Ghost Effect from Boltzmann Theory: Expansion with Remainder

Abstract

Consider the limit $\varepsilon \rightarrow 0$ of the steady Boltzmann problem

$$\begin{aligned} v\cdot \nabla _{x} \mathfrak {F}=\varepsilon ^{-1}Q[\mathfrak {F},\mathfrak {F}],\quad \mathfrak {F}\big |_{v\cdot n<0}=M_{w}\int _{v'\cdot n>0} \mathfrak {F}(v')|v'\cdot n|\textrm{d}v', \end{aligned}$$

(0.1)

where $M_{w}(x_0,v):=\frac{1}{2\pi (T_{w}(x_0))^2}\exp \big (-\frac{|v|^2}{2T_{w}(x_0)}\big )$ for $x_0\in \partial \Omega$ is the wall Maxwellian in the diffuse-reflection boundary condition. We normalize

$$\begin{aligned} T_{w}=1+O\left( |\nabla T_{w}|_{L^{\infty }}\right) . \end{aligned}$$

In the case of $|\nabla T_{w}|=O(\varepsilon )$, the Hilbert expansion confirms $\mathfrak {F}\approx (2\pi )^{-\frac{3}{2}}\textrm{e}^{-\frac{|v|^2}{2}}+\varepsilon (2\pi )^{-\frac{3}{4}}\textrm{e}^{-\frac{|v|^2}{4}}\big (\rho _1+T_1\frac{|v|^2-3}{2}\big )$ where $(2\pi )^{-\frac{3}{2}}\textrm{e}^{-\frac{|v|^2}{2}}$ is a global Maxwellian and $(\rho _1,T_1)$ satisfies the celebrated Fourier law

$$\begin{aligned} \Delta _xT_1=0. \end{aligned}$$

In the natural case of $|\nabla T_{w}|=O(1)$, for any constant $P>0$, the Hilbert expansion leads to

$$\begin{aligned} \mathfrak {F}\approx \mu +\varepsilon \left\{ \mu \left( \rho _1+u_1\cdot v+T_1\frac{|v|^2-3T}{2}\right) -\mu ^{\frac{1}{2}}\left( \mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right) \right\} , \end{aligned}$$

where $\mu (x,v):=\frac{\rho (x)}{(2\pi T(x))^{\frac{3}{2}}}\exp \big (-\frac{|v|^2}{2T(x)}\big )$, and $(\rho ,u_1,T)$ is determined by a Navier–Stokes–Fourier system with “ghost” effect

$$\begin{aligned} \left\{ \begin{array}{rcl} P&{}=&{}\rho T,\\ \rho (u_1\cdot \nabla _{x}u_1)+\nabla _{x} \mathfrak {p}&{}=&{}\nabla _{x}\cdot \left( \tau ^{(1)}-\tau ^{(2)}\right) ,\\ \nabla _{x}\cdot (\rho u_1)&{}=&{}0,\\ \nabla _{x}\cdot \left( \kappa \frac{\nabla _{x}T}{2T^2}\right) &{}=&{}5P(\nabla _{x}\cdot u_1), \end{array}\right. \end{aligned}$$

(0.2)

with the boundary condition

$$\begin{aligned} T\Big |_{\partial \Omega }= T_{w},\quad u_1\Big |_{\partial \Omega }:=(u_{1,\iota _1},u_{1,\iota _2},u_{1,n})\Big |_{\partial \Omega }=(\beta _0\partial _{\iota _1} T_{w},\beta _0\partial _{\iota _2} T_{w},0). \end{aligned}$$

(0.3)

Here $\kappa [T]>0$ is the heat conductivity, $(\iota _1,\iota _2)$ are two tangential variables and n is the normal variable, $\beta _0=\beta _0[T_{w}]$ is a function of $T_{w}$, $\tau ^{(1)}:=\lambda \left( \nabla _{x}u_1+(\nabla _{x}u_1)^t-\frac{2}{3}(\nabla _{x}\cdot u_1)\textbf{1}\right)$ and $\tau ^{(2)}:=\frac{\lambda ^2}{P}\left( K_1\big (\nabla _{x}^2T-\frac{1}{3}\Delta _x T\textbf{1}\big )+\frac{K_2}{T}\big (\nabla _{x}T\otimes \nabla _{x}T-\frac{1}{3}|\nabla _{x}T|^2\textbf{1}\big )\right)$ for some smooth function $\lambda [T]>0$, the viscosity coefficient, and positive constants $K_1$ and $K_2$. Tangential temperature variation creates non-zero first-order velocity $u_1$ at the boundary (0.3), which plays a surprising “ghost” effect [26, 27] in determining zeroth-order density and temperature field $(\rho ,T)$ in (0.2). Such a ghost effect cannot be predicted by the classical fluid theory, while it has been an intriguing outstanding mathematical problem to justify (0.2) from (0.1) due to fundamental analytical challenges. The goal of this paper is to construct $\mathfrak {F}$ in the form of

$$\begin{aligned} \mathfrak {F}(x,v)=\mu +\mu ^{\frac{1}{2}}\left( \varepsilon f_1+\varepsilon ^2 f_2\right) +\mu _{w}^{\frac{1}{2}}\left( \varepsilon f^{B}_1\right) +\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R \end{aligned}$$

(0.4)

for interior solutions $f_1$, $f_2$ and boundary layer $f^{B}_1$, where $\mu _w$ is $\mu$ computed for $T= T_{w}$, and derive equation for the remainder R with some constant $\alpha \ge 1$. To prove the validity of the expansion suitable bounds on R are needed, which are provided in the companion paper (Esposito 2023).

1 Introduction

The diffusive hydrodynamic limit of the Boltzmann equation in the low Mach number regime is described by the incompressible Navier–Stokes–Fourier equations under the extra assumption that the initial density and temperature profiles differ from constants at most for terms of the order of the Knudsen number. Such behavior has been proved in several papers and an overview is provided in [23] and [13], to which we refer for a partial list of references on the subject. We also stress that a similar result can be obtained starting from the compressible Navier–Stokes equations, which converge, in the low Mach number limit, to the solutions of the incompressible Navier–Stokes equations [18].

When the density and temperature do not satisfy the above mentioned assumptions, the limiting behavior of the Boltzmann equation deviates from the Navier–Stokes–Fourier equations. Such a discrepancy, called “ghost effect” [27], shows up in the macroscopic equations with the presence of some extra terms reminiscent of the limiting procedure such as some heat flow induced by the vanishingly small velocity field. Thus they are genuine kinetic effects which would be never detected in the standard hydrodynamic equations. Y. Sone has given the suggestive name of “ghost effects” to such phenomena. The meaning of the name is that the velocity field $u_1$ acts like a ghost since it appears at order $\varepsilon$ in the expansion and still affects $\rho$ and T at order 1. In [22] the local well-posedness of the time dependent equations is proven.

In this paper we confine our analysis to the stationary Boltzmann equation for a rarefied gas in a bounded domain with diffuse-reflection boundary data describing a non-homogeneous wall temperature with a gradient of order 1. In this situation the gradient of temperature along the boundary wall produces a flow called in literature thermal creep. For relevant physical background and discussion, we refer to [24].

We give a formal derivation of such new equations when the Mach number, proportional to the Knudsen number $\varepsilon$, goes to 0, and prove their well-posedness. In the companion paper [12] we study the much more involved problem of the rigorous proof of such a derivation. Here we construct the formal solution by a truncated expansion in $\varepsilon$ plus a remainder, both in the interior and in a boundary layer of size $\varepsilon$. In view of the control of the remainder, we carefully prepare the expansion by truncating at the second order in $\varepsilon$ in the bulk and at the first order in the boundary layer. Then a matching procedure allows to determine the boundary conditions for the limiting equations.

The explicit form of the equations for $(\rho , u_1, T, \mathfrak {p})$ is given in (0.2). The main difference between these equations and the incompressible ones is that $\nabla _{x}\cdot u_1$ is not anymore zero but is related to the gradient of the temperature. This is the analog of the constraint $\nabla _{x}\cdot u_1=0$ in the incompressible Navier–Stokes equations and is compensated by the Lagrangian multiplier $\mathfrak {p}$ in the equation for $u_1$. Moreover, in the equation for $u_1$ there are the usual Navier–Stokes terms involving $u_1$ and also a term $\tau ^{(2)}$ depending on the first and second gradient of the temperature. In particular, the “thermal stress” $\tau ^{(2)}$ is a new contribution different from the standard fluid theories. It is exactly this term that cannot be obtained from the compressible Navier–Stokes equation. The relevance of these equations, as also noted by Bobylev [5], is that they cannot be derived from the compressible Navier–Stokes equations. Let us notice that the particular solution corresponding to homogeneous initial condition for density and temperature is also solution of the incompressible Navier–Stokes equations.

We give also the proof of the existence of the solution to (0.2) under the assumption of small temperature gradient. The main difficulty in getting a rigorous proof of the hydrodynamic limit is the control of the remainder. This is achieved in [12].

Before stating the main results, we briefly introduce the history of the study of the ghost effect. Sone [25] and [19, 20] pointed out the new thermal effects in stationary situations. In [11], the equations from the Boltzmann equations in the time dependent case were formally derived, but without computing the transport coefficients. These equations were then discussed by Bobylev [5], who analyzed the behavior of the solutions in particular situations. He also showed that the thermodynamic entropy decreases in time. Finally, Sone and the Kyoto group exploited many other kinds of ghost effects in many papers [28, 29], both analytically and numerically and gave computations of the transport coefficients for the hard sphere case and for Maxwellian molecules. A detailed analysis can be found in [26] and [27] and references therein. Rigorous results in deriving the equations where obtained only in one-dimensional stationary cases [7, 8] and [1]. There are no rigorous results in the time dependent case, not even on the torus, but for [16] where the Korteweg theory is derived from the one-dimensional Boltzmann equation on the infinite line. We also refer to [15, 17] and the references therein.

1.1 Formulation of the Problem

We consider the stationary Boltzmann equation in a bounded three-dimensional $C^3$ domain $\Omega \ni x=(x_1,x_2,x_3)$ with velocity $v=(v_1,v_2,v_3)\in \mathbb {R}^3$. The density function $\mathfrak {F}(x,v)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} v\cdot \nabla _{x} \mathfrak {F}=\varepsilon ^{-1}Q[\mathfrak {F},\mathfrak {F}]&{}\quad \text {in }\Omega \times \mathbb {R}^3,\\ \mathfrak {F}(x_0,v)=\mathfrak {P}_{\gamma }[\mathfrak {F}]&{}\quad \text {for }x_0\in \partial \Omega \text { and }v\cdot n(x_0)<0. \end{array}\right. \end{aligned}$$

(1.1)

Here Q is the hard-sphere collision operator

$$\begin{aligned} Q[F,G]:=\frac{1}{2}\int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)\left( F(\mathfrak {u}_{*})G(v_{*})-F(\mathfrak {u})G(v)\right) \textrm{d}{\omega }\textrm{d}{\mathfrak {u}}, \end{aligned}$$

with $\mathfrak {u}_{*}:=\mathfrak {u}+\omega ((v-\mathfrak {u})\cdot \omega )$, $v_{*}:=v-\omega ((v-\mathfrak {u})\cdot \omega )$, and the hard-sphere collision kernel $q(\omega ,|\mathfrak {u}-v|):=q_0|\omega \cdot (v-\mathfrak {u})|$ for a positive constant $q_0$.

In the diffuse-reflection boundary condition

$$\begin{aligned} \mathfrak {P}_{\gamma }[\mathfrak {F}]:=M_{w}(x_0,v)\int _{v'\cdot n(x_0)>0} \mathfrak {F}(x_0,v')|v'\cdot n(x_0)|\textrm{d}{v'}, \end{aligned}$$

$n(x_0)$ is the unit outward normal vector at $x_0$, and the Knudsen number $\varepsilon$ satisfies $0<\varepsilon \ll 1$. The wall Maxwellian

$$\begin{aligned} M_{w}(x_0,v):=\frac{1}{2\pi (T_{w}(x_0))^2} \exp \left( -\frac{|v|^2}{2 T_{w}(x_0)}\right) \end{aligned}$$

for any $T_{w}(x_0)>0$ satisfies

$$\begin{aligned} \int _{v\cdot n(x_0)>0}M_{w}(x_0,v)|v\cdot n(x_0)| \textrm{d}v=1. \end{aligned}$$

The boundary condition in (1.1) implies that the total max flux across the boundary is zero.

1.2 Notation and Convention

Based on the flow direction, we can divide the boundary $\gamma :=\{(x_0,v): x_0\in \partial \Omega ,v\in \mathbb {R}^3\}$ into the incoming boundary $\gamma _-$, the outgoing boundary $\gamma _+$, and the grazing set $\gamma _0$ based on the sign of $v\cdot n(x_0)$. In particular, the boundary condition of (1.1) is only given on $\gamma _{-}$.

Denote the bulk and boundary norms

$$\begin{aligned} \Vert f\Vert _{L^r}:=\left( \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|f(x,v)|^r\textrm{d} v\textrm{d} x\right) ^{\frac{1}{r}},\qquad |f|_{L_{\gamma _\pm }^r}:=\left( \int _{\gamma _{\pm }}|f(x,v)|^r |v\cdot n|\textrm{d} v\textrm{d} x\right) ^{\frac{1}{r}}. \end{aligned}$$

Define the weighted $L^{\infty }$ norms for $T_M>0$, $0\le \varrho <\frac{1}{2}$ and $\vartheta \ge 0$ (see (4.7))

$$\begin{aligned} \Vert f\Vert _{L_{\varrho ,\vartheta }^\infty }:= & {} \underset{{(x,v)\in \Omega \times \mathbb {R}^3}}{\mathrm {ess\,sup}}\left( \langle {v}\rangle ^{\vartheta }\textrm{e}^{\varrho \frac{|v|^2}{2T_M}}|f(x,v)|\right) ,\\ |f|_{L^{\infty }_{\gamma _{\pm },\varrho ,\vartheta }}:= & {} \underset{{(x,v)\in \gamma _{\pm }}}{\mathrm {ess\,sup}}\left( \langle {v}\rangle ^{\vartheta }\textrm{e}^{\varrho \frac{|v|^2}{2T_M}}|f(x,v)|\right) . \end{aligned}$$

Denote the $\nu$-norm

$$\begin{aligned} \Vert f\Vert _{L_{\nu }^2}:=\left( \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\nu (x,v)|f(x,v)|^2\textrm{d} v\textrm{d} x\right) ^{\frac{1}{2}}. \end{aligned}$$

Let $\Vert \cdot \Vert _{W^{k,p}}$ denote the usual Sobolev norm for $x\in \Omega$ and $|\cdot |_{W^{k,p}}$ for $x\in \partial \Omega$. Let $\Vert \cdot \Vert _{W^{k,p}L^q}$ denote $W^{k,p}$ norm for $x\in \Omega$ and $L^q$ norm for $v\in \mathbb {R}^3$. The similar notation also applies when we replace $L^q$ by $L^{\infty }_{\varrho ,\vartheta }$ or $L^q_{\gamma }$.

Define the quantities (where $\mathcal {L}$ is defined in (2.2))

$$\begin{aligned} \overline{\mathscr {A}}:= & {} v\cdot \left( |v|^2-5T\right) \mu ^{\frac{1}{2}}\in \mathbb {R}^3,\quad \mathscr {A}:=\mathcal {L}^{-1} \left[ \overline{\mathscr {A}}\right] \in \mathbb {R}^3,\nonumber \\ \overline{\mathscr {B}}= & {} \left( v\otimes v-\frac{|v|^2}{3}\textbf{1}\right) \mu ^{\frac{1}{2}}\in \mathbb {R}^{3\times 3},\quad \mathscr {B}=\mathcal {L}^{-1}\left[ \overline{\mathscr {B}}\right] \in \mathbb {R}^{3\times 3},\nonumber \\ \kappa \textbf{1}:= & {} \int _{\mathbb {R}^3}\left( \mathscr {A}\otimes \overline{\mathscr {A}}\right) \textrm{d} v,\quad \lambda :=\frac{1}{T}\int _{\mathbb {R}^3}\mathscr {B}_{ij}\overline{\mathscr {B}}_{ij}\quad \text {for }i\ne j. \end{aligned}$$

(1.2)

Throughout this paper, $C>0$ denotes a constant that only depends on the domain $\Omega$, but does not depend on the data or $\varepsilon$. It is referred as universal and can change from one inequality to another. When we write C(z), it means a certain positive constant depending on the quantity z. We write $a \lesssim b$ to denote $a\le Cb$ and $a \gtrsim b$ to denote $a\ge Cb$.

In this paper, we will use o(1) to denote a sufficiently small constant independent of the data. Also, let $o_T$ be a small constant depending on $T_{w}$ satisfying

$$\begin{aligned} o_T=o(1)\rightarrow 0\quad \text { as }\quad |\nabla T_w|_{W^{3,\infty }} \rightarrow 0. \end{aligned}$$

(1.3)

In principle, while $o_T$ is determined by $\nabla T_{w}$ a priori, we are free to choose o(1) depending on the estimate.

1.3 Main Theorem

Throughout this paper, we assume that

$$\begin{aligned} |\nabla T_w|_{W^{3,\infty }}=o(1). \end{aligned}$$

(1.4)

Theorem 1.1

Under the assumption (1.4), for any given $P>0$, there exists a unique solution $(\rho ,u_1,T; \mathfrak {p})$ (where $\mathfrak {p}$ has zero average) to the ghost-effect (0.2) and (0.3) satisfying for any $s\in [2,\infty )$

$$\begin{aligned} \Vert u_1\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}+\Vert T-1\Vert _{W^{4,s}}\lesssim o_T. \end{aligned}$$

Also, we can construct $f_1$, $f_2$ and $f^{B}_1$ as in (2.31), (2.32), (2.48) such that

$$\begin{aligned} \Vert f_1\Vert _{W^{3,s}L^{\infty }_{\varrho ,\vartheta }}+|f_1|_{W^{3-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim & {} o_T,\\ \Vert f_2\Vert _{W^{2,s}L^{\infty }_{\varrho ,\vartheta }}+|f_2|_{W^{2-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim & {} o_T, \end{aligned}$$

and for some $K_0>0$ and any $0<r \le 3$

$$\begin{aligned} \left\| \textrm{e}^{K_0\eta }f^{B}_1\right\| _{L_{\varrho ,\vartheta }^\infty } + \left\| \textrm{e}^{K_0\eta }\frac{\partial ^r f^{B}_1}{\partial \iota _1^r}\right\| _{L_{\varrho ,\vartheta }^\infty } +\left\| \textrm{e}^{K_0\eta }\frac{\partial ^r f^{B}_1}{\partial \iota _2^r}\right\| _{L_{\varrho ,\vartheta }^\infty } \lesssim o_T. \end{aligned}$$

2 Asymptotic Analysis

In this section we construct a solution to (1.1) by a truncated expansion in $\varepsilon$ and determine the ghost effect equation in terms of the first terms of the expansion.

We seek a solution in the form

$$\begin{aligned} \mathfrak {F}(x,v)= & {} f+ f^{B}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\\= & {} \mu +\mu ^{\frac{1}{2}}\left( \varepsilon f_1+\varepsilon ^2 f_2\right) +\mu _{w}^{\frac{1}{2}}\left( \varepsilon f^{B}_1\right) +\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\nonumber \end{aligned}$$

where f is the interior solution

$$\begin{aligned} f(x,v):= \mu (x,v)+\mu ^{\frac{1}{2}}(x,v)\left( \varepsilon f_1(x,v)+\varepsilon ^2 f_2(x,v)\right) , \end{aligned}$$

(2.1)

and $f^{B}$ is the boundary layer term

$$\begin{aligned} f^{B}(x,v):= \mu _{w}^{\frac{1}{2}}(x_0,v)\left( \varepsilon f^{B}_1(x,v)\right) . \end{aligned}$$

Here R(x, v) is the remainder, $\mu (x,v)$ denotes a local Maxwellian which will be specified below and $\mu _{w}(x_0,v)=\mu (x_0,v)$ the boundary Maxwellian. The parameter $\alpha \ge 1$, will be equal to 1 in the companion paper [12].

We start to determine the first terms of the expansion. Inserting (2.1) into (1.1), at the lowest order of $\varepsilon$, we have

$$\begin{aligned} {\textbf {Order 0:}}\quad -Q[\mu ,\mu ]=0. \end{aligned}$$

This equation guarantees that $\mu$ is a local Maxwellian. Denote

$$\begin{aligned} \mu (x,v):=\frac{\rho (x)}{(2\pi T(x)\big )^{\frac{3}{2}}} \exp \left( -\frac{|v|^2}{2T(x)}\right) , \end{aligned}$$

where $\rho (x)>0$ and $T(x)>0$ will be determined later in terms of the solutions of the ghost equations. Notice that this local Maxwellian does not contain the velocity field since we are assuming the Mach number of order $\varepsilon$.

Linearized Boltzmann Operator Define the symmetrized version of Q

$$\begin{aligned} Q^{*}[F,G]:= & {} \frac{1}{2}\int \hspace{-0.5pc}\int _{\mathbb {R}^3\times \mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)\\{} & {} \hspace{1.5cm}\times \left( F(\mathfrak {u}_{*})G(v_{*})+F(v_{*})G(\mathfrak {u}_{*})-F(\mathfrak {u})G(v)-F(v)G(\mathfrak {u})\right) \textrm{d}{\omega }\textrm{d}{\mathfrak {u}}. \end{aligned}$$

Clearly, $Q[F,F]=Q^{*}[F,F]$. Denote the linearized Boltzmann operator $\mathcal {L}$

$$\begin{aligned} \mathcal {L}[f]:=-2\mu ^{-\frac{1}{2}}Q^{*}\left[ \mu ,\mu ^{\frac{1}{2}}f\right] :=\nu (v)f-K[f], \end{aligned}$$

(2.2)

where for some kernels $k(\mathfrak {u},v)$ (see [10, 14]),

$$\begin{aligned} \nu (v)= & {} \int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)\mu (\mathfrak {u})\textrm{d}{\omega }\textrm{d}{\mathfrak {u}},\\ K[f](v)= & {} \int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)\mu ^{\frac{1}{2}}(\mathfrak {u})\left( \mu ^{\frac{1}{2}}(v_{*})f(\mathfrak {u}_{*}) + \mu ^{\frac{1}{2}}(\mathfrak {u}_{*})f(v_{*})\right) \textrm{d}{\omega }\textrm{d}{\mathfrak {u}}\\{} & {} -\mu ^{\frac{1}{2}}(v)\int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)\mu ^{\frac{1}{2}}(\mathfrak {u})f(\mathfrak {u})\textrm{d}{\omega }\textrm{d}{\mathfrak {u}}. \end{aligned}$$

Note that $\mathcal {L}$ is self-adjoint in $L^2_{\nu }(\mathbb {R}^3)$. Also, the null space $\mathcal {N}$ of $\mathcal {L}$ is a five-dimensional space spanned by the orthogonal basis

$$\begin{aligned} \mu ^{\frac{1}{2}}\left\{ 1,v,\left( |v|^2-3T\right) \right\} . \end{aligned}$$

Denote $\mathcal {N}^{\perp }$ the orthogonal complement of $\mathcal {N}$ in $L^2(\mathbb {R}^3)$, and $\mathcal {L}^{-1}: \mathcal {N}^{\perp }\rightarrow \mathcal {N}^{\perp }$ the quasi-inverse of $\mathcal {L}$. Define the kernel operator ${\textbf {P}}$ as the orthogonal projection onto the null space $\mathcal {N}$ of $\mathcal {L}$, and the non-kernel operator ${\textbf {I}}-{\textbf {P}}$. Also, denote the nonlinear Boltzmann operator $\Gamma$ as

$$\begin{aligned} \Gamma [f,g]:=\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}} f,\mu ^{\frac{1}{2}} g\right] \in \mathcal {N}^{\perp }. \end{aligned}$$

2.1 Derivation of Interior Solution

Further inserting (2.1) into (1.1), we have

$$\begin{aligned} {\textbf {Order 1:}}{} & {} \quad v\cdot \nabla _{x}\mu -2Q^{*}\left[ \mu ,\mu ^{\frac{1}{2}}f_1\right] =0,\end{aligned}$$

(2.3)

$$\begin{aligned} {\textbf {Order}}\, \varvec{\varepsilon }{} {\textbf {:}}{} & {} \quad v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) -2Q^{*}\left[ \mu ,\mu ^{\frac{1}{2}}f_2\right] -Q^{*} \left[ \mu ^{\frac{1}{2}}f_1,\mu ^{\frac{1}{2}}f_1\right] =0. \end{aligned}$$

(2.4)

Inspired by the continuation of the expansion, we also require an additional condition that

$$\begin{aligned} {\textbf {Order}}\, \varvec{\varepsilon }^\textbf{2}{} {\textbf {:}}\quad \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_2\right) \right) \perp v\mu ^{\frac{1}{2}}. \end{aligned}$$

(2.5)

Note that we stop the bulk expansion at order $\varepsilon ^2$, so we do not need the orthogonality with $\mu ^{\frac{1}{2}}$ and $|v|^2\mu ^{\frac{1}{2}}$.

2.1.1 Equation (2.3)

Lemma 2.1

Equation (2.3) is equivalent to

$$\begin{aligned} \nabla _{x} P=\nabla _{x}(\rho T)=0 \end{aligned}$$

(2.6)

and for some $\rho _1(x)$, $u_1(x)$, $T_1(x)$,

$$\begin{aligned} f_1=-\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}+\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) . \end{aligned}$$

(2.7)

Proof

Equation (2.3) can be rewritten as

$$\begin{aligned} \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\mu \right) =-\mathcal {L}[f_1]. \end{aligned}$$

(2.8)

Then, by the orthogonality of $\mathcal {L}$ to $\mathcal {N}$, to satisfy (2.8) we must have

$$\begin{aligned} \int _{\mathbb {R}^3}\left( v\cdot \nabla _{x}\mu \right) \textrm{d}v=0,\quad \int _{\mathbb {R}^3}v\left( v\cdot \nabla _{x}\mu \right) \textrm{d}v={\textbf {0}},\quad \int _{\mathbb {R}^3}|v|^2\left( v\cdot \nabla _{x}\mu \right) \textrm{d}v=0. \end{aligned}$$

(2.9)

Note that

$$\begin{aligned} v\cdot \nabla _{x}\mu =\mu \left( v\cdot \frac{\nabla _{x}\rho }{\rho }+v\cdot \frac{\nabla _{x}T(|v|^2-3T)}{2T^2}\right) . \end{aligned}$$

(2.10)

Then the first and third conditions in (2.9) are satisfied by oddness. The second condition in (2.9) can be rewritten in the component form for $i\in \{1,2,3\}$ and summation over $j\in \{1,2,3\}$

$$\begin{aligned} \int _{\mathbb {R}^3}v_iv_j\mu \!\left( \frac{\partial _j\rho }{\rho }\!+\!\frac{\partial _jT(|v|^2\!-\!3T)}{2T^2}\right) \textrm{d}v\!=\! & {} \int _{\mathbb {R}^3}\delta _{ij}\frac{|v|^2}{3}\mu \!\left( \frac{\partial _j\rho }{\rho }\!+\!\frac{\partial _jT(|v|^2\!-\!3T)}{2T^2}\right) \textrm{d}v \qquad \quad \\\!=\! & {} \delta _{ij}\left( \rho T\cdot \frac{\partial _j\rho }{\rho }+5\rho T^2\cdot \frac{\partial _jT}{2T^2}-\rho T\cdot \frac{3\partial _jT}{2T}\right) \nonumber \\\!=\! & {} \delta _{ij}\left( T\partial _j\rho +\rho \partial _jT\right) =\delta _{ij}\partial _j(\rho T)=0.\nonumber \end{aligned}$$

(2.11)

Hence, (2.11) is actually (2.6).

Since $T\nabla _{x}\rho +\rho \nabla _{x}T=0$, we deduce $\frac{\nabla _{x}\rho }{\rho }=-\frac{\nabla _{x}T}{T}$. Thus, inserting this into (2.10), we have

$$\begin{aligned} v\cdot \nabla _{x}\mu =\mu \left( v\cdot \nabla _{x}T\right) \frac{|v|^2-5T}{2T^2}. \end{aligned}$$

(2.12)

Considering (2.8) and (2.12), we know

$$\begin{aligned} \mathcal {L}[f_1]=-\overline{\mathscr {A}}\cdot \frac{\nabla _{x}T}{2T^2}, \end{aligned}$$

and (2.7) holds.$\square$

2.1.2 Equation (2.4)

Lemma 2.2

Equation (2.4) is equivalent to

$$\begin{aligned} \nabla _{x}\cdot (\rho u_1)= & {} 0,\end{aligned}$$

(2.13)

$$\begin{aligned} \nabla _{x} P_1= & {} \nabla _{x}(T\rho _1+\rho T_1)=0,\end{aligned}$$

(2.14)

$$\begin{aligned} 5P(\nabla _{x}\cdot u_1)= & {} \nabla _{x}\cdot \left( \kappa \frac{\nabla _{x}T}{2T^2}\right) , \end{aligned}$$

(2.15)

and for some $\rho _2(x)$, $u_2(x)$, $T_2(x)$,

$$\begin{aligned} f_2\!=\!-\mathcal {L}^{-1}\!\left[ \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right] +\mathcal {L}^{-1}\left[ \Gamma [f_1,f_1]\right] + \mu ^{\frac{1}{2}}\!\left( \frac{\rho _2}{\rho }\!+\!\frac{ u_2\cdot v}{T}\!+\!\frac{T_2(|v|^2\!-\!3T)}{2T^2}\right) . \end{aligned}$$

(2.16)

Proof

Since the $Q^*$ terms in (2.4) are orthogonal to $\mathcal {N}$, we must have

$$\begin{aligned} \int _{\mathbb {R}^3}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \textrm{d}v=0,~~\int _{\mathbb {R}^3}v\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \textrm{d}v=0,~~\int _{\mathbb {R}^3}|v|^2\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \textrm{d}v=0. \end{aligned}$$

(2.17)

Using (2.7), the first condition in (2.17) can be rewritten as

$$\begin{aligned} \nabla _{x}\cdot \left( -\int _{\mathbb {R}^3}v\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\textrm{d}v +\int _{\mathbb {R}^3}v\mu \left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \textrm{d}v\right) =0. \end{aligned}$$

(2.18)

Since $\mathscr {A}$ is orthogonal to $\mathcal {N}$, the first term in (2.18) vanishes. Due to oddness, the $\rho _1$ and $T_1$ terms in (2.18) vanish. Hence, we are left with (2.13).

Similarly, the second condition in (2.17) can be rewritten as

$$\begin{aligned} \nabla _{x}\cdot \left( -\!\int _{\mathbb {R}^3}v\otimes v\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\textrm{d}v + \!\int _{\mathbb {R}^3}v\otimes v\mu \left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \textrm{d}v\right) \!=\!0. \end{aligned}$$

(2.19)

Due to the oddness of $\mathscr {A}$, the first term in (2.19) vanishes. For the same reason, the $u_1$ term in (2.19) also vanishes. Thus we are left with (2.14).

Finally, the third condition in (2.17) can be rewritten as

$$\begin{aligned} \nabla _{x}\cdot \left( -\int _{\mathbb {R}^3}v|v|^2\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\textrm{d}v + \int _{\mathbb {R}^3}v|v|^2\mu \left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \textrm{d}v\right) =0. \end{aligned}$$

(2.20)

Using the orthogonality of $\mathscr {A}$ to $\mathcal {N}$, we know

$$\begin{aligned} \int _{\mathbb {R}^3}v|v|^2\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\textrm{d} v = \int _{\mathbb {R}^3}\overline{\mathscr {A}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\textrm{d}v=\kappa \frac{\nabla _{x}T}{2T^2}, \end{aligned}$$

where $\kappa$ is defined in (1.2).

Due to oddness, the $\rho _1$ and $T_1$ terms in (2.20) vanish, so the $u_1$ term in (2.20) can be computed

$$\begin{aligned} \int _{\mathbb {R}^3}v|v|^2\mu \frac{ u_1\cdot v}{T}\textrm{d}v=5\rho T u_1=5P u_1. \end{aligned}$$

Hence, (2.20) becomes

$$\begin{aligned} \nabla _{x}\cdot \left( -\kappa \frac{\nabla _{x}T}{2T^2}+5P u_1\right) =0, \end{aligned}$$

which is equivalent to (2.15).

Equation (2.4) can be rewritten as

$$\begin{aligned} \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) -\Gamma [f_1,f_1]=-\mathcal {L}[f_2], \end{aligned}$$

and thus (2.16) holds.$\square$

2.1.3 Equation (2.5)

Lemma 2.3

We have the identity

$$\begin{aligned} \int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ (u_1\cdot v)\mu ^{\frac{1}{2}},\mathscr {A}\right] =-\int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2}\right) \overline{\mathscr {A}}+T\int _{\mathbb {R}^3}\mathscr {B}(u_1\cdot \overline{\mathscr {B}}). \end{aligned}$$

(2.21)

Proof

We follow the idea in [4]. Denote the translated quantities

$$\begin{aligned} \mu _s(x,v):=\frac{\rho (x)}{(2\pi T(x))^{\frac{3}{2}}} \exp \left( -\frac{|v-su_1|^2}{2T(x)}\right) ,\quad \mathcal {L}_s[f]:=-2\mu ^{-\frac{1}{2}}_s Q^{*}\left[ \mu _s,\mu ^{\frac{1}{2}}_sf\right] , \end{aligned}$$

and

$$\begin{aligned} \overline{\mathscr {A}}_s=\overline{\mathscr {A}}(v-su_1),\quad \mathscr {A}_s=\mathcal {L}_s^{-1}[\overline{\mathscr {A}}_s],\quad \overline{\mathscr {B}}_s=\overline{\mathscr {B}}(v-su_1),\quad \mathscr {B}_s=\mathcal {L}_s^{-1}[\overline{\mathscr {B}}_s]. \end{aligned}$$

Note that translation will not change the orthogonality, i.e. for any $s\in \mathbb {R}$

$$\begin{aligned} \int _{\mathbb {R}^3}\mathscr {B}_s\overline{\mathscr {A}}_s=\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\mathscr {A}_s=0. \end{aligned}$$

Taking s derivative, we know

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d} s}\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\mathscr {A}_s=0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \int _{\mathbb {R}^3}\frac{\textrm{d}\overline{\mathscr {B}}_s}{\textrm{d}s}\mathcal {L}^{-1}_s\left[ \overline{\mathscr {A}}_s\right] +\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\frac{\textrm{d}\mathcal {L}^{-1}_s}{\textrm{d} s}\left[ \overline{\mathscr {A}}_s\right] +\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\mathcal {L}^{-1}_s\left[ \frac{\textrm{d}\overline{\mathscr {A}}_s}{\textrm{d} s}\right] =0. \end{aligned}$$

(2.22)

For the first term in (2.22), due to oddness and orthogonality, we can directly verify that

$$\begin{aligned} \lim _{s\rightarrow 0}\int _{\mathbb {R}^3}\frac{\textrm{d}\overline{\mathscr {B}}_s}{\textrm{d} s}\mathcal {L}^{-1}_s\left[ \overline{\mathscr {A}}_s\right] =\int _{\mathbb {R}^3}\overline{\mathscr {B}}\left( \frac{u_1\cdot v}{2T}\right) \mathscr {A}. \end{aligned}$$

(2.23)

For the second term in (2.22), we have

$$\begin{aligned} \int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\frac{\textrm{d}\mathcal {L}^{-1}_s}{\textrm{d} s}\left[ \overline{\mathscr {A}}_s\right] = -\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\mathcal {L}^{-1}_s\frac{\textrm{d}\mathcal {L}_s}{\textrm{d} s}\mathcal {L}^{-1}_s\left[ \overline{\mathscr {A}}_s\right] =-\int _{\mathbb {R}^3}\mathscr {B}_s\frac{\textrm{d}\mathcal {L}_s}{\textrm{d} s}[\mathscr {A}_s]. \end{aligned}$$

Notice that for any g(v)

$$\begin{aligned} \lim _{s\rightarrow 0}\frac{\textrm{d}\mathcal {L}_s}{\textrm{d} s}[g]= & {} -2\lim _{s\rightarrow 0}\frac{\textrm{d}}{\textrm{d} s}\left( \mu ^{-\frac{1}{2}}_s Q^{*}\left[ \mu _s,\mu ^{\frac{1}{2}}_sg\right] \right) \\= & {} -2\left\{ -\frac{u_1\cdot v}{2T}\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ,\mu ^{\frac{1}{2}} g\right] +\mu ^{-\frac{1}{2}} Q^{*}\left[ \frac{u_1\cdot v}{T}\mu ,\mu ^{\frac{1}{2}} g\right] \right. \\{} & {} \qquad \left. +\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ,\frac{u_1\cdot v}{2T}\mu ^{\frac{1}{2}} g\right] \right\} \\= & {} -\frac{u_1\cdot v}{2T}\mathcal {L}[g]-2\mu ^{-\frac{1}{2}} Q^{*}\left[ \frac{u_1\cdot v}{T}\mu ,\mu ^{\frac{1}{2}} g\right] +\mathcal {L}\left[ \frac{u_1\cdot v}{2T}g\right] . \end{aligned}$$

Hence, we have

$$\begin{aligned}{} & {} \lim _{s\rightarrow 0}\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\frac{\textrm{d}\mathcal {L}^{-1}_s}{\textrm{d} s}\left[ \overline{\mathscr {A}}_s\right] = -\lim _{s\rightarrow 0}\int _{\mathbb {R}^3}\mathscr {B}_s\frac{\textrm{d}\mathcal {L}_s}{\textrm{d} s}\left[ \mathscr {A}_s\right] \\{} & {} =-\int _{\mathbb {R}^3}\mathscr {B}\left( -\frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}}+2\int _{\mathbb {R}^3}\mathscr {B}\mu ^{-\frac{1}{2}} Q^{*}\left[ \frac{u_1\cdot v}{T}\mu ,\mu ^{\frac{1}{2}} \mathscr {A}\right] -\int _{\mathbb {R}^3}\mathscr {B}\mathcal {L}\left[ \frac{u_1\cdot v}{2T}\mathscr {A}\right] \nonumber \\{} & {} =\int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}} + 2\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \frac{u_1\cdot v}{T}\mu ^{\frac{1}{2}},\mathscr {A}\right] -\int _{\mathbb {R}^3}\overline{\mathscr {B}}\left( \frac{u_1\cdot v}{2T}\right) \mathscr {A}.\nonumber \end{aligned}$$

(2.24)

For the third term in (2.22), we have

$$\begin{aligned} \lim _{s\rightarrow 0}\int _{\mathbb {R}^3}\overline{\mathscr {B}}_s\mathcal {L}^{-1}_s\left[ \frac{\textrm{d}\overline{\mathscr {A}}_s}{\textrm{d} s} \right] =\lim _{s\rightarrow 0}\int _{\mathbb {R}^3}\mathscr {B}_s\frac{\textrm{d}\overline{\mathscr {A}}_s}{\textrm{d} s}=\int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}}-2\int _{\mathbb {R}^3}\mathscr {B}(u_1\cdot \overline{\mathscr {B}}). \end{aligned}$$

(2.25)

Inserting (2.23), (2.24) and (2.25) into (2.22), we have

$$\begin{aligned} \int _{\mathbb {R}^3}\overline{\mathscr {B}}\left( \frac{u_1\cdot v}{2T}\right) \mathscr {A}+ & {} \int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}}+2\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \frac{u_1\cdot v}{T}\mu ^{\frac{1}{2}},\mathscr {A}\right] \\- & {} \int _{\mathbb {R}^3}\overline{\mathscr {B}}\left( \frac{u_1\cdot v}{2T}\right) \mathscr {A} +\int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}}-2\int _{\mathbb {R}^3}\mathscr {B}(u_1\cdot \overline{\mathscr {B}})=0. \end{aligned}$$

Hence, we know that

$$\begin{aligned} \int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \frac{u_1\cdot v}{T}\mu ^{\frac{1}{2}},\mathscr {A}\right] =-\int _{\mathbb {R}^3}\mathscr {B}\left( \frac{u_1\cdot v}{2T}\right) \overline{\mathscr {A}}+\int _{\mathbb {R}^3}\mathscr {B}(u_1\cdot \overline{\mathscr {B}}). \end{aligned}$$

This verifies (2.21).$\square$

Lemma 2.4

We have the identity

$$\begin{aligned}{} & {} \Gamma \left[ \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) , \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \right] \nonumber \\{} & {} =-\mathcal {L}\left[ \mu \left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) ^2\right] . \end{aligned}$$

(2.26)

Proof

The proof can be found in [3, (60)]. A different derivation can be achieved by considering the expansion with respect to $\varepsilon$ in $Q[\mu _F,\mu _F]=0$ where

$$\begin{aligned} \mu _F= & {} \frac{\rho _F}{(2\pi T_F)^{\frac{3}{2}}}\exp \left( -\frac{|v-u_F|^2}{2T_F}\right) \\= & {} \frac{(\rho +\varepsilon \rho _1+\varepsilon ^2\rho _2)}{\big (2\pi (T_0+\varepsilon T_1+\varepsilon ^2T_2)\big )^{\frac{3}{2}}}\exp \left( -\frac{|v-(\varepsilon u_1+\varepsilon ^2u_2)|^2}{2(T_0+\varepsilon T_1+\varepsilon ^2T_2)}\right) . \end{aligned}$$

$\square$

Lemma 2.5

Equation (2.5) is equivalent to

$$\begin{aligned} -\frac{P}{T}\nabla _{x}\cdot \left( -\frac{2}{3}|u_1|^2\textbf{1}+2(u_1\otimes u_1)\right) +\nabla _{x} \mathfrak {p}+\nabla _{x}\cdot \left( \tau ^{(1)}-\tau ^{(2)}\right) =0, \end{aligned}$$

where

$$\begin{aligned} \tau ^{(1)}:= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{\frac{1}{2}}\left( v\cdot \frac{\nabla _{x} u_1}{T}\cdot v\right) \right\} ,\\ \tau ^{(2)}:= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ v\cdot \nabla _{x}^2T\cdot \frac{\mathscr {A}}{2T^2}\right\} +\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}\frac{\mathscr {A}}{2T^2}\right) \cdot \nabla _{x}T\right\} \\{} & {} +\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2},\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right] . \end{aligned}$$

Proof

Equation (2.5) is equivalent to

$$\begin{aligned} \int _{\mathbb {R}^3}v\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_2\right) \right) \textrm{d}v=0. \end{aligned}$$

(2.27)

Using (2.16), (2.27) can be rewritten as

$$\begin{aligned}{} & {} \nabla _{x}\cdot \left( -\int _{\mathbb {R}^3}v\otimes v\mu ^{\frac{1}{2}}\mathcal {L}^{-1}\left[ \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right] +\int _{\mathbb {R}^3}v\otimes v\mu ^{\frac{1}{2}}\mathcal {L}^{-1}[\Gamma [f_1,f_1]]\right. \nonumber \\{} & {} \qquad \left. +\int _{\mathbb {R}^3}v\otimes v\mu \left( \frac{\rho _2}{\rho }+\frac{u_2\cdot v}{T}+\frac{T_2(|v|^2-3T)}{2T^2}\right) \right) =0. \end{aligned}$$

(2.28)

First Term in 2.28 For the first term in (2.28), by orthogonality, since $\mathcal {L}^{-1}$ is self-adjoint, using (2.7), we have

$$\begin{aligned}{} & {} -\int _{\mathbb {R}^3}v\otimes v\mu ^{\frac{1}{2}}\mathcal {L}^{-1}\left[ \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right] \\{} & {} =-\int _{\mathbb {R}^3}\left( v\otimes v-\frac{|v|^2}{3}\textbf{1}\right) \mu ^{\frac{1}{2}}\mathcal {L}^{-1}\left[ \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right] \nonumber \\{} & {} =-\int _{\mathbb {R}^3}\mathcal {L}^{-1}\left[ \overline{\mathscr {B}}\right] \left( \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right) \nonumber \\{} & {} =-\int _{\mathbb {R}^3}\mathscr {B}\left( \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right) ,\nonumber \\{} & {} =-\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( -\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}+\mu \left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \right) \right\} .\nonumber \end{aligned}$$

(2.29)

Due to oddness, the $\rho _1$ and $T_1$ terms in (2.29) vanish. Hence, the first term in (2.28) is actually

$$\begin{aligned} -\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( -\mu ^{\frac{1}{2}}\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}+\mu \frac{ u_1\cdot v}{T}\right) \right\} =-\tau ^{(1)}+\widetilde{\tau }^{(2)}+\widetilde{\varsigma }, \end{aligned}$$

where

$$\begin{aligned} \tau ^{(1)}:= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{\frac{1}{2}}\left( v\cdot \frac{\nabla _{x} u_1}{T}\cdot v\right) \right\} ,\\ \widetilde{\tau }^{(2)}:= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ v\cdot \nabla _{x}^2T\cdot \frac{\mathscr {A}}{2T^2}\right\} +\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{-\frac{1}{2}}v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}\frac{\mathscr {A}}{2T^2}\right) \cdot \nabla _{x}T\right\} , \end{aligned}$$

and

$$\begin{aligned} \widetilde{\varsigma }:= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{\frac{1}{2}}\left( v\cdot \frac{\nabla _{x}T}{T^2}\right) (u_1\cdot v)\right\} -\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\mu \right) \left( \frac{u_1\cdot v}{T}\right) \right\} \\= & {} \int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{\frac{1}{2}}\left( v\cdot \frac{\nabla _{x}T}{T^2}\right) (u_1\cdot v)\right\} -\int _{\mathbb {R}^3}\mathscr {B}\left\{ \mu ^{\frac{1}{2}}\left( v\cdot \frac{\nabla _{x}T}{2T^3}\right) \left( |v|^2-5T\right) \left( u_1\cdot v\right) \right\} \\= & {} \frac{\nabla _{x}T}{T^2}\cdot \int _{\mathbb {R}^3}\mathscr {B}\cdot \left\{ (u_1\cdot v)v\mu ^{\frac{1}{2}}\right\} -\frac{\nabla _{x}T}{2T^3}\cdot \int _{\mathbb {R}^3}\mathscr {B}\cdot \left\{ \overline{\mathscr {A}}\left( u_1\cdot v\right) \right\} . \end{aligned}$$

Second Term of 2.28 For the second term of (2.28), we have

$$\begin{aligned}{} & {} \int _{\mathbb {R}^3}v\otimes v\mu ^{\frac{1}{2}}\mathcal {L}^{-1}\left[ \Gamma [f_1,f_1]\right] = \int _{\mathbb {R}^3}\bar{\mathscr {B}}\mathcal {L} ^{-1}\left[ \Gamma [f_1,f_1]\right] =\int _{\mathbb {R}^3}\mathscr {B}\Gamma [f_1,f_1]\\{} & {} =\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ -\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2},-\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right] \nonumber \\{} & {} \quad +2\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ -\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2},\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1|v|^2-3T)}{2T^2}\right) \right] \nonumber \\{} & {} \quad +\!\int _{\mathbb {R}^3}\mathscr {B}\Gamma \!\left[ \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }\!+\!\frac{u_1\cdot v}{T}\!+\!\frac{T_1(|v|^2-3T)}{2T^2}\right) ,\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }\!+\!\frac{u_1\cdot v}{T}\!+\!\frac{T_1(|v|^2-3T)}{2T^2}\right) \right] .\nonumber \end{aligned}$$

(2.30)

For the first term in (2.30), denote

$$\begin{aligned} \overline{\tau }^{(2)}:=\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ -\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2},-\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right] . \end{aligned}$$

Then denote

$$\begin{aligned} \tau ^{(2)}:=\widetilde{\tau }^{(2)}+\overline{\tau }^{(2)}. \end{aligned}$$

For the second term in (2.30), using identity (2.21), we obtain

$$\begin{aligned} \overline{\varsigma }:= & {} 2\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ -\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2},\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \right] \\= & {} -\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \mathscr {A}\cdot \frac{\nabla _{x}T}{T^2},\mu ^{\frac{1}{2}}\left( \frac{ u_1\cdot v}{T}\right) \right] = -\frac{\nabla _{x}T}{T^3}\cdot \int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \mathscr {A},\mu ^{\frac{1}{2}}(u_1\cdot v)\right] \\= & {} \frac{\nabla _{x}T}{2T^3}\cdot \int _{\mathbb {R}^3}\mathscr {B}\cdot \overline{\mathscr {A}}(u_1\cdot v)-\frac{\nabla _{x}T}{T^2}\cdot \int _{\mathbb {R}^3}\mathscr {B}\cdot (u_1\cdot \overline{\mathscr {B}}). \end{aligned}$$

Then we have

$$\begin{aligned} \widetilde{\varsigma }+\overline{\varsigma }=0. \end{aligned}$$

For the third term in (2.30), direct computation using (2.26) and oddness reveals that

$$\begin{aligned}{} & {} -\!\int _{\mathbb {R}^3}\mathscr {B}\Gamma \left[ \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) ,\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{ u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \right] \\{} & {} =\int _{\mathbb {R}^3}\overline{\mathscr {B}}\mathcal {L}^{-1}\left[ \mathcal {L}\left[ \mu ^{\frac{1}{2}}\frac{(u_1\cdot v)^2}{T^2}\right] \right] =-\frac{2P}{3T}|u_1|^2\textbf{1}+\frac{2P}{T}(u_1\otimes u_1). \end{aligned}$$

Third Term of 2.28 For the third term of (2.28), due to oddness, $u_2$ terms vanish, and thus we have

$$\begin{aligned} \int _{\mathbb {R}^3}v\otimes v\mu \left( \frac{\rho _2}{\rho }+\frac{ u_2\cdot v}{T}+\frac{T_2(|v|^2-3T)}{2T^2}\right) =(T\rho _2+\rho T_2)\textbf{1}. \end{aligned}$$

$\square$

2.1.4 Ghost-Effect Equations

Collecting all above and rearranging the terms, we have

$$\begin{aligned} \mu (x,v)=\frac{\rho (x)}{\big (2\pi T(x)\big )^{\frac{3}{2}}} \exp \left( -\frac{|v|^2}{2T(x)}\right) \end{aligned}$$

and

$$\begin{aligned} f_1= & {} -\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}+\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) ,\end{aligned}$$

(2.31)

$$\begin{aligned} f_2= & {} -\mathcal {L}^{-1}\left[ \mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_1\right) \right) \right] +\mathcal {L}^{-1}\left[ \Gamma [f_1,f_1]\right] \nonumber \\{} & {} +\mu ^{\frac{1}{2}}\left( \frac{\rho _2}{\rho }+\frac{u_2\cdot v}{T}+\frac{T_2\big (|v|^2-3T\big )}{2T^2}\right) , \end{aligned}$$

(2.32)

where $(\rho ,0,T)$, $(\rho _1,u_1,T_1)$ and $(\rho _2,u_2,T_2)$ satisfy

• Order 1 equation:

  $$\begin{aligned} \nabla _{x} P=\nabla _{x}(\rho T)=0. \end{aligned}$$

  (2.33)

• Order $\varepsilon$ system:

  $$\begin{aligned} \nabla _{x}\cdot (\rho u_1)= & {} 0,\end{aligned}$$

  (2.34)
  $$\begin{aligned} \nabla _{x} P_1=\nabla _{x}(T\rho _1+\rho T_1)= & {} 0,\nonumber \\ \nabla _{x}\cdot \left( \kappa \frac{\nabla _{x}T}{2T^2}\right)= & {} 5P(\nabla _{x}\cdot u_1). \end{aligned}$$

  (2.35)

• Order $\varepsilon ^2$ system:

  $$\begin{aligned} \rho (u_1\cdot \nabla _{x} u_1)+\nabla _{x} \mathfrak {p}=\nabla _{x}\cdot \left( \tau ^{(1)}-\tau ^{(2)}\right) . \end{aligned}$$

  (2.36)

Here $u_k=(u_{k,1},u_{k,2},u_{k,3})$,

$$\begin{aligned} P:= & {} \rho T,\quad P_1:=T\rho _1+\rho T_1,\quad \mathfrak {p}:=T\rho _2+\rho T_2,\\ \overline{\mathscr {A}}:= & {} v\cdot \left( |v|^2-5T\right) \mu ^{\frac{1}{2}},\quad \mathscr {A}:=\mathcal {L}^{-1}[\overline{\mathscr {A}}]=\mathcal {L}^{-1}\left[ v\cdot \left( |v|^2-5T\right) \mu ^{\frac{1}{2}}\right] ,\nonumber \\ \kappa \textbf{1}:= & {} \int _{\mathbb {R}^3}\mathscr {A}\otimes \overline{\mathscr {A}}\textrm{d} v,\nonumber \end{aligned}$$

(2.37)

and

$$\begin{aligned} \tau ^{(1)}:= & {} \lambda \left( \nabla _{x} u_1+(\nabla _{x} u_1)^t-\frac{2}{3}(\nabla _{x}\cdot u_1)\textbf{1}\right) ,\\ \tau ^{(2)}:= & {} \frac{\lambda ^2}{P}\left( K_1\left( \nabla _{x}^2T-\frac{1}{3}\Delta _x T\textbf{1}\right) +\frac{K_2}{T}\left( \nabla _{x}T\otimes \nabla _{x}T-\frac{1}{3}|\nabla _{x}T|^2\textbf{1}\right) \right) \end{aligned}$$

for smooth functions $\lambda [T]>0$, and positive constants $K_1$ and $K_2$ [5, 20, 26].

We observe that (2.33), (2.34), (2.35) and (2.36) are a set of equations sufficient to determine $(\rho , u_1,T, \nabla _{x}\mathfrak {p})$ uniquely once suitable boundary conditions are specified:

$$\begin{aligned} \left\{ \begin{array}{rcl} \nabla _{x} P=\nabla _{x}(\rho T)&{}=&{}0,\\ \rho ( u_1\cdot \nabla _{x} u_1)+\nabla _{x} \mathfrak {p}&{}=&{}\nabla _{x}\cdot \left( \tau ^{(1)}-\tau ^{(2)}\right) ,\\ \nabla _{x}\cdot (\rho u_1)&{}=&{}0,\\ \nabla _{x}\cdot \left( \kappa \dfrac{\nabla _{x}T}{2T^2}\right) &{}=&{}5P(\nabla _{x}\cdot u_1). \end{array}\right. \end{aligned}$$

Notice that $\mathfrak {p}$ enters in the equations only through its gradient so we are free to choose a definite value by imposing $\int _{\Omega }\mathfrak {p}=0$.

Also, we are left with an additional requirement:

$$\begin{aligned} \nabla _{x} P_1=\nabla _{x}(T\rho _1+\rho T_1)=0. \end{aligned}$$

(2.38)

The higher-order terms of the expansion will be discussed in Section 3.

2.2 Normal Chart Near Boundary

In order to define the boundary layer correction, we need to design a coordinate system based on the normal and tangential directions on the boundary surface. Our main goal is to rewrite the three-dimensional transport operator $v\cdot \nabla _{x}$ in this new coordinate system. This is basically textbook-level differential geometry, so we omit the details.

Substitution 1: Spatial Substitution: For a smooth manifold $\partial \Omega$, there exists an orthogonal curvilinear coordinates system $(\iota _1,\iota _2)$ such that the coordinate lines coincide with the principal directions at any $x_0\in \partial \Omega$ (at least locally).

Assume $\partial \Omega$ is parameterized by $\textbf{r}=\textbf{r}(\iota _1,\iota _2)$. Let $|\cdot |$ denote the length. Hence, $\partial _{\iota _1}\textbf{r}$ and $\partial _{\iota _2}\textbf{r}$ represent two orthogonal tangential vectors. Denote $L_i=|\partial _{\iota _i}\textbf{r}|$ for $i=1,2$. Then define the two orthogonal unit tangential vectors

$$\begin{aligned} \varsigma _1:=\frac{\partial _{\iota _1}\textbf{r}}{L_1},\qquad \varsigma _2:=\frac{\partial _{\iota _2}\textbf{r}}{L_2}. \end{aligned}$$

Also, the outward unit normal vector is

$$\begin{aligned} n:=\frac{\partial _{\iota _1}\textbf{r}\times \partial _{\iota _2}\textbf{r}}{|\partial _{\iota _1}\textbf{r}\times \partial _{\iota _2}\textbf{r}|}=\varsigma _1\times \varsigma _2. \end{aligned}$$

Obviously, $(\varsigma _1,\varsigma _2,n)$ forms a new orthogonal frame. Hence, consider the corresponding new coordinate system $(\iota _1,\iota _2,\mathfrak {n})$, where $\mathfrak {n}$ denotes the normal distance to boundary surface $\partial \Omega$, i.e.

$$\begin{aligned} x=\textbf{r}-\mathfrak {n}n. \end{aligned}$$

Note that $\mathfrak {n}=0$ means $x\in \partial \Omega$ and $\mathfrak {n}>0$ means $x\in \Omega$ (before reaching the other side of $\partial \Omega$). Using this new coordinate system and denoting $\kappa _i$ the principal curvatures, the transport operator becomes

$$\begin{aligned} v\cdot \nabla _{x}=-(v\cdot n)\frac{\partial }{\partial \mathfrak {n}}-\frac{v\cdot \varsigma _1}{L_1(\kappa _1\mathfrak {n}-1)}\frac{\partial }{\partial \iota _1}-\frac{v\cdot \varsigma _2}{L_2(\kappa _2\mathfrak {n}-1)}\frac{\partial }{\partial \iota _2}. \end{aligned}$$

Substitution 2: Velocity Substitution: Define the orthogonal velocity substitution for $\mathfrak {v}:=(v_{\eta },v_{\phi },v_{\psi })$ as

$$\begin{aligned} \left\{ \begin{array}{l} -v\cdot n := v_{\eta },\\ -v\cdot \varsigma _1:=v_{\phi },\\ -v\cdot \varsigma _2:= v_{\psi }. \end{array}\right. \end{aligned}$$

Then the transport operator becomes

$$\begin{aligned} v\cdot \nabla _{x}= & {} v_{\eta }\frac{\partial }{\partial \mathfrak {n}}-\frac{1}{R_1-\mathfrak {n}}\left( v_{\phi }^2\frac{\partial }{\partial v_{\eta }}-v_{\eta }v_{\phi }\frac{\partial }{\partial v_{\phi }}\right) - \frac{1}{R_2-\mathfrak {n}}\left( v_{\psi }^2\frac{\partial }{\partial v_{\eta }} - v_{\eta } v_{\psi }\frac{\partial }{\partial v_{\psi }}\right) \\{} & {} +\frac{1}{L_1L_2}\left( \frac{R_1\partial _{\iota _1\iota _1}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_1(R_1-\mathfrak {n})} v_{\phi } v_{\psi } + \frac{R_2\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_2(R_2-\mathfrak {n})} v_{\psi }^2\right) \frac{\partial }{\partial v_{\phi }}\\{} & {} +\frac{1}{L_1L_2}\left( R_2\frac{\partial _{\iota _2\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_2(R_2-\mathfrak {n})} v_{\phi } v_{\psi } + \frac{R_1\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_1(R_1-\mathfrak {n})} v_{\phi }^2\right) \frac{\partial }{\partial v_{\psi }}\\{} & {} +\left( \frac{R_1 v_{\phi }}{L_1(R_1-\mathfrak {n})}\frac{\partial }{\partial \iota _1}+\frac{R_2 v_{\psi }}{L_2(R_2-\mathfrak {n})}\frac{\partial }{\partial \iota _2}\right) , \end{aligned}$$

where $R_i=\kappa _i^{-1}$ represent the radii of principal curvature.

Substitution 3: Scaling Substitution: Finally, we define the scaled variable $\eta =\frac{\mathfrak {n}}{\varepsilon }$, which implies $\frac{\partial }{\partial \mathfrak {n}}=\frac{1}{\varepsilon }\frac{\partial }{\partial \eta }$. Then the transport operator becomes

$$\begin{aligned} v\cdot \nabla _{x}= & {} \frac{1}{\varepsilon } v_{\eta }\frac{\partial }{\partial \eta }-\frac{1}{R_1-\varepsilon \eta }\left( v_{\phi }^2\frac{\partial }{\partial v_{\eta }} - v_{\eta } v_{\phi }\frac{\partial }{\partial v_{\phi }}\right) - \frac{1}{R_2-\varepsilon \eta }\left( v_{\psi }^2\frac{\partial }{\partial v_{\eta }} - v_{\eta } v_{\psi }\frac{\partial }{\partial v_{\psi }}\right) \\{} & {} +\frac{1}{L_1L_2}\left( \frac{R_1\partial _{\iota _1\iota _1}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_1(R_1-\varepsilon \eta )} v_{\phi } v_{\psi } + \frac{R_2\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_2(R_2-\varepsilon \eta )} v_{\psi }^2\right) \frac{\partial }{\partial v_{\phi }}\\{} & {} +\frac{1}{L_1L_2}\left( R_2\frac{\partial _{\iota _2\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_2(R_2-\varepsilon \eta )} v_{\phi } v_{\psi } + \frac{R_1\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_1(R_1-\varepsilon \eta )} v_{\phi }^2\right) \frac{\partial }{\partial v_{\psi }}\\{} & {} +\left( \frac{R_1 v_{\phi }}{L_1(R_1-\varepsilon \eta )}\frac{\partial }{\partial \iota _1}+\frac{R_2 v_{\psi }}{L_2(R_2-\varepsilon \eta )}\frac{\partial }{\partial \iota _2}\right) . \end{aligned}$$

2.3 Milne Problem with Tangential Dependence

To construct the Hilbert expansion in a general domain, it is important to study the Milne problem depending on the tangential variable $(\iota _1,\iota _2)$. Notice that, in the new variables, $\mu _{w}=\mu _{w}(\iota _1,\iota _2,\mathfrak {v})$. Set

$$\begin{aligned} \mathcal {L}_w[f]:=-2\mu _w^{-\frac{1}{2}} Q^{*}\left[ \mu _w,\mu _w^{\frac{1}{2}} f\right] =\nu _w f-K_w[f]. \end{aligned}$$

Let $\Phi (\eta ,\iota _1,\iota _2,\mathfrak {v})$ be solution to the Milne problem

$$\begin{aligned} v_{\eta }\frac{\partial \Phi }{\partial \eta }+\nu _w \Phi -K_w[\Phi ]=0, \end{aligned}$$

(2.39)

with in-flow boundary condition at $\eta =0$

$$\begin{aligned} \Phi (0,\iota _1,\iota _2,\mathfrak {v})=\mathscr {A}\cdot \frac{\nabla _{x} T}{2T^2}\quad \text {for } v_{\eta }>0, \end{aligned}$$

(2.40)

and the zero mass-flux condition

$$\begin{aligned} \int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\Phi (0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}=0. \end{aligned}$$

(2.41)

Theorem 2.6

Assume that $\nabla _{x}T\in W^{k,\infty }(\partial \Omega )$ for some $k\in \mathbb {N}$ and $|T|_{L_{\partial \Omega }^\infty }\lesssim 1$. Then there exists

$$\begin{aligned} \Phi _{\infty }(\iota _1,\iota _2,\mathfrak {v}):=\Phi _{\infty }(\iota _1,\iota _2,v):=\mu _{w}^{\frac{1}{2}}\left( \frac{\rho ^B}{\rho _w}+\frac{ u^B\cdot v}{ T_{w}}+\frac{T^B(|v|^2-3 T_{w})}{2 T_{w}^2}\right) \in \mathcal {N}, \end{aligned}$$

(2.42)

for $\rho _w:=P T_{w}^{-1}$ and some $(\rho ^B(\iota _1,\iota _2), u^B(\iota _1,\iota _2),T^B(\iota _1,\iota _2))$ such that

$$\begin{aligned} \int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v})\Phi _{\infty }(\mathfrak {v})\textrm{d} \mathfrak {v}=0, \end{aligned}$$

and a unique solution $\Phi (\eta ,\iota _1,\iota _2,\mathfrak {v})$ to (2.39) such that $\widetilde{\Phi }:=\Phi -\Phi _{\infty }$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} v_{\eta }\dfrac{\partial \widetilde{\Phi }}{\partial \eta }+\nu _w \widetilde{\Phi }-K_w\left[ \widetilde{\Phi }\right] =0,\\ \widetilde{\Phi } (0,\iota _1,\iota _2,\mathfrak {v})=\Phi (0,\iota _1,\iota _2,\mathfrak {v}) - \Phi _{\infty }(\iota _1,\iota _2,\mathfrak {v})~\text { for } v_{\eta }>0,\\ \int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}=0 \end{array}\right. \end{aligned}$$

(2.43)

and for some $K_0>0$ and any $0< r \le k$

$$\begin{aligned} |\Phi _{\infty }| + \left\| \textrm{e}^{K_0\eta }\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}\lesssim & {} |\nabla _{x}T|_{L^\infty _{\partial \Omega }},\end{aligned}$$

(2.44)

$$\begin{aligned} \left\| \textrm{e}^{K_0\eta } v_{\eta }\partial _{\eta }\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}+\left\| \textrm{e}^{K_0\eta } v_{\eta }\partial _{v_{\eta }}\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}\lesssim & {} |\nabla _{x}T|_{L^\infty _{\partial \Omega }},\end{aligned}$$

(2.45)

$$\begin{aligned} \left\| \textrm{e}^{K_0\eta }\partial _{v_{\phi }}\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}+\left\| \textrm{e}^{K_0\eta }\partial _{v_{\psi }}\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}\lesssim & {} |\nabla _{x}T|_{L^{\infty }_{\partial \Omega }},\end{aligned}$$

(2.46)

$$\begin{aligned} \left\| \textrm{e}^{K_0\eta }\partial _{\iota _1}^{r}\widetilde{\Phi }\right\| _{L^{\infty }_{\varrho ,\vartheta }}+\left\| \textrm{e}^{K_0\eta }\partial _{\iota _2}^{r}\widetilde{\Phi }\right\|\lesssim & {} |\nabla _{x}T|_{L^\infty _{\partial \Omega }}+\sum _{j=1}^{r}\left| \partial _{\iota _1}^j\nabla _{x}T\right| _{L^\infty _{\partial \Omega }}\nonumber \\{} & {} +\sum _{j=1}^{r}\left| \partial _{\iota _2}^j\nabla _{x}T\right| _{L^\infty _{\partial \Omega }}. \end{aligned}$$

(2.47)

Proof

Based on [2] and [30], we have the well-posedness of (2.39). Also, estimates (2.44), (2.45) and (2.46) follow. Hence, we will focus on (2.47). Let $W:=\frac{\partial \widetilde{\Phi }}{\partial \iota _i}$ for $i=1,2$. Then W satisfies

$$\begin{aligned} \left\{ \begin{array}{l} v_{\eta }\dfrac{\partial W}{\partial \eta }+\nu _w W-K_w[W]=-\dfrac{\partial \nu _w}{\partial \iota _i}\widetilde{\Phi }+\dfrac{\partial K_w}{\partial \iota _i}[\widetilde{\Phi }],\\ W(0,\iota _1,\iota _2,\mathfrak {v})=-\dfrac{\partial }{\partial \iota _i}\left( \mathscr {A}\cdot \dfrac{\nabla _{x} T}{2T^2}\right) -\dfrac{\partial \widetilde{\Phi }_{\infty }}{\partial \iota _i}(\iota _1,\iota _2,\mathfrak {v})\quad \text { for }~ v_{\eta }>0,\\ \displaystyle \int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v})W(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}=-\displaystyle \int _{\mathbb {R}^3} v_{\eta }\frac{\partial \mu _{w}^{\frac{1}{2}}}{\partial \iota _i}(\iota _1,\iota _2,\mathfrak {v})\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}. \end{array}\right. \end{aligned}$$

Multiplying $|\mathfrak {v}|^2\mu _{w}^{\frac{1}{2}}$ on both sides of (2.43) and integrating over $\mathbb {R}^3$ yield

$$\begin{aligned} \int _{\mathbb {R}^3} v_{\eta }|\mathfrak {v}|^2\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d}\mathfrak {v}=\int _{\mathbb {R}^3} v_{\eta }|\mathfrak {v}|^2\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\widetilde{\Phi }(\infty ,\iota _1,\iota _2,\mathfrak {v})\textrm{d}\mathfrak {v}=0, \end{aligned}$$

which, combined with the zero mass-flux of $\widetilde{\Phi }$, further implies

$$\begin{aligned} \int _{\mathbb {R}^3} v_{\eta }\frac{\partial \mu _{w}^{\frac{1}{2}}}{\partial \iota _i}(\iota _1,\iota _2,\mathfrak {v})\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}=0. \end{aligned}$$

Hence, W still satisfies the zero mass-flux condition. Also, notice that

$$\begin{aligned} \left\| \textrm{e}^{K_0\eta }\left( -\frac{\partial \nu _w}{\partial \iota _i}\widetilde{\Phi }+\frac{\partial K_w}{\partial \iota _i}[\widetilde{\Phi }]\right) \right\| _{L^{\infty }_{\varrho ,\vartheta }}\lesssim |\nabla _{x}T|_{L^{\infty }_{\partial \Omega }}. \end{aligned}$$

Therefore, based on [2], there exists a unique $W_{\infty }\in \mathcal {N}$ such that

$$\begin{aligned} |W_{\infty }|+\left\| \textrm{e}^{K_0\eta }(W-W_{\infty })\right\| _{L^{\infty }_{\varrho ,\vartheta }}\lesssim |\nabla _{x}T|_{L^{\infty }_{\partial \Omega }}+|\partial _{\iota _i} \nabla _{x}T|_{L^{\infty }_{\partial \Omega }}. \end{aligned}$$

In particular, since $\widetilde{\Phi }\rightarrow 0$ as $\eta \rightarrow \infty$, we must have $W_{\infty }=0$. Hence, (2.47) is verified for $r=1$. The $r>1$ cases follow inductively.$\square$

Let $\chi (y)\in C^{\infty }(\mathbb {R})$ and $\overline{\chi }(y)=1-\chi (y)$ be smooth cut-off functions satisfying

$$\begin{aligned} \chi (y)=\left\{ \begin{array}{ll} 1&{}\quad \text { if }|y|\le 1,\\ 0&{}\quad \text { if }|y|\ge 2, \end{array}\right. \end{aligned}$$

In view of the later regularity estimates (see the companion paper [12]), we define a cutoff boundary layer $f^{B}_1$:

$$\begin{aligned} f^{B}_1(\eta ,\iota _1,\iota _2,\mathfrak {v}):=\overline{\chi }(\varepsilon ^{-1} v_{\eta })\chi (\varepsilon \eta )\widetilde{\Phi } (\eta ,\iota _1,\iota _2,\mathfrak {v}). \end{aligned}$$

(2.48)

We can verify that $f^{B}_1$ satisfies

$$\begin{aligned}{} & {} v_{\eta }\frac{\partial f^{B}_1}{\partial \eta }+\nu _w f^{B}_1-K_w\left[ f^{B}_1\right] \\{} & {} = v_{\eta }\overline{\chi }(\varepsilon ^{-1} v_{\eta })\frac{\partial \chi (\varepsilon \eta )}{\partial \eta }\widetilde{\Phi }+\chi (\varepsilon \eta )\left( \overline{\chi }\left( \varepsilon ^{-1} v_{\eta }\right) K_w[\widetilde{\Phi }]-K_w[\overline{\chi }(\varepsilon ^{-1} v_{\eta })\widetilde{\Phi }]\right) , \end{aligned}$$

with

$$\begin{aligned} f^{B}_1(0,\iota _1,\iota _2,\mathfrak {v})=\overline{\chi }(\varepsilon ^{-1} v_{\eta })\left( -\mathscr {A}\cdot \frac{\nabla _{x} T}{2T^2}-\Phi _{\infty }(\iota _1,\iota _2,\mathfrak {v})\right) \quad \text { for } v_{\eta }>0. \end{aligned}$$

Due to the cutoff $\overline{\chi }$, $f^{B}_1$ cannot preserve the zero mass-flux condition, i.e.

$$\begin{aligned}{} & {} \int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v}) f^{B}_1(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}\nonumber \\{} & {} =\int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\overline{\chi }(\varepsilon ^{-1} v_{\eta })\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}\nonumber \\{} & {} =\int _{\mathbb {R}^3} v_{\eta }\mu _{w}^{\frac{1}{2}}(\iota _1,\iota _2,\mathfrak {v})\chi (\varepsilon ^{-1} v_{\eta })\widetilde{\Phi }(0,\iota _1,\iota _2,\mathfrak {v})\textrm{d} \mathfrak {v}\lesssim o_T\varepsilon . \end{aligned}$$

(2.49)

The zero mass-flux condition will be restored with the help of $f_2$ in (3.11).

2.4 Analysis of Boundary Matching

Considering the boundary condition in (1.1) and the expansion (0.4), we require the matching condition for $x_0\in \partial \Omega$ and $v\cdot n<0$:

$$\begin{aligned} \mu _{w}\big |_{v\cdot n<0}= & {} M_{w}\int _{v'\cdot n>0} \mu _{w}|v'\cdot n|\textrm{d}{v'},\end{aligned}$$

(2.50)

$$\begin{aligned} \mu _{w}^{\frac{1}{2}}\left( f_1+ f^{B}_1\right) \big |_{v\cdot n<0}= & {} M_{w}\int _{v'\cdot n>0} \mu _{w}^{\frac{1}{2}}\left( f_1+ f^{B}_1\right) |v'\cdot n|\textrm{d}{v'}+O(\varepsilon ). \end{aligned}$$

(2.51)

In order to guarantee (2.50), we deduce that

$$\begin{aligned} T(x_0)= T_{w}(x_0). \end{aligned}$$

(2.52)

This determines the boundary conditions for T.

In order to guarantee (2.51), due to (2.49), it suffices to require that at $\eta =0$

$$\begin{aligned} \mu _{w}^{\frac{1}{2}}(f_1+\Phi -\Phi _{\infty })\big |_{v\cdot n<0}=M_{w}\int _{v'\cdot n>0} \mu _{w}^{\frac{1}{2}}(f_1+\Phi -\Phi _{\infty })|v'\cdot n|\textrm{d}{v'}. \end{aligned}$$

(2.53)

Lemma 2.7

With the boundary condition (2.40) for (2.39), and for $x_0\in \partial \Omega$

$$\begin{aligned} u_1(x_0)=u^B,\quad T_1(x_0)=T^B, \end{aligned}$$

(2.54)

(2.53) is valid.

Proof

Using (2.31) and (2.42), we have for $x_0\in \partial \Omega$

$$\begin{aligned} f_1+\Phi -\Phi _{\infty }= & {} \mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}+\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }+\frac{u_1\cdot v}{T}+\frac{T_1(|v|^2-3T)}{2T^2}\right) \\{} & {} +\Phi -\mu ^{\frac{1}{2}}\left( \frac{\rho ^B}{\rho }+\frac{u^B\cdot v}{T}+\frac{T^B(|v|^2-3 T_{w})}{2T^2}\right) . \end{aligned}$$

With (2.54), we have

$$\begin{aligned} f_1+\Phi -\Phi _{\infty }=\left( \Phi +\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right) +\mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }-\frac{\rho ^B}{\rho }\right) . \end{aligned}$$

Since direct computation reveals that

$$\begin{aligned} \left. \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }-\frac{\rho ^B}{\rho }\right) \right| _{v\cdot n<0}=M_{w}\int _{v'\cdot n>0} \mu ^{\frac{1}{2}}\left( \frac{\rho _1}{\rho }-\frac{\rho ^B}{\rho }\right) |v'\cdot n|\textrm{d}{v'}, \end{aligned}$$

in order to verify (2.53), it suffices to require

$$\begin{aligned} \left. \left( \Phi +\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right) \right| _{v\cdot n<0}=M_{w}\int _{v'\cdot n>0} \left( \Phi +\mathscr {A}\cdot \frac{\nabla _{x}T}{2T^2}\right) |v'\cdot n|\textrm{d}{v'}. \end{aligned}$$

(2.55)

When (2.40) is valid, we know that

$$\begin{aligned} \left. \left( \Phi +\mathscr {A}\cdot \frac{\nabla _{x} T}{2T^2}\right) \right| _{v\cdot n<0}=0. \end{aligned}$$

(2.56)

Also, due to (2.41) and orthogonality of $\mathscr {A}$, we have

$$\begin{aligned} M_{w}\int _{\mathbb {R}^3} \mu ^{\frac{1}{2}}\Phi (v'\cdot n)\textrm{d}{v'}=M_{w}\int _{\mathbb {R}^3} \mu ^{\frac{1}{2}}\left( \mathscr {A}\cdot \frac{\nabla _{x} T}{2T^2}\right) (v'\cdot n)\textrm{d}{v'}=0, \end{aligned}$$

which, combined with (2.56), yields

$$\begin{aligned} M_{w}\int _{v'\cdot n>0} \mu ^{\frac{1}{2}}\left( \Phi +\mathscr {A}\cdot \frac{\nabla _{x} T}{2T^2}\right) |v'\cdot n|\textrm{d}{v'}=0. \end{aligned}$$

Then clearly (2.55) is true.$\square$

3 Construction of Expansion

In this section, we will present the detailed construction of ghost-effect solution, $f_1$, $f_2$ and $f^{B}_1$ based on the analysis in Section 2.4. Since the boundary conditions are tangled together, we divide the construction into several stages.

3.1 Construction of Boundary Layer $f^{B}_1$-Stage I

Since (2.40) involves $\nabla _{n}T$, which is not fully provided by $T_{w}$, we will have to split the tangential and normal parts of the boundary layer

$$\begin{aligned} f^{B}_1= f^{B}_{1,\iota _1}+ f^{B}_{1,\iota _2}+ f^{B}_{1,n},\quad \Phi =\Phi _{\iota _1}+\Phi _{\iota _2}+\Phi _{n}. \end{aligned}$$

Define

$$\begin{aligned} \Phi _{\iota _i}:=\left( \partial _{\iota _i} T_{w}\right) \mathcal {H}^{(i)}, \end{aligned}$$

where $\mathcal {H}^{(i)}$ for $i=1,2$ solves the Milne problem

$$\begin{aligned} \left\{ \begin{array}{l} v_{\eta }\dfrac{\partial \mathcal {H}^{(i)}}{\partial \eta }+\mathcal {L}_w\big [\mathcal {H}^{(i)}\big ]=0,\\ \mathcal {H}^{(i)}(0,\mathfrak {v})=-\dfrac{\mathscr {A}\cdot \varsigma _i}{2T^2}\quad \text { for } v_{\eta }>0,\\ \lim _{\eta \rightarrow \infty }\mathcal {H}^{(i)}(\eta ,\mathfrak {v})=\mathcal {H}^{(i)}_{\infty }\in \mathcal {N}, \end{array}\right. \end{aligned}$$

(3.1)

with the zero mass-flux condition

$$\begin{aligned} \int _{\mathbb {R}^3}v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v})\mathcal {H}^{(i)}=0. \end{aligned}$$

Denote

$$\begin{aligned} \Phi _{\iota _i,\infty }:=(\partial _{\iota _i} T_{w})\mathcal {H}^{(i)}_{\infty }. \end{aligned}$$

Since we lack the information of $\Phi _n$ at this stage, we are not able to determine the boundary condition $T_1=T^B$ yet. However, we can fully determine the boundary condition $u_1=u^B$. Denote $u_1=(u_{1,\iota _1},u_{1,\iota _2},u_{1,n})$ for the two tangential components $(u_{1,\iota _1},u_{1,\iota _2})$ and one normal component $u_{1,n}$. Due to (2.41), we have

$$\begin{aligned} u_{1,n}(x_0)=0. \end{aligned}$$

(3.2)

Due to oddness, the projection of $\mathcal {H}^{(i)}$ and $\mathcal {H}^{(i)}_{\infty }$ on $v\mu _{w}^{\frac{1}{2}}$ only has contribution on $(v\cdot \varsigma _i)\mu ^{\frac{1}{2}}$. Hence, from (2.54), we deduce

$$\begin{aligned} u_{1,\iota _1}(x_0)=\beta _{1}[ T_{w}]\partial _{\iota _1}T_w,\quad u_{1,\iota _2}(x_0)=\beta _{2}[ T_{w}]\partial _{\iota _2}T_w, \end{aligned}$$

where $\beta _{i}$ are functions depending on $T_{w}$. Due to isotropy, we know that $\beta _1=\beta _2$, and we denote it $\beta _0$. Hence, we arrive at

$$\begin{aligned} u_{1,\iota _1}(x_0)=\beta _0[T_{w}]\partial _{\iota _1}T_w,\quad u_{1,\iota _2}(x_0)=\beta _0[T_{w}]\partial _{\iota _2}T_w. \end{aligned}$$

(3.3)

Lemma 3.1

Under the assumption (1.4), for any $s\ge 1$, the boundary data of $u_1$ satisfies

$$\begin{aligned} |u_1|_{W^{3,\infty }}\lesssim o_T. \end{aligned}$$

Proof

Taking $\iota _i$ derivatives for $i=1,2$ on both sides of (3.1), using (1.4) and (2.44), we conclude that

$$\begin{aligned} |u_{1,\iota _1}|_{W^{3,\infty }}+|u_{1,\iota _2}|_{W^{3,\infty }}\lesssim o_T. \end{aligned}$$

Then using (3.2), we obtain the desired estimates.$\square$

Remark 3.2

Note that the boundary condition of $u_1$ only depends on $T_{w}$ and $\nabla T_{w}$ directly without referring to T in the bulk.

3.2 Well-Posedness of Ghost-Effect Equation

Based on our analysis above, the ghost-effect equation (0.2) will be accompanied with the boundary conditions (2.52), (3.2) and (3.3).

$$\begin{aligned} T(x_0)= T_{w},\quad u_{1,\iota _1}(x_0)=\beta _0[T_{w}]\partial _{\iota _1}T_w,\quad u_{1,\iota _2}(x_0)=\beta _0[T_{w}]\partial _{\iota _2}T_w,\quad u_{1,n}(x_0)=0. \end{aligned}$$

(3.4)

Theorem 3.3

Under the assumption (1.4), for any given $P>0$, there exists a unique solution $(\rho , u_1,T; \mathfrak {p})$ ($\mathfrak {p}$ has zero average) to the ghost-effect (0.2) with the boundary condition (3.4) satisfying for any $s\in [2,\infty )$

$$\begin{aligned} \Vert u_1\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}+\Vert T-1\Vert _{W^{4,s}}\lesssim o_T. \end{aligned}$$

Proof

Simplified Equations Denote $\overline{u}:=\rho u_1$. From the first and third equations in (0.2)

$$\begin{aligned} \nabla _{x}\cdot \overline{u}=\nabla _{x}\cdot (\rho u_1)=P\nabla _{x}\cdot \left( \frac{ u_1}{T}\right) =0, \end{aligned}$$

we have

$$\begin{aligned} \nabla _{x}\cdot u_1= u_1\cdot \frac{\nabla _{x}T}{T}. \end{aligned}$$

(3.5)

From the second equation in (0.2) and (3.5), we have

$$\begin{aligned}{} & {} -\frac{5}{3}\lambda [1]\Delta _xu_1+\nabla _{x}\mathfrak {p}=-\frac{5}{3}\left( \lambda [1]-\lambda [T]\right) \Delta _xu_1\\{} & {} -\nabla _{x}\cdot \left( \frac{\lambda ^2[T]}{P}\left( K_1[T]\left( \nabla _{x}^2T-\frac{1}{3}\Delta _x T\textbf{1}\right) +\frac{K_2[T]}{T}\left( \nabla _{x} T\otimes \nabla _{x} T-\frac{1}{3}|\nabla _{x} T|^2\textbf{1}\right) \right) \right) \nonumber \\{} & {} +\nabla _{x}\lambda [T]\cdot \left( \nabla _{x} u_1+(\nabla _{x} u_1)^t-\frac{2}{3}(\nabla _{x}\cdot u_1)\textbf{1}\right) +\lambda [T]\nabla _{x}\left( u_1\cdot \frac{\nabla _{x}T}{T}\right) -\frac{P}{T}u_1\cdot \nabla _{x} u_1.\nonumber \end{aligned}$$

(3.6)

Hence, we know

$$\begin{aligned}{} & {} -\frac{5}{3P}\lambda [1]\Delta _x\overline{u}+\nabla _{x} \mathfrak {p}=-\frac{5}{3P}\big (\lambda [1]-\lambda [T]\big )\Delta _x\overline{u}+\frac{5}{3P}\lambda [T]\Delta _x\big ((T-1)\overline{u}\big )\\{} & {} -\nabla _{x}\cdot \left( \frac{\lambda ^2[T]}{P}\left( K_1[T]\left( \nabla _{x}^2T-\frac{1}{3}\Delta _x T\textbf{1}\right) +\frac{K_2[T]}{T}\left( \nabla _{x} T\otimes \nabla _{x} T-\frac{1}{3}|\nabla _{x} T|^2\textbf{1}\right) \right) \right) \nonumber \\{} & {} +\nabla _{x}\lambda [T]\cdot \left( \nabla _{x}\left( P^{-1}T\overline{u}\right) +\big (\nabla _{x}(P^{-1}T\overline{u})\big )^t-\frac{2}{3}\big (\nabla _{x}\cdot (P^{-1}T\overline{u})\big )\textbf{1}\right) \nonumber \\{} & {} +\lambda [T]\nabla _{x}\left( P^{-1}\overline{u}\cdot \nabla _{x}T\right) -\overline{u}\cdot \nabla _{x}\left( P^{-1}T\overline{u}\right) .\nonumber \end{aligned}$$

Furthermore, from the fourth equations in (0.2)

$$\begin{aligned} \nabla _{x}\cdot u_1=\frac{1}{5P}\nabla _{x}\cdot \left( \kappa \frac{\nabla _{x}T}{2T^2}\right) =\frac{1}{5P}\frac{\kappa }{2T^2}\Delta _xT+\frac{1}{5P}\nabla _{x}\left( \frac{\kappa }{2T^2}\right) \cdot \nabla _{x}T, \end{aligned}$$

we have

$$\begin{aligned} \frac{1}{5P}\frac{\kappa }{2T^2}\Delta _xT=P^{-1}\overline{u}\cdot \nabla _{x}T-\frac{1}{5P}\nabla _{x}\left( \frac{\kappa }{2T^2}\right) \cdot \nabla _{x}T. \end{aligned}$$

Then we know

$$\begin{aligned} \Delta _xT=\frac{10T^2}{\kappa [T]}\left( \overline{u}\cdot \nabla _{x}T\right) -\frac{2T^2}{\kappa [T]}\nabla _{x}\left( \frac{\kappa [T]}{2T^2}\right) \cdot \nabla _{x}T. \end{aligned}$$

(3.7)

Setup of Contraction Mapping Collecting (3.5), (3.6) and (3.7), this is a system for the pair $(\overline{u},T)$. Then we can design a mapping $W^{3,s}\times W^{4,s}\rightarrow W^{3,s}\times W^{4,s}:(\widetilde{u},\widetilde{T})\rightarrow (\overline{u},T)$

$$\begin{aligned} \left\{ \begin{array}{rcl} -\frac{5}{3P}\lambda [1]\Delta _x\overline{u}+\nabla _{x}\mathfrak {p}&{}=&{}Z_1,\\ \nabla _{x}\cdot \overline{u}&{}=&{}0,\\ \Delta _xT&{}=&{}Z_3, \end{array}\right. \end{aligned}$$

where

$$\begin{aligned} Z_1:= & {} -\frac{5}{3P}\left( \lambda [1]-\lambda [\widetilde{T}]\right) \Delta _x\widetilde{u}+\frac{5}{3P}\lambda [\widetilde{T}]\Delta _x\big ((\widetilde{T}-1)\widetilde{u}\big )\\{} & {} -\nabla _{x}\cdot \left( \frac{\lambda ^2[\widetilde{T}]}{P}\left( K_1[\widetilde{T}]\left( \nabla _{x}^2\widetilde{T}-\frac{1}{3}\Delta _x\widetilde{T}\textbf{1}\right) \!+\!\frac{K_2}{\widetilde{T}}[\widetilde{T}]\left( \nabla _{x} \widetilde{T}\otimes \nabla _{x} \widetilde{T}\!-\!\frac{1}{3}|\nabla _{x}\widetilde{T}|^2\textbf{1}\!\right) \!\right) \!\right) \nonumber \\{} & {} +\nabla _{x}\lambda [\widetilde{T}]\cdot \left( \nabla _{x} \big (P^{-1}\widetilde{T}\widetilde{u}\big )+\big (\nabla _{x} \big (P^{-1}\widetilde{T}\widetilde{u}\big )\big )^t-\frac{2}{3}\left( \nabla _{x}\cdot \big (P^{-1}\widetilde{T}\widetilde{u}\big )\right) \textbf{1}\right) \nonumber \\{} & {} +\lambda [\widetilde{T}]\nabla _{x}\left( P^{-1}\widetilde{u}\cdot \nabla _{x}\widetilde{T}\right) -\widetilde{u}\cdot \nabla _{x}\big (P^{-1}\widetilde{T}\widetilde{u}\big ),\nonumber \\ Z_3:= & {} \frac{10\widetilde{T}^2}{\kappa [\widetilde{T}]}\big (\widetilde{u}\cdot \nabla _{x}\widetilde{T}\big )-\frac{2\widetilde{T}^2}{\kappa [\widetilde{T}]}\nabla _{x}\left( \frac{\kappa [\widetilde{T}]}{2\widetilde{T}^2}\right) \cdot \nabla _{x}\widetilde{T}. \end{aligned}$$

Boundedness and Contraction Based on [9] and [6, Theorem IV.5.8], noticing the compatibility condition

$$\begin{aligned} \int _{\partial \Omega }\overline{u}\cdot n=\int _{\Omega }(\nabla _{x}\cdot \overline{u})=0, \end{aligned}$$

we know that

$$\begin{aligned} \Vert \overline{u}\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}\lesssim \Vert Z_1\Vert _{W^{1,s}}+|\overline{u}|_{W^{3-\frac{1}{s},s}}. \end{aligned}$$

Based on standard elliptic estimates [21], we have

$$\begin{aligned} \Vert T-1\Vert _{W^{4,s}}\lesssim \Vert Z_3\Vert _{W^{2,s}}+|T|_{W^{4-\frac{1}{s},s}}. \end{aligned}$$

Under the assumption

$$\begin{aligned} \Vert \widetilde{u}_1\Vert _{W^{3,s}}+\Vert \widetilde{T}-1\Vert _{W^{4,s}}\lesssim 2 o_T, \end{aligned}$$

we directly obtain

$$\begin{aligned} \Vert Z_1\Vert _{W^{1,s}}\lesssim & {} o_T\left( \Vert \widetilde{u}\Vert _{W^{3,s}}+\Vert \nabla _{x}\widetilde{T}\Vert _{W^{3,s}}\right) ,\\ \Vert Z_3\Vert _{W^{2,s}}\lesssim & {} o_T\left( \Vert \widetilde{u}\Vert _{W^{3,s}}+\Vert \nabla _{x}\widetilde{T}\Vert _{W^{3,s}}\right) . \end{aligned}$$

Hence, we know that

$$\begin{aligned} \Vert \overline{u}\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}+\Vert T-1\Vert _{W^{4,s}}\lesssim o_T\left( \Vert \widetilde{u}\Vert _{W^{3,s}}+\Vert \nabla _{x}\widetilde{T}\Vert _{W^{3,s}}\right) +|\nabla T_w|_{W^{3,\infty }} \le 2 o_T. \end{aligned}$$

Hence, this mapping is bounded.

By a similar argument, for $(\widetilde{u}^{[k]},\widetilde{T}^{[k]})\rightarrow (\overline{u}^{[k]},T^{[k]})$ with $k=1,2$, we can show that

$$\begin{aligned}{} & {} \left\| \overline{u}^{[1]}-\overline{u}^{[2]}\right\| _{W^{3,s}}+\left\| \mathfrak {p}^{[1]}-\mathfrak {p}^{[2]}\right\| _{W^{2,s}}+\left\| T^{[1]}-T^{[2]}\right\| _{W^{4,s}}\\{} & {} \lesssim \left( \Vert \widetilde{u}\Vert _{W^{3,s}}+\Vert \nabla _{x}\widetilde{T}\Vert _{W^{3,s}}\right) \left( \left\| \widetilde{u}^{[1]}-\widetilde{u}^{[2]}\right\| _{W^{3,s}}+\left\| \nabla _{x}\widetilde{T}^{[1]}-\nabla _{x}\widetilde{T}^{[2]}\right\| _{W^{3,s}}\right) , \end{aligned}$$

which yields

$$\begin{aligned}{} & {} \left\| \overline{u}^{[1]}-\overline{u}^{[2]}\right\| _{W^{3,s}}+\left\| \mathfrak {p}^{[1]}-\mathfrak {p}^{[2]}\right\| _{W^{2,s}}+\left\| T^{[1]}-T^{[2]}\right\| _{W^{4,s}}\\{} & {} \lesssim o_T\left( \left\| \widetilde{u}^{[1]}-\widetilde{u}^{[2]}\right\| _{W^{3,s}}+\left\| \nabla _{x}\widetilde{T}^{[1]}-\nabla _{x}\widetilde{T}^{[2]}\right\| _{W^{3,s}}\right) . \end{aligned}$$

Hence, this is a contraction mapping.

In summary, we know that there exists a unique solution to (0.2) satisfying

$$\begin{aligned} \Vert \overline{u}\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}+\Vert T-1\Vert _{W^{4,s}}\lesssim o_T, \end{aligned}$$

and further

$$\begin{aligned} \Vert u_1\Vert _{W^{3,s}}+\Vert \mathfrak {p}\Vert _{W^{2,s}}+\Vert T-1\Vert _{W^{4,s}}\lesssim o_T. \end{aligned}$$

$\square$

Remark 3.4

Based on the first equation in (2.37), we have

$$\begin{aligned} \rho =PT^{-1}\in W^{4,s}. \end{aligned}$$

Then we have

$$\begin{aligned} P|\Omega |=\int _{\Omega }\rho (x)T(x)\textrm{d} x=\frac{1}{3}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu (x,v)\textrm{d} v\textrm{d} x. \end{aligned}$$

(3.8)

3.3 Construction of Boundary Layer $f^{B}_1$-Stage II

Now we can define the full boundary layer. Define

$$\begin{aligned} \Phi _{n}:=(\partial _{n}T)\mathcal {H}^{(n)}, \end{aligned}$$

where $\mathcal {H}^{(n)}$ solves the Milne problem

$$\begin{aligned} \left\{ \begin{array}{l} v_{\eta }\dfrac{\partial \mathcal {H}^{(n)}}{\partial \eta }+\mathcal {L}_w\left[ \mathcal {H}^{(n)}\right] =0,\\ \mathcal {H}^{(n)}(0,\mathfrak {v})=-\dfrac{\mathscr {A}\cdot n}{2T^2}\quad \text { for } v_{\eta }>0,\\ \lim _{\eta \rightarrow \infty }\mathcal {H}^{(n)}(\eta ,\mathfrak {v})=\mathcal {H}^{(n)}_{\infty }\in \mathcal {N}, \end{array}\right. \end{aligned}$$

with the zero mass-flux condition

$$\begin{aligned} \int _{\mathbb {R}^3}v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v})\mathcal {H}^{(n)}=0. \end{aligned}$$

Denote

$$\begin{aligned} \Phi _{n,\infty }:=(\partial _{n}T)\mathcal {H}^{(n)}_{\infty }. \end{aligned}$$

Here $\partial _nT$ comes from the ghost-effect equation (0.2) and is well-defined due to Theorem 3.3.

Finally, we have the full boundary layer from (2.48):

$$\begin{aligned} f^{B}_1(\eta ,\mathfrak {v})\!=\! & {} \overline{\chi }(\varepsilon ^{-1}v_{\eta })\chi (\varepsilon \eta )\left( \Phi _{\iota _1}(\eta ,\mathfrak {v})+\Phi _{\iota _2}(\eta ,\mathfrak {v})+\Phi _{n}(\eta ,\mathfrak {v})-\Phi _{\iota _1,\infty }-\Phi _{\iota _2,\infty }-\Phi _{n,\infty }\right) \\\!=\! & {} \overline{\chi }(\varepsilon ^{-1}v_{\eta })\chi (\varepsilon \eta )\left( \widetilde{\Phi }_{\iota _1}(\eta ,\mathfrak {v})+\widetilde{\Phi }_{\iota _2}(\eta ,\mathfrak {v})+\widetilde{\Phi }_{n}(\eta ,\mathfrak {v})\right) . \end{aligned}$$

Since the cutoff in $f^{B}_1$ is only defined in the normal direction, we can deduce tangential regularity estimates from Theorem 2.6:

Theorem 3.5

Under the assumption (1.4), we can construct $f^{B}_1$ such that for $i=1,2$, some $K_0>0$ and any $0<r\le 3$

$$\begin{aligned} \Vert \textrm{e}^{K_0\eta } f^{B}_1\Vert +\left\| \textrm{e}^{K_0\eta }\frac{\partial ^{r} f^{B}_1}{\partial \iota _i^{r}}\right\| \lesssim o_T. \end{aligned}$$

(3.9)

From (2.54) and (2.42), this fully determines the boundary condition of $T_1$:

$$\begin{aligned} T_1(x_0)=T^B. \end{aligned}$$

3.4 Construction of $(\rho _1,T_1)$

Theorem 3.6

Under the assumption (1.4), we can construct $(\rho _1,T_1)$ such that for any $s\in [2,\infty )$

$$\begin{aligned} \Vert f_1\Vert _{W^{3,s}L^{\infty }_{\varrho ,\vartheta }}+|f_1|_{W^{3-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim o_T. \end{aligned}$$

Proof

The boundary condition in (2.54) and Theorem 3.5 imply that

$$\begin{aligned} |T_1|_{W^{3,s}}\lesssim o_T. \end{aligned}$$

Then we can freely define a Sobolev extension for $T_1$ such that

$$\begin{aligned} \Vert T_1\Vert _{W^{3+\frac{1}{s},s}}\lesssim o_T. \end{aligned}$$

We choose the constant

$$\begin{aligned} P_1=0. \end{aligned}$$

Then we can deduce that

$$\begin{aligned}{} & {} \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\left( \mu ^{\frac{1}{2}}f_1+\mu _{w}^{\frac{1}{2}} f^{B}_1+\varepsilon \mu ^{\frac{1}{2}}f_2(x,v)\right) \textrm{d} x\textrm{d} v\\{} & {} =\int _{\Omega }\big (3\rho _1(x)T(x)+3T_1(x)\rho (x)+3\varepsilon \rho _2(x)T(x)+3\varepsilon \rho (x)T_2(x)\big )\textrm{d} x\nonumber \\{} & {} \quad +\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu _{w}^{\frac{1}{2}} f^{B}_1\textrm{d} x\textrm{d} v\nonumber \\{} & {} =\int _{\Omega }(3\rho _1(x)T(x)+3T_1(x)\rho (x))\textrm{d} x+\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu _{w}^{\frac{1}{2}} f^{B}_1\textrm{d} x\textrm{d} v\nonumber \\{} & {} =\int _{\Omega }3P_1\textrm{d} x+\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu _{w}^{\frac{1}{2}} f^{B}_1\textrm{d} x\textrm{d} v=\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu _{w}^{\frac{1}{2}} f^{B}_1\textrm{d} x\textrm{d} v,\nonumber \end{aligned}$$

(3.10)

where we have used $\int _{\Omega }\mathfrak {p}=\int _{\Omega }(T\rho _2+\rho T_2)=0$.

Then based on (2.38), we have

$$\begin{aligned} \rho _1=-T^{-1}(\rho T_1), \end{aligned}$$

and thus

$$\begin{aligned} \Vert \rho _1\Vert _{W^{3+\frac{1}{s},s}}\lesssim o_T. \end{aligned}$$

Note that $\rho _1$ is not necessarily equal to $\rho ^B$ on $\partial \Omega$. However, (2.53) can still hold due to (2.50).

Hence, we have shown that

$$\begin{aligned} \Vert f_1\Vert _{W^{3,s}L^{\infty }_{\varrho ,\vartheta }}+|f_1|_{W^{3-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim \Vert \rho _1\Vert _{W^{3,s}}+\Vert u_1\Vert _{W^{3,s}}+\Vert T_1\Vert _{W^{3,s}}+\Vert T\Vert _{W^{3,s}} \lesssim o_T. \end{aligned}$$

$\square$

Remark 3.7

We assume that the remainder R satisfies

$$\begin{aligned} \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu ^{\frac{1}{2}}R(x,v)\textrm{d} x\textrm{d} v=0. \end{aligned}$$

Hence, combining (3.8), (3.10) and (0.4), we know

$$\begin{aligned} \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mathfrak {F}(x,v)\textrm{d} v\textrm{d} x= & {} \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu (x,v)\textrm{d} v\textrm{d} x\\= & {} 3P|\Omega |+\varepsilon \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}|v|^2\mu _{w}^{\frac{1}{2}} f^{B}_1\textrm{d} x\textrm{d} v. \end{aligned}$$

3.5 Construction of $(\rho _2, u_2,T_2)$

Theorem 3.8

Under the assumption (1.4), we can construct $(\rho _2, u_2,T_2)$ such that for any $s\in [2,\infty )$

$$\begin{aligned} \Vert f_2\Vert _{W^{2,s}L^{\infty }_{\varrho ,\vartheta }}+|f_2|_{W^{2-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim o_T. \end{aligned}$$

Proof

Denote

$$\begin{aligned} Y(\iota _1,\iota _2):=-\varepsilon ^{-1}P^{-1}\int _{\mathbb {R}^3}v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v}) f^{B}_1(0,\mathfrak {v})\textrm{d} \mathfrak {v}. \end{aligned}$$

Due to (2.49), we have $|Y|\lesssim o_T$. Then we define $u_2$ via $u_2=\nabla _{x}\psi$ where $\psi$ solves

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _x\psi =\displaystyle -|\Omega |^{-1}\int _{\partial \Omega }Y(\iota _1,\iota _2)\textrm{d} s&{}\quad \text { in }~\Omega ,\\ \dfrac{\partial \psi }{\partial n}=Y&{}\quad \text { on }~\partial \Omega . \end{array}\right. \end{aligned}$$

Due to classical elliptic theory, we know that this equation is well-posed. In particular, due to (3.9), we know $Y\in W^{3,\infty }(\partial \Omega )$. Then we have $\psi \in W^{4,s}$ and thus $u_2\in W^{3,s}$ satisfying

$$\begin{aligned} \Vert u_2\Vert _{W^{3,s}}\lesssim o_T. \end{aligned}$$

From Theorem 3.3 and the third equation in (2.37), we know that

$$\begin{aligned} T\rho _2+\rho T_2\in W^{2,s}. \end{aligned}$$

We are free to take $\rho _2=0$ in $\Omega$, and thus $T_2$ is determined and satisfies

$$\begin{aligned} \Vert T_2\Vert _{W^{2,s}}\lesssim o_T. \end{aligned}$$

Hence, we have shown that

$$\begin{aligned} \Vert f_2\Vert _{W^{2,s}L^{\infty }_{\varrho ,\vartheta }}+|f_2|_{W^{2-\frac{1}{s},s}L^{\infty }_{\varrho ,\vartheta }}\lesssim \Vert f_1\Vert _{W^{2,s}L^{\infty }_{\varrho ,\vartheta }}+\Vert \rho _2\Vert _{W^{2,s}}+\Vert u_2\Vert _{W^{2,s}}+\Vert T_2\Vert _{W^{2,s}}\lesssim o_T. \end{aligned}$$

$\square$

Remark 3.9

Such choice of $u_2$ implies that on the boundary $\partial \Omega$

$$ u_2\cdot n=Y. $$

Hence, we know

$$\begin{aligned} \int _{\mathbb {R}^3}\left( \varepsilon ^2f_2+\varepsilon f^{B}_1\right) \mu _{w}(v\cdot n)=\varepsilon ^2P(u_2\cdot n)+\varepsilon \int _{\mathbb {R}^3}v_{\eta }\mu _{w}^{\frac{1}{2}}(\mathfrak {v}) f^{B}_1(0,\mathfrak {v})\textrm{d} \mathfrak {v}=0, \end{aligned}$$

(3.11)

and thus

$$\begin{aligned} \int _{\mathbb {R}^3}\left( \mu ^{\frac{1}{2}}+\varepsilon f_1+\varepsilon ^2 f_2+\varepsilon f^{B}_1\right) \mu _{w}^{\frac{1}{2}}(v\cdot n)=0. \end{aligned}$$

We restore the zero mass-flux condition of $\mu ^{\frac{1}{2}}+\varepsilon f_1+\varepsilon ^2 f_2+\varepsilon f^{B}_1$.

4 Remainder Equation

For sake of completeness, in this section we will present the remainder equation for R and report the main result in [12].

Now we begin to derive the remainder equation for R in (0.4), or equivalently the nonlinear Boltzmann equation (1.1). Denote

$$\begin{aligned} Q[F,F]= & {} Q_{\text {gain}}[F,F]-Q_{\text {loss}}[F,F]\\:= & {} \int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)F(\mathfrak {u}_{*})F(v_{*})\textrm{d}{\omega }\textrm{d}{\mathfrak {u}}\\{} & {} \quad -F(v)\int _{\mathbb {R}^3}\int _{\mathbb {S}^2}q(\omega ,|\mathfrak {u}-v|)F(\mathfrak {u})\textrm{d}{\omega }\textrm{d}{\mathfrak {u}}=\nu (F)F. \end{aligned}$$

Denote $\mathfrak {F}=\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R$, where

$$\begin{aligned} \mathfrak {F}_{\text {a}}:=\mu +\mu ^{\frac{1}{2}}(\varepsilon f_1+\varepsilon ^2 f_2)+\mu _{w}^{\frac{1}{2}}(\varepsilon f^{B}_1). \end{aligned}$$

We can split $\mathfrak {F}=\mathfrak {F}_+-\mathfrak {F}_-$ where $\mathfrak {F}_+=\max \{\mathfrak {F},0\}$ and $\mathfrak {F}_-=\max \{-\mathfrak {F},0\}$ denote the positive and negative parts, and the similar notation also applies to $\mathfrak {F}_{\text {a}}$ and R.

In order to study (1.1), we first consider an auxiliary equation (which is equivalent to (1.1) when $\mathfrak {F}\ge 0$)

$$\begin{aligned} v\cdot \nabla _{x} \mathfrak {F}+\varepsilon ^{-1}\left( Q_{\text {loss}}[\mathfrak {F},\mathfrak {F}]\!-\!Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) \!=\!\mathfrak {z} \int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\varepsilon ^{-1}\left( Q_{\text {loss}}[\mathfrak {F},\mathfrak {F}]\!-\!Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) , \end{aligned}$$

(4.1)

with diffuse-reflection boundary condition

$$\begin{aligned} \mathfrak {F}(x_0,v)=M_{w}(x_0,v)\int _{v'\cdot n(x_0)>0} \mathfrak {F}(x_0,v')|v'\cdot n(x_0)|\textrm{d}{v'} \quad \text { for }x_0\in \partial \Omega \text { and }v\cdot n(x_0)<0. \end{aligned}$$

Here $\mathfrak {z}=\mathfrak {z}(v)>0$ is a smooth function with support contained in $\{|v|\le 1\}$ such that $\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\mathfrak {z}=1$.

The auxiliary system (4.1) is equivalent to

$$\begin{aligned} v\cdot \nabla _{x} \mathfrak {F}-\varepsilon ^{-1}Q[\mathfrak {F},\mathfrak {F}]= & {} -\varepsilon ^{-1}\left( Q_{\text {gain}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) \\{} & {} +\mathfrak {z}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\varepsilon ^{-1}\left( Q_{\text {loss}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) , \end{aligned}$$

and due to orthogonality of Q, is further equivalent to

$$\begin{aligned} v\cdot \nabla _{x} \mathfrak {F}-\varepsilon ^{-1}Q[\mathfrak {F},\mathfrak {F}]= & {} -\varepsilon ^{-1}\left( Q_{\text {gain}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) \nonumber \\{} & {} +\mathfrak {z}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\varepsilon ^{-1}\left( Q_{\text {gain}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) . \end{aligned}$$

(4.2)

Remark 4.1

The extra terms

$$\begin{aligned} -\varepsilon ^{-1}\left( Q_{\text {gain}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) +\mathfrak {z}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\varepsilon ^{-1}\left( Q_{\text {gain}}[\mathfrak {F},\mathfrak {F}]-Q_{\text {gain}}[\mathfrak {F}_+,\mathfrak {F}_+]\right) . \end{aligned}$$

on the right hand side of (4.2) plays a significant role in justifying the positivity of $\mathfrak {F}$ (see [12]). Clearly, when $\mathfrak {F}\ge 0$, i.e. $\mathfrak {F}=\mathfrak {F}_+$, the above extra terms vanish and the auxiliary equation (4.2) reduces to (1.1).

Inserting $\mathfrak {F}=\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R:=\mu +\widetilde{\mathfrak {F}}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R$ into (4.2), we have

$$\begin{aligned}{} & {} v\cdot \nabla _{x} \left( \mu ^{\frac{1}{2}}R\right) + \varepsilon ^{-1}\mu ^{\frac{1}{2}}\mathcal {L}[R]\\{} & {} = \mathscr {S}+\varepsilon ^{-1}\left( 2Q^{*}\left[ \widetilde{\mathfrak {F}}_{\text {a}},\mu ^{\frac{1}{2}}R\right] +\varepsilon ^{\alpha }Q^{*}\left[ \mu ^{\frac{1}{2}}R,\mu ^{\frac{1}{2}}R\right] \right) \nonumber \\{} & {} \quad -\varepsilon ^{-\alpha }\left( \varepsilon ^{-1}Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \nonumber \\{} & {} \quad \left. -\varepsilon ^{-1}Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) \nonumber \\{} & {} \quad +\varepsilon ^{-\alpha }\mathfrak {z}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\left( \varepsilon ^{-1}Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \nonumber \\{} & {} \quad \left. -\varepsilon ^{-1}Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) ,\nonumber \end{aligned}$$

(4.3)

where

$$\begin{aligned} \mathscr {S}:=-\varepsilon ^{-\alpha }v\cdot \nabla _{x}\mathfrak {F}_{\text {a}}+\varepsilon ^{-\alpha -1}Q^{*}\left[ \mathfrak {F}_{\text {a}},\mathfrak {F}_{\text {a}}\right] . \end{aligned}$$

(4.4)

Hence, we know that the equation for the remainder R is

$$\begin{aligned} \left\{ \begin{array}{ll} v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}R\right) +\varepsilon ^{-1}\mu ^{\frac{1}{2}}\mathcal {L}[R]=\mu ^{\frac{1}{2}}S&{}\quad \text {in }\Omega \times \mathbb {R}^3,\\ R(x_0,v)=\mathcal {P}_{\gamma }[R](x_0,v)+h(x_0,v) &{}\quad \text {for }x_0\in \partial \Omega \text { and }v\cdot n(x_0)<0. \end{array}\right. \end{aligned}$$

(4.5)

where

$$\begin{aligned} \mathcal {P}_{\gamma }[R](x_0,v):=m_w(x_0,v)\int _{v'\cdot n(x_0)>0} \mu ^{\frac{1}{2}}(x_0,v')R(x_0,v')|v'\cdot n(x_0)|\textrm{d}{v'}, \end{aligned}$$

with

$$\begin{aligned} m_w(x_0,v):=M_{w}\mu _{w}^{-\frac{1}{2}}, \end{aligned}$$

satisfying the normalization condition

$$\begin{aligned} \mu _{w}^{\frac{1}{2}}(x_0,v)=m_w(x_0,v)\int _{v'\cdot n(x_0)>0} \mu _{w}(x_0,v')|v'\cdot n(x_0)| \textrm{d}{v'}=\frac{P}{\big (2\pi T_w(x_0)\big )^{\frac{1}{2}}}m_w(x_0,v). \end{aligned}$$

The source term S includes the nonlinear terms and the terms of the expansion coming from higher orders and h is a correction on the boundary condition.

Lemma 4.2

We have

$$\begin{aligned} h:= & {} \varepsilon ^{-\alpha }\left( \mathcal {P}_{\gamma }\left[ \mu ^{-\frac{1}{2}}\widetilde{\mathfrak {F}}_{\text {a}}\right] -\mu ^{-\frac{1}{2}}\widetilde{\mathfrak {F}}_{\text {a}}\right) \\= & {} \varepsilon ^{2-\alpha }\left( m_w\int _{v'\cdot n>0}\mu _{w}^{\frac{1}{2}}(v')f_2(v')|v'\cdot n|\textrm{d}{v'}-f_2\big |_{v\cdot n<0}\right) \nonumber \\{} & {} \quad -\varepsilon ^{1-\alpha }\left( m_w\int _{v'\cdot n>0} \mu _{w}^{\frac{1}{2}}\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }|v'\cdot n|\textrm{d}{v'}-\mu _{w}^{\frac{1}{2}}\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }\big |_{v\cdot n<0}\right) .\nonumber \end{aligned}$$

Proof

From (0.4), we know

$$\begin{aligned} h:=\varepsilon ^{-\alpha }\left( \mathcal {P}_{\gamma }\left[ \mu ^{-\frac{1}{2}}\widetilde{\mathfrak {F}}_{\text {a}}\right] -\mu ^{-\frac{1}{2}}\widetilde{\mathfrak {F}}_{\text {a}}\right) . \end{aligned}$$

Then due to (2.53) and (2.54), we know

$$\begin{aligned}{} & {} \varepsilon ^{-\alpha }\left( \mathcal {P}_{\gamma }\left[ \mu ^{-\frac{1}{2}}(\varepsilon f_1+\varepsilon f^{B}_1)\right] -\mu ^{-\frac{1}{2}}(\varepsilon f_1+\varepsilon f^{B}_1)\right) \\{} & {} =\varepsilon ^{1-\alpha }\left( \mathcal {P}_{\gamma }\left[ \mu ^{-\frac{1}{2}}\left( f_1+\widetilde{\Phi }-\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }\right) \right] -\mu ^{-\frac{1}{2}}\left( f_1+\widetilde{\Phi }-\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }\right) \right) \\{} & {} =-\varepsilon ^{1-\alpha }\left( m_w\int _{v'\cdot n>0} \mu _{w}^{\frac{1}{2}}\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }|v'\cdot n|\textrm{d}{v'}-\mu _{w}^{\frac{1}{2}}\chi (\varepsilon ^{-1}v_{\eta })\widetilde{\Phi }\Big |_{v\cdot n<0}\right) . \end{aligned}$$

Then the result follows by adding the $f_2$ contribution.$\square$

Lemma 4.3

We have

$$\begin{aligned} S:= & {} \mu ^{-\frac{1}{2}}\mathscr {S}+\varepsilon ^{-1}\mu ^{-\frac{1}{2}}\left( 2Q^{*}\left[ \widetilde{\mathfrak {F}}_{\text {a}},\mu ^{\frac{1}{2}}R\right] +\varepsilon ^{\alpha }Q^{*}\left[ \mu ^{\frac{1}{2}}R,\mu ^{\frac{1}{2}}R\right] \right) \\{} & {} -\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}}\left( \varepsilon ^{-1}Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \\{} & {} \left. -\varepsilon ^{-1}Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) \\{} & {} +\varepsilon ^{-\alpha }\mathfrak {z}\mu ^{-\frac{1}{2}}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\left( \varepsilon ^{-1}Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \\{} & {} \left. -\varepsilon ^{-1}Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) , \end{aligned}$$

where $\mathscr {S}$ is defined in (4.4). The detailed expression is

$$\begin{aligned} S=-\mathcal {L}^1[R]+\overline{S}, \end{aligned}$$

where

$$\begin{aligned} \mathcal {L}^1[R]:= & {} -2\varepsilon ^{-1}\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}(\varepsilon f_1),\mu ^{\frac{1}{2}}R\right] =-2\Gamma [f_1,R],\\ \overline{S}:= & {} S_0+S_1+S_2+S_3+S_4+S_5+S_6, \end{aligned}$$

for

$$\begin{aligned} S_0:= & {} 2\varepsilon ^{-1}\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}(\varepsilon ^2f_2),\mu ^{\frac{1}{2}}R\right] =2\varepsilon \Gamma [f_2,R],\\ S_1:= & {} 2\varepsilon ^{-1}\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}_w(\varepsilon f^{B}_1),\mu ^{\frac{1}{2}}R\right] =2\Gamma \left[ \mu ^{-\frac{1}{2}}\mu ^{\frac{1}{2}}_w f^{B}_1,R\right] ,\\ S_2:= & {} \varepsilon ^{\alpha -1}\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}R,\mu ^{\frac{1}{2}}R\right] =\varepsilon ^{\alpha -1}\Gamma [R,R],\\ S_3:= & {} \varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}}\frac{1}{R_1-\varepsilon \eta }\left( v_{\phi }^2\frac{\partial }{\partial v_{\eta }}- v_{\eta } v_{\phi }\frac{\partial }{\partial v_{\phi }}\right) \left( \mu _{w}^{\frac{1}{2}} f^{B}_1\right) \\{} & {} +\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}}\frac{1}{R_2-\varepsilon \eta }\left( v_{\psi }^2\frac{\partial }{\partial v_{\eta }}- v_{\eta } v_{\psi }\frac{\partial }{\partial v_{\psi }}\right) \left( \mu _{w}^{\frac{1}{2}} f^{B}_1\right) \\{} & {} -\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}}\frac{1}{L_1L_2}\left( \frac{R_1\partial _{\iota _1\iota _1}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_1(R_1-\varepsilon \eta )} v_{\phi } v_{\psi } + \frac{R_2\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _2}\textbf{r}}{L_2(R_2-\varepsilon \eta )} v_{\psi }^2\right) \frac{\partial }{\partial v_{\phi }}\left( \mu _{w}^{\frac{1}{2}} f^{B}_1\right) \\{} & {} -\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}}\frac{1}{L_1L_2}\left( \frac{R_2\partial _{\iota _2\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_2(R_2-\varepsilon \eta )} v_{\phi } v_{\psi } + \frac{R_1\partial _{\iota _1\iota _2}\textbf{r}\cdot \partial _{\iota _1}\textbf{r}}{L_1(R_1-\varepsilon \eta )} v_{\phi }^2\right) \frac{\partial }{\partial v_{\psi }}\left( \mu _{w}^{\frac{1}{2}} f^{B}_1\right) \\{} & {} -\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}}\left( \dfrac{R_1 v_{\phi }}{L_1(R_1-\varepsilon \eta )}\frac{\partial }{\partial \iota _1}+\frac{R_2 v_{\psi }}{L_2(R_2-\varepsilon \eta )}\frac{\partial }{\partial \iota _2}\right) \left( \mu _{w}^{\frac{1}{2}} f^{B}_1\right) \\{} & {} +\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}} v_{\eta }\overline{\chi }(\varepsilon ^{-1} v_{\eta })\frac{\partial \chi (\varepsilon \eta )}{\partial \eta }\left( \mu _{w}^{\frac{1}{2}}\widetilde{\Phi }\right) \\{} & {} +\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}}\chi (\varepsilon \eta )\left( \overline{\chi }(\varepsilon ^{-1} v_{\eta })K_w[\widetilde{\Phi }]-K_w[\overline{\chi }(\varepsilon ^{-1} v_{\eta })\widetilde{\Phi }]\right) ,\\ S_4:= & {} -\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}\left( \varepsilon ^2f_2\right) \right) \right) =-\varepsilon ^{2-\alpha }\mu ^{-\frac{1}{2}}\left( v\cdot \nabla _{x}\left( \mu ^{\frac{1}{2}}f_2\right) \right) ,\\ S_5:= & {} \varepsilon ^{3-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}f_2,\mu ^{\frac{1}{2}}f_2\right] +2\varepsilon ^{2-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}f_2,\mu ^{\frac{1}{2}}f_1\right] \\{} & {} +2\varepsilon ^{2-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu ^{\frac{1}{2}}f_2,\mu _{w}^{\frac{1}{2}} f^{B}_1\right] +2\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu _{w}^{\frac{1}{2}} f^{B}_1,\mu ^{\frac{1}{2}}f_1\right] \\{} & {} +\varepsilon ^{1-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu _{w}^{\frac{1}{2}} f^{B}_1,\mu _{w}^{\frac{1}{2}} f^{B}_1\right] +\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}} Q^{*}\left[ \mu -\mu _{w},\mu _{w}^{\frac{1}{2}} f^{B}_1\right] \\= & {} \varepsilon ^{3-\alpha }\Gamma [f_2,f_2]+2\varepsilon ^{2-\alpha }\Gamma [f_2,f_1]+2\varepsilon ^{2-\alpha }\Gamma \left[ f_2,\mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}} f^{B}_1\right] \\{} & {} +2\varepsilon ^{1-\alpha }\Gamma \left[ \mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}} f^{B}_1,f_1\right] +\varepsilon ^{1-\alpha }\Gamma \left[ \mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}} f^{B}_1,\mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}} f^{B}_1\right] \\{} & {} +\varepsilon ^{-\alpha }\Gamma \left[ \mu ^{-\frac{1}{2}}(\mu -\mu _{w}),\mu ^{-\frac{1}{2}}\mu _{w}^{\frac{1}{2}} f^{B}_1\right] , \end{aligned}$$

and

$$\begin{aligned} S_6:= & {} -\varepsilon ^{-\alpha }\mu ^{-\frac{1}{2}}\left( \varepsilon ^{-1} Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \\{} & {} \left. -\varepsilon ^{-1} Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) \\{} & {} +\varepsilon ^{-\alpha }\mathfrak {z}\mu ^{-\frac{1}{2}}\int \hspace{-0.5pc}\int _{\Omega \times \mathbb {R}^3}\left( \varepsilon ^{-1} Q_{\text {gain}}\left[ \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R,\mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right] \right. \\{} & {} \left. -\varepsilon ^{-1} Q_{\text {gain}}\left[ \left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+,\left( \mathfrak {F}_{\text {a}}+\varepsilon ^{\alpha }\mu ^{\frac{1}{2}}R\right) _+\right] \right) . \end{aligned}$$

Proof

This follows directly from (4.3).$\square$

We decompose

$$\begin{aligned} R={\textbf {P}}[R]+({\textbf {I}}-{\textbf {P}})[R] := \mu ^{\frac{1}{2}}(v)\left( p_{R}(x)+v\cdot {\textbf {b}}_{R}(x)+(|v|^2-5T)c_{R}(x)\right) +({\textbf {I}}-{\textbf {P}})[R], \end{aligned}$$

We further define the orthogonal split

$$\begin{aligned} ({\textbf {I}}-{\textbf {P}})[{R}] = \mathscr {A}\cdot {\textbf {d}}_{R}(x)+({\textbf {I}}-\overline{{\textbf {P}}})[R], \end{aligned}$$

where $({\textbf {I}}-\overline{{\textbf {P}}})[{R}]$ is the orthogonal complement to $\mathscr {A}\cdot {\textbf {d}}_{R}(x)$ in $\mathcal {N}^{\perp }$ with respect to $(\cdot ,\cdot )_{\mathcal {L}}=(\cdot ,\mathcal {L}[\cdot ])_{\mathcal {L}}$, i.e.

$$\begin{aligned} (\mathscr {A},({\textbf {I}}-\overline{{\textbf {P}}})[R])_{\mathcal {L}}=\langle \overline{\mathscr {A}},({\textbf {I}}-\overline{{\textbf {P}}})[R]\rangle =0. \end{aligned}$$

In summary, we decompose the remainder as (4.6),

$$\begin{aligned} R=\left( p+{\textbf {b}}\cdot v+c(|v|^2-5T)\right) \mu ^{\frac{1}{2}}+{\textbf {d}}\cdot \mathscr {A}+({\textbf {I}}-\overline{{\textbf {P}}})[R]. \end{aligned}$$

(4.6)

We can further define the Hodge decomposition ${\textbf {d}}=\nabla _{x}\xi +{\textbf {e}}$ with $\xi$ solving the Poisson equation

$$\begin{aligned} \left\{ \begin{array}{ll} \nabla _{x}\cdot \left( \kappa \nabla _{x}\xi \right) =\nabla _{x}\cdot (\kappa {\textbf {d}})&{}\quad \text {in }\Omega ,\\ \xi =0&{}\quad \text {on }\partial \Omega . \end{array}\right. \end{aligned}$$

We reformulate the remainder equation with a global Maxwellian in order to obtain $L^{\infty }$ estimates. Considering $\Vert \nabla _{x}T\Vert \lesssim o_T$ for $o_T$ defined in (1.3), choose a constant $T_M$ such that

$$\begin{aligned} T_M<\min _{x\in \Omega }T<\max _{x\in \Omega }T<2T_M\quad \text { and }\quad \max _{x\in \Omega }T-T_M= o_T. \end{aligned}$$

(4.7)

Define a global Maxwellian

$$\begin{aligned} \mu _M:=\frac{P}{(2\pi )^{\frac{3}{2}}T_M^{\frac{5}{2}}}\exp \left( -\frac{|v|^2}{2T_M}\right) . \end{aligned}$$

We can rewrite (4.5) as

$$\begin{aligned} \left\{ \begin{array}{ll} v\cdot \nabla _{x}R_M+\varepsilon ^{-1}\mathcal {L}_M[R]=S_M&{}\quad \text {in }\Omega \times \mathbb {R}^3,\\ R_M(x_0,v)=\mathcal {P}_M[R_M](x_0,v)+h_M(x_0,v)&{}\quad \text {for }x_0\in \partial \Omega \text { and }v\cdot n(x_0)<0, \end{array}\right. \end{aligned}$$

where $R_M=\mu _M^{-\frac{1}{2}}\mu ^{\frac{1}{2}}R$, $S_M=\mu _M^{-\frac{1}{2}}\mu ^{\frac{1}{2}}S$, $h_M=\mu _M^{-\frac{1}{2}}\mu ^{\frac{1}{2}} h$ and for $m_{M,w}:=\mu _M^{-\frac{1}{2}}\mu ^{\frac{1}{2}}(x_0,v)m_w(x_0,v)=M_w\mu _M^{-\frac{1}{2}}$

$$\begin{aligned} \mathcal {L}_M[R_M]:= & {} -2\mu _M^{-\frac{1}{2}} Q\left[ \mu ,\mu _M^{\frac{1}{2}} R_M\right] :=\nu _MR_M-K_M[R_M],\\ \mathcal {P}_M[R_M](x_0,v):= & {} m_{M,w}(x_0,v)\int _{v'\cdot n(x_0)>0} \mu _M^{\frac{1}{2}} R_M(x_0,v') |v'\cdot n(x_0)|\textrm{d}{v'}. \end{aligned}$$

Denote the working space X via the norm

$$\begin{aligned} \Vert R\Vert _X&\!:=\!&\varepsilon ^{-1}\Vert p\Vert _{L^2}+\varepsilon ^{-\frac{1}{2}}\Vert {\textbf {b}}\Vert _{L^2}+\Vert c\Vert _{L^2}+\varepsilon ^{-1}\Vert \xi \Vert _{L^2}+\varepsilon ^{-\frac{1}{2}}\Vert \xi \Vert _{H^2}+\varepsilon ^{-1}\Vert {\textbf {e}}\Vert _{L^2}\\{} & {} +\varepsilon ^{-1}\Vert ({\textbf {I}}-\overline{{\textbf {P}}})[R]\Vert _{L_{\nu }^2}+\Vert p\Vert _{L^6}+\Vert {\textbf {b}}\Vert _{L^6}+\Vert c\Vert _{L^6}+\varepsilon ^{-1}\Vert \xi \Vert _{L^6}+\Vert \xi \Vert _{W^{2,6}}\nonumber \\{} & {} +\Vert {\textbf {e}}\Vert _{L^6}+\Vert ({\textbf {I}}-\overline{{\textbf {P}}})[R]\Vert _{L^6}+|\mathcal {P}_{\gamma }[R]|_{L^2_\gamma }+\varepsilon ^{-\frac{1}{2}}|(1-\mathcal {P}_{\gamma })[R]|_{L^2_{\gamma _+}}\nonumber \\{} & {} +\left| \mu ^{\frac{1}{4}}(1-\mathcal {P}_{\gamma })[R]\right| _{L^4_{\gamma _+}}+\varepsilon ^{-\frac{1}{2}}|\nabla _{x}\xi |_{L^2_{\partial \Omega }}+\varepsilon ^{\frac{1}{2}}\Vert R_M\Vert _{L^{\infty }_{\varrho ,\vartheta }}+\varepsilon ^{\frac{1}{2}}|R_M|_{L^{\infty }_{\gamma ,\varrho ,\vartheta }}.\nonumber \end{aligned}$$

(4.8)

In the companion paper [12], we prove the following:

Theorem 4.1

Assume that $\Omega$ is a bounded $C^3$ domain and (1.4) holds. Then for any given $P>0$, there exists $\varepsilon _0>0$ such that for any $\varepsilon \in (0,\varepsilon _0)$, there exists a nonnegative solution $\mathfrak {F}$ to the equation (0.1) represented by (0.4) with $\alpha =1$ satisfying

$$\begin{aligned} \int _{\Omega }p(x)\textrm{d} x=0 \end{aligned}$$

(4.9)

and

$$\begin{aligned} \Vert R\Vert _{X}\lesssim o_T, \end{aligned}$$

(4.10)

where the X norm is defined in (4.8). Such a solution is unique among all solutions satisfying (4.9) and (4.10). This further yields that in the expansion (0.4), $\mu +\varepsilon \mu (u_1\cdot v)$ is the leading-order terms in the sense of

$$\begin{aligned} \left\| \mu ^{-\frac{1}{2}}[\mathfrak {F}-\mu ]\right\| _{L^2_{x,v}}\lesssim \varepsilon \end{aligned}$$

and

$$\begin{aligned} \left\| \int _{\mathbb {R}^3}[\mathfrak {F}-\mu -\varepsilon \mu (u_1\cdot v)]v\right\| _{L^2_{x}}\lesssim \varepsilon ^{\frac{3}{2}}, \end{aligned}$$

where $(\rho , u_1,T)$ is determined by the ghost-effect equations (0.2) and (0.3).