Problem. Given open, bounded set $\Omega\subset\mathbb R^d$ with smooth boundary $\partial\Omega$ and given smooth function $\varphi$ on $\partial\Omega$. As known, problem $$ \begin{cases} \Delta u=0, & \text{in $\Omega$,} \\ \\ \\ \lim_{\begin{matrix}x \to y \\ x \in \Omega \end{matrix}} u(x)=\varphi(y), & y\in\partial\Omega\setminus\{y_0\} \\ \end{cases} $$
has solution $u\in C^2(\Omega)$. Here boundary conditions are defined anywhere on $\partial\Omega$ except one point $y_0$.
- Show that solution is not unique.
- Show that bounded solution is unique.
For simplification we can use $d=2$ or $d=3$ ($d$ is the dimension of $\Omega$) and part of boundary, where $y_0$ locate, we can assume flat.
Can anyone give me some ideas?
$\bullet\;$ Take $\Omega=\{x\in\mathbb{R}^n\,\colon\,|x|<1,\,x_n>0\}$, $n\geqslant 2$. It is clear that a function $$ u(x)=x_n-\frac{x_n}{|x|^n} $$ solves the homogeneous bvp $$ \begin{cases}\tag{$\ast$} \Delta\,u=0,\quad x\in\Omega,\\ u\bigr|_{\partial\Omega\backslash \{0\}}=0. \end{cases} $$ $\bullet\;$ Let $u\in C^2(\Omega)\cap C\bigl(\overline{\Omega}\backslash\{0\}\bigr)$ be a bounded solution of a bvp $(\ast)$. Consider an odd extension of $u$ from $\,\Omega\,$ to a lower half of the unit ball $B=\{x\in\mathbb{R}^n\,\colon\,|x|<1\}$, namely, $$ u(x)= \begin{cases} u(x),\;\; x\in \overline{\Omega}\backslash \{0\},\\ u(x',-x_n),\;\; x\in B\,\colon\;x_n<0. \end{cases} $$ It is clear that $\,u\,$ is weakly harmonic in $\,B\backslash \{0\}$, and hence $u\in C^2(B\backslash \{0\})\cap C\bigl(\overline{B}\backslash\{0\})$ is a bounded harmonic function in $B\backslash\{0\}$ with a removable singularity at $x=0$, i.e., function $u\in C^2(B)\cap C\bigl(\overline{B})$ is harmonic in $B$, whence follows $\,u=0\,$ in $B\,$ by the maximum principle implying the uniqueness.