WITH THE TIDES OF A VISCOUS SPHEROID. 
5G9 
/3=fi'-\-/3 / , y=y'-j-y / , p—p’-\-p^ and find a', /3 1 , y , p any functions which satisfy the 
equations (40) throughout the sphere. 
Differentiate the three equations (40) by x, y, z respectively and add them together, 
and notice that 
df JV 
dy 
dz 
+ 2 ,:^j ) ^u i -^))+-( )+- 
dx 
dy 
dz 
= 0 , 
and that 
C K , d A . */_*. 
dx ^ dy ■*" dz ’ 
then we have v ~p = 0, of which y/ = 0 is a solution. 
Now if Y h be a solid harmonic of degree n, 
Hence 
V 2 r m Y u = m(2n -\-m +1) r m 3 V„ 
„dU „ r l tfU 
r ~—= v"- — 
dx 4(24 + 3) d® 
dc 
(41) 
Substituting from (41) in the equations of motion (40), and putting p=0, our equa¬ 
tions become 
Vfl va 
/ b + 1 ) XU 
d 
4 •'**-2iT5’- !i+ V4 U ^‘-fi= 0 
V ~{v/3' — &c.} = 0, V~{vy' — &c.} = 0 
(42) 
of which a solution is obviously 
a =- 
v 
1 [i+ 1 ,dU 
r + - 
4 dr 
/3' = &c., y' — k C. 
-4-— r 2;+5l/qj r -2i-i\l 
ic T 2t + 5 dad ; J 
> . 
(43) 
It may easily be shown that these values satisfy the equation of continuity, and 
thus together with p'= 0 they are the required values of a, (3', y,p, which satisfy 
the equations throughout the sphere. 
The next step is to find the surface stresses to which these values give rise. The 
formulas (13) of Part I. are applicable 
vt,'= v (ax J r /3'y-\-y'z) 
i(i + l)(2i + l) A 
-+±1),.4U= V U. 
4 24 + 5 4(24 + 5 ) 
4 D 2 
