T = K„ _(Kr) cos(2n-2) + 2 K . (Kr) cos(2n-l) 

 n zn-z zn-l 



+ K (Kr) cos 2 n (223) 



zn 



However, this type of function does not form a complete set and we cannot 

 express the diffraction potential which satisfies the body boundary 

 condition by means of the linear combination of these functions. In order 

 to find the missing term, we consider a distribution of wave sources with 

 density 1/2 G(x) e along the x-axis. Taking the Fourier transform 

 of O(x) such as 



oo 



a(k) = a 



(x) e ±kx dx (224) 



we have the velocity potential of the type e x> where 



X = - iKe Kz f ex P [iax-iA^-g bdl ~ ( _ a _ R) da 

 J „ /2~T 



-K 



A^c 



-K „°° 



[iOK-Va — K I y I ] — , „s , 

 — lzj-± a(-a-K) da 



/a -K 



oo /• o° / / l 



-, r ■ f /2 2 /2 2 .„ . /2 2 I | 



. 1 lax ~, „, , /u -a cos/u -a z+K sm/u -a -u y , 



+ - e a(-a-K)da J ~ 2 g— ~ 2 e du 



J -~ lot! u -a + K (225) 



Now we change the variables of integration to 



a + K = p /u -a 2 = Kv (226) 



80 



