# Laplace equation Dirichlet problem on punctured unit ball.

Let $\Omega = \{ x \in \mathbb{R}^n: 0<|x|<1 \}$ and consider the Dirichlet problem \begin{align} \Delta u &= 0 \\ u(0) &= 1 \\ u &= 0 ~~~\text{if} ~~|x|=1 \end{align} By considering the spherical average $$M_u(0, r) =\frac{1}{n w_n} \int_{|\xi|=1} u(r \xi) \, dS_\xi$$ show that the problem has no solution.

My attempt:

We have $$\lim_{r \to 1} M_u(0,r)=0$$ since $u=0$ on the boundary $|x|=1$ and $u$ must be continuous on $\Omega$. However, $u(0)=1$ so the Gauss Mean Value Theorem does not hold, and u is not harmonic at the origin.

I am uncomfortable with this argument. The origin is not even in the domain $\Omega$, so why should the mean value theorem hold when $x=0$? Why should we expect $u$ to be harmonic at $x=0$?

I can't think of any other proof for this problem, especially given that we are told to use the average over a sphere...

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Well, it depends on what exactly you mean by "the problem has no solution". If a solution to the Dirichlet problem is a continuous function on $\bar{\Omega}$ which is harmonic on $\Omega$, then by removal of singularities, a solution $u$ must be also harmonic at $x = 0$. You can also ask why should the mean value theorem hold, when the closed ball $\overline{B}(0,1)$ is not a subset of $\Omega$. –  levap Dec 21 '12 at 0:59
Right, I should have specified that $\lim_{r \to 1} M_u(0, r)=0$. The problem doesn't elaborate on what it means for the solution to not exist. What do you mean by removal of singularities in this case? We have $u(0)=1$ so there shouldn't be any singularity at the origin. –  B0112358 Dec 21 '12 at 1:02
A harmonic function on a punctured disc $\{0 < |x| < 1\}$ which is bounded around zero has a unique extension to a harmonic function on the disc $\{0 \leq |x| < 1\}$ - see here. If your solution $u$ is continuous on $\bar{\Omega}$, it must be bounded around zero and so $u$ must be also harmonic at zero. –  levap Dec 21 '12 at 1:10
No problem. You should also explain why you can take the limit as $r \to 1$. That is, why you can exchange the limit with the integral. –  levap Dec 21 '12 at 1:17
Honestly, I'm not sure why. I did my undergrad in physics, and many of the physics professors were astronomers so unfortunately we never really discussed stuff like this. –  B0112358 Dec 21 '12 at 1:26

One answer was given in the comments, but here is another one which might be closer to the "consider the average" suggestion. Namely, one can consider the average $M_u(0,r)$ as a function of $r$ and write a second-order linear differential equation for it (basically, Laplace's equation averaged over spheres). The solutions form a two-dimensional space, which is $\{c_1\log r+c_2:c_1,c_2\in\mathbb R\}$ in two dimensions and $\{c_1 r^{2-n}+c_2:c_1,c_2\in\mathbb R\}$ in higher dimensions. No element of this space can have distinct finite limits at $0$ and at $1$.