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< = 0 can easily be proved, in virtue of the
equations satisfied by u and v, to lead to
dz dz
P-- + Q-- = R.
dx dx
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
dPq dPQ dPp dPq dPQ dPp dPQ
--- = --- --- and --- = --- --- + ---;
dPx dPp dPx dPy dPp dPy dPy
these give
dPp dPq dPp dPq dPp dPQ
--- --- - --- --- = --- ---;
dPx dPy dPy dPx dPx dPy
by hypothesis dPp/dPx is not identically zero; therefore if the
Jacobian determinant of p and q in regard to x and y is zero
identically, so is dPQ/dPy, or Q does not contain y, so that q is
expressible as a function of p only. Conversely, such an expression
can be seen at once to make the Jacobian of p and q vanish
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