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

198 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 $$A =\int c dy$$, where the integration is to be performed upon $$c$$ considered as a function of $$y,$$ treating $$x$$ and $$z$$ as constants. Let $$B=\int a dz$$, $$C=\int b dx,$$ $$A'=\int b dz$$, $$B'=\int c dx$$, $$C'=\int a dy$$, 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 $

$=a.$