Prove that $n \cdot \frac{\partial u}{\partial n} = 0$ for divergence-free functions that vanishes on boundary.
I just read that if $u$ is a divergence-free (vectorial) function in $\Omega$ with $u = 0$ on $\partial \Omega$, then $$n \cdot \frac{\partial u}{\partial n} = 0$$ and $$\epsilon(u)n \cdot \frac{\partial u}{\partial n} = |\epsilon(u)|^2$$ on $\partial \Omega$, where $\epsilon(u) = \frac{1}{2}(\nabla u + (\nabla u)^T)$.
How can I prove this? I really don't see it. For the first, I think that it's related with the fact that $\frac{\partial u}{\partial \tau} = 0$ on $\partial \Omega$. Also, I've tried integration by parts, but it doesn't work.
Thanks in advance!