Salelt= 
it is possible by cross differentiating the first two equations, subtracting 
and combining the result with the third equation to obtain the result that 
(2) 72ttaa * 72ttss ~ Sts * 8 S55 
This provides the inhomogeneous solution for Z5(a, 6,t). The 
third equation then determines x5(a, 6,t) and the first two yield the same 
solution for P2: 
However Po does not satisfy the free surface condition, anda 
solution to the homogeneous equations must then be found that when added 
to this solution yields a proper form for Po: 
This provides solutions to the first three equations. When x, and 
2 
Z> are substituted into the last equation, it is seen that the vorticity can 
2 2k,6 2 2k,6 
be made zero by adding ay k),e tta, k, Ws € t to x,. 
The second order terms therefore become 
aja, w pte, (kj +k))6 
a) Se : ———): sin(kk, -k,)a - (w, -w))t + €5-€)) 
Ww, - W 
2 1 
aa, (k,-k,)6 
+ : (o, + wo, e sin((k, -k,)a - (wy - w,)t + €,-€ ) 
2k,6 2k, 6 
2 i 2 2 
ta) wikje tta, wok, e t 
aera (k,+k,)6 
mene a2 2 2 i, 2 z 
(27) Zo = (o, + W1%> + Ws Je cos((k, -k,)a - (5 w,)t + €5 - &) 
aja, (k-k))6 
- w,(w, + 0) e cog(k, - k))a- (w, -w,)t + E> - €1) 
5 2,6 a oe 2k,6 
= Weyl 1 (472 2 1 
Dy SED IG SN) ale =) 
(k) +k, )6 
- 2p aa, 050, € cod(k, - k))a- (w, - 0,)t + €5-€)) 
(k -k,)6 
+ 2p A] a,W,0) € cod(k, -k,)a-(w.) )t +P €,-€)) 
F, = a) a5, e(k2-ky) sin((k, - k)a- (w5 -w,)t + €,-€ )) 
(k,+k)) 
+a,a (a, -w je sid(k, - k,)a-(w,-@)tt€,-€) 
