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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
up vote 6 down vote accepted

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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