$\Omega = \{x\in\mathbb{R}^2: |x|>1\}$, then problem $\Delta u = 0 \mbox{ in } \Omega, u = u_0 \mbox{ in } S_1(0)$ is unique

Question

Let $\Omega = \{x\in\mathbb{R}^2: |x|>1\}$. Prove the unicity of the problem: $$\Delta u = 0 \mbox{ in } \Omega\\ u = u_0 \mbox{ in } S_1(0)$$ Where $u_0$ is continuous in $S_1(0)$ and we impose $u$ continuous and bounded in $\overline\Omega$

Suppose $u_1$ and $u_2$ are both solutions to this problem. Normally we'd like to come up with a proof of $w = u_1-u_2 = 0$.

We know $\Delta w = 0$ in $\Omega$ and $w = 0$ in $S_1(0)$. Maybe the maximum principle can be used here. We have a harmonic function and remember that its maximum must be attained in the boundary which is $S_1(0)$ and has value $0$.

Is this how we prove it? I don't think so, because no continuity and boundary conditions are used in my proof.

UPDATE:

I understood the answer below but I think the maximum principle can be used to make a cleaner proof. I know that the maximum principle requires a bounded domain but maybe that can be worked out because of the boundedness of $u$.

Maybe the fact that $u$ is harmonic and bounded imply something. At least for the entire $\mathbb{R}^2$, $u$ harmonic there would be $0$ because of Liouville's theorem

Answer 1

You have to be careful with statements like this. What does "$\Delta u = 0$" mean? Well, if $u \in C^2$, then it means $\frac{\partial^2}{\partial x^2}u + \frac{\partial^2}{\partial y^2} u = 0$. If $u$ is an arbitrary measurable function on $\Omega$, then we define "$\Delta u = 0$" to mean $\int u \Delta \phi = 0$ for all $\phi \in C_c^\infty(\Omega)$. However, for any $u$, $\Delta u = 0$ forces $u$ to be $C^2$ (actually, $C^\infty$ as well) and to satisfy $\frac{\partial^2}{\partial x^2}u + \frac{\partial^2}{\partial y^2} u = 0$, and also forces $u$ to satisfy the mean value property. My point is that, since continuity is guaranteed by $\Delta u = 0$, requiring $u$ to be continuous just gives information as to what happens at the boundary.

What does "$u = u_0$ in $S_1(0)$" mean? We are searching for a $u$ defined only $\Omega$. So "$u=u_0$ in $S_1(0)$" means, for example, that $\lim_{z \to \zeta} u(z) = u_0(\zeta)$ for each $\zeta \in S_1(0)$ and the limit is taken over $z$ in $\Omega$ approaching $\zeta$. The point is that if we extend the definition of $u$ to all of $\overline{\Omega}$, then this extended $u$ will be continuous.

Now, the maximum principle applies to harmonic functions that attain their maximum or minimum at a point. This happens for harmonic functions defined on the interior of a compact set that extend continuously to the boundary. $u$ does indeed extend continuously to $\partial \Omega = S_1(0)$, but $\overline{\Omega}$ is not compact. One could imagine $w$ not achieving its minimum or maximum at any point in $\Omega$.

Answer 2

WLOG all functions mentioned are real valued. Recall that $\ln |z|$ is harmonic in $\mathbb C\setminus \{0\}.$

Lemma: Suppose $v$ is continuous on $\{1\le |z|<\infty\}, $ with $v$ harmonic in $\{1<|z|<\infty\}.$ Assume $v=0$ on $\{|z|=1\}$ and $v(z)\to \infty$ as $|z|\to \infty.$ Then $v\ge 0$ in $\{1\le |z|<\infty\}.$

Proof: Choose $R>1$ such that $v\ge 0$ on $ \{|z|=R\}.$ Then $v\ge 0$ on the boundary of $\{1\le |z|\le R\}.$ By the minimum principle for harmonic functions, $v\ge 0$ on $\{1\le |z|\le R\}.$ Since $R$ is arbitrarily large, the lemma follows.

Thm: Suppose $u$ is continuous and bounded on $\{1\le |z|<\infty\}, $ with $u$ harmonic in $\{1<|z|<\infty\}.$ If $u=0$ on $\{|z|=1\},$ then $u\equiv 0.$

Proof: Let $\epsilon>0.$ Define $v(z) = \epsilon\ln |z| - u(z).$ By the lemma, $v\ge 0$ in $\{1\le |z|<\infty\}.$ I.e., $u(z) \le \epsilon \ln |z|$ in $\{1\le |z|<\infty\}.$ Since $\epsilon$ is arbitrary, $u\le 0$ in $\{1\le |z|<\infty\}.$ But the same argument applies to $-u.$ The theorem follows.

This proves your unicity result (although you should take care to state the result more precisely).