solving Problems in the Conduction of lie at. 607 



Starting from the solution of (1) 



A.&m.axe~ Ka2t 



which vanishes at #— 0, we are led by (3) to choose A so 

 that 



a A cos aa + h( A sin oca — r ) = ; 



i.e. A = 



a cos aa + It sin aa 

 This brings us to the solution of our problem in the form 



7iv C sin ax e~ KaH 7 ,, ,, /T> . • -. 



u = — I j—. da, over the path (P). (o) 



?7T i a cos oca + /i sin aa a 



It can be shown just as in § 2 that the value of v given 

 in (5) as a contour integral satisfies all the conditions of the 

 problem, for the roots of the equation 



a cos tw + h sin aa = (6) 



are all real *. and in the integral 



i 



sin ax du 



a cos aa + h sin aa a 



the path (F) can be replaced by the dotted part of fig. 3, so 

 that, when the radius tends to infinity, this integral vanishes. 

 Finally the solution obtained in (5) reduces as in § 2, by 

 Cauchy's Theorem, to a infinite series, and wje have 



v = hv +22 ; / \ 2^ e n i ' 



v [_l + na i (Ji(l + ha)+aa n 2 ) a n J 



the summation being taken over the positive roots of (6). 



4. If the end x — a is kept at temperature Qt, or radiation 

 takes place there into a medium at temperature Ct, the solu- 

 tions corresponding to § 2 (5) and § 3 (5) will be seen to be 



=- H 



. sm ax e KCL 

 V= , \ — — s — da. 



sm aa a 



3 



Kofif 



sin a# 



da, 



fci7rj a cos aa + h sin aa a' 



respectively, the integrals being taken over the standard 

 path (P). 



These can be reduced to infinite series as before, by intro- 

 ducing the path (Q) . 



* Cf. Carslaw, ' Fourier's Series and Integrals, ' § 133. 



