Temporal-spatial decays for the Stokes flows in half spaces

Abstract

In this paper, we prove new estimates on temporal-spatial decays in $L^1$ for solutions to the Stokes equations in the half spaces. Our approach is based on the Ukai’s solution formula of the Stokes equations and heat kernel estimates.

1 Introduction

In this paper, we consider the Stokes equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u -\Delta u +\nabla p=0 \ \ \ &{}\text {in} \ \ \mathbb {R}^n_+\times (0, \infty ),\\ \nabla \cdot u = 0 &{}\text {in} \ \ \mathbb {R}^n_+\times (0, \infty ),\\ u(x,t)=0 &{}\text {on} \ \ \partial \mathbb {R}^n_+\times (0, \infty ),\\ u(x,0)=u_0 &{}\text {in} \ \ \mathbb {R}^n_+.\\ \end{array}\right. } \end{aligned}$$

(1)

where

• $n\ge 2$, and $\mathbb {R}^n_+=\{x=(x',x_n): x'\in {\mathbb {R}}^{n-1}, x_n>0\}$ is the upper-half space of ${\mathbb {R}}^n$;

• $u=(u_1(x,t), u_2(x,t),\ldots , u_n(x,t))$ and $p=p(x,t)$ are unknown velocity vector and the pressure, respectively;

• the initial data $u_0=u_0(x)$ is assumed to satisfy a compatibility condition: $\nabla \cdot u_0 =0$ in ${\mathbb {R}}^n_+$ and the normal component of $u_0$ equals to zero on $\partial \mathbb {R}^n_+$.

The estimates of the solution to the Stokes equation play an important role in the study of the temporal and the spatial decays for weak solutions to the Navier–Stokes equation. See [6, 10, 12, 13, 15] and the references therein.

Denote by A the Stokes operator $-P\Delta$ in $\mathbb {R}^n_+$, where P is the projection: $L^r(\mathbb {R}^n_+)\rightarrow L^r_\sigma (\mathbb {R}^n_+)$, $1<r<\infty$. Then the solution to (1) can be written as $u(t,x)=e^{-tA} u_0$. In the whole space $\mathbb {R}^n$, the flow $e^{-tA}$ behaves similarly to the heat Laplacian semigroup $e^{t\Delta }$; hence,

$$\begin{aligned} \Vert e^{-tA} u_0\Vert _{L^q(\mathbb {R}^n)}\lesssim t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}\Vert u_0\Vert _{L^p(\mathbb {R}^n)} \end{aligned}$$

for all $1\le p\le q\le \infty$. However, this is not the case in the half-space $\mathbb {R}^n_+$. Precisely, in the half-space case, we have

$$\begin{aligned} \Vert e^{-tA} u_0\Vert _{L^q(\mathbb {R}^n_+)}\lesssim t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}\Vert u_0\Vert _{L^p(\mathbb {R}^n_+)} \end{aligned}$$

(2)

for all $1< p\le q\le \infty$ or $1\le p< q\le \infty$. See [10]. It is worth noticing that unlike the whole space case, (2) does not hold true for $p=q=1$. In fact, it was proved in [7] that there exists a function $u_0\in L^1(\mathbb {R}_+^n)$ such that $e^{-tA} u_0\notin L^1(\mathbb {R}_+^n)$. In order to deal with $L^1(\mathbb {R}_+^n)$ estimates, we need certain extra condition on the initial data $u_0$. We first recall the main result in [1, Theorem 1.1].

Theorem A. Assume that the initial date $u_0$ satisfies $\nabla \cdot u =0$ in $\mathbb {R}^n_+$ and

$$\begin{aligned} \int _{\mathbb {R}^{n-1}}u_0(y',y_n)dy'=0 \ \ \ a.e.\ \ \ y_n> 0. \end{aligned}$$

(3)

Then, there is a constant C independent of $u_0$ such that

$$\begin{aligned} \Vert e^{-tA}u_0\Vert _{{\mathcal {H}}^1(\mathbb {R}_+^n)}\le Ct^{-1}\int _{\mathbb {R}^n_+}y_n|y'||u_0(y)|dy \end{aligned}$$

(4)

and

$$\begin{aligned} \Vert e^{-tA}u_0\Vert _{{\mathcal {H}}^1(\mathbb {R}_+^n)}\le Ct^{-1/2}\int _{\mathbb {R}^n_+}|y'||u_0(y)|dy, \end{aligned}$$

(5)

where

$$\begin{aligned} {\mathcal {H}}^1(\mathbb {R}_+^n)=\inf \left\{ \Vert F\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}: F\in {\mathcal {H}}^1(\mathbb {R}^n), F|_{\mathbb {R}^n_+}=f\right\} \end{aligned}$$

with ${\mathcal {H}}^1(\mathbb {R}^n)$ being the Hardy space on $\mathbb {R}^n$. (See (19) for the precise definition of the Hardy space).

Then in [12] a stronger condition than (3) is considered. A vector $a=(a_1, a_2,\ldots , a_n)$ on $\mathbb {R}^n_+$, $n\ge 2$ satisfies the tangential parity condition if

• (i) for $1\le j,k\le n-1$ ($n\ge 3$) with $j \ne k$, $a_j(x_1,x_2,\ldots ,x_{n-1},x_n)$ is odd in $x_j$ and even in $x_k$; and $a_1(x_1,x_2)$ is odd in $x_1$ if $n=2$;

• (ii) for $1\le j\le n-1$, $a_n(x_1,x_2,\ldots ,x_{n-1},x_n)$ is even in each $x_j$.

Then it was proved that

Theorem B. [12] Assume that the initial date $u_0=(u^1_{0}, u^2_{0},\ldots , u^n_{0})$ satisfies $\nabla \cdot u =0$ in $\mathbb {R}^n_+$ and $u^n_{0}|_{\partial \mathbb {R}^n_+}=0$. Suppose further that $u_0$ satisfies the tangential parity condition. Then, for any $0<\alpha \le 1$ and $t>0$

$$\begin{aligned} \Vert e^{-tA}u_0\Vert _{L^1(\mathbb {R}_+^n)}\le Ct^{-\frac{\alpha }{2}}\int _{\mathbb {R}^n_+}|x|^{\alpha }|u_0(x)|dx. \end{aligned}$$

(6)

It is important to note that the method in [1] used to obtain the estimates (4) and (5) can not be applied to derive the estimate (6). This is a reason why in [12], the tangential parity condition is assumed to obtain the estimate (6). We would like to emphasize that the tangential parity condition is stronger than condition (3). It is worth noticing that such estimates (4) and (5) are not useful to study the long-time behaviour of solutions to the corresponding Navier–Stokes equation under the tangential parity condition. See for example [12].This show the motivation of the estimate (6).

The main aim of this paper is to prove the following general estimate.

Theorem 1.1

Assume that the initial date $u_0$ satisfies $\nabla \cdot u_0 =0$ in $\mathbb {R}^n_+$ and the condition (3). Then for any $t>0$, and $0< \alpha \le 1$ and $0\le \beta \le 1$,

$$\begin{aligned} \Vert e^{-tA}u_0\Vert _{{\mathcal {H}}^1(\mathbb {R}_+^n)}\le Ct^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |u_0(y)|dy. \end{aligned}$$

(7)

Some comments are in order:

• (a) Since the tangential parity condition is stronger than (3), Theorem 1.1 is still true under the tangential parity condition. Therefore it is also interesting to study the $L^1$ decay estimates for the solutions to the Navier–Stokes equations. However, we do not pursue this problem in this paper. Such a result should be elsewhere.

• (b) It is easy to see that the estimate (7) turns out to be the estimates (4) and (5) when we choose $(\alpha , \beta )=(1,1)$ and $(\alpha , \beta )=(1,0)$. Moreover, using the inequality $|y'|\le |y|$ and $y_n\le |y|$, the estimate (7) deduces, for $\alpha \in (0,2]$

  $$\begin{aligned} \Vert e^{-tA}u_0\Vert _{L^1(\mathbb {R}_+^n)}\le Ct^{-\frac{\alpha }{2}}\int _{\mathbb {R}^n_+}|x|^{\alpha }|u_0(x)|dx, \end{aligned}$$

  which extends the range of $\alpha \in (0,1]$ in (6) under the weaker assumption (3).

We would like to point out that the approaches in [1, 12] can not be applicable to derive the estimate (7). More importantly, although the decay estimates in [1, 12] hold true, there are gaps in their proofs. For example (please see the definitions of $\Lambda$ and E(t) in Sect. 3), both papers used the fact that $eE(t)u_0\in {\mathcal {H}}^1(\mathbb {R}^n)$, but this statement was not proved and it is not clear if this is true. See [1, p.17] and [12, p.1542]. Moreover, one of the key ingredients is proving the following $\partial _j \Lambda ^{-1}eE(t)u_0\in {\mathcal {H}}^1(\mathbb {R}^n)$. However, what was proved in [1, 12] is $\partial _j \Lambda ^{-1}E(t)u_0\in {\mathcal {H}}^1(\mathbb {R}^n)$. See the proof of [1, Lemma 3.3] and [12, Lemma 2.2]. Therefore, our approach fills in the gaps of the proofs of Theorem A and Theorem B in [1, 12].

The organization of the paper is as follows. In Sect. 2, we prove some kernel estimates which play an essential role in proving temporal-spatial decay estimates for the Stokes equation. The proof of Theorem 1.1 will be given in Sect. 3.

Throughout the paper, we always use C and c to denote positive constants that are independent of the main parameters involved but whose values may differ from line to line. We will write $A\lesssim B$ if there is a universal constant C so that $A\le CB$ and $A\sim B$ if $A\lesssim B$ and $B\lesssim A$.

2 Heat kernel estimates

In this section, we prove a number of kernel estimates which play an important role in the proof of the main result.

Lemma 2.1

Let $d\in {\mathbb {N}}$ and $\alpha >0$. Then we have

$$\begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx \lesssim t^{-\alpha } \end{aligned}$$

for all $y\in {\mathbb {R}}^d$ and $t>0$. As a consequence,

$$\begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{\frac{d+\alpha }{2}}}e^{-\frac{|x-y|^2}{c(s+t)}}dx \lesssim t^{-\alpha /2}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}&\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx\\&=\int _{|x-y|<t}\sup _{s>0}\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx\\&\ \ \ \ \ \ \ +\sum _{k=1}^\infty \int _{2^{k-1}t\le |x-y|<2^kt}\sup _{s>0}\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx. \end{aligned} \end{aligned}$$

(8)

It is obvious that

$$\begin{aligned} \begin{aligned} \int _{|x-y|<t}\sup _{s>0}\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx&\le \int _{|x-y|<t} \frac{1}{t^{d+\alpha }} dx \\&\le t^{-\alpha }. \end{aligned} \end{aligned}$$

It remains to estimate the sum in (8). To this end, we write

$$\begin{aligned} \begin{aligned} \sum _{k=1}^\infty \int _{2^{k-1}t\le |x-y|<2^kt}&\sup _{s>0}\frac{1}{(s+t)^{d+\alpha }}\Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}dx\\&\lesssim \sum _{k=1}^\infty \int _{2^{k-1}t\le |x-y|<2^kt}\frac{1}{|x-y|^{d+\alpha }}dx\\&\lesssim \sum _{k=1}^\infty (2^kt)^{-\alpha } \sim t^{-\alpha }. \end{aligned} \end{aligned}$$

This proves the first inequality.

The second inequality follows directly from the first one and the following inequality

$$\begin{aligned} e^{-\frac{|x-y|^2}{c(s+t)}}\lesssim \Big (1+\frac{|x-y|}{s+t}\Big )^{-(d+\alpha )}. \end{aligned}$$

This completes our proof. $\square$

Lemma 2.2

Let $\epsilon , \alpha , \gamma >0$. Then there exists C such that

• (i) for every $y\in \mathbb {R}^d$ and $t>0$,

  $$\begin{aligned} \int _{\mathbb {R}^d} \frac{1}{t^{d}}\Big (\frac{t}{t+|x-y|}\Big )^{d+\epsilon }dx \le C; \end{aligned}$$

• (ii) for every $M, t>0$,

  $$\begin{aligned} \int _0^\infty \frac{s^\alpha }{(s+t)^{\alpha +\gamma }}\Big (1+\frac{M}{s+t}\Big )^{-(\alpha +\gamma )}\frac{ds}{s}\le \frac{C}{(t+M)^\gamma }. \end{aligned}$$

Proof

(i) The proof of this item is simple and we omit the details.

(ii) We write

$$\begin{aligned} \begin{aligned} \int _0^\infty \frac{s^\alpha }{(s+t)^{ \alpha +\gamma }}\Big (1+\frac{M}{s+t}\Big )^{ -( \alpha +\gamma )}\frac{ds}{s}&= \int _0^{t+M}\cdots +\int _{t+M}^\infty \ldots \\&=E_1 + E_2. \end{aligned} \end{aligned}$$

For the term $E_1$, we have

$$\begin{aligned} \begin{aligned} E_1&\lesssim \int _0^{t+M}\frac{s^\alpha }{(s+t)^{ \alpha +\gamma }} \Big (\frac{s+t}{t+M}\Big )^{ \alpha +\gamma }\frac{ds}{s}\\&\sim \Big (\frac{1}{t+M}\Big )^{\gamma }. \end{aligned} \end{aligned}$$

For the term $E_2$, it is obvious that

$$\begin{aligned} \begin{aligned} E_2&\lesssim \int _{t+M}^\infty \frac{s^\alpha }{(s+t)^{\alpha +\gamma }}\frac{ds}{s}\\&\lesssim \int _{t+M}^\infty \frac{1}{s^\gamma }\frac{ds}{s}\\&\sim \Big (\frac{1}{t+M}\Big )^{\gamma }. \end{aligned} \end{aligned}$$

This completes our proof. $\square$

Lemma 2.3

Let $d\in {\mathbb {N}}$, $\alpha \in (0,1]$ and $\gamma \in [0,\alpha )$. Then for any $y\in {\mathbb {R}}^d$ and $t>0$, we have

$$\begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx \lesssim t^{-\alpha /2+\gamma /2}|y|^{\alpha }. \end{aligned}$$

Proof

We have

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx&\lesssim \int _{{\mathbb {R}}^d}\sup _{s: s+t<|y|^2}\ldots +\int _{{\mathbb {R}}^d}\sup _{s: s+t\ge |y|^2}\ldots \\&=: E +F. \end{aligned} \end{aligned}$$

For the term E, using Lemma 2.1,

$$\begin{aligned} \begin{aligned} E&\le \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|y|^{\alpha }}{(s+t)^{\alpha /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx\\&\le \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|y|^{\alpha }}{(s+t)^{\alpha /2}}e^{-\frac{|x-y|^2}{4(s+t)}} dx\\&\quad + \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2}}\frac{|y|^{\alpha }}{(s+t)^{\alpha /2-\gamma /2}} e^{-\frac{|x|^2}{4(s+t)}}dx\\&\le \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned} \end{aligned}$$

For the second term F, we have

$$\begin{aligned} \begin{aligned} F&\le \int _{{\mathbb {R}}^d}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&\ \ \ +\int _{{\mathbb {R}}^d}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x|^2+|y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx\\&=F_1 + F_2. \end{aligned} \end{aligned}$$

Using the inequality $e^u -1 \lesssim u^\alpha$ for $\alpha , u\in [0,1]$, we have

$$\begin{aligned} \begin{aligned} F_2&=\int _{{\mathbb {R}}^d}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big [e^{ \frac{ |y|^2}{4(s+t)}}-1\Big ]dx\\&\lesssim \int _{{\mathbb {R}}^d}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{|y|^{\alpha }}{(s+t)^{\alpha /2}}e^{-\frac{|x|^2+|y|^2}{4(s+t)}} dx\\&\lesssim \frac{|y|^{\alpha }}{t^{\alpha /2-\gamma /2}}, \end{aligned} \end{aligned}$$

where in the last inequality we used Lemma 2.1.

In order to take case of the term $F_1$, we write

$$\begin{aligned} \begin{aligned} F_1&= \int _{\Omega _{1}(y)}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}} \Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&\ \ \ \ \\&\quad +\int _{\Omega _{2}(y)}\sup _{s: s+t\ge |y|^2}\frac{1}{(s+t)^{d/2-\gamma /2}} \Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&=F_{11} + F_{12}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \Omega _{1}(y)=\{x: \langle x,y\rangle \ge 0\} \ \ \ \ \text {and} \ \ \ \ \Omega _{2}(y)=\{x: \langle x,y\rangle < 0\}. \end{aligned}$$

Hence, it suffices to show that

$$\begin{aligned} F_{11} + F_{12}\lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

To do this, we consider these two terms separately.

$\underline{\hbox {Estimate of }F_{11}}$

We have

$$\begin{aligned} \begin{aligned} F_{11}&\lesssim \int _{\Omega _{1}(y)}\sup _{s: s+t\ge \max \{|y|^2, \langle x,y\rangle /2 \}}\frac{1}{(s+t)^{d/2-\gamma /2}} \Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&\ \ \ +\int _{\Omega _{1}(y)}\sup _{s: |y|^2\le s+t<\langle x,y\rangle /2 }\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&=: G_1 + G_2. \end{aligned} \end{aligned}$$

Using $e^u -1 \lesssim u^\alpha$ for $\alpha , u\in [0,1]$,

$$\begin{aligned} \begin{aligned} G_1&=\int _{\Omega _{1}(y)}\sup _{s: s+t\ge \max \{|y|^2, \langle x,y\rangle /2 \}}\frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big [e^{\frac{\langle x,y\rangle }{2(s+t)}}-1\Big ]dx\\&\lesssim \int _{\Omega _{1}(y)}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{\langle x,y\rangle ^\alpha }{(s+t)^\alpha }e^{-\frac{|x|^2+|y|^2}{4(s+t)}}dx\\&\lesssim \int _{\Omega _{1}(y)}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{(|x||y|)^\alpha }{(s+t)^\alpha }e^{-\frac{|x|^2+|y|^2}{4(s+t)}}dx. \end{aligned} \end{aligned}$$

Owing to the following inequality

$$\begin{aligned} \frac{(|x||y|)^\alpha }{(s+t)^\alpha }e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\lesssim \frac{ |y| ^\alpha }{(s+t)^{\alpha /2}}e^{-\frac{|y|^2}{8(s+t)}}, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} G_1&\lesssim \int _{\Omega _{1}(y)}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|y|^\alpha }{(s+t)^\alpha }e^{-\frac{|x|^2}{8(s+t)}}dx\\&\lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}, \end{aligned} \end{aligned}$$

where in the last inequality we used Lemma 2.1.

For the term $G_2$, we have

$$\begin{aligned} \begin{aligned} G_2&\lesssim \int _{\Omega _{1}(y)}\sup _{s: s+t<\langle x,y\rangle /2 } \frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\quad + \int _{\Omega _{1}(y)}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{\langle x,y\rangle ^\alpha }{(s+t)^\alpha } e^{-\frac{|x|^2+|y|^2}{4(s+t)}} dx\\&=: G_{21}+G_{22}. \end{aligned} \end{aligned}$$

Similarly to the estimate of $G_1$, we have

$$\begin{aligned} G_{22}\lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

For the term $G_{21}$, we write

$$\begin{aligned} \begin{aligned} G_{21}&=\int _{\Omega _{1}(y)\cap \{|x|\le 2|y|\}}\sup _{s: s+t<\langle x,y\rangle /2 }\frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx \\&\ \ \ +\int _{\Omega _{1}(y)\cap \{|x|> 2|y|\}}\sup _{s: s+t<\langle x,y\rangle /2 }\frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\lesssim \int _{\Omega _{1}(y)\cap \{|x|\le 2|y|\}}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{\langle x,y\rangle ^{\alpha /2}}{(s+t)^{\alpha /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx \\&\ \ \ +\int _{\Omega _{1}(y)\cap \{|x|> 2|y|\}}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{\langle x,y\rangle ^\alpha }{(s+t)^\alpha } e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&=H_{1} +H_{2}. \end{aligned} \end{aligned}$$

Using Lemma 2.1,

$$\begin{aligned} \begin{aligned} H_1&\lesssim \int _{\Omega _{1}(y)\cap \{|x|\le 2|y|\}}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{(|x||y|)^{\alpha /2}}{(s+t)^{\alpha /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\lesssim \int _{{\mathbb {R}}^d}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{|y|^{\alpha }}{(s+t)^{\alpha /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned} \end{aligned}$$

For the term $H_2$, note that $|x-y|\sim |x|$ as $|x|>2|y|$. Hence, by Lemma 2.1,

$$\begin{aligned} \begin{aligned} H_2&\lesssim \int _{\Omega _{1}(y)\cap \{|x|> 2|y|\}}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{(|x||y|)^\alpha }{(s+t)^\alpha } e^{-\frac{|x-y|^2}{8(s+t)}}e^{-\frac{|x|^2}{c(s+t)}}dx\\&\lesssim \int _{\Omega _{1}(y)\cap \{|x|> 2|y|\}}\sup _{s>0 }\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|y|^\alpha }{(s+t)^{\alpha /2}} e^{-\frac{|x-y|^2}{8(s+t)}}dx\\&\lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}, \end{aligned} \end{aligned}$$

where in the second inequality we used the inequality

$$\begin{aligned} \frac{|x|^\alpha }{(s+t)^{\alpha /2}}e^{-\frac{|x|^2}{c(s+t)}}\lesssim 1. \end{aligned}$$

Collecting the estimates of $H_1, H_2, G_{21}, G_{22}$, we come up with

$$\begin{aligned} G_2 \lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

Taking this and the estimate of $G_1$ into account, we have

$$\begin{aligned} F_{11} \lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

$\underline{\hbox {Estimate of} F_{12}}$

We have

$$\begin{aligned} \begin{aligned} F_{12}&\lesssim \int _{\Omega _{2}(y)}\sup _{s: s+t\ge \max \{|y|^2, |\langle x,y\rangle |/2 \}}\frac{1}{(s+t)^{d/2-\gamma /2}} \Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&\ \ \ +\int _{\Omega _{2}(y)}\sup _{s: |y|^2\le s+t<|\langle x,y\rangle |/2 }\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx\\&=: M_1 + M_2. \end{aligned} \end{aligned}$$

We take care of $M_2$ first. Since $\langle x,y\rangle <0$ in this situation, we have

$$\begin{aligned} \begin{aligned} M_2&\le \int _{\Omega _{2}(y)}\sup _{s: |y|^2\le s+t<|\langle x,y\rangle |/2 }\frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\lesssim \int _{\Omega _{2}(y)\cap \{|x|\le 2|y|\}]}\sup _{s: |y|^2\le s+t<|\langle x,y\rangle |/2 }\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|\langle x,y\rangle |^{\alpha /2}}{(s+t)^{\alpha /2}}e^{-\frac{|x-y|^2}{4(s+t)}}dx\\&\ \ \ + \int _{\Omega _{2}(y)\cap \{|x|>|y|\}]}\sup _{s: |y|^2\le s+t<|\langle x,y\rangle |/2 }\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|\langle x,y\rangle |^{\alpha }}{(s+t)^{\alpha }}e^{-\frac{|x-y|^2}{4(s+t)}}dx. \end{aligned} \end{aligned}$$

At this stage, arguing similarly to the estimates of $H_1$ and $H_2$, we have

$$\begin{aligned} M_2\lesssim \frac{|y|^{\alpha }}{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

It remains to estimate $M_1$. We have

$$\begin{aligned} \begin{aligned} M_1&= \int _{\Omega _{2}(y)}\sup _{s: s+t\ge \max \{|y|^2, |\langle x,y\rangle |/2 \}}\frac{1}{(s+t)^{d/2-\gamma /2}} e^{-\frac{|x-y|^2}{4(s+t)}} \Big [ e^{ \frac{|\langle x, y \rangle |}{2(s+t)}}-1\Big ]dx. \end{aligned} \end{aligned}$$

This, along with the following inequality $e^u -1 \lesssim u^\alpha$ for $\alpha , u\in [0,1]$, yields

$$\begin{aligned} \begin{aligned} M_1&\lesssim \int _{\Omega _{2}(y)\cap \{|x|\le 2|y|\}}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|\langle x,y\rangle |^{\alpha /2}}{(s+t)^{\alpha /2}}e^{-\frac{|x-y|^2}{4(s+t)}} dx\\&\ \ \ +\int _{\Omega _{2}(y)\cap \{|x|> 2|y|\}}\sup _{s> 0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|\langle x,y\rangle |^{\alpha }}{(s+t)^{\alpha }}e^{-\frac{|x-y|^2}{4(s+t)}} dx. \end{aligned} \end{aligned}$$

At this stage, using the argument in the estimations of $H_1$ and $H_2$, we arrive at

$$\begin{aligned} M_1 \lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

This, along with the estimate of $M_2$, implies

$$\begin{aligned} F_{12} \lesssim \frac{|y|^\alpha }{t^{\alpha /2-\gamma /2}}. \end{aligned}$$

This completes our proof. $\square$

Lemma 2.4

For every $\alpha \in (0,1]$ and $\gamma \in [0,\alpha )$, we have

$$\begin{aligned} \int _{{\mathbb {R}}} \sup _{s>0} \frac{1}{(s+t)^{1/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x+y|^2}{4(s+t)}}\Big |dx\lesssim y^\alpha t^{-\alpha /2+\gamma /2} \end{aligned}$$

for all $y>0$.

Proof

We write

$$\begin{aligned}&\int _{{\mathbb {R}}} \sup _{s>0} \frac{1}{(s+t)^{1/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x+y|^2}{4(s+t)}}\Big |dx\\&\quad \le \int _{{\mathbb {R}}} \sup _{s>0} \frac{1}{(s+t)^{1/2-\gamma /2}}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx\\&\qquad +\int _{{\mathbb {R}}} \sup _{s>0} \frac{1}{(s+t)^{1/2-\gamma /2}}\Big |e^{-\frac{|x+y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx. \end{aligned}$$

The desired estimate then follows from Lemma 2.3. $\square$

Lemma 2.5

Let $d\in {\mathbb {N}}$, $\alpha \in (0,1]$, $\gamma \in [0,1]$ and $j=1,\ldots , d$. Then we have, for any $y\in \mathbb {R}^d$,

$$\begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |\partial _{x_j}\Big [e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big ]\Big |dx \lesssim t^{-\frac{\alpha +1-\gamma }{2}}|y|^{\alpha }. \end{aligned}$$

Proof

We have

$$\begin{aligned} \begin{aligned} \partial _{x_j}\Big [e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big ]&=\partial _{x_j}\Big [e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big ]+\partial _{x_j}\Big [e^{-\frac{|x|^2+|y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big ]\\&=-\frac{x_j-y_j}{2(s+t)}\Big [e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big ] + \frac{y_j}{2(s+t)}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\\&\ \ \ -\frac{x_j}{2(s+t)}\Big [e^{-\frac{|x|^2+|y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big ] \end{aligned} \end{aligned}$$

As a consequence,

$$\begin{aligned} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\Big |\partial _{x_j}\Big [e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big ]\Big |dx\le E_1 + E_2 + E_3, \end{aligned}$$

where

$$\begin{aligned} E_1= & {} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{|x_j-y_j|}{2(s+t)}\Big |e^{-\frac{|x-y|^2}{4(s+t)}}-e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\Big |dx, \\ E_2= & {} \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{|y_j|}{2(s+t)}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}dx, \end{aligned}$$

and

$$\begin{aligned} E_3=\int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{|x_j|}{2(s+t)}\Big |e^{-\frac{|x|^2+|y|^2}{4(s+t)}}-e^{-\frac{|x|^2}{4(s+t)}}\Big |dx. \end{aligned}$$

By making use of the following inequalities

$$\begin{aligned} \frac{|x_j-y_j|}{2(s+t)}e^{-\frac{|x-y|^2}{4(s+t)}} \lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{8(s+t)}}, \ \ \ \ \frac{|x_j-y_j|}{2(s+t)}e^{-\frac{|x|^2+|y|^2}{4(s+t)}} \lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x|^2+|y|^2}{8(s+t)}}, \end{aligned}$$

and

$$\begin{aligned} \frac{|x_j|}{2(s+t)}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x|^2+|y|^2}{8(s+t)}}, \ \ \ \ \frac{|x_j|}{2(s+t)}e^{-\frac{|x|^2}{4(s+t)}}\lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x|^2}{8(s+t)}}, \end{aligned}$$

the estimates of $E_1$ and $E_3$ can be obtained by the argument used in the proof of Lemma 2.3. Hence,

$$\begin{aligned} E_1+E_3\lesssim t^{-\frac{\alpha +1-\gamma }{2}}|y|^{\alpha }. \end{aligned}$$

It remains to estimate $E_2$. To do this, using the following inequality

$$\begin{aligned} e^{-\frac{|x|^2+|y|^2}{4(s+t)}}\lesssim \frac{(s+t)^{\frac{1-\alpha }{2}}}{(|x|^2+|y|^2)^{\frac{1-\alpha }{2}}}e^{-\frac{|x|^2+|y|^2}{8(s+t)}}\lesssim \frac{(s+t)^{\frac{1-\alpha }{2}}}{|y|^{1-\alpha }}e^{-\frac{|x|^2+|y|^2}{8(s+t)}}, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} E_2&\lesssim \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}}\frac{(s+t)^{\frac{1-\alpha }{2}}}{|y|^{1-\alpha }}\frac{|y_j|}{2(s+t)}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}dx\\&\lesssim \int _{{\mathbb {R}}^d}\sup _{s>0}\frac{1}{(s+t)^{d/2-\gamma /2}} \frac{|y|^{\alpha }}{2(s+t)^{\frac{\alpha +1}{2}}}e^{-\frac{|x|^2+|y|^2}{4(s+t)}}dx\\&\lesssim t^{-\frac{\alpha +1-\gamma }{2}}|y|^{\alpha }, \end{aligned} \end{aligned}$$

where in the last inequality we used Lemma 2.1. $\square$

Lemma 2.6

Let $\alpha \in [0,1]$ and $c_0, c_1>0$. Then we have

$$\begin{aligned} \int _{0}^\infty \frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} \Big [e^{-\frac{|y-z|^2}{c_0t}}-e^{-\frac{|y+z|^2}{c_0t}}\Big ] dz \lesssim t^{-\alpha /2 }|y|^{\alpha } \frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}} \end{aligned}$$

for every $s, t>0$, $x\in \mathbb {R}$ and $y>0$, and for some $c>0$.

Proof

In order to prove the lemma. We will employ the following simple inequality whose proof will be omitted: let $a>0$. Then there exists $c>0$ such that for any $s,t>0$ and $x,y\in \mathbb {R}$, we have

$$\begin{aligned} \int _{\mathbb {R}}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{as}}e^{-\frac{|y-z|^2}{at}} dz \lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}}. \end{aligned}$$

(9)

We consider two cases: $t<|y|^2$ and $t\ge |y|^2$.

Case 1: $t<|y|^2$

In this situation, we have

$$\begin{aligned} \begin{aligned} \int _{0}^\infty \frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} \Big [e^{-\frac{|y-z|^2}{c_0t}}-e^{-\frac{|y+z|^2}{c_0t}}\Big ] dz&\le \int _{\mathbb {R}}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}}e^{-\frac{|y-z|^2}{c_0t}} dz\\&\lesssim \frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}}\\&\lesssim \frac{y^{\alpha }}{t^{\alpha /2}}\frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}} \end{aligned} \end{aligned}$$

for some $c>0$, where we used (9) in the second inequality.

Case 2: $t\ge |y|^2$

In this case, we have

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} \Big |e^{-\frac{|y-z|^2}{c_0t}}-e^{-\frac{|y+z|^2}{c_0t}}\Big | dz&= \int _{\mathbb {R}_+\cap \{z: yz\ge 4t\}}\cdots + \int _{\mathbb {R}_+\cap \{z: yz< 4t\}}\ldots \\&=: E_1 +E_2. \end{aligned} \end{aligned}$$

For $E_2$, we write

$$\begin{aligned} E_2= \int _{\mathbb {R}_+\cap \{z: yz< 4t\}}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{(y+z)^2}{c_0t}}\Big [e^{\frac{4yz}{c_0t}}-1\Big ] dz. \end{aligned}$$

Using the inequality $e^u -1 \lesssim u^\alpha$ for $\alpha \in [0,1]$ and $u\lesssim 1$, we have

$$\begin{aligned} \begin{aligned} E_2&\lesssim \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{(y+z)^2}{c_0t}}\Big [\frac{yz}{t}\Big ]^{\alpha } dz\\&\lesssim \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{(y+z)^2}{2c_0t}}e^{-\frac{z^2}{2c_0t}}\Big [\frac{yz}{t}\Big ]^{\alpha } dz\\&\lesssim \frac{y^\alpha }{t^{\alpha /2}} \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{|y-z|^2}{2c_0t}} dz\\&\lesssim \frac{y^{\alpha }}{t^{\alpha /2}}\frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}} \end{aligned} \end{aligned}$$

for some $c>0$, where we used $e^{-\frac{(y+z)^2}{c_0t}}\le e^{-\frac{(y+z)^2}{2c_0t}}e^{-\frac{z^2}{2c_0t}}$ (since $y, z\ge 0$) in the second inequality and (9) in the last inequality.

For the term $E_1$, we notice that $yz>4t\ge 4y^2$, and hence $z>4y$. In this situation,

$$\begin{aligned} |y-z|\sim y+z \end{aligned}$$

so that

$$\begin{aligned} \Big |e^{-\frac{|y-z|^2}{c_0t}}-e^{-\frac{|y+z|^2}{c_0t}}\Big | \lesssim e^{-\frac{(y+z)^2}{c't}} \end{aligned}$$

for some $c'>0$.

Therefore,

$$\begin{aligned} \begin{aligned} E_2&\lesssim \int _{\mathbb {R}_+\cap \{z: yz\ge 4t\}}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{(y+z)^2}{c't}}\Big [\frac{yz}{t}\Big ]^{\alpha } dz\\&\lesssim \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{(y+z)^2}{2c't}}e^{-\frac{z^2}{2c't}}\Big [\frac{yz}{t}\Big ]^{\alpha } dz\\&\lesssim \frac{y^\alpha }{t^{\alpha /2}} \int _{\mathbb {R}_+}\frac{1}{\sqrt{st}}e^{-\frac{|x-z|^2}{c_1s}} e^{-\frac{|y-z|^2}{2c't}} dz\\&\lesssim \frac{y^{\alpha }}{t^{\alpha /2}}\frac{1}{\sqrt{s+t}}e^{-\frac{|x-y|^2}{c{(s+t)}}} \end{aligned} \end{aligned}$$

for some $c>0$, where we used the following estimate

$$\begin{aligned} e^{-\frac{z^2}{2c't}}\Big [\frac{yz}{t}\Big ]^{\alpha }\lesssim \frac{y^\alpha }{t^{\alpha /2}} \end{aligned}$$

in the third inequality and used (9) in the last inequality.

This completes our proof. $\square$

Lemma 2.7

Let $d\in \mathbb {N}$, $\alpha \in (0,1]$ and $a>0$. Then we have, for every $x,y\in \mathbb {R}^d$,

$$\begin{aligned}&\int _{0}^\infty \frac{\sqrt{s}}{(t+s)^{\frac{d}{2}}}\Big |\partial _j\Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |\frac{ds}{s} \lesssim \frac{|y|^{\alpha }}{t^{(d+\alpha )/2}} \Big [\Big (1+\frac{|x-y|}{\sqrt{t}}\Big )^{-(d+\alpha ) }\\&\quad +\Big (1+\frac{|x|}{\sqrt{t}}\Big )^{-(d+\alpha )}\Big ] \end{aligned}$$

Proof

We write

$$\begin{aligned} \begin{aligned} \int _{0}^\infty \frac{\sqrt{s}}{(t+s)^{\frac{d}{2}}}\Big |\partial _j\Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |\frac{ds}{s}&= \int _{\{s: s+t <|y|^2\}}\ldots + \int _{\{s: s+t \ge |y|^2\}}\ldots \\&=: E_1 + E_2. \end{aligned} \end{aligned}$$

Using the fact that

$$\begin{aligned} \Big |\partial _j\Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |\lesssim \frac{1}{\sqrt{s+t}}\Big [e^{-\frac{|x-y|^2}{c(t+s)}}+e^{-\frac{|x|^2}{c(t+s)}}\Big ] \ \ \ \hbox { for some}\ c>0, \end{aligned}$$

and Lemma  2.2 we have

$$\begin{aligned} \begin{aligned} E_1&\lesssim \int _{\{s: s+t <|y|^2\}} \frac{|y|^\alpha \sqrt{s}}{(t+s)^{\frac{d+\alpha +1}{2}}}\Big |e^{-\frac{|x-y|^2}{c(t+s)}}+e^{-\frac{|x|^2}{c(t+s)}}\Big |\frac{ds}{s}\\&\lesssim \int _{0}^\infty \frac{|y|^\alpha \sqrt{s}}{(t+s)^{\frac{d+\alpha +1}{2}}}\Big |e^{-\frac{|x-y|^2}{c(t+s)}}+e^{-\frac{|x|^2}{c(t+s)}}\Big |\frac{ds}{s}\\&\lesssim \frac{|y|^{\alpha }}{t^{(d+\alpha )/2}} \Big [\Big (1+\frac{|x-y|}{\sqrt{t}}\Big )^{-(d+\alpha )}+\Big (1+\frac{|x|}{\sqrt{t}}\Big )^{-(d+\alpha )}\Big ]. \end{aligned} \end{aligned}$$

For the second term, we note that

$$\begin{aligned} \partial _j \Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]= -\frac{2(x_j-y_j)}{a(t+s)}\Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ] -\frac{2y_j}{a(t+s)}e^{-\frac{|x|^2}{a(t+s)}}. \end{aligned}$$

Hence, for $s+t>|y|^2$, we consider two cases.

Case 1: If $|x|\le 2|y|$, then we have

$$\begin{aligned} \begin{aligned} \Big |\partial _j \Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |&\lesssim \frac{|y|}{t+s} e^{-\frac{|x-y|^2}{c(t+s)}} + \frac{|y_j|}{t+s}e^{-\frac{|x|^2}{c(t+s)}}\\&\lesssim \frac{1}{\sqrt{t+s}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}\Big [e^{-\frac{|x-y|^2}{c(t+s)}} + e^{-\frac{|x|^2}{c(t+s)}}\Big ]. \end{aligned} \end{aligned}$$

Case 2: If $|x|\ge 2|y|$, then we have $|x-y|\sim |y|$. If $|x||y|\ge (s+t)^2$, then we have we have

$$\begin{aligned} \begin{aligned} \Big |\partial _j \Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |&\lesssim \frac{|x-y|}{t+s}\frac{(|x||y|)^\alpha }{(t+s)^{\alpha }}e^{-\frac{|x-y|^2}{c(t+s)}} + \frac{|y_j|}{t+s}e^{-\frac{|x|^2}{c(t+s)}}\\&\lesssim \frac{|x-y|^{1+\alpha }}{(t+s)^{1+\alpha /2}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}e^{-\frac{|x-y|^2}{c(t+s)}} + \frac{|y_j|}{t+s}e^{-\frac{|x|^2}{c(t+s)}}\\&\lesssim \frac{1}{\sqrt{t+s}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}\Big [e^{-\frac{|x-y|^2}{2c(t+s)}} + e^{-\frac{|x|^2}{c(t+s)}}\Big ]. \end{aligned} \end{aligned}$$

Otherwise, if $|x||y|\ge (s+t)^2$, then we write

$$\begin{aligned} \begin{aligned} \Big |e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big |&=e^{-\frac{|x-y|^2}{a(t+s)}}\Big |1-e^{\frac{-\langle x,2x-y\rangle }{a(t+s)}}\Big |\\&\lesssim e^{-\frac{|x-y|^2}{a(t+s)}}\frac{(|2x-y||y|)^\alpha }{(t+s)^{\alpha }}\\&\sim e^{-\frac{|x-y|^2}{a(t+s)}}\frac{(|x-y||y|)^\alpha }{(t+s)^{\alpha }}, \end{aligned} \end{aligned}$$

where in the second inequality we used the following inequality $|1-e^{x}|\lesssim |x|^\alpha$ for all x with $|x|\lesssim 1$.

Therefore,

$$\begin{aligned} \begin{aligned} \Big |\partial _j \Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |&\lesssim \frac{|x-y|^{1+\alpha }}{(t+s)^{1+\alpha /2}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}e^{-\frac{|x-y|^2}{c(t+s)}} + \frac{|y_j|}{t+s}e^{-\frac{|x|^2}{c(t+s)}}\\&\lesssim \frac{1}{\sqrt{t+s}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}\Big [e^{-\frac{|x-y|^2}{2c(t+s)}} + e^{-\frac{|x|^2}{c(t+s)}}\Big ]. \end{aligned} \end{aligned}$$

So, we have proved that

$$\begin{aligned} \Big |\partial _j \Big [e^{-\frac{|x-y|^2}{a(t+s)}}-e^{-\frac{|x|^2}{a(t+s)}}\Big ]\Big |\lesssim \frac{1}{\sqrt{t+s}}\frac{|y|^\alpha }{(t+s)^{\alpha /2}}\Big [e^{-\frac{|x-y|^2}{2c(t+s)}} + e^{-\frac{|x|^2}{c(t+s)}}\Big ]. \end{aligned}$$

This, along with Lemma 2.2, implies

$$\begin{aligned} E_2\lesssim \frac{|y|^{\alpha }}{t^{(d+\alpha )/2}}\Big [\Big (1+\frac{|x-y|}{\sqrt{t}}\Big )^{-(d+\alpha )}+\Big (1+\frac{|x|}{\sqrt{t}}\Big )^{-(d+\alpha )}\Big ] \end{aligned}$$

as desired.

This completes our proof. $\square$

Lemma 2.8

Let $d\ge 2$, $\alpha \in (0,1]$ and $c, c'>0$. Then, for $x=(x',x_d), y=(y',y_d)\in \mathbb {R}^n$ and $j=1,\ldots , d-1$, we have

$$\begin{aligned} \begin{aligned} \int _{0}^\infty \frac{\sqrt{s}}{(t+s)^{\frac{d}{2}}}\Big |\partial _j&\Big [e^{-\frac{|x'-y'|^2}{c(t+s)}}-e^{-\frac{|x'|^2}{c(t+s)}}\Big ]\Big |e^{-\frac{|x_d-y_d|^2}{c'(t+s)}}\frac{ds}{s}\\&\lesssim \frac{|y'|^{\alpha }}{t^{(d+\alpha )/2}} \Big [\Big (1+\frac{|x-y|}{\sqrt{t}}\Big )^{-(d+\alpha )}+\Big (1+\frac{|x-(0',y_d)|}{\sqrt{t}}\Big )^{-(d+\alpha )}\Big ], \end{aligned} \end{aligned}$$

where $(0',y_d)=(0,\ldots ,0,y_d)\in \mathbb {R}^d$.

Proof

The proof of this Lemma is similar to that of Lemma 2.7 and we would like to leave the details for the interested reader. $\square$

3 Temporal-spatial decay estimates for the Stokes equations

This section is devoted to prove the temporal-spatial decay estimates for the Stokes equations in Theorem 1.1.

We first make some conventions of notations. For $x\in \mathbb {R}^n$, we denote its tangential components $(x_1,\ldots , x_{n-1})$ by $x'\in \mathbb {R}^{n-1}$, so that $x=(x',x_n)$. We set $\partial _j =\partial /\partial x_j$, $\nabla '=(\partial _1,\ldots ,\partial _{n-1})$, $\Delta '=\sum _{j=1}^{n-1}\partial ^2_j$, and $\Delta _n=\partial _n^2$.

The Riesz transforms $R_j$ $(j=1,\ldots , n)$, $S_j$ $(j=1,\ldots , n-1)$, and the operator $\Lambda$ are defined by

$$\begin{aligned} {\mathcal {F}}(R_jf)(\xi )=\frac{i\xi _j}{|\xi |}{\mathcal {F}}{\mathcal {f}}(\xi ),\\ {\mathcal {F}}(S_jf)(\xi )=\frac{i\xi _j}{|\xi '|}{\mathcal {F}}{\mathcal {f}}(\xi ),\\ {\mathcal {F}}(\Lambda f)(\xi )= |\xi '|{\mathcal {F}}{\mathcal {f}}(\xi ). \end{aligned}$$

We set $R'=(R_1,\ldots , R_{n-1})$, $S=(S_1,\ldots , S_{n-1})$, and define

$$\begin{aligned} Uf = rR'\cdot S(R'\cdot S +R_n)ef, \end{aligned}$$

where r is the restriction operator from $\mathbb {R}^n$ to $\mathbb {R}^n_+$, and e is the zero extension operator from $\mathbb {R}^n_+$ to $\mathbb {R}^n$ defined by

$$\begin{aligned} ef ={\left\{ \begin{array}{ll} f \ \ \ &{}\text {for} \ \ x_n\ge 0,\\ 0\ \ \ &{}\text {for} \ \ x_n= 0. \end{array}\right. } \end{aligned}$$

We also define the operators E(t) and F(t) by

$$\begin{aligned}&E(t)f(x) =\int _{\mathbb {R}^n_+} \left[ H_t(x-y)-H_t(x'-y',x_n+y_n)\right] f(y)dy,\ \end{aligned}$$

(10)

$$\begin{aligned}&F(t)f(x) =\int _{\mathbb {R}^n_+} \left[ H_t(x-y)+H_t(x'-y',x_n+y_n)\right] f(y)dy, \end{aligned}$$

(11)

where $G_t$ is the Gauss kernel $H_t(x)=(4\pi t)^{-n/2}e^{-|x|^2/4t}$. Note that E(t) and F(t) are the semigroups generated by the Dirichlet Laplacian and Neumann Laplacian, respectively.

We recall the following result in [19] regarding the solution formula of the Stokes equation (1).

Theorem 3.1

The solution to (1) can be expressed as

$$\begin{aligned}&u^n = UE(t)V_1u_0,\end{aligned}$$

(12)

$$\begin{aligned}&u' = E(t)V_2 u_0 - SUE(t)V_1u_0, \end{aligned}$$

(13)

where $V_1u_0 = -S\cdot u_0' + u_0^n$, and $V_2u_0=u_0' +Su_0^n$.

It is easy to see that the solution u defined in Theorem 3.1 is given as a restriction $r{\bar{u}}$ of ${\bar{u}} =({\bar{u}}^1,\ldots , {\bar{u}}^{n-1},{\bar{u}}^n):= ({\bar{u}}',{\bar{u}}^n)$ of the form

$$\begin{aligned}&{\bar{u}}^n = R'\cdot S(R'\cdot S+R_n)eE(t)V_1u_0, \end{aligned}$$

(14)

$$\begin{aligned}&{\bar{u}}' = eE(t)V_2 u_0 - SR'\cdot S(R'\cdot S+R_n)eE(t)V_1u_0. \end{aligned}$$

(15)

In order to prove Theorem 1.1 we need the following representations for ${\bar{u}}$ in [1, 12].

Proposition 3.2

Let ${\bar{u}}$ is given by (14) and (15). Then we have

$$\begin{aligned} {\bar{u}}^n= & {} \sum _{j=1}^{n-1}R_nR_jeE(t)u_0^j-\sum _{j=1}^{n-1}R_j^2eE(t)u_0^n+ \sum _{j,k=1}^{n-1}R_k^2\partial _j\Lambda ^{-1}eE(t)u_0^j\nonumber \\&+\sum _{j=1}^{n-1}R_nR_j\partial _j \Lambda ^{-1}eE(t)u_0^n, \end{aligned}$$

(16)

$$\begin{aligned} {\bar{u}}'= & {} {\widetilde{E}}(t)u_0' + R_n^2\nabla '\cdot \Lambda ^{-1} eE(t)u_0^n +\sum _{k=1}^{n-1}R'\cdot R_keE(t)u_0' \nonumber \\&-\sum _{k=1}^{n-1}R_n R_k \nabla ' \Lambda ^{-1} eE(t)u_0^k \sum _{k=1}^{n-1}R_k^2 \nabla ' \Lambda ^{-1}eE(t)u_0^n +R_nR'eE(t)u_0^n,\nonumber \\ \end{aligned}$$

(17)

where ${\widetilde{E}}(t)$ is an operator defined by

$$\begin{aligned} {\widetilde{E}}(t)f(x)=\int _{\mathbb {R}^n_+} \left[ H_t(x-y)-H_t(x'-y',x_n+y_n)\right] f(y)dy \end{aligned}$$

(18)

for all $x\in \mathbb {R}^n$ and $f\in L^1(\mathbb {R}^n_+)$.

We now recall the definition of the Hardy space ${\mathcal {H}}^1(\mathbb {R}^n)$. See [18]. A function $f\in L^1(\mathbb {R}^n)$ is said to be in Hardy space ${\mathcal {H}}^1(\mathbb {R}^n)$ if

$$\begin{aligned} \Vert f\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)} = \Big \Vert \sup _{t>0}|H_t*f\Vert \Big \Vert _{L^1(\mathbb {R}^n)}<\infty , \end{aligned}$$

(19)

where $H_t$ is the heat kernel of the Laplacian defined by

$$\begin{aligned}H_t(x)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{4s}}. \end{aligned}$$

In the sequel, we denote

$$\begin{aligned} H_{n-1, t}(x') =(4\pi t)^{-\frac{n-1}{2}}e^{-\frac{|x'|^2}{4t}}, \ \ \ x'\in \mathbb {R}^{n-1}, \end{aligned}$$

and

$$\begin{aligned} H_{1,t}(x_n)=(4\pi t)^{-\frac{1}{2}}e^{-\frac{x_n^2}{4t}}, \ \ \ x_n\in \mathbb {R}. \end{aligned}$$

Then we have

$$\begin{aligned} H_t(x)=H_{n-1,t}(x')H_{1,t}(x_n), \ \ \ x=(x',x_n)\in \mathbb {R}^n. \end{aligned}$$

In order to prove Theorem 1.1 we need the following auxiliary estimates.

Proposition 3.3

Assume that $a \in L^1(\mathbb {R}^n_+)$ satisfies

$$\begin{aligned} \int _{\mathbb {R}^{n-1}}a(y',y_n)dy'=0 \ \ \ a.e. \ \ y_n>0, \end{aligned}$$

(20)

then for any $1\le j\le n-1$, $t>0$, $\alpha \in (0,1]$ and $\beta \in [0,1]$,

$$\begin{aligned} \Vert {\widetilde{E}}(t)a\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)} \le Ct^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |a(y)|dy, \end{aligned}$$

(21)

$$\begin{aligned} \Vert \partial _j \Lambda ^{-1}eE(t)a\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\le Ct^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n}|y'|^\alpha y_n^\beta |a(y)|dy, \end{aligned}$$

(22)

and

$$\begin{aligned} \Vert R_jeE(t)a\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\le Ct^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n}|y'|^\alpha y_n^\beta |a(y)|dy. \end{aligned}$$

(23)

Proof

Proof of (21)

Denote by $a^*$ the odd extension of a from $\mathbb {R}^n_+$ to $\mathbb {R}^n$, i.e.

$$\begin{aligned} a^*(x',x_n)={\left\{ \begin{array}{ll} a(x',x_n) \ \ \ &{}\text {if} \ \ x_n\ge 0,\\ -a(x',-x_n) \ \ \ &{}\text {if} \ \ x_n< 0. \end{array}\right. } \end{aligned}$$

Then we have

$$\begin{aligned} {\widetilde{E}}(t)a(x) = e^{t\Delta }a^*(x), \end{aligned}$$

(24)

which implies that for $s>0$,

$$\begin{aligned} e^{s\Delta }{\widetilde{E}}(t)a(x)=e^{(s+t)\Delta }a^*(x). \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \Big \Vert&\sup _{s>0}|e^{s\Delta }{\widetilde{E}}(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)}\\&=\Big \Vert \sup _{s>0}|e^{(s+t)\Delta }a^*(x)|\Big \Vert _{L^1(\mathbb {R}^n)}\\&=\int _{\mathbb {R}^n}\sup _{s>0}\Big |\int _{\mathbb {R}^{n-1}}\int _0^\infty H_{n-1,s+t}(x'-y')[H_{1,s+t}(x_n-y_n)\\&\quad -H_{1,s+t}(x_n+y_n)]a(y',y_n)dy_ndy' \Big |dx. \end{aligned} \end{aligned}$$

Using the fact that $\int _{\mathbb {R}^{n-1}}a(y',y_n)dy'=0$,

$$\begin{aligned}&\Big \Vert \sup _{s>0}|e^{s\Delta }{\widetilde{E}}(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)}\nonumber \\&\quad =\int _{\mathbb {R}^n}\sup _{s>0}\Big |\int _{\mathbb {R}^{n-1}}\int _0^\infty F_{n-1,s+t}(x',y')G_{1,s+t}(x_n,y_n)a(y',y_n)dy_ndy' \Big |dx,\nonumber \\ \end{aligned}$$

(25)

where

$$\begin{aligned} \begin{aligned} G_{n-1, s+t}(x',y')&=H_{n-1,s+t}(x'-y')-H_{n-1,s+t}(x')\\&=\frac{1}{[4\pi (s+t)]^{\frac{n-1}{2}}}\Big [e^{-\frac{|x'-y'|^2}{4(s+t)}}-e^{-\frac{|x'|^2}{4(s+t)}}\Big ], \end{aligned} \end{aligned}$$

(26)

and

$$\begin{aligned} \begin{aligned} F_{1, s+t}(x_n,y_n)&=H_{1,s+t}(x_n-y_n)-H_{1,s+t}(x_n+y_n)\\&=\frac{1}{[4\pi (s+t)]^{1/2}}\Big [e^{-\frac{|x_n-y_n|^2}{4(s+t)}}-e^{-\frac{|x_n+y_n|^2}{4(s+t)}}\Big ]. \end{aligned} \end{aligned}$$

(27)

We consider two cases: $\beta >0$ and $\beta =0$.

Case 1: $\beta >0$

We have

$$\begin{aligned}&\Big \Vert \sup _{s>0} |e^{s\Delta }E(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)} \\&\quad \lesssim \int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |a(y)|dy \times \sup _{y'}\int _{\mathbb {R}^{n-1}} \sup _{s>0}|y'|^{-\alpha }|G_{n-1,s+t}(x',y')|dx'\\&\qquad \times \sup _{y_n}\int _{{\mathbb {R}}}\sup _{s>0}y_n^{-\beta }|F_{1,s+t}(x_n,y_n)|dx_n. \end{aligned}$$

Then applying Lemmas 2.3 and 2.4,

$$\begin{aligned} \begin{aligned} \Big \Vert \sup _{s>0}&|e^{s\Delta }E(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)}\lesssim t^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |a(y)|dy. \end{aligned} \end{aligned}$$

Case 2: $\beta =0$

Take $\gamma \in (0,\alpha )$. Then we have, from (25),

$$\begin{aligned} \begin{aligned}&\Big \Vert \sup _{s>0}|e^{s\Delta }E(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)}\\&\quad \lesssim \int _{\mathbb {R}^n_+} |y'|^\alpha |a(y)|dy\times \sup _{y'}\int _{\mathbb {R}^{n-1}} \sup _{s>0}|y'|^{-\alpha }(s+t)^{\gamma }|G_{n-1,s+t}(x',y')|dx'\\&\qquad \times \sup _{y_n}\int _{{\mathbb {R}}}\sup _{s>0}(s+t)^{-\gamma }|F_{1,s+t}(x_n,y_n)|dx_n. \end{aligned} \end{aligned}$$

Applying Lemma 2.4 to the first supremum and Lemma 2.1 for the second supremum, we find that

$$\begin{aligned} \begin{aligned} \Big \Vert \sup _{s>0}|e^{s\Delta }E(t)a(x)|\Big \Vert _{L^1(\mathbb {R}^n)}&\lesssim t^{-\frac{\alpha -\gamma }{2}-\frac{\gamma }{2}}\int _{\mathbb {R}^n_+} |y'|^\alpha |a(y)|dy\\&\lesssim t^{-\frac{\alpha }{2}}\int _{\mathbb {R}^n_+} |y'|^\alpha |a(y)|dy. \end{aligned} \end{aligned}$$

This completes the proof of (21).

Proof of (22) Note that $\Lambda ^{-1}=(-\Delta ')^{-\frac{1}{2}}$. Hence, by the subordination formula,

$$\begin{aligned} \Lambda ^{-1}=c\int _0^\infty \sqrt{u} e^{u\Delta '}\frac{du}{u}. \end{aligned}$$

It follows that, for $j=1,\ldots , n-1$,

$$\begin{aligned} \begin{aligned} e^{s\Delta }\partial _j \Lambda ^{-1}eE(t)a(x)&=c\int _0^\infty \sqrt{u} \partial _j e^{s\Delta }e^{u\Delta '}eE(t)a(x)\frac{du}{u}. \end{aligned} \end{aligned}$$

This, in combination with (20), implies

$$\begin{aligned} \begin{aligned} e^{s\Delta }\partial _j \Lambda ^{-1}eE(t)a(x)&=\int _0^\infty \int _{\mathbb {R}^{n}_+} \sqrt{u} \partial _j \left[ H_{n-1,s+u+t}(x',y')-H_{n-1,s+u+t}(x',0)\right] \\&\quad \times \int _{0}^\infty H_{1,s}(x_n,z_n)F_{1,t}(z_n,y_n)a(y',y_n)dz_ndy_ndy'\frac{du}{u}, \end{aligned} \end{aligned}$$

where we recall that $F_{1,t}(z_n,y_n)= H_{1,t}(z_n-y_n)-H_{1,t}(z_n+y_n)$.

By Lemmas 2.7 and 2.6,

$$\begin{aligned} \begin{aligned}&\int _0^\infty \sqrt{u} \left| \partial _j \left[ H_{n-1,s+u+t}(x',y')-H_{n-1,s+u+t}(x',0)\right] \right| \frac{du}{u}\\&\quad \lesssim \frac{|y'|^\alpha }{(s+t)^{\frac{n+\alpha -1}{2}}}\Big [\Big (1+\frac{|x'-y'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}+\Big (1+\frac{|x'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}\Big ] \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \int _{0}^\infty H_{1,s}(x_n,z_n)F_{1,t}(z_n,y_n)dz_n\lesssim \frac{|y_n|^\beta }{t^{\beta /2}} \frac{1}{\sqrt{s+t}}e^{-\frac{|x_n-y_n|^2}{c(s+t)}}. \end{aligned}$$

Consequently,

$$\begin{aligned} |e^{s\Delta }\partial _j \Lambda ^{-1}eE(t)a(x)|\lesssim & {} \int _{\mathbb {R}^{n}_+} \frac{|y'|^{\alpha }y_n^\beta }{t^{\beta /2}(s+t)^{\frac{n+\alpha }{2}}} |a(y',y_n)|\nonumber \\&\times \Big [\Big (1+\frac{|x'-y'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}+\Big (1+\frac{|x'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}\Big ]\nonumber \\&e^{-\frac{|x_n-y_n|^2}{c(s+t)}}dy_ndy'. \end{aligned}$$

(28)

From Lemma 2.1,

$$\begin{aligned}&\sup _y\int _{\mathbb {R}^{n-1}}\sup _{s>0}\frac{1}{(s+t)^{\frac{n-1}{2}+\frac{\alpha }{4}}} \Big [\Big (1+\frac{|x'-y'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}+ \Big (1+\frac{|x'|}{\sqrt{s+t}}\Big )^{-(n+\alpha -1)}\Big ]\\&\quad dx'\lesssim t^{-\frac{\alpha }{4}} \end{aligned}$$

and

$$\begin{aligned} \sup _y\int _{\mathbb {R}} \sup _{s>0}\frac{1}{(s+t)^{\frac{1}{2}+\frac{\alpha }{4}}}e^{-\frac{|x_n-y_n|^2}{c(s+t)}}dx_n\lesssim t^{-\frac{\alpha }{4}}. \end{aligned}$$

These two inequalities and (28) imply that

$$\begin{aligned} \Vert e^{s\Delta }\partial _j \Lambda ^{-1}eE(t)a\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\lesssim t^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |a(y',y_n)|dy \end{aligned}$$

Proof of (23)

Using the subordination formula,

$$\begin{aligned} R_j=c\int _0^\infty \sqrt{u} \partial _j e^{u\Delta }\frac{du}{u}. \end{aligned}$$

we have, for $j=1,\ldots , n-1$,

$$\begin{aligned} \begin{aligned} e^{s\Delta }R_j eE(t)a(x)&=c\int _0^\infty \sqrt{u} \partial _j e^{s\Delta }e^{u\Delta }eE(t)a(x)\frac{du}{u}. \end{aligned} \end{aligned}$$

We then apply (20) to write

$$\begin{aligned} \begin{aligned} e^{s\Delta }R_j eE(t)a(x)&=\int _0^\infty \int _{\mathbb {R}^{n}_+} \sqrt{u} \partial _j \left[ H_{n-1,s+u+t}(x',y')-H_{n-1,s+u+t}(x',0)\right] \\&\ \ \ \ \times \int _{0}^\infty H_{1,s+u}(x_n,z_n)F_{1,t}(z_n,y_n)a(y',y_n)dz_ndy_ndy'\frac{du}{u}. \end{aligned} \end{aligned}$$

Applying Lemma 2.6,

$$\begin{aligned} \int _{0}^\infty H_{1,s+u}(x_n,z_n)F_{1,t}(z_n,y_n)dz_n\lesssim \frac{|y_n|^\beta }{t^{\beta /2}} \frac{1}{\sqrt{s+u+t}}e^{-\frac{|x_n-y_n|^2}{c(s+u+t)}}. \end{aligned}$$

Therefore,

$$\begin{aligned} |e^{s\Delta }R_j eE(t)a(x)|= & {} \int _0^\infty \int _{\mathbb {R}^{n}_+} \frac{|y_n|^\beta }{t^{\beta /2}} \frac{\sqrt{u}}{\sqrt{s+u+t}} \Big |\partial _j \left[ H_{n-1,s+u+t}(x',y')\right. \\&\left. -H_{n-1,s+u+t}(x',0)\right] \Big | \times e^{-\frac{|x_n-y_n|^2}{c(s+u+t)}}|a(y',y_n)|dy_ndy'\frac{du}{u}. \end{aligned}$$

Then applying Lemma 2.8,

$$\begin{aligned} |e^{s\Delta }R_jeE(t)a(x)|\lesssim & {} \int _{\mathbb {R}^{n}_+} \frac{|y'|^{\alpha }y_n^\beta }{t^{\beta /2}(s+t)^{\frac{n+\alpha }{2}}} |a(y',y_n)|\nonumber \\&\times \Big [\Big (1+\frac{|x-y|}{\sqrt{s+t}}\Big )^{-(n+\alpha )}+\Big (1+\frac{|x-(0',y_n)|}{\sqrt{s+t}}\Big )^{-(n+\alpha )}\Big ]\nonumber \\&\quad e^{-\frac{|x_n-y_n|^2}{c(s+t)}}dy_ndy'. \end{aligned}$$

(29)

From Lemma 2.1,

$$\begin{aligned} \sup _y\int _{\mathbb {R}^{n}}\sup _{s>0}\frac{1}{(s+t)^{\frac{n}{2}+\frac{\alpha }{4}}}\Big [\Big (1+\frac{|x-y|}{\sqrt{s+t}}\Big )^{-(n+\alpha )}+\Big (1+\frac{|x-(0',y_n)|}{\sqrt{s+t}}\Big )^{-(n+\alpha )}\Big ] dx\lesssim t^{-\frac{\alpha }{4}}. \end{aligned}$$

This, along with (29), implies that

$$\begin{aligned} \Vert e^{s\Delta }R_j eE(t)a\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\lesssim t^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |a(y',y_n)|dy \end{aligned}$$

$\square$

We are now ready to give the proof of Theorem 1.1.

Proof of Theorem 1.1:

Using Proposition 3.2, we have

$$\begin{aligned} \begin{aligned} \Vert {\bar{u}}^n\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}&\le \sum _{j=1}^{n-1}|||R_n|||_{{\mathcal {H}}^1(\mathbb {R}^n)} \Vert R_jeE(t)u_0^j\Vert _{{\mathcal {H}}^1 (\mathbb {R}^n)}\\&\quad +\sum _{j=1}^{n-1}|||R_j|||_{{\mathcal {H}}^1 (\mathbb {R}^n)}\Vert R_jeE(t)u_0^n\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\\&\ \ \ +\sum _{j,k=1}^{n-1}|||R_k^2|||_{{\mathcal {H}}^1(\mathbb {R}^n)}\Vert \partial _j\Lambda ^{-1}eE(t)u_0^j\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\\&\ \ \ \ +\sum _{j=1}^{n-1}|||R_n|||_{{\mathcal {H}}^1(\mathbb {R}^n)}|||R_j|||_{{\mathcal {H}}^1(\mathbb {R}^n)}\Vert \partial _j \Lambda ^{-1}eE(t)u_0^n\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}. \end{aligned} \end{aligned}$$

At this stage, using Proposition 3.3 and the boundedness of the Riesz transforms $R_j$ on the Hardy space ${\mathcal {H}}^1(\mathbb {R}^n)$ for $j=1,\ldots , n$, we have

$$\begin{aligned} \Vert {\bar{u}}^n\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\lesssim t^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |u_0(y',y_n)|dy. \end{aligned}$$

Similarly,

$$\begin{aligned} \Vert {\bar{u}}'\Vert _{{\mathcal {H}}^1(\mathbb {R}^n)}\lesssim t^{-\frac{\alpha +\beta }{2}}\int _{\mathbb {R}^n_+}|y'|^\alpha y_n^\beta |u_0(y',y_n)|dy. \end{aligned}$$

This completes our proof. $\square$