THE UNIVERSITY OF MICHIGAN COLLEGE OF LITERATURE, SCIENCE, AND THE ARTS Department of Physics Technical Report SOME BOUNDARY VALUE PROBLEMS OF MATHEMATICAL PHYSICS R. D. Hazeltine ORA Project 01164 supported by: NATIONAL SCIENCE FOUNDATION GRANT NO. GP-7605 WASHINGTON, D.C. administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR November 1968

ACKNOWLEDGMENT All of the problems and many of the ideas in this report were suggested by Professor K. M. Case of The University of Michigan. The author is very grateful for Professor Case's advice and instruction.

TABLE OF CONTENTS Page LIST OF FIGURES iv ABSTRACT V I. INTRODUCTION1 II. CLASSICAL WEDGE PROBLEMS 5 1. The Laplace Equation 6 2. The Helmholtz Equation 20 III. A HALF-SPACE GREEN'S FUNCTION FOR ELASTIC WAVES 31 1. Integral Formulation 32 2. Solution of the Integral Equation 35 IV. LINEAR TRANSPORT THEORY 39 1. Introduction 39 2. General Formulation 41 3. The Infinite Space Green's Function 50 4. Half-Space Problems 53 5. Slab Problems 62 6. The i-Dependence 70 V. A CONJECTURE OF KAC 79 1. The T = oo Case 80 2. Factorization of V(k) 81 3. The Eigenvalue Problem 86 4. The Eigenvalues 88 5. The Eigenfunctions 90 VI. SOME QUANTUM FIELD THEORY PROPAGATORS 95 APPENDIX 103 REFERENCES AND ADDITIONAL FOOTNOTES 105 iii

LIST OF FIGURES Figure Page 1. The wedge of Section II.1. 5 2. Integration paths for Section II.2. 30 3. The region R of Section IV.2. The region R includes the entire plane except for the small neighborhoods of I enclosed by 7+. 48 4. An inverse Fourier transform integration contour for Section IV. 52 5. Integration contour for B (k), Section V. 0 - max(it/2, it/a). 83 iv

ABSTRACT In this thesis the following problems are solved: the homogeneous Laplace equation, with boundary conditions given on a radially finite wedge; the Helmholtz equation, both homogeneous and inhomogeneous, satisfied inside a radially infinite wedge; a half-space problem for the elastic wave equation; several infinite space, half-space, and slab problems, with point sources, in linear transport theory; an integral equation arising from the theory of Mathematical Statistics; and certain generalized wave equations from quantum field theory. Essential use is made throughout of the Fourier and related transforms, especially in combination with methods based on the theory of Cauchy integrals (including the Wiener-Hopf technique), and certain, usually elementary properties of generalized functions. V

I. INTRODUCTION The elementary theory of the Fourier transformation, depending as it does upon translation invariance, is not obviously applicable to problems involving boundaries. This report, in which a number of boundary value problems from various areas of Mathematical Physics are solved, illustrates several of the means by which the theory can be applied to such problems. Specifically, we obtain by the Fourier and related transforms the solutions to: wedge problems for the Laplace and Helmholtz equations (Section II); a half-space problem for the elastic wave equation (Section III); several infinitespace, half-space, and slab problems, with point sources, in linear transport theory (Section IV); an integral equation arising from the theory of Mathematical Statistics (Section V); and certain generalized wave equations from quantum field theory (Section VI). The specific technique used to solve each problem is briefly discussed at the beginning of the appropriate section. Here, it is convenient to summarize the notation and basic mathematical tools which will be generally relevant below. The Fourier transform of f(x) will be denoted by f(k) and defined by 00 ikx f(k) = e f(x)dx (1) -00 1 00 ikx - > f(x) - e f(k)dk 2P -M0 We will often have occasion to use a conventional decomposition of f(k): 1

2 f(k) = f+(k) + f_(k) (2) ~1-j ikx 0o ikx f+(k) = e f(x)dx; f (k)= e f(x)dx o - oo If f(x) has at most polynomial growth at infinity, we have the important fact that f (k) is analytic in the pper half k-plane _ ~lowe The f+(k) may have singularities on the real axis; for example, if f(x) = 1 -o < x < co then f (k) =-( +( ) k+io where the infinitesimal imaginary part specifies the interpretation of the pole in the usual way. Alternatively, we may consider equation (3) as defining the generalized function 1 1- =P- + i6(k) (4) k+io k where P denotes "principal value." The above remarks can easily be made rigorous by requiring f(x) to be a generalized function in the space S'of Gel'fand.( Then f(k) is also in S and the f+(k) always exist. On the occasions when we must assume f(x) to be in a different generalized function space, we will draw attention to the fact, although generally such matters will not be of crucial interest. Whenever f(k) = o(k ) for k co we can easily deduce the useful formulae

5 1 ff(k' )dk' f+(k) - - r^' (5) +(k) - 2ti k'-k-io () -00 + In general, subscripts will be used to denote functions which are defined and analytic in appropriate half-planes[the functions need not be Fourier transforms of any f(x)]. Superscripts, on the other hand, will denote the boundary values along the cuts of functions which are sectionally holomorphic in the (2) sense of Muskhelishvili. In particular, if I is any sufficiently smooth arc (or union of arcs) and () =1 p(t )dt' (z)- 2ti T t'-z (2) then for each t c ~, we define c-(t) = lim 0 (z) as z - t from the eft of ~ (6) right of ~ If p(t) satisfies a Htblder condition, the ~-(t) can be shown to exist, and the following useful formulae of Plemelj hold: + (t) + "(t) = - P f t' dt Jti t1-t (7) (t - (t)= p(t) As an example, observe that equation (5) could be stated as F (k) = + f+(k) (Im(k) = o) (8) where, for Im(k) / o, F(k) - 2i dk' (9) 27ti -oak'-k

4 and f(k) is suitably behaved. The second Plemelj formula, applied to f(k), is now merely the statement of (2).

II. CLASSICAL WEDGE PROBLEMS Two planes, intersecting along and terminating at the z-axis, constitute what we will call a wedge. In this section we solve some classical two-dimensional differential equations with simple boundary conditions specified on wedges. In Part 1 we find that solution of Laplace's equation which attains a given, constant value on a wedge of finite "width" (width is measured in the radial direction-see Figure 1), by the Wiener-Hopf technique. It is found that, in a space of generalized functions, an infinite set of solutions exists; of these, only the "least singular" solution is uniquely determined by the classical boundary conditions. The special case of the strip is also discussed in some detail. 8=wFTr=he wc8io r:O Figure 1. The wedge of Section II.1. Part 2 is concerned with the Helmholtz equation, satisfied inside a radially infinite wedge. The particular problems considered, which already have 5

6 known solutions, were chosen so as to clearly illustrate the main features of our method; this involves a modification of the Fourier transformation. 1. THE LAPLACE EQUATION We want to find qp(r,@) where, for each (r,@) not on the wedge, _ ar P (1) r 6r 2r 2 2 r and ~p(ra) = cp(r,-a) = Op for r < 1 (2) Here the wedge subtends an angle 2aQ, and we have conveniently assumed it to have unit width. With the orientation indicated in Figure 1, it is clear from symmetry that we can restrict our attention to the region 0 < 0 < rr, and that (assuming a i O,f), cp'(r) r) = 0 on Q = O and @ = ( (3) For a unique solution, an additional boundary condition must be specified:. the total charge per unit length on the wedge. (Since the wedge looks asymptotically like a line of charge, we could equivalently specify the behavior of p for large r. The fact that both cp and the charge must be given is essentially due to the fact that the wedge extends to infinity in the z-direction.) If we define D[f] = lim [f(G+e) - f(G-e)] = f(G+) - f(0-) then the charge density, q(r), on the wedge is clearly given by

7 q(r) - - D[p'] 0 r 1 (4) r -" and the specified quantity is 1 Q = 2 0 q(r)dr (5) Finally, we remark that with the exception of cp' on @ - a, 0 < r < 1, p and cp' are continuous for all 0 E (Or). In particular, cp'(r,a+) = p'(r,a-) for r > 1 (6) Equations (2) and (6) together constitute "mixed boundary conditions" and suggest use of the method of Wiener and Hopf.(6 The Least Singular Solution Our procedure for finding the cp(r,Q) satisfying (1) - (6) may be summarized in two steps: (i) a change of variable r-u = -In r, followed by Fourier transforming with respect to u*; solution of the resulting ordinary differential equation to obtain the O-dependence. (ii) use of the mixed boundary conditions to obtain a Wiener-Hopf equation for the transform variable dependence; solution of the Wiener-Hopf equation and inversion of the transformation of (i). Step (i): the angular dependence In terms of the variable u = -in r, our differential equation (1) takes the form Lwe write cp(e,G) = cp0u,Q) for convenience] *Equivalently, we could have taken the Mellin transform with respect to r.

8 2 2q acp acp ^ + =0 (7) u 2 a2 with boundary conditions cp(u,a) = cp 0 <u< o (8) 0 - - D[cp'] = o -oo < u < (9) cp'(u,O) = cp'(ur) = 0 -co < u < o (10) -2 D[cp'] du = Q (11) Using the convention specified in Section I, we take the Fourier transform of (7) to obtain (kg) - k2 p(k,) = 0 (12) a82 Recalling the continuity conditions discussed above, and using equation (10), it is a simple matter to solve (12): cosh kO g sO < 0 <oa cosh kg p(k,0) = A(k)= cosh k(-) (13) 1 — -- - a < < E cosh k(r-a) where A(k) is to be determined. Step (ii): the Wiener-Hopf equation With the definitions of p +(kQ) from Section I, and equation (8), we observe that icp p+(k,a)= 0 (14)

while from (9) we have p'(k,c+) ='(k,a-) (15) Now note from equation (4) that q+(k) = p+(k,a-) - c+(k,c+) is the transform of the charge density. Using (13) and (15) we find sinh k( T-a) sinh kg k sinh k3T q (k) kA(k) = +) [ si)h ko k,-+ 1 cosh k(Jr-a) cosh kjh k-cosh kJ (16) On the other hand, equation (13) implies that A(k) = c(k,a) = p+(k,a) + p (ka) Substituting this into (16), and using (14), we finally obtain the desired Wiener-Hopf equation: i(C q+(k) = H(k)[p (k,a) + kio (17) where k sinh kit H(k) k sinhk -- (18) H(k) cosh ka cosh k(-a) (18) Because of their known analyticity properties (cf., Section I), both of the unknown functions q+ and cp can be determined from equation (17)* by the WienerHopf technique, as follows: Suppose we can find functions h+(k) and h (k) such that *In deriving (17) from (7)-(11) we followed what is called, in Reference 6, Jones Method.

10 (a) H(k) k2h(k) h (k) (b) h+(k) is analytic and non-zero for Im(k) < 0 (c) h+(k) has at most polynomial growth at oo Then equation (17) implies that q+(k) h(k) = k h_(k)[ico + kp_(k )] (9) The function F(k) defined by q (k) h (k) Im(k) > 0 (20) F(k) kh_(k)[icp + k (k )] Im(k) < 0 (21) is easi-y seen to be entire: it is analytic for Im(k) i 0 and, by (19), continuous across the real axis. Now assuming q and cp to be generalized functions in the space S, and noting condition (c) above on the h+(k), we see that F(k) is bounded by a polynomial at infinity. Hence, by the "extended" Liouville (7) theorem, F(k) must itself be a polynomial: F(k) = B + B k +... + B k (22) o 1 n Assuming, for the present, the Bi to be known, equations (20)-(22) provide the desired solution to (19). Thus, the problem reduces to finding functions h,(k) which satisfy conditions (a), (b), and (c). For the H(k) of equation (18), this is not difficult. (7) Using known representations of the hyperbolic functions, we have 2 1. c 1.~-c 1. 1 rr-c k ik)( - i k -)r( - ik-) + i - ) (k) = 22 r(l-ik)r(l+ik) i r(1-ik)r(l+ik)

11 Noting that r(z) (which is nowhere zero), is analytic except for simple poles at z = o, -1, -2,..., we can immediately exhibit functions satisfying conditions (a) and (b): 1 a 1 2 -( r( - ik )(- - ik - ) + it r(l-ik) = 1 22) (2 ) r( + ik 1r( + ik h (k) = (24) where the X(ka) factors are to be chosen according to condition (c). Using Stirling's formula we find In h (k) - ln-ik[- In + In - ] - In k + k-o -a 2 so that h (k) will have exponential growth at infinity unless ik ( a- ik x(k)/ = -' ik- )ik = X(, —a) (25) This choice of X insures the proper asympotic behavior of h (k) also. In fact, it follows from (23)-(25) that 1 r h+(k) - k 2 for k - o in the lower half plane (26) ~_2upp lower We can determine the Bi, which fix the behavior of cp and q near k = o, only by appropriately restricting the class of functions which will be considered acceptable solutions. For the present, let us require, in the "classical" manner, that cp (k,a) be square-integrable, i.e., cp(k,ao) = o(k /2) k - c o (27)

12 The p(u,G) which results from this requirement will be square-integrable except for the trivial singularity unavoidably associated with equation (8) and will (8) be the "least singular" solution in the sense of Case.) Postponing discussion of the more singular solutions, we observe that (27) implies Bi = 0 for i > 0 (28) since, from (21) we have F(k) i___ T (k a) = F(k) (29) (k-io)2h (k) k-io which, with (26) and (27), gives (28). The remaining constant, B, is determined from the final boundary condition, equation (11): Q 00 2 - f [(u,) - cp'(u,a)]du = q+(o) (50) From (20), (22), and (23) we have B = =B q+(o) = h (o) = so that B - o 2 (31) Now it follows from (14) and (28)-(31) that A(k) = T(ka) = + i(io k-io 2(k-io) h (k) which equation, combined with (13), determines cp(k,0):

15 cosh kg -, 1 cosh ka 0 < Q < a cp(k,) = 1 + icp (k ) cosh k(-) - O (32) 2I +o k+io k-io cosh k(l-O) 2(k-io) h-(k) J osh k(r-a) -e Similarly [from (20)], q+(k) 2 h(k) (33) where the h+(k) are given by equations (23)-(25). Our problem in k-space is solved. As is clear from (23) —(25), the right-hand sides of (32) and (33) are meromorphic and well-behaved at infinity. Thus the inverse transform integral, 1 0 - iku - Cq(u,Q) -= f e cp(k,Q)dk 2t -00 involves only an elementary computation of residues (i.e., the contour can be closed by a semi-circle in the half-plane appropriate to the sign of u). In -u terms of the physical variable r = e, the results of this computation can be given in the following form: For 0 < a < T, - --- ) n - (r =2(x (_P) n + ) )]X(k,) ]r Q 2 ( 2 C l cp(r,O) = -( - r Z n+1,+1 cos[(n + )-]r 0 2 n=o n. 1 1 2c LI )- ](n - ) 2 2 a 2 for O< O <, 0 < r < l (34) = 0 - (+2-n2+n 2 + 2 in 2 + In r

14 or (n'(n+l)2 cos[(n+l)a]r[2 - -( n+l) 1 - (n+l) ]x(t(,0 1 7 cos[(n- l)Q r no,2 21 [- ) for O G< O< l < <oo (35) Here, k = -i(n+ ) t - i(n+l), X is given by (25) and y - In - + In n 2 C' n iT 7t-Ca -r-a For a < O < t, 0 < r < 1, cp(r,O) is clearly given by (34) with,,a replaced by T-~, it-Ca respectively. The charge density can be computed either from (33), or (34) with (4). By either method, the result is, for 0 < a < Tr and a -2 q(r) ( ) 2 n q (n + r_ [( 2)rX(PnCr 2) + o ^ n' 1 1 (-C6) 2r n=o n: -1 /(n + ) L i + (n 1 r[(n + - 2 ]x(qn,)r + +)-) 2 2 n + --— a -- (36) (t-a)r[ + (n + )( ) J 2 2 it-c where pn -i(n + -)- = -i(n+ - - n 2 a) n 2 rt-a' The Strip As we have implied, the cases OC = 0, C = jr and, for q(r), c = - are somewhat exceptional [this is clear from equation (32); note that for all three of these angles, the wedge reduces to a strip]. Nonetheless, the inverse transform can be found in essentially the same way as for the unexceptional a. We only give the result for a = 0, which is typical. C= 2' n

15 cp - Q [-(2 in 2 + in r) + r no (38) for a = 0 < < r <, < r <oo q(r) = r 1 /2 O<r<l (59) r /2 (1-r) / where we have used the fact that 00 (_l)n rn I.) 1/2 () r=(l-r)-1/2 (40) n r=o n. 1/ r [i-(n + )] Note that the factor of 2 in equation (5) is erroneous in the case a = 0; since the wedge is "closed," the integral of (5) automatically includes the charge on both wedge-planes. We have accounted for this in (37)-(38) and it is easily checked that (39) indeed satisfies 1 J q(r)dr =Q ( = 0) (41) The closed form of (39) suggests that we might have solved the strip problem in a much simpler way. That this is indeed the case can be seen in the following digression. The equations cp(re) = (constant) / q(r')lnjlr-dr'dr' (42) and e = _= O0 (43) a Q O

16 imply P q(r') dr' =0 0 < r < (44) o r -r Now assume that q(r) satisfies a Holder condition and let I I q(rl) t(z) = 2ti o r zdr' (45) (3) We note that ) (i) O(z) is analytic everywhere except on the line (0,1)o (ii) O(z) - for z ~ oo z (iii) ~ (r) exists [cf. equation (1.6)] for 0 < r < 1. Near the endpoints, ) < constant (z-c) where 0 < P < 1 and c = 0 or 1. (iv) By the Plemelj formulae and equation (44), D (r) + r (r) =0 0 < r < 1 The task of finding a O(z) which satisfies (i)-(iv) constitutes a "Hilbert Problem" Since the equation of statement (iv) is in this case particularly simple, the general method for solving such problems (which method we will use in later sections) is not needed here; it suffices to observe that the function P(z) Jz(l-z), where the br' —-l cut extends along the real axis from 0 to 1 and P(z) is any function continuous across this cut, satisfies (iv). By (i) and (iii), P(z) can have no singularities in the finite plane except possibly simple poles at 0 and 1, i.e.,

17 Q(z) P(Z) - _L p(z) - z(l-z) where Q(z) is an entire function. By (ii), Q(z) = constant, so we have D(z) = constant (46) z1/ (l-z) /2 and, again using the Plemelj formulae, q(r) constant (47) q(r) 1/2 )1/2 (47) r/2 (1-r) 1/2 as in equation (39). We remark that a generalization of this method can be effectively applied to the case of several strips aligned, say, along the real axis, with varying, given potentials on each strip. Behavior of q(r) Near the Endpoints When the wedge is not a strip, we cannot express q(r) in closed form. However, it is easy to obtain such expressions for qfr) valid near the extremities of the wedge. From equation (36) it is clear that near the vertex, q(r) (constant) Max 2(-) ] q(r) - (constant) Max r r r for r - O, 0< < < (48) (9) For r ~ 1 (near the edge), we use a well known Tauberian theorem: for -1 < r < 0,

18 [f (k) - k 1 for k -o] [f(u) - uB for u - 0] (49) Applying this to q+(k), by (33) and (26) we have -1/2 q(u) ~ u for u - 0 i.e., q(r) - (-in r)-1/2 (constant) -? for r 1 (50) So that near the edge, the charge distribution of the wedge is asymptotic to that of the strip, as we would expect. More Singular Solutions If we relax the requirement of equation (27), and thus allow F(k) to be a polynomial of degree greater than zero, the resulting p(k,@) no longer possesses an inverse Fourier transform in the ordinary, classical sense. However, we may still obtain useful solutions, provided we allow them to be generalized functions in the sense of Gel'fand.) Let us restrict our attention to the charge distribution. From (20) and (22) we have 1 00iku n q(u) = 2 e [B + B k +...+ B k ]h(k)dk (51) 21T-oo o 1 n Of course the integral does not exist, but in the generalized function space S' which we now explicitly use, the inverse Fourier transform always exists, and we have in general

19 F [k] = i d F- [] du where F [f] denotes the inverse Fourier transform of f(k). Thus q(u) = F [(B + Bk +...+ B k )h(k)] d d n = [B + Bi +..+ B (i ) ](u) (52) o 1n du n where q (u) is the least singular charge distribution obtained above (except that the constant - is no longer significant); q (u) is a regular generalized 2 O function in that all its singularities are integrable. Equation (52) is typical(8) ical: the general solution is a linear combination of the least singular solution and its derivatives. For clarity we specialize to the case of the strip (a = 0) and rewrite (52) in terms of r: A 2 1 d 1 d 2 d2 1 q( + A + A (r + r -) 4 (l - qvr) = ~r(l-r), 2 dr r(l-r)-+ A dr 2 rl drr +... (53) Two remarks are in order: (i) The above q(r) is not a regular generalized function, since the singularity at r = 1 is not integrable. Thus the derivatives with respect to r must be interpreted in the usual generalized function sense; e.g., dq f -r (r)dr = (q'(r), rt(r)) = -(q,[r]') (54) where ~(r) is one of an appropriately restricted class of test functions. (ii) The boundary conditions used to determine the least singular solution

20 are not sufficient-nor even applicable, since the integral of (5) no longer exists-to determine the more general solution of equation (53). Instead, some sort of "edge conditions" would seem to be called for. This concludes our discussion of Laplace's equation. 2. THE HELMHOLTZ EQUATION The Fourier transform, of course, owes its usefulness to the relation d ikx ikx d [e ]= ik [e ik] dx The observation that 2 1 a a 2 iKrsinhT a iKrsinhT r {-r r- - - ) [e ] - [e ] (1) r ar 2 6T suggests a new transform; this will simplify the Helmholtz operator in polar coordinates, much as the Fourier transform simplifies the differential operator in Cartesian coordinates. For cp(r,@), defined for all r c [-0o^ oo] (and sufficiently well-behaved), we define the "r-transform," (T,0Q), of p(r9,@) by 0\?0 iKirsinhi-'(T,Q) = cosh T f dr cp(r,@)e (2) -00 The inverse transform is given by \ 1 00 / \e -irrsinhT cp(r0,) -= dT 2(T,G)e r () 2T oo as can easily be seen by making the substitution k = K sinh r: we see that V(T@,) is simply cosh T * [cp(sinhTrQ)] Our purpose here is to examine the properties and usefulness of this trans

21 formation by means of a few examples. The Homogeneous Helmholtz Equation Inside a Wedge We find cp(r,~) such that ( -r -+ - 2 }p( r,) = 0(4) r ar Sr 2 2 r So for O < 0 < c 0 < r < oo and cp(r,O) = cp(r,a) = 0 (0 < a < 2Jt) for 0 < r < co (5) (Note that here the wedge is radially infinite and has a different angle and orientation from that used in the previous problem.) We first assume (4) and (5) to hold for all r, and formally apply the transformation (2). Using (1) we find that 2 2 (-+ —-4 —) 2 (TO) = 0 (6) ST 2 (TO) = (T, = (a)= (7) Clearly nT nT (TG) = Ae + Be sin n=l,2,... (8) We will see [cf., equation (11)] that both terms in (8) give the same result. Hence we choose B = 0 and apply the inverse transform (5) to obtain nit nt a -i0rsinhT cp(r,) = A sin - ~ f dT e e (9) a -oo With the change of variable T -- T + i - 2'

22 00- i- jT 2 - -- (T+i -) nT CX 2 -K rcoshT cp(r,) = A sin - @ e e dt (10) C^. JT -00o- 1 2 - KrcoshT In view of the rapidly decreasing behavior of e for large ITJ, it is clear from Cauchy's theorem that the integration path in (10) is equivalent to one along the real axis. That is n \ nr T a -KrcoshT cp(r,Q) = (constant) sin - @ f e e dT (11) a -oo (10) In this form, we recognize the integral and conclude p(r,0) = (constant) sin G K (Kr) n=l,2,... (12) n which is the familiar solution obtained by separation of variables. This simple example demonstrates the essential features of the T-transform: (i) It transforms the Helmholtz operator in polar coordinates into the Laplace operator in Cartesian coordinates [equation (6)]. (ii) It transforms a wedge into a strip [equation (7)]. (iii) It is not one-one; this is clear from equation (11). In general we can say that whenever the transformed function V is such that the manipulations of equations (10) and (11) are permissible, then the odd part of * will have vanishing inverse transform (since cosh T is even). The Inhomogeneous Equation In the above example no difficulties associated with the unphysical domain of r occurred. The necessity of allowing r to be negative does, however, re

23 quire us to exercise some care in dealing with inhomogeneous problems. To see this, consider the equation (V2 - 2 )cp(r,) = q(r) r > 0 (15) r,G 2 Multiplying by r and taking the transform we find itrsinhT 2 + (T r,0) = cosh T f dr r2 q(7)ei (14) T2 6o2 -00 where we have implicitly assumed (13) to hold for all r E[-00,o]. The significance of this assumption can be seen in the case of a point source. Placing this source at (x = r, y = 0) for convenience, we have q(7) = q b(x-r ) 6(y) The point is that q(7) i -q- 5(r-r )&(0) (15) r o 0 In fact q(F) = q 6(r cos @-r )5(r sin ~) (16) = q 5(r cos @-r ) 5(sin 0) + s 1 (r)] o'Irl Isin (17) The 6(r) term evidently gives no contribution to the integral in equation (14), while 00 6(sin 0) = Z 65(-nt) (18) n=-oo

24 so that (since r > 0) 000~~~~~0 which provides the correct source function to be inserted into equation (14). More generally, for a point source at (x,y ) = (r cos 0,r sin 0 ), we have 0 0' o 0 0 6(7- r ()= n 6(-9@-2nit) (r-r ) o r =-00 o o o 00 + 7 b[0-O-(2n+l)b] b(r+r)) (20) n= -oo 0 0 If we restrict out attentiion to e region 0 < < 2, this reduces to 6(r-r ) = - {(0-0 ) 6(r-r ) + 6(~- ~ i+ ) 5(r+r )} o ro o o o 0 for Q <, 0 < 0 < 2r (21) With these remarks we can find the Green's function for an infinite wedge() This satisfies (V - K )pG(r ) = q 5(r-r ) 0 < 0 < a (22) with the boundary condition G(r,O) = cG(r,a) = O (23) Transforming these equations according to (2), and using equations (14) and (21), we obtain

25 V G(TO) = f(T) 6(0-~ ) + f(-T) b(0-0 -r) 0 < 0 < a (24) 0 0 G(T,) = G (Ta) = O O< o < 2T where (choosing q = T K for convenience) f(T) = n K r cosh T e irsinh T 0 d iKrosinh T = -it d e (25) and we have assumed 0 < Jr, the modification for 0 > r being trivial [equation O O (21)]. Our task is to solve the equations (24); the solution to (22)-(25) will then be given by (3). The solution to (24) can clearly be written in the form G (T,;0~) =,(, Q;) ) + 1 (-T,Q; +t) (26) where 1 satisfies 1AVp, ) f(T) 5(~- ) 0 < < K 1(T,o) = (T,a) = (27) We find V1 by first determining the Green's function G satisfying V2G(T,G;T,0) = b(T-T') 6(0-0') (28) G(T,0;T,0') = G(T,a;T 1') = 0 (29) By Green's identity we have

26 f Gf b(0-G )dTdQ~-l(T;'') = / (G - - -)dr = 0 (30) V F1 n 1n where V is the strip 0 < K < a and r its boundary (G and I must of course be suitably behaved for large T). Renaming variables we have [since G(T,@;'T@) = G(r,O;r G )] 0 0 i(T,1 ) = J G(T,0;T,0o) f(r )dT (31) 1 G -oo 0 0.0 so that the problem is solved if we can find a G satisfying (28) and (29). But this is elementary: G is clearly the electrostatic potential due to a point charge between parallel conducting plates, and can be constructed in a wellknown way from an infinite sequence of "image" charges: 2Tr G(T,0;T0 o) = Re (F(z;z ) (32) where z = T + i~@ z T= + iO o o o and o-z z + i2na F(z,z ) = E in l - (33) O n=-co z - Z + i2na o Each term in (33) represents of course the field due to a point charge; the (7) charges are positioned so as to satisfy (29). It is not hard to show( that / \o 1 i2na F(z;z ) In z — o1 (34) i2na

27 sinh j- (z-z) = In (35) sinh -- (z-) 2 o0 Hence sinh 2 (z-z) G - I in -2 - (36) sinh - (z-z ) 2a o cosh - - C (-T )-cos - ((- ) 1 in 0 o0 a = In - (37) 2 cosh - (T-T )-COS - (0+0 a o aC o Substituting this expression into (31), and, as suggested by equation (25), integrating by parts, we obtain fO ir sinhT { sinh i (T-T ) 21 - irc sinhco - cosh - (T-T )-COS - (G-Q ) sinh - (T-T ) o ( -— r, —) dr (e8) T[ \\ cosh - (T-T )-cos a (G+ - ) or, with the change of variable = i(T-T ), - _100~ ~ - rsin ( ) e- ( eiKrsinh(T+i) sn12 i lo \cos O t-cos - (-Go ) a a' o sin- - rr — (d — d( (39) CS -os — os (+ ) a a o where the integration path must be understood as infinitesimally displaced to the right of the imaginary axis. In the corresponding expression for = (-T,G;

28 ~ +n), it is convenient to make the further substitution -o -~ +j, yielding 1( -TQ9@; +t) -= 1 i+ Teirosinh(T+iS) 01 o / 2i -io +jT f sin - (-At) sin (S- ).......... ---......~ - d~ (40) Los - (-It)-cos - (-Go -x) cos - (S-j)-cos - (+Q +r)J a a o a a o j Here, the integration path is such that Re (~) is infinitesimally less than Tr. Equations (39) and (40) provide, with (26), the solution to the problem (24) in T-space. Our prescription is to substitute this solution into (3); i.e., the Green's function for the wedge is 1 ~ -i rsinhr 9) = 0d iKrsinhT (41) G (r,) 2 -2x 1 (T; o) I+ 1(-T@;@+)e dT (41) Now for 5 c(Oji), it can be shown (cf. Appendix) that -iKrsinhT iKroSinh(T+i)dT 2K (flr-lr ) 42 f e, e ~d = 2K ( K r-r`42) f e= — oo O 0 where -r I g Jr + r - 2rr cos (43) Hence, combining equations (39) —(41), and inverting orders of integration, we finally obtain 1-100 sn sin sin cp (r,~) = -- 2i K (K r-r I ) -- d cos -, -cos - (Ge- ) cos - ~-cos - (9+~ ) a a o 0 aa

29 s inf sin - (4-k) 1 iioo+jr Jro ) a c cos - (-))-c-os a ((-~~ -) 0rri o ~rr: o sin - (i-t) ( -- -... d (44) Cos - ( - )-cos a (G+ o+j)J Equation (44) provides a solution to the problem for arbitrary wedge angle a C [0,2rt]. Its form is such that it is easily checked in the particular case a n = 1,2,..., since for such a the two expressions in braces in (44) are identical. Thus sin - PG(r. ) 2jti o Ko (ir-r I aTI r - cos - 5-cos - (-0 ) sin- T d~ cos - -cos (+G ) a ea o for - = n = 1,2,... (45) where r is the path indicated in Figure 2. It is easily seen that r can be deformed into the path F' as shown; thus we can evaluate the integral in (45) by residues. For definiteness, say a = T/3. Then the relevant poles of the first term in braces occur at =0-e0, 2 =f- -_-@ +2 i o o o' 2 fo 5 and the residue at n (n=o,l,2) is K (</ -r ); similarly for the second term 0 -o;smlry o h eodtr

50 (with its minus sign). Thus equation (45) gives the well-known solution by images. /r _ - Plane Figure 2. Integration paths for Section 11.2. Note that since the homogeneous equation, (4), possesses (an infinite set of) nontrivial solutions, we cannot expect the solution to (22) obtained above to be unique. In particular, for a < T, the inverse T-transform of i1(T,@), by itself, is a solution; this is already clear from equation (24). We do not pursue this point, but merely emphasize the fact that for a > j, and for other problems (such as the infinite space Green's function-see the Appendix), the inclusion of sources at negative r is essential.

III. A HALF-SPACE GREEN'S FUNCTION FOR ELASTIC WAVES A method for computing the elastic radiation from a small source in the (13) earth's interior has been presented by Case and Colwell. This method, which assumes the earth to be an infinite medium, could be modified to include effects due to the earth's surface simply by replacing the (known) infinite space Green's function which is used, by an appropriate half-space Green's function. It is the purpose of this section to derive a representation for the latter function. Our method is straightforward. We first formulate the problem in terms of an integral equation (following Case and Colwell), and then solve the integral equation by Fourier transforms. The only complication lies in the fact that, once we choose a definite orientation for our half-space, the matrices which occur are not tensors, i.e., not rotation invariant. Hence tensor theory arguments, with the computational simplifications they often afford, are not available to us. This section will consistently use the summation convention 3 a.b. = a.b. 11 ii 1 i i=1 and, augmenting the definitions of Section I, the notations a.= a;a= a, 1 ax. 1i 1 1 a = (al,a2); a = (n,a ) 51

52 We also define two- and three-dimensional Fourier transforms, by trivial extension of (I.1): A ik.r f(i,x3) = e f(7,x3)dr ik x f(k) = I e 3 3 k,x3)dx 1. INTEGRAL FORMULATION We consider first a general region V of r-space, and seek the solution f. (rr ) to -W p fi (r,r) kD ) km( a (rr) + 6j 6(r - r ) (1) ij o~ k ikm~~ mj ~o ij V r,r e V' 0O with the boundary condition n Di () f (r,r ) 0 (2) i ikm mjs s so r c av o's Here n. is (the ith component of) the inward normal to the region V and D () = + (. + ) (5) ikm( ik m + ( im k km i) ) The infinite space Green's function G.. satisfies (1) with V equal to all 1J space: -cpG. (rr') = aD, () Gi (rr') + bi (r - r') (4) all rr' -rV'

55 We now proceed in a standard way to multiply (4) by fi., (1) by Gi, and subtract. Using the easily verified identity, G kD f - D f..SDm G = k[G D f -f D G' i2 k ikm mj ij k ikm m i k [k ikm mj kj ikm G ] (5) we find G (r,r )..8(r - r ) - f (rr) - (r r ) + [Gk(rr) Ia I00%044V 0OV ^ o ijj k mo 0% A iO f kj(r ) Dikm() Gm (rr) ] = 0 (6) kj ^-^ ikm mV We now integrate (6) over V, apply the divergence theorem, and note equation (2). Upon renaming variables (and indices), and using the facts that G. (r,r ) G.(r,r) (7) and D m() Gj(rr) = - Dim(' ) G (r ) (8) ikm mj Al ikm N jm N ^ we obtain the desired equation f..(r,r ) = G..(r,r ) + S..(r, r) (9) fij(r ) = Gij o 1ijJ r c V where S. *(rr = dd r'n (r') f (r',r ) D (') G (r',r) (10) eij re o, s kofs Rj s o k&m m io se Here, of course

34 D k m( ) Gmi(r'r) [D () G r[D]) mi(re VI (11) ksm D-k mi'-s k (m mi r = s -I as A method for determining f. (r,r ) is clear from equation (9). Indeed, j13'' our problem clearly reduces to finding f j(r',r ); and, by taking the limit of ij "s — o (9) as r - r, we obtain an integral equation which may be solved for f (r'r; v ^ — s ~ j -S) o Specializing to the case in which V is the half-space x > 0, we denote the half-space Green's function by gij: 3 g..(r,r ) = G..(r,r ) + dr' g (7 0;rO )[D ( (5) G.(r' ]x ) = (12) A certain amount of care is required in taking the limit x3- 0 of (12), since the integrand is singular on aV. In fact, if we define 3.i(rr') - Dm () Gmi (r,r) (13) then it easily follows from the differential equation for G.., (4), that -7 E (,o+;?',o~) - i(7,o-;',o) - - 7) (14) Thus, in close analogy to (I.4), we define o(' ) [ [ i[(7,o+;7'r ) + 3i(rO-;r',0)] (15) -5Ii(' r, ~[.9 3, Now we let x - 0 in (12). Using (14) and (15), the result may be written as rO;r ) = (r,O;r ) + S.(r7, ) (16) 2~~~. ~._ S~ 1 O j

35 where 0-7 S. (rr ) - dr' gj(,;r) (',) (17)'a ^1'Q Aj 5'Q 311 (17) (12) and (16) are the basic equations by means of which our problem is to be solved. 2. SOLUTION OF THE INTEGRAL EQUATION (16) We define -ik x -7 1 o 5 5 -ik.r't F (kx) - dk, e 3- e-i r') (18) (13) Using the known fact that ik.r'k - k k e [kp k kij ij 2 2 2 2 2 (19 p k - k~ k - kt where 2 2 2 a_ P 2 wO p k = k - (20) i + 2B, t [ it requires only a straightforward computation to obtain 2 kj F - k2 [(k - s k ) I + 6 I + 2(I - k )[(kk - 6 k ) x (k. - jk )I + (kj - 6 k )6 I ~ (k - 6 k )63 I + I ] + 3(kk 2 k3)Iot k Sj t kj kit ) (21) +~~~~ ot 58j~k 53k)It 8jlltkjl

36 where -K Ix31 -7 t I (kx) = — e t t t -7 _ 3 it3 3 < - -K JX3I I (k, x ) = +ie t xe ) t 5 I(n ( 22), h- Ixiu e -t -x Ie Tt 1 3 t 3 23 53 2 2 2e( 3 t I (k,x ) = 2 ( e e ~I~~x ) k.9(~ te) + -e'Ke x 3 <;x 0 (22) 2 2~1 t 3 KI - Kt In (22), we have introduced the abbreviations 2 72 2(23) K:R k -k (25) 1~ t t Note that F (kx ) is independent of x": -ik x F ( kx5) dk e e i(k; O) (24) F1i~ 3 2 2-oo 3 Hence if we let F () = -[F i (io+) + F ( -)] (25) ~i 2[Fi ~i then the two-dimensional Fourier transform of (16) clearly takes the form [cf., equation (8)] 1 0A^- 7^ -7 ~ F (k)].(k, O'r) = G. (kO r) (26)

57 and we need merely find the inverse of the matrix IJAiJ, where 1 o. JA =2 i + F k) (27) Now it follows from (21) and (22) that F (k) a(2 (k - k) + b2 Si(k -8 k) (28) Ii' 2 5 i 5i 3 2 5i ~ 53 where 2 XK - (x + 24)K a(k) = -i - (29) aKK )(. + K )( + 24) XK - (x + 2))KI b(k) - i, (30) K t(K + K )(X + 24) () whence ""1 0 a k |AI| =t 0 1 ak (3.1) bk bk2 1 and we easily find 1 - abk abk k -ak ||Af1 2 — ^1.2 1 2 2 jAi=. 2/abk k 1 - a'bk1 -ak2 (2) 1- abk Lbk bk -bkl -bk2 1 Now gkj(S,O;ro) IIA-ki Gij(k,O;%) (5533)

38 and, since the two-dimensional Fourier transform of equation (12) is gij(,3;) = (kx;) - F(kx) g (kO;r) (34)'a 35^.0o ij 3'Lo Li 3 gj k O our problem is solved [cf., equations (21) and (33)]. The function Gij occurring in (33) and (34) is, of course, obtained from (19): A -1 -ik x ik.r k k k - k k Gij(kx;r ) = 1f dk e 3 3 e o + Geij(kex3ro) 2-00 3 2 2 2 2 2 c kk k k kik-0 K -x I -t Jx -x r(e 2 2 2 e2 e 2 -- K - < I.(Ix) - klj 2 - (35) ) 2vP \k - k't -2 Here, I. i= jl||.., with -ij'ij k k k k ik 11 12 1 i3 x IlIII = /k k k k ik a J( -k, 7 1 2 2 2 2 x 5 (36) ik ik - - 1 x 2 32x) 2 35 15 S.3 where -K Ix -xx I K 5 3 0 t 3 o3 1 t ~ J(k,x ) t= (57) 5 2 (2 2 I t(t - I) -7 The inverse transform with respect to k of (34) is obviously tedious and A will not be discussed here (gij may be directly useful for certain applications; cf., for example, Part 6 of Section IV). We merely draw attention to the fact that there is no suggestion, in equation (34), of an "image" source at -r. In fact, it is not hard to show from the differential equations (L) and (4), that no solution of (1) by images is possible.

IVo LINEAR TRANSPORT THEORY 1. INTRODUCTION In this section we solve a number of simple boundary value problems associated with the time-independent, one-speed transport equation with isotropic scattering [equation (2.1)]. One-dimensional solutions to this equation are already well-known(4); the extension here lies in the fact that although we do not consider boundary surfaces any more complicated than parallel planes, our boundary conditions will be such that the full three-dimensional form of the equation must be used. Essentially, we deal with point sources rather than plane sources. As in previous sections, the method used here relies on the Fourier transformation. We first obtain two basic equations which are valid for any configuration of boundaries and sources. The first of these, equation (2.6), gives the Fourier transform of the angular density in terms of a certain quantity in transform space, p (k), where the subscript V refers to the region enclosed by t9 PV the boundary surfaces; the second equation, (2.10), can, at least for the simple geometries considered here, be solved for pv(k). Thus (2o6) and (2.10) provide a general formulation by means of which the Fourier transforms of the solutions of all the problems considered can be found. The form of equation (2.10) obviously depends upon the particular region V, and on the boundary and/or external source contributions. Thus for the infinite medium the equation is trivial; for half-space problems we must use the Wiener-Hopf and related techniques; and for slab problems we must combine the 39

40 Wiener-Hopf technique with a Fredholm-like iterative procedure. It is not possible to express the inverse Fourier transform integrals in closed form; in particular, the dependence of the transformed solutions on the "transverse" transform variables, k, is very complicated. (Of course, by setting -7 k = 0, we obtain the known one-dimensional'.solutions, although, since our method is not the standard one, in somewhat unconventional form.) It is nonetheless found possible to express various significant physical quantities in fairly tractable form, generally by means of approximation techniques; these points are briefly discussed at the end of the section. We note here some notations which, in addition to those defined in Section I, will be used in this section: 7 A -7 1 fco -ikx - ik.r f(xk) = e f(k)dk = e f(r)d (1.1) is the two-dimensional Fourier transform of f(r), the latter being defined for all r in three-dimensional space. Here k (k,kz);r (y,z) (1.2) y z and f(k) is the three-dimensional transform, defined by trivial extension of (I.1). We will also use the function w, defined by!(t -? where Q refers to the y and z components of the three-dimensional, normalized, velocity vector, and p. = g:

41 = [ = ( cos (cp - 1 sin cp) (1.4) (cp is measured counter-clockwise from the positive y-axis). Note that 1 - ik.- = -i(k - ) (1.5) 2, GENERAL FORMULATION Let V be some (bounded or unbounded) region of three-dimensional space, with boundary S. Our problem is to find the function cp(rQ) for r c V, I I = 1S which satisfies (QSV + 1) cp(r,) - - p(r) + q( rQ) r c V V..., O/ A -. 4ft *% AWr* 01V cp(r,9) = (r ) r c S, 0 inward (2.1) Here p(r) -= d Cp(r,~) (2.2) while q(r ) for r c V and cp (r O) for r e S, 0 inward to V, are assumed to "V 19.AO S s SS s be given. To express (2.1) as an integral equation, let G(r - r'TQ), defined for all r, r', satisfy (-Q-V + 1) G(r - r1,O) = 6(r - r') (253) Then, by the conventional argument* (2,1), and (2o3) imply *cf., Section III.

42 cp(r,) = f dV' G(r' - r,)[- p(r') + q(r'S)] +.f n. dS G(r - r) cp (r ) -s s('r V S (2.4) A where n. is the inward normal to V. Note we have put the known function cp (r,) in the integrand, instead of cp(r,Q); this is justified by the easily verified fact that G(r - r ) = 0 for r c V, Q outward (2.5) Equation (2,4) holds of course only for r c V, the domain of definition of cp(r,s). We now extend this domain by assuming equation (2.4) to hold for all r, so that we may take its three-dimensional Fourier transform. [Note that it is not obvious how to extend the original differential equation (2.1) without contradicting the boundary conditions.] This takes the form c 1 + ik.r' 4TC 1 - ik.S2 (k) + +f n dS'e s cp (r' V) 4r 1 -k 1 - ik.S 1 - ik.0 i s'ls' (2.6) where ik-r' ~() - dV'e ~ ~ r P( ) (2_ ) V In deriving (2.6), we used the representation -ik (r' ) G(r - r ) 1 i3k (r (2.8) Y ~ 3 1 + ik.S0 (2Tt) I.which follows easily from (2,3), and made the convenient definition q(r,0) = o r/ V (2.9)

43 Upon integrating (2.6) over all directions Q, we obtain our basic equation (k) = [1 - (k)] pv(k) + B(k) +Q( (2.10) where d2 A(k) 1 - ik (2.11) is the three-dimensional dispersion function and ik.r' n 2 B( ) f dS e /I d - i p( (2.12) Q(k) f - (2.13) result of course from the (given) boundary and source contributions, respectively. We see from (2.6) that our problem in k-space is solved once we find the function p (k); this function will be determined from (2.10). Our method of solving (2.10) consists essentially in the observation that Pv f dk' (k')Av(k - k') (2.14) where i(k-k').~' A (k- k') = 1 f dV'e' u " (2.15) (2Tr) V so that (2.10) can be written, in general, as an integral equation for p(. The geometry of a particular problem enters solely through the kernel, Av(k; this will be, for the simple problems we consider, highly singular, i.e., a generalized function [when V = all space, for example, AV(k = 5()((k)], with the result that our solutions of (2.10) will depend more on analyticity argu

44 ments than on the Fredholm theory. Therefore, before proceeding further, we briefly examine the properties of A() as an analytic function of, say, k. The Dispersion Function The definition (2.11) may be written 1(ck 1 1 ( A( 4 1 d J dp 1 - ik t - iB cos (p - A) (2.16) 0 x where B -k = k + k (2.17) y z and k -l z A - tan -- (2.18) y The integrals are known* and we find, for real k and B, x~ A(k) = A (k,B) Im(k ) = 0 (2.19) 3 x x where 2 2 1 + ik B ic x A (k,B) 1 + c n - (2.20) x 2k +B 1- i k + B x x Note that for Im(k ) sufficiently large, x A(k) / A3(k,B) A(k) is in fact quite pathological as a function of k outside a certain neighx borhood of the real k axis; since our "basic" equations are true for real k x mple, Dwight *See, for example, Dwight No. l!46 and No. 580.001.

45 we always can, and will, use A (k,B), which can be taken to be defined by 5 x (2.20) for complex k also. x We will use the notation A(k x,B) A (k) where k k x A (k) has fairly simple analytic properties: its only singularities are branch points at k = + if where 2 t- UB + 1 (2.2l) and we will take the branch cuts, i, as extending to + io along the imaginery axis (cf., Figure )o Thus A (k) is analytic ii the plane cut along I and 35+ 2_. (14) Using well-known properties of the one-dimensional dispersion function, A (k) -A3(k,B0) (2.22) it is easy to show that A3(k) has only two simple zeroes. If we define v in 5 O the conventional way* A1(+ i/v ) = 0 (2023) *Note A1(i/v) = A(v), where A(v) is the function defined by Case.(14)

46 then the roots of A (k) are clearly given by 3(+ iK ) = 0 where 2 2 K \IB + 1/v (2.24) 0 0 It also follows from the one-dimensional theory that (i) c < 1= jK I > B and Im( ) = 0 (ii) c > 1 and |BI > Il/vO =i I K0 < B and Im(K ) = 0 (iii) c > 1 and B < Il/v I|=b | < B and Re( ) = 0 From the known fact that |v | > 1 for any c, we may conclude that the roots of A (k) are never located on its cuts, i.e., Im( ) = 0 i i | < P (2.25) o 0 The only other properties of A3(k) which we will need are both evident from (2.20): A3(k) = A(-k) (2.26) and A(k) - l1 (2027) Ikj oo Factorization of A 3(k) In Part 1 of Section II, we performed on the function H(k) a factorization appropriate to the Wiener-Hopf technique, essentially by inspection. Here we obtain an analagous representation for A (k) by a more complicated but conven(12) tional method,

47 It is clear that (any branch of) the function ln (k)(k + 32 L(k) -- In 3 )(k (2.28) (k + K 0 is analytic in the region R, as shown in Figure 3, and that L(k))- 0 (2.29) keR Hence, by Cauchy's theorem, L(k) - L (k) + L_(k) k e R (2,30) where L (k) =- L(k') dk' (2 31) 2ili k' - k + Here the contours y_ are close to, but not coincident with, the cuts I [see + + Figure 5; with this definition, the L (k) are well defined even for k e _ / R]o + -- + Equations (2,28) and (2,30) imply that 2 2 k + L (k)+L (k) O +A (k) = e +k E R (2.52) ^3(k): 2 2 k + l so that we may write A^(k) A (k) A (k) kcR (2.55) where the A (k) are given by k + iK L (k) A+(k) = — k + e (2.54) +' k + i

48 +i e R O k-Plane -o_ Figure 3. The region R of Section IV.2. The region R includes the entire plane except for the small neighborhoods of + enclosed by 7+. A-() k- i -L_(k) A1 (k) e - (2.35) k iKe o and clearly have the analyticity properties implied by the subscripts; equation (2.35) comprises a Wiener-Hopf factorization of A3(k). Note that A (k) -l (2.36) and, as follows easily from (2.26), ^ (-k) - ^ 1 (2.37)

49 Except for the path y_ [on which L (k) is not defined] and the obvious -- + poles, A+(k) is defined by equations (2.34) and (2.35) for all complex k. We now derive representations which are valid only in specified regions, but which we will eventually find useful. The argument consists wholly of applications of Cauchy's theorem. For k c R, it is evident that 2iri S — k 0 (2.38) 2ti 7 k' k + whence L+(k) = r+(k) k c R (2.39) where 1 In A(k') 1r (k) =- S dk' (2.40) 2Tci k' - k _7+ On the other hand, for k R (i.e., k E + or k c _), L+(k) + L_(k) 0 k / R (2.4i) 2 2 A(k) k + K k27k k R (2.42): A (k) k2 2 In particular, L+(k) = -L_(k) = r (k) for k c + (2.43) + -- ~+ +'

50 while k2 2 L+(k) = -L_(k) = I'(k) - n k (2.44) k + From (2.40) through (2.44) we deduce the relations k + iK (k) A_(k) = - e k c _ (2.45) k + i~ k + iK r (k) +(k) k -i e k + (2.46) + k+i5 + Incidentally and finally, we note the identities A (-i/z)e- r ) X(z) = v z- (B = 0) (2.47) o 0 e +(0-~/o)- r+( o) X(v) = (B = 0) (2~48) o where X(z) is the function defined by Casel) We are now prepared to apply our general method to the solution of specific transport problems, the simplest of which is clearly 3. THE INFINITE SPACE GREEN'S FUNCTION Here, the region V includes all space, there is a point source at r with ^Vo direction O G q(rs) = (r - r ) 6(Q - ) and no boundary contribution. Our basic equation (2.10) thus takes the form ik.r PG(k) = [1 - (k) ] (k) e =G [ _- )] G 1- ik-0

51 or ik.ro 1 e G i(k) 1 - ikQ (51) -O rO so that, from (2.6), ik'o ik-r e "% ^ 5(, - 0 ) c ei -o IGP G, 4_ (1 - ik. )(l - ik _2 _) +(k) (1 - ik.- (5.2) and we have merely (') to perform the inverse Fourier transformation. For the present, we examine only the inverse transformation with respect to k = ko x Using the notations discussed in III.1, and equation (2.19), we have -7 -7 A ~,k.kro ei -ik(x-X _ _ 1 x) (x,, -c - k-1o) e____f dk... A 2 -, t e 00 k (4T (k - )(k- w%) A3() o 0 s(0 k;) ='0 (303) where W c( 0o) (3.4) 0 0 This integral is easily reduced to a useful form. Assuming for definiteness that x > x, we close the contour in the lower half plane by means of a path which excludes the cut I (cf., Figure 4), and apply Cauchy's theorem. By the relation (2.25), our path will always enclose the pole of [A (k)]- at -iK; 5 0o we may have additional pole contributions from aco ), depending upon the sign of (4 ), since Im(o) > O 0 { 0 (355)

52 Oe~~I,e. An _. \ / \-i I ik\@ rO r - i! o( -i) \ I III/ \ ~ I / \ - / \ / oo) \ —' Figure 4. An inverse Fourier transform integration contour for Section IV. as is clear from (1.3), and provided that cc(c ) / I_. The result is that A -7 e ik'ro -i e"_o(X-Xo) cG(x,k;n) = - ( 1K + ))( i O o) ak - o o kIK 0 1 Cfd e-ik(x-xo) ic c - e c_ 2:r 4ff \ (k - a))(k - o> o)(k) +~) (~ 4r ( _) )'A (Cw) ( ~ (l0) 4 (c - C ) A(J + ( (1 ) ~o 8(, - Q )e ~ o) (3.6) for x >X, co, Co - R. 0 0

55 We remark that (i) if (wo ) E I_, which occurs, in particular, when k = 0, the only change in the above expression is that the first (second) term in brackets does not appear. (ii) the last term represents merely the uncollided source contl-ibution and is of little interest while (l4) (iii) the first and second terms give the so-called discrete and continuum "modes," respectively. Note that, by (2.25), the former always dominate for large x. Hence, for an asymptotic approximation, we could ignore the unwieldy integral term. (iv) we can most easily compute an analagous representation for p (x,k), Gr not from the definition (2.2), but directly from equation (3.1). A In fact the integrals over 2 of our solutions cp will in general be rather complicated, but it is an obvious feature of the present method that such integrals need never be explicitly evaluated. 4. HALF-SPACE PROBLEMS If the region V is the half-space lying to the right of the x = 0 plane, then* -7 -"7 1 00 ik xx d00 ik r' -7 4 P((k) =o xx e'xe ) = (k) (4.1) PV ~ () dx1 e + *Note that equation (4.1) could also have been obtained by noting that, for the half-space, t(k - k') t(k - k') Av(k k' Y Y z z V - J ) 2ri(k' - k - iO) x x. and using equation (1.5).

54 where k - k and we omit reference to the transverse variables k. (Throughout x this section subscripts will refer to regions of analyticity with respect to the k = k variable; k and k are always assumed to be real.) Our basic equax y z tion (2.10) becomes in this case P(k) = [1 - A3(k)] p(k) + B(k) + Q(k) (k real) (4.2) Decomposing p (k) according to (I.2), and using the factorization of A (k) provided by equations (2.34) and (2.355), we can write (4.2) in the form +(k) A+(k) + p_(k) A_(k) = [B(k) + Q(k)] A_(k) (k real) (4.3) Assuming that B and Q satisfy a Holder condition, equations of the form of (3) (4.3) can always be solved by a well-known generalization of the Wiener-Hopf technique. We let r+(k) A+(k) for Im(k) > 0 f(k) -- (4.4) -p_(k) A_(k) for Im(k) < 0 so that (4.3) is the statement f(k) - f-(k) = [B(k) + Q(k)] A_(k) (k real) (4.5) where we use the notation of (I.6). We see that f(k) is analytic in the plane cut along the real axis, and has a discontinuity across the cut given by (4.5). It follows that 1 = J [B(k') + Q(k')] A (k ) f(k)' -- dkt + A jim(k) 0] (4.6) ^T~l ~'~ K' -K

55 where the constant A is arbitrary. From (4.4) we have +(k) = 1 f [B(k' [ ) + (k')] A (k') A ~ -= / dk- + — (47) + 2ii A(k) L-00k' - k - iA + (k) 7) and the general half-space problem is solved. Note, from equation (2.36), that +(co) = A. Hence, for nonzero A, the density p(x,k) will have a S-function singularity at x = 0, and the choice A = 0 (4.8) clearly corresponds to taking the least singular solution as discussed in Section II.1. Aside from remarking that for B = Q = 0 there is no (nontrivial) least singular solution, we will generally confine out attention below to the case of equation (4,8), The Half-Space Albedo Problem The above remarks have their simplest application to the albedo problem, in which q(r,) = 0 => (k) = and p (Oyz,Q) = (y) 8(z) 6( - ) 2 > o, o > 0 ~S ~~~~O 0 BB(k) = - (4.9) w k 0 where cX is defined by (3.4). Note that o > 0 L= Im(o) < O (4.10) O O~~~~

56 We now have merely to substitute (4.9) into the formula (4.7), and perform an elementary integral. However, in this case it is perhaps more instructive to work directly from equation (4.3), which can here be written as (W - k) P (k) Ak) -A(k)[(o - k) P (k) + i] (4.11) w0 k a+ + 0 aThis equation, each side of which is analytic in an appropriate half-plane, is of the same form as (II.1.19). By precisely the same argument as was used in Section II, we conclude that both sides must equal a constant; and by setting k = o, and noting relation (4.10), we see that the constant is -iA (o ). Thus, -iA (w ) (k) = ( - ) Ak) (4.12) as we obviously would also have obtained from (4.7). With (4.12) and (2.34), we can now compute, for example iA (C- ) L+(k) - A o r -ikx ( k + i- e p (x,k) = 0 e a 2Tr — o' (k + i )(k - dk (4.13) 0 0 A By means of the same argument as was applied to p G, above, we find -r (-iK ) -K x p (x,k) = iA ( )( - K )e e -icw x'iA_(o o e -ik'x (k' + iP) -L+(k') dk' + e, e e dk? + x > 0 2j Tc'Y(k' + iK ) e A ) (4.14) where, again, the term in brackets does not occur when cd E. y' refers to O - - a path just outside the path 7, across which L (k) is discontinuous. Similarly, by (2.6),

-ic x A -7 5 (Pa(x,k,) = e ~ (Q - Q ) =O A()c (k+ i -)e-L+(k) -ikx A c f ~ () dk (4.15) 2 t - o 4T oo (k - w )(k - c)(k + )k (4.) and the integral can clearly be reduced to discrete and continuum modes, etc., in the usual manner. We will discuss the content of (4.15) in Part 6. A Half-Space Green's Function With the boundary condition of zero incidence at x = 0, cp ( 0,7) = o0 B( k) = 0 (4.16) and an isotropic point source at (x,0,0), q(r,,) = q~ 8(x - x ) s(y) 3(z) 1 1 A (k) Q(k) qeikxo0 A(k) (4.17) C our general half-space formula (4.7) yields ik'x e [A _(k') - A (k')] ikx ikx q 1 - - o q o!' (k) = 1 dk' + e + e pg+(k 2Tic A (k) dk k' - k - iO c c A (k) -Kz x A / \ ik'x 0 0 A (i1K ) oik 0 + q 1 l dk1 e A(k') c a^ A+(k)(iK - k) A (k) 2ri k k - iO - ok ~K (4. 18) where we have closed the contour in the upper half-plane [y1 refers, of course, to a path bordering on y+, analagously to y'_; by affixing appropriate super_-ripts o +

58 Af(k) A +(iK) Res[A_(k)] = Res[(] Afi (4.19) Aes (k)I = 5 o O ok i 0 The inverse transform of (4.18) contains little that is new: the first two terms are essentially the isotropic form of (3.1) and the third term is very similar to (4.12). The remaining integral term is clearly O(e ~) and, in the important case of large x, comparatively insignificant. We now consider this case in detail. The Milne Problem Here, we have the same boundary condition as above, but the source has been displaced to infinity, in such a way that the quantity -K X q - lim qe (4.20) X o x 0 ~~is finite and.~nos~nr~-zeroo is finite and nonzero. Using a certain amount of care with regard to the first two terms, we could read off the solution directly from (4.18); but it is somewhat more instructive to approach the problem from a different point of view, as follows. We replace the source at oo by the requirement that our solution, p (r), be O(e ~ ) for large x. This suggests the decomposition pm(r) = p() + p (r) (4.21) M ^_f 0 oo r(

59 Here po(r) = f(')e~ox (4.22) where f(r), which could be determined, for example, from the infinite space Green's function, is independent of x, and p(r) - (4.23) Our basic equation, (2.10), now may be written p(k) + - o(k) = [1 - A 3k) + (k(4.24) In view of the definition (4.22), the quantities p (k) and P (k) are not ob00 P00+ I — viously meaningful; indeed, in the generalized function space S' which we have hither-to implicitly used, they do not exist. But p (xr) is locally summable in x, and hence a well-defined generalized function in the space K' of Gel'fand.( It follows [assuming f(r) is suitably behaved] that the Fourier transform p (k) exists as a generalized function in the space Z', and in fact p(k) = 2T f(k) 6(k - i o) (k = k ) (4.25) Pooo0 Note that not only is the usual formula (I.1) meaningless here, but that the inverse transform A -I'2 00 ikx ( p^(x ) f e f(k) 6(k - i~ ) dk (4.26) might seem to give zero, since the "integral" is over real k, and is is not 0

6o (in general) real*. The correct formulae are, of course, obtained from the definition ( p,) = 2(P9) (4.27) where 4r(r) is a test function in K(Z). It happens in our case that p (k) exists in the classical sense: i(k-iK )x \ o.f(k) P (k) = f(k) _o e dx = ) (4.28) 00-'-/ v i too l(k- 1K 0 so we may conclude that p (k) = f(k)[2 (k - iK ) + k (4.29) 00+ 0'10 o k i0 (4 k2 i) 0 Now (4.24) may be written in the form [pI+(k) + p (k)] (k) = - A (k)[ (k) + p (k)] (430) or, using (4.29), [(k - iK ) " (k) + if(k)] A(k) = - A_(k)[(k - iK) p (k) - if(k)] 0 P(+ 0 - (4.51) = if(k) A (iK ) (4.32) *The point here is that po(k) is an analytic generalized function, the "support" of which is ambiguous. For example, it can be shown(l) that (n)(k)(-iK ) &(k - iK ) = Z- (in z') n

61 where we have obtained (4.32) by precisely the same Wiener-Hopf argument as was applied to (4.11). The result is, — - +(i A+(k) + (i) - iK (k) whence (+ k A A I) (k) To find f(k), we note from (4.18) that, for x < x, 0o o K X -K X -Px A q e 0 o eo p (xk)=i e e L J + O(e ) + (e ) (435) g C C 3A 6k iK 0 Thus (4.34) will be consistent with the application of (4.20) to (4.18) only if iq f(k) 0= 1 (4.36) c h3A k iK o Note that we could just as easily have obtained the above expression for f(k) from the infinite space, isotropic source, Green's function; the method used above for finding p is actually independent of that used for p. From ( 4. 7)( we find, in the usual way, Pm(x,k) =c 1 -f J m1 Ke + k- 2 (k - i ) A (k) a k i K 0 Note that we could just as easily have obtained the above expression for f(k) from the infinite space, isotropic source, Green's function; the method used above for finding is actually independent of that used for p5. m g From (4.37), we find, in the usual way, FA / * \/0 ~N x - ikx

62 It is clear that there exists a point x such that m ( (x,k) = 0(e ) (4.39) m m ioe, the asymptotically dominant component of p vanishes at x = x. The m~ m (14) definition (4.39) actually yields x < 0; it is conventionally denoted by m x = -z, z > 0 (4.40) m o o (z is called the extrapolated end-point.) Note that p (k), with which one would ordinarily compute p for x < 0, is not relevant here; since our original m differential equation (2.1) holds only in V, only p = + has physical significance. We have, from (4.38) through (4.40), -2K z A (iK )( - K )eL+(iKo O O + 0 0 e (4.41) 2K 0 -7 (14) which, with k = 0, is equivalent to the known one-dimensional formula. 5. SLAB PROBLEMS When the region V is the slab defined by V = (-Q < x < i, - < y < 00, -00 < z < o) we easily find - -7 - i(k-k')e A (k - k') (k -k' - k'e ) (501) ~~V 2~~ t irk - k' 8(k - k') ---- IkX — ) -\n = 8(k - k') -- -o k' - k- io (52)

63 where the equivalence of (5.1) and (5.2) is clear from (1.4) We restrict our attention to the case in which the inhomogeneous term, S(k) B(k) + Q(k) satisfies S(k) = S(-k) (5.3) so that [cf., equation (2.26)] it is permissible to assume p(k) = $'(-k) (5.4) Since, of course, any S(k) can be decomposed into odd and even parts, and the problem solved for each part in essentially the same way, (5o3) is not a serious restriction. Substituting this p(k) into (2.14), we find, from (5.2) and a few manipulations using (Io4), that —,' I-1 ikM -ikM pV() p(k) e J(k) -e J(-k) (5o5) where 1 oo -ik' + 2 Ti - o k' -k - io The form of equation (5.5) suggests the following decomposition of p(k): ( -ik ik P(k) e f(k) + e f(-k) (5.7)

64 and similarly -ik~ ik2 S (k) = e a( k) + e (-k) (5.8) [Note that (5.8) is automatic whenever any sources present are concentrated at the boundaries.] We easily find i o f+( k' )e -2ik'~ J (k)= f_(-k) +- J dk' f(k)e (59) 27+ 2[i o0 k' - k - io so that our basic equation (2.10) can be written in the form -ik~ ik ( ikFf 1 koo f+(k')e-2ik' f(k)e + f(-k)e = [1 - A(k)] (k) -- J dk' +( 3' ^ [+ 2Tii -co k' + k - io fo 1 0 f+(k')e-2ik'~ + e kf (-k) - -- dk' + 2 a) 2i - o k' - k - io -ik~ ik~ + e - a(k) + e a(-k) (5.10) ik~ -ik~ By considering the coefficients of e and e in (5.10), we conclude that if f(k) satisfies -2ik'_ 1 f+(k')e f(k) = [1 - A3(k)] f(k) - fo( dk' + (k) (5.11) then a solution to the symmetric slab problem is given by (5.7). Equation (5.11) is similar in form to (4.2); the ansatz (5.7) has, not surprisingly, reduced the slab problem to a half-space problem. We could solve (5.11) by our general half-space formula (4.7), f+(k) 2Ti A (k) L-d k' k' - Ak-]o (k 1 o A_(k') g(k') + A(k') I(k') 2ti A^ (k) - k' k' - k - io (5.1)

where f (ke-2ik'~ dk f,(k')e I (k) - = dk' (5o13) -2^* i' Ck' + k - io 5.15) except, of course, for the fact that I_(k') is not a known function. On the other hand, whenever f (k) is analytic in a region (Im(k) > - a; a > 0} it is evident that I_(k) is O(e ), i.e., small for large I. [We remark that the only important problem in which f (k) does not have this property is the socalled critical problem; that the method to be outlined is applicable even to this "worst case" is demonstrated explicitly below.] What is clearly called for is a perturbation expansion of f+(k), the zeroth approximation being given - 2nc~ by (5.12) with I = 0, and the nth correction term being O(e ). Thus we set f+(k)=: fZ )(k) (5.14) n=O where f(O)(k 1 O T(k?) A (k') f(0) (k) f dk' - (5.1.5) + 2(k i A (k) - dk k' - k - io and I(n-1)(k') A (k') f(n) = 1, (dk'1k n ( 6) + 2r i A (k) -m k' - k - io 1 ( where f(n-) (k)e-2ik -1 JW dk + I n -1 )(k,) = -J dk' n>l (5.17) — 2~ri _oo k' + k - io -

66 and, assuming the series (5.14) converges, the symmetric slab problem is solved. The iterative procedure of (5.14) through (5.17) is not as formidable as it may appear. In fact, the representation (5.14) can easily be reduced, in good approximation, to a simple power series, provided we stipulate (i) reasonably well behaved a(k); for example, the "symmetric slab albedo" problem, for which a(k) is given by (4.9), is easily tractable, (ii) a wide slab; explicitly, the power series obtained is correct to O(e ), the means of obtaining greater accuracy being clear in theory but cumbersome in practice. We illustrate these remarks by solving a problem the formulation of which is only slightly different. The Critical Problem Here we assume there is neither source nor boundary contribution, S(k) = O and seek that value of ~ such that there is a nonzero solution to the slab problem. The prescription of (5.14) through (5.17) is evidently not directly applicable; in order to obtain a nontrivial solution for a = 0, we must choose the constant A of equation (4.7) unequal to zero. With this minor modification, we have f+(k) = Z f(n)(k) n=0

67 where, f(o) (k) = A (518) + A (k) and the f n)(k) are given by (5.16) and (5.17). Note that f (k) is analytic for Im(k) > -P except for a pole at -iK o + 0 [We anticipate that for criticality, c > 1, so that K may be pure imaginery. Note that in this case the pole of (5.18) is to be interpreted as lying just below the real axis, as discussed in Section I.]. By induction, we assume that f(n-) (k) also has this property, whence (R(nn)o -21 o -2 (n-)(k) - - - e 0 +-(e )-2 (5.19) k - LK - io 0 where R(n-1) ) Res f(n )(k) (5.20) ~0. - iK 0o Now, from (5.16), (n-l) (e -2 o~ (n)( k ( K_ f (k =A- k)0Ri' )+ O(; 2 (59+21) + (k - i (i) A(k)+ 0e ) (5e2 and we observe that: (a), the induction hypothesis is fulfilled to 0(e ); and (b), the error term for each I(n) will be of the same form as that for I(): -2ik'~ Ti(o) A /d~ e -I( - fi A (k')(k' - k - io) (5.22) 7 + Hence a systematic means of obtaining more accurate solutions to our problem clearly lies in an appropriate asymptotic expansion of the integral (5.22).

68 With the abbreviation ~1 -i >1 -L (-iK ) R( )- Res = i(A - K )e (5.23) -iKL~ +(k it follows from (5.21) that (n) -2K ~ A (iK )(K)=-e 0 + o (n-i) ^'-R (K0i e ~ ~ R(K ) R (K.24) o 2iK 0 0 r R(n-1)(K) (5.25) 0 = R o)(K) = Ar R(K ) (5.26) 0 0 where (n) e2Ko A (iK ) r (n 2ilK = R(K0) (5.27) -2 R so that, to O(e ), (5.14) is the power series given by -2K A+(k) - A+'i m f+(k)= f( (k) - R( )( ) e k i r (528) A AR(K )e- 2 0A(k) - A+(iK A (k) 1 - r +(k)(k - iK ) Of course (5.29) follows from (5.28) in general only for Irl < 1. However, it is not hard to verify by direct substitution that the f (k) of (5.29) is in fact an appropriately approximate solution to (5.11) for any r / 1 [of course r must be given by (5.27)]. It is convenient to choose A = (1 - r), so that (1 - r) -2K o A (k) 1- A+(i f (k) - (rk) - R(K0) e [ 2 -Kj (5.50) + ~~- A()o I () k -

69 It is now clear [cf., our discussion of equation (4.7) above] that the density obtained from (5.30) and (5.7) must have the form $(xk) = (1 - r)[6(x + ~) + 6(x- ~)] + regular terms (5.31) and we conclude that the only nontrivial solutions to the homogeneous slab problem are singular, unless r = 1. We now consider this last case, which is the critical case, in detail. When r = 1 we have the (nonsingular) critical solution given by (5.7) and* A (k) - A (iK ) f +(k) = (const.) A (k + (5.32) AC(k)(k i~ ) + 0 Keeping in mind the fact that both multiplicative constants and, in our approximation, terms of 0(e ) are irrelevant, it is a simple matter to obtain the inverse (k = k) Fourier transform of the density as given by (5.7) and (5.32): x -K (X++) 1 oo -ikx -ik~ -K~(x-v) - Jf e [e i f (k)]dk c e + O(e ) (x > -Q) (5.33) and 1 0 ikx ik, - o ) - 2 f ef [e f +(-k)]dk e 0(e (x <) (5.34) whence p (x,k) = (const.) cosK |Ix (5.35) *Cf., equation (4.55).

70 The somewhat more interesting criticality condition on I is obtained of course from (5.23) and (5.27): -2K I o e A (i~ )( - K) Using (4.41), this can be written as -2K (~+zo) e = 1 (5.37) so that K0 must be pure imaginery, as we have assumed, and the critical value of i is Ir c 2IK| oI Both the results (5.55) and (5.58) are well-known.(l4) 6. THE k-DEPELNDENCE We have always been able, above, to bring the inverse Fourier transform with respect to kx of our solutions into fairly workable form [cf., for example, equations (3.6) and (4.538)] On the other hand, the dependence upon k = (k k ) y z of such functions as A+ is evidently too complicated for the inverse transforms with respect to k to be performed by any simple analytical procedure. The fact that useful information can be nonetheless obtained, without recourse to numerical techniques, from the solutions given above in (x,k) space, is here briefly illustrated. Our remarks are based on the identity

71 i- 00 (ik )n( ik )m - n f(xk) ff( md= f dry z f (r) (6.1) n,m=O from which it follows that an expansion of the (known) function f in powers of k and k immediately yields the moments of f with respect to y and z. These y z moments, which uniquely determine f(r), are generally of physical interest. In particular, for n = m = 0 we have f(x, ) = f(x) (6.2) where f(x) dr f(x, ) (6.3) is the average of f over r. Of course the relation "inverse" to (6.1), 0o ~ (-ik )n(-ik )m f ((r)= ynZ S dk fm(xk) (6.4) f(~) _ \ n' m (2e-) n,m=O would analagously furnish a power series for f(r), if we could compute the mo-"7 A ments with respect to k of f. Since, however, the latter computation is apparently no easier than computing f(r) directly, we do not pursue this point. The formula (6.2) can be immediately applied, for example, to equation (4.15). After some routine calculations [as in the derivation of equation (356)b] we find that the r-average of the albedo angular density is asymptotically given by -x/~ (v - 1)B e7(7~)-7(wO) -x/v cp(x,) = ( - e ( - t (o ) e + O(e ) (65)

72 where v is defined by (2.23), and () = r(-i/4) - r+(o) (B = 0) (6.6) The emergent angular density, P(O,,), is even easier to compute, since for x = 0 we may close the contour of (4.15) in the upper half k-plane, with the (exact) result that _ 7( )-7( 4 ) (OQ) = ) c(40- v )( - l)e o p(o,0) s( )- ( )l -- - - vZ)' —-- (6.7).~ ~o 4Tr (- 1))( - V Here we have used the representations (2.45) and (2.46). (Note that 0 < < < 1 i= -i/l. c _.) The corresponding density is given by P () =A_( o)! (6.8) a -- 0 B o - F F.(-i/.o) (. l r(e (B = ) (6.9) 0 0 Using (2.6) and (43.7) we similarly find, for the Milne problem v2 Y(v0)-Y() ~( o,T) - 1) (6.10) V *2t a (1 - v )(p + v )(v + 1) ak i/v 0 and r,( i/v ) q + 0 p(o) = i- (6.11) ak i/v 0

73 The Emergent Current We define J(r) - d Q cp(rS ) (6.12) The total emergent current for a half-space problem is clearly given by J (0) where x J(x) -= J(r)dr7 (6.13) and is very simple to compute. From the easily verified continuity equation V.J(r) (c - 1) p(r) (6.14) we find that J (x) = (c - 1) p(x) (6.15) dx x i.e., J(x) = (c - 1) fx p(x)dx (6.16) 2=A - o (c 1) x dxv (f dk e p(k,)} (6.17), - ikx c- - 0 1 ~ dk P(k O)-e (6.18) 2i -oo k - io whence J(o) - (l - c) (0) (6.19)

74 Using (6.19), we find, for the half-space albedo, |o(v - ) - 7(l) J =( c) -- " 1 e (6.20) x (1- ) o' and for the Milne problem i(l - c)v q O-v J (0) = - 00 e (6.21) ak i/ All the above results are equivalent to those obtained from the one-dimensional theory. [Note that 1 2 MA - v - - v = -i/k (6.22) (14) where A(v) is the function defined by Case.( ) Smewhat more interesting, therefore, are The Higher Moments For B > 0 but still small, we could proceed as follows, A general halfspace equation can be written as P (k) A (k) + (k) = S(k) (6.23) We expand the unknown functions 2 p (k) = +(k) + B P1+(k) + B p (k) +. (6.24) and the known functions A3(k) = A,(k) + B X (k) + B X4(k) +..+ S(k) = So(k) + B S,(k) + B S (k) +... (6.25) 01 2..

75 and find, by equating powers of B, po+(k) A (k) + p (k) So(k) Pl+(k) Al(k) + P1_(k) S1(k) p2+(k) A(k) + p2_(k) = S2(k)- p+(k) X2(k) (6.26) etc. In general, we must solve Pn+(k ) + _(k) = R(k) (k) ( (6.27) where R (k) is known in terms of SO,...,S,PO+..'P.(n2)+ and X2,...,X. Of course (6.27) is of a familiar form and its solution presents no difficulties: 1 B (k') A_(k') P+(k) =o dk^Ak (B 0) 6.8) n+ 2 ti A (k) -oo k' - k- io28) To obtain the X, we begin with (2.16), (\ crr dQ A3(k) = 1 - 4 J- if - - 4r.1 - ik4 - aB where 2 a ~ il - p cos(cp - A) (6.29) [cf., equation (2.18)]. For real k and B < 1, 11 - ikl| > jaBI so that we may expand 00 A3(c. 1 1 l f d ( B - )n (6 o) ^3 (k)l - 4-i 1 - ikZik *"~~-1~ n=O

76 The cp-integral is trivial and we find that (6.25) holds, with 2 -1 (i - (6ik1) The procedure of (6.24) through (6.31) may be applied, for example, to the half-space albedo problemo Here o 2 S(k) = 1 - ik~ - Bm; 0 = i ~ 1 (o 2 ) (6.) S(k) =; a i 1- p. cos(cpq - A) (6.32) l- ikp. -Ba o o o whence n o0 0 S (k) =n+ (6.33) (1 - ik4 ) Using the further result that 1 c X2(k) = [l (k) - ^ (6,34) 2k 1 + k we find 2 - cos(cp -) 1 R (k) = i -- - (6-35) 111o (k + i/(o)2 R2 () 2= 23 1 (k (k + i/. o) 2k 1 + k (6~36) so that equation (6.28) gives i A_(-i/p.) Po+(k) =(k + ~i/p,) A(k) (6.37) 0 +~~~~~~6.7

77 - i 2 cos(po - A) 1+=(k) - (k + i/% ) A+(k) k(6.38) 0 i(l - c ) cos - ) 1 A i A (-i/) (6.39) where 2 gk _ k + -c _ g(k) - 1 (k + 1) Al(k) and, of course, the A+ functions are to be evaluated at B = 0. Note that, because of the cosine factors in (6.38) and (6.39), the expansion (6.24) is of the same form as (6.l). In order to extend the half-space procedure of (6.24) through (6.31), it (14) is most convenient to return to the differential equation (2.1). Let 4 (xk ) - f dr z p(rQ) n2 fn(x) - dQ rn(x jQ (6.40) Similar functions could obviously be defined for the other types of moments; we consider only (6.40) for simplicity. By multiplying (2.1) by z and integrating over all r,* we obtain (assuming, for convenience, q 0), d c d + ) n(x,) f (x) + nQ n1(x, ) x e V x(x sQ) rns(X s ) Q inward (6. 41) *- us sum ht h oudryo i nepnet fr-7 *We must assume that the boundary of V is independent of r.

78 where, of course,,ns(, -f dr z s(r, (6.42) Hence, for each n we have only to solve a one-dimensional problem; this can be done by our general method of Part 2, with B = 0. Considering once again the example of the half-space albedo, we find f (k) n+(k,') /J Cc n+ n+ 31 (k — + (6.43) *n k,) =4- 1 - ikp 1 - ik where Qo+(k,') E +- ) (6.44) 0+4 k + i/p o 0 Q+( k) nQ (n (k,) n > 1 (6.45) z (n-l)+ and _ 1 dk'A (k') Qn+(k', f' (k): j i ddktQ (6.46) n+) 2fri l (k) k' - k- io - ik which formulae are completely equivalent to (6.37) through (6.39).

V. A CONJECTURE OF KAC Let p(x) - f= e e x dk (1) 2 T -00 where a is any positive number. For T < co, the integral equation T Xc(x) = f p(x-x' )c(x' )dx' (O < X < 1) (2) possesses discrete eigenvalues X and eigenfunctions pn (x). It has been conjectured that, for T large, -= 1 - + o(T-a) (3) n a T where pi is independent of T. The purpose of this section is to prove the stronger result nit a -a- -2a = 1- () + O(T -) o+(T )(4) n 2T -a- 1 and also to obtain the eigenfunctions p (x) to O(T ). The similarity between (2) and the critical problem of section IV is evident, and will be exploited below. The differences between these two problems, therefore, might well be made explicit here: (a) The function V(k) [equation (6)] which plays a role here corresponding to that of A(k) in Section IV, has, unlike A(k), both zeroes and branch points on the real axis, with the result that the Wiener-Hopf factorization of V(k) proceeds somewhat differently from Section IV. 1. 79

80 (b) We seek here, not just one critical width T, but an infinite sequence of eigenvalues x. Since the validity of a Neumann series does not in general extend beyond the first eigenvalue, the argument of Section IV. 6 is no longer convincing and we must determine our eigenvalues by a different procedure. With these two qualifications, the method of Section IV may be applied to the eigenvalue problem (2). For purposes of orientation, it is convenient first to briefly consider 1. THE T = o CASE When T = o, the Fourier transform of (2) may be written in the form V(k)p(k) = 0 (5) where V(k) - - p(k) = X - e- (6) V(k), which need not be defined for other than real k, has two real zeroes: V(~k ) = 0 o where 0 It follows that the two linearly independent solutions to (2) may be chosen as (k;o) = A(5(k-k ) + 6(k+k )) 1 o0 (8) (k;om) = B15(k-k ) - b(k+k ))

81 These satisfy the requirement cp(-k) = + ((k) (9) which may clearly be imposed for finite T also. The inverse transform of (8) yields gPl(x;o) = (constant) cos k x (10) 2 (x;oo) = (constant) sin k x Note that there exist solutions for each k of (7); the T = o case is charactero ized by a continuous spectrum: 0 X < 1 (11) 2. FACTORIZATION OF V(k) We now return our attention to the case of finite T, for the solution of which a Wiener-Hopf factorization of V(k) [equation (6)] is crucial. Observe that the function B(k) i V(k)(k2 ) where D is any real number, is continuous on the real k axis, and vanishes at o. Thus 1 ~ (k1'( B(k) - f dk' (13) 2iti -oo k'-k is analytic in the plane cut along the real axis, and its boundary values satisfy

82 B (k) - B (k) = i(k) k real (14) (14) is equivalent to 2 2 BB (k) V(k)(k+ ) +P e_ 2 2 - (k) (15) x(k2-k2) eB-(k) 0 so that with the choice (k2 k2) +, (k k) V(k) = (e (16) + (k) (k+ip) B-(k) V (k) = (k-i)e (17) we have V+(k) V(k) = V (18) v (k) where the V+(k) are analytic in appropriate half-planes, and V+(k) k k-*oo (19) Note (cf., our discussion in Section I. ) that the superscripts in the definitions (16), (17) are relevant only when k is real. It is convenient to derive here three properties of B (k) which will be required below. (i) B (k) can be analytically continued into a function cut, not along the real axis, but along the path r, below it (see Figure 5).

83 (2) <a a k- Plane Figure 5. Integration contour for B (k), Section V. G — max(n/2, n/a). We show that B (k) can be considered analytic above r as follows. Let k 2 12 2 -(kX- k2) 3 (20) Then B(k) = B(k) + B (k) (21) 2 where B+k 1 _ l(k') B (k) = 2 dk' ( (22) 2( 2r5i o k'+k-io ( 1 a (k'a ) + _]_lan

84 It is clear that the integration path for B (k) can be deformed down to the line 1 r() of Figure 5, by Cauchy's theorem. Similarly, that for B (k) can be de2 formed up to (-r ). Thus the two functions are cut along and (2) respectively (r - r(1) + r () (ii) a) If a is an integer, B (k) and its first a derivatives are finite near k = 0. b) If a = n + P, 0 < P < 1, n a positive integer, then B (k) and its d(n+l) +(k) k-1 near k = 0 first n derivatives are finite at k = 0, while (n+l) B (k) k near k = 0. Since property (a) is fairly self-evident, we prove only (b). Here, the general statement becomes clear upon consideration of the case n = 0. From (22) and property (i), + dB1 -1 d 1 2~i k(-) l(') -- (24) dk 2iti (i) 1 dk k-k We integrate by parts d+ d 1(k') dB1 -1 1(~) 1 dk' dkT ~ ~- 2 - + ^r iki)k (25) dk 2Tti k +2 i S (1) k'-k (25) F Similarly dB+ d(k') B2 1(0) 1 dk' - T + - f \(dk' (26) dk 2nik 2ri (2) k'+k -r whence

85 dBl dl dB 1 dk' 1 dk' fdk + —-- d dk' (27) 2k i d' 2i (2)) kk (2'7 d2 Now (for n = 0) - is singular at k = 0: dk dfi dP1 k-_1 (k 0) (28) dk from which it follows by a well-known theorem that + dB ~ a-l B k k O (29) dk For n > 1, we can continue to integrate by parts so long as the boundary (k=O) contribution is finite and then proceed just as in equations (28), (29). (iii) If b(k ) is defined by o 2ib(k ) = B+(-k )- B+k ) (30) 0 0 0 then b(k ) is realo o We have B (-k ) - B (k ) - dk' B(k ) +k- k- -io (31) o o 2iti-oo'+k -io k'-k -i o o Using equation (I.4) and the fact that B(ko) = B(-k0) (52) we find B+ o - B()=2 JdkB)ik (-k ) - B(k ) - ) ()dk' () B( k [tk' -k o 0 2r-x -o_,2,

86 so that the reality of b(k ) follows from that of ~(k). o Note also that b(k ) = O(k ) for k 0 O, as is evident from the definition. o o o 3. THE EIGENVALUE PROBLEM We begin our solution of (2) by closely following the argument of Section IV. Assume p(k) = p(-k) (34) in which case the Fourier transform of (2) takes the form ikT -ikT \cp(k) = p(k)(cp(k) - e J+(k) - e (-k)) (5) where -ik'T 1 o Cp+ (k')e J+(k) - dk' -- (36) 2 i -o k'-k-io and the ansatz.... -ikT ikT cp(k) = e " (k) + e 4(-k) (7) yields for V(k) the integral equation xi(k) = (k) (k) I_(k)) (38) where 1 +(ke-2ik'T I (k) -2i dk' + o (39) -00 k +k-io Equation (38) may be written in the form V(k) % (k) = -k i (k) - [X - V(k)]I (k) (k real) (40)

87 or, using equation (18), V(k) *+(k) = -V(k) i_(k) - [V (k) - V+(k)]I(k) (41) We define the function f(k) by V+(k) V+(k) Im(k) > 0 f(k) =V (k)[ (k) + I (k)] Im(k) < 0 (42) so that (41) implies f+(k) - f (k) = V (k) I (k) (45) whence 1 V +(k')I (k') f(k) = C + -- dk' -k- Im k 0 (44) 2mri - k'-k-io provided the integral exists. Now + v (k')I (k') f(k) C 1 ( ( ) +3(k) = v+k = vS+ dk' (45) (k) = V (k) V (k) 2niiV (k) - d k'-k-io provides a relatively simple integral equation for ~+(k). This is to be solved in the familiar way. Let c (n) +(k) = Z f (k) (46) + n=o + where, choosing C = 1, ( )(k) = v (47) and

88 (n) L _0 (i + (k) = 2r-iV(k) k'-k )io-)' where (n) -2ik'T ~(n- ) 1 0 +% (k')e I(n-1)(k) - 1 dkt — (49) 2iri -ob k'+k-io We will carry out the program of (46 )-(49.) below; but first it is necessary to determine 4. THE EIGENVALUES () It follows from (19) that the integral of (44) will exist only if I (k) = o(k1) (50) which implies 1 00 -2ikT 22i f dk (k)e = 0 (51) Equation (51) is our eigenvalue equation. Since it must hold independently of T, we have, for each n, 1 7dk (n) -2ikT (52) 2 dk ~ (k)e ( 2Ti1 -oo + so that (48) is meaningful. In particular, -2ikT 1 00 e i dk Vriky= ~ (53) 2iTi -0o V+(k) where the zeroes of V+(k) are to be considered as lying just below the real axis in the usual way. Because of property (i) noted above, (55) may be written in the form

89 -R(k )e-2ikoTR(-k )e2ikoT 0 0 -2ikT 1 e + 2-i dk Vk (4) a where 1 R(~k ) Res V k] (55) 0 ko V+(k) Now it follows easily from property (ii) that the integral term in the -EY- 1 (exact) eigenvalue equation (54) is O(T ). For example, in the case of property (ii b), we can integrate by parts n+l times, obtaining 1 e-2ikT -l n+l 1 -2ik'T d(nl) 1 f dk - =7 ( )n) fdk'e -2ikr (T6 2 -1 2i dk V (k) (-2iT (nl) V (k') (56) Now the integrand is singular at k=o, where it behaves like k (3 = a - n). Considering the contributions from rF and () separately) and using a wellknown( "Tauberian" theorem, we see that the integral behaves for large T like T, so that -2ikT fdk ~ TT (57) 2ti dV(k) T Hence for large T we have the eigenvalue equation R(k )e 2ikT + R(-k )e2koT 0 (58) o o which is correct to O(T ). Using (16), it can be written as

90 4ikoT _k +i + 4ikoT_ = o -B (ko) + B(-ko) e (59) k -iP o It is clear from our discussion of the T = co case that, for large T, the first (i.e., largest) eigenvalues will be close to 1. Now k = (+ In 1)1 o X = (l-X)l/ + O(-(1-X)1) (60) so that for sufficiently large T, k << P. Using also property (iii) we find, from (59) 4ikoT = 2i b(ko) + 2inr (61) But since b(k ) is O(k ) for small k, we have, for large T nit 2 k =- + O(T ) n = 1,2,... (62) on 2T Equation (60) now gives X 1 -(T) + o(T ) + O(T2 ) (63) n 2T the desired formula. 5. THE EIGENFUNCTIONS We find the eigenfunctions in the same approximation by the prescription of (46)-(49). First note that O (k) is analytic above r except for poles at ~k. By induction, we assume this is also true for 1+ (k) and let R(m-1)(+k ) = es ( (k) (64) o

91 With the observation that 1 _1 k' 1 k'+k-io k k k'+k-io ( and equation (52) we find (m-l) -2ik'T 2ik T If(k) = 2 ik; k' m (k')e R k(k )e2k M-^ -_-1 7, + ('e ~ 1o) ( l)(k) 2ik _-k k'+k-io k k+k -io 0o kR -1)(-k )e 2ik 0 0 o o (66) k-k -io 0 ~~~o_Here the integral along rF, which is easily seen to be at most O(T ), has been dropped. Another application of (52) yields r-i) 2k R' (k)e-2ikT I (k) = (67) (k- k)where the subscript in the denominator reminds us that the poles are to be interpreted as lying just above the real axis. Now (48) gives 2k Rm -l(k )e -2ikoT V (k') (k) fdk' (k) 2irilv +(k) -00 (k'-k-io)(k'2-k2) 0 - 2k R(m-l(k )e-2ikoT o o 0 0....2 --- (68) (k2-k2) as is clear from equation (16). We observe that (a) the induction hypothis i f (miswh in 5d(r) esis is fulfilled for +), and (b) ) is corsistent with equations (52) and (62). It is, however, clear from (68), and not surprising, that the series (46) will not converge for every eigenvalue:

92 R((k ) = e2ik R( (k) (69) o 0 R()(k )= 2i R(k) k ) = e-2imkT R(k (70) o o 0 2k R(k -2imk + (k ) + 5 m e 2imkoT (71) # -V+(k) (k2_k2) 0o Restricting ourselves therefore to the first eigenvalue [n=l in equation (62)], we have "convergence" in the sense of Cesaro: 00, (-1)n (72) nt-D-'oi (72) - 2 and the eigenfunction is k R(k ) 1 o o (k)*^ * (73) + __ (k) = V (k ) 2_k27) 0 It is now a simple matter to verify that this solution satisfies (51) [with -cz-1 the usual error of O(T )] for each k given by equation (62), and not merely 0 the first one. Furthermore, it is easy to check that, to O(T ), the +(k) of (73) is indeed a solution to equation (38), again for each k of (62), thus justifying the use of (72). Having *+(k) [_ (k) is of course not needed for |xJl T], we may take the inverse Fourier transform of equation (37) to find the even eigenfunctions of -Q- 1 (2). Dropping as usual contributions of O(T ), we easily find np (x) oc sin k T cos k x (74) n on on i.e., the only nontrivial solutions obtained in this way are for odd n. The solutions with n even are found by replacing equation (34) by

93 p(k) = -c(-k) (75) and proceeding analagously to equations (34)-(38). (The only changes are sign changes in these equations). The result is not surprising: (x) cos sink T s k x (76) n on on which vanishes for odd n. We conclude that the normalized eigenfunctions of equation (2) are approximately given by (P (X (x) 17 i (77) 2n+l 1 ( x 2T n__ 1 niT (x)) =^ sin - x (78) 2n = sin T where the error is O(T ). Note that these have the form of the T = oo solutions, except that k is chosen such that cP (~T) = 0. (80) We remark finally that another conjecture ) related to (3), is false. With nc given by (77) and (78), we define the functions f (x) = cp(x;T = 1) (81) and the notation 1 (fg) = I fn(x)g(x)dx (82) Then the conjecture we wish to consider may be written in the form

94 f (x)(f l)? /2 - ^ ^n ~? 2 Z )l "-=- C()(l-x ) (85) n Here C depends only on a and we have found [cf. equations (3) and (4)] nc n = (n2_) (84) n 2 Since (f f m) mn (85) (83) is equivalent to (fl) 2 / - C(a)(fm (l-x2) ). (86) m / Of course both sides vanish when m is even. For m = 2n+l, a trivial caluclation yields 2nl = 2(-l)n[(n + 1 ]-c l (87) n2n+l (12) But it is known that a2/2 1 2 /2 i(n + -)tx (f,(l- ) / = (1-x2) e d -1 a+1 n 21 T ( 1 r(a + 1) L 2 J[(n+)2] (88) 2 whence the invalidity of (83) is evident. In fact (83) is true only in the special case ca = 2.

VI. SOME QUANTUM FIELD THEORY PROPAGATORS We derive below generalized function representations for certain propagators of quantum field theory. Since most of our results have already been rge and(17) obtained by Gorge and Jauch 7 the significance of this section lies mainly in the method used. This consists of a systematic application of techniques due to Gel'fand.() It is well-known that the generalized function definitions k % ~ ik\t1 k (x~io) - x + e x_ (k -n) (1) n- 1 (x~-io) n x + (n (x) (n=1,2..) (2) can be extended to the case in which x is replaced by a quadratic form in more than one variable. Consider the Lorentz invariant form 2 2 2 2 p + m - p - p + m ~ ~ o The generalized functions u(p) (p2 + m ~ io) (3) exist in S' and we have, in analogy to (1), 2 2 2 2 ~ir 2 2 x (p + m ) i) = (p + m) + for k i -k, and, corresponding to (2), k- 1 ~( \ (2 2-k - ijr(-l) (k-l)( 22 m 2 uk(p) - (p + m ) + (k) 5 ( P + ) (5) 95

96 We will use in particular the functionals -k(p) 2 [uk(p) + -k(p)] = (p2 + m)k (6) _-k - k11 (-) h 2 6Bl(p m (7) u (P) - i(-l) u p) - u (p)] = 2n (k-l)2 2) where the right-hand-sides are given by 42 2-k (Uk, P) lm f d p(p +m ) c(p) (8) Y l I-mp2+m2l > ~ and -k-)1 002 (k-) (p, ~ +12+ (u P) k P dp [ +m - Here p - [.1 and - +2 2 2+2-! cp(p, +p +m -Q= /d cp(, J p +m -Q) (10) The definition (8) is an obvious four-dimensional extension of the ordinary principal value, while (9) results from the change of variable 2 2 2 Q = p +m - p., whence 2 hsi 5 (k)(p 2+ 2)o(p.,po P 2dpd2~dq r sjp dpddQ (k) -2 ) ( 112) T f e s ~t op, r, t (p2+m2 Q) 2 p +m -Q This is easily seen to reduce to (9).

97 Now consider the differential equations (O - m )kf(x) = (-1)k 4(x) (12) (- m2)k fo(x) = 0 (13) Here 2 2 ax o and we abbreviate 84(x) = (( x)o)(x )(x2)6(x3 The generalized functions discussed above lead naturally to particular, Lorentz invariant solutions of (12) and (13); these "canonical" solutions are precisely the propagators of quantum field theory. (12) and (13) have the Fourier Transforms [defined by four-dimensional extension of (Iol), with transform variables p, p ] 2 2k - (pm2)) f(p) 1 (14 (p 2+m2)k fO(p) = 0 (15) It is clear that the generalized functions u k, uk [equations (5), (6)] are solutions to (14), while the uk of (7) is a solution to (15). The uk can be j N"O expressed as linear combinations of uk and u k to which we therefore restrict our attention. We compute the inverse Fourier transforms of

98 f(p) = uk(p) and f~(p) u (p) by means of the known fact that, for Re(k) > 1, -2+ K [m(x2+io)1/2] -1 r=~ - 2+x J F [u]=i m _i _, (16)_ (2t) 2 r(-%) (x ~+io) 2 o2 where x = xx - x (16) can be written in a more useful form by using the identity -i --. 2 (1)t K (-iz) - e H (z) v 2 -v and equation (1). We find (2) K [m(x2)1/2 H1 )m(x21/2] K2 -2 *H(1) 2 1/2 2 K [.m x22+. /+ - ijr -(+2)[m( ) - I K <[m(x +io) ] =' + +lr -(- +L) (17) 2+[m(x ~io)] = 2 1/2(x+2) +2 (x )1/2(+2) (x _ 2 / whence 1u- +) 2X r ) x 2(1/2) 2, ( 1 L + (m jr2i J (X+2)[m( x 1/ u0(x) = lux + u2 I = i(2) (18) 2+x in(%-l) K [m(,x2)1/2 O(x) ie(X-l) u+ - m e 2+ + ] % x (2Ti) 22(% 2) r(- )L (X2)1/2(2- +) N_ ) [m(x2) 1/2 2 (x2)1/2(2+k) V19) (x

99 Equations (18) and (19) were derived by assuming Re(k) > -1, in which case the transform can be computed by actually performing the integrals. However, by considering the usual series expansions of the Bessel functions, and the known analytic properties of (x ), it is not hard to see that the expressions on the right-hand-sides are entire functions of k. Thus, by analytic continuation, we may take (18) and (19) to be true for all X. In particular, letting k-+ -k, we have our desired propagators: 6 6 2 2-k - (2-k) u () 4l(x) + ~m (x2) J [m(x2)1/2] (20) uk() 4t k - k-2 - 4jr2 (k-l): ()k-l 2[-k mK (x2) 1/2 N [m(x)1/ o (1 k-2 + Lk- 2 - u (x (21) -k2 22k(k-1)-L (x2)1/2(2-k) 2 (2)/2(2-k) [The reason for the extra term in u (x) is as follows. Upon expanding the Bessel function in (20) in a power series, we find o -(X+2) (x ) u (x) = N(\) - -l) + regular terms. (22) 2 -(x\-2) Both (x ) ( ) and (-X-l) have simple poles at X = -1, with known residues. The first term of (20) results from an application of L'hopital's ruleo] The case in which m = o is perhaps more interesting. We wish to find the canonical solutions to k g(x) = (-)k 54(x) (25) k o f m g (x) = o (24)

100 We denote these by v(k) (p), -k g0x) = v (x) -k and require them to correspond in some natural way to the u() (x) found above. -k Observing that, from (21), lim u (x) does not exist in any ordinary sense m+o -k 0 0 for k > 1 we might conclude that the v 2 v, etc., cannot be defined. In p-space, however, the situation appears somewhat differently. The limit lim 6(p2+m2) = 6(p2)" m-+o does exist, in the sense that there are well-defined generalized functions No, g (p) which satisfy 2~0 p g (p) = 0 In fact, there are two (linearly independent) such functions, which differ by (k-2) (constant) O (2)(p). Thus the difficulty is actually one of uniqueness; we must insure that our solutions to (23) and (24) are the canonical ones. This difficulty is easily surmounted. We require first that Re(\) > -2, and define v^(x) - lim u(x) Re(\) > - 2 (25) m-o u R) The limit exists and we find ~2 -k+2 v-(x)= i 22 r(= +2)(x2io (2+) (26) (2a)2r(-\.)

101 It is clear that the prescription vk(x) k v(x) (27) 1 + _k(x) =2 [v (x) + v (x)] (28) k~)(x) 2 -k -k v (x) i(-l [v k(x) - vk (x) (29) 0 would yield the m = o analogues of uk and u, and thus the desired canonical solutions to (23) and (24), provided the limit (27) exists. The fact that, for k > 1, this limit does not exist, requires only a slight and conventional modification of (27). 2- - (\+2) Specifically, it is known that (x +io) and F(\+2) have their only singularities-poles-at x = 0,1,2,..., and = -2 -3,..., respectively. Thus vX is well-behaved at h = -1, and we have, from (26), v (x) = i 2 (x +io) (30) (2r)2 For \ = -2, -395.., the formula (27) is useless and we use instead () limk a [(X+k)v(x)] k 2 1) K -k - + + Thus defined, vk is the so-called "regular part" of v at X =-k. It is clearly the constant term in the Laurent series for vk near \ = -k. 2 - (x+2) Using (26), and the known Laurent series for (x +io), we easily find -2k k-2 v- (x) = + (_l)k 2 (x2) [lnjx2 I~i( -x2)] (52)

102 Equations (28) —(2) furnish the desired propagators: 5(x2) v -(x) = 4 (33) l')k-2 -2k v (X) =( -1) 2 (x2k-2 e(-x2) k > 2 (4) v-(x) 2 (x2)-1 (35) -2k 2 r 0 2-2k 2k-2 2 -k(k) 2= (x) lnjx2I k > 2 (36) = (k-l)'(k-2): We remark that -k^ (. {k (x) l= im (37 v_ (x) U (x) while for vk(x), k > 2, we have only the correspondence of our definition: v x) = Regular part at x = -k oflim u(x k > 2 (8) -k M 3o) \X Lk>-2

APPENDIX We show that, for 0 < < T, c -iKrsinhT+iKr sinh(T+it) = 2K ( l A. e dT = 2K ( |rI'roI) (A.l.) — 00 0O as claimed in Section II.2 [equation (II.2.42)]. Beginning with the representation 1 CO oo ^-C~oo (<r- ) i dik e (r-r K (K|r-r |) = J dk (A.2) o0 2r -oo x -o y 2 2 2 k +k+ Kk x y which follows directly from the fact that K (i<7-r |) is the infinite space (12) Green's function for the two dimension Helmholtz equation() (V2-)K ( Kr'-r I) = -2r (7r-r ) (A.). 0 0 0 we "rotate" the integration variables: (k, k ) —(klk ) where x y 12 k = k cos @ - k sin @ x 1 2 k = k sin ~ + k cos 0 y 1 2 so that, in polar coordinates, the exponent is -ik (r-r ) -ikik r + ik r cos(0-0 ) + ik r sin(~ -0) (A.4) o 1 lo o 2o o We now perform the integral over k 3 -ikri 22 -iklr + ik lro cos(r-0) + ik + k+K ro sin(Q0-~) r —rol )= — dk {A) o ( o| 2 -cJ 1 21 k + K 103

104 and let k = s sinh T: K,(Ki-r i = 1 dT e-iKr sinh T + iKr sinh T cos(% -@) K (rro ) 1 - _dT e ~ o iKrocosh Tisin(Qo-O)I (A.6) The desired equation (A.1) now follows from the identity sinh(T+i~) = sinh T cos + i cosh T sin g the definition of I|-r | [equation (II.2.43)], and the fact that for E (0,j) sin = sin. We incidentally observe that according to equation (A.6), the T-transform of K ( Kl-7 [) is, for general @Q-, (T)) i= eirosinh[T + ia(0-G )] (A.7) where a = sgn[sin(G-G )] 0 Taking the T-transform of equation (A.3), we see that 4r should satisfy _+ 4r r(T (,) = -2Kr cosh T (ei0r T Z b(O-2nit) (a2 ~ei 0r ~ s n ) + e-iKrosinh T E 5[-(2n+ 1) (A.8) n provided equation (11.2.20) is correct; but the formula (A.8) is easily verified by performing the differentiations.

REFERENCES AND ADDITIONAL FOOTNOTES 1. I.M. Gel'fand and G.E. Shilov, Generalized Functions, Volume I, Academic New York, 1964. 2. N.I. Muskhelishvili, Singular Integral Equations, Noordhoff, Groningen, 1953. 3. Cf., for example, Reference 2. 4. H.M. MacDonald, Electric Waves, Cambridge University, Cambridge, 1902. (This reference gives an integral representation for the Green's function with arbitrary wedge angle. The homogeneous problem is of course elementary.) 5. A.S. Peters, Comm. Pure Appl. Math 5, 87 (1952). 6. Cf., for example, B. Noble, Methods Based on the Wiener-Hopf Technique, Pergamon, New York, 1958. 7. Cf., for example, E.C. Titchmarsh, The Theory of Functions, Oxford University, London, 1939. 8. K.M. Case, J. Math. Phys. 7, 12(1966)o 9. Cf., for example, A. Erdelyi, Asymptotic Expansions, Dover, New York, 1956. 10o Cf., for example, G.Nc Watson, Treatise on the Theory of Bessel Functions Cambridge, New York, 1944. 11. Cf. Reference 1, p. 185. 12. Cf., for example, P.M. Morse and H. Feshbach, Methods of Theoretical Physics, McGraw-Hill, New York, 1953. 13. K.M Case and J.F. Colwell, Geophysics XXXII, 1 (1967). 14. Cf., for example, K.M. Case and P.F. Zweifel, Linear Transport Theory, Addison-Wesley, Reading, Mass., 1967. 15. M. Kac, Probability and Statistics, The Harald Cramer Volume. 16. The method used here is similar to one used in Reference 14, Appendix 1.5. 17. V. Gorge and F. Rohrlich, J. Math. Phys. 8, 1748 (1967). 105

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