# weak solutions versus classical solutions

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

• I think that the test space must be $C_0^\infty(\Omega)$. – Tomás Jul 4 '13 at 17:51
• What is your definition of classical solution? – Tomás Jul 4 '13 at 18:34
• Yes you're right. I'hve corrected the mistake. – user55449 Jul 5 '13 at 8:08
• I've edited the answer. I hope that now it is more clear. – user55449 Jul 5 '13 at 8:16

## 1 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.