PARTIAL DIFFERENTIAL EQUATIONS. 
179 
Now X is a definite function of x x , x 2 , p 1} p 2 , q x , q. 2 , given by eliminating the 
differentials from the equations 
dp x = \dq x , dp 2 = hdq 2 , d(f> — \d\Jj, 
x l dq x A x. 2 dq 2 A difj = 0 . 
By means of the first two, the third becomes 
cty jty djr djr 
PPi d Pi d( h 
YPi u f h op* oq 2 
and the fourth 
Xl + X ty 1 + + I ® 2 + + ^q} dq - ~ 0 
3t|t 3-v|r 
OPl + 3?i 
The result of elimination is therefore 
= 0 , 
x x 
'xf* + ^ 
X 
A 
9 ?i/ 
+ 3A+A)(A- + A _ + A6A + A =« 
'dpi 3 qj\ dp 2 dqj \ 3 p 3 dq. 2 J\ dp, 3 q 1 
This shows the form of X as a function of x v x 2 , p x , p 2 , q v q 2 , not involving c v c*, 
c 3 , c 4 . Now this choice of X makes it possible to choose coefficients A, B, C, E, F, G, 
such that 
x | dp x A dp ).2 A d<f) — A {dp x — \dq x ) A B(c/p., — \dq 2 ) A G {d<f> — \d\jj ), 
x 1 dq l A % dq. 2 A dxfj = E (c/yq — Xc/^) A F (dp. 2 — XcZgq) A G (d<f> — Xdi//). 
Thus 
A{c£(yj>i, X) — XcZ(yq, X)} AB {d(p 2 , X) — hd(q 2 , X)} 
A C {d(<f>, X) — \d(xjj, X)} = x l d(p 1 , X) A x 2 d (p. 2 , X) A d(<f>, X) 
= multiples of bidifierentials of p x , p 2 , q x , q 2 
+ 3.q{^ d *i) + ^ (P* ®i) -h cZ (<^, n?i)} 
+ d (p lt x. 2 ) A x. 2 d (p 2i x 2 ) A d (</>, aq) j 
— gy d {y> a) 'P$d (aq, aq) | A d (y, x 2 ) yqd (*®n *a) j" 
A multiples of bidifierentials of p x , p 2 , q v q. 2 . 
2 a 2 
