Compressible Fluid past an Obstacle. 3 



of the cylinder. At the surface of the cylinder r=c, 

 d<j>/dr = 0, for all values of 0. 

 On the right hand of (6) 



dx dx 

 and from (7) 



d<f> dq 2 d<f> dq 2 _ d<f> dq 2 1 d<j> dq 2 



~ 2 ddd0 ; 



dy dy dr dr 

 j*-JJ*\\dr) ~* r\d0) J ■ 



(«) 



9 



U 



Also 



<>-2 



l+- 4 -^-cos20. 



(9) 





^-(l-5)«-*. ^ = -(l-4>in*; 



1 dq 2 4c 4 L 4c 2 0/J 



1 <7 7 2 4c- 2 . .. 

 U 2 rd0 r* 



Accordingly 

 V 2 <£ = 



^|-J(2-£)cos0 + cos30J. (10) 



The terms on the right of (10) are all of the form 

 r p cos ?i6, so that for the present purpose we have to solve 



raj d*6 ldd> , 1 J 2 6 n 

 T dr 2 r dr r z d0 z 



(ii) 



If we assume that <f> varies as r m cos n0, we see that 

 <m=p + 2, and that the complete solution is 



r r p+2 •} 



» = cos n g|Af* + Br-»+ (f + 2)1 ,_ >i , |, . (12) 



A aud B being arbitrary constants. In (10) we have to 

 deal with n = l associated with p— — 5 and —7, and with 

 n = 3 associated with p=—3. The complete solution as 

 regards terms in cos and cos '60 is accordingly 



</> = (Ar + Br- 1 )cos6''r-(Or 3 + Dr- 3 )cos3l9 



2U 3 c 2 r a ( c 2 c 4 \ cos 301 mo . 



The conditions to be satisfied at infinity require that, as in 

 (1), A = U, and that C = 0. We have also to make d<f>/dr 

 vanish when r = c. This leads to 



B = c 2 U + 



12a 2 ' 



D = 



UV 

 12a 2 * 



(14) 



B2 



