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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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