Schwere, Elektricität und Magnetismus:395

Bernhard Riemann: Schwere, Elektricität und Magnetismus
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VECTOR ANALYSIS.

from the nature of the formula itself. The most important discontinuities of scalars are those which occur at surfaces: in the case of vectors, discontinuities at surfaces, at lines, and at points, should be considered.

     74. From equation (3) we obtain

$\int \nabla (tu)\cdot d\rho =t''u''-t'u'=\int u\nabla t\cdot d\rho +\int t\nabla u\cdot d\rho ,\,$

where the accents distinguish the quantities relating to the limits of the line-integrals. We are thus able to reduce a line-integral of the form $\int \nabla t\cdot d\rho \,$ to the form $-\int t\nabla u\cdot d\rho \,$ with quantities free from the sign of integration.

     75. From equation (5) we obtain

$\iint \nabla \times (u\omega )\cdot d\sigma =\int u\omega \cdot d\rho =\iint u\nabla \times \omega \cdot d\sigma -\iint \omega \times \nabla u\cdot d\sigma ,\,$

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

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

$\iint \nabla u\times \nabla t\cdot d\sigma =\int u\nabla t\cdot d\rho =-\int t\nabla u\cdot d\rho .\,$

     76. From equation (4) we obtain

$\iiint \nabla \cdot \lbrack u\omega \rbrack dv=\iint u\omega \cdot d\sigma =\iiint \omega \cdot \nabla udu+\iiint u\nabla \cdot \omega du,\,$

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

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

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:—

$\iiint s\nabla t\cdot \nabla udv=\iint us\nabla t\cdot d\sigma \iiint u\nabla \cdot \lbrack s\nabla t\rbrack dv\,$ $=\iint ts\nabla u\cdot d\sigma -\iiint t\nabla \cdot \lbrack s\nabla u\rbrack dv.\,$

     77. From equation (6) we obtain

$\iiint \nabla \cdot \lbrack \tau \times \omega \rbrack dv=\iint \tau \times \omega \cdot d\sigma =\iiint \omega \cdot \nabla \times \tau dv-\iiint \tau \cdot \nabla \times \omega dv.\,$

A particular ease is

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

Integration of Differential Equations.

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

$\nabla u=0,\,$