One problem about harmonic functions 
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?
 A: $\bullet\;$ 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.
