Schwere, Elektricität und Magnetismus:397

VECTOR ANALYSIS.

<section begin=t1 /> ances within the space $u=a$,—then throughout the space

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Vorlage:IdtFor, if anywhere in the interior of the space $\mathrm{\nabla }u$ 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 $\mathrm{\nabla }u$ for this surface has the value zero in virtue of the general condition $\mathrm{\nabla }\cdot \mathrm{\nabla }u=0$. 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.

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

Vorlage:Idt81. If throughout a certain space (which need not be continuous, and which may extend to infinity,)

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and in all the bounding surfaces the normal component of $\mathrm{\nabla }u$ vanishes, and at infinite distances within the space (if such there are) $r^2\frac{du}{dr}=0$, where $r$ denotes the distance from a fixed origin, then throughout the space

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and in each continuous portion of the same

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Vorlage:IdtFor, if anywhere in the space in question $\mathrm{\nabla }u$ has a value different from zero, let it have such a value at a point $P$, and let $u$ be there equal to $b$. Imagine a spherical surface about the above-mentioned origin as center, enclosing the point $P$, and with a radius $r$. Consider that portion of the space to which the theorem relates which is within the sphere and in which $u<b$. The surface-integral of $\mathrm{\nabla }u$ for this space is equal to zero in virtue of the general condition $\mathrm{\nabla }\cdot \mathrm{\nabla }u=0$. That part of the integral (if any) which relates to a portion of the spherical surface has a value numerically not greater than $4\pi r^2\left(\frac{du}{dr}\right)$, where $\left(\frac{du}{dr}\right)$ denotes the greatest numerical value of $\frac{du}{dr}$ in the portion of the spherical surface considered. <section end=t1 />