Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/58

and $u={\text{constant}}$ for its boundary, which will be a single surface for a continuous aperiphractic space. Hence throughout the space

$\nabla u=\tau -\omega =0.$

88. If throughout an allelic space contained within finite boundaries but not necessarily continuous

$\nabla \times \tau =\nabla \times \omega$⁠and⁠$\nabla .\tau =\nabla .\omega ,$

and in all the bounding surfaces the normal components of $\tau$ and $\omega$ are equal, then throughout the whole space

$\tau =\omega .$

Setting $\nabla u=\tau -\omega ,$ we have $\nabla .\nabla u=0,$ throughout the space, and the normal component of $\nabla u$ at the boundary equal to zero. Hence throughout the whole space $\nabla u=\tau -\omega =0.$

89. If throughout a certain space (which need not be continuoas, and which may extend to infinity)

$\nabla .\nabla \tau =\nabla .\nabla \omega$

and in all the bounding surfaces

$\tau =\omega ,$

and at infinite distances within the space (if such there are)

$\tau =\omega ,$

then throughout the whole space

$\tau =\omega .$

This will be apparent if we consider separately each of the scalar components of $\tau$ and $\omega .$

Minimum values of the Volume-integral $\iiint u\omega .\omega dv.$

(Thomson's Theorems.)

90. Let it be required to determine for a certain space a vector function of position $\omega$ subject to certain conditions (to be specified hereafter), so that the volume-integral

$\iiint u\omega .\omega dv$

for that space shall have a minimum value, $u$ denoting a given positive scalar function of position.

a. In the first place, let the vector $\omega$ be subject to the conditions that $\nabla .\omega$ is given within the space, and that the normal component of $\omega$ is given for the bounding surface. (This component must of course be such that the surface-integral of $\omega$ shall be equal to the volume-integral $\int \nabla .\omega dv.$ If the space is not continuous, this must be true of each continuous portion of it See No. 57.) The solution is that $\nabla \times (u\omega )=0,$ or more generally, that the line-integral of $u\omega$ for any dosed curve in the space shall vanish.