[Math] How to solve the exterior Dirichlet problem for Laplace’s equation

harmonic functionspartial differential equations


Solve $\Delta u=0$ on $\Omega$, where $\Omega=\{x : \|x\|>1\}$. The conditions are $u=1$ on the boundary of $\Omega$, and $\lim_{x\to\infty}u(x)=0$.


The domain here is the exterior of unit ball in $\mathbb R^n$. If it was the interior, then the only harmonic function with $u=1$ on the boundary would be the constant one, $u\equiv 1$.

But here, $u\equiv 1$ is not an acceptable solution, as it does not satisfy $\lim_{x\to\infty}u(x)=0$.

Best Answer

Observe first that the problem is invariant under rotations, therefore solutions $u$ of the given problem are of the form $u(x)=f(|x|)$. Since \begin{eqnarray} \Delta u(x)&=&\sum_{i=1}^n\frac{\partial^2u}{\partial x_i^2}(x)=\sum_{i=1}^n\frac{\partial}{\partial x_i}\left(\frac{x_i}{|x|}f'(|x|)\right)=\sum_{i=1}^n\left[\frac{x_i^2}{|x|^2}f''(|x|)+\left(\frac{1}{|x|}-\frac{x_i^2}{|x|^3}\right)f'(|x|)\right]\cr &=&f''(|x|)+\frac{n-1}{|x|}f'(|x|), \end{eqnarray} we have to solve the ODE $$\tag{1} f''(t)+\frac{n-1}{t}f'(t)=0. $$ Integrating (1) we get $$ f(t)= \begin{cases} at^{2-n}+b & \text{ if } n \ge 3\\ a\ln t+b & \text{ if } n=2 \end{cases}, $$ where $a,b$ are real constants. Because of the conditions $f(1)=1$ and $\lim_{t \to \infty}f(t)=0$, there is no such $f$ for $n=2$. For $n \ge 3$, the solution is given by $$ f(t)=\frac{1}{t^{n-2}} \quad \forall t \ge 1. $$ Hence $$ u(x)=\frac{1}{|x|^{n-2}}\quad x \in \overline{\Omega}. $$