OgiZvie 



where j = 3 and 5, and $j , ^j , and Jij satisfy the following con- 

 ditions , respectively; 



$• + <|)j = 0; cE), = ni on z = d(x,y); $■ - v^i =0 on z = 0: 



(3-44) 



^i + ^i =0; ^i = mi on z = d(x,y); *: - v^.. =0 on z = 0; 



Jyy JzZ 'n ' ' Jz J 



(3-45) 

 Rjyy +«j^2 = 0; iij^ = on z =d(x,y); 



S2j - vflj = - {i/g)[2^j + 2xy*j + Xyy^j] on z = 0. 



Z * / 



(3-46) 



The quantities nj were defined previously, in (2-72), as the six 

 components of a generalized normal vector. Also, the quantities 

 mi were defined earlier, by (2-75). In the present notation, let 

 v(x,y,z) (see (2-74)) be defined by 



v(x,y,z) =V[x +x(x,y,z)]. 



Then rnj is again given by the previous formulas. Now it requires 

 just a bit of manipulating to show that the assumed solution above 

 indeed satisfies the body and free-surface boundary conditions; I 

 omit the proof. 



The above near-field solution must match the far-field solu- 

 tion, which has an inner expansion given in (3-40). In connection 

 with the latter, a comment was made earlier that the near-field 

 solution would have to represent a wave motion in which one com- 

 ponent grows linearly in amplitude as |y| ~* oo. Now we can see 

 that just such an interpretation must be given to the Qj functions, 

 for otherwise we cannot possibly find solutions to the problems set 

 above for J2j . The nonhomogeneous free-surface condition on ttj 

 can be compared to the free-surface condition that would result if a 

 pressure distribution were applied to the free surface. In fact, if 

 a pressure field were applied externally on z = 0, the pressure 

 being given by 



p(x,y,t) = ipC0Uej(t)[ 2$J^ + 2Xy^Jy + Xyy^j] » 



2 



then the potential function would have to be (ico) UJ2j^j(t), wi 



£ij(x,y,z) satisfying the conditions stated previously. This 



764 



