This is a follow-up question to Regularity of solution for Poisson equation on bounded domains.
We consider the Poisson equation $-\Delta u=f$ in $\Omega$ and $u=\varphi$ on $\partial\Omega$.
According to Theorem 6.14 of Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger there exists a solution $u\in C^{2,\alpha}(\overline{\Omega})$ if $\partial\Omega\in C^{2,\alpha}$, $f\in C^{\alpha}(\overline{\Omega})$, $\varphi\in C^{2,\alpha}(\overline{\Omega})$. They also say that one can weaken this condition. In the notes they do not really clarify what that means at least for me its not clear.
I am interested in weakening the regularity of the boundary but still keep a solution for which the second derivatives can be extended continuously up to the boundary.
Question
If $\partial\Omega$ is in $C^{2,\alpha}$ up to finitely many corners can the solution still be in $C^{2}(\overline{\Omega})$ or at least in $C^2(\overline{\Omega}\setminus\{x_1,…,x_n\})$ if $x_1,…,x_n\in \partial\Omega$ are the corner points?
Best Answer
In Theorem 6.14 of Gilbarg and Trudinger, it is in fact assumed that $\partial \Omega$ is of class $C^{2, \alpha}$. They discuss a variety of generalizations, but not of the global $C^{2, \alpha}$ estimate; rather, they generalize the existence theory implied by it.
The condition that $\partial \Omega$ is of class $C^{2, \alpha}$ is sort of sharp. Indeed, take a domain $\Omega$ which is, say, at least $C^{1, \alpha}$, with $y \in \partial \Omega$, and find a harmonic function in a neighborhood of $y$ which vanishes along $\partial \Omega$ and is positive on $\Omega$ (this is easy by solving some Dirichlet problem, for example). By basic regularity theory, $u \in C^{1, \alpha}(\bar{\Omega})$ locally, and by Hopf lemma the inward normal derivative of $u$ is strictly positive along $\partial \Omega$.
Let $\partial \Omega$ be locally given as a graph $\{x_n = g(x')\}$ with $\nabla' g(y') = 0$ for our designated point $y$ (and $\Omega = \{x_n > g(x')\}$ locally). Then we can take the equation $u(x', g(x')) = 0$ and differentiate it in the $x'$ variables, to get $\nabla' u(x) + \nabla' g(x') \partial_n u(x) = 0$. This gives the formula $$ \nabla' g(x') = \frac{\nabla' u(x)}{\partial_n u(x)} $$ The denominator is nonzero near $y$, as at $y$ itself we have $\partial_n u$ is the inward normal derivative (so: nonzero) and $\nabla' u(y) = 0$ as $\nabla'g(y') = 0$, while at nearby points it follows from the continuity of $\nabla u$.
But then if $u \in C^{2, \alpha}(\bar{\Omega}\cap B_r(y))$, this relation implies that $g$ is also $C^{2, \alpha}$. So we can conclude: (1) If a $C^{1, \alpha}$ domain has $\partial \Omega$ not $C^{2, \alpha}$ at a point, then there exists a harmonic function which is not $C^{2, \alpha}(\bar{\Omega})$ locally. This can be strengthened to say that any solution to the Poisson equation (or any equation! this isn't a PDE argument, this is the implicit function theorem) with nonzero normal at this point will fail to be $C^{2, \alpha}$ up to the boundary.
You can ask further questions here. For example, you can ask, well, if the normal derivative of $u$ does vanish at a point, is $u$ more regular there? There are theorems in that direction in the unique continuation literature. You can ask, what if the domain is not $C^{1, \alpha}$? Here it is useful to understand what happens on cones in the plane (i.e. consider the harmonic functions $\Re(z^\alpha)$ for $\alpha \geq 1/2$ on regions between two rays where they vanish): for cones with opening less than $\pi/2$, the harmonic function actually vanishes fast enough at the origin that it is $C^{2, \alpha}$ for some $\alpha$; for larger angles it is not $C^{2, \alpha}$ (except the special case of opening $\pi$). So you can say something if all of your corners are very acute.
For your second question, can you say that $u \in C^{2, \alpha}(\bar{\Omega}\setminus \{\text{bad set}\})$: sure. Use local boundary estimates, like Cor. 6.7 in Gilbarg-Trudinger. Everything in this theory is local.