DEPARTMENT OF PURE MATHEMATICS. 487 



hypothesis, that xl' =0 provides an integral of the equation (1), mei'ely requires 

 that the relation shall bo satisfied concurrently witli y = 0. Hence the Jacobian 



J, 





vanishes ; but it only needs to vanish concurrently with x^^ = 0. If J vanishes 

 identically (and this is the usual latent assumption in the fallacious proof), then 

 undoubtedly some functional relation, 



r(«, r,r/.)=0, 

 or, say, 



\|^=/(m, v), 



exists among the quantities n, v, ^ ; and the proposition then would hold. "When 

 the Jacobian vanishes, not identically but only concurrently with V' = ^> there is 

 no such functional relation; the proposition then does not hold. 



Moreover, it will be noticed that the proposition does not make a declaration 

 of a wider result that the equations 



^/. = 0,/(«,^.) = 0, 



are satisfied simultaneously ; it aims at the definite expression of -^t in terms of it 

 and i\ 



The range of the limitations upon the validity of the result can be indicated 

 by a different line of argument. As ?< = constant and r = constant .are two inde- 

 pendent integrals of the equations (2), not more than one (at the utmost) of the 

 Jacobians 





can vanish identicallj'. Suppose that the first of them does not vanish identically — 

 a hypothesis that imposes no condition. Ne.xt, let ,r = rt, y = ('>, s = c be a set of 

 simultaneous values satisfying the equation 



supposed to give an integral of the original partial equation. Usually there will 

 be a considerable choice in the selection of this set of values. If possible, let 

 a set of values a, b, c be chosen, such that u and r are regular functions of .r, y, s 



in the vicinity of a, b, c, and such also that J ( -^ ), which does not vanish iden- 



tically, does not vanisli for those values. This hypothesis does impose conditions. 

 When the hypothesis is justified, the two equations 



u = n{x,y,z), v=v {.i; y, z), 



can be resolved so as to express .v and y in terms of n, v, z in forms 



x-a=g iz-c,7i, v),y-b = h(z~c,v, v), 



where g and h are regular functions of their arguments. 



Now let these values of x and y be substituted in y^ {x, y, z), with a result 



where the s, that survives explicitly in (^ and is contained implicitly in ii and t> 

 is the z of the integral furnished by ■^ = 0, and therefore by (j) = 0. For this 

 integral we have 



dd> , d(f) /du du\ , d(b /dv , dv\ „ 

 'd(i) (du f)u\d(f>/dv , dv\ „ 





