Let $ \Omega $ be an open subset with compact closure of a Riemannian manifold $ M $. Let $ u \in H^1_{0}(\Omega) $ be a weak solution of the Dirichlet boundary problem:
$$ -\Delta u + qu = f \; \; \textrm{in} \; \Omega \; \; \; (1) $$
$$ u|_{\partial\Omega}=0 $$
In other words $ u $ satisfies the following equality for every $ \psi \in C^{\infty}_{0}(\Omega) $
$$ \int_{\Omega} \langle \nabla u,\nabla \psi \rangle + q u \psi = \int_{\Omega} f\psi \; \; \; (2)$$
Now from standard regularity results if $ q,f \in C^{\infty}(M) $ then $ u \in C^{\infty}(\Omega) $ but in general $ u \notin C^{\infty}(\overline{\Omega}) $. If $ \partial \Omega $ is smooth then $ u \in C^{\infty}(\overline{\Omega}) $ (and since $ u \in H^1_{0}(\Omega) $ we have in particular that $ u \in C^{\infty}_{0}(\overline{\Omega}) $ ) and by an integration by parts on (2) we can recover that u in actually a classical solution of (1) (in other words $ u $ satisfies equation (1) pointwise).
But for a general $\partial \Omega $ we have that $ u $ can have singularities on the boundary and we cannot apply the integration by parts. So i'm asking if it is true that in this case $ u $ is a classical solution of (1). Is it? The question seems me interesting since in this case we know that $ u \in C^{\infty}(\Omega) $.
Thanks
Best Answer
Assume that $u\in C^{\infty}(\Omega)$. Consider any open set $U\subset\Omega$ such that $\overline{U}\subset\Omega$ and $U$ has nice boundary. We have that $$\int_U\nabla u\nabla\psi+\int qu\psi=\int f\psi,\ \forall\ \psi\in C_0^\infty(U)$$
In particular, by appying Green formula, we get that $$\int\left(-\Delta u+qu-f\right)\cdot\psi=0,\ \forall\ \psi\in C_0^\infty(U)$$
The last equality implies that $-\Delta u(x)+q(x)u(x)=f(x)$ for all $x\in U$. For each fixed point $x\in \Omega$, we take $U=B_r(x)$ where $r>0$ is choosen in such a way that $\overline{B_r(x)}\subset\Omega$. we concude from the previous argument that $$-\Delta u(x)+q(x)u(x)=f(x),\ \forall\ x\in\Omega$$
To conclude, we have to prove that $u(x)=0$ in $\partial \Omega$. But to this happens, we need some regularity on the boundary of $\Omega$, for example, $\partial\Omega\in C^{0,1}$ is sufficient to conclude that $u\in C(\overline {\Omega})$. Indeed, the fact that $f\in C(\overline{\Omega})$ implies that $f\in L^p(\Omega)$ for all $p\in (1,\infty)$, therefore, $u\in W^{1,p}(\Omega)$ for all $p>1$. In particular, by using Sobolev embedding, we can conclude that $u\in C(\overline{\Omega})$, which would implies by the trace theorem that $u(x)=0$ in the boundary.