# The existence of harmonic function

If $\Omega$ is a connected bounded open set in $\mathbb R^n$ such that the boundary $\partial \Omega$ is smooth. Then can we find a function $u \in C(\Omega^c)$, such that $\Delta u=0$ in the complement of $\bar \Omega$ and $u=1$ on $\partial \Omega$ and $\mathop {\lim }\limits_{\left| x \right| \to \infty } u(x) = 0$?

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I suspect that this can be done, but only for $n\ge 3$. Indeed, for $n=2$ the fundamental solution of the Laplacian is infinite at infinity. –  Giuseppe Negro Jan 26 '13 at 1:52

This is discussed in any good book on potential theory, and the answer is what Giuseppe Negro wrote in the comment. For example, Classical Potential Theory by Armitage and Gardiner:

Theorem 6.7.1. If $\Omega$ is an unbounded open subset of $\mathbb R^n$, where $n\ge 3$, then $\infty$ is regular for $\Omega$.

Note that $\Omega$ in this context is $\mathbb R^n\setminus \overline{\Omega}$ in the question. Also, let's recall the definition of regular.

Definition 6.6.1. A point $y\in \partial^{\infty} \Omega$ is called regular (for $\Omega$) if $\lim_{x\to y}H_f(x)=f(y)$ for each $f\in C(\partial^\infty \Omega)$.

Here $\partial^\infty$ means the boundary of $\Omega$ taken in the topology of one-point compactification of $\mathbb R^N$, i.e., it includes $\infty$ when $\Omega$ is unbounded. And $H_f$ is the integral of $f$ against the harmonic measure on $\partial^\infty \Omega$.

Thus, if $n\ge 3$, then the Dirichlet problem for the function $f=1-\chi_{\{\infty\}}$ (which is continuous on $\partial^\infty \Omega$) has a classical solution, which is what you wanted.

And if $n=2$, then there is no such $u$. Indeed, applying a Möbius transformation of $\mathbb R^2$, we can map $\infty$ to a finite point, which is a removable singularity for bounded harmonic functions. By the maximum principle, $u_{\partial\Omega}=1$ implies $u\equiv 1$.

Remark. The preceding paragraph relies on the fact that Möbius transformations preserve harmonicity in two dimensions. Möbius maps are conformal in all dimensions, which implies that they preserve the functional $\int |\nabla u|^n$ under composition. Therefore they preserve the Euler-Lagrange equation for this functional. This equation is the Laplace equation only when $n=2$.

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Can you please expand a little the case $n=2$? The use of the maximum principle is somewhat unclear IMHO: the given initial condition is $u\equiv 1$ on $\partial \Omega$, while you are writing $u\equiv 0$ on $\partial \Omega$. Moreover, I could use more explanation on why the Möbius transformation argument does not push through in $n=3$ case. (By all means, that's a very nice answer.) –  Giuseppe Negro Jan 26 '13 at 10:33
@Guiseppe Thanks, I edited. –  user53153 Jan 26 '13 at 15:37

The object you are looking for is also known as the capacitary potential of the set $\Omega$, since it is the function whose energy $\int_{\Omega^c} |\nabla w|^2 dx$ achieves the capacity of the set $\Omega$. There is a weak-solution/variational formulation of the problem as well as the potential-theoretic one 5PM put up above.

Consider the Sobolev space $H^1(\Omega^c)$, with norm given by $$\|w\|_{H^1(\Omega^c)}^2 = \| \nabla w\|^2_{L^2(\Omega^c)} + \|w\|^2_{L^2(\Omega^c)}$$ Then you need only minimize the energy $\int_{\Omega^c} |\nabla w|^2 dx$ among all functions $w$ whose trace on $\partial \Omega$ is 1. When $n \geq 3$, the Sobolev embedding gives you the necessary condition for existence of a minimizer in this space (notice that the embedding just fails at $n=2$), and the usual theory for regular points of harmonic functions gives you the desired boundary conditions.

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