I have a question about Brownian motions and its heat kernel.

Using Dirichlet form theory, we can construct Brownian motions on manifolds, domains of Euclidean space under mild assumptions. For example, let $D$ be a domain of $\mathbb{R}^d$. Then, we can define the following bilinear form: \begin{equation*} \mathcal{E}(f,g)=\frac{1}{2}\int_{D}(\nabla f, \nabla g)\,dx,\quad f,g \in H^{1}(D), \end{equation*} where $H^{1}(D)$ is the Sobolev space of first order (with Neumann boundary condition). Furthermore, if this form is regular on $L^{2}(\bar{D},dx)$, we can construct reflecting Brownian motion on $\bar{D}$. I think Regularity assumption is very mild assumption. For example, when $D$ has continuous boundary, this form is regular.

**My question**

I am interested in estimate of heat kernel $p(t,x,y)$ of reflecting Brownian motion as above. In my research, **the existence is assured by this paper and references therein.** Furthermore, if the Sobolev inequality holds, we can obtain nice (**full time**) estimate like as
\begin{equation}
p(t,x,y) \le a_{1}t^{-d/2} \exp(-|x-y|^{2}/a_{2}t)\cdots(1),
\end{equation}
where $a_1, a_2$ are constants independent of $x,y,t$.

But I don't know nice heat kernel (upper) estimate when the Sobolev inequality is dropped. It is known that if $D$ is expressed as \begin{align*} D_{1}=\{(x_1,x_2)\mid x_{1}>1, 0<x_{2}<e^{-x_1} \}, \end{align*} we can prove Sobolev inequality does not hold. This is heat kernel of reflecting Brownian motion $\left\{X^{1}_{t} \right\}$ on $\bar{D}_{1}$ does not have nice estimate as (1).

On the other hand, if $0<t<1$, the sample path of $\left\{X^{1}_{t} \right\}$ is continuous, its heat kernel has nice estimate as (1). That is, I expected that heat kernel of $\left\{X^{1}_{t} \right\}$ has the following estimate: \begin{equation} p(t,x,y) \le a_{1}t^{-d/2} \exp(-|x-y|^{2}/a_{2}t), \end{equation} for all $x,y \in \bar{D}_{1}$ and $t\in (0,1)$, where $a_1, a_2$ are constants independent of $x,y,t$.

If you know related works on this question, please let me know.