125 
[655 
655] 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 
linator, that is, 
L And in the 
V. 
or any one of 
s meaning must 
f, r expressions 
nable, we have 
the next case 
reater difficulty, 
which, in virtue of (a, H) = 0, (b, H) = 0, becomes 
similarly, 
dH d(a,b,H) m 
dq K d{x, p, q)’ 
dH 0 
dq 0 
(«o, b 0 ) = 
Hence the terms in question are 
d (ft0} b 0 , -Ho) 
d O«,, p 0 , q 0 ) ' 
dH dH 0 
dq dq 0 (a ° 
h), 
dH dH 0 
dq dqâ { a> 
b), 
which are equal in virtue of (a, b) = (a 0 , b 0 ); and, similarly, the other conditions might 
be proved. But the proof would be more difficult in the case of a greater number 
of variables. 
the equations 
H = H 0 = const.) 
theorem that, 
;c. ; we find 
nd of opposite 
subtracting the 
> are the same 
ty in assuming 
Examples. Art. Nos. 71 to 79. 
71. The variables are taken to be x, y, z, p, q, r. As a first example, which will 
serve as an illustration of most of the preceding theorems, suppose pqr — 1 = H; the 
Hamiltonian system, with the adjoined equalities, is here 
dx _dy _ dz _dp _dq _dr _ ^ _ dV 
qr rp pq 0 0 0 3pqr' 
The integrals of the original system may be taken to be 
a =p, 
b = q, 
c = r, 
d = qy — px, 
e — rz — px, 
and there is of course the integral H — pqr—1, which is connected with the foregoing 
five integrals by the relation H = abc — 1. 
We form at once the equations 
(a, b) = 0, (a, c) = 0, (a, d) = — a, (a, e) = — a, 
(b, c) = 0, (6, d) = b, (b, e) = 0, 
(c, d)= 0, (c, e) = c, 
(d,e)= 0; 
hence it happens that no two of these integrals a, b, c, d, e give by the Poisson- 
Jacobi theorem a new integral. To show how the theorem might have given a new 
integral, suppose that the known integrals had been a = p + q, and e = rz — px, then 
(a, e) = —p: or the theorem gives the new integral a = p. 
e identically