Homework: Solve the poisson equation in the outer sphere Our teacher asked us to solve the poisson equation:
\begin{eqnarray}\left\{\begin{array}{ccc}\Delta u &= &0
\\ u|_{\partial \overline{B(0,R)}} & = & g \\
\lim_{|\vec{x}|\to\infty} u(\vec{x}) & = & 0\end{array}\right.\end{eqnarray}
I tried using the similar technique when the area is bounded. That is , choose an $\epsilon$ sphere near the point $M_0$ I needed and shrink it to zero. But integrating via another boundary: a big sphere is somehow hard since the rate of converging to zero is unknown. Also, the sphere $\overline{B(0,R)}$ is an obstacle fot the integration. Hope to find some hint, thanks!
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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We can choose the $\ds{z}$-axis in any direction and use spherical coordinates
$\ds{\pars{r > 0,\ 0 \leq \theta \leq \pi,\ 0 \leq \phi < 2\pi}}$. It turns out that the solution is $\ds{\phi}$-independent and it can be expressed as a linear combination of Legendre Polynomia
$\ds{\braces{{\rm P}_{\ell}\pars{\cos\pars{\theta}}\ \mid\
     \ell = 0,1,2,3,\ldots}}$.
\begin{align}
{\rm u}\pars{r,\theta}&
=\sum_{\ell = 0}^{\infty}\pars{A_{\ell}r^{\ell} + {B_{\ell} \over r_{\ell + 1}}}{\rm P}_{\ell}\pars{\cos\pars{\theta}}
\end{align}

Since $\ds{\lim_{r\ \to\ \infty}{\rm u}\pars{r,\theta} = 0}$, we set $\ds{A_{\ell} = 0\,,\forall\ \ell}$. Then,
  $$
{\rm u}\pars{r,\theta}
=\sum_{\ell = 0}^{\infty}{B_{\ell} \over r_{\ell + 1}}
{\rm P}_{\ell}\pars{\cos\pars{\theta}}
={B_{0} \over r} + \sum_{\ell = 1}^{\infty}{B_{\ell} \over r_{\ell + 1}}
{\rm P}_{\ell}\pars{\cos\pars{\theta}}
$$
  The boundary condition at $\ds{r = R}$ is given by:
  \begin{align}
g&={\rm u}\pars{R,\theta}
={{B_{0} \over R}}
+\sum_{\ell = 1}^{\infty}{B_{\ell} \over R_{\ell + 1}}
{\rm P}_{\ell}\pars{\cos\pars{\theta}}
\end{align}

Since $\ds{g}$ is a constant, we'll have
$\quad\ds{B_{\ell} = 0\,,\forall\ \ell \geq 1\quad\imp\quad B_{0} = gR}$

$$
\mbox{Then,}\quad
\color{#66f}{\large{\rm u}\pars{r,\theta}={R \over r}\,g}
$$

Since the $\ds{\tt OP}$ never answer my comment, I assumed that $\ds{g}$ is a constant.
