Does $\bigtriangleup u =0$ imply $u=0$?

Question

If $u \in H^2(\Omega) \cap H_0^1(\Omega)$ such that $\bigtriangleup u =0 $.

Is it necessarily that $u=0$?

Thank you!

Answer 1

$$ \int_{\Omega}u\Delta u \;dv = \int_{\Omega}\nabla\cdot(u\nabla u)-|\nabla u|^2 dv \\ = \int_{\partial\Omega}u\frac{du}{dn}ds-\int_{\Omega}|\nabla u|^2dv \\ = -\int_{\Omega}|\nabla u|^2dv $$ Therefore, if $\Delta u=0$, with $u\in H^2\cap H^1_0$, then $\int_{\Omega}|\nabla u|^2dx=0$, which forces $u$ to be constant, and that constant must be $0$ in order for $u\in H_0^1(\Omega)$ to hold.