I'm working on an exercise that requires me to prove the integration by parts formula
$$\int_{\Omega}u_{x_j}v\ dx=-\int_{\Omega}uv_{x_j}\ dx+\int_{\partial\Omega}\nu_j uv\ dS$$
by use of the divergence theorem
$$\int_{\Omega}\nabla F\ dx=\int_{\partial\Omega}\nu\cdot F dS$$
where $\nu$ denotes the outward pointing normal to $\partial\Omega$.
I don't exactly know how I would go about doing this, so any help/hints are highly appreciated!
Thanks in advance.
On
Let $F=(0,0,\ldots,0,uv,0,\ldots,0)$, where uv is the jth coordinate.
Then
$\nabla \cdot F=(uv)_{x_j}=u_{x_j}v + uv_{x_j} \ and \ F \cdot \nu =uv\nu^{j}$.
Now,
$\int_{\Omega}(u_{x_j}v+uv_{x_j})dx=\int_{\Omega} \nabla \cdot F dx=\int_{\partial \Omega} F \cdot \nu dS = \int_{\partial \Omega} uv\nu^{j} dS$
Hint: apply the divergence theorem to $$F=\begin{pmatrix}0\\\vdots\\0\\uv\\0\\\vdots\\0\end{pmatrix}$$ where the $uv$ is in the $j$-th entry.