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From the introductory part of Chapter 2 of Grisvard's book, we know that the PDE system \begin{align} -\Delta u &= 0 &\text{in}\ \Omega\subset \mathbb{R}^2\\ u &= g &\text{on}\ \Gamma\\ \end{align} where $g\in H^{1/2}(\Gamma)$, has a unique weak solution $u \in H^1(\Omega)$ which is also $H^2$ regular when $\Omega$ (open and bounded) has $C^{1,1}$ boundary $\Gamma = \partial \Omega$.

Question Is the solution $u \in H^1(\Omega)$ still $H^2$ regular if $\Gamma$ is only a subset of $\partial\Omega$?

Added Question Let $\Gamma_1 \subset \partial \Omega$ and define $\Gamma_2:=\partial \Omega \setminus \bar{\Gamma}_1$. Suppose I define the map $H^{1/2}(\Gamma_1) \ni g \mapsto f:=u|_{\Gamma_2}$ where $u$ is the solution of the PDE system $\Delta u = 0$ in $\Omega$ and $u = g$ on $\Gamma_1$. Does the solution of the new PDE system \begin{align} -\Delta v &= 0 &\text{in}\ \Omega\subset \mathbb{R}^2\\ v &= g &\text{on}\ \Gamma_1\\ v &= f &\text{on}\ \Gamma_2 \end{align} have an $H^2$ regularity?

I believe that, in this case, I now have an $H^2$ regular solution.

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No. Take $u(x,y)=\Re \sqrt{x+iy}$ then $u$ is harmonic in the slit plane, and satisfies homogeneous Dirichlet conditions in $\{x<0\}$ and homogeneous Neumann conditions on $\{x>0\}$. You can now construct a $C^{1,1}$ domain (take $\{x+iy: x^2+y^2<R^2, y>0\}$ and smooth out the corners) such that $u$ is zero on some part of the boundary. The function $u$ is $H^1$ but not $H^2$ (as it is not Hölder continuous for exponents greater than $\frac12$).

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  • $\begingroup$ Can you please point out what in particular does your counterexample violates for $u$ to be in $H^2$? Is it possible for $u$ to be $H^2$ regular in my problem above under additional assumptions? $\endgroup$ – Julienne Franz May 25 '18 at 12:15
  • $\begingroup$ This is essentially the same counterexample that I mentioned in your other question. $\endgroup$ – Hannes May 25 '18 at 12:38
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    $\begingroup$ $H^2$ embeds into all $W^{1,p}$, which embeds into $C^{0,1-2/p}$. For $p>4$, $\alpha=1-2/p>1/2$ so we would obtain $|x|^{1/2}=|u(x)-u(0)|\le C|x|^\alpha$, which is clearly wrong. The example is fairly typical for mixed boundary conditions (anywhere that Neumann and Dirichlet meet you will have such an example). But maybe you can show that nothing worse than this can happen (maybe you get something in $H^2$ after subtracting a multiple of my counterexample near each place where you switch between Dirichlet and Neumann?) $\endgroup$ – Kusma May 25 '18 at 13:29
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    $\begingroup$ @Kusma Many thanks for your explanation. I found a good discussion about this topic in Grisvard' 1992 book Singularities in Boundary Value Problems. Basically, $u$ splits into a regular and a singular solution where the singular solution is only in $H^1$. $\endgroup$ – Julienne Franz May 26 '18 at 4:48
  • $\begingroup$ @Hannes yes, I understand. I added another question above. I think in my new problem, I now have an $H^2$ regular solution. Is that correct? $\endgroup$ – Julienne Franz May 26 '18 at 5:25

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