140 Prof. Gr. Prasad on the Failure of 



Now, putting t = e~ v , we have 



>~ 3v cos vdv 



log; 



C t 2 cos (log I) dt= (e- 

 Jo V . */ J log L 



^ f /i 1 x i 1\ I 



— / — % cos I loo- - + tan St I r • 



Therefore, from (1), 



v^_n_^_ = * cos (log- + tan Sir- (*>) 



r <}?' \ 7 10 ( V r 3/ > 



B 2 V B 2 V d 2 V 



Hence it follows at once that ■=c—^ i ^v - * ? ^~r are all non- 



0*r Oy 2 oz 2 



existent. Thus S/ 2 Y has no meaning and, consequently, 



Poisson's equation fails at P. 



4. Case III. Let p = cos-. Then, proceeding as in 

 Case II., we find that 



-4ttI £ 2 eos- dt 



1 BV (^, Q _ Jo t 



~dr r 6 



(2) 



But, putting -=«, we have 

 t 



C T Q 1 7 r "cos ?? 7 



I r cos - a£ = « — T-dv, 



Jo * J 1 v 



which is numerically less than 2r 4 , since — is always 



positive and constantly diminishes as v increases. 

 Therefore, from (2), 



i ay 



r ~dr 



< 87rr. 



yv b 2 v b 2 v 



Hence it follows at once that <— „ , -^— r, , =r— 5- are all zero. 



B# 2 ' By 3« 2 



Thus V 2 V is zero and, consequently, Poisson's equation fails 

 at P unless we assign the value to p . It should be noted 

 that, in this case, p may be assigned any value without 

 affecting the value of V 2 V. 



