Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/146

(12) ${\color {White}xxx}a=K{\frac {dV}{dx}}+u,{\color {White}xxx}b=K{\frac {dV}{dy}}+v,{\color {White}xxx}c=K{\frac {dV}{dz}}+w$.

with the condition (1)

${\frac {du}{dx}}+{\frac {dv}{dy}}+{\frac {dw}{dz}}=0$,

then $V,u,v,w$ can be found without ambiguity from these four equations.

Corollary II. The general characteristic equation

${\frac {d}{dx}}K{\frac {dV}{dx}}+{\frac {d}{dy}}K{\frac {dV}{dy}}+{\frac {d}{dz}}K{\frac {dV}{dz}}+4\pi \rho =0$

where $V$ a finite quantity of single value whose first derivatives are finite and continuous except at the surface $S$, and at that surface fulfil the superficial characteristic

$l(K_{1}{\frac {dV_{1}}{dx}}-K_{2}{\frac {dV_{2}}{dx}})+m(K_{1}{\frac {dV_{1}}{dy}}-K_{2}{\frac {dV_{2}}{dy}})+n(K_{1}{\frac {dV_{1}}{dz}}-K_{2}{\frac {dV_{2}}{dz}})+4\pi \rho =0$

can be satisfied by one value of $V$, and by one only, in the following cases.

Case 1. When the equations apply to the space within any closed surface at every point of which $V=C$.

For we have proved that in this case $a,b,c$ have real and unique values which determine the first derivatives of $V$, and hence, if different values of $V$ exist, they can only differ by a constant. But at the surface $V$ is given equal to $C$, and therefore $V$ is determinate throughout the space.

As a particular case, let us suppose a space within which $\rho =0$ bounded by a closed surface at which $V=C$. The characteristic equations are satisfied by making $V=C$ for every point within the space, and therefore $V=C$ is the only solution of the equations.

Case 2. When the equations apply to the space within any closed surface at every point of which $V$ is given.

For if in this case the characteristic equations could be satisfied by two different values of $V$, say $V$ and $V'$, put $U=V-V'$, then subtracting the characteristic equation in $V'$ from that in $V$, we find a characteristic equation in $U$. At the closed surface $U=0$ because at the surface $V=V'$, and within the surface the density is zero because $\rho =\rho '$. Hence, by Case 1, $U=0$ throughout the enclosed space, and therefore $V=V'$ throughout this space.