655] 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 
123 
dr 
we find for -j— the equation 
CLOG 
d (c 0 , d 0 , e 0 ) (dr d(a, b, H) d (a, b, H)} ^ Q 
d(p 0 , q 0 , r 0 ) \dx d(r, p, q) d (x, p, q) j 
We have in the same way for 
dp 
dz 
the equation 
d(c 0 , d 0 , e 0 ) (dp d (a, b, H) + d(a, b, H)j + &c = Q 
d(p 0 , q 0 , r 0 )\dz d(p, r, q) d(z, r, q) j 
whence, adding, we obtain 
d (Cp, dp, e 0 ) (fdr _ dp\ d(a, b,Jl) d(a, b, H) d(a, b, H)} + &c = q 
d(p 0 , q 0) r 0 ) (U® dz) d (r, p, q) + d(x, p, q) d{r, z, q)\ 
where the terms denoted by the &c. are the like terms corresponding to the different 
permutations of the letters a, b, c, d, e. 
The equation may be simplified; we have identically 
da, 7 TT . db /TT . dH, 7X d (a, b, H) , d(a, b, H) 
-dq (b ’ a) -d q {a ’ b)= dim, P, q) + d(z, r. q) ' 
dH 
or, since {H, a) = 0, (H, b) = 0, the left-hand side is simply — -y— (a, b), and the 
UjQ 
equation becomes 
d(c 0) d 0 , e 0 ) \ fdr dp\d(a, b, H) dH , j 
d(p 0 , ff„ r.) [W* dz) d (r, p, q) dq W 0) ] + &C ' " 
68. This ought to give ^ — ~ 0; it will, if only 
(a, 6)1 = 0, 
d (po, do, &o) 
d (po, q 0 , r 0 ) 
which is thus the condition which has to be proved. By the Poisson-Jacobi theorem, 
(a, b) is a function of a, b, c, d, e: if we write 
_ d(a 0 , bp) + d(a 0 , b 0 ) d(a 0 , b 0 ) 
d (po, Xp) d (q 0 , y 0 ) + d(r 0 , z 0 ) ’ 
then (a 0 , bp) is the same function of a 0 , b 0 , c 0 , d 0 , e 0 ; but these are equal to a, b, c, d, e 
respectively, and we then have (a, b) = (a 0 , 6 0 ), and the theorem to be proved is 
2 
(d (Cp, dp, 6p) 
jc£ (p 0 , q 0 , r 0 ) 
= 0. 
But, substituting for (a 0 , 6 0 ) its value, the function on the left-hand side is, it 
is easy to see, the sum of the three functional determinants 
d (ao, bp, Cp, dp, 6p) d (ap, bp, Cp, dp, 6p) d(ap, bp, Cp, dp, 6o)