42 
Proceedings of Royal Society of Edinburgh. [jan. 7, 
and 
P= - / dt sin - t') = - f dt sin <*(t - $')P(£') . (17). 
J o Jo 
Since (/> ( t ), as we have seen, satisfies (13), P ( t ) must satisfy it 
also. Hence 
dP(t) _ d 2 P(t) 
^ dy dt 2 
. (18), 
for all values oi x, y , and t. How by differentiation of (17) we 
find, because P(0) = 0, and by (16), 
dP 
dt 
- f dt' sin u>t' -^P (t - t') = - / dt' sin - 1 ') . ( 19 ); 
J o dt J o 
and differentiating this, we find, because <£( 0 ) = (r + y + S )*?*" 1 
d 2 P (r + y + 6)» . /°* . ,d, u ^ 
— v u y sin wt - I dt sin co£ -=■ <£(2 - t ) 
Jo dt 
dt 2 
- - si „ * - Pdf sin «(< - t'ffP . (20). 
r J n dt L 
From this and the second form of (17) we find 
9 
dV d 2 P (r + y + by . 
= -sm J 
dy dt 2 r 
t 
whence, by (18) 
dV d 2 P (r + y 4 - by . 
q — ‘ 2 -sm a it 
u dy dt 2 r 
( 21 ); 
( 22 ); 
and therefore finally, by (3) above, we have, for the surface 
pressure, 
p j (x 2 + b 2 y + b \ * . /0QX 
P »=0) = C + \ + 62 1 sm “< • ( 23 )> 
as promised in (12) above. 
To work out our solution, remember that dV/dt is the 
velocity-potential of the motion ; and calling • this d>, we find, 
by (19), 
<E> = - j dt' sin (o{t — t')g>(t ) .... (24); 
J o 
