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Let $Q:=[0,1]^d$ and $g\in L^2(\Omega)$. Consider the PDE :

$$ \begin{cases} -\Delta f=g & \text{in $\Omega$} \\ f\equiv 0 & \mbox{on $\partial \Omega$.} \end{cases} $$

I know that if $\Omega$ is $C^2$ then we have : $f\in H^2(\Omega)$. The same holds true if $\Omega=\{z_n>0\}$.

Is this true for $\Omega=Q$ ?

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    $\begingroup$ Odd periodic continuation is your friend. $\endgroup$
    – Michael Renardy
    Commented Jul 11 at 0:44
  • $\begingroup$ @MichaelRenardy I know some what about that, can you please provide a complet answer. From what i can tell the proof using odd continuation is simple enought, but I havent seen it in any book, so i assume is something wrong with my understanding $\endgroup$
    – Alucard-o Ming
    Commented Jul 11 at 0:58
  • $\begingroup$ The cube is convex and on convex domains there is optimal $H^2 \cap H^1_0$ regularity for $L^2$ data. See for example the book of Grisvard. $\endgroup$
    – Hannes
    Commented Jul 11 at 7:18
  • $\begingroup$ Continue f and g as odd functions of period 2 in all directions. Your problem is then equivalent to the periodic problem, for which regularity is well known. $\endgroup$
    – Michael Renardy
    Commented Jul 11 at 12:36
  • $\begingroup$ I changed $\Omega=${$z_n>0$} to $\Omega=\{z_n>0\}$ (coded as Omega=\{z_n>0\}) so that there is proper horizontal spacing and no mismatch of fonts. Notice that if you must write $${}$$ $\Omega={}$something expressed other than in math mode,$${}$$then you can code the MathJax or LaTeX part as \Omega={} (with those curly braces) and then put no space between the MathJax or LaTeX part and the word that follows it, then because of those {curly braces} you will see the proper amount of horizontal space to the right of$\text{ “}{=}\text{”}.$ $\endgroup$
    – Michael Hardy
    Commented Jul 13 at 19:13

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