\section{Introduction}


One of the important applications of the Boltzmann theory is to derive hydrodynamic (fluid) equations as the mean free path $\varepsilon\rightarrow 0$. There has been explosive literature on such hydrodynamic limits both from physical and mathematical standpoints. Almost all of the fluid equations obtained from such derivations are compatible with standard fluid theory, and the role of such limits can be viewed as further justification
rather than new discovery of fluid equations from the Boltzmann theory.

A rare exception occurs in the study of the hydrodynamic limit of \eqref{large system-} for a steady gas. If $\abs{\nablaT_{w}}_{L^{\infty}}=O(\varepsilon)$, then a standard
Hilbert expansion leads, among the others, to the celebrated Fourier law for the infinitesimal temperature variation $T_1$ of order $\varepsilon$: $\Delta_x T_{1}=0$. On the other hand, for the more natural and general case of $\abs{\nablaT_{w}}_{L^{\infty}}=O(1)$, $\varepsilon^{0}$ order only results in $\nabla_{x}\big(\rho T\big)\equiv 0$. Even though $\abs{\nablaT_{w}}_{L^{\infty}}=O(1)$ is within the regime of the compressible Euler limit, the vanishing of the velocity at $\varepsilon^{0}$ order requires further expansion to determine uniquely $\rho $ and $T$. This process (see details of our companion paper \cite{AA024}) leads to \eqref{fluid system-}, which has been discovered by various researches as early as in 1960's. We refer to \cite{Sone1972, Bobylev1995, Kogan1958, Kogan.Galkin.Fridlender1976}.

It is important to note that \eqref{fluid system-} cannot be predicted by any classical fluid theory, which can be viewed as an exciting example of a new kinetic effect. Thanks to the mismatch of $\varepsilon$ orders, the infinitesimal first-order velocity $\varepsilon u_{1}$ acting like a ``ghost" (a term introduced by Y. Sone \cite{Sone1972}), has order $1$ impact in determining $\rho $ and $T$. Therefore, \eqref{fluid system-} is not even compatible with any classical continuum fluid theory, in which all fluid quantities are determined at the same level of order of $\varepsilon$.

Due to the challenging nature of Euler limit with $\abs{\nablaT_{w}}_{L^{\infty}}=O(1)$, it has remained as one of the intriguing outstanding open questions to
rigorously derive \eqref{fluid system-}. The main goal of this paper is to settle this open question in affirmative in a bounded domain. In order to achieve this result, we have to introduce new ideas and several technical novelties that we expect to be useful in many kinetic problems.


\subsection{Problem Formulation}


We consider the stationary Boltzmann equation in a bounded three-dimensional $C^3$ domain $\Omega\nix=(x_1,x_2,x_3)$
with velocity $v=(v_1,v_2,v_3)\in\mathbb{R}^3$. The density function $\mathfrak{F}(x,v)$ satisfies \eqref{large system-}.
Here $Q$ is the hard-sphere collision operator
\begin{align}
Q[F,G]:=&\int_{\mathbb{R}^3}\int_{\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})\Big(F({\mathfrak{u}}_{\ast})G(v_{\ast})-F({\mathfrak{u}})G(v)\Big)\mathrm{d}{\omega}\mathrm{d}{{\mathfrak{u}}},
\end{align}
with $\omega\in\mathbb{R}^3$ a unit vector,  ${\mathfrak{u}}_{\ast}:={\mathfrak{u}}+\omega\big((v-{\mathfrak{u}})\cdot\omega\big)$, $v_{\ast}:=v-\omega\big((v-{\mathfrak{u}})\cdot\omega\big)$, and the hard-sphere collision kernel $q(\omega,\abs{{\mathfrak{u}}-v}):=q_0\abs{\omega\cdot(v-{\mathfrak{u}})}$ for a positive constant $q_0$.

In the diffuse-reflection boundary condition in \eqref{large system-}, $n(x_0)$ is the unit outward normal vector at $x_0$, and the Knudsen number $\varepsilon$ satisfies $0<\varepsilon\ll 1$. The wall Maxwellian
\begin{align}
M_{w}(x_0,v):=\frac{1}{2\pi\big(T_{w}(x_0)\big)^2}
\exp\bigg(-\frac{\abs{v}^2}{2T_{w}(x_0)}\bigg),
\end{align}
for some $T_{w}(x_0)>0$ satisfies
\begin{align}
\int_{v\cdotn(x_0)>0}M_{w}(x_0,v)\abs{v\cdotn(x_0)}\mathrm{d}{v}=1.
\end{align}

Based on the flow direction, we can divide the boundary $\gamma:=\big\{(x_0,v):\ x_0\in\partial\Omega,v\in\mathbb{R}^3\big\}$ into the incoming boundary $\gamma_-$, the outgoing boundary $\gamma_+$, and the grazing set $\gamma_0$ based on the sign of $v\cdotn(x_0)$. In particular, the boundary condition of \eqref{large system-} is only given on $\gamma_{-}$.

Throughout this paper, in the low Mach number regime, we always assume \eqref{assumption:boundary} and the physically important hard-sphere collision kernel, even though our analysis can be easily extended to
kernels of hard potential with an angular cutoff. We intend to study the asymptotic behavior of $\mathfrak{F}$ as $\varepsilon\rt0$.


\subsection{Normal Chart near Boundary}\label{sec:geometric-setup}


We follow the approach in \cite{BB002} to define the geometric quantities and more details can be found in our companion paper \cite{AA024}. 

For 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 $\mathbf{r}=\mathbf{r}(\iota_1,\iota_2)$. Let the vector length be $L_i=\abs{\partial_{\iota_i}\mathbf{r}}$ and unit vector $\varsigma_i=L_i^{-1}\partial_{\iota_i}\mathbf{r}$.

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{align}
x=\mathbf{r}-\mnn.
\end{align}
Define the orthogonal velocity substitution for ${\mathfrak{v}}:=(v_{\eta},v_{\phi},v_{\psi})$ as
\begin{align}\label{aa 44}
-v\cdotn:=v_{\eta},\quad
-v\cdot\varsigma_1:=v_{\phi},\quad
-v\cdot\varsigma_2:=v_{\psi}.
\end{align}
Finally, we define the scaled variable $\eta=\dfrac{\mathfrak{n}}{\varepsilon}$, which implies $\dfrac{\partial}{\partial\mathfrak{n}}=\dfrac{1}{\varepsilon}\dfrac{\partial}{\partial\eta}$. 


\subsection{Asymptotic Expansion and Remainder Equation}


We seek a solution in the form 
\begin{align}\label{aa 08}
    \mathfrak{F}(x,v)=&f+f^{B}+\varepsilon^{\alpha}\mhR
    =\mu+\m^{\frac{1}{2}}\Big(\ef_1+\varepsilon^2f_2\Big)+\mb^{\frac{1}{2}}\Big(\ef^{B}_1\Big)+\varepsilon^{\alpha}\mhR,
\end{align}
where the interior solution is
\begin{align}\label{expand 1}
f(x,v):= \mu(x,v)+\m^{\frac{1}{2}}(x,v)\Big(\ef_1(x,v)+\varepsilon^2f_2(x,v)\Big),
\end{align}
and the boundary layer is
\begin{align}\label{expand 2}
f^{B}(\eta,\iota_1,\iota_2,{\mathfrak{v}}):= \mb^{\frac{1}{2}}(\iota_1,\iota_2,{\mathfrak{v}})\Big(\ef^{B}_1(\eta,\iota_1,\iota_2,{\mathfrak{v}})\Big).
\end{align}
Here $f_1$, $f_2$ and $f^{B}_1$ are constructed precisely in our companion paper \cite{AA024} (see Appendix \ref{sec:asymptotic} for definitions), $R(x,v)$ is the remainder, $\mu(x,v)$ denotes a local Maxwellian and $\mu_{w}(\iota_1,\iota_2,{\mathfrak{v}})=\mu(x_0,v)$ the boundary Maxwellian.

In \cite{AA024}, we proved the following
\begin{theorem}\label{thm:ghost}
    Under the assumption \eqref{assumption:boundary}, 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 equation \eqref{fluid system-} and \eqref{boundary condition} satisfying for any $s\in[2,\infty)$
    \begin{align}
    \nm{u_1}_{W^{3,s}}+\nm{\mathfrak{p}}_{W^{2,s}}+\nm{T-1}_{W^{4,s}}\lesssimo_{T}.
    \end{align}
    Also, we can construct $f_1$, $f_2$ and $f^{B}_1$ as in \eqref{final 34}\eqref{final 35}\eqref{final 14} such that 
    \begin{align}
    \nm{f_1}_{W^{3,s}L^{\infty}_{\varrho,\vartheta}}+\abs{f_1}_{W^{3-\frac{1}{s},s}L^{\infty}_{\varrho,\vartheta}}&\lesssimo_{T},\\
    \nm{f_2}_{W^{2,s}L^{\infty}_{\varrho,\vartheta}}+\abs{f_2}_{W^{2-\frac{1}{s},s}L^{\infty}_{\varrho,\vartheta}}&\lesssimo_{T},
    \end{align}
and for some $K_0>0$ and any $0<r\leq 3$
\begin{align}
    \lnmm{\mathrm{e}^{K_0\eta}f^{B}_1}+\lnmm{\mathrm{e}^{K_0\eta}\frac{\partial^rf^{B}_1}{\partial\iota_1^r}}+\lnmm{\mathrm{e}^{K_0\eta}\frac{\partial^rf^{B}_1}{\partial\iota_2^r}}&\lesssimo_{T}.
\end{align}
\end{theorem}

Based on the analysis in Section \ref{sec:remainder}, in order to show the validity of \eqref{aa 08}, it suffices to consider the remainder equation for $R$: 
\begin{align}\label{remainder}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}\left(\mhR\right)+\varepsilon^{-1}\m^{\frac{1}{2}}\mathcal{L}[R]=\mhS\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.2em} R(x_0,v)=\mathcal{P}_{\gamma}[R](x_0,v)+h(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
Here the linearized Boltzmann operator  $\mathcal{L}$ is defined in \eqref{att 11}, the boundary and bulk sources $h$ and $S$ are defined in \eqref{aa 32} and \eqref{aa 31} (In order to show positivity of $\mathfrak{F}$, we choose to add an artificial term $\ss_6$ to $S$),
\begin{align}
\mathcal{P}_{\gamma}[R](x_0,v):=m_w(x_0,v)\displaystyle\int_{{\mathfrak{u}}\cdotn(x_0)>0}
\m^{\frac{1}{2}}(x_0,{\mathfrak{u}})R(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}},
\end{align}
with $m_w(x_0,v):=M_{w}\mu_{w}^{-\frac{1}{2}}$
satisfying the normalization condition
\begin{align}
    \mu_{w}^{\frac{1}{2}}(x_0,v)=m_w(x_0,v)\displaystyle\int_{{\mathfrak{u}}\cdotn(x_0)>0}
    \mu_{w}(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}}=\frac{P}{\big(2\pi T_w(x_0)\big)^{\frac{1}{2}}}m_w(x_0,v).
\end{align}
Note that 
\begin{align}\label{oo 40}
v\cdot\nabla_{x}\left(\mhR\right)&=\m^{\frac{1}{2}}(v\cdot\nabla_{x}R)+\frac{1}{2}\m^{-\frac{1}{2}}(v\cdot\nabla_{x}\mu)R=\m^{\frac{1}{2}}(v\cdot\nabla_{x}R)+\frac{1}{2}\m^{\frac{1}{2}}\Big(v\cdot\nabla_{x}(\ln\mu)\Big)R\\
&=\m^{\frac{1}{2}}(v\cdot\nabla_{x}R)+\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}\right)R,\nonumber
\end{align}
where $\overline{\mathscr{A}}$ is defined in \eqref{final 22}.
Hence, we can rewrite \eqref{remainder} in the equivalent form
\begin{align}\label{remainder.}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}R+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x}T}{4T^2}\right)R+\varepsilon^{-1}\mathcal{L}[R]=S\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} R(x_0,v)=\mathcal{P}_{\gamma}[R](x_0,v)+h(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}


\subsection{Decomposition and Reformulation}


\subsubsection{Hodge Decomposition}


Note that the null space $\mathcal{N}$ of $\mathcal{L}$ is a five-dimensional space spanned by the orthogonal basis
\begin{align}\label{att 32}
\m^{\frac{1}{2}}\bigg\{1,v,\left(\abs{v}^2-3T\right)\bigg\}.
\end{align}
We denote $\mathcal{N}^{\perp}$ the orthogonal complement of $\mathcal{N}$ in $L^2(\mathbb{R}^3)$.
Define the kernel operator $\textbf{P}$
as the orthogonal projection onto $\mathcal{N}$, and the non-kernel operator $\textbf{I}-\textbf{P}$.
We decompose
\begin{align}
    R=&\textbf{P}[R]+(\textbf{I}-\textbf{P})[R]
    :=\m^{\frac{1}{2}}(v)\bigg(p_{R}(x)+v\cdot
    \textbf{b}_{R}(x)+\Big(\abs{v}^2-5T\Big)c_{R}(x)\bigg)+(\textbf{I}-\textbf{P})[R],
\end{align}
Note that we have chosen a different decomposition of $\textbf{P}[R]$ in contrast with the classical $(a_{R},\textbf{b}_{R},c_{R})$ decomposition in previous work \cite{Esposito.Guo.Kim.Marra2015} ($p_{R}=a_{R}+2T c_{R}$), for the convenience of the paper.
Moreover, we introduce the crucial orthogonal split
\begin{align}
(\textbf{I}-\textbf{P})[{R}]
&=\mathscr{A}\cdot\textbf{d}_{R}(x)+(\textbf{I}-\overline{\textbf{P}})[{R}],
\end{align}
where $\mathscr{A}$ is defined in \eqref{final 22}, $(\textbf{I}-\overline{\textbf{P}})[{R}]$ is the orthogonal complement to $\mathscr{A}\cdot\textbf{d}_{R}(x)$ in $\mathcal{N}^{\perp}$ with respect to $\brr{\ \cdot\ ,\ \cdot\ }=\brv{\ \cdot\ ,\mathcal{L}[\ \cdot\ ]}$
\begin{align}
\brr{\mathscr{A},(\textbf{I}-\overline{\textbf{P}})[{R}]}=\brv{\overline{\mathscr{A}},(\textbf{I}-\overline{\textbf{P}})[{R}]}=0.
\end{align}
From now on, when there is no confusion, we will omit the subscript ${R}$ and simply write $p,\textbf{b},c,\textbf{d}$.
In summary, we decompose the remainder as
\begin{align}
R&=\bigg(p+\textbf{b}\cdot v+c\Big(\abs{v}^2-5T\Big)\bigg)\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],\label{pp 01}
\end{align}
and from \eqref{remainder.}, $R$ satisfies
\begin{align}
v\cdot\nabla_{x}R+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)R+\varepsilon^{-1}\textbf{d}\cdot\overline{\mathscr{A}}+\varepsilon^{-1}\mathcal{L}\Big[(\textbf{I}-\overline{\textbf{P}})[R]\Big]+\lc^1[R]&=\barS.\label{pp 02}
\end{align}
We can further define the Hodge decomposition $\textbf{d}=\nabla_{x}\xi+\textbf{e}$
with $\xi$ solving the Poisson equation
\begin{align}\label{tt 01}
\left\{
\begin{array}{ll}
\nabla_{x}\cdot\left(\kappa\nabla_{x}\xi\right)=\nabla_{x}\cdot(\kappa\textbf{d})\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.2em}
\xi=0\ \ \text{on}\ \ \partial\Omega.
\end{array}
\right.
\end{align}
We can directly compute
\begin{align}\label{tt 13}
\nabla_{x}\cdot(\kappa\textbf{e})=\nabla_{x}\cdot(\kappa\textbf{d})-\nabla_{x}\cdot(\kappa\nabla_{x}\xi)=0.
\end{align}
This implies the crucial orthogonality: for any $\eta(x)$ such that $\eta=0$ on $\partial\Omega$, we have
\begin{align}\label{final 63}
\int_{\Omega}(\kappa\textbf{e})\cdot\nabla_{x}\eta\mathrm{d}{x}=-\int_{\Omega}\nabla_{x}\cdot(\kappa\textbf{e})\eta\mathrm{d}{x}=0.
\end{align}
Thus, taking $\eta=\xi$, we know
\begin{align}\label{energy 11}
\brr{\textbf{d}\cdot\mathscr{A},\textbf{d}\cdot\mathscr{A}}&=\br{\textbf{d}\cdot\mathscr{A},\mathcal{L}[\textbf{d}\cdot\mathscr{A}]}=\br{\textbf{d}\cdot\mathscr{A},\textbf{d}\cdot\overline{\mathscr{A}}}=\int_{\Omega}\kappa\abs{\nabla_{x}\xi+\textbf{e}}^2\mathrm{d}x\\
&=\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2\mathrm{d}x+\int_{\Omega}\kappa\abs{\textbf{e}}^2\mathrm{d}x+2\int_{\Omega}\kappa\left(\nabla_{x}\xi\cdot\textbf{e}\right)\mathrm{d}x\nonumber\\
&=\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2\mathrm{d}x+\int_{\Omega}\kappa\abs{\textbf{e}}^2\mathrm{d}x-2\int_{\Omega}\nabla_{x}\cdot(\kappa\textbf{e})\xi\mathrm{d}x=\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2\mathrm{d}x+\int_{\Omega}\kappa\abs{\textbf{e}}^2\mathrm{d}x.\nonumber
\end{align} 


\subsubsection{Reformulation with Global Maxwellian}


Due to the cubic velocity growth term in \eqref{remainder.}, we have to reformulate the remainder equation with a global Maxwellian in order to obtain $L^{\infty}$ estimates. Considering $\lnm{\nabla_{x}T}\lesssimo_{T}$ for $o_{T}$ defined in \eqref{def:oot}, choose constant $\displaystyleT_M: T_M<\min_{x\in\Omega}T<\max_{x\in\Omega}T<2T_M$ and $\displaystyle\max_{x\in\Omega}T-T_M=o_{T}$.
Define a global Maxwellian
\begin{align}\label{final 13}
    \mu_M:=\frac{P}{(2\pi)^{\frac{3}{2}}T_M^{\frac{5}{2}}}\exp\bigg(-\frac{\abs{v}^2}{2T_M}\bigg).
\end{align}
We can rewrite \eqref{remainder} as
\begin{align}\label{ll 01}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}\re_M+\varepsilon^{-1}\mathcal{L}_M[R]=S_M\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \re_M(x_0,v)=\mathcal{P}_M[\re_M](x_0,v)+h_M(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0,
\end{array}
\right.
\end{align}
where $\re_M=\mm^{-\frac{1}{2}}\mhR$, $S_M=\mm^{-\frac{1}{2}}\mhS$, $h_M=\mm^{-\frac{1}{2}}\m^{\frac{1}{2}} h$ and for $m_{M,w}:=\mm^{-\frac{1}{2}}\m^{\frac{1}{2}}(x_0,v)m_w(x_0,v)=M_w\mm^{-\frac{1}{2}}$
\begin{align}
    \mathcal{L}_M[\re_M]:=&-2\mm^{-\frac{1}{2}} Q^{\ast}\left[\mu,\mm^{\frac{1}{2}} \re_M\right]:=\nu_M\re_M-K_M[\re_M],\\
    \mathcal{P}_M[R_M](x_0,v):=&m_{M,w}(x_0,v)\displaystyle\int_{{\mathfrak{u}}\cdotn(x_0)>0}   \mm^{\frac{1}{2}}\re_M(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}}.
\end{align}


\subsubsection{Working Space}


Denote the working space $X$ via the norm
\begin{align}\label{ss 00}
\\
\xnm{R}:=&\varepsilon^{-1}\tnm{p}+\varepsilon^{-\frac{1}{2}}\tnm{\textbf{b}}+\pnm{c}{2}+\varepsilon^{-1}\tnm{\xi}+\varepsilon^{-\frac{1}{2}}\nm{\xi}_{H^2}+\varepsilon^{-1}\tnm{\textbf{e}}+\varepsilon^{-1}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&+\pnm{p}{6}+\pnm{\textbf{b}}{6}+\pnm{c}{6}+\varepsilon^{-1}\pnm{\xi}{6}+\nm{\xi}_{W^{2,6}}+\pnm{\textbf{e}}{6}+\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{6}\nonumber\\
&+\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}+\varepsilon^{-\frac{1}{2}}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\pnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{4}{\gamma_+}+\varepsilon^{-\frac{1}{2}}\tnms{\nabla_{x}\xi}{\partial\Omega}\nonumber\\
&+\varepsilon^{\frac{1}{2}}\lnmm{\re_M}+\varepsilon^{\frac{1}{2}}\lnmms{\re_M}{\gamma}.\nonumber
\end{align}


\subsection{Main Theorem}


\begin{theorem}\label{main}
Assume that $\Omega$ is a bounded $C^3$ domain and \eqref{assumption:boundary} 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 \eqref{large system-} represented by \eqref{aa 08} with $\alpha=1$ satisfying
\begin{align}\label{final 31}
    \int_{\Omega}p(x)\mathrm{d} x=0,
\end{align}
and
\begin{align}\label{final 21}
    \xnm{R}\lesssimo_{T},
\end{align}
where the $X$ norm is defined in \eqref{ss 00}. Such a solution is unique among all solutions satisfying \eqref{final 31} and \eqref{final 21}. This further yields that in the expansion \eqref{aa 08}, $\mu+\varepsilon\mu\big(u_1\cdotv\big)$ is the leading-order terms in the sense of
\begin{align}
    \int_{\Omega}\abs{\int_{\mathbb{R}^3}\Big(\mathfrak{F}-\mu\Big)}+\int_{\Omega}\abs{\int_{\mathbb{R}^3}\Big(\mathfrak{F}-\mu\Big)\left(\abs{v}^2-3T\right)}\lesssim\varepsilon,
\end{align}
and 
\begin{align}
    \int_{\Omega}\abs{\int_{\mathbb{R}^3}\Big(\mathfrak{F}-\mu-\varepsilon \mu\big(u_1\cdot v\big)\Big)v}\lesssim\varepsilon^{\frac{3}{2}},
\end{align}
where $(\rho,u_1,T)$
is determined by the ghost-effect equations \eqref{fluid system-} and \eqref{boundary condition}.
\end{theorem}

\begin{remark}
Our method also works for the slab $\Omega=\mathbb{T}^2\times[0,1]$ considered in \cite{Sone.Aoki.Takata.Sugimoto.Bobylev1996, Takata.Aoki2001}. In this case we can include higher-order boundary layers and take $\alpha=\frac{3}{2}$ to improve the estimates to $L^{\infty}$ convergence. 
\end{remark}

\begin{remark}
    There is no restriction on $\nabla T_{w}$ except smallness condition \eqref{assumption:boundary}. Moreover, our result is valid for all general smooth bounded domains (including non-convex domains), despite the presence of Knudsen boundary-layer approximation.
\end{remark}

\begin{remark}
    We stress that the boundedness of $\xnm{R}$ only implies that $\varepsilon^{-\frac{1}{2}}\nabla_{x}\xi$ is bounded in $L^2$, while we do not know that $\varepsilon^{-1}\nabla_{x}\xi$ is bounded in $L^2$, in contrast with the uniform bound in \cite{Esposito.Guo.Kim.Marra2015}, where the full $\varepsilon^{-1}\um{(\textbf{I}-\textbf{P})[R]}$ is bounded in $L^2$.
\end{remark}


\subsection{Methodology}


In the natural decomposition of the remainder $R=\textbf{P}[R]+(\textbf{I}-\textbf{P})[R]$, it is well-known that the $(\textbf{I}-\textbf{P})[R]$ part automatically enjoys functional analytic bound from basic energy estimate and coercivity of the linearized Boltzmann operator $\mathcal{L}$, so the major mathematical task is to control the macroscopic part $\textbf{P}[R]$. In a series of papers \cite{Guo2003, Guo2012}, an elliptic structure is discovered for $\textbf{P}[R]$, and $\textbf{P}[R]$ can be bounded via $(\textbf{I}-\textbf{P})[R]$ through the Boltzmann equation in high order Sobolev norms. More
applications can be found in \cite{Gressman.Strain2011}. Unfortunately, it is well-known that
Boltzmann solutions exhibit only limited regularity (or even singularity) in the presence of physical boundary conditions \cite{Kim2011, Guo.Kim.Tonon.Trescases2013} due to the characteristic nature of the grazing set in kinetic theory. To overcome this difficulty, a $L^{2}-L^{\infty }$ framework is established in \cite{Guo2010}, in which $\tnm{\textbf{P}[R]}$ is bounded via $\tnm{(\textbf{I}-\textbf{P})[R]}$ to bootstrap weighted $\lnm{R}$ bound via double-Duhamel principle along the
characteristics, thanks to the velocity mixing feature of $K$. An important advance is achieved in \cite{Esposito.Guo.Kim.Marra2013}, where $\tnm{\textbf{P}[R]}$ is estimated quantitatively in terms of $\tnm{(\textbf{I}-\textbf{P})[R]}$  via proper choices of test functions, which resembles a weak form of elliptic estimates in spirit.

For hydrodynamic limits of \eqref{large system-}, the basic energy estimate yields
\begin{align}
    \varepsilon^{-2}\um{(\textbf{I}-\textbf{P})[R]}^2=&-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}+\text{good terms}\\
    =&-\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}+\text{good terms}.\nonumber
\end{align}

In the case of $\lnm{\nabla_{x}T}=o(1)\varepsilon$, we have
\begin{align}\label{basicenergy}
    \varepsilon^{-1}\um{(\textbf{I}-\textbf{P})[R]}\lesssim o(1)\pnm{\textbf{P}[R]}{6}+1.
\end{align}
With a new $L^{6}$ gain for $\textbf{P}[R]$ in \cite{Esposito.Guo.Kim.Marra2015}, quantitative estimates in \cite{Esposito.Guo.Kim.Marra2013} lead to
\begin{align}\label{6bound}
    \pnm{\textbf{P}[R]}{6}\lesssim \varepsilon^{-1}\um{(\textbf{I}-\textbf{P})[R]}
\end{align}
where $c$ and $\textbf{b}$ are estimated via test functions of the form $\nabla_{x}\phi\cdot\overline{\mathscr{A}}$ and $\nabla_{x}\psi:\overline\b$, and the convergence to the incompressible Navier-Stokes-Fourier system as $\varepsilon\rightarrow 0$ in \cite{Esposito.Guo.Kim.Marra2015} is established. 

In the case of $\lnm{\nabla_{x}T}=o(1)$, however,
\begin{align}
    -\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}\approx o(1)\varepsilon^{-1}\pnm{\textbf{P}[R]}{6}^2
\end{align}
which is impossible to close via \eqref{6bound} with a severe loss of $\varepsilon^{-1}$. We note that the term $-\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}$ presents a generic major difficulty in compressible Euler
limits, and the main mathematical achievement of our contribution is to develop a systematic methodology to overcome this loss of $\varepsilon$ in the presence of boundary effects.\\

\paragraph{\underline{Step 1: Reduced Energy Estimate of $(\textbf{I}-\overline{\textbf{P}})[R]+\textbf{e}\cdot\mathscr{A}$}}
The first key idea is to use Hodge decomposition to split
\begin{align}
    (\textbf{I}-\textbf{P})[R]=\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]=\nabla_{x}\xi\cdot\mathscr{A}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R].
\end{align}
and such a construction of $\xi$ leads to a natural kinetic equation for the
combination $\nabla_{x}\big(c+\varepsilon^{-1}\xi\big)$ as
\begin{align}
    v\cdot\nabla_{x}R+\varepsilon^{-1}\big(\nabla_{x}\xi+\textbf{e}\big)\cdot\overline{\mathscr{A}}\approx \nabla_{x}\big(c+\varepsilon^{-1}\xi\big)\cdot\overline{\mathscr{A}}+\text{other terms},
\end{align}
along $\overline{\mathscr{A}}$ direction instead of $\nabla_{x} c$. This allows us to use $\varepsilon^{-1}\nabla_{x}\xi\cdot\mathscr{A}$ to cancel $-\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}$ exactly in the energy estimate, by a string of delicate and precise manipulations:
\begin{align}\label{string}
    \bbrx{\nabla_{x}T,\textbf{b} c}=&\bbrx{\nabla_{x}T\cdot\textbf{b},c}\approx\bbrx{\nabla_{x}\cdot\big(T\textbf{b}\big),c}\approx -\bbrx{\nabla_{x}\cdot\big(\kappa\textbf{d}\big),c}
    =-\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\xi\big),c}\\
    \approx& -\bbrx{\nabla_{x}\xi,\varepsilon^{-1}\kappa\textbf{d}}\approx-\bbrx{\nabla_{x}\xi,\varepsilon^{-1}\kappa\nabla_{x}\xi}=-\varepsilon^{-1}\iint_{\Omega\times\mathbb{R}^3}\kappa\abs{\nabla_{x}\xi}^2,\nonumber
\end{align}
which is cancelled with its counterpart in $\displaystyle\iint_{\Omega\times\mathbb{R}^3}R(\textbf{I}-\textbf{P})[R]$. To get the above chain of approximations we use the mass and energy conservation $\nabla_{x}\cdot\textbf{b}\approx 0$ and $\nabla_{x}\cdot\big(T\textbf{b}\big)+\nabla_{x}\cdot\big(\kappa\textbf{d}\big)\approx0$, and $\nabla_{x}\cdot\big(\kappa\textbf{e}\big)=0$
in the Hodge decomposition. Finally, we choose a \textit{new} test function $\nabla_{x}\phi\cdot\mathscr{A}$ with a smooth function $\phi(x)$ for \eqref{remainder}, such that
\begin{align}
\label{pp 03}
&-\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),c}+\varepsilon^{-1}\bbrx{\nabla_{x}\phi,\kappa\textbf{d}}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\\
=&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\barS}.\nonumber
\end{align}
A reduced energy estimate is then established as the building block of our analysis:

\begin{proposition}\label{thm:energy}
Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have
\begin{align}\label{ss 01}
    \varepsilon^{-\frac{1}{2}}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\varepsilon^{-1}\tnm{\textbf{e}}+\varepsilon^{-1}\um{(\textbf{I}-\overline{\textbf{P}})[R]}
    \lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
\end{align}
\end{proposition}

Remarkably, this new reduced energy
estimate, in comparison to \eqref{basicenergy}, sacrifices the control of $\nabla_{x}\xi\cdot \mathscr{A}$ in $(\textbf{I}-\textbf{P})[R]$ to cancel $-\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}$ via exact computations. The drawback is that now $\xi$ has to be controlled together with $\textbf{P}[R]$ through the Boltzmann equation. We notice 
that such manipulations \eqref{string} are not at all special, but generic for
even more general Maxwellian in the Euler scaling.\\

\paragraph{\underline{Step 2: Estimate of $p$ and $Z:=c+\varepsilon^{-1}\xi$}}
We can control $p\approx O(\varepsilon )$ directly from conservation laws as in \eqref{6bound}: 

\begin{proposition}\label{prop:p-bound}
Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary} and \eqref{final 31}, we have 
\begin{align}\label{oo 28=}
    \varepsilon^{-1}\tnm{p}+\pnm{p}{6}\lesssim o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
\end{align}
\end{proposition}

The construction of $\xi $ leads to a natural kinetic equation for the combination $v\cdot \nabla_{x}\left(c+\varepsilon^{-1}\xi\right),$ and the analog of \eqref
{6bound} can and \textit{only} can be obtained for $Z=c+\varepsilon^{-1}\xi$ via the same test function $\nabla_{x}\phi\cdot\mathscr{A}$: 

\begin{proposition}\label{prop:z-bound}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have 
    \begin{align}\label{oo 33=}
        \pnm{Z}{6}\lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{proposition}

In order to cope with the boundary layer interaction,
we have to further split $Z$ as regular part $Z^{R}$ and small singular part $Z^S$: $Z=Z^{S}+Z^{R}$ so that

\begin{proposition}\label{prop:z-splitting}
Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, we have $Z=Z^R+Z^S$ where 
\begin{align}
    \nm{Z^R}_{H^1_0}\lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha},\label{qq 06}\\
    \tnm{Z^S}\lesssim& o_{T}\varepsilon^{\frac{1}{2}}\xnm{R}+\varepsilon^{\alpha-\frac{1}{2}}\xnm{R}^2+o_{T}\varepsilon^{2-\alpha}.\label{qq 07}
\end{align}
\end{proposition}

\paragraph{\underline{Step 3: Dual Stokes Problem and Improved Estimate $\tnm{\textbf{b}}\lesssim \varepsilon^{\frac{1}{2}}$}}
We now obtain a surprising gain of $\varepsilon^{\frac{1}{2}}$ for $\textbf{b}$, which is necessary to justify the ghost effect contribution $\varepsilon u_{1}\cdot v\m^{\frac{1}{2}}$ beyond the first order of $\varepsilon$. 

\begin{proposition}\label{prop:b-bound}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have
    \begin{align}\label{oo 34=}
        \varepsilon^{-\frac{1}{2}}\tnm{\textbf{b}}+\pnm{\textbf{b}}{6}\lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{proposition}

The starting point is the choice of a \textit{new} test function $\nabla_{x}\psi:\mathscr{B}+\varepsilon^{-1}\psi\cdot v\m^{\frac{1}{2}}$ with a smooth function $\psi(x)$ satisfying $\nabla_{x}\cdot\psi=0$, $\psi\big|_{\partial\Omega}=0$ for \eqref{remainder}:
\begin{align}\label{pp 16}
&-\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}\\
&+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}+\br{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\nonumber\\
=&\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
&+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\bbr{\nabla_{x}\psi:\mathscr{B},\barS},\nonumber
\end{align}
which combines with the momentum conservation law. The key feature in \eqref{pp 16} is the absence of the $O(1)$ term $\varepsilon^{-1}(\textbf{I}-\overline{\textbf{P}})[R]$ so that a gain of $\varepsilon^{\frac{1}{2}}$ is possible by constructing a
solution to the following dual Stokes problem for $(\psi, q)$
\begin{align}\label{pp 17}
\\
\left\{
    \begin{array}{r}
    -\lambda\Delta_x\psi-\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)-\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big)+\bbrv{\nabla_{x}\psi:\mathscr{B},v\left(\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x} T}{2T^2}\right)}+\nabla_{x} q=\textbf{b}\abs{\textbf{b}}^{r-2}\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.2em}
    \nabla_{x}\cdot\psi=0\ \ \text{in}\ \ \Omega,\qquad\psi=0\ \ \text{on}\ \ \partial\Omega
    \end{array}\nonumber
\right.
\end{align}

\paragraph{\underline{Step 4: Dual Stokes-Poisson Problem and Estimate of $c$}}
In spite of the control of $c+\varepsilon^{-1}\xi$ in Proposition \ref{prop:z-bound}, to complete the control of $\textbf{P}[R]$, 
an independent estimate of $c$ is necessary.

\begin{proposition}\label{prop:c-bound}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have
    \begin{align}\label{oo 35=}
        \pnm{c}{6}\lesssim&o_{T}\varepsilon^{1-\alpha}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^{2}+\varepsilon^{2\alpha-2}\xnm{R}^{3}+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{proposition}

The starting point is the choice of a \textit{new} test function $\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\phi\big(\abs{v}^2-5T\big)\m^{\frac{1}{2}}$ where $\phi(x)$ is a smooth function satisfying $\phi\big|_{\partial\Omega}=0$ for \eqref{remainder}:
\begin{align}\label{pp 10}
&-\bbrx{\nabla_{x}\cdot(\kappa\nabla_{x}\phi),c}+\varepsilon^{-1}5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\br{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\\
=&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&+\varepsilon^{-1}\br{\phi\big(\abs{v}^2-5T\big)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\barS},\nonumber
\end{align}
which combines
with the energy conservation. In the combination of $\varepsilon^{-1}\times\text{\eqref{pp 16}}+\text{\eqref{pp 10}}$, we choose carefully the
following dual Stokes-Poisson system for $(\psi, q,\phi)$ to eliminate any
contributions from $\textbf{b}$ in \eqref{pp 10}
\begin{align}\label{pp 20}
\\
\left\{
    \begin{array}{r}
    -\lambda\Delta_x\psi-\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)-\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big)+\bbrv{\nabla_{x}\psi:\mathscr{B},\mathfrak{M}}+\nabla_{x} q=-5P\phi\nabla_{x} T\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.2em}
    \nabla_{x}\cdot\psi=0\ \ \text{in}\ \ \Omega,\qquad
    \psi=0\ \ \text{on}\ \ \partial\Omega,
    \end{array}
    \right.\nonumber
\end{align}
and
\begin{align}\label{pp 23}
    -\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big)=c\abs{c}^{r-2}\ \ \text{in}\ \ \Omega,\qquad
    \phi=0\ \ \text{on}\ \ \partial\Omega,
\end{align}
where
\begin{align}\label{pp 114}
    \mathfrak{M}:=v\left(\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x} T}{2T^2}\right)-2\m^{-\frac{1}{2}} Q^{\ast}\left[-\m^{\frac{1}{2}}\mathscr{A}\cdot\frac{\nabla_{x}T}{2T^2}+\mu\frac{u_1\cdot v}{T},v\mu\right].
\end{align}
Thanks to the precise cancellation of \eqref{final 05},  all but the boundary contribution
\begin{align}\label{final 75}
-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}
\end{align}
in $\varepsilon^{-1}\times\text{\eqref{pp 16}}+\text{\eqref{pp 10}}$ are under control. We note that $(1-\mathcal{P}_{\gamma})[R]$ is of $
O(\varepsilon^{\frac{1}{2}})$ and $\nabla_{x}\psi:\mathscr{B}$ is of $O(1)$, so there is still a loss of $\varepsilon^{\frac{1}{2}}$ here for closure.\\

\paragraph{\underline{Step 5: Boundary Layer $g^B$ and Interior $g$ Compensation}}
Through an extensive effort, the original loss of $\varepsilon^{-1}$ in $-\varepsilon^{-1}5P\br{\nabla_{x}T,\textbf{b} c}$ is now transferred to a
boundary loss of $\varepsilon^{-\frac{1}{2}}$ in $-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}$. With a gain of $%
\varepsilon^{\frac{1}{2}}$ in $L^{2}$ for any boundary layers, we carefully design a boundary layer $g^B$ and its interior counterpart $g$ to compensate such a loss with
\begin{align}
\big(g^B+g\big)\Big|_{\gamma _{+}}=-\big(\nabla_{x}\psi :\mathscr{B}\big)\Big|_{\gamma _{+}}.
\end{align}
In the control of the new error terms \eqref{pp 38} and \eqref{pp 45} from such a compensation $g^B+g$, the parity of $\nabla_{x}\psi:\mathscr{B}$ (hence $g^B$
and $g$) leads to two fundamental cancellations \eqref{final 04} and \eqref{final 06} to avoid any loss of $\varepsilon$. \\

\paragraph{\underline{Step 6: New Boundary Layer Estimates for Non-Convex Domain}}

\subparagraph{\underline{Hardy Inequality with $\varepsilon$ Gain}} 
One of the important challenges in the hydrodynamic limit of \eqref{large system-} is the necessary inclusion of
boundary layers $f^{B}_1$ and $g^B$. In fact, the determination of the ghost-effect equations depends on 
$f_{B}$ implicitly. Unfortunately, in \cite{AA003,AA007,AA009}, it is discovered that for non-flat domains the classical boundary layer theory in kinetic theory breaks down, due again to the characteristic and singular nature of the grazing set. Even though an alternative satisfactory theory has been established in convex domains \cite{AA004,AA013, BB002}, the non-convex case is completely open. In the proof of Lemma \ref{lemma:k1}, via the Hardy's inequality, the most difficult contribution in $\varepsilon^{-1}\br{h^B,R}$ (where $h^B$ denotes a generic quantity related to boundary layer) is treated as
\begin{align}
\varepsilon^{-1}\br{h^B,c\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}
=&\varepsilon^{-1}\br{h^B,Z\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}-\varepsilon^{-1}\br{h^B,\varepsilon^{-1}\xi\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}  \\
=&\varepsilon^{-1}\br{h^B,Z^R\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}
+\varepsilon^{-1}\br{h^B,Z^S\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}\nonumber\\
&-\varepsilon^{-2}\br{h^B,\xi\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}.\nonumber
\end{align}
Thanks to the fact that $Z^R\in H_0^1$ and $\xi\in H_{0}^{2}$, we express
\begin{align}
    \frac{Z^R}{\mathfrak{n}}=\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}Z^R,\quad \frac{\xi}{\mathfrak{n}}=\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\xi,
\end{align}
and apply Hardy's inequality \cite{Hardy1920, Masmoudi2011} along the normal $\mathfrak{n}$ direction to obtain
\begin{align}
    \nm{\frac{Z^R}{\mathfrak{n}}}_{L^2_{\mathfrak{n}}}\lesssim\nm{\partial_{\mathfrak{n}}Z^R}_{L^2_{\mathfrak{n}}},\quad \nm{\frac{\xi}{\mathfrak{n}}}_{L^2_{\mathfrak{n}}}\lesssim\nm{\partial_{\mathfrak{n}}\xi}_{L^2_{\mathfrak{n}}}.
\end{align}
Note that $\mathfrak{n}=\varepsilon\eta$ in the boundary layer scaling can be
absorbed by $h^B$ to produce extra $\varepsilon$ in $\mathfrak{n}h^B=\varepsilon\etah^B$:
\begin{align}
    \abs{\varepsilon^{-2}\br{h^B,\xi\left(\abs{v}^{2}-5T\right)\m^{\frac{1}{2}}}}\lesssim \Big(\varepsilon^{-\frac{1}{2}}\nm{\etah^B}_{L^2_{\mathfrak{n}}}\Big)\Big(\varepsilon^{-\frac{1}{2}}\nm{\partial_{\mathfrak{n}}\xi}_{L^2_{\mathfrak{n}}}\Big)
\end{align}
where $\varepsilon^{-\frac{1}{2}}\nm{\etah^B}_{L^2_{\eta}}$ is bounded thanks to the change-of-variable $\frac{\mathfrak{n}}{\varepsilon}=\eta$ which yields $\nm{\cdot}_{L^2_{\mathfrak{n}}}\lesssim\varepsilon^{\frac{1}{2}}\nm{\cdot}_{L^2_{\eta}}$, and $\varepsilon^{-\frac{1}{2}}\nm{\partial_{\mathfrak{n}}\xi}_{L^2_{\mathfrak{n}}}$ is bounded via an interpolation of
Hodge decomposition \eqref{tt 01} with $\varepsilon^{-1}\tnm{\xi}$.

\subparagraph{\underline{BV Estimate to Bound $\varepsilon ^{-1}\br{\partial_{v_{\eta}}f^{B}_1,g^B}$}}
It is well known from the boundary layer theory that the sharp pointwise bound near the grazing set holds
\begin{align}
    \abs{\partial_{v_{\eta}}f^{B}_1}\approx \abs{v_{\eta}}^{-1}.
\end{align}
With a cutoff $\abs{v_{\eta}}\geq\varepsilon$, $\nm{\partial_{v_{\eta}}f^{B}_1}_{L_{v_{\eta}}^{1}}\approx \abs{\ln \varepsilon}$, which creates a fatal logarithmic loss for closure, even with $\varepsilon $ in $\eta $ integration. To overcome such a $\abs{\ln \varepsilon}$ loss,  we
establish a new BV estimate for $f^{B}_1$ in Theorem \ref{boundary regularity}, which amounts to a
subtle but crucial gain in joint $(\eta, v_{\eta})$ integration with no
loss of $\abs{\ln \varepsilon}$ for the cutoff $\nm{\partial_{v_{\eta}}f^{B}_1}_{L_{v_{\eta}}^{1}}\lesssim 1$. For example, we can bound \eqref{final 36}-type terms
\begin{align}
    \varepsilon^{-1}\br{\partial_{\eta}f^{B}_1,g^B}\lesssim \varepsilon^{-1}\bnm{f^{B}_1}\lnm{g^B}\lesssim\lnm{g^B}^2+\varepsilon^{-2}\bnm{f^{B}_1}^2\lesssim o_{T}\tnm{c}^2+o_{T}.
\end{align}
We design an $\varepsilon$ cutoff of the standard boundary-layer Milne solution to achieve this goal. Such a cutoff leads to a non-local commutator of $K$ as in \eqref{mm 00}, which needs to be carefully controlled (see Lemma \ref{ss3-estimate}). Moreover, since it is extremely difficult to  go beyond the first order boundary layer approximation in general domains,  the
presence of boundary layer dictates $\alpha=1$ in \eqref{aa 08}. This can be relaxed for the flat case, for which higher-order boundary layers can
be constructed without any regularity obstruction.\\

\paragraph{\underline{Step 7: $L^2-L^6-L^{\infty}$ Framework}} 
We employ the $L^2-L^6-L^{\infty}$ framework in \cite{Esposito.Guo.Kim.Marra2015} to bootstrap $L^2$ to $L^6$ to $L^{\infty}$:

\begin{proposition}\label{thm:infinity}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, we have
    \begin{align}\label{ss 06}
    \varepsilon^{\frac{1}{2}}\lnmm{R}+\varepsilon^{\frac{1}{2}}\lnmms{R}{\gamma}\lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{proposition}

We note that the singular boundary layer term $\partial_{v_{\eta}}f^{B}_1$ in $S$ (see \eqref{mm 00}) is bounded by $o_{T}\varepsilon^{-1}$ in $L^{\infty}$ norm, which is under control based on \eqref{final 74}. \\

\paragraph{\underline{Step 8: Well-Posedness and Positivity}} 
The main effort of our paper is to obtain an a priori estimate for $\xnm{R}$. For the actual construction, we need to ensure positivity. Our method is different from those in \cite{Esposito.Guo.Marra2018} and \cite{Arkeryd.Nouri2007}. We prove it via a new artificial compensation \eqref{auxiliary system'}. The existence relies on the Fredholm alternative, which is subtle in $L^{\infty}$ space as in Proposition \ref{prop:linear}. We switch to the remainder around a suitable global Maxwellian to avoid the cubic velocity
growth from $\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x}T}{4T^2}$ in \eqref{pp 02}, and design a
huge artificial $\varepsilon^{-1}\textbf{P}_M[\re_M]$ in \eqref{linear remainder'} to overcome a severe boundary error terms, whose
$\varepsilon^{-1}$ dependence does not affect the compactness property. Finally,  since $\alpha=1$, the nonlinearity $\Gamma[R,R]$ has a strong effect, so we must employ the crucial gain of $L^6$ estimates in \cite{Esposito.Guo.Kim.Marra2015} to close the estimate.

Our analysis relies on several precise cancellations via accurate
computations based on new choices of test functions. In spite of this feature, Step 1-4 are completely generic, while Step 5-6 are specifically designed to overcome boundary difficulty, particularly the estimate of \eqref{final 75}.


\subsection{Literature Review}


The hydrodynamic limit of the Boltzmann equation has been the subject of many studies since the pioneering work by Hilbert, who introduced his famous expansion in the Knudsen number $\varepsilon$ in \cite{Hilbert1916, Hilbert1953} to answer the sixth of his famous questions \cite{Hilbert1900}. Mathematical results on the control of the remainder of the Hilbert expansion of the Boltzmann equation to show the convergence to the solutions of the compressible Euler equations for small Knudsen number $\varepsilon$, were obtained by Caflisch \cite{Caflisch1980}, and Lachowicz \cite{Lachowicz1987}, while Nishida \cite{Nishida1978}, Asano-Ukai \cite{Asano.Ukai1983} proved the convergence of the Boltzmann equation to the compressible Euler equations in the hydrodynamic limit by different methods. We also refer to \cite{Huang.Wang.Yang2013, Huang.Wang.Wang.Yang2016, Guo.Huang.Wang2021}. More recently, the convergence in the presence of singularities for the Euler equations have been obtained in \cite{Yu2005} and \cite{Huang.Wang.Yang2010, Huang.Wang.Yang2010(=)}. The relativistic Euler limit has been studied in \cite{Speck.Strain2011}.
We also refer to the recent progress on incompressible Euler limit \cite{Jang.Kim2021, Kim.La2022}.

On a longer time scale $\varepsilon^{-1}$, where diffusion effects become signiﬁcant, the problem can be faced only in the low Mach numbers regime (Mach number of order $\varepsilon$ or smaller) due to the lack of scaling invariance of the compressible Navier–Stokes equations. Hence the Boltzmann solution has been proved to be close to the incompressible Navier–Stokes–Fourier system. Mathematical results were given, among the
others, in \cite{Bardos.Ukai1991, Masi.Esposito.Lebowitz1989, Guo2006, Guo.Jang2010, Guo.Jang.Jiang2010} for smooth solutions. For weak solutions (renormalized solutions), after several partial steps \cite{Bardos.Golse.Levermore1991, Bardos.Golse.Levermore1993, Bardos.Golse.Levermore1998, Bardos.Golse.Levermore2000, Lions.Masmoudi2001, Masmoudi.Saint-Raymond2003}, the convergence of the renormalized solutions has been obtained by Golse and Saint-Raymond \cite{Golse.Saint-Raymond2004}.
For more references, see \cite{Saint-Raymond2009}.

Much less is known about the steady solutions, even for fixed Knudsen number, due to lack of $L^1$ and entropy estimates. Only entropy production from $\displaystyle\int Q[\mathfrak{F},\mathfrak{F}]\ln(\mathfrak{F})$ is available analytically. Our estimate for $\textbf{P}[R]$ can be viewed as a linear version of the entropy production estimate. For entropy production in the full nonlinear setting, we refer to \cite{Villani2002, Desvillettes.Villani2005, Desvillettes.Villani2001, Carlen.Carvalho1994}. In \cite{Guiraud1970, Guiraud1972}, steady solutions were constructed in convex domains near Maxwellians, and their positivity was left open. The only other results are for special, essentially one-dimensional geometry (see \cite{Arkeryd.Nouri2000} for results at ﬁxed Knudsen numbers and \cite{Arkeryd.Esposito.Marra.Nouri2010, Arkeryd.Esposito.Marra.Nouri2011, Esposito.Lebowitz.Marra1994, Esposito.Lebowitz.Marra1995} for results at small Knudsen numbers in certain special 2D geometry). We also refer to the recent review \cite{Esposito.Marra2020}. In recent papers \cite{Esposito.Guo.Kim.Marra2013, Esposito.Guo.Kim.Marra2015, BB002}, the steady solution to the Boltzmann equation close to Maxwellians was constructed, in 3D smooth domains, for a gas in contact with a boundary with a prescribed temperature profile modeled by the diffuse-reflection boundary condition. Also, the Navier-Stokes-Fourier limit was established in \cite{Esposito.Guo.Kim.Marra2015, BB002} for the diffusive
scaling of $\abs{\nabla_{x}T}=o(1)\varepsilon$ based on an improved $L^6-L^{\infty}$ framework. We also refer to the recent papers on the diffusive limit of the Boltzmann equation and related models \cite{Briant.Guo2016, Briant.Merino-Aceituno.Mouhot2019, Jiang.Masmoudi2022, Jang2009}. We list some recent developments along $L^p-L^{\infty}$ framework \cite{Esposito.Guo.Kim.Marra2013, Guo2010, Guo.Jang2010, Guo.Jang.Jiang2010, Guo.Kim.Tonon.Trescases2013, Guo.Kim.Tonon.Trescases2016, Kim2011, Kim2014, Speck.Strain2011, AA003}.

In a pioneering study of the unsteady ghost effect for 1D geometry $-\infty<x<\infty$ \cite{Huang.Wang.Wang.Yang2016}, convergence is established for a temperature variation along the normal direction (non-tangential direction) via delicate analysis in a Sobolev space with a crucial sign condition for $T'$.


\subsection{Appendix and Notation}


Due to the complexity of the problem, we put all the notation, including definitions of $\kappa$, $\sigma$, $\cdots$ etc, in Appendix for the convenience of
the reader.


\section{Boundary Layer Analysis}\label{sec: boundary-analysis}


In this section, we consider the Milne problem for $\g(\eta,\iota_1,\iota_2,{\mathfrak{v}})$ (see Section \ref{sec:geometric-setup}):
\begin{align}\label{Milne}
    \left\{
    \begin{array}{l}
    v_{\eta}\dfrac{\partial\g }{\partial\eta}+\nu_w\g- K_w\left[\g\right]=0,\\\rule{0ex}{1.2em}
    \g(0,\iota_1,\iota_2,{\mathfrak{v}})=\overline{h}(\iota_1,\iota_2,{\mathfrak{v}})\ \ \text{for}\ \ v_{\eta}>0,\\\rule{0ex}{1.5em}
    \displaystyle\int_{\mathbb{R}^3}v_{\eta}\mb^{\frac{1}{2}}(\iota_1,\iota_2,{\mathfrak{v}})\g(0,\iota_1,\iota_2,{\mathfrak{v}})\mathrm{d} {\mathfrak{v}}=0.
    \end{array}
    \right.
\end{align}
Here $\nu_w$ and $K_w$ are actually determined by the boundary Maxwellian $\mu_{w}$. In this section, we temporarily ignore the subscript $w$ for convenience. Also, when there is no confusion, we will not write $(\iota_1,\iota_2)$ explicitly.

\begin{theorem}\label{boundary well-posedness}
Assume $\overline{h}\in W^{k,\infty}_{\iota_1,\iota_2}$ for some $k\in\mathbb{N}$. Then there exists $\g_{\infty}({\mathfrak{v}})\in\mathcal{N}$ and a unique solution $\g(\eta,{\mathfrak{v}})$ to \eqref{Milne} such that $\mathfrak{G}:=\g-\g_{\infty}$ satisfies 
\begin{align}\label{Milne.}
        \left\{
        \begin{array}{l}
        v_{\eta}\dfrac{\partial\mathfrak{G} }{\partial\eta}+\nu \mathfrak{G}-K\left[\mathfrak{G}\right]=0,\\\rule{0ex}{1.2em}
        \mathfrak{G}(0,{\mathfrak{v}})=\overline{h}({\mathfrak{v}})-   \g_{\infty}({\mathfrak{v}}):=\mathfrak{h}({\mathfrak{v}})\ \ \text{for}\ \ v_{\eta}>0,\\\rule{0ex}{1.5em}
        \displaystyle\int_{\mathbb{R}^3}v_{\eta}\m^{\frac{1}{2}}({\mathfrak{v}})\mathfrak{G}(0,{\mathfrak{v}})\mathrm{d} {\mathfrak{v}}=0.
        \end{array}
        \right.
\end{align}
and for some $K_0>0$ and any $0<r\leq k$
\begin{align}
    \abs{\g_{\infty}}+\lnmm{\mathrm{e}^{K_0\eta}\mathfrak{G}}\lesssim& \lnmms{\overline{h}}{\gamma_-},\label{final 41}\\ \lnmm{\mathrm{e}^{K_0\eta}v_{\eta}\partial_{\eta}\mathfrak{G}}+\lnmm{\mathrm{e}^{K_0\eta}v_{\eta}\partial_{v_{\eta}}\mathfrak{G}}\lesssim& \lnmms{\overline{h}}{\gamma_-}+\lnmms{\nabla_{{\mathfrak{v}}}\overline{h}}{\gamma_-},\label{final 42}\\
    \lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\phi}}\mathfrak{G}}+\lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\psi}}\mathfrak{G}}\lesssim& \lnmms{\overline{h}}{\gamma_-}+\lnmms{\nabla_{{\mathfrak{v}}}\overline{h}}{\gamma_-},\label{final 43}\\
    \lnmm{\mathrm{e}^{K_0\eta}\partial_{\iota_1}^{r}\mathfrak{G}}+\lnmm{\mathrm{e}^{K_0\eta}\partial_{\iota_2}^{r}\mathfrak{G}}\lesssim& \lnmms{\overline{h}}{\gamma_-}+\sum_{j=1}^{r}\lnmms{\partial_{\iota_1}^j\overline{h}}{\gamma_-}+\sum_{j=1}^{r}\lnmms{\partial_{\iota_2}^j\overline{h}}{\gamma_-}.\label{final 44}
\end{align}
Also, if $\overline{h}$ is odd in $v_{\phi}$ and $v_{\psi}$, then $\g$, $\mathfrak{G}$ and $\g_{\infty}$ are also odd in $v_{\phi}$ and $v_{\psi}$.
\end{theorem}

\begin{proof}
    This is essentially Theorem 2.6 in the companion paper \cite{AA024}. Based on \cite{Bardos.Caflisch.Nicolaenko1986} and \cite{BB002}, we have the well-posedness of \eqref{Milne} as well as estimates \eqref{final 41}\eqref{final 42} and \eqref{final 43}. Then \eqref{final 44} follows from taking $\iota_i$ derivatives for $i=1,2$ on both sides of \eqref{Milne.} and study the resulting Milne problem.
    The oddness in Lemma \ref{lemma:m6} comes from the invariance of the collision kernel and the uniqueness. When we make the transformation $v_{\phi}\rightarrow-v_{\phi}$ or $v_{\psi}\rightarrow-v_{\psi}$, the equation in \eqref{Milne=} remains unchanged but the sign of the boundary data is flipped \cite{Glassey1996, Cercignani.Illner.Pulvirenti1994}.
\end{proof}

Note that $\partial_{\eta}\mathfrak{G}$ and $\partial_{v_{\eta}}\mathfrak{G}$ estimates require the extra weight $v_{\eta}$. In this section, we will focus on the BV regularity of $\mathfrak{G}$ without the weight.

For $f(\eta,\iota_1,\iota_2,{\mathfrak{v}})$, denote the semi-norm
\begin{align}
    \nm{f}_{\widetilde{\text{BV}}}:=\sup\bigg\{\iint_{\eta,{\mathfrak{v}}}f(\nabla_{\eta,{\mathfrak{v}}}\cdot\psi)\mathrm{d}\eta\mathrm{d}{\mathfrak{v}}:\ \psi\in C^1_c\ \text{and}\ \nm{\psi}_{L^{\infty}}\leq 1\bigg\},
\end{align}
and thus the BV norm can be defined as
\begin{align}
    \bnm{f}:=\pnm{f}{1}+\nm{f}_{\widetilde{\text{BV}}}.
\end{align}
It is classical that $W^{1,1}\hookrightarrow \text{BV}$, and Theorem \ref{boundary well-posedness} helps bound $L^1$ norm of $\mathfrak{G}$, $\dfrac{\partial\mathfrak{G}}{\partial\iota_1}$, $\dfrac{\partial\mathfrak{G}}{\partial\iota_2}$, $\dfrac{\partial\mathfrak{G}}{\partialv_{\phi}}$, $\dfrac{\partial\mathfrak{G}}{\partialv_{\psi}}$. Hence, in the following, we will focus on $L^1$ estimate of $\dfrac{\partial\mathfrak{G}}{\partial\eta}$ and $\dfrac{\partial\mathfrak{G}}{\partialv_{\eta}}$.


\subsection{$L$-Approximate Milne Problem}


In order to study the regularity of \eqref{Milne.}, we first consider an approximate Milne problem for $\ggg^L(\eta,{\mathfrak{v}})$ with $\eta\in[0,L]$ and ${\mathfrak{v}}\in\mathbb{R}^3$:
\begin{align}\label{Milne=}
    \left\{
    \begin{array}{l}
    v_{\eta}\dfrac{\partial\ggg^L }{\partial\eta}+\nu \ggg^L-K\left[\ggg^L\right]=0,\\\rule{0ex}{1.2em}
    \ggg^L(0,{\mathfrak{v}})=\mathfrak{h}({\mathfrak{v}})\ \ \text{for}\ \ v_{\eta}>0,\\\rule{0ex}{1.5em}
    \ggg^L(L,{\mathfrak{v}})=\ggg^L(L,\mathscr{R}[{\mathfrak{v}}]),
        \end{array}
        \right.
\end{align}
Here $\mathscr{R}[{\mathfrak{v}}]=(-v_{\eta},v_{\phi},v_{\psi})$ and $L>0$. 
This is justified since $\g_{\infty}\in\mathcal{N}$ with no $v_{\eta}$ component, then $\g_{\infty}$ satisfies both \eqref{Milne=} and specular-reflection boundary condition at $L$. Therefore, we deduce from the constructions of the $L$- approximate solutions in \cite{Bardos.Caflisch.Nicolaenko1986} and \cite{BB002} that: 
\begin{proposition}\label{lemma:m6}
There exists $\ggg^L_{L}({\mathfrak{v}})\in\mathcal{N}$ and a unique solution $\ggg^L(\eta,{\mathfrak{v}})$ to \eqref{Milne=} such that
\begin{align}
    \abs{\ggg^L_{L}}+\lnmm{\mathrm{e}^{K_0\eta}\big(\ggg^L-\ggg^L_L\big)}\lesssim \lnmms{\mathfrak{h}}{\gamma_-}.
\end{align}
Also, if $\mathfrak{h}$ is odd in $v_{\phi}$ and $v_{\psi}$, then $\ggg^L$ and $\ggg^L_L$ are also odd in $v_{\phi}$ and $v_{\psi}$.
\end{proposition}

We can define an extension of $\ggg^L$ by letting $\ggg^L=0$ for $\eta>L$ (for the convenience of assigning norms). Based on \cite{Bardos.Caflisch.Nicolaenko1986}, \cite{BB002} and \cite[Lemma 4.4]{AA004}, we have 
\begin{lemma}\label{lemma:m8}
We have
\begin{align}
    \lnmmss{\mathrm{e}^{K_0L}\ggg^L_L}\rt0\ \ \text{as}\ \ L\rightarrow\infty.
\end{align}
and 
\begin{align}
    \ggg^L\rightarrow\mathfrak{G}\ \ \text{weakly in $L^2([0,\infty)\times\mathbb{R}^3)$}\ \ \text{as}\ \ L\rightarrow\infty.
\end{align}
\end{lemma}


\subsection{$(L,\theta)$-Approximate Milne Problem}


In order to study \eqref{Milne=}, we further consider a truncated approximate Milne problem for $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ with $\eta\in[0,L]$ and ${\mathfrak{v}}\in\mathbb{R}^3$:
\begin{align}\label{Milne approximate}
    \left\{
    \begin{array}{l}
    v_{\eta}\dfrac{\partial\ggg^{L,\te} }{\partial\eta}+\nu \ggg^{L,\te}-K\left[\ggg^{L,\te}\right]=-\chi\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right],\\\rule{0ex}{1.5em}
    \ggg^{L,\te}(0,{\mathfrak{v}})=\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\mathfrak{h}({\mathfrak{v}})\ \ \text{for}\ \ v_{\eta}>0,\\\rule{0ex}{2.0em}
    \ggg^{L,\te}(L,{\mathfrak{v}})=\ggg^{L,\te}(L,\mathscr{R}[{\mathfrak{v}}]).
    \end{array}
    \right.
\end{align}
Here $0<\theta\ll1$. Clearly, the first equation of \eqref{Milne approximate} is equivalent to
\begin{align}\label{mm 04}
    v_{\eta}\dfrac{\partial\ggg^{L,\te} }{\partial\eta}+\nu \ggg^{L,\te}-\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right]=0.
\end{align}
Compared with \eqref{Milne.}, the key difference lies in the cutoff $\overline{\chi}\left(\theta^{-1}v_{\eta}\right)$. This term breaks the orthogonality, and thus we do not need to prescribe the mass-flux at $\eta=0$. Instead since
\begin{align}
    \int_{\mathbb{R}^3}v_{\eta}\m^{\frac{1}{2}}({\mathfrak{v}})\ggg^{L,\te}(L,{\mathfrak{v}})\mathrm{d}{\mathfrak{v}}=0,
\end{align}
by integrating ${\mathfrak{v}}\in\mathbb{R}^3$ on both sides of \eqref{mm 04}, we have
\begin{align}
    \int_{\mathbb{R}^3}v_{\eta}\m^{\frac{1}{2}}({\mathfrak{v}})\ggg^{L,\te}(0,{\mathfrak{v}})\mathrm{d}{\mathfrak{v}}=-\int_0^L\int_{\mathbb{R}^3}\m^{\frac{1}{2}}({\mathfrak{v}})\chi\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right](\eta,{\mathfrak{v}})\mathrm{d}\eta\mathrm{d}{\mathfrak{v}}.
\end{align}

\begin{lemma}\label{remark boundary}
    We have
    $\ggg^{L,\te}(\eta,{\mathfrak{v}})=0$ for any $\eta\in[0,L]$ and $-\theta\leqv_{\eta}<\theta$.
\end{lemma}
\begin{proof}
Note that the characteristics of the approximate Milne problem \eqref{mm 04} are straight lines $v_{\eta}=\text{const}$.
\begin{itemize}
    \item 
    For any $\eta\in[0,L]$ and $0<v_{\eta}\leq \theta$, we have
    \begin{align}
        \ggg^{L,\te}(\eta,{\mathfrak{v}})=\mathrm{e}^{-\frac{\nu}{v_{\eta}}\eta}\ggg^{L,\te}(0,{\mathfrak{v}})+\int_0^{\eta}\mathrm{e}^{-\frac{\nu}{v_{\eta}}(\eta-y)}\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right](y,{\mathfrak{v}})\mathrm{d} y.
    \end{align}
    Considering $\ggg^{L,\te}(0,{\mathfrak{v}})=\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\mathfrak{h}({\mathfrak{v}})=0$ and $\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right]=0$ for $0\leqv_{\eta}\leq \theta$, we have $\ggg^{L,\te}(\eta,{\mathfrak{v}})=0$ for any $\eta\in[0,L]$ and $0\leqv_{\eta}\leq \theta$. 
    \item
    For any $\eta\in[0,L]$ and $-\theta\leqv_{\eta}< 0$, we have
    \begin{align}
        \ggg^{L,\te}(\eta,{\mathfrak{v}})=\mathrm{e}^{\frac{\nu}{v_{\eta}}(L-\eta)}\ggg^{L,\te}(L,{\mathfrak{v}})+\int_{\eta}^L\mathrm{e}^{\frac{\nu}{v_{\eta}}(y-\eta)}\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right](y,{\mathfrak{v}})\mathrm{d} y.
    \end{align}
    Due to the specular reflection boundary condition at $\eta=L$, we have $\ggg^{L,\te}(L,{\mathfrak{v}})=0$ for $-\theta\leqv_{\eta}< 0$. Also, considering $\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right]=0$ for $0\leqv_{\eta}\leq \theta$, we have
    $\ggg^{L,\te}(\eta,{\mathfrak{v}})=0$ for any $\eta\in[0,L]$ and $-\theta\leqv_{\eta}< 0$.
\end{itemize}
\end{proof}

\begin{lemma}[$L^2$ Bound]\label{lemma:m1'}
Assume $\theta\ll L^{-8}$. Then there exists $\ggg^{L,\te}_{L}({\mathfrak{v}})\in\mathcal{N}$ and a unique solution $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ to \eqref{Milne approximate} satisfying
\begin{align}
    \lnmmss{\ggg^{L,\te}_L}+\tnm{\ggg^{L,\te}-\ggg^{L,\te}_L}\lesssim \tnms{\mathfrak{h}}{\gamma_-}.
\end{align}
Here the constant is independent of $\theta$ and $L$.
\end{lemma}
\begin{proof}
This proof follows from a similar argument as in \cite[Section 4.1.1]{BB002} with the source term 
\begin{align}
    \mathcal{S}^{L,\theta}:=-\chi\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right].
\end{align}
Based on \cite[(3.101)]{Glassey1996}, we have
\begin{align}
    \tnm{\mathcal{S}^{L,\theta}}\lesssim\theta^{\frac{1}{2}}\tnm{K\left[\ggg^{L,\te}\right]}\lesssim\theta^{\frac{1}{2}}\tnm{\ggg^{L,\te}}.
\end{align}
We split $\ggg^{L,\te}=:\mathfrak{w}^{L,\te}+\mathfrak{q}^{L,\te}$, where $\mathfrak{q}^{L,\te}\in\mathcal{N}$ and $\mathfrak{w}^{L,\te}\in\mathcal{N}^{\perp}$.

Following the similar argument as the proof of \cite[Lemma 4.1.5]{BB002} (and noticing that $\mathcal{S}^{L,\theta}$ is not necessarily in $\mathcal{N}^{\perp}$), we have
\begin{align}\label{mm 05'}
    \um{\mathfrak{w}^{L,\te}}^2\lesssim& \tnms{\mathfrak{h}}{\gamma_-}^2+\int_0^L\ggg^{L,\te}\mathcal{S}^{L,\theta}
    \lesssim\tnms{\mathfrak{h}}{\gamma_-}^2+\theta^{-\frac{1}{2}}\tnm{\mathcal{S}^{L,\theta}}^2+\theta^{\frac{1}{2}}\tnm{\mathfrak{w}^{L,\te}}^2+\theta^{\frac{1}{2}}\tnm{\mathfrak{q}^{L,\te}}^2,
\end{align}
which yields for $\theta\ll1$ 
\begin{align}
    \um{\mathfrak{w}^{L,\te}}\lesssim \tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}}\tnm{\ggg^{L,\te}}+\theta^{\frac{1}{4}}\tnm{\mathfrak{q}^{L,\te}}.
\end{align}
Following the similar argument as the proof of \cite[Lemma 4.1.7, Lemma 4.1.10]{BB002} and considering that $L$ is fixed, we obtain that there exists $q_L^{\theta}({\mathfrak{v}})\in\mathcal{N}$ such that
\begin{align}\label{mm 06}
    \abs{q_L^{\theta}}\lesssim&\tnms{\mathfrak{h}}{\gamma_-}+L \tnm{\mathcal{S}^{L,\theta}}+L\tnm{\mathfrak{w}^{L,\te}}
    \lesssim\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}} L\tnm{\ggg^{L,\te}}+\theta^{\frac{1}{4}}L\tnm{\mathfrak{q}^{L,\te}},
\end{align}
and
\begin{align}\label{mm 07'}
    \tnm{\mathfrak{q}^{L,\te}-q_L^{\theta}}\lesssim& \tnms{\mathfrak{h}}{\gamma_-}+\bigg[\int_0^L\bigg(\int_{\eta}^L\mathcal{S}^{L,\theta}(y)\bigg)^2\mathrm{d}\eta\bigg]^{\frac{1}{2}}+\bigg[\int_0^L\bigg(\int_{\eta}^L\mathfrak{w}^{L,\te}(y)\bigg)^2\mathrm{d}\eta\bigg]^{\frac{1}{2}}\\
    \lesssim& \tnms{\mathfrak{h}}{\gamma_-}+ L\tnm{\mathcal{S}^{L,\theta}}+ L\tnm{\mathfrak{w}^{L,\te}}
    \lesssim\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}} L\tnm{\ggg^{L,\te}}+\theta^{\frac{1}{4}}L\tnm{\mathfrak{q}^{L,\te}}.\nonumber
\end{align}
Hence, summarizing \eqref{mm 05'}, \eqref{mm 06} and \eqref{mm 07'}, letting $\ggg^{L,\te}_L=q_L^{\theta}$, we have 
\begin{align}\label{mm 08'}
    \abs{\ggg^{L,\te}_L}+\tnm{\ggg^{L,\te}-\ggg^{L,\te}_L}\lesssim
    \lesssim&\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}}L^2\abs{\ggg^{L,\te}_L}+\theta^{\frac{1}{4}}L\tnm{\ggg^{L,\te}-\ggg^{L,\te}_L},
\end{align}
which, when $\theta\ll L^{-8}$, yields
\begin{align}
    \abs{\ggg^{L,\te}_L}+\tnm{\ggg^{L,\te}-\ggg^{L,\te}_L}\lesssim&\tnms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{proof}

\begin{lemma}[$L^2$ Decay]\label{lemma:m1}
Assume $\theta\ll L^{-8}\mathrm{e}^{-4K_0L}$ with $0\leq K_0<1$. Then the unique solution $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ to \eqref{Milne approximate} satisfies
\begin{align}
    \tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)}\lesssim \tnms{\mathfrak{h}}{\gamma_-}.
\end{align}
Here the constant is independent of $\theta$ and $L$.
\end{lemma}
\begin{proof}
Based on \cite[(3.101)]{Glassey1996}, we have
\begin{align}
    \tnm{\mathrm{e}^{K_0\eta}\mathcal{S}^{L,\theta}}\lesssim\theta^{\frac{1}{2}}\tnm{K\left[\mathrm{e}^{K_0\eta}\ggg^{L,\te}\right]}\lesssim\theta^{\frac{1}{2}}\tnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}.
\end{align}
Following the similar argument as the proof of \cite[Lemma 4.1.9]{BB002}, we have
\begin{align}\label{mm 05}
    \um{\mathrm{e}^{K_0\eta}\mathfrak{w}^{L,\te}}\lesssim& \tnms{\mathfrak{h}}{\gamma_-}+\theta^{-\frac{1}{4}}\tnm{\mathrm{e}^{K_0\eta}\mathcal{S}^{L,\theta}}+\theta^{\frac{1}{4}}\tnm{\mathrm{e}^{K_0\eta}\mathfrak{q}^{L,\te}}\\
    \lesssim& \tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}}\tnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}+\theta^{\frac{1}{4}}\tnm{\mathrm{e}^{K_0\eta}\mathfrak{q}^{L,\te}}.\nonumber
\end{align}
Following the similar argument as the proof of \cite[Lemma 4.1.10]{BB002}, we have
\begin{align}\label{mm 07}
    \tnm{\mathrm{e}^{K_0\eta}\left(\mathfrak{q}^{L,\te}-q_L^{\theta}\right)}\lesssim& \tnms{\mathfrak{h}}{\gamma_-}+\bigg[\int_0^L\bigg(\int_{\eta}^L\mathrm{e}^{K_0y}\mathcal{S}^{L,\theta}(y)\bigg)^2\mathrm{d}\eta\bigg]^{\frac{1}{2}}+\bigg[\int_0^L\bigg(\int_{\eta}^L\mathrm{e}^{K_0y}\mathfrak{w}^{L,\te}(y)\bigg)^2\mathrm{d}\eta\bigg]^{\frac{1}{2}}\\
    \lesssim& \tnms{\mathfrak{h}}{\gamma_-}+L\tnm{\mathrm{e}^{K_0\eta}\mathcal{S}^{L,\theta}}+ L\tnm{\mathrm{e}^{K_0\eta}\mathfrak{w}^{L,\te}}\nonumber\\
    \lesssim&\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}} L\tnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}+\theta^{\frac{1}{4}}L\tnm{\mathrm{e}^{K_0\eta}\mathfrak{q}^{L,\te}}.\nonumber
\end{align}
Hence, summarizing \eqref{mm 05} and \eqref{mm 07}, we have 
\begin{align}\label{mm 08}
    \tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)}
    \lesssim&\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}}L^2\mathrm{e}^{K_0L}\abs{\ggg^{L,\te}_L}+\theta^{\frac{1}{4}}L\tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)},
\end{align}
which, when $\theta^{\frac{1}{4}}\ll L^{-2}\mathrm{e}^{-K_0L}$, yields
\begin{align}
    \tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)}\lesssim&\tnms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{proof}

\begin{lemma}[$L^{\infty}$ Bound and Decay]\label{lemma:m2}
There exists $\ggg^{L,\te}_{L}({\mathfrak{v}})\in\mathcal{N}$ and a unique solution $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ to \eqref{Milne approximate} satisfying
\begin{align}
    \lnmm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)}\lesssim \lnmms{\mathfrak{h}}{\gamma_-},
\end{align}
Here the constant is independent of $\theta$ and $L$.
\end{lemma}
\begin{proof}
This follows from \cite[Theorem 4.1.24]{BB002}.
\end{proof}

Summarizing Lemma \ref{lemma:m1'}, Lemma \ref{lemma:m1} and Lemma \ref{lemma:m2}, we arrive at the following:

\begin{proposition}\label{prop:Milne-approximate}
Assume $\theta\ll L^{-8}\mathrm{e}^{-4K_0L}$ with $0\leq K_0<1$. Then the unique solution $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ to \eqref{Milne approximate} satisfying
\begin{align}
    \lnmm{\ggg^{L,\te}}\lesssim \lnmms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{proposition}

\begin{remark}\label{rem:boundary}
Due to the presence of $\ggg^{L,\te}_L$, we cannot simply deduce that 
\begin{align}
    \tnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}+\lnmm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}\lesssim \lnmms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{remark}


\subsection{$L^1$ Estimate for $\partial_{\eta}\ggg^{L,\te}$}


\begin{lemma}\label{lemma:m3}
We have
\begin{align}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} }{1}\lesssim\pnm{\nu^2\ggg^{L,\te}}{1}+\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\lnmm{\ggg^{L,\te}}.
\end{align}
Here the constant is independent of $\theta$ and $L$.
\end{lemma}

\begin{proof}
For some $0<\delta\ll1$, considering \eqref{mm 04}, we have
\begin{align}\label{mm 11}
\\
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\geq\delta\}}}{1}\lesssim& \delta^{-1}\pnm{\nuv_{\eta}\partial_{\eta}\ggg^{L,\te} }{1}
    \lesssim\delta^{-1}\pnm{\nu^2\ggg^{L,\te}}{1}+\delta^{-1}\pnm{\nu K\left[\ggg^{L,\te}\right]}{1}\lesssim\delta^{-1}\pnm{\nu^2\ggg^{L,\te}}{1},\nonumber
\end{align}
and thus it suffices to bound 
\begin{align}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1}.
\end{align}
Taking $\eta$ derivative in \eqref{mm 04}, we obtain
\begin{align}\label{mm 09}
    v_{\eta}\partial_{\eta}\left(\partial_{\eta}\ggg^{L,\te} \right)+\nu \left(\partial_{\eta}\ggg^{L,\te} \right)=\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\partial_{\eta}\ggg^{L,\te} \right].
\end{align}
This equation holds for any $(\eta,{\mathfrak{v}})$, so we focus on the $\abs{v_{\eta}}\leq\delta$ part:
\begin{align}\label{mm 02}
    &v_{\eta}\partial_{\eta}\left(\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\right)+\nu \left(\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\right)=\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}K\left[\partial_{\eta}\ggg^{L,\te} \right]\\
    =&\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\geq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}+\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\leq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}.\nonumber
\end{align}
Here, ${\mathfrak{u}}=({\mathfrak{u}}_{\eta},{\mathfrak{u}}_{\phi},{\mathfrak{u}}_{\psi})$. By multiplying $\text{sgn}\left[\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\right]$ and integrating over $(\eta,{\mathfrak{v}})$ in \eqref{mm 02}, we obtain
\begin{align}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1}\lesssim&\int_{\mathbb{R}^3}\abs{\partial_{\eta}\ggg^{L,\te}(0)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}v_{\eta}\mathrm{d}{\mathfrak{v}}-\int_{\mathbb{R}^3}\abs{\partial_{\eta}\ggg^{L,\te}(L)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}v_{\eta}\mathrm{d}{\mathfrak{v}}\\
    &+\pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\geq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}}{1}\nonumber\\
    &+\pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\leq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d}{\mathfrak{u}}}{1}.\nonumber
\end{align}
Plugging in \eqref{Milne approximate} yields
\begin{align}\label{mm 10}
    \int_{\mathbb{R}^3}\abs{\partial_{\eta}\ggg^{L,\te}(0)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}v_{\eta}\mathrm{d}{\mathfrak{v}}\lesssim& \lnmms{\nu\ggg^{L,\te}(0)}{\gamma}+\lnmms{\chi\left(\theta^{-1}v_{\eta}\right)K\big[\ggg^{L,\te}\big](0)}{\gamma}\\
    \lesssim&\lnmms{\ggg^{L,\te}(0)}{\gamma}\lesssim\lnmm{\ggg^{L,\te}},\nonumber
\end{align}
and similarly
\begin{align}
    \int_{\mathbb{R}^3}\abs{\partial_{\eta}\ggg^{L,\te}(L)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}v_{\eta}\mathrm{d}{\mathfrak{v}}\lesssim\lnmm{\ggg^{L,\te}}.
\end{align}
Also, we know
\begin{align}
    &\pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\geq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}}{1}\\
    \lesssim&\nm{\int_{\abs{{\mathfrak{u}}_{\eta}}\geq\delta}k({\mathfrak{v}},{\mathfrak{u}})\mathrm{d} {\mathfrak{u}}}_{L^2_{{\mathfrak{v}}}}\nm{\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\geq\delta\}}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}
    \lesssim \delta^{-1}\nm{v_{\eta}\partial_{\eta}\ggg^{L,\te} }_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}\lesssim\delta^{-1}\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}.\nonumber
\end{align}
The remaining principal term is
\begin{align}\label{final 24}
    &\pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\int_{\abs{{\mathfrak{u}}_{\eta}}\leq\delta}k({\mathfrak{v}},{\mathfrak{u}})\partial_{\eta}\ggg^{L,\te} ({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}}{1}\\
    \lesssim&\left(\sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}k({\mathfrak{v}},{\mathfrak{u}})\mathrm{d} {\mathfrak{v}}\right)\pnm{\textbf{1}_{\{\abs{{\mathfrak{u}}_{\eta}}\leq\delta\}}\partial_{\eta}\ggg^{L,\te}({\mathfrak{u}})}{1}
    \lesssim\delta\pnm{\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1},\nonumber
\end{align}
where for ${\mathfrak{v}}'=(v_{\phi},v_{\psi})$ and ${\mathfrak{u}}'=(u_{\phi},u_{\psi})$, due to \cite[Lemma 3]{Guo2010}
\begin{align}\label{final 24'}
    \sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}k({\mathfrak{v}},{\mathfrak{u}})\mathrm{d} {\mathfrak{v}}\lesssim&\sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\abs{{\mathfrak{v}}-{\mathfrak{u}}}^{-1}\mathrm{e}^{-\frac{1}{8}\abs{{\mathfrak{v}}-{\mathfrak{u}}}^2}\mathrm{d} {\mathfrak{v}}\\
    \lesssim&\sup_{{\mathfrak{u}}}\int_{\mathbb{R}}\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\bigg(\int_{\mathbb{R}^2}\abs{{\mathfrak{v}}'-{\mathfrak{u}}'}^{-1}\mathrm{e}^{-\frac{1}{8}\abs{{\mathfrak{v}}-{\mathfrak{u}}}^2}\mathrm{d} {\mathfrak{v}}'\bigg)\mathrm{d}v_{\eta}\nonumber\\
    \lesssim&\sup_{{\mathfrak{u}}}\int_{\mathbb{R}}\textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}\mathrm{d}v_{\eta} \lesssim\delta.\nonumber
\end{align}
In summary, we have shown that
\begin{align}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1}\lesssim \delta^{-1}\pnm{\nu\ggg^{L,\te}}{1}+\delta^{-1}\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\delta\pnm{\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1}+\lnmm{\ggg^{L,\te}},
\end{align}
which yields
\begin{align}\label{mm 12}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} \textbf{1}_{\{\abs{v_{\eta}}\leq\delta\}}}{1}\lesssim \delta^{-1}\pnm{\nu\ggg^{L,\te}}{1}+\delta^{-1}\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}.
\end{align}
Combining \eqref{mm 11} and \eqref{mm 12}, we have
\begin{align}
    \pnm{\nu\partial_{\eta}\ggg^{L,\te} }{1}\lesssim \delta^{-1}\pnm{\nu^2\ggg^{L,\te}}{1}+\delta^{-1}\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\lnmm{\ggg^{L,\te}}.
\end{align}
\end{proof}


\subsection{$L^1$ Estimate for $\partial_{v_{\eta}}\ggg^{L,\te}$}


\begin{lemma}\label{lemma:m4}
We have
\begin{align}
    \pnm{\nu\partial_{v_{\eta}}\ggg^{L,\te} }{1}\lesssim&\pnm{\nu^2\ggg^{L,\te}}{1}+\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\lnmm{\ggg^{L,\te}}
    +\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}}.
\end{align}
Here the constant is independent of $\theta$ and $L$.
\end{lemma}

\begin{proof}
Taking $v_{\eta}$ derivative in \eqref{Milne approximate}, we obtain
\begin{align}\label{mm 03}
\\
    v_{\eta}\partial_{\eta}\left(\partial_{v_{\eta}}\ggg^{L,\te} \right)+\partial_{\eta}\ggg^{L,\te} +\nu\partial_{v_{\eta}}\ggg^{L,\te} +\partial_{v_{\eta}}\nu\ggg^{L,\te}=\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\partial_{v_{\eta}}K\left[\ggg^{L,\te}\right]+\theta^{-1}\overline{\chi}'\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right].\nonumber
\end{align}
By multiplying $\text{sgn}\left[\partial_{v_{\eta}}\ggg^{L,\te} \right]$ and integrating over $(\eta,v)$ in \eqref{mm 03}, 
we obtain
\begin{align}
    \pnm{\nu\partial_{v_{\eta}}\ggg^{L,\te} }{1}\lesssim&\abs{\int_{\mathbb{R}^3}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(0)}v_{\eta}\mathrm{d}{\mathfrak{v}}-\int_{\mathbb{R}^3}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(L)}v_{\eta}\mathrm{d}{\mathfrak{v}}}+\pnm{\partial_{\eta}\ggg^{L,\te}}{1}+\pnm{\partial_{v_{\eta}}\nu\ggg^{L,\te}}{1}\\
    &+\pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\partial_{v_{\eta}}K\left[\ggg^{L,\te}\right]}{1}+\pnm{\theta^{-1}\overline{\chi}'\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right]}{1}.\nonumber
\end{align}
Using Lemma \ref{remark boundary}, 
we have
\begin{align}
    \int_{\mathbb{R}^3}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(0)}v_{\eta}\mathrm{d}{\mathfrak{v}}=&\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(0)}v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}<0}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(0)}v_{\eta}\mathrm{d}{\mathfrak{v}}\\
    \leq&\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(0)}v_{\eta}\mathrm{d}{\mathfrak{v}}=\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\Big(\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\mathfrak{h}({\mathfrak{v}})\Big)}v_{\eta}\mathrm{d}{\mathfrak{v}}\nonumber\\
    \leq&\int_{v_{\eta}>0}\abs{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\theta^{-1}\overline{\chi}'\left(\theta^{-1}v_{\eta}\right)\mathfrak{h}}v_{\eta}\mathrm{d}{\mathfrak{v}}\nonumber\\
    \lesssim&\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}}.\nonumber
\end{align}
Due to specular-reflection boundary at $\eta=L$, we have
\begin{align}
    \int_{\mathbb{R}^3}\abs{\partial_{v_{\eta}}\ggg^{L,\te}(L)}v_{\eta}\mathrm{d}{\mathfrak{v}}=0.
\end{align}
Using Lemma \ref{lemma:m3}, we have
\begin{align}
    \pnm{\partial_{\eta}\ggg^{L,\te}}{1}\lesssim \pnm{\nu^2\ggg^{L,\te}}{1}+\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\lnmm{\ggg^{L,\te}}.
\end{align}
Direct estimate implies
\begin{align}
    \pnm{\partial_{v_{\eta}}\nu\ggg^{L,\te}}{1}\lesssim\pnm{\ggg^{L,\te}}{1},
\end{align}
and 
\begin{align}
    \pnm{\overline{\chi}\left(\theta^{-1}v_{\eta}\right)\partial_{v_{\eta}}K\left[\ggg^{L,\te}\right]}{1}\lesssim\pnm{\nu\ggg^{L,\te}}{1}.
\end{align}
The remaining principal term is
\begin{align}
    \pnm{\theta^{-1}\overline{\chi}'\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}\right]}{1}\lesssim&\theta^{-1}\pnm{\overline{\chi}'\left(\theta^{-1}v_{\eta}\right)\int_{\mathbb{R}^3}k({\mathfrak{v}},{\mathfrak{u}})\ggg^{L,\te}({\mathfrak{u}})\mathrm{d} {\mathfrak{u}}}{1}\\
    \lesssim&\theta^{-1}\left(\sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v_{\eta}}\leq\theta\}}k({\mathfrak{v}},{\mathfrak{u}})\mathrm{d} {\mathfrak{v}}\right)\pnm{\ggg^{L,\te}}{1}
    \lesssim\pnm{\ggg^{L,\te}}{1},\nonumber
\end{align}
where based on \eqref{final 24'}
\begin{align}
    \sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v_{\eta}}\leq\theta\}}k({\mathfrak{v}},{\mathfrak{u}})\mathrm{d} {\mathfrak{v}}\lesssim\theta.
\end{align}
In summary, we have shown that
\begin{align}
    \pnm{\nu\partial_{v_{\eta}}\ggg^{L,\te} }{1}\lesssim\pnm{\nu^2\ggg^{L,\te}}{1}+\nm{\nu\ggg^{L,\te}}_{L^1_{\eta}L^{2}_{{\mathfrak{v}}}}+\lnmm{\ggg^{L,\te}}+\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}}.
\end{align}
\end{proof}


\subsection{Limit $\theta\rt0$}


We formally take $\theta\rt0$ in \eqref{Milne approximate} and recover \eqref{Milne=}.

\begin{lemma}\label{lem:m6}
We have
\begin{align}
    \lnmmss{\ggg^{L,\te}_L-\ggg^L_L}+\tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^L\big)}\rt0\ \ \text{as}\ \ \theta\rt0.
\end{align}
\end{lemma}
\begin{proof}
This can be easily shown from the difference equation of $\ggg^{L,\te}-\ggg^L$:
\begin{align}\label{Milne approximate difference}
    \left\{
    \begin{array}{l}
    v_{\eta}\dfrac{\partial\big(\ggg^{L,\te}-\ggg^L\big) }{\partial\eta}+\nu\big(\ggg^{L,\te}-\ggg^L\big)-\overline{\chi}\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^{L,\te}-\ggg^L\right]=\chi\left(\theta^{-1}v_{\eta}\right)K\left[\ggg^L\right],\\\rule{0ex}{1.2em}
    \big(\ggg^{L,\te}-\ggg^L\big)(0,{\mathfrak{v}})=-\chi\left(\theta^{-1}v_{\eta}\right)\mathfrak{h}({\mathfrak{v}})\ \ \text{for}\ \ v_{\eta}>0,\\\rule{0ex}{1.5em}
    \big(\ggg^{L,\te}-\ggg^L\big)(L,{\mathfrak{v}})=\big(\ggg^{L,\te}-\ggg^L\big)(L,\mathscr{R}[{\mathfrak{v}}]).
    \end{array}
    \right.
\end{align}
Using the proof of Lemma \ref{lemma:m1'}, Lemma \ref{lemma:m1} and Lemma \ref{lemma:m2}, we have
\begin{align}
\\
    \lnmmss{\ggg^{L,\te}_L-\ggg^L_L}+\tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^L\big)}\lesssim\theta\tnms{\mathfrak{h}}{\gamma_-}+\theta^{\frac{1}{4}}L\tnm{\mathrm{e}^{K_0\eta}\big(\ggg^{L,\te}-\ggg^{L,\te}_L\big)}+\theta^{\frac{1}{4}}\mathrm{e}^{K_0L}\tnm{\ggg^{L,\te}_L}.\nonumber
\end{align}
Then for fixed $L$, the result naturally follows as $\theta\rt0$.
\end{proof}

\begin{corollary}\label{prop:Milne-approximate-improve}
Assume $\theta\ll L^{-8}\mathrm{e}^{-4K_0L}$ with $0\leq K_0<1$. Then the unique solution $\ggg^{L,\te}(\eta,{\mathfrak{v}})$ to \eqref{Milne approximate} satisfies
\begin{align}
    \tnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}+\lnmm{\ggg^{L,\te}}\lesssim \lnmms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{corollary}
\begin{proof}
    Combining Lemma \ref{lem:m6} and Proposition \ref{lemma:m6}, we obtain the $L^2$ bound. Then $L^{\infty}$ bound is from Proposition \ref{prop:Milne-approximate}.
\end{proof}

\begin{remark}\label{rem:boundary-uniform}
    Using H\"older's inequality and the utilizing the $L^2$ decay, the similar results as in Corollary \ref{prop:Milne-approximate-improve} also holds for any $r\in[1,\infty)$:
    \begin{align}
    \pnm{\mathrm{e}^{K_0\eta}\ggg^{L,\te}}{r}\lesssim \lnmms{\mathfrak{h}}{\gamma_-}.
\end{align}
\end{remark}

Next we turn to the BV estimates.
\begin{proposition}\label{lemma:m7}
We have
\begin{align}
    \bnm{\nu\ggg^L}\lesssim\lnmms{\mathfrak{h}}{\gamma_-}+\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}}.
\end{align}
\end{proposition}

\begin{proof}
Using Corollary \ref{prop:Milne-approximate-improve} and Remark \ref{rem:boundary-uniform}, we conclude that the estimates in Lemma \ref{lemma:m3} and Lemma \ref{lemma:m4} with $K_0=0$ are uniform in $\theta$. In combination with Theorem \ref{boundary well-posedness}, we have shown that 
\begin{align}\label{final 45}
    \nm{\ggg^{L,\te}}_{W^{1,1}}\lesssim\lnmms{\mathfrak{h}}{\gamma_-}+\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}},
\end{align}
uniformly in $\theta$, which yields a weakly $W^{1,1}$ convergent subsequence as $\theta\rt0$. 
Also, Proposition \ref{lem:m6} yields strong $L^{2}$ convergence as $\theta\rt0$.

Combining both of the convergence, we obtain the BV convergence of the subsequence $\ggg^{L,\te}\rightarrow\ggg^L$ as $\theta\rt0$, and thus, with the help of weak lower semi-continuity, we can pass to limit in \eqref{final 45} to obtain the desired result.
\end{proof}


\subsection{Limit $L\rightarrow\infty$}


Based on Lemma \ref{lemma:m8}, we take $L\rightarrow\infty$ in \eqref{Milne=} and recover the equation \eqref{Milne.}.

Finally, we can derive the BV estimates of $\mathfrak{G}$:
\begin{theorem}\label{boundary regularity}
We have
\begin{align}
    \bnm{\nu\mathfrak{G}}\lesssim\lnmms{\mathfrak{h}}{\gamma_-}+\int_{v_{\eta}>0}\abs{\partial_{v_{\eta}}\mathfrak{h} }v_{\eta}\mathrm{d}{\mathfrak{v}}+\int_{v_{\eta}>0}\abs{\mathfrak{h}}\mathrm{d}{\mathfrak{v}}.
\end{align}
\end{theorem}

\begin{proof}
    For any $C_c^{\infty}$ test function $\phi(\eta,{\mathfrak{v}})$ with compact support contained in $(\eta,{\mathfrak{v}})\in [0,L']\times B_r$, choosing $L>L'$ and using Lemma \ref{lemma:m8}, we have as $L\rightarrow\infty$
    \begin{align}
        \iint_{[0,L]\times\mathbb{R}^3}\frac{\partial\ggg^L}{\partial\eta}\phi=-\iint_{[0,L]\times\mathbb{R}^3}\ggg^L\frac{\partial\phi}{\partial\eta}\rightarrow \iint_{[0,L]\times\mathbb{R}^3}\mathfrak{G}\frac{\partial\phi}{\partial\eta}=\iint_{[0,L]\times\mathbb{R}^3}\frac{\partial\mathfrak{G}}{\partial\eta}\phi.
    \end{align}
    Hence, we know that the weak derivative $\dfrac{\partial\ggg^L}{\partial\eta}\rightarrow \dfrac{\partial\mathfrak{G}}{\partial\eta}$ in the sense of measure. Similarly, we can show that the weak derivative $\dfrac{\partial\ggg^L}{\partialv_{\eta}}\rightarrow \dfrac{\partial\mathfrak{G}}{\partialv_{\eta}}$ in the sense of measure. In addition, we know that
    \begin{align}
        \nm{\mathfrak{G}}_{\widetilde{\text{BV}}}\lesssim\nm{\ggg^L}_{\widetilde{\text{BV}}}.
    \end{align}
    Hence, using Proposition  \ref{lemma:m7}, our result follows.
\end{proof}


\section{Setup of Remainder Estimates}\label{sec:remainder}


Now we begin to estimate the remainder equation for $R$ in \eqref{aa 08}, or equivalently the nonlinear Boltzmann equation \eqref{large system-}. 
Denote 
\begin{align}
    Q[F,F]=&Q_{\text{gain}}[F,F]-Q_{\text{loss}}[F,F]\\
    :=&\int_{\mathbb{R}^3}\int_{\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})F({\mathfrak{u}}_{\ast})F(v_{\ast})\mathrm{d}{\omega}\mathrm{d}{{\mathfrak{u}}}-F(v)\int_{\mathbb{R}^3}\int_{\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})F({\mathfrak{u}})\mathrm{d}{\omega}\mathrm{d}{{\mathfrak{u}}}=\nu(F)F.\nonumber
\end{align}
Denote $\mathfrak{F}=\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR$,
where
\begin{align}
    \mathfrak{F}_{\text{a}}:=\mu+\m^{\frac{1}{2}}\big(\ef_1+\varepsilon^2f_2\big)+\mb^{\frac{1}{2}}\big(\ef^{B}_1\big).
\end{align}
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 \eqref{large system-}, we first consider an auxiliary equation
\begin{align}\label{auxiliary system}
    v\cdot\nabla_{x} \mathfrak{F}+\varepsilon^{-1}\Big(Q_{\text{loss}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)=\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{loss}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big),
\end{align}
with diffuse-reflection boundary condition
\begin{align}
    \mathfrak{F}(x_0,v)=M_{w}(x_0,v)\displaystyle\int_{{\mathfrak{u}}\cdotn(x_0)>0}
    \mathfrak{F}(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}} \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{align}
Here $\mathfrak{z}=\mathfrak{z}(v)>0$ is a smooth function with support contained in $\{\abs{v}\leq 1\}$ such that $\displaystyle\iint_{\Omega\times\mathbb{R}^3}\mathfrak{z}=1$.
The auxiliary system \eqref{auxiliary system} is equivalent to
\begin{align}
\\
v\cdot\nabla_{x} \mathfrak{F}-\varepsilon^{-1}Q[\mathfrak{F},\mathfrak{F}]
=&-\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{loss}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big),\nonumber
\end{align} 
and due to orthogonality of $Q$, is further equivalent to
\begin{align}\label{auxiliary system'}
\\
v\cdot\nabla_{x} \mathfrak{F}-\varepsilon^{-1}Q[\mathfrak{F},\mathfrak{F}]
=&-\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big).\nonumber
\end{align}

\begin{remark}
The extra terms
\begin{align}
    -\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big).
\end{align}
on the RHS of \eqref{auxiliary system'} plays a significant role in justifying the positivity of $\mathfrak{F}$ (see Section \ref{sec:well-posedness}).
Clearly, when $\mathfrak{F}\geq0$, i.e. $\mathfrak{F}=\mathfrak{F}_+$, the above extra terms vanish and the auxiliary equation \eqref{auxiliary system'} reduces to \eqref{large system-}.
\end{remark}


Denote $\mathfrak{F}_{\text{a}}=\mu+\widetilde{\mathfrak{F}}_{\text{a}}$ where
\begin{align}
    \widetilde{\mathfrak{F}}_{\text{a}}:=\m^{\frac{1}{2}}\big(\ef_1+\varepsilon^2f_2\big)+\mb^{\frac{1}{2}}\big(\ef^{B}_1\big).
\end{align}
Inserting $\mathfrak{F}=\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR=\mu+\widetilde{\mathfrak{F}}_{\text{a}}+\varepsilon^{\alpha}\mhR$ into \eqref{auxiliary system'}, we have
\begin{align}\label{wt 02}
    &v\cdot\nabla_{x} \left(\mhR\right)+\varepsilon^{-1}\m^{\frac{1}{2}}\mathcal{L}[R]\\
    =&\ \mathscr{S}+\varepsilon^{-1}\bigg(2Q^{\ast}\left[\widetilde{\mathfrak{F}}_{\text{a}},\mhR\right]+\varepsilon^{\alpha}Q^{\ast}\left[\mhR,\mhR\right]\bigg)\nonumber\\
    &-\varepsilon^{-\alpha}\bigg(\varepsilon^{-1}Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right]-\varepsilon^{-1}Q_{\text{gain}}\left[\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+,\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+\right]\bigg)\nonumber\\
    &+\varepsilon^{-\alpha}\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\bigg(\varepsilon^{-1}Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right]-\varepsilon^{-1}Q_{\text{gain}}\left[\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+,\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+\right]\bigg),\nonumber
\end{align}
where 
\begin{align}
    \mathscr{S}:=-\varepsilon^{-\alpha}v\cdot\nabla_{x}\mathfrak{F}_{\text{a}}+\varepsilon^{-\alpha-1}Q^{\ast}\left[\mathfrak{F}_{\text{a}},\mathfrak{F}_{\text{a}}\right].
\end{align}
Then we arrive at \eqref{remainder} with $h$ and $S$ defined in \eqref{aa 32} and \eqref{aa 31}.

\begin{lemma}\label{lemma:final 1}
    For $h$ and $S$ defined in \eqref{aa 32} and \eqref{aa 31}, we have the compatibility condition
    \begin{align}\label{compatibility}
    \iint_{\Omega\times\mathbb{R}^3}\m^{\frac{1}{2}} S+\int_{\gamma_-}\m^{\frac{1}{2}} h\mathrm{d}{\gamma}=0.
    \end{align}
\end{lemma}
\begin{proof}
We can directly verify that
\begin{align}\label{aa 36'}
    \int_{\gamma_-}h(x,v)\m^{\frac{1}{2}}(x,v)\mathrm{d}{\gamma}=&\varepsilon^{-\alpha}\left(\int_{\gamma_-}\mathcal{P}_{\gamma}\left[\m^{-\frac{1}{2}}\mathfrak{F}_{\text{a}}\right]\m^{\frac{1}{2}}\mathrm{d}{\gamma}-\int_{\gamma_-}\mathfrak{F}_{\text{a}}\mathrm{d}{\gamma}\right)\\
    =&\varepsilon^{-\alpha}\left(\int_{\gamma_+}\mathcal{P}_{\gamma}\left[\m^{-\frac{1}{2}}\mathfrak{F}_{\text{a}}\right]\m^{\frac{1}{2}}\mathrm{d}{\gamma}-\int_{\gamma_-}\mathfrak{F}_{\text{a}}\mathrm{d}{\gamma}\right)\nonumber\\
    =&\varepsilon^{-\alpha}\left(\int_{\gamma_+}\mathfrak{F}_{\text{a}}\mathrm{d}{\gamma}-\int_{\gamma_-}\mathfrak{F}_{\text{a}}\mathrm{d}{\gamma}\right)=\varepsilon^{-\alpha}\int_{\gamma}\mathfrak{F}_{\text{a}}(v\cdot n)\mathrm{d}v\mathrm{d} S_x.\nonumber
\end{align}
Using the orthogonality of $Q^{\ast}$ and the designed zero-mass condition of $\ss_6$, we verify that
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}S(x,v)\m^{\frac{1}{2}}(x,v)\mathrm{d}{v}\mathrm{d}{x}
    =&-\varepsilon^{-\alpha}\iint_{\Omega\times\mathbb{R}^3}\Big(v\cdot\nabla_{x}\mathfrak{F}_{\text{a}}\Big)\mathrm{d}{v}\mathrm{d}{x}
    =-\varepsilon^{-\alpha}\int_{\gamma}\mathfrak{F}_{\text{a}}(v\cdot n)\mathrm{d}v\mathrm{d} S_x.
\end{align}
Hence, \eqref{compatibility} is always guaranteed.
\end{proof}

\begin{proposition}\label{lemma:final 2}
    For $h$ and $S$ defined in \eqref{aa 32} and \eqref{aa 31}, we have
    \begin{align}\label{aa 36}
    \int_{\gamma_-}\m^{\frac{1}{2}} h\mathrm{d}{\gamma}=0,\quad \iint_{\Omega\times\mathbb{R}^3}\m^{\frac{1}{2}} S=0.
    \end{align}
\end{proposition}
\begin{proof}
    This is guaranteed by \eqref{aa 36'}, $u_1\cdot n=0$ and the choice of $u_2$ (see \eqref{aa 38}). Then the result follows from Lemma \ref{lemma:final 1}.
\end{proof}

\begin{remark}\label{remark 02}
Using Proposition \ref{lemma:final 2}, we can directly compute that for $x_0\in\partial\Omega$
\begin{align}\label{final 53}
    \textbf{b}(x_0)\cdot n=&\int_{\mathbb{R}^3}R(x_0)\m^{\frac{1}{2}}(v\cdot n)\mathrm{d} v=\int_{v\cdot n<0}h(x_0)\m^{\frac{1}{2}}(v\cdot n)\mathrm{d} v=0.
\end{align}
\end{remark}

\begin{lemma}[Green's Identity, Lemma 2.2 of \cite{Esposito.Guo.Kim.Marra2013}]\label{remainder lemma 2}
Assume $f(x,v),\ g(x,v)\in L^2_{\nu}(\Omega\times\mathbb{R}^3)$ and
$v\cdot\nabla_{x} f,\ v\cdot\nabla_{x} g\in L^2(\Omega\times\mathbb{R}^3)$ with $f,\
g\in L^2(\gamma)$. Then
\begin{align}
\iint_{\Omega\times\mathbb{R}^3}\Big(\left(v\cdot\nabla_{x} f\right)g+(v\cdot\nabla_{x}
g)f\Big)\mathrm{d}{x}\mathrm{d}{v}=\int_{\gamma}fg(v\cdot n)=\int_{\gamma_+}fg\mathrm{d}{\gamma}-\int_{\gamma_-}fg\mathrm{d}{\gamma}.
\end{align}
\end{lemma}

Using Lemma \ref{remainder lemma 2}, we can derive the weak formulation of \eqref{remainder.}. For any test function $\mathfrak{f}(x,v)\in L^2_{\nu}(\Omega\times\mathbb{R}^3)$ with $v\cdot\nabla_{x}\mathfrak{f}\in L^2(\Omega\times\mathbb{R}^3)$ with $\mathfrak{f}\in L^2(\gamma)$, we have
\begin{align}\label{weak formulation}
\\
    \int_{\gamma}R\mathfrak{f}(v\cdot n)-\iint_{\Omega\times\mathbb{R}^3}R\big(v\cdot\nabla_{x} \mathfrak{f}\big)+\iint_{\Omega\times\mathbb{R}^3}\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x}T}{4T^2}\right)R \mathfrak{f}+\varepsilon^{-1}\iint_{\Omega\times\mathbb{R}^3}\mathcal{L}[R]\mathfrak{f}=\iint_{\Omega\times\mathbb{R}^3}S\mathfrak{f}.\nonumber
\end{align}


\subsection{Preliminary Estimates}


\begin{lemma}[Lemma 2.3 of \cite{Guo2003}]\label{lemma:a1}
We have
\begin{align}
    \abs{\brv{\Gamma[f_1,f_2],g}}\lesssim&\Bigg\{\left(\int_{\mathbb{R}^3}\nu\abs{f_1}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\abs{f_2}^2\right)^{\frac{1}{2}}+\left(\int_{\mathbb{R}^3}\nu\abs{f_2}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\abs{f_1}^2\right)^{\frac{1}{2}}\Bigg\}\left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}},\\
    \tnm{\brv{\Gamma[f_1,f_2],g}}\lesssim& \Big(\sup_{x,v}\abs{\nu^3g}\Big)\min\Bigg\{\sup_x\left(\int_{\mathbb{R}^3}\abs{f_1}^2\right)^{\frac{1}{2}}\tnm{f_2}, \sup_x\left(\int_{\mathbb{R}^3}\abs{f_2}^2\right)^{\frac{1}{2}}\tnm{f_1}\Bigg\},\\
    \tnm{\Gamma[f_1,f_2]g}\lesssim& \Big(\sup_{x,v}\abs{\nu g}\Big)\min\Bigg\{\sup_x\left(\int_{\mathbb{R}^3}\abs{f_1}^2\right)^{\frac{1}{2}}\tnm{f_2}, \sup_x\left(\int_{\mathbb{R}^3}\abs{f_2}^2\right)^{\frac{1}{2}}\tnm{f_1}\Bigg\}.
\end{align}
\end{lemma}

\begin{lemma}[Lemma 5 of \cite{Guo2010}]\label{lemma:a2}
We have
\begin{align}
    \lnmm{\nu^{-1}\Gamma[f_1,f_2]}\lesssim\lnmm{f_1}\lnmm{f_2}.
\end{align}
\end{lemma}

\begin{lemma}\label{m-estimate}
    If $\abs{T-T_w}\lesssimz$ for some $0<z\ll1$, then we have
    \begin{align}
        \abs{\mu-\mu_w}\lesssim z\left(\abs{v}^2\mathrm{e}^{z\abs{v}^2}\right)\mu.
    \end{align}
\end{lemma}
\begin{proof}
    Notice that, since $\nabla_x P=0$
    \begin{align}
        \mu(x,v)=\frac{P}{\big(2\pi\big)^{\frac{3}{2}}\big(T(x)\big)^{\frac{5}{2}}}
        \exp\bigg(-\frac{\abs{v}^2}{2T(x)}\bigg),\quad \mu_w(x,v)=\frac{P}{\big(2\pi\big)^{\frac{3}{2}}\big(T_w(x)\big)^{\frac{5}{2}}}
        \exp\bigg(-\frac{\abs{v}^2}{2T_w(x)}\bigg).
    \end{align}
    Then we have $\mu-\mu_w=I+II$,
    where
    \begin{align}
        I:=&\frac{P}{\big(2\pi\big)^{\frac{3}{2}}}\bigg\{\frac{1}{T^{\frac{5}{2}}}-\frac{1}{T_w^{\frac{5}{2}}}\bigg\}\exp\bigg(-\frac{\abs{v}^2}{2T}\bigg),\\
        II:=&\frac{P}{\big(2\pi\big)^{\frac{3}{2}}}\frac{1}{T_w^{\frac{5}{2}}}\bigg\{\exp\bigg(-\frac{\abs{v}^2}{2T}\bigg)-\exp\bigg(-\frac{\abs{v}^2}{2T_w}\bigg)\bigg\}.
    \end{align}
    Since $\abs{T-T_w}\lesssimz$, we have
    \begin{align}
        \abs{I}=\frac{P}{\big(2\pi\big)^{\frac{3}{2}}}\frac{\abs{T_w^{\frac{5}{2}}-T^{\frac{5}{2}}}}{T^{\frac{5}{2}}T_w^{\frac{5}{2}}}\exp\left(-\frac{\abs{v}^2}{2T}\right)\lesssim \abs{T_w^{\frac{5}{2}}-T^{\frac{5}{2}}}\mu\lesssimz\mu.
    \end{align}
    Also, using mean-value theorem, for some $\delta\in(0,z)$, we know
    \begin{align}
        \abs{II}\lesssim& \mu\left\{1-\exp\left(\frac{\abs{v}^2}{2T}-\frac{\abs{v}^2}{2T_w}\right)\right\}\lesssim\mu\left\{1-\mathrm{e}^{z\abs{v}^2}\right\}
        \leq\mu\left(z\abs{v}^2\right)\mathrm{e}^{\delta\abs{v}^2}\lesssim z\left(\abs{v}^2\mathrm{e}^{z\abs{v}^2}\right)\mu.
    \end{align}
    Then our result naturally follows.
\end{proof}

Using a similar argument, we can derive the following estimates.
\begin{corollary}\label{m-estimate.}
    If $\abs{T-T_w}\lesssimz$ for some $0<z\ll1$, then we have
    \begin{align}
        \abs{\m^{\frac{1}{2}}-\m^{\frac{1}{2}}_w}\lesssim z\left(\abs{v}^2\mathrm{e}^{z\abs{v}^2}\right)\m^{\frac{1}{2}},\quad
        \abs{\m^{-\frac{1}{2}}-\m^{-\frac{1}{2}}_w}\lesssim z\left(\abs{v}^2\mathrm{e}^{z\abs{v}^2}\right)\m^{-\frac{1}{2}}.
    \end{align}
\end{corollary}

\begin{corollary}\label{m-estimate..}
    Let $\mu_{\#}$ be $\mu$ or $\mu_w$. If $\abs{T-T_w}\lesssimz$ for some $0<z\ll1$, then we have
    \begin{align}
        \tnm{\mu_{\#}^{-\frac{1}{2}} Q^{\ast}\Big[\mu-\mu_w,\mu_{\#}^{\frac{1}{2}}g\Big]}&=\tnm{\m^{\frac{1}{2}}\mu_{\#}^{-\frac{1}{2}} \Gamma\Big[\m^{\frac{1}{2}}-\m^{-\frac{1}{2}}\mu_w,\mu_{\#}^{\frac{1}{2}}\m^{-\frac{1}{2}} g\Big]}\lesssim z\tnm{\nu g},\\
        \lnmm{\mu_{\#}^{-\frac{1}{2}} Q^{\ast}\Big[\mu-\mu_w,\mu_{\#}^{\frac{1}{2}}g\Big]}&=\lnmm{\m^{\frac{1}{2}}\mu_{\#}^{-\frac{1}{2}} \Gamma\Big[\m^{\frac{1}{2}}-\m^{-\frac{1}{2}}\mu_w,\mu_{\#}^{\frac{1}{2}}\m^{-\frac{1}{2}} g\Big]}\lesssim z\lnmm{g}.
    \end{align}
\end{corollary}


\subsection{Estimates of Boundary Terms and Source Terms}\label{sec:bs}


In this section, we will give several types of estimates regarding $h$ and $S$. In the following, we will assume that the constant $r\in[2,6]$ and $g(x,v)$ is a given function.


\paragraph{\underline{Estimates of $h$}}


\begin{lemma}\label{h-estimate}
Under the assumption \eqref{assumption:boundary}, for $h$ defined in \eqref{aa 32}, we have
\begin{align}
    \tnms{h}{\gamma_-}&\lesssimo_{T}\varepsilon^{2-\alpha},\quad
    \jnms{h}{\gamma_-}\lesssimo_{T}\varepsilon^{1+\frac{3}{r}-\alpha},\quad
    \lnmms{h}{\gamma_-}\lesssimo_{T}\varepsilon^{1-\alpha},\quad \sup_{\iota_1,\iota_2}\int_{v\cdot n<0}\abs{h}\abs{v\cdot n}\mathrm{d} v\lesssimo_{T}\varepsilon^{2-\alpha}.
\end{align}
\end{lemma}

\begin{proof}
Note that from Theorem \ref{thm:ghost}
\begin{align}
    \lnmms{f_2}{\gamma}\lesssim\nm{f_2}_{W^{1,s}L^{\infty}_{\varrho,\vartheta}}\lesssim&o_{T}.
\end{align}
Then we know
\begin{align}
    \lnmms{\varepsilon^{2-\alpha}\left(m_w\displaystyle\int_{{\mathfrak{u}}\cdotn>0}
    \mb^{\frac{1}{2}}(v')f_2(v')\abs{{\mathfrak{u}}\cdotn}\mathrm{d}{{\mathfrak{u}}}-f_2\Big|_{v\cdot n<0}\right)}{\gamma_-}&\lesssim\varepsilon^{2-\alpha}\lnmms{f_2}{\gamma}\lesssimo_{T}\varepsilon^{2-\alpha}.
\end{align}
Then we obtain the similar estimates for $L^2_{\gamma_-}$ and $L^{\frac{2r}{3}}_{\gamma_-}$ norms. 

On the other hand, noticing that $\lnmms{\widetilde{\Phi}}{\gamma}\lesssimo_{T}$,
we have
\begin{align}
    &\lnmms{\varepsilon^{1-\alpha}\left(M_{w}\displaystyle\int_{{\mathfrak{u}}\cdotn>0}
    \mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\abs{{\mathfrak{u}}\cdotn}\mathrm{d}{{\mathfrak{u}}}-\mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\Big|_{v\cdot n<0}\right)}{\gamma_-}\\
    \lesssim&\varepsilon^{1-\alpha}\lnmms{\widetilde{\Phi}}{\gamma_-}\lesssimo_{T}\varepsilon^{1-\alpha}.\nonumber
\end{align}
Also, noticing that the cutoff $\chi\left(\varepsilon^{-1}v_{\eta}\right)$ implies a restriction to the domain $\abs{v_{\eta}}\leq\varepsilon$ and the $\gamma$ norm has an extra $v_{\eta}$, we have
\begin{align}
    &\jnms{\varepsilon^{1-\alpha}\left(M_{w}\displaystyle\int_{{\mathfrak{u}}\cdotn>0}
    \mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\abs{{\mathfrak{u}}\cdotn}\mathrm{d}{{\mathfrak{u}}}-\mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\Big|_{v\cdot n<0}\right)}{\gamma_-}\\
    \lesssim&\varepsilon^{1-\alpha}\jnms{\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}}{\gamma_-}\lesssimo_{T}\varepsilon^{1+\frac{3}{r}-\alpha},\nonumber
\end{align}
and the $L^2_{\gamma_-}$ estimate follows when $r=3$, and
\begin{align}
    &\int_{v\cdot n<0}\abs{\varepsilon^{1-\alpha}\left(M_{w}\displaystyle\int_{{\mathfrak{u}}\cdotn>0}
    \mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\abs{{\mathfrak{u}}\cdotn}\mathrm{d}{{\mathfrak{u}}}-\mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\Big|_{v\cdot n<0}\right)}\abs{v\cdot n}\mathrm{d} v\\
    \lesssim&\varepsilon^{1-\alpha}\int_{v\cdot n<0}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Phi}\abs{v_{\eta}}\mathrm{d} v\lesssim\varepsilon^{2-\alpha}.\nonumber
\end{align}
Then our estimates follow.
\end{proof}


\paragraph{\underline{Estimates of $\lc^1[R]$}}


\begin{lemma}\label{ssl-estimate}
Under the assumption \eqref{assumption:boundary}, for $\lc^1[R]$ defined in \eqref{final 51}, we have
\begin{align}
    \abs{\brv{\lc^1[R],g}}\lesssim o_{T}\left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\nu\abs{R}^2\right)^{\frac{1}{2}},
\end{align}
and thus
\begin{align}
    \abs{\br{\lc^1[R],g}}\lesssim o_{T}\um{g}\um{R}\lesssimo_{T}\um{g}\left(\tnm{\textbf{P}[R]}+\um{(\textbf{I}-\textbf{P})[R]}\right).
\end{align}
Also, we have
\begin{align}
    \tnm{\lc^1[R]}&\lesssimo_{T}\um{R},\quad
    \lnmm{\nu^{-1}\lc^1[R]}\lesssimo_{T}\lnmm{R}.
\end{align}
\end{lemma}

\begin{proof}
    The desired estimates follow from Lemma \ref{lemma:a1} and Lemma \ref{lemma:a2} since, by Theorem \ref{thm:ghost}, we have
    \begin{align}
        \lnmm{f_1}\lesssim\nm{f_1}_{W^{1,s}}\lesssimo_{T}.
    \end{align}
\end{proof}


\paragraph{\underline{Estimates of $S_0$}}


\begin{lemma}\label{ss0-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_0$ defined in \eqref{def-s0}, we have
\begin{align}
    \abs{\brv{S_0,g}}\lesssim o_{T}\varepsilon\left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\nu\abs{R}^2\right)^{\frac{1}{2}},
\end{align}
and thus
\begin{align}
    \abs{\br{S_0,g}}\lesssim o_{T}\varepsilon\um{g}\um{R}\lesssimo_{T}\varepsilon\um{g}\left(\tnm{\textbf{P}[R]}+\um{(\textbf{I}-\overline{\textbf{P}})[R]}\right).
\end{align}
Also, we have
\begin{align}
    \tnm{S_0}&\lesssimo_{T}\varepsilon\um{R},\quad
    \lnmm{\nu^{-1}S_0}\lesssimo_{T}\varepsilon\lnmm{R}.
\end{align}
\end{lemma}

\begin{proof}
    The desired estimates follow from Lemma \ref{lemma:a1} and Lemma \ref{lemma:a2} since, by  Theorem \ref{thm:ghost}, we have 
    \begin{align}
        \lnmms{f_2}{\gamma}\lesssim\nm{f_2}_{W^{1,s}L^{\infty}_{\varrho,\vartheta}}\lesssim&o_{T}.
    \end{align}
\end{proof}


\paragraph{\underline{Estimates of $S_1$}}


\begin{lemma}\label{ss1-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_1$ defined in \eqref{def-s1}, we have
\begin{align}
    \abs{\brv{S_1,g}}\lesssim \left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\nu\abs{f^{B}_1}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\nu\abs{R}^2\right)^{\frac{1}{2}},
\end{align}
and thus
\begin{align}
    \abs{\br{S_1,g}}\lesssim& o_{T}\um{g}\um{R}\lesssimo_{T}\um{g}\left(\tnm{\textbf{P}[R]}+\um{(\textbf{I}-\textbf{P})[R]}\right),\\
    \abs{\br{S_1,g}}\lesssim& o_{T}\um{f^{B}_1}\lnmm{g}\um{R}\lesssimo_{T}\varepsilon^{\frac{1}{2}}\lnmm{g}\left(\tnm{\textbf{P}[R]}+\um{(\textbf{I}-\textbf{P})[R]}\right).
\end{align}
Also, we have
\begin{align}
    \tnm{S_1}&\lesssimo_{T}\um{R},\quad
    \lnmm{\nu^{-1}S_1}\lesssimo_{T}\lnmm{R}.
\end{align}
\end{lemma}

\begin{proof}
    The desired estimates from Lemma \ref{lemma:a1} and Lemma \ref{lemma:a2} using Theorem \ref{thm:ghost}
    \begin{align}
        \lnmm{f^{B}_1}\lesssim&o_{T},\quad
        \um{f^{B}_1}\lesssimo_{T}\varepsilon^{\frac{1}{2}}.
    \end{align}
\end{proof}


\paragraph{\underline{Estimates of $S_2$}}


\begin{lemma}\label{ss2-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_2$ defined in \eqref{def-s2}, we have
\begin{align}\label{aa 39}
    \abs{\brv{S_2,g}}\lesssim \varepsilon^{\alpha-1}\left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^3}\nu\abs{R}^2\right),
\end{align}
and thus
\begin{align}\label{aa 40}
    \abs{\br{S_2,g}}\lesssim& \varepsilon^{\alpha-1}\um{g}\Big(\um{(\textbf{I}-\overline{\textbf{P}})R}\lnmm{R}+\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{3}\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{6}\Big)\\
    \lesssim&\varepsilon^{\alpha-1}\um{g}\xnm{R}^2.\nonumber
\end{align}
Also, we have
\begin{align}\label{aa 41}
    \tnm{S_2}&\lesssim\varepsilon^{\alpha-1}\Big(\um{(\textbf{I}-\overline{\textbf{P}})R}\lnmm{R}+\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{3}\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{6}\Big) \\
    &\lesssim\varepsilon^{\alpha-1}\xnm{R}^2,\nonumber\\
    \lnmm{\nu^{-1}S_2}&\lesssim\varepsilon^{\alpha-1}\lnmm{R}^2.\label{aa 42}
\end{align}
\end{lemma}

\begin{proof}
    Estimate \eqref{aa 39} follows from Lemma \ref{lemma:a1}. Then utilizing H\"older's inequality
    \begin{align}
        \abs{\int_{\mathbb{R}^3}\nuR(\textbf{I}-\overline{\textbf{P}})[R]}\lesssim&\um{(\textbf{I}-\overline{\textbf{P}})R}\lnmm{R},\\
        \int_{\mathbb{R}^3}\nu\abs{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}^2\lesssim&\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{3}\pnm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}{6},
    \end{align}
    we obtain \eqref{aa 40}. Then \eqref{aa 41} follows from duality. Finally, \eqref{aa 42} holds due to Lemma \ref{lemma:a2}.
\end{proof}


\paragraph{\underline{Estimates of $S_3$}}


\begin{lemma}\label{ss3-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_3$ defined in \eqref{mm 00}, we have
\begin{align}
    \pnm{S_3}{1}+\pnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}{1}+\pnm{\eta^2\left(\ss_{3b}+\ss_{3c}\right)}{1}&\lesssim o_{T} \varepsilon^{2-\alpha},\label{ss3-estimate1}\\
    \tnm{\br{v}^2S_3}+\tnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}+\tnm{\eta^2\left(\ss_{3b}+\ss_{3c}\right)}&\lesssim o_{T}\varepsilon^{1-\alpha},\label{ss3-estimate2}\\
    \pnm{S_3}{r}+\pnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}{r}+\pnm{\eta^2\left(\ss_{3b}+\ss_{3c}\right)}{r}&\lesssimo_{T}\varepsilon^{\frac{2}{r}-\alpha},\label{ss3-estimate3}\\
    \nm{S_3}_{L^{r}_{\iota_1\iota_2}L^1_{\mathfrak{n}}L^1_v}+\nm{\eta\left(\ss_{3b}+\ss_{3c}\right)}_{L^{r}_{\iota_1\iota_2}L^1_{\mathfrak{n}}L^1_v}&\lesssimo_{T}\varepsilon^{2-\alpha},\label{ss3-estimate4},
\end{align}
and
\begin{align}
    \nm{\ss_{3b}+\ss_{3c}}_{L^{r}_{x}L^1_v}+\nm{\eta\left(\ss_{3b}+\ss_{3c}\right)}_{L^{r}_{x}L^1_v}&\lesssimo_{T}\varepsilon^{1+\frac{1}{r}-\alpha},\label{ss3-estimate5}\\
    \abs{\br{\ss_{3a},g}}+\abs{\br{\eta\ss_{3a},g}}+\abs{\br{\eta^2\ss_{3a},g}}&\lesssim\varepsilon^{1-\alpha}\knm{\br{v}^2f^{B}_1}\jnm{\nabla_v g}\lesssimo_{T}\varepsilon^{2-\frac{1}{r}-\alpha}\jnm{\nabla_v g}.\label{ss3-estimate6}
\end{align}
Also, we have
\begin{align}\label{final 15}
    \lnmm{S_3}&\lesssimo_{T}\varepsilon^{-\alpha}.
\end{align}
\end{lemma}

\begin{proof} 
We start from the $L^{r}$ estimate \eqref{ss3-estimate3}, and then \eqref{ss3-estimate1} and \eqref{ss3-estimate2} will naturally follow. We first focus on $\ss_{3a}$. Notice that
    \begin{align}
    \frac{\partialf^{B}_1}{\partialv_{\eta}}(\eta,{\mathfrak{v}})=\varepsilon^{-1}\overline{\chi}'\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\widetilde{\Phi}(\eta,{\mathfrak{v}})+\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}.
\end{align}
From Theorem \ref{boundary well-posedness} and Theorem \ref{boundary regularity}, we have
\begin{align}
    \lnmm{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}\lesssim\varepsilon^{-1}\lnmm{v_{\eta}\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}\lesssim\varepsilon^{-1},
\end{align}
and
\begin{align}
    \pnm{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}{1}\lesssim 1.
\end{align}
Then by change of variable $\eta=\varepsilon^{-1}\mathfrak{n}$, we have
\begin{align}
    \pnm{\varepsilon^{-1}\overline{\chi}'\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\widetilde{\Phi}(\eta,{\mathfrak{v}})}{r}
    =&\left(\iint_{\Omega\times\mathbb{R}^3}\abs{\varepsilon^{-1}\overline{\chi}'\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\widetilde{\Phi}(\eta,{\mathfrak{v}})}^{r}\right)^{\frac{1}{r}}\\
    \lesssim&\varepsilon^{-1}\left(\iint_{\Omega\times\mathbb{R}^3}\abs{\overline{\chi}'\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\widetilde{\Phi}(\eta,{\mathfrak{v}})}^{r}\right)^{\frac{1}{r}}\lesssim\varepsilon^{\frac{2}{r}-1},\nonumber
\end{align}
and

\begin{align}\label{aa 43}
    \pnm{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}{r}
    =&\left(\iint_{\Omega\times\mathbb{R}^3}\abs{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}^{r}\right)^{\frac{1}{r}}\\
    \lesssim&\left(\lnmm{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}^{r-1}\iint_{\Omega\times\mathbb{R}^3}\abs{\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\frac{\partial\widetilde{\Phi}(\eta,{\mathfrak{v}})}{\partialv_{\eta}}}\right)^{\frac{1}{r}}\nonumber\\
    \lesssim&\big(\varepsilon^{-(r-1)}\varepsilon\big)^{\frac{1}{r}}\lesssim\varepsilon^{\frac{2}{r}-1}.\nonumber
\end{align}
Hence, we know
\begin{align}
    \pnm{\ss_{3a}}{r}\lesssim \varepsilon^{\frac{2}{r}-\alpha}.
\end{align}

Similarly $\ss_{3b}$ estimates follow from Theorem \ref{boundary well-posedness} and Theorem \ref{boundary regularity}. 

Noticing that
\begin{align}
    \ss_{3c}=&\varepsilon^{-\alpha}\m^{-\frac{1}{2}}\mb^{\frac{1}{2}}\chi(\varepsilon\eta)\bigg(\chi(\varepsilon^{-1}v_{\eta})K_w\Big[\widetilde{\Phi}\Big]-K_w\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Phi}\Big]\bigg),
\end{align}
which has one less $\varepsilon$-power but produces an $\varepsilon$-size cutoff $\chi(\varepsilon^{-1}v_{\eta})$. Clearly, the $\chi(\varepsilon^{-1}v_{\eta})K_w\Big[\widetilde{\Phi}\Big]$ can be estimated by a similar argument as \eqref{aa 43}. Then noticing that by the change of variable $w_{\eta}=\varepsilon^{-1}{\mathfrak{u}}_{\eta}$
\begin{align}
    \int_{\mathbb{R}^3}\abs{K_w\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Phi}\Big]}\mathrm{d} v=&\int_{\mathbb{R}^3}\abs{\int_{\mathbb{R}^3}k_w({\mathfrak{u}},v)\chi(\varepsilon^{-1}{\mathfrak{u}}_{\eta})\widetilde{\Phi}({\mathfrak{u}})\mathrm{d}{\mathfrak{u}}}\mathrm{d} v\\
    \lesssim&\sup_{{\mathfrak{u}}}\int_{\mathbb{R}^3}\abs{k_w({\mathfrak{u}},v)}\mathrm{d} v\abs{\int_{\mathbb{R}^3}\chi(\varepsilon^{-1}{\mathfrak{u}}_{\eta})\widetilde{\Phi}({\mathfrak{u}}) \mathrm{d}{\mathfrak{u}}}
    \lesssim\abs{\int_{\mathbb{R}^3}\chi(\varepsilon^{-1}{\mathfrak{u}}_{\eta})\widetilde{\Phi}({\mathfrak{u}})\mathrm{d}{\mathfrak{u}}}\lesssim\varepsilon,\nonumber
\end{align}
we can bound $K_w\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Phi}\Big]$ by a similar argument as \eqref{aa 43}. Hence, we complete the proof of \eqref{ss3-estimate3}.

Noting the rescaling $\eta=\varepsilon^{-1}\mathfrak{n}$ and the cutoff $\chi(\varepsilon^{-1}v_{\eta})$, using Theorem \ref{boundary well-posedness} and Theorem \ref{boundary regularity}, \eqref{ss3-estimate4} and \eqref{ss3-estimate5} follow from substitution in the integral.

Then we turn to \eqref{ss3-estimate6}. The most difficult term in $\abs{\br{\ss_{3a},g}}$ is essentially $\varepsilon^{1-\alpha}\abs{\br{\frac{\partialf^{B}_1}{\partialv_{\eta}},g}}$. Note that $\frac{\partialf^{B}_1}{\partialv_{\eta}}=0$ for $\abs{v_{\eta}}\leq \varepsilon$ due to the cutoff in $f^{B}_1$.
Integration by parts with respect to $v_{\eta}$ implies
\begin{align}
   \varepsilon^{1-\alpha}\abs{\br{\frac{\partialf^{B}_1}{\partialv_{\eta}},g}}\lesssim& \varepsilon^{1-\alpha}\abs{\br{f^{B}_1, \frac{\partial g}{\partialv_{\eta}}}}
   \lesssim\varepsilon^{1-\alpha}\jnm{f^{B}_1}\knm{\frac{\partial g}{\partialv_{\eta}}}.
\end{align}
From \eqref{aa 44} and $\dfrac{\partialx}{\partialv_{\eta}}\equiv\textbf{0}$, we know the substitution $(\mathfrak{n},\iota_1,\iota_2,v)\rightarrow (\mathfrak{n},\iota_1,\iota_2,{\mathfrak{v}})$ implies
\begin{align}
-\dfrac{\partialv}{\partialv_{\eta}}\cdotn=1,\quad 
-\dfrac{\partialv}{\partialv_{\eta}}\cdot\varsigma_1=0,\quad
-\dfrac{\partialv}{\partialv_{\eta}}\cdot\varsigma_2=0.
\end{align}
Hence, we know $\abs{\dfrac{\partialv}{\partialv_{\eta}}}\lesssim 1$,
and thus
\begin{align}
    \abs{\frac{\partial g}{\partialv_{\eta}}}\lesssim \abs{\nabla_v g}\abs{\frac{\partial v}{\partialv_{\eta}}}\lesssim\abs{\nabla_v g}.
\end{align}
Hence, we know that 
\begin{align}
   \varepsilon^{1-\alpha}\abs{\br{\frac{\partialf^{B}_1}{\partialv_{\eta}},g}}\lesssim&\varepsilon^{1-\alpha}\jnm{f^{B}_1}\knm{\nabla_v g}\lesssimo_{T}\varepsilon^{1+\frac{1}{r}-\alpha}\knm{\nabla_v g}.
\end{align}
Finally, \eqref{final 15} holds due to the cutoff $\overline{\chi}$.
\end{proof}

\begin{remark}\label{ss3-remark}
    Notice that the BV estimate in Theorem \ref{boundary regularity} does not contain exponential decay in $\eta$, and thus we cannot directly bound $\eta\ss_{3a}$ and $\eta^2\ss_{3a}$. Instead, we should first integrate by parts with respect to $v_{\eta}$ as in \eqref{ss3-estimate6} to study $f^{B}_1$ instead:
    \begin{align}
        \pnm{f^{B}_1}{r}+\pnm{\etaf^{B}_1}{r}+\pnm{\eta^2f^{B}_1}{r}&\lesssimo_{T}\varepsilon^{\frac{2}{r}-\alpha},\label{ss3-estimate7}\\
        \nm{f^{B}_1}_{L^{r}_{\iota_1\iota_2}L^1_{\mathfrak{n}}L^1_v}+\nm{\etaf^{B}_1}_{L^{r}_{\iota_1\iota_2}L^1_{\mathfrak{n}}L^1_v}&\lesssimo_{T}\varepsilon^{2-\alpha},\label{ss3-estimate8},\\
        \nm{f^{B}_1}_{L^{r}_{x}L^1_v}+\nm{\etaf^{B}_1}_{L^{r}_{x}L^1_v}&\lesssimo_{T}\varepsilon^{1+\frac{1}{r}-\alpha}.
    \end{align}
\end{remark}


\paragraph{\underline{Estimates of $S_4$}}


\begin{lemma}\label{ss4-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_4$ defined in \eqref{def-s4}, we have
\begin{align}
    \tnm{\br{v}^2S_4}&\lesssim o_{T}\varepsilon^{2-\alpha},\quad
    \pnm{S_4}{r}\lesssimo_{T}\varepsilon^{2-\alpha},\quad
    \lnmm{S_4}\lesssimo_{T}\varepsilon^{2-\alpha}.
\end{align}
Also, we have
    \begin{align}\label{final 25}
    \br{v\m^{\frac{1}{2}},S_4}=\varepsilon^{2-\alpha}\br{v, v\cdot\nabla_{x}\left(\mhf_2\right)}=0.
    \end{align}
\end{lemma}

\begin{proof}
    The desired estimates follow from Lemma \ref{lemma:a1} and Lemma \ref{lemma:a2} using Theorem \ref{thm:ghost}
    \begin{align}
        \lnmm{\nabla_{x} f_2}\lesssim\nm{f_2}_{W^{2,s}}\lesssimo_{T}.
    \end{align}
    Also, note that
    \begin{align}
    \br{v\m^{\frac{1}{2}},S_4}=\varepsilon^{2-\alpha}\br{v, v\cdot\nabla_{x}\left(\mhf_2\right)}=0,
    \end{align}
    due to the crucial orthogonality condition $\m^{-\frac{1}{2}}\left(v\cdot\nabla_{x}\left(\mhf_2\right)\right)\perp v\m^{\frac{1}{2}}$ in the companion paper \cite[(2.8)]{AA024}, which is equivalent to the second equation for the ghost-effect equations \eqref{fluid system-}.
\end{proof}


\paragraph{\underline{Estimates of $S_5$}}


\begin{lemma}\label{ss5-estimate}
Under the assumption \eqref{assumption:boundary}, for $S_5$ defined in \eqref{def-s5}, we have
\begin{align}
    \abs{\brv{S_5,g}}\lesssim o_{T}\varepsilon^{1-\alpha}\left(\int_{\mathbb{R}^3}\nu\abs{g}^2\right)^{\frac{1}{2}},
\end{align}
and thus
\begin{align}
    \abs{\br{S_5,g}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}-\alpha}\um{g},\quad
    \abs{\br{S_5,g}}\lesssimo_{T}\varepsilon^{2-\alpha}\lnmm{g}.
\end{align}
Also, we have
\begin{align}
    \tnm{S_5}&\lesssimo_{T}\varepsilon^{\frac{3}{2}-\alpha},\quad
    \lnmm{\nu^{-1}S_5}\lesssimo_{T}\varepsilon^{1-\alpha}.
\end{align}
\end{lemma}

\begin{proof}
    We obtain the desired estimates from Lemma \ref{lemma:a1} and Lemma \ref{lemma:a2} using Theorem \ref{thm:ghost}
    \begin{align}
        \lnmm{f_1}+\lnmm{f_2}\lesssim\nm{f_1}_{W^{2,s}}+\nm{f_2}_{W^{2,s}}\lesssim&o_{T},\quad
        \lnmm{f^{B}_1}\lesssimo_{T}.
    \end{align} 
    In particular, the estimate of $\mu-\mu_{w}$ follows from Lemma \ref{m-estimate} and Corollary \ref{m-estimate..} with $z=\varepsilon$, which provides an extra power of $\varepsilon$.
\end{proof}


\paragraph{\underline{Estimates of $\ss_6$}}


\begin{lemma}\label{ssp-estimate}
Under the assumption \eqref{assumption:boundary},  for $\ss_6$ defined in \eqref{def-sp} and any $m\in\mathbb{N}$, we have
\begin{align}
    \tnm{\br{v}^2\ss_6}&\lesssim o_{T}\varepsilon^{m-1-\alpha},\quad
    \pnm{\ss_6}{r}\lesssimo_{T}\varepsilon^{m-1-\alpha},\quad
    \lnmm{\ss_6}\lesssimo_{T}\varepsilon^{m-1-\alpha}.
\end{align}
\end{lemma}

\begin{proof}
Note that $\mathfrak{F}_{\text{a}}=\mu+O(\varepsilon)$, and thus $\mathfrak{F}_{\text{a}}\approx\mu$ when $\varepsilon$ is small. Also, for $w(v)=\mathrm{e}^{\varrho\frac{\abs{v}^2}{2T_M}}$ with $0\leq\varrho<\frac{1}{2}$, we have
\begin{align}
    \abs{\varepsilon^{\alpha}\mhR}= \abs{\varepsilon^{\alpha}\m^{\frac{1}{2}} w^{-1}\big(wR\big)}\lesssim\varepsilon^{\alpha}\m^{\frac{1}{2}} \mathrm{e}^{-\varrho\frac{\abs{v}^2}{2T_M}}\lnmm{R} \lesssim \varepsilon^{\alpha-\frac{1}{2}}\mu^{\frac{1}{2}+\varrho}\xnm{R}.
\end{align}
Hence, for $\varepsilon$ sufficiently small, as long as
\begin{align}
    \varepsilon^{\alpha-\frac{1}{2}}\mu^{\frac{1}{2}+\varrho}\xnm{R}\lesssim\mu
\end{align}
we will have
\begin{align}
    \mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\approx \mu+\varepsilon^{\alpha}\mhR\geq0,
\end{align}
and thus 
\begin{align}
    \overline{Q}:=Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right]- Q_{\text{gain}}\left[\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+,\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+\right]=0.
\end{align}
In other words, $\overline{Q}$ vanishes except in the region 
\begin{align}\label{oo 01}
    D:=\bigg\{(x,v): \varepsilon^{\alpha-\frac{1}{2}}\mu^{\frac{1}{2}+\varrho}\xnm{R}\gtrsim\mu\bigg\}.
\end{align}
The requirement in \eqref{oo 01} is equivalent to \begin{align}
    \mu^{\frac{1}{2}-\varrho}\lesssim\varepsilon^{\alpha-\frac{1}{2}}\xnm{R},
\end{align}
which is equivalently
\begin{align}
    \abs{v}\gtrsim \frac{T}{\frac{1}{2}-\varrho}\left(\abs{\ln(\varepsilon)}^{\frac{1}{2}}+\abs{\ln\xnm{R}}^{\frac{1}{2}}\right).
\end{align}
In addition, in the region $D$, we have
\begin{align}
    \mu\lesssim\varepsilon^{\frac{\alpha-\frac{1}{2}}{\frac{1}{2}-\varrho}}\xnm{R}^{\frac{1}{\frac{1}{2}-\varrho}}.
\end{align}
Hence, for any $m\in\mathbb{N}$, we can choose $\varrho$ with $\frac{1}{2}-\varrho\ll 1$ such that in the region $D$, we have
\begin{align}
    \mu\lesssim\varepsilon^{2m}\xnm{R}^{4m}.
\end{align}
and thus
\begin{align}\label{oo 02}
    \abs{\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR}\lesssim \varepsilon^{m}\xnm{R}^{2m}.
\end{align}

For the two groups of terms in $\ss_6$:
\begin{align}\label{oo 04}
    \varepsilon^{-\alpha-1}\Big(Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right]-Q_{\text{gain}}\left[\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+,\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+\right]\Big)
\end{align}
and 
\begin{align}\label{oo 05}
\\
    \varepsilon^{-\alpha-1}\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\bigg(Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right]-Q_{\text{gain}}\left[\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+,\left(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR\right)_+\right]\bigg),\nonumber
\end{align}
the effective region of $Q_{\text{gain}}$ integral is actually only on 
\begin{align}
    \abs{{\mathfrak{u}}_{\ast}}\gtrsim\abs{\ln(\varepsilon)}^{\frac{1}{2}}\abs{\ln\xnm{R}}^{\frac{1}{2}}\ \ \text{and}\ \  \abs{v_{\ast}}\gtrsim\abs{\ln(\varepsilon)}^{\frac{1}{2}}\abs{\ln\xnm{R}}^{\frac{1}{2}}.
\end{align}
Therefore \eqref{oo 02} is valid in both sets above so that
\begin{align}
    \lnmm{\ss_6}
    \lesssim&\varepsilon^{m-\alpha-1}\xnm{R}^{2m}.
\end{align}
\end{proof}


\subsection{Conservation Laws and Weak Formulation}


\subsubsection{Classical Conservation Laws}


\begin{lemma}\label{lem:conservation law}
    Under the assumption \eqref{assumption:boundary}, we have the conservation laws
    \begin{align}
    P\big(\nabla_{x}\cdot\textbf{b}\big)=&\brv{\m^{\frac{1}{2}},S}=\brv{\m^{\frac{1}{2}},S_3+S_4},\label{conservation law 1}\\
    P\nxp+\nabla_{x}\cdot\varpi=&\brv{v\m^{\frac{1}{2}},S}=\brv{v\m^{\frac{1}{2}},S_3+\ss_6},\label{conservation law 2}\\
    5P\Big(\nabla_{x}\cdot(\textbf{b} T)\Big)+\nabla_{x}\cdot(\kappa\textbf{d})=&\brv{\abs{v}^2\m^{\frac{1}{2}},S}=\brv{\abs{v}^2\m^{\frac{1}{2}},S_3+S_4+\ss_6},\label{conservation law 3}
    \end{align}
    where $\displaystyle\varpi:=\int_{\mathbb{R}^3}(v\otimes v)(\textbf{I}-\overline{\textbf{P}})[R]$.
\end{lemma}
\begin{proof}
    We multiply test functions $\m^{\frac{1}{2}},v\m^{\frac{1}{2}},\abs{v}^2\m^{\frac{1}{2}}$ on both sides of \eqref{pp 02} and integrate over $v\in\mathbb{R}^3$. Using the orthogonality of $\Gamma$ and the crucial property \eqref{final 25},
    the results follow. 
\end{proof}

\begin{lemma}\label{remark 01}
    Under the assumption \eqref{assumption:boundary}, we have for $r\in[2,6]$
    \begin{align}\label{tt 03}
    \nm{\xi}_{W^{2,r}}\lesssim& o_{T}\jnm{\textbf{b}}+o_{T}\varepsilon^{1+\frac{1}{r}-\alpha},\\ \jnm{\textbf{d}}\lesssim&o_{T}\jnm{\textbf{b}}+\jnm{\textbf{e}}+o_{T}\varepsilon^{1+\frac{1}{r}-\alpha},\label{tt 03'}\\
    \label{qq 33}
    \tnms{\nabla_{x}\xi}{\partial\Omega}\lesssim&\nm{\xi}_{H^2}\lesssimo_{T}\tnm{\textbf{b}}+o_{T}\varepsilon^{\frac{3}{2}-\alpha}.
    \end{align}
    In particular, we have
    \begin{align}
        \varepsilon^{-\frac{1}{2}}\nm{\xi}_{H^2}+\nm{\xi}_{W^{2,6}}+\varepsilon^{-\frac{1}{2}}\tnms{\nabla_{x}\xi}{\partial\Omega}\lesssimo_{T}\xnm{R}+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{lemma}
\begin{proof}
\eqref{conservation 
 law 1} and \eqref{conservation law 3} as well as \eqref{tt 01}, imply
\begin{align}\label{aa 35}
    \nabla_{x}\cdot(\kappa\nabla_{x}\xi)=&\nabla_{x}\cdot(\kappa\textbf{d})=\brv{\abs{v}^2\m^{\frac{1}{2}},S_3+S_4+\ss_6}-5P\Big(\nabla_{x}\cdot(\bbT)\Big)\\
    =&\brv{\abs{v}^2\m^{\frac{1}{2}},S_3+S_4+\ss_6}-5P\Big((\nabla_{x}\cdot\textbf{b})T+\textbf{b}\cdot\nabla_{x}T\Big)\nonumber\\
    =&\brv{\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4}+\brv{\abs{v}^2\m^{\frac{1}{2}},\ss_6}-5P(\textbf{b}\cdot\nabla_{x}T)\nonumber\\
    =&\brv{\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S}-5P(\textbf{b}\cdot\nabla_{x}T).\nonumber
\end{align}
By the standard elliptic estimate and Lemma \ref{ss4-estimate} and Lemma \ref{ssp-estimate}, we have for $1<r<\infty$ 
\begin{align}
    \nm{\xi}_{W^{2,r}}\lesssim&o_{T}\jnm{\textbf{b}}+\jnm{\brv{\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3}}+\jnm{\br{v}^2S_4}+\jnm{\ss_6}\\
    \lesssim& o_{T}\jnm{\textbf{b}}+o_{T}\varepsilon^{1+\frac{1}{r}-\alpha}+\jnm{\brv{\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3}}.\nonumber
\end{align}
Using \eqref{ss3-estimate5} and \eqref{ss3-estimate6}, we know
\begin{align}
    \jnm{\brv{\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3}}\lesssimo_{T}\varepsilon^{1+\frac{1}{r}-\alpha}. 
\end{align}
Then \eqref{tt 03} is verified. Hence, we have
\begin{align}
    \jnm{\textbf{d}}\lesssim\jnm{\nabla_{x}\xi}+\jnm{\textbf{e}}\lesssimo_{T}\jnm{\textbf{b}}+\jnm{\textbf{e}}+o_{T}\varepsilon^{1+\frac{1}{r}-\alpha}.
\end{align}
Then by Sobolev embedding and trace theorem
\begin{align}\label{qq 33,}
    \tnms{\nabla_{x}\xi}{\partial\Omega}\lesssim\nm{\xi}_{H^2}\lesssimo_{T}\tnm{\textbf{b}}+o_{T}\varepsilon^{\frac{3}{2}-\alpha}.
\end{align}

\end{proof}


\subsubsection{Weak Formulation with Test Function $\nabla_{x}\phi\cdot\mathscr{A}$}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for any smooth function $\phi(x)$, we have \eqref{pp 03}.
\end{lemma}
\begin{proof}
Letting test function $\mathfrak{f}=\nabla_{x}\phi\cdot\mathscr{A}$ in \eqref{weak formulation}, we obtain
\begin{align}
    &-\br{v\cdot\nabla_{x} \Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),R}+\varepsilon^{-1}\bbr{ \nabla_{x}\phi\cdot\mathscr{A}, \mathcal{L}[R]}+\iint_{\Omega\times\mathbb{R}^3}\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x}T}{4T^2}\right)R\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big)\\
    =&-\int_{\gamma}R \Big(\nabla_{x}\phi\cdot\mathscr{A}\Big)\mathrm{d}\gamma+\bbr{\nabla_{x}\phi\cdot\mathscr{A},S}.\nonumber
\end{align}
Using \eqref{pp 01}, oddness and the orthogonality of $\mathscr{A}$ with $\textbf{P}[R]$ and $(\textbf{I}-\overline{\textbf{P}})[R]$, we simplify
\begin{align}
&-\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),c}+\varepsilon^{-1}\bbrx{\nabla_{x}\phi,\kappa\textbf{d}}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\\
=&-\int_{\partial\Omega\times\mathbb{R}^3}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big)R (v\cdot n)+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\barS}.\nonumber
\end{align}
We rewrite the boundary condition in \eqref{remainder} as
\begin{align}\label{qq 109}
    R\textbf{1}_{\gamma}=\mathcal{P}_{\gamma}[R]\textbf{1}_{\gamma}+(1-\mathcal{P}_{\gamma})[R]\textbf{1}_{\gamma_+}+h\textbf{1}_{\gamma_-}.
\end{align}
Using orthogonality of $\mathscr{A}$ and $v\m^{\frac{1}{2}}$
\begin{align}
    \int_{\mathbb{R}^3}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big)\m^{\frac{1}{2}}(v\cdot n)=0,
\end{align}
we deduce \eqref{pp 03}.
\end{proof}

\begin{remark}
    Using \eqref{tt 01} and by integration by parts for $\varepsilon^{-1}\br{\nabla_{x}\phi,\kappa\textbf{d}}$, we can further rewrite for $Z=c+\varepsilon^{-1}\xi$
    \begin{align}\label{pp 03.}
    &-\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),Z}+\varepsilon^{-1}\bbrx{\nabla_{x}\phi,\kappa\textbf{e}}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\\
    =&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\barS}.\nonumber
    \end{align}
\end{remark}


\subsubsection{Weak Formulation with Test Function $\nabla_{x}\psi:\mathscr{B}$}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for any smooth function $\psi(x)$, we have 
    \begin{align}\label{pp 04}
    \\
    &-\br{\nabla_{x}\cdot\Big(\overline\b(\nabla_{x}\psi:\mathscr{B})\Big),\textbf{b}}+\varepsilon^{-1}\bbrx{\nabla_{x}\psi,\varpi}+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}+\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\nonumber\\
    =&\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}+\bbrv{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
    \end{align}
    where $\varpi$ is defined in Lemma \ref{lem:conservation law}.
\end{lemma}
\begin{proof}
Let the test function in \eqref{weak formulation} be chosen as $\mathfrak{f}=\nabla_{x}\psi:\mathscr{B}$ . We obtain
\begin{align}
    &-\br{v\cdot\nabla_{x} \Big(\nabla_{x}\psi:\mathscr{B}\Big),R}+\varepsilon^{-1}\bbr{ \nabla_{x}\psi:\mathscr{B}, \mathcal{L}[R]}+\iint_{\Omega\times\mathbb{R}^3}\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\dfrac{\nabla_{x}T}{4T^2}\right)R\Big(\nabla_{x}\psi:\mathscr{B}\Big)\\
    =&-\int_{\gamma}R \Big(\nabla_{x}\psi:\mathscr{B}\Big)\mathrm{d}\gamma+\bbr{\nabla_{x}\psi:\mathscr{B},S}.\nonumber
\end{align}
Using \eqref{pp 01}, oddness and orthogonality of $\mathscr{B}$ with $\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}$,
we obtain
\begin{align}
\\
&-\br{\nabla_{x}\cdot\Big(\overline\b(\nabla_{x}\psi:\mathscr{B})\Big),\textbf{b}}+\varepsilon^{-1}\bbrx{\nabla_{x}\psi,\varpi}+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}+\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\nonumber\\
=&-\int_{\partial\Omega\times\mathbb{R}^3}\Big(\nabla_{x}\psi:\mathscr{B}\Big)R (v\cdot n)+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}+\bbrv{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
Using \eqref{qq 109} and orthogonality of $\mathscr{B}$ with $\mathcal{P}_{\gamma}[R]$
\begin{align}
    \int_{\mathbb{R}^3}\Big(\nabla_{x}\psi:\mathscr{B}\Big)\m^{\frac{1}{2}}(v\cdot n)=0,
\end{align} 
we obtain \eqref{pp 04}.
\end{proof}


\subsubsection{Weak Formulation with Test Function $\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\phi\big(\abs{v}^2-5T\big)\m^{\frac{1}{2}}$}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for any smooth test function $\phi(x)$ satisfying $\phi\big|_{\partial\Omega}=0$, we have \eqref{pp 10}.
\end{lemma}
\begin{proof}
Multiplying $\phi(x)\in\mathbb{R}$ on both sides of \eqref{aa 35} and integrating over $x\in\Omega$, we obtain
\begin{align}\label{pp 06}
5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}-\bbrx{\nabla_{x}\phi,\kappa\textbf{d}}+\int_{\partial\Omega}\phi(\kappa\textbf{d})\cdot n=&\br{\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}.
\end{align}
Hence, adding $\varepsilon^{-1}\times$\eqref{pp 06} and \eqref{pp 03} to eliminate $\varepsilon^{-1}\brx{\nabla_{x}\phi,\kappa\textbf{d}}$ yields
\begin{align}\label{pp 08}
&-\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),c}+\varepsilon^{-1}5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}+\varepsilon^{-1}\int_{\partial\Omega}\phi(\kappa\textbf{d})\cdot n\\
&+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\nonumber\\
=&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&+\varepsilon^{-1}\br{\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\barS}.\nonumber
\end{align}
The assumption $\phi\big|_{\partial\Omega}=0$ completely eliminates the boundary term $\displaystyle\varepsilon^{-1}\int_{\partial\Omega}\phi(\kappa\textbf{d})\cdot n$ in \eqref{pp 08}. Hence, we have \eqref{pp 10}.
\end{proof}


\subsubsection{Weak Formulation with Test Function $\nabla_{x}\psi:\mathscr{B}+\varepsilon^{-1}\psi\cdot v\m^{\frac{1}{2}}$}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for any smooth function $\psi(x)$ satisfying $\nabla_{x}\cdot\psi=0$, $\psi\big|_{\partial\Omega}=0$, we have \eqref{pp 16}.
\end{lemma}
\begin{proof}
Multiplying $\psi(x)\in\mathbb{R}^3$ on both sides of \eqref{conservation law 2} and integrating over $x\in\Omega$, we obtain
\begin{align}\label{pp 12}
-P\bbrx{\nabla_{x}\cdot\psi, p}-\bbrx{\nabla_{x}\psi,\varpi}+\int_{\partial\Omega}\Big(Pp\psi+\psi\cdot\varpi\Big)\cdot n= &\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}.
\end{align}
Hence, adding $\varepsilon^{-1}\times$\eqref{pp 12} and \eqref{pp 04} to eliminate $\varepsilon^{-1}\brx{\nabla_{x}\psi,\varpi}$ yields
\begin{align}\label{pp 14}
&-\br{\nabla_{x}\cdot\Big(\overline\b(\nabla_{x}\psi:\mathscr{B})\Big),\textbf{b}}-\varepsilon^{-1}P\bbrx{\nabla_{x}\cdot\psi, p}+\varepsilon^{-1}\int_{\partial\Omega}\Big(Pp\psi+\psi\cdot\varpi\Big)\cdot n\\
&+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}+\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\nonumber\\
=&\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
&+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\bbr{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
The assumptions $\nabla_{x}\cdot\psi=0$ and $\psi\big|_{\partial\Omega}=0$ eliminates $\varepsilon^{-1}P\bbrx{\nabla_{x}\cdot\psi, p}$ and $\varepsilon^{-1}\displaystyle\int_{\partial\Omega}\Big(Pp\psi+\psi\cdot\varpi\Big)\cdot n$ in \eqref{pp 14}. Hence, we have
\begin{align}\label{pp 15}
&-\br{\nabla_{x}\cdot\Big(\overline\b(\nabla_{x}\psi:\mathscr{B})\Big),\textbf{b}}+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}+\br{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\\
=&\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
&+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\br{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
Using $\nabla_{x}\cdot\psi=0$, a standard simplification
\begin{align}
    \nabla_{x}\cdot\Big(\overline\b(\nabla_{x}\psi:\mathscr{B})\Big)=&\overline\b\cdot\nabla_{x}\big(\nabla_{x}\psi:\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big)\\
    &=\lambda\Delta_x\psi+\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\nonumber
\end{align}
further yields \eqref{pp 16}.
\end{proof}


\section{Basic Energy Estimates}


In this section, we plan to prove Proposition \ref{thm:energy}.


\subsection{Energy Structure}


\begin{lemma}\label{lem:final 9}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{qq 34}
    &\varepsilon^{-1}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+\varepsilon^{-2}\tnm{\textbf{e}}^2+\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2\\
    \lesssim&\varepsilon^{-1}\br{R-c\m^{\frac{1}{2}}\left(\abs{v}^2-5T\right)-\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),S}+o_{T}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
\end{lemma}

\begin{proof}
Taking $\mathfrak{f}=\varepsilon^{-1}R$ in the weak formulation \eqref{weak formulation}, we have
\begin{align}
    \varepsilon^{-1}\bbr{v\cdot\nabla_{x}R,R}+\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}+\varepsilon^{-2}\bbr{R,\mathcal{L}[R]}
    =\varepsilon^{-1}\bbr{R,S}.
\end{align}

\paragraph{\underline{Step 1: Weak Formulation}}
Using Green's identity in Lemma \ref{remainder lemma 2}, we obtain
\begin{align}
\frac{\varepsilon^{-1}}{2}\iint_{\Omega\times\mathbb{R}^3}v\cdot\nabla_{x}(R^2)=&\frac{\varepsilon^{-1}}{2}\int_{\partial\Omega\times\mathbb{R}^3}R^2(v\cdotn)=\frac{\varepsilon^{-1}}{2}\left(\int_{\gamma_+}\babs{R}^2\mathrm{d}\gamma-\int_{\gamma_-}\babs{\mathcal{P}_{\gamma}[R]+h}^2\mathrm{d}\gamma\right)\\
=&\frac{\varepsilon^{-1}}{2}\left(\int_{\gamma_+}\abs{R}^2\mathrm{d}\gamma-\int_{\gamma_-}\babs{\mathcal{P}_{\gamma}[R]}^2\mathrm{d}\gamma-\int_{\gamma_-}\babs{h}^2\mathrm{d}\gamma-2\int_{\gamma_-}\mathcal{P}_{\gamma}[R]h\mathrm{d}\gamma\right)\nonumber\\
=&\frac{\varepsilon^{-1}}{2}\left(\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2-\tnms{h}{\gamma_-}^2\right).\nonumber
\end{align}
Here, we use Proposition \ref{lemma:final 2} to conclude that
\begin{align}
    \int_{\gamma_-}\mathcal{P}_{\gamma}[R]h\mathrm{d}\gamma=0.
\end{align}
Also, we use the decomposition of $(\textbf{I}-\textbf{P})[R]=\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]$ and orthogonality to obtain
\begin{align}
\br{R,\mathcal{L}[R]}
=&\br{(\textbf{I}-\overline{\textbf{P}})[R],\mathcal{L}\Big[(\textbf{I}-\overline{\textbf{P}})[R]\Big]}+\bbr{\textbf{d}\cdot\mathscr{A},\mathcal{L}[\textbf{d}\cdot\mathscr{A}]}
\end{align}
where the middle term vanishes due to the orthogonality of $\mathscr{A}$ and $(\textbf{I}-\overline{\textbf{P}})[R]$. On the other hand, using the designed orthogonality \eqref{energy 11}, we have
\begin{align}
\bbr{\textbf{d}\cdot\mathscr{A},\mathcal{L}[\textbf{d}\cdot\mathscr{A}]}=\bbr{\textbf{d}\cdot\mathscr{A},\textbf{d}\cdot\overline{\mathscr{A}}}=\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2+\int_{\Omega}\kappa\abs{\textbf{e}}^2.
\end{align}
Noticing the coercivity property of $\mathcal{L}$ as \eqref{coercivity}, there exists a uniform $\delta_0>0$ such that
\begin{align}
    \br{(\textbf{I}-\overline{\textbf{P}})[R],\mathcal{L}\Big[(\textbf{I}-\overline{\textbf{P}})[R]\Big]}\geq \delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2.
\end{align}
Hence, we have
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}{R\mathcal{L}[R]}\geq\delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2+\int_{\Omega}\kappa\abs{\textbf{e}}^2.
\end{align}
In summary, using Lemma \ref{h-estimate}, we have shown that
\begin{align}\label{energy 01}
    &\frac{\varepsilon^{-1}}{2}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+\varepsilon^{-2}\delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\textbf{e}}^2\\
    \leq&-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}+ \frac{\varepsilon^{-1}}{2}\bbr{R,S}
    +\varepsilon^{-1}\tnms{h}{\gamma_-}^2\nonumber\\
    \leq&-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}+ \varepsilon^{-1}\bbr{R,S}
    +o_{T}\varepsilon^{3-2\alpha}.\nonumber
\end{align}

\paragraph{\underline{Step 2: Estimates of $\displaystyle\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}$}}
We decompose
\begin{align}
\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}&=\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)^2}
=\mathscr{H}_1+\mathscr{H}_2+\mathscr{H}_3+\mathscr{H}_4,
\end{align}
where
\begin{align}
\mathscr{H}_1&:=\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(\textbf{P}[R]\Big)^2}=\br{\m^{\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\bigg(p+v\cdot
\textbf{b}+\Big(\abs{v}^2-5T\Big)c\bigg)^2}\\
\mathscr{H}_2&:=\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(\textbf{d}\cdot\mathscr{A}\Big)^2},\\
\mathscr{H}_3&:=2\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\textbf{P}[R](\textbf{d}\cdot\mathscr{A})}=\br{\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{2T^2},\bigg(p +v\cdot
\textbf{b}+\Big(\abs{v}^2-5T\Big)c\bigg)(\textbf{d}\cdot\mathscr{A})},\\
\mathscr{H}_4&:=\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})+(\textbf{I}-\overline{\textbf{P}})[R]\Big)(\textbf{I}-\overline{\textbf{P}})[R]}.
\end{align}
Due to symmetry/oddness, $p^2$, $\textbf{b}\otimes\textbf{b}$, $c^2$ and $p c$ terms vanish. Due to orthogonality, $p \textbf{b}$ term also vanishes. Hence, recalling the precise computation $\displaystyle\int_{\mathbb{R}^3}\mu\abs{v}^2\Big(\abs{v}^2-5T\Big)^2=30PT^2$, we are left with
\begin{align}
\mathscr{H}_1&=2\br{\m^{\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},(v\cdot\textbf{b})\Big(\abs{v}^2-5T\Big)c}
=2\int_{\Omega}\frac{\nabla_{x}T}{4T^2}\cdot\int_{\mathbb{R}^3}\m^{\frac{1}{2}}\overline{\mathscr{A}}(v\cdot\textbf{b})\Big(\abs{v}^2-5T\Big)c\\
&=\frac{2}{3}\int_{\Omega}\frac{\nabla_{x}T}{4T^2}\cdot(\textbf{b} c)\int_{\mathbb{R}^3}\mu\abs{v}^2\Big(\abs{v}^2-5T\Big)^2
=5P\bbrx{\nabla_{x}T,\textbf{b} c}.\nonumber
\end{align}
Due to symmetry/oddness, we know $\mathscr{H}_2=0$. On the other hand, for $\mathscr{H}_3$,
due to symmetry, the $\textbf{b}$ term vanishes. Hence, we have
\begin{align}
\mathscr{H}_3&=\br{\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{2T^2},\bigg(p +\Big(\abs{v}^2-5T\Big)c\bigg)(\textbf{d}\cdot\mathscr{A})}\\
&=\br{\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{2T^2},p (\textbf{d}\cdot\mathscr{A})}
+\br{\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{2T^2},\Big(\abs{v}^2-5T\Big)c(\textbf{d}\cdot\mathscr{A})}=\brx{\textbf{d}\cdot\frac{\nabla_{x}T}{2T^2},p \kappa+c\sigma}.\nonumber
\end{align}
In summary, we have
\begin{align}\label{energy 02}
\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},R^2}=&5P\bbrx{\nabla_{x}T,\textbf{b} c}+\brx{\textbf{d}\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}\\
&+\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})+(\textbf{I}-\overline{\textbf{P}})[R]\Big)(\textbf{I}-\overline{\textbf{P}})[R]}.\nonumber
\end{align}
Inserting \eqref{energy 02} into \eqref{energy 01}, we obtain
\begin{align}\label{final 37}
    &\frac{\varepsilon^{-1}}{2}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+\varepsilon^{-2}\delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\textbf{e}}^2\\
    \leq&-\varepsilon^{-1}5P\bbrx{\nabla_{x}T,\textbf{b} c}-\varepsilon^{-1}\brx{\textbf{d}\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}\nonumber\\
    &-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})+(\textbf{I}-\overline{\textbf{P}})[R]\Big)(\textbf{I}-\overline{\textbf{P}})[R]}+ \varepsilon^{-1}\bbr{R,S}+o_{T}\varepsilon^{3-2\alpha}.\nonumber
\end{align}
Taking inner product of $c$ and \eqref{aa 35}, we have
\begin{align}
5P\bbrx{\nabla_{x}T,\textbf{b} c}&=-\bbrx{c,\nabla_{x}\cdot(\kappa\textbf{d})}+\br{c\m^{\frac{1}{2}}\left(\abs{v}^2-5T\right),S}.
\end{align}
Hence, we have
\begin{align}\label{energy 03}
    &\frac{\varepsilon^{-1}}{2}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+\varepsilon^{-2}\delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2+\varepsilon^{-2}\int_{\Omega}\kappa\abs{\textbf{e}}^2\\
    \leq&\varepsilon^{-1}\bbrx{c,\nabla_{x}\cdot(\kappa\textbf{d})}-\varepsilon^{-1}\brx{\textbf{d}\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}\nonumber\\
    &-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})+(\textbf{I}-\overline{\textbf{P}})[R]\Big)(\textbf{I}-\overline{\textbf{P}})[R]}+\varepsilon^{-1}\br{R-c\m^{\frac{1}{2}}\left(\abs{v}^2-5T\right),S}
    +o_{T}\varepsilon^{3-2\alpha}
    .\nonumber
\end{align}

\paragraph{\underline{Step 3: Estimates of $\displaystyle\bbrx{c,\nabla_{x}\cdot(\kappa\textbf{d})}$ and $\displaystyle\brx{\nabla_{x}\xi\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}$}}

Letting $\phi=\xi$ in \eqref{pp 03} and using \eqref{tt 01}, we obtain
\begin{align}\label{new 3}
&-\bbrx{c,\nabla_{x}\cdot(\kappa\textbf{d})}+\brx{\frac{\nabla_{x} T}{2T^2}\cdot\nabla_{x}\xi,\kappap+\sigma c}+\varepsilon^{-1}\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2\\
=&\bbrb{\nabla_{x}\xi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\xi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big)\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right),(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\nabla_{x}\xi\cdot\mathscr{A},S}.\nonumber
\end{align}
Subtracting $\varepsilon^{-1}\times$ \eqref{new 3} from \eqref{energy 03}, we arrive at
\begin{align}\label{energy 04}
    &\frac{\varepsilon^{-1}}{2}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+\varepsilon^{-2}\tnm{\textbf{e}}^2+\varepsilon^{-2}\delta_0\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2\\
    \leq&-\varepsilon^{-1}\brx{\textbf{e}\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}-\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})+(\textbf{I}-\overline{\textbf{P}})[R]\Big)(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}+\varepsilon^{-1}\br{(\nabla_{x}\xi\cdot\mathscr{A})\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &+\varepsilon^{-1}\bbrb{\nabla_{x}\xi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\varepsilon^{-1}\bbrb{\nabla_{x}\xi\cdot\mathscr{A},h}{\gamma_-}+\varepsilon^{-1}\br{R-c\m^{\frac{1}{2}}\left(\abs{v}^2-5T\right)-\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),S}+o_{T}\varepsilon^{3-2\alpha}.\nonumber
\end{align}
Note that the crucial term $\varepsilon^{-1}\displaystyle\int_{\Omega}\kappa\abs{\nabla_{x}\xi}^2$ on the LHS of \eqref{energy 03} is cancelled at this step. This is the cost paid to remove the difficult term $-\varepsilon^{-1}5P\bbrx{\nabla_{x}T,\textbf{b} c}$ from \eqref{final 37}.

We can further estimate each term on the RHS using H\"older's inequality, Lemma \ref{h-estimate} and Lemma \ref{remark 01}: 
\begin{align}
    &\abs{\varepsilon^{-1}\brx{\textbf{e}\cdot\frac{\nabla_{x}T}{2T^2},\kp+\sigma c}}\lesssim o_{T}\tnm{p}^2+o_{T}\tnm{c}^2+o_{T}\varepsilon^{-2}\tnm{\textbf{e}}^2\lesssimo_{T}\xnm{R}^2,
\end{align}
\begin{align}
    &\abs{\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big(2\textbf{P}[R]+2(\textbf{d}\cdot\mathscr{A})\Big)(\textbf{I}-\overline{\textbf{P}})[R]}}\\
    \lesssim&\ o_{T}\tnm{\textbf{P}[R]}^2+o_{T}\tnm{\nabla_{x}\xi}^2+o_{T}\tnm{\textbf{e}}^2+o_{T}\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2\lesssimo_{T}\xnm{R}^2,\nonumber
\end{align}
\begin{align}
    \abs{\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}}&\lesssimo_1^{-1}\nm{\xi}^2_{H^2}+o_1\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2\\
    &\lesssimo_1\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\xnm{R}^2+o_{T}\varepsilon^{3-2\alpha},\nonumber
\end{align}
\begin{align}
    \abs{\varepsilon^{-1}\br{(\nabla_{x}\xi\cdot\mathscr{A})\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right),(\textbf{I}-\overline{\textbf{P}})[R]}}&\lesssim \nm{\xi}^2_{H^1}+o_{T}\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2\\
    &\lesssimo_{T}\xnm{R}^2+o_{T}\varepsilon^{3-2\alpha},\nonumber
\end{align}
\begin{align}
    \abs{\varepsilon^{-1}\bbrb{\nabla_{x}\xi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}}&\lesssim o_2^{-1}\varepsilon^{-1}\tnms{\nabla_{x}\xi}{\partial\Omega}^2+o_2\varepsilon^{-1}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2\\
    &\lesssimo_2\varepsilon^{-1}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2+o_{T}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha},\nonumber
\end{align}
\begin{align}
    \abs{\varepsilon^{-1}\bbrb{\nabla_{x}\xi\cdot\mathscr{A},h}{\gamma_-}}\lesssim \varepsilon^{-1}\tnms{\nabla_{x}\xi}{\partial\Omega}^2+\varepsilon^{-1}\tnms{h}{\gamma_-}^2\lesssim o_{T}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha}.
\end{align}
Also, using the weighted $L^{\infty}$ estimate to control the large velocity, we have
\begin{align}
    &\abs{\varepsilon^{-1}\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2},\Big((\textbf{I}-\overline{\textbf{P}})[R]\Big)^2}}\\
    \lesssim&\abs{\varepsilon^{-1}\int_{\Omega}\int_{\abs{v}\leq\varepsilon^{-\frac{1}{4}}}\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}\right)\Big((\textbf{I}-\overline{\textbf{P}})[R]\Big)^2}+\abs{\varepsilon^{-1}\int_{\Omega}\int_{\abs{v}\geq\varepsilon^{-\frac{1}{4}}}\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}\right)\Big((\textbf{I}-\overline{\textbf{P}})[R]\Big)^2}\nonumber\\
    \lesssim&o_{T}\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\varepsilon^3\lnmm{R}^2\lesssimo_{T}\xnm{R}^2,\nonumber
\end{align}
Collecting all above, for $o_1,o_2\ll1$, we have \eqref{qq 34}.
\end{proof}


\subsection{Estimates of Source Terms}


In this section, we consider the source term in \eqref{qq 34}:
\begin{align}
    \varepsilon^{-1}\br{R-c\m^{\frac{1}{2}}\left(\abs{v}^2-5T\right)-\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),S}
    =&\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S}.
\end{align}

\begin{lemma}\label{lem:final 10}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{qq 36}
    \abs{\varepsilon^{-1}\br{R-c\m^{\frac{1}{2}}\Big(\abs{v}^2-5T\Big)-\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),S}}
    \lesssim&o(1)\Big(\varepsilon^{-2}\tnm{\textbf{e}}^2+\varepsilon^{-2}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2\Big)\\
    &+o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^4+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
\end{lemma}

\begin{proof}
Due to the orthogonality of $\Gamma$, using Lemma \ref{ssl-estimate}, Lemma \ref{ss0-estimate}, Lemma \ref{ss1-estimate}, Lemma \ref{ss5-estimate}, we know
\begin{align}
    &\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],\lc^1[R]+S_0+S_1+S_5}}\\
    =&\varepsilon^{-1}\abs{\bbr{\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],\lc^1[R]+S_0+S_1+S_5}}\nonumber\\
    \lesssim& o_{T}\varepsilon^{-2}\tnm{\textbf{e}}^2+o_{T}\varepsilon^{-2}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\um{R}^2+o_{T}\varepsilon^{3-2\alpha}.\nonumber
\end{align}
Also, using Lemma \ref{ss4-estimate} and Lemma \ref{ssp-estimate}, we have
\begin{align}
    &\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_4+S_6}}\\
    \lesssim&o_{T}\varepsilon^{-2}\tnm{\textbf{e}}^2+o_{T}\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\varepsilon^{-1}\tnm{p}^2+o_{T}\varepsilon^{-1}\tnm{\textbf{b}}^2+\varepsilon^{-1}\tnm{S_4}^2+\varepsilon^{-1}\tnm{S_6}^2\nonumber\\
    \lesssim&o_{T}\varepsilon^{-2}\tnm{\textbf{e}}^2+o_{T}\varepsilon^{-2}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\varepsilon^{-1}\tnm{p}^2+o_{T}\varepsilon^{-1}\tnm{\textbf{b}}^2+o_{T}\varepsilon^{3-2\alpha}.\nonumber
\end{align}
Also, we know
\begin{align}\label{final 09}
\\
    \abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_3}}
    \lesssim&\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}},S_3}}+\abs{\varepsilon^{-1}\br{\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_3}}.\nonumber
\end{align}
In the first term of \eqref{final 09}, using \eqref{ss3-estimate5} and \eqref{ss3-estimate6} with $r=2$, we obtain
\begin{align}
    &\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}},S_3}}\lesssim \abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}},\ss_{3a}}}+\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}},\ss_{3b}+\ss_{3c}}}\\
    \lesssim&o_{T}\varepsilon^{\frac{1}{2}-\alpha}\tnm{\nabla_v\Big(p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}\Big)}+\varepsilon^{-1}\nm{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}}_{L^2_xL^{\infty}_v}\nm{\ss_{3b}+\ss_{3c}}_{L^2_xL^1_v}\nonumber\\
    \lesssim&o_{T}\varepsilon^{-1}\tnm{p}^2+o_{T}\varepsilon^{-1}\tnm{\textbf{b}}^2+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
For the second term of \eqref{final 09}, we can directly bound
\begin{align}
    \abs{\varepsilon^{-1}\br{\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_3}}
    \lesssim&\varepsilon^{-1}\Big(\tnm{\textbf{e}}^2+\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)\tnm{S_3}\\
    \lesssim&o_{T}\varepsilon^{-2}\tnm{\textbf{e}}^2+o_{T}\varepsilon^{-2}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
In addition, due to the orthogonality of $\Gamma$, using Lemma \ref{ss2-estimate}, we know
\begin{align}
    &\abs{\varepsilon^{-1}\br{p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_2}}=\varepsilon^{-1}\abs{\bbr{\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R],S_2}}\\
    \lesssim& o(1)\varepsilon^{-2}\tnm{\textbf{e}}^2+o(1)\varepsilon^{-2}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2+\varepsilon^{2\alpha-2}\xnm{R}^4.\nonumber
\end{align}
In summary, we have
\begin{align}\label{qq 36,}
    \abs{\varepsilon^{-1}\br{R-c\Big(\abs{v}^2-5T\Big)-\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big),S}}
    \lesssim&o(1)\Big(\varepsilon^{-2}\tnm{\textbf{e}}^2+\varepsilon^{-2}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2\Big)\\
    &+o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^4+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
\end{proof}

\begin{proof}[Proof of Proposition \ref{thm:energy}]
Inserting \eqref{qq 36} into \eqref{qq 34}, our desired result follows.
\end{proof}


\subsection{Extended Energy Estimates}


\begin{lemma}\label{lem:final 11}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, we have
    \begin{align}\label{final 57}
        \pnm{\textbf{e}}{6}+\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{6}+\pnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{4}{\gamma_+}
        \lesssim&o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
    \end{align}
\end{lemma}
\begin{proof}
By interpolation, we obtain
\begin{align}
    \pnm{\textbf{e}}{6}\lesssim&\tnm{\textbf{e}}^{\frac{1}{3}}\lnmm{\textbf{e}}^{\frac{2}{3}}\lesssim o_{T}\xnm{R}^{\frac{1}{3}}\left(\varepsilon^{\frac{1}{2}}\lnmm{\textbf{e}}\right)^{\frac{2}{3}}\lesssim o_{T}\xnm{R}+o_{T}\varepsilon^{\frac{1}{2}}\lnmm{R},\label{tt 17}\\
    \pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{6}\lesssim&\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{\frac{1}{3}}\lnmm{(\textbf{I}-\overline{\textbf{P}})[R]}^{\frac{2}{3}}\lesssim o_{T}\xnm{R}^{\frac{1}{3}}\left(\varepsilon^{\frac{1}{2}}\lnmm{(\textbf{I}-\overline{\textbf{P}})[R]}\right)^{\frac{2}{3}}\label{tt 18}\\
    \lesssim& o_{T}\xnm{R}+o_{T}\varepsilon^{\frac{1}{2}}\lnmm{R},\nonumber\\
    \label{tt 19}
    \pnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{4}{\gamma_+}\lesssim&\tnm{(1-\mathcal{P}_{\gamma})[R]}^{\frac{1}{2}}\lnmms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{\frac{1}{2}}\lesssim o_{T}\xnm{R}^{\frac{1}{2}}\left(\varepsilon^{\frac{1}{2}}\lnmms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\right)^{\frac{1}{2}}\\
    \lesssim&o_{T}\xnm{R}+o_{T}\varepsilon^{\frac{1}{2}}\lnmms{R}{\gamma_+}.\nonumber
\end{align}
Then the desired result follows from \eqref{tt 17}\eqref{tt 18}\eqref{tt 19} and Proposition \ref{thm:energy}.
\end{proof}

\begin{lemma}[Ukai's Trace Theorem, Lemma 2.1 of \cite{Esposito.Guo.Kim.Marra2013} ]\label{remainder lemma 1}
Define the near-grazing set of $\gamma_+$ or $\gamma_-$ as
\begin{align}
\gamma_{\pm}^{\delta}=\left\{(x,v)\in\gamma_{\pm}:
\abs{n(x)\cdotv}\leq\delta\ \text{or}\ \abs{v}\geq\frac{1}{\delta}\ \text{or}\
\abs{v}\leq\delta\right\}.
\end{align}
Then for $p\in[1,\infty)$
\begin{align}
\abs{f\textbf{1}_{\gamma_{\pm}\backslash\gamma_{\pm}^{\delta}}}_{L^p}\leq
C(\delta)\bigg(\nm{f}_{L^p}+\nm{v\cdot\nabla_{x} f}_{L^p}\bigg).
\end{align}
\end{lemma}

\begin{proposition}\label{lem:final 12}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{qq 32}
    \tnms{\mathcal{P}_{\gamma}[R]}{\gamma}
    \lesssim&\tnm{p}+\tnm{\textbf{b}}+\tnm{c}+o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}+o_{T}\varepsilon^{1-\alpha}.
\end{align}
\end{proposition}

\begin{proof}
From Proposition \ref{thm:energy} and Lemma \ref{remark 01}, we have
\begin{align}
\tnm{R}\lesssim& \tnm{p}+\tnm{\textbf{b}}+\tnm{c}+\nm{\xi}_{H^1}+\tnm{\textbf{e}}+\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\\
\lesssim&\tnm{p}+\tnm{\textbf{b}}+\tnm{c}+o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.\nonumber
\end{align}
Multiplying $R$ on both sides of \eqref{remainder.}, we know
\begin{align}
\frac{1}{2}v\cdot\nabla_{x}\big(R\big)^2
&+R\bigg(\m^{-\frac{1}{2}}\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)R-\varepsilon^{-1}\xi(\nabla_{x}\cdot\overline{\mathscr{A}})+\varepsilon^{-1}\textbf{e}\cdot\overline{\mathscr{A}}+\varepsilon^{-1}\mathcal{L}\Big[(\textbf{I}-\overline{\textbf{P}})[R]\Big]-S\bigg)=0.
\end{align}
Taking absolute value and integrating over $\Omega\times\mathbb{R}^3$, with the help of Cauchy's inequality and Lemma \ref{remark 01}, we deduce
\begin{align}
\pnm{v\cdot\nabla_{x}\big(R\big)^2}{1}
\lesssim&\abs{\br{R, \m^{-\frac{1}{2}}\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)R-\varepsilon^{-1}\xi(\nabla_{x}\cdot\overline{\mathscr{A}})+\varepsilon^{-1}\textbf{e}\cdot\overline{\mathscr{A}}+\varepsilon^{-1}\mathcal{L}\Big[(\textbf{I}-\overline{\textbf{P}})[R]\Big]-S}}\\
\lesssim&\tnm{p}^2+\tnm{\textbf{b}}^2+\tnm{c}^2+o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha}+\abs{\bbr{R,S}}\nonumber\\
\lesssim&\tnm{p}^2+\tnm{\textbf{b}}^2+\tnm{c}^2+o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha},\nonumber
\end{align}
where we utilize the source term estimates in Section \ref{sec:bs}
\begin{align}
    \bbr{R,S}
    \lesssim& o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^4+o_{T}\varepsilon^{3-2\alpha}.
\end{align}
Then by Lemma \ref{remainder lemma 1}, we have
\begin{align}
\tnms{\textbf{1}_{\gamma\backslash\gamma^{\delta}}R}{\gamma}^2
\leq&
C(\delta)\left(\tnm{R}^2+\pnm{v\cdot\nabla_{x}\big(R\big)^2}{1}\right)\\
\lesssim&\tnm{p}^2+\tnm{\textbf{b}}^2+\tnm{c}^2+o_{T}\xnm{R}^2+\varepsilon^{2\alpha-2}\xnm{R}^2+o_{T}\varepsilon^{2-2\alpha}.\nonumber
\end{align}
Notice that
\begin{align}
    \mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}R\Big]=&\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}\Big(\mathcal{P}_{\gamma}[R]+\textbf{1}_{\gamma_+}(1-\mathcal{P}_{\gamma})[R]+\textbf{1}_{\gamma_-}h\Big)\Big]
\end{align}
Based on the definition, we can rewrite $\mathcal{P}_{\gamma}[R]=z_{\gamma}(x)\m^{\frac{1}{2}}$ for a suitable function
$z_{\gamma}(x)$ and
for $\delta$ small, we deduce
\begin{align}
\tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}\mathcal{P}_{\gamma}[R]\Big]}{\gamma}^2
=&\int_{\partial\Omega}\abs{z_{\gamma}(x)}^2
\left(\int_{\abs{v\cdotn(x)}\geq\delta,\delta\leq\abs{v}\leq\frac{1}{\delta}}\mu(v)\abs{v\cdotn(x)}\mathrm{d}{v}\right)\mathrm{d}{x}\\
\geq&\frac{1}{2}\left(\int_{\partial\Omega}\abs{z_{\gamma}(x)}^2\mathrm{d}{x}\right)\left(\int_{\mathbb{R}^2}\mu(v)\abs{v\cdotn(x)}\mathrm{d}{v}\right)=\frac{1}{2}\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}^2,\nonumber
\end{align}
where we utilized the fact that
\begin{align}
\int_{\abs{v\cdotn(x)}\leq\delta}\mu(v)\abs{v\cdotn(x)}\mathrm{d}{v}&\leq C\delta,\quad
\int_{\abs{v}\leq\delta\ or\
\abs{v}\geq\frac{1}{\delta}}\mu(v)\abs{v\cdotn(x)}\mathrm{d}{v}\leq C\delta.
\end{align}
Also, we have
\begin{align}
    \tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}\Big(\textbf{1}_{\gamma_+}(1-\mathcal{P}_{\gamma})[R]+\textbf{1}_{\gamma_-}h\Big)\Big]}{\gamma}\lesssim\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\tnms{h}{\gamma_-}.
\end{align}
Therefore, from
\begin{align}
\tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}R\Big]}{\gamma}\leq C
\tnms{\textbf{1}_{\gamma\backslash\gamma^{\delta}}R}{\gamma},
\end{align}
we conclude
\begin{align}
\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}
\leq&\tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}\mathcal{P}_{\gamma}[R]\Big]}{\gamma}
\leq\tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}R\Big]}{\gamma}+\tnms{\mathcal{P}_{\gamma}\Big[\textbf{1}_{\gamma\backslash\gamma^{\delta}}\Big(\textbf{1}_{\gamma_+}(1-\mathcal{P}_{\gamma})[R]+\textbf{1}_{\gamma_-}h\Big)\Big]}{\gamma}\\
\leq& C\tnms{\textbf{1}_{\gamma\backslash\gamma^{\delta}}R}{\gamma}+C\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+C\tnms{h}{\gamma_-}.\nonumber
\end{align}
Hence, using Lemma \ref{h-estimate} and Proposition \ref{thm:energy}, we conclude
\begin{align}\label{qq 32,}
\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}
\lesssim&\tnm{p}+\tnm{\textbf{b}}+\tnm{c}+o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}+o_{T}\varepsilon^{1-\alpha}.
\end{align}
\end{proof}


\section{$p$ and $Z$ Estimates}


\subsection{$L^{r}$ Estimates of $p$}


\begin{proof}[Proof of Proposition \ref{prop:p-bound}]
To establish \eqref{oo 28=}, thanks to energy estimates, it suffices to establish for $2\leqr\leq 6$
\begin{align}\label{oo 28}
    \pnm{p}{r}\lesssim&\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}+o_{T}\varepsilon^{1+\frac{2}{r}-\alpha}.
\end{align}
Denote a test function
\begin{align}
\psi=\m^{\frac{1}{2}}(v)\Big(v\cdot\phi(x)\Big),
\end{align}
where $\phi$ is defined via solving the Bogovskii problem
\begin{align}
\left\{
\begin{array}{l}
-\nabla_{x}\cdot\phi=\displaystyle p\abs{p}^{r-2}-\frac{1}{\abs{\Omega}}\int_{\Omega}p(x)\abs{p(x)}^{r-2}\mathrm{d} x\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.0em}
\phi=0\ \ \text{on}\ \ \partial\Omega.
\end{array}
\right.
\end{align}
Based on \cite{Bogovskii1979} and \cite[Theorem 2.4]{Acosta.Duran2017}, there exists a solution $\phi$ satisfying
\begin{align}
\nm{\psi}_{W^{1,\frac{r}{r-1}}L^{\infty}_{\varrho,\vartheta}}\lesssim\nm{\phi}_{W^{1,\frac{r}{r-1}}}\lesssim\pnm{p}{r}^{r-1}.
\end{align}
Taking $\mathfrak{f}=\psi $ in \eqref{weak formulation}, considering $\psi\big|_{\gamma}=0$, we obtain
\begin{align}
    -\bbr{R,v\cdot\nabla_{x}\psi }+\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}R,\psi }+\varepsilon^{-1}\bbr{\mathcal{L}[R],\psi}
    &=\bbr{S,\psi }.
\end{align}
Direct computation reveals that
\begin{align}
-\bbr{R,v\cdot\nabla_{x}\psi }+\br{\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}R,\psi }=-\br{\mhR,v\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)}.
\end{align}
We will split $R$ according to \eqref{pp 01} and analyze each term.
Due to oddness, we have
\begin{align}
-\br{(\textbf{b}\cdotv)\mu+\m^{\frac{1}{2}}(\textbf{d}\cdot\mathscr{A}),v\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)} =0.
\end{align}
On the other hand, using H\"{o}lder's inequality, we have
\begin{align}
\abs{\br{\m^{\frac{1}{2}}(\textbf{I}-\overline{\textbf{P}})[R],v\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)}}
&\lesssim\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\nm{\phi }_{W^{1,\frac{r}{r-1}}}\lesssim\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\pnm{p}{r}^{r-1}.
\end{align}
For $c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}$ part, due to orthogonality, we know
\begin{align}\label{kernel 8}
&-\br{\mu c\left(\abs{v}^2-5T\right),v\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)}
=-\br{\m^{\frac{1}{2}} c,\overline{\mathscr{A}}\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)}=0.
\end{align}
For $p\m^{\frac{1}{2}}$ part, considering \eqref{final 31} and $\displaystyle\int_{\mathbb{R}^3}\abs{v}^2\mu=3P$, we have
\begin{align}
-\br{p\mu ,v\cdot\nabla_{x}\left(\m^{-\frac{1}{2}}\psi \right)}
=&-\br{p\mu,v\cdot\nabla_{x}\Big(v\cdot\phi \Big)}
=-\frac{1}{3}\int_{\Omega}p\big(\nabla_{x}\cdot\phi \big)\int_{\mathbb{R}^3}\abs{v}^2\mu
=P\pnm{p}{r}^{r}.
\end{align}
Also, due to orthogonality, we have
\begin{align}\label{kernel 12}
\bbr{\mathcal{L}[R],\psi}&=\br{\mathcal{L}\Big[\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big],\psi}
=0 .
\end{align}
Collecting all above, we have shown that
\begin{align}
   \pnm{p}{r}^{r}\lesssim \pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\pnm{p}{r}^{r-1}+\abs{\bbr{S,\psi}}.
\end{align}
Due to orthogonality and oddness, we have
\begin{align}
    \bbr{S,\psi}=\bbr{S_3,\psi}+\bbr{\ss_6,\psi}.
\end{align}
Notice that $S_3$ contains derivatives of $f^{B}_1$ in the rescaled variable $\eta=\varepsilon^{-1}\mathfrak{n}$. We define a natural extension of $\partial_{\mathfrak{n}}\psi$ with zero value for $\mathfrak{n}>1$ (for the convenience of Hardy's inequality). Recall that $\psi=0$ on $\partial\Omega$. Using Hardy's inequality, Lemma \ref{ss3-estimate} for $\ss_{3b},\ss_{3c}$ and Remark \ref{ss3-remark} for $\ss_{3a}$, we have
\begin{align}\label{oo 30}
    &\abs{\bbr{S_3,\psi}}=\abs{\br{S_3,\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi}}=\varepsilon\abs{\br{\etaS_3,\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi}}\\
    \lesssim&\varepsilon\Big(\jnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}+\jnm{\etaf^{B}_1}\Big)\knm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi}\lesssim\varepsilon\Big(\jnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}+\jnm{\etaf^{B}_1}\Big)\knm{\partial_{\mathfrak{n}}\psi}\nonumber\\
    \lesssim&\varepsilon\Big(\jnm{\eta\left(\ss_{3b}+\ss_{3c}\right)}+\jnm{\etaf^{B}_1}\Big)\pnm{p}{r}^{r-1}\lesssimo_{T}\varepsilon^{1+\frac{2}{r}-\alpha}\pnm{p}{r}^{r-1}.\nonumber
\end{align}
Also, using Lemma \ref{ssp-estimate}, we know
\begin{align}
    \abs{\bbr{\ss_6,\psi}}\lesssim \pnm{\ss_6}{r}\pnm{\psi}{\frac{r}{r-1}}\lesssimo_{T}\varepsilon^{2-\alpha}\pnm{p}{r}^{r-1}.
\end{align}
In summary, we have shown
\begin{align}
    \pnm{p}{r}^{r}\lesssim&\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\pnm{p}{r}^{r-1}+o_{T}\varepsilon^{1+\frac{2}{r}-\alpha}\pnm{p}{r}^{r-1}+o_{T}\varepsilon^{2-\alpha}\pnm{p}{r}^{r-1}.
\end{align}
Hence, we have
\begin{align}\label{oo 28,}
    \pnm{p}{r}\lesssim&\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}+o_{T}\varepsilon^{1+\frac{2}{r}-\alpha}.
\end{align}
Then the desired estimate \eqref{oo 28} follows from Proposition \ref{thm:energy} and Lemma \ref{final 11}.
\end{proof}


\subsection{$L^{r}$ Estimates of $Z$}


\subsubsection{Estimates of $Z$}


\begin{proof}[Proof of Proposition \ref{prop:z-bound}]
To establish \eqref{oo 33=}, thanks to energy estimates, it suffices to establish for $2\leqr\leq6$
\begin{align}\label{oo 33}
    \pnm{Z}{r}\lesssim&\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.
\end{align}
In \eqref{pp 03.}, consider $\phi(x)$ satisfying
\begin{align}
\left\{
\begin{array}{l}
-\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi \big)=Z\abs{Z}^{r-2}\ \ \text{in}\ \
\Omega,\\\rule{0ex}{1.2em}
\phi =0\ \ \text{on}\ \ \partial\Omega.
\end{array}
\right.
\end{align}
Then we have
\begin{align}
    &\jnm{Z}^{r}+\varepsilon^{-1}\bbrx{\nabla_{x}\phi,\kappa\textbf{e}}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}\\
    =&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},S}.\nonumber
\end{align}
Using H\"{o}lder's inequality, we have
\begin{align}
    &\abs{\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}}+\abs{\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}}\\
    \lesssim&\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\nm{\phi }_{W^{2,\frac{r}{r-1}}}\lesssim\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\pnm{Z}{r}^{r-1},\nonumber
\end{align}
and
\begin{align}
    \abs{\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}}\lesssim&o_{T} \pnm{\kp+\sigma c}{r}\pnm{\phi }{\frac{r}{r-1}}\lesssimo_{T}\Big(\pnm{p}{r}+\pnm{c}{r}\Big)\pnm{Z}{r}^{r-1}.
\end{align}
Due to \eqref{final 63}, we have
\begin{align}
    \varepsilon^{-1}\bbrx{\nabla_{x}\phi,\kappa\textbf{e}}=0.
\end{align}
By Sobolev embedding and trace theorem in 3D, we have
\begin{align}
\abs{\bbrb{h,\mathscr{A}\cdot\nabla_{x}\phi}{\gamma_-}}&\lesssim\jnms{h}{\gamma_-}\knms{\nabla_{x}\phi }{\gamma_-}\lesssim\jnms{h}{\gamma_-}\pnm{Z}{r}^{r-1},\\
\abs{\bbrb{(1-\mathcal{P}_{\gamma})[R],\mathscr{A}\cdot\nabla_{x}\phi}{\gamma_+}}&\lesssim\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\knms{\nabla_{x}\phi }{\gamma_+}\lesssim\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\pnm{Z}{r}^{r-1}.
\end{align}
Finally, using Section \ref{sec:bs}, we have the source term estimates
\begin{align}
    \abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},S-S_2}}\lesssim&\tnm{\nabla_{x}\phi}\Big(o_{T}\um{R}+o_{T}\varepsilon^{\frac{3}{2}-\alpha}\Big)\lesssim \pnm{Z}{r}^{r-1}\Big(o_{T}\um{R}+o_{T}\varepsilon^{1-\alpha}\Big),\\
    \abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},S_2}}\lesssim&\varepsilon^{\alpha-1}\tnm{\nabla_{x}\phi}\xnm{R}^2\lesssim \varepsilon^{\alpha-1}\pnm{Z}{r}^{r-1}\xnm{R}^2.
\end{align}
In summary, we have shown 
\begin{align}
\pnm{Z}{r}^{r}
\lesssim& o_{T}\Big(\pnm{p}{r}+\pnm{c}{r}\Big)\pnm{Z}{r}^{r-1}+\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{r}\pnm{Z}{r}^{r-1}\\
&+\jnms{h}{\gamma_-}\pnm{Z}{r}^{r-1}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\pnm{Z}{r}^{r-1}\nonumber\\
&+\pnm{Z}{r}^{r-1}\Big(o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}\Big).\nonumber
\end{align}
Then the desired estimate \eqref{oo 33} follows from Proposition \ref{thm:energy} and Lemma \ref{final 11}.
\end{proof}


\subsubsection{Decomposition of $Z$}


We will introduce a decomposition $Z:=Z^R+Z^S$ such that that $\nm{Z^R}_{H^1_0}\lesssim\xnm{R}$ and $\tnm{Z^S}\lesssim\varepsilon^{\frac{1}{2}}\xnm{R}$. In other words, $Z^R$ is smoother and thus we can apply techniques like integration by parts or rewriting it with the fundamental theorem of calculus, while $Z^S$ is rougher but relatively small.

\begin{proof}[Proof of Proposition \ref{prop:z-splitting}]
The decomposition is rooted from the weak formulation. Equation \eqref{pp 03.} and \eqref{final 63} yield that for any test function $\phi(x)\in H^2_0(\Omega)$
\begin{align}\label{final 52}
\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),Z}
=&\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}\\
&-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}+\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
&+\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}-\bbr{\nabla_{x}\phi\cdot\mathscr{A},S}.\nonumber
\end{align}
In the following, we will split $Z$ and \eqref{final 52}.
\\
\paragraph{\underline{Construction of $Z^R$}}
By Lax-Milgram theorem, there exists a unique $Z^R\in H^1_0(\Omega)$ satisfying for any test function $\psi(x)\in H_0^1(\Omega)$,
\begin{align}\label{pp 125}
    \bbr{\kappa\nabla_{x}\psi,\nabla_{x} Z^R}
    =&-\br{\nabla_{x}\psi,\frac{\nabla_{x} T}{2T^2}(\kp+\sigma c)}\\
    &-\br{\mathscr{A}\cdot\nabla_{x}\psi,\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}+\bbr{\mathscr{A}\cdot\nabla_{x}\psi,S}.\nonumber
\end{align}
Taking $\psi=Z^R$, using Poincar\'e's inequality, we obtain
\begin{align}
\nm{Z^R}_{H^1_0}\lesssim& \tnm{\frac{\nabla_{x} T}{2T^2}(\kp+\sigma c)}+\tnm{\br{\mathscr{A},\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}(\textbf{I}-\overline{\textbf{P}})[R]}}+\tnm{\brv{\mathscr{A},S}}\\
\lesssim&o_{T}\tnm{p}+o_{T}\tnm{c}+o_{T}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}+o_{T}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^2+o_{T}\varepsilon^{1-\alpha}.\nonumber
\end{align}
Then the desired estimate \eqref{qq 06} follows from Proposition \ref{thm:energy}.
\\
\paragraph{\underline{Construction of $Z^S$}} Denote $Z^S=Z-Z^R$. Let $\psi(x)=\phi(x)\in H^2_0(\Omega)$ in \eqref{pp 125}. Integration by parts yields 
\begin{align}\label{qq 110.}
    \bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),Z^R}
    =&\br{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}(\kp+\sigma c)}\\
    &+\br{\mathscr{A}\cdot\nabla_{x}\phi,\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}-\bbr{\mathscr{A}\cdot\nabla_{x}\phi,S},\nonumber
\end{align}
and thus $Z^S\in L^2(\Omega)$ satisfies
\begin{align}\label{qq 110..}
\\
    \bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),Z^S}
    =&-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}+\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}.\nonumber
\end{align}
We choose the test function $\phi(x)$ satisfying
\begin{align}
\nabla_{x}\cdot(\kappa\nabla_{x}\phi) =Z^S,\quad \phi\Big|_{\partial\Omega}=0.
\end{align}
The standard elliptic estimate implies
\begin{align}
    \nm{\phi}_{H^2}\lesssim\tnm{Z^S}.
\end{align}
It is obvious that
\begin{align}
    \bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),Z^S}=\tnm{Z^S}^2.
\end{align}
On the other hand, we know
\begin{align}
    \abs{\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}}
    \lesssim&\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\nm{\phi}_{H^2}\lesssim \tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^2+o(1)\tnm{Z^S}^2.
\end{align}
The remaining boundary terms can be bounded as
\begin{align}
    \abs{\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}}
    \lesssim&\tnms{\nabla_{x}\phi}{\partial\Omega}\tnms{\bbrv{(v\cdot n)\mathscr{A}, h\textbf{1}_{\gamma_-}}}{\partial\Omega}
    \lesssimo(1)\tnm{Z^S}^2+\tnms{h}{\gamma_-}^2,
\end{align}
and
\begin{align}
    \abs{\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}}
    \lesssim&\tnms{\nabla_{x}\phi}{\partial\Omega}\tnms{\bbrv{(v\cdot n)\mathscr{A},(1-\mathcal{P}_{\gamma})[R]\textbf{1}_{\gamma_+}}}{\partial\Omega}\\
    \lesssim&o(1)\tnm{Z^S}^2+\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^2.\nonumber
\end{align}
In total, using Lemma \ref{h-estimate}, we have
\begin{align}
    \tnm{Z^S}\lesssim \um{(\textbf{I}-\overline{\textbf{P}})[R]}+\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+o_{T}\varepsilon^{2-\alpha}.
\end{align}
Then the desired estimate \eqref{qq 06} follows from Proposition \ref{thm:energy}.
\end{proof}


\section{Dual Stokes Problem and $\textbf{b}$ Estimates}


In this section, we plan to prove Proposition \ref{prop:b-bound}. To establish \eqref{oo 34=}, thanks to energy estimates and \eqref{ss 00}, it suffices to establish for $2\leqr\leq6$
    \begin{align}\label{oo 34}
    \jnm{\textbf{b}}^{r}\lesssim& \xnm{R}\jnm{\textbf{b}}^{r}+ \Big(\xnm{R}+1\Big)\Big(\jnm{\textbf{e}}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}\Big)\\
    &+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+o_{T}\varepsilon^{\alr-\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{r+1-r\alpha}.\nonumber
    \end{align}


\subsection{Combined Weak Formulation}


Assume $(\psi,q)\in\mathbb{R}^3\times\mathbb{R}$ (where $q$ has zero average) is the unique strong solution to the perturbed Stokes problem \eqref{pp 17}.
Considering that the perturbation terms $\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big)$ and $\bbrv{\nabla_{x}\psi:\mathscr{B},v\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}$ are small due to the smallness of $\nabla_{x} T$, we have the standard estimate \cite{Cattabriga1961}, Sobolev embedding for $2\leqr\leq 6$ and trace estimates
\begin{align}\label{qq 15}
    \nm{\psi}_{W^{2,\frac{r}{r-1}}}+\nm{q}_{W^{1,\frac{r}{r-1}}}+\nm{\psi}_{H^1}+\tnm{q}+\abs{\psi}_{W^{1,\frac{2r}{2r-3}}_{\partial\Omega}}+\knms{q}{\partial\Omega}\lesssim\knm{\textbf{b}\abs{\textbf{b}}^{r-2}}\lesssim \jnm{\textbf{b}}^{r-1}.
\end{align}
Multiplying $\textbf{b}$ on both sides of \eqref{pp 17} and integrating by parts for $\brx{\nabla_{x} q,\textbf{b}}$, we have
\begin{align} 
    -\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}\\
    -\bbrx{ q,\nabla_{x}\cdot\textbf{b}}+\int_{\partial\Omega}q(\textbf{b}\cdot n)&=\jnm{\textbf{b}}^{r},\nonumber
\end{align}
which, by combining \eqref{conservation law 1} and Remark \ref{remark 02}, implies
\begin{align}\label{pp 18}
    -\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+(\nabla_{x}\cdot\overline\b)(\nabla_{x}\psi:\mathscr{B}),\textbf{b}}+\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}\\
    -P^{-1}\br{q,\mhS_3+\mhS_4}&=\jnm{\textbf{b}}^{r},\nonumber
\end{align}
Inserting \eqref{pp 18} into \eqref{pp 16} to replace $-\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\bbr{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}$, we obtain
\begin{align}\label{pp 19}
&\jnm{\textbf{b}}^{r}+P^{-1}\br{q,\mhS_3+\mhS_4}+\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}\\
=&\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
&+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}-\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
&+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\bbr{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align} 

\begin{lemma}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{final 12}
  \jnm{\textbf{b}}^{r}\lesssim& \jnm{\textbf{e}}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\varepsilon^{r+1-r\alpha}\\
  &-P^{-1}\br{q,\mhS_3+\mhS_4}-\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\bbr{\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
\end{lemma}

\begin{proof}
We need to estimate each term in \eqref{pp 19} except the source terms. Using Lemma \ref{h-estimate}, we have
\begin{align}
&\abs{\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}}+\abs{\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}}\\
\lesssim& o(1)\knms{\nabla_{x}\psi}{\partial\Omega}^{\frac{r}{r-1}}+\jnms{\int_{v\cdot n>0}(1-\mathcal{P}_{\gamma})[R]\mathscr{B}(v\cdot n)}{\partial\Omega}^{r}+\jnms{\int_{v\cdot n<0}h\mathscr{B}(v\cdot n)}{\partial\Omega}^{r}\nonumber\\
\lesssim& o(1)\jnm{\textbf{b}}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+\jnms{h}{\gamma_-}^{r}\lesssim o(1)\jnm{\textbf{b}}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\varepsilon^{r+3-r\alpha}.\nonumber
\end{align}
In addition, \eqref{qq 15}, H\"older's inequality and Young's inequality yield
\begin{align}
\abs{\bbr{v\cdot\nabla_{x}(\nabla_{x}\psi:\mathscr{B}),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}}
\lesssim&o(1)\knm{\nabla_{x}^2\psi}^{\frac{r}{r-1}}+\jnm{\textbf{d}}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\\
\lesssim&o(1)\jnm{\textbf{b}}^{r}+\jnm{\textbf{d}}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r},\nonumber
\end{align}
and
\begin{align}
    &\abs{\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}}\\
    \lesssim&o_{T}\knm{\nabla_{x}\psi}^{\frac{r}{r-1}}+o_{T}\jnm{\textbf{d}}^{r}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}
    \lesssimo_{T}\jnm{\textbf{b}}^{r}+o_{T}\jnm{\textbf{d}}^{r}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
Collecting all above and using Lemma \ref{remark 01}, we have the desired result.
\end{proof}


\subsection{Estimates of Source Terms}


In this section, we will estimate the source term in \eqref{final 12}:
\begin{align}\label{final 58}
  -P^{-1}\br{q,\mhS_3+\mhS_4}-\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}+\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\bbr{\nabla_{x}\psi:\mathscr{B},\barS}.
\end{align}

\begin{lemma}
Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have
\begin{align}\label{final 11}
    &\abs{P^{-1}\br{q,\mhS_3+\mhS_4}}+\abs{\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}}+\abs{\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}}+\abs{\bbr{\nabla_{x}\psi:\mathscr{B},\barS}}\\
    \lesssim&o_{T}\jnm{\textbf{b}}^{r}+\xnm{R}\jnm{\textbf{b}}^{r}+ \Big(\xnm{R}+1\Big)\Big(\jnm{\textbf{e}}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}\Big)\nonumber\\
    &+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+\varepsilon^{\alr-\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{r+1-r\alpha}.\nonumber
\end{align}
\end{lemma}

\begin{proof}\ 
\paragraph{\underline{$\lc^1[R]$ Contribution}}
We need more detailed analysis than the estimates in Lemma \ref{ssl-estimate}. We split
\begin{align}
    \bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}=\br{\nabla_{x}\psi:\mathscr{B},2\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,R\right]}=K_1+K_2,
\end{align}
where
\begin{align}
    K_1:=&\br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,p\mu+(\textbf{b}\cdot v)\mu+\textbf{d}\cdot\mathscr{A}\m^{\frac{1}{2}}+\m^{\frac{1}{2}}(\textbf{I}-\overline{\textbf{P}})[R]\right]},\\
    K_2:=&\br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,c\left(\abs{v}^2-5T\right)\mu\right]}.
\end{align}
We can check that
\begin{align}
    \m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,p\mu\right]=p\mathcal{L}[f_1]=-p\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{2T^2}\right),
\end{align}
and thus by oddness of $\overline{\mathscr{A}}$ and $\mathscr{B}$, we have
\begin{align}
    \br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,p\mu\right]}=\brx{\bbrv{\nabla_{x}\psi:\mathscr{B},\overline{\mathscr{A}}},p\frac{\nabla_{x}T}{2T^2}}=0.
\end{align}
Also, considering that $\nm{f_1}_{\infty}\lesssimo_{T}$, H\"older's inequality and Young's inequality imply
\begin{align}
    &\abs{\br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,(\textbf{b}\cdot v)\mu+(\textbf{d}\cdot\mathscr{A})\m^{\frac{1}{2}}+\m^{\frac{1}{2}}(\textbf{I}-\overline{\textbf{P}})[R]\right]}}\\
    \lesssim& o_{T}\nm{\psi}_{W^{2,\frac{r}{r-1}}}\Big(\jnm{\textbf{b}}+\jnm{\textbf{d}}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)
    \lesssimo_{T}\jnm{\textbf{b}}^{r}+o_{T}\jnm{\textbf{d}}^{r}+o_{T}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
Hence, we have shown that
\begin{align}
    K_1\lesssim o_{T}\jnm{\textbf{b}}^{r}+o_{T}\jnm{\textbf{d}}^{r}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.
\end{align}
In addition, due to oddness, we have
\begin{align}\label{pp 81}
    K_2=&\br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\bigg[-\frac{\mathscr{A}\cdot\nabla_{x}T}{2T^2}\m^{\frac{1}{2}} +\mu\bigg(\frac{\rho_1}{\rho}+\frac{u_1\cdot v}{T}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\left(\abs{v}^2-5T\right)\mu\bigg]}\\
    =&\br{\nabla_{x}\psi:\mathscr{B},\m^{-\frac{1}{2}} Q^{\ast}\bigg[\mu\bigg(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\left(\abs{v}^2-5T\right)\mu\bigg]}\nonumber\\
    =&\br{\partial_{x_1}\psi_1\mathscr{B}_{11}+\partial_{x_2}\psi_2\mathscr{B}_{22}+\partial_{x_3}\psi_3\mathscr{B}_{33},\m^{-\frac{1}{2}} Q^{\ast}\bigg[\mu\bigg(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\left(\abs{v}^2-5T\right)\mu\bigg]}.\nonumber
\end{align}
Since $\m^{-\frac{1}{2}} Q^{\ast}\left[\mu\left(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\right),c\big(\abs{v}^2-5T\big)\mu\right]$ only depends on $\abs{v}^2$, for any $i=1,2,3$, using the definition of $\mathscr{B}_{ii}$, we have
\begin{align}\label{qq 16}
    &\brv{\mathscr{B}_{ii},\m^{-\frac{1}{2}} Q^{\ast}\bigg[\mu\bigg(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\big(\abs{v}^2-5T\big)\mu\bigg]}\\
    =&\brv{\lc^{-1}\left[\left(v_i^2-\frac{1}{3}\abs{v}^2\right)\m^{\frac{1}{2}}\right],\m^{-\frac{1}{2}} Q^{\ast}\bigg[\mu\bigg(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\big(\abs{v}^2-5T\big)\mu\bigg]}\nonumber\\
    =&\frac{1}{3}\brv{\lc^{-1}\left[\sum_{i=1,2,3}\left(v_i^2-\frac{1}{3}\abs{v}^2\right)\m^{\frac{1}{2}}\right],\m^{-\frac{1}{2}} Q^{\ast}\bigg[\mu\bigg(\frac{\rho_1}{\rho}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),c\big(\abs{v}^2-5T\big)\mu\bigg]}=0,\nonumber
\end{align}
since $\sum_{i=1,2,3}\left(v_i^2-\frac{1}{3}\abs{v}^2\right)=0$. 
Hence, we have $K_2=0$.

In total, using Lemma \ref{remark 01}, we have the $\lc^1$ contribution in \eqref{final 58}
\begin{align}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}}\lesssim& o_{T}\jnm{\textbf{b}}^{r}+o_{T}\jnm{\textbf{e}}^{r}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+o_{T}\varepsilon^{r+1-r\alpha}.
\end{align}

\paragraph{\underline{$S_3$ Contribution}}
We will apply a similar argument as \eqref{oo 30}.
Using $\psi\big|_{\partial\Omega}=0$ and Hardy's inequality, combined with \eqref{ss3-estimate5} and \eqref{ss3-estimate6}, we have
\begin{align}\label{pp 124}
    &\abs{\br{\varepsilon^{-1}\psi\cdot v\m^{\frac{1}{2}},S_3}}\\
    \leq&\varepsilon^{-1}\abs{\br{\left(\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y\right)\cdot v\m^{\frac{1}{2}},\ss_{3b}+\ss_{3c}}}+\varepsilon^{-1}\abs{\br{\left(\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y\right)\cdot \nabla_v(v\m^{\frac{1}{2}}),f^{B}_1}}\nonumber\\
    =&\varepsilon^{-1}\abs{\br{\left(\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y\right)\cdot v\m^{\frac{1}{2}},\mathfrak{n}\big(\ss_{3b}+\ss_{3c}\big)}}+\varepsilon^{-1}\abs{\br{\left(\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y\right)\cdot \nabla_v(v\m^{\frac{1}{2}}),\mnf^{B}_1}}\nonumber\\
    \leq&\knm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y}\nm{\eta\big(\ss_{3b}+\ss_{3c}\big)}_{L^r_xL^1_v}+\knm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\psi(y)\mathrm{d} y}\nm{\etaf^{B}_1}_{L^r_xL^1_v}\nonumber\\
    \leq&\knm{\partial_{\mathfrak{n}}\psi}\nm{\eta\big(\ss_{3b}+\ss_{3c}\big)}_{L^r_xL^1_v}+\knm{\partial_{\mathfrak{n}}\psi}\nm{\etaf^{B}_1}_{L^r_xL^1_v}
    \lesssim o(1) \jnm{\textbf{b}}^{r}+\nm{\eta\big(\ss_{3b}+\ss_{3c}\big)}_{L^r_xL^1_v}^{r}+\nm{\etaf^{B}_1}_{L^r_xL^1_v}^{r}.\nonumber
\end{align}
Similarly, we have
\begin{align}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_3}}\lesssim&\knm{\nabla_{x}\psi}\jnm{\ss_{3b}+\ss_{3c}}+\knm{\nabla_{x}\psi}\jnm{f^{B}_1}\\
    \lesssim& o(1)\jnm{\textbf{b}}^{r}+\nm{\ss_{3b}+\ss_{3c}}_{L^r_xL^1_v}^{r}+\nm{f^{B}_1}_{L^r_xL^1_v}^{r},\nonumber
\end{align}
\begin{align}
    P^{-1}\abs{\bbr{q,\mhS_3}}\lesssim o(1)\jnm{\textbf{b}}^{r}+\nm{\ss_{3b}+\ss_{3c}}_{L^r_xL^1_v}^{r}+\nm{f^{B}_1}_{L^r_xL^1_v}^{r}.
\end{align}
In total, using Lemma \ref{ss3-estimate}, the $S_3$ contribution in \eqref{final 58} is bounded by
\begin{align}
    o(1)\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{r+1-r\alpha}.
\end{align}

\paragraph{\underline{$S_1$ Contribution}}
Notice that
\begin{align}\label{pp 123}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_1}}\lesssim&\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,p\m^{\frac{1}{2}}+\textbf{b}\cdotv\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\right]}}\\
    &+\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}.\nonumber
\end{align}
Using \eqref{qq 15}, we have
\begin{align}
    \text{First Term in \eqref{pp 123}}\lesssim& \tnm{\nabla_{x}\psi}\lnmm{f^{B}_1}\Big(\tnm{p}+\tnm{\textbf{b}}+\tnm{\textbf{d}}+\um{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)\\
    \lesssim&o_{T}\jnm{\textbf{b}}^{r}+o_{T}\tnm{p}^{r}+o_{T}\tnm{\textbf{d}}^{r}+o_{T}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
On the other hand, we know
\begin{align}
    \text{Second Term in \eqref{pp 123}}\lesssim&\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,\big(Z^R-\varepsilon^{-1}\xi\big)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    &+\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,Z^S\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}.\nonumber
\end{align}
Similar to \eqref{pp 124}, using \eqref{qq 06}, we have
\begin{align}
    &\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,\big(Z^R-\varepsilon^{-1}\xi\big)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    =&\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\big(Z^R-\varepsilon^{-1}\xi\big)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\nonumber\\
    \lesssim&\varepsilon\tnm{\nabla_{x}\psi}\lnmm{\etaf^{B}_1}\tnm{\nabla_{x}Z^R-\varepsilon^{-1}\nabla_{x}\xi}\lesssim o_{T}\jnm{\textbf{b}}^{r}+o_{T}\Big(\varepsilon^{r}\nm{Z^R}_{H^1}^{r}+\nm{\xi}_{H^1}^{r}\Big)\nonumber\\
    \lesssim&o_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{r}\xnm{R}^{r}+o_{T}\varepsilon^{\alr}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}
Also,  using \eqref{qq 07}, we have
\begin{align}
    &\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,Z^S\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    \lesssim&\tnm{\nabla_{x}\psi}\lnmm{f^{B}_1}\tnm{Z^S}
    \lesssimo_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+o_{T}\varepsilon^{\alr-\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}
In total, for $2\leqr\leq6$, the $S_1$ contribution in \eqref{final 58}
\begin{align}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_1}}\lesssim&o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+o_{T}\varepsilon^{\alr-\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.
\end{align}

\paragraph{\underline{$S_2$ Contribution}}
We will need more detailed analysis than the estimates in Lemma \ref{ss2-estimate}. Notice that
\begin{align}\label{pp 103''}
    \bbr{\nabla_{x}\psi:\mathscr{B},S_2}=&\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}, \textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    &+\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}, (\textbf{I}-\overline{\textbf{P}})[R]\Big]}+\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[(\textbf{I}-\overline{\textbf{P}})[R], (\textbf{I}-\overline{\textbf{P}})[R]\Big]}.\nonumber
\end{align}
For the first term in \eqref{pp 103''}, by oddness, we have
\begin{align}\label{pp 104''}
    &\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    =&\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}\nonumber\\
    &+\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}\nonumber\\
    &+\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}.\nonumber
\end{align}
Using oddness, similar to \eqref{qq 16}, we know 
\begin{align}
    \varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}=0.
\end{align}
Also, due to $\frac{2r-3}{3r}+\frac{1}{3}+\frac{1}{r}=1$, considering the Sobolev embedding $\pnm{\nabla_{x}\psi}{\frac{3r}{2r-3}}\lesssim\nm{\psi}_{W^{2,\frac{r}{r-1}}}$, and interpolation $\pnm{\textbf{b}}{3}\leq\tnm{\textbf{b}}+\pnm{\textbf{b}}{6}\lesssim\xnm{R}$, we know
\begin{align}
    \abs{\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\lesssim& \varepsilon^{\alpha-1}\nm{\nabla_{x}\psi}_{L^{\frac{3r}{2r-3}}}\pnm{\textbf{b}}{3}\jnm{\textbf{b}}\\
    \lesssim&\varepsilon^{\alpha-1}\nm{\psi}_{W^{2,\frac{r}{r-1}}}\xnm{R} \jnm{\textbf{b}}\lesssim \varepsilon^{\alpha-1}\xnm{R}\jnm{\textbf{b}}^{r}.\nonumber
\end{align}
Similarly, using $\pnm{\textbf{d}}{3}\lesssim\xnm{R}$ due to interpolation, using \eqref{qq 15}, we have
\begin{align}
    \abs{\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\lesssim \varepsilon^{\alpha-1}\pnm{\textbf{d}}{3}\jnm{\textbf{b}}^{r}\lesssim \varepsilon^{\alpha-1}\xnm{R}\jnm{\textbf{b}}^{r},
\end{align}
and
\begin{align}
    \abs{\varepsilon^{\alpha-1}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}}\lesssim&\varepsilon^{\alpha-1}\jnm{\textbf{b}}^{r-1}\pnm{\textbf{d}}{3}\pnm{\textbf{d}}{r}\lesssim \varepsilon^{\alpha-1}\jnm{\textbf{b}}^{r-1}\xnm{R}\pnm{\textbf{d}}{r}\\
    \lesssim&  \varepsilon^{\alpha-1}\xnm{R}\Big(\jnm{\textbf{b}}^{r}+ \jnm{\textbf{d}}^{r}\Big).\nonumber
\end{align}
Hence, we know
\begin{align}
    \Babs{\text{First Term in \eqref{pp 103''}}}\lesssim& \varepsilon^{\alpha-1}\xnm{R}\Big(\jnm{\textbf{b}}^{r}+ \jnm{\textbf{d}}^{r}\Big).
\end{align}
On the other hand, noticing that $\lnmm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}\lesssim\lnmm{R}\lesssim \varepsilon^{-\frac{1}{2}}\xnm{R}$, we obtain
\begin{align}
    \Babs{\text{Second Term in \eqref{pp 103''}}}\lesssim&\varepsilon^{\alpha-1}\tnm{\nabla_{x}\psi}\lnmm{R}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\\
    \lesssim&\varepsilon^{\alpha-\frac{3}{2}}\tnm{\nabla_{x}\psi}\xnm{R}\um{(\textbf{I}-\overline{\textbf{P}})[R]}
    \lesssim\varepsilon^{\alpha-\frac{3}{2}}\pnm{\textbf{b}}{r}^{r-1}\xnm{R}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    \lesssim&\varepsilon^{\alpha-1}\xnm{R}\jnm{\textbf{b}}^{r}+\varepsilon^{\alpha-1-\frac{r}{2}}\xnm{R}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
Similarly,  noticing that $\lnmm{(\textbf{I}-\overline{\textbf{P}})[R]}\lesssim\lnmm{R}\lesssim \varepsilon^{-\frac{1}{2}}\xnm{R}$, we have
\begin{align}
    \Babs{\text{Third Term in \eqref{pp 103''}}}\lesssim \varepsilon^{\alpha-1}\xnm{R}\jnm{\textbf{b}}^{r}+\varepsilon^{\alpha-1-\frac{r}{2}}\xnm{R}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.
\end{align}
In total, using Lemma \ref{remark 01}, the $S_2$ contribution in \eqref{final 58} 
\begin{align}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_2}}&\lesssim  \varepsilon^{\alpha-1}\xnm{R}\Big(\jnm{\textbf{b}}^{r}+ \jnm{\textbf{d}}^{r}+\varepsilon^{-\frac{r}{2}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\Big)\\
    &\lesssim  \varepsilon^{\alpha-1}\xnm{R}\Big(\jnm{\textbf{b}}^{r}+ \jnm{\textbf{e}}^{r}+\varepsilon^{-\frac{r}{2}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+o_{T}\varepsilon^{r+1-r\alpha}\Big).\nonumber
\end{align}

\paragraph{\underline{Other Source Contribution}}
Using \eqref{qq 15}, Lemma \ref{ss4-estimate} and Lemma \ref{ssp-estimate}, we have
\begin{align}
    \abs{\varepsilon^{-1}\br{\psi\cdot v\m^{\frac{1}{2}},\ss_6}}+\abs{\bbr{\nabla_{x}\psi:\mathscr{B},\ss_6}}\lesssim&\varepsilon^{-1}\nm{\psi}_{W^{2,\frac{r}{r-1}}}\tnm{\ss_6}
    \lesssim o_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{2r-r\alpha},
\end{align}
and
\begin{align}
    P^{-1}\abs{\bbr{q,\mhS_4}}+\abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_4}}&\lesssim \Big(\nm{q}_{L^{\frac{r}{r-1}}}+\nm{\psi}_{W^{2,\frac{r}{r-1}}}\Big)\tnm{S_4}
    \lesssim  o_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{2r-r\alpha}.
\end{align}
Also, based on Lemma \ref{ss0-estimate} and Lemma \ref{ss5-estimate}, we know that
\begin{align}
    \abs{\bbr{\nabla_{x}\psi:\mathscr{B},S_0+S_5}}\lesssim&\nm{\psi}_{W^{2,\frac{r}{r-1}}}\Big(\tnm{S_0}+\tnm{S_5}\Big)
    \lesssimo_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{r}\xnm{R}^{r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.
\end{align}
In total, the $S_0$, $S_4$, $S_5$ and $\ss_6$ contributions in \eqref{final 58} are bounded by
\begin{align}
    o_{T}\jnm{\textbf{b}}^{r}+o_{T}\varepsilon^{r}\xnm{R}^{r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.
\end{align}
Collecting above estimates, we deduce \eqref{final 11}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:b-bound}]
Inserting \eqref{final 11} into \eqref{final 12}, for $\alpha\geq1$,
we obtain the desired result \eqref{oo 34}.
\end{proof}


\section{Dual Stokes-Poisson Problem, Boundary Layer, and $c$ Estimates}


In this section, we plan to prove Proposition \ref{prop:c-bound}. To establish \eqref{oo 35=}, thanks to energy estimates, it suffices to establish for $2\leqr\leq6$
    \begin{align}\label{oo 35}
    \\
    \jnm{c}\lesssim&\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+o_{T}\varepsilon^{1-\alpha}\xnm{R}+\varepsilon^{\alpha-1}\xnm{R}^{2}+\varepsilon^{2\alpha-2}\xnm{R}^{3}+o_{T}\varepsilon^{1-\alpha}.\nonumber
    \end{align}


\subsection{Combined Weak Formulation}


Adding $\varepsilon^{-1}\times$\eqref{pp 16} and \eqref{pp 10} yields
\begin{align}\label{pp 112}
    &5\varepsilon^{-1}P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}-\varepsilon^{-1}\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\varepsilon^{-1}\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}\\
    &+\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}-\bbrx{\nabla_{x}\cdot(\kappa\nabla_{x}\phi),c}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}\nonumber\\
    &+\bbr{\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B}+\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\nonumber\\
    =&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
    &-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
    &+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}+\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}-\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
    &+\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
Using oddness and $\nabla_{x}\cdot\psi=0$, similarly to \eqref{qq 16}, we can further simplify the $\textbf{P}[R]$ part in $\varepsilon^{-1}\bbr{\nabla_{x}\psi:\mathscr{B},\lc^1[R]}$:
\begin{align}\label{pp 113}
    &\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[\textbf{P}[R]\Big]}\\
    =&-\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},2\m^{-\frac{1}{2}} Q^{\ast}\bigg[-\m^{\frac{1}{2}}\mathscr{A}\cdot\frac{\nabla_{x}T}{2T^2}+\mu\bigg(\frac{\rho_1}{\rho}+\frac{u_1\cdot v}{T}+\frac{T_1(\abs{v}^2-3T)}{2T^2}\bigg),p\mu+(\textbf{b}\cdot v)\mu+c\left(\abs{v}^2-5T\right)\mu\bigg]}\nonumber\\
    =&-\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},2\m^{-\frac{1}{2}} Q^{\ast}\left[-\m^{\frac{1}{2}}\mathscr{A}\cdot\frac{\nabla_{x}T}{2T^2}+\mu\frac{u\cdot v}{T},(\textbf{b}\cdot v)\mu\right]}.\nonumber
\end{align}
We extract all $\varepsilon^{-1}$-order terms involving $\textbf{b}$ in \eqref{pp 112}:
\begin{align}
    5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}-\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\bbr{\nabla_{x}\psi\cdot \mathfrak{M},\textbf{b}}
\end{align}
for $\mathfrak{M}$ defined in \eqref{pp 114} which identifies the coefficients in front
of $\textbf{b}$ by combining the first term on the second line of \eqref{pp 112} and the third line of \eqref{pp 113}
\begin{align}
    \br{\nabla_{x}\psi:\mathscr{B},(\textbf{b}\cdot v)\left(\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)}-\br{\nabla_{x}\psi:\mathscr{B},2\m^{-\frac{1}{2}} Q^{\ast}\left[-\m^{\frac{1}{2}}\mathscr{A}\cdot\frac{\nabla_{x}T}{2T^2}+\mu\frac{u\cdot v}{T},(\textbf{b}\cdot v)\mu\right]}.
\end{align}
Assume $(\psi,q,\phi)\in\mathbb{R}^3\times\mathbb{R}\times\mathbb{R}$ (where $q$ has zero average) is the unique strong solution to the perturbed Stokes-Poisson problem \eqref{pp 20} and \eqref{pp 23}.
Since the perturbation terms $\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big)$ and $\bbrv{\nabla_{x}\psi:\mathscr{B},\mathfrak{M}}$ are small considering the smallness of $\nabla_{x} T$ and $u_1$, we have the standard estimate \cite{Cattabriga1961}, Sobolev embedding for $2\leqr\leq 6$ and trace estimate
\begin{align}\label{qq 17}
    \nm{\psi}_{W^{4,\frac{r}{r-1}}}+\nm{q}_{W^{3,\frac{r}{r-1}}}+\nm{\psi}_{H^3}+\nm{q}_{H^2}+\abs{\psi}_{W^{3,\frac{2r}{2r-3}}_{\partial\Omega}}+\abs{q}_{W^{2,\frac{2r}{2r-3}}_{\partial\Omega}}+\abs{\psi}_{H^1_{\partial\Omega}}\lesssim o_{T} \nm{\phi}_{W^{2,\frac{r}{r-1}}},
\end{align}
\begin{align}\label{qq 18}
    \nm{\phi}_{W^{2,\frac{r}{r-1}}}+\nm{\phi}_{H^1}+\knms{\phi}{\partial\Omega}\lesssim\knm{c\abs{c}^{r-2}}\lesssim\jnm{c}^{r-1}.
\end{align}
Multiplying $\textbf{b}$ on both sides of \eqref{pp 20} and integrating by parts in $\br{\nabla_{x} q,\textbf{b}}$, we have
\begin{align}
    -\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\bbr{\big(\nabla_{x}\psi:\mathscr{B}\big)\mathfrak{M},\textbf{b}}\\
    -\bbrx{q,\nabla_{x}\cdot\textbf{b}}+\int_{\partial\Omega}q(\textbf{b}\cdot n)&=-5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}},\nonumber
\end{align}
which, by combining \eqref{conservation law 1} and \eqref{final 53}, implies
\begin{align}\label{pp 21}
    -\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\bbr{\big(\nabla_{x}\psi:\mathscr{B}\big) \mathfrak{M},\textbf{b}}\\
    -P^{-1}\br{q,\mhS_3+\mhS_4}&=-5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}.\nonumber
\end{align}
Multiplying $c$ on both sides of \eqref{pp 23}, we have
\begin{align}\label{pp 24}
    -\bbrx{\nabla_{x}\cdot\big(\kappa\nabla_{x}\phi\big),c}=\jnm{c}^{r}.
\end{align}
Inserting \eqref{pp 21}\eqref{pp 24} into \eqref{pp 112} and using \eqref{pp 114} to replace $5P\bbrx{\phi,\nabla_{x} T\cdot\textbf{b}}-\bbrx{\lambda\Delta_x\psi,\textbf{b}}-\bbr{\overline\b\cdot\big(\nabla_{x}\psi:\nabla_{x}\mathscr{B}\big)+\big(\nabla_{x}\cdot\overline\b\big)\big(\nabla_{x}\psi:\mathscr{B}\big),\textbf{b}}+\bbr{\big(\nabla_{x}\psi:\mathscr{B}\big)\mathfrak{M},\textbf{b}}$ and  $-\bbrx{\nabla_{x}\cdot(\kappa\nabla_{x}\phi),c}$, we obtain
\begin{align}\label{pp 25}
    &\varepsilon^{-1}P^{-1}\br{q,\mhS_3+\mhS_4}\\
    &+\jnm{c}^{r}+\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}+\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}\nonumber\\
    =&\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}-\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
    &-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\nonumber\\
    &+\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}+\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    &-\br{\nabla_{x}\phi\cdot\mathscr{A},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right)(\textbf{I}-\overline{\textbf{P}})[R]}-\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}\nonumber\\
    &+\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
Note that $\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[\textbf{P}[R]\Big]}$ has been included in the definition of $\mathfrak{M}$ in \eqref{pp 114}, so we are left with $\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}$ in \eqref{pp 25}.

\begin{lemma}\label{lem:final 4}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{pp 28'}
\jnm{c}^{r}\leq&-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\\
&+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}\nonumber\\
&-\varepsilon^{-1}P^{-1}\br{q,\mhS_3+\mhS_4}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}-\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}\nonumber\\
&+\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\barS}.\nonumber
\end{align}
\end{lemma}

\begin{proof}
We need to estimate each term in \eqref{pp 25} except the source terms. 
Using \eqref{qq 18}, we have
\begin{align}
    \abs{\brx{\nabla_{x}\phi,\frac{\nabla_{x} T}{2T^2}\big(\kappap+\sigma c\big)}}\lesssim o_{T}\tnm{\nabla_{x}\phi}\Big(\tnm{p}+\tnm{c}\Big)\lesssim o_{T}\jnm{c}^{r}+o_{T}\tnm{p}^{r}\lesssimo_{T}\xnm{R}^{r}.
\end{align}
Also, using \eqref{qq 18} and Lemma \ref{h-estimate}, we have
\begin{align}
&\abs{\bbrb{\nabla_{x}\phi\cdot\mathscr{A},h}{\gamma_-}}+\abs{\bbrb{\nabla_{x}\phi\cdot\mathscr{A},(1-\mathcal{P}_{\gamma})[R]}{\gamma_-}}\\
\lesssim& o(1)\knms{\nabla_{x}\phi}{\partial\Omega}^{\frac{r}{r-1}}+\jnms{\int_{v\cdot n>0}(1-\mathcal{P}_{\gamma})[R]\mathscr{A}(v\cdot n)}{\partial\Omega}^{r}+\jnms{\int_{v\cdot n<0}h\mathscr{A}(v\cdot n)}{\partial\Omega}^{r}\nonumber\\
\lesssim& o(1)\jnm{c}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+\jnms{h}{\gamma_-}^{r}\lesssim o(1)\jnm{c}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\varepsilon^{2r-r\alpha}.\nonumber
\end{align}
In addition, \eqref{qq 18}, H\"older's inequality and Young's inequality yield
\begin{align}
\abs{\br{v\cdot\nabla_{x}\Big(\nabla_{x}\phi\cdot\mathscr{A}\Big),(\textbf{I}-\overline{\textbf{P}})[R]}}
\lesssim&o(1)\knm{\nabla_{x}^2\phi}^{\frac{r}{r-1}}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\\
\lesssim&o(1)\jnm{c}^{r}+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r},\nonumber
\end{align}
and
\begin{align}
    \abs{\br{\big(\nabla_{x}\phi\cdot\mathscr{A}\big)\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{2T^2}\right),(\textbf{I}-\overline{\textbf{P}})[R]}}
    \lesssim&o_{T}\knm{\nabla_{x}\phi}^{\frac{r}{r-1}}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r},\nonumber
\end{align}
Further, \eqref{qq 17}\eqref{qq 18}, H\"older's inequality and Young's inequality (and further integration by parts for $\nabla_{x}\xi$ in $\textbf{d}$ using $\xi\big|_{\partial\Omega}=0$) yield
\begin{align}
&\abs{\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}}\\
\leq&\abs{\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\nabla_{x}\xi\cdot\mathscr{A}}}+\abs{\varepsilon^{-1}\br{v\cdot\nabla_{x}\Big(\nabla_{x}\psi:\mathscr{B}\Big),\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}}\nonumber\\
\lesssim&\varepsilon^{-1}\nm{\psi}_{H^3}\Big(\tnm{\xi}+\tnm{\textbf{e}}+\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)
\lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{\xi}^{r}+o_{T}\varepsilon^{-r}\tnm{\textbf{e}}^{r}+o_{T}\varepsilon^{-r}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r},\nonumber
\end{align}
and
\begin{align}
    &\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}}\\
    \leq&\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\nabla_{x}\xi\cdot\mathscr{A}\Big)}}+\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\Big(\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big)}}\nonumber\\
    \lesssim&o_{T}\varepsilon^{-1}\nm{\psi}_{H^2}\Big(\tnm{\xi}+\tnm{\textbf{e}}+\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)
    \lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{\xi}^{r}+o_{T}\varepsilon^{-r}\tnm{\textbf{e}}^{r}+o_{T}\varepsilon^{-r}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
Finally, from \eqref{qq 17} and Lemma \ref{h-estimate}, we know
\begin{align}
\abs{\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},h}{\gamma_-}}
\lesssim& \tnms{\nabla_{x}\psi}{\partial\Omega}^{\frac{r}{r-1}}+\varepsilon^{-r}\tnms{h}{\gamma_-}^{r}
\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
In total, from \eqref{pp 25} and Proposition \ref{thm:energy} and Lemma \ref{lem:final 11}, we obtain the desired result.
\end{proof}


\subsection{Estimates of Source Terms}


In this subsection, we consider the source term contribution in \eqref{pp 28'}:
\begin{align}\label{pp 49}
    &-\varepsilon^{-1}P^{-1}\br{q,\mhS_3+\mhS_4}+\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}+\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}\\
    &+\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}+\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\overline{\ss}}.\nonumber
\end{align}

\begin{lemma}\label{lem:final 3}
Under the assumption \eqref{assumption:boundary}, for $\alpha\geq1$, we have
\begin{align}\label{pp 99''}
    &\abs{\varepsilon^{-1}P^{-1}\br{q,\mhS_3+\mhS_4}}+\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}}+\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}}\\
    &+\abs{\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3+\ss_6}}+\abs{\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3+S_4+\ss_6}}+\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\overline{\ss}}}\nonumber\\
    \lesssim
    &o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-\alr}.\nonumber
\end{align}
\end{lemma}

\begin{proof}
\ 
\paragraph{\underline{$\lc^1[R]$ Contribution}}
Lemma \ref{ssl-estimate} and \eqref{qq 17} imply
\begin{align}\label{pp 53}
    \abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},\lc^1[R]}}\lesssim& o_{T} \nm{\phi}_{H^{1}}\um{R}
    \lesssim o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.
\end{align}
Similarly, with the help of integration by parts for $\nabla_{x}\xi$ with $\xi\big|_{\partial\Omega}=0$, we have
\begin{align}
    &\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[(\textbf{I}-\textbf{P})[R]\Big]}}\\
    \lesssim&\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[\nabla_{x}\xi\cdot\mathscr{A}\Big]}}+\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big]}}\nonumber\\
    =&\abs{\varepsilon^{-1}\br{\nabla_{x}^2\psi:\mathscr{B},\lc^1\Big[\xi\cdot\mathscr{A}\Big]}}+\abs{\varepsilon^{-1}\br{\nabla_{x}\psi:\mathscr{B},\lc^1\Big[\textbf{e}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\Big]}}\nonumber\\
    \lesssim& \varepsilon^{-1}\nm{\psi}_{H^{2}}\Big(\tnm{\xi}+\tnm{\textbf{e}}+\um{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)\nonumber\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{\xi}^{r}+o_{T}\varepsilon^{-r}\tnm{\textbf{e}}^{r}+o_{T}\varepsilon^{-r}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
In summary, the $\lc^1$ terms in \eqref{pp 49} can be bounded by
\begin{align}
    o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{\textbf{e}}^{r}+o_{T}\varepsilon^{-r}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+o_{T}\xnm{R}^{r}.
\end{align}

\paragraph{\underline{$S_3$ Contribution}}
We will apply a similar argument as \eqref{oo 30}.
Using $\psi\big|_{\partial\Omega}=0$ and \eqref{qq 17}, with the help of the Taylor expansion with integral remainder and Hardy's inequality, we obtain
\begin{align}\label{pp 101}
    &\abs{\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},S_3}}\\
    =&\varepsilon^{-2}\abs{\br{\left(\psi(0)+\mathfrak{n}\partial_{\mathfrak{n}}\psi(0)+\int_0^{\mathfrak{n}}(\mathfrak{n}-y)\partial_{\mathfrak{n}\mn}\psi(y)\mathrm{d} y\right)\cdot v\m^{\frac{1}{2}},S_3}}\nonumber\\
    =&\varepsilon^{-2}\abs{\br{\left(\mathfrak{n}\partial_{\mathfrak{n}}\psi(0)+\int_0^{\mathfrak{n}}(\mathfrak{n}-y)\partial_{\mathfrak{n}\mn}\psi(y)\mathrm{d} y\right)\cdot v\m^{\frac{1}{2}},S_3}}\nonumber\\
    =&\varepsilon^{-2}\abs{\br{\partial_{\mathfrak{n}}\psi(0)\cdot v\m^{\frac{1}{2}},\mathfrak{n}\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}}+\varepsilon^{-2}\abs{\br{\left(\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\frac{\mathfrak{n}-y}{\mathfrak{n}}\partial_{\mathfrak{n}\mn}\psi(y)\mathrm{d} y\right)\cdot \nabla_v(v\m^{\frac{1}{2}}),\mathfrak{n}^2\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}}\nonumber\\
    =&\varepsilon^{-1}\abs{\br{\partial_{\mathfrak{n}}\psi(0)\cdot v\m^{\frac{1}{2}},\eta\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}}+\abs{\br{\left(\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\frac{\mathfrak{n}-y}{\mathfrak{n}}\partial_{\mathfrak{n}\mn}\psi(y)\mathrm{d} y\right)\cdot \nabla_v(v\m^{\frac{1}{2}}),\eta^2\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}}\nonumber\\
    \lesssim&\varepsilon^{-1}\nm{\partial_{\mathfrak{n}}\psi(0)\cdot v\m^{\frac{1}{2}}}_{L^2_xL^{\infty}_v}\nm{\eta\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\tnm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}\mn}\psi(y)\mathrm{d} y}\nm{\eta^2\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_xL^1_v}\nonumber\\
    \lesssim&\varepsilon^{-1}\nm{\partial_{\mathfrak{n}}\psi}_{W^{3,\frac{r}{r-1}}}\nm{\eta\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\nm{\partial_{\mathfrak{n}\mn}\psi}_{W^{2,\frac{r}{r-1}}}\nm{\eta^2\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_xL^1_v}\nonumber\\
    \lesssim& o_{T} \jnm{c}^{r}+\varepsilon^{-r}\nm{\eta\big(\ss_{3b}+\ss_{3c}\big)}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}^{r}+\varepsilon^{-r}\nm{\etaf^{B}_1}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}^{r}+\nm{\eta^2\big(\ss_{3b}+\ss_{3c}\big)}_{L^2_xL^1_v}^{r}+\nm{\eta^2f^{B}_1}_{L^2_xL^1_v}^{r}.\nonumber
\end{align}
Following a similar argument, the other $S_3$ contribution can be estimated
\begin{align}
    &\abs{\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}+\nabla_{x}\phi\cdot\mathscr{A},S_3}}\\
    =&\varepsilon^{-1}\abs{\br{\left(\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\phi\right)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_3}}+\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},S_3}}\nonumber\\
    \lesssim&\tnm{\nabla_{x}\phi}\nm{\eta\left(\ss_{3b}+\ss_{3c}+f^{B}_1\right)}_{L^2_xL^1_v}\lesssim o_{T} \jnm{c}^{r}+\nm{\eta\left(\ss_{3b}+\ss_{3c}\right)}_{L^2_xL^1_v}^{r}+\nm{\etaf^{B}_1}_{L^2_xL^1_v}^{r},\nonumber
\end{align}
and
\begin{align}
    \abs{\bbr{\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},S_3}}=&\varepsilon^{-1}\abs{\bbr{\left(\nabla_{x}\psi(0)+\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}\nabla_{x}\psi\right):\mathscr{B},S_3}}\label{final 54}\\
    \lesssim&\varepsilon^{-1}\nm{\partial_{\mathfrak{n}}\psi(0):\mathscr{B}}_{L^2_xL^{\infty}_v}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\tnm{\partial_{\mathfrak{n}}\nabla_{x}\psi}\nm{\eta\left(\ss_{3b}+\ss_{3c}+f^{B}_1\right)}_{L^2_xL^1_v}\nonumber\\
    \lesssim&\varepsilon^{-1}\nm{\partial_{\mathfrak{n}}\psi}_{W^{3,\frac{r}{r-1}}}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\nm{\partial_{\mathfrak{n}}\nabla_{x}\psi}_{W^{2,\frac{r}{r-1}}}\nm{\eta\left(\ss_{3b}+\ss_{3c}+f^{B}_1\right)}_{L^2_xL^1_v}\nonumber\\
    \lesssim& o_{T} \jnm{c}^{r}+\varepsilon^{-r}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}^{r}+\nm{\eta\left(\ss_{3b}+\ss_{3c}\right)}_{L^2_xL^1_v}^{r}+\nm{\etaf^{B}_1}_{L^2_xL^1_v}^{r}.\nonumber
\end{align}
Considering the Sobolev embedding $\lnm{q}\lesssim\nm{q}_{W^{3,\frac{r}{r-1}}}\lesssimo_{T}\jnm{c}^{r-1}$, H\"older's inequality, Young's inequality and \eqref{qq 17} and \eqref{qq 18} yield
\begin{align}
    \abs{\varepsilon^{-1}P^{-1}\br{q,\mhS_3}}\lesssim&\varepsilon^{-1}\lnm{q}\nm{\mhS_3}_{L^1}
    \lesssim \lnm{q}^{\frac{r}{r-1}}+\varepsilon^{-r}\nm{S_3}_{L^1}^{r}\\
    \lesssim& o_{T}\jnm{c}^{r}+\varepsilon^{-r}\nm{S_3}_{L^1}^{r}.\nonumber
\end{align}
In summary, using Lemma \ref{ss3-estimate}, the $S_3$ terms in \eqref{pp 49} can be bounded by
\begin{align}
    o_{T} \jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$S_1$ Contribution}}
Direct estimate yields
\begin{align}\label{pp 120.}
    &\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},S_1}}
    \lesssim \tnm{\nabla_{x}\phi}^{\frac{r}{r-1}}\lnmm{f^{B}_1}\um{R}^{r}
    \lesssim o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.
\end{align}
Also, similar to \eqref{pp 123}, we have
\begin{align}\label{pp 123=}
    \abs{\bbr{\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},S_1}}\lesssim&\abs{\br{\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,p\m^{\frac{1}{2}}+\textbf{b}\cdotv\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]\right]}}\\
    &+\abs{\br{\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}.\nonumber
\end{align}
Considering the Sobolev embedding $\nm{\psi}_{W^{1,\infty}}\lesssim\nm{\psi}_{W^{4,\frac{r}{r-1}}}\lesssim\jnm{c}^{r-1}$, we have
\begin{align}
    \text{First Term in \eqref{pp 123=}}\lesssim& \varepsilon^{-1}\lnm{\nabla_{x}\psi}\tnm{f^{B}_1}\Big(\tnm{p}+\tnm{\textbf{b}}+\tnm{\textbf{d}}+\um{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\tnm{p}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\tnm{\textbf{b}}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\tnm{\textbf{d}}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
On the other hand, we know
\begin{align}
    \text{Second Term in \eqref{pp 123=}}\lesssim&\varepsilon^{-1}\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,\big(Z^R-\varepsilon^{-1}\xi\big)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    &+\varepsilon^{-1}\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,Z^S\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}.\nonumber
\end{align}
Similar to \eqref{pp 124}, using \eqref{qq 06}, we have
\begin{align}
    &\varepsilon^{-1}\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,\big(Z^R-\varepsilon^{-1}\xi\big)\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    \lesssim&\lnm{\nabla_{x}\psi}\tnm{\etaf^{B}_1}\tnm{\nabla_{x}Z^R-\varepsilon^{-1}\nabla_{x}\xi}\lesssim o_{T}\jnm{c}^{r}+o_{T}\Big(\varepsilon^{\frac{r}{2}}\nm{Z^R}_{H^1}^{r}+\varepsilon^{-\frac{r}{2}}\nm{\xi}_{H^1}^{r}\Big)\nonumber\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\tnm{p}^{r}+\varepsilon^{-\frac{r}{2}}\tnm{\textbf{b}}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\tnm{c}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
Also, using \eqref{qq 07}, we have
\begin{align}
    &\varepsilon^{-1}\abs{\br{\nabla_{x}\psi:\mathscr{B},\Gamma\left[f^{B}_1,Z^S\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\right]}}\\
    \lesssim&\varepsilon^{-1}\lnm{\nabla_{x}\psi}\tnm{f^{B}_1}\tnm{Z^S}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\tnm{Z^S}^{r}\nonumber\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+o_{T}\varepsilon^{-\frac{r}{2}}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}
In total, we have the $S_1$ terms in \eqref{pp 49} is bounded by
\begin{align}
    &o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$S_2$ Contribution}}
Notice that 
\begin{align}\label{pp 103'}
    \varepsilon^{-1}\bbr{\nabla_{x}\psi:\mathscr{B},S_2}=&\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    &+\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},(\textbf{I}-\overline{\textbf{P}})[R]\Big]}+\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[(\textbf{I}-\overline{\textbf{P}})[R],(\textbf{I}-\overline{\textbf{P}})[R]\Big]}.\nonumber
\end{align}
For the first term in \eqref{pp 103'}, by oddness, we have
\begin{align}\label{pp 104'}
    &\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    =&\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}\nonumber\\
    &+\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}.\nonumber
\end{align}
For using oddness, similar to \eqref{qq 16}, we know
\begin{align}\label{final 05}
    \varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}=0.
\end{align}
Also, using $\tnm{\textbf{b}}\lesssim\varepsilon^{\frac{1}{2}}\xnm{R}$, we know
\begin{align}
    \abs{\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\lesssim& o_{T}\varepsilon^{\alpha-2}\lnm{\nabla_{x}\psi}\tnm{\textbf{b}}^2\\\lesssim&o_{T}\varepsilon^{\alpha-1}\jnm{c}^{{r-1}}\xnm{R}^2
    \lesssimo_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.\nonumber
\end{align}
Similarly, using $\tnm{\textbf{d}}\lesssim\varepsilon^{\frac{1}{2}}\xnm{R}$, we have
\begin{align}
    \abs{\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r},
\end{align}
and
\begin{align}
    \abs{\varepsilon^{\alpha-2}\br{\nabla_{x}\psi:\mathscr{B},\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
Hence, in total, we know
\begin{align}
    \Babs{\text{First Term in \eqref{pp 103'}}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
On the other hand, using $\um{(\textbf{I}-\overline{\textbf{P}})[R]}\lesssim\varepsilon\xnm{R}$ we have
\begin{align}
    \Babs{\text{Second Term in \eqref{pp 103'}}}\lesssim&\varepsilon^{\alpha-2}\lnm{\nabla_{x}\psi}\um{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\\
    \lesssim& o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.\nonumber
\end{align}
In a similar fashion, we obtain
\begin{align}
    \Babs{\text{Third Term in \eqref{pp 103'}}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
In total, \eqref{pp 103'} can be controlled
\begin{align}
\abs{\varepsilon^{-1}\bbr{\nabla_{x}\psi:\mathscr{B},S_2}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
Similarly, we know
\begin{align}\label{pp 103=}
    \bbr{\nabla_{x}\phi\cdot\mathscr{A},S_2}=&\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    &+\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},(\textbf{I}-\overline{\textbf{P}})[R]\Big]}+\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[(\textbf{I}-\overline{\textbf{P}})[R],(\textbf{I}-\overline{\textbf{P}})[R]\Big]}.\nonumber
\end{align}
For the first term in \eqref{pp 103=}, by oddness and orthogonality of $\mathscr{A}$, we have
\begin{align}\label{pp 104=}
    &\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    =&\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},\textbf{d}\cdot \mathscr{A}\Big]}.\nonumber
\end{align}
Using $\lnm{c}\lesssim\lnmm{R}\lesssim\varepsilon^{-\frac{1}{2}}\xnm{R}$ and $\tnm{\textbf{b}}\lesssim\varepsilon^{\frac{1}{2}}\xnm{R}$, we know
\begin{align}
    \abs{\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\lesssim& \varepsilon^{\alpha-1}\tnm{\nabla_{x}\phi}\lnm{c}\tnm{\textbf{b}}\\
    \lesssim& \varepsilon^{\alpha-1}\jnm{c}^{{r-1}}\xnm{R}^2\lesssimo_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.\nonumber
\end{align}
Similarly, we have
\begin{align}
    \abs{\varepsilon^{\alpha-1}\br{\nabla_{x}\phi\cdot\mathscr{A},\Gamma\Big[c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},\textbf{d}\cdot \mathscr{A}\Big]}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
Hence, in total, we know
\begin{align}
    \Babs{\text{First Term in \eqref{pp 103=}}}\lesssim o_{T}\jnm{c}^{r}+\varepsilon^{r\alpha-r} \xnm{R}^{2r}.
\end{align}
Similarly, we have
\begin{align}
    \Babs{\text{Second Term in \eqref{pp 103=}}}\lesssim&\varepsilon^{\alpha-1}\tnm{\nabla_{x}\phi}\lnmm{\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\\
    \lesssim&o_{T} \jnm{c}^{r}+\varepsilon^{r\alpha-r}\xnm{R}^{2r},\nonumber
\end{align}
and
\begin{align}
    \Babs{\text{Third Term in \eqref{pp 103=}}}\lesssim o(1) \jnm{c}^{r}+\varepsilon^{r\alpha-r}\xnm{R}^{2r}.
\end{align}
In total, we know that \eqref{pp 103=} can be controlled 
\begin{align}
\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A},S_2}}\lesssimo_{T} \jnm{c}^{r}+\xnm{R}^{2r}.
\end{align}
In summary, the $S_2$ contribution in \eqref{pp 49} can be bounded by
\begin{align}
o_{T} \jnm{c}^{r}+\varepsilon^{r\alpha-r}\xnm{R}^{2r}.
\end{align}

\paragraph{\underline{$S_4$ and $\ss_6$ Contribution}}
Based on Lemma \ref{ss4-estimate}, we have
\begin{align}
    \abs{\br{\varepsilon^{-1}\phi\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},S_4}}\lesssim& \varepsilon^{-1}\nm{\phi}_{W^{1,\frac{r}{r-1}}}\jnm{S_4}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Similarly, we have
\begin{align}
    \abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},S_4}}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha},
\end{align}
and
\begin{align}
    \abs{\varepsilon^{-1}P^{-1}\br{q,\mhS_4}}\lesssim& o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Based on Lemma \ref{ssp-estimate}, we have
\begin{align}
    \abs{\br{\varepsilon^{-2}\psi\cdot v\m^{\frac{1}{2}},\ss_6}}\lesssim \varepsilon^{-2}\tnm{\psi}\tnm{\ss_6}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Similarly, we have
\begin{align}
    \abs{\br{\varepsilon^{-1}\phi\big(\abs{v}^2-5T\big)\m^{\frac{1}{2}},\ss_6}+\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\ss_6}}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
In summary, the $S_4$ and $\ss_6$ terms \eqref{pp 49} can be bounded by
\begin{align}
o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$S_0$ and $S_5$ Contribution}}
Based on Lemma \ref{ss0-estimate}, we have
\begin{align}\label{pp 120}
    \\\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},S_0}}
    \lesssim& o_{T}\tnm{\nabla_{x}\phi}^{\frac{r}{r-1}}+o_{T}\tnm{\nabla_{x}\psi}^{\frac{r}{r-1}}+o_{T}\xnm{R}^{r}
    \lesssim o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.\nonumber
\end{align}
Also, 
\begin{align}
    \abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},S_5}}\lesssim&\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\varepsilon^{2-\alpha}\Gamma[f_i,f_j]+\varepsilon^{1-\alpha}\Gamma[f^{B}_1,f_i]}}.
\end{align}
Clearly, we have
\begin{align}
    \\\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\varepsilon^{2-\alpha}\Gamma[f_i,f_j]}}
    \lesssim& o_{T}\tnm{\nabla_{x}\phi}^{\frac{r}{r-1}}+o_{T}\tnm{\nabla_{x}\psi}^{\frac{r}{r-1}}+o_{T}\varepsilon^{r-r\alpha}
    \lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
Then following similar techniques as \eqref{pp 101} and \eqref{final 54}, we obtain
\begin{align}
    &\abs{\bbr{\nabla_{x}\phi\cdot\mathscr{A}+\varepsilon^{-1}\nabla_{x}\psi:\mathscr{B},\varepsilon^{1-\alpha}\Gamma[f^{B}_1,f_i]}}\\
    \lesssim&o_{T}\tnm{\nabla_{x}\phi}^{\frac{r}{r-1}}+o_{T}\tnm{\nabla_{x}\psi}^{\frac{r}{r-1}}+\varepsilon^{-r}\nm{f^{B}_1}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^2_v}^{r}+\tnm{\etaf^{B}_1}^{r}\lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
In summary, the $S_0$ and $S_5$ terms \eqref{pp 49} can be bounded by
\begin{align}
o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
We therefore deduce Lemma \ref{lem:final 3} by collecting the above estimates.
\end{proof}

Combining Lemma \ref{lem:final 3} and Lemma \ref{lem:final 4}, and applying Lemma \ref{h-estimate}, we conclude  
\begin{lemma}
Under the assumption \eqref{assumption:boundary}, we have
\begin{align}\label{pp 28}
\jnm{c}^{r}\lesssim&-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\\
&+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-\alr}.\nonumber
\end{align}
\end{lemma}

Now the remaining difficulty is to control the boundary term $\displaystyle\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}$.


\subsection{Auxiliary Function $\mathcal{G}$}


In order to handle the troublesome boundary term $\displaystyle\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}$, we plan to design an auxiliary function $\mathcal{G}(x,v)$ satisfying
\begin{align}\label{pp 46}
    \mathcal{G}\big|_{\gamma_+}=-\Big(\nabla_{x}\psi:\mathscr{B}\Big)\Big|_{\gamma_+}+\mathcal{G}_h\Big|_{\gamma_+},
\end{align}
and the almost zero mass-flux condition
\begin{align}\label{pp 116}
    \int_{\partial\Omega\times\mathbb{R}^3}\mathcal{G}\m^{\frac{1}{2}}(v\cdot n)=\mathcal{G}_m.
\end{align}
for some small $\mathcal{G}_h$ and $\mathcal{G}_m$ which will be specified later (see \eqref{final 61'} and \eqref{final 62'}). Taking the test function $\mathfrak{f}=\mathcal{G}$ in \eqref{weak formulation}, and considering that $\mathcal{L}$ is self-adjoint, we obtain
\begin{align}\label{pp 26}
    &\int_{\gamma}(\mathcal{G} R)(v\cdot n)+\br{-v\cdot\nabla_{x}\mathcal{G}+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\mathcal{G}+\varepsilon^{-1}\mathcal{L}[\mathcal{G}],R}=\bbr{\mathcal{G},S}.
\end{align}
Noting that \eqref{pp 116} yields
\begin{align}
    \int_{\partial\Omega\times\mathbb{R}^3}\Big(\mathcal{G}\mathcal{P}_{\gamma}[R]\Big)(v\cdot n)=\mathcal{G}_m\int_{\partial\Omega}\m^{-\frac{1}{2}}\mathcal{P}_{\gamma}[R],
\end{align}
with the help of \eqref{qq 109}, we know that \eqref{pp 26} becomes
\begin{align}\label{pp 27}
    -\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\bbrb{\mathcal{G}, h}{\gamma_-}+\bbrb{\mathcal{G}_h,(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\mathcal{G}_m\int_{\partial\Omega}\m^{-\frac{1}{2}}\mathcal{P}_{\gamma}[R]\\
    +\br{-v\cdot\nabla_{x}\mathcal{G}
    +\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\mathcal{G}+\varepsilon^{-1}\mathcal{L}[\mathcal{G}],R}&=\bbr{\mathcal{G},S}.\nonumber
\end{align}
Adding $\varepsilon^{-1}\times$\eqref{pp 27} and \eqref{pp 28} to eliminate $\displaystyle-\varepsilon^{-1}\bbrb{\nabla_{x}\psi:\mathscr{B},(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}$, we obtain
\begin{align}\label{pp 29}
    \jnm{c}^{r}\lesssim&-\varepsilon^{-1}\br{-v\cdot\nabla_{x}\mathcal{G}+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\mathcal{G}+\varepsilon^{-1}\mathcal{L}[\mathcal{G}],R}\\
    &+\varepsilon^{-1}\bbr{\mathcal{G},S}+\varepsilon^{-1}\bbrb{\mathcal{G},h}{\gamma_-}-\varepsilon^{-1}\bbrb{\mathcal{G}_h,(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\varepsilon^{-1}\mathcal{G}_m\int_{\partial\Omega}\m^{-\frac{1}{2}}\mathcal{P}_{\gamma}[R]\nonumber\\
    &+\jnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}+\jnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}^{r}+o_{T}\xnm{R}^{r}+\xnm{R}^{2r}+o_{T}\varepsilon^{1-\alpha}.\nonumber
\end{align}
Now the key is to estimate the auxiliary terms in \eqref{pp 29}
\begin{align}\label{pp 30}
    \varepsilon^{-1}\br{-v\cdot\nabla_{x}\mathcal{G}+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\mathcal{G}+\varepsilon^{-1}\mathcal{L}[\mathcal{G}],R}
\end{align}
and the interaction terms
\begin{align}\label{pp 51'}
    &\varepsilon^{-1}\bbr{\mathcal{G},S}+\varepsilon^{-1}\bbrb{\mathcal{G},h}{\gamma_-}-\varepsilon^{-1}\bbrb{\mathcal{G}_h,(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}-\varepsilon^{-1}\mathcal{G}_m\int_{\partial\Omega}\m^{-\frac{1}{2}}\mathcal{P}_{\gamma}[R].
\end{align}


\subsection{Construction of $\mathcal{G}$}


We define $\mathcal{G}:=g^B+g$ where $g^B$ is an auxiliary boundary layer and $g$ is an auxiliary interior solution. 

\paragraph{\underline{$g^B$ Construction}}
Using the substitution in Section \ref{sec:boundary-form}, define $g^B$ as a standard boundary layer similar to $f^{B}_1$:
\begin{align}\label{final 03}
    g^B(\eta,\iota_1,\iota_2,{\mathfrak{v}}):=\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\Big(\Psi(\eta,\iota_1,\iota_2,{\mathfrak{v}})-\Psi_{\infty}(\iota_1,\iota_2,{\mathfrak{v}})\Big):=\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)\chi(\varepsilon\eta)\widetilde{\Psi}(\eta,\iota_1,\iota_2,{\mathfrak{v}})
\end{align}
where
\begin{align}\label{pp 36'}
    \left\{
    \begin{array}{l}
    -v_{\eta}\dfrac{\partial\Psi}{\partial\eta}+\mathcal{L}_w\big[\Psi\big]=0\ \ \text{in}\ \ [0,\infty)\times\mathbb{R}^3,\\\rule{0ex}{1.0em}
    \Psi(0,\iota_1,\iota_2,{\mathfrak{v}})=\mathfrak{g}({\mathfrak{v}}):=-\Big(\nabla_{x}\psi(0,\iota_1,\iota_2,):\mathscr{B}(0,\iota_1,\iota_2,{\mathfrak{v}})\Big)\ \ \text{for}\ \ v_{\eta}<0,
    \end{array}
    \right.
\end{align}
satisfying (from the almost zero mass-flux condition \eqref{pp 116})
\begin{align}\label{pp 35}
    \int_{\mathbb{R}^3}\Psi(0,{\mathfrak{v}})\mu_w^{\frac{1}{2}}({\mathfrak{v}})v_{\eta}=0
\end{align}
and
\begin{align}
    \lim_{\eta\rightarrow\infty}\Psi(\eta,\iota_1,\iota_2,{\mathfrak{v}})=\Psi_{\infty}(\iota_1,\iota_2,{\mathfrak{v}})\in\mathcal{N}.
\end{align}

Let $\nm{\cdot}_{L^p_{\iota_1\iota_2}L^{\infty}_{\eta}L^{\infty}_{\varrho,\vartheta}}$ denote the $L^p$ norm in $\iota_1$ and $\iota_2$ and $L^{\infty}_{\varrho,\vartheta}$ norm in $\eta$ and ${\mathfrak{v}}$. The similar notation also applies to $\nm{\cdot}_{L^p_{\iota_1\iota_2}L^{\infty}_{\gamma,\varrho,\vartheta}}$
\begin{lemma}\label{lem:final 5}
    Under the assumption \eqref{assumption:boundary}, there exists $\Psi(\eta,\iota_1,\iota_2,{\mathfrak{v}})$ satisfying \eqref{pp 36'} and \eqref{pp 35} such that for some $K_0>0$
    \begin{align}\label{qq 24}
    \lnmm{\Psi_{\infty}}+\lnmm{\mathrm{e}^{K_0\eta}\widetilde{\Psi}}\lesssim& o_{T}\jnm{c}^{r-1},\\
    \label{qq 23'}
    \lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\phi}}\widetilde{\Psi}}+\lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\psi}}\widetilde{\Psi}}\lesssim& o_{T}\jnm{c}^{r-1},
    \end{align}
    \begin{align}
    \label{qq 23}
    \lnmm{\partial_{\iota_1}\Psi_{\infty}}+\lnmm{\partial_{\iota_2}\Psi_{\infty}}+\nm{\mathrm{e}^{K_0\eta}\partial_{\iota_1}\widetilde{\Psi}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\eta}L^{\infty}_{\varrho,\vartheta}}+\nm{\mathrm{e}^{K_0\eta}\partial_{\iota_2}\widetilde{\Psi}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\eta}L^{\infty}_{\varrho,\vartheta}}
    \lesssim&o_{T}\jnm{c}^{r-1},
    \end{align}
    \begin{align}\label{final 55}
        \bnm{\widetilde{\Psi}}\lesssimo_{T}\jnm{c}^{r-1}.
    \end{align}
    and for any $\eta$,
    \begin{align}\label{final 64}
    \int_{\mathbb{R}^3}\Psi(\eta,\iota_1,\iota_2 ,{\mathfrak{v}})\mu_w^{\frac{1}{2}}({\mathfrak{v}})v_{\eta}=0.
    \end{align}
    Then $g^B$ can be defined as \eqref{final 03}.
\end{lemma}

\begin{proof}
    For fixed $(\iota_1,\iota_2)$, using Theorem \ref{boundary well-posedness}, we have
    \begin{align}\label{qq 24=}
    \lnmm{\Psi_{\infty}}+\lnmm{\mathrm{e}^{K_0\eta}\widetilde{\Psi}}\lesssim& \lnmms{\mathfrak{g}}{\gamma_+},\\
    \label{qq 23'=}
    \lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\phi}}\widetilde{\Psi}}+\lnmm{\mathrm{e}^{K_0\eta}\partial_{v_{\psi}}\widetilde{\Psi}}\lesssim& \lnmms{\mathfrak{g}}{\gamma_+}+\lnmms{\partial_{v_{\phi}}\mathfrak{g}}{\gamma_+}+\lnmms{\partial_{v_{\psi}}\mathfrak{g}}{\gamma_+},
    \end{align}
    and
    \begin{align}
    \label{qq 23=}
    &\lnmm{\partial_{\iota_1}\Psi_{\infty}}+\lnmm{\partial_{\iota_2}\Psi_{\infty}}+\nm{\mathrm{e}^{K_0\eta}\partial_{\iota_1}\widetilde{\Psi}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\eta}L^{\infty}_{\varrho,\vartheta}}+\nm{\mathrm{e}^{K_0\eta}\partial_{\iota_2}\widetilde{\Psi}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\eta}L^{\infty}_{\varrho,\vartheta}}\\
    \lesssim& \nm{\mathfrak{g}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\gamma_+,\varrho,\vartheta}}+\nm{\partial_{\iota_1}\mathfrak{g}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\gamma_+,\varrho,\vartheta}}+\nm{\partial_{\iota_2}\mathfrak{g}}_{L^2_{\iota_1\iota_2}L^{\infty}_{\gamma_+,\varrho,\vartheta}}.\nonumber
    \end{align}
    Then for $2\leq r\leq 6$, due to trace theorem and Sobolev embedding 
    \begin{align}
    \tnms{\nabla_{x}^2\psi}{\partial\Omega}\lesssim\nm{\psi}_{H^3}\lesssim\nm{\psi}_{W^{4,\frac{r}{r-1}}}\lesssimo_{T}\jnm{c}^{r-1},
    \end{align}
    and
    \begin{align}
    \lnms{\nabla_{x}\psi}{\partial\Omega}\lesssim\nm{\psi}_{H^3}\lesssim\nm{\psi}_{W^{4,\frac{r}{r-1}}}\lesssimo_{T}\jnm{c}^{r-1},
    \end{align}
    from the definition of $\mathfrak{g}$ in \eqref{pp 36'}, we have the desired results. The estimate \eqref{final 55} follows from Theorem \ref{boundary regularity}. Also, \eqref{final 64} comes from direct integrating in \eqref{pp 36'}.
\end{proof}

\begin{lemma}\label{lem:final 6}
    Under the assumption \eqref{assumption:boundary}, we have that 
    $\Psi$, $\widetilde{\Psi}$ and $\Psi_{\infty}$ defined in \eqref{pp 36'} are all odd in $v_{\phi}$ and $v_{\psi}$. In particular, we have
    \begin{align}\label{pp 39}
    \Psi_{\infty}({\mathfrak{v}})=\big({\mathfrak{v}}\cdot\textbf{b}_{\infty}\big)\mu_w^{\frac{1}{2}}({\mathfrak{v}})=\Big(v_{\phi} b_{\infty,\iota_1}+v_{\psi} b_{\infty,\iota_2}\Big)\mu_w^{\frac{1}{2}}({\mathfrak{v}}),
    \end{align}
    for $\textbf{b}_{\infty}:=\Big(0,b_{\infty,\iota_1},b_{\infty,\iota_2}\Big)$ with some $b_{\infty,\iota_1}$ and $b_{\infty,\iota_2}$ only depending on $(\iota_1,\iota_2)$.
\end{lemma}
\begin{proof}
Since $\psi=(\psi_{\iota_1},\psi_{\iota_2},\psi_{\mathfrak{n}})=\textbf{0}$ on $\partial\Omega$, we know $\partial_{\iota_1}\psi=\partial_{\iota_2}\psi=0$. Also, due to $\nabla_{x}\cdot\psi=0$, we know $\partial_{\mathfrak{n}}\psi_n=0$. Hence,
\begin{align}
    \mathfrak{g}({\mathfrak{v}})=-\partial_{\mathfrak{n}}\psi_{\iota_1}(0)\mathscr{B}_{\mathfrak{n}\iota_1}(0)-\partial_{\mathfrak{n}}\psi_{\iota_2}(0)\mathscr{B}_{\mathfrak{n}\iota_2}(0).
\end{align}
Due to the oddness of $\mathfrak{g}({\mathfrak{v}})$ in $v_{\phi}$ and $v_{\psi}$ (since $\mathscr{B}_{\mathfrak{n}\iota_1}$ and $\mathscr{B}_{\mathfrak{n}\iota_2}$ are odd), using Theorem \ref{boundary well-posedness}, we know that the solution $\Psi(\eta,{\mathfrak{v}})$ is also odd in $v_{\phi}$ and $v_{\psi}$. Hence, for any two-variable function $\mathcal{M}$, we know
\begin{align}\label{final 01}
    \int_{\mathbb{R}^3}\mathcal{M}\big(\abs{v}^2,v_{\eta}\big)\Psi(\eta,{\mathfrak{v}})=0.
\end{align}
Denote
\begin{align}
    \textbf{P}[\Psi](\eta,{\mathfrak{v}}):=\bigg( p_{\Psi}(\eta )+{\mathfrak{v}}\cdot \textbf{b}_{\Psi}(\eta )+\left(\abs{{\mathfrak{v}}}^2-5\right) c_{\Psi}(\eta )\bigg)\mu_w^{\frac{1}{2}}({\mathfrak{v}}).
\end{align}
The oddness dictates that $p_{\Psi}=c_{\Psi}=0$. Also, for $\textbf{b}_{\Psi}:=\Big(b_{\Psi,\iota_1},b_{\Psi,\iota_2},b_{\Psi,\mathfrak{n}}\Big)$, \eqref{final 64} implies $b_{\Psi,\mathfrak{n}}=0$. Hence, we have
\begin{align}
    \textbf{P}[\Psi](\eta ,{\mathfrak{v}})=\Big(v_{\phi} b_{\Psi,\iota_1}(\eta )+v_{\psi} b_{\Psi,\iota_2}(\eta )\Big)\mu_w^{\frac{1}{2}}({\mathfrak{v}}).
\end{align}
Similarly, considering
\begin{align}
    \Psi_{\infty}({\mathfrak{v}})=\textbf{P}[\Psi_{\infty}](v):=\bigg( p_{\infty}+{\mathfrak{v}}\cdot \textbf{b}_{\infty}+\left(\abs{{\mathfrak{v}}}^2-5\right) c_{\infty}\bigg)\mu_w^{\frac{1}{2}}({\mathfrak{v}}),
\end{align}
we know $\textbf{b}_{\infty}=\Big(b_{\infty,\iota_1},b_{\infty,\iota_2},b_{\infty,\mathfrak{n}}\Big)$ satisfies $b_{\infty,\mathfrak{n}}=0$ from \eqref{pp 35}, and thus
\begin{align}
    \Psi_{\infty}({\mathfrak{v}})=\Big(v_{\phi} b_{\infty,\iota_1}+v_{\psi} b_{\infty,\iota_2}\Big)\mu_w^{\frac{1}{2}}({\mathfrak{v}}).
\end{align}
Similar to \eqref{final 01}, using \eqref{final 03}, we have the oddness of $g^B$, i.e. for any two-variable function $\mathcal{M}$, we know
\begin{align}\label{final 02}
    \int_{\mathbb{R}^3}\mathcal{M}\big(\abs{v}^2,v_{\eta}\big)g^B(\eta,{\mathfrak{v}})=0.
\end{align}
\end{proof}

\paragraph{\underline{$g$ Construction}}
Now we discuss the construction of $g$.
In order to fulfill \eqref{pp 46}, we need
\begin{align}
    g\Big|_{\partial\Omega}=\Psi_{\infty}=\big({\mathfrak{v}}\cdot \textbf{b}_{\infty}\big)\mu_w^{\frac{1}{2}}.
\end{align}

\begin{lemma}\label{lem:final 7}
    Under the assumption \eqref{assumption:boundary}, there exists an extension
    $\mathfrak{b}$ of $\textbf{b}_{\infty}$ satisfying
    \begin{align}\label{pp 47}
    \left\{
    \begin{array}{rll}
    \nabla_{x}\cdot\mathfrak{b}&=0&\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.0em}
    v\cdot\mathfrak{b}&={\mathfrak{v}}\cdot\textbf{b}_{\infty}&\ \ \text{on}\ \ \partial\Omega,
    \end{array}
    \right.
    \end{align}
    such that
    \begin{align}
        \nm{\mathfrak{b}}_{H^1}\lesssim o_{T}\jnm{c}^{r-1}.
    \end{align}
    Then $g$ can be defined as
    \begin{align}\label{final 10}
    g&=\m^{\frac{1}{2}}(v)\big(v\cdot\mathfrak{b} \big)
    \end{align}
\end{lemma}
\begin{remark}
    Since the change-of-variable $v\rightarrow{\mathfrak{v}}$ is orthogonal, we know $\m^{\frac{1}{2}}$ is invariant. Then thanks to \eqref{aa 44}, the boundary condition in \eqref{pp 47} is valid by applying a linear orthogonal transformation. Hence, we deduce that $\mathfrak{b}\cdotn=\textbf{b}_{\infty}\cdotn=0$. 
\end{remark}
\begin{proof}[Proof of Lemma \ref{lem:final 7}]
The construction of $\mathfrak{b}$ consists of two steps:

\subparagraph{$\overline{\textbf{b}}_{\infty}$ Construction:}
Let $\overline{\textbf{b}}_{\infty}(x)$ for $x\in\Omega$ be a Sobolev extension of $\textbf{b}_{\infty}$ satisfying $v\cdot\overline{\textbf{b}}_{\infty}={\mathfrak{v}}\cdot\textbf{b}_{\infty}$ on $\partial\Omega\times\mathbb{R}^3$ and
\begin{align}
    \nm{\overline{\textbf{b}}_{\infty}}_{H^1}\lesssim \nm{\textbf{b}_{\infty}}_{H^{\frac{1}{2}}_{\partial\Omega}}.
\end{align}

\subparagraph{$\overline\mathfrak{b}$ Construction:}
Since $\displaystyle\int_{\Omega}\nabla_{x}\cdot\overline{\textbf{b}}_{\infty}=\int_{\partial\Omega}\overline{\textbf{b}}_{\infty}\cdot n=\int_{\partial\Omega}\textbf{b}_{\infty,\mathfrak{n}}=0$, based on \cite[Theorem IV.5.2]{Boyer.Fabrie2013}, we know that there exists a unique solution $(\overline\mathfrak{b},\overline\mathfrak{p})\in H^1\times L^2$ (with $\overline\mathfrak{p}$ zero average) to the Stokes problem
\begin{align}
    \left\{
    \begin{array}{rll}
    -\Delta_x\overline\mathfrak{b}+\nabla_{x}\overline\mathfrak{p}&=0&\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.2em}
    \nabla_{x}\cdot\overline\mathfrak{b}&=-\nabla_{x}\cdot\overline{\textbf{b}}_{\infty}&\ \ \text{in}\ \ \Omega,\\\rule{0ex}{1.2em}
    \overline\mathfrak{b}&=0&\ \ \text{on}\ \ \partial\Omega,
    \end{array}
    \right.
\end{align}
satisfying
\begin{align}
    \nm{\overline\mathfrak{b}}_{H^1}+\tnm{\overline\mathfrak{p}}\lesssim \nm{\nabla_{x}\cdot\overline\textbf{b}_{\infty}}_{L^2}\lesssim \nm{\overline{\textbf{b}}_{\infty}}_{H^1}\lesssim \nm{\textbf{b}_{\infty}}_{H^{\frac{1}{2}}_{\partial\Omega}}.
\end{align}

\subparagraph{Summary of $g$ Construction:}
Hence, we know $\mathfrak{b}=\overline\mathfrak{b}+\overline{\textbf{b}}_{\infty}$ is the solution to \eqref{pp 47} and using \eqref{qq 23} it satisfies
\begin{align}
    \nm{\mathfrak{b}}_{H^1}\lesssim \nm{ \textbf{b}_{\infty}}_{H^{\frac{1}{2}}_{\partial\Omega}}\lesssim o_{T}\jnm{c}^{r-1}.
\end{align}

\end{proof}

\paragraph{\underline{Summary of $\mathcal{G}$ Construction}}
Therefore, we split \eqref{pp 30}
\begin{align}
    &\varepsilon^{-1}\br{-v\cdot\nabla_{x}\mathcal{G}+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)\mathcal{G}+\varepsilon^{-1}\mathcal{L}[\mathcal{G}],R}\\
    =&\varepsilon^{-1}\br{-v\cdot\nabla_{x}g^B+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)g^B+\varepsilon^{-1}\mathcal{L}\left[g^B\right],R}+\varepsilon^{-1}\br{-v\cdot\nabla_{x}g+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)g+\varepsilon^{-1}\mathcal{L}[g],R}.\nonumber
\end{align}
Similar to the derivation of $S_3$ in \eqref{mm 00}, direct computation using the substitution in Section \ref{sec:boundary-form} reveals 
\begin{align}\label{pp 38}
    &\varepsilon^{-1}\br{-v\cdot\nabla_{x}g^B+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)g^B+\varepsilon^{-1}\mathcal{L}\left[g^B\right],R}\\
    =&-\varepsilon^{-1}\bbr{\dfrac{1}{R_1-\varepsilon\eta}\bigg(v_{\phi}^2\dfrac{\partial }{\partialv_{\eta}}\bigg)g^B+\dfrac{1}{R_2-\varepsilon\eta}\bigg(v_{\psi}^2\dfrac{\partial }{\partialv_{\eta}}\bigg)g^B,R}\nonumber\\
    &+\varepsilon^{-1}\bbr{\dfrac{1}{R_1-\varepsilon\eta}\bigg(v_{\eta}v_{\phi}\dfrac{\partial}{\partialv_{\phi}}\bigg)g^B+\dfrac{1}{R_2-\varepsilon\eta}\bigg(v_{\eta}v_{\psi}\dfrac{\partial }{\partialv_{\psi}}\bigg)g^B,R}\nonumber\\
    &-\varepsilon^{-1}\bbr{\dfrac{1}{L_1L_2}\left(-\dfrac{\partial_{\iota_1\iota_1}\mathbf{r}\cdot\partial_{\iota_2}\mathbf{r}}{L_1(\varepsilon\kappa_1\eta-1)}v_{\phi}v_{\psi}
    +\dfrac{\partial_{\iota_1\iota_2}\mathbf{r}\cdot\partial_{\iota_2}\mathbf{r}}{L_2(\varepsilon\kappa_2\eta-1)}v_{\psi}^2\right)\dfrac{\partialg^B }{\partialv_{\phi}},R}\nonumber\\
    &-\varepsilon^{-1}\bbr{\dfrac{1}{L_1L_2}\left(\dfrac{\partial_{\iota_2\iota_2}\mathbf{r}\cdot\partial_{\iota_1}\mathbf{r}}{L_2(\varepsilon\kappa_2\eta-1)}v_{\phi}v_{\psi}
    +\dfrac{\partial_{\iota_1\iota_2}\mathbf{r}\cdot\partial_{\iota_1}\mathbf{r}}{L_1(\varepsilon\kappa_1\eta-1)}v_{\phi}^2\right)\dfrac{\partialg^B }{\partialv_{\psi}},R}\nonumber\\
    &-\varepsilon^{-1}\bbr{\left(\dfrac{v_{\phi}}{L_1(\varepsilon\kappa_1\eta-1)}\dfrac{\partial }{\partial\iota_1}+\dfrac{v_{\psi}}{L_2(\varepsilon\kappa_2\eta-1)}\dfrac{\partial }{\partial\iota_2}\right)g^B,R}\nonumber\\
    &+\varepsilon^{-1}\br{\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x}T}{4T^2}\right)g^B,R}-\varepsilon^{-2}\br{(v\cdot n)\frac{\partial\chi(\varepsilon\eta)}{\partial\eta}\widetilde{\Psi},R}\nonumber\\
    &-\varepsilon^{-2}\br{\chi(\varepsilon\eta)\bigg(\chi(\varepsilon^{-1}v_{\eta})K\Big[\widetilde{\Psi}\Big]-K\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Psi}\Big]\bigg),R}+\varepsilon^{-2}\br{\mathcal{L}\left[g^B\right]-\mathcal{L}_w\left[g^B\right],R}.\nonumber
\end{align}
Also, due to the direct computation
\begin{align}
    -\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\m^{-\frac{1}{2}}g\Big)=&-\m^{\frac{1}{2}} v\cdot\left(\m^{-\frac{1}{2}}\nabla_{x}g-\frac{1}{2}g\mu^{-\frac{3}{2}}\nabla_{x}\mu\right)\\
    =&-\m^{\frac{1}{2}} v\cdot\left(\m^{-\frac{1}{2}}\nabla_{x}g-g\mu^{-\frac{1}{2}}\left(\abs{v}^2-5T\right)\frac{\nabla_{x} T}{4T^2}\right)
    =-v\cdot\nabla_{x}g+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)g,\nonumber
\end{align}
and $g\in\mathcal{N}$ from \eqref{final 10}, we have
\begin{align}\label{pp 45}
    &\varepsilon^{-1}\br{-v\cdot\nabla_{x}g+\left(\m^{-\frac{1}{2}}\overline{\mathscr{A}}\cdot\frac{\nabla_{x} T}{4T^2}\right)g+\varepsilon^{-1}\mathcal{L}[g],R}=-\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\m^{-\frac{1}{2}}g\Big),R}.
\end{align}
Also, due to the cutoff $\overline{\chi}\left(\varepsilon^{-1}v_{\eta}\right)$ in $g^B$ definition \eqref{pp 36'}, we have
\begin{align}\label{final 61'}
    \mathcal{G}_h:=\chi\left(\varepsilon^{-1}v_{\eta}\right)\mathfrak{g}({\mathfrak{v}}),
\end{align}
with
\begin{align}\label{final 61}
    \tnms{\mathcal{G}_h}{\gamma_+}\lesssim o_{T}\varepsilon\jnm{c}^{r-1},
\end{align}
and
\begin{align}\label{final 62'}
    \mathcal{G}_m:=&\int_{\mathbb{R}^3}v_{\eta}\mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Psi}(0,{\mathfrak{v}})\mathrm{d} {\mathfrak{v}}=\int_{\mathbb{R}^3}v_{\eta}\mb^{\frac{1}{2}}\chi\left(\varepsilon^{-1}v_{\eta}\right)\mathfrak{g}(0,{\mathfrak{v}})\mathrm{d} {\mathfrak{v}},
\end{align}
with
\begin{align}\label{final 62}
    \abs{\mathcal{G}_m}\lesssimo_{T}\varepsilon^2\jnm{c}^{r-1}.
\end{align}


\subsection{Estimates of Auxiliary Boundary Layer $g^B$ Terms}\label{sec:gb estimate}


\begin{lemma}\label{lemma:k1}
    Let $h^B(\eta,\iota_1,\iota_2,v)$ with $\eta=\varepsilon^{-1}\mathfrak{n}$ be a boundary-layer-type quantity satisfying 
    \begin{align}\label{final 65}
        \tnm{h^B}+\nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}+\nm{\etah^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\lesssimo_{T}\jnm{c}^{r-1}.
    \end{align} 
    Then under the assumption \eqref{assumption:boundary}, we have
    \begin{align}
    \abs{\varepsilon^{-1}\bbr{h^B,R}}\lesssim&o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}. 
    \end{align}
\end{lemma}
\begin{remark}
    When applying Lemma \ref{lemma:k1}, we will let $h^B=g^B$, or its derivatives. 
\end{remark}
\begin{proof}[Proof of Lemma \ref{lemma:k1}]
We decompose
\begin{align}\label{qq 40}
    \varepsilon^{-1}\bbr{h^B,R}
    =I_1+I_2+I_3,
\end{align}
where
\begin{align}
\\
    I_1:=&\varepsilon^{-1}\bbr{h^B,(\textbf{I}-\overline{\textbf{P}})[R]},\quad
    I_2:=\varepsilon^{-1}\br{h^B,p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}},\quad
    I_3:=\varepsilon^{-1}\br{h^B,c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}}.\nonumber
\end{align}
We directly bound
\begin{align}
    \babs{I_{1}}\lesssim&\varepsilon^{-1}\tnm{h^B}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\lesssim o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.
\end{align}
Using the scaling $\eta=\varepsilon^{-1}\mathfrak{n}$, we can bound
\begin{align}\label{qq 41}
\\
    \babs{I_{2}}\lesssim& \varepsilon^{-1}\nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\mathfrak{n}}L^1_{v}}\Big(\tnm{p}+\tnm{\textbf{b}}+\tnm{\textbf{d}}\Big)
    \lesssim\varepsilon^{-\frac{1}{2}} \nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\Big(\tnm{p}+\tnm{\textbf{b}}+\tnm{\textbf{d}}\Big)\nonumber\\
    \lesssim& o_{T}\jnm{c}^{r}+o_{T}\Big(\varepsilon^{-\frac{r}{2}}\tnm{p}^{r}+\varepsilon^{-\frac{r}{2}}\tnm{\textbf{b}}^{r}+\varepsilon^{-\frac{r}{2}}\tnm{\textbf{d}}^{r}\Big)\lesssim o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.\nonumber
\end{align}
For $I_{3}$, notice that
\begin{align}\label{pp 41}
    c=Z-\varepsilon^{-1}\xi=\Big(Z^R-\varepsilon^{-1}\xi\Big)+Z^S:=& K_1+K_2.
\end{align}
We define a natural extension of $Z^R$ and $\xi$ with zero value for $\mathfrak{n}>1$ (for the convenience of Hardy's inequality). Then using \eqref{final 65}, Hardy's inequality, Lemma \ref{remark 01} and \eqref{qq 06}, we have
\begin{align}\label{pp 96=}
    \babs{I_{3, K_1}}\lesssim&\varepsilon^{-1}\br{h^B,\int_0^{\mathfrak{n}}\big(\partial_{\mathfrak{n}}Z^R-\varepsilon^{-1}\partial_{\mathfrak{n}}\xi\big)}
    \lesssim\varepsilon^{-1}\nm{\mathfrak{n}h^B}_{L^2_{\iota_1\iota_2}L^2_{\mathfrak{n}}L^1_{v}}\tnm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\big(\partial_{\mathfrak{n}}Z^R-\varepsilon^{-1}\partial_{\mathfrak{n}}\xi\big)}\\
    \lesssim&\nm{\etah^B }_{L^2_{\iota_1\iota_2}L^2_{\mathfrak{n}}L^1_{v}}\Big(\tnm{\partial_{\mathfrak{n}}Z^R}+\varepsilon^{-1}\tnm{\partial_{\mathfrak{n}}\xi}\Big)\lesssim\varepsilon^{\frac{1}{2}} \nm{\etah^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\Big(\nm{Z^R}_{H_0^1}+\varepsilon^{-1}\nm{\xi}_{H_0^1}\Big)\nonumber\\
    \lesssim& o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+\varepsilon^{\alr-\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}
Similar to the estimate of $I_{1}$, using \eqref{qq 07}, we have
\begin{align}\label{pp 96='}
    \babs{I_{3, K_2}}\lesssim& \varepsilon^{-1}\nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\tnm{Z^S}\lesssim\varepsilon^{-\frac{1}{2}} \nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\tnm{Z^S}\\
    \lesssim& \jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
We deduce Lemma \ref{lemma:k1} by collecting the above estimates.
\end{proof}

\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, we have
    \begin{align}\label{pp 44'}
    \babs{\text{\eqref{pp 38}}}\lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
\end{lemma}

\begin{proof}
Comparing the first term in \eqref{qq 40} with \eqref{mm 00}, we know
\begin{align}
    M:=\dfrac{1}{R_1-\varepsilon\eta}\bigg(v_{\phi}^2\dfrac{\partial }{\partialv_{\eta}}\bigg)g^B+\dfrac{1}{R_2-\varepsilon\eta}\bigg(v_{\psi}^2\dfrac{\partial }{\partialv_{\eta}}\bigg)g^B
\end{align}
is essentially $\ss_{3a}$ with $f^{B}_1$ replaced by $g^B$ and $\widetilde{\Phi}$ replaced by $\widetilde{\Psi}$. Hence, using the proof of Lemma \ref{ss3-estimate}, we have
\begin{align}
    \pnm{M}{r}&\lesssimo_{T}\varepsilon^{\frac{2}{r}-\alpha}\jnm{c}^{r-1}.\label{gg4}
\end{align}
Similar to \eqref{qq 40}, we split
\begin{align}\label{final 56}
    \text{First Term in \eqref{pp 38}}
    =&\varepsilon^{-1}\bbr{M,(\textbf{I}-\overline{\textbf{P}})[R]}+\varepsilon^{-1}\bbr{M,p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}}\\
    &+
    \varepsilon^{-1}\bbr{M,c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}}.\nonumber
\end{align}
For the first term in \eqref{final 56}, using \eqref{gg4}, we bound
\begin{align}
    \abs{\varepsilon^{-1}\bbr{M,(\textbf{I}-\overline{\textbf{P}})[R]}}\lesssim&\varepsilon^{-1}\tnm{M}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-\alr}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}.\nonumber
\end{align}
For the second term in \eqref{final 56}, we first integrate by parts with respect to $v_{\eta}$, and then use Lemma \ref{lem:final 5} with a similar argument as \eqref{qq 41} to obtain
\begin{align}
    &\abs{\varepsilon^{-1}\bbr{M,p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}}}\lesssim\varepsilon^{-1}\abs{\bbr{g^B,p\nabla_v\m^{\frac{1}{2}}+\textbf{b}\cdot \nabla_v(v\m^{\frac{1}{2}})+\textbf{d}\cdot\nabla_v\mathscr{A}}} \\
    \lesssim&\varepsilon^{-\frac{1}{2}}\nm{g^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\Big(\tnm{p}+\tnm{\textbf{b}}+\tnm{\textbf{d}}\Big)
    \lesssimo_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.\nonumber
\end{align}
For the third term in \eqref{final 56}, we first integrate by parts with respect to $v_{\eta}$ and then apply the splitting in \eqref{pp 41}. A similar argument as \eqref{pp 96=} and \eqref{pp 96='} with Lemma \ref{lem:final 5} imply
\begin{align}
    \abs{\varepsilon^{-1}\bbr{M,c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}}}\lesssim&o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Hence, we have shown that
\begin{align}
    &\Babs{\text{First Term in \eqref{pp 38}}}
   \lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{2r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Note that in the definition of $g^B$ \eqref{final 03}, the cutoff is only in the normal direction, so $g^B$ should satisfy the same tangential estimates as $\widetilde{\Psi}$ in \eqref{qq 23} and \eqref{qq 23'}. From Lemma \ref{lem:final 5}, we have for $h^B=g^B,\partial_{v_{\phi}}g^B,\partial_{v_{\psi}}g^B,\partial_{\iota_1}g^B,\partial_{\iota_2}g^B, \widetilde{\Psi}$
\begin{align}
    \tnm{h^B}+\nm{h^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}+\nm{\etah^B}_{L^2_{\iota_1\iota_2}L^2_{\eta}L^1_{v}}\lesssim &\jnm{c}^{r-1}.
\end{align}
Hence, applying Lemma \ref{lemma:k1} with $h^B=g^B,\partial_{v_{\phi}}g^B,\partial_{v_{\psi}}g^B,\partial_{\iota_1}g^B,\partial_{\iota_2}g^B, \widetilde{\Psi}$ and Lemma \ref{lem:final 5}, we have
\begin{align}
    &\Babs{\text{Second Term 
    to Sixth Term in \eqref{pp 38}}}\lesssimo_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
In addition, noticing that $\lnm{\frac{\partial\chi(\varepsilon\eta)}{\partial\eta}}\lesssim \varepsilon$, applying Lemma \ref{lemma:k1} with $h^B=\varepsilon^{-1}\frac{\partial\chi(\varepsilon\eta)}{\partial\eta}\widetilde{\Psi}$ and Lemma \ref{lem:final 5}, we have
\begin{align}
    &\Babs{\text{Seventh Term in \eqref{pp 38}}}\lesssimo_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Using the same idea as in the proof of Lemma \ref{ss3-estimate} and Lemma \ref{lem:final 5}, we obtain
\begin{align}
    \tnm{\chi(\varepsilon^{-1}v_{\eta})K\Big[\widetilde{\Psi}\Big]}+\tnm{K\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Psi}\Big]}\lesssim&\varepsilon\jnm{c}^{r-1},\\
    \nm{\chi(\varepsilon^{-1}v_{\eta})K\Big[\widetilde{\Psi}\Big]}_{L^2_{\mathfrak{n}}L^1_v}+\nm{K\Big[\chi(\varepsilon^{-1}v_{\eta})\widetilde{\Psi}\Big]}_{L^2_{\mathfrak{n}}L^1_v}\lesssim&\varepsilon^{\frac{3}{2}}\jnm{c}^{r-1}.
\end{align}
Hence, using Lemma \ref{lemma:k1} with $h^B=\chi\left(\varepsilon^{-1}v_{\eta}\right)K\left[\widetilde{\Psi}\right],K\left[\chi\left(\varepsilon^{-1}v_{\eta}\right)\widetilde{\Psi}\right]$, we obtain that
\begin{align}
    &\Babs{\text{Eighth Term in \eqref{pp 38}}}\lesssimo_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Using Corollary \ref{m-estimate.}, Corollary \ref{m-estimate..} and Lemma \ref{lem:final 5}, we know 
\begin{align}
    &\Babs{\text{Ninth Term in \eqref{pp 38}}}\lesssimo_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
\end{proof}


\subsection{Estimates of Auxiliary Interior Solution $g$ Terms}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, we have
    \begin{align}\label{pp 44''}
    \babs{\text{\eqref{pp 45}}}\lesssim&o_{T}\jnm{c}^{r}+o_{T}\xnm{R}^{r}.
    \end{align}
\end{lemma}

\begin{proof}
Oddness guarantees that in \eqref{pp 45}
\begin{align}
    &\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\m^{-\frac{1}{2}}g\Big),R}=\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\big(\mathfrak{b}\cdot v\big),R}\\
    =&\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\mathfrak{b}\cdot v\Big),p\m^{\frac{1}{2}}+\textbf{b}\cdot v\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}+(\textbf{I}-\overline{\textbf{P}})[R]}\nonumber\\
    =&\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\mathfrak{b}\cdot v\Big),p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}+(\textbf{I}-\overline{\textbf{P}})[R]}.\nonumber
\end{align}
Here, thanks to the oddness/parity of \eqref{pp 39} and the construction \eqref{pp 47} which yields the divergence-free $\nabla_{x}\cdot\mathfrak{b}=0$ for $\mathfrak{b}:=(\mathfrak{b}_{1},\mathfrak{b}_{2},\mathfrak{b}_{3})$, we observe the crucial cancellation of the $p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}$ contribution
\begin{align}\label{final 04}
    &\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\Big(\mathfrak{b}\cdot v\Big),p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}}\\
    =&\varepsilon^{-1}\br{\m^{\frac{1}{2}}\Big(v_1^2\partial_1\mathfrak{b}_{1}+v_2^2\partial_2\mathfrak{b}_{2}+v_3^2\partial_3\mathfrak{b}_{3}\Big),p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}}\nonumber\\
    =&\frac{\varepsilon^{-1}}{3}\br{\m^{\frac{1}{2}} \abs{v}^2\big(\nabla_{x}\cdot\mathfrak{b}\big),p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right) \m^{\frac{1}{2}}}=0.\nonumber
\end{align}
Also, using Lemma \ref{lem:final 7}, the $(\textbf{I}-\overline{\textbf{P}})[R]$ contribution can be controlled
\begin{align}
    \abs{\varepsilon^{-1}\br{\m^{\frac{1}{2}} v\cdot\nabla_{x}\big(\mathfrak{b}\cdot v\big),(\textbf{I}-\overline{\textbf{P}})[R]}}
    \lesssim&\varepsilon^{-1}\nm{\mathfrak{b}}_{H^1}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}^{r}.\nonumber
\end{align}
\end{proof}


\subsection{Estimates of Interaction Terms}


In this subsection, we consider the interaction terms in \eqref{pp 51'}.

\begin{lemma}
    Under the assumption \eqref{assumption:boundary} and $\xnm{R}\ll 1$, we have
    \begin{align}\label{pp 99'''}
    \babs{\text{\eqref{pp 51'}}}
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+\varepsilon^{2\alr-2r}\xnm{R}^{3r}+o_{T}\varepsilon^{r-r\alpha}.
    \end{align}
\end{lemma}

\begin{proof}
\ 
\paragraph{\underline{Boundary Terms Contribution}}
Naturally, using Lemma \ref{h-estimate}, the boundary term in \eqref{pp 51'} is controlled as
\begin{align}
    \abs{\varepsilon^{-1}\bbrb{\mathcal{G} ,h}{\gamma_-}}\leq&\abs{\varepsilon^{-1}\bbrb{g^B ,h}{\gamma_-}}+\abs{\varepsilon^{-1}\bbrb{g ,h}{\gamma_-}}
    \lesssim&\varepsilon^{-1}\tnms{\mathfrak{g}}{\gamma_-}\tnms{h}{\gamma_-}
    \lesssim o_{T}\jnm{c}^{r}+\varepsilon^{-r}\tnms{h}{\gamma_-}^{r}\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}.\nonumber
\end{align}
Also, using \eqref{final 61}, since $\mathcal{G}_h$ is restricted to $\abs{v_{\eta}}\lesssim\varepsilon$, we have
\begin{align}
    \abs{\varepsilon^{-1}\bbrb{\mathcal{G}_h,(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}}\lesssim& \varepsilon^{-1}\tnms{\mathcal{G}_h}{\gamma_+}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\lesssim o_{T}\jnm{c}^{r-1}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}\\
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+\varepsilon^{\frac{r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}
Using \eqref{final 62}, we know
\begin{align}
    \abs{\varepsilon^{-1}\mathcal{G}_m\int_{\partial\Omega}\m^{-\frac{1}{2}}\mathcal{P}_{\gamma}[R]}\lesssim \varepsilon^{-1}\abs{\mathcal{G}_m}\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}\lesssimo_{T}\varepsilon\jnm{c}^{r-1}\tnms{\mathcal{P}_{\gamma}[R]}{\gamma}\lesssim o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r}\xnm{R}^{r}.
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{\mathcal{G},\lc^1[R]}$ Contribution}}
We can decompose
\begin{align}\label{pp 97}
    \varepsilon^{-1}\bbr{\mathcal{G},\lc^1[R]}=&\varepsilon^{-1}\br{g^B,\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,\mhR\right]}+\varepsilon^{-1}\br{g,\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,\mhR\right]}.
\end{align}
For the first term in \eqref{pp 97}, using $\lnm{f_1}\lesssimo_{T}$, we can follow a similar argument as $g^B$ estimate in Section \ref{sec:gb estimate} to split $R$ and use Lemma \ref{lem:final 5} to bound
\begin{align}
    \abs{\varepsilon^{-1}\br{g^B,\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,\mhR\right]}}
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Since $g\in\mathcal{N}$, for the second term in \eqref{pp 97}, from Lemma \ref{lem:final 7}, we deduce from the orthogonality of $Q$
\begin{align}
    \varepsilon^{-1}\br{g,\m^{-\frac{1}{2}} Q^{\ast}\left[\m^{\frac{1}{2}} f_1,\mhR\right]}=0.
\end{align}
In total, we know 
\begin{align}
    \abs{\varepsilon^{-1}\bbr{\mathcal{G},\lc^1[R]}}
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g^B,S_0+S_5}$ Contribution}}
Using Lemma \ref{lem:final 5}, Lemma \ref{ss0-estimate} and Lemma \ref{ss5-estimate}, we have
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g^B,S_0+S_5}}\lesssim& \varepsilon^{-1}\tnm{g^B}\Big(\varepsilon\um{R}+o_{T}\varepsilon^{\frac{3}{2}-\alpha}\Big)\lesssimo_{T}\varepsilon^{-\frac{1}{2}}\jnm{c}^{r-1}\Big(\varepsilon\um{R}+o_{T}\varepsilon^{\frac{3}{2}-\alpha}\Big)\\
    \lesssim& o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{\frac{r}{2}}\xnm{R}^{r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g^B,S_4+\ss_6}$ Contribution}}
Using Lemma \ref{lem:final 5}, Lemma \ref{ss4-estimate} and Lemma \ref{ssp-estimate}, we have
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g^B,S_4+\ss_6}}\lesssim&\varepsilon^{-1}\tnm{g^B}\Big(\tnm{S_4}+\tnm{\ss_6}\Big)\lesssimo_{T}\varepsilon^{-\frac{1}{2}}\jnm{c}^{r-1}\Big(\tnm{S_4}+\tnm{\ss_6}\Big)\\
    \lesssim& o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{\frac{3r}{2}-r\alpha}.\nonumber
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g^B,S_1}$ Contribution}}
Using $\lnm{f^{B}_1}\lesssimo_{T}$, we can follow a similar argument as $g^B$ estimate in Section \ref{sec:gb estimate} to split $R$ and use  Lemma \ref{lem:final 5} to bound
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g^B,S_1}}
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-\alr}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g^B,S_3}$ Contribution}}
Using Lemma \ref{lem:final 5}, Lemma \ref{ss3-estimate}, we have
\begin{align}\label{final 36}
    \abs{\varepsilon^{-1}\br{g^B,S_3}}\lesssim&\varepsilon^{-1}\nm{g^B}_{L^{\infty}}\nm{S_3}_{L^1}
    \lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{-r}\nm{S_3}_{L^1}^{r}\lesssimo_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g^B,S_2}$ Contribution}}
We finally come to the most difficult term. Since $g^B$ is odd in $v_{\phi},v_{\psi}$, the estimate of $\abs{\varepsilon^{-1}\bbr{g^B,S_2}}$ follows a similar argument as estimating \eqref{pp 103'}. We split
\begin{align}
    \varepsilon^{-1}\bbr{g^B,S_2}=&\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    &+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},(\textbf{I}-\overline{\textbf{P}})[R]\Big]}
    +\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[(\textbf{I}-\overline{\textbf{P}})[R],(\textbf{I}-\overline{\textbf{P}})[R]\Big]},\nonumber
\end{align}
and
\begin{align}
    &\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},\textbf{P}[R]+\textbf{d}\cdot\mathscr{A}\Big]}\\
    =&\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}\nonumber\\
    &+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}, p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}\nonumber\\
    &+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}.\nonumber
\end{align}
Based on the oddness of $g^B$ in Lemma \ref{lem:final 6}, we have the crucial cancellation 
\begin{align}\label{final 06}
    \varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}},p\m^{\frac{1}{2}}+c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}=0.
\end{align}
Then using a similar argument as estimating \eqref{pp 103'} and Lemma \ref{lem:final 5}, we know 
\begin{align}
    &\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}}+\textbf{d}\cdot\mathscr{A}, p\m^{\frac{1}{2}}\Big]}}+\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}\\
    &+\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{b}\cdot v\m^{\frac{1}{2}}\Big]}}+\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},\textbf{d}\cdot \mathscr{A}\Big]}}\nonumber\\
    &+\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{P}[R]+\textbf{d}\cdot\mathscr{A},(\textbf{I}-\overline{\textbf{P}})[R]\Big]}}
    +\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[(\textbf{I}-\overline{\textbf{P}})[R],(\textbf{I}-\overline{\textbf{P}})[R]\Big]}}\nonumber\\
    \lesssim&\varepsilon^{\alpha-1}\lnmm{g^B}\Big(\varepsilon^{-\frac{1}{2}}\tnm{p}+\varepsilon^{-\frac{1}{2}}\tnm{\textbf{b}}+\varepsilon^{-\frac{1}{2}}\tnm{\textbf{d}}+\varepsilon^{-\frac{1}{2}}\tnm{(\textbf{I}-\overline{\textbf{P}})[R]}\Big)^2\nonumber\\
    \lesssim& o_{T} \jnm{c}^{r}+o_{T}\varepsilon^{\alr-r}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+o_{T}.\nonumber
\end{align}
Then all terms in $\abs{\varepsilon^{-1}\bbr{g^B,S_2}}$ have been bounded except
\begin{align}\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}+\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}.
\end{align}
Note that
\begin{align}\label{qq 01}
    &\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}=\varepsilon^{\alpha-2}\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b} c}\\
    =&-\varepsilon^{\alpha-2}\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}(Z^R-\varepsilon^{-1}\xi)}
    +\varepsilon^{\alpha-2}\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}Z^S}.\nonumber
\end{align}
For the first term in \eqref{qq 01}, noting \eqref{qq 24} and $\tnm{\textbf{b}}\lesssim\varepsilon^{\frac{1}{2}}\xnm{R}$ and using Hardy's inequality, \eqref{qq 06} and  Lemma \ref{lem:final 5}, we have
\begin{align}\label{qq 03}
    &\abs{\varepsilon^{\alpha-2}\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}(Z^R-\varepsilon^{-1}\xi)}}\\
    =&\varepsilon^{\alpha-2}\abs{\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}(Z^R-\varepsilon^{-1}\xi)}}\nonumber\\
    =&\varepsilon^{\alpha-2}\abs{\brx{\brv{\mng^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}(Z^R-\varepsilon^{-1}\xi)}}\nonumber\\
    =&\varepsilon^{\alpha-1}\abs{\brx{\brv{\etag^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b}\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}(Z^R-\varepsilon^{-1}\xi)}}\nonumber\\
    \lesssim&\varepsilon^{\alpha-1}\lnm{\etag^B}\tnm{\textbf{b}}\tnm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}(Z^R-\varepsilon^{-1}\xi)}\lesssim\varepsilon^{\alpha-\frac{1}{2}}\lnms{\mathfrak{g}}{\gamma_-}\xnm{R}\Big(\nm{Z^R}_{H^1}+\varepsilon^{-1}\nm{\xi}\Big)\nonumber\\
    \lesssim&o_{T}\jnm{c}^{r}+\xnm{R}^{r}\Big(o_{T}\varepsilon^{r\alpha-\frac{r}{2}}\xnm{R}^{r}+\varepsilon^{2r\alpha-\frac{3r}{2}}\xnm{R}^{2r}+o_{T}\varepsilon^{\frac{r}{2}}\Big).\nonumber
\end{align}
For the second term in \eqref{qq 01}, using  Lemma \ref{lem:final 5} and \eqref{qq 07}, we have
\begin{align}\label{qq 09}
    &\abs{\varepsilon^{\alpha-2}\brx{\brv{g^B,\Gamma\Big[v\m^{\frac{1}{2}},\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]},\textbf{b} Z^S}}\\
    \lesssim&\varepsilon^{\alpha-2}\lnm{g^B}\tnm{\textbf{b}}\tnm{Z^S}\lesssim\varepsilon^{\alpha-\frac{3}{2}}\lnms{\mathfrak{g}}{\gamma_-}\xnm{R}\tnm{Z^S}\nonumber\\
    \lesssim&o_{T}\jnm{c}^{r}+\xnm{R}^{r}\Big(o_{T}\varepsilon^{r\alpha-r}\xnm{R}^{r}+\varepsilon^{2r\alpha-2r}\xnm{R}^{2r}+o_{T}\Big).\nonumber
\end{align}
Inserting \eqref{qq 03} and \eqref{qq 09} into \eqref{qq 01}, we arrive at
\begin{align}
\\
    \abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{b}\cdot v\m^{\frac{1}{2}},c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}}
    \lesssim&o_{T}\jnm{c}^{r}+\xnm{R}^{r}\Big(o_{T}\varepsilon^{r\alpha-r}\xnm{R}^{r}+\varepsilon^{2r\alpha-2r}\xnm{R}^{2r}+o_{T}\Big).\nonumber
\end{align}
A similar argument justifies that
\begin{align}
\\
    &\abs{\varepsilon^{\alpha-2}\br{g^B,\Gamma\Big[\textbf{d}\cdot \mathscr{A},c\left(\abs{v}^2-5T\right)\m^{\frac{1}{2}}\Big]}}\lesssim
    o_{T}\jnm{c}^{r}+\xnm{R}^{r}\Big(o_{T}\varepsilon^{r\alpha-r}\xnm{R}^{r}+\varepsilon^{2r\alpha-2r}\xnm{R}^{2r}+o_{T}\Big).\nonumber
\end{align}
In summary, using Lemma \ref{remark 01}, we know
\begin{align}
    &\abs{\varepsilon^{-1}\bbr{g^B,S_2}}\lesssimo_{T} \jnm{c}^{r}+o_{T}\varepsilon^{\alr-r}\xnm{R}^{r}+\varepsilon^{\alr-r}\xnm{R}^{2r}+\varepsilon^{2r\alpha-2r}\xnm{R}^{3r}+o_{T}.
\end{align}

\paragraph{\underline{$\varepsilon^{-1}\bbr{g,\barS}$ Contribution}}
Based on Lemma \ref{lem:final 7}, due to orthogonality, we have
\begin{align}
    \varepsilon^{-1}\bbr{g,\barS}=\varepsilon^{-1}\bbr{g,S_3}+\varepsilon^{-1}\bbr{g,S_4}+\varepsilon^{-1}\bbr{g,\ss_6}.
\end{align}
Using Lemma \ref{lem:final 7}, Lemma \ref{ss4-estimate} and Lemma \ref{ssp-estimate}, we have
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g,S_4+\ss_6}}\lesssim&\varepsilon^{-1}\tnm{g}\Big(\tnm{S_4}+\tnm{\ss_6}\Big)\lesssimo_{T}\varepsilon^{-1}\jnm{c}^{r-1}\Big(\tnm{S_4}+\tnm{\ss_6}\Big)\\
    \lesssim& o_{T}\jnm{c}^{r}+\varepsilon^{-r}\tnm{S_4}^{r}+\varepsilon^{-r}\tnm{\ss_6}^{r}\lesssim o_{T} \jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
Considering the rescaling $\eta=\varepsilon^{-1}\mathfrak{n}$, we have
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g,S_3}}\lesssim&\varepsilon^{-1}\abs{\bbr{g(0)+\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}g,S_3}}
    \lesssim\varepsilon^{-1}\abs{\bbr{g(0),S_3}}+\varepsilon^{-1}\abs{\bbr{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}g(y)\mathrm{d} y,\mnS_3}}\\
    \lesssim&\varepsilon^{-1}\tnm{g(0)}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\tnm{\frac{1}{\mathfrak{n}}\int_0^{\mathfrak{n}}\partial_{\mathfrak{n}}g(y)\mathrm{d} y}\nm{\eta\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_xL^1_v}\nonumber\\
    \lesssim&\varepsilon^{-1}\tnm{g(0)}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}+\nm{g}_{H^1}\nm{\eta\big(\ss_{3b}+\ss_{3c}+f^{B}_1\big)}_{L^2_xL^1_v}\nonumber\\
    \lesssim& o_{T} \jnm{c}^{r}+\varepsilon^{-r}\nm{S_3}_{L^2_{\iota_i}L^1_{\mathfrak{n}}L^1_v}^{r}+\nm{\eta\big(\ss_{3b}+\ss_{3c}\big)}_{L^2_xL^1_v}^{r}+\nm{\etaf^{B}_1}_{L^2_xL^1_v}^{r}\lesssim o_{T} \jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.\nonumber
\end{align}
We conclude that
\begin{align}
    \abs{\varepsilon^{-1}\bbr{g,\barS}}
    \lesssim&o_{T}\jnm{c}^{r}+o_{T}\varepsilon^{r-r\alpha}.
\end{align}
Collecting the above estimates, we arrive at \eqref{pp 99'''}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:c-bound}]
Inserting \eqref{pp 44'}\eqref{pp 44''}\eqref{pp 99'''} into \eqref{pp 29}, we have the desired result.
\end{proof}


\section{$L^{\infty}$ Estimates}\label{sec:linfty estimate}


In this section, we plan to prove Proposition \ref{thm:infinity}.


\subsection{Stochastic Cycle}


We will rely on the reformulation \eqref{ll 01}. Due to our choice of $\mu_M$, we have
for any $a<1$
\begin{align}\label{ll 02}
    \mu_M\lesssim\mu\lesssim\mu_M^{a}.
\end{align}
From the first inequality in \eqref{ll 02}, we further know
\begin{align}\label{ll 03}
    \mm^{-\frac{1}{2}}\m^{\frac{1}{2}}\gs1.
\end{align}

\begin{definition}[Exit Time and Position]\label{exit}
For any $(x,v)\in\overline\Omega\times\mathbb{R}^3$ with $v\neq\textbf{0}$ and $(x,v)\notin\gamma_0$, define the backward exit time $t_b(x,v):=\inf\{t>0:x-\varepsilon tv\notin\Omega\}$. Also, define the backward exit position $x_b(x,v):=x-\varepsilon t_b(x,v)v\in\partial\Omega$.
\end{definition}

\begin{definition}[Stochastic Cycle]
For any $(x,v)\in\overline\Omega\times\mathbb{R}^3$ with $v\neq\textbf{0}$ and $(x,v)\notin\gamma_0$, let $(t_0,x_0,v_0)=(0,x,v)$. Define the first stochastic triple
\begin{align}
(t_1,x_1,v_1):=\Big(t_b(x_0,v_0),x_b(x_0,v_0),v_1\Big),
\end{align}
for some $v_1$ satisfying $v_1\cdotn(x_1)>0$.

Inductively, assume we know the $k^{th}$ stochastic triple $(t_k,x_k,v_k)$. Define the $(k+1)^{th}$ stochastic triple
\begin{align}
(t_{k+1},x_{k+1},v_{k+1}):=\Big(t_k+t_b(x_k,v_k),x_b(x_k,v_k),v_{k+1}\Big),
\end{align}
for some $v_{k+1}$ satisfying $v_{k+1}\cdotn(x_{k+1})>0$.
\end{definition}

\begin{definition}[Diffuse Reflection Integral]
Define $\mathcal{V}_{k}=\{v\in \mathbb{R}^3:v\cdotn(x_{k})>0\}$, so the stochastic cycle must satisfy $v_k\in\mathcal{V}_k$. Let the iterated integral for $k\geq2$ be defined as
\begin{align}
\int_{\prod_{j=1}^{k-1}\mathcal{V}_j}\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}:=\int_{\mathcal{V}_1}\ldots\bigg(\int_{\mathcal{V}_{k-1}}\mathrm{d}{\sigma_{k-1}}\bigg)\ldots\mathrm{d}{\sigma_1}
\end{align}
where $\mathrm{d}{\sigma_j}:=\mu_M(v_j)\abs{v_j\cdotn(x_j)}\mathrm{d}{v_j}$.
\end{definition}

We define a weight function scaled with parameters $0\leq\varrho<\dfrac{1}{2}$ and $\vartheta\geq0$,
\begin{align}\label{ktt 56}
w(v):=\br{\vv}^{\vth}\ue^{\varrho\frac{\abs{\vv}^2}{2T_M}},
\end{align}
\begin{align}\label{ktt 57}
\widetilde{w}(v):=\frac{\mu_M^{\frac{1}{2}}(v)}{w(v)}
=(2\pi)^{-\frac{3}{4}}T_M^{-\frac{5}{4}}\br{v}^{-\vartheta}\mathrm{e}^{-\left(\frac{1}{4T_M}+\frac{\varrho}{2T_M}\right)\abs{v}^2}
\end{align}
\begin{align}
\widetilde{w}_M(v):=\frac{1}{w(v)m_{M,w}(v)}
=(2\pi)^{\frac{1}{4}}T_w^{2}T_M^{-\frac{5}{4}}\br{v}^{-\vartheta}\mathrm{e}^{\left(\frac{1}{2T_w}-\frac{1}{4T_M}-\frac{\varrho}{2T_M}\right)\abs{v}^2}.
\end{align}

\begin{lemma}\label{ktt lemma 2}
For $\mathfrak{T}>0$ sufficiently large, there exist constants $C_1,C_2>0$ independent of $\mathfrak{T}$, such that for $k=C_1\mathfrak{T}^{\frac{5}{4}}$, and
$(x,v)\in\times\overline\Omega\times\mathbb{R}^3$,
\begin{align}\label{ktt 59}
\int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\{t_k(x,v,v_1,\ldots,v_{k-1})<\frac{\mathfrak{T}}{\varepsilon}\}}\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}\leq
\left(\frac{1}{2}\right)^{C_2\mathfrak{T}^{\frac{5}{4}}}.
\end{align}
\end{lemma}
\begin{proof}
This is a rescaled version of \cite[Lemma 4.1]{Esposito.Guo.Kim.Marra2013}. Choosing $\mathfrak{y}>0$ sufficiently small, we further define the non-grazing set for $1\leq j\leq k-1$ as 
\begin{align}
    \mathcal{V}_j^{\mathfrak{y}}=\Big\{v\in\mathcal{V}_j: v\cdot n(x_j)>\mathfrak{y}\Big\}\bigcup \Big\{v\in\mathcal{V}_j: \abs{v}<\frac{1}{\mathfrak{y}}\Big\}.
\end{align}
Then clearly, we have for some general $C>0$ independent of $j$, $\displaystyle\int_{\mathcal{V}_j\backslash\mathcal{V}_j^{\mathfrak{y}}}\leq C\mathfrak{y}$.
From the stochastic cycle, we know for $v\in\mathcal{V}_j^{\mathfrak{y}}$, if $t_k(x,v,v_1,\ldots,v_{k-1})<\frac{\mathfrak{T}}{\varepsilon}$, then there can be at most $\left[\dfrac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1$ number of $v_j\in\mathcal{V}_j^{\mathfrak{y}}$. Hence, we have
\begin{align}
    \int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\{t_k(x,v,v_1,\ldots,v_{k-1})<\frac{\mathfrak{T}}{\varepsilon}\}}\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}
    \leq&\sum_{m=1}^{\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1}\begin{pmatrix}k-1\\mu\end{pmatrix}\abs{\sup_j\int_{\mathcal{V}_j^{\mathfrak{y}}}\mathrm{d}\sigma_j}^m\abs{\sup_j\int_{\mathcal{V}_j\backslash\mathcal{V}_j^{\mathfrak{y}}}\mathrm{d}\sigma_j}^{k-m-1}.
\end{align}
Note that $\mathrm{d}\sigma_j$ is not exactly a probability measure, but the $(2\pi T_M)^{-\frac{1}{2}}$ multiple of a probability measure. Hence, we have
\begin{align}
    \abs{\sup_j\int_{\mathcal{V}_j\backslash\mathcal{V}_j^{\mathfrak{y}}}\mathrm{d}\sigma_j}^{k-m-1}
    \leq& \abs{\sup_j\int_{\mathcal{V}_j\backslash\mathcal{V}_j^{\mathfrak{y}}}\mathrm{d}\sigma_j}^{k-\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]-2}(2\pi T_M)^{-\frac{k}{2}-\frac{1}{2}\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]-1}\\
    \leq&\big(C\mathfrak{y}\big)^{k-\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]-2}(2\pi T_M)^{-\frac{k}{2}-\frac{1}{2}\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]-1}.\nonumber
\end{align}
Hence, considering that 
\begin{align}
    \begin{pmatrix}k-1\\mu\end{pmatrix}\leq\big(k-1\big)^m\leq\big(k-1\big)^{\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1},
\end{align}
we have
\begin{align}
    \int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\{t_k(x,v,v_1,\ldots,v_{k-1})<\frac{\mathfrak{T}}{\varepsilon}\}}\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}
    \leq&\left(\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1\right)\big(k-1\big)^{\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1}\big(C\mathfrak{y}\big)^{k-\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]-2}(2\pi T_M)^{-\frac{k}{2}}.
\end{align}
Let $k-2=N\left(\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1\right)$ for some $N$ to be determined later. Then if $\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]\geq1$, we have for given $T_M$
\begin{align}
    \int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\{t_k(x,v,v_1,\ldots,v_{k-1})<\frac{\mathfrak{T}}{\varepsilon}\}}\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}
    \leq&\left\{N\left(\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1\right)\big(C\mathfrak{y}\big)^N(2\pi T_M)^{-\frac{N}{2}}\right\}^{\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1}\\
    \leq& \left\{C_{N,\Omega}\mathfrak{T}\mathfrak{y}^{N-3}\right\}^{\left[\frac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1}.\nonumber
\end{align}
We choose $\mathfrak{y}=\left(\dfrac{1}{2C_{N,\Omega}\mathfrak{T}}\right)^{\frac{1}{N-3}}$, which yields $C_{N,\Omega}\mathfrak{T}\mathfrak{y}^{N-3}=\dfrac{1}{2}$. Then we know for $\mathfrak{T}$ sufficiently large $\left[\dfrac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\right]+1\approx C_{N,\Omega}\mathfrak{T}^{1+\frac{3}{N-3}}$
and $\dfrac{C_{\Omega}\mathfrak{T}}{\mathfrak{y}^3}\geq2$.

Finally, we choose $N=15$. Then our result follows.
\end{proof}

Based on \cite{Guo.Jang.Jiang2010}, using the second inequality in \eqref{ll 02}, we can further decompose $K_M=K_M^m+K_M^c$, where
\begin{align}
    K_M^m[\re_M]:=&\iint_{\mathbb{R}^3\times\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})\chi_m(\abs{{\mathfrak{u}}-v})\mu_M^{\frac{1}{2}}({\mathfrak{u}})\mu(v)\mu_M^{-\frac{1}{2}}(v)\re_M({\mathfrak{u}})\mathrm{d}\omega\mathrm{d}{\mathfrak{u}}\\
    &-\iint_{\mathbb{R}^3\times\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})\chi_m(\abs{{\mathfrak{u}}-v})\mu_M^{\frac{1}{2}}({\mathfrak{u}}^{\ast})\mu(v^{\ast})\mu_M^{-\frac{1}{2}}(v)\re_M(v^{\ast})\mathrm{d}\omega\mathrm{d}{\mathfrak{u}}\nonumber\\
    &-\iint_{\mathbb{R}^3\times\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})\chi_m(\abs{{\mathfrak{u}}-v})\mu_M^{\frac{1}{2}}(v^{\ast})\mu({\mathfrak{u}}^{\ast})\mu_M^{-\frac{1}{2}}(v)\re_M({\mathfrak{u}}^{\ast})\mathrm{d}\omega\mathrm{d}{\mathfrak{u}},\nonumber
\end{align}
and $K_M^c:=K_M-K_M^m$. Here $\chi_m=\chi(my)$ is a smooth cutoff function (see Section \ref{sec:boundary-form}) for any $m>0$. Based on \cite[Lemma 2.3]{Guo.Jang.Jiang2010}, when $m$ is sufficiently small, $K_M^m$ satisfies
\begin{align}\label{ll 04}
    \abs{K_M^m[\re_M]}\lesssim o(1)\nu_M\lnm{\re_M},
\end{align}
and $K_M^c$ has kernel $l_M({\mathfrak{u}},v)$
\begin{align}
    K_M^c[\re_M]=\int_{\mathbb{R}^3}l_M({\mathfrak{u}},v)\re_M({\mathfrak{u}})\mathrm{d}{\mathfrak{u}},
\end{align}
which satisfies the following:

\begin{lemma}\label{Regularity lemma 1'}
We have
\begin{align}
    \abs{l_M({\mathfrak{u}},v)}\lesssim\left(\abs{{\mathfrak{u}}-v}+\abs{{\mathfrak{u}}-v}^{-1}\right)\mathrm{e}^{-\frac{1}{4T_M}\abs{{\mathfrak{u}}-v}^2-\frac{1}{4T_M}\frac{\abs{\abs{{\mathfrak{u}}}^2-\abs{v}^2}^2}{\abs{{\mathfrak{u}}-v}^2}}.
\end{align}
Let $0\leq\varrho< \dfrac{1}{2}$ and $\vartheta\geq0$. Then for $\delta>0$ sufficiently small (depending on $T_M$) and any $v\in\mathbb{R}^3$,
\begin{align}
\int_{\mathbb{R}^3}\mathrm{e}^{\delta\abs{{\mathfrak{u}}-v}^2}\abs{l_M({\mathfrak{u}},v)}
\frac{\br{v}^{\vartheta}\mathrm{e}^{\varrho\frac{\abs{v}^2}{2T_M}}}{\br{{\mathfrak{u}}}^{\vartheta}\mathrm{e}^{\varrho\frac{\abs{{\mathfrak{u}}}^2}{2T_M}}}\mathrm{d}{{\mathfrak{u}}}
\lesssim \nu_M^{-1}.
\end{align}
\end{lemma}
\begin{proof}
This is a combination of  \cite[Lemma 3]{Guo2010} and \cite[Lemma 2.3]{Guo.Jang.Jiang2010}.
\end{proof}


\subsection{Mild Formulation}


Denote the weighted solution
\begin{align}
\overline{\rem}(x,v):=&w(v) \re_M(x,v),
\end{align}
and the weighted non-local operator
\begin{align}\label{final 66}
K^c_{M,w(v)}\left[\overline{\rem}\right](v):=&w(v)K^c_M\left[\frac{\overline{\rem}}{w}\right](v)=\int_{\mathbb{R}^3}l_{M,w(v)}({\mathfrak{u}},v)\overline{\rem}({\mathfrak{u}})\mathrm{d}{{\mathfrak{u}}},
\end{align}
where
\begin{align}
l_{M,w(v)}({\mathfrak{u}},v):=l_M({\mathfrak{u}},v)\frac{w(v)}{w({\mathfrak{u}})}.
\end{align}
Multiplying $\varepsilonw(v)$ on both sides of \eqref{ll 01}, we have
\begin{align}\label{ktt 64}
\\
\left\{
\begin{array}{l}
\varepsilonv\cdot\nabla_{x} \overline{\rem}+\nu_M \overline{\rem}=K^c_{M,w}\left[\overline{\rem}\right](x,v)+w(v)\widetilde S_M(x,v)\ \ \text{in}\ \ \Omega\times\mathbb{R}^3,\\\rule{0ex}{2em}
\overline{\rem}(x_0,v)=\displaystylew(v)m_{M,w}(x_0,v)\int_{{\mathfrak{u}}\cdotn>0}\widetilde{w}({\mathfrak{u}})\overline{\rem}(x_0,{\mathfrak{u}})\mathrm{d}{\mathfrak{u}}+w(v) h_M(x_0,v)\ \ \text{for}\ \ x_0\in\partial\Omega\ \
\text{and}\ \ v\cdotn<0,
\end{array}
\right.\nonumber
\end{align}
where
\begin{align}
    \widetilde S_M(x,v)=\varepsilon S_M(x,v)+K_M^m[\re_M].
\end{align}
We can rewrite the solution of the equation \eqref{ktt 64} along the characteristics by Duhamel's principle as
\begin{align}
\overline{\rem}(x,v)=& \mathrm{e}^{-\int_0^{t_1}\nu_M(v)}\bigg(w(v)h_M(x_1,v)+\int_{0}^{t_{1}}w(v) \widetilde S_M\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{\int_0^{s}\nu_M(v)}\mathrm{d}{s}\bigg)\\
&+\int_{0}^{t_{1}}K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
}\mathrm{d}{s}+\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)
}}{\widetilde{w}_M(v)}\int_{\mathcal{V}_1}\overline{\rem}(x_1,v_1)\widetilde{w}(v_1)\mathrm{d}{\sigma_1}.\nonumber
\end{align}
We can further rewrite the last term using \eqref{ktt 64} along the stochastic cycle by applying Duhamel's principle $k$ times as
\begin{align}\label{ktt 65}
\overline{\rem}(x,v)=& \mathrm{e}^{-\int_0^{t_1}\nu_M(v)}\bigg(w(v)h_M(x_1,v)+\int_{0}^{t_{1}}w(v) \widetilde S_M\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{\int_0^{s}\nu_M(v)}\mathrm{d}{s}\bigg)\\
&+\int_{0}^{t_{1}}K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)}\mathrm{d}{s}\nonumber\\
&+\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}\Big(G_{\ell}\big[x,v\big]+H_{\ell}\big[x,v\big]\Big)\widetilde{w}(v_\ell)
\bigg(\prod_{j=1}^{\ell}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_j)}\mathrm{d}{\sigma_j}\bigg)\nonumber\\
&+\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\prod_{j=1}^{k}\mathcal{V}_j}\overline{\rem}(x_k,v_k)\widetilde{w}(v_{k})
\bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg),\nonumber
\end{align}
where
\begin{align}\label{ktt 66}
G_{\ell}\big[x,v\big]:=&w(v_\ell)h_M(x_{\ell+1},v_{\ell})+\int_{t_{\ell}}^{t_{\ell+1}}\bigg(
w(v_\ell)\widetilde S_M\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell},v_{\ell}\Big)\mathrm{e}^{\int_{t_{\ell}}^{s}\nu_M(v_\ell)}\bigg)\mathrm{d}{s},\\
H_{\ell}\big[x,v\big]:=&\int_{t_{\ell}}^{t_{\ell+1}}\bigg(K^c_{M,w(v_{\ell})}\left[\overline{\rem}\right]\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell},v_{\ell}\Big)\mathrm{e}^{\int_{t_{\ell}}^{s}\nu_M(v_{\ell})}\bigg)\mathrm{d}{s}.\label{ktt 67}
\end{align}
We set $k=C_1\mathfrak{T}^{\frac{5}{4}}$ for $\mathfrak{T}$ defined in Lemma \ref{ktt lemma 2}. Then we need to estimate each term in \eqref{ktt 65}.


\subsection{Estimates of Source Terms and Boundary Terms}


\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for given $\mathfrak{T}>0$, we have
    \begin{align}\label{ktt 74}
    \text{Boundary Term Contribution}\lesssim \mathfrak{T}^{\frac{5}{4}}\ssm{w h_M}{\gamma_-}\lesssim \ssm{w h_M}{\gamma_-},
    \end{align}
    and
    \begin{align}\label{ktt 75}
    \text{Source Term Contribution}\lesssim \mathfrak{T}^{\frac{5}{4}}\lnm{\nu_M^{-1}w \widetilde S_M}\lesssim \lnm{\nu_M^{-1}w \widetilde S_M}.
    \end{align}
\end{lemma}

\begin{proof}
Since $t_1\geq0$, we have
\begin{align}\label{ktt 68}
\abs{w(v)h_M(x_1,v)\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}\leq \ssm{w h_M}{\gamma_-}.
\end{align}
Also,
\begin{align}\label{ktt 69}
\abs{\int_{0}^{t_{1}}w(v) \widetilde S_M\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)}\mathrm{d}{s}}&\leq \lnm{\nu_M^{-1}w \widetilde S_M}\abs{\int_{0}^{t_{1}}\nu_M(v)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)}\mathrm{d}{s}}\\
&\leq \lnm{\nu_M^{-1}w \widetilde S_M}.\nonumber
\end{align}
Then we turn to terms related to $G_{\ell}$ defined in \eqref{ktt 66}. Noting that $\dfrac{1}{\widetilde{w}_M}\lesssim 1$, we know
\begin{align}\label{ktt 70}
&\abs{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}w(v_\ell)h_M(x_{\ell+1},v_{\ell})\widetilde{w}(v_\ell)
\bigg(\prod_{j=1}^{\ell}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_j)}\mathrm{d}{\sigma_j}\bigg)}\\
\lesssim&\ssm{w h_M}{\gamma_-}\abs{\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}\widetilde{w}(v_\ell)\prod_{j=1}^{\ell}\mathrm{d}{\sigma_j}}
\lesssim\ssm{w h_M}{\gamma_-}\abs{\sum_{\ell=1}^{k-1}\int_{\mathcal{V}_{\ell}}\widetilde{w}(v_\ell)\mathrm{d}{\sigma_\ell}}\lesssim \mathfrak{T}^{\frac{5}{4}}\ssm{w h_M}{\gamma_-}.\nonumber
\end{align}
Similarly,
\begin{align}\label{ktt 71}
\\
&\abs{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}(v)}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}\bigg(\int_{t_{\ell}}^{t_{\ell+1}}
w(v_\ell)\widetilde S_M\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell},v_{\ell}\Big)\mathrm{e}^{\int_{t_{\ell}}^{s}\nu_M(v_\ell)}\mathrm{d}{s}\bigg)\widetilde{w}(v_\ell)
\bigg(\prod_{j=1}^{\ell}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_j)}\mathrm{d}{\sigma_j}\bigg)}\nonumber\\
\lesssim&\lnm{\nu^{-1}w \widetilde S_M}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}\bigg(\int_{t_{\ell}}^{t_{\ell+1}}\abs{
\nu_M(v_\ell)\mathrm{e}^{\int^{t_{\ell+1}}_{s}\nu_M(v_\ell)}\mathrm{d}{s}}\widetilde{w}(v_\ell)
\prod_{j=1}^{\ell}\mathrm{d}{\sigma_j}\bigg)\lesssim \mathfrak{T}^{\frac{5}{4}}\lnm{\nu_M^{-1}w \widetilde S_M}.\nonumber
\end{align}
Collecting all terms in \eqref{ktt 68}, \eqref{ktt 69}, \eqref{ktt 70} and \eqref{ktt 71}, we have the desired result.
\end{proof}


\subsection{Estimates of Multiple Reflections Terms}


We focus on the last term in \eqref{ktt 65}.
\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, for given $\mathfrak{T}>0$, we have for some $0<\delta\ll1$
    \begin{align}\label{ktt 76}
    \text{Multiple Reflection Term Contribution}\lesssim \delta \lnm{\overline{\rem}}.
    \end{align}
\end{lemma}

\begin{proof}
The last term in \eqref{ktt 65} can be decomposed based on accumulative time $t_{k+1}$:
\begin{align}
&\abs{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\prod_{j=1}^{k}\mathcal{V}_j}\overline{\rem}(x_k,v_k)\widetilde{w}(v_{k})
\bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}\\
\leq&\abs{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\Pi_{j=1}^{k}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\leq
\frac{\mathfrak{T}}{\varepsilon}\right\}}\overline{\rem}(x_k,v_k)\widetilde{w}(v_{k})
\bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}\nonumber\\
&+\abs{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\Pi_{j=1}^{k}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\geq
\frac{\mathfrak{T}}{\varepsilon}\right\}}\overline{\rem}(x_k,v_k)\widetilde{w}(v_{k})
\bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}:=J_1+J_2.\nonumber
\end{align}
Based on Lemma \ref{ktt lemma 2}, we have
\begin{align}\label{ktt 72}
J_1\lesssim&\lnm{\overline{\rem}}\abs{\int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\leq
\frac{\mathfrak{T}}{\varepsilon}\right\}}\bigg(\int_{\mathcal{V}_k}\widetilde{w}(v_{k})\mathrm{d}\sigma_k\bigg)
\bigg(\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}\bigg)}\\
\lesssim&\lnm{\overline{\rem}}\abs{\int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\leq
\frac{\mathfrak{T}}{\varepsilon}\right\}}
\bigg(\prod_{j=1}^{k-1}\mathrm{d}{\sigma_j}\bigg)}\lesssim \bigg(\frac{1}{2}\bigg)^{C_2\mathfrak{T}^{\frac{5}{4}}}\lnm{\overline{\rem}}.\nonumber
\end{align}
On the other hand, when $t_k$ is large, the exponential terms become extremely small, so we obtain
\begin{align}\label{ktt 73}
J_2\lesssim&\lnm{\overline{\rem}}\abs{\mathrm{e}^{-\nu_M(v)t_{1}}\int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\geq
\frac{\mathfrak{T}}{\varepsilon}\right\}}\bigg(\int_{\mathcal{V}_k}\widetilde{w}(v_{k})\mathrm{d}\sigma_k\bigg)
\bigg(\prod_{j=1}^{k-1}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}\\
\lesssim&\lnm{\overline{\rem}}\abs{\mathrm{e}^{-\nu_M(v)t_{1}}\int_{\Pi_{j=1}^{k-1}\mathcal{V}_j}\textbf{1}_{\left\{t_{k+1}\geq
\frac{\mathfrak{T}}{\varepsilon}\right\}}
\bigg(\prod_{j=1}^{k-1}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}\lesssim \mathrm{e}^{-\frac{\mathfrak{T}}{\varepsilon}}\lnm{\overline{\rem}}.\nonumber
\end{align}
Summarizing \eqref{ktt 72} and \eqref{ktt 73}, we have the desired result. 
\end{proof}


\subsection{Estimate of Non-Local Terms}


The only remaining terms in \eqref{ktt 65} are related to $K^c_{M,w(v)}$ and $H_{\ell}$ defined in \eqref{ktt 67}. 
\begin{lemma}
    Under the assumption \eqref{assumption:boundary}, we have for some $0<\delta\ll1$
    \begin{align}\label{ktt 93}
    K^c_{M,w}\ \text{term contribution}\lesssim& \delta\lnm{\overline{\rem}}+C(\delta)\varepsilon^{-\frac{1}{2}}\pnm{R}{6}+\ssm{w h_M}{\gamma_-}+\lnm{\nu_M^{-1}w \widetilde S_M}.
    \end{align}
\end{lemma}

\begin{proof}
We focus on
\begin{align}
\abs{\int_{0}^{t_{1}}K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_{s}^{t_1}\nu_M(v)}\mathrm{d}{s}}&\lesssim \lnm{K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)}.
\end{align}
Denote $X(s;x,v):=x-\varepsilon(t_1-s)v$. Notice that $K^c_{M,w(v)}$ defined in \eqref{final 66} is a non-local operator which contains an integral with respect to ${\mathfrak{u}}$. Define the back-time stochastic cycle from $(X,{\mathfrak{u}})$ as $(t_i',x_i',v_i')$ for $i=0,1,2,\cdots$ with $(t_0',x_0',v_0')=(0,X,{\mathfrak{u}})$. Then we can rewrite $K^c_{M,w}$ along the stochastic cycle as \eqref{ktt 65}
\begin{align}\label{ktt 77}
&\abs{K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)}=\abs{K^c_{M,w(v)}\left[\overline{\rem}\right](X,v)}\\
=&\abs{\int_{\mathbb{R}^3}l_{M,w(v)}(v,{\mathfrak{u}})\overline{\rem}(X,{\mathfrak{u}})\mathrm{d}{{\mathfrak{u}}}}\nonumber\\
\leq&\abs{\int_{\mathbb{R}^3}\int_{0}^{t_1'}l_{M,w(v)}(v,{\mathfrak{u}})K^c_{M,w({\mathfrak{u}})}\left[\overline{\rem}\right]\Big(X-\varepsilon(t_1'-r){\mathfrak{u}},{\mathfrak{u}}\Big)\mathrm{e}^{-\int_r^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}}\nonumber\\
&+\abs{\int_{\mathbb{R}^3}\frac{\mathrm{e}^{-\int_0^{t_1'}\nu_M({\mathfrak{u}})}}{\widetilde{w}({\mathfrak{u}})}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j'}l_{M,w(v)}(v,{\mathfrak{u}})H_{\ell}\big[X,{\mathfrak{u}}\big]\widetilde{w}(v_{\ell}')
\bigg(\prod_{j=1}^{\ell}\mathrm{e}^{-\int_{t_j'}^{t_{j+1}'}\nu_M(v_j')}\mathrm{d}{\sigma_j'}\bigg)\mathrm{d}{{\mathfrak{u}}}}\nonumber\\
&+\abs{\int_{\mathbb{R}^3}l_{M,w(v)}(v,{\mathfrak{u}})\Big(\text{boundary terms + source terms + multiple reflection terms}\Big)\mathrm{d}{{\mathfrak{u}}}}\nonumber\\
:=&I+II+III.\nonumber
\end{align}
Using estimates \eqref{ktt 74}, \eqref{ktt 75}, \eqref{ktt 76} from previous lemmas, we can bound $III$ directly
\begin{align}\label{ktt 78}
III\lesssim \ssm{w h_M}{\gamma_-}+\lnm{\nu_M^{-1}w \widetilde S_M}+\delta \lnm{\overline{\rem}}.
\end{align}
$I$ and $II$ are much more complicated. We can further rewrite $I$ as
\begin{align}
I=&\abs{\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\int_{0}^{t_1'}l_{M,w(v)}(v,{\mathfrak{u}})l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})\overline{\rem}\Big(X-\varepsilon(t_1'-r){\mathfrak{u}},{\mathfrak{u}}^{\ast}\Big)\mathrm{e}^{-\int_r^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}},
\end{align}
which will estimated in five cases:
\begin{align}
I:=I_1+I_2+I_3+I_4+I_5.
\end{align}

\paragraph{\underline{Case I: $I_1:$ $\abs{v}\geq N$}}
Based on Lemma \ref{Regularity lemma 1'}, we have
\begin{align}
\abs{\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}l_{M,w(v)}(v,{\mathfrak{u}})l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}}\lesssim\frac{1}{\nu_M(v)}\lesssim\frac{1}{N}.
\end{align}
Hence, we get
\begin{align}\label{ktt 87}
\babs{I_1}\lesssim\frac{1}{N}\lnm{\overline{\rem}}.
\end{align}

\paragraph{\underline{Case II: $I_2:$ $\abs{v}\leq N$, $\abs{{\mathfrak{u}}}\geq2N$, or $\abs{{\mathfrak{u}}}\leq
2N$, $\abs{{\mathfrak{u}}^{\ast}}\geq3N$}}
Notice this implies either $\abs{{\mathfrak{u}}-v}\geq N$ or
$\abs{{\mathfrak{u}}-{\mathfrak{u}}^{\ast}}\geq N$. Hence, either of the following is valid
correspondingly:
\begin{align}
\abs{l_{M,w(v)}(v,{\mathfrak{u}})}\leq& C\mathrm{e}^{-\delta N^2}\abs{l_{M,w(v)}(v,{\mathfrak{u}})}\mathrm{e}^{\delta\abs{v-{\mathfrak{u}}}^2},\\
\abs{l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})}\leq& C\mathrm{e}^{-\delta N^2}\abs{l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})}\mathrm{e}^{\delta\abs{{\mathfrak{u}}-{\mathfrak{u}}^{\ast}}^2}.
\end{align}
Based on Lemma \ref{Regularity lemma 1'} for $l_{M}$, we know
\begin{align}
\int_{\mathbb{R}^3}\abs{l_{M,w(v)}(v,{\mathfrak{u}})}\mathrm{e}^{\delta\abs{v-{\mathfrak{u}}}^2}\mathrm{d}{{\mathfrak{u}}}<&\infty,\quad
\int_{\mathbb{R}^3}\abs{l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})}\mathrm{e}^{\delta\abs{{\mathfrak{u}}-{\mathfrak{u}}^{\ast}}^2}\mathrm{d}{{\mathfrak{u}}^{\ast}}<\infty.
\end{align}
Hence, we have
\begin{align}\label{ktt 88}
\babs{I_2}\lesssim \mathrm{e}^{-\delta N^2}\lnm{\overline{\rem}}.
\end{align}

\paragraph{\underline{Case III: $I_3:$ $t_1'-r\leq\delta$ and $\abs{v}\leq N$, $\abs{{\mathfrak{u}}}\leq 2N$, $\abs{{\mathfrak{u}}^{\ast}}\leq 3N$}}
In this case, since the integral with respect to $r$ is restricted in a very short interval, the contribution is small 
\begin{align}\label{ktt 89}
\babs{I_3}\lesssim\abs{\int_{t_1'-\delta}^{t_1'}\mathrm{e}^{-(t_1'-r)}\mathrm{d}{r}}\lnm{\overline{\rem}}\lesssim \delta\lnm{\overline{\rem}}.
\end{align}

\paragraph{\underline{Case IV: $I_4:$ $t_1'-r\geq \abs{\ln(\delta)}$ and $\abs{v}\leq N$, $\abs{{\mathfrak{u}}}\leq 2N$, $\abs{{\mathfrak{u}}^{\ast}}\leq 3N$}}
In this case, $t_1'-r$ is significantly large, so $\mathrm{e}^{-(t_1'-r)}\leq\delta$ is very small. Hence, the contribution is small 
\begin{align}\label{ktt 89'}
\babs{I_4}\lesssim\abs{\int_{\abs{\ln(\delta)}}^{\infty}\mathrm{e}^{-(t_1'-r)}\mathrm{d}{r}}\lnm{\overline{\rem}}\lesssim \delta\lnm{\overline{\rem}}.
\end{align}

\paragraph{\underline{Case V: $I_5:$ $\delta\leq t_1'-r\leq \abs{\ln(\delta)}$ and $\abs{v}\leq N$, $\abs{{\mathfrak{u}}}\leq 2N$, $\abs{{\mathfrak{u}}^{\ast}}\leq 3N$}}
This is the most complicated case. Since $l_{M,w(v)}(v,{\mathfrak{u}})$ has
possible integrable singularity of $\dfrac{1}{\abs{v-{\mathfrak{u}}}}$, we need to
introduce the truncated kernel $l_{M,N}(v,{\mathfrak{u}})$ which is smooth and has compactly supported range such that
\begin{align}\label{ktt 79}
\sup_{\abs{v}\leq 3N}\int_{\abs{{\mathfrak{u}}}\leq
3N}\abs{l_{M,N}(v,{\mathfrak{u}})-l_{M,w(v)}(v,{\mathfrak{u}})}\mathrm{d}{{\mathfrak{u}}}\leq\frac{1}{N}.
\end{align}
Then we can split
\begin{align}\label{ktt 80}
l_{M,w(v)}(v,{\mathfrak{u}})l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})=&l_{M,N}(v,{\mathfrak{u}})l_{M,N}({\mathfrak{u}},{\mathfrak{u}}^{\ast})\\
&+\bigg(l_{M,w(v)}(v,{\mathfrak{u}})-l_{M,N}(v,{\mathfrak{u}})\bigg)l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})\nonumber\\
&+\bigg(l_{M,w({\mathfrak{u}})}({\mathfrak{u}},{\mathfrak{u}}^{\ast})-l_{M,N}({\mathfrak{u}},{\mathfrak{u}}^{\ast})\bigg)l_{M,N}(v,{\mathfrak{u}}).\nonumber
\end{align}
This means that we further split $I_5$ into
\begin{align}
I_5:=I_{5,1}+I_{5,2}+I_{5,3}.
\end{align}
Based on \eqref{ktt 79}, it is easy to check that
\begin{align}\label{ktt 82}
\babs{I_{5,2}}\lesssim&\frac{1}{N}\lnm{\overline{\rem}},\quad \babs{I_{5,3}}\lesssim\frac{1}{N}\lnm{\overline{\rem}}.
\end{align}
Therefore, the only remaining term is $I_{5,1}$. 
Considering $\abs{v},\abs{{\mathfrak{u}}},\abs{{\mathfrak{u}}^{\ast}}\leq 3N$, we know $\abs{\overline{\rem}}\lesssim_N\abs{\re_M}\lesssim_N\abs{R}$.
Also, since $l_{M,N}$ is bounded, we estimate
\begin{align}\label{ktt 85}
\babs{I_{5,1}}\lesssim_N&
\int_{\abs{{\mathfrak{u}}}\leq2N}\int_{\abs{{\mathfrak{u}}^{\ast}}\leq3N}\int_{0}^{t_1'-\delta}
\abs{R\Big(X-\varepsilon(t_1'-r){\mathfrak{u}},{\mathfrak{u}}^{\ast}\Big)}\mathrm{e}^{-\int_{r}^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}.
\end{align}
Note that we always have $X-\varepsilon(t_1'-r){\mathfrak{u}}\in\overline\Omega$. Hence, we define the change of variable ${\mathfrak{u}}\rightarrow y$ as
$y=(y_1,y_2,y_3)=X-\varepsilon(t_1'-r){\mathfrak{u}}$. Then the Jacobian
\begin{align}\label{ktt 81}
\abs{\frac{\mathrm{d}{y}}{\mathrm{d}{{\mathfrak{u}}}}}=\abs{\left\vert\begin{array}{ccc}
\varepsilon(t_1'-r)&0&0\\
0&\varepsilon(t_1'-r)&0\\
0&0&\varepsilon(t_1'-r)
\end{array}\right\vert}=\varepsilon^3(t_1'-r)^3\geq \varepsilon^3\delta^3.
\end{align}
Using H\"older's inequality, we estimate
\begin{align}\label{ktt 84}
\babs{I_{5,1}}\lesssim_N&\int_{\abs{{\mathfrak{u}}}\leq2N}\int_{\abs{{\mathfrak{u}}^{\ast}}\leq3N}\int_{0}^{t_1'-\delta}
\textbf{1}_{\{X-\varepsilon(t_1'-r){\mathfrak{u}}\in\overline\Omega\}}\abs{R\Big(X-\varepsilon(t_1'-r){\mathfrak{u}},{\mathfrak{u}}^{\ast}\Big)}\mathrm{e}^{-\int_{r}^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}\\
\leq&\bigg(\int_{\abs{{\mathfrak{u}}}\leq2N}\int_{\abs{{\mathfrak{u}}^{\ast}}\leq3N}\int_{0}^{t_1'-\delta}
\textbf{1}_{\{X-\varepsilon(t_1'-r){\mathfrak{u}}\in\overline\Omega\}}\mathrm{e}^{-\int_{r}^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}\bigg)^{\frac{5}{6}}\nonumber\\
&\times\bigg(\int_{\abs{{\mathfrak{u}}}\leq2N}\int_{\abs{{\mathfrak{u}}^{\ast}}\leq3N}\int_{0}^{t_1'-\delta}
\textbf{1}_{\{X-\varepsilon(t_1'-r){\mathfrak{u}}\in\overline\Omega\}}\abs{R\Big(X-\varepsilon(t_1'-r){\mathfrak{u}},{\mathfrak{u}}^{\ast}\Big)}^6\mathrm{e}^{-\int_{r}^{t_1'}\nu_M({\mathfrak{u}})
}\mathrm{d}{r}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{{\mathfrak{u}}^{\ast}}\bigg)^{\frac{1}{6}}\nonumber\\
\lesssim&\abs{\int_{0}^{t_1'-\delta}\frac{1}{\varepsilon^3\delta^3}\int_{\abs{{\mathfrak{u}}^{\ast}}\leq3N}
\int_{\Omega}\textbf{1}_{\{y\in\overline\Omega\}}\abs{R(y,{\mathfrak{u}}^{\ast})}^6\mathrm{e}^{-(t_1'-r)}\mathrm{d}{y}\mathrm{d}{{\mathfrak{u}}^{\ast}}\mathrm{d}{r}}^{\frac{1}{6}}
\lesssim\frac{1}{\varepsilon^{\frac{1}{2}}\delta^{\frac{1}{2}}}\pnm{R}{6}.\nonumber
\end{align}
Inserting \eqref{ktt 84} into \eqref{ktt 85}, we obtain
\begin{align}\label{ktt 86}
\babs{I_{5,1}}\lesssim_N \frac{1}{\varepsilon^{\frac{1}{2}}\delta^{\frac{1}{2}}}\pnm{R}{6}.
\end{align}
Combined with \eqref{ktt 82}, we know
\begin{align}\label{ktt 90}
\babs{I_5}\lesssim \frac{1}{N}\lnm{\overline{\rem}}+\frac{C(N)}{\varepsilon^{\frac{1}{2}}\delta^{\frac{1}{2}}}\pnm{R}{6}.
\end{align}

\paragraph{\underline{Summary of All Cases}}
Summarizing all five cases in \eqref{ktt 87}, \eqref{ktt 88}, \eqref{ktt 89}, \eqref{ktt 89'} and \eqref{ktt 90}, we obtain
\begin{align}
I\lesssim \bigg(\frac{1}{N}+\mathrm{e}^{-\delta N^2}+\delta\bigg)\lnm{\overline{\rem}}+C(N)\varepsilon^{-\frac{1}{2}}\delta^{-\frac{1}{2}}\pnm{R}{6}.
\end{align}
Choosing $\delta$ sufficiently small and then taking $N\gtrsim\delta^{-1}$ sufficiently large, we have
\begin{align}\label{ktt 91}
I\lesssim \delta\lnm{\overline{\rem}}+C(\delta)\varepsilon^{-\frac{1}{2}}\pnm{R}{6}.
\end{align}
By a similar but tedious computation, we arrive at
\begin{align}\label{ktt 92}
II\lesssim \delta\lnm{\overline{\rem}}+C(\delta)\varepsilon^{-\frac{1}{2}}\pnm{R}{6}.
\end{align}
Combined with \eqref{ktt 78}, we have
\begin{align}
&\abs{\int_{0}^{t_{1}}K^c_{M,w(v)}\left[\overline{\rem}\right]\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)}\mathrm{d}{s}}\\
&\lesssim\delta\lnm{\overline{\rem}}+C(\delta)\varepsilon^{-\frac{1}{2}}\pnm{R}{6}+\ssm{w h_M}{\gamma_-}+\lnm{\nu_M^{-1}w \widetilde S_M}.\nonumber
\end{align}
All the other terms in \eqref{ktt 65} related to $K^c_{M,w(v)}$ can be estimated in a similar fashion. Then our desired result follows.
\end{proof}


\subsection{Summary of $L^{\infty}$ Estimates}


\begin{proof}[Proof of Proposition \ref{thm:infinity}]
Summarizing all above and inserting \eqref{ktt 74}, \eqref{ktt 75}, \eqref{ktt 76} and \eqref{ktt 93} into \eqref{ktt 65}, we obtain for any $(x,v)\in\overline\Omega\times\mathbb{R}^3$,
\begin{align}\label{ctt 2}
\abs{\overline{\rem}(x,v)}\lesssim&\delta\lnm{\overline{\rem}}+C(\delta)\varepsilon^{-\frac{1}{2}}\pnm{R}{6}+\ssm{w h_M}{\gamma_-}+\lnm{\nu_M^{-1}w \widetilde S_M}.
\end{align}
For $\delta$ small, we have
\begin{align}
\abs{\overline{\rem}(x,v)}\lesssim&\varepsilon^{-\frac{1}{2}}\pnm{R}{6}+\ssm{w h_M}{\gamma_-}+\lnm{\nu_M^{-1}w \widetilde S_M}.
\end{align}
Then using \eqref{ll 03} and \eqref{ll 04}, we have
\begin{align}
\lnmm{\re_M}+\lnmms{\re_M}{\gamma}\lesssim&\lnmms{h_M}{-}+\varepsilon\lnmm{\nu^{-1}_MwS_M}+\lnm{\nu^{-1}_MwK_M^m[\re_M]}+\varepsilon^{-\frac{1}{2}}\pnm{R}{6}.
\end{align}
Since the cutoff $\chi_m$ in $K_M^m$ indicates that $\abs{{\mathfrak{u}}-v}\leq 2m$ and thus for sufficiently small $m>0$, $w(v)\simeq w({\mathfrak{u}})$ in the integral operator $K_M^m$. Hence, we know $wK_M^m[\re_M]\simeq K_M^m[w\re_M]$, and thus
\begin{align}
    \lnm{\nu^{-1}_MwK_M^m[\re_M]}\lesssim \lnm{\nu^{-1}_MK_M^m[w\re_M]}\lesssim o(1)\lnmm{\re_M}
\end{align}
Hence, we know
\begin{align}\label{final 74}
    \lnmm{\re_M}+\lnmms{\re_M}{\gamma}\lesssim&\lnmms{h}{-}+\varepsilon\lnmm{S}+\varepsilon^{-\frac{1}{2}}\pnm{R}{6}.
\end{align}

Using Lemma \ref{h-estimate}, Lemma \ref{ssl-estimate}, Lemma \ref{ss0-estimate}, Lemma \ref{ss1-estimate}, Lemma \ref{ss2-estimate}, Lemma \ref{ss3-estimate}, Lemma \ref{ss4-estimate}, Lemma \ref{ss5-estimate} and Lemma \ref{ssp-estimate}, we have the desired result.
\end{proof}


\section{A Priori Remainder Estimates}


\begin{theorem}\label{thm:apriori}
    Let $R$ be a solution to \eqref{remainder}. Under the assumption \eqref{assumption:boundary}, \eqref{final 31}, with $\alpha=1$ and $\xnm{R}\ll1$, for $h$ and $S$ defined in \eqref{aa 32} and \eqref{aa 31}, we have
    \begin{align}
        \xnm{R}\lesssimo_{T}.
    \end{align}
\end{theorem}

\begin{proof}
Let $\alpha=1$. Based on Lemma \ref{remark 01}, Proposition \ref{thm:energy} and Lemma \ref{final 11}, we have
\begin{align}\label{ss 01'}
    &\varepsilon^{-1}\tnm{\textbf{e}}+\varepsilon^{-1}\um{(\textbf{I}-\overline{\textbf{P}})[R]}+\pnm{\textbf{e}}{6}+\pnm{(\textbf{I}-\overline{\textbf{P}})[R]}{6}+\varepsilon^{-\frac{1}{2}}\nm{\xi}_{H^2}\\
    &+\varepsilon^{-\frac{1}{2}}\tnms{(1-\mathcal{P}_{\gamma})[R]}{\gamma_+}+\pnms{\mu^{\frac{1}{4}}(1-\mathcal{P}_{\gamma})[R]}{4}{\gamma_+}+\varepsilon^{-\frac{1}{2}}\tnms{\nabla_{x}\xi}{\partial\Omega}\nonumber\\
    \lesssim&o_{T}\xnm{R}+\xnm{R}^2+o_{T}.\nonumber
\end{align}
Collecting estimates from Proposition \ref{prop:p-bound}, Proposition \ref{prop:z-bound}, Proposition \ref{prop:b-bound} and Proposition \ref{prop:c-bound}, we have
\begin{align}\label{ss 02'}
    &\varepsilon^{-1}\tnm{p}+\pnm{p}{6}+\varepsilon^{-\frac{1}{2}}\tnm{\textbf{b}}+\pnm{\textbf{b}}{6}+\tnm{Z}+\tnm{c}+\pnm{c}{6}\\
    \lesssim&o_{T}\xnm{R}+\xnm{R}^{2}+\xnm{R}^{3}+o_{T}.\nonumber
\end{align}
Based on Proposition \ref{thm:infinity}, we have
\begin{align}\label{ss 06'}
\varepsilon^{\frac{1}{2}}\lnmm{R}+\varepsilon^{\frac{1}{2}}\lnmms{R}{\gamma}\lesssim&o_{T}\xnm{R}+\xnm{R}^2+o_{T}.
\end{align}
Based on Proposition \ref{lem:final 12}, using \eqref{ss 02'},  we have
\begin{align}\label{qq 32'}
    \tnms{\mathcal{P}_{\gamma}[R]}{\gamma}
    \lesssim&o_{T}\xnm{R}+\xnm{R}^{2}+\xnm{R}^{3}+o_{T}.
\end{align}
Combining \eqref{ss 01'}, \eqref{ss 02'}, \eqref{ss 06'} and \eqref{qq 32'}, we obtain that
\begin{align}\label{final 32}
    \xnm{R}\lesssim o_{T}\xnm{R}+\xnm{R}^2+\xnm{R}^{3}+o_{T}.
\end{align}
Hence, we have
\begin{align}
    \xnm{R}\lesssim \xnm{R}^2+\xnm{R}^{3}+o_{T}.
\end{align}
Then our desired result follows from an iteration/fixed-point argument, for $o_{T}$ and $R$ small.
\end{proof}


\section{Well-Posedness of Linear Remainder Equation}\label{sec:linear remainder}


In this section, we consider the well-posedness of the linear remainder equation \eqref{remainder} for given $S$ and $h$.

\begin{proposition}\label{prop:linear}
Assume $S$ and $h$ are given a priori satisfying \eqref{compatibility} and independent of $R$. Under the assumption \eqref{assumption:boundary}, for fixed $\varepsilon>0$, there exists a unique solution $R$ to the linear remainder equation \eqref{remainder} satisfying
    \begin{align}
    \xnm{R}\lesssim_{\varepsilon} \xnm{S}+\lnmms{h}{\gamma_-},
\end{align}
and
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mhR=3P\int_{\Omega}p(x)\mathrm{d}x=0.
\end{align}
\end{proposition}

Note that for fixed $\varepsilon>0$, $X$ norm is equivalent to $\widetilde X$ norm
\begin{align}
    \xnm{R}\approx\xxnm{R}:=\varepsilon^{-1}\nm{\re_M}_{L^2_{\nu_M}}+\varepsilon^{-\frac{1}{2}}\tnms{\re_M}{\gamma}+\varepsilon^{\frac{1}{2}}\lnmm{\re_M}+\varepsilon^{\frac{1}{2}}\lnmms{\re_M}{\gamma}.
\end{align}
Our goal is to justify the well-posedness of \eqref{remainder} in $\widetilde X$. \\

\paragraph{\underline{Reformulation with Global Maxwellian}}
Using the global Maxwellian \eqref{final 13}, we can rewrite \eqref{remainder} as
\begin{align}\label{global remainder}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}\widetilde{\re_M}+\varepsilon^{-1}\mathfrak{A}\left[\widetilde{\re_M}\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}\left[\widetilde{\re_M}\right]=\widetilde{S_M}\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \widetilde{\re_M}(x_0,v)=\widetilde{\mathcal{P}_M}\left[\widetilde{\re_M}\right](x_0,v)+\widetilde{h_M}(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0,
\end{array}
\right.
\end{align}
where $\widetilde{\re_M}:=\mm^{-\frac{1}{2}}\mhR$, $\widetilde{S_M}:=\mm^{-\frac{1}{2}}\mhS$ and $\widetilde{h_M}:=\mm^{-\frac{1}{2}}\m^{\frac{1}{2}} h$
\begin{align}
    \widetilde{\mathcal{L}_M}\left[\widetilde{\re_M}\right]:=-2\mm^{-\frac{1}{2}} Q^{\ast}\left[\mu_M,\mm^{\frac{1}{2}} \widetilde{\re_M}\right]:=\widetilde{\nu_M}\widetilde{\re_M}-\widetilde{K_M}\left[\widetilde{\re_M}\right],
\end{align}
\begin{align}\label{oo 31}
    \mathfrak{A}\left[\widetilde{\re_M}\right]:=2\mm^{-\frac{1}{2}} Q^{\ast}\left[\mu_M-\mu,\mm^{\frac{1}{2}} \widetilde{\re_M}\right],
\end{align}
\begin{align}
    \widetilde{\mathcal{P}_M}\left[\widetilde{\re_M}\right](x_0,v):=m_{M,w}(x_0,v)\displaystyle\int_{{\mathfrak{u}}\cdotn(x_0)>0}
    \mm^{\frac{1}{2}}\widetilde{\re_M}(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}},
\end{align}
for $m_{M,w}:=\mm^{-\frac{1}{2}}\m^{\frac{1}{2}}(x_0,v)m_w(x_0,v)$.
Note that
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}\widetilde{\re_M}=\iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mhR.
\end{align}
Notice that the notation here is slightly different that in \eqref{ll 01}. We will focus on the penalized version for $\theta>0$:
\begin{align}\label{linear remainder==}
\left\{
\begin{array}{l}
\theta\abs{v}^2\widetilde{\re_M}+v\cdot\nabla_{x}\widetilde{\re_M}+\varepsilon^{-1}\mathfrak{A}\left[\widetilde{\re_M}\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}\left[\widetilde{\re_M}\right]=\widetilde{S_M}\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \widetilde{\re_M}(x_0,v)=\widetilde{\mathcal{P}_M}\left[\widetilde{\re_M}\right](x_0,v)+\widetilde{h_M}(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0,
\end{array}
\right.
\end{align}
and then take limit $\theta\rt0$. \\

\paragraph{\underline{Non-Homogeneous Boundary Data}}
Due to the presence of $\widetilde{h_M}\neq0$, the solution set for $\widetilde{\re_M}$ is not a Banach space. We will first try to eliminate this $\widetilde{h_M}$ term. We introduce an auxiliary function $\widetilde{h}$ satisfying
\begin{align}\label{oo 32}
\left\{
\begin{array}{l}
\theta\abs{v}^2\widetilde{h}+v\cdot\nabla_{x}\widetilde{h}=0\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \widetilde{h}(x_0,v)=\widetilde{\mathcal{P}_M}\left[\widetilde{h}\right](x_0,v)+\widetilde{h_M}(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0,
\end{array}
\right.
\end{align}
There is no non-local term in \eqref{oo 32}, and thus we can show that there exists a unique $\widetilde{h}\in\widetilde{X}$ satisfying
\begin{align}
    \xxnm{\widetilde{h}}\lesssim\lnmms{\widetilde{h_M}}{\gamma_-}\lesssim_{\varepsilon}o_{T}.
\end{align}
and (using Lemma \ref{lemma:final 2})
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}\widetilde{h}=\int_{\gamma_-}\mm^{\frac{1}{2}}\widetilde{h_M}\mathrm{d}\gamma=\int_{\gamma_-}\m^{\frac{1}{2}} h\mathrm{d}\gamma=0.
\end{align}
Now under the substitution $\widetilde\re_M\rightarrow\re_M:=\widetilde\re_M-\widetilde h$, \eqref{linear remainder==} becomes 
\begin{align}\label{linear remainder.}
\\
\left\{
\begin{array}{l}
\theta\abs{v}^2\re_M+v\cdot\nabla_{x}{\re_M}+\varepsilon^{-1}\mathfrak{A}\left[{\re_M}\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}\left[{\re_M}\right]=S_M:=\widetilde{S_M}+\varepsilon^{-1}\mathfrak{A}\left[\widetilde h\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}\left[\widetilde h\right]\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \re_M(x_0,v)=\widetilde{\mathcal{P}_M}[\re_M](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.\nonumber
\end{array}
\right.
\end{align}
In particular, the boundary condition becomes homogeneous.
Hence, in order to study the well-posedness of \eqref{remainder} for $R$ in $\widetilde X$, it suffices to consider \eqref{linear remainder.} for $\re_M$ in $\widetilde X$.\\

\paragraph{\underline{Setup}}
For \eqref{linear remainder.}, using Lemma \ref{lemma:final 2},
we can directly verify that
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\mmhS_M=\iint_{\Omega\times\mathbb{R}^3}\mm^{\frac{1}{2}}\widetilde{S_M}=\iint_{\Omega\times\mathbb{R}^3}\mhS=0,
\end{align}
and thus direct integration in \eqref{linear remainder.} yields
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}\re_M=\iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mhR=3P\int_{\Omega}p=0.
\end{align}
Note that $\widetilde{\mathcal{P}_M}$ is not the natural diffuse-reflection boundary condition associated with $\mu_M$, and thus we do not have the desired energy structure. Hence, we need to split
\begin{align}
    \widetilde{\mathcal{P}_M}[\re_M]=\mathcal{P}_M[\re_M]+\mathfrak{B}[\re_M],
\end{align}
where
\begin{align}
    \mathcal{P}_M[\re_M]:=&P^{-1}\big(2\piT_M\big)^{\frac{1}{2}}\mm^{\frac{1}{2}}\int_{{\mathfrak{u}}\cdotn(x_0)>0}
    \mm^{\frac{1}{2}}{\re_M}(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}},\\
   \mathfrak{B}[\re_M]:=& \left(\mm^{-\frac{1}{2}}\m^{\frac{1}{2}}(x_0,v)m_w(x_0,v)-P^{-1}\big(2\piT_M\big)^{\frac{1}{2}}\mm^{\frac{1}{2}}\right)\int_{{\mathfrak{u}}\cdotn(x_0)>0}
    \mm^{\frac{1}{2}}{\re_M}(x_0,{\mathfrak{u}})\abs{{\mathfrak{u}}\cdotn(x_0)}\mathrm{d}{{\mathfrak{u}}}.
\end{align}
For the convenience of energy estimates, we further rewrite \eqref{linear remainder.} as
\begin{align}\label{linear remainder'}
\\
\left\{
\begin{array}{l}
\theta\abs{v}^2\re_M+v\cdot\nabla_{x}\re_M+\varepsilon^{-1}\mathfrak{A}\left[\re_M\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[\re_M]+\varepsilon^{-1}\textbf{P}_M[\re_M]=S_M+\varepsilon^{-1}\textbf{P}_M[\re_M]\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \re_M(x_0,v)=\mathcal{P}_M[\re_M](x_0,v)+\mathfrak{B}[\re_M](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.\nonumber
\end{array}
\right.
\end{align}
Here, $\textbf{P}_M$ and $\textbf{I}-\textbf{P}_M$ denote the projections onto $\mathcal{N}_M$ -- the nullspace  of $\mathcal{L}_M$, and its orthogonal complement $\mathcal{N}^{\perp}_M$.

Denote the linear operator
\begin{align}
    \mathcal{L}_{\varepsilon}[\re_M]:=\theta\abs{v}^2\re_M+v\cdot\nabla_{x}\re_M+\varepsilon^{-1}\mathfrak{A}\left[\re_M\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[\re_M]+\varepsilon^{-1}\textbf{P}_M[\re_M].
\end{align}

\begin{lemma}
    $\mathcal{L}_{\varepsilon}^{-1}$ is a bounded operator $\widetilde X\rightarrow\widetilde X$ with $\nm{\mathcal{L}_{\varepsilon}^{-1}}_{\widetilde X\rightarrow\widetilde X}\lesssim\varepsilon$.
\end{lemma}
\begin{proof}
Consider $\mathcal{L}_{\varepsilon}^{-1}[f]=g$, which is equivalent to 
\begin{align}\label{oo 54}
\left\{
\begin{array}{l}
\theta\abs{v}^2 g+v\cdot\nabla_{x} g+\varepsilon^{-1}\mathfrak{A}\left[g\right]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[g]+\varepsilon^{-1}\textbf{P}_M[g]=f\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} g(x_0,v)=\mathcal{P}_M[g](x_0,v)+\mathfrak{B}[g](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
We design an iteration with $g^0=0$ and
\begin{align}\label{oo 61}
\left\{
\begin{array}{l}
\theta\abs{v}^2 g^{n+1}+v\cdot\nabla_{x} g^{n+1}+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[g^{n+1}]+\varepsilon^{-1}\textbf{P}_M[g^{n+1}]=f-\varepsilon^{-1}\mathfrak{A}\left[g^{n}\right]\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} g^{n+1}(x_0,v)=\mathcal{P}_M[g^{n+1}](x_0,v)+\mathfrak{B}[g^{n}](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
We can show that $g^{n}$ is well-posed in $\widetilde X$. In the following, we will focus on showing that $g^n\rightarrow g$ in $\widetilde X$ by contraction mapping theorem.\\

\paragraph{\underline{$L^2$-Estimate}}
Note that
\begin{align}
    \lnmms{\mm^{-\frac{1}{2}}\m^{\frac{1}{2}}(x_0,v)m_w(x_0,v)-P^{-1}\big(2\piT_M\big)^{\frac{1}{2}}\mm^{\frac{1}{2}}}{\gamma_-}\lesssimo_{T},
\end{align}
and
\begin{align}
    \int_{\gamma_-}\mm^{\frac{1}{2}}\Big(\mm^{-\frac{1}{2}}\m^{\frac{1}{2}}(x_0,v)m_w(x_0,v)-P^{-1}\big(2\piT_M\big)^{\frac{1}{2}}\mm^{\frac{1}{2}}\Big)\mathrm{d}\gamma=0.
\end{align}
Also, since $\abs{T_M-T}\lesssimo_{T}$, using Lemma \ref{m-estimate}, we have
\begin{align}
    \abs{\mu_M-\mu}\lesssim o_{T}\left(\abs{v}^2\mathrm{e}^{o_{T}\abs{v}^2}\right)\mu_M.
\end{align}
Standard energy estimate in \eqref{oo 61} implies
\begin{align}\label{final 67}
    \varepsilon^{-1}\tnms{(1-\mathcal{P}_M)[g^{n+1}]}{\gamma_+}^2+\varepsilon^{-2}\nm{g^{n+1}}_{L^2_{\nu_M}}^2\lesssimo_{T}\varepsilon^{-1}\tnms{\mathcal{P}_M[g^{n}]}{\gamma_+}^2+o_{T}\varepsilon^{-2}\um{g^{n}}^2+\tnm{f}^2.
\end{align}
Next, we will try to bound $\tnms{\mathcal{P}_M[g^{n}]}{\gamma_+}$. Consider the cutoff function
\begin{align}
    \widetilde{\chi}(x,v):=\chi_0\big(\delta^{-1}(v\cdot n)\big)\chi(\delta^{-1}\mathfrak{n})
\end{align}
for some $0<\delta\ll1$, where $\chi$ are defined as in Section \ref{sec:boundary-form}, and $\chi_0\in C^{\infty}$ satisfying $\chi_0(y)=1$ if $y\geq1$ and $\chi_0(y)=0$ if $y\leq0$. 

Multiplying $\widetilde{\chi}$ on both sides of \eqref{oo 61} yields
\begin{align}\label{oo 55}
\\
\left\{
\begin{array}{l}
\theta\abs{v}^2 \widetilde{\chi} g^{n+1}+v\cdot\nabla_{x} \big(\widetilde{\chi} g^{n+1}\big)+\varepsilon^{-1}\widetilde{\chi}\widetilde{\mathcal{L}_M}[g^{n+1}]+\varepsilon^{-1}\widetilde{\chi}\textbf{P}_M[g^{n+1}]=f-\varepsilon^{-1}\widetilde{\chi}\mathfrak{A}\left[g^{n}\right]+v\cdot\nabla_{x}\widetilde{\chi} g^{n+1}\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \widetilde{\chi} g^{n+1}(x_0,v)=\widetilde{\chi}\mathcal{P}_M[g^{n+1}](x_0,v)+\widetilde{\chi}\mathfrak{B}[g^{n}](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.\nonumber
\end{array}
\right.
\end{align}
Notice that
\begin{align}
    \widetilde{\chi}\widetilde{\mathcal{L}_M}[g^{n+1}]=\widetilde{\mathcal{L}_M}[\widetilde{\chi} g^{n+1}]-(1-\widetilde{\chi})\widetilde{K_M}[g^{n+1}]+\widetilde{K_M}[(1-\widetilde{\chi})g^{n+1}]
\end{align}
and
\begin{align}
   \widetilde{\chi}\textbf{P}_M[g^{n+1}]= \textbf{P}_M[\widetilde{\chi} g^{n+1}]-(1-\widetilde{\chi})\textbf{P}_M[g^{n+1}]+\textbf{P}_M[(1-\widetilde{\chi})g^{n+1}].
\end{align}
Hence, \eqref{oo 55} is equivalent to
\begin{align}\label{oo 56}
\left\{
\begin{array}{l}
\theta\abs{v}^2 \widetilde{\chi} g^{n+1}+v\cdot\nabla_{x} \big(\widetilde{\chi} g^{n+1}\big)+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[\widetilde{\chi} g^{n+1}]+\varepsilon^{-1}\textbf{P}_M[\widetilde{\chi} g^{n+1}]=f+\mathcal{S}\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \widetilde{\chi} g^{n+1}(x_0,v)=\widetilde{\chi}\mathcal{P}_M[g^{n+1}](x_0,v)+\widetilde{\chi}\mathfrak{B}[g^{n}](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
where
\begin{align}
    \mathcal{S}:=&-\varepsilon^{-1}\widetilde{\chi}\mathfrak{A}\left[g^{n}\right]+v\cdot\nabla_{x}\widetilde{\chi} g^{n+1}+\varepsilon^{-1}(1-\widetilde{\chi})\widetilde{K_M}[g^{n+1}]-\varepsilon^{-1}\widetilde{K_M}[(1-\widetilde{\chi})g^{n+1}]\\
    &+\varepsilon^{-1}(1-\widetilde{\chi})\textbf{P}_M[g^{n+1}]-\varepsilon^{-1}\textbf{P}_M[(1-\widetilde{\chi})g^{n+1}].\nonumber
\end{align}
Since the overlapping region of $\widetilde{\chi}$ and $1-\widetilde{\chi}$ is very small, and $\nabla_{x}\widetilde{\chi}\lesssim\delta^{-1}$ in a small region, we know when $\varepsilon\ll\delta$
\begin{align}
    \br{\mathcal{S},\varepsilon^{-1}\widetilde{\chi} g^{n+1}}\lesssim \big(\varepsilon^{-2}\delta+\varepsilon^{-1}\big)\tnm{g^{n+1}}\lesssim \varepsilon^{-2}\delta\tnm{g^{n+1}}.
\end{align}
Then standard energy estimate in \eqref{oo 56} yields (noticing that there is no $\gamma_-$ contribution due to the cutoff)
\begin{align}
    \varepsilon^{-1}\tnms{\widetilde{\chi} g^{n+1}}{\gamma_+}^2+\varepsilon^{-2}\tnm{\widetilde{\chi} g^{n+1}}^2\lesssim \tnm{f}^2+\varepsilon^{-2}\delta\tnm{g^{n+1}}^2.
\end{align}
Then noticing that 
\begin{align}
    \mathcal{P}_M[g^{n+1}]\simeq&\int_{v'\cdot n>0}g^{n+1}(v')\mm^{\frac{1}{2}}(v')(v'\cdot n)\mathrm{d} v'\\
    =&\int_{v'\cdot n>2\delta}g^{n+1}(v')\mm^{\frac{1}{2}}(v')(v'\cdot n)\mathrm{d} v'+\int_{0<v'\cdot n<2\delta}g^{n+1}(v')\mm^{\frac{1}{2}}(v')(v'\cdot n)\mathrm{d} v',\nonumber
\end{align}
we have
\begin{align}\label{final 68}
    \tnms{\mathcal{P}_M[g^{n+1}]}{\gamma}^2\lesssim&\tnms{\int_{v'\cdot n>\delta}g^{n+1}(v')\mm^{\frac{1}{2}}(v')(v'\cdot n)\mathrm{d} v'}{\gamma_+}^2+\tnms{\int_{0<v'\cdot n<\delta}g^{n+1}(v')\mm^{\frac{1}{2}}(v')(v'\cdot n)\mathrm{d} v'}{\gamma_+}^2\\
    \lesssim&\tnms{\widetilde{\chi} g^{n+1}}{\gamma_+}^2+\delta\tnms{g^{n+1}}{\gamma}^2\lesssim\delta\Big(\varepsilon^{-1}\tnm{g^{n+1}}^2+\tnms{g^{n+1}}{\gamma}^2\Big)+\varepsilon\tnm{f}^2.\nonumber
\end{align}
Hence, combining \eqref{final 67} and \eqref{final 68}, we know
\begin{align}
\\
    \varepsilon^{-1}\tnms{\mathcal{P}_M[g^{n+1}]}{\gamma}^2+\varepsilon^{-1}\tnms{(1-\mathcal{P}_M)[g^{n+1}]}{\gamma_+}^2+\varepsilon^{-2}\nm{g^{n+1}}_{L^2_{\nu_M}}^2\lesssimo_{T}\varepsilon^{-1}\tnms{\mathcal{P}_M[g^{n}]}{\gamma}^2+o_{T}\varepsilon^{-2}\nm{g^{n}}_{L^2_{\nu_M}}^2+\tnm{f}^2.\nonumber
\end{align}
Then for fixed $\varepsilon$, when $\delta$ and $o_{T}$ are small, by contraction mapping theorem, we know the iteration \eqref{oo 61} is convergent $g^n\rightarrow g$ under the norm
\begin{align}
    \varepsilon^{-1}\tnms{\mathcal{P}_M[\ \cdot\ ]}{\gamma}^2+\varepsilon^{-1}\tnms{(1-\mathcal{P}_M)[\ \cdot\ ]}{\gamma_+}^2+\varepsilon^{-2}\nm{\ \cdot\ }_{L^2_{\nu_M}}^2.
\end{align}
Also, we have the uniform bound (for all $g^n$ and $g$)
\begin{align}
    \varepsilon^{-1}\tnms{\mathcal{P}_M[g]}{\gamma}^2+\varepsilon^{-1}\tnms{(1-\mathcal{P}_M)[g]}{\gamma_+}^2+\varepsilon^{-2}\nm{g}_{L^2_{\nu_M}}^2\lesssim\tnm{f}^2,
\end{align}
which yields
\begin{align}\label{final 71}
    \varepsilon^{-\frac{1}{2}}\tnms{g}{\gamma}+\varepsilon^{-1}\nm{g}_{L^2_{\nu_M}}\lesssim\tnm{f}.
\end{align}

\paragraph{\underline{$L^{\infty}$-Estimate}}
Noticing that $\textbf{P}_M$ is also a non-local operator similar to $\widetilde{K_M}$, we can follow the standard $L^{\infty}$ estimate in the proof of Proposition \ref{thm:infinity} (particularly \eqref{ktt 84}) to bound
\begin{align}\label{final 72}
    \lnmm{g}+\lnmms{g}{\gamma}\lesssim\varepsilon^{-\frac{3}{2}}\tnm{g}+\varepsilon\lnmm{\nu_M^{-1}f}\lesssim\varepsilon^{-\frac{1}{2}}\tnm{f}+\varepsilon\lnmm{\nu_M^{-1}f}.
\end{align}

\paragraph{\underline{Synthesis}}
Summarizing \eqref{final 71} and \eqref{final 72}, we know
\begin{align}\label{final 73}
    \xxnm{g}\lesssim\tnm{f}+\varepsilon^{\frac{3}{2}}\lnmm{f}\lesssim\varepsilon\xxnm{f}.
\end{align}
Hence, $\mathcal{L}_{\varepsilon}^{-1}$ is a bounded operator $\widetilde X\rightarrow\widetilde X$.
\end{proof}

Now we consider the solvability of \eqref{linear remainder.} or equivalently \eqref{linear remainder'}. Note that $\varepsilon^{-1}\mathcal{L}_{\varepsilon}^{-1}\textbf{P}_M[\re_M]$ is $O(1)$ from \eqref{final 73}. Hence, to prove the existence, we need to apply the Fredholm alternative instead of the contraction mapping principle. 

\begin{lemma}
    There exists a unique solution $\re_M\in\widetilde{X}$ to \eqref{linear remainder'} satisfying
    \begin{align}
        \xxnm{\re_M}\lesssim\xxnm{S_M}.
    \end{align}
\end{lemma}
\begin{proof}
Apply $\mathcal{L}_{\varepsilon}^{-1}$ on both sides of \eqref{linear remainder'}, we obtain
\begin{align}\label{temp 13}
    \re_M=\mathcal{L}_{\varepsilon}^{-1}[S_M]+\varepsilon^{-1}\mathcal{L}_{\varepsilon}^{-1}\textbf{P}_M[\re_M].
\end{align}
To solve \eqref{temp 13}, it suffices to solve
\begin{align}\label{temp 14}
    \textbf{P}_M[\re_M]=\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}[S_M]+\varepsilon^{-1}\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}\textbf{P}_M[\re_M],
\end{align}
and 
\begin{align}\label{temp 14'}
    (\textbf{I}-\textbf{P}_M)[\re_M]=(\textbf{I}-\textbf{P}_M)\mathcal{L}_{\varepsilon}^{-1}[S_M]+\varepsilon^{-1}(\textbf{I}-\textbf{P}_M)\mathcal{L}_{\varepsilon}^{-1}\textbf{P}_M[\re_M].
\end{align}

\paragraph{\underline{Solvability of $\textbf{P}_M[\re_M]$ and $(\textbf{I}-\textbf{P}_M)[\re_M]$}}
Rewrite \eqref{temp 14} as
\begin{align}\label{oo 09}
    \Big(I-\varepsilon^{-1}\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}\Big)\textbf{P}_M[\re_M]=\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}[S_M].
\end{align}
Direct estimate using \eqref{final 73} reveals that
\begin{align}
    \xxnm{\varepsilon^{-1}\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}[S_M]}&\lesssim\xxnm{S_M}.
\end{align}
Hence, it suffices to show the compactness of operator $\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}: \widetilde X\cap\mathcal{N}_M\rightarrow\widetilde X\cap\mathcal{N}_M$.

Our argument relies on the application of the Fredholm alternative. Considering that the remainder estimates in Theorem \ref{thm:apriori} implies the injection, as long as the compactness is guaranteed, we know the operator 
\begin{align}
    I-\varepsilon^{-1}\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}: \widetilde X\cap\mathcal{N}_M\rightarrow \widetilde X\cap\mathcal{N}_M
\end{align}
has a bounded inverse, and thus
there exists a unique solution $\textbf{P}_M[R]\in\widetilde X\cap\mathcal{N}_M$ to \eqref{oo 09} or equivalently \eqref{temp 14} satisfying 
\begin{align}
    \xxnm{\textbf{P}_M[\re_M]}\lesssim\xxnm{S_M}.
\end{align}
Also, from invertibility, there exists a unique solution $(\textbf{I}-\textbf{P}_M)[\re_M]\in\widetilde X\cap\mathcal{N}^{\perp}_M$ to \eqref{temp 14'} satisfying
\begin{align}
    \xxnm{(\textbf{I}-\textbf{P}_M)[\re_M]}\lesssim\xxnm{S_M}.
\end{align}
In the following, we will focus on proving the compactness.\\

\paragraph{\underline{Mild Formulation}}
Assume $\mathcal{L}_{\varepsilon}^{-1}[g]=f$ for $g\in\widetilde X\cap\mathcal{N}_M$. Notice that $g(x,v)$ is a linear combination of $\mathcal{N}^{\ast}:=\Big\{1,v,\abs{v}^2-5T\Big\}$, and thus due to linearity of $\mathcal{L}_{\varepsilon}^{-1}$, it suffices to consider the case $g(x,v)= r(x)q(v)$ for some $r(x)$ and $q(v)\in\mathcal{N}^{\ast}$, which means
\begin{align}\label{oo 14}
    \theta\abs{v}^2 f+v\cdot\nabla_{x} f+\varepsilon^{-1}\mathfrak{A}[f]+\varepsilon^{-1}\widetilde{\mathcal{L}_M}[f]+\varepsilon^{-1}\textbf{P}_M[f]=g=r(x)q(v).
\end{align}
Notice that both $\mathfrak{A}$ and $\textbf{P}_M$ are non-local operators. We abuse the notation and let
\begin{align}
    \mathcal{L}_M:=\widetilde{\mathcal{L}_M}+\mathfrak{A}+\varepsilon^{-1}\textbf{P}_M:=\nu_M-K_M.
\end{align}
For the non-local integral operator $K_M$, denote its kernel as $k_M$. Denote the weighted solution
\begin{align}
f_w(x,v):=&w(v) f(x,v),
\end{align}
for $w$ defined in \eqref{ktt 56} and the weighted non-local operator
\begin{align}
K_{M,w}[f_w](v):=&w(v)K_M\left[\frac{f_w}{w}\right](v)=\int_{\mathbb{R}^3}k_{M,w}({\mathfrak{u}},v)f_w({\mathfrak{u}})\mathrm{d}{{\mathfrak{u}}},
\end{align}
where
\begin{align}
k_{M,w}({\mathfrak{u}},v):=k_M({\mathfrak{u}},v)\frac{w(v)}{w({\mathfrak{u}})}.
\end{align}
Multiplying $\varepsilonw$ on both sides of \eqref{oo 14}, we have
\begin{align}\label{oo 24}
\varepsilonv\cdot\nabla_{x} f_w+\nu_M f_w=K_{M,w}[f_w]+\varepsilonw r q.
\end{align}
Similar to the derivation in Section \ref{sec:linfty estimate}, by Duhamel's principle, we can track along the characteristics in \eqref{oo 24} to rewrite $f_w$ in terms of $K_{M,w}[f_w]$ and $wr\tilde q$
\begin{align}\label{oo 17}
    f_w(x,v)=& \varepsilon\int_{0}^{t_{1}} w(v)r\Big(x-\varepsilon(t_1-s)v\Big)q\big(v\big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d}{s}\\
    &+\int_{0}^{t_{1}}\int_{\mathbb{R}^3}k_{M,w}(v,v')f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}+\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)
    }}{\widetilde{w}_M(v)}\int_{\mathcal{V}_1}f_w(x_1,v_1)\widetilde{w}(v_1)\mathrm{d}{\sigma_1}.\nonumber
\end{align}

\paragraph{\underline{Reflection Term Estimates}}
Similar to the proof of Proposition \ref{thm:infinity}, we can deduce that the boundary reflection term (i.e. the third term on the RHS of \eqref{oo 17})
\begin{align}\label{oo 25}
    &\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)
    }}{\widetilde{w}_M(v)}\int_{\mathcal{V}_1}f_w(x_1,v_1)\widetilde{w}(v_1)\mathrm{d}{\sigma_1}\\
    =&\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\sum_{\ell=1}^{k-1}\int_{\prod_{j=1}^{\ell}\mathcal{V}_j}\Big(G_{\ell}[x,v]+H_{\ell}[x,v]\Big)\widetilde{w}(v_\ell)
    \bigg(\prod_{j=1}^{\ell}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_j)}\mathrm{d}{\sigma_j}\bigg)\nonumber\\
    &+\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\prod_{j=1}^{k}\mathcal{V}_j}f_w(x_k,v_k)\widetilde{w}(v_{k})
    \bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg),\nonumber
\end{align}
where
\begin{align}
G_{\ell}[x,v]:=&\varepsilon\int_{t_{\ell}}^{t_{\ell+1}}
w(v_\ell)r\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell}\Big)\tilde q\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell},v_{\ell}\Big)\mathrm{e}^{\int_0^{s}\nu_M(v_\ell)}\mathrm{d}{s}\\
H_{\ell}[x,v]:=&\int_{t_{\ell}}^{t_{\ell+1}}\int_{r^3}k_{M,w}(v_{\ell},v_{\ell}')f_w\Big(x_{\ell}-\varepsilon(t_{\ell+1}-s)v_{\ell},v_{\ell}'\Big)\mathrm{e}^{\int_0^{s}\nu_M(v_{\ell})}\mathrm{d} v_{\ell}'\mathrm{d}{s}.
\end{align}
Based on the similar proof of Proposition \ref{thm:infinity}, we know for sufficiently large $\ell$
\begin{align}
    \lnm{\frac{\mathrm{e}^{-\int_0^{t_1}\nu_M(v)}}{\widetilde{w}_M(v)}\int_{\prod_{j=1}^{k}\mathcal{V}_j}f_w(x_k,v_k)\widetilde{w}(v_{k})
    \bigg(\prod_{j=1}^{k}\mathrm{e}^{-\int_{t_j}^{t_{j+1}}\nu_M(v_{j})}\mathrm{d}{\sigma_j}\bigg)}\lesssim\delta\lnm{f_w}.
\end{align}
Also, note that $G_{\ell}$ and $H_{\ell}$ have the similar structure as the first and second terms on the RHS of \eqref{oo 17}, and thus can be handled in a similar fashion. Hence, we just need to consider the bound of the first and second terms on the RHS of \eqref{oo 17}.\\

\paragraph{\underline{Non-Local Term Estimates}}
For $k_M$ term (i.e. the second term on the RHS of \eqref{oo 17})
\begin{align}\label{oo 26}
    \int_{0}^{t_{1}}\int_{\mathbb{R}^3}k_{M,w}(v,v')f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}
\end{align}
we can
introduce the truncated kernel $k_{M,N}(v,{\mathfrak{u}})$ which is smooth and has compactly supported range such that
\begin{align}
\sup_{\abs{v}\leq N}\int_{\abs{{\mathfrak{u}}}\leq
N}\abs{k_{M,N}(v,{\mathfrak{u}})-k_{M,w}(v,{\mathfrak{u}})}\mathrm{d}{{\mathfrak{u}}}\leq\frac{1}{N}.
\end{align}
Similar to the proof of Proposition \ref{thm:infinity}, we deduce that 
\begin{align}
    \lnm{\int_{0}^{t_{b}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\geq N\}}k_{M,w}(v,v')f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}}\lesssim\frac{1}{N}\lnm{f_w},
\end{align}
\begin{align}
    \lnm{\int_{0}^{t_{b}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{\abs{v'}\geq N\}}k_{M,w}(v,v')f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}}\lesssim\frac{1}{N}\lnm{f_w},
\end{align}
\begin{align}
\\
    \lnm{\int_{0}^{t_{1}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{\abs{v'}\leq N\}}\textbf{1}_{\{t_1-s\leq\delta\}}k_{M,w}(v,v')f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}}\lesssim\delta\lnm{f_w},\nonumber
\end{align}
and
\begin{align}
\\
    \lnm{\int_{0}^{t_{1}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{\abs{v'}\leq N\}}\textbf{1}_{\{t_1-s\geq\delta\}}\abs{k_{M,N}(v,{\mathfrak{u}})-k_{M,w}(v,{\mathfrak{u}})}f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}}\lesssim\frac{1}{N}\lnm{f_w}.\nonumber
\end{align}
When $N$ is large, all of these terms will only contribute $o(1)\lnm{f_w}$, which will not essentially change the compactness (which can be shown via contradiction). Hence, it suffices to consider 
\begin{align}
    \lnm{\int_{0}^{t_{1}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{\abs{v'}\leq N\}}\textbf{1}_{\{t_1-s\geq\delta\}}k_{M,N}(v,{\mathfrak{u}})f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d}{s}}.
\end{align}

\paragraph{\underline{Source Term Estimates}}
Consider the source term (i.e. the first term on the RHS of \eqref{oo 17})
\begin{align}
    \varepsilon\int_{0}^{t_{1}} w(v)r\Big(x-\varepsilon(t_1-s)v\Big)\tilde q\big(v\big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d}{s}.
\end{align}
Similar to the proof of Proposition \ref{thm:infinity}, we deduce that 
\begin{align}
    \lnm{\int_{0}^{t_{1}}\textbf{1}_{\{\abs{v}\geq N\}} w(v)r\Big(x-\varepsilon(t_1-s)v\Big) q\big(v\big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d}{s}}\lesssim \frac{1}{N}\lnm{r}
\end{align}
and
\begin{align}
    \lnm{\int_{0}^{t_{1}}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{t_1-s\leq\delta\}} w(v)r\Big(x-\varepsilon(t_1-s)v\Big) q\big(v\big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d}{s}}\lesssim \delta\lnm{r}.
\end{align}
When $N$ is large, all of these terms will only contribute $o(1)\lnm{r}$, which will not essentially change the compactness (which can be shown via contradiction). Hence, it suffices to consider 
\begin{align}
    \lnm{\int_{0}^{t_{1}}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{t_1-s\geq\delta\}} w(v)r\Big(x-\varepsilon(t_1-s)v\Big) q\big(v\big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d}{s}}\lesssim \delta\lnm{r}.
\end{align}

\paragraph{\underline{Summary of Simplification}}
Summarizing the analysis all above, after applying $\textbf{P}_M$ (i.e. multiplying $q(v)$ and integrating over $\mathbb{R}^3$) in \eqref{oo 17}, it suffices to prove the compactness of 
\begin{align}\label{oo 19}
    \int_{0}^{t_{1}}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{\abs{v'}\leq N\}}\textbf{1}_{\{t_1-s\geq\delta\}}k_{M,N}(v,{\mathfrak{u}})q(v)f_w\Big(x-\varepsilon(t_1-s)v,{\mathfrak{u}}\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v'\mathrm{d} v\mathrm{d}{s}
\end{align}
and
\begin{align}\label{oo 20}
    \int_{0}^{t_{1}}\int_{\mathbb{R}^3}\textbf{1}_{\{\abs{v}\leq N\}}\textbf{1}_{\{t_1-s\geq\delta\}}w(v)q(v)r\Big(x-\varepsilon(t_1-s)v\Big)\tilde q\Big(x-\varepsilon(t_1-s)v,v\Big)\mathrm{e}^{-\int_s^{t_1}\nu_M(v)
    }\mathrm{d} v\mathrm{d}{s}.
\end{align}

\paragraph{\underline{Compactness of \eqref{oo 19}}}
For \eqref{oo 19}, by substitution $v\rightarrow y=x-\varepsilon(t_1-s)v$ (or equivalently $v=\varepsilon^{-1}\frac{x-y}{t_1-s}$), the integral reduces to
\begin{align}\label{oo 21}
&\int_{\Omega}\int_{\abs{{\mathfrak{u}}}\leq N}\int_{0}^{\max\{t_1-\delta,0\}} q(\varepsilon^{-1}\tfrac{x-y}{t_1-s})k_{M,N}(\varepsilon^{-1}\tfrac{x-y}{t_1-s},v')w(v')f(y,v')\mathrm{e}^{-\int_{s}^{t_1}\nu_M(\varepsilon^{-1}\tfrac{x-y}{t_1-s})
}\mathrm{d}{s}\mathrm{d}{{\mathfrak{u}}}\mathrm{d}{y}.
\end{align}

On the one hand, \eqref{oo 14} naturally implies the $L^2$ bound
\begin{align}
    \tnm{f}\lesssim\xxnm{g},
\end{align}
which yields weak compactness of $f$ in $L^2$. 

On the other hand, note that $q$, $k_{M,N}$, $w$ terms are all smooth and uniformly bounded with respect to $(t,x)$ in $\abs{v'}\leq N$ and $t_1-s\geq\delta$. Hence, after the substitution, for fixed $(t,x)$
\begin{align}
    \int_{\abs{{\mathfrak{u}}}\leq N}\int_{0}^{\max\{t_1-\delta,0\}}q(\varepsilon^{-1}\tfrac{x-y}{t_1-s})k_{M,N}(\varepsilon^{-1}\tfrac{x-y}{t_1-s},v')w(v')\mathrm{e}^{-\int_{s}^{t_1}\nu_M(\varepsilon^{-1}\tfrac{x-y}{t_1-s})}\mathrm{d}{s}\mathrm{d}{{\mathfrak{u}}}
\end{align}
is a smooth $L^2$ function of $y$, and its $L^2$ norm is uniformly bounded with respect to $(t,x)$.

Therefore, for the weakly $L^2$ convergent sequence in terms of $f$, we can pass to the limit in the $y$ integral of \eqref{oo 21} to get strongly convergent sequence (uniformly in $(t,x)$). Hence, we have shown the strong compactness of \eqref{oo 19} in $L^{\infty}$.\\

\paragraph{\underline{Compactness of \eqref{oo 20}}}
Similarly, for \eqref{oo 20}, using substitution $v\rightarrow y=x-\varepsilon(t_1-s)v$ (or equivalently $v=\varepsilon^{-1}\frac{x-y}{t_1-s}$), the integral reduces to 
\begin{align}\label{oo 22}
    \int_{\Omega}\int_0^{\max\{t_1-\delta,0\}}w(\varepsilon^{-1}\tfrac{x-y}{t_1-s})q(\varepsilon^{-1}\tfrac{x-y}{t_1-s})\mathrm{e}^{-\int_s^t\nu_M(\varepsilon^{-1}\frac{x-y}{t_1-s})}r(y) q(\varepsilon^{-1}\tfrac{x-y}{t_1-s})\mathrm{d} s\mathrm{d} y.
\end{align}

On the one hand, $g\in\widetilde X$ implies 
\begin{align}
    \tnm{r}\lesssim\tnm{g}\lesssim\xxnm{g},
\end{align}
which yields weak compactness of $r$ in $L^2$. 

On the other hand,note that $q$, $w$ terms are all smooth and uniformly bounded with respect to $(t,x)$ in $\abs{v}\leq N$ and $t_1-s\geq\delta$. Hence, after the substitution, for fixed $(t,x)$
\begin{align}
    \int_0^{\max\{t_1-\delta,0\}}w(\varepsilon^{-1}\tfrac{x-y}{t_1-s})q(\varepsilon^{-1}\tfrac{x-y}{t_1-s})\mathrm{e}^{-\int_s^t\nu_M(\varepsilon^{-1}\frac{x-y}{t_1-s})} q(\varepsilon^{-1}\tfrac{x-y}{t_b-s})\mathrm{d} s
\end{align}
is a smooth $L^2$ function of $y$, and its $L^2$ norm is uniformly bounded with respect to $(t,x)$. 

Therefore, for the weakly $L^2$ convergent sequence in terms of $r$, we can pass to the limit in the $y$ integral of \eqref{oo 22} to get strongly convergent sequence (uniformly in $(t,x)$). Hence, we have shown the strong compactness of \eqref{oo 20} in $L^{\infty}$.\\

\paragraph{\underline{Summary of Compactness}}
In summary, we have shown that compactness of $f_w$ in $L^{\infty}$. Considering the weight function $w$ and \eqref{oo 14}, this is equivalent to the compactness of
$\textbf{P}_M\mathcal{L}_{\varepsilon}^{-1}: \widetilde X\cap\mathcal{N}_M\rightarrow \widetilde X\cap\mathcal{N}_M$ in $\widetilde X$.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:linear}]
For given $\theta$, we have shown that there exists a unique solution $R_{M,\theta}\in\widetilde X$ (we add subscript $\theta$ to highlight the dependence on $\theta$) to \eqref{linear remainder.} or equivalently \eqref{linear remainder'} satisfying
\begin{align}\label{oo 08'}
    \xxnm{R_{M,\theta}}\lesssim_{\varepsilon} o_{T}.
\end{align}
Note that this estimate is uniform in $\theta$.
In addition, direct integration on both sides of \eqref{linear remainder'} with multiplier $\m^{\frac{1}{2}}$ yields that $\widetildeR_{\theta}$ satisfies
\begin{align}\label{oo 07'}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}(v)R_{M,\theta}(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align}
Further we can recover the solution $\widetilde\re_M$ to \eqref{linear remainder==} by adding $\widetilde h$ such that
\begin{align}\label{oo 08}
    \xxnm{\widetilde{R_{M,\theta}}}\lesssim_{\varepsilon}o_{T},
\end{align}
and 
\begin{align}\label{oo 07}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}(v)\widetilde{R_{M,\theta}}(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align}
Since the estimates \eqref{oo 08} is uniform in $\theta>0$, we can find a subsequence that is weak-$\ast$ convergent in weighted $L^{\infty}$ and weak convergent in $L^2$ as $\theta_i\rt0$. By weak lower semi-continuity, we have 
\begin{align}
    \widetilde{R_{M,\theta_j}}\rightarrow \widetilde{\re_M}\ \ \text{in}\ \ \widetilde X,
\end{align}
and
\begin{align}
    \xxnm{\widetilde{\re_M}}\leq\liminf_{j\rightarrow\infty}\xxnm{\widetilde{R_{M,\theta_j}}}\lesssim_{\varepsilon} o_{T}.
\end{align}
Also, by passing to the limit in the weak formulation, we know $\widetilde\re_M$ or equivalently $R$ is a solution to the remainder equation \eqref{remainder} satisfying
\begin{align}
    \xxnm{R}\lesssim_{\varepsilon} o_{T},
\end{align}
and
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\m^{\frac{1}{2}}(v)R(x,v)\mathrm{d} x\mathrm{d} v=\iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\mm^{\frac{1}{2}}(v)\widetilde\re_M(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align}
By the a priori remainder estimates in Theorem \ref{thm:apriori}, we know that such $R$ is unique.
\end{proof}


\section{Proof of Main Theorem}


\subsection{Well-Posedness and Positivity}\label{sec:well-posedness}


\subsubsection{Well-Posedness of the Auxiliary Equation}


\paragraph{\underline{Setup of Iteration}}
Based on \eqref{remainder} and \eqref{wt 02}, we design an iterative scheme with $R^0=0$ and
\begin{align}\label{wt 01}
    &v\cdot\nabla_{x} \left(\mhR^{n+1}\right)-\varepsilon^{-1}\m^{\frac{1}{2}}\mathcal{L}[R^{n+1}]\\
    =&\ \mathscr{S}+\varepsilon^{-1}\bigg(Q^{\ast}\left[\mhR^n,\widetilde{\mathfrak{F}}_{\text{a}}\right]+Q^{\ast}\left[\widetilde{\mathfrak{F}}_{\text{a}},\mhR^n\right]+\varepsilon^{\alpha}Q^{\ast}\left[\mhR^n,\mhR^n\right]\bigg)\nonumber\\
    &+\varepsilon^{-\alpha}\varepsilon^{-1}\Big(Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n\right]-Q_{\text{gain}}\left[(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n)_+,(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n)_+\right]\Big)\nonumber\\
    &+\varepsilon^{-\alpha}\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\bigg(\varepsilon^{-1}Q_{\text{gain}}\left[\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n,\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n\right]-\varepsilon^{-1}Q_{\text{gain}}\left[(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n)_+,(\mathfrak{F}_{\text{a}}+\varepsilon^{\alpha}\mhR^n)_+\right]\bigg).\nonumber
\end{align}
This is equivalent to 
\begin{align}\label{remainder final}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}\left(\mhR^{n+1}\right)+\varepsilon^{-1}\m^{\frac{1}{2}}\mathcal{L}\left[R^{n+1}\right]=\mhS^n\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} R^{n+1}(x_0,v)=\mathcal{P}_{\gamma}\left[R^{n+1}\right](x_0,v)+h(x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
Here $S^n$ denotes $S$ with $R$ substituted by $R^n$. The sequence $\big\{R^n\big\}_{n=1}^{\infty}\subset X$ is guaranteed by Proposition \ref{prop:linear}.\\

\paragraph{\underline{Convergence}}
Applying the a priori remainder estimates in Theorem \ref{thm:apriori}, there exists $C_0>0$ such that letting $R=R^n$ on the RHS of \eqref{final 32}, we have
\begin{align}
    \xnm{R^{n+1}}\leq \frac{1}{2}\xnm{R^n}+C_0\left(\xnm{R^n}^2+\xnm{R^n}^{3}+o_{T}\right).
\end{align} 
Under the assumptions 
\begin{align}
    \xnm{R^n}\leq 3C_0o_{T},
\end{align}
for $o_{T}$ sufficiently small, by induction, we obtain
\begin{align}
    \xnm{R^{n+1}}\leq 3C_0o_{T}.
\end{align}
Hence, we obtain that the sequence $\big\{R^n\big\}_{n=1}^{\infty}$ is uniformly bounded. Also, based on Proposition \ref{prop:linear}, we have
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\m^{\frac{1}{2}}(v)R^{n+1}(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align}
By a similar proof, we know that this sequence is a contraction
\begin{align}
    \xnm{R^{n+1}-R^n}\lesssimo_{T}\xnm{R^n-R^{n-1}}.
\end{align}

\paragraph{\underline{Existence and Uniqueness}}
Using contraction mapping theorem, we know that $R^n\rightarrowR$ in $X$, and $R$ is a solution to \eqref{remainder} satisfying
\begin{align}
    \xnm{R}\lesssimo_{T},
\end{align}
and 
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\m^{\frac{1}{2}}(v)R(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align} 
The uniqueness follows from the a priori remainder estimates in Theorem \ref{thm:apriori}. Assume $R_1$ and $R_2$ are two distinct solutions to \eqref{remainder}. Then $\overline{R}=R_1-R_2$ satisfies 
\begin{align}
\left\{
\begin{array}{l}
v\cdot\nabla_{x}\left(\m^{\frac{1}{2}}\overline{R}\right)+\varepsilon^{-1}\m^{\frac{1}{2}}\mathcal{L}\left[\overline{R}\right]=\m^{\frac{1}{2}}\big(S^1-S^2\big)\ \ \text{in}\ \
\Omega\times\mathbb{R}^3,\\\rule{0ex}{1.5em} \overline{R}(x_0,v)=\mathcal{P}_{\gamma}\left[\overline{R}\right](x_0,v) \ \ \text{for}\ \ x_0\in\partial\Omega\ \ \text{and}\ \ v\cdotn(x_0)<0.
\end{array}
\right.
\end{align}
Here $S^k$ for $k=1,2$ denotes $S$ with $R$ substituted by $R^k$. Then using Theorem \ref{thm:apriori}, we obtain
\begin{align}
    \xnm{\overline{R}}\lesssim o_{T}\xnm{\overline{R}},
\end{align}
which naturally yields $\xnm{\overline{R}}=0$ and thus $\overline{R}=0$.


\subsubsection{Positivity of the Auxiliary Equation}


\paragraph{\underline{Mild Formulation of the Auxiliary Equation}} Now we focus on \eqref{auxiliary system},
which can be rewritten as
\begin{align}
    v\cdot\nabla_{x} \mathfrak{F}+\varepsilon^{-1}\nu(\mathfrak{F})\mathfrak{F}=\varepsilon^{-1}Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big).
\end{align}
Hence, using mild formulation to track the characteristics backward, we have
\begin{align}\label{oo 15}
    \mathfrak{F}(x,v)=&\mathrm{e}^{-\varepsilon^{-1}\nu t_b}\mathfrak{F}(x_b,v)\\
    &+\int_0^{t_b}\mathrm{e}^{-\varepsilon^{-1}\nu (t_b-s)}\left(\varepsilon^{-1}Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)\right)(s)\mathrm{d} s,\nonumber
\end{align}
for $t_b$ and $x_b$ defined in Definition \ref{exit}.\\

\paragraph{\underline{Boundary Term in the Mild Formulation}}
We know
\begin{align}
    \int_{\gamma_+}\mathfrak{F}(x_b,v)(v\cdot n)\mathrm{d} v=\int_{\gamma_+}\Big(\mathfrak{F}_{\text{a}}(x_b,v)+\varepsilon^{\alpha}\mhR(x_b,v)\Big)(v\cdot n)\mathrm{d} v.
\end{align}
By a similar argument as the $\ss_6$ estimates, we know that this integral is positive (since the negative part is only effective when $v$ is very large). Hence, $\mathcal{P}_{\gamma}[\mathfrak{F}]>0$ and thus $\mathfrak{F}\big|_{\gamma_-}\geq C\mu>0$. In \eqref{oo 15}, the characteristic will hit $\gamma_-$ when tracking backward, and thus we know
\begin{align}
    \mathrm{e}^{-\varepsilon^{-1}\nu t_b}\mathfrak{F}(x_b,v)\gs0.
\end{align}

\paragraph{\underline{Bulk Term in the Mild Formulation}}
Using the similar technique as the $\ss_6$ estimates, we have
\begin{align}
    \displaystyle\mathfrak{z}\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)\gtrsim -\varepsilon^{m-1}\mathfrak{z}.
\end{align}
Also, we know that $\mathfrak{F}\gtrsim \mu$ for $\abs{v}\lesssim \abs{\ln(\varepsilon)}^{\frac{1}{2}}$ and thus
\begin{align}
    \varepsilon^{-1}Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]=&\varepsilon^{-1}\int_{\mathbb{R}^3}\int_{\mathbb{S}^2}q(\omega,\abs{{\mathfrak{u}}-v})\mathfrak{F}_+({\mathfrak{u}}_{\ast})\mathfrak{F}_+(v_{\ast})\mathrm{d}{\omega}\mathrm{d}{{\mathfrak{u}}}\\
    \geq&\varepsilon^{-1}\int_{\mathbb{R}^3}\int_{\mathbb{S}^2}\textbf{1}_{\{\abs{{\mathfrak{u}}_{\ast}}\leq 1\}}\textbf{1}_{\{\abs{v_{\ast}}\leq 1\}}q(\omega,\abs{{\mathfrak{u}}-v})\mathfrak{F}_+({\mathfrak{u}}_{\ast})\mathfrak{F}_+(v_{\ast})\mathrm{d}{\omega}\mathrm{d}{{\mathfrak{u}}}\gtrsim \varepsilon^{-1}.\nonumber
\end{align}
Hence, considering that $\mathfrak{z}$ is compactly supported in $\{\abs{v}\leq 1\}$, we have
\begin{align}
\\
    \int_0^{t_b}\mathrm{e}^{-\varepsilon^{-1}\nu (t_b-s)}\left(\varepsilon^{-1}Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]+\mathfrak{z}\displaystyle\iint_{\Omega\times\mathbb{R}^3}\varepsilon^{-1}\Big(Q_{\text{gain}}[\mathfrak{F},\mathfrak{F}]-Q_{\text{gain}}[\mathfrak{F}_+,\mathfrak{F}_+]\Big)\right)(s)\mathrm{d} s\gtrsim \varepsilon^{-1}-\varepsilon^{m-1}\mathfrak{z}\geq0.\nonumber
\end{align}

\paragraph{\underline{Summary of Positivity}}
Combining the above results, when $\varepsilon$ is sufficiently small, we know that
\begin{align}
    \mathfrak{F}(x,v)\geq0.
\end{align}
In this case, \eqref{auxiliary system} reduces to \eqref{large system-}, and thus we know that the solution $\mathfrak{F}$ to \eqref{large system-} satisfies $\mathfrak{F}\geq0$
\begin{align}
    \xnm{R}\lesssimo_{T},
\end{align}
and
\begin{align}
    \iint_{\Omega\times\mathbb{R}^3}\abs{v}^2\m^{\frac{1}{2}}(v)R(x,v)\mathrm{d} x\mathrm{d} v=0.
\end{align}


\subsection{Ghost Effects}


Based on \eqref{aa 08}, we have
\begin{align}
    \m^{-\frac{1}{2}}\mathfrak{F}-\mhhf-\mhhf^{B}-\varepsilonR=0,
\end{align}
which yields
\begin{align}
    \tnm{\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilon\m^{-\frac{1}{2}}\mbhf^{B}_1-\varepsilonR}=\tnm{\varepsilon^2f_2}\lesssimo_{T}\varepsilon^2.
\end{align}
Due to the rescaling, we have
\begin{align}
    \tnm{\varepsilon\m^{-\frac{1}{2}}\mbhf^{B}_1}\lesssimo_{T}\varepsilon^{\frac{3}{2}},
\end{align}
and thus we know
\begin{align}\label{aa 09}
    \tnm{\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR}\lesssimo_{T}\varepsilon^{\frac{3}{2}}.
\end{align}
Let $\textbf{P}_{\m^{\frac{1}{2}}}$, $\textbf{P}_{v\m^{\frac{1}{2}}}$, $\textbf{P}_{(\abs{v}^2-3T)\m^{\frac{1}{2}}}$, $\textbf{P}_{\mathscr{A}}$ denote the $L^2$ projection onto the subspaces spanned by $\m^{\frac{1}{2}}$, $v\m^{\frac{1}{2}}$, $\big(\abs{v}^2-3T\big)\m^{\frac{1}{2}}$ or $\mathscr{A}$ (via taking inner product with these functions).

Applying these operators as well as $\textbf{I}-\overline{\textbf{P}}$ to $\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR$ and using \eqref{aa 09}, we have
\begin{align}
    \tnm{\textbf{P}_{\m^{\frac{1}{2}}}\Big[\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR\Big]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{v\m^{\frac{1}{2}}}\Big[\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR\Big]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{(\abs{v}^2-3T)\m^{\frac{1}{2}}}\Big[\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR\Big]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{\mathscr{A}}\Big[\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR\Big]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{(\textbf{I}-\overline{\textbf{P}})\Big[\m^{-\frac{1}{2}}\mathfrak{F}-\m^{\frac{1}{2}}-\ef_1-\varepsilonR\Big]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}}.
\end{align}
Then we have
\begin{align}
    \tnm{\textbf{P}_{\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\Big(\rho+\varepsilon\rho_1+\varepsilon(p-2Tc)\Big)\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{v\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\Big(\varepsilonu_1+\varepsilon\textbf{b}\Big)\cdot v\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{(\abs{v}^2-3T)\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\Big(T+\varepsilonT_1+\varepsilon c\Big)\left(\abs{v}^2-3T\right)\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{\mathscr{A}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\left(\varepsilon\left(-\frac{\nabla_{x}T}{2T^2}\right)-\varepsilon \textbf{d}\right)\cdot\mathscr{A}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{(\textbf{I}-\overline{\textbf{P}})\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-(\textbf{I}-\overline{\textbf{P}})[\varepsilonR]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}}.
\end{align}
Since $\xnm{R}\lesssimo_{T}$, we have $\varepsilon^{-\frac{1}{2}}\tnm{\textbf{b}}+\varepsilon^{-\frac{1}{2}}\tnm{\textbf{d}}+\varepsilon^{-1}\um{(\textbf{I}-\overline{\textbf{P}})[R]}\lesssimo_{T}$ and hence
\begin{align}
    \tnm{\textbf{P}_{\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\rho\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon,\\
    \tnm{\textbf{P}_{v\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-\varepsilonu_1\cdot v\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{\textbf{P}_{(\abs{v}^2-3T)\m^{\frac{1}{2}}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]-T\left(\abs{v}^2-3T\right)\m^{\frac{1}{2}}}\lesssim&o_{T}\varepsilon,\\
    \tnm{\textbf{P}_{\mathscr{A}}\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]+\varepsilon\left(\frac{\nabla_{x}T}{2T^2}\right)\cdot \mathscr{A}}\lesssim&o_{T}\varepsilon^{\frac{3}{2}},\\
    \tnm{(\textbf{I}-\textbf{P})\left[\m^{-\frac{1}{2}}\mathfrak{F}\right]}\lesssim&o_{T}\varepsilon^{\frac{3}{2}}.
\end{align}
Therefore, we know that $(\rho,\varepsilonu_1,T)$ indeed captures the leading-order terms of $\m^{-\frac{1}{2}}\mathfrak{F}$ in $L^2$ as $\varepsilon\rt0$.


