Page:On Faraday's Lines of Force.pdf/44

by that past of the whole effect at any point which is due to the distribution of conducting media, and not directly to the presence of the sources.

This quantity $Q$ is rendered a minimum by one and only one value of $p$, namely, that which satisfies the original equation.

THEOREM V.

If $a,b,c$ be three functions of $x,y,z$ satisfying the equation

${\frac {da}{dx}}+{\frac {db}{dy}}+{\frac {dc}{dz}}=0$,

it is always possible to find three functions \alpha, \beta, \gamma which shall satisfy the equations

${\frac {d\beta }{dz}}-{\frac {d\gamma }{dy}}=0$

${\frac {d\gamma }{dx}}-{\frac {d\alpha }{dz}}=0$

${\frac {d\alpha }{dy}}-{\frac {d\gamma }{dx}}=0$

Let $A=\int cdy$, where the integration is to be performed upon $c$ considered as a function of $y,$ treating $x$ and $z$ as constants. Let $B=\int adz$, $C=\int bdx,$ $A'=\int bdz$, $B'=\int cdx$, $C'=\int ady$, integrated in the same way.

Then

$\alpha =A=A'+{\frac {d\psi }{dx}}$,

$\beta =B=B'+{\frac {d\psi }{dy}}$,

$\alpha =C=C'+{\frac {d\psi }{dz}}$

Will satisfy the given equations ; for

${\frac {d\beta }{dx}}-{\frac {d\gamma }{dy}}=\int {\frac {da}{dy}}dz-\int {\frac {dc}{dz}}dx-\int {\frac {db}{dy}}dx+\int {\frac {da}{dy}}$,

and

$0=\int {\frac {da}{dx}}dx+\int {\frac {db}{dy}}dx+\int {\frac {dc}{dz}}dx$;

$\therefore {\frac {d\beta }{dz}}-{\frac {d\gamma }{dy}}=\int {\frac {da}{dx}}dx+\int {\frac {da}{dy}}dy+\int {\frac {da}{dz}}dz$

$=a.$