Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/54

76. From equation (4) we obtain

$\iiint \nabla .[u\omega ]dv=\iint u\omega -d\sigma =\iiint \omega .\nabla udv+\iiint u\nabla .\omega dv,$

where, as elsewhere in these equations, the surface-integral relates to the boundary of the volume-integrals.

From this, by substitution of $\nabla t$ for $\omega ,$ we derive as a particular case

$\iiint \nabla t.\nabla udv=\iint u\nabla t.d\sigma -\iiint u\nabla .\nabla tdv=\iint t\nabla u.d\sigma -\iiint t\nabla .\nabla udv,$

which is Green's Theorem. The substitution of $s\nabla t$ for $\omega$ gives the more general form of this theorem which is due to Thomson, viz.,

${\begin{aligned}\iiint s\nabla t.\nabla udv&=\iint us\,t.d\sigma -\iiint u\nabla .[s\nabla t]dv\\&=\iint ts\nabla u.d\sigma -\iiint t\nabla .[s\nabla u]dv.\end{aligned}}$

77. From equation (6) we obtain

$\iiint \nabla .[\tau \times \omega ]dv=\iint \tau \times \omega .d\sigma =\iiint \omega .\nabla \times \tau dv-\iiint \tau .\nabla \times \omega dv.$

A particular case is

$\iiint \nabla u.\nabla \times \omega dv=\iint \omega \times \nabla u.d\sigma .$

Integration of Differential Equations.

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

$\nabla u=0,$

then throughout the same space

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

79. If throughout any continuous space (or in all space) and in any finite part of that space, or in any finite surface in or bounding it,

$\nabla u=0,$

then throughout the whole space

$\nabla u=0,$and$u={\text{constant.}}$

This will appear from the following considerations:

If $\nabla u=0$ in any finite part of the space, $u$ is constant in that part.

If $u$ is not constant throughout, let us imagine a sphere situated principally in the part in which $u$ is constant, but projecting slightly into a part in which $u$ has a greater value, or else into a part in which $u$ has a less. The surface-integral of $\nabla u$ for the part of the spherical surface in the region where $u$ is constant will have the value zero: for the other part of the surface, the integral will be either greater than zero, or less than zero. Therefore the whole surface-integral for the spherical surface will not have the value zero, which is required by the general condition, $\nabla .\nabla u=0.$

Again, if $\nabla u=0$ only in a surface in or bounding the space in