Let $R$ be the shifted open unit cube \begin{equation} \Big(-\frac{1}{2}, \frac{1}{2} \Big)^3 \subset \mathbb{R}^3, \end{equation} and let $k \in \mathbb{C}$ be a constant with $\textrm{Re}(k) <0$. Fix an $r \in (0, 1)$, and let $\psi \in H^1(R, \mathbb{C})$ be a function that is $1$-periodic in each argument, i.e., $\psi (x + e_i) = \psi(x)$ for all $i \in \{ 1,2,3\}$, where $e_i$ is the ith standard basis vector, and which satisfies \begin{align} - \Delta \psi &= k \psi & \textrm{ in } & R \setminus \overline{B_{r}(0)} \\ \psi &= 1 & \textrm{ in } & \overline{ B_{r}(0)}, \end{align} where $B_{r}(0) \subset \mathbb{R}^3$ is the open ball of radius $r$.
Using standard results from the regularity theory of elliptic equations, I can show that $\psi \in C^{\infty}(R \setminus B_{r}(0))$. In particular, $\psi \in C^0 (R)$.
I would like to prove that $\psi$ is more regular (e.g, $\psi \in C^1(R)$).
By the very definition of $\psi$, we know that $\nabla \psi \equiv 0$ in the open ball $B_r(0)$. We also know that $\psi$ is smooth in $R \setminus \overline{B_r(0)}$. So, we are left to prove that $\nabla \psi = 0$ on $\partial B_r(0)$. However, I have no idea how to tackle this problem (in fact, I do not know if $\psi$ is continuously differentiable in $R$ at all).
I would be grateful for any suggestions!
generically $\psi \notin C^1$. You can already see that in one dimension: the solutions in 1d are exponential functions cut off at a certain height by a constant function (for $k$ real and negative).