(18) 



SOLUTION OF A LINEAR PARTIAL DIFFERENTIAL EQUATION. 555 



The two other parts of the proposition are established in exactly the same way.* 



We can now prove that 



Prop. II. Every non-singular solution o/(l) can he expressed in the form 



f(u,v) = (17). 



Since the functions u and v are independent, not more than one of the determinants 

 in the matrix 



U X Uy U z 

 V x Vy V z 



can vanish identically. 



1st, Let us suppose that none of the determinants in (18) vanish. Then, by means 

 of the equations (13), we can express any function \jj(x,y,z) in each of the three forms 



xi(x,u,v) , xt(y> u > v ) > x&w) ■ 



Hence the solution (2) of the differential equation (1) will be a derivative of (but not 

 necessarily equivalent to t) each of the three equations 



X i(x,u,v) = (190 , x&W) = (19 2 ) , joteM) = (19 3 ) . 



Now (19j) leads to relations of the forms x — g{u,v) = 0, or f(u,v) = ; (19 2 ) to relations 

 of the forms y — h(u,v) = 0, or f(u,v) = ; (19 3 ) to relations of the forms z — k(u,v) = 0, or 

 f(u,v) = 0. 



It follows, therefore, that any relation which is part of a solution of (1) is either a 



derivative of an equation of the form f(u,v) = 0,| or else it can be expressed in each of 



the three forms 



x — g(u,v) = 0, y — h(u,v) = 0, z — k(u,v) = 0, 



* The above is sufficient for our present purpose ; but it is interesting to notice that the condition (16) is not 

 sufficient if lg/^z=0. In that case we must add the further condition 



L | % wJ /%L = q . (16'). 



&=S<«i®) I Vy v 2 1 / lz 



It is obvious that, in general, Ig/lz will vanish under the conditions supposed, and that the condition (16') will not in 

 general be satisfied ; for, it is clear that if we set down an equation such as (1) at random, and connect (x,y,z) by the 

 relation P=0, the equation will not in general be satisfied. This leads to some curious analytical theorems which I 

 have partially verified in connection with the above theory. 



In like manner, in the case of a solution of (1) of the form y — h(u,v) = which makes lh/^z=0, in addition to the 

 condition Q = 0, we must have 



L 



y = h(u,v) 



l*=0. 



u z u x 



V, Vx 



Also in the case of a solution of the form a — k(u,v) = which makes 1 — lk/h = 0, in addition to K = 0, we must have 



u x 



L 



z=lc(u,v) 



v x v,\l\ V 



t Forgetfulness of this point seems to have been the real root of the many defective demonstrations that have been 

 given of Lagrange's Rule, beginning with his own. 



+ Any particular non-singular solution may come from all the three forms (19) in the form f(u,v) = 0; or it may 

 come from (19j) in the form x-g(u,v) = 0; and from the other two in the form f(u,v)=0; and so on. It should also be 

 noticed that the given solution may be only a derivative of, and not wholly equivalent to, f(u,v) = 0. I insist on these 

 elementary matters because they are so very much overlooked, especially by English mathematicians, the result having 

 been a plentiful crop of fallacies. 



