Under what condition is there solution to Laplace equation with following conditions:
Let be $\Omega \subset \mathbb{R}^n$ open with regular boundary and $f \in C(\partial \Omega)$. Find $u\in C^2(\mathbb{R}^n)$ that \begin{align} \Delta u(x) &= 0&x\in \mathbb{R}^n \\ u(x) &= f(x) &x\in \partial \Omega \end{align}
It is well known that there is solution $u\in C^2(\Omega)$. But I'm interested in $C^2$ solution in whole space. It does not have to exist.
Take for example $n=1$ $$ \Omega = (-2,-1)\cup(1,2)$$ $$f(-2)=f(2)=0$$$$ f(-1)=f(1)=1$$ Than the only candidate for solution is \begin{align} u(x) &= 1 &x\in [-1,1] \\ u(x) &= 2-|x| &x\in \mathbb{R}\setminus (-1,1) \\ \end{align} which does not have first derivative at points $-1,1$
I came up with condition that $\partial \Omega$ has to be retract of $\Omega^c$. By retract I mean that there is $$ H(x,t):\partial \Omega \times [0,1) \rightarrow \Omega^c $$$$ H(x,0) = x $$$$ H \text{ is diffeomorphism} $$ I can first solve Laplace equation in $\Omega$ then I know normal derivatives of $u$ at $\partial \Omega$ so I can look at Laplace equation in $\Omega^c$ as time equation $$ \frac{\partial^2 u}{\partial t^2} = - \nabla \cdot( A \nabla u ) $$ in $\partial \Omega \times [0,1)$. Where $A$ pops out when you pull the equation from $\Omega^c$ to $\partial \Omega \times [0,1)$.
Problem is that I know very little about equation $\partial_{tt} u = - \Delta u$. It looks like wave equation except the sign. To get a little grasp of this equation I tried to calculate solution in $\mathbb{R}\times [0,\infty)$ with Fourier transform. \begin{align} \partial_{tt} u &= - \partial_{xx} u \\ u(x,0) &= f(x) \\ \partial_t u(x,0) &= g(x) \end{align} after Fourier transform \begin{align} \partial_{tt} \hat u &= \omega^2 \hat u \\ \hat u(\omega,0) &= \hat f(\omega) \\ \partial_t \hat u(\omega,0) &= \hat g(\omega) \end{align} solution is $$ \hat u(\omega,t) = \frac{\sinh(\omega t)}{\omega} \hat g(\omega) + \cosh(\omega t) \hat f(\omega) $$ This gives quite crude restriction on $\hat g, \hat f$, they should decay faster than exponential. Than we can transform $\hat u$ back and obtain solution. I guess when I work on $\partial \Omega \times [0,1)$ for $\Omega$ bounded than I would not encounter such integration problems so $f,g$ could be any continuous functions. I don't know how to proof that.
The question is: Is my condition that $\partial \Omega$ is retract of $\Omega^c$ sufficient condition for existence of solution?
Apart from the geometry of $\partial \Omega$, the boundary values $f$ would have to be very exceptional. Indeed, a harmonic function on $\mathbb R^n$ is real-analytic on $\mathbb R^n$, and this confers a lot of rigidity on the values it can attain on $\partial \Omega$.
For example, let $\Omega$ be the unit disk in $\mathbb R^2$. Expand $f$ into Fourier series $f(\theta)\sim \sum_{n\in\mathbb Z} c_n e^{in\theta}$. (The series need not converge; so far this is a formal expansion.) Then in $\Omega$ the (unique) solution of the Dirichlet problem is given by $$u(r,\theta) = \sum_{n\in\mathbb Z} c_n r^{|n|} e^{in\theta} \tag{1}$$ If $u$ extends to a harmonic function on all of $\mathbb R^2$, then it must be represented by the series (1) on all of $\mathbb R^2$ (indeed, write $u$ as the real part of holomorphic function; that function must be entire). Which implies that the sequence $c_n$ is rapidly decaying on both ends: $$\lim_{|n|\to \infty} R^{|n|}c_n = 0 \quad \forall\ R>0 \tag{2}$$ This requires $f$ to be analytic and more. Conversely, if (2) holds, then (1) solves your problem in all of $\mathbb R$.
If $\Omega$ does not have a nice boundary, the rigidity is still there ($f$ has to be very special), but it's no longer feasible to describe what "special" is, without tautologies.