The Gauss-Green theorem for unbounded domain

Question

This question comes to me when I deal with the following PDE problem.

Suppose we have \begin{cases} -\Delta u=0 & x\in \mathbb R^N\setminus B(0,1)\\ u=0 & x\in\partial B(0,1)\\ u\to 0 & |x|\to\infty \end{cases} Then I am going to prove that $u\equiv 0$. This problem can be proven very quickly by using Maximum Principle.

But I got boring tonight and try to use energy method to prove this problem. I start with $$ 0=\int_{\mathbb R^N\setminus B(0,1)} \Delta u\, u\,dx = -\int_{\mathbb R^N\setminus B(0,1)} |\nabla u|^2\,dx+\int_{\partial B(0,1)} \nabla u\,u\,\nu d\sigma $$ and the last term is $0$ because of the boundary condition and we done.

Here I realize that I am using Gauss-Green theorem to do integration by parts on the unbounded domain and the integration over the boundary of $\mathbb R^N$ at "infinity" has been ignored by the condition $u\to 0$ as $|x|\to \infty$.

I remember I proved this result from my old classes... But now I can not justify it. I tried to do the following by taking

$$ 0=\int_{B(0,R)\setminus B(0,1)} \Delta u\, u\,dx = -\int_{B(0,R)\setminus B(0,1)} |\nabla u|^2\,dx+\int_{\partial B(0,1)} \nabla u\,u\,\nu d\sigma+\int_{\partial B(0,R)} \nabla u\,u\,\nu d\sigma $$ and I try to take $R\to \infty$.

But I don't see why $$ \lim_{R\to \infty} \int_{\partial B(0,R)} \nabla u\,u\,\nu d\sigma=0$$ even if $u$ vanish at infinity because we don't know the rate of vanishing and no information of $\nabla u$...

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Update:

Based on @Jose27's answer, my question has been well-solved. But in addition, I have a interested question here for case $N=2$.

It looks to me that for $N=2$, we need $\lim_{|x|\to \infty} u(x)$ to be exists, and hence we have $u$ is actually bounded.

Moreover, from Folland PDE book, page 115, proposition 2.74 I read that the following statement is equivalent if $u$ is harmonic outside $B(0,1)$, for $N=2$:

(a): $|u(x)|=o(\log|x|)$ as $x\to \infty$

(b): $|u(x)|=O(1)$ as $x\to\infty$

Quickly we have $(b)\implies (a)$. For converse, we notice that if $u$ satisfies $(a)$, then we have $\bar{u}$ is bounded and in turn $u$ is bounded as well.

Answer 1

You can prove that $u=o(1)$ at $\infty$ implies some stronger decay (at least for $N\geq 3$):

Consider the Kelvin transform of $u$, given by $v(x)=|x|^{2-N}u(x/|x|^2)$ defined (and harmonic) in the unit ball minus the origin. The decay of $u$ at $\infty$ implies that $v(x)|x|^{N-2} \to 0$ as $x\to 0$. By a result on isolated singularities of harmonic functions (see for example Han, Lin; "Elliptic Partial Differential Equations" Theorem 1.28) we get that $v$ is actually harmonic in $B(0,1)$. This implies that $|u(y)|\leq M|y|^{2-N}$ in the complement of $B(0,1)$.

Now just recall the inequality $$ |\nabla u(x_0)| \leq \frac{C_n}{s} \sup_{\partial B(x_0,s)} |u|. $$ Pick now $x_0\in \partial B(0,R)$ and $s=R/2$ to obtain that $|\nabla u (x_0)| \leq C/R^{N-1}$. Combining all this we obtain that $$ \left| \int_{\partial B(0,R)} \nabla u u \nu \right| \leq CR^{2-N} \to 0, \qquad \text{ as } R\to \infty. $$

For $N=2$ consider $v(x)=u(1/x)$, which is harmonic and bounded in $B(0,1)\setminus\{ 0\}$. As before $v$ extends to the unit ball. By specifying a harmonic conjugate with value $0$ at $0$, we consider $f$ the holomorphic function with $v$ as real part, and notice $f(0)=0$. Therefore $|v(x)|\leq |f(x)|\leq C_{r,u} |x|$ in $B(0,r)$. We conclude that $|u(z)|\leq C/|z|$ on the complement of $B(0,1/r)$ and, as before, $|\nabla u|\leq C/R^2$ on $\partial B(0,R)$ for $R$ large enough. We conclude that $$ \left| \int_{\partial B(0,R)} \nabla u u \nu \right| \leq C/R^2 \to 0, \qquad \text{ as } R\to \infty. $$ In fact in this case it's enough to ask that $\lim_{z\to \infty}u(z)$ exists, and the same argument gives that the integral vanishes in the limit.