-1
$\begingroup$

EDIT: Let $\Omega\subset \mathbb{R}^3$ be a bounded domain with smooth connected boundary. Let $f\colon \mathbb{R}^3\backslash \Omega \to \mathbb{R}$ be a continuous function which is harmonic in $\mathbb{R}^3\backslash \bar\Omega$ and constant on $\partial \Omega$.

Let us assume that $f$ decays at infinity as $O(1/r^2)$.

Is it true that $f\equiv 0$?

This question is a strengthened version of this one Harmonic functions in infinite domain in Euclidean space which contained an unnecessary assumption.

The motivation to ask this question comes from a classical question in electrostatics. Assume that the domain $\Omega$ is filled in with a conductor which is electrically neutral. Can the charge of the conductor redistribute in a non-trivial way so that the conductor will be in equilibrium?

$\endgroup$
12
  • 1
    $\begingroup$ $g(x)=|x|^{2-n} f(x/|x|^2)$ in $R^n$. $\endgroup$ Commented Sep 12, 2021 at 13:53
  • 3
    $\begingroup$ And that $g(x)=O(|x|^2)$ for $x \approx 0$ indicates that there is no singularity at $x=0$, hence $g$ is harmonic with constant boundary values. $\endgroup$ Commented Sep 12, 2021 at 13:59
  • $\begingroup$ @LeechLattice: If $\Omega$ is not a Euclidean ball then $g$ is not constant on the boundary. But in the Euclidean ball this answers the question. $\endgroup$
    – asv
    Commented Sep 12, 2021 at 14:31
  • 1
    $\begingroup$ @makt Even if $\Omega$ is not an Euclidean ball, if $f$ is nonzero at its boundary, $g$ will be bounded away from zero at its boundary (though not constant); in this case it's impossible to have $g(0)=0$ as $g$ is harmonic. $\endgroup$ Commented Sep 12, 2021 at 15:01
  • $\begingroup$ @LeechLattice: Looks like the final answer. $\endgroup$
    – asv
    Commented Sep 12, 2021 at 15:05

1 Answer 1

0
$\begingroup$

If $f$ is of one sign on $\partial\Omega$, then $f$ is of the same sign everywhere by the maximum principle. Now let $u(r)$ be the average of $f$ on any sphere centered at the origin which is large enough not to intersect with $\Omega$. Then we have $u''(r)+(2/r)u'(r)=0$, which is inconsistent with the decay you require unless $u=0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .