631. Prof. T. W. Havelock on the Stability of 
2n n? p™ 
(Makes ea Nios abit 9) tat 2 
«Py! = —{k(n—k)— 2(n—D} pe ee eRe i, 
k n n—k n 
_ ea Dk = (mk) p"} +p” Mn —k— kp") 
Q=ip’n (1+p")? a 
_ a, Pik (n= Eph =pe“H(n— hp") 
R=p’n (1+ p)? ? 
Var Oiges Waaciiigs ss 5's oo o C8) 
The equation for A is 
r —P, —Q’ —R’ |=0. (44) 
—P, X —hk’ Q' 
=@ oor r py 
—R Q —P,/ X 
Using the relation (37) we see that P}—P,= Rabon 
and we find that (44) reduces to a quadratic in 2, which 
can be solved in the form 
4n2=(L#+Mt)?—(P,—P,), . . . (43) 
where 
L=(P; + P,')? +4QQ’, 
M=(P,—Py')?+4RR’. . . . . (46) 
The condition for complete stability is that for the n values 
of & all the values of \2 must be real and negative, including 
possibly zero. From the symmetry of the coefficients (43) 
in kand n—k it is only necessary to examine the values of 
k from zero up to 3n if nis even, or ${n—1) if nis odd. 
7. We might examine now in detail the case when we 
take «/=x«, that is, when the vortices in the two rings are of 
equal strengths and opposite rotations; we shall state the 
results without giving the details of the algebraic analysis. 
It can be shown that when p satisfies the equation (37) the 
quantity QQ! increases in absolute value from £=0 to k=4n, 
while the quantity RR’ decreases in absolute value as k 
increases in this range. Further, except at k=O the 
quantity L of (46) is always negative, and thus the criterion 
for stability is reduced to M being negative. But when 
k=23n we have R=R/=0 ; and since P, is not in general 
equal to Py’, it follows that M is positive at k=3n. Hence, 
if n is even, the system is unstable for the mode k= $n at 
least. It can be seen that in general there are always some 
unstable modes in the neighbourhood of this mean value of & ; 
344 
