3
$\begingroup$

I have a question about Neumann heat kernels and its estimates.

Let $D$ be a domain of $\mathbb{R}^d$. We define the Dirichlet form $(\mathcal{E},\mathcal{F})$ on $L^{2}(D)$ as follows: \begin{align*} \mathcal{E}(f,g)&=\frac{1}{2}\int_{D}(\nabla f,\nabla g)\,dx,\quad f,g \in H^{1}(D), \end{align*} where $H^{1}(D)$ is the 1-st order Sobolev space with Neumann boundary condition. From general theory of Dirichlet forms, we can find $L^2$-semigroup $\{T_t\}_{t>0}$ associated with $(\mathcal{E},\mathcal{F})$. In the following, we assume $\{T_t\}_{t>0}$ admits an integral kernel $p_{t}(x,y)$.

If $D$ is bounded and the boundary of $D$ is sufficiently smooth, it is known that $p_{t}(x,y)$ has the following Gaussian estimate: \begin{align} (1)\quad p(t,x,y)\le c_{1}t^{-d/2}\exp(-|x-y|^2/c_{2}t),\quad dx-\text{a.e. } (x,y)\in D\times D,\,0<t\le 1. \end{align}

But if $D$ is unbounded, even if the boundary $\partial D$ is smooth, $p_{t}(x,y)$ does not necessarily have Gaussian estimate like (1).

My question

We assume $D$ satisfies the following assumption: there exist closed subsets $\{K_n\}_{n=1}^{\infty}$ of $\bar{D}$ such that

  • $K_{1} \subset K_2 \subset \cdots$.
  • each $K_{n} \cap \bar{D}$ is non empty bounded and its boundary is smooth.
  • $\bar{D}=\bigcup_{n=1}^{\infty}K_n$.

Then, can we show the following assertion?:

For each $n$,there exist some constants $c_{1,K_n}$, $c_{2,K_n}$ such that

\begin{align} (2)\quad p(t,x,y)\le c_{1,K_n}t^{-d/2}\exp(-|x-y|^2/c_{2,K_n}t),\quad dx-\text{a.e. } (x,y)\in K_n\times K_n,\,0<t \le 1. \end{align}

Roughly speaking, (2) is a kind of localized estimate of $p_t(x,y)$.

If you know papers related to (2), please let me know.

$\endgroup$
  • 1
    $\begingroup$ Two comments: (a) Your estimate (1) holds for small time only: the constants explode as $t \to \infty$. (b) I do not have any reference, but I believe the upper bound (1) also holds in unbounded domains, it is the lower bound which is problematic. Are you interested in a two-sided estimate, or just the upper bound? $\endgroup$ – Mateusz Kwaśnicki Aug 28 '17 at 9:01
  • 1
    $\begingroup$ Thank you for your reply. I am interested in upper estimates for Neumann heat kernels. Reply for your comment: (a) You're right. I was wrong. (b) Consider the following domain: $D=\{(x,y) \in \mathbb{R}^2 : 0<y<1/(1+x^2)\}$. Unfortunately, we can prove $H^{1}(D) \not \subset L^{q}(D)$ for each $q>2$. This implies $T_t$ is not ultracontractive. Therefore, (1) does not hold. $\endgroup$ – sharpe Aug 28 '17 at 10:05
  • 1
    $\begingroup$ Ah, I though you assume that the boundary is uniformly regular. $\endgroup$ – Mateusz Kwaśnicki Aug 28 '17 at 10:12
  • 1
    $\begingroup$ Sorry. What is the definition of uniformly regular? $\endgroup$ – sharpe Aug 28 '17 at 10:16
  • 1
    $\begingroup$ By "uniformly" I meant that the regularity constants (in particular, the localisation radius) do not depend on the boundary point, so in particular the set does not become too thin. Coming back to your original question, I would look into some papers by Grigor'yan and co-authors. For example here in Theorem 1.1 on-diagonal bounds are used to find off-diagonal bounds, and on-diagonal bounds are apparently simler. $\endgroup$ – Mateusz Kwaśnicki Aug 28 '17 at 12:04
1
$\begingroup$

Gaussian upper bounds for the Neumann heat kernel on a large class of unbounded domains are proved here.

Grigoryan cites a couple of papers by Guschin (in russian) claiming that he proved upper Gaussian estimates for the Neumann problem for parabolic equations on unbounded domains. See these notes.

$\endgroup$
  • 1
    $\begingroup$ P.S. I know that this does not answer your question directly, I'm just suggesting that you need some assumption on the boundary to get the estimate. Indeed, in your question the 'assumption' on D is satisfied by all open sets $\endgroup$ – Piero D'Ancona Aug 30 '17 at 6:47
  • $\begingroup$ Thank you for your reply. I modified assumptions for $D$. $\endgroup$ – sharpe Aug 30 '17 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.