340 T. H. Havelock. 
moving axes. Let ¢ be the surface elevation, and assume a frictional force in 
the liquid proportional to velocity, the frictional coefficient being ultimately 
made zero. 
The pressure condition at the free surface is 
<t —gC + v'd = constant, (1) 
and this gives 
Cr a 
Ae AP 0z p on °, o 
at 2=0, with xy =g/c? and u=y’/e. Assume for the velocity potential 
= [ do (f em KEN HID die 
21 ——T. 0 
t {i d0 I. F (c, 0) e+ de, (3) 
—T 0 
where @ = xcos 0 + ysin 0, and the real part of the expression is to be 
taken. The first term in (3) gives the velocity potential of the given source, 
namely m/r;, in a form valid for z+ f> 0. From the surface condition (2) 
we obtain 
18a, @)) 2 a 28 arene eee os) © 
2m kK — Ky sec? 0 + iu sec 0 
Hence we may write the solution in the form 
pa B— Bl cot oa {8 i, 6) 
. K, 
fy es 0 K — Ky sec” 0 + zusec 0 
where 
mae ty (e+fy?; re=2+ y+ (2 —f). 
It is to be understood that the limiting value of (5) is taken for > 0. 
We may now generalise by integration. We replace x and y by x — h and 
y — k respectively, and take o to be the surface density of source at a point 
(h, k, —f) on a surface 8 within the liquid. Thus the velocity potential is 
given by 
c aT [o} —K(f—zZ)+tco 
| ofS | sec? 00 | shoe eT Sela ai(6) 
T st 
. 2 
o K — Ky sec? 0 + iu sec 0 
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