# Definition of parabolic boundary by Lieberman

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: $$|X|=\max(|x|,|t|^{1/2}),\quad Q(X,r)=\{Y\in \mathbb{R}^{n+1}\colon |Y-X| where $$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 $$\partial D\times \{t=T\}$$?

## 1 Answer

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.$D \times (0,T)$" />

• 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. Commented Oct 23, 2021 at 8:12
• 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.
– sai
Commented Oct 25, 2021 at 5:06
• 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. Commented Oct 25, 2021 at 22:23
• 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. Commented Oct 25, 2021 at 22:35