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I want to see why

$$((u \cdot \nabla)w, w) = 0$$

holds, given $\nabla \cdot u = 0$.

The brackets are the $L^2$ norm and it supposed to follow from parts but I can't do it.

Thank you.

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up vote 3 down vote accepted

You're missing quite a number of assumptions there. If $w$ is a sufficiently smooth real-valued function on $\mathbb R^n$, $u$ is a sufficiently smooth real vector field on $\mathbb R^n$, and $u$ and/or $w$ decays sufficiently rapidly towards infinity, this follows by integration by parts (not: "from parts"):

$$ \begin{align} I &=((u \cdot \nabla)w, w)\\ &= \sum_i(u_i\partial_i w, w)\\ &=\sum_i\int u_i(\partial_i w)w\mathrm d^nx\\ &=\sum_i\left(\int \partial_i(u_iww)\mathrm d^nx-\int (\partial_i u_i)ww\mathrm d^nx-\int u_iw(\partial_iw)\mathrm d^nx\right)\\ &=\sum_i\lim_{t\to\infty}\left(\;\;\int\limits_{x_i=t} u_iww\mathrm dS_i-\int\limits_{x_i=-t} u_iww\mathrm dS_i\right)-\int(\nabla\cdot u)w^2\mathrm d^nx-I\\ &=-I\;, \end{align} $$

where the integrals over $\mathrm dS_i$ are over two bounding hyperplanes normal to the $i$-th axis and vanish in the limit if $u$ and $w$ decay sufficiently rapidly, and the second term vanishes since $\nabla\cdot u=0$.

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I understood everything except the two bounding hyperplanes. Thanks a lot for the really elegant proof! – user16697 Sep 26 '11 at 7:45
@QED: I've made the hyperplanes and the limiting process more explicit -- I hope it's clearer now? – joriki Sep 27 '11 at 10:42

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