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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!

Golbez
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    This is usually called the Laplace equation ($\Delta u=0$). Why is $u=g$ on the complement of unit ball? Did you mean $u=g$ on the boundary? If you know how to solve this PDE on the inside on the unit ball, the Kelvin transform gives the solution on the outside. The condition at infinity can be enforced by adding a multiple of $(1-R|x|^{-1})$. – Post No Bulls Dec 01 '13 at 06:45
  • @user111742 Oh,yes. The boundary condition is $g$, sorry for my mistake. By the way, we didn't mention this transformation on the class, it seems that I should prove this result before using it. Thanks for your comment! – Golbez Dec 01 '13 at 08:15
  • Is $g$ a constant ?. – Felix Marin Sep 04 '14 at 05:19

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

Felix Marin
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  • Thank you Felix. The $g$ here is not a constant, but a function depend on the position. Apologize for not responsing your comment, since I've got the answer for the question long ago. – Golbez Sep 12 '14 at 01:55