IEEE PRESS Series on Electromagnetic Waves ,
I
.
I
The IEEE PRESS Series on Electromagnetic Waves consists of new titl~s as well as reprints and revisions of recognized classics that maintain long-term archival significance in electromagnetic waves and applicatior/s.
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Series Editor Donald O. Dudley University of Arizona
Mathematical Foundations for Electromagnetic Theory
Advisory Board Robert E. Collin Case Western University
....
Akira Ishimaru University of Washington
Associate Editors Electromagnetic Theory, Scattering, and Diffraction Ehud Heyman Tel-Aviv University Differential Equation Methods Andreas C. Cangellaris UniversIty of Arizona
Donald G. Dudley University of Arizona, Tucson
Integral Equation Methods Donald R. Wilton Universi,ty of Houston Antennas, Propagation, and Microwaves David R. Jackson University of Houston
Series Books Published Collin, R. E., Field Theory of Guided Waves, 2nd ed., 1991 , Dudley, D.O., Mathematical Foundations for Electromagnetic Theory, 1994 Elliott, R. S., Electromagnetic.~: History, Theory, and Applications, 1993 Felsen, L. B., and Marcuvitz, N., Radiation and Scattering of Waves, 1994 Harrington, R. E, Field Computation by Moment Methods, 1993 Tai, C. T., Dyadic Green Functions in Electromagnetic Theory, 2nd ed., 1993 Tai, C. T., Generalized Vector and Dyadic Analysis: Applied Mathematics in Field Theory, 1991
IEEE PRESS Series on Electromagnetic Waves Donald G. Dudley, Series Editor
IEEE PRESS
I
lEEE Antenrias ancl Propagation Society, Sponsor
,IEEE PRESS 445 Hoes Lane, PO Box 1331 Piscataway, NJ 08855-1331 1994 Editorial Board William Perkins, Editor ill Chief J. D: Irwin S. V, Kartalopoulos P. Laplante A. LLaub M. Lightner
R. S. Blicq M.Eden D. M. Etter G. F. Hoffnagle R. F. Hoyt
J. M. F. Moura I. Pcden E. Sanchez-Sincncio L. Shaw D. J. Wells
Dedicated to Professor Robert S. Elliott
Dudley R. Kay, Director of Rook Puhlishing Carrie Briggs, Administrative Assistant Lisa Smey-Mizrahi. Review Coordinator
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Valerie Zaborski, Production Editor IEEE Antennas and Propagation Society, Spollsor AP·S Liaison to IEEE PRESS Robert J. Mailloux Rome Laboratory/ERI ,
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This book may be purchased at a discount from the publisher when ordered in bulle quantities. For more information contact: IEEE PRESS Marketing Attn: Special Sales P.O. Box 1331 445 Hoes Lane Piscataway, NJ 08855-1331
© 1994 by the Institute of Electrical and EI,ectronics Engineers, Inc. 345 East 47th Street, New York, NY 10017-2394 All rights reserved. No part of this book may be reproduced in any form, nor may it be stored in a retrieval system o~ transmitted in any fonn, without written permission from the publisher. Printed in the United Stales of America 10
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ISBN 0-7803-1022-5 IEEE Order Number: PC03715, Library of Congress Cataloging.in.Publication Data Dudley, Donal<. G. Mathematical found'ations for electromagnetic theory / Donald G. Dudley. p. ,;m. - (IEEE electromagnetic waves series) "IEEE Antennas and Propagation Society, sponsor." Includes bibliographical references and index. ISBN 0-7803-1022·5 I. Electro~agnetic theory-Mathematics. I. IEEE Antennas and Propagation Society." II. Title. III. Series.
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Contents
Preface
ix
1 Linear Analysis 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 A.I
1
Introduction I Linear Space I Inner Product Space 7 Normed Linear Space 10 Hilbert Space 15 Best Approximation 19 Operators in Hilbert Space 24 Method of Moments 33 Appendix-Proof of Projection Theorem Problems 38 References 43
2 The Green's Function Method 2.1 2.2 2.3 2.4 2.5 2.6
36
45
Introduction 45 Delta Function 45 Sturm-Liouville Operator Theory 50 Sturm-Liouville Problem of the First Kind 53 Sturm-Liouville Problem of the Second Kind 68' Stun:n-LiouviIle Problem of the Third Kind 77 Problems 94 References 97
I
:1
iI
I "
i
I
1
viii
Contents
3 The Spectral Representation Method 3.1 3.2 3.3 3.4 3.5
1'
99
Introduction 99 Eigenfunctions and Eigenvalues 99 Spectral Representations for SLP1 and SLP2 106 Spectral Representations for SLP3 111 Green's Functions and Spectral Representations 134 Problems 135 I References 138
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; :1
4 Electromagnetic Sources
139
4.1 "Introduction 139 4.2 Delta Function Transformations 139 4.3 Time-Harmonic Representations 143 4.4 The Electromagnetic Model 144 4.5 The Sheet Current Source 147 4.6 The Line Source 153. 4.7 The Cylindrical Shell Source 166 4.8 The Ring Source 168 4.9 Th'e Point Source 172 Problems 178 References 179
5 Electromagnetic Boundary Value Problems 181 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10
Introduction 181 SLP1 Extension to Three Dimensions 182 SLPI in Two Dimensions 191 SLP2 and SLP3 Extension to Three Dimensions 194 The Parallel Plate Waveguide 198 Iris in Parallel Plate Waveguide 206 Aperture Diffraction '216 Scattering by a Perfectly Conducting Cylinder 226 Perfectly Conducting Circular Cylinder 233 Dyadic Green's Functions 242 Problems 242 References 244
Index
246
Preface
This book is written for the serious student of electromagnetic theory. It is a principal product of my experience over the past 25 years interacting with graduate students in electromagnetics and applied mathematics at the University of Arizona. A large volume of literature has appeared since the latter days of World War II, written by researchers expanding the basic principles of electromagnetic theory and applying the electromagnetic model to many important practical problems. In spite of widespread and continuing interest in electromagnetics, the underlying mathematical principles used freely throughout graduate electromagnetic texts have not been systematically presented in the texts as preambles. This is in contrast to the situation regarding undergraduate electromagnetic texts, most of which contain preliminary treatments of fundamental applied mathematical principles, such as vector analysis, complex arithmetic, and phasors. It is my belief that there should be a graduate electromagnetic theory text with linear spac'es, Green's functions, and spectral expansions as mathematical cornerstones. Such a text should allow the reader access to the mathematics and the electromagnetic applications without the necessity for consulting a wide range of mathematical books written at a variety of levels. This book is an effort to bring the power of the mathematics to bear on electromagnetic problems in a single teXt. ' Since the mastery of the foundations for electromagnetics provided in this book can involve a considerable investment of time, I should like to indicate some of the potential rewards. When the student first begins a
x
,
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lfu~
study of electromagnetic theory at the graduate level, he/she is 'con'fron ted with a large array of series expansions and transforms with which to reduce the differential equations, and boundary conditions in a wide 'variety of canonical problems in Cartesian, cylindrical, and spherical coordin,ates. Often, it seems to the student that experience is the only way to determine specifically which expansions or transforms to use in a given problem. In addition, convergence properties seem quite mysterious. These isstles can be approached on a firm mathematical base through the foundations provided in this book. Indeed, the reader will find that different diff~rential operators with their associated boundary conditions lead to specificexpansions aRd transforms that are "rtatural" in a concrete mathematical sense for the problem being considered., My experience with graduate students has been that mastery of the foundations allows them to appreciate why certain expansions and transforms ar~ used in the study of canonical problems. Then, what is potentially more important, the foundations allow them to begin the more difficult task of formulating and solving problems on their own. I first became interested ih Green's functions and spectral representations during my graduate studies at UCLA in the I960s. I was particularly influenced by the treatment of the spectral representations of the delta function by Bernard Friedman [l], whose book at that time formed the cornerstone of the Applied Mathematics Program at UCLA in the Coilege of Engineering. Subsequently, examples of spectral representations began appearing in texts on electromagnetic theory, such as [2]-[4], and, more recently, [5]. However, no te~t specifically devoted to Green's functions and spectral expansions and their application to electromagnetic problems has been forthcoming. , . The material in this book forms a two-semester sequence for graduate students at the University of Arizona. The first three chapters contain the ; : I mathematical foundations, and are covered in a course offered every year to electrical engineering and applied mathematics graduate students with a wide range of interests. Indeed, the first three chapters in th~s book could be studied by applied mathematicians, physicists, and engineerswith no particular interest in the electromagnetic applications. The fourth and fifth chapters are concerned with the electromagnetics, and are ~overed in a course on advanced electromagnetic theory, offered biennially. In this book, I have presumed that the reader has a working knowledge of complex variables. In addition, in the last two chapters, I have assumed that the reader has studied an introductory treatment of electromagrietics at the graduate level, as can he found, for example. in the texts hy Harhngton !
Preface
xi
[6], Ishimaru [7], or Balanis [8]. I have therefore felt no necessity to include a chapter on Maxwell's equations or a chapter on analytic function theory, presupposing reader familiarity. Chapter I is an introduction to modem linear analysis. It begins with the notion of a linear space. Structure is added by the introduction of the inner product and the norm. With the addition of suitable convergence criteria, the space becomes a Hilbert space. Included in the discussion of Hilbert space are the concepts of best approximation and projection. The chapter concludes with a discussion of operators in Hilbert space. Emphasis is placed on the matrix representation of operations, a concept that leads naturally to the Method of Moments, one of the most popular techniques for the numerical solution to integral equations occurring in electromagnetic boundary value problems. Chapter 2 covers Green's functions for linear, ordinary, differential operators of second order. The chapter begins with a discussion of the delta functi9n. The Sturm-Liouville operator is introduced and discussed for three cases, which we title SLPl, SLP2, and SLP3. A clear distinction is made between self-adjoint and nonself-adjoint operators. In addition, the concepts of:limit point and limit circle cases are introduced and explored through exahlples applicable to electromagnetic problems. . Chapter 3 introduces the spectral representation of the delta function. The theory is applied by example to various operators and boundary Conditions. Included are important representations associated with the limit point and limit circle cases introduced in the previous chapter. A wide variety of spectral representations are presented in a form suitable for use in solving electromagnetic boundary value problems in multiple dimensions. These representations are augmented by further examples in the Problems. Chapter 4 contains a discussion of fundamental electromagnetic sources represented by delta functions. The sources are analyzed using spectral representations and Green'8 functions in Cartesian, cylindrical, and spherical conditions. A variety of useful alternative representations emerge. Included are sheet sources, line sources, ring sources, shell sources, and point sources. In Chapter 5, the ideas developed in the previous chapters are applied to a sample of electromagnetic boundary value problems. No attempt is made to produce an exhaustive collection. Rather, the purpose of the chapter is to demonstrate the power of the structure developed in the first three chapters. Static problems included involve the rectangular box and rectangular cylinder. Dynamic problems include propagation in a parallel plate waveguide, scattering by an iris obstacle in a parallel plate waveguide,
xii
Preface
aperture diffraction, and scattering by a conducting cylinder. Emphasis has been placed on the power of alternative representations by including useful alternatives in the examples on :the parallel plate waveguide and scattering from a conducting circular cylinder. . My graduate students over the past 25 years have had a major influence on this book. All have contributed through classroom and indivjdual discussions. Many too numerous to mention have made suggestions and corrections in early drafts. Specifically, I should like to acknowledge some special help. In the early 1980s; K. A. Nabulsi and Amal Nabulsi painstakingly typed a portion of my handwritten class notes. These typed notes were produced before the adv~nt of modem computational word processors, and formed the basis for my subsequent writing of Chapters 1-3 of i this book. Dr. Nabulsi, now a Professor in Saudi Arabia, sent me a gifted student, Muntasir Sheikh, for doctoral training. Mr. Sheikh has cr~tiGally read the entire book manuscript and offered suggestions and corrd:tibns. In addition, Charles Trantanella, Michael Pasik, and Jacob Adopley have carefully read portions of the manuscript. In the mid-1970s, I had the good fortune to be a part of the creation of the now greatly successful Program in Applied Mathematics at the University of Arizona. W. A. Johnson was my first student to graduate dlrough , . the program. Because of him, I became acquainted with three professors in the Department of Mathematics, C. L. DeVito, W. M. Greenlee, ~nd W. G. Faris. These four mathematicians have had a lasting influence pn the way I have come to consider many of the mathematical issues involved in electromagnetic theory. • ! Among my colleagues, there are several who have had a ~1arked influence on this book. R. E. Kleinman, University of Delaware, h~s consistently encouraged me to pursue my mathematical interests appiied to electromagnetic theory. L. B.Felsen, Polytechnic University, haS, influenced me in many ways, scientifically and personally. In addition, his comments concerning modem research applications led me to some important additions in Chapter 5: K. J. Langenberg, University of Kassel, has read in detail the first three chapters and offered important advice and criticism. R. W. Ziolkowski, University of Arizona, has taught a tourse using the material contained in Chapters 1-3 and offered many suggestions and corrections. I. Stakgold, University of Delaware, made me aware of the recent mathematicalliteratJre on limit point and limit circle problems. Many reviewers, anonymbus and known, have made comments that have led me to make changes ~nd additions. I would particularly like to mention Ehud Heyman, Tel Aviv University. whose comments concerning I
Chap. 1
xiii
References
alternative representations led me to strengthen this material in Chapter 5. I would also like to thank Dudley Kay and the staff at IEEE Press whose competence and diligence have been instrumental in the production phase of this book project. With Chalmers M. Butler, Clemson University, a distinguished educator and cherished friend, I have had the good fortune, to have a 20year running' discussion concerning methods of teaching electromagnetics to graduate students. Part of the fun has been that we have not always agreed. How:ever, one issue upon which there has been no disagreement is the importance of presenting electromagnetics to students in a structurklly organized manner, stressing the common links between wide ranges of problems. I have drawn strength, satisfaction, and pleasure from our association. My family has always seemed to understand my many interests, and this book has been a major one for more years than I should like to recall. It is with love and affection that I acknowledge my wife, Marjorie A. Dudley; my children, Donald L. Dudley and Susan D. Benson; and the memory of my former wife, Marjorie M. Dudley. Love truly does "make the world go 'round." Finally, it is with gratitude that I dedicate this book to my teacher, mentor, and friend, Robert S. Elliott, University of California at Los Angeles, a consummate scholar without whom none of this would have occurred.
REFERENCES [I] Friedman, B. (1956), Principles and Techniques of Applied Mathematics. New York: Wiley. [2] Jackson, J. D. (1962), Classical Electrodynamics.' New York: Wiley. [3] Collin, R. E. (1960, 2nd edition 1991), Field Theory of Guided Waves. New York: IEEE Press. [4] Felsen, L. B., and N. Marcuvitz (1973), Radiation and Scattering of Waves. Englewood Cliffs, NJ: Prentice-Hall. [5] Ishimaru, A. (1991), Electromagnetic Wave Propagation, Radiation, and Scattering. Englewood Cliffs, NJ: Prentice-Hall. [6] Harrington, R. F. (1961, reissued 1987), Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill. [7] Op. cit. Ishimaru. [8] Balanis, C. A. (1989), Advanced EnRineerinR ElectromaRlletics. New York: Wiley.
1 Linear Analysis
1.1 INTRODUCTION Fundamental to the study of many of the differential equations describing physical processes in applied physics and engineering is linear analysis. Linear analysis can be elegantly and logically placed in a mathematical structure called a linear space. We begin this chapter with the definition of a linear space. We then begin to add structure to the linear space by introducing ,the concepts of inner product and norm. Our study leads us to Hilbert space and, finally, to linear operators within Hilbert space. The characteristics of these operators are basic to the ensuing development of the differential operators and differential equations found in electromagnetic theory. Throughout this chapter, we shall be developing notions concerning vectors in a linear space. These ideas make use of both the real and complex number systems. A knowledge of the axioms and theorems governing real and complex numbers will be assumed in what follows. We shall use this information freely in the proofs involving vectors.
1.2 LINEAR SPACE Let a, b, c, be elements of a set S. These elements are called vectors. Let a, {J, he elements of the field of numhers F. In particular, let Rand
Linear Analysis
2
. Chap. 1
Sec. 1.2
3
Linear Space
I
,
.
C be the field of real and complex numbers, respectively. The set S is a linear space if the following rules for addition and multiplication apply: ,
I. Rules for addition among vectors in S:
= +
a. (a + b) + c a (lJ + c) I b. There exists a zero vector 0 such that a + 0 = 0 + a = a.i c. For every a E S, there exists -a E S such that a + (-a)
•
EXAMPLE 1.3 Consider C(O,I), the space of real-valued functions continuous on the interval (0, 1). For f and g in C(O, I) and a E R, we define addition and multiplication as follows:
(-a) +a = O. d. a +b = b+a
II. Rules for multiplication of vectors in
.
S by elements of F:
(f
a. a({3a) = (a{3)a =a c. a(a+b)=aa+ab: d. (a + {3)a = aa + {3a
+ g)(~) = f(~) + g(~) (Cif)(~)
b. la
EXAMPLE 1.1 follows:
EXAMPLE 1.2 Consider unitary space CIt. Vectors in the space are given by (1.1) and (1.2), where Cik and f3ko k = I, 2, ... ,11 are in C. Addition and multiplication are defined by (1.3) and (1.4) where a E C. Proof that CIt is a linear space follows the same lines as in Example 1.1. Note that C is a linear space, where we make the identification C = C I .
Consider Euclidean space R n • Define vectors a and bin R n as a = (CiI,CiZ,
,Cin)
b == (f3I, f3z,
, f3n)
(f
+ g)(~)
=
where Cik and fh, the components of vectors a and h, are in R, k =, 1,2, ... , n. Define addition and multiplication as follows: '
=
+b =
(Cil,···, an)
= (al
:= (CiCiI'
(1.3)
, Cin) , CiCi,,)
(IA)
where a E R. If we assume prior establishment of rules for addition and I)1ultiplication in the field of real numbers, it is easy to show that R n is a linear space. We must show that the rules in I and II are satisfied. For example, for addition rule d, a
+b=
(Cil
= ,({31
+ f31, ... , Ci" + f311) + CiJ, ••. , f311 + Ci n )
=b+a
We leave the completion of the proof for Problem 1.3.
+ (f3I,.··, f3n)
+ f3J' ... ,Cin + f3n)
Cia ;::: Ci(CiI'
+ g(~) g(~) + f(~) (g + f)(~)
= f(~)
(1.2)
I
a
(1.8)
•
In ordinary vector analysis over two or three spatial coordinates, we are often concerned with vectors that are parallel (collinear). This concept can be generalized in an abstract linear space. Let XI, X2, ... ,Xn be elements of a set of vectors in S. The vectors are linearly dependent if there exist ak E F, k I, 2, ... , n, not all zero, such that
=
(1.9)
If the only way to satisfy (1.9) is ak = 0, k Xk are linearly independent. The sum
= 1,2, ... , n, then the elements
(1.5)
We leave the satisfaction of the remainder of the rules in this example for Problem 1.2. Note that R is also a linear space, where we make the identification R = R I.
•
(1.7)
for all ~ E (0, 1). If we assume prior establishment of the rules for addition of two real-valued functions and multiplication of a real-valued function by a real scalar, it is easy to establish that C(O, 1) is a linear space by showing that the rules in I and II are satisfied. For example, for addition rule d,
;(1.1) ,
= af(~)
( 1.6)
is called a !inear comhination of the vectors Xk .
Linear Analysis
4
EXAMPLE 1.4 In R z, let Xl dependencd. We fonn
(l, 3), Xz
=
: Chap. I
(2,6). We test Xl and Xz f?r linear
from which we conclude that
:at+ 2a z=0 3at
+ 6a2 = 0
Sec. 1.2
5
Linear Space
In an abstract linear space S, it would be helpful to have a measure of how many and what sort of vectors describe the space. A linear space S has dimension n if it possesses a set of n independent vectors and if every set of n + 1 vectors is dependent. If for every positive integer k we can find k independent vectors in S, then S has infinite dimension. The set Xl, X2, ... ,Xn is a basis for S provided that the vectors in the set are " linearly independent, and provided that every XES can be written as a linear combination of the Xb viz.
I
These two equations are consistent and yield al = -2a2. Certainly, at ='az = 0 satisfies,this equation, but there is 'also an infinite number of nonzero possibilities. The vectors are therefore linearlY,dependent. Indeed, the reader can easily make a sketch to show that XI and Xz are collinear.
•
EXAMPLE 1.5 InC(O,I), let a set of vectors be defined by fk(~) = .j2 sin kTr~, k = 1,2, ... ,n. We test the vectdrs
!k
(1.12) The representation with respect to a given basis is unique. If it were not, then, in addition to the representation in (1.12), there would exist 13k E F, k = 1, 2, ... , n such that n
for linear dependence. We fonn
= L13kxk
X
n
L~kJ2sinkTr~ = 0
(1.10)
k=l
where ak That is,
E
R. The
fb
defined above, fonn an orthonormal set on ~
(1.13)
k=l Subtraction of (1.13) from (1.12) yields n
0= L(ak - 13k)xk
(0, I).
E,
(1.14)
k=l Since the Xk are linearly independent, we must have
, (1.11)
ak - 13k Multiplication of both sides of (1.1 0) by .j2 sin mn ~, m = I, 2, ... , n and integration over (0, I) give, with the help of (l.11), am = 0, m = I, 2, ... ,n. The vectors fk are therefore linearly independent.
•
In Example 1.5, we note that the elements !k are finite in number. We recognize them as a finite :subset of the countably infinite nu~ber of elements in the Fourier sine series fk, k = 1, 2, .... A question arises concerning the linear independence of sets containing a countably infinite number of: vectors. Let Xl, X2: . .. be an infinite set of vectors in S. The vectors are linearly independent if every finite subset of the vedtors is linearly independent. In Example 1.5, this requirement is realized, so that the infinite set of elements present in the Fourier sine series is linearly independent.' I
= 0,
k
= 1,2, ... , n
( 1.15)
which proves uniqueness with respect to a given basis. 'Finally, if S is n-dimensional, any set of n linearly independent vectors Xl, X2, .. X n forms a basis. Indeed, let XES. By the definition of dimension, the set X, Xl, X2, ... , X n is linearly dependent, and therefore, I I
,
n
ax
+ Lakxk = 0
(1.16)
k=l where we must have a
i
0. Dividing by a gives
(-a x=L -a k=l n
Therefore, the set Xl,
X2, ... , X'I
k)
is a basis.
Xk
(1. 17)
Linear Analysis
6
I
Chap. I
EXAMPLE 1.6 Consider Euclidean space R n • We shall show that the vectors e\ = (1,0, ... , 0),e2 = (0,1,.;.,0), , en = (0,0, ... ,1) satisfy the two requirements for a basis. First, the set el, , en is independent (Proble~ 1.8). Second, if Q'E R n ,
a
= (at, a2,
, an)
= a\ (I, 0,
, 0)
• EXAMPLE 1.7 It would be consistent with notation if the dimension of R n were, in fact, n. We now show that both r~quirements for dimension n are satisfied. First, since we have established an n-term basis for R n in Example 1.6, the space has a set of n independent vectors. Second, we must show that any set of n + I vedors is dependent. Let aj, a2, ... , an, ~n+ I be an arbitrary set of 11 + 1 vectors in, R n • We form the expression n+1
LYmam = 0
Expression (1.22) is a homogeneous set of n linear equations in n + 1 unknowns. The set is underdetermined, and as a result, always has a nontrivial solution [IJ. There is therefore at least one nonzero coefficient among Ym, m = 1, 2, ... , n + 1. The result is that the arbitrary set ai, ... , an, all + I is linearly dependent and the dimension of R II is n.
where we must show that there exist Ym E R, m = 1, 2, ... , 11 + 1, not aJI zero, such that (1.19) is satisfied. We express each of the members of the arbitrary set as a linear combination of the basis vectors, viz. m
=
1, 2, ... , n
+1
:(1.20) I '
Substitution of (1.20) into (1.19) and interchanging the order of the summations gives
t (~Yma~ml)
ek
=0
(1.21 )
0,
a. (x, y)
=
(y, x)
b. (x +y, z) ! c. (ax, 'y)
=
(x, z)
= a(x, y),
+ (y, z) -
a EC
d. (x, x) ::: 0, with equality if and only if x = 0
In a, the overbar indicates complex conjugate. Similar to the above is the real inner product space, which we produce by eliminating the overbar in a and requiring in c that a be in R. For the remainder of this section, we shall assume the complex case. We leave the reader to make the necessary specialization to the real inner product. EXAMPLE 1.8 that
We show from the definition of complex inner product space (0, y) =
°
Indeed, the result follows immediately if we substitute a
(1.23)
= 0 in rule c above.
•
EXAMPLE 1.9 Given the rules for the complex inner product in a-d, the following result holds: (1.24) (x, ay) = a(x, y)
(x,ay)
= (O'y,x) = O'(y, x) = a (y, x)
i ~
A linear space S is a complex inner product space if for every ordered pair (x, y) of vectors in S, there exists a unique scalar in C, symbolized (x, y), such that:
Indeed,
Since the ek are linearly independent,
n+l ' L" YmO'k(ml
1.3 INNER PRODUCT SPACE
:(1.19)
m=1
m=1
7
•
Therefore, any vector in the space can be expressed as a linear combination of the ek. A special case of this result is obtained by considering Euclidean space R 3 . The vectors el = (1, 0, 0), e2: = (0, 1, 0), e3 = (0, 0, 1) are a basis. These vectors are perhaps best known as the unit vectors associated with the Cartesian coordinate system.
.
Inner Product Space
+ a2 (0, 1, ... , 0) + ... + an (0, 0, ... , I) (1.18)
,
Sec. 1.3
k = 1.2, ... , n
~(1.22)
= a(x, y)
•
.j
Linear Analysis
8
EXAMPLE 1.10
Chap. 1
Given the rules: for the inner product space, we may shqw that n
n
('L akx~, y) = 'L adxk, y) k=]
(1.25)
Sec. 1.3
Inner Product Space
Herein, we refer to (1.28) as the CSB inequality. To prov~ the CSB inequality, we first note that for I(x, y) I = 0, there is nothing to prove. We may therefor~ assume y i= 0, with the result (y, y) i= 0, and ,define
k=1
The proof is left for Problem 1.9.
•
(x, y) a=-(y, y)
n
'L ak~k
=
(a, b) ,
j
II
(x, y}(y, x)
(y, y)
(y, y)
i l I' "i
= a(y,x}
11 ~i
k=l
Then, C n is a complex inner product space. To prove this, we must show that rules a-d for the complex inner product space are satisfied. For rule d, there are three parts to prove. First, we show that the inner product (a, a) is nonnegative. Indeed,
Ii I
,
I(x, y}1 2
,0 .26)
-I
.!i i
from which we have the result
EXAMPLE 1.11 In the space C~, with a and b defined in Example 1.2, de'fine an inner product by
r
9
= a(x, y}
= laI 2 (y, y}
(1.29)
With the help of rule d, we form
n
(a, a)
=
'L lakl
2
::::
O:s (x - ay, x - ay) = (x, x)
0
+ laI 2 (y, y}
k=l
)
Second, we ~how that (a, a)
= 0 implies that a = O. ,
= (x,x)
We have
n
=
0= (a, a)
'L lad k=l I
Since all the terms in the sum are nonnegative, ak = 0, k = 1,2, ... , 'n, and therefore a = O. Third, we must show that a = 0 implies (a, a) = O. wb leave the this for the reader. We also leave the reader to demonstrate that rules a-c ! inner product space are satisfied.
EXAMPLE 1.12 by
(x, y)
The set
(j, g) =
i
fl
f(~)g(~)d~
t
k
=
=0
0.30)
1, 2, ... is an orthogonal set if, for all members of the set,
ii=j
= 0,
(1.31 )
( 1.32)
•
One of the most importantinequalities in linear analysis follows from the basic rules for the complex inner product space. The C auchy-SchwarzBunjakowsky inequality is given by
Y}I:s ~~
".:
The set is an orthonormal set if
Then, C(a, (3) is a real inner product space. We leave the proof for Problem 1.10.
I(x,
Zk,
(Zi, Zj)
Let f and g be iwo vectors in C(a, (3). Define an inner product :
2
from which the result in 0.28) follows. Two concepts used throughout this book involve the notions of orthogonality and orthonormality. The concepts are generalizations of the ideas introduced in Example 1.5. Two vectors x and yare orthogonal if
ror
•
_ l(x,y}1 (y, y)
- a(x, y} - a(y, x}
(1.:28)
where Oij
I
i
=j
0,
i
i= j
=( '
(1.33)
An orthogonal set is called proper if it does not contain the zero vector. We can show that a proper orthogonal set of vectors is linearly independent. Indeed, we form
Linear Analysis
10
Chap. 1
Sec. 1.4
II
11
Normed Linear Space
Using the CSB inequality, we obtain
LQHk =0 k=)
Taking the inner product of both sides with Zi gives Taking the square root of both sides yields the result in c.
n
(LakZb Zi)
= (0, Zi)
EXAMPLE 1.13 we can show that
k=)
From the basic rules for the norm and the definition in (1.34),
Using (1.25) and (1.31), we obtain adZi, Zi}
=
(1.36)
°
from which'we conclude that ai= 0, i = 1,2, ... , n and the set is li!learly independent. Further, if the index n is arbitrary, the countably infinite set Zb k = 1,2, ..., is linearly independent.
Indeed,
1.4 NORMED LINEAR SPACE
from which the result in (1.36) folIows.
IIx
A linear space S is a normed linear space if, for every vector XES; there is assigned a unique number IIx1l E R such that the following rules apply:
IIxll
2
L
lakl 2
(1.37)
•
(1.35)
EXAMPLE 1.15 For the real linear space C(a, fJ), with inner product defined by (1.27), the norm of a vector f in the space is
re~uire
(1.38)
+ y)
fJ) is therefore a normed linear space.
•
One of the useful consequences of the normed linear space is that it provides a measure of the "closeness" of one vector to another. We note from rule a that IIx - yll = 0 if and only if x = y. Therefore, closeness can be indicated by the relation IIx - y II < E. This observation brings us to the notion of convergence. Among the many forms of convergence, there
Since the real part of a complex ~umber is less than or equal to its magnitude, :::::
- y, x' - y)
•
(1.34)
= (x, x) + (y, y) + (x, y) + (y, x) = IIxf + IIYll2 + 2Re(x, y} IIx + )'11 2
= (x + y, x + y) + (x = 2(x, x} + 2(y, y}
Unitary space is therefore a normed linear space.
The space C(a,
(x +,y,x
2
k=1
It is easy to show that the norm defined by (1.34) meets the ments in rules a, b, and c above. We leave the reader to show that a: and b are satisfied. For c, for x and y in S, we have
IIx + Yll2 =
YII
n
Using (1.34), we find that the c'SB inequality in (1.28) can be writt~n :
Y}! ::::: IIxlillyli
+ IIx -
Iiall =
Although there are many possible definitions of norms, we use exclusively the norm induced by the inner product, defined by
I(x,
2
EXAMPLE 1.14 For unitary space en, with inner product defined by (1.26), the norm of a vector a in the space is easily found to be
IIx II ::: 0, with equality if and only if x = 0 b. lIaxli = lalllxll, a E F c. IIx) + x211 ::::: IIx)1I + 1IX:211 (triangle inequality) a.
IIxll=~
+YII
+ 1I.v11 2 + 21(x, y}1 I.
li '
Linear Analysis
12
,
~
,
.
Chap. I
are two fonns whose relationship is crucial to placing finn "boundaries" on the linear space. The type of boundary we seek is one that assur¢s that the limit of a sequence in the linear space also is contained in the space. In a normed linear space 5, a sequence of vectors {xd~l con}'etges to a vector x E 5 if, given an € > 0, there exists a number N such that IIx - Xk II < E whenever k > N. We write Xk -+ x or
=x
Sec. 1.4
Nonned Linear Space
13
We can show that convergence implies Cauchy convergence. Indeed, let x E 5 be defined as the limit of a sequence, as in (1.39). Then, by the triangle inequality, IIx m
-
XII II ~ IIxl1l
IIx and therefore, for min(m, n) > N,
!
Note that if Xk -+ x, IIx - xkll -+ O. Fundamental to studies of approximation of one vector by. another vector, to be studied later in this chapter, is the notion of continuity of the inner prod~ct. We show that if {xd~l is a sequence in 5 converging to x E 5, then
j
(Xk, h) -+ (x, h)
I !''''
I J I
t
i
!
!
(l.40)
where h is any vector in 5. To prove (1.40), it is sufficient to show tha,t
or
x,
(Xli; -
h) -+
0
(1.41 )
By the fonn of the CSB inequality in (1.35), we have I(Xk -x,h}1
But, since;n -+
2
~
Ilxk
_x11 2 11h11 2
X,
IIxk -xlI-+ 0
so that (1.41) is verified. We remark that another useful way of writing (1.40) is as follows: lim (Xk, h)
k---'>oo
= (k---'>oo lim :q, h)
(1.42)
This relationship indicates that, given Xb the order of application of the . limit and the inner product with h can be interchanged. In 5, a sequence {Xk }~l converges in the Cauchy sense if; given an E > 0, there exists a number N such that IIx m - Xn II < E whenever min(m, n) > N. We write lim IIx n,.n--. 00 l1l
-
XII II
=0
xnll ~
'I
Ij
E
2:
.,1
which proves the assertion. Unfortunately, the converse is not always true. The interpretation is that it is possible for two members of the sequence to become arbitrarily close without the sequence itself approaching a limit in S. A nonned linear space is said to be complete if every Cauchy sequence in the space convergcs to a vector in the space. The concept of completeness is an important one in what is to follow. Although it is beyond the scope of this book to include a detailed treatment, we shall give a brief discussion. In real analysis, the space of rational numbers is defined [2] as those numbers that can be written as p / q, where p and q are integers. It is a standard exercise [3],[4] to produce a sequence of rational numbers that has the Cauchy property and yet fails to converge in the space. (We consider an example in Problem 1.15.) This incompleteness is caused by the fact that in between two rational numbers, no matter how close, is an infinite number of irrational numbers; often, a sequence of rationals can converge to an irrational. The solution to this problem is a procedure due to Cantor [5] whereby the irrationals are appended to the rationals in such a manner so as to produce a complete linear space called the space of real numbers R. We shall assume henceforth that R is complete, and direct the reader to the literature in real analysis for details. EXAMPLE 1.16 We can show that Euclidean space R n is complete. For vectors a and b in the space, defined by (1.1) and (1.2), we define an inner product by n
(a, b) = Ladh
(1.44)
k=!
Let am, m
=
i
1.2.... be a Cauchy sequence in R n , where t
(1.43)
Xn II
Since Xk -+ x, there exists a number N such that for n > N
( 1.39)
lim Xk
(400
+ IIx -
x II
-
_ (ml u
am -
j
(m)
.CJ 2
• ...
i
il il
(ml)
,a"
14
Linear Analysis
Then,
lIam
-
apli =
It
\aim J -
Sec. 1.5
[akmJ -
111m - Ikll 2
aiPJrj2 ~ E
lim
m,k"""""'oo
I
te~s in the sum are nonnegative, we must h~ve
ak pJ I ~ E,
15
Hilbert Space
1
k=1
for min(m, p) > N. Since all the
' Chap. I
i
k = I, 2, ... , n
-7
a.
•
111m - lkil
= 0
which proves that the sequence is Cauchy. However, it is apparent that the sequence Ik converges to the Heaviside function H (~), defined by
for min(m, p) > N. Since the space of real numbers R is complete, tHen as ! m -700, k = 1,2, ... , n and therefore am
(I kI)
~ max ; '
H(~)
But, H
=
I
0,
~
I,
~>O
(1.46)
(n rf- C(-I ,I). Therefore, the space C(a, f3) is not complete. y
t
EXAMPLE 1.17 We can show that the normed linear space C(a, f3) with norm given by (1.38) is incomplete. We shall consider a well-known [6],[7] Cauchy sequence that fails to converge to a vector in the space. Without loss of gene.rality, let (a, f3) be (-1,1). Consider the sequence 1
-I ~ ~ ~ 0
O~~~t
(l.45)
t~~~I where k = 1,2, .... This sequence is continuous, and therefore is in the .linear space C(-I,I). We show that this sequence is Cauchy. We form the difference between two members of the sequehce. For m > k, 0,
1m -
Ik
=
-I ~ ~ ~ 0
(m - k)~,
O~~~~
(l-kn
..!.. < t <.!.
0,:
m -
<; -
Fig. 1-1
t~~~I
Ikll 2 =
11k
Two members /k and 1m of sequence in (1.45) for the case ,m > k.
•
k
I-I.
We display the two sequence members /k and 1m in Fig. Note that .the difference 1m - Ik is always less than unity. It follows that unity is an upper bound on (fm - Ik)2. We therefore must have
111m -
11m
[II (fm(~) - Ik(~))2d~ ~ ~
Although the above result has been obtained for m > k, an interchange of /11 and k gives the general result '
1.5 HILBERT SPACE A linear space is a Hilbert space if it is complete in the norm induced by the inner product. Therefore, in any Hilbert space, Cauchy convergence implies convergence. From Example 1.16, Euclidean space R n is complete in the norm induced by the inner product in (1.44). Therefore, R n is a Hilbert space. In a similar manner, it can be shown that unitary space ell is complete. However, from Example 1.17, C(a, fJ) is incomplete. In a manner similar to the completion of the space of rational numbers, C can
Linear Analysis
16
~hap. 1
be completed. The result is £2(ex, (3), the Hilbert space of real functions f (~) square integrable on the i~terval (ex, (3), viz.
i j2(~)d~ f3
<
00
;0 .47)
Sec. 1.5
Hilbert Space
17
The next member of the sequence is generated by (1.50)
and Z2
e2= - -
In (1.47), the integration is to be understood in the Lebesgue sen~e [8]. Although the Lebesgue theory is essential to the understanding of the proof of completeness, in this book sJch proofs will be omitted. For discussions of the issues involved, the reader is directed to [9] ,[ 10]. In linear analysis, we are often concerned with subsets of vectors in a linear~pace. One such subset is called a linear manifold. If S is a linear space and ex, f3 E F, then M is a linear manifold in S, provided that exx + f3y are in M whenever x and y are in M. It is easy to show that M is a linear space. The proof is left for Problem 1.16. EXAMPLE 1.18 In R 2 , the set of all vectors in the first quadrant is not a linear manifold. Indeed, for x and y in the first quadrant, it is easy to find a, fJ E R such that ax + f3y is not in the first quadrant.
•
IIz211
(1.51 )
For the third member, we form (1.52)
and Z3
e3= - -
IIz311
(1.53)
This process continues until the final member of the sequence is produced by . n-1 Zn
= X n - L(xn , edek
(1.54)
j
k=J
and
1
'1
Zn
en = - IIZn
II
( 1.55)
We leave the reader to show that the sequence {el, , en} possesses the orthonormal property. In addition, each ek, k = I, , n is a linear combination of XI, ... ,Xk. We conclude that any linear combination
•
I'
The results in Example 1.19 raise an interesting issue. Can th~ same linear manifold be generated by more than one sequence of vector~? We shall show that indeed this is the case. We consider the Gram-Schmidt orthogonaiization process. Let {Xl, ... , x n } be a linearly independent sequence of vectors generating the linear manifold M c H.. The GramSchmidt process is a constructive procedure for generating an orthonormal sequence leI, ... , en} from theindependent sequence. To begin, let ZI
= XI ZI
.
II z. Iii
n
Lf3k x k k=1
The original sequence and the orthornormal sequence obtained from it therefore generate the same linear manifold. Given the sequence {I, r, r 2, ... } E £2 (-I, I), we use the Gram-Schmidt procedure to produce an orthonormal sequence. Indeed, we define an inner product by EXAMPLE 1.20
i
:(1.48)
and 'el = - -
is also a linear combination
1.49)
(f.
~) == [~f(r)g(r)dr
,', il
1
Linear Analysis
18
Then,
=
ZI('r) el(r)
Z2(r)
=r
~ 1/11111
.....c
(r,
Chap. I
m
This process continues for as mahy terms in the orthonormal sequence as we wish to calculate. We remark that the members of the sequence so produced are proportional to the orthogonal sequence of Legendre polynomials, whose fi~stfew members an: , Po(r) = 1
1
=r 2
P2 (i)=2:(3r -1)
P3 (r)
=
1 3 2:(Sr - 3r)
Xm = L O'kZk (1.56) k=1 This sum generates a linear manifold M c 1-l. Different members of the linear manifold are produced by assigning various values to the sequence of coefficients {O'k }Z~\. We should like to determine what choice of coefficients results in X m being the "best" approximation to x. Specifically, let us make X m "close" to x by adjusting the coefficients to minimize IIx - xmll. We expand the square of the norm as follows:
Ilx
1 4 2 + 3) 8 1 3 5 Ps(r) = -(63r -70r + lSr) P4(r) = -(3Sr - 30r
-xm 11
2
= (x -Xm,X -xm ) = (x,x) + (xm,x m) m
= IIxI1 2 + L lad -
8 The Legendre functions, orthogonal but not orthonormal, are constructed in such a way that Pn(±l) = ±l. ,
M. We conclude
Within the stru:cture of the Hilbert space, it is possible to generalize th~ concepts of approximation of vectors and functions. Let x be a vector in a Hilbert space 'H and let {ZklZ~1 be an orthonormal set in 1-l. We form the sum
J4S/8 (r 2 - 1/3)
,
E
1.6 BEST APPROXIMATION
j3fi r
,PI (r)
19
= l/h
Z3(~) = r 2 - 1/3 e3(r) =
Besl Approximation
1-l, xk --* X E 1-l. But M is closed, and therefore x that M is a Hilbert space.
1
l/h}(l/h) = r
e2(r) =
Sec. 1.6
I
•
We next discuss a characteristic associated with linear manifolds that plays a central role in approximation theory. A linear manifold M IS said to be closed if it contains the limits of all sequences that can be constructed from the members of M. It is easy to demonstrate that not all' linear manifolds are closed. For example, the space C(O', (3) is a linear m~nifold since a linear combination of two continuous functions is a continuous function. However, in Example 1.17, we have given a sequence of vectors in C that fails to converge to a vector in C. The linear manifold C is therefore not closed. An interesting result occUrs if a closed linear manifold is contained in a Hilbert space. Specifically~ if 1-l is a Hilbert space and M is a closed linear manifold in 1-l, then M is a Hilbert space. Indeed, let {Xk }~1 be a Cauchy sequence in M. Then; since M is contained in the Hilbert space
k=1
(xm,x) - (x,x m ) m
m
LO'd x , Zk) - LO'k(X, Zk) k=1 k=\ '
Completing the square, we obtain m
Ilx - xm l1 2 = IIxI1 2 + L
m
(O'k - (x,
zd) (O'k
- (x, Zk) -
k=\
L !(x, zdl
2
k=1
(1.57) Since the sum in (1.57) containing the coefficients O'k is nonnegative, the norm-squared (and hence the norm) is minimized by the choice
k=I,2, ... ,m
(1.58)
Expression (1.58) defines the Fourier coefficients associated with orthonormal expansions. Note that once we have made the selection given in (1.58), we can define an error vector em by
I
em
=x =X -
Xm m
L(x, Zk)Zk k=1
(1.59)
Linear Analysis
20
Chap. 1
Sec. 1.6
Best Approximation
21
I
Taking the inner product with any member Zj of the orthonormal we find that Zj)
= (x, Zj)
-
I
I)x,
Zk)(Zk, Zj)
k=l
'
= 0,
j
= 1,2, ... ,m
d.60)
Since Xm is a linear combination of members of the sequence {Zj };~;, we must have (em, XIII) = 0 (1.61) i
We summarize these results as follows: a. The vector x E 'H. has been decomposed into a vector XIII E J0c 'H. plus an error vector em, viz. ' j
Indeed, let (ek}~l be a sequence in
.
X
= XIII + em
d.62) !
b. The error vector em is orthogonal to the approximation vector XIII' ,
I
EXAMPLE 1.21 We consider the Fourier sine series in Example 1.5. Let f(~) £2(0, 1) with inner product (j, g)
t f(~)g(Od~
=:.,Jo
E
(1.63)
Let {.j2 sin krr ~ J~I be an orthonormal set generating a linear manifold M c H. Then, by (1.56) and (1.58), the "best" approximating function fm E M to f(~) is given by : m
fm(~) = :I>k.j2sinbr~
(1.64)
MJ.. converging to a vector e in 'H..
~h~
(ek, x)
m
(em,
sequ~nce,
for all
X
=0
E·M. But, by continuity of the inner product, lim (ek, x)
k->oo
= (e, x) = 0
so that e E Ml., which is therefore closed. The closed linear manifold M J..·IS called the orthogonal complement to M. We now can produce a result called the Projection Theorem. Let X be any vector in the Hilbert space 'H., and let M C 'H. be a closed linear manifold. Then, there is a unique vector Yo E M c 'H. closest to x in the sense that Ilx - Yo II :::: Ilx - y II for all y in M. Furthermore, the necessary , and sufficient condition that Yo is the unique minimizing vector is that e = x - Yo is in M J... The proof of this important theorem is deferred to Appendix A.l at the end of the chapter. The vector Yo is called the projection of x onto M. The vector e is called the projection of x onto MJ... The ideas inherent to the projection theorem have a well-known interpretation in two- and three-dimensional vector spaces, as in the following example. EXAMPLE 1.22 Let a = (al. (2) be any vector in R 2. Let M be the set of all vectors b in R 2 with second component equal to zero, viz. b
= (131.0)
Since linear combinations of vectors in M are also in M, the set M is a linear manifold. In addition, the manifold is closed. Indeed, let b 1, b2 , ••• be a sequence in M converging to b E H, where
'k=1
where
bk ak
=
fal f(17)./isinbr17d17
which is the classical Fourier result.
(1.65)
I
= (13l kJ , 0)
kJ
Since bk ---+ bE H, 13i ---+ 131. Therefore, bE M. By the projection theorem, among all vectors b E M, the vector fj closest to a can be obtained from
•
The above results for the approximation of a vector X E 'H. by a vector M C 11 can be generalized. We shall need the concept of a manifqld that is orthogonal to a given man'ifold. If M is a linear manifold, theh the vector e E 'H. is a member of a set MJ.. if it is orthogonal to every vector in M. The set .MJ.. is a linear manifold since linear combinations of vectors in MJ.. are also orthogonal to vectors in M. In fact, MJ.. is also closed.
Xm E
,
(a - fj, b) = 0
where
fj
= (~I, 0)
Solving this equation, employing the usual definition of inner product for R2 , we obtain
Linear Analysis
22
,
.
~hap. I
Since f31 is arbitrary, ~ I = al. This result can be visualized (Fig. 1-2) by drawing the vector a and noting that the vector b lies along the x-axis. The "best" b is then obtained by dropping a perpendicular from the tip of a to the x-axis. The r~sult is a vector b, called the projection ora onto the x-axis. Note that the error vbctor e is such that it is orthogonal to any ~ector along the x-axis and that a = b 11- e, as required by the projection theorem, I
23
Best Approximation
Substituting (1.66) and rearranging, we have M
L (Xj (Yj, yd = (y, Yk),
t
(Yl,YM)
Illustration of the projection theorem in R 2 , as given in Example 1.22. I
! !
• In (1.56)-(1.58), we introduced best approximation in terms of orthonormal expansion functions generating a linear manifold. With the aid of the projection theorem, we next generalize the concept of best approximation to include expansion functions that are linearly independent but not necessarily orthogonal. Let Y E 11, and let {Yj }~l be a linearly indepenI dent sequence of vectors in 11. We form the sum M
~~::~~~jr~~j r~~:~~~j . . .
(L66)
j=l
We wish to approximate Y with y by suitable choice of the coefficients (Xj' We have already indicated in Example 1.19 (Problem 1.17) that linear combinations of the type in (1.66) form a linear manifold M. In fact, since the limit of sequences of vectors in M must necessarily be in M, the manifold is closed and therefore meets the requirements of the projection theorem. Since Yj EM, j = 1, 2, ... , M, the projection theorem gives
y, Yk) =0,
:
:
(YM, YM)
(XM
-
:
(1.69)
(y, YM)
Inversion of this matrix yields the coefficients (Xj, j = 1, 2, ... , M. These coefficients then determine yin (1.66). The square matrix on the left-hand side of (1.69) is the transpose of a matrix called the Gram matrix. In addition to its appearance in best approximation, it also finds use in proofs of linear independence (Problem 1.21). Note that the result in (1.69) is a generalization of the Fourier coefficient result in (1.58). Indeed, for cases where the independent sequence Yj, j = 1, 2, ... , M is orthogonal, the matrix in (1.69) diagonalizes. Inversion then produces the Fourier coefficient result. One of the classic problems of algebra is the approximation of a function by a polynomial. This problem is easily cast as best approximation in the following example. EXAMPLE 1.23 In the Hilbert space .c 2(0, 1), consider the approximation of a function fer) by a polynomial. Let {rn-l}~=1 be a sequence in .c2(O, 1). The sequence is linearly independent. Indeed, by the fundamental theorem of algebra, the equation N
L.B"r n - 1 =0 n=1
Y= I>:¥jYj
(y -
( 1.68)
We write (1.68) in matrix form as follows:
r
....
= 1,2, ... , M
k
j=l
(Yl,Yl) (Yl, Y2)
y
Fig. 1-2
Sec. 1.6
k = 1, 2, ... , M
0:67)
can have at most N - 1 roots. Therefore, the only solution valid for all r is.B" = 0, n = 1,2, ... , N. We wish to approximate fer) by
E
(0, I)
N
fer) = Lan rn - t
(1.70)
n=l
All possible suc~ linear combinations form a closed linear manifold so that the projection theorem applies. Comparing (1.66), we identify (1.71)
Linear Analysis
24
'Chap. I
Sec. 1.7
Operators in Hilbert Space
25
The solution is given formally by
Define an inner product for £2 (0, :1) by
x=A-Iz
Jol
f(r)g(r)dr
I
(1.72)
The matrix operation is linear. Indeed, given XI, X2, ordinary matrix methods
We then have
1
1 m-l
= o ,r
(Ym, Yn)
r
1 1
(y, Ym)
=
n-Id
m
r
=
1
m+n-
1
i
!
1
r - f(r)dr
I
(1.74)
I
Substitution of (1.73) and (1.74) into (1.69) gives 1
"2
1
1
[1
'I ]
3"
: (1.75)
1 N+I
Inversion of this matrix equation yields the best approximation.
I
j
1
1.7 OPERATORS IN HILBERT SPACE Consider the following transfonnation in R 3 :
!
= CXII~I + CX12~2 + CX13~3 ~2 = CX21 ~ I + CX22~2 + CX23~3 ~3 = CX31~1 + CX32~2 + CX33~3 ~I
I
A(cxx]
, d.73)
•
+ {>x2) = cxAxl + f3Ax2
Ir, l'
~2
~3 f
x =:± [~I
~2
~3 f
where T indicates matrix transpose. We then have
"
Ax
where "
i ;
=z
+ f3Z2
The domain of the operator L is the set of vectors x for which the mapping is defined. The range of the operator L is the set of vectors y resulting from the mapping. The operator L is linear if the mapping is such that for any XI and X2 in the domain of L, the vector CXIXI + CX2X2 is also in the domain and L (CXIX]
+ CX2X2) = CX] LXI + cx2Lx2
0.77)
A linear operator L with domain V L C 1i is bounded if there exists a real number Y such that (1. 78) IILull ~ Yllull for all u
E
VL.
EXAMPLE 1.24 Let Roo be the space of all vectors consisting ~f a countably infinite set of re~1 numbers (components), viz. a
where ak
E
R. If b
=
(ai, a2, a3 ...)
(1.79)
= (tiI, tho th. ...), define an inner product for the space by , 00
z = [~I
= CXZI
and Z2, we have by
The concepts of linearity and inversion for matrices can be generalized to linear operators in a Hilbert space. An operator L is a mapping that assigns to a vector x 'E S another vector Lx E S. We write Lx = y 0.76)
Using the usual matrix notation, we let
I
Zl,
(a,b) =
i
Ladh
(1.80)
k=1
Let the norm for the space be induced by the inner product. We restrict Roo to those vectors with finite norm. Define the right shift operator A R in Roo by
ARa = (0, ai, a2,
...)
The right shift operator A R is linear. The proof is easy and is omitted. In addition, A R is bounded. Indeed,
Sec. 1.7
Linear Analysis
26
i· Therefore, the operator A R is boundetl in Roo. Indeed, the least upper bound on : is unity.
y
EXAMPLE 1.25 On the complex Hilbert space £2 (0, I), we consider the! following integral equation:
10t
:
u(~')k(~, ~/)d( = /(0
A linear operator L with domain V L C 7{ is continuous if given an > 0, there exists a 0 > such that, for every Uo E V L, II Luo ~ Lu II < E, for all u E VI. satisfying lIuo - u II < o. We can interpret this definition to mean that when an operator is continuous, Luo is close to Lu whenever Uo is close to u. There is an important theorem on interchange of operators and limits that follows immediately from the above definition. A linear operator L with domain V L C 7{ is continuous if and only if for every sequence {un }~1 E V L converging to Uo E V L ,
°
E
:.
(1.81)
, This equation can be written in operator notation as follows: Lu
Luo
=/
L =
1
(.
1f
Ik(t
)k(~, ~/)d(
(1.82)
II Lu o - LU n II <
~')12d~d( <
00
(1.83)
where
111 u(~/)k(~, ~/)d~f
:: 1 1
lu(S')!2d(
= lIull211
(1.84)
f Ik(~. nI2d~'
n~N
E,
and
Luo
n
This property of the kernel k(~, is:called the Hilbert-Schmidt property, and the operator L it generates is called a H,ilbert-Schmidt operator. To show thatLis bounded, we form
1/(01 2 =
lim LU n
n~oo
The proof is in two parts. First, we suppose that L is continuous and E > is given. We may select a 0 according to the definition o(continuity and suppose that lIuo - unll < o. Since Un is a member of a converging sequence, Iluo -'unll < 0 for all n ~ N. Therefore,
We shall show that the operator L is bounded if 1
= L n---+oo lim Un =
°
where L is the linear operator given by 1
27
Operators in Hilbert Space
=
lim LU n
11->00
This first part of the proof shows that, if an operator is continuous, the operator and the limit can be interchanged. In the second part of the proof, we must show that if the operator and limit can be interchanged, the operator is continuous. This part is not essential to our development and is omitted. The interested reader is referred to [12 J. . We now give a theorem linking the boundedness and continuity of operators. A linear operator L with domain V L C 7{ is bounded if and only if it is continuous. The proof is in two parts. In the first part, we show that if the operator is bounded, it is continuous. Indeed, if L is bounded and Uo E V L ,
IILuo - Lull
Ik(~, nI2d~'
= IIL(uo ~ YIIuo -
It follows that:
for all u
E
V L . Then, given any
E
ull
> 0, it is easy to find a 0 :> Osuch that
IILuo - Lull < and finally,
u)1I
E
whenever IIL~II::::
lIuo - ull < 0
Mllull
where M is the bound on the double integral.
•
Indeed, the choice 0 = E/ JI is sufficient. In the second part of the proof, we must show that if an operator is continuous, it is bounded. This part is
I
Sec. 1.7
Linear Analysis
28
not essential to our development and is omitted. The interested reader is referred to [13]. It is straightforward to show that the differential operator L = d/d~ is unbounded. The proof is by contradiction. We suppose that d/d~ is bounded. Then, it is continuous. Therefore, for any Un ~ u, we rhust have Lu = L lim Un = lim LU n n->oo n->oo 1
Un =
11
, Il (IltTlJo I>kLZk. Zj) = (J, Zj), ,
we have lim Un
n->oo
L n->oo lim Un But, lim LU n
n---+oo
l'
=
1,2, ...
"'J [(xIJ (X2 = [(J,ZI)J (J, Z2)
...
.. .
j(~) = -I·W(~) +
I
where
k(~, n = ~(l -
l' k(~, nu(nd~' O::s~::sf
$'),
nl-n
~'
::s ~ ::s I
(1.88)
(1.89)
(1.90)
In operator notation,
(1.85)
j,
~ E
(0, I)
(1.91)
where / is the identity operator. The operator L - Ji-/ is bounded. We leave the proof for .Problem 1.26: We wish to obtain the matrix representation and thereby solve th.e mtegral equatIOn. We define the inner product for the space as in (1.63). For baSIS functions, we choose
Since boundedness implies continuity, 'n
(L - Ji-l)u =
n
= L n->oo lim L(XkZk = lim L L(XkZk k=l n->oo k=l
Zn = sinmr~,
By the linearity of the operator L, we then have
n = 1,2, ...
(1.92)
Then, operating on any member of the basis set, we obtain
:6.86)
(I, - p.l)z" = -1£ sin
mr~ +
1k(~, 1
() silll17r( d(
I
L i
On the real Hilbert space .c 2 (0, I), consider the integral equa-
EXAMPLE 1.26 tion
I~
I;
~f this matrix can be inverted to give the coefficients (Xl, (X2, . : . , substitution ' mto (1.85) completes the determination of u.
n
il
Ij
I
j
" "
lim L(XkZk
ij
Equa~ion (1.87) is a matrix equation that can be written in standard matrix notation as follows:
z·) J '
,
"
: n->oo k=l
Lu
11
II
=0
We expand U in the basis as follows:
=
= 1,2, ...
(1.87)
= (I
zJ')
(Lz 2 ,Zt) (Lz 2, Z2)
= n---+oo lim (-7f sinl17f~)
II
,Il
= 1,2, ...
and this limit is undefined. We therefore have arrived at a contradiction, and we conclude that d / d~ is ~nbounded. i Giveil the concepts of continuity and boundedness of a linear operator, we can show that a bounded linear operator is uniquely determined by a matrix. Indeed, let {Zk}~l be a basis for H. Let L be a bounded linear operator with Lu = 1
U
j
By ~ontinuity of the inner product and the rules for inner products, We obtam
=0
and therefore,
.il )1
k=l
lim "" (XdLzk. Il~OO~ k=l
11,
29
We take the inner product of both sides of (1.86) with a member of the basis set to gi~e
But, if we choose Un as a member of the sequence cosmr~
Operators in Hilbert Space
(1.93)
But, using (1.90), we find that
1 1
o
Sec. 1.7
Linear Analysis
30
' k(;,nsinmr(d;' = (1-;)
1·~
1 1
(sinnrr(d(
+;
; (I-nsinnrr,(d(
~
0
/1l)Zn, Zm) =
=
,
[(n~)2 - /1] (Zn, Zm)
~ [(n~)2 - /1] 8
11m
(1.94)
where the inner product is the usual inner product for £2 (0, I) and 8nm ha~ been defined in (1.33). The matrix representation in (1.88) therefore diagonalize~, and i the inversion yields I Uk
=
2 1 (krr)2 -
/1,
(j,
zd,
k = 1,2, ...
(1.95)
[x, y]
Ix
1= J(Lx, x)
(1.97)
(1.98)
We.empha~i~e that the operator L must be positive for (1.9~b to satisfy the baSIC d~finltlOns of a norm. Indeed, the energy inner product and norm defi.ned In (1.97) and (1.98) must be shown in each case to satisfy the rules for Inner p~od~cts and norms. For positive-definite operators, we can prove the follOWIng Important relationship between norms: 1
IIxil :s -Ixl c (1.96)
• In the above example of representation of an operator by a matrix, the choice of the basis functions resulted in diagonalization of the matrix and, therefore, trivial matrix inversion.I There are many operators, however,' that do not have properties that result ~n this diagonalization. These concepts are better understood after a study of operator properties and resulting Greeh's functions and spectral representations in the next two chapters. , An important collection of operators for which there are e~tabl:ished convergence criteria are nonnegative, positive, and positive-definite ,operators. The reader is cautioned that there is little uniformity of notation concerning these operators in the literature. For the purposes herein, an operator L is nonnegative if (Lx, x) 2: 0, for all x E 'DL. An operator is positive if (Lx, x) > 0, for all x#-O in 'DL. An operatoris positive-definite if (Lx, x) 2: c2 11x1l 2 , for c > 0 and x E 'D L . An operator L is symmetric if (Lx, x) (x, Lx). It is easy to show that nonnegative, positive, and positive-definite operators are symmetric. In fact, any operator haviqg the property that (Lx, x) is real is symmetric. Indeed, '
(1.99)
Indeed,
Therefore,
I
IIxIl2:s~lxF c
1.
i
Taking the square root of both sides yields the desired result. Amon~ the many forms ofconvergence criteria, there are several types that are particularly useful in numerical methods in electromagnetics. For a sequence {un} C H, Un converges to U is written ' (1.100) and means that lim
n---+oo
lIu n - u'l = 0
The statement Un converges in energy to
=
== (Lx, x) = (Lx, x)
= (Lx, y)
With ~his inner product definition, 'DL becomes a Hilbert space HL. The assocIated energy norm in H L is given by
Substitution of (1.95) into (1.85) yields the final result, viz.
(x, Lx)
31
..A speci~1 inner product and norm [17], associated with positive and posltlve-defimte operators, are useful in relating convergence criteria. Define the energy inner product with respect to the operator L by
,
After some elementary integrations, we obtain the general matrix element in ,the square matrix in (1.88), viz.
«L -
Operators in Hilbert Space
U
(1.101)
is written
e
Un~U
(1.102)
and means that lim
n---+oo
IUn -
ul = 0
(1.103)
I
Linear Analysis
32 I
Chap. I
.
The statement Un converges weakly to
U
is written
w
d.104)
:U n - + U
I
lim I(u n -
U,
g)1
11-+00
=0
~
B. Convergence implies vveak convergence. C. Convergence in energy implies LU n ~ f. The weak convergence is for those g, defined by (1.105), in 'HL. If, however, II Lunll
fin 'H. is bounded, then LU n ~ ,
We first prove Property A. We have
= I(L(u n - u), ~n - u)1 ~ IIL(u n - u)lIl1u n - ull = IILu n - Luililu n - ull ~ (IiLu nll + IILulI) lIu n -ull
Since, by hypothesis, II Lunll is'bounded and Un -* u, a limiting operation gives li~ IUn - up = 0 n~oo
To prove Property B, we use the CSB inequality to give
for any g in 'H. Taking the limit yields the desired result, viz. -
U,
c
ul
yields the desired result.
00
1.8 METHOD OF MOMENTS The purpose of this section is to introduce the Method of Moments in a general way and develop various special cases. Emphasis is on convergence and error minimization. If L is a linear operator, an approximate solution to Lu = f is given by the following procedure. For L an operator in 'H, consider Lu -
f =0
(1.106)
where U E V L , f E R L . Define the linearly independent sets {cPdk=l C V Land {wd k=l' where cPk and Wk are called expansion functions and weighting functions, respectively. Define a sequence of approximants to u by , n
Un
=L
akcPk.
1l
= 1, 2, . . .
(1.107)
k=l
A matrix equation is formed in (1.106) by the condition that, upon replacement of U by Un, the left side shall be orthogonal to the sequence {Wk}. We have . (Lu n - f, wm ) = 0, m = 1,2, ... , n (1.108) Substitution of (1.107) into (1.108) and use of (1.25) gives the matrix equation of the Method of Moments [18] ,[ 19], viz.
I(u n - u, g)1 ~ lIu n - ulIlIgll
lim I(u n
1 lIu n -ull ~ -Iun -
,
D. If L is positive-definite, convergence in energy implies convergence.
2
for g E 'H L. This procedure proves the first half of Property C. The proof of the second half is based on the Hilbert space 'HL being dense in 'H and is omitted. (See [17, p. 24-25].) To prove Property D, we write
Taking the limit as n -*
A. If II Lunll is bounded, convergence implies convergence iD energy. ,
l
33
(1.105)
It is straightforward to show the following relationships among the types of convergence:
un
Method of Moments
I
and means that for every g E 1-t
IU -
Sec. 1.8
n
L ak(LcPk. w
g)1 = 0
m)
=
(f, wm ),
m=I,2, ... ,n
(1.109)
k=l
n~oo,
I
To prove Property C, we have I(Lu n - f, g)1
= I(L(u n ~ u), g)1 = I[u n -
u, g]l
~ IUn
-c-
ulhl
where we have used the CSB inequality on Hilbert space 'HL. By hypothesis, we have convergence in energy. Therefore, lim I(Lu" - f, g)1
n-+oo
=0
Note that the exact operator equation (1.106) in a Hi Ibert space'H has been transformed in(o an approximate operator equation on Hilbert space C n , viz. '
Ax =b
(1.110)
where, in usual:matrix form, x
= (a]
a2
(1.111)
.
"
Ii Linear Analysis
34
b
= ( (j, WI)
(i.112)
. (j, W2)
'A
Chap. I
n
Un
I
:
= (j,
Substitution into (1.107) gives
(1.113)
We note the following interesting result. If the operator L is boun~ed, if Wk=
m=1,2, ... ,n
(Ll15)
k=l
If nothing is known about the mathematical properties of the operator L other than its linearity, nothing'in general can be said concerning the convergence of the approximants Un to the solution u. Unfortunately, most of the interesting and practical problems in electromagnetics involve operators that are neither positive nor positive-definite. Therefore, most of the large body of solutions to electromagnetic problems by the Method of Moments lack any sort of mathematical convergence criteria. If, however, the operator L is positive, we may define the Hilbert space 7-{L with the norm given by (1.98). We then write (1.115) as folI9ws:
k=1
m=1,2, ... ,n
I
'
We further assume that the sequence {
ak=[ll,¢kl
~1.117) i
(Ll18)
which is the Fourier series expansion in 7-{L of Un with Fourier coefficients given by (1.117). Therefore, lim
n--->oo
I
Un -
U
I= 0
(1.119)
By Property C, the resu It in (1.119) implies that LU n ~ f. Uilfortunately, nothing can be said about the nearness of Un to u. If, however, L is positivedefinite, Property D states that the approximants converge, vi~. lim
n-...oo
lIu
lI -
llll = 0
(1.120)
In the Galerkin procedure, if the operator L is positive and the sequence {¢d is complete in 7-{L, the method is calIed the Rayleigh-Ritz method. For a classical treatment of the Rayleigh-Ritz method, the reader should consult [17J,[20]. For the more general operators often encountered in electromagnetics, a positive operator can be produced by the following procedure. Consider
Lu
=f
(1.121)
Let the adjoint! operator L * be defined by I
(Lu, v)
= (u, L*v)
(Ll22)
.
,
for u E D L , V E V L*. Then, if the adjoint L * exists, multiplication of both sides of (1.121) by L * produces
L*Lu
G1.116)
.
= L[u, ¢k]
n
L ad
35
Method of Moments
I
= [amd
where T denotes transpose and amk are the individual matrix eleJ;Tlents, I . given by , amk' = (L
Lak(L
Sec. 1.8
= L* f
(1.123)
for any f E DL*. Provided that Lu = 0 has none but the trivial solution, it is easy to show that the operator L * L is positive. Indeed, (L * Lu, u) = 2 II Lu 11 > 0, unless Lu = O. But, Lu = 0 implies u = O. . The Method of Moments applied to L * L gives n
L adL * L¢k, Will} = (L * f, Will), k=J
m
= 1,2, .... ,n
(1.124)
i
I
Linear Analysis
36
Chap. I
Sec. A.I
\
The Galerki:l specialization follows immediately, viz. ,
.
.'
n CXk (L *L¢k,
L
¢m)
= (L* f, ¢m),
m
=
1, 2, ... , n
(1,.125)
k=1
Since L * L is positive, if the sequence {¢k} is complete in VL*L, (1.125) is the Rayleigh-Ritz method and convergence in energy Un ~ U is assured, viz. (1.126) lim Un - U = 0 n--->oo
I
I
Appendix-Proof of Projection Theorem
In proving the Projection Theorem, we begin by noting that the first equality in (A. 1) makes sense. Indeed, IIx - y 11 is bounded below by zero, and therefore has a greatest lower bound. We next show that there exists at least one vector Yo closest to x. We begin by asserting that there exists a vector Yn E M such that by the definition of infimum, (A.2) Taking the limit as n --*
where the energy norm is with respect to the operator L * L. By properties of the adjoint, (1.125) can also be written n
LcxdL¢k, L¢m}
= (f, L¢m),
m=I,2, ...
,11
0.127)
k=1
which is the result in the Method of Least Squares, more usually derived [20] by minimization of
lim IILu n n--->oo .
fII = 0
In this Appendix, we prove the Projection Theorem, considered in S:ection 1.6. We restate the theorem here for convenience. Let x be any ve~tor in the Hilbert space 1t, and let M c 1t be a closed linear manifold. !Then, there is a unique vector Yo E M c 1t closest to x in the sense that I inf
'lix - YII
=
IIx - Yn II = 8
Ilx - yoll
(A.3)
Therefore, we can always define a sequence {Yn} E M such that IIx - Yn II converges to 8. In (1.36), if we replace x by x - Yn and Y by x - Ym, we obtain [21], afte~ some rearrangement,
llYn - Ym 1
2
= 211x - Ynl1 2 + 211x ,
Ym 11 2
-
411x -
~(Yn + Ym)1I 2 2
(AA)
+ Ym)/2 E M, and we may assert that ,:i
A.1 APPENDIX-PROOF OF PROJECTION THEOREM ,
yEM,
lim
n--->oo
q.128)
so that LU n --* f. Unless the operator L * L is positive-definite, n9 thin g can be saidconceming the convergence of Un to u.
8=
we find that
00,
Since M is a linear manifold, (Yn
It is easy to show that (1.126) implies that
37
i (A.l) I
where inf is the greatest lower bound, or infimum. In other words, Yo is closest to x in the sense that IIx - Yoll ~ IIx - yll for all Y in M. Furthermore, the necessary and sufficient condition that Yo is the pnique minimizing vector is that e = x - Yo is in M 1-. The vector Yo is called the projection of x onto M. The vector e is called the projection of x ontb M 1- •
i"
Therefore,
In the limit as m, n --* 00, the right side goes to 28 2 + 28 2 - 48 2 = 0, and we conclude that the sequence {Yn} is Cauchy. Since 1t is a Hilbert space and M is closed, M is a Hilbert space and Cauchy convergence implies convergence. Therefore, Yn --* Yo E M. We next show that Yo is unique [22]. Suppose it is not unique. Then, , we must have at least two solutions Yo and Yo satisfying IIx, - Yo II IIx - Yoll = 8. Then,
lIyo - Yo 11 2 ~ 211x - Yoll 2 + 211x - Yoll 2 - 411x ~ 28 2 + 28 2 Therefore, Yo
= Yo.
-
48 2
=0
~(y~ + yo) 11 2 2 (A.6)
I
Problems
38
Chap. 1
Finally, we show that e =x - Yo E Ml-. We must show that e is orthogonal to every vector in M. Suppose that there exists a vector Z E M that is not orthogonal to e. Then, we would have [23] (e, z)
= (x -
Yo,z)
= A =I 0,
zEM
(A,7)
==
Yo
+
A \I Z11 2 Z
I
e\
= (1,0, ... ,0), e2 = (0, 1, ... ,0), ... , en = (0,0, ...',1)
is linearly independent. Is the same conclusion valid in C n ? Is the set of' vectors a basis for C n ? n
n
k=1
k=\
(l:= akXb y} = L adxb y} 1.10. Show that C(a, (3) with inner product defined by (1.27) is a real inner product.
".
IIx - zoll2
= (x -
space.
A' A Yo - - z2· x - Yo - - z2)
IIzll " 2 A = Ilx - Yo II - IIzll 2 (x
1.11. Consider the linear space of real continuous twice differentiable functions
IIz11
A
- Yo, z) -
IIz11 2(z, x
- Yo)
1~12
+ 11211 2
IAI 2 = IIx - Yo II - IIzll 2 2
(A.9)
Therefore,
1
Ilx - zoll .
1.8. Show that in R n the set of vectors
(A;8)
Then,
I
39
Problems
1.9. Given the basic definition of an inner product space, show that
We define a vector Zo EM such that Zo
Chap. 1
<
Ilx -
Yo II
over the interval (0, I). As a candidate for an inner product for the space, consider J"(r)gl/(r)dr
where f and g are members ofthe space and "primes" indicate'differentiation. Determine whether (j, g} is a legitimate inner product.
1.12. Prove the following corollary to the CSB inequality in (1.35):
(..\.10)
which, by (A. I ), is impossible.
1 1
=
(j, g}
=
I(x, y}1
IIxlillyll
if and only if x and yare linearly dependent.
1.13. Prove the following identity:
PROBLEMS 1.1. 1.2. 1.3. 1.4.
Using the rules defining a linear space, show that Oa = 0 and -la = -\-a. Show that Rn is a linear space. As an extension to Example 1.4; in R 2 , let Xl = (l, 3), X2 = (2,6.00000001). Show that XI and X2 are linearly independent. Comment on what might occur in solving this problem on a computer with eight-digit accuracy. (This problem is indicative of the difficulties that can arise in establishing linear independence in numerical experiments in finite length arithmetic.)
1.5. If XI, X2, • •• , X n is a linearly dependent set, show that at least one member of the set can be written as a linear combination of the other members. ~~.
IIx - yll
1.14. Consider a complex inner product space with norm induced by the inner product. If X and yare members of the space, prove that
Show that C(O,I) is a linear space.
1.6. Show that if 0 is a member of the set Xl,
Ilixil - lIylll .:::
I
X2, •.• , X n ,
the set is linearly depen-
i
= 2' (lix + iYll2
(x, y} - (y, x}
and (X, y}
+ (y, x}
=
- IIx - ;YIl2)
~ (li x + Yll2 -
2 IIx _ Y1I )
1.15. Given the following sequence in the space of rational numbers: n
X
n
= {; (k
1 - I)!
!
1.7. Show that if a set containing nvectors is linearly dependent, and if m additional vectors are added to the set, the resulting set of n + m vectors is linearly i dependent. i 1
First, show that the sequence is Cauchy; next, show that the seqiJence does not converge in the space. (Indeed, it converges to e; the details can be found in [4J.)
I
40
Problems
'
C;:hap.l
Chap. 1
Problems
41
,
1.16. Show that if S is a linear space, a linear manifold M c S is also a!linear space. 1.17. Let Xb k = 1,2, ... , n be a linearly independent sequence of vectors ,in the Hilbert space 'H.. Define M to be the set of all linear combinations of the n vectors. Prove that M is a linear manifold. 1.18. Let Roo be the space of all vectors consisting of a countably infinite set pf real numbers (components), viz.
consider the following integral equation:
j
Ct
11] -
u(O In
~I
-ct
d~
Ja2_~2
=
f(1])
r
a
=
(ai, a2, ...)
where ai E R. Let M be the set of vectors in Roo with only a finite nOmber of the countably infinite number of components different from zero. Show that M is a linear manifold. Show that M is not closed. Hint: Consider the I concept of closed as applied specifically to the sequence of vectors I Xn
=
This integral equation occurs in diffraction by a slit in a perfectly conducting screen. It can be shown [14] that the operator
is bounded. Solve the integral equation by using the Chebyshev polynomials Tn (~/a) as a basis for L2( -a, a) and obtaining the matrix representation for, L. Hint: The following are useful integral relations [15]:
(I,,~, ~, ... , n~,O, 0, ...) :2 3
-Tf
In(2/a)To(~/a),
n>O
1.19. The Legendre functions Pn (~.), n = 0, 1, 2, ... , are orthogonal on ~ E (-1, I),but they are not orthonormal. Create a sequence of orthonormalI ized Legendre functions Pn (n, n = 0, I, 2, . . . .
m =/= n
m=n=/=O
1.20. Given that in R 3 , XI = (I, 2, 0), X2 = (0, I, 2), X3 = (1, 0, I). (a) Prove that {Xl, X2, X3} is a linearly independent set of vectors. (b) From the linearly indepen~ent set, produce the first two members of the associate orthonormal set !using the Gram-Schmidt procedure.
1.21. Show that the determinant of the Gram matrix in (1.69) is nonzero if and only ' if the sequence of vectors {Yk }t,l is linearly independent [II]. 1.22. Let Roo be the space described in Problem 1.18. If b = (f3I, f32, ...), ;define i an inner product for the space by
m =n =0 1.24. Let L = d / d ~ , and consider the sequence of partial sums n
Un
=L k=1
= I>kf3k
lim Un n->oo
k=l
Let the norm for the space be induced by the inner product. We rest&t Roo to those ,vectors with finite norm. Define the operator A in Roo by
Aa
=: (ai, ~a2' ~a3' ...)
1.25. Let L =
d/d~,
and consider the sequence of partial sums n
Un
1.23. On the Hilbert space L2( -a, a), with inner product (f, g) =
f
. -ct
.f
Ja 2 -
=
1
L 2"k coskTf~ k=l
d~
Ct
=-In(2sinTf~) 2
Show that limn->oo LU n is undefined. The problem is that L is unbounded. This result is an example of the fact that a Fourier series cannot always be differentiated term by term.
Test the operator A for boundedness. I
1
-k coskTf~
It is well-known [16] that
00
(a, b)
n=O
~
2
Using well-known series summation results [161, show that, although L is unbounded, in this case the operator and limit can be interchanged.
i Problems
42
,
.
Chap. 1
1.26. Show that the operator in (1.91) with kernel defined in (1.90) is bounded.
(u,~) =
References
43
1.31. Consider the real Hilbert space £2(0,1) with inner product '
I
1.27. Consider Hilbert space £2 (-1, 1) with inner product ,
Chap. 1
11 u(Ov(~)d~
where fU;), g(O
where all functions are real-valued. Consider the following function
f
1f(~)g(~)d~ 1
(j, g) =
I
(0:
E
£2(0, 1). Suppose that
f(~) = 1 - i
2
(a) By the method of best approximation, approximate f(~) by i(~), where
i(~) is a linear combination constructed from the orthonormal set
(JEkcoskn~}t=o in £2(0, 1). In the orthonormili set, and 2 for k # O. (b) Calculate the norm of the error IIf(O - i(~)II.
Ek
is 1 for k
=
0
1.32. Consider Euclidean space R 4 . Define vectors a and b in R 4 by a = (aI,
, a4)
b = (fJI,
, f34)
Define an inner product for the space by 2rr
(u, v) = ,
where members of the space
1 0
~re
4
u(Ov(~)d~
(a, b) =
L.:>kf3k k=l
Consider those vectors a in R 4 restricted by
complex functions.
(a) Show that the sequence '
is an orthogonal sequence. (b) Produce an orthonormal (O.N.) sequence from the orthogonal sequence. (c) Using the members of the O.N. sequence contained on -N ~ ~ ~ N, where N is a positive integer, find the best approximation to f(~)
.
2a) ,
=;sm (~-2-
aER
in the sense given in Section 1.6, Best Approximation. 1.30. Consider the real Hilbert space £2(-1, 1). For define an inner product
(j, g) =
1 -1 /
f(~)g(~)
f(o.
g(O E £2('-1, 1),
d~
JI=T2
Determine whether this dcfini'tion results in a legitimate inner product.
(a) Show that all vectors with this restriction form a linear manifold M. (b) Find all vectors b that are members of M 1- •
REFERENCES [I] Stewart, G.w. (1973), Introduction to Matrix Computations. New York: Academic Press, 54-56. [2] Hardy, G.H. (1967), A Course in Pure Mathematics. London: Cam~ bridge University Press, 1-2. [3] De Lillo, N.J. (1982), Advanced Calculus with Applications. New York: Macmillan, 32. ' [4] Stakgold, I.: (1967), Boundary Value Problems of Mathematical Physics. Vol, I. New York: Macmillan, tal-102. [5] Mac Duffee; c.c. (1940), Introduction to Ahstract Algehra. New York: Wiley. chapter VI.
References
44
[6] Greenberg, M;D. (1978), Fhundations ofApplied Mathematics.! Englewood Cliffs, NJ: Prentice-Hall, 319-320. [7] Helmberg, G. (1975), Int~oduction to Spectral Theory in Hilbert Space. Amsterdam: N0 rth-Holland,20--2l. [8] Ibid., Appendix B. , [9] Shilov, G.E. (1961), An Introduction to the Theory of Linear Spaces. Englewood qiffs, NJ: Prentice-Hall, sect. 83-85. I [10] Stakgold, 1. (1979), Green's Functions and Boundary Vatue 'Problems. New York: Wiley-~nterscience, 36-41. : [11] Lue)1berger, D.G. (1969),: Optimization by Vector Space Methods. New iork: Wiley, 5~57. ', [12] op.cit. Helmberg, 73. [13] Akhiezer, N.L, and LM. Glazman (1961), Theory ofLinear Operators ' in Hilbert Space. New York: Frederick Ungar, 33,39. [14] Dudley, D.G. (1985), Error minimization and convergence in n~mer ical methods, Electromagnetics 5: 89-97. [15] Butler, C.M., and D.R. Wilton (1980), General analysis of narrow strips and slots, IEEE Trans. Antennas Propagat. AP-28: 42-48., [16] Abramowitz, M., and LA. 'Stegun (Eds.) (1964), Handbook of Mathematical Functions, National Bureau of Standards, Applied Mathematics Series, 55, Superintendent of Documents, U.S. Government Printing Office, Washingtqn, DC 20402, 1005. , [17] Mikhlin, S.G. (1965), The Problem of the Minimum of a Quadratic Functional. San Francisco: Holden-Day. [18] Harrington, R.E (1968), field Computation by Moment Methods. New York: Macmillan. [19] Harrington, R.E (1967), "Matrix methods for field problems," hoc. IEEE 55: 13~149. [20] op.cit. Stakgold (1967), v61. 2, sect. 8.10. [21] Krall,A.M. (1973), Linear Methods ofApplied Analysis. ReadIng, MA: Addison-Wesley, 181-182. [22] Ibid., !182. [23] op.cit. Akhiezer and Glazman, 10. ,
l
r
\
, Chap. 1
2 The Green's Function Method , ,
,
2.1 INTRODUCTION In this chapter, we begin our study of linear ordinary differential equations of second order. Our goal is to develop a procedure whereby we can solve the differential equations using fundamental solutions called Green's functions. We begin >yith a brief discussion of the delta function. We follow with a description of the Sturm-Liouville operator L and Its properties. We de~ne three types of Sturm-Liouville problems and investigate their propertIes. In all three types, we examine the role of the operator L and its ' adjoint operator L *. These operators are used to define the Green's func- : ~ion and the adjoint Green's function, respectively. Our study culminates: 10 a procedure fo~ applying the Green's function and/or the adjoint Green's, function in solving the differential equation Lu = f. 2.2 DELTA FUNCTION The concept of the delta function arises when we wish to fix attention on ' the value of a function f(x) at a given point xo. Mathematically, we seek an operator T that transforms a function f (x), continuous at xo, into f (xo), the value of the function at xo. In equation form, we require T such that T [f(x)] = f(xo)
(2.1)
:
!
The Green's Function Method
46
Chap. 2
E
< X < Xo
+E
lim : (2.2)
otherwise Note that, regardless of the value of E, the area under the pulse is 'unity. Indeed, if (a, b) is any interval containing (xo - E, Xo + E),
l
b
.
P€ (x - xo)dx
=
a
l
XO
+€ I
Xo-€
-
2E
dx
=I
Delta Function
Taking the limit as
We begin by considering the pulse function P€ (x - xo), defined by ;
Xo -
Sec. 2.2
€-+o
l
E ---+
0, we have
b
= f(xo),
f(x)P€(x - xo)dx
a
Xo
EXAMPLE 2.1 Let f(x) = x 2, Xo continuous at x = 0 and we have
. (2.3) lim
,->0
1" a
=
0, and Xo
E (a,
= ,->0 lim[~ 2E
f(x)p,(x - xo)dx
An important property of the pulse function is that it is even about Xo, viz.
I'
I
l
I'
\
b
f(x)p€(x - x.o)dx
a
:
= -I
2E
l
+€
b
f(x)P€(x - xo)dx
Since f(O)
(2.6)
~Pe(X-Xo)
,->0
1"
f(x)p,(x - xo)dx = lim - 1 ,->02E
a
Since
f
(rr/3)
=
so that
-1-----+----:---+----+-----" x
2-1
+8
Pulse function p,(x - xo) and function f(x).
(i +
• E (a,
b). In this case,
dx
E) - sin
(i -
E)]
I~ =
•
Expression (2.7) fonns the cornerstone of our definition of the delta function, as follows:
1
I·
2 X dX]·
1/2, we have again verified (2.7).
h
f(x)
Fi~.
1+' cosx 1-'
1
f(x)8(x - xo)dx
a
Xo
1
= !~ 12E [sin
,--------=,
1/28
_,
= 0, w,e have verified (2.7).
'
=f
b). In this case, f(x) is
l'
EXAMPLE 2.2 Let f(x) = cosx, Xo = rr/3, and Xo f (x) is continuous at x = rr /3 and we have lim
f is the mean value of f(x)
(2.7)
2
(2.5)
f(x)dx
Xo-€
By the mean value theorem for ihtegrals [I], if on the interyal x E (xo - E, Xo + E),
l
xo
(a, b)
= ,-.0 lim(E ) = 0 3
(2.4)
This property can be proved by interchanging x and Xo in (2.2). The details are left for the problems. Multiplying the pulse function by f (x) and integrating over any interval containing the pulse gives (Fig. 2-1)
E
The integration followed by the limiting operation in (2.7) transfonns f(x) to f (xo), the value of the function at Xo.
,
~
47
=
lim €-+o
l
l
b
f(x)P€(x - xo)dx
(2.8)
= f(xo)
(2.9)
a
b
f(x)8(x - xo)dx
for any Xo in the interval (a, b). Note in (2.2) that as E becomes smaller, the pulse function becomes narrower and higher while maintaining unit area. If the limit in (2.7) could be taken under the integral, we would have
8(x -xo) 4: lim [p, (x -xo)] €-+O
(2.10)
I
i 48
The Green's Function Method
Sec. 2.2
Since this limit does not exist, the interchange of limit and integration in (2.8) is not valid. We have therefore placed an "s" over the equality in (2.10) to indicate symbolic equality only. The delta function 8 (x - xo) has two remarkable properties. Symbolically, it is a function that is zero everywhere except at x = xu, where it is undefined. Second, when integrated against a function f that is continuous at xo, it yields the value of the function at xo. We note that (2.9) defines the operator T we were seeking in (2.1). Indeed, comparing (2.1) and (2.9) yields b (2.11 ) T = (.)8(x - xo)dx
1
....
a.
where (.) indicates the position of the function upon which T operates. From the basic definition of the delta function in (2.9), we obtain some additional relations. If we set Xo equal to zero, we find
l
b
f(x)8(x)dx
= f(O)
Delta Function
49
Suppose we wish to solve the equation Lu
=f
(2.15)
where L is a diffe~ential operator. Formally, the solution is given by multiplying both sides of (2.15) by the inverse operator, viz.
or (2.16) Since L is a differential operator, we shall assume that its inverse is an integral operator with kernel g(x, ~), so that u(x)
=
f
g(x,
~)f(~)d~
(2.17)
i
(2.12) I
Substitution into (2.15) gives
! I
Also, if in (2.9) we set f(x)
\
l
a
= 1, we obtain b
,
8(x - xo)dx
=1
f(x)
=
f
I
(2.13)
=L
Finally, from (2.4) and (2.8), we conclude symbolically that
= i
8(x - xo) :b 8(xo - x)
L[u(x)]
(~.14) I I
In concluding our development of the delta function and its properties, we remark that there are certain difficulties with the definitions. Indeed, any function that is zero everywhere except at one point must producci zero when Riemann integrated over any interval containing the point. The result in (2.13), for example, is therefote unacceptable in the Riemann sense. To interpret the integral, it seems that we must return to the basic defirtition in (2.8). The mathematical acceptability of integrals involving the delta function have, however, been fdrmalized in the Theory of Distributions, introduced by Schwartz [2]. In the theory, the delta function is called a generalized function, and the integral in (2.9) is said to exist in the distributional sense. Although the theory is beyond the scope of this book, the interested reader can find introductory treatments in [3],[4]. ' The central role played by'the delta function in the solution to certain differential equations becomes apparent in the following argument.
I I
f
g(x,
~)f(~)d~
Lg(x,
~)f(~)d~
(2.18)
where we have assumed, without proof, that we can move the operator L inside the integral. But, from the properties of the delta function, we have
f(x)
for x
E (a, b).
=
l
b
8(x -
~)f(S)d~
(2.19)
Comparing (2.18) and (2.19), we identify (2.20)
Presumably, if we can solve (2.20), then the solution to (2.15) is given explicitly by (2.17). The kernel g(x, ~) is called the Green's function for the problem. It is the purpose of this chapter to formalize and structure the introductory ideas above. The result will be the solution to a class of linear ordinary differential equations of second order by the Green's function method.
I
The Green's Function Method
50
Chap. 2
d 2u(x) du(x)' ao(x)--2-+at(x)--+a2(x)u(X)-AU(X) dx dx
Sturm-Liouville Operator Theory
EXAMPLE 2.3
2.3 STURM-LIOUVILLE OPERATOR THEORY Consider the following linear, ordinary, differential equation of order:
Sec. 2.3
s~cond
Consider Bessel's equation of order v, given by 1 I -u " - -u X
(v 2
+ x- 2 - k 2) u = f
(:2.21)
where A is, in general, a complex parameter independent of x. The fun6tions ao, at, and a2 are real and assumed to have the following properties [5],[6]:
ao =-1
= -O/x) = (v/x)2
al
a. a2, datJdx, and d2ao/dx2 are continuous in a ::'S x ::'S b ".. b. ao f:. 0 in a < x < b
a2
To transform to Sturm-Liouville form, we use (2.23)-(2.25) and obtain
In (2.21), we also require that u(x) be twice differentiable and that i f (x) be piecewise continuous. We may always recast this differential eqJation in Sturm-Liouville form, as follows:
q(x) =
I,
(2.22)
w(x)
so that 1 I I --(xu) X
(2.23)
= h p [fX at (t) dt]
(2.24)
w(~) = _ p(x)
(2.25)
.
i
ao(t)
ao(x) I
We may verify these transformations by substituting (2.23)-(2.25) ihto (2.22) to produce (2.21). The details are left for the problems. We r~wtite (2.22) in operator notation as follows: I
=f
2
x
EXAMPLE 2.4 ferent form, viz.
=x
2 2) k u =f
+ (V-x 2 -
•
-x 2U "
-
xu I -
[k ( x) 2 -
A = _v 2
ao
= _x 2
al
= -x
a2 = _(kx)2
L
1 'd[ p(x)d] + q(x) =-w(x) dx
Using (2.23)-(2.25), we obtain
(2.27)
dx
For the rem~inder of this chapter, without loss of generality, L will always mean the Sturm-Liouville operator in (2.27). I
v 2) u =
f•
(2.30)
We note that (2.30) is obtained simply by multiplying both sides of (2.28) by x 2 . In this case, we identify
(2.26)
where we identify the Sturm-Liouville operator L, viz.
(2.29)
Consider Bessel's equation, given by (2.28), in a slightly dif-
I
(L - A)U
V
2"
p(x) = x
The necessary coefficient transformations are given by [7]
p(x)
(2.28)
where "prime" indicates differentiation with respect to x. Comparing to (2.21), we identify
= f(x),
d [ p(x)-dU(X)] + . q(x)u(x) - AU(X) = f(x) - -1- w(x) dx dx
51
q(x) = _(kx)2
p(x) = x
1
w(x) = x
The Green's Function Method
52
,
i Chap. 2
so that in Sturm-Liouville form,
.
-x(xu')' - (kx)2 u
+ v 2u = f
.!
,
We note, in particular, that the weighting function w(x) differs from that in ;Example 2.3. ' , ,
•
It might appear that the distinction between (2.29) and (2.31) is tri;ial since the latter can be obtained from the former by dividing by x 2 . However, the difference in the weighting functions between (2.29) and (2.31) changes the Hilbert.space, and makes a major difference in spectral representations associated ~ith the radial portion of the Helmholtz equation in cylindrical ' coordinates [8], as we shall find in' Chapter 4. ,
.
EXAMPLE 2.5 follows:
Consider Legendre's equation on the interval x -(1 - x 2 )u"
+ 2xu' -
n(n
Stunn-Liouville Problem of the First Kind
E (-I, I),
For the first form of the Sturm-Liouville problem, we consider,(L - A)ll = over a finite interval x E (a, b) and for real A and real f. For -00 < a < b < 00, consider the Hilbert space L2(a, b) with real inner product
f
(u, v)
L).)l
l
b
u(x)v(x)w(x)dx
(2.34)
=
f,
a<x
(2.35)
where as
+ 1)11 = f
LA and where
L A = n(n
=
for al1 u, v E Ida, b). We define the Sturm-Liouville Problem ofthe First Kind, abbreviated SLP I, as follows:
,We identify I
53
2.4 STURM-LIOUVILLE PROBLEM OF THE FIRST KIND
,
,
Sec. 2.4
+ 1)
=
L - A
= __l_!- [P(X)!-] + q(x) w(x) dx
dx
(2.36)
(2.37)
We impose the following restrictions [9]:
ao =,-(1 _x 2 )
\
a. p, pi, q, ware real and continuous for a ::::: x ::::: b al
=,2x
a2
=0
b. p(x) > 0, w(x) >
c. A is real and independent of x
:Using (2.23)-(2.25), we obtain q(x)
In addition, we require u(x) E V L C Ida, b), where V L is the domain of the operator L. Because we are dealing with second-order differential operators, the domain is restricted to those functions that are twice differentiable. Finally, we require that u(x) satisfy two boundary conditions as follows:
=0
p(x)=l-x
2
w(x) = 1
'so that in Sturm-Liouville form,
- [(I -
x 2 )u']' - n(n
°for a ::::: x ::::: b
+ I)u
=
f
(2.33)
• The Sturm-Liouville form of the second order differential equati,on, given by (2.22), plays a central role in the solution of electromagnetic boundary value problems. We distinguish three forms of the SturmLiouville problem, which we consider in the next three sections.
B I (u)
= a = allu(a) + a12u'(a) + aI3u(b) + aI4u' (b)
(2.38)
B2(U)
= f3 = a2l 11 (a) + a22 u' (a) + a23u(b) + a24u'(b)
. (2.39)
where, for SLP I, a, f3, and aij are real. Typically, in (2.38), if a is nonzero, the boundary condition is said to be inhomogeneous. If a = 0, the boundary condition is homoger'eous. . There are several important special cases contained in the boundary conditions in (2.38): and (2.39). A boundary condition is unmixed if it involves conditions bn u(x) at one boundary only. If SLPI involves an
chap. 2
The Green's Function Method
54
unmixed condition at one end of the boundary and an unmixed condition at the other end, we refer to. this caSe as SLPI with unmixed conditions.'. The most general case of unmixed bqundary conditions is al3 = al4 = a21 = a22 = 0, so that , BI (u) = a = allu(a)
B2(U)
= f3
:::i: a23u(b)
, + a\2u (a)
(2.40)
+ a24u'(b)
(2.41) I
The relations in (2.38) and (2.39) are said to be initial conditions, if dll' = a22 = 1 and:all other aij coeffidents are zero, so that .
I
I
Bl(U)
= a = u(a)
(2.42)
B2(U}
= f3 = u'(a)
(2.43)
The two conditions in (2.38) and (2.39) are periodic if the value of the function u(x) at one boundary is identical to the value at the other boundary, and if the value of the derivative u' (x) at one boundary is identical to the I value at the other boundary. To produce the periodic conditions, we require all = -a13= a22 = -a24 = I land all other coefficients zero, so that
\;
(2.45)
:
k
E
E
(0, 1): I
R
with two homogeneous unmixed boundary conditions
= u(1) = 0
•
The operator Lin SLPI has aformal adjoint, which we constn:lctby the following procedure. For u,.V E £2(a, b), we form i (Lu, v) ::::::
l
a
b
{ - -1- d
(Lu, v)
= Ja{b u(x) {-
I d [dV(X)] W(x) dx p(x)~
['p(x)-dU(X)] w(x) dx' dx .
: + q(x)u(x) } v(x)w(x)dx
+ q(x)v(x) } w(x)dx
_ {P(X) [V(X) d~~X) _ u(x) d~~)]}
b
l
a
(2.47)
We write this result in inner product notation as (Lu, v) = (u, L *v)
+ J(u, v)
I:
(2.48)
where J (u, v) is called the conjunct and is given by J(u, v)
= -p(u'v -
uv')
(2.49)
The operator L *, produced in the integration by parts operation, is called the formal adjoint to L. We note that
=L
(2.50)
When (2.50) is true, we say that L is formally self-adjoint. We conclude, in general, that the Sturm-Liouville operator for SLPI is formally selfadjoint. In our search for a solution, or solutions, to (2.35), we shall first assume that the boundary conditions in (2.38) and (2.39) are homogeneous. We then follow with the extension to the inhomogeneous case. Accordingly, if u(f) is to be a solution to (2.35), we require the following I restrictions: a. u
We identify p(x) = w(x) = 1, q(x) = 0, A. = k 2 , a = 0, b = I, a = f3 = O. In the boundary conditions in (2.3~) and (2.39), all coefficients aij = 0, except all = a23 = 1. We find that all requirements for SLPI are satisfied.
55
Integrating by parts twice, we obtain
(2.44)
Consider the following differential equation on x
u(O)
Sturm-Liouville Problem of the First Kind
L*
= u(b) u'(a) = u'(b) u(a)
EXAMPLE 2.6
Sec. 2.4
E
£2(a,b)
b. u E V L
c. u satisfies two boundary conditions, B j (u)
= 0, B2(U) = 0
These restrictions define a linear manifold M L C £2(a, b). The proof is left for the problems. We next consider the function vex) in (2.48). We place the following restrictions on v (x):
a. v
E £2(a,b)
b. v
E
VL*
I
(2.46) I
II i
c. v satisfies two ad;oint hnundary conditions, Rt(v) = 0, RI(I') =
°
The Green's Function Method
56
!
Chap. 2
Since vex) is unspecified in the ~riginal problem statement in (2.35), we are free to choose the adjoint boundary conditions in any manner we wish, consistent with the integration by parts operation in (2.48). We define the adjoint boundary conditions to by those conditions Bt(v) = 0, B~(v? = 0 that when coupled with the boundary conditions on u(x), result 1\1 the vanishing of the conjunct, viz. ,
Sec. 2.4
Substitution into the conjunct gives J(u, v)
I:
In this case, (2.51) is satisfied if we choose adjoint boundary conditions
1
:
In =
(2.51 )
0
These res,trictions on vex) defin~ a linear manifold ML* C L2(a, b). At present, it is not clear that it is ,Possible to define the a?j~int boun?~ry conditions such that (2.51) is satisfied. We next show explicitly the adJomt boundary condition result for the unmixed, initial, and periodic cases. We have defined the unmixed boundary case in (2.40) and (2.41). For the homogeneous case, they become
B2(U) = IX23,U(b)
la =
+ IX24U'(b) =
a
.
IX
23 v(b)'+ v' (b)] - p(a)u(a) IX24 '
[~v(a) + V1(d)] IX \2
(2.61)
+ IX24V'(b)
= 0
(~.56)
We note that in the unmixed boundary case, the boundary conditiohs on vex) in (2.55) and (2.56) are identical to those on u(x) in (2.52) and (2.53!. Therefore, for the case of unmixed boundary conditions, the linear ?1 am fold ML is the same as the line~r manifold ML*. A formally self-adjoint operator with ML = ML* is said to be self-adjoint. We shall. find s~bse quently that self-adjoint problems possess remarkable properties. I For homogeneous initial conditions, we have u(a) = 0
I ~.57)
/l'(a) =
(2.58)
0
(2.62)
- u(a) [p(a)v'(a) - p(b)v'(b)]
In this case, (2.51) is satisfied if we choose adjoint boundary conditions Bt(v)
'
(t·
B;(v) = IX2jv(b)
-pCb) [u'(a)v(b) - u(a)v'(b)]
= p(a)v(a) -
p(b)v(b)
= p(a)v'(a) -
p(b)v'(b)
=0
(2.63)
=0
(2.64)
(2.54)
In this case, (2.51) is satisfied if we choose the following adjoint boundary . ! conditions: . I 55 ) Bt(v) = IXII v(a) + IX\2v'(a) = 0
..
0
= u'(a) [p(a)v(a) - p(b)v(b)]
(2.53)
0
I
=:p(b)u(b) [
B;(v) = v'(b) =
+ pea) [u'(a)v(a) - u(a)v'(a)]
We use these expressions in the ~onjunct to eliminate u' (a) and u' (b), ~iz. b
(2.60)
h
i
J(J, v)
(2.52)
I
0
We note that for initial conditions, the boundary conditions on vex) in (2.60) and (2.61) are not the same as those on u (x) in (2.57) and (2.58). Therefore, M L '# ML*, and the initial condition case is never self-adjoint. For periodic:conditions, we substitute (2.44) and (2.45) into the conjunct to give
l
I
Bt(v) = v(b) =
b
J(i,t, v)
J(u, v)
(2.59)
= -pCb) [u'(b)v(b) - u(b)v'(b)]
I
\
57
Stunn-Liouville Problem of the First Kind
Bi(v)
We note that for the general form of L in (2.37) and for periodic' conditions, the boundary conditions on vex) in (2.63) and (2.64) are notthe same as those on u(x) in (2.44) and (2.45). However, if the operator L is such that pea) = pCb), the conditions are identical and the problem becomes self-adjoint. To produce the solution to SLPI by the Green's function method, we define two auxiliary problems: the Green's function problem and the adjoint Green's function problem. The Green's function problem is defined as follows: •
,
L).g(x,O =
I
8(x -~) w(x)
a < ~ < b
,
(2.65)
B)(g) =
0
(2.66)
B2(g) =
0
(2.67)
\
The Green's Function Method
58
Chap. 2
I
where w(x) is the weight functiori defined in (2.25) and (2.27) and appearing in the inner product definition in (2.34). We note that, by definition, the boundary conditions on g are identical to the boundary conditions on u. The adjoint Green's function problem is defined as follows: LAh(x,~)
=
8(x - ~) ~(x) ,
(2.68)
Bt(h)
=0
B;(h)
=0
,
Sec. 2.4
Sturm-Liouville Problem of the First Kind
°
Bt(h) = 0, B2(h) = reduces the second term in (2.74) to zero. The extension to the inhomogeneous case, however, is now available. We simply apply the given b
Consider the following differential equation on x E (a, b):
-uti - AU
(2.70)
We note tliat, by definition, the boundary conditions on h are identic~1 ~o the boundary conditions on v. We also note that the boundary conditions associated with the Green's function and the adjoint Green's function are always homogeneous. The solution to SLPI by th~ Green's function method is obtained by taking the inner product of LAu with h, viz. ,
l
u(a) u'(a)
In this case, p(x)
:=:: w(x)
u(O=
l
b
f(x)h(x,Odx+a
=
leu, h)
dU(X) -p(x) [ ~h(x, 0
-
dh(X,O] u(x) dx
,
u(~) = (f, h) -
b
leu, h)
,
I
(2.73)
Q
or, explicitly,
u(~) =
i
b
,
+
f(x)h(x, {
~)w(x?dx
, [dU(X) : p(x) ~h(x,~) - u(x)
dh(X'~)]}IX=b dx
X=Q
(2.74)
Equation (2.74) is the formal solution to SLP1, provided that we can ~olve the adjoint Green's function pr6blem, given in (2.68)-(2.70). Fot homogeneous boundary conditions BI (Il) = 0, B2(1l) = 0. the selection
dx
where we have applied the boundary conditions u(a) adjoint boundary conditions h(b,
~) =
dh(b, dx
~) =
u(~) =
I
b
f(x)h(x,
-f3h(a,~)
=
a, u'(a)
=
f3 and the
°
Note that for homogeneous boundary conditions, a ishes and
(2.72)
Substitution of (2.35) and (2.68) Into (2.71) gives
dh(a,~)
a
I
where the integrations indicated by the inner products are with respect to x. From (2.49), the conjunct 1 (u, h) is given by :
=a = f3
= 1, and (2.74) yields
')
\
=f
with boundary conditions
<:2.71)
(
59
= f3 = 0, the conjunct van-
~)dx
•
We should note that in (2.74) and in Example 2.7, the solution is given in terms of the variable ~, withx as the dummy variable of integration. This notation causes no difficulty since ~ simply refers to a point of evaluation of U on the interval (a, b). We shall subsequently obtain the solution Il in terms of x with ~ as the dummy integration variable by a simple interchange of x and ~. The reader is cautioned, however, to withhold performing this step until after explicit evaluation of the adjoint Green's function. This evaluation is the next subject for discussion. We now show that it is never necessary to find the adjoint Green's function h(x,~) directly from (2.68)-(2.70). Indeed, if we determine the Green's function g(x, ~), defined by (2.65)-(2.67), the adjoint Green's function follows immediately. The details follow. We form (LAI(x,
~), hex, ()} =
(I(x,
~). L)..h(x, ~')} + 1(1(, h) 1:::
(2.75)
Sec. 2.4
The Green's Function Method
60
Stunn-Liouville Problem of the First Kind
a. Region I: a < x <
where integrations in the inner products are with respect to x. ApplicatIon of (2.66), (2.67), (2.69), and (2.70) reduces the conjunct to zero, viz~
61
~
b. Region 2: ~ < x < b
I
J(q' h) [ : : = 0 Using this result and (2.65) and (2.68) in (2.75) gives h(~,()
= g«(,~)
A simple variable change gives ".
h(x,~)
= g(~, x)
b.76)
We conclude that, if we can find the Green's function g(x, 0, the adjoint Green's function h (x, 0 follows immediately by an interchange of:x and ~. Substitution of h(x,~) into (2.74) completes the solution to SLP~. A further simplification occurs if the Green's function problem i:s selfadjoint. In this case, the operator and boundary conditions for the Green's function and the adjoint Green'i> function are identical, and we must have h = g. Therefore, h(x,O
= g(x,~) = g(~:, x)
(self-adjoint case)
~2.77)
When g(x, 0 = g(~, x), the Green's function g is said to be symmetric. Substituting (2.77) into (2.74) gives, for the self-adjoint case, ,
u(~) =
i +
,
b
~)w(x)dx
{P(X)[~g(X'O-U(X) du(x) ,
{
__ I ~ [ p( x~] _ 8(x - 0 )d- +q(X)-A } g(x,~)---w (x )dx
r=b
dg(x,~)'
dx
I
]} x=a
.<2. 78 ) I
We note that in the self-adjoint case, we have produced the useful result that it is unnecessary to consider any aspect of the adjoint problem. Indeed, (2.78) involves the Green's function g(x, ~) rather than the adjoint dreen's function hex, ~). The only remaining step in the solution to SLP 1 is the specifiC' determination ot g(x, ~). The differential equation that describes the Green's function is given by (2.65). We' write this equation for x =I- ~ as follows:
x
w(x)
(2.80)
We multiply by w(x) and integrate over the region (~ - E,~' + E) to give -
'
j(x)g(x,
Since the equation is of second order, the solution in Region I will contain two as yet undetem1ined coefficients. In Region 2, the solution will contain two additio~al undetermined coefficients. These four coefficients require four constramts on the Green's function for their determination. The conditions B I (g~ = 0 and B2 (g) = 0 provide two constraints. The remaining two are provIded by conditions joining together the two regions at x = ~. Recall that in seeking a solution u(x) to (L - A)U = j, we have required that u(x) be twice differentiable. In the solution to (2.65), however, we' r~lax thi~ requiremen~ so that the Green's function is required to be simply dIfferentIable on the Interval a < x < b. This relaxation is logical since the s~cond differentiation of the Green's function produces a singularity' functIOn 8 (x - ~) at x = ~. Since a differentiable function is continuous, the third constraint on the Green's function is that it must be continuous at x = ~. For the fourth constraint, we write explicitly the Sturm-Liouville operator in (2.65), viz. '
l
HE
~-E
g d(d P- ) dx + l~+E (q -
-d X
dX
~-E
A)gwdx = I
(2.81 )
In t~e second integ~al, since q, g, and ware continuous, (q - A)gW is contmuous over the Interval [~ - E, ~ + E]. Since the interval is closed and bounded, the continuous function (q - A)gW is bounded on the interval. Let M be the upper bound on I(q - A)gWj. Then,
I~~:E (q _ A)gWdX! :::: 2EM
I
:(2.79) This homogeneous second-order equation can be solved in the following ~~~:
I
and therefore,
IJ~-E
lim {HE (q - A)gWdxl E--+O
=0
Performing the integration in the first integral in (2.81) and taking the limit as E --+ 0 gives the fourth constraint, viz. dg , dx ~+
I
-
dg dx ~- =
-
1
p(~)
(2.82)
The Green's Function Method
62
,
.
CHap; 2
where we have used the continuity of p(x) atx = ~. Typically, our notation : ~- indicates the limit as x -+ ~ from below. We collect the characteristics of the Green's function g(x, ~) that allow for its determination as foll?ws:
LAg =0,
x
=0
(~.85)
= gl~+
(2.86)
I
dg
I
dx ~+ -,dx ~- =
Sturm-Liouville Problem of the First Kind
g(x.O
=
0,
x <
( C sin kx
- p(~)
(2.92)
x>~
+ Dcosk~
= 0
Jump gives k(Ccosk~
We summarize the procedure fordetermining the Green's function as, fol-
+ D cos kx,
s
The remai~i~g two coefficients C, D are evaluated by applying the continuity and Jump condItIOns. Continuity gives C sink~
(2.87)
1
63
This result can be verified by substituting (2.91) into (2.90) to show that the differ- , entia( equation is satisfied. We apply the two boundary conditions, given explicitly by (2.89). The result is A = B = 0, so that
d83) ,
BI (g) = 0
gl~dg
~
(2.84)
B 2 (g)
....
=1=
Sec. 2.4
I
-
Dsink~)
=-1
Solving simultaneously gives
lows: 1. Solve (2.83) for x < ~ and for x > ~. The result will contain;four as yet undetermined coefficients. 2. Apply the boundary conditions indicated in (2.84) and (2.85). These two conditions will result in determination of two ofr the . four coefficients. i 3. Apply the continuity condition (2.86) and the jump condition (2.87). These two conditions will result in the determination of the final two ~oefficients.
C
k
I
I
~/,.
.
(.
",
= _ cosk~
D = sink~ k
Substitution into (2.92) and application of a trigonometric identity yielcis
I
We demonstrate the procedure in several examples. EXAMPLE 2.8
Consider the follo~ing Green's function problem onx E (0, b):
d 2g dx 2
---eg=8(x-~)
(~.88)
g(O,~) ~ dg~; ~) = 0
(2.8,9)
g(x, S) =
d2
_-!2 dx
=
I
(2.93)
x>~
'
EXAMPLE 2.9
•
Consider the following Green's function problem on XE (0, a):
d 2g - dx 2
-
2
k g
= 8(x - ~)
(2.94)
g(O, ~) = g(a,~) = 0
(2.95)
Proceeding as in Example 2.8, we obtain
;
k 2g = 0,
x
.
f= ~
(2.90)
A solution to (2.90) can be written for the two regions bisected by x
g(x,~)
k
x < ~
It is instructive to verify that (2.93) satisfies the requirements for the Green's function given in (2.83)-(2.87). The details are left for the problems.
We shall solve for the Green's function g(x, ~) by using (2.83)-(2.87). For x. f= ~,
(2.88) becomes
(
0, sink(s - x)
A sin kx
+ B cos kx ,
x < ~
Csinkx
+ Dcoskx.
x > ~
= ~ as follows: (2.91)
1:) _(
g (x,<; -
~. sinkx + Bcoskx, ~sink(a - x)
+ Dcosk(a -
x <
x),
s
(2.96)
x > ~
~ote that th~ form of sol uti on for x > ~ is chosen so that the arguments for the slOe and cosme are equal to zero at the boundary x = a. This form satisfies (2.90).
The Green's Function Method
64
Chap. 2
while making the evaluation of the urdetermined coefficients C and D easier. (The reader should verify that selection of the form C sin kx + D cos kx would yield the same result; but the process of coefficient evaluation would be more complicated.) Applying the boundary conditions given in (2.95), we obtain B = D = 0, imd
I
Alsinkx, , Csink(a - x),
x < ~
Sec. 2.4
Sturm-Liouville Problem of the First Kind
6S
Self-Adjoint Green's Function Problem I. Write the solution in the form given by (2.78). 2. Substitute the boundary conditions Bl (u) (2.78).
=a
'
B ( ) 2
u
= f3
I'nto
~2.'97)
3. Substitute the boundary conditions B I (g) = 0 B ( ) = 0 . (2.78). ' 2 g Into
The remaining two coefficients C, D are evaluated by applying the continuIty and . I ' jump conditions. The results are . I I
4. Solv~ the .Green's function problem given by (2.65)"':'(2.67) and substitute Into (2.78). 5. Interchange the variables x and ~ in (2.78).
g(x,O=
x>~
sin k~ C=-k sin ka
.... A
=
,(2.98)
_si_n_k(_a_-_~:...:...) k sin ka
(2.99)
EXAMPLE 2.10
Substitution into (2.97) yields
1k \ g(x,O=-k' sm a
0
sin kx,
x < ~
-x)sink~,
x > ~
sin k(a 1
sin'k(a
We ~em~rk again that, in the self-adjoint case, there is no necessity for consldenng a.ny aspect of the adjoint problem. We next illustrate these procedures With some examples. Consider the following differential equation on x
-u " - k. 2 u = (~.l 00)
1
•
We next summarize the steps for solving the differential equation L)."u = f by the Green's function method. We distinguish two cases.
Nonself-Adjoint Green's Function Problem 1. Write the solution in the form given by (2.74). 2. Substitute the boundary conditions Bl (u) = a, B2(U) = f3 jnto (2.74). 3. Substitute the adjoint bbundary conditions Bj(h) = 0, Bi(ll) = 0 into ( 2 . 7 4 ) . ' , ,
4. So~ve the Green's function problem given by (2.65)-(2.67)'. 5. Obtain the adjoint Green's function through (2.76) and substitute into (2.74). . 6. Interchange the variables x and ~ in (2.74).
(0, b):
f
with boundary conditions
u(O)
Note that the Green's function derived in (2.100) is symmetric, g(x,~) = g(~, x), a result that we anticipate from the unmixed boundary conditions in (2.95) and the self-adjoint property in (2.77).
E
u' (0)
= ex = fJ
These boundary conditions define an initial value problem which we kno . ' . w IS never self-adjoint. We therefore use (2.74) and obtain
l
u(~)=.
b
o
f(x)h(x,~)dx+ex
dh(O '~)-fJh(O~) dx'
(2.101)
where the adjoint $reen's function equation is given by
d 2h - dx 2
-
eh
= 8(x -~)
(2.102)
and where we have used the adjoint boundary conditions
h(b, 1::) <;
=
dh(b, ~) dx
=0
(2.103) ,
As shown in (2;76), w~ can obtain the adjoint Green's function h(x,~) directly from the Green s functIOn problem, given in this case by
d 2g - dx 2
-
g(O,O
e g = 8(x -~)
(2.104)
= dg(O, ~) = 0
(2.105)
dx
·I.· .~
l ~
The Green's Function Method
66
Chap. 2
I
x < ~
g(x,~)
=
I
sink(~ -x)
:
k
'
67
Sturm-Liouville Problem of the First Kind
EXAMPLE 2.11
But, we have obtained this Green's (unction in (2.93), Example 2.8, as follows: 0,;
Sec. 2.4
Consider the following differential equation on x _utI =
E
(0, 1):
f
(2.106) with boundary conditions
x>~
u(O)
=a
Application of (2.76) yields the adjoint Green's function, viz. u(l) = 0
h(X'()~1
x>~
0, sin k(x k
(2.107)
~)
The associated Green's function problem is
x<~
".
From (2.107), we also obtain h(O,~)
sink~
= --k-
dh(O, ~) = cos k~ dx Substitution of (2.107)-(2.109) int~ (2.101) gives
l
J
I
u(~)
=
~
1o
f(x)
(2.108) C2.109)
Since the bounda~ conditions are unmixed, the Green's function ~roblem is selfadjoint. We therefore use (2.78). After application of the boundary coilditions on u(x) and g(x, ~), 't'e have u(~)
sin k(x - ~) sin k~ , dx +fJ-- +acosk~ k k
=
, 1 o
.
10r
fcn
sinkC~-x)
k
"
sinkx
d~+fJ-k-+acoskx
,
g(x,~)
,
1 [ . u(x) = "kIm e- rkx
1r 0
. ]
sin kx f(~)e'k~d~ +fJ-- +acoskx k
(~.Ill) \ i
Since real differentiation and the imaginary part operator can be interchan?ed, it is now straightforward to show tha~ this solution satisfies the original differential equation and the boundary conditions. The details are left for Problem 2.1 I.
•
=
dg(O,~)
+ a----'------
(2.112)
dx
I
(1 - nx,
x<~
x)~,
x>~
(1 -
(2.110)
It is important to assure that our solution in (2.110) satisfies the differential equation and the boundary conditions. To do so, it is necessary to twice differentiate (2:.110). This differentiation requires some care. We note that the integral in (2.110) involves a variable upper limit. To differentiate, we make use of a theorem [10], as follows: "The derivative of the definite integral of a continuous function with res~ect to the upper limit of integration is equal to the value of the integrand functlon at this upper limit." In (2.110), howev¢r, the variable x occurs not only in the ,upper limit, but also under the integral sign. To remedy this problem, we write (2.110) as follows: : ! '
~)dx
Using the procedure for Green's function evaluation, we find that.
An interchange of x and ~ yields the final solution, viz. u(x) =
f(x)g(x,
and
n
dg(O, = 1_ ~ dx Substitution into C2.112) followed by an interchange of variables yields
l U(nd~ + [I x
u(x) = (I - x)
x
(I -
nf(~)d~ + a(l -
x)
We leave it to the reader to show that this solution satisfies the differential equation and the boundary conditions.
•
In this section, we have defined the requirements for SLPI and have given a procedure for its solution by the Green's function method. Note that the parameter A and the forcing function f (x) were constrained to be real. In many of the interesting problems of electromagnetic theory. Aand lex) are complex. We consider this case in the next section.
!
The Green's Function Method
68
,Chap. 2
2.5 STURM-LIOUVILLE PROBLEM OF THE SECOND KIND For the second form of the Sturm-Liouville problem, we consider (L A)U = f over a finite intervalx E (a, b) and for complex A and complex f. Since we are now dealing with complex quantities, the Hilbert space {da, b) is now defined with complex inner product ' (U, v) =,
I
h
Sec. 2.5
Sturm-Liouville Problem of the Second Kind
(Lu, v)
l = l =
h {
a b
a
for allu, v E {da, b). We idefine the Sturm-Liollville Problen~ of the SeconiJ-Kind, abbreviated SLP2, as follows:
=
f,
-(X)
< a < x < b <
(X)
(2.119)
= (1/, Lv) + J(u, v)
I
(2.115)
i
+ q(x)v(x) } w(x)dx
_ dv(x)
[V(X)~ - u(x)~ ]} du(x)
- {P(X)
(2,114)
+ q(x)u(x) } v(x)w(x)dx
d [dV(X)] u(x) { - -1 - p(x)-w(x) dx dx
,
where
Lii
1 -d d [dU(X)] --p(x)-w(x) x dx
i
LAu
=
In a similar manner to the SLPI development, we may s~ow that the operator Lin SLP2 is formally self-adjoint. Indeed, for u, v E £:2(a, b), we find that :
(i 113)
u(x)v(x)w(x)dx
Lu
,
69
[b a
jh a
(2.120) and where
where we have used Lv = Lv and where, for SLP2, L
= __1_. ~ [p(X)~] + q(x) w(x) dx
dx
(2.116)
We impose, the following restrictions ,
,
°for
S x S b c. A is complex and independent of x
b. p(x) > 0, w(x) >
In addition, we require
a
I
u(x) E, V L). C £2(a, b), where V L). is the domain
of the operator LA' Because we are dealing with second-order differential operators, the domain is restricted to those functions that are twin; differentiable. Finally, we require that U (x) satisfy two boundary conditions as follows: I
I ,
i
BI(u)
= a = allu(a) +~12u'(a) +a13u(b) +aI4u'(b)
(2.117)
B2(U)
= fJ = a21u(a) + ~22u'(a) + a23u(b) + a24 u '(b)
(2.118)
i
~
(2.121)
We shall require the same restrictions on u and v as those developed for SLPI. For homogeneous boundary conditions, we plac~ the following restrictions on u(x):
a. p, p', q, ware real and continuous for a S x S b '
J(u, v) = -p(u'v - uv')
where the aij are real. Beca~se of the generalization of A and f(x) ~o include complex values, we anticipate that the solution u (x) to (2.1 i 4) WIll be a complex function. Since ~ (x) is complex, (2.117) and (2.118) :will, in general, generate complex values of a and fJ. We note, however, that the i , operator L is real; that is,
a. u
E £2(a, b)
b. u
E
C.
VL
u satisfies two boundary conditions, B] (u)
= 0, B2(U) = a
We place the following restrictions on vex): a. v
E £2(a, b)
b. v
E
VL*
= 0, Bi(v) = 0 We define the adjoint boundary conditions to be those conditions Bi (v) = 0 Bi(v) = a that, when coupled with the boundary conditions on u(x), result c. v satisfies two adjoint boundary conditions, Bi(v)
in the vanishing of the conjunct, viz. J(u, v)
1:=0
(2.122)
The Green's Function Method
70
,,
,
qhap.2
We note in (2.121) that the conjunct involves V, rather than v. Howevyr, the conditions on v are easily produced from the conditions on V. We define the conjugate adjoint boundary conditions by
Sec. 2.5
EXAMPLE 2.12 We consider characteristics of the following differential equation on x E (0, 1): (L - A)U
I
+ afzv'(a) +ai3v(b) + ai4v'(b) = 0 aZI v(a) + azzv'(a) +az3 v (b) + aZ4 v'(b) = 0
aflv(a)
q.123)
ail v(a) + aizv'(a) +'ai3v(b) + ai4 v'(b) = 0
(2.125)
+ azzv'(a) +az3v(b) + aZ4v'(b) = 0
(2.126)
...·aZlv(a)
'
We conclude that, because the aij 's are real in SLP2, the conjugate adjoint boundary conditions and the adjoint boundary conditions are identical. We next consider specific conditions that result in L being self-apjoint in SLP2. For the unmixed bountlary case, we again have 'i.
"I,
t~l'
'.
,: ~.:,
••.
allu(a)
+ al2u'(a) = 0
(2.127)
anu(b)
+ aZ4u'(b) = 0
(2.128)
We use these expressions in the 'conjunct to eliminate u'(a) and u'(b), viz.
~I
J(u, v)
\b = p(b)u(b) ,
a .
Z3 [a v(bJ, aZ4'
+ V'(b)]
- p(a)u(a)
'
[~v(a) +v'(a)] al2 i
(2.129)
The conjunct in (2.129) vanishes provided
+ al2v'(a) = 0 anv(b) + a24v'(b) = 0
all v(ai
which, in SLP2, always implies'that
+ al2v'(a) = 0 anv(b.) + a24v'(b) = 0
all v (a)
with L
=
2
(2.130) (2.1,31 )
I (2.132) I
(2.133)
We note that, in SLP2, the unmixed boundary case yields boundary conditions on v(x) identical to those on u(x). Therefore, the linear manifold ML is the same as the linear manifold Mu. We conclude that unmixed boundary conditions in SLP2 yield a self-adjoint operator just as in~LPl. We remark that this result depends on the restriction to real aij' We shall find in the next chapter that this restriction has a dramatic effect pn the eigenvalues of the operator L.
=f
2
_d ; dx and with boundary conditions
(4. 124)
where the a~. 's are chosen so that (2.122) is satisfied. Taking the complex conjugate ofh.123) and (2.124) and noting that the a;j 's are real, we obtain
,
Stunn-Liouville Problem of the Second Kind
u'(O) - au(O)
u(l)
=0 =a
~
,
where A and f are complex and Cl is real. We note that the problem meets all of the requirements for SLP2. In addition, the boundary conditions are unmixed. We therefore conclude that the operator L is self-adjoint. We stress that a different result would have been obtained if Cl E C. Indeed, for u, v E £2(0; b); we have (Lu, v)
=
(Ii, Lv) - u'(l)v(l) - Ii(O) [V'(O) - ClV(O)]
where we have applied the boundary conditions on lI. The conjugate adjoint boundary conditions that reduce the conjunct to zero are V'(O) - ClV(O)
" '* t,
=0
I
v(l) = 0
Taking the complex conjugate, we have V'(O) - &v(O) v(1)
=0 =0
We conclude that the conditions on v are not the same as the conditions on lI, and therefore the operator L is no longer self-adjoint. We shall investigate the distinction between real and complex Cl in this example again in the next chapter.
•
We next con.sider homogeneous initial conditions. Following a similar procedure to that in (2.127)-(2.133), we find that the initial condition case is not self-adjoint in SLP2. The details are left for the problems. For periodic conditions in SLP2, a similar procedure shows that the operator L is self-adjoint, provided that p(a) = p(b). Again, the details are left for the problems. The solution procedure for SLPI has been given in (2.71)-(2.74). We now show that the solution to SLP2 follows along similar lines, with modification to accommodate complex A and f(x). We take the inner product of LAu with the adjoint Green's function h and integrate by parts twice to give
":-;:
72 j
The Green's Function Method
Chap. 2
Sec. 2.5
Stunn-;-Liouville Problem of the Second Kind
73
.
(LAu, h) = (u,
where
L~h) + leu, h) [::
i
(4.134) , I I
i~ = L - i.
(2.135)
B2(1~) =
B~(h)
f3 in c0!1junction with the conjugate adjoint boundary conditions
= 0, B;(h) = O.
As in the case of SLP1, we can show that it is never ne~essary to find the conjugate adjoint Green's function directly. Indeed, we form
and
(LAg(x, , leu, h)
=
-p(x)
[dU(X)~h(x, 0
-
U(X)
dh(X,O] dx
(2.136)
w(x)
,
,
I
- leu, h)
(2.138)
IX=b X=(J
or, explicitly, u(O
=
l
(J
,+
b
,
p(x)
h(x,~)
= g(~,x)
LAg(X,~)
[dU(X) - :
dll(x, ~h(~,~) - u(x) dx
~)] Ilx=b
X=(J
(2.141)
i3;(h) = 0
(2.142)
Similar to the SLPI case, the selection of the boundary condition~ proceeds as follows. We first assume that the boundary conditions on u are homogeneous. Then, Bi(h) and B;(h) are chosen such that the conju~ct in (2.138) vanishes. The extension to the inhomogeneous case, however, is nowavailahle. We simply apply the given boundary conditions 81 (u), = ex,
(2.144)
w(x)
(2.145)
(2.147)
LAh(x,~) = 8(x
(2.148)
B2(g)
=0
.
=0 =0
B 1(g)
(2.140)
8(x -~)
= ---
(2.139)
We note that (2.139) is the solution to SLP2, provided that we can determine the conjugate adjoint Green's function h(x, ~). Taking the co~plex conjugate of both sides of (2.137), we obtain •
B~(h)
= g(~',~)
We shaH always proceed to find the Green's function g(x, ~), and then produce the ~onjugate adjoint Green's function h(x, ~) from (2.144). Substitution of h (x, ~) into (2.139) completes the solution to SLP2. A further simplification occurs when the operator L is self-adjoint. ~h~ ,
f(x)h(x, ~)w(x)dx
. I
II(~, () or, with a variable change,
I
= (f, h)
(2.143)
from which
(2.137) ,
Substitution of (2.137) into (2.134) gives u(O
~), hex, (») = (g(x, ~), L~h(x, (») + leg, h) [ : :
I
The adjoint Green's function pr0blem is given by ". : 8(x - t) L*h = --_sA :
i
(2.146)
and - ~) _ w(x) B](h) = 0
(2.149)
=0
(2.150)
B2(h)
The fact that the boundary conditions on h are identical to the boundary conditions on g is deduced as follows. First, the conditions on g and hare identical from the self-adjoint property of L; second, the conditibns on h and hare alwa£,s 1dentical in SLP2. We conclude from (2.145)-(2.150) that g(x, 0 = h(x" ~), and therefore, using (2.144), we find that' (self-adjoint case) Suhstitution into (2.139) gives, for the sclf-ndjoint case,
(2.151)
The Green's Function Method
74 .
u(~) ~
1 b
f(x)g(x,
~)w(~)dx
{P(X)[~g(X'~)-U(X) du(x) .
Sec. 2.5
Sturm-Liouville Problem of the Second Kind
EXAMPLE 2.13
:
a :
+
Chap. 2
Consider the following differential equation on x (L - A)U =
dg(x,~)
dx
IX=b ]} x=a
•
(0, a):
f
with boundary conditions
,
We shall summarize the steps for solving SLP2 by the Green's function i method. We distinguish two cases.
u'(O)
:
L.:Write the solution in tHe form given by (2.139). 2. Su'Jstitute the boundary conditions B1 (u) = a, B2(1l) =f3 into (2.139). 3. Substitute the adjoint boundary conditions Bi (h) = 0, Bi (h) = into ( 2 . 1 3 9 ) . ' ' 4. Solve the Green's function problem given by (2.65)-(2.67). 5. Obtain the conjugate adjoint Green's function h through (2.144) and substitute into (2.139). 6. Interchange the variables x and ~ in (2.139).
°
wher~.1 and A are .complex. The probl.em is of class SLP2. Since th~ boundary ~OndItlOns are unmIxed, the operator L IS self-adjoint with respect to the complex mner product (u, v) =
u(~) =
It is interesting and extremely useful to note that for L self-adjoint, the procedure for obtaining the solution to SLP1 and SLP2 is identical. Indeed, we have proved that in both of these cases, we may obtain the solution in terms of the Green's function g(x, ~) rather than the adjoint Green's function (SLPl) or the conjugate adjoint Green's function (SLP2). Spec,ifically, (2.78) and (2.151) are identical. It is only in the cases of nonself:.adjoint operators where we use the adjoint Green's function (SLP I) or the conjugate adjoint Green's function (SLP2), respectively. We i1Iustrate these ideas in the following examples.
u(x)jj(x)dx .
l l f(~)g(x, ~)d~ a
f(x)g(x, ndx
The self-adjoint property produces a symmetric Green's function, so that a
Self-Adjoint Green's Function Problem
°
r
~
Because of the self-adjoint property of L, the form of the solution is given b (2.152). In this case, y
u(x) =
1. Write the solution in tbe form given by (2.152). 2. Substitute the boundary conditions BI(U) = a, B2(u) ='{3 into (2.152). 3. Substitute the boundary conditions B1 (g) = 0, B2(g) = into (2.152). 4. Solve the Green's function problem given by (2.65)-(2.67) and substitute into (2.152): 5. Interchange the variables x and ~ in (2.152).
=0
u'(a) = 0
I
Nonself-Adjoint Green's Function Problem
E
d2 L = - -2 dx
(2.152) I I
75
where we require the solution to the Green's function problem
d 2g t ---Ag=8(x" dx 2
)
with boundary cond'itions dg(O,~)
dx
=
dg(a,~)
-0
dx-
We form g= .
I
A cos .Jf..x,
x<~
Bcos.Jf..(a-x),
x>~
a
where we have applied the boundary conditions at x = 0 and x = to eliminate two coefficients. Application of the continuity and jump conditions at x = ~ yields A = _ cos .Jf..(a - ~) .,fi. sin .,fi.a B
=_
cos .,fi.~ .,fi. sin .,fi.a
I
The Green's Function Method
76
Chap. 2
The Green's function therefore is ~iven by ,
g(x,~)
.
1
0,
x < ~
cos J)..~ cos J)..(a - x),
x>~
cos J)..x cos J)..(a -
1
= --=---=::J).. sin J)..a
, As expected from the self-adjoint property, the Green's function is symm~tric. EXAMPLE 2.14
Consider the iollowing differential equation on x
E
•
(0, h):
Sec. 2.6
Stunn-Liouville Problem of the Third Kind
77
In our study of SLPI and SLP2 problems, SLPI could properly be considered as SLP2, with the specialization that all quantities are real. We now take that point of view and classify all problems so far studied in this chapter as SLP2. Green's function problems classified as SLP2 do not nearly exhaust all of the cases of practical interest. There are many problems of interest in electrornagnetics that do not satisfy the requirements of SLP2. Such problems are classified SLP3 and are considered in the next section.
2.6 STURM-LIOUVILLE PROBLEM OF THE THIRD KIND
". with boundary conditions
In defining the ,~tlmn-Liolll'ille Problem of the Third Kind, abbreviated SLP3, we again consider the following differential equation:
u(O) = a u'(O)
= f3
I
It is identical to k ,a, f3 , f are complex. The problem is of class SLP2. h were f S' h b' d ry Example 2.10 except for the extension to complex k, a, f3,... mce t e ~un a conditions define an initial value problem, it is not self-adJomt. We there~ore use (2.139) and find that dii(O, ~) u(~)= f(x)ii(x,~)dx+a d -f3h(O,O o x b
LJ..u
a<x
(2.153)
where AEC
and where L
fo
= f,
1d [p(x)-d] + q(x) = -w(x) dx
dx
(2.154) •
(2.155)
In SLP2, we demanded that the interval (a, b) be finite and that the coefficients in (2.155) satisfy the following conditions:
where the conjugate adjoint Gree~ 's function equation is given by 2
d ii _ --' -eh =8(x-0 dx
a. p, p', q, ware real and continuous for a :::: x :::: b
2
b. p(x) > 0, w(x) >
a fora
:::: x:::: b
and where we have used the conjugate adjoint boundary conditions : h(h 0 '.
=
dii(b,~) dx
=
,
0
i !
We note that the conjugate adjoint Green's function problem is identical to the adjoint Green's function problem in Example 2.10. The solution therefore proceedS identically, and we produce the following result: ,
.u(x) = 10r o
f(~)
sink(~
. k
- x)
sinkx d~ + f3- k
+ a cos
If the interval (a, b) is not finite, or if any of the above conditions on the coefficients is violated, the problem is SLP3. In the mathematical literature, SLP2 problems are teoned regular Stuon-Liouville problems, while SLP3 problems are teoned singular [11]. We consider the SLP3 problem in Hilbert space Ida, b) with inner product (j, g)
k
=
Ib
f(x)g(x)w(x)dx
(2.156)
x
Although the form of solution is the same as in Example 2. I0, complex produces a complex solution lI(x~.
i
kt, f3, f I
•
There are several classes of SLP3 problems that are important in electromagnetic applications, defined by the following situations: 1. The interval is semi-infinite. In this problem, there is a singular point as x ~ 00.
Chap. 2
The Green's Function Method
78
2. The interval is finite, but p (x) = 0 at an endpoint. In this problem, there is a singular point at the endpoint where p (x) vanishes. 3. The interval is (-00,00). In this problem, there are singula'rpoints as x -+ ±oo. 4. The interval is semi-infinite, and p(x) vanishes at the finite endpoint. In this problem" there are singular points as x -+ 00 and at the finite endpoint. We shall classify singular problems by considering the homogeneous equa' tion associated with (2.153), \liz. , ,(2'.157) According to Weyl's theorem[12]:
.
.
i
2. For every A with Im(A) i= 0, there exists at least one u E 42(a, b). . i We omit the proof of this theorem and refer the reader to [12]. The theorem effectively divides singular problems into two mutually exclusiyecases
[13]: 1. The limit circle case: All solutions u are in £2 (a, b) for all A. 2. The limit point case: There is either one solution or no solutions in £2(a, b), according to the following: I
.
u
fa,
a. If Im(A) i= 0, there exists exactly one solution in £2 b). b. If Im(A) = 0, there is either one solution or no sol\ltions in £2(a, b).
:
;
Slunn-Liouville Problem of the Third Kind
79
of which is absolutely square integrable over (0, 00). Therefore, the'solutions are : not in £z (0, 00) and we have the limit point case.
EXAMPLE
2.1~
•
Consider Bessel's equation of order zero in £z(O, a):
'[-~~ (x~) - AJ x dx dx
U
= 0,
O
(2.159)
The en~point. x = a is ~ regular point. Since p(x) = x = 0 at x = 0, the endpoint x = 0 IS a smgular pomt. To determine the limit point or limit circle case we examine two linearly independent solutions to (2.159) for A = 0, namely, III '= j and Uz = log x. Although IlZ is 10garithmicaIly singular at x = 0, both U I and Uz are absolutely square integrable over (0, a). Therefore, both solutions are in £z(O, a), and we have the limit circle case. .
.
1. If for a particular value of A, every u that is a solution to (2.157) is in fda, b), then for all A, every u is in fda, b).
;
Sec. 2.6
EXAMPLE 2.17
Consider Bessel's equation of order zero in £z (0, 00):
(x~) -AJU =0 [ -~~ x dx dx
•
(2.160)
Both end~oint~ a:e si~gular ~oi.nts: In such cases, we pick an interior point x = ~ and examme !lmlt pomt or !lmlt CIrcle conditions on two intervals: ~ < x < 00 and 0 < x < ~. From Example 2.16, we have the limit circle case on 0 < x < ~. For ~ < x < 00, the endpoint x = ~ is regular and the endpoint x ~ 00 is singular. Further, neither U I = I nor IlZ = log x is absolutely square integrable over (~, 00), and therefore neither is in £z(~, 00). We conclude that we have the l~m~t P?int case. We say that Bessel's equation of order zero in £z(O, 00) is in the ., !lmIl Circle case at x = 0 and the limit point case as x ~ 00 .
•
I
We note that we can determine the limit point or limit circle case by examination of the solutions to (2.157) at a single value of A. If all solutions u are in £2(a, b), the limit circ\e case applies. If not, by exclusion, the limit point case a p p l i e s . ' ,
The. method of construction of the Green's function for SLP3 problems IS dIrectly related to the limit point and limit circle classifications. We shall proceed by considering a few examples, and follow with some conclusions and generalizations. .
EXAMPLE 2.15
EXAMPLE 2.18 Consider the foIlowing Green's function problem on the interval x E (0, 00):
Consider th¢ foIlowing differential equation in £z (0, 00): ) ' dZ - - -A ( , dx z
U
=0
(2.158)
The endpoint x = 0 is a regulat point. The endpoint x ~ 00 is a singular point. To determine the limit point or limit circle case, we examine solutions to (2.158) for A = O. Two linearly independent solutions are U I = I and Uz = :x "neither
(2.161 )
with the boundary condition g(O,~)
=
°
The Green's Function Method
80
.
Chap. 2
I '
In the beginning, we shall not assign a boundary condition as x ---+ 00. However, the method for dealing with this 1eficiency will emer.ge as we proce~d. i From Example 2.15, we have the limit point case. We begin the const~uctlOn. of the Green's function by considering solutions to the homogeneous equation for f =f. ~ , viz. d 2g
- - -}.,g dx 2
(~.162)
=0,
I
Possible forms of solution to this equation are sin v1cx, cos v1cx, exp(i v1c~), an~ exp(-iv1cxj. Since the problem is a limit point probl~m,.we know from ~eyl ~ Theorem that, if Im(}.,) =f. 0, there is exactly one solutIOn In £2 (0, 00). Our ~ask is to fincht. (We shall have no need to consider the case where Im(A~ = ~Ince we can always approach this case by taking a limit as Tm(}.,) ---+ 0.) SIn~e ~e~th~r sin v1cx nor' cos v1cx is absolutely square integrable over (0, 00), neIther IS In £2(0,00). Consider the two exponential solution forms. We have
?
r
00
• le- i J):X I2dx
io and
:
1
00
=
1
00
e 2(ImJ):)Xdx
0
1
~2.163) , I
=
e- 2(linJ):)xdx
(2.164)
Which of these two exponential solution forms is in £2 (0, 00) depends on "Yhether Im(v1c) is negative or positive. Since}., is a parameter specified in the problem statement, we shall choose for definiteness (2.165)
tm(.fA) < 0 I
With this choice
1 ie00
o
i J):Xj 2dx <
00
(2.166)
i
,
.
and we conclude that exp( -i v1cx) is the one solution to (2.162) in £2(0, do)., We now proceed with the construction of the Green's function in the usual manner. •
Wewnte
'. x
. ,~)
g(
=
I
II
A sin .fA1, + C cos .fAx, B e -iJ):x +' D e ,J):x , ,
Application of the boundary condition at x
Stunn-Liouville Problem of the Third Kind
81
In SLPI or SLP2 problems, we would next apply a second boundary condition to eliminate another coefficient. In the limit point case in SLP3, however, we replace the second boundary condition with the requirement that the solution be in £2(0,00). Since sih v1cx, as used in (2.168), has support only on (0, ~), the only part of the solutionin (2.168) that is not in £2(0, 00) is expUv1cx). We therefore choose D = 0 and obtain
_I
g(x,O-
A sin .fAx,
Be-
.r, 1VAx
,
x<~
(2.169)
x>~
We next apply the continuity and jump conditions at x and obtain
=
r
~ in the usual manner
e-iJ):~
A=-v1c
(2.170)
sin v1c~ B=--...::.v1c Substitution of these constants into (2.170) gives
(2.171)
I
00
\e i J):Xj 2dx
Sec. 2.6
x < ~
I
(2.167)
x>~
= 0 results in C = 0, with the,result x < ~ x>~
(2.168)
x<~
(2.172)
x>~ I'
where v1c is constrained by (2.165). We note that the Green's function derived in (2.172) is symmetric, g(x, ~) = g(~, x).
•
Example 2.18 suggests the following procedure for dealing with Green's functions associated with problems in the limit point case at one boundary, say x = b, and regular at the other boundary. First, we write the solution to the Green's function problem in the usual manner, in terms of four undetermined coefficients, as in (2.167). To determine one of the four coefficients, we apply the boundary condition at the regular endpoint. Next, to determine a second coefficient, we apply the requirement that the solution on the interval S < x < b must be in £2(a, b). The remaining two coefficients are determined in the usual manner by the contin:uity and jump conditions at x = S. Mathematically, for the limit point case, we may show that for Im(A) # 0, the single solution to L))I = 0 in £2(a, b) is always obtained simply by invoking the £2 requirement. In addition, if an unmixed boundary condition is applied at the regular endpoint, this condition, together with the £2 requirement, renders the problem self-adjoint. No boundary condition
J
I
The Green's Function Method
82
C;:hap.2
is required at the limit point boundary. The proof of this crucial re~ult is contained in: a review paper by Hajmirzaahmad and Krall [14]. As w~ have shown, the self-adjoint property'results in a symmetric Green's function. There is an alternate method leading to the determination of the Green's function in Example 2.18 [15]. Indeed, if we invoke in (j.168) the physically reasonable condition that the Green's function vanishes as x -+ 00, we produce the same result as we do by invoking the [2(0, 00) requirement. That is, we can invoke a limit condition lim
x-+~
g(x,~)
=0
in place ~f the second "boundari' condition in the statement of the pr~blem. This is an appealing procedure since such an unmixed limit condition can be viewed as an extension of the regular boundary condition g(b,~)
=0
Sec. 2.6
Sturm-Liouville Problem of the Third Kind
where u, f, A are complex. We choose the inner product
1'")0 u(x)v(x)dx
(U, v) =
From Example 2.18, we know that this problem is singular in the limit point case as x -* 00. We therefore invoke the limit condition ' lim u(x)
x-'>oo
=
°
The problem is self-adjoint and the Green's function is symmetric. We therefore use (2.152) which, specialized to this case, yields u(g)
=
1
00
o
f(x)g(x,
~)dx + [dU(X) g(x, g) _ u(x) dg(x, ~)] IX=OO b
b
(2.174)
x=o
We apply the boundary condition and limit condition on u(x) and choose
Indeed, consider the Green's function problem
g(O, g) =
lim g(x,~)
x-.>oo
=
°
°
and find that
with boundary conditions
1
00
u(O
g(O, ~) = g(b, ~) = 0
The result in (2.172) can be obtained by solving this problem and then taking the limit as b -+ 00. Th~ details are left for the problems. In summary, for the case of a regular unmixed boundary condition at x = a and the limit point case at x = b, the [2 requirement takes the place of a boundary condition at x = b. Furthermore, the problem is selfadjoint. In,the case where b -+ 00, we may use the alternate procedure of applying a limit condition in place of the £2 requirement. We remark that it is sufficient to have a procedure that picks out the one [2 solution required in the mathematical proofs, such as those in [14]. Invoking the limit condition is such a procedure. We consider these ideas further in the following example. . EXAMPLE 2.19
83
Consider the following differential equation on x E (0,00): I
-:u" - AU = f with boundary condition u(o) = 0
(2.173)
=
Finally, after interchanging x and
~,
f(x)g(x,
~)dx
we obtain
1 f(~)g(x,g)d~ 00
lI(x)
where the Green's function g(x,
=
~)
is given by (2.172).
•
We note that the result in (2.174) is an extension to the result for self-adjoint operators in SLP2. Specifically, the arguments for the SLP2 unmixed boundary case given in (2.122)-(2.133) carry over to the SLP3 limit point case at infinity, provided again that the aij 's are constrained to be real. We may establish this result simply by observing that the arguments in (2.122)-(2.133) are not altered by taking the limit as a ---+' -00 or b -+ 00, or both., Since the problem is self-adjoint, the Green's' function is symmetric, an4 the result in (2.174) is assured before solving for the specific Green's function. We shall illustrate this important point in an additional example.
The Green's Function Method
84
EXAMPLE 2.20
Consider the foll~wing differential equation on x E (- 00,(0): (L - A)U
=
y,
Im(v}.:)
<
°
Sec. 2.6
Stunn-Liouville Problem of the Third Kind
85
and where we have invoked the two limit conditions. We note that, consistent with the limit ~oin~ case and Im(A) # 0, our two limit conditions have produced exactly one solutton In L2(-00,;) and exactly one solution in L2(;, 00). Applying the continuity and jump conditions at x = ;, we obtain
where eifi~
d2 L i = - -2 dx
A=--
2iv}.:
, This problem is in the limit point case,as x ~ 00 and as x ~ -00. Our procedure : in dealing with limit points at both ends of the interval along the real line ,is to , pick an interior point x = ~. Since Im(A) # 0, there is exactly one s~luti~~ to , LAu = in L2( -00, ~) and exactly one solution in L2(~, 00). These two solu~tons to the hom~geneous equation fonn t~e building blocks for the construction of the Green's function. Hajmirzaahmad and Krall [14] prove the following: For the , Stunn-LiouviI:e operator L with the.limit point case at both ends of the interval,
°
I. No boundary conditions need be invoked.
e-ifi~
B=--
2iv}.:
Therefore, g(x,;)
=
_1_/
e-ifi(x- O ,
2iv}.:
e
-ifi(~-x)
,
x>; x<;
or, more compactly,
2. L is self-adjoint.
e-ifilx-~I
Again, in lieu of the L2 requirement, we shall invoke limiting conditions, o~e at : each end of the interval, viz.
lim u(x)
x--.+-oo
=
lim u(x) =
x~oo
° °
f:
!(Og(x,
Od~
i
EXAMPLE 2.21
Consider the following Green's function problem on x
E
(0,00):
[i C(X
(x
dg )] -Ag dx
=
8(x -;) x '
,(2.176)
°
From the results in Ex~mple 2.1 7, we have the limit circle case at x = and the limit point case as x ~ 00. We begin our construction of the Green's function in the usual manner by considering the homogeneous equation
(L - A)g(X,~) = 8(x -~)
lim g(x,;)
(2. I75)
2iv}.:
•
-~x
where the Green's function must satisfy
x-+-oo
= ---
We have established in the above paragraphs a procedure for deriving the Green's function in limit point problems. We now tum to a consideration of the limit circle case. We begin with two examples.
We assume that u, A,! are complex.: Since L is self-adjoint, the Green's fun~tion is symmetric. The solution to the differential equation is therefore given by , U(x) =
g(x,;)
= x-+oo lim g(x,O =
°
_1x [~(xd8)] dx dx
_
A8 = 0,
We write the solution for the Green's function as
x>; x < ~
which is Bessel's equation of order zero. Solutions can be constructed from linear combinations of the Bessel function Jo(v}.:x), the Neumann function Yo(.v'Ax), and the two Hankel functions Hcil)(v}.:x) and Hci 2 )hfi:x). We write
+ CYo(v}.:x), )(v}.:x) + DHcil)(v}.:x),
A Jo(v}.:x)
where
Irri( v}.:)
<
°
BHci
2
x
<;
x>;
(2.177)
The Green's Function Method
86
Chap: 2
We may evaluate one of the coefficients in (2.177) by following our procedure for dealing with limit points. We therefore invoke the following limiting condition as "
,
.
x~
Stunn-Liouville Problem of the Third Kind
_ A dJo(~x)] [ BdHri2)(~X) dx dx
I
x--> OQ
(2,178)
=0
The asymptotic fonns of the two Hankel functions are given by [16)
87
The jump condition gives
!
I
00:
lim g(x,~)
I
Sec. 2.6
i
1 X=~
(2.182)
~
Performing the indicated derivatives and solving (2.181) and (2 182)' I ously for A gives . slmu tane-
A [Jo(J.iOH t(2)(v'AO
-
JI(v'AOHri2>Cv'A~)J = Hri2)(~~) ~~
(2.183)
Bya well-known Wronskian relationship [17], we have I
These asymptotic forms show that if we constrain Im~ < 0, H(~I) (~x) diverges as x ~ 00. We therefore set D = 0 and obtain
g=
I
AJo(v'Ax}+CYo(v'Ax), 2 B Hri )(J):.x) ,
x <
~
(2.179)
Using this relation i~ (2.183), we obtain I
7f
Substitution of this result into (2.181) gives . I
Therefore,
yt
Yo(t) = - In -2 - ... .
7f
i
where In y is Euler's constant. Since we have the limit circle case, we know a priori that both of these functions are square integrable over (0, ~). Therefore, invoking . I the requirement that the solution b~ in £2 (0, 0 does not evaluate a coef~cient, as was the case for limit points. We do have, however, a condition that "Y e can invoke from physical principles. Bessel's equation with forcing function atI -': = ~ normally results from considerations of the radial dependence in problems in cylindrical coordinates. In such problems, we shall find in Chapter 4 that, :based on physical grounds, the solution must remain finite as x ~ O. The Neumann function does not meet such a condition at x = 0, and we therefore set C =: 0 and i obtain '
g=
The continuity condition at x
I
A Jo(v'Ax) ,
x < ~
2 B Hri )(v'Ax),
x>~
=~
yields
2
A = 2i Hri )(v'AO
== 1 + ... . 2
= -/:17ft
x>~
Detennining the remaining three coefficients requires three conditions. The continuity and jump conditions at x = ~ will provide two conditions. To produce the third, we consider the limit circle case at x = O. The leading terms in the expansion of the Bessel and Neuma~n functions are given by [16] Jo(t)
2 J 1 (t)Hri )(t) - Jo(t)Hi 2)(t)
(2.180) \
I
I I
(2.181)
g(x,O
= !!-I 2i
x < ~ x > ~
(2.184)
We note A useful . r that . the Green's function is symmetric ' g(x , ~1:) -- g (I:) s x sp.ecla IzatlOn of the result in (2.184) can be obtained by taking the l'imi~ as 1:' 0 WIth the result s ~ , (2.185)
•
We. have noted in t~e a~ove example that the Green's function is sym~etr~c. We ar~ ,I~d to InqUIre if the operator that produced the Green's f~nctI~n IS self-adJOInt. .Consider first the case where we have the SturmLlOu.vIlle operator L WIth the limit circle case at x = a and a re ular unmixed boundary condition at x = b. This case has been by K~per, K wo.n~, a~d Zettl [18], who have proved the following: The operator L IS self-adjOint If U E 'D t , and if:
e1arific~
Sec. 2.6
The Green's Function Method
88
1. u satisfies an unmixed condition at the regular boundary x
2. u exists and is finite as
= b.
x < ~
Ajo(kx),
x-+a
89
and therefore set D = O. With these conditions, we find that
u---+ a, the limit circle boundary.
In addition, they show that the finiteness condition is mathematically equivalent to the condition lim [p(x)u'(x)] = 0
g = I
The continuity cOQdition at x
1
(2)
Bh o (kx),
=~
(2.192)
x>~
11 I
yields
!
(This equivalence is important in making the connection between the physically appealing finiteness condition and the classical Weyl theory.) This important result has been extended [14] to show that the self-adjoint property is r~tained when the regular point at x = b is replaced by a limit point, as in the 'previous example. We consider another example. ! • EXAMPLE 2.22
Stunn-Liouville Problem of the Third Kind
(2.193), The jump condition gives
[
B dh62 )(kx) _ A djO(kx)] dx dx
I r=~
Consider the following Green's function problem O,ll x E
~2
(2.194)
Solving for A in (2.193) and substituting into (2.194), we have
(0,00):
kEC
(2.186) (2.195) I
This problem is in the limit circle case at x = 0 and the limit point case as x ~ 00. We therefore invoke a finiteness condition at x = 0 and the fOllowing , , limit condition as x ~ 00: i lim g(x,~)
:rhe ex,rression in s~uare brackets in (2.195) is one of many Wronskian expressions mvolvmg the spheflcal Bessel functions. In this case, we find [20] that
=0
11(2).,
Equation (2.186) is the spherical Bessel equation of order zero [191. Its solution is given by linear combinations of spherical Bessel, spherical Neumann, and spherical Hankel functions. We therefore write Ajo(kx)+ Ch62) (kx), Bh62) (ki)
+ Dh61)(kx),
11(2)'. 0 Jo
o Jo -
x~oo
x < ~ x>~
where all arguments are with respect to argument. Using (2.196) in (2.195) gives
i
B
(2.187)
,
i
=
=
z and
i
2"
(2.196)
z
differentiation is with respect to
-ikjo(k~)
(2.197)
Substitution into (2.193) yields
I I
(2.198)
where sinkx
jo(kx) = - kx ,
(2.188)
e ikx
h~I)(kx) = -.-
(2.189)
lkx
:
,
Substitution of (2.197) and (2.198) into (2.192) yields the Green's function
e- ikx
h(2)(kx) = - o ikx
:
=
-ik
1jo(kx)h62)(k~). jo(k~)h62) (kx),
x < ~ x > ~
(2.199)
(2.190) 1
To preserve finiteness as x ~ 0, We set C = O. To satisfy the limit condition at infinity, we adopt the constraint ; Im(k) < 0
g
(2.191)
An alternate form can be obtained by using (2.188) and (2.190). We have g
=
_1_1 e-ik~
sin kx,
x < ~
c- ikx
sin ki;,
x > ~
kxi;
(2.200)
The Green's Function Method
90
Chap. 2
I
Sec. 2.6
Sturm-Liouville Problem of the Third Kind
.
hm
~->O
g(x,~)
e- ikx
= --
i (2.201)
x
i .
.
•
In dealing with the singular point in limit circle cases, we ~ave not been able to evaluate a coefficient by invoking the requirement that the solution be in £2. Instead, we have invoked a condition based on physical grounc!~. The self-adjoint property of L results in a Green's function that I is symmetric. The above examples all concern operators that are self-adjoint. We next present an example wherethe operator manifold contains two u'nmixed I conditions, but the operator is not self-adjoint. EXAMPLE 2.23
Consider the following differential equation on x
"-u" - AU = f
E
We note that the houndary conditions on h(x, 0 are identical to the boundary conditions on !lex). If we recall that the boundary conditions on the Green's function g(x,~) are always identical to the boundary conditions on u(x), we find that (2.209) We therefore have
=
dg 2
---Ag=8(x-~) dx 2
(2.212)
: (2.202) lim
(2.203)
aEC
, (2.204)
:Iim u(x) = 0
r--+ oo
In addition, since a is complex, the problem is not an extension to a finite interval self-adjoint problem. (See Example 2.12 for a discussion.) We therefore;pr9ceed using (2.139). Applying the boundary and limit conditions given in (2.103) and (2.204), we obtain u(~) =
f(x)h(x,
o
[ ~)«x + u(O) ah(O,~)
dh(O d~
0]
i
I .
'(2.205) I
We write
I
A cos .Ji.x
(2.213)
= 0
+ B sin .Ji.x,
x <
C e- i fix,
~
lim h(x,~)
=0
where we have applied the limit condition as x ---*
00
and have chosen
Im(.Ji.) < 0
I
Applying the boundary condition at x
g
=
A (cos .Ji.x
+
::x
= 0, we obtain x<~
sin .Ji.x) ,
Ce- ifix ,
(2.206)
x>~ ~
results in
e-i.fi~
A =-=-i.Ji.+a
If we now choose
dh(O t)
_
d~" =ah(O,O
•(2.207)
(2.214)
x>~
Invoking the continuity and jump conditions at x = x400
we ohtain
g=
g(x,~)
I
where we have chosen
Ii
(2.21 \)
~O,oo):
x->oo
J(oo,
(2.210)
f(x)g(x, Odx
where we must solve
and where we assume that u, A, f are complex. The boundary x = 0 is a regular point. As x ---* 00, we have the limit point case. We therefore apply the limit condition
,,
1
00
!l(0
where u'(O) ±::: au(O),
(2.208)
f(x)h(x, Odx
I
We shall uSe this important result in Chapter 4.
,
1
u(~) =
We note that in the limit as ~ ---*P, we produce the result
91
00
C=
cos JI~
+ vA~ sin JI~
iJI+a
(2.215)
,......
I I
The Green's Function Method
92
Sec. 2.6
Chap. 2
Sturm-Liouville Problem of the Third Kind
93
on two intervals ~ < x < 00 and 0 < x < ~. For A = 0, the homogeneous equation (L - A)U = () has the two independent solutions
Therefore, the G~een's function is given by
I
1
g(x,~)=. I
[e-i~~ (co.s v0.x + 5x sin v0.x) ,
x < ~
,
x > ~
'ex) e-i ~x (cos v0.~ + JI sin v0.~
JI + ex
U, =x "
and (2.222)
(2.216) We note that, although the operator in this problem is not self-adjoint, we still have g(x,
Consider the interval ~ < x < 00. We first examine whether UI is in £2(~, 00). We have
lOO x"x"xdx = loo x
n = g(~, x)
2Re (t')+ldx
u,
and, therefore, interchanging x and ~ i~ (2.210), we have
u(x) =
(2.221)
By (2.219), this integral diverges and is not in £2(~, 00). We therefore have the limit point case as x ---+ 00 and assign the limit condition
1if(~)g(X,~)d~
lim g(x,~) = 0
(2.223)
x-->oo
We remark that if an operator is self-adjoint, the Green's function is symmetfic. However, if the operator is not self-adjoint, it does not necessarily follow that the Green's function is not symmetric. This seemingly small distinction has a mar~ed effect on characteristics of eigenvalues, as will be discussed in the next chapter. "
'.
c~rcle!
Consider the following Green's function problem associated,
,
lim g(x,
8(x - ~)
= ---
x-->o
(2.217)
x
,
l~ x"x"xdx = l~ x 2Re (,,)+ldx n,
~ith Bessel's equation of order v in £2(0, 00): (L-A)g
u,
This integral exists when 2Re(v) + 1 > -lor Re(v) > -1. Replacing v by -v, we find that U2 is in £2(0, provided that Re(-v) > -lor Re(v) < 1. Combining these two results, we find that both solutions are in £2(0, ~), provided that -I < Re(v) < 1. We therefore have the limit circle case as x ---+ 0 for -1 < Re( v) < I and the limit point case otherwise. In either case, we shall demand satisfaction of the physically motivated limit condition
In certain cases, detennination of the limit point or limit is dependent on parameters in the differential equation. We illustrate this f~ct with the follo~ing example.
~XAMPLE 2.24
The situation on the interval 0 < x < ~ is more delicate. We first examine whether is in £2(0, ~). We have
where
_~x [~ (x'!!-)] + xv2 dx dx 2
=
g(X,')~~
(2.218)
~nd where v and A are complex param~ters. We assume that Re(v) > 0
(2.219)
1
00
uvxdx
I
H: 2) (v0.~)J,,( v0.x),
x< ~
H: 2) (v0.x )J,,( v0.n
x> ~
(2.225)
Im(v0.) < 0 That this is the solution for g(x, ~) can be determined by construction in the usual manner. We defer the details until Example 3.6 in the next chapter.
We define the inner product for the sp~ce to be
=
(2.224)
where
I
(u, v)
finite
The Green's function problem defined by (2.217) and (2.218) with limiting conditions given by (2.223) and (2.224) yields the solution
I
L
~)
(2.220)
We note that both endpoints are singular points. Proceeding as in Example 2.17; we pick an interior point x = ~ and ex'amine limit point or limit circle condition~ !
,
I
•
The limit point ~nd limit circle cases in SLP3 problems are a subject of continuing interest to mathematicians. We have only considered the
94
Problems
Chap. 2
portion of the theory of interest to os in our application to electromag'netic boundary value problems. For ani in-depth discussion of limit point ~nd limit circle cases, as well as a weiI-compiled bibliography, the readerI is referred to [14] and [21]. , We have now completed our discussion of the solution to linear, o~di nary, second-order differential equ~tions by the Green's function methp& In the next chapter, we shall discuss an alternate method where we shall ge~ ~ermine the solution to (L - A)U = 'j by finding the spectral representation I 'associated with the differential operator L. t
I
•
I
!
Problems
95
2.8. Solve the following differential equation:
--=dg,--(O_,~'--) _ dg(L, dx dx
_u" = f(x),
2.2. 2.3.
I
a E
R
u'(L) = 0
(a) Show that, if a solution exists, it is unique. (b) Construct the solution by the Green's function method.
By substituting (2.23)-(2.25) into (2.22), verify that the Sturm-Liouville differential equation is transform~d into the general form in (2.21). I I
The Chebyshev differential equation is defined on the interval x as follows: , -(1 - xz)~" + xu' - nZu = f
E
2.10. Given the differential equation Lu = f on the interval x real and with the following boundary conditions:
E
(0, L) with f(x)
(-I,' I), a> 0
u(O) = au'(O), u'(L) = 0
Transform to Sturm-Liouville foim.
Show that this problem is self-adjoint. 2.11. Show that the solution given in (2.111) satisfies the differential equation and the boundary conditions in Example 2.10. ' 2.12. Solve the following , SLPI differential equation on the interval x E (0, L):
Z.4. Transform the Laguerre differential equation
I
x E (0, L)
u(O) = a,
For the pulse function p( (x - xo~, defined in (2.2), show that p(x -xo) = p(xo -x)
i,
-
with f(x) real and with boundary conditions
PRO'BLEMS
2.1.
n _0
2.9. Consider the following differential equation:
i
! "'t.
Chap. 2
!
-xu" - (1-x)u' -nu = f
,
to Sturm-Liouville form.
I
2.5. In SLPI, the following are restrictions on u(x):
u'(O)
(a) u E Li(a, b); (b) u E'DL ; (c) u satisfies two boundary conoitions, B I (u) = 0, Bz(u) = O.
Show that these restrictions define a linear manifold M
L
2.13. Consider the following differential equation:
C Lz(a, b).
dZu - dx z =
dZg(x, ~) = 8(x dx z
~)=
dg(1, dx
=0
u'(L)=f3
2.6. Solve the following differential equation:
g(O,
_u" -eu = f
~)
-~)
f
du(1) u(O) = - dx
= 0
du(O) - - =a, dx'
2.7. Verify that (2.93) satisfies the requirements for the Green's function givep in (2.83)-(2.87). ;
I
a E R
where x E (0, I) and where f is a real-valued function of x. Solve the differential equation hy the Green's function method.
Problems
96
Chap. 2
i.14. For SLP2, show that the operator L is not self-adjoint for the case of initial :. conditions. , 2.15. For SLP2, show that the operator L is self-adjoint for periodic conditions, i provided p.(a) = p.(b). . ~.16. Repeat Problem 2.12 for the case where k E C and f(x) is complex so ihat j the problem becomes SLP2. ' 2.17. Repeat Pr~blem 2.9 for the case where f(x) is complex so that the probilem becomes SLP2. 2.18. Consider the following SLP2 Green's function problem:
".
d
2
g
- - -)"g
dx 2
0
= 8(x -
= g(2n,
dg(O,O
dg(2n, ~)
dx,
dx
=
~I -
n)J
2J):: sin J)::n
(L - ),,)u
=f d2
L = - -2 dx C and where the following boundary conditions apply: u'(O) - au(O) = 0,
a E C
u(1)=O
(a) Show that the operator L is not self-adjoint. i ' (b) Find the Green's function g(x, ~), the adjoint Green's function h(i, ~), .
=0
2.21. Consider the Green's function problem
with boundary conditions
= g(b, 0 = 0
REFERENCES
2.19. Consider the following SLP3 boundary-value problem, defined on the interval , x E (0, 1):
.
lim [u(x)] .\"-----+00
0
,cos [J)::(lx -
)
where k E C and Im(k) < O. Obtain the solution in tenns of the appropriate Green's function g(x, 0 by demanding that
g(O,O
Show that the solution is
E
97
Obtain the result in (2.172) by solving this problem and then taking'the limit : as h ~ 00.
g(O,~)
where)"
References
:
with periodic boundary conditions
g(x,O
Chap. 2
and the conjugate adjoint Green's function (c) Solve the differential equation.
hex, O·
2.20. Consider the following SLP3 boundary-value prohlem, defined on the int~rval x E (0, (0):
-u" .
eu = .r
u'(O) = 0
[I] Taylor, A.E. (1955), Advanced Calculus. Boston: Ginn, 52-53. [2] Schwartz, L. (1950), TheOl'ie des Distributions, Vol. I. Paris: Hermann. [3] Stakgold, I. (1967), Boundary Value Problems of Mathematical Physics, Vol. I. New York: Macmillan, 18-58. [4] Papoulis, A. (1962), The Fourier Integral Gnd Its Applications. New York: McGraw-Hill, 269-282. [5] Arfken, G. (1985), Mathematical Methods for Physicists, 3rd edition. New York: Academic, 497. [6] Krall, A.M. (1973), Linear Methods of Applied Analysis. Reading, MA: Addison.:.Wesley, 132. [7] op.cit. Arfken, 498-499. [8] Felsen, L.B., and N. Marcuvitz (1973), Radiation and Scattering of Waves. Englewood Cliffs, NJ: Prentice-Hall, 323-328. [9J Hellwig, G. (1964), Differential Operators of Mathematical Physics. Reading, MA: Addison-Wesley, 39-40. [10] op.cit. Taylor, 55. [11] op.cit. Stakgold, 268-269. [12] Ibid., 297-301. [13] Stakgold. I. (1979), Green's Functions and Boundarv Value Problems. New York: Wiley, 438.
,
:...
,
98
References
Chap. 2 .
P4] Hajmirzaahmad, M., and A.M. Krall (1992), Singular second~ ~r der operators: The maximal and minimal operators, and selfadJOIl1t operators in between, SIAM Review 34: 614-634. , [15] Friedman, B. (1956), Principles and Techniques of Applied Mathematics. New York: Wiley, 230-231. [16] Harrington, R.E (1961, reissued 1987), Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill,461-463. [17] op.cit. Arfken, 605. [18] Kaper, H.G., M.K. Kwong, and A. Zettl (1986), Characteriza~ions: of the F:riedrichs extensions of singular Sturm-Liouville expresSIOns, SIAM J. Math. Anal. 17: 772-.777. [19] op.cit., Arfken, 622-636. [20] Hansen, J.E. (Ed.) (1988), Spherical Near-Field Antenna MeasU1~e ments. London: Peter Perigrinus, 314-315.
3 The Spectral Representation Method
[21] op.cit. Hellwig, 221-276.
3.1 INTRODUCTION In this chapter, we continue our discussion of linear ordinary differential equations of second order. We begin with a discussion of eigenvalues and eigenfunctions. We follow with a description of the method of solution for self-adjoint SLP2 problems in terms of the eigenfunctions. We include a discussion of the determination of the eigenfunctions directly from the Green's function for the problem. This determination leads to a spectral representation of the delta function specific to a particular operator and its domain. We next consider problems on unbounded intervals (SLP3). We are able to expand our analysis to produce the appropriate spectral representations for many important unbounded interval problems. We conClude the chapter by emphasizing the connection between solutions by the Green's function method, and by the spectral representation method.
3.2 EIGENFUNCTIONS AND EIGENVALUES A complex number f.1 is called an eigenvalue of the linear operator L if there exists a nonzero vector v in the domain of L such that
Lv = f.1v
(3.1)
The vector v is called an eigenfunction of the linear operator L. We remark that, although an eigenfunction is by definition nonzero, it can be associated with a zero eigenvalue. 99
The Spectral Representation Method
100
Sec. 3.2
So far, it is unclear whether or not there are any eigenfunctions:and eigenvalues a~sociated with a specific operator. However, if eigenvalues and eigenfunctions do exist, they have remarkable properties. We 'first show that if UI, U2, , Un are eigenfunctions corresponding to dim~rei1t eigenvalues >-:1, A2, , An, associated with the operator L, then {ud,is a linearly independent sequence. Our proof is by induction. Let n = I iand examine alUI = 0 Since U 1 is an eigenfunction, u I. f. 0, and therefore al = O. We ~ow suppose that the linearly independent assertion is true for n - I and examine ".
'
I
n
101
Eigenfunctions and Eigenvalues
ator L is self-adjoint, however, we may show that its eigenvalues are real. Indeed, let fL be an eigenvalue associated with the eigenfunction v: Then, fL(V, v) = (fLV, v) = (Lv, v)
But, since L is self-adjoint, fL(V, v) = (v, Lv)
=
(v, fLV) = ji(v, v)
Therefore, (fL - ji)(v, v) = 0
Since v
f. 0,
(v, v) > 0 and we must have
Lakuk = 0 k=l'
which implies that It E R. We next establish that eigenfunctions of a self-adjoint operator corresponding to different eigenvalues are orthogonal. Indeed, let
Operating on both sides with L ---: An, we obtain n,
n-I
0= (L - An) L akUk = L ak(Ak - An)Uk k=1
k=1
Since the n ~ I length sequence has been supposed independent, where Am
k = I, 2, ... , n - I
f. An Then,
which implies that I
Since L is self-adjoint,
'k = I, 2, ... , n - I
The expression
n'
But, we have established that An is real. Therefore,
LJakuk = 0 k=l
then reduces to
i I I
I
I I
which implies that an = O. What we have shown is that if the (n --: 1)length sequence is independent, then the n-length sequence is indepen~ent. We have established the result for n = I, and therefore it must be true for n = 2. By induction, it must therefore be true for arbitrary n. Further, since n is arbitrary, the countably infinite sequence UI, U2, ... is linearly independent.. , In the above proof of linear independence of the eigenfunctions, we assumed nothing about the operator L except linearity. If the linear ~per-
and (Am - An)(U m , Un) = 0
Since Am f. An, we must have (u m , un) = O. We next state a central result for Hilbert space L2(a, b). It can be shown that the eigenfunctions Uk of a self-adjoint operator form an orthogonal basis in L2(a, b). Therefore, any U E L2(a, b) can be expanded:
I
I
I I
The Spectral Representation Method
102
Crap. 3
From the self-adjoint property, the conjunct J is zero. Subs~ituting (3.4) and (3.7) into (3.10), we obtain
The equality, is interpreted in the: sense that
,
n .
lim lIu- Lakukll
=0
I The proof of this property involves the theory of integral equation~ and n-+oo
(f, un)
'k=!
is beyond the scope of this book.. The interested reader is referred tbI the literature [ 1 ] . ; : The fact that the eigenfunctions form an orthogonal basis allows us to solve the self-adjoint SLP2 problem in terms of the eigenfunctions Un of the operator L. We begin by noting that the eigenfunctions can always be normalized, so that we shall assume that they are orthonormal. Therefore, if . (3.2) u h Lanu n .
I
n
=
(u, (An - i)u n ) = (An - A)(U, Un) an
an: = (u, un)
I' ,
,/
(An - A)a n
= (f, un)
(3.11)
An - A Substituting (3.11) into (3.2) gives the solution in terms of the eigenvalues and eigenfunctions, viz. _ "i - J(f, un) ---U n n An - A
U -
f
with associated boundary conditi,ons ~! (u) = 0
f
(3:4)
I !
(3.5) I
B2(U) = 0
where
f
and A are complex and where (3.14)
and U'(O)
= u'(a) = 0
(3.7)
(3.15)
The problem is of class SLP2. The operator L with the given unmixed boundary conditions is self-adjoint. The associated eigenproblem is given by
(3.6)
We assume that the operator L is self-adjoint. We associate with this SLP2 problem the following eigenproblem:
(3.12)
(3.13)
where the index n runs over all of the eigenfunctions. On the interval x E (a, b), consider, the following SLP2 problem: i (L '- A)U =
!
EXAMPLE 3.1 Using the eigenfunction~igenvalue method, we wish to solve the following differential equation on x E (0, a): (L - A)U =
'(3.3)
=
Therefore,
then by (1.58), the Fourier coeffiCients are given by .till
103
Eigenfunctions and Eigenvalues
Sec. 3.2
(3.16) with U~(O) = u~(a)
=0
(3.17)
The orthonormal solutions to the eigenproblem are given by En )1/2
U ll
= ( -;;
nJfX
n = 0, I, ...
cos 7'
(3.18)
where
_ (nIT)2
An and where
Ell
(3.19)
a
is Neumann's number, given by En
=
11, 2,
n=O n:;fO
(3.20)
, The Spectral Representation Method
104
Ch~p.
3
Sec. 3.2
The solution in (3.18) can be easily verified by substitution into (3.16). We note that the factor :,jEn/a normalizes the; eigenfunctions. Substitution of (3.18)',and . (3.19) in (3.12) yields 00
' \ ' En
ra / (X) ', nrr x' , Jo COS - d x
U(X) = ~ - :
n=O
a
(n; )2 _
i
a
/lIT X COS - -
A
(3.21)
a
•
, I
In Example 3.1, we have solved the differential equation in (3.1\3)i (3.15) by the eigenfu~ction-eigenyalue method. The method is also callep , the spectrQ! representation method. The reader should compare the s01ution in Example 3.1 to the solution by Green's function methods, given in Example 2.13. It appears that the: Green's function method might always i be preferred since the spectral representation method contains a summation that must be performed before the mathematics can be reduced to a , numerical answer. There are, however, many reasons why the spectral rep: resentation is important. First, consider (3.21) in a slightly different form, ! viz.
105
Eigenfunctions and Eigenvalues
U = Lallu ll
(3.24)
all = (u, u ll )
(3.25)
I
We say that (3.25) is a transformation of the function u(x) E £2(a, b) into coefficients all' Conversely, (3.24) is the inverse transformation of the coefficients all into the function u(x). We represent this transformation relationship by u (x) ¢::=} all' We now show that if L is self-adjoint and
I
I
n7fX
00
u(x) = L
, where All =
[
a
Ell 2
(II:) _A
i
J.
0
Lu
¢::=}
Alla ll
. (3.27)
Indeed, forming the transformation defined in (3.25), we have (Lu, u ll )
=
(u, LUll)
= AIl(U, u ll ) = Allall
. (3.28)
Having established the basic result in (3.27), we reconsider the original problem stated in (3.4)-(3.6), which we repeat here for convenience, viz.
I
(L - A)u =
(3.29)
with associated b?undary conditions
a Q
then,
(3.22)
All cos--
11:=0
(3.26)
, n7fx' I I(x ) cos(--)dx
(3,23)
a
We interpret (3.22) as a Fourier sum over the natural modes of the system with modal coefficients All' We note that it is the interaction between the forcing function I(x) and the modes in (3.23) that determines the modal coefficients. Second, in &aling with multidimensional problems in later chapters, we shall encounter partial differential equations whose solutions are rarely expressible in terms of closed-form Green's function$. However, proper combination of the spectral representation and Green ~s function methods results in a poweiful tool to solve many partial differeqtial equations appearing in electromagnetic problems. , There is a variation on the procedure used to produce the result in (3.12). This variation results in amore direct approach to the solutio~ bf differential equations by the eigenfunction-eigenvalue method. In addition, the variation is very useful in the solution to the partial differer,ltial equations considered in later chapters. We proceed as follows. Suppo~e L is self-adjoint with associated orthonormal eigenfunctions u". Then,
Bt(u)=O
(3.30)
B2(U) = 0
: (3.31)
where we assume that L is self-adjoint. If U ll are eigenfunctions and All are eigenvalues of L, then I , (3.32) and (3.33) If we define
I
(x)
¢::=}
f31l
(3.34)
then (3.29) transforms into (3.35) Solving for all' we obtain (f, u ll ) f31l all = - - - = - - -
Ail - A
All - A
Substitution of this result into n.24) yields n.12).
(3.36)
I
The Spectral Representation Method
106
Chap. 3
3.3 SPECTRAL REPRESENTATIONS FOR SLP1 AND SLP2
I
107
Spectral Representations for SLPI and SLP2
Therefore, the specific forcing function 8(x - ~)/w(x) must produce the Green's function g(x,~, A), We therefore obtain
'
,
Sec. 3.3
.
In Example 3.1, we assumed that we had somehow obtained the eigenfunctions given by (3.18). In this, section, we shall present a method. of bbtaining the eigenfunctions and eigenvalues of a self-adjoint operator directly from the Green's function for the problem. ~ince SLPI is a spe~ial case contained' in SLP2, we shall confine our attention to SLP2. Note that the solution in (3.12) in terms of the eigenfunctions and eigenvalues is parametrically dependent on A, viz.
1
-. 2nl
i
g(x,~,
8(x-~) A)dA = - - - -
(3.39)
w(x)
C
The solution to the contour integral in (3.39) for a specific Green's function associated with a specific operator L and boundary conditions is called the spectral representation of the delta function [3] for the operator L. We shall demonstrate the utility of this result in an example. Im(A)
"'.
A-plane
Consider
J
u(x,A)dA
-><-t------l<--'J~' Ra( A)
fCR
where C R is a circle of radius R centered at the origin in the complex A-plane (Fig. 3-1). We have
J
'.
u(x, A)dA = -
fCR
L(f, un)un i ,n
CR
A
~\ n I
where the sum is over those eigerwalues An contained within the cir~le: The singularities of the integrand are simple poles with residue of unitX at all A = An within the contour. We note that since L is self-adjoint, the poles must lie on the real axis in the A-plane. Taking the limit as R -+ 00, we enclose all of the singularities and obtain by the Residue Theorem [,2] lim
J
u(x, A)d>" = -2ni
R-+oo fCR
L(f. n
,
Un)U n
(3.37)
J
2m fc
u(x, A)dA
=-
Circular contour of radius R centered at the origin in the complex A-plane. The x indicate possible simple pole locations at A = An.
EXAMPLE 3.2
Consider the following operator defined on x E (0, a): d2
L = - -2 dx u'(O)
I
g (x. ~. A) = - -----=,-------=,-
fi sin fia
First, consider the case x <
(3.38) I
where C is the contour at infinity obtained in the limiting operatio~ in (3.37). There is an important special case to the result in (3.38). ;We note that the general forcing function .r(x) produces the response u(x, IA).
= u'(a) = 0
(3.41)
The Green's function associated with LA with these boundary conditions has been previously derived in Example 2.13. We repeat it here for convenience,viz.
~,
I
cos.fix cos .fi(a - 0,
x < ~
cos .fi~ cos .fi(a - x),
x>~
g(X • .,.A)-
Substitution into (3.39) gives 8(x
-0
= -
(3.42)
so that _ _ cos
t
f(x)
(3.40)
with boundary conditions
,
l
where the sum is now over all of the eigenfunctions. The summatton is simply the Fourier expansion of the forcing function in terms of ~he eigenfunctions. Therefore, we find, that _1_.
Fig. 3-1
1
fix cos fiCa - ~)
(3.43)
"" vAsin v Aa
i
2IT; c
cos fix cos fiCa fi sin fia
0
dA
(3.44)
The Spectral Representation Method
108
Chap. 3
109
Spectral Representations for SLPI and SLP2
Sec. 3.3
In order to solve this closed contour integral by the residue theorem, we first inves~ tigate the singularities of the integran<;1. Since,JI is a multiple-valued function, 'we might expect that the integrand contains a branch cut with a branch point at A = O. We may show, however, that although ,JI has a branch cut, g(x, g, A) d?es not. Indeed, define (3.45) 2lf>¢>0
Im(A)
A-plane
1,11
------1'io-....- .....~
I
so that
Fig. 3-2
Ji = IAI 1/ 2/
If
> if!.- > 0
2
(3.46)
This definition of Aresults in a branch cut in ,JI along the positive-real axis in th 7 A-plane (Fi/r.· 3-2). In fact, lim Jj = IA1 1/ 2 ¢-->O
.
Polar representation of A in the complex A-plane showing branch cut for ,JI along positive-real axis (thick line).
We note that peA) =1= {} anywhere, and that q(A) = 0 wherever .JIa = 0, ±If, ±2lf, .... We conclude that q(A) = 0 whenever
12
lim h
Re(A)
= -IAI /
¢-->2"
Applying this result to (3.43), we find that '. hm g(x, g, A) = ¢-->O . hm ¢-+2" g(x, g, A) =
A" -_
c~sIAII/2xcosIAI1/2(a-0 • 1'1 1/ 2 1'1 1/ 2 ~n A A a
(lIlf)2
a
= 0, 1,2, ...
Differentiating the denominator, we find that
cos(-IAI1/ 2x)cos[-IAI1/ 2(a - 0] -
11
,
1/ 2 ) 1'1 1/ 2 sm . ( - 1'1 A A a
I
q (A)
r.; ,JIa) ="2a ( cosvAa + sin,JIa
Some minor algebraic manipulation shows that lim g(x, g, A) = lim g(x, g, A) ¢-->o
from which
¢-->2"
11=0,1, ...
We conclude that, although there is a branch cut in .JI along the positive-real axis, the Green's function is continuous there. We next consider the possible location and order of poles of g(x, g, A). Since the numerator of the Green's functiort in (3.43) is finite throughout the complex A-plane, it is sufficient to search for poles caused by the denominator. Let : I I(A) = ,JI sin.JIa
peA)
- q(A)
where E" is Neumann's number, defined in (3.20). We have shown that the singularities at An are simple poles. For the residues, we have Res{f(A); An} =
8(x -
peA) = I
p(A n )
Res{.f(A); A,,) = --;--(') q A" I
g) =
L cos - a 00
lIlf X
n=O
q(A) = Jisinha
:
a
Returning to (3.44), we now have
where
By a well-known theorem of complex analysis [4], I(A) has a simple pole at A = An if p(A n) =1= 0, q (An) = 0, and q' (A,,) =1= O. In addition, the residue at the simple pole location is given by
(_I)n En
00
~
n En
n=O
a
[8pt] = L..,,(-I) -
fllf
cos -(a - ORes{f()..); An) a lIlf X COS -
a
lIlf
cos - ( a -
a
g)
and finally, 00 En fllf X fllf I; 8(x-0= L-cos--cos--
n=O
a
a
a
(3.47)
,
The Spectral Representation Method
110
Chap. 3
Equation (3.47) is the spectral representation of the delta function associated ";,,ith the operator L :;= _d 2 / dx 2 with boundary conditions u ' (0) = u' (a) = O. Recall that our result is for x < ~. To produce the result for x > ~, we interchange x ,and ~ in (3.47). Since this interchange produces no change in the result, (3.47) holds for all x and ~ on the interval (0, a). The reader is cautioned that the series o~ the right side of (3.47) does not converge. It is, however, an extremely useful symb~lic equality, as has been discussed in Section 2.2. Indeed, we may show that (3.47) is merely a disguised form of the Fourier cosine series. For any sex) E £2(0~ a), we have, from (2.9), I
1 a
"
sex)
=
8(x -
~)s(~)d~
(3~48)
Substituting (3.47), we obtain, after an interchange of integration and summation, 00 . (E:: nrr x sex) = Lany -;; cos -a-
n=O
where an =
lo
as({). fin- cos nrr ~
-d~
o
a a We note that we have produced the ,eigenfunctions and eigenvalues inferreq in (3.18) and (3.19) directly from the Green's function associated with the operator L and its boundary conditions.
•
In the example spectral representation in (3.47), each term in the sum consists of the product of the orth6normal eigenfunction as a function of x with the same orthonormal eigenfanction as a function of ~. We may show that this result can be generalizec;l to all self-adjoint operators on SLP2. Indeed, if the forcing function in (3.12) is the delta function, we have' i 8(x'-~)
,
( w(x') ,lIn(X ))x'
g(x,~,A)=L'
An
/l'
-A
lI/l(X)
, where, as indicated, the inner product is with respect to x'. Performing the , I ' integration gives _'""
g(x,~) -,1-, n
lIn(X)U;;-(~)
(3:49)
A - A
-~) l' - - - = - - . LU/l(x)ulI(~) w(x)
2]'[1
n
Spectral Representations for SLP3
i
dA
---,
CA/l-A
111
or 8(x-~) _ - - - = LUn(X)lI/l(~) w(x) II
Equation (3.50) gives the general spectral representation for selfadjoint operators on SLP2. The procedure for solving self-adjoint SLP2 problems by the spectral representation method can now be summarized as follows: ' I. For a given self-adjoint operator L and given boundary conditions, solve the Green's function problem LAg(x, ~, A) = 8(x-~)/w(x). 2. Substitute the Green's function g(x,~, A) into (3.39) and solve for the spectral representation of the delta function. The resulting form should appear as in (3.50). 3. Substitute the normalized eigenfunctions and their eigenvalues, obtained in the spectral representation, into (3.12) to produce the solution u(x). In addition to Example 3.2, we have included in the problems several common examples to illustrate the method.
3.4 SPECTRAL REPRESENTATIONS FOR SLP3 The spectral representation of the delta function takes on a different character when the interval along the real axis becomes unbounded. Consider the problem in Example 3.2, defined on the interval x E (0, a). The Green's function g (x , ~, A) was shown to have simple poles at A = (mr / a)2. Note that as a becomes larger, the poles become closer together. In the limit as a -+ 00, the poles become arbitrarily close. This behavior leads us to inquire into the form of the singularity along the positive-real axis in the A-plane in this limiting case. We shall illustrate the obtaining of the spectral representation with an example. EXAMPLE 3.3
Consider the following SLP3 differential equation:
-U" - AU = u(O)
/l
: This form of the Green's functidn is called the bilinear series form [5]. Substitution of (3.49) into (3.39) gives /s(x
Sec. 3.4
I,
(3.51)
=0
(3.52)
This problem is in the limit point case as x -+ 00. We therefore assign the limit condition (3.53) lim u(x) = 0 x->oo
The associated (Jrc<:,n's fllnction prnhlcm is
The Spectral Representation Method
112 d 2g
Chap. 3
dx 2
= 0
lim g(x,:~) = 0
J
x->oo
Jc.+c, +cp+c,
:We have previously obtained this Greep's function in (2.172) and repeat it here fOf
e-~,;I~ sin v'Ax,
x < ~
e-,i,;Ix Sin v'A~ ,
x>~
lim lim
"'.
To produce the restriction in (3.55), we define 21T < <jJ < 41T
(3,56)
[
X
cp
so that v'A
<jJ
= IAI 1/ 2ei t/>/2,
1T < -
< 21T
(3.57)
2 ! The angular definition in (3.56) defines a Riemann sheet of the A-pla~e. The
lim
J>- =
_IAI
=
lim g(x,~, A)
->2](
,
lim g(x,~, A) =
->4](
g(x,~, A)dA =
-21Ti8(x
-~)
(3.59)
A)dA = i
2rr+r
P
1
4](-r
. (
1/2 ir/J/2 )
.
sm p e x -i(pl/'e'~/'~) ir/JdA. 1/2ir/J/2 e e'l' P e
Since the integrand on the right side is bounded as p -+ 0 and is bounded and lim lim (
r ->0 p->O 1Cp
g(x,~, A)dA =
Im(A)
r
-+ 0, the integral
0
(3.60)
A-plane
1 2 /
Unlike the situation in Example 3.2, however, the branch cut in JI produces: a branch cut in g(x, ~,A) along the positive-real axis. Indeed, for x < ~, i
I:
( g,,,,
angular restriction in (3.57) indicates that Imv'A < 0 everywhere on thiS sheet. i We shall refer to this sheet as the pro/?er Riemann sheet: We note that ?nce (3.55) has been invoked, any result with ImJI > 0 would vIOlate the reqUirement f~r the proper Riemann sheet. The definition of A i~ (3.56) result~ in a branch c~t In JI along the positive-real axis in the A-plane (Fig. 3-2). In thiS case, , ->2](
(3.58)
where we have assumed that, even in the presence of the branch cut, the integral around the circle of infinite radius centered at the origin produces the delta-function contribution. Consider Cpo Letting A = p exp(i<jJ), we have
I~v'A < 0
•• "
(
r->o R->oo lc.
where
A = IAlei ,
g(x,~,A)dA=O
We examine the contributions from the various portions of the contour in the limit as p -+ 0, R -+ 00, and r -+ O. Consider CR. From (3.39), we have
:convenience, viz.
"
113
We next produce the spectral representation by considering the closed contour in the complex A-plane shown in Fig. 3-3. Since the contour excludes the branch cut and since g(x, ~, A) has no other singularities, Cauchy's theorem gives
,
---Ag=8(x-~)
g(O,'~)
Spectral Representations for SLP3
Sec. 3.4
----t----+-~~--
....- - - . R e ( , \ )
eiIAI'/'~ sin IA1 1/ 2x
1'1 1/ 2 A e-iIAI'/'~ sin IA1 1/ 2x
1'1 1/ 2 A ! .
Since the exponential changes sign, g (x, ~, A) is discontinuous across the pOsl,tlvereal axis. The result for x > ~ is the same, except that x and ~ are intercha~ged.
Fig.3-3
Contour for evaluation of the spectral representation for Example 3.3.
Chap. 3
, The Spectral Representation Method
114
Sec. 3.4
:
Consider the integral along C 1 an~ C2. On CI, let
Spectral Representations for SLP3
115
Substituting (3.64), we obtain, after an interchange of integrations, 00
2 f(x)=:n:
A = re i (4rr-f)
On C2 , let
where
1 1 f(~) sink~d~
F(k)sinkxdk
(3.65)
0
00
F(k) =
(3.66)
We indicate the Fourier sine transform relationship symbolically by
<==>
f(x)
F(k)
(3.67)
We now return to the solution to the differential equation considered In (3.51)(3.53), repeated here for convenience. Consider L2 (0, 00) with inner product Taking the limits, we have
. . .1
!tm !tm !tm
1
00
. 1 :
(u, v) =
o·
1/2 smr x -ir'f2~d 1/2 e r
g(x,~, A)dA =
p ..... OR ..... oof ..... O C,+C2
00.
+
1 1
sm r
o
= 2I·
00
1/2
In
x
e
ir'/2~ d
-u" - AU = f, r
r sin(rl/2x) sin(rl/2~)
o
r
1/2
'1
+ 2i
sin(r 1/ 2x) sin(rl/2~)
00
r l/2
o
lim u(x)
(3.62)
dr
x ..... 00
2 u(x) = :n:
(3.63)
dr = 0
=0
(3.69) (3.70) (3.71 )
/ 'rt
00
1
U(k)sinkxdk
(3.72)
0
where
U(k) = (u,sinkx) = so that
dr dk = 2r l (2
sUbstit~te into (3.63) and produce the following spectral representation: , -2 8(x -~) == :n:
I'
C
From Example 2.18 and the discussion following, this problem is self-adjoint. Using the results in (3.66) and (3.67), we expand u(x) as follows:
We let
We
AE
u(O) = 0
Combining the results in (3.58)-(3.62), we obtain
-2:n:i8(x -~)
(3.68)
Consider
r
00
u(x)v(x)dx
1 00
(3.64)
sinkx sink~dk
0
have
I
f(x) =
[00
.10
8(x -
~).f(~)d~
(3.73)
Expression (3.72) is the equivalent to (3.24), except in this case we have an integral, rather than a sum. We note that sin kx plays the role that the eigenfunction u" plays in (3.24). In (3.73), U (k) is similar to the Fourier coefficient in (3.25), where sin kx plays the same role as the eigenfunction u" in (3.25). To further investigate the similarity to the eigenfunction U", we note that d 2 sinkx
dx 2
In a similar manner to that in Example 3.2, we now show that (3.64) i:s merely a disguised form of the Fourier sine transform. Indeed, for f(x) E [2(0\ 00), we
1°Ou(~)sink~d~
=
2. k smkx
(3.74)
so that sin kx appears to be an eigenfunction of the self-adjoint operator -d 2 /dx 2 with eigenvalue k 2. However, sin kx is not in L2(0, 00), and therefore cannot be an eigenfunction. We shall adopt the notation of Friedman [61 and call sin kx an improper eigenfunction with improper eigenvalue k 2 . Fortunately. the procedure we
.'"1'
I
I
The: Spectral Representation Method
Chap.J
adopted in (3.24)-(3.36) for solving differential equations by the eigenfunctioneigenvalue method can be extended to apply to improper eigenfunctions. We begin by showing that if II
{::::::>
Sec. 3.4
Dividing both sides by (k 2 I
-
I,II(X)
(3.75)
U
, 117
Spectral Representations for SLP3
i
A) and taking the inverse transform, we obtairi
21
= rr
0
00
, (3.81)
F(k) . sm kxdk k-A
-2--
•
then 2
d u.{::::::> k 2 U 2
(3.76) .
- -
dx
Indeed, following (3.28), we form d 2u
( - - , sinkx) = (u, ch 2
d2 sin'kx
.
2) + J(II, smkx)
dx.
\00 0
~ k'U + (-:: ,ink< +ku ,"'kX)[
(3.77) .
,
Using the conditions in (3.70) and (3.71), we have 2 u smkx) . du. k (_d_ =.k 2 U - I·1m -sm·x
dx2'
x-->oo
dx
(3.~8) !
We note that the improper eigenfunction sin kx does not vanish in the limit as x :-+ 00. This behavio,r is contrary to what we , found when dealing with eigenfunctions I ' on finite intervals. Fortunately, in electromagnetic problems, when we have I
,
lim II(X)
'
X-->OO
=0
In the above example, we have indicated explicitly the various limiting operations involved inevaluating the integral of the Green's function around the closed contour indicated in Fig. 3-3. In subsequent discussions, we shall, whenever appropriate, simplify the contour (Fig. 3-4) such that the limiting operations have already taken place. The contour in Fig. 3-4 is to be interpreted as follows. The contour segments C 1 and C2 are straight lines that are the result of limits as we approach the branch cut fr~m below and above, respectively. By Cauchy's Theorem, if there are no singularities inside the contour, then the integral along CR is the negative of the integral along C\ + C2. Therefore, we have [ CI+C2 g(x,~,A)dA
=- [
.
g(x,~,A)dA CR
(3.82)
w(x)
In (3.82), we have assumed that there is no contribution obtained from the integral along C p (Fig. 3-3) in the limit as p -+ O. We use this abbreviated method in the following example. Im(A)
then also .
2ni8(x -~)
= -----
A-plane
du(x)
hm -.-- =0
x-->OO
:dx
I
For example, if u is a component of the electric field, then au/ax is acomponenl of the magnetic field; if the E-field vanishes at infinity, then the H -field also vanish,es. Therefore, in the usual cases in electromagnetics, we obtain ,
(3.79)
---+------t7l--------.. Re(A)
which establishes (3.76). As a footnbte, we remark that there are mathematical theorems that generalize this result to classes of functions possessing certain continuity and absolute integrability properties. The interested reader is referred
to [7].
:
We now are able to solve the original differential equation in (3.69) usinglhe spectral representation. Taking the Fohrier sine transform of both sides of (3.69)1 I we obtain
(k 2 -
A)U(k) = F(k)
(3.80)
Fig. 3-4
Simplified contour for evaluation of the spectral representation of the delta function for SLP3 problems.
Chap! 3 .
The Spectral Representation Method
118
EXAMPLE 3.4
i:
Consider Hilbert space L2( -00,00) with inner product (u, v) =
(3.83)
u(x)v(x)dx
2
I
(3.84)
2
with limiting conditions I
X-4-00
(3.85)
= x---+oo lim u(x) = 0
u(x) i
;
X-4-00
=
(3.86)
Imv"): < 0
lim g(x,!;)
where U(k) =
g(x, ~)
==
I
(3.88)
2i v"):
The singularities of g (x, ~) involve th~ branch cut associated with this branch cut using (3.56) and (3.57), and find for x > ~
,JX.
We define
eiIAII/2(X-~)
lim g(X,~,:A) =
1'-->2"
.
-
(3.89)
2'1'1 1/2 I 1\
1'-->4"
:Therefore (Fig. 3-4), there is a branch ~ut in g(x,
2rri8(~ -~) = Joo
2ilAII/2
, dA -
Jo
2ijAI I /2 dA
100
: (3;91)
iWe let (3.92)
-~) = - f~ e-ik(x-Odk + loo eik(t-Odk
: Replacing k by -k in the fir~t integral give~ the final re~ult, viz.
(3.96)
ikx u(x)e-ikxdx = (u, e )
AEC
(3.97)
(3.99)
lim u(x) = 0
(3.100)
X---l>-OO
(3.101)
lim u(x) = 0
x .....
00
We begin by showing that if u (x)
{::::=}
U (k)
-u"(x)
{::::=}
eU(k)
then
(3.102)
Indeed, . ~~kx lkx (-u", e ) = (u. ---2-) dx
and find that 2rr8(x
U(k)eikxdk
-00
We may lise the Fourier transform to solve the following differential equation by the spectral representation method:
['Xl eiIAII/2(x-O
e- i \A\I/2(x-O
(3.95)
In (3.96) and (3.97), we identify exp(ikx) as an improper eigenfunction with improper eigenvalue k 2 • We indicate the Fourier transform relationship symbolically by u(x) {::::=} U(k) (3.98)
~, A) along the positive-real i xiS .
:Using (3.82), we o b t a i n .
fO
i:
(3~90),
2'1'1 1/2 I 1\
(3.94)
u(~)8(x - !;)d~
-u" - AU = j,
e-iIAII/2(x-~)
• lim g(x, ~,A) =
i: 2rr
The solution to this problem was given in (2.175), viz. ; e-iJ):lx-~1
=
u(x) = - I
(3.87)
X-4(X)
eik(x-Odk
Substitution of (3.94) followed by a change in the order of integration yields
I
=0
1-0000
which means that we can again reverse the interchange and reclaim the result in (3.94). We next use the spectral representation in (3.94) to produce the Fourier tran.~form. We write
In Example 2::20, we considered the foilowing Green's function problem:
. d 2g ---Ag=8(x-n dx 2
119
We have obtained this result for the case x > ~. However, this restriction can be removed. Indeed, to obtain the case x < ~, we merely interchange x and ~ in (3.94). However, 8(x - ~) = 8(~ - x)
.
lim g(x,!;)
I
2rr
L = _d /dx
lim u(x)
Spectral Representations for SLP3 8(x - !;) = -
We seek the spectral representation for the self-adjoint operator
,
Sec. 3.4
(3.93) = eU(k)
+ J(u, eikx) 100
+ (_u'e- ikx
-00 - iklle-
ikx
)
[00
(3.103)
The Spectral Representation Method
120
Chap. 3
= k2U(k) _
Spectral Representations for SLP3
(u'e-ikx)!oo -00
From the discussion associated with (3.78), we know that in the usual cases in electromagnetics, (3.100) and (3.101) imply that lim u'(x) = 0
g(x,O
=
71' 2i
I
HJ2\ ../i.~)Jo(../i.x),
x < ~
Ht) (../i.x )Jo(../i.o,
x > ~
Im../i. < 0
e
(-u", ikx ) = k 2U(k)
(3.104)
and obtain
(e - A)U(k) =
-271' < ¢ < 0
(3.112)
so that F(k)
Dividing both sides by (k 2 - A) and taking the inverse Fourier transform, weihave I
F(k) u(x) = -I ' foo _ _ eikxdk 271' -00 k 2 - A
,I (3,105) :
We note that the result in (3.102) could also be obtained by twice differentiating (3.96), provided that we can intercha(Jge differentiation and integration on the right side. Our method of proof provides ajustification of this interchange in this case.
•
"
We next provide two examples leading to solutions involving Bessel functions. These examples will be useful in problems in cylindrical coordinates to be considered in later chapters. ' Consider the following differential equation on x E (0,00):, (L - A)U =
(3.111)
To assure the condition in (3.111), we restrict A as follows:
which proves (3.102). We now take :the Fourier transform of both sides of (3.99)
EXAMPLE 3.5
(3.1 10)
where
x~±oo
and therefore, --.
121
We seek the spectral representation of the operator in (3.107) with the Ii 't' d' . . . mt 109 co.n Itrons gIven 10 (3.108) and (3.109). The Green's function ass~ciated with thIS op~rator has been obtained previously in (2.184) and is repeated here for convel1lence, as follows:
Using the conditions in (3.100) and (3.101), we have (-u", e ikx )
Sec. 3.4
,
f
h = IAI I/2 e i ¢/2,
-71' <
~
< 0
(3.113)
We.~ay show t.ha~ this definition produces a branch cut in g(x,~, A) along the positIve-real aXIs In the A-plane . Consider the case x > "I: . A pproac h,'109 the .. positive-real axis from above, we have
¢~~" HJ2)(../i.x)Jo(h~)=
HJ2)(e-i"p.. 11/2x)Jo(e-i"IAII/2~)
(3.114)
But, using a well-known Bessel function identity [8], we have Jo(e-i"j)1.11/2~) = Jo(iAII/2~)
(3.115)
and using a well-known Hankel function identity [9], we have HJ2\e- i" P. I1/2x ) = -HJ 1)(IAI 1/ 2x)
(3.116)
so that
(3.106)
(3.117) On the other hand, approaching the positive-real axis from below, we have
where
L = -
~ [:x (x :x ) ]
(3.107)
0
(3.108)
lim u(x) finite
(3.109)
.t--.O
(3.118)
I
From Example 2.21, this problem is in the limit circle case as x ---+ 0 and the limit point case as x ---+ 00. Furthermore, the operator L is self-adjoint. We invoke,
lim u(x) = x->oo
J~ Hri2)(../i.x).lo(../i.~) = H~2)(IAII/2x).lo(IAII/2~)
We not~ ~hat (3.117) and (3.118) indicate ajump in the Green's function as w~ cross the POSitive .real axis. 1;0 produce the spectral representation, we again consider the contour 10 Fig. 3-3 and write (3.119)
:The Spectral Representation Method
122
Chap. 3
Sec. 3.4
Spectral Representations for SLP3
In a manner similar to Example 3.3, we may show that the contribution alorig C p vanishes as r ~ and p ~ 0. Vfe leave this for the reader to verify. Along C 1 + C2, we have
°
sex) =
123
lOO S(k)Jo(kx)kdk
(3.127)
Symbolically, we write sex)
<==>
S(k)
A very useful relation is obtained by noting that for II (x) E £2(0,00),
~ [~(xdll)] =
x
dx
dx
U(k) {~~ [xdJo(kx)]} kdk 1roo x dx dx 0 (3.128)
= loor-eu(k)]Jo(kx)kdk
Along C R, we have
. I'1m \1m
1 (' g
l'~o R~oo
Therefore,
8 A) d A = - 2· Jrl (x-
x,~,
CR
O
Taking the appropriate limits in (3.119) and substituting (3.120) and (3.121), we obtain 8(x -
~)
=
Letting k =
roo Jo(IAII/2x)Jo(IAII/20dA
<==>
2
-k U(k)
(3.129)
We note that the result in (3.128) depends on the interchange of differentiation and integration. This operation can be justified by using the procedure followed in (3.76)-(3.79) and in (3.102)-(3.104). The details are left for the problems.
•
(3.142)
210'
x 12 1A1 / ,
~
~ [;~r (x ~~;)]
(3.121 )
X
EXAMPLE 3.6
we find that
1
00
8(x- 0 =. -
x
'0
Jo(kx)Jo(kOkdk
'1
00
s(x)t(x)xdx
Im(k) <
(3.123)
which is the required spectral representation. Although this representation has been obtained with the restriction x > ~, the restriction can now be removed in the same manner as in Example 3.3 because of the symmetry of the Green's function. The representation in (3.123) leads to the FOllrier-Bessel Trall.~f(}rm of order zero. Indeed, consider a Hilbert space £2(0, 00) with inner product
(05, t) =
We wish to find the spectral representation of the operator
(3.124)
on £2(0, 00). By examining (2.22), we identify p(x) = x and w(x) Green's function diff~rential equation associated with LA is given by -x
[~ (x::)] - (kx)2 g -
-x sex) =
1 =1 o
s(~)
8(x -
x
0
~d~
(3.125)
00
s(x)Jo(kx)xdx
q.126) !
[~ (x dll)] dx dx
(kx)2
II -
For A = 0, two independent solutions are given by
Substituting (3.123) into (3.125), we produce the Fourier-Bessel transform :pair , S(k)
Ag =x8(x
-~)
J3.130)
=
I/x. The ,
(3.131 )
where we have identified 8(x - Olw(x) = x8(x - O. We investigate limit point and limit circle conditions as x ~ 0 and as x ~ 00 by examining solutions to the homogeneous equation
For any sex) E £2(0,00), we have 00
°
III
= H~2)(kx)
112
= Ho (kx)
and
(I)
All = 0
The Spectral Representation Method
124
Chap. 3
'Let ~ be an arbitrary interior point on the interval x E (0, (0). As x -+ 0, both UI and Uz are kgarithmically singula~. The singularity is weak enough, however, that they are both in .cz(O, ~). We therefore have the limit circle case as x ---+ o. :The solution Uz diverges exponentially as x -+ 00, and thus is not in .cz(~, (0). :We therefore have the limit point case as x -+ 00. We invoke the limit condit~ons !
Sec. 3.4
Spectral Representations for SLP3
The evaluation of the coefficients A and B proceeds in a manner similar to that in Example 2.21. Invoking the continuity and jump conditions at x = ~, we find that Jr A = -H(Z)(k/::) 2i v ., Jr B = 2i lv(k~)
(3.132)
lim u(x) = 0
125
.l'---lo-OO
(3.133)
lim u(x) finite
Substitution into (3.139) gives
X-4>O
The limit conditions associated with the Green's function are I
(3.134)
limg(x,O=O
H~Z)(k~)Jv(kx),
x < ~
H~Z)(kx)lv(kn
x>~
:(3.140)
x->oo
lim g(x,
~)
(3.135)
finite
where
X->O
Define a parameter v by v = eirr/ZJ).. = i J)..
--," t,T
Then, for x
(3.136)
Im(k)<0
(3.141)
Re(v) > 0
(3.142)
The last step in the determination of the Green's function involves making the transformation from v to A in accordance with (3. I 36). If we define
i= ~, we have (3.131)
We identify (3.137) as Bessel's equa'tion of order v and argument kx [10]. The Green's function can be composed of linear combinations of various Bessel flmctions as follow~: x < ~
( kx)V lv(kx) ----+ -V , 2 v!
Therefore, if we choose Re( v) > 0, Lv diverges as x -+ 0 and we must set C := The Green's function can now be written as follows: :
x>~
ImJ).. < 0
7f
-
2
(3.145)
¢+Jr
Jr
2
2
> - - >--
o.
(3.139)
(3.146)
The angular range in (3.146) is consistent with the restriction on v in (3.142). We therefore have
2i
(-v)!
BH:Z)(kx),
(3.144)
This result implies that
g(x,~, A) =!!.-I H/~(k~)li,fi..(kx),
2V(kx)-V
x < ~
¢ 0> - > -Jr 2
Substituting (3.144) into (3.136) gives
where lv, Lv,Hl~Z), and H~I) are linearly independent solutions to Bessel's equation of order v [10]. For Im(k) < 0; H~I) diverges as x -+ 00 [11]. Therefore, D = O. We may set C = 0 by using the following argument. For x -+ 0, we have
Al,.(kx),
(3.143)
then
(3.138)
x>~
Lv(kx) ----+
0> ¢ > -2Jr
H(Z) (kx)I re(k/::) i,fi..
IvA" ,
x < ~
x > ~
(3.147)
where the branch cut in .j). lies along the positive-real axis and is explicitly determined by (3.143). . O~r next step is to determine the spectral representation of x<5(x - ~) by tntegratmg over the Green's function with respect to A in a similar manner to that performed in Example 3.5. We first consider the case x < ~. We find that the
The Spectral Representation Method
126
Chap. 3
branch cut in -II, defined in (3.143), produces a branch cut in g(x,~, A) along the positive-real axis. Indeed,
Sec. 3.4
Spectral Representations for SLP3 ,
The representation in (3.154) leads to the Kantorovich-Lehedev Transform [14],[ 15]. Indeed, consider a Hilbert space £2(0, 00) with inner product 1
lim H.(2) (k~)Ji.fi(kx)-I'12(k~)LiI>-I'/2(kx)
o
lim H.(2) (k~)J ..fi(kx) = Hii~'/2 (kOJil >-II/2 (kx)
,.fi
1>->0
'
1
00
(s, t) =
,.fi
1>->-2rr
....
1
00
sex) =
d~ s(~)x8(x - ~)-
f(x) = (2)
_.
-irr(iI>-I'/2) H(2) il>-I'/2
(3.149)
Substituting (3.149) into (3.148) and combining integrals gives
1
00
-4x8(x -
0 =
(3.150)
(3.151) Substitution into (3.150) gives 4x8(x
_~) = i
'
(3.152)
Let
(3.153) Then, x8(x -
~) =
1
too
I) F(f3)HA 2)(kx)fJdf3
(3.158)
iOO
F(f3) H?) (kx)f3df3
ioo
(3.159)
(x)
{:::=:}
F (f3)
(3.160)
If we apply the operator L in (3.130) to both sides of (3.159), we produce the useful relationship
(3.161)
roo e-irr (iI>-II/2) sin(irrIAII/2)Hii~I'/2(kOHii~'/2(kx)dA
10
{i -
(3.157)
The details of producing (3.159) from (3.158) are left for Problem 3.6. We indicate the Kantorovich-Lebedev transform relationship by
f
But for any vEe [13],
dt f(x)Hfi2)(kx)~ x
Alternately, we can manipulate (3.158) to produce
4
irr(iI>-I'12j H(2) (kl:) Jo e. il>-I'/2 " . [LiI>-I'/l (kx) - eirr(iI>-I'/l) J il >-I'/2 (kx) ] dA
~ too (e- i2rr 41 0
f(x) = -11-
(00
(3.156)
Substituting (3.154) into (3.156), we produce the Kantorovich-Lebedev transform pair
o
H_ il >-1112 - e
(3.155)
~
o
F(f3) =
But [12],
dx s(x)t(x)x
For any sex) E £2(0,00), we have
,
Since J v and Lv are linearly independ~nt for any VEe, v:e concl~de t.hat there i~ a jump in the Green's function across the positi~e-re.al aXIS, resultl.ng ~n a branch cut. The appropriate contour is the one shown m FIg. 3-4. Substltutmg (3.147) into (3.82), we find for x < ~
.'
127
I
i
4. 10 (e- i2rr /l ~ I) Hfi2) (kx) Hfi2) (kOf3dfJ
,
(3.154)
:We note that this result is not altered by interchanging x and ~. TherefOre,OUj.re:striction x < ~ can be removed. Equation (3.154) gives the spectral repres~nta, Ion of the delta function for the operator defined in (3.130)-(3.133). 1
The interchange of differentiation and integration used to produce (3.161) can be justified in the same manner as in the procedure in (3.76)-(3.79) and in (3.102)(3.104). The details are left for the problems. The Kantorovich-Lebedev transform is useful in solving certain electromagnetic problems in cylindrical coordinates, as we shall discover in the next chapter.
•
In the mathematical literature, the spectral contribution resulting from pole contributions, such as in (3.47), is called the discrete spectrum, whereas the contribution from the branch cut, such as in (3.64), is called the contin//o//s spectrum. We next inquire if it is possihle to have both a continiJous
r
:j
I
1
i
.I1
:)
~l
"1
.,1
,I
The Spectral Representation Method
128
Chap. 3
and discrete spectrum associated with an operator. The example we c~oose involves an operator that is not self-adjoint. The theory of nonself-adjoint operators is both difficult and incomplete. However, in the simple ex~mple to follow, we are able to obtain the spectral representation in a straightforward manner. We consider th~ spectral representation of the operator L = with boundary and limiting conditions
Sec. 3.4
Spectral Representations for SLP3
The result for x > ~ is obtained by interchanging X and ~. In addition to the bran~h cut on. the positive-real axis, g(x,~, A) has an isolated singularity at the locatIOn AO, given by solving
with the result
EXAMPLE 3.7 _d 2 / dx 2 ,
(~.162)
lim u(x) = 0
,
X-HXl
u'(O)
= au(O),
Since a is complex, the operator function problem is given by
I,
(3.163)
Re(a) < 0
129
AO = _a 2 We now show that this singularity is on the proper Riemann sheet. Indeed . . ' we have r;:: -' F . .v AO - Ia. rom t h'IS relatIOnship, we easily find that the relation Im.,f).. < 0 Impltes that Re(a) < 0, as assumed in the problem statement. We now show that this singularity is a simple pole. For x < ~, we write
is nonself-adjoint. The associated Green's I
(3.166) where
hex, ~, A)
limg(x,O=O x->oo We have previously obtained this Green's function in Example 2.23. We repeat the result given in (2.216) for convenience, viz.
e-ifix(cos.fi~
+
i
e-i fi~ (cos v I,AX
+
c;,
sin.,f)..n vA a. 1,) .,f).. SIn v AX ,
=
II (A)
dg~ ~) = ag(O, 0
We note that
h (x, ~, A)
I i.fi+a
= e-ifi~ (cos.fix + ~ sin .fix) .fi
(3.167)
(3.168)
is regular at AO. Consider II (A) = peA) q(A)
x>~
q.I64)
where peA)
x < ~
=
I and q(A)
= i.fi +a
We have I
where
I
!m.,f).. < 0
(t 165)
The restriction in (3.165) can again be assured by defining .,f).. as in (3.56) and (3.57) so that, once again (Fig. 3-2), lim' .fi = -IAI I / 2 ¢->2rr lim .fi = IA\I/2 ¢->41r The branch cut in .fi along the positive-real axis results in a branch cut in the same location in g(x, ~,A). Indeed, for x < ~, we obtain in (3.164) . ltm R(x,~, A) ¢->2rr
=a
e
-
iIAI'/2~
a. 1/2 . . 1/2 (cos \AI 1/2 x + ~/2 SIn IAI x) /IA\ IAI
_iIAI'1/2~
. _ e 1/2 a.. 1/2 ltm R(X.~, A) . 1/2 (cos IA! X + ~/2 SIn IA\ x) 1>--4rr a + 'IAI IAI
I 2a We conclude that II (A) has a simple pole at AO with residue ,
2
q (-a) = -
Reslfl (A); AO! = p(AO) = 2a q'(AO) To obtain the spectral representation, we integrate the Green's function g(x, ~, A) around the closed contour shown in Fig. 3-5. For x > ~, we have
fg(X'~' A)dA = 2rriRes(g(x,~, A); Ao! = 2rrieax[cos(ia~) = 2rri(2a)e,,(x H l
i sin(iaOJReslfl(A); AO! (3.169)
.. : Tht Spectral Representation Method
130
Chapi 3
I Im(A)
___
A-plane
+----~--___{~-------.
Re(A)
Sec. 3.4
Spectral Representations for SLP3
131
In Problem 5.6 given at the end of Chapter 5, the spectral representation in (3.172) will be used to characterize a source over a flat surface characterized by a surface impedance. We shall discover that we may associate the first term with a surface wave bound to the surface and the second term with radiation carried away from the surface. We remark that we assumed Re(a) < 0 in the problem statement. In addition, we found that ImJI < 0 implies Re(a) < 0, as required. If the problem had stated that Re(a) :::: 0, the pole at AO = _(X2 would have been on the improper Riemann sheet and the discrete spectral term in (3.172) would be missing. In a similar manner to that in Examples 3.2 and 3.3, we now cast (3.172) in a form that we shall call the impedance transform. For u(x) E £2(0, (0), we have
".
1
00
u(x) =
8(x -
~)u(~)d~
(3.173)
Substituting (3.172) and interchanging the order of integration, we obtain, I
!lex) = -2ae
Fig. 3-5: Contour for evaluation of the spectral representation for Example 3.4. Contour includes the simple pole (x) at AO'
""'t
... ,
ax
21
00
Uo + -
1T
k
0
where :The integral around the closed contour;(Fig. 3-5) consists of the integral around thy 'circle of radius R plus the integrals along either side of the branch cut. Therefore, 'in the limit as R ~ 00, we obtain '
1'
, 1. -8(x-O+-. hm i
I
21T I R--->oo
d' 2 a(x+O g(x,~,)..) 1 \ = ae
(3.170)
C, +C2
:Evaluating the integrals along CI and C2 in a similar manner to the proces's in Example 3.3, we obtain lim R--->oo
l'
= 2i
C,+C2
[00
Jo
112
:
j1 / 2d)"
a. 1/2 IA ~ -: 1)..1 1/ 2 SIO 1)..1 ~)-a2=-+-I)..-1
(3.171)
21
-
1T
+a
(3.174)
00
Uo =
1
u(x)eaXdx
(3.175)
a
00
U(k) =
+-
(3.176) sinkx)dx k Equations (3.175) and (3.176) comprise the impedance transform, yielding the spectral coefficients Uo and U(k). Equation (3.174) is the inverse impedance transform. We call (3.175) the zeroth-order impedance transform, while (3.176) is the kth-order impedance transform. In (3.174), we identify
o
u(x)(coskx
0
00
,
(coskx + '
a
as an eigenfunction of the operator -d 2 /dx 2 with boundary conditions given in (3.162) and (3.163). In addition, coskx
We let k = IAI I/2 and obtain for the spectral representation
-0 = _2ae a(xHJ +
1
k
(cos l)..i l/2 x + l)..a\1/2 sin 1)..11/2 x )
. (cos IAI
8(x
2
a k dk U(k)(coskx + - sinkx)-2--2
a .
2
k dk k2
k sinkx)(cosk~ + k smk~) a2 +
, is an improper eigenfuflction of the same operator with the same boundary conditions. We define an in~er product for the space by
(3.172)
The first term on the right side gives the discrete spectral contribution, whil~ ~~e second term gives the continuous spectral contribution. Although the delt.a functl9 n representation has been obtained with the restriction x > ~, we no~e 10 (3.164) that the case x < ~ can be obtained by interchanging x and ~. Smce such an interchange leaves the result in (3.172) unaltered, the restriction can he rem~ved.
a .
+ k smkx
1
00
(u, v) =
u(x)v(x)dx
With this definition, we may write (3.175) and (3.176) as follows:
Uo = (II, ea.t )
(3.177)
132
Repre~entation
,The Spectral
ii.
+ k smkx)
U(k) = (u: coskx
Method
Chap. 3
,
(3.178)
Sec. 3.4
"
Spectral
ii .
Repre~entations
(-u ,coskx+-smkx) = (u, k
for SLP3
2 d - - 2 (coskx
dx
133
+ ii- sinkx») k
We identify
,
as an adjoint eigenfunction of the operator ditions given by
-d 2 /dx 2
a . d a +[-u (coskx+- smkx)+u-(coskx+k dx k with adjoint boundarY con,
v'(O) = iiv(O)
lim v(x) = 0
(3.179)
x---.+oo;
= e(u, coskx
= eU(k)
lim u(x) = 0
".
cos kx
,
k
sinkx~J 1o
(3.185) where we have used (3.181) and (3.182). In addition, we have used the fact that, in the usual cases in electromagnetics,
In addition,
x-->oo
a
+ k sin kx
implies that
is an improper adjoint eigenfunction pfthe same operator with the adjoint boundary condition given in (3.179). We next use the impedance transform to solve the following SLP3 prohletn:
, !
+ ~ sinkx)
00
-u" - A~ = f,
AEC
lim u'(x) = 0
x-->oo
Symbolically, we indicate this transform by
(3.180) .
We now apply the zeroth-order transform to (3.180) to give
with boundary condition
Re(a) < 0
u'(O) = au,(O),
This problem is in the limit point case as x -+ condition
lim u(x)
x-->oo
00.
We therefore invoke the limiting
=0
_(a
(3.181)
+ A)Uo =
Fo
where Fo is the zeroth-order transform of f(x). Rearranging, we have ,
Uo =
(3,182)
In order to solve the differential equation in (3.180) by the use of the impedarice transform, we require the zeroth-order and kth-order impedance transform of _d 2 u/ dx 2 . For the zeroth order, we, have I '
2
F
o -2 a +A
(3.187)
Similarly, for the kth-order transform, we find that
(e - A)U(k) =
F(k)
where F(k) is the kth-order transform of f(x). Rearranging, we have (3.188)
(3.183) where we have used (3.181), (3.182), and
lim e
x-+oo
Substituting these results into (3.174) gives the final result, viz. ax
= 0
Symbolically, we indicate this zeroth-order transform by
,,0 -a 2U0 -u===}
.
Similarly, for the kth-order transform, we have '
2dk 2ae Fo + -2£00 F(k) (cos kx + I sin kx) -k _ k2, k2 +a 2 a 2 + A' :n: 0 -A ax
u(x) =
(3.184)
(3.189)
•
We have now concluded our presentation of the spectral representation method. This method and the Green's function method together
1'-,.~
, The Spectral Representation Method
134
Chap. 3
comprise a powerful tool for the solution of many of the partial differential equations found in electromagnetic radiation, scattering, and diffraction. In subsequent chapters, we shall study electromagnetic source representations, and then develop the solution methods for a large class of electromagnetic problems. We shall conclude this chapter with a brief discussion of the connection between the Green's function method and the spectral representation method.
3.5 GREEN'S FUNCTIONS AND SPECTRAL REPRESENTATIONS There is ~n important connection between the Green's function method and the spectral representation method. Indeed, consider the result in Example 3.3, given in (3.81), viz.
2 £00 F(k) u(x) = - , - - sinkxdk rr . 0 k 2 - A
(3.190)
Problems
135
as we could easily verify by the calculus of residues. Since (3.193) requires an integration and (3.194) does not, it would appear that the spectral representation is not as useful in practice. Its utility, however, becomes clear as soon as we begin considering partial rather than ordinary differential equations in subsequent chapters.
PROBLEMS 3.1. For the operator L = -d 2 /dx 2 and boundary conditions u(O) = u(a) = 0, begin with the Green's function for LA and show that the spectral n;presentation of the delt
f'--
2 I17r x I17r ~ 8(x -~) = - Gsin--sina 11=1 a a
10 f(~)sink~d~ 00
F(k) =
(3.191)
Ifwe substitute (3.191) into (3.190) and interchange the order of integration, we obtain
u(x) =
00 f(~) [2-.10. 00 sinkkx sin' k~ dk] d "t
10o
2
rro
(3.192)
-/I.
_ ~ 10 g(x,~) rr 0
00 sin kx sin k~ dk k2
-
(3.193)
'
/I.
We compare this result with the Green's function obtained in Exampl~ 2.18, given in (2.172), viz.
g(x,~)=
1 f7I
VA
I
e-i..fi~ sin~x, ."
J,"
e-l"AxsinVA~,
x <
k2
-
A
2v1.
I
=f
3.2. For the operator L = _d 2 / dx 2 and boundary conditions u (0) = u (2rr) and u' (0) = II' (2rr), the Green's function for LA was found in Problem 2.18. The result is repeated here for reference, viz. t
1
g ( x,d = - - - - - -
v'A"(~
- x - rr),
x<~
cos v'A"(x - ~ - rr),
x>~
[ cos
Beginning with this Green's function, show that the spectral representation of the delta function is given by 1 1 00 8(x -~) = -2 + - L(cosl1xcosl1~ +sinl1xsinl1~) rr rr 11=1 By using Euler's identity, show that an alternate representation is given by
~
(31. 194 )
x>~
I
Although (3.193) and (3.194) appear very different, they are different representations of the same Green '5 function. Indeed, comparing (3.193) and (3.194), we must have
00 sin kx sin k~ rr ! ---dk=--
~I
with the operator L and the boundary conditions given above.
2v'A" sin v'A"rr
We identify the term in square brackets as the Green's function
-I
II
Use this spectra,' representation to obtain the solution to the differential equation LAli
where
10o
Chap. 3
1
8(x - ~) = 2rr
L 00
11=-00
Show that this alternate representation leads to the complex Fourier series f(x) =
If 1 If L
all
00
11=-00
e-i..fi~ sin ~x,
x<~
e-i..fix sin ~~,
x>~
(3.195)
.
e11l(X-O
_e illx
2rr
2Jf
all =
o
f(x)
- e -;nx dx
2rr
ii
Chap. 3
Problems
136
_d 2/ d~2 with boundary condition u' (0) = 0 and
3.3. For the operator L limiting condition
begin with the Green's function for L>.. and show that the spectral representation of the delta function is giver by
21
8(x x
f
dx
s(x) =
10
dx
f
=
fo
1
00
~ [~(xdU)] = (OO U(k) {~~ [x_dJO(_kx)]} kdk x
Jv(kx)J,,(k~)kdk
0
S(k) =
3.4. In Example 3.5, it was shown that
~l;
1
00
0
Show that this representation leads to the Fourier-Bessel Transform of order v, given by the pair
with the operator L and the boundary conditions given above.
,....
137
--'- =
cos kx cos k~dk
0
L>..u =
I ,
Note: This result can be obtained directly from Example 3.6 by identifying A above with in(3.131) in Example 3.6. By integrating the Green's function around the closed contour in the complex A-plane, as described in (3.119), show that the spectral representation of the delta function for the operator L and boundary conditions given above is given by
00
Use this spectral representation to obtain the solution to the differential equation
".
Problems
e
lim u(x) = 0 X-+OO
8(x -~) = ,n
Chap. 3
x dx
{OO S(k)Jv(kx)kdk
10
Produce the following Fourier-Bessel transform pair:
dx
oo
[~(x~)] +x { -~ ,x dx dx
[-k 2 U(k)]Jo(kx)kdk
V
and therefore,
~x [~ (x dU)] ~ dx dx
s(x)J,,(kx)xdx
2
2 }
sex)
~k
2
S(k)
3.6. In the development of the Kantorovich-Lebedev transform, carefully complete the steps necessary to produce (3.159) from (3.158).
2
-k U(k)
I
•
I.
This result involves the interchange of differential operator and mtegr;lllO~. Justify this result by following the procedure used in (3.76)-(3.79) and m (3.102)-(3.104). '
3.5. Consider the following Green's function problem associated with Bessel's equation of order v: : 8(x -~) (L ~ A)g = --'--'x
3.7. Justify the result in (3.161) by following the procedure used in (3.76)-(3.79) and (3.102)-(3.104).
3.8. In the development of the Kantorovich-Lebedev transform in Example 3.6, we considered the operator Im(k) < 0 where lim u(x) = 0
where
x-+oo
lim u (x) finite x-+o
In [14J, Stakgold considers the Kantorovich-Lebedev problem for a slightly different operator, viz.
with limiting condition lim g(x,O = 0
x-+oo
where Ais complex. By invoki~g the condition that g must be finite as Xi I show that : x <
~)
x >
~)
-+ 0,
/.1>0
His resulting spectral representation is given by (3.196)
References
138
Chap. 3
where K)..(y) is the modified Bessel function of the second kind, Alh order [16]. By making the appropriate transformation between k and p., show that the spectral representation in (3.154) transforms to (3.196). '
REFERENCES [1] Stakgbld, 1. (1979), Green's Functions and Boundary Value Prohlems. New York: Wiley Interscience, 414-415. [2] Carrier, G.P., M. Krook, and C.E. Pearson (1966), Functimls of a : Complex Variable. New York: McGraw-Hili, 56-59. [3] FrIedman, B. (1956), Principles and Techniques of Applied Mathematics. New York: Wiley, 213-222. ~ [4] Churchill, R.Y. (1960), Complex Variahles and Applications. New York: McGraw-Hili, 159~161. [5] Arfken, G. (1985), Mathematical Methods for Physicists, 3rd Edition. New York: Academic, 529-530. [6] op.cit. Friedman, 234-237. [7] Sneddon, LN. (1972), Th~ Use of Integral Tran~forms. New York: McGraw-Hill,46-49. [8] Abramowitz, M., and LA Stegun (Eds.) (1964), Handhook ofMathematical Functions, National Bureau of Standard, Applied Mathematics Series, 55, Superintendent of Documents, U.S. Government '. Printing Office, Washington, DC 20402, 361, eq. 9.1.35. [9] Ibid., eq. 9.1.39. [10] Ibid., 358, sect. 9.1.1. I [11] Harrington, R.P. (1961, reissued 1987), Time-Harmonic Etect~'Omagnetic Fields. New York McGraw-Hill, 461-462. [12] op.cit. Abramowitz and Stegun, eq. 9.1.6. [13] Ibid., eq. 9.1.4. [14] Stakgold, 1. (1967), Boundary Value Problems of Mathematical Physics. New York: Macmillan, 317. [15] Felseh, L.B., and N. Marcuvitz (1973), Radiation and Scattering of Waves. Englewood-Cliffs, NJ: Prentice-Hall, 323-326. [16] op.cit. Harrington, 363.
i--
-~
'~
til II
4 Electromagnetic Sources,
4.1 INTRODUCTION In this chapter, we introduce electromagnetic source representations. We begin by collecting expressions for the delta function in cylindrical and spherical coordinates. We follow with a discussion of time-harmonic representations of functions and vectors. We next introduce the electromagnetic model in the time domain, and then specialize to the time-harmonic case. We begin our study of electromagnetic sources with a consideration of the sheet current source. We then, in sequence, study the line source, the ring source, and the point source. Throughout, we shall utilize the Green's function method and the spectral representation method, developed in Chapters 2 and 3, respectively, in order to obtain alternative representations of the fields from various sources.
4.2 DELTA FUNCTION TRANSFORMATIONS In what follows, we shall require delta function representations in two and three dimensions. Depending on the particulars of the analysis, it will be convenient to express the results in different coordinate systems. In particular, we consider rectangular, polar, cylindrical, and spherical coordinates. We begin with a point source in two dimensions (Fig. 4-1), located at the point Q(r'. v'). We represent this point source hy 8(x - x')8(y - y/) 110
Electromagnetic Sources
140
Sec. 4.2
Delta Function Transformations
141
Integrating (4.5) over the xy-plane gives
y
LL 2Jr
I =
L
oo
h(p)8(p)pdpd¢
oo
=
(4.6)
[2rrph(p)]8(p)dp
Equation (4.6) is reduced to an identity by the choice Fig. 4-1 x'
".
Point source in two dimensions, located at (x', y').
I
h(p) = -2rrp
(4.7)
o(x)o(y) = 8(p) 2rrp
(4.8)
Substitution into (4.5) gives
and seek a corresponding representation in polar coordinates. Since the polar coordinate point corresponding to (x', y ') is (p', ¢'), we wri te . I
o(x - x')o(y - y') =
i
fl (p, ¢)o(p -
p')o(¢ - ¢')
I
:(4.1)
The function f1 (p, ¢) allows fpr the possibility of an additional factor other than the delta functions that might be introduced by the Jacobian in the coordinate transformatio~. Integrating both sides of (4.1) over ~he xy-plane gives ' 1=
r rIO fJ(p,'¢)O(p Jo Jo .
The proper transformations in three dimensions between rectangular and cylindrical coordinates require no further analysis since the z-cobrdinate remains the same in both systems. We have o(x - x')8(y _ y')o(z _ z') = o(p - p')8(¢ - ¢')8(z - z')
2Jr
p')o(¢ - ¢')pdpd¢
I
:
8(x)8(y)8(z) = 8(p)8(z) 2rrp
(4.2)
(4.10)
In transformations to spherical coordinates, there are three cases of interest. We begin with a point source (Fig. 4-2) at the location Q (x', y', z') and write
Equation (4~2) is reduced to an identity by the choice (4.3)
!rep, ¢) = -
(4.9)
P
p
8(x -x')8(y- y')8(z- z') = her, e, ¢)8(r-r')8(e -e')8(¢-¢') (4.11)
Integrating both sides over all space gives
Substitution into (4.1) gives , ., o(p - p')o(¢ - ¢') o(x - x )o(y - y) = - - - - - -
p
2Jr
, (4.4)
Equation (4.4) gives the polar representation of a point source in a plane as long as the source is at a location other than the origin. For a point source at the origin, we have o(x)o(y). In transforming this representation to polar coordinates, we note that the origin in polar coordinates is given by p =:= 0, independent of ¢. We say that the coordinate ¢ is if.?norahle at the or!gin [I] and write o(x)8(y) = h(p)8(p)
; (4.5)
I=
Jr
r r roo 13 (r, e, ¢)8 (r Jo Jo Jo ' :
r')8 (e - e')8 (¢ - ¢')r 2 sin e4rded¢
(4.12)
Equation (4.12) is reduced to an identity by the choice
13 (r, e, ¢) =
1 r
2 .
SIn
e
(4.13)
Therefore, 8(x - x')8(y - y')8(z - z') =
8(r
r')8 (e e')8 (A. . 'I' 2 r slOe
A.') 'I'
(4.14)
Sec. 4.3
Electromagnetic Sources
142
Time-Harmonic Representations
143
Equation (4.20) is reduced to an identity by the choice
Z
P(x, y, Z) I I I I Q(x',y',Z') I I I I I I
8
XF---.,....----;----+---I
~J
)
".
x
Fig. 4-2
=
1 4-Z nr
~
8(r) 8(x)8(y)8(z) = - 2 . 4nr
Y
Point source in three dimensions, located at (x', y',
8(x)8(y)8(z - z') = /4(r, e)8(r - r')8(e)
z').
(4.15)
In subsequent considerations of the electromagnetic model, we shall be dealing with quantities that vary harmonically with time t. The timeharmonic representation is useful in the determination of the cosinusoidal steady-state behavior of the electromagnetic fields. The representation can also be directly extended to give the response for more general forms of source input [2]. An excellent treatment of time-harmonic representations is given in [3]. We shall include herein only the major results. A timeharmonic function /(t) has the form
Integrating over all space gives
rr
{27r (Xl Jo Jo /4(r, e)8(r o
=
rr roo [2nr
Jo Jo
2
/(t) =
2
- r')8(e)r sinedrded¢
sin e/4Cr, e)]8(r - r')8(e)drd()
.
(4.16)
Focos(wt
+ ¢)
e)
1
= 2 2' e nr sm
Substitution into (4.15) gives
(4.24)
(1. 1.1) '
(4·18)
In the case where the point source is at the origin (0,0,0), the coordihates are both ignorable. We have
e and ¢
I
8(x)o(y)8(z) = /5(r)8(r)
(4.19) I
Integrating over all space gives
1 = {2rr {rr
Jo =
I i
roo /5(r)8(r)r 2 sin ()drd()d¢
Jo Jo .
~oo [4nr 2 /s(r)]8(r)dr
where
F I
.., 8(r - r')8(e) 8(x)8(y)8(z - z) = 2 2 ' () nr sm
(4.23)
where w is radian fr~quency and where Fo and ¢ are real and time-independent. We write this function in terms of the real-part operator as follows:
Equation (4.16) is reduced to an ldentity by the choice /4(r,
(4.22)
4.3 TIME-HARMONIC REPRESENTATIONS
In the case where the point source is located along the z-axis, the coordinate ¢ is ignorable. We have
1= J
(4.21)
Substitution into (4.19) gives
I I I I I I I " I "" I
I I
~
her)
= Foe i
(4.25)
That (4.24) is equivalent to (4.23) can be demonstrated by substituting (4.25) into (4.24) arid performing the real-part operation. The details are left to the reader. We may show the following relation for derivatives:
d/
-
dt
=
.
Re(iwFe"lJt )
(4.26)
The proof is straightforward and is left for the problems. The real-part operator has the following useful properties [4]: For z, Zl, Z2 E C and a, t E R, Re(z\)
+ Re(Z2) =
Re(zi
+ Z2)
(4.27)
(4.20) Re(az) = nRe(z)
(4.2R)
,
144 1
~ [Re(z)]
at
= Re
Electromagnetic Sources
Chap. 4
(az) at
(4.29)
/ Re(z)d/ = Re ( /
Zdt)
Sec. 4.4
The Electromagnetic Model
av
'1x1t=-+:1
(4.37)
'1·V=p
(4.38)
'1· S= Pm
(4.39)
at
(4.30)
'These real-part relationships are e~sily verified. The details are left for the problems. An additional relationship involving the real-part operator is ~he following: If A, BEe and are time-independent, and if Re(Ae iw1 )
= Re(Be iw1 ),
ap '1.:1=-at
(4.40)
'1.M = _ aPm all t
(4.41)
at
(4.31)
where the symbols are!defined as follows:
then
".
A= B
(4.32)
Indeed, let t = O. Then Re(A) = Re(B). Let wt = n/2. Then Re(iA) == Re(i B), which implies that Im(A) = Im(B). Since A and B are tirhe~ independent, the above equality 9f their real parts and their imaginary . . parts must be true for all t, and thus A = B. The complex representation bf time-harmonic functions can be tended to vectors. Indeed, let £(t) be a real, time-varying vector with I , time-harmonic components, viz.
eX"
£(t) = xEx cos(wt +
+ yEy cos(wt +
(4.$3)
x,
where y, i are unit vectors in the three Cartesian coordinate directi~ms I and Ex, Ey, E z,
..,
145
£ electric field intensity (volts/meter) 1t magnetic field intensity (amps/meter) V electric flux density (coulombs/meter2 )
S magnetic flux density (webers/meter2 ) :1 electric current density (amps/mete~)
M magnetic current density (volts/mete~) P electric charge density (coulombs/meter3 ) Pm magnetic charge density (webers/meter3)
In performing dimensional analyses, it is helpful to have the following equalities: . coulomb=amp. second weber=volt . second
,
ohm=volt/amp .
where E = x Exei,px
+ YEyei,py + iEzei,p,
(4.~5)
The equivalence of (4.34) and (4.33) can be verified by substituting (4.35) into (4.34) and performing the real-part operation. The details are left for I the reader. .
All field and source quantities vary with both space and time. Typically, for the electric field, we have £(x, y, z, t), which we write in shorthand notation as fer, t). The magnetic charge PIIl(r, t) and the magnetic current M(r, t) have not been shown to exist in nature, but can be defined as equivalent sources in equivalence theorems involving the electric fields [5]. We shall assume that all quantities vary time-harmonically. Typically,
4.4 THE ELECTROMAGNETIC MODEL
fer, t)
I
The macroscopic model for the behavior of electromagnetic fields is given by the following set of equations: ' i
as
'1x£=,---M
at
(4)6)
= Re [E(r, w)e iW1 ]
(4,42)
:t
(4.43)
Then, in (4.36), we have
V x [Re(Ee
ieJJ1 )]
= -
[Re(Be iW1 )]
-
Re(Me iw1 )
Electromagnetic Sources
146
Chap.
4
X
(E/ wt )] =i= -Re [(iwB + M)e
iUJt
But, by a well-known vector identity, (4.,45) i
J(z) =
Substituting (4.45) into (4.44) and applying (4.31) and (4.32), we obtain
Vx E
~
-iwB - M
". A similar procedure in (4.37)-(4.41) yields
l'
xlso8(z)
1 ::~
"I
x
I
iI
V x E = -iwf,LH - M
(4.55)
V x H = iWEE + J
(4.56)
(4. 46) 1
VxH=iwD+J
(4.47)
V·D=p
(4:48)
V·B = Pm
(4.49)
V· J = -iwp
(4;50)
V· M = -iwPm
(4Js 1)
i
In general, the permittivity is complex and is given by [6] E
=
a
Ed
I
We remark that, in keeping with usual practice, we have used the same symbols for electric charge density in both the time and frequency domains, land similarly for magnetic charge density. Which domain we are consideting will be clear in context. ' In a simple medium, the electric flux density D is simply related to the electric field intensity E by
+-;-
where a is the conductivity in mhos/m and Ed is the permittiyity of a perfect dielectric (a = 0). The complex permittivity accounts for losses in the medium. For the case where the losses are negligible, the complex permittivity reduces to the perfect dielectric permittivity Ed. Since the source, given by (4.54), varies only with z, and since there are no scattering objects present, we conclude that a/ax = a/ay = O. Substituting this result and (4.54) into Maxwell's equations, setting M = 0, and expanding in Cartesian coordinates, we obtain dHy
Similarly~
(4.57)
lW
---;;; = J.r08(Z)
where E is the permittivity of the medium in farads/meter. the magnetic field, '
(4.54)
where is a unit vector in the x-direction and lso is a constant surface current density in amps/m. We write Maxwell's curl equations by substituting (4.52) into (4.47) and (4.53) into (4.46) to give
I ~I
147
Consider a planar electric current sheet in the xy-plane (Fig. 4-3). We assume linear polarization in the x-direction and no variations in either x or y. The mathematical representation of this current sheet involves the electric current density J(z) in amps/m 2 , given by
]
V x (Ee iwt ) = (V x E)e irtJt
The Sheet Current Source
4.5 THE SHEET CURRENT SOURCE
Using the real-part operator relationships in (4.26)-(4.29), we obtain!
Re [V
Sec. 4.5
.
+ lWE Ex
(4.58)
Ey
(4.59)
for
where f,L is the permeability of the tnedium in henrys/meter. The basic units i offarads can be deduced from the units ofD and E in (4.52), and simil,arly • for the units of henrys in (4.53). I' , I Equations (4.46)-(4.53) constitute the electromagnetic model in simi pie media. We shall use the model in what follows to develop represe1ntaI ' i • tions of electromagnetic sources. In Chapter 5, we shall use the model to develop solutions to some example boundary value problems.
dHx --;;; =
. lWE
0= iWEE z
(4.60)
dEy . --;;; = 1W f,L H.t
(4.61)
dE
.
x -= -lWltH\,. dz
(4.62) (4.63)
148
Electromagnetic Sources
Chap. 4
Sec. 4.5
i
z
II
The Sheet Current Source
The problem is therefore completely described by (4.65) and (4.66), together with appropriate conditions as Z -7 ±oo. We take the derivative of (4.65) with respect to Z and substitute (4.66) to obtain the following set: 2
,
149
d Ex dz 2
.
+k
2
Ex
.
= IW/-Llso8(z)
(4.70)
J-~~------_y
1 dE x Hy = - - - - iW/-L dz
(4.71 )
where the wavenumber k is given by ".
(4.72)
x
Fig. 4-3
From (4.60) and (4.63), we find
(4.73) and
th~t
E z ~ Hz = 0 t
with
Electric current sheet 'located in the x y-plane and linearly polarized in the x-direction. Current sheet extends over : entire xy-plane.
(4.74) (4.64) !
The remaining equations decouple ~nto two independent sets. The first se( is given by , , I ' . dE (4.65) - -x == -lw/-LHy I I dz I . dHy (4.q6) ----;jz = l.108(z) + IWEE x 1
The wavenumber kd is the wavenumber that would be present in a perfect dielectric (0' = 0). In the engineering literature, S is called the loss tangent [7]. We note that, if the second-order, linear, ordinary differential equation in (4.70) can be solved for the electric field Ex, the magnetic field Hy can be found by the simple differentiation indicated in (4.71). For limiting conditions, we demand that, for k E C,
I
The second set is given by
,-
(4.75)
To solve the differential equation in (4.70), we let
dE
--y
dz
== iW/-LHx
(4.76)
(4.<57) Substitution into (4.70) yields
dHx . (4.J8) - - = IWEE y dz Note that the first set contains Ex, Hy , and lso, while the second set contains f. y , Hx , and no sources. Since the ~econd set is source-free throughout all space and is not coupled in any manner to the first set, we must conclude that the only solution to (4.67) and,(4.68) is the trivial solution, viz. I
lim Ex = 0
z-->±oo
'
Ey = Ht = 0
(4.69)
d 2g
-- k 2 g = 8(z) dz 2
(4.77)
lim g = 0
(4.78)
with limiting conditions z--> ±oo
We recognize this as a Green's function problem in SLP3. In general, the Green's function is associated with the delta-function source 8 (z -
n.
I , ·;:I
II
;;:t: . .
,
150
Electromagnetic Sources
Chap. 4
This source would produce a Green's function g(z, {). In the case in (4.77), { = 0 and we obtain g(z, 0). The solution to this Green's function problem has been given in Example 2.20, viz.
j"
.'"
g(z, 0) =
e~iklzl 2ik '
Im(k) < 0
Sec. 4.5
The Sheet Current Source
G(fJ, 0) =
G(fJ, 0) =
WJ1,
Ex (z) = . - 2kJof oe-iklzl
1 g(z, 0) = 27f
We may ~ormalize this result by letting (4.81)
ik 7f
WJ1,
.,
J" 't"
ExCz)=--
We then have
ExC~) = e-iklzl
(4.82)
e -ikz , _e ikz ,
, :1~
e z>O
(4.88)
2
roo e ifJz L oo f32 _ k2 df3
(4.89)
100 -00
ifJz
e 2df3 k - f3
(4.90)
2
'k -iklzl _ ~ - 7f
/00 -00
eiflz f32 _ k 2 df3,
Im(k) < 0
(4.91)
r "
(4.83)
z
,t"
"
f3 - k
Since the solution to (4.70) is unique, we may equate the two results in (4.82) and (4.90) to give the following useful relationship:
Substitution of this result into (4.71) yields the accompanying magnetic field, viz.
~. .J"
1 2
Substituting into (4.76) and applying (4.81), we find that
2k Jso = - -
~>'
(4.87)
Taking the inverse Fourier transform yields the alternative solution form !
~: ,.
g(Z, O)e-ifJzdz
Taking the Fourier transform of (4.77) and rearranging gives
(4.79)
Substitution into (4.76) gives
L:
151
We remark that the result in (4.91) can also be obtained by contour integration techniques, as we show in the following example.
where the intrinsic impedance TJ is given by EXAMPLE 4.1 WJ1,
TJ=
T
(4.84)
The solution to (4.77) with limiting conditions in (4.78) has been obtained by the Green's function method. Alternately, the solution can also be obtained by spectral methods. From the result in Example 3.4, the spectral representation of the delta function for the operator -d 2/ d Z2 with limiting conditions given by (4.78) is given by 8(z -
n =_1 /00 eifJ(z-l;)df3 .27f
-00
(4.85)
This spectral representation defines the Fourier transform. Applying this transform to the Green's function, we have g(z, 0) =
~ /00 27f
-00
G(fJ,O)eifJzdfJ
(4.86)
We shall show that eif!z 7f _ _-dR = _e-iklzl . 2 -00 f32 - k fJ ik '
1
00
fm(k) <
0
Consider
i
e;f!z
CdC,
,
-2--2 df3 f3 - k
(4.92)
around a closed contour (Fig. 4-4) along the real axis from - R to R and a semicircle of radius R through the upper half-plane. We constrain R such that R > Ik I. The denominator of (4.92) has simple poles at f3 = ±k. We have Im(k) < O. It can be shown that this selection of the sign of the imaginary part, together with the definition of k in (4.72), implies that Re(k) > O. The details are left for the problems. Therefore, the pole at f3 = -k is enclosed by the contour, and the residue theorem gives
i.
• (.' . d
C,
eif!z
- 2- - 2 dfJ
fJ - k
= 2niRes
[eif!Z - 2--.2;
fJ - k
-k
]
(4,93)
1'"
~52
Electromagnetic Sources
Sec. 4.6
Chap.
The Line Source
153
I
I
4.6 THE LINE SOURCE
Im(.B) .B-plane
Consider a line current source located along the z-axis (Fig. 4-5) and extending from z = -00 to z = 00. We represent the current density associated with this source by J(p) = ZI08(x)8(y)
,---'-----"t---l--
"--~
Re(.B)
"
Fig.4-4
,
.
Contour for the evaluation of the contour integral ,in (4.92).
where 10 is a constant current in amps. We begin our study of the fields produced by this current source by considering the problem in cylindrical coordinates. From (4.8), the cylindrical coordinate representation of the current J(p) is given by J(p) = ZIo 8(p)
Evaluating the residue and splitting th¢ contour integral into two pieces gives'
f
R
-R
eifJz
fJ2 - k 2dfJ
+
1 ,
C
R
27rp
eifJz
Jr, d - _e-,kz fJ2 - k 2 fJ - ik
(4.94)
iWe now show that
'.
lim
R->oo
1
1
CR
eifJz
fJ2
-
I
k,2 dfJ = 0,
(4.95)
2>0
(4.99)
(4.100)
Since the current source is independent of
Ep dH --z dp
Indeed. on C R, let (4.96)
= Hp = 0
(4.101)
=
(4.102)
-iWEE",
~ [~ (PE",)] = -iw~Hz
Then,
(4.103)
z
t ,RJr < --:----:c-:c2
R2
(4.'97)
-lkl
°
In the last inequality, we have used the: fact that zR sin () > 0, < () < Jr to bound the integrand with unity. Therefore, in the limit as R -+ 00, the integral aro~nd CR approaches zero. Taking the limitin (4.94) gives I
-1---------- Y
I
oo
'~
f
eifJz
:
Jr
'
d - _e-,kz _oofJ 2 -k 2 fJ- ik '
2>0
I
(4J98)
Equation (4.98) gives the result for 2 > 0. For 2 < 0, we close the coniour C 1 along a semi-circle through the lower half-plane. The details are left for' the problems.
•
x Fig.4-5
Electric line current located along the z-axis and extending z = -00 to z = 00.
from
154
Electromagnetic Sources
Chap. 4
(4.104) (4.105)
I
where J
== 10 8(p)
z.
(4.106)
2Jrp
We note that (4.102) and (4.103) are source-free and independent of (4.104) and (4.105). Therefore, (4.107) .... Hz = Erp = 0 We conclude that (4.104) and (4.105), together with appropriate boundary and/or limiting conditions, compl~tely characterize the proble.m. We ~ul~ tiply (4.104) by p, take the derivative with respect to p, multIply by lip, and then divide by iWfl to obtain '
; _1 [~(p dE,z)] = ~ [~(PHrp)] iWflP dp dp P dp
[d (p_z dE)] + e Ez = IWflI0-2. 8(p) dp Jrp 1 dE z H
, I
(4.109) I (4.110)
.
'
and obtain
~ [~ (p d p dp
g)'] + e g=
dp.
with the limiting condition given in (4.113) and a finiteness condition at the origin. The solution was given in (2.184) and is repeated here with some trivial changes in notation, viz.
I
2 Hci )(kp') Jo(kp) , Hci 2)(kp) Jo(kp') ,
Jr g(p, p') = ----: ,21
(4.1 P)
lim g =0
Jr (2) g(p,O) = 2i Ho (kp)
E = _ Wfllo H(2)(k ) z 4 0 P
8(p - p') p
(4.114)
(4.117)
Substitution of this result into (4.110) gives
'kl H
(4.118)
The solution to (4.114) with limiting condition (4.113) at infinity and a finiteness condition at the origin has been obtained by the Green's function method. Alternately, the solution can also be obtained by spectral methods. From Example 3.5, the spectral representation of the differential operator in (4.114) with the given limiting and finiteness conditions is given by roo
10
JO(Ap) JOCAp')AdA
(4.119)
As found in Example 3.5, this representation gives the Fourier-Bessel transform pair
f(p) =
To solve (4.112), we first conside~ a result from Example 2.21. In that example, we considered
(4.116)
Substituting (4.116) into (4.111) and solving for E z , we obtain
(4.1\2)
(4.113)
p->oo
(4.115)
p> p'
I
F(A) = _ 8(p) p
p < p'
Taking the limit as p' ~ 0 yields the solution to (4.112), viz.
with limiting condition
~p [~(pdg)] 4- k 2g = dp dp
155
8(p - p') = p
To solve the differential equation in (4.109), we let
2Jr E z g = - iWfllo
The Line Source
(4.108)
Substitution of (4.108) into (4.105) produces the following set:
-1 P dp
Sec. 4.6
i i
oo
f(p)JO(Ap)pdp
. (4.120)
F(A)Jo(Ap)AdA
(4.121)
oo
Taking the Fourier-Bessel transform of both sides of (4.114) gives (-A
2
+ k 2)G(A, p') =
-JO(Ap')
(4.122)
where G(A, p') is the Fourier-Bessel transform of g(p, p'). Solving for G and taking the inverse Fourier-Bessel transform gives
, _ rx:> JO(Ap).lO(Ap')
g(P,P)-10
A2 -k 2
AdA
(4.123)
Electromagnetic Sources
156
Chap',4
To produce the solution to (4.112), we take the limit as pi -+ 0 and obtain g(p, 0) =
!i
Sec. 4.6
The Line Source
157
Set 1: TM z
I OO
o
Substitution into (4.111) yields E = _iwfl!O roo Jo(Ap) AdA z 2n A2 - k 2 0
1
o
P
'JT
10
. aE z ax = lWflHy
(4.135)
aHx -~- - -a- = I 08(x)8(y)
y
uX
i
A2 -k 2
(4.134)
aHy
(4.1 25 )
Comparing (4.125) to (4.117), we obtain the following integral representation of ~e Hankel function [8J:' : H(2)(k ) = 2i roo JO(Ap) AdA
aEz = -iWflH ay x
-
JoU,-p) 2 2 AdA A - k
aHz - - = -iWEE
Taking the Fourier-Bessel transfonn gives the relation
A2 _ k2 =
aE y ax
(00 (2) 2i 10 H o (kp)Jo(Ap)pdp I
We have obtained two representations of the electric field E z produced by a line current located along the i-axis. These representations are given by (4.117) and (4.125). Further representations can be obtained by considering the same problem in Cartesian coordinates. Since a/az = 0, Maxwell's curl equations reduce to I (4.128)
I
aE z
ax: = iWf.lHy
aE y aEx . - - - - ' - = -IWflHz ax ay
aH
_z = iWEE x ay
(4.129)
.
z
(4)31) I
I
(4.139)
Set.1 is .labeled T ~z. since it contains no magnetic field component in the z-dlrectlOn. In a sImilar manner, Set 2 is T E z . We note that the two sets are not coupl~d and that Set 2 is source-free. Therefore, the only solution to Set 2 consIsts of the null fields, viz. Hz = Ex = E y = 0
(4.140)
To solve ~or the T M z fields, we differentiate (4.134) with respect to y, (4.135) wIth respect to x, add the result, and substitute (4.136) to obtain
a2 E z a2 E z
-a 2
x
+ -a 2 +k y
2
.
E z = lWf.l 108(x)8(y)
(4.141)
(4.142)
•
I
H = _1_ aEz y iwf.l ax
(4.143)
aH' __ z = -iWEE ax ' y
(4.132)
-aHy - -aHx = I08(x)8(y) + iWEE z
(4.133)
lim Ez(x, y) = 0 x->±oo
(4.144)
These six equations can be grouped into two independent sets, as follows:
lim Ez(x, y) = 0 y->±oo .
(4.145)
ax
.,
aEx ay
(4.138)
y
(4.130) I
(4.136)
(4.137)
- - - - - = -lwf.lH
n
. aE z '= -lwllH -a' r x y ,
+ iWEE z
aHz - - = iWEE ay x ax
1
i
Set 2: T E z
(4.p6)
ay'
, 'i
We consider (4.141), together with the limiting conditions
.'
~1
Electromagnetic Sources
158
Chap. 4
To reduce' (4.141), we combine the spectral representation and Green's function methods. First, from Example 3.4, the spectral represendtion of a2 /ax 2 with limiting condition in (4.144) is given by , 8(x - x')
/00 eik,,(.~-x )dkx 2rr -00
== - 1
I
I
~4.146)
i Multiplying both sides of (4.146) by Ez(x', y) and integrating over (-00, (0) gives the spatial Fourier transform pair ! Ez(X, y) = - 1 2rr
". Ez(k x , y)
/00 Ez(kx , y)e' x·xdk x -00
i:
J
'k
A
Ez(x, y)e-ikxXdx
~4.147)
(4.148)
We therefore take the Fouriertr*nsform of both sides of(4.141) and produce d 2 EZ d 2 y
where and where Ez(x,
i
A
+ k;E z = k~
iWfJ- 108(y)
= Jk 2 - k;
(4.149)
(4.150)
y) and Ez(kx \ y) are Fourier transform pairs. We let
Sec. 4.6
The Line Source
159
Equation (4.155) gives another form of solution to the line source ~roblem. If we co~pare (4.117) and (4.155), we produce the following
mtegral representatIOn of the Hankel function: , 1 e-i(k 2-k;)J/2Iyl H( ) [k(x 2 + = ikxXdk o rr (k2 _ k;)1/2 e x
2
-1-0000
i)1/2J
(4.156)
where Im(k 2
-
k;) 1/2 < 0
(4.157)
Taking the transform of both sides of (4.156) yields e- i (k 2-k;)J/2Iy/ 1 2 2 ikxx (k2 _ kD I /2 = '2 Hci ) [k(x + i)1/2Jedx
100 -00
(4.158)
We note in (4.141), (4.144), and (4.145) that the x and y differential operators and their manifolds are identical. We therefore could have taken the Fourier transform with respect to y. The result can be immediately obtained by interchanging x with y and k x with kyo We now have three representations of the electric field from the line current source along the z-axis, given by (4.117), (4.125), and (4.155). A fourth representation can be obtained by taking the Fourier transforms of (4.141) with respect to x and y to obtain
. (-kx2- k y2+ k 2) E-z = IWfJ-1o
I
J4.151)
(4.159)
! I
so that (4.152)
where Ez(x, y) and Ez(k x , k y ) are two-dimensional Fourier transform pairs. Solving for Ez and taking the two-dimensional inverse· transform gives
1
Ez(x, y) = iWfJ-1o ( 2rr
)2
00 100 ei(kxx+kyY) 1-00 -00 k 2 - k 2 _ k2 dkxdk x
y
(4.160)
Y
Comparing (4.117) and (4.160), we obtain the following double-integral representation of the Hankel function: Im(k y) < 0
(4.154)
(2) [
Ho
2
2
k(x +y)
1/ 2]
1 =irr 2
/00 /00 ei(kxx+kyY) -00 _ook2_k2_k2dkxdky x y
Substituting into (4.151), solv;ing for transform, we obtain
E z ( x,y) = ,
Ez' and taking the inverse Fourier
/00 -e-ik\"lyl . ,... 0 - - e1kxxdkx 4rr -00 k v
(vII I - - I-
Taking the two-dimensional transform, we obtain
k2 (4.155)
(4.161)
.
_
1
k2x - kY2
= ,
I
~4 /00
/00 H(2) [k(x2 + y2)1/2] e-i(k,x+k,v)dx d y
-00-00
0
.
(4.162)
,
I
Electromagnetic Sources
160
Chap. 4
(4.163) where we have used (4.4) to m~ke the delta function coordinate traJ;1s~or mation. We expand Maxwell's curl equations in cylindrical coordin~tes and obtain
(~.164)
....
Sec. 4.6
The Line Source
where we have used a/az = 0 in making the expansions. We have arranged the six equations above such that the first three are T Ez and the second three are T M z . We note that the T E z and T Mz fields are not coupled, and that the T E z fields are source-free. We therefore conclude that the T E z fields are zero. For the T M z set, we multiply (4.168) by p and differentiate with respect to p to give
a(
ap
(4.171)
a
p a
Using (4.170) and (4.171) in (4.169) to eliminate HI/> and H p yields the following: \72 E + k 2E _. I 8(p - p')8( z z - / WfJ. 0 -----:..--:.....:...
(4.167)
i
(4.170)
la 2E z aHp . - - - = -/wfJ.--
(~.165)
(4.168)
aEz ) a p ap = iwfJ. ap (pHI/»
Next, we differentiate (4.167) with respect to
!
(4.166)
161
p
(4.172)
where, from (4.167) and (4.168), we have HI/> = _1_aEz iwfJ. ap
(4.169)
H
p
(4.173)
= __I_aEz
(4.174)
i WfJ.p a
z
and where
~P [~ (p~)] + p21 a
: \72 I
-
pI/> -
(4 175)
,.
The boundary and limiting conditions associated with E z are as follows: };:,----t-----;-r----i~ Y
. ¢i'A',p .' ,
,,
x·
,
y' !
(4.176)
lim E z = finite
(4.177)
p--+O
Ez
x
Fig. 4·6
lim E z = 0
p--+oo
Electric line current located at (x', y') and extending from z = -00 to Z = 00.
aE z a
I1/>=1/>0 =
I
Ez
II/> =1/>0
aEz 1>=1/>0 = a
+2rr
I 4> =1/>0 +2rr
(4.178) (4.179)
I
Electromagnetic SourCf1s
162
I! Chap. 4
where ¢o is any fixed angle. I.,et
163
Fourier series expansion or the Kantorovich-Lebedev transform. We shall
Ez
i Wf.L 10
Substitution of (4.180) and (4.175) into (4.172) yields the two-dim~nsional Green's function problem
2 g 1 -a + g kg 2 , -1 [ -a I (a p- )] + = - 8(p - p') 8(¢-¢) p ap ap p2 a¢2 p
A..I A..')
~ an ( p,p ,¢ y2;e {1 int/J = n~oo I
lim g = finite
(4.183)
all(p, p', ¢') = (2:rr g(p, ¢, p', ¢I); 1 e-int/Jd¢
p--+o
10
(4.184)
I
ag ag a¢ t/J=t/Jo = a¢ t/J=t/Jo+2:rr
(4.185)
= (g,
2rc
U Il
(4.191)
)
where U ll is the ~ormalized eigenfunction (4.192):
In order to separate the p-operator from the ¢-operator, we multiply both sides of (4.181) by p2 and obtain
a ( p aapg )] p [ ap
+
2 a g a¢2
and where we have defined the complex inner product . 2 g = -p8(p + ~kp)
p ")8(¢ - ¢)
(4.186)
A perhaps more descriptive way of writing (4.186) is as follows:
(L p
+ Lt/J)g =
p8(p - p')8(¢ - ¢')
L = -p
[~ (p~)] . ap ap
a
2
Lt/J = - a¢2
(kp)2
(U, v)
= fa2:rr u(¢)"V(¢)d¢
(4.193)
We symbolize the transform from g to an given in (4.191) by (4.187)
I
,
where p
(4.190)
The coefficient all is given by (4:182)
g I·t/J,=t/Jo - g 1t/J=t/Jo+2:rr
')
'(4.181)
. lim g = 0 . p--+ 00
senes. Using the complex Fourier expansion in Problem 3.2, we expand the Green's function as follows:
g ( P''/-''P ''/-'
1
I
The Line Source
co~sider both choices in tum. We begin by choosing the complex Fourier
g=---
".
Sec. 4.6
'(4.188) ,I I
~4.189)
The operator L p , with boundary and limiting conditions given in :(4.182) and (4.183), is called the Kantorovich-Lebedev operator and leads to the Kantorovich-Lebedev transform considered in Example 3.6. The operator Lt/J with periodic boundary conditions given in (4.184) and (4.185) produces the complex Fourier series considered in Problem 3.2. To solve lfor the Green's function g(p, ¢, p', ¢), we have the choice of applying a cbmplex I I
(4.194) Since the operator Lt/J is self-adjoint, we use the procedure in (3.24)-(3.27) and fi nd that (4.195) Also, 8(¢ _ ¢')
===> ;
1 e-int/J'
2rc
(4.196)
Using these results to transform (4.187), we obtain (4.197)
164
Electromagnetic Sources
Chap. 4
Substituting (4.188) and dividing:both sides by p2 gives
2
_I~,p [~(p~)] + k 2, _ 8p 8p
n p2
j bn =
Sec. 4.6
The Line Source
A coordinate transformation x -+ x - x', y -+ y - y' gives, in cylindrical coordinates, .
8(p - p') P
(4.198) : (4.205)
where
where
,
an
bn = --==---
(4.199)
fle-int/,'
y2ir
For limiting conditions associated with bn , we choose lim bn = 0
(4.200)
=
(4.201)
P-V:JO
lim bn
p--+o'
finite
The reader should verify that this choice of limiting conditions is consi~tent with the limiting conditions associated with g given in (4.182) and (4: li83). We now solve (4.198) by the Green's function method. This particular Green's function problem has been previously considered in Example i 3.6. Using the results therein, we mar: immediately write b = n
~ 2i
I
H;2)(~P')J.(kp),
P < p'
HP)(kp)Jn(kp'),
p > p'
Substituting (4.202) into (4.199), into (4.190) yields I'
,
• ,
_
~
g(p, ¢, p , ¢) - 4i
00
n~oo
(4.102)
Ip - p'l = J(x
:
I
HP)(kp)Jn(kp'),
p> p' (4.207) which is the Addition Theorem for the Hankel function. We next obtain an alternative representation of the solution for the electric field by applying the Kantorovich-Lebedev transform ,to (4.186). If g(p, ¢, p', ¢') and G(f3, ¢, p', ¢') are Kantorovich-Lebedev transform pairs, then the transform applied to (4.186) gives '
-d¢2 + f32 G =
f
4. n=-oo
ein(t/>-t/>')
I
f3
(4.208)
G(f3, ¢, ¢') = G(f3, ¢, p', ¢') HJ2) (kp')
(4.209)
I i
:HP)(kp')In(kp),
p < p'
HP) (kp)Jn(kp'),
p> p'
,:
-H(2)(kp')8(¢ _ ¢')
where we have used the transform in (3.157) and the derivative transform in (3.161). We defin,e
p < p,' p > )
Finally, this result substituted intQ (4.180) gives E = _ wf-L1o
+ p'2 -
p < p'
d 2G
!
H,~2)(kp')Jn(kp),
y')2 = J p2
2pp' cos(¢ _ ¢') . (4.206) Since (4.204) and (4.205) must yield the same result, we may equate them to give
~olving for an, and substituting the r~sult
ein(t/>-t/>')
i- x')2 + (y -
(4.203)
z
165
and obtain 2
I
We recall from (4.117) that the electric field from a line source the origin is given by
locat~d
at
A
d G -d¢2
(4.204)
+ f3
2
A
,
G = -8(¢ - ¢)
(4.210)
with the boundary conditions "
,.....
G(f3, ¢o, ¢ ) = G(f3, ¢o
+ 2n, ¢
I
)
dG(f3, ¢o, ¢') dG(f3, ¢o + 2n, ¢') = d¢ d¢
(4.211) (4.212)
166
Electromagnetic Sources
cha~. 4
Sec. 4.7
The Cylindrical Shell Source
167
From Problem 2.18, the solution to this Green's function problem is given by : ' ~
z
cos[fJ(I¢ - ¢'I - JT)]
G =
(4.213) 2fJ sin JT fJ , We substitute (4.213) into (4.209), solve for G, and use the inj{erse Kantorovich-Lebedev transformgiven in (3.159) to obtain
~
l-
Fig. 4-7 ioo
HJ2)(kp)HJ2\kp') cos[fJ(l¢ - ¢'I -JTil g(p,¢,p,¢)- 8 ' . fJ '. dfJ ioo. Sin JT ~ , . (4.~14) Substituting into (4.180) and solving for E z gives the electric field at (p, ¢) caused by an electric line current source at (p', ¢'), viz. ,
,
I
__
i~Jllo
.
HJ2) (kp) HJ2\kpl) cos[fJ(l¢ - ¢'I - JT)] , . dfJ ioo SInJTfJ ; , (4.215) Equation (4.215) gives an alternative representation to (4.204) for the ~Iec tric field. By comparison with (4.205), we obtain the following integral representation alternative to the Hankel function addition theorem in (4.207): Ez = - 8
[-ioo
(2) . 1 [-ioo HJ2\kp)HJ2) (kp') cos[fJ(l¢ - ¢'I- JT)] Ho (kip - p'!) = -2' . fJ ' dfJ 1 ioo SInJT (4.216) For further discussion of the Kantorovich-Lebedev transform, the reader is referred to [9]. The transfolll1 is particularly useful in the solution to electromagnetic problems involving conducting wedges [10]. I
Electric cylindrical shell current located symmetrically about the z-axis and extending from z = -00 to Z = 00.
~ [_d p dp
(p_dE_z )] dp
1l 2JT
o
0
OO
10 [o(p_pl)] pdpd¢ = 10 2JTp
z.
Therefore, using (4.109) and
+ k2Ez
= iWIl'/O-O-(p p_/) 2JTp HrJ> = _1_dE z iWJl dp
(4.218) (4.219)
Let
2JT E z
g=---
(4.220)
iWJl10
so that
~ [~ (p d g )] + k 2g =
__ o(_p
p dp dp with the limiting condition
p_/) P
(4.221)
lim g(p, p') = 0
p---+oo
Consider a circularly-cylindrical'shell current source (Fig. 4-7), loqated symmetrically about the z-axis and extending over Z E (-00, (0) .. We represent the current by
The factor 2JT has been included so that the total current is 10, viz.
Y
x
This problem is independent of both ¢ and (4.110), we have
4.7 THE CYLINDRICAL SHELL SOURCE
';1 o(p - p') J =Z 0 - - - 2JTp
)---H-------l~
I I
(4}17)
and a finiteness condition at the origin. The solution to this Green's function problem has been given in (4.115), viz. I 'JT g(p, p ) = --:21
I
HJ2) (kp') Jo(kp),
p < p'
HJ2) (kp) JO(kp'),
p> p'
(4.222)
Substituting into (4.220) and solving for E z , we obtain
p < p' (4.223)
p> p'
i
168
Electromagnetic Sources
i
Chap.4
Sec. 4.8
,
,j
l
lim E z = ~ Wf..J., 0 Hci 2) (kp) p'--+o . 4 .
169
z
Note that
;
The Ring Source
(4.224)
.which reproduces the result for the line current at the origin, give~ in (4.117). The reader should compare the development of the cylindrical : shell source in this section to the treatment in [II]. }---------l~
Y
4.8 THE RING SOURCE Consider a magnetic ring source M, located symmetrically about the z-his (Fig. 4-8»We describe the sourc~ by
=
M
Po 8(p - pi) 8(z - z')¢ ,p
a
I
aEz
az
ap
aHp
-- az
. ,= -M", -lwf..J.,H", ~
ap
.
= lWEE",
aE '
a
az
a ] --(pH",) pap
aE z = iWEap
(4.233)
Subtracting (4.232) from (4.233) and substituting (4.228), we obtain (4.234) But,
~ [~~(PH )]
I
ap
(4.227)
,
[I
p ap
'"
_ ~~ (p aH",) _ H", -
p ap
ap
p2
(4.235)
Substituting (4.235) into (4.234), we obtain
(4.228) I '(4.236)
aHz
'" = lWf..J., . Hp -I
a ap
az,
aE p
-- - -
x
(4.225)
where ¢ is a \Init vector in the ¢~direction and Po is a magnetic current moment in volt· meters. Since the problem is symmetric in ¢, we must have a/a¢ = O. With this restriction, :Maxwell's curl equations in cylindrical , coordinates reduce to aH! --'" = lWE . EP --(pH",) = iWEE z pap :
Fig. 4-8 Magnetic dng current source located symmetrically about the z-axis.
(4.229)
where (4.237)
(4.230)
i
and
--(pE",) = -iwf..J.,Hz pap ,
(4.231)
•We have arranged the six equations above so that the first three are T E", and : the second three are T M",. We note that the T E", and T M", fields are not . coupled, and that the T M", fields are source-free. We therefore conclude i that the T M", fields are zero. For the T E", set, we differentiate (4.226) with respect to Z and (4.227) with respect to p to obtain
M", = Po 8(p - pi) 8(z - Z') p
Once we have solved (4.236), the electric field components can be obtained from (4.226) and (4.227), viz. E
I_aH", az
iWE
p -
2
a H",
.
aE p
- - - = lWE-az az 2
(4.232)
(4.238)
I
a
E z = -.--(pH",) lWEp ap
(4.239) (4.240)
Electromagnetic Sources
170
Sec. 4.8
Chap. 4
By suitably normalizing H tP , we ma~ obtain the following Green's function problem: '
(Lp+Lz-k
2) g,=: 8(p P- p') 8(z-z),
The Ring Source
171
. V!e may satisfy the limiting conditions in (4.245) and (4.246) by requlflng
(4.241)
lim H finite
(4.252)
=0
(4.253)
p ...... O
lim H
p ...... oo
where
The solution to (4.251) with conditions in (4.252) and (4.253) has been previously considered in Problem 3.5. We find that
(4.242) l
;
I
'
H =
(4.243):
,, i
and where '-. !
I
H = tP
lim' g finite
(4.245)
lim g = 0
(4.21 6)
p ...... O
p ...... oo
lim g = 0 (4.247) i z...... ±oo i lAs shown in P~oblem 3.5, the operator L p in (4.242) with limiting conditions in (4.245) and (4.246) leads to the Fourier-Bessel transform of order ;one. As shown in Example 3.4, the operator L z in (4.243) with limiting 'conditions in (4.247) leads to the Fourier transform. We therefore have the choice of applying either of these transforms. We shall consider each in iturn, beginning with the Fourier transform. Using the Fourier transform over the z-variable in (4.241), we obtain i
pop
op
;
(e -1) P
p2
G = e- ik ,Z,8(p - p')
,
G
e-ik,z'
_~~ (p OH) pop
op
_ (k 2 ~ ~) H p , p2
I
1
00
dkzeik,(z-z'l
H?l(kpp')J, (kpp),
8(p - p') P
z,
p',
z') =
1
g(p,
z,
p', z')J, (Ap)pdp
(4.256)
z,
p', z') =
1
C(A,
z,
p', z')J 1 (Ap)AdA
(4:257)
with inverse 00
g(p,
From Problem 3.5, w~ have the following Fourier-Bessel transform pair:
(p~) +~] g [ -~~ fJ dp dp p2
{::=}
(4.2~I)
P < 'p'
H?l(kpp)JI (kpp'),
-00
A2 C
(4:258)
Applying (4.256) and 1(4.258) in (4.241), we obtain "
2
~
~
d H , - dz 2 -
fJ
2 ~
H = 8(z - z')
(4.259)
where ~ G H=--Jl (Ap')
=
(4.254)
00
C(A,
and obtain I
P > P'
P > P' (4.255) We may obtain an alternative representation by taking the FourierBessel tran~form of order one in (4.241). From the results in Problem 3.5, we may wnte the following Fourier-Bessel transform pair: ' •
(4.248)
(4.250)
H--i
H?l(kpp)JI(kpp'),
WE Po
4
P
where g(p, z, p', z') and G(p, k z , p', z') are Fourier transform pairs and' k2 d k2 - k2 (4.249) p' z We let
2i
Substituting (4.254) into (4.250) and the result into (4.244), we obtain
The associated limiting conditions .are
_
p < p'
I
iWEPO
_~~ (p oG)
l H,(2 (k p p')J, (kpp),
I
(4.244)
g='--.!!.L
'!!-l
(4.260)
and (4.261)
172
Electromagnetic Sources
Chap, 4 I
Sec. 4,9
The Point Source
173
To satisfy the limiting condition in (4.247), we require
.,I
lim
z-",±oo
dt! "
.
I
I
Ii
z
= 0
We have previously considered this Green's function problem in Example 2.20. The result applied to (4.25?) is as follows: e-iPlz~z'l
H
'I
H =
!'
,
2if3
Im(f3) < 0
II
(4 j 263) Fig. 4-9
,I
Substituting (4.263) into (4.260) and the result into (4.257), we obtain
i
"'. g=
~oo
-iPlz-z'l , e .
o
JI(Ap)JI(Ap)Ad'A
2zf3
,
H¢ = -iWE Po
~o
-iPlz-z'l
e. .
JI ('AP')JI (Ap )'Ad'A
2zf3
J = iIoe8(r)
A=
iA z
' (4.271)
(41266)
(4.272) We let (4.273) and obtain
- (V
of Joe are amp' meters and where
8(r)
I~,."'
Sin~e the cu~rent is in the z-direction, the magnetic vector potential is also z-dIrected, VIZ. ' The z-component of (4.268) yields a differential equation for determining the vector potential A z , viz.
Consider an electric point source located at the origin (Fig. 4-9) and polarized in the z-direction. We ~ssume that the source radiates in empty space and that there is a small amount of loss present so that we may apply limiting conditions on the fields :as we approach ±oo in x, y, and z. We may describe the source in terms of its current moment Joe as follows: :
UI~its
E = -iw [A + kl2 V(V. A ) ] : (4.270)
(4.265)
4.9 THE POINT SOURCE
where the
x
(4.264)
Finally, the magnetic field can be obtained by substituting (4.264) into (4.244), viz. oo
Electric current point source located at the origin.
2
+ e) g =
8(r)
(4.274)
yve
== 8(x)8(y)8(z)
first consider the solution to the point source problem in Cartesian coordmates. Expanding (4.274), we have '
The fields from such a source can be described conveniently by use Of the magnetic vector potential A, as follows: ;
,. '.' '.
(4.275)
)
,
,
..
'
The magnetic and electric fields are obtained from the vector potential by the following two relationships:
H=
I
-v x A J1-
(4~269)
From t~e resuI~s in Exampe 3.4, the spectral representation of each of the t~ree dlfferentl~1 operators in (4.275), with limiting conditions at ±oo, ~1e1.d.s the FO~~ler transform. Typically, for the operator -a 2 /ax 2 with IImltmg condItIons lim g =0
x-..±oo
i
I, I
I
Electromagnetic Sources
174 we have
8(x - X')
= ,_1_ 2rr
1
Chap. 4
Sec. 4.9
175
Using (4.273), we obtain the magnetic vector potential A z as folIo ws..
00
eikx(x-x')dkx
(4.276)
Illoe (2rr)2
Az = --
-00
Taking the Fourier transform in (4.275) over x, y, _ (_k x2 _ k,2y _ k z2
The Point Source
z, we obtain
+ k 2) G =
I
where g {=:::::> G is a triple Fourier transform pair over (x, y, z), (k x , k y , k z )· Solving for G, we obtain
11 00
00
-00
-00
_
1
g _ - - -3 (2rr )
100 100 100 , _
ei(kxx+kyy+kzz) . , dk,dk\.dk z _ k 2 - k x2 - k y2 - k Z2 .• .
00 00 00, _
(4.277)
_~~ (a g )
\
I
)
/00 /00 100, ei(kxx+kyy+kzz) -00 -00 -00 k 2 _ k'} _ k; _ k;dkxdkydk z
ag -az 2
-
(~ +e dz 2
k 2 ) {; Y
= 8 (z)
(4?78)
G =
(4.279)
where
,B =
J
k2 -
I
--2
(2rr)
100 /00 ~-ifJlzl -00'
21,B
2rrp
(4.284)
(4.285)
CU.. , z)
1
=
g(p, z)Jo()..p)pdp
(4.286)
and
f=Jk 2
-)..2
(4.287)
The solution to (4.285) with limiting conditions z~±oo
C=
0
is the same as (4.263), viz. _
e-iflzl
Im(f) < 0
(4.288)
Taking the inverse Fourier-Bessel transform yields
1
k'} - k;
_._ei(k,x+k"Y)dkxdk y
-00
8(p)8(z)
(4.280)
(4.281)
Taking the inverse Fourier transform over k x , k y yields
g=
kg =
where
2rrG = 2j'f'
e-itllzl
2i{3'
(4.283)
00
where g {=:::::> {; is a double Fourier transform pair over (x, y), (k), k y). The one-dimensional Green's function problem in (4.279) is identital to that posed in (4.152) and (4.153). The solution is given by :
Im(,B) < 0
y
2 -
lim
A
2
2 -
d G 2 8(z) - -2- f G = dz 2rr
An alternative form in Cartesian coordinates can be obtained by t~klng the Fourier transform over any two Cartesian variables. In (4.275), if we take the Fourier transform over x and y, we obtain
k2 x
x
The spectral representation of the p-operator produces the Fourier-Bessel transform of order .zero, while the spectral representation of the z-operator ~rodu.ces .the Founer transform. We shall derive three forms of solution 111 cyllndncal coordinates. For the first form, we apply the Fourier-Bessel transform to (4.284) and obtain
Using (4.273), we obtain, for the magnetic vector potential,
__ !LlOe Az (2rr)3
2i,B
We next ~onsi~er t~e point source problem in cylindrical co~rdinates. We not~ that, 111 ~}'Imdncal coordinates, the Green's function problem in (4.274) ~s a functl~n of p and z and independent of
pap Pap
Taking the inverse transform over k x , k y , k z yields
e-ifJlzl . --e'(k,x+k"y)dk dk
1
00
g(p, z) = -2 rr 0
e-iflzl
- .-Jo()..p»)"d)" 21 f
(4.289)
Using (4.273), we produce the magnetic vector potential (4.282)
Illoe
1
2rr
()
Az = -
00
e-if!zl
- .-Jo()..p»)"d)" 21 f
(4.290)
,
Electromagnetic Sources
176
Chap. 4
The second form of solution is obtained by taking the Fourier transform of (4.284) with respect to Z1 with the result ,
Sec. 4.9
The Point Source
177
where we have used (4.22) for the spherical coordinate representation of the delta function at the origin. We have obtained the solution to this Green's function problem previously in Example 2.22. Using (2.201), we obtain
(4.291)
e- ikr
g = 4rrr where i
Again using (4.273), we obtain the magnetic vector potential, as follows:
(4.292)
ikr ) A z -_ -/LIoe (e-r 4rr
and r = Jk 2 -
k~
(4.293)
We now use (2.185) and write the solution to (4.291) as follows: , rr (2 2rrG = 2i HO )(rp) (4.294)
1 g = - . 8rrl
(4.300)
~e may exhibit five identities from the alternative representations of the pomt source. Comparing (4.299) and (4.277), we obtain e- ikr 1_ foo foo foo ei(k,x+k"y+k,z) r 2rr 2 -00 -00 -00 k2 _ k2 _ k2 _ k2dkxdkydkz (4.301) ,
, Taking the inverse Fourier transform, we obtain
(4.299)
x
y
z
From (4.299) and (4.282), we have
:, 0 H( 2) (rp)eik,Zdk z ,
e- ikr
(4.195)
/00
1 foo foo e-ifllzl
-- = -
i
-00
r
Using (4.273), we have, for the rrtagnetic vector potential,
rr
-00 -00
_ _ ei(k,x+kyy)dk dk
2if3
x
y
(4.302)
where
A
z
=
/LIoe foo H(2\rp)e ik ,Zdk 8rr i
-00
0
z
(4.296)
The third form of solution is obtained by taking the Fourier-Bessel transform with respect to p and the Fourier transform with respect to z, with the result
:
+ k:< -
2
(4.303)
From (4.299) and (4.289), we have
e- ikr -- = 2 r
1
k2 )g = 2rr Solving for 9 and taking the iqverse Fourier-Bessel transform and the inverse Fourier transform gives ().
f3 = Jk 2 - k x2 - k y2
100 e- if1z / .-Jo().p))'d)' 21f
0
(4.304)
where f =
Jk 2 _).2
(4.305)
From (4.299) and (4.295), we have
ikr e1 foo - - = -: H(2\rp)e ik ,Zdk r 21 -00 0 z
(4.'29,7)
(4.306)
where r = Jk 2 -
k~
(4.307)
From (4.299) and (4.297), we have
ikr
(4.298)
e- = -J foo r
rr
--'XJ
100 0
eik,z Jo().p) ).2+k~-k2
).d)'dk
(4.308)
I
i
I
Problems
178
dhap.4
PROBLEMS
Chap. 4
179
References
Show that the magnetic field radiated by this source is given by
4.1. Given th~ time-hannonic representation of f(t) in (4.23), show that
,
'
df'
----:. = Re(iwFe
p
,
,wt
dt
<:
p'
p> p'
)
4.2. Verify the four relations for the real-part operator, given in (4.27)-(14.30). Hint: To prove the two relations for derivatives and integrals, begin wi1th the ' basic definition of a derivative and a Riemann integral.
Show that, despite the presence in the sum of the multiplicative factor 11, the series converges as 11 ---+ ±oo.
z
4.3. One of the important theorems of electromagnetic theory is the principle of duality [12],[13]. Using duality, make the necessary changes in (4.80)-~4.83) to obtain the fields produced by the magnetic sheet source M(z) = xM,,08(z)
where M"o is a constant magnetic surface current density in volts/m. 4.4. From (4.72), the wavenumber with loss is given by
i
,
Show that the requirement Im(k) < 0 implies that Re(k) > O. Hint: Write ik in tenns of its real and imaginary parts, viz.
ik = a
+ if3 =
Fig.4-10
"
I
4.5. In (4.98), we obtained
J
-00
eiflz ,
7f,
---df3 - _e- 1kz f32 - k~ - ik '
z> 0
By contour integration and the calculus of residues, obtain the res~lt for z < O. Hint: In Example 4.1, we closed the contour on a semi-circle through the uppe~ half of the f3-plane. For z < 0, close the contour through the lower half of the f3-plane. ' 4.6. An interesting variation [14] on the line source problem examined in Section 4.3 is the line source located a~ (x', y') parallel to the z-axis and polarized in the p-direction (Fig. 4-10). Such a source can be represented by ,
J=
PIo 8(p p
Line source parallel to z-axis and p-polarized.
ikdv'l=TS
Solve for a and f3 by squaring both sides and discarding the extraneous root. Note that Im(k) < 0 implies Re(a) > O. From the sign of a, it is then possible to infer th.e sign of f3.
OO
x
p') 8(
REFERENCES [1] Friedman, B. (1956), Principles and Techniques of Applied Mathematics. New York: Wiley, 290-293. [2] Collin, R.E. (1991), Field Theory of Guided Waves, 2nd edition. New York: IEEE Press, 2-3. [3] Shen, L.C., and J.A. Kong (1987), Applied Electromagnetism, 2nd edition. Boston: PWS Engineering, chapter 1. [4] Harrington, R.E (1961, reissued 1987), Time-Harmonic Electromagnetic Fields. New York: McGraw-Hili, 16-18. [5] Ibid., 106-116. [6] Paul, c.R., and S.A. Nasar (1987), Introduction to Electromagnetic Fields. New York: McGraw-Hili, 298-305. [7] Ibid. [8] Gradshteyn, I.S., and I.M. Ryzhik (1980), Table of Integrals, Series, and Products!. San Diego: Academic, 678, #6.532(4). i
I
'I
180
References
C;hap.4
[9] Felsen, L.B., and N. Marcuvitz (1973), Radiation and Scattering of Waves. Englewood Cliffs, NJ: Prentice-Hall, 323-328. [10] [11] [12] [13]
Ibid., 639-645. op.cit. Harrington, 227-228. op.cit. Harrington, 98-100, Balanis, c.A. (1989), Advanced Engineering Electromagnetics. New York: Wiley, 310-312. [14] Butler, C.M. (1991), Vector potential Green's function for a radiallydirected line dipole in open space, Microwave and Optical Technrlogy I Let(~rs 4(11): 430-433. :
5 Electromagnetic Boundary Value Problems
5.1 INTRODUCTION In this chapter, we consider electromagnetic boundary value problems. We apply the concepts developed in the first four chapters, concentrating on the application of the mathematical ideas to a representative set of electromagnetic examples. Our principal objective is structure. Once the reader understands how the concepts in linear spaces, coupled with the theories of Green's functions and spectral expansions, can be applied to examples, it should become <).pparent how the methods are used to approach 'the study of electromagnetic propagation, scattering, and diffraction in an organized, logical manner. : ' We begin by: extending the Green's function method to three dimensions. We next consider the case where the three-dimensional geometry is independent of one spatial coordinate so that the problem reduces to two dimensions. We then present a series of examples. We shall find that we may constmct solutions in two and three dimensions by using combinations of the one-dimensional Green's functions and one-dimensional spectral representations discussed in Chapters 2 and 3, respectively. One of the important solution characteristics that emerges in the examples is the fact that there exist alternative representations for the solutions. In particular, we exhibit alternative representations for the fields in a parallel plate waveguide and the fields scattered by a perfectly conducting 1!l1
I I
Electromagnetic Boundary Value Problems
182
I
Chap. 5
cylinder. These solutions not only have important physical interpretations, but also are useful in different portions of the frequency spectrum. \
5.2 SLP1 EXTENSION TO THREE DIMENSIONS
Sec. 5.2
SLPI Extension to Three Dimensions
n
at points on S. In addition, Vu I.'I . indicates the normal derivative of u (r) evaluated at points on S. The condition in (5.4) has two important special cases. If (X2 = 0 and (X I = 1, we have B(u)
We begin by considering the ne~ative Laplacian operator L = three-dimensional closed and bounded region V. This region is surrounded by a surface S whose parts mayor may not be contiguous. For ex~mple (Fig. 5-1), the surface S might ~onsist of an external surface Se and two internal surfaces SI and S2 with
2 V on a
183
= uls =
(5.5)
(X
Equation (5.1), coupled with boundary condition (5.5), is called the Dirichlet problem, and (5.5) is referred to as the inhomogeneous Dirichlet boundary condition [11. (If (X = 0, the boundary condition is homogeneous.) If (XI = 0 and (X2 = 1, we have
I
....
B(u) = Vuls'
In this case, the region V consists of the volume internal to Se but external to SI and S2. By convention, the unit normal vector points outward from V. Our interest is in the three-dimensional partial differential equation
n
(5.1) ~
where
I
l
f
is a real function and where L>..
= L - A = - V2 -
The functional dependence of u and
f
A,
n=
(5.6)
(X
Equation (5.1), coupled with boundary condition (5.6), is called the Neumann problem, and (5.6) is referred to as the inhomogeneous Neumann boundary condition. The general case in (5.4) is called the mixed problem. We point out that ,it is perfectly reasonable to have one type of boundary condition (Dirichlet, Neumann, or mixed) on a portion of the surface Sand a different type ofiboundary condition (Dirichlet, Neumann, or mixed) on the remainder. ;
,(5.2)
A
n
is
u = u(r)
1=
fer)
where rEV. In Cartesian coordinates, for example, u (r) stands for u(x, y, z). Let u(r) and vCr) be members of a Hilbert space 7i with inner product (u, v)
,
.
==
r u(r)v(r)dV
lv
i
'
v
(5.3)
for all u, v E 7i. The three-dimensional problem involving (5.1) dan be stated as follows: Given the partial differential equation in (5.1) and a suitable boundary condition involving u (r) and its normal derivative on the surface,S, determine u(r) throughout V. We shall require that u(r) have the following specification on S:
(Lu, v) = [(-V 2 U)VdV
(5.7)
I
: (5.4) ! i
where the coefficients (X, (XI. (X2 are real and where'-; is the outgoing n'ormal from the surface S. Our notation II Is indicates the [unction II (r) evaluated
In the one-dimensional case, the adjoint was found by integrating by parts twice. The extension to three dimensions can be obtained by integrating by parts over all three coordinates comprising V. A more direct and com-
184 i
Electromagnetic Boundary Value Problems
Chap. 5
pact method, however, employ~ Green's theorem [2]. In the case of the Laplacian operator, Green's theqrem is given by 2
2
[(-V U)VdV = [U(-V V)dV
+ ~ (-vVu + uVv)' iidS
, where
th~ conjunct l(u, v) l(u, v)
+ l(u, v) I
185
EXAMPLE 5.2 Consider the homogeneous Neumann problem. Then, B(u) = Vuls '11 = 0, and (5.12) gives :
0= iUVV"ldS
(5.9)
To satisfy this relationship, we choose B*(v) = Vvl . n = O. Since the boundary condition on v is identical to the boundary conditio~ on u, the operator L for the Neumann problem is self-adjoint.
•
s
Is is:given by
Is=
SLPI Extension to Three Dimensions
(5.8)
We write this result in inner pr09uct notation as (Lu, v) = ('u, L*v)
Sec. 5.2
L
(-vVu
To produce the solution to (5.l), we define two auxiliary problems: the Green's function problem and the adjoint Green's function problem. The Green's function problem is defined as follows:
+ uVv)· lidS
(5.10)
The operator L * produced by Green's theorem in (5.9) is the formal adjoint to L. We observe in (5.8) that
LAg(r, r') = o(r - r')
(5.13)
B(g) = 0
(5.14)
I
i
L* = L
(5.11 ) :
V2
,
.
and therefore, the operator L = is formally self-adjoint. As was the case in Chapter 2, we shall assume initially that the b<;mndary condition on u is homogene~us, B(u) = O. We now choose the boundary condition on v to be that condition B*(v) = 0 which, when coupled with the boundary condition on u, results in the vanishing of the conjunct, viz. i l(u, v)
:
I= s
0
(15.12) ~
where LA is defined in (5.2). We note that, by definition, the boundary condition on g is identical to the homogeneous boundary condition on u. The adjoint Green's function problem is defined as follows: LAh(r, r') = o(r - r')
(5.15)
B*(h)=O
(5.16)
We note that, by definition, the boundary condition on h is identical to the boundary condition on v. In the same manner as in Section 2.4, the solution to (5.1) is obtained by taking the inner product of LAu with h, viz.
In general, the boundary conditions associated with v are different: from those associated with u. When they are the same, however, the operator L is self-adjoint.
(LAu, h) = (u, LAh)
+ l(u, h)
Is
(5.17)
where the integrations are with respect to the unprimed coordinates. Substitution of (5.1) and (5.15) into (5.17) gives u(r') = (t, h) - l(u, h)
Is
(5.18)
or, explicitly, u(r')
= [f(r)h(r, r')dV + ~ [her, r')Vu(r) -
u(r)Vh(r, r')]. iidS (5.19)
I
186
'
Chap. 5
Electromagnetic Boundary Value Problems
We note that (~.19) is the solution to (5.1), provided that we can determin~ the adjoint Gr~en's function h(r, r'). In a manner similar to the Green's function method developed in Section 2.4, we can show that it is never necessary to find the adjoint Green's function directly. Indeed, We form I (L).g(r, r'), h(r, r"») = (g(r,
~'), L).h(r, r"») + J(g, h)
i
Is
,
(5·fO) I
Sec. 5.2
of the terms in the conjunct will survive. The final step in the solution involves the interchange of the primed and unprimed coordinates, in the same manner as in Section 2.4. We demonstrate these concepts in the following example. EXAMPLE 5.3 Consider a rectangular box (Fig. 5-2) with dimensions a, b, c. It is required to find the solution to - v 2 u = j in the region V inside the box, where it is given that u satisfies the Dirichlet condition B(u) = u == 0 on the boundary. The formulation of the problem is as follows:
Is
We are given the boundary conditions on g. We choose the boundary conditions on h so that i I
,
-v2u
Then, substitution of (5.13), (5.15), and (5.21) into (5.20) gives
(5.27)
u(x, 0, z)
= u(x, b, z) = 0
(5.28)
/I(x, y, 0)
= u(x, y, c) = 0
,(5.29)
We know that the operator L = with Dirichlet boundary conditions is selfadjoint. We may therefore use (5.25) rather than (5.19), viz. u(r') =
(5.22)
h(r, r') = g(r', r)
Therefore, the adjoint Green's function is given simply by interchanging r and r' in the expression for the Green's function g(r, r'). In cases where L is self-adjoint, the boundary conoitions on h are the same as those on g, and we must have
= g(r, r') = g(r', r)
(5.23)
I
+
1
[g(r, r')Vu(r) - u(r)Vg(r,
S'
r')] . tld)S I
I
I'
Iv j(r)g(r, r')dV
j(r)g(r, r')dV
(5.30)
where we require the solution to a2 ( - ax 2 -
2 a a 2 -
y
2 a ) , , , , az 2 g(x, y, z, x, y ,z) = 8(x - x')8(y - y )8(z - z')
(5.31)
gl =gl =0
(5.32)
gl =gl =0
(5.33)
=gl z=c =0
(5.34)
x=O
x=a
y=O
, (5.24) For the case presently under consideration, where the boundary conditi~ns 'on u are homogeneous, the term involving the conjunct in (5.24) vanishes, ,and we are left with u(r') =
Iv
with
;Therefore, the Green's function is symmetric. For the self-adjoint case,:We may substitute (5.23) into (5.19) to obtain
lv
(5.26)
0
or, with a change in variables,
u(r') = ( j(r)g(r, r')dV
in V
V2
h(r', r"~ = g(r", r')
h(r, r')
= j,
u(O, y, z) = u(a, y, z) =
(5.~1)
J(g,h) Is = 0
".
'
187
SLPI Extension to Three Dimensions
gl
lz=o
where we have chosen the boundary conditions on g to be identical to the boundary conditions on /I. We begin the solution to (5.31) by invoking the spectral 'representation of 8(x - x'). As we found in Problem 3.1, this representation produces the orthonormal eigenfunctions
(5.25) u",(x) =
iTo extend the , results to the inhomogeneous case, we simply apply the I :inhomogeneotls boundary conditions to (5.24), with the result that some
y=b
. ff
and leads to the Fourier sine series, viz.
mIT x
- sm-a a
(5.35)
Electromagnetic Boundary Value Problems
188
dap.5
Sec. 5.2
SLPI Extension to Three Dimensions
189
Equation (5.42) is a partial differential equation whose solution yields am, To solve (5.42), we invoke the spectral representation of 8(y - y"). The operator (_a 2 /8 y 2) with boundary conditions am (0, Z, x', y', z') = a"Jb, z, x', y', z') = 0
c a
results in orthonormal eigenfunctions that we may again obtain from the results in Problem 3.1, viz.
x Fig.5-2
z
Rectangular box problem.
00
".
"""'
, y , ,Z,) = L...., am ( y, z, x,"y",Z) II", () g(x, y, Z, x, X
(5.36)
v,,(y)
Ii . /lITY
= -
sm-(5.43) b b These eigenfunctions lead to a Fourier sine series representation of am, viz.
: m=l 00
where am(y,z,x',y',Z') =
~a g(x,y,Z,x',y',z')II",(x)dx
a",(y, Z, x', y', z')
= L.Bm,Jz, x', y', Z')V,,(y)
(5.44)
,,=1
(5.37) where
In a manner similar to (3.24) and (3.2~), we note that (5.37) transforms the Gr~e~ 's function g into the coefficient am, viz. I (5.38) Also, (5.36) provides the inverse tra~sformation of am into g, viz.
8 {::= am
II
,
'
,
.Bm,,(Z,x,y,z)=
l
0
b I
'
I
am(y,z,x,y,z)v,,(y)dy
(5.45)
Since the operator
,
82 Ly 8y2
(5.39)
is self-adjoint, we proceed in the same manner as in the above treatment of Lx. With respect to the transformation given in (5.45), we have
Since the operator
82 Lx = - 8x 2 is self-adjoint, (3.27) and (3.28) give
Lxg
==>
mIT )2 ( ~ a",
(5.40)
We also easily establish that
l
We use these relations to transform (5.42), with the result
a
8(x - Xi)lIm(x)dx = IIm(X')
d2
2 )
- ( dz 2 - Ym"
i so that
8(x - x')
==>
IIm(X')
(5.41)
.
(5.42)
,
"
,
(5.46)
where
2 = (mIT)2 (/lIT)2 a + b
Using (5.38), (5.40), and (5.41) to transform (5.31), we obtain
82+ 82- (mIT " , , , ') - [- )2] a m(y,z,x,y,z)=I1",(x)8(y-y)8(z-z· 8y2 az 2 a
"
.Bm,,(Z, x, y ,z) = " m(x )V,,(y )8(z - Z)
Ymn
, (5.47)
We let fom,,(Z. x', y', Z')
= fi",,,(z.
Z')II",(X')V,,(v')
(5.48)
rl
'I!:j '~
Electro~agnetic Boundary
190
Value Problems
Chap. 5
191
SLPI in Two Dimensions
Sec. 5.3
I I
: !
5.3 SLP1 IN TWO DIMENSIONS
and obtain the ordinary differential equation
.,
d2 ( - dz 2
I
2:) f3mn ,
+ Ymit
,
(5.49)
= 8(z - z)
:We associate the following boundary ~onditions with (5.49): Pmn Iz=o
I
•
I
~ Pmn Iz=c = 0
(5.50)
I
I
.It is easy to show that satisfaction of (5.50) satisfies (5.34). The Green 's func~ion problem posed by (5.49) with boundary conditions in (5.50) can be solved by the standard Green's function methods developed in Chapter 2. The details are,left I for the problem~. The results are as follows: •.
.
:, ,. I . f3mn(Z,Z) = ,Ym" SInh Ym"c
I
I
sinhYmn(c-z')sinhYmn z ,
z < z'
sinh 'Ymn (c - z) sinh ymnz'
z > Z'
(5,51)
Substitution of (5.51) into (5.48), (5.48) into (5.44), and (5.44) into (5.36) yields the required Green's function, viz. '
,g = :
ff
um(x)vn(y~um(x')vn(y')'l
m=ln=1
YmnsInhYmn c
sinh Ymn(c - z') sinh ymnZ,
Z < z',
sinhYmn(c-z)sinhYmnZ',
z>z' (5.52) iWe note in (5.52) that the Green's function is symmetric, g(r, r') = g(r', r), as predicted by the self-adjoint property of the negative Laplacian operator. Substitution of (5.52) into (5.30), followed by an interchange of the primed and unprimed coordinates, yields the final solution, viz.
ff Um(~)Vn(Y) /0r
vn(y') rUm(X')[sinhYmll(C-Z)
·f(x', /, z') sinh ymnz'dz'
+ sinh YmllZ
b
u(x,y,z)=
m=1 n=1 Ymn SInh Ymll C
10
t
In many of the interesting problems in e1ectromagnetics, the assumption is made that the fields are independent of one of the three spatial coordinates, with the result that the problem to be solved is two-dimensional. To solve two-dimensional problems, we modify the Green's function method developed for three dimensions in the previous section. The starting point is again the application of Green's theorem to the negative Laplacian operator, as in (5.8), viz.
i (-
,Pu)vdV =
i
u( - V 2 v)dV
+
Is (-vVu + uVv) . izd S. (5.54)
In the two-dimensional case, the Laplacian is two-dimensional. For example, if the problem is independent of z, we would write the negative Laplacian as - V;,.. In addition, the volume V and the surface S become degenerate, in the sense that the integration over one of the coordinates involved in both the volume and the surface integrals is trivial. For example, if the problem is independent of z, the integrations over z in all three integrals in (5.54) will cancel. We illustrate these ideas in the following example. EXAMPLE 5.4 Consider a rectangular cylinder (Fig. 5-3) with cross-sectional dimensions a and b. It is required to find the solution to - 'i,7;yU = f in the region V interior to the cylinder, where it is assumed that f is independent of z. Since the geometry is also independent of z, the solution u will be z-independent. Suppose it is given that homogeneous Dirichlet boundary conditions apply on the surfaces bounding the cylinder, except for the surface at y = b, where it is given, that the inhomogeneous bourl,dary condition Uly=h = a
10
applies, where a is a real constant. We state the problem as follows: -\1;)'u = 11(0, y)
f
(5.55)
= u(a, y) = u(x, 0) = 0
(5.56) (5.57)
u(x, b) = a (5.53) In the production of the Green's function, we chose to begin with a spectral expan'c sion over the x-coordinate, followed by a spectral expansion over the y-coordinate. In a manner similar to many of the multiple-dimension cases considered in Chapc ter 4, alternate representations are possible. Other forms could be obtained by expanding spectrally over y and z or over x and z. Another possibility is to exp:and spectrally over all three coordinates.
•
a
v y
Fig. 5-3
Rectangular cylinder problem.
x
b
r,
~' [
~
r
Ii, i
ElectrQmagnetic Boundary Value Problems
192
Chap. 5
The associated Green's function problem is as follows:
-V;yg = S(x - x ' )8(y glx=o I
I
Sec. 5.3
y')
(5.58)
= g!x=a'= gly=o = gly=b = 0
where, consistent with the discussion associated with (5.13) and (5.14), we have chosen the boundary conditions associated with g to be the same as the homogeneous form of the boundary conditions associated with II. The homogen'eous Dirichlet boundary condition case as~ociated with (5.55) is a self-adjoint problem. We therefore formulate in terms of the Green's function g (x, y, x', y'). In this case, (5.54) becomes
00
""
g(x,y,x ,y) = Lam(y,x ,y )um(x) " m=1 ,
I
am(y, x, y ) =
2u)gdxdydZ ~
-00
,
,
u(x,y)=
b loa loo
0
i
(5.66)
mJrx
- sin--
a
Using the same progedure as in (5.37)-(5.41), we obtain
2
8 (x - x') =?
i: i 0
U m (x')
Using these relations to transform (5.58), we obtain
[-:;2 + f] ('naJr
I
0
a (Vg· y)ly=/XdZ
(5.61) I !
y')
= am(h, x',
y')
= 0
(5.68)
Application of these Dirichlet boundary conditions satisfies the boundary condition requirements in (5.59) at y = 0 and y = b. The solution to (5.67) is available immediately from the result previously obtained in (5.51), viz.
I
_
am -
-I
'• I ' loa ag (x, l7, X,} I , ' ) IiJ !(x,y)g(x,y,x,y)dxdy-a a ax 0 Y : (5.63)
(5.67)
am(y, x', y') = u m(x')8(y - y')
We associate the following boundary conditions with (5.67):
am(O, x',
Since the integrands in all three integrals are independent of z, the z-integrations cancel and we. have a b b loa (-V; ' u)gdxdy = ,u(-V;yg)dxdy+a loa ag dx (5.62) . y 0 0 0 ay y=b . loo 0 I where we have used a/az = 0 to reduce the Laplacian to two dimensions. S';1bstituting (5.55) and (5.58) and performing the delta function integration, we o~tain
Io
f§a
With respect to the transformation given in (5.65), we have
fOO lob loa u(-V 2g)dxdydZ +a
(5.65)
az
az
-00
I
all
By hypothesis, there are no variations with respect to z, and these partial deriv~tives vanish. The integral at z ~ -00 vanishes in a similar manner. The only remaining surface integral contribution is over the surface at y = b, and (5.60) reduces to the following:
f oo lob0 'loa0 (-V
,
g(x, y, x, y )um(x)dx
a g (mJr)2 --a 2 =? am x a
: • ag Vg· n = I
a
0
um(x) =
Vu' n = -
,
i
(5.64)
where
(5.60)
The surface integral S consists of integrals over the surfaces bounding the cylinder at x = 0, x = a, y = 0, and y = ,b, plus the integrals over the cylinder cross sections at z ~ -00 and z ~ 00. Application of the boundary conditions specified in (5,56) and (5.59) reduces all surface integrals to zero, except over the surface at y == b and over the surfaces at z ~ -00 and z ~ 00. Consider the integral at z ~ 00. We have
•
To complete the solution, we must solve the Green's function problem given in (5.58) and (5.59). In the same manner as in Example 5.3, the spectral representation in the x-direction leads to a Fourier sine series, viz.
(5.59)
Iv (~V2u)gdV = Iv u(-V 28;)dV + i (-gVu + uVg)· lidS
,.
193
SLPI in Two Dimensions
I
(x') mrr sinh mrrb a a U
m
I
sinh mrr(b-y') sinh a
~ a '
y
I
. h mrr(b-y) . h ~ sm a sm a'
.
y>y
(5.69)
I
Substitution in (5.64) yields the Green's function
(
, 2 sin mrrx sin mrrx' ')_ -- L a sinh mrrb 00
a
g x, y, x, Y
m=1
mrr
a
a
a
h mrr(b-y') . h ~ sm - - a - sm a" . h mrr(b-y) . h mrry' sm a sm -a-'
I
!
y
y>y (5.70)
I
Chap. 5
Electromagnetic Boundary Value Problems
194
Equation (5.70) gives the Green's fu~ction required in the first integral in (5.'63). In the second integral in (5.63), we require agjay evaluated at y = b. Perfom1ing the required differentiation in (5.70) yields ag(x, b, x, y ) 2 L -=.:..-c... = - I
00
I
a
ay
111=1
'.
.
u(x,y) =
a..
(5,71)
21 l
I
lo
b
0
• mTCX sm . mTCX' 'sl'n'h mTC(b-y') . sl'nh I1lTCy , sm a a a a . L : mTC . h mTCb, ' m=l a sm -asi'nh mTC(b-y) sinh I1lTCy a (, ' 00
l'
l
a
o
mn x mn x' l L sin - - sin - - d x ' = a a a a ()
2
-
00
a
AEC
,
,
Define the three-dimensional inner product
,i:
0
(f, g) =
(5.77)
The Green's function problem associated with (5.74) is given by Y
LAg = 8(r - r')
Y> Y
a
8(x - x')dx' = a (5.73)
(LAu, h)
= Iv [(- v 2 -
A) u] hdV = Iv(- V 2 u)hdV + Iv (-Au)hdV
(5.79) To produce the adjoint operator L~ and the conjunct J(u, h), We again use Green's theorem. In the case of the Laplacian operator, we have
Iv (-V u)hdV = Iv u(-V h)dV + 1(-hVu + uVh) . ndS
mnx'
8(x - x ) = - ~ sin - - sin - a m=1 a a
It is also important to show that the solution in (5.72) satisfies the original differential equation in (5.55). The details ,are left for the Problems.
2
I
(5.80)
where S is the surface bounding V and It is the unit normal to S in the direction outward from V. But,
•
5.4 SLP2 AND SLP3 EXTENSION TO THREE DIMENSIONS In this section, we consider complex !, complex A, and admit the possibility of unbounded regions. We produce an SLP2 and SLP3 exten, sian to three dimensions. We again confine our attention to the thre~ dimensional negative Laplacian operator and consider the partial diffe~en ; tial equation
(5.78)
Extending our analysis in Sections 2.5 and 2.6, we solve (5.74) by taking the inner product with the adjoint Green's function her, r'), as follows:
2
mnx
Iv !(r)g(r)dV
:
where we have used the spectral representation of the delta function 2 ~
(5.75)
and where L is the Laplacian operator
dxdy.r(x;y')
m=l
,
A,
(5.76)
(5.72) It is easy to show that (5.72) satisfies the boundary conditions at x = 0, x =i= a, and Y = O. The details are left for the reader. We now show that (5.72) satisfies the inhomogeneous boundary condition at y = b required by (5.57). Indeed, at y = b, the second term vanishes and we have u(x, b) = a
=L -
a
2 00 mnx mnx' sinh I1lTCy , ~-Lsin--sin-,'-. m:bdx +o a m=1 a a smh -aa a
where
h mIry
mnx . mnx sm - a sm - - sm - - ---"-~ a a sinh I1lTCb
195
(5.74)
LA ,
Substitution of (5.70) and (5.71) intc) (5.63), followed by an interchange of the prime and unprimed coordinates, yields the following result:
"
SLP2 and SLP3 Extension to Three Dimensions
Sec. 5.4
(5.81 ) and
Iv (-Au)hdV = Iv u(-Xh)dV = (u, -Xh)
(5.82)
Substituting (5.81) in (5.80) and then (5.80) and (5.82) into (5.79), we obtain (LAu, h) = (u,
L~h) + J(u, h) Is
(5.83)
I
Electromagnetic Boundary Value Problems
196
Chap.' 5 i
where
L~u J(u, h)
=,'
-1
I= s
-
\;r2 - X
(5'.84) i
(71Y'u - llY'71) . ,idS
(5;.85)
i
s
We note that (5.83) is the three-dimensional extension of (2.134). We may solve for u in (5.83) by considering the adjoint Green's function problem i given by L~h ~ j'
8(r - r')
(5.86)
, Substitution of (5.74) and (5.86) Into (5.83) yields u(r') = (j, h) - J(u, h)
Is
d87) I
!
or, explicitly,
u(r') =
r f(r)71(r, r')dV + I [71(r, r')Y'u(r) -
lv ,~
,
u(r)Y'71(r, r')] . ft~S (5.88)
We note that (5.88) is the solution to (5.74), provided that we can determine the conjugate adjoint Green:'s function 71(r, r'). Taking the complex conjugate of both sides of (5.86),;we obtain a partial differential equ~tion for 71, viz. i '
L~h(r, r')
= L).,,71(r, r') = 8(r -
d.89)
r')
As in Chapter 2, we can sh~w that it is never necessary to fin~ the . conjugate adjoint Green's function directly. Indeed, we form I
(L)."g(r, r'), h(r, r")) == (g(r, r'), ,
L~h(r, r")) + J(g, h) I
s
(5 90)
r I I
'
, We are given the boundary conditions on g. We choose the boundary : conditions on 71 so that J(g, h)
Is= 0
SLP2 and SLP3 Extension to Three Dimensions
197
Therefore, the conjugate adjoint Green's function is given simply by interchanging r andr' in the expression for the Green's function g(r, r'). In cases where the Green's function is symmetric, g(r, r') = g(r', r) and 71(r, r')
= g(r, r') = g(r', r)
(symmetric case)
(5.93)
We shall find that the Green's function is symmetric in many of the interesting cases to follow. It is certainly symmetric in cases wherethe operator L is self-adjoint. In addition, we shall find symmetry, as we have previously found in Chapter 2, when examining many problems contai~ing limit points and limit citcles. For the symmetric Green's function case, we may substitute (5.93) into (5.88) to obtain u(r') =
Iv f(r)g(r, r')dV + ~ [g(r, r')Y'u(r) -
u(r)Y'g(r; r')]. 'IdS
(5.94) We shall summarize the steps for solving (5.74) by the above-described extension to the one-dimensional Green's function method. We distinguish two cases, dependent on whether or not the Green 's fun~tion is symmetric. Nonsymmetric Green's Function Case
1. Write the solution in the form given by (5.88). 2. Substitute the boundary conditions for u on the surface S into (5.88). 3. Substitute the conjugate adjoint boundary conditions for 71 on the surface S into (5.88). 4. Solve the Green's function problem given by (5.78) with boundary conditions on S the same as the homogeneous form of the boundary conditions on u. 5. Obtain the conjugate adjoint Green's function Ii through (5~92) and substitute into (5.88). 6. Interchange the variables rand r' in (5.88).
(5.91)
Symmetric Green's Function Case
: Then, substitution of (5.78), (5.86), and (5.91) into (5.90) gives
1. Write the solution in the form given by (5.94).
71(r', r") = g(r", r')
2. Substitute the boundary conditions for u on the surface S into (5.94).
I
; or, with a change in variables,
71(r, r') = g(r', r)
Sec. 5.4
(5.92)
3. Substitute the boundary conditions for g on the surface S into (5.94).
i
,
.
1
I 198
Electromagnetic Boundary Value Problems
Chap. 5
Sec. 5.5
The Parallel Plate Waveguide
199
I
4. Solve the Green's function problem given by (5.78) with bound~ry conditions on S the same asthe homogeneous form of the boundary conditions on u and substitute into (5.94). 5. Interchange the variables r. and r' in (5.94).
u-+oo
f
x
5.5 THE PARALLEL PLATE WAVEGUIDE In this section, we consider the pr9pagation of electromagnetic waves in a parallel plat~ waveguide. We consider a waveguide of uniform cross 'section with no scattering objects. : Suppose that the waveguide is formed from two parallel, perfec'tly conductingplates (Fig. 5-4), separ~ted by a distance d and extending from ~oo to 00 in the y-direction and th~ z-direction. Assume that the meditim: petween the parallel piates is free space. We shall choose the source of th~ electromagnetic field to be independent of y. Since the structure is also ! independent of y, we must have
!ay = 0
(5.~5)
Fig. 5-4
Parallel plate waveguide.
( V2xz
,
We begin with Maxwell's curl equ\ltions, given in (4.55) and (4.56). If EO is the permittivity of free space, we have ;
I
,
V x E = -'-M - iWJloH
(5.cF)
'
+ k 2) H olex y = '1WEoMy + -olz - -
ox
Ex
az
aE aE ; . - x - - z = -M - lWJloH az ax y y
(5.99)
--y = Jz + iWEoE z ax
(5.100)
aE y -oz oHx
'
= Mx
(5.101)
+ lWJloHx
- - - - - = ly
oz ox, : aE y
-ax = -M . 7
-
(aHy + lx) oz
(OHy -
_.1_
oz
ox
UVEO
lz)
.
+ 1WEoEy
(5.102)
iW/l 0 H.
(5.103)
•
(5.104) (5.105) (5.106)
where (5.107) and
k2 =
2
(5.108)
W JloEo
A similar procedure :applied to (5.101)-(5.103) yields Set 2: T E z t
_. 1 oMz ( V21+k2)E xz! y - lWJlo y - - -
•
aHzo
= __._1_
Ez =
I
~e expand these two equations in Cartesian coordinates and group them into two sets as follows: Set 1: T M z aHy : . (5.98) - - = -lx - lWEOE x
Set 2: T E z
u-+oo
UuEO
(5.96)
aH
_t_ _t_
We note that the transverse magnetic (T M z ) set is not coupled to the transverse electric (T E z ) set. It is therefore possible to excite one set independent of the other by appropriate selection of the J and M sources. We produce s~cond-order partial differential equations governing each set by th~ followmg procedure. We differentiate (5.98) with respect to z, (5.100) WIth respect to x, add the result, and substitute (5.99) to obtain the JolIowing: Set I: TM z
I
V x H == J + iWEOE
a
.
ox
(OE
y H t = -I - - - - M
.
iw/J..o
H_. - - -I ,
iW/l0
az
oMx + -oz
) x
y (fJE --+M. )
ax
.
(5.109) (5.110) (5.111 )
Electrom~gnetic Boundary Value Problems
200
I
Chap. 5
yve remark that the sources My, Jx, Jz excite the T ~z set, whereas tre, sources J y , Mx , M z e~cite the T Ez,set. We shaIl consIder the T M z c~se. ; To excite the T M z fields only, we set J y = Mx = M z = 0 SUice, : these sources excite T E z modes. We are left with three options to exc~te the T M z modes, namely, My, Jx , or, Jz. Since My enters into (5.104) in t:he least complicated manner of the three, we choose My and set Jx = Jz =:0. This choice results in E y = Hx = Hz = 0 and
(V~z ".
+k
2
)Ry = iWEOMy
1 aHy iWEo az 1 aHy Ez = - iWEO ax
Ex =
~---
(5.112) (5.10)
lim Hy(x,z)
aHy(O, z) ax
=0
The Parallel Plate Waveguide
Iv [(- V
2
-
k
2
)
Hy ] hdV =
.
aHy(a, z) = 0 ax
Iv Hy ( - V
k 2 ) hdV
-
We anticipate, and will verify, that the Green's function will be symmetric, and use (5.93) to write
Iv [(-V
2
-
k
2
)
Hy] gdV =
Iv Hy ( - V
2
k 2 ) gdV
-
+ t(HyVg-gVHy).izdS
,
= ±ag -
(5.116)
~yHy·nAI j
ay
y---+±oo
I y
aH=±
ay
y---+±oo
= 0
y---+±oo
I
=0
y---+±oo
where we have used (5.95). The contributions on all remaining surfaces vanish with application of (5.115), (5.116), (5.118), and (5.119). The result IS
(5.11?)
(5.123)
i
lim g (x;, z, x ", Z ) = 0
z---+±oo
,(5.122)
The volume V is that region between the paraIlel plates. The surface S consists of the surfaces of the two paraIlel plates and the cross-sectional planar surfaces at z -,> ±oo and y -,> ±oo. The surface integral contributions at y -,> ±oo vanish since
(5.115)
Equation (5.116) arises from the fact that E z is tangential to the perfectly conducting waveguide surfaces at ~ = 0 and x = a, and therefore m~st vanish. Substituting this information in (5.114) yields (5.116). The Green's function problem associated with (5.112), (5.115), and (5.116) is as foIlows::
2
+ t(HyVh-hVHy).izdS(5.121)
Vg·nAI
~
201
Explicitly,
(5.11:4)
At present, the only restriction on the source My is that it is independ:nt, 6f y. Associated with the Laplacian operator in (5.112) are the foIlowmg boundary and limiting conditions on the field Hy(x, z): z---+±oo
Sec. 5.5
(5.11,8),
i
ag(O,z,x',z') ~ ag(a,z,x',z') =0 (5.11'9) ax ax To solve for the T M z fields, we shall solve (5.112) by the Green's function method given in Section 5.4. In the present case, however, tre problem is independent of y. We introduce this simplification by beginnirg with (5.83), which gives, in this case, i I
(5.120) i
I
Because of the absence of y-variations, the dy portion of the volume integrals cancel and the del-operator reduces to V xz ' with the result
i: la [(-
V;z
~ k 2 ) Hy] gdxdz =
i: la
Hy ( - v;z - k 2 ) gdxdz
. (5.124) Substituting (5.112) and (5.117) and performing the right-side iptegration, we obtain
Hv(x', z') =
-iWEo
i
Mv(x, z)g(x, z, x'. z')dxdz
Electromagnetic Boundary Value Problems
, 202
Chap. 5
where A indicates the area occupied by the source. An interchange of primed and u~primed coordinates gives '
Hy(x, z) =
I
-iWEo
i
My(X', z')g(x, z, x', z')dx'dz'
(5.125)
where we have again assumed that the Green's function is symmetric. Equation (5.125) gives the Hy-field everywhere inside the parallel plates, provided we Can solve for the Green's function g, which we consider next. To solve for the Green's function, defined by (5.117)-(5.119)', we expand in terms of the spectral representation over the x-coordinate, viz. 00
g(x, z, x',
z') =;=
L
f3n(Z, x', z')un(x)
(5.126)
n=O
Sec. 5.5
Let
f3n Yn = - ( ') Un X With this definition, (5.129) becomes
(::2
a
a
La g(x, z, x', z')un(x)dx
Yn =
=
Un (x')
(n;y
(5.135)
-iWEo f:
[1
, My(x', Z')
(En) e-ik:lz-z'l cos nrrx'dX'd Z'] cos nrrx a
a (5.136) The electric fields Ex and E z associated with (5.136) can be calculated from (5.113) and (5.114), respectively. We examine the modal structure of (5.136) by specializing the source My as follows: '
Using these relations to transform (5.117), we obtain
k; =k2 _
(En) e-ik,lz-z'l nrr X nrr X', L 2'k cos--cos-n=O a z a a 1
Hy(x,z)
x', z') = -u,(x')8(z - z')
(5.134)
W,e note th~t the Green 's ~unction is symmetric, as anticipated. Substituting thiS result mto (5.125) gives the magnetic field Hy , viz. ,
.
===}
Im(kz} < 0
2ik z
00
g(X,Z,X ,Z) =
n=O
,
(5.133)
These conditions!are consistent with those in (5.118). The solution to this: one-dimensional Green's function differential equation has been obtained previously in Example 2.20. Applied to (5.132), we find that I
(5.128)
Using the procedure in (5.37)-(5.41), we have
where
i
Substituting (5.1 34) into (5.131) and the result into (5.126) gives
g ===} f3n
(::' H;) ~,(z,
(5.132)
lim Yn = 0
(5.'127)
With respect to the transformation in (5.128), we have
8(x - x')
z')
z-->±oo
"
,
-8(z -
e-ik,lz-z'l
En is Neumann's number. The coefficient f3n is given by f3n(Z, x', z') =
i ) Yn =
1
nrrx un(x):::;:'~n - cos-and
+k
(5.131)
We associate the fOllowing limiting conditions with (5.132):
where the normalized eigenfunction Un (x) is given by
;
203
The Parallel Plate Waveguide
(5~ 129)
A
My(x', z') = M sy (x')8(z' - e)
(5.137)
;vhere Msy is a magnetic surface current in volts/m. Substituting (5.137) mto (5.136) and performing the indicated z' -integration, we obtain •
(5.130)
a
2lk z
Hy(x, z) =
-lWEO
LB n=O
e-ik,lz-ll ~n
00
n
.
2lk z
nrrx cos-a a
-
(5.138)
i
204
Electromagnetic Boundary Value Problems
Chap. 5
The Parall~l Plate Waveguide
Sec. 5.5
20S
I
where Bn is a modal coefficient, given by
Bn =
fI
a
associated with -a;2 /az 2 subject to the limiting condition in (5.118) leads to the Fourier transform pair
'
, En nJrx , ' Msy(x) - cos --dx o a a
l
(5.139):
The representation of the Hy-field jn (5.138) shows the decomposition of the field into the familiar T E M (n ~ 0) and T M (n > 0) modes described: in the undergraduate texts ([3],[4J', for example). By making different choices of Ms~, we may adjust the coefficient Bn associated with dch mode. For example, we may excite only the TEM mode by choosing i[
".
·If
M.~y
= :
,
(5.140)
-Mo a
"Z) = 1-00 (Xl g (x, G( x, k z' x,
1
g(x,z,x',z')
{:::::::::>
_a2 g az 2
1
n#O n=O
,
M,o,
(5.1~l)
G(x,kz,x',z')
(5.146)
(5.147)
and easily find that
I
{:::::::::>
k2G
(5.1 48)
z
8(z - z')
Bn =
(5.145)
00
" ) =--. I G( x, k z,x,z ' ') eik'Zdk z g(x,z,X,Z 2Jr -00 We represent the transform pair by
where Mo is a constant. This choice gives
0,
z, x,"Z) e -ik,zd Z
{:::::::::>
e-ik,z'
(5.149)
Applying (5.145) and (5.147)-(5.149) to (5.117), we obtain 2
_ (d , dx 2
Substituting into (5.138), we obtairi
+ k 2)
G = e-ik,z' 8(x - x')
(5.150)
x
I'
If
Hy(x,Z)=-iwEo -Mo : a
e-iklz-il . 21k z
where (5.14,2)
i
=le.
which is a pure T EM wave traveling away from the source location z For Z < the wave travels right to left; for Z > the wave travels left,to we may select right. If we are interested specifically in the region Z > the constant Mo to produce a unit left-to-right T EM wave. Indeed, the choice :
e,
e,
M 0=
- .. iWE 0 (
(5.151) We let
0= Geik,z'
e,
and obtain
d2 - ( -dx + kx2) G = 2 A
eike)-l
(5.143)
7i,jO,k z
,
(5.153)
8 (x - x )
with boundary conditions inferred from (5.119), viz.
I
produces
Hy(x, z) = e:- ikz ,
(5.152)
Z>
e
dOl
(5.14~)
, In (5.126), we chose to expand the Green's function in a spect~al expansion over the x-coordinate. This expansion led to the Fourier cosine series, and produced a solution for the magnetic field H y in terms of the waveguide modes. An alternative representation is possible. We shall' begin by expanding the Green's function in a spectral expansion over:z, rather than x. In (5.117), the spect~al representation of the delta function
dx x=o
=
dOl
=0
(5.154)
dx x=a
The solution to this differential equation and associated boundary conditions has been obtained previously in Example 2.13, viz.
G A
I = - k x sinkxa
1
cos kxx cos k.r (a - x'), coskrx' coskx(a - x)
x < x'
x> x'
;(5.1 55)
j
206
Electromagnetic Boundary Value Problems
•
Chap. 5
Sec. 5.6
Using (5.152) and taking the inverse Fourier transfonn, we have I
g(x,
Z,
I
.1
-00
----1
k x sm kxa
-R-e-gj-On-1
cos kxx cos kAa - x'), coskxx'coskAa-x)
x < x'
(5.156)
x>x'
i
!
,
(E ) e-ik,lz-z'l nrrx nrrx' g(x,z,x,z)= L -' . cos--. cos-.....
"
00
n=O
!
11
a
, 21k z
a
a
(~.157) I
Although these two representations lead to the same Green 's fu~ction g(x, z, x', z'), their fonns are quite different. In (5.157), the cross-sectional waveguide modes are displayedexplicitly. In (5.156), we find no e~plicit modal display. However, (5.156) is the starting point for constructing waveguide ray representations [5]. These ray representations are particularly useful in cases where thejfrequency is so high that a large number of modes can propagate in the waveguide. In addition, Felsen and Kamel have shown that the Green's fun~tion fonns in (5.156) and (5.157) c'an be combined to produce what are c,alled hybrid ray-mode formulations; The hybrid fonns effectively exhibit .the useful features in both the modal and ray fonnulations. The details can be found in [5]. ! In this section, we have studied fonnulations describing the Jropagation of the T E M and T M modes in a parallel plate waveguide: We leave the production of a modal series describing the T E modes f¢r the problems. In the next section, we shall consider an obstacle in a paratlel plate waveguide. We shall demonstrate the decomposition of the fields into incident, transmitted, and reflected waves.
L_i"'oo
____I a
1-'0'£:0
This represdntation of the Green's function is an alternative to the representation given in (5.135), which we repeat here for convenience, vi~. i
.
x
_t......__
1 jooeikz(Z-Z')dk z ,--.----=-
_
x, z) - - . 2rr
207
Iris in Parallel Plate Waveguide
Fig. 5-5
b
t__ ~z
Iris in parallel plate waveguide.
through an aperture, located at Z = 0, x E (0, b). We note that the insertion of the iris does not change the y-independence of the waveg~ide geometry. Therefore, Maxwell's equations again separate into a T E z set and a T Me set, in the same manner as in Section 5.5. We shall cause excitation of the T M z set hy placing a constant magnetic sheet current source at z = -d in Region I, viz. (5.158) My = Mo8(z + d) Such a choice will produce a TEM wave incident from left to right in Region 1. As we shall discover in the ensuing analysis, this TEM wave, when encountering the iris, will cause TEM and TM waves to scatter from the iris into both Region I and Region 2. Using (5.112)-(5.114), we write the equations describing the T M z fields in Region 1 and Region 2 as follows: Region 1: 2 (5.159) (\7;Z + k )RY I = iWEoMy
1 aRyl E xl = - - . - - 1WEO
EZI
az
1 aRyl 1WEO ax
=-.---
Region 2:
I
I
°
5.6 IRIS IN PARALLEL PLAtE WAVEGUIDE
(\7;Z
In Section 5.5, we solved for the fields in a parallel plate waveguide. In this section,we add an iris to the waveguide interior. The waveguide has cross-sectio~al dimension a (Fig: 5-5), and contains an infinitesimally thin, perfectly conducting iris connected to the top plate at x = a, Z = and extending perpendicular to the plate and into the waveguide interior. The iris effectively divides the interior of the waveguide into two regions: Region 1, z < 0; Region 2, Z > 0. The regions are connected electromagnetically
1 aRy 2 E x 2 = --.- - 1WEO az
°
2
+ k2 ) R y 2 =
1 aRy 2 iWEo ax
E'2 = - - C
(5.160) (5.161)
(5.162) (5.163) (5.164)
We first consider Region 1. Using the Green's function method and anticipating that the Green's function will be symmetric, we adapt (5.122) to the
.. 208
209
Iris in Parallel Plate Waveguide
Sec. 5.6
Electromagnetic Boundary Value Problems
(5.172), the boundary condition is over the iris and the aperture. We therefore have a Green's function problem (Fig. 5-6) for a parallel plate waveguide extending from Z -+ -00 and terminating in a perfect conductor at z = 0. Before commenting on this choice, we substitute (5.167)-(5.172) into (5.165) and obtain
present case as follows:
where gl is the Green's function
i~
-(V;z +k 2 )gl
{[(-V 2 -k 2 )Hyl ]gldV= J~{ H yl (-V 2 -k 2 )gldV'
J~
Region 1, governed by
b 8(x -
x')8(z - z')
-foo Jo{b
(5.166)
(gl aHYI)1
I
,
with boundary and limiting conditions yet to be determined. The volume VI consists of Region 1. The surface SI consists of the following parts: I
:
1. The surfaces.of the two parallel plates in Region 1 (z < 0, x ~ and z < 0, x= a). 2. The cross-sectional planar surfaces at z -+ -00, x E (0, a) lmd , I Y -+ ±oo, x E (0, a). ' I 3. The surface of the iris and, aperture, z = 0, x E (0, a).
9
(5.173) Because of the invariance with y, the integrations with respect to y cancel in all terms. In addition, the del-operator reduces to V xz and we have
i:L
G [(
-V;z - k
2
)
Hyl ] gldxdz =
i
I
By the same reasoning as in the previous section, the surface integrals at Y -+ ±oo vanish. In addition, we have the following boundary and limiting conditions governing the Hyl-fields: '
I
lim Hyl =
z--> -00
°
°
(5.168)
x E (b, a)
(5.169)
_a_H.::....yl:....;(_O,_z...:...) _ aHyl(a, z) _ ax ax aHyl(X, O)
:
az
:
-~--=O,
(5.167)
lim gl(x,z,x',z') =0 z~-oo
ag l (x, 0, x', z') az ,=0,
+
2
H yl (-V;z - k ) gldxdz
My(x, z)gl (x,
b
1 o
,
,
gl(x,O,x,z)
Z,
x', z')dxdz
aHyl(X,O) az
dx
Anticipating the symmetry of the Green's function, we interchange the primed and unprimed coordinates and then substitute (5.158) and (5.160) to obtain x
(5.170)
t
u-+oo
(5.171) u-+oo
xE(O,a)
•
a
(5.172)
We note that the Green's function boundary and limiting conditions 'are the same as those associated with the Hyl-field, with one important excep~ tion. In (5.169), the boundary condition on HI'! is over the iris. whereas in
I
i
H yl (x', z') = -iWEo
I
agl(O,z,x',z') : agl(a,z,x',z') ax = ax =0
G
(5.174) Substituting (5.159) and (5.166), doing the delta-function integrations, and rearranging, we obtain
i
.We choose the following boundary and limiting conditions for the Green's ifunction gl:
i:L
- f(gla~llodX
I
,
I
dxdy
z=o
az
-00
IT-+
Fig. 5-6
00
-..z
Green's function problem for Region 1.
210
Electromagnetic Boundary Value Problems
Hy1(x, z)
= -iWfoMo {b
'lo
la
Chap. 5
'
Our choice of the boundary condition in (5.172) deserves some Somment. If we had chosen this condition to be the same as that governing the H -field derivative in (5.169), we would have produced a Green's function problem with an unspecified bouhdary condition over x E (0, b), z =:= 0. Our choice completes the specification of the Green's function in Region I, and has the added advantage .of simplifying the problem solution by eliminating a portion of the surface integral. . in We now consider Region 2., In a manner similar to our treatment I Region I, we' obtain
r
[(_V 2 _k 2 ) Hyz]gZdV'=
r Hyz(-V
l~
2
°
z-+oo
(5.175) I
agz(o, z, x', Z') __ agz(a, z, x', Z') __ ax ax
I I
l~
211
Iris in Parallel Plate Waveguide
We choose the following boundary and limiting conditions for the Green's function gz: (5.181) lim gz(x, z, x', Z') =
gl(X, Z,X ' , -d)dx' - iWfo
gl(X,Z,xl,O)Etl(xl,O)dx '
Sec. 5.6
agz(x, 0, x', Z') = az
°
°
, x E (0, a)
(5.183)
We therefore have a Green's function problem (Fig. 5-7) for a parallel plate waveguide extending to Z ---+ 00 and terminating in a perfect conductor at z = 0. We substitute (5.178)-(5.183) into (5.176) and obtain
{[(-VZ-e)Hyz]gZdV=
l~
r Hyz(-VZ-e)gZdV
l~
+
-k 2 )gZdV
/00 lo{b (gZ aHyz) az -00
+ { (H yzVgz-gzVHyz)·,'id5 lsz .
(5.176)
(5.182)
dxdy
II
z=O
(5.184) Again, the integrations with respect to y cancel in all terms and the deloperator reduces to V xz ' We have
where gz is the Green's function in Region 2, governed by
-(V;z
+ e)gZ =
8(x - x ' )8(z -
z')
(5.177)
The volume Vz consists of Region 2. The surface 5z consists of the following parts:
1. The surfaces of the two parallel plates in Region 2 (z > 0, x=O and z > 0, x = a). , 2. The cross-sectional planar surfaces at z ---+ 00, x E (0, a) aQd y ---+ ±oo, x E (0, a). 3. The surface of the iris and aperture, z = 0, x E (0, a).
(5.185) Substituting (5.162) and (5.177), doing the delta-function integrations, and rearranging, we obtain I
I
Hvz(x,z)=.
i
0
b
I
I
gz(x,O,x,z)
aHyz(x,O) . dx az
,
Again, the suIface integrals at y -4 ±oo vanish. In addition, we hav~ the following boundary and limiting conditions governing the Hy2-fields: lim Hyz =
z-+oo
°
aHyz(O, z) aHyz(a, z) =0 ax . ax aHyz(x, O) -~-- = 0, x E (b,a) az -~--=
x
t
(1' ..... 00
(5. i78) (1'-+00
(5.179)
a
I
I
(5.180)
Fig.5-7
Green's function problem for Region 2.
....-------~-----.. z
!
, Electromagnetic Boundary Value Problems
212
Chap. 5
Anticipating the symmetry of the Green's function, we interchange the primed and unprimed coordinates and substitute (5.163) to obtain b
I
io
I
We define ") fJ n ( Z,X, Z
= an(z,x',Z') un(x')
(5.195)
k;) fJn(Z, x', Z') = 8(z - z')
(5.196)
I
H y2(X, z) = iWEo ( g2(X, Z, x', 0)E x 2(X', O)dx'
:
(5.186)
and obtain
-(::2 +
,
We next consider the determination of the Green's functions. In Regibn I, we wish to solve (5.166) with boundary conditions given by (5.170)(5.172). As we found in Exm:,rtple 3.2, the spectral representation of (-a 2 Iax 2 ) with Neumann boundary conditions given in (5.171) results in the Fourier cosine series. We therefore expand the Green's function
We assign the boundary and limiting conditions lim fJn(Z, x', z') =
z--+ -00
....
.
gl (x, Z, x', z')
00
=L
an(z, x', z')un(x)
213
Iris in Parallel Plate Waveguide
Sec. 5.6
(5.187)
dfJn (0, x', z') dz =0,
,
a
(5.197)
xE(O,a)
(5.198)
n=O
where Un (x) is the orthonormal ~jgenfunction n:rrx = -n cos--
(5.188)
~a gl (x, Z, x', z')un(x)dx
(5.189)
un(x)
:
~a
a
Satisfaction of these two conditions results in the satisfaction of the con- • ditions in (5.170) and (5.172). The differential equation in (5.196) with! conditions in (5.197) and (5.198) is solved by the standard Green's function' methods developed in Chapter 2. The details are left for the problems. The, result is .
and where an(z, x', z') =
•
Equation (5.189) defines the transformation
1
fJn(Z,X',z')=-;-k ,
I
ax
=}
(n:rr)2 an
a
.
8(x - x')
(5.1 91)
z'
e' ,z coskzz,
Im(k z ) <
=}
un(x')
i
k;) an(z, x', z') = Un (x')8(z - Z')
Z'
(5.199),
Z > Z'
a
gl (x,
I
I
z, x , Z ) =
n:rr x n:rr x' L -.cos - - cos - lkza a a En
00
n=O
(5.'192)
We use (5.190)-(5.192) to transform (5.166), with the result
-(::2 +
'k'
Substituting (5.199) into (5.195), solving for an, and substituting the result into (5.187) gives
Using the procedure in (5.37)-(5.41), we establish that 2
z<
eik,z cosk z'
where (5.:190)
_a g12
z
I
(5.;193)
g2 ( X, Z,
" _ X, Z) -
00 '"'"
L
En n:rrx n:rrx , -:--k cos - - cos - -
n=O I
(5.' 194) I
eik,z cos kl z' ,
.k ' e' ,z coskzz,
z<
Z'
Z > Z'
(5.200) To obtain the Green's function g2 for Region 2, we note that the geometry . in Fig. 5-7 can be obtained from Fig. 5-6 by reflection through the z = a plane. We therefore can obtain g2 from gl by replacing z by -z and z' by -z', with the result
I
where
I
zG
a
G
I
kzZ, '
z> z'
e-' ,z cos kzz,
Z < Z'
e-ik,z 'k'
COS
(5.201)
214
Electromagnetic Boundary Value Problems
Chap. 5
We note that the anticipated symmetry of the Green's functions occurs for both gI and g2. In the solution for the fields, we shall need the Green's functions with Z' = O. We obtain 00 E. nrrx nrrx ' gI (x, z, x', 0) = L ~elkzz cos - - cos - n=O lkza a a
g2(X,
Z,
I 00 En" X, 0) = L :-----'e- lkzz n~l~a
nrrx
nrrx'
COS - - COS - -
a
a
Sec. 5.6
-d <
z<0
215
Therefore, in the; limit, we produce a unit-strength electric field, incident from left to right land reflecting from a perfect conductor at Z = '0. As a final ~tep in the problem formulation, we note in (5.186) and: (5.207) that :
(5.202)
(5.203)
Z/E(O,b)
H yl (x, z) =
2 cos kz
.
- [(uEo
~
b
g\ (x,
Mo' ~
I
fob gj(x, Z, X', O)ExJ(x' , 0)4.:x'
. 0 '
(5.205) I
As a normalization, we choose
Mo = _2e ikd
(b
10
HyI(X,Z) =
2 cos kz
. -lWEO
~
fo
Hyl(X,Z) =
2coskz I]
k ~ Ell ik nrrx --L...J-k e zZcos-~ 11=0 zO a
(5.206)
k 00 E n ' nrrx H y2(X,Z) = - L-e-lk,Zcos-~ 11=0 kza a
b->O
y
1(x,
z\ = {
0
0
b
I' nrrx ' I EA(X)cos--dx a
(5.211) b
nrrx ' EA(XI)cos--dx ' a
(5.212)
l
nrrx a
l
b
nrrx ' EA(XI)cos--dx ' 0 a
(5.213)
or
~~ (e ikZ + e- ikZ )
Substituting into (5.160) gives lim ExJ(x, z) = eik<. - e- ikz b-+O
k
11=0 a k z
(5.207)
This normalization is used to produce a unit-strength incident electric field. Indeed, consider the limiting cas~ where the size of the aperture shrinks to zero. We have lim H
E
1 = L~-cos--
.0
l
l
These two expressions give the magnetic fields everywhere in the two regions. We note, however, that the electric field E A in the aperture is as yet unknown. We have, however, one additional boundary condition that we have not utilized, namely, the continuity of the tangential magnetic field in the aperture. We invoke this continuity by equating (5.211) and (5.212) in the aperture and obtain 00
I I I g\(x,z,x,O)Ex\(x,O)dx
(5.209) :
(5.210)
g2(X, z, x', O)EA(xl)dx '
Substitution of (5.202) and (5.203) gives
and obtain : b
I I I X ,O)EA(x )dx
I
We note that in (5.202)-(5.204), care must be taken to select the p~oper cases for the Green's functions gJ and g2. In (5.202), Z < Z'; in (5.203), Z > Z'; and, in (5.204), z > Z' = '-d. ' The magnetic current source in Region 1, given by (5.158), has been chosen to produce a TEM wave incident from left to right. We may show this by substituting (5.204) into the first term in (5.175) to give Hyl(X, z) = __ e- ikd coskz- iWEo
Z,
0
H v2 (x,z) = iWEo
(5.204)
l
I
..
(5.208) :
which is a statement that the tangential electric field is continuous in the : aperture. We symbolize this aperture field by E A(X') and write (5.186) and . (5.207) as
We also need the Green's function in Region 1 evaluated with Z' = :-d. Confining..ouf interest to regions to the right of the source, we obtain i I ~ En nrr X nrr X' -ik.d gI(X, Z, X, -d) = L...J -.- cos -.- cos - - e . coskzz, : n=O lkza :0 0
Iris in Parallel Plate Waveguide
(5.214) where I ~ En k nrrx nrrx ' Q(X, X) = L...J - - COS - - COS-n=O a kz a a
(5.215)
Expression (5.214) is an integral equation whose solution yields the aperture field Ell. Once E,1 is known. the result can be substituted into
,
216
Electromagnetic Boundary Value Problems
Chap. 5
Sec. 5.7
217
Aperture Diffraction
I
..
(5.211) and (5.212) to yield the magnetic fields. The corresponding electric fields can be obtained by substituti9n of these results into (5.160), (5.161), (5.163), and (5.164). I Unfortunately, the integral equation in (5.214) cannot be solved ~n alytically. A popular method for finding an approximate solution for ~he aperture field EA is the Method of Moments (MOM), introduced in Section 1.8. Although the approximate solution to (5.214) is beyond the central i theme of this book, a few comments are in order. The kernel Q(x, x') for the integral equation in (5.214) is logarithlni~ cally singular. Therefore, care must be taken in dealing with the limit as x' ---+ x. For a discussion of the issues involved, the reader is referred to [6]-[8]. The series contained in Q(x, x') is slowly converging. For m~th ods to speed the convergence, the reader is referred to [6]-[10]. Fina~ly, the aperture field possesses an edge singularity [6],[11] in its beha~ior as x ---+ b. This singularity must be considered in evaluations involving MOM. For a discussion, the reader is referred to [6],[8],[12].
x
t
/ kp 1/ r
® Region 1
a
~~XE~O
Aperture in a perfectly conducting screen.
oE z
. ox = IWJLoHy
oEx
5.7 APERTURE DIFFRACTION
Region 2
+--y
L Fig. 5-8
P(x,y)
-
oy
oEy
- -
ox
= M
z
(5.221)
+ iw"oHz f-'"
(5.222)
The set comprised of (5.217), (5.218), and (5.222) is excited by M z and is decoupled from the unexcited set comprised of (5.219)-(5.221). We therefore have H x = H y = E z = O. Differentiating (5.217) with respect to y, (5.218) with respect to x, adding, and then substituting (5.222) gives the following set:
(V Xy
+ k2 )
Hz
= iWEOMz
10H Ex = - -z oy
iWEo
10H E y = - - -z
oH . . - -z == 1WEoE oy :
oHz -- = ox oHy
-- ox
-iWEOE
oHx
~
oy
y
.
= 1WEoEz
oE z -:::::; -iwJLoHx oy
,
iWEo
(5.217)
x
(5.2~8)
(5.219) i (5.220)
ox
(5.223)
(5.224) (5.225)
where k is given by (5.108). We next specialize the above equations to Regions 1 and 2 as follows: Region 1:
(VXy
+ k2 )
Hz]
= iWEoMz
10Hz! Ex! = - . - - 1WEO
oy
(5.226) (5.227)
,
.
I
'218
Electromagnetic Boundary Value Problems E)'1
.
: 1 aHzI =--.--!Wfo ax
Region 2:
(VXy
+k2 ) H 2 = ° z
!
1 aHz2
E x 2= - . - - 1WfO ay
.
E 2 y
: 1 aHz2 =;--iWfo ax
!
Chap. 5 I
i (5.228) ,I (5.229)
Ivl
, + , i
(HzIVg I - gIVHzJ)·
I
'
i
ndS!
°
2 2 2 2 {[(-V -k )Hz IJgldV= { H ZI(-V -k )gldV
l~
lv,
(5.233)
-+
: By the same reasoning as in the treatment of the parallel plate waveguide, the surface integrals at z -+ ±oo vanish. In addition, we have the following boundary and limiting conditions governing the Hzl-fields:
lim H,j = 0,
x-+±oo
aHz! (x, 0) ay
-
= 0, .
°
(5.234)
Y E (-00,0)
(5.235)
x f/ (-a/2, a /2)
(5.236)
(5.239)
We note that the Green's function boundary and limiting conditions are the same as those associated with the HZI-field, with one exception. At y = 0, the boundary condition on HzI is over the screen, whereas the boundary condition on gl is over the screen and the aperture. Therefore, the Green's function problem is for the half-space y < with a perfectly conducting surface at y = 0. We substitute (5.234)-(5.239) into (5.232) and obtain
-
: with boundary and limiting conditions to be determined. The volume VI ! is Region 1. The surface SI consists of the following parts: ' i 1. The surface of the screen and aperture at y = 0. 2. The planar surface at y ~ -00. 3. The planar surfaces at x -+ ±oo, y < 0. ±oo, y < 0. 4. The planar surfaces at Z
, (5.238)
agl(x,O,x',y') =0 ay
(5.231)
where gl is the Green's function ip. Region 1, governed by
lim HZI = y-+-oo
Y E (-00,0)
lIm gt(x,y,x,y)=O,
x-+±oo
(5.232)
I
.
(5.230)
(_V
L
Inspection of (5.227) indicates that the condition in (5.236) is equivalent to the vanishing of the tangential electric field on the surface of the screen. We choose the following boundary and limiting conditions for the Green's function gl: I ' (5.237) lim gl(x,y,x ,y) =0 y-+-oo
. We first coosider Region 1. Using the Green's function method and anticipating symmetry of the Green's function, we adapt (5.165) to the present , case as follows: I : 2 2 2 2 - k ) Hz! ] gldV 7= Hzl . - k ) gldV
lv, [( _V
219
Aperture Diffraction
Sec. 5.7
J-00 OO
jal2
aHZI g l -- dxd z -a12 ay
(5.240)
Because of the invariance with z, the integrations with respect to Z cancel in all terms. In addition, the del-operator reduces to V xy and we have
1°001:
[(-V;y -k2)dHzI]gldxdy
/00 HZI (-V;y - e) gtdxdy -00 -00 o
aH I gl-Z-dx -a12 ay . (5.241) Substituting (5.226) and (5.233), doing the delta-function integrations, and rearranging, we obtain
=
j
HZI (x',
Y') =
-iWfo
1 A
I
+j
jal2
Mz(x, y)gl (x, y, x', y')dxdy ,
al 2
I ,aHzI(x,O) g\(x,O,x,y) a dx -a12 y
Anticipating the symmetry of the Green's function, we interchange the primed and unprimed coordinates and substitute (5.227) to obtain
, I
Electromagnetic Boundary Value Problems
220
HZl (x, y)
= -iWEo .
1
M;(x' , y')gl (x, y, x', y')dx ' dy'
A
+ iWEo .
Chap. 5
I I
alz
I
gl'(X, y, x', O)E xl (x', O)dx' j -aIZ'
(5.242) I
We now consider Region, 2. In a manner similar to Re&ion! 1" we . , obtain
Iv [(+
Z
V -
k Z) HzzJ gzdV
2
= fV
Hzz (- V
Z
Z k ) gzdV
-
2
,
".
+
1 S2
,
(HzzVgz - gzVHzz), ndS' (S.243)
where gz is' the Green's functior in Region 2, governed by
- (V;y + k Z) gz = 8(x -
x ' )8(y -
y')
i
(6.244)
with boundary and limiting co~ditions to be determined. The volum y Vz is Region 2. The surface Sz consists of the following parts:
1. The surface of the
scre~n and aperture at y
= O.
2. The planar surface at y' ~ 00. 3. The planar surfaces at x ~ ±oo, y > O. 4. The planar surfaces at Z ~ ±oo, y > O.
aHzz(x, 0)
ay
~ 0,
(0,00)
x ¢ (-aI2, a12)
(5.247) ,
We choose' the following boundary and limiting conditions for the Green's I function gz: (5.248) lim g~(X,y,x',y')=O y-+oo : E
agz(x,O,x',y') =0
tJy
+j
(0,00)
(:5.249) i
I , (15 .250 )
Hzz(-VZ-kZ)gZdV
oo jalZ ( gZ_Z_ aH Z)[ : dxdz -aiZ tJy z~o (5.251) Again, the integrations with respect to Z cancel in all terms and the deloperator reduces to V xy . We have
~oo
-00
L:[(-
Z V;z - k ) HzzJ gzdxdy
=
~ooo
joo Hzz ( - V;z ,., - k Z) gzdxdy -00
+
jalz ( gZ--' tJHzz) [ dx -aiZ tJy, z=O
(5.252) Substituting (5.229) and (5.244), doing the delta-function integrations, and rearranging, we obtain
jalZ I ' aHzz(x,O) . gz (x, 0, x , y ) . dx -aiZ ay I ,
I
Anticipating the, symmetry of the Green's function, we interchange thd, primed and unprimed coordinates and substitute (5.230) to obtain
alz Hzz(x, y) = -iWEo gz(x, y, x', O)Exz(x' , O)dx' j -aiZ ,
(5.246)
:
Y
lV2
: I'
i
lim gz(x,y,x',y') =0, x-+±oo
{[(-VZ~kZ)HzZJgZdV= {
lV2
(5.:245) E
221
We therefore have a Green's function problem for a half-space y > 0 with a perfectly conducting surface at y = O. We substitute (5.245)-(5.250) into (5.243) and obtain
I
iim Hzz = 0 y'-+oo
Y
Aperture Diffraction
Hzz (x , y ) = -
Again, the surface integrals at Z ~ ±oo vanish. In addition, we have the following boundary and limiti~g conditions governing the Hzz-fields:
lim Hzz, = 0, x-+±oo '
Sec. 5.7
(5.253)
At this point in the development, we have two remaining tasks, namely. the specification of boundary conditions at points in the aperture and the selection of a specific magnetic current source. We first require that th~ tangential electric field in the aperture be continuous. We symbolize the aperture electric field by E A (x) and write x E (-aI2, a12) We next specialize the source M z to be a line source located ~t x'
(5.254)
= ~, y' =
7], VIZ.
, ' , Mz(x ,y ) = Mo8(x -
'
~)o(y -
7])
(5.255)
Electromagnetic Boundary Value Problems
222
Chap. 5
We substitute (5.254) into both, (5.242) and (5.253). Also, we substitute , (5.255) into (5.242) and perform the indicated delta-function integrations to give "i I
H Z1 (x, y) = -iwEoMogl (x,
Y,~, ry)+iWEo ,
f
Sec. 5,7
223
Aperture Diffraction
ox
The spectral representation of (-0 2 1 2 ) with limiting conditions in (5.261) leads to the Fourier transform, as given in Example 3.4. In this case, we have
a/2'
-a/2
(5.264)
gl (x, Y, x', O)E A(x')dx ' (5.256)
02 g2
2
- - - {:=:::}
Hz2(X, y) = -iWEo
f
ox 2
a/ 2
: g2(X, y, x', O)EA(xl)dx'
(5.257)
8(x - x')
k G2
(5.265)
e- ikxx '
(5.266)
x
{:=:::}
-a/2
Expressions (5.256) and (5.257) give the magnetic fields everywhere, provided that we know the Green's functions gl and g2 and provided we can find the aperture electric field EA. We shall derive the Green's functions subsequently. The aperture field EA is obtained by requiring the tangential magnetic field in the aperture to be continuous, viz.
x
E
Applying these, relationships to (5.260) gives (5.267,) where
(-aI2, a12)
(5.268)
Substituting (5.256) and (5.257) into (5.258) yields the integral equation
Mog] (x,
O,~, ry) =
L a
:
: l
[gI(X, 0, x', 0) + g2(X, 0, x', 0)] E A (x'~d~'
: (5.259) Once the Green's functions have been determined, an approximate sdlution to the integral equation (using Method of Moments, for example) yields an approximation to the aperture field EA. The aperture field can t11en be substituted into (5.256) and (5.257) to give the magnetic fields everyy.'here. Once the magnetic fields are known, the electric fields can be obtained by differentiation in (5.227), (5.228), (5.230), and (5.231). We next cdnsider , the Green's functions gl and g2. We may determine g2 from (5.244) and (5.248)-(5.250), which we reproduce here for convenience, viz. '
(
- 02 -2 - -0
ox
2
oy2
(5.269)
I
2)'
- kg2
= 8(x -
lim g2(X, y, x', y') = 0,
x-+±oo
og2(X, 0, x', y')
x I )8(y - y ' )
Y
-=-~------'-- =
oy
E
(0,00)
i ($.261)
! 0
lim g2(X, y, x', )") = 0 )""400
(5.260)
(5.262) (5.263)
The boundary and limiting conditions associated with (5.267) are as follows: ' x ,_0_,_x'_,,-y_') = 0 _d_G_2(_k_ (5.270)
dy
~
I'
lim G2(X, y, x ,y ) = 0
(5.271)
y-+oo
Invoking these conditions is consistent with the conditions on g2. The solution to (5.267) with the above associated conditions can be inferred from the Green's function problem discussed previously in (5.196)-(5. 199). If in (5.199) weletz -+ _y,Z'-+ -y',andk z -+ ky,weobtain
1_1 e-ikyy' coskyy,
y
e-ik"Ycoskyy',
y>y
G __ 2-
iky
Expanding the cosine terms into exponentials gives the following useful alternate form:
G2 =
_?_ [e-ik"IY-y' + e-ik,,(Y+.v'l] 21k I
(5.272)
y
Suhstituting this result into (5.269) and then taking the inverse Fourier transform, we have
224
Electromagnetic Boundary Value Problems
I g2 = - . 4rrl
/00 e-ik)'ly-y'1 + e-ik)'(y+y') e,kxC , r-00 ky !
x
Chap. 5
1
)dkx
(5.273)
Sec. 5.7
225
Aperture Diffraction
where (p', ¢/~ marks the position of the line source (Fig. 5-9). We may locate the line source at a distance remote from the aperture by letting p' become very large, in which case
From (4.156), we have the following identity:
e-ikvlYleik,x Hci2)( k y/x 2 +y2 ):=\ -rrI /00 -00 k dkr
(5.:274) Furthermore, using the large argument approximation for the Hankel function, given in Example 2.21, we have
y
Therefore
1
g2 =
-'.
~i {Hci2)[kJ(x -
+Hci2)[kJ(x -
x ' )2
x ' )2
+(y -
yl)2]
~5.275)
+(y +y')2]}
Referring to the development in Section 4.6, we recognize (5.27,5) as a descriptio~ of the radiation from a line source located at x = x', = y', plus a line source at the image location x = x', y = - Y I, with respect to the ground plane. ' Consider g\. We may obtain the solution for gl directly from the above solution for g2 by replaCing y by - y and y by - Y Such replacement does not change the solution, and we therefore have
H(2) Ik [p' - p cos(¢ =f= ¢')]} '" / 2i e- ikp'eikpcoS(Of') o rrkp' Using (5.280) in (5.278), we obtain
J
1
gi (p, ¢, p', ¢') '" :i /
I
I R--e-' b i 'k P2e''k XCOS'I'-1. ' cos(kysin¢) I = --;41 rrkp' 1
(5.281) Let
(5.276) Mo=
The above determination of the Green's functions completes the fdrmula• : I hon of the problem.; ! In many applications, the source M z is located at a distance far enough from the aperture so that a plane wave approximation can be invoked. To accomplish this, we consider the Green's function in the first term in (5.256), viz. '
g\(x,
y,~, 1]) =
:i [Hci
2 )
[~J(X - ~)2 + (y _1])2]
+kci ) [kJ(X -~)2 +(y +1])2] 2
I
Rewriting this expression in cylindrical coordinates, we have g\ (p, ¢, p', 4/) =
~i [ ~ci2\ [kJ p2 + p~ +Hci2~ [kJ p2 + p'2 -
rr~ple-ikpl [eikPCOS(-I) + ~ikPCOS(H')]
I.
, g2 = gl
+cP l)]
[
-WEoRbi -ikPI]-1 --e rrkp'
(5.282)
-4 x
t P(p,
p
(5.277)
Region 1
I (5.278)
Region 2
+~y
p'
;
2pp' cos(¢ - ¢';]
2 pp ' cos(¢
(5.280)
O(p',
Fig. 5-9
Aperture in a perfectly conducting screen, cylindrical coordinates.
226
Electro?1agnetic Boundary Value Problems
Chap. 5
Sec. 5.8
We substitute (5.275), (5.276), (5.281), and (5.282) into (5.256) and (5.257) and obtain Hz1(x, y) =
2eikxcos'
P(p)
cos(kysin¢')
l ja Hci) kJ(x - x')2 + y2 EA(x')dx' 21] -a12 . (5.283) ja H2(X, y) = - Hci) kJ(x - x')2 + y2 EA(x')dx' 21] -a12
+ -k
Z
2
l
k
227
Scattering by a Perfectly Conducting Cylinder
2 [
]
I
=-------'--------"-+-_ x
I
2
2 [
]
where we have used
(5.i84)
Fig. 5-10
k
W~o
=-
Electric current J z exciting a perfectly conducting cylinder.
1]
We note that in the limit as the ap~rture length approaches zero, we haye lim Hz! (x, y) =
2e ikx COs¢'
a~O
cos(ky sin ¢')
!
=
eikxCOS¢', (eikYSin¢'
+ e-ikYSin¢')
(5.285)
We require the fields at a point pep). We begin with Maxwell's curl equations, viz. v x H = ZJz + iWfOE ' (5.288)' V x E = -iwJloH
which represents a unit magnitude plane wave approaching the screen at angle ¢' and reflecting according 'to Snell's law of reflection. Continuity of the tangential H -field in the aperture gives the integral equation 2 al2 : _-.!2eikxcos¢1 = Hci2)(klx - x'I)EA(x')dx' (5.2~6) k i-al2 I , ; .which completes the problem formulation for the case of plane wave inci~ :dence. i Again, the integral equation in (5.286) cannot be inverted analytic~lly. 'The aperture field EA therefore must be determined approximately us:ing ,numerical methods. We note, however, that for the aperture size a small 2 I enough, the Hankel function Hci ) (k Ix - x'l) can be approximated by In Ix x'l, in which c'ase an analytical solution is possible in terms of Chebyshey 'polynomials. The resulting integra! equation is considered in Problem 1.23. ,The details are, given in [13],[14]. ,
The curl operator in (5.288) and (5.289) is composed of components transverse to the z-direction, plus a z-component, viz.
r
I
V = V(
a ~ = dZ
0
(5.n7)
A
az
, (5.290)
V = V(
We divide the electric and magnetic fields into transverse and z-components, viz. (5.291) H = H( +iBz (5.292) E = E( +zE z Substituting (5.290)-(5.292) into (5.288) and (5.289), we obtain
+ V(
V( X E(
!Consider an electric current J z that excites a surface current on a perfeCtly conducting cylinder of uniform cross section (Fig. 5- I0). We assume that :the cylinder geometry and the source are independent of z, so that
a
+ z-
In this case, because of (5.287), we have
V( x H(
i5.8 SCATIERING BY A PERFECTLY CONDUCTING CYLINDER
(5.289)
x
zHz =
+ V( x
ZJz + iWfOE
zE z = -iwJloH
(5.293) (5.294)
Equating transverse components and z-components on either side of (5.293) and (5.294), we produce the following two sets: Set I: T M z (5.295) V( X H( = (Jz + ;wfoE z )
z
V( x i.E; = -;o)/loH(
(5.296)
Chap. 5
Electrbmagnetic Boundary Value Problems
, 228
Set 2: T E z
(5.2~7)
Vt x E( = -ziwILoHz Vt x zHz = :
1
.
I
+iWEoEz) = V
t X
".
(5:299)
, and thus Set 1 becomes
V t2 E z +;k 2 E z = iW/l,oJz
= --._l_ Vt lWILo
x iE z =
-._l_ z x VtE z
(5.301)
where k is defined in (5.108). The procedure is now to solve the partial differential equation in (5.301) to yield E z . The result can then be substituted into (5.302) to produce the magn~tic fields H t • Anticipating the symmetry of the Green's function, we adapt (5,122) to the present case and obtain
r g ('1
lv
2
+ k 2 )EzdV
=
fv
Ez
lim g = 0
(V 2 + k 2 )gdV + Is (gVE z -
EzVg);ridS
Therefore, the Green's function problem is for two-dimensional free space. We substitute (5.305)-(5.307) into (5.303) and obtain 2
+k 2 )EzdV=
r Ez (V
2
lv
+e)gdV+
r gVEz.ndS
l~
We note that we did not require
glSe = 0 Although such a requirement would eliminate the surface integral, we would be unable to find an analytical solution for the Green's function, except in the special case where the cross section is circular. (We shall consider the circular case subsequently.) Because of the invariance with z, the integrations with respect to z cancel in all terms. In addition, from (5.290), the del-operator reduces to Vt and we have
r g (V
2
+;k 2 ) EzdA =
r E z (Vt + k
t 1A lA
2
2
)
gdA
I
+
r gVtE, z ' nds
lSe
(5.308)
where Sc is the ~rc-Iength integration around the cross section of the cylind~r and A is the planar area external to Sc' We shall consider the ds integration in some detail subsequently. Substitution of (5.301) and (5.304) into (5.308) gives, after some rearrangement,
Ez(p')
(5,303)
The volume V consists of all spate exterior to the cylinder. The surf~ce S is the surfac~ of the cylinder Sc plus the surface at infinity. By the ~ame reasoning as in the case of the parallel plate waveguide, the surface integrals at Z -+ ±oo vanish. The Green's function g is governed by
(5.307)
p~oo
(5.302)
lWILo
(5.306)
We choose the following condition for the Green's function g:
r g(V
: where we have used a well-known vector identity to expand the double~curI. ; But, (5.300)
Ht
lim E z = 0
lv V t2 (iE z )]
229
P-H)O
I
V t x zEz
= [Vt (V t . zEz) -
Scattering by a Perfectly Conducting Cylinder
(5j298)
iWEOE t
Set 1 is excited by the source lz; while Set 2 is unexcited. We therefore have E t = Hz = O. We take the, curl of (5.296) and substitute (5.2~5) to obtain '
-iwILoZ (Jz
Sec. 5.8
=
-iWILo
rg(p, p')lz(p)dA + her g(p, p')V;Ez(p) . nds
lA
(5.309)
However, from (5.302), we have
Z x VtE z = iWILoH t so that
I
(5)04)
I I
with boundary and/or limiting conditions to be determined. The condi'tions I on E z are as follows:
J'
But,
z x (z x VtE z) = -VtEz where we have used the vector triple product identity and the fact that
i.. VtE z = 0
230
Electromagnetic Boundary Value Prohlems
Chap. 5
Taking the inner product with the normal vector ii gives
i
in· "VtE z =
'iwJ.Lon· H t x
z=
iwJ.Loz· (n x H t ) = iWJ.Lolsz
(5.310)
{yhere lsz is the equivalent surface current in the z-direction in amps/m. Substituting this result into (5.309) and interchanging the primed and un~ primed coordinates, we obtain ' i
Ez(p) = -iwJ.Lo
i
g(p,
pl)lz(~I)dA' + iwJ.Lo
!.
g(p, pl)J.dpl)ds '
:i
Hci 2) (kip -
(5.312)
p'l)
231
Scattering by a Perfectly Conducting Cylinder
We shall parametrize with respect to the polar angle
(5.311) The Green's function problem given in (5.304) and (5.307) has been previously solved in (4.116), followed by the coordinate transformation indicated in (4.205). We have, including 2n from (4.10),
g(p, p') =
Sec. 5.8
(2)
Ez(p) = --4:-, Ho (kip - Pol)
w:qfa"Hci I
+
[
2
'(klp - p'l) J,,(p') (:;:)
2 .' .2] 1/2 +(:;,)
dqi (5.318):
and for p
E Sco
We shall specialize the source lz t9 be a line source of strength 10 amps~ located at p' = Po, viz.
(5.3~3)
lz = I08(p' - Po)
(5.319):
Substituting (5.312) and (5.313) into (5.311), we have
1
i WJ.Lo (2) I I,; i wJ.Lo/o (2) Ez(p) = - . Ho (kip - Pol) + - . Ho (kip - p I) lsz(p)lis 41 . 41 Sc I , (5.3l4) This equation gives the electric field E z everywhere exterior to the cylinder, provided that we can determine the surface current J.~z on the surface! of the cylinder. We accomplish this by forming an integral equation. We let p approach a general point on the surface of the cylinder p ESc. Since E z = 0 on the cylinder surface, we .have :,
P
I'
,
Is)
(dx '2 + dy'
2) 1/2
x' = p' cos
Y = p sm
E Sc
(5.3 Equations (5.314) and (5.315) complete the formulation of the proble~. Considerable care must be taken in the evaluation of the arc-length integral in (5.314) and (5.315). In Cartesian coordinates, the differential element can be represented by
ds' =
As in the previous problems in this chapter, we must solve the integral equation to determine the unknown quantity under the integral, in this case, lsz. In general, numerical methods must be employed to obtain an approximation to the solution. In the case where the cylindrical cross section is circular, however, we may invert the integral equation in (5.319) analytically. Indeed, consider a perfectly conducting circular cylinder of radius a (Fig. 5-11). In this case,
(5.316)
I
dx ' I' I = -p sm
I
232
Electromagnetic Boundary Value Problems
Chap. 5
P(p)
Sec. 5.9
Substituting thi~ result and (5.320) into (5.314), we obtain the following expansion for the electric field E z : ;
y
r
(J)II I Ez(p) = ~~
+ -=-------.L---I--:.-.L---l~ I
Fig.5-ll
Electric current Jz exciting a perfectly condu~t ing circular cylinder.' .
We shall eD;!ploy the addition theorem for the Hankel function from (4.2q7), viz.
H(2) (kl
o
_ 'I) =
p
P
~
in(t/J-ip')
i....J e
n=-oo
I
I
HP) (kp')Jn(kp), k k I Hn(2) (p)J n( p),
P < pi I i p> pi .
(5.321) Substituting (5.320) and (5.321) into (5.319) and rearranging, we have! .
,
.
J=J
p' (kPo) J. (ka)]e''.
IDe -"% H
:
~ .1=00 [10'" Jd a ,,4>')e-"" ad4>' H;"(ka)J. (ka) ] e"4 (5.322) To obtain (5.322), since p ESe, we have chosen the p < pi case in (5.321). We recognize each side of (5.322) as a complex Fourier series on (0, 2,rr). We equate coefficients and rearrange to give .
211' . . , : loe-int/Jo H(2) (kpo) ,1,1) -lnt/J dA,I '-n Jsz (a, '+' e '+' (2)
1 o
233
Pcrfeclly Conducting Circular Cylinder
I
aHn (ka)
We shall use this result to find the el~ctric field E z by noting that, in (5.318),
I
2 00 Hri ) (kip - Pol) - n~oo ein(t/J-t/Jo)
)!
. In(ka) H(2)(k )H(2)(k PO P H,~2) (ka) n n . (5.325) which is the classical ¢-directed eigenfunction expansion for the electric field [15]. In the above example of the circular cylinder, we were able to invert the integral equation analytically. This event occurred because the surface Sc was a coordinate surface, in this case, p = a. In cases where the cylinder does not conform to a complete coordinate surface, the integral equation in (5.315) must be inverted numerically. We include a specific case, th~ rectangular cylinder, in the problems.
5.9 PERFECTLY CONDUCTING CIRCULAR CYLINDER In the previous section, we derived the fields associated with scattering from a perfectly conducting circular cylinder by beginning with the conducting cylinder of arbitrary cross section. We obtained an integral equation in (5.319). For the case of circular cross section, we were able to invert the integral equation and obtain an expression for the electric field E z in (5.325). It is, however, possible to proceed more directly. In this section, we derive the fields scattered from a perfectly conducting cylinder of circular cross section when the excitation is an electric current line source. We are able to verify the result obtained in (5.325). Next, we obtain an alternative representation, useful in describing scattering in the form of creeping waves. We again consider the geometry in Fig. 5-11. The source is given explicitly in (4.163) and the differential equation describing the Ez-field in (4.180) and (4.181), which we repeat here for convenience, viz.
~ [~ p
(p ag )] ap ap
2
+ ~ a g + k2g = _ lJ(p p2 a¢2
g= -
Ez i(j}/lOfo
p')lJ(¢ - ¢'). p
(5.326)
(5.327)
234
Electromagnetic Boundary Value Problems
Cha~., 5
From the results in Problem 3.2, we may expand the Green's function g! in terms of the spectral representation with respect to ¢, viz. .
Sec. 5.9
Solving for C and substituting into (5.336), we have
I
A [In(kp) -
g ( p,
..
;."
'f',
P
I
,/,1) _
,'f'
-
~ nf:::oo an
: (
I
,/,1)
p, p ,'v
in¢ yfl 2;e
235
Perfectly Conducting Circular Cylinder
I
b, =
I
(5.328)
C n HP)
p < pi
(kp)] ,
B H,~2) (kp),
(5.337)
P > pi
where We write this transformation
Cn
In(ka)
=
(5.329) I
and easily find that
'.
a2 g - a¢2
o(¢ -
,
2
{=::}
~
,
a
,,/,I
_e- lI1 ", 2rr
= pi allows us to evaluate
Invoking the continuity and jump conditions at p the coefficients A and B, as follows:
rr A = _H(2)(k p ')
(5.330)
II an
(5.338)
H,F)(ka)
2i
B = ~ [In(kp') -
(5.331 )
(5.339)
n
(5.340)
C n H,;2) (kpl)]
where we have used the Bessel function identity
Applying (5.329)-(5.331) to (5.326), we obtain
J H(2)' - J 'H(2) = ~ n n n n irr x
(5.332)
I
(5.341)
where the prime indicates differentiation with respect to x. Therefore, where
_jl;l
rr
- e -in¢' bn 2rr
an -
(5.3~3) I
:
So far, the development is identicaI"to that in (4.190)-(4.198), except that tj1e boundary and limiting conditions are now
bnl:
n
b = 2i
= 0
(5.33,4)
lim'bn =0
(5.3~5)
H~2)(kp') [In(kp)
H,~2)(kp) [In(kpl) -
g=
:i n1:;oo
ein(¢-¢') [
+
,
We write the solution for bn as a linear combination of Bessel and Hankel fhnctions, as follows: I
p < pi
The limiting condition in (5.335) results in D = O. At p = a, we have:
A.ln(ka)
+ C 1I,;2) (ka)
= 0
-cnHP)(kp)H~2)(kp')
I
H,;2) (kp')Jn(kp),
p < pi ]
H,~2) (kp) I n(kp'),
p > pi
(5.343)
Use of (5.327) and the addition theorem given in (4.207) again produces the result in (5.325), which we display here for reference, viz.
(5.336)
P > pi
(5.342)
p > pi
cnHP)(kpl)] ,
Substitution of (5.342) into (5.333) and the result into (5.328) gives
p=a p-400
p < pi
- cnH,;2)(kp)] ,
Ez(p) = - wflo1o 4
I
Hci 2) (kip - Pol) -
t
n=-oo
ein(¢-t/Jo)
)!
. In(ka) H(2)(k )H(2)(kPO P H,;2) (ka) n n
(5.344)
236
Electromagnetic Boundary Value Problems
C~ap. 5
It is instructive to consider the important special case of a plane wave incident on the cylinder. In (5.344), the first term in the brackets is the incident field, given by , Einc
z
= _ woJiolo H(21(kl
4
0
P
_
1\)
P
(5.345)
We expand Ip - pll in cylindrica(coordinates and obtain
Ip - p I I =, [2 p
+p
12
i
I
-,2pp cos(¢ - ¢I)
]
1/2
: ; (5.34~)
iThe plane-wave case is produced by allowing the line source to be very f~r removed from the cylinder. Mathematically, pi > > p and !
';1
, Ip~ p/l ~ p' [ 1- ( 2
)
Sec. 5.9
In obtaining the fields for T M propagation between parallel plates in Section 5.5, we found that there was an alternative representation for the Green's function, useful at high frequencies. In the case under consideration here, we may again obtain a useful alternative representation. We begin by writing the differential equation describing the Green's function in (5.326) in the form that separates the p-operator from the ¢-operator. This form is given in (4. I87)-(4.189) and is repeated here for convenience, viz. (5.352) (L p + L,p)g = p8(p - pl)8(¢ - ¢I) where L
= _ iW/Lo/o 4i
z
j
I
(5.347)
2i
e-ik[p'-pcos(,p-,p'l]
rrkpl
z
= _ iW/LOloj 2i e-ikp' e-ikpcos,p 4i rrkpl
iw/L010 4i
j
2i rrkpl
e_ikP,]-1
(kp)2
(L p
-
(5.348)
(5.349)
(5.350)
(5.351)
(5.354)
= - a¢2
A)G = p8(p - pi)
(5.355)
GI
=0
(5.356)
lim G = 0
(5.357)
p=a
p~oo
The reader should carefully compare this problem to the problem in Example 3.6. The only difference is in the boundary condition at the lower end of the interval., In Example 3.6, we had a finiteness condition at p = O. In this case, we Have a Dirichlet condition at p = a. We still have the limit I point case as p ~ 00, but the condition at p = a is regular. We write the solution as
G=I
AJv(kp)
+ C HS21(kp),
BHS 21(kp)
+
DHS1)(kp),
p < pi
(5.35In P > pi
where \!
so that
(5.353)
We require the spectral representation of the operator L p • The Green's function problem associated with this spectral representation is
To produce a unit magnitude plane wave from left to right, we adjust the intensity 10 as follows: '
10 = [
[~ ap (p~)] ap -
a2
iWe let the incident wave arrive froni left to right along the x-axis (Fig. 5-11), so that ¢I = rr and Einc
= -p
L,p
:where we have discarded terms in p/ pi higher than first order, and wher~ we :have used the first two terms in the Taylor series expansion for ../(1 + x). !Substituting (5.347) into (5.345), and using the large argument approxi~a'tion for the Hankel function given in Example 2.21, we obtain ' E inc
p
cos(¢ - ¢I) ] 1/2
~ pi _ pcos{¢ _ ¢I)
237
Perfectly Conducting Circular Cylinder
= i.fi.
(5.359)
in the same manner as in (3. I36). Application of the limiting condition in (5.357) results in D = O. From this point, the solutiorl for the Greeti's
C~ap. 5
Electromagnetic Boundary Value Problems
!
;function follows the development in (5.337)-(5.342). The result is
I' H~2)(kp') [Jv(kp) ~ HS 2)(kp) [JV(kpl)
G = 2i
C vHS
2 )(kp)] ,
~ cvH~2)(kpl)] ,
p < pi p > pi
I
2i
iv'A
[J.,v'A (kp) _
lim
(5.360)
where the brarich cut in ..;).Iies along the positive real axis and is expliCitly (determined byI (3.143). i , ' O u r next step is to determine the spectral representation of p8 (p ! pi) by integrating the Green's function with respect to A by the methods developecl,in Chapter 3. We first consider the case p < p'. We have G = !!...-H(2) (kp')
Jiv'A(ka) H(2) (k )] H(2) (k) iv'A P iv'A a
239
We substitute (5.365) into (5.363); in addition, we substitute (5.366) and (5.367) into (5.362). After some routine algebra, we find that
!
!
Jr
Perf~ctly Conducting Circular Cylinder
Sec. 5.9
(5.361 )
(5.368)
G = lim G
Therefore, there is no branch cut in G along the positive- real axis. Since G has no branch cut singularities, the spectral representation of the delta function is given by (3.39), viz. I -p8(p - p) = -I.
2Jr/
f
(5.369)
G(p, p I ,A)dA
where the only possible singularities in G are poles. Our analysis of the pole contributions is based on the treatments in [17] and [18]. We write the expression for G given in (5.361) as (5.370)
G = Gl +G2 ,
where we have used (5.338) and (5.359). We define the branch cut associated with..;). by using (3.143) and (3.144), and produce a cut along the positive-real axis in the A-plane. Following the development in Example 3.6, we now investigate whether tpe branch cut in ..;). produces a branch cut in G. As we approach the positive- real axis from above and below, we have, r e s p e c t i v e l y , ' ,
where (5.371)
02 =
_!!...-
(5.377)
Jiv'A(ka) H.(2) (kp)H.(2) (kp')
2i H(2) (ka)
iv'A
,v'A
,v'A
There are no poles contained in G 1; there are, however, a countably infini,te number of simple poles [18] in G2 whenever
H/~(ka) = 0
(5.373)
Therefore, by Cauchy's Theorem, only the second term. in (5.361) contributes to the contour integral in (5.369), and we have r = i IA1
12 /
(5.364)
But, from [16], we have
p
8(p - pi) =
f
-~
4
iv'A
'
Using the residue theorem, we obtain J = r
~2,[N(I) + H(2)] r r
(5.365)
p8(p - pi) =
I
Jr.
~ fi:(ka)H.(2)f1(kp)H.(2~(kp')Res
2/ L
r=l
!
J.
ly~p
'yAp
where i (5.3 67) 1
,I
[+); H
iA
Ar ]
(5.375)
!
and, from (3.149),
'yAp
,
(5.366) l'
(5.374)
Jiv'A(ka) H.(2) (kp)H.(2) (kp')dA H(2) (ka) 1v'A ,v'A
Res[f(z); z]
~40
Electromagnetic Boundary Value Problems
Chap. 5
signifies the residue of f(z) evaluated at Z, and where the sum is over the zeros evaluated in (5.373). Using the relationship between A and 11 in (5.359), we obtain I
Sec. 5.9
241
Perfectly Conducting Circular Cylinder
We substitute into (5.381) and find that (5.384)
00
po(p - p') =
:L ¢p(kp)¢p(kp')
(5.376)
1'=1
Using (5.327), ;we produce the electric field
where
1 2
¢p(p) =
Jv(ka) rri I' p. .2.- [H S2) (ka)]
1 /
V
I
av
H(2)(kp)
(5.377)
v"
""
Expression·(5.376) gives the required spectral representation for the delt~ function. We note that the result is symmetric with respect to p and, p~. Therefore, the restriction p < p' can be removed. Using the methods in Chapter 3, we may develop the spectral iep:resentation in (5.376) into a Fouri~r expansion useful for solving (5.3~2). For f(p) E .c2(a, 00), we have ' i •
1
1
00
f(p) =
00
j(p')o(p - p')dp' =
f(p') [p'o(p - p')]
dp~'
(5.378)
- ' ; I ~ cos[vp(I - <1>'1- rr)]", (k )'" (k ') E z - !W/lO 0 L 2' '+'p P '+'1' P VI' sm vprr : 1'=1
We again specialize to the case where a plane wave is incident from left to right along the x -axis (Fig. 5-11). Let <1>' = rr,
(5.386)
-rr < ::::: rr
so that (5.387)
cos[vp(I - <1>'1 - rr)] = cos vp
Let p' become large enough so that the Hankel function can be approximated by
a I Since p' 0 (p - p') = po (p - p'), we may substitute (5.376) and obtain
,
00
=
f(p)
L (Xp¢p(kp)
00
,
'
(XI' =
1
• dp f(p)¢p(kp)-
P We next use this Fourier expansion to solve (5.352). Let a
g =
'
00
,
1'=1
'
L (Xp(p', <1>, ')¢(kp)
(5.388)
i i
(5.379)
Then,
(5.380)
2 W/Lorrl0R£;i -'k' ~. J v" (ka) HS )(kp) cos VI' - - e I P L! v" ---:._-=--!:-P _ 2 rrkp' 1'=1 aav [HS 2\ka)]t'p sin vprr (5.389) To produce a unit plane wave incident, we use (5.350) and produce
1'=1 where
(5.385)
Ez =
I
(5.3~1)
I
(5.390)
Substitution into (5.352) gives
(dd;2
+ V~) (Xp
=;= -¢p(kp')o( - <1>')
(5.3~2)
This Green's function problem with periodic boundary conditions has ,been solved in Problem 2.18. The result:applied here is I
(XI' =
cos[vp(I - <1>'1 - rr)] , . ¢p(kp ) 2vp sm vprr
(5.383)
The reader may wish to compare this result with [17, eq. (129)] by using (5.365) and (5.373). In [17], James has used the classic residue series approach to produce the representation for the electric field that we give in (5.390). We have used the alternative spectral representation, a method also used in [18]. The alternative spectral representation is useful for obtaining solutions at high frequencies where summing the series in (5.344) requires a large numher of terms for convergence. The alternative representation is
242
Problems
Chap. 5
particularly suited for the so-called:shadow region [19] behind the cylinder, away from the side directly illumi~ated by the incoming plane wave. H~re, the field is givyn in the form of cr~eping waves [17], [20]. For a thorough discussion of the zeros of the Hankel function needed in (5.373), the rdder is referred to [18] and [21].
Chap. 5
243
Problems
5.2. Show that the solution in (5.72) satisfies the differential equation in (5.55); 5.3. Beginning with the T E z equation set in (5.109)-(5.111), derive a modal seri,es dual to the modal series describing the T M z modes in (5.138).
5.4. Using the Green's function method, show that (5.199) is the solution to (5.196) with the boundary conditions in (5.197) and (5.198).
5.5. In the problem describing the scattering from a perfectly conducting cylinder 5.10 DYADIC GREEN'S FUNCTIONS I
J'
.
In the electromagnetic problems in this chapter, the geometry and soJrce in each case have been independent of one coordinate dimension. niese two-dimeill;ional problems have been chosen as models to illustrate :the use of spectral expansions and Green's functions. Indeed, many of ithe interesting and useful problems in electromagnetic theory can be mod~led in two spatial dimensions. Additional two-dimensional examples directly using the methods developed in this book can be found in [7]-[9]. I . There are, however, many electromagnetic problems where it is !not feasible to assume that the problem is independent of one spatial dimension. In these three-dimensional cases, the analysis in this book may be dire6tly and elegantly extended using a dyadic form of Green's theorem. The dyadic method is presented in detail in the book by Tai [22]. Dyadic analysis is based on the formulation of dyadic spectral ~ep resentations of the delta function ahd the derivation of problem-dependent dyadic Green's functions. The interested reader is referred to [22] !for a description of the procedures, as well as application to some classical problems, such as waveguide propagation, scattering from cylinders, hnd interactions with plane stratified media. In addition, the book by Collin [23] provides a logical, systematic presentation of dyadic Green's functibns and their use in electromagnetics. . Dyadic analysis can be appHed to boundary value problems where the solution depends on inverting an integral equation. In these cases, the reader is cautioned that the analysis of the singularities associated with dyadic kernels in integral equations is a delicate matter. For a discussi6n, : ' i I the reader is referred to [23],[24]. ' I .
given in Section 5.8, assume that the cylinder cross section is rectangular. For this specific case, specialize the expression for the electric field in (5.318) and the fonn of the integral equation in (5.319).
5.6. Consider a y-directed magnetic current source My above an impedance plane (Fig. 5-12). Let the current source be independent of y, so that
a ay
-=0
Assume that the boundary condition at the impedance plane is given by z
lim (E ) = icvL Hy
x-->o
where L > 0 is the inductance of the impedance sheet. (a) Show that the only nonzero field components are Hy , Ex, E z· (b) Using Green's theorem, formulate an expression for the magnetic field H v . Solve explicitly for the Green's function by two methods: I. Use a spectral expansion in z, followed by a closed form solution in
x. 2. Use a spectral expansion in x, followed by a closed form solution in z. Note: This spectral expansion utilizes the impedance transform derived in Chapter 3. (c) Specialize the solution to the case where the magnetic current source is a line source on the x-axis at a distance d above the impedance plane. x
PROBLEMS
5.1.
Using the Green's function method, show that (5.51) is 6the solution to (5.49) with the boundary conditions in (5.50).
Impedance plane
Fig. 5-12
~z
Magnetic current M" above an impedance pl~ne.
,
I !
References
244
Ch~p.5
REFERENCES [1] Dennemeyer, R (1968), Introduction to Partial Differe/?tial i Equations. New York: McGraw-Hill, 92-93. [2] Stratton, J.A. (1941), Electromagnetic Theory. New Yprk: I McGraw-Hill, 165. .. I [3] Ramo, S., J.R. Whinnery, and T. Van Duzer (1984), Fields and Waves in Communication Electronics. New York: Wiley, 396-399. [4] Cheng, D.K. (1989), Field and Wave Electromagnetic~. Reading, MA: Addison-Wesley, 534-5,39. [5] Felsen, L.B., and A.H. Kam~1 (1981), Hybrid ray-mode formula~ion of parallel plate waveguide Green's functions, IEEE Trans. Antennas Propagat. AP-29: 637-649. ; , [6] Wilton, D.R, and C.M. Butl~r (1990), Effective methods for solving integral and integrodifferentiill equations, Electromagnetics 1: 289308; reprinted in Hansen, R.t. (Ed.), Moment Methods in Antennas and Scattering. Norwood, r;vt:A: Artech House, 58-77. !, . [7] Zhang, S.1., and D.G. Dudley (1987), Scattering by a radiallinejoi'nep to a coaxial waveguide, 1. Electromagnetic Waves and Appl. I: 299'-321. I [8] Lee, R, D.G. Dudley, and K.F. Casey (1989), Electromagnetic qoupiing by a wire through a circular aperture in an infinite planar scrben, J. Electromagnetic Waves and Appl. 3: 281-305. I [9] Wright, V.B., D~G. Dudley, ~nd K.F. Casey (1990), Electromagnetic coupling by a wire through a cavity-backed circular aperture iri an infinite screen, J. Electromagnetic Waves and Appl. 4: 375-400.: [10] Collin, R.E. (1991), Field Theory of Guided Waves, 2nd i edition. New York: IEEE Press, 811-821. . , . I ![ 11] Mlttra, R'., and S.W. Lee (1971), Analytical Techniques in the Theory of Guided Waves. New Yor~: Macmillan, sect. 1-3. [12] Butler, C.M. (1984), Current induced on a conducting strip which resides on the planar interfac~ between two semi-infinite half-spaces, IEEE Trans. Antennas Propagat. AP-32: 226-231. i[ 13] Dudley, D.G. (1985), Error rrlinimization and convergence in numer, ical methods, Electromagnetfcs 5: 89-97. [14] Butler, C.M., and D.R. Wilton (1980), General analysis of narrow strips and slots, IEEE Trans. Antennas Propagat. AP-28: 42-48. I
Chap. 5
References
245
[15] Harrington, R.F. (1961), Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill, 232-238. [16] Abramowitz, M., and LA. Stegun (Eds.) (1964), Handbook of Mathematical Functions, National Bureau of Standards, Applied Mathematics Series, 55, Superintendent of Documents, U.S. Government Printing Office, Washington, DC 20402, 358, eq. 9.1.3 and 9.1.4. [17] James, G.L. (1980), Geometric Theory ofDiffraction for Electromagnetic Waves. Stevenage, UK: Peter Peregrinus, 74-76. [18] Felsen, L.B., and N. Marcuvitz (1973), Radiation and Scattering of Waves. Englewood Cliffs, NJ: Prentice-Hall, 326-327, 710-718, [19] Ishimaru, A. (1991), Electromagnetic Wave Propagation, Radiation, and Scattering. Englewood Cliffs: NJ, Prentice-Hall, 330-333. [20] Franz, W. (1954), Uber die greensche funktionen des zylinders und der kugel, Z. Natwforsch 9: 705-716. [21] Wait, J.R (1988), Electromagnetic Radiation from Cylindrical Structures Stevenage, UK: Peter Peregrinus, 54-60. [22] Tai, c.T. (1993), Dyadic Green's Functions in Electromagnetic Theory, 2nd edition. New York: IEEE Press. [23] op.cit. Collin, chapter 2. [24] Johnson, W.A., A.Q. Howard, and D.G. Dudley (1979), On the irrotational component of the electric Green's dyadic, Radio Science 14, 961-967.
247
Index
Index A Addition, rules for; for vectors, 2 Addition Theoremfor the Hankel I ' , function, 165 Adjoint boundary conditions, 69 Adjoint Green's function problem, 57 Adjoint operator, 35, 45, 54 Alternative repres~ntation, 150-51, 155-56, 158-59, 164, 166, ! 171-72, l7fl-77, 204, 237 Aperture diffraction, 216-26 Approximate operator equation, 33 Approximation theory, 18
B I.
BasIs, 5 Bessel's equation, 51, 79 Bessel function expansions, 86 Best approximation, 19-24,42
c Cauchy convergence, 13, 15 Cauchy-Schwarz-Bunjakowsky inequality, 8-9 Cauchy's theorem, 113 Chebyshev polynomials, 41, 226 Collinear vectors, 3 Complete normed linear space, 13 Components of vectors, 2 Conjugate adjoint, 70-73 boundary conditions, 70 Green's function, 72 Conjunct, 55
Continuity condition, 62 • Continuity of inner product, 12 Continuous spectrum, 127 , Convergence, 12-13,31-32 Cauchy, 12 in energy, 31 weak, 32 , Creeping waves, 233, 242 Cylindrical shell source, 166-68
I
o : Delta function, 45--49 spectral representation of, 107 transformations, 139--43 Dimension,S Dirichlet boundary condition, 183, 237 Dirichlet problem, 183 Discrete spectrum, 127 Distributions, theory of, 48 Domain, 25 Duality, principle of, 178 Dyadic analysis, 242
E Eigenfunction-eigenvalue method. See spectral representation method Eigenfunctions, 99-105 I improper, 115 Eigenvalues, 99-105 improper, 115 Electromagnetic boundary value problems, 181-82 aperture diffraction, 216--26 dyadic Green's functions. 242
iris in parallel plate waveguide, 206-16 parallel plate waveguide, 198-206 perfectly conducting circular cylinder, 233--42 scattering by perfectly conducting cylinder, 226-33 SLPI extension to three dimensions, 182-90 SLPI in two dimensions, 191-94 SLP2 and SLP3 extension to three dimensions, 194-98 Electromagnetic model, 144--46, 146 time-harmonic representations, 145--46 Electromagnetic sources, 139 cylindrical shell source, 168-72 delta function transformations, 139--43 line source, 153-66 point source, 172-77 sheet current source, 147-52 Energy inner product, 31 Energy norm, 31 Euclidean space, 2 Expansion functions, 33
F
I
Formal adjoint, 5~ Formally self-adjbint, 55 Fourier-Bessel transform, 122, 137, 171 of order one, 170 Fourier coefficients, 19 Fourier sine series, 4 Fourier transform, 119
G Galerkin's method, 34, 35 Galerkin specialization, 36 Generalized function, 48 Gram matrix, 23 '
Gram-Schmidt orthogonalization process, 16-17, 34 Greatest lower bound, 36 Green's functions dyadic, 242 and spectral representations, 134-35 Green's function method, 45 delta function, 45--49 Sturm-Liouville operator theory, 50-52 Sturm-Liouville problem of the first kind, 53 Sturm-Liouville problem of the second kind, 68-77 Sturnl-Liouville problem of the third kind, 77-94 Green's function problem. 57 Green's theorem, 184, 195
H Hankel function, 156, 159,226 addition theorem for, 165 asymptotics for, 86 Helmholtz equation, 52 , Hilbert-Schmidt operator, 26 Hilbert-Schmidt property, 26 Hilbert space, I, 15-19 operators in, 24-33 Homogeneous boundary condition, 53" 69 Hybrid ray-mode formulations, 206
I
Impedance transform, 131, 243 Improper eigenfunction, lIS Improper eigenvalue, 115 Infimum, 36 Inhomogeneous boundary condition, 53 Inhomogeneous Dirichlet boundary condition. 183
248 Irhomogeneous Neumann boundary , condition, 183 Initial conditions. 54 Inner product sp~ce, 7
j ,
Jump condition, 62
K Kantorovich-.!;-ebedevoperator, 162-63 Kantorovich-Lebedev transfonn, 126, 137, 165-66 kth-order impedance transfonn, 131
L Laplacian operator, 195 Least squares, method of, 36 Lebesgue theory, 16 Legendre polynomials, 18 Legendre's equation, 52 Limit circle, 78 Limit condition, 82 Limit point, 78 : Linear analysis, 1 best approximation, 19-24 Hilbert space, 15-19 inner product space, 7-10 linear space, 1-7 method of moments, 33-36 nonned linear space, 10-15 operators in Hilbert space, 24-33 proof of projection theorem, 36-38 Linear combination, 3 Linear dependence, 3,4 Linear independence, 3,4 Linear manifold, 16 Linear space, 1-7 nonned, 10-15 Line source, 153-66 Loss tangent, 149
Indp
M Maxwell's curl equations, 156 Maxwell's equations, 147 Method of Least Squares, 36 Method of Moments (MOM), 33-36j 216 : Mixed problem, 183 Modal coefficients, 104 Multiplication, rules for, for vectors, 2
N Natural modes, 104 Neumann problem, 183 Neumann's number, 103-4 Nonnegative operator, 30 Nonself-adjoint Green's function problem, 64, 74 Nonself-adjoint operators, 128 Nonsymmetric Green's function case, 197 Nonned linear space, 10-15 Nonn induced by the inner product, 10
o Operator, 25 bounded, 25 continuous. 27 differential. 28 Hilbert-Schmidt, 26 in Hilbert space, 24-33 right shift, 25 Orthogonal complement, 21 Orthogonal vectors, 9 Orthononnalized Legendre functions~ 40 Orthonormal set, 4, 9
p Parallel plate waveguide, 198-206 iris in, 206-16 Parallel vectors, 3
Index
249
Perfectly conducting circular cylinder, 233-42 Periodic boundary conditions, 54 Point source, 172-77 Positive-definite operator, 30 Positive operator, 30 Projection, 21, 36 Projection theorem, 21 proof of, 36-38 Proper Riemann sheet, 112, 129 Proper orthogonal set, 9 Pulse function, 46
Singular point, 77 SLPI. See Stunn-Liouville Problem of the First Kind SLP2. See Stunn-Liouville Problem of the Second Kind SLP3. See Stunn-Liouvillt Problem of the Third Kind Snell's law of reflection, 226 Spatial Fourier transfonn pair, 158 Spectral representation method. 99-105 eigenfunctions and eigenvalues, 99-105 .
R
Green's functions and spectral representations, 134-35 spectral representati~ns for SLP3, III spectral representations for SLP I and SLP2, 106-11 Spectral representations of delta function, 107 for SLPI, 106-11 for SLP2, 106-11 for SLP3, 111-34 Spherical Bessel equation, 88 Spherical Bessel function, 88 Spherical Hankel function, 88 Spherical Neumann function, 88 Stunn-Liouville fonn, 50 Sturm-Liouville operator, 45, 50 Sturm-Liouville operator theory, 50-52 Stuml-Liouville Problem of the First Kind (SLP I), 53-67 extension to three dimensions, 182-90 spectral representations for, 106-1 I in two dimensions. 191-94 Stunn-Liouville Problem of the Second Kind (SLP2), 68-77 extension to three dimensions. 194-98 spectral representations for, 106-1 1
Range, 25 , Rational numbers, 13 Rayleigh-Ritz m~thod, 35, 36 Ray representations, 206 Real inner product space, 7 Rectangular box, 187-90 Rectangular cylinder, 191-94 Residue theorem, 106 Riemann sheet, I 12 Right shift operator, 25 Ring source, 168-72
s Scattering by perfectly conducting cylinder, 226-33 Self-adjoint Green's function problem, 65, 74 Self-adjoint operator, 56 formally, 56 Self-adjoint property, 75 Set orthonormal, 4, 9 proper, 9 Shadow region, 242 Sheet current source, 147-52 Simple medium, 146
Index
250
Stunn-Liouville Problem of the Third Kind (SLP3), 77-94 extension to three dimensions, 194-98 spectral representations for, 111-34 Symbolic equality, 48 Symmetric Green's function Case, 197-98 Symmetric operator, 30
T ".
Three dimensions SLPI: extension to, 182-90 SLP2 extension to, 194-98 SLP3 extension to, 194-98. Time-harmonic representatio~s, 143-44 Transpose of a matrix, 23 Two dimensions, SLPI in, 191-94
u Unitary space, 3 Unmixed conditions, 54
v Vectors, I components of. 2 orthogonal, 9 rules for addition among, 2 rules for multiplication of, ~
w Weighting functions, 33 Weyl's theorem, 78 Weyl theory, 88
z i
.
Zeroth-order impedance trar~sfonn, 131 t
