568 
ME, G. H. DARWIN ON PROBLEMS CONNECTED 
The solution of this has the same form as in the first approximation, viz,: equation 
(31), with a 0 written for a, and Y' for Y. 
We shall have occasion hereafter to use the velocity of flow resolved along 1 the radius 
vector, which may be called p. Then 
Hence 
P o 
Po~ a 0 f + A)~ +7oy 
1 [ i~(i + 2) cr — i(y — l)r 2 1 Y' 
lv 
2 ( 1 - 1 ) 
cos vt 
Then observing that Y ' -r-r i is independent of r, we have as the surface value 
a ;+1 i(2i+l) Y' 
Po = 
lv r 
—r COS Vt 
(33) 
(34) 
Secondly, 
clpJ , at, woo? i(i + 2) dY . . 0 „ , 
+UV '“ U+ 1^'2h^dr Sm ^ = 0 ’ &C, > &C .( 3o ) 
This, again, may clearly be solved in the same way, and we have 
, __ wvcP i(i + 2) [ 
a ° = ¥d ‘ —i) t 
and 
, _ wvct" i(i -f- 2) (" i?(i + 2)<x~— i(i~ — l)?’ 2 
p0 = Thd ‘ 2 b-i)l 2 (i- 1 ) 
and its surface value is 
Y . 
— sm vt 
i(i + 2) __ 0 (i + l)(2i + 3) . 
2b-1)^ 2(2i + l) r 
dY _ i_ 
dx 2i +1 
2i+3_ 
dx 
(Yr ~ l J ) l sin vt 
Thirdly, let 
p Q '=wva i+3 
d(i + 2)(2i + 1) 
[21v(i-l)f 
Y . 
— sin vt 
r l 
wv Y » 
lv 2(2i+1) 
sin vt 
(36) 
(37) 
(38) 
(39) 
So that U is a solid harmonic of the i ih degree multiplied by a simple time harmonic. 
Then the rest of the terms to be satisfied are given in the following equations :— 
. . . . (40) 
-£+ to. = 0 , -f+ to. =0 
dij dz J 
These equations have to be satisfied throughout a sphere subject to no surface 
stresses. The procedure will be exactly that explained in Part I., viz.: put a=a'-fa / , 
d P . _ 3 
■v + uV ~a — 
dx 
b +1) (2 i +3 +2 ir~ i+i ~(lJr~ 2i ~ l ) 
dx' 
