Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/56

where $r$ denotes the distance from a fixed origin, then throughout the space

$\nabla u=0,$

and in each continuous portion of the same

$u={\text{constant.}}$

For, if anywhere in the space in question $\nabla u$ has a value different from zero, let it have such a value at a point $P,$ and let $u$ be there equal to $b.$ Imagine a spherical surface about the above-mentioned origin as center, enclosing the point $P,$ and with a radius $r.$ Consider that portion of the space to which the theorem relates which is within the sphere and in which $u<b.$ The surface integral of $\nabla u$ for this space is equal to zero in virtue of the general condition $\nabla .\nabla u=0.$ That part of the integral (if any) which relates to a portion of the spherical surface has a value numerically not greater than $4\pi r^{2}\left({\frac {du}{dr}}\right)',$ where $\left({\frac {du}{dr}}\right)'$ denotes the greatest numerical value of ${\frac {du}{dr}}$ in the portion of the spherical surface considered. Hence, the value of this part of the surface-integral may be made less (numerically) than any assignable quantity by giving to $r$ a sufficiently great value. Hence, the other part of the surface-integral (viz., that relating to the surface in which $u=b,$ and to the boundary of the space to which the theorem relates) may be given a value differing from zero by less than any assignable quantity. But no part of the integral relating to this surface can be negative. Therefore no part can be positive, and the supposition relative to the point $P$ is untenable.

This proposition also may be generalized by substituting $\nabla .[t\nabla u]=0$ for $\nabla .\nabla u=0,$ and $tr^{2}{\frac {du}{dr}}$ for $r^{2}{\frac {du}{dr}}=0.$

82. If throughout any continuous space (or in all space)

$\nabla t=\nabla u,$

then throughout the same space

$t=u+{\text{const.}}$

The truth of this and the three following theorems will be apparent if we consider the difference $t-u.$

83. If throughout any continuous space (or in all space)

$\nabla .\nabla t=\nabla .\nabla u,$

and in any finite part of that space, or in any finite surface in or bounding it,

$\nabla t=\nabla u,$

then throughout the whole space

$\nabla t=\nabla u,$⁠and⁠$t=u+{\text{const.}}$