Compressible Fluid past an Obstacle. 

 In the present problem 



„ a d 2 2d 1 d ( . n d\ 



ar- /• (//• rsin 6 dv\ ddj 

 while P„ satisfies 



sin 



so that 



reduces to 



V 2 *=^P„ . . • i 



^ + 2d4>_n { ,^r l , rK 



dr 2 r dr r 1 



(18) 



(19) 

 (20) 

 (21) 



The solution, corresponding to the various terms of (17), is 

 thus 



a- rP+2F * m~) 



9 ( p +t)(p+'d)-n(7i+i)' ' • ; v J 



With use o£ {22), (6) gives 



IP J _c 6 P 1 c^ 3c 3 P 3 3c«P, 3c 9 PQ 

 9 a 2 L 5r 5 + 24r 8 10r 2 10r 5 + 176r 8 J 



+ A ? 'P 1 + Br- 2 P 1 + Cr 3 P 3 +Dr^P 3 , . . . (23) 



A, B, C, D being- arbitrary constants. The conditions at 

 infinity require A = U, C = 0. The conditions at the surface 

 of the sphere give 



and. thus is completely determined to the second approxi- 

 mation. 



The P's which occur in (23) are of odd order, and are 

 polynomials in /x( = cos 6) of odd degree. Thus d<f>/dr is odd 

 (in fi) and d<f>/dd = sin x even function of /jl. Further, 



^ 2 = even function -I- sin 2 6 x even function = even function, 

 dq 2 /dr = even function, 

 dq 2 /d6 = sin #Xodd function. 



Accordingly 



d$ dq 2 1 d$ dq 2 , c c 



-j- -, - + n-jk ttt = odd tunction or /x, 

 <ir dr r J du dd n 



and can be resolved into a series of P's of odd order. Thus 

 not only is there no resultant force discovered in the second 



