Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/55

Rh which $$\nabla. \nabla u = 0, u$$ will be constant in this surface, and the surface will be contiguous to a region in which $$\nabla. \nabla u = 0$$ and $$u$$ has a greater value than in the surface, or else a less value than in the surface. Let us imagine a sphere lying principally on the other side of the surface, but projecting slightly into this region, and let us particularly consider the surface-integral of $$\nabla u$$ for the small segment cut off by the surface $$\nabla u = 0.$$ The integral for that part of the surface of the segment which consists of part of the surface $$\nabla u = 0$$ will have the value zero, the integral for the spherical part will have a value either greater than zero or else less than zero. Therefore the integral for the whole surface of the segment cannot have the value zero, which is demanded by the general condition, $$\nabla. \nabla u = 0.$$

80. If throughout a certain space (which need not be continuous, and which may extend to infinity) and in all the bounding surfaces  and (in case the space extends to infinity) if at infinite distances within the space $$u = a,$$—then throughout the space  For, if anywhere in the interior of the space $$\nabla u$$ has a value different from zero, we may find a point $$P$$ where such is the case, and where $$u$$ has a value $$b$$ different from $$a,$$—to fix our ideas we will say less. Imagine a surface enclosing all of the space in which $$u < b.$$ (This must be possible, since that part of the space does not reach to infinity.) The surface-integral of $$\nabla u$$ for this surface has the value zero in virtue of the general condition $$\nabla. \nabla u = 0.$$ But, from the manner in which the surface is defined, no part of the integral can be negative. Therefore no part of the integral can be positive, and the supposition made with respect to the point $$P$$ is untenable. That the supposition that $$b > a$$ is untenable may be shown in a similar manner. Therefore the value of $$u$$ is constant.

This proposition may be generalized by substituting the condition $$\nabla. [t\nabla u] = 0$$ for $$\nabla. \nabla u = 0, t$$ denoting any positive (or any negative) scalar function of position in space. The conclusion would be the same, and the demonstration similar.

81. If throughout a certain space (which need not be continuous, and which may extend to infinity) and in all the bounding surfaces the normal component of $$\nabla u$$ vanishes, and at infinite distances within the space (if such there are) $$r^2 \frac{du}{dr} = 0,$$