WGWEERWG i.it..hr,i 
 
 «i.*s COria ANTENNA ^ 
 
 It G55-te 
 
 BORATORY 
 Technical Report No. 37 
 
 
 ON THE SOLUTION OF A CLASS OF WIENER-HOPF 
 
 INTEGRAL EQUATIONS IN FINITE 
 
 AND INFINITE RANGES 
 
 CONPENCE ROOM 
 
 *»v:.V 
 
 .**** 
 
 Raj Mittra 
 
 15 May 1959 
 
 Contract No. AF33(616)-6079 
 Project No. 9-113-6278) Task 40572 
 
 Sponsored by: 
 WRIGHT AIR DEVELOPMENT CENTER 
 
 ELECTRICAL ENGINEERING RESEARCH LABORATORY 
 
 ENGINEERING EXPERIMENT STATION 
 
 UNIVERSITY OF ILLINOIS 
 
 URBANA, ILLINOIS 
 
ANTENNA LABORATORY 
 Technical Report No, 37 
 
 ON THE SOULUTION OF A CLASS OF WIENER-HOPF INTEGRAL EQUATIONS 
 IN FINITE AND INFINITE RANGES 
 
 by 
 Raj Mittra 
 
 15 May 1959 
 
 Contract No. AF33 l616) -6079 
 Project No, 9- (13 -6278) Task 40572 
 
 Sponsored by : 
 WRIGHT AIR DEVELOPMENT CENTER 
 
 Electrical Engineering Research Laboratory 
 
 Engineering Experiment Station 
 
 University of Illinois 
 
 I rbana , Illinois 
 
A <^ CONTENTS 
 
 Page 
 
 L„ Introduction 1 
 
 2„ Solution of the Semi-Infinite Range Integral Equation 2 
 
 2„1 Derivation of the Simultaneous Set of Equations 2 
 
 2 2 Solution of the Set of Equations for the Unknown Coefficient A. 5 
 
 on 
 2„3 Extension to the Case When the Difference of Order of F (p ) 
 
 and F _(p^) is more than one, 7 
 
 3 The Integral Equation with a Finite Range 10 
 
 3„1 Solution of the Simultaneous Set of Equations Obtained for the 
 
 Finite Range Case 12 
 
 3 P 2 Inversion of the Matrix of the Equation (34a) 13 
 
 4, Calculation of the Determinant and the Cof actors of the Determinant 
 of the Matrix M 16 
 
 5 The Fourier Transform Method of Formulation 20 
 
 5„1 Fourier Transform Formulation 20 
 
 5„2 Development of a Set of Simultaneous Equations 26 
 
 5„3 Iterative Method of Solution 30 
 
 6. The Eigenvalue Problem 33 
 
 6„1 The Eigenvalue Matrix Equation 33 
 
 6,2 Upper Limit Infinity 37 
 
 7„ Numerical Example 40 
 
 8„ Applications 42 
 
 9„ Conclusions 43 
 
 References 44 
 
1. INTRODUCTION 
 
 The subject of investigation in this report is the Wiener-Hopf Integral 
 Equation 
 
 g(t) 
 
 o 
 for a T either finite or infinite. It is assumed that Kit! admits of an 
 
 expression of the form 
 
 -k It I 
 
 K Itl S C e r 
 I I m 
 
 The method of solution for the semi-infinite range (T -oo) i s discussed first. 
 The method is then extended to T finite. Two basically different approaches 
 are used. The first one leads to a set of simultaneous equations which can 
 be solved by analytic means. 
 
 The second one uses the technique of Fourier transforms in a complex 
 plane and is more suitable for an iterative method of solution for the 
 finite T case (for T ^&o f the Wiener-Hopf technique yields the answer 
 directly) 
 
 The eigenvalue problem corresponding to the above kernel is also 
 discussed. Suitable numerical examples are included and certain applications 
 in the fields of Information Theory and Electromagnetic Wave propagation are 
 also pointed out. 
 
SOLUTION OF THE SEMI -INFINITE RANGE INTEGRAL EQUATION 
 
 Consider the integral equation 
 
 f (t) K | t-T| dT t > (1) 
 
 where K|t-T| is of the type 
 
 n -k ! t-T| 
 
 K|t-T|= 2 C e r , Re(k )>0 (2) 
 
 r r 
 
 C = constant coefficients 
 r 
 
 Equation (1) is the well known Wiener-Hopf Integral Equation and a 
 
 solution of the equation can be obtained for a general K(t) by the use 
 
 of the method of Integral Transforms in the complex plane . For the 
 
 particular type of Kernel which appears in (2), however, the equation can 
 
 be solved directly by using the theory of residues of a complex variable. 
 
 The first step for solving the problem is similar to the one used by 
 
 2 
 Laning and Battin , A set of simultaneous equations are obtained when 
 
 their method is used and these equations are solved here using the technique 
 
 of residues. 
 
 2.1 Deri vation of the Simultaneous Set of Equations 
 
 The function g(t) can be expanded in a series of exponentials 
 
 -i t 
 
 g(t) = Z D e r , t>0 
 
 r 
 
 Hence the problem in Equation (1) can be reduced to that of solving for 
 
 -it 
 a g(t) of the form e and superposing the contribution due to the differ- 
 ent terms in the series. 
 
 An alternative approach is the following. Operate on both sides of 
 (1) with the operator L, where 
 
 l - n ,p 2 - kr 2 ) , ,4 
 r=i 
 
Digitized by the Internet Archive 
 in 2013 
 
 http://archive.org/details/onsolutionofclas37mitt 
 
Then, using 
 
 2 a "kit" 1 "! 
 (p -k ) e = -2k 6(t-T), (6(t) = Dirac delta) 
 
 T r 
 
 one obtains, 
 
 n 2 2 n n (n ^ 2 2 
 
 L g(t) = n (p -k ) g(t) = -2 2 k C n (p -k )f(t) (3) 
 
 i S n r r i m 
 
 g= 1 50= 1 m= 1 
 
 where the superscript (n) in the continued product implies that the factor 
 corresponding to m = n is to be omitted. Equation (3) is the differential 
 equation satisfied by f(t)„ The equation is clearly inhomogeneous and can 
 be written in the form 
 
 F. (P 2 ) 
 
 -f(t) = g(t) < t < ~> (4) 
 
 F 2 (P ) 
 
 2 J 2 
 It can be easily shown that F (-p )/F r (-p ) is the Fourier transform of the 
 
 kernel function K(t) , i.e. 
 
 ,°o . 2 
 
 F. <-p ) 
 
 K| t | e' ipt dt^ J -^- 
 
 F 2 (-p ) 
 
 Returning to C4) , one knows that (4) has a particular solution say f (t) , 
 
 * ' P 
 
 and a general solution of the integral equation can be written in the 
 
 form 
 
 n-1 -a . t 
 
 f(t) = f (t) + A 6ft) + S A. e 1 . t > (5) 
 
 p o 1 
 
 v i=l 
 
 i 2 
 
 where a. s are the roots of F (p ). From (3) one sees that the order 
 
 1 2 2 
 
 of the polynomial F m p is (n-1), so F (p ) has symmetrical roots, 
 
 p = «- a , + a + .... + a -1 
 — 1 — 2 — n 
 
 * For the case, the degree of F (p ) is one less than that of F„(p ) „ 
 
 When the difference in the degrees is more than one, the method discussed 
 in the next section is applicable 
 
Only half of these roots, however, the ones with the negative sign, will 
 give solutions with exponential decay as t — » + oo„ The others, with a 
 positive real part in the exponent are not admissible as solutions. Also 
 it can be shown (see for instance Reference 2, p. 416, that the solution 
 of the integral equation must contain the Dirac-6 included in (5). 
 Let g (t), o < t < oo , be defined as 
 
 g 1 (t) = [ f (T) K(it-Ti) dr 
 
 n-1 ~ aT ] 
 f(T) - A 6(T) - S A. e * > K(lt-T| ) dT (6) 
 
 i=l X J 
 
 using (1) and (2), (6) can be written as 
 
 g x (t) - g(t) 
 
 r -a t -k |t-T| 
 
 (2 A. e )(SC e V )dT 
 
 i l r 
 
 r 
 "o 
 
 -k t 
 
 -ASCe r t > (7) 
 
 o r 
 
 Upon evaluation of the integral in the r h,s. of (7), the following is 
 obtained, 
 
 -k.t 
 
 g. (t) - gft) = S D. e 
 
 1 i i 
 
 where D s can be calculated in terms of A , A. ' and the C s 
 i o i s r 
 
 -at 
 
 The coefficients of e can be shown to be equal to zero since 
 
 I . 2 
 
 a. s are the roots of F, (p ). Hence the right hand side contains terms 
 
 -k t 
 
 involving the exponentials e only 
 
 The left hand side of (8) is known if the particular solution f (t) 
 
 P 
 and hence g it) is known From (8), remembering that the coefficients of 
 
 -a , t 
 e are zero, it follows that 
 
 -k.t 
 
 g.lt) -g(t)= SC, P. e X (9) 
 
 x i l. l 
 
where P. s can be obtained from the knowledge of g (t). 
 
 1 -k t 
 
 r 
 Equating the coefficients of e the following set of n - equations 
 
 are obtained for the n-unknowns A , A, ... A 
 
 1 n-1, 
 
 n-1 A . 
 
 A o 4- 2 — 4r- = P . r = 1, 2, „„.n (10) 
 
 , a -k r 
 J-l J r 
 
 where P s in Equation (10) are considered as known, 
 r 
 
 The solution of (10) will now be derived using the theory of residues, 
 2 2 Solution of the Set of Equation for the Unknown Coefficients A ^ 
 Consider first of all the solution of a set of equations 
 
 n-1 A (m) 
 
 ^ + * ^T-IT - P 6 ". m = l, 2, ...n (11) 
 
 . a .-k m r 
 J=l J r 
 
 If A and A. s can be solved for all m's, then A , A , ... etc are 
 obtained by superposition, i.e., using 
 
 A Q = S A (m) , A - S A. (m) (12) 
 
 m=l J m=l J 
 
 Attention will therefore be devoted presently to the solution of (11) 
 For this purpose consider a contour integral of the form, 
 
 where Qtz) satisfies the following conditions, 
 
 1 lim Q(z) = Q (constant) 
 z — >°° o 
 
 2 Q(z) has (n-1) poles atz-a . a n . . . a 
 
 1 2 n-1 
 
 3. Q(z) has (n-1) zeros at z = k , r = 1, 2, m-1 , m+l,--n, 
 excluding n - m 
 Such a choice of Q(z) is readily found as, 
 
 n tm) 
 Q(z) = Q H (z-k ) 
 o r 
 
 r=^l 
 
 ; (Q = constant) (13) 
 
 n-1 o 
 
 n (z-a ), 
 
 r 
 
 r-1 
 
Now let the contour in the integral 
 
 Q(z) 
 z-k 
 
 dz 
 
 be chosen large enough to include all the singularities of Q(z). If one 
 lets the contour C recede to infinity, say by taking an infinitely large 
 circle for C, one has: 
 
 n-1 Q( . 
 
 q ^2 dz - 2 7T jQ = 2ff j 2 ^L (z-a ) 
 
 o z-k o . z-k p 
 
 r p=l r 
 
 2 7T JQ(z) 
 
 zaa 
 
 (14) 
 
 z=k 
 
 Since Q(z) has zeros for z = k , r ^ m, (14) can be written as 
 
 Q(z) 
 
 n-1 Q(z)(z-a ) 
 
 q - 2 P 
 
 o . (z-k ) 
 p=l r 
 
 zsa 
 
 n 
 
 z=k 
 
 m 
 
 Using the expression of Q(z) given in (13) in the Equation (15), the 
 following results, 
 
 (15) 
 
 n (m) ( k -k ) 
 
 1 n r=l m r 
 
 = q < — — ->6 
 
 a -k 
 
 P r 
 
 v. 
 
 n-l 
 
 n (k-a ) 
 
 , m r 
 
 (16) 
 
 Comparing (16) with (11), it is seen that if one puts the following 
 relations in the equation (II), 
 
 n (m) (k -k > 
 
 Ifc.1 m r 
 
 o A n-1 
 
 n (k -a ) 
 V s ! 
 
 Q 
 
 m 
 
 = A 
 
 (m) 
 
 (17a) 
 
 (17b) 
 
 -Q 
 
 n (B) (a -k ) 
 D.1 P - r 
 
 n tp) (a =a ) 
 1*1 P r 
 
 = A< m > 
 
 (17c) 
 
The set of 'n 1 equations are identically satisfied. Q in (17) can be 
 
 O , X 
 
 calculated from the knowledge of P . The unknowns in (11), viz A , 
 . m o 
 
 A .... A etc., are then obtained using (17b) and (17c). 
 
 1 n 
 
 Since the a s and k s are known, the solution of (11) is complete 
 p r 
 
 Also, as was pointed out earlier, the solution of (10) can be constructed 
 from the solutions of (11) with the use of (12). The solution of the 
 Integral Equation (1) is then obtained by substituting the coefficients 
 
 Ai , and A in (5) „ 
 
 2 2 
 
 2.3 Extension to the Case When the Difference of Order of F (p, ) and F„ (p ) 
 
 is more than one 
 In the previous section the derivation of a simultaneous set of equaf 
 
 tions for A , A n ... etc was based on the assumption, ;that the order of 
 
 ° 1 2 2 
 
 the polynomial F (p ) is one less than that of F (p ). In this section 
 
 this condition will be relaxed and the process for formulating a new set 
 
 of equations will be outlined. The method of solution will also be 
 
 presented, 
 
 2 2 
 
 Let, ' g 1 and } n' be the order of F (p ) and F (p ) respectively where 
 
 J. ^j 
 
 g < n. As has been pointed out by Laning and Battin (Ref 2.2 p„ 416), the 
 solution f(t) will now contain the derivatives of the Dirac Delta and f(t) 
 will be of the form 
 
 s_1 ( . q 
 
 f(t) = A 6(t) + 2 (-1) X A b{\\ + 2 A e^ 1 * t ? (18) 
 1=1 1=1 
 
 S - n-q > 1 , 6 = i-th derivative of 6 
 
 Following exactly the same method as was used for the derivation of 
 (10) one now obtains the set of equations, 
 
 / 
 s-1 q A . 
 
 A f 2 A.k 1 + 2 — = P (19) 
 
 o . . i r . , a -k r ' 
 
 1=1 i=l i r 
 
 where P ? s are the same as in (10) 
 r 
 
 Again consider solving the equation similar to (11), viz , 
 
(m) + S 2 A (m) k ± + 
 o . , i r 
 
 i=l 
 
 * Cm) 
 
 q a. 
 z x 
 
 a.-k 
 1=1 1 r 
 
 P 6 
 m r 
 
 m 
 
 (20) 
 
 by using the method of residues Let M(z) be a function of the complex 
 variable z, which satisfies 
 
 1. 
 
 lim < M(z) - 2 l.z 1 
 z-* - 3 i=l 1 
 
 (constant) 
 
 2„ M(z) has q poles at z = a , i = 1, 2, ., „, q 
 
 3. M(z) has (n-1) zeros at z = k , r = 1, 2, „ , n, r/m 
 A choice of M(z) which staisfies the above conditions is 
 
 n (m) f ,- 
 
 M(z) ■ M 
 
 i=l 
 
 (z-k.) 
 
 l 
 
 o q 
 
 n (z-a ) 
 i=l i 
 
 (21) 
 
 One can find the constants L and L in terms of M , k and a. from 
 
 i o o r i 
 
 the equation (21), 
 
 Now consider the integral 
 
 f mi/ ^ s_1 L.Z 
 
 c r i=l r 
 
 where C is an infinitely large circle. 
 
 Since the integrand behaves as L /z as z-^oo, one has 
 
 o' 
 
 (22) 
 
 j J 2 Mp. (z _a . ) 
 o i. , z-k l 
 1=1 r 
 
 Equation (23) can be rewritten as 
 
 s-1 
 - 2 L.k 1 + M(z) 
 
 z=a 
 
 i=l 
 
 l r 
 
 z=k 
 
 Y = 1, 2, ... n 
 
 s-1 
 
 f 
 
 L f 2 L.k - 2 - 
 o i r 
 
 i=l l-l 
 
 M(z) (z-a, ) 
 
 z=a 
 
 a,-k 
 l r 
 
 = M(z) 
 
 Dl 
 
 z=k 
 
 ra 
 
 (23) 
 
 (24) 
 
Hence, comparing the r h.s of (24) and (20) and using the expression 
 
 for M(z) in (21), one sets M(z) 
 
 z=k 
 
 = P , i.e. 
 m 
 
 m 
 
 M =P 
 o m 
 
 n (k -a.) 
 
 i=l m X 
 
 n (m !k -k.) 
 i=i m x 
 
 (25) 
 
 As was explained above the constants L , L , 
 
 once M is known, 
 o 
 
 Again comparing (24) with (20) one has, 
 
 L , ,. are determined 
 (s-1) 
 
 L =A (m) , L. =A, fm) , i = 1, 2, ..., (s-1) 
 o o 11 
 
 -M(z)(z-a ) 
 
 r 
 
 s M 
 
 z=a 
 
 / (m) 
 
 r = 1, 2, 
 
 (26) 
 
 The expression (26) yields all the unknowns in (20). The solution 
 of (19) is obtained by using superposition. The method of solution has 
 
 thus been extended in this section for the case when the degree of the 
 
 2 2 
 
 polynomial in F (p ) differs by more than one from the degree of F (p ) 
 
 , 2 / , 2 
 Recall here that F (,-p )/F (-p ) is the Fourier Transform of the kernel 
 
 1 ^ 
 
 K |t I . 
 
10 
 3. THE INTEGRAL EQUATION WITH A FINITE RANGE 
 
 The solution of the integral equation, 
 
 .T 
 g(t) = f(T) Kl t-TjdT (27) 
 
 which is similar to (1) except for the upper limit, will be detailed in 
 
 this section, K|tj is again assumed to be expressible as a series of 
 
 -k. I t| 
 exponentials e i as in (2). 
 
 The method formulating the problem in terms of a set of simultaneous 
 equations, is similar to the one used in the previous section. The solution 
 of these equations, however, is considerably more involved. 
 
 The Differential Equation for f (t) and the particular solution fp(t) 
 are the same as obtained above in (24), if of course g(t) and K(t) are 
 the same. The range however is different, i.e. 
 
 F (p ) 
 
 — — f (t) = g(t) < t < t (28) 
 
 F 2 (P ) 
 
 Hence a general solution of (28) is of the form 
 
 n-1 -a.t n-1 a.(t-T) 
 f(t) = f (t) + 2 A. e 1 + £ B. e * (29) 
 
 P i=l X i=l X 
 
 where a . s are the same as defined in the previous section, 
 l 
 
 2 2 
 
 It is assumed that the order of F (p ) is one less than that of F (p ) 
 
 -L £ 
 
 The method can be extended for the case when this difference is more than 
 one, as was discussed before. It can also be shown that the solution of 
 the Integral Equation must contain Dirac - 6's i.e. the complete, expression 
 for f(t) which satisfies (27) is, 
 
 n-1 -a.t a (t-T) 
 
 f(t) = f (t) + A 6(t) + B 6 (t-T) + 2 A.e 1 + 2 Be 1 (30) 
 
 P o o i i 
 
 i=l i=l 
 
11 
 
 The problem in hand is to find the 2n coefficients, A , B , A s and 
 . o o i 
 
 B, s. 
 
 l 
 
 One again defines a g (t) as 
 
 g 1 (t) = f (t) K|t-r|dT 
 
 — 'o 
 
 and it is possible to verify, by following the lines used in deriving (29) 
 
 that 
 
 n -k t n k (t-T) 
 g x (t) - g(t) = S C r P r e r + 2 C r R r e r (31) 
 
 R=l r=l 
 
 Substituting (30) in (27) one obtains after carrying out the integration 
 (for details see Laning, ibid. p. 320) 
 
 r C } r fn-1 -a.T n-1 a. (t-T) 
 
 if(T)-f (t) I K|t-T|dT = J 2 A.e + 2 B.e 1 ^K|t-T|dT 
 
 _a .t 
 
 / n-1 A. B.e v -k t 
 
 = S f A + 2 ~. 2 -~ — z ) C e r (32) 
 
 \ o , a .-k . a.+k / r 
 
 r\ 1=1 l r l ir/ 
 
 -a T 
 
 / B, A.e 1 > k (t-T) 
 
 + 2 B +2 ~i— - 2 n X . . C e r 
 
 \ o „ Q . -k a.+k r 
 
 r \ iir i ir/ 
 
 Since (31) and (32) should be identical for all values of (t), one 
 
 -k t 
 gets the following set of equations by equating the coefficients of e r 
 
 , k (t-T). 
 and e r 
 
 _q t 
 n-1 A. n-1 B.e 
 
 A + 2 n 1 . - S ~ — r— = P (33a) 
 
 o J , a . -k a.+k r 
 i=l l r i=l l r 
 
 -a T 
 n-1 B„ n-1 A.e 1 
 
 B + 2 — 2 -| ■ = Q (33b) 
 
 i=l i r i=l i r 
 
 r = 1, 2,. . . n 
 
 UWVEKSITY Of nUNOli 
 UiftMtf 
 
12 
 
 (33a) and (33b) are the desired set ( of 2n equations for the unknowns 
 
 A , A. , B , B . 
 o 1 o i 
 
 It is to be noted that (33) reduces to the form (8 0) when T— ><=*». 
 
 3.1 Solution of the Simultaneous Set of Equations Obtained for the Finite 
 
 Range Case 
 
 The solution of the set of 2n equations for A , A , B , and B., are 
 
 o i o i 
 
 discussed in this section. As a first step to solving (33), form a new 
 set of equations by adding and subtracting (33a) and (33b) and obtaining 
 
 F o + 2 F J 
 
 
 -a T 
 
 
 
 
 1 
 
 l 
 e 
 
 a.+k 
 l r 
 
 )■ 
 
 P 
 r 
 
 + Q r 
 
 a -k 
 l r 
 
 
 -a T 
 
 
 
 
 1 
 
 e x N 
 
 )■ 
 
 P 
 
 r 
 
 - Q 
 r 
 
 a.-k 
 
 a +k 
 
 J + S J ; — = P - Q (34b) 
 
 o i I a -k a +k / r r 
 i \ l r i r / 
 
 where 
 
 F - A + B , J = A - B 
 o o o o o o 
 
 F . = A . + B . , J = A -B 
 ill l i l 
 
 The advantage of using (34a) and (34b) in preference to (33a) and 
 (33b) becomes immediately clear if one notices that the set (34a) and 
 (34b) are uncoupled. This is because the matrix corresponding to (34a) 
 is the same as that of (34b). Hence, it is only necessary to invert the 
 (nxn) matrix corresponding to one set say (34a), rather than the 2n x 2n) 
 matrix of (33a) and (33b) „ 
 
 The inversion of the matrix (34a) is not straight forward, however, 
 since the residue method is not directly applicable. 
 
 It is nevertheless possible to construct the elements of the inverted 
 matrix for (34a) , when the cof actors and the determinant of the matrix for 
 the set (10) is known. This is detailed in the following section. 
 
13 
 
 3.2 Inversion of the Matrix of the Equation (34a) 
 
 The matrix \ M 1 for the equation (34a) is, 
 
 [ 
 
 M i = 
 
 1, 
 
 1 e 
 
 -c^T 
 
 V k i VV' lV k i W 
 
 1 e 
 
 -a T 
 2 
 
 -a T-, 
 n-1 
 
 \ (— — - ' ) 
 r"'v.-r k i °n-i-v 
 
 -a 
 
 1, 
 
 1 e 
 
 IT 
 
 1 e 
 
 -a T 
 r 
 
 V k 2 I V k 2 
 
 «V k 2> ! a n" k 2 
 
 1, 
 
 1 e 
 
 -a t 
 
 -a T 
 n-1 
 
 (a -k ) a +k 
 In In 
 
 a -k a + k 
 
 n-1 n n-1 n 
 
 (35) 
 
 The matrix corresponding to (10), say Tm] is 
 
 [„]. 
 
 a -k 
 1 1 
 
 1. T 
 
 a -k 
 1 r 
 
 a -k 
 r r 
 
 a -k 
 r r 
 
 n-1 1 
 
 1, 
 
 a -k 
 1 r 
 
 a -k 
 n-1 n 
 
 (36) 
 
 Let the cofactors of the determinant of the matrix [Ml be designated 
 by the symbol A 
 
 pq 
 
 Then 
 
 A = A (a , a . .a a , . . .a k n , k , . . .k . , k . , . . .k ) (37) 
 pq pq 1 2 q-2 q n-1 1 2' p-1' p+1 n 
 
14 
 
 i.e. the expression for A involves all a. . ' s and k 's excluding o. 
 
 pq 1 r q-1 
 
 and k and this can be easily verified from the form of (36) . 
 P 
 The determinant A for Tm1 however involves all the a- .' s and k ' s. 
 
 Suppose now A is known, then, A corresponding to TlVI'l can be 
 pq pq L J 
 
 expressed in terms A , as follows, 
 
 pq 
 
 n-1 -a. T 
 / „ l. 
 
 A = A + S e 1 , A (-a, ) 
 pq pq =1 ' T>Q ij 
 
 n-2 n-1 -(a. +a . ) T 
 
 + S 7 s' e l l X 2 A (-a. , -a. ) 
 
 h-vv 1 pq ^ l2 
 
 T 
 n-3 n-2 n-1 -(a. +a , +a ) 
 
 + Z S 7 2/ e x l x 2 x 3 A (-a -a , -a . ) 
 
 H' 1 W 1 W 1 pq ij l2 l3 
 
 -(a +a +. . . a ) 
 
 + e X l X 2 1 n-l A (-a -a. , -a i ) 
 
 pq 1 i 2 n-1 
 
 (38) 
 
 where the prime on the summation symbol implies that the term corresponding 
 to (q-1) is to be omitted. 
 
 Also, A (- a . , - a ) implies that the signs of the particular set of 
 
 pq i 1 i 2 
 
 a' s appearing in the argument are to be changed in the expression for A 
 keeping the signs of the other a's unchanged. As an example, if 
 
 then 
 
 A (-a ) = (k -k ) (k -k ) (-a -a ) (a -a ) 
 pq 1 2 3 12 12 2 3' 
 
 and 
 
 A P q ( - Q l'-V - (k 2" 1 V (k r k 2 )( - Q r a 2 )(a 2 +a 3 ) 
 and so on. 
 
15 
 
 Equation (38) is a rather long and involved one, but it is a straight- 
 forward process to show its validity. Furthermore, in a practical problem 
 it is only necessary to compute a few of the terms in the series (38). This 
 is because the higher order terms have an exponential decay with the order. 
 
 Equation (38) provides a means for obtaining the cofactors of the 
 matrix [M' ] when the cofactor of the matrix TmI is known. The cofactors 
 of [m] can be obtained by a method outlined in the next section. 
 
 In order to find the inverse of the matrix [M' 1 , one still needs the 
 expression for the determinant A" of [m"] in addition to the expression 
 for the cofactors. The expression is very similar to that of A' and is 
 
 pq 
 
 given in the following. 
 
 -a T 
 n-1 i 
 
 A - A + 2 e A (-a . ) 
 
 i 1= l 
 
 n-2 n-1 -<V^i>T 
 
 + S S e A(-a. , -a . ) 
 
 i 
 1=1 1 2 =1 1+ 1 1 2 
 
 -fa +a +. . .+a )T 
 
 + e A (-a. -a, , , . .a ) (39) 
 
 V X 2 Vl 
 
 where the symbols have the same meaning as used in connection with (38). 
 Using (38) and (39) it is possible to construct the element N 
 
 pq 
 
 which is p, qth element of the inverse of Tm' 1 by using the well known 
 expression 
 
 n = -H3. 
 
 qp A' 
 
 The problem of inverting the matrix [M* ] and solving the set of 
 Equations (34a) and (34b) can thus formally be considered as solved. 
 It has been tacitly assumed in the above that A and A can be 
 
 pq 
 
 obtained by some reasonably simple means. The actual method for calculating 
 them is discussed in the following section. 
 
16 
 
 CALCULATION OF THE DETERMINANT AND THE COFACTORS OF THE DETERMINANT OF 
 THE MATRIX M 
 
 In the previous chapter it was shown that the elements of the matrix 
 Tm p can be developed from the knowledge of the determinant and the 
 cofactors of the determinant of Tm] . The matrix I'm] is associated with 
 the set of Equation (10). 
 
 Consider first of all the task of expressing the determinant A of [Ml 
 in a suitable form. The determinant is rewritten here for convenience 
 
 A = 
 
 *W~ " ( VrV 
 
 i i 
 
 (a -k ) " (a -k ) 
 
 12 n-1 2' 
 
 1 1 
 
 (d -k ) 
 
 n-1 n 
 
 (a -k ) 
 
 1 n 
 
 (39) 
 
 The determinant can be easily transformed to the form of a double 
 alternant (for the definition of an alternant see Reference 3), by subtract- 
 ing the first row from the rest. After this operation, when the common 
 factors are taken out, there results, 
 
 A = 
 
 I 
 
 ! 
 
 1 
 
 i 
 
 ( VV (a 2 - ki ) 
 
 (a -k ) (a -k ) 
 v 1 2 v 2 2 
 
 n-1 1 
 
 (a . -k ) 
 n-1 1 
 
 (a -k ) (a - k ) 
 
 In 2 n 
 
 (a -k ) 
 n-1 n 
 
 xF 
 
 (40) 
 
where 
 
 17 
 
 F = 
 
 n (k -k ) 
 r=2 r X 
 n-1 
 
 n (a -k) 
 
 i r 1 
 r=l 
 
 Now consider the (n-l)th order determinant V where 
 
 V = 
 
 (VV (a 2 - k 2 ) 
 
 a -k 
 
 1 n 
 
 2 n 
 
 n-1 Z 
 
 (a -k ) 
 n-1 n 
 
 (41) 
 
 The determinant V is a double alterbant and can be conveniently expressed 
 as a ratio of products as follows 
 
 V= , "- 1 t h (a a ..... a )• fc*(-k_, -k ..., -k ) 
 (-1) 1 2 n-1 2 3 n_ 
 
 n-1 , n 
 
 n (a =k ) 
 
 r in 
 n=l , m=2 
 
 (42) 
 
 where t, 2 is a difference product defined as 
 
 -i 
 
 1-1, q 
 
 I (x x . x ) = ' n ( x k ~ x «) 
 
 q k=l,i=2 k X 
 
 (43) 
 
 From (40) 
 
 A = V-F 
 
 1 
 
 i v 
 
 & * ( °1- "V ' ' Vl' l * ( - k l'- k 2 •••• "V 
 
 n-1 ,n 
 
 n (a -k ) 
 
 i , r m 
 r=l , m=l 
 
 (44) 
 
18 
 
 Equation (44) gives the desired expression for the determinant A of the 
 matrix [m] . The expression for the cofactors are also determined in a 
 similar manner. 
 
 Consider first of all the cofactors A , , where 
 
 ml 
 
 Xi = <-« 
 
 mH-1 
 
 a -k 
 1. 1 
 
 a -k 
 1, m-1 
 
 I 
 
 i 
 a i " k m i 
 
 1 m+l 
 
 a -k 
 1 n 
 
 a -k 
 n-1 1 
 
 a -k 
 n-1 m-1 
 i 
 
 i 
 
 1 
 
 -k 
 
 a- n-1 m+l 
 
 a -k 
 n-1 n 
 
 (45) 
 
 The determinant in expression (45) is a double alternant. Hence, 
 can be expressed as 
 
 A 
 
 _„■* l*«V V •• Vl> *<•*!• " k 2 •' - k n )(m> 
 
 ml 
 
 n-1 , n 
 
 n ( a „-k„) 
 r=l , p=l 
 
 Pi^m 
 
 (46) 
 
 r p 
 
 where the superscript on the second product implies that k is omitted 
 
 m 
 
 from the argument. 
 
 Next consider A , q ^ 1 
 mq 
 
 A mq - ("I) 
 
 m«-q 
 
 V k i VrV 
 
 i ! 
 
 1 1 
 
 a ,-k a -k 
 1 m-1 q-1 m-1 
 
 a -k 
 
 a -k 
 
 1 m*l q-1 m+l 
 
 (a -k ) 
 1 n 
 
 a -k ' 
 
 q+1 1 
 
 a -k 
 q+1 m-1 
 
 a -k 
 q+1 m+l 
 
 i i 
 
 a - k 
 n-1 1 
 
 a -k . 
 
 n-1 m-1 
 
 a -k 
 n-1 m+l 
 
 a -k 
 n-1 n 
 
 (47) 
 
19 
 
 The determinant appearing in (47) is of the type A, and the method 
 
 used above for expressing A is applicable here. The first step is to 
 
 subtract the first row from the others and take out the common factors. 
 
 The resulting determinant is a double alternant and can be expressed 
 conveniently. 
 
 Omitting the details, the final expression for A , q ^ 1 is given 
 below 
 
 ^(a , a .... a i ) (q) ;*<-*, -k ..,, -k ) (m) 
 A , ,^m+q v 1' 2 n-r 1' 2 n 
 
 A mq = t" 1 ) ^l~n— ' (48) 
 
 n (a r -k e ) 
 
 r=l,i=l 
 r^q,i/m 
 
 Equations (44), (46) and (48) are the desired expressions for the 
 determinant and the cofactors of the determinant A of the matrix I'm] . 
 
20 
 
 5. THE FOURIER TRANSFORM METHOD OF FORMULATION 
 In the previous chapter the method of solving the equation, 
 
 g(t) = f(t) K|t-T|)dT < t < T 
 Jo 
 
 g(t) specified for < t < T 
 involved as a first s^ep, the formulation in terms of a set of simultaneous 
 equations. This was possible because the kernel of the Integral Equation 
 could be expressed in a suitable series of exponentials. In this section 
 there is discussed an alternative method of formulating the problem using 
 the technique of Fourier Transformation in the complex plane. The advantage 
 of this method is that the formulation is useful for a general KJt| , the 
 only requirement being that the existence of the transform for the function 
 K|t|. It is also assumed that the asymptotic behavior of g(t) is such 
 that the transform can be defined when gCt) is extrapolated beyond the 
 range it is specified. 
 
 It is also demonstrated \^.iat when the kernel K(t-T) can be expressed 
 in the form of a series of exponentials, as was the case assumed before, 
 the simultaneous set of equations can also be derived from the present 
 formulation. The exact solution is then possible by the method described 
 in the previous section. For a general kernel K(t-T), however, the solution 
 is obtained approximately by a series of iterations. An insight into this 
 approach is gained by first solving the particular case of the series form 
 of K(t-T). This is pointed out in the discussion given in connection with 
 the iterative method of solution. 
 5.1 Fourier Transform Formulatio n 
 
 The formulation in terms of the Fourier Transforms is discussed in 
 the following. 
 
 The Fourier Transform H(oo) of a function h(t) is usually defined as 
 
 H(«") = 
 
 10Jt 
 
 e h(t)dt, w real 
 
 -oo 
 
21 
 
 The inverse transform of H(oj) is then given by, 
 
 h(t) 
 
 27T 
 
 - icot „ v 
 e H(co)doo 
 
 For the purpose of this work, however, the transforms will be defined in 
 
 terms of a complex variable s = s + is , 
 
 The definitions of Fourier Transform of h(t) in the complex plane 
 
 which are suitable here are given below. 
 
 -a, t 
 
 1 2 
 
 Let h(t) be a function such that f (t) e , a > is in class L 
 
 " a 9 t 1 2 
 
 in the interval < t < °o, and f (t) e , a < 0, is in L for -°o < t < 0, 
 
 then 
 
 r 
 
 .OO 
 
 F (s) = 
 
 + 
 
 „, . ist 
 
 f (t) e dt, s = s +is , s > a 
 
 (49) 
 
 and 
 
 F (s) 
 
 f(t) e 1St dt, 
 
 s < a 
 2 2 
 
 (50) 
 
 The inversion formulas are 
 
 and 
 
 1_ 
 277 
 
 1_ 
 27T 
 
 oo+ia. 
 
 -<30+ia. 
 
 r 00 ^ 
 
 F + (s) e 1S ds = f (t), t > 
 
 ist 
 F (s) e ds = f (t) t < 
 
 J -oo+i 
 
 "2 
 
 =0, t > 
 Hence the Fourier Transform for h(t) is given by 
 
 (51) 
 
 '.T. [h(t)] = H (s) + H_(s) 
 
22 
 
 Now consider the Fourier Transform of a function f (t) defined as 
 
 f x (t) = f(t) 
 
 o < t < T 
 
 (52) 
 
 = 
 
 otherwise 
 
 In general f(t) can be expressed as 
 
 f(t) = q(t) + r(t) 
 
 o < t < T 
 
 where 
 
 q(t) =ZA e 
 n 
 
 -c t 
 n 
 
 Re(c ) > 
 n — 
 
 (53) 
 
 r(t) =Sfe eS (t " T) 
 n 
 
 Re(dn) > 
 
 (54) 
 
 Note that q(t) and r(t) decrease exponentially for 't positive and 
 
 negative respectively. 
 
 The transform for f (t) , F (s) can be expressed as 
 
 F 1 (s) = F.T. of f x (t) 
 
 C 
 
 -X) 
 
 -ist , v 
 e f (t) dt 
 
 q(t) + r(t) 
 
 -co 
 
 -ist 
 e dt 
 
 (55) 
 
 Rewriting (53) one has 
 
 F 1 (s e 1St dt 
 
 q(t) e iSt dt 
 
 (56) 
 
 i -ist 
 + r(t) e dt - 
 
 r(t) e" 1St dt 
 
 -oo 
 
23 
 
 Using the shift theorem for transforms it is possible to write, 
 
 q(t) e" iSt dt = e" lsT V (s) (57) 
 
 + 
 T 
 
 where 
 
 similarly 
 
 Also let 
 
 and 
 
 V (s) = q(t+T) e~ 1St dt 
 
 >T 
 
 , v -ist -is* . . ,■_„* 
 
 r(t) e dt = e W (s) (58) 
 
 -ist 
 W (s) = r(t+T) e dt 
 
 ^30 
 
 jx> 
 
 Q (s) = q(t) e 1St dt, (59) 
 
 -ist 
 R (s) = / r(t) e dt (60) 
 
 Putting (57), (58), (59) and (60) in (56), there results, 
 
 F,(s) = Q (s) -e" 1St V (s) 
 1 + + 
 
 e 1S W (s) - R (s) 
 
 (61) 
 
 Note that the integral in (55) was broken up into (56) because of the 
 behavior of q(t) and r(t) given in (53) and (54). The range was changed 
 from o < t < T to o < t < <x> and T < t < -T5, so that the transforms 
 could be defined in terms Q (s), V (s) and W (s) , R_(s) which are analytic 
 
24 
 in the complex s-plane in the ranges shown below. 
 
 Q (s) and V (s) analytic for Im(s) > 
 
 + + 
 
 W (s) and R (s) analytic for Im(s) < 
 
 Now consider the transform of a function g(t) which is known in the 
 range o < t < T 
 Then let 
 
 g(t) - g x (t) + g 2 (t) o < t < T (62) 
 
 where say 
 
 -v t 
 
 g, (t) = S P e n Re(v ) > (63) 
 
 1 n n — 
 
 and 
 
 6 (t-T) 
 
 g_(t) = S Q e n ' Re (6 ) > (64) 
 
 2. n n — 
 
 Also let g(t) be extrapolated for all t in the form, 
 
 g(t) = g x (t) U(t) + g 2 (t) U(T-t) + g 3 (t) U(t-T) + g 4 (t) U(-t) (65) 
 
 In (65) the unknowns are g~(t) and g 4 (t) and U(t) = unit step, i.e. 
 
 U(t) =0 t < 
 
 = 1 t > 1 
 
 Let the F.T. of g(t) be G(s), then 
 
 00 OO 
 
 e" 1St g(t) dt = 
 
 - OO 
 
 e~ ist g(t) dt 
 
 (66) 
 
 ,^x -ist 
 + g(t) e dt 
 
 -oO 
 
 Using (65) and (66), the following is obtained by obvious application 
 
of the shift theorem. 
 
 25 
 
 ~, x ~ / ^ tisT , -isT 
 
 G(s) = G 1+ (s) + e I x _(s) + e J 2 + (s) + G 4_ (s > 
 
 (67) 
 
 where 
 
 .00 
 
 e 1St g 2 (t) U(T-t)dt = I e iat g 2 (t)dt 
 
 -00 
 
 -00 
 
 -ist -ist ,. mXJ , -isT 
 
 / e g (t+T)dt = e I, (s) 
 
 1- 
 
 -00 
 
 and similarly 
 
 I 2+ (s) = / e _ist g 3 (t+T)dt 
 
 Also 
 
 and 
 
 F.T. 
 
 F.T. 
 
 s 1 (t) U(t) 
 
 g 4 (t) U(-t) 
 
 = G. (s) 
 
 1 + 
 
 = G 4 _(s). 
 
 (68) 
 
 Also if K(t) — *- o as |t|— *°o, then the transform of K(t) say ^(s) is, 
 
 (s) = 
 
 . . I . 1 -ist,, 
 K t e dt 
 
 (69; 
 
 -00 
 
 The equations (61), (67) and (69) give the transforms of f(t), g(t) and 
 K(|t|) respectively under the assumptions mentioned. 
 
 If now the F.T. is taken of the original Integral Equation, 
 
 J f(t) k|t-T|dT 
 
 O 
 
 o < t < T 
 
 by first writing it in the form 
 
26 
 
 g (t) U(t) + g 2 (t) U(T-t) + g 3 (t) U(t-T) + g 4 (t) U(-t) 
 
 OO 
 
 = J f (t) k |t-T|dT 
 
 J -OO 
 
 One gets using the expressions for the appropriate transforms developed 
 above, 
 
 -isT f 
 G (s) + G. (s) + e \l ■ (s) + I (s) 
 
 1+ 4- 1- 2 + 
 
 [Q (s) - R_(s) + e" lsT |V-(s) - V (s)\] (s) 
 
 (70) 
 
 The unknowns in (70) are G„ (s) , I (s) , Q (s) , R (s) , W (s) and V (s) . 
 
 4- 2+ + + 
 
 Regrouping terms in (70) there results, 
 
 G. (s) + G. (s) - Q (s) - R (syf^s) 
 1+ 4- + " ^ 
 
 = e~ isT [ |w_(s) - V + (s)j K(s) - I x _(s) - I 2+ (s)] (71) 
 
 From the definition of the various transforms, it is possible to 
 verify that the only terms dependent on T are the ones appearing on the 
 r.h s. of (71). Since (71) is to be valid for all T, the left and right 
 sides of (71) must separately be equal to zero. The two equations which 
 are obtained using this condition are, 
 
 JQ + (s) - R_(s)} J^(s) = G 1+ (s) + G 4 _(s) (72) 
 
 |w_(s) - V + (s)jJc(s) = ^(s) + I 2+ (s) (73) 
 
 \e known q 
 are JL ( s ) and ^1 ( s ^ • 
 
 and 
 
 The known quantities in (72) arte J^ (s) and G, (s) and in (73) , they 
 
 "1- 
 The equations can be solved if there exists a common strip of 
 
 analyticity of the transformed functions involved. 
 
 5.2 Development of a Set of Simultaneous Equations 
 
 Consider the case when there exists an expansion of K(t) in the form 
 
27 
 
 of a series 
 
 M -k lt| 
 K(|t|) = S C r e r 
 
 r=i 
 
 Re(k r ) > 
 
 and when 
 
 n -k t n k (t-T) 
 
 g(t) =SCPe r + SCQe r 
 ° r r r r 
 
 It was shown in Chapter 1, that such a g(t) is obtained when the 
 particular solution f (t) is filtered out of the integral equation by 
 calculating 
 
 f (t) - f (t) ). K t-T|dT 
 P 
 
 Then from (62) , 
 
 g. (t) = S C P^ e 
 1 - r r 
 
 Y=l 
 
 -k t 
 r 
 
 and 
 
 Using 
 
 g 2 (t) = S C r Q r e 
 Y =l 
 
 k (t-T) 
 r 
 
 r 
 
 oo 
 
 J. 
 
 -k t -ist . 1 
 
 re dt = 
 
 e k +is 
 
 one has the expressions for G 1 (s) and I-,_( s ), a s , 
 
 C P 
 r r 
 G. (s) = S- ~ 
 
 1+ k +is 
 
 (74) 
 
 and 
 
 -k T 
 
 C Q e r 
 
 I ( S ) = H-*-?—. 
 
 1- k -is 
 
 (75) 
 
Also 28 
 
 •J& 
 
 C k 
 r r 
 
 "- 1 , 2 2 
 n (s +a r ) 
 
 4 ?=1 ' 
 
 „ 2 2 
 
 k +3 
 
 r 
 
 n 2 2 
 
 n (s +k 
 
 r= 1 
 
 F.T. (K t|) = /T (s) = 22 
 
 where d is a constant and a.r s are the zeros of /£, (s) . Equation (72) and 
 (73) for this case becomes 
 
 n ; L . 2 2 
 
 > n (S ^r > C P 
 Q + (s) - R_( 3 )|,d ^ = 2 r ^"f- + G 4 _(s) (76) 
 
 n (s +k r 
 
 r=l 
 
 and 
 
 n-1 2 2 
 , n (S +a r > C Q -k T 
 
 w (s) - v (s)Ld — = ^tt^tI e r + l o (s > (77) 
 
 + n k tis 2+ 
 
 n (s 2 +k 2 ) 
 r=l 
 
 From the nature of the differential equation of which 1 (t) is the 
 
 P 
 particular solution one knows that the complementary solution f(t) - f (t) 
 
 has the form, 
 
 n-1 -a T n-1 -a (t-T) 
 
 A 6(t) + 2 A e r +B 6 (t-T) + 2 B e r 
 o r o . . r 
 
 r=l i=l 
 
 Therefore Q (s) can be assumed to be of the form 
 
 + 
 
 A 
 
 Q (s) = A + 2 —A- (78) 
 
 + o a -is 
 
 Similarly 
 
 B 
 W (s) = B + 2 ' r . (79) 
 
 O CL +1S 
 
 r 
 
 The unknowns are A , A s , B and B s . 
 
 o l o i 
 
 It is possible to express the remaining quantities in the l.h s. 
 of (76) and (77), i.e. R_(s) and V (s) , in terms of the coefficients 
 
29 
 
 A s and B s. For this purpose refer to (57) and (58) where the definitions 
 
 of R (s) and V (s) were given. It is seen that 
 
 and 
 
 F.T." 1 R (s) = r(t) U(-t) 
 
 -1 
 
 F.T. W_(s) = r(t+T) U(-t) 
 If R (s) is chosen to be 
 
 (80) 
 
 (81) 
 
 R (s) = S 
 
 B e 
 r 
 
 -a . T 
 
 x 
 
 a. +is 
 r r 
 
 (82) 
 
 The Equations (80) and (81) are consistent as is verified by inverting 
 R_(s), and using (79). 
 Similarly 
 
 -1 
 
 and 
 
 Hence 
 
 F.T. V (s) = q(t+T) 
 
 + 
 
 F.T " Q (s) = q(t) 
 
 + 
 
 a T 
 i 
 
 A e 
 
 V (s) = A + 2 -£ r- 
 
 + o a -is 
 
 (83) 
 
 Substituting (78} , (79), (80) and (81) in (76) and (77), one gets 
 after dividing both sides hyjCis) 
 
 -a T 
 f A Be 1 ^ 1 CP G (s) 
 
 A o + s 5^Ii ~ 2 -^T" } ^^(^ S FTi7 + ^"Ts7 
 
 and 
 
 B A e 
 
 r r 
 
 B + S —I— - £ 
 
 o a +is 
 
 -a T -k T 
 
 1 ;- C rV r I 2 + (S) 
 
 ,/& 
 
 (84) 
 
 (85) 
 
 Notice that lA/Ms) has zeros for i' = + k . Hence if one sets 
 / s — r 
 
 i = k in (84), m = 1, 2, . „.„n, one obtains the set of equations, 
 
-a T 
 
 A B e , C P 
 
 A o + ^ a -k " ^ a + k ~ T~(k -i J M^TT 
 
 r m rm I m s ^ c J 
 
 30 
 
 m=l, 2, ... n (86) 
 
 s=ik 
 m 
 
 This is because all the terms in the summation on the right hand side 
 
 go to zero except the one containing l/k -i in the denominator. Notice 
 
 / m s 
 
 that G (s) can not have a pole for s = -ik since it is regular in the 
 4- m 
 
 lower half of the s - plane. 
 
 Similarly, from (37) one obtains by setting i = -k or s = ik , 
 
 s m m 
 
 -a T -k T 
 
 B Ae r CQe m 
 
 r r mm 
 
 B + S — - - S : = T-P7-: 7T r-n m = 1, 2, ... n (87) 
 
 o a -k a +k [ £(s)(k +i 1 ' 
 
 r m r m lj^ m s J . , 
 
 s=ik 
 m 
 
 By comparison one notices that (86) and (87) have the same form as 
 (33a) and (33b) in Chapter 3. In particular the matrix associated with 
 these equations is exactly the same. Methods discussed in Chapter 3 can 
 therefore be used to solve (86) and (87) . 
 5.3 Iterative Method of Solution 
 
 It was shown in the discussion above that the transformed equations 
 (72) and (73) can be reformulated in terms of a set of simultaneous equations 
 similar to the ones obtained in Chapter 3. This was shown to be possible 
 when the kernel K(t-T) is expressible in a suitable series of exponentials. 
 However, this might not be a convenient thing to do for all kernels. It 
 is therefore useful to find alternative methods for solving (72) and (73). 
 An iterative method for solving the above equations is presented here. 
 
 The method of solution is based on the following reasoning. Suppose 
 one starts with an approximation of q(t), say q (t) . Let q (t) be the 
 solution of the problem for T - °o. q (t) is therfore obtainable by 
 using Wiener-Hopf method. The following steps are carried out for the 
 
 iterative procedure. 
 
 (1) 
 1. Predict q (t+T 
 
 solution of the equation corresponding to T = °o, 
 
 1. Predict q (t+T) - approximate q(t+T) using q ' (t) which is the 
 
31 
 
 2. Find V (s) , the transform of q (t+T) , substitute in (73) and solve 
 
 (1) + 
 for W (s) , the transform of r(t+T) (see the definitions of V and W in 
 
 + - 
 
 (57) and (58), by using Wiener-Hopf technique 
 
 3. Find V ' (t) by inverting W (s) and substitute in (73). Next solve 
 
 (2) 
 
 for q (t) by applying Wiener-Hopf method to (72) . 
 
 4. Repeat steps (1), (2) and (3) until the solution converges. 
 
 No proof is included here regarding the convergence of the process. 
 However, it is possible to make the following observations. If q(t+T) is 
 small for t > o, i.e. if the solution q(t) decays with increasing t and 
 if T is large enough to make q(t+T) small, then the solution r (t+T) is a 
 good approximation of the actual r(t). Also if r(t) is fairly small for 
 t < o, then from the definition of R_(s) it is again seen that q.. (t) 
 obtained from (72) is a close approximation of q(t). The process is then 
 expected to converge fast. 
 
 Although it might not be practically convenient to express K 1 1— t | 
 in a series of exponentials, it is nevertheless useful for making certain 
 statements about the convergence of the iterative process by considering 
 a formal series expansion of K|t-r| in terms of exponentials. From the 
 discussion in Chapter 4 (see Equation (42)) that 
 
 JO, t a. (t-T) 
 
 f (t) - f (t) = A 6(t) + 2 A e 1 + B + S B, e 1 
 P ° i = l ± ° i=l l 
 
 From the above expression it is seen that q(t+T) for t > T and 
 r(t) for t < are small provided T is large or the a . ' s are sufficiently 
 large. It is therefore observed that if the kernel has a reasonably fast 
 decay with the increase of its argument, and the particular solution is 
 filtered out from f(t) in the beginning, it is expected that the conver- 
 gence of the iterative process will also be reasonably fast. 
 
 It is not necessary that the kernel be of an exponentially decaying 
 type. The essential condition for a rapid convergence of the process is 
 that q(t) and r(t-T) tend asymptotically to the corresponding semi-infinite 
 range solution at t = T. It is assumed in the above that the particular 
 solution has been filtered out from the general one. Considerations such 
 
32 
 
 as these apply to kernels of the type 
 
 Ho (2) Mt-T 
 
 which appear in the formulation of electro-magnetic wave problems, say for 
 instance in the problem of diffraction of e„m. waves by a perfectly 
 conducting strip. 
 
33 
 
 6. THE EIGENVALUE PROBLEM 
 
 In the preceding chapters the discussion was centered around solving 
 the integral equation 
 
 f T 
 
 gl"t) = f(T) K(lt-Tj) dT < t < T 
 Jo 
 
 g(t) specified in the range < t < T and methods were presented for 
 solving the equation when 
 
 n -k |t| 
 Kit! = S C e r 
 r=l r 
 
 In a class of problems in probability theory the eigenvalue problem 
 
 is encountered. The origin of the Equation (88) and a transform method 
 
 4 
 o f solution has been discussed by Youla in a recent paper. The purpose 
 
 here is to show that the eigenvalue problem can be formulated in terms of 
 
 a matrix without resorting to the transform method. The direct method of 
 
 formulation is simple and involves comparatively less amount of work. 
 
 6.1 The Eigenvalue Matrix Equation 
 
 Assume <j>(t) in the form 
 
 n -y t v t 
 
 (h(t) =E(Ae m +Be m ) 0<t<T (89) 
 
 " , m m ■ 
 
 m=l 
 
 Such a form is dictated by the fact that the integral 
 
 r T -y t 
 
 e m K |t--r| dT 
 Jo 
 
 -*y t 
 
 yields e multiplied by a factor, plus some other terms of the type 
 
 -k t k t 
 
 e and e If by adjusting n 's and B s, the coefficients of 
 
 m m 
 
34 
 
 k t 
 -k t r 
 
 r and e in the integrated expression of 
 e 
 
 J 
 
 (j>(T) K |t-T|dT, 
 O 
 
 can be made equal to zero, then it will be possible to obtain the solution 
 of the eigenvalue problem. 
 
 When (89) is substituted in (88) , the evaluation of the resulting 
 expression involves integrals of the type 
 
 r T -k |t-T| 2k e pt -V e^ + ( f -^ r 
 
 | e P T e r d T , -Jl— - S— — (90) 
 
 2 , 2 p-t-k p-k 
 p -k r r 
 
 •/ o r 
 
 Hence 
 
 n -> t rt /- T 
 
 X d)(t) =H 2 A e m + 2 B e m l= <j>(t) K|t-T| dT 
 
 m=l /; 
 
 - / o 
 
 T T 1 I 
 
 n r V t n -k t-T n / -Y t -k |t-T| 
 
 = S / Ae m SCe r dT+ZI B e m 2 C e r dT 
 
 / m r m r 
 
 m=l ) r=l m=l / 
 
 J o - / o 
 
 -k t k t 
 
 k A C -> t A C e r e r -(k +Y )T 
 
 = -2 2 _LJL£ o e m - 2 -JL* 2 A C - v r m 
 
 «, 2 , 2 -y +k m r k +y 
 
 mr V -k mr m r mr r m 
 
 m r 
 
 Y t -k t k t-(k -Y )t 
 
 kBCe m BCe r B C e r 
 
 _ 2 j. r m r _ m r _ m r 
 
 v, 2 , 2 k +y ~ k -y (91) 
 
 mr y -k mr r m mr r m 
 
 m r 
 
 -Yt 
 
 If (91) is to be true for all *t ', the coefficients of e and 
 
 r m t 
 
 e in both sides of (91) must agree for all <m f and the coefficients 
 
 -k t k t 
 
 of e and e must vanish for all *r *. 
 
 k t 
 
 Applying the second condition, one has from (91) Coeff of e is 
 
34 
 
 k t 
 -k t r 
 
 r and e in the integrated expression of 
 e 
 
 j 
 
 <j)(T) K |t-T|dT, 
 O 
 
 can be made equal to zero, then it will be possible to obtain the solution 
 of the eigenvalue problem. 
 
 When (89) is substituted in (88) , the evaluation of the resulting 
 expression involves integrals of the type 
 
 r T -k |t-T| 2k e St -V e V + < f - k r> J 
 
 [ e P T e ' ' iT ^r___e^ -_ (90) 
 
 2 , 2 p+k p-k 
 p -k r r 
 
 ^ o r 
 
 Hence 
 
 n -Vt 7t /- T 
 
 r 6(t) = r f S A e m + S B e, m 1 = <h(T) K|t-T| dT 
 \ Y \ i- , m m J / t ii 
 
 m=l ) 
 
 •S o 
 
 T T I 
 
 n /* V t n -k t-T n / -Y t -k | "t— T J 
 
 = E / Ae m 2Ce r dt + Z / B e m E C e r dT 
 
 / m r m r 
 
 m=l ) r=l m=l / 
 
 «/ o So 
 
 -k t k t 
 
 k A C -/ t A C e r e r -(k +Y )T 
 
 „^ rmrm v m r _ r m 
 
 = -2 i — - -7. e - Zy — =rr— ; - Z A C : r: 
 
 .2,2 -y +k m r k +y 
 
 mrV -k mr m r mr r m 
 m r 
 
 Y t -k t k t-(k -Y )T 
 
 kBCe" 1 BCe r B C e r 
 
 _ 2 j r m r - 2 m r _, m r 
 
 V, 2 , 2 k +y k -Y < 91 > 
 
 mr V -k mr r y m mr r m 
 
 ^m r 
 
 -yt 
 
 If (91) is to be true for all ft', the coefficients of e and 
 
 y m t 
 
 e in both sides of (91) must agree for all 'm* and the coefficients 
 -k t k t 
 
 of e and e must vanish for all i r^. 
 
 k t 
 
 Applying the second condition, one has from (91) Coeff of e r is 
 
3o 
 
 equal to zero for all r or, 
 
 A B 
 
 m r m m r m 
 -k L 
 Coeff of e is equal to zero for all r , or (92) 
 
 -Y t y t 
 
 A e m B e m 
 2 J? + 2 J5 = 
 
 k +y k +y 
 
 m r m m r m 
 
 -k t 
 r 
 Here the common factor e has been canceled from the second equation 
 
 in (92) . Equation (92) is a homogeneous set of 2n' equations for the 
 
 coefficients A and B . ■%* , -^ . 
 
 mm -T t f x 
 
 m ni t i 
 
 Also by equating the coefficients of e or c ' . for each m in 
 
 both sides of (91), the following is obtained. 
 
 k C 
 i = r 2 2 ^ r r = 1, 2, . .. n (93) 
 
 K r T ~k 
 
 m r 
 
 The routine for calculating \ and the unknowns in the expansion of the 
 
 eieenfunction, viz, A , B and y is the following. 
 
 ° m m m 
 
 Noting that Equation (93) is an algebraic equation of nth order for 
 
 2 
 y ' and that the coefficients of this equation in value \, the following 
 m 
 
 steps can be outlined. 
 
 2 2 2 2 
 1 Find 'n' roots of r , viz Y , 'V »'. . V ' from (93), Note that the 
 
 m 1 2, n 
 
 expression for these roots involve X.' 
 
 2. Substitute Y , Y , ... etc. in (92). Since (92) is a homogeneous 
 
 equation for A 's and B 's, the determinant must be equal to zero. Using 
 
 A = where A is the determinant of (92) , solve the transcendental equations 
 
 for X, yielding an infinite number of solutions X , X , .... . 
 
 3 For each X, say X calculated (m = 1, . . . . n) from the expressions 
 
 P mp 
 
 in step 1. 
 
 4. Using these 'Y ' s solve (92) for A ( s and B 's corresponding to X 
 mp mp mp p 
 
 Note that the coefficients can be determined only within a multiplicative 
 
36 
 
 th 
 factor. The expression for the p eigenfunction is then 
 
 -7 t +7 t 
 
 <$> = S A e mp + S B e m P 
 T p m mP m mp 
 
 (94) 
 
 The difficulty in actually working through the above procedure is encountered, 
 
 first, in solving for the n roots of X in step 1, since the coefficients 
 
 m 
 
 involve the unknown X. Even more difficult is the solution of A(X) = 0, 
 since this equation is transcendental. Only approximate solutions can 
 therefore be expected. 
 Example: 
 
 Consider the case K|t-T| = e 
 Equation (93) is then, 
 
 -k t-T 
 
 j t,e v 
 
 2k 
 
 X T 2 2 
 
 r - k 
 
 or 
 
 2 2 
 
 y + 2kX - k 
 
 or 
 
 = 
 
 y= (k 2 -2^x)^ 
 
 (95) 
 
 From (92; one has, 
 
 k-r k+r 
 
 - = o 
 
 Ae 
 
 -YT 
 
 k+y 
 
 Be 
 
 7T 
 
 k-r 
 
 = 
 
 (96) 
 
 or 
 
 or 
 
 A 
 
 i 
 
 k-y 
 
 -7t 
 
 e 
 
 k+y 
 
 k+y 
 
 k-r 
 
 e 
 
 e"* T 
 
 (k-f) 2 
 
 (k+r) 2 
 
 ft 
 
 e = + 
 
 k-y 
 
 = 
 
 = 
 
 - k+y 
 
 (97) 
 
37 
 
 Substituting from (95) in (97) 
 
 (K 2 -2k\)* T k - (k 2 -2k X)^ , Qjn 
 
 — 9, 1 
 
 k +(k -2kX) 5 
 
 Solutions of (98) which is a transcendental equation for X. are the 
 eigenvalues of the problem. 
 
 For a particular \ say X , 
 
 y = (k 2 -2kX )■ 
 P P 
 
 from (95) and from (96) 
 
 Hence 
 
 k+y 
 
 B = __i A 
 
 p fTk i 
 
 -7 1 k+r y t 
 
 4> = A \ e P + Z7-T- eP l 0<t<T (99) 
 
 Y p p L y -k J 
 
 P 
 which rearranged reads, 
 
 <j> * — sinh X t + coshT^ t (100) 
 
 y P r ' p p 
 
 P 
 
 where the constant factor has been dropped. 
 
 4 
 The same solution was obtained by Youla although by using a different 
 
 method of formulation. 
 
 6.2 Upper Limit Infinity 
 
 It is of interest to consider the case when T, the upper limit of the 
 Integral Equation is replaced by infinity. The formulation presented in 
 this section can still be used but there results considerable simplification 
 of the process of obtaining \ and <)>. This is discussed in the following 
 
 It is to be noticed in the first instance that because the range of 
 the integral equation is from to oo , continuous eigenvalues will result 
 
38 
 
 due to the type of kernel considered. In order for the integral 
 
 -k t -y t y t 
 
 (2 C e r ) (2 A e r + Be r ) dt 
 r r 
 
 to converge it is essential that if Re y = y (y >0) then 
 
 r rl rl 
 
 y , < k, , k real. It has been assumed that k, < k„ < ... < k and 
 nl 1 r 1 2 n 
 
 Y, - < \„- < ... < Y , • It is to be noted that since v , > 0, all -v s 
 
 11 21 nl 'rl 'r 
 
 are admissible. 
 
 Since the y s involve the eigenvalue X, this sets a limit to the 
 values that X can take. Consider a simple example in which k|t-T|= e 
 Then the integral equation is, 
 
 k t-T 
 
 j.^ ^ I , n -k t-T 
 
 <j>(t) = X (t) e ' IdT, k real 
 
 (101) 
 
 Assuming § (t) to be of the form, 
 
 <j,(T) = A e" Vt + B e^ 
 
 (102) 
 
 one gets by substituting (102) in (101), 
 
 iUe^+Be-^) = -: 2k 
 
 Ae 
 
 - Y t 
 
 Ae 
 
 -kt 
 
 - 2k 
 
 Be 
 
 yt 
 
 Be 
 
 -kt 
 
 Y 
 
 2 -k 2 ^ +k v 2 -k 2 '^ +k 
 
 V 
 
 (103) 
 
 Hence 
 
 2k 
 
 X ~ " v2 2 
 
 (104) 
 
 and 
 
 A B 
 
 + 
 
 Y+k ~Y+k 
 
 = 
 
 (105) 
 
 from (104) 
 
 k -2kX 
 
 or 
 
 2 . 
 k -2k\ 
 
 (106) 
 
39 
 
 In (105), A can be arbitrarily chosen and solving for B one gets, 
 
 B »* A 
 
 +k 
 
 Note that (105) can be satisfied for all y by the choice of B/A. 
 Therefore, the eigenvalue equation (L05) is satisfied for a continuous 
 range of X. The solution 4>('t) is then 
 
 (107) 
 
 cj>(t) = A 
 
 - v/k -2kX t v/k"-2k\-k V k~-2\k t 
 
 e " + — e 
 
 k *2kX+k 
 
 (108) 
 
 Note that if Re X< then Re(^y)>k hence the negative values of Re(X) 
 are inadmissible. 
 
 Equation (101) was solved by the Wiener-Hopf technique and appears as 
 an example in the book by Morse and Feshbach . 
 
 When the series representation of the kernel contains more than one 
 term say p , then it is found that the problem reduces to finding (p+1) 
 coefficients from 'p 1 simultaneous equations. This can be done by choosing 
 one of the coefficients arbitrarily and solving the 'p' equations. 
 
 The eigenvalue spectrum is continuous, hence the allowed values of 
 X are arbitrary, except for the restriction on Re(X). This restriction is 
 imposed from the convergence condition, i.e. the restriction on Re (■y) , 
 which was discussed earlier. 
 
40 
 
 NUMERICAL EXAMPLE 
 
 In this Chapter a numerical example is worked out to demonstrate the 
 application of the theory discussed in Chapters '2 and 3. The equation 
 considered is of the form (1) 
 
 Let the kernel K |t-T | of the form 
 
 .L I - - |t-x| n -2 It-T -3|t-T 
 K t-T = e ' ' -T- 2e ' ' -e ' ' 
 
 i.e. k x =1, k 2 =2, k 3 .3 
 
 2 
 F 0*> ) , the Fourier transform of K(,t) is 
 
 F (co ) = 2 
 
 2 2 2 
 co +1 co +4 co +9 
 
 i 2 
 
 ci s are obtained from the roots of F (co ) , and are 
 
 a = 0,95837 
 
 a = 3.44949 
 2 
 
 The matrix [mJ which corresponds to the case of T = oo is then, 
 
 [«]■ 
 
 1 -1.84628 
 1 -0 64866 
 1 -0 39345 
 
 0.40825 
 0.68990 
 2 22474 
 
 The cofactors and the determinant for [mJ and M calculated using 
 the expressions in Chapters 3 and 4 and verified by direct calculation 
 are, 
 
41 
 
 *L1 ■ 
 
 -1 
 
 17167 
 
 *21 - 
 
 -3 
 
 94688 
 
 *si - 
 
 -1 
 
 00893 
 
 \2 = 
 
 -1 
 
 53485 
 
 ^3 = 
 
 
 
 25522 
 
 ^2 = 
 
 1 
 
 81650 
 
 ^3 = 
 
 -1 
 
 45283 
 
 *32 = 
 
 -0 
 
 28165 
 
 ^33 = 
 
 1 
 
 19762 
 
 \l 
 
 = -i 
 
 28444 
 
 4i 
 
 = 4 
 
 17186 
 
 4 
 
 = -1 
 
 05800 
 
 \2 
 
 = -1 
 
 53485 
 
 4 
 
 = 
 
 27322 
 
 4 
 
 = 1 
 
 81650 
 
 4 
 
 = -1 
 
 51622 
 
 4 
 
 = -0 
 
 28165 
 
 a.„ 
 
 = 1 
 
 24221 
 
 are 
 
 A = 1.76627 
 
 A = 1,82941 
 
 The inverse of Tm] and Tm 1 constructed from the above data 
 
 [»] 
 
 -1 
 
 -0 66336 
 -0.86898 
 -0.14450 
 
 2.23458 
 
 1 02844 
 
 -0.82254 
 
 -0.57122 
 
 -0.15946 
 
 0,67805 
 
 l"']- 1 
 
 -0.70211 
 
 -0 83899 
 
 0.14935 
 
 -2,28044 
 
 0.99294 
 
 -0.82880 
 
 -0.57833 
 
 -0,15396 
 
 0.67902 
 
 Once the inverses [ M] and [m'] are known the Integral Equations 
 for both T = o=> and a finite T can be obtained by the methods outlined 
 in Chapters 2 and 3 
 
42 
 8 APPLICATIONS 
 
 The Wiener-Hopf Equation for the finite and infinite ranges occur in 
 the problems associated with the theory of linear prediction and the design 
 of optimum filters in the field of Information Theory, 
 
 In the field of Electromagnetic Theory, the Wiener-Hopf Equation 
 occurs in numerous problems of wave propagation. 
 
 Examples are the problems of diffraction by a slit, of finite width 
 in an infinite plane, or the diffraction by an infinite stack of equispaced 
 metal planes Several other examples are also scattered around in the 
 literature on wave propagation and boundary value problems. 
 
43 
 9 . CONCLUS IONS 
 
 In this report, methods have been developed for the solution of 
 finite and infinite range Integral Equations of Wiener-Hopf type and the 
 eigenvalue problem associated with the kernel. The assumption in a 
 part of the work is that the kernel K|t-T| can be conveniently expressed 
 as a series of exponentials. Iterative methods are presented in another 
 part for the case where the series expansion is inconvenient. 
 
 Examples are worked out to illustrate the methods. Applications 
 in the fields of Information Theory and E.M. wave propagation are 
 pointed out. Actual application to a particular problem in field theory 
 viz. the diffraction by a stack of finite-sized metal planes is left 
 for discussion in a future report. 
 
44 
 REFERENCES 
 
 1. Morse and Feshbach Methods of Theoretical Physics , Pt . I, p 978, 
 McGraw Hill 
 
 2 Laning and Battin, Random Processes in Automatic Control , Chapt 7, 
 McGraw Hill. 
 
 3. Muir and Metzler, A T reatise on the Theory of Determinants , Longmans 
 Green & Co. 
 
 n i* 
 
 4 Youla, The Solution of a Homogeneous Wiener-Hopf Integral Equation, 
 IRE Trans, on Information Theory, Vol. IT-3 , No. 3, Sept, 1957 
 
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 Technical Report No, 16 , Robert L Carrel, 15 January 1957, 
 
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 Raytheon Manufacturing Co. 
 Waltham 54, Massachusetts 
 
 Canoga Corporation 
 
 5955 Sepulveda Boulevard 
 
 M/F Contract AF08 C'603)-4327 
 
 P.O, Box 550 
 
 Van Nuys, California 
 
UNIVERSITY OF ILLINOIS-URBANA 
 
 Q621.365IL655TE «« 
 
 TECHNICAL REPORTSURBANA 
 37 1959 
 
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 3 0112 008079326