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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.

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    $\begingroup$ pass to polar coordinates $\endgroup$
    – Gustave
    Commented Nov 7, 2021 at 10:44
  • $\begingroup$ 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? $\endgroup$
    – Eric
    Commented Nov 7, 2021 at 11:05

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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.

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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.

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