBLEMS

67

directly b y the Wiener-Hopf technique (§2.5 below) o r we can convert into (2. 55) or (2.56) as follows. Write (2.63) as co + ib

..... - - f

1 lim y O iJ y 2 1T iJ2

- lo + ib

or

(

y-le - ixx - yy

iJx2 +

7c

2

) f

+ ib

- co

8e - ikx cos 0 ,

sin

0

( x < 0) ,

(2.64)

(x < 0) .

(2.65)

. f e(;)e'''' d; dr:t.

y-1e - ,C<X

1

:;;

f e(;)ei"� d; dr:t.

= _ 2 ik <Xl

iJ 2

0

- oo + ib

.

- 00 =

<

2i7c sin

8e - ikx cos 0 ,

We can now interchange orders of integration and evaluate the inner integral by ( 1 . 60), ( 1 . 6 1 ) (cf. 2.60) to find (2.55), ( 2 . 5 6 ) . Further comments o n the formulation o f problems in terms of integral equations are given in exs. 2.6, 2.7, 2.8. 2.5

Solution o f the integral equations

In §2 .4 we have obtained integral equations of several different types which we analyse in turn. A fundamental result we shall need is the Fourier inverse of (1 .60) , ( 1 .6 1 ) :

li

00

J H&l l {k(X2 + y2) 1/2}eiC<X dx

- 00

=

y-le -Y I Y I ,

First of all consider the integral equation (cf. 2 .49) co

J k(x - $)h(�) d� = f(x) ,

o

where k, f are given, co

g

(0 � x < (0 ) ,

(2.67)

is unknown. Extend this equation by writing

J k(x - �)h(�) d �

e (x),

=

o

( -00

<

x

< 0),

(2 .68)

where e(x) is an unknown function defined by this equation. Apply a Fourier transform to (2.67 ) , (2.68 ) : multiply throughout by ( 2n ) -1/ 2 exp (iIXx) and integrate with respect to x from -00 to 00 . Then 00

co

: f eic<x f k(x - �)h(�) d� dx

F+ (IX) + E_(IX) = 2 1/2 ( ) =

- co co

0

co

: f h(�) f k(x - �)ei"x dx d�,

( 2 ) 1/2

o

- 00

.

68

THE

where

F+(oc)

W I E N E R-H O P F T E C H N I Q U E

co

=

f .

: f e (x)eic<x dx.

1 f(x)e'rxx dx ( 2 7T) 1 / 2

E_(oc) = ( 2 ) 1/2

o

, 0

- co

( 2.69)

In the inner integral change variable to y = (x - �). The double integral separates into two independent integrals and we find

(2.70)

where

H+ (oc)

=

co

: f

( 2 ) 1/2

o

co

h(�) eic
K(oc) =

f k(y)eic
- co

(2.71)

Note that K(oc) has been defined in such a way that there is no factor (27T) -1/2 in front of the integral, in order to simplify the form of ( 2.70) . This equation is of the Wiener-Hopf type (2.25) if it holds in a strip T_ < T < T+ of the oc-plane. In the concrete example we are considering, namely ( 2.49),

F+ (oc)

=

=

K(oc)

=

2 _ _ ( 27T) 1/2

co

f eix(c< - k

o

cos El )

dx

2 i . , (27T) 1/2 (oc - k cos 0) i

co

I Hb1) (klxi l eic<X dx

- co

=

( 2.72a)

(T > k 2 cos 0),

2 y-1 ,

Hence (2.70) becomes

2 ( 27T ) 1/2 ' (oc

i 2 = + ( 2.73 ) H (oc). k cos 0) E_ (oc) Y + This equation corresponds to ( 2.24 ) in Jones ' s method. By definition h(x) = ocp tloy on y = 0, x > o. Since I cptl < exp (k2 x cos 0) as x� + 00 , H+ is regular in T > k 2 cos 0. Also, as x � - oo e (x)

=

_

co

co

i f HiP(kl x - �i l h(�) d � ,...., 0 1 f (� - x) -1/2eik( g -X)h(�) d�; o

l e (x) 1

co

<

0

0 2ek2X f e - k2gh (�) d � , o

(x �

-

(0 )

.

( 2.74)

i.e. E_( oc ) is regular in T < k2• On using these results and inspecting ( 2.73 ) we see that this equation holds in k2 cos 0 < T < k 2 . Deal

BASIC PROCEDURE S :

HALF -PLAN E PROBLEMS

69

with (2.73) as with (2 .24) in Jones ' s method. Multiply by (oc - k) I/ 2 and rewrite as P(oc)

i k) I/2 - (k cos 0 - k) I/2 } + E_(oc) (oc k cos 0 { 2 i(k cos 0 - k) I/2 2 (2.75) - (oc + k) I/2 H+ (oc) - (2 1T ) 1 / 2 ' (oc k cos 0) 2

. ( 21T ) 1 / 2 oc

=

_

_

This defines a function which is regular in the whole of the oc-plane. From the physics of the problem it is known that h(x) ,..."" X-1/2 as x � +0. Hence H+(oc) ,..."" oc-1/2 as oc � 00 in an upper half-plane. From the definition (2.74), e (x) """" 0 as x � -0, where 0 is some constant, and so E_ (oc) "" oc-1 as oc � 00 in a lower half-plane. Hence from (2.75), P(oc) � ° as oc � 00 in any direction in the oc-plane and from Liouville ' s theorem P(oc) is identically zero. Thus

H+(oc)

1

=

( 21T ) 1/2

.

i(k cos 0 - k) I/2 (0I: + k) I/2 , (k 2 cos 0 < T < k 2 ) · (oc k cos 0) _

(2.76)

To obtain the potential at any point of space, return to (2.44) , (2.45) , (2.46) :

CPt

=

=f

q (x,y)

00

ti I Hb1 l[k {(x - '; ) 2 + y2} 1/2]h( '; ) d '; , o

(2 .77)

where the upper sign applies for y � 0, the lower for y � 0, and

q (x,y)

=

(

e - ikX COS 0 - ikY Sin 0 + e - ikx cos 0 +iky sin 0 , ° ,

(y � 0), (y � 0).

To complete the solution substitute the following expression in (2.77) :

h( '; )

=

oo +ib

: J H+ (oc)e-irx." doc, +ib

( 2 ) 1/2

- 00

where this integral is automatically zero for � < 0. The limits of integration in (2.77) can now be extended to ( - 00 , 00 ) . Interchange orders of integration and change variable in the inner integral by setting y = (x - '; ) . Evaluate the inner integral by (2.66) . We find

CP t = q (x,y)

=f

:

oo + ib

( 2 ) 1/2

- co

J y-le -irxx =f YVH+ (oc) doc. + ib

( Alternatively this could be obtained by applying a convolution

70

TH E W I E N E R-H O P F T E C H N I Q U E

theorem to (2 .77 ) . ) Insert H+ (oc) from (2 .7 6) and use the result ( 1 .l3a ) i.e. (k cos 0 - k)1/2 = -i(k - k cos 0) 1 / 2 . This gives

q(x,y)

00

-x

+ ib

exp ( ioc =f yy) d oc. (oc - k) 1 /2 (OC - k cos 0 ) - 00 + ib (2 .7 8) Shift the contour parallel to itself to a line T = a where -k 2 < a < k 2 cos 0. We cross a pole at oc = k cos 0 and the contribution from this pole produces a term -exp (-ik cos 0 + iky sin 0) in y � 0, and + exp (-ik cos 0 - iky sin 0) in y � o. The solution is then identical with that found by Jones ' s method i.e. CP t = cP ; + cP where CP i = exp (-ik cos 0 - iky sin 0), the incident wave, and cP is given by ( 2. 32 a ) .

CP t =

=f

1

1/2 21T (k - k cos 0)

f

x

x x

Further details regarding solution of this type of integral equation be found in E. T. Copson [2], B. B. Baker and E. T. Copson [ 1 J . We treat the other integral equations of §2.4 in less detail. Consider next (cf. (2.56)) will

( 02 + k2 f k(x - �)eW d� - {g(x), (x 0), (2 .79) ox2 ) h(x), (x > 0), where the integral equation is given in the first line, e(�) being unknown, and h(x) is defined by the second line as the value of the left-hand side when x > Apply a Fourier transform in x: multiply both sides by exp (iocx) and integrate with respect to x in - 00 , 00 On the left-hand side the double differentiation with respect to x can be dealt with by integration by parts. Interchange orders of integra o

<

_

- 00

o.

(

)

.

tion and change variable as before. These steps give

G_(oc) + H+(oc) = (k2 - oc2 )E_(oc)K(oc) . In the particular case (2.56) 2ik sin 0 G_(oc) - 21T ) 1 /a ( _

fe o

- 00

ix(ex - k cos 0 )

and K(oc) is given by (2.72b). Thus

(2.80)

2k sin 0 dx - 21T 1 / 2 ( ) (OC k cos 0) _

_

,

k sin 0(21T)-1 /2 (OC - k cos 0) -1 + !H+ (oc) = - (oc2 - k2 ) 1 /2E_(oc). ( 2.81 ) A detailed analysis as before shows that this equation holds in -k 2 < T < k 2 cos 0. Apart from this, the equation is almost the same as (2.24 ) obtained in Jones ' s method, since E_(oc) is the same as

BASIC PROCEDURES :

HALF -PLANE

PROBLEMS

71

D_(rx) and h(x) is proportional to orfo/0'Y on 'Y O . (See the derivation of the integral equations in §2.4.) There is therefore no need for further analysis of this equation. We have incidentally illustrated a point that we mention again later. In Jones's method the unknown functions (e.g. H+ (rx) and E_(rx) in (2.81 )) are treated symmetrically, and their physical significance is evident by definition. In the integral equation the symmetry is not quite so self-evident though it appears when transforms are applied. Finally consider (2.55), =

02 - lim � y� O 'Y

o

J k{(x - �) : 'Y}e (�) d�

- 00

=

g(x),

(x

<:

0),

(2.82)

where we have written the kernel as k { (x - �) : 'Y} to emphasize that this is given as a function of 'Y although we shall eventually let 'Y tend to zero. As before extend this to general x by defining the left-hand side for x > 0 as a function h(x) . Apply a Fourier transform as before and assume that orders of the limiting process, differentiation, and integration, can be interchanged. This can be justified by uniform convergence. We find, using an obvious notation, (2.83) In the cases in which we are interested it will be found that we can now perform the differentiation and go to the limit ; e.g. in (2 .55) we find

Hence (2.83) is identical with (2.80 ) . We have deliberately avoided solving the integral equations (2.59), (2.63) obtained by transform methods, which are equivalent to the integral equations solved above. It will be clear to the reader that if we apply Fourier transforms to those equations we shall obtain results equivalent to those obtained above. But if we do this we are merely "unwinding" the process by which the integral equations were obtained. There is no reason for obtaining integral equations by transforms and then applying transforms in reverse to undo the process. This is the point of Jones ' s method, that the Wiener-Hopf equations are obtained by working directly with functions obtained by applying transforms to the partial differential equation, without the intermediate step of formulating integral equations.

72

THE

2.6

WIEN ER-H O P F T E CHNI Q U E

Discussion of the solution

With some care it may be possible to provide a rigorous justifi cation of the methods described in the preceding sections. We should then know that the solution obtained is the only solution satisfying the conditions of the problem. (The original work of N. Wiener and E. Hopf on the solution of integral equations was of this type. ) When dealing with partial differential equations it may be easier to obtain a solution by a more or less formal procedure. It is then necessary to verify that the formal solution satisfies the conditions of the problem. If we know that the problem has a unique 8olution we can conclude that our solution is correct. The solution obtained by the Wiener-Hopf technique will usually be in the form of a contour integral which cannot be evaluated explicitly. In order to show how a result of this type can be verified consider the contour integral solution of the Sommerfeld problem (cf. (2.32a)) :

CP t

=

e - ikz cos El - iky sin El ± 1

=f 217

«> + ia

(k - k cos 0) 1 /2

J

- oo +ia

(oc

_

e - ilXZ - y l lll doc , k) 1 / 2 (OC k cos 0) _

(2.84a)

where -k 2 < a < k 2 cos 0. First of all we prove that this solution satisfies the steady-state wave equation. We use standard theorems dealing with the uniform convergence of infinite integrals, and the conditions under which such integrals can be differentiated under the integral sign (e.g. H. S. Carslaw, The theory of Fourier integrals and series, Macmillan (1930), pp. 196-201 ) . For I y l > 0 the integrand in (2.84a) decreases exponentially as Re oc -- ±oo in the strip -k 2 < 'T < k 2 cos 0. The integrals obtained by applying the operators o2 /ox2 , 02/oy2 under the integral sign converge uniformly for I Y I > o. Hence when forming second derivatives of the integral it is permissible to interchange orders of differentiation and integration for I y l > 0, -00 < x < 00 . Similarly if the oc-plane is cut by lines parallel to the y-axis from k k 1 + ik 2 to k 1 + i oo and -k to -k 1 -ioo , and the contour is deformed on to the two sides of the appropriate cut depending on whether x > 0 or x < 0, it will be found that it is permissible to differentiate under the integral sign for Ixl > 0, 0 � y < 00 , -00 < y � O. We can also show that for x > 0, 02cp/OX2 , 02cp/oy2 are continuous across y = o. (In fact for this special case the integral and its even order deriva tives are identically zero on y = 0.) As a consequence of these results it is easily verified that (2.84a) satisfies the steady-state wave =

BASIC PROCEDURES :

equation in the whole of the x-y plane cut along the line y - 00 < x � O. Consider next

ocfo 1 - = - (k - k cos 0) 1 / 2 oy 21T

73

H A L F - P L A NE P R O BL E M S

f

CXl + ia

- CXl + ia

(IX + k) 1/2e - iCXX - y lvl dlX ' IX - k cos 0

=

0,

(2.84b)

obtained by differentiating the integral in (2.84a) . The integral is uniformly convergent for all x, y such that (X2 + y2 ) 1/2 ;;? e where e is arbitrarily small. For y = 0, x < ° we can complete the contour by an infinite semi-circle in the upper half-plane. The only singularity of the integral is a simple pole at IX = k cos 0 and this gives exactly (2.6) in condition (i) of §2. 1 . Similarly we can verify that conditions (ii) and (iii) of §2 .1 are satisfied. From theorem B, §1 .4, since the integrand of the integral in (2.84a) is regular in -k 2 < T < k 2 cos 0 we have that I cfo l < exp (-k 2 + c5)x as x ---+ + 00 , and I cfo l < exp (k 2 cos 0 - c5)x as x ---+ - 00 for arbitrarily small c5 (cf. condition (iv), §2 . 1 ) . Finally consider condition (v), §2 . 1 . I f y = 0, x > ° in (2.84b) the contour can be deformed in the lower half-plane. Cut the IX-plane by a line parallel to the imaginary axis from -k = -k 1 - ik 2 to -k 1 - ioo. (2.84b) becomes

ocfo 1 1 / 2eikx -irr/4 - = - (k - k cos 0) oy 1T

f

CXl

o

U

u1 / 2e -ux du . . - �(k + k cos 0)

OX-1/ 2 as x ---+ +0. The other conditions

Hence from ( 1 .74), ocf% y in (v) §2 . 1 can be verified in a similar way. (A detailed verification of a formal solution obtained by a Wiener Hopf technique is given in J. Bazer and S. N. Karp [1].) r-..J

For completeness w e note that (2.84a) can b e evaluated b y results given in § 1 . 6 to give a well-known expression in terms of Fresnel functions. If we compare (2. 84a) with ( 1 .62), remembering that (k - k cos 0) 1/2 = i(k cos 0 - k) 1/2, we find rpt e - ikx cos 8 -iky sin 8 =f i(27T)-lI, ( 2.85) where I is given in terms of Fresnel integrals by ( 1 .65), and the minus sign holds for y :> 0, the plus for y <: o. On using the result F ( v ) + P ( ) = �/2 exp (i7Tj4) we find for y :> 0 i.e. 0 <: () <: 7T, rpt 7T -1 /2e - i rr/4 [e -ikr cos ( O - 8 )P{ - (2kr) 1/2 cos ! (() - 0)} + + e -ikr cos (0 + 8 )P{ (2kr) 1 /2 cos ! (() + 0)}], (2.86) where P ( v ) is the Fresnel integral defined by ( 1 .66). For y <: 0 we must remember that (2. 84a) and ( 1 . 64) are expressed in terms of I Y I s o that if () is used to denote the actual angle that the radius vector makes with the line y 0, x > 0, () being negative for y < 0, then () in ( 1 .65) must =

-

v

=

=

74

T H E W I ENE R-H O P F T E C H N I Q U E

be replaced by ( - 0) if Y < O. We must also take the lower sign in (2.85) and it is found that the combination of these two factors means that rfot is again given by (2. 86) for y <: 0 i.e. -7r <: 0 <: O. The total field in the whole space is given by (2.86) provided that we take -7r <: 0 <: 7r.

The edge condition (v) of §2 .1 is the condition which ensures uniqueness of the solution obtained by the Wiener-Hopf technique in the problem we have just considered. We wish to solve (2 .29) , namely J ( oc ) = (oc + k) -1/2 <1>� (0) + H+ (oc) = -(oc - k) I/2 D_ - H_(oc), (2 .87) where both H+ and H_ ,....." oc l as oc ---+ 00 in appropriate half-planes. Instead of assuming (v) in §2.1 suppose quite generally that ocf% y ,....." xl' as x ---+ + 0 and cfo ,....." XV as x ---+ -0 on y O. Then (oc - k) I/2 D_ ,......., OC - (1/2) - v (oc + k) -1/2<1>� (0) ,....." OC - ( 3/2) - 1' as oc ---+ 00 in appropriate half-planes. By the extended form of Liouville ' s theorem in §1 .2 we deduce from (2 .87) that J(oc) = Pn ( oc ) where Pn ( oc) is a polynomial of degree less than or equal to n where n is the integral part of the lesser of ( ! - fl), ( - t - 'V), providing that this quantity is greater than or equal to zero. (Po(oc) = 0 = constant.) Thus to prove J (oc) = 0 it is sufficient to assume either 'V finite and fl > --J i.e. ocf% y < 0 lx - ( 3/2) + e as x ---+ 0 on y = 0 or fl finite, 'V > - t i.e. cfo < 02X- (1 /2 ) + e as x ---+ -0 on y = ±O (8 is an arbitrarily small positive quantity) . We have in fact assumed fl = -t, 'V = 0 i.e. we have assumed conditions on both cfo and ocf% y which are amply sufficient to prove J ( oc) = O. Also we assumed cfo ,....." 0 as x ---+ -0 (since cfot ,....." 0, cfoi """" 0) whereas cfo ,....." XI/2 as we can prove from the final solution. This is not important so long as we do not relax the condition too much as discussed in the next paragraph. It would be possible to provide a more rigorous procedure by working in terms of inequalities e.g. to write I cfo l < 0 instead of cfo ,....." O. But this is unnecessary here. If we relax the conditions on cfo and ocf% y we no longer obtain a unique solution for ·the Sommerfeld problem. Thus suppose we assume ocf% y ,....." 0 IX-3/2 as x ---+ + 0 on y = 0, and cfo ,......., 0 2X-1 / 2 as x ---+ -0 on y = ±O. Then Liouville ' s theorem gives J ( oc ) = 0 where 0 is a constant. As in the analysis leading to (2.32a) we find =

-

<1> + (0) = -(oc + k) I/2H+(oc) + O(oc + k) I / 2 , o cfo = cfo o =f 2 1 2 ( 1T ) /

f

co +ia

- co +ia

exp (-iocx - yly l )

(oc

_

k) I/2

doc,

BASIC PROCEDURES :

H AL F - P L ANE P R O B L E M S

75

where CPo is given by (2.32a) and upper and lower signs refer to � 0, y � ° respectively. On y = 0, using ex. 2 .4, we find

y

cP

=

{ CPo =F Oi7T-1/2 (

_ x ) -1/2 eikx ,

(x < 0),

CPo

(x > 0). ' cP If we impose our previous condition f'.I const. as x -+ -0 then of course we must take 0 = 0.

A different kind of non-unique solution has appeared in an elasticity problem investigated by W. T. Koiter [2] , pp. 369-370, where the additional term corresponds to a concentrated force at the corner of a wedge. At first sight it might seem that we can obtain a non-unique solution by a procedure of the following type. Multiply (2.87) by (ex + p) where p is some complex constant. Liouville's theorem gives (ex + p )J ( ex ) = constant 0, say, and =

� (O) = - (ex + k) 1 /2H+(ex) + O (ex + k) 1 / 2 (ex + p) - l . But since � (0) is regular in T > -k 2 we must have 1m p � -k 2 . Similarly by examining D_ we find that 1m p � k 2 cos 0 . Hence o ° and we arrive back at our previous solution. =

The purpose of the foregoing discussion is to show the part played by the edge conditions (v) of §2. 1 in determining the uniqueness of the solution when using the Wiener-Hopf technique. From a more general point of view the question of the necessary edge conditions for uniqueness of solution is connected with existence theorems. The subject has a considerable literature. We mention papers by : C. J. Bouwkamp [1], p. 45 (a summary) ; Philip8 Re8. Rep. 5 ( 1 950) , 40l . D. S . Jones [1] ; Pmc. Lond. Math. Soc. (3) 2 ( 1 952), 440 ; Quart . J. Me ch. Appl. Math. 5 ( 1 952), 363. A. W. Maue, Z. Phys . 1 26 ( 1 949), 60l . J . Meixner, Ann. Phys . 6 ( 1 949) , 2 . It would take us too far afield to reproduce the arguments of any of these authors in detail. They all lead to essentially the same results from different viewpoints. In scalar problems it would seem to be sufficient from our point of view to assume k = ° in the immediate neighbourhood of edges and consider the corresponding problem for Laplace ' s equation. For the problem considered above we have oCP t/oy = ° on y = 0, x < ° and it is well known that in the static case o CPt/oy ,....., X-1/ 2 on y = 0, x-+ +0, and CPt is finite near the edge. (Physically we can consider the flow of incompressible fluid past the edge of a sheet on which the normal velocity is zero. ) Similarly if we had CP t ° on y 0, =

=

76

THE

W I E N E R- H O P F T E C H N I Q U E

X < ° we should have CPt ""'" Xl /2 on y = 0 , X -+ +0, and OCPt/OY ""'" ( _ X)-1/2 on Y = 0, x -+ -0. (Physically we can consider the electro·

static field and charge near the sharp edge of a perfectly conducting sheet at zero potential. ) In electromagnetic problems convenient edge conditions are that the component of current density normal to the edge vanishes at the edge as r1/ 2 and the component tangential to the edge vanishes as r-l/ 2 , where r is the distance from the edge. The charge density varies as r-1/ 2 near the edge. Discontinuities in field components are related to current and charge densities by n X

n .

(B+ - B_ ) ( E+ - E_)

= =

Ok, w,

where 0 is the magnitude of the current, k is a unit vector giving its direction, and w is the charge density. These edge conditions are extremely important since it is possible to produce solutions of problems which satisfy all the boundary conditions except that they have the wrong edge singularities. 2.7

Comparison o f methods

We first of all compare Jones's method and the integral equation method. The integral equation method needs : choice of Green's functions, formulation of the integral equation, application of transforms. Jones's method is more direct since the Wiener-Hopf equation is obtained directly from transforms applied to the partial differential equation. The Green's function method for formulating integral equations is cumbersome. It is sometimes not completely obvious which Green's function should be chosen; examples have occurred in the literature where this has caused confusion or made problems seem more compli. cated than need be. Also functions may be introduced whose Fourier transforms are required e.g. the Hankel functions as in (2.66) ; these complicated functions are avoided altogether in Jones's method-the required transforms are obtained in the process of solution. In each Wiener-Hopf equation of type (2.25) there are two unknown functions. In Jones's method these appear in a completely symmetrical way and the physical ·significance of each of the unknown functions is immediately obvious. In the integral equation approach the symmetry is lost in the integral equation itself, though it reappears when the integral equation is transformed. The main advantage of the integral equation method of approach seems to be that it is very easy to recognize problems that can be solved by the Wiener-Hopf technique since the corresponding integral equations have a semi ·infinite range and the kernels are of the form k(x - ,; ) i.e. a function of (x - ,;) only (cf. (2.49), ( 2 . 55), (2.56) ) . Sometimes when using Jones's method in complicated problems it may

BASIC PROCEDURE S :

H A L F - P LANE

P R O BL E M S

77

not b e s o immediately obvious that the transform equations can be reduced to the Wiener-Hopf form. It may be possible that a solution can be obtained by means of the Wiener-Hopf technique from an integral equation formulated by a Green's function procedure, whereas it may not be possible to obtain the same solution by Jones's method. But no such case has so far occurred in connexion with partial differential equations, to my knowledge. We have probably mentioned a sufficient number of points to show the reader why we have used Jones's method in most of this book. At first sight Jones's method and the dual integral equation method of §2.3 seem quite dissimilar except for the use of the fundamental factorization (2. 1 ) . Jones's method uses analytic continuation to solve an equation of type (2.25) valid in a strip in the (X-plane. The dual integral equation method does not use analytic continuation, and it uses only functions specified on a contour in the (X-plane. The relation ship between the two methods will be clarified in later sections (e.g. §§4.3, 6 . 1 ) . We content ourselves with stating that in this book the relative advantages of the two methods are : Jones's method enables us to give a careful discussion in a routine manner of the regions in which functions are regular, the effect on the solution of edge conditions, and so on. The dual integral equation method will enable us to solve easily problems involving general boundary conditions, e.g. when atl ay = f(x) instead of atf ay = 0 on y = 0, x < 0 (cf. §2.8) but questions of uniqueness and rigour will not be so obvious as in Jones's method. 2.8

Boundary conditions specified by general functions

We have now completed our discussion of basic procedures applied to the Sommerfeld problem. In this section and the next we consider two of the generalizations mentioned in §2. 1 . These will be sufficient to enable us to deal with the other cases mentioned, some of which are outlined in exercises at the end of this chapter. At the end of §2.2 it was proved that in the problem considered, cp = 0 on y = 0, x > o. Hence the problem solved in §2.2 is equivalent to the following : find a solution of \12 cp + k2 cp = 0 in y ;;::: 0, - 00 < x < 00 such that cp represents outgoing waves at infinity; also on y = 0 (cf. condition (i) of §2. 1 ) , ocp/oy = ik sin 0 exp (- ikx cos 0) . cp = 0, x>0 An obvious generalization is : find a solution of \1 2 cp + P cp = 0 such that cp represents outgoing waves at infinity; also on y = 0 cp = f(x), (x > 0) ocp/oy = g(x) , (x < 0). (2.88) Assume that If(x) 1 < 0 1 exp ( T_X ) as x-+ + 00 , I g(x) 1 < 0 2 exp ( T+X ) as x -+ - 00 , with -k 2 � T_ < T+ � k 2 . There are no sources or incident waves in y ;;::: o. We give three methods of solution.

78

T H E W I E NE R - H O P F T E C H NI Q U E

Perhaps the simplest method is to use the dual integral equation approach of §2.3. As in §§2.2, 2.3 apply a Fourier transform in x and find in y � 0,

co + ia A(oc) e - ilXX - YY doc, cp = (2 ) 1/2 - co + ia

f

:

= A(oc) e -YY

(2 . 8 9 )

where T < a < T+ . Application of the boundary conditions on y = 0 gives the dual integral equations _

oo + ia (2.90a ) (x > 0), A(oc) e - ilXX doc = f(x), (2 ) 1/2 a co +i co +ia (2.90b) (x < 0). yA(oc) e - ilXX doc = -g(x), (2 ) 1/2 co a - +i As in §2 .3 replace x by (x + �) in the first equation, and by (x - �) in the second. Multiply respectively by 0/11 1 (�), vV 2 (�) given by (2.40) and integrate with respect to � from 0 to infinity. On using (2.38) with N+(oc) = (oc + k) -1/ 2 , N_(oc) = (oc - k) -1/ 2 as in §2.3, we

f

:

f

:

find

oo + ia

f

1

(2 n ) I/2 - co + ia =

n-l /2e i rr/4

oo +ia

1

(2n ) I / 2

(oc - k) -1 /2A(oc) e - irxx doc

f

co

f �-1 /2ei;;kf(x + �) d�,

(x > 0),

(2.91a)

(x < 0).

(2.91b)

o

(oc - k) 1/2A(oc) e - irxx doc

- co + ia

co

= _ n-l/2e - i rr/4

f �-1/2ei;;kg(x - �) d�, o

Multiply the first equation by exp (ikx) and differentiate with respect to x. Then, for x > 0,

:

( 2 ) 1 /2

oo + ia

f

(oc - k) I / 2A (oc) e - ilXX doc

- oo + ia =

n-l/ 2iei rr/ 4e - ikX !:.... dx

co

f �-1/2e;k(xHlj (x

o

+

�)

M.

(2.92)

BASIC PROCEDURES :

79

H A L F - P L A N E P R O B L E MS

The left-hand sides of ( 2.91 b ), (2.92) are identical. Hence we can use the Fourier inversion theorem to find

A(oc)

=

2-1 / 2n-1I oc - k) -1 / 2e3irr/4

00

X

:U f $-1/2eik(uH>j (u

+

o

00

{ f ei(rx - k)u du 0

o

f

$) d$ + - 00

eirxu du

00

f $-1/2eik
0

(2.93) The value of cp at any point is obtained by substituting in (2.89) .

This gives

00

00

{ f M(u,x,y)e - iku du d� f $-1/2eik(uH)! (u + $) d$

cp = fn-3/2e3i rr/ 4

o

+ where

0

00

o

f M(u,x,y) du f $-1/2eik
- 00

+

(2.94a)

0

f

oo + ia

M(u ,x,y )

=

- 00

\' N

A

=

{

2n1/ 2ei rr/4 (u

( 2 .94b )

""

+ ia

y = 0 since then (ex. 2.4) x) -1/2e ik( , -x) , ( u - x) > 0,

The solution simplifies slightly

M(u,x,O)

- k) -1/2eirx(U - x) - yy d N •

_

if

o

,

lu - x) < 0.

( 2.95 )

The nature of the solution will be clarified by di:41113sion of more general equations in Chapter VI. We next obtain the same solution by Jones's method. A" in §2.2 . write (2.06a) <1>+ (0) + <1>_(0) = A(oc),

<1> � (O) + <1>'-- (0)

Eliminate A(oc) :

=

(2.96b)

-yA(oc) .

(2.97) <1>� (0) + <1>'-- (0) = -y{<1>+(O) + <1>_(0)} . The functions <1>� , <1>_ are unknown; <1>'-- , <1>+ are known. In fact 00

<1>-;- (0)

=

: f

!( $ ) eirx< d $ (2 ) 1 / 2 o

1 _ (0) = 2n ( ) 1/ 2

mo l 'l'

o

f g( $) e,rx' d$.

- 00

.

e

( 2 .98)

80

T H E W I E N E R-H O P F T E C H N I Q U E

Rearrange (2 .97) by writing (cf. (2.29» ( rx + k) -1/2<1> � (0) + H+ ( rx ) = - ( rx - k) 1/ 2<1>_(0) - H_ ( rx ) , where H+ and H_ are obtained from J( rx)

=

H+ ( rx) + H_(rx) = { ( rx + k) -1/2<1>'-- (0) + (rx - k) 1/2<1>+(0)}, by using theorem B , §1.3 to decompose the function on the right hand side into the sum of two functions regular in suitable upper and lower half-planes. For example

� f

oo + ;a

H_ ( rx) = 2 i

{( ' + k) -1/2<1> '_ ( ,,o)

- oo + ia

+ ('

_

+

k) 1 / 2<1>+( ,,0)}

,

d, rx '

where T_ < a < T+ and (1m rx) = T < a. As in §2.2 we assume that sufficient conditions are given to prove that J( rx ) is an analytic function regular over the whole rx-plane vanishing at infinity so that it must be identically zero. Then <1>_(0) = - ( rx - k) -1 / 2H_ ( rx ) , and

f

oo + ia

A( rx)

=

<1>+(0) + ( 2ni) -1 (rx - k) -1 / 2

{( ' + k) -1/2<1>'-- ( ,,0)

- oo + ia

+ ( ' - k) 1 / 2<1>+( ',o)} Since 1m (' - rx )

>

I

oo + ia

(' + k) -1/2 d'

- oo + �

- 00 + ia

rx '

(2.99)

00

0

I g(�) ei'< d� I eiu( t -a) du.

- 00

The integral in , is (ex. 2.4(d»

f

d'

° we can write the first half of the integral as

-(2 n ) -3/ 2 (rx - k) -lj2

oo + ia

,

+

0

(2.100)

(' + k) -1 / 2ei � ( HU) d,

{

, o = 2 nl/ 2e - irr/4 _ i(H u)k 2 -1 , ( � - U) / e

(� + u) > 0, (� + u ) < 0,

when 1m ( ' + k) > 0, a condition which is assumed to be satisfied in the above analysis. Thus (2.100) reduces to 00

_ 2-1/2n-1 (rx _ k) -1/2e - i rr/4 I e - iu(H a) du o

-u I g(�)e - ik« _ � - U)-1/2 d�,

- 00

BASIC PRO C E D URE S :

HALF-PLANE

which is readily proved to be equivalent to the term involving (2.93) . The remaining part of (2.99) is

oo + ia <1>+ (0) + (2m) -1 (oc - k) -1/2

- 00

=

(2m) -1 ( oc - k) -1/2

81

PROBLEMS

J+ ia ( ' - k)1/2<1>+ ( ,,o) ,

d,

g

in

oc '

1m ( ' - oc) > 0,

oo + ib

J+ib (' - k) 1/2<1>+ ( ',0) ,

d,

oc ' 1m ( ' - oc) < 0, (2. 101)

- 00

where we have shifted the contour parallel to itself over the pole at , oc. This is written (cf. (2. 100)) =

00 00 oo +� ( 21T) -3/2 (OC - k) -1/2 I ( ' - k) 1 / 2 d, I f W eirx< d� I e -iu, , -rx) duo 0 0 - oo +ib

In this case we cannot repeat our previous procedure exactly since the integral in , obtained by changing orders of integration is divergent. However we can write the integral as 00

J

� J J

eiku f(�) d 0 oo + ib X ( ' - k) -1/2 eim -u) d, d� duo (2.102) - 00 +i b

i(21T)-3/ 2 ( OC - k) -1/2 ei(rx -k)u o

00

The integral in , is (ex. 2.4c)

oo + ib

J+ib( ' - k)-1/2ei'« -u) d , { 2�/2ei7T/4(�

- 00

=

°

_

U ) -1/2eik(; - U ) , ,

(� - u) > 0, (� - u) < 0.

The result of substituting this in (2. 102) is readily proved to be identical with the term in (2.93) involving f. We finally obtain the same results by a "superposition method" which is practically the same as the second method just described,

82

THE W I E N E R-HOPF T E CH N I Q U E

except that it does not make explicit use of theorem B , §1 .3. have

f (x)

=

� ( ',O) =

--;-/2 (27T)

We

oo + ib

f

- 00

<1>+ ( ',O) e -i ,X d"

(2.103a)

+ ib

o

: fg(�)ei'< d� : � ( " O) e -i 'X d', ( 2 ) ] /2 f ( 2 ) 1/ 2

- co

oo + ia

g(x)

=

(2. 103b)

- co + ia

where in the first equation b > T_ , and in the second a < T+. Solve two separate problems : (a) a �/ ay = exp ( -i 'x ) on y 0, x < 0 : � ° on y = 0, x > 0. If we use Jones's method exactly as in §2.2, we find =

=

where

i 1 , 1m (oc - ') < 0. . (27T) 1/2 (oc k) 1/2 ( ' + k) 1/2 (OC ') (b) a�/ ay = ° on y = 0, x < 0 : � exp ( -i'x) on y 0, x > 0.

A (oc, ') 1

=

_

_

=

=

Jones ' s method gives

� (O) = -yA2 (oc, '), say : <1>_(0) = - i(27T) 1 / 2 (OC - ,) -1 + A 2 (oc, '),

where

Now superimpose solutions using ( 2. 103). This yields oo + �

A(oc)

=

oo + m

: f � ( ,,0)A 1 (OC, ') d' + (2:)1/2 f +( ',0)A 2(oc, ') d"

(2 ) 11 2

- oo + �

- oo + m

where T_ A 2 we find that this is identical with (2.99) (remember result

(2.101)).

BASIC PRO C E D UR E S :

2.9

HAL F - PLAN E

83

PROBLEMS

Radiation-type boundary conditions

We deal briefly with one example of problems where radiation type boundary conditions occur. Examples of a similar type, with references, are given in ex. 2.12. The problem treated here is equivalent to diffraction of an electromagnetic wave by a semi-infinite metallic sheet of finite conductivity which has been treated in detail by T. B. A. Senior [1 ] . We choose our conventions so that our results are directly comparable with those of Senior. (See the end of this section. Note that we take the plane to lie in 'Y = 0, x > 0, not 'Y = 0, x < 0 as in the first part of this chapter. ) We wish t o solve V2CPt + PCPt = 0 in - 00 < x, 'Y < 00 with incident wave CPi = exp ( - ikx cos 0 - ik'Y sin 0 ) . Suppose that an imperfectly absorbent sheet lies in the plane '!J = 0, x > 0, the conditions on the sheet being

CPt

=

± i(j(ocpFJ'Y),

'Y

=

± O,

(0 � x < 00 ) .

(2.104)

Write CPt = CPi + cpo A Fourier transform can be applied to cpo As in §2.2 we have = A(IX) exp (-Y'Y), 'Y � 0, and = B(IX) exp (Y'Y), 'Y � O. In the usual notation these give

<1>+ ( +0) + <1>_(0) A (IX), (2. 105a) (2. 105b) � ( +O) + � (O) -yA(IX), <1>+ ( - 0) + <1>_(0) = B(IX), (2. 106a) (2.106b) � (-O) + � (O) = yB(IX) . Elimination of A(IX), B(IX) gives (2. 107a) � ( +O) + � (O) = -y{+ ( +O) + _(O)}, (2.107b) � ( -O) + � (O) = y{+ ( -O) + _ (O) } . If we write CPt = CPi + cP in (2.104) and apply a Fourier transform =

=

we find

i(27T ) -1/2(1 - (jk sin 0)(1X - k cos 0 ) -1 + <1>+ ( +0) - i(j � ( + 0)

=

i(27T )-1/ 2(1 + (jk sin 0)(1X - k cos 0)-1 + <1>+ ( - 0) + i(j� ( -0)

=

0, (2.108a)

Eliminate + ( +0) between (2. 107a) and (2. 108a) : <1>+( -0) between (2.107b) and (2. 108b) . Then � (O) + y_(O) = - ( 1 + i(jy)� ( +0) +

o.

(2. 108b) eliminate

+ i( 2 7T)-1/ 2y( 1 - (jk sin 0)(1X - k cos 0)-1 ,

� (O) - y _(O)

=

- ( 1 + i(jy)� (-O) -

- i(27T) -1/ 2y(1 + (jk sin 0)(1X - k cos 0) -1.

84 THE WIEN ER-H OPF TECHN I Q U E Two independent Wiener-Hopf equations are obtained by adding and subtracting these equations : 2<1> � (0) -(1 + ic5y){ � ( +O) + � ( -O)} (2.109a) - 2i(21T)-1/ 2 c5yk sin 0(1X - k cos 0)-1 , 2y _(0) = - (1 + ic5y){ � ( +O) - � ( -O)} + (2. 109b) + 2i(21T)-1/ 2y(1X - k cos 0)-1 . The logic of the procedure by which (2.109a, b) are obtained should be clear. The writing down of (2.105), (2. 106) is routine. We wish to obtain equations involving only functions whose regions of regularity are known, so we eliminate A (IX), B(IX) . Next use the boundary conditions : these lead to (2. 108) . In (2.107), (2.108) we have four equations involving six unknown functions <1>_(0) , � (O), +(±O), � ( ±O). (2.108) involve only +(±O), � (±O) . So clearly we must try to obtain two Wiener-Hopf equations involving _(0), � (0) and two "plus" functions, using (2.108) to eliminate two of the four "plus" functions from (2.107). There are many ways of carrying out the manipulations, but the reader will readily show that the final Wiener-Hopf equations are always equivalent to (2.109). The main object ofthis section is to show in detail how the Wiener Hopf equations (2.109) are obtained. Now that this has been accom plished we merely summarize the remaining steps in the solution. Both of the equations in (2.109) are similar to : (2. 1 10) p(1X - k cos 0)-1 + F_( IX ) = K(IX) G+ (IX), where p is some constant. The procedure is : (a) Prove that the equation holds in a strip' k 2 cos 0 < T < k 2 . (In the Sommerfeld problem of §2. 1 the strip was - k 2 < T < k 2 cos 0 because in that case the sheet was chosen in y 0, x < 0.) (b) Factorize K(IX) 1 + ic5y in the form K+ (IX)K_(IX) (ex. 2. 10) . (c) Rewrite (2.1 10) as F_( IX ) 'p P(IX) = + k cos 0 K_(IX) K_(k cos 0) IX K_(IX) 1 P . (2. 1 1 1 ) . K+(IX) G+ (IX) k cos 0) K_(k cos 0) (IX (d) Investigate the behaviour of (2.1 1 1 ) a s IX tends t o infinity in appropriate half-planes so that Liouville's theorem can be applied to determine P(IX) . One important difference between this problem and the other =

=

=

-

=

{ I

I

_

}

85 problems considered in this chapter is that previously the factoriza tion step (b) could be carried out by inspection (namely ( oc2 - k2 ) 1 / 2 = (oc - k) I/2 (OC + k) 1/2 ) whereas in the present case the factorization of K(oc) = 1 + ib(oc2 - P) I/2 = K+ (oc)K_(oc) is much more difficult. Details are given in ex. 2.10. A detailed examination of (2. 109a, b) along these lines would take us too far afield. We shall examine several similar cases in Chapter III, and details for the present case will be found in T. B. A. Senior [1], who gives explicit expressions for the distant field. Senior's integral equation approach is of course more cumbersome than Jones's method. Finally we compare our notation with that used by Senior. He uses the terminology of electromagnetic theory as explained in ex. 2 . 1 1 , the results of which we now assume. The problem treated here is equivalent to the case of E-polarization. In the following fl is the permeability, s the dielectric constant. We have CPt = E . , ocp tfoy = ifl wHw ' The boundary condition used by Senior, corre sponding to (2. 104), is E. = �'fJZHw on Y = ±O, where 'fJ is the conductivity and Z = fll/2 S-1/2 . Thus on comparing our notation with that of Senior, -'fJZ(iflw) -1 ib i.e. b 'fJlk , where k = W (flS) I/2 , and k is the constant in the wave equation V 2cp + Pcp = O. Senior also writes I 1 (x) = (E. ) y = + o - (E. ) y = - o , i.e. 1 1 (oc) - «1>+ ( + 0) - «1>+ ( - 0), where we have used a bar to denote the transform of 1 1' Also I 2 (x) = (Hw ) y = + o - (Hw ) y = - o , 1 . I.e. ,*, l+ ( +0) - ffi I- 2 (oc) = - {ffi ,*, /+ (-0) } . �fl W Noting that oc8 = 0, ,8 = - oc where the superscript 'S' refers to Senior's notation, it will be found that (2. 108) , (2.109) are identical with Senior ' s (27) and (14). Although many of the problems in this chapter can be solved by other means, some of the virtues of methods based on the Wiener Hopf technique should already be apparent. In particular the procedures used to solve half-plane problems with general boundary conditions and with radiation type boundary conditions are direct extensions of the method for the much easier Sommerfeld problem. In the next chapter we consider examples involving more complicated geometrical configurations. BASIC

PROCEDURES :

HALF - PLANE

=

.

PROBLEMS

86

T H E W I E N E R-H O P F T E C H N I Q U E

Miscellaneous Examples and Results II

The conventions used in this book are to some extent arbitrary and in the following various other possibilities are illustrated in connex ion with the Sommerfeld half-plane problem. For convenience we quote two equations from §2.2 : 2.1

-ty{
=

=

Most of the details below are left to the reader. 0 lies in y (i) If the half-plane on which iJt/ iJy instead of - 00 < x < 0, then (2.20), (2.21) become =

-ty{
=

=

=

0, 0

<

k 2 cos e <

e)-I .

x < T

<

00

k2 '

(ii) If the half-plane litJs in x 0, - 00 < y < 0, with iJt/ iJx 0, a Fourier transform in y is used. The incident wave is given by (2.3) but -t7T < e < +t7T. (2.20) is unchanged in form but holds in -k 2 < T < k sin e. ( 2 . 2 1 ) becomes 2 =

=

(27T)-1/2k cos e(oc - k sin e)-I . (iii) If the time factor is exp ( + iwt) then k has a negative imaginary part and y = (oc2 - k2 ) 1/2 + i ( k2 - oc2 ) 1/2. The incident wave, if situated as in Fig. 2. 1 , is i exp (ikx cos e + iky sin e), 0 < e < 7T. (2.20) is unchanged but - 0 then the contour of integration in the inversion formula is the real axis, indented at at oc oc -k, oc -k cos e, and + k.
=

=

=

=

below above wave incident from the lower half-plane can be obtained by =

=

(iv) A taking 7T < e < 27T, or by setting i exp ( ikx cos e + iky sin e ) , o < e < 7T. (v) The sign of oc can be reversed in the Fourier transform (i.e. we take the lower sign in ( 1 . 5 2) , § 1 . 5 ) . Define =

=

00

.l.e -'!xx. dx' (27T) 1/2 f 1

--

- 00

,

'I'

00

=

1 f .I.e .

__

(27T) 1/2

o

'I'

- ,!XX

dx , etc.

The result is essentially to interchange upper and lower halves of the oc-plane. (vi) The Laplace transform can be used instead of the Fourier transform. Define III (s,y)

00

J e -8X dx

- 00

00

=

J e - 8X dx,

o

BASIC PR O C E D UR E S :

HALF-PLANE

PROBLEMS

87

and similarly for cllN where cllp , cllN are valid in right and left half-planes respectively. The partial differential equation gives

-A(s) exp ( -iKY) , 0) ; (y .;;; 0) , where K = (S2 + k2 ) 1/2 with K = k when s = O. The results for Fourier and Laplace transforms will be found to correspond exactly if a: == is, y == -iK, (1m a:) == (Re s), cll+ (a:) == (2:".)-1/2cllp (S), etc. for k > o. 2 2.2 Suppose we wish to solve (rpt) .,,,, + ( rpt ) 1/1I + k 2 rpt = s(x,y) with a half·plane in y = 0 , - 00 < x .;;; 0 on which orpt! oy O. The term s(x,y) represents an arbitrary source distribution and there is now no (y

;;;.

=

wave incident from infinity. We use the method of superposition . Set

co

S (fJ,�) =

co

L J J s(x,y)ei(,B",HlI) dx dy - co -

00

ic + co id + co

s( x,y)

=

L J J ic - co id - co

S(M) e - i(,BxHlI) dfJ d{.

First of all solve the special case (lpt)",,,,

+ (lpt ) yy + k21pt = exp { -i(fJx + �y)}. Introduce 11' defined by lpt = 11' + (k2 - fJ2 - �2 ) - 1 exp { -i(fJx + �y)}. Then we have to solve lpxx + lpyy + k21p = 0 with boundary condition 01p/ oy = i�(k2 - fJ2 - �2 ) -1 exp ( -ifJx) on y = 0, x .;;; O. From (2 . 6), ( 2.3 1 ), if -k 2 < 1m ex < 1m f1 < k 2 ' and A = + l (y > 0 ), - 1 (y < 0) , e -i"'''' - yllll

da:. (a: - k) l/2 (a: - fJ ) By superposition we can set rpt = rp i + rp, where rpi

ic + co

=

LJ ic -

00

ic - co

id+ co

dfJ

J id - co

(a)

id - 00

ia+ co

x

J ia - co

e -i"'''' - yllll da:. (a: - k) l/2 (a: - fJ)

(b)

This gives the answer to the problem. In the special case of a line source, s(x,y) = - 47TO ( X - xo)o(y - Yo ) . Hence S(fJ ,�) = - 2 exp (ifJxo + i�yo) . The integral in � in (a) and (b)

88

T H E W I E N E R-H O P F T E C H N I Q U E

can be evaluated by residues and the integral obtained from (a) can then be evaluated from ( 1 .60), ( 1 . 6 1 ) . This gives

rpi

=

rp

=

R2 = (x ia + co

7TiHh1)(kR), w + co

-

�:o f

f

dP

ie - co

ia - oo

doc

- xO ) 2 + (y - YO )2. e -iocx - rlv le -i/3xo - lllvo l

( oc _ k)I/2 ( P

_

k)I/2 (OC + P )

,

where -k2 < a < -c < k2 II = ( P2 - k2) 1 /2 and we have changed the sign of p. This is symmetrical in (x,y), (xo ,Yo ) and in oc, p as we should expect. Other applications of a similar type of analysis are given by P. C. Clemmow [2] who deals with reduction of the double integral to a single integral, and with asymptotic evaluation of more general double integrals. The half-plane problem for a line source has also been treated by R. F. Harrington [ 1]. '

Suppose that instead of iJrptl iJy 0 on y 0, x < 0, we have 0 for the Sommerfeld half-plane problem of §2. 1 . If rpt rpi + rp,
rpt

=

=

=

=

=

corresponding to (2.24) turns out to be

is

i

A (oc) = ( 27T ) 1 /2 (oc _ k cos 0) • co

J tq - 1e ± i"t dt

2.4

o

=

(k

+ k cos 0) 1/2 , ( -k2 < (oc + k ) I /2

r(q)oc - qe ±iqrr/2,

T

< k 2 cos 0 ) .

(0 < R e q < 1 ) , (t > 0) ,

(t < 0) ,

where upper and lower signs go together, (1m oc) > 0 for the upper sign and ( 1m oc) < 0 for the lower. The first is a standard result and the second is its Fourier inverse. We can deduce the following useful special cases : co

J 1; - 1/2e ± i«" U) 'dl;

o

J ( _I;)-1/2e 'F io(" U) dl;

- 00

ia + co

J ia -

co

=

e ± irr/4 7T1/2 (oc ±

k)-1/2 ,

1m (k

±

oc)

>

0, (a)

e ± irr/4 wl/2(oc ±

k) - 1/2 ,

1m (k ±

oc)

>

0, (b)

o

=

I; >

O} ,

1; < 0 1m (k ± oc)

>

0,

(c)

B A S I C PR O C E D UR E S :

ia + co

f

(ex ± k)-1/2e ±iCt.< dex

ia - 00

HALF -PLAN E PROBLEMS

�

}

89

>O ' 0 1m (k ± ex) > 0, (d)

o

� <

2-rTl/2e 'f irr/ 4 ( _ �) - 1/2e - i
where upper and lower signs go together.

2 . 5 Another method of solving the general equation (2.97) is the following. Multiply through by (ex + k)-1/2 and take the inverse Fourier transform, replacing ex by say fJ for convenience:

ia + co

�

f

{(fJ + k) -1/2 0), (2TT) I/2 ia - 00 ia + co 1 (fJ - k)1/2 0 respectively. If we multiply through by (2TT) -1/2 exp (iexx) and integrate with respect to x from 0 to 00 , the right-hand side becomes (ex + k) -1/2
f

f

these results. The basis of the method j ust described is the following procedure for writing F(ex) in the form F+(ex) + F_(ex) . Define j(x) by

1 j(x) = (2TT) I/2 Then

co

=

1 f j(x)e'Ct.. x dx

ia+ co

f ia -

Ji'_(ex ) =

--

(2TT) I/2

F(ex)e -iCt.x dex.

00

o

o

: f j(x)eiCt.X dx.

(2 ) 1/ 2

- co

When integral equations were established by the Green's function method in §2 .4 , the space was divided into the two regions o < y < 00 and - 00 < y < O. Alternatively it is possible to apply Green's theorem to the whole space, excluding the line - 00 < x < 0 occupied by the half-plane, e.g. we can set x > - R, y = -0 (i.e. B = -TT). Then let R 00, r O. This will obviously give an integral equation in terms of the discontinuity in

=

�

�

90

T H E WI E N E R- H O P F T E C H N I Q U E

2 .7 It is sometimes possible to obtain the integral equation for a problem in a direct way by a physical argument. Consider a half-plane in y 0, - 00 < x .;;; ° under two conditions. (i) If there is a charge distribution on y = 0, x < 0, the potential at (x,y) due to a unit line charge at is R2 = (x + y2, and b y superposition the total potential at (x,y) is o (x,y) = 1Ti rp =

u(x) (�,o) -rriH�l)(kR),

�)2

f H�l)(kR)u(�) d� .

- 00

The potential �s continuous on crossing y = 0, x < ° but the deriva tive of the potential in the y-direction is discontinuous. (ii) If there is a distribution of y-oriented dipoles on y 0, x < 0, by superposition the total potential at (x,y) is o

v(x)

rp (x,y)

=

= 1)-->0 1Ti �01} f Hbl) (kR)v(�) d�, lim

- 00

The potential is discontinuous across the layer but the y-derivative of the potential is continuous. As an example consider the Sommerfeld problem of §2. 1 . Using the notation rpt = rp i + rp, orp/ oy is continuous across y = ° and we can use representation (ii) above. This gives o

( orpoy) 1I�0

=

lim lim -rri �

oy 01}

1/-->0 1)-->0

f Hbl)(kR)v (�) d� ik =

- 00

sin 0e

-ikx

cos

8.

This agrees with (2.54), apart from a factor of proportionality relating and e. Similarly for the absorbent half-plane on which rpt ° we should use (i) above.

v

=

2.8 It is possible to obtain the integral equations of §2 .4 by applying transforms in y instead of x. As an example consider case (2b) of §2.4. Set rpt rp i + rp and apply a transform in y. On integrating by parts, since rp is discontinuous on y = 0, =

f 02rp2 e"1.1/. d 00

1

--

(21T)1/2

_ .-

- 00

y =

oy

. �or.(21T)-1/2 2e(x) - or.2 ([> (or.) ,

-

where 2e(x) = rpt(x, + O) - rpt(x, -0), 00 < x < 0, and of course = ° for x > 0. On solving the ordinary differential equation for ([> by ex. 1 . 15, and inverting,

e(x)

0 ia + L f ior.y-le - ilXl/ fe (�)e-Y l x -el d� dor.. ia - oo 00

rp

=

- 00

BASIC PR O C E D URE S :

HAL F - PLAN E P R O B L E M S

I f we use the boundary condition o/ oy find, by an obvious manipulation, -ik sin

0e-ikxcos 8

d2 = lim � o y--.. 2 rr dy2

o

=

ik sin

91

0 exp ( -ikx cos 0), we

ia + ; fe ( ) fy-le-YIX -ol-iIXY drx d;, ia- oo co

- co

(x < 0) .

This is equivalent to (2.64), by ( 1 . 60), ( 1 . 6 1 ) .

2 . 9 The half-plane problems o f this chapter (mixed boundary conditions on y = 0, - 00 < x < 0, 0 < x < 00 ) can also be solved by introducing unknown functions on x = 0, - 00 < y < 0, 0 < y < 00 . 2 . 1 0 I n order t o perform the factorization (2. 1 1 2) write =

In K+ (rx) + In K_(rx) d d In K+( rx) + In K_ (rx) drx drx =

=

In { I + i<5(rx2 - k2 ) 1/2},

i<5rx { 1 + i<5(rx2 - k2 ) 1/2 }-1 (rx2 - k2 ) -1/2

1 1 i + 2(rx - id) + 2 (rx + id) 2<5

(

1 1 rx - id + rx + id

)

1 (rx2 - k2)1/2

(a)

where d 2 = {( 1 /<52) - k2}. This is obtained by multiplying the fraction in the previous line top and bottom by { I - i<5(rx2 - k2 ) 1/2} and separ ating into partial fractions. Now use the result obtained at the end of § 1 . 3, that if ! ± (rx)

=

7T-1 (rx2 - k2 ) -1/2 arc cos ( ± rx/k),

then !+ (rx), !_(rx) are regular in 'T

- k ' 'T < k respectively, and 2 2 !+ (rx) + L (rx) = (rx2 - k2 )-1/2. >

From this we deduce by inspection that if p is any complex number and

where the upper sign holds for p in the upper half-plane and the lower sign for p in the lower half-plane, then F+ (rx), F_(rx) are regular in upper and lower half-planes respectively, and (rx - p)-1(rx2 - k2 ) -1/2. From (a) , using the fact that if rx = -id then ( rx2 - k2 )1/2 we can show that F+(rx ) + F_(rx)

=

A similar procedure can be used for

K(rx)

=

1

+ ic(rx2 - k2 )-I/2 .

=

( -i/<5),

92

T H E W I E N E R- H O P F T E C H N I Q U E

The results can be deduced from the above or we can proceed inde pendently to find

1 !!. { f+ (IX) - f+(p) + f+ (IX) - f+ ( } (c) = ) K IX ( + dlX 2 (IX + k) 2 IX IX + 82)1/2 . 2 (k where In order to find K+ (IX) expressions (b) and (c) have to be integrated. It is not possible to do this explicitly but formulae which are practical for numerical work may be obtained by the substitution arc cos (IX/k) which gives, for example, fIX + (IX)id dlX 7Ti2. f d + ikdOcos f The denominator can be expanded as a series in ( ik/d ) cos if d is large. Similar formulae are given by T. Senior [1] and Heins and � In

-P)

_

_

-P

P

_

P =

0

a

,

=

-

arc coB ( alk)

0

0'

=

0 A. E.

B. A.

H. Feshbach [2]. Senior's formulae can be obtained by writing

1 IX2 ! d2 152 {I - 15(k2 I(2)1/2 + 1 + 15(k; I(2)1/2} , 1 O dO f 1 - 15k sin O ' =

_

_

arc cos ( a lk)

2. 1 1 We remind the reader of the connexion between the scalar wave equation and certain two-dimensional cases of the electromagnetic equations curl E = curl = -i8W E,

H

i,uwH

where 8 is the dielectric constant and is the permeability. If the fields are independent of the z-co-ordinate then these separate into two inde pendent sets : 0 and (a) TM or mag n t c waves when = x =

,u

ei

transverse

Hz E E1I aH1I aHx -iewEz· ax ay =

aEz i,uwHx ay Then E. satisfies a2Ez +. a2Ez + k2E ax2 ay2 The boundary condition on a perfectly conducting plate lying in a plane parallel to the z-axis is Ez O. When the plate lies in a plane a x b, y currents are induced in the --

=

=

__

<

<

•

__

=

0,

- 00

= 0, =

z-direction. The current is given by

Iz = (Hx)+o - (Hx

Lo

=

00 ,

aEz) (aEz) } �,uw { ( uy + 0 uY _ o , a x b,

= .

1

-,,

-

-,,

<

<

BASIC PRO C E D URE S :

HAL F - PLANE

93

PR O B L E M S

where the symbols ± O refer to ± O, i.e. the two sides of the conducting plane in o. This case is sometimes referred to as or (b) or waves when H", = HlI = 0 and

y= y = parallelTEpolarization transverseE-polarization. electric E. = aH. = uw. E aH. -iewE", -ax ay Then H. satisfies the two-dimensional steady-state wave equation with 2• When perfect conductor lies in, say, a < x < b, y k-2 ep,w the boundary condition is E", i.e. aH.f ay < < 11

-- =

=

00

00

z

a

=

= =

0

0, O.

Electric currents occur in the conductor in the x-direction and we have

It is often useful to represent these fields in terms of a scalar which is a solution of the two-dimensional wave equation, in the following way : H = ( )

. a fay

This case is sometimes referred to as or H On using the above terminology we see that the half-plane problem of §2. 1 is equivalent to diffraction of an H-polarized wave falling on a perfectly conducting half-plane. Similarly the problem of ex. 2.3 is equivalent to diffraction of an E-polarized wave falling on a perfectly conducting half-plane.

perpendicular polarization

-polarization.

2.12 We summarize below some of the cases considered in the literature. The following headings are used : (i) the equation solved, (ii) the boundary conditions, (iii) the function K(cx) which has to be factorized, (iv) references.

(A)

V2t + k2t p( atf ay) - t r( atf ay) - st

(i) (ii)

q

t

(iii) K(cx)

=

i + ,

= (py

i

+ q) (

(iv) The special case

q

0,

=

= 0, =

=

0,

y :> Y =

0,

Y

0,

=

- oo < x <

0,

o

El

<x <

- 00

( -ikx cos - iky sin

)

(See ex. 2. 10).

ry + S -I.

00 ,

<x <

exp

00 .

El l .

o.

= 0 or s = 0 is considered by T. B. A. Senior

[ 1] (cf. §2.9) . Application of the general case to propagation of radio waves over a land-sea interface has been made by P. C. Clemmow [2] (see note in (B) (iv) below) . A. E . Heins and H. Feshbach [2] have

94

T H E W I E N E R- H O P F T E CH N I Q U E

considered the corresponding problem in acoustics where the boundary conditions are due to partially sound-absorbent materials. (B)

(i) V2rpt + k2rpt

=

>

y

0,

= 0,

y < 0,

orp,! oy

= 0,

y = 0,

(ii)

( orpt!

iJy!+O

=

p( orpt! OYL o

rp, contmuous rpi

K(oc)

(iii)

- 00

0,

V2rpt + K2rpt

- 00

°

},

exp ( -ikx cos 0

=

(oc2

_

k2 ) 1/2 +

-

p(oc2

00 .

<x < 00 .

<x <

- 00

Y = 0,

=

00 ,

<x <

< x < 0.

iky sin 0 ) . _

K2 ) 1/2.

(iv) P. C. Clemmow [2] discusses an application to the propagation of radio waves past a land-sea interface. Clemmow shows that in this case, if we are interested only in the field at sea-level due to a trans mitter on the ground, the far-field can be expressed in terms of K(oc) alone and we do not need K+(oc), K_(oc). This holds also if we apply (A) above to the problem. (C)

- oo < x < 00 . Y ';;; 0, ° < x < 00 , orp! oy = 0, Y = 0, - 00 < x < 0. orp! oY = prp, Y = 0, K(oc) 1 - p(oc2 + k2)-1/2.

V2rp - k2rp

(i) (ii)

=

0,

=

(iii)

(iv) T. R. Greene and A. E . Heins [ 1] and A. E. Heins [ 1 0] discuss an application to surface waves on water. (D)

(i) ( ii) rp

V2rp = 1,

=

+ ierp (y

x < 0)

= 0,

K(oc)

(iii)

Y

0,

= 1

.;;;; 0, :

+

- 00

orpf oy - irp

i(oc2

<x <

=

0, (y

00 .

=

0,

- ie) 1/2.

x >

0) .

(iv) G. F. Carrier and W. H . Munk [ 1] discuss the diffusion of tides in permeable rock. There is a periodic tidal fluctuation of pressure on y = 0, x < ° and the problem is to find the fluctuation of the free surface level of the water in the rock, where the free surface is assumed to lie in y 0, x > 0, approximately. =

(E)

(i) (ii) rp

V2rp = 1,

-

y

2n(

= 0,

orp! ox) x >

= 0,

y ;;;. 0, orp! oy

°

- 00

= 0,

y

<x < =

0,

00 .

x < 0.

(iii) (iv) These equations describe convection of heat from a flat plate in y = 0, x > 0, at constant temperature, in an inviscid stream of constant velocity flowing in the x-direction. If we set tp = rp exp then tp satisfies tp�� + tpw - 2tp 0.

n

( -nx)

=

BASIC PRO C E D URE S :

HAL F - PLAN E P R O B L E M S

95

2.13 The oscillating quarter-plane airfoil at supersonic speeds has been considered by J. W. Miles [1]. The problem can be reduced to solution of

- 00 < x, y < with boundary conditions, on

;;;. 0,

Z = °

o"" oz = /(x,y), (known) , x ;;;. 0, orp/ ox - iprf> = 0, x ;;;. 0, x < 0, rp = 0, If we apply a Fourier transform

(1)(rt.,{3 )

z

00 ,

00

00

- 00

- 00

y ;;;. 0, y < 0, - 00 < y < 00 .

= 2� f I rpei(a.x+{J'V) dx dy,

it appears that if (1) -+ ° as z -+ 00 the rt. and {3 planes must be cut so that y ({32 + k2 - rt.2 ) 1/2 has a positive real part for all rt., {3 considered. On using the arguments in § 1 . 5 and ex. 1 . 1 8 this means that if we set {3 0 = (rt.2 - k2 ) 1/2 , 1m {3 0 > 0, the {3 plane must be cut from + {3 0 to inflnity in the upper half-plane and -{3 0 to infinity in the lower half plane. But the rt.-plane must be cut from ± ((32 + k2 ) 1 /2 to inflnity in the lower half-plane. When the cuts are fixed the solution is straight forward. There are easier methods for solving this particular problem but the Wiener-Hopf technique may be useful when considering generalizations. =

2.14 Consider diffraction of the three-dimensional electromagnetic wave Ei = (A,Il,V) exp { -ik(ax + by + cz)} by a semi-inflnite per fectly conducting plane in y = 0, x < 0, - 00 < z < 00. The notation means that the x, y, z components of Ei are given by multiplying the exponential factor by A, Il , v. These constants must be chosen so that the wave is plane. Also a = sin (Jo cos rpo, b = sin (Jo sin rp o ' c = cos (Jo. All quantities are proportional to exp ( -ikcz) and this is the only way in which they depend on z. If we write Et Ei + E, E = (E." Ell ' E.) exp ( -ikcz), each of the components of E satisfies rp.,., + rpllll + k2 sin2 (Jorp = 0. Denote the transform of E., with respect to x in (0, 00 ) by �.,+ , etc. The method of §2.2 gives the. simultaneous Wiener Hopf equations =

r!?J;- = -y28,+ + rt.kc�: + i(21T)-1 /2 (rt. -ka) -l {Art.kc -vy2}. - r!?J;- k2 (I -c2 )�: + rt.kc�,+ + i(21T)-1f2 (rt. -ka)-1{Ak2 ( I -c2 ) + vrt.kc}. where

=

These equations can be solved exactly. D. S. Jones [1], [3], cf. E. T. Copson [4] .

Appropriate references are

96

THE

W I E N E R-H O P F T E C H N I Q U E

2 . 1 5 Consider diffraction of waves in an elastic medium by a half plane. The strains and stresses can be expressed in terms of two potentials , 'P satisfying the steady-state equations (cf. I. N. Sneddon [1], p. 444) :

where e2 = (A + 2p)lp, 02 = pip : A, p are Lame's elastic constants and p is the density. The strains and stresses are given by a",,,, = -k2A + 2p(", ,,, + 'P "", ), u = '" + 'PII a'llll = -k2A + 2p(1I1I - 'P "", ), 'P = ", V 11 a"", = p(2"", - l{J",,,, + 'P'IIII ) . Suppose that a plane wave in an infinite elastic medium falls on the half-plane - 00 < x < 0, y = ° on which a'llll = a"", = 0. Write t = i + , 'Pt = 'Pi + 'P, where i = a exp ( -ikx cos El - iky sin El ) , 'Pi = b exp ( -iKx cos El - iKy sin El ) . The method o f §2.2 gives

where

�;t (0) + �;' (O) �i;, (0) + � ';( O )

= =

2p(yK2 )-1{ ( (X2 - IK2 ) 2 - (X2y r}F_, 2p( rK2)-1 {( (X2 - IK2 ) 2 - (X2y r}E_,

y ( (X2 k2 ) 1/2 U_( + O) - U_( -O) = E_ =

_

r

( (X2 - K2 ) 1/2 V_( + O) - V_( -O)

=

=

F_.

In the above equations �;' (O) and �';' (O) are known. The difficult step in the solution is the factorization of

K( (X )

=

( (X2 - tK2 ) 2 - (X2 ( (X2 - k2 ) 1/2 ( (X2 - K2 ) 1/2 .

(a)

The (X-plane can be cut from +k to +K and from -k to -K. Details are given in A. W. Maue [1] who has obtained the above results by an integral equation method. (Cf. A. W. Maue, Z. Angew. Math. Meeh., 34 ( 1 954) , pp. 1-12, where a transient problem involving a half-plane is solved by the method of conical flows and reduction to a Hilbert problem. The factorization which occurs in the Hilbert problem is identical with that required in (a) above.) The problem of diffraction by a half-plane on which u = v = ° can be formulated in a similar way. 2 . 1 6 The diffraction of waves by a half-plane in a viscous medium has been considered by J. B. Alblas [1]. using the linearized Navier Stokes equations. These equations are effectively of fourth order but the solutions can be expressed in terms of two potentials satisfying second order equations (cf. ex. 2 . 1 5 ) . The factorization required is

k complex,

Z

real.

Alblas discusses the far-field and the field near the half-plane (the boundary layer) .

BASIC PROCEDURES :

H AL F - PL A N E PR O B L E M S

x x

97

2.17 The boundary-layer problem for two-dimensional flow o f a viscous fluid past a plate lying in y = 0 , > 0 has been considered by J. A. Lewis and G. F. Carrier [1]. The flow as ...... - 00 is assumed to be uniform with velocity U in the positive x-direction. Oseen's linearized equations can be converted into the following equation for a stream function Ij! :

(::2 + a�2) ( a:2 + a�2 - :x)

Ij!

=

O.

(a)

This equation is of Laplace type and in order to apply the Wiener Hopf technique it is convenient to introduce a parameter k (real and positive) and to write the Fourier transform of (a) as ·

(� �2 (�2 'Yl 2 ) dy ·/ ) dy2 -

where

_

'Y

=

0,

The parameter k is set equal to zero in the final solution. The results are of the correct form but disagree with the Blasius solution near the half-plane by a factor of proportionality. A semi-empirical correction factor is suggested in the paper quoted. 2 . 1 8 All the problems solved in this chapter have been steady state, with time factor exp ( -iwt). Theoretically it is easy to derive the solution of transient problems since a Laplace transform in t (parameter s) applied to the wave equation produces the steady-state wave equation with k2 = - (S2/C2 ). Hence if we solve a suitable steady state problem, replace k by (is/c) , and take the inverse Laplace trans form, we obtain the solution of a transient problem. So far as I know, the only Wiener-Hopf example where this has been carried through in detail is given by D. S. Jones [5] . Transient problems involving the heat conduction equation can be solved in a similar way.

CHAP T E R I I I

FURTHER WAV E PROBLEMS

3.1

Introduction

The main features of various methods of exact solution using the Wiener-Hopf technique have been explained in some detail in connexion with the examples solved in the last chapter. In this chapter we apply the same basic technique to solve problems with more complicated geometry. One of the main objects is to illustrate practical difficulties by examining concrete · examples in detail. As far as possible the end-results are presented in a form suitable for numerical computation, although numerical results are not quoted. For simplicity we use only one of the three techniques described in Chapter II, namely Jones's method. The reader should be able to follow the examples in this chapter after reading §§2.1, 2.2 and certain background material from Chapter 1. The corresponding solutions by the integral equation method are developed in some of the exercises and in most of the references. A feature common to most of the problems considered in this chapter is the presence of a duct or waveguide. We remind the reader of some results concerning the steady-state wave equation in - OC) < x < 00 , - 00 < Z < 00 , - b .;;; y .;;; b, when it is assumed that wave propaga tion is in the x-direction and is independent of z. The problem is two dimensional and '" satisfies (3.1) where w e first o f all assume that k i s real. I f the boundary condition on y ±b is '" 0 then the only permissible solutions of the equation are ( 3 . 2) n 1, 2, 3 . . . . '" = fn(z) !;Iin (n-rr/ 2b ) (y - b), =

=

=

Substituting in the wave equation, we find fn(x) where

=

A neY"x + Bne - Y"x,

(3.3)

A n' Bn are constants and (3.4)

If (n1T/2b) > k then Yn is real, and one wave is exponentially increasing; the other exponentially decreasing, as x increases. These are called 98

99

F U R T H E R W A V E P R O BL E M S

Yn i s imaginary. Since a time

attenuated ( -iwt) (rm/2b) k

factor exp

modes. If < then is assumed we write

(3. 5)

Kn travelling progressive where

i s real and positive. The corresponding waves are called or modes :

=

{A n exp ( -iKnx) + Bn exp (iKnx)} sin

(n7T/ 2b)(y - b).

The term in B n represents a wave travelling to the right in the duct and the term in A n represents a wave to the left. The coefficients A n and Bn are called the amplitudes of the travelling waves. For a given only a finite number of progressive modes exist. Thus if 0 no travelling waves can exist : if < < only one type of travelling wave can exist, corresponding to /<1 in the above notation : and so on. In a similar way, when the conditions on the wall of the duct are 0, = the only permissible solutions are of the form

7T/2b k 7T/b

o/ oy

=

.;;; k .;;; (7T/2b),k,

y ±b, gn ( x ) 2.... (n7T/2b)(y - b), n go(x) A oe - ikx Boeikx • cos

=

=

0, 1 ,

( 3. 6)

We always have the progressive solution

+

=

(3.7)

The other solutions correspond exactly with ( 3 . 3 ) , ( 3. 4 ) and may b e either attenuated o r progressive modes. Next consider propagation in the rectangular duct z .;;; - 00 < x < 00 . Suppose that satisfies

-c ';;; c,

-b .;;; y .;;; b,

IXIX l1li zz K2 (3. 2 ) hnp (x) (n7T/2b)(y - b) (p7T/2c)(z - c). +

Corresponding to

+

+

= O.

(3.8)

and ( 3 . 6) we have

sin cos

=

sin cos

On substituting in the wave equation we find where now

hn p (x) Yn p

= =

A n p exp ( Yn px) + B np exp - Yn p x),

(

{( n7T/2b j2 (p7T/2c) 2 +

_.

K2}1 I 2 .

Again a finite number of the waves may be progressive : the remainder will be attenuated. Suppose next that in the problem considered the waves have all the same z-factor. Then in the course of the solution we can forget about the z-facto£ providing that inst·ead of K2 we take 2} and consider the corresponding two -dimensional {K2 problem involving 3 . 1 instead of ( 3.8). Once the solution of this problem is obtained, the solution of the original problem is found by multiplying by the factor in z and expressing in terms of K. The above comments hold for real I t will be necessary later to assume that has a small positive imaginary part. In this case the

k2

=

(p7T/2c) ( ) k

k, K. k

1 00

T H E W I E N E R- H O P F T E C H N I Q U E

strict distinction between attenuated and progressive waves no longer holds-all progressive waves are attenuated to some extent. However if the imaginary part of k is small and we define y" by ( 3.4) as above, there will be a sharp difference of degree at a certain value of n-above this value the modes will be strongly attenuated and will have little progressive character : below this value waves will exist which have only small attenuation. We shall still be justified in referring to these as attenuated and progressive modes respectively. In this connexion we remind the reader of a result in ex. 1 . 3 . If k kl + ik ' k l > 0, k > 0 2 2 and � is real, then (3.9) =

The physical meaning of this is that if k is complex then all modes are attenuated at least as rapidly as exp ( - k2 I x l ) in the appropriate x-direction. 3.2

plane wave incident on two semi-infinite parallel planes

A

Suppose that the plane wave cfoi exp (-ikx cos 0 iky sin 0 ) is incident on two semi-infinite parallel planes of zero thickness in y = ±b, x � O. (See Fig. 3.1.) The boundary condition on the =

-

y

a ¢t/ay= 0

b

b

FIG. 3 . 1.

plates is assumed to be ocfotf(Jy = 0, y ±b, x < o. If we set cfot = cfoi + cfo we find (cf. §2.1 ) (i) ocf% y = ik sin 0 exp (-ikx cos 0 =f ikb sin 0), y = ±b, x < o. (ii) ocf% y is continuous on y = ±b, - 00 < x < 00 . (iii) cfo is continuous on y = ±b, x > O . (iv) cfo = 0(1), ocf% y = 0(rl/ 2 ) as r ---+ O, where r is the distance from (x,b) to (O,b) or (x,-b) to (O, -b), respectively, with x > o. (v) For any fixed y, - 00 < y < 00, I cfo l < 0 1 exp (-k 2X) as x ---+ + 00 due to the presence of a diffracted wave. For any fixed y, =

101 1

(y � b)

(y ::S;; -b) : (3.1Oa) C-b

::S;;

y ::S;; b).

(3.1Ob)

As in §2.2 an essential step in the procedure is the elimination of the unknown functions A - D in favour of functions whose regions of regularity are known. In complicated examples the best method for carrying out this elimination to obtain the equations of Wiener Hopf type may not be immediately obvious. Although the present example is relatively straightforward we give two methods of procedure. In the first, the boundary conditions given in the prob lems are introduced at an early state to help in the elimination of A - D : in the second, the functions A-D are eliminated and then the boundary conditions are introduced afterwards. The first method is shorter and is used below : the second method is more systematic and is explained in ex. 3 . l . Since o

=

=

Be -yb - Oeyb ,

(3.12a)

Be -yb + Oeyb ,

(3.12b)

Beyb + Oe -yb,

(3.12c)

- Beyb + Oe - Yb,

(3.12d)

From (iii ) ,

=

2F_(-b),

(3.13a) (3.13b)

1 02 T H E W I E N E R -H O P F T E C H N I Q U E where from (v) the F_ are functions regular in (J < k 2 cos 0. We find (3.14) We have also, on taking the derivative of (3.lOb) and setting y = ±b , «l> � (b) + «l>'-- (b) = y(-Be-yb + Oeyb ) , (3.15a) «l> � (-b) + «l>'-- (-b) = y(-Beyb + Oe -yb ) . (3.15b) From (i) «l> '-- ( ±b ) k sin 0e 'f ikb sin 0 ( 27T ) -1 /2 (OC - k cos 0 ) -1 . (3.16) Define =

«l> � (b ) + «l> � ( - b ) = S�

: «l> � ( b) - «l> � (-b) = D� , (3.17a) F_(b) - F_(-b) = D_ : F_(b) + F_(-b) = S_ . (3.17b) Add and subtract the two equations in (3.14) and the two equations in (3.15). Introduce notation (3.17) and use result (3. 16) . Eliminate (B + 0) and (B - 0) between the resulting equations. This gives , 2k sin 0 cos (kb sin 0) ( 1 + e -2yb ) D_, (3. 1Sa) S+ + 21T 1 /2 ( ) (OC k cos 0) - - 1' 2ik sin 0 sin (kb sin 0) , D+ - 21T) 1/2 - ( 1 e -2 Yb ) S_. (3. 1Sb) ( (OC k cos 0) - 1' These are equations to which the standard Wiener-Hopf technique can be applied. -Suppose we can write !(1 + e -2 yb ) = e -yb cosh yb = K(oc) = K+ (oc)K_ ( oc) , (3.1 9a) ( 2 y b) -I (1 - e- 2 yb ) = (yb)-le - yb sinh yb = L(oc) = L+ (oc) L_ (oc), (3.1 9 b) _

_

_

_

where K+ , L+ are regular in T > - k 2 ; K_, L_ are regular in T < [.; 2 ; I K+I , I K_I are asymptotic to constants, and I L+ I , I L_I are asymp totic to l ocl-1/2 as oc tends to infinity in appropriate half-planes. Explicit expressions . for these quantities are given below. Rewrite (3.1Sa) as 1 2k sin 0 cos (kb sin 0) S '+ (oc + k) I/2K+(oc) + (21T) 1 /2 (OC - k cos 0) (oc + k) I/2K+ (oc) -

{

-

(k cos 0 + k):/2K+(/C cos 0)

2k sin 0 cos (kb sin 0) . (21T) 1 /2 (OC - k cos 0)(k cos 0 + k) 1 / 2K+(k cos 0)

}

1 03 From (iv), D_ = 0( \ oc l -1 ) and S� = 0( \ oc l -1/2 ) as oc -+ 00 in appro priate half-planes. Hence all terms in the above equation tend to zero as oc -+ 00 in appropriate half-planes. There is a common strip of regularity, -k2 < T < k 2 cos 8 . On applying Liouville's theorem, each side of the equation equals zero. Hence D_ = k sin 8 cos (kb sin 8) ( 2 7T)1/ 2 (k + k cos 8) 1/2K+(k cos 8)(oc - k) 1 / 2K_(oc)(oc - k cos 8) ( 3. 2 0 ) In an exactly similar way, from (3. I8b), ik sin 8 sin (kb sin 8) S_ = 27T) 1 /2 b(k + k cos 8)L+(k cos 8)(oc - k) L_(oc}(oc - k cos 8) • ( ( 3.21 ) We next obtain explicit factorizations. (Cf. J. F. Carlson and A. E. Heins [1], A. E. Heins [1].) Consider first K+, K_ defined in (3.I9a) . The factor cosh yb is an integral function to which the infinite product theory of ex. 1 .9 and §1.3 can be applied. From the general theory (cf. ex. 1 .9) F U R T H E R WAVE PR O B L E M S

00

cosh yb = cos kb II { I - (oc/OCn _ t ) 2 } n�l 00 cos kb = II ( 1 - k2b� _ ! ), n�l

where ocn - ! = i {(I/ b� _ t ) - k2 }1/2 Combine these infinite products : 00

bn - ! = b / {(n - t) 7T } .

cosh yb = II {( I - k2bL t) + oc2 bL t } = H(oc) , n�l Decompose H(oc) in the form

say.

GO

II {(I - k2bL t ) 1 / 2 =F i ocbn - t } e ± ic
H ± (oc)

=

=

X

00

II {(I - k2b�_l) 1/2 =F iocbn_ t}e± ic
n= l

(3.22a)

1 04 T H E W I E N E R- H O P F T E C H N I Q U E where X 1 (Ot) is an arbitrary function which will now be chosen so that K+ and K_ have polynomial behaviour at infinity. As in exs. 3.4, 3.6 when l Oti ---+ 00 the behaviour of K+, K_ is independent of the term (k2 b�_ t ) in the infinite product. Thus, using results in ex. 1 .10, as l Oti ---+ 00 in a lower half-plane, 00 K_(Ot) ,....., exp {X 1 (Ot) + ibOt71'-1 ln (-2Ot/k) } II (1 + iOtb n_t ) e - iab ,,_t , n=l

,....., A exp [X 1 (Ot) + ibOt71'-1 { 1 - 0 + In ( 71'/2bk) + it71'}] , where A is a constant independent of Ot, and 0 0·5772 . . . , the Euler constant. Choose (3.22b) Xl (Ot) = -ibOt71'-1 {I - 0 + In (71'/2bk) } + tOtb, =

and then both K+ and K_ are asymptotic to constants as l Oti in upper and lower half-planes respectively. In a similar way we can decompose (3.1 9b) in the form

L ± (Ot) where

= exp {=r= X2(Ot) - T ± (Ot)} II {I - Pb�)1/2 =r= iOtbn}e 00

n=l

---+

00

ib,

± a "

(3.23a)

- ibOt71'-1{1 - 0 + In ( 271'/bk) } + lOtb. (3 . 23b) In this case I L+ I , I L_I ""'" I OtI -1/2 as I OtI ---+ oo in appropriate half planes. Note that

b n = (b/n71')

X 2 ( Ot)

=

Now return to the solution (3.20), (3.2 1 ) . From (3.14), (3.17b), B

=

-t(8_ - D_) e-yb

and we readily deduce oo + i-r cp = - 2 1/ 2 (8_ cosh yy + D_ sinh yy) e -yb - iax dOt, ( ) - (:J) + i'T (-b < y < b), (3.25a) OO + iT (8_ sinh yb ± D_ cosh yb) e - yl v l - iax dOt, ( 2 ) 1/2 - CX) +i 'T ( Iyl � b), (3.25b)

: f

=

: f

where the upper sign refers to y � b, the lower to y < -b.

1 05 In the region - b � y � b, x < 0 close the contour in the upper half-plane. For simplicity assume that 0 < k < (7T/2b) . Then S_ and D_ have poles at at = k cos 0 and S_ has a pole at at = k. These are the only real poles. On evaluating the residues, the pole at at k cos 0 gives a contribution - exp (-ikx cos 0 - iky sin 0) which cancels the incident wave. The pole at at k gives the travel ling wave in the duct : sin (kb sin 0) e - ikx . (3.26) kb sin 0 L+ (k cos 0) L_( k) All other poles give terms which decrease exponentially as x -+ 00 There are no branch points for this region of the x-y plane. The far field can be evaluated by the method given in § 1 . 6 . Use cylindrical co-ordinates (r,O). Then, as an example, in the region o � 0 < 7T - 0, if 0 is not too near (7T - 0), we find on using ( 1 .71) that the field as r -+ 00 is given by rP '"" 2 1 /2 (k7T) -1/ 2eti rr sin to sin t o(cos 0 + cos 0) -1 (kb sin 0) cos (kb sin 0) X K+(k cos 0)K+(k cos 0) + i sin (kb sin 0) sin (kb sin 0) . . (3.27) + 2kb cos to cos tOL+(k cos 0) L+(k cos 0) This is symmetrical in 0 and 0 as we should expect. Formulae for the practical computation of the K and L factors occurring in (3.26) and (3.27) are given in ex. 3.3. F U R T H E R W A VE P R O B L E M S

=

=

-

{COS

3.3

.

}

Radiation from two parallel semi-infinite plates

In §3.2 we discussed the progressive waves excited in a duct formed by two parallel semi.infinite plates by a plane wave incident from outside the duct. In this section we reverse the situation and discuss the waves radiated from a pair of plates by progressive waves travelling towards the open end from inside the duct. As in §3.2 assume that the plates lie in y = ±b, x � o. Consider two cases together. (a ) orPloy 0 on the plates. Assume that the progressive wave (3.28) rPi = exp (iKx) cos (N7T/2b)(y - b) , k > (N7T/2b), for some given N, K = {k2 - (N7T/2b) 2 }1/2 , is the only wave travelling towards the open end of the duct. N may be any of the numbers 0, 1 , 2, . . . , with N < 2bk/7T. (b) rPt = 0 on the plates. Assume that the only progressive wave inside the duct, travelling towards the open end of the duct, is rP, = exp (iKx) sin (N7T/2b)(y - b), ( 3.29) =

T H E W I E N E R-H O P F T E C H N I Q U E 1 06 where K is defined as before and again N is a given number but this time N = 1 , 2, 3 . . . , with N < 2bk/7T, i.e. zero is excluded. In the regions Y � b, Y � -b, only radiated waves exist and we can set = DeY1l, Y � -b. Y�b Wt = A e- Y1I, In - b � Y � b set rp t rp i + rp where rp can be represented by a Fourier transform and -b � y � b. We have removed the incident wave which is the only wave which increases exponentially as x -+ - 00 inside the duct. All other waves inside the duct are attenuated at least as rapidly as exp ( - k 2 Ixi ) as x -+ - oo (cf. (3.9)). Outside the duct only radiated waves exist and as in §2.2 these are attenuated as exp (-kl'lxi ) as I xl -+ 00 . Hence in -k 2 < T < k 2 ' (3.30a) Wt+(b + 0) + Wdb + 0) = A e -yb , (3.30b) W+(b - 0) + W_(b - 0) Be -yb + Ceyb , b (3.30c) W;+ (b + 0) + W;_ (b + 0) = _ yAe -y , b -yb _ e Cey ), (3.30d) W� ( b - 0) + W�(b - 0) = y( B with similar equations for y = -b. Consider cases (a) and (b) separately. (a) If orp t/oy = 0 on the plates with rpi given by (3.28), then orp t/oy = orp/oy on y = ±b, - 00 < x < 00, A = B - Ce2yb , and W; + (b + 0) W� ( b - 0) = W� (b), say W;_ (b + 0) = W� (b - 0) 0 W t+(b + 0) - W + (b - 0 ) i(27T)-l/ 2 (oc + K) -l . Introduce the unknown function, Wd b + 0) - W_(b - 0) 2 F_(b) . Subtract 13.30b) from (3.30c), and write down the common form of (3.30c, d) . (3.3Ia) ti(27T)-l/2 (oc + k)-l + F_ (b) = - Ceyb , (3.3Ib) W� (b) = -y(Be -yb - Ceyb ) . In an exactly similar way, conditions at y = -b give Ii COS N7T(27T)-l/2 (OC + K) l + F_ ( - b ) = -Beyb , (3.3Ic) b -yb ey Ce = ), W� (-b) -y(B where (3.3Id) 2 F_ ( -b ) = Wt_(- b - 0) - W_ ( -b + 0) . =

=

=

=

=

=

-

1 07 Use notation (3 .17 ) and follow exactly the same procedure as in §3.2. Then (3.31a-d) give - 1' ( 1 + C 2 yb ){D_ + ti(27T) -1/ 2 (OC + K) -I ( l - cos N7T) } , S '+(3.32a) b y D'+- 1' (1 - e 2 ){S_ + !i ( 27T) -1 /2 (OC + K) -I ( l + cos N7T) } . (3.32b) Thus there are two subdivisions (i) If N is even 4> has even symmetry about y 0 and ( 1 - cos N7T) o . By Liouville ' s theorem we can prove from the first of the above equations that S,+- , D_ are zero. The remaining equation gives, on writing (2yb) -1 { 1 - exp ( -2yb)) L+(oc)L_(oc) defined in §3.2, and rearranging in the usual way, (oc + k) -I {L+ (oc)}-I D'+- (oc) - i2b(27T) -1/ 2 (OC + K) -I (K + k)L_(-K) -2b(oc - k)L_(oc) S_(oc) - i2b(27T) -1/ 2 (OC + K) -I {(OC - k)L_(oc) + F U R T H E R W A VE P R O B L E M S

=

=

-

=

=

=

=

+

(K + k)L_(-K) } .

By the usual arguments both sides are zero, and

D,+- (oc)

=

i2b(27T) -1/ 2 (K + k)L_(-K) (oc + k)(oc + K) -I L+(oc) . (3.33)

It has already been noted that S'+- (oc) O. Hence from the definitions (3.1 7a), '+- (b) = - ,+- (- b) = tD'+- . From the equations at the beginning of this section we can deduce A - D (e.g. (3.30c, d) and the corresponding equations for. y -b) . It is found that =

=

../.. 't'

4> t

=

=

OO + iT

(oc + k)L+ (oc) cosh yy - irxx d e . oc, sinh yb y(oc + K) - 00 + i -r ( -b � y � b), (3.34) oo + i'T ib (oc + k)L+ (oc) y( b - Ivi ) - irxx d e - 27T (K + k)L_(-K) oc, y(oc + K) - 00 + i'T ( \y l � b) .

ib k)L_ (-K) ') (K + �7T

J

J

(ii) If N is odd, 4> has odd symmetry about y 0 and (1 + cos N7T) o. Corresponding to (3.33) - (3.34) we find S ,+- (oc) - 2(27T) -1/2 (K + k) I/2K_(-K)(oc + k) I / 2K+ (oc)(oc + K) -I , (3.35a) =

=

=

D'+- (oc)

=

0,

(3.35b)

1 08 cP

T H E W I E N E R-H O P F T E C H N I Q U E

-

=

1 (K + k) I/2K_(-K) 27T

(-b

� y � b),

(3.36) co + i'T 2 I/ 1 ( at +-:-k) (at) (b:f V) - i<xx -: i-----:K= ey CPt = ± (K + k) I / 2K_i -K) dat, 27T y( at + K co 'T + i where the upper sign refers to y � b, the lower to y � -b. (b) The argument in this case is very similar. Instead of (3.32) we find

f

S+

=

D+

=

_ y-l (l + e -2yb ) X {D'-- + (N7TJ4b)(27T) -1/2 (at + K) -li( l - COS N7T)}, (3.37a) -y-l (l e -2yb ) X {S'-- + (N7TJ4b)(27T) -1/2 (at + K ) -li( l + COS N7T)}. _

(3.37b)

Again there are two cases : (i) N even, odd symmetry in cP about y = 0. We find

D+(at) S+(at)

= =

-iN7T(27T) -1/2 L_(-K)L+(at) ( at + K) -l ,

(3.38a)

0,

(3.38b)

( - b � y � b), (3.39)

where the upper sign refers to y � b, the lower to y � -b. (ii) N odd, even symmetry in cP about y = 0. We find

S+(at)

=

K+(at) N7T K_(-K) T . ( 27T ) 1 /2{ K + k) I/2 . (at + K)(at + k) I/2 '

(3.40a)

D+(at)

=

0,

(3.40b)

F U R T H E R WAVE PR O B L E M S

f

oo + �

109

. Kt-(oc) cosh yy i<xx e doc, . (oc + K)(oc + k) 1/2 cosh yb - oo + i-r ( -b � y � b), (3.41 ) CO + iT _ N K+ (oc) K_( -K) ../.. ey( b - I yl ) - i<xx doc, 'l' t - 4 b . (K + k) 1/ 2 (oc + K) ( oc + k) 1/ 2 - 00 + i T ( I y l > b) . A similar analysis can be applied to the results of all four cases, i.e. equations (3.34), (3.36), 13.39) , (3.41 ) . These are of the form oo + i 'T f(oc)e -yz - i<xx doc, CPt or cP = - 00 + i T where -k 2 < T < k 2 ' and z is some constant, Z ;? O . (A) In the region I y l ;? b we use the theory of §1 .6. The apparent pole at oc -K in the explicit formulae is also a zero of the numer ator in each case since L+ (-K) = 0 if N is even and K+ ( -K) = 0 if N is odd. Hence there are no poles. We have for the distant field, considering y > b, r � 00 in direction e where polar co-ordinates are taken from (O,b) as origin (from equation ( 1 .71 ) ) N K_(-K) cP = 4 . b (K + k) 1/2

f

f

=

(3.42) (B) In the region -b � y � b, x � +00, close the contour in a lower half-plane. It will be found that the pole at oc = -K gives in each case -exp (iKx) {sin or cos } (n7Tj2b)(y - b) i.e. a term which exactly cancels the incident wav�. The remaining integral has a branch point at oc = -k and the distant field can be dealt with by the theory of §1 .6, though this is unnecessary in practice since the field as x -+ +00 must be continuous with that found in (A) . (0) In the region - b � y � b, x < 0, we can close the contour in the upper half-plane. It is found that there are no branch points in this region and an infinite number of poles at the roots of L_(oc) = 0 or K_(oc) = O. (L+ (oc)(sinh yb) -l = {ybL_(oc)}- l exp (-yb), etc.) In case (a) (i) there is also a root at oc = k. Denote the roots by oc = ocn • Then (3.43) If we let k2 � 0 it will be found that a finite number of the OCn are real, the remainder imaginary. Thus there are a finite number of propagated modes and an infinite number of attenuated modes. Of course it will be found that the propagated waves travel to the

no

THE WIEN ER-H OPF TE CHN I Q U E

left in the duct and the attenuated modes tend to zero exponentially as x � - 00 in the duct. We consider one result in more detail, namely when oCP t/oy = 0 on the walls of the duct and the only progres sive waves are exp ( ± ikx) . This is case (a) (i) with N = 0, K = k. The solution is (3.34) and in conjunction with (3.43) this gives the reflected wave ( - b � Y � b, x < 0). The coefficient of exp ( -ikx) is the reflection coefficient R and we find (ex. 3.3) R = - I Rl e2ikl , where I R I = e -bk , (3.44a) ( l/b) = 1T-l {l - 0 + In (21T/bk)} co

- (bk) -l 2: {sin-1 (bk/n1T) - (bk/n1T) ) . n�l The distant field, from (3.42), is CPt "-' ( 2k1T) -1/2e - !j 1T 2kbL+ (k)L+ ( -k cos O) rl/2eikr • (x r cos 0, Iy l - b = r sin 0) . =

3.4

(3.44b)

(3.45)

Radiation from a cylindrical pipe

Use cylindrical co-ordinates (p,z) and suppose that there is a rigid circular pipe in p = a, - 00 < z � O. CPt satisfies the steady-state wave equation in cylindrical co-ordinates 1 0 oCPt 02 cp t 2 _ (3.46) - 3"" P :l + """"5"2 + k CPt - 0, p up up uZ where for simplicity we assume symmetry in the radial direction. Consider the waves that can exist in 0 � P � a, z < O. As in §3.1 this is equivalent to considering waves in a tube 0 � p � a, - 00 < Z < 00 . By separation of variables we set CP t = f( p)g(z) and find 1 d df(p ) d2g ( z ) --p+ �2f(p ) = 0, -- - (�2 - P)g(z) = 0 2 dz p dp dp (3.4 7) i.e. 2 f(p ) AJo(�p ) + B Yo(�p ) , g(z) exp { ± ( �2 - P ) 1/ Z} In these equations � is the separation parameter. Since the pipe is rigid, o CP t/op 0 on p = a. Also CP t must be finite on p = O. Hence we must have B = 0 and =

=

=

(3.48)

III

F U R T H E R W A V E P R O BL E M S

Suppose that the roots of J l ( fl) = 0, apart from the root fl = 0, are given by fli(i = 1 ,2,3 . . . ) where fl 1 = 3·832, fl 2 = 7·016, . . . etc. Then any wave in the tube can be expressed in the form 00

CPt = aoeikz + L a.JO(flip ja) exp {- (fl7 - Pa2 ) 1 / 2 (zja)} + i�l 00 + boe - ikz + L biJ O (flip ja) exp {+ (fl7 - k2a2 ) 1/2 (zja)}. (3.49) i�l Return now to the problem of the semi-infinite pipe. Suppose that a wave CP i = exp (ikz) is incident on the mouth of the pipe in o � p � a from z = - 00 , where 0 < ka < fl 1 = 3·832 . . . , so that only the fundamental wave propagates. Thus inside the pipe we must have (cf. (3.49)) 00

CPt = eikz + Re - ikz + L biJ O(fl ip ja) exp {(fll - Pa2 ) 1 /2 (zja)}. (3.50) i�l Also as z � + 00 in 0 � P � a, CP t "-' Z-l exp (ikz). Hence if for o � p � a we write CPt = CPi + cP , then cp( p ,oc) the Fourier transform of cP with respect to z in 0 � p � a, is regular in -k 2 < T < k 2 . For p � a, using spherical polar co-ordinates, as

r -+

00 outside the pipe. (3.51)

Hence t(p ,oc), the Fourier transform of CPt with respect to z in p � a is regular in -k2 < T < k2 • The remaining boundary condition we require is oCP tjop = ocpjop = 0, (3.52) p = a, -00 < z � O. cP satisfies the same equation as CPt ' namely (3.46) . If we apply a Fourier transform in z in the usual way we find 1 d d t 1 d d - d- p - - y2 <1> t = 0 , - - p - y2<1> = 0, ( p � a) p p dp p dp dp (0 � p � a), (3.53) 2 where y = (oc2 - P) 1/ . Suitable solutions are -

t

=

=

A(oc)Ko( Yp ) , B(oc) lo(yp ),

( p � a), (0 � p � a) .

(3.54a) (3.54b)

where in (3.54b) we choose the solution which is finite as p � 0 and in (3.54a) the solution which tends to zero or represents an outgoing wave at infinity. Note that Ko(-iKp ) = !7riH�l ) (Kp ) lo(- i Kp) = J O ( Kp), (3.55)

T H E W I E N E R- H O P F T E C H N I Q U E 1 12 where y = (oc2 - P) 1/ 2 = -i(k2 - OC2 ) 1 / 2 = -iK, in our standard· notation. By Jones's procedure, (3.54) give (3.56a) «1> t+ (a) + «1> t-(a) = A(oc)Ko(ya), = (3.56b) B(oc) lo(ya), «1>+ (a) + «1>_ (a) «1>� (a) + «1> '_ (a) = yA(oc)K� (ya) = yB(oc) l� (ya), (3.56c) where in the third equation we have used the fact that orfolop = orfolop on p = a, - 00 < Z < 00 , and we have written K� (u) = dKo(u)/du = -K l ( U) l� (u) = dlo(u)/du = 1 1 ( U ) . Note that K l (-iKp ) = -inHil ) (Kp ) 1 1 (-iKp ) = -iJ 1 (Kp ), (3.57a ) (3.57b) 1 . (u)K; (u) - K. (u) l; (u) = _ u-1 • Eliminate A(oc) and B(oc) from (3.56) : «1> t+ (a) + «1> d a) = Ko (ya)y-l {K� (ya)}-l {«1>� (a) + «1> '- (a)}, «1>+ (a) + «1>_ (a) = lo (ya)y-l {l� (ya)}-l {«1>� (a) + «1>'- (a)}. Subtract these equations. Use the results «1>'- (a) = 0 and

Set Then _ 1_ i � Ko(ya) l� (ya) - lo(ya)K�(ya) «1> ' (a) __ + F_ _ -Y + (2n ) 1/2 ' oc + k l� (ya)Kb (ya) = {y 2aKb (ya) lb (ya)}-1«1>� (a) . (3.58) Set K(oc) = - 2Kb (ya) lb (ya) = 2K 1 (ya)1 1 (ya) (3.59 ) niHil ) ( K a) Jil ) ( Ka), where the numerical factors have been inserted for convenience . Decompose this function in the form K(oc) = K+(oc)K_(oc) by the theory of §1 . 3, theorem C. Then for - k 2 < c < k 2 ' y = ( ,2 -' k2 ) 1/2 , oo + ic 1 In {2K 1 (ya)1 1 (ya)} y d':" In K+ (oc) = (1m oc > c) , 2 nt ':,y - oc c oo + i oo + ic 1 In {2K 1 (ya)1 1 (ya)} y d':" In K_(oc) = - (1m oc < c) . 2 nt. , - oc - 00 + i c

{

_

}

=

•

f

f

1 13

F U R T H E R W A VE P R O B L E M S

The following results will be useful later (3.60) Note that although In {2K l (ya)I l (ya)} ,......, -In l Oti as Ot � CO in the strip, the integrals are convergent if taken in the sense T + iT

I

lim

T-.+ oo

{

} dOt.

- T + iT

We could insert an extra factor y so that the logarithm tends to zero at infinity, but this is not necessary. Let k 2 tend to zero and let Ot tend to a real number �, -k < � < k. For simplicity consider only procedure (a) of §1.3, following theorem C. Then, on setting (k2 - � 2 ) 1/2 = K, (3.61a) In K+ (Ot) t In {7TiHp ) (Ka)J 1 ( Ka)} + tll (�), (�) , t ( a)} (3.61b) a)J ( 7TiH n K n K_(Ot) K { ll p) t I I 1 where, on using ( 1 .30 ) with F(z) = In {2 K l (Z)I l (Z )} we have =

=

-

2 �a + 7Ti P For convenience set �

=

) (Z)J l (z)} f In2{7TiH�l 2 2 . 2 ka

o

a (k - � )

_

z

(k2a2

z dz Z2 ) 1/2 . _

k cos 0. Then

(3.6 2) Il(k cos 0) p(k cos 0) + iq (k cos 0), say, where p and q are real and ka 2ka cos 0 arc tan {-J l (z)/ Y l (Z)} z dz C\ , p(k cos �) = P Z2 - k2a2 sin2 0 . (k2a2 - Z2 ) 1 / 2 7T =

f o

( 3.63a )

(3.63b)

1 14 T H E W I E N E R- H O P F T E CH N I Q U E Return to the solution of (3.58) which we now write as 1 i + F_ = ( 21T )l/2 (oc + k)

-

(oc2

_

2 k2 )aK+ (oc)K_(oc)

<1> '+ (a) .

In the usual way this can b e rewritten in the form

( 21T ) -1 /2i(oc + k ) -1 {(oc - k)K_(oc) + 2 kK_( - k) } + (oc - k)K_(oc)F_ 2kK_( -k) 2 , i . (a) + 21T 1 / 2 ( ) · (oc + k) (oc + k)aK+( oc)

<1> +

From the general theory (ex. (1 .12)) we have 1 K ± (oc) 1 1 oc 1 -1/2 as oc tends to infinity in appropriate half-planes. Assuming the usual edge conditions, 1 F_ I l ocl -1 and 1 � 1 1 oc 1 -1/2 as oc -+ 00 in appropriate half-planes. Hence the above equation defines an integral function which by Liouville's theorem is identically zero. Thus � ( oc ) = ika( � 1T )-1/2K+(oc)K_ (-k) . r-.;

r-.;

r-.;

On using (3.54), (3.56c) we have finally

( p � a), (3.64a) (0 � p � a) . (3.64b) There are three regions to consider (a) 0 � p � a , :Ii -+ + 00. Deform the contour into the lower half plane. There is a branch point at oc -k. The small semi-circle round this branch point gives a contribution -exp (ikx) which exactly cancels the incident wave. The asymptotic behaviour of �t in this region is most easily obtained as a limiting case of (c) below. (b) 0 � p � a, :Ii � - 00 . Close the contour in the upper half plane. There are no branch points and only simple poles at the zeros of yI 1 (ya). The zero at oc = k gives a contribution -K_(-k)K+( k)e - ikZ :""' {K+ (k)}2e - ikZ = Re - ikZ, (3.65) =

=

since we have already defined the reflection coefficient R by (3.50) as the coefficient of exp (-ikz) as :Ii -+ - 00 in 0 � p � a. Write R = - I RI exp (2ikl) where I RI is the magnitude of the reflection

�

1 15 coefficient and l is the "end correction" . In (3.61a), if � __ k the first term on the right hand side vanishes and we have K+ (k) = exp {W ( k ) } = exp {!p(k) + !iq ( k ) } , F U R T H E R W A VE P R O B L E M S

where p and q are real and defined by (3.63) with 0 there is no need for the principal value sign and

2ka I R I - exp { 7T -

ka

J o

= O.

In this case

}

arc tan {-J 1 (Z)/ Y 1 (Z) } , dz Z (Pa2 - z2) 1 / 2

(3.66)

a

(c) P > a. We consider the far field as (p2 + Z2)1 /2 __ 00 . Write p = r sin e, Z = r cos e where (r,e) are spherical polar co-ordinates. Let r __ 00 in (3.64a) and use the asymptotic expansion Ko( Y p) '"'-' (7T/2yp) 1/2 exp (-yp) which holds as r __ oo for any fixed e. Then (3.64a) takes the form (1 .56) considered in §1.6. Use of the asymp totic expansion ( 1 .71) gives CP t '"'-' -

aK_( -k)K+( -k cos e) eikr 7T sin e Ht1 ) ( ka sin e) . r eikr eikr K+ ( k ) aiJ1 (ka sin e ) . (e) = f sin e r ' K+ (k cos e) . r

(3.67)

where we have used the results (3.59), (3.60) and the function f (e) has been defined in (3.51 ) . This gives the variation of the far-field with e. We easily find (remember (3.59), (3.65)) f(O)

=

- !ika2

f(7T)

=

tika2 R.

(3.68)

These equations differ in sign from H. Levine and J. Schwinger [1], equation (III, 12, 13), and P. M. Morse and H. Feshbach [1], equation (11 .4.33), but the sign in the references seems to be in error. This completes our discussion of the radiation from a circular pipe. Problems involving k > 3·832 when more than one progressive mode can exist in the pipe, and problems when the incident mode is not axially symmetrical can be handled by a straightforward extension of the above analysis (cf. the general case for the two dimensional duct in §3.3 and L. A. Vajnshtejn [3] , [4] , [5]) .

116 T H E W IE N E R- H O P F T E C H N I Q U E Next consider a plane wave � i = exp { -ikr (cos e cos eo + sin e sin eo cos ("P "Po))}, (3.69) incident on a circular pipe in p a, Z � 0 from direction (eo '''P o) where we use (r,e,"P) for spherical polar co-ordinates, and we shall use ( p , "P , z ) for cylindrical polars, axial symmetry being no longer present. We use the well-known result 00 � ( _ i)m e- im0Jm (v) (3.70) e -iv cos 0 m = - oo to write (3.69) in the form 00 �i e - ikz cos 00 � ( _ i)me - im('P-'Po) Jm (k p sin eo). (3.71 ) m = - oo Write � t = � i + �, where � satisfies 1 0 o� 1 02 � 02 � 3"2 + k2 � O. (3.72) + uZ -P '3'""" u p" + 2 u"P u p" P '3'""" P '5'"2 The condition o�tlo p 0 on the wall of the pipe gives O� i o� = op op 00 -k sin e oe ikz cOS Oo L ( _ i)me -im('P -'PolJ:n (kp sin eo) m = - oo on p = a, -00 < z � o. Apply a finite transform in "P and an infinite transform in z :

-

=

=

=

=

=

-

=

2".

ip (n) =

J �ei:n'P d"P o

�

=

00

(n,oc)

=

: J ip(n)ei<xz d - 00

z

( 2 ) 1/2

ip (n)

=

1

�

.

_

�(n)e-·n'P , L... 11. = - 00 oo +ib (n,oc) e - iCXZ doc. ( 2 ) 1 12 - oo +ib (3.73)

2 1T

: J

The equation for (n,oc) is 1 d d(n,oc) n2 - 2 + Y2 (n,oc) - O. p P d p . dp p This gives ( p � a) , (n,oc) A n (oc)Kn (yp), (0 � p � a). Bn (oc) ln (y p ), The boundary conditions on p = a give +(n, a + 0) + _(n, a + 0) = A n (oc)Kn (ya) , +(n, a - 0) + _(n, a 0) = Bn (oc) ln (ya), � (n) + � (n) yA n (oc)K� (ya) = yBn (oc) I� (ya) .

(

=

=

=

-

)

_

(3.740,) (3.74b) (3.75a) (3.75b) (3.75c)

�

1 17

F U R T H E R W A VE P R O B L E M S

Introduce

(fU n, a + 0) - _(n, a - 0) = D_(n), and use the result 2". 0 k sin e o n <1> '- (n) = - (2 n ) 1/2 e, 'l' dlll-r e,"(<x - k cos V o l dz o - 00 00 x 2: ( _ i) me - i m( 'I' - 'I' o lJ:n (ka sin eo) m = - oo = ( _ 1 ) nin + l( 2 n ) 1 /2k sin eo J�(ka sin eo)(oc - k cos eo) -lein'l'o . ---

f'

f'

0

Subtract (3.75b) from (3.75a) and use (3.75c) to find

1 D_(n) = 2 y aI�( ya)K�(y oc) k sin e 0 J'n (ka sin e 0 ) n'l' . e' o ' (n) + ( _ 1 ) ni n+1 ( 2n ) 1/2 X <1> + oc - k cos eo

{

.

}

This equation can be solved by the standard Wiener-Hopf technique. We deal with the case n = 0 in detail. The equation holds in -k 2 < T < k 2 cos e o . Introduce K(oc) defined in (3.59). Then rearrange and separate in the usual way. This gives 2i(2 n ) 1/2k sin e o J l (ka sin eo) 1 . . D_(O) = a(k + k cos eo)K+( k cos eo) (oc - k cos eo)(oc - k)K_(oc) (3.76) From (3.73), (3.74) we have (0 ::::;; p ::::;; a), (3.77) where -k 2 < b < k 2 cos eo. We can use (3.75a, b) to express Bn ( oc) in terms of D_( n). The term in the infinite series .with n 0 is =

i k sin eo J 1 (ka sin eo) - - ;: . (k + k cos eo)K+( k cos eo) oo + ib yK 1 (ya) Io(yp ) e - i<xz doc, X (oc _ k cos eo)(oc _ k)K_(oc) 00 + ib -

f

T H E W I E N E R-H O P F T E C H N I Q U E lIS where we have used (3.76). As z � - 00 in 0 � p � a, close the contour in the upper half-plane. The pole at IX = k cos 0 0 gives -Jo(ka sin ( 0 ) exp ( - ikz cos ( 0 ) which exactly cancels the term in n = 0 in the incident wave. The pole at IX = k gives a contribution 2J] (ka sin Oo)K+(k) - ikz f (Oo) - ikz e e , = ka sin OoK+ (k cos ( 0 ) f(O) where f (O) is the angle distribution factor defined in (3.51) and found explicitly in (3.67). The agreement between these formulae is an expression of the reciprocity principle that the amplitude of the progressive wave produced in a tube by an incident wave of unit intensity at angle 0 with the axis of the tube is proportional to the wave radiated at angle 0 by a progressive wave of unit amplitude incident on the open end of the tube from inside the tube. The remaining terms in the infinite series (3.77) can be found in exactly the same way but we do not pursue this here.

3.5

Semi-infinite strips parallel to the walls of a duct

The problems to be examined in this section and the next differ from those in previous sections in that they will be concerned with only the interior of a duct 0 � y � 2 b , - 00 < z < 00 . One impor tant result is that the integrands of integrals expressing the potential at any point will possess only poles and no branch points. Consider the duct 0 � y � 2 b , -00 < Z < 00 with ocptfCJy 0 on y 0, 2b, and a strip in y = e , 0 � Z < 00 with CPt = 0 on the strip. Suppose there is a wave CPi = exp (ikz) incident from z - 00 . Write CP t CP i + cp o Apply a Fourier transform to the equation for cP and use the condition that d <'Pldy = 0 on y = 0, 2b. In the usual way this gives (0 � y � e), <'P(y) = A(IX) cosh yy = B(IX) cosh y( 2b - y), ( e � y � b). On y e we have that <'P ± (y), <'P � (y) are continuous but <'P � (y) is discontinuous. Also =

=

=

=

=

co

.1 f

<'P+(e) - - ( 21T 1/ 2 ) _

__

o

.

1

_. eiZ(o< +k) dz - - _� ( 21T )1/ 2 (IX + k) .

Thus -i( 21T)-1/ 2 (1X + k)-l + <'P_(e)

_

___

A(IX) cosh ye = B(IX) cosh y(2b - e ) , (3.7Sa) ( 3.7 S b ) <'P� ( e - 0) + <'P� ( e) yA (IX) sinh ye , <'P � ( e + 0) + <'P� ( e) = -yB(IX) sinh y( 2 b - e ) . ( 3.7 S c ) =

=

1 19

F U R T H E R W A VE P R O B L E MS

Introduce

(3.79a) «l> � (c + 0) - «l> � ( C - 0) D � . Subtract (3.78b) from (3.78c) and eliminate A, B by means of (3.78a) . We find ' 1 y sinh 2 yb . D+ «l> ( C ) (2n)�_ 1/2 (oc +_ k) cosh yc cosh y( 2b - c) Define (3.79b) K(oc) = 2 y b cosh yc cosh y( 2b - c) {sinh 2 y b }-1 = K+(oc)K_(oc), (3.79c) where / K+ / , / K_/ ,-....., / OC/ 1/2 as oc -+ 00 in appropriate half-planes. Explicit formulae are easily obtained by the infinite product theory of §1 .3. (See (3.96) below for the case b c) . Rearrange (3.79b) as : 2k i K+(oc)D,+ . + 1/2 2b (oc + k) ( 2n ) (oc + k)K_(-k) 2k i 1 (oc - k)«l>_(c) (OC - k) . = + 2n ) 1 / 2 ' K_(oc) ( (oc + k) K_( oc) + K_(-k) Apply the Wiener-Hopf technique in the usual way. Then 2k i I i K_(oc) . «l>_(c ) = 2n 1/2 . ( ) (oc + k) + ( 2n ) 1/ 2 ' (oc2 - k2 ) K_( k ) Hence from (3.78a), 2k i K_(oc) A (oc) cosh yc = B(oc) cosh y( 2b - c) = . . ( 2n ) 1/ 2 (oc2 k2 ) K_ ( - k) (3.80a) 2b 1 _ 4kb cosh cosh c) y( yc . - ( 2n�.) 1/ 2 ' y . sinh 2 yb K+ (oc)K_(-k) (3.80b) The potential at any point is given by =

{

_

_

_

_

.

__

}

•

=

{

}

-

_

_

A.. 'f' t

=

. e,kz

_

1 + ( 2n ) 1/ 2

--

_

oo + ib

J A ( oc) cosh "ye - tc<. z doc

- oo + ib

I

: J B(oc) cosh y(2b

'

(0 � y � c), (3.81)

oo + ib

=

eikz +

( 2 ) 1/ 2

-

y) e - iaz doc,

(c � y � 2b).

- oo + ib

If z < 0 complete the contour in the upper half-plane and use forms given by (3.80b) for A, B. The only singularities are simple

1 20 THE WIENER-H OPF T E CHNI Q U E poles at IX. k, and values of IX. in the upper half-plane defined by 2 yb m7Ti(m 1 , 2, 3 . . . ) i.e. IX. i {(m7T/2b) 2 - k2 }1 / 2 iY m say, (3.82) where we assume k < (7T/2b) so that I' m is real and positive. Then either of the integrals in (3.81) gives for z < 0, 0 :::;; y :::;; 2b, =

=

=

=

A..

_

'f' t - e

ikz

=

'

1 e - ikz {K+( k) }2 2k � 1 n7TC n7TY cos 2b cos -2b eYnz • (i ) K K+ (k) nL. + Y Y n �l n

-

(3.83)

If z > 0 we close the contour in the lower half-plane and use forms given by (3.S 0a) for A, B. For 0 :::;; Y :::;; c, c :::;; Y :::;; 2b there are simple poles at IX. -k and values of IX. in the lower half-plane given by yc (n + t)7Ti and y( 2b - c) (n + t)7Ti respectively. The pole at IX. -k exactly cancels the incident wave and the remaining terms give eigenfunction expansions in terms of cos (n + t)7TY/C and cos (n + t)7T(2b - y)/(2b - c), respectively, which are appro priate for boundary conditions oCPt/oy 0 on y 0, 2b and CPt = 0 on y c. We next assume that the strip is resistive so that instead of CPt 0 on the strip we have CPt ± i(jocpt/oy on y c ± 0 respec tively. This gives, on writing CPt CPi + cP as before, + (c ± 0) =t= i(j� (c ± 0) = -i( 2 7T)-1/2 (1X. + k)-l . (3.84) =

=

=

=

=

=

=

=

=

=

=

Instead of (3.78) we have

+(c - 0) + _(c) = A(IX.) cosh yc, '+ (c - 0) + :"' (c) yA(IX.) sinh yc, +(c + 0 ) + _(c) B(IX.) cosh y ( 2b - c), � (c + 0) + :"' (c) -yB(IX.) sinh y( 2b - c) . =

=

=

_(c) _(c) where

-

+

B

and use of (3.84) (cf. §2.9) gives 1 i(27T)-1/2 (1X. + k)-l + 1'- coth yc :"' (c) (3.85a) + L(IX. : c HI' sinh YC}-l'+ (C - 0), i(27T)-1/2 (1X. + k)-l 1'-1 coth y( 2b - c):"' (c) - L(IX. : 2b - cHI' sinh y( 2b - C) }-l� (C + 0), (3.85b)

Elimination of A,

=

=

L( IX. : d)

=

cosh I'd

-

+

ifJY sinh yd .

121

F U R T H E R W A VE P R O B L E M S

These are simultaneous Wiener-Hopf equations which cannot be solved exactly in the general case by the standard procedure (cf. §4.4) . However if c = b, (3.85) can be reduced to two independent Wiener-Hopf equations. On subtracting (3.85b) from (3.85a) we can readily prove that in this special case � ( b ) = 0, � ( b + 0) - � (b - 0) . These results are obvious physically from the sym metry of the problem. If we next add (3.85a, b) with c = b, we find, using notation (3.79a), =

I }

{

i 2 y sinh yb (3.86) ( b ) - 2n 1/2 oc. cosh yb + i(jy sinh y b _ +k ( ) An equation which is similar to this, apart from notation, is con sidered in ex. 3.13. If (j = 0, (3.86) reduces to the special case of (3.79b) with c = b . A similar analysis holds for an infinite circular duct containing a semi-infinite circular tube. Consider the case where there are no resistive losses. Suppose that the infinite duct occupies 0 � p � a, -00 < z < 00 with ocp t/o p = 0 on p = a; suppose that CPt = 0 on a cylinder lying in P = b, 0 � z < 00. Assume a wave CP i = exp ( ikz ) incident from z = 00 and write CPt CP i + cp o If we apply a Fourier transform in z we find (cf. §3.4) ,

D+ =

-

.

=

-

(O � p � b), ( oc. ) = A(oc.)Io( Yp ), B(oc.) {Io(yp )K� (ya) - Ko(y p )I� (ya) } , (b � p � a), where we have used the boundary conditions : finite on p = 0, and d jdp = 0 on p = a. The boundary conditions on p = b give -i(2 n )-1 / 2 (oc. + k)-l + _(oc.) ----: A (oc.) Io (yb) B(oc.){Io(yb)K� (ya) - Ko(yb)I� (ya)}, � ( b + 0) + � ( b ) = yB(oc.){I� (y b)K� (ya) - K� (yb) I� (ya)}, '+ (b - 0) + � (b) = yA (oc.)I� (yb). Eliminate � ( b ) by subtracting the last two equations, and set =

=

� (b + 0) - � (b - 0) = D � (oc.). Eliminate A and B, and use (3.57b). Then K(oc.)D� (oc.) = {_ ( oc. ) - i(2n ) -1/ 2 (oc. + k)-l } ,

(3.87)

where

K(oc.) = bl o ( yb){I� (ya) }-l {Io(y b )K� (ya) - Ko(yb)I� (ya)}. This is now in standard form (cf. (3.79b)) and the solution proceeds as before. It might seem that K(oc.) has branch points at oc. ±k =

12 2 T H E W I E N E R- H O P F T E CH N I Q U E because of the presence of K� (ya), Ko(y b) but this is not the case since yK 1 (ya) yl 1 (ya) In (ly) + a function with no branch point!', Ko(yb) = - Io (yb) In ( !y) + a function with no branch points. =

Hence the logarithmic terms in K(oc.) cancel, and K(oc.) has only simple zeros and poles. The factorization can therefore be performed by an infinite product decomposition. (See L. L. Bailin [1], N . Marcuvitz [ 1 ] . These references contain numerical tables.) 3.6

A strip across a duct

Consider the duct 0 � y � 2b, - 00 < z < 00 with a strip o � y � b at z O. (Finite width in the x-direction can be dealt with as already indicated in §3. 1 .) Suppose that the boundary condition on all boundaries is oCP t/on = 0 and that a wave CP i = exp (ikz) is incident from z - 00 . Set CPt CP i + cp o Then ocp/oz = - ik on z = 0, 0 � y � b. Apply a Fourier transform in z to the equation for cP in the region b � y � 2b. Using the condition that ocp/oy = 0 on y 2b we find CP(y,Ot) A (oc.) cosh y (2b - y) . Hence on y = b, in the usual notation, CP+ (b , oc.) + CP_ (b , oc.) A(oc.) cosh yb, (3.88a) =

=

=

=

=

=

CP '+ (b , oc.) + CP � (b , oc.)

=

-yA ( oc.) sinh yb.

(3.88b)

Elimination of A (oc.) gives where we have simplified the notation slightly by omitting reference to 'b ' . Apply a Fourier transform in z in the region 0 � y � b, z � 0, noting that

where the suffix zero refers to the value on z = O. We know that (ocp/oz)o = -ik but (cp) o is unknown. Denote (cp)o by f(y) . Then the partial differential equation becomes

123 To eliminate the unknown function f(y), change the sign of oc and add the resulting equation to (3.90). This gives d 2 {+ (y,oc) + +(y,-OC)}/dy2 - y 2 {+(y,oc ) + +(y, -oc)} = - (2 1T)- 1/22ik. FURTHER WAVE PRO B L E M S

The required solution of this equation, with d + /dy

0 on y = 0, is +(y,oc) + +(y,- oc) = B(oc) cosh yy + (21T)-1/ 2 2iky -2 . (3.9 1 ) Differentiate with respect t o y , eliminate B(oc) between the resulting equation and (3.91), and set y = b . We obtain + (oc) + +(-oc) = y-l coth yb{� (oc) + � ( - oc)} + + (21T)-1 /22iky-2 . (3.92) In exactly the same way, in 0 � Y � b, - 00 < z � 0, we find _(oc) + _(-oc) = y-l coth yb{� (oc) + � (-oc) } (3.93) - ( 21T)-1 / 22iky-2 . Equations (3.89), (3.92), (3.93) are the basic equations for the problem. Add (3.92), (3 .93) : =

{+( oc) + _(oc)} + {+(-oc) + _(-oc)} = y-l coth yb[{ '+ (oc ) + � (oc)} + {� ( - oc) + � ( -oc)}] . From (3.89) : {+(oc) + _ (oc) } + {+(-oc) + _(- oc) } = _ y-l coth yb[{� (oc) + � (oc)} + { '+ ( - oc) + � ( - oc)}] . Add and subtract these two equations and apply the Wiener-Hopf technique. Then � (oc) = -� (-oc). (3.94) Eliminate + (oc) from (3.89), (3.92) by subtraction and use (3.94). We find

This may be compared with (3.79b) with b

y b coth yb

=

=

c.

(-k 2 < T < k 2 ) . (3.9 5) Set

K(oc) = K+(oc)K_(oc), b bp = - . p1T

(3.96a) (3.96b)

124 T H E W I E N E R-H O P F T E C HN I Q U E This decomposition has been obtained in ex. 3.6. We can rewrite (3.95) as ik 1 oc - k (-oc) - 21T 1/2 ' K_(oc)
{ I

f

{

I }

I }

On changing the signs of z and oc we see directly that, as implied by the first equation in (3.94), tfo ( y, -z ) = - tfo ( y, z ) . (3.9 8) To obtain tfo for z � 0, b � Y � 2b, close the contour in (3.97) in a lower half-plane and evaluate by residues. It is convenient to use (3.96a) and the result K+(-oc) = K_(oc) to rearrange (3.97) as oo + iT ik K_(oc) b tfo ( y, z ) = 21T K+ (k) y2 cosh y b. Y sinh ybK_(oc) - oo + iT X cosh y( 2b - y ) e - i",z doc. Remembering that tfo t = tfo i + tfo and tfo ( y , -z) = - tfo(y,z) we obtain (z � 0), (3.99a) tfo t Poeikz + S( y , z) , ik ik (z � 0), ( 3.99b) = e z + (1 - P o) e - z - S( y ,z), where, if Y 21 = { (p1Tlb) 2 - k2 }l/2 ,

f{

=

}

1 25

F U R T H E R WAVE PR O B L E M S

We have indicated how (3.99a, b) are derived for b � 'Y � 2b. Exactly the same expressions are found in 0 � 'Y � b by a similar method. The results of this section have been obtained by G. L. Baldwin and A. E. Heins [1]. We have already noted that (3.95) and (3.79b) with b = c are of similar form. The equivalence of the problems leading to these two equations has been shown by S. N. Karp and W. E. Williams [1] by means of a direct argument not involving the Wiener-Hopf technique. A connexion between the problem solved above and a set of linear simultaneous algebraic equations is indicated in ex. 4.12. Various methods have been suggested for approximate solution of problems like those solved exactly in this Chapter. It should now be possible to check exact against approximate solutions, and perhaps to develop improved approximate methods. In connexion with diffraction problems some interesting investigations of this type are given in L. A. Vajhnstejn [1], [2], [6]. Miscellaneous Examples and Results I I I 3.1 In complicated examples the following procedure may be pre ferred to that used in §3.2. We use it to obtain (3. 1 8) . First define our standard notation. If '" is continuous on y ±b, x > 0, we define

S+

=

(a) 0, we define If however ", is discontinuous on y
S<Wf == S

=

=

=

t (S
D<
=

=

=

=

=

=

- y{
=

=

Add and subtract these and introduce the above notation S'+ + -y(D+ + D<�h -y(A - D}e - yb• D'+ + D� - y(S+ + s
S� =

=

=

= =

(c) (d)

126

T H E W I E N E R-H O P F T E CH N I Q U E

Similarly (3. lOb) give
= Be hb + Ce ±rb, = - y Be hb + yCe ± yb ,

where upper and lower signs go together. Add and subtract in pairs and eliminate (B + C) and (B - C) . Introducing our standard notation, we find S+ D'+

+ s'- = y coth yb(D+ + D
=

=

2y( C - B) cosh yb , 2y(C + B) sinh yb.

(e) (f)

Clearly (c) and (e) go together. The functions S'+ , D+ , D�) , Dt = 0 on the planes instead of at/ ay = o. Show that corresponding to (3. 1 8) we have S+ + (27T)-1/22i cos (kb sin 0){ IX - k cos 0}-1 = _ y-l ( l + e - 2rb ) D '- ,

D+ + (27T)-1/22 sin (kb sin 0){1X - k cos 0}-1

=

-

y-l ( l - e - 2rb )S,- .

Hence (cf. (3.20), (3.2 1 ) ) D'- (27T)-1 /2i cos (kb sin 0 ) (k

s'-

+ k cos 0)1/2(1X - k)1/2 {K+ (k cos 0)K_(IX)(1X - k cos 0n-I , / - (27T) 1 2b 1 sin (kb sin 0){L+(k cos 0)L_(IX)(1X .:... k cos 0)}-1. X

The far field and the travelling waves inside the duct can be deduced as in §3.2. As an example suppose that t7T < kb < 7T and define K = {k2 - (7T/2b)2}1/2. In - b <: y <: b, x < 0 close the contour for in the upper half-plane. It will be found that there are two poles at IX = k cos 0, K. The residue from the first pole exactly cancels the incident wave and the residue from the second gives the travelling wave 2b cos (kb sin 0)(k + k cos 0)1/2 (K - k)1 /2 . . hm --'-::=--,-:-'-----'--::-:----' 7T K - k cos 0 CX--> K K+ (k cos 0) -

-

-

-

} e_. IX {K_(IX) -K

--

'KX

cos

The coefficient of exp ( - iKX) is the "transmission coefficient" . limit is easily computed by writing lim

CX--> K

{� K_(IX) }

=

K+ (K) lim

CX--> K

{� K(IX) }

where we have used the result that for

IX

=

=

K+ (K) JC(K)

K, Y

=

(K)7T 2Kb2i '

= K+

- ti7Tb-1.

(1rY)

- . 2b The

127

F U R T H E R W A V E P R O BL E M S

The solution o f this problcm b y means o f an integral equation approach is given by A. E. Heins [1]. 3.3 Equations (3.22), (3 . 23) can be transformed into expressions which are convenient for numerical work as follows. If 0 < kd < 1 we write

( 1 - k2d2 ) 1 /2 - irJ.d where tan

'P

=

=

{ I - (k2 - rJ.2 )d2}1/2 exp ( - i'P)'

rJ.dj( l - k2d2 ) 1/2 , o r equivalently

this latter form being convenient when rJ. If 0 < kb < 7T so that kbn = (kbjnrr) <

L+ (rJ.)

=

=

k. 1 for all n, (3.23) give

{(yb)-l sinh ybp /2 exp [ - irJ.b + irJ.brr-l {l - 0 + In (2rrjbk)} 00

- ybrr-1 arc cos (rJ.jk) + i L (rJ.b n - 'Pn ) ) ' n =l

where

If rJ., k are real and -k < rJ. < k we can set rJ. = k cos JI. (JI. real), and then y = -ik sin JI., arc cos (rJ.jk) = JI. and an elegant expression is obtained for L+ (k cos A). In particular

L+ (k)

=

exp

[ -ikb + ikbrr-1{ 1 - 0 + In (2rrjbk)} 00

+

+ i L { (bkjnrr) - sin-1 (bkjnrr)}]. n =l

From

(3.22), with 0 < kb < irr, we find

00

- ybrr-1 arc cos (rJ.jk) + i nL (rJ.bn - l - 'Pn - l ) ) , =l bn - l' 'Pn - l are defined as for b n, 'Pn with (n - i) in place of n. If kb < � and we define K = {k2 - (rrj2b) 2}1/2 , it is convenient to write the first term of the infinite product in (3 . 22a) as {( I - k 2bi ) 1/2 /2 irJ.b1 /2 } = -i(rJ. + K)(2bjrr) and we find K+ (rJ.) = -i {cosh ybj(rJ.2 - K2 )}1/2 (rJ. + K ) exp [ -trJ.b where

trr

<

00

- ybrr-1 arc cos (rJ.jk) + irJ.brr-l{ 3 - 0 + In (7Tj2kb)} + i L (rJ.b n - t - 'Pn - t ) ) . n =2 The term under the square root is positive for all rJ. in the range 0 < rJ. < k (rJ., k feal). When the Laplace transform is used, the appropriate formulae have been given by D. S. Jones [4 ) .

128

THE

W I E N E R-H O P F T E CH N I Q U E

3.4 Consider the asymptotic behaviour as ex tends to infinity of the following expression (cf. ( l . 1 9 ) ) ,

K+ (ex)

+ (ex/exn)}e - c
co

{K(OWi2e - X(C<) II { I

=

Suppose that

+b

(a)

n=l

a as n ->- 00 . (b) c fln = Compare the behaviour of K+ (ex) as ex tends to infinity with that of

exn

a

=

J(ex)

n

IT

=

n=l

+

O (n-l )

( + an_ex+_b) e -rxian 1

K+( ex)/J(ex)

i.e.

=

=

e -Cc
IT (an + b ) IT ( exn

n=l

a

r �+ �+ 1

l

a

a

{ � ( � L) }

{K(OWi2e -x(c<) exp ex x

(� + )j (

n=l

a

)

(c)

-

exn + ex an + b + ex) .

(d)

Denote the last infinite product by Q(ex). An algebraic rearrangement gives

{ + exn n b} an + b ex an + b + ex K(ex) . I fn(ex) I On-2 ex) exl + Q (ex) II {I + fn(ex)} Q(ex)

=

co

_

II 1

n=l

a

_

-

=

co

II { I

n =l

+ fn(ex)},

say.

Assume that i s non-zero for all ex , (1m ex) > - (1m exl ) where exl is the smallest root of Then in view of (b) it is easy to prove that < where 0 is a constant independent of ex, for all ex such that (1m :> - (1m ) 6. Hence the infinite product for Q(ex) is uniformly convergent and lim

<X----+ co

=

co

=

lim

n = l <x----+ 00

1,

(As in J. Bazer and S. N. Karp [ 1], if the exn are imaginary it is possible to prove that this result holds for - !7T + 6 ' < arg ex < i7T - 6 ' but this is not necessary here.) Now return to (c). Inserting the asymptotic expression for J(ex), we find

K+ (ex)

�

{

exp -X(ex)

+. ex-

a

( 1 - 0) -

(-ex + -b + -1 ) (-ex + -b + ) 2 a

a

hI

a

a

1

+

3. 5 We reduce the infinite product (a) of ex. 3.4 to a form suitable for computation for the special case exn = (k2 - <5�)li2 = i( <5� - k2) 1/2, fln = iBn' where <5n, 6n are sequences of real numbers. Assume that k and ex are real. exn is real for 1 .;;; n .;;; N, say, and imaginary for n > N.

F U R T H E R W A VE P R O B L E M S

129

Denote the real values by exn = I
K+ (ex) K_(ex) where

'Y

=

GO

L

=

e - 2x(<x)

IT I
n � l l
{"Pn - (ex/en )},

tan

"P n

{

+ 2i

�

L �l

n

ex or sin "P n Yn

= -

n�N+l

O n multiplying the above expression b y K(ex)

K+ (ex)

=

{ / { fl (I<; - ex2) }T/2 fl (I
e - x(<x K(ex)

The modulus

=

K+ (ex)K_(ex) w e have

exp

{ � C:) } i

- i 'Y .

[K+ (ex) [ has a particularly simple form.

Consider the decomposition of

3 .6

K(ex) where

(�) - 2i'Y} , en

c+d

=

=

(cd/a)y sinh ya{sin yc sinh yd}-l

K+ (ex)K_(ex)

a. As in ex. 1.9 and §3. 2

(yb) -1 sinh yb

=

Define

G+(ex,b) and write

=

GO

II {( I - k2b;) + ex2b;},

n �l

=

00

II {( I - k2b;,jI/2 - iexbn}ei<xbn,

n=l

If the three infinite products are combined into one we obtain 00 {( 1 k2a2n ) 1/2 - i exan} . (a) K+ ( ex ) - e -Xl(<X) II I - k2c;.j1 / 2 - iexcn}{( l - k2d;) 1!2 - iexdn} n � l {( _

The arbitrary function Xl (ex) is chosen so that K+ (ex) has algebraic behaviour as ex -+ w in an upper half-plane. We examine the asymptotic behaviour of K+ (ex) for large ex. The terms k2a; etc. can be ignored (ex. 3.4) and

K+ (ex)

�

{ r( l - iexlZ1T-1 )}-l r ( 1 - iexC7T-1 ) r ( 1 - iexd1T-1 ) exp { -Xl (ex)}.

On using Stirling's fonnula we deduce that if we choose

Xl (ex) = iex1T-1 (a In a - c In c - d In d), then K+ (ex) --- ex1/2 as ex -+ w in an upper half-plane.

1 30

THE WIEN ER-H OPF TECHNIQUE

A slightly different, though equivalent, expansion is obtained by writing instead of (a),

K+ (or.)

=

e

_

( )

X2 "

n °O {( I - k2a �n_l )1/2 - ior.a2 n _ l}{( 1 - k2a �n ) - i or.a 2 n} . { ( I _ k2C�)1/2 _ ior.cn}{ ( l _ k2d � ) - ior.dn} n=l

It is then found that we must choose =

i or.7T-1(a In ia - c In c - d In d) . This is particularly convenient when c = d = �a since then X 2 (or.) = O. In this case we have K(or.) = yb coth yb = K+ (or.)K_(or.) where X 2 (or.)

K+ (or.) =

00

I - k2 b�_ t) 1/2 - i or.bn _

J1 {( {( I - k2b�)1/2 - ior.bn} t}

A result which is required later is that if we set and let k --+ 0, then

or. =

k in this formula (b)

3.7 The decomposition of K(or.) = 2Il (ya)Kl (ya) = 7TiH�l ) (Ka)Jl ( Ka) where y = (or.2 - k2 ) 1/2 = -iK, has been discussed by H. Levine and J. Schwinger [ 1 ] who use methods similar to (a), (b) at the end of § 1 . 3, D. S. Jones [5] who uses a variant of method (b), and L. Vajnshtejn [3] who uses method (c). When these authors shift contours in methods (b) and (c) the presence of the zeros of Il (ya) tends to make manipulations complicated. It would seem to be simpler to proceed as follows : Set

K(or.) = 2e -ya{(ya) -lIl (ya)}{yaeyaKl (ya)}. The factor exp ( -ya) has the simple decomposition given in ( 1 .35) . (ya)-lIl (ya) can be dealt with by the infinite product theory already

developed. The only factor which needs to be decomposed by the integral formula is ya exp (ya)Kl (ya) and this function has no zeros in the complex plane. We can obtain some idea of the complexity of the integrand in the decomposition formulae ( 1 .32), ( 1.33) by considering We have

Hence

K1 ( -z) = K1 ( e -i1Tz) K1 (iz) = -t7TH�2) ( Z ) z

)

-K1 (z) + 7TiI1 (z), K1 ( - iz) = -i7TH�l ) (z).

=

i In [K(z)fi{Ki (z) + 7T2Ii (z)}] - li7T - i arc tan {K1 (z)/7TI1 (z)}, F(iz) - .F( -iz) = - 2i arc tan { Y1 (z)/J1 (z)}. F(z) - F(

-

The simplicity of this second result suggests the use of method (c) of §1.3 and in fact elegant formulae o f this type are given by L. A . Vajnshtejn [3] . All three authors quoted above give approximate. formulae.

F U R T H E R W A VE

131

PR O B L E M S

The only published table of values for K+(ex) are given by D . S . Jones [5] who gives tables for ex = k, 1<1 ' 1< 2 where I< i = (7c2 - ri ) 1/2 and ri is the ith root of the equation Jo ( ria) = O. The tables are for ka = 0(0, 25) 10. 3 . 8 Integral equation formulations . There are broadly speaking three methods of procedure and most of the important features have already been mentioned in Chapter II. We give a brief summary to introduce the reader to the literature. General references are P. M. Morse and H. Feshbach [ 1 ] and A. E. Heins [ 10] . It has already been emphasized that when we use Jones's method the discussion of integral equations is unnecessary. (i) The easiest procedure in some cases is the physical method of ex. 2 . 7 . Thus suppose there are two semi-infinite plates in y = ± b, - 00 < x .;;; 0 with

=

where

o

J {H�1) (kR1 )f1 (;) + Hbl) (kR2)f2 ( m d;,

- co

Ri = (x - ; ) 2 + (y - b) 2 R� = (x - ; ) 2 + (y + b) 2 . Let (x, y ) tend to (x, ± b) and obtain two simultaneous integral equations o

J {H�l) (k lx - ; [ )f1 ( ;) + H�1) (kR)f2 (;)} d; +
=

0,

rri J {H�1) (kR)f1 m +Hb1) (kl x - ; [ )f2 (;)} d; +
=

rri

- co

o

- co

R2

=

( - 00 < x .;;; 0), 0 , ( - 00 <

x .;;; 0),

(x - ;) 2 + 4 b 2 . If we add and subtract these and set s (;) , f1 ( ;) - f2 ( ;) = d( ;) we obtain two independent integral equations, each holding in - 00 < x .;;; 0 : where

f1 ( ;) + f2 ( ; )

-rri

=

o

J {H�l) (k lx - ; 1 ) ± H�l) (k R )} �W) d; +
- co

=

O.

These are of the Wiener-Hopf type and can be solved in the usual way, remembering the result (cf. ( 1 .60), ( 1 . 6 1 ) ) :

i

co

J H�1) {k(a2 + x2 )1/2}ei<XX dx

- co

=

2y-1e r i a l -

.

A. E . Heins [1], part I, has used this approach to solve the problem of ex. 3 . 2 . See also L. Lewin [1], L. A. Vajnshtejn [1].

1 32

T H E W I E N E R-H O P F T E C H N I Q U E

(ii) Integral equations are readily established by applying Green's formula to the whole space excluding the region occupied by the thin plates (cf. ex. 2.6). This has been used by H. Levine and J. Schwinger [ 1 ] for radiation from a circular pipe, and W. Chester [ 1 ] for radiation from parallel plates. (iii) The space can be split into subspaces and Green's formula can be applied separately to each of the subspaces. This is the procedure used in § 2 .4. In the problem of two parallel plates with
b , ( b ) - b .;;; y .;;; b, ( c ) y .;;; - b . Green's functions for (a) and ( c ) present no difficulty. For (b) a suitable Green's function would have

27T

= -

b

2: y - l sm. 00

n�l

n

nrr

2b

-

. n7T (y - b) sm (Yo - b)e - y"lx -xol , 2b -

where Yn = {(n7T/2b) 2 - k2p/2 . Although this type of Green's function is frequently quoted in the literature it is not very useful since we are interested ultimately only in the Fourier transform of this with respect to x and to compute a closed form of this result we should need to sum an infinite series. It is much easier to find the transform of the Green's function by applying a Fourier transform in x to

-

The required transform of the Green's function is the solution of this equation such that

2 a ( a2 k2 )-1'Y1 ( a ) + p'Y 2" ( a ) q
where p,

q

K( a )

K+ ( a ) /K_ ( a ) and �et F( a )K_( a ) = E+ (rt.) + E_(rt.)

are given constants. Write

K_( a )'Y:i ( a ) a+k

=

=

=

K( a )

=

{K_ ( a )'Y:i (rt.) - K1 ( -k)'Y:i ( -k)} K1 ( -k)'Y:i ( -k) . + a+k a+k

Then the first equation can be solved by the Wiener-Hopf technique . in the usual way, and we find

1 33

F UR T H E R W A V E P R O B L E M S

Similarly the second equation can be solved to give

where L(ex) L_(ex)/L+ (ex). If we set ex k in (a) and ex = -k in (b) we obtain two simultaneous linear algebraic equations for the unknown constants (Dl (k), 'Y1 ( -k). The solution can then be completed. =

=

3.10 Examine the radiation from a coaxial waveguide with infinite central conductor and semi-infinite outer conductor i . e . solve the steady state wave equation in p ;> a, - 00 < Z < 00 with a cylinder in p = b > a, - 00 < z ..;; 0, and wave incident from z - 00 in a ..;; p ..;; b (cf. N. Marcuvitz [ 1], p. 208 ) . =

3.11

( i ) Examine the radiation from the parallel plate duct y = ± b , = ± B, - 00 < x < 00

- 00 < x ..;; 0 enclosed i n parallel walls y

(B � b). (ii) Examine the radiation from the circular duct p a, - 00 < z ..;; 0 enclosed in the tube p = A , - 00 < Z < 00 (A � a ) . In the limiting case A , B 00 we obtain the problems of §§3 . 3, 3 . 4 . The case o f infinite A , B i s much more complicated than the case of finite A, B due to the presence of branch points. It would be interesting to know how far the results for finite (but large) A, B could be used to approximate to the results for infinite space e.g. near the mouth of the duct. (The transition from a function with poles to a function with branch points is examined from another point of view in V. Kourganoff [ 1], §27, where references are given to the original work of S. Chand rasekhar. ) =

->-

3.12 The reflection o f an electromagnetic wave falling on an infinite set of staggered semi-infinite plates lying in y na, na cot A ..;; x < 00 , (n = . . . - 2, - 1, 0, + 1 , + 2 . . . ) has been treated by J. F. Carlson and A . E. Heins [1], A. E. Heins and J. F. Carlson [ 1 ] , and A. E. Heins [ 1 1]. From the point of view of Jones's method it is natural to make use of the periodicity of the structure in y, to consider the problem in the strip 0 ..;; y ..;; a, - 00 < x < 00 (cf. A. E . Heins [ 1 2] ) . The solution involves the infinite product decomposition of =

cosh ya - cos ( kp - exb),

( a, k, p, b given constants) .

3.13 Sound propagation in a lined duct. Consider transmission of sound inside the duct 0 ..;; y ..;; b, - 00 < x < 00 with a
=

b,

=

where

K(ex)

=

y yb{y sinh

=

sinh yb - ik1] cosh b} l.

y

-

1 34

THE

Define

W I E N E R- H O P F T E C H N I Q U E

I-' m by the equation 1Tl-' m tan 1Tl-' m

As 'YJ -+ 0,

1-' 0 � 1T-1 ( - ik'YJb) 1/2

:

P m ""

=

-ikb'YJ.

m - ikb'YJ(�)-l , (m

=

1, 2, . . . ) .

The infinite product decomposition of K(cx) gives 00

L+ (cx) = (cx+ k)(cx+ ko )-l II [ ( I _ k2A � ) 1/2 - icxA m][( I -k2B� )1/2 -icxB m]-l , m=l

where A m = b/(mw), B m = b/(l-' m1T), ·Wiener-Hopf equation gives

As z -+

([>+ (b)

k�

=

k2 - (1T1-'0/b) 2 . Solution of the

=

-( 21T)-1/2k'YJK_( -k)K+ (cx)/(cx + k). - 00 , t ,.., exp (ikz) + R exp ( -ikz) and i t i s easily deduced

that the reflection coefficient R is given by

R

= =

-ik'YJ( 4k2b)-lK_( -k)K+ (k) k - k0 __ II [( 1 - k2A� ) 1/2 - ikA m ]2 [(1 - k 2 B� )1/ 2 - ikB m]-2 . k + ko m = l 00

This problem has been treated in detail by an integral equation method by A. E. Heins and H. Feshbach [1] : see also P. M. Morse and H . Feshbach [1], p. 1 522. The problem of the semi-infinite resistive strip parallel to the walls of a duct, and centrally placed (§3.5, (3.86)), is equivalent to the above problem except that then we are generally interested in small 15 = 'YJ-1 instead of small 'YJ. A similar problem involving a resistive strip has been considered by V. M. Papadopoulos [1] but the solution of his Wiener-Hopf equation (his equation ( 17)) is unnecessarily complicated . Consider transmission of sound inside the circular duct a, - 00 < z < 00 , with ot/ op = 0 on p = a, z < 0 and ot/ op = ik'YJt for p = a, z > 0, and incident wave i = exp (ikz).

3 .14 o <: p <:

3 . 1 5 I t i s convenient t o summarize some standard notation for radiation-type boundary conditions. For sound waves falling on partially absorbent material we have o/ On = idcp, where (Re d ) is positive but (1m d ) may be positive or negative (P. M. Morse and H . Feshbach [1], p. 1 522). The positive direction o f the normal i s taken into the material. For electromagnetic waves we generally write o/ on = id1 for E-polarization, = -id2 0/ on for H-polarization, where 151 ' 15 are constants with positive real and imaginary parts (cf. ex. 2 2. 1 1 for terminology and T. B. A. Senior [1] for references). If a time factor exp ( + iwt) is used then the sign of i must be changed throughout. 3.16 A. E. Heins [2], [3], [ 4 ], M. Weitz and J. B. Keller [1], have considered various cases of reflection of water waves in water of finite

1 35

F U R T H E R W A VE P R O B L E M S

depth in - b .;;; y .;;; 0, - 00 < x < 00 , - 00 < z "'''' + 'Y'Y - k2 = 0, (k real). Typical problems are (i) iJ/ iJy

iJ/ iJy

=

=

0 on y

p o n y 0 on y iJ/ iJy = p on y iJ/ iJy = 0 on y

(ii) iJ/ iJy

=

=

=

=

=

=

-b, 0, x

- 00

<x <

.;;; 0

00 .

iJ/ iJy

- 00

=

00 .

q on y

< x < 00 . 0, - 00 < x < 00 . -a, 0 ';;; x < 00 , ( - b

-b,

<

<

satisfies

=

0, x

>

o.

-a < 0).

In both cases the infinite product decomposition of the following function is required :

p, sinh p,a - A cosh p,a,

3.17 Laplace-type equations. We consider solution of 'i/2 = 0 by means of the Wiener-Hopf technique. The two-dimensional examples which follow can be solved by conformal mapping. The justification for discussing the Wiener-Hopf solutions is two-fold. The same tech nique can be used for axially symmetrical problems : and the same procedure can be applied to more complicated equations where it may not be very obvious how to obtain the solution by conformal mapping. (i) In some cases it is possible to solve Laplace's equation directly without considering the limiting case k 0 of a solution of 'i/2 + k2 = o. As an example consider the solution of 'i/ 2 0 in 0 .;;; y .;;; b, V on y = 0, - 00 < x < 00 with = 0 on y b, - 00 < Z < 00 , o < Z < 00 , iJ/ iJy = 0 on y = 0, - 00 < Z < o. In order to apply the Wiener-Hopf technique replace = V by --+

=

=

=

V exp ( -ez) =

y

=

0,

o

<

z <

(e > 0).

00 ,

e o. The Wiener-Hopf solution is i IX)-l , - (21T )-1/2 Vb-1K_ ( - ie)K+(IX)(e

We shall ultimately let

on

=

--+

-

where K(IX) = IXb coth IXb = K+ (IX)K_(IX). This function has been factorized in ex. 1 . 1 1 . We can evaluate the contour integral for by residues. We obtain, for z > 0,

=

v

{

Sin e (b - y) ez e - + Ksin eb

If we let

=

(ii) with 0
--+

e

V

--+

{I

0,

�

_

n

sin

}

e - (nrrZ/b) .

n�l

1 r ( n + ! ) sin nwy e - (nrrZ/b) b 1T3 J2 L-. n r (n + 1 ) n�l

_ '¥. _ __

b

( - ie) L-.� ( K_I() -n1Ti/b)(n1T/b)(n1T(b--eby))

}

,

z

>

O.

Consider the solution of 'i/ 2 t = 0 in 0 .;;; y .;;; b, - 00 < z < 00 , iJtf iJy = 0 on y = b, - 00 < z < 00 , iJtf iJy = 0 on y = 0, < oo , and t = O on y = 0, - 00 < Z < O, such that t ->- Az + B + 00 , where A is given but B is to be determined. We first solve

136

THE W I E N E R - H O P F T E C H N I Q U E

the corresponding problem for V 2
Wiener-Hopf technique we find

2kbO

oo + ia

=

cosh y(b - y) e - icxz d ex. sinh yb .

J K_(ex)y - oo + ia

If we let k 0, then y l exl and the integral is divergent . However the correct procedure becomes clear if we evaluate by residues. For z > 0, -+

-+

where Yn = {(nor/b)2 - k2}112. We set 2kO iA and as k -+ 0 we adj ust so that A remains constant. Then on using (b) in ex. 3 . 6 we find that for z > 0, =

o

{

2b In 2 or

_

�

(rVIrY) e - (nTrZ1bl } . ' b

r (n + � .,(d12 L nr n + t)1 ) cos ( n=l

(iii) Consider the limiting case k 0 for transmission in a duct with a strip across it (§3. 6 ) . As in (ii) the troublesome terms are those involving the progressive waves. From ( 3 . 99), ( 3 . 100), -+

•

•

•

•

•

},

z :> 0, z .;;; 0,

where { . . . } represents exponentially decreasing terms. We multiply by 0, let ikO A , and as k -+ 0 adjust 0 so that A remains constant. Then on using (b) of ex. 3.6 we find =

=

=

(ik)-lA cos kz + A{k-l sin kz + (2b/or) In 2 cos kz} + A{ . . . } + O(k), (ik)-lA cos kz + A{k-l sin kz - (2b/or) In 2 cos kz} + A{ . . . } + O(k),

z

:>

0,

z .;;; O.

If we let k 0 the first term tends to infinity. However if we subtract (ik)-lA cos kz (which satisfies V
0: and

=

=

=

-+

-+

=

=

=

1 37

F U R T H E R W A VE P R O B L E M S

(vi) Next in th-'J static semi-infinite (§3 . 3 ( 3 . 3 4 ) ) .

{

consider a two-dimensional example where the potential case is logarithmic at infinity, namely radiation from two plates in Y ±b , x .;;; 0, with orpt/ oy 0 on the plates For clarity suppose that

rpt = B eikX + Re - ikX + =

00

�

Rn coS

+ ia f(IX)e -iax -YIIII d IX, - 00 + ia

B

=

=

f

(n:Y ) eYnx} ,

I YI

>

I Y I < b, x < O, (a)

b, (b)

where we have used the asymptotic expansion concrete examples we find that as k 0,

( l . 7 1 ) for ( l .56). In

--->-

R

R n = ikTn + 0(k2 ), ( c ) - 1 + ik{p In k + q} + 0(k2 ) (d) f( -k cos 0) sin 0 fo + O(k), say,

=

=

where fo is a constant independent of =

O.

e.g. from ( 3 . 34), ( 3 . 44), ( 3 . 45) :

-

- 1 - (2ikb/7r) { 1 C + In (27r/bk) + trri} , (e) where C is Euler's constant 0·5772 . . . . In (a), (b) set 2ikB A, let 00 in such a way that A remains constant, and use (c), (d) . k 0, B Then for small k, 00 rp t = A{x + t(p In k + q) + t 2: Tn cos (nrry/b)enrrx/b + 0(k2 )}, n=l I Y I < b, x < O. (f) -1/2eikre tirr 00 . - , as r rpt -iAf0 (7r/2) 1/2 (kr) fo

=

--->-

(b /7r)

R

--->-

=

�

--->-

If the far field is expressed in terms of Hankel functions this gives (ex. l .2 1 ) as r --->- 00 . Now choose a fixed large r and let Hb1l( kr) for small (kr) we have

k -+ O . If we use the expansion of

(g) Afo{In (tkr) + C - irri }. 00 is orpt/ or = Afo/r. The velocity of flow The velocity of flow as r as x - 00 in the duct is orpt/ ox = A. Hence the equation of continuity g i es 7rfo = b. If we impose the condition rpt (Ab/7r) In r + 0 ( 1 ) as 00 we must subtract (Ab/7r){In tk + C - t7ri} from (f) and (g) . r The coefficient of In k in the modified form of (f) must then be zero which gives p = 2fo = (2b/7r) . Then finally rpt A [x + {lq + (b/7r) ( In 2 - C + trri ) }] as x --->- 00 , Iyl < b.

rpt

v

�

--->-

--->-

�

--->-

--->-

In the particular case (e) , rpt (Ab/7r) In r, as �

rp ,

--->-

-

00 outside the duct, r A [x - (b/7r){1 + In (7r/b)}], as x --->- 00 , --->-

-

I Y I < b.

1 38

T H E W I E N E R-H O P F T E C H N I Q U E

(vii) In contrast with the two-dimensional example in (vi), axially symmetrical problems in infinite space usually present no difficulties when we let k --+ O. Thus for radiation from a circular pipe (§3 .4) we can deduce directly from ( 3 . 64)-(3.67) that on multiplying rpt by 0, setting ikOa2 = 2A, and letting k --+ 0,

rp t ,...., -A/r, . r --+ 00 outside the pipe, rp t --+ (4A/a2) ( z - l ) , z --+ - 00 inside the pipe, where l is obtained by setting k = 0 in ( 3 . 66) i.e. l a

=

1

-

n

4 u-2 ln {1/2Kl(U)Il (u)} du · J

=

0·6 1 3 3 . . . .

o

3.18 The infinities we have eliminated in ex. 3 . 1 7 (vi) are by no means peculiar to the Wiener-Hopf technique. They are inherent in the transition from the steady-state wave equation to Laplace's equation in two dimensions. Thus the solution of the two-dimensional steady state wave equation for a line source at the origin is

rp

=

wiH�l) (kr)

"'"

wi

- 2 ln (ikr) - 20, as k 0, (a) is Euler's constant and r = (x2 + y2 ) 1 /2 . The solution of --+

where a Laplace's equation with a line source is

rp

=

- 2 ln r .

(b)

To make (a) reduce to (b) as k 0 we need to subtract (wi - 2 ln ik - 20) which is similar to the procedure used in ex. 3 . 1 8 (vi). It is sometimes convenient to avoid difficulties of this type by working in terms of the derivatives u = iJrp/ iJx, v iJrp/ iJy instead of rp. --+

=

3 . 1 9 The biharmonic equation. ( i ) Consider the determination of stresses in a two-dimensional elastic strip in - b .;;; y .;;; b, - 00 < x < 00 with boundary conditions

y

- 00 < x < 0,

=

- 00 < x < 00 .

± b, v

< x < 00,

(a)

y

=

±b.

(b)

where rp satisfies V 4rp = O. If we apply a Fourier transform in obtain, on using symmetry conditions and (a) above,

x we

=

0,

o

rp such that iJ2rp/ iJx2

Introduce the Airy stress function a ",,,,

=

iJ2rp/ iJy2

cI> (()(,y)

=

.

.

a

1l1l =

B( ()(){()(y sinh ()(y -

( 1 + ()(b coth ()(b ) cosh

()(y}.

If we apply a Fourier transform in x to the stresEHltrain relations it is found that the transform of V can be expressed in terms of the transform of rp as follows :

()(2E V( ()( )

=

(1

- v2 ) d3cI> (y)/dy3 + ()(2 (v2 - V - 2) dcI> (y)/dy,

139

FURTHER WAVE PROBLEMS

where E , v are constants. I f we set allll - p exp (ex) for y - 00 < x < 0 w e obtain the Wiener-Hopf equation

±b,

=

IX (sinh IXb cosh IXb + IXb) G - ( IX) , 2 sinh2 IXb ( 27T)1/2(e + iIX) where 11'+( IX) is the transform of all 11 in (0, 00 ) and G_ ( IX) =E( 1 _ v2 )-1 V_(IX) . The function which has to be decomposed in the form K+( IX )K_( IX ) has F+ ( IX) +

P

_

no branch points and the standard infinite product theory can be applied, though the zeros of the numerator are complex (cf. the approxi mate method of §4.5) 4> can be expressed as a contour integral which can be evaluated by residues and the final solution is obtained by taking the limit e --+ o. A similar example is considered in W. T. Koiter [2] . (ii) Consider the flow o f a highly viscous fluid i n - b <: y <: b , - 00 < x < 00 with a semi·infinite plate i n y 0 , 0 < x < 00 . The velocity components can be expressed in terms of a stream function "P t which satisfies the biharmonic equation . The velocity components are given by u - O"Pt/ oy, v = 0"Pt/ ox . Set tpt tpi + "P with "Pi = iB{(y/b) - i ( y/b) 3} , where "Pt = 0 on y = 0, B on y = b. Then "P satisfies the boundary conditions =

=

"P = 0 on y = 0, tp � o"P/ oy 02"P/ oy2 = 0, - 00 < Z < 0, otp/ oy

=

=

=

on y b, - 00 < Z < 00 . - i ( B/b) , 0 < Z < 00 , on y = O .

0

=

This can b e formulated as a Wiener-Hopf problem involving the factorization approximate solution for this example and for two similar problems involving deflexion of plates has been given by W. T. Koiter [ 1] .

An

3 .20 A supersonic jet i n a subsonic stream. (The application o f the Wiener-Hopf technique to this problem was suggested by Prof. D. C . Pack, the Royal College o f Science and Technology, Glasgow, and the formulation given below was carried out in collaboration with S. C. Lennox, the Heriot-Watt College, Edinburgh. Relevant references, which do not however use the Wiener-Hopf technique, are : "Supersonic flow of a two-dimensional j et", S. I. Pai, J. Aero . Sci., 19 ( 1 952), pp. 6 1-65, and a note on this paper by E . B. Klunker and K. C. Harder, p. 427 of the same issue.) Suppose that a two-dimensi<:mal supersonic jet lies in a region -b <: y <: b, - 00 < x < 00, with a subsonic stream in I y l > b. Assume that the jet issues from rigid walls in y ±b, 00 < x < 0, and that the supersonic flow and the subsonic stream are in contact along y = ±b, x > O. The boundary conditions on these lines are : o4>/ oy is continuous but the pressure, 04>/ ox, is discontinuous. More precisely we assume =

-

( 04)/ oX ) y = b + O - m( 04>/ oX ) II = b - O

=

Oe -ex,

x >

0,

where m, � are constants and e is a positive constant. We shall ultimately let e tend to zero. In - b <: y UJ - 4>1111 0, =

1 40

T H E W I E N E R- H O P F T E C H N I Q U E

but in Iyl > b the equation will b e assumed to b e xx + yy + k2 = 0, where we shall ultimately let k O. To avoid logarithmic potentials at infinity we work in terms of velocities u aI ax, v = aI ay with corresponding Fourier transforms in x given by U, V. We have -+

=

i.e. Then

where y

=

V+(ot,y) + V_(ot,y)

=

V+(ot,y) + V_(ot,y)

=

(ot2 - k2 ) 1/2. The boundarY conditions are U+(ot, b + 0) - m U+(ot, b - 0)

=

6(s

- iyot-1Ae -1'II , ( y > b), -iO sin oty, (0 ..;; y ..;; b) ,

- iot)-l.

The Wiener-Hopf equation is 6(s

- iot )-l + { U_(ot, b + 0) - m U_(ot, b - OJ} =

i (oty-l - m cot otb) V+(ot,b) .

If k 0, then (ot/y) (ot/ l otl J which is a typical situation arising in connexion with Laplace's equation. We do not pursue the solution further. It is left to the reader to show that by application of a Fourier transform in x the following integral equation obtained by Klunker and Harder, can be reduced to the above Wiener-Hopf problem with 6 0: -+

-+

loco cit. ,

=

v(x) + v(x + 2p)

=

- !!.

'IT

00

f {xV(�)- � - x +v(�)2p - �} d�,

o

Suppose next that there are rigid walls at y = ± B, B aI ay = v = O. The Wiener-Hopf equation is then 6(s

(x ;;;. 0). >

b, on which

- iot )-l + { U_( ot, b + 0) - mU_(ot, b - OJ} = i (oty-l coth yB - m cot otb) V+( ot,b) .

When B is finite the function to be factorized has no branch points and the infinite product method is applicable. It is probable that the above method can be applied to the subsonic jet in a subsonic or supersonic stream, and to axially symmetrical jets.

CHAPTER IV

E X TENS I O NS AND L I M I TAT I O NS O F THE METH O D 4.1

Introduction

Chapters II and III have been concerned mainly with the solution of specific examples. In this chapter we discuss a number of more general topics. In §4.2 it is shown that the complex variable problem solved by the Wiener-Hopf technique is a special case of the Hilbert problem. In §4.3 we examine the criteria for successful application of the Hilbert problem (or the Wiener-Hopf technique) to a typical mixed boundary value problem, involving any co-ordinate of a separable system. §4.4 deals with a generalization of the basic Wiener-Hopf equation where there is a set of simultaneous equations for several unknown functions. A major difficulty in applying the Wiener-Hopf technique in practice is often the fundamental factorization K(rx) = K+ (rx)K_(rx) . §4.5 discusses the position when we can find a function K*(rx) with a known factorization such that K*(rx) and K(rx) are approxi mately equal to each other in the strip in which the Wiener-Hopf equation holds. In §4.6 it is shown that certain problems involving Laplace's equation in polar co-ordinates can be solved by the Wiener-Hopf technique in conjunction with a Mellin transform. We draw the reader's attention to various topics in connexion with the solution of simultaneous linear algebraic equations in exs. 4.10-4. 13 at the end of the chapter. 4.2

The Hilbert prob1emt

The fundamental equation which appears in the Wiener-Hopf method is · (4. 1 ) where A , B , 0 are known analytic functions, «1>+ (rx) , 7_(rx) are unknown, and the equation holds in a strip T_ < T < T+ , - 00 < Cf < 00 in the complex rx-plane. In this section we show that the t Also called the Riemann problem, or the Riemann-Hilbert problem.

141

1 42 T H E W IE N E R-H O P F T E C H N I Q U E solution of (7.1) can be regarded as a special case of the Hilbert problem : to find certain analytic functions' when the boundary values of these functions on a set of smooth contours and arcs of arbitrary shape satisfy (4.2) a(t)«l\(t) + b(t)'Y_( t ) + C (t) = 0, where t is the complex number specifying any point on the contour. In this equation + (t), 'Y_(t) are the (unknown) boundary values of analytic functions + (at), 'Y_(at) which are to be determined. The functions a(t), b(t), c(t) are given and need not be related to analytic functions. We first specify the non-homogeneous Hilbert problem more precisely as follows. (N. 1. Muskhelishvili [1], pp. 86, 92, 235.)

FIG. 4.1.

Consider the complex at-plane. Let S+ be a connected region bounded by smooth contours and arcs 00 , 0 1 > . . . , 0 1> ' not intersecting each other, the first of which encloses all the others (see Fig. 4 . 1 ) . The contour 00 may be absent in which case S+ is an infinite region. The contours and arcs may go to infinity but in this case some modi fications are needed in the following statements. By 0 denote the union of 00 , 0 1 > . . . , 0 1> . Denote by S- that part of the plane which is the complement of S+ + 0 , and by S' that part not belonging to 0 (i.e. S' S+ + S-) . Let E(at) be a function of the complex variable at such that (i) E(at) is regular everywhere in S' and E(at) r-...I atm as at -+ 00, where m is an integer. (ii) E(at) tends to a finite limiting value if at tends to a point on 0 from either side of 0 , other than the ends of any arcs. (iii) At the end of any arc, say at = d, =

(0 < fl < 1 ) . Each part of 0 has a definite positive and negative side. For arcs these may be defined arbitrarily. For contours the positive side is the side adjoining S+ . If at lies on 0 denote its value by t. Denote by

143 E+ (t) , E-(t) the boundary values of the function E(oc) as oc tends to the point t on 0 from positive and negative sides of 0 respectively. We can now state the non-homogeneous Hilbert problem : find a function E(oc) with the above properties such that E X T E N S I O N S A N D L 1; M I T A T I O N S

O F M E TH O D

E+ (t) = G(t)E-(t) + g(t) ,

(4.3) where G(t), g(t) are functions given on 0 and G(t) is a non-vanishing function on 0 satisfying everywhere on 0 the Holder condition

/ G(t 2 ) - G ( t l ) / < A / t 2 - t l / " where A , Y are positive constants. We now show that the solution of the Wiener-Hopf equation (4.1 ) can be reduced to the solution of a Hilbert problem. In (4. 1 ) we assume (a) The equation holds in a strip L < T < T+ , -00 < (] < 00 . In this strip A (oc) , B(oc) , O(oc) are regular functions of oc . (b) + (oc) is regular in T > T_ and / <1>+ / < O l / OC/ P as oc -+ oo in this half-plane. '¥_(oc) is regular in T < T+ and / '¥_ / < 0 2 / OC/q as oc -+ 00 in this half-plane. Choose any T V T_ < T I < T+ such that no zeros of A (oc) , B(oc) lie on the line T T V - 00 < (] < 00 . Call this line the contour 0 and denote the complex number representing a point on 0 by the letter t. Then on 0, (7.1 ) becomes (4.4) A ( t )+ ( t ) + B (t )'¥_( t ) + O t t) = 0, where now + ( t ) is the limiting value, as oc tends to the point t, of an analytic function +(oc) where + (oc) is regular for T � T I ' and similarly for '¥_(oc) . Also A (t), B(t) are continuous, differentiable and non-zero on O. We see that this is a restricted form of the Hilbert problem. The two functions + (oc), '¥_(oc) together make up the function E(oc) defined above. The. contour 0 of the Hilbert problem is now very simple-an infinite straight line parallel to the real axis. In com paring Wiener-Hopf and Hilbert problems we need to be careful when considering conditions at infinity. Apply a transformation, say oc = (w - a )-I , to bring the point at infinity into the finite part of the plane-in this case to w = a. For the Wiener-Hopf problem in the w-plane we have a contour passing through w a, and as w -+ a on one side of the contour, / <1>+ / < O l /W - a/- p , and as w -+ a on the other side, / '¥_ / < 0 2 / W - a/ -q • On the other hand for the Hilbert problem considered at the beginning of this section there is no contour through the point w = a in the w-plane but E(oc) has a pole of order m at this point. However the essential difference between (4.3) and (4.4) is that in =

=

1 44 T H E W I E N E R- H O P F T E C H N I Q U E (4.3), G(t) need only satisfy a Holder condition whereas in (4.4), A (t) and B(t) must satisfy much more restrictive conditions. In (4.4) A (t) and B(t) are derived from functions which are regular in a strip in the oc-plane, which in turn is the basic feature used to solve the original equation (4. 1 ) by the Wiener-Hopf method. The function G(t) in (4.3) need have no connection, directly, with analytic functions. If the function G(t) in (4.3) is regular in a strip of finite width including 0, the Wiener-Hopf method can be used to solve the Hilbert problem. But in the general case it is necessary to use a different method. We outline briefly the solution of the Hilbert problem for contours in order to show the connexion with the Wiener-Hopf technique. For details and a rigorous exposition the reader is referred to N. 1. Muskhelishvili [1]. The major difficulty in solving (4.3) is the determination of suitable K+ (t), K-(t) such that (4.5) K+ (t)G(t) = K-(t), where K+ (t), K-(t) are limiting values of functions regular in S+ , S respectively. Muskhelishvili calls this the corresponding homo geneous Hilbert problem. The first steps of his solution are reminis cent of the theorem in ex. 1 .12. When t vanishes over any of the closed contours Ok' In G(t) will vary by an integral multiple of 27Ti (cf. ex. 1 .5) . We write Go(t) = H-(t)G(t)jH+ (t) where H ± (t) are multiplying factors chosen so that In Go(t} is one-valued as well as satisfying a Holder condition on 0, and H ± (t) are the limiting values of functions H ± (oc) regular in S ± respectively. Explicit forms for H ± (oc) are given by N. 1. Muskhelishvili [1], p. 88. Introduce Kf1= (t) = H ± (t)K ± (t) where the upper and lower signs go together. Then (4.5) gives (4.6) where now Kf1= (t) are unknown, Go(t} is known. Take logarithms of both sides : (4.7) In Ko (t) In Kt (t) In Go(t). At this stage it is no.t permissible to use theorem B of §1 .3 to deter� mine Kf1= (t) which is the Wiener-Hopf procedure. Instead of Muskhelishvili uses the Plemelj formulae which can be stated as follows. Consider the Hilbert problem for contours. Define -

Q(oc)

=

=

� 27T� J a

cp(s)oc ds,

s

-

(4 8 ) .

for any complex oc not on O. It is assumed that cp (s) satisfies a Holder condition on O. Then Q(oc) is regular everywhere in the

1 45 IX-plane except on O. Denote the limiting value of Q(IX) as IX tends to a point t of 0 from S+ by Q+ ( t ) and similarly for Q- ( t ) . Define direction on 0 so that S+ lies on the left of 0 when 0 is traversed in the positive direction. In f4,.8 ) let oc tend to a point t on 0 from S+ , and from S-. By indenting the contour suitably we find cp(s) Q+ (t) = LI.(t) + � P (4.9a) ds, 21T� S - t 2 '1' E X T ENS I O N S A N D L I M I T A T I O N S O F

M E TH O D

J

c

Q- ( t )

=

1 -icp(t) + 2 1T�. P

J scp(s)- t ds,

c

(4.9b)

where the P sign indicates that the integrals are now Cauchy principal values (cf. ex. 1 .24 ) . These are the Plemelj formulae. By subtraction, Q+ ( t ) - Q- ( t ) cp(t). (4.10) Hence we have the theorem : If cp(t) is given on 0 and satisfies a Holder condition then a function Q(oc) whose limiting values on 0 satisfy (4. 10) is given by (4.8). This theorem can be used to solve the Hilbert problem (equation (4.3)) in exactly the same way as theorem B of §1.3 was used to solve the corresponding Wiener-Hopf problem. We first of all solve the subsidiary equation (4.6) . In simple cases it may be possible to do this by inspection. In the general case we use the above theorem. Define =

In Ko(oc)

=

1 --: 2 1T�

J

c

In Go(s) -- ds . S - oc

(4. 1 1 )

Then the limiting values as oc tends to 0 from different sides of 0 determine In Kr; ( t ) and In Ket (t) . Hence we can find K+ (t ) , K- ( t ) . Multiply (4.3) b y K+ (t ) : K+ ( t )E+ ( t ) - K- ( t )E- (t ) = K+ (t)g(t ) . It may be possible to solve this by inspection but in the general case we again use (4.8)- (4.10). Define

1 K(oc)E(oc) = -. 2 1T�

J

C

K+ (s)g(s) ds. S - IX

(4 . 1 2)

Then K+ (t)E+ (t) and K- ( t )E- ( t ) are the limiting values of this expression as oc tends to 0 from opposite sides of O. This completes the solution of (4.3 ) . Formally the above solution is exactly what we should obtain by

1 46 T H E W I E N E R- H O P F T E C H N I Q U E using the Wiener-Hopf technique in a strip of finite width and then letting the width of the strip tend to zero. We note that the applications of the Hilbert problem made by N. 1. Muskhelishvili are quite different from the applications in this book. Muskhelishvili is interested in cases where the solution of the partial differential equation in two independent variables (x,y) can be expressed directly as z = x + iy, cp(x,y) = F( z) , and the Hilbert problem is posed in terms of the complex variable z . In contrast to this, we are interested in cases when the solution of such an equation can be expressed in the form cp(x,y)

=

J f(IX)K(x,y : IX) dIX,

a

and the Hilbert problem is formulated on the contour 0 in the IX-plane. There are three ways in which the Hilbert problem is more general than the Wiener-Hopf problem : (a) The Hilbert problem is formulated on a contour and not in a strip of finite width. Hence it can be used to solve directly problems involving functions whose transforms exist only on a contour in the IX-plane instead of in a strip. Strictly speaking nearly all the problems considered in this book are of this type. In previous chapters we have written k = k l + ik 2 where k 2 is finite, to ensure that transforms exist in a strip of finite width, although we have ultimately let k 2 tend to zero, in which case the transforms usually exist only on a contour. The solution for the case k 2 0 could have been obtained directly by solving a Hilbert problem. (b) The function G(t) occurring in the Hilbert problem need be defined only on a contour and it need satisfy only a Holder condition on the contour. There is no need for G(t) to be derived from a function of a complex variable which is analytic in a strip in the complex plane. (c) The Hilbert problem can be solved for an arbitrary numb�r of contours and arcs· of arbitrary shape. The Hilbert problem has been mentioned in some detail because almost certainly it will be useful in future researches. But in spite of the generality of the solution of the Hilbert problem we have not used it in the main part of this book. We have preferred the Wiener Hopf technique because it seems more routine when used in con junction with transforms for the particular kind of problem solved in this book. The method that we have used proceeds by a very straightforward analysis in the complex plane: branch cuts are =

147 clearly defined and we see automatically where indentations are required in the limiting case k 2 -+ O. E X T E N S I O N S A N D L I M I TA T I O N S O F M E T H O D

4.3

General considerations

The problems solved in previous chapters form an impressive collection, especially if we think of how few problems of this type can be solved by any other means. On the other hand the range of problem which can be solved is limited. The methods described so far for the exact solution of problems apply essentially to two-part mixed boundary value problems when the partial differential equation possesses a separation-of-variables solution exp (iocx), the complex variable oc being the parameter of separation. Different conditions involving c/> and oc/>/on are specified on - oo < x < O and O < x < 00 . (The Mellin transform case is essentially the same and can b e obtained by change of variable. The separation-of-variables solution is p-s and the mixed boundary conditions are specified on 0 :::;; p < a, and a < p < 00 . ) Ideally we should like to be able to solve two-part mixed boundary value problems for any co-ordinate of any separable system. In this section we aim to clarify the relation of the Wiener-Hopf technique to this very general objective. Unfortunately it will become only too clear that we should not expect the Wiener-Hopf technique to work for any co-ordinate system. Consider first of all a concrete example. Suppose that 0 :::;; x < 00 , 0 :::;; y < 00 , V 2 c/> + Pc/> = 0, oc/>/ox = 0 on x = 0, (0 :::;; y < (0 ) , c/> 1 , (0 :::;; x < a), oc/>/oy = 0, (a < x < (0), on y = O. (A related physical problem is diffraction of a normally incident wave in infinite two-dimensional space, by a strip in y = 0, - a :::;; x :::;; a.) The natural transform to use is a cosine transform : _

=

( 2 ) 1/2 (oc,y) = ;:

00

J c/>(x,y) cos ocx dx

o

c/>

=

00

( ) J cos ocx dx. 2 1/2 ;:

o

(4.13)

Without going through the details (cf. §§5.2, 5.6) we state that the problem can be reduced to solution of the following equation which can be regarded as a generalization of (4. 1 ) : (4. 14) ( 2/7T) 1 /2 OC-1 sin oca + 2 (OC) = _ y-1 �(OC), where 00 a 1 /2 2 1/2 oc/> cos ocx dx c/> cos IX.X dx. 1' (oc) = 2 ( oc) = 7T oy o a (4.15a)

( ) J-

(� ) J

1 48 T H E W E I N E R- H O P F T E O H N I Q U E cp and Cocp/ay) are taken on y = O . � (ex) is an integral function of ex and 2 ( ex ) can be written 2 (ex) ei<xa + ( ex ) + e -i<>a+ ( -ex), (4.15b) =

+ (ex)

=

1 __/ 2 (2 1T ) 1

00

f A..'f'ei<x(x -a) dx.

a Owing to the occurrence of + (-ex) in conjunction with <1>+ (ex) , it is found that (4.14) is more complicated than (4. 1 ) and cannot be solved exactly by the Wiener-Hopf technique. It can be solved approximately for large ka by the method of §§5.5, 5.6. Another approach is to try two-dimensional polar co-ordinates (4. 16) If k = 0 i.e. if we are dealing with Laplace's equation, problems of this type can be solved by a Mellin transform (§4.6). But if we apply a Mellin transform to (4. 16) for k #- 0 we find

d 2<1> (s)/d(}2 + s2 <1> ( S ) + k2 <1> ( s + 2)

=

O.

(4. 17)

In this case it is the separation-of-variables solution which is at fault-the transform does not fit the partial differential equation. Further progress could be made if it were possible to solve (4. 17) but this is deceptively simple. The next step is to enquire whether there is a generalization of the Mellin transform which does fit the partial differential equation. This is the Lebedev-Kontorovich transform (A. Erdelyi et al. [1], Vol. II) :

(fl)

=

f cpKI'(Ar) dr 00

o

-;:

.

f

c + i c.o

cp

=

�

i

flCP(fl) II' (Ar) d fl ,

c - i oo

(4. 18)

A>

where A = - ik. On applying (4.18) to (4.16) the problem can e reduced to an equation similar to (4. 14) but involving the following unknown functions (cf. ex. 5.12) :

2 (fl)

=

f cpKI' (Ar) d; , 00

a

(4. 19)

1 49 where cp and (ocp/o()) are taken on () = o. In this case
cp =

I ,u(t)K(t,x)L(t,y) dt,

c

(4.20)

where ,u(t) is an arbitrary function and 0 is in general some contour in the complex at-plane. We emphasize at the outset that there is no necessity here for ,u, K, L to be regular or analytic functions of t. The functions need only be defined for special values of at, namely at on 0, and it is for this reason that we use a separate letter t to denote these special values of at. Suppose that by suitable choice of K, L we can satisfy all boundary conditions except on y = Yo where

cp = f(x),

ocp/oy = g(x),

We define where

L' (t,yo)

=

[o L (t , y)/o Y]iI= vo .

(a � x < p l .

1 50 T H E W I E N E R-H O P F T E C H N I Q U E From (4.20), on applying the boundary conditions on Y = Yo' we find the dual integral equations

I K(t)X(t)K(t,x) dt = f(x),

c

I

X (t)K(t,x) dt = g(x),

c

(p

x � b) ,

(4.21a)

(a � x < p) .

(4.21b)

<

We next assume that there is an inversion theorem

I K(t,x)p(t) dt = p(x),

(a

c

� x � b),

(4.22a)

b

I k(t,x)p(x) dx = P(t) ,

(t on G) .

(4.22b)

a

Suppose that the left-hand sides of (4.21a) and (4.21b) equal e (x) and h(x) for a � x < p, p < x � b, respectively. Invert by (4.22). This gives

K(t) X (t) = F(t) + E(t) where E(t)

=

p

I

: X (t)

=

G(t) + H(t),

(t on G),

(4.23)

H(t) = I k(t,x)h(x) dx,

(4.24)

b

k(t,x) e (x) dx

a

p

and there are similar definitions for F(t), G(t) . Elimination of X (t) gives E(t) = K(t)H(t) + M(t), where M(t) = K(t)G(t) - F(t) . (4.25) E, H are unknown and all other functions are known. This equation is formally very similar to the equation (4.2) of the Hilbert problem or (4.1 ) for the equivalent Wiener-Hopf problem. Before examining (4.25) we digress to show that the same equation can be obtained directly from the transform point of view by Jones's method. Define (cf. (4.22)) b

(t,y) = I k(t,x)cp(x,y) dx,

(t on G).

a

Apply this transform to the partial differential equation, solve the ordinary differential equation which results, and insert boundary conditions on all boundaries except y = Yo. Our statement in connexion with (4.20), (4.21) that K, L are chosen so that all boundary conditi�are satisfied except on y = Yo means that we shall find (t,y) = fl(t)L(t,y) ,

151 where fl is an arbitrary function and L is known. Form d/dy and set y = Yo in the expressions for and d/dy. Insert the mixed boundary conditions and use the notation for E, F, G, H defined above. Elimination of fl(t) then gives (4.25) . Although both E(t) and H(t) are unknown in (4.25) , the fact that they are defined by (4.24) provides additional information which is sufficient to specify e (x), h(x) completely. If we wish to try to apply the Wiener-Hopf technique, we define E(rx) , H(rx) for the general complex variable rx by (cf. (4.24)) E X T E N SI O N S A N D L I M I T A T I O N S O F M E TH O D

E(rx)

=

p

I k(rx,x)e(x) dx

H (rx)

a

=

b

I k(rx,x)h(x) dx.

p

(4.26)

Suppose that the contour 0 (on which rx = t) divides the rx-plane into two parts, say S+, S-. Then if E(rx) is regular in say S-, H(rx) in S+, we can extend our notation to write (4.25) in the form

E_(t)

=

K(t)H+(t) + M(t),

where now E_(t), H+(t) are (unknown) values of analytic functions E_(rx), H+(rx) which are regular in S-, S+ respectively. The problem is now to determine the analytic functions E_(rx), H+(rx) , which is precisely the Hilbert problem (4.2) . However we emphasize that E(rx) will b e regular in say S- only in very special circumstances. If we set E(t) = E l (t) + iEz(t) where E l and Ez are real, then it is well known that the specification of El alone is sufficient to determine functions E_(rx) , E+(rx) regular in S- and S+, and unique apart from an arbitrary constant. This is true because E l (t) defines a Dirichlet boundary value problem in say S-. Since the problem is two-dimensional the solution of the Dirichlet problem, say U( a,r} , determines a conjugate function V( a,T) such that U( a,T) + i V( a,T) is a function of (a + iT) . If ( a,T) tends to a point on the boundary of the region then U( a,T) will tend to E l (t) but V( a,T) will not in general tend to E 2 (t) . This means that when E l (t) is specified, E 2 (t) cannot be specified arbitrarily if E_(rx) is to exist. In the general case it will therefore not be possible to identify E(rx), H(rx), defined by (4.26), with E_(rx), H+ (rx) regular in S-, S+ respectively. It will be necessary to examine the structure of E(rx), H(rx) in detail. Some examples are considered in Chapter V. Two features illustrated by this investigation are : (i) Any two-part boundary value problem can be formulated as an equation of type (4.25). The all-important factor which deter mines whether this equation is equivalent to a Hilbert problem is the

152 T H E WIE N E R- H O P F T E C H N I Q U E behaviour of the integrals (4.26) considered as functions of a complex variable. (ii) From the Hilbert problem point of view, the function K(t) need not have any direct connexion with analytic functions of a complex variable : it need be defined only on a contour in the complex plane and it need satisfy only a Holder condition. In the Wiener-Hopf method K(ex) must be analytic in a strip in the complex ex-plane but from a general viewpoint this would seem to be accidental rather than fundamental. In particular cases it may be possible to deduce the solution of dual integral equations almost by inspection without going through the reduction to a Hilbert or a Wiener-Hopf problem. Thus consider the solution of the following equations (see §2.3 (2.34) (2.35), cf. P. C. Clemmow [1], p. 296. Note that Clemmow takes a time factor exp (iwt) so that his contour is indented differently.) : 1 n (2 ) 1/2

1 __ ( 2n ) 1/2

J A(t)e .

c

J (t2 - k2)1/2A(t)e

c

- ,tJ;

dt

- itx

dt

=

=

0,

(x > 0),

-ik sin 0e - ikx cos e

(4.27a) '

(x < 0), (4.27b) - k and

where the contour 0 is the real axis indented above at t below at t k cos 0 and t k. We assume that an analytic function A (ex) can be derived such that its limiting value on 0 gives A (t) which satisfies the above equation. As previously remarked it is not obvious that such an A(ex) exists but we can either prove the existence of A (ex) by general arguments as above, or merely find A ( ex) and show a posteriori that it satisfies the required conditions. In the first integral complete the contour by an infinite semi-circle below O. Then the first equation is satisfied by choosing (4.28) ( T < 0), A (ex) '¥_(ex), =

=

=

=

r

where '¥_(ex) is regular below the contour and has algebraic behaviour at infinity. In the second integral complete the contour by an infinite semi-circle above O. Then a suitable form for A (ex) is 1 / +(ex) - k sin 0 (4.29) , (2 n ) 1/2 (ex2 - k2 ) 1/ 2 · +(k cos 0) . (ex - k cos 0) since the pole at ex = k cos 0 (contour indented below) gives the right-hand side of (7 .33b) and there are no other poles or branch points of (ex2 - k2 )A(ex) above O. The limiting value of A (ex) in A(ex) -

1 53 (7.34) as T � -0 must be the same as in (7.35) as T � + 0. It is easily seen that we need to write (oc2 - k2 ) 1/ 2 = (oc - k) 1/ 2 (OC + k) 1/ 2 and choose + (oc) (oc + k) 1/ 2 . Then E X T E N S I O N S A N D L I M I TA T I O N S O F M E T H O D

=

1 1 1 k sin 0 . A (oc) = . (21T) 1/ 2 (k + k cos 0) 1/ 2 (oc - k) 1/ 2 (oc � k cos 0) This agrees with (2.31 ). There are a group of methods due to L. A. Vajnshtejn ([1], [3], [4], [5]), S . N. Karp [1], and P. C. Clemmow [I], which are essentially the same as the method illustrated in the last paragraph. All these methods are similar in that they start with a pair of dual integral equations obtained by separation of variables, or solution of an integral equation or a physical argument. Instead of reduction to a Hilbert or Wiener-Hopf problem the equations are solved by "function-theoretic" methods depending on the regions of regularity, and poles, of the functions under the integral sign. Instead of working with general j, g on the right-hand sides of their equations Vajnshtejn-Karp-Clemmow work with explicit j, g so that the poles of unknown functions are known as in the last paragraph. The justification is essentially a posteriori (cf. J. Bazer and S. N. Karp [1].) The V-K-C methods may be convenient in particular cases where they give an elegant solution without invoking the reduction to a Hilbert or Wiener-Hopf problem but they are not essentially different from the procedures used in this book and we do not describe them in further detail.

•

4.4

Simultaneous Wiener-Hopf equations

In this section we consider the situation which arises when, instead of the single Wiener-Hopf equation (4. 1 ) in two unknown functions there are n simultaneous equations in 2n unknown functions. Many of the examples in Chapter III were in fact simultaneous Wiener-Hopf equations with n = 2 but in all cases except one it was possible to reduce these twin equations to independent single equations of the Wiener-Hopf type. The exception was (3.85a, b). These equations are of the form

= {M(oc) + i t5 }t (oc) + F(oc), N( oc )'Fi ( OC ) = - {N(oc) + i t5}i (oc) + G(oc),

'F1 (oc) - M(oc)'Fi (oc)

(4.30a)

'F1 (oc) +

(4.30b)

where (j is a (small) constant and M(oc), N(oc) possess only simple poles and no branch points. It is possible to obtain an exact solution when M(oc) = N(oc), or when (j = 0, but not in the general case.

1 54

T H E W I E N E R-H O P F T E C H N I Q U E

In Chapter III we deliberately restricted ourselves (except in the one case we have just mentioned) to problems which reduced to single Wiener-Hopf equations. But it is easy to suggest extensions of the problems in Chapter III which lead to twin Wiener-Hopf equations. Thus consider a plane wave incident on two parallel partially reflecting half-planes (cf. §2.2 ). For simplicity suppose that a wave exp (-ikx) is incident on two half-planes in y ±b, - 00 < x < O. Suppose that =

ocp t /oy = =F i oCPt, Y = -b ± 0,

for - 00 < x < O. From symmetry oCP t/oy 0 on y = O. It is left to the reader to show that the Wiener-Hopf equations are =

� ( b ) + o ( ex - k) -1 {y tanh yb - io}_(b - 0) + y tanh y b + ( b ), � ( b ) - o (ex - k) -1 = ( -y + i o)_( b + 0) - y + ( b ). =

These are similar to ( 4 . 30 ) but in this case the function corresponding to N( ex ) possesses branch points. Other generalizations of the problems of §§3.2, 3.4 which result in simultaneous Wiener-Hopf equations are : (i) A plane wave incident on two parallel staggered plates in y b, - 00 < z < -h, and y - b, - 00 < z < +h. (ii) A plane wave incident on three parallel plates in y = 0, ±b, -00 < z < O. (iii) Radiation from a coaxial guide a :'( p :'( b, - 00 < z < o. Before considering the general theory we examine the solution of the equation. =

=

(4.31 ) where 8 1 is a parameter introduced for convenience, as will appear later. We assume that the only singularities of M ( ex ), L ( ex ) are simple poles, though K( ex ) may have branch points. In the usual way we assume that the equation holds in T < T < T+ , - 00 < 0 < 00 , and that all functions are regular in this strip and have suitable behaviour as ex � 00 in appropriate half-planes. We set _

K( ex ) = K+( ex )K_( ex ) : L( ex ) = L+( ex )L_( ex ) : M( ex ) = M+(ex )M_(ex ) . Divide (4.31 ) throughout by K_ ( ex )M+( ex ) and suppose that L( ex ) � 1s ' M+ ( ex ) = L ( ex ) + 8L. = 1 ex - fJ s

1 55

EXTENSIONS AND LIMITATIONS OF M E THOD

where {3s are the zeros of M+(IX) and the poles of L(IX) in the lower half plane. We can write (4.32a) where

'Y (1X) A - (IX) = l- (IX) 1 + K_(IX) We write

�

{

l_ s _ 'Y 1 (1X) K_(IX)

� - {3. 8 IX _

_

'Y 1 ({3s) K_({3s)

}

( 4.32b ) .

(4.32c)

N (IX) {M+ (IX)K_(IX)}-1 = G+(IX) + G_(IX), say.

Separation of (4.31 ) by means of the Wiener-Hopf technique then gives

�

'Y 1 ({3. ) l. m+ K+(IX) -V 1 (IX) + 8 1 M+(IX) � . {3 • K_({3 . ) IX , 8= 1 _

=

M+(IX)G+ (IX). (4.33)

If we are given a second equation in the form

P(IX)'Y 1 (1X) + 8 2Q(IX)
then an exactly similar procedure gives lTJ' P_(lX b: 1 (IX) + 8 2 R_(IX)

where

� IX q.

�

0

_

8=1

Q(IX)/I!_(IX)

=

•

q+(IX) +

= 00

=

S(IX),

R_(IX) T_(IX),

L: IX 8=1

q.

0

•

(4.34)

(4.35)

'

S(IX){R_(IX)P+ (IX)}-1 = T+(IX) + T_(IX). The functions

equations (4.31 ) , (4.34) are therefore given by (4.33 ) , (4.35), where the constants 'Y 1 ({3. ), t (O . ) are so far unknown. Twin sets of infinite simultaneous linear algebraic equations for these constants are obtained by setting IX = 0.. in (4.33) and IX = /1.. in (4.35), r = 1, 2, . . . . The equations can be written 00

a;c.. + 8 1 L: 0 8=1 "

� rJ.

=

A.. ,

(r = 1 , 2, . . . ) , (r = 1 , 2, . . . ) ,

where

156

THE

W I E N E R-H O P F T E C H N I Q U E

with appropriate expressions for ar , br , A r, Br . We have also set 'YJ s = -/3s so that () s and 'YJ8 both lie in an upper half-plane. In general it would appear that these equations must be solved numeri cally. If 8 1 or 8 2 is small, or if 8 1 and 8 2 are both small, it may be convenient to solve the equations iteratively. The scope of the above method for solution of twin Wiener-Hopf equations is of course strictly limited since the functions L, M, Q, R must not possess branch points. When the method is applicable it will be advantageous to arrange that 8 and 8 2 are as small as 1 possible. Thus in the case of (4.30) if we divide (4.30a) by M(Ot), (4.30b) by N(Ot), and add, we find

2 U(Ot)'P'1 (0t)

=

where

i t5 V(Ot)S+ (Ot) + { I + i t5 U ( Ot ) }D+ ( Ot ) + + F(Ot)jM(Ot) + G( Ot ) jN ( Ot ),

S+ (Ot) <1>t (Ot) + <1>t (Ot) 2 U(Ot) = {M( Ot ) }-l + {N ( Ot ) }-l =

D+ (Ot) = <1>t (Ot) - <1>t (Ot), 2 V(Ot) = {M(Ot)}-l - {N(Ot)}-1 .

This is a convenient equation when 15 is small. From (4.31 ), (4.34) the second equation must involve only S+(Ot), <1> 1 (Ot), 'P'i (Ot), and this can be derived from (4.30) in only one way. We now consider some general theory. Instead of (4.1 ) suppose that we are given a set of n equations n

n

L A s ( )<1> + ( Ot ) + sL Brs ( Ot )'P's- ( Ot ) + 0r ( Ot ) s �l r Ot s �1

=

0, (r

=

1 , 2, . . . , n),

(4.36) where there are 2n unknown functions <1>: (Ot), 'P's- ( Ot ), regular in upper and lower half-planes, T > T T < T+ , respectively, with T < T+. The equations hold in the strip L < T < T+. The above equations can be written in matrix notation as _,

_

A«I»+ + B'I'- + C = 0, (4.37) where A, B are (n X n) square matrices and «1»+ , '1'-, C are (n X 1 ) matrices. A, B "are �ssumed non-singular in T_ < T < T+ . The following set of equations of type (4.37) can be solved exactly: (4.38) K+ «I»+ + K-'I'- + L = 0,

where all equations are assumed to hold in a common strip and the general elements K:};. , Kr--; of the matrices K+ , K- are regular in upper and lower half-planes respectively. Then the standard Wiener-Hopf procedure can be used to solve each of the n simulta neous equations exactly, and the <1>: , 'P's- can then be obtained by solving two n X n sets of simultaneous linear algebraic equations.

E XT E N S I O N S A N D LIMITATI O N S

OF METHOD

157

It may be possible to reduce (4.37) to form (4.38) by premultiply ing by some n X n matrix D whose elements are functions of at (i.e. by taking linear combinations of the equations in (4.37 ) , multiplied b y suitable functions o f at ) . Then (4.37 ) would become

DA«I>+ + DB'I'- + DC

=

0, and on comparing with (4.38) it is clear that D must be chosen so that =

K+

DB = K-. (4.39) Suppose that A is a non-singular matrix so that we can write D = K+ A-l . Eliminate D from (4.39) by using this result in the second equation. Then we must have DA

K+A-I B

=

K-.

(4.40 )

The existence of K+ and K- satisfying this equation is obviously a

sufficient condition for solution of the original equation ( 4.37 ) providing of course that certain conditions concerning behaviour at infinity, etc. are satisfied. However it is by no means obvious that, given any equation ( 4.37 ) with general A, B and non-singular A , then matrices K+ and K- can always be found to satisfy (4.40), so that ( 4.37 ) can always be solved in this way. However we showed in §4.2 that the Wiener-Hopf problem was a special case of the Hilbert problem : similarly the Wiener-Hopf problem for several unknown functions is a special case of the Hilbert problem for several unknown functions. This latter problem has been treated in N. 1. Muskheli shvili [1], Chap. 18, where it is proved that the matrices K+ , K exist under quite general conditions. (The references quoted by Muskhelishvili go back to Plemelj .) However although the proof of the existence of K+ , K- given by Muskhelishvili provides in theory a method for constructing K+ , K- this does not seem to have been carried through in any concrete example. An interesting suggestion has been made by A. E. Heins [6]. If we know the eigenvalues of the matrix K then in general we can find a matrix L such that K exp {L } . =

The elements of K and L are functions of at . If we decompose each element of L additively,

lij( at )

=

lit ( at ) + lij ( at ),

then in a n obvious notation we can write

K+

=

exp {L+}

It would seem difficult to obtain concrete results for specific examples by this method.

1 58

T H E W I E N E R -H O P F T E C H N I Q U E

In the remainder of this section we content ourselves with the reduction of (4 .37 ) to a set of Fredholm equations in two different ways. There is no loss of generality in considering, instead of (4 .37 ) , +

«1>+

K'I'- + M

=

O.

(4.41 )

Assume for simplicity that the strip in which this holds is such that T < T < T+. As in §5.3, ( 5. 2 8 ) , if t is a real number, _

l.2 «1> ,:I- (t) t'¥; (t)

00

f «1> :I- (s) --t 1 ds f - ----; P '¥; (s) 1

= -.

27T� P

=

•

- 00

2 7T�

S

00

ds

-- ,

(i

s-t

- 00

= 1 to n ) .

Write these in matrix notation in an obvious way :

+ l«l> 2 (t)

1

= -.

27T�

00

P

f «I>+ (s) --t S

- 00

ds

-

(4.42)

- 00

Take values of «I> + (t) from (4.41) for values of at on the line T at = t and substitute in the first equation in (4.42) . Then 00

1

t {K(t)'I'- (t) + M(t)} = -. P 2 7T�

f

- 00

=

0,

ds . {K(s)'I'- (s) + M(s) } s-t

Multiply through by [K(t)] -l and add to the second equation in (4.42) . This gives 00

. 'I'- (t) - 27T�

- 00

=

ds

f {[K(t)]-lK(s) - I}'I'- (s) s-t [K(t)] -l {� P M(s) � - l.M(t) } 27T� f s-t 2 1

00

- 00

(4.43)

where the integral on the left is no longer a Cauchy principal value integral since the integrand is now well-behaved at s = t. This is a system of Fredholm integral equations of the second kind. The above method is based on the Plemelj treatment of the homogeneous Hilbert problem (cf. N. 1. Muskhelishvili [1], p. 387 ) .

E X T E N S I O N S A N D L I M I TATI O N S O F M E T H O D

Another method of reduction is the following. We have n

<1>,+ ( IX) + 8:L�1KrilX)'¥8- (IX) Suppose that Then

Krt <1>r+

+

Mr(lX) = 0,

Krr(lX) = Kr-:; (IX) /Krt (IX) . +

K;-;. '¥r-

n

( r = 1 to n ) .

Krt (IX) .

Multiply through by

Krt 8:L'�1 Krs '¥8-

+

159

+

K;!;. Mr = 0,

(4.44)

where the dash indicates that the term r = s i s omitted. In the following the limits s = 1 to n on the summation sign will be omitted. Set 00

Krt (lX) :L' Krs( IX)'¥8- ( IX )

=

1

2 TTZ.

-

ic

+

f

ic - co

d'

Krt m :L' KrSm'¥8- m '" - IX r

where the first term is regular in an upper half-plane and the second in a lower. If we apply the Wiener-Hopf technique in the usual way we find from (4.44) that

K;'¥; Krt <1>,+

1 - ---2 TTZ-;

d" f Krt m :L' Krsm'¥;m '" - IX + N;(IX)

id+ 00

r -

id - 00 ic + 00

1 +2 TTZ.

f Krt m:L'Krsm'¥; m � '" - IX + N,+ (IX)

=

=

Pr(IX) , (4.46a)

- Pr(IX) ,

ic - oo

(4.46b) where we have set Krt Mr = N,+ + N; and the Pr(lX) are arbitrary polynomials determined by conditions at infinity. (4.46a) holds for r = 1 to n and this gives a set of n simultaneous integral equations for the '¥ ; . If these are found the <1>r+ are then obtained from (4.46b) . As in the analysis leading to (4.43), if we let = 0, denote the real value of , by s and let IX tend to a real number t, then (4.46a) can be reduced to the system

d

K; (t)'¥; (t) - !K;!;.(t) :L' Krs(t)'¥; (t) -� 2m

00

f Krt ( ) :L' {Kr8(S)'¥; (s)

- 00

s

-

( r = 1 to n. )

Krs ( t)'¥; (t) } � s-t

160

THE

WIEN ER-H O P F T E CHNI Q U E

If the matrix K is diagonal the exact solution can be obtained immediately from these equations and therefore we should expect that this form will be useful for iterative solution if the diagonal terms are predominant. 4.5

Approximate factorization

All the problems solved in this book depend on the fundamental

factorization K(oc.) K+ (oc.)K_(oc.) . If we have to use the integral formula of theorem C, § 1 .3, to perform this factorization then the practical details of finding numerical solutions tend to become complicated. For concreteness consider =

(4.47) valid in T_ < T < T+ , - 00 < set

(J

< 00 . It may happen that we can

(4.48) K*(oc.) + Q(oc.), where a simple factorization exists for K*(oc.), and K(oc.) is very nearly equal to K*(oc.) in the strip in which (4.47) is true. Assume

K(oc.)

=

that in this case we can find the solution of

K*(oc.)+. (oc.)

=

'Y! (oc.) + F(oc.) .

(4.49)

It is then natural to compare <1>+ , +. and 'Y_, 'Y! . Two slightly different situations arise : (a) We may wish to know whether the solution of (4.49) can be taken as an approximation to the solution of (4.47 ) . (b) It may b e possible to write (4.48) in the form

K(oc.)

=

K*(oc.) + eq (oc.) ,

(4.50)

where e is a small parameter, and we may be interested in the perturbation of the solution of (4.49) caused by the presence of the term in e. In problems of type (a) it is easy to check whether K(oc.) and K*(oc.) are approximately �qual since it is necessary only to compare th e numerical values of these functions on some line T = T V - 00 < (J < 00, where T_ < T l < T+ . The factors K+(oc.) , K_(oc.) and the final solution can be expressed as integrals along a contour consisting of this line. If K(oc.), K*(oc.) are approximately equal on this line then the final solutions will be approximately equal. It is particularly important to notice that there is no need for the behaviour of K(oc.) and K*(oc.) to be similar in the complex plane, off the line T = T l ' - 00 < (J < 00 . (Apart from the argument just given we recall from §4.2 that the analytic nature of K(oc.) is in a sense

E X T E N SI O N S A N D LIMITA T I O N S

OF METHOD

161

accidental. If we use the Hilbert problem formulation then K(oc) need be defined only on a contour in the oc-plane and need not even be defined off this contour. ) In this connexion a n instructive example has been given by W. T. Koiter [1], part IIb. He compares problems involving the following two functions, both of which possess simple factorizations : K (oc)

=

oc-1 tanh oc

K*(oc)

=

(oc2 + 1 ) -1/2.

The behaviour of these functions is similar in a narrow strip, - a < T < a, - 00 < (] < 00 . They both tend to unity as oc tends to zero and to / oc/ -1 as oc tends to infinity. A comparison of numerical values indicates that they agree to within 9 % on the real axis. However the behaviour off the real axis is completely different. K(oc) has an infinite number of poles and zeros : K*(oc) has no zeros or poles, but it possesses two branch points. Consider

oc-1 tanh oc +(oc) (oc2 + 1 ) -1 /2 t (oc)

=

=

i(27T) -1/2 OC-1 + 'Y_ (oc), i(27T) -1/ 2 OC-1 + 'Y : (oc),

0 < T < t7T,

The solution of these equations is straightforward and it is left to the reader to verify that (cf. W. T. Koiter [1]) :

(x < 0),

- 00

cp(x)

=

{I

1p* (x)

=

1

cp*(x)

=

where

-

-

(4.51a) exp (-7TX) }-1/ 2 ,

E[y(-x)] ,

(x > 0),

(x < 0),

(7TX)-1/ 2 exp ( -x) + E[yx], E(s)

8

=

27T-1/2 I e - t2 dt,

(4.51b) (4.52a)

(x > 0) ,

(4.52b)

the error function.

o

If these expressions are compared it is found that cp(x) , cp * (x) both ,....." (7TX) -1/2 as x --+ 0, and they both tend to unity as x --+ 00 . The difference in their values does not exceed 3 % for any x. The expres sions for 1p(x) , 1p*(x) both vary as { I - 2 ( _ X) 1 / 2 7T-1/ 2 } as x --+ -0 but as x --+ - 00 , 1p(x) ,....." 27T-1 exp (t7TX), 1p*(x) ,....., (-7TX) -1/ 2 exp (x) . The numerical agreement is within 5 % for 0 > x > -t but becomes progressively worse as x --+ -00 (though then 1p(x) and 1p* (x) are small) .

162

THE

W I E N ER-H O P F T E CHN I Q U E

It is of interest that we can find explicit formulae for the kernels of the corresponding integral equations :

1 k(x) = 27T

00

I

- 00

.

oc.-1 tanh oc.e - '001: doc. =

1 1 + exp (-f7Tlxl ) , In 7T 1 - exp (-!n l xi )

-

00

k*(x)

=

� I (oc.2 + 1 )-1/2e - iOOl: doc. � Ko( l xi ) .

2

=

- 00

As Ixl ---+ 00, k(x) ,--, 27T-1 exp (-t7Tl xi ), k*(x) ,--, ( 27Tl xi ) -1/2 exp ( - I x i ) . W . T. Koiter has suggested that in order to check and improve the accuracy of any solution obtained by replacing K(oc.) by K*(oc.) , a second approximate solution should be carried out, using

oc.4 + Ooc.2 + D (4.53) oc.4 + Eoc.2 + D where 0 , D, E are chosen so that K * *(oc.) and K(oc.) agree as closely as possible. It is assumed that K(oc.) is an even function of oc. and that K(oc.), K*(oc.) agree as oc. ---+ 0 and oc. ---+ 00, so that this is the simplest K**(oc.)

=

K*(oc.)

'

rational function which can be used. The object of choosing the correction factor as a rational function is merely that it can be factorized easily, so that the split can still be performed in a straight forward way. By this means W. T. Koiter has reduced the difference between the solutions for K(oc.) and K * *(oc.) to less than 1 %. W. T. Koiter [ 1] has also made some applications of these ideas to the approximate factorization of functions which occur in problems involving the biharmonic equation, e.g. sinh2 oc. - oc.2 sinh oc. cosh oc. - oc. = K l (oc.) = K (oc.) 2 2 2 oc. (sinh oc. cosh oc. - oc.) 2 oc. sinh oc. These can be approximated by A(oc.2 + W) -1/ 2 or aoc.-1 tanh (boc.) to within 1 0 %, and by more complicated expressions of the form (4.53) to within 1 %. But for the above functions an exact decomposition is possible by means of the infinite product theory, though the roots involved are complex. It seems to me that in this case the crude approximations provide an excellent method for obtaining a rough answer. But if an accurate answer is required it is doubtful whether a refined approximation of type (4.53) is to be preferred to a straight forward though laborious solution based on the infinite product decomposition. We consider next problem (b) stated at the beginning of this section. Suppose that we require a solution of the following equation as a function of e for small e: [K*(oc.) + eq(oc.)] +(oc.) = 'Y_(oc.) + A(oc. + k cos 0) -1 , (4.54)

EXTEN SIONS AND LIMITATIO N S O F

METHOD

1 63

where k2 cos 0 < T < k 2 ' say, and the solution is known for s = O. We set K*(oc) = K� (oc)K"'- (oc), a known decomposition, divide (4.54) throughout by K"'- (oc) and use theorem B of § 1 . 3 to obtain the additive decomposition of q(oc) + (oc)/K"'- (oc) . Then separation by the Wiener-Hopf technique gives

A

= (oc + k cos 0)K+- (k cos 0) .

This is an integral equation for +(oc) and holds for any T V - Cf) < (] < Cf) , with k 2 cos 0 < C < T l < k 2 . Iterative solution gives successive approximations :

�p(oc) = A {(oc + k cos 0)K+- (k cos 0)K :\,- (oc)}-1 , <; ) (oc) = �) (oc) As

271'iK:\,- (k cos 0)K:\,- (oc) (4.55) We examine two concrete examples (i) K*(oc) = (oc2 - k2 ) 1 / 2 , q( oc) = 1. Then the integral in (4.55) is convergent and the iterative procedure is presumably satisfactory. (ii) K*(oc) (oc2 - k2 ) -1/2 , q(oc) 1 . The integrand in (4.55) behaves as l ocl -1 as l oci � Cf) in the strip and the integral is divergent. In this case the iterative solution is not satisfactory. Some light is thrown on the difficulty which has arisen in this second case, by the following remarks : =

=

( A) . The exact decomposition for the above examples has been investigated in ex. 2.10. On examining case (ii) it appears that

K+ (oc)

=

K:\,- (oc) + s ln s S+(oc) + . . . ,

( 4.56 )

i.e. the first correction term is of order s In s, not of order s. (B) . If we apply the product decomposition theorem C of § 1 .3 to (4.50 ) we have c + i co

In K+ (oc)

=

1 -. 2 71'�

f

c -i oo

In [K *

i')- +oc sqW ] d,.

In example (ii) above for finite s , large " the term sq W is more important than K t ( ') and this suggests the presence of terms in In s.

1 64

TH E W I E N E R-H O P F T E CH N I Q U E

(C) . Suppose that we can write

K(oc)

=

Kt (oc) { 1 + I3s+ (oc)}K '!:. (oc) { l + I3s_(oc)} = K * (oc) { 1 + I3 (s+(oc) + s_(oc)) + t3 2s+(oc)s_(oc)} , (4.57)

where 13 is small. Then on comparing with (4.50) it is natural to identify 13 with 8, neglect 13 2 in (4.57) and find s+(oc ), s_(oc) such that

s+(oc) + s_(oc)

=

q(oc)jK*(oc) .

This can be done explicitly for both the above examples by means of ( l .35) . For case iii) we find (4.58) But this result is wrong since the leading correction term should be of order 8 ln 8. We cannot assume that if K(oc) can be expanded in a power series in 8 then so can K+ (oc) , K_(oc) . If we multiply (4.58) by the corresponding expression for K_(oc) we find that for a given finite 8 the result does not approximate to K(oc) as l oci --+ 00 in the strip. This is in contrast to the examples considered in the first half of this section where the approximate K*(oc) agreed with the exact K(oc) over the whole range of oc in the strip. The question of finding a valid second approximation by means of relatively simple formulae in problems involving K(oc) of the form given in (ii) above would seem to merit further investigation. Several of the K(oc) which have appeared in previous chapters are of this type. 4.6

Laplace's equation in polar co-ordinates

Various problems involving Laplace's equation in two-dimensional polar and spherical polar co-ordinates can be solved by the Wiener Hopf technique in conjunction with the Mellin transform :

J

co

$(s)

=

o

2� J

a + i co

cpr" - l dr .

cp

=

$(s)r- S ds,

where s is a complex variable, s Laplace's equation are

=

(J

+ iT. The basic forms of

(two-dimensional polars),

1 � . ocp � .2 ocp __ sm 0 or · or + sin 0 00 00

(4.59)

a - i oo

_ -

0,

(spherical polars) .

(4.60a) (4.60b)

EXTENSIONS A N D

LIMITATIO N S

OF

METHOD

1 65

From one point of view there is no need to use the Mellin transform at all. The substitution r = exp (u) converts the above equations into

02 cp 0 2 cp = 0, ou2 + oe 2 1 O . ocp 0 2cp ocp = 0. sm e + U+ 2 sin e oe ou O oe These can be solved by a Fourier or Laplace transform in u. (This corresponds to the fact that the Mellin and Laplace transforms are related by an exponential substitution, cf. ( 1 .49).) However in practice it is convenient to use the Mellin transform and we proceed to show how this can be done. We write <1>p(8,e)

=

1

f

cp(r,e)r" - l dr

<1>N (8 ,e )

o

=

00

f cp(r,e)rS-1 dr,

(4.6 1 )

1

where subscripts ' P', 'N ' are used instead of ' + ' ' - ' to remind the reader that <1>p, <1>N are regular in right- and left-hand half-planes instead of upper and lower half-planes as for the Fourier transform. We need the following results derived from example (2) following theorem A in §1 .3, and from the Abelian theorem quoted at the end of §1.6 (In (r) R::> (r - 1 ) when r R::> 1 ) : If / cp / < A r - " as r -+ 0 , <1>p(8) is regular in a > a_ . If / cp / N (8) is regular in a < a+ . If cp ( 1 - r) < as r -+ 1 - 0, then / <1>p(8) / 1""-' / 8 / - 0 - 1 as 8 -+ 00 . If cp (r - 1 )'7 as r -+ 1 + 0 , then / <1>N (8) / 1""-' / 8 / - 17 1 as 8 -+ 00 . -

I""-'

-

I""-'

In the last two lines 8 -+ 00 in right and left half-planes respectively. Consider a concrete example. Suppose that there is a cone at uniform potential V in ° < r < 1 , e = e. Application of a Mellin transform to (4.60b) gives

d . d <1>(8) Sill e � + 8(8 - 1 )<1>(8) = 0, sin e de <1>(8) = A (8)Ps_ 1 (cos e) + B(8) Ps_ 1 ( -cos e). The functions Ps_ 1 (cos e), Ps_ 1 ( -COS e) are singular at e 1

respectively. Hence we write

<1>(8) = A (8)Ps_ 1 (cos e) B(8) Ps_ 1 ( - cos e), =

0 < e < e, e < e < 1T.

= 1T ,

0,

(4.62a) (4.62b)

1 66

T H E WIE N E R-H O P F T E C H N I Q U E

We apply boundary conditions on e O . Use notation (4.6 1 ) and denote the corresponding transforms of acpjae by �(s,e), � (s,e). Let �(s, ° + 0) denote the value o f �(s, e) as e --+ ° through values of e greater than 0, etc. (s) is continuous on e ° and from (4.62) we write =

=

Then

A(s)Ps_ I (cos

O)

B(s)Ps_ I ( -cos

=

p(s,0) + N (s,0) � (s, 0 + 0) + �(s,0)

O)

=

O(s), say.

(4.63) O(s), = O( S )P; _ I (-COS 0) I Ps_ I (- c os O ) ,

=

(4.64a)

� (s, ° - 0) + �(s,0)

=

O(S)P; _ I (COS 0 ) I Ps_ I ( cos O ) . (4.64b)

These can be reduced by the standard procedure of §2.2 to the Wiener-Hopf equation

-Fp(s,0)

where

=

K( S ) {N ( S ,0) + VIs},

K(s) sin 1TS{1T sin 0Ps_ I ( cos 0)Ps-I ( -cos 0) }-I , 2 Fp(s,0) = � (s, ° + 0) - CI>� (s, ° - 0), =

(4.65) (4.66 )

and we have used the results

P;_ l (-COS 0)P; _ I (COS

O)

- Ps_ I (cos 0)P; _ I (-COS O ) 2 sin 1Tsl1T sin 0, =

1

p(s,0)

=

V f rS-1 dr == VIs. o

K(oc) is an integral function of s and can be decomposed by the infinite product theory. Details are given in S. N. Karp [2] . The solution can then be completed in the usual way. A similar example is discussed in J. Bazer and S. N. Karp [1] . To illustrate the procedure we consider the special case of a disk, ° :::::; r :::::; 1 , ° = 1T( 2. It is known that I

Ps- l (o) = 1TI/2 {r(t + ts) r(l - tS )}-I .

Then (4.66) can be reduced to

K(s)

=

2 r(t + ts) r(l - ts) { r(ts) r(i -,- is)}-I .

(4.67)

We have the additional information : (i) As r --+ ° the charge density on the disk is finite. Hence Fp(s,O) is regular for (] > - 1 . (ii) p(s,O) exists and is regular for (] > 0.

EXTENSIONS AND LIMITATI O N S O F M E TH O D

1 67

(iii ) As r -H/J, 14>1 < Or-I . Hence Cl>N(S,O) is regular for a < 1 . (iv) As r .- l + 0 , 4> ,-.....- const. Hence I Cl>N(S,O) I ,-.....- l sl-1 as s .- 00 in a left half-plane. From (i)-(iii) and (4.67) we conclude that (4.65) holds in the strip ° < a < 1 . The required factorization of (4.67) is then K(s) = Kp (s)KN(S) where On applying the Wiener-Hopf technique, (4.65) gives the solution

Hence ..I. 't'

=

f

c + i oo

r( t - is) PS_1 ( l cos O \ ) s d r s. s r ( 1 - .12 S ) • P8- 1 (0) c - i oo

V_

_

' 1/2 2 7TZ7T

T o obtain the charge distribution o n the disk w e calculate r-1ca 4>/o O ) on 0 = + 0 . On using the result

P; - l (O) =

we find

-

2 7Tt {r (tS ) r (t

-

is) }-l ,

Further details concerning application of the Wiener-Hopf technique in connection with Laplace ' s equation in polar co-ordinates will be found in J. Bazer and S. N. Karp [1], S. N. Karp [1], . [2] , and D. S . Jones [7], [8] . Miscellaneous Examples and Results IV In this book we have been concerned with solution of partial differential equations and in general it has not been necessary to formulate problems in terms of integral equations. However a great deal of the literature is concerned with integral equations and for reference we quote the classic example considered by N. Wiener and E. Hopf [ l]. This is Milne's integral equation which occurs in connexion with radiation and neutron transport problems : 4.1

f(x) =

'"

f f(�)k(x - �) d�,

o

(x

>

0)

k(�)

=

i

'"

I A - I e - A dA.

1<1

(a)

168

THE

W IE N E R-H O P F T E C H N I Q U E

If we suppose that the right-hand side of this equation defines a function g(x) for x < 0, application of a double-sided Laplace transform gives

K(s)

1

=

Fp(s)K(s)

=

J k(x)e-S'" dx

=

ro

.

- ro

GN(s), 1

-

(b)

ts-1 In

l + s . 1 s

--

K(s) has a double zero at the origin and In K(s) has branch points at s = ± 1 . Equation (a) is usually deduced from (c)

J Iji (Z,Il) dll

+1

lji o

=

='=

f(z) in (a) above .

-1

The Wiener-Hopf technique is often applied to the following equation which can also be obtained from (c) :

Fp(s)K(s) =

-

t

o

1 m d�. 1 + s�

J Fp(

-1

-

(d)

On comparing (d) and (b) it is clear that the right-hand side of (d) is a different representation of GN(s). In fact (d) can be obtained by apply ing a Laplace transform in (0, 00 ) to (a). This gives

F (s)K(s)

ro

=

ro

J f(�) J e S"k(y) dy d�.

o

Ii

It is left to the reader to show that the right-hand side of this expression can be reduced to the right-hand side of (d). There is an extensive literature, e.g. N . Wiener and E . Hopf [ 1 ] , E. Hopf [ 1 ] , G. Placzek [ 1 ] , G. Placzek and W. Seidel [ 1 ] , C . Mark [ 1 ] , R . E . Marshak [ 1 ] . Brief treatments are given in E . C. Titchmarsh [ 1 ] , I. N. Sneddon [ 1 ] . Two books which deal with other methods of solution of the equation in addition to the Wiener-Hopf method are V. Kourganoff [ 1 ] , and B. Davison [ 1 ] . 4.2 We summarize briefly the conditions for uniqueness of solution of the equation

f(x)

ro

=

J k([ x - y[ ) f(y) dy + g(x),

o

(0

<:

x

<

00 ) ,

where f is unknown, g and k are known. For the homogeneous equation, i.e. g(x) = 0, this has been examined by N. Wiener and E. Hopf [ 1 ] . (See E . C. Titchmarsh [ 1 ] , p. 340. ) For the non-homogeneous case the theory has been extended by V. Fock [ 1 ] , [2] . In the following remarks

EXTENSIONS AND

LIMITATI O N S O F M E T H O D

169

integrals are t o b e understood in the Lebesgue sense. Suppose that k 1 (x) = k(x) exp (ex) is of limited variation and absolutely integrable in (0, w ) for some e > 0. Define - 00

which is even and regular in the strip -e < 1m ex < + e ; I exK(ex) I is bounded at infinity in the strip ; In {I - K(ex)} tends to zero as l exl -1 at infinity in the strip. (i) The homogeneous equation, i.e. g(x) 0. (a) If {I - K(ex)} ° has no real roots or singular points and if "'0 is the imaginary part of the complex root or singular point nearest to the real axis, then the homogeneous equation has no solutions satisfying the inequality I f(x) I < a exp hx), (A) where a is a constant. (b) If {I - K(ex)} = ° has 2n real roots, the homogeneous equation has exactly n solutions, say fr (x) , which are linearly independent and satisfy equation (A) . (ii) The non-homogeneou8 equation, i.e. g(x) =1= 0, where g(x) is absolutely integrable in (0, w ) and of limited variation. (a) If {I - K(ex)} = ° has no real roots there exists exactly one solution which is bounded and tends to zero at infinity. (b) If {I - K(ex)} = ° has 2n roots whose multiplicity does not exceed 8, and if g(x) satisfies two additional conditions, namely x S-1g (x) In x is absolutely integrable and =

=

J g(x) fr(x) dx = 0,

o

(r

=

1 to n) ,

where the fr (x) have been defined in (i) (b) , then the non-homogeneous equation has exactly one solution which tends to zero at infinity. 4.3 When solving integral equations by the usual Wiener-Hopf method it is assumed that there exists a common strip of regularity for all transforms in the complex plane. It is possible to extend the theory to the case where Fourier transforms exist only on a common line parallel to the real axis in the complex plane by reducing the problem to a Hilbert problem. The normal Wiener-Hopf theory applies when the kernel k(x) of the integral equation decays exponentially, I k(x) 1 < a exp ( -e I X I ) , e > 0, as Ix l --- w . The case considered in this example occurs when k(x) decreases algebraically, I k(x) 1 < Olxl-p , p > 0, as I x l -+ w . The following is based on J. A. Sparenberg [1], [2]. Consider

f(x) =

fl

( )

J a2 +f(xt;

o

_

t; ) 2

dt; + g(x),

x > 0,

7Tfl <

a.

(a)

170

T H E W I E N E R-H O P F T E C H N I Q U E

Introduce a function h(x), equation for x < O. f(x) in 00 < x < 00 gives

x

<

0, defined by the right-hand side of this 0 for x < O. A Fourier transform

g(x)

=

-

=

F+(ex)K(ex) + H_(ex) = G+ (ex), K(ex) = 1 - fl7TG,-l exp ( - a l ex l ) .

(b)

This holds only for real ex since the Fourier transform o f the kernel holds for only real ex. We can treat this as a Hilbert problem. Define In

K+ (ex)

=�

27T�

00

dx J x K(x) - ex In

- 00

,

1m

ex

>

(c)

0,

with a suitable indentation of the contour if 1m ex -+ O. Define by a similar formula with a minus sign and 1m ex .;;; O. Then

F+ (ex)

=

00

1

2?TiK+(ex)

J K_(x)G+(x(x)- ex) dx

- 00

K_(ex)

.

Complications arise if 7Tfl > a since then K(ex) has two zeros on the real axis, say x = ±p. This suggests that f(x) is finite and oscillatory as x -+ + 00 due to the presence of terms of the form exp ( ±ipx) . Following this hint we introduce

v(x) = f(x) - Ae- ip," - Beip," ,

(x > 0),

where A, B are constants which will be determined by the condition that v (x ) -+ 0 as x -+ + 00 . Introduce into (a) . For simplicity we consider the homogeneous case, g(x) O. Then for x > 0, =

. . J v(x) + Ae-'P," + Be'P,"

00

=

fl

o

a2

v(�) d�

+ (x - � ) 2

+

00

fl

J Ae-ip+ <(x

+

a2

Beipl; _

0

�)2

d�.

The terms in A, B cancel as x -+ + 00 . Suppose that the right-hand side of this equation defines a function h(x) for x < O. A Fourier transform gives

In

{

}

i A B + K(ex). (27T ) 1/2 ex - p ex + p --

--

order to avoid difficulties from the zeros of K(ex) we define

=

� m

00

dx J x� L(X� - ex

o

l

,

1m

ex

>

O.

E X TENSIONS AN D LIMITATI O NS

O F M E TH O D

171

Suppose that the arbitrary separation polynomial is a constant Then separation gives Bi Ai o V+ (O() = 2 2 (0() ) L (0( p +

O.

_

In order that V+ (O() should have no poles at 0( = ±p we need to choose ( 21T) lf20i (21T)1/20i A = B = . 2pL+ ( -p) 2pL+ (p)

The solution is now determined. 4.4 J. A. Sparenberg [2] has considered integral equations of the following form

f {b of( ';) + bd' ( ';)}k(x ro

a o f(x) + ad' (x) 4.5

=

o

Show that

-

.;) d';.

ro

( .; )

tA

=

f ��) -

o

cos h "2

d , 'f} ) 'f}

( .;

>

0),

can be reduced to the Wiener-Hopf equation
Show that integral equations of the form 1

f(x) + g(x)

=

A

f x .; d';, f f( ';) d';,

f(x) + g(x)

=

A

(0

+

o

or

f( '; )

1

,.;;; x

,.;;;

1),

( 1 ,.;;; x < 00 ),

x + .;

are of the Wiener-Hopf type. (Set x = exp ( -y), .; Cf. ex. 4.5. See E . C. Titchmarsh [ 1 ] , p. 342.)

=

exp ( -'f} ) .

4.7 E. C. Titchmarsh [ 1], p. 335, has solved a particular p air of dual integral equations by a method related to various topics in this book. We use the method to solve the general equations ro + ic

1 f

.

K(O() A (O() e -'cxx dO(

I'i::\fl2 (21T)

- ro + ic ro + ic

1 f

(21TjI12

- ro + ic

.

A (O() e -WZ dO(

= f (x),

x > 0,

(a)

=

x < 0,

(b)

g (x),

1 72

THE WIEN ER-H OPF T E C H N I Q U E

where

c

is a real number,

T_

T+ .

< c < A (Ol) W c,

is regular in this strip and

K ( Ol ) is regular and non-zero in the strip . Suppose that we can write K ( Ol ) = K_(Ol)/K+ (Ol) in the usual way, and introduce B(Ol) = A (Ol)/K-i-(Ol) . Multiply (a) by exp (iw1x), 1m I > and integrate with respect to x from 0 to 00 . Similarly multiply (b) by exp (iw 2x), 1m w 2 and integrate with respect to x from - 00 to o. The equations become i (2 Tr ) 172

co + io

f

K_ ( Ol )

- co + ic co + ic

i (2 Tr ) I/2

B

co

( Ol )

� =

Ol - wI

B

f

< c,

f eiW1Xf(x) dx o

-� (2Tr)

co + ib

f

f

WI

11'+(w I ) '

say, (c )

o

eiw,Xg(x) dx K+ ( Ol ) ( Ol ) � Ol - w 2 - co Now choose some constant b such that 1m I the line of integration in (c) to T = b. This gives - co + ·ic

=

=

G_(w 2 ), say, (d)

c < W
T

+ and move

IC(Ol)B(Ol) � = - (2Tr) 1/ 2K_(w l )B(w l ) + F+ ( wl ) · (e) Ol -

- co + ib

The function on the left-hand side is regular in the lower half-plane 1m W I and shift b. Similarly choose a so that T_ a < 1m w 2 the contour in (d) to T = a. Then

<

--;(2Tr ) /2

< c,

<

co + ia

f

K+ (Ol)B(Ol) � = _ (2Tr ) I/2K+ ( W 2 )B(w 2 ) + G_(w 2 ) · (f) Ol - w 2

- co + ia

The left-hand side and therefore the right-hand side is regular in the upper half-plane 1m w 2 > a. From the point of view adopted in this book the most direct method of solution at this point would be to replace W I and W 2 by Ol, denote the left-hand side of (e) by 'Y_(Ol), the left-hand side of (f) by cI>+ (Ol), and then eliminate B(Ol) to obtain a Wiener-Hopf problem. Titchmarsh proceeds as follows : Since the function on the right in (f) is regular in the upper half-plane T > T_ , so is Hence

co + ic

f { B(Ol)

- co +ic

B(Ol) - (2Tr)-1/2 {K+ (0l)}-I G_ (0l). �

}

G_(Ol) - 0, � (2Tr) 1/2K+ (0l) Ol - W

Similarly from (e)

(1m

W < c).

(g)

(1m

W < c).

(h)

co + ic

-

} dOl f {B( Ol) - (2Tr)F+(Ol) I/2K_(0l) Ol - W co + ic { B(W) (2Tr��( W )} =

- 2m

-

,

E X TENSION S AN D LIM ITATIONS

O F M E TH O D

1 73

In this case there is exactly one pole, at oc = w, when the contour is completed in the lower half-plane. The function B(w) is now obtained by eliminating the integral involving B(oc) between (g) and (h). This is the required solution, and it can be reduced to the general solution in §6.2 (cf. ex. 6 . 1 ) . In practice it may not be possible to carry out the procedure exactly as described because of convergence difficulties but artifices to deal with these can be invented along the lines indicated in §6.2. 4 . 8 The method of §4.6 has been used to solve the problem of steady flow of incompressible fluid past a thin aerofoil by D. S. Jones [7]. The method has been extended to deal with flow past an oscillating aerofoil in D. S. Jones [8]. An interesting feature is the way in which the Kutta-Joukowsky condition is applied at the trailing edge. 4.9 The method of §4.6 can be used to solve problems involving stresses in a wedge of elastic material when mixed boundary value conditions are specified on the faces of the wedge, cf. W. T. Koiter [2], and ex. 3. 19. 4.10 The Wiener-Hopf technique can be used to solve the set of simultaneous linear algebraic equations : co

2: kn_mx m = b n >

(n = 0, 1, 2 . . . ) ,

m�O

(a)

where the known coefficients kn_m are functions of only the difference (n - m). We extend the equations by writing the left-hand side equal to the (unknown) constants Cn for n = - 1, - 2, . . . . Multiply the equation for any given n by ocn and sum over n from - w to + w . Inter change orders of summation on the left-hand side. It is found that co

= 2: bn ocn + n�O

-1

2: cn ocn ,

n � - co

or

(b)

where the subscripts indicate that B+(oc), X+ (oc) are regular inside some circle of convergence, whereas O_(oc) is regul ar outside a certain circle in the oc-plane. Assume that K(oc) is a regular function of oc in r+ < l oc i < r_, with 'plus' functions regular in l oc i < r_, 'minus' functions regular in l oc i > r+. Then (b) holds in a strip in the complex plane and the Wiener-Hopf technique can be applied. We write K(oc) = K+ (oc)K_(oc). Divide (b) by K_(oc) and write B+(oc)/K_(oc) Then (b) can be written

=

H+ (oc) + H_( oc ) .

J (oc) = K+ (oc)X+ (oc) - H+(oc)

=

O_(oc)/K_(oc) + H_(oc) .

This defines a function which is regular in the whole of the oc-plane and X+ (oc) can be determined by Liouville's theorem in the usual way. The xn can be found by expanding X+ (oc) as a power series in oc.

1 74

THE

W IE N E R- H O P F T E CH N I Q U E

This solution has been obtained in a slightly different way by A. N. Fel'd [1]. As an application he considers the propagation of a wave in a waveguide 0 .;;; x .;;; a, 0 .;;; y .;;; b, 00 < Z < 00, with cylindrical posts of (small) radius r, the centres lying in y = d, Z = 0, D, 2D, . . . , each post extending over 0 .;;; x .;;; a. (The solution of this example is not such a feat as might appear at first sight since once it is realized that the solution is of the form x m = pp m where P, p are constants, the values of P, p can be obtained by direct sp.bstitution in the original simultaneous equations. Nevertheless the example provides an instruc tive application of the theory. ) -

4 . 1 1 Some o f the problems we have solved b y means o f the Wiener Hopf technique can be formulated in terms of simultaneous linear algebraic equations. In such cases the linear equation formulation is often the more natural procedure. We first of all reduce a typical Wiener-Hopf problem to a set of linear simultaneous algebraic equations. Consider (a)

T_ T T+,

where this equation holds in < < - 00 < (1 < 00 , and K(cx) is regular and non-zero in the strip. It is necessary to assume that K(cx) has no branch points in the upper half-plane. Suppose that in the upper K(cx) has simple poles CX l ' CX 2' . . . and simple zeros fl l ' fl 2' half-plane and that these are interlaced : .

.

•

I l < I fl l < I 1 < I fl21 < . . . cxl

l

CX 2

Suppose that K(cx) can be expanded in the form

(b) Substitute in (a), set H(cx) =

H+

( cx ) + H_(cx), and rearrange in the form

This defines an integral function which we assume to be zero . Then (c)

fln

fln F-(fln ) H(fln )'

K(fln ) +(fln) F-(fln ) - H_ (fln ) H+ (fln ).

in (a), by definition If we set cx = is in an upper half-plane. Then =

l .e.

= 0 and G

=

is finite since

E XTENSIONS AND LIMITATION S O F METHOD

175

O n setting ex = Pm in (c) we obtain a set o f linear simultaneous algebraic equations for G+ (exn) : co

L

n = l Pm

� ctn anG+ (exn ) = H+( Pm ) '

(m = 1 , 2, 3, . . . ) .

The procedure can obviously b e worked backwards. To solve co

� ctn xn = c m>

L

n = l Pm

(m = 1, 2 , 3, . . ) ,

(d)

.

we need to find a function K(ex) which can be written in the form (b) and such that K( Pm ) = o. Then if H(ex) is a function such that H( Pm ) = c m and we can solve the Wiener-Hopf problem (a), the solution of (d) is given by Xn = anG+ (exn).

Consider a wave exp ( i kz ) incident from z = - 00 in a duct .;;; b, - 00 < z < 00 with iJ
4.12 .;;; Y

1 + Ao +

co

co

2: Bn cos (n - t)rry/b, n=l

2: A n cos nrry/b n=l

co

co

ik( 1 - A o ) + 2: YnA n cos nrry/b = - 2: Yn- t Bn cos (n - t)rry/b, n=l n =l where A n' Bn are unknown constants and Y p = {(prr/b)2 k2 }l/2 . By eliminating say A n show that the equations for Bn can be written in the form (d) of ex. 4 . 1 1 . Hence find the Bn. (exn = iYn t , n = 1 , 2 , . . . : PI = k, Pm = iYm_l ' m = 2, 3, . . . . K(ex) = Y tanh yb, H(ex) = 4 kby- 1 -

tanh yb. ) The simultaneous equations are a generalization of a set considered by E. H. Linfoot and W. M. Shepherd, Quart. J. Math. ( Oxford) 1 0 ( 1939), 84-98. Cf. W. Magnus and F. Oberhettinger, Oomm. Pure and Applied Math. 3 ( 1 950), 393-410.) An equivalent problem is that of a strip across a duct and we can use the above method to find the solution (3.99), ( 3 . 100) of §3.6. 4 . 1 3 It has been shown by W. Magnus, that the integral equation ro

I H�I) (klx - �i l f( � ) d�

o

=

g(x),

Z. Phys. 1 7 7 ( 19 4 1 ) , 168-179, (0 <

x < 00 ) ,

(a)

can be solved by setting ro

g( x) = 2: ( -i) mJ m (kx)a m >

m=O

co

f(x) = trrcirr/4 2: ( - i ) m ( 2m + l )x-IJm H(kx)Om· m=O

(b)

176

THE

WIEN ER-H O P F T E C HN I Q U E

The unknown O m are related to the known a m by En

i {

1 + n+ -n + m + t

Om

� -!} +

=

an > EO

=

1, E n

=

2 (n ;;;. 1 ) .

(c) We can deduce (c) and the solution of (c) from (a), (b) by a method related to the Wiener-Hopf technique. Application of a Fourier transform to (a) gives m=O

(d) where F+( O( ) , H_( O( ) are unknown. Introduce (cf. § 1 . 6 ) IX - k cos {3, -ik sin (3. Write G+({3) in place of G+ (IX) etc. Since H_({3) has no singularity at {3 = 0 ( 0( - k) an examination of cuts and symmetries in the (3-plane shows that H_({3) H_( -(3). In the (3-plane (d) is =

I'

=

=

=

2(k sin (3)-l F+((3)

=

G+({3) + H_({3).

(e)

Change the sign of {3 and subtract the resulting equation from (e). Then (f) 2 (k sin (3)-l{F+( (3) + F + ( - {3)} = G+( {3) - G+( -(3).

The Fourier transforms of (b) give (A. Erdelyi et al. [1] ) : G+(O( )

=

:

00

1'-1 ( 0( + y)- mk ma m (2 ) 1/2 m = O

L

(g)

Replace 0( by - k cos {3, remembering that by analytic continuation ( 0( _ik1/2 exp (li{3). Then (f) gives + 1')1/2 =

00

2: ( - l ) m cos m{3 a m

m=O

=

00

21T 2: ( - l ) m cos (m + !){3 0 m ' m=O

I f we multiply by cos n{3 and integrate in (0,1T) we obtain (c). But if we multiply by cos (n + !){3 and integrate in (0,1T) we obtain the solution of W. MagnJls directly : om

1

= -

2�

oo a { I I -!} . + n l n m -n m + + + + L

n=O

One unsatisfactory feature o f the above analysis i s that we have had to assume the forms of the expansions for f( x) and g(x) (cf. (g ) ) . It would be of great interest if one could deal with branch points by extending the above method to apply to the general Wiener-Hopf equation.

E X TEN SIONS AND LIMITATIONS O F METHOD

177

4.14 Consider the potential distribution due t o a two-dimensional perfectly conducting strip in 0 < x < a charged to a uniform potential V. Suppose that there is a charge distribution 1 ( x) per unit length on the strip. By superposition the potential at any point ( x,y) is a

I 1(1;) In {(x - 1; )2 + y2} dl;. This gives the following integral equation for 1( 1;) : - 2 I1(1;) In ( I x - 1; 1 ) dl; (0 < x < a) . V, '"

=

-

o

a

=

o

Differentiate with respect to x :

J I;1(1;)- x dl; a

=

0,

(0

< x < a) .

(a)

o

To solve this equation let the right-hand side equal the unknown function g(x) for x > o. Apply a Mellin transform. Interchange orders of integration on the left and set x I;rJ . Then =

a

00

00

I 1(1;)1;8-1 dl; I rJS - 1( 1 - rJ )-l drJ I g(x)xs - 1 dx,

i.e.

o

=

0

This is a Wiener-Hopf equation which can From the point of view adopted in this formulate the integral equation (a). The applying a Mellin transform directly to the in polar co-ordinates (cf. §4.6).

a

be solved in the usual way. book it is unnecessary to problem can be solved by partial differential equation

CHAPTER V

SOME APPRO XIMATE METHODS 5.1

Introduction

All the problems considered in previous chapters have been highly idealized. In connection with the two-dimensional wave equation, for example, only parallel, infinitely thin, semi-infinite plates were considered. In this chapter we discuss some approximate methods which can be used to deal with plates of finite thickness and finite length. From a mathematical point of view we can say that the typical two-dimensional wave-equation problem considered so far has involved two-part mixed boundary value conditions on infinite parallel boundaries, e.g. on boundaries, say, y ±b, - 00 < x < 00 we have been given different conditions involving cp and ocp/oy on - 00 < x < 0, and ° < x < 00 . Ideally we should like to be able to solve problems involving the two-dimensional wave equation with : (i) Boundaries along x constant as well as y constant, i.e. boundaries perpendicular to each other as well as parallel to each other. (ii) m-part mixed boundary value problems when the boundary is divided into m parts and different conditions involving cp and ocp/oy are specified on alternate segments. We consider below only thtee part problems. In this chapter we make some limited progress towards these objectives. The basic Wiener-Hopf equation. which has appeared in the exact solutions considered in previous chapters is =

=

=

A { oc)
=

0,

where A, B, 0 are known analytic functions, <1>+ , 'Y_ are unknown functions regular in upper and lower half-planes respectively, and the equation holds in a strip T_ < T < T+ , - 00 < (J < 00 in the complex oc-plane. The problems in this chapter can be reduced to an equation of the type A { oc) + { oc) + E { oc) + { -oc) + B { oc)'Y_{ oc) + D { OC ) I { OC ) + O { oc) = 0,

(5.1 )

where A , . . . E are known, <1>+ , 'Y_ are unknown, and <1> 1 is an un known integral function. Equation (5.1) cannot be solved exactly by 178

S O M E APPROXIMATE

METHOD S

179

the Wiener-Hopf technique. It will be convenient in some cases to write-E(IX) = B(IX)b(lX) so that (5. 1 ) can be written

A (IX)
(5.2)

The nature of the function b(lX) will play an important part in the approximate solution. §5. 2 deals with formulation of specific problems as equations of type (5. 1 ) . The remainder of this chapter is divided into two parts. One deals with "perpendicular boundaries" (case (i) above) . The other deals with three-part problems (case (ii) above). The reader who is interested mainly in applications may prefer to skip §§5.3, 5.5 on a first reading, whereas -the mathematically minded reader may prefer to read these first. §§5.3, 5.5 require a good understand ing of the basic decomposition theorem B of § 1 . 3 : If f (lX) is regular in T _ < T < T+ , - 00 < a < 00 , with suitable behaviour as I a l -+ 00 in the strip then we can write

f

ic + oo

f(lX)

=

� 2m

ie - co

f m d, - � 2m , - IX

id+ 00

f

id - oo

f ( ' ) d" , - IX

f+(IX) + f- (IX), respectively, (5.3) where L < C < T < d < T+ , and f+ (IX) is regular for T > T_, f_(IX) for T < T+ . =

In a book of this size devoted to one particular technique it is impossible to give the reader a sense of proportion. The Wiener Hopf technique is only one method out of many for solution of field problems, and from another point of view it is only one method out of many for solution of singular integral equations. Even though a problem can be solved exactly or approximately by the Wiener-Hopf technique there may be some alternative method of obtaining sufficiently accurate results in an easier way. These remarks apply particularly to the approximate methods discussed in this chapter. A discussion of various methods for solving the type of problem in which we are interested will be found in P. M. Morse and H. Feshbach [1]. More specialized references are N. Marcuvitz [ 1 ] , and L. Lewin [1] for waveguide problems, and C. J. Bouwkamp [ 1 ] for diffraction theory. In connexion with the topics in this chapter it would seem that the Wiener-Hopf procedure is the only technique available at the moment for the thick-plate problem of §5.4, and it is convenient for dealing with problems involving strips or apertures of large width as in §5.6. For problems like the flanged pipe (§5.2, (5.14) ) , the duct with a step (ex. 5 . 1 ) , and diffraction by slits and strips of small

1 80

T H E W I E N E R- H O P F T E C H N I Q U E

width, the most natural approach is to try to find the field distribu tion across a finite aperture or obstacle in the system whereas the Wiener-Hopf technique concentrates attention on semi-infinite apertures or obstacles ; on the other hand, in many such problems the approximate methods of this chapter will be convenient for certain ranges of parameters. A note is necessary on the use of the word "approximate" in the title of the chapter. In the problem considered in §5.4 the solution is expressed in terms of constants which satisfy a given infinite set of simultaneous linear algebraic equations. The constants must be determined numerically by solving a finite number of the linear equations, so that the constants and hence the solution are known only approximately. For small enough wave numbers , i.e. long enough wavelengths the solution can be determined as accurately as desired by solving a large enough set of equations. On the other hand for wave numbers greater than a certain value the solution of n X n sets of equations for various n may not converge as n --+ CIJ so that it may not be possible to obtain a satisfactory approximate solution in this way. The solutions of §§5.5- 5.6 are approximate in a different sense. At an early stage of the calculation an integral is replaced by the first term of an asymptotic expansion for the integral so that the solutions are inherently approximate. It may be possible to ascribe an order of magnitude to the error. But some degree of error is inevitably present and the size of the error will increase as the wave number decreases. The essential feature distinguishing the two types of approximation involves the nature of certain functions of a complex variable which occur in the Wiener-Hopf equations. In the first case these functions possess only simple poles : in the second case branch points occur. We shall return to this point at the end of this chapter. 5.2

Some problems which cannot be solved exactly

In this section we consider the formulation of some problems which cannot be solved exactly by the Wiener-Hopf technique. Only very simple examples are considered and these are treated from various points of view in order to illustrate several ways in which equations of form (5.2) can occur. The symbols (oc), +(oc) , etc. in this section sometimes refer to transforms with respect to x, and sometimes with respect to y. Similarly dashed quantities will sometimes denote derivatives with respect to y, sometimes with respect to x. The different meanings will be clear from the context. Suppose first that the plane wave exp (-ikx) travelling from right to left parallel to the x-axis is incident normally on the end of a

181

S O M E APP R O X I M A T E M E TH O D S

plate of finite thickness lying in - b � 'Y � b , - (f) < x � o. There is symmetry about 'Y 0 and we can write CPt CPi + cpo In the usual way, by applying a Fourier transform in x to the two dimensional wave equation,

flat

=

<1>(Ot)

=

=

A (Ot) exp (-Y'Y),

'Y ?: b .

(5.4)

In -b � 'Y � b, x ?: 0, on integrating b y parts, 00

0 2 cp

J ox2 eirxx dx

=

-

o

(oxocp ) 0 + iOt(A.)'I' 0 - (27T) 1/2Ot2<1>+ (Ot) .

Hence in this region the partial differential equation becomes

d 2<1>+(Ot)/d'Y2 - y 2<1>+(Ot)

( 27T)-1/ 2 (Ocp/OX)o - (27T) -1/ 2iOt(cp)o. (5.5) Assume that oCPt/on = 0 on the plate so that ocp/ox = ik on x 0, -b � 'Y � b. In this case (cp)o is unknown. To eliminate (cp)o from (5.5) change the sign of Ot and add the resulting equation to (5.5). =

=

This gives

d 2 {<1>+(Ot) + <1>+( - Ot)}/d'Y2 - y 2 {<1>+(Ot) + <1>+{ - Ot ) }

(27T) -1/ 22ik. Solve this equation and find, on using the symmetry about 'Y 0, =

=

(0 � 'Y � b). (5.6) Insert <1> = <1>+ + <1>_ in (5.4) with 'Y = b . Differentiate (5.4) with respect to 'Y, set <1>' <1>� + <1> '_ on 'Y b. Eliminate A and use the fact that <1>� = 0 to find =

=

<1>+(Ot) + <1>_(Ot)

=

_ y-l<1> � (Ot) .

(5. 7a)

Similarly from (5.6), on 'Y = b,

<1>+(Ot) + <1>+(-Ot)

=

- (27T)-1 / 2 2iky-2 + y-l coth yb {<1>� (Ot) + <1>� (-Ot)}.

(5.7b)

Eliminate <1>+ (Ot) between (5.7a, b).

2 y-l (1 - r 2yb ) -1 <1>� (Ot) + [{<1>_ (Ot) - <1>+(-Ot)} + + y-l coth yb<1>� (-Ot)] - (27T) -1/ 2ik y-2 = O. (5.8) This is in form (5.2) with b(Ot) y-l coth yb, i.e. b(Ot) has only simple =

poles and no branch points. Equation (5.8) contains the complete solution of the problem.

1 82

T H E W IE N E R-H O P F T E C H N I Q U E

An alternative formulation can be obtained by applying trans forms in the y-direction and matching functions across x o. We have in the usual way =

(oc) Define, on x

+ (oc)

=

=

1 __ 12

( 2n ) /

=

x ;? o.

A(oc) exp (- yx),

0,

f

00

A. eilX(y -b) dy 'I'

b

1 ( oc)

=

1

(5.9) b . f Y 1 2 cpe'IX dy.

( 2n ) /

-b

(5. lOa)

l (OC) is an integral function of x and +(oc) has been defined so that it has polynomial behaviour as l oci � 00 in an upper half-plane. From the symmetry about y = 0,

(5. lOb) (oc) eilXb+(oc ) + l (OC) + e - ilXb+(-oc). Since ocp/ox ik on x 0, -b <; y <; b, b 2ik ik I l (OC) ( 2 n ) 1 / 2 e'IXY dy = (2 n ) 1 / 2 oc sin ocb. -b Differentiate (5.9) and eliminate A between the resulting equation and (5.9). Set x 0 and insert the above notation, which gives _ y{eilXb+ (oc) + l (OC) + e - ilXb +(-oc)} eilXb 't- (oc) + 2ik(2 n ) 1 /2 oc-1 sin ocb + e - ilXb 't- (-oc) . (5. 1 1 ) =

=

=

=

f.

=

=

Next consider x <; 0 , y ;? b . Apply a Fourier transform in y and eliminate the unknown value of cp on y b as in the analysis leading to ( 5.6) . We find =

+ (oc)

+

<1>+ ( - oc)

=

B exp (yx),

(x <; 0),

where <1>+ is defined in (5. lOa) . Differentiate with respect to x, set 0, and eliminate B to find

x

=

y{+(oc)· + +)-oc)} 't- (oc) + 't- (-oc). (5.12) Multiply (5.1 1 ) by. exp (-iocb) and add to (5.12) to eliminate + (oc) =

and obtain

2 <1>� (oc) - y(1 - r 2ilXb ) {+( _ oc) + i y-1 cot ocb 't- (-oc) } + O. (5. 1 3) + y e - ilXb l (OC) + (2 n ) -1/ 22ikoc-1 sin rx b r ilXb This is in form (5.2) with b(oc) = i y-1 cot ocb so that in this case b(oc) has branch points at oc ±k. =

=

1 83

S O ME APPRO X IMATE ME TH O D S

We might have expected on general grounds that the coefficient of

� (-oc) in (5.8) should have only poles whereas in (5.13) branch points should be present. In (5.8) � (-oc) was introduced by a boundary of finite length whereas in (5.13) the boundary introducing � (-oc) goes to infinity. Consider next a semi-infinite duct with a flange. Suppose that the duct lies in -b � 'Y � b, - 00 < x � 0, and the flange in X = 0, I'YI � b. Assume that a wave exp (-ikx) is incident from the right on the duct, from the half-space x � 0, - 00 < 'Y < 00 . Suppose that oCPt/ox 0 on the flange and o CP t/o'Y = 0 on the walls of the duct. Proceed as above, and find =

(0 � 'Y � b) .

Differentiate each o f these equations with respect t o 'Y , set 'Y = b , eliminate A, B and then + (oc) , to obtain, on using the result

'- (oc,b) = 0, 2 y-l (1 - e -2yb )-1 � (oc) + {+ (-oc) - _ (oc)} + + y-l� ( - OC) + (21T) -1/22iky-2 = O. (5.14) This is an equation of type (5.2) with b(oc) = y-l . Since � (- oc) was introduced by a semi-infinite boundary, b(oc) has branch points as we should expect. To conclude this section, consider a typical three-part problem. Suppose that V 2 cp + k2 cp = 0 in - 00 < 'Y < 00, x � 0, with

ocp/ox 1 , (x = 0, - b < y < b) , cP = 0, (x = 0, - 00 < 'Y < -b, b < 'Y < 00 ) . Apply a Fourier transform in 'Y in the usual way and find (oc) A (oc) exp (-yx) as in (5.9 ) . Differentiate with respect to x, eliminate A (oc), and set x = O. cP is symmetrical about 'Y = 0 and we can use =

=

the notation (5.lOa, b). Insertion of the boundary conditions on x 0 gives =

eilXb� (oc) + y l (OC) + e- ilXb � (- oc)

=

- (21T) -1/22oc-1 sin ocb.

(5. 15) This equation is of type (5.2) with b(oc) = o. We finally formulate the problem by means of transforms in the x-direction. The procedure leading to (5.6) for eliminating unknown functions on x = 0 gives

+ (�,'Y) - + (-oc,'Y) + (oc,'Y) + + (-oc,'Y)

= =

A e-Y1I , - ( 21T)-lj22 y-2 +

B

cosh yY,

(y � b),

(0 � y � b).

I S4

T H E W IE N E R-H O P F T E C H N I Q U E

Differentiate each of these equations with respect to 'Y, set 'Y = b, eliminate A, B, and then «P+(Ot,b). Write «P+(Ot) instead of «P+(Ot,b) , etc. We find

y-I ( 1 + coth yb)«P't- (Ot) - 2 «P+ (- Ot ) -

- y-I ( 1 - coth yb)«P� (- Ot) - (27T) -1/22 y- 2

=

o.

(5 . 1 6 )

This is of form (5.2) ; the function b(Ot) has branch points at ot = ±k. Of the five formulations in this section, namely (5.S), ( 5 . 13 ) , ( 5 . 14 ), ( 5 . 15 ), (5.16), the only cases that can b e solved satisfactorily at the moment by methods based on the Wiener-Hopf technique are (5.S) for small b and (5.15) for large b. The remainder of this chapter is devoted to the corresponding special cases of (5.2) . One of the objects of this section is to indicate that a practical method for solution of the general equation (5.2) should prove to be of consider able value. 5.3

General theory of a special equation

In this section we examine the following special case of (5.2) where no term involving the integral function «P I (ot) is present : valid in a strip T_ < T < T+ , Ot-plane. Set

- 00

<

(]

<

00 ,

K_(Ot) C (Ot)jB(Ot)

of the complex =

E+ (Ot) + E_ (Ot). (5. 1 S)

Introduce these in (5. 17) and rearrange : K+ (Ot)«P+ (Ot) + E+ (Ot) =

- K_(Ot)'Y_(Ot) - E_ (Ot) - b(Ot)K_ (Ot)«P+ ( -Ot).

(5. 1 9)

The last term is not regular in either an upper or a lower half-plane, in general. In many examples b(Ot) possesses only simple poles in the lower half-plane T < T+ and we deal with this special case first. b(Ot) can then be written

(5.20) where b. is the residue of b(Ot) at ot = fl. and the sum is over all the

185

SOME APPRO XIMATE METH O D S

poles in the lower half-plane ( n may b e infinite) . Introduce (5.20) into (5.19) and add extra terms to both sides :

6n oc. b P. K_(P.)W+( -P.) •

K+(oc.)W+(oc.) + E+(oc.) + =

-K_(oc.)'F_(oc.) - E_(oc.) - b_(oc.)K_(oc.) W+ ( -oc.)

- � b. { oc.

P.

-

}

(5.2 1 )

o.

(5.22a)

K_(oc.)W+ (-oc.) - K_(P.)W+(-P. ) .

The additional terms ensure that the left- and right-hand sides are regular in upper and lower half-planes respectively. Apply the Wiener-Hopf technique and assume that both sides are then identically zero, from conditions at infinity. Then

K+(oc.)W+(oc.) + E+(oc.) +

86n oc. b P. K_(P.)W+(-P.) •

=

This equation holds for all oc.. Set oc. -Pr , r 1 to n. A set of simultaneous linear algebraic equations is obtained for the unknown constants W+(-P.) x. ' say : =

=

=

K+(-Pr )xr + E+(-Pr )

n

- L Pr b •

=1

•

+

P•

K_(P.)x.

=

(r

0,

=

When the x. are determined the solution is complete. For reference, the analogous solution for the equation

A (oc.){W+(oc.) + a(oc.)'F_(- oc.)} + B(oc.)'F_ (oc.) m

a(oc.)

=

a+(oc.) +

+

O (oc.)

=

0,

a.

L -oc. - oc. ,

8=1

(5.22b)

(5.23a) (5.23b)

•

is given by the following equations in which Y.

1 to n) .

=

'F_(-oc..) : (5.23c)

where (5.23d) When these results are used to solve concrete problems it can be shown that the xr' Yr are related to physical quantities occurring in the problem (cf. ex. 5.2).

186

THE

b (x) in (5.17) .

W I E N E R- H O P F T E C H N I Q U E

Next consider general gives

The standard formula (5.3)

f b WK_ W
� + oo

U + oo

� - oo

U - oo

< c <

<

<

<

>

(5.24) into (5.19) and separate in the usual way, assuming that each half of the equation is identically zero. Then

ic + b ( -,) d , f WK,W
1 . K+( at)
=

0,

(5.25a)

f b WK,W
(5.25b)

ic - co

1 . K_(at)'Y_(at) + E_(at) - 2 7T�

id + 00

id - 00

The first equation is an integral equation for + (at) is found, 'Y_ (at) can be found from the second equation. If the only singularities of at) in the lower half-plane are simple poles the contour in (5.25a) can be completed in the lower half-plane and the integral can be evaluated by residues. With representation (5.20) for b(at) it is found that this gives exactly (5.22a) so that the integral equation method merely reproduces results obtained by the more direct procedure described at the beginning of this section. In the general case (5.25a) can be reduced to a Fredholm integral equation in the following way. For simplicity suppose that we can take c 0 and denote the value of , on the real axis by s . Suppose that at tends to a point on the real axis which we denote by t . Then the contour must be indented below, and on using ex. 1 .24, equation (5.25a) reduces to

b(

=

+ 1

-. 27T�

p

f b (S)K_(S)�+ ( -S) 00

- a)

s -

ds

= 0,

( - 00 <

t

< (0 ) ,

(5.26)

where P denotes that the integral must be understood as a Cauchy

1 87

S OME APPRO XIMATE METH O D S

principal value. The singularity in the kernel can be eliminated as follows. From Cauchy's theorem

(ex in the lower half-plane), (5.27)

- co

where we assume that «1>+ ( + ex) tends to zero uniformly as ( + ex) tends to infinity in an upper half-plane, so that the above integral can be converted into a contour integral by completing the contour in the lower half-plane. Let ex tend to a real number t and indent the contour above to find that (5.27) reduces to ( - 00

- co

<

t < (0 ) . (5.28)

Substitute for t«1>+( -t) in (5.26) and find K+ (t) «1>+ (t) + E+( t)

f {b(s)K_(s) - b(t)K_ (t) }",'1'+(-s) ds + -. co

1

27TZ

- co

S

-t

=

0,

( - 00 < t < (0 ),

(5.29)

where the kernel is now finite at s t so that the integral is no longer a Cauchy principal value. The result in this section from (5.24) onwards will not be used in the remainder of this chapter, but have been included to indicate one way in which it may be possible to solve more complicated cases. =

5.4

Diffraction by a thick semi-infinite strip

We consider a single example in detail to illustrate the theory at the beginning of the last section. Suppose that a wave CP i = exp ( -ikx cos 0 - iky sin 0) is incident on a plate lying in x � 0, -b � y � b, with o CP t/on = 0 on the plate. The results given below are due to D. S. Jones [4] who treats also the case CPt 0 on the plate. Our method of determining the basic equations is somewhat different from that of D. S. Jones. Set =

y � b, CPt = cP + e - ikx cos 8 - iky sin 8 + e - ikx cos 8 + ik(y - 2b) sin 8 , (5.30) x � 0; and y � b, -b � y � b, cP , so that cP ---+ 0 as 14 I Y I ---+ 00, and ocp/on = 0 on the plate. Define =

in the usual way

1 88

TH E WIEN ER-H O P F T E C H N I Q U E

+(IX,y) is regular in T > k 2 cos e , <1>_ in T < k 2' and the common strip is k 2 cos e < T < k 2 • From (5.30), � is continuous on y ±b, and <1>+ is continuous on y = - b . On y = + b, with the usual notation that cp(x, b + 0) denotes the value of cp(x,y) as y--+ b from values of y just greater than b, etc. (5.30) gives =

cp(x, b + 0) + 2e - ikx cos EJ - ilcb sin EJ cp(x, b - 0) , (x > 0) , 2i exp (-ikb sin 0) . +(IX, b - 0) . (5.3 1 ) I .e. + (x, b + 0) + 27T) 1 / 2 ( IX k cos 0 =

=

_

Apply a Fourier transform in x to the two-dimensional steady state wave equation. In -b < y < b, x ? 0,

where (cp)o denotes the value of cp on x = 0, -b < y < b. The partial differential equation, in X ? 0, -b < y < b, therefore transforms into

d 2<1>+(IX,y) /dy2 - y 2<1>+(IX,y)

=

-ilX(27T) -1 / 2 (cp) O .

(5.32)

Eliminate (cp)o by changing the sign of IX and adding the resulting equation to (5.32). This gives

d2 2 dy2 {+ (IX,y) + +( -IX,y)} - y {+ (IX,y) + +( - IX,y)} = 0, -b

<

y < b.

From this we deduce

A e - )'b + Beyb , - yA e - yb + yBeyb ,

+(IX, b - 0) + +(- IX, b - 0) � (IX,b) + � (-IX, b)

=

=

+ (IX,- l, ) + +(- IX,- b) = A eyb + Be -yb , � (IX,-b) + � ( -IX,-b) = -yA eyb + yBe -yb .

(5.33a) (5.33b) (5.34a ) (5.34b)

Introduce the notation (5.35 )

1 89

SOME APPROXIMATE METH O D S

and similarly for dashed quantities. Elimination of A , B between (5.33), ( 5.34) then gives

S � (oc) + S � ( - oc) = y coth y b { D+(oc) + D+ (-oc)}, (5.36a ) D � (oc) + D � ( - oc) = y tanh y b {S+ (oc) + S+(- oc) } . ( 5.3 6 b ) Application of transforms in y � b, y � - b , gives (y � b), (y �

(5.37a)

- b) .

(5.37b)

Differentiate with respect to y, eliminate C, D between the resulting equations and (5.37), and set y = b + 0, y -b in the appropriate equations. This gives, on remembering that '-- (oc,b) = '_ (oc, -b) =O, =

� (oc,b) = -y{+(oc, b + 0) + _(oc,b)}, � (oc, -b) = y{+ (oc, -b) + _(oc, -b)}.

(5.38a) (5.38b)

Add and subtract, use (5.31 ) , and introduce notation (5.35 ) . This gives

, -y { D+(oc) D_(oc) - (21T2i)1/2 expoc (-ikbk cossin00) } 2i exp (-ikb sin 0) } . D ,+ (oc) = -y {S+ (oc) + S_(oc ) - 1T 1/2 (2 ) oc cos 0 S+ (oc)

+

=

_

_

( 5.39a )

'

( 5.39b )

Equations ( 5.36 ) , ( 5.39 ) contain the solution of the problem. These split into two independent pairs of equations. Eliminate D+ (oc) between (5.36a), (5.39a) . This gives

S� (oc) + S� (-oc)

=

{

y coth y b D+ ( -OC) - D_(oc) - y-lS� (OC) +

}

+

2i exp (-ikb sin 0) . (5.40) (21T) 1/2 oc - k cos 0

Set

( 5.41 )

where K+ , K_ have been defined by ( 3.19a ) , ( 3.22 ) . Then ( 5 .40 ) can be rearranged as

S� (oc) (oc + k) 1 /2K+ (oc)

-;--..,.:.,.:,, .,...., <,=---:-:-

=

2i exp (-ikb sin 0) 1 /2K- (oc) . oc - k cos 0 (oc - k) ( 21T) 1 /2 + (oc - k) 1 /2K_(oc) { D+ ( - oc) - D_(oc)} --

+

- y-l tanh y b(oc - k) 1 /2K_(oc) S � ( -oc).

(5.42)

190

TH E WIEN ER-H O P F T E C H N I Q U E

This is in form (5.19) with K_(ex) in that equation replaced by (ex - k) 1 / 2K_(ex) and b(ex) = y-l tanh yb. Introduce notation (5.43) Yr = { (r1T/2b) 2 - k 2P /2 Y o = - ik. The poles and residues of y-l tanh y b in the lower half-plane are (cf. the notation in (5.20» :

(0 ) .

(r = 0 to

The solution o f (5.42) can now b e written down from (5.22), or it is easy to work out the solution from first principles. It is left to the reader to check that the conditions at the corners (O,±b), namely I cfo l ,....., const. , I grad cfol ,....., r-1 /3 , ensure that the separation polynomial is identically zero when the Wiener-Hopf technique is applied. We find 2i exp (-ikb sin 0)(k cos 0 - k) 1/2K_(k cos 0) S� (ex) ex - k cos 0 (ex + k) 1/2K+ (ex) (21T) 1 /2

� � ( -iY 2S+ 1 - k )1/2K_( ---:iY 2S+1 )S� ( iY 2S+ 1 ) = O.

+b 6

Y 2s+ 1 (ex + � Y 2S+ 1 )

(5.44)

We introduce simplifying notation which also makes the results comparable with D. S. Jones [4]. Set (5.45a) .1 2s+ 1 = 1T (2b) -1/2 (28 + 1 )( Y 28+ 1 - ik) -1/2 {K+ (iY 28+ 1 )}-1 , in El x 2s+ 1 = lnl /2b-l /2eikb s X e - i7T/4 (k cos 0 - k) -1 /2 {K_(k cos 0) }-1 (28 + 1 ) -lS+ (iY S+ ) ' 2 1

(5.45b)

Then (5.44) becomes

{

i S'+ ( ex ) - A( ex ) ex k cos 0 -:_

_

}

i1T � (28 + 1 ) 2X 2s+ 1 b 6 Y 28+ 1 (ex + iY 2s+ 1 ) .1 2 s+ 1 '

(5.46a)

where

A ( ex) = (2/1T ) 1 /2e -ikb sjn El ( k cos 0 - k) 1/2K_( k cos 0)(ex + k) 1/2K+ (ex) . (5.46b) An

infinite set of simultaneous linear algebraic equations for the

X 2s+ 1 are obtained by setting ex = i Y 2r+ l (r = 0, 1 , 2, . . . ) in this equation :

2b 1 1 x 1T Jl, 2r+ l 2r+ l - Y 2r+ l + "k cos 0 .

_

.

_

�

b

OO . (28 + 1 ) 2X S+ L Y 2 � 1 ( Y 2s+ 1 + Y 22� 11) .1 2S+ 1

. =0

..,

•

"

(5.47a)

191

S O M E APPR O X IMATE M E TH O D S

We shall later neglect small order terms. Set

Y 2s+ 1 = ( 2s + 1 ) ( n/2b){1 + O(k 2b 2 )} .

(5.47b)

Then (5.47a) becomes

(5.47c) Equations (5.36b) , (5.39b) can be solved in exactly the same way. Eliminate S+ (oc) and introduce a split defined in (3.19b), (3.23) :

(yb) -l e -yb sinh y b = L+ (oc)L_(oc). In this case we find b(oc) = y-l coth yb with poles and residues in the lower half-plane given by

b o = - (2kb) -1

f30 = -k = -iyo f3. = - iY 2.

The solution is

2i exp (-ikb sin 0)(k cos 0 - k)L (k cos 0) D�(oc) b(oc + k)L+ (oc) - (2 n ) I/ 2 oc - k cos 0 + L_(-k)D� (k) (i + k) L_ (- iY 2s )D � (iY 2S ) O. _ i � Y 2s (5.48) = b(oc + k) Y 2sb(oc + iY 2s ) 6 _

--

Again introduce simplifying notation

fl2s Y 2s

= =

(5.49a) i2 1 / 2nsb-l (iY 2S + k) -I {L+( iY 2s )}-I , 8 si _ i (4bs) -lnl / 2eikb n (k cos 0 - k) -I {L_( k cos 0)}-1 D� (iY 2 S ) ' (5.49b)

Then (5.48) becomes

D ,+ (oc) = B(oc)

where

{

i OC - k cos

t::\ �

2bQ D � (k) +n . OC + k co • 4 nS2'!l2s � , (5.50a) . b 8 � 1 Y 2s ( OC + � Y 2s )fl 2s

}

-

L

B(oc) = ( 2 /n ) I/2e - ikb sin 8 (k cos 0 - k)L_(k cos 0)b(oc

+

k)L+ (oc), (5.50b)

Q = ( n/2 ) 1 / 2 ( n/2b 2 )L_( _ k) eikb Sin 8 (k - k cos 0) -I {L_ (k cos 0)}-1 . (5.50c)

1 92

T H E W I E N E R- H O P F T E C H N I Q U E

An infinite set of simultaneous linear equations for the Y 28 is obtained by setting oc = iY 2r (r 0, 1 , 2 . . . ) in (5.50a) . Write =

(r � 1 ) ,

(5.51a)

and neglect small order terms. Then (5.50a) gives

_

-

_

1 i7Tk Y8 4k � 2 ' b · k - k cos 0 + 8L 8 = 1 f-l 2

(r � 1 ) ,

(5.51b)

( r = 0).

(5.51 c)

In order to find the field in, say, Y � b, we use (5.37a) . To determine 0 in this equation, since we have found S+ ( oc ) , D+(oc), we differentiate (5.37a) with respect to y and set y b. An inverse transform gives =

The far field can be found by the asymptotic expansion methods of §1 .6. The values of S+ , D+ are given by (5.46a), (5.50a) . If we set x = r cos 0, y - b = r sin 0, the saddle point occurs at oc = -k cos 0, but we cannot use the standard formula ( 1 .7 1 ) since the pole at oc = k cos 0 may be near the saddle point oc = -k cos ° (cf. ( 1 .72 ) ) . T o overcome this difficulty, apply the method associated with (1 .73) , and write (5.52) in the form ia -t' 00

: f [y-1{S+ (oc) + D+ (oc)} - !( oc)]e - Y(II -bl - iOCX doc

tfo = - 2(2 )1/2

ia - oo

-

ia + 00

1_

__

2 ( 27T) 1/2

f !( oc)e - 'Y(II - bl - irxx doc '

ia - oo

where

! ( oc)

I

e ik

sin

= (27T)2i1/2 · (k( occos- 0k) -/2( k)-/2 k-cosb 0) I

OC

0

The first integral has now no pole near the saddle point and can be

193

SOME APPROXIMATE METH O D S

evaluated asymptotically by ( 1 .7 1 ) . The second integral is given by ( 1 .62), ( 1 .65). As r --+ 00 , 0 < () < 7T - 0 ,

[

_ e lirr HS � ( - k cos () ) + D� ( -k cos ()) ) (k cos 0 - k)1 / 2 sin ()e -ikb sin 8 i (kr) - !e' kr + G(r,()) , + 1 2 (27T) / · (k cos () + k) 1 /2 ( COS () + cos 0 ) (5.53a) where G( r , ()) = _ 7T-l / 2e3irr/4e - ik b sin 8 [ _ e-ikr cos (0 - 8 ) F { (2kr)1 / 2 COS H() - 0)} + e-ikr cos (B + 8 ) F{ (2kr)I/ 2 cos t(() + 0)}] , (5.53b)

cP

'""

]

F (v) = Write

.

00

f exp (iu2 ) duo

v

(5.53c) (5.54)

where G(r,()) is proportional to the field due to diffraction by a semi. infinite plate of zero thickness, CP l is the correction term that should be added to G to give the far field due to diffraction by a semi. infinite duct, and CP 2 is the further contribution due to the flat end of the plate. CP 2 comes from the parts of S� , D� that do not involve ( ex. - k cos 0) -1 , i.e. (2 8 + 1 ) 2x 28+ 1 � . CP 2 - { _ 7T (2b) -1 e - i rr/4A ( _ k cos ()) L. ( k cos () ) 1L' 28+ 1 8 � O Y 2s+ 1 �Y 2 S+ 1 _

2:00

47T82y 28 (5.55) . ( k cos () ) ft 28 8 � 1 Y 2 . �Y 2s - ( bj7T ) e + i rr/4 B( -k cos () )QD� (k)(k - k cos ())-1 }(kr) -1 / 2eikr . We now neglect terms of small order in (kb). In addition to (5.47b), (5.51a) we need the results that for small ex., _

L ± ( ex.)

(2b) -le -irr/4B ( - k cos ())

=

1 + O(kb In kb)

K ± (ex.) = 1 + O(kb In kb) .

Then (5.46b), (5.50b, c ) , (5.51c) give, for small kb, A(-k cos ()) = -i(2j7T) 1 /2 2k sin to sin t(),

B(-k cos ())

Q D� (k)

= = =

_ (2j7T) 1 /2k2b(1 - cos 0)(1 - cos ()), ( 7Tj2) 1 /2 (7Tj2b 2 )(k - k cos 0 )-1 , -ikb(2j7T) 1 / 2 .

The notation has been chosen so that x2&+1 ' Y 28' A.2s+ 1 > ft 28 are purely numerical. Hence in (5.55) the first and last terms are of order (kb) :

1 94

T H E W I E N E R-H O P F T E C H N I Q U E

the middle term is of order (Pb 2 ) and will be ignored. In the series, (k cos 0) can be ignored compared with i Y2 s+1 ' Then (5.55) gives, for small (kb), CP 2 = {e - irr/4 (2/7T ) 1 /2kp sin to sin to + e - irr/4t(2/7T) 1 / 2kb}(kr) -1 / 2eikr , (5.56a) co

p = (4/ 7T )b L: X 2S+ 1 A2-; � 1'

where

(5.56b)

s�O

The series in (5.56b) is a purely numerical constant which we now proceed to determine. From (3.22a, b), (5.45a), it is easily shown that for small (kb)

Am = 2 m r ( tm + t)m1 / 2 exp {tm ( 1

- In

2m ) } ,

(5.57a)

: Aa = 4·22 : .1.5 = 5·51 : .1.7 = 6·55. (5.57b) The first four equations in (5.47a), assuming x 2s+ 1 = 0 for 8 > 3

i.e. Al = 2·33 are

2 ' 759x 1 + 0 ' 355x + 0 ' 302x5 + 0· 267x7 = 1 ·0000 a 0'2 14x 1 + 4·457 x + 0· 227x5 + 0·214x7 = 0 · 3333 a 0· 143x 1 + 0 · 1 7 8xa + 5·69 1x5 + 0 · 1 7 8x7 = 0· 2000 0 ' 107x1 + 0 ' 142xa + 0 ' 151x5 + 6 ' 7 0 3x 7 = 0 · 1 429.

We find an approximate solution of the infinite system by solving the first n equations in n unknowns, assuming x2r+ 1 = 0, r ;? n . Denote the resulting roots by X&�� l ' We find : xiI ) = 0 ·3624. X�2) = 0 · 0577. xi2) = 0 ' 3550 X�3 ) = 0 · 0566 : xhS) = 0·0246. xiS) = 0'3524 X�4) = 0·0560 : xh4) = 0·0242 : X�4 ) = 0·0140. xi4 ) = 0 ·35 1 2 Successive approximations alter the values of the unknowns by only a few per cent. The first unknown is much larger than the others. Since iJcp/ay ,...., r-1 /a at the corner of the plate we have 1 «l>� (oc,d) 1 ,...., l ocl -2/a as oc __ 00 in appropriate half-planes i.e. from . (5.45b), X2r+1 A (2 r + 1 ) -5 /a as f __ 00. A numerical check can be obtained from the above results : ,....,

X�4)/X�4) = 2 - 3 1 X�4)/X�4) = 1 ·73

(5/ 3) 5 / 3 = 2·34. (7 / 5) 5 / 3 = 1 ·75.

By using the asymptotic expansion of the gamma-function, (5.57a) gives Am � (2mn) 1 /2 for large m . This gives .1.6 � 5'60, ;"7 � 6·63

195

S O M E APPR O X IMATE M E TH O D S

which should b e compared with the exact values in (5.57b). We can now examine the value of the series in (5.56b ) . Set

The above numerical results give

8 1 = 0 . 1 55 5 : 8 2 = 0 . 166 1 83 = 0 . 169 1 : 84 = 0 . 1705 • These indicate that 84 gives an underestimate for 8 = 800- Using the previous approximations, another estimate for 8 is given by n-2 00 8� = L X�� lA2-;.l:-1 + X��)- lA;':'l L ( 2n - 1 ) 1 3/6 ( 28 + 1 ) -1 3/6 .

s �n - 1

S�O

The second series can be summed by means of the result

00 L ( 28 + 1 )-Z

S�O

=

,(z)(1 - 2-Z),

where ,(z) is the Riemann zeta-function. The numerical results quoted above give

0 . 1830 : 8� = 0 . 178 6 : 83 = 0 . 1770 : 8� = 0 . 1763 • These indicate that 8� gives an overestimate of 8 8 00 • Hence we 8i.

=

=

can say with some confidence, though the argument is not rigorous, that 0 · 171 < 8 < 0 · 176. D. S. Jones [4] has shown rigorously that 8 = 0· 175 ± 0 · 014. If we take 8 = 0 · 175 then (5.56b) gives p = 0·22b. (5.58) We can now provide a physical interpretation of CP 2 . Make the following substitution (equivalent to a change of origin) r = r' + q cos ()'

() = 0' (qfr') sin ()' in G (r,()) defined by (5.53b). For large r' and small (kb), ignoring small -

order terms,

(() ± e) = r' cos W ± e) + q cos e , (2kr) 1/2 cos i(() ± e) = ( 2kr') 1/2 cos iW ± e) + + iq(2k/r') 1 / 2 cos tW F(z + bz) = F(z) - (bz) exp (iz2 ). r cos

=f

e ),

To the order t o which we are working G(r, (} ) = eikq cos El G(r' , (} ' ) -

- ( 2 f7T) 1/ 2e - ;rr/4kq sin i()' sin ie(kr' )-1/2eikr' .

(5.59)

1 96

T H E W I E N E R-H O P F T E C H N I Q U E

The total field is G(r,fJ) + CP l + CP2 ' from (5.54). If we choose q = p = 0 · 22b, the last term in (5.59) will cancel the first term in (5.56a), for CP 2 ' and we find

e ikp cos 8 G(r',fJ') + CP l (r,fJ) + e - i ,,/4t(2/7T) 1 / 2 kb(kr') -1/ 2e ikr ' . CP l is a small term and to our order of approximation CP l ( r,fJ) CP l (r' ,fJ' ) . The first two terms then represent the field of a semi infinite duct extending from x = -00 to x = p 0·22b. The last cP

=

=

=

term is equivalent to the potential produced by a line source of strength -i(27T)-l kb, assuming a line source of unit strength produces a field 7TiH�l ) (kr'). This line source is due to the fact that there would be a wave of fundamental frequency propagated along the corresponding semi-infinite duct. For the condition CPt = 0 on the plate, for small thicknesses of plate, there will be no propagated wave and this term will not occur. These results are independent of o and therefore hold for any field constructed from a spectrum of . plane waves. The technique of this section has been applied to a semi-infinite solid cylindrical rod by D. S. Jones [5]. Propagation of waves in a two-dimensional duct with a step has been considered by W. E. Williams [2] (cf. exs. 5.1-5.4) . The corresponding problem for a cylindrical duct has been considered by V. M. Papadopoulos [2] . (The reader may prefer the method used in these references to the method used in this section. Cf. ex. 5.2.) From a numerical point o f view the success o f the method depends on whether it is practical to perform the factorizations and solve the simultaneous linear equations. One of the main reasons for discuss ing the numerical details in the above example is to show that in this particular case a suitable solution can be obtained by solving a comparatively small number of the simultaneous linear equations. Also the first unknown is much larger than the others and it is possible to obtain crude results by considering only the first equation and assuming that all the unknowns are zero except the first. It would appear that these remarks are often true in other cases. 5.5

General theory of another special equation

As explained in the next section by means of specific examples, the following equation occurs in connection with three-part mixed boundary value problems :

eirtq
(5.6 0 )

1 97

S O M E APPR O XIMATE M E TH O D S

where A is a constant and the equation holds in the strip -k 2 < T < k 2 • CP+(ot), cp_(ot), cp l (ot) are unknown functions.

cp+ (ot) cp- (ot)

=

=

1 _ ( 21T ) 1 / 2

_

1 __2 (21T) 1 /

f ..I.. (x)eiex(x-q) dx co

q

'f'

f ..I.. (x)eicx(X -P) dx p

- co

(5.6 1 )

'f'

CP+ (ot) is assumed regular in T > -k 2 , cp_(ot) in T < k 2 , and cp l (ot) is an integral function.

" 1 e -,exqcp 1 (ot) - -( 21T) 1/ 2

fq ,p(x)e' x -q dx "

ex(

P

)

"

f

0

1 ,p(u + q ) e'exu du (2 1T ) 1 / 2 p -q

--

'

so that the left-hand side is regular in a lower half-plane and has algebraic behaviour at infinity as ot ---+ 00 in a lower half-plane T < k 2 • Similarly exp (-iotp)cp l (ot) is regular in an upper half-plane and has algebraic behaviour as ot ---+ 00 in the upper half-plane T > - k 2 . It is assumed that K(ot) is regular in -k2 < T < k2, and has branch points at ot = ±k. Suppose that we can write Assume 0 < 0 < i1T so that k 2 cos 0 > O. It will then prove convenient to rearrange (5 . 60) so as to apply the Wiener-Hopf technique in a strip -k 2 < T < k2 cos 0. (If i1T < 0 < 1T then k2 cos 0 < 0 and we should apply the Wiener-Hopf technique in a strip k2 cos 0 < T < k 2 . In either case the object is to obtain relations holding in a symmetrical strip -k 2 l cos 0 1 < T < k 2 1 cos 0 1 . The reason for this will become clear below. ) Multiply (5.60) by exp (-iotq ) {K+(ot)}-l and rearrange in the form

{

I

}

A e -ik cos 0 q CP+(ot) 1 1 • 2 K+ (ot) (21T) / ot - k cos 0 K+ (ot) K+ ( k cos 0 ) + + U+ (ot) + V+ (ot) = - e - iexqK_(ot)cp l (ot) - U_(ot) - V_(ot) + e -ik cos El q A . (5.62) + 2 1 / ( 21T ) ' (ot - k cos 0)K+(k cos 0) --

1 98

T H E W I E N E R- H O P F T E C H N I Q U E

In this equation we have written

U+(Ot) + U_(Ot) = eia(P -ql (f>- (Ot) /K+ (Ot} , V+(Ot) + V_(Ot) = A(27T)-1/2eia(p -ql - 'ik cos 0P/ {(0t - k cos 0)K+ (Ot)}. These decompositions cannot be performed by inspection and it is necessary to use the general theorem B of §1 .3 (cf. (5.3) above) . In a similar way, multiply (5.60) by exp ( - iOtp ){K_(Ot)}-l and rearrange as

e - ik cos 0p A
(5.63)

where

R+ (Ot) + R_(Ot) = eia(q- Pl -k 2 . The other sides are regular in T < k 2 cos 0 . Assume that behaviours at infinity are such that Liouville's theorem can be applied in the usual way to prove that each side of each equation equals zero. We are interested in the left-hand sides of these equations. It will appear that it is convenient, for brevity, to introduce the notation (5.64a) -k 2 • 'Y_ is regular in T < k 2 cos 0 . On equating the left-hand sides of (5.62), (5.63) to zero, introducing explicit expressions for the U+ , V+ , R+ , S+ by using the general decomposition theorem (e.g. (5.3» and using notation (5.64), we obtain

+

ic - co e -ik cos 0 q A · 2 1 (27T) / (Ot - k cos 0)K+(k cos 0)

= 0,

S O ME APPROXIMATE

1 99

M E TH O D S

In these equations -k2 < d < k 2 cos 0 , -k 2 < c < k 2 cos 0. In the first equation T > c ; in the second T < d. From the assumption ° < 0 < in we can choose a so that -k 2 cos 0. < a < k 2 cos 0 and take d - c = a. In the first equation above replace , by (- , ) . I n the second equation replace IX. b y ( - IX. ) . This gives, remembering that K+ ( - IX.) = K_( + IX.), =

ia - co e - ik cos 8 q A + n ) 1/2 ' (IX. - k cos 0)K+(k cos 0) (2

where now T >

-a

=

0,

in both equations. Define

Dt ( lX. )

= 'Y t ( lX. )

- 'Y_ ( IX.), -

(5.65) where in this case the star denotes that expressions are regular in (1m IX.) > -k 2 cos 0 except for simple poles at IX. = k cos 0 . Then adding and subtracting the above equations gives

Dt ( lX. ) K+ (IX.)

1

+ 2ni

f

ia+ co

ei � (q -p) Dt ( O d, + ' ( + IX.)K_W a C() i e - ik cos 8 q A + (2n) 1 / 2 (IX. - k cos 0)K (k cos 0) - °. + --

.

(5 .66b)

Each of these equations is of the same type and we obtain an approximate solution by a method due to D. S. Jones [2] . Consider first the asymptotic evaluation of integrals of the form 1

f � f3 HWei,z d"

ia+ co = ' a C() i -

(5.67a)

200

THE

W I E N E R- H O P F T E C H N I Q U E a

< k 2 • Suppose that we can write H m = ( ' - k) T+ Ym = ( ' - k) TH{f(p) + ( ' - p) f '( p } + . . . },

where - (1m fJ ) <

(5.67b) (5.67c)

where r is one of the numbers - 1 , 0, + 1 , + 2 , . . . , and f( '} can be expanded as a Taylor series about an arbitrary number p which will be chosen later. f(p} is non-zero and finite. Since - (1m fJ) < a, the poje at , = -fJ lies below the contour. On the other hand fJ can lie close to , = - k and this is the reason why the factor ( ' + fJ} -1 is kept separate from H ( ,) : otherwise it would be sufficient to expand ( ' + fJ} -lH m in the form (5.67c). In (5.67a) it is assumed that Hm possesses no branch points other than , k in (1m ') � a and that 1 is a large positive number so that the contour can be deformed into the upper '-plane. Cut the '-plane from k = k l + ik 2 to k l + ioo by a straight line parallel to the imaginary axis and deform the contour on to the two sides of the cut. When the contour is deformed in this way the contributions from any poles crossed in the process must be included : these are ignored here. Deform the contour and set , - k iu. On the right-hand side of the cut ( ' - k} 1/2 = U1 /2 exp (in/4) and u goes from 0 to 00 . On the left-hand side ( ' _k} 1 /2 = U1/2 exp ( - 3in/4) and u goes from + 00 to o. Then (5.67a, c) give =

=

1 ,.....,

2eirr/4eikli r [f(p} Wr (z} + + f '(p) {il-1 Wr+ l (z} + (k - p) Wr (z} } + . . . ] .

(5.68a)

In this equation Z = -i(k + fJ}l and we have used the notation (5.68b) where Wk ,m is a Whittaker function (cf. ex. 5.9). The expansion used by D. S. Jones corresponds to the choice p = k in (5.68a) . Then (5.69) This is the expansion used in the remainder of this section. An improved expansion is obtained by choosing p in (5.68a) so that the coefficient of f '(p} vanishes , i.e. Z

=

-i(k + fJ}l.

This gives an expansion similar to (5.69) except that f(k} is replaced by f(p}. (There is some scope for investigation of improvements on (5.69) . This asymptotic formula is the factor which limits the

201

S O M E APPRO XIMATE M E TH O D S

accuracy o f the method o f this section. An alternative approach would be to write, in (5.67a) , 1

,+

(3

HW

=

( ' - k)"H ( ' - k)"H f( - (3 ) + { lW - f ( - (3 )}, , + (3 , + (3

where we have used the notation (5.67b). The second term on the right is regular at , + (3 = 0 and the first gives an integral which can be expressed in terms of the Wr function. It may also be prac tical to include several terms of an asymptotic expansion.) Now return to the problem of finding an asymptotic solution of

A f

ia+ co Ft (ex) Ft (ex) ei , ( q -p) d, + 27Ti K+ (ex) ( ' + ex)K_ W ia - 00 A e - ik cos 0 q , (5.70) . = (27T) 1/2 (ex - k cos 0)K+ (k cos 0) where Ft will be set equal to s t , Dt in turn, so that from (5 .64), (5.65), (5.66), Ft has the form e - ik cos 0 q + AA e - ik cos 0p A Ft (ex) = F+(ex) . 1 2 1/2 (ex + k cos 0) . (27T) / ex k cos 0 (27T) _

(5.7 1 ) A is a constant which later we shall take as - l or + 1 . F+(ex) is regular in T > - k 2 • We should expect that F+ (ex) will have a branch point at ex -k but for large 1 this is sufficiently far from the point ex + k to enable us to write in the above asymptotic expansion: (5.72a) if {K_W}-l '""" hr ( , - k) 'H as ' � k, r as k � then H k) ' F+W /K_ W '""" F+(k)hr ( ' . Define (5.72b) Er 2eirr/4eikl1 - r - ii rhr · 1, (5.67a) , (5.69) give, on writing ( q - p ) =

=

=

=

say;

ia - co

(5.72c)

202

T H E W I E N E R- H O P F T E C H N I Q U E

where

Er [Wr{ -i(k ± k cos 0)l} - Wr {-i(k + lX)l} ] , (5.73) 27Ti(1X =j= k cos 0) and upper and lower signs refer to R I > R 2 respectively. The last R 1. 2 ( IX )

=

two results are obtained by splitting the integrand into partial fractions. On introducing (5.7 1 ) into (5.70) and using the above expressions for the resulting integrals we find :

F+(IX)jK+(IX) = (27T) -1/ 2A e - ik coS 0 Q{ P 1 (1X) + AR 2 (1X)} -(27T) -1/ 2AA e - ik cos 0p { P 2 (1X) + AR 1 (1X)} - AT(IX)F+(k),

(5.74)

where, in addition to our previous definitions, (5.75) Upper and lower signs refer to P 1 ' P 2 respectively. The value of F+(k) can be found by setting IX = k in (5.74). When this value of F+(k) is substituted back into (5.74) we find, on remembering that A2 = 1 ,

F+(IX) K+(IX)

where

=

A (1X)} (1X) (27T) 1/2 {G 1 - AG 2 AA K+(k) T(IX) - 27T) 1/2 · 1 {G (k) - AG 2 (k)}, + AT(k)K+(k) 1 (

(5.76a) (5.76b)

G 2 (1X) = e - ik cos 0P P 2 (1X) - e - ik cos 0 Q R 2 (1X).

(5.76c)

Next return to (5.66). From definitions (5.64), (5.65), for S,!!- (IX) we take A = - 1 , F+(IX) = «l\(IX) +
(27T)-1/2AG 1 (IX)K+(IX) + A K+(k) T(IX)K+(IX) (k) + T(k)K+(k)G 1 (k)}. (5.77) + (27T) 1/2 { I --;- T2(k)K� (k) } {G 2 If in this equation G 1 is replaced by G 2 , and G 2 by G1 ' and the sign of IX is changed, the resulting expression gives
=

.

In specific problems it will be possible to simplify these formulae to some extent by neglecting some terms completely and using the asymptotic forms of Wr in other terms. In any case the Wr which are of importance in applications can be expressed in terms of tabulated functions (ex. 5.9). An application of this analysis is given in the next section.

2 03

S O M E APPR O X I MATE M E TH O D S

5.6

Diffraction by strips and slits of finite width

Consider diffraction of the incident wave rP i = exp (-ikx cos 0 iky sin 0) in the following four cases : (i) By the strip p < x < q, y = 0, with o rP tfCJ y = 0 on the strip, (ii) By the strip p < x < q, y = 0, with rP t = 0 on the strip, (iii) By the slit p < x < q, y = 0 with orPtloy = 0 on y = 0, off the slit, (iv) By the slit p < x < q, y slit.

In cases (i) and (ii) write rP t

rP t = e - ikx cos 0 - iky sin 0 rP t = rP ,

±

=

0 with rP t = 0 on y

=

0, off the

rP i + rP. In cases (iii) and (iv) write e - ikx cos 0 +iky sin 0 + rP , (y � O) , (y < 0 ) ,

=

with the upper sign for case (iii) , the lower for case (iv). The object of these substitutions is to ensure that transforms are regular in -k 2 < T < k 2 ' since rP tends to zero as exp ( -k 2 I xl ) as Ixl -- 00 in all cases. Apply a Fourier transform in x to the· partial differential equation for rP and find in the usual way (y < O ) . B exp (yy), (y � 0 ) ; Differentiate with respect to y , eliminate A and B and let y tend to

=

A exp ( -yY),

plus and minus zero respectively. This gives

y

=

J.

where

G(Ot) = (2 n ) -1/2 (Ot - k cos 0) -I {ei( IX - k cos 0 )q

_

ei(IX - k cos 0 )p} .

204

THE WIEN ER-H OPF T E CH N I Q U E

Hence the addition of (5.78), setting

=

2'1,\(0) , (5.79)

=

This is of form (5.60) . Similarly in case (ii) � is continuous on '!J 0 and a�/a'!J is con tinuous on '!J 0 except for p � x � q. Also =

=

=

=

=

=

=

=

=

=

=

In case (iv), � is continuous on y

=

=

0 and
=

=

0,

=

=

2 k sin GG(Ot) .

Subtract (5.78) , �ite
eic
=

=

2'¥� (0) etc. to find

-k sin GG( Ot) .

(5.82)

The equations for all four cases are of form (5.60) . The similarity of (5.79), (5.82) corresponding to cases (i) , (iv), and of (5.80) , (5.81 ) corresponding to cases (ii), (iii) , is obviously connected with Babinet's principle. An approximate solution in each of the above cases can be obtained by using the theory in §5.5. Consider case (iv). Comparing the notation in (5.60) and (5.82) we have

=

Hence

K_(Ot) From (5.72a), From (5.72b),

r

=

-1

'¥'-- (O), =

-k sin G .

e - i TT/4 (k - Ot) I/2 = e irr/4 (Ot k) I /2 . h_ l = e - iTT/4 •

=

_

SOME APPRO XIMATE

M E TH O D S

205

Consider the field in y � O. Using the notation defined in connex Ion with (5.82) , the transform of the potential in y � 0 is given by

(J) (y) = B exp yy = -y -l {e iaq 'P' 't (O) + 0/;(0) + eiap 'P''- ( O )}e l'v . ia + 00 1>( x , y) = y-l {eiaQ 'P' 't (O) + 'P'i(O) + ia - 00 + eiap 'P' '- ( O )} eYV - iax d('J. . In these equations 'P'i(O) = - A G ( ('J. ). 'P''t (0), 'P" _ (O) can be com puted in a straightforward way from (5.77) , using definitions (5.72) - (5.76) . For convenience we assume that the slit lies in - d � x � d so that q = d, p = -d. -

(2:)1/2 f

When the various approximations are substituted into the integral it is found that the potential can be split into two parts :

1>(tot.)

=

1>(sep.) + 1>(int.).

1>(sep.) is the sum of the diffracted waves produced by each half plane separately i.e. acting as though the other half-plane were absent. This is a natural first approximation. ia+ co ei(a - k cos 8 )d -ik sin 0 r 'a. x -y[v [ d('J. 1> (sep ) 1 . - 2n (k + k cos 0) / 2 (('J. - k cos 0)(k - ('J. ) 1 /2 ia - 00 ia + co e - i(a - k cos 8 )d ik sin 0 . + 2n(k - k cos 0) 1 / 2 ( ('J. - k cos 0)(k ('J. 1 / 2 e - ,ax - y [v [ d('J. . + ) ia - 00 The exact value of 1> (sep.) can be found in terms of Fresnel integrals by using ( 1 .62), (1 .65) and the corresponding expressions obtained by changing the signs of ('J., x and replacing 0 by ( n - 0). 1>(int.) is an interaction term. Using the notation of ( 5.72) - ( 5.76)

f

f

it is given by

A 1>(int.) = 2n

-

id - iad { ikd cos 8 R ( ('J. ) e a + e - ikd cos 8 R (- ('J. ) e e f 1 K_( ('J. ) 2 K+( ('J. ) ia e - iad } . x eiad e -'a -y[v[ d('J. , T(-('J.) T(('J.)

ia+ co

- 01

where

--

--

co

--

K_( ('J. )

-0

2

-

--

K+( ('J. )

K+(k) T k)K )G (k) , T2 (k)K't ( k) {G2 (k) + ( +(k 1 } and O2 is obtained by interchanging G and G 2 . On examining the 1 forms of R 1 (('J.), R 2 ( ('J.), T( ('J.) it is found that the far-field due to 0 = 1 1

_

206 THE W I E N E R- H O P F T E C H N I Q U E cp(int.) can b e found by a straightforward asymptotic expansion as in §1 .6, ( 1 .71 ) . Define (r,e) by x = r cos e, Iyl r sin e, so that e =

is measured from the positive x-axis in a clockwise direction. (0 is positive measured from the positive x-axis in an anti-clockwise direction. ) From (1 . 71 ), as r --)- 00,

cp(int.)

r-..J

-k2 (2 n ) -1/ 2 sin 0 sin ee - i 1r/4 (kr) -1 / 2eikr X [{K+ (k cos e )}-l {eikd(Cos El - cos .B) R ( _ k cos e) 1 - 0 le - ikd Cos B T(-k cos e) } + {K_(k cos e) }-I {e - ikd(cos El - cos B) R 2 (k cos e) - 0 2eikd Cos B T(k cos e) }] .

When explicit expressions are substituted in this equation it is found that the W-1 functions which occur can be replaced by their asymptotic expansions if (kd) is large, provided that e, 0 are not nearly equal to 0 or n :

kd --)- 00 , W_ l {-2i(k ± k cos c5)d} nl/ 2 {- 2i(k ± k cos c5)d }-I , if c5 is not nearly equal t o 0 or n . It is then found that considerable r-..J

simplifications occur, and after some manipulation the interaction field as r --)- 00 turns out to be

1 i (2 n ) 1/ 2 · 1 + (4nikd )-le4ikd · (4nkd) 1 / 2 . (kr)1 / 2 X [ { ( sin te cos t o ) -l e - ikd( cos B - cos El ) + (cos te s in t o ) - l eikd(Cos B - cos El ) } ei1r/4 (4nkd) -1/ 2e2ikd{( sin te sin to )-le - ikd(Cos H cos El) + + (cos te cos t o )-leikd(CoS H cos El ) }]. _

This is identical with the result found by S . N. Karp and A. Russek [1] who used a completely different argument. (Note that they used a time-factor exp ( + iwt) so that in comparisons the sign of i must be changed. Also their cp, CPo are related to our e, 0 by e + cp = t n , o - CP o = tn .) The argument used by S . N. Karp and A. Russek is essentially physical. It is assumed that the field diffracted by, say, the left-hand half-plane can be regarded as produced by a line source at the edge of the left half-plane when considering the inter action effect caused by diffraction of this field by the right-hand half-plane. There is a discussion of this physical argument, and of numerical results, in the paper of S. N. Karp and A. Russek. The present method and the Karp- Russek method are both related to the work of K. Schwartz child and E. N. Fox. References and some discussion can be found in B. Baker and E. T. Copson [1] , and S. N. Karp and A. Russek [1 ].

SOME APPROXIMATE

207

M E TH O D S

The exposition we have given may have obscured the laborious nature of the present method, whereas the Karp-Russek method is relatively elementary and direct. On the other hand the Wiener Hopf method can be extended to many other problems in a straight forward way. In would be interesting to know whether the Karp Russek method could be extended to deal with more complicated problems, assuming the results for corresponding semi-infinite obstacles. Three papers apply the analysis in the last two sections to more complicated problems. D. S. Jones [2] and W. E. Williams [1] examine diffraction by a waveguide of finite length with CPt = 0, oCPt/on = 0, respectively, on the walls of the guide. When the separation of the plates tends to zero these reduce to the strip problems considered in this section. W. E . Williams [3] examines diffraction by a circular waveguide of finite length. Various complicating features can appear in these problems. Resonance can occur within the guide : in this case the factor {I - T2 (k)K; (k) } in (5.77) is small. Poles may occur at 0( = k or near 0( = k in addition to the poles at 0( = ±k cos 0 : the expansions of K+( O(), K_(O() may have expansions containing significant terms of the form (0( - k) S In (0( - k) : if either of these complications occur it will be necessary to modify the general formula (5.77 ) . It may b e possible to improve the treatment o f the present section by taking further terms in the asymptotic expansion or by using a different type of asymptotic expansion. However one basic feature underlying the whole of this chapter is that if the singularities of certain critical functions are simple poles a solution can be given which is satisfactory for small wave numbers : if branch points occur an asymptotic expansion method will give a solution suitable for large wave numbers. If it is desired to find two approximate solutions, one for large and one for small wave numbers, it is at present necessary to formulate the problem in two different ways by using the possibilities outlined in §5.2 or by using some completely different method (cf. ex. 5.12 for the strip problem) . Most of the topics considered in this chapter offer considerable scope for further investigation. Miscellaneous Examples and Results V 5 . 1 Suppose that a wave d > o. Set
-

=

208

T H E W I E N E R- H O P F T E C H N I Q U E

in this type of problem the basic equations can be obtained by other methods, and the problem can be formulated in terms of a different, complementary, set of equations. W. E. Williams [2] has obtained (e), (g) below by using a Laplace transform and a variant of the method of ex. 5.2. He gives a discussion of the physical interpretation of the results together with numerical values of the reflection coefficient for various (kDjTr) , (kdjTr) including the case where more than one wave propagates in the duct. Apply Fourier transforms in the z-direction. In 0 .;;; y .;;; d, z .;;; 0, the partial differential equation becomes d2 (fUr:x ,y)jdy2 - y2 (fUr:x,y)

=

(2Tr)-1/2ik + (2Tr)-1/2ir:x (cp) 0 '

(a)

Eliminate (cp) o as in §5.4 to obtain cI Ur:x) +
=

- (2Tr)-1/2 2iky-2 + y-l coth yd{cI> '-- ( r:x )+ cI>'-- ( - r:x)} .

(b)

In d .;;; y .;;; D cI>+ (r:x) + cI>_(r:x)

_ y-l coth y(D - d)cI>'-- ( r:x ) .

(c)

Eliminate cI>_( r:x ) and set _y-l sinh yD{sinh yd sinh y(D - d)}-l = M+ ( r:x)ilC ( r:x ) . Separation of the resulting equations as in §5.4 gives , M_ (r:x)cI> _ (r:x)

=

-

1 1 cI>'-- ( - k) + + ' ' f2 1 _ (k) 2kd(r:x 2Tr) ( k) k) ( r:x M+ }d+ (k) 00

+

where Introduce and set r:x

=

Yn xn

(d)

, cI> _ ( - iYn ) ' L... iYnd( r:x - iYn ) M+ ( iYn ) n�l

""

=

=

1

,

(e)

{(nTrjd) 2 - k2 }1/2 , Y o = - ik. 1 - i (2Tr) /2 cI>'-- ( - iYn )M+ (k) ,

(f)

-iY m in (e) . Then

(m

=

0, 1 , 2 . . . ) .

(g)

If only one mode is propagated in the guide these can be reduced to a set of real equations. Introduce J+ ( r:x ) = (k + r:x )M+ ( r:x ) where J+( iYn ) is now real, multiply by (k + iY m ) ' and separate real and imaginary parts. It will be found that the imaginary part is a series which is independent of m. The first equation of the set (i.e. m = 0) can be used to eliminate this imaginary part in the remaining equations. By solving the resulting (real) equations for m = 1, 2, 3, . . . we can find x ' x 2 ' in terms of xo ' Substitution in the first equation then givesl Xo (cf. W. E. Williams [2] ) . •

•

•

S O M E APPR O X IMA T E M E TH O D S

209

5 . 2 We give an alternative method for obtaining the results of ex. 5. 1 which illustrates the physical significance of the cI>'_ ( -iYn ) ' Proceed as in ex. 5. 1 up to equation (a) . Suppose that (
(
=

00

�L

n�O

e

�in cos (n1Tyjd)

d

in

=

f (
(h )

o

Introduce into (a) and solve the resulting equation with cI>� (0l,0) = O. Differentiation with respect to y and elimination of an unknown function of Ol then gives, on y = d,

This replaces (b) . Eliminate cI>_ (0l) between (i) and (c) and separate in the usual way to find

(j ) To complete the solution we need a relationship between cI> '_ (Ol) and in which can be found in various ways, e.g. (i) The left-hand side of (i) is regular in a lower half-plane. Hence the residues at Ol = - k, -iYn ' on the right-hand side must cancel, i.e. in = (21T) 1/2 y,;;- l cos n1TcI>� ( - iYn ) , (n > 0) : 10 = (21T) 1/2ik-1 cI>� ( - k) + d. (k) (ii) The solution of (a) such that cI>� (0l,0) = 0 is cI>_ (Ol,y)

=

_ (21T)-1/2iky-2 + ( 2 1T)-1/2y--1iOl

y

f (
o

+

A cosh yy.

(1)

Differentiate with respect to y, set y = d, and substitute the resulting value of A into (1) . The right-hand side of the equation for cI>_ which is obtained in this way must be regular in a lower half-plane, and cancellation of residues at Ol = - le, - iYn again gives (k) . (iii) Apply a finite cosine transform in 0 .;;; y .;;; d, z .;;; 0 and solve the resulting ordinary differential equation, assuming 0/ oy = 0 on y = 0, and oj oy known on y = d. This gives, for n » 1, with oj oz = 0 on z = 0,

210

THE WIE N ER-H O P F T E C H N I Q U E

Set z 0 and this gives (k) for n ;;;. 1. A similar procedure gives the case n = O. If now we set ex = - iYm in (j ) and use (k), equations are obtained for the in ' These are equivalent to (g) . The advantage of the present method is that the
-

5.3 In ex. 5.2 (rp) o was expanded in a series such that orp/ oy = 0 on y = 0, d. It is logical to try an alternative expansion such that oN oy = 0 on y 0, rp = 0 on y = d i.e. =

(rp) o

=

00

J (rp)o d

� 2 >n cos {(n + !)rry/d}

gn

n�O

=

o

cos {(n + !)rry/d} dy. (m)

I n place o f (i) we find

Cancellation of residues at ex (

-

=

-

iYn + t gives (cf. (k) )

I ) ngn = (27T ) 1/2Y;-f t (n + ! ) {7r/d)

-

iYn H ) - {(n + !) {7r/d) Yn H}-lik. (0 )

Eliminate
N_(ex)
=

-

27T � ik - 2 i 1/ · L. ( ex ) G ( - I )n (n + ! )gnH0) (ex), 2 1/2 2 ( 7T) ( 7T) d2 _

n �O

H�) (ex) + H0) (ex) G+ (ex) + G_ (ex) N+ (ex)/N_(ex)

=

-

�)

ex ( ex2 + Y; H )-l {y tanh y(D - d)}-lN+ ( ex ) ,

_y-2 tanh yd{tanh y(D - d)}-lN+ ( ex ) , - sinh yD{cosh yd sinh y(D - d)}-l .

On setting ex = - iY';'+ t in (p) and using ( 0 ) we obtain a set of equations for the gn. Of course from a practical point of view this formulation is much more complicated than the previous one because of the splits required for H('!!.) , G_ . The main point we wish to make, however, is that theoretically two complementary formulations are possible. 5 .4 An approach which gives the two formulations in a direct and symmetrical way is the following. Proceed as in ex. 5 .1 up to (b) . Deal with this equation in one of two ways.

SOME APPR O XIMATE

METHOD S

21 1

(i) Rewrite (b) as

cI>_(()() - A_(()() + (27T)-1/2i(()( - k)-1 = - cI>_ ( ()() + A+ (()() + (27T)-1/2* + k)-l , where -

(q)

Write where

Then Equation (q) can be separated, and if this value of A _ ( ()() is inserted the resulting equation is exactly (i) . (ii) Rewrite (b ) as (r )

where

B+ ( ()() + B_(()() Write

= y tanh yd { cI>_( ()() + cI>_( - ()() + (27T)-1/22iky-�} .

b+ ( ()() + b_ (()(), say. B_( oc ) = b_(oc) + b+ ( - oc) .

y tanh yd{ cI>_( ()() + (27T)-1/2iky-2} =

Then

Equation (r) can be separated and if this value of B_(oc) is inserted the resulting equation is exactly (n) . 5 . 5 A rigid-walled duct occupying 0 .;; y .;; D, - w < z < w is filled with a medium ( 1 ) for which V24>f + k24>i = 0 except for the space 0 .;; y .;; d, z :> 0 which is filled with a medium ( 2 ) for which V24>� + K24>� = O. Suppose that at the interface between the two media 4>i = 4>� and c o4>i/ on = o4>M on. Assume a wave 4>i = exp ( i kz ) incident from z = - w and write 4>i = 4>i + 4>1 ' 4>� = 4>i + 4> 2 ' so that 4> 2 = 4> 1 ' 04> 2/ on = c o4>1 / on + (c - l )ik on z = 0, 0 .;; y .;; d. Expand 4>1 and 04>1/ oZ on z = 0, 0 <: y .;; d as cosine series (cf. ex. 5.2) :

2�

(4)1 ) 0 = d L e nI in cos (n"Y /d) n-O

( 04)1 / oZ) 0 =

00

� L e:,.gn cos (n7Ty/d ) . n-O

I t will b e found that the procedure o f ex. 5. 2 gives two sets o f equations corresponding to (i), (k), each involving both in and gn' Linear com binations of in and gn are related to cI>� _ ( - iYn) and cI>� + (i rn) where r; = {(n7T/d) 2 - K2}. E quation (c ) is still true and the equations can be manipulated to give two relations corresponding to (j ) . From these simultaneous linear algebraic equations for the in' gn can be deduced. But the splits required for the solution seem complicated (cf. ex. 5 . 3 ) .

212

T H E W I E N E R-H O P F T E C H N I Q U E

5.6 A plane wave exp (ikz) is incident in a duct 0 .;;; y .;;; 2b, - 00 < z < 00 on a plate of finite thickness lying in c .;;; y .;;; d, o .;;; z < 00, where 0 < c < d < 2b. iJ
Two generalizations of the problem of the strip across a duct

(§3 .6) give rise to equations of type (5. 1 7 ) , namely (i) the strip of

arbitrary width, (ii) the strip of fmite resistivity. For simplicity instead of the problem in §3.6 consider the following. Find a solution of the steady-state wave equation in 0 .;;; y .;;; 2b, 0 .;;; z < 00 with incident wave
( 2) 1/2 f iJ2

Then

:;;.

o

()

2 1 /2 ik - or.2 'Y (or.) . cos or.x dx = iJx2 :;;.

S O M E APPR O X IMATE

213

METHO D S

A cosine transform reduces the partial differential equation to an ordinary differential equation for 'Y(()(), the solution of which gives (5.6) directly. In the notation of (5.68b) it can be shown that ru + 1 )Z-1 as z 00 , l arg zl < 7T, 1 W 1 ( - iy ) = 27T /2e-iYy-l/2 F( y'y ) , W o ( - iy) = 7T1/2 + 27Tl/2ie-iYyl/2F( y'y ) ,

5.9

Wj - t(z)

�

where

--+

F(v)

00

=

I e u2 duo i

v

5.10 It will be possible to extend the methods of this chapter to diffraction of waves by a two· dimensional obstacle of finite length and finite width, for large length and small width (cf. the discussion in D. S. Jones [4], p. 169). A similar problem is that of a plane wave incident from any angle on a solid cylindrical rod of large length and small diameter. This is a possible method of approach to antenna problems. 5 . 1 1 Consider a plane electromagnetic wave in infinite space, falling on a semi·infinite slab of dielectric material lying in - 00 < x .;;; 0, - b .;;; y .;;; b (cf. §5.4 and ex. 5 . 5 ) .

5 .12 Various problems involving diffraction o f waves b y disks and strips can be reduced to the solution of infinite sets of simultaneous linear algebraic equations by using the Kontorovich-Lebedev transform (see A. Erdelyi et al. [ 1 ] , Vol. II ) : 00

'Y(p,) =

J

J

c + i oo

dr IjJ(r)Kp (Ar) -:;:

IjJ(r)

o

=

�

7Tt

p,'Y(p,)I p (Ar) dp, .

(a)

c - i co

For simplicity the analysis is carried out on the assumption that A has a positive real part, but we set A = real, at the end of the work, corresponding to a time factor exp ( - i w t ) . The analysis and results for disks are based on a paper by A. Leitner and C. P. Wells [ 1 ] ; they use the Vajnshtejn-Karp-Clemmow procedure. The method used here is the same as that in §5.4. We give the analysis in some detail since this would seem to be the first case where a transform other than the related Fourier-Laplace-Mellin trio has been used to give significant results by a Wiener-Hopf method . I t i s necessary t o consider the properties o f I p (Ar), K p (Ar) a s functions of p, for fixed (Ar). Asymptotically (b) I p (Ar) I p, /Arl ?> 1. { r( 1 + p,) }-I (lAr)p{ 1 + O (p,-I ) }. I p (Ar) decreases rapidly as Re p, --+ + 00 but increases as Re p, --+ - 00 .

-ik, k

�

Kp (Ar )

=

i7T

1 _ p (Ar) - I (Ar) P . SIn 7Tp,

�

i(iAr) Hr( ±p,),

Re p,

--+

± 00.

(c)

214

T H E W I E N E R- H O P F T E C H N I Q U E

Hence K",(Ar) increases rapidly as I Re ,u l --- 00 in both right and left half-planes_ On the imaginary axis, ,u iT, =

T ---

± 00 _

From these results it is seen that if we write (d) then

ID p (,u )

a

�

I f(r) (r/a)", - l dr,

,u --- 00 ,

Re ,u

o

> 0,

(e )

assuming that f(r) const_ as r ___ 0 so that the integral converges for Re ,u > 0_ ID p (,u ) is regular and bounded in the right-hand half-plane Re ,u > O. In this notation �

a

ff(r)K",(Ar) �

= WVa) - "' r(,u )IDp( - ,u ) + (i).a )"'r( - ,u )IDp(,u)} .

(f)

o

Next consider the integral

00

I

=

f g(r)K",(Ar) --:;:dr

.

a

(g)

It is assumed that g(r) is such that I is convergent at both upper and lower limits for all ,u. Then I represents an integral function of ,u and from (b), (c) we can write 00

00

(i).a ) - '"

r(

�,u) f g(r)K",(Ar) � = 'YN (,u), a

(h)

where 'Y p, 'YN are regular and tend to zero in right and left half-planes . respectively. For problems involving circular disks use spherical co-ordinates (r,B,,/» . Assume axial symmetry so that the steady-state wave equation is

1

0 .

sin B oB Set r1/2

=

sm B

o ° ° + + k2r2 = O. r2 oB or or

1p. The equation for 1p is 1 0 . 01p ° 01p - sm - r - --

sin B oB

B

oB

+

or

r

or

- i1p + k2r21p = O.

SOME APPR O X IMATE

215

METHODS

Application of transform (a) gives 1 d do/ -- - sin 0 - + ( 11.2 - 1)0/ sin 0 dO dO r

=

0

,

(i) o/(p,) = AP I-' - t (cos 0) + BPI-' - t( - cos 0) . Consider one example in detail. A rigid disk lies in 0 = 1-7T , 0 « r « a. A wave
=

=

=

o/(p,)

=

A PI' _t (cos 0) , 0 « 0

«

!7T ; -A PI' -t( - cos 0) , i7T

«

0 « 7T.

(j )

Differentiate these equations with respect to 0 , eliminate A , and apply !7T. Then, if suffix zero refers to 0 !7T, boundary conditions on 0 =

ik

=

00

J r1/2K,A).r) dr + J (��) oKI-' ().r) d; a

o where

J f(r)KI-'().r) � , a

= - 2K (p, )

(k)

0

a

so that Also

K(p,) = -iP� - !t (O)/P l-'- t (O) L p(p,)/ LN (P, ) , 1 2 {PI-' _ t(O)}-l = 7T- / r ( ! + !p,) r(! - !p,) {P� _ t(O)}-l = - !7T-1/2 r ( t + !p, ) r ( t - ip,) =

(1 )

L p(p,) = r (i + !p,)/ r (l + !p, ) LN(P, ) r ( i- - ip,) / r( ! - -lp, )

(m)

=

and Lp , LN are regular and non-zero for Re p, > -t, < ! , respectively. In the following manipulations we must remember that ( (m - p,) r( - p,) -m!

2 _m!

( 1)m lim - p,) r ( -ip, ) = -- . 1'-> 2m In view of (f), (h) it is convenient to rewrite (k) as (cf. (5. 1 7 ) ) :

lim I'->m

=

- 1)m

(2m

_p,)}-l f rl /2K,A).r) dr + LN (P, )o/N (P, ) a

ik LN (P, ) (i),a) - I-'{ r (

o

Set L p(p,) r ( p,){ r ( p, ) } ( !),a ) - 2 1-'cI> p( - p,) = [ L p(p,) r (p,){r ( -p, )}-1 ( i),a) - 2 1-'cI> p ( - p,) - u - V] + [u + V],

_

l

-

(0)

216

T H E W I E N E R- H O P F T E C H N I Q U E

where Un

=

n r(.l.n + Q) 4, . 2 (n! ) 2 r(in + t) ,

__

r(n + l) (2n + l) Vn - -:= -=.:--= { r(2n + �-)}2 . n! -

--

[ U + V] is regular in Re ft > - 1 , and the first bracketed term on the / right of (0) is regular in Re ft < i, assuming that '" � const. i.e. f(r) � r1 2 as r ->- o. Next examine

I r1/2K,A).r) dr a

J

=

=

o

{

(iAa) -1'+2n+(3 /2) Tr � 1 21 /2 . / 3 2 sin Trft L.- � r( - ft + n + 1 ) ( - ft + 2n + i ) ). n�O - l'(ft

+

(iAa)I'+ 2n+(3 /2) n

+

l ) (ft

+

2n

+

}

i) ,

where we have used (c), substituted the series expansion of 11" and integrated term by term. This is valid for - Re ft < i < Re ft. The series defines the analytic continuation of the integral for all ft. At first sight it would seem that there are poles at ft = ±n , but the reader will readily prove that this is not so. The only poles occur at ft = ± (2n + 1) with residues Write

where 21/2 1 ).3/2 · l'(2n + I) '

- ( - l )n _ Wn -

and W is regular in Re ft > -J. We can now separate (n) by the Wiener-Hopf technique to give, since p (ft) � ft-1 as Re ft ->- + 00 , (q) This agrees with A. Leitner and C. P. Wells [1], p. 28. Introduce

ikTr-1/2a3/2X (ft)

=

Lp(ft)p(ft) ·

(r)

SOME APPR O XIMATE

Then (q) gives

X ( p,) + _

217

METH O D S

2:00 - . -(tAa) 2n X(n) (n! ) 2 p, + n 2:00 2n + .3. ( 1 Aa ) 4nH n

n=l

2

2

"2"

{ r ( 2n + tW · (p, + 2n n=O

00

X(2n

+ .3.) 2

a

I ) n (lA ) 2n + i ) (p, + 2n

( '" _- "2"1 7T1/2 L r 2n n=O

i)

+

(

-

+

!-)

.

( ) s

From the form o f this result it is clear that X(p,) can b e expanded as a series

X(p,) =

�+

p, +

2"

00

'" X

1=2

m (P, ) (iAa) m .

(t)

If this is substituted in (s) and the coefficient of each power of (tAa) is equated to zero, the X m can be found explicitly by means of recursive formulae. 4 2 16 x (P, ) 5 (p, + 1 ) 1 5 (p, + I) 97T(p, + i ) 3 1 16 x (P, ) 7 + etc. 4 7 (p, + 2) 27 (p, + 1 ) 9 45 (p, + ¥ ) ,

_

_

_

From (a), (j ) , for 0

f1 I2rf> =

'P

.;;;

0

.;;;

i7T,

c + i oo

a

f f

1

= ;;

p,

c - i oo

!(p)K",(Ap)

0

l - (c O 0) �"'l (O) I",(Ar) dp, .

dp P

P

S

To evaluate this integral for r > a it is necessary to substitute (c) for K",(lp) and change the sign of p, in the part involving I _ ",(Ap), remember ing that P", - i is an even function of p, . This gives a

:i f p, f !(p)I", (Ap) � P��:��) O) K", (Ar) dp, c + i oo

r1/2rf> =

0

c - i oo

c + i oo

= 7T-2ka3/2

f

( i Aa)"'{ r (p,)}-l r (i - ip, ) r( t + ip, ) X P", - 1 (cos O)K",(Ar)X(p,) dp" where we have used (d), (r) . The contour can be completed in a right half-plane. The only singularities are poles at p, = 2n + i, and these give c - i oo

rf>

=

00

27T lkIa3 '"

.

( _ I )n(Aa )2n

L 1 . 3. 5 . . . (4n + 1 ) n-O

X(2n + i )

( )

2 2A I / X P 2n+ l (cos O) � K2n+ (S/2 )(Ar) .

218

T H E W I E N E R- H O P F T E C H N I Q U E

The far field is obtained by using the asymptotic expansion

{ 2,1./(1IT)}1/2K2n + (3 /2) ( ,1.r)

�

r-1 e -AY,

and 'this term comes outside the summation sign. The field depends only on X(2n + i ) which can be computed from (t) . In order to evaluate the field to (,1.a)4 we have, for instance,

1 = 31 75 (,1.a)2 271T -2 (,1.a)3 + -(,1.a)4 . 5 72 2 ... = 5-1 - -32(,1.a)2 .7 _

_

4

+

_

+ . . .

+

The scattering cross-section is given by the left-hand side of the following equation. On substituting the above expression for 4> we find +1 2 Re i4> r2 d( COS O ) -1 1 6 k4a6 311 - 1 + (ka)4 + . . . . ( ka)2 + 25 6 1 25 27 1T

;f

{ ( :�)}

= - [ -8

--

]

This agrees with the existing literature, e.g. C. J. Bouwkamp [ 1]. This concludes our discussion of the problem. In exactly the same way a solution can be found for the case 4>t 0 instead of a4>t! an = 0 on the disk. The method can also be used to deal with the biconical antenna-as shown by J. A. Meier and A. Leitner [ 1] . A similar type of analysis can be used for the two-dimensional strip . The equation is o a4> 024> + + k2r24> = O. r�r a02 or =

Application of transform (a) gives

d2(D + p,2(D d02

=

0

(D

=

A sin p, O + B cos p, O.

The solution proceeds as before. The factorization corresponding to (m) can again be expres5!ed in terms of gamma-functions but it so happens that double poles occur in the split (0) . It will be found that the separation of these double poles involves differentiations of the type

These introduce a familiar parameter of the form

=

0+

(tkb) - i1T/2, We shall not pursue the analysis here. p

In

o

= 0·57 72 . . . .

219

S O M E APPR O X IMATE M E TH O D S

The solutions of these problems given in the literature usually involve complicated evaluation of integrals. One of the interesting features of the present method is that once the machinery is established the solution needs little more than a few elementary properties of gamma functions. 5.13 Consider a wave exp ( -ikz) incident on a circular disk in z = 0, .;;; p .;;; a. By applying Fourier transforms in the z-direction and introducing unknown functions on p a, - 00 < z < 00 , formulate this o

=

problem as a generalized Wiener-Hopf equation of type (5. 1 7 ) . This formulation should be suitable for large k. A similar formulation can be obtained for the two-dimensional strip. 5.14 If we try to generalize the method of §5.6 to a circular disk of radius a it would appear that corresponding to the Fourier transform in x we should use a Hankel transform in the radial co-ordinate p. To obtain a decomposition corresponding to (5. 10) we should presumably write (a < P < 00 ) .

The Fourier inverse of ( 1 . 60), ( 1 . 6 1 ) gives

5.15

ii

00

f H�1) {k(X2 + y2 )1/2}eicxx dx = y-1e - yllil .

- 00

In this result replace x by x

ii

00

- X,

and then set X = � cos v, Iyl

f H�1) {k(X2 + �2 - 2x� cos v)1/2}eicxx dx = y-1ei«",

- 00

=

� sin v :

cos v + iy sin v )

Consider the integral equation 00

ii f f(�)[H�l ) {k l x - m + o x > 0, + pH�1) {k(x2 + �2 - 2x� cos v W /2] d� = g(x), where p is a constant. A Fourier transform gives y-l {F+(or.) + pF+(or. cos v + iy sin v)} = G+(or.) + H_(or.) .

This equation can be regarded as a generalization of ( 5. 1 7 ) which is the case v 1T. The case v i1T has been considered to some extent by R. Jost [1] in connexion with a problem in quantum mechanics. He does not obtain an explicit solution suitable for calculation. Similar integral equations appear in connexion with diffraction problems involving wedges. In this type of problem it is natural to make the substitution or. = -k cos fJ of § 1 .6. =

=

CHAP T E R VI

THE G E N ERAL S O L UTION OF THE BAS I C WIENER-HOPF PRO B L E M 6.1

Introduction

So far we have been concerned mainly with the solution of specific examples. In §6 2 we obtain the general solution of the basic Wiener-Hopf problem underlying all the examples which have been solved exactly in previous chapters. As indicated in Chapter II, problems which can be solved exactly by the Wiener-Hopf technique can be reduced to the solution of one of three equivalent subsidiary problems : (i) Solution of the complex variable equation .

where this equation holds in the strip 'T < 'T < 'T+, 00 < (J < 00 , and h(x) , e(x) , the transforms of H+(cx), E_(cx) , are the unknown functions. (ii) Solution of certain types of integral equation, e.g. _

co

J k(x - �) h(�) d � = l(x)

(a)

0

or In

-

(b) {d 2/dx2 + k2 }

,

(x > 0) ,

(6.2)

J l(x - �)h(�) d� = 1(x),

(x > 0).

(6.3)

co

0

these equations h(�) is unknown. (iii) Solution of certain dual integral equations :

1 (217 ) 1/2

ia + co

J

-ill - co

1 (2 17 ) 1 /2

K(cx)A (cx)e - iex", dcx = 1(x) ,

(x > 0),

(6.4a)

(x < 0),

(6.4b)

ia + co

I

A ( cx ) e -iax dcx = g(x) ,

ia - c:o

where A ( cx ) is unknown. Problems (i)-(iii) are equivalent under certain conditions and the 220

SOLUTION

OF THE

WIEN ER-H OPF PR O B L E M

221

notation has been chosen t o indicate the relationship between the various functions in the three problems. Thus

- GO

(6.5)

For convenience we have defined

K(rt.)

GO

=

J k(x)eiax dx,

- GO

(6.6)

i.e. there is no factor (21T) -1/2 in front of the integral. To show that (iii) is equivalent to (i) suppose that the left -hand sides of (6.4a) , (6.4b) are equal to the unknown functions e(x) , h(x) for x < 0, x > ° respectively. Then a Fourier transform gives Elimination of A(rt.) gives (6 . 1 ) . Conversely if in (6.1) we define then from ( 6 . 1 )

A (rt.) K(rt.)A(rt.)

=

=

H+(rt.) + G_(rt.),

(6.8a)

F+ (rt.) + E_(rt.) .

(6.8b)

The Fourier inverses of (6.8a), (6 .8b) give (6.4b ) , (6 .4a) for x < 0, x > 0, respectively. These manipulations assume of course that the

appropriate Fourier transforms exist. In §6.2 we solve the dual integral equations (6 .4) by the 'multi plying factor method' of §2 .3. This is probably the easiest formal method for finding the general solution of the three problems. A direct solution of problem (i) is sketched in ex. 6 . 1 . A result which is related to the investigation in §6.2 is that any Wiener-Hopf integral equation of type (6.2) can be decomposed into repeated Volterra integral equations (ex. 6 .4) . In exs. 6.5 and 6.8 the results in §6 .2 and ex. 6.4 are used to reduce certain singular integral equations of the first kind (or related equations) to Fredholm integral equations of the second kind. These techniques can be applied to approximate solution of the three part problems considered in §§5.5, 5.6 but it would seem that this method of approach is not so useful as the complex variable methods of Chapter V. However there are some interesting and effective applications of the same basic ideas in connection with dual integral equations involving Bessel functions, applied to boundary value problems involving disks. These applications will be discussed elsewhere.

222 6.2

T H E W I E N ER-H O P F T E C H N I Q U E

The exact solution of certain dual integral equations

In this section we solve the dual integral equations (6.4) by the method of §§2.3, 2.8. As in §2.3 replace x by (x + �) in (6.4a) and by (x - �) in (6.4b) , with � > 0 in each case. Multiply by .At- m , .;V+( �) respectively where these functions will b e defined below. Integrate each equation with respect to � from 0 to 00. Introduce N_(oc) , N+ (oc) defined by 00

: J .At_me -i<x� d�

(2 ) 1 /2

=

N_(oc)

o

(6.9a)

00

: J .At+(�)ei<x� d�

(2 ) 1 /2 By inversion

.;V_(�)

N+(oc) .

ib + 00

: J

=

(2 ) 1 / 2

ib - oo

: J

'w +

.;V+W

=

o

=

(2 ) 1 / 2

N_(oc ) ei<x e doc (6.9b)

00

N+(oc) e - i<x � doc,

ic - r:LJ

where these integrals are zero for � < O.' Using these results, (6.4) reduce to

ia - r:LJ

00

=

1

J

ia +

(21T) 1 / 2

00

: J .At- Wf(x + �) d�,

(2 ) 1 / 2

(x > 0), (6.lOa)

o

N+(oc)A(oc) e - iax doc

ia - 00

00

=

: J .At+(�)g(x - �) d�,

(2 ) 1 / 2

(x <

0) . (6. lOb)

o

The natural method of procedure would be to determine N+ , N_ in these equations so that

(6.1 1 )

S O LUTION O F T H E WIEN ER-H OPF PROBLEM

223

but, as indicated in §2.3 in a special case, convergence difficulties indicate that a slight modification in this procedure is required. Consider two cases separately (i) I K(Ot) 1 "'-' I Ot I -1 as I Ot I - 00 in the strip, (ii) IK(Ot) 1 ,......, l Oti as I OtI - 00 in the strip. (i) In case (i) assume that we can write K(Ot) = K+(Ot)K_(Ot) where I K+ (Ot) I I Ot I -1/2 , I K_(Ot) 1 "'-' I OtI -1/ 2 , as Ot --+ 00 in appropriate half-planes. For convenience we define here several functions that we need later : ,.....,

ia+ 00 K+(Ot) e - ic 0) = 0, (u < 0), ( 2n ) 1 / 2 ia - 00 (6. 12a) ia+ 00 1 K_(Ot) eic 0) = 0, (u < 0), ( 2n ) 1/2 ia - co (6. 12b) ia + 00 1 i e - ic 0) (u < 0), (6. 1 2c) ia+ 00 i e''XU dOt - 2n 1 / 2 K_(Ot)(Ot - Ot 2 ) ( ) ia - 00 = .A'_(u), = 0, (u > 0) ( u < 0) , (6.12d) where Ot 1 , Ot 2 represent arbitrary complex constants in lower and 1

f

=

f

f

f

l

.

upper half-planes respectively. Each of these equations has an inverse analogous to one of (6.9a) . We use the shorthand :

. d ±. �t- - ±e 'F 'C
.

�;} - ± e ± 'C<2Y

d . e 'F "'2Y dy

(6.13)

where upper and lower signs go together and the appropriate variable to be used in place of y (namely x or r; ) will be obvious from the context. Comparing the equation K(Ot) = K+ (Ot)K_(Ot) with (6. 1 1 ) and remembering that IK+ I , I K_I "'-' I OtI -1/ 2 as I Otl _ 00 in appropriate half· planes we see that suitable expressions for N+ , N_, which of course do not satisfy (6. 1 1 ) , are

N_(Ot) = {K_(Ot)(Ot - Ot 2 )}-1, (6.14) N+ (Ot) = K+ (Ot) where Ot 2 is an arbitrary constant in the upper half-plane, inserted

to overcome convergence difficulties. Comparing (6.9b), (6.12a, d),

224

T H E W I E N E R- H O P F T E C H N I Q U E

we see that .Y_(�) = iJt_(�), .Y+W = 2+(�) so that (6. 10) become ia + 00

: f (oc - oc2)-lK+ (oc)A(oc)e - iax doc

(2 ) 1/ 2

ia - co

=

00

: f Jt_( �) f(x + �) d�,

(2 ) 1/ 2

ia + 00

(x > 0) ,

(6.15a)

o

: f K+(oc)A(oc)e - iax doc

(2 ) 1/ 2

ia - 00

=

00

: f

(2 ) 1/2

2+ (�)g(x

o

- �) d �, (x < 0). (6.15b)

Multiply the first equation by exp ( ioc 2x ) and differentiate with respect to x : ia - 00

=

:

( 2 ) 1/2

00

�2

f Jt_( � )f(x + �) d�,

(x > 0) , (6 .16)

o

using notation (6 .13). The left-hand sides of (6 .15b), (6 .16) are identical and we can invert to find

K+ (oc)A(oc)

=

00

Lf

eia'l �2

00

f Jt-( �) f ('YJ + �) d� d'YJ

0

o

+

o

00

Lf f - 00

+

eicx'l

0

2+ ( � )g( 'YJ

- � ) d � d'YJ . (6.17)

This is the solution of the dual integral equations. Quantities which. are of interest are the left-hand sides of (6 .4a), (6.4b) when x < 0, x > ° respectively. Inserting (6 . 17) in (6.4a) and using notation (6 . 12) gives, for x < 0,

e ( x) =

00

00

L f 2_( 'YJ - x) f Jt_(� - 'YJ )f(�) d� d'YJ + L f - ( 'YJ - x) f +( 'YJ - �)g(�) d� d'YJ . o

�2

o

'I

2

+

'I

- 00

2

(6.18)

SOLUTION OF THE

W I E N E R-H O P F P R O B L E M

225

I f w e insert ( 6 . 1 7 ) in (6 .4b) w e cannot interchange orders o f integra tion directly. We use the following device to avoid divergence, considered for a typical case ,

ia+ 00 drx e - 'ax K+(rx) ia - oo

f ·

--

co

f F('Y} )e'.lX� d'Y}

0

= ( 21T ) 1/2e - '.IX,x

d e' ,x dx lX .

x

f F( 'Y} ) 1+ (x - 'Y} ) d'Y} ,

(6. 1 9 )

o

where we have interchanged orders o f integration and used notation (6.12c). Insertion of (6 . 1 7 ) in (6 .4b) now gives, for x > 0, co

x

h e x)

=

L f»t f 1+ (x - 'Y} ) f 1_(� - 'Y} )f(�) d� d'Y} + L f»t f 1+(x - 'Y} ) f .:£'+('Y} - �)(J(�) d� d'Y} . f»2

o

+

� �

o

- co

- co

(6 .20)

(ii) If /K(rx) / ,...,., / rx/ as / rx/ ..--+ 00 in the strip, assume /K+( rx) / , /K_(rx) / ,...,., / rx/ 1 / 2 as / rx/ ..--+ 00 in appropriate half-planes. This time we define 1

( 21T) 1/2

ia+ 00

e - 'IX. U drx f K+(rx) l iaco

=

.:£'+ (u) , (u > 0 )

= 0,

ia+ co 1 = 0, -- e'IXU drx = .:£'- (u) , (u > 0) (2 1T ) 1/2 K_(rx) ia - 00 ia+ co i K+(rx) - i U e IX d rx 1/2 (21T) rx - rx 1 ia - 00 = 1+ (u) , (u > 0 ) = 0, (u < 0), ia+ co i K-(rx)- U e'IX drx - 1T) 1/2 rx - rx 2 (2 ia - co = 1_ (u), = 0, (u < 0). (u > 0 )

f l . f

f

(u < 0), (6.21a)

(u < 0), (6.21b)

---

.

(6.21c)

(6.21d)

226

T H E W I E N E R-H O P F T E C H N I Q U E

Then choose (cf. (6. 14»

(Ot - Ot 1 ) -lK+(Ot)

=

N+ (Ot)

Instead of (6.15) we find

ia - co =

f

ia + 00

f .!l'-mf(x + �) d�, : 2 1 (2 ) / o 00

(x > 0), (6.22a)

K+ (Ot) . A(Ot) e - '<xx dOt 2 1 (2 7T ) / Ot - Ot 1 ia - 00 00 1

--

=

-

�

(2 ) 1 / 2

f .A+(�)g(x - �) d�,

(x < 0). (6.22b)

o

Multiply the second equation by exp (iOt 1x) and differentiate with respect to x to find

ia - co (x < 0). (6.23) We can now invert (6.22a), (6.23). This gives (cf. (6 .17, 18, 20»

K+ (Ot)A(Ot)

=

� f ei<x'I f .!l'-(�)f( 'Yj + �) d� d'Yj 00

00

2

+ 00

o

0

+

L f ei<X'Ig)t f .A+ (�)g( 'Yj - �) d� d'Yj . o

- 00

00

00

0

e(X) = L �2 f .A_('Yj - x) f .!l'_(� - 'Yj )f(�) d� d'Yj + o 'I o f'I .A+('Yj - �)g(�) d� d'Yj, f + L �2 .A_( rJ - x) �t x

(6 .24a)

- 00

(x < 0 ) ,

(6.24b)

SOLUTION

h(x)

=

L f !l'+(X - r; ) f !l'_(� - r; )fW d� dr; o

+

227

O F T H E W I E N E R- H O P F P R O B L E M co

:r

'1

o

'1

L f !l'+(X - r; ) !l)t f vIt+(r; - �)g(�) - co

+

M

d r; ,

(x > o).

- co

(6.24c)

This concludes our discussion of case (ii).

These general solutions have not been used in previous chapters for various reasons. They are unwieldy since they involve fourfold integrals. And the known functions F+(oc) , G_(oc) in (6 . 1 ) (or f(x) , g (x) in (6.4)) are usually given explicitly and have a simple form : in concrete examples it is usually easy to solve the Wiener-Hopf equation from first principles. However in several important cases the functions !l' ± (u) , Jt ± (u) which occur in the general solution can be evaluated explicitly and this means that the general solution involves only twofold integrals. Thus for the Sommerfeld half 2 2 -1/ 2 plane problem of Chapter . II, with K(oc) = ( oc - k ) , we can choose

K+(oc)

=

(oc

+

k) -1/2 : K_(oc) = (oc - k)-1/2 : oc 1

=

- k : oc2 =

+ k.

This gives (ex. 2.4) :

Jt+ (u)

vIt_(u )

=

=

ia + co � (oc ( 2n ) 1 / 2 ia - 00

f

- 2; 1 / 2 ( )

+

ia + co

f ia

(oc

k) -1/2e - ic 0) , (6.25a) - k) -1/2e + ic
- co

=

2 1 / 2e - i TT/4u-1 / 2eiku , ( u > 0).

(6.25b)

As an application of these results consider the solution of co

ii J Hb1 ) (k l x - � I lh(�) d� = f(x) ,

(x > 0).

(6.26)

o

Suppose that the left-hand side of this equation is equal to the (unknown) function e (x) for x < O. Application of a Fourier transform as in §2.5 gives (6 . 1 ) with G_ ( oc ) 0 , K(oc) = (oc2 - k2 ) -1/ 2 . =

228

T H E W I E N E R- H O P F T E C H N I Q U E

The solution of (6 .26) is therefore given by (6.20) with (J(�) On using the notation (6.13) and the results (6.25) we have

h(x)

=

l o d - - e"kx 1T dx

-

=

O.

x

J (x - 1])-1/2e -2'Ok

'I

a

co

X

� J f(�)(� - 1])-1/2eikl; d� d1].

d

(6.27)

'I

In §2.5 we solved (2.49), i.e. (6 .26) with f(x) exp ( i kx cos 0). If this value of f(x) is substituted in (6.27) it is found that the inner integral can be evaluated and the inner differentiation can then be performed. This gives (2kX)1/2 cos i =

h(x)

=

21T-1/2e - irr/4 tan l2 0 eikx !:..- e - i(k+ k cos 8 )x dx

-

J :xp (iy2) dy:

a

It is left to the reader to show that the solution in Chapter II, which is given by inverting (2.76), is equivalent to this result. Miscellaneous Examp les and Results VI 6.1 Show that, by using the decomposition theorem B of § 1 . 3, the function H+ {rx) in (6. 1 ) is given by

where

In these integralso T_ < If we use the results

{' - rx)-l

(' - rx)-l

=

< T < T+.

C

E_{rx).

-

00

i f ei(' -e<)'l d1],

co

1m (' - rx) > 0,

o

= i f e - ia-e<)'l o

Obtain a similar formula for

d1],

1m ( ' - rx) < 0,

SOLUTION OF THE

W I E N E R- H O P F P R O B L E M

229

and replace F+(0, G_(O by their expressions in terms of j(x), g(x), it will be found that the inverses of H+(rx), E_(rx) obtained above can be reduced to the results ( 6. 1 8) , ( 6.20), ( 6.24) for h(x), e(x). of

6.2

By applying a Fourier transform in (0, 00 ) show that the solution x

J 2'+ (x - ;)g(;) d; = j(x),

o

is

g(x)

=

+ 2- 2)l 21'

x > 0,

(a)

x

f o4+ (x - ;) j(;) d;, o

where we have used the notation (6. 12), (6. 1 3) , i.e. o4+ (u) is defined by (6. 12c) where K+ (rx) is the inverse of (6. 12a), namely

Show that the solution of co

J 2'_(; - x)g(;) d; = j(x),

x

is

x > 0,

(b)

co

g(x) =

L J 04_(; - x) j(;) d;, 2)2-

x

where o4_(u) is defined by (6. 1 2d), K_(rx) being the inverse of ( 6 . 1 2b), namely =

(Suppose that

1 (21' ) 1/2

--

co

co

J 2'-( u)e - "XU. duo

o

J 2'_(; - x )g(;) d; = e(x),

x < 0,

(c)

o

where e(x) is unknown. Apply (b) and (c) . Then

a

Fourier transform in ( - 00 , 00 ) to

If this is inverted the second term on the right will give no contribution t o g ( x), x > 0.)

230

THE

W I E N E R- H O P F T E C H N I Q U E

6.3 In case (i) of §6.2, using notation (6. 12a, b), the inverse of (6.6) gives ia + 00

k(x - �)

=

L f K+(ex)K_(ex) e -io« x -<) dex ia - co

=

1 __ (27T)3/2

ia + 00

00

f K+(ex) f !l'-(y)e -io
ia - 00

= As

00

0

L f !l'_(u - x)!l'+(u - �) du. max (x.�)

a special case show that

trriH�l) (klx - � i l

=

00

f (u max (x. <)

e - ik(x H)

(a)

e2iku

xJ1/2 (u

_

_

W/2

du .

6.4 We can use the result in ex. 6.3 to show that any Wiener-Hopf type equation can be reduced to a repeated Volterra equation. Substi tute (a) from ex. 6.3 in (6.2) : 00

x

f h(�) f !l'_(u - x)!l'+ (u - �) du d� x

o

+

+

00

00

f h(�) f!l'_(u - x)!l'+ (u - �) du d�

x

�

=

27T f(x),

x > O.

Interchange orders of integration in each repeated integral. It is found that then the integrals can be combined to give 00

u

f !l'_(u - x) f !l'+ (u - �) h (�) d� du

x

=

27T f( x ) ,

x > O.

0

But this is equivalent to the repeated Volterra equations u ,

f !l'+(u - �)h (�) d�

=

p(u),

=

27Tf(x) ,

u > 0,

o

00

f !l'

II!

_

(u

-

x )p (u ) du

x

> O.

If these are solved by means of ex. 6.2 it will be found that the expres sion for h (x), the solution of (6.2), is identical with that found in §6.2 , namely (6.20) with g (�) O. =

S O L U TI O N O F THE

W I E N E R -H O P F P R O B L E M

231

I was led t o the above analysis by a paper o f E . T. Copson [5] which deals with the problem of the electrified disk. This problem is in fact of the Wiener-Hopf type although the Wiener-Hopf technique is not used in Corson's paper. A related paper is that of D. S. Jones [6] . I have made extensive applications of this method in connection with problems involving parallel disks, disks in tubes, etc . It is not necessary to use the Wiener-Hopf technique explicitly. It is appropriate to denote by 'Copson's method' the technique of interchange of orders of integration to produce repeated Volterra integral equations (cf. 1 . N. Sneddon, Elements of Partial Differential Equations, McGraw-Hill ( 1 957), p. 1 7 9 ) . Consider the three-part problem

6.5

1 ( 27T) 1/2

J K(Ol)A (Ol)e-'cxx. dOl u(x), (x J A (Ol)e -'''x. dOl s(x) (p < x < 00

ia +

=

> q),

(a)

q),

(b)

(x (x), 1JI(x) for x q respectively. Take (b) and (c) together and replace x by (X + q). Then on setting B(Ol) A (Ol) exp ( -iOlq),

� ( 27T) /2 1 ( 27T) 1/2

J

ia +

ia - 00 ia + 00

=

00

K(Ol)B(Ol)e - icxX dOl

=

J B (Ol)e _ '·cxX dOl {r/>S(XX ++ =

ia - 00

(

u(X + q), q),

(X

>

(d)

0),

( (p - q) < X < 0), (X < (p - q) ) .

q),

(e)

Consider case (i) of §6.2 where I K(Ol) I I Oll-l as l Oll ->- 00 in the strip . Then ( 6.20) gives, on interchanging orders of integration in the second repeated integral, and rearranging, �

1JI (X + q) +

00

J r/>(p - Z)N(X,Z) dZ

=

G(X),

(X

> 0),

(f)

o

where G (X) is known and o

N ( X Z) ,

=

-

L P}t J

-d-Z

.A+ (X - 'I} ) !l'+ ( 'I} + . Z + d) dn, d

=

(q - p l .

(g)

232

THE

W I E N E R- H O P F T E C HN I Q U E

In exactly the same way, by considering (a) and (b ) , we can obtain a similar relation with the roles of and !p interchanged. This gives two integral equations of Fredholm type for the unknown functions and !po It may be possible to solve these iteratively but we shall not con sider this here. The above analysis can be regarded as a generalization of a special result obtained by a different method by E. N. Fox, Phil. Trans. Roy. Soc. (A) 241 ( 1 948), 7 1- 103. To indicate the relation between the present method and that of Fox, invert (e) which is equal to !p(X + q ) for X > 0, and insert the resulting expression for B(rt.) in (d) . This gives 00

f {!p(X + q) !c ( Y - X) + (p - X)!C( Y + X

o

+

d)} dX = F ( Y ),

Y

>

0,

(h)

where F ( Y ) is a known function. Suppose that we can write

k ( Y + X + d)

00

=

f k( Y - Z)M( Z,X) dZ.

(i)

o

If we substitute this expression in (h) and interchange orders of inte gration, we find 00

f

o

(

!C ( Y - X) !p(X + q) +

00

f (p - Z)M(X, Z) dZ} dX = F( Y) . 0

This is now a straightforward Wiener-Hopf equation for the quantity in brackets, which is identical with the expression on the left of (f) if we have M(X , Z) "'" N(X , Z) . This indicates that

k( Y + X + d) =

00

-

L J !C Y (

o

d+X - Z)� + l

J

.4+ ( Z

+

'YJ )..'l'+ (X

+ d - 'YJ ) d'YJ dZ.

(j )

0

In the special case discussed by Fox we have .4+ ( u ) · = i2'+( u)

=

21/2eirr/4u-l/2eiku.

The integral in (g) is elementary and can be evaluated explicitly. We find that (j ) reduces to Fox's result : H �l ){k(X +

1 Y + d)} = -

1T

00

J

o

H �l) ( lc I X

+ y - Z I ) Zeik(Z +YY++ d)d {d-Z +

}

1/2

dZ.

The problem considered above is of the same type as that considered in §5.5 (cf. ex. 6.6). It would appear that for the problem of diffraction

S O LU TI O N

O F THE

233

W I E N ER-H O P F P R O B L E M

by slits and strips the method o f §§5.5, 5.6 i s t o be preferred t o that given above. (The two methods are of course related.) 6 . 6 Show that the solution of the integral equations (a)-(c) of ex. 6.5 is equivalent to the solution of the generalized Wiener-Hopf equation

K(IX){eic
=

Show that the following two problems are equivalent : (i) Solution of an integral equation

6.7

00

f {k(x - �) + l (x + mh(�) d� = f(x),

(x

o

>

0) .

(ii) Solution of the generalized Wiener-Hopf equation

K(IX)H+ (IX) + L(IX)H+ ( - IX) = F+ (IX) + E_(IX).

6.8

We reduce the equation 1

f k(x - �)h(�) d�

=

f(x),

(0

o

<

x

<

1),

(a)

to an integral equation of Fredholm type. For simplicity suppose that 00 , the transform of k(x), i.e. K(IX), is even, [ K(IX) [ [ IX[-l as [ IX[ and K_( IX) = K+ ( - IX). Then in definitions (6. 1 2 ), if 1X 2 = - IXI ' �

2'+(u) = 2'_(u) = 2' ( u ) , say

--+

.4+ (u) = .4_(u) = .4(u), say.

From ex. 6.2 the following are inverse formulae :

J 2'(x - �)g(�) d� x

f(x) =

o

' J .4 (x

1 ' d g(x) = - etfix - e - 'fix dx 21T

(b)

x

- �) f(�)

d�,

o

where fJ is a complex constant in the upper half-plane, By simple change of variable we have also

f(x)

J 2'(� - x)g( �) d, 1

=

x

g (x)

=

1 d - - e -''fix dx 21T

e''fix

J .4(� - x) f(�) d�. 1

x

(0)

234

T H E W IE N E R-H O P F T E C H N I Q U E

We now consider (a) . From ex. 6.3 we can set

- �)

k(x

=

�

J

.!l'(U - X)Z(U ma.x (x.e) 217

Substitution in (a) gives, for 0 u

-

�) du +

co

�f

+

1

1

1

<

Z(u

x

<

-

x) Z(u

- � ) duo

1,

I .!l'(u - x) I Z(u - �)h(�) d� +

x

0

+

co

1

I h(�) I Z(u - x) .!l'(u

o

Set

-

�) du d�

=

217 j(x).

(d)

1

u

I .!l' (U - �)hW d�

=

say.

x(u) ,

(e)

o

Apply the inversion formula in (c) to (d) . In the second integral in (d) replace h(�) by its expression in terms of X (u) obtained by inverting (e) by (b) . Then co

. e -ipx .!:.. eiPx f JI (u - x) J .!l'(' - u)Nm d, du - -..:. 4772 dx 1 a; 1 _ e-ipa; :x eipa; J JI(u - x) j(u) du, a; 1 v e-i ei f Nm (1) .4(v - �)xW d�} Z(' - v) dv fJ v ! { f 1

X (x )

=

where

(f)

=

o

=

.!l'(,

1

0

1

- 1 ) f .4( 1 - �)X(�) d� - f XWI(�) d�,

.

0

o

with

fJ(V 1

I(�)

=

E l -E =

-

�) e- 'Pv .!:.. {eiPv.!l'(, dv

-

v ) } dv

f .4(u)e-iP(uH) � {eiP(uH) .!l'(, - u - m

o

d

duo

(g)

SOLUTION

O F THE

W I E N E R-H O P F P R O B L E M

235

From the symmetry of the expression in curly brackets we can replace

d/du by d/d; . We can then verify that B -i{3�

:; Bi/30 f Jt(u) 2'( { - u 1-;

-

; ) du = - Jt ( I - ; ) .2"(� - 1 ) + 1(; ) . (h)

o

Substitution of 1(;) from this expression into (g) gives, on then setting

v = ; + u,

N(C)

f X(;) B - i/30 :; Bi/30 f Jt(v - ;)2'(� - v) dv. 1

1

= -

o

e

On using this result in (f) we find, for (0 < x < 1),

f X(;)K(x,;) d; 1

X(x) + where

o

K(x,;) = k(x,�)

=

=

_ B - i{3x

:x Bi{3x f Jt (; - x) f(;) d;, 1

(i)

x

00

J k(x,�)k(;,�) d�,

1

(j )

{ f Jt(U - x).2"(� - u) dU} '

� B - i{3x � BiflX 27T dx

1

(k)

It is obvious that alternatively we could have obtained a Fredholm equation for h( ; ) by inverting (d) twice by (b) and (c ) . But the kernel of the resulting integral equation is not so elegant as (j ) . As an example consider

!i Then

1

J H�l) (k l x - ; i lh(;) d� = f(x),

o

(0 <

x < 1) .

K+(or.) = Bi1T/4 (k + or.)-1/2 : K_(or.) = Bi1T/4 (k _ or.) -1/2 = B - i1T/4 (or. _ k)-1/2 . Choose or. 2 = - or.1

=

k. Then (cf. (6. 12), (6.25) ), 2'(u) = Jt (u) = 21/2U-1 /2Biku .

The integration in (k) is elementary and we find from (j ) , (k) , if f1

- 1/2 _ � Bik( � -x) ��(��/�1 )�--� ' 7T ( I _ x)1 2 ( � _ x) K(x,;)

=

k,

236 where

T H E W I E N E R-H O P F T E C H N I Q U E

F(;) =

(1

ex;

- ;) I

u-1eiu duo

2k(1 - �)

6.9 Show that the problems in exs. 6.5, 6.8 are equivalent since equations (a)-(c) of ex. 6.5 can be reduced to a single integral equation of type (a) in ex. 6.8 and conversely. Show also that the solution of the integral equation (a) in ex. 6.8 is equivalent to the solution of a generalized Wiener-Hopf equation of the type given in ex. 6.6 (cf. R. Latter [1]).

[Note added in page proof: Approximate methods based on the Wiener Hopf technique can be considered under three headin gs : (i) Problems which can be formulated exactly in terms of a generalized Wiener-Hopf complex variable equation (p. 178) ; the approximations are made when trying to solve this equation. The problems in Chapter V are of this type. Examples which have been treated in the literature from an integral equation point of view are J. B. Alblas [2], H. Levine [2], [3]. (See p. 242 for the last two references.) The integral equation method and Jones's method for obtaining the complex variable equation are equivalent and most of the comments in §2. 7, p. 76, apply (though these refer to problems which can be solved exactly by means of the Wiener-Hopf technique) . (ii ) Problems which are usually formulated in terms o f integral equations or dual integral equations ; these are converted into equivalent integral equations of a more convenient form, e.g. Fredholm integral equations of the second kind. The examples of this kind given in this book (exs. 6.5, 6.8) are of type (i) above and in my ' opinion they are more conveniently formulated and solved by the methods of Chapter V (cf. exs. 6.6, 6.9). But this is not necessarily true in all cases (cf. p. 221, bottom) . (iii) Problems not directly amenable to the Wiener-Hopf technique which are converted into a form suitable for application of the technique by some approximation or transformation, or by virtue of the method used in formulating the problem. There are a large number of possibilities, e.g. H. Levine [ 1], H. Levine and T. T. Wu [1], [2] (the first and last references are given on p. 242). No examples of this type are given in this book. It is hoped that examples of types (ii) and (iii) will be included in a book being written' by Prof. I. N. Sneddon and myself which will be devoted mainly to methods other than the Wiener-Hopf technique for the solution of mixed boundary value problems. The above remarks were prompted by an interesting discussion with Dr. H. Levine at the International Congress of Mathematicians, Edinburgh, 1958.]

BI BLIO O H A P H Y (Items marked * arc not directly connoeto
J. B . ALBLAS [ 1 ] On the diffraction of sound waves in a viscous medium, Appl. Sci. Res . A6 ( 1 9 5 7 ) , 2 3 7-262. [2] On the generation of water waves by a vibrating strip, Appl. Sci. Res . A7 ( 1 95 8 ) , 2 24-2 36. W. S . AMENT [ I ] Application of a Wiener-Hopf technique to certain diffrac tion problems, U.s. Naval Res. Lab . Report No. 4334, Washington, D . C . , ( 1 954). L . L . BAILIN [ I ] An analysis of the effect of the discontinuity in a bifurcated circular guide upon plane longitudinal waves, J. Res. Nat. Bur. Stand. 47 ( 1 9 5 1 ) , 3 1 5-33 5 . B . B . BAKER and E . T . COPSO N [ I ] The Mathematical Theory of Huyghens' Principle, 2nd Ed. , Oxford University Press ( 1 950). G. L . BALDWIN and A. E . HEINS [ I ] On the diffraction of a plane wave by an infinite plane grating, Math. Scand. 2 ( 1 954), I03- 1 I 8 . J. BAZER and S . N. KARP [ I ] O n a steady-state potential flow through a conical pipe with a circular aperture, J. Rat. Mech. and Ana.lysis 5 ( 1 956), 2 7 7-322. (See also EM-66 ( 1 954) which has in addition an application to diffraction theory . ) J. BLASS [ I ] The extension o f the Wiener-Hopf technique t o radiation problems involving boundaries of elliptic cross-section, Dissertation, Poly technic Institute of Brooklyn ( 1 9 5 1 ) . C. J. BOUWKAMP [ I ] Diffraction theory, Rep. Progr. Phys. 1 7 ( 1 95 4 ) , 35- 1 00. (This is a shortened version of EM-50 ( 1 953 ) . ) J . F. CARLSON and A . E . HEINS [ I ] The reflection o f electromagnetic waves by an infinite set of plates I, Quart. Appl. Math. 4 ( 1 946), 3 1 3-32 9 . G. F . CARRIER [ I ] A generalization o f the Wiener-Hopf technique, Quart. Appl. Math. 7 ( 1 949), 1 05- 1 0 9 . [2] The mechanics o f the Rijke tube, Quart. Appl. Math. 1 2 ( 1 954), 3 8 3-395. [3] Sound transmission from a tube with flow, Quart. Appl. Math. 1 3 ( 1 956), 457-46 1 . G . F . CARRIER and W . H . MUNK [ I ] O n the diffusion o f tides into permeable rock, Proc. of Symposia in Applied Math. V, McGraw-Hill ( 1 954), 89-96. W. CHESTER [I] The propagation of sound waves in an open-ended channel, Phil. Mag. ( 7 ) 41 ( 1 950), 1 I-33. P. C. CLEMMOW [ I ] A method for the exact solution o f a class o f two -dimen sional diffraction problems, Proc. Roy . Soc. A205 ( 1 95 1 ) , 2 86--30 8 . [ 2 ] Radio propagation over a flat earth across a boundary separating two different media, Phil. Trans. Roy . Soc. 246 ( 1 953), I-55. E . T . COPSON [ 1 ] * Theory of Functions of a Complex Variable, Oxford Univer sity Press ( 1 935). [2] On an integral equation arising in the theory of diffraction, Quart. J. Math. 1 7 ( 1 946), 1 9-34.

238

THE

W I E N E R-H O P F T E C H N I Q U E

[3] * An integral equation method of solving plane diffraction problems, Proc. Roy. Soc. A1 86 ( 1 946), 100- 1 1 8. [4] * Diffraction by a plane screen, Pmc. Roy . Soc. A202 ( 1 950), 2 7 7-284. [5] * The electrified disk, Proc. Edin. Math. Soc. ( 3 ) 8 ( 1 94 7 ) , 1 4- 1 9 . J. CREASE [ 1 ] Long waves on a rotating earth in the presence of a semi infinite barrier, J. Fluid Mech. 1 ( 1 956), 86-96. B . DAVISON [ 1 ] Neutron Transport Theory, Oxford University Press ( 1 9 5 7 ) . R. C. DIPRIMA [ 1 ] O n the diffusion of tides into permeable rock o f finite depth, Quart. Appl. Math. 15 ( 1 95 7 ) , 329-339. J. P. ELLro= [ 1 ] Milne's problem with a point source, Proc. Roy. Soc . A228 ( 1 955), 424-433. A. ERDELYI [ 1] * Asymptotic Expansion8, Dover ( 1 95 6 ) . A. ERDELYI e t a l . [ 1 ] * Tables of Integral Tran8form8, Vols. I and I I , McGraw Hill ( 1 954). A. N. FEL'D [ 1 ] An infinite system of linear algebraic equations connected

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V. FOCK [ 1 ] Sur certaines equations integrales d e physique mathematique, G.R. Acad. Sci. U. R.S.S. 36 ( 1 942), 1 33-1 3 6 . [ 2 ] On some integral equations of mathematical physics, (in Russian) Mat. Sborn. 14 ( 1 944), 3-50 . R. C. GAST [ 1 ] Diffraction b y two parallel half-planes with source excitation, Tech. Rep. No . 25, Mathematics Department, Carnegie Institute of Tech nology ( 1 956). T . R. GREENE and A. E . HEINS [ 1 ] Water waves over a channel of infinite depth, Quart. Appl. Math. 1 1 ( 1 953), 201-2 1 4 . R. F. HARRINGTON [ 1 ] A current element near the edge o f a conducting half-plane, J. Appl. PhY8 . 24 ( 1 95 3 ) , 547-550. A. E . HEINS [1] The radiation and transmission properties of a pair of semi infinite parallel plates I, Quart. Appl. Math. 6 ( 1 948), 1 5 7- 1 6 6 ; II ibid . , 2 1 5-220. [2] Water waves over a channel of finite depth with a dock, Amer. J. Math. 70 ( 1 948), 730-748. [3] Water waves over a channel of finite depth with a submerged plane barrier, Ganad. J. Math. 2 ( 1 950), 2 1 0-222. [4] Some remarks on the coupling of two ducts, J. Math. Phys. 30 ( 1 9 5 1 ) , 1 64-1 69. [5] A note on a singular integral equation, Proc. Gamb. Phil. Soc. 46 ( 1 950), 268-2 7 1 . [ 6 ] Systems o f Wie.ner-Hopf equations, Proc. of Symposia i n Applied Math. II, McGraw-Hill ( 1 950), 7 6-8 1 . [ 7 ] A note o n a pair o f dual integral equations, (short summary only) Bull. Amer. Math. Soc. 56 ( 1 950), 1 7 2 . [ 8] Sur les couples d'equations integrales, G . R. Acad. Sci., Paris 230 ( 1 950), 1 732-4. [9] On gravity waves, Pmc. of Symposia in Applied Math. I V, McGraw-Hill ( 1 953), 75-86. [ 1 0] The scope and limitations of the method of Wiener and Hopf, Gommun. Pure Appl. Math. 9 ( 1 95 6 ) , 447-466. [ 1 1] The reflection of electromagnetic waves by an infinite set of plates III, Quart. Appl. Math. 8 ( 1 950), 2 8 1-2 9 1 .

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F. SMITHIES [ 1 ] Singular integral equations, Proc. Lond. Math. Soc. ( 2 ) 46 ( 1 940), 409-466. I. N. SNEDDON [ 1 ] Fourier Transforms, McGraw-Hill ( 1 9 5 1 ) . J. A . SPARENBERG [ 1 ] Application o f the theory o f sectionally holomorphic functions to Wiener-Hopf type integral equations, Koninkl. Ned. Akad. Wetenschap. A59 ( 1 956), 29-34. [2] The homogeneous first order integro-differential equation of the Wiener Hopf type, Rep. No. 1 3 , Technische Hogeschool, Delft ( 1 956). [3J On a shrink-fit problem, Appl. Sci. Res . A7 ( 1 958), 109-120. J. A. SPARENBERG, T . C. BRAAKMAN and C. W. BENTHEM [ 1 ] Discussion of a Wiener-Hopf type integro-differential equation, Appl. Sci. Res. B6 ( 1 95 7 ) , 3 1 2-322. E . C. TITCHMARSH [ 1 ] Theory of Fourier Integrals, Oxford University Press ( 1937). [2] * Theory of Functions, 2nd Ed., Oxford University Press ( 1 9 3 9 ) . C. J. TRANTER [ 1 ] * Integral Transforms i n Mathematical Physics, Methuen ( 1951) . L. A. VAJNSHTEJN [ I J Rigorous solution of the problem of an open-ended parallel-plate waveguide, Izv. Akad. Nauk, Ser. Fiz. 1 2 ( 1 948), 144- 1 6 5 . [ 2 J O n the theory of diffraction b y two parallel half-planes, Izv . Akad. Nauk, Ser. Fiz. 12 ( 1 948), 1 66- 180. [3J Theory of symmetric waves in a cylindrical waveguide with an open end, Zh. Tekh. Fiz. 1 8 ( 1 948), 1 543- 1 564. [4J The theory of sound waves in open tubes, Zh. Tekh. Fiz. 19 ( 1 949 ) , 9 1 1-9 30. [5] Radiation of asymmetric electromagnetic waves from the open end of a circular waveguide. Dokl. Akad. Nauk 74 ( 1 950), 485-488. [6] Diffraction at the open end of a circularly polarized cylindrical wave guide whose diameter is much greater than the wavelength, Dokl. Akad. Nauk 74 ( 1 950), 909-9 1 2 . (The above six reports have been translated into English b y J. Shmoys in EM-63 ( 1 954) , ) G . N. WATSON [ I J * Bessel Functions, 2nd Ed., Cambridge University Press ( 1 944). M. WEITZ and J . B . KELLER [1] Reflection of water waves from floating ice in water of finite depth, Commun. Pure Appl. Math. 3 ( 1 950), 305- 3 1 8 . N. WIENER [ 1 ] The Extrapolation, Interpolation, and Smoothing of Station ary Time Series, Wiley ( 1 949 ) . N. WIENER and E . HOPF [ 1 ] Uber eine Klasse singularer Integralgleichungen, S. B. Preuss. Akad. Wiss. ( 1 9 3 1 ) , 696-706 . W. E. WILLIAMS [ 1 ] Diffraction by two parallel planes of finite length, Proc. Camb. Phil. Soc. 50 ( 1 954), 309-3 1 8 . [ 2 J Step discontinuities in waveguides, EM- 7 7 ( 1 955). [3] Diffraction by a cylinder of finite length, Proc. Camb. Phil. Soc. 52 ( 1 956), 322-335.

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(I am indebted to H . Levine and L . Lewin for most of the following references. added in page proof. Relevant pages in this book are indicated at the end of each title. I would be grateful for notice of omissions from the bibliography and for reprints of papers. past and future. ) F. BERZ [ 1 ] * Reflection and refraction of microwaves at a set of parallel metallic plates. Proc. lnat. Elect. Engrs 98 pt. III ( 1 9 5 1 ) . 47-55. (Ex. 3 . 12. p . 1 3 3 and ex. 4. 1 1 . p . 1 74. ) G. F. CARRIER and R. C. DIPRIMA [ 1 ] On the unsteady motion of a viscous fluid past a semi-infinite flat plate. J. Math. Phys. 35 ( 1 956). 359-383. (Ex. 2 . 1 7. p. 9 7 . ) G. GRUNBERG [ 1 ] * Theory o f the coastal refraction o f electromagnetic waves. (in Russian) Akad. Nauk SSSR. Zh. Eksp. Teo. Fiz. 14 ( 1 944). 84-1 1 1 . (Ex. 2 . 12(A). p. 9 3 . ) A. E . HEINS and R. C. MACCAMY [ 1 ] * A function-theoretic solution of certain integral equations I. Quart. J. Math. (Oxford) 9 ( 1 958). 1 32-143. ( §4.3. • p. 147.) C. LAFLEUR and V. NAMIAS [ 1 ] Sur la resolution de l'equation de Wiener Hopf basee sur l'utilisation des proprietes formelles des fonctions «5+ et «5_. Acad. Roy. Belg . Bull. Ol. Sci. (5) 40 ( 1 954). 787-790. G. E . LATTA [1] A note on the Wiener-Hopf technique for solving integral equations. Technical Report No. 56. Applied Mathematics and Statistics Laboratory. Stanford University ( 1 956). (Ex. 4.3. p. 1 6 9 . ) H . LEVINE [ 1 ] O n the theory o f sound reflection in a n open-ended cylindrical tube, J. Acoust. Soc. Amer. 26 ( 1 954). 200-2 1 1 . ( §3.4. p. 1 1 0. and p. 236.) [2] Diffraction by an infinite slit. Technical Report No. 6 1 . Applied Mathematics and Statistics Laboratory. Stanford University ( 1 9 5 7 ) . ( §5.6. p . 2 0 3 . and p. 2 3 6 . ) [3] Skin friction o n a strip of finite width moving parallel to its length. J. Fluid Mech. 3 (1957). 1 45-1 58. (p. 236.) H. LEVINE and T . T . Wu [2] The scattering cross-section of a row of circular cylinders. Technical Report No . 73. Applied Mathematics and Statistics Laboratory. Stanford University ( 1 958). (p. 236.) V. NAMIAS [ 1 ] Utilisation des proprietes formelles des fonctions impulsives «5+ et «5_ pour la discussion de l'equation de Wiener-Hopf. Acad. Roy. Belg . Bull. Cl. Sci. (5) 41 ( 1 955). 435-440. L . A. VAJNSHTEJN [7] Diaphragms in waveguides. (in Russian) Zh. Tekn. Fiz . 25 ( 1 955). 841-846. ( §3 . 6. p. 122 . ) [ 8 1 Diffraction o f electromagnetic waves b y a grid o f parallel conducting strips. (in Russian) Zh. Tekn. Fiz . 25 ( 1 955). 847-852. ( §3 . 6 . p. 122 . )

INDEX

( The letters ff. following a page number generally denote that an example has been worked in some detail. For notation and conventions see the summary on p. x. References in the text to papers in the biblio graphy are indexed under the author's name., Abelian theorems 36, 4 6 , 5 7 Absorbent boundaries (see Resistive boundaries) Aerofoil, quarter.infinite 9 5 ; two · dimensional 1 73 ALBLAS, J. B . 9 6 , 236 AMENT, W. J. 47 Analytic continuation 7ff. , 38ff. Antenna, biconical 2 1 8 ; cylindrical rod 2 1 3 Approximate factorization 1 60ff. Approximate methods 1 7 8ff. , 236 (see also Wiener-Hopf complex variable equation, generalized) ; for integral equations 2 3 1 ff. , 23 3ff. ; for strips and slits 20 3ff. ; for thick plates 1 8 7ff. ; meaning of 'approximate' 1 8 0 ; miscellaneous formulations 1 8 1ff. Asymptotic expansions, for contour integrals 33ff. , 46; for transforms 3 6 ; in approximate solutions 200ff. ; (see also Far.field) BAILIN, L. L. 122 BAKER, B . B . and COPSON, E . T . 7 0 , 206 BALDWIN, G. L. and HEINS, A. E . 1 2 5 BAZER, J. and KARP, S . N. 73, 1 2 8, 1 53, 166, 1 6 7 Biharmonic equation, (see also Elastic medium) approximate factorization for 1 6 2 ; stresses in strip 1 3 8 Boundary conditions, specified by arbitrary functions 7 7ff. ; (see also Resistive boundaries) Boundary layer on flat plate , 96, 9 7 BOUWKAMP, C. J. 75, 1 7 9, 2 1 8 Branch cuts 8ff. ; special cases 9, 19, 38, 39 Branch points and poles (see Poles and branch points) CARLSON, J. F. and HEINS, A. E . 1 03, 1 33

CARRIER, G. F. and MUNK, W. H. 94 CARSLAW, H. S . 72 CHESTER, W. 20, 1 3 2 CHUBCHILL, R . V. 2 7 Circular waveguide (see Duct, cylin· drical) CLEMMOW, P. C. 36, 8 8, 9 3 , 94, 1 52 (see also Vajnshtejn-Karp-Clem mow method) Cone, Laplace's equation 1 6 5 ; wave equation 2 1 8 Contour integrals, for sum and pro · duct decompositions 1 8ff. ; special type 3 1 ff. Convection of heat from flat plate 94 Conventions x, 1 0, 28, 86 COPS ON, E. T. vii, 5, 6, 70, 95, 2 3 1 Copson's method 2 3 1 Cosine transform 147, 2 1 2 Cylindrical pipe (see Duct, cylin· drical) DAVISON, B. 1 6 8 Dielectric slab 2 1 1 , 2 1 3 Diffraction, (see Disc, Duct, Half· planes, Strips and slits) methods other than the Wiener-Hopf for 125, 1 7 9 Dirac delta function 4 3 Disc, Laplace's equation 1 6 6 , 23 1 ; wave equation, high frequency 2 1 9 , - low frequency 2 1 3ff., 2 3 1 Divergent integrals 64 DOETSCH, G. 2 7 , 3 6 Dual integral equations, exact solution of general cases 7 7ff. , 222ff. ; for boundary conditions specified arbi· trarily 7 7ff. ; formulation in terms of 4, 58, 1 5 0 ; for three-part problems 23 1 ; general comments 48, 7 7 , 22 1 , 2 3 6 ; half· plane prob. lem solved by 5 8ff. ; involving Bessel functions viii, 22 1 ; multi· plying factor method for 5 8ff. , 7 7ff. , 222ff. ; reduction to complex variable equation 1 50, 1 5 1 , 221

244

IND EX

Duct, cylindrical, finite length 207 ; infinite length, resistive wall 1 34, semi-infinite tube in 12 1 ; propagated and attenuated modes 1 l 0 ; semi-infinite, electromagnetic waves 1 3 2 , - radiation from l 1 1ff_ , - radiation from coaxial 1 3 3 , 1 54, - wave incident on 1 1 6ff. Duct, two-dimensional, finite length 207 ; infinite length with semi infinite obstacle, dielectric slab 2 1 1 , - resistive wall 1 33, - set of posts 174, - solution via simultaneous linear algebraic equa tions 1 7 5 , - strip parallel to wall 1 1 8ff. , - strip of finite thickness parallel to wall 2 1 2 ; infinite length with step 1 9 6 , 207ff. ; infinite length with strip across 12 2ff. , 2 1 2 ; propagated and attenuated modes 98ff. ; semi infinite (see Half-planes, two parallel) Edge conditions 5 1 , 74ff. ; references 75 Elastic medium, (see also Biharmonic equation) diffraction by half-plane in 9 6 ; stresses in wedge with mixed boundary conditions 1 7 3 Electromagnetic waves, boundary conditions 1 34 ; diffraction by half-plane 9 5 ; diffraction by semi infinite cylinder 1 3 2 ; edge con ditions 7 6 ; terminology 85, 92 End correction, cylindrical tube 1 1 5 ; thick semi-infinite strip 1 9 6 ERDELYI, A. 36, 46 ERDELYI, A. et al. 148, 1 7 6, 2 1 3 Factorization (see Product factoriza tion) Far-field, examples of calculation of 105, 1 l 0, 1 1 5 , 1 92, 206 FEL'D, A. N. 1 7 4 FocK, V. 1 6 8 Fourier integral Iff. , 21ff. ; in com plex plane 2 3 ; generalized 42 Fourier transforms 2; basic example 53ff. Fox, E. N. 232 Fox's integral identity 232 Fredholm integral equations, first kind (see Wiener-Hopf integral equations) ; second kind 221, 23 1ff. , 233ff. , 2 3 6 Fresnel functions 3 3 , 4 5 , 73, 2 1 3 Function-theoretic methods 1 5 3

Gamma functions 4 0 ; factorization by means of 4 1 , 1 6 7 , 1 7 1 , 2 1 5 Green's function 6 Iff. GREENE, T. R. and HEINS, A. E . 9 4 Half-planes, infinite set o f staggered 1 3 3 ; single infinitely thin, basic (Sommerfeld) diffraction problem 48ff. , - diffraction of electro magnetic waves by 95, - fin ite resistivity 83, - generalizations of basic problem 5 1ff. , - in elastic or viscous medium 96, - solution by simultaneous linear algebraic equations 1 7 5 ; single of finite thickness, 1 8 1 , 1 87ff. , dielectric material 2 1 3 ; three parallel 1 5 4 ; two parallel, radia tion from 1 05ff. , - wave falling on 1 00ff. , 1 2 5 , 126, - with flange 1 8 3 ; two parallel, staggered 1 54 Hankel functions, asymptotic behav iour 4 5 ; integral representation 3 2 HARRINGTON, R. F . 88 HEINS, A. E. 94, 103, 1 2 7 , 1 3 1 , 1 3 3 , 1 34, 1 57 , 1 7 1 HEINS, A. E . and CARLSON, J . F . 1 3 3 HEINS, A. E . and FESHBACH, H. 92, 93, 1 3 4 Hilbert problem 4 9 , 141ff. ; integral equation application 1 69 HoPF, E. 1 6 8 IIJIMA, T. 1 32 Imperfect conductivity (see Resistive boundaries) Indentation of contour 30, 44, 8 6 ; for transform i n time 44 Infinite product 1 5 , 40, 103, 129, 1 30 ; asymptotic behaviour of 1 04, 128, 1 2 9 ; computation of 127, 1 2 8 ; for gamma functions 41 Institute of Mathematical Sciences, New York, reports 237 Integral equation method, general remarks vii, 49, 7 1 , 76, 236 Integral equations (see Fredholm, Volterra, Wiener-Hopf) Integral functions 6, 1 7 8, 1 9 7 Jets 1 3 9 JONES, D . S . vii, viii, 48, 52, 75, 9 5 , 9 7 , 1 2 7 , 1 30, 1 3 1 , 1 67 , 1 7 3, 1 8 7 , 1 9 � 1 9 � 1 9 � 1 9 9 , 20� 207, 2 3 1 Jones's method, detailed application to a special example 52ff. ; for approximate solutions 1 7 8ff. , 2 3 6 ; for boundary conditions specified arbitrarily 79f,f. ; general remarks

IN D E X

vii, 48, 7 1 , 76, 98; routine applica tions 100ff. JOST, RES 2 1 9

k = k t ± ik., 10, 29, 38, 44, 45, 86 KARP, S. N. 1 66, 167 (see also Vajnshtejn-Karp-Clemmow method) KARP, S. N. and RUSSEK, A. 2 0 6 KARP, S . N. and WILLIAMS, W. E . 125 KLUNKER, E . B . and HARDER, K. C . 139 KOlTER, W. T . 75, 1 39, 1 6 1 , 1 62 , 1 7 3 KOURGANOFF, V. 1 33 , 1 68 Laplace transform 1 0, 2 1 , 52, 8 6 ; complete equivalence with Fourier transform 22, 2 6 Laplace's equation, a s limit o f wave equation 1 3 5ff. ; in j et problem 1 4 0 ; in polar coordinates 1 64ft'. , 177 Laplace-type equations 1 3 5 (see also Biharmonic equation) Oseen's equations 97 LATTER, R . 2 3 6 Lebedev-Kontorovich transform 1 4 8 , 2 1 3ff. Lebesgue integration, reason for not using 23 LEITNER, A. andWELLS , C. P. 2 1 3 , 2 1 6 LENNOX, S . C . 1 39 LEVINE, H. 2 3 6 , 242 LEVINE, H. and SCHWINGER, J. 2 0 , 42, 1 1 5 , 1 30, 1 3 2 LEVINE, H. and Wu, T . T . 2 3 6 LEWIN, L . 1 3 1 , 1 79, 242 LEWIS , J. A. and CARRIER, G. F . 97 LINFOOT, E. H. and SHEPHERD , W. M. 1 75 Liouville's theorem 6, 38, 57, 69, 103, 1 1 4 MAGNUS, W. 1 7 5 MAGNUS, W. and OBERHETTINGER, F. 1 75 MARCUVITZ, N. 122, 133, 1 7 9 MARK, C. 1 6 8 MARSHAK, R. E . 1 68 MAUE, A. W. 75, 9 6 MEIER, J . A. and LEITNER, A . 2 1 8 MEIXNER, J . 7 5 Meromorphic functions 3 9 MILES, J . W. 9 5 Milne's integral equation 1 67 Mixed boundary value problem 1 ; solution of a specific example 77ff. ; three-part 1 83 , 203ff. , 2 3 1 ff. ; two-part, general theory 147

241)

MORSE, P. M. and FESHDAUH, H. 02, 1 1 5, 1 3 1 , 1 34, 1 79 Multiplying factor method 5811'. , 7 8ff. , 221ff. MUSKHELISHVILI, N. I. 142, 1 44, 1 46 , 158 Navier-Stokes equations, linearized 96 Notation x, 5 4 ; i n certain approxi mate methods 1 82 , 1 9 7 Numerical evaluation o f infinite products 1 2 7 , 1 2 8 Numerical tables viii, 1 2 2 , 1 3 1 OBERHETTINGER, F . 3 6 Oseen's equations 9 7 PACK, D . C. viii, 1 3 9 PAl, S . I. 1 3 9 PAPADOPOULOS , V. M . 1 34, 1 9 6 PEARSON, J. D . 1 32 Perpendicular boundaries 1 7 8, 1 8 1 , 1 8 7ff. , 2 07ff. PLACZEK, G. 1 6 8 PLACZEK, G. and SEIDEL, W. 1 68 Plemelj formulae 47, 145 POL, B . VAN DER and BREMMER, H. 27, 36 Polar coordinates, Laplace's equation 1 64ff. , 1 7 7 ; steady-state wave equation 148, 2 1 4ff. ; wedge of elastic material 1 7 3 Poles and branch points 4 7 , 1 33 , 207 Product factorization 1 4ff. ; approxi mate 160ff. ; general theorems 1 5 , 4 1 ; involving gamma functions 4 1 , 1 67 , 1 7 1 , 2 1 5 ; of exp ( - yd ) , 2 0 ; o f 1 + i6y, 1 + ic Y -I, 9 1 , 1 6 3 ; o f 1 ± exp ( - 2yb ) , 1 02ff. ; ofKt (ya)It ( ya) , 1 1 2 , 1 3 0 ; transfor mation ex = - k cos fJ for 47 Quantum mechanics, problem in 2 1 9 Quarter-plane aerofoil

reference

to

95

Radiation condition at infinity 2 , 3 1 Radiation-type boundary conditions (see Resistive boundaries) Radio-wave propagation over land sea interface 9 3 , 94 Reflection coefficient in duct 1 1 0, 1 1 4, 120, 1 2 4 Regularity, of functions defined by integrals l lff. ; of transforms in complex plane 1 2 Resistive boundaries, half-plane 83ff. ; notation 1 3 4 ; strip across duct 2 1 2 ; strip parallel t o walls o f duct 1 2 0 Rod, solid cylindrical, finite length 2 1 3 ; semi-infinite 1 9 6

246

INDEX

Scattering cross-section of disc 2 1 8 SCHWINGER, J_ vii Semi-infinite plane ( see Half-planes) SENIOR, T_ R A_ 83, 85, 92, 93, 1 34 Separation of variables 1 , 147, 149 Sinlultaneous linear algebraic equations, from simultaneous Wiener Hopf equations 1 5 5 ; in approxi mate solutions 185, 1 90ff- , 208, 2 1 7 ; in duct problem 1 7 5 ; in half-plane problem 176; solution of special sets 1 7 3, 1 74ffSimultaneous Wiener-Hopf equa tions, general theory 1 53ff- ; in electromagnetic cylindrical duct problem 1 32 ; in electromagnetic half-plane problem 9 5 ; in two dinlensional duct problems 1 2 1 , 1 54 Slits ( see Strips and slits) SNEDDON, L N_ viii, 2, 96, 1 68, 23 1 , 236 Sommerfeld, half-plane problem 48f[ ; radiation condition 3 1 Sound waves ( see Duct, Half-planes, etc _ ) Source distribution 44, 5 2 , 87 SPARENBERG, J_ A_ 1 69, 1 7 1 Steady-state wave equation 2 8 Steepest descent 34 Strips and slits of finite thickness, semi-infinite length 1 8 1 , 1 87ff_ , and finite length 2 1 3 , in duct 2 1 2 Strips and slits o f finite width, La place 's equation 1 7 7 ; wave equa tion, high frequency 1 83, 203ff_ , 2 1 9 , -- low frequency 2 1 8 Sum decomposition, function with poles 3 9 ; general formulae for concrete cases 1 8 ; general theorem 1 3 , -- applications of 80, 1 79, 1 86, 198, 228 ; in approxinlate solutions 1 79, 1 86 ; of Y = (0(.2 - k2)lj2 20, 4 7 ; transforma tion 0(. = -k cos fl for 47 Superposition, method of 8 1 , 87 Tables ( see Numf)rical tables) Tides in permeable rock 94 Time factor exp ( - iwt) 28 TITCHMARSH, E_ C. 2, 5, 8, 1 1 , 39, 4� 4 1 , 42, 1 6� 1 7 1 Titchmarsh's method 1 7 1 Transformation 0(. = - k cos fl, applied to diffraction by half-plane 1 7 6 ; for evaluation of integrals 3 3 ; for sum decompositions 47 ; in wedge problems 2 1 9 Transient problems 2 8 , 9 7 TRANTER, C. J. 2

Uniqueness 7 2 ; and edge conditions 5 1 , 74; of solution of integral equations 69, 1 68 VAJNSHTEJN, L. A. 20, 1 1 5, 125, 1 3 0, 131, 1 32, 1 5 3 Vajnshtejn-Karp-Clemmow method 49, 1 53, 2 1 3 Verification o f solution 7 2 Viscous medium 96, 9 7 , 1 3 9 Volterra integral equations 2 2 9 , 230, 233 Water waves 9 4 , 1 34 Watson, G. N. 36, 46 Wave equation 27ff. Waveguide ( see Duct, two -dinlen sional; Duct, cylindrical ; Half planes, two parallel) Wedges, diffraction by 2 1 9 ; of elastic material 173 WEITZ, M. and KELLER, J. B _ 1 34 Whittaker functions 200, 2 1 3 WIDDER, D . V . 27, 3 6 WIENER, N. and HOPF, E . 4 1 , 1 67, 1 68 Wiener-Hopf complex variable equa tion (for exact solutions) , basic procedure 3 6ff. ; derived by Jones's method 55, 79, 84, 1 0 1 , 125, 1 50, -- from an integral equation 68, 1 68, -- from dual integral equa tions 1 50, 1 5 1 , 22 1 ; general solutions 7 9ff. , 89, 222ff., 228; reduction to linear algebraic equa tions 1 74 ; solutions for simple cases 55ff. , 68ff. Wiener-Hopf complex variable equa tion, generalized (for approxinlate solutions) 1 7 8 ; formulations of 1 8 1ff. , 1 87ff. , 203ff. , 233, 2 3 6 ; solutions of 1 84ff., 1 9 6ff_ Wiener-Hopf equations, general con siderations 1 5 1 Wiener-Hopf integral equations, formulation by Green's functions 6 1ff_ , 89, 1 32, -- by physical reasoning 90, 1 3 1 , 1 7 7 , -- by transforms 4, 65ff. ; Milne's integral equation 167; reduction of more general types to Fredholm equations of second kind 232, 233ff. ; reduction to repeated Vol terra equations 2 3 0 ; solution by the Wiener-Hopf technique 67ff. , 1 6 7ff. , 2 2 7 , -- via the Hilbert problem 1 6 9 ; special equations 1 7 1 , 1 7 7 ; uniqueness of solution 69, 1 68 WILLIAMS, W. E. viii, 1 96, 207, 208

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