570 
MR. G. H. DARWIN ON PROBLEMS CONNECTED 
Then remembering' that 
XV: 
1 r r^-r^3| (r -c/-i U) ^ 
2i +1 dx 
We have 
r° 
d? _ i(i + l)(2i + 1) [ A dJ] , 4?’ 3 n 3 ^CJ 4b 2 
’dx ~~ 4(2i + 5) 
i(i + l) 
dx 2i + l clx 2 i +1 dx 
. r Zi + 3±( r -V-l U) 
4(2* + 
|[(2i+5)r*f-4^« f |( } -^U)}.(44) 
Again, by the properties of homogeneous functions, 
„l r^-l\x=v(xj+yj+zj\/-va' 
dr 
dx J dy dzj 
■ (t +1)^ + 2) 4,cZU i(i + 2) 2 . +5 j _2*_ 1 tt'v 
4 cfo~2* + 5 
(45) 
Also p'=0. 
Then adding (44) and (45) together, we have for the component of stress parallel to 
the axis of x across any of the concentric spherical surfaces, 
Fr= —p'x-\-v 
b’ + l) 2 .dXJ , i 
J \ d? 
1 iv~ 1 r + Xh 
by (13), Part I. 
dx 2i + 5 
' F5 ~(r 2i 1 [J) by (44) and (45). 
And at the surface of the sphere, where r = a, 
F _(wfl)_W +2 
— i+1 
ALJ 7 
dx 
i 
2i + 5" 
+. za i+ * 
tS v 
dx 
U) 
(46) 
The quantities in square brackets are independent of r, and are surface harmonics of 
orders i—l and i 4-1 respectively. 
Let 
Where 
A;-]. — ■ 
(* + l) 
a i+ ~ 
V. — /+1 
F=-A / _ 1 -A f - +1 
dJJ 
dx 
Ai+1 2i + 5 
(V 
i +2 
i 
!• • (47) 
J 
Also let the other two components G and H of the surface stress due to a.', /3', y , p 
be given by 
G= B 2+1 , H=-G_ 1 -C ;+1 .(47) 
Then by symmetry it is clear that the B’s and C’s only differ from the As in having 
y and 2 in place of x. 
