Showing $ u =0 $ for a particular laplace equation.

Question

Given open $\Omega\subset \mathbb{R}^n $ with smooth enough boundary, if for $x_0 \in \Omega$ there exists $ u \in C^2(\Omega\setminus\{x_0\}) $( we may as well take $u$ as smooth as we want) that satisfies \begin{equation*} \begin{cases} \Delta u = 0 & \text{ in } \Omega\setminus\{x_0\} \\ u = 0 & \text{ in } \partial\Omega \end{cases} \end{equation*} Then for some $ y \in \Omega\setminus\{x_0\}$ if $u(y) = 0 $ then is it true that $ u = 0 $ everywhere in $\Omega$ ? I can see that $ \Delta u = \sum_\alpha C_\alpha D^\alpha \delta_{x_0} $ in sense of distribution, hence perhaps $ u(x) = \sum_\alpha C_\alpha D^\alpha G(x,x_0) $ where $G$ is the green's function. Probably a good use of maximum principle would do it but I am stuck. Any help would be greatly appreciated. Thank you.

Answer 1

First assume that for any sequence $x_n\in \Omega\setminus \{x_0\}$ with $x_n\to x_0$, we have that $$\liminf u(x_n)>0$$

hence, by the maximum principle $u(x)>0$ in $\Omega\setminus \{x_0\}$. On the other hand, if for any sequence $x_n\in \Omega\setminus \{x_0\}$ with $x_n\to x_0$, we have that $$\limsup u(x_n)<0$$

then, by the maximum principle $u(x)<0$ for $x\in \Omega\setminus \{x_0\}$. To conclude, if $u\in C(\overline{\Omega})$ and $u(x_0)=0$ then $u$ must be zero everywhere.

For some references, take a look here in pages 7 and 8.

Answer 2

For a bounded domain $\Omega$ with $\,\partial\Omega\in C^1$, it is obviously true if $\,u\in C(\overline{\Omega})$ and $x_0\,$ is a removable singular point. Of course, it cannot be true whenever a singularity $x_0\,$ be non-removable, e.g., take $\,\Omega=\{x\in\mathbb{R}^2\,\colon\,\,|x|<1\}\,$ with $\,x_0=0$, and consider an example of $\,u(x)= x_1 x_2\bigl(1-|x|^{-4}\bigr)$.  It is clear that $\,u\in C^2(\Omega\backslash \{x_0\})\,$ and $\,\Delta u=0\,$ in $\,\Omega\backslash \{x_0\}$, while $\,u|_{\partial\Omega}=0$.  But $\,u=0\,$ on the lines $\,x_1=0\,$ and $\,x_2=0\,$ inside $\,\Omega\backslash \{x_0\}\,$, while $u\not\equiv 0$.