Consider the following evolution equation

$$u_t=\Delta u$$ in a bounded and regular open subset $\Omega$ of $\mathbb{R}^N$, with smooth initial conditions $u_0\geq 0$ and homogeneous Dirichlet boundary conditions.

It is known that this equation has a smooth global solution $u$. By differentiating $E(t):=\int_\Omega u^2 dx$, using integration parts and Poincaré's inequality we can prove that

$$|u(t)|_{L^2(\Omega)}\leq e^{ -\frac{1}{c^2}t}|u(0)|_{L^2(\Omega)}$$ where $c$ is the Poincaré constant. This means that the solution decays exponentially in $L^2(\Omega)$.

My question is: Does this decay result holds in $L^p(\Omega)$ for $p>2$ or maybe even $L^\infty(\Omega)$? The fact that $p=2$ is very important here because after the integration by parts we get the term $\int_\Omega |\nabla u|^2dx$ which we know how to estimate in terms of $\int_\Omega u^2 dx$ via Poincaré's inequality.

  • $\begingroup$ Welcome to MathOverflow! By semigroup and spectral theory the estimate remains true if you multiply the right handside by an appropriate constant $C_p \ge 1$ (at least for $p < \infty$). Does this suffice for your purposes? $\endgroup$ Jul 29, 2019 at 19:35
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    $\begingroup$ The decay is exponential in $L^\infty$ norm, too. This is a consequence of ultracontractivity, introduced by Davies and Simon (I think here). The $L^\infty$ norm of $u(t + 1)$ is bounded by a constant times the $L^2$ norm of $u(t)$, which we already know decays exponentially with $t$. $\endgroup$ Jul 29, 2019 at 20:58
  • $\begingroup$ @JochenGlueck Thank you! do you have any reference where I can see this with details? $\endgroup$ Jul 29, 2019 at 22:19
  • $\begingroup$ Actually, I think @MateuszKwaśnicki's comment provides a much simpler approach to the question than my comment about spectral theory. In fact, one does not even need the general theory of ultracontractivity. It follows, for instance, from Proposition 1.1. in [Arendt, Bénilan: Wiener Regularity and Heat Semigroups on Spaces of Continuous Functions (1999)] that the solution $u$ is bounded by the solution of the heat equation on $\mathbb{R}^N$ (with the same initial condition), so the solution semigroup of the equation on $\Omega$ maps $L^2$ into $L^\infty$. $\endgroup$ Aug 3, 2019 at 9:47
  • $\begingroup$ @JochenGlueck: Of course, you are right: in this case ultracontractivity is just a fancy word for "for a fixed $t > 0$, the heat kernel is a bounded function". $\endgroup$ Aug 3, 2019 at 14:02

1 Answer 1


You have the very powerful probabilist theorem $$ u(t,x) = \mathbb{E}_x(u(0,B_{t \wedge T})) $$ where $B_t$ is a Brownian process starting at $x$ and $T$ is the stopping time $T=\inf{s : B_s \notin \Omega}$ which gives the solution of the heat equation. In particular $$ \|u(t,.)\|_{L^{\infty}} \leq \|u(0,)\|_{L^{\infty}} \sup_{x\in \Omega} \mathbb{P}_x[\forall s\leq t, B_s\in \Omega ] \leq \|u(0,)\|_{L^{\infty}} C e^{-\gamma t}$$ for some $C>0$ and $\gamma >0$.

  • $\begingroup$ instead of this probabilist approach one can use directly the maximum principle to obtain the same (or at least) similar result... (I seem to recall). $\endgroup$
    – Math604
    Aug 8, 2019 at 0:19

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