dPu dPu dPu dPv dPv dPv P--- + Q--- = R--- = 0 and P--- + Q--- = R--- = 0. dPx dPy dPz dPx dPy dPz
Then if F(x, y, z) = 0 be a relation satisfying the original differential equations, this relation giving rise to
dPF dPF dPz dPF dPF dPz dPF dPF dPF --- + --- --- = 0 and --- + --- --- = 0, we have P--- + Q--- = R--- = 0. dPx dPz dPx dPy dPz dPy dPx dPy dPz
It follows that the determinant of three rows and columns vanishes whose first row consists of the three quantities dPF/dPx, dPF/dPy, dPF/dPz, whose second row consists of the three quantities dPu/dPx, dPu/dPy, dPu/dPz, whose third row consists similarly of the partial derivatives of v. The vanishing of this so-called Jacobian determinant is known to imply that F is expressible as a function of u and v, unless these are themselves functionally related, which is contrary to hypothesis (see the note below on Jacobian determinants). Conversely, any relation [phi](u, v) = 0 can easily be proved, in virtue of the equations satisfied by u and v, to lead to
The solution of this partial equation is thus reduced to the solution of the two ordinary differential equations expressed by dx/P = dy/Q = dz/R. In regard to this problem one remark may be made which is often of use in practice: when one equation u = a has been found to satisfy the differential equations, we may utilize this to obtain the second equation v = b; for instance, we may, by means of u = a, eliminate z--when then from the resulting equations in x and y a relation v = b has been found containing x and y and a, the substitution a = u will give a relation involving x, y, z.
_Note on Jacobian Determinants._--The fact assumed above that the vanishing of the Jacobian determinant whose elements are the partial derivatives of three functions F, u, v, of three variables x, y, z, involves that there exists a functional relation connecting the three functions F, u, v, may be proved somewhat roughly as follows:--
The corresponding theorem is true for any number of variables. Consider first the case of two functions p, q, of two variables x, y. The function p, not being constant, must contain one of the variables, say x; we can then suppose x expressed in terms of y and the function p; thus the function q can be expressed in terms of y and the function p, say q = Q(p, y). This is clear enough in the simplest cases which arise, when the functions are rational. Hence we have