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From [Grisvard, Thm. 2.4.2.7, p. 126], the BVP \begin{align} -\Delta u &= 0 & \text{in}\ \Omega\\ -\frac{\partial u}{\partial n} + au &= g & \text{on}\ \Gamma\\ \end{align} where $a>0$ and $g \in H^{1/2}(\Gamma)$ has unique solution $u$ with $H^2$ regularity for open bounded domain $\Omega \subset \mathbb{R}^2$ with boundary $\Gamma$ of class $C^{1,1}$.

Question Let $a_1, a_2 >0$, $g_1 \in H^{1/2}(\Gamma_1)$ and $g_2\in H^{1/2}(\Gamma_2)$ where $\Gamma_1 \subset \Gamma$ and $\Gamma_2 = \Gamma\setminus \bar{\Gamma}_1$. Does the unique solution $u \in H^1(\Omega)$ of the BVP \begin{align} -\Delta u &= 0 & \text{in}\ \Omega\\ -\frac{\partial u}{\partial n} + a_1u &= g_1 & \text{on}\ \Gamma_1\\ -\frac{\partial u}{\partial n} + a_2u &= g_2 & \text{on}\ \Gamma_2 \end{align} have higher regularity (in particular, is $u\in H^2(\Omega)$) for domain $\Omega$ of class $C^{1,1}$?

I know $u \in H^2(\Omega)$ if $\Gamma_1$ and $\Gamma_2$ are both closed, i.e., $\Omega$ is an annular domain. But, I do not know if I still have an $H^2$ regularity for the case given above.

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Without further assumptions, no. Consider the case $a_1=a_2=0$ and $\Omega$ smooth. If $u\in H^2(\Omega)$, then $\partial u/\partial n$ is in $H^{1/2}(\Gamma)$ by the trace theorem. However, $g_1\in H^{1/2}(\Gamma_1)$, $g_2\in H^{1/2}(\Gamma_2)$ does not imply $g\in H^{1/2}(\Gamma)$ (a step function is not in $H^{1/2}$).

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  • $\begingroup$ Thank you for you answer. Can you please state what must be the additional assumption for $u$ to be in $H^2$, and would it be possible for the case $a_1, a_2 >0$? $\endgroup$ – Julienne Franz May 25 '18 at 1:00

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