364 
Proceedings of Royal Society of Edinburgh, [july 15 , 
It will still be convenient occasionally to use (1). We proceed to 
find the complete solution of the problem before us, consisting of 
expressions for u, v, iv, p satisfying (22) . . . (25) for all values of 
x, y, z, t ; and the following initial and boundary conditions : — 
when t = 0: u, v, w, to be arbitrary functions ) 
of x, y, z, subject only to (1) j 
u = 0, v = 0, w — 0, for y = 0 and all values of x, z, t ( 
u = 0, v = 0, w— 0, for y = b „ „ j 
■ ( 26 ); 
• (27)- 
33. First let us find a particular solution u, v, w, p, which shall 
satisfy the initial conditions (26), irrespectively of the boundary 
conditions (27), except as follows : — 
V = 0, when t — 0 and y = 0 
V = 0, when t = 0 and y = b 
(28). 
Next, find another particular solution, u, it), p, satisfying the 
following initial and boundary equations : — ■ 
it = 0, = 0, tt) = 0, when t — 0 . . . . (29), 
it -Ml = 0 , t> + V = 0 , tt) + w = 0 , when y = 0 ) 
and when y = b) 
(30). 
The required complete solution will then be 
m = U + u, v = + z^ = U) + w. . . . (31). 
34. To find u, V, W, remark that, if y. were zero, the complete 
integral of (21) would be 
l = arb. func. ( x - /3yt) ; 
and take therefore as a trial for a type-solution with ju not zero, 
£ 
»p i \mx + {n -mpt)y+ qz\ 
( 32 ); 
where T is a function of t, and t denotes Substituting 
accordingly in (21), we find 
r TV 
— = — y[m 2 + {n- m/St) 2 + g 2 ]T .... (33); 
whence, by integration, 
-pt[m2+n2+q2-nnipt+^ pw] 
By the second of (21) and (32) we find 
( 34 ). 
