$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ #[Chapter 7. Laplace equation](id:chapter-7) ##[7.1. General properties of Laplace equation](id:sect-7.1) > 1. [Existence and unicity](#sect-7.1.1) ###1. [Existence and unicity](id:sect-7.1.1) Consider problem \begin{align} & \Delta u-cu =f && \text{in }\mathcal{D}, \label{eq-7.1.1}\\\\[3pt] & u=0 && \text{on }\ \Gamma\_- \label{eq-7.1.2}\\\\[3pt] & \partial\_\nu u -\alpha u =0 && \text{on }\ \Gamma\_+\label{eq-7.1.3} \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-7.1.4} \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-7.1.5} \end{equation} we conclude that $u=0$. So, if (\ref{eq-7.1.4}) is fulfilled but (\ref{eq-7.1.5}) fails problem (\ref{eq-7.1.1})--(\ref{eq-7.1.3}) 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 1.](id:thm-7.1.1)** If (\ref{eq-7.1.4}) is fulfilled but (\ref{eq-7.1.5}) fails problem (\ref{eq-7.1.1})--(\ref{eq-7.1.3}) is uniquely solvable. Assume now that (\ref{eq-7.1.5}) 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-7.1.6}\\\\[3pt] & \partial\_\nu u =h && \text{on }\ \Gamma\label{eq-7.1.7} \end{align} does not exist unless \begin{equation} \int\_\mathcal{D}f\,dxdy+\int\_\Gamma h \,ds=0. \label{eq-7.1.8} \end{equation} One can prove that under assumption (\ref{eq-7.1.8}) the solution exists (sorry, we do not have analytic tools for this). **[Theorem 2.](id:thm-7.1.2)** If (\ref{eq-7.1.5}) is fulfilled problem (\ref{eq-7.1.6})--(\ref{eq-7.1.7}) has a solution iff (\ref{eq-7.1.8}) is fulfilled and this solution is unique modulo constant. --------------- [$\Leftarrow$](../Chapter6/S6.5.html) [$\Uparrow$](../contents.html) [$\Rightarrow$](./S7.2.html)