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Let $U \subset \mathbb{R}^n$ be a bounded domain, and consider the following problem : $$\left\{ \begin{array}{lcr} -\Delta u = 0 & & \text{in } U, \\ u = g & & \text{on } \partial U, \end{array} \right.$$ where $g$ is a given function and $u \in C^2(U) \cap C(\overline{U})$.

Does the existence of Green's function for $U$ imply the existence of a solution to the above Dirichlet problem ?

I know that the existence of a solution to the above Dirichlet problem depends both on the regularity of $\partial U$ and on the choice of $g$. On the other side, Green's function is defined as $G(x,y) = \Psi(x-y) - \phi^x(y)$, $x,y \in U$ and $x \neq y$, where $\Psi$ is the fundamental solution to Laplace's equation (and thus independent of $g$) and $\phi^x$ satisfies $$\left\{ \begin{array}{lcr} \Delta \phi^x = 0 & & \text{in } U, \\ \phi^x = \Psi(y-x) & & \text{on } \partial U, \end{array} \right.$$ which is also independent of $g$. If $u \in C^2(\overline{U})$ solves the Dirichlet problem, then $$u(x) = - \int_{\partial U} g(y) \frac{\partial G}{\partial \nu}(x,y) dS(y) + \int_U f(y)G(x,y) dy, \hspace{3mm} x \in U.$$

So, I'd say no : the existence of Green's function for $U$ does not imply the existence of a solution to the above Dirichlet problem. Yet, I need someone to confirm this.

And now, the other way around...

Does the existence of a solution to the above Dirichlet problem implies the existence of Green's function for $U$ ?

Thanks in advance.

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Your question requires some clarifications to be answered precisely. Your first statement: I know that the existence of a solution to the above Dirichlet problem depends both on the regularity of $\partial U$ and on the choice of $g$ seems to imply that you are considering a general open set $U$.

In that case, you should explain what the identity $u=g$ on $\partial \Omega$ means (in the sense of traces for regular boundaries, but in general?).

Next, you define a Green function by introducing $\phi^x$, and further down, you talk of a normal derivative of $G$, $\partial_n G$ : should we therefore understand that in fact the domain $U$ has a regular boundary, where a normal exists ?

If you are happy with a regular boundary (e.g. satisfying an exterior cone condition in dimension $\geq3$, and less in dimension $2$, see the related question here), the classical Dirichlet problem is uniquely solvable for any $g\in C^0(\bar{U})$, see Gilbarg & Trudinger Chapter 2 for example. Then, provided that the Green formula you wrote makes sense (a normal exists etc.) you have constructed a Green function as well.

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  • $\begingroup$ I'm considering a general open set $U$ that does not satisfy the exterior sphere condition at every point of its boundary. I'm looking for a solution $u \in C^2(U) \cap C(\overline{U})$ such that $\Delta u = 0$ in $U$ and $u_{\left| \partial U \right.} = g$ for some $g \in C(\overline{U})$. If the Barrier Postulate is satisfied, then there exists a solution to the Dirichlet problem, which I assume can be written in terms of $g$ and Green's function for $U$. Now, does the existence of Green's function for $U$ implies the existence of a solution to Dirichlet problem ? Thanks. $\endgroup$
    – Gatz'
    Oct 23 '13 at 19:33
  • $\begingroup$ @Gatz': The Barrier Postulate as you call it implies the unique solvability of the Dirichlet Problem, Thm 2.14 in Gilbard-Trudinger, i.e. it is solvable and the solution is unique. $\endgroup$
    – username
    Oct 24 '13 at 5:52
  • $\begingroup$ I understand. But does the existence of Green's function for $U$ is equivalent to the existence of a barrier for the Dirichlet problem at every point of the boundary $\partial U$ ? My concern was whether there can be an open set $U$ (that does not satisfy the exterior sphere condition) for which Green's function exists but for which the associated Dirichlet problem has no solution ? $\endgroup$
    – Gatz'
    Oct 24 '13 at 7:25

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