1
$\begingroup$

Let $R > 0$. Determine the radial solution of the problem

\begin{align} - \Delta u(x) & = 1 \text{ if $|x| < R$}\\ u(x) & = 0 \text{ if $|x| = R$} \end{align}

We know the fundamental solution of the Laplace equation in $\mathbb{R}^n$ for n>2:

$\Phi(x) = \frac{1}{(n-2) \cdot w_n} \cdot |x|^{2-n}$,

where $w_n$ denotes the surface area of the unit sphere in $\mathbb{R}^n$.

Additionally we know following theorem:

Suppose $f \in C^2_c(\mathbb{R}^n)$ and let $u = \Phi \ast f$. Then $u \in C^2(\mathbb{R}^n)$ and $- \Delta u = f$ in $\mathbb{R}^n$.

My first guess is, that we have to choose $f = \mathbb{1}_{B(0,R)}$. But this function is not even continuous. And just tacking $f = \mathbb{1}_{\mathbb{R}^n}$ seems to make no sense to me.

Can anyone give me a hint how to approach this problem?

$\endgroup$
  • $\begingroup$ You should use the green's function for a ball e.g. math.stackexchange.com/q/1464667/80734 $\endgroup$ – Calvin Khor Apr 22 at 9:29
  • $\begingroup$ Thanks for the hint. I saw this solution in Evans book on pdes. In my course, this question is explicitly asked before the introduction to Green's function. So I would like to solve the problem without this specific function. $\endgroup$ – mathlettuce Apr 22 at 9:39
  • 1
    $\begingroup$ You are asked to find a radial solution. So express $\Delta$ in spherical coordinates, and you will find an ODE. $\endgroup$ – Calvin Khor Apr 22 at 9:43
2
$\begingroup$

Recipe -

  1. Show that if $u(x) = U(r)$, $r=|x|$, then $$ \Delta u=\frac{\partial^{2} u}{\partial r^{2}}+\frac{N-1}{r} \frac{\partial u}{\partial r}= U'' + \frac{N-1}rU'.$$
  2. Now solve $ U'' + \frac{N-1}rU' = -1$ on $r\in[0,R]$ subject to $ U(R)=0$ and (since $u$ is radial and differentiable at $0$) $U'(0)=0$.
  3. Conclude.

A similar problem - Poisson equation inside a ball $B(0, 1)$

$\endgroup$

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.