Ogitvie 



Using Green's theorem, we can write: 



*,(x;y,z) = ^J [^log r' - *, Ijdog r')] d^', 



where r^ = [ (y-y') + (z-z') ] , and the integration is carried out in 

 the cross section, with (y',z') ranging over the body contour, the 

 free-surface contour, and a closing contour at infinity. The last of 

 these contours contributes nothing and can be ignored. We assumed 

 that ^1 is known on the free surface, and, from the body boundary 

 condition, we know 9$| /9N on the hull. If we let the field point, 

 (x;y,z), approach the hull surface, we obtain an. integral equation, 

 with ^1 unknown on the hull and 8$|/9N unknown on the free 

 surface. This is not quite the usual form for an integral equation, 

 but it should be possible to solve it approximately by essentially 

 standard numerical methods. Then the Green's -theorem integral 

 can be used to express $, at all points in that cross section. Thus, 

 the solution of an integral equation in one dimension allows the 

 potential to be found. 



This procedure has not used the information contained in the 

 free- surface conditions. Usually, we look on the free surface con- 

 ditions as complications that cause tremendous difficulty in the 

 finding of solutions. Now we take an opposite point of view: 

 Supposing that we have solved the above problem at some x, we use 

 the kinematic conditions to predict the value of 2, just downstream: 



;(x + Ax,y) = ;(x,y) + Ax ;^(x,y) + . . . 



= ;(x,y) +^[^l^(x,y,;(x,y)) - $,y^y] +... . 



Similarly, we predict the value of $, on the free surface just 

 downstream: 



l^(x+Ax,y) 



'z 



where the right-hajid side is evaluated at (x,y , ^(x,y)) , and the 

 dynamic boundary condition is used to evaluate ^i^. 



Now we are ready to start over. Presumably having solved 

 the problem at some x, we have used the free- surface conditions to 

 formulate the equivalent problem at x + Ax. The most serious 

 difficulty m.ay very well be in starting the whole process , and there 

 seems to be no elegant prescription for carrying out that essential 

 first step; in some problems, it is possible that a linearized solution 

 may suffice for a start, but this is not certain. Another serious 

 difficulty may be the stability of the method. 



744 



