Hölder kernel estimates for Robin operators and Dirichlet-to-Neumann operators

Abstract

Consider the elliptic operator

$$\begin{aligned} A = - \sum _{k,l=1}^d \partial _k \, c_{kl} \, \partial _l + \sum _{k=1}^d a_k \, \partial _k - \sum _{k=1}^d \partial _k \, b_k + a_0 \end{aligned}$$

on a bounded connected open set $\Omega \subset \mathbb {R}^d$ with Lipschitz boundary conditions, where $c_{kl} \in L_\infty (\Omega ,\mathbb {R})$ and $a_k,b_k,a_0 \in L_\infty (\Omega ,\mathbb {C})$, subject to Robin boundary conditions $\partial _\nu u + \beta \, {\mathop {\mathrm{Tr \,}}}u = 0$, where $\beta \in L_\infty (\partial \Omega , \mathbb {C})$ is complex valued. Then we show that the kernel of the semigroup generated by $-A$ satisfies Gaussian estimates and Hölder Gaussian estimates. If all coefficients and the function $\beta$ are real valued, then we prove Gaussian lower bounds. Finally, if $\Omega$ is of class $C^{1+\kappa }$ with $\kappa > 0$, $c_{kl} = c_{lk}$ is Hölder continuous, $a_k = b_k = 0$ and $a_0$ is real valued, then we show that the kernel of the semigroup associated to the Dirichlet-to-Neumann operator corresponding to A has Hölder Poisson bounds.

The chain condition

Let $\Omega \subset \mathbb {R}^d$ be open and connected. We say that $\Omega$ satisfies the chain condition if there exists a $c > 0$ such that for all $x,y \in \Omega$ and $n \in \mathbb {N}$ there are $x_0,\ldots ,x_n \in \Omega$ such that $x_0 = x$, $x_n = y$ and $|x_{k+1} - x_k| \le \frac{c}{n} \, |x-y|$ for all $k \in \{ 0,\ldots ,n-1 \}$. Obviously in general $\Omega$ does not satisfy the chain condition.

Proposition A.1

Let $\Omega \subset \mathbb {R}^d$ be open bounded connected and Lipschitz. Then $\Omega$ satisfies the chain condition.

The proof requires some preparation. Let $\Omega \subset \mathbb {R}^d$ be open bounded connected and Lipschitz. If $T > 0$ and $\gamma :[0,T] \rightarrow \Omega$ is a Lipschitz curve, then $\gamma$ is differentiable almost everywhere. We define the length of $\gamma$ by $\ell (\gamma ) = \int _0^T |\gamma '(t)| \, dt$. Define the geometric distance $d :\Omega \times \Omega \rightarrow [0,\infty )$ by d(x, y) is the infimum of $\ell (\gamma )$, where $T > 0$ and $\gamma :[0,T] \rightarrow \Omega$ is a Lipschitz curve with $\gamma (0) = x$ and $\gamma (T) = y$. Obviously $|x-y| \le \ell (\gamma )$ and hence $|x-y| \le d(x,y)$.

We first consider a special Lipschitz chart.

Lemma A.2

Let $U \subset \mathbb {R}^d$ be an open set and $\Phi$ be a bi-Lipschitz map from an open neighbourhood of $\overline{U}$ onto an open subset of $\mathbb {R}^d$ such that $\Phi (U) = E$ and $\Phi (\Omega \cap U) = E^-$. Then there are $c_1,c_2 > 0$ such that $d(x,y) \le c_1 \, |x-y|$ and $|x-y| \le c_2$ for all $x,y \in \Omega \cap U$.

Proof

Let $L \in \mathbb {R}$ be larger than both the Lipschitz constant for $\Phi$ and $\Phi ^{-1}$. Further, let $x,y \in \Omega \cap U$. Define $\gamma :[0,1] \rightarrow \Omega$ by $\gamma (t) = \Phi ^{-1} ( (1-t) \, \Phi (x) + t \, \Phi (y) )$. Then $\gamma (0) = x$ and $\gamma (1) = y$. Moreover, the curve $\gamma$ is Lipschitz continuous and $|\gamma '(t)| \le L \, |\Phi (y) - \Phi (x)| \le L^2 \, |y-x|$ for almost every $t \in [0,1]$. So $d(x,y) \le \ell (\gamma ) \le L^2 \, |x-y|$. Also $|x-y| \le L \, |\Phi (x) - \Phi (y)| \le 2 L$. $\square$

We next show that the geometric distance is equivalent with the induced Euclidean distance on $\Omega$.

Lemma A.3

There exists a $c > 0$ such that $|x-y| \le d(x,y) \le c \, |x-y|$ for all $x,y \in \Omega$.

Proof

By a compactness argument there are $N \in \mathbb {N}$ and for all $k \in \{ 1,\ldots ,N \}$ there are open $U_k \subset \mathbb {R}^d$ and a bi-Lipschitz map $\Phi _k$ from an open neighbourhood of $\overline{U_k}$ onto an open subset of $\mathbb {R}^d$ such that $\Phi _k(U_k) = E$ and $\Phi _k(\Omega \cap U_k) = E^-$; and moreover, $\Gamma \subset \bigcup _{k=1}^N U_k$. For all $k \in \{ 1,\dots ,N \}$ fix $w_k \in \Omega \cap U_k$. Again by compactness there are $N' \in \{ N+1,N+2,\ldots \}$ and for all $k \in \{ N+1,\ldots ,N' \}$ there are $w_k \in \Omega$ and $r_k > 0$ such that $B(w_k,r_k) \subset \Omega$ and

$$\begin{aligned} \overline{\Omega } \subset \bigcup _{k=1}^N U_k \cup \bigcup _{k=N+1}^{N'} B(w_k,r_k) . \end{aligned}$$

By Lemma A.2 there are $c_1,c_2 \ge 1$ such that $d(x,y) \le c_1 \, |x-y|$ and $|x-y| \le c_2$ for all $k \in \{ 1,\ldots ,N \}$ and $x,y \in \Omega \cap U_k$. Without loss of generality we may assume that $2 r_k \le c_2$ for all $k \in \{ N+1,\ldots ,N' \}$. For simplicity write $U_k = B(w_k,r_k)$ for all $k \in \{ N+1,\ldots ,N' \}$. Then $d(x,y) \le c_1 \, |x-y|$ and $|x-y| \le c_2$ for all $k \in \{ N+1,\ldots ,N' \}$ and $x,y \in U_k$.

We next prove that the geometric distance d is bounded on $\Omega$. Define $M = 2 c_2 + \max \{ d(w_k,w_l) : k,l \in \{ 1,\ldots ,N' \} \}$. Let $x,y \in \Omega$. Then there are $k,l \in \{ 1,\ldots ,N' \}$ such that $x \in U_k$ and $y \in U_l$. Hence, $d(x,y) \le d(x,w_k) + d(w_k,w_l) + d(w_l,y) \le M$. Therefore, d is bounded by M.

Finally, suppose that there is no $c > 0$ such that $d(x,y) \le c \, |x-y|$ for all $x,y \in \Omega$. Then for all $n \in \mathbb {N}$ there are $x_n,y_n \in \Omega$ such that $d(x_n,y_n) > n \, |x_n - y_n|$. It follows that $|x_n - y_n| \le \frac{M}{n}$ for all $n \in \mathbb {N}$. The sequence $(x_n)_{n \in \mathbb {N}}$ is bounded since $\Omega$ is bounded. Passing to a subsequence if necessary, we may assume that the sequence $(x_n)_{n \in \mathbb {N}}$ is convergent. Let $x = \lim _{n \rightarrow \infty } x_n$. Then $\lim _{n \rightarrow \infty } y_n = x$ and $x \in \overline{\Omega }$. Since $\overline{\Omega } \subset \bigcup _{k=1}^N U_k \cup \bigcup _{k=N+1}^{N'} B(w_k,r_k)$, there exists a $k \in \{ 1,\ldots ,N' \}$ such that $x \in U_k$. Because $U_k$ is open there exists an $N_0 \in \mathbb {N}$ such that $x_n \in U_k$ and $y_n \in U_k$ for all $n \in \mathbb {N}$ with $n \ge N_0$. Finally, choose $n \in \mathbb {N}$ such that $n \ge \max \{ N_0,c_1 \}$. Then

$$\begin{aligned} n \, |x_n - y_n| < d(x_n,y_n) \le c_1 \, |x_n - y_n| \le n \, |x_n - y_n| . \end{aligned}$$

This is a contradiction. $\square$

Now we are able to prove the proposition.

Proof of Proposition A.1

Let $c > 0$ be as in Lemma A.3. Let $x,y \in \Omega$ and $n \in \mathbb {N}$. Since the case $x=y$ is trivial, we may assume that $x \ne y$. There exist $T > 0$ and a Lipschitz curve $\gamma :[0,T] \rightarrow \Omega$ such that $\gamma (0) = x$, $\gamma (1) = y$ and $\ell (\gamma ) \le 2 d(x,y)$. For all $k \in \{ 1,\ldots ,n-1 \}$ let

$$\begin{aligned} t_k = \min \{ t \in [0,T] : \ell (\gamma |_{[0,t]}) = \frac{k \, \ell (\gamma )}{n} \} , \end{aligned}$$

which exists by continuity. Set $x_k = \gamma (t_k)$. Further define $x_0 = x$ and $x_n = y$. Then

$$\begin{aligned} |x_{k+1} - x_k| \le d(x_{k+1},x_k) \le \frac{\ell (\gamma )}{n} \le \frac{2 d(x,y)}{n} \le \frac{2 c}{n} \, |x-y| \end{aligned}$$

for all $k \in \{ 0,\ldots ,n-1 \}$, as required. $\square$