On regularity of a Kinetic Boundary layer

Hongxu Chen Department of Mathematics, The Chinese University of Hong Kong, Shatin, N.T., Hong Kong, email: [email protected]

Abstract.

We study the nonlinear steady Boltzmann equation in the half space, with phase transition and Dirichlet boundary condition. In particular, we study the regularity of the solution to the half-space problem in the situation that the gas is in contact with its condensed phase. We propose a novel kinetic weight and establish a weighted $C^{1}$ estimate under the spatial domain $x\in[0,\infty)$ , which is unbounded and not strictly convex. Additionally, we prove the $W^{1,p}$ estimate without any weight for $p<2$ .

1. Introduction

1.1. Background

The regularity issue for the boundary value problem of the Boltzmann equation has been a challenging problem due to the singularity of the characteristic near the boundary and the nonlocal nature of the collision operator. Significant progress has been achieved in a convex bounded domain by Guo-Kim-Tonon-Trescases in [19], where the boundedness and strict convexity of the domain play crucial roles in the analysis (see [11] for a discussion on curvature). In this paper, we are interested in the regularity estimate in a domain that lacks both boundedness and strict convexity, specifically, we consider the half-space kinetic boundary layer problem. Other domains with similar properties are discussed in Remark 2.

We consider the phase transition problem in the kinetic theory of gas. This problem can be modeled by the steady Boltzmann equation in the half space with slab-symmetry and suitable boundary conditions, and converging to some Maxwellian equilibrium at the far field. Let $F(x,v)$ be the mass density of the gas particle at distance $x>0$ and velocity $v\in\mathbb{R}^{3}$ . Then the half-space problem is formulated as

\begin{cases}v_{1}\partial_{x}F(x,v)=Q(F,F)(x,v),\ v\in\mathbb{R}^{3},\ x>0,\\ F(x,v)\to M_{1,u,1}(v),\ \ \text{ as }x\to\infty.\end{cases}

(1.1)

The Maxwellian equilibrium in (1.1) is denoted as $M_{\rho,u,T}(v)$ . It is characterized by the constant parameters $\rho\in\mathbb{R}^{+}$ , $u\in\mathbb{R}$ , and $T\in\mathbb{R}^{+}$ , which represent the mass density, flow velocity, and temperature, respectively. The equilibrium is expressed as:

M_{\rho,u,T}(v)=\frac{\rho}{(2\pi T)^{3/2}}\exp\left(-\frac{(v_{1}-u)^{2}+v_{2}^{2}+v_{3}^{2}}{2T}\right).

(1.2)

In (1.1), $Q(F,F)$ is the Boltzmann collision operator. We consider the hard sphere model with an angular cut-off kernel, where $Q$ is defined as

Q(F,F)(x,v):=\int_{\mathbb{R}^{3}\times\mathbb{S}^{2}}(F(x,v^{\prime})F(x,v_{*}^{\prime})-F(x,v)F(x,v_{*}))|(v-v_{*})\cdot\omega|\mathrm{d}\omega\mathrm{d}v_{*}.

(1.3)

In (1.3) $v^{\prime},v_{*}^{\prime}$ represent the post-collision velocity, which can be determined through $\omega\in\mathbb{S}^{2}$ and the conservation of momentum and energy:

	$\displaystyle v^{\prime}+v_{}^{\prime}=v+v_{},\ \ \ \|v^{\prime}\|^{2}+\|v^{\prime}_{}\|^{2}=\|v\|^{2}+\|v_{}\|^{2},$
	$\displaystyle v^{\prime}=v-((v-v_{})\cdot\omega)\omega,\ v_{}^{\prime}=v_{}+((v-v_{})\cdot\omega)\omega.$

The well-posedness and asymptotic behavior of the linear half-space problem (linearized version of (1.1)), also known as Milne problem, have been studied in the pioneering work by Bardos-Caflisch-Nicolaenko [1]. The Milne problem serves as a fundamental boundary layer problem with specified incoming boundary condition. This theory has found significant applications in the hydrodynamic limit problem of the kinetic equation when there exhibits a mismatch between the kinetic and fluid boundaries, as demonstrated in [32, 13, 33].

For the nonlinear half-space problem (1.1) with Dirichlet boundary conditions, not all Dirichlet data are admissible and the admissible conditions depend on the far field Maxwellian. When $u$ does not take the singular values $\{0,+\sqrt{5/3},-\sqrt{5/3}\}$ , Ukai-Yang-Yu devised a penalization method to construct a unique solution in [30, 31], and the solution is proved to converge exponentially to $0$ as $x\to\infty$ . In this study, the convergence rate and the condition on the admissible boundary data depend on the parameter $u$ . When $u$ takes the singular value $u=0$ , Golse studied the well-posedness and boundary admissible condition in [15].

In our paper, we focus on the case of $0<|u|\ll 1$ with the Dirichlet boundary condition, which corresponds to the phase transition in the condensation-evaporation problem. For more comprehensive details on the boundary admissible condition of these problems, we refer readers to [2] and the reference therein. We also refer readers to [29, 28] for extensive numerical computation on these topics.

To begin, we denote the standard global Maxwellian by $M:=M_{1,0,1}$ with $M_{\rho,u,T}$ defined in (1.2). We apply a change of variable $\xi:=v-(u,0,0),$ and set the Boltzmann equation $F$ (1.1) as a perturbation around the Maxwellian

F(x,v)=(M+\sqrt{M}f)(x,v-(u,0,0)).

The equation of the perturbation $f$ , with $\xi=v-(u,0,0)$ , reads

\begin{cases}(\xi_{1}+u)\partial_{x}f(x,\xi)+\mathcal{L}f(x,\xi)=\Gamma(f,f)\\ f(x,\xi)\to 0\text{ as }x\to\infty.\end{cases}

(1.4)

Here $\mathcal{L}f$ is the linear Boltzmann operator, which is defined as

\mathcal{L}f:=\nu(\xi)f-K(f)=-\frac{Q(M,\sqrt{M}f)}{\sqrt{M}}-\frac{Q(\sqrt{M}f,M)}{\sqrt{M}}.

(1.5)

$\Gamma(f,f)$ is the nonlinear Boltzmann operator defined as

\Gamma(f,g):=\frac{Q(\sqrt{M}f,\sqrt{M}g)+Q(\sqrt{M}g,\sqrt{M}f)}{2\sqrt{M}}.

(1.6)

The properties of these Boltzmann operators are summarized in Lemma 1.

Now we discuss boundary conditions of (1.4). We assume that the gas is in contact with its condensed phase as in [3]. At $x=0$ , we impose a Dirichlet boundary condition:

f(0,\xi)=f_{b}(\xi),\ \ \ \xi_{1}+u>0.

(1.7)

For other boundary conditions such as diffuse boundary condition and specular boundary condition, the well-posedness and asymptotic behavior are obtained in [16, 12], and recently, with continuity, in [20, 21] under different functional spaces. We refer to [23, 24] for more recent progress on the range of Mach number.

Recently, for $u$ closed to the singular value $0$ , Bernhoff and Golse [3] proposed an elegant approach to establish the well-posedness and asymptotic stability under certain assumptions on the boundary data(also see [27]). In this result, the solution $f$ is proved to exhibit slab symmetry and converge exponentially as $x\to\infty$ , with the convergence rate being uniform in $|u|\ll 1$ . We document this well-posedness result in Theorem 3 in Section 2.

1.2. Main result

While the well-posedness of (1.4) is well understood in [3] and other literature, the regularity of such a boundary layer problem remains open. In this paper, we tackle this problem by constructing the weighted $C^{1}$ estimate and $W^{1,p},p<2$ estimate without weight.

It is well-known that with the presence of the boundary, the regularity of the Boltzmann equation possesses singularity due to the non-local nature of the Boltzmann operator. As demonstrated in [26, 17, 25, 18], in a general convex domain, the spatial derivative $\nabla_{x}f(x,\xi)$ generates singularity as $1/(n(x_{\mathbf{b}})\cdot\xi)$ , where $x_{\mathbf{b}}$ corresponds to the backward exit position defined as

\displaystyle x_{\mathbf{b}}(x,\xi)=x-t_{\mathbf{b}}(x,\xi)\xi,\ t_{\mathbf{b}}(x,\xi)=\sup\{s\geq 0,\ x-s\xi\in\Omega\}.

In particular, in [19], Guo-Kim-Tonon-Trescases proposed a kinetic weight that can compensate the singularity and capture the convexity of domain at the same time. This innovative approach successfully overcomes the challenges posed by the non-local properties of the Boltzmann equation and establishes a weighted $C^{1}$ estimate. The introduction of kinetic weight has significantly advanced the understanding of the Boltzmann regularity, see [8, 6, 5, 10].

In our specific problem, it is evident that $\partial_{x}f$ exhibits a singularity as $1/(\xi_{1}+u)$ . However, due to the lack of strict convexity and boundedness in the domain, we need to introduce a new kinetic weight to address this singularity. The proposed kinetic weight is defined as follows:

Definition 1 (Kinetic weight).

We define

\tilde{\alpha}(x,\xi):=\sqrt{|\xi_{1}+u|^{2}+(c\nu_{0})^{2}x^{2}},

(1.8)

where $\nu_{0}$ corresponds to the lower bound of the collision frequency defined in (2.8). $0<c<1$ is a constant that will be specified later in (2.13).

We introduce a cut-off function for the kinetic weight. Define $\chi:[0,\infty)\rightarrow[0,\infty)$ which stands for a non-decreasing smooth function such that

\chi(s)=s\ \text{for}\ s\in[0,1/2],\ \chi(s)=1\ \text{for}\ s\in[2,\infty),\ and\ |\chi^{\prime}(s)|\leq 1\ \text{for}\ s\in[0,\infty).

(1.9)

Then we introduce a cut-off function to the kinetic weight:

\alpha(x,\xi):=\chi(\tilde{\alpha}(x,\xi)).

(1.10)

With the extra cutoff function, we directly have

\alpha(x,\xi)\leq 1,\ \ \ 1\leq\frac{1}{\alpha(x,\xi)}.

(1.11)

Remark 1.

On the boundary $x=0$ , clearly we have

\displaystyle\alpha(0,\xi)\leq|\xi_{1}+u|,\ \ \frac{1}{|\xi_{1}+u|}\leq\frac{1}{\alpha(0,\xi)}.

(1.12)

Thus the singularity $\frac{1}{|\xi_{1}+u|}$ can be compensated by $\alpha(x,\xi)$ when $|\xi_{1}+u|\ll 1$ .

From (1.4), we can control $\partial_{x}f(x,\xi)$ using a trivial weight $\xi_{1}+u$ . At the boundary $x=0$ , $\partial_{x}f$ is expected to be discontinuous and possess a singularity as $\frac{1}{\xi_{1}+u}$ . Away from the boundary, $f$ is expected to be continuous; however, the trivial weight $\xi_{1}+u$ does not provide information regarding this continuity. Our weight (1.8) is non-zero away from the boundary, and controlling $\alpha\partial_{x}f$ implies the desired estimate for $\partial_{x}f$ when $x>0$ , as illustrated in Theorem 1.

Remark 2.

The kinetic weight proposed in [19] is almost invariant along the characteristic thanks to the strict convexity and boundedness. It is important to note that even under a flat and unbounded domain, our weight (1.8) still possesses favorable properties similar to [19] in the following two aspects.

I. Due to the presence of the spatial variable $x$ in (1.8), our weight $\alpha(x,\xi)$ does not remain invariant along the characteristic. In fact, the rate of change along the characteristic of these weights can be explicitly computed (see Lemma 4 and Lemma 5). Importantly, it turns out that the rate of change depends on the choice of $c$ . Meanwhile, the collision frequency $\nu$ from the linear Boltzmann operator $\mathcal{L}$ is bounded below by $\nu_{0}$ according to Lemma 1. By selecting an appropriate $c$ , the growth factor $e^{2c\nu_{0}s}$ in Lemma 5 along the characteristic can be compensated by the damped factor $e^{-\nu s}$ .

II. The extra spatial variable $x$ in (1.8) plays a key role in the regularity estimate of the non-local Boltzmann operator. For example, in (1.23), the integral over $\mathrm{d}\xi^{\prime}$ does not blow up; instead, it becomes a function of the spatial variable $x$ . This allows us to proceed the computation and focus on the $\mathrm{d}s$ integral. We refer more detailed computation to Lemma 6.

We anticipate that this type of kinetic weight can be adapted to study the regularity issue in unbounded domains with flat boundaries. For instance, the initial boundary value problem of the Boltzmann equation in [22], [4] with domain $\mathbb{T}^{2}\times\mathbb{R}^{+}$ , and in [7] with domain $\mathbb{R}^{2}\times(-1,1)$ .

In the following result, we establish the regularity estimate of the solution to the boundary layer problem (1.4). We denote the exponential weight in velocity $\xi$ as

\begin{split}&w(\xi):=e^{\theta|\xi|^{2}},\ \theta<\frac{1}{4},\ \ \ w_{\tilde{\theta}}(\xi):=e^{\tilde{\theta}|\xi|^{2}}\text{ for }\tilde{\theta}\ll\theta.\end{split}

(1.13)

We define the grazing set as

\mathcal{D}:=\{(0,\xi)|\xi\in\mathbb{R}^{3},\ \ \xi_{1}+u=0\}.

(1.14)

We denote the inner product and orthongonal matrix as

\displaystyle\langle f\rangle:=\int_{\mathbb{R}^{3}}f(\xi)\mathrm{d}\xi,

(1.15)

\mathcal{R}:=\begin{pmatrix}1&0&0\\ 0&-1&0\\ 0&0&-1\\ \end{pmatrix}.

(1.16)

Theorem 1.

Assume $0<|u|<r$ for some $r\ll 1$ . Assume that for some $\varepsilon\ll 1$ , the boundary data $f_{b}$ (1.7) satisfy

f_{b}\circ\mathcal{R}=f_{b},\ f_{b}\in C(\mathbb{R}^{3})\text{ and }\|wf_{b}\|_{L^{\infty}_{\xi}}\leq\varepsilon,\ \ \mathcal{R}\text{ defined in }\eqref{matrix},

(1.17)

and

\langle(\xi_{1}+u)Y_{1}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=\langle(\xi_{1}+u)Y_{2}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=0.

(1.18)

Here $Y_{1}[u]$ , $Y_{2}[u]$ , and $\mathfrak{R}_{u,\gamma}$ are defined in Appendix A.

We further assume that (1.4) with boundary condition (1.7) has a unique solution $f$ such that for some $C>0$ and $\gamma\ll 1$ ,

\displaystyle f\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D}),\ \|wf(x,\xi)\|_{L^{\infty}_{\xi}}\leq C\varepsilon e^{-\gamma x}.

(1.19)

Then for some $C_{\infty}>0$ , the unique solution $f$ satisfies the weighted $C^{1}$ estimate:

\|w_{\tilde{\theta}}\alpha\partial_{x}f\|_{L^{\infty}_{x,\xi}}\leq C_{\infty}\varepsilon e^{-\gamma x}.

(1.20)

For $1\leq p<2$ and $\gamma_{0}<\gamma$ , the solution $f$ also satisfies the $W^{1,p}$ and $H^{1}_{loc}$ estimate:

\displaystyle\|w_{\tilde{\theta}/2}e^{\gamma_{0}x}\partial_{x}f\|_{L^{p}_{x,\xi}}\leq C_{p}\varepsilon\text{ for some }C_{p}>0,

(1.21)

\displaystyle\int_{\delta}^{\infty}\int_{\mathbb{R}^{3}}w_{\tilde{\theta}}(\xi)e^{2\gamma_{0}x}|\partial_{x}f|^{2}\mathrm{d}\xi\mathrm{d}x\leq C_{\delta}\text{ for any }\delta>0\text{ and some }C_{\delta}>0.

(1.22)

Remark 3.

The continuity and $w$ -weighted $L^{\infty}$ estimate assumption (1.19) will be justified in Theorem 2.

Bernhoff and Golse construct a unique solution $f$ using polynomial weight $(1+|\xi|^{3})$ (see Theorem 3) in [3]. In this study, they focus on the well-posedness analysis of a penalized problem, which subsequently establishes the well-posedness of the original problem (1.4) by eliminating the penalization. To remove the penalization, additional assumptions (1.18) are introduced. The assumptions (1.18) also imply that the incoming boundary forms a manifold of co-dimension two for given $u$ , as stated in [3].

In our paper, we do not incorporate these additional conditions in our analysis. For precise definitions of the functions $Y_{1}[u]$ , $Y_{2}[u]$ , and $\mathfrak{R}_{u,\gamma}$ in (1.18), we refer readers to the Appendix A and also to [3].

Remark 4.

For the $W^{1,p}$ estimate with $p<2$ , the exponential weight $w_{\tilde{\theta}}(\xi)$ and exponential decay $e^{-\gamma x}$ in the weighted $C^{1}$ estimate (1.20) guarantee the integrability as $x,|\xi|\to\infty$ . For a general convex, smooth, and bounded domain, [9] establishes a $W^{1,p}$ estimate for $p<3$ by employing control over the backward exit time $t_{\mathbf{b}}\lesssim|n(x_{\mathbf{b}})\cdot\xi|/|\xi|^{2}$ . This control allows for a gain in local integrability with respect to the singularity $1/(n(x_{\mathbf{b}})\cdot\xi)$ . We notice that in our paper, we do not have the benefit of such an additional control since our domain is unbounded and not strictly convex.

Remark 5.

The $H^{1}_{loc}$ estimate (1.22) coincides with the regularity of the Milne problem, as stated in Theorem 3.3 in [1]:

\displaystyle\int_{\delta}^{\infty}\int_{\mathbb{R}^{3}}(1+|\xi|)e^{2\gamma x}|\partial_{x}f|^{2}\mathrm{d}\xi\mathrm{d}x\leq C_{\delta}.

Bardos-Caflisch-Nicolaenko derived this upper bound through an $L^{2}$ energy estimate to the equation of $\partial_{x}f$ . In our Theorem 1, we deal with the nonlinear problem, and we derive (1.22) through direct computation using the weighted $C^{1}$ estimate (1.20). In this computation, the extra weight in (1.20) generates extra singularity $\frac{1}{|\alpha(x,\xi)|^{2}}$ in integration, which is controllable away from $x=0$ , as demonstrated in Section 1.3.

The assumption (1.19) is justified in the following theorem.

Theorem 2.

There exists $\varepsilon$ , $r$ , $\gamma$ and $C$ such that if the boundary data $f_{b}$ (1.7) satisfies

f_{b}\circ\mathcal{R}=f_{b},\ f_{b}\in C(\mathbb{R}^{3})\text{ and }\|wf_{b}\|_{L^{\infty}_{\xi}}\leq\varepsilon,\ \ \mathcal{R}\text{ defined in }\eqref{matrix},

\langle(\xi_{1}+u)Y_{1}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=\langle(\xi_{1}+u)Y_{2}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=0.

for each $0<|u|\leq r$ . Then (1.4) with boundary condition (1.7) has a unique solution $f$ that is continuous away from the grazing set: $f\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D})$ , and

f(x,\mathcal{R}\xi)=f(x,\xi)\text{ for a.e }(x,\xi)\in\mathbb{R}_{+}\times\mathbb{R}^{3}.

Moreover, $f$ satisfies the uniform-in- $u$ decay estimate

\|wf(x,\xi)\|_{L^{\infty}_{\xi}}\leq C\varepsilon e^{-\gamma x}.

Remark 6.

The $w$ -weighted $L^{\infty}$ estimate (1.19) is crucial in constructing the weighted $C^{1}$ regularity (1.20) in Theorem 1. As stated in (2.7) in Lemma 1, the nonlinear operator $\Gamma$ shares a similar structure with the linear operator $Kf$ (2.2) by employing the weight $w$ . This allows us to control the contribution of both $K$ and $\Gamma$ in the weighted $C^{1}$ estimate using a uniform strategy.

1.3. Difficulty and main idea

1. Continuity proof by penalization. To begin, we justify the continuity and $w$ weighted $L^{\infty}$ estimate assumption (1.19) in Theorem 2. Following the approach in [3], we introduce a linearized penalized problem, as described in Proposition 5. The construction of the penalized operator is based on solving the eigen-value problem (3.1). In [3], the corresponding eigen-function $\phi_{u}$ is proved to be bounded in $L^{\infty}$ with polynomial weight $(1+|\xi|)^{s}$ . To prove $\|wg\|_{L^{\infty}_{x,\xi}}$ , the first step is to control the eigen-function $\phi_{u}$ in $L^{\infty}$ with exponential weight. We use a modified linear Boltzmann operator,

\displaystyle\mathcal{L}_{\theta}\phi_{u}

\displaystyle=-\frac{Q(M,\sqrt{M}e^{-\theta|\xi|^{2}}\phi_{u})+Q(\sqrt{M}e^{-\theta|\xi|^{2}}\phi_{u},M)}{\sqrt{M}e^{-\theta|\xi|^{2}}}=\nu(\xi)\phi_{u}-K_{\theta}\phi_{u}.

The modified linear operator $K_{\theta}$ shares a structure similar to the original operator $K$ (see Lemma 3). The eigen-value problem associated with the modified operator can be solved in the same spirit. We obtain the eigen-function to the original problem as $e^{-\theta|\xi|^{2}}\phi_{u}$ with the desired weight(see Lemma 8 for detail).

Subsequently, we employ a fixed-point argument in the $w$ -weighted $L^{\infty}$ space to establish the well-posedness and the continuity, for the linearized penalized problem in Proposition 6, and after, for the nonlinear penalized problem, in Proposition 8.

2. Regularity

2-a. Weighted $C^{1}$ estimate. The construction of the derivative involves fixed-point argument to the penalized problem (3.11) with an a-priori weighted $C^{1}$ estimate. In the following, we illustrate the main difficulty and idea in the a-priori estimate(Lemma 12).

The difficulty of the singularity $\frac{1}{\xi_{1}+u}$ is addressed by the proposed kinetic weight $\alpha(x,\xi)$ . To control its rate of change along the characteristic(Lemma 5), we set $c=1/8$ and thus this growth rate can be controlled by the collision frequency $\nu$ in (1.5).

The derivative along the characteristic is given by the piece-wise formula (4.5)-(4.10). The main difficulty comes from the contribution of the linear operator $K(\partial_{x}g)$ in (4.8). By employing the Grad estimate (2.2) to $K(\partial_{x}g)$ , we see a $\mathrm{d}\xi^{\prime}$ integration in this term as

\displaystyle\int^{t}_{0}\mathrm{d}se^{-\nu(t-s)}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi^{\prime})\mathrm{d}\xi^{\prime}.

From a standard technique, we iterate along the characteristic with velocity $\xi^{\prime}$ again and obtain a double Duhamel’s formula(in (4.15)):

\displaystyle\int_{0}^{t}\mathrm{d}se^{-\nu(t-s)}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\int^{s}_{0}\mathrm{d}s^{\prime}e^{-\nu(s-s^{\prime})}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\partial_{x}g(x-(\xi_{1}+u)(t-s)-(\xi_{1}^{\prime}+u)(s-s^{\prime}),\xi^{\prime\prime}).

To control this term, we expect to utilize the estimate on $\|g\|_{\infty}$ or gain a smallness contribution from the time integration. To achieve our goal, we split the $\mathrm{d}s^{\prime}$ integral into two regions: $s-s^{\prime}\leq\varepsilon$ and $s-s^{\prime}>\varepsilon$ . When $s-s^{\prime}>\varepsilon$ , we observe the following change of variable

\displaystyle\frac{\partial_{\xi^{\prime}_{1}}[g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})]}{-(s-s^{\prime})}=\partial_{x}g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime}).

With the lower bound of $s-s^{\prime}$ , we convert the $x$ -derivative $\partial_{x}g$ into a velocity $\xi_{1}$ derivative $\partial_{\xi_{1}}[g]$ . Via an integration by part, we remove the derivative in $\partial_{\xi_{1}}[g]$ and control the $g$ using the $L^{\infty}$ estimate mentioned in the previous paragraphs(Theorem 2). On the other hand, the case $s-s^{\prime}<\varepsilon$ corresponds to a small time contribution, we need to incorporate the nonlocal operator $K(\partial_{x}g)$ with the kinetic weight $\alpha$ . By introducing the kinetic weight $\alpha$ in the Duhamel’s formula and isolating $\|\alpha\partial_{x}g\|_{L^{\infty}_{\xi}}$ , we arrive at the following integral:

\displaystyle\int_{t-\varepsilon}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}\frac{1}{\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})}.

(1.23)

This type of estimate was first studied in [19] when the domain is bounded and strictly convex. In our case, the integrability in $\mathrm{d}\xi^{\prime}$ arises from the additional $x^{2}$ term in (1.8). The $\mathrm{d}\xi^{\prime}$ integration can be explicitly computed as a function of $\xi,s$ (see (2.23) and (2.24) for detail):

\displaystyle\int_{t-\varepsilon}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)}\ln\Big{(}1+\Big{|}\frac{1}{x-(\xi_{1}+u)(t-s)}\Big{|}^{2}\Big{)}.

We aim to control this integral by $\frac{1}{\alpha(x,v)}$ to close the weighted $C^{1}$ estimate $\|\alpha\partial_{x}g\|_{\infty}.$ Despite the lack of strict convexity and boundedness, our setting has one beneficial aspect - the spatial variable is one-dimensional. Along the characteristic, the spatial variable can be explicitly computed as $x-(\xi_{1}+u)(t-s)$ . We tackle this integration by carefully comparing the scale of $\frac{1}{x-(\xi_{1}+u)(t-s)}$ with the scale of $\frac{1}{\alpha(x,v)}$ (approximately $\frac{1}{\sqrt{|\xi_{1}+u|^{2}+x^{2}}}$ in (1.8)). We refer the detailed analysis and conclusion of this integral to Lemma 6.

The contribution of the nonlinear operator (4.10) can be controlled in the same spirit. The corresponding integral reads

	$\displaystyle\int^{t}_{0}\mathrm{d}se^{-\nu(t-s)}\partial_{x}\Gamma(g,g)(x-(\xi_{1}+u)(t-s),\xi)\mathrm{d}\xi$
	$\displaystyle\lesssim\\|wg\\|_{L^{\infty}_{x,\xi}}\int_{0}^{t}\mathrm{d}e^{-\nu(t-s)}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\|\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi^{\prime})\|\mathrm{d}\xi^{\prime}.$

Here the $w$ weight plays an important role to have the kernel $\mathbf{k}(\xi,\xi^{\prime})$ in the integration, as emphasized in Remark 6. The smallness of $\|wg\|_{L^{\infty}_{x,\xi}}$ plays the same role as the small time contribution discussed in the previous paragraph. In fact, the large time integration includes an additional growing factor in the time scale $t$ , as stated in Lemma 6. Consequently, we set $t$ to be large but fixed, and absorb it using the smallness of $\|wg\|_{L^{\infty}_{x,\xi}}$ .

2-b. $W^{1,p}$ estimate. The $W^{1,p}$ estimate without weight in (1.21) is derived after obtaining the weighted $C^{1}$ estimate. In the previous paragraph we establish the weighted $C^{1}$ estimate to the penalized problem $g$ . The solution $f$ to (1.4) is set to be (see Section 3.4 for detail)

f(x,\xi)=e^{-\gamma x}g(x,\xi).

The exponential decay factor $e^{-\gamma x}$ and the velocity weight in $\|w_{\tilde{\theta}}(\xi)\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}$ provide the integrability as $x,|\xi|\to\infty$ . In (1.21), the constraint of $p<2$ arises from the local integrability of the term $\frac{1}{\alpha^{p}(x,\xi)}$ . The local integral of $\frac{1}{\alpha^{p}(x,\xi)}\sim\frac{1}{(|\xi_{1}+u|^{2}+x^{2})^{p/2}}$ corresponds to a two-dimensional integration involving the variables $\xi_{1}$ and $x$ . Therefore, this integral over $x\in(0,1)$ is bounded for $p<2$ , and the integral over $x\in(\delta,1)$ is bounded by some constant $C_{\delta}$ for $\delta>0$ and $p=2$ . This justifies the $H^{1}_{loc}$ estimate in (1.22).

Outline. In Section 2, we provide the properties of the kinetic weight and Boltzmann operator as preliminary material. In Section 3, we introduce the penalized problem and analyze its continuity as well as the exponential weighted $L^{\infty}$ estimate. This analysis allows us to justify the assumption (1.19) and conclude Theorem 2 after removing the penalization. In Section 4, we apply the properties of kinetic weight to conclude Theorem 1 by establishing the weighted $C^{1}$ and $W^{1,p}$ estimate for $p<2$ .

2. Preliminary

First, we document the well-posedness result of (1.4) in [3].

Theorem 3 (Theorem 2.1 in [3]).

There exists $\varepsilon$ , $r$ , $\gamma$ and $C$ such that if the boundary data $f_{b}$ (1.7) satisfies

f_{b}\circ\mathcal{R}=f_{b}\text{ and }\|(1+|\xi|)^{3}f_{b}\|_{L^{\infty}_{\xi}}\leq\varepsilon,

(2.1)

for each $0<|u|\leq r$ . Then (1.4) with boundary condition (1.7) has a unique solution that satisfies

f(x,\mathcal{R}\xi)=f(x,\xi)\text{ for a.e }(x,\xi)\in\mathbb{R}_{+}\times\mathbb{R}^{3}

and the uniform decay estimate

\|(1+|\xi|)^{3}f(x,\xi)\|_{L^{\infty}_{\xi}}\leq C\varepsilon e^{-\gamma x}

if and only if the boundary data $f_{b}$ satisfies further conditions

\langle(\xi_{1}+u)Y_{1}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=\langle(\xi_{1}+u)Y_{2}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=0.

Here $Y_{1}[u]$ , $Y_{2}[u]$ , and $\mathfrak{R}_{u,\gamma}$ are defined in Appendix A.

2.1. Properties of the Boltzmann operator in (1.5) and (1.6).

Lemma 1.

The linear Boltzmann operator $K(f)$ in (1.4) is given by

\displaystyle Kf(x,\xi)=\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})f(x,\xi^{\prime})\mathrm{d}\xi^{\prime}.

(2.2)

The kernel $\mathbf{k}(\xi,\xi^{\prime})=\mathbf{k}_{1}(\xi,\xi^{\prime})+\mathbf{k}_{2}(\xi,\xi^{\prime})$ is given by the Grad estimate [14]:

	$\displaystyle\mathbf{k}_{1}(\xi,\xi^{\prime})$	$\displaystyle=C_{\mathbf{k}_{1}}\|\xi-\xi^{\prime}\|e^{-\frac{\|\xi\|^{2}+\|\xi^{\prime}\|^{2}}{4}},$		(2.3)
	$\displaystyle\mathbf{k}_{2}(\xi,\xi^{\prime})$	$\displaystyle=C_{\mathbf{k}_{2}}\frac{1}{\|\xi-\xi^{\prime}\|}e^{-\frac{1}{8}\|\xi-\xi^{\prime}\|^{2}-\frac{1}{8}\frac{(\|\xi\|^{2}-\|\xi^{\prime}\|^{2})^{2}}{\|\xi-\xi^{\prime}\|^{2}}}.$		(2.4)

The kernel satisfies

|\mathbf{k}(\xi,\xi^{\prime})|\lesssim\mathbf{k}^{\varrho}(\xi,\xi^{\prime}),\ \ \mathbf{k}^{\varrho}(\xi,\xi^{\prime}):=e^{-\varrho|\xi-\xi^{\prime}|^{2}}/|\xi-\xi^{\prime}|.

For $\nu$ and $\Gamma$ in (1.4), we have

\nu(\xi)\gtrsim 1+|\xi|,\quad|\nabla_{\xi}\nu(\xi)|\lesssim 1,

(2.5)

\begin{split}\Big{\|}\frac{w}{1+|\xi|}\Gamma(f,g)(x,\xi)\Big{\|}_{L^{\infty}_{x,\xi}}&\lesssim\|wf\|_{L^{\infty}_{x,\xi}}\times\|wg\|_{L^{\infty}_{x,\xi}},\end{split}

(2.6)

\begin{split}|\nabla_{x}\Gamma(f,g)(x,\xi)|&\lesssim[1+|\xi|]\|wf\|_{L^{\infty}_{x,\xi}}|\nabla_{x}g(x,\xi)|+\|wf\|_{L^{\infty}_{x,\xi}}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})|\nabla_{x}g(x,\xi^{\prime})|\mathrm{d}\xi^{\prime}\\ &+\|wg\|_{L^{\infty}_{x,\xi}}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})|\nabla_{x}f(x,\xi^{\prime})|\mathrm{d}\xi^{\prime}.\end{split}

(2.7)

Proof.

The proof of (2.3) – (2.5) can be found in [14]. The proof of (2.6) can be found in [17].

We only prove (2.7). We take $x$ -derivative to have

	$\displaystyle\|\nabla_{x}\Gamma(f,g)\|$	$\displaystyle=\|\Gamma(\nabla_{x}f,g)+\Gamma(f,\nabla_{x}g)\|$
		$\displaystyle\leq\\|wf\\|_{L^{\infty}_{x,\xi}}\Big{\|}\Gamma(e^{-\theta\|\cdot\|^{2}},\nabla_{x}g)\Big{\|}+\\|wg\\|_{L^{\infty}_{x,\xi}}\Big{\|}\Gamma(\nabla_{x}f,e^{-\theta\|\cdot\|^{2}})\Big{\|}.$

The $\Gamma$ terms above have the same structure as the linear operator $\mathcal{L}$ in (1.5) with replacing $\mu$ by a different exponent $\sqrt{\mu}e^{-\theta|\xi|^{2}}$ . Then (2.7) follows the expression of $Kf$ in (2.3) and (2.4) with replacing $1/8$ by another coefficient. For ease of notation, we keep the same $\mathbf{k}(\xi,\xi^{\prime})$ in (2.7).

∎

Lemma 2.

In the linear operator (1.5), the collision frequency $\nu(\xi)$ satisfies

\nu_{0}[1+|\xi|]\leq\nu(\xi)\leq\nu_{1}[1+|\xi|].

(2.8)

The kernel of $\mathcal{L}$ is

\text{Ker}L=\text{Span}(X_{+},X_{-},X_{0},\xi_{2}\sqrt{M},\xi_{3}\sqrt{M}),

where

X_{\pm}=\frac{1}{\sqrt{30}}(|\xi|^{2}+\sqrt{15}\xi_{1})\sqrt{M},\ \ X_{0}=\frac{1}{\sqrt{10}}(|\xi|^{2}-5)\sqrt{M}.

(2.9)

Denote

\langle\phi\rangle:=\int_{\mathbb{R}^{3}}\phi(\xi)\mathrm{d}\xi.

The family $(X_{+},X_{0},X_{-},\xi_{2},\xi_{3})$ is orthonormal in $L^{2}(\mathbb{R}^{3};\mathrm{d}\xi)$ , and is orthogonal for the bilinear form $(f,g)\to\langle\xi_{1}fg\rangle$ , with

\langle\xi_{1}X_{\pm}^{2}\rangle=\pm\sqrt{\frac{5}{3}},\ \ \langle\xi_{1}X_{0}^{2}\rangle=\langle\xi_{1}X_{2}^{2}\rangle=\langle\xi_{1}\xi_{3}^{2}\rangle=0.

Proof.

The proof is standard and we omit it. ∎

Lemma 3.

Let $0\leq\theta<\frac{1}{4}$ , denote $\mathbf{k}_{\theta}(\xi,\xi^{\prime}):=\mathbf{k}(\xi,\xi^{\prime})\frac{e^{\theta|\xi|^{2}}}{e^{\theta|\xi^{\prime}|^{2}}}.$ Then we have

\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\frac{e^{\theta|\xi|^{2}}}{e^{\theta|\xi^{\prime}|^{2}}}\mathrm{d}\xi^{\prime}\lesssim\frac{C}{1+|\xi|}.

(2.10)

And there exists $C_{\theta}>0$

\mathbf{k}_{\theta}(\xi,\xi^{\prime})\lesssim\frac{e^{-C_{\theta}|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}.

(2.11)

The derivative on $\mathbf{k}(\xi,\xi^{\prime})$ shares similar property: for $0<\tilde{\theta}\ll\theta,$

|\nabla_{\xi}\mathbf{k}(\xi,\xi^{\prime})|\frac{e^{\tilde{\theta}|\xi|^{2}}}{e^{\tilde{\theta}|\xi^{\prime}|^{2}}}\lesssim\frac{[1+|\xi|^{2}]e^{-C_{\theta}|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|^{2}}.

(2.12)

Proof.

The proof of (2.10) and (2.11) can be found in [17]. We provide the proof of (2.12).

Taking the derivative to (2.3) and (2.4) we have

\displaystyle\nabla_{v}\mathbf{k}_{1}(\xi,\xi^{\prime})

\displaystyle=\frac{\xi-\xi^{\prime}}{|\xi-\xi^{\prime}|}\mathbf{k}_{1}(\xi,\xi^{\prime})-\xi\mathbf{k}_{1}(\xi,\xi^{\prime}),

	$\displaystyle\nabla_{v}\mathbf{k}_{2}(\xi,\xi^{\prime})$	$\displaystyle=\frac{\xi-\xi^{\prime}}{\|\xi-\xi^{\prime}\|^{2}}\mathbf{k}_{2}(\xi,\xi^{\prime})$
		$\displaystyle-\mathbf{k}_{2}(\xi,\xi^{\prime})\Big{[}\frac{\xi-\xi^{\prime}}{4}+\frac{\xi(\|\xi^{\prime}\|^{2}-\|\xi\|^{2})\|\xi-\xi^{\prime}\|^{2}-(\|\xi^{\prime}\|^{2}-\|\xi\|^{2})^{2}(\xi-\xi^{\prime})}{4\|\xi-\xi^{\prime}\|^{4}}\Big{]}.$

Since $|\xi^{\prime}|^{2}-|\xi|^{2}=|\xi^{\prime}-\xi|^{2}+2\xi\cdot(\xi^{\prime}-\xi)$ , we have

	$\displaystyle\big{\|}\|\xi^{\prime}\|^{2}-\|\xi\|^{2}\big{\|}\lesssim\|\xi^{\prime}-\xi\|^{2}+\|\xi\|\|\xi^{\prime}-\xi\|,$
	$\displaystyle\big{\|}\|\xi^{\prime}\|^{2}-\|\xi\|^{2}\big{\|}^{2}\lesssim\|\xi^{\prime}-\xi\|^{4}+\|\xi\|^{2}\|\xi^{\prime}-\xi\|^{2}.$

This leads to

	$\displaystyle\Big{\|}\nabla_{\xi}\mathbf{k}(\xi,\xi^{\prime})\frac{e^{\tilde{\theta}\|\xi\|^{2}}}{e^{\tilde{\theta}\|\xi^{\prime}\|^{2}}}\Big{\|}$	$\displaystyle\lesssim\big{[}\frac{1+\|\xi^{\prime}\|^{2}+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}[\mathbf{k}_{1}(\xi,\xi^{\prime})+\mathbf{k}_{2}(\xi,\xi^{\prime})]\frac{e^{\tilde{\theta}\|\xi\|^{2}}}{e^{\tilde{\theta}\|\xi^{\prime}\|^{2}}}$
		$\displaystyle\lesssim\big{[}\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}\mathbf{k}^{C_{\theta}}(\xi,\xi^{\prime})$
		$\displaystyle=\big{[}\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}\frac{e^{-C_{\theta}\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}\lesssim\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}\frac{e^{-c\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}$

for some $c<C_{\theta}$ . In the second line we have applied (2.10). In the last line we used that for $c<C_{\theta}$ , we have

\displaystyle e^{-C_{\theta}|\xi-\xi^{\prime}|^{2}}\lesssim\frac{e^{-c|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|},\ e^{-C_{\theta}|\xi-\xi^{\prime}|^{2}}\lesssim\frac{e^{-c|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|^{2}}.

For ease of notation, we conclude (2.12) with coefficient $C_{\theta}$ .

∎

2.2. Properties of kinetic weight $\alpha$ in (1.8).

A key property of the kinetic weight (1.8) is that $\tilde{\alpha}$ grows exponentially fast with a factor $c$ along the characteristic. We take

c:=\frac{1}{8},

(2.13)

so that the exponential growth can be controlled by a faster exponential decay from the collision frequency $\nu(\xi)$ in (2.8).

Lemma 4 (Velocity Lemma).

Along the characteristic, whenever $x\geq 0$ and $x-(\xi_{1}+u)s\geq 0$ , for $\tilde{\alpha}(x,\xi)$ defined in (1.8), we have

e^{-c\nu_{0}s/2}\tilde{\alpha}(x-s(\xi_{1}+u),\xi)\leq\tilde{\alpha}(x,\xi)\leq e^{c\nu_{0}s/2}\tilde{\alpha}(x-s(\xi_{1}+u),\xi).

(2.14)

Proof.

We take spatial derivative to $\tilde{\alpha}(x,\xi)$ and have

	$\displaystyle\|(\xi_{1}+u)\partial_{x}\tilde{\alpha}(x,\xi)\|$
	$\displaystyle=\Big{\|}(\xi_{1}+u)\frac{(c\nu_{0})^{2}x}{\tilde{\alpha}(x,\xi)}\Big{\|}=\frac{\|(c\nu_{0})^{1/2}(\xi_{1}+u)(c\nu_{0})^{3/2}x\|}{\tilde{\alpha}(x,\xi)}$
	$\displaystyle\leq\frac{(c\nu_{0})(\xi_{1}+u)^{2}+(c\nu_{0})^{3}x^{2}}{2\tilde{\alpha}(x,\xi)}=\frac{c\nu_{0}(\tilde{\alpha}(x,\xi))^{2}}{2\tilde{\alpha}(x,\xi)}=\frac{c\nu_{0}\tilde{\alpha}(x,\xi)}{2}.$		(2.15)

Since

\displaystyle\frac{\mathrm{d}}{\mathrm{d}s}\tilde{\alpha}(x-s(\xi_{1}+u),\xi)

\displaystyle=-(\xi_{1}+u)\partial_{x}\tilde{\alpha}(x-s(\xi_{1}+u),\xi),

by (2.15), we have

\displaystyle\frac{\mathrm{d}}{\mathrm{d}s}\tilde{\alpha}(x-s(\xi_{1}+u),\xi)

\displaystyle\leq\frac{c\nu_{0}\tilde{\alpha}(x-s(\xi_{1}+u),\xi)}{2}.

By Gronwall’s inequality, we conclude (2.14).

∎

With the extra cut-off function, the weight (1.10) shares similar property as demonstrated in the following lemma:

Lemma 5.

Along the characteristic, when $x\geq 0$ and $x-(\xi_{1}+u)s\geq 0$ , for $\alpha(x,\xi)$ defined in (1.10), we have

e^{-2c\nu_{0}s}\alpha(x-s(\xi_{1}+u),\xi)\leq\alpha(x,\xi)\leq e^{2c\nu_{0}s}\alpha(x-s(\xi_{1}+u),\xi).

(2.16)

Proof.

First we prove that $\chi(s)$ in (1.9) has the following property:

s\chi^{\prime}(s)\leq 4\chi(s).

(2.17)

By (1.9), when $s\geq 2$ , $s\chi^{\prime}(s)=0\leq 4\chi(s)$ . When $s\leq\frac{1}{2}$ , we have $\chi(s)=s,$ and thus $s\chi^{\prime}(s)=s\leq 4\chi(s)=4s$ . When $\frac{1}{2}<s<2$ , since $\chi^{\prime}(s)\leq 1$ , we have $s\chi^{\prime}(s)\leq s<2=4\chi(1/2)\leq 4\chi(s)$ . Then we conclude (2.17).

Then we compute

	$\displaystyle\|(\xi_{1}+u)\partial_{x}\alpha(x,\xi)\|$
	$\displaystyle=\|(\xi_{1}+u)\chi^{\prime}(\tilde{\alpha}(x,\xi))\partial_{x}\tilde{\alpha}(x,\xi)\|$
	$\displaystyle=\chi^{\prime}(\tilde{\alpha}(x,\xi))\|(\xi_{1}+u)\partial_{x}\tilde{\alpha}(x,\xi)\|\leq\frac{c\nu_{0}}{2}\chi^{\prime}(\tilde{\alpha}(x,\xi))\tilde{\alpha}(x,\xi)$
	$\displaystyle\leq 2c\nu_{0}\chi(\tilde{\alpha}(x,\xi))=2c\nu_{0}\alpha(x,\xi).$

In the third line, we applied the computation in (2.15). In the last line, we used (2.17). Then by

\frac{\mathrm{d}}{\mathrm{d}s}\alpha(x-s(\xi_{1}+u),\xi)=-(\xi_{1}+u)\partial_{x}\alpha(x-s(\xi_{1}+u),\xi)

and Gronwall’s inequality, we conclude (2.16).

∎

The next lemma addresses the integral (1.23) mentioned in Section 1.3, which consists of the weight $1/\alpha$ and $K(f)$ in Lemma 3.

Lemma 6.

Let $t\gg 1$ , $T\leq t$ and $x-T(\xi_{1}+u)\geq 0$ . For $T>1$ , we have

\displaystyle\int^{t}_{t-T}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}\frac{1}{\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})}\lesssim\frac{t}{\alpha(x,\xi)}.

(2.18)

For $T\leq 1$ , we have

\displaystyle\int^{t}_{t-T}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}\frac{1}{\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})}\lesssim\frac{\sqrt{T}+T\ln(t)}{\alpha(x,\xi)}.

(2.19)

In result, for $\varepsilon\ll 1$ such that $\varepsilon\ln(t)\ll 1$ and $x-\varepsilon(\xi_{1}+u)\geq 0$ ,

\displaystyle\int^{t}_{t-\varepsilon}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}\frac{1}{\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})}\lesssim\frac{\sqrt{\varepsilon}+\varepsilon\ln(t)}{\alpha(x,\xi)}.

(2.20)

Remark 7.

We note that there is a time-growing factor $t$ in the upper bound. In the steady problem, we will fix $t$ to be large and fixed, and this factor will be damped by either $e^{-\nu(\xi)t}$ or $\gamma$ (e.g. see (3.10) and (4.14)).

Proof.

We only consider the case $\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})=\tilde{\alpha}(x-(t-s)(\xi_{1}+u),\xi^{\prime})$ . For the other case, from (1.9), we have $\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})\geq\frac{1}{2}$ , this leads to

	$\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{1}_{\alpha\neq\tilde{\alpha}}\frac{e^{-C\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|\alpha(x-(t-s)(\xi_{1}+u),\xi^{\prime})}$
	$\displaystyle\lesssim\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C\|\xi-\xi^{\prime}\|}}{\|\xi-\xi^{\prime}\|}\lesssim\min\{1,T\}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.$

The last inequality follows from (1.11).

Then we focus on

\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi-\xi^{\prime}|^{2}}}{|\xi-\xi^{\prime}|}\frac{1}{\sqrt{(\xi_{1}^{\prime}+u)^{2}+(c\nu_{0})^{2}(x-(t-s)(\xi_{1}+u))^{2}}}.

Step 1: integral over $\mathrm{d}\xi^{\prime}$ .

First we compute the $\mathrm{d}\xi^{\prime}$ integral. We use a notation

\eta:=\xi+(u,0,0).

(2.21)

We apply a change of variable $\xi^{\prime}+(u,0,0)\to\xi^{\prime}$ . Denote $\xi_{\parallel}=(0,\xi_{2},\xi_{3})$ , then the $\mathrm{d}\xi^{\prime}$ integral reads

	$\displaystyle\int_{\mathbb{R}^{3}}\frac{e^{-C\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}\frac{1}{\sqrt{(\xi_{1}^{\prime}+u)^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi^{\prime}$
	$\displaystyle=\int_{\mathbb{R}^{3}}\frac{e^{-C\|\eta-\xi^{\prime}\|^{2}}}{\|\eta-\xi^{\prime}\|}\frac{1}{\sqrt{\|\xi_{1}^{\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim\iint\frac{e^{-C\|\eta_{\parallel}-\xi^{\prime}_{\parallel}\|^{2}}}{\|\eta_{\parallel}-\xi^{\prime}_{\parallel}\|}\int_{\mathbb{R}}\frac{e^{-C\|\eta_{1}-\xi^{\prime}_{1}\|^{2}}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim\int_{\mathbb{R}}\frac{e^{-C\|\eta_{1}-\xi^{\prime}_{1}\|^{2}}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle=\int_{\mathbb{R}}\mathbf{1}_{\|\xi_{1}^{\prime}\|\leq 2\|\eta_{1}\|}+\mathbf{1}_{\|\xi_{1}^{\prime}\|>2\|\eta_{1}\|}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim\int_{0}^{2\|\eta_{1}\|}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle+\int_{\mathbb{R}}\frac{e^{-C\|\xi_{1}^{\prime}\|^{2}/4}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim 1+\Big{[}\int_{0}^{1}+\int_{0}^{2\|\eta_{1}\|}\Big{]}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}.$		(2.22)

In the second last line we used that for $|\xi_{1}^{\prime}|>2|\eta_{1}|$ ,

\displaystyle|\eta_{1}-\xi_{1}^{\prime}|\geq|\xi_{1}^{\prime}|-|\eta_{1}|\geq|\xi_{1}^{\prime}|-\frac{|\xi_{1}^{\prime}|}{2}=\frac{|\xi_{1}^{\prime}|}{2}.

In the last line we used

\displaystyle\int_{1}^{\infty}\frac{e^{-C|\xi_{1}^{\prime}|^{2}/4}}{\sqrt{|\xi^{\prime}_{1}|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\lesssim 1.

The integral in (2.22) reads

	$\displaystyle\int_{0}^{2\|\eta_{1}\|}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle=\ln\Big{(}\sqrt{\|\xi_{1}^{\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}+\|\xi_{1}^{\prime}\|\Big{)}\Big{\|}_{0}^{2\|\eta_{1}\|}$
	$\displaystyle=\ln\Big{(}\sqrt{4\|\eta_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}+2\|\eta_{1}\|\Big{)}$
	$\displaystyle-\ln\Big{(}\sqrt{(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}\Big{)}$
	$\displaystyle=\ln\Big{(}\sqrt{1+\Big{\|}\frac{2\eta_{1}}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}^{2}}+\Big{\|}\frac{2\eta_{1}}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}\Big{)}.$		(2.23)

Similarly

	$\displaystyle\int_{0}^{1}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle=\ln\Big{(}\sqrt{1+\Big{\|}\frac{1}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}^{2}}+\Big{\|}\frac{1}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}\Big{)}.$		(2.24)

Step 2: integral over $\mathrm{d}s$ .

Then we compute the $\mathrm{d}s$ integral:

\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\eqref{xi_integral}\lesssim\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}[1+\eqref{ln_bdd_eta}+\eqref{ln_bdd_1}].

The contribution of the constant is bounded as

\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\lesssim\min\{1,T\}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)},

where we have used (1.11).

Then we consider the contribution of (2.23) and (2.24). We split the discussion into $\xi_{1}+u<0$ and $\xi_{1}+u>0$ .

Step 2-1: case of $\xi_{1}+u<0$ . We have $x\leq x-(\xi_{1}+u)(t-s)$ .

Contribution of (2.23).

If $|\eta_{1}|/(c\nu_{0}x)\leq 1$ , then $|\eta_{1}|/[c\nu_{0}(x-(\xi_{1}+u)(t-s))]\leq 1$ for all $t-T\leq s\leq t$ , this leads to (2.23) $\lesssim 1$ and thus

\displaystyle\int_{t-T}^{t}\mathrm{d}s\mathbf{1}_{|\eta_{1}|/(c\nu_{0}x)\leq 1}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\lesssim\min\{1,T\}\leq\frac{\min\{1,T\}}{\alpha(x,\xi)}.

Then we consider $|\eta_{1}|/(c\nu_{0}x)>1$ . From (2.21), we have $\eta_{1}=\xi_{1}+u$ . Then

\displaystyle\alpha(x,\xi)

\displaystyle\leq\sqrt{(\xi_{1}+u)^{2}+(c\nu_{0})^{2}x^{2}}\leq|\xi_{1}+u|+c\nu_{0}x<2|\xi_{1}+u|.

Thus we have

\displaystyle\frac{1}{|\xi_{1}+u|}

\displaystyle\leq\frac{2}{\alpha(x,\xi)}.

(2.25)

We bound (2.23) by

	$\displaystyle\eqref{ln_bdd_eta}$	$\displaystyle\lesssim 1+\ln\Big{\|}\frac{\eta_{1}}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}$
		$\displaystyle\lesssim 1+\|\ln\|\eta_{1}\|\|+\|\ln\|x-(\xi_{1}+u)(t-s)\|\|.$		(2.26)

The contribution of $|\ln|\eta_{1}||$ can be bounded as

	$\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\|\ln\|\eta_{1}\|\|\mathrm{d}s$	$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln\|\eta_{1}\|\|}{[1+\|\xi\|]}=\mathbf{1}_{\|\xi_{1}+u\|\geq 1}+\mathbf{1}_{\|\xi_{1}+u\|<1}$
		$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln(2\|\xi\|+2)\|}{[1+\|\xi\|]}+\frac{\min\{1,T\}}{\|\xi_{1}+u\|}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.$		(2.27)

Here for $|\xi_{1}+u|<1$ we used $\ln|\xi_{1}+u|\leq\frac{1}{|\xi_{1}+u|}$ . For the other case we used $|u|\ll 1$ and $|\xi_{1}+u|\geq 1$ to derive $|\xi_{1}+u|<2|\xi_{1}|$ , and thus

\displaystyle\ln(|\xi_{1}+u|)\leq\ln(2|\xi_{1}|)\leq\ln(2|\xi|+2).

Then we compute the contribution of the last term of (2.26), which reads

	$\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\|\ln\|x-(\xi_{1}+u)(t-s)\|\|\mathrm{d}s$
	$\displaystyle=\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\geq 1}+\mathbf{1}_{x-(\xi_{1}+u)(t-s)<1}\mathrm{d}s$
	$\displaystyle\leq\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\geq 1}e^{-\nu(\xi)(t-s)/2}\ln(-(\xi_{1}+u)/(c\nu_{0})-(\xi_{1}+u)(t-s))\mathrm{d}s$
	$\displaystyle+\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)<1}e^{-\nu(\xi)(t-s)/2}\|\ln(-(\xi_{1}+u)(t-s))\|\mathrm{d}s$
	$\displaystyle\lesssim\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}[\|\ln\|\xi_{1}+u\|\|+\|\ln(t-s)\|+\|\ln(t-s+\frac{1}{c\nu_{0}})\|]\mathrm{d}s$		(2.28)
	$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln\|\xi_{1}+u\|\|}{[1+\|\xi\|]}+\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\alpha(x,\xi)}\lesssim\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\alpha(x,\xi)}.$

In the third line, we used $x<\frac{|\eta_{1}|}{c\nu_{0}}=-\frac{(\xi_{1}+u)}{c\nu_{0}}$ for $\xi_{1}+u<0$ . In the last line, we applied the same computation (2.27) for the first term in (2.28). For the rest terms in (2.28), we used the following computation:

	$\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}[\|\ln(t-s)\|+\|\ln(t-s+\frac{1}{c\nu_{0}})\|]\mathrm{d}s=\int_{t-T}^{t}\mathbf{1}_{t-s\geq 1}+\mathbf{1}_{t-s\leq 1}\mathrm{d}s$
	$\displaystyle\lesssim\mathbf{1}_{T\geq 1}\int_{t-T}^{t-1}e^{-\nu(\xi)(t-s)/2}\|t-s+\frac{1}{c\nu_{0}}\|\mathrm{d}s+\int_{0}^{\min\{1,T\}}\|\ln(s)\|+\|\ln(s+\frac{1}{c\nu_{0}})\|\mathrm{d}s$
	$\displaystyle\lesssim\mathbf{1}_{T\geq 1}[1+\frac{1}{c\nu_{0}}]+\int_{0}^{\min\{1,T\}}\|\ln(s)\|\mathrm{d}s$
	$\displaystyle\lesssim\min\{1,T\}+\min\{1,\|T\|+\|T\ln(T)\|\}$
	$\displaystyle\lesssim\min\{1,\|T\|+\|T\ln(T)\|\}\leq\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\alpha(x,\xi)}.$		(2.29)

We conclude

\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s\lesssim\frac{\min\{1,|T|+|T\ln(T)|\}}{\alpha(x,\xi)}.

(2.30)

Contribution of (2.24).

If $1/(c\nu_{0}x)\leq 1$ , we have (2.24) $\lesssim 1$ and thus

\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\mathbf{1}_{1/(c\nu_{0}x)\leq 1}\eqref{ln_bdd_1}\mathrm{d}s\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.

Then we suppose $c\nu_{0}x\leq 1$ . If $c\nu_{0}x<|\eta_{1}|$ , we bound (2.24) similarly as (2.26):

\displaystyle\eqref{ln_bdd_1}

\displaystyle\lesssim 1+|\ln(x-(\xi_{1}+u)(t-s))|.

(2.31)

We follow the same computation as in (2.28) to conclude

\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\mathbf{1}_{1/(c\nu_{0}x)\leq 1,c\nu_{0}x<|\eta_{1}|}\eqref{ln_bdd_1}\mathrm{d}s\lesssim\frac{\min\{1,|T|+|T\ln(T)|\}}{\alpha(x,\xi)}.

If $c\nu_{0}x\geq|\eta_{1}|$ , we have

\displaystyle\alpha(x,\xi)

\displaystyle\leq\sqrt{|\xi_{1}+u|^{2}+(c\nu_{0})^{2}x^{2}}\leq 2c\nu_{0}x,

which leads to

\displaystyle\frac{1}{c\nu_{0}x}

\displaystyle\leq\frac{2}{\alpha(x,\xi)}.

(2.32)

Then we use the bound (2.31) and further compute

	$\displaystyle\int_{t-T}^{t}\mathrm{d}se^{-\nu(\xi)(t-s)/2}\|\ln(x-(\xi_{1}+u)(t-s))\|=\int_{t-T}^{t}[\mathbf{1}_{x-(\xi_{1}+u)(t-s)\leq 1}+\mathbf{1}_{x-(\xi_{1}+u)(t-s)\geq 1}]$
	$\displaystyle\leq\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\leq 1}e^{-\nu(\xi)(t-s)/2}\|\ln(x)\|\mathrm{d}s$
	$\displaystyle+\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)>1}e^{-\nu(\xi)(t-s)/2}\ln(x+c\nu_{0}x(t-s))\mathrm{d}s$
	$\displaystyle\lesssim\|\ln(x)\|\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}[1+\ln(1+c\nu_{0}(t-s))]\mathrm{d}s$
	$\displaystyle\lesssim\|\ln(x)\|\min\{1,\|T\|+\|T\ln(T)\|\}\lesssim\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\|x\|}$
	$\displaystyle\lesssim\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\alpha(x,\xi)}.$

In the second last line, for the first inequality we applied the same computation (2.29), for the second inequality we used $x\leq\frac{1}{c\nu_{0}}$ . In the last inequality, we used (2.32).

We conclude

\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_1}\mathrm{d}s\lesssim\frac{\min\{1,|T|+|T\ln(T)|\}}{\alpha(x,\xi)}.

(2.33)

Step 2-2: case of $\xi_{1}+u>0$ . In such case $x-(\xi_{1}+u)(t-s)<x$ .

Contribution of (2.23).

When $|\eta_{1}|/(c\nu_{0}x)\geq 1$ , we have $1/|\xi_{1}+u|\lesssim 1/\alpha(x,\xi)$ . Then we use the same bound (2.26) for (2.23), where the contribution of the first two terms are independent of $x$ , and thus can be bounded using the same computation in (2.27). We only need to compute the contribution of $|\ln(x-(\xi_{1}+u)(t-s))|$ . We split the integral into $x-(\xi_{1}+u)(t-s)\geq 1$ and $x-(\xi_{1}+u)(t-s)\leq 1$ . For the first case, we have

	$\displaystyle\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\geq 1}e^{-\nu(\xi)(t-s)/2}\|\ln(x-(\xi_{1}+u)(t-s))\|\mathrm{d}s$
	$\displaystyle\leq\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\geq 1}e^{-\nu(\xi)(t-s)/2}\|\ln((\xi_{1}+u)/(c\nu_{0})-(\xi_{1}+u)(t-s))\|\mathrm{d}s$
	$\displaystyle\lesssim\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\|\ln(\xi_{1}+u)\|\mathrm{d}s+\int_{t-T}^{t}\mathbf{1}_{1/(c\nu_{0})-(t-s)\geq 0}e^{-\nu(\xi)(t-s)/2}\|\ln(1/(c\nu_{0})-(t-s))\|\mathrm{d}s$
	$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln(\xi_{1}+u)\|}{[1+\|\xi\|]}+\int_{0}^{\min\{1/(c\nu_{0}),T\}}\|\ln(s)\|\mathrm{d}s$
	$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln(\xi_{1}+u)\|}{[1+\|\xi\|]}+\min\{1,\|T\|+\|T\ln(T)\|\}\lesssim\frac{\min\{1,\|T\|+\|T\ln(T)\|\}}{\alpha(x,\xi)}.$		(2.34)

In the last inequality we used the same computation in (2.27).

For the second case $x-(\xi_{1}+u)(t-s)\leq 1$ , without loss of generality, we assume $x-(\xi_{1}+u)T<1$ , otherwise, the integration vanishes. We use a change of variable $y=x-(\xi_{1}+u)(t-s)$ with $\mathrm{d}y=(\xi_{1}+u)\mathrm{d}s$ and $x-(\xi_{1}+u)T\leq y\leq\min\{1,x\}$ , then

	$\displaystyle\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\leq 1}e^{-\nu(\xi)(t-s)/2}\|\ln(x-(\xi_{1}+u)(t-s))\|\mathrm{d}s$
	$\displaystyle\leq\int_{x-(\xi_{1}+u)T}^{\min\{1,x\}}e^{-\nu(\xi)\frac{x-y}{\xi_{1}+u}/2}\|\ln(y)\|\frac{1}{\xi_{1}+u}\mathrm{d}y\leq\frac{1}{\xi_{1}+u}\int^{\min\{1,x\}}_{x-(\xi_{1}+u)T}\|\ln(y)\|\mathrm{d}y$
	$\displaystyle\leq\frac{1}{\xi_{1}+u}\int_{0}^{\min\{1,(\xi_{1}+u)T\}}\|\ln(y)\|\mathrm{d}y$		(2.35)
	$\displaystyle\lesssim\frac{1}{\xi_{1}+u}\min\{1,\|(\xi_{1}+u)T\|+\|(\xi_{1}+u)T\ln((\xi_{1}+u)T)\|\}$		(2.36)
	$\displaystyle\lesssim\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}.$		(2.37)

In the third line, we use the fact that the integral domain is on $0<y\leq 1$ , and the length of the integral domain is bounded by $\min\{1,x\}-[x-(\xi_{1}+u)T]\lesssim\min\{1,(\xi_{1}+u)T\}$ . In the last line, in the case of $(\xi_{1}+u)T>1$ , we have $\frac{1}{\xi_{1}+u}<T$ , and thus from (2.25),

\displaystyle\frac{1}{\xi_{1}+u}\lesssim\min\{T,\frac{1}{\alpha(x,\xi)}\}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.

In the case of $(\xi_{1}+u)T\leq 1$ , for $T>1$ , by (2.25) we bound (2.35) $\lesssim\frac{1}{\xi_{1}+u}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.$ For $T\leq 1$ , from $(\xi_{1}+u)T\leq 1$ we have $|\ln((\xi_{1}+u)T)|\lesssim\frac{1}{\sqrt{(\xi_{1}+u)T}}$ and

	$\displaystyle\frac{(\xi_{1}+u)T+\|(\xi_{1}+u)T\ln((\xi_{1}+u)T)\|}{\xi_{1}+u}$
	$\displaystyle=T+\|T\ln((\xi_{1}+u)T)\|=\mathbf{1}_{(\xi_{1}+u)\geq 1}+\mathbf{1}_{(\xi_{1}+u)<1}$
	$\displaystyle\lesssim T+\mathbf{1}_{\xi_{1}+u\geq 1}\frac{T}{\sqrt{(\xi_{1}+u)T}}+\mathbf{1}_{\xi_{1}+u<1}[T\ln(T)+T\|\ln(\xi_{1}+u)\|]$		(2.38)
	$\displaystyle\lesssim T+\sqrt{T}+\|T\ln(T)\|+\frac{T}{\xi_{1}+u}\lesssim\frac{\sqrt{T}}{\alpha(x,\xi)}\lesssim\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}.$

In the last line, we used (2.25) and $T\leq 1$ .

Then we focus on the scenario that $|\eta_{1}|/(c\nu_{0}x)\leq 1$ , if $|\eta_{1}|\leq(c\nu_{0})(x-(\xi_{1}+u)(t-s))$ for all $t-T\leq s\leq t$ , then we have (2.23) $\lesssim 1$ and thus

\displaystyle\int_{t-T}^{t}\mathbf{1}_{|\eta_{1}|\leq(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\eqref{ln_bdd_eta}\mathrm{d}s

\displaystyle\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.

If $|\eta_{1}|>(c\nu_{0})(x-(\xi_{1}+u)(t-s))$ for some $t-T<s<t$ , then there is a unique $t^{\prime}$ such that $|\eta_{1}|=(c\nu_{0})(x-(\xi_{1}+u)(t-t^{\prime}))$ . Denote $x^{\prime}=x-(\xi_{1}+u)(t-t^{\prime})$ . Then

\displaystyle\frac{|\eta_{1}|}{(c\nu_{0})(x^{\prime}-(\xi_{1}+u)(t^{\prime}-s))}>1\text{ for all }t-T\leq s<t^{\prime},

\displaystyle\frac{|\eta_{1}|}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\leq 1\text{ for all }t^{\prime}\leq s\leq t.

By the observation above, we split the $s$ -integral into two parts. The first part is bounded as

\displaystyle\int_{t^{\prime}}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s\lesssim\int_{t^{\prime}}^{t}e^{-\nu(\xi)(t-s)/2}\mathrm{d}s\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.

For the second part, we have $\frac{\eta_{1}}{(c\nu_{0})x^{\prime}}\geq 1$ , and thus

\displaystyle\frac{1}{|\xi_{1}+u|}\lesssim\frac{1}{\alpha(x^{\prime},\xi)}.

Then the integral for the second part reads

\displaystyle\int_{t-T}^{t^{\prime}}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s=e^{-\nu(\xi)(t-t^{\prime})/2}\int_{t-T}^{t^{\prime}}e^{-\nu(\xi)(t^{\prime}-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s.

Note that

\displaystyle\eqref{ln_bdd_eta}

\displaystyle=\ln\Big{(}\sqrt{1+\Big{|}\frac{2\eta_{1}}{(c\nu_{0})(x^{\prime}-(\xi_{1}+u)(t^{\prime}-s))}\Big{|}^{2}}+\Big{|}\frac{2\eta_{1}}{(c\nu_{0})(x^{\prime}-(\xi_{1}+u)(t^{\prime}-s))}\Big{|}\Big{)}.

Following the same computation as (2.34), (2.37) for the case $|\eta_{1}|/(c\nu_{0}x)>1$ , we replace $x$ by $x^{\prime}$ and conclude that

	$\displaystyle\int_{t-T}^{t^{\prime}}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s\lesssim e^{-\nu(\xi)(t-t^{\prime})/2}\frac{\min\{1,\sqrt{T}\}}{\alpha(x^{\prime},\xi)}$
	$\displaystyle\lesssim e^{-\nu_{0}(t-t^{\prime})/2}e^{2c\nu_{0}(t-t^{\prime})}\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}\leq\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}.$

In the second line, we have used Lemma 5 with $2c=\frac{1}{4}$ .

Combining with (2.34) and (2.37), we conclude that

\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}\mathrm{d}s\lesssim\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}.

(2.39)

Contribution of (2.24).

If $1/(x-(\xi_{1}+u)(t-s))\leq 1$ for all $s$ , then (2.24) $\lesssim 1$ so that the contribution of (2.24) is bounded by $\min\{1,T\}/\alpha(x,\xi)$ . Thus we only consider the integral over $x-(\xi_{1}+u)(t-s)\leq 1$ :

\displaystyle\int_{t-T}^{t}\mathbf{1}_{x-(\xi_{1}+u)(t-s)\leq 1}e^{-\nu(\xi)(t-s)/2}|\ln(x-(\xi_{1}+u)(t-s))|\mathrm{d}s.

(2.40)

In such a case, we have $x-(\xi_{1}+u)T<1$ . We apply a change of variable $y=x-(\xi_{1}+u)(t-s)$ , with $x-(\xi_{1}+u)T\leq y\leq\min\{x,1\}$ . Then

\displaystyle\eqref{s_integral}

\displaystyle\leq\int_{x-(\xi_{1}+u)T}^{\min\{x,1\}}e^{-\nu(\xi)\frac{x-y}{\xi_{1}+u}/2}|\ln(y)|\frac{1}{\xi_{1}+u}\mathrm{d}y.

(2.41)

If $x>2$ , we apply (2.36) to have

	$\displaystyle\eqref{s_integral}$	$\displaystyle\leq e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}\frac{1}{\xi_{1}+u}\int_{x-(\xi_{1}+u)T}^{\min\{x,1\}}\|\ln(y)\|\mathrm{d}y$
		$\displaystyle\lesssim e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}\frac{1}{\xi_{1}+u}\int_{0}^{\min\{1,(\xi_{1}+u)T\}}\|\ln(y)\|\mathrm{d}y$
		$\displaystyle\lesssim e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}\frac{\min\{1,(\xi_{1}+u)T+\|(\xi_{1}+u)T\ln((\xi_{1}+u)T)\|\}}{\xi_{1}+u}.$

For $(\xi_{1}+u)T\geq 1$ we have $\frac{1}{\xi_{1}+u}\leq T$ , from $\frac{e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}}{\xi_{1}+u}\lesssim 1$ , we further have

\displaystyle\frac{e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}}{\xi_{1}+u}

\displaystyle\lesssim\min\{1,e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}T\}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.

For $(\xi_{1}+u)T<1$ and $T>1$ , from $\frac{e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}}{\xi_{1}+u}\lesssim 1$ we have

\displaystyle\eqref{s_integral}\mathbf{1}_{T>1}

\displaystyle\lesssim\frac{e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}}{\xi_{1}+u}\mathbf{1}_{T>1}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}\mathbf{1}_{T>1}.

For $(\xi_{1}+u)T<1$ and $T\leq 1$ , from $\frac{e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}}{\sqrt{\xi_{1}+u}}\lesssim 1$ we have

	$\displaystyle e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}\frac{(\xi_{1}+u)T+\|(\xi_{1}+u)T\ln((\xi_{1}+u)T)\|}{\xi_{1}+u}$
	$\displaystyle\lesssim e^{-\frac{\nu(\xi)}{2(\xi_{1}+u)}}[T+\frac{T}{\sqrt{(\xi_{1}+u)T}}]\lesssim\sqrt{T}\lesssim\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)}.$

Then we consider $x\leq 2$ . We further discuss two cases. The first case is $|\eta_{1}|/(c\nu_{0}x)\leq 1/(2tc\nu_{0})$ , which implies $x>2t|\eta_{1}|$ . Then

	$\displaystyle\alpha(x,\xi)$	$\displaystyle\leq\sqrt{(\xi_{1}+u)^{2}+(c\nu_{0})^{2}x^{2}}\leq(\xi_{1}+u)+c\nu_{0}x$
		$\displaystyle\leq\frac{x}{2t}+c\nu_{0}x\lesssim x,$

here we used $t\gg 1$ . Thus we derive $\frac{1}{x}\lesssim 1/\alpha(x,\xi).$

Since $0\leq s\leq t$ , we have

	$\displaystyle x-(\xi_{1}+u)(t-s)$	$\displaystyle=\frac{x}{2}+\frac{x}{2}-(\xi_{1}+u)(t-s)$
		$\displaystyle\geq\frac{x}{2}+t\|\eta_{1}\|-t\|\eta_{1}\|=\frac{x}{2}.$

Then we have

	$\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\|\ln(x-(\xi_{1}+u)(t-s))\|\mathrm{d}s$
	$\displaystyle\leq\min\{1,T\}\max\{\|\ln(x)\|,\|\ln(x/2)\|\}\lesssim\frac{\min\{1,T\}}{x}\leq\frac{\min\{1,T\}}{\alpha(x,\xi)},$

where we have used $x\leq 2$ .

The other case is $|\eta_{1}|/(c\nu_{0}x)>1/(2tc\nu_{0})$ , which implies $2t|\eta_{1}|>x$ . We have

\displaystyle\alpha(x,\xi)

\displaystyle\leq(\xi_{1}+u)+(c\nu_{0})x\leq(\xi_{1}+u)(1+2tc\nu_{0}),

which leads to

\displaystyle\frac{1}{\xi_{1}+u}

\displaystyle\leq\frac{1+2tc\nu_{0}}{\alpha(x,\xi)}.

In this case we apply (2.41) with (2.35) and (2.36) to have

\displaystyle\eqref{cov_y}

\displaystyle\leq\eqref{third}\leq\frac{1}{\xi_{1}+u}\min\{1,|(\xi_{1}+u)T|+|(\xi_{1}+u)T\ln((\xi_{1}+u)T)|\}.

When $(\xi_{1}+u)T>1$ , we have $\frac{1}{\xi_{1}+u}<T$ and

\displaystyle\eqref{cov_y}

\displaystyle\lesssim\frac{1}{\xi_{1}+u}\lesssim\min\{T,\frac{t}{\alpha(x,\xi)}\}\lesssim\frac{\min\{T,t\}}{\alpha(x,\xi)}.

When $(\xi_{1}+u)T\leq 1$ , for $T>1$ we use (2.35) to bound

\displaystyle\eqref{cov_y}\mathbf{1}_{T>1}

\displaystyle\lesssim\frac{\mathbf{1}_{T>1}}{\xi_{1}+u}\lesssim\frac{\mathbf{1}_{T>1}t}{\alpha(x,\xi)}.

For $T<1$ , we follow the computation in (2.38) to have

	$\displaystyle\eqref{cov_y}\lesssim T+\sqrt{T}+\|T\ln(T)\|+T\mathbf{1}_{\xi_{1}+u\leq 1}\|\ln(\xi_{1}+u)\|$
	$\displaystyle\lesssim\sqrt{T}+T\mathbf{1}_{\frac{1}{\xi_{1}+u}\geq 1}\ln(\frac{1}{\xi_{1}+u})$
	$\displaystyle\lesssim\sqrt{T}+T\ln(\frac{1+2tc\nu_{0}}{\alpha(x,\xi)})\lesssim\sqrt{T}+T\ln(t)+T\|\ln(\alpha(x,\xi))\|$
	$\displaystyle\lesssim\frac{\sqrt{T}+T\ln(t)}{\alpha(x,\xi)}.$

In the last inequality we used $\alpha(x,\xi)\lesssim 1$ to have $|\ln(\alpha(x,\xi))|\lesssim\frac{1}{\alpha(x,\xi)}$ .

We conclude that

\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_1}\mathrm{d}s\lesssim\frac{\mathbf{1}_{T>1}t+\mathbf{1}_{T\leq 1}[\sqrt{T}+T\ln(t)]}{\alpha(x,\xi)}.

(2.42)

Step 3: conclusion

In summary, in the case of $\xi_{1}+u<0$ in Step 2-1 and the contribution of (2.23) in the case of $\xi_{1}+u>0$ in Step 2-2, we collect (2.30), (2.33) and (2.39) to have

\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\eqref{ln_bdd_eta}+\mathbf{1}_{\xi_{1}+u<0}\eqref{ln_bdd_1}\mathrm{d}s

\displaystyle\lesssim\frac{\min\{1,\sqrt{T}\}}{\alpha(x,\xi)},

which satisfies (2.18) for $T>1$ and (2.19) for $T\leq 1$ .

For the contribution of (2.24) in the case of $\xi_{1}+u>0$ in Step 2-2, we use (2.42) to have

\displaystyle\int_{t-T}^{t}\mathbf{1}_{\xi_{1}+u>0}\eqref{ln_bdd_1}\mathrm{d}s

\displaystyle\lesssim\frac{\mathbf{1}_{T>1}t+\mathbf{1}_{T\leq 1}[\sqrt{T}+T\ln(t)]}{\alpha(x,\xi)},

which also satisfies (2.18) and (2.19).

∎

The proof of Lemma 6 directly implies the following result:

Lemma 7.

\int^{t}_{0}\mathrm{d}se^{-\nu(\xi)(t-s)}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{e^{-C|\xi^{\prime}|^{2}}}{\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}\lesssim\frac{t}{\alpha(x,\xi)},

(2.43)

\int^{t}_{0}\mathrm{d}se^{-\nu(\xi)(t-s)}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}w^{-1}(\xi^{\prime})\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\frac{e^{-C|\xi^{\prime}-\xi^{\prime\prime}|^{2}}}{|\xi^{\prime}-\xi^{\prime\prime}|\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime\prime})}\mathrm{d}\xi^{\prime\prime}\lesssim\frac{t}{\alpha(x,\xi)}.

(2.44)

Proof.

Proof of (2.43). The integral over $\mathrm{d}\xi^{\prime}$ is bounded as

\displaystyle 1+\int_{0}^{1}\frac{1}{\sqrt{|\xi^{\prime}_{1}|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}=1+\eqref{ln_bdd_1}.

Here, we applied the same computation as (2.22). Then follow Step 2 in the proof of Lemma 6 to conclude the lemma.

Proof of (2.43). Again following the computation of (2.22), with $\eta=\xi^{\prime}+(u,0,0)$ , the $\mathrm{d}\xi^{\prime}\mathrm{d}\xi^{\prime\prime}$ integral is bounded as

	$\displaystyle\int_{\mathbb{R}^{3}}e^{-\theta\|\eta\|^{2}}\mathrm{d}\eta\int_{\mathbb{R}}\frac{e^{-C\|\eta_{1}-\xi_{1}^{\prime\prime}\|^{2}}}{\sqrt{\|\xi_{1}^{\prime\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime\prime}$
	$\displaystyle\lesssim\int_{\mathbb{R}}\frac{e^{-C_{1}\|\xi_{1}^{\prime\prime}\|^{2}}}{\sqrt{\|\xi_{1}^{\prime\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime\prime}\lesssim 1+\eqref{ln_bdd_1}.$

Here we choose $C_{1}=\frac{\varepsilon}{2}C$ for some $\varepsilon>0$ such that $\theta>\varepsilon C$ and

	$\displaystyle-\theta\|\eta_{1}\|^{2}-C\|\eta_{1}-\xi^{\prime\prime}_{1}\|^{2}$	$\displaystyle\leq-\theta\|\eta_{1}\|^{2}-\varepsilon C\|\eta_{1}-\xi^{\prime\prime}_{1}\|^{2}\leq(-\theta-\varepsilon C)\|\eta_{1}\|^{2}+2\varepsilon C\eta_{1}\xi^{\prime\prime}-\varepsilon C\|\xi^{\prime\prime}\|^{2}$
		$\displaystyle\leq(\theta-\varepsilon C+2\varepsilon C)-(\varepsilon C-\frac{\varepsilon C}{2})\|\xi^{\prime\prime}\|^{2}.$

∎

3. Continuity and exponential decay in $\xi$ .

In this section, we conclude justify the assumption (1.19) and conclude Theorem 2.

We start from proving the continuity and $w$ -weighted $L^{\infty}$ estimate of the linearized penalized problem (3.11) in Proposition 6, then we will move onto the nonlinear penalized problem (3.31) in Proposition 8. At the end of this section, we will make use of the penalized problem to recover the solution the boundary layer problem (1.4).

To define the penalized problem (3.11), we denote $\prod_{+}$ as the orthogonal projection on $X_{+}$ :

\prod_{+}g=\langle gX_{+}\rangle X_{+}.

Here, $X_{+}$ is defined in (2.9), and the inner product is defined in (1.15).

Denote $\phi_{u}$ to be the eigen-function of the following eigen-value problem:

\begin{cases}&\mathcal{L}\phi_{u}=\tau_{u}(\xi_{1}+u)\phi_{u},\\ &\langle(\xi_{1}+u)\phi_{u}^{2}\rangle=-u.\end{cases}

(3.1)

3.1. Continuity and $w$ -weighted $L^{\infty}$ estimate of the eigen-value problem (3.1).

For $u\neq 0$ , we define

\psi_{u}:=\frac{\phi_{u}-\phi_{0}}{u}.

(3.2)

For $u$ near $0$ , it was shown in [3] that there exists solution $\phi_{u}\in L^{2}_{\xi}\cap L^{\infty,s}_{\xi}$ to (3.1):

Proposition 4 (Proposition 3.1 in [3]).

There exists $r>0$ and real analytic function $u\to\tau_{u}\in\mathbb{R}^{3}$ with $|\tau_{u}|\sim u$ and real analytic map $u\to\phi_{u}\in\mathcal{H}\cap dom(\mathcal{L})$ with $0<|u|<r$ such that $\phi_{u}$ satisfies (3.1). Furthermore, there exists a positive constant $C_{s}$ such that for each $s\geq 0$ , $\phi_{u}$ satisfies

\|(1+|\xi|)^{s}\phi_{u}\|_{L^{\infty}_{\xi}}\leq C_{s}

(3.3)

for all $s\geq 0$ uniformly in $u\in(-r,0)\cup(0,r)$ .

Remark 8.

In [3], the linearization reads $F=M+Mf$ , and their statement regarding (3.1) reads as following. The eigen-function of the following eigen-value problem

\begin{cases}&\frac{-Q(M\phi_{u},M)-Q(M,M\phi_{u})}{M}=\tau_{u}(\xi_{1}+u)\phi_{u}\\ &\int_{\mathbb{R}^{3}}(\xi_{1}+u)\phi_{u}^{2}M\mathrm{d}\xi=-u\end{cases}

satisfies

\displaystyle\|(1+|\xi|)^{s}\phi_{u}\sqrt{M}\|_{L^{\infty}_{\xi}}\leq C_{s}.

By substituting $\phi_{u}\sqrt{M}$ by $\phi_{u}$ , this statement is exactly Proposition 4.

First we show that the eigen-function in (3.1) is continuous and decays exponentially in $\xi\in\mathbb{R}^{3}$ .

Lemma 8.

For the eigen-function $\phi_{u}$ in Proposition 4, we have

\phi_{u}\in C(\mathbb{R}^{3}).

(3.4)

Recall the exponential weight in (1.13), the eigen-function in (3.1) further satisfies

\|w\phi_{u}\|_{L^{\infty}_{\xi}}<\infty.

(3.5)

Proof.

Proof of (3.4). From (3.1), the solution $\phi_{u}$ satisfies

\displaystyle\phi_{u}

\displaystyle=\frac{1}{\nu(\xi)-\tau_{u}(\xi_{1}+u)}K\phi_{u}.

(3.6)

Since $|u|<r$ for some small $r$ , on RHS we have $\nu(\xi)-\tau_{u}(\xi_{1}+u)>\frac{\nu(\xi)}{2}$ . We use the contradiction argument to show that $\phi_{u}$ is continuous.

Suppose $\phi_{u}$ is not continuous at some $\xi=\xi^{0}$ , then at $\xi=\xi^{0}$ , RHS of (3.6) reads

\displaystyle\frac{1}{\nu(\xi^{0})-\tau_{u}(\xi^{0}_{1}+u)}\int_{\xi^{\prime}\in\mathbb{R}^{3}}\mathbf{k}(\xi^{0},\xi^{\prime})\phi_{u}(\xi^{\prime})\mathrm{d}\xi^{\prime}.

The first term is continuous in $\xi^{0}$ due to $\tau_{u}\ll 1$ . For the second term, note that from (3.3), $\phi_{u}\in L^{\infty}_{\xi}$ , the second term is differentiable at $\xi^{0}$ from (2.12):

\displaystyle\Big{|}\int_{\xi^{\prime}\in\mathbb{R}^{3}}\nabla_{\xi}\mathbf{k}(\xi^{0},\xi^{\prime})\phi_{u}(\xi^{\prime})\mathrm{d}\xi^{\prime}\Big{|}\lesssim[1+\xi_{0}]^{2}\int_{\mathbb{R}^{3}}\frac{e^{-C|\xi^{\prime}-\xi^{0}|^{2}}}{|\xi^{\prime}-\xi^{0}|^{2}}\mathrm{d}\xi^{\prime}<\infty.

Since both terms are continuous, we conclude (3.4) by contradiction.

Proof of (3.5). For $\theta<\frac{1}{4}$ , we consider a variant of perturbation around the Maxwellian as

\displaystyle F(x,\xi)=M+\sqrt{M}e^{-\theta|\xi|^{2}}f.

Then we denote the corresponding linear Boltzmann operator as

\displaystyle\mathcal{L}_{\theta}f

\displaystyle=-\frac{Q(M,\sqrt{M}e^{-\theta|\xi|^{2}}f)+Q(\sqrt{M}e^{-\theta|\xi|^{2}}f,M)}{\sqrt{M}e^{-\theta|\xi|^{2}}}=\nu(\xi)f-K_{\theta}f.

(3.7)

With the extra weight $e^{\theta|\xi|^{2}}$ and $\theta<\frac{1}{4}$ , $K_{\theta}f$ can be expressed as

\displaystyle K_{\theta}f(\xi)=\int_{\mathbb{R}^{3}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})f(\xi^{\prime})\mathrm{d}\xi^{\prime}=\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\frac{e^{\theta|\xi|^{2}}}{e^{\theta|\xi^{\prime}|^{2}}}f(\xi^{\prime})\mathrm{d}\xi^{\prime},

here $\mathbf{k}_{\theta}(\xi,\xi^{\prime})$ is given in Lemma 3, which has the form in (2.3), (2.4) with different coefficients in the exponent( due to (2.10) and see Lemma 3 in [17] for the proof).

Therefore, $\mathcal{L}_{\theta}$ is still self-adjoint, and $K_{\theta}$ is still a bounded operator from $L^{2}_{\xi}$ to $L^{\infty,1/2}_{\xi}$ , and from $L^{\infty,s}_{\xi}$ to $L^{\infty,s+1}_{\xi}$ . Here

L^{\infty,s}(\mathbb{R}^{3}):=\{\phi\in L^{\infty}(\mathbb{R}^{3})|(1+|\xi|)^{s}\phi\in L^{\infty}(\mathbb{R}^{3})\}.

The kernel of $\mathcal{L}_{\theta}$ is given by $(1,\xi_{i},|\xi|^{2})\sqrt{M}e^{\theta|\xi|^{2}}\in L^{2}_{\xi}$ . Then applying the same argument in Proposition 3.1 in [3], for $|u|\ll 1$ , there exists $\phi_{\theta}\in L^{2}_{\xi}\cap L^{\infty,s}_{\xi}$ for any $s>0$ to the following eigen-value problem

\begin{cases}&\mathcal{L}_{\theta}\phi_{\theta}=\tau_{u}(\xi_{1}+u)\phi_{\theta},\\ &\int_{\mathbb{R}^{3}}(\xi_{1}+u)|\phi_{\theta}|^{2}\mathrm{d}\xi=-u.\end{cases}

Then for some constant $C=C(\theta,u)$ , $\phi=e^{-\theta|\xi|^{2}}\phi_{\theta}$ is an eigen-function of the following problem

\begin{cases}&\mathcal{L}\phi=\tau_{u}(\xi_{1}+u)\phi,\\ &\langle(\xi_{1}+u)|\phi|^{2}\rangle=C(\theta,u).\end{cases}

Through rescaling, we conclude that the eigen-function $\phi_{u}$ in (3.1) satisfies (3.5).

∎

In the following lemma, we bound $\psi_{u}$ in the $w$ -weighted $L^{\infty}$ norm. Here we recall the definition of $\psi_{u}$ in (3.2).

Lemma 9.

For the eigen-function $\phi_{u}$ in (3.1) in Proposition 4, we further have

\|w\psi_{u}\|_{L^{\infty}_{\xi}}=\big{\|}w\frac{\phi_{u}-\phi_{0}}{u}\big{\|}_{L^{\infty}_{\xi}}\lesssim 1,

(3.8)

where the inequality in (3.8) does not depend on $u$ .

Proof.

Following the proof of Proposition 3.1 in [3], for $z$ near $0$ , there exists $\phi_{z}$ and $\lambda(z)=u(z)z$ which are analytic in $z$ such that

\displaystyle T(z)\phi(z):=(\mathcal{L}-z\xi_{1})\phi(z)=\lambda(z)\phi(z).

Denoting $\dot{\phi}_{z}$ as the derivative with respect to $z$ , then we have

\displaystyle\mathcal{L}\dot{\phi}_{z}-z\xi_{1}\dot{\phi}_{z}-\xi_{1}\phi_{z}=\lambda(z)\dot{\phi}_{z}+\dot{\lambda}(z)\phi_{z},

which is equivalent to

\displaystyle[\nu(\xi)-z(\xi_{1}+u)]\dot{\phi}_{z}=K\dot{\phi}_{z}+(\dot{\lambda}(z)+\xi_{1})\phi_{z}.

Since $\phi_{z}$ is analytic in $z$ (also see Proposition 4), we have $\|\dot{\phi}_{z}\|_{L^{2}_{\xi}}\lesssim 1$ . Combining with $\|[1+\xi]\phi_{z}\|_{L^{\infty}_{\xi}}\lesssim 1$ from Proposition 4, and the fact that $K$ is bounded from $L^{2}_{\xi}$ to $L^{\infty,1/2}_{\xi}$ , we conclude that

\displaystyle\|\dot{\phi}_{z}\|_{L^{\infty}_{\xi}}\lesssim\big{\|}\frac{K\dot{\phi}_{z}}{\nu(\xi)}\big{\|}_{L^{\infty}_{\xi}}+\|[1+\xi]\phi_{z}\|_{L^{\infty}_{\xi}}\lesssim 1.

(3.9)

From the Taylor’s theorem, $\frac{\phi_{u}-\phi_{0}}{u}=\dot{\phi}_{\bar{u}}$ for some $\bar{u}\in[0,u]$ .

To prove (3.8), we apply the same argument in the proof of Lemma 8. With the modified linear operator $\mathcal{L}_{\theta}$ defined in (3.7), we apply the same computation of (3.9) to the eigen-value problem $[\mathcal{L}_{\theta}-z\xi_{1}]\phi_{\theta}(z)=\lambda(z)\phi_{\theta}(z)$ and deduce that

\displaystyle\|\dot{\phi}_{\theta}(z)\|_{L^{\infty}_{\xi}}\lesssim 1.

Then we conclude (3.8) from the Taylor’s theorem and the fact that $\dot{\phi}(z)=e^{-\theta|\xi|^{2}}\dot{\phi}_{\theta}(z)$ .

∎

3.2. Continuity and $w$ -weighted $L^{\infty}$ estimate of the linearized penalized problem

With $\psi_{u}$ defined in (3.2), we define

\mathbf{p}_{u}g=-\langle(\xi_{1}+u)\psi_{u}g\rangle\phi_{u},\ \ \mathbf{P}_{u}g=-\langle\psi_{u}g\rangle(\xi_{1}+u)\phi_{u}.

Now we define the linearized penalized problem in the following proposition.

Proposition 5 (Proposition 5.3 and Proposition 5.6 in [3]).

Define the linearized penalized collision operator as

\mathcal{L}^{p}g:=\mathcal{L}g+2\gamma\prod_{+}((\xi_{1}+u)g)+2\gamma\mathbf{p}_{u}g-\gamma(\xi_{1}+u)g.

(3.10)

Here $\gamma$ is a small constant $\gamma\ll 1$ .

Let $Q\in L^{2}_{x,\xi}$ . There exists a unique solution $g$ to the following linearized penalized problem

\begin{cases}&(\xi_{1}+u)\partial_{x}g+\mathcal{L}^{p}g=Q,\ x>0,\ \xi\in\mathbb{R}^{3}\\ &g(0,\xi)=g_{b}(\xi),\ \xi_{1}+u>0.\end{cases}

(3.11)

Moreover, this solution satisfies the $L^{2}$ -estimate

\|\nu g\|_{L^{2}_{x,\xi}}\leq C_{2}\big{[}\|Q\|_{L^{2}_{x,\xi}}+\|\nu g_{b}\|_{L^{2}_{\xi}}\big{]}.

(3.12)

If we further assume $Q$ and $g_{b}$ satisfy

\displaystyle(1+|\xi|)^{3}g_{b}\in L^{\infty}_{\xi},\ \ (1+|\xi|)^{2}Q\in L^{\infty}_{x,\xi},\ \ Q(x,\cdot)\perp\ker\mathcal{L},

then the solution further satisfies the $L^{\infty}$ -estimate

\displaystyle\|(1+|\xi|)^{3}g\|_{L^{\infty}_{x,\xi}}\leq C\big{[}\|(1+|\xi|)^{3}g_{b}\|_{L^{\infty}_{\xi}}+\|(1+|\xi|)^{2}Q\|_{L^{\infty}_{x,\xi}}+\|g\|_{L^{2}_{x,\xi}}\big{]}.

(3.13)

Remark 9.

Here we note that the $L^{\infty}$ estimate in (3.13) is controlled by the $L^{2}$ estimate $\|g\|_{L^{2}_{x,\xi}}$ , and thus it can be further bounded using (3.12):

\displaystyle\|(1+|\xi|)^{3}g\|_{L^{\infty}_{x,\xi}}\leq C\big{[}\|(1+|\xi|)^{3}g_{b}\|_{L^{\infty}_{\xi}}+\|e^{\delta x}(1+|\xi|)^{2}Q\|_{L^{\infty}_{x,\xi}}\big{]}.

Denote

\displaystyle-K^{p}g

\displaystyle=-Kg+2\gamma\prod_{+}((\xi_{1}+u)g)+2\gamma\mathbf{p}_{u}g,

(3.14)

so that $\mathcal{L}^{p}g=[\nu(\xi)-\gamma(\xi_{1}+u)]g-K^{p}g$ . With the $w$ -weighted estimate for the eigen-function problem in Lemma 8, we have the following property for $K^{p}$ :

Lemma 10.

\|K^{p}g\|_{L^{2}_{x,\xi}}\lesssim\|g\|_{L^{2}_{x,\xi}}.

(3.15)

\|w(\xi)K^{p}g(\xi)\|_{L^{\infty}_{x,\xi}}\lesssim\|wg\|_{L^{\infty}_{x,\xi}}.

(3.16)

Proof.

Proof of (3.15). Clearly $\|Kf\|_{L^{2}_{x,\xi}}\lesssim\|f\|_{L^{2}_{x,\xi}}$ . For the rest, the contribution of $\prod_{+}(\xi_{1}+u)g$ is bounded as

	$\displaystyle\\|\chi_{+}\langle(\xi_{1}+u)g,\chi_{+}\rangle\\|_{L^{2}_{x,\xi}}^{2}$	$\displaystyle\lesssim\iint_{\mathbb{R}^{+}\times\mathbb{R}^{3}}e^{-\frac{\|\xi\|^{2}}{4}}\big{(}\int_{\mathbb{R}^{3}}\|g(x,\xi^{\prime})\|^{2}\mathrm{d}\xi^{\prime}\big{)}\big{(}\int_{\mathbb{R}^{3}}(\xi_{1}^{\prime}+u)^{2}e^{-\frac{\|\xi^{\prime}\|^{2}}{4}}\mathrm{d}\xi^{\prime}\big{)}\mathrm{d}x\mathrm{d}\xi$
		$\displaystyle\lesssim\iint_{\mathbb{R}^{+}\times\mathbb{R}^{3}}\|g(x,\xi^{\prime})\|^{2}\mathrm{d}x\mathrm{d}\xi^{\prime}=\\|g\\|^{2}_{L^{2}_{x,\xi}}.$

The contribution of $\mathbf{p}_{u}g$ is bounded as

	$\displaystyle\\|\phi_{u}\langle(\xi_{1}+u)\psi_{u}g\rangle\\|_{L^{2}_{x,\xi}}^{2}$
	$\displaystyle\lesssim\iint_{\mathbb{R}^{+}\times\mathbb{R}^{3}}\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}^{2}w^{-2}(\xi)\big{(}\int_{\mathbb{R}^{3}}\|g(x,\xi^{\prime})\|^{2}\mathrm{d}\xi^{\prime}\big{)}\\|w\psi_{u}\\|_{L^{\infty}_{x,\xi}}^{2}\big{(}\int_{\mathbb{R}^{3}}(\xi^{\prime}_{1}+u)^{2}w^{-2}(\xi^{\prime})\mathrm{d}\xi^{\prime}\big{)}\mathrm{d}x\mathrm{d}\xi$
	$\displaystyle\lesssim\iint_{\mathbb{R}^{+}\times\mathbb{R}^{3}}\|g(x,\xi^{\prime})\|^{2}\mathrm{d}x\mathrm{d}\xi^{\prime}\\|g\\|_{L^{2}_{x,\xi}}^{2}.$

Here we have applied Hölder inequality to $\langle(\xi_{1}+u)\psi_{u}g\rangle$ and used Lemma 8 and Lemma 9.

Proof of (3.16). The contribution of $K$ in $K^{p}$ can be controlled using Lemma 3:

\displaystyle\|wKg\|_{L^{\infty}_{x,\xi}}\leq\|wg\|_{L^{\infty}_{x,\xi}}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\frac{w(\xi)}{w(\xi^{\prime})}\mathrm{d}\xi^{\prime}\lesssim\|wg\|_{L^{\infty}_{x,\xi}}.

The contribution of $\prod_{+}(\xi_{1}+u)g$ is bounded as

\displaystyle w(\xi)\chi_{+}\langle(\xi_{1}+u)g,\chi_{+}\rangle\lesssim\|wg\|_{L^{\infty}_{x,\xi}}\langle(\xi_{1}+u)e^{-\theta|\xi|^{2}},\chi_{+}\rangle\lesssim\|wg\|_{L^{\infty}_{x,\xi}}.

(3.17)

The contribution of $\mathbf{p}_{u}g$ can be bounded using Proposition 4 and Lemma 8:

	$\displaystyle\|w(\xi)\phi_{u}(\xi)\langle(\xi_{1}+u)\psi_{u}g\rangle\|\leq\\|w\phi_{u}\\|_{L^{\infty}_{x,\xi}}\\|wg\\|_{L^{\infty}_{x,\xi}}\|\langle(\xi_{1}+u)\psi_{u}e^{-\theta\|\xi\|^{2}}\rangle\|$
	$\displaystyle\lesssim\\|w\phi_{u}\\|_{L^{\infty}_{x,\xi}}\\|wg\\|_{L^{\infty}_{x,\xi}}\\|\psi_{u}\\|_{L^{2}_{\xi}}\lesssim\\|wg\\|_{L^{\infty}_{x,\xi}}.$		(3.18)

Then we conclude the lemma.

∎

In the following proposition, we improve the result in Proposition 5 by establishing the continuity and $w$ -weighted $L^{\infty}$ estimate.

Proposition 6.

Suppose all conditions in Proposition 5 are satisfied. We assume $g_{b}(\xi)\in C(\mathbb{R}^{3})$ , and $Q$ is continuous away from $\mathcal{D}$ defined in (1.14), then the unique solution $g$ of (3.11) is continuous away from $\mathcal{D}$ . If we further assume $\|wg_{b}\|_{L^{\infty}_{\xi}}<\infty$ and $\big{\|}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}<\infty$ , the solution in Proposition 5 satisfies

\|wg\|_{L^{\infty}_{x,\xi}}\leq C\big{[}\|wg_{b}\|_{L^{\infty}_{\xi}}+\big{\|}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}+\|g\|_{L^{2}_{x,\xi}}\big{]}.

(3.19)

Remark 10.

By the $L^{2}$ bound (3.12), for some $\delta\ll 1$ , (3.19) can be further bounded as

\|wg\|_{L^{\infty}_{x,\xi}}\leq C\big{[}\|wg_{b}\|_{L^{\infty}_{\xi}}+\big{\|}e^{\delta x}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}\big{]}.

(3.20)

Proof.

We consider a variant of (3.11):

\begin{cases}&(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-\lambda K^{p}g=Q\\ &g(0,\xi)=g_{b}(\xi),\ \xi_{1}+u>0.\end{cases}

(3.21)

It is straightforward to apply the same argument in [3] to show that (3.21) is well-posed and satisfies both estimates (3.12) and (3.13) uniformly in $\lambda$ . To obtain the $L^{\infty}_{\xi}$ estimate with exponential weight $w(\xi)$ , we will prove the a-priori estimate in Step 1, and in Step 2, we will use fixed point argument to justify that the solution $g$ indeed satisfies (3.19) and is continuous away from $\mathcal{D}$ .

Step 1. A-priori estimate. In this step we prove the following statement: suppose the solution to (3.21) satisfies $\|wg\|_{L^{\infty}_{x,\xi}}<\infty$ , then we have

\|wg\|_{L^{\infty}_{x,\xi}}\lesssim\|wg_{b}\|_{L^{\infty}_{\xi}}+\big{\|}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}+\|g\|_{L^{2}_{x,\xi}}.

(3.22)

Here the inequality does not depend on $\lambda$ .

We denote $G(\xi):=\|g(x,\xi)\|_{L^{\infty}_{x}}$ . Applying the Duhamel’s principle and taking $\sup$ in $x$ , we have

\displaystyle G(\xi)

\displaystyle\leq|g_{b}(\xi)|+\frac{|K^{p}G(\xi)|}{\nu(\xi)-\gamma|\xi_{1}+u|}+\frac{\|Q(x,\xi)\|_{L^{\infty}_{x}}}{\nu(\xi)-\gamma|\xi_{1}+u|}.

(3.23)

To estimate $wG(\xi)$ , we first estimate $w|K^{p}G|$ . We apply Lemma 3 to compute the contribution of $K$ in $K^{p}$ as

	$\displaystyle w(\xi)KG(\xi)=w(\xi)\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})G(\xi^{\prime})\mathrm{d}\xi^{\prime}$
	$\displaystyle=\int_{\|\xi^{\prime}\|>N\text{ or }\|\xi^{\prime}-\xi\|<\frac{1}{N}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})w(\xi^{\prime})G(\xi^{\prime})\mathrm{d}\xi^{\prime}+\int_{\|\xi^{\prime}\|\leq N\text{ and }\|\xi^{\prime}-\xi\|\geq\frac{1}{N}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})w(\xi^{\prime})G(\xi^{\prime})\mathrm{d}\xi^{\prime}$
	$\displaystyle\leq o(1)\\|wG\\|_{L^{\infty}_{\xi}}+\int_{\|\xi^{\prime}\|\leq N}G(\xi^{\prime})\mathrm{d}\xi^{\prime}\lesssim o(1)\\|wG\\|_{L^{\infty}_{\xi}}+C_{N}\\|G\\|_{L^{2}_{\xi}}.$		(3.24)

In the last line, in the first inequality, we used $\mathbf{k}_{\theta}(\xi,\xi^{\prime})w(\xi^{\prime})\lesssim_{N}1$ when $|\xi^{\prime}|\leq N,\ |\xi^{\prime}-\xi|\geq\frac{1}{N}$ ; when $|\xi^{\prime}-\xi|<\frac{1}{N}$ , we directly have $\int_{|\xi^{\prime}-\xi|<\frac{1}{N}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})\mathrm{d}\xi^{\prime}\lesssim o(1)$ ; when $|\xi^{\prime}|>N$ , we further split the case into $|\xi|<\frac{N}{2}$ and $|\xi|>\frac{N}{2}$ , for the first case we have $|\xi-\xi^{\prime}|>\frac{N}{2}$ , and thus $\int_{|\xi^{\prime}-\xi|>\frac{N}{2}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})\mathrm{d}\xi^{\prime}\lesssim o(1)$ ; for the second case we have $\mathbf{1}_{|\xi|>\frac{N}{2}}\int_{\mathbb{R}^{3}}\mathbf{k}_{\theta}(\xi,\xi^{\prime})\mathrm{d}\xi^{\prime}\lesssim o(1)$ from (2.10). In the last inequality, we have used the Hölder inequality in the bounded space $|\xi^{\prime}|\leq N$ .

The contribution of $\gamma\prod_{+}((\xi_{1}+u)g)$ and $\gamma\mathbf{p}_{u}g$ can be controlled using (3.17) and (3.18), with the extra constant $\gamma\ll 1$ :

\displaystyle\gamma\|wG\|_{L^{\infty}_{\xi}}\leq o(1)\|wG\|_{L^{\infty}_{\xi}}.

(3.25)

Combining (3.23), (3.24) and (3.25), we conclude that

\displaystyle\|wG\|_{L^{\infty}_{\xi}}

\displaystyle\lesssim\|wg_{b}\|_{L^{\infty}_{\xi}}+\|G\|_{L_{\xi}^{2}}+\big{\|}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}.

(3.26)

The $L^{2}(\mathrm{d}\xi;L^{\infty}_{x})$ estimate $\|G\|_{L^{2}_{\xi}}$ follows from (5.19) in [3]:

\displaystyle\|G\|_{L^{2}_{\xi}}\leq 2\|g_{b}\|_{L^{2}_{\xi}}+C\|Q\|_{L^{2}(\mathrm{d}\xi;L^{\infty}_{x})}+\|g\|_{L^{2}_{x,\xi}},

where we can further control the RHS as

\displaystyle\|G\|_{L^{2}_{\xi}}\lesssim\|wg_{b}\|_{L^{\infty}_{\xi}}+\big{\|}\frac{w}{[1+|\xi|]}Q\big{\|}_{L^{\infty}_{x,\xi}}+\|g\|_{L^{2}_{x,\xi}}.

(3.27)

Combining (3.27) and (3.26), we conclude (3.22).

Step 2. Fixed point argument.

For fixed boundary data $g_{b}(\xi)$ , we denote $\mathcal{L}_{\lambda}^{-1}$ to be the solution operator associated with (3.21), i.e, the solution to (3.21) is $g=\mathcal{L}_{\lambda}^{-1}(Q)$ .

We define a Banach space as

\mathcal{X}:=\Big{\{}g(x,\xi):\|wg\|_{L^{\infty}_{x,\xi}}<\infty,\ \|g\|_{L^{2}_{x,\xi}}<\infty,\ \ g\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D})\Big{\}},

(3.28)

with the associated norm

\|g\|_{\mathcal{X}}:=\|wg\|_{L^{\infty}_{x,\xi}}+\|g\|_{L^{2}_{x,\xi}}.

We start from $\lambda=0$ and define an operator as

T_{\lambda}g=\mathcal{L}_{0}^{-1}(\lambda K^{p}g+Q).

When $g\in\mathcal{X}$ , from the assumption, we have that both $g,Q\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D})$ . The continuity of $K^{p}g$ follows from the fact that $Kg$ is continuous and $\phi_{u},\chi_{+}$ are continuous functions from Lemma 8. Since $g_{b}\in C(\mathbb{R}^{3})$ , without the contribution of $K^{p}$ acting on $T_{\lambda}g$ , we conclude that $T_{\lambda}g=\mathcal{L}_{0}^{-1}(\lambda K^{p}g+Q)\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D})$ .

Again, without the contribution of $K^{p}T_{\lambda}g$ , $\|wT_{\lambda}g\|_{L^{\infty}_{x,\xi}}$ can be directly controlled by (3.26), which is bounded. Thus we can apply the a-priori estimate (3.22) with the $L^{2}$ estimate (3.12) to have

	$\displaystyle\\|T_{\lambda}g\\|_{L^{2}_{x,\xi}}+\\|wT_{\lambda}g\\|_{L^{\infty}_{x,\xi}}$
	$\displaystyle\lesssim\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\\|\lambda K^{p}g+Q\\|_{L^{2}_{x,\xi}}+\big{\\|}\frac{w}{[1+\|\xi\|]}Q\big{\\|}_{L^{\infty}_{x,\xi}}+\lambda\\|wK^{p}g\\|_{L^{\infty}_{x,\xi}}+\\|T_{\lambda}g\\|_{L^{2}_{x,\xi}}$
	$\displaystyle\lesssim\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\big{\\|}\frac{w}{[1+\|\xi\|]}Q\big{\\|}_{L^{\infty}_{x,\xi}}+\lambda\\|g\\|_{L^{2}_{x,\xi}}+\lambda\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|Q\\|_{L^{2}_{x,\xi}}.$		(3.29)

Here we have used Lemma 10. From the assumption that $\|Q\|_{L^{2}_{x,\xi}},\|\frac{w}{[1+|\xi|]}Q\|_{L^{\infty}_{x,\xi}},\|wg_{b}\|_{L^{\infty}_{\xi}},\|g\|_{\mathcal{X}}<\infty$ , we conclude that $T_{\lambda}g\in\mathcal{X}.$

Given $g_{1},g_{2}\in\mathcal{X}$ , then $g_{T}:=T_{\lambda}(g_{1}-g_{2})$ satisfies

\begin{cases}&(\xi_{1}+u)\partial_{x}g_{T}+[\nu(\xi)-\gamma(\xi_{1}+u)]g_{T}=-\lambda K^{p}(g_{1}-g_{2})\\ &g_{T}(0,\xi)=0,\ \xi_{1}+u>0.\end{cases}

Again we apply the a-priori estimate (3.22). With (3.12) and Lemma 10, we have

$\displaystyle\\|g_{T}\\|_{\mathcal{X}}$	$\displaystyle=\\|wg_{T}\\|_{L^{\infty}_{x,\xi}}+\\|g_{T}\\|_{L^{2}_{x,\xi}}$
	$\displaystyle\lesssim\lambda\\|K^{p}(g_{1}-g_{2})\\|_{L^{2}_{x,\xi}}+\lambda\\|wK^{p}(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+\\|g_{T}\\|_{L^{2}_{x,\xi}}$
	$\displaystyle\lesssim\lambda\\|g_{1}-g_{2}\\|_{L^{2}_{x,\xi}}+\lambda\\|w(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+\lambda\\|K^{p}(g_{1}-g_{2})\\|_{L^{2}_{x,\xi}}$
	$\displaystyle\leq C\lambda\\|g_{1}-g_{2}\\|_{\mathcal{X}}.$	(3.30)

Then we choose $\lambda_{*}=\lambda_{*}(C)$ to be small enough such that $\lambda_{*}C\leq\frac{1}{2}$ , then for $0\leq\lambda\leq\lambda_{*}$ , from the Banach fixed point theorem we conclude that $T_{\lambda}$ has a fixed point, i.e, there exists $g\in\mathcal{X}$ such that $T_{\lambda}g=g$ , which is equivalent to $g=\mathcal{L}_{0}^{-1}(\lambda Kg+Q)$ , and

\displaystyle(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-\lambda K^{p}g=Q.

Therefore, the solution to (3.21) is continuous away from $\mathcal{D}$ and satisfies $\|g\|_{\mathcal{X}}<\infty$ for $0\leq\lambda\leq\lambda_{*}$ .

Next we define

T_{\lambda_{*}+\lambda}g=\mathcal{L}^{-1}_{\lambda_{*}}(\lambda Kg+Q).

Since the estimates (3.12) and (3.22) are uniform in $\lambda$ , we can apply a similar fixed-point argument in (3.30) to conclude that there exists a fixed point $T_{\lambda+\lambda_{*}}g=g$ for $g\in\mathcal{X}$ . Step by step, we can conclude that $\mathcal{L}_{1}^{-1}(Q)$ , the solution to (3.11), is in $\mathcal{X}$ and thus is continuous away from $\mathcal{D}$ .

∎

3.3. Continuity and $w$ -weighted $L^{\infty}$ estimate of the nonlinear penalized problem

With the continuity and $w$ -weighted $L^{\infty}_{x,\xi}$ estimate of the eigenfunction and the linearized penalized problem in Lemma 8 and Proposition 6, we are ready to prove the continuity and $w$ -weighted $L^{\infty}_{x,\xi}$ estimate of the nonlinear penalized problem. The nonlinear problem is given by

\begin{cases}&(\xi_{1}+u)\partial_{x}g+\mathcal{L}^{p}g=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(g-h\phi_{u},g-h\phi_{u}),\\ &h(x)=-e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Gamma(g-h\phi_{u},g-h\phi_{u})\rangle(x+z)\mathrm{d}z,\\ &g(0,\xi)=f_{b}(\xi)+h(0)\phi_{u}(\xi),\ \ \xi_{1}+u>0.\end{cases}

(3.31)

The well-posedness of the nonlinear problem is already established in [3]:

Proposition 7 (Proposition 6.1 in [3]).

Suppose the boundary data $f_{b}(\xi)$ satisfy (2.1). There exists a unique solution $(g,h)$ to (3.31) such that

g(x,\mathcal{R}\xi)=g(x,\xi)

and for $\varepsilon\ll 1$ ,

\|(1+|\xi|)^{3}g\|_{L^{\infty}_{x,\xi}}+\|h\|_{L^{\infty}_{x}}\leq C\varepsilon.

In the following proposition, we construct the continuity and $w$ -weighted $L^{\infty}_{x,\xi}$ estimate of the nonlinear problem.

Proposition 8.

Suppose the boundary data (2.1) further satisfies (1.17). Then the solution in Proposition 7 satisfies

h(x)\in C(\mathbb{R}^{+}),\ \ g(x,\xi)\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D}),\ \ \|wg\|_{L^{\infty}_{x,\xi}}+\|h\|_{L^{\infty}_{x}}\leq C\varepsilon.

Proof.

For some $C>0$ specified later, we denote a Banach space as

	$\displaystyle\mathcal{X}_{1}:=$	$\displaystyle\big{\{}(g,h):\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|h\\|_{L^{\infty}_{x}}\leq 2C\varepsilon,\ g(x,\mathcal{R}\xi)=g(x,\xi),$
		$\displaystyle\ \ \text{ and }h\in C(\mathbb{R}^{3}),\ g\in C(\mathbb{R}^{+}\times\mathbb{R}^{3}\backslash\mathcal{D})\big{\}},$		(3.32)

with norm given by

\|(g,h)\|_{\mathcal{X}_{1}}:=\|wg\|_{L^{\infty}_{x,\xi}}+\|h\|_{L^{\infty}_{x}}.

Given $(\tilde{g},\tilde{h})\in\mathcal{X}_{1}$ , we focus on the following linear problem:

\begin{cases}&(\xi_{1}+u)\partial_{x}g+\mathcal{L}^{p}g=e^{-\gamma x}(\mathbf{\mathbf{missing}}{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u}),\\ &h(x)=-e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(x+z)\mathrm{d}z,\\ &g(0,\xi)=f_{b}(\xi)+h(0)\phi_{u}(\xi),\ \ \xi_{1}+u>0.\end{cases}

(3.33)

The well-posedness of the above system is given by Proposition 5, with $w$ -weighted estimate given in Proposition 6. Then we denote $\mathcal{S}$ as the map from $(\tilde{g},\tilde{h})$ to the solution $(g,h)$ : $\mathcal{S}((\tilde{g},\tilde{h}))=(g,h)$ .

By the continuity of $\tilde{h}$ and $\tilde{g}$ from $(\tilde{g},\tilde{h})\in\mathcal{X}_{1}$ , we conclude the continuity of $h(x)$ from the continuity of $\phi_{u}$ in Lemma 8.

Since $f_{b}$ is continuous from (1.17) and $\phi_{u}$ is continuous in Lemma 8, $g(0,\xi)$ is continuous. Then the continuity of $g$ follows from Proposition 6, with the continuity of $\phi_{u}$ , $g(0,\xi)$ , and the assumption that $(\tilde{g},\tilde{h})\in\mathcal{X}_{1}$ .

Applying (2.6), we have the estimate for the nonlinear term $\Gamma$ as

	$\displaystyle\Big{\\|}\frac{w}{[1+\|\xi\|]}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\Big{\\|}_{L^{\infty}_{x,\xi}}\lesssim\\|w(\tilde{g}-\tilde{h}\phi_{u})\\|_{L^{\infty}_{x,\xi}}^{2}$
	$\displaystyle\lesssim\\|w\tilde{g}\\|^{2}_{L^{\infty}_{x,\xi}}+\\|w\phi_{u}\\|_{L^{\infty}_{x,\xi}}^{2}\\|\tilde{h}\\|_{L^{\infty}_{x}}^{2}\lesssim\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}^{2}.$		(3.34)

In the last inequality we have applied Lemma 8.

Thus we bound $h(x)$ in (3.33) as

\displaystyle\|h\|_{L^{\infty}_{x}}

\displaystyle\lesssim\Big{\|}\frac{w}{[1+|\xi|]}\Gamma\Big{\|}_{L^{\infty}_{x,\xi}}\Big{\langle}\psi_{u}\frac{[1+|\xi|]}{w}\Big{\rangle}\lesssim\|\psi_{u}\|_{L^{2}_{\xi}}\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}\lesssim\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}.

(3.35)

Here we have used Proposition 4.

For the boundary condition, we have

\displaystyle g_{b}=f_{b}+h(0)\phi_{u},

then from Lemma 8 and (3.35), we have

\displaystyle\|wg_{b}\|_{L^{\infty}_{\xi}}\leq\|wf_{b}\|_{L^{\infty}_{\xi}}+\|w\phi_{u}\|_{L^{\infty}_{\xi}}\|h\|_{L^{\infty}_{x}}\lesssim\|wf_{b}\|_{L^{\infty}_{\xi}}+\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}.

(3.36)

Now we apply (3.20) to $g$ in (3.33), and we take $\delta=\gamma$ to have

	$\displaystyle\\|wg\\|_{L^{\infty}_{x,\xi}}$	$\displaystyle\lesssim\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\Big{\\|}e^{(\delta-\gamma)x}\frac{w}{[1+\|\xi\|]}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\Big{\\|}_{L^{\infty}_{x,\xi}}$
		$\displaystyle\lesssim\\|wf_{b}\\|_{L^{\infty}_{\xi}}+\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}^{2}.$		(3.37)

In the second line we have used (3.36) and (3.34), and applied the following computation for $\mathbf{P}_{u}$ :

	$\displaystyle\Big{\\|}\frac{w}{[1+\|\xi\|]}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(\xi_{1}+u)\phi_{u}\Big{\\|}_{L^{\infty}_{x,\xi}}$
	$\displaystyle\lesssim\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\Big{\\|}\frac{w}{[1+\|\xi\|]}\Gamma\Big{\\|}_{L^{\infty}_{x,\xi}}\Big{\langle}\psi\frac{[1+\|\xi\|]}{w(\xi)}\Big{\rangle}\lesssim\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\\|\psi\\|_{L^{2}_{\xi}}\Big{\\|}\frac{w}{[1+\|\xi\|]}\Gamma\Big{\\|}_{L^{\infty}_{x,\xi}}\lesssim\Big{\\|}\frac{w}{[1+\|\xi\|]}\Gamma\Big{\\|}_{L^{\infty}_{x,\xi}}.$		(3.38)

Combining (3.35) and (3.37), we conclude that for some $C>0$ ,

\displaystyle\|(g,h)\|_{\mathcal{X}_{1}}

\displaystyle\leq C\|wf_{b}\|_{L^{\infty}_{\xi}}+C\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}.

Then we let $\varepsilon$ in (1.17) be small enough such that $4C^{2}\varepsilon<1$ . Then if $\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}\leq 2C\varepsilon$ , we have

\displaystyle\|(g,h)\|_{\mathcal{X}_{1}}\leq C\varepsilon+4C^{3}\varepsilon^{2}\leq 2C\varepsilon.

(3.39)

Combining the continuity of $g$ , $h$ and (3.39), we conclude that $(g,h)\in\mathcal{X}_{1}$ .

It remains to prove the contraction property. Given $(\tilde{g}_{1},\tilde{h}_{1}),(\tilde{g}_{2},\tilde{h}_{2})\in\mathcal{X}_{1}$ as the source terms in (3.33), we denote the corresponding solutions as $(g_{1},h_{1}),(g_{2},h_{2})$ . The equations for $g_{1}-g_{2},h_{1}-h_{2}$ read

\begin{cases}&(\xi_{1}+u)\partial_{x}(g_{1}-g_{2})+\mathcal{L}^{p}(g_{1}-g_{2})=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Sigma,\\ &\Sigma=\Gamma(\tilde{g}_{1}-\tilde{g}_{2}-(\tilde{h}_{1}-\tilde{h}_{2})\phi_{u},\tilde{g}_{1}-\tilde{h}_{1}\phi_{u})+\Gamma(\tilde{g}_{2}-\tilde{h}_{2}\phi_{u},\tilde{g}_{1}-\tilde{g}_{2}-(\tilde{h}_{1}-\tilde{h}_{2})\phi_{u}),\\ &(h_{1}-h_{2})(x)=-e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Sigma\rangle(x+z)\mathrm{d}z,\\ &(g_{1}-g_{2})(0,\xi)=(h_{1}-h_{2})(0)\phi_{u}(\xi),\ \ \xi_{1}+u>0.\end{cases}

(3.40)

Applying (2.6) and the same computation in (3.34), the source term is bounded as

	$\displaystyle\Big{\\|}\frac{w}{[1+\|\xi\|]}\Sigma\Big{\\|}_{L^{\infty}_{x,\xi}}$	$\displaystyle\lesssim\\|w[\tilde{g}_{1}-\tilde{g}_{2}-(\tilde{h}_{1}-\tilde{h}_{2})\phi_{u}]\\|_{L^{\infty}_{x,\xi}}\big{[}\\|w(\tilde{g}_{1}-\tilde{h}_{1}\phi_{u})\\|_{L^{\infty}_{x,\xi}}+\\|w(\tilde{g}_{2}-\tilde{h}_{2}\phi_{u})\\|_{L^{\infty}_{x,\xi}}\big{]}$
		$\displaystyle\lesssim[\\|(\tilde{g}_{1},\tilde{h}_{1})\\|_{\mathcal{X}_{1}}+\\|(\tilde{g}_{2},\tilde{h}_{2})\\|_{\mathcal{X}_{1}}]\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}\lesssim\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}.$

Then we apply the same computation in (3.35) to bound $h_{1}-h_{2}$ as

\displaystyle\|h_{1}-h_{2}\|_{L^{\infty}_{x}}

\displaystyle\lesssim\Big{\|}\frac{w}{[1+|\xi|]}\Sigma\Big{\|}_{L^{\infty}_{x,\xi}}\lesssim\varepsilon\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\|_{\mathcal{X}_{1}}.

(3.41)

We denote the incoming boundary as $(g_{1}-g_{2})_{b}$ , then it is bounded similarly as (3.41):

\displaystyle\|w(g_{1}-g_{2})_{b}\|_{L^{\infty}_{\xi}}

\displaystyle\lesssim\|h_{1}-h_{2}\|_{L^{\infty}_{x}}\lesssim\varepsilon\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\|_{\mathcal{X}_{1}}.

Now we apply (3.20) with $\delta=\gamma$ and similar computation in (3.37) to bound $g_{1}-g_{2}$ as

\displaystyle\|w(g_{1}-g_{2})\|_{L^{\infty}_{x,\xi}}

\displaystyle\lesssim\|w(g_{1}-g_{2})_{b}\|_{L^{\infty}_{\xi}}+\Big{\|}\frac{w}{[1+|\xi|]}(\mathbf{I}-\mathbf{P}_{u})\Sigma\Big{\|}_{L^{\infty}_{x,\xi}}\lesssim\varepsilon\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\|_{\mathcal{X}_{1}}.

(3.42)

Combining (3.41) and (3.42) we conclude that for some $C_{1}>0$ ,

\displaystyle\|(g_{1}-g_{2},h_{1}-h_{2})\|_{\mathcal{X}_{1}}

\displaystyle\leq C_{1}\varepsilon\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\|_{\mathcal{X}_{1}}.

(3.43)

Last we take $\varepsilon$ to be small such that $C_{1}\varepsilon<1$ . We conclude the proposition from the Banach fixed-point theorem.

∎

3.4. Proof of Theorem 2

Finally, the solution of the boundary layer problem (1.4) is constructed by assuming further condition (1.18) on the boundary data. The solution reads

f(x,\xi)=e^{-\gamma x}g(x,\xi)-e^{-\gamma x}h(x)\phi_{u}(\xi),

where $g,h$ are the unique solution constructed in Proposition 7. With the continuity of $g,h,\phi_{u}$ and $w$ -weighted estimate of $g,\phi$ established in Lemma 8 and Proposition 8, we conclude Theorem 2.

4. Weighted $C^{1}$ estimate and $W^{1,p}$ estimate for $p<2$ without weight

In the section, we will conclude Theorem 1 by applying the weight $\alpha$ in (1.8) with its property in Section 2.2.

We begin with proving the existence of derivative to the damped transport equation

\begin{cases}&(\xi_{1}+u)\partial_{x}f(x,\xi)+\nu f(x,\xi)=Q\\ &f(0,\xi)|_{\xi_{1}+u>0}=f_{b}(\xi).\end{cases}

(4.1)

Lemma 11.

Suppose $|Q|<\infty$ and $\partial_{x}Q$ exists. Then $\partial_{x}f(x,\xi)$ exists.

Proof.

First we consider the case of $\xi_{1}+u=0$ . Then $f(x,\xi)$ can be expressed as

\displaystyle f(x,\xi)=e^{-\nu(\xi)t}f(x,\xi)+\int^{t}_{0}\mathrm{d}se^{-\nu(\xi)(t-s)}Q(x,\xi).

The difference quotient reads

\displaystyle\frac{f(x+\varepsilon,\xi)-f(x,\xi)}{\varepsilon}=e^{-\nu(\xi)t}\frac{f(x+\varepsilon,\xi)-f(x,\xi)}{\varepsilon}+\int^{t}_{0}\mathrm{d}se^{-\nu(\xi)(t-s)}\frac{Q(x+\varepsilon,\xi)-Q(x,\xi)}{\varepsilon},

thus

\displaystyle(1-e^{-\nu(\xi)t})\frac{f(x+\varepsilon,\xi)-f(x,\xi)}{\varepsilon}=\int^{t}_{0}\mathrm{d}se^{-\nu(\xi)(t-s)}\frac{Q(x+\varepsilon,\xi)-Q(x,\xi)}{\varepsilon}.

From the assumption that $\partial_{x}Q$ exists, we can pass $\varepsilon$ to $0$ to conclude that $\partial_{x}f(x,\xi)$ exists for $\xi_{1}+u=0$ .

Next, we consider the case of $\xi_{1}+u\neq 0$ . Then from the equation (4.1), we have $|\partial_{x}f|=\Big{|}\frac{Q-\nu f}{\xi_{1}+u}\Big{|}<\infty$ from $\|wf\|_{L^{\infty}_{x,\xi}}<\infty$ . Thus $\partial_{x}f(x,\xi)$ exists.

∎

4.1. Weighted $C^{1}$ estimate of the linearized penalized problem

Before we construct the derivative of the linearized penalized problem, we first establish an a-priori estimate in the following lemma:

Lemma 12 (A-priori weighted $C^{1}$ estimate).

Let $g$ be the solution of the linearized problem (3.11), suppose $\partial_{x}g$ exists and satisfies

\displaystyle(\xi_{1}+u)\partial_{x}(\partial_{x}g)+\mathcal{L}^{p}(\partial_{x}g)=\partial_{x}Q,\ x>0,\ \xi\in\mathbb{R}^{3}.

If we further assume $\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}<\infty$ , then we have

	$\displaystyle\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}\lesssim\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\\|w_{\tilde{\theta}}(\xi)Q(0,\xi)\\|_{L^{\infty}_{\xi}}$
	$\displaystyle+\underbrace{\sup_{x,\xi}\Big{[}w_{\tilde{\theta}}\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-(\nu(\xi)-\gamma(\xi_{1}+u))(t-s)}\|\partial_{x}Q(x-(\xi_{1}+u)(t-s),\xi)\|\Big{]}}_{\eqref{aprior_deri}_{*}}.$		(4.2)

Proof.

First we let $1\ll t$ be large and fixed, so that

\frac{1}{t}\ll 1,\ \ e^{-\nu_{0}t/4}\ll 1\text{ and }te^{-\nu_{0}t/4}\ll 1.

(4.3)

Recall the definition of $K^{p}$ in (3.14). We denote

\bar{\nu}(\xi):=\nu(\xi)-\gamma(\xi_{1}+u)\geq\frac{\nu(\xi)}{2}\geq\frac{\nu_{0}(\xi)}{2}:=\frac{\nu_{0}[1+|\xi|]}{2}.

(4.4)

By method of characteristic, we express $g$ as

	$\displaystyle g(x,\xi)$	$\displaystyle=[\mathbf{1}_{\frac{x}{\xi_{1}+u}>t}+\mathbf{1}_{\xi_{1}+u<0}]e^{-\bar{\nu}(\xi)t}g(x-(\xi_{1}+u)t,\xi)$
		$\displaystyle\ +\mathbf{1}_{0<\frac{x}{\xi_{1}+u}\leq t}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}g(0,\xi)$
		$\displaystyle\ +\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}K^{p}g(x-(\xi_{1}+u)(t-s),\xi)$
		$\displaystyle\ +\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}Q(x-(\xi_{1}+u)(t-s),\xi).$

From the assumption that $\partial_{x}g$ exists, we take derivative to the above formula to have

$\displaystyle\partial_{x}g(x,\xi)$	$\displaystyle=[\mathbf{1}_{\frac{x}{\xi_{1}+u}>t}+\mathbf{1}_{\xi_{1}+u<0}]e^{-\bar{\nu}(\xi)t}\partial_{x}g(x-(\xi_{1}+u)t,\xi)$	(4.5)
	$\displaystyle\ -\mathbf{1}_{0<\frac{x}{\xi_{1}+u}\leq t}\frac{\bar{\nu}(\xi)}{\xi_{1}+u}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}g(0,\xi)$	(4.6)
	$\displaystyle\ -\mathbf{1}_{0<\frac{x}{\xi_{1}+u}\leq t}\frac{1}{\xi_{1}+u}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}K^{p}g(0,\xi)$	(4.7)
	$\displaystyle\ +\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}K^{p}\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi)$	(4.8)
	$\displaystyle\ -\mathbf{1}_{0<\frac{x}{\xi_{1}+u}\leq t}\frac{1}{\xi_{1}+u}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}Q(0,\xi)$	(4.9)
	$\displaystyle\ +\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\partial_{x}Q(x-(\xi_{1}+u)(t-s),\xi).$	(4.10)

We estimate (4.5) - (4.10) term by term.

For (4.5), we apply (4.4) and to have

\displaystyle|\eqref{p_x_1}|

\displaystyle\lesssim e^{-\nu_{0}t/2}\frac{\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x-(\xi_{1}+u)t,\xi)}\lesssim e^{-\nu_{0}t/4}\frac{\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.

In the last inequality, we applied Lemma 5 with $c=1/8$ .

For (4.6), we use (1.12) to have

	$\displaystyle\|\eqref{p_x_2}\|$	$\displaystyle\lesssim\frac{1}{\alpha(0,\xi)}e^{-\frac{\nu_{0}x}{2(\xi_{1}+u)}}\frac{[1+\|\xi\|]}{w(\xi)}\\|wg_{b}\\|_{L^{\infty}_{\xi}}$
		$\displaystyle\lesssim\frac{1}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}e^{\frac{\nu_{0}x}{4(\xi_{1}+u)}}e^{-\frac{\nu_{0}x}{2(\xi_{1}+u)}}\\|wg_{b}\\|_{L^{\infty}_{\xi}}\lesssim\frac{\\|wg_{b}\\|_{L^{\infty}_{\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.11)

In the second line we applied Lemma 5 with $s=x/(\xi_{1}+u)$ and $c=\frac{1}{8}$ .

For (4.7), we first consider the contribution of $Kg$ . Applying Lemma 3, we have

	$\displaystyle\frac{1}{\xi_{1}+u}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}\frac{1}{w_{\tilde{\theta}}(\xi)}\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\frac{w_{\tilde{\theta}}(\xi)}{w_{\tilde{\theta}}(\xi^{\prime})}w_{\tilde{\theta}}(\xi^{\prime})g(0,\xi^{\prime})\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim\frac{1}{\alpha(0,\xi)}e^{-\frac{\nu_{0}x}{2(\xi_{1}+u)}}\\|wg\\|_{L^{\infty}_{x,\xi}}\lesssim\frac{1}{\alpha(x,\xi)}\\|wg\\|_{L^{\infty}_{x,\xi}},$

in the last inequality we applied the same computation as (4.11).

For the contribution of $\gamma\prod_{+}((\xi_{1}+u)g)$ and $\gamma\mathbf{p}_{u}g$ in (4.7), we have

	$\displaystyle\frac{1}{w_{\tilde{\theta}}(\xi)}\frac{e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}}{\xi_{1}+u}\gamma w_{\tilde{\theta}}(\xi)\big{[}\prod_{+}((\xi_{1}+u)g(0))+\mathbf{p}_{u}g(0)\big{]}$
	$\displaystyle\lesssim\frac{1}{w_{\tilde{\theta}}(\xi)}\frac{\gamma}{\xi_{1}+u}e^{-\bar{\nu}(\xi)\frac{x}{\xi_{1}+u}}\\|wg\\|_{L^{\infty}_{x,\xi}}\lesssim\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$

Here we have applied the same computation in (3.17) and (3.18).

Thus we conclude the estimate for (4.7) as

\displaystyle|\eqref{p_x_3}|

\displaystyle\lesssim\frac{\|wg\|_{L^{\infty}_{x,\xi}}}{\alpha(x,\xi)}.

(4.9) is directly bounded as

|\eqref{p_x_5}|\lesssim\frac{1}{w_{\tilde{\theta}}(\xi)\alpha(0,\xi)}e^{-\frac{\nu_{0}x}{2(\xi_{1}+u)}}\|w_{\tilde{\theta}}Q(0)\|_{L^{\infty}_{\xi}}\lesssim\frac{\|w_{\tilde{\theta}}Q(0)\|_{L^{\infty}_{\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.

The contribution of (4.10) is already included in (4.2).

For (4.8), first we compute the contribution of $\gamma\prod_{+}((\xi_{1}+u)g)$ and $\gamma\mathbf{p}_{u}g$ as

	$\displaystyle\gamma\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}[1+\|\xi\|](t-s)/2}\Big{\|}\chi_{+}(\xi)\langle(\xi_{1}+u)\chi_{+}\partial_{x}g\rangle(x-(\xi_{1}+u)(t-s))$
	$\displaystyle\quad\quad\quad\quad\quad+\phi_{u}(\xi)\langle(\xi_{1}+u)\psi_{u}\partial_{x}g\rangle(x-(\xi_{1}+u)(t-s))\Big{\|}$
	$\displaystyle\lesssim\frac{\gamma}{w_{\tilde{\theta}}(\xi)}\times\big{[}\\|w_{\tilde{\theta}}\chi_{+}\\|_{L^{\infty}_{\xi}}+\\|w_{\tilde{\theta}}\phi_{u}\\|_{L^{\infty}_{\xi}}\big{]}\\|\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}$
	$\displaystyle\times\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}[1+\|\xi\|](t-s)/2}\int_{\mathbb{R}^{3}}\frac{w^{-1/2}(\xi^{\prime})}{\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim\frac{\gamma t\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.12)

In the last line we have applied Lemma 7. In the second line we used Lemma 8. In the third line, for the integration $\langle(\xi_{1}+u)\chi_{+}\partial_{x}g\rangle$ and $\langle(\xi_{1}+u)\psi\partial_{x}g\rangle$ , we used Lemma 9 to have

	$\displaystyle\big{\|}\langle(\xi_{1}+u)\chi_{+}\partial_{x}g\rangle+\langle(\xi_{1}+u)\psi_{u}\partial_{x}g\rangle\big{\|}=\int_{\mathbb{R}^{3}}[(\xi^{\prime}_{1}+u)\chi_{+}(\xi^{\prime})\partial_{x}g(\xi^{\prime})+(\xi_{1}^{\prime}+u)\psi_{u}(\xi^{\prime})\partial_{x}g(\xi^{\prime})]\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim[\\|w\psi_{u}\\|_{L^{\infty}_{\xi}}+\\|w\chi_{+}\\|_{L^{\infty}_{\xi}}]\\|\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}\int_{\mathbb{R}^{3}}\frac{w^{-1/2}(\xi^{\prime})}{\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}\mathrm{d}\xi^{\prime}.$

We conclude that

	$\displaystyle\eqref{p_x_1}+\eqref{p_x_2}+\eqref{p_x_3}+\eqref{p_x_5}+\eqref{p_x_6}+\eqref{p_x_4}_{\gamma\mathbf{p}_{u}g\text{ and }\gamma\prod_{+}((\xi_{1}+u)g)}$
	$\displaystyle\lesssim\frac{[e^{-\nu_{0}t/4}+\gamma t]\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}+\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}Q(0)\\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.13)

For the contribution of

K\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi)=\int_{\mathbb{R}^{3}}\mathbf{k}(\xi,\xi^{\prime})\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi^{\prime})

in (4.8), we expand $\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi^{\prime})$ along $\xi^{\prime}$ using (4.5) - (4.10).

In the expansion of $\partial_{x}g(x-(\xi_{1}+u)(t-s),\xi^{\prime})$ , the contribution of the $\eqref{p_x_1}-\eqref{p_x_6}$ except $Kg$ are bounded by (4.13) with replacing $(x,\xi)$ by $(x-(\xi_{1}+u)(t-s),\xi^{\prime})$ :

	$\displaystyle\frac{1}{w_{\tilde{\theta}}(\xi)}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}[1+\|\xi\|](t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}w_{\tilde{\theta}}(\xi)\mathbf{k}(\xi,\xi^{\prime})$
	$\displaystyle\times\frac{[e^{-\nu_{0}t/4}+\gamma t]\\|\alpha w_{\tilde{\theta}}\partial_{x}g\\|_{L^{\infty}_{x,\xi}}+\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}Q(0)\\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}}{w_{\tilde{\theta}}(\xi^{\prime})\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}$
	$\displaystyle\lesssim\frac{t\Big{\{}[e^{-\nu_{0}t/4}+\gamma t]\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}+\\|wg_{b}\\|_{L^{\infty}_{\xi}}+\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}Q(0)\\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}\Big{\}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.14)

In the last line, first we applied (2.11), then we applied Lemma 6.

We focus on the contribution of the $Kg$ in (4.8), which induces a double Duhamel formula. Denote $y=x-(\xi_{1}+u)(t-s)$ , this formula equals

	$\displaystyle\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})$
	$\displaystyle\times\int^{s}_{\max\{0,\mathbf{1}_{\xi_{1}^{\prime}+u>0}(s-\frac{y}{\xi_{1}^{\prime}+u})\}}\mathrm{d}s^{\prime}e^{-\bar{\nu}(\xi^{\prime})(s-s^{\prime})}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\partial_{x}g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime}).$		(4.15)

We split the $s^{\prime}$ -integral into

\displaystyle\underbrace{\mathbf{1}_{s-s^{\prime}\leq\varepsilon}}_{\eqref{s'_integral}_{1}}+\underbrace{\mathbf{1}_{s-s^{\prime}>\varepsilon}}_{\eqref{s'_integral}_{2}}.

(4.16)

The contribution of (4.16)₁ in (4.15) is bounded as

	$\displaystyle\frac{1}{w_{\tilde{\theta}}(\xi)}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})\frac{w_{\tilde{\theta}}(\xi)}{w_{\tilde{\theta}}(\xi^{\prime})}$
	$\displaystyle\times\int^{s}_{s-\varepsilon}\mathrm{d}s^{\prime}e^{-\nu_{0}(\xi^{\prime})(s-s^{\prime})/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\frac{w_{\tilde{\theta}}(\xi^{\prime})\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi^{\prime\prime})\alpha(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})}$
	$\displaystyle\lesssim\frac{1}{w_{\tilde{\theta}}(\xi)}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})\frac{w_{\tilde{\theta}}(\xi)}{w_{\tilde{\theta}}(\xi^{\prime})}\frac{[\sqrt{\varepsilon}+\varepsilon\ln(t)]\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}}{\alpha(y,\xi^{\prime})}$
	$\displaystyle\lesssim\frac{t[\sqrt{\varepsilon}+\varepsilon\ln(t)]\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}\lesssim\frac{1}{t}\frac{\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.17)

In the third line we applied (2.20) in Lemma 6 with (2.11) for the $\mathrm{d}s^{\prime}$ integral. In the last line, we applied (2.18) with (2.11) for the $\mathrm{d}s$ integral. In the last inequality we take $\varepsilon=\frac{1}{t^{4}}$ , which is small since $t\gg 1$ , and that

\sqrt{\varepsilon}=\frac{1}{t^{2}},\ \ t[\sqrt{\varepsilon}+\varepsilon\ln(t)]\lesssim\frac{1}{t}+\frac{\ln t}{t^{3}}\lesssim\frac{1}{t}.

(4.18)

For the contribution of (4.16)₂ in (4.15), without loss of generality, we assume $s>\varepsilon$ . Otherwise, we bound (4.16)₂ by (4.17). Then we observe the following chain rule:

\displaystyle\frac{\partial_{\xi^{\prime}_{1}}[g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})]}{-(s-s^{\prime})}=\partial_{x}g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime}).

(4.19)

In such case $s-s^{\prime}\leq\varepsilon$ , we only integrate over the $\xi^{\prime}$ such that $\frac{y}{\xi_{1}^{\prime}+u}>\varepsilon$ , we denote

\displaystyle V:=\{\xi^{\prime}\in\mathbb{R}^{3}:\frac{y}{\xi^{\prime}_{1}+u}>\varepsilon\},\ \tilde{V}:=\{\xi^{\prime}\in\mathbb{R}^{3}:\frac{y}{\xi^{\prime}_{1}+u}=\varepsilon\}.

We apply (4.19) and an integration by part for $\mathrm{d}\xi^{\prime}_{1}$ to compute this contribution as

	$\displaystyle\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\int_{V}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})$
	$\displaystyle\times\int^{s-\varepsilon}_{\max\{0,\mathbf{1}_{\xi_{1}^{\prime}+u>0}(s-\frac{y}{\xi_{1}^{\prime}+u})\}}\mathrm{d}s^{\prime}e^{-\bar{\nu}(\xi^{\prime})(s-s^{\prime})}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\frac{\partial_{\xi^{\prime}_{1}}[g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})]}{-(s-s^{\prime})}$
	$\displaystyle=\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\int_{V}\mathrm{d}\xi^{\prime}\int^{s-\varepsilon}_{\max\{0,\mathbf{1}_{\xi_{1}^{\prime}+u>0}(s-\frac{y}{\xi_{1}^{\prime}+u})\}}\mathrm{d}s^{\prime}$
	$\displaystyle\times\partial_{\xi^{\prime}_{1}}[\mathbf{k}(\xi,\xi^{\prime})\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})e^{-\bar{\nu}(\xi^{\prime})(s-s^{\prime})}]\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\frac{g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})}{-(s-s^{\prime})}$		(4.20)
	$\displaystyle+\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\int_{V}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})$
	$\displaystyle\times\partial_{\xi^{\prime}_{1}}[\frac{y}{\xi_{1}^{\prime}+u}]e^{-\frac{\bar{\nu}(\xi^{\prime})y}{(\xi^{\prime}_{1}+u)}}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\frac{g(0,\xi^{\prime\prime})}{y/(\xi_{1}^{\prime}+u)}$		(4.21)
	$\displaystyle+\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\int_{\tilde{V}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})$
	$\displaystyle\times\int^{s-\varepsilon}_{s-\varepsilon}\mathrm{d}s^{\prime}e^{-\bar{\nu}(\xi^{\prime})(s-s^{\prime})}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\frac{g(y-(\xi^{\prime}_{1}+u)(s-s^{\prime}),\xi^{\prime\prime})}{-(s-s^{\prime})}.$		(4.22)

For the last term, we apply $\|wg\|_{L^{\infty}_{x,\xi}}<\infty$ to have $\eqref{kk_geq_e_3}=0$ .

For $\partial_{\xi^{\prime}_{1}}[\mathbf{k}(\xi,\xi^{\prime})\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})]$ in (4.20), we use $2\tilde{\theta}<\theta$ to have

	$\displaystyle\frac{1}{w_{\tilde{\theta}}(\xi)}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\int^{s}_{\max\{0,\mathbf{1}_{\xi_{1}^{\prime}+u>0}(s-\frac{y}{\xi_{1}^{\prime}+u})\}}\mathrm{d}s^{\prime}e^{-\nu_{0}(\xi^{\prime})(s-s^{\prime})/2}$
	$\displaystyle\times\frac{w_{\tilde{\theta}}(\xi)}{w_{2\tilde{\theta}}(\xi^{\prime})}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\Big{\|}\partial_{\xi^{\prime}_{1}}\mathbf{k}(\xi,\xi^{\prime})\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})+\mathbf{k}(\xi,\xi^{\prime})\partial_{\xi^{\prime}_{1}}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\Big{\|}\frac{w_{2\tilde{\theta}}(\xi^{\prime})}{w_{2\tilde{\theta}}(\xi^{\prime\prime})}\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}}{\varepsilon}$
	$\displaystyle\lesssim\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}}{\varepsilon w_{\tilde{\theta}}(\xi)}\lesssim t^{4}\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.23)

In the last inequality we applied (1.11) and (4.18). To get the first inequality in the last line, we have applied (2.10) and (2.12) to have

	$\displaystyle\frac{w_{\tilde{\theta}}(\xi)}{w_{2\tilde{\theta}}(\xi^{\prime})}\Big{\|}\partial_{\xi^{\prime}_{1}}\mathbf{k}(\xi,\xi^{\prime})\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})+\mathbf{k}(\xi,\xi^{\prime})\partial_{\xi^{\prime}_{1}}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\Big{\|}\frac{w_{2\tilde{\theta}}(\xi^{\prime})}{w_{2\tilde{\theta}}(\xi^{\prime\prime})}$
	$\displaystyle\lesssim\frac{[1+\|\xi\|^{\prime}]^{2}}{w_{\tilde{\theta}}(\xi^{\prime})}\Big{\|}\frac{e^{-C_{\theta}\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|^{2}}\frac{e^{-C_{\theta}\|\xi^{\prime}-\xi^{\prime\prime}\|^{2}}}{\|\xi^{\prime}-\xi^{\prime\prime}\|}+\frac{e^{-C_{\theta}\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}\frac{e^{-C_{\theta}\|\xi^{\prime}-\xi^{\prime\prime}\|^{2}}}{\|\xi^{\prime}-\xi^{\prime\prime}\|^{2}}\Big{\|}\in L^{1}(\mathrm{d}\xi^{\prime}\mathrm{d}\xi^{\prime\prime}).$

For the other term $\partial_{\xi^{\prime}_{1}}e^{-\bar{\nu}(\xi^{\prime})(s-s^{\prime})}$ in (4.20), recall the definition of $\bar{\nu}$ in (4.4), we apply (2.5) to bound it by (4.23) using the same computation.

Then we compute (4.21). The $\xi_{1}^{\prime}$ derivative on $y/(\xi_{1}^{\prime}+u)$ can be combined with the extra $1/(y/(\xi_{1}^{\prime}+u))$ , we apply (1.12) and Lemma 5 to have

\displaystyle\Big{|}\partial_{\xi_{1}^{\prime}}[\frac{y}{\xi^{\prime}_{1}+u}]\frac{\xi_{1}^{\prime}+u}{y}\Big{|}=\frac{1}{|\xi_{1}^{\prime}+u|}\leq\frac{1}{\alpha(0,\xi^{\prime})}\lesssim\frac{e^{\frac{\nu_{0}y}{4(\xi_{1}^{\prime}+u)}}}{\alpha(y,\xi^{\prime})}.

Then we proceed the computation as

	$\displaystyle\|\eqref{kk_geq_e_2}\|$	$\displaystyle\lesssim\frac{1}{w_{\tilde{\theta}}(\xi)}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})\frac{w_{\tilde{\theta}}(\xi)}{w_{\tilde{\theta}}(\xi^{\prime})}$
		$\displaystyle\times e^{-\frac{\nu_{0}y}{4(\xi_{1}^{\prime}+u)}}\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}}{\alpha(y,\xi^{\prime})}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})\frac{w_{\tilde{\theta}}(\xi^{\prime})}{w_{\tilde{\theta}}(\xi^{\prime\prime})}\lesssim\frac{t\\|wg\\|_{L^{\infty}_{x,\xi}}}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.$		(4.24)

In the last inequality we applied (2.18) in Lemma 6 with (2.11).

Collecting (4.17), (4.23) and (4.24) we conclude

\displaystyle|\eqref{kk}|

\displaystyle\lesssim\frac{\frac{1}{t}\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}+t^{4}[\|wg_{b}\|_{L^{\infty}_{\xi}}+\|wg\|_{L^{\infty}_{x,\xi}}]}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.

This, together with (4.14), leads to the estimate

\displaystyle\eqref{p_x_4}

\displaystyle\lesssim\frac{[\frac{1}{t}+te^{-\nu_{0}t/4}+t^{2}\gamma]\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}+t^{4}[\|wg_{b}\|_{L^{\infty}_{\xi}}+\|wg\|_{L^{\infty}_{x,\xi}}+\|w_{\tilde{\theta}}Q(0)\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}]}{w_{\tilde{\theta}}(\xi)\alpha(x,\xi)}.

(4.25)

(4.13) and (4.25) leads to the estimate for $\partial_{x}g(x,\xi)$ :

\displaystyle|\partial_{x}g(x,\xi)|

\displaystyle\lesssim\eqref{4_bdd}.

Applying (4.3) to a fixed $t$ , we choose a smaller $\gamma$ in (3.10) such that $t^{2}\gamma\ll 1$ , we conclude the lemma by

\displaystyle\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}

\displaystyle\lesssim t^{4}[\|wg_{b}\|_{L^{\infty}_{\xi}}+\|wg\|_{L^{\infty}_{x,\xi}}+\|w_{\tilde{\theta}}Q(0)\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}].

∎

In the following lemma, we construct the derivative to the linearized penalized problem.

Lemma 13.

Suppose $\partial_{x}Q$ exists and satisfies $\eqref{aprior_deri}_{*}<\infty$ , $\|w_{\tilde{\theta}}Q\|_{L^{\infty}_{x,\xi}}<\infty$ , then the derivative of the unique solution to linearized penalized problem in (3.11) exists, and it is given by

\displaystyle(\xi_{1}+u)\partial_{x}(\partial_{x}g)+\mathcal{L}^{p}(\partial_{x}g)=\partial_{x}Q.

Moreover, the derivative satisfies (4.2).

Proof.

We start by considering the derivative to the following equation:

\begin{cases}&(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-\lambda K^{p}\tilde{g}=Q,\\ &g(0,\xi)=g_{b}(\xi),\ \ \xi_{1}+u>0.\end{cases}

(4.26)

\displaystyle(\xi_{1}+u)\partial_{x}(\partial_{x}g)+[\nu(\xi)-\gamma(\xi_{1}+u)](\partial_{x}g)-\lambda K^{p}(\partial_{x}\tilde{g})=\partial_{x}Q.

(4.27)

We define a Banach space as

\displaystyle\mathcal{X}_{2}:=

\displaystyle\Big{\{}g(x,\xi):\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}<\infty,\ \ \|wg\|_{L^{\infty}_{x,\xi}}<\infty,\ \ \|g\|_{L^{2}_{x,\xi}}<\infty\Big{\}},

with associated norm

\displaystyle\|g\|_{\mathcal{X}_{2}}

\displaystyle:=\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}+\|g\|_{\mathcal{X}}.

Here $\|g\|_{\mathcal{X}}$ is defined in (3.28).

For given $\tilde{g}\in\mathcal{X}_{2}$ , from Proposition 6 there exists a unique solution $g$ to (4.26). By Lemma 11, the derivative of $g$ exists and is given by (4.27).

We first prove that $\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}<\infty$ so that we can apply the a priori estimate (4.2). From the proof of Lemma 12, $\partial_{x}g$ can be expressed by (4.5) - (4.10) with replacing $K^{p}g$ by $\lambda K^{p}(\tilde{g})$ .

For (4.5), when $\xi_{1}+u=0$ , (4.5) can be absorbed by LHS. When $\xi_{1}+u\neq 0$ , we have a uniform-in- $x$ bound for $\partial_{x}g$ by the equation of $g$ in (3.11):

	$\displaystyle\\|w_{\tilde{\theta}}(\xi)\partial_{x}g(x,\xi)\\|_{L^{\infty}_{x}}\leq\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}Q\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}K^{p}(\tilde{g})\\|_{L^{\infty}_{x,\xi}}}{\|\xi_{1}+u\|}$
	$\displaystyle\lesssim\frac{\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w_{\tilde{\theta}}Q\\|_{L^{\infty}_{x,\xi}}+\\|w\tilde{g}\\|_{L^{\infty}_{x,\xi}}}{\|\xi_{1}+u\|}.$

Here we have applied Lemma 10. From the uniform-in- $x$ bound, we can multiply $w_{\tilde{\theta}}(\xi)$ to (4.5) - (4.10), and take $sup$ in $x$ to the LHS, so that $w_{\tilde{\theta}}(\xi)\times\eqref{p_x_1}$ can be absorbed by the LHS.

Since $\alpha(x,\xi)\leq 1$ , we further multiply $\alpha(x,\xi)$ to (4.5) - (4.10), and $\alpha(x,\xi)w_{\tilde{\theta}}(\xi)\times\eqref{p_x_1}$ can also be absorbed by the LHS.

Without the contribution of $K^{p}g$ in (4.26), we can apply the same argument as the a priori estimate in Lemma 12, here we do not need to estimate (4.5), (4.7) and (4.8). Then we derive that (4.27) can be bounded by the same estimate as (4.2). Hence, for some $C_{\alpha}$ ,

\displaystyle\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}

\displaystyle\leq C_{\alpha}\Big{[}\|wg\|_{L^{\infty}_{x,\xi}}+\|w_{\tilde{\theta}}Q(0)\|_{L^{\infty}_{\xi}}+\eqref{aprior_deri}_{*}+\eqref{kp_deri_bdd}\Big{]},

(4.28)

where (4.29) corresponds to the contribution of $\lambda K^{p}(\partial_{x}\tilde{g})$ :

	$\displaystyle\lambda w_{\tilde{\theta}}(\xi)\Big{[}K^{p}\tilde{g}(0,\xi)+\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu(\xi)(t-s)}K^{p}\partial_{x}\tilde{g}(x-(\xi_{1}+u)(t-s),\xi)\mathrm{d}s\Big{]}$		(4.29)
	$\displaystyle\lesssim\lambda\\|w\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\lambda\gamma t\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}$
	$\displaystyle+\lambda\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu(\xi)(t-s)}\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\frac{w_{\tilde{\theta}}(\xi)\mathbf{k}(\xi,\xi^{\prime})}{w_{\tilde{\theta}}(\xi^{\prime})\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}$
	$\displaystyle\lesssim\lambda\gamma\\|w\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\lambda\gamma t\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\lambda t\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}.$		(4.30)

In the second line we have applied Lemma 10 to $K^{p}\tilde{g}$ for the first term, for the second term, we applied the same computation as (4.12) to compute the contribution of $\gamma\prod_{+}((\xi_{1}+u)g)+\gamma\mathbf{p}_{u}((\xi_{1}+u)g)$ . In the last line we have used Lemma 6 with (2.11).

Combining (4.28), (4.30), $\|g\|_{\mathcal{X}}<\infty$ from (3.29) and the assumption that $\eqref{aprior_deri}_{*}<\infty$ , we conclude that $g\in\mathcal{X}_{2}$ .

Next, for given $\tilde{g}_{1},\tilde{g}_{2}$ , we denote the associated solution as $g_{1},g_{2}$ . Then we apply (4.28), (4.30) and (3.30) to have

	$\displaystyle\\|g_{1}-g_{2}\\|_{\mathcal{X}_{2}}=\\|w_{\tilde{\theta}}\alpha\partial_{x}(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+\\|g_{1}-g_{2}\\|_{\mathcal{X}}$
	$\displaystyle\leq C_{\alpha}\\|w(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+[\lambda\gamma t+\lambda t]C_{\alpha}\\|w_{\tilde{\theta}}\alpha\partial_{x}(\tilde{g}_{1}-\tilde{g}_{2})\\|_{L_{x,\xi}^{\infty}}+C_{\alpha}\lambda\gamma\\|w(\tilde{g}_{1}-\tilde{g}_{2})\\|_{L^{\infty}_{x,\xi}}+C\lambda\\|\tilde{g}_{1}-\tilde{g}_{2}\\|_{\mathcal{X}}$
	$\displaystyle\leq C_{\alpha}\lambda[\gamma t+t]\\|w_{\tilde{\theta}}\alpha\partial_{x}(\tilde{g}_{1}-\tilde{g}_{2})\\|_{L^{\infty}_{x,\xi}}+C_{\alpha}\\|g_{1}-g_{2}\\|_{\mathcal{X}}+C_{\alpha}\lambda[\gamma+C]\\|\tilde{g}_{1}-\tilde{g}_{2}\\|_{\mathcal{X}}$
	$\displaystyle\leq C_{\alpha}\lambda[\gamma t+t]\\|w_{\tilde{\theta}}\alpha\partial_{x}(\tilde{g}_{1}-\tilde{g}_{2})\\|_{L^{\infty}_{x,\xi}}+C_{\alpha}\lambda[\gamma+2C]\\|\tilde{g}_{1}-\tilde{g}_{2}\\|_{\mathcal{X}}.$

Then we choose $\lambda$ to be sufficiently small such that all coefficients are bounded as

\displaystyle C_{\alpha}\lambda[\gamma t+t]<1\ \ \ C_{\alpha}\lambda[\gamma+2C]<1.

By Banach fixed point theorem, there exists a unique $g\in\mathcal{X}_{2}$ satisfying

\begin{cases}&(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-\lambda K^{p}g=Q,\\ &g(0,\xi)=g_{b}(\xi),\ \ \xi_{1}+u>0,\\ &(\xi_{1}+u)\partial_{x}(\partial_{x}g)+[\nu(\xi)-\gamma(\xi_{1}+u)](\partial_{x}g)-\lambda K^{p}(\partial_{x}g)=\partial_{x}Q.\end{cases}

From the well-posedness of (4.1), next we consider the derivative to the solution of the following problem,

\begin{cases}&(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-\lambda K^{p}g=\lambda K^{p}(\tilde{g})+Q,\\ &g(0,\xi)=g_{b}(\xi),\ \ \xi_{1}+u>0,\\ &(\xi_{1}+u)\partial_{x}(\partial_{x}g)+[\nu(\xi)-\gamma(\xi_{1}+u)](\partial_{x}g)-\lambda K^{p}(\partial_{x}g)=\lambda K^{p}(\partial_{x}\tilde{g})+\partial_{x}Q.\end{cases}

With given source $\tilde{g}\in\mathcal{X}_{2}$ , from previous argument we know that $\partial_{x}g$ exists and $g\in\mathcal{X}_{2}$ . It is straightforward to show that one can obtain the same uniform in $\lambda$ a-priori estimate (4.2). Thus (4.28) still holds. Then we apply the same fixed-point argument to show that there exists a unique $g$ satisfying

\begin{cases}&(\xi_{1}+u)\partial_{x}g+[\nu(\xi)-\gamma(\xi_{1}+u)]g-2\lambda K^{p}g=Q\\ &g(0,\xi)=g_{b}(\xi),\ \ \xi_{1}+u>0,\\ &(\xi_{1}+u)\partial_{x}(\partial_{x}g)+[\nu(\xi)-\gamma(\xi_{1}+u)](\partial_{x}g)-2\lambda K^{p}(\partial_{x}g)=\partial_{x}Q.\end{cases}

Step by step, we construct solution with coefficient $1$ for $K^{p}(\partial_{x}g)$ , then we conclude the lemma.

∎

4.2. Weighted $C^{1}$ estimate of nonlinear penalized problem

Next, we construct the derivative to the nonlinear problem (3.31).

Proposition 9.

The derivative to the solution of (3.31) in Proposition 8 exists. Moreover, we have the weighted $C^{1}$ estimate

\displaystyle\|\alpha\partial_{x}h\|_{L^{\infty}_{x,\xi}}+\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}

\displaystyle\leq C\|wg\|_{L^{\infty}_{x,\xi}}+C\|wg_{b}\|_{L^{\infty}_{\xi}}+C\|h\|_{L^{\infty}_{x}}.

(4.31)

Proof.

We denote a Banach space as

\begin{split}\mathcal{X}_{3}&:=\Big{\{}(g,h):\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L_{x,\xi}^{\infty}}+\|\alpha\partial_{x}h\|_{L^{\infty}_{x,\xi}}+\|(g,h)\|_{\mathcal{X}_{1}}\leq 2C\varepsilon\Big{\}},\end{split}

(4.32)

with the associated norm defined as

\begin{split}\|(g,h)\|_{\mathcal{X}_{3}}&=\|w_{\tilde{\theta}}\alpha\partial_{x}g\|_{L^{\infty}_{x,\xi}}+\|\alpha\partial_{x}h\|_{L^{\infty}_{x,\xi}}+\|(g,h)\|_{\mathcal{X}_{1}}.\end{split}

Here, $\mathcal{X}_{1}$ is defined in (3.32). Clearly we have $\|(g,h)\|_{\mathcal{X}_{1}}\leq\|(g,h)\|_{\mathcal{X}_{3}}$ .

Given $(\tilde{g},\tilde{h})\in\mathcal{X}_{3}$ , we consider the following linear problem with given sources as $\tilde{g},\tilde{h}$ :

\begin{cases}&(\xi_{1}+u)\partial_{x}g+\mathcal{L}^{p}g=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u}),\\ &h(x)=-e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(x+z)\mathrm{d}z,\\ &g(0,\xi)=f_{b}(\xi)+h(0)\phi_{u}(\xi),\ \ \xi_{1}+u>0,\\ &(\xi_{1}+u)\partial_{x}(\partial_{x}g)+\mathcal{L}^{p}(\partial_{x}g)=\partial_{x}\Big{[}e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\Big{]},\\ &\partial_{x}h(x)=-\partial_{x}\Big{[}e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(x+z)\mathrm{d}z\Big{]}.\end{cases}

$h(x)$ can also be expressed as

	$\displaystyle h(x)$	$\displaystyle=-e^{-\gamma x}\int_{x}^{\infty}e^{(\tau_{u}-2\gamma)(z-x)}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(z)\mathrm{d}z$
		$\displaystyle=-e^{(\gamma-\tau_{u})x}\int_{x}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(z)\mathrm{d}z.$

Then we take derivative to have

\displaystyle\partial_{x}h(x)=(\gamma-\tau_{u})h(x)+e^{-\gamma x}\langle\psi_{u}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})\rangle(x).

(4.33)

From the equation of $h$ in (4.33), we apply (1.11) to have

\displaystyle\|\alpha\partial_{x}h\|_{L^{\infty}_{x,\xi}}

\displaystyle\leq\|\partial_{x}h\|_{L^{\infty}_{x}}\lesssim\|h\|_{L^{\infty}_{x}}+\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}\lesssim\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}.

(4.34)

In the second inequality, we applied (3.35) to the source term $\Gamma$ . In the last inequality, we used $\|h\|_{L^{\infty}_{x}}\lesssim\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}$ from (3.35).

The well-posedness of $\partial_{x}g$ is guaranteed by Lemma 13, and $\partial_{x}g$ satisfies (4.2). The boundary term in (4.2) can be controlled as

\displaystyle\|wg_{b}\|_{L^{\infty}_{\xi}}+\|w_{\tilde{\theta}}(\xi)Q(0,\xi)\|_{L^{\infty}_{\xi}}\lesssim\|wf_{b}\|_{L^{\infty}_{\xi}}+\|w\phi_{u}\|_{L^{\infty}_{\xi}}\|h\|_{L^{\infty}_{x}}+\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}\lesssim\varepsilon+\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}^{2}.

(4.35)

Here we applied (3.34) and (3.38) with $w_{\tilde{\theta}}(\xi)\lesssim\frac{w(\xi)}{1+|\xi|}$ to $Q(0)=(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})(0)$ in the first inequality. In the second inequality, $\varepsilon$ comes from the assumption $\|wf_{b}\|_{L^{\infty}}\leq\varepsilon$ in Proposition 8, and we used (3.35) to $h$ .

Then we evaluate $\eqref{aprior_deri}_{*}$ with $Q=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})$ . Note that

\displaystyle\partial_{x}Q

\displaystyle=-\gamma e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})+e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\partial_{x}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u}).

(4.36)

The contribution of the first term on RHS of (4.36) is bounded as

	$\displaystyle w_{\tilde{\theta}}(\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu(\xi)(t-s)}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})(x-(\xi_{1}+u)(t-s),\xi)$
	$\displaystyle\lesssim\big{\\|}\frac{w}{[1+\|\xi\|]}(\mathbf{I}-\mathbf{P}_{u})\Gamma\big{\\|}_{L^{\infty}_{x,\xi}}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu(\xi)(t-s)}\lesssim\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}^{2}.$		(4.37)

Here we have applied the same computation in (3.37) with $w_{\tilde{\theta}}(\xi)\lesssim\frac{w(\xi)}{[1+|\xi|]}$ .

For $\eqref{aprior_deri}_{*}$ , the contribution of $I\partial_{x}\Gamma$ in the second term of RHS of (4.36) is bounded as

\displaystyle w_{\tilde{\theta}}(\xi)\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}|\partial_{x}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})(x-(\xi_{1}+u)(t-s),\xi)|.

We apply (2.7). The contribution of the first term in (2.7) is bounded as

	$\displaystyle\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}[1+\|\xi\|]\\|w(\tilde{g}-\phi_{u}\tilde{h})\\|_{L^{\infty}_{x,\xi}}$
	$\displaystyle\times w_{\tilde{\theta}}(\xi)\alpha(x,\xi)\|[\partial_{x}\tilde{g}-\partial_{x}\tilde{h}\phi_{u}](x-(\xi_{1}+u)(t-s),\xi)\|$
	$\displaystyle\leq\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x}}]\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}[1+\|\xi\|]e^{\nu_{0}(t-s)/4}$
	$\displaystyle\leq\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x}}]\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/4}[1+\|\xi\|]$
	$\displaystyle\lesssim\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x}}].$		(4.38)

In the third line we have used $\|w_{\tilde{\theta}}(\xi)[1+|\xi|]\phi_{u}\|_{L^{\infty}_{\xi}}\lesssim\|w\phi_{u}\|_{L^{\infty}_{\xi}}\lesssim 1$ and applied Lemma 5.

The contribution of the second and third term in (2.7) are bounded as

	$\displaystyle w_{\tilde{\theta}}(\xi)\alpha(x,\xi)\\|w(\tilde{g}-\phi_{u}\tilde{h})\\|_{L^{\infty}_{x,\xi}}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}$
	$\displaystyle\times\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}\mathbf{k}(\xi,\xi^{\prime})\|\partial_{x}[\tilde{g}-\tilde{h}\phi_{u}](x-(\xi_{1}+u)(t-s),\xi^{\prime})\|$
	$\displaystyle\leq\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x}}]\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}$
	$\displaystyle\times\int_{\mathbb{R}^{3}}\frac{w_{\tilde{\theta}}(\xi)\mathbf{k}(\xi,\xi^{\prime})}{w_{\tilde{\theta}}(\xi^{\prime})\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim t\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x}}].$		(4.39)

In the fourth line we have used Lemma 8 to have $\phi_{u}(\xi^{\prime})\leq\frac{\|w\phi_{u}\|_{L^{\infty}_{\xi}}}{w_{\tilde{\theta}}(\xi^{\prime})}$ . In the last line applied Lemma 6 with (2.11).

For $\eqref{aprior_deri}_{*}$ , the contribution of $\mathbf{P}_{u}\partial_{x}\Gamma$ in the second term of the RHS of (4.36) is controlled as

	$\displaystyle\alpha(x,\xi)\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\bar{\nu}(\xi)(t-s)}\\|w_{\tilde{\theta}}(\xi_{1}+u)\phi_{u}\\|_{L^{\infty}_{\xi}}\\|w\psi_{u}\\|_{L^{\infty}_{\xi}}$
	$\displaystyle\times\int_{\mathbb{R}^{3}}w^{-1}(\xi^{\prime})\big{\|}\partial_{x}\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})(x-(\xi_{1}+u)(t-s),\xi^{\prime})\big{\|}\mathrm{d}\xi^{\prime}.$

Again we apply (2.7). The contribution of the first term in (2.7) is bounded as

	$\displaystyle\alpha(x,\xi)\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\\|w\psi_{u}\\|_{L^{\infty}_{\xi}}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}$
	$\displaystyle\times\int_{\mathbb{R}^{3}}w^{-1}(\xi^{\prime})[1+\|\xi\|^{\prime}]\\|w(\tilde{g}-\tilde{h}\phi_{u})\\|_{L^{\infty}_{x,\xi}}[\partial_{x}\tilde{g}-\partial_{x}\tilde{h}\phi_{u}](x-(\xi_{1}+u)(t-s),\xi^{\prime})\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim\alpha(x,\xi)\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x,\xi}}]$
	$\displaystyle\times\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}\frac{w^{-1/2}(\xi^{\prime})}{\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime})}\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim t\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x,\xi}}].$		(4.40)

In the last line we have used Lemma 7.

The contribution of the second and third term in (2.7) are bounded as

	$\displaystyle\alpha(x,\xi)\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}$
	$\displaystyle\times\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime}w^{-1}(\xi^{\prime})\int_{\mathbb{R}^{3}}\mathrm{d}\xi^{\prime\prime}\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})[\|\partial_{x}\tilde{g}\|+\|\partial_{x}\tilde{h}\|](x-(\xi_{1}+u)(t-s),\xi^{\prime\prime})$
	$\displaystyle\lesssim\alpha(x,\xi)\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x,\xi}}]$
	$\displaystyle\times\int^{t}_{\max\{0,\mathbf{1}_{\xi_{1}+u>0}(t-\frac{x}{\xi_{1}+u})\}}\mathrm{d}se^{-\nu_{0}(\xi)(t-s)/2}\int_{\mathbb{R}^{3}}w^{-1}(\xi^{\prime})\mathrm{d}\xi^{\prime}\int_{\mathbb{R}^{3}}\frac{\mathbf{k}(\xi^{\prime},\xi^{\prime\prime})}{\alpha(x-(\xi_{1}+u)(t-s),\xi^{\prime\prime})}\mathrm{d}\xi^{\prime\prime}$
	$\displaystyle\lesssim t\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{1}}[\\|w_{\tilde{\theta}}\alpha\partial_{x}\tilde{g}\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}\tilde{h}\\|_{L^{\infty}_{x,\xi}}].$		(4.41)

In the last line we have applied (2.44) in Lemma 7.

Collecting (4.37), (4.38), (4.39), (4.40) and (4.41), we conclude that the contribution of $\eqref{aprior_deri}_{*}$ is bounded as

\displaystyle\eqref{aprior_deri}_{*}\mathbf{1}_{Q=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Gamma(\tilde{g}-\tilde{h}\phi_{u},\tilde{g}-\tilde{h}\phi_{u})}\lesssim t\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{3}}^{2}.

(4.42)

It has been proved in (3.39) that given $\|(\tilde{g},\tilde{h})\|_{\mathcal{X}_{1}}\leq 2C\varepsilon$ , then $\|(g,h)\|_{\mathcal{X}_{1}}\leq 2C\varepsilon$ . This, combining with the estimate of $\partial_{x}h$ in (4.34), the estimate for $\partial_{x}g$ in (4.2), (4.35) and (4.42), implies that for some $C_{1}=C_{1}(t)>0$ ,

	$\displaystyle\\|(g,h)\\|_{\mathcal{X}_{3}}$	$\displaystyle\leq\\|(g,h)\\|_{\mathcal{X}_{1}}+\eqref{p_x_h_bdd}+\eqref{p_x_bdr_term}+\eqref{deri_gamma_bdd}$
		$\displaystyle\leq C_{1}\varepsilon+C_{1}\\|(\tilde{g},\tilde{h})\\|_{\mathcal{X}_{3}}^{2}\leq C_{1}\varepsilon+4C_{1}C^{2}\varepsilon^{2}\leq 2C\varepsilon.$

Here we have taken $C$ in (4.32) to be $C=C_{1}$ and let $\varepsilon$ be small enough such that $C_{1}C^{2}\varepsilon^{2}=C_{1}^{3}\varepsilon^{2}\leq C\varepsilon$ . Therefore, we conclude that $(g,h)\in\mathcal{X}_{3}$ .

Next we prove the contraction property. Given $(\tilde{g}_{1},\tilde{h}_{1}),(\tilde{g}_{2},\tilde{h}_{2})\in\mathcal{X}_{3}$ , we denote the corresponding solutions as $(g_{1},h_{1}),(g_{2},h_{2})$ . Then $(g_{1}-g_{2},h_{1}-h_{2})$ satisfies (3.40), and the derivative satisfies

\begin{cases}&(\xi_{1}+u)\partial_{x}(\partial_{x}(g_{1}-g_{2}))+\mathcal{L}^{p}(\partial_{x}(g_{1}-g_{2}))=\partial_{x}\big{[}e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Sigma\big{]},\\ &\partial_{x}[h_{1}-h_{2}](x)=-\partial_{x}\Big{[}e^{-\gamma x}\int_{0}^{\infty}e^{(\tau_{u}-2\gamma)z}\langle\psi_{u}\Sigma\rangle(x+z)\mathrm{d}z\Big{]},\end{cases}

here $\Sigma$ is defined in (3.40).

To estimate $h_{1}-h_{2}$ , we apply the same computation as (4.34), with (3.41), we obtain

	$\displaystyle\\|\alpha\partial_{x}(h_{1}-h_{2})\\|_{L^{\infty}_{x,\xi}}\leq\\|\partial_{x}(h_{1}-h_{2})\\|_{L^{\infty}_{x,\xi}}\lesssim\\|h_{1}-h_{2}\\|_{L^{\infty}_{x}}+\\|e^{-\gamma x}\langle\psi\Sigma\rangle\\|_{L^{\infty}_{x,\xi}}$
	$\displaystyle\lesssim\big{\\|}\frac{w}{[1+\|\xi\|]}\Sigma\big{\\|}_{L^{\infty}_{x,\xi}}\lesssim\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}.$		(4.43)

To estimate $\partial_{x}(g_{1}-g_{2})$ , we apply (4.2). First we compute the boundary term with $Q(0)=(\mathbf{I}-\mathbf{P}_{u})\Sigma(0)$ as

	$\displaystyle\\|w(g_{1}-g_{2})_{b}\\|_{L^{\infty}_{\xi}}+\\|Q(0)\\|_{L^{\infty}_{\xi}}\lesssim\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\\|h_{1}-h_{2}\\|_{L^{\infty}_{x}}+\\|(\mathbf{I}-\mathbf{P}_{u})\Sigma(0)\\|_{L^{\infty}_{\xi}}$
	$\displaystyle\lesssim\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}+\big{\\|}\frac{w}{[1+\|\xi\|]}\Sigma\big{\\|}_{L^{\infty}_{x,\xi}}+\big{\\|}\frac{w}{[1+\|\xi\|]}\mathbf{P}_{u}\Sigma\big{\\|}_{L^{\infty}_{x,\xi}}\lesssim\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}.$		(4.44)

Here we applied the same estimate for $h_{1}-h_{2}$ in (3.41) and the same estimate for $\Sigma$ in (3.42).

For $\eqref{aprior_deri}_{*}$ , we express the derivative as

-\gamma e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Sigma+e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})(\partial_{x}\Sigma).

(4.45)

We apply the same estimate as (4.37) to the first term in (4.45), and apply the same estimate as (4.42) to the second term in (4.45), this yields

	$\displaystyle\eqref{aprior_deri}_{*}\mathbf{1}_{Q=e^{-\gamma x}(\mathbf{I}-\mathbf{P}_{u})\Sigma}\lesssim[\\|(\tilde{g}_{1},\tilde{h}_{1})\\|_{\mathcal{X}_{1}}+\\|(\tilde{g}_{2},\tilde{h}_{2})\\|_{\mathcal{X}_{1}}]\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{1}}$
	$\displaystyle+t[\\|(\tilde{g}_{1},\tilde{h}_{1})\\|_{\mathcal{X}_{3}}+\\|(\tilde{g}_{2},\tilde{h}_{2})\\|_{\mathcal{X}_{3}}]\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{3}}$
	$\displaystyle\lesssim t\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{3}}.$		(4.46)

Finally, we collect (4.43), (4.2), (4.44), (4.46) and (3.43) to conclude that

	$\displaystyle\\|(g_{1}-g_{2},h_{1}-h_{2})\\|_{\mathcal{X}_{3}}=\\|w_{\tilde{\theta}}\alpha\partial_{x}(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+\\|\alpha\partial_{x}(h_{1}-h_{2})\\|_{L^{\infty}_{x}}+\\|(g_{1}-g_{2},h_{1}-h_{2})\\|_{\mathcal{X}_{1}}$
	$\displaystyle\lesssim\\|w(g_{1}-g_{2})\\|_{L^{\infty}_{x,\xi}}+t\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{3}}$
	$\displaystyle\leq C_{2}\varepsilon\\|(\tilde{g}_{1}-\tilde{g}_{2},\tilde{h}_{1}-\tilde{h}_{2})\\|_{\mathcal{X}_{3}}.$

Here $C_{2}$ depends on $t$ . With small enough $\varepsilon$ such that $C_{2}(t)\varepsilon<1$ , we conclude the proposition with Banach fixed point theorem.

∎

4.3. Proof of Theorem 1

The unique solution to (1.4) is constructed as

f(x,\xi)=e^{-\gamma x}g(x,\xi)-e^{-\gamma x}h(x)\phi_{u}(\xi).

With $h(x)$ and $g(x)$ satisfying (4.31), we take $x$ -derivative and have

	$\displaystyle\\|w_{\tilde{\theta}}\alpha\partial_{x}f\\|_{L^{\infty}_{\xi}}$	$\displaystyle\lesssim e^{-\gamma x}[\\|wg\\|_{L^{\infty}_{x,\xi}}+\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\\|h\\|_{L^{\infty}_{x}}]$
		$\displaystyle+e^{-\gamma x}[\\|w_{\tilde{\theta}}\alpha\partial_{x}g\\|_{L^{\infty}_{x,\xi}}+\\|w\phi_{u}\\|_{L^{\infty}_{\xi}}\\|\alpha\partial_{x}h\\|_{L^{\infty}_{x}}]\lesssim e^{-\gamma x}\varepsilon.$

We conclude the theorem with some $\bar{C}>0$ .

In summary, we have

\displaystyle\|e^{\gamma x}w_{\tilde{\theta}}\alpha\partial_{x}f\|_{L^{\infty}_{x,\xi}}\lesssim\varepsilon.

(4.47)

This verifies (1.20).

Then we prove (1.21). For $\gamma_{0}<\gamma$ , (4.47) leads to

	$\displaystyle\\|w_{\tilde{\theta}/2}e^{\gamma_{0}x}\partial_{x}f\\|_{L^{p}_{x,\xi}}=\Big{(}\int_{0}^{\infty}\int_{\mathbb{R}^{3}}e^{p\tilde{\theta}\|\xi\|^{2}/2}e^{p\gamma_{0}x}\|\partial_{x}f(x,\xi)\|^{p}\mathrm{d}\xi\mathrm{d}x\Big{)}^{1/p}$
	$\displaystyle\lesssim\\|e^{\gamma x}w_{\tilde{\theta}}\alpha\partial_{x}f\\|_{L^{\infty}_{x,\xi}}\Big{(}\int_{0}^{\infty}\int_{\mathbb{R}^{3}}\frac{e^{-(\gamma-\gamma_{0})px}w^{-p}_{\tilde{\theta}/2}(\xi)}{\alpha^{p}(x,\xi)}\mathrm{d}\xi\mathrm{d}x\Big{)}^{1/p}.$		(4.48)

From the definition of $\alpha(x,\xi)$ in (1.10), when $\xi_{1}\geq 1$ or $x\geq 1$ , we have $\alpha(x,\xi)\gtrsim 1$ . Then we proceed the computation as

$\displaystyle\eqref{W1p_first_bdd}$	$\displaystyle\lesssim\varepsilon\Big{(}\int_{\mathbb{R}^{2}}\mathrm{d}\xi_{2}\mathrm{d}\xi_{3}w^{-p}_{\tilde{\theta}/2}(\xi_{2})w^{-p}_{\tilde{\theta}/2}(\xi_{3})\int_{0}^{\infty}\int_{\mathbb{R}}\frac{e^{-(\gamma-\gamma_{0})px}w^{-p}_{\tilde{\theta}/2}(\xi_{1})}{\alpha^{p}(x,\xi)}\mathrm{d}\xi_{1}\mathrm{d}x\Big{)}^{1/p}$
	$\displaystyle\lesssim\varepsilon\Big{(}\int_{\|\xi_{1}\|\geq 1\text{ or }x\geq 1}e^{-(\gamma-\gamma_{0})px}w_{\tilde{\theta}/2}^{-p}(\xi_{1})\mathrm{d}\xi_{1}\mathrm{d}x+\int_{0}^{1}\int_{0}^{1}\frac{1}{\big{[}\|\xi_{1}+u\|^{2}+(c\nu_{0})^{2}x^{2}\big{]}^{p/2}}\mathrm{d}\xi_{1}\mathrm{d}x\Big{)}^{1/p}$
	$\displaystyle\lesssim\varepsilon\Big{(}1+\int_{0}^{1}\int_{0}^{1}\frac{1}{\big{[}\|\xi_{1}\|^{2}+x^{2}\big{]}^{p/2}}\mathrm{d}\xi_{1}\mathrm{d}x\Big{)}^{1/p}.$	(4.49)

In the last inequality we apply the change of variable $\xi_{1}+u\to u$ and $c\nu_{0}x\to x$ . Note that the integral domain becomes $(x,\xi_{1})\in[0,1]\times[0,1]\subset\{(x,\xi_{1})|x^{2}+\xi_{1}^{2}\leq 4\}$ , then we apply the polar coordinate $x=r\cos\theta,\ \xi_{1}=r\sin\theta$ to have

	$\displaystyle\eqref{W1p_second_bdd}$	$\displaystyle\lesssim\varepsilon\Big{(}\int_{0}^{2\pi}\int_{0}^{2}\frac{1}{r^{p}(\|\cos^{p}\theta\|+\|\sin^{p}\theta\|)}r\mathrm{d}r\mathrm{d}\theta\Big{)}^{1/p}$
		$\displaystyle\lesssim\varepsilon\Big{(}\int_{0}^{2\pi}\frac{1}{\|\cos^{p}\theta\|+\|\sin^{p}\theta\|}\mathrm{d}\theta\Big{)}^{1/p}\lesssim\varepsilon.$

In the last line, we have used $p<2$ . We conclude the $W^{1,p}$ estimate (1.21).

To prove (1.22), similar to the computation in (4.49), we only focus on the integration over $[\delta,1]\times[0,1]$ , we have

	$\displaystyle\int_{\delta}^{\infty}\int_{\mathbb{R}^{3}}w_{\tilde{\theta}}(\xi)e^{2\gamma_{0}x}\|\partial_{x}f\|^{2}\mathrm{d}\xi\mathrm{d}x$
	$\displaystyle\lesssim\varepsilon+\varepsilon\int_{0}^{1}\int_{\delta}^{1}\frac{1}{\|\xi_{1}\|^{2}+x^{2}}\mathrm{d}\xi_{1}\mathrm{d}x=\varepsilon+\varepsilon\int_{\delta}^{1}\frac{1}{x}\arctan(\frac{1}{x})\mathrm{d}x\lesssim C_{\delta}\varepsilon.$

We conclude the theorem.

Appendix A Definition of (1.18)

In the appendix we give the definition of the addition assumptions (1.18). To be specific, $Y_{1}[u]$ and $Y_{2}[u]$ are defined in (A.2), and $\mathfrak{R}_{u,\gamma}$ is defined in (A.3).

The original version of the linearized penalized operator (3.10) has extra coefficients to be determined:

\mathcal{L}^{p}g=\mathcal{L}p+\alpha\prod_{+}((\xi_{1}+u)g)+\beta\mathbf{p}_{u}g-\gamma(\xi_{1}+u)g.

(A.1)

In order to remove the extra penalization terms, [3] proposed the following lemma:

Lemma 14 (Lemma 4.3 in [3]).

Let

\displaystyle\mathcal{A}:=\begin{bmatrix}\alpha&0&-u\beta\langle\psi_{u}X_{+}\rangle\\ 0&0&-\beta\langle\phi_{u}X_{0}\rangle\\ \alpha\langle\psi_{u}X_{+}\rangle&\frac{1}{u}\tau_{u}&\tau_{u}-\beta\langle\psi_{u}\phi_{u}\rangle\end{bmatrix}

For $|u|\ll 1$ , $\mathcal{A}$ has three different eigenvalues. Let $(l_{1},l_{2},l_{3})$ be a real basis of left eigenvectors of $\mathcal{A}$ , define

\begin{split}Y_{1}[u](\xi):=&(X_{+}(\xi),X_{0}(\xi),\psi_{u}(\xi))\cdot l_{1}(u)\\ Y_{2}[u](\xi):=&(X_{+}(\xi),X_{0}(\xi),\psi_{u}(\xi))\cdot l_{2}(u).\end{split}

(A.2)

Then for $g$ satisfying (3.11) with $\mathcal{L}^{p}$ given by (A.1), one has

\displaystyle\langle(\xi_{1}+u)X_{+}g\rangle=\langle(\xi_{1}+u)\psi_{u}g\rangle=0\Leftrightarrow\begin{cases}\langle(\xi_{1}+u)Y_{1}[u]g\rangle|_{x=0}=0\\ \langle(\xi_{1}+u)Y_{2}[u]g\rangle|_{x=0}=0\end{cases}.

Taking $\alpha=\beta=2\gamma$ as in (3.10), we obtain a unique solution $(g,h)$ to the nonlinear penalized problem (3.31) as in Proposition 7 and Proposition 8. Define

\mathfrak{R}_{u,\gamma}[f_{b}](\xi):=g(0,\xi),\ \ \xi\in\mathbb{R}^{3}.

(A.3)

Then removing the penalization in the nonlinear penalized problem (3.31) is equivalent to imposing the condition in Lemma 14:

\displaystyle\langle(\xi_{1}+u)Y_{1}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=\langle(\xi_{1}+u)Y_{2}[u]\mathfrak{R}_{u,\gamma}[f_{b}]\rangle=0,

which is exactly (1.18).

Acknowledgement. The author thanks Professor Chanwoo Kim for suggesting the problem and engaging in stimulating discussion. HC is supported by NSF 2047681 and GRF grant (2130715) from RGC of Hong Kong. HC thanks the host from the Chinese University of Hong Kong, and thanks Professor Kung-Chien Wu for helpful discussion.

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	$\displaystyle\Big{\|}\nabla_{\xi}\mathbf{k}(\xi,\xi^{\prime})\frac{e^{\tilde{\theta}\|\xi\|^{2}}}{e^{\tilde{\theta}\|\xi^{\prime}\|^{2}}}\Big{\|}$	$\displaystyle\lesssim\big{[}\frac{1+\|\xi^{\prime}\|^{2}+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}[\mathbf{k}_{1}(\xi,\xi^{\prime})+\mathbf{k}_{2}(\xi,\xi^{\prime})]\frac{e^{\tilde{\theta}\|\xi\|^{2}}}{e^{\tilde{\theta}\|\xi^{\prime}\|^{2}}}$
		$\displaystyle\lesssim\big{[}\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}\mathbf{k}^{C_{\theta}}(\xi,\xi^{\prime})$
		$\displaystyle=\big{[}\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}+\|\xi-\xi^{\prime}\|+[1+\|\xi\|]\big{]}\frac{e^{-C_{\theta}\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}\lesssim\frac{1+\|\xi\|^{2}}{\|\xi-\xi^{\prime}\|}\frac{e^{-c\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}$

	$\displaystyle\|(\xi_{1}+u)\partial_{x}\alpha(x,\xi)\|$
	$\displaystyle=\|(\xi_{1}+u)\chi^{\prime}(\tilde{\alpha}(x,\xi))\partial_{x}\tilde{\alpha}(x,\xi)\|$
	$\displaystyle=\chi^{\prime}(\tilde{\alpha}(x,\xi))\|(\xi_{1}+u)\partial_{x}\tilde{\alpha}(x,\xi)\|\leq\frac{c\nu_{0}}{2}\chi^{\prime}(\tilde{\alpha}(x,\xi))\tilde{\alpha}(x,\xi)$
	$\displaystyle\leq 2c\nu_{0}\chi(\tilde{\alpha}(x,\xi))=2c\nu_{0}\alpha(x,\xi).$

	$\displaystyle\int_{\mathbb{R}^{3}}\frac{e^{-C\|\xi-\xi^{\prime}\|^{2}}}{\|\xi-\xi^{\prime}\|}\frac{1}{\sqrt{(\xi_{1}^{\prime}+u)^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi^{\prime}$
	$\displaystyle=\int_{\mathbb{R}^{3}}\frac{e^{-C\|\eta-\xi^{\prime}\|^{2}}}{\|\eta-\xi^{\prime}\|}\frac{1}{\sqrt{\|\xi_{1}^{\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi^{\prime}$
	$\displaystyle\lesssim\iint\frac{e^{-C\|\eta_{\parallel}-\xi^{\prime}_{\parallel}\|^{2}}}{\|\eta_{\parallel}-\xi^{\prime}_{\parallel}\|}\int_{\mathbb{R}}\frac{e^{-C\|\eta_{1}-\xi^{\prime}_{1}\|^{2}}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim\int_{\mathbb{R}}\frac{e^{-C\|\eta_{1}-\xi^{\prime}_{1}\|^{2}}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle=\int_{\mathbb{R}}\mathbf{1}_{\|\xi_{1}^{\prime}\|\leq 2\|\eta_{1}\|}+\mathbf{1}_{\|\xi_{1}^{\prime}\|>2\|\eta_{1}\|}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim\int_{0}^{2\|\eta_{1}\|}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle+\int_{\mathbb{R}}\frac{e^{-C\|\xi_{1}^{\prime}\|^{2}/4}}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle\lesssim 1+\Big{[}\int_{0}^{1}+\int_{0}^{2\|\eta_{1}\|}\Big{]}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}.$		(2.22)

	$\displaystyle\int_{0}^{2\|\eta_{1}\|}\frac{1}{\sqrt{\|\xi^{\prime}_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}}\mathrm{d}\xi_{1}^{\prime}$
	$\displaystyle=\ln\Big{(}\sqrt{\|\xi_{1}^{\prime}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}+\|\xi_{1}^{\prime}\|\Big{)}\Big{\|}_{0}^{2\|\eta_{1}\|}$
	$\displaystyle=\ln\Big{(}\sqrt{4\|\eta_{1}\|^{2}+(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}+2\|\eta_{1}\|\Big{)}$
	$\displaystyle-\ln\Big{(}\sqrt{(c\nu_{0})^{2}(x-(\xi_{1}+u)(t-s))^{2}}\Big{)}$
	$\displaystyle=\ln\Big{(}\sqrt{1+\Big{\|}\frac{2\eta_{1}}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}^{2}}+\Big{\|}\frac{2\eta_{1}}{(c\nu_{0})(x-(\xi_{1}+u)(t-s))}\Big{\|}\Big{)}.$		(2.23)

	$\displaystyle\int_{t-T}^{t}e^{-\nu(\xi)(t-s)/2}\|\ln\|\eta_{1}\|\|\mathrm{d}s$	$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln\|\eta_{1}\|\|}{[1+\|\xi\|]}=\mathbf{1}_{\|\xi_{1}+u\|\geq 1}+\mathbf{1}_{\|\xi_{1}+u\|<1}$
		$\displaystyle\lesssim\min\{1,T\}\frac{\|\ln(2\|\xi\|+2)\|}{[1+\|\xi\|]}+\frac{\min\{1,T\}}{\|\xi_{1}+u\|}\lesssim\frac{\min\{1,T\}}{\alpha(x,\xi)}.$		(2.27)

On regularity of a Kinetic Boundary layer

Abstract.

1. Introduction

1.1. Background

1.2. Main result

Definition 1 (Kinetic weight).

Remark 1.

Remark 2.

Theorem 1.

Remark 3.

Remark 4.

Remark 5.

Theorem 2.

Remark 6.

1.3. Difficulty and main idea

2. Preliminary

Theorem 3 (Theorem 2.1 in [3]).

2.1. Properties of the Boltzmann operator in (1.5) and (1.6).

Lemma 1.

Proof.

Lemma 2.

Proof.

Lemma 3.

Proof.

2.2. Properties of kinetic weight α\alpha in (1.8).

Lemma 4 (Velocity Lemma).

Proof.

Lemma 5.

Proof.

Lemma 6.

Remark 7.

Proof.

Lemma 7.

Proof.

3. Continuity and exponential decay in ξ\xi.

3.1. Continuity and ww-weighted L∞L^{\infty} estimate of the eigen-value problem (3.1).

Proposition 4 (Proposition 3.1 in [3]).

Remark 8.

Lemma 8.

Proof.

Lemma 9.

Proof.

3.2. Continuity and ww-weighted L∞L^{\infty} estimate of the linearized penalized problem

Proposition 5 (Proposition 5.3 and Proposition 5.6 in [3]).

Remark 9.

Lemma 10.

Proof.

Proposition 6.

Remark 10.

Proof.

3.3. Continuity and ww-weighted L∞L^{\infty} estimate of the nonlinear penalized problem

Proposition 7 (Proposition 6.1 in [3]).

Proposition 8.

Proof.

3.4. Proof of Theorem 2

4. Weighted C1C^{1} estimate and W1,pW^{1,p} estimate for p<2p<2 without weight

Lemma 11.

Proof.

4.1. Weighted C1C^{1} estimate of the linearized penalized problem

Lemma 12 (A-priori weighted C1C^{1} estimate).

Proof.

Lemma 13.

Proof.

4.2. Weighted C1C^{1} estimate of nonlinear penalized problem

Proposition 9.

Proof.

4.3. Proof of Theorem 1

Appendix A Definition of (1.18)

Lemma 14 (Lemma 4.3 in [3]).

References

2.2. Properties of kinetic weight $\alpha$ in (1.8).

3. Continuity and exponential decay in $\xi$ .

3.1. Continuity and $w$ -weighted $L^{\infty}$ estimate of the eigen-value problem (3.1).

3.2. Continuity and $w$ -weighted $L^{\infty}$ estimate of the linearized penalized problem

3.3. Continuity and $w$ -weighted $L^{\infty}$ estimate of the nonlinear penalized problem

4. Weighted $C^{1}$ estimate and $W^{1,p}$ estimate for $p<2$ without weight

4.1. Weighted $C^{1}$ estimate of the linearized penalized problem

Lemma 12 (A-priori weighted $C^{1}$ estimate).

4.2. Weighted $C^{1}$ estimate of nonlinear penalized problem