Schwere, Elektricität und Magnetismus:400

Bernhard Riemann: Schwere, Elektricität und Magnetismus
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VECTOR ANALYSIS.

     Setting $\nabla u=\tau -\omega \,$, we have $\nabla \cdot \nabla u=0\,$ 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 continuous, and which may extend to infinity)

$\nabla \cdot \nabla \tau =\nabla \cdot \nabla \omega \,$

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 $\iiint u\,\omega \cdot \omega dv\,$. (Thomson’s Theorems.)

     90. Let it he 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

$\iiint u\,\omega \cdot \omega dv\,$

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 $\nabla \cdot \omega \,$ 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 $\int \nabla \cdot \omega dv\,$. 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 closed curve in the space shall vanish.

     The existence of the minimum requires that

$\iiint u\,\omega \cdot \delta \omega dv=0\,,$

while $\delta \omega \,$ is subject to the limitation that

$\nabla \cdot \delta \omega =0,\,$

and that the normal component of $\delta \omega \,$ at the bounding surface vanishes. To prove that the line-integral of $u\omega \,$ vanishes for