$\renewcommand{\Re}{\operatorname{Re}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$

### General properties of Laplace equation

#### Existence and unicity

Consider problem \begin{align} & \Delta u-cu =f && \text{in }\mathcal{D}, \label{eq-1}\\[3pt] & u=0 && \text{on }\ \Gamma_- \label{eq-2}\\[3pt] & \partial_\nu u -\alpha u =0 && \text{on }\ \Gamma_+\label{eq-3} \end{align} where $\mathcal{D}$ is a connected bounded domain, $\Gamma$ its boundary (smooth), consisting of two non-intersecting parts $\Gamma_-$ and $\Gamma_+$, and $\nu$ a unit interior normal to $\Gamma$, $\partial_\nu u:=\nabla u\cdot \nu$ is a normal derivative of $u$, $c$ and $\alpha$ real valued functions.

Then \begin{gather*} - \int_\mathcal{D} fu\,dxdy = -\int_\mathcal{D} (u\Delta u-cu^2)\,dxdy=\\[3pt] \int_\mathcal{D} (|\nabla u|^2+cu^2)\,dxdy + \int _\Gamma u\partial_\nu u\,ds=\\[3pt] \int_\mathcal{D} (|\nabla u|^2+cu^2)\,dxdy + \int _{\Gamma_+} \alpha u^2\,ds \end{gather*} as we can integrate over $\Gamma_+$. Therefore assuming that \begin{equation} c\ge 0,\qquad \alpha \ge 0 \label{eq-4} \end{equation} we conclude that $f=0\implies \nabla u=0$ and then $u=\const$ and unless \begin{equation} c\equiv 0,\quad \alpha\equiv 0, \qquad \Gamma_-=\emptyset \label{eq-5} \end{equation} we conclude that $u=0$.

So, if (\ref{eq-4}) is fulfilled but (\ref{eq-5}) fails problem (\ref{eq-1})--(\ref{eq-3}) has no more than one solution (explain why). One can prove that the solution exists (sorry, we do not have analytic tools for this).

Theorem. If (\ref{eq-4}) is fulfilled but (\ref{eq-5}) fails problem (\ref{eq-1})--(\ref{eq-3}) is uniquely solvable.

Assume now that (\ref{eq-5}) is fulfilled. Then $u=C$ is a solution with $f=0$. So, problem has no more than one solution modulo constant. Also \begin{equation*} \int_\mathcal{D}f\,dxdy= \int_\mathcal{D}\Delta u\,dxdy=-\int_\Gamma \partial_\nu u \,ds. \end{equation*} and therefore solution of \begin{align} & \Delta u =f && \text{in }\ \mathcal{D},\label{eq-6}\\[3pt] & \partial_\nu u =h && \text{on }\ \Gamma\label{eq-7} \end{align} does not exist unless \begin{equation} \int_\mathcal{D}f\,dxdy+\int_\Gamma h \,ds=0. \label{eq-8} \end{equation} One can prove that under assumption (\ref{eq-8}) the solution exists (sorry, we do not have analytic tools for this).

Theorem. If (\ref{eq-5}) is fulfilled problem (\ref{eq-6})--(\ref{eq-7}) has a solution iff (\ref{eq-8}) is fulfilled and this solution is unique modulo constant.