Does null Dirichlet condition implies null Neumann condition?

Question

Let $\Omega \subset \mathbb{R}^N$ be an open and bounded set, and suppose that $u \in H_0^1(\Omega)$. This means that $ u\rvert_{\partial \Omega} = 0$ (Dirichlet condition).

But, in that case, do we also have that $u$ verifies $\nabla u \cdot \vec{n}\rvert_{\partial \Omega} = 0$? (Neumann condition)

Answer 1

Short answer: No.

Counterexample in 1D: Take $\Omega=(0,1)$ and $u(x)=x(1-x)$. This function is in $H^1_0(\Omega)$ but $u’(0)=1$.

As indicated in the comments, in general, $\nabla u\cdot n|_{\partial\Omega}$ may not even be defined. For example, $u(x)=x(1-x)\sin(\log x)$ is in $H^1_0(\Omega)$ but $u’(x)$ doesn’t converge as $x\to0$.