Let $B\subset \mathbb R^2$ be given as a unit ball. Let $\omega\subset W^{2,2}(B)\cap L^\infty$ be given. (So we are only in 2d space)
Let $\Gamma\subset B$ be a closed Lipschitz curve such that $\mathcal H^1(\Gamma)<\infty$. Let $\Gamma_n\subset B$ be a sequence of closed Lipschitz curve such that $$ d_\mathcal H(\Gamma,\Gamma_n)\to 0 $$ and $$ \limsup_{n\to\infty}\mathcal H^1(\Gamma_n)<\infty. $$
Let function $u$ and $u_n$ be the solution such that $$ \begin{cases} -\Delta u=0&x\in\Omega\setminus \Gamma\\ u=\omega&x\in\partial B \end{cases} \text{ and } \begin{cases} -\Delta u_n=0&x\in\Omega\setminus \Gamma_n\\ u_n=\omega&x\in\partial B \end{cases} $$ So the boundary value on curve is free.
My question: can we have $$ \int_B|\nabla u_n|^2dx\to \int_B|\nabla u|^2dx $$ If not, would it be helpful if we increase the regularity of $\Gamma$ and $\Gamma_n$? i.e., change them to close $C^2$ curve?