326 Dr. R. A. Houstoun on the Transverse 



upwards. Let a be the radius o£ the section. Then $, the 

 velocity potential, must satisfy the equation 



P + P = (1) 



at all points within the semicircle, the condition 



g=0 ... (2) 



at all points on the semicircle and 



B 2 <£ d<£ n /Q . 



at all points on the surface. 



The problem cannot be solved by transforming conform- 

 ally the case of triangular section solved by Kirchhoff *, 

 because when the triangle is transformed into the semicircle, 

 the surface condition no longer holds, 



Suppose that the time-factor is e iat . Then the surface 

 condition becomes 



Let -^r be the stream function. Then, omitting the time 

 factor, we may assume 



c/) + ^ = A + A 1 (^ + z < ?/)+A 2 (^ + ^) 2 



If we write the constants A n in the form Pi» + zQ», we 

 obtain 



(/> = P + X r n (P n cos nO — Q n sin n6), 



i/r = Q + £]V(Q» cos nO + Pn sin n6). 



This satisfies (1). 



Expressing the surface condition in polars we obtain 



o^=?|* for = 0. 



r ov 



and ^=-^11 for 6 = tt. 



r o" 



Substituting the expression for $, we find that the surface 

 condition is satisfied, provided that 



fiQ»=-ir«P.- 1 /^ (4) 



* Ibid. § 255. 



