Gradient Decay in the Boltzmann Theory of Non-isothermal Boundary

Abstract

We consider the Boltzmann equation in a convex domain with a non-isothermal boundary of diffuse reflection. For both unsteady/steady problems, we construct solutions belonging to $W^{1,p}_x$ for any $p<3$. We prove that the unsteady solution converges to the steady solution in the same Sobolev space exponentially quickly as $t \rightarrow \infty$.

1 Introduction

One of the central questions in kinetic theory has been finding the effective control of derivatives of solutions to the Boltzmann equation

$$\begin{aligned} \partial _t F + v\cdot \nabla _x F = Q(F,F) \ \ \text {in} \ \ {{\mathbb {R}}}_+ \times \Omega \times {{\mathbb {R}}}^3 \end{aligned}$$

(1.1)

in the presence of a physical boundary $\partial \Omega$. Here we consider the hard sphere collision operator, where Q(F, F) takes the form

$$\begin{aligned} \begin{aligned} Q(F_1,F_2)&:= Q_{\text {gain}}(F_1,F_2)-Q_{\text {loss}}(F_1,F_2) \\&=\int _{{{\mathbb {R}}}^3}\int _{{{\mathbb {S}}}^2}|(v-u) \cdot \omega | \big [F_1(u')F_2(v')-F_1(u)F_2(v) \big ]\textrm{d}\omega \textrm{d}u, \end{aligned}\nonumber \\ \end{aligned}$$

(1.2)

with $u'=u+[(v-u)\cdot \omega ]\omega$, $v'=v-[(v-u)\cdot \omega ]\omega$ with $\omega \in {{\mathbb {S}}}^2$.

A particular motivation for this question could be justifying one of the major a priori assumptions of the celebrated theory of Devillettes–Villani [6], namely, that of uniform in time and space smoothness of Boltzmann solutions. While there has been much progress on the regularity estimate, such a desired uniform estimate is far less satisfactory. In this paper, we focus on the diffuse reflection boundary condition

$$\begin{aligned} F(\cdot , x,v)|_{n(x)\cdot v<0} = M_W(x,v)\int _{n(x)\cdot u>0} F(\cdot ,x,u) \{n(x)\cdot u\} \textrm{d}u ,\ \ x\in \partial \Omega , \nonumber \\ \end{aligned}$$

(1.3)

where the outward normal at the boundary $\partial \Omega$ is denoted by n(x). This condition implies an instantaneous equilibration of bounce-back particles to a thermodynamic equilibrium of the wall temperature $T_W(x)$ at the boundary point $x\in \partial \Omega$:

$$\begin{aligned} M_W(x,v) = M_{1, 0, T_W (x)}(v) :=\frac{1}{2\pi [T_W(x)]^2}e^{-\frac{|v|^2}{2T_W(x)}}. \end{aligned}$$

(1.4)

When the wall temperature is non-constant (non-isothermal boundary), any existing steady solutions of the Boltzmann equation with the diffuse reflection boundary condition (1.5)–(1.6) do not have a local Maxwellian, and hence do not have a (local) minimizer of entropy:

$$\begin{aligned} v\cdot \nabla _x F_s&= Q(F_s,F_s), \end{aligned}$$

(1.5)

$$\begin{aligned} F_s(x,v)|_{n(x)\cdot v<0}&= M_W(x,v) \int _{n(x) \cdot u>0} F_s(x,u)\{n(x)\cdot u\} \textrm{d}u, \ \ x\in \partial \Omega . \end{aligned}$$

(1.6)

Esposito–Guo–Kim–Marra established the construction of steady solutions and their dynamical stability in $L^\infty$ when the data are small in [7]. On the other hand, the diffuse reflection boundary condition solely stabilizes the dynamics even without the intermolecular collision [13, 14].

So far, any available results regarding regularity of the Boltzmann equation with the diffuse reflection boundary condition stop before a $H^1$ threshold, and hence the following question is open:

$$\begin{aligned} \nabla _x F(t,x,v) \in L^2 (\Omega \times {{\mathbb {R}}}^3) . \end{aligned}$$

(1.7)

In [12], the second author and collaborators prove that dynamic solutions belong to $W^{1,p}_x$ for $p<2$, while such control grows exponentially in time. In [3, 12], they use a kinetic distance function and prove a weighted estimate beyond $H^1_x$. All these studies left the $H^1$ threshold problem unsolved (also see [5]).

In this paper, we pass over this $H^1$ threshold for the first time, which would mark a resolution of a major open problem in a boundary regularity theory of kinetic theorem. Moreover, we prove that both steady/unsteady solutions belong to $W^{1,p}_x$ for $p<3$. Furthermore, we prove that such $W^{1,p}_x$-control of unsteady solutions decays exponentially in time. Therefore the result would make a surprising breakthrough beyond a standing open problem

Theorem 1

(Informal statement of the Main Theorem) Assume that the domain is strictly convex (see (2.2)) and the boundary is $C^3$. Suppose $\sup _{x \in \partial \Omega }|T_W(x)-T_0|\ll 1$ for some constant $T_0>0$ and $T_W(x) \in C^1(\partial \Omega )$. Then any steady solution $F_s (x,v)$ of a finite total mass satisfies

$$\begin{aligned} \frac{1}{\sqrt{M_{1,0,T_0} (v)}} |\nabla _x F_s(x,v) | \in L^p (\Omega \times {{\mathbb {R}}}^3) \ \ \text {for all } \ p<3 . \end{aligned}$$

(1.8)

Here, a global Maxwellian of constant temperature $T_0$ is denoted by $M_{1,0,T_0} = \frac{1}{2\pi [T_0]^2}e^{-\frac{|v|^2}{2T_0}}$.

Suppose that an initial datum

$$\begin{aligned} F_0 (x,v) = F_s (x,v) + \sqrt{M_{1, 0, T_0} (v) } f_0(x,v) \end{aligned}$$

has the same total mass as the steady solution $F_s (x,v)$, and $\nabla _x f_0(x,v) \rightarrow 0$ pointwisely, fast enough as $|v| \rightarrow \infty$. Finally, we assume a smallness condition $\Vert e^{\theta |v|^2} f_0 \Vert _\infty \ll 1$ for some $\theta >0$. Then, for any $p<3$,

$$\begin{aligned} \nabla _x F(t,x,v)\rightarrow \nabla _x F_s(x,v) \ \text {in} \ L^p (\Omega \times {{\mathbb {R}}}^3) \ \ \text {exponentially fast as} \ t\rightarrow \infty . \end{aligned}$$

(1.9)

Our result, apart from its proof, has two implications. First, it shows that the generic solutions of the steady/unsteady Boltzmann equations are regular globally as $W^{1,3-}$. Second, it shows that the gradient of solutions are asymptotically stable, which demonstrates the versatility of the method of [7] and [12].

Notations. Throughout this paper we will use the following notations:

$$\begin{aligned}&f\lesssim g \Leftrightarrow \text { there exists } 0<C<\infty \text { such that } 0 \le f\le Cg; \end{aligned}$$

(1.10)

$$\begin{aligned}&f\thicksim g \Leftrightarrow \text { there exists } 0<C<\infty \text { such that } 0 \le \frac{f}{C}\le g\le Cf; \end{aligned}$$

(1.11)

$$\begin{aligned}&f \ll g\Leftrightarrow \text { there exists a small constant } c>0 \text { such that } 0 \le f \le cg; \end{aligned}$$

(1.12)

$$\begin{aligned}&\Vert f\Vert _\infty = \Vert f\Vert _{L^\infty (\Omega \times {{\mathbb {R}}}^3)}, \ |f|_\infty = \Vert f\Vert _{L^\infty (\partial \Omega \times {{\mathbb {R}}}^3)}, \ \Vert f\Vert _p = \Vert f\Vert _{L^p(\Omega \times {{\mathbb {R}}}^3)}. \end{aligned}$$

(1.13)

Moreover, $f = o(1) \Leftrightarrow |f|\ll 1$ and $\langle v\rangle = \sqrt{1+|v|^2}$.

2 Results

2.1 Basic Setting

In this paper we assume the domain is a bounded open subset of ${{\mathbb {R}}}^3$ defined as $\Omega = \{x \in {{\mathbb {R}}}^3: \xi (x) <0\}$ and the boundary is defined as $\partial \Omega =\{x\in {{\mathbb {R}}}^3: \xi (x)=0\}$ via a $C^3$ function $\xi : {{\mathbb {R}}}^3 \rightarrow {{\mathbb {R}}}$. Equivalently, we assume that, for all $q\in \partial \Omega$, there exists a $C^3$ function ${ \eta _q}$ and $0<\delta _1 \ll 1$ such that

$$\begin{aligned} \eta _q: B_+(0; \delta _1) \ni \textbf{x}_q := (\textbf{x}_{q,1},\textbf{x}_{q,2},\textbf{x}_{q,3}) \rightarrow { {{\mathbb {R}}}^3}, \end{aligned}$$

(2.1)

where the map is one-to-one and onto to the image ${\mathcal {O}}_q := \eta _q (B_+ (0;\delta _1))$ when $\delta _1$ is sufficiently small. Moreover, $\eta _q (\textbf{x}_q) \in \partial \Omega$ if and only if $\textbf{x}_{q,3}=0$ within the image of $\eta _q$. We refer to [8] for the construction of such $\xi$ and $\eta _q$. We further assume that the domain is strictly convex in the following sense:

$$\begin{aligned} \sum _{i,j=1}^3\zeta _i \zeta _j\partial _i\partial _j \xi (x) \gtrsim |\zeta |^2 \ \text { for all } x \in {\bar{\Omega }} \text { and } \zeta \in {{\mathbb {R}}}^3. \end{aligned}$$

(2.2)

Without loss of generality, we may assume that $\nabla \xi \ne 0$ near $\partial \Omega$. Then the outward normal can be defined as

$$\begin{aligned} n(x):=\frac{\nabla \xi (x)}{|\nabla \xi (x)|}. \end{aligned}$$

(2.3)

The boundary of the phase space is

$$\begin{aligned} \gamma :=\{(x,v)\in \partial \Omega \times {{\mathbb {R}}}^3\}. \end{aligned}$$

We decompose $\gamma$ as

$$\begin{aligned} \gamma _+&:= \{(x,v)\in \partial \Omega \times {{\mathbb {R}}}^3: n(x)\cdot v>0\}, \text { (the outgoing set) }\\ \gamma _-&:= \{(x,v)\in \partial \Omega \times {{\mathbb {R}}}^3: n(x)\cdot v<0\}, \text { (the incoming set)} \\ \gamma _0&:= \{(x,v)\in \partial \Omega \times {{\mathbb {R}}}^3: n(x)\cdot v=0\}. \text { (the grazing set)} \end{aligned}$$

2.2 Main Result

In this paper, we investigate the regularity of the dynamical problem (1.1). We also analyze the rate at which the gradient of the solution to the dynamical problem (1.1) approaches the gradient of the solution to the steady problem (1.5). Without loss of generality, we assume the wall temperature has a small fluctuation around 1($T_0=1$ in Theorem 1). We denote $\mu$ as the global Maxwellian:

$$\begin{aligned} \mu := \frac{1}{2\pi }e^{-\frac{|v|^2}{2}} = \sqrt{M_{1,0,1}(v)}. \end{aligned}$$

(2.4)

Then we consider the steady problem (1.5) as a perturbation around the Maxwellian

$$\begin{aligned} F_s(x,v)=\mu (v)+ \sqrt{\mu (v)} f_s(x,v) , \end{aligned}$$

(2.5)

and consider the dynamical problem (1.1) as a perturbation around the steady problem (1.5)

$$\begin{aligned} F(t,x,v) = F_s(x,v) + \sqrt{\mu (v)} f(t,x,v). \end{aligned}$$

(2.6)

The equation for f in (2.6) reads as

$$\begin{aligned} \partial _t f +v\cdot \nabla _x f +\nu (v) f = K(f)+\Gamma (f_s,f)+\Gamma (f,f_s)+\Gamma (f,f), \end{aligned}$$

(2.7)

where we denote the linear Boltzmann operator as

$$\begin{aligned} Lf=\nu (v) f - K(f)=-\frac{Q(\mu ,\sqrt{\mu }f)}{\sqrt{\mu }}-\frac{Q(\sqrt{\mu }f,\mu )}{\sqrt{\mu }} \end{aligned}$$

(2.8)

and the nonlinear Boltzmann operator as

$$\begin{aligned} \Gamma (f,g) = Q(\sqrt{\mu }f,\sqrt{\mu }g)/\sqrt{\mu }. \end{aligned}$$

(2.9)

The properties of these Boltzmann operators can be found in Lemma 3. Here $f_s$ is defined in (2.5).

The initial condition is given as

$$\begin{aligned} f_0(x,v)=f(0,x,v)=(F(0,x,v)-F_s(x,v))/\sqrt{\mu }. \end{aligned}$$

(2.10)

On the boundary $\gamma _-$, F and $F_s$ are given by (1.3) and (1.6), hence the boundary condition of f is given by

$$\begin{aligned} f(t,x,v)|_{n(x)\cdot v<0} = \frac{M_W(x,v)}{\sqrt{\mu (v)}}\int _{n(x)\cdot u>0} f(t,x,u)\sqrt{\mu (u)}\{n(x)\cdot u\} \textrm{d}u. \nonumber \\ \end{aligned}$$

(2.11)

From the equation of f in (2.7), we define the initial condition for $\partial _t f(0,x,v)$ as

$$\begin{aligned} \partial _t f(0,x,v)&: = -v\cdot \nabla _x f(0,x,v)-L(f(0))+\Gamma (f_s,f(0))\nonumber \\&\quad +\Gamma (f(0),f_s)+\Gamma (f(0),f(0)). \end{aligned}$$

(2.12)

It is well-known that the Boltzmann equation possesses a generic singularity at the boundary. We adopt the following weight of [12]:

Definition 1

We define a kinetic distance

$$\begin{aligned} \begin{aligned} \alpha (x,v)&: = \chi ( {\tilde{\alpha }}(x,v) ) , \ \ {\tilde{\alpha }}(x,v) : = \sqrt{ |v \cdot \nabla _x \xi (x)|^2 - 2 \xi (x) (v \cdot \nabla _x^2 \xi (x) \cdot v)}, \\&\quad \ (x,v) \in {\bar{\Omega }} \times {{\mathbb {R}}}^3, \end{aligned} \nonumber \\ \end{aligned}$$

(2.13)

where $\chi : [0,\infty ) \rightarrow [0,\infty )$ stands for a non-decreasing smooth function such that

$$\begin{aligned}{} & {} \chi (s) = s \ \text {for} \ s \in [0, 1/2 ], \ \chi (s) = 1 \ \text {for} \ s \in [ 2, \infty ), \ and \ \nonumber \\ {}{} & {} | \chi ^\prime (s) | \le 1 \ \text {for} \ s \in [0,\infty ). \end{aligned}$$

(2.14)

Remark 1

We note that $\alpha \equiv 0$ on the grazing set $\gamma _0$. An important property of this kinetic weight is that $\alpha (x,v)$ is almost invariant along the characteristic, see Lemma 4.

Convexity is a necessary condition for studying the regularity of the Boltzmann equation. It is well-known that singularity propagates in non-convex domains [15], and in such cases, we can only expect bounded variation regularity at best [11]. The convexity condition ensures the positivity of the kinetic weight in (2.13). This weight has been instrumental in the development of the study of the regularity of the Boltzmann equation, leading to progress in recent years [1, 2, 4].

In this paper, we first establish a weighted $C^1$ estimate for the dynamical problem that converges exponentially fast to the weighted $C^1$ estimate to the steady problem.

Theorem 2

Assume that the domain is convex (2.2), the boundary is $C^3$, $\sup _{x \in \partial \Omega }|T_W(x)-T_0|\ll 1$ for some constant $T_0>0$, the initial condition (2.10) is continuous away from $\gamma _0$ and satisfies that

$$\begin{aligned}{} & {} \iint _{\Omega \times {{\mathbb {R}}}^3}f(0,x,v)\sqrt{\mu (v)} \textrm{d}x \textrm{d}v =0, \end{aligned}$$

(2.15)

$$\begin{aligned}{} & {} \Vert wf(0)\Vert _\infty + |wf(0)|_\infty \ll 1, \ \ w(v):=e^{\theta |v|^2} \text { for some } 0<\theta <1/4,\nonumber \\ \end{aligned}$$

(2.16)

and that the compatibility condition is

$$\begin{aligned} f(0,x,v)|_{n(x)\cdot v<0} = \frac{M_W(x,v)}{\sqrt{\mu (v)}} \int _{n(x)\cdot u>0}f(0,x,u)\sqrt{\mu (u)}\{n(x)\cdot u\} \textrm{d}u\nonumber \\ \end{aligned}$$

(2.17)

Then there exists a unique solution f to the dynamical problem (2.6) and (2.11). Moreover, f is continuous away from $\gamma _0$, and, for some $\lambda \ll 1$, it satisfies

$$\begin{aligned} e^{\lambda t}\Big [\Vert wf(t)\Vert _{\infty } +|wf(t)|_\infty \Big ]\lesssim \Vert wf(0)\Vert _\infty + |wf(0)|_\infty . \end{aligned}$$

(2.18)

If we further assume that $T_W(x)\in C^1(\partial \Omega )$, the initial condition $\partial _t f(0,x,v)$ in (2.12) and $\alpha \nabla _x f(0,x,v)$ are continuous away from $\gamma _0$ and satisfy

$$\begin{aligned}{} & {} \Vert w_{{\tilde{\theta }}}(v)\alpha (x,v)\nabla _x f(0)\Vert _\infty <\infty , \ \ w_{{\tilde{\theta }}}(v) = e^{{\tilde{\theta }}|v|^2} \text { for some }{\tilde{\theta }} \ll 1, \end{aligned}$$

(2.19)

$$\begin{aligned}{} & {} \Vert w \partial _t f(0,x,v)\Vert _\infty <\infty \end{aligned}$$

(2.20)

and the compatibility condition

$$\begin{aligned} \partial _t f(0,x,v)|_{\gamma -} = \frac{M_W(x,v)}{\sqrt{\mu (v)}}\int _{n(x)\cdot u>0} \partial _t f(0,x,u)\sqrt{\mu (u)}\{n(x)\cdot u\} \textrm{d}u. \end{aligned}$$

(2.21)

Then, for all $t>0$ and $\lambda \ll 1$ in (2.18), $f(t,x,v)\in C^1({\bar{\Omega }}\times {{\mathbb {R}}}^3 \backslash \gamma _0)$ and satisfies

$$\begin{aligned} \begin{aligned}&e^{\lambda t}\Vert w_{{\tilde{\theta }}}(v)\alpha (x,v)\nabla _x f(t)\Vert _\infty \\&\lesssim \Vert w_{{\tilde{\theta }}}(v)\alpha (x,v) \nabla _x f(0)\Vert _\infty + \Vert w\partial _t f(0)\Vert _\infty \\&\qquad + \sup _t e^{\lambda t}\{\Vert wf(t)\Vert _\infty + |wf(t)|_\infty \} . \end{aligned} \end{aligned}$$

(2.22)

Remark 2

According to [12], the continuity assumption on $\alpha \nabla _x f(0)$ and the compatibility assumption (2.21) imply that f belongs to $C^1$ away from the grazing set. The additional assumption (2.20) is needed to control the contribution of the time derivative $\partial _t f$ in the characteristic formula (4.4)–(4.11).

Here, we can observe that the additional assumption (2.20) is not actually a very restrictive one. In fact, it is very similar to the assumption (2.19). To see this, we first note that the last four terms on the right-hand side of (2.12) can be bounded in $L^\infty$. Now, on the boundary $\partial \Omega$, let $\tau _1, \tau _2$, and n(x) be the basis vectors. We can decompose the first term as follows:

$$\begin{aligned} v\cdot \nabla _x f(0,x,v)&= (n(x)\cdot v) \partial _n f(0,x,v) + \sum _{i=1,2} (\tau _i \cdot v) \partial _{\tau _i} f(0,x,v). \end{aligned}$$

We can see that the assumption $(n(x)\cdot v) \partial _n f(0,x,v)$ is bounded in $L^\infty$ is almost the same as (2.19). This is because both assumptions cancel out the singularity, and we have $|n(x)\cdot v| \thicksim \alpha (x,v)$.

Remark 3

In this paper, we enhance the weighted $C^1$ estimate of the dynamical problem by making it uniform in time. Additionally, we prove that the weighted $C^1$ norm of the dynamical problem converges to the weighted $C^1$ norm of the steady problem with an exponential rate.

The weighted $C^1$ estimate (2.22) allows us to study the $W^{1,p}$ estimate without any weight. The following theorem is the formal version of Theorem 1, which proves a $W^{1,p}$ estimate for both the steady and dynamical problems

Theorem 3

Assume that the domain is convex (2.2), the boundary is $C^3$, $\sup _{x \in \partial \Omega }|T_W(x)-T_0|\ll 1$ for some constant $T_0>0$, $T_W(x)\in C^1(\partial \Omega )$. Then there is a unique steady solution to (2.5) and (1.6), and moreover, $f_s$ belongs to $W^{1,p}(\Omega \times {{\mathbb {R}}}^3)$ for $p<3$:

$$\begin{aligned} \Vert \nabla _x f_s\Vert _p \lesssim \Vert T_W-T_0\Vert _{C^1(\partial \Omega )}. \end{aligned}$$

(2.23)

Assuming that all conditions in Theorem 2 are satisfied so that we have the decay (2.22), the dynamical problem f(t) in (2.6) further belongs to $W^{1,p}(\Omega \times {{\mathbb {R}}}^3)$ for $p<3$ and

$$\begin{aligned} \begin{aligned}&e^{\lambda t}\Vert \nabla _x f(t)\Vert _p \lesssim e^{\lambda t}\Vert w_{{\tilde{\theta }}}(v)\alpha (x,v)\nabla _x f(t)\Vert _\infty <\infty . \end{aligned} \end{aligned}$$

(2.24)

Remark 4

It is worth mentioning that the $W^{1,p}$ estimate we establish does not involve any weight. As shown in (2.24), our result proves that the $W^{1,p}$ norm of the dynamical problem converges to the $W^{1,p}$ norm of the steady problem with exponential rate.

Remark 5

It is important to note that Theorem 3 only establishes an interior $W^{1,p}$ estimate without any weight for $p<3$. We don’t expect the boundary regularity result up to $W^{1,p}$ for $p<3$. Indeed, Kim and his collaborators construct a counterexample of such boundary regularity in Lemma 12 of [12], in which they prove $\int _0^1 \int _{\gamma _-}|\nabla _x f(s,x,v)|^2 \textrm{d}\gamma \textrm{d}s = +\infty$ for free transport equation.

However, the interior estimate fails for the second-order derivative. We refer detailed discussion to Appendix.

Outline. The structure of this paper is straightforward. In Sect. 3, we present several lemmas that serve as preliminary results. In Sect. 4, we establish the weighted $C^1$ estimate and prove Theorem 2. Section 5 is dedicated to applying Theorem 2 to obtain the $W^{1,p}$ estimate in Theorem 3.

3 Preliminaries

In this section, we will begin by introducing and reparametrizing the backward exit position as stochastic cycles. Next, we will present several lemmas that describe the properties of stochastic cycles, Boltzmann operators, and kinetic weight.

3.1 Stochastic Cycles and Reparametrization

We mainly define the stochastic cycles and the reparametrization, then deduce another representation of the boundary condition in (3.12). Denote $t_{\textbf{b}}(x,v),x_{\textbf{b}}(x,v)$ as the backward exit time and backward exit position:

$$\begin{aligned} t_{\textbf{b}}(x,v) := \sup \{ s>0:x-sv \in \Omega \},\quad {x_{\textbf{b}}(x,v):= x- t_{\textbf{b}}(x,v) v.} \end{aligned}$$

(3.1)

Definition 2

We define a stochastic cycle as $(x^0,v^0)= (x,v) \in {\bar{\Omega }} \times {{\mathbb {R}}}^3$, and, inductively,

$$\begin{aligned}&x^1:= x_{\textbf{b}}(x,v), \ v^1 \in \{v^1\in {{\mathbb {R}}}^3:n(x^1)\cdot v^1>0\} , \end{aligned}$$

(3.2)

$$\begin{aligned}&v^{k}\in \{v^{k}\in {{\mathbb {R}}}^3:n(x^k)\cdot v^k>0\}, \ \ \text {for} \ k \ge 1, \end{aligned}$$

(3.3)

$$\begin{aligned}&x^{1} := x_{\textbf{b}}(x^k, v^k) , \ t_{\textbf{b}}^{k}:= t_{\textbf{b}}(x^k,v^k) \ \ \text {for} \ n(x^k) \cdot v^k\ge 0 . \end{aligned}$$

(3.4)

Choose $t\ge 0.$ We define $t^0=t$ and

$$\begin{aligned} t^{k} = t- \{ t_{\textbf{b}}+ t_{\textbf{b}}^1 + \cdots + t_{\textbf{b}}^{k-1}\}, \ \ \text {for} \ k \ge 1. \end{aligned}$$

(3.5)

Remark 6

Here $x^{k+1}$ depends on $(x,v,x^1,v^1,\cdots , x^k,v^k)$, while $v^k$ is a free parameter whose domain (3.3) only depends on $x^k$.

Recall (2.1). Since the boundary is compact and $C^3$, for fixed $0<\delta _1 \ll 1$, we choose a finite number of $p \in \tilde{{\mathcal {P}}} \subset \partial \Omega$ and $0<\delta _2\ll 1$ such that ${\mathcal {O}}_p=\eta _p( B_+(0; \delta _1)) \subset B(p;\delta _2) \cap {\bar{\Omega }}$ and $\{{\mathcal {O}}_p \}$ forms a finite covering of $\partial \Omega$. We further choose an interior covering ${\mathcal {O}}_0 \subset \Omega$ such that $\{ {\mathcal {O}}_p\}_{p \in {\mathcal {P}}}$ with ${\mathcal {P}} = \tilde{{\mathcal {P}}}\cup \{0\}$ forms an open covering of $\bar{\Omega }$. We define a partition of unity

$$\begin{aligned} \textbf{1}_{\bar{\Omega }} (x)= \sum _{p \in {\mathcal {P}}} \iota _p(x) \text { such that }0 \le \iota _p(x) \le 1, \ \ \iota _p(x) \equiv 0 \ \text {for} \ x \notin {\mathcal {O}}_p. \end{aligned}$$

(3.6)

Without loss of generality (see [16]) we can always reparametrize $\eta _p$ such that $\partial _{\textbf{x}_{p,i}} \eta _p \ne 0$ for $i=1,2,3$ at $\textbf{x}_{p,3}=0$, and an orthogonality holds as

$$\begin{aligned} \partial _{\textbf{x}_{p,i}}\eta _p \cdot \partial _{\textbf{x}_{p,j}}\eta _p =0 \ \ \text {at} \ \ \textbf{x}_{p,3}=0 \text { for } i\ne j \text { and } i,j\in \{1,2,3\}. \end{aligned}$$

(3.7)

Futhermore at $\textbf{x}_{p,3}=0$, the $\textbf{x}_{p,3}$ derivative gives the outward normal

$$\begin{aligned} {n_p(\textbf{x}_p) = \frac{\partial _{\textbf{x}_{p,3}}\eta _p}{\langle \partial _{\textbf{x}_{p,3}}\eta _p,\partial _{\textbf{x}_{p,3}}\eta _p\rangle }.} \end{aligned}$$

(3.8)

For simplicity, we denote

$$\begin{aligned} \partial _i \eta _p(\textbf{x}_p): = \partial _{\textbf{x}_{p,i}} \eta _p. \end{aligned}$$

(3.9)

Definition 3

For $x \in {\bar{\Omega }}$, we choose $p \in {\mathcal {P}}$ as in (2.1). We define

$$\begin{aligned} T_{\textbf{x}_p}&=\left( \begin{array}{ccc} \frac{\partial _1 \eta _p(\textbf{x}_p)}{\sqrt{g_{p,11}(\textbf{x}_p) }} &{} \frac{\partial _2 \eta _p(\textbf{x}_p)}{\sqrt{g_{p,22}(\textbf{x}_p) }} &{} \frac{\partial _3 \eta _p(\textbf{x}_p)}{\sqrt{g_{p,33}(\textbf{x}_p) }} \\ \end{array} \right) ^t, \end{aligned}$$

(3.10)

with $g_{p,ij}(\textbf{x}_p) =\langle \partial _i \eta _p(\textbf{x}_p),\partial _j \eta _p(\textbf{x}_p)\rangle$ for $i,j\in \{1,2,3\}$. Here $A^t$ stands as the transpose of a matrix A. Note that when $\textbf{x}_{p,3}=0$, $T_{\textbf{x}_p} \frac{\partial _i \eta _p(\textbf{x}_p)}{\sqrt{g_{p,ii}(\textbf{x}_p) }} = e_i$ for $i=1,2,3$ where $\{e_i\}$ is a standard basis of ${{\mathbb {R}}}^3$.

We define

$$\begin{aligned} \textbf{v}_j(\textbf{x}_p) = \frac{\partial _j \eta _p(\textbf{x}_p)}{\sqrt{g_{p,jj}(\textbf{x}_p) }} \cdot v. \end{aligned}$$

(3.11)

We note that from (3.7), the map $T_{\textbf{x}_p}$ is an orthonormal matrix when $\textbf{x}_{p,3}=0$. Therefore both maps $v \rightarrow \textbf{v} (\textbf{x}_p )$ and $\textbf{v} (\textbf{x}_p ) \rightarrow v$ have a unit Jacobian at $\textbf{x}_{p,3}=0$. This fact induces a new representation of boundary integration of diffuse boundary condition in (2.11): For $x \in \partial \Omega$ and $p \in {\mathcal {P}}$ as in (2.1),

$$\begin{aligned} \begin{aligned}&\int _{n(x)\cdot v>0}f(t,x,v)\sqrt{\mu (v)}\{n(x)\cdot v\}\textrm{d}v \\&\quad = \int _{\textbf{v} _{p ,3}>0}f(t, \eta _{p } (\textbf{x}_{p } ), T^t_{\textbf{x} _{p }} \textbf{v} ( \textbf{x}_p) )\sqrt{\mu (\textbf{v} (\textbf{x}_p))}\textbf{v} _{ 3}(\textbf{x}_p) \textrm{d}\textbf{v} (\textbf{x}_p). \end{aligned} \end{aligned}$$

(3.12)

We have used the fact of $\mu (v )=\mu (|v |)=\mu (|T^t_{\textbf{x}_{p } }\textbf{v} (\textbf{x}_p) |)=\mu (|\textbf{v} (\textbf{x}_p) |)=\mu (\textbf{v} (\textbf{x}_p) )$ and $\textbf{x}_{p,3}=0$.

Now we reparametrize the stochastic cycle using the local chart defined in Definition 2.

Definition 4

Recall the stochastic cycles (3.3). For each cycle $x^k$ let us choose $p^k \in {\mathcal {P}}$ in (2.1). Then we denote

$$\begin{aligned} \begin{aligned} \textbf{x}^k_{p^k}&:= (\textbf{x}^k_{p^k,1}, \textbf{x}^k_{p^k,2},0) \text { such that } \eta _{p^k} (\textbf{x}^k_{p^k}) = x^k, \ \ \text {for} \ k \ge 1, \\ \textbf{v}^k_{p^k,i}&:= \frac{\partial _i \eta _{p^k}(\textbf{x}_{p^k}^k)}{\sqrt{g_{p^k,ii}(\textbf{x}_{p^k}^k) }} \cdot v^k , \ \ \text {for} \ k \ge 1. \end{aligned} \end{aligned}$$

(3.13)

From (3.8), the outward normal at $x^k$ can also be denoted as

$$\begin{aligned} {n(x^k)} { = n_{p^k}(\textbf{x}_{p^k}^k). } \end{aligned}$$

(3.14)

Finally, we define

$$\begin{aligned} \partial _{\textbf{x}^{k}_{p^{k},i}}[ a( \eta _{p^{k}} ( \textbf{x}_{p^{k} }^{k} ), {v}^{k} ) ] := \frac{\partial \eta _{p^{k}}(\textbf{x}^{k}_{p^{k},i})}{\partial \textbf{x}^{k}_{p^{k},i}} \cdot \nabla _x a ( \eta _{p^{k}} ( \textbf{x}_{p^{k} }^{k} ), v^{k}) , \ \ i=1,2. \nonumber \\ \end{aligned}$$

(3.15)

3.2 Properties of Stochastic Cycles, Boltzmann Operators, and Kinetic Weight

The derivative of $x_{\textbf{b}}(x,v),t_{\textbf{b}}(x,v)$ is given by the following lemma:

Lemma 1

(Lemma 2.1 in [3]) The derivative of $t_{\textbf{b}}$ to $x_j$ and $v_j$ reads as

$$\begin{aligned}{} & {} \frac{\partial t_{\textbf{b}}^{1}}{\partial x_j^{1}} = \frac{1}{ \textbf{v}^{2}_{p^{2},3} } \frac{ \partial _{3} \eta _{{p}^{2}}(\textbf{x}_{p^2}^2)}{\sqrt{g_{{p}^{2},33}(\textbf{x}_{p^2}^2)}} \cdot e_j, \end{aligned}$$

(3.16)

$$\begin{aligned}{} & {} \frac{\partial t_{\textbf{b}}^{1}}{\partial v_j^{1}}=-\frac{t_{\textbf{b}}^{1}e_j}{\textbf{v}^{2}_{p^{2},3}}\cdot \frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}}. \end{aligned}$$

(3.17)

This leads to

$$\begin{aligned} \begin{aligned} \nabla _x t_{\textbf{b}}&= \frac{n(x_{\textbf{b}})}{n(x_{\textbf{b}}) \cdot v},\ \ \nabla _v t_{\textbf{b}}= - \frac{t_{\textbf{b}}n(x_{\textbf{b}})}{n(x_{\textbf{b}}) \cdot v},\\ \nabla _x x_{\textbf{b}}&= Id_{3\times 3} - \frac{n(x_{\textbf{b}}) \otimes v}{n(x_{\textbf{b}}) \cdot v},\ \ \nabla _v x_{\textbf{b}}= - t_{\textbf{b}}Id + \frac{ t_{\textbf{b}}n(x_{\textbf{b}}) \otimes v}{n(x_{\textbf{b}}) \cdot v}. \end{aligned} \end{aligned}$$

(3.18)

For $i=1,2$, $j=1,2$, $k=0,1$(recall $x^0=x$ in Definition 2),

$$\begin{aligned} \frac{\partial \textbf{x}_{p^{k+1},i}^{k+1}}{\partial x^{k}_j}= & {} \frac{1}{\sqrt{g_{p^{k+1}, ii} (\textbf{x}^{k+1}_{p^{k+1} } )}}\nonumber \\{} & {} \left[ \frac{\partial _{i} \eta _{p^{k+1}} (\textbf{x}^{k+1}_{p^{k+1} } ) }{\sqrt{g_{p^{k+1},ii}(\textbf{x}^{k+1}_{p^{k+1} } ) }} - \frac{\textbf{v}_{p^{k+1}, i}^{k+1}}{\textbf{v}_{p^{k+1}, 3}^{k+1}} \frac{\partial _{3} \eta _{p^{k+1}} (\textbf{x}^{k+1}_{p^{k+1} } ) }{\sqrt{g_{p^{k+1},33}(\textbf{x}^{k+1}_{p^{k+1} } )}} \right] \cdot e_j, \end{aligned}$$

(3.19)

$$\begin{aligned} \frac{\partial \textbf{x}^{2}_{p^{2},i}}{\partial {\textbf{x}^{1 }_{p^{1 },j}}}= & {} \frac{1}{\sqrt{g_{p^{2}, ii} (\textbf{x}^{2}_{p^{2} } )}} \left[ \frac{\partial _{i} \eta _{p^{2}} (\textbf{x}^{2}_{p^{2} } ) }{\sqrt{g_{p^{2},ii}(\textbf{x}^{2}_{p^{2} } ) }} - \frac{\textbf{v}^{2}_{p^{2}, i}}{\textbf{v}^{2}_{p^{2}, 3}} \frac{\partial _{3} \eta _{p^{2}} (\textbf{x}^{2}_{p^{2} } ) }{\sqrt{g_{p^{2},33}(\textbf{x}^{2}_{p^{2} } )}} \right] \cdot \partial _j \eta _{p^{1}}(\textbf{x}^{1}_{p^{1} } ) , \nonumber \\ \end{aligned}$$

(3.20)

$$\begin{aligned} \frac{\partial \textbf{x}^{2}_{p^{2},i}}{\partial v_j^{1}}= & {} -t_{\textbf{b}}^{1}e_j \cdot \frac{1}{\sqrt{g_{p^{2},ii}(\textbf{x}_{p^{2}}^{2}))}} \Big [\frac{\partial _i \eta _{p^{2}}(\textbf{x}^{2}_{p^{2} } )}{\sqrt{g_{p^{2},ii}(\textbf{x}^{2}_{p^{2} } )}}- \frac{\textbf{v}_{p^{1},i}^{1}}{\textbf{v}_{p^{2},3}^{2}}\frac{\partial _3 \eta _{p^{2}}(\textbf{x}^{2}_{p^{2} } )}{\sqrt{g_{p^{2},33}(\textbf{x}^{2}_{p^{2} } )}} \Big ].\nonumber \\ \end{aligned}$$

(3.21)

The properties of the collision operator $\nu (v)$, Kf and $\Gamma (f,f)$ in (2.7) are summarized in Lemmas 3 and 2.

Lemma 2

The linear Boltzmann operator K(f) in (2.7) is given by

$$\begin{aligned} Kf(x,v)=\int _{{{\mathbb {R}}}^3}\textbf{k}(v,u)f(x,u)\textrm{d}u. \end{aligned}$$

(3.22)

For some $\varrho >0$, the kernel $\textbf{k}(v,u)$ satisfies

$$\begin{aligned} |\textbf{k} (v,u)|&\lesssim \textbf{k}_\varrho (v,u), \ | \nabla _ u\textbf{k} (v,u)| \lesssim \langle u\rangle \textbf{k}_\varrho (v,u)/|v-u|, \nonumber \\ \textbf{k}_\varrho (v,u)&:= e^{- \varrho |v-u|^2}/ |v-u|. \end{aligned}$$

(3.23)

If $0<\frac{{\tilde{\theta }}}{2}<\varrho$, if $0<{\tilde{\varrho }}< \varrho - \frac{{\tilde{\theta }}}{2}$,

$$\begin{aligned} \textbf{k}(v,u) \frac{e^{{\tilde{\theta }} |v|^2}}{e^{{\tilde{\theta }} |u|^2}} \lesssim \textbf{k}_{{\tilde{\varrho }}}(v,u) , \end{aligned}$$

(3.24)

where $\textbf{k}$ and $\textbf{k}_{{\tilde{\varrho }}}$ are defined in (3.23).

Moreover, the derivative on $\textbf{k}(v,u)$ shares a similar property; for some ${\tilde{\varrho }}<\varrho -\frac{{\tilde{\theta }}}{2}$,

$$\begin{aligned} | \nabla _v \textbf{k}(v,u)| \frac{e^{{\tilde{\theta }}| v|^2}}{e^{{\tilde{\theta }}| u|^2}} \lesssim \frac{[1+|v|^2]\textbf{k}_{{\tilde{\varrho }}}(v,u)}{| v-u|} . \end{aligned}$$

(3.25)

Proof of (3.24)

From the standard Grad estimate in [9], the $\textbf{k}$ in (3.23) equals $\textbf{k}_1(v,u)+\textbf{k}_2(v,u)$, where

$$\begin{aligned} \textbf{k}_1(v,u)&= C_{\textbf{k}_1}|u-v|e^{-\frac{|v|^2+|u|^2}{4}}, \end{aligned}$$

(3.26)

$$\begin{aligned} \textbf{k}_2(v,u)&= C_{\textbf{k}_2}\frac{1}{|u-v|}e^{-\frac{1}{8}|u-v|^{2}-\frac{1}{8} \frac{(|u|^{2}-|v|^{2})^{2}}{|u-v|^{2}}}. \end{aligned}$$

(3.27)

Then, for some $\frac{1}{8}\ge \varrho >0$,

$$\begin{aligned} \begin{aligned} \textbf{k}(v,u) \frac{e^{{\tilde{\theta }} |v|^2}}{e^{{\tilde{\theta }} |u|^2}} \lesssim \frac{1}{|v-u| } \exp \left\{ - {\varrho } |v-u|^{2} - {\varrho } \frac{ ||v|^2-|u|^2 |^2}{|v-u|^2} + {\tilde{\theta }} |v|^2 - {\tilde{\theta }} |u|^2 \right\} . \end{aligned} \end{aligned}$$

Let $v-u=\eta$ and $u=v-\eta$, the exponent equals

$$\begin{aligned}{} & {} - \varrho |\eta |^{2}-\varrho \frac{||\eta |^{2}-2v\cdot \eta |^{2}}{ |\eta |^{2}}-{\tilde{\theta }} \{|v-\eta |^{2}-|v|^{2}\} \\{} & {} \quad =-2 \varrho |\eta |^{2}+ 4 \varrho v\cdot \eta - 4 \varrho \frac{|v\cdot \eta |^{2}}{|\eta |^{2}}-{\tilde{\theta }} \{|\eta |^{2}-2v\cdot \eta \} \\{} & {} \quad =(-2 \varrho -{\tilde{\theta }} )|\eta |^{2}+(4 \varrho +2{\tilde{\theta }} )v\cdot \eta - 4 \varrho \frac{\{v\cdot \eta \}^{2}}{|\eta |^{2}}. \end{aligned}$$

If $0<{\tilde{\theta }} <2 \varrho$ then the discriminant of the above quadratic form of $|\eta |$ and $\frac{v\cdot \eta }{|\eta |}$ is

$$\begin{aligned} (4 \varrho +2{\tilde{\theta }} )^{2}-4 (-2 \varrho -{\tilde{\theta }} )(- 4 \varrho ) =4{\tilde{\theta }} ^{2}- 16 \varrho ^2<0. \end{aligned}$$

Hence, the quadratic form is negative definite. We thus have, for $0<{\tilde{\varrho }}< \varrho - \frac{{\tilde{\theta }}}{2}$, the following perturbed quadratic form is still negative definite: $-(\varrho - {\tilde{\varrho }})|\eta |^{2}-(\varrho - {\tilde{\varrho }})\frac{||\eta |^{2}-2v\cdot \eta |^{2}}{|\eta |^{2}}-{\tilde{\theta }} \{|\eta |^{2}-2v\cdot \eta \} \le 0,$ i.e,

$$\begin{aligned} - {\varrho } |v-u|^{2} - {\varrho } \frac{ ||v|^2-|u|^2 |^2}{|v-u|^2} + {\tilde{\theta }} |v|^2 - {\tilde{\theta }} |u|^2 \le -{\tilde{\rho }}|v-u|^2, \end{aligned}$$

we conclude (3.24). $\square$

Proof of (3.25)

Taking the derivative to (3.26) and (3.27), we have

$$\begin{aligned} \nabla _v \textbf{k}_1(v,u)&= \frac{v-u}{| v-u|}\textbf{k}_1(v,u) - v \textbf{k}_1(v,u),\\ \nabla _v \textbf{k}_2(v,u)&= \frac{v-u}{| v-u|^2} \textbf{k}_2(v,u) \\&\quad - \textbf{k}_2(v,u)\Big [\frac{v-u}{4} + \frac{v(| u|^2-| v|^2)| v-u|^2 - (| u|^2-| v|^2)^2 (v-u) }{4| v-u|^4} \Big ] . \end{aligned}$$

Since $|u|^2 - |v|^2 = |u-v|^2 + 2v\cdot (u-v)$, we have

$$\begin{aligned}&\big | |u|^2 - |v|^2\big | \lesssim |u-v|^2 + |v||u-v| , \\&\big | |u|^2 - |v|^2\big |^2 \lesssim |u-v|^4 + |v|^2 |u-v|^2. \end{aligned}$$

This leads to

$$\begin{aligned} \Big |\nabla _v \textbf{k}(v,u) \frac{e^{{\tilde{\theta }}| v|^2}}{e^{{\tilde{\theta }}| u|^2}} \Big |&\lesssim \big [\frac{1+|u|^2+|v|^2}{| v-u|}+| v-u|+\langle v\rangle \big ] [\textbf{k}_1(v,u)+\textbf{k}_2(v,u) ] \frac{e^{{\tilde{\theta }}| v|^2}}{e^{{\tilde{\theta }}| u|^2}} \\&\lesssim \big [\frac{1 +|v|^2}{| v-u|}+| v-u|+\langle v\rangle \big ] \textbf{k}_{{\tilde{\varrho }}}(v,u) \\&=\big [\frac{1 +|v|^2}{| v-u|}+| v-u|+\langle v\rangle \big ] \frac{e^{-{\tilde{\varrho }}|v-u|^2}}{|v-u|} \lesssim \frac{1+|v|^2}{|v-u|} \frac{e^{-c|v-u|^2}}{|v-u|} \end{aligned}$$

for some $c<{\tilde{\varrho }}$. In the second line we have applied (3.24). In the last line we used that, for $c<{\tilde{\varrho }}$, we have

$$\begin{aligned}&e^{-{\tilde{\varrho }}|v-u|^2} \lesssim \frac{e^{-c|v-u|^2}}{|v-u|}, \ e^{-{\tilde{\varrho }}|v-u|^2} \lesssim \frac{e^{-c|v-u|^2}}{|v-u|^2}. \end{aligned}$$

Since $c<{\tilde{\varrho }}<\varrho -\frac{{\tilde{\theta }}}{2}$, for ease of notation, we conclude (3.25) with coefficient ${\tilde{\varrho }}$. $\square$

Lemma 3

With ${\tilde{\theta }}\ll \theta$, K(f) is bounded as

$$\begin{aligned} \begin{aligned} \Vert w_{{\tilde{\theta }}}(v)K(f)(t,x,v)\Vert _\infty&\lesssim e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty ,\\ | w_{{\tilde{\theta }}}(v)K(f)(t,x,v)|_\infty&\lesssim e^{-\lambda t} \sup _t e^{\lambda t}| wf(t)|_\infty . \end{aligned} \end{aligned}$$

(3.28)

Moreover, for $\nu$ and $\Gamma$ in (2.7), we have

$$\begin{aligned} \Big \Vert \frac{w}{\langle v\rangle }\Gamma (f,g)(t,x,v) \Big \Vert _\infty\lesssim & {} \Vert wf(t)\Vert _\infty \times \Vert wg(t)\Vert _\infty ,\nonumber \\ \Big | \frac{w}{\langle v\rangle }\Gamma (f,g)(t,x,v) \Big |_\infty\lesssim & {} | wf(t)|_\infty \times |wg(t)|_\infty , \end{aligned}$$

(3.29)

$$\begin{aligned} \nu\gtrsim & {} \sqrt{|v|^2+1}, \ \ |\nabla _v \nu | \lesssim 1. \end{aligned}$$

(3.30)

$$\begin{aligned} | \nabla _x \Gamma (f,g)(t,x,v) |\lesssim & {} \langle v\rangle \Vert wf(t) \Vert _\infty | \nabla _x g(t,x,v) | + \Vert wf(t)\Vert _\infty \nonumber \\{} & {} \int _{{{\mathbb {R}}}^3} \textbf{k}(v,u) |\nabla _x g(t,x,u)| \textrm{d}u \nonumber \\{} & {} + \Vert wg\Vert _\infty \int _{{{\mathbb {R}}}^3} \textbf{k}(v,u) |\nabla _x f(t,x,u)| \textrm{d}u. \end{aligned}$$

(3.31)

Proof

Proof of (3.28). By the property of K(f) in (3.26) and (3.27), we apply Lemma 2 to have

$$\begin{aligned}&|w_{{\tilde{\theta }}}(v) K(f)(t,x,v)| =\Big |\int _{{{\mathbb {R}}}^3} \textbf{k}(v,u)\frac{w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)} w_{{\tilde{\theta }}}(u) f(t,x,u) \textrm{d}u \Big | \\&\quad \lesssim \Vert w_{{\tilde{\theta }}}f(t)\Vert _\infty \Big | \int _{{{\mathbb {R}}}^3} \textbf{k}_{{\tilde{\varrho }}}(v,u) \textrm{d}u \Big | \le \Vert wf(t)\Vert _\infty . \end{aligned}$$

Proof of (3.29). By the definition of $\Gamma (f,g)$ in (2.9), we have

$$\begin{aligned}&|\Gamma (f,g)(t,x,v)| \\&\quad \lesssim \iint _{{{\mathbb {R}}}^3\times {{\mathbb {S}}}^2}|(v-u)\cdot \omega |\sqrt{\mu (u)}[|f(t,x,v')||g(t,x,u')|+|f(t,x,v)||g(t,x,u)|] \\&\quad \lesssim w^{-1}(v)\Vert wf(t)\Vert _\infty \Vert wg(t)\Vert _\infty \iint _{{{\mathbb {R}}}^3\times {{\mathbb {S}}}^2} |(v-u)\cdot \omega |\sqrt{\mu (u)}w^{-1}(u) \textrm{d}u\\&\quad \lesssim w^{-1}(v) \langle v\rangle \Vert wf(t)\Vert _\infty \Vert wg(t)\Vert _\infty . \end{aligned}$$

Here we have used $|u'|^2 + |v'|^2 = |u|^2 + |v|^2$.

Proof of (3.30). The proof is standard, we refer to [10].

Proof of (3.31). Taking the derivative we have

$$\begin{aligned} |\nabla _x \Gamma (f,g)|&= |\Gamma (\nabla _x f,g) + \Gamma (f,\nabla _x g) | \nonumber \\&\le \Vert wf(t)\Vert _\infty \Big |\Gamma (e^{-\theta |\cdot |^2},\nabla _x g) \Big | + \Vert wg(t)\Vert _\infty \Big |\Gamma (\nabla _x f, e^{-\theta |\cdot |^2}) \Big |. \end{aligned}$$

The $\Gamma$ terms above share the same form as the linear operator L in (2.8). Then (3.31) follows the expression of $\nu f$ and Kf in (3.22), with replacing $\mu$ by a different exponent $\sqrt{\mu }e^{-\theta |v|^2}$. For ease of notation, we keep the same $\textbf{k}(v,u)$ in (3.31). $\square$

The properties of the kinetic weight $\alpha$ are summarized in the next two lemmas.

Lemma 4

For a $C^3$ convex domain, the kinetic weight $\alpha ,{\tilde{\alpha }}$ in (2.13) is almost invariant along the characteristic:

$$\begin{aligned} \alpha (x,v)\thicksim \alpha (x-sv,v), \ {\tilde{\alpha }}(x,v)\thicksim {\tilde{\alpha }}(x-sv,v) \ \text {as long as } x-sv \in {\bar{\Omega }} . \end{aligned}$$

(3.32)

Here we recall the notation $\thicksim$ is defined in (1.11).

From the definition of ${\tilde{\alpha }}$ in (2.13), we have an upper bound for $\alpha (x,v)$ as

$$\begin{aligned} \alpha (x,v)\le \min \{1,{\tilde{\alpha }}(x,v)\}, \ \alpha (x,v)\lesssim |v|. \end{aligned}$$

(3.33)

Thus

$$\begin{aligned} {\tilde{\alpha }}(x,v)\sim |n(x_{\textbf{b}}(x,v))\cdot v|, \ \ \frac{1}{|n(x_{\textbf{b}}(x,v))\cdot v|}\lesssim \frac{1}{\alpha (x,v)} . \end{aligned}$$

(3.34)

Proof

(3.33) follows from the definition of $\chi$ in (2.14).

Since $x_{\textbf{b}}(x,v)\in \partial \Omega$, we have $|n(x_{\textbf{b}}(x,v))\cdot v|\sim {\tilde{\alpha }}(x_{\textbf{b}}(x,v),v)$. By (3.32), we further have $|n(x_{\textbf{b}}(x,v))\cdot v|\sim {\tilde{\alpha }}(x,v) \ge \alpha (x,v)$, then we conclude (3.34).

Then we focus on (3.32).

First we prove that $\chi (s)$ has the following property:

$$\begin{aligned} s\chi '(s)\le 4\chi (s). \end{aligned}$$

(3.35)

From the definition in (2.14), when $s\ge 2$, $s\chi '(s)=0\le 4\chi (s)$. When $s\le \frac{1}{2}$, we have $\chi (s)=s,$ and thus $s\chi '(s)=s \le 4\chi (s)=4s$. When $\frac{1}{2}<s<2$, since $\chi '(s)\le 1$, we have $s\chi '(s)\le s<2=4\chi (1/2)\le 4\chi (s)$. Then we conclude (3.35).

From the definition of ${\tilde{\alpha }}(x,v)$ in (2.13), we directly compute that

$$\begin{aligned}&|v\cdot \nabla _x \alpha (x,v)| = \chi '({\tilde{\alpha }}(x,v))|v\cdot \nabla _x {\tilde{\alpha }}(x,v)| \nonumber \\&\quad =\chi '({\tilde{\alpha }}(x,v)) \frac{|2v\cdot \nabla \xi (x)[v\cdot \nabla ^2 \xi \cdot v] - 2v\cdot \nabla \xi (x)[v\cdot \nabla ^2 \xi \cdot v]- 2v \{ v\cdot \nabla ^3 \xi \cdot v \} \xi (x)| }{{\tilde{\alpha }}(x,v)} \nonumber \\&\quad = 2\chi '({\tilde{\alpha }}(x,v))\frac{|v\{v\cdot \nabla ^3 \xi \cdot v\}\xi (x)|}{{\tilde{\alpha }}(x,v)} \lesssim \chi '({\tilde{\alpha }}(x,v)) \frac{|v|^3|\xi (x)|}{{\tilde{\alpha }}(x,v)} \nonumber \\&\quad \lesssim \chi '({\tilde{\alpha }}(x,v))\frac{|v|({\tilde{\alpha }}(x,v))^2}{{\tilde{\alpha }}(x,v)} = |v|\chi '({\tilde{\alpha }}(x,v)){\tilde{\alpha }}(x,v)\lesssim |v|\chi ({\tilde{\alpha }}(x,v))=|v|\alpha (x,v). \end{aligned}$$

(3.36)

In the last line, we have used the convexity (2.2) in the first inequality, and used (3.35) in the second inequality. Then for some C, we have

$$\begin{aligned} -C |v|\alpha (x,v)\le v\cdot \nabla _x \alpha (x,v) \le C |v|\alpha (x,v). \end{aligned}$$

Since

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}s} \alpha (x-sv,v)=-v\cdot \nabla _x \alpha (x-sv,v), \end{aligned}$$

by Grönwall’s inequality, we conclude that

$$\begin{aligned} e^{-C|v|s}\alpha (x-sv,v)\le \alpha (x,v)\le e^{C|v|s}\alpha (x-sv,v). \end{aligned}$$

Since $x-sv\in {\bar{\Omega }}$, we have $|v|s\le C_{\Omega }$ and conclude $\alpha (x,v)\sim \alpha (x-sv,v)$.

Following the same computation in (3.36) with replacing $\chi '$ by 1, for ${\tilde{\alpha }}(x,v)$ we have

$$\begin{aligned} |v\cdot {\tilde{\alpha }}(x,v)|\lesssim |v|{\tilde{\alpha }}(x,v). \end{aligned}$$

Again by Grönwall’s inequality we can conclude that ${\tilde{\alpha }}(x,v)\sim {\tilde{\alpha }}(x-sv,v)$.

We conclude (3.32). $\square$

The following nonlocal-to-local estimate for the kinetic weight (2.13) is essential in the analysis of the regularity:

Lemma 5

Recall the definition of $\alpha$ in (2.13), then for $C>0$ and $\varrho >0$, we have

$$\begin{aligned} \begin{aligned} \int ^{t}_{t- t_{\textbf{b}}(x,v)} \int _{{{\mathbb {R}}}^{3}}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho |v-u|^{2}}}{|v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s&\lesssim \frac{1}{\alpha (x,v)}. \end{aligned} \end{aligned}$$

(3.37)

For $\varepsilon ,\delta \ll 1$, recall the notation o(1) in (1.12),

$$\begin{aligned}{} & {} \int ^{t}_{t- \varepsilon } \int _{{{\mathbb {R}}}^{3}}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho |v-u|^{2}}}{|v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \lesssim \frac{o(1)}{\alpha (x,v)}, \end{aligned}$$

(3.38)

$$\begin{aligned}{} & {} \int ^{t}_{t-t_{\textbf{b}}(x,v)} \int _{|u|<\delta }\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho |v-u|^{2}}}{|v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \lesssim \frac{o(1)}{\alpha (x,v)}. \end{aligned}$$

(3.39)

Proof

We prove the following estimate: when $t-t_{\textbf{b}}(x,v)\le t-t_1 \le t-t_2 \le t$, for $0<\beta <1$ and some $C_1>0$,

$$\begin{aligned}&\int ^{t-t_1}_{t-t_2} \int _{{{\mathbb {R}}}^{3}}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho | v-u| ^{2}}}{| v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \lesssim \frac{| e^{-C_1\langle v\rangle t_1 }-e^{-C_1\langle v\rangle t_2}| ^{\beta }}{\alpha (x,v)}, \end{aligned}$$

(3.40)

$$\begin{aligned}&\int ^{t-t_1}_{t-t_2} \int _{|u|<\delta }\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho | v-u| ^{2}}}{| v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \lesssim o(1)\frac{| e^{-C_1\langle v\rangle t_1 }-e^{-C_1\langle v\rangle t_2}| ^{\beta }}{\alpha (x,v)}, \end{aligned}$$

(3.41)

Clearly Lemma 5 follows from (3.40) and (3.41).

In the proof we assume $\alpha (x-(t-s)v,u)={\tilde{\alpha }}(x-(t-s)v,u)$. For the other case, from (2.13), we have $\alpha (x-(t-s)v,u)\gtrsim 1$, which combined with $\frac{e^{-\varrho |v-u|^2}}{|v-u|}\in L^1_u$, gives that

$$\begin{aligned}&\int ^{t-t_1}_{t-t_2} \int _{{{\mathbb {R}}}^{3}}\textbf{1}_{\alpha (x-(t-s)v,u)\gtrsim 1}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho |v-u|^{2}}}{|v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \\&\quad \lesssim \int ^{t-t_1}_{t-t_2} \int _{{{\mathbb {R}}}^{3}}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho |v-u|^{2}}}{|v-u| }\textrm{d}u\textrm{d}s \\&\quad \lesssim \int ^{t-t_1}_{t-t_2} e^{-C\langle v\rangle (t-s)} \textrm{d}s \lesssim |e^{-C\langle v\rangle t_1}-e^{-C\langle v\rangle t_2}|\lesssim \frac{|e^{-C\langle v\rangle t_1}-e^{-C\langle v\rangle t_2}|}{\alpha (x,v)}. \end{aligned}$$

In the last inequality, we used (3.33). Now we prove (3.40).

Step 1. We claim that for $y\in {\bar{\Omega }}$ and $\varrho >0$,

$$\begin{aligned}{} & {} \int _{{{\mathbb {R}}}^3}\frac{e^{-\varrho | v-u|^2}}{| v-u| } \frac{1}{ \alpha (y,u)} \textrm{d}u \lesssim 1+ | \ln | \xi (y)| | + | \ln | v| | . \end{aligned}$$

(3.42)

$$\begin{aligned}{} & {} \int _{|u|<\delta }\frac{e^{-\varrho | v-u|^2}}{| v-u| } \frac{1}{ \alpha (y,u)} \textrm{d}u \lesssim o(1)[1+ | \ln | \xi (y)| | + | \ln | v| | ]. \end{aligned}$$

(3.43)

Recall (3.11), we set $\textbf{v}= \textbf{v} (y)$ and $\textbf{u}= \textbf{u} (y)$. For $| u| \gtrsim {| v| }$, we have

$$\begin{aligned} \big [ | \textbf{u}_3 (y)|^2+| \xi (y)| | u|^2 \big ]^{1/2} \gtrsim \big [| \textbf{u}_3 (y)|^2+| \xi (y)| | v|^2 \big ]^{1/2}. \end{aligned}$$

(3.44)

Thus, when $\frac{|v|}{4}\le |u|\le 4|v|$, we have

$$\begin{aligned} {\int _{\frac{| v| }{4} \le | u| \le 4| v| }}\lesssim&\ \iint \frac{e^{-\varrho | \textbf{v}_\parallel -\textbf{u}_\parallel |^2}}{| \textbf{v}_\parallel -\textbf{u}_\parallel | }\textrm{d}\textbf{u}_\parallel \int _0^{ 4| v| }\frac{ \textrm{d}\textbf{u}_3}{\big [ | \textbf{u}_3|^2+| \xi (y)| | v|^2\big ]^{1/2}} \nonumber \\ \le&\ \int _0^{4| v| }\frac{\textrm{d}\textbf{u}_3}{\big [ | \textbf{u}_3|^2+| \xi (y)| | v|^2\big ]^{1/2}}\nonumber \\ =&\ \ln \Big ( \sqrt{| \textbf{u}_3|^2+| \xi (y)| | v|^2}+| \textbf{u}_3| \Big )\Big | _{0}^{4| v| } \nonumber \\ =&\ln (\sqrt{16| v|^2 + | \xi (y)| | v|^2}+ 16 | v|^2)- \ln (\sqrt{ | \xi (y)| | v|^2})\nonumber \\ \lesssim&\ln | v| + \ln | \xi (y)| . \end{aligned}$$

(3.45)

If $| u| \ge 4| v|$, then $| u-v|^2 \ge \frac{| v|^2}{4}+\frac{| u|^2}{4}$ and hence the exponent is bounded as

$e^{- \varrho | v-u|^2} \le e^{- \frac{\varrho }{8} | v|^2}e^{- \frac{\varrho }{8} | u|^2} e^{- \frac{\varrho }{2} | v-u|^2}$. This, together with (3.44), implies

$$\begin{aligned} {\int _{| u| \ge 4| v| }} \lesssim&\ e^{- \frac{\varrho }{8}| v|^2} \iint \frac{e^{-\frac{\varrho }{2}| \textbf{v}_\parallel -\textbf{u}_\parallel |^2}}{| \textbf{v}_\parallel -\textbf{u}_\parallel | }\textrm{d}\textbf{u}_\parallel \int _0^{\infty }\frac{ e^{-\frac{\varrho }{8}| \textbf{u}_3|^2} }{\big [ | \textbf{u}_3|^2+| \xi (y)| | v|^2\big ]^{1/2}} \textrm{d}\textbf{u}_3 \nonumber \\ \lesssim&\ e^{- \frac{\varrho }{8}| v|^2} \int _0^{\infty }\frac{ e^{-\frac{\varrho }{8}| \textbf{u}_3|^2} }{\big [ | \textbf{u}_3|^2+| \xi (y)| | v|^2\big ]^{1/2}} \textrm{d}\textbf{u}_3 \nonumber \\ \lesssim&\ e^{- \frac{\varrho }{8}| v|^2} + e^{- \frac{\varrho }{8}| v|^2} \int _0^{1}\frac{\textrm{d}\textbf{u}_3}{\big [ | \textbf{u}_3|^2+| \xi (y)| | v|^2\big ]^{1/2}} \nonumber \\ =&\ e^{- \frac{\varrho }{8}| v|^2}+ e^{- \frac{\varrho }{8}| v|^2}\ln \Big ( \sqrt{| \textbf{u}_3|^2+| \xi (y)| | v|^2}+| \textbf{u}_3| \Big )\Big | _{0}^{1}\nonumber \\ =&\ e^{- \frac{\varrho }{8}| v|^2} \Big \{1+ \ln (\sqrt{1 + | \xi (y)| | v|^2} + 1 )- \ln (\sqrt{ | \xi (y)| | v|^2} ) \Big \} \nonumber \\ \le&\ e^{- \frac{\varrho }{8}| v|^2} \{\ln | v| + \ln | \xi (y)| \} . \end{aligned}$$

(3.46)

When $| u| \le \frac{| v| }{4}$, we have $| v-u| \ge \big | | v| -| u| \big | \ge | v| -\frac{| v| }{4}\ge \frac{| v| }{2}$, which leads to

$$\begin{aligned} {\int _{| u| \le \frac{| v| }{4}}}&\lesssim \frac{ {e^{- \frac{\varrho }{4}| v|^2}} }{| v| }\int _{| \textbf{u}_3| +| \textbf{u}_\parallel | \le \frac{| v| }{2}} \frac{ \textrm{d}\textbf{u}_3 \textrm{d}\textbf{u}_\parallel }{\Big [| \textbf{u}_3|^2+| \xi (y)| | \textbf{u}_\parallel |^2 \Big ]^{1/2}}\\&\lesssim | v| {e^{- \frac{\varrho }{4}| v|^2}}\int _{| {\tilde{\textbf{u}}}_3 | \le \frac{1}{2}} \int _{| {\tilde{\textbf{u}}}_\parallel | \le \frac{1}{2}} \frac{ \textrm{d}{\tilde{\textbf{u}}}_\parallel \textrm{d}{\tilde{\textbf{u}}}_3 }{\Big [| {\tilde{\textbf{u}}}_3|^2+| \xi (y)| | {\tilde{\textbf{u}}}_\parallel |^2 \Big ]^{1/2}}, \end{aligned}$$

where we have used $| v| {\tilde{u}} = u$. Using the polar coordinate ${\tilde{\textbf{u}}}_{1}=| {\tilde{\textbf{u}}}_\parallel | \cos \rho , {\tilde{\textbf{u}}}_{2}=| {\tilde{\textbf{u}}}_\parallel | \sin \rho$, we proceed with the computation as follows:

$$\begin{aligned}&{\int _{| u| \le \frac{| v| }{4}}} \lesssim \ | v| {e^{- \frac{\varrho }{4}| v|^2}}\int _0^{ \frac{1}{2}} \textrm{d}{\tilde{\textbf{u}}}_3 \int _{0}^{2\pi }\int _{0}^{\sqrt{1/2}} \frac{ | {\tilde{\textbf{u}}}_\parallel | \textrm{d}| {\tilde{\textbf{u}}}_\parallel | \textrm{d}\rho }{\Big [| {\tilde{\textbf{u}}}_3|^2+| \xi (y)| | {\tilde{\textbf{u}}}_\parallel |^2 \Big ]^{1/2}} \nonumber \\&\lesssim | v| {e^{- \frac{\varrho }{4}| v|^2}} \int _0^{ \frac{1}{2}} \textrm{d}{\tilde{\textbf{u}}}_3 \int _{0}^{ {1/2}} \frac{ \textrm{d}| {\tilde{\textbf{u}}}_\parallel |^2 }{\Big [| {\tilde{\textbf{u}}}_3|^2+| \xi (y)| | {\tilde{\textbf{u}}}_\parallel |^2 \Big ]^{1/2}} \nonumber \\&= \ | v| {e^{- \frac{\varrho }{4}| v|^2}} \int _0^{ \frac{1}{2}} \textrm{d}{\tilde{\textbf{u}}}_3 \frac{1}{| \xi (y)| } \Big ( \sqrt{ | {\tilde{\textbf{u}}}_3|^2+\frac{| \xi (y)| }{2} } - | {\tilde{\textbf{u}}}_3| \Big ) \nonumber \\&= \ \frac{| v| {e^{- \frac{\varrho }{4}| v|^2}}}{| \xi (y)| } \nonumber \\&\quad \times \bigg \{\frac{1}{2}| {\tilde{\textbf{u}}}_3| \sqrt{| {\tilde{\textbf{u}}}_3|^2+\frac{| \xi | }{2}}+\frac{| \xi | }{4}\log \Big (\sqrt{| {\tilde{\textbf{u}}}_3|^2+\frac{| \xi | }{2}}+| {\tilde{\textbf{u}}}_3| \Big )-\frac{1}{2}| {\tilde{\textbf{u}}}_3| ^{1/2} \bigg \}\bigg | _{| {\tilde{\textbf{u}}}_3| =0}^{| {\tilde{\textbf{u}}}_3| =1/2} \nonumber \\&= \frac{| v| e^{-C| v|^2}}{| \xi | }\bigg \{\frac{1}{4}\sqrt{\frac{1}{4}+\frac{| \xi | }{2}}+\frac{\xi }{4}\log \Big ( \sqrt{\frac{1}{4}+\frac{| \xi | }{2}}+\frac{1}{2}\Big )-\frac{| \xi | }{4}\log \Big (\sqrt{\frac{| \xi | }{2}} \Big )-\frac{1}{8} \bigg \} \nonumber \\&\lesssim \frac{| v| e^{-C| v|^2}}{| \xi | }\Big [| \xi | \log (| \xi | )+| \xi | \log \Big (1+\sqrt{1+| \xi | }\Big ) \Big ] \nonumber \\&\lesssim 1+\log (| \xi (y)| ). \end{aligned}$$

(3.47)

Collecting terms from (3.45), (3.46), and (3.47), we prove (3.42).

The proof of (3.43) is the same, and uses the following estimate:

$$\begin{aligned}&\int _{|u|<\delta } \frac{e^{-\varrho |v-u|^2}}{|v-u|} \frac{1}{\alpha (y,u)} \textrm{d}u \\&\quad \lesssim \iint _{|\textbf{u}_\parallel |<\delta } \frac{e^{-\varrho | \textbf{v}_\parallel -\textbf{u}_\parallel |^2}}{| \textbf{v}_\parallel -\textbf{u}_\parallel | }\textrm{d}\textbf{u}_\parallel \int _0^{\delta }\frac{ \textrm{d}\textbf{u}_3}{\big [ | \textbf{u}_3|^2+| \xi (y)| | \textbf{u}_\parallel |^2\big ]^{1/2}} \\&\quad \lesssim o(1)\int _0^{\delta }\frac{ \textrm{d}\textbf{u}_3}{\big [ | \textbf{u}_3|^2+| \xi (y)| | \textbf{u}_\parallel |^2\big ]^{1/2}}. \end{aligned}$$

Step 2. We prove the following statement: for $x\in \partial \Omega$ , we can choose $0 < {\tilde{\delta }}\ll _\Omega 1$ such that

$$\begin{aligned}&{\tilde{\delta }}^{1/2} | v\cdot \nabla \xi ( x-(t-s) v) | \gtrsim _\Omega \ | v| \sqrt{-\xi ( x-(t-s) v)}, \nonumber \\&\qquad \qquad \qquad \qquad \text {for} \ s\in \Big [t-t_{\textbf{b}}(x,v), t-t_{\textbf{b}}(x,v)+ { {\tilde{t}}} \Big ]\cup \Big [t- { {\tilde{t}}}, t \Big ] , \end{aligned}$$

(3.48)

$$\begin{aligned}&{\tilde{\delta }}^{1/2} \times {\tilde{\alpha }}(x,v) \lesssim _\Omega | v| \sqrt{-\xi ( x-(t-s) v)}, \ \ \text {for} \ s\in \Big [ t-t_{\textbf{b}}(x,v)+ { {\tilde{t}}}, t- { {\tilde{t}}} \Big ]. \end{aligned}$$

(3.49)

Here ${\tilde{t}} =\min \{ \frac{t_{\textbf{b}}(x,v)}{2}, {\tilde{\delta }}\frac{{\tilde{\alpha }}(x,v)}{| v|^2}\}$. We note that when ${\tilde{t}}<{\tilde{\delta }}\frac{{\tilde{\alpha }}(x,v)}{| v|^2}$, (3.49) vanishes.

If $v=0$ or $v\cdot \nabla \xi (x)\le 0$, since $x\in \partial \Omega$, (3.48) and (3.49) hold clearly with $t_{\textbf{b}}(x,v)={\tilde{t}}=0$. We may assume $v\ne 0$ and $v\cdot \nabla \xi (x)>0$. Since $v\cdot \nabla \xi (x)> 0$, from (3.32), we have $| v\cdot \nabla \xi (x_{\textbf{b}}(x,v) )| >0$. Then we must have $v\cdot \nabla \xi (x_{\textbf{b}}(x,v) ) <0$, otherwise $v\cdot \nabla \xi (x_{\textbf{b}})>0$ implies $x_{\textbf{b}}$ is not the backward exit position defined in (3.1). By the mean value theorem there exists at least one $t^{*} \in (t-t_{\textbf{b}}(x,v) , t)$ such that $v\cdot \nabla \xi (x- t^*v)=0$. Moreover, the convexity in (2.2) leads to $\frac{d}{ds}\big (v\cdot \nabla \xi (x-(t-s)v)\big )=v\cdot \nabla ^{2}\xi (x-(t-s)v )\cdot v\ge C_{\xi }| v| ^{2},$ and therefore $t^{*}\in (t-t_{\textbf{b}}(x,v), t)$ is unique.

Let $s \in \big [t-{\tilde{t}},t\big ]$ for $0<{\tilde{\delta }}\ll 1$ and ${\tilde{t}}\le {\tilde{\delta }}\frac{{\tilde{\alpha }}(x,v)}{| v|^2}$. Then, from the fact that $v\cdot \nabla _x \xi (x-(t-\tau )v)$ is non-decreasing function in $\tau$ and $x\in \partial \Omega$,

$$\begin{aligned} | v|^2 (-1) \xi (x-(t-s)v)= & {} \int ^t_s | v|^2 v\cdot \nabla _x \xi (x-(t-\tau )v) \textrm{d}\tau \nonumber \\ {}\le & {} {\tilde{\delta }} {\tilde{\alpha }}(x,v) | v\cdot \nabla _x \xi (x)| . \end{aligned}$$

(3.50)

By (3.32) $| v\cdot \nabla _x \xi (x)| \le {\tilde{\alpha }}(x,v) \le C_\Omega {\tilde{\alpha }}(x-(t-s)v,v)\le C_\Omega \{ | v \cdot \nabla \xi (x-(t-s)v)| + \Vert \nabla _x^2 \xi \Vert _\infty | v| \sqrt{- \xi (x-(t-s)v)} \},$ we choose ${\tilde{\delta }}\ll C_\Omega \Vert \nabla _x^2 \xi \Vert _\infty ^{-2}$ and absorb the $|v|^2 |\xi (x-(t-s)v)|$ term in

$$\begin{aligned}&{\tilde{\delta }} {\tilde{\alpha }}(x,v) \times C_\Omega \{ | v \cdot \nabla \xi (x-(t-s)v)| + \Vert \nabla _x^2 \xi \Vert _\infty | v| \sqrt{- \xi (x-(t-s)v)} \} \\&\quad \le {\tilde{\delta }} \times \{C_\Omega \Vert \nabla _x^2 \xi \Vert _\infty | v| \sqrt{- \xi (x-(t-s)v)}\}^2 + {\tilde{\delta }}\times \{C_\Omega |v\cdot \nabla \xi (x-(t-s)v)|\}^2 \end{aligned}$$

by the left hand side of (3.50). This gives (3.48) for $s \in \big [t- {\tilde{t}} ,t\big ]$. The proof for $s \in \big [t-t_{\textbf{b}}(x,v), t-t_{\textbf{b}}(x,v)+ {\tilde{t}} \big ]$ is same.

For (3.49), we assume ${\tilde{t}} = {\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2}$, otherwise (3.49) vanishes. Since $v\cdot \nabla _x \xi (x-(t-s)v)>0$ when $s>t_*$, we have that $\xi (x-(t-s)v)$ is increasing in s when $s>t_*$, thus $| v|^2 (-1) \xi (x-(t-s)v)\ge | v|^2 (-1) \xi \Big (x-{\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} v\Big )$ for $s\in \big [t- t^*, t- {\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} \big ]$. By an expansion, for $s^*:=t-{\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2}$,

$$\begin{aligned} \begin{aligned}&| v|^2 (-1) \xi \Big (x-{\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} v\Big ) \\&\quad =| v|^2 ( v\cdot \nabla _x \xi (x )){\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} -\int ^t_{s^*}\int ^t_{\tau }| v|^2 v\cdot \nabla _x^2 \xi (x-(t-\tau ^\prime ) v) \cdot v\textrm{d}\tau ^\prime \textrm{d}\tau . \end{aligned} \nonumber \\ \end{aligned}$$

(3.51)

The last term of (3.51) is bounded by $\Vert \nabla _x^2 \xi \Vert _\infty {\tilde{\delta }}^2 \big (\frac{{\tilde{\alpha }}(x,v)}{| v|^2}\big )^2 | v| ^4 \le \Vert \nabla _x^2 \xi \Vert _\infty {\tilde{\delta }}^2 ({\tilde{\alpha }}(x,v))^2$. Since $v \cdot \nabla _x \xi (x) = {\tilde{\alpha }}(x,v)$ when $x\in \partial \Omega$, for ${\tilde{\delta }} \ll \Vert \nabla _x^2 \xi \Vert _\infty ^{-1/2}$, the right hand side of (3.51) is bounded below by $\frac{{\tilde{\delta }}}{2} ({\tilde{\alpha }}(x,v))^2$. This completes the proof of (3.49) when $s\in \big [t- t^*, t- {\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} \big ]$. The proof for the case of $s\in \big [ t - t_{\textbf{b}}(x,v)+ {\tilde{\delta }} \frac{{\tilde{\alpha }}(x,v)}{| v|^2} , t- t^* \big ]$ is same.

Step 3. From (3.42), for the proof of (3.40), it suffices to estimate that

$$\begin{aligned} \begin{aligned}&\int ^t_{t-t_{\textbf{b}}(x,v)} \textbf{1}_{t-t_2\ge s \ge t-t_1 } e^{-C \langle v\rangle (t-s)} \big | \ln | \xi (x-(t-s)v)| \big | \textrm{d}s \\&\quad + \int ^t_{t-t_{\textbf{b}}(x,v)} \textbf{1}_{t-t_2\ge s \ge t-t_1 } e^{-C \langle v\rangle (t-s)}\big (1+ \big | \ln | v| \big | \big )\textrm{d}s . \end{aligned} \end{aligned}$$

(3.52)

We bound the second term of (3.52) as

$$\begin{aligned} (1+| \ln | v| | )\int ^{t-t_2}_{t-t_1}e^{-C\langle v\rangle (t-s)} \lesssim (1+| \ln | v| | )\langle v\rangle ^{-1} | e^{-C\langle v\rangle t_2}-e^{-C\langle v\rangle t_1}| . \nonumber \\ \end{aligned}$$

(3.53)

For the first term of (3.52), we first assume $x\in \partial \Omega$. For utilizing (3.48) and (3.49), we split the first term of (3.52) as

$$\begin{aligned} \underbrace{ \int ^t_{t- {\tilde{t}}} + \int ^{t -t_{\textbf{b}}(x,v)+ {\tilde{t}}}_{t -t_{\textbf{b}}(x,v) }}_{(3.54)_1} + \underbrace{ \int _{t -t_{\textbf{b}}(x,v)+ {\tilde{t}}} ^{t- {\tilde{t}}}}_{(3.54)_2}. \end{aligned}$$

(3.54)

Without loss of generality, we assume $t-t_2\in [t-{\tilde{t}},t]$, $t-t_1\in [t-t_{\textbf{b}}(x,v)+{\tilde{t}},t-{\tilde{t}}]$. For the first term $(3.54)_1$ we use a change of variables $s \mapsto -\xi (x-(t-s)v)$ in $s \in [t-t_{\textbf{b}}(x,v), t-t^*]$ and $s \in [t-t^*,t ]$ separately with $\textrm{d}s = | v\cdot \nabla _x \xi (x-(t-s)v)| ^{-1} \textrm{d}| \xi |$. From (3.50), we deduce $| \xi (x-(t-s)v) | \le {\tilde{\delta }}\frac{{\tilde{\alpha }}^2(x,v)}{| v|^2}$. Then applying Hölder’s inequality with $\beta +(1-\beta )=1$ and using (3.48), we get

$$\begin{aligned}&(3.54)_1 \textbf{1}_{\{t-t_2\in [t-{\tilde{t}},t]\},t-t_1\in [t-t_{\textbf{b}}(x,v)+{\tilde{t}},t-{\tilde{t}}]} \nonumber \\&\quad \lesssim \Big (\big [\int ^{t-t_2}_{t-{\tilde{t}}} e^{-C\langle v\rangle (t-s)/\beta } \textrm{d}s \big ]^{\beta }+\big [\int ^{t-t_{\textbf{b}}(x,v)+{\tilde{t}}}_{t-t_1} e^{-C\langle v\rangle (t-s)/\beta } \textrm{d}s \big ]^{\beta }\Big ) \nonumber \\&\qquad \times \Big [ \int _0^{{\tilde{\delta }}\frac{{\tilde{\alpha }}^2(x,v)}{| v|^2}} | \ln | \xi | | ^{1/(1-\beta )} \frac{\textrm{d}| \xi | }{ {\tilde{\delta }}^{-1/2}| v| \sqrt{ | \xi | }} \Big ]^{1-\beta } \nonumber \\&\quad \lesssim | e^{-C\langle v\rangle t_2/\beta }-e^{-C\langle v\rangle t_1/\beta }| ^\beta \frac{1}{| v| ^{1-\beta }}, \end{aligned}$$

(3.55)

where we have used $t-{\tilde{t}}>t-t_1, \quad t-t_2>t-t_{\textbf{b}}(x,v)+{\tilde{t}}$ and $\frac{| \ln | \xi | | ^{1/(1-\beta )}}{\sqrt{\xi }} \in L^1_{loc}(0,\infty )$ for $\beta <1$ in the last line.

On the other hand, since $\Vert \xi \Vert _\infty \lesssim 1$, we only consider the contribution of $|\xi |\lesssim 1$ in $\ln (|\xi |)$. From (3.49),

$$\begin{aligned} (3.54)_2&\le \int ^{t-t_2}_{t-t_1} e^{-C \langle v\rangle (t-s)} \Big | \ln \Big ({\tilde{\delta }} \frac{ ({\tilde{\alpha }}(x,v))^2}{| v|^2}\Big ) \Big | \textrm{d}s \nonumber \\&\le 2 \int ^{t-t_2}_{t-t_1} e^{-C \langle v\rangle (t-s)} \{ | \ln {\tilde{\delta }} | + | \ln {\tilde{\alpha }}(x,v)| + | \ln | v| | \} \textrm{d}s \nonumber \\&\le 2 | e^{-C\langle v\rangle t_2}-e^{-C\langle v\rangle t_1}| \langle v\rangle ^{-1/2}\{ | \ln {\tilde{\delta }} | + | \ln {\tilde{\alpha }}(x,v)| + | \ln | v| | \}, \end{aligned}$$

(3.56)

where we have used a similar estimate of (3.53).

Now as assume $x\notin \partial \Omega$. We find ${\bar{x}}\in \partial \Omega$ and ${\bar{t}}$ so that

$$\begin{aligned} x = {\bar{x}} - ({\bar{t}} - t)v \text { and } {\bar{t}}>t. \end{aligned}$$

Then clearly, $x-(t-s)v = {\bar{x}}-({\bar{t}}-s)v.$ Since ${\bar{x}}\in \partial \Omega$, applying the same computation as (3.55) and (3.56), the first term of (3.52) is bounded by

$$\begin{aligned}&\int ^{{\bar{t}}}_{{\bar{t}}-t_{\textbf{b}}({\bar{x}},v)} \textbf{1}_{t-t_2\ge s \ge t-t_1} e^{-C\langle v\rangle ({\bar{t}}-s)}| \ln | \xi ({\bar{x}}-({\bar{t}}-s)v)| | \textrm{d}s \nonumber \\&\quad \lesssim (3.55) + | e^{-C\langle v\rangle t_2}-e^{-C\langle v\rangle t_1}| \langle v\rangle ^{-1/2}\{ | \ln {\tilde{\delta }} | + | \ln {\tilde{\alpha }}({\bar{x}},v)| + | \ln | v| | \}. \end{aligned}$$

(3.57)

Here we used ${\tilde{\alpha }}({\bar{x}},v) \thicksim {\tilde{\alpha }}(x,v)$ from (3.32) to obtain the same upper bound as (3.55). Using ${\tilde{\alpha }}({\bar{x}},v) \thicksim {\tilde{\alpha }}(x,v)$ again, we conclude that

$$\begin{aligned} (3.57) \lesssim (3.55) + (3.56). \end{aligned}$$

Then collecting (3.52), (3.53), (3.55) and (3.56), we conclude that for $C_1=C/\beta$,

$$\begin{aligned}&\int ^{t-t_1}_{t-t_2} \int _{{{\mathbb {R}}}^{3}}\frac{e^{-C\langle v\rangle (t-s)} e^{-\varrho | v-u| ^{2}}}{| v-u| \alpha ( x-(t-s)v, {u}) }\textrm{d}u\textrm{d}s \nonumber \\&\quad \lesssim | e^{-C_1\langle v\rangle t_1 }-e^{-C_1\langle v\rangle t_2}| ^{\beta }\big [ \langle v\rangle ^{-1/2}\big ( 1+ | \ln | v| | + | \ln {\tilde{\alpha }}(x,v)| \big ) + \frac{1}{| v| ^{1-\beta }}\big ] . \end{aligned}$$

(3.58)

Then by (3.33), ${\tilde{\alpha }}(x,v)\lesssim |v|$, $\alpha (x,v)\lesssim \min \{1,|v|\}$, and thus $1\le \frac{1}{\alpha (x,v)}$, we bound

$$\begin{aligned} \frac{|\ln |v||}{\langle v\rangle ^{1/2}}&= \textbf{1}_{|v|\ge 1} + \textbf{1}_{|v|<1}\lesssim 1+\textbf{1}_{|v|< 1} \frac{|\ln |v||}{\langle v\rangle ^{1/2}} \\&\lesssim 1+\textbf{1}_{|v|<1}\frac{1}{|v|}\lesssim \frac{1}{\alpha (x,v)},\\ \frac{|\ln {\tilde{\alpha }}(x,v)|}{\langle v\rangle ^{1/2}}&=\textbf{1}_{{\tilde{\alpha }}(x,v)\ge 1} + \textbf{1}_{{\tilde{\alpha }}(x,v)<1} \lesssim \textbf{1}_{|v|\gtrsim 1}\frac{|\ln |v||}{\langle v\rangle ^{1/2}} \\&\quad + \textbf{1}_{\alpha (x,v)\lesssim 1}\frac{|\ln \alpha (x,v)|}{\langle v\rangle ^{1/2}}\lesssim \frac{1}{\alpha (x,v)},\\ \frac{1}{|v|^{1-\beta }}&=\textbf{1}_{|v|\ge 1} + \textbf{1}_{|v|<1} \le 1+\frac{1}{(\alpha (x,v))^{1-\beta }}\lesssim \frac{1}{\alpha (x,v)} . \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} (3.58)\lesssim \frac{| e^{-C_1\langle v\rangle t_1 }-e^{-C_1\langle v\rangle t_2}|^{\beta }}{\alpha (x,v)}, \end{aligned}$$

So (3.40) follows. (3.41) follows by the extra o(1) from (3.43). $\square$

4 Uniform-in-Time Weighted $C^1$ Estimate

In this section, we prove Theorem 2. We will take ${\tilde{\theta }} \ll \theta$ in (2.19), and thus ${\tilde{\theta }}$ satisfies the condition in Lemma 2.

First, we note that, under the conditions in Theorem 2, the dynamical stability (2.18) has been established in [7], and the steady problem (2.5) is well-posedness with $L^\infty$ bound

$$\begin{aligned} \Vert wf_s\Vert _\infty + |wf_s|_\infty \lesssim \Vert T_W-T_0\Vert _{L^\infty (\partial \Omega )} . \end{aligned}$$

(4.1)

Moreover, [3] constructed the weighted $C^1$ estimate of the steady problem

$$\begin{aligned} \Vert w_{{\tilde{\theta }}}(v) \alpha (x,v) \nabla _x f_s (x,v)\Vert _\infty&\lesssim \Vert T_W-T_0\Vert _{C^1(\partial \Omega )}. \end{aligned}$$

(4.2)

Here, $\alpha (x,v)$ is the kinetic weight defined in (2.13). We sketch the proof of (4.2) in Sect. 4.5.

Remark 7

In Sect. 4.5, we will only focus on the a priori estimate. The existence can be justified through the sequential argument, we refer detail to [3].

For $(x,v)\in \Omega \times {{\mathbb {R}}}^3$, we apply method of characteristic to (2.7) and obtain

$$\begin{aligned} f(t,x,v)&= \textbf{1}_{t\le t_{\textbf{b}}(x,v)} e^{-\nu t} f_0(x-tv,v) \\&\quad + \textbf{1}_{t>t_{\textbf{b}}(x,v)} e^{-\nu t_{\textbf{b}}(x,v)} f(t-t_{\textbf{b}},x_{\textbf{b}},v) \\&\quad + \int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (v)(t-s)} \int _{{{\mathbb {R}}}^3} \textbf{k}(v,u) f(s,x-(t-s)v,u) \textrm{d}u \textrm{d}s\\&\quad + \int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (v)(t-s)} h(s,x-(t-s)v,u) \textrm{d}s. \end{aligned}$$

Here we denote all the $\Gamma$ term in (2.7) as

$$\begin{aligned} h(t,x,v) = [\Gamma (f_s,f)+\Gamma (f,f_s)+\Gamma (f,f)](t,x,v). \end{aligned}$$

(4.3)

We take spatial derivative to get

$$\begin{aligned} \nabla _x f(t,x,v)&=\textbf{1}_{t\le t_{\textbf{b}}(x,v)} e^{-\nu t} \nabla _x f_0(x-tv,v) \end{aligned}$$

(4.4)

$$\begin{aligned}&\quad - \textbf{1}_{t>t_{\textbf{b}}} \nu \nabla _x t_{\textbf{b}}(x,v) e^{-\nu t_{\textbf{b}}(x,v)} f(t-t_{\textbf{b}},x_{\textbf{b}},v) \end{aligned}$$

(4.5)

$$\begin{aligned}&\quad - \textbf{1}_{t>t_{\textbf{b}}} e^{-\nu t_{\textbf{b}}} \nabla _x t_{\textbf{b}}\partial _t f(t-t_{\textbf{b}},x_{\textbf{b}},v) \end{aligned}$$

(4.6)

$$\begin{aligned}&\quad + \textbf{1}_{t>t_{\textbf{b}}} e^{-\nu t_{\textbf{b}}} \sum _{i=1,2} \nabla _x \textbf{x}_{p^1,i}^1 \partial _{\textbf{x}_{p^1,i}^1} f(t-t_{\textbf{b}},x_{\textbf{b}},v) \end{aligned}$$

(4.7)

$$\begin{aligned}&\quad + \int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textbf{k}(v,u) \nabla _x f(s,x-(t-s)v,u) \textrm{d}u \textrm{d}s \end{aligned}$$

(4.8)

$$\begin{aligned}&\quad +\textbf{1}_{t>t_{\textbf{b}}} e^{-\nu t_{\textbf{b}}}\nabla _x t_{\textbf{b}}\int _{{{\mathbb {R}}}^3} \textbf{k}(v,u) f(t-t_{\textbf{b}},x_{\textbf{b}},u) \textrm{d}u \end{aligned}$$

(4.9)

$$\begin{aligned}&\quad + \int _{\max \{0,t-t_{\textbf{b}}\}}^t e^{-\nu (t-s)} \nabla _x h(s,x-(t-s)v,u) \textrm{d}s \end{aligned}$$

(4.10)

$$\begin{aligned}&\quad + \textbf{1}_{t>t_{\textbf{b}}} e^{-\nu t_{\textbf{b}}} \nabla _x t_{\textbf{b}}h(t-t_{\textbf{b}},x_{\textbf{b}},u) . \end{aligned}$$

(4.11)

Here h in (4.10) and (4.11) is defined in (4.3). In (4.7), $\textbf{x}_{p^1,i}^1$ is defined in Definition 4 and $x_{\textbf{b}}=x^1=\eta _{p^1}(\textbf{x}_{p^1}^1)$.

Remark 8

With a compatibility condition it is standard to check the piece-wise formula (4.4)–(4.11) is actually a weak derivative of f and continuous across $\{t=t_{\textbf{b}}(x,v)\}$(see [11] for the details). We will only focus on the a priori estimate to the weighted $C^1$ estimate $\Vert w_{{\tilde{\theta }}}(v)\alpha \nabla _x f\Vert _\infty .$ For existence one can apply iteration sequence, we refer detail to [12].

Before proving Theorem 2, we need some preparations for (4.6), (4.7), (4.10) and (4.11), they are presented in Sects. 4.1, 4.2 and 4.3 respectively. Then, in Sect. 4.4, we prove Theorem 2.

4.1 Preparation for (4.6).

The following corollary gives a decay-in-time estimate for the temporal derivative $\partial _t f(t)$ in (4.6):

Corollary 4

Assume the compatibility condition (2.17) and (2.21) hold for $f_0(x,v)$ and $\partial _t f_0(x,v)$ defined in (2.12).

If $\partial _t f_0(x,v)$ satisfies (2.20) and is continuous away from $\gamma _0$, then $\partial _t f$ is continuous away from the grazing set $\gamma _0$. Moreover, we have the $L^\infty$ estimate

$$\begin{aligned} \Vert w\partial _t f(t)\Vert _\infty + |w\partial _t f(t)|_\infty&\lesssim e^{-\lambda t} \Vert w\partial _t f_0\Vert _\infty . \end{aligned}$$

(4.12)

To prove Corollary 4 we first cite a result from [7] regarding the decay-in-time $L^\infty$ estimate for the following linear problem:

$$\begin{aligned} \begin{aligned} \partial _t f+v\cdot \nabla _x f + L(f) =g ,&\\ f|_{\gamma _-} = P_\gamma f + r.&\end{aligned} \end{aligned}$$

(4.13)

Here $P_\gamma f$ is defined as the projection

$$\begin{aligned} P_\gamma f = \sqrt{\mu (v)} \int _{n(x)\cdot u>0} f(u) \sqrt{\mu (u)}\{n(x)\cdot u\} \textrm{d}u. \end{aligned}$$

(4.14)

Proposition 5

(Proposition 6.1 and 7.1 in [7]) Let $\Vert wf_0\Vert _\infty + |\langle v\rangle wr|_\infty + \big \Vert \frac{w}{\langle v\rangle }g\big \Vert _\infty < \infty$ and $\iint \sqrt{\mu }g=\int _\gamma r\sqrt{\mu }=\iint f_0\sqrt{\mu }=0$, then there exists a solution to (4.13) that satisfies

$$\begin{aligned} \Vert wf(t)\Vert _\infty + |wf(t)|_\infty\lesssim & {} e^{-\lambda t}\left\{ \Vert wf_0\Vert _\infty +\sup _{s\le t} e^{\lambda s}\Big \Vert \frac{w}{\langle v\rangle }g(s) \Big \Vert _\infty \right. \nonumber \\{} & {} \left. + \sup _{s\le t}e^{\lambda s}|\langle v\rangle wr(s)|_\infty \right\} . \end{aligned}$$

(4.15)

If $f_0|_{\gamma _-}=P_\gamma f_0 + r_0$, $f_0$, r and g are continuous away from $\gamma _0$, then f(t, x, v) is continuous away from $\gamma _0$.

Proof of Corollary 4

First, we consider the linear problem

$$\begin{aligned} \partial _t \partial _t f + v\cdot \nabla _x \partial _t f+L(\partial _tf) = \partial _t g&\\ \partial _t f|_{\gamma _-} = P_{\gamma } \partial _t f + \partial _t r, \end{aligned}$$

with following the initial condition satisfying (2.21):

$$\begin{aligned} \partial _t f_0|_{\gamma _-} = P_\gamma \partial _t f_0 + \partial _t r_0. \end{aligned}$$

Assuming $\partial _t r$ and $\partial _t g$ are continuous away from $\gamma _0$ and $\iint \sqrt{\mu }\partial _t g=\int _\gamma \partial _t r\sqrt{\mu }=\iint \partial _t f_0\sqrt{\mu }=0$, by Proposition 5, there exists a unique f and $\partial _t f$ such that

$$\begin{aligned} \Vert w\partial _t f(t)\Vert _\infty + |w \partial _t f(t)|_\infty&\lesssim e^{-\lambda t}\{\Vert w \partial _t f_0\Vert _\infty +\sup _{s\le t} e^{\lambda s} \Big \Vert \frac{w}{\langle v\rangle } \partial _t g(s)\Big \Vert _\infty \nonumber \\&\quad + \sup _{s\le t}e^{\lambda s}|\langle v\rangle w \partial _t r(s)|_\infty \}. \end{aligned}$$

(4.16)

In order to apply (4.16), we consider the following iteration sequence:

$$\begin{aligned} \partial _t f^{\ell +1} + v\cdot \nabla _x f^{\ell +1}+L( f^{\ell +1}) = \Gamma (f_s, f^\ell )+\Gamma ( f^\ell ,f_s)+\Gamma ( f^\ell , f^\ell ).\nonumber \\ \end{aligned}$$

(4.17)

The temporal derivative of (4.17) reads as

$$\begin{aligned}{} & {} \partial _t (\partial _t f^{\ell +1})+v\cdot \nabla _x \partial _t f^{\ell +1}+L( \partial _t f^{\ell +1}) \nonumber \\{} & {} \quad =\partial _t [ \Gamma (f_s, f^\ell )+\Gamma ( f^\ell ,f_s)+\Gamma ( f^\ell , f^\ell )], \end{aligned}$$

(4.18)

with boundary condition and initial condition given by

$$\begin{aligned} \partial _t f^{\ell +1}|_{\gamma _-}= & {} P_\gamma \partial _tf^{\ell +1}+\frac{M_W(x,v) - \mu }{\sqrt{\mu }}\int _{n(x)\cdot u>0}\partial _t f^{\ell }\sqrt{\mu }\{n(x)\cdot u\} \textrm{d}u,\\ \partial _t f^{\ell +1}(0,x,v)= & {} \partial _t f_0(x,v). \end{aligned}$$

The initial sequence is defined to be the initial condition $\partial _t f^0(t,x,v) = \partial _t f_0(x,v)$.

We first check all assumptions for (4.16) hold and then apply (4.16) to the iteration sequence (4.18).

We check the condition $\iint \partial _t f_0\sqrt{\mu }=0$. From the definition of $\partial _t f_0(x,v)$ in (2.12), we apply Green’s identity to get that

$$\begin{aligned}&\iint v\cdot \nabla _x f_0 \sqrt{\mu } \\&\quad = \int _{n(x)\cdot v>0} [n(x)\cdot v] f_0 \sqrt{\mu } + \int _{n(x)\cdot v<0} [n(x)\cdot v] f_0 \sqrt{\mu } \\&\quad = \int _{n(x)\cdot v>0} [n(x)\cdot v] f_0 \sqrt{\mu } + \int _{n(x)\cdot v<0} [n(x)\cdot v]\sqrt{\mu } \frac{M_W(x,v)}{\sqrt{\mu (v)}} \\&\qquad \int _{n(x)\cdot u>0} f_0(x,u)[n(x)\cdot u] \sqrt{\mu (u)}\\&\quad = \int _{n(x)\cdot v>0} [n(x)\cdot v] f_0 \sqrt{\mu } - \int _{n(x)\cdot u>0} f_0(x,u)[n(x)\cdot u] \sqrt{\mu (u)}=0. \end{aligned}$$

In the third line, we applied the compatibility condition (2.17) for $f_0$ and used

$$\begin{aligned} \int _{n(x)\cdot v<0} M_W(x,v)[ n(x)\cdot v]=-1. \end{aligned}$$

It is clear from the mass conservation that $\iint [-L(f_0)+\Gamma (f_s,f_0)+\Gamma (f_0,f_s)+\Gamma (f_0,f_0)]\sqrt{\mu } =0$. These estimates lead to the desired condition $\iint \partial _t f_0(x,v) \sqrt{\mu } = 0$.

For $\partial _t r$ and $\partial _t g$, we have that

$$\begin{aligned} \int _{v\in {{\mathbb {R}}}^3} \frac{M_W(x,v)-\mu }{\sqrt{\mu }}\sqrt{\mu } [n(x)\cdot v] = 0, \end{aligned}$$

and again from mass conservation,

$$\begin{aligned} \iint \partial _t [ \Gamma (f_s, f^\ell )+\Gamma ( f^\ell ,f_s)+\Gamma ( f^\ell , f^\ell )] \sqrt{\mu } =0. \end{aligned}$$

The conditions for (4.16) are satisfied. We apply (4.16) to (4.18) and get that

$$\begin{aligned}&\Vert w\partial _t f^{\ell +1}(t)\Vert _\infty + |w\partial _t f^{\ell +1}(t)|_\infty \nonumber \\&\quad \lesssim e^{-\lambda t}\Big \{\Vert w\partial _t f_0\Vert _\infty + \sup _{s\le t}e^{\lambda s}\Big |\langle v\rangle w \frac{M_W(x,v)-\mu }{\sqrt{\mu }}\int _{n(x)\cdot u>0}\partial _t f^{\ell }(s)\sqrt{\mu }\{n(x)\cdot u\}\Big |_\infty \nonumber \\&\qquad + \sup _{s\le t}e^{\lambda s}\Big [\Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f_s,f^{\ell })(s)\Big \Vert _\infty +\Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f^{\ell },f_s)(s)\Big \Vert _\infty \nonumber \\&\qquad +\Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f^{\ell },f^{\ell })(s)\Big \Vert _\infty \Big ]\Big \} . \end{aligned}$$

(4.19)

For the third line in (4.19), with $w=e^{\theta |v|^2}$, by (3.29) and (4.1), we have that

$$\begin{aligned} \Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f^\ell ,f_s)(s)\Big \Vert _\infty&= \Big \Vert \frac{w}{\langle v\rangle }\Gamma (\partial _t f^\ell (s),f_s)\Big \Vert _\infty \lesssim \Vert wf_s\Vert _\infty \Vert w\partial _t f^\ell (s) \Vert _\infty \\&= o(1)\Vert w\partial _t f^\ell (s)\Vert _\infty . \end{aligned}$$

Similarly,

$$\begin{aligned} \Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f_s,f^\ell )(s)\Big \Vert _\infty \lesssim o(1) \Vert w\partial _t f^\ell (s) \Vert _\infty , \end{aligned}$$

and, by (2.18),

$$\begin{aligned} \Big \Vert \frac{w}{\langle v\rangle }\partial _t \Gamma (f^\ell ,f^\ell )(s)\Big \Vert _\infty&\lesssim \Big \Vert \frac{w}{\langle v\rangle }\Gamma (\partial _t f^\ell ,f^\ell )(s)\Big \Vert _\infty + \Big \Vert \frac{w}{\langle v\rangle }\Gamma (f^\ell ,\partial _t f^\ell )(s)\Big \Vert _\infty \\&\lesssim \Vert wf^\ell (s)\Vert _\infty \Vert w\partial _t f^\ell (s)\Vert _\infty =o(1)\Vert w\partial _t f^\ell (s)\Vert _\infty . \end{aligned}$$

To deal with the boundary term in the second line of (4.19), with $\theta <\frac{1}{4}$, there exists $\delta$ such that $\theta <\frac{1}{4}-\delta$. Then we have

$$\begin{aligned}&\langle v\rangle w\frac{M_W(x,v)-\mu }{\sqrt{\mu }} \\&\quad =\langle v\rangle e^{\theta |v|^2} \frac{e^{-|v|^2/2T_W(x)}/T_W(x) - e^{-|v|^2/2}}{e^{-|v|^2/4}} \\&\quad =\langle v\rangle e^{(\theta +\frac{1}{4})|v|^2}\times \Big [\frac{1}{T_W(x)}\big (e^{-\frac{|v|^2}{2T_W(x)}} - e^{-\frac{|v|^2}{2}} \big ) + e^{-\frac{|v|^2}{2}}\bigg (\frac{1}{T_W(x)}-1 \bigg ) \Big ]\\&\quad \lesssim \langle v\rangle e^{(\frac{1}{2}-\delta )|v|^2}\Big [ e^{-\frac{|v|^2}{2}}\Vert T_W-1\Vert _{L^\infty (\partial \Omega )} + \frac{|v|^2}{2} \big |\frac{1}{T_W(x)} -1 \big |e^{-\frac{|v|^2}{2\inf \{ T_W(x)\}}} \Big ] \\&\quad \lesssim \Vert T_W-1\Vert _{L^\infty (\partial \Omega )} \langle v\rangle [1+|v|^2] e^{(\frac{1}{2}-\delta -\frac{1}{2\inf \{T_W(x)\}})|v|^2} \\&\quad \lesssim \Vert T_W-1\Vert _{L^\infty (\partial \Omega )} \frac{1}{\delta } e^{(\frac{1}{2}-3\delta -\frac{1}{2\inf \{T_W(x)\}})|v|^2} = o(1). \end{aligned}$$

In the fourth line, we applied mean value theorem. In the last line we used the Taylor expansion of $e^{x}$ to get that $|v|^2 \le \frac{e^{\delta |v|^2}}{\delta }$, and thus

$$\begin{aligned} \langle v\rangle [1+|v|^2] \lesssim 1+|v|^4\lesssim \frac{e^{2\delta |v|^2}}{\delta }. \end{aligned}$$

Also we take $\Vert T_W-1\Vert _{L^\infty (\partial \Omega )}$ to be small enough such that $\frac{\Vert T_W-1\Vert _{L^\infty (\partial \Omega )}}{\delta } \ll 1$ and $\frac{1}{2}-3\delta - \frac{1}{2\inf \{T_W(x)\}}<0$.

Then we control the boundary term in the second line (4.19) as

$$\begin{aligned} \Big | \langle v\rangle w \frac{M_W(x,v)-\mu }{\sqrt{\mu }}\int _{n(x)\cdot u>0}\partial _t f^{\ell }(s)\sqrt{\mu }\{n(x)\cdot u\}\Big |_\infty \lesssim o(1) | w\partial _t f^\ell (s)|_\infty . \end{aligned}$$

Hence we obtain the $L^\infty$ estimate for $\partial _t f^{\ell +1}$ in (4.19) as

$$\begin{aligned}&\Vert w\partial _t f^{\ell +1}(t)\Vert _\infty + |w\partial _t f^{\ell +1}(t)|_\infty \nonumber \\&\quad \lesssim e^{-\lambda t}\{\Vert w\partial _t f_0\Vert _\infty + o(1)\sup _{s\le t}e^{\lambda s}\Vert w\partial _t f^\ell (s)\Vert _\infty + o(1)\sup _{s\le t}e^{\lambda s}| w\partial _t f^\ell (s)|_\infty \}. \end{aligned}$$

(4.20)

This is equivalent to the fact that, for some $C>1$ and $c\ll 1$ and $c\times C\ll 1$, we have

$$\begin{aligned}&\Vert w\partial _t f^{\ell +1}(t)\Vert _\infty + |w\partial _t f^{\ell +1}(t)|_\infty \\&\quad \le C e^{-\lambda t}\{\Vert w\partial _t f_0\Vert _\infty + c\sup _{s\le t}e^{\lambda s}\Vert w\partial _t f^\ell (s)\Vert _\infty + c\sup _{s\le t}e^{\lambda s}| w\partial _t f^\ell (s)|_\infty \}. \end{aligned}$$

Next we use induction to prove that, for all $\ell \ge 0$,

$$\begin{aligned} \Vert w\partial _t f^{\ell }(t)\Vert _\infty + |w\partial _t f^{\ell }(t)|_\infty&\le 2Ce^{-\lambda t}\Vert w\partial _t f_0\Vert _\infty . \end{aligned}$$

Since the initial sequence is given by $\partial _t f^0(t,x,v)=\partial _t f_0(x,v)$, we have $\Vert w\partial _t f^{1}(t)\Vert _\infty + |w\partial _t f^{1}(t)|_\infty \le Ce^{-\lambda t}\Vert w\partial _t f_0\Vert _\infty$.

Assume that the induction hypothesis holds; i.e,

$$\begin{aligned} \Vert w\partial _t f^{\ell }(t)\Vert _\infty + |w\partial _t f^{\ell }(t)|_\infty \le 2Ce^{-\lambda t}\Vert w\partial _t f_0\Vert _\infty . \end{aligned}$$

Applying (4.20) to $\partial _t f^{\ell +1}$ yields that

$$\begin{aligned}&\Vert w\partial _t f^{\ell +1}(t)\Vert _\infty + |w\partial _t f^{\ell +1}(t)|_\infty \\&\quad \le C e^{-\lambda t}\{\Vert w\partial _t f_0\Vert _\infty + 2Cc\Vert w\partial _t f_0\Vert _\infty \} \le 2C e^{-\lambda t}\Vert w\partial _t f_0\Vert _\infty . \end{aligned}$$

The induction is valid and we conclude that, for all $\ell$,

$$\begin{aligned}&\sup _t e^{\lambda t}\Vert w\partial _t f^{\ell +1}(t)\Vert _\infty + \sup _t e^{\lambda t}|w\partial _t f^{\ell +1}(t)|_\infty \lesssim \Vert w\partial _t f_0\Vert _\infty . \end{aligned}$$

With the uniform-in-$\ell$ estimate, it is standard to take the difference $\partial _t f^{\ell +1}-\partial _t f^{\ell }$ to conclude that $\partial _t f^{\ell }$ is Cauchy sequence in $L^\infty$ and then pass the limit to conclude Corollary 4. $\square$

4.2 Preparation for (4.7).

For the estimate of the derivative to (4.7) at the boundary, we will use the boundary condition (2.11) with some properties from the convexity (2.2) in the next two lemmas.

Lemma 6

Given a $C^2$ convex domain in (2.2),

$$\begin{aligned} \begin{aligned}&|n_{p^{j}} (\textbf{x}_{p^{j}} ^{j}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )| \thicksim |x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) |^2, \ \ j=1,2, \\&{|\textbf{v}^{1}_{p^{1}, 3}| } / {|\textbf{v}^{1}_{p^{1} }|} \thicksim |x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) |. \end{aligned} \end{aligned}$$

(4.21)

For $j'=1,2$,

$$\begin{aligned} \bigg | \frac{\partial [ n_{p^{j}} (\textbf{x}_{p^{j}} ^{j}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )]}{\partial {\textbf{x}_{p^{2},j^\prime }^{2}}} \bigg | \lesssim \Vert \eta \Vert _{C^2} |x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2})| , \ \ j=1,2. \end{aligned}$$

(4.22)

Proof

First we prove (4.21). By Taylor’s expansion, for $x,y\in \partial \Omega$ and some $0\le t\le 1$,

$$\begin{aligned} \xi (y)-\xi (x)&=0-0=\nabla \xi (x)\cdot (y-x) + \frac{1}{2}(y-x)^T \nabla ^2 \xi (x+t(y-x)) (y-x). \end{aligned}$$

Thus from (2.3)

$$\begin{aligned} |n(x)\cdot (x-y)| \sim (y-x)^T \nabla ^2 \xi (x+t(y-x)) (y-x). \end{aligned}$$

From the convexity (2.2), we have

$$\begin{aligned} |n_{p^{j}} (\textbf{x}_{p^{j}} ^{j}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )|\ge C_\Omega |x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) |^2. \end{aligned}$$

Since $\xi$ is $C^2$ at least,

$$\begin{aligned} |\{x^{1}-y\}\cdot n(x^{1})|\le \Vert \xi \Vert _{C^2}|x^{1}-y|^2. \end{aligned}$$

Also, notice that

$$\begin{aligned} |n_{p^{1}} (\textbf{x}_{p^{1}} ^{1}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )|=|\textbf{v}^{1}_{p^{1}, 3}| (t^{2}-t^{1}), \end{aligned}$$

and thus

$$\begin{aligned} \begin{aligned} \frac{|\textbf{v}^{1}_{p^{1}, 3}| }{|\textbf{v}^{1}_{p^{1} }|}\ge&\frac{1}{|\textbf{v}^{1}_{p^{1} }|}\frac{C_\Omega }{|t^{1}-t^{2}|}\Big |x^{1}-x^{2} \Big |^2\\ =&C_\Omega |x^{1}-x^{2}| = C_\Omega |x^{1}-\eta _{p^{2}}(\textbf{x}_{p^{2}}^{2})|. \end{aligned} \end{aligned}$$

By the same computation, we can conclude that

$$\begin{aligned} \frac{|\textbf{v}^{1}_{p^{1}, 3}| }{|\textbf{v}^{1}_{p^{1} }|} \le C_\xi |x^{1}-\eta _{p^{2}}(\textbf{x}_{p^{2}}^{2})| . \end{aligned}$$

Then we prove (4.22). For $j=1$, $j'=1,2$ we apply (3.7) and (3.14) to obtain

$$\begin{aligned} \begin{aligned}&\bigg | \frac{\partial [ n_{p^{1}} (\textbf{x}_{p^{1}} ^{1}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )]}{\partial {\textbf{x}_{p^{2},j^\prime }^{2}}} \bigg |\le |n_{p^{1}} (\textbf{x}_{p^{1}} ^{1}) \cdot \partial _{j'}\eta _{p^{2}}(\textbf{x}_{p^{2}} ^{2})| \\&\quad =\Big |n_{p^{1}} (\textbf{x}_{p^{1}} ^{1}) \cdot \partial _{j'}\eta _{p^{1}}(\textbf{x}_{p^{1}} ^{1})+n_{p^{1}} (\textbf{x}_{p^{1}} ^{1}) \cdot \big [\partial _{j'}\eta _{p^{2}}(\textbf{x}_{p^{2}} ^{2})-\partial _{j'}\eta _{p^{1}}(\textbf{x}_{p^{1}} ^{1})\big ]\Big |\\&\quad \le 0+ \Vert \eta \Vert _{C^2} |x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2})|. \end{aligned}\nonumber \\ \end{aligned}$$

(4.23)

For $j=2$, we again apply (3.7) and (3.14) to get that

$$\begin{aligned} \begin{aligned} \bigg | \frac{\partial [ n_{p^{2}} (\textbf{x}_{p^{2}} ^{2}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2}) )]}{\partial {\textbf{x}_{p^{2},j^\prime }^{2}}} \bigg |\lesssim&|n_{p^{2}} (\textbf{x}_{p^{2}} ^{2}) \cdot \partial _{j'}\eta _{p^{2}}|+\Vert \eta \Vert _{C^2}|x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2})| \\ =&\Vert \eta \Vert _{C^2}|x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2})|. \end{aligned} \end{aligned}$$

$\square$

Lemma 7

The following map is one-to-one:

$$\begin{aligned} v^{1} \in \{ n(x^{1}) \cdot v^{1} >0: x_{\textbf{b}}(x^{1},v^{1}) \in B(p^{2}, \delta _2)\} \mapsto (\textbf{x}^{2}_{p^{2},1}, \textbf{x}^{2}_{p^{2},2}, t_{\textbf{b}}^{1}), \end{aligned}$$

(4.24)

with

$$\begin{aligned} \det \left( \frac{\partial (\textbf{x}^{2}_{p^{2},1}, \textbf{x}^{2}_{p^{2},2}, t_{\textbf{b}}^{1})}{\partial v^{1}}\right) = \frac{1}{\sqrt{ g_{p^{2},11}(\textbf{x}^{2}_{p^{2}}) g_{p^{2},22}(\textbf{x}_{p^{2}}^{2}) }} \frac{|t_{\textbf{b}}^{1}|^3}{ |n(x^{2}) \cdot v^{1}| }. \end{aligned}$$

(4.25)

Proof

Combining (3.17) and (3.21), we conclude that

$$\begin{aligned} \begin{aligned}&\det \left( \frac{\partial (\textbf{x}^{2}_{p^{2},1}, \textbf{x}^{2}_{p^{2},2}, t_{\textbf{b}}^{1})}{\partial v^{1}}\right) \\&\quad = |t_{\textbf{b}}^{1}|^3\det \left( \begin{array}{c} -\frac{1}{\textbf{v}^{1}_{p^{2},3}} \frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \\ \frac{1}{\sqrt{g_{p^{2},11}(\textbf{x}_{p^{2}}^{2})}} \Big [\frac{\partial _1 \eta _{p^{2}}}{\sqrt{g_{p^{2},11}}}\Big |_{x^{2}}- \frac{\textbf{v}_{p^{1},1}^{1}}{\textbf{v}_{p^{2},3}^{1}}\frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \Big ] \\ \frac{1}{\sqrt{g_{p^{2},22}(\textbf{x}_{p^{2}}^{2})}} \Big [\frac{\partial _2 \eta _{p^{2}}}{\sqrt{g_{p^{2},22}}}\Big |_{x^{2}}- \frac{\textbf{v}_{p^{1},2}^{1}}{\textbf{v}_{p^{2},3}^{1}}\frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \Big ] \\ \end{array} \right) \\&\quad = -|t_{\textbf{b}}^{1}|^3\frac{1}{\textbf{v}^{1}_{p^{2},3}} \frac{1}{\sqrt{g_{p^{2},11}(\textbf{x}_{p^{2}}^{2})g_{p^{2},22}(\textbf{x}_{p^{2}}^{2})}} \frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \\&\qquad \cdot \bigg (\Big [\frac{\partial _1 \eta _{p^{2}}}{\sqrt{g_{p^{2},11}}}\Big |_{x^{2}}- \frac{\textbf{v}_{p^{1},1}^{1}}{\textbf{v}_{p^{2},3}^{1}}\frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \Big ]\times \Big [\frac{\partial _2 \eta _{p^{2}}}{\sqrt{g_{p^{2},22}}}\Big |_{x^{2}}- \frac{\textbf{v}_{p^{1},2}^{1}}{\textbf{v}_{p^{2},3}^{1}}\frac{\partial _3 \eta _{p^{2}}}{\sqrt{g_{p^{2},33}}}\Big |_{x^{2}} \Big ] \bigg )\\&\quad =\frac{1}{\sqrt{g_{p^{2},11}(\textbf{x}_{p^{2}}^{2})g_{p^{2},22}(\textbf{x}_{p^{2}}^{2})}}\frac{|t_{\textbf{b}}^{1}|^3}{\textbf{v}^{1}_{p^{2},3}}=~(4.25), \end{aligned} \end{aligned}$$

where we have used  (3.7).

Now we prove the map (4.24) is one-to-one. Assume that there exists v and ${\tilde{v}}$ satisfy $x_{\textbf{b}}(x^{1},v)=x_{\textbf{b}}(x^{1},{\tilde{v}})$ and $t_{\textbf{b}}(x^{1},v)=t_{\textbf{b}}(x^{1},{\tilde{v}})$. We choose $p\in \partial \Omega$ near $x_{\textbf{b}}(x^{1},v)$ and use the same parametrization. Then by an expansion, for some ${{\bar{v}}\in \{a{\tilde{v}}+(1-a)v: a\in [0,1]\},}$

$$\begin{aligned}0=\left( \begin{array}{c} \textbf{x}_{p,1}(x^{1},{\tilde{v}}) \\ \textbf{x}_{p,2}(x^{1},{\tilde{v}})\\ t_{\textbf{b}}(x^{1},{\tilde{v}}) \\ \end{array} \right) -\left( \begin{array}{c} \textbf{x}_{p,1}(x^{1},v) \\ \textbf{x}_{p,2}(x^{1},v) \\ t_{\textbf{b}}(x^{1},v) \\ \end{array} \right) =\left( \begin{array}{c} \nabla _v \textbf{x}_{p,1}({ x^{1}},{\bar{v}}) \\ \nabla _v \textbf{x}_{p,2}({ x^{1}},{\bar{v}}) \\ \nabla _v t_{\textbf{b}}({ x^{1}},{\bar{v}}) \\ \end{array} \right) ({\tilde{v}}-v). \end{aligned}$$

This equality can be true only if the determinant of the Jacobian matrix equals zero. Then (4.25) implies that $t_{\textbf{b}}(x^{1},{\bar{v}})=0$. But this implies $x^{1}=x_{\textbf{b}}(x^{1},{\bar{v}})$ and hence $n(x^{1})\cdot {\bar{v}}=0$ which is out of our domain. $\square$

4.3 Preparation for (4.10) and (4.11).

Applying Lemma 5, we have the following estimate for (4.10) and (4.11) respectively:

Lemma 8

Recall the definition of h in (4.3). We have an $L^\infty$ control for h as

$$\begin{aligned} |w_{{\tilde{\theta }}}(v)h(t,x,v)|_\infty\lesssim & {} e^{-\lambda t}\big [(\sup _t e^{\lambda t}| wf(t)|_\infty )^2 + \sup _t e^{\lambda t}| wf(t)|_\infty | wf_s|_\infty \big ] \nonumber \\\lesssim & {} e^{-\lambda t} \sup _t e^{\lambda t}|wf(t)|_\infty . \end{aligned}$$

(4.26)

The derivative along the characteristic is bounded as

$$\begin{aligned}&\Big |\int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\nu (t-s)} \nabla _x h(s,x-(t-s)v,v) \textrm{d}s \Big | \nonumber \\&\quad \lesssim e^{-\lambda t}\Big [\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \frac{\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)} +\Vert wf_s\Vert _\infty \frac{\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s) \Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \nonumber \\&\qquad + \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \frac{\sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\Big ] \nonumber \\&\quad =\frac{o(1)e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\big [\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty + \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty \big ]. \end{aligned}$$

(4.27)

Proof

First we prove (4.26). From the definition of h(t, x, v) in (4.3), we apply (3.29) with $w_{{\tilde{\theta }}}(v)\lesssim \frac{w(v)}{\langle v\rangle }$ to get that

$$\begin{aligned} |w_{{\tilde{\theta }}}(v)h(t,x,v)|_\infty&\lesssim |wf(t)|_\infty \times |wf_s|_\infty + |wf(t)|_\infty ^2 \\&\lesssim e^{-\lambda t}\big [\sup _t e^{\lambda t}|wf(t)|_\infty \times |wf_s|_\infty + \sup _t e^{2\lambda t} |wf(t)|_\infty ^2 \big ]. \end{aligned}$$

This concludes the first inequality in (4.26). The second inequality in (4.26) follows from $|wf_s|_\infty \ll 1$ and $e^{\lambda t}|wf(t)|_\infty \ll 1$ from (2.18) and (4.1).

Then we prove (4.27). First we consider the contribution of $\Gamma (f,f_s)$ and compute that

$$\begin{aligned} \Big |\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}\nabla _x \Gamma (f,f_s)(s,x-(t-s)v,v) \textrm{d}s \Big |. \end{aligned}$$

We bound $\nabla _x\Gamma (f,f_s)$ by (3.31), the contribution of the first term on RHS of (3.31) is bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}}\textrm{d}s e^{-\nu (t-s)} \Big [{\langle v\rangle }\Vert wf(s)\Vert _\infty |\nabla _x f_s(x,v)|\Big ] \\&\quad \le \int ^t_{\max \{0,t-t_{\textbf{b}}\}}\textrm{d}s e^{-\nu (t-s)} \langle v\rangle \Big [ e^{-\lambda s} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \frac{\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\Big ] \\&\quad \lesssim \frac{e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)/2} \langle v\rangle \textrm{d}s \\&\quad \lesssim \frac{e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

In the second line, we used $\lambda \ll 1$ and $1\lesssim \nu (v)$ from (3.30), so that $e^{-\nu (t-s)}e^{-\lambda s}\le e^{-\lambda t}e^{-\nu (t-s)/2}$.

The contribution of the second term of the RHS of (3.31) is bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}} \textrm{d}s e^{-\nu (t-s)} e^{-\lambda s}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u)\frac{\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s \Vert _\infty }{w_{{\tilde{\theta }}}(u)\alpha (x-(t-s)v,u)} \\&\quad \lesssim \frac{e^{-\lambda t}\sup _t e^{\lambda t} \Vert wf(t)\Vert _\infty \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{ w_{{\tilde{\theta }}}(v)}\\&\qquad \int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\frac{\nu (t-s)}{2}} \int _{{{\mathbb {R}}}^3} \frac{\textbf{k}(v,u)w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)\alpha (x-(t-s)v,u)} \textrm{d}u \textrm{d}s \\&\quad \lesssim \frac{e^{-\lambda t}\sup _t e^{\lambda t} \Vert wf(t)\Vert _\infty \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)}\\&\qquad \int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\frac{\nu (t-s)}{2}} \int _{{{\mathbb {R}}}^3} \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{\alpha (x-(t-s)v,u)} \textrm{d}u \textrm{d}s\\&\quad \lesssim \frac{ e^{-\lambda t}\sup _t e^{\lambda t} \Vert wf(t)\Vert _\infty \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

In the third line, we have applied (3.24) Lemma 2. In the last line, we have applied Lemma 5.

By a similar computation, the contribution of the third term of the RHS of (3.31) is bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}} \textrm{d}s e^{-\nu (t-s)} e^{-\lambda s}\Vert wf_s\Vert _\infty \int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u)\frac{e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s) \Vert _\infty }{w_{{\tilde{\theta }}}(u)\alpha (x-(t-s)v,u)} \\&\quad \lesssim \frac{e^{-\lambda t}\Vert wf_s\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{ w_{{\tilde{\theta }}}(v)}\\&\qquad \int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\frac{\nu (t-s)}{2}} \int _{{{\mathbb {R}}}^3} \frac{\textbf{k}(v,u)w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)\alpha (x-(t-s)v,u)} \textrm{d}u \textrm{d}s \\&\quad \lesssim \frac{ e^{-\lambda t}\Vert wf_s\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

Thus the contribution of $\Gamma (f,f_s)$ in the h of (4.27) corresponds to the second line in (4.27).

Then we consider the contribution of $\Gamma (f_s,f)$ and bound

$$\begin{aligned} \Big |\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}\nabla _x \Gamma (f_s,f)(s,x-(t-s)v,v) \textrm{d}s \Big |. \end{aligned}$$

Again by (3.31), the contribution of the first term on RHS is bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}\langle v\rangle \Big [\Vert wf_s\Vert _\infty e^{-\lambda s} \frac{\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \Big ] \textrm{d}s \\&\quad \lesssim \frac{e^{-\lambda t}\Vert wf_s\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

The contribution of the second and third terms on RHS of (3.31) are bounded using the same computation for $\nabla _x \Gamma (f,f_s)$. Thus the contribution of $\Gamma (f_s,f)$ in h of (4.27) corresponds to the second line in (4.27).

Last we consider the contribution of $\Gamma (f,f)$ and bound

$$\begin{aligned} \Big |\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}\nabla _x \Gamma (f,f)(s,x-(t-s)v,v) \textrm{d}s \Big |. \end{aligned}$$

By (3.31), the contribution of the first term on RHS is bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)}e^{-2\lambda s} \langle v\rangle \Big [\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \frac{\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \Big ] \textrm{d}s \\&\quad \lesssim \frac{e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

The contribution of the second and third terms on RHS of (3.31) are bounded as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\nu (t-s)} e^{-2\lambda s}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \int _{{{\mathbb {R}}}^3} \textbf{k}(v,u)\frac{\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s) \Vert _\infty }{w_{{\tilde{\theta }}}(u)\alpha (x-(t-s)v,u)} \textrm{d}u \textrm{d}s \\&\quad \lesssim e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty w_{{\tilde{\theta }}}^{-1}(v) \\&\qquad \times \int ^t_{\max \{0,t-t_{\textbf{b}}\}}e^{-\nu (t-s)/2} \int _{{{\mathbb {R}}}^3} \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{\alpha (x-(t-s)v,u)}\\&\quad \lesssim \frac{ e^{-\lambda t}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

Thus the contribution of $\Gamma (f,f)$ in h of (4.27) corresponds to the third line of (4.27).

We conclude the first inequality in (4.27). The second inequality follows by

$\Vert wf_s\Vert _\infty \ll 1$ and $\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \ll 1$ from (4.1) and (2.18). $\square$

4.4 Proof of Theorem 2

We prove Theorem 2 by estimating each term in (4.4)–(4.11).

For fixed and small $\varepsilon \ll 1$, we denote that

$$\begin{aligned} {\mathcal {B}}&:= \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty + \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_0\Vert _\infty \nonumber \\&\quad + \sup _t e^{\lambda t}|w\partial _t f(t)|_\infty + \sup _t e^{\lambda t}|wf(t)|_\infty + \varepsilon ^{-1}\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty . \nonumber \\ \end{aligned}$$

(4.28)

Here we note that $\sup _t e^{\lambda t}| w\partial _t f(t)|_\infty$ is bounded in Corollary 4,

$\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty , \ \sup _t e^{\lambda t}|wf(t)|_\infty$ are bounded in (2.18) and $\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty$ is bounded in (4.2).

We will prove the following estimate:

$$\begin{aligned} |\nabla _x f(t,x,v)| \lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times [{\mathcal {B}}+ o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty ].\nonumber \\ \end{aligned}$$

(4.29)

We present the proof of the theorem in four steps.

In Step 1, we estimate (4.4), (4.5), (4.6), (4.9), (4.10), and (4.11).

In Step 2, we estimate (4.7) using the boundary condition (2.11) with the representation (3.12). We will apply the characteristic (4.4) - (4.11), so that the estimate in Step 1 can be utilized.

In Step 3, we estimate (4.8). We will apply the characteristic (4.4)–(4.11), and thus the estimates in Step 1 and Step 2 can be utilized.

Finally, in Step 4, we summarize all the estimates and conclude the theorem.

In fact, Theorem 2 follows directly from (4.29) as illustrated in Step 4.

Step 1: estimate of (4.4), (4.5), (4.6), (4.9), (4.10) and (4.11).

From $\lambda \ll 1$ and $\nu (v)\gtrsim 1$, (4.4), (4.5), (4.6), (4.9) are bounded as

$$\begin{aligned}&|(4.4)| + |(4.5)| + |(4.6)| + |(4.9)| \nonumber \\&\quad \lesssim e^{-\lambda t}\frac{w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)} \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_0 \Vert _\infty + \frac{w^{-1}(v)\nu }{\alpha (x,v)}e^{-\lambda (t_{\textbf{b}}+t-t_{\textbf{b}})} e^{\lambda (t-t_{\textbf{b}})}| wf(t-t_{\textbf{b}})|_\infty \nonumber \\&\qquad + \frac{w^{-1}(v)}{\alpha (x,v)} e^{-\lambda (t_{\textbf{b}}+t-t_{\textbf{b}})} e^{\lambda (t-t_{\textbf{b}})} | w\partial _tf(t-t_{\textbf{b}})|_\infty + \frac{w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}e^{-\lambda (t_{\textbf{b}}+t-t_{\textbf{b}})} | wf(t-t_{\textbf{b}})|_\infty \nonumber \\&\quad \lesssim { \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times {\mathcal {B}}. } \end{aligned}$$

(4.30)

${\mathcal {B}}$ is defined in (4.28). Here we applied (3.18) and (3.34) to $\nabla _x t_{\textbf{b}}$ and (3.28) to (4.9).

(4.10) and (4.11) are bounded using (4.26) and (4.27):

$$\begin{aligned}&|(4.10)| + |(4.11)| \nonumber \\&\quad \lesssim \frac{o(1)e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\big [\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty + \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty \big ]\nonumber \\&\qquad +\frac{e^{-\nu t_{\textbf{b}}}e^{-\lambda (t-t_{\textbf{b}})} \sup _t e^{\lambda t} |wf(t)|_\infty }{\alpha (x,v)w_{{\tilde{\theta }}}(v)}\nonumber \\&\quad \lesssim { \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s} \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty ]. } \end{aligned}$$

(4.31)

Step 2: estimate of (4.7).

(4.7) corresponds to the boundary condition (2.11). We use the representation in (3.12) and apply (3.19) with (3.34) to bound it as

$$\begin{aligned}&|(4.7)| \nonumber \\&\quad \lesssim \frac{e^{-\nu t_{\textbf{b}}}|v| \big |\partial _{\textbf{x}_{p^1,j}^1} M_W(\eta _{p^1}(\textbf{x}_{p^1}^1),v)\big |}{\sqrt{\mu (v)}\alpha (x,v)}\nonumber \\&\qquad \times \Big |\int _{n(x^1)\cdot v^1>0} f(t^1,x^1,v^1)\sqrt{\mu (v^1)}[n(x^1)\cdot v^1]\textrm{d}v^1 \Big | \end{aligned}$$

(4.32)

$$\begin{aligned}&\quad +\frac{e^{-\nu t_{\textbf{b}}}|v|M_W(\eta _{p^1}(\textbf{x}_{p^1}^1),v)}{\sqrt{\mu (v)}\alpha (x,v)}\nonumber \\&\qquad \times \Big |\int _{\textbf{v}^{1}_{p^{1},3}>0} \underbrace{\partial _{\textbf{x}^{1}_{p^{1},j}} [f(t^1, \eta _{p^{1}} ( \textbf{x}_{p^{1} }^{1} ), T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}}) ]}_{(4.33)_*}\sqrt{\mu (\textbf{v}^{1}_{p^{1}})} \textbf{v}^{1}_{p^{1},3} \textrm{d}\textbf{v}^{1}_{p^{1}}\Big |. \end{aligned}$$

(4.33)

Here we recall that $x^1, v^1$ and $t^1$ are defined in Definition 2.

The second term in (4.32) is directly bounded by

$$\begin{aligned} e^{-\lambda t^1} \sup _t e^{\lambda t}|wf(t)|_\infty = e^{-\lambda (t-t_{\textbf{b}})} \sup _t e^{\lambda t}|wf(t)|_\infty . \end{aligned}$$

(4.34)

For the first term in (4.32), we directly apply the derivative to (1.4) to bound it as

$$\begin{aligned} \frac{|v| \Big | \partial _{\textbf{x}_{p^1,j}^1}M_W(\eta _{p^1}(\textbf{x}_{p^1}^1),v)\Big |}{\sqrt{\mu (v)}}&\lesssim \Vert \eta \Vert _{C^1}\Vert T_W\Vert _{C^1(\partial \Omega )}[1+|v|]\frac{M_W(x,v)}{\sqrt{\mu (v)}}\lesssim w^{-1}_{{\tilde{\theta }}}(v) . \end{aligned}$$

Here we used ${\tilde{\theta }}\ll 1$ and $\Vert T_W-1\Vert _{L^\infty (\partial \Omega )} \ll 1$.

Thus combining with the extra $e^{-\lambda (t-t_{\textbf{b}})}$ from (4.34), we bound

$$\begin{aligned} (4.32)&\lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times \sup _t e^{\lambda t}|wf(t)|_\infty \lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times {\mathcal {B}}. \end{aligned}$$

(4.35)

Now we consider (4.33). First we consider the velocity derivative in $(4.33)_*$ given by

$$\begin{aligned} \begin{aligned}&\left( \partial _{\textbf{x}^{1}_{p^{1},j}} T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}}\right) \cdot \nabla _v f(t^1 ,\eta _{p^{1}} ( \textbf{x}_{p^{1} }^{1} ), T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}})\\&\quad = \sum _{l,m} \frac{\partial }{ \partial {\textbf{x}^{1}_{p^{1}, j}} }\left( \frac{\partial _m \eta _{p^{1}, l} (\textbf{x}^{1}_{p^{1}})}{\sqrt{g_{p^{1},mm}(\textbf{x}^{1}_{p^{1}})} }\right) \textbf{v}_{p^{1},m}^{1}\partial _{v_l}f(t^1,\eta _{p^{1}} ( \textbf{x}_{p^{1} }^{1} ), T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}})\\&\quad = \sum _{ m,n} (4.37)_{mn} \textbf{v}_{p^{1},m}^{1} \partial _{\textbf{v}^{1}_{p^{1},n}} [f(t^1, \eta _{p^{1}} ( \textbf{x}_{p^{1} }^{1} ) , T^t_{\textbf{x}^{1}_{p^{1}}}\textbf{v}^{1}_{p^{1}})], \end{aligned}\nonumber \\ \end{aligned}$$

(4.36)

where

$$\begin{aligned} (4.37)_{mn} :=\sum _l \frac{\partial }{ \partial {\textbf{x}^{1}_{p^{1}, j}} }\left( \frac{\partial _m \eta _{p^{1},l} (\textbf{x}^{1}_{p^{1}})}{\sqrt{g_{p^{1},mm}(\textbf{x}^{1}_{p^{1}})} }\right) \frac{\partial _n \eta _{p^{1},l} (\textbf{x}^{1}_{p^{1}})}{\sqrt{g_{p^{1},nn}(\textbf{x}^{1}_{p^{1}})} } . \end{aligned}$$

(4.37)

Then we apply an integration by parts with respect to $\partial _{\textbf{v}_{p^{1}}^{1}}$ to derive that

$$\begin{aligned} \begin{aligned}&\Big |\int _{\textbf{v}^{1}_{p^{1},3}>0} f(t^1,\eta _{p^{1}} ( \textbf{x}_{p^{1}}^{1} ), T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}}) \sum _{m,n} (4.37)_{mn} \partial _{\textbf{v}^{1}_{p^{1},n}} \big [\textbf{v}^{1}_{p^{1}, m} \textbf{v}^{1}_{p^{1},3}\sqrt{\mu (\textbf{v}^{1}_{p^{1}})}\big ] \textrm{d}\textbf{v}^{1}_{p^{1}} \Big | \\&\quad \lesssim e^{-\lambda (t-t_{\textbf{b}})} e^{\lambda (t-t_{\textbf{b}})} | wf(t-t_{\textbf{b}})|_\infty . \end{aligned}\nonumber \\ \end{aligned}$$

(4.38)

Here we notice that

$$\begin{aligned} f(t^1,\eta _{p^{1}} ( \textbf{x}_{p^{1} }^{1} ), T^t_{\textbf{x}^{1}_{p^{1}}} \textbf{v}^{1}_{p^{1}}) \sum _{m,n} (4.37)_{mn}\textbf{v}^{1}_{p^{1}, m} \textbf{v}^{1}_{p^{1},3} \sqrt{\mu (\textbf{v}^{1}_{p^{1}})}\equiv 0\end{aligned}$$

when $\textbf{v}^{1}_{p^{1},3}=0$ for ${| wf(t-t_{\textbf{b}})|_\infty }<\infty$.

We conclude that the contribution of (4.36) in (4.33), i.e, the partial derivative with respect to the velocity variable, is bounded by

$$\begin{aligned}&\frac{e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}\sup _t e^{\lambda t}| wf(t)|_\infty \le \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times {\mathcal {B}}. \end{aligned}$$

(4.39)

Next we consider the spatial derivative in $(4.33)_*$, which is given by

$$\begin{aligned}&\int _{n(x^1)\cdot v^1 > 0} \partial _{\textbf{x}_{p^1,j}^1}f(t^1, \eta _{p^1}(\textbf{x}_{p^1}^1),v^1) \sqrt{\mu (v^1)} [n(x^1)\cdot v^1] \textrm{d}v^1. \end{aligned}$$

(4.40)

Here we rewrite the velocity variable using the Cartesian coordinate $v^1$.

Note that from (3.15), we have $\partial _{\textbf{x}_{p^1,j}^1} f(t^1,\eta _{p^1}(\textbf{x}_{p^1}^1),v^1) = \partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \nabla _x f(t^1,\eta _{p^1}(\textbf{x}_{p^1}^1),v^1)$. Since $n(x^1)\cdot v^1>0$, we can expand $\nabla _x f(t^1,\eta _{p^1}(\textbf{x}_{p^1}^1),v^1)$ along the characteristic using (4.4)–(4.11).

The contribution of (4.4), (4.5), (4.6), (4.9), (4.10) and (4.11) in the expansion of $\nabla _x f(t^1,\eta _{p^1}(\textbf{x}_{p^1}^1),v^1)$ can be computed similarly as (4.30) and (4.31), where we just need to replace t by $t^1$ and (x, v) by $(x^1,v^1)$. Thus the contribution of these six terms in (4.40) are bounded by

$$\begin{aligned}&\int _{n(x^1)\cdot v^1>0} \frac{e^{-\lambda (t-t_{\textbf{b}})}}{\alpha (x^1,v^1)}[n(x^1)\cdot v^1]\sqrt{\mu (v^1)}[{\mathcal {B}} + o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty ] \textrm{d}v^1 \nonumber \\&\quad \lesssim e^{-\lambda (t-t_{\textbf{b}})} [{\mathcal {B}} + o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty ], \end{aligned}$$

(4.41)

where we have used (2.13) and (3.34) to cancel the singularity of $\frac{1}{\alpha (x^1,v^1)}$ as follows:

$$\begin{aligned}&\frac{|n(x^1)\cdot v^1|}{\alpha (x^1,v^1)} \sqrt{\mu (v^1)} = \textbf{1}_{\alpha (x^1,v^1) \ge \frac{1}{2}} + \textbf{1}_{\alpha (x^1,v^1) < \frac{1}{2}} \nonumber \\&\quad \le 2|n(x^1)\cdot v^1| \sqrt{\mu (v^1)} + \frac{|n(x^1)\cdot v^1|}{{\tilde{\alpha }}(x^1,v^1)} \lesssim [|n(x^1)\cdot v^1| + 1]\sqrt{\mu (v^1)}. \end{aligned}$$

(4.42)

Next we consider the contribution of (4.8) in (4.40), which reads as

$$\begin{aligned}&\Big |\int _{n(x^1)\cdot v^1>0} \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1]\partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \nonumber \\&\quad \times \int _{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}}^{t^1} \textrm{d}s e^{-\nu (v^1)(t^1-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v^1,u) \nabla _x f(s,x^1-(t^1-s)v^1,u) \Big |. \end{aligned}$$

(4.43)

We separate the $\textrm{d}s$ integral into $t^1-s<\varepsilon$ and $t^1-s\ge \varepsilon$. When $t^1-s<\varepsilon$, we denote the above term as

$$\begin{aligned} (4.43)_1&:= \Big |\int _{n(x^1)\cdot v^1>0} \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \nonumber \\&\quad \times \int _{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1),t^1-\varepsilon \}}^{t^1} \textrm{d}s e^{-\nu (v^1)(t^1-s)}\nonumber \\&\quad \int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v^1,u) \nabla _x f(s,x^1-(t^1-s)v^1,u) \Big |\nonumber \\&\lesssim \int _{n(x^1)\cdot v^1>0} \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \int _{\max \{t^1-t_{\textbf{b}}(x^1,v^1),t^1-\varepsilon \}}^{t^1} \textrm{d}s e^{-\nu (v^1)(t^1-s)} \nonumber \\&\quad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v^1,u)e^{-\lambda s}}{\alpha (x^1-(t^1-s)v^1,u)} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

(4.44)

Note that from (3.30), we have $\nu (v^1) \thicksim \sqrt{|v^1|^2+1}$, together with $\lambda \ll 1$, we derive $e^{-\nu (v^1)(t^1-s)}e^{-\lambda s^1}\le e^{-\nu (v^1)(t^1-s)/2} e^{-\lambda t^1}$. We further bound

$$\begin{aligned} (4.44)&\lesssim e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \int _{n(x^1)\cdot v^1>0 } \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \nonumber \\&\quad \times \int _{\max \{t^1-\varepsilon ,t^1-t_{\textbf{b}}(x^1,v^1)\}}^{t^1} \textrm{d}s e^{-\nu (v^1)(t^1-s)/2} \int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v^1,u)}{\alpha (x^1-(t^1-s)v^1,u)} \nonumber \\&\lesssim o(1) e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \int _{n(x^1)\cdot v^1>0} \sqrt{\mu (v^1)}\frac{[n(x^1)\cdot v^1]}{\alpha (x^1,v^1)}\textrm{d}v^1 \nonumber \\&\lesssim o(1) e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

(4.45)

Here we have applied Lemma 5 to get the third line, and we applied (4.42) in the last line.

When $t^1-s\ge \varepsilon$, we further split the $v^1$ integration into $n(x^1)\cdot v^1<\delta$ and $n(x^1)\cdot v^1 > \delta$. For $n(x^1)\cdot v^1<\delta$, we have

$$\begin{aligned}&\Big |\int _{n(x^1)\cdot v^1<\delta } \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \nonumber \\&\qquad \times \int _{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}}^{t^1-\varepsilon } \textrm{d}s e^{-\nu (v^1)(t^1-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v^1,u) \nabla _x f(s,x^1-(t^1-s)v^1,u) \Big | \nonumber \\&\quad \lesssim e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \int _{n(x^1)\cdot v^1< \delta } \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \nonumber \\&\qquad \times \int _{\max \{t^1-t_{\textbf{b}}(x^1,v^1)\}}^{t^1-\varepsilon } \textrm{d}s e^{-\nu (v^1)(t^1-s)/2} \int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v^1,u)}{\alpha (x^1-(t^1-s)v^1,u)} \nonumber \\&\quad \lesssim e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \int _{n(x^1)\cdot v^1<\delta } \sqrt{\mu (v^1)}\frac{[n(x^1)\cdot v^1]}{\alpha (x^1,v^1)}\textrm{d}v^1 \nonumber \\&\quad \lesssim o(1) e^{-\lambda t^1} \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

(4.46)

Here we applied Lemma 5 in the second last line.

For $n(x^1)\cdot v^1 > \delta$ and $t^1-s\ge \varepsilon$, we use the following change of variable

$$\begin{aligned} \nabla _x f(s,x^1-(t^1-s)v^1,u) = -\frac{\nabla _{v^1}[f(s,x^1-(t^1-s)v^1,u)]}{t^1-s}. \end{aligned}$$

(4.47)

In such case, without loss of generality, we assume $t^1>\varepsilon$, otherwise, we directly bound (4.43) by (4.45). Then in (4.43), we only integrate over the $v^1$ such that $t_{\textbf{b}}(x^1,v^1)>\varepsilon$. Then we denote

$$\begin{aligned} V: = \{v^1\in {{\mathbb {R}}}^3: n(x^1)\cdot v^1> \delta , \ t_{\textbf{b}}(x^1,v^1)>\varepsilon \}. \end{aligned}$$

We also denote the corresponding hypersurface as ${\tilde{V}}$. Since $n(x^1)\cdot v^1>\delta$, by (3.18), ${\tilde{V}}$ is piecewise smooth.

In such case, we denote (4.43) as

$$\begin{aligned} (4.43)_2&:= \Big |\int _{V} \textrm{d}v^1 \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \\&\quad \times \int _{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}}^{t^1-\varepsilon } \textrm{d}s e^{-\nu (v^1)(t^1-s)}\\&\quad \int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v^1,u) \frac{-\nabla _{v^1} f(s,x^1-(t^1-s)v^1,u)}{t^1-s} \Big |\nonumber . \end{aligned}$$

Here we applied (4.47).

Then we apply an integration by part in $v^1$ to get that

$$\begin{aligned} \begin{aligned}&(4.43)_2 \\&\quad = \Big |\int _{V} \nabla _{v^1}[\sqrt{\mu (v^1)}[n(x^1)\cdot v^1]]\partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \\&\qquad \times \int ^{t^1-\varepsilon }_{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}} \textrm{d}s e^{-\nu (v^1)(t^1-s)} \int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v^1,u) \frac{f(s,x^1-(t^1-s)v^1,u)}{t^1-s} \\&\qquad + \int _{V} \sqrt{\mu (v^1)}[n(x^1)\cdot v^1]\partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \times \int ^{t^1-\varepsilon }_{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}} \textrm{d}s \\&\qquad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u \nabla _{v^1}[ e^{-\nu (v^1)(t^1-s)}\textbf{k}(v^1,u)] \frac{f(s,x^1-(t^1-s)v^1,u)}{t^1-s} \\&\qquad +\int _{V} \sqrt{\mu (v^1)}[n(x^1)\cdot v^1]\partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \textbf{1}_{t^1-t_{\textbf{b}}(x^1,v^1)>0}\\&\qquad \times \nabla _{v^1} t_{\textbf{b}}(x^1,v^1) e^{-\nu (v^1)t_{\textbf{b}}(x^1,v^1)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v^1,u) \frac{f(t^1-t_{\textbf{b}}(x^1,v^1),x_{\textbf{b}}(x^1,v^1),u)}{t_{\textbf{b}}(x^1,v^1)} \\&\qquad - \int _{{\tilde{V}}} \sqrt{\mu (v^1)}[n(x^1)\cdot v^1] \partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \textbf{n} \int ^{t^1-\varepsilon }_{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}} \textrm{d}s \\&\qquad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u e^{-\nu (v^1)(t^1-s)}\textbf{k}(v^1,u) \frac{f(s,x^1-(t^1-s)v^1,u)}{t^1-s}\Big | . \end{aligned}\nonumber \\ \end{aligned}$$

(4.48)

In the last term $\textbf{n} = \frac{-\nabla _{v^1} t_{\textbf{b}}(x^1,v^1)}{|\nabla _{v^1} t_{\textbf{b}}(x^1,v^1)|}$. Then the last term is bounded using $\Vert w f(s)\Vert _{\infty }<\infty$ by

$$\begin{aligned}&\varepsilon ^{-1} e^{-\lambda t^1} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty . \end{aligned}$$

(4.49)

Here we applied $e^{-\nu (v^1)(t^1-s)}e^{-\lambda s^1}\le e^{-\nu (v^1)(t^1-s)/2} e^{-\lambda t^1}$, $\textbf{k}(v^1,u)\in L^1_u$ and $t^1-s\ge \varepsilon$.

Again by the three conditions above, the first term on the RHS of (4.48) is bounded as

$$\begin{aligned}&\varepsilon ^{-1} e^{-\lambda t^1} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty . \end{aligned}$$

(4.50)

For the second term in (4.48), we use (3.25) with ${\tilde{\theta }} \ll \theta$, and $|v^1|^2 w^{-1}_{{\tilde{\theta }}}(v^1)\lesssim 1$ to get that

$$\begin{aligned} |\nabla _{v^1}\textbf{k}(v^1,u)| |f(s,u)|&\le \Vert wf(s)\Vert _\infty w^{-1}_{{\tilde{\theta }}}(v^1)|\nabla _{v^1} \textbf{k}(v^1,u)| \frac{w_{{\tilde{\theta }}}(v^1)}{w_{{\tilde{\theta }}}(u)} \\&\lesssim \Vert wf(s)\Vert _\infty w^{-1}_{{\tilde{\theta }}}(v^1) [1+|v^1|^2 ] \frac{\textbf{k}_{{\tilde{\varrho }}}(v^1,u)}{|v^1-u|}\\&\lesssim e^{-\lambda s} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \frac{\textbf{k}_{{\tilde{\varrho }}}(v^1,u)}{|v^1-u|} \in L^1_u. \end{aligned}$$

Thus the contribution of $\nabla _{v^1} \textbf{k}(v^1,u)$ in the second term on the RHS of (4.48) is bounded as

$$\begin{aligned}&\varepsilon ^{-1} \int _{n(x^1)\cdot v^1>0} \nu (v^1)\sqrt{\mu (v^1)} [n(x^1)\cdot v^1] \int ^{t^1-\varepsilon }_{\max \{0,t^1-t_{\textbf{b}}(x^1,v^1)\}} \textrm{d}s e^{-\nu (v^1)(t^1-s)} e^{-\lambda t^1} \nonumber \\&\quad \lesssim \varepsilon ^{-1}e^{-\lambda t^1} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty . \end{aligned}$$

(4.51)

The contribution of $\frac{|\nabla _{v^1} e^{-\nu (v^1)(t^1-s)}|}{t^1-s}\lesssim e^{-\nu (v^1)(t^1-s)}$ in the second term on the RHS of (4.48) can be bounded by (4.51) by the same computation. Here we have used (3.30).

For the third term we apply (3.18) to have $\big | \frac{\nabla _{v^1}t_{\textbf{b}}(x^1,v^1)}{t_{\textbf{b}}(x^1,v^1)} \big | \lesssim \frac{1}{\alpha (x^1,v^1)}$, and thus the third term in (4.48) is bounded by

$$\begin{aligned}&e^{-\lambda t^1} \sup _t e^{\lambda t}| wf(t)|_\infty . \end{aligned}$$

(4.52)

Collecting (4.45) and (4.49) – (4.52) we conclude that the contribution of (4.8) in (4.40) is bounded by

$$\begin{aligned}&e^{-\lambda t^1}\big [{\mathcal {B}}+ o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ] . \end{aligned}$$

(4.53)

Here we recall ${\mathcal {B}}$ is defined in (4.28).

Next, we consider the contribution of (4.7) in (4.40). We use the boundary condition (2.11) with the representation (3.12) at $x^2$, then such contribution reads as

$$\begin{aligned}&\int _{n(x^1)\cdot v^1>0} \sum _{j'=1,2} \frac{\partial \textbf{x}_{p^2,j'}^2}{ \partial \textbf{x}_{p^1,j}^1} e^{-\nu (v^1)t_{\textbf{b}}^1} \sqrt{\mu (v^1)} [n(x^1)\cdot v^1] \textrm{d}v^1 \nonumber \\&\quad \times \Big [\frac{\partial _{\textbf{x}_{p^2,j'}^2} M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1)}{\sqrt{\mu (v^1)}} \int _{n(x^2)\cdot v^2>0} f(t^2,x^2,v^2) \sqrt{\mu (v^2)}[n(x^2)\cdot v^2] \textrm{d}v^2 \nonumber \\ \end{aligned}$$

(4.54)

$$\begin{aligned}&\quad + \frac{M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1)}{\sqrt{\mu (v^1)}} \int _{\textbf{v}_{p^2,3}^2>0} \partial _{\textbf{x}_{p^2,j'}^2} [f(t^2,\eta _{p^2}(\textbf{x}_{p^2}^2),T^t_{\textbf{x}_{p^2}^2}\textbf{v}_{p^2}^2)] \sqrt{\mu (\textbf{v}_{p^2}^2)} \textbf{v}^2_{p^2,3} \textrm{d}\textbf{v}_{p^2}^2 \Big ] . \end{aligned}$$

(4.55)

Here we have applied (3.15) to get that $\partial _j \eta _{p^1}(\textbf{x}_{p^1}^1) \nabla _x \textbf{x}_{p^2,j'}^2 = \frac{\partial \textbf{x}_{p^2,j'}^2}{ \partial \textbf{x}_{p^1,j}^1}$.

From (3.20), the first term (4.54) is directly bounded as

$$\begin{aligned} (4.54)&\lesssim \Vert \eta \Vert _{C^1(\partial \Omega )} \sup _t e^{\lambda t} | wf(t)|_\infty e^{-\lambda t_{\textbf{b}}^1} e^{-\lambda (t^1-t_{\textbf{b}}^1)} = e^{-\lambda t^1}\sup _t e^{\lambda t}| wf(t)|_\infty , \end{aligned}$$

where we have used $t^2 = t^1-t_{\textbf{b}}^1$.

For the second term, we apply the change of variable in Lemma 7 to get that

$$\begin{aligned} (4.55)&= \sum _{p^2\in {\mathcal {P}}} \iint _{|\textbf{x}_{p^2}^2|<\delta _1} \textrm{d}\textbf{x}_{p^2,1}^2 \textbf{x}_{p^2,2}^2 \int _0^{t^1} \textrm{d}t_{\textbf{b}}^1 \sqrt{g_{p^2,11} g_{p^2,22}} e^{-\nu (v^1)t_{\textbf{b}}^1} \iota _{p^2}(\eta _{p^2}(\textbf{x}_{p^2}^2)) \\&\quad \sum _{j^\prime =1,2} \frac{\partial \textbf{x}^{2}_{p^{2},j^\prime }}{\partial {\textbf{x}^{1 }_{p^{1 },j}}} \frac{n_{p^{1}} (\textbf{x}_{p^{1}}^{1}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{t_{\textbf{b}}^{1}} \frac{ n_{p^{2}}(\textbf{x}^{2}_{p^{2}}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{|t_{\textbf{b}}^{1}|^4} \\&\quad \times M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1) \int _{\textbf{v}_{p^2,3}^2>0} \partial _{\textbf{x}_{p^2,j'}^2} [f(t^1-t_{\textbf{b}}^1,\eta _{p^2}(\textbf{x}_{p^2}^2),T^t_{\textbf{x}_{p^2}^2}\textbf{v}_{p^2}^2)]\\&\quad \sqrt{\mu (\textbf{v}_{p^2}^2)} \textbf{v}^2_{p^2,3} \textrm{d}\textbf{v}_{p^2}^2. \end{aligned}$$

We apply the integration by parts with respect to $\partial _{\textbf{x}^{2}_{p^{2}, j^\prime }}$ for $j^{\prime }=1,2$. Then we derive that

$$\begin{aligned} (4.55)&= \sum _{p^{2} \in {\mathcal {P}}}\iint \int ^{t^{1}}_0 \int _{\textbf{v}_{p^2,3}^2>0} f(t^1-t_{\textbf{b}}^1,\eta _{p^2}(\textbf{x}_{p^2}^2),T^t_{\textbf{x}_{p^2}^2}\textbf{v}_{p^2}^2)\sqrt{\mu (\textbf{v}_{p^2}^2)} \textbf{v}^2_{p^2,3} \textrm{d}\textbf{v}_{p^2}^2 \nonumber \\&\quad \times \sum _{j'=1,2}\Big \{ \partial _{\textbf{x}_{p^{2},j^\prime }^{2}} \Big [ e^{- \nu (v^{1}) t_{\textbf{b}}^{1}} \iota _{p^{2}} ( \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2} ) ) M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1)\Big ] \cdots \end{aligned}$$

(4.56)

$$\begin{aligned}&\quad + \partial _{\textbf{x}_{p^{2},j^\prime }^{2}} \Big [ \frac{\partial \textbf{x}^{2}_{p^{2},j^\prime }}{\partial {\textbf{x}^{1 }_{p^{1 },j}}} \sqrt{g_{p^{2},11}g_{p^{2},22} } \Big ] \cdots \end{aligned}$$

(4.57)

$$\begin{aligned}&\quad + \partial _{\textbf{x}_{p^{2},j^\prime }^{2}} \Big [ \frac{n_{p^{1}} (\textbf{x}_{p^{1}}^{1}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{t_{\textbf{b}}^{1}} \cdot \frac{ n_{p^{2}}(\textbf{x}^{2}_{p^{2}}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{|t_{\textbf{b}}^{1}|^4}\Big ]\cdots \Big \}. \end{aligned}$$

(4.58)

Here we note that for $\iota _{p^{2} } ( \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2} ) )=0$ when $|\textbf{x}_{p^{2}}^{2}|= \delta _1$ from (3.6), such contribution of $|\textbf{x}_{p^{2}}^{2}|= \delta _1$ vanishes.

From (4.21) and (4.22), we bounds the terms in (4.56)–(4.58) as follows:

$$\begin{aligned}&|n(x^1)\cdot v^1| = \frac{|n(x^1)\cdot (x^2-x^1)|}{t_{\textbf{b}}^1}\lesssim \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^2}{t_{\textbf{b}}^1} , \nonumber \\&\Big | \frac{n_{p^{1}} (\textbf{x}_{p^{1}}^{1}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{t_{\textbf{b}}^{1}} \frac{ n_{p^{2}}(\textbf{x}^{2}_{p^{2}}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}}^{2}) ) }{|t_{\textbf{b}}^{1}|^4} \Big |\lesssim \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^4}{|t_{\textbf{b}}^1|^5},\nonumber \\&\Big | \frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } [n(x^1)\cdot v^1]\Big | \lesssim \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|}{t_{\textbf{b}}^1}, \nonumber \\&\Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } [n(x^2)\cdot (x^1-\eta _{p^2}(\textbf{x}_{p^2}^2))] \Big | \lesssim |x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|,\nonumber \\&\Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } v^1 \Big | = \Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } [x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)] \Big |/t_{\textbf{b}}^1 \lesssim \frac{\Vert \eta \Vert _{C^1}}{t_{\textbf{b}}^1} , \nonumber \\&\Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } e^{-\nu (v^1)t_{\textbf{b}}^1} \Big | \lesssim \frac{\Vert \eta \Vert _{C^1} e^{-\nu (v^1)t_{\textbf{b}}^1}}{t_{\textbf{b}}^1}, \nonumber \\&\Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } [M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1)] \Big | \lesssim \Vert \eta \Vert _{C^1}[1+|v^1|^2]M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1) \nonumber \\&\quad + |v^1|\Big |\frac{\partial }{\partial _{\textbf{x}_{p^2,j'}^2} } v^1 \Big |M_W(\eta _{p^2}(\textbf{x}_{p^2}^2),v^1) \nonumber \\&\lesssim \Vert \eta \Vert _{C^1}\Big [1 + \frac{1}{t_{\textbf{b}}^1} \Big ] e^{-\frac{|v^1|^2}{4}}\lesssim \Big [1 + \frac{1}{t_{\textbf{b}}^1} \Big ] e^{-\frac{|x^1 - \eta _{p^2}(\textbf{x}_{p^2}^2)|^2}{4|t_{\textbf{b}}^1|^2}},\nonumber \\&\bigg | \frac{\partial }{\partial {\textbf{x}_{p^{2},j^\prime }^{2}}} \bigg ( \sum _{j^\prime =1,2} \frac{\partial \textbf{x}^{2}_{p^{2},j^\prime }}{\partial {\textbf{x}^{1 }_{p^{1 },j}}} \sqrt{g_{p^{2},11}g_{p^{2},22} } \bigg ) \bigg |\nonumber \\&\quad \lesssim \Vert \eta \Vert _{C^2} \Big \{1+ \frac{|\textbf{v}^{2}_{p^{2} ,\parallel }|}{|\textbf{v}^{2}_{p^{2}, 3}|^2 } | \partial _3 \eta _{p^{2}} ( \textbf{x}^{2 }_{p^{2 }} ) \cdot \partial _j \eta _{p^{1}} ( \textbf{x}^{1 }_{p^{1 }} )| \Big \}\nonumber \\&\quad \lesssim \Big \{1+ \frac{|\textbf{v}^{2}_{p^{2} }|}{|\textbf{v}^{2}_{p^{2}, 3}|^2 } |x^{1} - \eta _{p^{2}} ( \textbf{x}^{2 }_{p^{2 }} ) | \Big \} \lesssim \frac{1}{|\textbf{v}^{2}_{p^{2}, 3}|}=\frac{|t_{\textbf{b}}^{1}|}{|n_{p^{2}}(\textbf{x}_{p^{2}}^{2}) \cdot (x^{1} - \eta _{p^{2}} (\textbf{x}_{p^{2}} ^{2})|} \nonumber \\&\qquad \lesssim \frac{|t_{\textbf{b}}^1|}{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^2}. \end{aligned}$$

(4.59)

For the last estimate (4.59), we also used (3.20) and (4.23).

Thus the integrand of (4.56) is bounded as

$$\begin{aligned}&\Big [1+ \frac{1}{t_{\textbf{b}}^1} \Big ] |v^1|e^{-\frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2) |^2}{4|t_{\textbf{b}}^1|^2}} \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^4}{|t_{\textbf{b}}^1|^5} \\&\quad \lesssim \Big [ \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^4}{|t_{\textbf{b}}^1|^6}+\frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^4}{|t_{\textbf{b}}^1|^5}\Big ] e^{-\frac{\big |x^1-\eta _{p^2}(\textbf{x}_{p^2}^2) \big |^2}{4|t_{\textbf{b}}^1|^2}}, \end{aligned}$$

the integrand of (4.57) is bounded as

$$\begin{aligned}&\frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^4}{|t_{\textbf{b}}^1|^5}\frac{|t_{\textbf{b}}^1|}{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^2} e^{-\frac{\big |x^1-\eta _{p^2}(\textbf{x}_{p^2}^2) \big |^2}{4|t_{\textbf{b}}^1|^2}} \lesssim \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^2}{|t_{\textbf{b}}^1|^4}e^{-\frac{\big |x^1-\eta _{p^2}(\textbf{x}_{p^2}^2) \big |^2}{4|t_{\textbf{b}}^1|^2}} , \end{aligned}$$

the integrand of (4.58) is bounded as

$$\begin{aligned}&\frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^3}{|t_{\textbf{b}}^1|^5} e^{-\frac{\big |x^1-\eta _{p^2}(\textbf{x}_{p^2}^2) \big |^2}{4|t_{\textbf{b}}^1|^2}}. \end{aligned}$$

With $e^{-\nu (v^1)t_{\textbf{b}}^1}e^{-\lambda (t^1-t_{\textbf{b}}^1)} \le e^{-\nu (v^1)t_{\textbf{b}}^1/2}e^{-\lambda t^1}$, we derive that

$$\begin{aligned} \begin{aligned}&|(4.55)|\\&\quad \lesssim e^{-\lambda t^1}\sup _t e^{\lambda t} |wf(t)|_\infty \iint \int _0^{t^{1}} e^{-\nu (v^1) t_{\textbf{b}}^{1}/2}e^{-\frac{\big |x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})\big |^2}{4|t_{\textbf{b}}^{1}|^2}} \\&\qquad \times \Big [ \frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^3}{|t_{\textbf{b}}^{1}|^5}+ \frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^2}{|t_{\textbf{b}}^{1}|^4}+ \frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^4}{|t_{\textbf{b}}^{1}|^6}\Big ] \\&\quad \lesssim e^{-\lambda t^1}\sup _t e^{\lambda t} |wf(t)|_\infty \\&\qquad \int _0^{\infty } \frac{e^{-\nu (v^1) t_{\textbf{b}}^{1}/2}}{|t_{\textbf{b}}^{1}|^{1/2}}\iint \Bigg [\frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{3/2}} + \frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{1/2}}\Bigg ]\\&\quad \lesssim e^{-\lambda t^1}\sup _t e^{\lambda t} |wf(t)|_\infty , \end{aligned}\nonumber \\ \end{aligned}$$

(4.60)

where we have used that

$$\begin{aligned} \begin{aligned}&\Big [ \frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^3}{|t_{\textbf{b}}^{1}|^5}+ \frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^2}{|t_{\textbf{b}}^{1}|^4}+\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^4}{|t_{\textbf{b}}^{1}|^6}\\&\qquad +\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^4}{|t_{\textbf{b}}^{1}|^5}\Big ] e^{-\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^2}{4|t_{\textbf{b}}^{1}|^2}}\\&\quad \le \Big [\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{11/2}}{|t_{\textbf{b}}^{1}|^{11/2}}+\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{9/2}}{|t_{\textbf{b}}^{1}|^{9/2}} + \frac{|x^1-\eta _{p^2}(\textbf{x}_{p^2}^2)|^{7/2}}{|t_{\textbf{b}}^1|^{7/2}}\Big ] \\&\qquad \times \Big [\frac{1}{|t_{\textbf{b}}^{1}|^{1/2}} \frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{3/2}} + \frac{1}{|t_{\textbf{b}}^{1}|^{1/2}} \frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{1/2}} \Big ] e^{-\frac{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^2}{4|t_{\textbf{b}}^{1}|^2}}\\&\quad \lesssim \frac{1}{|t_{\textbf{b}}^{1}|^{1/2}} \frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{3/2}} + \frac{1}{|t_{\textbf{b}}^{1}|^{1/2}} \frac{1}{|x^{1} - \eta _{p^{2} } (\textbf{x}^{2}_{p^{2}})|^{1/2}}. \end{aligned} \end{aligned}$$

Collecting (4.41) (4.53), and (4.60) we conclude that

$$\begin{aligned} (4.40)&\lesssim e^{-\lambda t^1}\big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ]. \end{aligned}$$

This, together with (4.39), conclude that

$$\begin{aligned} (4.7)&\lesssim (4.35)+(4.33) \lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ]. \end{aligned}$$

(4.61)

Step 3: estimate of (4.8). Last we compute (4.8). We apply the characteristic (4.4)–(4.11) to expand $\nabla _x f(s,x-(t-s)v,u)$ along u.

The contribution of (4.4)–(4.7) and (4.9)–(4.11) in the expansion of $\nabla _x f(s,x-(t-s)v,u)$ can be computed using the same computation in (4.30) and (4.61), by replacing x by $x-(t-s)v$, v by u, t by s. This leads to a bound for these seven terms as

$$\begin{aligned}&\int ^t_{\max \{0,t-t_{\textbf{b}}\}} e^{-\nu (t-s)} \int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u) \frac{e^{-\lambda s}w^{-1}_{{\tilde{\theta }}}(u)}{\alpha (x-(t-s)v,u)}\nonumber \\&\qquad \times \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ] \nonumber \\&\quad \lesssim \frac{e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}\times \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ] . \end{aligned}$$

(4.62)

Here we have used $e^{-\nu (t-s)}e^{-\lambda s}\le e^{-\lambda t} e^{-\nu (t-s)/2}$ and applied Lemma 5 with Lemma 2.

Then we consider the contribution of (4.8), denoting $y=x-(t-s)v$, such contribution reads as

$$\begin{aligned}&\int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u)\nonumber \\&\quad \times \int ^s_{\max \{0,s-t_{\textbf{b}}(y,u)\}} \textrm{d}s'e^{-\nu (u)(s-s')} \int _{{{\mathbb {R}}}^3} \textrm{d}u'\textbf{k}(u,u')\nabla _x f(s',y-(s-s')u,u') . \end{aligned}$$

(4.63)

We split the $\textrm{d}s'$ integral into $s-s'<\varepsilon$ and $s-s'\ge \varepsilon$.

When $s-s'<\varepsilon$, we apply Lemmas 5 and 2 to get that

$$\begin{aligned}&(4.63)_1:=\Big |\int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u)\nonumber \\&\qquad \times \int ^s_{\max \{0,s-t_{\textbf{b}}(y,u),s-\varepsilon \}} \textrm{d}s' e^{-\nu (u)(s-s')} \int _{{{\mathbb {R}}}^3} \textrm{d}u'\textbf{k}(u,u')\nabla _x f(s',y-(s-s')u,u') \Big |\nonumber \\&\quad \lesssim \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u) \int ^s_{\max \{s-t_{\textbf{b}}(y,u),s-\varepsilon \}} \textrm{d}s' e^{-\nu (u)(s-s')} \nonumber \\&\qquad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u'\textbf{k}(u,u') \frac{e^{-\lambda s'}w^{-1}_{{\tilde{\theta }}}(u')}{\alpha (y-(s-s')u,u')}\sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \nonumber \\&\quad \lesssim \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}e^{-\lambda s}\int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v,u)}{w_{{\tilde{\theta }}}(u)} \int ^s_{\max \{s-t_{\textbf{b}}(y,u),s-\varepsilon \}} \textrm{d}s' e^{-\nu (u)(s-s')/2} \nonumber \\&\qquad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u' \frac{\textbf{k}_{{\tilde{\varrho }}}(u,u')}{\alpha (y-(s-s')u,u')}\sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \nonumber \\&\quad \lesssim o(1)e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)\times \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f\Vert _\infty \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)/2}\int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{\alpha (y,u)} \nonumber \\&\quad \lesssim o(1)\frac{e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}\sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

(4.64)

Here we note that we applied (3.38) in the fifth line and (3.37) in the last line.

When $s-s'\ge \varepsilon$, we further split the u integration into $|u|>\delta$ and $|u|<\delta$. For $|u|<\delta$, by a similar computation as to that of (4.64), we have that

$$\begin{aligned}&\Big |\int _{\max \{0,t-t_{\textbf{b}}\}}^{t-\varepsilon } \textrm{d}s e^{-\nu (t-s)}\int _{|u|<\delta } \textrm{d}u \textbf{k}(v,u)\\&\qquad \times \int ^s_{\max \{0,s-t_{\textbf{b}}(y,u),s-\varepsilon \}} \textrm{d}s' e^{-\nu (u)(s-s')} \int _{{{\mathbb {R}}}^3} \textrm{d}u'\textbf{k}(u,u')\nabla _x f(s',y-(s-s')u,u') \Big |\\&\quad \lesssim e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)\times \sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f\Vert _\infty \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)/2}\int _{|u|<\delta } \textrm{d}u \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{\alpha (y,u)} \\&\quad \lesssim o(1)\frac{e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}\sup _{s\le t} e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

In the last line we used (3.39).

For $s-s'\ge \varepsilon$ and $|u|>\delta$, we apply the following change of variable:

$$\begin{aligned} \nabla _x f(s',y-(s-s')u,u') = -\frac{\nabla _{u}[f(s',y-(s-s')u,u')]}{s-s'}. \end{aligned}$$

(4.65)

Without loss of generality, we assume $s>\varepsilon$, otherwise, we directly bound (4.63) by (4.64). In such case, we only integrate over the u such that $t_{\textbf{b}}(y,u)>\varepsilon$. Then we denote that

$$\begin{aligned} U:=\{u\in {{\mathbb {R}}}^3: |u|>\delta , \ t_{\textbf{b}}(y,u)>\varepsilon \}. \end{aligned}$$

Then by Lemma 10, we have $|n((x_{\textbf{b}}(y,u)))\cdot u|\gtrsim \varepsilon |\delta |^2$. With (3.18), the corresponding hypersurface, denoted as ${\tilde{U}}$, is piecewise smooth.

We use the change of variable (4.65) to express such a case as

$$\begin{aligned}&(4.63)_2:= \Big | \int _{\max \{0,t-t_{\textbf{b}}\}}^{t} \textrm{d}s e^{-\nu (t-s)}\int _{U} \textrm{d}u \textbf{k}(v,u) \\&\quad \times \int ^{s-\varepsilon }_{\max \{0,s-t_{\textbf{b}}(y,u)\}} \textrm{d}s'e^{-\nu (u)(s-s')} \int _{{{\mathbb {R}}}^3} \textrm{d}u'\textbf{k}(u,u') \frac{-\nabla _u f(s',y-(s-s')u,u')}{s-s'} \Big | . \end{aligned}$$

Now we apply an integration by part in $\textrm{d}u$ to obtain

$$\begin{aligned} \begin{aligned}&(4.63)_2= \Big |\int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \int ^{s-\varepsilon }_{\max \{0,s-t_{\textbf{b}}(y,u)\}} \textrm{d}s' \\&\quad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u'\nabla _u[e^{-\nu (u)(s-s')}\textbf{k}(v,u)\textbf{k}(u,u')] \frac{f(s',y-(s-s')u,u')}{s-s'} \\&\quad + \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u) \\&\quad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u' \textbf{k}(u,u')\nabla _u t_{\textbf{b}}(y,u)e^{-\nu (u)t_{\textbf{b}}(y,u)} \frac{f(s-t_{\textbf{b}}(y,u),x_{\textbf{b}}(y,u),u')}{t_{\textbf{b}}(y,u)} \\&\quad + \int _{\max \{0,t-t_{\textbf{b}}\}}^{t} \textrm{d}s e^{-\nu (t-s)}\int _{{\tilde{U}}} \textrm{d}u \textbf{n}_u \textbf{k}(v,u) \int ^{s-\varepsilon }_{\max \{0,s-t_{\textbf{b}}(y,u)\}} \textrm{d}s' e^{-\nu (u)(s-s')} \\&\quad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u' \textbf{k}(u,u') \frac{f(s',y-(s-s')u,u')}{s-s'} \Big |. \end{aligned}\nonumber \\ \end{aligned}$$

(4.66)

In the last term, $\textbf{n}_u = \frac{-\nabla _u t_{\textbf{b}}(y,u)}{|\nabla _u t_{\textbf{b}}(y,u)|}$. Then the last term is bounded as

$$\begin{aligned}&\frac{\varepsilon ^{-1} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty }{w_{{\tilde{\theta }}}(v)} \int ^t_{\max \{0,t-t_{\textbf{b}}\}} \textrm{d}e^{-\nu (t-s)} \int _{{\tilde{U}}} \textrm{d}u \textbf{k}(v,u)\frac{w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)} \nonumber \\&\qquad \times \int ^{s-\varepsilon }_{s-t_{\textbf{b}}(y,u)} \textrm{d}s' e^{-\nu (u)(s-s')}\int _{{{\mathbb {R}}}^3} \textrm{d}u' \textbf{k}(u,u') e^{-\lambda s'}\frac{w_{{\tilde{\theta }}}(u)}{w_{{\tilde{\theta }}}(u')} \nonumber \\&\quad \lesssim \frac{\varepsilon ^{-1}e^{-\lambda t} \sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

(4.67)

In the last line we used $\alpha (x,v)\lesssim 1$ from (3.33), and we used (3.24).

The first term in (4.66) is bounded as

$$\begin{aligned}&\sup _t e^{\lambda t}\Vert wf(t)\Vert _\infty \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \int ^{s-\varepsilon }_{\max \{0,s-t_{\textbf{b}}(y,u)\}} \textrm{d}s' \nonumber \\&\qquad \times \Big |\int _{{{\mathbb {R}}}^3} \textrm{d}u'\nabla _u[e^{-\nu (u)(s-s')}\textbf{k}(v,u)\textbf{k}(u,u')] \frac{e^{-\lambda s'}w^{-1}(u')}{s-s'} \Big |\nonumber \\&\quad \lesssim \varepsilon ^{-1}e^{-\lambda t} w^{-1}_{{\tilde{\theta }}}(v)\sup _s e^{\lambda s}\Vert wf(s)\Vert _\infty \lesssim \frac{e^{-\lambda t}\times {\mathcal {B}}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)} . \end{aligned}$$

(4.68)

In the last inequality, we used (3.33). To get the first inequality in the last line, we have used ${\tilde{\theta }} \ll \theta$ so that $w^{-1}(u')\le w^{-1}_{2{\tilde{\theta }}}(u')$, $|u|^2 w_{{\tilde{\theta }}}^{-1}(u)\lesssim 1$, then we (3.24) and (3.25) to control the $\textbf{k}$ terms in the third line as

$$\begin{aligned}&|\nabla _u [\textbf{k}(v,u)\textbf{k}(u,u')] | w^{-1}(u') \le |\nabla _u [\textbf{k}(v,u)\textbf{k}(u,u')] | w^{-1}_{2{\tilde{\theta }}}(u')\\&\quad \lesssim |\nabla _u \textbf{k}(v,u)| w^{-1}_{2{\tilde{\theta }}}(u) \textbf{k}(u,u')\frac{w_{2{\tilde{\theta }}}(u)}{w_{2{\tilde{\theta }}}(u')} + \textbf{k}(v,u) w^{-1}_{2{\tilde{\theta }}}(u) |\nabla _u \textbf{k}(u,u')|\frac{w_{2{\tilde{\theta }}}(u)}{w_{2{\tilde{\theta }}}(u')} \\&\quad \lesssim w^{-1}_{{\tilde{\theta }}}(v)w^{-1}_{{\tilde{\theta }}}(u) |\nabla _u \textbf{k}(v,u)| \frac{w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)} \textbf{k}_{{\tilde{\varrho }}}(u,u') \\&\qquad + \textbf{k}(v,u) w^{-1}_{{\tilde{\theta }}}(u) [1+|u|^2]w^{-1}_{{\tilde{\theta }}}(u) \frac{\textbf{k}_{{\tilde{\varrho }}}(u,u')}{|u-u'|} \\&\quad \lesssim w^{-1}_{{\tilde{\theta }}}(v)[1+|u|^2] w^{-1}_{{\tilde{\theta }}}(u) \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{|v-u|} \textbf{k}_{{\tilde{\varrho }}}(u,u') + w^{-1}_{{\tilde{\theta }}}(v) \textbf{k}(v,u) \frac{w_{{\tilde{\theta }}}(v)}{w_{{\tilde{\theta }}}(u)} \frac{\textbf{k}_{{\tilde{\varrho }}}(u,u')}{|u-u'|} \\&\quad \lesssim w^{-1}_{{\tilde{\theta }}}(v) \frac{\textbf{k}_{{\tilde{\varrho }}}(v,u)}{|v-u|} \textbf{k}_{{\tilde{\varrho }}}(u,u') + w^{-1}_{{\tilde{\theta }}}(v)\textbf{k}_{{\tilde{\varrho }}}(v,u) \frac{\textbf{k}_{{\tilde{\varrho }}}(u,u')}{|u-u'|} \in L^1_{u,u'}. \end{aligned}$$

The contribution of $\nabla _u e^{-\nu (u)(s-s')}$ can be controlled by (3.30) and the extra $\frac{1}{s-s'}$.

For the second term in (4.66), we apply (3.18), $e^{-\nu (u)t_{\textbf{b}}(y,u)} e^{-\lambda (s-t_{\textbf{b}}(y,u))} \le e^{-\lambda s}$ and $e^{-\nu (t-s)}e^{-\lambda s}\le e^{-\nu (t-s)/2}e^{-\lambda t}$ to obtain a bound as

$$\begin{aligned}&\sup _t e^{\lambda t}| wf(t)|_\infty \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v,u)}{\alpha (y,u)} \nonumber \\&\qquad \times \int _{{{\mathbb {R}}}^3} \textrm{d}u' \textbf{k}(u,u') e^{-\nu (u)t_{\textbf{b}}(y,u)}e^{-\lambda (s-t_{\textbf{b}}(y,u))} w^{-1}(u') \nonumber \\&\quad \lesssim \sup _t e^{\lambda t}| wf(t)|_\infty \int _{\max \{0,t-t_{\textbf{b}}\}}^t \textrm{d}s e^{-\nu (t-s)}\int _{{{\mathbb {R}}}^3} \textrm{d}u \frac{\textbf{k}(v,u)}{\alpha (y,u)w_{{\tilde{\theta }}}(u)} \int _{{{\mathbb {R}}}^3} \textrm{d}u' \textbf{k}_{{\tilde{\varrho }}}(u,u') e^{-\lambda s} \nonumber \\&\quad \lesssim \frac{e^{-\lambda t}w_{{\tilde{\theta }}}^{-1}(v)\sup _t e^{\lambda t}| wf(t)|_\infty }{\alpha (x,v)} \lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}\times {\mathcal {B}}. \end{aligned}$$

(4.69)

Here we recall ${\mathcal {B}}$ is defined in (4.28).

Collecting (4.64) and (4.67) – (4.69) we conclude that

$$\begin{aligned} (4.63)&\lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ]. \end{aligned}$$

(4.70)

Combining (4.62) and (4.70) we conclude that

$$\begin{aligned} (4.8)&\lesssim \frac{e^{-\lambda t}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)} \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ] . \end{aligned}$$

(4.71)

Step 4: conclusion. To conclude the theorem, we combine (4.30), (4.61) and (4.71) to get that

$$\begin{aligned} |\nabla _x f(t,x,v)|&\lesssim \frac{e^{-\lambda t}w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)} \times \big [{\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty \big ] , \end{aligned}$$

and thus

$$\begin{aligned} |e^{\lambda t}w_{{\tilde{\theta }}}(v)\alpha \nabla _x f(t)|&\lesssim {\mathcal {B}}+o(1)\sup _{s\le t}e^{\lambda s}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(s)\Vert _\infty . \end{aligned}$$

Since the inequality holds for all $t>0$, we take the supremum in $s\le t$ to have

$$\begin{aligned} \sup _{t}e^{\lambda t}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(t)\Vert _\infty&\lesssim {\mathcal {B}}. \end{aligned}$$

We recall ${\mathcal {B}}$ is defined in (4.28), which is bounded, in particular, $\sup _t e^{\lambda t}|w\partial _t f(t)|_\infty$ is bounded by $\Vert w\partial _t f_0\Vert _\infty$ in Corollary 4. With fixed $\varepsilon >0$, we finish the proof.

4.5 Proof of (4.2).

From (1.5) and (1.6), the equation for $f_s$ in (2.5) reads as

$$\begin{aligned} \begin{aligned}&v\cdot \nabla _x f_s +\nu f_s = K(f_s)+\Gamma (f_s,f_s) , \\&f_s(x,v)|_{n(x)\cdot v<0}=\frac{M_W(x,v)}{\sqrt{\mu (v)}}\int _{n(x)\cdot u>0}f_s(x,u) \sqrt{\mu (u)} \{n(x)\cdot u\}\textrm{d}u+r(x,v). \end{aligned}\nonumber \\ \end{aligned}$$

(4.72)

Here the remainder term is

$$\begin{aligned} r(x,v):=\frac{M_W(x,v)-\mu (v)}{\sqrt{\mu (v)}}. \end{aligned}$$

(4.73)

Denoting $g=K(f_s)+\Gamma (f_s,f_s)$, we apply method of characteristic to (4.72) with fixed $t\gg 1$ to get that

$$\begin{aligned} \nabla _x f_s(x,v)&= \textbf{1}_{t\ge t_{\textbf{b}}} e^{-\nu (v)t_{\textbf{b}}(x,v)} \nabla _x [f_s(x_{\textbf{b}}(x,v),v) ] \end{aligned}$$

(4.74)

$$\begin{aligned}&\quad - \textbf{1}_{t\ge t_{\textbf{b}}} \nu (v) \nabla _x t_{\textbf{b}}(x,v) e^{-\nu (v)t_{\textbf{b}}(x,v)} f_s(x_{\textbf{b}}(x,v),v) \end{aligned}$$

(4.75)

$$\begin{aligned}&\quad +\textbf{1}_{t<t_{\textbf{b}}}e^{-\nu (v)t} \nabla _x [ f_s(x-tv,v)] \end{aligned}$$

(4.76)

$$\begin{aligned}&\quad +\int _{\max \{0,t-t_{\textbf{b}}\}}^t e^{-\nu (v)(t-s)} \nabla _x g ( x-(t-s) v,v) \textrm{d}s \end{aligned}$$

(4.77)

$$\begin{aligned}&\quad - \textbf{1}_{t\ge t_{\textbf{b}}} \nabla _x t_{\textbf{b}}e^{-\nu (v)t_{\textbf{b}}}g(x-t_{\textbf{b}}v,v) . \end{aligned}$$

(4.78)

Compared with (4.4)–(4.11), the difference is that there is no temporal derivative, no initial condition, while there is an extra term (4.76), and the boundary condition has an extra term (4.73). On the boundary, $f_s$ is bounded in (4.1). Thus we only need to consider the contribution of (4.76) and r in (4.74).

With $t\gg 1$, we bound (4.76) as

$$\begin{aligned} |(4.76)|&\lesssim \frac{o(1)\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty }{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

For (4.74), by a similar computation as (4.32), such extra contribution reads as

$$\begin{aligned}&\frac{e^{-\nu t_{\textbf{b}}}|v| \partial _{\textbf{x}_{p^1,j}^1} M_W(\eta _{p^1}(\textbf{x}_{p^1}^1),v)}{\sqrt{\mu (v)}\alpha (x,v)} \\&\quad \lesssim \frac{e^{-\nu t_{\textbf{b}}}|v| }{\sqrt{\mu (v)}\alpha (x,v)} \Vert \eta \Vert _{C^1}\Vert T_W\Vert _{C^1(\partial \Omega )}[1+|v|]M_W(x,v) \lesssim \frac{\Vert T_W-T_0\Vert _{C^1(\partial \Omega )}}{w_{{\tilde{\theta }}}(v)\alpha (x,v)}. \end{aligned}$$

Following the same computation as in Sect. 4.4 for the rest of the terms, we conclude (4.2).

5 Uniform-in-time $W^{1,p}$-estimate for $p<3$

We aim to utilize Theorem 2 to obtain the $\alpha$-weighted $C^1$ estimate (2.22), and then deduce a $W^{1,p}$ estimate without weight for $p<3$. Since the weight $\alpha$ in Theorem 2 behaves as $\alpha (x,v)\thicksim |n(x_{\textbf{b}}(x,v))\cdot v|$, we address the singularity of $1/|n(x_{\textbf{b}}(x,v))\cdot v|$ in the $W^{1,p}$ estimate through Lemma 11.

We will use the change of variable in Lemma 9 and some properties of the convex domain in Lemma 10 to deduce Lemma 11.

Lemma 9

For any g,

$$\begin{aligned} \begin{aligned} \iint _{\Omega \times {{\mathbb {R}}}^3} g(y,v) \textrm{d}y \textrm{d}v&= \int _{\gamma _+}\int _0^{t_{\textbf{b}}(x,v)}g(x-sv,v)|n(x)\cdot v|\textrm{d}s \textrm{d}v \textrm{d}S_x \\&= \int _{\gamma _-} \int _0^{t_{\textbf{b}}(x,-v)}g(x+sv,v)|n(x)\cdot v|\textrm{d}s \textrm{d}v \textrm{d}S_x. \end{aligned} \nonumber \\ \end{aligned}$$

(5.1)

Proof

First, we prove that

$$\begin{aligned} (s,x,v) \in (0,t_{\textbf{b}}(x,v)) \times \gamma _+ \rightarrow (x-sv,v)\in \Omega \times {{\mathbb {R}}}^3 \end{aligned}$$

(5.2)

is a bijection, with a Jacobian given by

$$\begin{aligned} \Big |\det \Big (\frac{\partial (x-sv,v)}{\partial (s,x,v)}\Big )\Big | = |n(x)\cdot v|. \end{aligned}$$

(5.3)

Clearly the map is well-defined, since $x-sv\in \Omega$ for $0<s<t_{\textbf{b}}(x,v)$.

To show the map is injective we suppose $(x_1-s_1v_1,v_1) = (x_2-s_2v_2,v_2)$, then $v_1=v_2=v$, and thus $x_1-s_1v = x_2-s_2v$, which implies $x_1-x_2 = (s_1-s_2)v$. If $s_1\ne s_2$, then $x_1\ne x_2$, and $x_1,x_2$ are in the same characteristic. $(x_1,v)\in \gamma _+$ and $x_2\in \partial \Omega$ implies $x_2 = x_1 - t_{\textbf{b}}(x_1,v)v=x_{\textbf{b}}(x_1,v)$, which further implies $(x_2,v)\in \gamma _-\cup \gamma _0$. This contradicts to the fact that $(x_2,v)\in \gamma _+$. By a contradiction argument, we conclude that $s_1=s_2$ and $x_1=x_2$.

To show the map is surjective, we define that

$$\begin{aligned} t_{\textbf{f}}(x,v) := \sup \{s>0:x+sv\in \Omega \}, \ \ x_{\textbf{f}}(x,v):=x+t_{\textbf{f}}(x,v)v. \end{aligned}$$

(5.4)

For any $(y,v)\in \Omega \times {{\mathbb {R}}}^3$, we take $x=y+t_{\textbf{f}}(y,v)v\in \partial \Omega$. Then $x-t_{\textbf{f}}(y,v)v = y$ and $(x,v)\in \gamma _+$. Since $y\in \Omega$, we then have $t_{\textbf{f}}(y,v)<t_{\textbf{b}}(x,v)$. Thus, (5.2) is surjective and we conclude that it is also bijective.

Then we compute the Jacobian of the map. For $x\in \Omega$, by (2.1) $x = \eta _{p}(\textbf{x}_p)$, then we evaluate the following determinant as:

$$\begin{aligned}&\Big | \det \Big (\frac{\partial (\eta _p(\textbf{x}_p)-sv, v)}{\partial (\textbf{x}_{p,1},\textbf{x}_{p,2},v)}\Big )\Big | \\&\quad = \Big | \det \left( \begin{array}{cccc} v &{} \partial _1 \eta _{p}(\textbf{x}_p) &{} \partial _2 \eta _p(\textbf{x}_p) &{} -s Id \\ 0 &{} 0&{} 0&{} Id \end{array} \right) \Big | \\&\quad = \Big | \det \left( \begin{array}{ccc} v&\partial _1 \eta _{p}(\textbf{x}_p)&\partial _2 \eta _p(\textbf{x}_p) \end{array} \right) \Big | \\&\quad = |v\cdot (\partial _1 \eta _p(\textbf{x}_p)\times \partial _2\eta _p(\textbf{x}_p))| = |v\cdot n(x)||(\partial _1 \eta _p(\textbf{x}_p)\times \partial _2\eta _p(\textbf{x}_p))|. \end{aligned}$$

Here we have applied (3.7).

Since the surface measure equals $\textrm{d}S_x = |\partial _1 \eta _p(\textbf{x}_p) \times \partial _2 \eta _p(\textbf{x}_p)|\textrm{d}\textbf{x}_{p,1}\textrm{d}\textbf{x}_{p,2}$, we conclude (5.3) and the first line of (5.1). The proof of the second line of (5.1) is similar. $\square$

Lemma 10

In a convex and $C^2$ domain (2.2), $t_{\textbf{b}}(x,v)$ is bounded as

$$\begin{aligned} t_{\textbf{b}}(x,v) \lesssim \frac{|n(x_{\textbf{b}}(x,v))\cdot v|}{|v|^2} . \end{aligned}$$

(5.5)

For $x\in \partial \Omega$, we have

$$\begin{aligned} |n(x)\cdot v| \thicksim |n(x_{\textbf{b}}(x,v))\cdot v|. \end{aligned}$$

(5.6)

Proof

The proof is similar to [10]. First we prove (5.5). It suffices to prove $t_{\textbf{b}}(x,v)\lesssim \frac{|n(x_{\textbf{b}}(x,v))\cdot v|}{|v|^2}$ when $(x,v)\in \gamma _+$. When $x\notin \partial \Omega$, (5.5) follows from the fact that $t_{\textbf{b}}(x,v)\le t_{\textbf{b}}(x_{\textbf{f}}(x,v),v)$, where $x_{\textbf{f}}(x,v)$ is defined in (5.4). Then for $(x,v)\in \gamma _+$, we have $\xi (x)=0=\xi (x-t_{\textbf{b}}(x,v)v)$, then

$$\begin{aligned} 0&=\xi (x-t_{\textbf{b}}(x,v)v)=\xi (x)+\int _0^{t_{\textbf{b}}(x,v)} [-v\cdot \nabla _x \xi (x-sv)] \textrm{d}s \\&= [-v\cdot \nabla _x \xi (x-t_{\textbf{b}}(x,v)v)]t_{\textbf{b}}(x,v)+\int _0^{t_{\textbf{b}}(x,v)}\int _{t_{\textbf{b}}(x,v)}^s \{v\cdot \nabla _x^2 \xi (x-\tau v)\cdot v\}\textrm{d}\tau \textrm{d}s. \end{aligned}$$

This leads to

$$\begin{aligned} -[v\cdot \nabla _x \xi (x_{\textbf{b}}(x,v))]t_{\textbf{b}}(x,v) = \int _0^{t_{\textbf{b}}(x,v)} \int _s^{t_{\textbf{b}}(x,v)} \{v\cdot \nabla _x^2 \xi (x-\tau v)\cdot v\} \textrm{d}\tau \textrm{d}s . \end{aligned}$$

From the convexity (2.2), we have $-[v\cdot \nabla _x \xi (x_{\textbf{b}}(x,v))]t_{\textbf{b}}(x,v)\gtrsim (t_{\textbf{b}}(x,v))^2|v|^2$, thus we conclude (5.5).

Then we prove (5.6). By the definition of the ${\tilde{\alpha }}$ in (2.13), for $x\in \partial \Omega$ and $x_{\textbf{b}}(x,v)\in \partial \Omega$ such that $\xi (x)=0,\xi (x_{\textbf{b}}(x,v))=0$, we have

$$\begin{aligned} {\tilde{\alpha }}(x,v)\thicksim |n(x)\cdot v|, \ \ {\tilde{\alpha }}(x_{\textbf{b}}(x,v),v)\thicksim |n(x_{\textbf{b}}(x,v))\cdot v|. \end{aligned}$$

Then, by the velocity lemma (3.32), we conclude that $|n(x_{\textbf{b}}(x,v))\cdot v|\thicksim |n(x)\cdot v|$. $\square$

Now we control the singularity of $|n(x_{\textbf{b}}(x,v))\cdot v|^{-p}$ from the following lemma:

Lemma 11

In a convex domain (2.2), for $p<3$, we have

$$\begin{aligned} \iint _{\Omega \times {{\mathbb {R}}}^3} \frac{w^{-p}_{{\tilde{\theta }}}(v)}{|n(x_{\textbf{b}}(x,v))\cdot v|^p}\textrm{d}x\textrm{d}v \lesssim 1. \end{aligned}$$

(5.7)

Proof

We directly compute the LHS of (5.7) as

$$\begin{aligned}&\iint _{\Omega \times {{\mathbb {R}}}^3} \frac{w^{-p}_{{\tilde{\theta }}}(v)}{|n(x_{\textbf{b}}(x,v))\cdot v|^p}\textrm{d}x\textrm{d}v \\&\quad \lesssim \int _{\gamma _+}\int _0^{t_{\textbf{b}}(x,v)} w^{-p}_{{\tilde{\theta }}}(v) \frac{|n(x)\cdot v|}{|n(x_{\textbf{b}}(x-sv,v))\cdot v|^p} \textrm{d}s \textrm{d}v \textrm{d}S_x \\&\quad \lesssim \int _{\gamma _+}w^{-p}_{{\tilde{\theta }}}(v) \frac{|n(x)\cdot v|t_{\textbf{b}}(x,v)}{|n(x_{\textbf{b}}(x,v))\cdot v|^p} \textrm{d}v \textrm{d}S_x\\&\quad \lesssim \int _{\gamma _+} w^{-p}_{{\tilde{\theta }}}(v) \frac{|n(x)\cdot v| |n(x_{\textbf{b}}(x,v))\cdot v|}{|v|^2|n(x_{\textbf{b}}(x,v))\cdot v|^p} \textrm{d}s \textrm{d}v \textrm{d}S_x \\&\quad \lesssim \int _{\gamma _+} w^{-p}_{{\tilde{\theta }}}(v) \frac{1}{|v|^2|n(x)\cdot v|^{p-2}} \textrm{d}v \textrm{d}S_x \lesssim 1. \end{aligned}$$

In the second line, we apply the change of variable (5.1). In the fourth line, we apply (5.5). In the last line, we apply (5.6). In the last inequality, for fixed $x\in \partial \Omega$ and fixed n(x), we used the spherical coordinate

$$\begin{aligned} v_{\parallel ,1}=r\sin \theta \cos \phi , \ v_{\parallel ,2} = r\sin \theta \sin \phi , \ n(x)\cdot v = r\cos \theta . \end{aligned}$$

Here $v_{\parallel ,i}$ is defined to be the projection of v to $\tau _i$, where $(\tau _1,\tau _2,n(x))$ is a basis of ${{\mathbb {R}}}^3$ such that $\tau _1\cdot \tau _2=0$ and $\tau _i\cdot n(x)=0$. Then, for $p<3$, we apply the spherical coordinate to get that

$$\begin{aligned} \int _{{{\mathbb {R}}}^3} \frac{w^{-p}_{{\tilde{\theta }}}(v)}{|v|^2 |n(x)\cdot v|^{p-2}}\textrm{d}v&\lesssim \int _0^{\pi } \frac{1}{|\cos \theta |^{p-2}} \textrm{d}\theta \int _0^{\infty } \textrm{d}r w_{{\tilde{\theta }}}^{-p}(r) \frac{r^2}{r^{p} } \lesssim 1. \end{aligned}$$

$\square$

We are ready to prove Theorem 3.

Proof of Theorem 3

Recall the weighted $C^1$ estimate for $f_s$ in (4.2):

$$\begin{aligned} \Vert w_{{\tilde{\theta }}}(v)\alpha \nabla _x f_s\Vert _\infty <\infty . \end{aligned}$$

Then

$$\begin{aligned} \Vert \nabla _x f_s \Vert _p&= \Big (\iint _{\Omega \times {{\mathbb {R}}}^3} |\nabla _x f_s|^p \textrm{d}x\textrm{d}v \Big )^{1/p} \\&\lesssim \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty \Big [\iint _{\Omega \times {{\mathbb {R}}}^3}\Big (\frac{w_{{\tilde{\theta }}}^{-1}(v)}{\alpha (x,v)} \Big )^{p} \textrm{d}x \textrm{d}v\Big ]^{1/p} \\&\lesssim \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty \Big ( \iint _{\Omega \times {{\mathbb {R}}}^3} \frac{w^{-p}_{{\tilde{\theta }}/2}(v)}{|n(x_{\textbf{b}}(x,v))\cdot v|^p} \textrm{d}x \textrm{d}v\Big )^{1/p} \\&\lesssim \Vert w_{{\tilde{\theta }}}\alpha \nabla _x f_s\Vert _\infty <\infty . \end{aligned}$$

In the last line, we have applied Lemma 11. In the third line we have used (3.34) and ${\tilde{\alpha }}(x,v)\lesssim |v|$ to get that

$$\begin{aligned} \frac{w^{-1}_{{\tilde{\theta }}}(v)}{\alpha (x,v)}&= \frac{1}{w_{{\tilde{\theta }}/2}(v){\tilde{\alpha }}(x,v)} \frac{{\tilde{\alpha }}(x,v)}{w_{{\tilde{\theta }}/2}(v)\alpha (x,v)} \\&\lesssim \frac{1}{w_{{\tilde{\theta }}/2}(v)|n(x_{\textbf{b}}(x,v))\cdot v|} \frac{\langle v\rangle }{w_{{\tilde{\theta }}/2}(v)} \lesssim \frac{1}{w_{{\tilde{\theta }}/2}(v)|n(x_{\textbf{b}}(x,v))\cdot v|} . \end{aligned}$$

We conclude (2.23).

Similarly, when all conditions in Theorem 2 are satisfied, we have

$$\begin{aligned} e^{\lambda t}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(t)\Vert _\infty <\infty , \end{aligned}$$

so

$$\begin{aligned} \sup _{t} e^{\lambda t}\Vert \nabla _x f(t) \Vert _p&=\sup _t \Big (\iint _{\Omega \times {{\mathbb {R}}}^3} |e^{\lambda t}\nabla _x f(t)|^p \textrm{d}x\textrm{d}v \Big )^{1/p} \\&\lesssim \sup _t e^{\lambda t}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(t)\Vert _\infty \Big [\iint _{\Omega \times {{\mathbb {R}}}^3}\Big (\frac{w_{{\tilde{\theta }}}^{-1}(v)}{\alpha (x,v)} \Big )^{p} \textrm{d}x \textrm{d}v\Big ]^{1/p} \\&\lesssim \sup _t e^{\lambda t}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(t)\Vert _\infty \Big ( \iint _{\Omega \times {{\mathbb {R}}}^3} \frac{w^{-p}_{{\tilde{\theta }}/2}(v)}{|n(x_{\textbf{b}}(x,v))\cdot v|^p} \textrm{d}x \textrm{d}v\Big )^{1/p} \\&\lesssim \sup _t e^{\lambda t}\Vert w_{{\tilde{\theta }}}\alpha \nabla _x f(t)\Vert _\infty <\infty , \end{aligned}$$

and we conclude (2.24). $\square$

Remark on the second order derivative

Here we illustrate our attempts for the second order derivative in the interior and highlight the obstacle. We investigate the second order derivative to the steady problem (4.72). Taking one more derivative to the (4.74)–(4.78) with $g=K(f_s)+\Gamma (f_s,f_s)$, we have

$$\begin{aligned} \partial _{x_i}\partial _{x_j} f_s(x,v)&= \textbf{1}_{t<t_{\textbf{b}}} e^{-\nu (v)t}\partial _{x_i}\partial _{x_j}f_s(x-tv,v) \end{aligned}$$

(A.1)

$$\begin{aligned}&-\textbf{1}_{t\ge t_{\textbf{b}}} \nu (v)\partial _{x_i}t_{\textbf{b}}(x,v) e^{-\nu (v)t_{\textbf{b}}(x,v)}\partial _{x_j}[f_s(x_{\textbf{b}}(x,v),v)] \end{aligned}$$

(A.2)

$$\begin{aligned}&+\textbf{1}_{t\ge t_{\textbf{b}}} e^{-\nu (v)t_{\textbf{b}}(x,v)} \partial _{x_j}[\partial _{x_i}[f_s(x_{\textbf{b}}(x,v)),v]] \end{aligned}$$

(A.3)

$$\begin{aligned}&-\textbf{1}_{t\ge t_{\textbf{b}}}\nu (v) \partial _{x_i}\partial _{x_j}t_{\textbf{b}}(x,v) e^{-\nu (v)t_{\textbf{b}}(x,v)} f_s(x_{\textbf{b}}(x,v),v) \end{aligned}$$

(A.4)

$$\begin{aligned}&+ \textbf{1}_{t\ge t_{\textbf{b}}}(\nu (v))^2 \partial _{x_j}t_{\textbf{b}}(x,v)\partial _{x_i}t_{\textbf{b}}(x,v) e^{-\nu (v)t_{\textbf{b}}(x,v)}f_s(x_{\textbf{b}}(x,v),v) \end{aligned}$$

(A.5)

$$\begin{aligned}&-\textbf{1}_{t\ge t_{\textbf{b}}}\nu (v)\partial _{x_j} t_{\textbf{b}}(x,v)e^{-\nu (v)t_{\textbf{b}}(x,v)}\partial _{x_i}[f_s(x_{\textbf{b}}(x,v),v)] \end{aligned}$$

(A.6)

$$\begin{aligned}&+ \int _{\max \{0,t-t_{\textbf{b}}\}}^t e^{-\nu (t-s)} K \partial _{x_i}\partial _{x_j} f_s(x-(t-s)v,v) \textrm{d}s \end{aligned}$$

(A.7)

$$\begin{aligned}&+\int _{\max \{0,t-t_{\textbf{b}}\}}^t e^{-\nu (t-s)} \partial _{x_i}\partial _{x_j} \Gamma (f_s,f_s)(x-(t-s)v,v) \textrm{d}s \end{aligned}$$

(A.8)

$$\begin{aligned}&-\textbf{1}_{t\ge t_{\textbf{b}}}\partial _{x_i}t_{\textbf{b}}e^{-\nu (v)t_{\textbf{b}}} K \partial _{x_j} f_s(x-t_{\textbf{b}}v,v) \end{aligned}$$

(A.9)

$$\begin{aligned}&- \textbf{1}_{t\ge t_{\textbf{b}}}\partial _{x_i}t_{\textbf{b}}e^{-\nu t_{\textbf{b}}}\partial _{x_j}\Gamma (f_s,f_s)(x-t_{\textbf{b}}v,v) \end{aligned}$$

(A.10)

$$\begin{aligned}&-\textbf{1}_{t\ge t_{\textbf{b}}}\partial _{x_i} \partial _{x_j} t_{\textbf{b}}e^{-\nu (v)t_{\textbf{b}}} (Kf_s+\Gamma (f_s,f_s))(x-t_{\textbf{b}}v,v) \end{aligned}$$

(A.11)

$$\begin{aligned}&+\textbf{1}_{t\ge t_{\textbf{b}}}\nu (v) \partial _{x_i}t_{\textbf{b}}\partial _{x_j}t_{\textbf{b}}e^{-\nu (v) t_{\textbf{b}}} (Kf_s+\Gamma (f_s,f_s))(x-t_{\textbf{b}}v,v) \end{aligned}$$

(A.12)

$$\begin{aligned}&+\textbf{1}_{t\ge t_{\textbf{b}}}\partial _{x_j}t_{\textbf{b}}e^{-\nu (v)t_{\textbf{b}}} \big \{\partial _{x_i}[Kf_s(x_{\textbf{b}}(x,v),v)] \nonumber \\&\quad + \partial _{x_i}[\Gamma (f_s,f_s)(x_{\textbf{b}}(x,v),v)]\big \}. \end{aligned}$$

(A.13)

We consider (A.9), which reads as

$$\begin{aligned}&\partial _{x_i} t_{\textbf{b}}e^{-\nu t_{\textbf{b}}} \int _{{{\mathbb {R}}}^3} \textrm{d}u\textbf{k}(v,u)\partial _{x_j} f_s(x_{\textbf{b}}(x,v),u) . \end{aligned}$$

(A.14)

On the boundary, we express the normal derivative as

$$\begin{aligned} v\cdot \nabla _x f_s = (n\cdot v)\partial _n f_s + \sum _{i=1,2}(\tau _i\cdot v)\partial _{\tau _i}f_s, \end{aligned}$$

where $\tau _1,\tau _2,n$ is a basis. Thus using the equation (4.72) the normal derivative exhibits a singularity as

$$\begin{aligned} \partial _{n}f_s(x,u) = \frac{-\sum _{i=1,2}(\tau _i\cdot u)\partial _{\tau _i}f_s-Lf_s+\Gamma (f_s,f_s)}{n(x)\cdot u}. \end{aligned}$$

Whenever $e_j\cdot n(x_{\textbf{b}}(x,v))\ne 0$, $\partial _{x_j}f(x_{\textbf{b}}(x,v),u)$ exhibits a singularity, which leads to

$$\begin{aligned} \mathrm{(A.9)} = \mathrm{(A.14)}&\thicksim \partial _{x_i}t_{\textbf{b}}e^{-\nu t_{\textbf{b}}} \int _{{{\mathbb {R}}}^3} \textrm{d}u \textbf{k}(v,u) \frac{1}{|n(x_{\textbf{b}}(x,v))\cdot u|}. \end{aligned}$$

Due to the fact that $\frac{1}{|n(x_{\textbf{b}}(x,v))\cdot u|}\notin L^1_u$, in the presence of the Boltzmann operator, the argument fails to control any second order derivatives.