﻿66 Solution of Problems in Diffraction. 



If, however, we remove the small semicircle then we easily 

 find that 



1 i cosn(«7r — — ti f ) T f 

 ~ 1 V ■ — J„ (mr f ) K {mr)dn 



ITT J A , Sill J17T ' S n\ / 



= i { J (m/) K (wr) + 2 2 J, (mr') K„ (mr) x 



7T L ,4-1 



cosn(0-0 ! )\. . (11) 



The path A r is that with the circle removed. There is a 

 well-known addition theorem for ~K (mr) which gives 



K (mR) = J Cmr') K (mr) + 22 J„ (m/) K n (mr) 



n = l 



XGOsn(0-0'). (12) 

 From (11) and (12) there results 



K (mR) = I f C0SW Cy-^--^) J (m/) K (mr) ^ (13) 

 v « Ja' sin mr " v y v y v 7 



This equation gives a solution for the unbounded space. 

 It is necessary now to add terms which shall satisfy the 

 boundary equations while introducing no new singularities. 

 After many trials I have found that the following equation 

 satisfies the conditions completely : — 



Y=- cos n(ir — — b')— cosn(ir — a — 0') — 



i J A , L su 



sin 7icc 



. , sum (a — 6*) 1 



— cos n(ir — 0) r^ '- 



v y sm no. J 



mr 



J. (mrOK. (,..., ^ _ _ 



Sill ?27T v y 



If this is tested term by term it will be found to satisfy 

 the differential equation and the boundary conditions. With 

 some laborious work the trigonometry can be simplified, and 

 the final result appears as 



tt «.->• C sin w(« — 0) sin«0' T , ,. Tr , w /irN 



V = 2z I ~ J^(mr / )K n (mr)dn . (15) 



J A , sin ?l(X n \ • * \ • v / 



r>r' and a>0>0'. 



Since in (15) there is no pole at the origin, then A and A' 

 are identical. 



By Cauchy's residue theorem the whole solution is now 



