Do $u\in H^1(\Omega)$ and $\Delta u\in L^2(\Omega)$ imply $u\in H^2(\Omega)$?

Suppose $\Omega$ is a bounded open domain in $\mathbb R^n$ with regular boundary. If $u\in H^1(\Omega)$ and $-\Delta u\in L^2(\Omega)$, do we have $u\in H^2(\Omega)$ for sure?

Here we say $\Delta u\in L^2(\Omega)$ by meaning $u$ has weak derivatives $\partial_{x_i}^2 u$ for $i=1,\cdots,n$ and $$ \Delta u:=\sum_{i=1}^n \partial_{x_i}^2u\in L^2(\Omega). $$ There is a related problem, and some comments refer to Theorem 8.12 in Gilbarg and Trudinger's book. I think their theorems are essentially proving the following result:

Suppose $\Omega$ is a bounded open domain in $\mathbb R^n$ with regular boundary. If $u\in H_0^1(\Omega)$ and $-\Delta u\in L^2(\Omega)$, then $u\in H^2(\Omega)$.

The proof can be accomplished using the Lax-Milgram theorem and the regularity of the weak solution. However, the original problem does not assume $u$ has zero trace, and this result cannot be directly applied.

In Helffer's Spectral Theory and Its Applications, Equation (4.4.5) at Page 38 defines the space $$ W(\Omega) = \{u\in H^1(\Omega):-\Delta u\in L^2(\Omega)\} $$ and claims that there is NO $W(\Omega)\subset H^2(\Omega)$. It seems that the answer to the original problem is negative, but I am not sure where to find a counterexample. Any helpful comments or answers are appreciated.

Solution 1:

No. The main problem is the regularity of the data on the boundary. Suppose for the sake of being concrete that $\partial \Omega$ is smooth. Then pick boundary data $g \in H^{1/2}(\partial \Omega) \backslash H^1(\partial \Omega)$. From weak well-posedness theory there exists a unique $u \in H^1(\Omega)$ such that $u \vert_{\partial \Omega} = g$ in the trace sense and $-\Delta u =0$ in $\Omega$. The latter requires that $u \in C^\infty(\Omega)$, and yet the regularity of $u$ on the boundary is locked into place by that of $g$. Now, if we had that $u \in H^2(\Omega)$ then trace theory would require that $u\vert_{\partial \Omega} \in H^{3/2}(\partial \Omega) \subset H^1(\partial \Omega)$, which is impossible by the choice of $g$. Thus, while $u$ is smooth in $\Omega$, its second derivatives are not square integrable up to the boundary.

Solution 2:

For a specific example consider the three-quarter disc, $\Omega=\{(r, \theta):0 < r < 1, 0 < \theta < \frac{3 \pi}{2}\}$ and $$u(r, \theta) = r^{\alpha} \sin(\alpha\theta)$$ with $\alpha = \frac{2}{3}$. It holds that $\Delta u = 0$ which is clearly $L^2$, but if you check, you find that $u\in H^1(\Omega)$ and $u\not\in H^2(\Omega)$.

I think this and a whole family of functions like it are great examples as they arise as 'singular' solutions to the Poisson problem. I used a similar example recently to show how irregularities arise from the corners of an otherwise smooth domain. You should checkout this fantastic book on irregularties of elliptic problems by Grisvard.

Edit:

As mentioned by @Glitch, the above example does not have a regular boundary, but does have regular BCs. However, it is quite simple to construct the reverse.

Consider the unit disc $D = \{(x,y):x^2+y^2<1\}$ and in polar coordinates $(0 \le \theta < 2\pi)$ $$u(r, \theta) = r^{\frac12}\sin(\frac12 \theta).$$ Again, this has $u\in H^1(D)$ but $u\not\in H^2(D)$.

Indeed, it is the solution to the Laplace problem \begin{alignat}{2} -\Delta u =&\, 0 \quad &&\textrm{in }D, \\ u =& \sin(\frac12 \theta) \quad &&\textrm{on }\partial D. \end{alignat} Clearly this problem is posed in a smooth domain, but with irregular boundary data.