Px [mu] dPy / dPy
and hence
dP[psi] dP[phi] dP[psi] dP[phi]
------- ------- - ------- ------- = 0;
dPx dPy dPy dPx
this shows that, as functions of x and y, [psi] is a function of [phi]
(see the note at the end of part i. of this article, on Jacobian
determinants), so that we may write [psi] = F(z, [phi]), from which
[sigma] dPF dP[psi] dPF dPF dP[phi] dPF [sigma] dPF
------- = -------; then ------- = --- + ------- ------- = --- + ------- . [mu]Z = --- + [sigma]Z
[mu] dP[phi] dPz dPz dP[phi] dPz dPz [mu] dPz
dP[psi] dPF
or ------- - [sigma]Z = ---;
dPz dPz
in virtue of [psi](x, y, z) = f(z), and [psi] = F(z, [phi]), the
function [phi] can be written in terms of z only, thus dPF/dPz can be
written in terms of z only, and what we required to prove is proved.
Consider lastly a simple type of differential equation containing
_two_ independent variables, say x and y, and one dependent variable
z, namely the equation
dPz dPz
P--- + Q--- = R,
dPx dPy
where P, Q, R are functions of x, y, z. This is known as Lagrange's
linear partial differential equation of the first order. To integrate
this, consider first the ordinary differential equations dx/dz = P/R,
dy/dz = Q/R, and suppose that two functions u, v, of x, y, z can be
determined, independent of one another, such that the equations u = a,
v = b, where a, b are arbitrary constants, lead to these ordinary
differential equations, namely such that
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
derivat
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