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.
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.
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 $\nabla\cdot u=0$.