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 $\lim_{x\to\infty}u(x)=0$.

  • $\begingroup$ How many independent variables are there in the PDE? $\endgroup$ – doraemonpaul Nov 2 '12 at 15:02

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 $$ u(x)=\frac{1}{|x|^{n-2}}\quad x \in \overline{\Omega}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.