3D hard sphere Boltzmann equation: explicit structure and the transition process from polynomial tail to Gaussian tail

Yu-Chu Lin Yu-Chu Lin, Department of Mathematics, National Cheng Kung University, Tainan, Taiwan yuchu@mail.ncku.edu.tw , Haitao Wang Haitao Wang, School of Mathematical Sciences, Institute of Natural Sciences, MOE-LSC, IMA-Shanghai, Shanghai Jiao Tong University, Shanghai, China haitallica@sjtu.edu.cn and Kung-Chien Wu Kung-Chien Wu, Department of Mathematics, National Cheng Kung University, Tainan, Taiwan and National Center for Theoretical Sciences, National Taiwan University, Taipei, Taiwan kungchienwu@gmail.com

Abstract.

We study the Boltzmann equation with hard sphere in a near-equilibrium setting. The initial data is compactly supported in the space variable and has a polynomial tail in the microscopic velocity. We show that the solution can be decomposed into a particle-like part (polynomial tail) and a fluid-like part (Gaussian tail). The particle-like part decays exponentially in both space and time, while the fluid-like part corresponds to the behavior of the compressible Navier-Stokes equation, which dominates the long time behavior and exhibits rich wave motion. The nonlinear wave interactions in the fluid-like part are precisely characterized and therefore we are able to distinguish the linear and nonlinear wave of the solution. It is notable that although the solution has polynomial tail in the velocity initially, the transition process from the polynomial to the Gaussian tail can be quantitatively revealed due to the collision with the background global Maxwellian.

Key words and phrases:

Boltzmann equation, Maxwellian, polynomial tail.

2020 Mathematics Subject Classification: 35Q20; 82C40.

1. Introduction

1.1. The model

The Boltzmann equation is a fundamental model in the collisional kinetic theory, which describes the evolution of a phase space distribution function of moderately dilute gas. Precisely, the Boltzmann equation reads

(1)

\left\{\begin{array}[]{l}\partial_{t}F+\xi\cdot\nabla_{x}F=Q(F,F)\text{,}\\[11.38109pt] F(0,x,\xi)=F_{0}(x,\xi)\text{,}\end{array}\right.\quad(t,x,\xi)\in{\mathbb{R}}^{+}\times{\mathbb{R}}^{3}\times{\mathbb{R}}^{3}\text{,}

where $F\left(t,x,\xi\right)$ is the distribution function for particles at time $t>0$ , position $x=\left(x_{1}\text{, }x_{2}\text{, }x_{3}\right)\in\mathbb{R}^{3}$ and microscopic velocity $\xi=\left(\xi_{1}\text{, }\xi_{2}\text{, }\xi_{3}\right)\in\mathbb{R}^{3}$ , and initial data $F_{0}\left(x,\xi\right)\geq 0$ is given. Here the Boltzmann collision operator $Q\left(\cdot,\cdot\right)$ is given by

Q\left(G,F\right)=\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|q\cdot n\right|\left[G\left(\xi_{\ast}^{\prime}\right)F\left(\xi^{\prime}\right)-G\left(\xi_{\ast}\right)F\left(\xi\right)\right]dnd\xi_{\ast}

where $q=\xi-\xi_{\ast}$ is the relative velocity and the post-collisional velocities satisfy

\xi^{\prime}=\xi-\left[\left(\xi-\xi_{\ast}\right)\cdot n\right]n\text{, \ \ \ }\xi_{\ast}^{\prime}=\xi_{\ast}+\left[\left(\xi-\xi_{\ast}\right)\cdot n\right]n\text{.}

It is well known that the global Maxwellians are the steady solutions to the Boltzmann equation. In the perturbation regime near the Maxwellian, we look for the solution in the form of

(2)

F=\mathcal{M}+f\text{, \ \ \ }F_{0}(x,\xi)=\mathcal{M}+\varepsilon f_{0}\text{,}

where $\varepsilon>0$ sufficiently small, with a perturbation function $f$ to $\mathcal{M}$ . Here the global Maxwellian $\mathcal{M}$ is normalized as

\mathcal{M=}\frac{1}{\left(2\pi\right)^{3/2}}\exp\left(-\frac{\left|\xi\right|^{2}}{2}\right)\text{.}

Substituting (2) into (1), the perturbation function $f$ satisfies the equation

(3)

\left\{\begin{array}[]{l}\partial_{t}f+\xi\cdot\nabla_{x}f=\mathcal{L}f+Q\left(f,f\right),\vspace{3mm}\\ f|_{t=0}=F_{0}(x,\xi)-\mathcal{M=}\varepsilon f_{0}\text{, \ \ }\left(x,\xi\right)\in\mathbb{R}^{3}\times\mathbb{R}^{3}\text{,}\end{array}\right.

where

\mathcal{L}f=Q\left(\mathcal{M},f\right)+Q\left(f,\mathcal{M}\right)\text{.}

In fact, $\mathcal{L}$ can be split into

\mathcal{L}f=-\nu+\mathcal{K}\text{,}

with

\nu\left(\xi\right)=\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|q\cdot\omega\right|\mathcal{M}\left(\xi_{\ast}\right)d\omega d\xi_{\ast}\sim\left(1+\left|\xi\right|\right)\text{,}

\mathcal{K}f=\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|q\cdot n\right|\left[\mathcal{M}\left(\xi_{\ast}^{\prime}\right)f\left(\xi^{\prime}\right)+f\left(\xi_{\ast}^{\prime}\right)\mathcal{M}\left(\xi^{\prime}\right)-f\left(\xi_{\ast}\right)\mathcal{M}\left(\xi\right)\right]dnd\xi_{\ast}\text{.}

Note that this kind of perturbation (2) allows the initial data to have a polynomial tail in the microscopic velocity, which is different from the standard perturbation, $F=\mathcal{M}+\sqrt{\mathcal{M}}f$ , where initial data is assumed to have a Gaussian tail.

It is known that there are extensive studies on the standard perturbation, including the global existence, time-asymptotic behavior, and even the pointwise structure, see [7, 16, 27, 28] and the references therein. It would be very interesting to see that can we still obtain the precise space-time structure of the solution for initial data with a polynomial tail?

Moreover, since the perturbation setting describes the collisions between a small amount of released particles and the ambient particles that have reached thermal equilibrium, the physical intuition suggests that the distribution of the released particles will also approach thermal equilibrium over a long period, namely, it will become close to a Gaussian in terms of the microscopic velocity. It is a challenge problem that is it possible to give a quantitative description of the transition process?

The main goal of this paper is to answer the above two questions. Specifically, we will construct a pointwise estimate of solution to (3) with respect to all variables, space, time, and velocity. The estimate not only exhibits the wave motion in space-time, but also reveals how the solution transitions from a polynomial tail to a Gaussian tail.

1.2. Review of previous works

Concerning the polynomial tail perturbation for collisional kinetic equation, there has been substantial progress recently. For the torus case, it was initiated by Gualdani-Mischler-Mouhot [17] for the Boltzmann equation with hard sphere. It was then generalized by [8] for Landau equation, by [1, 18] for the Boltzmann equation without angular cutoff, and by [5] for the Boltzmann equation with soft potentials. For the whole space case, we refer to [9, 6] for the non-cutoff Boltzmann equation and [13] for the cutoff Boltzmann equation. These works mainly focused on global existence and large time decay of the solution, whereas our study aim at providing a more quantitative description of the structure of the solution.

Next, we review some space-time pointwise results closely related to the current study. It was initiated by Liu [22] for 1D viscous conservation laws, then developed to multi-dimensional compressible Navier-Stokes equation [10, 12, 23, 24]. There are two key ingredients. The first is the construction of Green’s function for the linearized system, where rich wave phenomena, such as dissipative Huygens wave, diffusion wave, Riesz wave are identified. The other one is the careful estimate of nonlinear wave couplings between the above basic wave patterns. As is known, the long time behavior of the Boltzmann equation is governed by macroscopic fluid. There are some parallel results for Boltzmann equation with standard Gaussian tail. The result for 1D hard sphere Boltzmann equation was constructed by Liu-Yu [25]. As the nonlinear interaction is strong in 1D, the authors need to extract the so called “kinetic Burger equations” to close the nonlinear problem. Later, [26, 27] obtained the explicit structure of Green’s function for the linearized Boltzmann equation with hard sphere in 3D. Recently, [20] constructed the explicit structure of the relativistic Boltzmann equation for “hard ball”, an exponentially sharp ansatz similar to structure in [10] was justified. In these works, the nonlinear interactions have been estimated to the extent necessary to close the nonlinear ansatz.

The transition from polynomial tail to Gaussian tail is related to the decay estimates for large velocities in the Boltzmann equation. For space homogeneous case, there are extensive studies on $L^{1}_{v}$ moments or pointwise decay, both for polynomial and exponential weight, see [2, 3, 11, 15] and references therein. For space inhomogeneous case, the results are relatively fewer. Generation of polynomial moments in $L^{1}_{v}$ or pointwise sense was established under suitable moment bound conditions for hard potential with or without cutoff, see [4, 17, 19]. Different from the conditional results, our result provides a dynamic process for Gaussian tail generation in the perturbation regime.

1.3. Notations

Before stating our main results, we introduce some notations used in this paper. We denote $w_{\beta}\left(\xi\right)=\left\langle\xi\right\rangle^{\beta}=(1+|\xi|^{2})^{\beta/2}$ , $\beta\in{\mathbb{R}}$ . For the microscopic variable $\xi$ , we denote

|g|_{L_{\xi}^{p}}=\Big{(}\int_{{\mathbb{R}}^{3}}|g|^{p}d\xi\Big{)}^{1/p}\hbox{ if }1\leq p<\infty\hbox{,}\quad\quad|g|_{L_{\xi}^{\infty}}=\sup_{\xi\in{\mathbb{R}}^{3}}|g(\xi)|\hbox{,}

and the weighted norms can be defined by

|g|_{L_{\xi,\beta}^{p}}=\Big{(}\int_{{\mathbb{R}}^{3}}\left|\left\langle\xi\right\rangle^{\beta}g\right|^{p}d\xi\Big{)}^{1/p}\hbox{ if }1\leq p<\infty\hbox{,}\quad\quad|g|_{L_{\xi,\beta}^{\infty}}=\sup_{\xi\in{\mathbb{R}}^{3}}\left|\left\langle\xi\right\rangle^{\beta}g(\xi)\right|\hbox{.}

The $L_{\xi}^{2}$ inner product in ${\mathbb{R}}^{3}$ will be denoted by $\left\langle\cdot,\cdot\right\rangle_{\xi}$ , i.e.,

\left\langle f,g\right\rangle_{\xi}=\int f(\xi)\overline{g(\xi)}d\xi\hbox{.}

For the Boltzmann equation with hard sphere, the natural norm in $\xi$ is $|\cdot|_{L_{\sigma}^{2}}$ , which is defined as

|g|_{L_{\sigma}^{2}}^{2}=\left|\left\langle\xi\right\rangle^{\frac{1}{2}}g\right|_{L_{\xi}^{2}}^{2}\hbox{.}

For the space variable $x$ , we have similar notations, namely,

|g|_{L_{x}^{p}}=\left(\int_{{\mathbb{R}^{3}}}|g|^{p}dx\right)^{1/p}\hbox{ if }1\leq p<\infty\hbox{,}\quad\quad|g|_{L_{x}^{\infty}}=\sup_{x\in{\mathbb{R}^{3}}}|g(x)|\hbox{.}

Finally, with $\mathcal{X}$ and $\mathcal{Y}$ being two normed spaces, we define

\left\|g\right\|_{\mathcal{XY}}=\left|\left|g\right|_{\mathcal{Y}}\right|_{\mathcal{X}}\hbox{,}

and for simplicity, we denote

\left\|g\right\|_{L^{2}}=\left\|g\right\|_{L_{\xi}^{2}L_{x}^{2}}=\left(\int_{\mathbb{R}^{3}}\left|g\right|_{L_{x}^{2}}^{2}d\xi\right)^{1/2}\hbox{.}

For any two functions $f(x,t)$ and $g(x,t)$ , we define the space-time convolution as

f(x,t)\ast_{x,t}g(x,t)=\int_{0}^{t}\int_{{\mathbb{R}}^{3}}f(x-y,t-\tau)g(y,\tau)dyd\tau.

For any two real numbers $a$ and $b$ , we define $a\wedge b=\min\{a,b\}$ .

For simplicity of notations, hereafter, we abbreviate $\leq C$ to $\lesssim$ , where $C$ is a constant depending only on fixed numbers.

1.4. Main theorem and significant points of our results

In order to achieve our goal, we introduce the decomposition $f=f_{1}+\sqrt{\mathcal{M}}f_{2}$ . Here $f_{1}$ corresponds to the part with only polynomial tail, while $\sqrt{\mathcal{M}}f_{2}$ has Gaussian tail. Heuristically, the closer a distribution function to a Gaussian, the closer the behavior resembles macroscopic hydrodynamics. Therefore, one may expect $f_{1}$ behaves like particle, and $f_{2}$ behaves like fluid.

The following coupled system designed for $f_{1}$ and $f_{2}$ is to realize one’s intuition:

(4)

\left\{\begin{array}[]{l}\partial_{t}f_{1}+\xi\cdot\nabla_{x}f_{1}=-\nu\left(\xi\right)f_{1}+\mathcal{K}_{s}f_{1}+Q\left(f_{1},f_{1}\right)+Q\left(f_{1},\sqrt{\mathcal{M}}f_{2}\right)+Q\left(\sqrt{\mathcal{M}}f_{2},f_{1}\right),\vspace{3mm}\\ \partial_{t}f_{2}+\xi\cdot\nabla_{x}f_{2}=Lf_{2}+\mathcal{K}_{b}f_{1}+\Gamma\left(f_{2},f_{2}\right),\end{array}\right.

with initial data

f_{1}\left(0,x,\xi\right)=f_{1,0}\left(x,\xi\right)=\varepsilon f_{0}\left(x,\xi\right)\hbox{, \ \ \ }f_{2}\left(0,x,\xi\right)=f_{2,0}\left(x,\xi\right)=0\hbox{,}

where

Lg=\frac{1}{\sqrt{\mathcal{M}}}\left[Q\left(\mathcal{M},\sqrt{\mathcal{M}}g\right)+Q\left(\sqrt{\mathcal{M}}g,\mathcal{M}\right)\right]

is the so called linearized Boltzmann collision operator,

\Gamma\left(f_{2},f_{2}\right)=\frac{1}{\sqrt{\mathcal{M}}}Q\left(\sqrt{\mathcal{M}}f_{2},\sqrt{\mathcal{M}}f_{2}\right)

is the nonlinear Boltzmann collision operator, and $\mathcal{K=K}_{s}+\mathcal{M}^{1/2}\mathcal{K}_{b}$ with

\mathcal{K}_{s}:=\chi_{\{\left|\xi\right|\geq R\}}\mathcal{K}\hbox{, \ \ \ \ }\mathcal{K}_{b}:=\mathcal{M}^{-1/2}\chi_{\{\left|\xi\right|<R\}}\mathcal{K}

for a constant $R>0$ large enough, $\chi_{\{\cdot\}}$ being the indicator function.

It is noted that similar decomposition was also employed in [6] for Boltzmann equation without angular cutoff. Based on this decomposition, we have the main theorem as follows:

Theorem 1.

Let $\beta>4$ be sufficiently large. Assume that $f_{0}\in L_{\xi,\beta}^{\infty}L_{x}^{\infty}$ is compactly supported in the variable $x$ for all $\xi$

f_{0}\left(x,\xi\right)\equiv 0\text{ for }\left|x\right|\geq 1\text{, }\xi\in\mathbb{R}^{3}\text{.}

Then for any fixed $\delta>0$ , there exists $\varepsilon>0$ small enough such that the solution $f$ of $(\ref{Linearized})$ exists for $t>0$ and it can be decomposed as $f=f_{1}+\sqrt{\mathcal{M}}f_{2}$ , where $f_{1}$ and $f_{2}$ satisfy $\left(\ref{decom-System}\right)$ with the following estimates:

For $f_{1}$ , we have

\left|\mathbf{1}_{\{\left<\xi\right>\leq t\}}f_{1}\right|\lesssim\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\overline{c}_{0}\left[\left<\xi\right>^{2}+(t+|x|)\right]}

and

\left|\mathbf{1}_{\{\left<\xi\right>>t\}}f_{1}\right|\lesssim\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\overline{c}_{0}\left[\left<\xi\right>t+(t+|x|)\right]}

for some positive constant $\overline{c}_{0}$ .

For $f_{2}$ , we have

	$\displaystyle\left\|f_{2}\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle\lesssim\varepsilon\left[\begin{array}[]{l}\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+\left(1+t\right)^{-3/2}e^{-\frac{\|x\|^{2}}{\widehat{D}\left(1+t\right)}}\\ \\ +\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+e^{-\frac{t+\|x\|}{\widehat{c}_{\delta}}}\end{array}\right]$
		$\displaystyle\quad+\varepsilon^{2}\mathbf{1}_{\{\left\|x\right\|\leq(\mathbf{c}+\delta)t\}}\left[\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]$

for some positive constants $\hat{D},\hat{D}_{0},\hat{c}_{\delta}$ . Here $\mathbf{c}=\sqrt{5/3}$ is the sound speed associated with global Maxwellian $\mathcal{M}$ .

Several remarks on the main theorem are in order:

•

The main result is the combinations of Theorems 9, 10, and 15. In Theorem 9, we obtain the global wave structure, which is accurate in the time-like region but only shows polynomial decay for $f_{2}$ in the space-like region. In Theorem 10, the estimate in space-like region is further improved to be exponentially sharp. Theorem 15 describes the dynamic process of transition from polynomial tail to Gaussian tail for $f_{1}$ .
•

The result shows that the polynomial tail part $f_{1}$ decays exponentially in both space and time, while Gaussian tail part $\sqrt{\mathcal{M}}f_{2}$ exhibits rich wave phenomena and dominates the solution at large time. This is consistent with our intuition: the polynomial and Gaussian tail parts are associated with particle-like and fluid-like behaviors, respectively.
•

In the estimate of $f_{2}$ , we see that the terms consist of $\varepsilon$ and $\varepsilon^{2}$ orders. The $\varepsilon$ order terms represent linear waves, such as Huygens, diffusion, and Riesz waves, as given by the Green’s function. The $\varepsilon^{2}$ order terms arise from nonlinear interactions between these basic waves. They consist of polynomial versions of Huygens and diffusion waves, primarily concentrate inside the acoustic wave cone. Compared to the linear waves, the nonlinear waves not only have a $\varepsilon^{2}$ order magnitude, but also decay faster by $(1+t)^{-1/2}$ than their linear counterparts. However, previous works [10, 12, 20, 23, 24] only showed that they have the same decay rates. Thus, our results provide a more accurate description for the nonlinear effect, based on sharper estimates of nonlinear wave couplings.
•

The polynomial tail of the solution is fully captured by $f_{1}$ part. The pointwise estimate of $f_{1}$ in velocity variable thus shows how the polynomial tail transitions to a Gaussian tail. Specifically, $f_{1}$ only exhibits a polynomial tail initially, as time evolves it immediately acquires an exponential tail. As time continues to evolve, the particles with velocity $\left\langle\xi\right\rangle<t$ will become Gaussian tail, while the non-Gaussian part is of the order $e^{-t^{2}}$ and any moments generated by the non-Gaussian part are bounded by $e^{-t^{2}}$ . Therefore, as time tends to infinity, the distribution function will eventually transit to a Gaussian tail; but however, the transition cannot be completed in any finite time.
•

The assumption that the initial data has compact support in space is unessential. It is not hard to generalize to the case where the initial data decays polynomially in space, but in this case, the space-like behavior should be modified accordingly. We do not pursue this, as our focus is on the quantitative description of wave motion and the transition to a Gaussian tail.

1.5. Ideas and strategies

We now outline the ideas and strategies for the proof of the main theorem.

1.5.1. Global wave structure

Let us begin with Theorem 9, the global wave structure of the solution. Consider the coupled system (4). Let $\mathbb{S}^{t}$ and $\mathbb{G}^{t}$ be the solution operators to the damped transport equation and (standard) linearized Boltzmann equation respectively, namely, $g(t)\equiv\mathbb{S}^{t}g_{0}$ solves

\partial_{t}g+\xi\cdot\nabla_{x}g+\nu\left(\xi\right)g=0\hbox{,}\quad g(0,x,\xi)=g_{0}\,,

and $g(t)\equiv\mathbb{G}^{t}g_{0}$ solves

\partial_{t}g+\xi\cdot\nabla_{x}g-Lg=0\hbox{,}\quad g(0,x,\xi)=g_{0}\,.

Then by Duhamel principle, solutions $f_{1}$ and $f_{2}$ to (4) satisfy the following coupled integral system

(5)

\left\{\begin{aligned} &f_{1}(t)=\varepsilon\mathbb{S}^{t}f_{0}+\int_{0}^{t}\mathbb{S}^{t-\tau}\mathcal{K}_{s}f_{1}(\tau)d\tau+\int_{0}^{t}\mathbb{S}^{t-\tau}\Bigl{[}Q(f_{1},f_{1})+Q(f_{1},\sqrt{\mathcal{M}}f_{2})+Q(\sqrt{\mathcal{M}}f_{2},f_{1})\Bigr{]}(\tau)d\tau,\\ &f_{2}(t)=\int_{0}^{t}\mathbb{G}^{t-\tau}\mathcal{K}_{b}f_{1}(\tau)d\tau+\int_{0}^{t}\mathbb{G}^{t-\tau}\Gamma(f_{2},f_{2})(\tau)d\tau.\end{aligned}\right.

The essential step for constructing global wave structure is to find out the accurate ansatz for the solution.

We neglect the nonlinear effects for the time being. In the equation for $f_{1}$ , if $R$ is chosen sufficiently large, the integral operator $\mathcal{K}_{s}$ can be regarded as a perturbation of the damped transport operator. This results in an exponential decay of $f_{1}$ in both space and time. We then substitute the estimate of $f_{1}$ into the integral equation for $f_{2}$ , requiring consideration of $\int_{0}^{t}\mathbb{G}^{t-\tau}\mathcal{K}_{b}f_{1}(\tau)d\tau$ . The explicit structure of Green’s function $\mathbb{G}$ is constructed in [26, 27] (we stated it in Lemma 8), showing rich wave structures, including a dissipative Huygens wave (acoustic wave described by a moving heat kernel), a diffusion wave (thermal wave described by a stationary heat kernel), a Riesz wave (related to vorticity of macroscopic fluid, described by a polynomial analogue of diffusion wave confined in the wave cone), and a space-time exponential decay term. The estimate of $f_{2}$ is given by the convolution of the Green’s function and the source term $\mathcal{K}_{b}f_{1}$ , inheriting a structure similar to that of the Green’s function (see Lemmas 19-21). This indicates that $f_{1}$ can be viewed as particle-like wave, while $f_{2}$ can be viewed as fluid-like wave.

Next, we incorporate nonlinear effects. In designing the coupled system, we intentionally placed all nonlinear terms involving $f_{1}$ in the first equation of (4), as it describes the particle-like behavior. The term $\Gamma(f_{2},f_{2})$ is included in the second equation, as it is associated with the fluid-like behavior. The key term is $\int_{0}^{t}\mathbb{G}^{t-\tau}\Gamma(f_{2},f_{2})(\tau)d\tau$ , which accounts for the nonlinear wave interactions. A fundamental property is that the nonlinear operator $\Gamma$ is purely microscopic, and when acted upon by the Green’s operator, it gains an extra $\frac{1}{2}$ -order time decay. Substituting the linear estimates leads to the convolution estimates:

\frac{\bigl{(}\mbox{Diffusion+Huygens+Riesz+Exponential decay}\bigr{)}}{\sqrt{1+t}}\underset{(x,t)}{\ast}\bigl{(}\mbox{Diffusion+Huygens+Riesz+Exponential decay}\bigr{)}^{2}.

The main effort is to provide sharp estimates of these convolutions. Without delving into the details, we provide a heuristic explanation of the interaction process here. We illustrate this with the convolution of two Huygens waves.

\int_{0}^{t}\int_{{\mathbb{R}}^{3}}(1+t-s)^{-5/2}e^{-\frac{(\lvert x-y\rvert-(t-s))^{2}}{D_{0}(t-s)}}(1+s)^{-4}e^{-\frac{(\lvert y\rvert-s)^{2}}{C_{1}(s+1)}}dyds,

where $\mathbf{c}=1$ for simplicity. The following figure is to explain the interaction process:

Refer to caption — Figure 1. Interaction between two Huygens waves

Here, the source $(1+s)^{-4}e^{-\frac{(\lvert y\rvert-s)^{2}}{C_{1}(s+1)}}$ is plotted as a forward cone with thickness $\sqrt{s}$ , and the propagator $(1+t-s)^{-5/2}e^{-\frac{(\lvert x-y\rvert-(t-s))^{2}}{D_{0}(t-s)}}$ is plotted as a backward cone with thickness $\sqrt{t-s}$ . The space is represented in 2D in the figure. The interaction essentially occurs in the following space-time region:

\left\{(y,s)\big{|}\,\lvert\lvert x-y\rvert-(t-s)\rvert\leq O(1)\sqrt{t-s}\mbox{ and }\lvert\lvert y\rvert-s\rvert\leq O(1)\sqrt{s}\right\}.

Inside this region, the exponential term in Huygens wave is not effective, and the decay is mainly due to time decay factor. The key point is the sharp estimate of the volume for this space-time region. In Section 6.2, we first provide some heuristic calculations. In fact, our rigorous estimates are greatly motivated by the heuristics. We identify the strong interaction region (which appears in the regions $D_{4}$ and $D_{5}$ of Section 6) and perform very careful estimates there. The results match those obtained by the heuristic argument.

Through these sharp convolution estimates, we propose an appropriate ansatz, which is exponential decay for $f_{1}$ and polynomially sharp for $f_{2}$ as the main focus here is the region inside wave cone, where only polynomial decay can be expected. Justifying the ansatz involves even more complicated convolution estimates, but the underlying idea remains similar. Additionally, the damped transport operator $\mathbb{S}^{t}$ is used to compensate the loss of velocity decay from the nonlinear operator $\Gamma$ in the justification for $f_{1}$ (see Lemmas 4 and 5). Our result distinguishes between the linear and nonlinear parts of the solution, significantly improving upon previous results in [10, 12, 20, 23, 24], where the nonlinear couplings and the linear part have the same decay rate.

1.5.2. Exponential decay outside the acoustic wave cone

In Theorem 9, we obtain space-time exponential decay for $f_{1}$ , but for $f_{2}$ , we only achieve a polynomial decay estimate. This is because we designed a polynomial-type ansatz to facilitate the control of the nonlinear part of $f_{2}$ . This ansatz is accurate for the structure inside the acoustic wave cone. However, since the initial data has compact support in space and $f_{2}$ corresponds to a fluid structure that propagates at a finite speed, we expect the solution to decay exponentially outside the wave cone.

To observe the behavior of $f_{2}$ outside the cone, we multiply a suitable weight function on $f_{2}$ and prove an $L^{\infty}$ bound of the weighted solution through regularization and energy estimates. This approach for obtaining the space asymptotic behavior of the Boltzmann equation was developed in our previous work [21], and here it is adapted to handle $f_{2}$ .

It is worth mentioning that in our previous work, we proved exponential decay outside a wave cone $\lvert x\rvert<Mt$ for a sufficiently large $M$ . Here, through careful calculation of the micro projection, we show that $f_{2}$ indeed decays space-time exponentially outside the wave cone $\lvert x\rvert<(\mathbf{c}+\delta)t$ for any positive $\delta$ (see Theorem 10), where $\mathbf{c}$ is the sound speed. This result is more consistent with physical reality.

1.5.3. The transition from polynomial tail to Gaussian tail

In the decomposition $f=f_{1}+\sqrt{\mathcal{M}}f_{2}$ , the latter already exhibits Gaussian tail. Therefore, studying the transition process is equivalent to examining the generation of the Gaussian tail for $f_{1}$ . The mechanism for generating velocity decay comes from $\displaystyle e^{-\nu(\xi)t}$ in the damped transport operator $\mathbb{S}^{t}$ :

\mathbb{S}^{t}g_{0}=e^{-\nu(\xi)t}g_{0}(x-\xi t,\xi).

At first glance, it seems that $f_{1}$ can gain arbitrary velocity decay as time evolves. However, the upper bound for velocity decay is limited by the coupling between $f_{1}$ and $\sqrt{\mathcal{M}}f_{2}$ , specifically by the term

\int_{0}^{t}\mathbb{S}^{t-\tau}\Bigl{[}Q(f_{1},\sqrt{\mathcal{M}}f_{2})+Q(\sqrt{\mathcal{M}}f_{2},f_{1})\Bigr{]}(\tau)d\tau

in the second equation of (5). The velocity weight will ultimately be slowed down by $\sqrt{\mathcal{M}}f_{2}$ as it decay at most as a Gaussian. We design a suitable weighted function

e^{\kappa\langle\xi\rangle\min\{\langle\xi\rangle\wedge t\}}

to capture this feature. By carefully analyzing the commutator between this weight function and the $\mathcal{K}_{s},\,Q$ operators, we complete the description of dynamic transition process (see Theorem 15).

It is interesting to note that the mechanism for gaining velocity weight and the limitation of the maximal generation both stem from collisions with the global Maxwellian. This is entirely consistent with our physical intuition.

1.6. Organization of the paper

The rest of this paper is organized as follows: In Section 2, we first prepare some basic properties of the integral operator $\mathcal{K}$ , and present some estimates for the damped transport equation and the linearized Boltzmann equation. In Section 3, we construct the global wave structures of the solution, fully utilizing sharp nonlinear wave interactions. In Section 4, we apply the weighted energy estimate to prove that the solution indeed decays exponentially in space-time outside the wave cone. In Section 5, we provide a quantitative description of how solution approaches a Gaussian tail in terms of the microscopic velocity. Finally, we present the proof of all kinds of wave interactions in Section 6.

2. Preliminaries

To begin with, we study some essential properties of the collision frequency $\nu\left(\xi\right)$ , the operator $\mathcal{K}$ and the collision operator $Q$ . It is well known that there exist two positive constants $\nu_{0}$ and $\nu_{1}$ such that

\nu_{0}\left\langle\xi\right\rangle\leq\nu\left(\xi\right)=\int_{\mathbb{R}^{3}}\int_{S^{2}}\left|q\cdot n\right|\mathcal{M}\left(\xi_{\ast}\right)dnd\xi_{\ast}\leq\nu_{1}\left\langle\xi\right\rangle

for all $\xi\in\mathbb{R}^{3}$ . In [5, Lemma 2.1 and Lemma 6.2], the following estimates of the collision kernel have been proved.

Lemma 2 ([5]).

For any $\beta>4$ , there exists a constant $C_{\beta}>0$ depending only on $\beta$ such that

\int_{\mathbb{R}^{3}}\int_{S^{2}}\left|\xi-\xi_{\ast}\right|\frac{w_{\beta}\left(\xi\right)}{w_{\beta}\left(\xi_{\ast}^{\prime}\right)}e^{-\frac{1}{4}|\xi^{\prime 2}}dnd\xi_{\ast}\leq\frac{C}{\beta}\nu\left(\xi\right)+C_{\beta}\frac{\nu\left(\xi\right)}{\left(1+\left|\xi\right|\right)^{2}}

and

\int_{\mathbb{R}^{3}}\int_{S^{2}}\left|\xi-\xi_{\ast}\right|\frac{w_{\beta}\left(\xi\right)}{w_{\beta}\left(\xi^{\prime}\right)w_{\beta}\left(\xi_{\ast}^{\prime}\right)}dnd\xi_{\ast}\leq\frac{C}{\beta}\nu\left(\xi\right)+C_{\beta}\frac{\nu\left(\xi\right)}{\left(1+\left|\xi\right|\right)^{2}}\hbox{.}

where $C>0$ is a universal constant independent of $\beta$ . Here $w_{\beta}\left(\xi\right)=\left\langle\xi\right\rangle^{\beta}$ .

Using this lemma, one can get the estimates for the operators $\mathcal{K}$ and $Q$ .

Lemma 3 ([5]).

For any $\beta>4$ , there exists a constant $C_{\beta}>0$ depending only on $\beta$ such that

\left\langle\xi\right\rangle^{\beta}\left|\mathcal{K}f\right|\leq\left(\frac{C}{\beta}\nu\left(\xi\right)+C_{\beta}\frac{\nu\left(\xi\right)}{\left(1+\left|\xi\right|\right)^{2}}\right)\left|f\right|_{L_{\xi,\beta}^{\infty}}

and

\left|\left\langle\xi\right\rangle^{\beta}Q\left(f,g\right)\right|\leq C_{\beta}\nu\left(\xi\right)\left|f\right|_{L_{\xi,\beta}^{\infty}}\left|g\right|_{L_{\xi,\beta}^{\infty}}\hbox{.}

Moreover, for any $R>0$ ,

\chi_{\{\left|\xi\right|\geq R\}}\left\langle\xi\right\rangle^{\beta}\left|\mathcal{K}f\right|\leq\left(\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}\right)\nu\left(\xi\right)\left|f\right|_{L_{\xi,\beta}^{\infty}}\,.

In order to study the first equation of (4), we introduce the damped transport operator $\mathbb{S}^{t}$ , that is, $\mathbb{S}^{t}g_{0}$ is the solution of the equation:

\partial_{t}g+\xi\cdot\nabla_{x}g+\nu\left(\xi\right)g=0\hbox{,}\quad g(0,x,\xi)=g_{0}\,.

Lemma 4.

Let $\beta>4$ and $M>0$ . Assume that $U\left(t,x,\xi\right)$ and $h(t,x,\xi)$ satisfy

\left|\nu\left(\xi\right)^{-1}U\left(t,x,\xi\right)\right|_{L_{\xi,\beta}^{\infty}}\leq Ae^{-\frac{t+\left|x\right|}{c}}

and

\left|h\left(t,x,\xi\right)\right|_{L_{\xi,\beta}^{\infty}}\leq Be^{-\frac{t+\left|x\right|}{c}}

for some constants $A,c>0$ with $\frac{\nu_{0}}{2}>\frac{1}{c}$ . If $g$ satisfies the integral equation

g\left(t,x,\xi\right)=\int_{0}^{t}\mathbb{S}^{t-\tau}\left(\mathcal{K}_{s}h+U\right)\left(\tau,x,\xi\right)d\tau\hbox{,}

then

\left|g(t,x,\xi)\right|_{L_{\xi,\beta}^{\infty}}\leq 2(A+\eta B)e^{-\frac{t+\left|x\right|}{c}}\hbox{,}

where

\eta=\eta(\beta,R)=\left(\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}\right)\,.

Proof.

Let $y=x-\left(t-\tau\right)\xi$ , we have

$\displaystyle\left\|\left\langle\xi\right\rangle^{\beta}g\left(t,x,\xi\right)\right\|$	$\displaystyle=$	$\displaystyle\left\|\int_{0}^{t}e^{-\nu\left(\xi\right)\left(t-\tau\right)}\left\langle\xi\right\rangle^{\beta}\left[\mathcal{K}_{s}h\left(\tau,x-\xi\left(t-\tau\right),\xi\right)+U\left(\tau,x-\xi\left(t-\tau\right),\xi\right)\right]d\tau\right\|$
	$\displaystyle\leq$	$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(\nu\left(\xi\right)\right)^{-1}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\langle\xi\right\rangle^{\beta}\left[\left\|\mathcal{K}_{s}h\left(\tau,y,\xi\right)\right\|\right]d\tau$
		$\displaystyle+\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(\nu\left(\xi\right)\right)^{-1}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\langle\xi\right\rangle^{\beta}\left\|U\left(\tau,y,\xi\right)\right\|d\tau$
	$\displaystyle\leq$	$\displaystyle\eta\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\cdot\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\|h\left(\tau,y,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}}d\tau$
		$\displaystyle+\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\cdot\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\|\nu^{-1}(\xi)U\left(\tau,y,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}}d\tau$
	$\displaystyle\leq$	$\displaystyle 2(A+\eta B)e^{-\frac{t+\left\|x\right\|}{c}}\text{.}$

The proof is completed. $\hfill\square$

Lemma 5.

Let $\alpha>0$ and $\rho>0$ . Assume that $V\left(t,x,\xi\right)$ satisfies

\left|\nu\left(\xi\right)^{-1}V\left(t,x,\xi\right)\right|_{L_{\xi,\beta}^{\infty}}\leq\left(1+t\right)^{-\alpha}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}\,.

If $g$ satisfies the integral equation

g\left(t,x,\xi\right)=\int_{0}^{t}\mathbb{S}^{t-\tau}V\left(\tau,x,\xi\right)d\tau\hbox{,}

then

\left|g(t,x,\xi)\right|_{L_{\xi,\beta}^{\infty}}\leq C_{\alpha,\rho}\left(1+t\right)^{-\alpha}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}

for some constant $C_{\alpha,\rho}>0$ .

Proof.

The proof is similar to Lemma 4. It suffices to verify that

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha,\rho}\left(1+t\right)^{-\alpha}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\rho}\,.$

Let $0\leq\tau\leq t$ . If $\left|y\right|>\left|x\right|/2$ , then

\left(1+\frac{\left|y\right|^{2}}{1+\tau}\right)^{-\rho}\leq\left(1+\frac{\left|x\right|^{2}}{4\left(1+t\right)}\right)^{-\rho}\leq 4^{\rho}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}\hbox{,}

and thus

e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left|x-y\right|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left|y\right|^{2}}{1+\tau}\right)^{-\rho}\leq 4^{\rho}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}\hbox{.}

If $\left|y\right|\leq\left|x\right|/2$ , then

e^{-\frac{\nu_{0}}{2}\left|x-y\right|}\leq e^{-\frac{\nu_{0}\left|x\right|}{4}}\leq C_{\rho}\left(1+\left|x\right|^{2}\right)^{-\rho}\leq C_{\rho}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}\hbox{,}

and so

e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left|x-y\right|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left|y\right|^{2}}{1+\tau}\right)^{-\rho}\leq C_{\rho}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\rho}\hbox{.}

Therefore,

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle\left(C_{\rho}+4^{\rho}\right)\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}\left(C_{\rho}+4^{\rho}\right)\left(1+t\right)^{-\alpha}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\rho}\hbox{.}$

∎

Lemma 6.

Let $\alpha>0$ and $\rho>0$ . Assume that $W\left(t,x,\xi\right)$ satisfies

\left|\nu\left(\xi\right)^{-1}W\left(t,x,\xi\right)\right|_{L_{\xi,\beta}^{\infty}}\leq\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-\rho}\,.

If $g$ satisfies the integral equation

g\left(t,x,\xi\right)=\int_{0}^{t}\mathbb{S}^{t-\tau}W\left(\tau,x,\xi\right)d\tau\hbox{,}

then

\left|g(t,x,\xi)\right|_{L_{\xi,\beta}^{\infty}}\leq C_{\alpha,\rho}\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-\rho}

for some constant $C_{\alpha,\rho}>0$ .

Proof.

The proof is similar to Lemma 4 and it suffices to verify that

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha,\rho}\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}$

We consider two cases $\left|x\right|\geq\mathbf{c}t$ and $\left|x\right|\leq\mathbf{c}t$ .

Case 1: $\left|x\right|\geq\mathbf{c}t$ . We split $\mathbb{R}^{3}$ into two parts

\{y\in\mathbb{R}^{3}:\left|\left|y\right|-\mathbf{c}\tau\right|>\frac{\left|x\right|-\mathbf{c}t}{2}\}\hbox{ and }\{y\in\mathbb{R}^{3}:\left|\left|y\right|-\mathbf{c}\tau\right|\leq\frac{\left|x\right|-\mathbf{c}t}{2}\}\hbox{.}

If $\left|\left|y\right|-\mathbf{c}\tau\right|\leq\frac{\left|x\right|-\mathbf{c}t}{2}$ , then

	$\displaystyle\left\|x-y\right\|$	$\displaystyle\geq$	$\displaystyle\left\|\left\|x\right\|-\left\|y\right\|\right\|=\left\|\left\|x\right\|-\mathbf{c}\tau+\mathbf{c}\tau-\left\|y\right\|\right\|\geq\left\|\left\|x\right\|-\mathbf{c}\tau\right\|-\left\|\left\|y\right\|-\mathbf{c}\tau\right\|$
		$\displaystyle\geq$	$\displaystyle\left\|x\right\|-\mathbf{c}t-\frac{\left\|x\right\|-\mathbf{c}t}{2}=\frac{\left\|x\right\|-\mathbf{c}t}{2}\hbox{,}$

and thus

e^{-\frac{\nu_{0}}{2}\left|x-y\right|}\leq e^{-\frac{\nu_{0}\left|\left|x\right|-\mathbf{c}t\right|}{4}}\leq C_{\rho}\left(1+\left(\left|x\right|-\mathbf{c}t\right)^{2}\right)^{-\rho}\leq C_{\rho}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}

If $\left|\left|y\right|-\mathbf{c}\tau\right|>\frac{\left|x\right|-\mathbf{c}t}{2}$ , then

\left(1+\frac{\left(\left|y\right|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}\leq\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4\left(1+t\right)}\right)^{-\rho}\leq 4^{\rho}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}

Therefore,

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{\left(1+\tau\right)}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle\left(C_{\rho}+4^{\rho}\right)\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}\left(C_{\rho}+4^{\rho}\right)\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}$

Case 2: $\left|x\right|\leq\mathbf{c}t$ . If $0\leq\tau\leq\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}$ , then

t-\tau\geq t-\left(\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}\right)=\frac{\mathbf{c}t-\left|x\right|}{2\mathbf{c}}\geq 0\hbox{,}

and thus

e^{-\nu_{0}\left(t-\tau\right)}\leq e^{-\frac{\nu_{0}}{2\mathbf{c}}\left(\mathbf{c}t-\left|x\right|\right)}\leq C_{\rho}\left(1+\left(\mathbf{c}t-\left|x\right|\right)^{2}\right)^{-\rho}\leq C_{\rho}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}

It implies that

			$\displaystyle\int_{0}^{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\rho}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}C_{\rho}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}$

If $\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|\leq\frac{\mathbf{c}\tau+\left|x\right|}{2}$ , then

\mathbf{c}\tau-\left|y\right|\geq\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\geq 0\hbox{,}

and so

\left(1+\frac{\left(\left|y\right|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}\leq\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{16\left(1+t\right)}\right)^{-\rho}\leq 16^{\rho}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}

If $\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|>\frac{\mathbf{c}\tau+\left|x\right|}{2}$ , then

\left|x-y\right|\geq\left|y\right|-\left|x\right|\geq\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{,}

and thus

e^{-\frac{\nu_{0}}{2}\left|x-y\right|}\leq e^{-\frac{\nu_{0}}{8}\left(\mathbf{c}t-\left|x\right|\right)}\leq C_{\rho}\left(1+\left(\mathbf{c}t-\left|x\right|\right)^{2}\right)^{-\rho}\leq C_{\rho}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-\rho}\text{.}

Consequently,

			$\displaystyle\int_{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{\left(1+\tau\right)}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle\left(16^{\rho}+C_{\rho}\right)\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}\left(16^{\rho}+C_{\rho}\right)\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}$

Combining all the discussion, there exists a constant $C_{\alpha,\rho}>0$ such that

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha,\rho}\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\hbox{,}$

as desired. ∎

For the second equation of decomposition (4), we list some basic properties of the linearized Boltzmann collision operator $L$ and nonlinear operator $\Gamma$ as below.

It is well known that the null space of $L$

\ker\left(L\right)=\text{span }\{\chi_{0},\chi_{1},\chi_{2},\chi_{3},\chi_{4}\}\text{,}

is a five-dimensional vector space, where

\chi_{0}=\mathcal{M}^{1/2},\text{ }\chi_{i}=\xi_{i}\mathcal{M}^{1/2},\text{ }\chi_{4}=\frac{1}{\sqrt{6}}\left(\left|\xi\right|^{2}-3\right)\mathcal{M}^{1/2}\text{, \ }i=1\text{, }2\text{, }3\text{.}

Let $\mathrm{P}_{0}$ be the orthogonal projection with respect to the $L_{\xi}^{2}$ inner product onto $\mathrm{\ker}(L)$ , and $\mathrm{P}_{1}\equiv\mathrm{Id}-\mathrm{P}_{0}$ . That is, for any $g\in L^{2}_{\xi}$ ,

\mathrm{P}_{0}g=\sum_{i=0}^{4}\left<\chi_{i},g\right>_{\xi}\chi_{i}\,,\quad\mathrm{P}_{1}g=g-\mathrm{P}_{0}g\,.

The solution of the wave propagation is connected to the operator $\mathrm{P}_{0}(\xi\cdot\omega)\mathrm{P}_{0}$ for $\omega\in\mathbb{S}^{2}$

(6)

\left\{\begin{array}[]{l}\mathrm{P}_{0}\xi\cdot\omega E_{j}=\lambda_{j}E_{j}\,,\\[5.69054pt] \lambda_{0}=\mathbf{c}\,,\quad\lambda_{1}=-\mathbf{c}\,,\quad\lambda_{2}=\lambda_{3}=\lambda_{4}=0\,,\mathbf{c}=\sqrt{\frac{5}{3}}\,,\\[5.69054pt] E_{0}=\sqrt{\frac{3}{10}}\chi_{0}+\sqrt{\frac{1}{2}}\omega\cdot\overline{\chi}+\sqrt{\frac{1}{5}}\chi_{4}\,,\\[5.69054pt] E_{1}=\sqrt{\frac{3}{10}}\chi_{0}-\sqrt{\frac{1}{2}}\omega\cdot\overline{\chi}+\sqrt{\frac{1}{5}}\chi_{4}\,,\\[5.69054pt] E_{2}=-\sqrt{\frac{2}{5}}\chi_{0}+\sqrt{\frac{3}{5}}\chi_{4}\,,\\[5.69054pt] E_{3}=\omega_{1}\cdot\overline{\chi}\,,\\[5.69054pt] E_{4}=\omega_{2}\cdot\overline{\chi}\,.\end{array}\right.

where $\overline{\chi}=(\chi_{1},\chi_{2},\chi_{3})$ , and $\{\omega_{1},\omega_{2},\omega\}$ is an orthonormal basis of ${\mathbb{R}}^{3}$ .

Lemma 7 ([26]).

The collision operator $L$ consists of a multiplicative operator $\nu(\xi)$ and an integral operator $K$ , namely,

Lf=-\nu(\xi)f+Kf\,,

where $\nu(\xi)\sim(1+|\xi|)$ and

\left|Kg\right|_{L_{\xi,\eta+1}^{\infty}}\leq C\left|g\right|_{L_{\xi,\eta}^{\infty}}\text{, }\eta\in\mathbb{R}\text{,}

for some universal constant $C>0$ .

The nonlinear operator $\Gamma$ has the following estimate

\left|\nu^{-1}(\xi)\Gamma(g,h)\right|_{L_{\xi,\eta}^{\infty}}\leq C\left|g\right|_{L_{\xi,\eta}^{\infty}}\left|h\right|_{L_{\xi,\eta}^{\infty}}\text{, }\eta\in\mathbb{R}\text{,}

for some universal constant $C>0$ .

It is vital to get the space-time structure of the solution of the linearized Boltzmann equation

\partial_{t}h+\xi\cdot\nabla_{x}h=Lh\text{,}

and we denote $\mathbb{G}^{t}$ as its solution operator.

Lemma 8 ([26, 27]).

The solution $h=\mathbb{G}^{t}h_{0}$ of the equation

\partial_{t}h+\xi\cdot\nabla_{x}h=Lh\hbox{, }h\left(0,x,\xi\right)=h_{0}\left(x,\xi\right)\hbox{.}

satisfies

\left|\mathbb{G}^{t}h_{0}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left|x\right|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left|x\right|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+|x|}{c_{0}}}\end{array}\right]\ast_{x}\left|h_{0}\right|_{L_{\xi,\beta}^{\infty}}

for any $\beta>3/2$ and some constants $\mathcal{C}_{1}$ , $c_{0}$ , $D_{0}>0$ , where $h_{0}\in L_{\xi,\beta}^{\infty}(L_{x}^{\infty}\cap L_{x}^{1})$ . Moreover, if $\mathrm{P}_{1}h_{0}=0$ , then

\left|\mathbb{G}^{t}h_{0}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\left[\begin{array}[]{l}\left(1+t\right)^{-5/2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-2}e^{-\frac{\left|x\right|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left|x\right|\leq\mathbf{c}t\}}\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+|x|}{c_{0}}}\end{array}\right]\ast_{x}\left|h_{0}\right|_{L_{\xi,\beta}^{\infty}}.

3. Global wave structure

In this section we will prove the global wave structure of the equation $(\ref{Linearized})$ which is described in the following theorem.

Theorem 9.

Let $\beta>4$ be sufficiently large. Assume that $f_{0}\in L_{\xi,\beta}^{\infty}L_{x}^{\infty}$ is compactly supported in the variable $x$ for all $\xi$

f_{0}\left(x,\xi\right)\equiv 0\text{ for }\left|x\right|\geq 1\text{, }\xi\in\mathbb{R}^{3}\text{.}

Then there exists $\varepsilon>0$ small enough such that the solution $f$ of $(\ref{Linearized})$ exists for $t>0$ and it can be written as $f=f_{1}+\sqrt{\mathcal{M}}f_{2}$ , where $f_{1}$ and $f_{2}$ satisfy $\left(\ref{decom-System}\right)$ with

\left|f_{1}\right|_{L_{\xi,\beta}^{\infty}}\leq 2\mathcal{C}_{1}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}\left(t+\left|x\right|\right)}\text{,}

and

	$\displaystyle\quad\left\|f_{2}\right\|_{L_{\xi,\beta}^{\infty}}$
	$\displaystyle\leq\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]$
	$\displaystyle\quad+2\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]$

for some positive constants $\mathfrak{B},\mathcal{C}_{1},\mathfrak{C},\hat{D}_{0},\hat{c}_{0}$ .

Proof.

To clarify the space-time structures of the solutions $f_{1}$ and $f_{2}$ to $\left(\ref{decom-System}\right)$ , we design an iteration $\{f^{n}\}$ with $f^{n}=f_{1}^{n}+\sqrt{\mathcal{M}}f_{2}^{n}$ , as follows:

(7)

\left\{\begin{array}[]{l}\partial_{t}f_{1}^{n+1}+\xi\cdot\nabla_{x}f_{1}^{n+1}+\nu\left(\xi\right)f_{1}^{n+1}=\mathcal{K}_{s}f_{1}^{n}+U(f_{1}^{n},f_{2}^{n})\vspace{3mm}\\ \partial_{t}f_{2}^{n+1}+\xi\cdot\nabla_{x}f_{2}^{n+1}=Lf_{2}^{n+1}+\mathcal{K}_{b}f_{1}^{n+1}+\Gamma(f_{2}^{n},f_{2}^{n})\end{array}\right.

with

f_{1}^{n+1}\left(0,x,\xi\right)=\varepsilon f_{0}\left(x,\xi\right)\text{, }f_{2}^{n+1}\left(0,x,\xi\right)=0\text{, }\ f_{1}^{0}\left(t,x,\xi\right)=f_{2}^{0}\left(t,x,\xi\right)=0\text{,}

where

U(f_{1}^{n},f_{2}^{n})=Q\left(f_{1}^{n},f_{1}^{n}\right)+Q\left(f_{1}^{n},\sqrt{\mathcal{M}}f_{2}^{n}\right)+Q\left(\sqrt{\mathcal{M}}f_{2}^{n},f_{1}^{n}\right)\text{.}

We shall study the space-time structures of $f_{1}^{n}$ and $f_{2}^{n}$ . Now we rewrite (7) as integral forms:

\left\{\begin{array}[]{l}f_{1}^{n+1}=\varepsilon\mathbb{S}^{t}f_{0}+\int_{0}^{t}\mathbb{S}^{t-\tau}\left[\mathcal{K}_{s}f_{1}^{n}(\tau)+U(f_{1}^{n},f_{2}^{n})(\tau)\right]d\tau=:f_{1,1}^{n+1}+f_{1,2}^{n+1}\text{,}\vspace{3mm}\\ f_{2}^{n+1}=\int_{0}^{t}\mathbb{G}^{t-\tau}\mathcal{K}_{b}\left(f_{1}^{n+1}\right)(\tau)d\tau+\int_{0}^{t}\mathbb{G}^{t-\tau}\Gamma(f_{2}^{n},f_{2}^{n})(\tau)d\tau=:f_{2,1}^{n+1}+f_{2,2}^{n+1}\text{.}\end{array}\right.

First, one can see that $f_{1}^{1}=f_{1,1}^{1}$ , $f_{2}^{1}=f_{2,1}^{1}$ and $f_{1,2}^{1}=f_{2,2}^{1}=0$ . Since $f_{0}$ is compactly supported in the unit ball with respect to the variable $x$ uniformly for all $\xi$ , we have

\left|f_{1}^{1}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+|x|)}

for some constant $\mathcal{C}_{1}$ . In view of Lemmas 3, 8 and Lemmas 19 – 21, we get

	$\displaystyle\left\|f_{2}^{1}\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle=\left\|\int_{0}^{t}\mathbb{G}^{t-\tau}\mathcal{K}_{b}\left(f_{1}^{1}\right)(\tau)d\tau\right\|_{L_{\xi,\beta}^{\infty}}$
		$\displaystyle\leq 2\mathcal{C}_{1}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
		$\displaystyle\cdot\mathcal{C}_{1}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{c_{0}}}\end{array}\right]\ast_{x,t}e^{-\frac{\nu_{0}}{2}(t+\|x\|)}$
		$\displaystyle\leq 2\mathcal{C}_{2}\mathcal{C}_{1}^{2}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
		$\displaystyle\cdot\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]$
		$\displaystyle\leq\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]\text{.}$

In fact, by definition of $f_{1,1}^{n}$ , we have $f_{1,1}^{n}=f_{1,1}^{1}$ for all $n$ and so

\left|f_{1,1}^{n}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+|x|)}\text{.}

Now we choose $\beta>4$ and $R>0$ sufficiently large such that the constant $\eta\left(\beta,R\right)$ defined in Lemma 4 satisfies

\eta\left(\beta,R\right)=\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}<\frac{1}{8}\text{.}

Fixing such $\beta$ and $R$ , we shall prove that if $\varepsilon>0$ is sufficiently small, we have

\left|f_{1,2}^{n}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+|x|)}\text{,}

\left|f_{2,1}^{n}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathfrak{B}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left|x\right|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left|x\right|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+|x|}{\widehat{c}_{0}}}\end{array}\right]\text{,}

\left|f_{2,2}^{n}\right|_{L_{\xi,\beta}^{\infty}}\leq 2\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]\text{,}

for all $n\in\mathbb{N}$ by induction on $n$ , here the large constant $\mathfrak{C}$ $>0$ will be determined later.

By induction hypothesis and Lemma 3, we have

$\displaystyle\left\|\nu^{-1}U\left(f_{1}^{n},f_{2}^{n}\right)\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle\leq$	$\displaystyle C\left(\left\|f_{1}^{n}\right\|_{L_{\xi,\beta}^{\infty}}^{2}+2\left\|f_{1}^{n}\right\|_{L_{\xi,\beta}^{\infty}}\left\|f_{2}^{n}\right\|_{L_{\xi,\beta}^{\infty}}\right)$
	$\displaystyle\leq$	$\displaystyle C\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left(4\mathcal{C}_{1}+\mathfrak{B}\left(1+2\mathfrak{CB}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\right)$
		$\displaystyle\cdot\mathcal{C}_{1}\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+\|x\|)}\text{.}$

By Lemma 4,

$\displaystyle\left\|f_{1,2}^{n+1}\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle=$	$\displaystyle\left\|\int_{0}^{t}\left\langle\xi\right\rangle^{\beta}\mathbb{S}^{t-\tau}\left[\mathcal{K}_{s}f_{1}^{n}(\tau)+U(f_{1}^{n},f_{2}^{n})(\tau)\right]d\tau\right\|$
	$\displaystyle\leq$	$\displaystyle\left[4\eta\left(\beta,R\right)+2C\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left(4\mathcal{C}_{1}+\mathfrak{B}\left(1+2\mathfrak{CB}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\right)\right]$
		$\displaystyle\cdot\mathcal{C}_{1}\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+\|x\|)}$
	$\displaystyle\leq$	$\displaystyle\mathcal{C}_{1}\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+\|x\|)}\text{,}$

after we choose $\varepsilon>0$ sufficiently small such that

2C\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left(4\mathcal{C}_{1}+2\mathfrak{CB}\left(1+\mathfrak{B}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\right)<1/2\text{.}

Therefore,

\left|f_{1,2}^{n+1}\right|_{L_{\xi,\beta}^{\infty}}\leq\mathcal{C}_{1}\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+|x|)}\text{,}

and so

\left|f_{1}^{n+1}\right|_{L_{\xi,\beta}^{\infty}}\leq 2\mathcal{C}_{1}\varepsilon\|f_{0}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}(t+|x|)}\text{.}

Likewise $f_{2,1}^{1}$ , we have

$\displaystyle\left\|f_{2,1}^{n+1}\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle=$	$\displaystyle\left\|\int_{0}^{t}\mathbb{G}^{t-\tau}\mathcal{K}_{b}\left(f_{1}^{n+1}\right)(\tau)d\tau\right\|_{L_{\xi,\beta}^{\infty}}$
	$\displaystyle\leq$	$\displaystyle 2\mathcal{C}_{1}\varepsilon\\|f_{0}\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
		$\displaystyle\cdot\mathcal{C}_{1}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{c_{0}}}\end{array}\right]\ast_{x,t}e^{-\frac{\nu_{0}}{2}(t+\|x\|)}$
	$\displaystyle\leq$	$\displaystyle 2\mathcal{C}_{2}\mathcal{C}_{1}^{2}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
		$\displaystyle\cdot\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]$
	$\displaystyle\leq$	$\displaystyle\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]\text{.}$

Also, there exits a constant $\mathcal{C}_{3}>1$ depending on $\widehat{D}_{0}$ and $\widehat{c}_{0}$ such that

			$\displaystyle\left\|f_{2,1}^{n+1}\right\|_{L_{\xi,\beta}^{\infty}}$
		$\displaystyle\leq$	$\displaystyle 2\mathcal{C}_{2}\mathcal{C}_{1}^{2}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
			$\displaystyle\cdot\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-\frac{3}{2}}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]$
		$\displaystyle\leq$	$\displaystyle 2\mathcal{C}_{3}\mathcal{C}_{2}\mathcal{C}_{1}^{2}\left[\left(2\pi\right)^{\frac{3}{4}}e^{\frac{R^{2}}{4}}\left(\frac{C\nu_{1}}{\beta}\left(1+R\right)+C_{\beta}\right)\right]$
			$\displaystyle\cdot\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\left(1+t\right)^{-\frac{3}{2}}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]$
		$\displaystyle\leq$	$\displaystyle\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\left(1+t\right)^{-\frac{3}{2}}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]\text{.}$

Since $f_{2,2}^{n+1}$ satisfies the equation

\partial_{t}f_{2,2}^{n+1}+\xi\cdot\nabla_{x}f_{2,2}^{n+1}=Lf_{2,2}^{n+1}+\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\text{ with }f_{2,2}^{n+1}\left(0,x,\xi\right)=0\text{, }

we further decompose $f_{2,2}^{n+1}$ into two parts as $f_{2,2}^{n+1}=h_{1}^{n+1}+h_{2}^{n+1}$ where $h_{1}^{n+1}$ , $h_{2}^{n+1}$ satisfy the equations

\left\{\begin{array}[]{l}\partial_{t}h_{1}^{n+1}+\xi\cdot\nabla_{x}h_{1}^{n+1}+\nu\left(\xi\right)h_{1}^{n+1}=\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\text{,}\vspace{3mm}\\ \partial_{t}h_{2}^{n+1}+\xi\cdot\nabla_{x}h_{2}^{n+1}+\nu\left(\xi\right)h_{2}^{n+1}=Kf_{2,2}^{n+1}\text{,}\end{array}\right.

with $h_{1}^{n+1}\left(0,x,\xi\right)=h_{2}^{n+1}\left(0,x,\xi\right)=0$ . By induction hypothesis and (3),

$\displaystyle\left\|f_{2}^{n}\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle=$	$\displaystyle\left\|f_{2,1}^{n}+f_{2,2}^{n}\right\|_{{}_{L_{\xi,\beta}^{\infty}}}$
	$\displaystyle\leq$	$\displaystyle\left[\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}+2\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\right]$
		$\displaystyle\cdot\left[\left(1+t\right)^{-2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}+\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}\right]$
	$\displaystyle\leq$	$\displaystyle 2\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\left(1+t\right)^{-2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}+\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}\right]\text{,}$

so that

$\displaystyle\left\|\nu^{-1}\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\right\|_{L_{\xi,\beta}^{\infty}}$	$\displaystyle\leq$	$\displaystyle C\left\|f_{2}^{n}\right\|_{L_{\xi,\beta}^{\infty}}^{2}$
	$\displaystyle\leq$	$\displaystyle 8C\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}$
		$\displaystyle\cdot\left[\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}+\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\right]\text{.}$

In view of Lemmas 5 and 6, we have

			$\displaystyle\left\|\left\langle\xi\right\rangle^{\beta}h_{1}^{n+1}\right\|$
		$\displaystyle=$	$\displaystyle\left\|\left\langle\xi\right\rangle^{\beta}\int_{0}^{t}\mathbb{S}^{t-\tau}\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\left(\tau\right)d\tau\right\|$
		$\displaystyle\leq$	$\displaystyle 8C^{2}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}$
			$\displaystyle\cdot\left[\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\left(1+t\right)}\right)^{-2}+\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-3}\right]$
		$\displaystyle\leq$	$\displaystyle\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\left(1+t\right)}\right)^{-2}+\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-3}\right]\text{.}$

As for $h_{2}^{n+1}$ , we have

\left|\left\langle\xi\right\rangle^{\beta}h_{2}^{n+1}\right|=\left|\left\langle\xi\right\rangle^{\beta}\int_{0}^{t}\mathbb{S}^{t-\tau}K\left(f_{2,2}^{n+1}\right)\left(\tau,x,\xi\right)d\tau\right|\text{.}

It follows from Lemma 7 that

\left|\nu\left(\xi\right)^{-1}\left\langle\xi\right\rangle^{\beta}Kf_{2,2}^{n+1}\right|_{L_{\xi}^{\infty}}\leq\frac{C}{\nu_{0}}\left|f_{2,2}^{n+1}\right|_{L_{\xi,\beta-1}^{\infty}}\text{.}

In view of Lemma 8 and (3), together with convolution estimates in Section 6.2, to be more specific, Lemmas 22–27, we obtain

$\displaystyle\left\|f_{2,2}^{n+1}\right\|_{L_{\xi,\beta-1}^{\infty}}$	$\displaystyle=$	$\displaystyle\left\|\int_{0}^{t}\left\langle\xi\right\rangle^{\beta-1}\mathbb{G}^{t-\tau}\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\left(\tau\right)d\tau\right\|$
	$\displaystyle\leq$	$\displaystyle\mathcal{C}_{1}\nu_{1}\left[\begin{array}[]{l}\left(1+t\right)^{-\frac{5}{2}}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-2}e^{-\frac{\left\|x\right\|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{c_{0}}}\end{array}\right]\ast_{x,t}\left\|\nu^{-1}\Gamma\left(f_{2}^{n},f_{2}^{n}\right)\right\|_{L_{\xi,\beta}^{\infty}}$
	$\displaystyle\leq$	$\displaystyle\mathcal{C}_{1}\nu_{1}8C\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\begin{array}[]{l}\left(1+t\right)^{-\frac{5}{2}}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-2}e^{-\frac{\left\|x\right\|^{2}}{D_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{c_{0}}}\end{array}\right]$
		$\displaystyle\ast_{x,t}\left[\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}+\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\right]$
	$\displaystyle\leq$	$\displaystyle 8C^{2}\mathcal{C}_{1}\nu_{1}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}$
		$\displaystyle\cdot\left[\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}+\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\right]\text{.}$

Therefore, using Lemmas 5 and 6,

			$\displaystyle\left\|h_{2}^{n+1}\right\|_{L_{\xi,\beta}^{\infty}}$
		$\displaystyle\leq$	$\displaystyle C\cdot\frac{C}{\nu_{0}}8C^{2}\mathcal{C}_{1}\nu_{1}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}$
			$\displaystyle\cdot\left[\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]$
		$\displaystyle\leq$	$\displaystyle\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]$

after we choose $\mathfrak{C}>0$ sufficiently large. Combining (3) and (3) gives

\left|f_{2,2}^{n+1}\right|_{L_{\xi,\beta}^{\infty}}\leq 2\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]\text{.}

Consequently, we get the desired estimates for $f_{1,2}^{n}$ , $f_{2,1}^{n}$ , $f_{2,2}^{n}$ by induction on $n$ .

Finally, it is straightforward to prove that $\left\{\left(f_{1}^{n},f_{2}^{n}\right)\right\}$ is a Cauchy sequence in $L_{\xi,\beta}^{\infty}\left(L_{x}^{2}\cap L_{x}^{\infty}\right)$ . As a consequence, $\left(f_{1}^{n},f_{2}^{n}\right)$ converges to $\left(f_{1},f_{2}\right)$ in $L_{\xi,\beta}^{\infty}\left(L_{x}^{2}\cap L_{x}^{\infty}\right)$ and $\left(f_{1},f_{2}\right)$ solves the problem (4) with

\left|f_{1}\right|_{L_{\xi,\beta}^{\infty}}\leq 2\mathcal{C}_{1}\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}e^{-\frac{\nu_{0}}{2}\left(t+\left|x\right|\right)}\text{,}

and

	$\displaystyle\quad\left\|f_{2}\right\|_{L_{\xi,\beta}^{\infty}}$
	$\displaystyle\leq\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\left[\begin{array}[]{l}\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|^{2}}{\widehat{D}_{0}\left(1+t\right)}}\\[5.69054pt] +\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\|x\|}{\widehat{c}_{0}}}\end{array}\right]$
	$\displaystyle\quad+2\mathfrak{C}\left(\mathfrak{B}\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)^{2}\left[\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\right]\,.$

The proof of Theorem 9 is completed. ∎

4. Exponential decay outside the wave cone

In Theorem 9, the estimate for $f_{1}$ is exponentially sharp, while the estimate of $f_{2}$ is only polynomially sharp. The reason is that, to facilitate the closure of nonlinearity, we focused on the structure inside sound wave cone, so we chose a polynomial ansatz. However, since our initial data is compactly supported in space and $f_{2}$ represents the fluid part with an essentially finite propagation speed. Therefore, we expect a faster decay in the space-like region.

In this section, we will improve the behavior of $f_{2}$ outside the sound wave cone. Indeed, we can prove that $f_{2}$ has exponential decay both in space and time there. The result is stated as follows.

Theorem 10.

Under the same assumption of Theorem 9, for any $0<\delta\ll 1$ , if $\varepsilon>0$ is sufficiently small, there exists a large positive constant $D$ depending on $\delta$ , such that for $\left|x\right|>\left(\mathbf{c}+\delta\right)t$ ,

\left|f_{2}\right|_{L_{\xi,\beta}^{\infty}}\leq C_{\delta}\varepsilon e^{-\frac{\left\langle x\right\rangle+t}{D}}\text{,}

where the constant $C_{\delta}>0$ is independent of time and $C_{\delta}\rightarrow\infty$ as $\delta\rightarrow 0$ .

To attain this end, we consider the weighted nonlinear equation corresponding to $\left(\ref{decom-System}\right)$ . That is, let $u=f_{w}:=wf=u_{1}+\sqrt{\mathcal{M}}u_{2}$ , $u_{i}=wf_{i}$ $\left(i=1\text{, }2\right)$ , where the weight function is given by

w\left(t,x\right)=\exp\left(\frac{\left\langle x\right\rangle-Mt}{\ell}\right)\text{,}

$\mathbf{c}<M<\mathbf{c}+1$ and sufficiently large $\ell>0$ . Note that by Theorem 9, $f_{1}$ decays in space and time exponentially, so that $u_{1}$ can be controlled if $\ell$ is large enough, that is, there exists $c_{0}>0$ such that

\|u_{1}\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\text{, }\|u_{1}\|_{L_{\xi,\beta}^{\infty}L_{x}^{2}}\lesssim\varepsilon e^{-c_{0}t}.

In view of $\left(\ref{decom-System}\right)$ , $u_{2}$ satisfies the equation

\left\{\begin{array}[]{l}\partial_{t}u_{2}+\xi\cdot\nabla_{x}u_{2}-w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)u_{2}=Lu_{2}+\mathcal{K}_{b}u_{1}+\Gamma(f_{2},u_{2})\vspace{3mm}\\ u_{2}\left(0,x,\xi\right)=0\text{.}\end{array}\right.

After choosing $\ell>0$ large such that $M\ell^{-1}$ is small, we have

\widetilde{\nu}\left(t,x,\xi\right)=\nu\left(\xi\right)+w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\geq\frac{3}{4}\nu\left(\xi\right)\text{.}

Under this situation, we are ready to estimate $u_{2}$ . Let $T>0$ be a finite number. Denote

C_{u_{2},T}^{\infty}=\varepsilon^{-1}\sup_{0\leq t\leq T}\left\|u_{2}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\text{, \ \ \ \ }C_{u_{2},T}^{2}=\varepsilon^{-1}\sup_{0\leq t\leq T}\left\|u_{2}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{2}}\text{.}

Also denote

\begin{array}[]{ccc}C_{f_{2}}^{\infty}=\varepsilon^{-1}\sup\limits_{t\geq 0}\left(1+t\right)^{3/2}\left\|f_{2}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\text{,}&&C_{f_{2}}^{2}=\varepsilon^{-1}\sup\limits_{t\geq 0}\left(1+t\right)^{3/4}\left\|f_{2}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{2}}\text{,}\end{array}

which are finite due to Theorem 9.

To estimate $u_{2}$ , we design a Picard-type iteration: the zeroth order approximation $u_{2}^{\left(0\right)}$ is defined as

\partial_{t}u_{2}^{\left(0\right)}+\xi\cdot\nabla_{x}u_{2}^{\left(0\right)}+\widetilde{\nu}\left(\xi\right)u_{2}^{\left(0\right)}=\mathcal{K}_{b}u_{1}+\Gamma(f_{2},u_{2})\text{, \ }u_{2}^{\left(0\right)}\left(0,x,\xi\right)=0\text{,\ }

and define the $j$ th order approximation $u_{2}^{\left(j\right)}$ , $j\geq 1$ , inductively as

\partial_{t}u_{2}^{\left(j\right)}+\xi\cdot\nabla_{x}u_{2}^{\left(j\right)}+\widetilde{\nu}\left(\xi\right)u_{2}^{\left(j\right)}=Ku_{2}^{\left(j-1\right)}\text{, \ }u_{2}^{\left(j-1\right)}\left(0,x,\xi\right)=0\text{.}

Thus, the wave part and the remainder part are defined respectively as follows:

W_{w}^{(m)}=\sum_{j=0}^{m}u_{2}^{\left(j\right)}\text{, \ }\mathcal{R}_{w}^{\left(m\right)}=u_{2}-W_{w}^{\left(m\right)}\text{,}

$\mathcal{R}_{w}^{\left(m\right)}$ solving the equation

\partial_{t}\mathcal{R}_{w}^{\left(m\right)}+\xi\cdot\nabla_{x}\mathcal{R}_{w}^{\left(m\right)}+\widetilde{\nu}\left(\xi\right)\mathcal{R}_{w}^{\left(m\right)}=K\mathcal{R}_{w}^{\left(m\right)}+Ku_{2}^{\left(m\right)}\text{, \ }\mathcal{R}_{w}^{\left(m\right)}\left(0,x,\xi\right)=0\text{.}

In fact, $m=6$ is enough to get the $H_{x}^{2}$ regularization estimate for the remainder part. Following a similar argument as [21], we have

Lemma 11.

Let $\beta>4$ be large enough. Then for $0\leq t\leq T$ ,

\left\|u_{2}^{\left(j\right)}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\lesssim\varepsilon e^{-c_{0}t}+\varepsilon^{2}C_{u_{2},T}^{\infty}\left(1+t\right)^{-\frac{3}{2}}\text{, \ \ }\left\|u_{2}^{\left(j\right)}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{2}}\lesssim\varepsilon e^{-c_{0}t}+\varepsilon^{2}C_{u_{2},T}^{2}\left(1+t\right)^{-\frac{3}{2}}\text{,}

for all integers $j\geq 0$ .

Lemma 12.

For $0\leq t\leq T$ ,

\left\|u_{2}^{\left(6\right)}\right\|_{L_{\xi}^{2}H_{x}^{2}}\lesssim\varepsilon e^{-c_{0}t}+\varepsilon^{2}C_{u_{2},T}^{2}\left(1+t\right)^{-\frac{3}{2}}\text{.}

Proposition 13.

(the $H_{x}^{2}$ regularization estimate for $\mathcal{R}_{w}^{\left(6\right)}$ ) Let $0<\delta\ll 1$ . Then for $M=\mathbf{c}+25\delta$ and large $\ell>0$ , the corresponding weighted remainder $\mathcal{R}_{w}^{\left(6\right)}$ satisfies

\left\|\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}H_{x}^{2}}\leq C_{\delta}\left(\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}\right)

for some positive constant $C_{\delta}$ , $C_{\delta}\rightarrow\infty$ as $\delta\rightarrow 0$ .

Proof.

Note that

w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)=-\frac{M}{\ell}+\frac{\xi\cdot x}{\ell\left\langle x\right\rangle}\text{,}

\left|\partial_{x_{i}}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right]\right|\leq\frac{2\left|\xi\right|}{\ell\left\langle x\right\rangle}\text{, \ }\left|\partial_{x_{i}x_{j}}^{2}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right]\right|\leq\frac{6\left|\xi\right|}{\ell\left\langle x\right\rangle^{2}}

for $1\leq i$ , $j\leq 3$ . Let $0<\delta\ll 1$ . Consider the quantity

B\left(t\right):=\left\|\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}L_{x}^{2}}^{2}+\frac{\delta^{2}}{4\mathfrak{D}^{4}}\sum_{i=1}^{3}\left\|\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}L_{x}^{2}}^{2}+\left(\frac{\delta^{2}}{4\mathfrak{D}^{4}}\right)^{2}\sum_{i,j=1}^{3}\left\|\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}L_{x}^{2}}^{2}\text{,}

where the constant $\mathfrak{D}=\left(\int_{\mathbb{R}^{3}}\sum_{j=0}^{4}\left|\xi\right|\chi_{j}^{2}d\xi\right)^{1/2}>1$ . The direct computation gives

	$\displaystyle\frac{1}{2}\frac{d}{dt}\left\\|\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}$	$\displaystyle=$	$\displaystyle\int_{\mathbb{R}^{3}}\left\langle\left(-\frac{M}{\ell}+\frac{\xi\cdot x}{\ell\left\langle x\right\rangle}\right)\left(\mathrm{P}_{0}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\mathcal{R}_{w}^{\left(6\right)}\right),\left(\mathrm{P}_{0}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\mathcal{R}_{w}^{\left(6\right)}\right)\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{n}}\left\langle L\mathcal{R}_{w}^{\left(6\right)},\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}+\int_{\mathbb{R}^{3}}\left\langle Ku_{2}^{\left(6\right)},\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx\text{,}$

			$\displaystyle\frac{1}{2}\frac{d}{dt}\left\\|\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}$
		$\displaystyle=$	$\displaystyle\int_{\mathbb{R}^{3}}\left\langle\left(-\frac{M}{\ell}+\frac{\xi\cdot x}{\ell\left\langle x\right\rangle}\right)\left(\mathrm{P}_{0}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right),\left(\mathrm{P}_{0}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right)\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle L\left(\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right),\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle\mathcal{R}_{w}^{\left(6\right)}\partial_{x_{i}}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right],\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx+\int_{\mathbb{R}^{3}}\left\langle K\partial_{x_{i}}u_{2}^{\left(6\right)},\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx\text{,}$

			$\displaystyle\frac{1}{2}\frac{d}{dt}\left\\|\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}$
		$\displaystyle=$	$\displaystyle\int_{\mathbb{R}^{3}}\left\langle\left(-\frac{M}{\ell}+\frac{\xi\cdot x}{\ell\left\langle x\right\rangle}\right)\left(\mathrm{P}_{0}\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right),\left(\mathrm{P}_{0}\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}+\mathrm{P}_{1}\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right)\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle L\left(\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right),\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}+\int_{\mathbb{R}^{3}}\left\langle\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\cdot\partial_{x_{j}}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right],\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle\partial_{x_{j}}\mathcal{R}_{w}^{\left(6\right)}\cdot\partial_{x_{i}}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right],\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle\mathcal{R}_{w}^{\left(6\right)}\partial_{x_{i}x_{j}}^{2}\left[w^{-1}\left(\partial_{t}w+\xi\cdot\nabla_{x}w\right)\right],\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx$
			$\displaystyle+\int_{\mathbb{R}^{3}}\left\langle K\left(\partial_{x_{i}x_{j}}^{2}u_{2}^{\left(6\right)}\right),\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\rangle_{\xi}dx\text{.}$

Using the fact that in (6)

\mathrm{P}_{0}\left(\frac{x\cdot\xi}{\left|x\right|}\right)\mathrm{P}_{0}g=\lambda_{1}\left\langle g,E_{1}\right\rangle_{\xi}E_{1}+\lambda_{2}\left\langle g,E_{2}\right\rangle_{\xi}E_{2}\text{, }x\neq 0\text{,}

where $\lambda_{1}=-\lambda_{2}=\mathbf{c}$ , we have

			$\displaystyle\frac{1}{2}\frac{d}{dt}B\left(t\right)$
		$\displaystyle=$	$\displaystyle-\frac{1}{\ell}\left(M-\mathbf{c}-\delta-3\delta-\frac{81\delta^{3}}{4\mathfrak{D}^{4}}\right)\left\\|\mathrm{P}_{0}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}-\left(\nu_{0}-\frac{1}{\ell}-\frac{\mathfrak{D}^{2}}{4\ell\delta}-\frac{3\delta}{\ell\mathfrak{D}^{2}}-\frac{81\delta^{3}}{4\ell\mathfrak{D}^{6}}\right)\left\\|\mathrm{P}_{1}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\sigma}^{2}L_{x}^{2}}^{2}$
			$\displaystyle+\frac{\delta^{2}}{4\mathfrak{D}^{4}}\sum_{i=1}^{3}\left\{-\frac{1}{\ell}\left(M-\mathbf{c}-\delta-6\delta\right)\left\\|\mathrm{P}_{0}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}-\left(\nu_{0}-\frac{1}{\ell}-\frac{\mathfrak{D}^{2}}{4\ell\delta}-\frac{6\delta^{3}}{4\ell\mathfrak{D}^{6}}\right)\left\\|\mathrm{P}_{1}\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\sigma}^{2}L_{x}^{2}}^{2}\right\}$
			$\displaystyle+\left(\frac{\delta^{2}}{4\mathfrak{D}^{4}}\right)^{2}\sum_{i,j=1}^{3}\left\{-\frac{1}{\ell}\left(M-\mathbf{c}-4\delta\right)\left\\|\mathrm{P}_{0}\partial_{x_{i}x_{j}}^{2}\mathcal{R}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}-\left(\nu_{0}-\frac{1}{\ell}-\frac{\mathfrak{D}^{2}}{4\ell\delta}-\frac{3\delta}{\ell\mathfrak{D}^{2}}\right)\left\\|\mathrm{P}_{1}\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\sigma}^{2}L_{x}^{2}\xi}^{2}\right\}$
			$\displaystyle+C\left\\|u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}\left\\|\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}+\frac{\delta^{2}}{4\mathfrak{D}^{4}}\sum_{i=1}^{3}\left[C\left\\|\partial_{x_{i}}u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}\left\\|\partial_{x_{i}}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}\right]$
			$\displaystyle+\left(\frac{\delta^{2}}{4\mathfrak{D}^{4}}\right)^{2}\sum_{i,j=1}^{3}\left[C\left\\|\partial_{x_{i}x_{j}}^{2}u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}\left\\|\partial_{x_{i}x_{j}}^{2}\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}\right]$
		$\displaystyle\leq$	$\displaystyle 3C\left[\left\\|u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}+\frac{\delta^{2}}{4\mathfrak{D}^{4}}\sum_{i=1}^{3}\left\\|\partial_{x_{i}}u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}+\left(\frac{\delta^{2}}{4\mathfrak{D}^{4}}\right)^{2}\sum_{i,j=1}^{3}\left\\|\partial_{x_{i}x_{j}}^{2}u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{2}}^{2}\right]^{1/2}\sqrt{B\left(t\right)}$
		$\displaystyle\leq$	$\displaystyle 3C\left\\|u_{2}^{\left(6\right)}\right\\|_{L_{\xi}^{2}H_{x}^{2}}\sqrt{B\left(t\right)}$

if we choose $M=$ $\mathbf{c}+25\delta$ , and then choose $\ell>0$ sufficiently large such that $\nu_{0}-\frac{166}{\ell}-\frac{\mathfrak{D}^{2}}{4\ell\delta}>0$ . Under this choice, we have

\sqrt{B\left(t\right)}\leq\int_{0}^{t}3C\left\|u_{2}^{\left(6\right)}\right\|_{L_{\xi}^{2}H_{x}^{2}}d\tau\lesssim\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}

by Lemma 12, so that

\left\|\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}H_{x}^{2}}\leq C_{\delta}\left(\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}\right)\text{,}

for some positive constant $C_{\delta}$ , $C_{\delta}\rightarrow\infty$ as $\delta\rightarrow 0$ . The proof of this proposition is complete. ∎

Therefore, the Sobolev inequality implies that

\left\|\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}L_{x}^{\infty}}\lesssim C_{\delta}\left(\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}\right).

Combining this with Lemma 11, we have

$\displaystyle\left\\|u_{2}\right\\|_{L_{\xi}^{2}L_{x}^{\infty}}$	$\displaystyle\leq$	$\displaystyle\left\\|W^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{\infty}}+\left\\|\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{\infty}}$
	$\displaystyle\lesssim$	$\displaystyle\left\\|W^{\left(6\right)}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}+\left\\|\mathcal{R}_{w}^{\left(6\right)}\right\\|_{L_{\xi}^{2}L_{x}^{\infty}}$
	$\displaystyle\lesssim$	$\displaystyle\varepsilon e^{-c_{0}t}+\varepsilon^{2}C_{u_{2},T}^{\infty}\left(1+t\right)^{-\frac{3}{2}}+C_{\delta}\left(\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}\right)\,.$

According to the wave-remainder decomposition, $u_{2}=W^{\left(7\right)}+\mathcal{R}_{w}^{\left(7\right)}$ and

\mathcal{R}_{w}^{\left(7\right)}=\int_{0}^{t}\exp\left(-\int_{\tau}^{t}\widetilde{\nu}\left(r,x-\left(t-r\right)\xi,\xi\right)dr\right)\left(K\mathcal{R}_{w}^{\left(6\right)}\right)\left(\tau\right)d\tau\text{.}

Since

\left\|\mathcal{R}_{w}^{\left(7\right)}\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\lesssim\int_{0}^{t}e^{-\frac{3}{4}\nu_{0}\left(t-\tau\right)}\left\|\mathcal{R}_{w}^{\left(6\right)}\right\|_{L_{\xi}^{2}L_{x}^{\infty}}\left(\tau\right)d\tau\text{,}

we get

\left\|u_{2}\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\lesssim\varepsilon e^{-c_{0}t}+\varepsilon^{2}C_{u_{2},T}^{\infty}\left(1+t\right)^{-\frac{3}{2}}+C_{\delta}\left(\varepsilon+\varepsilon^{2}C_{u_{2},T}^{2}\right)\,.

By finite steps of bootstrap argument, we obtain the desired $L_{\xi,\beta}^{\infty}L_{x}^{\infty}$ estimate for $u_{2}$ . Similarly, by Lemma 11, Proposition 13, and the bootstrap argument, we obtain $L_{\xi,\beta}^{\infty}L_{x}^{2}$ estimate for $u_{2}$ as well. We summarize the estimates for $u_{2}$ as below.

Proposition 14.

Let $\beta>4$ be large enough and let $0<\delta\ll 1$ . Then for $M=\mathbf{c}+25\delta$ and large $\ell>0$ , the corresponding $u_{2}$ satisfies

C_{u_{2},T}^{2}\leq C_{1}\left[1+\varepsilon C_{u_{2},T}^{2}+C_{\delta}\left(1+\varepsilon C_{u_{2},T}^{2}\right)\right]

C_{u_{2},T}^{\infty}\leq C_{2}\left[1+\varepsilon C_{u_{2},T}^{\infty}+C_{\delta}\left(1+\varepsilon C_{u_{2},T}^{2}\right)\right]

for some positive constants $C_{1}$ and $C_{2}$ dependent on $\|f_{0}\|_{L_{\xi,\beta}^{\infty}\left(L_{x}^{\infty}\cap L_{x}^{2}\right)}$ but independent of $T$ and $\delta$ .

Now, we are ready to prove Theorem 10. For any fixed $0<\delta\ll 1$ , we take $M=\mathbf{c}+25\delta$ , and consider the weight function

w\left(x,t\right)=\exp\left(\frac{\left\langle x\right\rangle-Mt}{\ell}\right)\,\text{,}

with $\ell>0$ being chosen large. In view of Proposition 14, choosing $\varepsilon>0$ sufficiently small gives

\left|u_{2}\right|_{L_{\xi,\beta}^{\infty}}\leq CC_{\delta}\varepsilon\text{,}

for some positive constant $C$ dependent on $\|f_{0}\|_{L_{\xi,\beta}^{\infty}\left(L_{x}^{\infty}\cap L_{x}^{2}\right)}$ .

Note that $\left|x\right|>\left(M+\delta\right)t=\left(\mathbf{c}+26\delta\right)t$ , we have

	$\displaystyle\left\langle x\right\rangle-Mt$	$\displaystyle=$	$\displaystyle\frac{\frac{\delta}{2M}\left\langle x\right\rangle}{2+\frac{\delta}{2M}}+\frac{2\left\langle x\right\rangle}{2+\frac{\delta}{2M}}-Mt$
		$\displaystyle>$	$\displaystyle\frac{\frac{\delta}{2M}\left\langle x\right\rangle}{2+\frac{\delta}{2M}}+\frac{\frac{\delta}{2}t}{2+\frac{\delta}{2M}}\geq\frac{\delta\left(\left\langle x\right\rangle+t\right)}{4M+\delta}\text{.}$

Then for $\left|x\right|>\left(\mathbf{c}+26\delta\right)t$ ,

e^{\frac{\delta\left(\left\langle x\right\rangle+t\right)}{\left(4M+\delta\right)\ell}}\left|f_{2}\right|_{L_{\xi,\beta}^{\infty}}\leq\left|u_{2}\right|_{L_{\xi,\beta}^{\infty}}\leq CC_{\delta}\varepsilon

which implies that

\left|f_{2}\right|_{L_{\xi,\beta}^{\infty}}\leq CC_{\delta}\varepsilon e^{-\frac{\delta\left(\left\langle x\right\rangle+t\right)}{\left(4M+\delta\right)\ell}}\text{.}

The proof of Theorem 10 is complete.

5. The transition from polynomial tail to Gaussian tail

In this section, we study the behavior of $f_{1}$ in the microscopic variable as $t$ increases. We provide a quantitative description of how the velocity variable transitions from a polynomial tail to a Gaussian tail. The result is as follows:

Theorem 15.

Let $\beta>4$ , $R>0$ be sufficiently large and $0<\kappa<\min\{1/4$ , $\nu_{0}/2\}$ . Assume that $f_{0}$ satisfies the same condition as in Theorem 9. If $\varepsilon>0$ is sufficiently small, then there exists a constant $\overline{C}_{\beta}>0$ only depending on $\beta$ such that

\left|f_{1}\left(t,x,\xi\right)\right|\leq\overline{C}_{\beta}\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\kappa\rho\left(\xi,t\right)}\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}

for all $t\geq 0$ , $x\in\mathbb{R}^{3}$ , $\xi\in\mathbb{R}^{3}$ , where $\rho\left(\xi,t\right)=\left\langle\xi\right\rangle\left(\left\langle\xi\right\rangle\wedge t\right)$ . Consequently, for any fixed $t>0$ , then

|\mathbf{1}_{\{\left<\xi\right>\leq t\}}f_{1}|\lesssim\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\kappa\left<\xi\right>^{2}}

and

|\mathbf{1}_{\{\left<\xi\right>>t\}}f_{1}|\lesssim\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\kappa t^{2}}\,.

Firstly, we need some estimate for the weight function $e^{\kappa\rho\left(\xi,t\right)}$ . For simplicity, we define

(20)

\rho\left(\xi,t\right)=\overline{\rho}\left(\left|\xi\right|,t\right)\hbox{, }\xi\in\mathbb{R}^{3}\hbox{, }t\geq 0\hbox{,}

with $\overline{\rho}\left(z,t\right)=\left(1+z^{2}\right)^{1/2}\left(\left(1+z^{2}\right)^{1/2}\wedge t\right)$ , $t\geq 0$ . Now, we give an inequality regarding the function $\overline{\rho}$ .

Lemma 16.

Let $\overline{\rho}\left(z,t\right)$ be a function defined by

\overline{\rho}\left(z,t\right)=\left(1+z^{2}\right)^{1/2}\left(\left(1+z^{2}\right)^{1/2}\wedge t\right)

for $z\in\mathbb{R}$ and $t\geq 0$ . Then

\overline{\rho}\left(a,t\right)+\overline{\rho}\left(b,t\right)\geq\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)

for all $a$ , $b\geq 0$ .

Proof.

If either $a=0$ or $b=0$ , it is trivial. By symmetry, we may assume that $b>a>0$ and so $\sqrt{a^{2}+b^{2}}>b>a>0$ . In the following we discuss the inequality in four cases.

Case 1: $t\geq\sqrt{a^{2}+b^{2}}>b>a>0$ . Then

\overline{\rho}\left(a,t\right)+\overline{\rho}\left(b,t\right)=2+a^{2}+b^{2}=1+\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)\geq\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)\hbox{.}

Case 2: $\sqrt{a^{2}+b^{2}}>b>a>t$ . Then

	$\displaystyle\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)$	$\displaystyle=$	$\displaystyle\left(1+a^{2}+b^{2}\right)^{1/2}t\leq\left(1+a^{2}+1+b^{2}\right)^{1/2}t$
		$\displaystyle\leq$	$\displaystyle\left(1+a^{2}\right)^{1/2}t+\left(1+b^{2}\right)^{1/2}t=\overline{\rho}\left(a,t\right)+\overline{\rho}\left(b,t\right)\hbox{.}$

Case 3: $\sqrt{a^{2}+b^{2}}>t>b>a$ . Then

\overline{\rho}\left(a,t\right)+\overline{\rho}\left(b,t\right)=2+a^{2}+b^{2}\geq\left(1+a^{2}+b^{2}\right)^{1/2}t=\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)\hbox{.}

Case 4: $\sqrt{a^{2}+b^{2}}>b>t>a$ . Then

			$\displaystyle\left[\overline{\rho}\left(a,t\right)+\overline{\rho}\left(b,t\right)\right]^{2}-\left[\overline{\rho}\left(\sqrt{a^{2}+b^{2}},t\right)\right]^{2}$
		$\displaystyle=$	$\displaystyle\left[1+a^{2}+\left(1+b^{2}\right)^{1/2}t\right]^{2}-\left(1+a^{2}+b^{2}\right)t^{2}$
		$\displaystyle=$	$\displaystyle\left(1+a^{2}\right)^{2}+2\left(1+a^{2}\right)\left(1+b^{2}\right)^{1/2}t-a^{2}t^{2}\geq 2\left(1+a^{2}\right)t^{2}-a^{2}t^{2}\geq 0\hbox{.}$

As $a=b>0$ , it is a consequence of Case 1-Case 3. Gathering all the cases, the proof is complete. $\hfill\square$

According to this lemma, we can prove the following weighted estimate regarding $Q$ and $\mathcal{K}$ .

Lemma 17.

Let $\beta>4$ , $\kappa>0$ and let $\rho\left(\xi,t\right)$ be defined by $(\ref{weight-exponent})$ . Then there exists a constant $C_{\beta}>0$ depending only on $\beta$ such that

(21)

\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|\left(\xi-\xi_{\ast}\right)\cdot n\right|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}\left\langle\xi^{\prime}\right\rangle^{\beta}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{*}\leq\left(\frac{C}{\beta}+\frac{C_{\beta}}{\left\langle\xi\right\rangle^{2}}\right)\nu\left(\xi\right)\hbox{.}

Moreover, if $0<\kappa<\frac{1}{4}$ , we have

(22)

\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|\left(\xi-\xi_{\ast}\right)\cdot n\right|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}}e^{-\frac{1}{4}|\xi^{\prime 2}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{*}\leq\left(\frac{C}{\beta}+\frac{C_{\beta}}{\left\langle\xi\right\rangle^{2}}\right)\nu\left(\xi\right)\hbox{.}

Proof.

Let $\eta=\xi^{\prime}-\xi$ and $\omega=\xi_{\ast}-\xi^{\prime}$ . Then $\eta\bot\omega$ , $\xi^{\prime}=\xi+\eta$ , $\xi_{\ast}=\xi+\eta+\omega$ , $\xi_{\ast}^{\prime}=\xi+\omega$ . By change of variables,

			$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left\|\left(\xi-\xi_{\ast}\right)\cdot n\right\|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}\left\langle\xi^{\prime}\right\rangle^{\beta}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{*}$
		$\displaystyle\leq$	$\displaystyle\int_{\mathbb{R}^{3}}\int_{\{\omega\hbox{ }:\hbox{ }\eta\bot\omega\}}\frac{1}{\left\|\eta\right\|}\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi+\omega\right\rangle^{\beta}\left\langle\xi+\eta\right\rangle^{\beta}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi+\omega,t\right)}e^{\kappa\rho\left(\xi+\eta,t\right)}}d\omega d\eta=:\mathrm{I}\hbox{.}$

We split $\xi$ into two parts

\xi=\xi_{\|}+\xi_{\bot}\hbox{, }\xi_{\bot}=\frac{\left(\xi\cdot\eta\right)\eta}{\left|\eta\right|^{2}}\hbox{, }\xi_{\|}\|\omega\hbox{, }\xi_{\bot}\bot\omega\hbox{, }\left|\xi\right|^{2}=\left|\xi_{\|}\right|^{2}+\left|\xi_{\bot}\right|^{2}\hbox{.}

Then

	$\displaystyle\mathrm{I}$	$\displaystyle=$	$\displaystyle\int_{\mathbb{R}^{3}}\frac{\left\langle\xi\right\rangle^{\beta}e^{\kappa\left[\rho\left(\xi,t\right)-\rho\left(\xi+\eta,t\right)\right]}}{\left\langle\xi+\eta\right\rangle^{\beta}\left\|\eta\right\|}\int_{\{\omega\hbox{ }:\hbox{ }\eta\bot\omega\}}\frac{e^{-\kappa\left(1+\left\|\xi_{\\|}+\omega\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\left(\left(1+\left\|\xi_{\\|}+\omega\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\wedge t\right)}}{\left(1+\left\|\xi_{\\|}+\omega\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{\beta/2}}d\omega d\eta$
		$\displaystyle=$	$\displaystyle\int_{\mathbb{R}^{3}}\frac{\left\langle\xi\right\rangle^{\beta}e^{\kappa\left[\rho\left(\xi,t\right)-\rho\left(\xi+\eta,t\right)\right]}}{\left\langle\xi+\eta\right\rangle^{\beta}\left\|\eta\right\|}\int_{\mathbb{R}^{2}}\frac{e^{-\kappa\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\left[\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\wedge t\right]}}{\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{\beta/2}}d\varpi d\eta\hbox{.}$

Making a change of variable by $\varpi=\sqrt{1+\left|\xi_{\bot}\right|^{2}}z$ gives

			$\displaystyle\int_{\mathbb{R}^{2}}\frac{e^{-\kappa\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\left[\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\wedge t\right]}}{\left(1+\left\|\varpi\right\|^{2}+\left\|\xi_{\bot}\right\|^{2}\right)^{\beta/2}}d\varpi$
		$\displaystyle=$	$\displaystyle\left(1+\left\|\xi_{\bot}\right\|^{2}\right)^{-\frac{\beta-2}{2}}\int_{\mathbb{R}^{2}}\frac{e^{-\kappa\left(1+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\left(1+\left\|z\right\|^{2}\right)^{1/2}\left[\left(1+\left\|\xi_{\bot}\right\|^{2}\right)^{1/2}\left(1+\left\|z\right\|^{2}\right)^{1/2}\wedge t\right]}}{\left(1+\left\|z\right\|^{2}\right)^{\beta/2}}dz$
		$\displaystyle=$	$\displaystyle\frac{2\pi}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\int_{1}^{\infty}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle\zeta\left[\left\langle\xi_{\bot}\right\rangle\zeta\wedge t\right]}}{\zeta^{\beta-1}}d\zeta\hbox{.}$

If $\left\langle\xi_{\bot}\right\rangle\geq t$ , then

	$\displaystyle\frac{2\pi}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\int_{1}^{\infty}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle\zeta\left[\left\langle\xi_{\bot}\right\rangle\zeta\wedge t\right]}}{\zeta^{\beta-1}}d\zeta$	$\displaystyle=$	$\displaystyle\frac{2\pi}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\int_{1}^{\infty}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle t\zeta}}{\zeta^{\beta-1}}d\zeta$
		$\displaystyle\leq$	$\displaystyle\frac{2\pi}{\beta-2}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle t}}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}=\frac{2\pi}{\beta-2}\frac{e^{-\kappa\rho(\xi_{\bot},t)}}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\hbox{.}$

If $\left\langle\xi_{\bot}\right\rangle<t$ , then $\left\langle\xi_{\bot}\right\rangle\zeta<t$ for $\zeta<t/\left\langle\xi_{\bot}\right\rangle$ and thus

	$\displaystyle\frac{2\pi}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\int_{1}^{\infty}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle\zeta\left[\left\langle\xi_{\bot}\right\rangle\zeta\wedge\tau\right]}}{\zeta^{\beta-1}}d\zeta$	$\displaystyle=$	$\displaystyle\frac{2\pi}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\left[\int_{1}^{\frac{t}{\left\langle\xi_{\bot}\right\rangle}}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle^{2}\zeta^{2}}}{\zeta^{\beta-1}}d\zeta+\int_{\frac{t}{\left\langle\xi_{\bot}\right\rangle}}^{\infty}\frac{e^{-\kappa\left\langle\xi_{\bot}\right\rangle\zeta t}}{\zeta^{\beta-1}}d\zeta\right]$
		$\displaystyle\leq$	$\displaystyle\frac{2\pi}{\beta-2}\frac{1}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\left(e^{-\kappa\left\langle\xi_{\bot}\right\rangle^{2}}+e^{-\kappa t^{2}}\right)\leq\frac{4\pi}{\beta-2}\frac{e^{-\kappa\rho(\xi_{\bot},t)}}{\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\hbox{.}$

Therefore,

\mathrm{I}\leq\frac{4\pi}{\beta-2}\int_{\mathbb{R}^{3}}\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi+\eta\right\rangle^{\beta}\left\langle\xi_{\bot}\right\rangle^{\beta-2}}\frac{e^{\kappa\left[\rho\left(\xi,t\right)-\rho\left(\xi+\eta,t\right)-\rho(\xi_{\bot},t)\right]}}{\left|\eta\right|}d\eta\hbox{.}

Observe that $\left|\xi_{\bot}\right|^{2}+\left|\xi+\eta\right|^{2}=\left|\xi_{\bot}\right|^{2}+\left|\xi\right|^{2}+2\xi\cdot\eta+\left|\eta\right|^{2}=\left|\xi_{\bot}\right|^{2}+\left|\xi\right|^{2}+2\xi_{\bot}\cdot\eta+\left|\eta\right|^{2}=\left|\xi\right|^{2}+\left|\xi_{\bot}+\eta\right|^{2}$ . Hence,

	$\displaystyle\rho\left(\xi,t\right)-\rho\left(\xi+\eta,t\right)-\rho(\xi_{\bot},t)$	$\displaystyle=$	$\displaystyle\overline{\rho}\left(\left\|\xi\right\|,t\right)-\overline{\rho}\left(\left\|\xi+\eta\right\|,t\right)-\overline{\rho}(\left\|\xi_{\bot}\right\|,t)$
		$\displaystyle\leq$	$\displaystyle\overline{\rho}\left(\sqrt{\left\|\xi_{\bot}\right\|^{2}+\left\|\xi+\eta\right\|^{2}},t\right)-\overline{\rho}\left(\left\|\xi+\eta\right\|,t\right)-\overline{\rho}(\left\|\xi_{\bot}\right\|,t)\leq 0\hbox{,}$

the last inequality being valid due to Lemma 16. It follows

\mathrm{I}\leq\frac{4\pi}{\beta-2}\int_{\mathbb{R}^{3}}\frac{1}{\left|\eta\right|}\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi+\eta\right\rangle^{\beta}\left\langle\xi_{\bot}\right\rangle^{\beta-2}}d\eta=\frac{4\pi}{\beta-2}\int_{\mathbb{R}^{3}}\frac{1}{\left|\xi^{\prime}-\xi\right|}\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi^{\prime}\right\rangle^{\beta}\left\langle\xi_{\bot}\right\rangle^{\beta-2}}d\xi^{\prime}\hbox{.}

According to the estimate in the proof of [5, Lemma 2.11], there exists a constant $C_{\beta}>0$ depending only on $\beta$ such that

\mathrm{I}\leq\frac{4\pi}{\beta-2}\int_{\mathbb{R}^{3}}\frac{1}{\left|\xi^{\prime}-\xi\right|}\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi^{\prime}\right\rangle^{\beta}\left\langle\xi_{\bot}\right\rangle^{\beta-2}}d\xi^{\prime}\leq\left(\frac{C}{\beta}+\frac{C_{\beta}}{\left\langle\xi\right\rangle^{2}}\right)\nu\left(\xi\right)\hbox{,}

where $C>0$ is a universal constant. This completes the proof of (21).

For the estimate of (22), applying the same argument as (21), one gets

	$\displaystyle\quad\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left\|\left(\xi-\xi_{\ast}\right)\cdot n\right\|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}}e^{-\frac{1}{4}\|\xi^{\prime 2}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{*}$
	$\displaystyle\leq\frac{4\pi}{\beta-2}\int_{\mathbb{R}^{3}}\frac{1}{\left\|\xi^{\prime}-\xi\right\|}\frac{\left\langle\xi\right\rangle^{\beta}}{e^{\frac{1}{4}\|\xi^{\prime 2}}\left\langle\xi_{\bot}\right\rangle^{\beta-2}}d\xi^{\prime}\,.$

Then one can modify the argument in [5, Lemma 2.11] (in fact, it is easier) to conclude our result. ∎

Corollary 18.

Let $\beta>4$ , $\kappa>0$ and let $\rho\left(\xi,t\right)$ be defined by $(\ref{weight-exponent})$ . Then there exists a constant $C_{\beta}^{{}^{\prime\prime}}>0$ depending only on $\beta$ such that

e^{\kappa\rho\left(\xi,t\right)}\left\langle\xi\right\rangle^{\beta}\left|Q\left(g,h\right)\right|\leq C_{\beta}^{{}^{\prime\prime}}\nu\left(\xi\right)\left|e^{\kappa\rho\left(\xi,t\right)}g\right|_{L_{\xi,\beta}^{\infty}}\left|e^{\kappa\rho\left(\xi,t\right)}h\right|_{L_{\xi,\beta}^{\infty}}\hbox{.}

Moreover, for any $R>0$ , and $0<\kappa<\frac{1}{4}$ , we have

\chi_{\{\left|\xi\right|\geq R\}}e^{\kappa\rho\left(\xi,t\right)}\left\langle\xi\right\rangle^{\beta}\left|\mathcal{K}f\right|\leq\left(\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}\right)\nu\left(\xi\right)\left|e^{\kappa\rho(\xi,t)}f\right|_{L_{\xi,\beta}^{\infty}}\,.

Proof.

We only prove the estimate of $Q$ since the estimate of $\mathcal{K}$ is similar. By definition of $Q$ ,

$\displaystyle e^{\kappa\rho\left(\xi,t\right)}\left\langle\xi\right\rangle^{\beta}\left\|Q\left(g,h\right)\right\|$	$\displaystyle\leq$	$\displaystyle\left\|e^{\kappa\rho\left(\xi,t\right)}g\right\|_{L_{\xi,\beta}^{\infty}}\left\|e^{\kappa\rho\left(\xi,t\right)}h\right\|_{L_{\xi,\beta}^{\infty}}$
		$\displaystyle\cdot\left[\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left\|\left(\xi-\xi_{\ast}\right)\cdot n\right\|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}\left\langle\xi^{\prime}\right\rangle^{\beta}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{\ast}\right.$
		$\displaystyle\left.+\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left\|\left(\xi-\xi_{\ast}\right)\cdot n\right\|\frac{1}{\left\langle\xi_{\ast}\right\rangle^{\beta}e^{\kappa\rho\left(\xi_{\ast},\tau\right)}}dnd\xi_{\ast}\right]\hbox{.}$

In view of Lemma 17,

\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left|\left(\xi-\xi_{\ast}\right)\cdot n\right|\frac{\left\langle\xi\right\rangle^{\beta}}{\left\langle\xi_{\ast}^{\prime}\right\rangle^{\beta}\left\langle\xi^{\prime}\right\rangle^{\beta}}\frac{e^{\kappa\rho\left(\xi,t\right)}}{e^{\kappa\rho\left(\xi_{\ast}^{\prime},t\right)}e^{\kappa\rho\left(\xi^{\prime},t\right)}}dnd\xi_{\ast}\leq C_{\beta}^{\prime}\nu\left(\xi\right)

for some $C_{\beta}^{\prime}>0$ . By straightforward computation,

			$\displaystyle\int_{\mathbb{R}^{3}}\int_{\mathbb{S}^{2}}\left\|\left(\xi-\xi_{\ast}\right)\cdot n\right\|\frac{1}{\left\langle\xi_{\ast}\right\rangle^{\beta}e^{\kappa\rho\left(\xi_{\ast},\tau\right)}}dnd\xi_{\ast}$
		$\displaystyle\leq$	$\displaystyle 4\pi\left[\int_{\left\|\xi_{\ast}\right\|\leq\left\|\xi\right\|}\left\|\xi-\xi_{\ast}\right\|\frac{1}{\left\langle\xi_{\ast}\right\rangle^{\beta}}d\xi_{\ast}+\int_{\left\|\xi_{\ast}\right\|>\left\|\xi\right\|}\left\|\xi-\xi_{\ast}\right\|\frac{1}{\left\langle\xi_{\ast}\right\rangle^{\beta}}d\xi_{\ast}\right]$
		$\displaystyle\leq$	$\displaystyle\left(4\pi\right)^{2}\left(\frac{2\left\|\xi\right\|}{\beta-3}+\frac{2}{\beta-4}\frac{1}{\left\langle\xi\right\rangle^{\beta-4}}\right)\hbox{.}$

Hence,

e^{\kappa\rho\left(\xi,t\right)}\left\langle\xi\right\rangle^{\beta}\left|Q\left(g,h\right)\right|\leq C_{\beta}^{{}^{\prime\prime}}\nu\left(\xi\right)\left|e^{\kappa\rho\left(\xi,t\right)}g\right|_{L_{\xi,\beta}^{\infty}}\left|e^{\kappa\rho\left(\xi,t\right)}h\right|_{L_{\xi,\beta}^{\infty}}

for some constant $C_{\beta}^{{}^{\prime\prime}}>0$ only depending upon $\beta$ , as desired. ∎

Proof of Theorem 15.

In virtue of $\left(\ref{decom-System}\right)$ , $f_{1}$ can be expressed as

f_{1}=\mathbb{S}^{t}f_{0}+\int_{0}^{t}\mathbb{S}^{t-\tau}\left[K_{s}f_{1}+Q\left(f_{1},f_{1}\right)+Q\left(f_{1},\sqrt{\mathcal{M}}f_{2}\right)+Q\left(\sqrt{\mathcal{M}}f_{2},f_{1}\right)\right]\left(\tau\right)d\tau\hbox{.}

Let $T>0$ be any number and denote

u\left(t,x,\xi\right)=\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}f_{1}\left(t,x,\xi\right)\hbox{.}

Multiplying $\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}$ on both sides of the integral equation gives

$\displaystyle\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}f_{1}$	$\displaystyle=$	$\displaystyle\varepsilon\mathbb{S}^{t}\left(\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}f_{0}\right)+\int_{0}^{t}\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}\mathbb{S}^{t-\tau}Q\left(f_{1},f_{1}\right)\left(\tau\right)d\tau$
		$\displaystyle+\int_{0}^{t}\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}\mathbb{S}^{t-\tau}\left[K_{s}f_{1}+Q\left(f_{1},\sqrt{\mathcal{M}}f_{2}\right)+Q\left(\sqrt{\mathcal{M}}f_{2},f_{1}\right)\right]\left(\tau\right)d\tau$
	$\displaystyle=$	$\displaystyle:\mathrm{I}+\mathrm{II}+\mathrm{III}\hbox{.}$

At first, it is easy to see

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\left|\mathrm{I}\right|\leq\varepsilon\left\|\left\langle\xi\right\rangle^{\beta}e^{-\nu_{0}\left\langle\xi\right\rangle t+\kappa\left\langle\xi\right\rangle t}\left|f_{0}\right|\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\leq\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\hbox{,}

since $\kappa\rho\left(\xi,t\right)\leq\kappa\left\langle\xi\right\rangle t$ .

As for $\mathrm{II}$ and $\mathrm{III}$ , we can find

e^{-\nu\left(\xi\right)\left(t-\tau\right)}e^{\kappa\rho\left(\xi,t\right)}\leq e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\rho\left(\xi,\tau\right)}

for all $\xi$ and $0\leq\tau\leq t$ . To see this, for $\left\langle\xi\right\rangle>t$ , we have $\rho\left(\xi,t\right)=\left\langle\xi\right\rangle t$ and thus

e^{-\nu\left(\xi\right)\left(t-\tau\right)}e^{\kappa\rho\left(\xi,t\right)}=e^{-\nu\left(\xi\right)\left(t-\tau\right)}e^{\kappa\left\langle\xi\right\rangle t}\leq e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\left\langle\xi\right\rangle\tau}=e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\rho\left(\xi,\tau\right)}\hbox{;}

for $\left\langle\xi\right\rangle\leq t$ , we have $\rho\left(\xi,t\right)=\left\langle\xi\right\rangle^{2}$ , so that

	$\displaystyle e^{-\nu\left(\xi\right)\left(t-\tau\right)}e^{\kappa\rho\left(\xi,t\right)}$	$\displaystyle\leq$	$\displaystyle e^{-\nu_{0}\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\left\langle\xi\right\rangle^{2}}=e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\left\langle\xi\right\rangle\left(\left\langle\xi\right\rangle-t\right)+\kappa\left\langle\xi\right\rangle\tau}$
		$\displaystyle\leq$	$\displaystyle e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\rho\left(\xi,\tau\right)}\hbox{.}$

Therefore,

\left|\mathrm{II}\right|\leq\int_{0}^{t}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}e^{\kappa\rho\left(\xi,\tau\right)}\left\langle\xi\right\rangle^{\beta}\left|Q\left(f_{1},f_{1}\right)\left(\tau\right)\right|d\tau\hbox{,}

\left|\mathrm{III}\right|\leq\int_{0}^{\infty}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,\tau\right)}\left|\left[K_{s}f_{1}+Q\left(f_{1},\sqrt{\mathcal{M}}f_{2}\right)+Q\left(\sqrt{\mathcal{M}}f_{2},f_{1}\right)\right]\left(\tau\right)\right|d\tau\hbox{.}

By Corollary 18, we have

(24)

\left|\mathrm{II}\right|\leq C_{\beta}^{\prime\prime}\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}^{2}\int_{0}^{t}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\nu\left(\xi\right)d\tau\leq\frac{C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}\left(\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\right)^{2}\hbox{.}

Regarding $\mathrm{III}$ , in view of Theorem 9 and Corollary 18, together the fact that

e^{\kappa\rho\left(\xi,\tau\right)}\sqrt{\mathcal{M}}=e^{\kappa\left\langle\xi\right\rangle\left(\left\langle\xi\right\rangle\wedge\tau\right)}e^{-\frac{\left|\xi\right|^{2}}{4}}\leq e^{\kappa\left\langle\xi\right\rangle^{2}-\frac{\left|\xi\right|^{2}}{4}}\leq e^{1/4}\hbox{,}

we have

			$\displaystyle\int_{0}^{\infty}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,\tau\right)}\left\|Q\left(f_{1},\sqrt{\mathcal{M}}f_{2}\right)+Q\left(\sqrt{\mathcal{M}}f_{2},f_{1}\right)\right\|d\tau$
		$\displaystyle\leq$	$\displaystyle 2C_{\beta}^{\prime\prime}\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\cdot\sup_{0\leq t\leq T}\left\\|e^{\kappa\rho\left(\xi,\tau\right)}\sqrt{\mathcal{M}}f_{2}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\int_{0}^{t}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\nu\left(\xi\right)d\tau$
		$\displaystyle\leq$	$\displaystyle\left(\frac{\widetilde{C}_{\beta}\varepsilon}{\nu_{0}-\kappa}\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\hbox{.}$

By Corollary 18,

			$\displaystyle\int_{0}^{\infty}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,\tau\right)}\left\|K_{s}f_{1}\right\|d\tau$
		$\displaystyle\leq$	$\displaystyle\left(\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}\right)\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\int_{0}^{\infty}e^{-\left(\nu_{0}-\kappa\right)\left\langle\xi\right\rangle\left(t-\tau\right)}\nu\left(\xi\right)d\tau$
		$\displaystyle\leq$	$\displaystyle\frac{\nu_{1}}{\nu_{0}-\kappa}\left(\frac{C}{\beta}+\frac{C_{\beta}}{R^{2}}\right)\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\leq\frac{1}{4}\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\hbox{,}$

after choosing $\beta>4$ and $R>0$ sufficiently large. Hence,

(25)

\left|\mathrm{III}\right|\leq\left(\frac{1}{4}+\frac{\widetilde{C}_{\beta}\varepsilon}{\nu_{0}-\kappa}\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\hbox{.}

Combining (23), (24) and (25), we have

	$\displaystyle\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}$	$\displaystyle\leq$	$\displaystyle\varepsilon\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}+\left(\frac{1}{4}+\frac{\widetilde{C}_{\beta}\varepsilon}{\nu_{0}-\kappa}\left\\|f_{0}\right\\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\right)\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}$
			$\displaystyle+\frac{C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}\left(\sup_{0\leq t\leq T}\left\\|u\right\\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\right)^{2}\hbox{.}$

We may assume that $\frac{C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}>1$ . Choosing $\varepsilon>0$ sufficiently small such that

\frac{\widetilde{C}_{\beta}\varepsilon}{\nu_{0}-\kappa}\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}<\frac{1}{4}\hbox{ and }\left(\frac{4C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}\right)^{2}\varepsilon\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}<1\hbox{,}

we obtain

\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\leq 2\varepsilon\left\|f_{0}\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}+\frac{2C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}\left(\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\right)^{2}\hbox{.}

Since $\left\|u\left(0,x,\xi\right)\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}=\left\|f_{1}\left(0,x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}=\varepsilon\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}$ ,

\sup_{0\leq t\leq T}\left\|u\right\|_{L_{\xi}^{\infty}L_{x}^{\infty}}\leq\left(\frac{4C_{\beta}^{\prime\prime}\nu_{1}}{\nu_{0}-\kappa}\right)\varepsilon\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}\hbox{,}

for any finite $T>0$ . Consequently,

\left\langle\xi\right\rangle^{\beta}e^{\kappa\rho\left(\xi,t\right)}\left|f_{1}\left(t,x,\xi\right)\right|=\left|u\left(t,x,\xi\right)\right|\leq\overline{C}_{\beta}\varepsilon\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}

for some constant $\overline{C}_{\beta}>0$ depending only on $\beta$ , i.e.,

\left|f_{1}\left(t,x,\xi\right)\right|\leq\overline{C}_{\beta}\varepsilon\left\langle\xi\right\rangle^{-\beta}e^{-\kappa\rho\left(\xi,t\right)}\left\|f_{0}\left(x,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}L_{x}^{\infty}}

for all $t\geq 0$ , $x\in\mathbb{R}^{3}$ , $\xi\in\mathbb{R}^{3}$ . ∎

6. Some convolution estimates

In this section, we will compute the interactions between different wave patterns, which are essential for determining the precise space-time structure of the solution. Although these estimates appear complicated, there is a clear physical picture behind them (see Section 6.2 for some illustrations). The proofs in fact aim to translate this heuristic picture into refined convolution estimates.

To facilitate the estimates, we decompose space-time domain into the following $5$ regions:

	$\displaystyle D_{1}$	$\displaystyle=\left\{\|x\|\leq\sqrt{1+t}\right\}\,,$
	$\displaystyle\ D_{2}$	$\displaystyle=\left\{\mathbf{c}t-\sqrt{1+t}\leq\|x\|\leq\mathbf{c}t+\sqrt{1+t}\right\}\,,$
	$\displaystyle D_{3}$	$\displaystyle=\left\{\|x\|\geq\mathbf{c}t+\sqrt{1+t}\right\}\,,$
	$\displaystyle D_{4}$	$\displaystyle=\left\{\sqrt{1+t}\leq\|x\|\leq\frac{1}{2}\mathbf{c}t\right\}\,,$
	$\displaystyle D_{5}$	$\displaystyle=\left\{\frac{1}{2}\mathbf{c}t\leq\|x\|\leq\mathbf{c}t-\sqrt{1+t}\right\}\,.$

6.1. Linear interaction

Lemma 19 (Diffusion wave convolved with exponential decay).

\left(1+t\right)^{-3/2}e^{-\frac{\left|x\right|^{2}}{D_{0}\left(1+t\right)}}\ast_{x,t}e^{-\frac{t+\left|x\right|}{c_{0}}}\lesssim\left(1+t\right)^{-3/2}e^{-\frac{\left|x\right|^{2}}{\widehat{D}\left(1+t\right)}}+e^{-\frac{t+\left|x\right|}{\widehat{c}}}\hbox{,}

for some constants $\widehat{c}$ and $\widehat{D}>0$ .

Lemma 20 (Huygens wave convolved with exponential decay).

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}\ast_{x,t}e^{-\frac{t+\left\|x\right\|}{c_{0}}}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}\left(1+t\right)}}+e^{-\frac{t+\left\|x\right\|}{\widehat{c}}}\hbox{,}$

for some constants $\widehat{c}$ and $\widehat{D}>0$ .

Lemma 21 (Riesz wave convolved with exponential decay).

	$\displaystyle B$	$\displaystyle=$	$\displaystyle\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}\ast_{x,t}e^{-\frac{t+\left\|x\right\|}{c_{0}}}$
		$\displaystyle\lesssim$	$\displaystyle\mathbf{1}_{\{\left\|x\right\|<\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\left\|x\right\|}{\widehat{c}_{0}}}+\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}\hbox{,}$

for some constants $\widehat{c}_{0},\widehat{D}_{0}>0$ .

The proof of Lemma 19 is easy and hence we omit it.

Proof of Lemma 20 (Huygens wave convolved with exponential decay).

We rewrite

A=\int_{0}^{t}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left|y\right|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left|x-y\right|}{c_{0}}}dyd\tau\hbox{.}

We discuss the integral $A$ in each domain $D_{i}$ $\left(1\leq i\leq 5\right)$ for which $\left(x,t\right)$ belongs to.

Case 1: $\left(x,t\right)\in D_{1}$ . Direct computation gives

$\displaystyle A$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{t}{2c_{0}}-\frac{\left\|x-y\right\|}{c_{0}}}dyd\tau+\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|\leq\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle+\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{2c_{0}}}+e^{-\frac{\mathbf{c}^{2}t}{32D_{0}}}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle+\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{2c_{0}}}+e^{-\frac{\mathbf{c}t}{32D_{0}}}+\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau\hbox{.}$

Note that if $\frac{t}{2}\leq\tau\leq t$ , $\left|y\right|>\frac{\mathbf{c}\tau}{2}$ , then

\left|x-y\right|\geq\left|y\right|-\left|x\right|\geq\frac{\mathbf{c}t}{4}-\sqrt{1+t}\geq\frac{\mathbf{c}t}{8}+\frac{\mathbf{c}t}{8}-\sqrt{1+t}\geq\frac{\mathbf{c}t}{8}

for $t\geq 40$ , so that

			$\displaystyle\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\mathbf{c}t}{8c_{0}}}\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\mathbf{c}t}{8c_{0}}}\int_{\frac{t}{2}}^{t}\left(1+\tau\right)^{-2+\frac{5}{2}}d\tau\lesssim e^{-\frac{\mathbf{c}t}{16c_{0}}}\hbox{,}$

for $t\geq 40$ . Consequently,

A\lesssim e^{-\frac{t}{2c_{0}}}+e^{-\frac{\mathbf{c}t}{32D_{0}}}+e^{-\frac{\mathbf{c}t}{16c_{0}}}\lesssim e^{-\frac{t+\left|x\right|}{\widehat{c}}}

for all $t\geq 0$ , where $\frac{1}{\widehat{c}}=\frac{1}{2}\min\{\frac{1}{2c_{0}},\frac{\mathbf{c}}{32D_{0}},\frac{\mathbf{c}}{16c_{0}}\}$ .

Case 2: $\left(x,t\right)\in D_{2}$ .

	$\displaystyle A$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{t}{2c_{0}}}e^{-\frac{\left\|x-y\right\|}{c_{0}}}dyd\tau+\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t\right)^{-2}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{2c_{0}}}+\left(1+t\right)^{-2}\lesssim\left(1+t\right)^{-2}\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D\left(1+t\right)}}\hbox{.}$

Case 3: $\left(x,t\right)\in D_{3}$ . We split the integral $A$ into four parts

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}+\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\right)\left(\cdots\right)dyd\tau+\int_{\frac{t}{2}}^{t}\left(\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}+\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\right)\left(\cdots\right)dyd\tau$
		$\displaystyle=$	$\displaystyle:A_{11}+A_{12}+A_{21}+A_{22}\hbox{.}$

Note that if $\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\left|x-y\right|\geq\left|x\right|-\left|y\right|\geq\frac{\left|x\right|-\mathbf{c}\tau}{2}=\frac{\left|x\right|-\mathbf{c}t}{2}+\frac{\mathbf{c}\left(t-\tau\right)}{2}\hbox{.}

if $\left|y\right|>\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\left|y\right|-\mathbf{c}\tau>\frac{\left|x\right|-\mathbf{c}\tau}{2}=\frac{\left|x\right|-\mathbf{c}t}{2}+\frac{\mathbf{c}\left(t-\tau\right)}{2}\hbox{.}

It immediately follows that

	$\displaystyle A_{11}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}e^{-\frac{t}{2c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}e^{-\frac{t}{2c_{0}}-\frac{1}{2c_{0}}\left(\frac{\left\|x\right\|}{2}-\frac{\mathbf{c}t}{4}\right)}dyd\tau\lesssim e^{-\frac{t}{2c_{0}}-\frac{1}{2c_{0}}\left(\frac{\left\|x\right\|}{2}-\frac{\mathbf{c}t}{4}\right)}\lesssim e^{-\frac{t}{2c_{0}}-\frac{\left\|x\right\|}{8c_{0}}}\hbox{,}$

A_{12}\lesssim\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}e^{-\frac{t}{2c_{0}}}e^{-\frac{\left|x-y\right|}{c_{0}}}\lesssim e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}e^{-\frac{t}{2c_{0}}}\hbox{,}

A_{22}\lesssim\int_{\frac{t}{2}}^{t}\int_{\left|y\right|\geq\frac{\left|x\right|+\mathbf{c}\tau}{2}}\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}e^{-\frac{\left(t-\tau\right)+\left|x-y\right|}{c_{0}}}dyd\tau\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}\hbox{.}

As for $A_{21}$ ,

$\displaystyle A_{21}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\int_{\frac{t}{2}}^{t}\left(1+\tau\right)^{-2+\frac{5}{2}}d\tau\lesssim\left(1+t\right)^{\frac{3}{2}}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\lesssim\left(1+t\right)^{\frac{3}{2}}e^{-\frac{\sqrt{1+t}}{4c_{0}}}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{4c_{0}}}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{4c_{0}}}\hbox{,}$

since $\left|x\right|-\mathbf{c}t\geq\sqrt{1+t}$ . Now we discuss two cases: (i) $\left|x\right|-\mathbf{c}t\leq\frac{1}{2}\left(1+t\right)$ , and (ii) $\left|x\right|-\mathbf{c}t\geq\frac{1}{2}\left(1+t\right)$ . For case (i),

\left|x\right|-\mathbf{c}t\geq\frac{2\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\hbox{,}

which follows that

e^{-\frac{\left|x\right|-\mathbf{c}t}{4c_{0}}}\lesssim e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{2c_{0}\left(1+t\right)}}\hbox{.}

For case (ii), it is easy to see

\left(1+t\right)\leq 2\left(\left|x\right|-\mathbf{c}t\right)

and thus

\left|x\right|=\left|\left|x\right|-\mathbf{c}t+\mathbf{c}t\right|\leq\left(\left|x\right|-\mathbf{c}t\right)+\mathbf{c}t\leq\left(1+2\mathbf{c}\right)\left(\left|x\right|-\mathbf{c}t\right)<4\left(\left|x\right|-\mathbf{c}t\right)\hbox{,}

so that

e^{-\frac{\left|x\right|-\mathbf{c}t}{4c_{0}}}=e^{-\frac{\left|x\right|-\mathbf{c}t}{8c_{0}}}e^{-\frac{\left|x\right|-\mathbf{c}t}{8c_{0}}}\lesssim e^{-\frac{\left(1+t\right)}{16c_{0}}}e^{-\frac{\left|x\right|}{32c_{0}}}\lesssim e^{-\frac{t}{16c_{0}}}e^{-\frac{\left|x\right|}{32c_{0}}}\hbox{.}

Hence,

A_{21}\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{2c_{0}\left(1+t\right)}}+e^{-\frac{t}{16c_{0}}}e^{-\frac{\left|x\right|}{32c_{0}}}\hbox{.}

Combining all above estimates, we have

A\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{\widehat{D}\left(1+t\right)}}+e^{-\frac{t+\left|x\right|}{\widehat{c}}}

for some constants $\widehat{c}$ and $\widehat{D}>0$ .

Case 4: $\left(x,t\right)\in D_{4}$ . We split the integral into two parts

A=\left(\int_{0}^{\frac{2}{3}t}+\int_{\frac{2}{3}t}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:A_{1}+A_{2}\hbox{.}

For $A_{1}$ ,

A_{1}\lesssim\int_{0}^{\frac{2}{3}t}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left|x-y\right|}{c_{0}}}e^{-\frac{t}{3c_{0}}}dyd\tau\lesssim e^{-\frac{t}{3c_{0}}}\lesssim e^{-\frac{t}{6c_{0}}}e^{-\frac{\left|x\right|}{3c_{0}\mathbf{c}}}

since $\sqrt{1+t}\leq\left|x\right|\leq\mathbf{c}t/2$ .

For $A_{2}$ , we decompose $\mathbb{R}^{3}$ into two parts

A_{2}=\int_{\frac{2}{3}t}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}\tau}{2}}+\int_{\left|y\right|>\frac{\left|x\right|+\mathbf{c}\tau}{2}}\right)\left(\cdots\right)dyd\tau=:A_{21}+A_{22}\hbox{.}

If $\frac{2}{3}t\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\mathbf{c}\tau-\left|y\right|\geq\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t}{3}-\frac{\mathbf{c}t}{4}=\frac{\mathbf{c}t}{12}\geq\frac{\mathbf{c}t-\left|x\right|}{12}\hbox{.}

If $\frac{2}{3}t\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\left|x-y\right|\geq\left|y\right|-\left|x\right|\geq\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t}{12}\geq\frac{\mathbf{c}t-\left|x\right|}{12}\hbox{.}

Hence,

	$\displaystyle A_{21}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{2}{3}t}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{144D_{0}\left(1+t\right)}}\hbox{,}$

and

	$\displaystyle A_{22}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{2}{3}t}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+t\right)^{-2}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}e^{-\frac{\mathbf{c}t-\left\|x\right\|}{24c_{0}}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\mathbf{c}t-\left\|x\right\|}{24c_{0}}}\hbox{.}$

Since $t\geq 1$ and $\mathbf{c}t-\left|x\right|>\frac{\mathbf{c}t}{2}\geq\frac{\mathbf{c}}{4}\left(1+t\right)$ for $\left(x,t\right)\in D_{4}$ ,

\mathbf{c}\left(1+t\right)\leq 4\left(\mathbf{c}t-\left|x\right|\right)\hbox{,}

and thus

\left|x\right|=\left|\left|x\right|-\mathbf{c}t+\mathbf{c}t\right|\leq\mathbf{c}t-\left|x\right|+\mathbf{c}t\leq 5\left(\mathbf{c}t-\left|x\right|\right)\hbox{,}

which implies that

e^{-\frac{\mathbf{c}t-\left|x\right|}{24c_{0}}}=e^{-\frac{\mathbf{c}t-\left|x\right|}{48c_{0}}}e^{-\frac{\mathbf{c}t-\left|x\right|}{48c_{0}}}\lesssim e^{-\frac{\mathbf{c}\left(1+t\right)}{192c_{0}}}e^{-\frac{\left|x\right|}{240c_{0}}}\lesssim e^{-\frac{\mathbf{c}t}{192c_{0}}}e^{-\frac{\left|x\right|}{240c_{0}}}\hbox{.}

Therefore,

A_{22}\lesssim e^{-\frac{\mathbf{t}}{96c_{0}}}e^{-\frac{\left|x\right|}{192c_{0}}}\hbox{.}

Combining this with $A_{1}$ and $A_{11}$ , we get the desired estimate

A\lesssim e^{-\frac{t}{6c_{0}}}e^{-\frac{\left|x\right|}{3c_{0}\mathbf{c}}}+\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{144D_{0}\left(1+t\right)}}+e^{-\frac{\mathbf{t}}{96c_{0}}}e^{-\frac{\left|x\right|}{192c_{0}}}\hbox{.}

Case 5: $\left(x,t\right)\in D_{5}$ . We split the integral $A$ into three parts

A=\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}}+\int_{\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:A_{1}+A_{2}+A_{3}\hbox{.}

It immediately follows that

A_{1}\lesssim\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left|x-y\right|}{c_{0}}}e^{-\frac{t}{2c_{0}}}dyd\tau\lesssim e^{-\frac{t}{2c_{0}}}\lesssim e^{-\frac{t}{4c_{0}}-\frac{\left|x\right|}{4\mathbf{c}c_{0}}}\hbox{,}

and

$\displaystyle A_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-2}e^{-\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{D_{0}\left(1+\tau\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\int_{\frac{t}{2}}^{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{\mathbb{R}^{3}}e^{-\frac{\left\|x-y\right\|}{c_{0}}}e^{-\frac{t-\tau}{2c_{0}}}e^{-\frac{1}{2c_{0}}\left(\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}\right)}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\mathbf{c}t-\left\|x\right\|}{4c_{0}\mathbf{c}}}\hbox{.}$

For $A_{3}$ , we decompose $\mathbb{R}^{3}$ into two parts

A_{3}=\int_{\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}\tau}{2}}+\int_{\left|y\right|>\frac{\left|x\right|+\mathbf{c}\tau}{2}}\right)\left(\cdots\right)dyd\tau=A_{31}+A_{32}\hbox{.}

If $\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\mathbf{c}\tau-\left|y\right|\geq\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

If $\frac{t}{2}+\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|+\mathbf{c}\tau}{2}$ , then

\left|x-y\right|\geq\left|y\right|-\left|x\right|>\frac{\mathbf{c}\tau-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

Hence,

$\displaystyle A_{3}$	$\displaystyle=$	$\displaystyle A_{31}+A_{32}$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{16D_{0}\left(1+t\right)}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle+\int_{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}\tau}{2}}\left(1+t\right)^{-2}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\mathbf{c}t-\left\|x\right\|}{8c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{16D_{0}\left(1+t\right)}}+\left(1+t\right)^{-2}e^{-\frac{\mathbf{c}t-\left\|x\right\|}{8c_{0}}}\hbox{.}$

Since $\sqrt{1+t}\leq\mathbf{c}t-\left|x\right|\leq\frac{\mathbf{c}t}{2}\leq\frac{\mathbf{c}}{2}\left(1+t\right)$ , we have

\mathbf{c}t-\left|x\right|\geq\frac{2\left(\mathbf{c}t-\left|x\right|\right)^{2}}{\mathbf{c}\left(1+t\right)}\hbox{,}

and thus

e^{-\frac{\mathbf{c}t-\left|x\right|}{4c_{0}\mathbf{c}}}\lesssim e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{2c_{0}\mathbf{c}^{2}\left(1+t\right)}}\hbox{, \ \ }e^{-\frac{\mathbf{c}t-\left|x\right|}{8c_{0}}}\lesssim e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{4c_{0}\mathbf{c}\left(1+t\right)}}\hbox{.}

Consequently,

A_{2}\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{2c_{0}\mathbf{c}^{2}\left(1+t\right)}}\hbox{, \ \ }A_{3}\lesssim\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{16D_{0}\left(1+t\right)}}+\left(1+t\right)^{-2}e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{4c_{0}\mathbf{c}\left(1+t\right)}}\hbox{.}

Combining this with $A_{1}$ , we get the desired estimate. ∎

Proof of Lemma 21 (Riesz wave convolved with exponential decay). .

We compute the integral for two cases: $\left|x\right|>\mathbf{c}t$ and $\left|x\right|\leq\mathbf{c}t$ .

Case 1: $\left|x\right|>\mathbf{c}t$ . Let $0\leq\tau\leq t$ . If $\left|x\right|>\mathbf{c}t$ and $\left|y\right|<\mathbf{c}\tau$ , then

\left|x-y\right|\geq\left|x\right|-\left|y\right|\geq\left|x\right|-\mathbf{c}t+\mathbf{c}t-\left|y\right|\geq\left|x\right|-\mathbf{c}t\geq 0\hbox{.}

Hence,

			$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}1_{\{\left\|y\right\|<\mathbf{c}\tau\}}\left(1+\tau\right)^{-3/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3/2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}\left(1+\tau\right)^{-3/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3/2}dyd\tau$
		$\displaystyle=$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:B_{1}+B_{2}\hbox{.}$

It immediately follows that

B_{1}\leq e^{-\frac{t}{2c_{0}}}e^{-\frac{\left|x\right|-\mathbf{c}t}{2c_{0}}}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}e^{-\frac{\left|x-y\right|}{2c_{0}}}\left(1+\tau\right)^{-3/2}dyd\tau\lesssim e^{-\frac{t}{2c_{0}}}e^{-\frac{\left|x\right|-\mathbf{c}t}{2c_{0}}}\hbox{.}

As for $B_{2}$ ,

$\displaystyle B_{2}$	$\displaystyle=$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\int_{\frac{t}{2}}^{t}\left(\int_{\left\|y\right\|\leq\frac{\left\|x\right\|}{2}}+\int_{\left\|y\right\|>\frac{\left\|x\right\|}{2}}\right)e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}\left(1+\tau\right)^{-3/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3/2}dyd\tau$
	$\displaystyle\leq$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}e^{-\frac{\left\|x\right\|}{8c_{0}}}\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|}{2}}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{4c_{0}}}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|}{2}}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}e^{-\frac{\left\|x\right\|}{8c_{0}}}+\left(1+t\right)^{-3}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\lesssim\left(1+t\right)^{-3}e^{-\frac{\left\|x\right\|-\mathbf{c}t}{2c_{0}}}\hbox{.}$

since $\left|x\right|>\mathbf{c}t$ .

Now, we consider the cases: $0\leq\left|x\right|-\mathbf{c}t\leq 2\mathbf{c}\left(1+t\right)$ and $\left|x\right|-\mathbf{c}t\geq 2\mathbf{c}\left(1+t\right)$ . For $0\leq\left|x\right|-\mathbf{c}t\leq 2\mathbf{c}\left(1+t\right)$ , we have

\left|x\right|-\mathbf{c}t\geq\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{2\mathbf{c}\left(1+t\right)}\hbox{,}

so that

e^{-\frac{\left|x\right|-\mathbf{c}t}{2c_{0}}}\leq e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{4\mathbf{c}c_{0}\left(1+t\right)}}\hbox{.}

For $\left|x\right|-\mathbf{c}t\geq 2\mathbf{c}\left(1+t\right)$ , it follows that

\left|x\right|-\mathbf{c}t=\frac{\left|x\right|}{2}+\frac{\left|x\right|}{2}-\mathbf{c}t\geq\frac{\left|x\right|}{2}+\frac{\mathbf{c}t}{2}\geq\frac{\left|x\right|}{2}+\frac{t}{2}\hbox{,}

and thus

e^{-\frac{\left|x\right|-\mathbf{c}t}{2c_{0}}}\leq e^{-\frac{\left|x\right|+t}{4c_{0}}}\hbox{.}

Consequently,

			$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}1_{\{\left\|y\right\|<\mathbf{c}\tau\}}\left(1+\tau\right)^{-3/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3/2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4\mathbf{c}c_{0}\left(1+t\right)}}+e^{-\frac{\left\|x\right\|+t}{4c_{0}}}\hbox{.}$

Case 2: $\left|x\right|<\mathbf{c}t$ . Direct computation gives

$\displaystyle B$	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{2c_{0}}}\int_{0}^{t/2}\int_{\mathbb{R}^{3}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}\int_{t/2}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|}{2}}e^{-\frac{\left(t-\tau\right)+\left\|x-y\right\|}{c_{0}}}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-3/2}e^{-\frac{\left\|x\right\|}{4c_{0}}}\int_{t/2}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|}{2}}e^{-\frac{t-\tau}{c_{0}}}e^{-\frac{\left\|x-y\right\|}{2c_{0}}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{2c_{0}}}+\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t+\left\|x\right\|}{4cc_{0}}}+\mathbf{1}_{\{\left\|x\right\|<\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3/2}\hbox{.}$

To sum up,

B\lesssim\mathbf{1}_{\{\left|x\right|<\mathbf{c}t\}}\left(1+t\right)^{-3/2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-3/2}+e^{-\frac{t+\left|x\right|}{\widehat{c}_{0}}}+\left(1+t\right)^{-2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{\widehat{D}_{0}\left(1+t\right)}}

for some constants $\widehat{D}_{0}$ , $\widehat{c}_{0}>0$ . ∎

6.2. Nonlinear wave interaction

Lemma 22 (Diffusion wave convolved with Diffusion wave).

			$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}*_{x,t}\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

Lemma 23 (Space-time exponential decay convolved with Huygens wave).

			$\displaystyle e^{-\frac{t+\|x\|}{c_{0}}}*_{x,t}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

Lemma 24 (Riesz wave convolved with Huygens wave).

	$\displaystyle I$	$\displaystyle=$	$\displaystyle\mathbf{1}_{\{\left\|x\right\|\leq\mathbf{c}t\}}\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\ast_{x,t}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

Lemma 25 (Huygens wave convolved with Huygens wave).

	$\displaystyle J$	$\displaystyle=$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}*_{x,t}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

Lemma 26 (Huygens wave convolved with diffusion wave).

	$\displaystyle K$	$\displaystyle=$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}*_{x,t}\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

Lemma 27 (Diffusion wave convolutied with Huygens wave).

	$\displaystyle L$	$\displaystyle=$	$\displaystyle\left(1+t\right)^{-2}e^{-\frac{\left\|x\right\|^{2}}{D_{0}\left(1+t\right)}}\ast_{x,t}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

We omit the proof of Lemma 22 and Lemma 23. Before we proceed to the detailed proof of Lemmas 24-27, let us present some heuristic calculations to help understand the mechanism of nonlinear wave interactions. We use two examples for illustration: the convolution of two Huygens waves (Lemma 25) and convolution of diffusion wave with a Huygens wave (Lemma 27).

Heuristic estimate for Lemma 25.

Set $\mathbf{c}=1$ for simplicity. Then

(26)

J=\int_{0}^{t}\int_{{\mathbb{R}}^{3}}(1+t-s)^{-5/2}e^{-\frac{(\lvert x-y\rvert-(t-s))^{2}}{D_{0}(t-s)}}(1+s)^{-4}\Bigl{(}1+\frac{\left(\left|y\right|-s\right)^{2}}{1+s}\Bigr{)}^{-2}dyds.

We can interpret $(1+t-s)^{-5/2}e^{-\frac{(\lvert x-y\rvert-(t-s))^{2}}{D_{0}(t-s)}}$ as a receiver located at $(x,t)$ that can only receive signals along the wave cone concentrated on $\lvert x-y\rvert=(t-s)$ with thickness $\sqrt{t-s}$ . Similarly, $(1+s)^{-4}\Bigl{(}1+\frac{\left(\left|y\right|-s\right)^{2}}{1+s}\Bigr{)}^{-2}$ can be viewed as a sender located at $(0,0)$ that sends signals along the wave cone concentrated on $\lvert y\rvert=s$ with thickness $\sqrt{s}$ .

During the interaction process, the interaction becomes strong in the following space-time region (See Figure 2)

(27)

E=\left\{(y,s)\big{|}\,\lvert\lvert x-y\rvert-(t-s)\rvert\leq O(1)\sqrt{t-s}\mbox{ and }\lvert\lvert y\rvert-s\rvert\leq O(1)\sqrt{s}\right\}.

Inside this region, the space decay terms in the convolution are of order $O(1)$ , and the decay is mainly from time factor. The key point is the sharp estimate of the volume for the strong interaction region.

If $\lvert x\rvert>t$ , as $s$ increases from $0$ to $t$ , the receiving and sending wave cones are almost disjoint, so the interaction is very weak.

Now we focus the case where $O(1)\sqrt{t}<\lvert x\rvert<t-O(1)\sqrt{t}$ . The interaction process starts at $s=(t-\lvert x\rvert)/2$ and ends at $s=(t+\lvert x\rvert)/2$ (See Figure 2). To satisfy the condition in (27), one has

\left\{\begin{aligned} &\lvert\lvert x-y\rvert-(t-s)\rvert\leq O(1)\sqrt{t-s},\\ &\lvert\lvert y\rvert-s\rvert\leq O(1)\sqrt{s}.\end{aligned}\right.

Let $r=\lvert y\rvert$ , $\theta$ be the angle between $x$ and $y$ , and set $O(1)=1$ . The above constraints are equivalent to

(28)

\left\{\begin{aligned} &\lvert\sqrt{\lvert x\rvert^{2}+r^{2}-2r|x|\cos\theta}-(t-s)\rvert\leq\sqrt{t-s},\\ &\lvert r-s\rvert\leq\sqrt{s},\end{aligned}\right.

The first equation of (28) implies

\frac{\lvert x\rvert^{2}+r^{2}-(t-s)^{2}-(t-s)}{2r\lvert x\rvert}-\frac{(t-s)^{3/2}}{r\lvert x\rvert}\leq\cos\theta\leq\frac{\lvert x\rvert^{2}+r^{2}-(t-s)^{2}-(t-s)}{2r\lvert x\rvert}+\frac{(t-s)^{3/2}}{r\lvert x\rvert}

This leads to

\Delta\cos\theta\approx\frac{(t-s)^{3/2}}{r\lvert x\rvert}.

For $s$ large, $r\approx s$ , $\Delta r\approx\sqrt{s}$ . Then inside the strong interaction region, (26) is approximated by

	$\displaystyle J$	$\displaystyle\approx\int_{\frac{t-\lvert x\rvert}{2}}^{\frac{t+\lvert x\rvert}{2}}(1+t-s)^{-5/2}(1+s)^{-4}\underset{r^{2}dr}{\underbrace{s^{2}\sqrt{s}}}\,\underset{\sin\theta d\theta}{\underbrace{\frac{(t-s)^{3/2}}{s\lvert x\rvert}}}ds$
		$\displaystyle\lesssim\int_{\frac{t-\lvert x\rvert}{2}}^{\frac{t+\lvert x\rvert}{2}}(1+t-s)^{-1}(1+s)^{-5/2}ds\frac{1}{\lvert x\rvert}$
		$\displaystyle\lesssim\Bigl{(}\int_{\frac{t-\lvert x\rvert}{2}}^{\frac{t}{2}}(1+t)^{-1}(1+s)^{-5/2}ds+\int_{\frac{t}{2}}^{\frac{t+\lvert x\rvert}{2}}(1+t-s)^{-1}(1+t)^{-5/2}ds\Bigr{)}\frac{1}{\lvert x\rvert}$
		$\displaystyle\lesssim\Bigl{[}(1+t)^{-1}(1+t-\lvert x\rvert)^{-3/2}+(1+t)^{-5/2}\ln\frac{t}{t-\lvert x\rvert}\Bigr{]}\frac{1}{\lvert x\rvert}.$

When $\sqrt{t}\leq\lvert x\rvert\leq t/2$ , it is bounded by

(1+t)^{-5/2}\lvert x\rvert^{-1}\lesssim(1+t)^{-2}\Bigl{(}1+\frac{\lvert x\rvert^{2}}{1+t}\Bigr{)}^{-3/2}.

When $t/2\leq\lvert x\rvert\leq t-\sqrt{t}$ , it is bounded by

t^{-2}(1+t-\lvert x\rvert)^{-3/2}+(1+t)^{-5/2}\frac{t^{2}}{(t-\lvert x\rvert)^{2}}(1+t)^{-1}\lesssim t^{-5/2}\Bigl{(}1+\frac{(t-\lvert x\rvert)^{2}}{1+t}\Bigr{)}^{-1}.

This is the desired estimate in Lemma 25. ∎

Heuristic estimate for Lemma 26.

Set $\mathbf{c}=1$ , and one has

K=\int_{0}^{t}\int_{\mathbb{R}^{3}}\left(1+t-s\right)^{-5/2}e^{-\frac{\left(\left|x-y\right|-(t-s)\right)^{2}}{D_{0}\left(1+t-s\right)}}\left(1+s\right)^{-3}\left(1+\frac{\left|y\right|^{2}}{1+s}\right)^{-3}dyds.

Similar as heuristics argument of Lemma 25, we view $(1+t-s)^{-5/2}e^{-\frac{\left(\left|x-y\right|-(t-s)\right)^{2}}{D_{0}\left(1+t-s\right)}}$ as a receiver and $\left(1+s\right)^{-3}\left(1+\frac{\left|y\right|^{2}}{1+s}\right)^{-3}$ as a sender, to have the following figure representing the interaction:

For $O(1)\sqrt{t}\leq\lvert x\rvert\leq t-O(1)\sqrt{t}$ , the interaction occurs at $s\approx t-\lvert x\rvert$ and its duration is of the order $\sqrt{s}$ . So the volume of the interaction region in space-time is approximated by

\underset{\mbox{vol of space}}{\underbrace{s^{3/2}}}\cdot\underset{\mbox{vol of time}}{\underbrace{\sqrt{s}}}

The contribution of the integrand during the interaction is approximated by

(1+t-s)^{-5/2}(1+s)^{-3}\Big{|}_{s\approx t-\lvert x\rvert}.

So we conclude

	$\displaystyle K$	$\displaystyle\approx(1+t-s)^{-5/2}(1+s)^{-3}s^{2}\Big{\|}_{s\approx t-\lvert x\rvert}\approx(1+\lvert x\rvert)^{-5/2}(1+t-\lvert x\rvert)^{-1}$
		$\displaystyle\lesssim\begin{cases}(1+t)^{-1}(1+\lvert x\rvert)^{-5/2}\lesssim(1+t)^{-2}\Bigl{(}1+\frac{\lvert x\rvert^{2}}{1+t}\Bigr{)}^{-3/2},&\sqrt{t}\leq\lvert x\rvert\leq t/2,\\ (1+t)^{-5/2}(1+t-\lvert x\rvert)^{-1}\lesssim(1+t)^{-5/2}\Bigl{(}1+\frac{(t-\lvert x\rvert)^{2}}{1+t}\Bigr{)}^{-1},&t/2\leq\lvert x\rvert\leq t-\sqrt{t}.\end{cases}$

This yields the desired estimate in Lemma 26. ∎

Other convolutions can be estimated heuristically in a similar manner. It turns out that the convolutions in Lemmas 25 and 26 are the dominated ones among all the nonlinear wave couplings.

We now begin the rigorous proofs, transforming the previous heuristic calculations into refined (and complex) convolution estimates!

Proof of Lemma 24.

(Riesz wave convolved with Huygens wave).

Case 1: $\left(x,t\right)\in D_{1}$ . Direct computation gives

$\displaystyle I$	$\displaystyle=$	$\displaystyle\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-4}\int_{\frac{t}{2}}^{t}\int_{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}+\left(1+t\right)^{-4}\left(1+t\right)^{3/4}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

Case 2: $\left(x,t\right)\in D_{2}$ . We split the integral $I$ into two parts

I=\left(\int_{0}^{\frac{t}{4}}+\int_{\frac{t}{4}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:I_{1}+I_{2}\hbox{.}

For $I_{2}$ , one can see that

$\displaystyle I_{2}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{\frac{t}{4}}^{t}\int_{\mathbb{R}^{3}}\mathbf{1}_{\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{\frac{t}{4}}^{t}\left(1+t-\tau\right)^{-1/4}d\tau\lesssim\left(1+t\right)^{-4}\left(1+t\right)^{3/4}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $I_{1}$ , we further decompose $\mathbb{R}^{3}$ into two parts

I_{1}=\left(\int_{0}^{\frac{t}{4}}\int_{\left|y\right|\leq\frac{2}{3}\left|x\right|}+\int_{0}^{\frac{t}{4}}\int_{\left|y\right|>\frac{2}{3}\left|x\right|}\right)\left(\cdots\right)dyd\tau=I_{11}+I_{12}\hbox{.}

If $\left|y\right|\leq\frac{2}{3}\left|x\right|$ , then we have

\left|x-y\right|\geq\left|x\right|-\left|y\right|\geq\frac{\left|x\right|}{3}\hbox{,}

and thus

$\displaystyle I_{11}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

If $\left|y\right|>\frac{2}{3}\left|x\right|$ and $0\leq\tau\leq\frac{t}{4}$ ,

\left|y\right|-\mathbf{c}\tau\geq\frac{2}{3}\left|x\right|-\mathbf{c}\tau\geq\frac{\left|x\right|}{3}+\frac{1}{3}\left(\mathbf{c}t-\sqrt{1+t}\right)-\frac{\mathbf{c}t}{4}\geq\frac{\left|x\right|}{3}+\frac{1}{24}\mathbf{c}t

for $t\geq 40$ . We mention here that $\left(x,t\right)\in D_{2}$ with $t\geq\frac{2+2\sqrt{1+\mathbf{c}^{2}}}{\mathbf{c}^{2}}(\doteqdot 3.15)$ implies that $\left|x\right|\geq\sqrt{1+t}$ . Hence for $t\geq 40$ ,

$\displaystyle I_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\int_{\left\|y\right\|>\frac{2}{3}\left\|x\right\|}\mathbf{1}_{\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}$
		$\displaystyle\cdot\left(1+\tau\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\frac{1}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\int_{\left\{\left\|y\right\|>\frac{2}{3}\left\|x\right\|\right\}\cap\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}$
		$\displaystyle\cdot\left(1+\tau\right)^{-\frac{5}{2}}\left(1+\frac{\left\|x-y\right\|^{2}}{1+t-\tau}\right)^{-\frac{1}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\left(1+\tau\right)^{-\frac{5}{2}}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle(1+t)^{-1/2}\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

and for $0\leq t\leq 40$ ,

	$\displaystyle I_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{4}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle C\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

Thereupon,

I_{1}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{,}

and so

I\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 3: $\left(x,t\right)\in D_{3}$ . We split the integral $I$ into two parts

I=\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:I_{1}+I_{2}\hbox{.}

For $I_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts as follows:

I_{1}=\int_{0}^{\frac{t}{2}}\left(\int_{\left|y\right|\leq\frac{2}{3}\left|x\right|}+\int_{\left|y\right|>\frac{2}{3}\left|x\right|}\right)\left(\ldots\right)dyd\tau=:I_{11}+I_{12}\hbox{.}

If $\left|y\right|\leq\frac{2}{3}\left|x\right|$ , then

\left|x-y\right|\geq\frac{\left|x\right|}{3}\hbox{,}

and thus

	$\displaystyle I_{11}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

If $\left|y\right|>\frac{2}{3}\left|x\right|$ and $0\leq\tau\leq\frac{t}{2}$ , we have

\left|y\right|-\mathbf{c}\tau\geq\frac{2}{3}\left|x\right|-\frac{\mathbf{c}t}{2}=\frac{1}{6}\left|x\right|+\frac{1}{2}\left(\left|x\right|-\mathbf{c}t\right)\geq\frac{1}{6}\mathbf{c}t\hbox{,}

so that

$\displaystyle I_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|>\frac{2}{3}\left\|x\right\|}\mathbf{1}_{\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}$
		$\displaystyle\cdot\left(1+\tau\right)^{-\frac{5}{2}}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\frac{1}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\int_{\left\{\left\|y\right\|\geq\frac{2}{3}\left\|x\right\|\right\}\cap\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\tau\right)^{-\frac{5}{2}}\cdot\left(1+\frac{\left\|x-y\right\|^{2}}{\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-\frac{5}{2}}\cdot\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle(1+t)^{-1/2}\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\lesssim(1+t)^{-1/2}\left(\frac{1+t}{2}+\frac{\left\|x\right\|^{2}}{2}\right)^{-\frac{3}{2}}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}$

since $\left|x\right|\geq\sqrt{1+t}$ . Therefore,

I_{1}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

For $I_{2}$ , we decompose $\mathbb{R}^{3}$ into two parts

I_{2}=\int_{\frac{t}{2}}^{t}\left(\int_{\left|y\right|-\mathbf{c}\tau\leq\frac{\left|x\right|-ct}{2}}+\int_{\left|y\right|-\mathbf{c}\tau>\frac{\left|x\right|-ct}{2}}\right)\left(\cdots\right)dyd\tau=I_{21}+I_{22}\hbox{. }

If $\left|y\right|-\mathbf{c}\tau\leq\frac{\left|x\right|-\mathbf{c}t}{2}$ , we have

\left|x-y\right|\geq\left|x\right|-\left(\frac{\left|x\right|-\mathbf{c}t}{2}\right)-\mathbf{c}\tau\geq\frac{\left|x\right|-\mathbf{c}t}{2}\hbox{,}

and thus

$\displaystyle I_{21}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\left\|x\right\|-\mathbf{c}t\right)^{2}\right)^{-\frac{3}{2}}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}(1+t-\tau)^{-1/2}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{3}{2}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{\frac{t}{2}}^{t}(1+t-\tau)^{-1/2}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

If $\left|y\right|-\mathbf{c}\tau>\frac{\left|x\right|-\mathbf{c}t}{2}$ , then

$\displaystyle I_{22}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\int_{\frac{t}{2}}^{t}\int_{\{\left\|x-y\right\|\leq\mathbf{c}\left(t-\tau\right)\}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\int_{\frac{t}{2}}^{t}(1+t-\tau)^{-1/2}\ln\left(1+t-\tau\right)d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

Therefore,

I_{2}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

Combining all the estimates, we get desired estimate

I\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

Case 4: $\left(x,t\right)\in D_{4}$ . In this region $\sqrt{1+t}\leq\left|x\right|\leq\frac{\mathbf{c}t}{2}$ . We split the integral $I$ into three parts:

	$\displaystyle I$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{\left\|y\right\|\leq\mathbf{c}\tau\}}dyd\tau$
		$\displaystyle=$	$\displaystyle\left(\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}+\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}+\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:I_{1}+I_{2}+I_{3}\hbox{.}$

For $I_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts

I_{1}=\int_{0}^{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}\left(\int_{\left|y\right|\leq\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}}+\int_{\left|y\right|>\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}}\right)=:I_{11}+I_{12}\hbox{.}

If $0\leq\tau\leq\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}$ and $\left|y\right|\leq\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}$ , then

\mathbf{c}\left(t-\tau\right)-\left|x-y\right|\geq\mathbf{c}\left(t-\tau\right)-\left|x\right|-\left|y\right|\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{,}

so that

	$\displaystyle I_{11}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t+\left\|x\right\|\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\leq\mathbf{c}\tau}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\left(1+t\right)^{3/4}\lesssim\left(1+t\right)^{-13/4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

If $0\leq\tau\leq\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}$ and $\left|y\right|>\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}$ , then

\left|y\right|>\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{,}

and thus

$\displaystyle I_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-4}\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{\mathbf{c}}}\int_{\mathbb{R}^{3}}(1+\tau)^{-1/2}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}(1+\tau)^{-1/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{2}+\frac{1+t}{2}\right)^{-\frac{3}{2}}\left(1+t\right)^{\frac{7}{2}}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}$

since $\mathbf{c}t-\left|x\right|\geq\frac{\mathbf{c}t}{2}$ . Therefore,

I_{1}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

For $I_{2}$ , we use the spherical coordinates to obtain

$\displaystyle I_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}$
		$\displaystyle\cdot\left(1+\tau\right)^{-2}\left(1+\frac{\left\|r\right\|^{2}}{\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{r\leq\mathbf{c}\tau\}}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}z\left(1+\tau\right)^{-2}$
		$\displaystyle\cdot\left(1+\frac{r^{2}}{\left(1+\tau\right)}\right)^{-\frac{3}{2}}\frac{r}{\left\|x\right\|}\mathbf{1}_{\{r\leq\mathbf{c}\tau\}}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\left(1+t-\tau\right)^{-4+\frac{3}{2}}\left(1+\tau\right)^{-2}\cdot r\left(1+\frac{r^{2}}{\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{r\leq\mathbf{c}\tau\}}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\left(1+t-\tau\right)^{-\frac{5}{2}}\left(1+\tau\right)^{-1}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}\left(1+\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-\frac{3}{2}}\left(1+\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-1}$
	$\displaystyle\lesssim$	$\displaystyle(1+t)^{-1/2}\left(1+\left\|x\right\|\right)^{-3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

by setting $z=\sqrt{\left|x\right|^{2}+r^{2}-2r\left|x\right|\cos\theta}$ and $\sin\theta d\theta=\frac{z}{r\left|x\right|}dz$ .

For $I_{3}$ , we decompose $\mathbb{R}^{3}$ into two parts

I_{3}=\int_{t-\frac{\left|x\right|}{2\mathbf{c}}}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}\right)=:I_{31}+I_{32}\hbox{.}

If $t-\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\left|x\right|-\left|y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\left|x\right|}{4}\hbox{.}

If $t-\frac{\left|x\right|}{2\mathbf{c}}\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\left|x\right|}{4}\hbox{.}

Hence,

	$\displaystyle I_{31}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-\frac{5}{2}}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle(1+t)^{-1/2}\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}\hbox{,}$

since $\left|x\right|\geq\sqrt{1+t}$ , and

	$\displaystyle I_{32}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}\hbox{.}$

Therefore,

I_{3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}\hbox{.}

Combining all the estimates yields

I\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 5: $\left(x,t\right)\in D_{5}$ . In this region $\mathbf{c}t/2\leq\left|x\right|\leq\mathbf{c}t-\sqrt{1+t}$ . Now we split the integral $I$ into three parts

	$\displaystyle I$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{\left\|y\right\|\leq\mathbf{c}\tau\}}dyd\tau$
		$\displaystyle=$	$\displaystyle\left(\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}+\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}+\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:I_{1}+I_{2}+I_{3}\hbox{.}$

For $I_{1}$ , the estimate is the same as the $I_{1}$ of Case 4, so

I_{1}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

For $I_{2}$ , we use the spherical coordinates to obtain

$\displaystyle I_{2}$	$\displaystyle=$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}$
		$\displaystyle\cdot\left(1+\tau\right)^{-2}\left(1+\frac{r^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{r\leq\mathbf{c}\tau\}}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle=$	$\displaystyle\frac{1}{\left\|x\right\|}\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}z\left(1+\tau\right)^{-2}$
		$\displaystyle\cdot r\left(1+\frac{r^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}\mathbf{1}_{\{r\leq\mathbf{c}\tau\}}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}\left(1+t-\tau\right)^{-\frac{5}{2}}\left(1+\tau\right)^{-1}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}\left(1+\frac{3\left\|x\right\|}{4\mathbf{c}}-\frac{t}{4}\right)^{-\frac{3}{2}}\left(1+\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-1}$
	$\displaystyle\lesssim$	$\displaystyle\left\|x\right\|^{-1}(1+t)^{-1/2}\left(1+\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}+\frac{\left\|x\right\|}{\mathbf{c}}-\frac{t}{2}\right)^{-1}\left(1+\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-1}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}$

by setting $z=\sqrt{\left|x\right|^{2}+r^{2}-2r\left|x\right|\cos\theta}$ and $\sin\theta d\theta=\frac{z}{r\left|x\right|}dz$ .

For $I_{3}$ , we decompose $\mathbb{R}^{3}$ into two parts

I_{3}=\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)\right)}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}\right)\left(\cdots\right)dyd\tau=:I_{31}+I_{32}\hbox{.}

If $\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)\right)\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\mathbf{c}t+\left|x\right|}{8}=\frac{\mathbf{c}t-\left|x\right|}{8}+\frac{|x|}{4}\hbox{.}

If $\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)\right)\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{8}+\frac{|x|}{4}\hbox{.}

Hence,

$\displaystyle I_{31}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\int_{\|y\|\leq\mathbf{c}\tau}\left(1+t-\tau\right)^{-2}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\left(1+t-\tau\right)^{-2}(1+\tau)^{-1/4}d\tau$
	$\displaystyle\lesssim$	$\displaystyle(1+t)^{-1/4}\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}\lesssim\left(1+t\right)^{-9/4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

On the other hand,

	$\displaystyle I_{31}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\|x\|^{2}\right)^{-2}\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\int_{\|y\|\leq\mathbf{c}\tau}\left(1+t-\tau\right)^{-2}\left(1+\tau\right)^{-2}\left(1+\frac{\left\|y\right\|^{2}}{D_{0}\left(1+\tau\right)}\right)^{-\frac{3}{2}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle(1+t)^{-17/4}\hbox{.}$

By interpolation, one has

I_{31}\lesssim(1+t)^{-13/4}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\,.

For $I_{32}$ ,

$\displaystyle I_{32}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\left(1+\left\|x\right\|-\mathbf{c}\left(t-\tau\right)\right)^{-3}(1+\tau)^{-1/2}$
		$\displaystyle\cdot\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-4}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{1+t-\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\left(1+\left\|x\right\|-\mathbf{c}\left(t-\tau\right)\right)^{-3}\left(1+t-\tau\right)^{-4+\frac{5}{2}}\left(1+\tau\right)^{-\frac{1}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\lesssim(1+t)^{-1/2}\left(1+\left\|x\right\|\right)^{-3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}\hbox{,}$

so that

I_{3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

To sum up,

I\leq\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{\left(1+t\right)}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

∎

Proof of Lemma 25.

(Huygens wave convolved with Huygens wave).

Case 1: $\left(x,t\right)\in D_{1}\cup D_{2}$ . Direct computation gives

$\displaystyle J$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle+\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau+\left(1+t\right)^{-4}\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\hbox{.}$

Case 2: $\left(x,t\right)\in D_{3}$ . We split the integral $J$ into four parts

	$\displaystyle J$	$\displaystyle=$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(\int_{\left\|y\right\|-\mathbf{c}\tau\leq\frac{\left\|x\right\|-\mathbf{c}t}{2}}+\int_{\left\|y\right\|-\mathbf{c}\tau>\frac{\left\|x\right\|-\mathbf{c}t}{2}}\right)\left(\cdots\right)dyd\tau+\int_{\frac{t}{2}}^{t}\left(\int_{\left\|y\right\|-\mathbf{c}\tau\leq\frac{\left\|x\right\|-\mathbf{c}t}{2}}+\int_{\left\|y\right\|-\mathbf{c}\tau>\frac{\left\|x\right\|-\mathbf{c}t}{2}}\right)\left(\cdots\right)dyd\tau$
		$\displaystyle=$	$\displaystyle J_{11}+J_{12}+J_{21}+J_{22}\hbox{.}$

Note that if $\left|y\right|-\mathbf{c}\tau\leq\frac{\left|x\right|-\mathbf{c}t}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\left|x\right|-\left|y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}t}{2}\hbox{.}

Hence,

	$\displaystyle J_{11}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|-\mathbf{c}\tau\leq\frac{\left\|x\right\|-\mathbf{c}t}{2}}\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{D_{0}\left(1+t\right)}}\hbox{.}$

For $J_{12}$ , we have

$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|-\mathbf{c}\tau>\frac{\left\|x\right\|-\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}\left(1+\tau\right)^{-2}\left(1+\left(\left\|x\right\|-\mathbf{c}t\right)^{2}\right)^{-2}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-4}\hbox{,}$

and

$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|-\mathbf{c}\tau>\frac{\left\|x\right\|-\mathbf{c}t}{2}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\left(1+\left\|y\right\|-\mathbf{c}\tau\right)^{-4}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\int_{\mathbf{c}\tau+\frac{\left\|x\right\|-\mathbf{c}t}{2}}^{\infty}\left(1+r-\mathbf{c}\tau\right)^{-4}r^{2}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\int_{\frac{\left\|x\right\|-\mathbf{c}t}{2}}^{\infty}\left(1+r\right)^{-4}\left(r+\mathbf{c}\tau\right)^{2}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\left[r^{-1}+\mathbf{c}\tau\cdot r^{-2}+\left(\mathbf{c}\tau\right)^{2}r^{-3}\right]_{r=\frac{\left\|x\right\|-\mathbf{c}t}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left[\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-1}+\ln\left(2+t\right)\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-2}+t\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-3}\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-1}\left[1+\ln\left(2+t\right)\left(1+\sqrt{1+t}\right)^{-1}+t\left(1+\sqrt{1+t}\right)^{-2}\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-1}\hbox{,}$

which implies that

$\displaystyle J_{12}$	$\displaystyle=$	$\displaystyle\left(J_{12}\right)^{\frac{1}{3}}\left(J_{12}\right)^{\frac{2}{3}}\lesssim\left[\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-4}\right]^{\frac{1}{3}}\left[\left(1+t\right)^{-5/2}\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-1}\right]^{\frac{2}{3}}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{5}{3}}\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{8}{3}}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $J_{21}$ and $J_{22}$ , it follows that

$\displaystyle J_{21}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|-\mathbf{c}\tau\leq\frac{\left\|x\right\|-\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{2D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}\hbox{,}$

$\displaystyle J_{22}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|-\mathbf{c}\tau>\frac{\left\|x\right\|-\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

Gathering all the estimates, we can conclude that

J\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 3: $\left(x,t\right)\in D_{4}$ . We split the integral into four parts

	$\displaystyle J$	$\displaystyle=$	$\displaystyle\left(\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}+\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}+\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau$
		$\displaystyle=$	$\displaystyle:J_{1}+J_{2}+J_{3}+J_{4}\hbox{.}$

For $J_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts

J_{1}=\int_{0}^{\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}}\left(\int_{\left|y\right|\leq\frac{\mathbf{c}t-\left|x\right|}{2}}+\int_{\left|y\right|>\frac{\mathbf{c}t-\left|x\right|}{2}}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:J_{11}+J_{12}\hbox{.}

If $0\leq\tau\leq\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}$ , $\left|y\right|\leq\frac{\mathbf{c}t-\left|x\right|}{2}$ , then

\mathbf{c}\left(t-\tau\right)-\left|x-y\right|\geq\mathbf{c}\left(t-\tau\right)-\left|x\right|-\left|y\right|\geq\frac{\mathbf{c}t-\left|x\right|}{4}\geq\frac{\mathbf{c}t}{8}\hbox{.}

Hence,

	$\displaystyle J_{11}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\int_{\left\|y\right\|\leq\frac{\mathbf{c}t-\left\|x\right\|}{2}}\left(1+t-s\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}\hbox{,}$

and

$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\int_{\left\|y\right\|>\frac{\mathbf{c}t-\left\|x\right\|}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-2}\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}\left(1+\tau\right)^{-2}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\lesssim\left(1+t\right)^{-1}\left(1+\left\|x\right\|\right)^{-3}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

since $\sqrt{1+t}\leq\left|x\right|\leq\frac{\mathbf{c}t}{2}$ , so that

J_{1}\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

For $J_{2}$ and $J_{3}$ ,

$\displaystyle J_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}\right)^{-4}\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+t\right)^{-4}\left(1+t\right)^{3}\lesssim\left(1+t\right)^{-7/2}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

$\displaystyle J_{3}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\frac{1}{4}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)^{-5/2}\left(1+t\right)^{-4}\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-13/2}\left(1+t\right)^{3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

due to Lemma 5.4 of [24].

For $J_{4}$ , we decompose $\mathbb{R}^{3}$ into two parts

J_{4}=\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}t}{2}}+\int_{\left|y\right|>\frac{\left|x\right|+\mathbf{c}t}{2}}\right)\left(\cdots\right)dyd\tau=:J_{41}+J_{42}\hbox{.}

If $\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}t}{2}$ , then

\mathbf{c}\tau-\left|y\right|\geq\frac{\mathbf{c}t}{2}+\frac{1}{4}\left(\mathbf{c}t+\left|x\right|\right)-\frac{\left|x\right|+\mathbf{c}t}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\geq\frac{\mathbf{c}t}{8}\hbox{.}

If $\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|+\mathbf{c}t}{2}$ , then

	$\displaystyle\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)$	$\displaystyle\geq$	$\displaystyle\left\|y\right\|-\left\|x\right\|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left\|x\right\|+\mathbf{c}t}{2}-\left\|x\right\|-\mathbf{c}t+\frac{\mathbf{c}t}{2}+\frac{1}{4}\left(\mathbf{c}t+\left\|x\right\|\right)$
		$\displaystyle\geq$	$\displaystyle\frac{\mathbf{c}t-\left\|x\right\|}{4}\geq\frac{\mathbf{c}t}{8}\hbox{.}$

Hence,

$\displaystyle J_{41}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}(1+t)$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5}\lesssim\left(1+t\right)^{-2}\left(1+\left\|x\right\|\right)^{-3}\lesssim\left(1+t\right)^{-7/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

$\displaystyle J_{42}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{2D_{0}\left(1+t\right)}}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{2D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{2D_{0}\left(1+t\right)}}\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\left(1+t-\tau\right)^{-\frac{5}{2}+\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{t}{C}}\lesssim\left(1+t\right)^{-7/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

for some $C>0$ , so that

J_{4}\lesssim\left(1+t\right)^{-7/2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

As a result,

J\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}e^{-\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{2D_{0}\left(1+t\right)}}

for $\left(x,t\right)\in D_{4}$ .

Case 4: $\left(x,t\right)\in D_{5}$ . We split the integral $J$ into five parts

J=\left(\int_{0}^{\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}}+\int_{\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{\frac{|x|}{2\mathbf{c}}+\frac{1}{4}(t+\frac{|x|}{\mathbf{c}})}+\int_{\frac{|x|}{2\mathbf{c}}+\frac{1}{4}(t+\frac{|x|}{\mathbf{c}})}^{\frac{t}{2}+\frac{1}{4}(t+\frac{|x|}{\mathbf{c}})}+\int^{t}_{\frac{t}{2}+\frac{1}{4}(t+\frac{|x|}{\mathbf{c}})}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:\sum_{i=1}^{5}J_{i}\hbox{.}

For $J_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts

J_{1}=\int_{0}^{\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}}\left(\int_{\left|y\right|\leq\frac{\mathbf{c}t-\left|x\right|}{2}}+\int_{\left|y\right|>\frac{\mathbf{c}t-\left|x\right|}{2}}\right)\left(\cdots\right)dyd\tau=:J_{11}+J_{12}\hbox{.}

If $0\leq\tau\leq\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}$ , $\left|y\right|\leq\frac{\mathbf{c}t-\left|x\right|}{2}$ , then

\mathbf{c}\left(t-\tau\right)-\left|x-y\right|\geq\mathbf{c}\left(t-\tau\right)-\left(\left|x\right|+\left|y\right|\right)\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

If $0\leq\tau\leq\frac{t}{4}-\frac{\left|x\right|}{4\mathbf{c}}$ , $\left|y\right|>\frac{\mathbf{c}t-\left|x\right|}{2}$ , then

\left|y\right|-\mathbf{c}\tau\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

Hence,

	$\displaystyle J_{11}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\int_{\left\|y\right\|\leq\frac{\mathbf{c}t-\left\|x\right\|}{2}}\left(1+t\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{D_{0}\left(1+t\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{D_{0}\left(1+t\right)}}\int_{0}^{\frac{\mathbf{c}t-\left\|x\right\|}{4}}\left(1+\tau\right)^{-4+\frac{5}{2}}d\tau\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{D_{0}\left(1+t\right)}}\hbox{.}$

For $J_{12}$ , we have

$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\int_{\left\|y\right\|>\frac{\mathbf{c}t-\left\|x\right\|}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-2}\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\mathbf{c}t-\left\|x\right\|\right)^{2}\right)^{-2}\int_{0}^{\frac{\mathbf{c}t-\left\|x\right\|}{4}}\left(1+\tau\right)^{-2}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-4}\hbox{,}$

and

$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\int_{\frac{\mathbf{c}t-\left\|x\right\|}{2}}^{\infty}\left(1+r-\mathbf{c}\tau\right)^{-4}r^{2}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{\mathbf{c}t-\left\|x\right\|}{4}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-1}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-1}\hbox{,}$

which implies that

	$\displaystyle J_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(J_{12}\right)^{\frac{1}{3}}\left(J_{12}\right)^{\frac{2}{3}}\lesssim\left[\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-4}\right]^{\frac{1}{3}}\left[\left(1+t\right)^{-5/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-1}\right]^{\frac{2}{3}}$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{5}{3}}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-2}\lesssim\left(1+t\right)^{-\frac{8}{3}}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

Therefore, we get

J_{1}\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{D_{0}\left(1+t\right)}}+\left(1+t\right)^{-\frac{8}{3}}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

For $J_{2}$ , we use the spherical coordinates to obtain

$\displaystyle J_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}rz\frac{1}{\left\|x\right\|}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\int_{0}^{\infty}\int_{0}^{\infty}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}rz\frac{1}{\left\|x\right\|}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\int_{0}^{\infty}\left(1+t-\tau\right)^{-\frac{5}{2}+\frac{3}{2}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}\frac{r}{\left\|x\right\|}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{5}{2}+\frac{3}{2}}\left\|x\right\|^{-1}\int_{\frac{t}{4}-\frac{\left\|x\right\|}{4\mathbf{c}}}^{\frac{t}{2}}\left(1+\tau\right)^{-4+\frac{3}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-\frac{3}{2}}\lesssim\left(1+t\right)^{-2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{\frac{1}{2}}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{,}$

by setting $z=\sqrt{\left|x\right|^{2}+r^{2}-2r\left|x\right|\cos\theta}$ and $\sin\theta d\theta=\frac{z}{r\left|x\right|}dz$ .

For $J_{3}$ , we use the spherical coordinates again to obtain

$\displaystyle J_{3}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{\frac{3\|x\|}{4\mathbf{c}}+\frac{t}{4}}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{\frac{3\|x\|}{4\mathbf{c}}+\frac{t}{4}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}rz\frac{1}{\left\|x\right\|}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left\|x\right\|^{-1}\int_{\frac{t}{2}}^{\frac{3\|x\|}{4\mathbf{c}}+\frac{t}{4}}\int_{0}^{\infty}\left(1+t-\tau\right)^{-\frac{5}{2}+\frac{3}{2}}\left(1+\frac{\left(r-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}rdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\int_{\frac{t}{2}}^{\frac{3\|x\|}{4\mathbf{c}}+\frac{t}{4}}\left(1+t-\tau\right)^{-1}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left[\ln\left(1+\frac{t}{2}\right)-\ln\left(1+\frac{\mathbf{c}t-\|x\|}{4\mathbf{c}}\right)\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $J_{4}$ , similar to $J_{3}$ , we have

$\displaystyle J_{4}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\int_{\frac{3\|x\|}{4\mathbf{c}}+\frac{t}{4}}^{\frac{\|x\|}{4\mathbf{c}}+\frac{3t}{4}}\left(1+t-\tau\right)^{-1}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left[\left(1+\frac{3(\mathbf{c}t-\|x\|)}{4\mathbf{c}}\right)^{-1}\left(1+\frac{(\mathbf{c}t-\|x\|)}{2\mathbf{c}}\right)\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $J_{5}$ , we decompose $\mathbb{R}^{3}$ into two parts

J_{5}=\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}t}{2}}+\int_{\left|y\right|>\frac{\left|x\right|+\mathbf{c}t}{2}}\right)\left(\cdots\right)dyd\tau=:J_{51}+J_{52}\hbox{.}

If $\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|+\mathbf{c}t}{2}$ , then

\mathbf{c}\tau-\left|y\right|\geq\frac{\mathbf{c}t}{2}+\frac{1}{4}\left(\mathbf{c}t+\left|x\right|\right)-\frac{\left|x\right|+\mathbf{c}t}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

If $\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|+\mathbf{c}t}{2}$ , then

\displaystyle\left|x-y\right|-\mathbf{c}\left(t-\tau\right)

\displaystyle\geq

\displaystyle\left|y\right|-\left|x\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|+\mathbf{c}t}{2}-\left|x\right|-\mathbf{c}t+\frac{\mathbf{c}t}{2}+\frac{1}{4}\left(\mathbf{c}t+\left|x\right|\right)\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

Hence,

$\displaystyle J_{51}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|+\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\left(\mathbf{c}t-\left\|x\right\|\right)$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-2}\hbox{,}$

$\displaystyle J_{52}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|+\mathbf{c}t}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{2D_{0}\left(1+\mathbf{c}t-\left\|x\right\|\right)}}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{2D_{0}\left(1+t-\tau\right)}}\left(1+t\right)^{-4}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)}{2D_{0}}}\int_{\frac{t}{2}+\frac{1}{4}\left(t+\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\left(1+t-\tau\right)^{-\frac{5}{2}+\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)}{2D_{0}}}\hbox{.}$

Gathering all the estimates, we have

J\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}\hbox{.}

∎

Proof of Lemma 26.

(Huygens wave convolved with diffusion wave).

Case 1: $\left(x,t\right)\in D_{1}$ . Direct computation gives

$\displaystyle K$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-3}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}\left(1+\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-3}\left(1+\tau\right)^{\frac{3}{2}}dyd\tau+\left(1+t\right)^{-3}\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}+\left(1+t\right)^{-3}\left(1+t\right)\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

Case 2: $\left(x,t\right)\in D_{2}$ . We split the integral $K$ into two parts

K=\left(\int_{0}^{\frac{3}{4}t}+\int_{\frac{3}{4}t}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:K_{1}+K_{2}\hbox{.}

For $K_{1}$ , it is easy to see

	$\displaystyle K_{1}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{3}{4}t}\int_{\mathbb{R}^{3}}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{3}{4}t}\left(1+t\right)^{-5/2}\left(1+\tau\right)^{-3}\left(1+\tau\right)^{\frac{3}{2}}d\tau\lesssim\left(1+t\right)^{-5/2}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $K_{2}$ , we decompose $\mathbb{R}^{3}$ into two parts

K_{2}=\int_{\frac{3}{4}t}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|}{2}}+\int_{\left|y\right|>\frac{\left|x\right|}{2}}\right)\left(\cdots\right)dyd\tau=:K_{21}+K_{22}.

It readily follows that

$\displaystyle K_{22}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\int_{\frac{3}{4}t}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\int_{\frac{3}{4}t}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\hbox{.}$

If $\frac{3}{4}t\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|}{2}$ , then

	$\displaystyle\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)$	$\displaystyle\geq$	$\displaystyle\left\|x\right\|-\left\|y\right\|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left\|x\right\|}{2}-\frac{\mathbf{c}t}{4}\geq\frac{1}{2}\left(\mathbf{c}t-\sqrt{1+t}\right)-\frac{\mathbf{c}t}{4}$
		$\displaystyle\geq$	$\displaystyle\frac{\mathbf{c}t}{8}+\left(\frac{\mathbf{c}t}{8}-\frac{\sqrt{1+t}}{2}\right)\geq\frac{\mathbf{c}t}{8}$

for $t\geq 11$ . Hence, there exists a universal constant $C>0$ such that

$\displaystyle K_{21}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{t}{C}}\int_{\frac{3}{4}t}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}e^{-\frac{t}{C}}\int_{\frac{3}{4}t}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+\tau\right)^{\frac{3}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{t}{C}}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}$

for $t\geq 11$ , and

	$\displaystyle K_{21}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\int_{\frac{3}{4}t}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle C\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}$

for $0\leq t\leq 11$ . Therefore we can conclude that

K\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-3}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 3: $\left(x,t\right)\in D_{3}$ . We split the integral $K$ into four parts

	$\displaystyle K$	$\displaystyle=$	$\displaystyle\int_{0}^{\frac{t}{2}}\left(\int_{\left\|y\right\|\leq\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\right)\left(\cdots\right)dyd\tau+\int_{\frac{t}{2}}^{t}\left(\int_{\left\|y\right\|\leq\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\right)\left(\cdots\right)dyd\tau$
		$\displaystyle=$	$\displaystyle K_{11}+K_{12}+K_{21}+K_{22}\hbox{.}$

If $\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\left|x\right|-\left|y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}=\frac{\left|x\right|-\mathbf{c}t+\mathbf{c}\tau}{2}\hbox{.}

Hence,

	$\displaystyle K_{11}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}+\left(\mathbf{c}\tau\right)^{2}}{4D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-3}\left(1+\tau\right)^{\frac{3}{2}}dyd\tau\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}\hbox{,}$

$\displaystyle K_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+\tau+\left\|y\right\|^{2}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\int_{\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}^{\infty}\left(1+r\right)^{-6}r^{2}drd\tau\lesssim\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}}\left(1+\left\|x\right\|-\mathbf{c}t+\mathbf{c}\tau\right)^{-3}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\left\|x\right\|-\mathbf{c}t\right)^{-2}\lesssim\left(1+t\right)^{-7/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{,}$

	$\displaystyle K_{21}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}}^{t}\int_{\left\|y\right\|\leq\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}+\left(\mathbf{c}\tau\right)^{2}}{4D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}e^{-\frac{t}{C}}\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+\tau\right)^{-\frac{3}{2}}d\tau\lesssim e^{-\frac{t}{C}}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}$

for some $C>0$ , and

$\displaystyle K_{22}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}+\mathbf{c}^{2}t^{2}}{1+t}\right)^{-3}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}+t\right)^{-3}\int_{\frac{t}{2}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

It implies that

K\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 4: $\left(x,t\right)\in D_{4}$ . First note that $\sqrt{1+t}\leq\left|x\right|\leq\mathbf{c}t/2$ in this region. Now we split the integral $K$ into three parts

K=\left(\int_{0}^{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}+\int_{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}^{t-\frac{\left|x\right|}{2\mathbf{c}}}+\int_{t-\frac{\left|x\right|}{2\mathbf{c}}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:K_{1}+K_{2}+K_{3}\hbox{.}

For $K_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts

K_{1}=\int_{0}^{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}\left(\int_{\left|y\right|\leq\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}}+\int_{\left|y\right|>\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}}\right)\left(\cdots\right)dyd\tau=:K_{11}+K_{12}\hbox{.}

If $0\leq\tau\leq\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}$ , $\left|y\right|\leq\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}$ , then

\mathbf{c}\left(t-\tau\right)-\left|x\right|-\left|y\right|\geq\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

If $0\leq\tau\leq\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}$ , $\left|y\right|>\frac{\mathbf{c}\left(t-\tau\right)-\left|x\right|}{2}$ , then

\left|y\right|>\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

Hence, we have

	$\displaystyle K_{11}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{16D_{0}\left(1+t\right)}}\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{\left\|y\right\|\leq\frac{\mathbf{c}\left(t-\tau\right)-\left\|x\right\|}{2}}\left(1+\tau\right)^{-3}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-3}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{16D_{0}\left(1+t\right)}}\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}\left(1+\tau\right)^{-3+\frac{3}{2}}d\tau\lesssim\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{4D_{0}\left(1+t\right)}}\hbox{,}$

$\displaystyle K_{12}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{\left\|y\right\|\geq\frac{\mathbf{c}\left(t-\tau\right)-\left\|x\right\|}{2}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\int_{0}^{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{\frac{\mathbf{c}\left(t-\tau\right)-\left\|x\right\|}{2}}^{\infty}r^{-6}r^{2}drd\tau\lesssim\left(1+t\right)^{-5/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{,}$

so that

K_{1}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

For $K_{2}$ , we use the spherical coordinates to obtain

$\displaystyle K_{2}$	$\displaystyle=$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle=$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}r^{2}\frac{zdz}{r\left\|x\right\|}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t-\frac{\left\|x\right\|}{2\mathbf{c}}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+\left\|x\right\|\right)^{-5/2}e^{-\frac{\left(c\tau-\left(\mathbf{c}t-z\right)\right)^{2}}{D_{0}\left(1+t\right)}}\left(1+t\right)^{-3}\left(1+\frac{r^{2}}{1+t}\right)^{-3}r\frac{zdz}{\left\|x\right\|}drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\frac{1}{\left\|x\right\|}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+\left\|x\right\|\right)^{-5/2}\sqrt{1+t}\left(1+t\right)^{-3}\left(1+\frac{r^{2}}{1+t}\right)^{-3}rzdzdr$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|\right)^{-7/2}\left(1+t\right)^{-\frac{5}{2}}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+\frac{r^{2}}{1+t}\right)^{-3}rzdzdr$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|\right)^{-7/2}\left(1+t\right)^{-\frac{5}{2}}\left[\int_{0}^{\left\|x\right\|}\int_{\left\|x\right\|-r}^{\left\|x\right\|+r}zr\left(1+\frac{r^{2}}{1+t}\right)^{-3}dzdr+\int_{\left\|x\right\|}^{\infty}\int_{r-\left\|x\right\|}^{r+\left\|x\right\|}zr\left(1+\frac{r^{2}}{1+t}\right)^{-3}dzdr\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|\right)^{-7/2}\left(1+t\right)^{-\frac{5}{2}}\left[\int_{0}^{\left\|x\right\|}r^{2}\left\|x\right\|\left(1+\frac{r^{2}}{1+t}\right)^{-3}dr+\int_{\left\|x\right\|}^{\infty}r^{2}\left\|x\right\|\left(1+\frac{r^{2}}{1+t}\right)^{-3}dr\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|\right)^{-7/2}\left(1+t\right)^{-\frac{5}{2}}\left[\left\|x\right\|\left(1+t\right)^{\frac{3}{2}}+\left\|x\right\|\left(1+t\right)^{\frac{3}{2}}\left(1+\frac{\left\|x\right\|}{\sqrt{1+t}}\right)^{-3}\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|\right)^{-5/2}\left(1+t\right)^{-1}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

by setting $z=\sqrt{\left|x\right|^{2}+r^{2}-2r\left|x\right|\cos\theta}$ and $\sin\theta d\theta=\frac{z}{r\left|x\right|}dz$ .

For $K_{3}$ , we split $\mathbb{R}^{3}$ into two parts

K_{3}=\int_{t-\frac{\left|x\right|}{2\mathbf{c}}}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}\right)\left(\cdots\right)dyd\tau=:K_{31}+K_{32}\hbox{.}

If $t-\frac{\left|x\right|}{2\mathbf{c}}<\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\left|x\right|-\left|y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\left|x\right|}{4}\hbox{.}

If $t-\frac{\left|x\right|}{2\mathbf{c}}<\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\left|x\right|}{4}\hbox{.}

Hence,

$\displaystyle K_{31}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\|x\|^{2}\right)^{-\frac{5}{2}}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\|x\|^{2}\right)^{-\frac{5}{2}}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\left(1+\tau\right)^{-\frac{3}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

$\displaystyle K_{32}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t-\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t-\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-3}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\int_{t-\frac{\left\|x\right\|}{2\mathbf{c}}}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+t-\tau\right)^{\frac{5}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-3}\hbox{,}$

so that

K_{3}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

To sum up,

K\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 5: $\left(x,t\right)\in D_{5}$ . We split the integral $K$ into three parts

K=\left(\int_{0}^{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}+\int_{\frac{t}{2}-\frac{\left|x\right|}{2\mathbf{c}}}^{\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)}+\int_{\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:K_{1}+K_{2}+K_{3}\hbox{.}

For $K_{1}$ , the estimate is the same as the term $K_{1}$ of Case 4 and so

K_{1}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

For $K_{2}$ , we use the spherical coordinates to obtain

$\displaystyle K_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}\int_{\mathbb{R}^{3}}\left(1+t\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}\int_{0}^{\infty}\int_{0}^{\pi}\left(1+t\right)^{-5/2}e^{-\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t\right)}}\left(1+\tau\right)^{-3}\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}\left(1+t\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t\right)}}z\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-3}\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}\frac{r}{\left\|x\right\|}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-3}\int_{\frac{t}{2}-\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}\int_{0}^{\infty}\int_{\left\|\left\|x\right\|-r\right\|}^{\left\|x\right\|+r}e^{-\frac{\left(c\tau-\left(\mathbf{c}t-z\right)\right)^{2}}{D_{0}\left(1+t\right)}}z\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}rdzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-3}\left(1+t\right)^{\frac{1}{2}}$
		$\displaystyle\cdot\left[\int_{0}^{\left\|x\right\|}\int_{\left\|x\right\|-r}^{\left\|x\right\|+r}z\left(1+\frac{r^{2}}{1+\mathbf{c}t-\left\|x\right\|}\right)^{-3}rdzdr+\int_{\left\|x\right\|}^{\infty}\int_{r-\left\|x\right\|}^{r+\left\|x\right\|}z\left(1+\frac{r^{2}}{1+\mathbf{c}t-\left\|x\right\|}\right)^{-3}rdzdr\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-3}\left[\int_{0}^{\left\|x\right\|}r^{2}\left\|x\right\|\left(1+\frac{r^{2}}{1+\mathbf{c}t-\left\|x\right\|}\right)^{-3}dr+\int_{\left\|x\right\|}^{\infty}r^{2}\left\|x\right\|\left(1+\frac{r^{2}}{1+\mathbf{c}t-\left\|x\right\|}\right)^{-3}dr\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-3}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-3}\left[\left\|x\right\|\left(1+\mathbf{c}t-\left\|x\right\|\right)^{\frac{3}{2}}+\left\|x\right\|\left(1+\mathbf{c}t-\left\|x\right\|\right)^{3}\left(1+\left\|x\right\|\right)^{-3}\right]$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-\frac{3}{2}}\lesssim\left(1+t\right)^{-2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{\frac{1}{2}}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

For $K_{3}$ , we decompose $\mathbb{R}^{3}$ into two parts

K_{3}=\int_{\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)}^{t}\left(\int_{\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}+\int_{\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}}\right)\left(\cdots\right)dyd\tau=K_{31}+K_{32}\hbox{.}

If $\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|\leq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|x-y\right|-\mathbf{c}\left(t-\tau\right)\geq\left|x\right|-\left|y\right|-\mathbf{c}\left(t-\tau\right)\geq\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

If $\frac{3}{2}\left(t-\frac{\left|x\right|}{\mathbf{c}}\right)\leq\tau\leq t$ , $\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}$ , then

\left|y\right|>\frac{\left|x\right|-\mathbf{c}\left(t-\tau\right)}{2}\geq\frac{\mathbf{c}t-\left|x\right|}{4}\hbox{.}

Hence, we use the spherical coordinates to obtain

$\displaystyle K_{31}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{0}^{\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\int_{0}^{\pi}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t-\tau\right)}}e^{-\frac{\left(\sqrt{\left\|x\right\|^{2}+r^{2}-2r\left\|x\right\|\cos\theta}-\mathbf{c}\left(t-\tau\right)\right)^{2}}{2D_{0}\left(1+t-\tau\right)}}$
		$\displaystyle\cdot\left(1+\tau\right)^{-3}\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}r^{2}\sin\theta d\theta drd\tau$
	$\displaystyle\lesssim$	$\displaystyle e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t\right)}}\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{0}^{\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\int_{\left\|x\right\|-r}^{\left\|x\right\|+r}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(z-\mathbf{c}\left(t-\tau\right)\right)^{2}}{2D_{0}\left(1+t-\tau\right)}}z\left(1+\tau\right)^{-3}$
		$\displaystyle\cdot\left(1+\frac{r^{2}}{1+\tau}\right)^{-3}\frac{r}{\left\|x\right\|}dzdrd\tau$
	$\displaystyle\lesssim$	$\displaystyle\frac{1}{\left\|x\right\|}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t\right)}}\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\left(1+t-\tau\right)^{-5/2}\left(1+\tau\right)^{-3}\left(1+t-\tau\right)^{\frac{3}{2}}\left(1+\tau\right)d\tau$
	$\displaystyle\lesssim$	$\displaystyle\frac{1}{\left\|x\right\|}e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t\right)}}\left(\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}+\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\right)\left(1+t-\tau\right)^{-1}\left(1+\tau\right)^{-2}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\frac{e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t\right)}}}{\left\|x\right\|}\left[\left(1+\frac{3\left\|x\right\|}{4\mathbf{c}}-\frac{t}{4}\right)^{-1}\left(1+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)^{-1}+\left(1+t\right)^{-2}\ln\left(1+\frac{3\|x\|}{4\mathbf{c}}-\frac{t}{4}\right)\right]$
	$\displaystyle\lesssim$	$\displaystyle\frac{e^{-\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{32D_{0}\left(1+t\right)}}}{1+t}\left(1+t\right)^{-3/2}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-x\right)^{2}}{1+t}\right)^{-1}\hbox{,}$

by setting $z=\sqrt{\left|x\right|^{2}+r^{2}-2r\left|x\right|\cos\theta}$ and $\sin\theta d\theta=\frac{z}{r\left|x\right|}dz$ . As for $K_{32}$ , direct computation gives

$\displaystyle K_{32}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\left\|y\right\|^{2}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}+\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\right)\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}(\cdot\cdot\cdot)dyd\tau$
	$\displaystyle=$	$\displaystyle K_{321}+K_{322}\,,$

we then have

$\displaystyle K_{321}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\left\|y\right\|^{2}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)}^{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}\left(1+t-\tau\right)^{-5/2}\left(1+\left\|x\right\|-\mathbf{c}\left(t-\tau\right)\right)^{-3}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\mathbf{c}t-\left\|x\right\|\right)^{-2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

and

$\displaystyle K_{322}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\int_{\left\|y\right\|>\frac{\left\|x\right\|-\mathbf{c}\left(t-\tau\right)}{2}}\left(1+t-\tau\right)^{-5/2}e^{-\frac{\left(\left\|x-y\right\|-\mathbf{c}\left(t-\tau\right)\right)^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\left\|y\right\|^{2}\right)^{-3}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{1}{2}\left(t+\frac{3}{2}\left(t-\frac{\left\|x\right\|}{\mathbf{c}}\right)\right)}^{t}\left(1+\left\|x\right\|-\mathbf{c}\left(t-\tau\right)\right)^{-6}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left\|x\right\|+\mathbf{c}t\right)^{-5}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-\frac{7}{2}}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

It implies that

K_{3}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}+\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

As a consequence,

K\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\mathbf{c}t-\left|x\right|\right)^{2}}{1+t}\right)^{-1}+\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

∎

Proof of Lemma 27.

(Diffusion wave convolution with Huygens wave).

Case 1: $\left(x,t\right)\in D_{1}$ . We split the integral into two parts to obtain

$\displaystyle L$	$\displaystyle=$	$\displaystyle\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle+\left(1+t\right)^{-4}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau+\left(1+t\right)^{-4}\int_{t/2}^{t}\left(1+t-\tau\right)^{-\frac{1}{2}}d\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

Case 2: $\left(x,t\right)\in D_{2}$ . We split the integral into two parts:

L=\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:L_{1}+L_{2}\hbox{.}

First one can see

	$\displaystyle L_{2}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}$

As for $L_{1}$ , we further decompose $\mathbb{R}^{3}$ into two parts:

L_{1}=\int_{0}^{\frac{t}{2}}\left(\int_{\left|y\right|\leq\frac{2}{3}\left|x\right|}+\int_{\left|y\right|>\frac{2}{3}\left|x\right|}\right)\left(\ldots\right)dyd\tau=:L_{11}+L_{12}.

If $\left|y\right|\leq\frac{2}{3}\left|x\right|$ , then we have

\left|x-y\right|\geq\left|x\right|-\left|y\right|\geq\frac{\left|x\right|}{3}\hbox{,}

and thus

			$\displaystyle L_{11}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\leq\frac{2}{3}\left\|x\right\|}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-2}\int_{0}^{\frac{t}{2}}\left(1+\tau\right)^{-4}\left(1+\tau\right)^{\frac{5}{2}}d\tau\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-2}\hbox{.}$

If $\left|y\right|\geq\frac{2}{3}\left|x\right|$ and $0\leq\tau\leq\frac{t}{2}$ , then

\left|y\right|-\mathbf{c}\tau\geq\frac{2}{3}\left|x\right|-\mathbf{c}\tau\geq\frac{2}{3}\left(\mathbf{c}t-\sqrt{1+t}\right)-\frac{\mathbf{c}t}{2}\geq\frac{\mathbf{c}t}{12}+\frac{\mathbf{c}t}{12}-\frac{2}{3}\sqrt{1+t}\geq\frac{\mathbf{c}t}{12}

for $t\geq 40$ . Hence,

			$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\geq\frac{2}{3}\left\|x\right\|}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}\left(1+\tau\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-9/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}$

whenever $t\geq 40$ , and

			$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\geq\frac{2}{3}\left\|x\right\|}\left(1+t-\tau\right)^{-2}\left(1+\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}\right)^{-2}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle C\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}$

whenever $0\leq t\leq 40$ . Therefore, we get

L_{12}\lesssim\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Combining all the estimates, we have

L\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-2}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-1}\hbox{.}

Case 3: $\left(x,t\right)\in D_{3}$ . We split the integral into two parts

L=\left(\int_{0}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\cdots\right)dyd\tau=:L_{1}+L_{2}\hbox{.}

For $L_{1}$ , we further decompose the space domain into two parts:

L_{1}=\int_{0}^{\frac{t}{2}}\left(\int_{\left|y\right|\leq\frac{2}{3}\left|x\right|}+\int_{\left|y\right|>\frac{2}{3}\left|x\right|}\right)\left(\cdots\right)dyd\tau=:L_{11}+L_{12}\hbox{.}

If $\left|y\right|\leq\frac{2}{3}\left|x\right|$ , then

\left|x-y\right|\geq\frac{\left|x\right|}{3}\hbox{,}

and thus

	$\displaystyle L_{11}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

If $\left|y\right|\geq\frac{2}{3}\left|x\right|$ and $0\leq\tau\leq\frac{t}{2}$ , we have

\left|y\right|-\mathbf{c}\tau\geq\frac{2}{3}\left|x\right|-\frac{\mathbf{c}t}{2}=\frac{1}{6}\left|x\right|+\frac{1}{2}\left(\left|x\right|-\mathbf{c}t\right)\hbox{,}

so that

$\displaystyle L_{12}$	$\displaystyle\lesssim$	$\displaystyle\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\geq\frac{2}{3}\left\|x\right\|}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-2}\left(1+\tau+\left(\left\|x\right\|-\mathbf{c}t\right)^{2}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\left\|x\right\|-\mathbf{c}t\right)^{2}\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+\left(\left\|x\right\|-\mathbf{c}t\right)^{2}\right)^{-2}(1+t)^{-1/2}\lesssim\left(\frac{1+t}{2}+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{2}\right)^{-2}(1+t)^{-1/2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}$

since $\left|x\right|-\mathbf{c}t\geq\sqrt{1+t}$ for $\left(x,t\right)\in D_{3}$ . Therefore, we obtain

L_{1}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}

Next for $L_{2}$ , we decompose $\mathbb{R}^{3}$ into two parts

L_{2}=\int_{\frac{t}{2}}^{t}\left(\int_{\left|y\right|-\mathbf{c}\tau\leq\frac{\left|x\right|-ct}{2}}+\int_{\left|y\right|-\mathbf{c}\tau>\frac{\left|x\right|-ct}{2}}\right)\left(\cdots\right)dyd\tau=:L_{21}+L_{22}\hbox{.}

If $\left|y\right|-\mathbf{c}\tau\leq\frac{\left|x\right|-\mathbf{c}t}{2}$ , we have

\left|x-y\right|\geq\left|x\right|-\left(\frac{\left|x\right|-\mathbf{c}t}{2}\right)-\mathbf{c}\tau\geq\frac{\left|x\right|-\mathbf{c}t}{2}\hbox{,}

	$\displaystyle L_{21}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{2D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{,}$

and

	$\displaystyle L_{22}$	$\displaystyle\lesssim$	$\displaystyle\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}$

Therefore,

L\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(1+t\right)^{-5/2}\left(1+\frac{\left(\left|x\right|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-2}\hbox{.}

Case 4: $\left(x,t\right)\in D_{4}\cup D_{5}$ . Observe that

\frac{\sqrt{1+t}}{2\mathbf{c}}\leq\frac{\left|x\right|}{2\mathbf{c}}\leq\frac{t}{2}\hbox{.}

We split the integral into three parts:

L=\left(\int_{0}^{\frac{\left|x\right|}{2\mathbf{c}}}+\int_{\frac{\left|x\right|}{2\mathbf{c}}}^{\frac{t}{2}}+\int_{\frac{t}{2}}^{t}\right)\int_{\mathbb{R}^{3}}\left(\ldots\right)dyd\tau=:L_{1}+L_{2}+L_{3}\hbox{.}

For $L_{1}$ , we decompose $\mathbb{R}^{3}$ into two parts $\{\left|y\right|\leq\frac{3}{4}\left|x\right|\}$ and $\{\left|y\right|>\frac{3}{4}\left|x\right|\}$ . If $\left|y\right|\geq\frac{3}{4}\left|x\right|$ and $0\leq\tau\leq\frac{\left|x\right|}{2\mathbf{c}}$ , then

\left|y\right|-\mathbf{c}\tau\geq\frac{1}{4}\left|x\right|\hbox{.}

Thus,

$\displaystyle L_{1}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|\leq\frac{3\left\|x\right\|}{4}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
		$\displaystyle+\left(1+\left\|x\right\|^{2}\right)^{-2}\int_{0}^{\frac{t}{2}}\int_{\left\|y\right\|>\frac{3\left\|x\right\|}{4}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\frac{3}{2}}+\left(\frac{\left\|x\right\|^{2}}{2}+\frac{1+t}{2}\right)^{-2}(1+t)^{-1/2}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|}{1+t}\right)^{-\frac{3}{2}}\hbox{.}$

As for $L_{2}$ and $L_{3}$ , it immediately follows that

$\displaystyle L_{2}$	$\displaystyle\lesssim$	$\displaystyle\int_{\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{t}{2}}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}\left(1+\tau\right)^{-4}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-2}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-2}\int_{\frac{\left\|x\right\|}{2\mathbf{c}}}^{\frac{t}{2}}\left(1+\tau\right)^{-2}\int_{\mathbb{R}^{3}}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}dyd\tau$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|}{2\mathbf{c}}\right)^{-3}\left(1+t\right)^{\frac{3}{2}}$
	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-5/2}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-2}\hbox{,}$

and

	$\displaystyle L_{3}$	$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-4}\int_{\frac{t}{2}}^{t}\int_{\mathbb{R}^{3}}\left(1+t-\tau\right)^{-2}e^{-\frac{\left\|x-y\right\|^{2}}{D_{0}\left(1+t-\tau\right)}}dyd\tau$
		$\displaystyle\lesssim$	$\displaystyle\left(1+t\right)^{-7/2}\lesssim\left(1+\left\|x\right\|^{2}\right)^{-\frac{3}{2}}(1+t)^{-1/2}\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left\|x\right\|}{1+t}\right)^{-\frac{3}{2}}\hbox{,}$

since $\sqrt{1+t}\leq\left|x\right|\leq\mathbf{c}t$ . Therefore, we have

L\lesssim\left(1+t\right)^{-2}\left(1+\frac{\left|x\right|}{1+t}\right)^{-\frac{3}{2}}\hbox{.}

∎

Acknowledgments:

This work is partially supported by the National Key R&D Program of China under grant 2022YFA1007300. Y.-C. Lin is supported by the National Science and Technology Council under the grant NSTC 112-2115-M-006-006-MY2. H.-T. Wang is supported by NSFC under Grant No. 12371220 and 12031013, the Strategic Priority Research Program of Chinese Academy of Sciences under Grant No. XDA25010403. K.-C. Wu is supported by the National Science and Technology Council under the grant NSTC 112-2628-M-006-006 -MY4 and National Center for Theoretical Sciences.

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$\displaystyle\left\|\left\langle\xi\right\rangle^{\beta}g\left(t,x,\xi\right)\right\|$	$\displaystyle=$	$\displaystyle\left\|\int_{0}^{t}e^{-\nu\left(\xi\right)\left(t-\tau\right)}\left\langle\xi\right\rangle^{\beta}\left[\mathcal{K}_{s}h\left(\tau,x-\xi\left(t-\tau\right),\xi\right)+U\left(\tau,x-\xi\left(t-\tau\right),\xi\right)\right]d\tau\right\|$
	$\displaystyle\leq$	$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(\nu\left(\xi\right)\right)^{-1}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\langle\xi\right\rangle^{\beta}\left[\left\|\mathcal{K}_{s}h\left(\tau,y,\xi\right)\right\|\right]d\tau$
		$\displaystyle+\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(\nu\left(\xi\right)\right)^{-1}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\langle\xi\right\rangle^{\beta}\left\|U\left(\tau,y,\xi\right)\right\|d\tau$
	$\displaystyle\leq$	$\displaystyle\eta\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\cdot\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\|h\left(\tau,y,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}}d\tau$
		$\displaystyle+\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\cdot\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left\|\nu^{-1}(\xi)U\left(\tau,y,\xi\right)\right\|_{L_{\xi,\beta}^{\infty}}d\tau$
	$\displaystyle\leq$	$\displaystyle 2(A+\eta B)e^{-\frac{t+\left\|x\right\|}{c}}\text{.}$

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left\|y\right\|^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle\left(C_{\rho}+4^{\rho}\right)\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}\left(C_{\rho}+4^{\rho}\right)\left(1+t\right)^{-\alpha}\left(1+\frac{\left\|x\right\|^{2}}{1+t}\right)^{-\rho}\hbox{.}$

	$\displaystyle\left\|x-y\right\|$	$\displaystyle\geq$	$\displaystyle\left\|\left\|x\right\|-\left\|y\right\|\right\|=\left\|\left\|x\right\|-\mathbf{c}\tau+\mathbf{c}\tau-\left\|y\right\|\right\|\geq\left\|\left\|x\right\|-\mathbf{c}\tau\right\|-\left\|\left\|y\right\|-\mathbf{c}\tau\right\|$
		$\displaystyle\geq$	$\displaystyle\left\|x\right\|-\mathbf{c}t-\frac{\left\|x\right\|-\mathbf{c}t}{2}=\frac{\left\|x\right\|-\mathbf{c}t}{2}\hbox{,}$

			$\displaystyle\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{\left(1+\tau\right)}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle\left(C_{\rho}+4^{\rho}\right)\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}\left(C_{\rho}+4^{\rho}\right)\left(1+t\right)^{-\alpha}\left(1+\frac{\left(\left\|x\right\|-\mathbf{c}t\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}$

			$\displaystyle\int_{0}^{\frac{t}{2}+\frac{\left\|x\right\|}{2\mathbf{c}}}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\sup_{y\in\mathbb{R}^{3}}e^{-\frac{\nu_{0}}{2}\left(t-\tau+\left\|x-y\right\|\right)}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\left\|y\right\|-\mathbf{c}\tau\right)^{2}}{1+\tau}\right)^{-\rho}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\rho}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\int_{0}^{t}e^{-\frac{\nu\left(\xi\right)}{2}\left(t-\tau\right)}\nu\left(\xi\right)\left(1+\tau\right)^{-\alpha}d\tau$
		$\displaystyle\leq$	$\displaystyle C_{\alpha}C_{\rho}\left(1+\tau\right)^{-\alpha}\left(1+\frac{\left(\mathbf{c}t-\left\|x\right\|\right)^{2}}{1+t}\right)^{-\rho}\hbox{.}$