558 PROFESSOR CHRYSTAL ON LAGRANGE'S RULE FOR THE 



(in the memoir last quoted) est exposee qu'en passant et presque sans demonstration, j'ai 

 cru qu'il serait avantageux aux progres du calcul integral de la presenter de nouveau de 

 La maniere la plus directe et avec toute la generalite dont elle est susceptible." 



The demonstration, which seems intended to prove that every possible solution is of 

 the Lagrangian form, may be summarised as follows. We have to determine p and q as 

 functions of (x,y,z), so that the two equations 



Pp + Q,q = R > pdx + qdy = dz 



may be simultaneously satisfied. As a consequence of these we have 



q(Pdz- 1 Rdx)+q(Qdz-'Rdy) = (A). 



Let u(x,y,z) = a, v(x,y,z) = b be the integrals of the system 



Pdz-Rdx = 0, Qdz-~Rdy = 0; 



then, if we replace the variables (x,y,z) by u,v, and any one of the others, say z, we 

 can readily show that 



H?dz — Hdx — ~R(v y du — u tJ dv)/T , 

 Qdz— ~Rdy = ~R(u x dv — v x du)/T , 



where T u y v x — u x v y . 

 Hence (A) becomes 



R 



f^{(pv y —qv x )du + (qu x —pu y )dv} = . . . (B). 



Lagrange then rejects the factor R/T, and writes (B) in the form 



du+ qUx ~ pUy dv = (C). 



pvy—qv x 



Then he reasons thus. Since (C) contains only the two differentials du and dv, it can 

 only subsist if the coefficient of dv be a function of u and v alone. Consequently, if 

 we substitute for x and y their values in terms of (u,v,z) as determined by the equations 

 u(x s y,z) ■■= u, v(x,y,z) = v, the variable z must disappear of itself, and we must have 



{qu x -pu y )l(pv y -qv x )=f(u,v) ... .(D); 



(C) then becomes du+f(u,v)dv = ; the integration of which gives us a relation of the 

 form F(u,v) = 0. 



This demonstration is utterly unsound. So far from demonstrating that all solutions 

 have the Lagrangian form, it does not even apply to solutions that actually have that 

 form. 



In the first place the equation (B) might be satisfied by R/T = 0. If, for example, 

 we take the equation x 2 p — x 2 q= (z — x—y) 2 , and consider, the solution z — x — y = 0, 

 Then p = 1, q = 1 ; and we may take 



u = x + y + x(z — x — y)/(z — 2x — y) ; 

 v = x(z—x — y)/(z — 2x — y) : 



