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Let $\mathbb H^+$ denote the upper half plane in $\mathbb R^2$. Consider the following equation \begin{equation}\label{pf0} \begin{aligned} \begin{cases} \Delta u=0\,\quad &\text{on $\mathbb H^+$}, \\ u=f\,\quad &\text{on $\partial \mathbb H^+$,}\\ \end{cases} \end{aligned} \end{equation} where $f \in L^2(\mathbb R)$. My question is what is the right Sobolev space where the classical solution to this equation belongs to.

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  • $\begingroup$ If $H^+=\{(x,t): t>0\}$ then $\|u(\cdot,t)\|_2 \le \|f\|_2$ for every $t>0$ but $u$ does not belong to $L^2(H^+)$ and the same holds for $p \neq 2$. Usually one looks at the solution using $t$ as a parameter. $\endgroup$
    – Giorgio Metafune
    Commented Dec 9, 2020 at 12:12
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    $\begingroup$ Yes, the problem is ill-posed if you consider classical Sobolev spaces. Indeed, even the trace operator $H^{s} \to H^{s-1/2}$ is ill-defined for $s = 1/2$. There are anisotropic Sobolev spaces that have been defined that would do the job (See homepage.univie.ac.at/piotr.chrusciel/papers/Dissertationes/…) but I have to think more about this.. $\endgroup$
    – Romain Gicquaud
    Commented Dec 9, 2020 at 12:16

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