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EDIT: Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with smooth boundary. Let $f\colon \mathbb{R}^n\backslash \Omega \to \mathbb{R}$ be a continuous function which is harmonic in $\mathbb{R}^n\backslash \bar\Omega$ and vanishes on $\partial \Omega$. Let us also assume that $f$ vanishes at infinity.

What assumptions on the decay rate at infinity imply that $f\equiv 0$?

The case $n=3$ is of special interest to me.

ADD: The motivation of my question comes from the very classical problem from electrostatics ($n=3$) which is probably solved. Assume the domain $\Omega$ is filled with a conductor and electrified with a charge. All the charged is necessarily accumulated on the surface of the domain. The potential of the created electric field in the space is a harmonic function outside if the domain and is constant on the boundary. It decays at infinity like $1/r^2$. Is such potential unique? Equivalently, is distribution of charge on the surface is unique?

Probably this is a very well studied question, but I am not a specialist.

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  • $\begingroup$ You'd at least have to say something about the boundary conditions on the surface of the ball ... $\endgroup$ Sep 11, 2021 at 13:40
  • $\begingroup$ @MichaelEngelhardt: well, there is a bounded smooth domain containing the ball, such that on its boundary the function vanishes. $\endgroup$
    – asv
    Sep 11, 2021 at 14:04
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    $\begingroup$ Without boundary conditions there is no rate that suffice. In $n = 3$ for example $\partial_1^m \frac1r$ is harmonic and decays like $r^{-1-m}$. $\endgroup$ Sep 11, 2021 at 14:06
  • $\begingroup$ @WillieWong: You are right, thanks. I edited the question. $\endgroup$
    – asv
    Sep 11, 2021 at 14:13
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    $\begingroup$ Any harmonic function which vanishes on $\partial\Omega$ AND at $\infty$ is evidently zero. This follows from the Maximum Principle. $\endgroup$ Sep 11, 2021 at 15:40

1 Answer 1

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Assuming "vanish at infinity" means that

for every $\epsilon > 0$ there exists a sufficiently large ball $B_\epsilon$ such that $\big|f|_{\mathbb{R}^n \setminus B_\epsilon}\big| < \epsilon$

then you can just apply the maximum principle to $B_\epsilon \setminus \Omega$ and conclude that $f$ is bounded by $\epsilon$ on that set.

Take $\epsilon \to 0$ you get that $f$ must vanish identically.

No decay rate needed.

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