From Wikisource
Jump to: navigation, search
This page has been proofread, but needs to be validated.

in positive directions. To the right hand sides of the equations (34) we may apply transformation (35) with these values of \pi_{ba}, d\Omega-being infinitely small of the fourth order and it being allowed to confine ourselves to quantities of this order.

On the left hand sides of (34), however, we must take into consideration, the surface being of the third order, that the values of \pi_{ba} change from point to point. Let \mathrm{x}_{1},\dots\mathrm{x}_{4} be the changes which x_{1},\dots x_{4} undergo when we pass from P to any other point of the surface. Then we must write for the value of the coefficient at this last point


We thus have

\int u'_{a}d\sigma=\sum(b)\pi_{ba}\int u_{b}d\sigma+\sum(b)\int u_{b}\sum(c)\frac{\partial\pi{}_{ba}}{\partial x_{c}}\mathrm{x}_{c}d\sigma

It will be shown presently that the last term vanishes. This being proved, it is clear that the relations (34) follow from (33); indeed, multiplying equations (33) by\pi_{1a},\dots\pi_{4a} respectively and adding them we find

\int u'_{a}d\sigma=v'_{a}d\Omega

§ 30. The proof for

\sum(b)\int u_{b}\sum(c)\frac{\partial\pi{}_{ba}}{\partial x_{c}}\mathrm{x}_{c}d\sigma=0 (36)

rests on the relations

\frac{\partial\pi{}_{ba}}{\partial x_{c}}=\frac{\partial\pi{}_{ea}}{\partial x_{b}} (37)

which follow from

\pi{}_{ba}=\frac{\partial x'_{b}}{\partial x_{b}},\ \pi{}_{ea}=\frac{\partial x'_{a}}{\partial x_{e}},

The integral which occurs in (36) differs from

\int u_{b}d\sigma (38)

by the infinitely small factor under the sign of integration

\sum(c)\frac{\partial\pi{}_{ba}}{\partial x_{c}}\mathrm{x}_{c}

Now we have calculated in § 26 integrals like (38) by taking together each time two opposite sides, one of which \Sigma_{1} passes through P while the second \Sigma_{2} is obtained from the first by a shift in the