# Function with divergence, curl and normal trace on boundary equals zero is zero

Let $u\in H^1(\Omega)$ with $\nabla\times u=0$ in $\Omega\subset\mathbb{R}^3$ (open bounded domain), $u\times n=0$ on $\partial\Omega$ (where $n$ is a a normal vector to $\partial\Omega$), $\operatorname{div}(u)=0$ in $\Omega$ and $u\cdot n=0$ on $\partial\Omega$.

Prove that $u=0$.

Thanks.

For $u\in H^1(\Omega)$, you can write : $$||u||_{H^1(\Omega)}\leq C \{ ||u||_{L^2(\Omega)}+ ||div\ u||_{L^2(\Omega)} + ||curl\ u||_{L^2(\Omega)} + ||u\cdot n||_{H^{\frac{1}{2}}(\partial\Omega)}\}$$ or, $$||u||_{H^1(\Omega)}\leq C \{ ||u||_{L^2(\Omega)}+ ||div\ u||_{L^2(\Omega)} + ||curl\ u||_{L^2(\Omega)} + ||u\times n||_{H^{\frac{1}{2}}(\partial\Omega)}\}.$$ Now if $$u =div\ u=curl\ u = 0 \quad\mbox{ in }\Omega$$ and $$u\cdot n =0 \quad \mbox{ or, }\quad u\times n =0 \quad\mbox{ on }\partial\Omega$$ then following the above estimates you can conclude $$u = 0 \quad\mbox{ in }\Omega.$$
I could not comment in the above answer by tuhin, but strangely it assumes $u=0$ to prove $u=0$!!!!
This result is actually far from trivial. It is not necessarily true if you only assume $\Omega$ to be just open and bounded. However, if you assume $\partial\Omega$ to be $C^{2},$ the result is true.
The estimate, simplified to the case of the question is $$\lVert u \rVert_{H^1(\Omega)} \leq C \left( \lVert \operatorname*{curl}u \rVert_{L^2(\Omega)} + \lVert \operatorname*{div}u \rVert_{L^2(\Omega)} + \lVert n \times u \rVert_{H^{1/2}(\partial\Omega)} + \lVert n \cdot u \rVert_{L^1(\partial\Omega)} \right).$$