WITH THE TIDES OF A VISCOUS SPHEROID. 
571 
We now have got in (43) values of a, /3', y, which satisfy the equations (40) 
throughout the sphere, together with the surface stresses in (47) to which they 
correspond. Thus (43) would be the solution of the problem, if the surface of the 
sphere were subject to the surface stresses (47). It only remains to find a p /3, y /5 to 
satisfy the equations 
— ~+vV 2 a = 0, —~ )/ + &C. = 0, — ^+&C. = 0 .... (48) 
dx ‘ ay dz ' ' 
throughout the sphere, which is not under the influence of bodily force, but is subject 
to surface stresses of which + + C;_ 1 + C; +1 are the components. 
r Fhe sum of the solution of these equations and of the solutions (43) will clearly be 
the complete solution; for (43) satisfies the condition as to the bodily force in (40), 
and the two sets of surface actions will annul one another, leaving no surface action. 
For the required solutions of (48), Sir W. Thomson’s solution given in (15) and (16) 
of Part I. is at once applicable. 
We have first to find the auxiliary functions T,_ : , <b; corresponding to A/_ 1; B;_ 1? 
C/_ l5 and T/, <f>; +2 corresponding to A/ +1 , B,- +1 , C,- +1 . It is easy to show that 
^-3=0, <P; +2 = 0, 
and 
= — a 
i+2. 
2i + 5 
*r" + iv 5MU )}+| 
+ 
dz 
— a i+2 l)(2t + 3) ^j 
2i + 5 
* a; 11 ( 7 "tl)" 
— r- l+l a l+ --— 
9 
d 
dx 
u. 
We have next to substitute these values of the auxiliary functions in Thomson’s 
solution (15), Part I. It will be simpler to perform the substitutions piece-meal, and 
to indicate the various parts which go to make up the complete value of eq by accents 
to that symbol. 
First. For the terms in oq depending on A,-_ 1 , Tq_ 2 , 4>;, we have 
, _ 1 _ J_ 1 
_c 4 \ i(i+1) 3 dU 
“ ul4(i-l)(i-2) dx 
_« 4 (t + 1) 2 dU 
v 4(i — 1) dx 
d<D. 1 . . , 
' + . -A i-yr 1 1 
\2i-l) dx ' %-*>■ 
r a+ 1) 2 du 
2(i—2) dx 
(49) 
(Note that i — 2 divides out, so that the solution is still applicable when ? = 2). 
