My motivation in this problem is to show that one can approximate the eigenvalues of the Helmholtz equation on an L-shaped domain with Dirichlet boundary conditions (famously has no closed-form solutions) by solving a system of equations that govern the solution of a Schrodinger equation with sufficiently large potential. Are there any issues with my problem here? I am mostly concerned if my ansatz in equation (2) makes sense. In 1D for a finite potential well problem, one can use the fact that the solutions must be $C^1$... but is this the case for the 2D finite potential well problems? I suppose my solution here works if we have for a bounded smooth domain $\Omega \subset \mathbb R^2$ and piecewise constant $V$, are the Schrodinger equation solutions in $C^1(\Omega)$. Is that true?
Problem:
*Consider the regions $S = (0,2)^2$ and the L-shaped region $L = S \setminus (1,2)^2$. Solve the time-independent Schrodinger equation with Dirichlet boundary conditions for the Schrodinger operator $\hat H_{V_0} = -\Delta + V(x,y;V_0)$ with $V(x,y;V_0) = V_0 \chi_{S\setminus L} (x,y) $ where $\chi_{S\setminus L} (x,y)$ is the indicator function on ${S\setminus L}$ and we consider large $V_0 \gg 1$: \begin{equation}(1)\qquad \hat H_{V_0} u_{k,\ell} = \lambda_{k,\ell}(V_0) u_{k,\ell},\quad \left.u\right\vert_{(x,y) \in \partial S} = 0\end{equation}
My Attempt:
We label the following quadrants of $S$: \begin{equation} \begin{aligned} {\operatorname{I}} &= [0,1)\times[0,1), \qquad {\operatorname{II}} &= [0,1) \times [1,2) \\ {\operatorname{III}} &= [1,2) \times [0,1), \qquad {\operatorname{IV}} &= [1,2)\times [1,2) \end{aligned} \end{equation} Any solution on a quadrant solves (1), and we have from the boundary conditions on each quadrant that solutions are of the form: \begin{equation} \begin{aligned}(2)\qquad u_{k,\ell}(x,y) &= \begin{cases} A_{k,\ell} \sin( \xi^{\operatorname{I}}_k x) \sin(\eta^{\operatorname{I}}_\ell y), & (x,y) \in {\operatorname{I}} \\ B_{k,\ell} \sin( \xi^{\operatorname{II}}_k x) \sin(\eta^{\operatorname{II}}_\ell (2-y)), & (x,y) \in {\operatorname{II}} \\ C_{k,\ell} \sin( \xi^{\operatorname{III}}_k (2-x)) \sin(\eta^{\operatorname{III}}_\ell y), & (x,y) \in {\operatorname{III}} \\ \sinh( \xi^{{\operatorname{IV}}}_k (2-x)) \sinh(\eta^{{\operatorname{IV}}}_\ell(2- y)), & (x,y) \in {\operatorname{IV}} \end{cases}\\ \lambda_{k,\ell} &= (\xi^{i}_k)^2 + (\eta^i_\ell)^2 = -(\xi^{{\operatorname{IV}}}_k)^2 - (\eta^{{\operatorname{IV}}}_\ell)^2 + V_0,\ \text{for } i \in \{\operatorname{I},\operatorname{II},\operatorname{III}\}. \end{aligned} \end{equation} After requiring the solution to be in $C^1(S)$ and piecing together along the lines $x=1$ and $y=1$, we then have the following system of 11 equations for 11 unknowns (for fixed $k$ and $\ell$): \begin{align} A\xi^{\operatorname{I}} \sin( \xi^{\operatorname{I}})\sin(\eta^{\operatorname{I}} y) &= C\sin(\xi^{\operatorname{III}}) \sin(\eta^{\operatorname{III}} y) \label{eq:s1}\\ A \xi^{\operatorname{I}} \cos(\xi^{\operatorname{I}})\sin(\eta^{\operatorname{I}} y) &= - C \xi ^{\operatorname{III}} \cos( \xi^{\operatorname{III}} )\sin(\eta^{\operatorname{III}} y) \label{eq:s2} \\ B \sin( \xi^{\operatorname{II}} ) \sin( \eta^{\operatorname{II}}(2-y)) &= \sinh( \xi^{{\operatorname{IV}}} ) \sinh(\eta^{{\operatorname{IV}}}(2-y)) \label{eq:s3}\\ B \xi^{\operatorname{II}} \cos(\xi^{\operatorname{II}})\sin(\eta^{\operatorname{II}} (2 -y))&= - \xi^{{\operatorname{IV}}} \cosh (\xi^{{\operatorname{IV}}}) \sinh( \eta^{{\operatorname{IV}}} (2 - y))\label{eq:s4} \\ A \sin(\eta^ I)\sin(\xi^{\operatorname{I}} x) & = B\sin(\eta^{\operatorname{II}})\sin(\xi^{\operatorname{II}} x) \label{eq:s5} \\ A \eta^{\operatorname{I}} \cos(\eta^{\operatorname{I}})\sin(\xi^{\operatorname{I}} x) &= -B\eta^{\operatorname{II}}\cos(\eta^{\operatorname{II}})\sin(\xi^{\operatorname{II}} x) \label{eq:s6} \\ C \sin(\xi^{\operatorname{III}}(2-x))\sin(\eta^{\operatorname{I}}) &= \sinh (\xi^{{\operatorname{IV}}}(2-x)) \sinh( \eta^{{\operatorname{IV}}} ) \label{eq:s7}\\ C \eta^{\operatorname{I}} \sin( \xi^{\operatorname{III}}(2-x)) \cos(\eta^{\operatorname{I}}) &= - \eta^{{\operatorname{IV}}}\sinh(\xi^{{\operatorname{IV}}}(2-x))\cosh(\eta^{{\operatorname{IV}}}) \label{eq:s8} \\ (\xi^{\operatorname{I}})^2 + (\eta^{\operatorname{I}})^2 &= -(\xi^{{\operatorname{IV}}})^2 - (\eta^{{\operatorname{IV}}})^2 + V_0 \label{eq:s9} \\ (\xi^{\operatorname{II}})^2 + (\eta^{\operatorname{II}})^2 &= -(\xi^{{\operatorname{IV}}})^2 - (\eta^{{\operatorname{IV}}})^2 + V_0 \label{eq:s10}\\ (\xi^{\operatorname{III}})^2 + (\eta^{\operatorname{III}})^2 &= -(\xi^{{\operatorname{IV}}})^2 - (\eta^{{\operatorname{IV}}})^2 + V_0. \label{eq:s11} \end{align} Since these functions are all analytic, this system has a solution.