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.