jw ON A CLASS OF INTEGRAL EQUATIONS AND ITS APPLICATIONS TO THE THEORY OF LINEAR ANTENNAS 4 't, r by rDario Castellanos A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the University of Michigan 1964 Doctoral Comnittee: Professor Chl|ao-Min Chu, Chairman Professor La berto Cesari Professor Nicholas D. Kazarinoff Professor Charles B. Sharpe Associate Professor Herschel Weil RL-334 = RL-334

ADDENDUM TO ON A CLASS OF INTEGRAL EQUATIONS AND ITS APPLICATIONS TO THE THEORY OF LINEAR ANTENNAS.

PREFACE 1J / An exact olution of Hallen's integral equation has long been a goal of many researchers in the theory of linear antennas. This thesis presents a method of solution for a class of integral equations which, in / / particular, yields a closed form solution to Hallen's equation. The germ gf the idea leading to the solution is to be found in P. M. Horse and H. Feshbach's "Methods of 4' Theoretical Physcs in the chapter on integral equations. Professor Feshbach was most kind in communicating his solution of the infinite system of equations, and his assistance is kindly acknowledged. The contents of Professor E. D. Rainville's course "Special Functions" form most of the foundation of this thesis, and to him I express my most sincere gratitude. To Professor C. M. Chu, chairman of the doctoral committee, I am especially indebted. His constant encouragemenand guidance provided the continuing impetus for te completion of this works I also w ih to acknowledge the assistance of Professors L. Cesari and N. D. Kazarinoff of the Mathematics Department, and C. B. Sharpe and H. Weil of the Electrical Engineering Department. From them I obtained many valuable suggestions during the preparation of the manuscript. ii

Finally I wish to thank Dr. S. G. Gallo and The Creole Foundation for the financial assistance they provide4during my completion of the doctor's degree. Ann Arbor, Michigan July i964. iii

An error has been brought to my attention by Dr. Olov Einarsson of the Radiation Laboratory, who has pointed out that one cannot hope to solve the system of equations (3.1-7) of Chapter I when a or b is infinite. If one proceeds as indicated in the thesis it will always be found that (-b C = exp(-y2 )H(y)g(y)dy, (3.1-6) is independent of the limits a and b. However, this equation gives us the m'th coefficient in the expansion of the function g(y) a < x, b '0 elsewhere in series of Hermite polynomials; and Cm= / exp(y2)Hm(y)h(y)dy (3.1-12) -00oo is the m'th coefficient in the expansion of 1 (f(s) h(y)=- exp(-isy) ) ds (3.1-11) 27r K(s) -w in series of Hermite polynomials. Since the Hermite polynomials form a complete set it would follow that g(y)=h(y) almost everywhere, which is certainly incorrect. This shortcoming of the procedure can, however, be overcome by operating directly on the system of equations (3. 1-7) and the corresponding one for the equation with infinite limits without making any attempt to eliminate the C's. That is, one has f(k)()= E (-)n anCn+ n=o 1

where b 2 C = eY H (y)g(y)dy m m and 00 (k) (-)naC' -n n+k n=o where /"Io _y2 C' = ee H (y)h(y)dy. m / m '-co From this one can write (-)na C (-) a C n a k n n+k n=o n=o or explicitly z CO-alCl+a2C2-. = C -aC+a2C-... C1-alc2+a2c3-* C 11 C2 23 C2-alC3+a2C4-* = C2-alC3+a2C 4 ' If one multiplies the second equation by (-in7r) 1 (b-a)l ' and adds it to the first equation, this gives (-inT) (-in+r) +aC 0C+ (b-a)l! 2..1 2 2 where a2, at,.. are the new coefficients. We can now multiply the third equation by, (-in7T)2 a 2 (b-a) 2! and add it to the first to give 2

(-in7r) (-in7r)2 -a"C + CO (b —a)l C1+ (b'a) C2-a'3 C3+ ' (b-a) 2! (-in )CT+ (-inr)2 - a" Ct+ = C'+ C'+ ' - a~ C '+. C0 (b-a)' 1 (b-a)22! C2 3 3 It is obvious that by continuing this procedure one can again recover Equation (3. 1-18) 00 2 —1 exp(-s2/4)exp(-b n)rs f(s)ds,Ib 2 2 n27r2 i/ Sp(-y2)g(y)ep (-2n -Ya+ (b)2) dy b-a (b-a) Ca z It should be noted that from an infinite set of relations between the C's and the C' 's we have obtained one relation at the best. It would seem a more satisfactory procedure to begin the above scheme with the k'th equation, i. e. multiply the (k+l)th equation by a certain constant and add it to the k'th. Multiply the (k+2)th equation by a certain constant and add it to the k' th, etc. In this manner one would obtain one such relation for each k and hope to determine a g(y) which would satisfy each of them. I must confess that if I proceed in this fashion I find myself unable to solve the resulting set. It is my belief that because of the form of the solution f(s) 5 -(b-a)2 g(y)= exp(-iy) ) ds s(s 3 2 and its limiting behavior as b -- oo, a -- -oo giving h(y), that this g(y) is at least a part of the complete solution. If an independent proof were to be found, the above expression should appear naturally in the complete solution. 3

Furthermore when applied to the problem of the radiating antenna the solution yields an expansion for the current whose structure is in agreement with the travelling wave argument advanced by Hallen. Quantitatively, however, the solution falls short of b.ing in agreement with experiment. Recently P. C. Waterman in a Technical Memorandum entitled "Exact Theory of Scattering by Conducting Strips" (AVCO Research and Development Division, Wilmington, Mass.) has shown that the wave scattered by a strip J (a;r) is 0 determined by the equation / ikx'cos a (;r))-(-1/2) sia| dkxe H (kR')-S (;r), _ a where S is a linear operator which depends only in the solution for the infinite a p plane AO (O ikx'cos a -(1/2) isin a i kdx'e H (kR') -OD It is seen that in this equation the scattering by the finite strip is described as the solution of an operator equation involving the solution of the infinite plane. This is a confirmation of our general belief that the solution of the equation with finite limits can be expressed in terms of the solution of the equation with infinite limits, Inasmuch as the scattering by a strip is determined by a solution of the equation a H ((kl x-g )J (g) AE a:x a. -)-a DC/cfw Dario Castellanos The University of Michigan Radiation Laboratory 30 March 1965 4

ABSTRACT In this stay a systematic method of solution is presented for a class ofiehn integral equations. The most significant result is a closed form solution, valid under very weak restrictions, for the equations rb f(x) = K(x-y) g(y)y, a < x < b and pb f(x) + g(x)= K(x-y) g(y) dy, a < x < b v Ja - if the interval (a,b) istiite. If the interval (a,b) is infinite the solution is given in series of orthogonal polynomials with explicit coefficients. Integral equations of the first kind whose kernels are generating functions for polynomial sets are also treated, b f(x) = K(x, y) g(y) dy, a < x < b & K(x, y)= (y)X x where n(y) is polynomial of degree n in y.: A general sblution is also obtained for the equation J f(x) = e g(y) dy with I generally complex. The special choice a = O, b = o, and \= 1 leading to Laplace's integral equation is illustrated by two examples.

A striking application of these results is a closed form solution of Hallen's integral equation r a radiating antenna. The current wave is obtained as the superposition of traveling waves predicted by Hallen. The current consists of two parts: a series of TM waves representing waveguide modes inside a hollow tube antenna and a superposition of travelipg waves on the surface of the antenna. One further application of the method concerns the problem of electromagnetic back-scattering from a cylindrical wire. The solution to this problem possesses all the qualitative features one would naturally expect, but contrary to common belief the efirsurfaces are seen to play a major role in the determination of the current distribution. The solution, consequently, affords a very poor representation of the physical facts. The difficulties created by the presence of the end surfaces for the scattering problem do not exist for the radiating antenna when the antenna is assumed to be a hollow tube. I,.

Table of Contents Preface............................................... i jE - 1 Chapter One A Class Section Section Section Section Section of 1. 2. 3. 3. 3. Section 3. Section 3. Integral Equations......................... 1 Statement of the Problem.................. 1 Some Results from fhe Theory of Distributions.............................. 2 4 ~ee Kernel K(x-y).................. -. 4 1. Th9 interval (a,b) is finite........... 4 11. Expression of the Solution as an Infinite Integral........................... 4 12. Expression of the Solution in Series of Orthogonal Polynomials................. 19 13. Expression of the Solution as a Contour Integral...........................27 2. The Interval (a,b) is infinite..........29 Kernels which are Generating Functions or Polynomials Sets....................... 31 he Kernel exp(- Axy)......................41 Chapter Two Section Section 3. 4. Section 5. Radiation from a Linear Antenna........................ ection. Formulation of Hallen' Interal Euat Section 1. Formulation of Hallen's Integral Equation.. 48 48 iv

Section 2. Solution of Hallen's Integral Equation....52 Section 3. Niture of the Solution....................57 j Chapter Three Scattering by a Thin Wire..............................66 Section 1. Formulation of the Integral Equation.....66 Section 2. Solution of the Integral Equation.........68 Section 3. Nature of the Solution....................69 4 Section 4. The Expression for the Back-Scattering Crdss Section......................73 Appendix Fundamentals of the Theory of Distributions..............77 1. V

List of Illustrations Chapter One Figure 1. 1. Figure 2. 1. Figure 2. 2. Figure 3. 1. ontour C............................... 28 Chapter Two Cylindrical Antenna.....................53 Contour r............................60 Chapter Three Orientation of the Incident Z-Vector %fnd the Position of the Cylindrical Wire.............................7 ^;~ ~~~~~~~~~~~~~~~c~6 vi

CHAPTER ONE A CLASS OF INTEGRAL EQUATIONS. geZion 1. Statement of the problem. We wish to consider a class of Fredholm integral equations of the first kind f(x) = K(x,y)g(y)dy a<x<b (1-1) at where f(x) and K(x,y) are given functions satisfying appropriate cor -tions and g(y) is to be determined. f(x) is defined in the interval a<x<b and K(x,y) in the square a<x,y<b. As is well known, the central difficulty in the study of equations of this type is the following: if K(x,y) is continuous, Eq.(1-1) maps the set of all piecewise continuous functions into a more restrictive set, since all functions f(x) obtained in this manner are certainly continuous. If K(x,y) is differentiable every piecewise continuous function and in fact every integrable function is mapped.nto a differentiable function. Hence, in general, the integral equation with continuous f(x) cannot be solved by A continuous function g(y). If f(x) belongs to a more general class of functions we may expect Eq.(1-1) to be solvable only if K(x,y) deviates from continuity in some way. 1

2 In light of the above remarks, the method to be expounded in the sequel possesses a definite advantage. For instanceihen the kernel is of the form K(x-y) it is only necessary in order for the method to be applicable that g(x) be integrable and bounded, provided that f and K satisfy suitable conditions. For other types of kernels the conditions on g(x) become mbre severe, but it is never necessary that the kernel be continuous. 4 The techniques to be presented were developed in vraom ---.. order to solve certain integral equations which azrew of importance in the^ttudy of distribution of currents in linear antennas. Section 2. Some results from the theory of distributions. In what follows we shall have occasion to use some results of the theory of distributions. The pertinent lemmas are stated in this section and are proved in the appendix. Lemma 1. In the sense of convergence of distributions lim j exp(ix() d& = 2w 6(xx - -v Lena2. I2f, I 1 2 f )exp(-x /4t), t>0 2/FF

3 Then in the sense of convergence of distributions 1 fm(x) = 6(x) Lemma 3. If f(x) * g(x) denotes the convolution of two functions f(x) and g(x) f(x) * g(x) = f(&)' g(x-)) d& and if D be a differential operator, then 4 Dftg) = Df*g = f*Dg Lemma 4. If f Tv then f * g+f*g in each of the following cases: a) The functionals fv are concentrated in one and the same bounded set. b) g is concentrated in a bounded set. c) The supports of the functionals f and g are bounded on the same side in a manner independent of v. Lemma 7. With the Fourier transform of the functional f denoted by the symbols f or F[f] we have exp(bx) = 2 6(s-ib) for any co lex b. Lemma 8. Te Fourier transform of the convolution of two functions f(x) * g(x) = f(&) g(x-E) d& is given by F f(x) gCx) f(a)g(a).

4 Section 3. Th_ kernel K(x-y). To obtain1 a solution of the equation f(x) = K(x-y)g(y)dy a<x<b (3-1) a we will consider two separate cases: the interval (a,b) I is finite and the interval (a,b) is infinite. Section 3.1. The interval (a,b) is finite. Section 3.11. %Expression of the solution as an infi eat... integral. We will construct a solution of Eq.(3-1) as a function of the solution of the same equation with both limits infinite. The essence of, the method will be the determination of an entity related to Eq.(3-1), which is independent of the limits of the integral equation, this quantity being then evaluated in terms of the solution of the equation with infinite limits. We will assume that f, K and g satisfy the following conditions ci-1) f(x) ha a Maclaurin expansion (ii-l) f(x) an be continued analytically'for all x and J 2 is such that its analytic continuation belongs to L ( —,-) (iii-l) K(x-y) has a formal (not necessarily convergent) series expansion K(x-y) = n kn(y)x

5 (iv-l) K(x) Acan be continued analytically for all x and is such that its analytic continuation is L(-",). (v-l) K(x) j o(Ix-n) as I|x|l for all n>O. 2 (vi-l) f(s)/K(s) belongs to L ( —,-). (vii-l) g(x) is bounded in (a,b), and the interval can be broken up into a finite number of open partial intervals, in each of which g(x) is monotdnic. 2 a-o (viii-l) exp(-s /if/K is bounded and integrable in every interval (a,b). Conditions (i-l) to (viii-l) will later be relaxed by using,4istributions. In Eq.(3-1) let us expand exp(t2)K(t) in a series of Hermite polynomials( exp(tZ)K(t) =n a H (t) (3.1-1) n-o n n The coefficients an in this expansion are given by a 1 J H (t)K(t)dt (3.1-2) 2nn!i J.. The convergence of this integral is guaranteed by conditions iv-l) and (v-l). Replac4g t by x-y in Eq.(3.1-1),there follows K(x-y) = exg(-y2) n a exp(2xy-xz) H (x-y) (3.1-3) 9 n=O0 n n (2) By use of the generating function2 (-)nH (y) xk exp(2xy-x2)H (x-y) nk (3.1-4) n k=O

6 Eq.(3. 1-3) becomes -. ()nH k(y) xk K(x-y) exp(-y2) z a n+k n=O k= k! Introducing this expansion into Eq.(3-1), and making use of condition (i-1), there follows b f(k)(0) = Z (-)na exp(-y2)H n+(y)g(y)dy < (3.1-5) n=O a Some words ae necessary regarding the validity of Eq.(3.1-5), which was obtained by means of formal series; especially, s-ince similar arguments are to be uset again. It is necessary when studying the summation of series to satisfy requirements of absolute convergence for rearrangement of infinite series and uniform convergence for term by term integration. It is to be remembered, however, that such arguments are only pertinent when one is concerned with the sum of the series. Our only need was for establishing the relationship among the coefficients expressed ~i-Eq.(3.1-5), and the sum of the series did not enter iro the discussion.. To jus$tify these remarks one can replace the infinite series by finite sums which agree with the infinite expressions through the term in xs to obtain b (k) xk s Xk n E f(0)-( = ) x Ez a (-)n exp(-y2)H n (y)g(y)dy. k:=0 k! k=O k! n=O n a

7 Hence 0 f(O) = a (-)" exp(-y2)Hnk(y)g(y)dy, k=0,1,2,...,s. a As s is arbitrary it follows that the above relation is true for all k. Since the use of formal Series leads to correct results in the case considered above, we shall continue, when,4 necessary, to use them in what follows leaving the arguments in the present section as a guide to any further proofs...t Returning now to Eq.(3.1-5), we let b Cm = exp(-y2)iH (y)g(y)dy (3.1-6) a and Eq.(3.1-5) takes the form f()() ( = (-)n a Cn+k; for each k. (3.1-7) n=0 n n k Eqs.(3.1-7) are a set of equations determining Cn. This system of quations was solved by Professor Feshbach in order to slve Eq.(3-1) with a —, b. His solution Ai reproduced here by permission. Let us write out equations (3.1-7)

8 f(O)Y-= C - aC, f() (0) = C1 - aLC2 f(2) (0O = Cz - aC3 - -.I + a2C2-. ~ + a2C3-*. * + aZC4-.... In the above ao was set equal to one. To find CO one multiplies the expression for r)(0) by a, and adds the first two equations. This 4 automatically Eliminates C1. This gives a new coeftSrreirrt for C2 which can beeliminated by multiplying the expression for f (0) by a suitable coefficient. In this manner one can successively eliminate all the C's except Co. A similar procedure Obviously works for any Cn. Each Cn is thus obtained as a linear combination of f() () C = z f ( On+s ) T n s=0 (3.1-8) where the coefficients T are given by S r2+ 4+.* (rl+r2+r3+..,,) r) r2 Ts = Z (-) 1 a a2 a3... (3.1-9) r1! r2! r3!.. summed over all combinations of the integers r. such that 2. rl + 2r2 + 3r3 + 4r4 +... = s.

9 It is seen that the C's are determined only by the values of the derivatives of f(x) at the origin _a J e cojficients a of Eq.(3.1-l), but neither of these depends on the limits of the integral equation. In other words, we will obtain the system of equations (3.1-7) regardless of the limits in Eq.(3-1). Because of conditions (ii-1) and (iv-l) we see that we can determine C without carrying )ut the difficult inversion (i.1-8) to (3.1-9), provided that we can solve the integral equation f(x) = K(x-y)h(y)dy -o<x<- (3.1-10) This equation has been considered by Titchmarsh(4). Titchmarsh proves the following theorem Theorem 1. Let f(x) belong to L2( —,-), and K(x) to L( —,-). Then in order that there should be a solution h(x) of L2( —,-), it is necessary and sufficient that T/Rz should belong to L2( —,-). Sinc the conditions of theorem 1 are satisfied by hypothesis it follows that h(x) is given by h(x) -- j iexp(-isx) ds (3.1-11) 2w S) h~xf

10 and Cm H m xp(-yZH(y)h(y)dy. (3.1-12) In order to express Cm as a single integral let us expand exp(-isx) in a series of Hermite polynomials. We have i exp(-isx) = I,. n=0 '()ninsnxn n! 4 (3.1-13) aslNsu,,''.. but() bu t n/2] n! H 2(x) x E - - k=O k! 2n(n-2k)! e3. 1-1.4) where [n/2] denotes the greatest integer in n/2. Replacing (3.1-14) in (3.1-13) we have [n/2] (-)ninsnn! H (x) exp(-isx) = Z EZ n-2k n=O k=O n! 2 k!(n-2k)! (_)nin( )kn+2k H (x) n=O k 2n+k k! n! 1 exp(-/4) (-)" i" s" H (x) exp(-/4) 2 n n=O 2n n! (3.1-15) and from Eq.(3.1-ll)

11 nf(s) co H (x)7 h(x): n - exp(-s2/4)s ds n=O 2n n! 2w S Ts) so that, usjg the orthogonality of the Hermite polynomials C xp(-s2/4)s' - - - ds (3.1-16) m 2,w ts) This solution is, of course, equivalent to (3.1-8). Combining Eqs. (3.1-6) and (3.1-16) we have 4 (-) im exp(-s 2/)sm f) s ds = exp(-y2)H (y) gy)dy(3.1-17) 2 K -l s Let us multiply both sides of Eq.(3.1-17) by (-inw/(b-a))m/m! and sum over m. This gives 1 exp(-s2/4)exp(-nws)..) ds 2 - J _ b-a k(s) b 2 2 = exp(-y2)g(y)exp(-2n + n dy (3.1-18) b-a (b-a) a where we made use of Eq.(3.1-4) with n = 0. The in rchange of the order of summation and integration tan be justified without difficulty. For the right hand side it follows by condition (vii-l) that g(x) is integrable (6) and bounded. The series (3.1-4) converges uniformly for all y. The interchange is then permissible by a well known theorem of analysis (7). Similarly for the left hand side it suffices to recall that the exponential

12 series cOnverqg uniformly for all values of its argument. Hence by condition (viii-l) and a pertinent theorem(8), term by term i egration is permissible. iH.non t.1-18) can be written in the form < exp(-2i ny ) exp(-y2)g(y)dy b-a a b-a n2r2 exp b-a) / - exp.(.. exp(-s2/4)exp(- ns ) f(s) ds. 2 7 (b-a) b-a K( - This is just tIW Fourier coefficient in the expansioa..f exp(-y2)g(y) in a Fourier series in the interval (a,y) -, g(y)= e)2) -- exp ( 2ni - n2r2 ) exp(- s2 n(rs fs f(s) b-a) b-a (b-a)2 b-a s) Since condition (vii-1) corresponds to Dirichlet conditions (9), the above series is uniformly convergent and it follows that g(y) - e(2) jexp(-s2/4) f(s) 2V'(b-a) - K(s) ex2 exp i2n (.n2F +12i ds 2(b-a) The expression Lin brackets will be recognized as lthe third theta fu tion defined by(10) 3(},q) n — q exp(i2nZ) hence g(y) eXp(y2) exp(-s2/4) 03( a Y T r8 -2 /b-a) )d 2'(r(b-a) -. K(s) "b-a 2(b-a) (3.1-19)

13 eWith the notat ion, = exp(rit), Im(r)> 0, (3.1-20) and 83 (. ) = 83 (1(t) the third theta function is known to satisfy the functional equation(11) 93(Z|r) = ( -T)-1/2 exp- ) (3.1-21) Use of equation 3.1-21) in equation (3.1-19) yields g(y)= - s exp(-is3 (b-a) ( -siy),e (ba2) ds K K(s) (3.1-22) which is the desired solution. We wish now to show that by appealing to the theory of distributions it is possible to relax considerably conditions i-1 to viii-l and still preserve Eq. (3.1-22) as a solution. We will assume, then, that f, K and g of Eq. (3-l) satisfy the following conditions (i-2) f(x) can be continued analytically for all x and is such that its analytic continuation belongs to L (-v,00) (ii-2) K(x-y) h a formal expression K(x-y) = kn(y)xn nlO (iii-2) K(x) can be continued analytically for all x and is such that its analytic continuation isL(- ), c ). (iv-2) K(x) = o((xt -n) as lxt- 00 for all n>O. (v-2) G(y,s) of Eq. (3.1-23) is integrable and bounded in

14 the interval a y<b. Consider the equation I ( -2 f(s) exp(-kx), K(x-y)G(y,s)dy, a< xb (3.1-23) 2y — 0 -,a where fVs) denotes the Fourier transform of f(x) and s is to be interpreted as a parameter. Since the left hand side of E1. (3.1-23) has a Maclaurin expansion it is possible to proceed as before to Eq. (3.1-7) with Cm, exp(-y2) (y)G(y,s)dy (3.1-24V Since exp(-isx) can be continued analytically for all x, we will determine C, by solving the equation 1 f) exp(-isx) - K(x-y)h(y)dy (3.1-25) By use of lemmas 7 and 8 of section 2, we have h(04= F1 f(s) ( 1 L i(S) J (3.1-26) where F-1 dFnots the inverse Fourier transform. It follows then that (F-1 ^ - iS~; | f ~ ) -} f -~s ) ( ) K(>) 2L /% —j ' J L K(d) J2 K(S)

15 =lr. 1 f a)-fl I x)d. -2W7,-) 00 _ _ _ _ 1 f ) ~inx s0 tha t F-i f ~~( d-s x) 4' 2 V (x) dx Li!2f.8 L(LIL *xp(-isx). Ar*) 4 ( 3. 1- L7 The above derivation of gq.(3.,1-27) is based on Parseval's formula for distributions (g,014) =2 v'(f *Y)I which is discussed in the appe —dix. Hence h (x) = fL i4 ep(.'isx). 2v ") By u se o~f 3q. (3, 115) it follow at once that and from Bq. (3. 1- 24) 4 f (s) exp(-s2/4)(-)m gm 2 V -— K'( 8s) b =fexp(-y 2))Jim(y) G(y a) dy a

16 raii.BZq. (3.1-17) we obtain G(y.s)= 1 exp( Y)- (b-) iy ) (3 (3.1-)28 as the solution of Eq. (3.1-23). Let us now integrate both sides of Eq. (3.1-23) with / respect to a from -p to V jb Y 1 /-isx. b= f(s)exp(-isx)dsJ K(x-y) (y,s)ds dy (3.1-29) -Y ~ The interchange of the order of integration follows by Fubini's theorem since the left hand side is integrable. Let t ing V 2 J'G(ys)ds and assuming that the infinits integral r 00 f() J G(y, )dsd a - (a,b) is finite f(x) =J k(x-y) J G(ys)ds dy &a -(> In other words the solution of Eq. (3-1) is

17 2ir - D e 3(b- ).1- iy) ieCp ( -Ob-a))d a which is equa~ on (3.1-22).a It is not dif ficult to extend the previous results to equations of the second kind f (X) + g (X) K(x-Y)g(y)dy,, ~cx~b. (3. 1- 30) Considering g(x) as a testing function vanisthing outside the interval' (a,b) we can writ'e f (X) that + (g (x-Y) i.<3 (y)) (,(x-y),g(y) is f (X) = wx - Y) - (x - Y), g (Y) (3. 1- 31) Lot us consider the related equation ( 2 4t G(y t) dy ( 3 * 1-32) where t is a perameter greater than zero. Lett ing It K, (x-y) = x(x-y) - 1 exp(-(x -Y)2/4) (3.1-33) and assuming that f,, K, and G satisfy conditions i-2 to v-2,, we have from Zq. (3.1-22)

18 -l G(y,t) = exp(-iys) 93 ((b-a)( - y) (b-) ds. 27r Ks- 3 (3.1-34),a Af ti" result exp(-s2t) fexp(isx) ( exp(-x2/4t)) dx in Eq.(3.1-34) gives G(y.,t)= 2- exp(-iys)., (s) e b-a) ( fiy) e- (b-a)2d K(s) -exp(2t) 2 (3. 1 -S) as the solution %f q. (3.1-32). Taking the limit of Eq. (3.1-32) as t approaches zero gives, in view of lmMM 2 f(x) = (K(x-y) - (x-y),g(y)) where g(y) = lira G(y,t) t -3 0 It follows then from Bq.(3.1-35) that the solution of Eq. (3.1-30) il g (y e (-iys)L) 063 (b-a) ( 2 iy), C-(ba) d (3.1-36) which is the required result

19 Section 3.12. Epression of the solution in series of orthogonal polymils. It is posble to obtain the solution of Eq. (3-1) in series of orthogonal polynomials. This form of the solution has the advantage that it is not necessary to obtain the analytic continuation off f(x) in order to determine g(x). One other point in its favor will be seen in the next section. f, K, and g will be assumed to satisfy the folldWing /' conditions (i-3) f(x) has a Maclaurin expansion (ii-3) K(x-y) has a formal expansion K(-y) = kn(y)x" (iii-3) K(x) can be continued analytically for all x and is such that its analytic continuation is L(-m,w). (iv-3) -n K(x) = o(Ix/ ) as Ix-' ~ for all n0O. (v-3) exp(-y2) g(y)/w(y) can be expanded in a uniformly convergent seri4 n=O where j n) is a set of polynomials orthogonal in the interval (a,b) with respect to the weighting function w(y)2 0.

20 Conditions (i-3) to (iv-3) are sufficient to go from Eq.(3.1) to Eqj 3.1-5). With Cm defined by Eq. (3.1-6) we b C f exp(-y2) H(y)g(y)dy (3.1-6) a Multiplying both sides of this equation by (-t)'/2^m! and summing from m - 0 to r, we obtain 4 exp(-y2)exp(-yt-_ t 0 2g (y)dy n'. or ep(2 ) C(-)" tm /b exp(t) <N - 4 m~O 2a ' a Zxpanding the exponentials in ing powers of t we have 's O/2J mrO n=- 2('-2n) n rN-O comparing coe cients Em/ b 'C n ym exp( n-O 2m (-2n)a *a exp'(-yt) (exp(-y2)g(ycdy t in a power series and collect-:(J y exp(-y2)g(y)dy! a (3.1-37) (-y2) g(y)dy We now expend exp(-y2)g(y)/w(y) in a uniformly convergent

21 series ao exp(-y2)g(y)/V(y) - bn. (Y) nj (3.1-38) where n(y) are polynomials orthogonal in the interval (a,b) with respect to the weighting function w(y)>0. The existence of such a set is guaranteed by application of the Gram-Schmidt orthogonalization process to the sequence of functions (w(y)) 1/21 (w(y)) /2y (w(y') /y2.,, ( ~... in the interval (a,b). We shall assume that the set n(Y is comp]O We will assume that the polynomials Pn(y) are given by an equation of the form K (Y) = L O(k, m) y m=0 (3.1 -39) Multiplying both sides of Eq. (31.-37) by q((k,m) and summing over m from 0 to k we find n (k. )Cm-2n n a:- k(y)exp (-y2)g(y) m=0 n=O 20(a-2n): n' a dy 1 (3, substitution o Eq.(3.1-38) in Eq.(3.1-40) yields bk = (km)Cm-2n ml b k m=O n=O 2m(m-2n) n (3. o~~~(.( )C3.1-40).1-41)

22 where gk = a w(Y)k (y)dy a (3.1-42) h pdfyAnomiaisl8 t (y) are such thatif(y) is an even function of y when n is even and odd when n is odd, thenltn(y) is given by an equation of the form Cn/27 01 (y) - O(n,m)y -2m (=n(y) = (3.1-43) m=O ' In this case we hav from Zq. (3.1-37) G/ /2 'i (tm)Ck 2m-n2n(k2m)m): 7Z 2- 2k2m(k-2m-2n). n: m=O n=O b e P k(y) e:p(-y2)g(y)dy If 1 -A44 and substituting, as before, Eq.(3.1-38) in Eq.(3.1-44) we find k/2=O [/2nk nr-=O n=O 2(kam)Ck22n(k-2mn): 2 (k-2m-2n): n: (3.1-45) with gk given by Eq.(3.1-42). We have iown then that the solution of Eq.(3-1) is 1. ' ^ given by =o Ok Lm/2n g(y) =exp(y2 )w(y) I k=O m=O n=O <<k.M).m! CM,-2 n (y) 2k2 (m-2n)!n: (3.1-46)

23 where k k (y) = Z C~(km) y me (3.1-39) are polynomials orthogonal in the interval (a,b) with respect to the weighting function w(y)> O. Cm is given by Eq.(3.1-8). If the polynomial set is such thatln(y) is an even function of y when n is even and odd when n is odd, then with [n/27 C' tn (y) -= B(nm)Yn-2 the solution takes the form g e [k/) k/2-m~ g(y)=exp(y2)W(y) kI = n ' (3.1-43) 1B(km)Ck-2m-2n(k-2m)' k (Y) gk -2^(k-2m-2n):t: (3.1-47) If instead of conditions (i-3) to (iv-3) f, K and g satisfy condit ns (i-1) to (vi-i) and also condition (v-3), then Cm s given by Eq.(3.1-16) 00 Cm ( )mm exp(-s2)(2s)m f(2s) ds - X(K2s) (3.1-16)

24 where we replaced s by 2s. Substitution of this result in Eq. (3.1-46) gives -4 2- 9 d k g(y)=exp(y2)w(y) I > q(k,m)m! (-)mimjj(y) k=O m=O 2mm n m-2n (-)m (2s) exp-s2) f(2s).. (-2n).(2) - ds n'. '(m-2n). K 2s) 4 (3.1- 40 but (12) t/2j Hm(8) = n=O n m-2n (-)m (2s)2n n' (m-2n). (3.1-49) and it follows immediately that r0 exp(-s2)Hm() (2s) ds 2mm: ~y -0O X(2s) is the rth coetficient in the expansion of f(2s)/K(2s) in a series of Hermite polynomials f(2s) k(2s) (3.1- 50) and Eq. (3.1-48) becomes

25 'gy)-e= 2),w(y) T,'4("')'exp(yk=O k z c((k, m) me.( -)n i hin4tk (y) m0Oa% (3. 1 -51) Similarly, by sub~jtituting Eq.(3.l-16) in Eq.(3.1-47) we find g (y) =exp (y 2) w(y) cO [k /2] kaO m=O k Y ()m+k k 0 (ke m) (k -2m) -.. lip 1 2k-m(k-2m)'. C'F [o k/2 -m] 00n=O _ n k2).2)k-2m-2n n! (k-2zu-2n)! 2.f (2s) d dxp- K12 S) (3. 1-52) It but again j- 2m) /] z= H - 2m (S (_) n(k -2mn),(2s) I~2 n~ (k-2m-2n)! and Eq. (3.1-52) becomes g(y)=exp(y 2)w(y) k=O m=O 0 km ( 2).( )m+k.khk-2 Y 4 4 %~('3. 1- 53) where hm is given by Eq.(3.l-50). The possibility of the expansion (3.1-50) is guaranteed by condition (i1*(3

26 In the event that f(x) does not have a Maclaurin expansion as assumed by condition (i-1), one can, at least in ~,rincle, replace f(x) by a sequence f of C functions converging to f; solve the corresponding integral equation for fy and take the limit of the resulting solution. As illustration let a = -1, p = +1. One possible choice fort (y) are the Legendre polynomials (14) 4 P (y) = n k=0 k ' (n-2k) ' (3.1- 54) The factorial defined as function (a)n used in Eq. ( 3,1- 54) is (a)n = a(a + 1) (a + 2)...(a + n - 1), nl and (a) = 1 a X 0. For the Legendre polynomials we have 2 2k + 1, w(y) = 1 m (1- k-2m ' (k,m) -=. -2 m' (k-2m)

27 and Eq. (3.1-46) becomes Ck/23 g(y) 2p3 )9 kk () c +-) (2) k-n 2k ^ hk22mPk(Y) k=0 m=O m: (3.1-55) corresponding to the solution of the equation f(x) = k(x-y)g(y)dy. -1 (3.156) Solutions in series of orthogonal polynomials can also be obtained for equations of the second kind by using the limit process employed to derive Eq.(3.1-36) Section 3.13. Expression of the solution as a contour integral. In many cases it is possible to transform the solution in series of orthogonal polynomials into a contour integral We do this following a method due to Watson. ( The solution of Eq.(3.1) in series of polynomials is given in its most general form by Eq.(3.1-38) g(y) - w(y) exp(y2) n=O bn tn (Y) (3.1-5 7) bn will be assumed such that if n is allowed to vary continuously, then b is an analytic function of n. n

28 Also 'n(y) will be assumed defined when the index n varies continuously and to be an analytic function of n. We will show tha4 in this case 9(y) can be represented by the following integral g(y) = w(y)exp(y2) i f b exp(ir) (y)d 2 C c pin r-a (3.1-58) where C is a contour that starts at a-i& in the z-plane, goes below the real axis to z = - - and then above the real axis to a+ i. C -/2 1 2 3 FIGURE 1.1: CONTOUR C Since bz and B1(y) are analytic functions of z, the only singularities of the integrand are poles at those values of z inside C for which sin r z = 0, those are z = 0, 1, 2,.... Since the contour C is described in the clockwise direct ion, the integral (3.1-58) is equal to the negative sum of the residues at the poles, so that g(y)=w(y)exp(y2) (-2ri) f bn()n tn(Y) 7rcos rr n

29 =w(y)exp(y2) 7 'n(y) n=o which is Eq. (:l-57). Equation (3.1-58) can be used to advantage with the principle of deformation of contours to transform the solution into another one more directly accessible to numerical computation. 4 Section 3.2. The interval (ab) is infinite. Equations (3.146) and (3.1-47) are, of course, also valid if the ~i:erval (a,b) is infinite, provided that g(y) is a sufficiently well behaved function. Condition (v-3) is no longer sufficient to guarantee term by term integration when the interval is infinite, but we will assume g(y) to be such that term by term integration is permissible. One example of the conditions g(.y) may satisfy is given in reference (6), page 178, Theorem III. When a =, b =X we may select for n (y) the Laguerre polynomials Ln (y), (16) orthogonal in the interval (0,X ) with respect to the weighting function y exp(-y), Re(lq) > -1. n k (Y) I -)(l+C)n y n k=0 k (n-k)( 'l+ (32-1) k (3.2-1)

30 For this set we have(17) s = I' _ (A = Ir(1( ) (l-+ )n n! nl w(y) = y exp(-y), Re(() -l 9q(k,m) = (m (1 (+ )k m' (k-m) ' (l+o<)m and Eq.. (3.1-51) becpmes 0c g(y)= exp (y2-y)yq r (l++) K=o k m= m=0 i" k! hm Lk (Y) (k-m) '(l(t)m (3.2-2) corresponding to the solution of the equation f(x) = K (x-y)g(y)dy -O 0<x <( (3.2-3) In particular for~(= 0, Eq.(3.2-2) takes the form oo A k=O m i hmk L(Y) k00 rn g(y) = exp(y2-y) (3.2-4) where ( ) is the binomial coefficient. In both Eqs.(3.2-2) and (3.2-4) h is given in Eq. (3.1- 50r. An equation of the type

31 b f(x) K(xy) g(y) dy (3.2-5) a can be reducedjto the form (3-1) by putting x = exp ( ), y = exp (-? ), and writing f(exp,) * ( ), -exp(- t )g(exp(- Q)) - ( ), and K(exp( - )) ( - ) to obtain b 0(<?) = a K d'\ )+( >1)d where a = -ln ~, b s -In b. A similar set of substitutions can be used for the equation b f(x) = / K(x/y)g(y)dy. (3.2-6) Section 4. Kernels which are generatig functions for polynomial sets. A class of equations directly a men able to the previous techniques consists of those integral equations whose kernels are generating functions for a set of polynomials. k (x,y) n=0 n(y) X (4-1) where (n(Y) is a polynomial of degree n in y. If we substitute Eq. (4-1) in the equation

32 b f(x) = K (x,y)g(y)dy (4-2) we n j f(n) (0) =1 n(y)g(y)dy (4-3) i where we have assumed that f(x) possesses a Maclaurin expansion. Let us also assume that g(y)/w(y) can be expanded in a series of polinomials 7 (y) orthogonal in the n interval (a,b) with respect to the weighting function w(y) > 0. ao g(y) = w(y) E bmm(y) m=O (4-4) Substituting Eq. (4-4) in Eq. (4-3) and assuming term by term integration permissible, we have fn() j 2 bm w(y) n(y) m(y)dy m=O a (4-5) We now appeal to a theorem in the theory of polynomial sets. Theorem 2. Let tn(x) be a polynomial of degree n in x and assume there is one such polynomial for each n. Let w(x) >O on a<x<b. Then a necessary and sufficient

33 condition that the set!An(x)I be orthogonal with respect to w(x) over the interval a<x<b is that b f w(x)xwn(x)dx = 0, k=0,l,..., (n-). This is a well know result. (18) Since Yn(y) is a polynomial of degree n, it follows from theorem 2 that the integral in Eq.(4-5) vanishes identically for m>n, n b f(n) (0) bm J w(y) Fn (y) m (y) dy (4-6) m=0 a This relation can be written in the form f(0) b __ f w(y) V0(y) 0(y)dy (4-7 a (n) I (b() 0) m b m a w(y) n(Y)m) (Y)dy n l w(y) (Y) ^(y) dy bn Giving successive values to n, Eq.(4-7) provides an iterative scheme for determining the b's of Eq.(4-4). If in Eq. (4-6) we let f() (0) = w(y) n(y) (y)dy (4-8)

34 Then as is obvious by inspection of Eq. (4-7) b0 = 1, b = 0 m)l. Hence if we have iterated Eq.(4-7) N times and have J obtainea a set of relations m b = arm f(n)(0), m = 0 1,..., N. (4-9) m n=O then by letting f(n)(0) take the values specified by, 4 Eq. (4-8) we must obtain b 1. 1, bm = 0, ml. If this result is not obtained, a mistake was made in the computations. This device can be used to check the numerical accuracy of Eqs.(4-9). It should be noticed that the integral in Eq.(4-6) w(y) n(y) l(y) dy (4-10) a is g-1 times the mth coefficient in the expansion of 9 (y) in a series of polynomials {m(y), where gm is given by Eq.(3.1-42). Hence in order to do the integral (4-10) it is, perhaps,' the simplest procedure, to expand +n(y) in series of lm,(y) and pick out the coefficients. Such expansions are given for a number of classical polynomials in Professor Rainville's "Special FunctionS." (See References).

35 We shall see now that it is quite easy to solve the system of equations (4-6) when the polynomials n(y) possejs a gene ating function of the type A(t)G(xt) = z n=O (x) tn, X(n) (4-11) where Xis some function of n. We shall assume that A(t) and G(t) have the formal expansions l>CL - A (t) = C( n I n= ris A(tj-1 30 =0 Bnt (4-12) (4-13) G(t) nt G(t) = n=0 ntna Consider the integral equation b f(x) - j a G(xy)g(y)dy (4-14) Multiply both sides of this equation by A(x) A(x)f (x) = b a A(x)G(xy)g (y)dy (4-15) Let A(x)f (x) = / >m) C xm m=O0 m (4-16) then from Eqs. (4-11) and (4-16)

36 b c = f fmn(y)g(y)dy a (4-1-7) Let - ~ n=o (4 -13) Substitution of Eq. (4-18) in Eq.'(4-17) gives in view of theorem 24 n=0 b a W(y)Y)~ which are equations (4-6). Now from Eqs. (4-13) and (4,14) there follows f (n) (0)b ly, n 'a yn g(y) dy (4 - 19) Let the Polynomnialst tk(y) be k of the forms / (y) -.4-1 ni O CX\((k, n) yn (4 - 20) MU It 'ply both sidJes of Eq. (4,z- 19) biy (A(k, n) arid s umi o ve r An I L1K;~lI -= b -, ~-~k n ~ n)(0 I >(y)gSi(y)jy '~ n a (4 - 21) Insertingy Eq. (4-18) in EF-q. (4 -21) we have by virtue- of

37 the orthogonality of the polynomials n(y) bk=L- 2 gk r b a (k, nf ( O) Fn n: (4-22) where w(y) k (y)dy. (4-23) 4 From Eqc. (4-16) and (4-12) we have 00 f(x) = 2A(X)] I (m)CmX n= 0 = (m) mxn+m n= n=O n mr m=O r(m) Cm3n-m xn so that n f(n) () = m=O X (m)Cm n' nn-m (4-24) and Eq.(4-22) becomes bk = - 1 f gk n n! k n =0 m=0 (m) Dn-m 0(k, n)Cm I (4-25) which is the solution of Eqs. (4-6). If the polynomials 7 k(y) are of the form

38 1(y)= ZIP 13(k, n) yk2 ri=O (4-26) t hen we hav e f rom E~q. (4 -19) [k /i nQO Ik-2n (k-2n) b k a Use of Eq. (4-18) gives as b~ef ore 4 b:k = 1 Ii/2 0= ~n) f(k -2 n) (0) ~k - 2n (k-2n) 8 (4 -27) with gkgiven by Eq. (4-23). F ina 1 y, f rom Eq. (4 -24) C/27j k-2n 1 0= m=0 f3 (k, n) '>\ (m) 5k-2n -mCm <k - 2n (4 -28) We have thus shown that the solution of the equation f (X) = a K(x,y)g (y)dy i.(4 -2 9) with K(x, y) == xn (4 -30) and ~~y) satisfying Eqs. (4-11) to (4-13) is given by

39 g0 g(y) = w(y)) n=O bn n (Y) (4-31) where the Ik(y) are polynomials orthogonal in the interval (a,b) with respect to the weighting function w(y)> 0, and bk is given by kL k n b 1 g9 n=0 m=0 \(m) n-m (k,n)f(m) (0) (n (4-32) - if k nn=0 k4 - (4-33) or b = 1 k -- 9k Lk/2) k-2n n=0 m=0 (k.n) k(m) 1k-2a-m f) (2n k-2n if =0II k (Y) = n=O (4- 34) (4-35) P (k, n) y-2n Many polynomial sets have generating functions of the form (4-11). Thus for example, the Hermite polynomials jo 2 2 Hn(x)tn exp(-t 2)exp(2xt)=exp(2xt-t ) -= n(_t n=0 n ( (4-36) and A(n) = 1/nI, 02n = 1/n:. 2n+l = 0, yn=2n/n

40 In Eq. (4-25) replace m by n-2m, then k LD/23 bk L (n-2m) i2m (km ) -n-2m k n=O n= since 2m+l 0 Hence for this case k [n/2i 9k ni m=O 4 o((k,n)n: Cn-2m (n-2m): m! 2" (4-37) which is, of course, Eq.(3.1-41). A similar substitution can be made for Eq. (4-34). For the Laguerre polynomials(19) C0 exp(t)0F1(-; 1+c;: -xt) = (L nt) t" L(+ (1+q)n (4-38) On=( -)"/n, 4=(-)n/(l+c )n n', A(n) =l/(l+ o)n and from Eq.(4-32) k n bk n= m=L yk n-O= m=O (-)m q(k,n) (1+( )n n: Cm (1+C )m (n-m) (4-39) The substitution could just as easily have been made in Eq.(4-34).

41 Other polynomial sets with generating functions of the form (4-11) are m odnomia (20) Berse_ ---i po nomial, 00 t 7 Bn(x) tn exp(xt) = Z - (4-40) exp(t)-l nt0 n* Euler polynomials, (21) 2.p(x). En(x) tn (4-41) -Xp(xt) exp(t)+l n0 n etc. Section 5. The kernel exp( xy). As an incidental result to the derivation of Eqs. (3.1-41) and (3.1-45)we have found an interesting way to solve the integral equation f(x) = { exp(-Axy)g(y)dy (5-1) with the parantterAgenerally complex. We shall assume that f(x) has a Taylor series expansion about some point x0. ~o (n) (x) f(x) = f( (O) (x-xO)n. (5-2) n! Also g(y) will be assumed to have the uniformly convergent series expansion

42 g(y) = w(y)exp(A,-x y)7 n=o b t (y) (5 -3) ~(Yy)~are polynomials orthogonal in the interval1 (a,b) with respect to the weighting function wl(y)70O. In Eq. (5 -l) replace x, by x, + x b f (X+xO) exp(-xy) exp,( -?\x0y) g(y) dy 0 (5 -40 Expan~ding thaL exponential in x in a power series A~d making use of Eq4 -2) we have (b yn exp(-,Axoy)g(y)dy [1 0 (5 5) We shall assume the polynomials 1 Y to be given by an equation of the form k 'tk,(Y) = n N (Ik, n) yn (5 *-6) Multiplying Eq.(5-5) byc~(k,n) and summing over n from I k zA r n=oA = (y) exp(-,A x,,y) g(y)d y a (5 -7) Use of Eq. (5 -3) in Eq. (5 -7) gives finally

43 k bk = 1 gk n=O (-)nO((k,n) f(n) (>) (5-8) so that n q(Y) = w(y)exp(AOxy) z n k) (6- ) n=O k-=O (nk) (-kf(k) Ak gn is the solution of Eq.(5-1) If the polynomial set is such that k(y) is given by an equation of the form 4 Lk/3 n —, f (k,n) yk-2n (5-10) then we modify Eq.(5-5) to read kf( -2 n (x I-2n b a yk-2n exp(-A x y)g(y)dy Y0 and an identical procedure leads to (_)k f(k-2n) (x) (kn) Nk-2n 9k 'L n=O (5-11) The solution of Eq. (5-1) is then g(Y)=w(y)e-xp(,xgY) n= k O )n ( (nk) f(n-2k) (xo) n (Y) - - - -- - - gn "- 2k ---- (5-12)

44 A special case of Eq. (5-1) corresponding to, =i and theAnterval (a,b) finite f(x) = exp(-ixy)g(y)dy (5-13) is of central importance in the study of antenna (22) synthesis. 22 Another special case is provided by the choice Sc -- > = 1, a =0, b = W, which gives Laplace's intergral (23) equation 23 00 f(s) = O exp(-st)F(t)dt (5-14) Selecting as the set /+n(y) the Laguerre polynomials L(n (y) we have from Eq.(5-9) and the results on page 30 ac n F(t) t exp(-(l-snt) f(k) (So) () r( +d) n=O k=O (l+)k: (5-15) as the solution of Eq.(5-14) Suppose for example that m f(s) = (5-16) (s+a)m+l then m 00 k k +m sm (m+l)k ( ) sk (s+a)1 k=O ak+m+l k

45 and f (k+ml) (0)= ak+m+1 (5 -17) Then from Eq. (5 -15) with 0(= 0, so 0 2~ 0 ~'~~k (k) (0) L (t) (1)n~kf I n-i 00 = exp ( -t).4 OD0 2,f (k!) n.* 4 (1) n+k+rm f (k +) (0) Ln+k+m J(-t) (i2 k+M n (0t)Z =exp(t n=0 k=0 (1) n+m f (k+m) (0) Ln+m (t) (1)2 (n -k) l k-fmr With Eq. (5 -17), F (t) takes the f orm 00 n F(t) =exp(-t) 0= k=0 (1) n+mn (m+I) k (- k Ln+m (t) ak+m+l1 (1) k+m (n-k)! (5- 19) but (1) (1) (1+m) k +m m and Eq. (5-18) becomes m+1 F~)=xp at __ (1)n+m Ln+m (t) m! n! n n (-)k a -k k=O k

46 =exp(-t) ill rn+1 a n=O 4, Use of the result (24) (1 ~ n+m (t) (1 - 1/a), m! n n=o (n+m)! Ln~ (t) xn m. n. 11-4' 1-x ') i-x ) -x 4 gives alm+l == exp -( -at) L (at) = exp(-at),F.(-m;l;at) (5 -193) or by use of Kummer's first formula(25) exp(-z) 1F1(a;b;z) = 1Fi(b-a,-b;-z) equation (5-19) becomes F(t) = F1 (1 + m; 1; -,at). Thus if1 de'notes the inverse Laplace transform we have -1 s m (s + a) = 1F1 (1 + mn; 1; - at) (5 -2 Cl) This result is well known. (26)

47 Suppose now that f(s) = - 9 S / (5-21) then with so = 1 Sn f(s + 1) = -.. (s + 1)n, but this is the example we just considered with a = 1. 4, It follows then from Eqs. (5-15) and (5-19) that.t, -%r exp (-t) FCt) I exp (-t) Ln(t) that is '1 = Ln (t), (5-22) n (N 5 -s/f also a known result. (27) We now proceed to consider some applications of the above methods to physical problems.

CHAPTER TWO RADIATION FROK A LINEAR ANTENNA. J Section 1. Formulation of Hallen's integral equation. We will consider the problem of electromagnetic radiation from a perfectly conducting cylindrical antenna excited across a small air gap by an external a-c source. It is well known that the electric field vector can be derived frik a scalar and a vector potential according to the relation 0 E = - V l - iaA (1-1) If the potentials are subject to the Lorentz condition V A + iaU e 0 (1-2) then E can be derived from a vector potential alone 2 2 E ( (V A) (1-3) C where c = 1/=r -e r(r) For the radiated field E the vector potential -(r) A is related to the induced current density by means of the equation (r) _ ', t - C A - 4wd (1-4) r being the distance from the origin of coordinates to the field point, and r' the corresponding distance to 48

49 the source point. We shall assume the z-axis chosen along the axis of the -na, tfe length of which extends from -1 to +1 (Fig. 2.1). At z = 0 the antenna is fed across a small air gap by an external a-c voltage 2V exp(ia)t). Because o of the assumption of infinite conductivity, the current only flows along the surface of the antenna, and we may assume that it isL symmetrically distributed with respect to the axis of the cylinder. The current density vector has thus only a z-component, and the vector potential is parallel to the z-axis at all points. The end surfaces of the cylinder make an exception and should consequently be assumed small. We can, however, surmount the difficulty by assuming that the antenna consists of a tube thus eliminating the end surfaces. On the surface of the antenna E = 0 because of the z assumption of infinite conductivity. Furthermore, for a straight antenna V. A / c A /SE. We have then for z the z-component of Eq. (1-3) 2 _ z 2 + k A = 0 (1-5) 52 z where k w C /u. The solutions of Eq. (1-5) are of the type coskz and sin kz.

50 As the antenna we have considered consists of two halves between which there exists a difference of potential we ehmid not Sxpect the coefficients of cos kz and sin kz terms to be the same for both halves of the antenna. Instead it is the vector potential which is to be symmetrical with respect to the air gap when the antenna is fed at the middle. Consequently we shall put A Z (= cos kz + A2 sin k z )exp(iwt), where A. and A areonstants. For the scalar potential of the antenna we have from Eq. (1-2) iao = - c a A /<)z z i.e., 2 id = c k(A, sin kz + A2 cos kz ) exp(iwt) where the upper sign corresponds to z > 0 and the lower one to z < 0. Hence the potential has two values = + icA exp(iwt) at z = 0. We know already, however, z that there, - + V exp(iwt) and A2 is thus defined. The o- 2 vector potenti4L obtains then the form A (A cos kz - 7V sin k zl) exp(iwot). Z 1 C 0 O It is possible to complement the last term in the parenthesis by a corresponding real cosine term to give an exponential function, the cosine part of which is taken from the first term. A obtains then the alternative z

51 form V A = (A' cos kz + -- exp(-ik Izl)) exp(iwot) (1-6) z C whrm A' is a new constant. The vector potential A may be expressed in terms z of the antenna current with the help of Eq. 1-4). Let K be the radial coordinate in a 'cylindrical coordinate system whose polar axis coincides with the axis of the antenna and with origin in the mid point of -,he tenna. Let "a" be the rpdius of the antenna. The surface current I(z) is then given by the limit of (J 2ra) A P as AD-_e and J ---- - Equation (1-4) becomes a surface integral upon replacing J dv by I(f)d a d$/27r a, where the integration over / has, so to speak, already been performed. For the time dependence exp(iwt) the retardation factor Ir - rt) takes the form exp iw(t - c The magnitude of the vector potential A on the surface of the antenna is given by 2 A =- I() exp(-ikr) d d (1-7) z 4 ( 2d7 r where r is the distance from the source point on the surface of the antenna, to the field point al o on the surface of the antenna (Fig.2.1) r (z- ) + 4a sin2 (1-8) 2

52 From Eqs. (1-6) and (1-7), the integral equation of a cylindrical transmitting antenna fed at the middle is obtained in Fe form 41 _ ei'kr) 4-(V exp(-ikjlz) + A cos kz) =d exe( ) r I 2 0 where A is a new constant and - This is Hallen's integral equation. (28) It has been argued by Hallen that to Eq. (1-9) 4 one must add the boundary condition I () = I(-_) =_0 to specify the constant A. We shall not follow his suggestion for, as we shall see, it is the constant A that is determined by the solution and not the other way around; that is I(') is independent of A. Section 2. Solution of Hallen's integral euation. In Eq.(1-9) replace z by lx and f by ly, then I 27 4v 1 exp-ikrl) -(V exp(-ikllxl)+ A cos klx) dyI(ly) =p d)o -I (2-1) where now 2 Ja 2 1 r= (x-y) + 4 sin (2-2) 2 d2 Multiply both sides of Eq. (2-1) by -2 + (kl), to obtain 4wV 2) d2 2 Y_ 1 _ _ (-2ikl)5(x) = (d + (kl)) dy(ly) exp(lr dx2 ) r / o (2-3) where < (x) is the delta function. The factor (-2ikl) accounts for the jump discontinuity in the first derivative of exp(-ikllx() at x = 0.

153 z 4 z fy FIGURE 2.1 CYLINDRICAL ANTENNA A

54 Let us as in page 14, consider the related equation Trr 2 exp (- ik-r)d exp(isx) = 2 + (k)2) dyG(y,s) dx o x(2-4) where s is a parameter. By letting G(y,s) = 0 for iyl >1 we have by virtue of lemma 3 of chapter one I z7r in 2 1 exp(-ikfr exp(isx) =4V k jdyG(y,s) ( +(k)2) ) (2-5) but this is Eq,(3.1-23) with f(s) = 2r, a = -1, b =_, s replaced by -s an (2-6) -Tr 2 2 a 2 1 ~ 7 2 21 exp(ik (x-y) +4 sin2 K(x-y)-4v ( + ) - - 2 a 2 1 4V x ( x-2 4 - - 0 'x k (x- V) +4 sin ~ - C Z:; In order to satisfy the requirement that K(x) = o((xl n) as f x! -- -n for all n >0, we require k to have a small negative imaginary part, corresponding to the medium having finite non-zero conductivity k = p - iq p,q>0. (2-7) for substitution in Eq. (3.1-28) we need the Fourier transform of Eq.(2-6). (29) To obtain this consider the following integral ( 2 2 t -a --- -I = — b-b F / 2 l b b a 2b2 K ar (t2 + z 2 (-b a + b 2 j (bt) 2 /2v tdt = - K (taZ,, -/ -72 0 1 1 In this equation let V = = a = ik, b = s, z = a and since J l(z) = (2/7 z) cos z, and K (z) = (T/2z) exp(-z), 2 2 there follows

55 00 eXp(-ikR t+ -- 2k) ) Cos st "P — dt = K aS2-(Q )(d vt a a or 22 exp (ist) exp(-i )dt = 2 K a 8 - (ki)2 \t2 + a - OC Hence by Fourier's theorem exp(-ikiV t+ a ) 1 /2 2 2 - a exp(-ist)K as - (k ds t + a Replace t by x-y, a by 2 (a/1) sin 2 6 and integra- with respect to 0 from 0 to 2T 4a 2.. 1 exp(-iki x-y)2 + 4 sin2 2 2 a2 --- - dj ) (x-y) + 4 sin -i o2 zTF_ (2-8) Y. 2 -a n-1 2 2 = 2- j exp(-is(x-y)) {CO 2 0 sin 2 s - (kj) djds Consider now the following integral (30 I (z)K (z) = - K, (2z cos&) cos6edO that is 0 7(,/2 I (z) K (z) = - K,, (2z cos 6 ) c6sptd& (2-9) 1 - 2 2 2 Let 0 = 0, 6 = 2 - 2' z = (a/-) (s2-(kg)2), and Eq. (2-9) becomes I /a 2 - (k )2 K (' / - (kj) ) = T /2 )2' - K(2 - sin s - (kj) ) d$ (2-10) 2r

56 Substitution of this result in Bq.(2-8) yields 27 /r 2 1 sin.. r exp(-ik2 (x-y)2 + 4 sin 2 / id (x-y) + 4 sin 2 O /. oO 1_ 2 (]cj 2 )K 2' 2- exp(-is(x-y)) 2I ( - (k)2 )Ko( - (k) 2ds ( i 2-11) _, (2-11) By use of the result, proved in the appendix, that P (ds ) f = F P(ix) f.s I where P(x) is a polynomial in x, we have a 2 1 2 2 1 exp(-ik (-)2 + 4 sin2 + (kj) ) "- - -- -- 2 2r -2 (x-y) + 4 sin 2 1 2 2a a =-s -<< (2-; 4 -d f -I \ - (k) ) ds L2) Hence from Eq. (2-6) we have.-_ isn (k) -2 a 2 -2 K(s) ~ ' ' s - (k ) K ( - (kI 0 o (2-13) which is the desired transform. From Eq.(3.1-28) there follows G(y,s) = exp(iys) 2V kE 39 (s + 2iy, exp(-4)) o 0 3 __ _ 2 2 a 2 2 a 2Q ir((k9) - s ) I(T s2-(k) )K( 0( - )) (2-14)

57 where we made use of the fact that the third theta function is an even function of its argument. Let us nowintegrate both sides of Eq.(2-4) with respect to s from -2 to v and take the limit as l/-so. Then by use of lemma 1, we have lI d2 2 1 exp(-ik r) 5 (x) = 8 V A 2 + (k )) dy (y, s)s d dr o dx -I -oo 0 - which is equation (2-3). Hence i2kV x(isz) 3 (2i + exp(-4))ds I( (s - k)I - k )K (a s k ) -000 0 (2-15) which is the desired result. Section 3. Nature of the solutiop. We now examine, in some detail, the nature of the results given in section 2. The solution of Hallen's integral equation is given from Eq.(2-15) by I( ) - ~o exp(isz) 3 (st + 2i z/,. exp(-4)) ds o J (s2 - k2)I (a - k )KO(a _s - k ) -00 (3-1) Let us first note that because of the complex nature of k (Eq.(2-7)), the integrand in (3-1) has no singularities

58 in the real axis. The singularities of the integrand are located in the s-plane at s - k and a = -k. Use qf the rdation 93 (s + 2i 4, exp(-4)) = 1 + 2 exp(-4n2) cos 2n(2i + sl) n~/ in Eq.(3-1), gives i(2kV exp(isz) ds ( -1 0 07K iV k exp(is (-+z))exp(-4rn(L+z) )exp(is(2n-l)Q)25 + (2 - k2 )I iV 1 No (s -k ) (a s - k )K (a s - k =-/ oC) 0rl 0 0 0 00 This expression is seen to be of the form I(z) = i (z) + / i(1 + z) + i (1 - z) (3-3) where _ - / i (2)k V! exp (isz) ds 0 ) (2 -k)I (a )K (a k 0 - 00 oo (3-4) iV k I exp(is(l+z))exp(-4rr-(l+z))expS(C~n-,)~ i (l+z) - -xp(-4n(n-l)) 2 2 n r/=, 2.... o (s - k )I (a s kK (a -k O O (3-5)

59 That is i (z) = i (-z) is the outgoing current wave and o o the i are reflected traveling current waves of different n orders as functions of the distances 1 + z, 1 - z respectively from the end points, i (1 + z) traveling in the positive and i (1 - z) in the negative z-direction. The expression for i (z) was obtained independently by Hallen by considering the infinitely long antenna. Note from Eqs.(3-4) and (3-5) that 4 exp (-4n(n-l))exp(-4r) i2V0k exp(is ( +(2-l)j) ds i( = exp(-4n( +n-))i(+ (2n - 1)1) (3-6) approach. Since Eq4'(3-4) does not change when s is replaced by -sHe and z by -z it suffices to is convenient to replace z byEq(3-6) to the exponential term 00 i (o k exp(-is+z() ds 0 2 2 2 2 (s2 k2)I (a s -k)K - o (3-7)

60 Let us now replace the path of integration by another path (Fig.2.2) s-plane r s ",F 2 Figure 2.2: CONTOURr The rotation of the left half of the path of integration in Eq.(3-7) is permissible because of the factor exp(-is Iz ) in the numerator, and the asymptotic behavior of I K () O O when residues are provided for the poles. Since I (z)K (z) + I, (z)K (z)= 1/z(33 there follows 0 1 0 I K 1 a 1 +1 \ (3-8) 2 2 I K7 (s - k )I K k fo o ] 2 o K a s - k K has no zeros in the lower half of the complex plane. a 2

61 where f are the zeros of J (z). Now(3)I' (z) = I (z) 0 o' 00 1 and so in the immediate neighborhood of the poles a1 = 1 2 s (s -s) s -k I m m o Consequently if the contribution of the poles is determined by the calculus of residues, the expression for the outgoing wave takes the form i (z) = V 5k exp(-iSm Z) - ' O m. 0e / 2iV k $ (iI ) d (3-10) + o ) f 2k 2 2 2 + 0 k eiel -ilz' dsa 0 (a (2 - k )I (a s2 - k)K (a s2 - k It is a simple matter to determine the physical significance of the infinite series above. When k<j /a, then s in Eq.(3.9) is imaginary (Fig.2). The terms of the m series are then 4irV exp(- a -a2k2) i = iak - -r2- (3-11) om i 2 k jo a2k2 and are exponentially attenuated with distance. When k > /a thei ~ c) 1 2 2 m = - - (C (3-12) C lC where 0 = c /a. Also The poles c wY s 2 c in question then lie on the real axis between k and the origin. The terms of the series then are of the form

62 V o izi 2 i (z) = - exp(-i V - c om Nc c (3-13) where n I Xo 2 2 V (3-14) The series terms in both cases represent guided waves within the hollow cylindrical antenna, and wc has the meaning of a cutoff frequency. Equation (3-10) gives the traveling wave partly as an external antenna current expressed by the &aist term, and partly as a series of internal waves in wave-guides. It was pointed out before, that Hallen's integral equation was rigorously valid when the antenna consisted of a thin metal tube. It is for this reason that the sum of both the internal and the external solution is obtained as the answer to the problem. Hallen has shown that the integral in Eq. (3-10) can be put in the form 4r V t (kz) exp(-iklzl) (3-15) % 10 where (kz) is a non-periodic complex amplitude function given by

63 / i"\ (kz) = 2 Tr 2 + 2 — exp(ikIz|) ) oI du u H H)(aku)2 exp(-iklzl cos 1)-, sin 6 H(1) (aksii o exp(-ik z ) )2 dd nx) l 7/2 f+ i exp(-k!zlcot ) + a.k/i H (1) (ak/sins) 2 (3-16) 4 where H( (z) is a Hankel function of the first kind. o The last two 4itegrals have both finite ranges of integration and finite integrands and are suitable for numerical evaluation as functions of kz. The first term which is a constant has an important physical significance. It forms the limiting value of the real part of ~ when z-40, and this term alone in Eq. (3-15) gives the conductance at the feeding point of an infinitely long antenna, i.e., the characteristic wave conductance of an antenna of arbitrary length, 2Tr n 4 _ du o I R o C u l- u I H (( aku) 0 0 (3-17) The imaginary part of, according to Eq. (3-16) does not approach a finite limit as z approaches zero. The reason for this lies in the fact that as the gap

64 between the two halves of the antenna is reduced, the mutual capacitance increases without bound. In reality the air gap never anishes since even when the separation equals a the capacitance is still very small. The absence of a finite limit when the antenna is considered without the feeding line makes it impossible to assign to the antenna a well defined susceptance B 4 whereas it may be ascribed a well defined conductance G From Eqs.(3-3), (3-6), (3-10) and (3-16) it follows that the solution of Haflen's integral equation is 0 m o I(z) - V k exp(-isk|zt) x/ n L s + 7 7 exp -4n v,&(-z)+n-l] 2J V - exp {-ism|(jCz)+(2n-l)X1 f + 4 V 'f(kz) exp (-ikjz{) + IV exp -4n[L (Q+z)+n-1] ]'l(k (lz)+kl (2n-1) )exp -ik [ (;+z)+ (2n-1l + 0 n exp -4n[(J~-z)+n-l] t(k(j-z)+k.(2n-l))exp -ik[(C-z)+~(2n-1)j (3-18)

65 where i/ 2 to__ e k and f are the zeros of J (z). The first term corresponds, as we have seen, to an outgoing internal wave in the tube antenna. The second and third terms given by the doubly infinite sums represent a series of reflected wave-guide currents of different orders as functions of the distance i + z, 2 - z respectively from tie end' ints, the former traveling in the positive and the latter in the negative z-direction. The fourth term represents the outgoing external current wave in the antenna. The fifth and sixth terms represent reflected external currents of different orders as functions of the distances J + z, i - z respectively from the end points, one set traveling in the positive and one in the negative z-direction. The factor exp(-ikc?(2n-l)) in the last two terms corresponds to the total distance 2 (2n-l) over which th6 wave has traveled before the last reflection.

CHAPTER THREE J SCATTERING BY A THIN WIRE. Section 1. Formulation of the integral geuation. We wish now to consider the problem of electromagnetic back scattering from a perfectly conducting cylindrical wire excited by a plane electromagnetic wave with harmonic time dependence. From Bq. (1-7) of cfapter two we have for the magnitude of the vector potential on the surface of the wire 2 f A- 4 L 21 exp(kr)d (1-1) 0 -z 4r 2r r where as before k - W/C and r is the distance from the source point on the surface of the wire, to the field point also on the surface of the wire r = z - )2 + 4a2sin2 1 (1-2) The wire must be; assumed thin enough so that the assumption, used in arriving at q. (1-1), that the current is symmetrically distributed with respect to the axis of the cylinder is satisfied. On the surface of the wire, the tangential component of the scattered field E must be equal to the negative 66

67 z 4 I I z =R - - I I I I x z = -j I 0 I ii I, I I I I I I I I I I I/ / / 4 Incident E-vector / / / / 7,) -— I --- -.-.-,/ I - C2a - Cylindrical Wire FIGURE 3. 1 ORIENTATION OF THE INCIDENT E-VECTOR AND THE POSITION OF THIE CYLINDRICAL WIRE

68 of the tangential component of the incident field E, that is a r -2= (2 + k2) df d Z i d 2 r 4 (1-3) With the direction of the incident E-vector indicated in Fig. (3.1) we obtain at once 17r o i(_ 2 ikr E sintcosTexp(ikzcose) = -4k I(. dd o, 4rk d2 2( ) ]d r dz ) c (1-4) where qo = Vu/c and E is the amplitude of the incident field. Eq.(1-4) was derived on the assumption that the end surfaces play a negligible role and its solution will, accordingly, bear the same limitation. Section 2.. Solution of the integral equation. In Eq. (1-4) replace z by lx and T by ly 2 r E singcos exp (ikif x coso) = 2 (~- 2) I J Y)O: rdjd 47rk x dx (2-1) 2 r = /(x - y)2 + 4 a sin2 - We have already solved this equation in section 2 of chapter two. In fact it follows from Eq. (2-4) that I(z) is ob

69 tained from G(y,s) in Eq. (2-14) upon replacing J by kl cos, V by TE 1 sin 0 cos, and y by z/l; that is 4E co2smexp(ikzcos9)9 (kaco89+2i, exp(-4)) I (z) =. (2) rok sin 9 J (aksin9) H (aksin9) 0 0 0 (2-2) where we made use of the relations I (iaksin 9in= J (aksin ), K ( iksin 8). ( ) 0 0 0 2 0 Eq.(2-2) is the desired solution. t Section 3. Nature of the solution. For the scattering problem we have obtained for the current 4E cos' exp(ikzcos 0) 83 (kQcos 9 + 2i Z, exp(-4)) i(z) = (2I (2) rO ksin 0 J (aksin 9) H2 (aksin 8) 0 0 0 (3-1) The imaginary part of k will henceforth be taken equal to zero. Use of the relation oo 3 (klcos + 2i zexp(-)) 1 + 2 exp(-4n )cos2n(kpcos 9 + 2i 7Z / in Eq.(3-1) gives

70 4E 2-cos'~f"'exP(ikz con 9) I (Z) =. 2 %k sin 9 J. (a~ksin 9) H (aksin 9) 4E cosItA1 e~xp (-4n (n-l) exp (-4n I(+z))exp(ikUI+z+(2n-l)_J))com 9 + (akin (2) r~k sin9 J 0(ki 9) H (aksin 9) +4E cos'/L exp(-4n(nl1))exp(A4n ) (e-z))exp(-ik((~-z+(2n-1)~))cos 8 rjk sin 9 Jo (aksin H() (aksin 9) This expression is qplpo of the form I(Z) =i(z) + ni(l + Z) + in (1 - Z) (3-3) where 4E0 cos ^th"exp (ikzcos 8) i (Z) =(2) (3-4) 0%k sinG9J (aksin 9) H (aksin 0) 4E coslh exp(-4n(n-1))exp(-4rr (.Q+z) ) exp (ik (~+Z+ (2n-l).) i(2 +Z) =-(2CS9) n ~k sin 8 J (a~ksin 8) H 2 (ak si n 9) 000 (3-5) 1 4E cost kexp,(-4n (n-l) ) exp (-4n7r(J-z) ) exp.)(-ik (2-z+2n-l).P) i lZ) (2) n ~k sin 9J (aksin9) H 2)(aksinE)) o (3-6) That is, i 0(z) is the induced primary current wave and the i nare the reflected traveling currents of different orders

71 as functions of the distances 1 + z, 1 - z respectively from the end points. Ume expredsion for I(z) is given to a very good approximation by keeping terms to order n = 1 only, 4E cos t f I(z) = I ) (2) exp(ikzcos *) ro k sin 9 J (aksin Q) H (aksin 9) + exp -4(l1 cos k(z+2A)cos! +iexp 4 ( 14) sin k (+21)cos 1 L- L + expl-4 (l-)j Yos k(z-2Q)cos 9 + iexp -4(1-J) sin k(z-2k)cos e (3-7) This solution possesses the following features: I(z) is composed of a traveling wave plus standing waves. Away from the ends the standing waves are nearly space harmonic; but contrary to what is commonly assumed, for normal incidence, 8 = r/2, the standing waves are no longer harmonic. In fact, for normal incidence the z-component of the impinging fiqld is uniform in space and there is, indeed, no a priori reason for the belief that the standing waves in this case must be space harmonic. One more fact of this solution is yet to be discussed, namely the behavior of the current at the ends of the wire. For this we return to Eq. (3-1)

72 4E cosPexp(ikzcos )983(kicos 9 + 2i, exp(-4)) I(z) = - 3(2 n, k sin 9 JO (akin 9) H (aksin 8) The zeros of the third theta function 83 (z) are located at(35) z = lT + Tr r - mT - nmr, where m and n are any integers and t is defined in E. (3.1-20) of chapter one. For Eq. (3-1) we have U = i4/7r and I(z) vanishes at 1 - z = 1(2-4n-i7r4( - m + k1 cos 9)) (3-8) since the range of variation of z is the interval (-',*), n can only take the values 0 and 1, and m must be such that 27rcos 8 = (2m-l) A/l. (3-9) Equation (3-9) assures the vanishing of the imaginary part of Eq.(3-8). For angles such that Eq.(3-9) is satisfied I(z) vanishes at -1 and +1. For other angles the non-vanishing of the current at the ends indicates the presence there of a displacement current in order to satisfy the equation of continuity. In fact it is well known that the radiation field of an electric dipole vanishes along the axis of the dipole,

73 but in the near zone there is an appreciable longitudinal component of E; and indeed it is this component, and not the ~adiatiow field, which determines the distribution of current in a vertical antenna above the earth. (36) On the other hand, experiment has shown that the current at the ends of the wire is always very small contrary to the large values predicted by Eq. (3-7). 4 This emphasizes the fact that although Eq. (3-1) is the exact solution of Eq.(1-4), the neglecting of the end surfaces affords a very poor representation of the physical situation. This point was emphasized by L. Brillouin. ( Brillouin alleged that the end surfaces played a fundamental role in the scattering by a thin cylinder and consequently should be taken into account in the formulation. When this is done, the analysis leads to coupled integal equations which are considerably more difficult to solve. Other reseirchers have compensated for the limitations of Eq.(1-4) by assuming current distributions suggested by the experimental results. Section 4. expression for the back-scattering cross section. According to definition, the back-scattering cross

74 section d(9, t) is given by 4rR2 f E cos tl2 d(9, ) - --....2 o 0 (4-1) where E9 denotes the field produced by the induced current I(z) at a large distance R in the direction opposite that of the incident waver that is, i ksin 9 exp(-ik R E - 0R 9 WR o Hence i 4 I (z) exp(ikzcos 9) dz -- c- (4-2) d(ej ) = o2k sin 9 cos2t o 4rE2 0 I(z) exp(ikzcos 9) dz (4-3) For obtaining numerical values of d(9,f), it is convenient to eliminate the t-dependence by introducing the average value of d(9,f) over all values of rT corresponding to a random distribution of dipoles. Denoting the 'latter by d(0), we have Tr 1 d e) = dO (ie,, > ) d' t 0 (4-4) From Eqs. (4-3) and (4-4) there follows

75 (a) 312 2L2 2 27 J (aksin ) H( (aksin 9) |2 O o exp(2ik x cos 9) 83 (kecos 9 + 2ix, exp(-4))dx L -" (4-5) Use of Eq.(3-7) gives to a very good approximation 24 dL () _ v 3k2 -i \2 272 2J2aksin 9) H(2) (aksin 9) 2.IrO (4-6) - 2 sin(2klcos ) + 2 2 k-cos 9 2 i2 i 4 + k cos j which is the expression sought. If ak <<1, then J (ak sin 9) is nearly unity and for the Hankel function we have for small values of z8) H(2)(z) = 1 - 2 log z (4-7) 0 2 where log y = 0.5772... is the Euler-Mascheroni constant. The expression for the scattering cross section takes the form

76 2 8 ()2 + (log ak sin e)2 22 i sin(2kcos 0) 2 + 22 L kcos9 4 + k22 cos2 L This expression is similar to that obtained by other research workers, ' but it predicts values much smaller than those indicated by experiment. This is not a surprising fat since we have seen that the end surfaces play a dominant role in the shaping of the current distribution.

APPENDIX FUNDAMENTALS OF THE THEORY OF DISTRIBUTIONS In the main body of the thesis we had occasion to use several results from the theory of distributions. It is the purpose of this appendix to give rigorous justification to these assertions. Our approach is patterned after the Gelfand-S)kilov theory as found in their book (41) Les Distributions. ( By a class K of testing functions is meant a set of finite functions in(x), i.e., functions that vanish outside a bounded set (corresponding to each function ~((x)) and which have continuous derivatives of all orders. Adding the testing functions, or multiplying them by complex numbers, we will always obtain testing functions; in other words K is a vector space. The sequence of testing functions 1 ( ),,...,.. (x),. is said to converge to zero in K if they all vanish outside the same bounded set, and if they, together with their derivatives of all orders, converge uniformly to zero in 77

78 the ordinary sense. By a continuous linear functional f in the space K is meant a definite rule that assigns to each testing function a certain complex number (f, s) satisfying the following conditions: a) For any two complex numbers a, a2 and any two testing functions 1 (x), ^2 (x) we have the equality (f' c1 + a2 ) = a1 (f, () + a2 (f, (linearity of the functional f); b) If the sequence Yf testing functions 1' i2 ' '''' /'i' ' tends toward zero in K, the sequence of numbers (f, 1) (f ). * ** 0 (f. 9 0)~ *.. tends toward zero. (Continuity of the linear functional f). From here on we will designate as distributions every continuous linear functional in the space of testing functions. If a function f(x) is alsolutely integrable in every finite interval this function is said to te Locally integrable. With each such function we can make correspond to each testing function P(x) the number ox) (f, (D) = f(x) (x)dx (A-l)

79 where f denotes the complex conjugate of f and the integration ranges over the bounded set outside of which 9 (x) vanishes. It s easy to show that the conditions (a) and (b) above are satisfied. Distributions of this type are known as regular distributions. Any other type is known as a singular distribution (in particular the 5-function defined by (6(x), (p(x)) s 4(0) for any testing function). A distribution f is said to vanish in the neighborhood U of the point x if for each testing function 4(x) vanishing outside I4the equality (f, 4 (x)) = 0 holds. Hence, the distribution f, corresponding to the ordinary function f, vanishes in the neighborhood U of x0 if the function f(x) itself vanishes almost everywhere in this neighborhood. The singular distribution 6 (x-x1) vanishes in a certain neighborhood of any point x # xl. If the distribution f does not vanish in any neighborhood of the point x, x is called an essential point of the functional f. The union of all the essential points is called the support of the distribution. The support of a distribution f, corresponding to an ordinary function (continuous or piecewise continuous) is the closure of the set over which f(x) # 0. The support of the distribution 6(x-xo) is made of the single point x. If the set F o o

80 contains the support of the functional f, the functional f is said to be concentrated in the set F. aXj;4 initbn, a distribution f vanishes in the open set G if it vanishes in a certain neighborhood of each point of this set. Given two distributions f and g, we define their sun as the linear functional defined by 4 (f + g, io ) = (f, I ) + (g, I )..1 The product of a distribution by a number a is defined by (cf, ) = a(f, ) =(f,a 9). The multiplication by an infinitely differentiable function a(x) is defined by (a(x)f, i),= (f,a(x) ( ). (A-2) By definition, the sequence of distributions fl f2, f3..., f,... converges to the distribution f if for any testing function ) (x) lim (f ~ ) = (f <). 1J —., o,

81 o0 In the same fashion, the series of distributions X h "Zo n is said to converge to the distribution g if the sequence of partial sumsA L i h converges to the distribution g in the sense indicated above. Let f be a continuous linear functional over the space K of testing functions: theidistribution g given by the formula (g, ) = (f,- P ) is called the derivative 4 of the functional f and is designated by the notationf' or df/dx. Lemma 1. In the sense of convergence of distributions L; lim exp(ix7)d~ 2r 6(x). ai, Proof. It is well known that j fi (x)dx = s in 'x dx 1 -oo -o00 -c~C Also for bba>O b TT fJ x Y J - x J The same result being valid for a< b <. Finally, the magnitude

82 x C y ais uniformly bounded with respect to a and b, for any., Let us now consider the sequence of primitives F,(x) = f1t ()dp It follows immediately from the definition of the sequence f L (x), that the function F > (x) has for limit as v -boe a constant equal to zero for x < 0, and unity for x >0, and is also uniformly bounded with respect to - in every finite interval. It follows then that the sequence F, (x) has for limit, in the sense of distributions, the step function ac(x), 0 for xO<0 and 1 for x>0, hence the sequence of functions f (x) = F' (x) has for limit, in the sense of distributions, the function a' (x) (a (x), fic)) = (a(x), - f )) = - ' (Xx) dx = (0) 0 and by definition of the delta function ca' (x) = 6(x)

83 so that, 1 sin 5 x im — ~.. = b (x). 1 sinjlx The function.. - -- can in turn be represented as the 7[ x result of integrating with respect to, from - J to L, the function (1/27r)exp(i fx). We have then: In the sense of convergence of distributions 4 -A 2 Lema 2. If f (x) = 2 -7t exp(-x /4t), t > O, then in the sense of convergence of distributions lim f (x) - b(x) Proof. Sirnce f t(x) > 0, then for any a and b b o f-(x)dx^ 24- exp(-x2/4t)dx - 1 Making the change of variables x/V7 3= y, we have for a<0< b lim f (x)dx = lim exp(-y /4)dy = 1 - -1 t t

84 also, for any b>0O p(-x2/4t)dx < 2 E exp(-x2/4t) x2t dx.... 2t b d t 22. b b = exp(-b2/4t) -- 0, as t —0. The integrals over the intervals (b,co), b > 0 tend to zero, and an analogous result holds for tie intervals (- c, a), a < f. Constructing as in lemma 1 the sec -nce of primitives Ft(x) = t ft(f)d -I shows, in view of the results above, that the functions ft (x) form a sequence converging to the delta function. That is 1 2 2 t -p exp (-x /4t) - 6 (x). Convolution of distributions. In classical analysis one frequently uses the operation of convolution of two functions f(x) and g(x) f(x)*g(x) = f( )g(x- )d. (A-3)

85 The definition of convolution in the domain of distributions is made starting with the concept of direct product of two distributions wyLch is now introduced. Let a distribution f(x) be defined over the space of testing functions K of one variable x and g(y) a distribution defined over the space of testing functions K of I Y one variable y. Starting from these distributions we 4 wish to define a distribution h(z) over the space of testing functions K of two independent variables x and z y. We proceed in theFfollowing way: We denote the testing functions for the distribution h(z) by C (x,y). Let us fix x and consider o (x, ) as a function of y. This is obviously a testing function in the space K. Let us apply to it the functional g(y) to obtain a certain function I(x). This function is infinitely differentiable, since (X +x)-^x) (x +Ax.v) _- (x.v) )((y).(.Y) tx - (g(y), Ax ' x gx Where we used tle fact that the sequence.(x + x.v) - (x.v) Ax converges to 9 ( (x,y)/~ x in the sense of convergence in the space K and g(y) is continuous. It is evident also that Y

86 /(x) is finite. Hence, rf (x) is a testing function in the space K and we can apply to it the functional f(x). x TherefQr the ejression (f(x),(g(y), ~(x,y))) (A-4) is well defined. This is a certain functional over the space K. From the continuity of the functionals X(x) z and g(y) one caA deduce the continuity of this functibnal. This functional is <ignated by h(z) = f(x) X g(y) and is called the direct product of the functional f(x) by the functional g(y). The direct product has a particularly simple appearance when it is applied to a testing function 2(x,y), product of two testing functions (l(x) and 2 (y). In this case, acceding to the definition (f(x) X g(y), ~1(x) )2 (y)) (f(x), (g(y), ~l(x) Y2(y))) (A-5) - (f(x), 1(x)(g(y), (Y))) - (f(x), (1(x))(g () Y2 (y)). If f(x) and g(x) are two absolutely integrable functions

87 over the line and h(x) = f(x)*g(x) their convolution, the functional defined by the function h(x) (absolutely integrable) can b1 expressed in the following fora (hh (,(x)) = h(x) (x)dx = ( g(x- )d ((x)dx = f( i) g(r) (( f + r)dddr. In other words, the desired result is no other than the result of the appliation of the functional f(x)g(y), which one can consider as the direct product of the functions f(x) and g(y), over the function Y(x + y). Quite naturally, the general definition of the convolution of distributions f and g is given by (f*g, f) = (f(x) X g(y), ({(x + y)). (A-6) We now show that the direct product of two distributions is commutative, f(x) X g(y) = g(y) X f(x). (A-7) For the proof let us note that by virtue of the continuity of both members of Eq. (A-7), it is sufficient to prove the equality over a space of testing functions which is everywhere dense. We will consider then the dense set of

88 functions of the form Y (X) nj (y) J2' j,' where (j (x), Ij(y) (j - 1,2,.., L j U 1,2,...) are testing functions in their corresponding variables. We have then (f(x) x g (y), > ^(x)r(y)) = j (f(x) X g(Y), ij (xX) t-(y) )., =?L(f(x), y(X))(9g(y)r.j(y)) In the same fashion (g (y) f f(x),. (x) j (y)) (g (y), Tj (y)) (f(x) j i (x) ) and the result follows at once. aL 3. If D be a differential operator then D(f*g) = Dfg = f*Dg. Proof. We have (D(f*g), ) = (f*g,D* (0)

89 where, for example D* (-) D if D is a homogeneous differential operator of order V. Furthermore, according to the definition of convolution (f*g,D* c ) (g(y), (f(x),D* D (x + y))) = (g(y), (Dtf(x), ((x + y))) = (Df*g, ()! hence D(f*g) = Df*g and by virtue of the commutativity of the convolution D(f*g) = D(g*f) = Dg*f = f*Dg. Let us now establish the following lemma, relating to the continuity of the convolution. Lemma 4. If f, f then fi *g -- f*g in each of the following cases: a) The functionals f, are concentrated in one and the same bounded set. b) g is concentrated in a bounded set.

90 c) The supports of the functionals f, and g are bounded on one and the saae side in a manner independent of V Proof. ) According to the definition of convolution, for any testing function t (f,*g, ) = (f (y),(g(x), 9p(x + y))) For the case (a) the function (g(x), 9 (x + y)) can be modified to a functiofvanishing outside the set where all the f (y) are concentrated. Hence (f *g, ) ( (f,, (y)) — (f.) ( (f*g, C) and f *g -- f*g. For the case (b), Y (y) = (g(x), C(x + y)) is a testing function and the proof proceeds as above. For case (c), if we suppose, to fix the ideas, that the support of the functionals f, and g are bounded on the

91 left, then the function t (Y) = (gc ) (x + y)) has a support bounded on the right. We can then change it to a testing function vanishing outside the set where the functionals f, (x) are concentrated and the proof proceeds as before. 4 Fourier transfore of distributions. In classical anlysis there exists a one-to-one correspondence between the class of functions possessing a Fourier transform and their corresponding transforms. This correspondence preserves linear operations and the properties of convergence. An analogous coreespondence can be established among the linear functionals defined over these spaces. This correspondence is established in such a way that it makes correspond the classical Fourier transform to functionals corresponding to absolutely in4 tegrable functiois. Let f(x) be an absolutely integrable function and g(6) its Fourier transform. For each testing function '(x) and its Fourier transform 3'(6) we have Parseval's formula

92 (f, L f(x)(x) f) x ) exp(ix)d dx -00 -oo -00 -oo00 -oo This formula is valid when f(x) and $ (x), and hence their Fourier transforms g(d) and ((d), are of integrable square. Parseval 's formula shows that g (d) considered as a distribution, acts on a tel*ng function according to the formula (g,z) = 27r(f, f), (A-8) This formula is used to define the distribution g over the space Z of testing functions P corresponding to any given distribution f over the space K. The functional g defined by Bq. (A-8), is called the Fourier transform of the functional f and is denoted by the symbols f or F[f]. ' The usual rules of differentiation of Fourier transforms carry over for Fourier transforms of distributions: d [P(dx)f F[P(ix)f (A-9) F[P()f] - P(-is)f (A-10)

93 where P(x) is a polynomial in x. For the proof it is sufficient to consider the case d d P( x) = 'dx) dx We have then ((ixf),+) = 2r(ixf, P) = 2r(f,-ix f) = (f,(-ix )) 4 (f d y2) = df 4 -which gives equation (A-9). Equation (A-10) is established in an analogous manner. Lemma 5. T = 1. Proof. According to the definition (b, () = 27r(6,,) = 27r ((0) = (c/)do = (1, ') -hence hence 6 = 1.

94 Lemma 6. if P(x) in a polynomial dd P(x) = [P(x)'&l] = P(-i -l and (1 )= 2r (l, F) =~ I f (Ux) dx = 21j (x)exp(.-ixO)dx = 27r f'(O') = 27r (b5, h) hence x-. d P (x) = 2T P (-i dsib6(S) Lemma 7. The Fourier transform of the exponential function exp (bx) is 2wb,(s-ib). Proof.I. We will make use of the convergence in the complex plane of the series exp (bx) c"9 n n b x C n

95 applying the Fourier transform operator term by term to this series, we obtain bn n - = - b n exp(bx) = n x n zO = 2r bn (-id )" 6(s) 7- ds /7:: and b n d n (2Tr n (-i-s) "(s), J7 = o Tf(s)) brO I1 = 0 d (s). (s)) -ds S ) n^ n. 37 — o oo = 2 (6 (s), i (n )(s) n f --- and the series /.- f (s) n. Z?= O converges in the functional sense to the function i1(s + ib) Hence 0 ((s)) = b (-id )nb (s) (s)) = 2Tr(6(), (s + ib)) _n ds J-7 ~C = (2tn (s - ib), I (s)) where the translated b-function, b(s + h), is defined by

96 the equality (E (s + h, u(s)) = (6 (s), s (s - h)) = f(-h) (A-ll) for all (complex) h. Hence exp(bx) = 2v 6 (s-ib) (A-12) for any complex bf, Fourier transform of a direct product. The Fourier transform of the functional f, acting over the space K of testing functions ( (x) of several independent variables x = (x1, x2..., x ), is defined 1 ~ n as the functional g acting over the space Z of testing functions 1r (x), s = (s s s..., s ), by the formula (g, t ) =4 (27 )n(f, ( ) (A-13) where f = " is the Fourier transform of the function (x). The functional g is also designated as f or F[f]. Let f(x) and g(y) be given distributions over the variables x and y respectively, and let f( ) and g(r) be

97 their Fourier transforms. The Fourier transform of the direct product f(x) \ g(y) is given by the formula f X g = f( ) g() (A-14) that is, it is equal to the direct product of the Fourier transforms of the functionals f and g. For the proof it is sufficient to consider testing functions 7 (x,y) of the form J-I In this case (f % (2)2(fg, Z I) f) (2r)2(f) - (g, ) - f = (f,,Z x) = (f X I 9 R p ) When the convolution is given by ah expression of the form f (x) *g (x) = f ( )g(x- )d (A-15) then the ordinary Fourier transform

98 f (x) * (x) exp (ix d)dx can be considered as a distribution depending on a parameter O' defined over the space of testing functions exp(ix <&). According to Eq. (A-6) we have then 4 (f*g,exp(Ixd)) = (f(x) A g(y),exp(i Cr(x + y)) (f x) X g(y), exp ix a) exp (iy ) ) = (f(x), exp (ix a') )(g (), exp (ixc = f( C)g( ), where we used Eq. (A-5). We have then Lemma 8. The Fourier transform of the convolution f(x)*g(x) = f( )g(x- )d is jiven by F[f(x)*g(x) ] = f( )g( ).

99 This result, commonly used in classical analysis, is known as the convolution theorem for Fourier transforms.! 4jr

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