# Prove that the problem $-\Delta u = 0$ with $u = g$ on the boundary has no solution for a given $g$.

This was an exercice in my functional analysis course: Let $$\Omega = B(0,1)\backslash \{0\} \subset \mathbb R^N, N \ge 3.$$ We define on $$\partial\Omega$$ $$g(x) = \begin{cases} 1 & \text{if }~ x= 0,\\ 0 & \text{if }~ x \neq 0, \end{cases}$$ and I have to prove that the problem $$\begin{cases} -\Delta u = 0 & \text{on }~ \Omega,\\ u = g & \text{on }~ \partial \Omega. \end{cases}$$ has no solution. I have absolutly no idea how to do that. I thought at first I could argue by contradiction and then use the strong maximum principle but I have not found anything that allows me to conclude.

• pass to polar coordinates Commented Nov 7, 2021 at 10:44
• It doesn't seem to me that we can find any harmonic function $u$ that satisfies the Dirichlet problem. Like you said you can get a contradiction that $u$ would be constantly zero if $u$ was ever a solution, right?
– Eric
Commented Nov 7, 2021 at 11:05

You haven't said it but I guess the exercise is to prove no solution to your problem exists. I'm also going to go ahead and assume that $$u \in C^2(\Omega) \cap C(\overline{\Omega})$$ - I expect that this is also probably assumed in your problem.
Let us assume, for the sake of contradiction, that there exist a solution $$u \in C^2(\Omega) \cap C(\overline{\Omega})$$. Then we must have that $$u$$ is radially symmetric. Indeed, fix $$e\in \mathbb \partial B(0,1)$$, and let $$v(x) = u(x) - u(x_\ast) \qquad x \in B(0,1)$$ where $$x_\ast$$ denotes the reflection of $$x$$ across the hyperplane $$\{ x \cdot e = 0 \}$$. Then you can check that $$v$$ satisfies $$\Delta v = 0$$ in $$\Omega$$ and $$v=0$$ on $$\partial \Omega$$. Thus, by uniqueness, $$v\equiv 0$$ in $$\overline{B(0,1)}$$. This implies that $$u$$ is symmetric with respect to $$\{ x \cdot e = 0 \}$$. Since $$e$$ was arbitrary, we must have that $$u$$ is radially symmetric.
So far we have shown that $$u(x) = w( \vert x \vert)$$ for some function $$w : (0,1) \to \mathbb R$$. Plugging this into the PDE we have $$w''(r)+ \frac{n-1}{r} w'(r) = 0 \qquad \text{in } (0,1)$$ with boundary conditions $$w(0)=1$$, $$w(1)=0$$. We can solve the above ODE (using that $$n>2$$) to get $$w(r) = A r^{2-n} +B,$$ with $$A,B \in \mathbb R$$. The boundary conditions imply that $$A=B=0$$ (if $$A \neq 0$$ then $$w$$ diverges at 0). Thus, $$u \equiv 0$$. But this clearly doesn't satisfy the boundary conditions.
Alternately, by the weak max maximum principle $$0\leqslant u \leqslant 1$$. Then, by standard elliptic regularity theory we will have that if $$u$$ exists then $$u \in C^\infty ( \overline{\Omega})$$. By continuity, $$\Delta u = 0$$ in $$B(0,1)$$. Since $$u(0)=1$$, the strong max. principle implies $$u$$ is constant which is a contradiction.
Such a solution $$u$$ would be a harmonic function in $$B(0,1)\setminus\{0\}$$ satisfying $$\lim_{x\to 0}u\!\left(x\right)/\lvert x \rvert^{2-N}=0$$. As we established in your other question asked today, minutes earlier than this one, such a function must be identical throughout $$B(0,1)\setminus\{0\}$$ to the solution $$v$$ of the Dirichlet problem on $$B(0,1)$$. But that clearly has $$v(0)=0$$ in this instance. Contradiction.