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I have a question concerning the definition of parabolic boundary in the book "Second Order Parabolic Differential Equations" (2nd edition, 1996) by Gary M. Lieberman. He uses the following notation: $$|X|=\max(|x|,|t|^{1/2}),\quad Q(X,r)=\{Y\in \mathbb{R}^{n+1}\colon |Y-X|<r,s<t\},$$ where $X=(x,t),Y=(y,s)$.

He defines on page 7 for an arbitrary bounded domain $\Omega\subset \mathbb{R}^{n+1}$ the parabolic boundary as the points $X\in \partial \Omega$, where $\partial \Omega$ denotes the topological boundary, such that for any $\epsilon >0$ the cylinder $Q(X,\epsilon)$ contains points which do not belong to $\Omega$.

Now he claims that in the special case $\Omega=D\times (0,T)$, where $D$ is a bounded domain in $\mathbb{R}^n$, the parabolic boundary consists of $B\Omega=D\times \{t=0\}$ (bottom), $C\Omega=\partial D\times \{t=0\}$ (corner), $S\Omega=\partial D\times (0,T)$ (side). Could anyone explain to me why he does not include the set $\partial D\times \{t=T\}$?

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Given that the domain is $\Omega = D \times (0,T)$ and the definition of the parabolic boundary of this domain is $\mathcal{P}(\Omega) = \{ X \in \partial \Omega \ | \ \forall \ \epsilon > 0, \ Q(X,\epsilon) \not\subset \Omega \} $:

$\mathcal{P}(\Omega)$ is a backwards-in-time cylinder anchored at the point $X = (x,t)$. It extends from $t$ to $(t-\epsilon)$.

  • Using this cylinder on the lower face $B \Omega = D \times \{ t = 0 \}$ and the boundary of this face $C\Omega = \partial D \times \{ t=0 \}$ with $\epsilon > 0$, we see no intersection with $\Omega$. Thus, this is a part of $\mathcal{P}(\Omega)$.

  • At $S \Omega = \partial D \times (0,T) $, the cylinder has a partial intersection with the domain, but there are some points outside the domain. Thus, this is a part of $\mathcal{P}(\Omega)$.

  • For the domain $(D \times {t=T})$, any cylinder anchored on this face that has a finite $\epsilon$ extends into the domain $\Omega$. Thus, this face does not belong to the parabolic boundary $\mathcal{P}(\Omega)$.

I hope the schematic helps.Schematic for Parabolic boundary of <span class=$D \times (0,T)$" />

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  • $\begingroup$ Thanks for your answer. This was already clear to me, but what can you say about the set $\partial D \times \{t=T\}$? I guess it should also be part of the parabolic boundary. $\endgroup$
    – C. Phil
    Oct 23, 2021 at 8:12
  • $\begingroup$ I apologise for not reading the question well.. I concur with you that by this logic, we expect $\partial D \times \{t=T\}$ to be in $\mathcal{P}(\Omega)$. I am still working on why not.. I have been on the lookout for a few references that can explain this. And found: sites.pitt.edu/~armin/PDEIIWS16/pde2.pdf where Def 1.4.1 on pg 16 clearly states that this set does not belong to $\mathcal{P}(\Omega)$. Def 2.1.2 on pg 26 shows that both statements must be equivalent (exercise). I hope this is useful. Will get back once I have a complete answer. $\endgroup$
    – sai
    Oct 25, 2021 at 5:06
  • $\begingroup$ Thank you very much for the reference. One should note carefully that in Liebermann's book he considers domains of the form $\Omega= D\times (0,T)$ whereas in your reference (also in Evans book or the book of Haim Brezis) they consider domains of the form $\Omega=D\times (0,T]$. Thus in the later case the top of the cylinder is naturally excluded. $\endgroup$
    – C. Phil
    Oct 25, 2021 at 22:23
  • $\begingroup$ I'm also not sure whether there is a typo in your reference (cf. diva-portal.org/smash/get/diva2:914466/FULLTEXT01.pdf or Evans, Sect. 2.3 or Ladyzhenskaja's book), since $\partial D\times \{t=T\}$ should be included by the given definition. $\endgroup$
    – C. Phil
    Oct 25, 2021 at 22:35

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