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There are many places in the literature where the positivity of some semigroups is treated. However I did not know anyone which states and proves the strong positivity even for the basic semigroups like the Neumann laplacian semigroup.

Here is a simplified mathematical problem:

$$ \begin{cases}\dfrac{\partial u}{\partial t}(t,x)-\Delta u(t,x)+u(t,x)=0, & (t,x)\in (0,T)\times\Omega \\ \dfrac{\partial u}{\partial\nu}(t,x)=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega \end{cases}, $$

where $\Omega\subset\mathbb{R}^N$ is an open, bounded set with smooth boundary, and $f\in L^{\infty}(\Omega)^+=\{g\in L^{\infty}(\Omega)\ |\ g(x)\geq 0,\ \text{a.e. on}\ \Omega\}$.

If we denote by $S(t)$ the semigroup generated by $-\Delta+I$ with Neumann b.c. on $L^2(\Omega)$ then $S(t)$ is a positive semigroup, i.e. $u(t,\cdot)=S(t)f\in L^{\infty}(\Omega)^+$ for any $t\in [0,T]$. See for example W. Arendt - Heat Kernels (Theorem 3.3.1) https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/arendt/downloads/internetseminar.pdf.

In some articles I found, without proof or references that if $f\in L^{\infty}(\Omega)^+$ and $f\neq 0$ then $u(t,\cdot)\in \text{int}(L^{\infty}(\Omega)^+)$, i.e. there is some constant $c(t,u_0)>0$ such that $u(t,x)>c(t,u_0)$ a.e. on $\Omega$. How can we prove that?

It looks like a parabolic Harnack-type inequality is needed here...

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    $\begingroup$ $\Omega$ should also be connected for such a statement to possibly be true, right? $\endgroup$
    – Hannes
    Oct 18, 2022 at 7:07
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    $\begingroup$ As pointed out in the other comments, one possibility is to use the strong maximum principle (see Ch. 2, Sec. 5 in the book of Friedman "Partial differential equations of parabolic type"). A functoipnal analytic approach, which uses "irreducuibility" and form methods, can be found in the book by Ouhabaz "Analysis of the heat equation in domains", see Chapter 4. $\endgroup$ Oct 18, 2022 at 7:58
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    $\begingroup$ For rather general situations (e.g. rough boundary conditions where Hopf's boundary lemma doesn't apply) combining irreducibility with ultracontractivity and a bit of abstract nonsense does the trick. I'll explain the details in an answer as soon as I have a bit more time left. $\endgroup$ Oct 18, 2022 at 8:42
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    $\begingroup$ @bogdan Yes, true. One has to add also a regularizing effect of the semigroup for positive $t$ . In this case the solution is continuous, positive at every point and then the infimum is positive. However, this argumet does not give any quantittave bound. $\endgroup$ Oct 18, 2022 at 17:18
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    $\begingroup$ $u$ is continuous only for $t>0$, unless $f$ is continuous, too. The approach I have in mind is based on regularity up to the boundary and I need $\partial \Omega \in C^2$. The semigroup is analytic in the whole scale of $L^p$ with domain included into $W^{2,p}$. Choosing $p>N/2$, by Sobolev embedding one obtains thjat $u(t, \cdot)$ is continuous. A standard reference is the bool by A. Lunardi "Analytic semigroups and optimal regularity in parabolic problems". the approach described by @Jochen Glueck is surely more efficient. $\endgroup$ Oct 19, 2022 at 7:22

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As other users have indicated in the comments, for sufficiently smooth domains one can get it by combining, for instance, elliptic regularity with Hopf's boundary point lemma (and then go from the elliptic to the parabolic case by, for instance, a semigroup argument).

However, the same result remains true in much more general situations; for the Neumann Laplace, for instance, it suffices if the domain is connected and has the extension property for the Sobolev space $H^1$. Here is general operator theoretic argument that can be used in such situations:

Let $X$ and $Y$ be an $L^p$- and an $L^q$-space or, more generally, Banach lattices and let $T: X \to Y$ be a bounded linear operator which is positive in the sense that $Tf \ge 0$ whenever $f \ge 0$.

  1. A vector $0 \le f \in X$ is called a quasi-interior point of $X_+$ if for every non-zero functional $0 \le \varphi \in X'$ one has $\langle \varphi, f \rangle > 0$.

  2. Example: If $0 \le f \in X = L^p(\Omega,\mu)$ for a $\sigma$-finite measure space $(\Omega,\mu)$ then the following holds: (a) if $1 \le p < \infty$, then $f$ is quasi-interior point if and only if $f(\omega) > 0$ for almost all $\omega \in \Omega$. (b) If $p=\infty$, then $f$ is a quasi-interior point if and only if $f \ge \varepsilon 1$ for some $\varepsilon$. This distinction between the cases $p=\infty$ and $p < \infty$ will be very important at the end of our argument.

  3. Proposition. Assume that there exists $0 \le g \in X$ such that $Tg$ is a quasi interior point of $Y_+$. Then for every quasi-interior point $f$ of $X_+$ the vector $Tf$ is a quasi-interior point of $Y_+$.

  4. Proof of the proposition. This follows easily from a duality argument, since the assumption on $Tg$ implies that the dual operator $T'$ does not have any non-zero positive functionals in its kernel. For more details see for instance (warning: self-promotion ahead!) Proposition 2.21 in this article by Martin Weber and myself (there the notion almost interior point is used; in Banach lattices this coincides with the notion quasi-interior point).

  5. How to apply it to the heat equation with Neumann boundary conditions? The solution semigroup $S$ is irreducible and analytic on $L^2$. Hence, it even has the property that $S(t)f$ is a quasi-interior point of $L^2$ for every $t > 0$ and every non-zero function $f \ge 0$; see for instance Theorem C-III-3.2(b) in this classical book on positive semigroups. Moreover, as we assumed $\Omega$ to have the extension property, the Neumann semigroup $S$ is ultracontractive, so $S(t)$ maps $L^2$ into $L^\infty$ for each $t > 0$. Moreover, the constant function $1$, which is a quasi-interior point in the positive cone of $L^\infty$, is a fixed point of the semigroup, i.e., $S(t) 1 = 1$ for all times $t$. Hence, we can now apply the proposition: for every $t > 0$ and every non-zero $0 \le f \in L^2$ the vector $S(t/2)f$ is a quasi-interior point of $L^2$, so by the proposition $S(t)f = S(t/2)S(t/2)f$ is a quasi-initerior point in $L^\infty$.

  6. Generalizations: A similar type of argument can be used (if one combines it with some additional tools) for more general second order differential operators with rough coefficients on rough domains, and e.g. also with Robin boundary conditions - see (warning: self-promotion, again!) this article by Wolfgang Arendt, Tom ter Elst and myself.

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  • $\begingroup$ Thanks for the answer! I have a little question: Why we can't just apply the parabolic Harnack inequality (G.Lieberman - Second order parabolic differential equations , page 192, corollary 7.42)? $\endgroup$
    – Bogdan
    Oct 19, 2022 at 3:31
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    $\begingroup$ @Bogdan: The parabolic Harnack inequality gives you information inside the domain, but I doesn't apply to points on the boundary. $\endgroup$ Oct 19, 2022 at 9:18
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    $\begingroup$ Note that the Harnack inequality is formulated in terms of the action of the differential operator - it does not take any boundary conditions into account. Hence, the inequality cannot hold for boundary points, since the strict positivity on the boundary is, for instance, clearly not true for Dirichlet boundary conditions (as pointed out by MaoWao in a comment). $\endgroup$ Oct 19, 2022 at 9:22
  • $\begingroup$ Now I fully understand. Thank you very much for your interest! $\endgroup$
    – Bogdan
    Oct 19, 2022 at 11:50

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