Characterize where the Dirichlet Problem for the Laplacian is always solvable

Conway's 1978 textbook Functions of One Complex Variable I gives an unsatisfying characterization of the regions for which the Dirichlet Problem can always be solved, and then comments no cleaner characterization is known. Has any progress been made since then? And what simpler characterization is known today, if one is known?

Here is the problem definition:

An open connected set $G\subseteq \mathbb{C}$ is called a Dirichlet Region if for each continuous function $f:\partial_\infty G\rightarrow \mathbb{R}$ there is a continuous function $u:G^- \rightarrow \mathbb{R}$ such that $u$ is harmonic in $G$ and $u(z)=f(z)$ for all $z$ in $\partial_\infty G$.

(The notation $\partial_\infty G$ refers to the boundary of $G$ in $\mathbb{C}\cup\{\infty\}$, and $G^-$ denotes the closure of $G$ in $\mathbb{C}\cup\{\infty\}$.)

The characterization given in the book is:

Given $a \in \partial_\infty G$, a barrier for $G$ at $a$ is a family $\{\psi_r: r>0\}$ of functions such that:
1. $\psi_r$ is well-defined and superharmonic on $B(a;r) \cap G$ with $0\leq \psi_r(z) \leq 1$
2. $\lim_{z\rightarrow a}\psi_r(z) = 0$, and
3. $\lim_{z\rightarrow w} \psi_r(z) = 1$ for $w$ in $G \cap \{w:|w-a|=r\}$.

An open connected set $G$ is a Dirichlet Region iff there is a barrier for $G$ at each point of $\partial_\infty G$.

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I'm not sure why you are not satisfied. I assume there are large classes of regions, simple to describe, for which the Dirichlet problem can be satisfied with any continuous boundary data. I also assume there are some classes, simply described, for which some continuous boundary data causes failure (although 0 boundary data would seem to always work). Exactly classifying the "always" regions strikes me as subtle. Note that your result seems to be identical with Theorem 2.14 in Gilbarg and Trudinger, in the section on Perron's method. –  Will Jagy May 24 '11 at 22:57
If the boundary is $C^{1,\alpha}$ there is always a solution. –  Will Jagy May 24 '11 at 23:38
I like the fact that there is always a unique solution when $\partial_\infty G$ is a simple closed curve on $\mathbb{C}\cup\{\infty\}$ as follows from Theorem 2.24 in Markushevich: Theory of Functions of a Complex Variable Vol. 3. –  GH from MO May 24 '11 at 23:55
This isn't an actual characterization but after Theorem 2.14 in Gilbarg and Trudinger, they also go on to remark that the boundary value problem is solvable (in 2 dimensions still, where we can use complex analysis) in any bounded domain where each connected component of the complement consits of of more than a single point. –  Spencer May 25 '11 at 9:09
Thanks to those who have offered elegant sufficient conditions. But as Will suggested, I am indeed interested in exactly classifying the "always" regions. My interest is from mathematical logic. When I come across an unwieldy characterization which has resisted simplification for a long time, I often wonder if that unwieldiness is unavoidable. In logical terms, what I mean by unwieldiness can be made precise: the definition of a Dirichlet Region is syntactically $\Pi^1_2$, and on its face the barrier characterization is also syntactically $\Pi^1_2$; i.e. just as complicated. –  Linda Brown Westrick May 25 '11 at 20:45

http://eom.springer.de/r/r080680.htm contains some characterizations of domains where the Dirichlet problem is solvable. I believe that a key term in searching for references is "Wiener's criterion."

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I assume you are asking about strong solutions (so u is actually $C^2(G)\cap C(\partial_\infty G))$. In this case, the characterization via barriers, or equivalently, as Will says, using Perron's method, cannot be improved upon, I think.

Here's what I remember, please correct me if there are flaws in the argument.

Define a regular point as a point $a$ on the boundary of $G$ such that a barrier exists at $a$ (with respect to G).Conway's characterization is saying domains with boundaries consisting of only regular points are Dirichlet regions.

Now for the converse: if a region is a Dirichlet region, it must have a boundary of regular points. Suppose there is a domain $G$ which is a Dirichlet region. Let $y$ be a point on the boundary of the domain. Consider a continuous function $f:\partial_\infty G \rightarrow \mathbb{R}$ such that $f(y)=0$, and $f>0$ for all other parts of the boundary.The solution $u$ of the Dirichlet problem with $f$ as data is, by the strong maximum principle, a barrier at $y$. Hence $y$ is a regular point.

The question of the boundary regularity necessary for solvability of the Dirichlet problem has indeed been studied, and the answer may vary depending on the specific notion of solvability (strong solution? weak solution? solution a.s.?). Gilbarg and Trudinger, H\"ormander, Maz'ya have all written nice books on this and related topics.

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Different page, someone named James McKernan at Santa Barbara (at the time), exterior cone suffices, his Theorem 5.5: The Dirichlet problem can be solved for any region $U$ such that each boundary point is the end point of a line segment whose other points are exterior to $U.$ So, a variant of the cardioid where the spine touching the origin oscillates would be an example, appropriate boundary data uncertain. math.ucsb.edu/~mckernan –  Will Jagy May 25 '11 at 4:51
@Will Jagy Check out "Partial Differential Equations" Second Edition by Jost pages 74-75, the end of the "alternating method of H.A. Schwarz" section. He gives an explicit example of a domain with an inward cusp where the solution to Dirichlet problem is not continuous up to the boundary. The relevant pages are available on google books. –  Yakov Shlapentokh-Rothman May 25 '11 at 16:39
@Linda The Jost example is in three dimensions. I don't have Conway's book, but I imagine that his interest stems from the Riemann mapping theorem. Hence he probably only considers two dimensional regions. –  Yakov Shlapentokh-Rothman May 25 '11 at 22:06
@Nilima Yes, I would be very interested in a simple characterization that works only for the plane. Geometric conditions are often the right kind of simple. Now to be more specific about "simple": most of math can be formalized in Second Order Arithmetic; see en.wikipedia.org/wiki/Second-order_arithmetic for its syntax. The formula which says "G is a Dirichlet Region" has two set quantifiers (for all continuous $f$, there exists a harmonic $u$ such that...). Reduction to a single set quantifier would be useful, and an arithmetic characterization would be especially simple. –  Linda Brown Westrick May 25 '11 at 22:36
Yakov, excellent. Ahlfohrs says that the best result from Perron type arguments is this: The Dirichlet problem can be solved for any region whose complement is such that no component reduces to a point. He says that this provides a quick proof of Riemann mapping if taken as an axiom. The next section is multiply (but finitely) connected regions, concentric slit regions and so on. I don't think the original question by Linda has a nice answer. –  Will Jagy May 26 '11 at 1:48