Ogt lute 
bation potential satisfies the following 
[LJ egies nero 
[H] Oo = Up! (x) on y=. = ibe; 
Jy 
a = © on Za = He 5 
[A ] Be UGA =e AKO 
on z= 0 
[B] § Mact® 
The last condition, [B] » is of course the rigid-wall condition which 
replaces the free-surface condition. The dynamic boundary condition 
on the free surface, [A] » serves only for the determination of the 
free-surface shape, $ , after the potential problem is solved. In the 
body boundary condition, we have stated a separate condition for the 
bottom of the wedge, for we do not need or want to restrict ourselves 
to a ''thin'' slender-body over the entire body length. 
The above problem can be solved precisely, by mapping, for 
example. We do not need that complete solution, however. Let 
@(x,y,z) be the solution of this 2-D problem which has the property 
2UHb' 2 
&(x, y, z) pe dee se log ts +x 
us 
ray a 2 
| i as lie . ee 
Then the perturbation potential, ¢(x,y,z), is given by 
O(%sVo2)) oS4 0b Yee) to PaGe)ass 
where F (x) is given by [8] 
'F (x) dle ae fg (2) sgn (x- £) log 2|x- | +7 H,(K(x- &) ) 
+ (2 + sgn (x6) )=RY, (K |x- nf , 
1504 
