Laplace equation in unbounded domain

Question

Let consider the Laplace equation in unbounded domain in $\Omega = \{ x \in \mathbb{R}^n ; |x|＞1\}$

$$-\Delta f(x) =0 \ \ \text{in} \ \Omega$$

I want to prove if $f \in C^2 (\overline{\Omega})$ satisfies the above equation and $$\int_{\Omega}|f(x)|^2 dx + \int_{\Omega} |\nabla f(x)|^2 dx \ ＜ \ \infty$$

then $f=0$.

I could prove in case of $n=1$ or $|f(x)|=0 \ \text{on} \ |x|= 1$ or $|\nabla f(x)|=0 \ \text{on} \ |x|= 1$ with the use of integration by parts or the maximum principle.

But in general case, I have no idea.

Any advice would be appreciated.

Answer 1

I think your question needs revision. Here is what I think the revised statement should be:

Proposition: Suppose that $u\in C^2(\mathbb R^n)\cap L^2(\mathbb R^n)$ satisfies $\Delta u = 0$ in $\mathbb R^n \setminus B_1$, $u=0$ on $\partial B_1$. Then $u=0$ in $\mathbb R^n \setminus B_1$.

I will give two counter-examples which justify why I think these changes are appropriate. Then I will prove this proposition.

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Why can we only conclude $u=0$ in $\mathbb R^n \setminus B_1$? This was answered by Dan Doe in the comments. In fact, any smooth function $v:B_{1/2} \to \mathbb R$ can be smoothly extended so that $v=0$ in $\mathbb R^n \setminus B_1$. Hence, there is absolutely no way to say anything about $u$ in $B_1$.

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Why do we need $u=0$ on $\partial B_1$? Let $u(x) = \vert x \vert^{2-n}$ be the fundamental solution. Then $\Delta u=0$ in $\mathbb R^n \setminus B_1$ (it is harmonic everywhere but $0$). Then $u^2 = \vert x \vert^{4-2n}$ which is integrable in $\mathbb R^n \setminus B_1$ provided that $n>4$. Moreover, $\vert \nabla u \vert^2 = (n-2)^2 \vert x \vert^{2-2n}$ which is integrable provided that $n>2$. Hence, for $n>4$ the fundamental solution contradicts your statement.

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Proof of Proposition: Let $\eta \in C^1(B_{2r}\setminus B_1)$ be nonnegative such that $\eta =0$ on $\partial B_{2r}$. Since $u$ vanishes on $\partial B_{1}$, integration by parts gives that \begin{align*} 0 &= \int_{B_{2r}\setminus B_1} \eta^2 u \Delta u \, dx \\ &= - \int_{B_{2r} \setminus B_1} \nabla (\eta^2 u) \cdot \nabla u \, dx \\ &=- 2\int_{B_{2r} \setminus B_1} \eta u \nabla \eta \cdot \nabla u \, dx - \int_{B_{2r} \setminus B_1} \eta^2 \vert \nabla u \vert^2 \, dx. \end{align*} Rearranging, applying the Cauchy-Schwarz inequality then applying the Hölder inequality gives \begin{align*} \int_{B_{2r} \setminus B_1} \eta^2 \vert \nabla u \vert^2 \, dx &=- 2\int_{B_{2r} \setminus B_1} \eta u \nabla \eta \cdot \nabla u \, dx \\ &\leqslant 2\int_{B_{2r} \setminus B_1} \eta \vert u \vert \vert \nabla \eta \vert\cdot \vert \nabla u \vert \, dx \\ &\leqslant 2\bigg ( \int_{B_{2r} \setminus B_1} \eta^2 \vert \nabla u\vert^2 \, dx\bigg ) ^{1/2} \bigg (\int_{B_{2r} \setminus B_1} u^2 \vert \nabla \eta \vert^2\, dx \bigg )^{1/2} \end{align*} This implies the following (small) variant on the classical Caccioppoli inequality: \begin{align*} \int_{B_{2r} \setminus B_1} \eta^2 \vert \nabla u \vert^2 \, dx &\leqslant 4\int_{B_{2r} \setminus B_1} u^2 \vert \nabla \eta\vert^2 \, dx . \end{align*}

Fix $\varepsilon >0$ small and choose $\eta$ such that $$\eta (x) = \begin{cases} 1, &\text{if } x\in B_r \setminus B_1, \\ \frac 1 r (2r-\vert x \vert^2 ), &\text{if } x \in B_{2r} \setminus B_r. \end{cases} $$ Then, since $\vert \nabla \eta \vert = \frac1r$ in $B_{2r} \setminus B_r$\begin{align*} \int_{B_{r} \setminus B_1} \vert \nabla u \vert^2 \, dx \leqslant\int_{B_{2r} \setminus B_1} \eta^2 \vert \nabla u \vert^2 \, dx \leqslant 4\int_{B_{2r} \setminus B_1} u^2 \vert \nabla \eta\vert^2 \, dx\leqslant \frac 4 r\| u \|_{L^2(\mathbb R^n)}^2.\end{align*} Sending $r\to \infty$, we obtain $\| \nabla u \|_{L^2(\mathbb R^n \setminus B_1)}=0$, so $u$ is a constant. Then, since $u$ is in $L^2$ it must be zero.