basis is an orthonormal basis of $\mathbb{R}^{3}$ which is defined by

$\begin{split}&\mathbf{{e}}^{k}_{0}:=\frac{v^{k}}{|v^{k}|},\\\
&\mathbf{{e}}^{k}_{\perp,1}:=\mathbf{e}^{k}_{0}\times\frac{\partial_{2}\eta_{p^{k}}(x^{k})}{\sqrt{g_{{p}^{k},22}(x^{k})}}\bigg{/}\Big{|}\mathbf{e}^{k}_{0}\times\frac{\partial_{2}\eta_{p^{k}}(x^{k})}{\sqrt{g_{{p}^{k},22}(x^{k})}}\Big{|},\\\
&\mathbf{{e}}^{k}_{\perp,2}:=\mathbf{e}^{k}_{0}\times\mathbf{{e}}^{k}_{\perp,1},\end{split}$
(6.10)

for each $k$ and $(n(x^{k}),v^{k})$.

From uniform nondegeneracy (6.8), we study the following Jacobian with respect
to speed-direction coordinate,

$\det\frac{\partial
X(s;t,x,v)}{\partial(|\mathbf{v}_{p^{k}}^{k}|,\hat{\mathbf{v}}_{p^{k},1}^{k},\hat{\mathbf{v}}_{p^{k},2}^{k})}.$

Since perturbation of speed $|\mathbf{v}_{p^{k}}^{k}|=|v|$ does not change
bouncing points. Using this advantage, we can obtain (see (3.32) of [25])

$\displaystyle\left[\begin{array}[]{ccc}\frac{\partial
X(s)}{\partial|v|}\cdot\mathbf{e}_{0}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{0}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{0}^{k}\\\ \frac{\partial
X(s)}{\partial|v|}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{\perp,1}^{k}\\\ \frac{\partial
X(s)}{\partial|v|}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial
X(s)}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{\perp,2}^{k}\end{array}\right]=\left[\begin{array}[]{c|cc}-(t-s)&{\bf(*)_{1,2}}\\\
\hline\cr\mathbf{0}_{2,1}&{\bf(**)_{2,2}}\end{array}\right]_{3\times 3}.$

To check full rank property of above matrix, we study determinant of lower
right ${\bf(**)_{2,2}}$ which is generated by directional derivative
$\frac{\partial X(s)}{\partial\hat{\mathbf{v}}_{p^{k},i}^{k}}$. To study
${\bf(**)_{2,2}}$ efficiently, we define specular matrix (Definition 3.3 in
[25]).

######  Definition 6.2 (Specular Matrix).

For fixed $k\in\mathbb{N}$ and a $C^{1}$-map $Y:(y_{1},y_{2})\mapsto
Y(y_{1},y_{2})\in\Omega$, assume (6.9) with
${x}^{k}={x}^{k}(t,Y({y}_{1},{y}_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$ and
${v}^{k}={v}^{k}(t,Y({y}_{1},{y}_{2}),$
$|v|,\hat{v}_{1},\hat{v}_{2})$. We define the $4\times 4$ specular transition
matrix $\mathcal{S}^{k,p^{k},Y}=\mathcal{S}^{k,p^{k},Y}(t,y_{1},y_{2},\\\
|v|,\hat{v}_{1},\hat{v}_{2})$ as

$\mathcal{S}^{k,p^{k},Y}:=\left[\begin{array}[]{c|c}\mathcal{S}_{1}^{k,p^{k},Y}&0_{2\times
2}\\\
\hline\cr\mathcal{S}_{2}^{k,p^{k},Y}&\mathcal{S}_{3}^{k,p^{k},Y}\end{array}\right]_{4\times
4},$ (6.12)

where

$\begin{split}\mathcal{S}_{1}^{k,p^{k},Y}&:=\left[\begin{array}[]{cc}\partial_{1}\eta_{{p}^{k}}\cdot\mathbf{e}^{k}_{\perp,1}&\partial_{2}\eta_{{p}^{k}}\cdot\mathbf{e}^{k}_{\perp,1}\\\
\partial_{1}\eta_{{p}^{k}}\cdot\mathbf{e}^{k}_{\perp,2}&\partial_{2}\eta_{{p}^{k}}\cdot\mathbf{e}^{k}_{\perp,2}\end{array}\right]_{2\times
2},\\\
\mathcal{S}_{2}^{k,p^{k},Y}&:=\left[\begin{array}[]{cc}\Big{(}\sum_{\ell=1}^{3}\partial_{1}\big{[}\frac{\partial_{\ell}\eta_{{p}^{k}}}{\sqrt{g_{{{p}^{k}},\ell\ell}}}\big{]}\hat{\mathbf{v}}^{k}_{{{p}^{k}},\ell}\Big{)}\cdot\mathbf{e}^{k}_{\perp,1}&\Big{(}\sum_{\ell=1}^{3}\partial_{2}\big{[}\frac{\partial_{\ell}\eta_{{p}^{k}}}{\sqrt{g_{{{p^{k}}},\ell\ell}}}\big{]}\hat{\mathbf{v}}^{k}_{{{p}},\ell}\Big{)}\cdot\mathbf{e}_{\perp,1}^{k}\\\
\Big{(}\sum_{\ell=1}^{3}\partial_{1}\big{[}\frac{\partial_{\ell}\eta_{{p^{k}}}}{\sqrt{g_{{{p^{k}}},\ell\ell}}}\big{]}\hat{\mathbf{v}}^{k}_{{{p^{k}}},\ell}\Big{)}\cdot\mathbf{e}^{k}_{\perp,2}&\Big{(}\sum_{\ell=1}^{3}\partial_{2}\big{[}\frac{\partial_{\ell}\eta_{{p^{k}}}}{\sqrt{g_{{{p^{k}}},\ell\ell}}}\big{]}\hat{\mathbf{v}}^{k}_{{{p^{k}}},\ell}\Big{)}\cdot\mathbf{e}^{k}_{\perp,2}\end{array}\right]_{2\times
2},\\\
\mathcal{S}_{3}^{k,p^{k},Y}&:=\left[\begin{array}[]{cc}\Big{[}\frac{\partial_{1}\eta_{p^{k}}}{\sqrt{g_{p^{k},11}}}-\frac{\partial_{3}\eta_{p^{k}}}{\sqrt{g_{p^{k},33}}}\frac{\hat{\mathbf{v}}^{k}_{p^{k},1}}{\hat{\mathbf{v}}^{k}_{p^{k},3}}\Big{]}\cdot\mathbf{e}^{k}_{\perp,1}&\Big{[}\frac{\partial_{2}\eta_{p^{k}}}{\sqrt{g_{p^{k},22}}}-\frac{\partial_{3}\eta_{p^{k}}}{\sqrt{g_{p^{k},33}}}\frac{\hat{\mathbf{v}}^{k}_{p^{k},2}}{\hat{\mathbf{v}}^{k}_{p^{k},3}}\Big{]}\cdot\mathbf{e}^{k}_{\perp,1}\\\
\Big{[}\frac{\partial_{1}\eta_{p^{k}}}{\sqrt{g_{p^{k},11}}}-\frac{\partial_{3}\eta_{p^{k}}}{\sqrt{g_{p^{k},33}}}\frac{\hat{\mathbf{v}}^{k}_{p^{k},1}}{\hat{\mathbf{v}}^{k}_{p^{k},3}}\Big{]}\cdot\mathbf{e}^{k}_{\perp,2}&\Big{[}\frac{\partial_{2}\eta_{p^{k}}}{\sqrt{g_{p^{k},22}}}-\frac{\partial_{3}\eta_{p^{k}}}{\sqrt{g_{p^{k},33}}}\frac{\hat{\mathbf{v}}^{k}_{p^{k},2}}{\hat{\mathbf{v}}^{k}_{p^{k},3}}\Big{]}\cdot\mathbf{e}^{k}_{\perp,2}\end{array}\right]_{2\times
2},\end{split}$

where $\eta_{p^{k}}$ and $g_{p^{k}}$ are evaluated at
$x^{k}(t,Y({y}_{1},{y}_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$. We also define the
$4\times 4$ specular matrix
$\mathcal{R}^{k,p^{k},Y}=\mathcal{R}^{k,p^{k},Y}(t,y_{1},y_{2},|v|,\hat{v}_{1},\hat{v}_{2})$
as

$\mathcal{R}^{k,p^{k},Y}:=\mathcal{S}^{k,p^{k},Y}\frac{\partial(\mathbf{x}^{k}_{{p}^{k},1},\mathbf{x}^{k}_{{p}^{k},2},\hat{\mathbf{v}}^{k}_{{p}^{k},1},\hat{\mathbf{v}}^{k}_{{p}^{k},1})}{\partial(y_{1},y_{2},\hat{v}_{1},\hat{v}_{2})},$
(6.13)

where
$\mathbf{x}^{k}_{{p}^{k}}=\mathbf{x}^{k}_{{p}^{k}}(t,Y({y}_{1},{y}_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$,
$\mathbf{v}^{k}_{{p}^{k}}=\mathbf{v}^{k}_{{p}^{k}}(t,Y({y}_{1},{y}_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$.
Here, $\hat{v}_{i}$ is directional component in global Eulerian coordinate
$\hat{v}_{i}=\frac{v_{i}}{|v|}$.

Once we perform triple iteration, we can generate extra variables $y_{1}$ and
$y_{2}$. Explicitly, we have

$\displaystyle\left[\begin{array}[]{ccc}\mathbf{e}_{0}^{k}&\mathbf{e}_{\perp,1}^{k}&\mathbf{e}_{\perp,2}^{k}\end{array}\right]^{-1}\frac{\partial
X(s;t,Y_{x}(y_{1},y_{2}),|v|,\hat{v}_{1},\hat{v}_{2})}{\partial(|v|,\hat{v}_{1},\hat{v}_{2},y_{1},y_{2})}$
$\displaystyle=\underbrace{\left[\begin{array}[]{ccccc}\frac{\partial
X}{\partial|v|}\cdot\mathbf{e}_{0}^{k}&\frac{\partial
X}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{0}^{k}&\frac{\partial
X}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{0}^{k}&\frac{\partial X}{\partial
y_{1}}\cdot\mathbf{e}_{0}^{k}&\frac{\partial X}{\partial
y_{2}}\cdot\mathbf{e}_{0}^{k}\\\ \frac{\partial
X}{\partial|v|}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial
X}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial
X}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial
X}{\partial y_{1}}\cdot\mathbf{e}_{\perp,1}^{k}&\frac{\partial X}{\partial
y_{2}}\cdot\mathbf{e}_{\perp,1}^{k}\\\ \frac{\partial
X}{\partial|v|}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial
X}{\partial\hat{v}_{1}}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial
X}{\partial\hat{v}_{2}}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial
X}{\partial y_{1}}\cdot\mathbf{e}_{\perp,2}^{k}&\frac{\partial X}{\partial
y_{2}}\cdot\mathbf{e}_{\perp,2}^{k}\end{array}\right]}.$

Using the specular basis (6.10) and the specular matrix (6.13), the
underbraced term becomes

$\displaystyle{\left[\begin{array}[]{c|c}-(t-s)&\begin{subarray}{c}-|{v}^{k}|\nabla_{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}}t^{k}+\nabla_{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}}\mathbf{x}^{k}_{{p}^{k},\ell}\partial_{\ell}\eta_{{p}^{k}}\cdot\mathbf{e}_{0}^{k}\\\
-(t^{k}-s)|{v}^{k}|\sum_{j=1}^{2}\bigg{(}\sum_{\ell=1}^{3}\frac{\partial}{\partial{\mathbf{x}^{k}_{{p}^{k},j}}}\Big{[}\frac{\partial_{\ell}\eta_{{p}^{k}}}{\sqrt{g_{{p}^{k},\ell\ell}}}\Big{]}\hat{\mathbf{v}}^{k}_{{p}^{k},\ell}\bigg{)}\nabla_{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}}\mathbf{x}^{k}_{{p}^{k},j}\end{subarray}\\\
\hline\cr\begin{subarray}{c}0\\\ 0\end{subarray}&*_{2\times
4}\end{array}\right]_{3\times 5}},$

where the lower right $2\times 4$-submatrix equals

$\begin{split}&\left[\begin{array}[]{cccc}\mathcal{R}_{1,1}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{1,2}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{1,3}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{1,4}^{k,{p}^{k},Y_{x}}\\\
\mathcal{R}_{2,1}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{2,2}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{2,3}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{2,4}^{k,{p}^{k},Y_{x}}\end{array}\right]\\\
&-(t^{k}-s)|{v}^{k}|\left[\begin{array}[]{cccc}\mathcal{R}_{3,1}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{3,2}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{3,3}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{3,4}^{k,{p}^{k},Y_{x}}\\\
\mathcal{R}_{4,1}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{4,2}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{4,3}^{k,{p}^{k},Y_{x}}&\mathcal{R}_{4,4}^{k,{p}^{k},Y_{x}}\end{array}\right].\end{split}$
(6.17)

Here $\mathcal{R}^{k,p^{k},Y_{x}}_{i,j}$ is defined in (6.13) with
$x^{k}=x^{k}(t,Y_{x}(y_{1},y_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$ and
$v^{k}=v^{k}(t,Y_{x}$
$(y_{1},y_{2}),|v|,\hat{v}_{1},\hat{v}_{2})$. From the exact form of
underbraced submatrix, choosing $2\times 2$ submatrix in the underbraced
matrix is equivalent to choosing two independent variables
$\\{\zeta_{1},\zeta_{2}\\}$ among $\\{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}\\}$.
(Here, $(y_{1},y_{2})$ is introduced to represent a local spatial coordinate
of a manifold which is constructed by $\partial_{|v|},\partial_{\hat{v}_{i}}$
in its second Duhamel iteration step.)

Step 3 From the existence of
$\\{\zeta_{1},\zeta_{2}\\}\subset\\{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}\\}$
which gives nondegeneracy of $2\times 2$ submatrix of (6.17), we can prove
full rank property. We recall Lemm 3.7 of [25]. Away from arbitrary small time
intervals near $C^{1}$ function
$\psi_{i}^{k}:B_{\varepsilon}(t,x,v)\rightarrow\mathbb{R}$ (for $i=1,2,3$), we
obtain nondegeneracy. The following Lemma is same as Lemma 3.7 in [25]. We
obtain the following Lemma by replacing $\mathbf{e}_{3}$ into
$\hat{\varphi}(\varphi)$ WLOG.

######  Lemma 6.3.

[$S_{\varphi}$ version for Lemma 3.7 in [25]] Let $x\in S_{\varphi}$. We also
assume at most $M$ bounces during $[t-1,t]$ and

$\begin{split}\frac{1}{N}\leq|v|\leq N,\
\frac{1}{N}\leq|v\cdot\hat{\varphi}(\varphi)|,\
\frac{1}{N}\leq|n(x^{1}(t,x,v))\cdot\hat{\varphi}(\varphi)|,\\\ \text{and}\
{|n(x^{k}(t,x,v))\cdot v^{k}(t,x,v)|}\geq\delta_{1}>0,\quad 1\leq k\leq
M.\end{split}$ (6.18)

where $\hat{\varphi}(\varphi)$ is a unit vector that is perpendicular to the
cross-section $S_{\varphi}$. Let a $C^{1}$-map $Y_{x}:(y_{1},y_{2})\mapsto
Y_{x}(y_{1},y_{2})\in\bar{\Omega}$ with $Y_{x}(0,0)=x$ and
$\|Y\|_{C^{1}_{x,y_{1},y_{2}}}\lesssim 1$. We assume that

$\Big{|}\Big{(}\frac{\partial Y_{x}(0,0)}{\partial y_{1}}\times\frac{\partial
Y_{x}(0,0)}{\partial y_{2}}\Big{)}\cdot
R_{x^{1}(t,x,v)}v^{1}(t,x,v)\Big{|}>\delta_{3}>0.$ (6.19)

For $k\in\mathbb{N}$ with $t^{k}\geq t-1$, there exists $\varepsilon>0$ and
finitely many $C^{0,\gamma}$-functions
$\psi^{k}_{i}:B_{\varepsilon}(t,x,v)\rightarrow\mathbb{R}$ with
$\|\psi^{k}_{i}\|_{C^{0,\gamma}_{t,x}}\lesssim 1$, and there exists a constant
$\epsilon_{\delta_{1},\delta_{3},N,\Omega,\delta_{*}}>0$ and
$\\{\zeta_{1},\zeta_{2}\\}\subset\\{\hat{v}_{1},\hat{v}_{2},y_{1},y_{2}\\}$
such that

$\displaystyle\text{if }\
\min_{i}|s-\psi_{i}^{k}(t,Y_{x}(y_{1},y_{2}),v)|>\delta_{*}$ $\displaystyle\ \
\ \ \text{and}\
(s;t,Y_{x}(y_{1},y_{2}),v)\in[\max\\{t-1,t^{k+1}\\},\min\\{t-\frac{1}{N},t^{k}\\}]\times
B_{\varepsilon}(t,x,v),$ $\displaystyle\text{then}\ \det\left(\frac{\partial
X(s;t,Y_{x}(y_{1},y_{2}),|v|,\hat{v}_{1},\hat{v}_{2})}{\partial(|v|,\zeta_{1},\zeta_{2})}\right)>\epsilon_{\delta_{1},\delta_{3},N,\Omega,\delta_{*}}>0.$

Before we apply above Lemma to the following Proposition, we again note that
the second and third conditions of (6.18) are written as

$\begin{split}\frac{1}{N}\leq|v_{3}|,\quad\frac{1}{N}\leq|n(x^{1}(t,x,v))\cdot
e_{3}|,\end{split}$

in [25], where the choice of $e_{3}$ can be changed into any other fixed
direction vector in $\mathbb{S}^{2}$ WLOG. Since we are dealing with $x\in
S_{\varphi}$, we choose $\hat{\varphi}(\varphi)$ instead of $e_{3}$ which is
more suitable in our domain.

Modifying Theorem 3.9 of [25] into $\mathcal{Z}^{\varepsilon}$ version
($\mathcal{Z}^{\varepsilon}$ is obtained in Theorem 5.7), we obtain the
following Proposition.

######  Proposition 6.4.

[$\mathcal{Z}^{\varepsilon}$ version for Theorem 3.9 in [25]] Fix any
arbitrary $(t,x,v)\in[T,T+1]\times\overline{\Omega}\times\mathbb{V}^{N}$.
There exists $\delta_{2}>0$ and $C^{1}$-function
$\psi^{\ell_{0},\vec{\ell},k}$ for $k\leq M$ where
$\psi^{\ell_{0},\vec{\ell},k}$ is defined locally around
$(T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell})$
with
$(\ell_{0},\vec{\ell})=(\ell_{0},\ell_{1},\ell_{2},\ell_{3})\in\\{0,1,\cdots,\lfloor\frac{1}{\delta_{2}}\rfloor+1\\}\times\\{-\lfloor\frac{N}{\delta_{2}}\rfloor-1,\cdots,0,\cdots,\lfloor\frac{N}{\delta_{2}}\rfloor+1\\}^{3}$
and $\|\psi^{\ell_{0},\vec{\ell},k}\|_{C^{1}}\leq
C_{N,\Omega,\varepsilon,\delta_{2}}<\infty$.

Moreover, if

$(X(s;t,x,v),u)\in\mathcal{Z}^{\varepsilon},$ (6.20)

$(s,u)\in[T+(\ell_{0}-1)\delta_{2},T+(\ell_{0}+1)\delta_{2}]\times
B(\delta_{2}\vec{\ell};2\delta_{2}),$ (6.21)

$s^{\prime}\in\big{[}t^{k+1}(T+\delta_{2}\ell_{0};X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell})+\frac{1}{N},\
t^{k}(T+\delta_{2}\ell_{0};X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell})-\frac{1}{N}\big{]},$
(6.22)

and

$|s^{\prime}-\psi^{\ell_{0},\vec{\ell},k}(T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell})|>N^{2}(1+\|\psi^{\ell_{0},\vec{\ell},k}\|_{C^{1}})\delta_{2},$
(6.23)

then

$\big{|}\partial_{|u|}X(s^{\prime};s,X(s;t,x,v),u)\times\partial_{\hat{u}_{1}}X(s^{\prime};s,X(s;t,x,v),u)\big{|}>\epsilon_{\Omega,N,\varepsilon,\delta_{2}}.$
(6.24)

Here $\epsilon_{\Omega,N,\varepsilon,\delta_{2}}>0$ does not depend on
$T,t,x,v$.

There exists $\delta_{3}>0$ and $C^{1}$-functions

$\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{1},\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{2},\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{3},$
(6.25)

for $k^{\prime}\leq M$ where
$\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{n}$ is defined locally
around
$(T+\delta_{3}m_{0};X(T+\delta_{3}m_{0};T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell}),\delta_{3}\vec{m})$
for some
$(m_{0},\vec{m})=(m_{0},m_{1},m_{2},m_{3})\in\\{0,1,\cdots,\lfloor\frac{1}{\delta_{3}}\rfloor+1\\}\times\\{-\lfloor\frac{N}{\delta_{3}}\rfloor-1,\cdots,0,\cdots,\lfloor\frac{N}{\delta_{3}}\rfloor+1\\}^{3}$
with $0<\delta_{3}\ll 1$.

Moreover, if we assume (6.20), (6.21), (6.22), (6.23),

$\begin{split}&(X(s^{\prime};s,X(s;t,x,v),u),u^{\prime})\in\mathcal{Z}^{\varepsilon},\
\text{and}\\\
&\quad\quad\Big{|}\Big{(}\frac{\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)}{|\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)|}\Big{)}\cdot
u^{\prime}\Big{|}>\frac{1}{N}>0,\end{split}$ (6.26)
$\begin{split}s^{\prime\prime}\in&\
\Big{[}t^{k^{\prime}+1}(T+\delta_{3}m_{0};X(T+\delta_{3}m_{0};T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell}),\delta_{3}\vec{m})+\frac{1}{N}\\\
&\ \ \ \
,t^{k^{\prime}}(T+\delta_{3}m_{0};X(T+\delta_{3}m_{0};T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell}),\delta_{3}\vec{m})-\frac{1}{N}\Big{]},\end{split}$
(6.27)

and

$\begin{split}&\min_{n=1,2,3}|s^{\prime\prime}-\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{n}(T+\delta_{3}m_{0};X(T+\delta_{3}m_{0};T+\delta_{2}\ell_{0},X(T+\delta_{2}\ell_{0};t,x,v),\delta_{2}\vec{\ell}),\delta_{3}\vec{m})|\\\
&>N^{2}(1+\max_{n=1,2,3}\|\psi^{\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}}_{n}\|_{C^{1}})(\delta_{3}),\end{split}$
(6.28)

then for each $\ell_{0},\vec{\ell},k,m_{0},\vec{m},k^{\prime}$ we can choose
two distinct variables
$\\{\zeta_{1},\zeta_{2}\\}\subset\\{|u|,\hat{u}_{1},\hat{u}_{1}^{\prime},\hat{u}_{2}^{\prime}\\}$
such that $(|u^{\prime}|,\zeta_{1},\zeta_{2})\mapsto
X(s^{\prime\prime};s^{\prime},X(s^{\prime};s,X(s;t,x,v),u),u^{\prime})$ is
one-to-one locally and

$\Big{|}\det\left(\frac{\partial
X(s^{\prime\prime};s^{\prime},X(s^{\prime};s,X(s;t,x,v),u),u^{\prime})}{\partial(|u^{\prime}|,\zeta_{1},\zeta_{2})}\right)\Big{|}>\epsilon^{\prime}_{\Omega,N,\varepsilon,\delta_{2},\delta_{3}}.$
(6.29)

Here $\epsilon^{\prime}_{\Omega,N,\varepsilon,\delta_{2},\delta_{3}}>0$ does
not depend on $T,t,x,v$.

###### Proof.

Main proof is nearly the same as the proof of Theorem 3.9 in [25]. Proof of
Theorem 3.9 in [25] comes from Lemma 3.7 in [25]. Similarly, Proposition 6.4
can be proved using Lemma 6.3. Instead of providing full proof, we just
explain why it is possible to apply Lemma 6.3 and Theorem 5.7.
Let us combine Theorem 5.7 and Lemma 6.3. First, when we assume Theorem 5.7,
we should check that $(x,v)\in\mathcal{Z}^{\varepsilon}$ satisfies (6.18) in
Lemma 6.3. The first and fourth conditions of (6.18) hold obviously by the
statement of Theorem 5.7, by the definition of $\mathbb{V}^{N}$ and choosing
$\delta_{1}=C^{*}_{\varepsilon,L}$ of Theorem 5.7. Let us check the second and
third conditions.
For the second condition, we recall Lemma 4.16 in which we excluded
$R_{\perp}^{\varepsilon}$. If we consider Lemma 4.16 in general $S_{\varphi}$
cross-section, excluding $R_{\perp}^{\varepsilon}$ exactly corresponds to
$\frac{1}{N}\leq|v\cdot\hat{\varphi}(\varphi)|$ for some proper
$\varepsilon\ll 1$ and $N\gg 1$. The third condition of (6.18) is also similar
because it is easy to observe

$|v\cdot\hat{\varphi}(\varphi)|=0\Longleftrightarrow|n(x^{1}(t,x,v))\cdot\hat{\varphi}(\varphi)|=0,\quad
x\in\overline{S_{\varphi}},$

by $z$-axisymmetric structure. In conclusion,
$(x,v)\in\mathcal{Z}^{\varepsilon}$ satisfies all conditions in (6.18).
Now, above argument is applied to $(X(s;t,x,v),u)\in\mathcal{Z}^{\varepsilon}$
and $(X(s^{\prime};s,X(s;t,x,v),u)$
$,u^{\prime})\in\mathcal{Z}^{\varepsilon}$ by (6.20) and (6.26). Moreover,
condition (6.26) implies (6.19) putting $x^{0}=X(s^{\prime};s,X(s;t,x,v),u)$.
Therefore, the proof of Theorem 3.9 of [25] can be directly applied to prove
Proposition 6.4. Note that $\delta_{1}$ of Lemma 6.3 is chosen by
$\delta_{1}=C^{*}_{\varepsilon,L}$ by Theorem 5.7. So choosing $L=NT_{0}=N$
(we choose time interval $T_{0}=1$ here), we get the dependence of
$\epsilon^{\prime}_{\Omega,N,\varepsilon,\delta_{2},\delta_{3}}$. We finish
the proof. ∎

##  7\. Triple Duhamel iterations

Note that the kernel $\mathbf{k}(v,u)$ satisfies (see [16, 12])

$\mathbf{k}(v,u)\lesssim\big{(}\frac{1}{|v-u|}+|v-u|\big{)}e^{-\frac{1}{8}|v-u|^{2}-\frac{1}{8}\frac{||v|^{2}-|u|^{2}|^{2}}{|v-u|^{2}}}.$

Now we can write an equation for $f$ from (1.1) and (1.16):

Motivated by characteristic $(X(s;t,x,v),V(s;t,x,v))$, we define mild solution
of (1.21):

$\begin{split}f(t,x,v)&=e^{-\int_{0}^{t}\nu(V(\tau;t,x,v))d\tau}f_{0}(X(0;t,x,v),V(0;t,x,v))\\\
&\quad+\int_{0}^{t}e^{-\int_{s}^{t}\nu(V(\tau;t,x,v))d\tau}\int_{u\in\mathbb{R}^{3}}k(V(s;t,x,v),u)f(s,X(s;t,x,v),u)duds\\\
&\quad+\int_{0}^{t}e^{-\int_{s}^{t}\nu(V(\tau;t,x,v))d\tau}\Gamma(f,f)(s,X(s;t,x,v),V(s;t,x,v))ds.\\\
\end{split}$ (7.1)

In this section, we also briefly explain triple Duhamel expansion which was
explained in [25]. First, let $f$ solves linearized Boltzmann equation (7.2)

$\begin{split}&\partial_{t}f+v\cdot\nabla f+\nu f=Kf,\quad
f(t,x,v)=f(t,x,R_{x}v)\ \ \text{on $x\in\partial\Omega$}.\end{split}$ (7.2)

Let $h:=wf$ where $w=(1+|v|)^{\beta},\ \beta>5/2$. Then the linearized
Boltzmann equation becomes

$\begin{split}&\partial_{t}h+v\cdot\nabla_{x}h+\nu h=K_{w}h,\quad
K_{w}h:=wK(\frac{h}{w}).\end{split}$

Let us define,

$\begin{split}E(v,t,s)&:=\exp\Big{\\{}-\int_{s}^{t}\nu(V(\tau;t,x,v))\Big{\\}}.\\\
\end{split}$

Along the backward in time trajectory,

$\begin{split}&\frac{d}{ds}\Big{(}E(v,t,s)h(s,X(s;t,x,v),V(s;t,x,v))\Big{)}\\\
&={E(v,t,s)}\big{[}K_{w}h\big{]}(s,X(s;t,x,v),V(s;t,x,v)).\end{split}$

By integrating from $0$ to $t$, we obtain the first Duhamel expansion,

$\begin{split}h(t,x,v)=&E(v,t,0)h_{0}(X(0),V(0))\\\
&+\int^{t}_{0}E(v,t,s)\int_{\mathbb{R}^{3}}k_{w}(u,V(s))h(s,X(s;t,x,v),u)\mathrm{d}u\mathrm{d}s,\end{split}$
(7.3)

where we abbreviated $X(s)=X(s;t,x,v)$. Recall the standard estimates (see
Lemma 4 and Lemma 5 in [19])

$\int_{\mathbb{R}^{3}}|k_{w}(v,u)|\mathrm{d}u\leq C_{K}\langle v\rangle^{-1}.$
(7.4)

From double and triple iteration, we get

$\displaystyle h(t,x,v)$ $\displaystyle=$ $\displaystyle
E(v,t,0)h_{0}(X(0),V(0))$
$\displaystyle+\int^{t}_{0}E(v,t,s)\int_{u}k_{w}(u,V(s))E(u,s,0)h_{0}(X(0),V(0))duds$
$\displaystyle+\int^{t}_{0}E(v,t,s)\int_{u}k_{w}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})$
$\displaystyle\quad\times\int_{u^{\prime}}k_{w}(u^{\prime},V(s^{\prime}))h(s^{\prime},X(s^{\prime}),u^{\prime})du^{\prime}ds^{\prime}duds$
$\displaystyle=$ $\displaystyle E(v,t,0)h_{0}(X(0),V(0))$
$\displaystyle+\int^{t}_{0}E(v,t,s)\int_{u}k_{w}(u,V(s))E(u,s,0)h_{0}(X(0),V(0))duds$
$\displaystyle+\int^{t}_{0}E(v,t,s)\int_{u}k_{w}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})$
$\displaystyle\quad\times\int_{u^{\prime}}k_{w}(u^{\prime},V(s^{\prime}))E(u^{\prime},s^{\prime},0)h_{0}(X(0),V(0))du^{\prime}ds^{\prime}duds$
$\displaystyle+\int^{t}_{0}E(v,t,s)\int_{u}k_{w}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})\int_{u^{\prime}}k_{w}(u^{\prime},V(s^{\prime}))E(u^{\prime},s^{\prime},s^{\prime\prime})\quad\quad$
$\displaystyle\quad\times\int_{u^{\prime\prime}}k_{w}(u^{\prime\prime},V(s^{\prime\prime}))h(s^{\prime\prime},X(s^{\prime\prime}),u^{\prime\prime})du^{\prime\prime}ds^{\prime\prime}du^{\prime}ds^{\prime}duds,$

where we abbreviated notations (similar for $V$),

$\begin{split}X(s)&:=X(s;t,x,v),\quad
X(s^{\prime}):=X^{\prime}(s^{\prime};s,X(s;t,x,v),u),\\\
X(s^{\prime\prime})&:=X(s^{\prime\prime};s^{\prime},X^{\prime}(s^{\prime};s,X(s;t,x,v),u),u^{\prime}).\end{split}$
(7.6)

Now, we choose $m({N})$ so that

$k_{w,m}(u,v):=\mathbf{1}_{\\{|u-v|\geq\frac{1}{m},\ |u|\leq m\\}}k_{w}(u,v),$
(7.7)

satisfies $\int_{\mathbb{R}^{3}}|k_{w,m}(u,v)-k_{w}(u,v)|\
\mathrm{d}u\leq\frac{1}{{N}}$ for sufficiently large ${N}\geq 1$. From (7.7)
and uniform lower bound of collision frequency, we can rewrite (7) as

$\displaystyle h(t,x,v)$ $\displaystyle=E(v,t,0)h_{0}(X(0),V(0))$
$\displaystyle\quad+\int^{t}_{0}E(v,t,s)\int_{u\in\mathbb{V}^{N}}k_{w,m}(u,V(s))E(u,s,0)h_{0}(X(0),V(0))duds$
$\displaystyle\quad+\int^{t}_{0}E(v,t,s)\int_{u\in\mathbb{V}^{N}}k_{w,m}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})$
$\displaystyle\quad\quad\times\int_{u^{\prime}}k_{w}(u^{\prime},V(s^{\prime}))E(u^{\prime},s^{\prime},0)h_{0}(X(0),V(0))du^{\prime}ds^{\prime}duds$
$\displaystyle\quad+\int^{t}_{0}E(v,t,s)\int_{u}k_{w,m}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})\int_{u^{\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime},V(s^{\prime}))E(u^{\prime},s^{\prime},s^{\prime\prime})$
$\displaystyle\quad\quad\times\int_{u^{\prime\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime\prime},V(s^{\prime\prime}))h(s^{\prime\prime},X(s^{\prime\prime}),u^{\prime\prime})$
$\displaystyle\quad+O_{\Omega}(\frac{1}{N})\sup_{0\leq s\leq
t}\|h(s)\|_{L^{\infty}}$ $\displaystyle\lesssim
e^{-\frac{\nu_{0}}{2}t}\|h_{0}\|_{L^{\infty}}+O_{\Omega}(\frac{1}{N})\sup_{0\leq
s\leq t}\|h(s)\|_{L^{\infty}}$
$\displaystyle\quad+\int^{t}_{0}E(v,t,s)\int_{u}k_{w,m}(u,V(s))\int^{s}_{0}E(u,s,s^{\prime})\int_{u^{\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime},V(s^{\prime}))E(u^{\prime},s^{\prime},s^{\prime\prime})$
$\displaystyle\quad\quad\times\int_{u^{\prime\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime\prime},V(s^{\prime\prime}))h(s^{\prime\prime},X(s^{\prime\prime}),u^{\prime\prime}).$
(7.8)

Let us analyze (7.8). We define following sets for fixed
$\delta,n,\vec{n},k,m,\vec{m},k^{\prime}$, where Proposition 6.4 does not
hold,

$\begin{split}R_{1}&:=\\{u\ |\ (X(s;t,x,v),u)\notin\mathcal{Z}^{\varepsilon}\
\text{and}\ u\notin B(\vec{n}\delta;2\delta)\\},\\\ R_{2}&:=\\{s^{\prime}\ |\
|s-s^{\prime}|\leq\delta\\},\\\ R_{3}&:=\\{s^{\prime}\ |\
|s^{\prime}-\psi_{1}^{n,\vec{n},k,m,\vec{m},k^{\prime}}(n\delta,X(n\delta;t,x,v),\vec{n}\delta)|\lesssim_{N}\delta\|\psi_{1}\|_{C^{1}}\\},\\\
R_{4}&:=\\{u^{\prime}\ |\
(X(s^{\prime};s,X(s;t,x,v),u),u^{\prime})\notin\mathcal{Z}^{\varepsilon}\
\text{and}\ u^{\prime}\notin B(\vec{m}\delta;2\delta)\\},\\\
R_{5}&:=\\{u^{\prime}\ |\
\Big{|}\Big{(}\frac{\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)}{|\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)|}\Big{)}\cdot
u^{\prime}\Big{|}\leq\delta\\},\\\ R_{6}&:=\\{s^{\prime\prime}\ |\
|s^{\prime}-s^{\prime\prime}|\leq\delta\\},\\\ R_{7}&:=\\{s^{\prime\prime}\ |\
\min_{r=1,2}|s^{\prime\prime}-\psi_{r}^{n,\vec{n},k,m,\vec{m},k^{\prime}}(m\delta,X(m\delta;n\delta,X(n\delta;t,x,v),\vec{m}\delta),\vec{n}\delta)|\lesssim_{N}\delta\min_{r=1,2}\|\psi_{r}\|_{C^{1}}\\}.\\\
\end{split}$

Therefore, (7.8) can be controlled by

$\displaystyle\lesssim\sum_{n=0}^{[t/\delta]+1}\sum_{|\vec{n}|\leq
N}\sum_{m=0}^{[t/\delta]+1}\sum_{|\vec{m}|\leq
N}\sum_{k=1}^{M}\sum_{k^{\prime}=1}^{M}\int^{(n+1)\delta}_{(n-1)\delta}\
\int^{t^{k}-\delta}_{t^{k+1}+\delta}\int^{t^{k^{\prime}}-\delta}_{t^{k^{\prime}+1}+\delta}e^{-{\nu_{0}}(t-s^{\prime\prime})}$
$\displaystyle\quad\quad\times\int_{u,u^{\prime},u^{\prime\prime}\in\mathbb{V}^{N}}\
|h(s^{\prime\prime},X(s^{\prime\prime}),u^{\prime\prime})|\
\mathbf{1}_{R_{1}^{c}\cap R_{2}^{c}\cap R_{3}^{c}\cap R_{4}^{c}\cap
R_{5}^{c}\cap R_{6}^{c}}$ (7.9) $\displaystyle\quad+B+R.$

Here $B$ term corresponds to when the trajectory locates near bouncing points
and $R$ corresponds to where $(u,s^{\prime},u^{\prime},s^{\prime\prime})$
belong to at least one of $R_{1}\sim R_{7}$. Exact definition and estimates
for $B$ and $R$ are given as following

$\begin{split}B&\leq\int_{0}^{t}\int_{0}^{s}\int_{0}^{s^{\prime}}e^{-{\nu_{0}}(t-s^{\prime\prime})}\int_{u\in\mathbb{V}^{N}}k_{w,m}(u,v)\int_{u^{\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime},u)\\\
&\quad\quad\times\int_{u^{\prime\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime\prime},u^{\prime})h(s^{\prime\prime},X^{\prime\prime}(s^{\prime\prime}),u^{\prime\prime})\mathbf{1}_{|s^{\prime}-t^{k}|\leq\delta\
\text{or}\ |s^{\prime\prime}-t^{k^{\prime}}|\leq\delta}\\\ &\leq
C_{N}\delta\sup_{0\leq s\leq t}\|h(s)\|_{\infty},\\\
R&\leq\int_{0}^{t}\int_{0}^{s}\int_{0}^{s^{\prime}}e^{-{\nu_{0}}(t-s^{\prime\prime})}\int_{u\in\mathbb{V}^{N}}k_{w,m}(u,v)\int_{u^{\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime},u)\\\
&\quad\quad\times\int_{u^{\prime\prime}\in\mathbb{V}^{N}}k_{w,m}(u^{\prime\prime},u^{\prime})h(s^{\prime\prime},X^{\prime\prime}(s^{\prime\prime}),u^{\prime\prime})\
\mathbf{1}_{R_{1}\cup R_{2}\cup R_{3}\cup R_{4}\cup R_{5}\cup R_{6}\cup
R_{7}}\\\ &\leq C_{N}\delta\sup_{0\leq s\leq t}\|h(s)\|_{\infty}.\\\
\end{split}$ (7.10)

In particular, for about $R$ estimate, deriving smallness
$(\delta+\varepsilon)$ from $R_{2},R_{3},R_{5},R_{6},R_{7}$ is easy. To derive
smallness from $R_{1}$ and $R_{4}$, we use (5.42) in Theorem 5.7 to obtain

$\int_{u}k_{w,m}(u,V(s))\mathbf{1}_{R_{1}}\lesssim\delta,\quad\int_{u^{\prime}}k_{w,m}(u^{\prime},V(s^{\prime}))\mathbf{1}_{R_{4}}\lesssim\delta,$

choosing $\varepsilon$ sufficiently small depending on $N\gg 1$ in (5.42).
For (7.9), we are away from two sets $B$ and $R$. Under the condition of
$(u,s^{\prime},u^{\prime},s^{\prime\prime})\in R_{1}^{c}\ \cap\ R_{2}^{c}\
\cap R_{3}^{c}\ \cap R_{4}^{c}\ \cap R_{5}^{c}\ \cap R_{6}^{c}\ \cap
R_{7}^{c}$, indices $n,\vec{n},k,m,\vec{m},k^{\prime}$ are determined so that

$\begin{split}t\ &\in\ [(n-1)\delta,(n+1)\delta],\\\ (X(s;t,x,v),u)\ &\in\
\mathcal{Z}^{\varepsilon},\\\ u\ &\in\ B(\vec{n}\delta;2\delta),\\\
(X(s^{\prime};s,X(s;t,x,v),u),u^{\prime})\ &\in\ \mathcal{Z}^{\varepsilon},\\\
u^{\prime}\ &\in\ B(\vec{m}\delta;2\delta),\\\
\Big{|}\Big{(}\frac{\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)}{|\partial_{y_{1}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)\times\partial_{y_{2}}Y_{X(s^{\prime};s,X(s;t,x,v),u)}(0,0)|}\Big{)}\cdot
u^{\prime}\Big{|}&\geq\delta.\end{split}$ (7.11)

We can apply Proposition 6.4 which gives local time-independent lower bound of

$\Big{|}\det(\frac{\partial(X(s^{\prime\prime}))}{\partial(|u^{\prime}|,\zeta_{1},\zeta_{2})})\Big{|}\
\geq\ \epsilon^{\prime}_{\delta}.$

Note that
$\\{\zeta_{1},\zeta_{2}\\}\subset\\{|u|,\hat{u}_{1},\hat{u}^{\prime}_{1},\hat{u}^{\prime}_{2}\\}$
are chosen variables in Theorem 6.4 and
$\\{\zeta_{3},\zeta_{4}\\}\subset\\{|u|,\hat{u}_{1},\hat{u}^{\prime}_{1},\hat{u}^{\prime}_{2}\\}$
are unchosen variables. Let us use $\mathcal{P}$ to denote projection of

$B(\vec{n}\delta;2\delta)\cap(\mathcal{Z}^{\varepsilon})_{X(s;t,x,v)}\times
B(\vec{m}\delta;2\delta)\cap(\mathcal{Z}^{\varepsilon})_{X(s^{\prime};s,X(s;t,x,v),u)},$

into $\mathbb{R}^{3}$ which corresponds to
$(|u^{\prime}|,\zeta_{1},\zeta_{2})$ components. Note that both
$(\mathcal{Z}^{\varepsilon})_{X(s;t,x,v)}$ and
$(\mathcal{Z}^{\varepsilon})_{X(s^{\prime};s,X(s;t,x,v),u)}$ are defined using
definition (4.46). If we choose sufficiently small $\delta$, there exist small
$r_{\delta,n,\vec{n},k,m,\vec{m},k^{\prime}}$ such that there exist one-to-one
map $\mathcal{M}$,

$\displaystyle\mathcal{M}$ $\displaystyle:$
$\displaystyle\mathcal{P}\Big{(}B(\vec{n}\delta;2\delta)\cap(\mathcal{Z}^{\varepsilon})_{X(s;t,x,v)}\times
B(\vec{m}\delta;2\delta)\cap(\mathcal{Z}^{\varepsilon})_{X(s^{\prime};s,X(s;t,x,v),u)}\Big{)}$
$\displaystyle\quad\quad\mapsto
B(X(s^{\prime\prime};s^{\prime},X(s^{\prime};s,X(s;t,x,v),u),u^{\prime}),r_{\delta,n,\vec{n},k,m,\vec{m},k^{\prime}}).$

Now, we perform change of variable for in (7.9) to obtain

$\begin{split}\text{\eqref{main2}}&\leq\sum_{n=0}^{[t/\delta]+1}\sum_{|\vec{n}|\leq
N}\sum_{m=0}^{[t/\delta]+1}\sum_{|\vec{m}|\leq
N}\sum_{k=1}^{M}\sum_{k^{\prime}=1}^{M}\int^{(n+1)\delta}_{(n-1)\delta}\
\int^{t^{k}-\delta}_{t^{k+1}+\delta}\int^{t^{k^{\prime}}-\delta}_{t^{k^{\prime}+1}}e^{-{\nu_{0}}(t-s^{\prime\prime})}\\\
&\quad\quad\times\int_{u^{\prime\prime}}\mathrm{d}u^{\prime\prime}\int_{\hat{u}_{2},\zeta_{3},\zeta_{4}}\
\mathbf{1}_{|u|\leq N,|u^{\prime}|\leq N,|u^{\prime\prime}|\leq N}\
\mathrm{d}\hat{u}_{2}\mathrm{d}\zeta_{3}\mathrm{d}\zeta_{4}\\\
&\quad\quad\times\int_{|u^{\prime}|,\zeta_{1},\zeta_{2}}\
\mathrm{d}|u^{\prime}|\mathrm{d}\zeta_{1}\mathrm{d}\zeta_{2}\
|h(s^{\prime\prime},X(s^{\prime\prime}),u^{\prime\prime})|\
\mathrm{d}s\mathrm{d}s^{\prime}\mathrm{d}s^{\prime\prime}\\\
&\leq\sum_{n=0}^{[t/\delta]+1}\sum_{|\vec{n}|\leq
N}\sum_{m=0}^{[t/\delta]+1}\sum_{|\vec{m}|\leq
N}\sum_{k=1}^{M}\sum_{k^{\prime}=1}^{M}\int^{(n+1)\delta}_{(n-1)\delta}\
\int^{t^{k}-\delta}_{t^{k+1}+\delta}\int^{t^{k^{\prime}}-\delta}_{t^{k^{\prime}+1}+\delta}e^{-{\nu_{0}}(t-s^{\prime\prime})}\\\
&\quad\quad\times\int_{\hat{u}_{2},\zeta_{3},\zeta_{4}}\ \mathbf{1}_{|u|\leq
N,|u^{\prime}|\leq N,|u^{\prime\prime}|\leq N}\
\mathrm{d}\hat{u}_{2}\mathrm{d}\zeta_{3}\mathrm{d}\zeta_{4}\\\
&\quad\quad\times\int_{u^{\prime\prime}}\int_{B(X(s^{\prime\prime}),r_{\delta,n,\vec{n},i,k,m,\vec{m},j,k^{\prime}})}\
|h(s^{\prime\prime},x,u^{\prime\prime})|\
\frac{1}{\epsilon^{\prime}_{\Omega,N,\varepsilon,\delta}}\mathrm{d}x\mathrm{d}u^{\prime\prime}\mathrm{d}s\mathrm{d}s^{\prime}\mathrm{d}s^{\prime\prime}\\\
&\leq\
C_{\Omega,N,\varepsilon,\delta}\int_{0}^{t}e^{-{\nu_{0}}(t-s^{\prime\prime})}\
\int_{\Omega}\int_{|u^{\prime\prime}|\leq
N}|h(s^{\prime\prime},x,u^{\prime\prime})|\
\mathrm{d}u^{\prime\prime}\mathrm{d}x\mathrm{d}s^{\prime\prime}\\\ &\leq\
C_{t,\Omega,N,\varepsilon,\delta}\int_{0}^{t}\|h(s)\|_{L^{2}_{x,v}}ds.\\\
\end{split}$ (7.12)

We collect (7.12), (7.9), (7.10), and choose sufficiently large $N$ and
(corresponding) sufficiently small $\varepsilon,\delta$ to obtain

$\|h(t)\|_{L^{\infty}}\lesssim
e^{-\frac{\nu_{0}}{2}t}\|h(0)\|_{L^{\infty}}+C_{t}\int_{0}^{t}\|h(s)\|_{L^{2}}ds.$
(7.13)

##  8\. Linear and nonlinear decay : Proof of main theorem

This part is also standard and similar as [26] and [16]. Let us briefly
explain key steps. We first start with linear $L^{2}_{x,v}$ decay.

###  8.1. Linear $L^{2}$ decay

We use a standard coercivity estimate (Proposition 1.4 in [25]):

######  Proposition 8.1.

Let $f$ solves linearized Boltzmann equation (7.2) with the specular BC.
Furthermore, we assume (1.46). Then there exists $C>0$ such that for all
$N\in\mathbb{N}$,

$\int^{N+1}_{N}\|\mathbf{P}f(t)\|_{2}^{2}\mathrm{d}t\leq
C\int^{N+1}_{N}\|(\mathbf{I}-\mathbf{P})f(t)\|_{\nu}^{2}\mathrm{d}t.$ (8.1)

######  Remark 8.2.

In [16], Guo obtained the coercivity estimate (8.1) by using contradiction
method. When we consider a periodic cylindrical domain with an annulus cross-
section, we derive the same coercivity estimate through a constructive method
in Appendix, Proposition 9.6.

######  Corollary 8.3.

Assume the same conditions of Proposition 8.1. Then we have following estimate

$\sup_{0\leq t}e^{\lambda
t}\|f(t)\|_{L^{2}_{x,v}}^{2}\lesssim\|f_{0}\|_{L^{2}_{x,v}}^{2},$ (8.2)

for some $\lambda>0$.

###### Proof.

From Green’s identity (or $L^{2}$ energy estimate) for (7.2) in a time
interval $[0,N]$,

$\|f(N)\|_{2}^{2}+\int^{N}_{0}\iint_{U\times\mathbb{R}^{3}}fLf\leq\|f(0)\|_{2}^{2}.$

From (7.2), for any $\lambda>0$, we get

$\big{[}\partial_{t}+v\cdot\nabla_{x}\big{]}(e^{\lambda t}f)+L(e^{\lambda
t}f)=\lambda e^{\lambda t}f,$

Green’s identity (or $L^{2}$ energy estimate) yields

$\|e^{\lambda
t}f(N)\|_{2}^{2}+\underbrace{\int^{N}_{0}\iint_{U\times\mathbb{R}^{3}}e^{2\lambda
s}fLf}_{(I)}-{\lambda\int^{N}_{0}\iint_{U\times\mathbb{R}^{3}}|e^{\lambda
s}f(s)|^{2}}\leq\|f(0)\|_{2}^{2}.$ (8.3)

Let us consider $(I)$ in (8.3). From semi-positivity of linearized Boltzmann
operator $L$, $(I)$ is bounded from below as

$\displaystyle(I)\geq\delta_{L}\int^{N}_{0}\iint_{U\times\mathbb{R}^{3}}\langle
v\rangle|e^{\lambda
s}(\mathbf{I}-\mathbf{P})f|^{2}\geq\delta_{L}\int^{N}_{0}\|e^{\lambda
s}(\mathbf{I}-\mathbf{P})f\|_{\nu}^{2}.$

By time translation, we apply coercivity estimate (8.1) to obtain

$\begin{split}(I)&\geq\frac{\delta_{L}}{2}\int^{N}_{0}\|e^{\lambda
s}(\mathbf{I}-\mathbf{P})f\|_{\nu}^{2}+\frac{\delta_{L}}{2C}\int^{N}_{0}\|e^{\lambda
s}\mathbf{P}f\|_{2}^{2}\geq\frac{\delta_{L}}{2C}\int^{N}_{0}\|e^{\lambda
s}f\|_{2}^{2},\end{split}$

where $\|\cdot\|_{\nu}=\|\cdot\sqrt{\nu}\|_{L^{2}_{x,v}}$. Therefore, we
derive

$e^{2\lambda
N}\|f(N)\|_{2}^{2}+\Big{(}\frac{\delta_{L}}{2C}-\lambda\Big{)}\int^{N}_{0}\|e^{\lambda
s}f(s)\|_{2}^{2}ds\leq\|f_{0}\|_{2}^{2}.$ (8.4)

For a time interval $[N,t]$, it is not hard to get

$\|f(t)\|_{2}^{2}\leq\|f(N)\|_{2}^{2}.$ (8.5)

Choosing $\lambda\ll 1$, from (8.4) and (8.5), we conclude that

$e^{\lambda t}\|f(t)\|_{2}^{2}=e^{\lambda(t-N)}e^{\lambda
N}\|f(N)\|_{2}^{2}\leq 2\|f_{0}\|_{2}^{2},$

and obtain (8.2). ∎

###  8.2. Nonlinear $L^{\infty}$ decay

Now, we are ready to derive nonlinear $L^{\infty}_{x,v}$ decay from previous
$L^{2}_{x,v}$ decay.

###### Proof of Theorem 1.8.

From (7.13), for $T\leq t<T+1$,

$\begin{split}\sup_{s\in[T,t]}\|h(s)\|_{\infty}&\lesssim\
e^{-\nu_{0}(t-T)}\|h(T)\|_{\infty}+\int_{T}^{t}\|f(s)\|_{2}\mathrm{d}s.\end{split}$

We assume that $m\leq t<m+1$ and define
$\lambda^{*}:=\min\\{\nu_{0},\lambda\\}$, where $\lambda$ is some constant
from Corollary 8.3. We use (7.13) repeatedly for each time step, $[k,k+1),\
k\in\mathbb{N}$ and Corollary 8.3 to perform $L^{2}-L^{\infty}$ bootstrap,

$\begin{split}\|h(t)\|_{\infty}&\lesssim
e^{-m\nu_{0}}\|h(0)\|_{\infty}+\sum_{k=0}^{m-1}e^{-k\nu_{0}}\int_{m-1-k}^{m-k}\|f(s)\|\mathrm{d}s\\\
&\lesssim
e^{-m\nu_{0}}\|h(0)\|_{\infty}+\sum_{k=0}^{m-1}e^{-k\nu_{0}}\int_{m-1-k}^{m-k}e^{-\lambda(m-1-k)}\|f(0)\|\mathrm{d}s\lesssim
e^{-\lambda^{*}t}\|h(0)\|_{\infty}.\\\ \end{split}$

For nonlinear problem from Duhamel principle,

$\begin{split}h&:=U(t)h_{0}+\int_{0}^{t}U(t-s)w\Gamma(\frac{h}{w},\frac{h}{w})(s)\mathrm{d}s,\\\
\|h(t)\|_{\infty}&\lesssim
e^{-\lambda^{*}t}\|h(0)\|_{\infty}+\Big{\|}\int_{0}^{t}U(t-s)w\Gamma\big{(}\frac{h}{w},\frac{h}{w}\big{)}(s)\mathrm{d}s\Big{\|}_{\infty},\end{split}$
(8.6)

where $U(t)$ is a semi-group for linearized Boltzmann equation. Using
Duhamel’s principle again, we have

$U(t-s)=G(t-s)+\int_{s}^{t}G(t-s_{1})K_{w}U(s_{1}-s)\mathrm{d}s_{1},$

where $G(t)$ is semi-group for the system

$\begin{split}&\partial_{t}h+v\cdot\nabla_{x}h+\nu
h=0,\quad\text{and}\quad|G(t)h_{0}|\leq e^{-\nu_{0}t}|h_{0}|.\end{split}$

For the last term in (8.6), using standard nonlinear estimate
$|w\Gamma(f,f)|\lesssim\langle v\rangle\|wf\|_{\infty}^{2}$, we obtain (see
[16, 26, 25] for detail)

$\begin{split}&\Big{\|}\int_{0}^{t}U(t-s)w\Gamma\big{(}\frac{h}{w},\frac{h}{w}\big{)}(s)\mathrm{d}s\Big{\|}_{\infty}\\\
&\leq\Big{\|}\int_{0}^{t}G(t-s)w\Gamma\big{(}\frac{h}{w},\frac{h}{w}\big{)}(s)\mathrm{d}s\Big{\|}_{\infty}+\Big{\|}\int_{0}^{t}\int_{s}^{t}G(t-s_{1})K_{w}U(s_{1}-s)w\Gamma\big{(}\frac{h}{w},\frac{h}{w}\big{)}(s)\mathrm{d}s_{1}\mathrm{d}s\Big{\|}_{\infty}\\\
&\leq Ce^{-\lambda^{*}t}\Big{(}\sup_{0\leq
s\leq\infty}e^{\lambda^{*}s}\|h(s)\|_{\infty}\Big{)}^{2}.\end{split}$

Therefore, for sufficiently small $\|h_{0}\|_{\infty}\ll 1$, we have uniform
bound

$\sup_{0\leq t\leq\infty}e^{\lambda^{*}t}\|h(t)\|_{\infty}\ll 1,$

and hence global decay and uniqueness. Also note that positivity of $F$ is
standard by sequence $F^{\ell}$ which solves

$\begin{split}&\partial_{t}F^{\ell+1}+v\cdot\nabla
F^{\ell+1}=Q_{+}(F^{\ell},F^{\ell})-\nu(F^{\ell})F^{\ell+1},\ \
F|_{t=0}=F_{0},\\\ &F^{\ell+1}(t,x,v)=F^{\ell+1}(t,x,R_{x}v)\ \ \ \text{on}\ \
\partial U.\end{split}$

From $F_{0}\geq 0$ and $F^{\ell}\geq 0$, we have $F^{\ell+1}\geq 0$. ∎

##  9\. Appendix: $L^{2}$ coercivity in an axisymmetric domain

In Appendix, we consider the linearized hard-sphere Boltzmann equation

$\partial_{t}f+v\cdot\nabla_{y}f+Lf=g,$ (9.1)

with specular boundry condition

$\displaystyle f(t,x,R_{y}v)=f(t,y,v)\text{ on }y\in\partial U,$ (9.2)

where $R_{y}:=(\mathbf{I}-2n(y)\otimes n(y))$. We consider a periodic
cylindrical domain with annulus cross-section:

$\displaystyle U:=\Omega\times[0,H]\subset\mathbb{R}^{3},\quad\partial
U:=\partial\Omega\times(0,H),$ (9.3)

where $\Omega:=\\{(x,y)\in\mathbb{R}^{2}:0<R_{1}<x^{2}+y^{2}<R_{2}\\}$ is an
annulus and $[0,H]$ is a periodic interval. It can be parametrized by using a
global triply orthogonal coordinate system: for $y\in U$,

$\displaystyle
y=\eta(\theta,z,r)=(r\cos\theta,r\sin\theta,z)=(\eta_{1}(\theta,z,r),\eta_{2}(\theta,z,r),\eta_{3}(\theta,z,r)),$
(9.4)

where $0<R_{1}<r<R_{2},\;0\leq\theta\leq 2\pi,\;0\leq z\leq H$.

The main object of Appendix is to control the physical macroscopic quantities:

$\displaystyle a(t,y)$
$\displaystyle=\int_{\mathbb{R}^{3}}f(t,y,v)\sqrt{\mu}dv,$ $\displaystyle
b(t,y)$ $\displaystyle=\int_{\mathbb{R}^{3}}vf(t,y,v)\sqrt{\mu}dv,$
$\displaystyle c(t,y)$
$\displaystyle=\int_{\mathbb{R}^{3}}\frac{|v|^{2}-3}{2}f(t,y,v)\sqrt{\mu}dv.$

We provide a constructive proof of the coercivity estimate in the presence of
the specular reflection boundary condition. Firstly, we introduce the mass and
energy ($a$ and $c$) estimates in Proposition 9.5. On the other hand, when it
comes to the momentum component $b$, we use the global triply orthogonal
coordinate system (9.4) to transform the momentum part such as (9.9). Then, we
divide it into the tangential momentum and normal momentum. Unlike mass and
energy, momentum is not fully conserved, so it needs to be handled using the
delicate geometric method. The proof consists of construction of correction
term $\zeta$ to derive the Laplace-Beltrami operator (9.24).

###  9.1. Transformed equations and macroscopic parts

In Appendix, we only consider the periodic cylindrical domain with an annulus
cross-section. However, Lemma 9.1, Lemma 9.3, and Lemma 9.4 in section 9.1
hold for a domain with global triply orthogonal coordinate system of a single
chart as well.
Within Appendix, we use variables $(\theta,z,r)$ as $x_{1},x_{2}$ and $x_{3}$
respectively. To distinguish derivatives, we use the following derivative
symbols.

$\partial_{y_{i}}:=\frac{\partial}{\partial
y_{i}},\quad\partial_{i}:=\frac{\partial}{\partial
x_{i}},\quad\nabla_{y}:=(\partial_{y_{1}},\partial_{y_{2}},\partial_{y_{3}}),\quad\nabla:=(\partial_{1},\partial_{2},\partial_{3}).$

We also define

$\begin{split}g_{ij}:=\langle\partial_{i}\eta,\partial_{j}\eta\rangle,\quad
D_{i}:=\frac{1}{\sqrt{g_{ii}}}\partial_{i},\quad
D:=(D_{1},D_{2},D_{3}).\end{split}$ (9.5)

In the periodic cylindrical domain with annulus cross-section, it is directly
computed that

$\displaystyle g_{11}=r^{2},\quad g_{22}=g_{33}=1,\;\text{and}\quad
g_{ij}=0,\quad\text{whenever }i\neq j.$ (9.6)

We define

$\mathbf{v}:=\begin{bmatrix}\mathbf{v}_{{1}}\\\ \mathbf{v}_{{2}}\\\
\mathbf{v}_{{3}}\end{bmatrix}=Q^{T}\begin{bmatrix}v_{{1}}\\\ v_{{2}}\\\
v_{{3}}\end{bmatrix},\quad
Q:=\begin{bmatrix}D_{1}\eta_{1}&D_{2}\eta_{1}&D_{3}\eta_{1}\\\
D_{1}\eta_{2}&D_{2}\eta_{2}&D_{3}\eta_{2}\\\
D_{1}\eta_{3}&D_{2}\eta_{3}&D_{3}\eta_{3}\\\ \end{bmatrix},$ (9.7)

where $Q$ is orthonormal matrix, $QQ^{T}=Q^{T}Q=I$.

Given the global triply orthogonal coordinate system (9.4), we define
transformed local mass $a_{x}$ and local energy $c_{x}$ by,

$a_{x}:=a\circ\eta\quad\text{and}\quad c_{x}:=c\circ\eta.$ (9.8)

For the local momentum vector $b_{x}$, it is crucial to take account of
geometry: we define

$\begin{split}b_{x_{i}}&:=\int_{\mathbb{R}^{3}}\mathbf{v}_{i}f\sqrt{\mu(\mathbf{v})}\mathrm{d}\mathbf{v}=\int_{\mathbb{R}^{3}}D_{i}\eta\cdot
vf\sqrt{\mu(\mathbf{v})}\mathrm{d}\mathbf{v}=D_{i}\eta\cdot b.\end{split}$
(9.9)

Therefore, we get that, as (9.7),

$\begin{bmatrix}b_{x_{1}}\\\ b_{x_{2}}\\\
b_{x_{3}}\end{bmatrix}=Q^{T}\begin{bmatrix}b_{{1}}\\\ b_{{2}}\\\
b_{{3}}\end{bmatrix}.$ (9.10)

As the mapping in the velocity space is isometric (see $Q^{T}$ is an
orthonormal matrix), we note that

$|\mathbf{v}|=|v|,\quad|b_{x}|=|b|,\quad\mu=\frac{1}{\sqrt{(2\pi)^{3}}}e^{-\frac{|v|^{2}}{2}}=\frac{1}{\sqrt{(2\pi)^{3}}}e^{-\frac{|\mathbf{v}|^{2}}{2}}.$
(9.11)

In the coordinates, now we rewrite the transport operator in terms of
$\mathbf{v}$ and local derivatives $\nabla$. Since

$\begin{split}\partial_{i}&:=\frac{\partial}{\partial
x_{i}}=\sum_{j}\frac{\partial y_{j}}{\partial x_{i}}\frac{\partial}{\partial
y_{j}},\end{split}$

we have

$\begin{split}\begin{bmatrix}\partial_{1}\\\ \partial_{2}\\\
\partial_{3}\end{bmatrix}=\begin{bmatrix}\sqrt{g_{11}}&0&0\\\
0&\sqrt{g_{22}}&0\\\ 0&0&\sqrt{g_{33}}\\\
\end{bmatrix}Q^{T}\begin{bmatrix}\partial_{y_{1}}\\\ \partial_{y_{2}}\\\
\partial_{y_{3}}\end{bmatrix},\quad\begin{bmatrix}D_{1}\\\ D_{2}\\\
D_{3}\end{bmatrix}=Q^{T}\begin{bmatrix}\partial_{y_{1}}\\\ \partial_{y_{2}}\\\
\partial_{y_{3}}\end{bmatrix}.\end{split}$ (9.12)

Therefore, identities (9.7) and (9.12) yield that

$\begin{split}v\cdot\nabla_{y}\
&=\begin{bmatrix}\mathbf{v}_{1}&\mathbf{v}_{2}&\mathbf{v}_{3}\end{bmatrix}Q^{T}Q\begin{bmatrix}D_{1}\\\
D_{2}\\\ D_{3}\end{bmatrix}=\mathbf{v}\cdot
D{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}.}\end{split}$
(9.13)

We also note that, from (9.7) and (9.10),

$\begin{split}b\cdot
v&=\begin{bmatrix}b_{x_{1}}&b_{x_{2}}&b_{x_{3}}\end{bmatrix}Q^{T}Q\begin{bmatrix}\mathbf{v}_{1}\\\
\mathbf{v}_{2}\\\
\mathbf{v}_{3}\end{bmatrix}=b_{x}\cdot\mathbf{v}.\end{split}$ (9.14)

Together with $a_{x}$ and $c_{x}$ defined in (9.8), we can rewrite the
macroscopic part $\mathbf{P}f$ as

$\begin{split}\mathbf{P}f:=\Big{(}a_{x}+b_{x}\cdot\mathbf{v}+\frac{|\mathbf{v}|^{2}-3}{2}\Big{)}\sqrt{\mu}.\end{split}$
(9.15)

Now we consider some important properties of
$D_{i}:=\frac{1}{\sqrt{g_{ii}}}\partial_{i}$. By the normalization in (9.5)
and (9.6),

$\langle
D_{i}\eta,D_{j}\eta\rangle=\delta_{ij}.{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}}$
(9.16)

We define $D_{ij}$ and $\Gamma_{D,ij}^{k}$ by

$\begin{split}D_{ij}&:=D_{i}D_{j},\quad\Gamma_{D,ij}^{k}:=\langle
D_{ij}\eta,D_{k}\eta\rangle.\end{split}$ (9.17)

Notice that we calculated the Christoffel symbol in the domain $U$

$\displaystyle\Gamma_{D,33}^{1}=\Gamma_{D,33}^{2}=0,\quad\Gamma_{D,22}^{1}=\Gamma_{D,22}^{3}=0,\quad\Gamma_{D,11}^{3}=-\frac{1}{r},\quad\Gamma_{D,11}^{2}=0.$
(9.18)

######  Lemma 9.1.

(1) We have

$\Gamma_{D,ij}^{k}=-\Gamma_{D,ik}^{j},\quad\Gamma_{D,ij}^{j}=0,\quad\text{for
all}\ \ i,j,k\in\\{1,2,3\\}.\\\ $ (9.19)

(2) We have the following commutator:

$\begin{split}D_{i}D_{j}-D_{j}D_{i}=\Gamma_{D,jj}^{i}D_{j}-\Gamma_{D,ii}^{j}D_{i},\quad
i\neq j.\end{split}$ (9.20)

(3) When all $i,j,k\in\\{1,2,3\\}$ are distinct,

$\Gamma_{D,ij}^{k}=0.$ (9.21)

###### Proof.

(1) We get (9.19) directly from (9.16) by taking derivatives.

(2) First we compute that, for $i\neq k$,

$\begin{split}\partial_{i}\big{(}\frac{1}{\sqrt{g_{kk}}}\big{)}&=-\frac{1}{\sqrt{g_{kk}}^{3}}\langle\partial_{ik}\eta,\partial_{k}\eta\rangle=-\sqrt{\frac{g_{ii}}{g_{kk}}}\langle\frac{1}{\sqrt{g_{kk}g_{ii}}}\partial_{ki}\eta,\frac{1}{\sqrt{g_{kk}}}\partial_{k}\eta\rangle\\\
&=-\sqrt{\frac{g_{ii}}{g_{kk}}}\langle\frac{1}{\sqrt{g_{kk}}}\partial_{k}\big{(}\frac{1}{\sqrt{g_{ii}}}\partial_{i}\eta\big{)},\frac{1}{\sqrt{g_{kk}}}\partial_{k}\eta\rangle\\\
&=-\sqrt{\frac{g_{ii}}{g_{kk}}}\langle
D_{ki}\eta,D_{k}\eta\rangle=-\sqrt{\frac{g_{ii}}{g_{kk}}}\Gamma_{D,ki}^{k}=\sqrt{\frac{g_{ii}}{g_{kk}}}\Gamma_{D,kk}^{i},\\\
\end{split}$ (9.22)

where we used (9.19) at the last equality. Therefore we conclude (9.20) as

$\begin{split}D_{i}D_{j}-D_{j}D_{i}&=\frac{1}{\sqrt{g_{ii}}}\partial_{i}\big{(}\frac{1}{\sqrt{g_{jj}}}\partial_{j}\big{)}-\frac{1}{\sqrt{g_{jj}}}\partial_{j}\big{(}\frac{1}{\sqrt{g_{ii}}}\partial_{i}\big{)}\\\
&=\frac{1}{\sqrt{g_{ii}}}\partial_{i}\big{(}\frac{1}{\sqrt{g_{jj}}}\big{)}\partial_{j}-\frac{1}{\sqrt{g_{jj}}}\partial_{j}\big{(}\frac{1}{\sqrt{g_{ii}}}\big{)}\partial_{i}\\\
&=\frac{1}{\sqrt{g_{jj}}}\Gamma_{D,jj}^{i}\partial_{j}-\frac{1}{\sqrt{g_{ii}}}\Gamma_{D,ii}^{j}\partial_{i}\\\
&=\Gamma_{D,jj}^{i}D_{j}-\Gamma_{D,ii}^{j}D_{i}.\\\ \end{split}$

(3) When $i,j,k$ are all distinct, using (9.19) and (9.20), we derive that

$\begin{split}\Gamma_{D,ij}^{k}&=-\Gamma_{D,ik}^{j}=-\langle
D_{j}\eta,D_{ik}\eta\rangle\\\ &=-\langle
D_{j}\eta,D_{ki}\eta+\Gamma_{D,kk}^{i}D_{k}\eta-\Gamma_{D,ii}^{k}D_{i}\eta\rangle=-\Gamma_{D,ki}^{j}.\end{split}$

Now using above relation successively we derive that

$\Gamma_{D,ij}^{k}=-\Gamma_{D,ki}^{j}=\Gamma_{D,jk}^{i}=-\Gamma_{D,ij}^{k}.$

Therefore we conclude (9.21). ∎

######  Definition 9.2.

Define the Laplacian operator in the normalized $D$-derivatives:

$\Delta_{D}:=\big{(}D_{1}^{2}+D_{2}^{2}+D_{3}^{2}\big{)}.$ (9.23)

We also define the Laplace-Beltrami operator in the coordinates:

$\Delta_{bel}:=\Delta_{y}=\sum_{i,j}\frac{1}{\sqrt{|g|}}\partial_{i}\big{(}\sqrt{|g|}g_{ij}\partial_{j}\big{)}.$
(9.24)

######  Lemma 9.3.

Two operator $\Delta_{D}$ and $\Delta_{bel}$ have the following relation:

$\begin{split}\Delta_{bel}&=\Delta_{D}-\sum_{i}\sum_{k\neq
i}\Gamma_{D,kk}^{i}D_{i}.\\\ \end{split}$ (9.25)

###### Proof.

Using (9.1), we derive that

$\begin{split}\Delta_{bel}&=\frac{1}{\sqrt{|g|}}\sum_{i=1}^{3}\partial_{{i}}\big{(}\sqrt{|g|}\frac{1}{g_{ii}}\partial_{{i}}\big{)}=\frac{1}{\sqrt{|g|}}\sum_{i=1}^{3}\partial_{{i}}\big{(}\sqrt{g_{jj}g_{kk}}D_{i}\big{)}\\\
&=\sum_{i=1}^{3}D_{i}^{2}+\frac{1}{\sqrt{|g|}}\sum_{i=1}^{3}\Big{(}\sqrt{g_{kk}}\partial_{x_{i}}\sqrt{g_{jj}}+\sqrt{g_{jj}}\partial_{x_{i}}\sqrt{g_{kk}}\Big{)}D_{i}\\\
&=\sum_{i}D_{i}^{2}+\sum_{i=1}^{3}\Big{(}\langle\frac{\partial_{ij}\eta}{\sqrt{g_{ii}g_{jj}}},\frac{\partial_{j}\eta}{\sqrt{g_{jj}}}\rangle+\langle\frac{\partial_{ik}\eta}{\sqrt{g_{ii}g_{kk}}},\frac{\partial_{k}\eta}{\sqrt{g_{kk}}}\rangle\Big{)}D_{i}\\\
&=\sum_{i}D_{i}^{2}+\sum_{i=1}^{3}\Big{(}\langle\frac{1}{g_{jj}}\partial_{j}\big{(}\frac{1}{\sqrt{g_{ii}}}\partial_{i}\eta\big{)},\frac{\partial_{j}\eta}{\sqrt{g_{jj}}}\rangle+\langle\frac{1}{g_{kk}}\partial_{k}\big{(}\frac{1}{\sqrt{g_{ii}}}\partial_{i}\eta\big{)},\frac{\partial_{k}\eta}{\sqrt{g_{kk}}}\rangle\Big{)}D_{i}\\\
&=\sum_{i}D_{i}^{2}+\sum_{i=1}^{3}\Big{(}\langle
D_{ji}\eta,D_{j}\eta\rangle+\langle D_{ki}\eta,D_{k}\eta\rangle\Big{)}D_{i}\\\
&=\Delta_{D}-\sum_{i=1}^{3}\Big{(}\langle D_{jj}\eta,D_{i}\eta\rangle+\langle
D_{kk}\eta,D_{i}\eta\rangle\Big{)}D_{i}=\Delta_{D}-\sum_{i=1}^{3}\big{(}\Gamma_{D,jj}^{i}+\Gamma_{D,kk}^{i}\big{)}D_{i}\\\
&=\Delta_{D}-\sum_{i}\sum_{k\neq i}\Gamma_{D,kk}^{i}D_{i}.\end{split}$

∎

######  Lemma 9.4.

We have the following differentiation identity:

$D_{i}\mathbf{v}_{j}=\sum_{k}\Gamma_{D,ij}^{k}\mathbf{v}_{k}.$

###### Proof.

From definition (9.7),

$\begin{split}D_{i}\mathbf{v}_{j}&=D_{i}\langle D_{j}\eta,v\rangle=\langle
D_{ij}\eta,v\rangle\\\
&=\langle\sum_{k}\Gamma_{D,ij}^{k}D_{k}\eta,v\rangle=\sum_{k}\Gamma_{D,ij}^{k}\langle
D_{k}\eta,v\rangle=\sum_{k}\Gamma_{D,ij}^{k}\mathbf{v}_{k}.\end{split}$

∎

###  9.2. Mass and Energy estimates

The main purpose of this section is to control macroscopic quantities:

$\displaystyle a(t,y)=\int_{\mathbb{R}^{3}}f(t,y,v)\sqrt{\mu}dv,\quad
c(t,y)=\int_{\mathbb{R}^{3}}\frac{|v|^{2}-3}{2}f(t,y,v)\sqrt{\mu}dv.$

######  Proposition 9.5.

Let $U$ be a domain with global triply orthogonal coordinate system of a
single chart. Assume that $f$ solves linearized Boltzmann equation (9.1) with
the specular boundary condition $f(t,y,v)=f(t,y,R_{y}v)$ on $y\in\partial U,$
where $R_{y}:=(I-2n(y)\otimes n(y))$. Then we have the following mass and
energy estimate:

$\begin{split}\|a\|_{2}^{2}&\lesssim\frac{d}{dt}G(t)+\|b\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\|c\|_{2}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon\|b\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2},\quad\varepsilon\ll
1.\end{split}$ (9.26)

###### Proof.

A starting point of the proof is the weak formulation:

$\begin{split}&\iint_{\partial U\times\mathbb{R}^{3}}f^{b}\Psi^{b}(v\cdot
n)dAdv-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla_{y})\Psi dydv\\\
&=-\iint_{U\times\mathbb{R}^{3}}\Psi\partial_{t}fdydv-\iint_{U\times\mathbb{R}^{3}}\Psi
Lfdydv+\iint_{U\times\mathbb{R}^{3}}\Psi gdydv\\\
&\quad+\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f(v\cdot\nabla_{y})\Psi
dydv.\end{split}$ (9.27)

To estimate the mass part $a$, we pick
$\Psi_{a}=\sum_{i=1}^{3}\phi_{i}(v)\psi_{i}(y)$ with

$\begin{split}\phi_{i}(v)&=(|v|^{2}-\beta_{a})v_{i}\sqrt{\mu},\\\
\psi_{i}(y)&=\partial_{y_{i}}u_{a}.\end{split}$ (9.28)

Here, $u_{a}$ solves

$-\Delta u_{a}=a(x),\quad\partial_{n}u_{a}|_{\partial U}=0,$ (9.29)

and the constant $\beta_{a}=10$ is chosen so that

$\int_{\mathbb{R}^{3}}(|v|^{2}-\beta_{a})(\frac{|v|^{2}-3}{2})v_{i}^{2}\mu
dv=0,\quad i=1,2,3.$ (9.30)

Note that the Poisson equation with Neumann boundary condition is uniquely
solvable as we have the total mass conservation $\int_{U}ady=0$. When the
solution uniquely well-defined, we use the notation

$u_{a}=-(-\Delta)_{N}^{-1}a.$ (9.31)

For the boundary contribution in (9.27), we rewrite the integral in the
$x$-coordinate. Using $|\mathbf{v}|=|v|$ in (9.11) and
$v\cdot\nabla_{y}=\mathbf{v}\cdot D$ by (9.13), we have that

$\begin{split}\iint_{\partial U\times\mathbb{R}^{3}}f^{b}\Psi_{a}^{b}(v\cdot
n)=\sum_{i=1}^{3}\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\mathbf{v}_{3}(|\mathbf{v}|^{2}-\beta_{a})\mu\mathbf{v}_{i}D_{i}u_{a}=0.\end{split}$
(9.32)

This term always vanishes because when $i=1,2$ velocity integration is odd;
while when $i=3$ the Neumann boundary condition implies $D_{3}u_{a}|_{\partial
U}=0$.

For the second term in LHS of (9.27), using (9.30) and (9.29), we derive that

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{a}\\\
&=-\iint_{U\times\mathbb{R}^{3}}\sqrt{\mu}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}(v\cdot\nabla)\Psi_{a}\\\
&=-\sum_{i,j=1}^{3}\iint_{U\times\mathbb{R}^{3}}a{\mu}(|v|^{2}-\beta_{a})v_{i}v_{j}\partial_{y_{j}}^{2}u_{a}dy\\\
&=-C\sum_{i=1}^{3}\int_{U}a\partial_{y_{i}}^{2}u_{a}dy=C\|a\|^{2}_{2}.\end{split}$
(9.33)

Here,
$C=\int_{\mathbb{R}^{3}}\mu(|v|^{2}-\beta_{a})v_{i}^{2}dv=\int_{\mathbb{R}^{3}}\mu(|v|^{2}-10)v_{i}^{2}dv=-5\neq
0$ for all $i=1,2,3.$

For the RHS of (9.27), from the elliptic estimate to (9.29), we use
$\|u_{a}\|_{H^{2}}^{2}\lesssim\|a\|_{2}^{2}$. Then we could conclude that

$\begin{split}&\text{RHS of (\ref{glo IBP
re})}+\frac{d}{dt}\iint_{U\times\mathbb{R}^{3}}f\Psi_{a}dydv\\\
&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{a}dydv\Big{|}+\varepsilon\|u_{a}\|_{H^{2}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}\\\
&\lesssim\underbrace{\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{a}dydv\Big{|}}_{(**)}+\varepsilon\|a\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.34)

Here, we set $G(t)=\iint_{U\times\mathbb{R}^{3}}f\Psi_{a}dydv$ which satisfies
that $G(t)\lesssim\|f(t)\|_{2}^{2}$.

To estimate the part $(**)$, we device another test function:

$\varphi_{a}=-(-\Delta)^{-1}_{N}(\partial_{t}a)\sqrt{\mu}.$ (9.35)

Note that $(-\Delta)^{-1}_{N}(\partial_{t}a)$ is uniquely well-defined as the
mass conservation
$\int_{U}a(t)=\int_{U}\int_{\mathbb{R}^{3}}f_{0}\sqrt{\mu}=0$ for all $t\geq
0$. When this test function combined with $\partial_{t}f$, we get

$\begin{split}\iint_{U\times\mathbb{R}^{3}}\partial_{t}f\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}a)&=\|\nabla\Delta_{N}^{-1}(\partial_{t}a)\|^{2}_{2}.\end{split}$

On the other hand, considering other terms, except $\partial_{t}f$,

$\begin{split}\|\nabla\Delta_{N}^{-1}(\partial_{t}a)\|^{2}_{2}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}a)\Big{(}-v\cdot\nabla_{y}f+g-Lf\Big{)}\Big{|}\\\
&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{P}f+(\mathbf{I}-\mathbf{P})f)\
\nabla\cdot\big{(}v\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}a)\big{)}\Big{|}\\\
&\quad+\
\Big{|}\iint_{U\times\mathbb{R}^{3}}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}a)\big{(}g-Lf\big{)}\Big{|}\\\
&\lesssim\varepsilon\|\nabla\Delta_{N}^{-1}(\partial_{t}a)\|^{2}_{2}+\varepsilon\|a\|_{2}^{2}+\Big{(}\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|b\|^{2}_{2}+\|g\|_{2}^{2}\Big{)},\\\
\end{split}$ (9.36)

where we used Poincare inequality and the Young’s inequality with sufficiently
small $\varepsilon\ll 1$. Therefore we get

$\|\nabla\Delta_{N}^{-1}(\partial_{t}a)\|^{2}_{2}\lesssim\varepsilon\|a\|_{2}^{2}+\Big{(}\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|b\|^{2}_{2}+\|g\|_{2}^{2}\Big{)}.$
(9.37)

Using (9.37) and $\|\cdot\|_{2}\leq\|\cdot\|_{\nu}$ , we can estimate $(**)$
of (9.34) by

$\begin{split}\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{a}dydv\Big{|}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{a}dydv\Big{|}+\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f\partial_{t}\Psi_{a}dydv\Big{|}\\\
&\lesssim\|b\|_{2}^{2}+\|\nabla\Delta_{N}^{-1}(\partial_{t}a)\|^{2}_{2}+\|(\mathbf{I-P})f\|_{2}^{2}\\\
&\lesssim\varepsilon\|a\|_{2}^{2}+\|b\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.38)

Combining (9.27), (9.32), (9.33), (9.34), and (9.38), we gain

$\begin{split}\|a\|_{2}^{2}&\lesssim\frac{d}{dt}G(t)+\|b\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2}.\\\
\end{split}$ (9.39)

Similar as above mass estimate, in energy $c$ estimate, we pick
$\Psi_{c}=\sum_{i=1}^{3}\phi_{i}(v)\psi_{i}(y)$ where,

$\begin{split}\phi_{i}(v)&=(|v|^{2}-\beta_{c})v_{i}\sqrt{\mu},\\\
\psi_{i}(y)&=\partial_{y_{i}}u_{c},\\\ \end{split}$ (9.40)

when each are read on $k$-chart and constant $\beta_{c}=5$ is chosen so that

$\int_{\mathbb{R}^{3}}(|v|^{2}-\beta_{c})v_{i}^{2}\mu dv=0,\quad i=1,2,3,$
(9.41)

and $u_{c}$ solves,

$-\Delta u_{c}=c(y),\quad\partial_{n}u_{c}|_{\partial U}=0.$ (9.42)

Note that we have energy conservation $\int_{U}cdy=0$. Similar as (9.32), we
use oddness integral in $\mathbf{v}_{3}$ and Neumann condition of $u_{c}$ to
get

$\begin{split}\iint_{\partial U}f^{b}\Psi_{c}^{b}(v\cdot
n)=\sum_{i=1}^{3}\iint
f^{b}\mathbf{v}_{3}(|\mathbf{v}|^{2}-\beta_{c})\mu\mathbf{v}_{i}D_{i}u_{c}=0.\end{split}$
(9.43)

For transport term of LHS, similar as (9.33),

$\begin{split}-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{c}dydv&=-\sum_{i,j=1}^{3}\iint_{U\times\mathbb{R}^{3}}c{\mu}(|v|^{2}-\beta_{c})\frac{|v|^{2}-3}{2}v_{i}v_{j}\partial_{y_{j}}^{2}u_{c}dydv\\\
&=-C\sum_{i=1}^{3}\int_{U}c\partial_{y_{i}}^{2}u_{c}dy=C\|c\|^{2}_{2}.\end{split}$
(9.44)

Here, the constant
$C=\int_{\mathbb{R}^{3}}\mu(|v|^{2}-\beta_{c})\left(\frac{|v|^{2}-3}{2}\right)v_{i}^{2}dv=\int_{\mathbb{R}^{3}}\mu(|v|^{2}-5)\left(\frac{|v|^{2}-3}{2}\right)v_{i}^{2}dv=5\neq
0$ for all $i=1,2,3.$

For the RHS of (9.27), similar as (9.34),

$\begin{split}\text{RHS of (\ref{glo IBP
re})}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\Psi_{c}\partial_{t}fdydv\Big{|}+\varepsilon\|u_{c}\|_{H^{2}}^{2}+\|(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad\varepsilon\ll
1,\\\
&\lesssim\frac{d}{dt}G(t)+\underbrace{\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{c}dydv\Big{|}}_{(**)}+\varepsilon\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\end{split}$ (9.45)

where $|G(t)|\lesssim\|f(t)\|_{2}^{2}$. To estimate time part $(**)$, we apply
another test function,

$\varphi_{c}=\big{(}\frac{|v|^{2}-3}{2}\big{)}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}c).$
(9.46)

Note that $\Delta^{-1}_{N}(\partial_{t}c)$ is well defined by energy
conservation. When this test function combined with $\partial_{t}f$,

$\begin{split}\iint_{U\times\mathbb{R}^{3}}\partial_{t}f\big{(}\frac{|v|^{2}-3}{2}\big{)}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}c)&=\|\nabla\Delta_{N}^{-1}(\partial_{t}c)\|^{2}_{2}.\end{split}$

On the other hand, considering other terms, except $\partial_{t}f$,

$\begin{split}\|\nabla\Delta_{N}^{-1}(\partial_{t}c)\|^{2}_{2}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\big{(}\frac{|v|^{2}-3}{2}\big{)}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}c)\Big{(}-v\cdot\nabla_{y}f+g-Lf\Big{)}\Big{|}\\\
&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{P}f+(\mathbf{I}-\mathbf{P})f)\
\nabla\cdot\big{(}v\big{(}\frac{|v|^{2}-3}{2}\big{)}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}c)\big{)}\Big{|}\\\
&\quad+\
\Big{|}\iint_{U\times\mathbb{R}^{3}}\big{(}\frac{|v|^{2}-3}{2}\big{)}\sqrt{\mu}\Delta^{-1}_{N}(\partial_{t}c)\big{(}g-Lf\big{)}\Big{|}\\\
&\lesssim\varepsilon\|\nabla\Delta_{N}^{-1}(\partial_{t}c)\|^{2}_{2}+\|b\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\end{split}$ (9.47)

where we used Poincare inequality and sufficiently small $\varepsilon\ll 1$
comes from Young’s inequality. Therefore we get

$\|\nabla\Delta_{N}^{-1}(\partial_{t}c)\|^{2}_{2}\lesssim\|b\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.48)

Using (9.48) and $\|\cdot\|_{2}\leq\|\cdot\|_{\nu}$, we can estimate $(**)$ by

$\begin{split}\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{c}dydv\Big{|}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{c}dydv\Big{|}+\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f\partial_{t}\Psi_{c}dydv\Big{|}\\\
&=\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f\partial_{t}\Psi_{c}dydv\Big{|}\\\
&\lesssim\varepsilon\|\nabla\Delta_{N}^{-1}(\partial_{t}c)\|^{2}_{2}+\|(\mathbf{I-P})f\|_{2}^{2}\\\
&\lesssim\varepsilon\|b\|_{2}^{2}+\varepsilon\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2},\end{split}$
(9.49)

where we used

$\begin{split}\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{c}&=\sum_{i=1}^{3}\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}{\mu}(|v|^{2}-\beta_{c})v_{i}\partial_{i}u_{c}\\\
&=\sum_{i=1}^{3}\iint_{U\times\mathbb{R}^{3}}b_{i}{\mu}v_{i}^{2}(|v|^{2}-\beta_{c})\partial_{i}u_{c}=0,\\\
\end{split}$ (9.50)

because of our choice (9.41).

Combining (9.43), (9.44), (9.45), and (9.49), we gain

$\begin{split}\|c\|_{2}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon\|b\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2},\quad\varepsilon\ll
1.\\\ \end{split}$ (9.51)

From (9.39) and (9.51), we prove Proposition 9.5. ∎

###  9.3. Momentum $b$ Estimate

The main difficulty comes from estimating $\|b\|_{2}$. In contrast with the
mass and energy estimates, momentum $b$ is not entirely conserved and
therefore we ought to distinguish the “tangential” momenta and “normal”
momentum to the boundary as (9.9). In this section we use the global triply
orthogonal coordinate system (9.4).

For the momentum estimate, we express the weak formulation (9.27) in the
coordinate system of (9.4). Let us use $\mathbb{R}^{3}_{+}$ for the domain of
$\eta$, namely $\eta^{-1}(U)$. We use a set of test functions
$\Psi=\phi(\mathbf{v})\psi(x)$, which depends on the coordinates and
transformed velocity (9.7). Now we expand interior integral of the LHS of
(9.27) as

$\displaystyle-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi
dydv$
$\displaystyle=-\sum_{j,m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\mathbf{v}_{j}\mathbf{v}_{m}\phi(\mathbf{v}){\mu}d\mathbf{v}\Big{)}b_{x_{m}}D_{j}[\psi(x)]\sqrt{|g|}dx$
$\displaystyle\quad-\sum_{j,m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\mathbf{v}_{j}\mathbf{v}_{m}D_{j}[\phi(\mathbf{v})]\mu
d\mathbf{v}\Big{)}b_{x_{m}}\psi(x)\sqrt{|g|}dx.$ (9.52)

For the second term (9.52), we use the integration by parts in velocity,
(9.17), and Lemma 9.4:

$\displaystyle-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi
dydv$
$\displaystyle=-\sum_{j,m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\mathbf{v}_{j}\mathbf{v}_{m}\phi(\mathbf{v}){\mu}\Big{)}b_{x_{m}}D_{j}\psi(x)\sqrt{|g|}dx$
(9.53) $\displaystyle\quad+\sum_{j,m}\sum_{\ell\neq
m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}_{j}\mathbf{v}_{\ell}{\mu}\Big{)}b_{x_{m}}\Gamma^{\ell}_{D,jm}\psi(x)\sqrt{|g|}dx$
(9.54) $\displaystyle\quad+\sum_{j,m}\sum_{\ell\neq
j}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}_{m}\mathbf{v}_{\ell}{\mu}\Big{)}b_{x_{m}}\Gamma^{\ell}_{D,jj}\psi(x)\sqrt{|g|}dx.$
(9.55)

Using oddness in velocoity integration, we can expand as following.

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi
dydv\\\
&=-\sum_{j=1}^{3}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\mathbf{v}^{2}_{j}\phi(\mathbf{v}){\mu}\Big{)}b_{x_{j}}D_{j}\psi(x)\sqrt{|g|}dx\quad(j=m)\\\
&\quad-\sum_{j\neq
m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\mathbf{v}_{j}\mathbf{v}_{m}\phi(\mathbf{v}){\mu}\Big{)}b_{x_{m}}D_{j}\psi(x)\sqrt{|g|}dx\quad(j\neq
m)\\\ &\quad+\sum_{j=\ell\neq
m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}^{2}_{j}{\mu}\Big{)}b_{x_{m}}\Gamma^{j}_{D,jm}\psi(x)\sqrt{|g|}dx\quad(j=\ell,\
\ell\neq m)\\\ &\quad+\sum_{j\neq\ell,\ell\neq
m}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}_{j}\mathbf{v}_{\ell}{\mu}\Big{)}b_{x_{m}}\Gamma^{\ell}_{D,jm}\psi(x)\sqrt{|g|}dx\quad(j\neq\ell,\
\ell\neq m)\\\ &\quad+\sum_{m=\ell\neq
j}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}^{2}_{m}{\mu}\Big{)}b_{x_{m}}\Gamma^{m}_{D,jj}\psi(x)\sqrt{|g|}dx\quad(m=\ell,\
\ell\neq j)\\\ &\quad+\sum_{m\neq\ell,\ell\neq
j}\int_{\mathbb{R}^{3}_{+}}\Big{(}\int_{\mathbb{R}^{3}}\phi(\mathbf{v})\mathbf{v}_{m}\mathbf{v}_{\ell}{\mu}\Big{)}b_{x_{m}}\Gamma^{\ell}_{D,jj}\psi(x)\sqrt{|g|}dx.\quad(m\neq\ell,\
\ell\neq j)\\\ \end{split}$ (9.56)

####  9.3.1. Normal momentum $b_{x_{3}}$ estimate.

We try three types of test functions, Symmetric, Non-symmetric 1, and Non-
symmetric 2.
(Symmetric case) We try test function
$\Psi_{3,1}=\phi_{3}(\mathbf{v})\psi_{3}(x)$, where

$\begin{split}\phi_{3}(\mathbf{v})&=(\mathbf{v}^{2}_{3}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{3}(x)&=\zeta_{3}(x)D_{3}u_{3},\end{split}$ (9.57)

where $\zeta_{3}(x)$ to be determined and $u_{3}$ solves Dirichlet problem,

$-\Delta_{bel}u_{3}=b_{x_{3}},\quad u_{3}|_{\partial U}=0.$ (9.58)

Above $\phi_{3}$ yields only $\mathbf{v}^{2}_{3}\sqrt{\mu}$-terms in (9.56).
With multiplying proper normalization constants to each $\phi_{i}$’s, LHS of
(9.27) becomes

$\begin{split}&\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{3,1}^{b}(v\cdot
n)dAdv-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla_{y})\Psi_{3,1}dydv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}\underbrace{b_{x_{3}}\zeta_{3}D^{2}_{3}u_{3}}_{main}+b_{x_{3}}D_{3}\zeta_{3}D_{3}u_{3}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{1}}\Gamma^{1}_{D,33}-b_{x_{2}}\Gamma^{2}_{D,33}+b_{x_{3}}\Gamma^{3}_{D,11}+b_{x_{3}}\Gamma^{3}_{D,22}\Big{)}\zeta_{3}D_{3}u_{3}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{3}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{3}(x)\sqrt{g_{11}g_{22}}dA.\end{split}$ (9.59)

(Non-symmetric case 1) We try test function
$\Psi_{3,2}=\sum_{i=1,2}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=\mathbf{v}_{i}\mathbf{v}_{3}\sqrt{\mu},\\\
\psi_{i}(x)&=\zeta_{3}(x)D_{i}u_{3},\end{split}$ (9.60)

for $i=1,2$. Above $\phi_{1}$ and $\phi_{2}$ yield only
$\mathbf{v}_{3}\mathbf{v}_{1}\sqrt{\mu}$ and
$\mathbf{v}_{2}\mathbf{v}_{3}\sqrt{\mu}$ terms in (9.56) respectively. With
multiplying proper normalization constants to each $\phi_{i}$’s, LHS of (9.27)
becomes

$\begin{split}&\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{3,2}^{b}(v\cdot
n)dAdv-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla_{y})\Psi_{3,2}dydv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{1}}\zeta_{3}D_{31}u_{3}+b_{x_{1}}D_{3}\zeta_{3}D_{1}u_{3}+b_{x_{3}}\zeta_{3}D_{11}u_{3}+b_{x_{3}}D_{1}\zeta_{3}D_{1}u_{3}\big{)}\sqrt{|g|}dx\\\
&\quad-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{2}}\zeta
D_{32}u_{3}+b_{x_{2}}D_{3}\zeta_{3}D_{2}u_{3}+b_{x_{3}}\zeta_{3}D_{22}u_{3}+b_{x_{3}}D_{2}\zeta_{3}D_{2}u_{3}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}2b_{x_{3}}\Gamma^{1}_{D,33}+2b_{x_{1}}\Gamma^{3}_{D,11}+b_{x_{3}}\Gamma^{1}_{D,22}+b_{x_{1}}\Gamma^{3}_{D,22}+\underbrace{b_{x_{2}}\Gamma^{3}_{D,12}+b_{x_{2}}\Gamma^{1}_{D,32}}_{=0\
\text{by}\ (\ref{geo cond})}\Big{)}\zeta_{3}D_{1}u_{3}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}2b_{x_{2}}\Gamma^{3}_{D,22}+2b_{x_{3}}\Gamma^{2}_{D,33}+\underbrace{b_{x_{1}}\Gamma^{3}_{D,21}+b_{x_{1}}\Gamma^{2}_{D,31}}_{=0\
\text{by}\ (\ref{geo
cond})}+b_{x_{2}}\Gamma^{3}_{D,11}+b_{x_{3}}\Gamma^{2}_{D,11}\Big{)}\zeta_{3}D_{2}u_{3}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\int_{\mathbb{R}^{3}}f^{b}\phi_{1}(\mathbf{v})\mathbf{v}_{3}\Big{)}\underbrace{\psi_{1}(x)}_{=0}\sqrt{g_{11}g_{22}}dA+C\int_{x_{3}=0}\Big{(}\int_{\mathbb{R}^{3}}f^{b}\phi_{2}(\mathbf{v})\mathbf{v}_{3}\Big{)}\underbrace{\psi_{2}(x)}_{=0}\sqrt{g_{11}g_{22}}dA,\end{split}$
(9.61)

where last two terms are zero by Dirichlet boundary condition of $u_{3}$.

(Non-symmetric case 2) We try test function
$\Psi_{3,3}=\sum_{i=1,2}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=(\mathbf{v}^{2}_{i}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{i}(x)&=\zeta_{3}(x)D_{3}u_{3},\end{split}$ (9.62)

for $i=1,2$. Above $\phi_{1}$ and $\phi_{2}$ yield only
$\mathbf{v}^{2}_{1}\sqrt{\mu}$ and $\mathbf{v}^{2}_{2}\sqrt{\mu}$ terms in
(9.56) respectively. With multiplying proper normalization constants to each
$\phi_{i}$’s, LHS of (9.27) becomes

$\begin{split}&\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{3,3}^{b}(v\cdot
n)dAdv-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla_{y})\Psi_{3,3}dydv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{1}}\zeta_{3}D_{13}u_{3}+b_{x_{1}}D_{1}\zeta_{3}D_{3}u_{3}\big{)}\sqrt{|g|}dx\\\
&\quad-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{2}}\zeta
D_{23}u_{3}+b_{x_{2}}D_{2}\zeta_{3}D_{3}u_{3}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{2}}\Gamma^{2}_{D,11}-b_{x_{3}}\Gamma^{3}_{D,11}+b_{x_{1}}\Gamma^{1}_{D,22}+b_{x_{1}}\Gamma^{1}_{D,33}\Big{)}\zeta_{3}D_{3}u_{3}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{1}}\Gamma^{1}_{D,22}-b_{x_{3}}\Gamma^{3}_{D,22}+b_{x_{2}}\Gamma^{2}_{D,11}+b_{x_{2}}\Gamma^{2}_{D,33}\Big{)}\zeta_{3}D_{3}u_{3}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{1}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{1}(x)\sqrt{g_{11}g_{22}}dA+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{2}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{2}(x)\sqrt{g_{11}g_{22}}dA.\end{split}$ (9.63)

####  9.3.2. Condition for $\zeta_{3}(x)$

We add (9.59), (9.61), and $(-1)\times$(9.63). This is equivalent to applying
test function

$\Psi_{b,3}:=\Psi_{3,1}+\Psi_{3,2}-\Psi_{3,3},$ (9.64)

to (9.56). Also using (9.20),

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{b,3}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{b,3}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\zeta_{3}\big{(}D_{1}^{2}+D_{2}^{2}+D_{3}^{2}\big{)}u_{3}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\Big{(}2\Gamma_{D,33}^{1}\zeta_{3}-D_{1}\zeta_{3}+\Gamma_{D,22}^{1}\zeta_{3}\Big{)}D_{1}u_{3}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\Big{(}2\Gamma_{D,33}^{2}\zeta_{3}-D_{2}\zeta_{3}+\Gamma_{D,11}^{2}\zeta_{3}\Big{)}D_{2}u_{3}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\Big{(}2\Gamma_{D,11}^{3}\zeta_{3}+2\Gamma_{D,22}^{3}\zeta_{3}-D_{3}\zeta_{3}\Big{)}D_{3}u_{3}\sqrt{|g|}dx\\\
&+\sum_{i=1,2}\int_{\mathbb{R}^{3}_{+}}b_{x_{i}}\Big{(}-\Gamma_{D,33}^{i}\zeta_{3}+D_{i}\zeta_{3}\Big{)}D_{3}u_{3}\sqrt{|g|}dx\\\
&+\sum_{i=1,2}\int_{\mathbb{R}^{3}_{+}}\Big{(}\Gamma_{D,22}^{3}\zeta_{3}+\Gamma_{D,11}^{3}\zeta_{3}-D_{3}\zeta_{3}\Big{)}b_{x_{i}}D_{i}u_{3}\sqrt{|g|}dx.\end{split}$
(9.65)

If we match index $3\rightarrow r,1\rightarrow\theta,2\rightarrow z$ and pick

$\displaystyle\zeta_{3}=\zeta_{r}=\frac{1}{r}$ (9.66)

$\zeta_{3}(x)$ satisfies

$\begin{cases}D_{1}\zeta_{3}&=\Gamma_{D,33}^{1}\zeta_{3}\\\
D_{2}\zeta_{3}&=\Gamma_{D,33}^{2}\zeta_{3}\\\
D_{3}\zeta_{3}&=\Gamma_{D,11}^{3}\zeta_{3}+\Gamma_{D,22}^{3}\zeta_{3},\\\
\end{cases}\quad\text{and}\quad c\leq|\zeta_{3}|\leq C,$ (9.67)

for uniformly positive $c$ and $C$ due to (9.4), (9.5) and (9.18). Then above
system becomes,

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{b,3}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{b,3}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\zeta_{3}\big{(}D_{1}^{2}+D_{2}^{2}+D_{3}^{2}\big{)}u_{3}\sqrt{|g|}dx+\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\zeta_{3}\sum_{i=1}^{3}\sum_{k\neq
i}\Gamma_{D,kk}^{i}D_{i}u_{3}\sqrt{|g|}dx\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{3}}\zeta_{3}\Delta_{bel}u_{3}\sqrt{|g|}dx=\int_{U}\zeta_{3}b^{2}_{x_{3}}dy,\end{split}$
(9.68)

where we used Lemma 9.3 and (9.58).

With $\Psi_{b,3}$, which is defined in (9.64), the RHS of (9.27) is controlled
by

$\begin{split}\text{RHS of (\ref{glo IBP
re})}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\Psi_{b,3}\partial_{t}fdy\Big{|}+\varepsilon\|u_{3}\|_{H^{2}}^{2}+\|(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad\varepsilon\ll
1.\\\
&\lesssim\frac{d}{dt}G(t)+\underbrace{\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{b,3}dy\Big{|}}_{(**)}+\varepsilon\|b_{x_{3}}\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\end{split}$ (9.69)

where $|G(t)|\lesssim\|f(t)\|_{2}^{2}$. Here, we used
$\|u_{3}\|_{H^{2}}^{2}\lesssim\|b_{x_{3}}\|_{2}^{2}$ and
$\|\cdot\|_{2}\leq\|\cdot\|_{\nu}$.

To estimate time part $(**)$, we apply another test function,

$\varphi_{3}=\mathbf{v}_{3}\sqrt{\mu}\Delta^{-1}_{D}\partial_{t}b_{x_{3}}.$
(9.70)

When this test function combined with $\partial_{t}f$, we get

$\begin{split}\iint_{U\times\mathbb{R}^{3}}\partial_{t}f\mathbf{v}_{3}\sqrt{\mu}\Delta^{-1}_{D}\partial_{t}b_{x_{3}}&=\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|^{2}_{2}.\end{split}$

On the other hand, considering other terms, except $\partial_{t}f$,

$\begin{split}\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|^{2}_{2}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{v}_{3}\sqrt{\mu}\Delta^{-1}_{D}\partial_{t}b_{x_{3}}\Big{(}-v\cdot\nabla
f+g-Lf\Big{)}\Big{|}\\\
&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{P}f+(\mathbf{I}-\mathbf{P})f)\
\nabla\cdot\big{(}v\mathbf{v}_{3}\sqrt{\mu}\Delta^{-1}_{D}\partial_{t}b_{x_{3}}\big{)}\Big{|}\\\
&\quad+\
\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{v}_{3}\sqrt{\mu}\Delta^{-1}_{D}\partial_{t}b_{x_{3}}\big{(}g-Lf\big{)}\Big{|}\\\
&\lesssim\varepsilon\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|^{2}_{2}+\Big{(}\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}\Big{)},\\\
\end{split}$ (9.71)

from Poincare inequality and sufficiently small $\varepsilon\ll 1$ comes from
Young’s inequality. Therefore we get

$\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|^{2}_{2}\lesssim\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.72)

Using (9.72), we can estimate $(**)$ in (9.69) by

$\begin{split}\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{b,3}dy\Big{|}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{b,3}dy\Big{|}+\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f\partial_{t}\Psi_{b,3}dy\Big{|}\\\
&\lesssim\big{(}\|c\|_{2}\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|_{2}\big{)}+\big{(}\varepsilon\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|^{2}_{2}+\|(\mathbf{I-P})f\|_{2}^{2}\big{)}\\\
&\lesssim\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2},\end{split}$
(9.73)

where we used

$\begin{split}&\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{b,3}\\\
&=\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{3,1}+\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{3,2}-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{3,3}\\\
&\leq\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{3}^{2}-\beta_{b})\sqrt{\mu}\zeta_{3}D_{3}\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\\\
&\quad+\underbrace{\sum_{i=1}^{2}\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b_{x}\cdot\mathbf{v}+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{i}\mathbf{v}_{3})\sqrt{\mu}\zeta_{3}D_{3}\Delta_{D}^{-1}\partial_{t}b_{x_{3}}}_{=0\
\text{odd in velocity}}\\\
&\quad-\sum_{i=1}^{2}\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{i}^{2}-\beta_{b})\sqrt{\mu}\zeta_{3}D_{3}\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\\\
&\lesssim\|c\|_{2}\|\nabla\Delta_{D}^{-1}\partial_{t}b_{x_{3}}\|_{2}\\\
&\lesssim\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.74)

Combining (9.68), (9.69), and (9.73), we gain

$\begin{split}\|b_{x_{3}}\|_{2}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2}.\\\
\end{split}$ (9.75)

####  9.3.3. Tangential momentum $b_{x_{1}}$ estimate

(Symmetric case) We try test function
$\Psi_{1,1}=\phi_{1}(\mathbf{v})\psi_{1}(x)$, where

$\begin{split}\phi_{1}(\mathbf{v})&=(\mathbf{v}^{2}_{1}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{1}(x)&=\zeta_{1}(x)D_{1}u_{1},\end{split}$ (9.76)

where $\zeta_{1}(x)$ to be determined and $u_{1}$ solves Neumann problem,

$-\Delta_{bel}u_{1}=\frac{1}{\zeta_{1}}b_{x_{1}}-\langle\frac{1}{\zeta_{1}}b_{x_{1}}\rangle:=\widetilde{\frac{1}{\zeta_{1}}b_{x_{1}}},\quad\partial_{n}u_{1}|_{\partial
U}=0,\quad\text{where}\quad\langle
f\rangle:=\fint_{U}f=\frac{1}{|U|}\int_{U}f.$ (9.77)

Above $\phi_{1}$ yields only $\mathbf{v}_{1}^{2}\sqrt{\mu}$ terms in (9.56).
With multiplying proper normalization constants to each $\phi_{i}$’s, LHS of
(9.27) becomes

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{1,1}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{1,1}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{1}}\zeta_{1}D^{2}_{1}u_{1}+b_{x_{1}}D_{1}\zeta_{1}D_{1}u_{1}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{2}}\Gamma^{2}_{D,11}-b_{x_{3}}\Gamma^{3}_{D,11}+b_{x_{1}}\Gamma^{1}_{D,22}+b_{x_{1}}\Gamma^{1}_{D,33}\Big{)}\zeta_{1}D_{1}u_{1}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{1}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{1}(x)\sqrt{g_{11}g_{22}}dA.\end{split}$ (9.78)

(Non-symmetric case 1) We try test function
$\Psi_{1,2}=\sum_{i=2,3}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=\mathbf{v}_{i}\mathbf{v}_{1}\sqrt{\mu},\\\
\psi_{i}(x)&=\zeta_{1}(x)D_{i}u_{1},\end{split}$ (9.79)

for $i=2,3$. Above $\phi_{2}$ and $\phi_{3}$ yield only
$\mathbf{v}_{1}\mathbf{v}_{2}\sqrt{\mu}$ and
$\mathbf{v}_{3}\mathbf{v}_{1}\sqrt{\mu}$ terms in (9.56) respectively. With
multiplying proper normalization constants to each $\phi_{i}$’s, LHS of (9.27)
becomes

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{1,2}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{1,2}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{2}}\zeta_{1}D_{12}u_{1}+b_{x_{2}}D_{1}\zeta_{1}D_{2}u_{1}+b_{x_{1}}\zeta_{1}D_{22}u_{1}+b_{x_{1}}D_{2}\zeta_{1}D_{2}u_{1}\big{)}\sqrt{|g|}dx\\\
&\quad-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{3}}\zeta_{1}D_{13}u_{1}+b_{x_{3}}D_{1}\zeta_{1}D_{3}u_{1}+b_{x_{1}}\zeta_{1}D_{33}u_{1}+b_{x_{1}}D_{3}\zeta_{1}D_{3}u_{1}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}2b_{x_{1}}\Gamma^{2}_{D,11}+2b_{x_{2}}\Gamma^{1}_{D,22}+\underbrace{b_{x_{3}}\Gamma^{2}_{D,13}+b_{x_{3}}\Gamma^{1}_{D,23}}_{=0\
\text{by}\ (\ref{geo
cond})}+b_{x_{1}}\Gamma^{2}_{D,33}+b_{x_{2}}\Gamma^{1}_{D,33}\Big{)}\zeta_{1}D_{2}u_{1}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}2b_{x_{3}}\Gamma^{1}_{D,33}+2b_{x_{1}}\Gamma^{3}_{D,11}+b_{x_{3}}\Gamma^{1}_{D,22}+b_{x_{1}}\Gamma^{3}_{D,22}+\underbrace{b_{x_{2}}\Gamma^{3}_{D,12}+b_{x_{2}}\Gamma^{1}_{D,32}}_{=0\
\text{by}\ (\ref{geo cond})}\Big{)}\zeta_{1}D_{3}u_{1}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{2}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{2}(x)\sqrt{g_{11}g_{22}}dA+C\int_{x_{3}=0}\Big{(}\int_{\mathbb{R}^{3}}f^{b}\phi_{3}(\mathbf{v})\mathbf{v}_{3}\Big{)}\underbrace{\psi_{3}(x)}_{=0}\sqrt{g_{11}g_{22}}dA,\end{split}$
(9.80)

where we used oddness in velocity integration and Neumann boundary condition.

(Non-symmetric case 2) We try test function
$\Psi_{1,3}=\sum_{i=2,3}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=(\mathbf{v}^{2}_{i}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{i}(x)&=\zeta_{1}(x)D_{1}u_{1},\end{split}$ (9.81)

for $i=2,3$. Above $\phi_{2}$ and $\phi_{3}$ yield only
$\mathbf{v}^{2}_{2}\sqrt{\mu}$ and $\mathbf{v}^{2}_{3}\sqrt{\mu}$ terms in
(9.56) respectively. Multiplying proper normalization constants to each
$\phi_{i}$’s, LHS of (9.27) becomes

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{1,3}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{1,3}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{2}}\zeta_{1}D_{21}u_{1}+b_{x_{2}}D_{2}\zeta_{1}D_{1}u_{1}\big{)}\sqrt{|g|}dx\\\
&\quad-\int_{\mathbb{R}^{3}_{+}}\big{(}b_{x_{3}}\zeta_{1}D_{31}u_{1}+b_{x_{3}}D_{3}\zeta_{1}D_{1}u_{1}\big{)}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{1}}\Gamma^{1}_{D,22}-b_{x_{3}}\Gamma^{3}_{D,22}+b_{x_{2}}\Gamma^{2}_{D,11}+b_{x_{2}}\Gamma^{2}_{D,33}\Big{)}\zeta_{1}D_{1}u_{1}\sqrt{|g|}dx\\\
&\quad+\int_{\mathbb{R}^{3}_{+}}\Big{(}-b_{x_{1}}\Gamma^{1}_{D,33}-b_{x_{2}}\Gamma^{2}_{D,33}+b_{x_{3}}\Gamma^{3}_{D,11}+b_{x_{3}}\Gamma^{3}_{D,22}\Big{)}\zeta_{1}D_{1}u_{1}\sqrt{|g|}dx\\\
&\quad+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{1}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{1}(x)\sqrt{g_{11}g_{22}}dA+C\int_{x_{3}=0}\Big{(}\underbrace{\int_{\mathbb{R}^{3}}f^{b}\phi_{2}(\mathbf{v})\mathbf{v}_{3}}_{=0\
\text{oddness in}\
\mathbf{v}_{3}}\Big{)}\psi_{2}(x)\sqrt{g_{11}g_{22}}dA.\end{split}$ (9.82)

####  9.3.4. Condition for $\zeta_{1}(x)$

We add (9.78), (9.80), and $(-1)\times$(9.82). This is equivalent to applying
test function

$\Psi_{b,1}:=\Psi_{1,1}+\Psi_{1,2}-\Psi_{1,3},$ (9.83)

to (9.56). Using (9.20),

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{b,1}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{b,1}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\zeta_{1}\big{(}D_{1}^{2}+D_{2}^{2}+D_{3}^{2}\big{)}u_{1}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\Big{(}2\Gamma_{D,22}^{1}\zeta_{1}+2\Gamma_{D,33}^{1}\zeta_{1}-D_{1}\zeta_{1}\Big{)}D_{1}u_{1}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\Big{(}2\Gamma_{D,11}^{2}\zeta_{1}-D_{2}\zeta_{1}+\Gamma_{D,33}^{2}\zeta_{1}\Big{)}D_{2}u_{1}\sqrt{|g|}dx\\\
&+\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\Big{(}2\Gamma_{D,11}^{3}\zeta_{1}-D_{3}\zeta_{1}+\Gamma_{D,22}^{3}\zeta_{1}\Big{)}D_{3}u_{1}\sqrt{|g|}dx\\\
&+\sum_{i=2,3}\int_{\mathbb{R}^{3}_{+}}b_{x_{i}}\Big{(}-\Gamma_{D,11}^{i}\zeta_{1}+D_{i}\zeta_{1}\Big{)}D_{1}u_{1}\sqrt{|g|}dx\\\
&+\sum_{i=2,3}\int_{\mathbb{R}^{3}_{+}}\Big{(}\Gamma_{D,22}^{1}\zeta_{1}+\Gamma_{D,33}^{1}\zeta_{1}-D_{1}\zeta_{1}\Big{)}b_{x_{i}}D_{i}u_{1}\sqrt{|g|}dx.\end{split}$
(9.84)

Similar to the normal momentum estimate, if we match index $3\rightarrow
r,1\rightarrow\theta,2\rightarrow z$ and pick

$\displaystyle\zeta_{1}=\zeta_{\theta}=\frac{1}{r},$ (9.85)

$\zeta_{1}(x)$ satisfies:

$\begin{cases}D_{1}\zeta_{1}&=\Gamma_{D,22}^{1}\zeta_{1}+\Gamma_{D,33}^{1}\zeta_{1}\\\
D_{2}\zeta_{1}&=\Gamma_{D,11}^{2}\zeta_{1}\\\
D_{3}\zeta_{1}&=\Gamma_{D,11}^{3}\zeta_{1}\\\ \end{cases},\quad\text{and}\quad
c\leq|\zeta_{1}|\leq C,$ (9.86)

for uniformly positive $c$ and $C$ from (9.4), (9.5) and (9.18). Then above
system becomes nearly same as (9.68) and using Lemma 9.3, the LHS of (9.27) is

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{b,1}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{b,1}^{b}(v\cdot n)dAdv\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\zeta_{1}\big{(}D_{1}^{2}+D_{2}^{2}+D_{3}^{2}\big{)}u_{1}\sqrt{|g|}dx+\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\zeta_{1}\sum_{i=1}^{3}\sum_{k\neq
i}\Gamma_{D,kk}^{i}D_{i}u_{1}\sqrt{|g|}dx,\\\
&=-\int_{\mathbb{R}^{3}_{+}}b_{x_{1}}\zeta_{1}\Delta_{bel}u_{1}\sqrt{|g|}dx\\\
&=\int_{U}b_{x_{1}}^{2}dy-\frac{1}{|U|}\int_{U}\zeta_{1}b_{x_{1}}\int_{U}\frac{1}{\zeta_{1}}b_{x_{1}}.\\\
\end{split}$ (9.87)

With $\Psi_{b,1}$, which is defined in (9.83), the RHS of (9.27) is controlled
by

$\begin{split}\text{RHS of (\ref{glo IBP
re})}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\Psi_{b,1}\partial_{t}fdy\Big{|}+\varepsilon\|u_{1}\|_{H^{2}}^{2}+\|(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\quad\varepsilon\ll
1.\\\
&\lesssim\frac{d}{dt}G(t)+\underbrace{\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{b,1}dy\Big{|}}_{(**)}+\varepsilon\|b_{x_{1}}\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\end{split}$ (9.88)

where $|G(t)|\lesssim\|f(t)\|_{2}^{2}$. Here, we used
$\|u_{1}\|_{H^{2}}^{2}\lesssim\|b_{x_{1}}\|_{2}^{2}$ and
$\|\cdot\|_{2}\leq\|\cdot\|_{\nu}$.

To estimate time part $(**)$ in (9.88), we apply another test function,

$\varphi_{1}=\mathbf{v}_{1}\sqrt{\mu}\Delta^{-1}_{N}\widetilde{\partial_{t}b_{x_{1}}}.$
(9.89)

When this test function is combined with $\partial_{t}f$, we get

$\begin{split}&\iint_{U\times\mathbb{R}^{3}}\partial_{t}f\mathbf{v}_{1}\sqrt{\mu}\Delta^{-1}_{N}\widetilde{\partial_{t}b_{x_{1}}}=\int_{U}\partial_{t}b_{x_{1}}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\\\
&=\int_{U}\nabla\cdot\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}+\langle\partial_{t}b_{x_{1}}\rangle\underbrace{\int_{U}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}}_{=0}\\\
&=\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|^{2}_{2}.\end{split}$

On the other hand, considering other terms except $\partial_{t}f$,

$\begin{split}\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|^{2}_{2}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{v}_{1}\sqrt{\mu}\Delta^{-1}_{N}\widetilde{\partial_{t}b_{x_{1}}}\Big{(}-v\cdot\nabla
f+g-Lf\Big{)}\Big{|}\\\
&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{P}f+(\mathbf{I}-\mathbf{P})f)\
\nabla\cdot\big{(}v\mathbf{v}_{1}\sqrt{\mu}\Delta^{-1}_{N}\widetilde{\partial_{t}b_{x_{1}}}\big{)}\Big{|}\\\
&\quad+\
\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{v}_{1}\sqrt{\mu}\Delta^{-1}_{N}\widetilde{\partial_{t}b_{x_{1}}}\big{(}g-Lf\big{)}\Big{|}\\\
&\lesssim\varepsilon\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|^{2}_{2}+\Big{(}\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}\Big{)},\\\
\end{split}$ (9.90)

where sufficiently small $\varepsilon\ll 1$ comes from Young’s inequality and
we used Poincare inequality. Therefore we get

$\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|^{2}_{2}\lesssim\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.91)

Using (9.91), we can estimate $(**)$ in (9.88) by

$\begin{split}\Big{|}\iint_{U\times\mathbb{R}^{3}}f\partial_{t}\Psi_{b,1}dy\Big{|}&\lesssim\Big{|}\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{b,1}dy\Big{|}+\Big{|}\iint_{U\times\mathbb{R}^{3}}(\mathbf{I-P})f\partial_{t}\Psi_{b,1}dy\Big{|}\\\
&\lesssim\|c\|_{2}\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|_{2}+\big{(}\varepsilon\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|^{2}_{2}+\|(\mathbf{I-P})f\|_{2}^{2}\big{)}\\\
&\lesssim\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2},\end{split}$
(9.92)

where we used

$\begin{split}&\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{b,1}\\\
&=\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{1,1}+\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{1,2}-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f\partial_{t}\Psi_{1,3}\\\
&\leq\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{1}^{2}-\beta_{b})\sqrt{\mu}\zeta_{1}D_{1}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\\\
&\quad+\underbrace{\sum_{i=2}^{3}\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b_{x}\cdot\mathbf{v}+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{i}\mathbf{v}_{1})\sqrt{\mu}\zeta_{1}D_{1}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}}_{=0\
\text{odd in velocity}}\\\
&\quad-\sum_{i=2}^{3}\iint_{U\times\mathbb{R}^{3}}\Big{(}a+b\cdot
v+c\frac{|v|^{2}-3}{2}\Big{)}\sqrt{\mu}(\mathbf{v}_{i}^{2}-\beta_{b})\sqrt{\mu}\zeta_{1}D_{1}\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\\\
&\lesssim\|c\|_{2}\|\nabla\Delta_{N}^{-1}\widetilde{\partial_{t}b_{x_{1}}}\|_{2}\\\
&\lesssim\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I}-\mathbf{P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.93)

Combining (9.86), (9.87), (9.88), and (9.92), we gain

$\begin{split}\int_{U}b_{x_{1}}^{2}dy&\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}+\int_{U}\zeta_{1}b_{x_{1}}\int_{U}\frac{1}{\zeta_{1}}b_{x_{1}},\\\
\end{split}$ (9.94)

where $|G(t)|\lesssim\|f(t)\|_{2}^{2}$.

####  9.3.5. Tangential momentum $b_{x_{2}}$ estimate.

This is nearly similar as what we computed for $b_{x_{1}}$ and $u_{1}$ case.
We define

$-\Delta_{bel}u_{2}=\frac{1}{\zeta_{2}}b_{x_{2}}-\langle\frac{1}{\zeta_{2}}b_{x_{2}}\rangle:=\widetilde{\frac{1}{\zeta_{2}}b_{x_{2}}},\quad\partial_{n}u_{2}|_{\partial
U}=0.$ (9.95)

(Symmetric case) $\Psi_{2,1}=\phi_{2}(\mathbf{v})\psi_{2}(x)$, where

$\begin{split}\phi_{2}(\mathbf{v})&=(\mathbf{v}^{2}_{2}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{2}(x)&=\zeta_{2}(x)D_{2}u_{2}.\end{split}$ (9.96)

(Non-symmetric case 1) We try test function
$\Psi_{2,2}=\sum_{i=1,3}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=\mathbf{v}_{i}\mathbf{v}_{2}\sqrt{\mu},\\\
\psi_{i}(x)&=\zeta_{2}(x)D_{i}u_{2},\end{split}$ (9.97)

for $i=1,3$.
(Non-symmetric case 2) We try test function
$\Psi_{2,3}=\sum_{i=1,3}\phi_{i}(\mathbf{v})\psi_{i}(x)$, where

$\begin{split}\phi_{i}(\mathbf{v})&=(\mathbf{v}^{2}_{i}-\beta_{b})\sqrt{\mu},\quad\text{where}\quad\int_{\mathbb{R}^{3}}(\mathbf{v}^{2}_{i}-\beta_{b})\mu
d\mathbf{v}=0,\\\ \psi_{i}(x)&=\zeta_{2}(x)D_{2}u_{2},\end{split}$ (9.98)

for $i=1,3$. We just change role of index $1$ and $2$ to gain. If we match
index $3\rightarrow r,1\rightarrow\theta,2\rightarrow z$ and pick

$\displaystyle\zeta_{2}=\zeta_{z}=1,$ (9.99)

$\zeta_{2}(x)$ satisfies:

$\begin{cases}D_{1}\zeta_{2}&=\Gamma_{D,22}^{1}\zeta_{2}\\\
D_{2}\zeta_{2}&=\Gamma_{D,11}^{2}\zeta_{2}+\Gamma_{D,33}^{2}\zeta_{2}\\\
D_{3}\zeta_{2}&=\Gamma_{D,22}^{3}\zeta_{2}\\\ \end{cases}\quad\text{and}\quad
c\leq|\zeta_{2}|\leq C,$ (9.100)

with uniform positive constants $c$ and $C$ due to (9.5) and (9.18).
Considering above three test function cases, by defining

$\Psi_{b,2}=\Psi_{2,1}+\Psi_{2,2}-\Psi_{2,3},$ (9.101)

we get

$\begin{split}&-\iint_{U\times\mathbb{R}^{3}}\mathbf{P}f(v\cdot\nabla)\Psi_{b,2}dydv+\iint_{\partial
U\times\mathbb{R}^{3}}f^{b}\Psi_{b,2}^{b}(v\cdot n)dAdv\\\
&=\int_{U}b_{x_{2}}^{2}dy-\frac{1}{|U|}\int_{U}\zeta_{2}b_{x_{2}}\int_{U}\frac{1}{\zeta_{2}}b_{x_{2}}.\\\
\end{split}$ (9.102)

With $\Psi_{b,2}$, similar as (9.94), we have

$\begin{split}\int_{U}b_{x_{2}}^{2}dy&\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{2}^{2}+\|c\|_{2}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}+\int_{U}\zeta_{2}b_{x_{2}}\int_{U}\frac{1}{\zeta_{2}}b_{x_{2}},\\\
\end{split}$ (9.103)

where $|G(t)|\lesssim\|f(t)\|_{2}^{2}$.

######  Proposition 9.6.

Let $U$ be an annular cylinder domain (9.3) which is periodic in vertical
$z$-direction. We assume that $f$ solves the linearized Boltzmann equation
(9.1) with specular boundary condition (9.2). then we have coercivity estimate

$\begin{split}\|\mathbf{P}f\|_{\nu}^{2}&\lesssim\frac{d}{dt}G(t)+\|(\mathbf{I-P})f\|_{\nu}^{2}+\|g\|_{2}^{2},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2},\\\
\end{split}$ (9.104)

under the assumption for angular momentum and momentum with respect to
$z$-axis:

$\displaystyle\int_{U\times\mathbb{R}^{3}}\\{(y-y_{0})\times z\\}\cdot
vf(t,y,v)\sqrt{\mu}dydv=0\text{ and
}\int_{U\times\mathbb{R}^{3}}v_{3}f(t,y,v)\sqrt{\mu}dydv=0,$ (9.105)

for all $t\geq 0$.

###### Proof.

From (9.75), (9.94), and (9.103), we have

$\begin{split}\|b\|_{L^{2}_{y}}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{L^{2}_{y}}^{2}+\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}\\\
&\quad+\int_{U}\zeta_{1}b_{x_{1}}\int_{U}\frac{1}{\zeta_{1}}b_{x_{1}}+\int_{U}\zeta_{2}b_{x_{2}}\int_{U}\frac{1}{\zeta_{2}}b_{x_{2}},\quad|G(t)|\lesssim\|f(t)\|_{2}^{2}.\\\
\end{split}$ (9.106)

WLOG, we set $y_{0}=(0,0,0)$ and $z=(0,0,1)$ in (9.105). Then, we can rewrite
(9.105) by using the global triply orthogonal coordinate system (9.4):

$\displaystyle\begin{split}&\int_{U\times\mathbb{R}^{3}}\\{(y-y_{0})\times
z\\}\cdot vf(t,y,v)\sqrt{\mu}dydv\\\
&\quad=\int_{U\times\mathbb{R}^{3}}(r\sin\theta,-r\cos\theta,0)\cdot
vf(t,y,v)\sqrt{\mu}dydv=\int_{U}\frac{1}{\zeta_{1}}b_{x_{1}}=0,\\\
&\int_{U\times\mathbb{R}^{3}}v_{3}f(t,y,v)\sqrt{\mu}dydv\\\
&\quad=\int_{U}(0,0,1)\cdot
vf(t,y,v)\sqrt{\mu}dydv=\int_{U}\frac{1}{\zeta_{2}}b_{x_{2}}=0.\end{split}$
(9.107)

Therefore, the last two terms in the RHS of (9.106) become 0, and then we
obtain

$\displaystyle\|b\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{L^{2}_{y}}^{2}+\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.108)

Combining with Proposition 9.5 and (9.108), we have

$\displaystyle\begin{split}\|a\|_{L^{2}_{y}}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon\|a\|_{L^{2}_{y}}^{2}+\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\|c\|_{L^{2}_{y}}^{2}&\lesssim\frac{d}{dt}G(t)+\varepsilon^{2}\|a\|_{L^{2}_{y}}^{2}+\varepsilon\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.109)

From the estimate for $a$ in (9.109), if $\varepsilon<1/2$, we derive the
following as:

$\displaystyle\|a\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.110)

Using the above estimate and the estimate for $c$ in (9.109), we can further
estimate for $c$

$\displaystyle\|c\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\varepsilon\|c\|_{L^{2}_{y}}^{2}+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},$

which implies

$\displaystyle\|c\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.$
(9.111)

Furthermore, from (9.108), (9.110), and (9.111), we get

$\displaystyle\begin{split}\|a\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},\\\
\|b\|_{L^{2}_{y}}^{2}\lesssim\frac{d}{dt}G(t)+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2}.\end{split}$
(9.112)

Using the fact that there exists $C_{1}>0$ such that
$\|\mathbf{P}f\|_{\nu}^{2}\leq C_{1}\|\mathbf{P}f\|_{2}^{2}$, one obtains that

$\displaystyle\|\mathbf{P}f\|_{\nu}^{2}\leq
C_{1}\|\mathbf{P}f\|_{2}^{2}=C_{1}(\|a\|_{L^{2}_{y}}^{2}+\|b\|_{L^{2}_{y}}^{2}+\|c\|_{L^{2}_{y}}^{2})\lesssim\frac{d}{dt}G(t)+\|\sqrt{\nu}(\mathbf{I-P})f\|_{2}^{2}+\|g\|_{2}^{2},$

where the second inequality comes from (9.111) and (9.112). ∎

######  Remark 9.7.

In the proof of Proposition 9.6, two terms in (9.107)

$\displaystyle\int_{U}\frac{1}{\zeta_{1}}b_{x_{1}}\quad\text{and}\quad\int_{U}\frac{1}{\zeta_{2}}b_{x_{2}}$

are corresponding to angular momentum and momentum with respect to the
$z$-axis. Without the assumption (9.105), this explicitly shows that an
estimate like (9.108) cannot be obtained. To ensure that the solution $F$ to
the Boltzmann equation (1.1) converges to the global Maxwellian $\mu$, it is
necessary to assume (9.105).

######  Remark 9.8.

To the best of our knowledge, this result is the first constructive proof of
(hypo)coercivity of the linearized Boltzmann equation in the presence of
specular reflection boundary, which is not trivial (e.g. a half space). The
claim of [35] could not hold as the angular momentum preserves in any axis-
symmetric domains.

######  Remark 9.9.

In our proof we heavily rely on the orthogonal coordinate of a single chart.
In general, we do not even have a global single chart of coordinate system.

## Acknowledgement

GK and DL are partly supported by the National Research Foundation of
Korea(NRF) grant funded by the Korean government(MSIT)(No.
NRF-2019R1C1C1010915 and No. RS-2023-00212304). CK is partly supported by NSF-
DMS 1900923, NSF-CAREER 2047681, the Simons fellowship in Mathematics,and the
Brain Pool fellowship funded by the Korean Ministry of Science and ICT. He
thanks Yan Guo’s comments about [35] and kind hospitality of POSTECH during a
stay in August/2022.

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