SOLUTION OF A LINEAR PARTIAL DIFFERENTIAL EQUATION. 559 



and we have for the equation (B) of Lagrange's theory 



(z — u) 2 ( — du + dv) = ; 

 and for the equation (C) 



— du+dv . 



Now (B) is satisfied by the solution z — x — y = 0, that is, z — u = 0, on account of the 

 occurrence of the factor (2 - u) 2 ; but Lagrange's equation (C) is not satisfied by the 

 solution at all. 



There is a still more fatal objection. The statement which I have italicised in 

 Lagrange's demonstration is quite erroneous, as may be easily verified in the simplest 

 cases. Consider, for example, the same linear equation as before, along with the obvious 

 solution z— 2x — y = 0. In this case p = 2, q = 1 ; and we may take 



u=x+y; 



v = (z - 2x - y)/x(z - x - y) - l/(x + y) . 



The factor R/T in this case reduces to — x\z — x — yf, and does not furnish the 

 solution in question. The equation (C) reduces to 



[2(v + l/u){l/(z-u) + v + l/u}-2vju— v*]du-dv, 



and instead of Lagrange's conclusion (D) we get 



gux-puy 1 , D , 



V v y -q. v x ' %(v + l/u){ll(z-u) + v + l/u}-2v/u-v 2 K h 



from which z does not disappear. It is perfectly true that the solution in question, viz., 

 z — 2x — y = 0, or, as it may be written, 



(v+l/w)(z-uy 2 /{l + (v + lju)(z-u)} =0 , 



the effective derivative of which for the present purpose is v+l/u = 0, is a Lagrangian 

 solution, and does satisfy the equation (C), as may be readily verified ; but this does not 

 happen in the way required by Lagrange's demonstration. The fallacy seems to have 

 arisen from forgetfulness of the fact that (C) does not require to be satisfied identically, 

 but only as a derivative of the relation which constitutes the solution. Lagrange's 

 proof is, therefore, utterly defective. 



Lagrange gives two other demonstrations of his rule, one in his Legons sur le Calcul 

 des Fonctions (Lecon 20 e ), the other in his Theorie des Fonctions (chap. xvi.). The 

 latter is merely a verification that every linear equation admits of a Lagrangian solution 

 f(u,v) — 0, and need not be considered here. The former is another example of the 

 fatality which has pursued the attempts at a complete demonstration. It runs as 

 follows : — 



Let yp-(x,y,z) = be a solution of Vp + Qq = R ; then 



P^ x +Q^„ + Ity* = (A). 



VOL. XXXVI. PART II. (NO. 20). 4 P 



