556 PROFESSOR CHRYSTAL ON LAGRANGE'S RULE FOR THE 



and is therefore such that it makes 



P = 0, Q = 0, R = 0, 



that is to say, it is a singular solution. 



2nd, Let us suppose that one of the determinants (18) vanishes identically, then by (7) 



we must have 



P=0, or Q=0, or R=0. 



That is to say, we have to prove Lagrange's Rule for the special equations 



Q? = R (20); 



Pp=R (21); 



Pl> + Qq = (22). 



Let us consider (20). The corresponding auxiliary system is 



dx/0 = dy/Q,(x,y,z) = dz/R(x,y,z) . 



One integral of this is x = a ; the other is given by 



dy/Q,(a,y,z) = dz/~R(a,y,z) . 



Since neither Q nor R vanish identically, this last equation has no integral of the 

 form y = const, or z = const. Hence its integral may be written v(a,y,z) = b, where v 

 contains both y and z. In the present case, therefore, we may put 



u=x, v=v(x,y,z). 



We can, therefore, express \)j(x,y,z) in each of the tvw forms x*(y> u > v )> Xs{ z > u > v )- 



It follows, therefore, precisely as before, that any part of a solution of (20) is either 

 a derivative of an equation of the form f(u,v) = 0, or else is expressible in each of the 

 two forms y — h(u,v) = 0, z — k(u,v) = 0, and therefore makes Q = 0, R = ; that is to say, 

 is a singular solution of (20). Similar reasoning applies to (21) and to (22). 



The above reasoning is obviously immediately applicable to the case of n independent 

 variables. It therefore affords a rigorous general demonstration of Lagrange's Rule, so 

 far as that rule is correct. It would be in vain to attempt to extend the demonstration 

 to singular solutions, because solutions of this kind can be produced for which Lagrange's 

 Rule is not true. See below, p. 557. 



It may be well to illustrate the foregoing theory by a few examples. 



Ex. 1. Let us consider the solution 



z— x— y=0 

 of the equation 



x 2 p — x 2 q = (z — x — y)' 2 . 

 Here we may take 



u=x + y , v=x(z — x — y)/(z — 2x — y). 

 We have 



yjr(x,y^)=z-x-y. 

 and 



Xi(x,u,v)^exvI(v - x) ; 



X2(y> u > v )^( u - y) v K v - u +y); 



Xz(z,u,v)^z-u. 



