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\begin{document}
\title{Lifespan of Solution to MHD Boundary Layer Equations with Analytic Perturbation of General Shear Flow}
\author{{\bf Feng Xie}\\[1mm]
\small School of Mathematical Sciences, and LSC-MOE,\\[1mm]
\small Shanghai Jiao Tong University,
Shanghai 200240, P.R.China\\[1mm]
{\bf Tong Yang}\\[1mm]
\small Department of Mathematics,
City University of Hong Kong,\\[1mm]
\small Tat Chee Avenue, Kowloon, Hong Kong
}
\date{}
\maketitle

\begin{center}
 Dedicated to Professor Philippe G. Ciarlet on the Occasion of
his 80th Birthday
\end{center}

\bigskip

\begin{abstract} In this paper, we consider the lifespan of solution to the
	MHD boundary layer system as an analytic perturbation of general shear flow.
	By using the cancellation mechanism in the system observed in
	\cite{LXY1}, the lifespan of solution is shown to have a lower bound in the order of
	$\varepsilon^{-2+}$ if the strength of the perturbation is of the order
	of $\varepsilon$. Since there is no restriction on the strength of the
	shear flow and the lifespan estimate is larger than the one obtained for the
	classical Prandtl system in this setting, it reveals the stabilizing effect
	of the magnetic field on the electrically conducting fluid near the boundary.
\end{abstract}
\footnotetext[1]{{\it E-mail address:}
tzxief@sjtu.edu.cn (F. Xie)}
\footnotetext[2]{{\it E-mail address:} matyang@cityu.edu.hk(T. Yang)}

\vskip 2mm

\noindent {\bf 2000 Mathematical Subject Classification}: 76N20, 35Q35, 76N10, 35M33.

\vskip 2mm

\noindent {\bf Keywords}: MHD boundary  layer, analytic perturbation, lifespan estimate, shear flow.

\section{Introduction}
Consider the high Reynolds number limit to the MHD system near a no-slip boundary, the following MHD boundary layer system was derived in \cite{LXY1}
when both of the Reynolds number and the magnetic Reynolds number have the
same order in two space dimensions. Precisely, consider
the MHD system in  the domain $\{(x,Y)|x\in\mathbb{R}, Y\in\mathbb{R}_+\}$
with $Y=0$ being the boundary,
\begin{align}
\label{V1}
\left\{
\begin{array}{ll}
\partial_t u^\epsilon+(u^\epsilon\partial_x+v^\epsilon\partial_Y)u^\epsilon+\partial_xp^\epsilon-(h^\epsilon\partial_x+g^\epsilon\partial_Y)h^\epsilon=\epsilon(\partial_x^2 u^\epsilon+\partial_Y^2 u^\epsilon),\\
\partial_t v^\epsilon+(u^\epsilon\partial_x+v^\epsilon\partial_Y)v^\epsilon+\partial_Yp^\epsilon-(h^\epsilon\partial_x+g^\epsilon\partial_Y)g^\epsilon=\epsilon(\partial_x^2 v^\epsilon+\partial_Y^2 v^\epsilon),\\
\partial_t h^\epsilon+(u^\epsilon\partial_x+v^\epsilon\partial_Y)h^\epsilon-(h^\epsilon\partial_x+g^\epsilon\partial_Y)u^\epsilon=\kappa\epsilon(\partial_x^2 h^\epsilon+\partial_Y^2 h^\epsilon),\\
\partial_t g^\epsilon+(u^\epsilon\partial_x+v^\epsilon\partial_Y)g^\epsilon-(h^\epsilon\partial_x+g^\epsilon\partial_Y)v^\epsilon=\kappa\epsilon(\partial_x^2 g^\epsilon+\partial_Y^2 g^\epsilon),\\
\partial_x u^\epsilon+\partial_Y v^\epsilon=0,\qquad \partial_x h^\epsilon+\partial_Y g^\epsilon=0,
\end{array}
\right.
\end{align}
where both the viscosity and resistivity coefficients are denoted
by a small positive parameter $\epsilon$, $(u^\epsilon, v^\epsilon)$
and $(h^\epsilon, g^\epsilon)$ represent the velocity and
 the magnetic field respectively.
 The no-slip boundary condition is imposed on the velocity field
\begin{align}
\label{BCV}
(u^\epsilon, v^\epsilon)|_{Y=0}={\bf{0}},
\end{align}
and the perfectly conducting boundary condition is given for the magnetic
 field
\begin{align}
\label{BCM}
(\partial_Yh^\epsilon, g^\epsilon)|_{Y=0}={\bf{0}}.
\end{align}
Formally, when $\epsilon=0$,  (\ref{V1}) is reduced into the following incompressible ideal MHD system
\begin{align}
\label{ILE0}
\left\{
\begin{array}{ll}
\partial_t u^0_e+(u^0_e\partial_x+v^0_e\partial_Y)u^0_e+\partial_xp^0_e-(h^0_e\partial_x+g^0_e\partial_Y)h^0_e=0,\\
\partial_t v^0_e+(u^0_e\partial_x+v^0_e\partial_Y)v^0_e+\partial_Yp^0_e-(h^0_e\partial_x+g^0_e\partial_Y)g^0_e=0,\\
\partial_t h^0_e+(u^0_e\partial_x+v^0_e\partial_Y)h^0_e-(h^0_e\partial_x+g^0_e\partial_Y)u^0_e=0,\\
\partial_t g^0_e+(u^0_e\partial_x+v^0_e\partial_Y)g^0_e-(h^0_e\partial_x+g^0_e\partial_Y)v^0_e=0,\\
\partial_x u^0_e+\partial_Y v^0_e=0,\qquad \partial_x h^0_e+\partial_Y g^0_e=0.
\end{array}
\right.
\end{align}
Since the sovability of the system (\ref{ILE0}) requires only the
 normal components of the velocity and magnetic fields $(v_e^0, g^0_e)$ on the boundary
\begin{align}
\label{IBE}
(v^0_e, g^0_e)|_{Y=0}={\bf 0},
\end{align}
in the limit from (\ref{V1}) to (\ref{ILE0}), a Prandtl-type boundary layer can be derived to resolve the mis-match of the tangential components between the viscous flow $(u^\epsilon, h^\epsilon)$ and invicid flow $(u^0, h^0)$ on the  boundary $\{Y=0\}$. And this system governing the fluid behavior in the
leading order of approximation near
the boundary is  derived in \cite{GP,LXY1,LXY2}:
\begin{align}
\label{1.1}
\left\{
\begin{array}{ll}
\partial_tu_1+u_1\partial_xu_1+u_2\partial_yu_1=b_1\partial_xb_1+b_2\partial_yb_1+\partial^2_yu_1,\\
\partial_tb_1+\partial_y(u_2b_1-u_1b_2)=\kappa\partial_y^2b_1,\\
\partial_xu_1+\partial_yu_2=0,\quad \partial_xb_1+\partial_yb_2=0
\end{array}
\right.
\end{align}
in $\mathbb{H}=\{(x,y)\in\mathbb{R}^2| y\geq0\}$ with the fast variable $y=Y/\sqrt{\epsilon}$. Here, the trace of the horizontal ideal MHD flow (\ref{ILE0}) on the boundary $\{Y=0\}$ is
 assumed to be a constant vector so that the pressure term $\partial_xp^0_e(t,x,0)$ vanishes by the Bernoulli's law.

Consider the system $(\ref{1.1})$ with initial data
\begin{align}
\label{ID}
u_1(t,x,y)|_{t=0}=u_0(x,y),\qquad\qquad b_1(t,x,y)|_{t=0}=b_0(x,y),
\end{align}
and the boundary conditions
\begin{align}
\label{1.2}
\left\{
\begin{array}{ll}
u_1|_{y=0}=0,\\
u_2|_{y=0}=0,
\end{array}
\right.
\qquad\hbox{and}\qquad
\left\{
\begin{array}{ll}
\partial_yb_1|_{y=0}=0,\\
b_2|_{y=0}=0.
\end{array}
\right.
\end{align}
And the far field state is denoted by $(\bar{u}, \bar{b})$:
\begin{align}
\label{1.3}
\lim_{y\rightarrow+\infty}u_1=u^0_e(t,x,0)\triangleq\bar{u},\qquad \lim_{y\rightarrow+\infty}b_1=h^0_e(t,x,0)\triangleq\bar{b}.
\end{align}
First of all, a  shear flow $(u_s(t,y),0,\bar{b}, 0)$ is a trivial
solution to the system  \eqref{1.1} with $u_s(t,y)$ solving
\begin{align}
\label{GS}
\left\{
\begin{array}{ll}
\partial_tu_s(t,y)-\partial_y^2u_s(t,y)=0, \quad (t, y)\in\mathbb{R}_+\times\mathbb{R}_+,\\
u_s(t,y=0)=0, \qquad \lim_{y\rightarrow\infty}u_s(t,y)=\bar{u},\\
u_{s}(t=0, y)=u_{s0}(y).
\end{array}
\right.
\end{align}
In the following discussion, we assume the shear flow $u_s(t,y)$ has the following properties:
\begin{align*}
{\bf (H)}\quad \|\partial_y^iu_s(t,\cdot)\|_{L^\infty_y}\leq \frac{C}{\langle t \rangle^{i/2}}\quad (i=1,2),\quad \int_0^\infty|\partial_yu_s(t,y)|dy<C,\quad\|\theta_\alpha\partial_y^2u_s(t,\cdot)\|_{L^2_y}\leq \frac{C}{\langle t \rangle^{3/4}},
\end{align*}
for some generic constant $C$.
\begin{re}
	\label{RK2}
	The assumption (H) on the shear flow holds
	for  a large class of initial data $u_{s0}$.
	For example, it holds for the initial data $u_{s0}=\chi(y)$ with $\chi(y)\in C^\infty(\mathbb{R})$, $\chi(y)=0$ for $y\leq 1$ and $\chi(y)=\bar{u}$ for $y\geq 2$ considered in \cite{ZZ} for the Prandtl system. Note that here we do
	not assume the smallness of the shear flow. In addition, it also holds when  $u_{s0}(y)=\frac{1}{\sqrt{\pi}}\int_0^{y}\exp(-\frac{z^2}{4})dz$
	 considered in \cite{IV} for the Prandtl system where the almost global solution is
	 obtained. Note that for the classical Prandtl equations, such shear flow
	 in the form of Guassian error
	 function yields a time decay damping term in the time evolution equation of
	 $u_1$, however, it does not leads to  any damping effect in the MHD boundary
	 layer system \eqref{1.1}.
\end{re}

To define the function space of the solution considered in this paper,
the following Gaussian weighted function $\theta_\alpha$ will be used:
\begin{align*}
\theta_{\alpha}(t,y)=\exp{(\frac{\alpha z(t,y)^2}{4})},\quad \hbox{with}\quad z(t,y)=\frac{y}{\sqrt{\langle t\rangle}},\quad \langle t\rangle=1+t\quad\hbox{and}\quad \alpha\in[1/4,1/2].
\end{align*}
With this and
\begin{align*}
M_m=\frac{\sqrt{m+1}}{m!},
\end{align*}
define  the Sobolev weighted semi-norms by
\begin{align}
\label{NM1}
&X_m=X_m(f,\tau)=\|\theta_{\alpha}\partial_x^mf\|_{L^2}\tau^mM_m,\quad
D_m=D_m(f,\tau)=\|\theta_{\alpha}\partial_y\partial_x^mf\|_{L^2}\tau^mM_m,\nonumber\\
&Z_m=Z_m(f,\tau)=\|z\theta_{\alpha}\partial_x^mf\|_{L^2}\tau^mM_m,\quad
Y_m=Y_m(f,\tau)=\|\theta_{\alpha}\partial_x^mf\|_{L^2}\tau^{m-1}mM_m.
\end{align}
Then the following space of analytic functions
in the tangential variable $x$ and Sobolev weighted in the normal variable $y$
is defined by
\begin{align*}
X_{\tau, \alpha}=\{f(t,x,y)\in L^2(\mathbb{H}; \theta_{\alpha}dxdy): \|f\|_{X_{\tau, \alpha}}<\infty\}
\end{align*}
with $\tau>0$ and the norm
\begin{align*}
\|f\|_{X_{\tau, \alpha}}=\sum_{m\geq 0}X_m(f,\tau).
\end{align*}
In addition,  the following two semi-norms will also be used:
\begin{align*}
\|f\|_{D_{\tau, \alpha}}=\sum_{m\geq 0}D_m(f,\tau)=\|\partial_yf\|_{X_{\tau, \alpha}},\quad
\|f\|_{Y_{\tau, \alpha}}=\sum_{m\geq 1}Y_m(f,\tau).
\end{align*}
Here, the summation over $m$ is considered in the $l^1$ sense that is similar
to the definition used in \cite{IV,ZZ} rather than in the $l^2$ sense used in \cite{KV}. With the above notations, we are now
 ready to state the main Theorem as follows.

\begin{thm}
\label{THM}
For any $\lambda\in [3/2, 2)$, there exists a small positive constant
$\varepsilon_*$ depending on $2-\lambda$.
Under the assumption (H) on the backgroud shear flow
$(u_s(t,y), 0, \bar{b},0)$ with $\bar{b}\neq 0$, assume the initial data $u_0$ and $b_0$  satisfy
\begin{align}
\label{THM1}
\|u_0-u_s(0,y)\|_{X_{2\tau_0,1/2}}\leq \varepsilon,
\quad
\|b_0-\bar{b}\|_{X_{2\tau_0,1/2}}\leq \varepsilon,
\end{align}
for some given $\varepsilon\in (0, \varepsilon_*]$. Then there exists a unique solution $(u_1, u_2, b_1, b_2)$ to the MHD boundary layer equations (\ref{1.1})-(\ref{1.3}) such that
\begin{align*}
(u_1-u_s(t,y),  b_1-\bar{b})\in X_{\tau,\alpha},\  \alpha\in[1/4,1/2],
\end{align*}
with analyticity radius $\tau$ larger than $\tau_0/4$ in the time interval $[0, T_\varepsilon]$.
And the lifespan $T_\varepsilon$ has the following low bound estimate
\begin{align}
\label{THM3}
T_\varepsilon\geq {C}\varepsilon^{-\lambda},
\end{align}
where the constant ${C}$ is  independent of $\varepsilon$.
\end{thm}

As is well-known that the leading order characteristic boundary layer for the
incompressible Navier-Stokes equations
with no-slip boundary condition is described by the classical Prandtl
equations derived by Prandtl  \cite{P} in 1904. In the two space dimensions,
under the monotonicity
assumption on the tangential velocity in the normal direction, Oleinik firstly
obtained the local existence of classical solutions
by using the Crocco transformation, cf.  \cite{O} and
Oleinik-Samokhin's classical
book \cite{OS}. Recently, this well-posedness result was re-proved by using an energy method  in the framework of
 Sobolev spaces in \cite{AWXY} and \cite{MW1} independently by observing the cancellation mechanism in the convection terms. And by imposing an additional
favorable condition on the pressure,
a global in time weak solution was obtained in
\cite{XZ}.

When the monotonicity condition is violated, singularity formation or separation of the boundary
layer is well expected and observed. For this, E-Engquist constructed a finite
time blowup solution to the Prandtl
equations in \cite{EE}. Recently, when the background shear flow has a non-degenerate critical point, some interesting ill-posedness (or instability) phenomena of solutions to both linear and nonlinear classical Prandtl equations around
shear flows are studied, cf.
\cite{GD,GN,GN1}. All these results show that the monotonicity assumption on the tangential velocity plays a key role for well-posedness theory
except in the frameworks of analytic functions and Gevrey regularity classes. Indeed, in the framework of
analytic functions, Sammartino and Caflisch \cite{SC,CS} established
the local well-posedness theory of the Prandtl system in three space dimensions and also justified the Prandtl ansatz in this setting by applying
the abstract Cauchy-Kowalewskaya (CK) theorem initated by Asano's unpublished work.
 Later, the analyticity requirement in the normal variable $y$ was removed by
Lombardo, Cannone and Sammartino in \cite{LCS} because of the viscous effect in the normal direction.

Recently, Zhang and Zhang obtained the lifespan of small analytic solution to the classical Prandtl equations with small analytic initial data in \cite{ZZ}. Precisely, when the strength of background shear flow is of the order
 of $\varepsilon^{5/3}$ and the perturbation is of the order of $\varepsilon$, they showed that  the classical Prandtl
system has a unique solution with a lower bound estimate on the lifespan in the order of  $\varepsilon^{-4/3}$.
Furthermore, if the initial data is a small analytic perturbation of the Guassian error function (\ref{GS}),  an
almost global existence for the Prandtl boundary layer equations is proved in \cite{IV}.

On the other hand, to study the high Reynolds number limits for  the MHD equations (\ref{V1}) with no-slip boundary condition on the velocity (\ref{BCV}) and perfect conducting boundary condition (\ref{BCM}) on the magnetic field,
one can apply the Prandtl ansatz to derive the boundary layer system  (\ref{1.1}) as
 the leading order description on the flow near the boundary. For this, readers can
  refer to \cite{GP,LXY1,LXY2,LXY3,XY} about  the formal derivation of (\ref{1.1}), the well-posedness theory
  of the system  and the justification of the Prandtl ansatz locally in time.

This paper is about  long time existence of solutions to (\ref{1.1})-(\ref{1.3}). Precisely, we will show that if the initial data is a small perturbation of a shear flow analytically in the order of  $\varepsilon$, then there exists a unique solution to (\ref{1.1})-(\ref{1.3}) with the lifespan $T_\varepsilon$ of the order of $ \varepsilon^{-2+}$.  Compared with the
estimate on the lifespan of solutions to the classical Prandtl system studied in  \cite{ZZ}, the lower bound estimate
is larger and there is no requirement on the smallness of the background shear flow because the mechanism in the
system is used due to the non-degeneracy of the tangential  magnetic field. However, it is not known
whether one can obtain a global or
almost global in time solution like the work on the Prandtl system  when the background shear velocity is taken to be a Gaussian error function in \cite{IV}. We mention that even though  Lin and Zhang showed the almost global existence of solution to MHD boundary layer equations with zero Dirichlet boundary condition on the  magnetic field in \cite{LZ} when
the components of both the background velocity and magnetic fields are Guassian error functions, it is not clear wheather
 the system (\ref{1.1}) holds with zero Dirichlet boundary condition even in formal derivation.

The analysis on the lifespan of the perturbed system in this paper relies on the introduction of some new unknown functions that capture the cancellation of some  linear terms. Unlike the work
in \cite{IV} on the  Prandtl system for which the cancellation
yields  a damping  term in the time evolution of the perturbation of the tangetial velocity field, there
is no such damping effect observed for the  MHD boundary layer system.

Finally, the rest of the paper is organized as follows. After giving  some preliminary estimates,  a
uniform estimate on the solution will be proved in the next section.  Based on this uniform estimate,  a low bound of the lifespan of  solution is  derived  in Section 3. The uniqueness part is done in Section 4. Throughout the paper,  constants denoted by  $C$, $\bar{C}, C_0, C_1$ and $C_2$ are generic and independent of the small parameter $\varepsilon$.

\section{Uniform Estimate}

We first list the following two priliminary estimates on the functions in the norms defined in the previous section. The first
estimate indeed is from  Lemma 3.3 in \cite{IV} (also see \cite{H}).
\begin{lem}
\label{LEM2.1}
(Poincar\'e type inequality with Gaussian weight) Let $f$ be a function such that $f|_{y=0}=0\ (or\ \partial_yf|_{y=0})$ and $f|_{y=\infty}=0$. Then, for $\alpha\in[1/4,1/2], m\geq 0$ and $t\geq 0$, it holds that
\begin{align}
\label{2.1}
\frac{\alpha}{\langle t\rangle}\|\theta_{\alpha}\partial_x^mf\|_{L^2_y}^2\leq\|\theta_{\alpha}\partial_y\partial_x^mf\|_{L^2_y}^2.
\end{align}
\end{lem}

The second lemma is  used  in \cite{IV} and we include it here with a short proof for convenience of readers.

\begin{lem}\label{LEM2.2}
Let $f$ be a function such that $f|_{y=0}=0\ (or\ \partial_yf|_{y=0})$ and $f|_{y=\infty}=0$.  Then
\begin{align}
\label{LL}
\sum_{m\geq 0}\frac{\|\theta_{\alpha}\partial_y\partial_x^mf\|^2_{L^2}}{\|\theta_{\alpha}\partial_x^mf\|_{L^2}}\tau^mM_m\geq \frac{\alpha^{1/2}\beta}{2\langle t\rangle^{1/2}}\|f\|_{D_{\tau,\alpha}}+\frac{\alpha(1-\beta)}{\langle t\rangle}\|f\|_{X_{\tau,\alpha}},
\end{align}
for $\beta\in(0, 1/2)$.
\end{lem}
\begin{pf}
In fact, by Lemma \ref{LEM2.1}, one has
\begin{align*}
\frac{\|\theta_{\alpha}\partial_y\partial_x^mf\|^2_{L^2}}{\|\theta_{\alpha}\partial_x^mf\|_{L^2}}\geq & \frac{\beta}{2}\frac{\|\theta_{\alpha}\partial_y\partial_x^mf\|^2_{L^2}}{\|\theta_{\alpha}\partial_x^mf\|_{L^2}}+\frac{2-\beta}{2}\frac{\alpha^{1/2}}{\langle t\rangle^{1/2}}\|\theta_{\alpha}\partial_y\partial_x^mf\|_{L^2}\\
\geq &\frac{\beta}{2}\frac{\|\theta_{\alpha}\partial_y\partial_x^mf\|^2_{L^2}}{\|\theta_{\alpha}\partial_x^mf\|_{L^2}}+\frac{\beta\alpha^{1/2}}{2\langle t\rangle^{1/2}}\|\theta_{\alpha}\partial_y\partial_x^mf\|_{L^2}+\frac{\alpha(1-\beta)}{\langle t\rangle}\|\theta_{\alpha}\partial_x^mf\|_{L^2}\\
\geq& \frac{\beta\alpha^{1/2}}{2\langle t\rangle^{1/2}}\|\theta_{\alpha}\partial_y\partial_x^mf\|_{L^2}+\frac{\alpha(1-\beta)}{\langle t\rangle}\|\theta_{\alpha}\partial_x^mf\|_{L^2}.
\end{align*}
Multiplying the above inequality by $\tau^mM_m$  and summing up in $m\geq 0$ give (\ref{LL}).
\end{pf}

 We are now ready to study a uniform estimate on the solution. For this, we first
rewrite the solution to (\ref{1.1})-(\ref{1.3}) as a perturbation $(u, v, b, g)$ of the $(u_s(t,y), 0, \bar{b}, 0)$ by denoting
\begin{align}
\label{EXP}
\left\{
\begin{array}{ll}
u_1=u_s(t,y)+u,\\
u_2=v,
\end{array}
\right.
\qquad\qquad
\left\{
\begin{array}{ll}
b_1=\bar{b}+b,\\
b_2=g.
\end{array}
\right.
\end{align}
Without loss of generality, take $\bar{b}=1$
and $\kappa=1$. Then  (\ref{1.1}) yields
\begin{align}
\label{1.4}
\left\{
\begin{array}{ll}
\partial_tu+(u_s+u)\partial_xu+v\partial_y(u_s+u)-(1+b)\partial_xb-g\partial_yb-\partial_y^2u=0,\\
\partial_tb-(1+b)\partial_xu-g\partial_y(u_s+u)+(u_s+u)\partial_xb+v\partial_yb-\partial_y^2b=0.
\end{array}
\right.
\end{align}
And the initial and boundary data of $(u, v)$ and $(b, g)$ are given by
\begin{align}
\label{NID}
u(t,x,y)|_{t=0}=u_0(x,y)-u_s(0,y),\qquad b(t, x,y)|_{t=0}=b_0(x,y)-1,
\end{align}
\begin{align}
\label{NBC}
\left\{
\begin{array}{ll}
u|_{y=0}=0,\\
v|_{y=0}=0,
\end{array}
\right.
\qquad\hbox{and}\qquad
\left\{
\begin{array}{ll}
\partial_yb|_{y=0}=0,\\
g|_{y=0}=0,
\end{array}
\right.
\end{align}
with the corresponding far field condition
\begin{align}
\label{NFC}
\lim_{y\rightarrow+\infty}u=0,\qquad \lim_{y\rightarrow+\infty}b=0.
\end{align}

It suffices to establish the long time existence of solutions to (\ref{1.4})-(\ref{NFC}). In this section, we focus on the uniform a priori estimate on  the solution to (\ref{1.4}) in the analytical framework defined in Section 1.

Integrating equation $(\ref{1.4})_2$ over $[0,y]$ gives that
\begin{align}
\label{3.1}
\partial_t\int_0^ybd\tilde{y}+v(1+b)-(u_s+u)g=\partial^2_y\int_0^ybd\tilde{y},
\end{align}
where the boundary conditions that $\partial_yb|_{y=0}=v|_{y=0}=g|_{y=0}=0$ are used.\\
Define
\begin{align*}
\psi(t,y)=\int_0^ybd\tilde{y},
\end{align*}
one has
\begin{align}
\label{3.2}
\partial_t\psi+v(1+b)-(u_s+u)g=\partial^2_y\psi.
\end{align}
Now introduce  new unknown functions by taking care of the cancellation mechamism in the system as obseved in \cite{LXY1} as follows
\begin{align}
\label{3.3}
\tilde{u}=u-\partial_yu_s\psi,\qquad \tilde{b}=b.
\end{align}
Then $(\tilde{u}, \tilde{b})$ satisfies the following equations.
\begin{align}
\label{3.4}
\left\{
\begin{array}{ll}
\partial_t\tilde{u}-\partial_y^2\tilde{u}+(u_s+u)\partial_x\tilde{u}+v\partial_y\tilde{u}-(1+b)\partial_x\tilde{b}-g\partial_y\tilde{b}-2\partial_y^2u_s\tilde{b}+v\partial_y^2u_s\psi=0,\\
\partial_t\tilde{b}-\partial_y^2\tilde{b}-(1+b)\partial_x\tilde{u}-g\partial_y\tilde{u}+(u_s+u)\partial_x\tilde{b}+v\partial_y\tilde{b}-g\partial_y^2u_s\psi=0.
\end{array}
\right.
\end{align}
Here we have used the following fact that $u_s$ is the solution to the heat equation. That is,
\begin{align*}
\partial_tu_s-\partial_y^2u_s=0,\qquad \partial_t\partial_yu_s-\partial_y^3u_s=0.
\end{align*}
By a direct calculation, the boundary conditions of $(\tilde{u}, \tilde{b})$ are given by
\begin{align}
\label{BC}
\tilde{u}|_{y=0}=0,\qquad \partial_y\tilde{b}|_{y=0}=0,
\end{align}
\begin{align}
\label{FBC}
\tilde{u}|_{y=\infty}=0,\qquad \tilde{b}|_{y=\infty}=0.
\end{align}
We then turn to show the existence of solution $(\tilde{u}, \tilde{b})$ to (\ref{3.4})-(\ref{FBC}) with the
corresponding initial data.
\begin{align}
\label{IIII}
\tilde{u}(0,x,y)=u(0,x,y)-\partial_yu_s(0,y)\int_0^yb(0,x,\tilde{y})d\tilde{y},\qquad \tilde{b}(0,x,y)=b(0,x,y).
\end{align}
Note that
\begin{align}
\label{IIIIE}
\|\tilde{u}(0,x,y)\|_{X_{2\tau_0, \alpha}}\leq \|u(0,x,y)\|_{X_{2\tau_0, \alpha}}+C\|b(0,x,y)\|_{X_{2\tau_0, \alpha}},
\end{align}
for $\alpha\in[1/4, 1/2]$.

Moreover, once the existence of solution $(\tilde{u}, \tilde{b})$ to (\ref{3.4})-(\ref{IIII}) is obtained,  one can define $(u, b)$ by
\begin{align}
\label{RS}
u(t,x,y)=\tilde{u}(t,x,y)+\partial_yu_s(t,y)\int_0^y\tilde{b}(t,x,\tilde{y})d\tilde{y},\qquad b(t,x,y)=\tilde{b}(t,x,y).
\end{align}
It is straightforward to check that $(u, b)$ is a solution to (\ref{1.4})-(\ref{NFC}) with the following estimates
\begin{align*}
\|u\|_{X_{\tau, \alpha}}\leq \|\tilde{u}\|_{X_{\tau, \alpha}}+C\|\tilde{b}\|_{X_{\tau, \alpha}},\qquad  \|b\|_{X_{\tau, \alpha}}=\|\tilde{b}\|_{X_{\tau, \alpha}}.
\end{align*}
Therefore, we only need to estimate  the solution $(\tilde{u}, \tilde{b})$ to (\ref{3.4})-(\ref{IIII})
in the analytic norms as shown in the next two subsections.

\subsection{A priori estimate on velocity field}
For  $m\geq 0$,  by applying the tangential derivative operator $\partial_x^m$ to $(\ref{3.4})_1$ and multiplying it  by $\theta_{\alpha}^2\partial_x^m\tilde{u}$, the integration over  $\mathbb{H}$ yields
\begin{align}
\label{3.5}
\int_{\mathbb{H}}\partial_x^m(\partial_t\tilde{u}-\partial_y^2\tilde{u}+(u_s+u)\partial_x\tilde{u}+v\partial_y\tilde{u}-
(1+b)\partial_x\tilde{b}-g\partial_y\tilde{b}-2\partial_y^2u_s\tilde{b}+v\psi\partial_y^2u_s)\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy=0.
\end{align}
We now  estimate each term in $(\ref{3.5})$ as follows. Firstly, note that
\begin{align}
\label{3.6}
&\int_\mathbb{H}\partial_t\partial_x^m\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy\nonumber\\
= &\frac12\frac{d}{dt}\int_\mathbb{H}(\partial_x^m\tilde{u})^2\theta_{\alpha}^2dxdy-\int_\mathbb{H}(\partial_x^m\tilde{u})^2\theta_{\alpha}\frac{d}{dt}\theta_{\alpha}dxdy\\
= &\frac12\frac{d}{dt}\|\theta_{\alpha}\partial_x^m\tilde{u}\|^2_{L^2}+\frac{\alpha}{4\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{u}\|^2_{L^2},\nonumber
\end{align}
and
\begin{align*}
-\int_\mathbb{H}\partial_y^2\partial_x^m\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
= \|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2+\int_\mathbb{H}\partial_y\partial_x^m\tilde{u}\partial_y(\theta_{\alpha}^2)\partial_x^m\tilde{u}dxdy.
\end{align*}
The boundary  term vanishes because  of the boundary condition  $\partial_x^m\tilde{u}|_{y=0}=0$. Furthermore,
\begin{align*}
&\int_\mathbb{H}\partial_y\partial_x^m\tilde{u}\partial_y(\theta_{\alpha}^2)\partial_x^m\tilde{u}dxdy
=-\frac12\int_\mathbb{H}(\partial_x^m\tilde{u})^2\partial_y^2(\theta_{\alpha}^2)dxdy\\
=&-\frac{\alpha}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}\partial_x^m\tilde{u}\|^2_{L^2}-\frac{\alpha^2}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{u}\|^2_{L^2},
\end{align*}
where we have used
\begin{align*}
\partial_y^2(\theta_{\alpha}^2)=\frac{\alpha}{\langle t\rangle}\theta_{\alpha}^2+\frac{\alpha^2}{\langle t\rangle}z^2(t,y)\theta_{\alpha}^2.
\end{align*}
Consequently,
\begin{align}
\label{3.7}
-\int_\mathbb{H}\partial_y^2\partial_x^m\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
= \|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2-\frac{\alpha}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}\partial_x^m\tilde{u}\|^2_{L^2}-\frac{\alpha^2}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{u}\|^2_{L^2}.
\end{align}
For the nonlinear terms in (\ref{3.5}), we have
\begin{align*}
\int_{\mathbb{H}}\partial_x^m((u_s+u)\partial_x\tilde{u})\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xu\partial_x^{j+1}\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
\triangleq R_1
\end{align*}
and
\begin{align*}
|R_1|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xu\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xu\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$, by (\ref{RS}), one has
\begin{align*}
&\|\partial^{m-j}_xu\|_{L^2_xL^\infty_y}=\|\partial^{m-j}_x(\tilde{u}+\partial_yu_s\psi)\|_{L^2_xL^\infty_y}\\
\leq &\|\partial^{m-j}_x\tilde{u}\|_{L^2_xL^\infty_y}+\|\partial_yu_s\partial^{m-j}_x\psi\|_{L^2_xL^\infty_y}\\
\leq &C\|\theta_{\alpha}\partial^{m-j}_x\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j}_x\partial_y\tilde{u}\|^{1/2}_{L^2}+C\langle t\rangle^{-1/4}\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|_{L^2},
\end{align*}
where in the last inequality, we have used
$$
\|\partial_yu_s\|_{L_y^\infty}\leq \frac{C}{\sqrt{\langle t\rangle}},
$$
according to the assumption (H). Moreover,
\begin{align*}
&\|\partial^{m-j}_x\psi\|_{L^2_xL^\infty_y}=\|\int_0^y\partial^{m-j}_x\tilde{b}d\tilde{y}\|_{L^2_xL^\infty_y}\\
=&\|\int_0^y\theta_{\alpha}\partial^{m-j}_x\tilde{b}\exp(-\frac{\alpha}{4}z^2)d\tilde{y}\|_{L^2_xL^\infty_y}
\leq C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|_{L^2}.
\end{align*}
And
\begin{align*}
\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^\infty_xL^2_y}\leq C\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+2}_x\tilde{u}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$, we have
\begin{align*}
\|\partial^{m-j}_xu\|_{L^\infty_{xy}}\leq &\|\partial^{m-j}_x\tilde{u}\|_{L_{xy}^\infty}+\|\partial_yu_s\partial^{m-j}_x\psi\|_{L_{xy}^\infty}\\
\leq &C\|\theta_{\alpha}\partial^{m-j}_x\tilde{u}\|^{1/4}_{L^2}\|\theta_{\alpha}\partial^{m-j}_x\partial_y\tilde{u}\|^{1/4}_{L^2}
\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|^{1/4}_{L^2}\|\theta_{\alpha}\partial^{m-j+1}_x\partial_y\tilde{u}\|^{1/4}_{L^2}\\
&+C\langle t\rangle^{-1/4}\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
Hence,
\begin{align}
\label{NE1}
\frac{|R_1|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}(X^{1/2}_{m-j}D^{1/2}_{m-j}+\langle t\rangle^{-1/4}\bar{X}_{m-j})Y^{1/2}_{j+1}Y^{1/2}_{j+2} \right.\\
&\left.+\sum_{j=[m/2]+1}^{m}(X^{1/4}_{m-j}X^{1/4}_{m-j+1}D^{1/4}_{m-j}D^{1/4}_{m-j+1}+\langle t\rangle^{-1/4}\bar{X}_{m-j}^{1/2}\bar{X}_{m-j+1}^{1/2})Y_{j+1}\right\}.\nonumber
\end{align}
From now on, we use $X_i, D_i, Y_i$ to denote the semi-norms of function $\tilde{u}$ defined in (\ref{NM1}), and
 $\bar{X}_i, \bar{D}_i$ and $\bar{Y}_i$ for the corresponding semi-norms for $\tilde{b}$. Note that
\begin{align*}
\int_{\mathbb{H}}\partial_x^m(v\partial_y\tilde{u})\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xv\partial_x^{j}\partial_y\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
\triangleq R_2
\end{align*}
and
\begin{align*}
|R_2|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_xv\|_{L^2_xL^\infty_y}=&\|\int_0^y\partial^{m-j+1}_xud\tilde{y}\|_{L^2_xL^\infty_y}\\
\leq&\|\int_0^y\partial^{m-j+1}_x\tilde{u}d\tilde{y}\|_{L^2_xL^\infty_y}+\|\int_0^y\partial_yu_s(\int_0^{\tilde{y}}\partial^{m-j+1}_x\tilde{b}ds)d\tilde{y}\|_{L^2_xL^\infty_y}\\
\leq&\|\int_0^y\partial^{m-j+1}_x\tilde{u}d\tilde{y}\|_{L^2_xL^\infty_y}+\|\int_0^y\partial_yu_s d\tilde{y}\|_{L^\infty_y}\|\int_0^{\tilde{y}}\partial^{m-j+1}_x\tilde{b}ds\|_{L^2_xL^\infty_y}\\
\leq&C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|_{L^2}+C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2},
\end{align*}
where we have used $\int_0^\infty|\partial_yu_s(t,y)|dy<C$ by the assumption (H). Note that
\begin{align*}
\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^\infty_xL^2_y}\leq C\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+1}_x\partial_y\tilde{u}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_xv\|_{L^\infty_{xy}}\leq & \|\int_0^y\partial^{m-j+1}_x\tilde{u}d\tilde{y}\|_{L^\infty_{xy}}+\|\int_0^y\partial_yu_s(\int_0^{\tilde{y}}\partial^{m-j+1}_x\tilde{b}ds)d\tilde{y}\|_{L^\infty_{xy}}\\
\leq & \|\int_0^y\partial^{m-j+1}_x\tilde{u}d\tilde{y}\|_{L^\infty_{xy}}+\|\int_0^y\partial_yu_s d\tilde{y}\|_{L^\infty_{y}}\|\int_0^{\tilde{y}}\partial^{m-j+1}_x\tilde{b}ds\|_{L^\infty_{xy}}\\
\leq & C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|_{L^2_yL^\infty_x}+C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2_yL^\infty_x}\\
\leq &C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{u}\|^{1/2}_{L^2}+C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
Hence,
\begin{align}
\label{NE2}
\frac{|R_2|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}(\langle t\rangle^{1/4}Y_{m-j+1}+\langle t\rangle^{1/4}\bar{Y}_{m-j+1})D^{1/2}_{j}D^{1/2}_{j+1} \right.\\
&\left.+\sum_{j=[m/2]+1}^{m}(\langle t\rangle^{1/4}Y_{m-j+1}^{1/2}Y_{m-j+2}^{1/2}+\langle t\rangle^{1/4}\bar{Y}_{m-j+1}^{1/2}\bar{Y}_{m-j+2}^{1/2})D_{j}\right\}.\nonumber
\end{align}
Recall  $b=\tilde{b}$ so that
\begin{align*}
R_3\triangleq\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_x\tilde{b}\partial_x^{j+1}\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy,
\end{align*}
and
\begin{align*}
|R_3|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_x\tilde{b}\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_x\tilde{b}\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_x\tilde{b}\|_{L^2_xL^\infty_y}
\leq C\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|^{1/2}_{L^2}\|\partial^{m-j}_x\partial_y\tilde{b}\|^{1/2}_{L^2},
\end{align*}
and
\begin{align*}
\|\partial^{j+1}_x\tilde{b}\|_{L^\infty_xL^2_y}\leq C\|\partial^{j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\partial^{j+2}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_x\tilde{b}\|_{L^\infty_{xy}}
\leq C\|\partial^{m-j}_x\tilde{b}\|^{1/4}_{L^2}\|\partial^{m-j}_x\partial_y\tilde{b}\|^{1/4}_{L^2}\|\partial^{m-j+1}_x\tilde{b}\|^{1/4}_{L^2}\|\partial^{m-j+1}_x\partial_y\tilde{b}\|^{1/4}_{L^2}.
\end{align*}
Therefore,
\begin{align}
\label{NE3}
\frac{|R_3|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}\bar{X}_{m-j}^{1/2}\bar{D}_{m-j}^{1/2}\bar{Y}_{j+1}^{1/2}\bar{Y}_{j+2}^{1/2}\right.\\
&\left.+\sum_{j=[m/2]+1}^{m}\bar{X}_{m-j}^{1/4}\bar{X}_{m-j+1}^{1/4}\bar{D}_{m-j}^{1/4}\bar{D}_{m-j+1}^{1/4}\bar{Y}_{j+1}\right\}.\nonumber
\end{align}
 Note that
\begin{align*}
\int_{\mathbb{H}}\partial_x^m(g\partial_y\tilde{b})\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xg\partial_x^{j}\partial_y\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy
\triangleq R_4
\end{align*}
and
\begin{align*}
|R_4|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}\leq  C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2},
\end{align*}
and
\begin{align*}
\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|_{L^\infty_xL^2_y}\leq C\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+1}_x\partial_y\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^\infty_{xy}}\leq &
C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2_yL^\infty_x}\\
\leq & C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
As a consequence, we have
\begin{align}
\label{NE4}
\frac{|R_4|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq  \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}\langle t\rangle^{1/4}\bar{Y}_{m-j+1}\bar{D}^{1/2}_{j}\bar{D}^{1/2}_{j+1}
+\sum_{j=[m/2]+1}^{m}\langle t\rangle^{1/4}\bar{Y}^{1/2}_{m-j+1}\bar{Y}^{1/2}_{m-j+2}\bar{D}_{j}\right\}.
\end{align}
And
\begin{align*}
|\int_\mathbb{H}\partial_y^2u_s\partial_x^m\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy|
\leq&\|\partial_y^2u_s\|_{L^\infty_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
\leq&
C\langle t\rangle^{-1}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2},
\end{align*}
that is,
\begin{align}
\label{NE5}
\frac{|\int_\mathbb{H}\partial_y^2u_s\partial_x^m\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy|}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq C\langle t\rangle^{-1}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2},
\end{align}
where we  have used $\|\partial_y^2u_s\|_{L_y^\infty}\leq \frac{C}{\langle t\rangle}$ by the assumption (H). We now consider
\begin{align*}
R_5\triangleq\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{H}\partial^{m-j}_xv \partial_y^2u_s   \partial_x^{j}\psi\theta_{\alpha}^2\partial_x^m\tilde{u}dxdy.
\end{align*}
Note that
\begin{align*}
|R_5|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}  \|\partial^{j}_x\psi\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}  \|\partial^{j}_x\psi\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{m}_x\tilde{u}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$, we have
\begin{align*}
\|\partial^{m-j}_xv\|_{L^2_xL^\infty_y}
\leq C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|_{L^2} + C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2},
\end{align*}
and
\begin{align*}
\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}\leq \frac{C}{\langle t\rangle^{3/4}},
\end{align*}
provided that $\alpha<1$ by the  assumption (H). And
\begin{align*}
\|\partial^{j}_x\psi\|_{L^\infty_{xy}}\leq C\langle t\rangle^{1/4} \|\theta_{\alpha}\partial^{j}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$, we have
\begin{align*}
\|\partial^{m-j}_xv\|_{L^\infty_{xy}}
\leq& C\langle t\rangle^{1/4} \|\theta_{\alpha}\partial^{m-j+1}_x\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{u}\|^{1/2}_{L^2}\\
&+C\langle t\rangle^{1/4} \|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
And
\begin{align*}
\|\partial^{j}_x\psi\|_{L^2_xL^\infty_y}\leq C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial_x^j\tilde{b}\|_{L^2}.
\end{align*}
Hence,
\begin{align}
\label{NE6}
\frac{|R_5|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{u}\|_{L^2}}\leq & \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}(\langle t\rangle^{-1/4}Y_{m-j+1}+\langle t\rangle^{-1/4}\bar{Y}_{m-j+1})\bar{X}^{1/2}_{j}\bar{X}^{1/2}_{j+1} \right.\\
&\left.+\sum_{j=[m/2]+1}^{m}(\langle t\rangle^{-1/4}Y_{m-j+1}^{1/2}Y_{m-j+2}^{1/2}+\langle t\rangle^{-1/4}\bar{Y}_{m-j+1}^{1/2}\bar{Y}_{m-j+2}^{1/2})\bar{X}_{j}\right\}.\nonumber
\end{align}
Combining the estimates (\ref{3.6})-(\ref{NE6})  and summing over $m\geq 0$ give
\begin{align}
\label{NEU}
&\frac{d}{dt}\|\tilde{u}\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}+\frac{\alpha(1-2\alpha)}{4\langle t\rangle}\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}z\partial_x^m\tilde{u}\|^2_{L^2}}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}\nonumber\\
&-\frac{\alpha}{2\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}}-\frac{C}{\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}}
\leq\dot{\tau}(t)\|\tilde{u}\|_{Y_{\tau,\alpha}}\\&+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)(\|\tilde{u}\|_{Y_{\tau,\alpha}}+\|\tilde{b}\|_{Y_{\tau,\alpha}}),\nonumber
\end{align}
where we have used the fact that for any positive sequences $\{a_j\}_{j\geq 0}$ and $\{b_j\}_{j\geq 0}$,
\begin{align*}
\sum_{m\geq 0}\sum_{j=0}^{m}a_jb_{m-j}\leq \sum_{j\geq 0}a_j\sum_{j\geq 0}b_j.
\end{align*}
Choosing $\alpha\leq1/2$ in (\ref{NEU}) yields
\begin{align}
\label{NEUU}
&\frac{d}{dt}\|\tilde{u}\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}-\frac{\alpha}{2\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}}-\frac{C}{\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}}\nonumber\\
\leq& \dot{\tau}(t)\|\tilde{u}\|_{Y_{\tau,\alpha}}+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)\nonumber\\
&\times(\|\tilde{u}\|_{Y_{\tau,\alpha}}+\|\tilde{b}\|_{Y_{\tau,\alpha}}).
\end{align}
\subsection{A priori estimate on  magnetic field}
Similarly,
for $m\geq 0$,  by applying the tangential derivative operator $\partial_x^m$ to $(\ref{3.4})_2$ and  multiplying it by $\theta_{\alpha}^2\partial_x^m\tilde{b}$, the integration  over  $\mathbb{H}$ gives
\begin{align}
\label{3.8}
\int_{\mathbb{H}}\partial_x^m(\partial_t\tilde{b}-\partial_y^2\tilde{b}-(1+b)\partial_x\tilde{u}-g\partial_y\tilde{u}+
(u_s+u)\partial_x\tilde{b}+v\partial_y\tilde{b}-g\partial_y^2u_s\psi)\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy=0.
\end{align}
We now estimate $(\ref{3.8})$ term by term as follows. Firstly,
\begin{align}
\label{3.19}
&\int_\mathbb{H}\partial_t\partial_x^m\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
= \frac12\frac{d}{dt}\int_\mathbb{H}(\partial_x^m\tilde{b})^2\theta_{\alpha}^2dxdy-\int_\mathbb{H}(\partial_x^m\tilde{b})^2\theta_{\alpha}\frac{d}{dt}\theta_{\alpha}dxdy\\
= &\frac12\frac{d}{dt}\|\theta_{\alpha}\partial_x^m\tilde{b}\|^2_{L^2}+\frac{\alpha}{4\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{b}\|^2_{L^2}.\nonumber
\end{align}
And
\begin{align*}
-\int_\mathbb{H}\partial_y^2\partial_x^m\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
=\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2+\int_\mathbb{H}\partial_y\partial_x^m\tilde{b}\partial_y(\theta_{\alpha}^2)\partial_x^m\tilde{b}dxdy,
\end{align*}
where we have used the boundary condition $\partial_y\partial_x^m\tilde{b}|_{y=0}=0$. Moreover,
\begin{align*}
&\int_\mathbb{H}\partial_y\partial_x^m\tilde{b}\partial_y(\theta_{\alpha}^2)\partial_x^m\tilde{b}dxdy
=-\frac12\int_\mathbb{H}(\partial_x^m\tilde{b})^2\partial_y^2(\theta_{\alpha}^2)dxdy\\
=&-\frac{\alpha}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}\partial_x^m\tilde{b}\|^2_{L^2}-\frac{\alpha^2}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{b}\|^2_{L^2}.\nonumber
\end{align*}
Hence,
\begin{align}
\label{3.7}
-\int_\mathbb{H}\partial_y^2\partial_x^m\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
= \|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2-\frac{\alpha}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}\partial_x^m\tilde{b}\|^2_{L^2}-\frac{\alpha^2}{2}\frac{1}{\langle t\rangle}\|\theta_{\alpha}z\partial_x^m\tilde{b}\|^2_{L^2}.
\end{align}
Similar to Subsection 2.1, the nonlinear terms can be estimated as follows.
Firstly,
\begin{align*}
\int_{\mathbb{H}}\partial_x^m((1+b)\partial_x\tilde{u})\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{H}\partial^{m-j}_x\tilde{b}\partial_x^{j+1}\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
\triangleq R_6,
\end{align*}
and
\begin{align*}
|R_6|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_x\tilde{b}\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_x\tilde{b}\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_x\tilde{b}\|_{L^2_xL^\infty_y}
\leq C\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j}_x\partial_y\tilde{b}\|^{1/2}_{L^2},
\end{align*}
and
\begin{align*}
\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|_{L^\infty_xL^2_y}\leq C\|\theta_{\alpha}\partial^{j+1}_x\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+2}_x\tilde{u}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_x\tilde{b}\|_{L^\infty_{xy}}\leq
C\|\theta_{\alpha}\partial^{m-j}_x\tilde{b}\|^{1/4}_{L^2}\|\theta_{\alpha}\partial^{m-j}_x\partial_y\tilde{b}\|^{1/4}_{L^2}
\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/4}_{L^2}\|\theta_{\alpha}\partial^{m-j+1}_x\partial_y\tilde{b}\|^{1/4}_{L^2}.
\end{align*}
Hence,
\begin{align}
\label{NE7}
\frac{|R_6|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{b}\|_{L^2}}\leq \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}\bar{X}_{m-j}^{1/2}\bar{D}_{m-j}^{1/2}Y_{j+1}^{1/2}Y_{j+2}^{1/2}+\sum_{j=[m/2]+1}^{m}\bar{X}^{1/4}_{m-j}\bar{X}^{1/4}_{m-j+1}\bar{D}^{1/4}_{m-j}\bar{D}^{1/4}_{m-j+1}Y_{j+1}\right\}.
\end{align}
Moreover,
\begin{align*}
\int_{\mathbb{H}}\partial_x^m(g\partial_y\tilde{u})\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xg\partial_x^{j}\partial_y\tilde{u}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
\triangleq R_7,
\end{align*}
and
\begin{align*}
|R_7|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}\leq C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2},
\end{align*}
and
\begin{align*}
\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|_{L^\infty_xL^2_y}\leq C\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{u}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+1}_x\partial_y\tilde{u}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^\infty_{xy}}\leq C \langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
Therefore,
\begin{align}
\label{NE8}
\frac{|R_7|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{b}\|_{L^2}}\leq \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}\langle t\rangle^{1/4}\bar{Y}_{m-j+1}D_{j}^{1/2}D_{j+1}^{1/2}+\sum_{j=[m/2]+1}^{m}\langle t\rangle^{1/4}\bar{Y}_{m-j+1}^{1/2}\bar{Y}_{m-j+2}^{1/2}D_{j}\right\}.
\end{align}
Denote
\begin{align*}
R_8\triangleq\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xu\partial_x^{j+1}\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy,
\end{align*}
then
\begin{align*}
|R_8|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xu\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xu\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}.
\end{align*}
Similar to the estimation on $R_1$, we can obtain
\begin{align}
\label{NE9}
\frac{|R_8|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{b}\|_{L^2}}
\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}(X_{m-j}^{1/2}D_{m-j}^{1/2}+\langle t\rangle^{-1/4}\bar{X}_{m-j})\bar{Y}_{j+1}^{1/2}\bar{Y}_{j+2}^{1/2}\right.\\
&\left.+\sum_{j=[m/2]+1}^{m}(X_{m-j}^{1/4}X_{m-j+1}^{1/4}D_{m-j}^{1/4}D_{m-j+1}^{1/4}+\langle t\rangle^{-1/4}\bar{X}^{1/2}_{m-j}\bar{X}^{1/2}_{m-j+1})\bar{Y}_{j+1}\right\}.\nonumber
\end{align}
And
\begin{align*}
\int_{\mathbb{H}}\partial_x^m(v\partial_y\tilde{b})\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
=\sum_{j=0}^m(\begin{array}{ll}m\\j\end{array})\int_{\mathbb{H}}\partial^{m-j}_xv\partial_x^{j}\partial_y\tilde{b}\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
\triangleq R_9.
\end{align*}
Thus
\begin{align*}
|R_9|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|_{L^\infty_xL^2_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xv\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{j}_x\partial_y\tilde{b}\|_{L^2}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}.
\end{align*}
Similar to  the estimation on  $R_2$, we have
\begin{align}
\label{NE10}
\frac{|R_9|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{b}\|_{L^2}}
\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}(\langle t\rangle^{1/4}Y_{m-j+1}+\langle t\rangle^{1/4}\bar{Y}_{m-j+1})\bar{D}_{j}^{1/2}\bar{D}_{j+1}^{1/2}\right.\\
&\left.+\sum_{j=[m/2]+1}^{m}(\langle t\rangle^{1/4}Y_{m-j+1}^{1/2}Y_{m-j+2}^{1/2}+\langle t\rangle^{1/4}\bar{Y}_{m-j+1}^{1/2}\bar{Y}_{m-j+2}^{1/2})\bar{D}_{j}\right\}.\nonumber
\end{align}
Denote
\begin{align*}
R_{10} \triangleq  \int_\mathbb{H}\partial_x^m(g\partial_y^2u_s\psi)\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy
=\sum_{j=0}^m\int_\mathbb{H}\partial_x^{m-j}g\partial_y^2u_s\partial_x^j\psi\theta_{\alpha}^2\partial_x^m\tilde{b}dxdy.
\end{align*}
Then
\begin{align*}
|R_{10}|\leq& \sum_{j=0}^{[m/2]}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}  \|\partial^{j}_x\psi\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}\\
+&\sum_{j=[m/2]+1}^{m}(\begin{array}{ll}m\\j\end{array})\|\partial^{m-j}_xg\|_{L^\infty_{xy}}\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}  \|\partial^{j}_x\psi\|_{L^2_xL^\infty_y}\|\theta_{\alpha}\partial^{m}_x\tilde{b}\|_{L^2}.
\end{align*}
For $0\leq j\leq [m/2]$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^2_xL^\infty_y}
\leq C\langle t\rangle^{1/4}\|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|_{L^2},
\end{align*}
and
\begin{align*}
\|\theta_{\alpha}\partial_y^2u_s\|_{L^2_y}\leq \frac{C}{\langle t\rangle^{3/4}},
\end{align*}
provided that $\alpha<1$ by the assumption (H). Moreover,
\begin{align*}
\|\partial^{j}_x\psi\|_{L^\infty_{xy}}\leq C\langle t\rangle^{1/4} \|\theta_{\alpha}\partial^{j}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{j+1}_x\tilde{b}\|^{1/2}_{L^2}.
\end{align*}
For $[m/2]+1\leq j\leq m$,
\begin{align*}
\|\partial^{m-j}_xg\|_{L^\infty_{xy}}
\leq C\langle t\rangle^{1/4} \|\theta_{\alpha}\partial^{m-j+1}_x\tilde{b}\|^{1/2}_{L^2}\|\theta_{\alpha}\partial^{m-j+2}_x\tilde{b}\|^{1/2}_{L^2},
\end{align*}
and
\begin{align*}
\|\partial^{j}_x\psi\|_{L^2_xL^\infty_y}\leq C\langle t\rangle^{1/4}\|\theta_\alpha\partial_x^j\tilde{b}\|_{L^2}.
\end{align*}
Consequently,
\begin{align}
\label{NE11}
\frac{|R_{10}|\tau^mM_m}{\|\theta_{\alpha}\partial_x^m\tilde{b}\|_{L^2}}
\leq& \frac{C}{(\tau(t))^{1/2}}\left\{\sum_{j=0}^{[m/2]}\langle t\rangle^{-1/4}\bar{Y}_{m-j+1}\bar{X}_j^{1/2}\bar{X}_{j+1}^{1/2}\right.\\
&\left.+\sum_{j=[m/2]+1}^{m}\langle t\rangle^{-1/4}\bar{Y}^{1/2}_{m-j+1}\bar{Y}^{1/2}_{m-j+2}\bar{X}_j\right\}.\nonumber
\end{align}
From the estimates (\ref{3.19})-(\ref{NE11}), summing over $m\geq 0$ yields
\begin{align}
\label{NEB}
&\frac{d}{dt}\|\tilde{b}\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}}+\frac{\alpha(1-2\alpha)}{4\langle t\rangle}\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}z\partial_x^m\tilde{b}\|^2_{L^2}}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}}-\frac{\alpha}{2\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}}\nonumber\\
\leq&\dot{\tau}(t)\|\tilde{b}\|_{Y_{\tau,\alpha}}+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)\nonumber\\
&\times(\|\tilde{u}\|_{Y_{\tau,\alpha}}+\|\tilde{b}\|_{Y_{\tau,\alpha}}).
\end{align}
Similarly, by choosing $\alpha\leq1/2$, we have
\begin{align}
\label{NEBB}
&\frac{d}{dt}\|\tilde{b}\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}}-\frac{\alpha}{2\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}}\nonumber\\
\leq& \dot{\tau}(t)\|\tilde{b}\|_{Y_{\tau,\alpha}}+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)\nonumber\\
&\times(\|\tilde{u}\|_{Y_{\tau,\alpha}}+\|\tilde{b}\|_{Y_{\tau,\alpha}}).
\end{align}
\section{The Proof of Estimate of Lifespan in Theorem \ref{THM}}
By the uniform {\it a priori} estimates obtained in Section 2, we now estimate
the low bound on lifespan of the solution.  Consider
$(\ref{NEUU})+K\times(\ref{NEBB})$ with $K> 1$ to be determined later,
\begin{align}
\label{NEUB}
&\frac{d}{dt}(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})+\sum_{m\geq 0}\tau^mM_m(\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}
+K\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}})\nonumber\\
&-\frac{\alpha}{2\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}}-(C+\frac{K\alpha}{2})\frac{1}{\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}}\\
\leq&\left(\dot{\tau}(t)+\frac{C_0(K+1)}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)\right)\nonumber\\
&\times(\|\tilde{u}\|_{Y_{\tau,\alpha}}+K\|\tilde{b}\|_{Y_{\tau,\alpha}}).\nonumber
\end{align}
Choose the function $\tau(t)$ satisfies the following ODE.
\begin{align}
\label{ODE}
\frac{d}{dt}(\tau(t))^{3/2}+\frac{3C_0(K+1)}{2}\left(\langle t\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}})\right)=0.
\end{align}
From (\ref{NEUB}) and (\ref{ODE}), one has
\begin{align}
\label{NEUBU}
\frac{d}{dt}(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})&+\sum_{m\geq 0}\tau^mM_m\left(\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}
+K\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}}\right)\nonumber\\
&-\frac{\alpha}{2\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}}-(C+\frac{K\alpha}{2})\frac{1}{\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}} \leq
0.
\end{align}
By  lemma \ref{LEM2.2}, we have
\begin{align}
\label{ESTU}
\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{u}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{u}\|_{L^2}}\geq \frac{\alpha^{1/2}\beta_1}{2\langle t\rangle^{1/2}}\|\tilde{u}\|_{D_{\tau,\alpha}}+\frac{\alpha(1-\beta_1)}{\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}},
\end{align}
and
\begin{align}
\label{ESTB}
\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_{\alpha}\partial_x^m\partial_y\tilde{b}\|_{L^2}^2}{\|\theta_\alpha\partial_x^m\tilde{b}\|_{L^2}}\geq \frac{\alpha^{1/2}\beta_2}{2\langle t\rangle^{1/2}}\|\tilde{b}\|_{D_{\tau,\alpha}}+\frac{\alpha(1-\beta_2)}{\langle t\rangle}\|\tilde{b}\|_{X_{\tau,\alpha}},
\end{align}
for $\beta_1, \beta_2\in(0,1/2)$.

From (\ref{NEUBU}), (\ref{ESTU}) and (\ref{ESTB}), it follows that
\begin{align}
\label{ESTT}
\frac{d}{dt}(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})&+\frac12(\alpha(1-2\beta_1))\frac{1}{\langle t\rangle}\|\tilde{u}\|_{X_{\tau,\alpha}}+
\frac12(\alpha(1-2\beta_2)-\frac{2C}{K})\frac{1}{\langle t\rangle}K\|\tilde{b}\|_{X_{\tau,\alpha}}\nonumber\\
&\qquad\qquad+\frac{\alpha^{1/2}\beta_1}{2\langle t\rangle^{1/2}}\|\tilde{u}\|_{D_{\tau,\alpha}}+\frac{K\alpha^{1/2}\beta_2}{2\langle t\rangle^{1/2}}\|\tilde{b}\|_{D_{\tau,\alpha}}\leq 0.
\end{align}
Choose
\begin{align*}
\alpha=\frac12-\delta,\quad \beta_1=\frac{\delta}{2},\quad \beta_2=\frac{\delta}{2},\quad K=\frac{4C}{\delta},
\end{align*}
where $0<\delta<1/4$ is sufficiently small to be determined later, then
\begin{align*}
\alpha(1-2\beta_1)=\frac12-\frac32\delta+\delta^2,
\end{align*}
and
\begin{align*}
\alpha(1-2\beta_2)-\frac{2C}{K}=\frac12-2\delta+\delta^2.
\end{align*}
Then, there exist small positive constants $\eta_1=\delta$ and $\eta_2=\frac{\delta}{8}$  such that
\begin{align}
\label{ESTTT}
\frac{d}{dt}(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})+\frac{1/4-\eta_1}{\langle t\rangle}\left(\|\tilde{u}\|_{X_{\tau,\alpha}}+
K\|\tilde{b}\|_{X_{\tau,\alpha}}\right)+\frac{\eta_2}{\langle t\rangle^{1/2}}(\|\tilde{u}\|_{D_{\tau,\alpha}}+K\|\tilde{b}\|_{D_{\tau,\alpha}})\leq 0.
\end{align}
It implies that
\begin{align}
\label{ESTTTT}
\frac{d}{dt}(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})\langle t\rangle^{1/4-\eta_1}&+\frac{1/4-\eta_1}{\langle t\rangle^{3/4+\eta_1}}\left(\|\tilde{u}\|_{X_{\tau,\alpha}}+
K\|\tilde{b}\|_{X_{\tau,\alpha}}\right)\nonumber\\
&+\frac{\eta_2}{\langle t\rangle^{1/4+\eta_1}}(\|\tilde{u}\|_{D_{\tau,\alpha}}+K\|\tilde{b}\|_{D_{\tau,\alpha}})\leq 0.
\end{align}
As a consequence,
\begin{align}
\label{ESTTTTL}
(\|\tilde{u}\|_{X_{\tau,\alpha}}+K\|\tilde{b}\|_{X_{\tau,\alpha}})\langle t\rangle^{1/4-\eta_1}&+\int_0^t\frac{\eta_2}{\langle s\rangle^{1/4+\eta_1}}(\|\tilde{u}(s)\|_{D_{\tau,\alpha}}+K\|\tilde{b}(s)\|_{D_{\tau,\alpha}})ds\nonumber\\
&\qquad\leq (\|\tilde{u}(0)\|_{X_{\tau,\alpha}}+K\|\tilde{b}(0)\|_{X_{\tau,\alpha}})\leq C(1+K)\varepsilon,
\end{align}
where we have used (\ref{IIIIE}).
Then, by noting  that $K=\frac{4C}{\delta}$, one has
\begin{align}
\label{T1}
&\frac{3C_0}{2}(K+1)\int_0^t\langle s\rangle^{-1/4}(\|\tilde{u}(s)\|_{X_{\tau,\alpha}}+\|\tilde{b}(s)\|_{X_{\tau,\alpha}})ds\nonumber\\
=&\frac{3C_0}{2}(\frac{4C}{\delta}+1)\int_0^t\langle s\rangle^{-1/4}(\|\tilde{u}(s)\|_{X_{\tau,\alpha}}+\|\tilde{b}(s)\|_{X_{\tau,\alpha}})ds\nonumber\\
\leq &\frac{3CC_0\varepsilon}{2}(\frac{4C}{\delta}+1)^2\int_0^t\langle s\rangle^{-1/2+\eta_1}ds
\leq 3CC_0\varepsilon(\frac{4C}{\delta}+1)^2\langle t\rangle^{1/2+\eta_1},
\end{align}
and
\begin{align}
\label{T2}
&\frac{3C_0}{2}(K+1)\int_0^t\langle s\rangle^{1/4}(\|\tilde{u}(s)\|_{D_{\tau,\alpha}}+\|\tilde{b}(s)\|_{D_{\tau,\alpha}})ds\nonumber\\
=&\frac{3C_0}{2}(\frac{4C}{\delta}+1)\int_0^t\langle s\rangle^{1/4}(\|\tilde{u}(s)\|_{D_{\tau,\alpha}}+\|\tilde{b}(s)\|_{D_{\tau,\alpha}})ds\nonumber\\
=&\frac{3C_0}{2}(\frac{4C}{\delta}+1)\frac{8}{\delta}\int_0^t\langle s\rangle^{1/2+\eta_1}\frac{\eta_2}{\langle s\rangle^{1/4+\eta_1}}(\|\tilde{u}(s)\|_{D_{\tau,\alpha}}+K\|\tilde{b}(s)\|_{D_{\tau,\alpha}})ds\nonumber\\
\leq&(\frac{4C}{\delta}+1)^2\frac{12CC_0}{\delta}\langle t\rangle^{1/2+\eta_1}\varepsilon.
\end{align}
On the other hand,  (\ref{ODE}) implies that
\begin{align}
\label{ODES}
\tau(t)^{3/2}
=\tau(0)^{3/2}-\frac{3C_0(K+1)}{2}\int_0^t(\langle s\rangle^{-1/4}(\|\tilde{u}\|_{X_{\tau,\alpha}}+\|\tilde{b}\|_{X_{\tau,\alpha}})+\langle s\rangle^{1/4}(\|\tilde{u}\|_{D_{\tau,\alpha}}+\|\tilde{b}\|_{D_{\tau,\alpha}}))ds.
\end{align}
From (\ref{T1}), (\ref{T2}) and (\ref{ODES}), one has
\begin{align*}
\tau(t)^{3/2}\geq \tau_0^{3/2}-\max\{3CC_0(\frac{4C}{\delta}+1)^2\langle t\rangle^{1/2+\eta_1}\varepsilon,\quad (\frac{4C}{\delta}+1)^2\frac{12CC_0}{\delta}\langle t\rangle^{1/2+\eta_1}\varepsilon \},
\end{align*}
for all $t\geq 0$.

Choose $\delta=\frac{1}{\ln(1/\varepsilon)}$. It is straightforward  to show that
\begin{align*}
\tau(t)\geq \frac{\tau_0}{4},
\end{align*}
in the time interval $[0, T_\varepsilon]$, where $T_\varepsilon$ satisfies
\begin{align}
\label{LS}
T_\varepsilon= \bar{C}\left(\frac{1}{\varepsilon(\ln(1/\varepsilon))^3}\right)^{2-4/(\ln(1/\varepsilon)+2)}-1.
\end{align}
This gives the estimate on the lifespan of solution stated  in (\ref{THM3}).

\section{The Proof of Uniqueness Part in Theorem \ref{THM}}
Assume there are two solutions $(\tilde{u}_1, \tilde{b}_1)$ and  $(\tilde{u}_2, \tilde{b}_2)$ to (\ref{3.4}) with the same initial data $(\tilde{u}_0, \tilde{b}_0)$, which satisfies $\|(\tilde{u}_0, \tilde{b}_0)\|_{X_{2\tau_0}, \alpha}\leq \varepsilon$. And the tangential radii of analyticity of $(\tilde{u}_1, \tilde{b}_1)$ and  $(\tilde{u}_2, \tilde{b}_2)$ are $\tau_1(t)$ and $\tau_2(t)$, respectively.

Define $\tau(t)$ by
\begin{align}
\label{U1}
\frac{d(\tau(t))^{3/2}}{dt}+\frac{3C_0(K+1)}{2}&\left(\langle t\rangle^{-1/4}(\|\tilde{u}_1\|_{X_{\tau_1(t),\alpha}}+\|\tilde{b}_1\|_{X_{\tau_1(t),\alpha}})\right.\nonumber\\
&\left.+\langle t\rangle^{1/4}(\|\tilde{u}_1\|_{D_{\tau_1(t),\alpha}}+\|\tilde{b}_1\|_{D_{\tau_1(t),\alpha}})\right)=0,
\end{align}
with initial data
\begin{align}
\label{U2}
\tau(0)=\frac{\tau_0}{8}.
\end{align}
By the estimates obtained  in Section 2, there exists a time interval $[0, T_0]$ with $T_0\leq T_\varepsilon$  such that
\begin{align}
\label{U3}
\frac{\tau_0}{16}\leq \tau(t)\leq \frac{\tau_0}{8}\leq \frac{\min\{\tau_1,\tau_2\}}{2}
\end{align}
for all $t\in[0,T_0]$.

Set  $U=\tilde{u}_1-\tilde{u}_2$ and $B=\tilde{b}_1-\tilde{b}_2$. Then
\begin{align}
\label{U4}
\partial_tU-\partial_y^2U+(u_s+u_1)\partial_xU&+(v_1-v_2)\partial_y\tilde{u}_1-(1+b_1)\partial_xB-(g_1-g_2)\partial_y\tilde{b}_1\nonumber\\
&\qquad\qquad-2\partial_y^2u_s B+(v_1-v_2)\partial_y^2u_s\psi_1+R_{s1}=0,
\end{align}
and
\begin{align}
\label{U5}
\partial_tB-\partial_y^2B-(1+b_1)\partial_xU-(g_1-g_2)\partial_y\tilde{u}_1&+(u_s+u_1)\partial_xB+(v_1-v_2)\partial_y\tilde{b}_1\nonumber\\
&\qquad-(g_1-g_2)\partial_y^2u_s\psi_1+R_{s2}=0,
\end{align}
with the source terms $R_{s1}$ and $R_{s2}$ given by
\begin{align}
\label{U6}
R_{s1}=(u_1-u_2)\partial_x\tilde{u}_2+v_2\partial_yU-(b_1-b_2)\partial_x\tilde{b}_2-g_2\partial_yB+v_2\partial_y^2u_s(\psi_1-\psi_2),
\end{align}
and
\begin{align}
\label{U7}
R_{s2}=-(b_1-b_2)\partial_x\tilde{u}_2-g_2\partial_yU+(u_1-u_2)\partial_x\tilde{b}_2+v_2\partial_yB-g_2\partial_y^2u_s(\psi_1-\psi_2).
\end{align}
Note that the initial data and the boundary conditions are
\begin{align}
\label{DEI}
U(t,x,y)|_{t=0}=0,\qquad B(t, x,y)|_{t=0}=0,
\end{align}
and
\begin{align}
\label{DE}
\left\{
\begin{array}{ll}
U|_{y=0}=0,\\
U|_{y=\infty}=0,
\end{array}
\right.
\qquad\hbox{and}\qquad
\left\{
\begin{array}{ll}
\partial_yB|_{y=0}=0,\\
B|_{y=\infty}=0.
\end{array}
\right.
\end{align}
Similar to Section 2, we have
\begin{align}
\label{U8}
&\frac{d}{dt}\|U\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_\alpha\partial_y\partial_x^m U\|^2_{L^2}}{\|\theta_\alpha\partial_x^m U\|_{L^2}}-\frac{\alpha}{2\langle t\rangle}\|U\|_{X_{\tau,\alpha}}-\frac{C}{\langle t\rangle}\|B\|_{X_{\tau,\alpha}}\\
\leq&\dot{\tau}(t)\|U\|_{Y_{\tau,\alpha}}+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}_1\|_{X_{\tau(t),\alpha}}+\|\tilde{b}_1\|_{X_{\tau(t),\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}_1\|_{D_{\tau(t),\alpha}}+\|\tilde{b}\|_{D_{\tau_1(t),\alpha}})\right)\nonumber\\
&\times (\|U\|_{Y_{\tau,\alpha}}+\|B\|_{Y_{\tau,\alpha}})\nonumber\\
&+\frac{C_0}{(\tau(t))^{1/2}}(\|\tilde{u}_2\|_{Y_{\tau,\alpha}}+\|\tilde{b}_2\|_{Y_{\tau,\alpha}})
(\langle t\rangle^{-1/4}(\|U\|_{X_{\tau,\alpha}}+
\|B\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|U\|_{D_{\tau,\alpha}}+\|B\|_{D_{\tau,\alpha}}))\nonumber,
\end{align}
and
\begin{align}
\label{U9}
&\frac{d}{dt}\|B\|_{X_{\tau,\alpha}}+\sum_{m\geq 0}\tau^mM_m\frac{\|\theta_\alpha\partial_y\partial_x^m U\|^2_{L^2}}{\|\theta_\alpha\partial_x^m U\|_{L^2}}-\frac{\alpha}{2\langle t\rangle}\|B\|_{X_{\tau,\alpha}}\\
\leq&\dot{\tau}(t)\|B\|_{Y_{\tau,\alpha}}+\frac{C_0}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}_1\|_{X_{\tau(t),\alpha}}+\|\tilde{b}_1\|_{X_{\tau(t),\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}_1\|_{D_{\tau(t),\alpha}}+\|\tilde{b}_1\|_{D_{\tau(t),\alpha}})\right)\nonumber\\
&\times (\|U\|_{Y_{\tau,\alpha}}+\|B\|_{Y_{\tau,\alpha}})\nonumber\\
&+\frac{C_0}{(\tau(t))^{1/2}}(\|\tilde{u}_2\|_{Y_{\tau,\alpha}}+\|\tilde{b}_2\|_{Y_{\tau,\alpha}})
(\langle t\rangle^{-1/4}(\|U\|_{X_{\tau,\alpha}}+
\|B\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|U\|_{D_{\tau,\alpha}}+\|B\|_{D_{\tau,\alpha}}))\nonumber.
\end{align}
Then, we have
\begin{align}
\label{U10}
&\frac{d}{dt}(\|U\|_{X_{\tau,\alpha}}+K\|B\|_{X_{\tau,\alpha}})+\sum_{m\geq 0}\tau^mM_m\left(\frac{\|\theta_\alpha\partial_y\partial_x^m U\|^2_{L^2}}{\|\theta_\alpha\partial_x^m U\|_{L^2}}+K\frac{\|\theta_\alpha\partial_y\partial_x^m B\|^2_{L^2}}{\|\theta_\alpha\partial_x^m B|_{L^2}}\right)\nonumber\\
&-\frac{\alpha}{2\langle t\rangle}\|U\|_{X_{\tau,\alpha}}-\frac{2C+K\alpha}{2\langle t\rangle}\|B\|_{X_{\tau,\alpha}}\nonumber\\
\leq&\left(\dot{\tau}(t)+\frac{C_0(1+K)}{(\tau(t))^{1/2}}\left(\langle t\rangle^{-1/4}(\|\tilde{u}_1\|_{X_{\tau(t),\alpha}}+\|\tilde{b}_1\|_{X_{\tau(t),\alpha}})+\langle t\rangle^{1/4}(\|\tilde{u}_1\|_{D_{\tau(t),\alpha}}+\|\tilde{b}_1\|_{D_{\tau(t),\alpha}})\right)\right.\nonumber\\
&\left.\times(\|U\|_{Y_{\tau,\alpha}}+K\|B\|_{Y_{\tau,\alpha}})\right)\\
&+\frac{C_0(1+K)}{(\tau(t))^{1/2}}(\|\tilde{u}_2\|_{Y_{\tau,\alpha}}+\|\tilde{b}_2\|_{Y_{\tau,\alpha}})
(\langle t\rangle^{-1/4}(\|U\|_{X_{\tau,\alpha}}+
\|B\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|U\|_{D_{\tau,\alpha}}+\|B\|_{D_{\tau,\alpha}}))\nonumber.
\end{align}
From (\ref{U1}), one has
\begin{align}
\label{U11}
\dot{\tau}(t)+\frac{C_0(K+1)}{(\tau(t))^{1/2}}&\left(\langle t\rangle^{-1/4}(\|\tilde{u}_1\|_{X_{\tau,\alpha}}+
\|\tilde{b}_1\|_{X_{\tau,\alpha}})\right.\nonumber\\
&\left.+\langle t\rangle^{1/4}(\|\tilde{u}_1\|_{D_{\tau,\alpha}}+\|\tilde{b}_1\|_{D_{\tau,\alpha}})\right)(\|U\|_{Y_{\tau,\alpha}}+\|B\|_{Y_{\tau,\alpha}})
\leq 0,
\end{align}
because $\tau(t)\leq \tau_1(t)$ and the norms
$X_{\tau, \alpha}$ and $D_{\tau, \alpha}$  are increasing in $\tau$.

By the inequalities (\ref{ESTU}), (\ref{ESTB}) and (\ref{U11}), one has
\begin{align}
\label{U12}
&\frac{d}{dt}(\|U\|_{X_{\tau,\alpha}}+K\|B\|_{X_{\tau,\alpha}})+\frac{\alpha(1-2\beta_1)}{2\langle t\rangle}\|U\|_{X_{\tau,\alpha}}
+\frac{\alpha(1-2\beta_2)-\frac{2C}{K}}{2\langle t\rangle}K\|B\|_{X_{\tau,\alpha}}\nonumber\\
&+\frac{\alpha^{1/2}\beta_1}{2\langle t\rangle^{1/2}}\|U\|_{D_{\tau,\alpha}}+\frac{\alpha^{1/2}\beta_2}{2\langle t\rangle^{1/2}}K\|B\|_{D_{\tau,\alpha}}\\
\leq &\frac{C_0(1+K)}{(\tau(t))^{1/2}}(\|\tilde{u}_2\|_{Y_{\tau,\alpha}}+\|\tilde{b}_2\|_{Y_{\tau,\alpha}})
(\langle t\rangle^{-1/4}(\|U\|_{X_{\tau,\alpha}}+
\|B\|_{X_{\tau,\alpha}})+\langle t\rangle^{1/4}(\|U\|_{D_{\tau,\alpha}}+\|B\|_{D_{\tau,\alpha}}))\nonumber,
\end{align}
for $\beta_1, \beta_2\in (0, 1/2)$.
Since
\begin{align}
\label{U13}
\|\tilde{u}_2\|_{Y_{\tau,\alpha}}\leq \frac{1}{\tau}\|\tilde{u}_2\|_{X_{2\tau,\alpha}}\leq \frac{1}{\tau}\|\tilde{u}_2\|_{X_{\tau_2,\alpha}}
\leq \frac{C(1+K)}{\tau}\varepsilon\langle t\rangle^{-1/4+\eta_1}
\end{align}
and
\begin{align}
\label{U14}
\|\tilde{b}_2\|_{Y_{\tau,\alpha}}\leq \frac{1}{\tau}\|\tilde{b}_2\|_{X_{2\tau,\alpha}}\leq \frac{1}{\tau}\|\tilde{b}_2\|_{X_{\tau_2,\alpha}}
\leq \frac{C(1+K)}{\tau}\varepsilon\langle t\rangle^{-1/4+\eta_1},
\end{align}
we have
\begin{align}
\label{U15}
\frac{C_0(1+K)}{(\tau(t))^{1/2}}(\|\tilde{u}_2\|_{Y_{\tau,\alpha}}+\|\tilde{b}_2\|_{Y_{\tau,\alpha}})\leq \frac{2(1+K)^2CC_0\varepsilon}{(\tau(t))^{3/2}\langle t\rangle^{1/4-\eta_1}}.
\end{align}
Notice that $t\in [0,T_\varepsilon]$ with $T_\varepsilon=\varepsilon^{-2+\delta_0}$, where $\delta_0$ is a fixed
small positive constant. As in Section 3, we can choose $\alpha=1/2-\delta, \beta_1=\beta_2=\frac{\delta}{2}, K=\frac{4C}{\delta}$ and
$\delta=1/\ln(1/\varepsilon)$, then $\eta_1$ can be chosen to be $\delta$.  Let $\varepsilon$ suitably small to have
\begin{align*}
\frac{\alpha(1-2\beta_1)}{2}>\frac{2(1+K)^2CC_0\varepsilon\langle t\rangle^{1/2+\eta_1}}{(\tau(t))^{3/2}},\qquad
\frac{\alpha(1-2\beta_1)-2C/K}{2}>\frac{2(1+K)^2CC_0\varepsilon\langle t\rangle^{1/2+\eta_1}}{(\tau(t))^{3/2}K},
 \end{align*}
and
\begin{align*}
\frac{\alpha^{1/2}\beta_1}{2}>\frac{2(1+K)^2CC_0\varepsilon\langle t\rangle^{1/2+\eta_1}}{(\tau(t))^{3/2}},\qquad
\frac{\alpha^{1/2}\beta_2}{2}>\frac{2(1+K)^2CC_0\varepsilon\langle t\rangle^{1/2+\eta_1}}{(\tau(t))^{3/2}K}.
  \end{align*}
 (\ref{U12}) and (\ref{U15}) imply that
\begin{align}
\label{U16}
\frac{d}{dt}(\|U\|_{X_{\tau,\alpha}}+K\|B\|_{X_{\tau,\alpha}})+\eta_3(\|U\|_{X_{\tau,\alpha}}
+K\|B\|_{X_{\tau,\alpha}})\leq 0
\end{align}
for suitably small $\eta_3>0$ and any $t\in [0,T_\varepsilon]$.  It implies uniqueness of solution to (\ref{3.4})  in the time interval $[0, T_\varepsilon]$.\\

\noindent{\bf Acknowledgement:} Feng Xie' research  was supported by National Nature Science Foundation of
China 11571231, the China Scholarship
Council and Shanghai Jiao Tong University SMC(A). Tong Yang's research was supported by internal research funding of City University of Hong Kong, 7004847.

redpajama_set_name: RedPajamaArXiv


text:
Q: How do i change the background color in python of this combined png image? I have the following code that works great except, however it produces a black background image, and i would like it to be white.  I have tried changing bg_color but anything I change it to still keeps the black background.  Is there also a way that i can make a box around each image that i am combining?
direction = 'vertical'
bg_color=(1,1,1,1)
aligment='center'

images = [Image.open(x) for x in ['LVISF2_ABoVE2019_0801_R2003_084168.TXT_ZG.png', 'LVISF2_ABoVE2019_0801_R2003_084168.TXT_RH95.png']]
widths, heights = zip(*(i.size for i in images))

if direction=='horizontal':
    new_width = sum(widths)
    new_height = max(heights)
else:
    new_width = max(widths)
    new_height = sum(heights)
new_im = Image.new('RGB', (new_width, new_height), color=bg_color)
offset = 0
for im in images:
    if direction=='horizontal':
        y = 0
        if aligment == 'center':
            y = int((new_height - im.size[1])/2)
        elif aligment == 'bottom':
            y = new_height - im.size[1]
        new_im.paste(im, (offset, y))
        offset += im.size[0]
    else:
        x = 0
        if aligment == 'center':
            x = int((new_width - im.size[0])/2)
    elif aligment == 'right':
        x = new_width - im.size[0]
        new_im.paste(im, (x, offset))
        offset += im.size[1]

new_im.save('test4.jpg')

A: turns out my original png background was not white, it was transparent So i had to change the transparent background to white, and then I was able to change the background color.
Here is how i changed the original png backgrounds to white
images1 = ['LVISF2_ABoVE2019_0801_R2003_084168.TXT_ZG.png', 'LVISF2_ABoVE2019_0801_R2003_084168.TXT_RH95.png']

for i in images1:
    im = Image.open(r'{0}'.format(i))
    fill_color = (255,255,255)
    im = im.convert("RGBA")
    if im.mode in ('RGBA', 'LA'):
        background = Image.new(im.mode[:-1], im.size, fill_color)
        background.paste(im, im.split()[-1]) # omit transparency
        im = background
    im.convert("RGB").save(r'newimage_{0}.jpg'.format(i))

redpajama_set_name: RedPajamaStackExchange


The Good Ground Farmers Market will be held at 84 Montauk Highway (across from the Hampton Bays Fire Dept) on Thursdays from 3 - 7 pm, starting May 23rd through August 29th, 2019! The market will have all locally grown, caught, and hand-prepared products.
Please call the Library to register!
Join us for ECI Community Yoga at its new home in 2019 - stay tuned!
ECI will host a free preview screening of the upcoming new documentary Grow Food on Thursday, August 30th, 2018, 7 - 9 pm at the Hampton Bays Public Library.
New Classes in Organic Gardening!
The Ecological Culture Initiative will host a matinee screening of the award winning documentary INHABIT: A Permaculture Perspective on Saturday, April 21st, 2018, 2:30 - 5 pm at the Hampton Bays Public Library. Running time is 1hr 32 minutes. After the movie, the ECI Staff will hold a question and answer session.
Price: $40 per person - limited to 40 people - UPDATE: SOLD OUT!!!
Join the staff of ECI for a mid-winter gathering featuring teas from locally grown or foraged sources, as well as savory organic baked goods. Learn how to make teas from native plants that have medicinal properties.
Price: $20 per person - SOLD OUT!!!!
The Ecological Culture Initiative will host a screening of the award winning documentary Inhabit: A Permaculture Perspective on Tuesday, January 23rd, 2018, 7 - 9 pm at the Hampton Bays Public Library. Running time is 1hr 32 minutes. After the movie, the ECI Staff will hold a question and answer session.
Please join us for the groundbreaking of the Heritage Gardens at St. Joseph's Villa, a new ECI Teaching Garden for learning ecosystem gardening techniques, habitat restoration, localized food production, composting, rain harvesting & much more!
Please join the staff of ECI for an evening of gourmet permaculture and conviviality. Bonfire, baked goods, LI cheese pumpkin pies, mulled cider, and more!
Price: $50 per person - limited to 50 people Update: SOLD OUT!!!!


text:
HEALTH ‑- COUNTIES ‑- CITIES AND TOWNS ‑- BOARDING HOMES FOR THE AGED ‑- PLACES OF REFUGE ‑- REGULATION AND LICENSING
AGO 1958 No. 156 - Feb 11 1958
HEALTH ‑- BOARDING HOMES FOR THE AGED ‑- PLACES OF REFUGE ‑- REGULATION AND LICENSING.
COUNTIES ‑- BOARDING HOMES FOR THE AGED ‑- PLACES OF REFUGE ‑- REGULATION AND LICENSING.
CITIES AND TOWNS ‑- BOARDING HOMES FOR THE AGED ‑- PLACES OF REFUGE ‑- REGULATION AND LICENSING.
Local authorities may not enact or enforce local ordinances or regulations pertaining to the licensure and/or inspection of places of refuge or boarding homes for the aged, as defined by chapter 253, Laws of 1957 (chapter 74.32 RCW).
Honorable Bernard Bucove, M.D.
State Director of Health
214 General Administration Building
Olympia, Washington Cite as: AGO 57-58 No. 156
You have requested the opinion of this office on the following question:
In view of the enactment of chapter 253, Laws of 1957 (chapter 18.20 RCW), may cities, towns or counties enact or enforce laws, ordinances or regulations pertaining to the licensure and/or inspection of boarding homes for the aged or places of refuge?
It is our opinion that this question must be answered in the negative.
Prior to the enactment of chapter 253, Laws of 1957, the licensure and regulation of establishments providing housing for aged persons was vested in local authorities by chapter 74.32 RCW pertaining to places of refuge.
RCW 74.32.010 defined a place of refuge as "any . . . building or dwelling for housing the aged, infirm, or imbeciles, wherein three or more persons, not related by blood or marriage to the . . . operator, . . . are lodged or boarded more than fifteen days in any calendar month." RCW 74.32.020 made it illegal for any person to maintain a place of refuge as defined without obtaining a license from the county commissioners or governing body of the city or town. These official bodies were empowered by RCW 74.32.090 to make rules and regulations pertaining to the construction and operation of such institutions.
The legislature, however, in enacting chapter 253, Laws of 1957, repealed chapter 74.32 RCW in its entirety and provided in its stead a comprehensive statute relating to the licensure and inspection of boarding homes for the aged. This act provides for the enactment of state‑wide [[statewide]]rules and regulations by the state board of health after consultation with an advisory council constituted by law. The licensure, enforcement and inspection, of all such homes are vested in the state department of health or in such department and a local health department jointly as provided in § 10 of the act (RCW 18.20.100). This section reads as follows:
"Where it is determined by the director together with the jurisdictional health officer, that a city, county, city-county health department or health district is qualified to carry out the provisions of this act, he shall authorize such political subdivision or agency to administer and enforce this act, and the rules and regulations promulgated hereunder.
"Any such authorization may be withdrawn by the director after thirty days' notice in writing to the authorized department should the director determine that the authorized department is unwilling or unable to carry out the duties and responsibilities hereunder." (Emphasis supplied.)
A comparison of the provisions of chapter 253, Laws of 1957, with chapter 117, Laws of 1951 (chapter 18.51 RCW), relating to the regulation of nursing homes, reveals that the administrative provisions of the two enactments are essentially identical. Compare RCW 18.51.020 with 18.20.100; RCW 18.51.090 with 18.20.110.
A previous opinion of this office dated June 12, 1951, addressed to J. A. Kahl, M.D., Acting State Director of Health (AGO 51-53-69) [[Opinion No. 51-53 No. 69]], a copy of which is enclosed, held that the state has preempted the field of regulation of nursing homes to the exclusion of local authorities. Inasmuch as the provisions interpreted by that opinion are substantially identical with the corresponding sections of chapter 253, the reasoning of this earlier opinion would [[Orig. Op. Page 3]] likewise apply to our answer to your query. Accordingly, the appropriate part of that opinion is quoted below in its entirety.
"It is a fundamental principle of law that municipal ordinances are inferior and subordinate to the general laws of the state. Indeed, it is provided by the constitution in section 11, Article XI, that
"'Any county, city, town or township, may make and enforce within its limits all such local, police, sanitary, and other regulations as are not in conflict with general laws.'
"It is an equally fundamental principle of law that the mere fact that the state has legislated on a subject does not of necessity deprive a municipality of the power to deal with the subject by ordinance. State ex rel. Isham v. Spokane, 2 Wn. (2d) 392, 98 P. (2d) 306. Concurrent jurisdiction by the state and a municipality may exist. Nevertheless, if the state has asserted its sovereign jurisdiction over a given subject matter and there is no room for concurrence, the municipal regulations or ordinances must give way. State ex rel. Webster v. Superior Court, 67 Wash. 37, 120 Pac. 861; L.R.A. 1915-C 287, Ann. Cas. 1913-D 78;Seattle Electric Co. v. Seattle, 78 Wash. 203, 138 Pac. 892. Therefore, inquiry must be directed toward the question as to whether or not the legislature intended that a local authority should exercise its powers over a particular subject matter.
"In passing chapter 117, Laws of 1951, the state has acted in the field of regulation of nursing homes, and the preservation of public health in this aspect has become of general concern to the state. Can local authorities operate under or enact regulations governing the same subject matter? The opinion of this office is to the contrary.
"An examination of the statute does not disclose that a nursing home in order to be invalid operation may also have to conform to local health ordinances. There is no positive provision that local authorities may also issue licenses for nursing homes. Generally, it is true that local ordinances may enlarge upon the provisions of a statute to a greater extent than the statute requires. State ex rel. Isham v. Spokane, supra. However, section 3 of the act makes provision for cities, counties, or district health departments to participate in the control of nursing homes. This express reference [[Orig. Op. Page 4]] to local enforcement of the provisions of chapter 117, Laws of 1951, can only mean that the legislature intended that the local authorities act solely in this manner. There is no room for concurrent jurisdiction. To hold otherwise would invoke a lack of uniformity of control over the operation of nursing homes. It is our opinion that the state has preempted the field of regulation of nursing homes to the exclusion of local authorities except as provided in the statute, and therefore, local authorities may not pass regulations which are more stringent than those provided for in chapter 117, Laws of 1951."
We therefore conclude that local authorities may not enact or enforce local ordinances or regulations pertaining to the licensure and/or inspection of places of refuge or boarding homes for the aged as defined by chapter 253, Laws of 1957.
We hope that the foregoing satisfactorily answers your inquiry.
LIVER J. NEIBEL, JR.

redpajama_set_name: RedPajamaCommonCrawl


text:
ou U-121 a été le nom de plusieurs sous-marins allemands ou U-boot durant la Première et la Seconde Guerre mondiale :
 L'Unterseeboot 121, un sous-marin de type UE 2, a été construit en 1918 et a participé à la première bataille de l'Atlantique durant la Première Guerre mondiale et qui s'est rendu à la France pour être coulé comme cible le .
 Durant la première Guerre mondiale, l'Allemagne a possédé deux autres sous-marins avec un nom similaire:
 UB-121, un sous-marin de type UB III lancé en 1918 et qui s'est rendu en 1918.
 L'Unterseeboot 121, un sous-marin de type II.B a été construit en 1940 et a participé à la Seconde Guerre mondiale et qui a été sabordé le  à la suite de l'opération Regenbogen.

Sources

Sous-marin de la Marine allemande
Nom de bateau
Bateau des années 1910
Bateau des années 1940

redpajama_set_name: RedPajamaWikipedia


The Department of Education's Office of Educational Technology (OET) has released new guidance aimed at bringing educators and ed tech entrepreneurs onto the same page.
The guide, written by educators and researchers for developers, startups, and entrepreneurs, aims to bring attention to the real, persistent problems that actually need solutions.
The free guide, titled "Ed Tech Developer's Guide: A Primer for Developers, Startups and Entrepreneurs," delves into what schools need most and what digital tools educators and administrators would like to see created.
According to eSchool News, the problems educators need digital solutions for deal with many of the following topics: mastery of academic skills, fostering and measuring of non-cognitive skills, embedding of formative assessments, engaging families, supporting college and career exploration, providing job-embedded professional development, improving educator productivity, increasing access for all students, and closing achievement gaps.
What's cool about this guidance is it's a win-win for all. Educators and administrators get a say in suggesting the tools that will ideally be coming to their classrooms, and entrepreneurs and startups get free insight into the minds of educators.
One common issue with the ed tech surge is that there appears to be a lack of communication between developers and schools. The guidance makes sure to stress that, despite the helpful tips it provides, it's very important that the ed tech world maintains communication with schools even when they're not pitching an idea.


text:
David Love Grant was born on 23rd May 1898 at Bellshill in Lanarkshire. He was working as a coal miner when he enlisted with the Royal Marines on 26th May 1915 at Glasgow and was given the short service number 18644.
After completing his initial training at the Deal Depot David was transferred to the Portsmouth Division in January 1916. He served aboard HMS Renown and HMS Agincourt before being transferred to the Royal Naval Division, 2nd RMLI, on 8th July 1917.
David would see action with the Division at both Passchendaele and Welsh Ridge before he was taken prisoner during the enemy's Spring Offensive at the Battle of Bapaume. Held captive for three days he managed to escape and return to the British line on the 27th March.
Transferred to the 1st Reserve RMLI on 9th April, he would be questioned regarding his absence 3 days later by Colonel Anstruther.
By kind permission of The National Archives
With his name cleared, David returned to the Portsmouth Division and was posted to HMS Arrogant where he would see out the war. He was invalided from service on 31st March 1919.
return to Detailed Personnel
©2015- RoyalNavalDivision.info


Desi, la petite chienne blanche (amie de moi et de Perrine), veut célébrer avec toi. Elle porte son chapeau de fête.
We are working hard trying *all* the gifts; playing dominos and the "casse-tête"... The duck is truly wonderful; The noise he makes makes Chloé die with laughter and I think she'll sleep with it.
I gave her the gifts today because it's her birthday... Actually she was very happy because we only celebrate tomorrow due to conflicting family schedules.
Thank you so much! YOu should receive a little word for the young lady soon!
Tell Chloé that if you come visit I'll see that you meet Desi. You'll want to pack her in your suitcase and take her back to France with you.
Did I get the French right? If I make mistakes you have to tell me; it's the only way I'll learn.