II (V 2 (f> 2 64> 2 -<j> 2 V 2 (6(j> 2 )} dxdz = {()) 2n <S(f> 2" <t) 2 (6<|) 2ri )} ds = ° 



D 2 S J US 2F 



and from the linear free surface condition on S„„, Equation (5), and the condition 



zr 



at t = T, we have 



J J (5 *2*2n-*2 6 *2n ) dzdt = "J J < 6 *2*2n-*2 6 *2n> dxdt 



0S J ° S 2F 



(10) 



= lj j (6 *2*2tt-*2 6 *2tt ) dtdx " | J (6 *2 < f , 2t-*2 5 *2t ) t=x 



dx = 



S 2F ° S 2F 



so that the last integral of Equation (9) vanishes. Since 6<j>.. , 6<j>„, 6d)_ , and 6h 



1 z zn 



are arbitrary, we obtain from Equations (9) and (10) the corresponding time- 

 dependent, free surface boundary value problem represented by Equations (2)- (4) 

 together with the Laplace equation for <j> . Therefore, solving the variational problem 

 with the functional J of Equation (6) is equivalent to solving the Laplace equation 

 with the boundary conditions as set forth in Equations (2)- (4), provided we assume 

 that 



» 2t = at t = x 



LINEAR PROBLEM 

 If we assume a linear free surface condition in both S.. and S„ and keep the 



exact boundary condition [Equation (3)] on the body surface, 



= h / vV+1 (ID 



t x 



then we may use 



