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 $\nabla u$...


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.


1 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.

  • $\begingroup$ Honestly I didn't expect a this long answer. I thought this just some "easy prove" calculus fact but not a highly related PDE property... Indeed your answer is definitely make a perfect sense! Thanks so much. I read this problem from some online source and now I think that author should write more details rather then just "integration by parts"... $\endgroup$
    – spatially
    Dec 27, 2014 at 15:58
  • 1
    $\begingroup$ I also update my post for $N=2$ case, please have a look, thank you! $\endgroup$
    – spatially
    Dec 27, 2014 at 21:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.