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

42 and $$u = \text{constant}$$ for its boundary, which will be a single surface for a continuous aperiphractic space. Hence throughout the space 88. If throughout an allelic space contained within finite boundaries but not necessarily continuous and in all the bounding surfaces the normal components of $$\tau$$ and $$\omega$$ are equal, then throughout the whole space  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) and in all the bounding surfaces  and at infinite distances within the space (if such there are)  then throughout the whole space  This will be apparent if we consider separately each of the scalar components of $$\tau$$ and $$\omega.$$

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 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.