text:
\section{Introduction}

\subsection{Formulation}

We consider the motion of an incompressible viscous fluid, subject to the influence of gravity and surface tension forces, in a moving domain
\begin{equation}\label{omegat}
\Omega(t)= \left\{ x \in \mathbb{R}^3 \mid  -b< x_{3} < h(t, x_{1}, x_{2}) \right\}.
\end{equation}
The lower boundary of $\Omega(t)$ is assumed to be rigid and given with the constant $b>0$, but the upper boundary is a free surface that is the
graph of the unknown function $ h:\mathbb{R}_+\times\Rn{2}\to \Rn{}$. The fluid is described by its velocity and pressure, which are given for
each
$t\ge0$ by $ u(t,\cdot):\Omega(t) \to \Rn{3} $ and $ p(t,\cdot):\Omega(t) \to \Rn{}$, respectively.  For each $t>0$, $(u, p, h)$ solve the
following problem for the  incompressible Navier-Stokes equations
\begin{equation}\label{NS}
\begin{cases}
\partial_t u + u \cdot \nabla u + \nabla p-\varepsilon  \Delta u  = 0& \text{in }
\Omega(t) \\
\nabla\cdot{u}=0 & \text{in }\Omega(t) \\
p {\bf n}-  2  \varepsilon  S u {\bf n} =  {g} h {\bf n}-\sigma H {\bf n} & \text{on } \{x_3=h(t,x_1,x_2)\} \\
\partial_t h = u\cdot{\bf N} &
\text{on }\{x_3=h(t,x_1,x_2)\} \\
u_3 = 0,\quad (Su (- e_3))_i=-\kappa u_i,\ i=1,2 & \text{on } \{x_3=-b\}%
\end{cases}%
\end{equation}
for $Su = {1 \over 2} \( \nabla u + \nabla u^t\)$ the symmetric part of the gradient of $u$ and ${\bf n}={\bf N}/|{\bf N}|$ the outward unit normal of the
free surface with ${\bf N}=(-\partial_1h,-\partial_2h,1)^t$. $\varepsilon>0$ is the viscosity, $g>0$ is the strength of gravity, $\sigma>0$ is the surface tension
coefficient and $\kappa$ is the friction coefficient. Finally, $H$ is twice the mean curvature of the free surface given by the formula
\begin{equation}
H =\nabla\cdot\left(\frac{\nabla h }
{\sqrt{1+|\nabla h|^2}}\right).
\end{equation}
The kinematic boundary condition, the fourth equation in \eqref{NS}, implies that the free surface is adverted with the
fluid, and the dynamic boundary condition, the third equation, states the balance of stress tensor on the free surface. We have imposed the
Navier
slip boundary condition on the lower fixed boundary. Note that in \eqref{NS} we have shifted the gravitational forcing to the boundary and
eliminated the constant atmospheric pressure, $p_{atm}$, in the usual way by adjusting the actual pressure $\bar{p}$ according to $p = \bar{p} +
g x_3 - p_{atm}$.

The initial surface is given by the graph of the function $h(0)=h_0: \Rn{2}\rightarrow \mathbb{R}$, which yields the initial domain $\Omega(0)$
on which the initial velocity $u(0)=u_0: \Omega(0) \rightarrow \mathbb{R}^3$ is specified. It  will  be assumed that $h_0 > -b$ and that $(u_0,
h_0)$ satisfy certain compatibility conditions.

The movement of the free boundary and the subsequent change of the domain create numerous mathematical difficulties. To circumvent these, as
usual, we will transform the free boundary problem under consideration to a problem with a fixed domain and fixed boundary. Consider a  family of
diffeomorphism $\Phi(t, \cdot)$ of the form
\begin{equation}\label{diff}
\begin{split}
\Phi(t, \cdot) : \,   \Omega= \mathbb{R}^2 \times (-b, 0) & \rightarrow   \Omega({t}) \\
(y,z) &\rightarrow    (y, \varphi(t, y, z)).
\end{split}
\end{equation}
$\varphi$ is chosen so that $\partial_{z} \varphi>0$ which ensures that $\Phi(t, \cdot)$ is a diffeomorphism.
For the fluid domain under consideration, $\varphi$ can be chosen as
\begin{equation}\label{eqphi}
\varphi(t,y,z)=    z + \eta(t, y, z),
\end{equation}
where $\eta$ is a chosen extension of $h$ onto $\{z\le 0\}$ defined by
\begin{equation}\label{eqeta}
\hat{\eta}(t, \xi, z)= \( 1+\frac{z}{b}\)\exp{(A|\xi|z)} \hat{h}(t, \xi).
\end{equation}
Here $\hat{\cdot}$ stands for the horizontal Fourier transform with respect  to the $y$ variable. It is verified in Proposition \ref{diffeoprop}
that for given $h_0$ if the  number $A>0$ is chosen sufficiently small, then
\begin{equation}\label{Adeb}
\partial_{z} \varphi(0,y,z)\ge c_0 >0\text{ in }\Omega.
\end{equation}

Then one can reduce the problem into the fixed domain $ \Omega  $ by setting
\begin{equation}\label{vdef}
v(t,y,z)= u(t, \Phi(t,y,z)), \ q(t,y,z)= p(t, \Phi(t,y,z) )\text{ in }\Omega.
\end{equation}
Set
$$ \partial^\varphi_{i} =  \partial_{i}  - { \partial_{i} \varphi \over \partial_{z} \varphi}
\partial_{z}, \quad i=t, \, 1,\, 2, \quad \partial^{\varphi}_{3}= \partial^{\varphi}_{z}=  {1 \over \partial_{z} \varphi} \partial_{z} $$
such that
$$   \partial_{i}u \circ \Phi(t, \cdot)= \partial^{\varphi}_{i}v, \quad i= t, \, 1, \, 2, \, 3.
     $$
Then by the change of coordinates \eqref{diff}, the problem \eqref{NS} becomes
\begin{equation}\label{NSv}
\begin{cases}   \partial^{\varphi}_{t}v +  v \cdot \nabla^\varphi  v + \nabla^\varphi q - \varepsilon   \Delta^\varphi v =0 &\text{in }\Omega
\\ \nabla^\varphi  \cdot  v = 0  &\text{in }\Omega
\\q {\bf n} - 2 \varepsilon S^\varphi v   {\bf n} = gh  {\bf n}-\sigma H{\bf n} &\text{on }\{z=0\}
\\\partial_{t}h = v \cdot {\bf N} &\text{on }\{z=0\}
\\v_3 = 0,\quad (S^\varphi v e_3)_i= \kappa v_i,\ i=1,2 &\text{on }\{z=-b\}
\\ (v,h)\mid_{t=0}= (v_0,h_{0}).
\end{cases}
\end{equation}
Here we have naturally written $ \(\nabla^\varphi  \)_i =  \partial^{\varphi}_{i}  , \ \Delta^\varphi=  \partial^{\varphi}_{i}\partial^{\varphi}_{i} , \  \nabla^\varphi\cdot  v=  \partial^{\varphi}_{i} v_{i} $
and $S^\varphi v= {1 \over 2}\big( \nabla^\varphi v + (\nabla^\varphi v)^t)$.
Note that $\nabla^\varphi\cdot  S^\varphi v= {1 \over 2} \Delta^\varphi v$ for vector fields satisfying $\nabla^\varphi\cdot  v=0$.

\subsection{Previous works}

Free boundary problems in fluid mechanics have been studied intensively in the mathematical community. There are a huge amount of mathematical
works, and we only mention briefly some of them below. We may refer to the references cited in these works for more proper survey of the
literature. For the incompressible Navier-Stokes equations, we refer to, for instance, Beale \cite{beale_1}, Hataya \cite{H}, Guo and Tice
\cite{GT13,GT_per,GT_inf} for the well-posedness without surface tension, and Beale \cite{beale_2}, Tani \cite{Ta}, Tanaka and Tani \cite{TT} for
the well-posedness with surface tension. Those well-posedness results are strongly based on the regularizing effect of the viscosity, and the
solutions are shown to be global for the small initial data \cite{H,GT_per,GT_inf,beale_2,TT}. Note that the surface tension only has a
regularizing effect on the free surface, and its effect serves to enhance the decay rate, see \cite{GT_per,GT_inf} for more discussions. For the
incompressible Euler equations, the problem becomes much more difficult. The early works were focused on the irrotational fluids, which began
with
the pioneering work of Nalimov \cite{N} of the local well-posedness without surface tension for the small initial data and was generalized to the
general initial data by the breakthrough of Wu \cite{Wu1,Wu2} for the case without surface tension and by Beyer and G\"unther \cite{BG} for the
case with surface tension. For the irrotational inviscid fluids, certain dispersive effects can be used to establish the global well-posedness
for the small initial data; we refer to  Wu \cite{Wu3,Wu4}, Germain, Masmoudi and Shatah \cite{GMS1}, Ionescu and Pusateri \cite{IP} and Alazard
and Delort \cite{AD} for the case without surface tension, and Germain, Masmoudi and Shatah \cite{GMS2} and Ionescu and Pusateri \cite{IP2} for
the case with surface tension. For the general incompressible Euler equations without the irrotational assumption, only local well-posedness
results could be found. The first local well-posedness in 3D was obtained by Lindblad \cite{Lindblad05} for the case without surface tension and
by Coutand and Shkoller
\cite{CS07} for the case with (and without) surface tension. We also refer to the results of Shatah and Zeng \cite{SZ} and Zhang and Zhang
\cite{ZZ}.

Various approaches are used to prove those well-posedness results mentioned above, depending on whether viscosity or surface tension is presented
or not. It is then very natural and interesting to study the asymptotic behavior of vanishing these two parameters in the equations. The
vanishing viscosity limit for the Navier-Stokes equations is a classical issue. When there is no boundary, the problem has been well studied; we
refer to Swann \cite{Swann71}, Kato \cite{Kato72}, DiPerna and Majda \cite{DM1,DM2}, Constantin \cite{Con86} and Masmoudi \cite{M} for example.
However, in the presence of boundaries, the situation is more complicated and the problem becomes challenging due to the possible formation of
boundary layers. In a fixed domain with the no-slip boundary condition, there is formation of boundary layers in the vicinity of the boundary and
the solution $u^\varepsilon$ of the Navier-Stokes equations is expected to behavior like $u^\varepsilon\sim u^0+U(t,y,z/\sqrt{\varepsilon})$ (we assume the boundary
is locally given by $z=0$), where $u^0$ is the solution of the Euler equations satisfying only the impermeable boundary condition and $U$ is some
profile. In view of this small scale behavior, it is impossible in general  to get uniform strong estimates in any Sobolev spaces containing
normal derivatives. Consequently, the vanishing viscosity problem with the no-slip boundary condition is widely open except the works by Asano
\cite{Asano} and Sammartino and Caflisch \cite{SC} in the framework of analytic initial data and and the work by Maekawa \cite{Mae14} for the
initial vorticity located away from the boundary and a recent work by Guo and Nguyen \cite{GN} for a steady flow over a moving plane. However,
when the no-slip boundary condition is replaced by the Navier slip boundary condition,
the situation becomes  better. Indeed, now the solution is expected to behavior like $u^\varepsilon\sim u^0+\sqrt{\varepsilon}U(t,y,z/\sqrt{\varepsilon})$; the
amplitude of the boundary layer is  weaker, and one can hope to get an uniform estimates involving one normal derivative. In this case, the
vanishing viscosity limit in 3D has been justified rigorously in Iftimie and Planas \cite{Iftimie-Planas}, Iftimie and Sueur
\cite{Iftimie-Sueur}, Masmoudi and Rousset \cite{MasRou12} and Xiao and Xin \cite{XiaoXin13}. Furthermore,  for some special types of Navier
boundary conditions or boundaires,  uniform estimates in higher order Sobolev spaces can be obtained, see  Xiao and Xin \cite{Xin} and Beir\~ao
da Veiga and Crispo \cite{BC11}.

Going back to the free-surface incompressible Navier-Stokes equations, since the dynamic boundary condition can be viewed as the same type of
slip boundary conditions, one has the hope to establish the vanishing viscosity limit. For the case without surface tension and there is no
boundary below the fluid, Masmoudi and Rousset \cite{MasRou} justified the inviscid limit by using the framework of their earlier work
\cite{MasRou12} and some additional techniques. Later, Elgindi and Lee \cite{EL14} discussed the same problem for the case with surface tension,
however, some key points in their arguments are not clear to us. On the other hand, for the free boundary problems, it is also interesting to
show the vanishing surface tension limit. This is supposed to be somewhat simpler than the inviscid limit problem since the equation on the free
boundary is defined without boundary. Yet, one needs to develop the well-posedness which is uniform with respect to surface tension, and
generally this is nontrivial. These have been done for the irrotational Euler equations, see \cite{AM1,AM2} and references therein; for the
general Euler equations, a priori uniform estimates have been derived in \cite{SZ}; while for the Navier-Stokes equations with small initial
data, we refer to the recent work of Tan and
Wang \cite{TW14}. The purpose of this paper is to derive the uniform estimates of solutions of the system \eqref{NS}
(equivalently, \eqref{NSv}) on a time interval independent of both viscosity and surface tension coefficients. These allow us to justify the
vanishing viscosity and surface tension limits by the strong compactness argument. As a byproduct, we can get a unified local well-posedness of
the
incompressible free-surface Euler equations with or without surface tension by the inviscid limits.

\section{Main results}\label{sec main}

\subsection{Statement of the results}

We shall use Sobolev conormal spaces on $\Omega$ as \cite{MasRou12,MasRou}. Set
$$  Z_{i}= \partial_{i}, \, i=1, \, 2, \ Z_{3}= z(z+b) \partial_{z},$$
which are tangent to $\partial\Omega$. The Sobolev conormal space $H^k_{co}$  is defined as
$$ H^k_{co}(\Omega)= \left\{ f \in L^2(\Omega), \   Z^\alpha  f  \in L^2 (\Omega), \  \alpha\in\mathbb{N}^3,\ | \alpha| \leq k \right\},$$
where $ Z^\alpha = Z_{1}^{\alpha_{1}} Z_{2}^{\alpha_{2}} Z_{3}^{\alpha_{3}}  $, with norm defined as
$$ \norm{  f  }_{k} =\sum_{\alpha\in\mathbb{N}^3\atop|\alpha| \leq k}\norm{ Z^\alpha f }_{L^2}, \ \norm{  f  } = \norm{ f  }_0 = \norm{  f
}_{L^2}.
  $$
Similarly, $W^{k, \infty}_{co}$  is defined as
$$ W^{k, \infty}_{co}(\Omega)= \left\{ f \in L^\infty(\Omega), \   Z^\alpha  f  \in L^\infty (\Omega), \  \alpha\in\mathbb{N}^3, \  | \alpha|
\leq k \right\}$$
with norm
$$ \norm{  f  }_{k, \infty} =\sum_{\alpha\in\mathbb{N}^3\atop|\alpha| \leq k} \norm{ Z^\alpha f }_{L^\infty}.$$
$H^k$ and $W^{k,\infty}$ will denote for the usual Sobolev spaces on $\Omega$, and $|\cdot|_{s}$ and $|\cdot|_{s,\infty}$ stand for the standard
Sobolev norms on $\mathbb{R}^2$.
To estimate the time derivatives of the solution, as \cite{EL14}, we introduce also the {\it spatial-time} Sobolev conormal norms on $\Omega$ as:
\begin{equation}\label{xym}
\norm{  f  }_{{\mathbb X}^{m,k}}^2 =\sum_{\ell=0}^m \norm{\partial_t^\ell f }_{m+k-\ell}^2, \  \norm{  f   }_{{\mathbb X}^{m}}= \norm{  f   }_{{\mathbb X}^{m,0}}, \text{ and
}\norm{  f  }_{{\mathbb Y}^{m}} =\sum_{\ell=0}^m \norm{\partial_t^\ell f }_{m-\ell,\infty},
\end{equation}
 and the {\it spatial-time} Sobolev norms on $\mathbb{R}^2$ :
\begin{equation}\label{xymb}
\abs{  f  }_{{\mathbb X}^{m,s}} =\sum_{\ell=0}^m \abs{\partial_t^\ell f }_{m+s-\ell}^2, \   \abs{  f  }_{{\mathbb X}^{m}}= \abs{  f  }_{{\mathbb X}^{m,0}},\text{ and }\abs{  f
}_{{\mathbb Y}^{m}} =\sum_{\ell=0}^m \abs{\partial_t^\ell f }_{m-\ell,\infty}.
\end{equation}
Note that in these definitions $k$ and $m$ are assumed to be non-negative integers, but $s$ is allowed  to be any real number, typically, halfs.
In the following we will abuse the notation $\frac{m}{2}$ for $\[\frac{m}{2}\]$.

The aim of this paper is to get a local well-posedness result for strong  solutions of  \eqref{NSv} in Sobolev conormal spaces
which is valid on an interval of time independent of ${\varepsilon,\sigma} \in (0,1]$.  Note that such a result will also  imply  the  local
well-posedness for the Euler equation with or without surface tension. As it is well-known that
when there is no surface tension a Taylor sign condition on the free boundary is needed
to get local well-posedness for the Euler  equation; when there is surface tension, no such condition is needed.
By the change of coordinates \eqref{diff}, the Taylor sign condition reads as
\begin{equation}\label{Taylor2}
    - \partial_{z}^\varphi q +g\geq c_{0}>0  \text{ on }\{z=0\}.
\end{equation}

In the below, ${\bf N}$ and ${\bf n}$ are extended to $\bar{\Omega}$ by
 \begin{equation}{\bf N}(t,y,z)=(-\partial_1\varphi(t,y,z),-\partial_2\varphi(t,y,z),1)^t\text{ and }{\bf n}={\bf N}/|{\bf N}|.
 \end{equation}
 Note that ${\bf N}(t,y,0)$ is indeed the outward normal to the free surface defined before, and ${\bf N}(t,y,-b)=e_3$.
Define $\Pi=  \mbox{I} - {\bf n}\otimes {\bf n}$, and let $\chi$ be a smooth function which takes the value zero in the vicinity of
$\{z=0\}$ and one in the vicinity of $\{z=-b\}$. Let $\norm{\cdot}_{L^p_TX}$ be the norm of the space $L^p([0,T];X)$. Then the main result in
this paper is stated as follows.
  \begin{theoreme}
    \label{main}
Let $m \geq 14$. Assume that the initial data $(h ^{\varepsilon,\sigma}_0, v ^{\varepsilon,\sigma}_0)$ is given such that
\begin{align}\label{hypinit}
  &\abs{h^{\varepsilon,\sigma}(0)}_{{\mathbb X}^m}^2+ \sigma \abs{h ^{\varepsilon,\sigma}(0)}_{{\mathbb X}^{m,1}}^2+ \varepsilon  \abs{h ^{\varepsilon,\sigma}(0)}_{{\mathbb X}^{m,\frac{1}{2}}}^2
   \\\nonumber&\quad+ \norm{v ^{\varepsilon,\sigma}(0)}_{{\mathbb X}^m}^2 + \norm{\partial_{z} v ^{\varepsilon,\sigma}(0)}_{{\mathbb X}^{m-1}}^2
+ \norm{ \partial_{z} v^{\varepsilon,\sigma}(0)}_{{\mathbb Y}^{\frac{m}{2}+2}}^2 +\varepsilon  \norm{\partial_{zz} v^{\varepsilon,\sigma}(0) }_{L^\infty}^2  \leq R_{0},   \end{align}
  and that initially the compatibility condition
   $ \Pi^{\varepsilon,\sigma} \(S^{\varphi^{\varepsilon,\sigma}} v^{\varepsilon,\sigma} {\bf n}^{\varepsilon,\sigma}-\kappa \chi v^{\varepsilon,\sigma}\)= 0$ on $\{z=0,-b\}$, the diffeomorphism \eqref{Adeb} and the Taylor sign
   condition \eqref{Taylor2} on $\{z=0\}$ hold.
There exists $T>0$ and $C>0$  such that for every  $\varepsilon,\sigma\in (0, 1]$, the unique solution $(v^{\varepsilon,\sigma}, h^{\varepsilon,\sigma})$  of  \eqref{NSv} satisfies the
estimate:
\begin{align}\label{mainborne1}
&\abs{h^{\varepsilon,\sigma} }_{L^\infty_T {\mathbb X}^{m-1,1}}^2+ \sigma \abs{h ^{\varepsilon,\sigma} }_{L^\infty_T{\mathbb X}^{m-1,2}}^2+ \varepsilon  \abs{h ^{\varepsilon,\sigma} }_{L^\infty_T{\mathbb X}^{m-1 , {3\over 2}}}^2
\\&\quad+ \norm{v ^{\varepsilon,\sigma} }_{L^\infty_T{\mathbb X}^{m-1,1}}^2 + \norm{\partial_{z} v ^{\varepsilon,\sigma} }_{L^\infty_T{\mathbb X}^{m-2}}^2  + \norm{ \partial_{z} v^{\varepsilon,\sigma}
}_{L^\infty_T{\mathbb Y}^{\frac{m}{2}+2}}^2+\varepsilon  \norm{\partial_{zz} v^{{\varepsilon,\sigma}} }_{L^\infty_T L^\infty}^2\nonumber\\&\quad+  \abs{\partial_t^m h^{\varepsilon,\sigma} }_{L^4_T L^2
}^2+ \sigma \abs{\partial_t^m h^{\varepsilon,\sigma} }_{L^4_T H^1 }^2+ \varepsilon  \abs{\partial_t^m  h ^{\varepsilon,\sigma} }_{L^4_TH^{ \frac{1}{2}}}^2+ \norm{ \partial_t^m v^{\varepsilon,\sigma} }_{L^4_T L^2 }^2+ \norm{
\partial_{ z} v^{\varepsilon,\sigma} }_{L^4_T {\mathbb X}^{m-1} }^2\nonumber
 \\\nonumber&\quad+\sigma^2 \abs{h ^{\varepsilon,\sigma} }_{L^2_T {\mathbb X}^{m-1,\frac{5}{2}} }^2 +\varepsilon  \norm{\nabla v ^{\varepsilon,\sigma} }_{L^2_T {\mathbb X}^{m-1,1} }^2+ \varepsilon \norm{\nabla
 \partial_z v ^{\varepsilon,\sigma} }_{L^2_T {\mathbb X}^{m-2} }^2   \leq   C.\end{align}
\end{theoreme}

It should be remarked that the reason for the $L^4$-in-time estimates of $\norm{\partial_{z} v^{\varepsilon,\sigma}(t)}_{{\mathbb X}^{m-1}} $ rather than in $L^\infty$
stated in
Theorem \ref{main} is related  to the boundary  control of the vorticity for the Navier-Stokes system ($i.e.$, viscous boundary layers). Also,
owing
to the less regularity of the pressure as will be shown in our proof later, which is due to the presence of surface tension, we can only prove
the $L^4$-in-time estimates of the highest time
derivative term $\abs{\partial_t^m h^{\varepsilon,\sigma}(t) }_{L^2 }+ \sqrt{\sigma} \abs{\partial_t^m h^{\varepsilon,\sigma}(t) }_{H^1 }+ \sqrt{\varepsilon } \abs{\partial_t^m h^{\varepsilon,\sigma}(t) }_{H^\frac{1}{2} }+ \norm{
\partial_t^m v^{\varepsilon,\sigma}(t) }_{L^2 }$. As immediate consequences of the uniform estimates of Theorem \ref{main}, one establishes easily, by
standard compactness arguments,  the justification of the inviscid limit, vanishing surface tension limit,  and any their combinations. In
particular, we can obtain a local well-posedness of the free surface Euler equations, independent of surface tension, by the inviscid limit.
Furthermore, for the Euler equations, by using the equation for the vorticity one can
improve those $L^4$-in-time estimates to be in $L^\infty$ and also recover the standard Sobolev regularity.

\subsection{Strategy of the proof}

The main step in the proof of Theorem \ref{main} is to derive the a priori uniform estimates on a time interval small but independent of
${\varepsilon,\sigma}\in(0,1]$ for a sufficiently smooth solution of the equations \eqref{NSv}. Our approach is
strongly motivated by the strategy of Masmoudi and Rousset \cite{MasRou} where the vanishing viscosity limit was justified for the problem
without surface tension. However, there are several new difficulties arising in the presence of surface tension. Indeed, it has been already
known for the free-surface incompressible Euler equations that the problem with surface tension is more difficult than the problem without
surface tension in certain sense, see Coutand and Shkoller \cite{CS07} for some discussions. This is even so when one
wants to show the inviscid limit for the free-surface incompressible Navier-Stokes equations. When showing the inviscid limit for each fixed
$\sigma>0$, one finds that surface tension only serves to provide the improved regularity of the free surface. Indeed, the presence of
surface tension makes the problem more difficult: first, the required nonlinear estimates are more difficult to close due to the mean curvature
term; second, the less regularity of the pressure makes the arguments much more involved. Note that the less regularity of the free surface for
the problem without surface tension was get around in \cite{MasRou} by using Alinhac good unknowns \cite{Alinhac}. In this paper we will still
use Alinhac good unknowns also without taking advantage of the improved regularity of the free surface provided by any fixed surface tension;
nevertheless, we will develop some ideas to overcome the difficulties caused by the presence of  surface tension as illustrated below. As a
consequence, we will be able to deal with both the vanishing viscosity and surface tension limits.

For notational convenience we shall suppress the subscripts $\varepsilon$ and $\sigma$ below. Let $\mathcal{N}(T)$  be the quantity appearing in the
left hand side of \eqref{mainborne1}, while $\mathcal{Q}(T)$  be the first two lines in \eqref{mainborne1}. The crucial point is thus to get the
closed a priori estimates of $\mathcal{N}(T)$ on a small time interval independent of $\varepsilon$ and $\sigma$ in terms of the initial data. We start
with the basic physical energy identity:
\begin{align}\label{intro1}
& \frac{1}{2} \frac{d}{dt} \(  \int_{\Omega} |v|^2 \, d\mathcal{V}_{t} +   \int_{z=0} g|h|^2+2\sigma\(\sqrt{1+|\nabla_y h|^2}-1\) \,dy  \)
 \\\nonumber&\qquad\qquad\qquad\qquad\qquad\qquad+ 2 \varepsilon  \int_{\Omega} |S^\varphi v|^2\, d\mathcal{V}_{t}+2 \kappa \varepsilon\int_{z=-b}|v|^2\, dy=0.
 \end{align}
Here $d\mathcal{V}_{t}$ stands for the volume element induced by the change of variable \eqref{eqphi}: $d\mathcal{V}_{t}=\partial_{z}\varphi \, dydz$.

To get the estimates for higher order conormal estimates, one then applies $Z^\alpha$ for $1\le |\alpha|\le m$ to the equations \eqref{NSv}.
Since the operators $\partial_i^\varphi$ involve $\nabla\varphi$, the estimate of the commutator between $Z^\alpha$ and $\partial_i^\varphi$ needs a
control of $\norm{Z^\alpha \nabla\varphi}\lesssim \abs{Z^\alpha h}_{1/2}$. In the absence of the surface tension this yields a loss of $1/2$
derivative, and the difficulty was overcome in Masmoudi and Rousset \cite{MasRou} by using a
crucial cancellation observed by Alinhac \cite{Alinhac}. The idea is to use the good unknowns $V^\alpha= Z^\alpha v - {\partial_{z}^\varphi v}
Z^\alpha \eta$ and $Q^\alpha=  Z^\alpha q - {\partial_{z}^\varphi q} Z^\alpha \eta$, and some cancellation occurs when considering the equations
for $V^\alpha$ and $Q^\alpha$ which allows one to derive an $L^2$ type energy estimates similar to  \eqref{intro1}:
\begin{align}\label{intro2}
&\frac{1}{2} \frac{d}{dt}\( \int_{\Omega} |V^\alpha|^2 \, d\mathcal{V}_{t} +  \int_{z=0} (g - \partial_{z}^\varphi q ) |Z^\alpha h|^2+\sigma\abs{\nabla_y Z^\alpha h}^2 dy\)+
2 \varepsilon  \int_{\Omega} |S^\varphi V^\alpha|^2\, d\mathcal{V}_{t}
\\\nonumber&\quad =-\int_{z= 0}  \sigma Z^\alpha H \sum_{|\alpha'|=1}C_{\alpha}^{\alpha'}Z^{ \alpha'}{\bf N}\cdot Z^{\alpha-\alpha'} v
\,dy-\int_{\Omega} \mathcal{C}^\alpha (d) Z^\alpha q \, d\mathcal{V}_{t} +  {\sum}_0 +  {\sum}_\sigma,
\end{align}
where, using the symmetric commutator notation $\[ \cdot ,  \cdot , \cdot \]$ defined by \eqref{symcom},
\begin{align}\label{intro11}
  \partial_z\varphi \mathcal{C}^\alpha (d)   =  \[  Z^\alpha ,  {\bf N} , \cdot\partial_{z}v  \]  +   \[  Z^\alpha ,  \partial_{z}
  \eta,\partial_1v_1+\partial_2v_2\].
     \end{align}
Here ${\sum}_0$ denotes the terms that can be controlled in a similar way as the case without surface tension \cite{MasRou}, and ${\sum}_\sigma$
stands for  the terms related to surface tension, after some delicate arguments, that can be controlled well with the improved energy estimates
$\sigma\abs{  Z^\alpha h}_1^2$. The first two terms in the right hand side of \eqref{intro2} are singled out in order to indicate the main
difficulties for the case with surface tension. Note that the regularity $\sigma\abs{ Z^\alpha h}_1^2$ in the energy is not enough to control the
first term. Indeed, one needs $\sigma\abs{Z^\alpha h}_{3/2}$, $i.e.,$ there is a loss of $1/2$ derivative. To improve the regularity of $h$, one
then resorts to the ``dissipation" estimate by using the dynamic boundary condition: $-\sigma H  =q-gh-  2 \varepsilon S^\varphi v  \,{\bf n} \cdot
\,{\bf n}$ on $\{z=0\}$, which requires a control of $ \abs{Z^\alpha q}_{-1/2} $.  On the other hand, it seems that the only way to control the
pressure $q$ is through the elliptic equation with the Neumann boundary conditions
\begin{equation}\label{intro3}
\begin{cases}
\Delta^\varphi q = - \nabla^\varphi \cdot (v\cdot \nabla^\varphi v)& \text{in }\Omega
\\\nabla^\varphi q\cdot{\bf N} = -\partial_t v\cdot {\bf N}-(v_y\cdot\nabla_y) v\cdot {\bf N}+\varepsilon   \Delta^\varphi v \cdot{\bf N} &\text{on }\{z=0\}
\\\nabla^\varphi q\cdot{\bf N} =  \varepsilon   \Delta^\varphi v \cdot{\bf N} &\text{on }\{z=-b\}.
\end{cases}
\end{equation}
It should be noted  that on the boundary $\{z=0\}$ one could not use the Dirichlet boundary condition $q=-\sigma H +gh+  2 \varepsilon S^\varphi v
\,{\bf n} \cdot \,{\bf n}  $ since there is no control for $-\sigma H$ yet, which is in contrast to
the case $\sigma=0$ where the Dirichlet boundary condition is successfully used in  \cite{MasRou}. Note that  the appearance of $\partial_t v$ in
\eqref{intro3} forces one to include the time derivatives in $Z^\alpha$. Then elliptic estimates for \eqref{intro3} provide a control of
$\norm{\nabla q}_{{\mathbb X}^{m-1}}$, which then yields a bound for $\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}$. This implies in particular that we need to separate
the estimates of \eqref{intro2} into two cases: $\alpha_0\le m-1$ and $\alpha_0=m$. For the former case, the main conclusion is that
\begin{align}\label{intro4}
&\norm{v(t)}_{{\mathbb X}^{m-1,1}}^2+\abs{h(t)}_{{\mathbb X}^{m-1,1}}^2 +\sigma\abs{h(t)}_{{\mathbb X}^{m-1,2}}^2+\varepsilon  \abs{h(t)}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2
\\\nonumber&\quad \le C_0+\Lambda\(\mathcal{Q}(T)\)\(t+\int_0^t  \abs{\partial_t^m h}_{-\frac{1}{2}}^2+\sigma \abs{\partial_t^m h}_{1}^2 +\norm{\partial_t^m v}^2+
\norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 \).
\end{align}
The crucial step is to treat  \eqref{intro2} for the case $\alpha_0=m$, which can be rewritten as
\begin{align}\label{intro5}
&\frac{1}{2} \frac{d}{dt}\( \int_{\Omega} |V^m|^2 \, d\mathcal{V}_{t} +  \int_{z=0} (g - \partial_{z}^\varphi q ) |\partial_t^m h|^2+\sigma\abs{\nabla_y \partial_t^m h}^2 dy\)+ 2 \varepsilon
\int_{\Omega} |S^\varphi V^m|^2\, d\mathcal{V}_{t}
\\\nonumber&\quad =-\int_{z= 0}  \sigma \partial_t^m H m\partial_t {\bf N}\cdot \partial_t^{m-1} v\,dy-\int_{\Omega} m\partial_t {\bf N}\cdot \partial_t^{m-1}\partial_z v \partial_t^m q \, dydz +
{\sum}_0 +  {\sum}_\sigma+  {\sum}_q.
\end{align}
Here ${\sum}_q$ stands for the terms related to $q$, which are easier to estimate after some delicate arguments than the second term on the
right hand side above. The main difficulties now are that there are no any estimates for $\partial_t^mq$ so that it is difficult to bound  the second
term, and hence that the regularity $\sigma\abs{ \partial_t^m h}_1^2$ in the energy can no longer be improved through the dissipation so that it is
difficult to bound the first term. Our approach  to
overcome these difficulties is to integrate by parts over $t$ and $z$ in an appropriate order to obtain a crucial cancelation. More precisely,
we integrate by parts in $z$ first and then in $t$ the second term to obtain
\begin{align}\label{intro6}
&-\int_{\Omega} m\partial_t {\bf N}\cdot \partial_t^{m-1}\partial_z v \partial_t^m q \, dydz
\\&\nonumber\quad=-\int_{z=0} m\partial_t {\bf N}\cdot \partial_t^{m-1} v \partial_t^m q \, dy +\int_{\Omega} m \partial_z(\partial_t {\bf N} \partial_t^m q)\cdot \partial_t^{m-1} v \, dydz
\\&\nonumber\quad=-\int_{z=0} m\partial_t {\bf N}\cdot \partial_t^{m-1} v \partial_t^m q \, dy +\frac{d}{dt}\int_{\Omega} m\partial_z(\partial_t{\bf N} \partial_t^{m-1} q)\cdot \partial_t^{m-1} v
dydz +{\sum}_0  .
\end{align}
Note that one can not integrate by parts in $t$ for the first term in the last line of \eqref{intro6} since we can not control $\partial_t^{m} v$ on
$\{z=0\}$. The crucial observation is that there is a cancelation between this term and the first term in the right hand side of \eqref{intro5}.
Indeed, by the kinematic boundary condition, it holds that
\begin{align}\label{intro7}
&-\int_{z= 0}  \sigma \partial_t^m H m\partial_t {\bf N}\cdot \partial_t^{m-1} v\,dy-\int_{z=0} m\partial_t {\bf N}\cdot \partial_t^{m-1} v \partial_t^m q \, dy
 \\\nonumber&\quad=-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \(g\partial_t^mh +2\varepsilon \partial_t^m \( S^\varphi v  \,{\bf n} \cdot \,{\bf n}\)\)\,
 dy={\sum}_0.
\end{align}
The last difficulty in closing the conormal estimate for the case $\alpha_0=m$ is due to the second term in the last line of \eqref{intro6}: the
control of $\norm{\nabla q}_{{\mathbb X}^{m-1}}$ (especially involves the dissipation) is not pointwise in time but only $L^2$. To get the estimates, we
are forced to integrate in time twice and the main conclusion is that
\begin{align}
&\(\int_0^t\(\norm{\partial_t^m v }^2+\abs{\partial_t^m h }_{0}^2 +\sigma \abs{\partial_t^m h }_{1}^2+ \varepsilon   \abs{\partial_t^m h}_{\frac{1}{2}}^2 \)^2\)^\frac{1}{2}
\\\nonumber&\quad\le  C_0+   \Lambda(\mathcal{Q}(T))\(t + \int_0^t  \abs{\partial_t^mh}_{0}^2 + \sigma     \abs{\partial_t^mh}_{1}^2
+ \norm{\partial_t^m v} ^2 + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 + \varepsilon  \abs{\partial_t^m h}_{\frac{1}{2}}^2  \)^\frac{1}{2} .
\end{align}

The next step is to derive the conormal estimates for $\partial_z v$. As in \cite{MasRou}, the main idea is to first introduce the equivalent quantity
$S_{{\bf n}}= \Pi \(S^\varphi v   {\bf n}-\kappa\chi v\)$, which satisfies a convention-diffusion type equation with the homogeneous Dirichlet boundary
condition. The
main difficulty is that $\mathcal{Q}(T)$ controls only $\sqrt{\varepsilon} \norm{\partial_{zz}v}_{L^\infty}$. Since we have included the time
derivatives, so if one followed the arguments of \cite{MasRou}, which only involves the spatial derivatives, to estimate the commutator resulting
from the viscosity term, one would need  to control $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{{\mathbb Y}^{k}}$ for some high $k\ge 1$. Our key
observation here  is that since in the vicinity of the boundary the solution behaves like $v(t,x)\sim v^0(t,x)+\sqrt{\varepsilon}U(t,y,z/\sqrt{\varepsilon})$,
so it is possible to control $\varepsilon \partial_{zz}v $ (and even $\varepsilon \partial_{zzz}v$!). This can be done easily by using the first equation in
\eqref{NSv}. We can then perform the $L^2$ type energy estimates to conclude that
\begin{align}
\norm{\partial_z v(t)}_{{\mathbb X}^{m-2}}^2\le  C_0+   \Lambda(\mathcal{Q}(T))\(t + \int_0^t  \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2   \) .
\end{align}
Note that  the $m-2$ order estimate above cannot be improved to be $m-1$ due to the appearance of
$(\nabla^\varphi)^2 q$ in the source term in the equation for $S_{{\bf n}}$, which is only in ${\mathbb X}^{m-2}$. To get a better estimate, following
\cite{MasRou,MasRou12,Xin}, one would proceed with the vorticity  $\omega=\nabla^\varphi\times v$ instead of $S_{{\bf n}}$. $\omega$ again satisfies
a convention-diffusion type equation, but the main difficulty is that it does not vanish on the boundary and the boundary value is at a low
regularity. To split the difficulty, for $|\alpha|=m-1$ we set
$ Z^\alpha \omega= \omega^\alpha_{nh}+ \omega^{\alpha}_{h}$, where $\omega^\alpha_{nh}$ satisfies the nonhomogeneous equation with the
homogeneous boundary condition which can be handled by performing the $L^2$ type energy estimates and $\omega^\alpha_{nh}$ satisfies the
homogeneous equation with the nonhomogeneous boundary condition. Note that $\omega^\alpha_{nh}$ solves exactly the same problem as \cite{MasRou}
so that one can apply their estimates directly, though, the proof is quite involved. Combining these two estimates, one concludes  that
\begin{align}
\(\int_0^t \norm{\partial_z v(t)}_{{\mathbb X}^{m-1}}^4\)^\frac{1}{2} \le  C_0+   \Lambda(\mathcal{Q}(T))\(t + \int_0^t  \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2   \) .
\end{align}

The last step is to derive the $L^\infty$ estimates of $\partial_zv$ and $\sqrt{\varepsilon}\partial_{zz}v$. In this step, one needs to estimate a low number of
derivatives of $v$, say $\frac{m}{2}+2$, while the boundary is $H^m$ with  $m$ being as large as needed, it is convenient to use a normal
geodesic coordinate system in the vicinity of the boundary so that the Laplacian has the simplest expression. Note that $\sqrt{\varepsilon}\partial_{zz}v$
can be controlled in the same way as \cite{MasRou}, however, for $\partial_zv$ we will employ a different argument. Again, it is more convenient to
estimate the equivalent quantity $S_{\bf n}$. After some computations, we find that $\rho$, an equivalent quantity of $S_{\bf n}$ in the new coordinates,
solves
\begin{equation} \label{intro10}
\partial_{t}  \rho + w \cdot \nabla   \rho - \varepsilon  \partial_{zz}   \rho= \mathcal{H}
\end{equation}
for some source term $\mathcal{H}$ and a vector field $w$ with $w_3=0$ on the boundary. The main difficulty in the analysis of \cite{MasRou} is a
commutator between $Z^\alpha$ and $\varepsilon\partial_{zz}$ which is difficult to
control when applying the maximum principle. The main idea in \cite{MasRou} is to rewrite \eqref{intro10} into a one-dimensional Fokker-Planck
type equation and then use the explicit representation of the solution. Our observation here is that since we have included the time derivatives
in $Z^\alpha$, it is easy to estimate the commutator by, roughly speaking, replacing the $\varepsilon\partial_{zz}$ with $\partial_t$ by using the equations. Main
conclusion in this step is that
\begin{align}
\norm{\partial_zv(t)}_{{\mathbb Y}^{\frac{m}{2}+2}}^2+\varepsilon\norm{\partial_{zz}v(t)}_{L^\infty}^2\le  C_0+   \Lambda(\mathcal{Q}(T))t .
\end{align}

Combining the estimates in all these steps, we then derive the desired estimates $\mathcal{N}(T)\le C_0$ for some $T$ sufficiently small but
independent of $\varepsilon$ and $\sigma$. Note that the Taylor sign condition and the condition that $\Phi(t,\cdot)$ is a diffeomorphism can be easily
justified due to our estimates of time derivatives. Finally we remark that for each fixed $\sigma>0$, the Taylor sign condition is no longer
needed for the inviscid limit problem. This can be seen from \eqref{intro1} and \eqref{intro2}: even $g-\partial_z^\varphi q\le 0$, one can use the
Sobolev interpolation to get the estimates of $h$ for each $\sigma>0$.

We will set the conventions for notation to be used later.  $\mathbb{N} = \{ 0,1,2,\dotsc\}$ stands  for the collection of non-negative integers,
$\mathbb{N}^{1+d} = \{ \alpha = (\alpha_0,\alpha_1,\dotsc,\alpha_d) \}$ is used to emphasize that the $0-$index term is related to temporal
derivatives,
while  $\mathbb{N}^d$ is related for spatial derivatives.  For $\alpha \in \mathbb{N}^{1+d}$,  $Z^\alpha = \partial_t^{\alpha_0} Z_1^{\alpha_1}\cdots
Z_d^{\alpha_d}.$
The Einstein convention of summing over repeated indices will be used.
Throughout the paper $C>0$ will denote a generic constant that does
not depend on the data, the surface tension coefficient $\sigma$ and
the  viscosity coefficient $\varepsilon$, but can depend on the
other parameters of the problem, $g,\kappa$, $m\ge 14$ and $\Omega$.  We refer to
such constants as ``universal''.  Such constants are allowed to change from line to line.  We will
employ the notation $a \lesssim b$ to mean that $a \le C b$ for a
universal constant $C>0$. Throughout the paper, the notation $\Lambda(\cdot, \cdot)$ stands for a continuous increasing function in all its
arguments, independent of $\varepsilon$ and $\sigma$ and  that may change from line to line, and $\Lambda_0=\Lambda( \frac{1}{c_0})$.

The rest of the paper is organized as follows. We collect some analytic tools related to Sobolev conormal spaces, the properties of Poisson
extension and some geometric estimates in Appendixes \ref{aa1}, \ref{aa2} and \ref{aa3}, respectively. In Section \ref{sectionconorm} we study
the equations satisfied by $(Z^\alpha v, Z^\alpha q, Z^\alpha h)$ and present the estimates of the commutators. Section \ref{sectionpressure} is
devoted to derive the pressure estimates using elliptic regularity in Sobolev conormal spaces, and Section \ref{sectionprelimeta} contains the
smoothing regularity estimates of $h$ due to viscosity and surface tension. In Section \ref{secconormal}, the conormal estimates of the solution
are derived, and the conormal estimates for normal derivatives are given in Section \ref{secnormal}. In Section \ref{secinfty}, we prove the
needed $L^\infty$ estimates for normal derivatives.
Finally, the proof of Theorem \ref{main} is given in Section \ref{finalsec}.

\section{Equations satisfied by $(Z^\alpha v,Z^\alpha q,Z^\alpha h)$}\label{sectionconorm}

\subsection{A commutator estimate}

In order to perform higher order conormal estimates, one needs to compute the equations satisfied by $(Z^\alpha v,Z^\alpha q,Z^\alpha h)$, which
requires  to commute $Z^\alpha$ with each term in the equations \eqref{NSv}. It is thus useful to establish the following general expressions and
estimates for commutators to be used often later.

We will not commute $Z^{\alpha}$ with $\partial^{\varphi}_t $ directly. For $i=  1, \, 2,\, 3$, set
\begin{equation}\label{com1}
Z^\alpha \partial^{\varphi}_{i}f=\partial^{\varphi}_{i} Z^\alpha f-\partial^{\varphi}_{z}f \partial^{\varphi}_{i} Z^\alpha \eta+\mathcal{C}^\alpha_{i}(f),
\end{equation}
where the commutator $\mathcal{C}^\alpha_{i}(f)$ is given for $\alpha \neq 0$ and $i \neq 3$ by
\begin{equation}\label{Cialpha}
\mathcal{C}^\alpha_{i}(f)= \mathcal{C}^\alpha_{i, 1}(f)+ \mathcal{C}^\alpha_{i,2}(f)+\mathcal{C}^\alpha_{i, 3}(f)
\end{equation}
with
\begin{align}
\label{Cialpha1} \mathcal{C}^{\alpha}_{i,1}&=-\[ Z^\alpha, {\partial_{i} \varphi \over \partial_{z} \varphi }, \partial_{z} f  \], \\
\label{Cialpha2} \mathcal{C}^{\alpha}_{i,2}&=-\partial_{z} f \[ Z^\alpha, \partial_{i} \varphi, {1 \over \partial_{z} \varphi}\]     -
\partial_{i} \varphi\partial_zf\[Z^{\alpha-\alpha'}, { 1 \over (\partial_{z} \varphi)^2}\]Z^{\alpha'}\partial_{z} \eta, \\
\label{Cialpha3} \mathcal{C}^\alpha_{i, 3}&= - {\partial_{i} \varphi \over \partial_{z} \varphi} [Z^\alpha, \partial_{z}]f
 + {\partial_{i} \varphi \over (\partial_{z} \varphi)^2}\, \partial_{z} f\,  [Z^\alpha, \partial_{z}] \eta,
\end{align}
for any $|\alpha'|=1$. Note that for $i= 1, \, 2$ ,  $\partial_{i}\varphi=\partial_{i}\eta$ and that for $\alpha \neq 0$,
$Z^\alpha\partial_{z}\varphi= Z^\alpha \partial_{z}\eta$. For $i=3$, similar decomposition for the commutator holds (basically, it suffices to
replace $\partial_{i}\varphi$ by $1$ in the above expressions). Since $\partial^{\varphi}_{i}$ and $\partial^{\varphi}_{j}$ commute, it holds that
\begin{equation}\label{com122}
Z^\alpha \partial^{\varphi}_{i} f =\partial^{\varphi}_{i}(Z^\alpha f-\partial^{\varphi}_{z}f  Z^\alpha \eta)+\partial^{\varphi}_{z}\partial^{\varphi}_{i}f Z^\alpha \eta+\mathcal{C}^\alpha_{i}(f).
\end{equation}
It was first observed by Alinhac \cite{Alinhac} that the highest order term of $\eta$ (which is difficult to control for the case $\sigma=0$)
will be canceled when one uses the good unknown $Z^\alpha f-\partial^{\varphi}_{z}fZ^\alpha \eta$, which allows one to perform high order energy estimates.

Since the expressions as $f/\partial_{z} \varphi$ will appear often later, we shall first state a general estimate. It is assumed that
$\partial_{z} \varphi \geq \frac{{c_{0}}}{2}$ and $|h|_{2, \infty} \leq {1 \over c_{0}}$.
\begin{lem}
For every $k\in \mathbb{N}$, it holds that
\begin{equation} \label{quot}
\norm{ {f \over \partial_{z} \varphi}}_{{\mathbb X}^k} \leq \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+1}}  + \norm{ f }_{{\mathbb Y}^{\frac{k}{2}}}
\)\(\abs{h}_{{\mathbb X}^{k,{1 \over 2}}} +\norm{f}_{{\mathbb X}^k} \).
\end{equation}
\end{lem}
\begin{proof}
Since $\partial_{z} \varphi= 1+ \partial_{z} \eta$, so
$$ {f \over \partial_{z} \varphi}=   f  -   f {  \partial_{z} \eta \over  1+  \partial_{z} \eta } = f - fF(\partial_{z} \eta),$$
where $F(x)= x/(1+ x)$ is smooth and bounded together with all  its derivatives on $1+x \geq c_{0}>0$ and $F(0)=0$. Consequently, the product
estimate \eqref{gues} implies that
$$\norm{{f \over \partial_{z} \varphi } }_{{\mathbb X}^k} \lesssim  \norm{f   }_{{\mathbb X}^k}+  \norm{ f }_{{\mathbb X}^k}\norm{ F(\partial_{z} \eta)
}_{{\mathbb Y}^{\frac{k}{2}}}+\norm{ f }_{{\mathbb Y}^{\frac{k}{2}}} \norm{ F(\partial_{z} \eta) }_{{\mathbb X}^k} .$$
Notice that
$$ \norm{ F(\partial_{z} \eta)}_{{\mathbb Y}^{\frac{k}{2}}} \le  \Lambda\( {1 \over c_{0}}, \norm{\nabla \eta}_{{\mathbb Y}^{\frac{k}{2}}}\) ,  $$
and \eqref{gues} implies again,
$$ \norm{F(\partial_{z} \eta) }_{{\mathbb X}^k} \lesssim \Lambda\( {1 \over c_{0}}, \norm{\nabla \eta }_{{\mathbb Y}^{\frac{k}{2}}}\) \norm{ \partial_{z}
\eta}_{{\mathbb X}^k}.$$
Hence the estimate \eqref{quot} follows from these,  \eqref{etainfty} and \eqref{etaharm}.
\end{proof}

Next lemma deals with the estimates of the commutators $\mathcal{C}^\alpha_{i}(f)$.
\begin{lem}\label{comi}
For $1 \leq | \alpha |\leq m$, $i=   1, \, 2, \, 3$,  it holds that
\begin{equation}\label{comiest}
\norm{\mathcal{C}_{i}^\alpha(f) }     \leq \Lambda\( {1 \over c_{0}}, |h |_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{  \partial_z f }_{{\mathbb Y}^{\frac{m}{2}}} \)
\(\norm{\partial_z f }_{{\mathbb X}^{m-1}} + \abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}}\).
\end{equation}
\end{lem}
\begin{proof}
We only present the proof for $i=1, \, 2$, and the last case is similar and slightly easier.

First, for $\mathcal{C}_{i,1}^\alpha$, it follows from  the commutator estimate \eqref{comsym} that
$$\norm{  \mathcal{C}^\alpha_{i,1} } \lesssim  \norm{  Z \({\partial_{i }  \varphi \over \partial_{z} \varphi } \)
}_{{\mathbb Y}^\frac{m-2}{2}}\norm{Z\partial_{z} f }_{{\mathbb X}^{m-2}} +   \norm{ Z \partial_{z} f }_{{\mathbb Y}^{\frac{m-2}{2}}} \norm{Z\({\partial_{i }  \varphi \over
\partial_{z} \varphi } \) }_{{\mathbb X}^{m-2}}.$$
Consequently,  \eqref{quot} yields that
$$\norm{  \mathcal{C}^\alpha_{i,1} } \leq  \Lambda\({1\over c_{0}},\norm{\nabla \varphi}_{{\mathbb Y}^\frac{m}{2}} + \abs{h}_{{\mathbb Y}^{\frac{m-1}{2}+1}} +
\norm{  \partial_{z} f }_{{\mathbb Y}^{\frac{m}{2}}}\)\( \norm{ \partial_{z} f }_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}} +\norm{ \partial_{i}\eta
}_{{\mathbb X}^{m-1}} \).$$
It then follows from \eqref{eqphi}, \eqref{etainfty} and \eqref{etaharm} that
\begin{equation}\label{Calpha1}
\norm{\mathcal{C}^\alpha_{i,1} } \leq \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m}{2}+1}}+\norm{\partial_z f }_{{\mathbb Y}^{\frac{m}{2}}}\) \(\norm{\partial_z
f }_{{\mathbb X}^{m-1}} + \abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}}\).
\end{equation}

Next, for the first term in $\mathcal{C}_{i,2}^\alpha$, one can use similar arguments: \eqref{comsym}
   and \eqref{eqphi} yield  that
$$\norm{ \partial_{z} f \[ Z^\alpha, \partial_{i} \varphi, {1 \over \partial_{z} \varphi}\] }
\lesssim  \norm{ \partial_{z} f }_{L^\infty} \( \norm{Z\partial_{i }   \varphi }_{{\mathbb X}^{m-2}} \norm{  {Z\partial_z\eta \over (\partial_{z} \varphi)^2}
}_{{\mathbb Y}^{\frac{m-2}{2}}}+  \norm{  Z  \partial_{i }   \varphi  }_{{\mathbb Y}^\frac{m-2}{2}}\norm{ {Z\partial_z\eta \over (\partial_{z} \varphi)^2}}_{{\mathbb X}^{m-2}}
\)
$$
and hence by using \eqref{quot}, \eqref{etaharm} and \eqref{etainfty}, one can show that
$$ \norm{ \partial_{z} f \[ Z^\alpha, \partial_{i} \varphi, {1 \over \partial_{z} \varphi}\] }
  \le    \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{   \partial_z f }_{L^\infty} \)\abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}}.$$
By \eqref{com} instead of \eqref{comsym}, the same estimate holds for the second term in $\mathcal{C}_{i, 2}^\alpha$. Hence,
\begin{equation}\label{Calpha2}
\norm{\mathcal{C}^\alpha_{i, 2} }
\leq\Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{   \partial_z f }_{L^\infty} \)\abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}}.
\end{equation}

It remains to estimate $\mathcal{C}_{i, 3}^\alpha$. Notice that
\begin{equation}\label{idcom}
\[Z^\alpha, \partial_{z}\] f= \sum_{| \beta | \leq m-1} c_{\beta } \partial_{z} ( Z^\beta f)
\end{equation}
for some harmless smooth bounded functions $c_{\beta}.$ This yields, by using again \eqref{etaharm},
\begin{align}\label{Calpha3}
\norm{ \mathcal{C}^\alpha_{i, 3}}&\leq \Lambda\( {1 \over  c_{0}}, \norm{\partial_{i} \varphi}_{L^\infty}+\norm{\partial_{z} f }_{L^\infty}\)\(
\norm{\partial_{z} f }_{{\mathbb X}^{m-1}} + \norm{\partial_{z} \eta }_{{\mathbb X}^{m-1}} \)
\\\nonumber&\leq \Lambda\( {1 \over  c_{0}}, \abs{h}_{1,\infty} + \norm{\partial_z f }_{L^\infty}\)\( \norm{\partial_z f }_{{\mathbb X}^{m-1}}+   \abs{h}_{{\mathbb X}^{m-1
,\frac{1}{2}}}\).
\end{align}

Consequently, the estimate \eqref{comiest} follows by collecting \eqref{Calpha1}, \eqref{Calpha2} and \eqref{Calpha3}.
\end{proof}

\subsection{Interior equations}

We shall now derive the equations  in the domain $\Omega$ satisfied by the good unknowns $V^\alpha = Z^\alpha v - \partial^{\varphi}_{z}v\, Z^\alpha \eta  $ and
$Q^\alpha = Z^\alpha q - \partial^{\varphi}_{z}q\, Z^\alpha \eta.$
  \begin{lem}
  \label{lemValpha}
  For $1 \leq | \alpha | \leq m$, it holds that
   \begin{eqnarray}
  \label{eqValpha} & &  \partial^{\varphi}_{t} V^\alpha + v \cdot \nabla^\varphi V^\alpha + \nabla^\varphi Q^\alpha- 2 \varepsilon \nabla^\varphi \cdot S^\varphi
  V^\alpha
     \\
    \nonumber   & &  \quad  = (\partial^{\varphi}_{z}v  \cdot \nabla^\varphi v)
        Z^\alpha \eta-\mathcal{C}^\alpha(\mathcal{T}) - \mathcal{C}^\alpha(q)+\varepsilon  \mathcal{D^\alpha}\big( S^\varphi v \big) + \varepsilon
        \nabla^\varphi \cdot \big( \mathcal{E}^\alpha (v)\big)
       , \\
   \label{divValpha}
     & &    \nabla^\varphi \cdot V^\alpha =- \mathcal{C}^\alpha (d),
       \end{eqnarray}
       where the commutators $\mathcal{C}^\alpha(q)$, $\mathcal{C}^\alpha(d)$, $\mathcal{E}^\alpha(v)$ and $\mathcal{C}^\alpha (\mathcal{T})$
       satisfy the estimates:
       \begin{align}
       \label{Cq} &\norm{\mathcal{C}^\alpha (q)} \leq \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{  \partial_z q
       }_{{\mathbb Y}^{\frac{m}{2}}} \) \( \norm{ \partial_z q }_{{\mathbb X}^{m-1}} + |h|_{{\mathbb X}^{m-1,{1 \over 2 }}}\), \\
        \label{Cd} &  \norm{\mathcal{C}^\alpha (d)} \leq \Lambda\( {1 \over c_{0}}, |h |_{{\mathbb Y}^{\frac{m}{2}+1}}   +\norm{   \partial_z v
        }_{{\mathbb Y}^{\frac{m}{2}}} \) \( \norm{ \partial_z v }_{{\mathbb X}^{m-1}} + |h|_{{\mathbb X}^{m-1,{1 \over 2 }}}\),    \\
  \label{CE} &\norm{\mathcal{E}^\alpha(v)}\leq\Lambda\( {1 \over c_{0}}, |h |_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{  \partial_z v }_{{\mathbb Y}^{\frac{m}{2}}} \)
   \( \norm{ \partial_z v}_{{\mathbb X}^{m-1}} + |h|_{{\mathbb X}^{m-1,{1 \over 2 }}}\),
  \\\label{CT}  &\norm{ \mathcal{C}^\alpha (\mathcal{T}) }  \leq \Lambda \( {1 \over c_{0}}, |h|_{{\mathbb Y}^{\frac{m}{2}+1}} +  \norm{   v
  }_{{\mathbb Y}^{\frac{m}{2}}}  +  \norm{  \partial_z v }_{{\mathbb Y}^{\frac{m}{2}}}   \)  \( \norm{v}_{{\mathbb X}^{m-1}}+\norm{ \partial_z v}_{{\mathbb X}^{m-1}}+
  \abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}\),
 \end{align}
and $\mathcal{D}^\alpha (S^\varphi v ) $ is given by
$$\mathcal{D}^\alpha\big( S^\varphi v \big)_{ij}= 2 \,  \mathcal{C}_{j}^\alpha \big( S^\varphi v)_{ij}. $$

  \end{lem}
  It should be noted  that the commutator $\mathcal{D}^\alpha(S^\varphi v)$ will be estimated later by using the integration by parts.
  \begin{proof}
First, the equations  \eqref{eqValpha}--\eqref{divValpha}  follows from applying $Z^\alpha$ to the equations \eqref{NSv}. Indeed,  \eqref{com122}
implies that
  \begin{equation}
  \label{comp1}
   Z^\alpha \nabla^\varphi q= \nabla^\varphi Q^\alpha  +\partial_{z}^\varphi \nabla^\varphi q   Z^\alpha \eta
   + \mathcal{C}^\alpha(q),
   \end{equation}
   where $  \mathcal{C}^\alpha(q)= \( \mathcal{C}_{1}^\alpha(q) ,\mathcal{C}_{2}^\alpha (q) ,
   \mathcal{C}_{3}^\alpha(q)\)^t$, and
\begin{equation}
\label{div1}
Z^\alpha  \nabla^\varphi \cdot v  = \nabla^\varphi\cdot V^\alpha + \partial_{z}^\varphi \nabla^\varphi\cdot v Z^\alpha \eta
 + \mathcal{C}^\alpha(d),
\end{equation}
where
$ \mathcal{C}^\alpha(d) =  \sum_{i=1}^3 \mathcal{C}^\alpha_{i}(v_i).$

Next, note that
   \begin{equation}
   \label{transportW} \partial^{\varphi}_{t} + v \cdot \nabla^\varphi
   =  \partial_{t}+ v_{y} \cdot \nabla_{y}v +  V_{z} \partial_{z},\end{equation}
 where $V_{z}$ is defined by
   \begin{equation}
   \label{Wdef}
    V_{z}= {1\over \partial_{z} \varphi} v_{z}\ \text{ with }v_z=    v\cdot {\bf N}- \partial_{t} \varphi =  v\cdot{\bf N}- \partial_{t} \eta .
   \end{equation}
By using \eqref{transportW}, one can thus get that
\begin{align}  \label{T1}    Z^\alpha \( \partial^{\varphi}_{t} + v \cdot \nabla^\varphi   \)v
&    =   \(\partial_{t}+ v_{y} \cdot \nabla_{y} +  V_{z} \partial_{z} \) Z^\alpha v + \big(v\cdot Z^\alpha {\bf N}-
\partial_{t} Z^\alpha \eta \big) \partial^{\varphi}_{z} v
\\\nonumber&\quad- \partial^{\varphi}_{z} Z^\alpha \eta \big( v \cdot {\bf N} - \partial_{t} \eta\big)
 \partial^{\varphi}_{z}v
+ \mathcal{C}^\alpha(\mathcal{T}) \\
\nonumber & =  \big( \partial^{\varphi}_{t}+ v \cdot \nabla^\varphi\big)Z^\alpha v - \partial^{\varphi}_{z}v \big( \partial^{\varphi}_{t}+ v \cdot \nabla^\varphi) Z^\alpha \eta
 +  \mathcal{C}^\alpha(\mathcal{T})
 \\
\nonumber & =  \big( \partial^{\varphi}_{t}+ v \cdot \nabla^\varphi\big)V^\alpha + \partial^{\varphi}_{z} \( \partial^{\varphi}_{t}+ v \cdot \nabla^\varphi\)v Z^\alpha \eta
-\partial^{\varphi}_{z} v \cdot \nabla^\varphi v Z^\alpha \eta
 +  \mathcal{C}^\alpha(\mathcal{T}),
 \end{align}
 where the commutator $\mathcal{C}^\alpha(\mathcal{T})$ is defined by
 \begin{align}\label{calphat}
 \mathcal{C}^\alpha(\mathcal{T})= &  \[Z^\alpha, v_{y}\]\partial_{y} v+ \[Z^\alpha, V_{z},\partial_{z}v\]+
   \[Z^\alpha, v_{z}, {1 \over \partial_{z} \varphi}\] \partial_{z} v+ {1 \over \partial_{z}\varphi} \[Z^\alpha, v
   \]\cdot{\bf N}\partial_{z}v,\\\nonumber
       & +v_z\partial_zv\[Z^{\alpha-\alpha'}, { 1 \over (\partial_{z} \varphi)^2}\]Z^{\alpha'}\partial_{z} \eta+ V_{z} \[Z^\alpha,
       \partial_{z}\]v+{ v_z\partial_zv \over (\partial_{z} \varphi)^2}\[Z^{\alpha},\partial_z\]  \eta
 \end{align}
 for any $|\alpha'|=1$.

  It remains to  compute $ \varepsilon Z^\alpha \Delta^\varphi v=  2\varepsilon Z^\alpha \nabla^\varphi \cdot  (S^\varphi v).$ Note that
   $$ 2  Z^\alpha \nabla^\varphi \cdot  (S^\varphi v) =  2 \nabla^\varphi \cdot \big( Z^\alpha \, S^\varphi v \big) - 2 \big( \partial^{\varphi}_{z}\, S^\varphi
   v \big) \nabla^\varphi
     ( Z^\alpha \eta) + \mathcal{D^\alpha}\big( S^\varphi v \big)$$
with $ \mathcal{D}^\alpha\big( S^\varphi v \big)_{i}= 2 \,  \mathcal{C}_{j}^\alpha \big( S^\varphi v)_{ij},$ and
    \begin{equation}
    \label{comS} 2 Z^\alpha  \big( S^\varphi v \big) =  2 S^\varphi \big( Z^\alpha v \big) - \partial^{\varphi}_{z} v \otimes \nabla^\varphi Z^\alpha \eta
     - \nabla^\varphi Z^\alpha \eta
  \otimes \partial^{\varphi}_{z} v  + \mathcal{E}^\alpha(v)\end{equation}
     with $\big( \mathcal{E}^\alpha (v)\big)_{ij}= \mathcal{C}^\alpha_{i}(v_{j})+ \mathcal{C}^\alpha_{j}(v_{i}).$
Hence one deduces that
     \begin{align}
     \label{deltaexp}
       \varepsilon Z^\alpha \Delta^\varphi v  & = 2 \, \varepsilon\, \nabla^\varphi \cdot  S^\varphi(Z^\alpha v )  - 2 \varepsilon \nabla^\varphi \cdot \Big(  \partial^{\varphi}_{z}
       v \otimes \nabla^\varphi Z^\alpha \eta
     - \nabla^\varphi Z^\alpha \eta  \otimes \partial^{\varphi}_{z} v \Big)\\
    \nonumber      &\quad -
      2  \varepsilon \big( \partial^{\varphi}_{z}\, S^\varphi v \big) \nabla^\varphi
     ( Z^\alpha \eta)  +   \varepsilon  \mathcal{D^\alpha}\big( S^\varphi v \big) + \varepsilon  \nabla^\varphi \cdot \big( \mathcal{E}^\alpha( v)\big)
     \\
    \nonumber      & =2 \, \varepsilon\, \nabla^\varphi \cdot  S^\varphi V^\alpha -
      2  \varepsilon \big( \partial^{\varphi}_{z} \nabla^\varphi\cdot S^\varphi v \big)
      Z^\alpha \eta   +   \varepsilon  \mathcal{D^\alpha}\big( S^\varphi v \big) + \varepsilon  \nabla^\varphi \cdot \big( \mathcal{E}^\alpha(
      v)\big).\end{align}
Consequently, these and  \eqref{NSv}  imply \eqref{eqValpha}--\eqref{divValpha}.

Now we estimate these commutators. First, thanks to Lemma \ref{comi},  the estimates \eqref{Cq}--\eqref{CE} hold.
To estimate the  commutator $\mathcal{C}^\alpha(\mathcal{T})$ defined by \eqref{calphat}, one needs to bound $v_{z}$ and  $V_{z}$. It follows
from  \eqref{etainfty} that
 \begin{equation}\label{vz1}
 \norm{v_{z}}_{{\mathbb Y}^{\frac{m}{2}}} +\norm{V_{z}}_{{\mathbb Y}^{\frac{m}{2}}}  \leq \Lambda\({1 \over c_{0}},  \norm{v }_{ {\mathbb Y}^{\frac{m}{2}}}+  \norm{\nabla
 \eta }_{ {\mathbb Y}^{\frac{m}{2}}} + \norm{  \partial_{t} \eta }_{ {\mathbb Y}^{\frac{m}{2}}} \)\leq \Lambda\({1 \over c_{0}},  \norm{v }_{ {\mathbb Y}^{\frac{m}{2}}}+
 \abs{  h }_{ {\mathbb Y}^{\frac{m}{2}+1}} \).\end{equation}
And  \eqref{gues}, \eqref{etaharm}, and \eqref{etainfty} yield that
     \begin{align}
 \label{Zpetitvz}
      \norm{Z v_{z}}_{{\mathbb X}^{m-2}} &  \lesssim   \norm{Z \partial_{t} \eta }_{{\mathbb X}^{m-2}}+ \(1+  \norm{\nabla \eta}_{{\mathbb Y}^{\frac{m-2}{2}}}
      \)\norm{v}_{{\mathbb X}^{m-1}} + \norm{v}_{{\mathbb Y}^{\frac{m-2}{2}}} \norm{\nabla \eta }_{{\mathbb X}^{m-1}}\\
   \nonumber & \le
       \Lambda\( \norm{v}_{{\mathbb Y}^{\frac{m-2}{2}}} +  \abs{h}_{{\mathbb Y}^{\frac{m}{2}}}   \) \(\norm{v}_{{\mathbb X}^{m-1}} +\abs{h }_{{\mathbb X}^{m ,-\frac{1}{2}}} \).\end{align}
With \eqref{Zpetitvz} and \eqref{vz1} in hand, by using \eqref{gues}, \eqref{quot}, \eqref{etaharm} and  \eqref{etainfty}, one can obtain that
\begin{align}\label{Zvzm-2prov}
\norm{ Z V_{z}}_{{\mathbb X}^{m-2}} &\lesssim \norm{Z\( {1\over \partial_{z} \varphi} \)  v_{z}}_{{\mathbb X}^{m-2}} +  \norm{{1\over \partial_{z} \varphi} Zv_{z}
}_{{\mathbb X}^{m-2}}
\\\nonumber&\leq \Lambda\({1 \over c_{0}},  \norm{v }_{ {\mathbb Y}^{\frac{m}{2}}}+  \abs{  h }_{ {\mathbb Y}^{\frac{m}{2}+1}} \)\(\norm{v}_{{\mathbb X}^{m-1}} +\abs{h
}_{{\mathbb X}^{m ,-\frac{1}{2}}} \).
\end{align}
Consequently, one may use \eqref{com}, \eqref{comsym}, \eqref{idcom} and \eqref{quot} combined with \eqref{vz1}--\eqref{Zvzm-2prov}
and also again \eqref{etaharm}, \eqref{etainfty}, similarly as in the proof of Lemma \ref{comi}, to conclude the estimate \eqref{CT}.
  \end{proof}

\subsection{Boundary conditions}

We shall now also compute the boundary conditions satisfied   by $(Z^\alpha v,Z^\alpha q,Z^\alpha h)$ when $\alpha_{3} = 0$ (for $\alpha_{3}\neq
0$,
$Z^\alpha v= 0$ on the boundary). As a preliminary, one has one has the following.

\begin{lem} \label{lembord}
For $k\in \mathbb{N}$:
\begin{equation}\label{dzvb}
\abs{\nabla v }_{{\mathbb X}^{k }}   \leq    \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{k}{2}+1}}   + \norm{  \nabla v }_{{\mathbb Y}^{\frac{k}{2}}} \)  \(
\abs{  v }_{{\mathbb X}^{k, 1}}+\abs{h }_{{\mathbb X}^{k, 1}} \)
\end{equation}
and
\begin{equation}\label{dzvb2}
\abs{\nabla v }_{{\mathbb X}^{k,s}}\le\Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{k}{2}+2}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+1}}\)\(\abs{v
}_{{\mathbb X}^{k,s+1}}+\abs{h }_{{\mathbb X}^{k, s+1}} \)\text{ for }s=-\frac{1}{2},\frac{1}{2}.
\end{equation}
\end{lem}
\begin{proof}
Note that it suffices to prove the estimates for $\partial_z v$. Since $\nabla^\varphi \cdot  v=0$, thus
\begin{equation}\label{eqdzvn}
\partial_{z} \varphi \(\partial_{1} v_{1}+ \partial_{2} v_{2}\)+\partial_{z} v \cdot  {\bf N}=0.
\end{equation}
Then for $s=-\frac{1}{2},\frac{1}{2}$,  \eqref{gues2} implies that
\begin{align}\label{dzvnb1}
\abs{\partial_{z}v \cdot {\bf n}}_{{\mathbb X}^{k,s}} &\le  \Lambda\( {1 \over c_{0}}, \abs{\nabla \eta}_{{\mathbb Y}^{\frac{k}{2}+1}}   + \norm{  \nabla_y v
}_{{\mathbb Y}^{\frac{k}{2}+1}} \)
 \(\abs{v}_{{\mathbb X}^{k,s+1}}+  \abs{\nabla \eta}_{{\mathbb X}^{k,s}}\)
 \\\nonumber&\le   \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}}   + \norm{  \nabla_y v }_{{\mathbb Y}^{\frac{k}{2}+1}} \)
 \(\abs{v}_{{\mathbb X}^{k,s+1}}+  \abs{h}_{{\mathbb X}^{k,s+1}}\).
 \end{align}
where the second inequality follows from Lemma \ref{propeta} and the trace estimate \eqref{trace}.

To bound $\Pi \partial_{z}v $, we shall use the boundary conditions in \eqref{NSv} which yield
\begin{equation} \label{bordpi}
\Pi\(S^\varphi v {\bf n}-\kappa\chi v\)=0.
\end{equation}
To compute $\Pi \(S^\varphi v{\bf n}\)$, one can use the local basis $(\partial_{y^1}, \partial_{y^2}, \partial_{y^3})$ in $\Omega_{t}$ induced by
\eqref{diff}. The induced riemannian metric is given by $g_{ij}= \partial_{y^i}\cdot \partial_{y^j}$, whose inverse denoted by $g^{ij}$. It
follows
from the definition and \eqref{vdef} that $(\partial_{y^i}u)(t,  \Phi(t, \cdot))= \partial_{i}v.$ Hence,
\begin{equation}\label{Su}
2 Su {\bf n}={\bf n}\cdot \nabla u+\nabla u_k{\bf n}_k= \partial_{{\bf n}}u + g^{ij} \partial_{y^j} u \cdot {\bf n}  \partial_{y^i}= \partial_{{\bf n}}u + g^{ij}  \partial_{j}
v \cdot {\bf n}  \partial_{y^i}.
\end{equation}
Note also that
\begin{equation}\label{eqdzvPi0}
\partial_{{\bf n}}u   =\frac{{\bf N}}{\abs{{\bf N}}}\cdot  \nabla^\varphi v
= {\abs{{\bf N}} \over \partial_{z} \varphi} \partial_{z}v- \partial_{1} \varphi \partial_{1} v - \partial_{2}\varphi \partial_{2} v,
\end{equation}
one then gets from \eqref{bordpi} that
\begin{equation}\label{eqdzvPi'}
\Pi \partial_{z}v={\partial_{z} \varphi \over \abs{{\bf N}}}\(\partial_{1}\varphi\Pi \partial_{1} v + \partial_{2}\varphi \Pi \partial_{2} v-g^{ij}
\partial_{j}v \cdot {\bf n} \Pi \partial_{y^i}-\kappa\chi\Pi v \).
\end{equation}
Hence, using \eqref{gues2} again shows that
\begin{align}\label{dzvnb2}
\abs{\Pi \partial_{z}v }_{{\mathbb X}^{k,s}} &\le  \Lambda\( {1 \over c_{0}}, \abs{\nabla \varphi}_{{\mathbb Y}^{\frac{k}{2}+1}}   + \norm{  \nabla  v
}_{{\mathbb Y}^{\frac{k}{2}+1}} \)
 \(\abs{v}_{{\mathbb X}^{k,s+1}}+  \abs{h}_{{\mathbb X}^{k,s+1}}+  \abs{\partial_{z}v \cdot {\bf n}}_{{\mathbb X}^{k,s}} \)
 \\\nonumber&\le   \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}}   + \norm{  \nabla  v }_{{\mathbb Y}^{\frac{k}{2}+1}} \)
 \(\abs{v}_{{\mathbb X}^{k,s+1}}+  \abs{h}_{{\mathbb X}^{k,s+1}}\).
 \end{align}
where in the second inequality \eqref{dzvnb1} has been used.

Consequently, we conclude the estimate \eqref{dzvb2} by combining \eqref{dzvnb1} and \eqref{dzvnb2}. And \eqref{dzvb} follows similarly by using
\eqref{gues}
instead of \eqref{gues2}.
   \end{proof}

We now study the dynamic boundary condition on $\{z=0\}$ and the Navier slip boundary condition on $\{z=-b\}$.
  \begin{lem}
  \label{lembordV}
  For $1 \leq |\alpha | \leq m$ such that $\alpha_{3}= 0$, it holds that on $\{z=0\}$
    \begin{align}\label{bordV}
  &2 \varepsilon S^\varphi V^\alpha\, {\bf N}  -\( Z^\alpha q  -  g  Z^\alpha h+\sigma Z^\alpha H \){\bf N}
   \\ \nonumber&\quad= - 2 \varepsilon S^\varphi v \Pi Z^\alpha {\bf N}  - 2 \varepsilon Z^\alpha h \partial^{\varphi}_{z}\big( S^\varphi v \big){\bf N}+\varepsilon
   \mathcal{C}^\alpha(\mathcal{B}_\varepsilon) ,
  \end{align}
  where the commutator $\mathcal{C}^\alpha(\mathcal{B}_\varepsilon)$ satisfies the estimate:
  \begin{equation}
  \label{CalphaB}
 \abs{ \mathcal{C}^\alpha(\mathcal{B}_\varepsilon) }_{0} \leq    \Lambda\( {1 \over c_{0}}, \abs{ h }_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{  \nabla v
 }_{{\mathbb Y}^{\frac{m}{2}}} \) \( \abs{  v }_{{\mathbb X}^{m-1,1}} + \abs{ h}_{{\mathbb X}^{m-1,1}}\)
    .
  \end{equation}
Similarly, on $\{z=-b\}$ one has that
  \begin{equation}\label{bord222}
  V^\alpha_3 = 0,\quad (S^\varphi V^\alpha e_3)_i=  \kappa V^\alpha_i-\mathcal{E}^\alpha(v)_{i3},\ i=1,2,  \end{equation}
  and
\begin{equation}\label{eeest}   \abs{\mathcal{E}^\alpha(v) }_{0} \leq  \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{m}{2}+1}}   +\norm{  \nabla v
}_{{\mathbb Y}^{\frac{m}{2}}} \) \( \abs{ v }_{{\mathbb X}^{m-1,1}} +  \abs{h}_{{\mathbb X}^{m-1,1}}\).\end{equation}
  \end{lem}

 \begin{proof}
 Applying $Z^\alpha$ to the dynamic boundary condition and using \eqref{comS}, one gets
 \begin{align}
 \label{bordV1}
  \nonumber&\varepsilon \( 2  S^\varphi \( Z^\alpha v \)  -  \partial^{\varphi}_{z} v \otimes \nabla^\varphi Z^\alpha \varphi
     -  \nabla^\varphi Z^\alpha v \otimes \partial^{\varphi}_{z} v   + \varepsilon \mathcal{E}^\alpha(v) \) {\bf N}- \( Z^\alpha q - g Z^\alpha h+\sigma Z^\alpha H\) {\bf N}
     \\&\quad =-
     \( 2 \varepsilon S^\varphi v - (q- gh+\sigma H )I \) Z^\alpha {\bf N}   - \[Z^\alpha,2\varepsilon S^\varphi v-(q-gh+\sigma H)I, {\bf N}\].\nonumber
     \\&\quad =- 2 \varepsilon\(  S^\varphi v - S^\varphi v {\bf n} \cdot {\bf n} I \) Z^\alpha {\bf N}   - 2\varepsilon \[Z^\alpha, S^\varphi v-S^\varphi v {\bf n} \cdot {\bf n} I,
     {\bf N}\]
     \\&\quad\equiv- 2 \varepsilon  S^\varphi v \Pi  Z^\alpha {\bf N}   - 2\varepsilon \[Z^\alpha,   S^\varphi v \Pi  , {\bf N}\].\nonumber\end{align}
This yields \eqref{bordV} with the commutator $\mathcal{C}^\alpha(\mathcal{B}_\varepsilon)$ defined by
   \begin{equation}
 \mathcal{C}^\alpha(\mathcal{B}_\varepsilon) =- \mathcal{E}^\alpha(v)  {\bf N}- 2  \[Z^\alpha,   S^\varphi v \Pi  , {\bf N}\].
  \end{equation}
Similarly, applying $Z^\alpha$ to the Navier slip boundary condition and noting that $V^\alpha=Z^\alpha v$, one shows \eqref{bord222}.

Now, it follows from \eqref{comsym}, \eqref{etainfty} and Lemma \ref{lembord} that
   \begin{align}\label{CBe1}
  \abs{\[Z^\alpha,   S^\varphi v \Pi  , {\bf N}\]}_{0}
  &  \lesssim  \abs{ Z(   S^\varphi v \Pi  )}_{{\mathbb X}^{m-2}} \abs{Z {\bf N}}_{{\mathbb Y}^{\frac{m-2}{2}}}+ \abs{Z(   S^\varphi v \Pi  )}_{{\mathbb Y}^{\frac{m-2}{2}}}
  \abs{Z {\bf N}}_{{\mathbb X}^{m-2}}
    \\\nonumber& \le \Lambda\( {1 \over c_{0}}, \norm{  \nabla \eta}_{{\mathbb Y}^{\frac{m}{2}}}   + \norm{  \nabla v }_{{\mathbb Y}^{\frac{m}{2}}} \) \(
    \abs{\nabla v }_{{\mathbb X}^{m-1}} + \abs{h}_{{\mathbb X}^{m-1,1}}\)
    \\\nonumber&   \le \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m}{2}+1}}   + \norm{  \nabla v }_{{\mathbb Y}^{\frac{m}{2}}} \) \(  \abs{  v
    }_{{\mathbb X}^{m-1,1}}  + |h|_{{\mathbb X}^{m-1,1}}\).
      \end{align}
On the other hand, following the proof of Lemma \ref{comi} and using again Lemma \ref{lembord}, one has
\begin{align}\label{CBe2}\abs{\mathcal{E}^\alpha(v)}_{0}& \leq  \Lambda\( {1 \over c_{0}}, \norm{\nabla\eta}_{{\mathbb Y}^{\frac{m}{2}}}   +\norm{  \nabla
v }_{{\mathbb Y}^{\frac{m}{2}}} \) \(  \abs{\nabla v}_{{\mathbb X}^{m-1}} +  \abs{\nabla \eta}_{{\mathbb X}^{m-1}}\)
 \\\nonumber&\leq  \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{m}{2}+1}}   +\norm{  \nabla v }_{{\mathbb Y}^{\frac{m}{2}}} \) \( \abs{ v }_{{\mathbb X}^{m-1,1}}
 +  \abs{h}_{{\mathbb X}^{m-1,1}}\).
 \end{align}
Consequently, the estimates \eqref{eeest} and \eqref{CalphaB} follows.
 \end{proof}

Finally, we study the kinematic boundary condition on $\{z=0\}$.
\begin{lem}
\label{lembordh}
  For $1 \leq |\alpha | \leq m$ such that $\alpha_{3}= 0$, it holds that on $\{z=0\}$
  \begin{equation}
  \label{bordh}
  \partial_{t} Z^\alpha h + v_y \cdot\nabla_y Z^\alpha h - V^\alpha \cdot {\bf N} =-\partial^{\varphi}_zv\cdot{\bf N} Z^\alpha h+\mathcal{C}^\alpha(h)
  \end{equation}
  where the commutator $\mathcal{C}^\alpha(h)$ satisfies
  \begin{equation}
  \label{bordhC}
 \abs{ \mathcal{C}^\alpha(h)}_0\leq \Lambda\( {1 \over c_{0}},      \abs{h}_{{\mathbb Y}^{\frac{m}{2}+1}} + \norm{v}_{{\mathbb Y}^{\frac{m}{2}}}\) \(
 \abs{h}_{{\mathbb X}^{m-1,1}} +\abs{ v}_{{\mathbb X}^{m-1}}
\) .
\end{equation}
Moreover,
\begin{equation}\label{chtilde}
\mathcal{C}^\alpha(h)= \sum_{|\alpha'|=1}C_{\alpha}^{\alpha'}Z^{\alpha-\alpha'} v_y\cdot \nabla_{y} Z^{ \alpha'}h +
\tilde{\mathcal{C}}^\alpha(h),
\end{equation}
 and $\tilde{\mathcal{C}}^\alpha(h)$  satisfies the estimate:
  \begin{equation}
  \label{bordhC1}
 \abs{ \tilde{\mathcal{C}}^\alpha(h)}_1\leq \Lambda\( {1 \over c_{0}},      \abs{h}_{{\mathbb Y}^{\frac{m}{2}+2}} +  \abs{v}_{{\mathbb Y}^{\frac{m}{2}+1}}\) \(
 \abs{h}_{{\mathbb X}^{m-1,2}} +\abs{ v}_{{\mathbb X}^{m-2,1}}
\) .
\end{equation}
\end{lem}
\begin{proof}
Applying $Z^\alpha$ to the kinematic boundary condition yields
\begin{equation}
  \partial_{t} Z^\alpha h + v_y \cdot\nabla_y Z^\alpha h -Z^\alpha v \cdot {\bf N}=    {\mathcal{C}}^\alpha(h)
  \end{equation}
where
\begin{equation}
{\mathcal{C}}^\alpha(h)=- \[ Z^\alpha, v_{y}, \nabla_{y}h\].
  \end{equation}
This yields \eqref{bordh}, and the estimate \eqref{bordhC} follows from \eqref{comsym}.
One may further single out the highest order derivative terms according to \eqref{chtilde}, where $\tilde{\mathcal{C}}^\alpha(h)$ is defined by
\begin{equation}\label{tildech} \tilde{\mathcal{C}}^\alpha(h) = -\sum_{\beta+\gamma=\alpha\atop \beta\neq0,|\gamma|\ge 2}C_{\beta,\gamma} Z^{\beta} v_y\cdot
\nabla_{y} Z^{ \gamma}h  .\end{equation}
And the estimate \eqref{bordhC1} follows by using \eqref{gues}.
\end{proof}

\section{Pressure estimates}\label{sectionpressure}

In view of the equation \eqref{eqValpha}, one needs to estimate the pressure $q$. The first equation in \eqref{NSv} implies that
\begin{equation}\label{eqq} \Delta^\varphi q = - \nabla^\varphi \cdot (v\cdot \nabla^\varphi v) \text{ in }\Omega.\end{equation}
 Moreover, the dynamic boundary condition gives
\begin{equation}\label{boundary0d} q  =  2 \varepsilon S^\varphi v  \,{\bf n} \cdot \,{\bf n} + gh-\sigma H \text{ on }\{z=0\}.
\end{equation}
Projecting the first equation in \eqref{NSv} along ${\bf N}$ onto $\{z=0\}$ and $\{z=-b\}$ yields
\begin{equation}\label{boundary0}\nabla^\varphi q\cdot{\bf N} = -\partial_t v\cdot {\bf N}-(v_y\cdot\nabla_y) v\cdot {\bf N}+\varepsilon   \Delta^\varphi v \cdot{\bf N} \text{ on }\{z=0\}
\end{equation}
and
\begin{equation}\label{boundaryb}\nabla^\varphi q\cdot{\bf N} =  \varepsilon   \Delta^\varphi v \cdot{\bf N} \text{ on }\{z=-b\}.\end{equation}
Here in \eqref{boundaryb} one has used the fact that ${\bf N}=e_3$ and $v_3=0$ on $\{z=-b\}$.

Note that to solve the pressure, one has two choices of boundary conditions on $\{z=0\}$, $i.e.,$ \eqref{boundary0d} and \eqref{boundary0}.
Without surface tension, one can use the elliptic problem \eqref{eqq}, \eqref{boundary0d} and \eqref{boundaryb} to establish the regularity
estimates for $q$. The subtlety lies in that the energy dissipation estimates of \eqref{NSv} in the case without surface tension  provide the
needed
estimates for those boundary terms. When there is surface tension, however, the energy dissipation estimates do not provide enough
estimates for the boundary term $ -\sigma H$ (which is of one half regularity less). This would suggest that the elliptic problem \eqref{eqq},
\eqref{boundary0d} and \eqref{boundaryb} is not the right choice for estimating the pressure $q$ in the case with surface tension. Our way to get
around this difficulty is to use instead the elliptic problem \eqref{eqq}, \eqref{boundary0} and \eqref{boundaryb}. It is then noticed that this
approach forces one to estimate the time derivatives of $v$, that is, one needs to perform energy estimates for the time derivatives of the
solution.
However, there is an essential difficulty arising: when doing energy estimates with time derivatives up to $m$ order,  we can only obtain the
estimates of time derivatives of $q$ up to $m-1$ order due to the presence of $\partial_t v$ in \eqref{boundary0}. Thus the energy estimates cannot be
closed since it seems that $m$ order time derivative of $q$ is involved. We will explain this and our way to overcome it in more details in
Section
\ref{secconormal2}.

It follows from the definition of $\partial_i^\varphi$ that
 \begin{equation}
 \label{graddiv} \nabla^\varphi \cdot v= {1 \over \partial_{z}\varphi}  \nabla\cdot \big( P v \big) , \quad  \nabla^\varphi f= {1 \over
 \partial_{z} \varphi} P^* \nabla f,
  \quad P= \left( \begin{array}{ccc}  \partial_{z} \varphi & 0 & 0 \\ 0 & \partial_{z} \varphi & 0 \\ -\partial_{1} \varphi  & - \partial_{2}
  \varphi & 1 \end{array}
   \right).\end{equation}
And then $\Delta^\varphi$ can be expressed as
\begin{equation}
\label{deltaphi} \Delta^{\varphi}f = \nabla^\varphi \cdot (\nabla^\varphi f) = {1 \over \partial_{z} \varphi} \nabla \cdot \big( E \nabla f
\big)\end{equation}
 with the matrix $E$ defined by
 $$ E =  \frac1{\partial_{z} \varphi} PP^* \equiv \left( \begin{array}{ccc} \partial_{z} \varphi & 0 & {- \partial_{1} \varphi} \\
  0 & \partial_{z} \varphi & -\partial_{2} \varphi  \\ -\partial_{1} \varphi & -\partial_{2} \varphi &
   {1 + (\partial_{1} \varphi)^2} + (\partial_{2} \varphi)^2 \over \partial_{z} \varphi \end{array} \right).
  $$
   Note that $E$ is symmetric positive and that if
   $ \| \nabla_y \varphi \|_{L^\infty} \leq { 1 \over c_{0}} $ and
 $  \partial_{z} \varphi \geq c_{0}>0$ then
   there exists  $\delta(c_{0})>0$ such that
   \begin{equation}
   \label{Eminor}
   E X \cdot X \geq \delta |X|^2, \quad \forall X \in \mathbb{R}^3.
   \end{equation}
   Moreover,
   \begin{equation}
   \label{ELinfty}
\norm{ E}_{\tilde{{\mathbb Y}}^{k}} \leq \Lambda\({1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+1}}\).
   \end{equation}
One can write
   $$ E= \mbox{Id}  + \tilde{E}, \quad
\tilde  E= \left( \begin{array}{ccc}  \partial_{z}\eta & 0 & {- \partial_{1} \eta} \\
  0 &  \partial_{z}\eta  & -\partial_{2} \eta  \\ -\partial_{1} \eta & -\partial_{2} \eta &
   { (\partial_{1} \eta)^2} + (\partial_{2} \eta)^2  - \partial_{z} \eta  \over \partial_{z} \varphi \end{array} \right), $$
where
 \begin{equation}
 \label{EHs} \norm{ \tilde{E}}_{\tilde{{\mathbb X}}^{k}} \leq \Lambda\({1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{\frac{k}{2}+1}} \) \abs{h}_{{\mathbb X}^{k,\frac{1}{2}}} .
 \end{equation}
Here $\tilde{{\mathbb X}}^k$ and $\tilde{{\mathbb Y}}^k$ are referred to the usual {\it spatial-time} Sobolev spaces as defined similarly as \eqref{xym}.

Since $ {\bf N}=  P^\ast e_3$ on $\{z=0,-b\}$, the equations \eqref{eqq}--\eqref{boundaryb} can be rewritten as
 \begin{alignat}{3}
 \label{eqq12} &- \nabla \cdot \big( E \nabla q\big) =F:=  \partial_{z}\varphi \nabla^\varphi v \cdot \nabla^\varphi v  &&\quad\text{in
 }\Omega,\\
\label{boundary0d12}& q  =G^1:=  2 \varepsilon S^\varphi v  \,{\bf n} \cdot \,{\bf n} + gh-\sigma H &&\quad\text{on }\{z=0\},
\\\label{boundary012}&{E\nabla q\cdot e_3} =G^2:= -\partial_t v\cdot {\bf N}-(v_y\cdot\nabla_y) v\cdot {\bf N}+\varepsilon   \Delta^\varphi v \cdot{\bf N} &&\quad\text{on
}\{z=0\},
\\\label{boundaryb12}&{E\nabla q\cdot e_3} =G^3:= \varepsilon   \Delta^\varphi v \cdot{\bf N} &&\quad\text{on }\{z=-b\}.
\end{alignat}

We shall now prove the estimates for the pressure $q$.
  \begin{prop}\label{proppressure}
The following estimates hold:
    \begin{align}\label{qpressureest}
  &    \norm{   q }_{{\mathbb X}^{k}}+\norm{ \nabla q }_{{\mathbb X}^{k}}+\norm{ \partial_{zz} q}_{{\mathbb X}^{k-1}}
  \\\nonumber&  \quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\abs{h}_{{{\mathbb X}^{{\frac{k+7 }{2}}}}}  +\norm{  v
  }_{{\mathbb Y}^{\frac{k+3}{2}}}  +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{v}_{{\mathbb X}^{\frac{k+7}{2}}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{k+5}{2}}}  \)
  \\\nonumber&\qquad\times\(\abs{h}_{{\mathbb X}^{k+1,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{k,2}}+\norm{v}_{{\mathbb X}^{k+1}}  + \norm{\nabla v}_{{\mathbb X}^{k}} +
  \varepsilon\abs{  v }_{{{\mathbb X}^{{k}, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\), \text{ for } k\ge 3;
  \end{align}
\begin{align} \label{qinfty}
&\norm{   q }_{{\mathbb Y}^{k}}+\norm{ \nabla q }_{{\mathbb Y}^{k}}+ \norm{ \partial_{zz} q }_{{\mathbb Y}^{k-1}}
\\\nonumber&\quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{k}{2}+4}}  +\abs{h}_{{\mathbb X}^{k+4}}+\norm{  v
}_{{\mathbb Y}^{\frac{k+5}{2}}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+3}} +  \norm{v}_{{\mathbb X}^{k+4}}  + \norm{\nabla v}_{{\mathbb X}^{k+3}}   \), \text{ for }k\ge 1.
\end{align}
\end{prop}
     \begin{proof}
Multiplying the equation \eqref{eqq12} by $q$  and then integrating by parts over $\Omega$, using
\eqref{boundary012} and \eqref{boundaryb12}, one obtains
  \begin{equation}
  \(E \nabla q,\nabla q\)_{\Omega} =\(F, q\)_{\Omega}+\(G^2, q\)_{z=0}-\(G^3, q\)_{z=-b}.
  \end{equation}
Here $\(\cdot,\cdot\)_{\Omega},\ \(\cdot,\cdot\)_{z=0}$ and $ \(\cdot,\cdot\)_{z=-b}$ denote the $L^2$ inner products on $\Omega$, $\{z=0\}$ and
$\{z=-b\}$, respectively. It follows from the trace estimate $\abs{q}_{\frac{1}{2}}\lesssim \norm{  q }_{H^1}$ and Cauchy's inequality that
  \begin{equation}
  \label{rhoH1'0}
  \norm{\nabla q }^2 \leq {1 \over \eta}\Lambda_0  \(\norm{F}^2+\abs{G^1}_{0}^2+\abs{G^2}_{-\frac{1}{2}}^2+\abs{G^3}_{-\frac{1}{2}}^2\)+\eta\norm{  q
  }_{H^1}^2
  \end{equation}
  for any $\eta>0.$
 By the Poincar\'e inequality \eqref{poin} and \eqref{boundary0d12}, one can get by taking $\eta$
  sufficiently small that
    \begin{equation}
  \label{rhoH1'}
 \norm{  q }_{H^1}  \leq  \Lambda_0  \(\norm{F} +\abs{G^1}_{0} +\abs{G^2}_{-\frac{1}{2}} +\abs{G^3}_{-\frac{1}{2}} \).
  \end{equation}
Note that if one uses solely the problem \eqref{eqq}, \eqref{boundary0d} and \eqref{boundaryb} to estimate $p$, then one needs
$\abs{G^1}_{\frac{1}{2}}$.
This half less regularity requirement enables us to control the surface tension term by the energy dissipation estimates.

   Next, applying $Z^\alpha$ with $|\alpha|=k$ to the equation \eqref{eqq12} and using \eqref{boundary012}--\eqref{boundaryb12} lead to
  \begin{equation}
  \(Z^{\alpha}(E \cdot \nabla q),\nabla Z^{\alpha} q\)_{\Omega} =\(Z^{\alpha} F, Z^{\alpha} q\)_{\Omega}+\(Z^{\alpha} G^2, Z^{\alpha}
  q\)_{z=0}-\(Z^{\alpha} G^3 Z^{\alpha} q\)_{z=-b}.
  \end{equation}
Then as for \eqref{rhoH1'}, one can derive after using \eqref{boundary0d12} that
\begin{align}
       \label{elprov3}
      \norm{   q }_{{\mathbb X}^k}+\norm{ \nabla  q }_{{\mathbb X}^k}\leq \Lambda_0  & \(
      \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}\right.
        \\\nonumber& \ + \norm{ E\cdot \[Z^\alpha , \nabla\] q } +  \norm{\[Z^\alpha,  E\]\cdot \nabla q}\Big).
      \end{align}
We then estimate the commutators in the right hand side of \eqref{elprov3}. First, \eqref{idcom} implies that
 \begin{equation}\label{al11}
  \norm{  E\cdot \[Z^\alpha , \nabla\] q} \lesssim    \norm{ E}_{L^\infty}   \norm{  \nabla q }_{{\mathbb X}^{k-1}}\lesssim \abs{h}_{ {\mathbb Y}^{1}} \norm{\nabla q
  }_{{\mathbb X}^{k-1}} . \end{equation}
However, the other communtator needs more attentions according to $1\le k\le 3$ or $k\ge 4$. Indeed, for $1\le k\le 3$, direct estimates by
controlling the $\nabla q$ terms in $L^2$ and $E$ terms in $L^\infty$ yield
    \begin{align} \label{al1122}
\norm{ \[Z^\alpha,  E\]\cdot \nabla q}    \lesssim      \norm{ {E}}_{\tilde{\mathbb Y}^{k}} \norm{\nabla q }_{{\mathbb X}^{k-1}}
\lesssim      \abs{h}_{ {\mathbb Y}^{k+1}} \norm{\nabla q }_{{\mathbb X}^{k-1}} .
      \end{align}
Plugging \eqref{al11}--\eqref{al1122} into \eqref{elprov3}, by an induction argument and \eqref{rhoH1'}, one can deduce that
\begin{align}
\label{elprov31112}
     \nonumber \norm{   q }_{{\mathbb X}^k}+\norm{ \nabla  q }_{{\mathbb X}^k}&\leq \Lambda\({1 \over c_{0}},\norm{ {E}}_{\tilde{\mathbb Y}^{3}}  \)  \(
     \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\norm{\nabla q }_{{\mathbb X}^{k-1}}\)
      \\&\leq \Lambda\({1 \over c_{0}},\abs{ h}_{{\mathbb Y}^{4}}  \)  \(
      \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}\).
              \end{align}
Since the equation \eqref{eqq12}   gives
      \begin{equation}
      \label{eqdzzrho} \partial_{zz}  q ={1 \over E_{33}} \(    F -  \partial_{z}\( \sum_{j<3}E_{3,j}\partial_{j} q  \)-
       \sum_{i<3, \, j} \partial_{i}\( E_{i j} \partial_{j} q\) \),\end{equation}
in a similar way and by \eqref{elprov31112}, one can also obtain
 \begin{align}\label{q11}
\norm{ \partial_{zz} q}_{{\mathbb X}^{k-1}} &\leq \Lambda\({1 \over c_{0}},  \norm{ E }_{\tilde{\mathbb Y}^k}\)\(\norm{  F}_{{\mathbb X}^{k-1}}+\norm{\nabla q }_{{\mathbb X}^{k}}\)
 \\\nonumber&\leq \Lambda\({1 \over c_{0}},| h|_{{\mathbb Y}^{4}} \)\(
 \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}\).
 \end{align}
It then follows from \eqref{elprov31112} and \eqref{q11} that for $1 \le k\le 3$,
  \begin{align}\label{qclaim}
  &    \norm{   q }_{{\mathbb X}^k}+\norm{ \nabla q }_{{\mathbb X}^k}+\norm{ \partial_{zz} q}_{{\mathbb X}^{k-1}}
  \\\nonumber&  \quad\leq \Lambda \({1 \over c_{0}}, \abs{ h}_{{\mathbb Y}^{4}} +\abs{ h}_{{\mathbb Y}^{\frac{k+3}{2}}}+\abs{ h}_{{\mathbb X}^{\frac{k+3
  }{2},\frac{1}{2}}}+\norm{F}_{{\mathbb X}^{\frac{k+3}{2}}} +\abs{G^1}_{{{\mathbb X}^{\frac{k+3}{2}}}}+\abs{G^2}_{{{\mathbb X}^{\frac{k+3 }{2},-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{\frac{k+3
  }{2},-\frac{1}{2}}}}\)
   \\\nonumber& \qquad\times\(\abs{ h}_{{\mathbb X}^{k,{1 \over 2}}}+
   \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}\).
  \end{align}

We now claim that \eqref{qclaim} holds for all $k\ge 1$. This will be proved by induction. Assume that $k\ge 4$
and that \eqref{qclaim} holds for $k-1$. The commutator estimate \eqref{com} yields that
\begin{align}
   \label{eses112}
   \norm{ \[Z^\alpha,  E\]\cdot \nabla q} &  \lesssim  \norm{Z{E}}_{{\mathbb X}^{k-1}} \norm{ \nabla q }_{{\mathbb Y}^{\frac{k-1}{2}}} +
   \norm{Z{E}}_{{\mathbb Y}^{\frac{k-1}{2}}} \norm{\nabla q }_{{\mathbb X}^{k-1}}  \\
     \nonumber &   \le     \Lambda \({1 \over c_{0}}, \abs{ h}_{{\mathbb Y}^{\frac{k+3}{2}}}+\norm{ \nabla q
     }_{{\mathbb Y}^{\frac{k-1}{2}}}\)\(\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}+\norm{\nabla q }_{{\mathbb X}^{k-1}}\).
      \end{align}
Plugging \eqref{eses112} and \eqref{al11} into \eqref{elprov3} and using \eqref{eqdzzrho} again lead to
\begin{align}
       \label{elprov31212}
      \norm{   q }_{{\mathbb X}^k}+\norm{ \nabla  q }_{{\mathbb X}^k}+\norm{  \partial_{zz} q}_{{\mathbb X}^{k-1}} \leq  & \Lambda \({1 \over c_{0}}, \abs{
      h}_{{\mathbb Y}^{\frac{k+3}{2}}}+\norm{ \nabla q }_{{\mathbb Y}^{\frac{k-1}{2}}}\) \Big(\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}+\norm{\nabla q }_{{\mathbb X}^{k-1}}
        \\\nonumber&\qquad \left. + \norm{F}_{{\mathbb X}^k}+\abs{G^1}_{{{\mathbb X}^{k}}}+\abs{G^2}_{{{\mathbb X}^{k,-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{k,-\frac{1}{2}}}}\).
      \end{align}
To remove the dependence of $\Lambda$ on $\norm{ \nabla q }_{{\mathbb Y}^{\frac{k-1}{2}}}$, one applies the anisotropic Sobolev embedding estimate
\eqref{emb} and the induction assumption to obtain
\begin{align}\label{gogogo}
&\norm{ \nabla q }_{{\mathbb Y}^{\frac{k-1}{2}}}\lesssim  \norm{ \partial_{z} \nabla q }_{{\mathbb X}^{\frac{k+1}{2}}} \norm{   \nabla q }_{{\mathbb X}^{\frac{k+3}{2}}}
\\\nonumber&\quad\le \Lambda \({1 \over c_{0}}, \abs{ h}_{{\mathbb Y}^{4}} +\abs{ h}_{{\mathbb Y}^{\frac{\frac{k+3}{2}+3}{2}}}+\abs{ h}_{{\mathbb X}^{\frac{\frac{k+3}{2}+3
}{2},\frac{1}{2}}}+\norm{ F}_{{\mathbb X}^{\frac{\frac{k+3}{2}+3}{2}}} \right.
\\\nonumber&\qquad\qquad\left.+\abs{G^1}_{{{\mathbb X}^{\frac{\frac{k+3}{2}+3}{2}}}}+\abs{G^2}_{{{\mathbb X}^{\frac{\frac{k+3}{2}+3
}{2},-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{\frac{\frac{k+3}{2}+3 }{2},-\frac{1}{2}}}} \)
\\\nonumber&\qquad  \times\( \abs{ h}_{{\mathbb X}^{\frac{k+3}{2},{1 \over 2}}}+  \norm{F}_{{\mathbb X}^{\frac{k+3}{2}}}
+\abs{G^1}_{{{\mathbb X}^{\frac{k+3}{2}}}}+\abs{G^2}_{{{\mathbb X}^{\frac{k+3}{2},-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{\frac{k+3}{2},-\frac{1}{2}}}} \)
\\\nonumber&\quad\le \Lambda \({1 \over c_{0}}, \abs{ h}_{{\mathbb Y}^{4}} + \abs{ h}_{{\mathbb Y}^{\frac{k+3}{2}}}+\abs{ h}_{{\mathbb X}^{\frac{k+3}{2},{1 \over 2}}}+
\norm{F}_{{\mathbb X}^{\frac{k+3}{2}}} +\abs{G^1}_{{{\mathbb X}^{\frac{k+3}{2}}}}+\abs{G^2}_{{{\mathbb X}^{\frac{k+3}{2},-\frac{1}{2}}}}+\abs{G^3}_{{{\mathbb X}^{\frac{k+3}{2},-\frac{1}{2}}}} \)
.
\end{align}
Here one has used the fact that $\frac{k+3}{2}\le k-1$ since $k\ge 4$, which allows one to use the induction assumption. Plugging the estimate
\eqref{gogogo} into \eqref{elprov31212} and using the induction assumption to estimate $\norm{\nabla q }_{{\mathbb X}^{k-1}}$, one thus concludes
 \eqref{qclaim} for all $k\ge 4$.

We now estimate the right hand side of \eqref{qclaim} for $k\ge 3$.  It follows by the product estimate \eqref{gues} that
\begin{align}\label{Fest1}
 \norm{F}_{{\mathbb X}^{k}}&=
 \norm{\partial_{z}\varphi \nabla^\varphi v \cdot \nabla^\varphi v }_{{\mathbb X}^{k}}
 \\\nonumber&\leq   \Lambda \( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{\frac{k}{2}+1}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}}}   \)
  \( \abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}  + \norm{\nabla v}_{{\mathbb X}^{k}} \).
  \end{align}
Similarly, since $k\ge 3$,
\begin{align}\label{Fest2}
\norm{F}_{{\mathbb X}^{\frac{k+3 }{2}}}
 &  \leq  \Lambda \( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{ \frac{k+3 }{4} +1}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k+3 }{4} }}   \)
  \( \abs{h}_{{\mathbb X}^{\frac{k+3 }{2} ,\frac{1}{2}}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{k+3 }{2}  }} \)
 \\\nonumber&\le \Lambda \( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{\frac{k}{2}+1}}+\abs{h}_{{\mathbb X}^{\frac{k }{2}+2}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}}} +
 \norm{\nabla v}_{{\mathbb X}^{{\frac{k +3}{2}}}}  \).
  \end{align}
By \eqref{gues} and Lemma \ref{lembord},
   \begin{align}\label{G1est1}
 \abs{ G^1 }_{{\mathbb X}^{k}} &= \abs{2 \varepsilon S^\varphi v  \,{\bf n} \cdot \,{\bf n} + gh-\sigma H}_{{\mathbb X}^{k}}
   \\\nonumber&\leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}}}  \) \( \varepsilon \abs{ \nabla v
   }_{{\mathbb X}^{k}}
    +  | h |_{{\mathbb X}^{k}}+ \sigma | h |_{{\mathbb X}^{k,2}}\)  \\\nonumber
     & \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}}}  \) \( \varepsilon \abs{   v  }_{{\mathbb X}^{k,1}}
    +  \abs{h}_{{\mathbb X}^{k}}+ \sigma \abs{h}_{{\mathbb X}^{k,2}}\) .
    \end{align}
 Similarly, since $k\ge 3$,  the trace estimate implies that
    \begin{align}\label{G1est2}
\abs{G^1}_{{{\mathbb X}^{\frac{k+3 }{2} }}}
     & \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k+3 }{4}+2}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k+3 }{4}}}  \) \( \varepsilon \abs{   v
     }_{{\mathbb X}^{\frac{k+3 }{2},1}}
    +  \abs{h}_{{\mathbb X}^{\frac{k+3 }{2}}}+ \sigma \abs{h}_{{\mathbb X}^{\frac{k+3 }{2},2}}\)
     \\\nonumber
     & \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}} +  \abs{h}_{{\mathbb X}^{\frac{k+7 }{2}}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}}}
     +\norm{   v  }_{{\mathbb X}^{\frac{k  }{2}+3}}+\norm{ \nabla  v  }_{{\mathbb X}^{\frac{k  }{2}+2}}\).
    \end{align}

To estimate the most delicate term $G^2$, we start with $\varepsilon\Delta^\varphi v\cdot{\bf N}$. Note that
    \begin{align}
\Delta^\varphi v\cdot{\bf N}=2\(\nabla^\varphi\cdot  S^\varphi v\)\cdot{\bf N} = 2\nabla^\varphi\cdot \( S^\varphi v {\bf N} \)- 2   S^\varphi v: \nabla^\varphi
{\bf N}
    \end{align}
and
    \begin{align}
 \nabla^\varphi\cdot \( S^\varphi v {\bf N} \)=\partial_1 \( S^\varphi v {\bf N} \)_1+\partial_2 \( S^\varphi v {\bf N} \)_2+ \frac{1}{\partial_z\varphi}\partial_z\( S^\varphi v {\bf N}
 \)\cdot{\bf N} .
    \end{align}
Hence by the estimate \eqref{gues2} with $s=-\frac{1}{2},\frac{1}{2}$, one can get
\begin{align}\label{g3es1}
\abs{\Delta^\varphi v\cdot{\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}&\lesssim \abs{\frac{1}{\partial_z\varphi}\partial_z\( S^\varphi v {\bf N}
\)\cdot{\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}+\abs{S^\varphi v {\bf N}}_{{{\mathbb X}^{{k},\frac{1}{2}}}}+ \abs{S^\varphi v: \nabla^\varphi {\bf N} }_{{{\mathbb X}^{{k},-\frac{1}{2}}}}
 \\\nonumber
     & \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+1}}+\norm{\partial_z\( S^\varphi v {\bf N}
     \)\cdot{\bf N}}_{{\mathbb Y}^{\frac{k}{2}+1}} \) \\\nonumber
     & \quad\times\(\abs{ \partial_z\( S^\varphi v {\bf N} \)\cdot{\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}+\abs{\nabla v
     }_{{{\mathbb X}^{{k},\frac{1}{2}}}}+\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\).
\end{align}
But
    \begin{align}
\partial_z\( S^\varphi v {\bf N} \)\cdot{\bf N}=\partial_z\( S^\varphi v {\bf N}\cdot{\bf N} \)-  S^\varphi v {\bf N}  \cdot \partial_z{\bf N},
    \end{align}
and recalling the matrices $P$ and $E$,
    \begin{align}
 \partial_z\( S^\varphi v {\bf N}\cdot{\bf N} \)&=\partial_z\( \nabla^\varphi v {\bf N}\cdot{\bf N} \)=\partial_z\(  \frac{1}{\partial_z\varphi}P^\ast \nabla v P^\ast e_3\cdot P^\ast
 e_3 \)
 \\\nonumber&
  = \nabla^\varphi (\partial_z v) {\bf N}\cdot{\bf N}  - \partial_j v_i\partial_z(E_{3i}P_{3j}), \end{align}
one can then deduce from \eqref{g3es1} that
\begin{align}\label{g3es2}
\abs{\Delta^\varphi v\cdot{\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}& \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\norm{\nabla
v}_{{\mathbb Y}^{\frac{k}{2}+1}}+\norm{\nabla^\varphi (\partial_z v) {\bf N}\cdot{\bf N} }_{{\mathbb Y}^{\frac{k}{2}+1}} \) \\\nonumber
     & \quad\times\(\abs{ \nabla^\varphi (\partial_z v) {\bf N}\cdot{\bf N} }_{{{\mathbb X}^{{k},-\frac{1}{2}}}}+\abs{\nabla v
     }_{{{\mathbb X}^{{k},\frac{1}{2}}}}+\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\).
\end{align}
Note further that
    \begin{align}
 &  \nabla^\varphi (\partial_z v) {\bf N}\cdot{\bf N}  = \nabla^\varphi \(\partial_z v\cdot {\bf N}\)\cdot{\bf N}-\({\bf N}\cdot\nabla^\varphi\){\bf N}\cdot \partial_z v, \end{align}
and one can compute by using $\nabla^\varphi\cdot v = 0$ that
    \begin{align}
 &   \nabla^\varphi \(\partial_z v\cdot {\bf N}\)\cdot{\bf N}= -\nabla^\varphi \(\partial_z\varphi(\partial_1v_1+\partial_2 v_2)\)\cdot{\bf N}.\end{align}
It follows from these, \eqref{g3es2} and Lemma \ref{lembord} that
 \begin{align}\label{gg1es1}
\abs{\Delta^\varphi v\cdot{\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}& \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\norm{\nabla
v}_{{\mathbb Y}^{\frac{k}{2}+2}}  \)  \(\abs{\nabla v }_{{{\mathbb X}^{{k},\frac{1}{2}}}}+\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\)
\\\nonumber& \leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+2}}  \)  \(\abs{  v }_{{{\mathbb X}^{{k},
\frac{3}{2}}}}+\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\).
\end{align}
Next, one easily has
 \begin{align}\label{gg1es2}
\abs{(v_y\cdot\nabla_y) v\cdot {\bf N}}_{{{\mathbb X}^{{k},-\frac{1}{2}}}}\leq   \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{
v}_{{\mathbb Y}^{\frac{k}{2}+2}}  \)  \(\abs{  v }_{{{\mathbb X}^{{k}, \frac{1}{2}}}}+\abs{h}_{{{\mathbb X}^{{k},\frac{1}{2}}}}\).
\end{align}
Finally, we estimate the remaining time derivative term $\partial_t v\cdot{\bf N}$. One first has
 \begin{align}\label{gggesrt00}
 \abs{\partial_t v\cdot{\bf N}}_{{{\mathbb X}^{k-1,\frac{1}{2}}}} & \leq\Lambda\( {1 \over c_{0}}, \abs{\nabla h }_{{\mathbb Y}^{\frac{k-1}{2}+1 }}  + \norm{ \partial_t v
 }_{{\mathbb Y}^{\frac{k-1}{2}+1}}    \)
   \(\abs{\partial_t v}_{{\mathbb X}^{k-1,\frac{1}{2}}}   + \abs{h}_{{\mathbb X}^{k-1, {3 \over 2 }}}\)
\\\nonumber&\leq\Lambda\( {1 \over c_{0}}, \abs{  h }_{{\mathbb Y}^{\frac{k+3}{2} }}  + \norm{  v }_{{\mathbb Y}^{\frac{k+3}{2}}}    \)
   \(\abs{ v}_{{\mathbb X}^{k,\frac{1}{2}}}   + \abs{h}_{{\mathbb X}^{k-1, {3 \over 2 }}}\).
\end{align}
It then suffices to estimate $\abs{\partial_t^{k }(\partial_t v\cdot{\bf N})}_{-\frac{1}{2}} $. However, this will lead to some difficulties since $k$ can be $m-1$, and
energy estimates yield only $\partial_t^{m} v\in L^2(\Omega)$ which cannot ensure the control of the
$H^{-\frac{1}{2}}\{z=0\}$ norm of $\partial_t^m v$. The key observation is that $\partial_t^m v\cdot {\bf N}$ is indeed in $H^{-\frac{1}{2}}(\{z=0\})$. The way of achieving this
is to
use the Alinhac good unknown, returning back to $k$, $V^{k+1}=\partial_t^{k+1}v-\partial^{\varphi}_zv\partial_t^{k+1}\eta$. Indeed, since $\nabla^\varphi\cdot V^{k+1}=-
C^{k+1}(d)$ with $C^{k+1}(d)$ defined as in \eqref{divValpha} with $Z^\alpha=\partial_t^{k+1}$,  by Lemma \ref{-1/2b}, and using the estimate
\eqref{Cd} and \eqref{etaharm}, one gets that
 \begin{align}
\abs{V^{k+1}\cdot {\bf N}}_{-\frac{1}{2}}&\leq   \Lambda\( {1 \over c_{0}} \)  \(\norm{ V^{k+1} }+\norm{ \nabla^\varphi\cdot V^{k+1} } \)
\\\nonumber&\leq   \Lambda\( {1 \over c_{0}},\norm{\nabla v}_{L^\infty} \)  \(\norm{\partial_t^{k+1} v}  +   \norm{\partial_t^{k+1}\eta} +\norm{ C^{k+1}(d) }
\)
\\\nonumber&\leq\Lambda\( {1 \over c_{0}}, |h |_{{\mathbb Y}^{\frac{{k+1}}{2}+1}}   + \norm{ \nabla v }_{{\mathbb Y}^{\frac{{k+1}}{2}}} \)
   \(\norm{\partial_t^{k+1} v}  + \abs{h}_{{\mathbb X}^{{k+1},-{1 \over 2 }}}+ \norm{\nabla v}_{{\mathbb X}^{k}}\).
\end{align}
This in turn implies
 \begin{align}\label{tttt}
\abs{\partial_t^{k+1} v \cdot {\bf N}}_{-\frac{1}{2}}&\leq   \abs{V^{k+1}\cdot {\bf N}}_{-\frac{1}{2}}+\abs{  {\bf N}\cdot \partial^{\varphi}_zv\partial_t^{k+1}\eta}_{-\frac{1}{2}}
\\\nonumber&\leq\Lambda\( {1 \over c_{0}}, |h |_{{\mathbb Y}^{\frac{{k+3}}{2} }}   + \norm{ \nabla v }_{{\mathbb Y}^{\frac{{k+1}}{2}}} \)
   \(\norm{\partial_t^{k+1} v} + \abs{h}_{{\mathbb X}^{{k+1},-{1 \over 2 }}} + \norm{\nabla v}_{{\mathbb X}^{k}} \).
\end{align}
Hence, it follows from \eqref{gggesrt00}, \eqref{tttt} and the trace estimate $|\cdot|_{-\frac{1}{2}}\le |\cdot|_{\frac{1}{2}}\lesssim \norm{\cdot}_{H^1}$ that
 \begin{align}\label{gggesrt}
&\abs{\partial_t v\cdot{\bf N}}_{{{\mathbb X}^{k,-\frac{1}{2}}}} \leq   \abs{\partial_t v\cdot{\bf N}}_{{{\mathbb X}^{k-1,\frac{1}{2}}}}+\abs{\partial_t^{k }(\partial_t v\cdot{\bf N})}_{-\frac{1}{2}}
\\\nonumber&\quad\leq\Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{k+3}{2}  }}  + \norm{  v }_{{\mathbb Y}^{\frac{k+3}{2}}}  + \norm{ \nabla v
}_{{\mathbb Y}^{\frac{k+1}{2}}} \)
   \(\norm{v}_{{\mathbb X}^{k+1}}  + \norm{\nabla v}_{{\mathbb X}^{k}} + \abs{h}_{{\mathbb X}^{k+1,-{1 \over 2 }}}\).
\end{align}
Therefore, we conclude from \eqref{gg1es1}, \eqref{gg1es2} and \eqref{gggesrt} that
 \begin{align}\label{G2est1}
\abs{G^2}_{{{\mathbb X}^{{k},-\frac{1}{2}}}} \leq &  \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}}  + \norm{  v }_{{\mathbb Y}^{\frac{k+3}{2}}} +\norm{\nabla
v}_{{\mathbb Y}^{\frac{k}{2}+2}}  \)
\\\nonumber&\times\(\norm{v}_{{\mathbb X}^{k+1}}  + \norm{\nabla v}_{{\mathbb X}^{k}} + \abs{h}_{{\mathbb X}^{k+1,-{1 \over 2 }}}+\varepsilon\abs{  v }_{{{\mathbb X}^{{k},
\frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\).
\end{align}
Similarly, since $k\ge 3$, due to the trace estimates, one can get
 \begin{align}\label{G2est2}
\abs{G^2}_{{{\mathbb X}^{{\frac{k+3}{2}},-\frac{1}{2}}}} &\leq \Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k+3}{4}+3}}  + \norm{  v
}_{{\mathbb Y}^{\frac{\frac{k+3}{2}+3}{2}}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{k+3}{4}+2}}  \)
\\\nonumber&\quad\times\(\norm{v}_{{\mathbb X}^{\frac{k+3}{2}+1}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{k+3}{2}}} + \abs{h}_{{\mathbb X}^{\frac{k+3}{2}+1,-{1 \over 2
}}}+\varepsilon\abs{  v }_{{{\mathbb X}^{{\frac{k+3}{2}}, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{{\frac{k+3}{2}},\frac{3}{2}}}}\)
\\\nonumber&\leq\Lambda\( {1 \over c_{0}},  \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\abs{h}_{{{\mathbb X}^{{\frac{k+7 }{2}}}}} + \norm{  v }_{{\mathbb Y}^{\frac{k+3}{2}}}
+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{v}_{{\mathbb X}^{\frac{k+7}{2}}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{k+5}{2}}}  \)  .
\end{align}

Consequently, plugging the estimates \eqref{Fest1}--\eqref{Fest2}, \eqref{G1est1}--\eqref{G1est2} and  \eqref{G2est1}--\eqref{G2est2}, along with
doing the the same estimates for $G^3$ as for $G^2$, into \eqref{qclaim}, we can obtain that for $k\ge 3$,
  \begin{align}\label{qkxes}
  &    \norm{   q }_{{\mathbb X}^{k}}+\norm{ \nabla q }_{{\mathbb X}^{k}}+\norm{ \partial_{zz} q}_{{\mathbb X}^{k-1}}
  \\\nonumber&  \quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{k}{2}+3}} +\abs{h}_{{{\mathbb X}^{{\frac{k+7 }{2}}}}} +\norm{  v
  }_{{\mathbb Y}^{\frac{k+3}{2}}}  +\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+2}} +\norm{v}_{{\mathbb X}^{\frac{k+7}{2}}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{k+5}{2}}}  \)
  \\\nonumber&\qquad\times\(\abs{h}_{{\mathbb X}^{k+1,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{k,2}}+\norm{v}_{{\mathbb X}^{k+1}}  + \norm{\nabla v}_{{\mathbb X}^{k}} +
  \varepsilon\abs{  v }_{{{\mathbb X}^{{k}, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{{k},\frac{3}{2}}}}\).
  \end{align}
This proves the estimate \eqref{qpressureest}.

To prove \eqref{qinfty}, one can use the anisotropic Sobolev embedding estimate \eqref{emb} and the trace estimates to have that for $k\ge 1$, by
\eqref{qpressureest},
  \begin{align}
  &\norm{   q }_{{\mathbb Y}^{k}}+\norm{ \nabla q }_{{\mathbb Y}^{k}}\lesssim \norm{ \nabla q }_{{\mathbb X}^{k+2}}+\norm{ \partial_{zz} q}_{{\mathbb X}^{k+1}}
  \\\nonumber&  \quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{k }{2}+4}} +\abs{h}_{{{\mathbb X}^{{\frac{k+9}{2}}}}} +\norm{  v
  }_{{\mathbb Y}^{\frac{k+5}{2}}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k }{2}+3}} +\norm{v}_{{\mathbb X}^{\frac{k+9}{2}}}   + \norm{\nabla v}_{{\mathbb X}^{\frac{k+7}{2}}}  \)
  \\\nonumber&\qquad\times\(\abs{h}_{{\mathbb X}^{k+3,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{k+2,2}}+\norm{v}_{{\mathbb X}^{k+3}}  + \norm{\nabla v}_{{\mathbb X}^{k+2}} +
  \varepsilon\abs{  v }_{{{\mathbb X}^{{k+2}, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{{k+2},\frac{3}{2}}}}\)
    \\\nonumber&  \quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{k}{2}+4}}  +\abs{h}_{{\mathbb X}^{k+4}}+\norm{  v
    }_{{\mathbb Y}^{\frac{k+5}{2}}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{k}{2}+3}} +  \norm{v}_{{\mathbb X}^{k+4}}  + \norm{\nabla v}_{{\mathbb X}^{k+3}}   \).
  \end{align}
  Note that one has used the fact that $k+2\ge 3$ so that \eqref{qpressureest} can be used with $k+2$. This, together with
 \eqref{eqdzzrho} again, proves \eqref{qinfty}.
\end{proof}

\section{Smoothing estimates of $h$}\label{sectionprelimeta}

We first show the smoothing regularity estimates of $h$ coming from viscosity.
\begin{prop}\label{heps}
For every $m\in \mathbb{N}$,  $\varepsilon \in(0,1)$, it holds that
\begin{equation}\label{hepses1}
\varepsilon\abs{h(t)}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2\leq   \varepsilon\abs{h(0)}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2+\int_{0}^t \Lambda\(\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+2}}
+\norm{ v }_{{\mathbb Y}^{\frac{m}{2}+2}}  \)\( \varepsilon\abs{  v} _{{\mathbb X}^{m-1, \frac{3}{2}}}^2 +  \varepsilon\abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2 \)
\end{equation}
and
\begin{equation}\label{hepses2}
\varepsilon\abs{h(t)}_{{\mathbb X}^{m,\frac{1}{2}}}^2\leq   \varepsilon\abs{h(0)}_{{\mathbb X}^{m,\frac{1}{2}}}^2+\int_{0}^t \Lambda\(\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+2}}  +\norm{ v
}_{{\mathbb Y}^{\frac{m}{2}+2}}  \)
\( \varepsilon\abs{  v} _{{\mathbb X}^{m,\frac{1}{2}}}^2 +  \varepsilon\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}^2 \).
\end{equation}
\end{prop}
\begin{proof}
We prove only the estimate \eqref{hepses2}, and the estimate \eqref{hepses1} follows in the same way.
Apply $Z^\alpha$ with $\alpha\in \mathbb{N}^{1+2}$, $|\alpha|\le m$, to the kinematic boundary condition to get that on $\{z=0\}$:
\begin{equation}\label{heq1}
\partial_{t} Z^\alpha h + v  \cdot \nabla_y  Z^\alpha h -Z^\alpha v_{3}  + \[ Z^\alpha, v_{y} \]\cdot \nabla_y h=0.
\end{equation}
Then applying further $\Lambda^\frac{1}{2}$, the tangential Fourier multiplier, to \eqref{heq1} gives
\begin{equation}\label{heq2}
\partial_{t} \Lambda^\frac{1}{2} Z^\alpha h + v  \cdot \nabla_y \Lambda^\frac{1}{2} Z^\alpha h -\Lambda^\frac{1}{2} Z^\alpha  v_{3} + \[\Lambda^\frac{1}{2}, v_{y} \]\cdot
\nabla_y Z^\alpha h + \Lambda^\frac{1}{2} \(\[ Z^\alpha, v_{y} \]\cdot \nabla_y h\)=0.
\end{equation}
A standard energy estimate on the equation \eqref{heq2} yields
\begin{align}\label{heq3}
  \frac{d}{dt} \abs{Z^\alpha h}_{ \frac{1}{2} }^2&  \lesssim \norm{\nabla_{y} v}_{L^\infty }  \abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}^2
\\ \nonumber & \quad +  \(   \abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}+\abs{\[\Lambda^\frac{1}{2}, v_{y} \]\cdot \nabla_y Z^\alpha h}_{0} +\abs{\[ Z^\alpha, v_{y} \]\cdot
\nabla_y h}_{\frac{1}{2}}\) \abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}.
\end{align}
Due to the commutator estimate \eqref{comr2}, one has
\begin{equation}\label{heq4}
\abs{\[\Lambda^\frac{1}{2}, v_{y} \]\cdot \nabla_y Z^\alpha h}_{0}\lesssim  \norm{\nabla_{y} v  }_{L^\infty} \abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}
+  \norm{\nabla_{y} h  }_{L^\infty} \abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}.
\end{equation}
And the estimate \eqref{gues2} leads to
\begin{align}\label{heq5}
\abs{\[ Z^\alpha, v_{y} \]\cdot \nabla_y h}_{\frac{1}{2}}&\le \sum_{|\beta|+|\gamma|=\alpha\atop |\beta'|=1}\abs{Z^{\beta-\beta'}Z^{\beta'} v_{y}
\cdot \nabla_y Z^{\gamma} h}_{\frac{1}{2}}
\\\nonumber&\lesssim\norm{ v }_{{\mathbb Y}^{\frac{m}{2}+2}}\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}+\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+2}}\abs{  v} _{{\mathbb X}^{m,\frac{1}{2}}}.
\end{align}
Hence, plugging the estimates \eqref{heq4} and \eqref{heq5} into \eqref{heq3} and summing over $|\alpha|\le m$, by Cauchy's inequality, one can
deduce
that
\begin{equation}\label{heq6}
  \frac{d}{dt}  \abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}^2\lesssim\(1+\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+2}}+\norm{ v }_{{\mathbb Y}^{\frac{m}{2}+2}}\)\( \abs{  v} _{{\mathbb X}^{m,\frac{1}{2}}}^2 +
 \abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}^2\).
\end{equation}
Integrating the inequality \eqref{heq6} directly in time yields \eqref{hepses2}.
\end{proof}

Next, we show the smoothing estimates of $h$ due to surface tension.
\begin{prop}\label{hsig}
For every $m\in \mathbb{N}$,  ${\varepsilon,\sigma} \in(0,1)$, it holds that
\begin{align}\label{hsiges}
 \sigma^2\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}^2\le& \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m+3}{2}}}   + \norm{  \nabla v
 }_{{\mathbb Y}^{\frac{m+1}{2}}} \)
\(\abs{ q}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2 +\abs{ h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2 \right.
 \\\nonumber&\qquad\qquad\quad\left.   +\varepsilon^2 \abs{  v }_{{\mathbb X}^{m-1,\frac{3}{2}}}^2+\varepsilon^2\abs{h }_{{\mathbb X}^{m-1,\frac{3}{2}}}^2   \).
\end{align}
\end{prop}
\begin{proof}
Apply $Z^\alpha$ with $\alpha\in \mathbb{N}^{1+2}$, $|\alpha|\le m,\alpha_0\le m-1$, to the dynamic boundary condition to have
\begin{equation}
-\sigma Z^\alpha H=Z^\alpha  q  -2 \varepsilon Z^\alpha( S^\varphi v  \,{\bf n} \cdot \,{\bf n})- gZ^\alpha h .
\end{equation}
Note that
\begin{align}
     Z^\alpha H &=\nabla_y\cdot \(Z^\alpha\(\frac{\nabla_y h}{\sqrt{1+|\nabla_y h|^2}}\) \)
  \\& \nonumber=\nabla_y\cdot \( \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}+\nabla_y h Z^\alpha\(\frac{1}{\sqrt{1+|\nabla_y h|^2}}\)
  +\[Z^\alpha, \nabla_y h, \frac{1}{\sqrt{1+|\nabla_y h|^2}}\] \)
      \end{align}
and
\begin{align}
Z^\alpha\(\frac{1}{\sqrt{1+|\nabla_y h|^2}}\)&=Z^{\alpha-\alpha'} Z^{\alpha'}\(\frac{1}{\sqrt{1+|\nabla_y h|^2}}\)
=-Z^{\alpha-\alpha'} \(\frac{\nabla_y h\cdot \nabla_y Z^{\alpha'} h}{\sqrt{1+|\nabla_y h|^2}^3}\)
\\\nonumber& =- \frac{\nabla_y h\cdot \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}^3}-\[Z^{\alpha-{\alpha'}},\frac{\nabla_y h}{\sqrt{1+|\nabla_y
h|^2}^3}\]\cdot \nabla_y Z^{\alpha'} h
\end{align}
for any $|\alpha'|=1$. Hence,
  \begin{align}\label{sigsur1}
 Z^\alpha H  = \sigma \nabla_y\cdot \( \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha
 h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h
  +\mathcal{C}(\mathcal{B}_\sigma^\alpha) \),
    \end{align}
where
\begin{align}\label{sigsur2}
 \mathcal{C}(\mathcal{B}_\sigma^\alpha)=-\[Z^{\alpha-\alpha'},\frac{\nabla_y h}{\sqrt{1+|\nabla_y h|^2}^3}\]\cdot \nabla_y Z^{\alpha'} h
 \nabla_y h+\[Z^\alpha, \frac{1}{\sqrt{1+|\nabla_y h|^2}}, \nabla_y h\]
\end{align}
for any $|\alpha'|=1$. It follows that
\begin{equation}\label{fkdal}
-\sigma   \nabla_y\cdot \( \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y
h|^2}^3}\nabla_y h  \) =\sigma   \nabla_y\cdot  \mathcal{C}(\mathcal{B}_\sigma^\alpha)+Z^\alpha  q  -2 \varepsilon Z^\alpha( S^\varphi v  \,{\bf n}
\cdot \,{\bf n})- gZ^\alpha h .
\end{equation}

Then apply further $\Lambda^\frac{1}{2}$ to \eqref{fkdal} to get
\begin{align}
&-\sigma   \nabla_y\cdot \( \frac{ \nabla_y \Lambda^\frac{1}{2} Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y \Lambda^\frac{1}{2}
Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h  \)
 \\\nonumber&\quad=\sigma   \nabla_y\cdot \( \[\Lambda^\frac{1}{2}, \frac{ 1}{\sqrt{1+|\nabla_y h|^2}}-1\]\nabla_y  Z^\alpha h- \[\Lambda^\frac{1}{2},
 \frac{\nabla_y h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \cdot \]\nabla_y  Z^\alpha h \)
   \\\nonumber&\qquad+\Lambda^\frac{1}{2}\( \sigma   \nabla_y\cdot  \mathcal{C}(\mathcal{B}_\sigma^\alpha)+Z^\alpha  q  -2 \varepsilon Z^\alpha( S^\varphi v
   \,{\bf n} \cdot \,{\bf n})- gZ^\alpha h\).
\end{align}
It follows from a standard energy estimate for this elliptic equation that
\begin{align}
& \sigma^2\int_{z=0}  \( \frac{  |\nabla_y \Lambda^\frac{1}{2} Z^\alpha h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y h\cdot \nabla_y \Lambda^\frac{1}{2}
Z^\alpha h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy
 \\\nonumber&\quad=- \sigma^2\( \[\Lambda^\frac{1}{2}, \frac{ 1}{\sqrt{1+|\nabla_y h|^2}}-1\]\nabla_y  Z^\alpha h- \[\Lambda^\frac{1}{2}, \frac{\nabla_y
 h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \cdot \]\nabla_y  Z^\alpha h, \nabla_y \Lambda^\frac{1}{2} Z^\alpha h \)
 \\\nonumber&\qquad-\sigma^2  \(\Lambda^\frac{1}{2}\mathcal{C}(\mathcal{B}_\sigma^\alpha),\nabla_y \Lambda^\frac{1}{2} Z^\alpha h \)+\(\Lambda^\frac{1}{2}\(Z^\alpha
 q  -2 \varepsilon Z^\alpha( S^\varphi v  \,{\bf n} \cdot \,{\bf n})- gZ^\alpha h\), \sigma \Lambda^\frac{1}{2} Z^\alpha h \)  .
\end{align}
Since for any vector $\mathbf{a}\in\mathbb{R}^2$,
\begin{equation}\label{sigsur3}
\frac{  |\mathbf{a}|^2 }{\sqrt{1+|\nabla h|^2}}-  \frac{ |\nabla_y h\cdot \mathbf{a}|^2  }{\sqrt{1+|\nabla_y h|^2}^3}
\ge   \frac{ 1  }{\sqrt{1+|\nabla_y h|^2}^3} |\mathbf{a}|^2,
\end{equation}
so Cauchy's inequality, together with \eqref{cont2D} and \eqref{gues2}, since $|\alpha_1|+|\alpha_2|\ge 1$, give
\begin{align}\label{la1}
 \sigma^2\abs{\nabla_yh}_{{\mathbb X}^{m-1,\frac{5}{2}}}^2&\le \Lambda_0
\( \abs{ q}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2 +\varepsilon^2\abs{S^\varphi v  \,{\bf n} \cdot \,{\bf n}}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2   +\abs{ h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2\right.
 \\\nonumber&\quad  +\sigma^2\abs{ \mathcal{C}(\mathcal{B}_\sigma^\alpha)}_\frac{1}{2}^2+\sigma^2\abs{ \[\Lambda^\frac{1}{2}, \frac{ 1}{\sqrt{1+|\nabla_y
 h|^2}}-1\]\nabla_y  Z^\alpha h}_0^2
  \\\nonumber&\quad \left. +\sigma^2\abs{ \[\Lambda^\frac{1}{2}, \frac{\nabla_y h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \cdot \]\nabla_y  Z^\alpha
  h}_0^2\)
 \\&\nonumber\le \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m+3}{2}}}   + \norm{  \nabla v }_{{\mathbb Y}^{\frac{m+1}{2}}} \)
\(\abs{ q}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2 +\abs{ h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2\right.
   \\&\nonumber\left.\quad+\sigma^2 \abs{h}_{{\mathbb X}^{m-1,\frac{3}{2}}}^2 +\varepsilon^2\abs{S^\varphi v  \,{\bf n} \cdot \,{\bf n}}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2\).
\end{align}
Using the similar arguments in the previous section, one can have
$$
\abs{S^\varphi v  \,{\bf n} \cdot \,{\bf n}}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2  \leq    \Lambda\( {1 \over c_{0}}, \abs{h }_{{\mathbb Y}^{\frac{m-1}{2}+2}}   + \norm{
\nabla v }_{{\mathbb Y}^{\frac{m-1}{2}+1}} \)  \( \abs{  v }_{{\mathbb X}^{m-1,\frac{3}{2}}}
   +\abs{h }_{{\mathbb X}^{m-1,\frac{3}{2}}} \).
   $$
Then \eqref{hsiges} follows from \eqref{la1}, the Sobolev interpolation and Young's inequality.
\end{proof}

\section{Conormal estimates}\label{secconormal}

We shall derive a priori estimates on a time interval $[0, T^{\varepsilon,\sigma}]$ on which it is assumed that
\begin{equation}\label{apriori}
\partial_{z} \varphi \geq \frac{{c_{0}}}{2} , \  |h|_{2, \infty} \leq {1 \over c_{0}}\text{ and }  g  -   \partial_{z}^\varphi q   \geq
{c_{0}\over
2}\text{ on } \{z=0\}.
\end{equation}
Note in particular that this will allow one to use Lemmas \ref{mingrad} and \ref{Korn}.

To derive the higher order energy estimates, we shall use the good unknown $ V^\alpha= Z^\alpha v-\partial^{\varphi}_{z}v Z^\alpha \eta$, $\alpha\neq0$. A
key point is that the control of $V^\alpha$ and $Z^\alpha h$ will yield a control of $Z^\alpha v$:
\begin{equation}\label{equiv1}
\norm{Z^\alpha v} \lesssim \norm{V^\alpha} +  \Lambda\({1 \over c_{0}} , \norm{\nabla v}_{L^\infty}\) \abs{Z^\alpha h}_{-\frac{1}{2}},\ \norm{V^\alpha} \lesssim
\norm{Z^\alpha v} +  \Lambda\({1 \over c_{0}} , \norm{\nabla v}_{L^\infty}\) \abs{Z^\alpha h}_{-\frac{1}{2}}.
\end{equation}

Define
\begin{equation}\label{deflambdainfty1}
\Lambda_{\infty}(t)= \Lambda \( {1 \over c_{0}},  \abs{h(t)}_{{\mathbb X}^{\frac{m}{2}+5}} +\norm{v(t) }_{{\mathbb X}^{\frac{m}{2}+5}}+\norm{\partial_z v (t)
}_{{\mathbb X}^{\frac{m}{2}+4}} +\norm{\partial_z v(t) }_{{\mathbb Y}^{\frac{m}{2}+2}}+ \varepsilon^{1 \over 2} \norm{\partial_{zz}v(t)}_{L^\infty} \).
\end{equation}
It will be shown then that those functions $\Lambda(\cdot,\cdot)$ defined in the previous three sections can be bounded by $\Lambda_\infty$ for
sufficiently large
$m$, and also the elliptic estimates of $q$ and the smoothing estimates of $h$ will be restated along the way. First, taking $k=\frac{m}{2}$
for $m\ge 6$ in the estimate \eqref{qinfty} yields
\begin{align}\label{qminfty}
&\norm{   q }_{{\mathbb Y}^{\frac{m}{2}}}+ \norm{ \nabla q }_{{\mathbb Y}^{\frac{m}{2}}}+ \norm{ \partial_{zz} q }_{{\mathbb Y}^{\frac{m}{2}-1}}
\\\nonumber&\quad\le \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{m}{4}+4}}  +\abs{h}_{{\mathbb X}^{\frac{m}{2}+4}}+\norm{  v
}_{{\mathbb Y}^{\frac{m}{4}+\frac{5}{2}}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{m}{4}+3}} +  \norm{v}_{{\mathbb X}^{\frac{m}{2}+4}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{m}{2}+3}}
\)\leq \Lambda_\infty,
\end{align}
while taking $k=m-1$ in the estimate \eqref{qpressureest} gives
\begin{align}\label{qpressurem}
  &    \norm{   q }_{{\mathbb X}^{m-1}}+\norm{ \nabla q }_{{\mathbb X}^{m-1}}+\norm{ \partial_{zz} q}_{{\mathbb X}^{m-2}}
  \\\nonumber&  \quad\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{m+5}{2}}}  +\abs{h}_{{\mathbb X}^{\frac{m}{2}+3}} +\norm{  v
  }_{{\mathbb Y}^{\frac{m}{2}+1}} +\norm{\nabla v}_{{\mathbb Y}^{\frac{m+3}{2} }} +  \norm{v}_{{\mathbb X}^{\frac{m}{2}+3}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{m}{2}+2}}   \)
  \\\nonumber&\qquad\times\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-1,2}}+\norm{v}_{{\mathbb X}^{m}}  + \norm{\nabla v}_{{\mathbb X}^{m-1}} +
  \varepsilon\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\)
   \\\nonumber&  \quad\leq \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-1,2}}+\norm{v}_{{\mathbb X}^{m}}  + \norm{\partial_z v}_{{\mathbb X}^{m-1}} +
   \varepsilon\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\).
  \end{align}
Here one has required $m\ge 6$ so that by Sobolev's inequality
$$  \abs{h}_{{\mathbb Y}^{\frac{m}{4}+4}}  \le\abs{h}_{{\mathbb Y}^{\frac{m+5 }{2} }} \lesssim \abs{h}_{{\mathbb X}^{\frac{m}{2}+5}},
$$
and that by the anisotropic Sobolev embedding estimate \eqref{emb}
 $$\norm{  v }_{{\mathbb Y}^{\frac{m+3}{2}+1}}\lesssim \norm{  v }_{{\mathbb X}^{\frac{m}{2}+5}}+\norm{\partial_z  v }_{{\mathbb X}^{\frac{m}{2}+4}}.$$
It can be checked easily that all the functions $\Lambda( \cdot,\cdot)$ defined in Propositions \ref{heps} and \ref{hsig} and Lemmas
\ref{lemValpha},
\ref{lembordV} and \ref{lembordh} are bounded by $\Lambda_{\infty}$, due to \eqref{qminfty}. Moreover, Proposition \ref{heps} implies
that
\begin{equation}\label{mheps1}
\varepsilon\abs{h(t)}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2\leq   \varepsilon\abs{h(0)}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2+\int_{0}^t \Lambda_\infty\( \varepsilon\abs{  v} _{{\mathbb X}^{m-1,
\frac{3}{2}}}^2 +  \varepsilon\abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2 \)
\end{equation}
and
\begin{equation} \label{mheps2}
\varepsilon\abs{h(t)}_{{\mathbb X}^{m,\frac{1}{2}}}^2\leq   \varepsilon\abs{h(0)}_{{\mathbb X}^{m,\frac{1}{2}}}^2+\int_{0}^t \Lambda_\infty
\( \varepsilon\abs{  v} _{{\mathbb X}^{m,\frac{1}{2}}}^2 +  \varepsilon\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}^2 \).
\end{equation}
Proposition \ref{hsig}, together with \eqref{qpressurem} and the trace estimate $\abs{q}_{{\mathbb X}^{m-1,\frac{1}{2}}}\lesssim\norm{   q
}_{{\mathbb X}^{m-1}}+\norm{ \nabla q }_{{\mathbb X}^{m-1}}$, yields that
\begin{align}\label{mhsig1}
 \sigma^2\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}^2\le& \Lambda_\infty
\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}^2+\sigma^2 \abs{h}_{{\mathbb X}^{m-1,2}}^2+ \norm{v}_{{\mathbb X}^{m}}^2  + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 +  \varepsilon^2\abs{  v
}_{{{\mathbb X}^{m-1, \frac{3}{2}}}}^2+\varepsilon^2\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}^2 \).
\end{align}
Since $\abs{h}_{{\mathbb X}^{m-1,2}}^2\le \abs{h}_{{\mathbb X}^{m-1}}^\frac{1}{5}\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}^\frac{4}{5}$, one may improve \eqref{mhsig1} by
using Young's inequality to get
\begin{align}\label{mhsig}
 \sigma^2\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}^2\le& \Lambda_\infty
\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}^2+ \norm{v}_{{\mathbb X}^{m}}^2  + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 +  \varepsilon^2\abs{  v }_{{{\mathbb X}^{m-1,
\frac{3}{2}}}}^2+\varepsilon^2\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}^2 \).
\end{align}

Note that in the following we will use frequently these $L^\infty$ bounds involved in $\Lambda_\infty$.

\subsection{Basic $L^2$ estimate}\label{secconormal0}

We start with the estimates of $(v,h)$ itself, that is, the case $\alpha=0$.
  \begin{prop}
  \label{basicL2}
  For any smooth solution of \eqref{NSv}, it holds that
\begin{align}\label{l2estimate}
 &\norm{v(t)}^2 +g\abs{h(t)}_0^2+\sigma\abs{h(t)}_1^2 + \varepsilon \int_{0}^t \norm{ \nabla v}^2
\\\nonumber&\quad \le \Lambda_0  \(\norm{v_{0}}^2+ \abs{h_0}_0^2+\sigma\abs{h_0}_1^2+ \int_{0}^t \norm{v}^2\).
 \end{align}
  \end{prop}

\begin{proof}
Standard energy identity yields
\begin{equation}\label{l20}
\frac{1}{2} \frac{d}{dt} \int_{\Omega} |v|^2 \, d\mathcal{V}_{t} +   \varepsilon  \int_{\Omega} |S^\varphi v|^2\, d\mathcal{V}_{t}=
\int_{z=0} \(2 \varepsilon S^\varphi v - q \mbox{I} \) {\bf N} \cdot v \, dy-\int_{z=-b} \(2 \varepsilon S^\varphi v - q \mbox{I} \) e_3 \cdot v \, dy.
\end{equation}
The Navier slip boundary condition implies that
\begin{equation}\label{l21}
-\int_{z=-b}\(2 \varepsilon S^\varphi v - q \mbox{I} \) e_3 \cdot v \, dy=-\int_{z=-b}  2 \varepsilon (S^\varphi v   e_3)_i  v_i \, dy=-2 \kappa
\varepsilon\int_{z=-b}|v|^2\, dy.
\end{equation}
While the dynamic boundary condition and the kinematic boundary condition give
\begin{align}\label{l22}
\int_{z=0} \(2 \varepsilon S^\varphi v - q \mbox{I} \) {\bf N} \cdot v \, dy& =  -  \int_{z=0} (g h-\sigma H) {\bf N}  \cdot v \, dy=  -  \int_{z=0} (g h-\sigma
H) \partial_t h \, dy
\\\nonumber&=   - \frac{1}{2} \frac{d}{dt} \int_{z=0} g|h|^2+2\sigma\(\sqrt{1+|\nabla h|^2}-1\) \,dy.
\end{align}
Consequently,
\begin{align}\label{l23}
& \frac{1}{2} \frac{d}{dt} \(  \int_{\Omega} |v|^2 \, d\mathcal{V}_{t} +   \int_{z=0} g|h|^2+2\sigma\(\sqrt{1+|\nabla h|^2}-1\) \,dy  \)
 \\\nonumber&\qquad\qquad\qquad\qquad\qquad\qquad+ 2 \varepsilon  \int_{\Omega} |S^\varphi v|^2\, d\mathcal{V}_{t}+2 \kappa \varepsilon\int_{z=-b}|v|^2\, dy=0.
 \end{align}

Note that
$$\sqrt{1+|\nabla h|^2}-1\ge \frac{1}{2}\frac{1}{\sqrt{1+({1 \over c_0})^2}}|\nabla h|^2$$
due to \eqref{apriori}, and the trace estimate
$$\abs{v}_{L^2(\{z=-b\})}^2\lesssim \norm{\nabla v}\norm{v}+\norm{v}^2.$$
Hence, \eqref{l2estimate} follows from \eqref{l23}, Lemma \ref{Korn}  and Cauchy's inequality.
\end{proof}

\subsection{Estimate of $(Z^\alpha v, Z^\alpha h)$ for $\alpha_0\le m-1$}\label{secconormal1}

Next, we derive the energy estimates of $(Z^\alpha v, Z^\alpha h)$ for $1 \leq | \alpha |\leq m$ and $\alpha_0\le m-1$, that is, except the cases
$\alpha=0$ or $\alpha_0=m$.

\begin{prop}\label{conormv1}
Any smooth solution of \eqref{NSv} satisfies the estimate
\begin{align}\label{conormales1}
\nonumber&\norm{v(t)}_{{\mathbb X}^{m-1,1}}^2+\abs{h(t)}_{{\mathbb X}^{m-1,1}}^2 +\sigma\abs{h(t)}_{{\mathbb X}^{m-1,2}}^2 +\varepsilon  \abs{h(t)}_{{\mathbb X}^{m-1,
\frac{3}{2}}}^2+\int_0^t\varepsilon  \norm{\nabla v}_{{\mathbb X}^{m-1,1}}^2+\sigma^2\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}^2
\\&\quad \le \Lambda_0\( \norm{v(0)}_{{\mathbb X}^{m-1,1}}^2+\abs{h(0)}_{{\mathbb X}^{m-1,1}}^2 +\sigma\abs{h(0)}_{{\mathbb X}^{m-1,2}}^2+ \varepsilon \abs{h(0)}_{{\mathbb X}^{m,
\frac{1}{2}}}^2\)
\\\nonumber&\qquad+\int_0^t\Lambda_\infty \(\abs{h}_{{\mathbb X}^{m-1,1}}^2+\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}^2+\sigma \abs{h}_{{\mathbb X}^{m,1}}^2+ \varepsilon  \abs{h}_{{\mathbb X}^{m-1,
\frac{3}{2}}}^2+\norm{v}_{{\mathbb X}^{m}}^2+  \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 \).
\end{align}
\end{prop}
\begin{proof}
The energy identity for the equations \eqref{eqValpha}--\eqref{divValpha} yields
\begin{equation} \label{conormen}
\frac{1}{2} \frac{d}{dt}\int_{\Omega}\abs{V^\alpha }^2 d\mathcal{V}_{t} + 2 \varepsilon \int_{\Omega} \abs{S^\varphi V^\alpha}^2\, d\mathcal{V}_{t}
\\=\mathcal{I}_0^\alpha+\mathcal{I}_b^\alpha+ \mathcal{R}_{C}^\alpha+\mathcal{R}_{S}^\alpha,
\end{equation}
where
\begin{align}
\label{I0}&\mathcal{I}_0^\alpha=\int_{z= 0}\( 2 \varepsilon S^\varphi V^\alpha  - Q^\alpha \mbox{I} \) {\bf N} \cdot V^\alpha\, dy
\\
\label{Ib}&\mathcal{I}_b^\alpha=-\int_{z= -b}\( 2 \varepsilon S^\varphi V^\alpha  - Q^\alpha \mbox{I} \) e_3 \cdot V^\alpha\, dy,
\\\label{RC} & \mathcal{R}_{C}^\alpha=\int_{\Omega}\( \( \partial^{\varphi}_{z}v  \cdot \nabla^\varphi v Z^\alpha \eta-  \mathcal{C}^\alpha(\mathcal{T})-
\mathcal{C}^\alpha(q)\) \cdot V^\alpha-\mathcal{C}^\alpha (d) Q^\alpha \) \, d\mathcal{V}_{t},
\\
\label{RS}&\mathcal{R}_{S}^\alpha=\int_{\Omega} \( \varepsilon \mathcal{D}^\alpha(S^\varphi v)  + \varepsilon \nabla^\varphi \cdot \big( \mathcal{E}^\alpha (v)
\) \cdot V^\alpha\, d\mathcal{V}_{t}.
\end{align}

We first estimate $\mathcal{I}_0^\alpha$. The boundary condition \eqref{bordV} implies
\begin{align}\label{I0'}
\mathcal{I}_0^\alpha& =  \int_{z= 0} \big( 2 \varepsilon S^\varphi V^\alpha  - Z^\alpha q \mbox{I} +\partial^{\varphi}_z q Z^\alpha\eta I\big) {\bf N} \cdot V^\alpha\,dy
\\& \nonumber =  \int_{z= 0}  - (g-\partial^{\varphi}_zq) Z^\alpha h {\bf N}\cdot V^\alpha\, dy+\int_{z= 0}  \sigma Z^\alpha H {\bf N}\cdot
V^\alpha\, dy+ \int_{z=0}  \varepsilon \mathcal{C}^\alpha(\mathcal{B}_\varepsilon)\cdot V^\alpha \, dy
\\\nonumber  &  \quad- \int_{z= 0} 2 \varepsilon S^\varphi v\Pi Z^\alpha {\bf N} \cdot V^\alpha \, dy
-\int_{z=0} 2 \varepsilon Z^\alpha h\, \partial^{\varphi}_{z} \big( S^\varphi v\big) {\bf N}  \cdot V^\alpha \, dy.
\end{align}
By \eqref{CalphaB}, the third term in the right hand side of \eqref{I0'} can be bounded by
\begin{equation}\label{m11}
\abs{\int_{z=0}  \varepsilon \mathcal{C}^\alpha(\mathcal{B}_\varepsilon)\cdot V^\alpha \, dy}\le \Lambda_{\infty}\varepsilon \(\abs{v }_{{\mathbb X}^{m-1,1}}  + \abs{h
}_{{\mathbb X}^{m-1,1}}\)\abs{V^\alpha}_0.
 \end{equation}
Due to \eqref{cont2D}, it holds that
\begin{align}\label{m12}
\abs{ \int_{z= 0}2 \varepsilon  S^\varphi v\Pi Z^\alpha {\bf N} \cdot V^\alpha \, dy}&\le 2\varepsilon\abs{  Z^\alpha \nabla h }_{-{1 \over 2 }} \abs{S^\varphi v\Pi
V^\alpha}_{{1 \over 2}}
\\\nonumber&
\le \Lambda_{\infty}  \varepsilon \abs{h}_{{\mathbb X}^{m-1,{3 \over 2}}} \abs{V^\alpha}_{\frac{1}{2}}.
\end{align}
Note that $\Lambda_\infty$ involves $\sqrt{\varepsilon}||\partial_{zz} v||_{L^\infty}$, one has
\begin{align}\label{m13}
\abs{ \int_{z=0} 2 \varepsilon Z^\alpha h\, \partial^{\varphi}_{z} \( S^\varphi v\) {\bf N}  \cdot V^\alpha \, dy}
&\leq 2 \varepsilon  \abs{Z^\alpha h}_{0}\norm{\partial^{\varphi}_{z} \( S^\varphi v\) {\bf N} }_{L^\infty} \abs{V^\alpha}_{0}
\\\nonumber&\leq \Lambda_\infty\varepsilon^{\frac{1}{2}}  |h|_{{\mathbb X}^{m-1,1}} \abs{V^\alpha}_{0}.
\end{align}

For the first gravity term, one may use the boundary condition \eqref{bordh} to rewrite it as
 \begin{align}\label{gravity0}
  &\int_{z= 0}  - (g-\partial^{\varphi}_zq) Z^\alpha h {\bf N}\cdot
   V^\alpha\,dy
 \\\nonumber&\quad= \int_{z=0}  - (g-\partial^{\varphi}_zq) Z^\alpha h \(\partial_{t} Z^\alpha h+ v_y \cdot \nabla_{y}Z^\alpha h +\partial^{\varphi}_z v \cdot {\bf N}
 Z^\alpha h- \mathcal{C}^\alpha(h)\)\,dy.
    \end{align}
Integrating by parts in $t$ and using \eqref{qminfty} lead to
\begin{equation}\label{gravity1}
  \int_{z=0} -\(g-\partial^{\varphi}_zq\) \, Z^\alpha h  \partial_{t} Z^\alpha h\,dy
     \le -\frac{1}{2}  \frac{d}{dt}  \int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{Z^\alpha h }^2\,dy+\Lambda_\infty \abs{h}_{{\mathbb X}^{m-1,1}}^2.
\end{equation}
The integration by parts in $y$ gives
\begin{equation}\label{gravity2}
\abs{ \int_{z=0}  \(g-\partial^{\varphi}_zq\) Z^\alpha h v_y \cdot \nabla_{y}Z^\alpha h\, dy}\le   \Lambda_\infty\abs{h}_{{\mathbb X}^{m-1,1}}^2.
\end{equation}
Due to \eqref{bordhC}, one has
\begin{equation}\label{gravity3}
\abs{ \int_{z=0} \(g-\partial^{\varphi}_zq\) Z^\alpha h \(\partial^{\varphi}_z v \cdot {\bf N} Z^\alpha h- \mathcal{C}^\alpha(h)\)\, dy}
\leq  \Lambda_\infty    \abs{h}_{{\mathbb X}^{m-1,1}}\( \abs{h}_{{\mathbb X}^{m-1,1}} + \abs{ v}_{{\mathbb X}^{m-1}}\).
\end{equation}
Hence, in light of the estimates \eqref{gravity1}--\eqref{gravity3}, one may conclude from \eqref{gravity0} that
 \begin{align}\label{m14}
 &\int_{z= 0}  - (g-\partial^{\varphi}_zq) Z^\alpha h {\bf N}\cdot
   V^\alpha\,dy
 \\\nonumber&\quad\le - {1 \over 2} {d \over dt}  \int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{Z^\alpha h }^2\,dy
+\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,1}}^2 +  \abs{ v}_{{\mathbb X}^{m-1}}^2\).
    \end{align}

To deal with the second term involving surface tension, one has by \eqref{sigsur1} that
  \begin{align}\label{sigmaes0}
   \int_{z= 0}  \sigma Z^\alpha H {\bf N}\cdot
   V^\alpha\,dy=\int_{z= 0}\sigma \nabla_y\cdot \( \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha
   h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h
  +\mathcal{C}(\mathcal{B}_\sigma^\alpha) \){\bf N}\cdot  V^\alpha\,dy,
    \end{align}
where $ \mathcal{C}(\mathcal{B}_\sigma^\alpha)$ is defined by \eqref{sigsur2}. By Lemma \ref{lembordh}, one may deduce
  \begin{align}\label{sigmaes1}
   \int_{z= 0}  \sigma\nabla_y\cdot\mathcal{C}(\mathcal{B}_\sigma^\alpha) {\bf N}\cdot  V^\alpha\,dy
  \nonumber &\le \sigma\abs{  \nabla_y\cdot\mathcal{C}(\mathcal{B}_\sigma^\alpha)}_0\abs{\partial_{t} Z^\alpha h + v_y \cdot\nabla_y Z^\alpha h
  + \partial^{\varphi}_z v \cdot {\bf N}
 Z^\alpha h- \mathcal{C}^\alpha(h)}_0
  \\& \le \Lambda_\infty\sigma \abs{h}_{{\mathbb X}^{m-1,2}}\(\abs{ h}_{{\mathbb X}^{m ,1}}+\abs{h}_{{\mathbb X}^{m-1,2}}+\abs{h}_{{\mathbb X}^{m-1,1}}+\abs{v}_{{\mathbb X}^{m-1}}\)
  \\\nonumber& \le \Lambda_\infty \sigma \abs{h}_{{\mathbb X}^{m-1,2}}\(\abs{ h}_{{\mathbb X}^{m ,1}} +\abs{v}_{{\mathbb X}^{m-1}}\).
    \end{align}
To study the other two terms, one rewrite it as, by using the boundary condition \eqref{bordh} again,
 \begin{align} & \int_{z= 0}
   \sigma \nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y
   h|^2}^3}\nabla_y h\){\bf N}\cdot V^\alpha  \,dy
   \\\nonumber&\quad = \int_{z=0}   \sigma \nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y
   Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \)\(\partial_{t} Z^\alpha h+v_y \cdot \nabla_{y}Z^\alpha h\)\,dy
    +\mathcal{R}_{\mathcal{B}_\sigma^1}^\alpha,
    \end{align}
where
\begin{equation}
\mathcal{R}_{\mathcal{B}_\sigma^1}^\alpha= \int_{z=0}   \sigma \nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}+
\frac{\nabla_y h\cdot \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \) \(\partial^{\varphi}_zv\cdot{\bf N} Z^\alpha h-   \mathcal{C}^\alpha(h) \) dy.
 \end{equation}
It follows from an integration by parts, \eqref{chtilde} and \eqref{bordhC1} that
 \begin{align}\label{sigmaes2}
 \mathcal{R}_{\mathcal{B}_\sigma^1}^\alpha&\lesssim \sigma\abs{   \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y
 Z^\alpha h}{\sqrt{1+|\nabla h|^2}^3}\nabla_y h  }_{0}\abs{\nabla_y\(\partial^{\varphi}_zv\cdot{\bf N} Z^\alpha h-\tilde{\mathcal{C}}^\alpha(h)\)}_{0}
  \\\nonumber&\quad + \sigma \abs{\nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha
  h}{\sqrt{1+|\nabla h|^2}^3}\nabla_y h \)}_{-\frac{1}{2}}\abs{Z^{\alpha-\alpha_1} v_y\cdot \nabla_{y} Z^{ \alpha_1}h }_{\frac{1}{2}}
 \\&\nonumber \le \Lambda_\infty  \(\sigma\abs{h}_{{\mathbb X}^{m-1,2}}\(\abs{h}_{{\mathbb X}^{m-1,2}}+\abs{v}_{{\mathbb X}^{m-2,1}}\) +\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}
 \abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}} \).
    \end{align}
Integrating by parts in both $y$ and $t$, one finds that
\begin{align}\label{sigmaes3}
&\int_{z=0}   \sigma \nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha
h}{\sqrt{1+|\nabla h|^2}^3}\nabla_y h \)\partial_{t} Z^\alpha h\,dy
\\\nonumber&\quad=-\frac{1}{2}\frac{d}{dt}\int_{z=0}  \sigma\( \frac{  |\nabla_y Z^\alpha h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y
h\cdot \nabla_y Z^\alpha h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy+\mathcal{R}_{\mathcal{B}_\sigma^2}^\alpha,
   \end{align}
where
\begin{align}\label{sigmaes4}
\mathcal{R}_{\mathcal{B}_\sigma^2}^\alpha
&=\frac{1}{2}\int_{z=0}  \sigma\(\partial_t\( \frac{ 1  }{\sqrt{1+|\nabla_y h|^2}}\)|\nabla_y Z^\alpha h|^2-  \partial_t\(\frac{ 1}{\sqrt{1+|\nabla_y
h|^2}^3}\) |\nabla_y h\cdot \nabla_y Z^\alpha h|^2\right.  \\\nonumber&\qquad\qquad\qquad\left. - \frac{\nabla_y h\cdot \nabla_y Z^\alpha
h}{\sqrt{1+|\nabla_y h|^2}^3} \nabla_y \partial_{t} h\cdot \nabla_y Z^\alpha h \)\,dy
\\\nonumber&\le \Lambda_\infty \sigma \abs{h}_{{\mathbb X}^{m-1,2}}^2.
   \end{align}
Similarly, the integration by parts twice yields
  \begin{align}\label{sigmaes5}
  - \int_{z=0}   \sigma \nabla_y\cdot \(  \frac{ \nabla_y Z^\alpha h}{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y Z^\alpha
  h}{\sqrt{1+|\nabla h|^2}^3}\nabla_y h \)  v_y \cdot \nabla_{y}Z^\alpha h \, dy\le \Lambda_\infty \sigma \abs{h}_{{\mathbb X}^{m-1,2}}^2.
    \end{align}
Hence, by the estimates \eqref{sigmaes1}, \eqref{sigmaes2}--\eqref{sigmaes5}, one may conclude from \eqref{sigmaes0} that
 \begin{align}\label{m15}
 \int_{z= 0}  \sigma Z^\alpha H {\bf N}\cdot
   V^\alpha\, dy
 \le& -\frac{1}{2}\frac{d}{dt}\int_{z=0}  \sigma\( \frac{  |\nabla_y Z^\alpha h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y h\cdot \nabla_y
 Z^\alpha h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy
 \\\nonumber&+\Lambda_\infty    \(\sigma\abs{h}_{{\mathbb X}^{m-1,2}} \abs{h}_{{\mathbb X}^{m,1}}+\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}} \abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}} \).
    \end{align}
Note also that Lemma \ref{sobbord} implies that
\begin{equation} \label{equiv12}
\abs{V^\alpha}_0\lesssim \abs{v}_{{\mathbb X}^{m-1,1}}+\Lambda_\infty\abs{h}_{{\mathbb X}^{m-1,1}}\text{ and }\abs{V^\alpha}_\frac{1}{2}\lesssim
\abs{v}_{{\mathbb X}^{m-1,\frac{3}{2}}}+\Lambda_\infty\abs{h}_{{\mathbb X}^{m-1,\frac{3}{2}}}.
\end{equation}
Consequently, plugging the estimates \eqref{m11}--\eqref{m13}, \eqref{m14} and \eqref{m15} into \eqref{I0'}, by \eqref{equiv12} and Cauchy's
inequality, one may finish the
estimates of $\mathcal{I}_0^\alpha$ as:
 \begin{align}\label{I0estimate}
\mathcal{I}_0^\alpha
\le& -\frac{1}{2}\frac{d}{dt}\int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{Z^\alpha h }^2+ \sigma\( \frac{  |\nabla_y Z^\alpha h|^2 }{\sqrt{1+|\nabla_y
h|^2}}-  \frac{ |\nabla_y h\cdot \nabla_y Z^\alpha h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy
 \\\nonumber&+\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,1}}^2 +    \abs{ v}_{{\mathbb X}^{m-1}}^2 + \sigma   \abs{h}_{{\mathbb X}^{m-1,2}}
 \abs{h}_{{\mathbb X}^{m,1}}+\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}} \abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}}    \right.
   \\\nonumber& \qquad\quad\left.+
    \varepsilon\(\abs{v }_{{\mathbb X}^{m-1,1}}^2
  +  \abs{h}_{{\mathbb X}^{m-1,{3 \over 2}}}^2+ \abs{h}_{{\mathbb X}^{m-1,\frac{3}{2}}} \abs{v}_{{\mathbb X}^{m-1,\frac{3}{2}}}\) \).
    \end{align}

It also follows from \eqref{eeest} and \eqref{equiv12} that $\mathcal{I}_b^\alpha$ admits the following bound:
\begin{align}\label{Ibestimate}
\mathcal{I}_b^\alpha&=-\int_{z= -b}\( 2 \varepsilon S^\varphi V^\alpha e_3 \)_i \cdot V_i^\alpha\, dy=-\int_{z= -b}2\varepsilon\( \kappa
V^\alpha_i-\mathcal{E}^\alpha(v)_{i3}\)  \cdot V_i^\alpha\, dy
 \\\nonumber&\le \Lambda_\infty \varepsilon \( \abs{V^\alpha}_0+\abs{ v }_{{\mathbb X}^{m-1,1}} +  \abs{h}_{{\mathbb X}^{m-1,1}}\)\abs{V^\alpha}_0
  \\\nonumber&\le \Lambda_\infty \varepsilon \( \abs{ v }_{{\mathbb X}^{m-1,1}}^2 +  \abs{h}_{{\mathbb X}^{m-1,1}}^2\).
\end{align}

Next, the commutator $\mathcal{R}_C$ is estimated by using \eqref{CT}, \eqref{Cq}, \eqref{Cd}, \eqref{equiv1}, \eqref{qpressurem} and
 \eqref{qminfty} as
  \begin{align}\label{RCestimate}
  \mathcal{R}_{C}^\alpha &\le \Lambda_\infty\(\( \norm{ Z^\alpha \eta}+\norm{\mathcal{C}^\alpha(\mathcal{T})}+\norm{\mathcal{C}^\alpha(q)}\) \norm{
  V^\alpha}+\norm{\mathcal{C}^\alpha (d)}\norm{Q^\alpha}\)
  \\\nonumber& \le \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}+ \norm{v}_{{\mathbb X}^{m-1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}+ \norm{\nabla
  q}_{{\mathbb X}^{m-1}}\)\(\norm{v}_{{\mathbb X}^{m-1,1}}+ \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\)
  \\\nonumber&\quad+\Lambda_\infty \(\norm{\nabla v}_{{\mathbb X}^{m-1}}+ |h|_{{\mathbb X}^{m-1,{1 \over 2 }}}\)\(\norm{q}_{{\mathbb X}^{m-1,1}}+ \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\)
   \\\nonumber &\le \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-1,2}}+\norm{v}_{{\mathbb X}^{m}}  + \norm{\nabla v}_{{\mathbb X}^{m-1}} +
   \varepsilon\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\)
  \\\nonumber&\quad\times \(\norm{v}_{{\mathbb X}^{m-1,1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}+ \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\).
  \end{align}

 It remains to estimate the commutator $\mathcal{R}_{S}^\alpha$. First, it follows from the integration by parts, \eqref{CE}
 and \eqref{eeest} that
\begin{align}\label{RSe}
&\int_{\Omega} \varepsilon \nabla^\varphi( \mathcal{E}^\alpha(v) \big) \cdot V^\alpha d \mathcal{V}_{t}=- \int_{\Omega} \varepsilon\, \mathcal{E}^\alpha(v) \cdot \nabla
V^\alpha d \mathcal{V}_{t}+ \int_{z=0}\varepsilon\,  \mathcal{E}^\alpha(v) {\bf N} \cdot V^\alpha \, dy
\\\nonumber&\quad\leq \Lambda_{\infty}\varepsilon\(\( \norm{\nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\)  \norm{\nabla V^\alpha} + \(
\abs{h}_{{\mathbb X}^{m-1,1}} + \abs{v}_{{\mathbb X}^{m-1,1}}\)\abs{V^\alpha}_0\).
\end{align}
Next, for the first term, one actually has to estimate
\begin{align}\label{RSi}
\mathcal{R}_{Si}^\alpha  & =  \varepsilon \int_{\Omega} \mathcal C^\alpha_{j}(S^\varphi v)_{ij}  V^\alpha_{j} d \mathcal{V}_{t} \\
&\nonumber=\varepsilon \int_{\Omega} \mathcal C^\alpha_{j, 1}(S^\varphi v)_{ij}  V^\alpha_{j} d \mathcal{V}_{t}+ \varepsilon \int_{\Omega} \mathcal C^\alpha_{j,
2}(S^\varphi v)_{ij}  V^\alpha_{j} d \mathcal{V}_{t}+  \varepsilon \int_{\Omega} \mathcal C^\alpha_{j, 3}(S^\varphi v)_{ij}  V^\alpha_{j} d \mathcal{V}_{t} \\
&\nonumber: = \mathcal{R}_{Si}^{\alpha,1} + \mathcal{R}_{Si}^{\alpha,2} + \mathcal{R}_{Si}^{\alpha,3}
\end{align}
due to \eqref{Cialpha}. For $\mathcal{R}_{Si}^{\alpha,1}$, by \eqref{Cialpha1}, it suffices to estimate terms like
$$\varepsilon  \int_{\Omega}  Z^\beta \big( {\partial_{j} \varphi \over \partial_{z} \varphi}\big) \big(Z^{\tilde{\gamma}} \partial_{z}(S^\varphi
v)_{ij} \big)V^\alpha_{j} d \mathcal{V}_{t},$$
where $\beta $ and $\tilde\gamma$ are such that $\beta \neq 0, \, \tilde \gamma \neq 0$ and $|\beta | + |\tilde \gamma|=m.$ By using
\eqref{idcom}, one can reduce the problem to the estimate of
$$ \varepsilon  \int_{\Omega}  c_{\gamma} Z^\beta \big( {\partial_{j} \varphi \over \partial_{z} \varphi}\big)  \partial_{z}\big( Z^\gamma (S^\varphi
v)_{ij} \big)V^\alpha_{j} d \mathcal{V}_{t}$$
with $\beta$ as before (thus $| \beta | \leq m-1$) and $| \gamma | \leq | \tilde \gamma|\leq m-1.$ The integration by parts shows that it
suffices
to estimate three types of terms:
$$  \mathcal{I}_{1}=  \varepsilon\int_{\Omega} Z^\beta \big( {\partial_{j} \varphi \over \partial_{z} \varphi } \big)\,Z^\gamma (S^\varphi
v)_{ij}\,\partial_{z} V_{j}^\alpha d \mathcal{V}_{t},$$
$$\mathcal{I}_{2}=\varepsilon\int_{\Omega} \Big( \partial_{z}Z^\beta \big( {\partial_{j} \varphi \over \partial_{z} \varphi }  \big) \Big)Z^\gamma
(S^\varphi v)_{ij} V_{j}^\alpha d \mathcal{V}_{t},$$
and
$$ \mathcal{I}_{3}= \varepsilon \int_{z=0} Z^\beta \big( {\partial_{j} \varphi \over \partial_{z} \varphi } \big)\, Z^\gamma (S^\varphi v)_{ij}
V_{j}^\alpha dy.$$
For $\mathcal{I}_1$ and $\mathcal{I}_2$,  since $\beta \neq 0$, it follows from \eqref{gues}, \eqref{quot} and Lemma \ref{propeta} that
\begin{equation}\nonumber
\abs{\mathcal{I}_{1}} \leq\Lambda_{\infty}\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,{1 \over 2}}} \) \norm{ \nabla V^\alpha}
\end{equation}
and
$$\abs{ \mathcal{I}_{2}} \le\Lambda_{\infty}\varepsilon\(   \norm{ \nabla v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,{3 \over 2}}} \) \norm{   V^\alpha}.$$
By \eqref{gues}, $\beta \neq 0$ and Lemma \ref{lembord}, it holds that
$$\abs{\mathcal{I}_{3}}\leq\Lambda_\infty\varepsilon\( \abs{h}_{{\mathbb X}^{m-1,1} } +  \abs{\nabla v}_{{\mathbb X}^{m-1}} \)\abs{V^\alpha}_0 \leq\Lambda_\infty\varepsilon\( \abs{h}_{{\mathbb X}^{m-1,1}
} +  \abs{ v}_{{\mathbb X}^{m-1,1}} \)\abs{V^\alpha}_0 .$$
Consequently, one can get from the previous three estimates that
\begin{align}\label{RSi1}
\abs{\mathcal{R}_{Si}^{\alpha,1}}\leq\Lambda_{\infty}\varepsilon&\(\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^\alpha}
+\(   \norm{ \nabla v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,{3 \over 2}}} \) \norm{   V^\alpha}\right.
\\\nonumber&\left.+ \( \abs{h}_{{\mathbb X}^{m-1,1} } + \abs{ v}_{{\mathbb X}^{m-1,1}}  \)\abs{V^\alpha}_0 \) .
\end{align}
 The estimate of $\mathcal{R}_{Si}^{\alpha,2}$ is straightforward, one gets from the definition \eqref{Cialpha2} that
\begin{equation}\label{RSi2}
\abs{\mathcal{R}_{Si}^{\alpha,2}}\leq\Lambda_{\infty}\varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}\norm{V^\alpha}.
\end{equation}
To estimate $\mathcal{R}_{Si}^{\alpha,3}$, one derives from \eqref{Cialpha3} and \eqref{idcom} that
$$\varepsilon \abs{\int_{\Omega}  {\partial_{i}\varphi \over (\partial_{z}\varphi)^2} \partial_{z}\big( S^\varphi v)  [Z^\alpha, \partial_{z}]\varphi
V_{j}^\alpha d\mathcal{V}_{t}} \leq\Lambda_{\infty}\varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}\norm{V^\alpha}.$$
Note that one has used again in the previous two estimates the fact that $\Lambda_\infty$ involves $
\varepsilon^{\frac{1}{2}}\norm{\partial_{zz}v}_{L^\infty}$. For the term
$$\varepsilon \int_{\Omega}{\partial_{i}\varphi \over \partial_{z} \varphi}  V^\alpha \, [Z^\alpha, \partial_{z}](S^\varphi v)  d\mathcal{V}_{t}, $$
performing an integration by parts and using a similar arguments as for $\mathcal{R}_{Si}^{1}$ show that
$$\abs{\int_{\Omega}{\partial_{i}\varphi \over \partial_{z} \varphi}  V^\alpha \, [Z^\alpha, \partial_{z}](S^\varphi v)  d\mathcal{V}_{t}}\leq\Lambda_\infty \varepsilon
\norm{\nabla v}_{{\mathbb X}^{m-1}}\( \norm{V^\alpha} + \norm{\nabla  V^\alpha}\).$$
Consequently,
\begin{equation}\label{RSi3}
\abs{\mathcal{R}_{Si}^{\alpha,3}}\leq\Lambda_\infty\( \varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}\norm{V^\alpha}+\varepsilon\norm{\nabla v}_{{\mathbb X}^{m-1}}\( \norm{V^\alpha}
+ \norm{\nabla  V^\alpha}\) \) .
\end{equation}
It then follows from \eqref{RSi1}--\eqref{RSi3} that
\begin{align}
\varepsilon\abs{ \int_{\Omega} \mathcal D^\alpha(S^\varphi v) \cdot V^\alpha d \mathcal{V}_{t}}&\leq\Lambda_\infty\(\(\varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\varepsilon
\abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}+\varepsilon\norm{\nabla v}_{{\mathbb X}^{m-1}}\)\norm{V^\alpha}\right.
\\\nonumber&\left.\quad +\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^\alpha}+ \( \abs{h}_{{\mathbb X}^{m-1,1} } +
\abs{ v}_{{\mathbb X}^{m-1,1}} \)\abs{V^\alpha}_0 \).
\end{align}
This, together with \eqref{RSe}, \eqref{equiv1} and \eqref{equiv12}, implies that
\begin{align}\label{RSestimate}
\mathcal{R}_{S}^\alpha\leq&\Lambda_\infty\(\(\varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\varepsilon \abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}+\varepsilon\norm{\nabla
v}_{{\mathbb X}^{m-1}}\)\(\norm{v}_{{\mathbb X}^{m-1,1}}+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\)\right.
\\\nonumber&\left.\qquad +\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^\alpha}+ \abs{h}_{{\mathbb X}^{m-1,1} }^2 +
\abs{  v}_{{\mathbb X}^{m-1,1}}^2\).
\end{align}

We can now finish the proof of the proposition. By the estimates \eqref{I0estimate}--\eqref{RCestimate} and \eqref{RSestimate},  the trace
estimates $\abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}}\lesssim \norm{v}_{{\mathbb X}^{m-1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}$ and $\abs{v}_{{\mathbb X}^{m-1,1}}\lesssim \norm{\nabla
v}_{{\mathbb X}^{m-1,1}}^\frac{1}{2}\norm{v}_{{\mathbb X}^{m-1,1}}^\frac{1}{2}+\norm{v}_{{\mathbb X}^{m-1,1}} $, using Cauchy's inequality, one may deduce from \eqref{conormen} that
\begin{align}\label{alphaend1}
&\frac{1}{2} \frac{d}{dt} \mathcal{E}^\alpha + 2 \varepsilon \int_{\Omega} \abs{S^\varphi V^\alpha}^2\, d\mathcal{V}_{t}
\\\nonumber&\quad \le\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,1}}^2 + \sigma   \abs{h}_{{\mathbb X}^{m-1,2}}
\abs{h}_{{\mathbb X}^{m,1}}+\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}\(\norm{v}_{{\mathbb X}^{m-1,1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}\) \right.
  \\\nonumber&\qquad\qquad+\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}^2+ \norm{\nabla v}_{{\mathbb X}^{m-1}}^2 + \norm{v}_{{\mathbb X}^{m}}^2 + \varepsilon  \abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}^2
  \\\nonumber&\qquad\qquad+ \varepsilon\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}} \(\norm{v}_{{\mathbb X}^{m-1,1}}+\norm{\nabla
  v}_{{\mathbb X}^{m-1}}+\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\)
\\\nonumber&\left.\qquad \qquad+\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^\alpha}  \),
\end{align}
where
\begin{equation}
\mathcal{E}^\alpha:=\int_{\Omega}\abs{V^\alpha }^2 d\mathcal{V}_{t} +\int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{Z^\alpha h }^2+ \sigma\( \frac{  |\nabla_y Z^\alpha h|^2
}{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y h\cdot \nabla_y Z^\alpha h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy.
\end{equation}
It follows from \eqref{equiv1}, the Taylor sign condition in \eqref{apriori} and \eqref{sigsur3} that
$$\norm{Z^\alpha v}^2+\abs{Z^\alpha h}_0^2+\sigma\abs{Z^\alpha h}_1^2 \leq \Lambda\({1\over c_{0}}\)  \mathcal{E}^\alpha.$$
One can use the Korn inequality of Lemma \ref{Korn} and \eqref{equiv1} to get that
$$\norm{\nabla V^\alpha}^2 \leq \Lambda\({1\over c_{0}}\) \(\int_{\Omega} \abs{S^\varphi V^\alpha}^2\, d\mathcal{V}_{t} + \norm{v}_{{\mathbb X}^{m-1,1}}^2+
\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2\).$$
On the other hand, by the definition of $V^\alpha$,
$$\varepsilon \norm{\nabla Z^\alpha v}^2\le \varepsilon \norm{\nabla V^\alpha}^2+\Lambda_\infty\(\varepsilon\norm{h}_{{\mathbb X}^{m-1,\frac{3}{2}}}^2+\norm{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2\).$$
Then integrating \eqref{alphaend1} in time, using the trace estimate $\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}\lesssim \norm{\nabla v }_{{{\mathbb X}^{m-1,
1}}}+\norm{  v }_{{{\mathbb X}^{m-1, 1}}}$ and Cauchy's inequality, together with \eqref{l2estimate}, one deduces that
\begin{align}\label{alphaend2}
&\norm{v(t)}_{{\mathbb X}^{m-1,1}}^2+\abs{h(t)}_{{\mathbb X}^{m-1,1}}^2 +\sigma\abs{h(t)}_{{\mathbb X}^{m-1,2}}^2+\varepsilon\int_0^t  \norm{\nabla v}_{{\mathbb X}^{m-1,1}}^2
\\\nonumber&\quad\le \Lambda_0\( \norm{v(0)}_{{\mathbb X}^{m-1,1}}^2+\abs{h(0)}_{{\mathbb X}^{m-1,1}}^2 +\sigma\abs{h(0)}_{{\mathbb X}^{m-1,2}}^2\)
\\\nonumber&\qquad+ \int_0^t \Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,1}}^2 + \sigma   \abs{h}_{{\mathbb X}^{m-1,2}}
\abs{h}_{{\mathbb X}^{m,1}}+\sigma\abs{h}_{{\mathbb X}^{m-1,\frac{5}{2}}}\(\norm{v}_{{\mathbb X}^{m-1,1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}\)  \right.
  \\\nonumber&\qquad\qquad\left.+\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}^2+ \norm{\nabla v}_{{\mathbb X}^{m-1}}^2 + \norm{v}_{{\mathbb X}^{m}}^2 + \varepsilon  \abs{h}_{{\mathbb X}^{m-1,
  \frac{3}{2}}}^2\).
\end{align}
This, together with \eqref{mheps1}, \eqref{mhsig} and Cauchy's inequality, leads to \eqref{conormales1}.
\end{proof}

\subsection{Estimate of $(\partial_t^m v, \partial_t^m h)$}\label{secconormal2}

We now derive the energy estimates of $(\partial_t^m v, \partial_t^m h)$, that is, the case $\alpha_0=m$.
\begin{prop}\label{conormvm}
Any smooth solution of \eqref{NSv} satisfies the estimate
 \begin{align}\label{alphaend2m1ll}
&\int_0^t\(\norm{\partial_t^m v }^2+\abs{\partial_t^m h }_{0}^2 +\sigma\abs{\partial_t^m h }_{1}^2+ \varepsilon  \abs{h }_{{\mathbb X}^{m , \frac{1}{2}}}^2 \)^2+\int_0^t\(\varepsilon\int_0^s
\norm{\nabla \partial_t^m v}_{0}^2\)^2
\\\nonumber&\quad\le t\Lambda_0\( \norm{\partial_t^m v(0)}^2+\abs{\partial_t^m h(0)}_{0}^2 +\sigma\abs{\partial_t^m h(0)}_{1}^2+ \varepsilon  \abs{h(0)}_{{\mathbb X}^{m ,
\frac{1}{2}}}^2\)^2
\\\nonumber&\qquad+ t\(\int_0^t\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m }}^2 + \sigma     \abs{h}_{{\mathbb X}^{m,1}}^2
  + \norm{v}_{{\mathbb X}^{m}}^2 + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2+ \varepsilon  \norm{\nabla v}_{{\mathbb X}^{m-1,1}}^2+ \varepsilon  \abs{h}_{{\mathbb X}^{m , \frac{1}{2}}}^2  \)\)^2
   \\\nonumber &\qquad +\int_0^t\Lambda_\infty \(\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}^2+\sigma \abs{h}_{{\mathbb X}^{m-1,2 }}^2+\norm{v}_{{\mathbb X}^{m-1,1 }}^2+\norm{\partial_z
   v}_{{\mathbb X}^{m-2 }}^2\)  \\\nonumber&\qquad\qquad \times\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}^2+\sigma \abs{h}_{{\mathbb X}^{m,1}}^2 +\norm{v}_{{\mathbb X}^{m}}^2  +
   \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 +  \varepsilon^2\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}^2+\varepsilon^2\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}^2\).
\end{align}

\end{prop}

\begin{proof}
In the current case, \eqref{conormen} can be restated as:
\begin{equation}\label{conormenm}
\frac{1}{2} \frac{d}{dt}\int_{\Omega} \abs{V^m }^2 d\mathcal{V}_{t} + 2 \varepsilon \int_{\Omega} \abs{S^\varphi V^m}^2\, d\mathcal{V}_{t} \\
= \mathcal{I}_0^m+\mathcal{I}_b^m+ \mathcal{R}_{Q}^m+ \mathcal{R}_{C}^m+\mathcal{R}_{S}^m,
\end{equation}
where
\begin{align}
\label{Im0} &\mathcal{I}_0^m=  \int_{z= 0} \( 2 \varepsilon S^\varphi V^m  - Q^m \mbox{I} \) {\bf N} \cdot V^m\, dy,\\
\label{Imb} &\mathcal{I}_b^m= - \int_{z= 0} \( 2 \varepsilon S^\varphi V^m  - Q^m \mbox{I} \) e_3 \cdot V^m\, dy, \\
\label{RQm} & \mathcal{R}_{Q}^m=   - \int_{\Omega}    \mathcal{C}^m (d) \partial_t^m q  \, d\mathcal{V}_{t},\\
\label{RCm} & \mathcal{R}_{C}^m=   \int_{\Omega}\( \( \partial^{\varphi}_{z}v  \cdot \nabla^\varphi v \partial_t^m \eta-  \mathcal{C}^m(\mathcal{T})- \mathcal{C}^m(q)\)
\cdot V^m+\mathcal{C}^m (d) \partial^{\varphi}_zq\partial_t^m\eta \) \, d\mathcal{V}_{t},\\
\label{RSm} & \mathcal{R}_{S}^m=   \int_{\Omega} \( \varepsilon \mathcal{D}^m(S^\varphi v)  + \varepsilon \nabla^\varphi \cdot  \mathcal{E}^m (v) \) \cdot
V^m\, d\mathcal{V}_{t}.
\end{align}
Here $V^m=\partial_t^m v-\partial^{\varphi}_z v\partial_t^m\eta,\ Q^m=\partial_t^m q-\partial^{\varphi}_z v\partial_t^m\eta$, and $\mathcal{C}^m(\cdot)$ are those commutators $\mathcal{C}^\alpha(\cdot)$
for the case $\alpha_0=m$. Note that we have singled out the term $\mathcal{R}_{Q}^m$ from $\mathcal{R}_{C}^m$.

We first estimate $\mathcal{I}_0^m$, which can be rewritten as (similar to \eqref{I0'})
\begin{align}\label{I0'm}
\mathcal{I}_0^m&   =  \int_{z= 0}  - (g-\partial^{\varphi}_zq) \partial_t^m h {\bf N}\cdot V^m\, dy+\int_{z= 0}  \sigma \partial_t^m H {\bf N}\cdot V^m\, dy+ \int_{z=0}  \varepsilon
\mathcal{C}^m(\mathcal{B}_\varepsilon)\cdot V^m \, dy
\\\nonumber&  \quad- \int_{z= 0} 2 \varepsilon S^\varphi v\Pi \partial_t^m {\bf N} \cdot V^m \, dy-\int_{z=0} 2 \varepsilon \partial_t^m h\, \partial^{\varphi}_{z} \big( S^\varphi v\big) {\bf N}
\cdot V^m \, dy.
\end{align}
Following the analysis in \eqref{m11}--\eqref{m13}, one can bound the last three terms in \eqref{I0'm} by
\begin{equation}\label{m11m}
\Lambda_{\infty}\(\varepsilon \(\abs{v }_{{\mathbb X}^{m-1,1}}  + \abs{h }_{{\mathbb X}^{m-1,1}}\)\abs{V^m}_0 +\varepsilon \abs{ h}_{{\mathbb X}^{m,\frac{1}{2}}} \abs{V^m}_{\frac{1}{2}}+\varepsilon^\frac{1}{2}
\abs{ h}_{{\mathbb X}^{m}}\abs{V^m}_{0}\).
\end{equation}
As \eqref{m14}, one deduces
\begin{align}\label{m11m1}
& \int_{z= 0}  - (g-\partial^{\varphi}_zq) \partial_t^m h {\bf N}\cdot V^m\,dy
\\\nonumber&\quad\le - \frac{1}{2}  \frac{d}{dt}  \int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{\partial_t^m h }^2\,dy
+\Lambda_\infty\(\abs{h}_{{\mathbb X}^{m}}^2  +  \abs{ v}_{{\mathbb X}^{m-1}}^2 \).
\end{align}
However, as explained in Section \ref{sec main}, one can not use the arguments leading to \eqref{m15} to estimate $\int_{z= 0}  \sigma \partial_t^m H
{\bf N}\cdot V^m\, dy$ since there is one half
regularity loss for $\partial_t^m h$ so that it is difficult to control the following term, after using the kinematic boundary condition,
\begin{equation}\label{surout}
 -\int_{z= 0} \sigma \partial_t^m  H\,  m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v \, dy.
\end{equation}
The crucial observation here is that, this term will be cancelled out from estimating the term $\mathcal{R}^m_Q$ defined by \eqref{RQm}.
So the estimates of this term will be postponed till we estimate $\mathcal{R}^m_Q$.

Similarly as \eqref{Ibestimate}, $\mathcal{I}_b^m$ admits the bound
\begin{equation}\label{Ibestimatem}
\mathcal{I}_b^m \le \Lambda_\infty \varepsilon \( \abs{V^m}_0+\abs{ v }_{{\mathbb X}^{m-1,1}}  +  \abs{h}_{{\mathbb X}^{m-1,1}}\)\abs{V^m}_0.
\end{equation}

We now estimate $\mathcal{R}^m_Q$. Note carefully that there is no any estimates of $\partial_t^m q$, so one needs to integrate by parts in $t$.
To continue, one needs more explicit expression of $\mathcal{C}^m (d)$. Indeed, we will use a variant of \eqref{Cialpha}. It follows from
the divergence free condition that
$$\partial_z\varphi\(\partial_1v_1+\partial_2v_2\)+\partial_zv\cdot{\bf N}=0.$$
Applying $\partial_t^m$ to the above and using the definition of $\mathcal{C}^m(d)$, one gets that
 \begin{align}\label{com1m}
  \partial_z\varphi \mathcal{C}^m(d)   =  \[ \partial_t^m ,  {\bf N} , \cdot\partial_{z}v  \]  +   \[ \partial_t^m ,  \partial_{z} \eta,\partial_1v_1+\partial_2v_2\].
     \end{align}
 Moreover, to integrate by parts in $t$, one needs to single out in $\mathcal{C}^m(d) $ the highest $m-1$ order time
 derivatives terms and use the following decomposition
\begin{align}
 \partial_z\varphi\mathcal{C}^m (d)=  \mathcal{C} ^{m} (d)_1+\mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3+\mathcal{C} ^{m} (d)_4+\mathcal{C} ^{m}
 (d)_5
     \end{align}
   with
\begin{align}
\mathcal{C} ^{m} (d)_1= & m\partial_t  {\bf N}  \cdot \partial_t^{m-1} \partial_{z} v,\quad \mathcal{C} ^{m} (d)_2=m\partial_t \partial_{z} \eta
\partial_t^{m-1}(\partial_1v_1+\partial_2v_2), \\
\mathcal{C} ^{m} (d)_3
  = & m \partial_t^{m-1}{\bf N}\cdot \partial_t \partial_{z} v,\quad
\mathcal{C} ^{m} (d)_4
  = m \partial_t^{m-1}\partial_z\eta \partial_t (\partial_1v_1+\partial_2v_2),\\
 \mathcal{C} ^{m} (d)_5
 =   & \sum_{\ell=2 }^{m-2} C_m^\ell\(\partial_t^\ell {\bf N}  \cdot\partial_t^{m-\ell}\partial_{z} v+\partial_t^\ell \partial_{z} \eta
 \cdot\partial_t^{m-\ell}(\partial_1v_1+\partial_2v_2)\).
     \end{align}
Accordingly,
 \begin{equation}\label{rmq0}
\mathcal{R}^m_Q= -\int_{\Omega} \(\mathcal{C} ^{m} (d)_1+\mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3+\mathcal{C} ^{m} (d)_4+\mathcal{C} ^{m}
(d)_5\) \partial_t^m q   \, dydz.
  \end{equation}
The fifth term in \eqref{rmq0} can be easily treated by the integration by parts in $t$ as
 \begin{align}\label{rmq11}
 -\int_{\Omega}  \mathcal{C} ^{m} (d)_5 \partial_t^m q   \, dydz
 =- \frac{d}{dt}\int_{\Omega}   \mathcal{C} ^{m} (d)_5 \partial_t^{m-1} q\,dydz+\mathcal{R}_5^m,
  \end{align}
  with
   \begin{align}\label{rmq11'}
\mathcal{R}_5^m=\int_{\Omega} \partial_t \mathcal{C} ^{m} (d)_5   \partial_t^{m-1} q\,dydz \le \Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\norm{\partial_z
v}_{{\mathbb X}^{m-2}}\)\norm{ \partial_t^{m-1} q}.
  \end{align}
Integrate by parts in $t$ to write the fourth term as
 \begin{equation}\label{rmq12}
-\int_{\Omega}  \mathcal{C} ^{m} (d)_4\partial_t^m q   \, dydz
=- \frac{d}{dt}\int_{\Omega}   \mathcal{C} ^{m} (d)_4 \partial_t^{m-1} q\,dydz+\mathcal{R}_4^m,
\end{equation}
where, by further integrating by parts in $z$ and the trace theory,
 \begin{align}\label{rmq12'}
  \mathcal{R}_4^m&=\int_{\Omega}m \(\partial_t^{m}\partial_z\eta \partial_t (\partial_1v_1+\partial_2v_2) +  \partial_t^{m-1}\partial_z\eta \partial_t^2 (\partial_1v_1+\partial_2v_2) \)
  \partial_t^{m-1} q\,dydz
 \\\nonumber& = \int_{z=0}   m \partial_t^{m}h  \partial_t\( \partial_1 v_1+\partial_2 v_2\)\partial_t^{m-1} q  \, dy -\int_{\Omega}   m\partial_t^{m}\eta\partial_z\( \( \partial_1
 v_1+\partial_2 v_2\)\partial_t^{m-1} q \) \, dydz \\\nonumber&\quad +\int_{\Omega}m
  \partial_t^{m-1}\partial_z\eta \partial_t^2 (\partial_1v_1+\partial_2v_2)  \partial_t^{m-1} q\,dydz
\\\nonumber& \le  \Lambda_\infty\(\abs{\partial_t^m h}_{-\frac{1}{2}}\abs{\partial_t^{m-1} q}_{\frac{1}{2}}+\norm{\partial_t^{m} \eta}\norm{\partial_t^{m-1} q}_{H^1}+\norm{\partial_t^{m-1}\partial_z
\eta}\norm{\partial_t^{m-1} q} \)
\\\nonumber& \le \Lambda_\infty \abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}} \norm{\partial_t^{m-1} q}_{H^1}  .
 \end{align}
Similarly, integrate by parts in both $t$ and $y$ to bound the second and third terms by
 \begin{align}\label{rmq13}
 -\int_{\Omega}  \(\mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3\) \partial_t^m q   \, dydz
 =- \frac{d}{dt}\int_{\Omega}   \(\mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3\)  \partial_t^{m-1} q\,dydz+\mathcal{R}_{2,3}^m,
  \end{align}
  where
   \begin{align}\label{rmq13'}
\mathcal{R}_{2,3}^m &=\int_{\Omega}  \partial_t \(\mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3\)  \partial_t^{m-1} q\,dydz
\\\nonumber&
\le \Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\norm{  v}_{{\mathbb X}^{m}}+\norm{\nabla  v}_{{\mathbb X}^{2}}\)\norm{\partial_t^{m-1} q}_{H^1}.
  \end{align}
Finally, we turn to the most delicate term, the one involving $\mathcal{C}^m(d)_1$ in \eqref{rmq0}. Integrate by parts in $z$ first to get
 \begin{equation}\label{rmq14}
-\int_{\Omega}   \mathcal{C}^m (d)_1  \partial_t^{m} q\,dydz
 =-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \partial_t^{m} q \, dy+\int_{\Omega}   m  \partial_z\( \partial_t^{m} q\partial_t  {\bf N}\)\cdot \partial_t^{m-1} v  \,
 dydz.
  \end{equation}
 Then integrate by parts in $t$ to obtain
\begin{equation}
 \int_{\Omega}   m  \partial_z\( \partial_t^{m} q\partial_t  {\bf N}\)\cdot \partial_t^{m-1} v  \, dydz
= \frac{d}{dt}\int_{\Omega}   m \partial_z\( \partial_t^{m-1} q\partial_t  {\bf N}\)\cdot \partial_t^{m-1} v  \, dydz+\mathcal{R}^m_1,
\end{equation}
with
\begin{align}
\mathcal{R}^m_1&= -\int_{\Omega}   m \partial_z\( \partial_t^{m-1} q\partial_t  {\bf N}\)\cdot \partial_t^{m} v + m \partial_z\( \partial_t^{m-1} q\partial_t^2  {\bf N}\)\cdot \partial_t^{m-1} v
\, dydz
\\&\le   \Lambda_\infty \norm{  v}_{{\mathbb X}^{m}}
\norm{\partial_z\partial_t^{m-1} q} .\nonumber
  \end{align}
Note carefully that we integrate by parts in $z$ first rather than in $t$ since there is no estimates of $\partial_t^{m}v$ on the boundary.
This also indicates the difficulty in controlling the first term in the right hand side of \eqref{rmq14} since one can no longer integrate by
parts in $t$. Recall here that there was also
one term out of control, that is, \eqref{surout}.
Our crucial observation is that there is a  cancelation between them since $ q=g h-\sigma H+2\varepsilon S^\varphi v  \,{\bf n} \cdot \,{\bf n} $ on
$\{z=0\}$. This motivates us to estimate together the first term in \eqref{rmq14} and the second surface tension term in \eqref{I0'm}, by the
kinematic boundary condition,
  \begin{align}\label{rmq15}
  & \int_{z= 0}   \sigma \partial_t^m H    {\bf N} \cdot V^m \, dy-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \partial_t^{m} q \, dy
  \\\nonumber&\quad= \int_{z= 0} \sigma \partial_t^m  H \({\bf N} \cdot V^m+  m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v\)\, dy
   \\\nonumber&\qquad-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \(g\partial_t^mh +2\varepsilon \partial_t^m \( S^\varphi v  \,{\bf n} \cdot \,{\bf n}\)\)\, dy
  \\\nonumber&\quad= \int_{z= 0} \sigma \partial_t^m  H \(\partial_t^{m+1}h+v_y\cdot\nabla_y \partial_t^m h+\partial^{\varphi}_zv\cdot{\bf N} \partial_t^m h- \tilde{\mathcal{C}}^m(h) \)\,
  dy
  \\\nonumber&\qquad-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \(g\partial_t^mh +2\varepsilon \partial_t^m \( S^\varphi v  \,{\bf n} \cdot \,{\bf n}\)\)\, dy,
    \end{align}
where $\tilde{\mathcal{C}}^m(h)$ is the commutator $\tilde{\mathcal{C}}^\alpha(h)$ defined by \eqref{tildech} for the case $\alpha_0=m$.
 Note that the last term in \eqref{rmq15} can be estimated as follows, thanks to Lemma \ref{lembord},
   \begin{align}\label{rmq151}
  & -\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \(g\partial_t^mh +2\varepsilon \partial_t^m \( S^\varphi v  \,{\bf n} \cdot \,{\bf n}\)\)\, dy
  \\\nonumber&\quad\le \Lambda_\infty \abs{\partial_t^{m-1} v}_{\frac{1}{2}}\(\abs{\partial_t^mh}_{-\frac{1}{2}}+\varepsilon\abs{\partial_t^m \( S^\varphi v  \,{\bf n} \cdot \,{\bf
  n}\)}_{-\frac{1}{2}}\)
  \\\nonumber&\quad\le \Lambda_\infty \abs{\partial_t^{m-1} v}_{\frac{1}{2}}\(\abs{\partial_t^mh}_{-\frac{1}{2}}+\varepsilon\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}+\varepsilon\abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}\).
    \end{align}
The integration by parts and \eqref{bordhC1} yield
 \begin{align}\label{rmq152}
 &\int_{z= 0} \sigma \partial_t^m  H \( \partial^{\varphi}_zv\cdot{\bf N} \partial_t^m h- \tilde{\mathcal{C}}^m(h) \)\, dy
 \\\nonumber&\quad\le \sigma\abs{  \partial_t^m\( \frac{ \nabla_y  h}{\sqrt{1+|\nabla_y h|^2}} \)}_{0}\abs{\nabla_y\(\partial^{\varphi}_zv\cdot{\bf N} \partial_t^m
 h-\tilde{\mathcal{C}}^m(h)\)}_{0}
 \\&\nonumber\quad \le \Lambda_\infty   \sigma\abs{h}_{{\mathbb X}^{m,1}}\(\abs{\partial_t^m h}_{1}+\abs{h}_{{\mathbb X}^{m-1,2}}+\abs{v}_{{\mathbb X}^{m-2,1}} \).
    \end{align}
It follows from \eqref{sigsur1} and \eqref{sigsur2} that
  \begin{align}\label{sigmaes0m}
  & \int_{z= 0}  \sigma Z^\alpha H \(\partial_t^{m+1}h+v_y\cdot\nabla_y \partial_t^m h\)\,dy
  \\\nonumber&\quad= \int_{z= 0}\sigma\nabla_y \cdot\( \frac{ \nabla_y \partial_t^{m}h }{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y
  \partial_t^{m}h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h
  +\mathcal{C}(\mathcal{B}_\sigma^m) \)\(\partial_t^{m+1}h+v_y\cdot\nabla_y \partial_t^m h\)\,dy,
    \end{align}
where
\begin{align}\label{sigsur2m}
 \mathcal{C}(\mathcal{B}_\sigma^m)=-\[\partial_t^{m-1},\frac{\nabla_y h}{\sqrt{1+|\nabla_y h|^2}^3}\]\cdot \nabla_y \partial_t h \nabla_y h+\[\partial_t^m,
 \frac{1}{\sqrt{1+|\nabla h|^2}}, \nabla_y h\].
\end{align}
Similarly as \eqref{sigmaes3}--\eqref{sigmaes5}, one can deduce that
\begin{align}\label{rmq153}
&\int_{z= 0}\sigma \nabla_y\cdot \( \frac{ \nabla_y \partial_t^{m}h }{\sqrt{1+|\nabla_y h|^2}}- \frac{\nabla_y h\cdot \nabla_y
\partial_t^{m}h}{\sqrt{1+|\nabla_y h|^2}^3}\nabla_y h \)\(\partial_t^{m+1}h+v_y\cdot\nabla_y \partial_t^m h\)\,dy
\\\nonumber&\quad\le-\frac{1}{2}\frac{d}{dt}\int_{z=0}  \sigma\( \frac{  |\nabla_y \partial_t^{m} h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y
h\cdot \nabla_y \partial_t^{m} h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy+\Lambda_\infty   \sigma\abs{h}_{{\mathbb X}^{m,1}}^2.
   \end{align}
Integrate by parts in both $t$ and $y$ to have
  \begin{align}\label{rmq154}
\nonumber \int_{z= 0}\sigma
 \nabla_y\cdot\( \mathcal{C}(\mathcal{B}_\sigma^m)\) \partial_t^{m+1}h\,dy&\le - \frac{d}{dt}\int_{z= 0}\sigma
  \mathcal{C}(\mathcal{B}_\sigma^m)\cdot \nabla_y  \partial_t^{m}h\,dy+
  \int_{z= 0}\sigma
 \partial_t \mathcal{C}(\mathcal{B}_\sigma^m) \cdot\nabla_y\partial_t^{m}h\,dy
 \\&\le - \frac{d}{dt}\int_{z= 0}\sigma
  \mathcal{C}(\mathcal{B}_\sigma^m)\cdot \nabla_y  \partial_t^{m}h\,dy+
 \Lambda_\infty   \sigma\abs{h}_{{\mathbb X}^{m,1}}^2.
    \end{align}
One easily has
  \begin{align}\label{rmq155}
  \int_{z= 0}\sigma\nabla_y \cdot\(  \mathcal{C}(\mathcal{B}_\sigma^m) \) v_y\cdot\nabla_y \partial_t^m h \,dy\le \Lambda_\infty
  \sigma\abs{h}_{{\mathbb X}^{m-1,2}}\abs{h}_{{\mathbb X}^{m,1}}.
    \end{align}
Hence, by the estimates \eqref{rmq151}, \eqref{rmq152}, \eqref{rmq153}--\eqref{rmq155}, one may conclude from \eqref{rmq15} that
\begin{align}\label{rmq15m}
  & \int_{z= 0}   \sigma \partial_t^m H    {\bf N} \cdot V^m \, dy-\int_{ {z=0}}   m\partial_t  {\bf N}  \cdot \partial_t^{m-1} v  \partial_t^{m} q \, dy
  \\\nonumber&\quad\le-\frac{1}{2}\frac{d}{dt}\int_{z=0}  \sigma\( \frac{  |\nabla_y \partial_t^{m} h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y
  h\cdot \nabla_y \partial_t^{m} h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy- \frac{d}{dt}\int_{z= 0}\sigma
  \mathcal{C}(\mathcal{B}_\sigma^m)\cdot \nabla_y  \partial_t^{m}h\,dy
    \\\nonumber&\qquad+
 \Lambda_\infty  \( \sigma\abs{h}_{{\mathbb X}^{m,1}}^2 +\sigma\abs{h}_{{\mathbb X}^{m,1}} \abs{v}_{{\mathbb X}^{m-2,1}} + \abs{\partial_t^{m-1}
 v}_{\frac{1}{2}}\(\abs{\partial_t^mh}_{-\frac{1}{2}}+\varepsilon\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}+\varepsilon\abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}\)\).
\end{align}
This in particular finishes the estimates of the second surface tension term in \eqref{I0'm} and $\mathcal{R}^m_Q$, which can be stated as
follows:
\begin{align}\label{msigQ}
  & \int_{z= 0}   \sigma \partial_t^m H    {\bf N} \cdot V^m \, dy+\mathcal{R}^m_Q
  \\\nonumber&\quad\le-\frac{1}{2}\frac{d}{dt}\int_{z=0}  \sigma\( \frac{  |\nabla_y \partial_t^{m} h|^2 }{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y
  h\cdot \nabla_y \partial_t^{m} h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy- \frac{d}{dt}\mathcal{G}^m
    \\\nonumber&\qquad+
 \Lambda_\infty  \( \sigma\abs{h}_{{\mathbb X}^{m,1}}^2 +\sigma\abs{h}_{{\mathbb X}^{m,1}} \abs{v}_{{\mathbb X}^{m-2,1}} + \abs{\partial_t^{m-1}
 v}_{\frac{1}{2}}\(\abs{\partial_t^mh}_{-\frac{1}{2}}+\varepsilon\abs{h}_{{\mathbb X}^{m,\frac{1}{2}}}+\varepsilon\abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}\)\)
    \\\nonumber&\qquad+\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\norm{  v}_{{\mathbb X}^{m}}+\norm{\nabla  v}_{{\mathbb X}^{2}}\)\norm{\partial_t^{m-1} q}_{H^1},
\end{align}
where
\begin{align}
\mathcal{G}^m=&  \int_{\Omega} \( \mathcal{C} ^{m} (d)_2+\mathcal{C} ^{m} (d)_3+\mathcal{C} ^{m} (d)_4+\mathcal{C} ^{m} (d)_5\) \partial_t^{m-1} q
\, dydz
\\\nonumber&-\int_{\Omega}   m \partial_z\( \partial_t^{m-1} q\partial_t  {\bf N}\)\cdot \partial_t^{m-1} v  \, dydz+\int_{z= 0}\sigma
  \mathcal{C}(\mathcal{B}_\sigma^m)\cdot \nabla_y  \partial_t^{m}h\,dy.
\end{align}

It remains to estimates the commutators $\mathcal{R}_C^m$ and $\mathcal{R}_S^m$. It follows from \eqref{CT}, \eqref{Cq}, \eqref{Cd},
\eqref{equiv1}, \eqref{qpressurem} and \eqref{qminfty} that
  \begin{align}\label{RCestimatem}
  \mathcal{R}_{C}^m &\le\Lambda_\infty\( \( \norm{ \partial_t^m \eta}+\norm{   \mathcal{C}^m(\mathcal{T})}+ \norm{ \mathcal{C}^m(q)}\) \cdot V^m +  \norm{
  \mathcal{C}^m (d)} \norm{ \partial_t^m\eta} \)
  \\\nonumber& \le \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}+ \norm{v}_{{\mathbb X}^{m-1}}+\norm{\nabla v}_{{\mathbb X}^{m-1}}+ \norm{\nabla
  q}_{{\mathbb X}^{m-1}}\)\(\norm{V^m}+\norm{\partial_t^m\eta}\)
   \\\nonumber &\le \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-1,2}}+\norm{v}_{{\mathbb X}^{m}}  + \norm{\nabla v}_{{\mathbb X}^{m-1}} +
   \varepsilon\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\)
  \\\nonumber&\quad\times \(\norm{v}_{{\mathbb X}^m}+\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}\).
  \end{align}
Similarly as \eqref{RSestimate}, it holds that
\begin{align}\label{RSestimatem}
\mathcal{R}_{S}^m\leq&\Lambda_\infty\(\(\varepsilon^{\frac{1}{2}}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\varepsilon \abs{h}_{{\mathbb X}^{m-1, \frac{3}{2}}}+\varepsilon\norm{\nabla
v}_{{\mathbb X}^{m-1}}\)\(\norm{v}_{{\mathbb X}^m}+\abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}\)\right.
\\\nonumber&\left.\qquad +\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^m}+ \(\abs{h}_{{\mathbb X}^{m-1,1} }  +  \abs{
v}_{{\mathbb X}^{m-1,1}}\)\(\abs{h}_{{\mathbb X}^{m} }  +  \abs{  v}_{{\mathbb X}^{m}}\)\).
\end{align}

We can now finish the proof of the proposition. As a consequence of \eqref{m11m}, \eqref{m11m1}, \eqref{Ibestimatem}, \eqref{msigQ},
\eqref{RCestimatem}, \eqref{RSestimatem} and Cauchy's inequality, one may deduce from \eqref{conormen} that
\begin{align}\label{alphaend1m}
&\frac{1}{2} \frac{d}{dt} \mathcal{E}^m+ \frac{d}{dt}\mathcal{G}^m + 2 \varepsilon \int_{\Omega} \abs{S^\varphi V^m}^2\, d\mathcal{V}_{t}
\\\nonumber&\quad \le\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m }}^2 + \sigma     \abs{h}_{{\mathbb X}^{m,1}}^2
 + \norm{\nabla v}_{{\mathbb X}^{m-1}}^2 + \norm{v}_{{\mathbb X}^{m}}^2 + \varepsilon  \abs{v}_{{\mathbb X}^{m}}^2+ \varepsilon  \abs{h}_{{\mathbb X}^{m , \frac{1}{2}}}^2 \right.
  \\\nonumber&\qquad\qquad \left.+ \varepsilon\abs{  v }_{{{\mathbb X}^{m , \frac{1}{2}}}} \(\norm{v}_{{\mathbb X}^{m }}+\norm{\nabla v}_{{\mathbb X}^{2}}+\abs{h}_{{{\mathbb X}^{m ,\frac{1}{2}}}}\)
+\varepsilon\(  \norm{ \nabla v}_{{\mathbb X}^{m-1}}+  \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }} \) \norm{ \nabla V^m}  \),
\end{align}
where
\begin{equation}
\mathcal{E}^m:=\int_{\Omega}\abs{V^m }^2 d\mathcal{V}_{t} +\int_{z=0} \( g- \partial^{\varphi}_{z} q\) \abs{\partial_t^m h }^2+ \sigma\( \frac{  |\nabla_y \partial_t^{m} h|^2
}{\sqrt{1+|\nabla_y h|^2}}-  \frac{ |\nabla_y h\cdot \nabla_y \partial_t^{m} h|^2  }{\sqrt{1+|\nabla_y h|^2}^3}   \)\,dy.
\end{equation}
Similarly as \eqref{alphaend2}, by the trace estimates
$$\abs{v}_{{\mathbb X}^{m,\frac{1}{2}}}\lesssim \norm{v}_{{\mathbb X}^{m}}+\norm{\nabla v}_{{\mathbb X}^{m}}\text{ and }
\abs{v}_{{\mathbb X}^{m}}\lesssim \norm{\nabla v}_{{\mathbb X}^{m}}^\frac{1}{2}\norm{v}_{{\mathbb X}^{m}}^\frac{1}{2}+\norm{v}_{{\mathbb X}^{m}}, $$
using Cauchy's inequality and
\eqref{mheps2}, one can then deduce that
\begin{align}\label{alphaend2m}
&\norm{\partial_t^m v(t)}^2+\abs{\partial_t^m h(t)}_{0}^2 +\sigma\abs{\partial_t^m h(t)}_{1}^2+ \varepsilon  \abs{h(t)}_{{\mathbb X}^{m , \frac{1}{2}}}^2 +\varepsilon\int_0^t  \norm{\nabla \partial_t^m
v}_{0}^2
\\\nonumber&\quad\le \Lambda_0\( \norm{\partial_t^m v(0)}^2+\abs{\partial_t^m h(0)}_{0}^2 +\sigma\abs{\partial_t^m h(0)}_{1}^2+ \varepsilon  \abs{h(0)}_{{\mathbb X}^{m ,
\frac{1}{2}}}^2\)-\mathcal{G}^m
\\\nonumber&\qquad+ \int_0^t\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m }}^2 + \sigma     \abs{h}_{{\mathbb X}^{m,1}}^2
 + \norm{v}_{{\mathbb X}^{m}}^2+ \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 + \varepsilon  \norm{\nabla v}_{{\mathbb X}^{m-1,1}}^2+ \varepsilon  \abs{h}_{{\mathbb X}^{m , \frac{1}{2}}}^2  \).
\end{align}
Note that
\begin{align}\label{ges}
 -\mathcal{G}^m\le \Lambda_\infty\(\(\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}+\abs{v}_{{\mathbb X}^{m-1,1 }}+\norm{\partial_z v}_{{\mathbb X}^{m-2
 }}\)\norm{\partial_t^{m-1}q}_{H^1}+\sigma\abs{h}_{{\mathbb X}^{m-1,2 }}\abs{h}_{{\mathbb X}^{m,1 }}\).
\end{align}
In contrast to the previous case, the difficulty here is that $\norm{\partial_t^{m-1}q}_{H^1}$ and hence $-\mathcal{G}^m$  are not in $L^\infty([0,T])$
but only in $L^2([0,T])$.
Our basic idea is to integrate in time twice. Indeed, we take the square and then integrate in time to have, by Cauchy's inequality,
 \begin{align}\label{alphaend2m1}
&\int_0^t\(\norm{\partial_t^m v }^2+\abs{\partial_t^m h }_{0}^2 +\sigma\abs{\partial_t^m h }_{1}^2+ \varepsilon  \abs{h }_{{\mathbb X}^{m , \frac{1}{2}}}^2 \)^2+\int_0^t\(\varepsilon\int_0^s
\norm{\nabla \partial_t^m v}_{0}^2\)^2
\\\nonumber&\quad\le t\Lambda_0\( \norm{\partial_t^m v(0)}^2+\abs{\partial_t^m h(0)}_{0}^2 +\sigma\abs{\partial_t^m h(0)}_{1}^2+ \varepsilon  \abs{h(0)}_{{\mathbb X}^{m ,
\frac{1}{2}}}^2\)^2+\int_0^t\abs{\mathcal{G}^m}^2
\\\nonumber&\qquad+ t\(\int_0^t\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m }}^2 + \sigma     \abs{h}_{{\mathbb X}^{m,1}}^2
  + \norm{v}_{{\mathbb X}^{m}}^2 + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2+ \varepsilon  \norm{\nabla v}_{{\mathbb X}^{m-1,1}}^2+ \varepsilon  \abs{h}_{{\mathbb X}^{m , \frac{1}{2}}}^2  \)\)^2.
\end{align}
It follows from \eqref{ges} and \eqref{qpressurem} that
\begin{align}\label{hles}
 \int_0^t\abs{\mathcal{G}^m}^2\le& \int_0^t\Lambda_\infty\(\(\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}^2+\abs{v}_{{\mathbb X}^{m-1,1 }}^2+\norm{\partial_z v}_{{\mathbb X}^{m-2
 }}^2\)\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-1,2}}\right. \right.
 \\\nonumber&\left.\left.+\norm{v}_{{\mathbb X}^{m}}  + \norm{\partial_z v}_{{\mathbb X}^{m-1}} +  \varepsilon\abs{  v }_{{{\mathbb X}^{m-1,
 \frac{3}{2}}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}\)^2+\sigma^2\abs{h}_{{\mathbb X}^{m-1,2 }}^2\abs{h}_{{\mathbb X}^{m,1 }}^2\)
 \\\nonumber\le& \int_0^t\Lambda_\infty \(\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}^2+\sigma \abs{h}_{{\mathbb X}^{m-1,2 }}^2+\abs{v}_{{\mathbb X}^{m-1,1 }}^2+\norm{\partial_z v}_{{\mathbb X}^{m-2 }}^2\)
 \\\nonumber& \times\(\abs{h}_{{\mathbb X}^{m,-{1 \over 2 }}}^2+\sigma \abs{h}_{{\mathbb X}^{m,1}}^2 +\norm{v}_{{\mathbb X}^{m}}^2  + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2 +
 \varepsilon^2\abs{  v }_{{{\mathbb X}^{m-1, \frac{3}{2}}}}^2+\varepsilon^2\abs{h}_{{{\mathbb X}^{m-1,\frac{3}{2}}}}^2\)  .
\end{align}
We thus conclude \eqref{conormales1} by plugging the estimate \eqref{hles} into \eqref{alphaend2m1}.
 \end{proof}

\section{Normal derivative estimates}\label{secnormal}

In view of the conormal estimates in Propositions \ref{conormv1} and \ref{conormvm} in Section \ref{secconormal}, the next main step is to
estimate
$\norm{\partial_{z}v}_{{\mathbb X}^{m-1}}$.

Recall the definition of $\Lambda_\infty$ from \eqref{deflambdainfty1} and all the facts of the $L^\infty$ controls elaborated in the beginning of
Section \ref{secconormal}. Note that $\Lambda_{\infty}$ involves only $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{L^\infty}$. For the case without
surface tension \cite{MasRou} that involves only the spatial derivatives, this is sufficient for deriving the normal derivative estimates since
in
such situation applying the product or commutator estimates to control the commutators resulting from the viscosity term needs only the control
of $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{L^\infty}$. However, in the current case that involves the time derivatives, following the arguments
of \cite{MasRou} would require the control of $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{{\mathbb Y}^{k}}$ for some $k\ge 1$. Recall from Proposition 9.8 in
\cite{MasRou} that deriving the bound of $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{L^\infty}$ requires a crucial use of the heat kernel and the first
order compatibility condition $S_{{\bf n}}|_{t=0}=0$ on the boundary. Hence, to control $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{{\mathbb Y}^{k}}$, it seems to
involve much more delicate use of the various properties of the heat kernel for the time differentiated problems; furthermore, it should require
more compatibility conditions of initial data.

Our key observation here is that since in the vicinity of the boundary the solution behaves as $v(t,x)\sim
v^0(t,x)+\sqrt{\varepsilon}U(t,y,z/\sqrt{\varepsilon})$, it indicates that there may be better control of $\varepsilon \partial_{zz}v $ (and even $\varepsilon
\partial_{zzz}v$!)
in Sobolev conormal spaces. This is indeed the case as shown in the following lemma.
\begin{lem}\label{dzzlem}
It holds that
\begin{align}\label{dzzves11}
\varepsilon \norm{\partial_{zz}v}_{{\mathbb Y}^{\frac{m}{2}+1}}+\varepsilon \norm{\partial_{zzz}v}_{{\mathbb Y}^{\frac{m}{2}}}\le \Lambda_\infty
\end{align}
and for $k\le m-1$:
\begin{align}\label{dzzves221}
\varepsilon \norm{\partial_{zz}v}_{{\mathbb X}^{k}}
\le  \Lambda_\infty& \( \varepsilon\(\norm{\partial_z   v}_{{\mathbb X}^{k,1}}+\abs{ h}_{{\mathbb X}^{k,\frac{3}{2}}}\)  +\norm{\partial_zv}_{{\mathbb X}^{k}}+\norm{v}_{
{{\mathbb X}^{k+1}}}+\abs{h}_{{\mathbb X}^{k+1,-\frac{1}{2}}}+\norm{\nabla  q}_{{\mathbb X}^{k}}\)
\end{align}
and
\begin{align}\label{dzzves22'}
\varepsilon \norm{\partial_{zzz}v}_{{\mathbb X}^{k-1}}
\le  \Lambda_\infty& \( \varepsilon\(\norm{\nabla v}_{{\mathbb X}^{k-1,2}}+\abs{ h}_{{\mathbb X}^{k-1,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}} \right.
\\&\nonumber\   +\norm{\partial_zv}_{{\mathbb X}^{k}} +\norm{v}_{ {{\mathbb X}^{k,1}}} +\norm{\nabla  q}_{{\mathbb X}^{k-1,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{k-1}}\Big) .
\end{align}
\end{lem}
\begin{proof}
It follows from the first equation in \eqref{NSv}, \eqref{transportW} and \eqref{deltaphi} that
\begin{align}\label{eqdzzv}
\varepsilon\partial_{zz}  v ={1 \over E_{33}}  \( -  \varepsilon \sum_{j<3} \partial_{z} \(E_{3,j}\partial_{j} v\)   -\varepsilon\sum_{i<3, \, j} \partial_{i}\( E_{i
j} \partial_{j} v\) +\partial_z\varphi\(\partial_t v+v_y\cdot\nabla_y v\)+v_z\partial_z v +\partial_z\varphi \nabla^\varphi  q  \).
\end{align}
This implies that, since $v_z=0$ on the boundary,
\begin{align}\label{dzzves}
\varepsilon \norm{\partial_{zz}v}_{{\mathbb Y}^{\frac{m}{2}+1}}
\le  \Lambda& \(\frac{1}{c_0},\varepsilon\(\norm{\nabla \nabla_y v}_{{\mathbb Y}^{\frac{m}{2}+1}}+\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+3}}\)+\abs{h}_{{\mathbb Y}^{\frac{m}{2}+2}}
\right. \\\nonumber&\qquad\ \left.+\norm{\partial_zv}_{{\mathbb Y}^{\frac{m}{2}+1}}+\norm{v}_{{\mathbb Y}^{\frac{m}{2}+2}}+\norm{\nabla  q}_{{\mathbb Y}^{\frac{m}{2}+1}}\)\le
\Lambda_\infty .
\end{align}
Applying $\partial_z$ to \eqref{eqdzzv}, using the estimate \eqref{dzzves}, one deduces
\begin{align}\label{dzzves2}
\varepsilon \norm{\partial_{zzz}v}_{{\mathbb Y}^{\frac{m}{2}}}
&\le  \Lambda \(\frac{1}{c_0},\varepsilon\(\abs{ h}_{{\mathbb Y}^{\frac{m}{2}+3}}+\norm{\nabla^2\nabla_y  v}_{{\mathbb Y}^{\frac{m}{2}}}\)+\abs{h}_{{\mathbb Y}^{\frac{m}{2}+2}}
\right. \\\nonumber&\qquad\qquad \left.+\norm{\partial_zv}_{{\mathbb Y}^{\frac{m}{2}+1}}+\norm{v}_{{\mathbb Y}^{\frac{m}{2}+2}}+\norm{\partial_z \nabla
q}_{{\mathbb Y}^{\frac{m}{2}}}+\norm{ \nabla  q}_{{\mathbb Y}^{\frac{m}{2}}}\)\le \Lambda_\infty ,
\end{align}
where one has used the fact that, since $v_z=0$ on the boundary,
\begin{align}\label{dzzves23}
\nonumber\norm{ \partial_z(v_z\partial_z  v)}_{{\mathbb Y}^{\frac{m}{2}}}&\le \norm{ \partial_z  v_z \partial_z  v }_{{\mathbb Y}^{\frac{m}{2}}}+\norm{ v_z\partial_{zz}  v
}_{{\mathbb Y}^{\frac{m}{2}}}=\norm{ \partial_z  v_z \partial_z  v }_{{\mathbb Y}^{\frac{m}{2}}}+\norm{ \frac{1}{z(z+b)}v_zZ_3 \partial_z v}_{{\mathbb Y}^{\frac{m}{2} }}
\\&\lesssim \norm{ \partial_z  v_z}_{{\mathbb Y}^{\frac{m}{2}}} \norm{\partial_z  v }_{{\mathbb Y}^{\frac{m}{2}}}+\norm{ \partial_zv_z}_{{\mathbb Y}^{\frac{m}{2} }}\norm{ \partial_z
v}_{{\mathbb Y}^{\frac{m}{2}+1 }}.
\end{align}
Combining \eqref{dzzves} and \eqref{dzzves23} proves the estimates \eqref{dzzves11}.

The estimates \eqref{dzzves221}--\eqref{dzzves22'} follow by a similar argument as that for \eqref{dzzves11}. Indeed, it follows from
\eqref{eqdzzv} and
\eqref{gues} that for $k\ge m-1$,
\begin{align}
\nonumber\varepsilon \norm{\partial_{zz}v}_{{\mathbb X}^{k}}
\le  \Lambda_\infty  \( \varepsilon\(\norm{\nabla \nabla_y v}_{{\mathbb X}^{k}}+\abs{ h}_{{\mathbb X}^{k,\frac{3}{2}}}\)+\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}  +\norm{\partial_zv}_{{\mathbb X}^{k}}+\norm{v}_{
{{\mathbb X}^{k+1}}}+\abs{h}_{{\mathbb X}^{k+1,-\frac{1}{2}}}+\norm{\nabla  q}_{{\mathbb X}^{k}}\) .
\end{align}
This yields \eqref{dzzves221}. Now applying $Z_i$, $i=1,2,3$ to \eqref{eqdzzv} and then using \eqref{gues} again lead to
\begin{align}\label{dzzvesh12}
\varepsilon \norm{\partial_{zz}v}_{{\mathbb X}^{k-1,1}}
\le  \Lambda_\infty& \( \varepsilon\(\norm{\nabla \nabla_y v}_{{\mathbb X}^{k-1,1}}+\abs{ h}_{{\mathbb X}^{k-1,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{k-1,\frac{3}{2}}} \right.
\\&\nonumber\left. \ +\norm{\partial_zv}_{{\mathbb X}^{k-1,1}}+\norm{v}_{ {{\mathbb X}^{k,1}}}+\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}+\norm{\nabla  q}_{{\mathbb X}^{k-1,1}}\) .
\end{align}
Then applying $\partial_z$ to \eqref{eqdzzv} and using the estimate \eqref{dzzves11} and the similar observation as in \eqref{dzzves23} give
\begin{align}\label{dzzvesh13}
\varepsilon \norm{\partial_{zzz}v}_{{\mathbb X}^{k-1}}
\le  \Lambda_\infty& \( \varepsilon\(\norm{\nabla^2 \nabla_y v}_{{\mathbb X}^{k-1}}+\abs{ h}_{{\mathbb X}^{k-1,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{k-1,\frac{3}{2}}}\right.
\\&\nonumber\ \left.  +\norm{\partial_zv}_{{\mathbb X}^{k}}+\norm{v}_{ {{\mathbb X}^{k-1,1}}}+\abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}+\norm{\partial_z\nabla  q}_{{\mathbb X}^{k-1}}+\norm{ \nabla
q}_{{\mathbb X}^{k-1}}\) .
\end{align}
Combining \eqref{dzzvesh13} and \eqref{dzzvesh12} thus proves the estimate \eqref{dzzves22'}.
\end{proof}

Lemma \ref{dzzlem} then allows one to derive the normal derivative estimates.

\subsection{Estimate of  $\norm{\partial_z v}_{{\mathbb X}^{m-2}}$}

As in Section 8 of \cite{MasRou}, one defines
\begin{equation}\label{Sn}
S_{{\bf n}}= \Pi \(S^\varphi v   {\bf n}-\kappa\chi v\).
\end{equation}
The main advantages of this quantity are that
\begin{equation}\label{Snb}
S_{{\bf n}}=0\text{ on }\{z=0,-b\}
\end{equation}
and that the following estimates hold:
\begin{lem}\label{lemdzS}
For every $k \geq 0$:
\begin{equation}\label{dzvk}
\norm{ \partial_{z} v}_{{\mathbb X}^{k}} \le \Lambda_\infty \(\norm{ S_{{\bf n}}}_{{\mathbb X}^{k}} + \abs{h}_{{\mathbb X}^{k,\frac{1}{2}}} +  \norm{ v}_{{\mathbb X}^{k,1}}\)
\end{equation}
and
\begin{equation}\label{dzzvk}
\norm{\partial_{zz} v}_{{\mathbb X}^{k}} \le \Lambda_\infty\( \norm{\partial_z S_{{\bf n}} }_{{\mathbb X}^{k}} + \abs{h}_{{\mathbb X}^{k,\frac{3}{2}}} + \norm{v}_{{\mathbb X}^{k,2}}\).
\end{equation}
\end{lem}
\begin{proof}
We start with the estimate \eqref{dzvk}. The normal component of $\partial_{z}v$ is given by, due to the divergence free condition,
\begin{equation}\label{eqdzvnk}
\partial_{z} v \cdot  {\bf n}=-\frac{\partial_{z} \varphi}{\abs{{\bf N}}} \(\partial_{1} v_{1}+ \partial_{2} v_{2}\).
\end{equation}
Then it follows from \eqref{gues} and Lemma \ref{propeta} that
 \begin{equation}\label{vnorm}
\norm{\partial_{z}v \cdot {\bf n} }_{{\mathbb X}^{k}} \le \Lambda_\infty \( \norm{ v   }_{{\mathbb X}^{k,1}} + \abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}\).
\end{equation}
It thus suffices to estimate the tangential components of $\partial_{z} v $. Recall from the derivation of \eqref{eqdzvPi'} that
\begin{equation}\label{eqdzvPin}
\Pi \partial_{z}v={\partial_{z} \varphi \over \abs{{\bf N}}}\(\partial_{1}\varphi\Pi \partial_{1} v + \partial_{2}\varphi \Pi \partial_{2}
v+2S_{{\bf n}}-g^{ij} \partial_{j}v \cdot {\bf n} \Pi \partial_{y^i} -\kappa\chi\Pi v\),
\end{equation}
which follows from the derivation of \eqref{eqdzvPi'}. Hence, by \eqref{gues} and Lemma \ref{propeta}, one has
\begin{equation} \label{vnormm}
\norm{\Pi \partial_{z}v }_{{\mathbb X}^{k}} \le \Lambda_\infty \( \norm{ v   }_{{\mathbb X}^{k,1}} + \abs{h}_{{\mathbb X}^{k,\frac{1}{2}}}+\norm{S_{{\bf n}}}_{{\mathbb X}^{k}}+\norm{\partial_{z}v \cdot
{\bf n} }_{{\mathbb X}^{k}} \).
\end{equation}
Thus the estimate \eqref{dzvk} follows from combining \eqref{vnorm} and \eqref{vnormm}.

The estimate \eqref{dzzvk} follows from applying $\partial_z$ to \eqref{eqdzvnk} and \eqref{eqdzvPin}, employing a similar argument as
for \eqref{dzvk} and using \eqref{dzvk}.
\end{proof}

In light of Lemma \ref{lemdzS}, we then turn to estimate $S_{{\bf n}}$ instead of $\partial_{z} v $. It follows from the first equation in \eqref{NSv}
that
\begin{equation}\label{eqS0}\partial^{\varphi}_{t} \nabla^\varphi v+ \(v \cdot \nabla^\varphi\) \nabla^\varphi v  + \(\nabla^\varphi v\)^2 + \(\nabla^\varphi\)^2 q  -\varepsilon
\Delta^\varphi \nabla^\varphi v = 0.
\end{equation}
Taking the symmetric part of \eqref{eqS0} yields
\begin{equation}\label{eqS}
\partial^{\varphi}_{t} S^\varphi v+ \(v \cdot \nabla^\varphi\) S^\varphi v  + \frac{1}{2}\((\nabla^\varphi v)^2+ ((\nabla^\varphi v)^t)^2\) +  \(\nabla^\varphi\)^2 q
-\varepsilon \Delta^\varphi (S^\varphi v)= 0.
\end{equation}
Consequently,
\begin{equation}\label{eqPiS}
\partial^{\varphi}_{t}  S_{{\bf n}}+ \(v \cdot \nabla^\varphi\) S_{{\bf n}}  -\varepsilon \Delta^\varphi (S_{{\bf n}})= F_{S}
\end{equation}
where
\begin{equation}\label{FSdef}
F_{S}= F_{S}^1 + F_{S}^2+ F_{S}^3+ F_{S}^4
\end{equation}
with
\begin{align}
\label{FS1}& F_{S}^1 =   - \frac{1}{2}  \Pi\( (\nabla^\varphi v)^2+ ((\nabla^\varphi v)^t)^2\) {\bf n} + \( \partial_{t} \Pi + v \cdot \nabla^\varphi \Pi
\)S^\varphi v  {\bf n} +    \Pi S^\varphi v \( \partial_{t} {\bf n} + v \cdot \nabla^\varphi {\bf n}\)
\\&\qquad+\kappa \(\partial^{\varphi}_{t}+  v \cdot \nabla^\varphi\) (\chi \Pi v),\nonumber   \\
\label{FS4} &F_{S}^2 =- \Pi\( \(\nabla^\varphi\)^2 q \){\bf n},  \\
\label{FS2}& F_{S}^3 =    - \varepsilon \(\Delta^\varphi \Pi  \)S^\varphi v  {\bf n}  - \varepsilon  \Pi S^\varphi v \Delta^\varphi {\bf n}, \\
\label{FS3}& F_{S}^4 =   - 2 \varepsilon  \partial^{\varphi}_{i} \Pi \partial^{\varphi}_{i} \( S^\varphi v  {\bf n} \)  -  2 \varepsilon  \Pi \( \partial^{\varphi}_{i}\( S^\varphi v \)  \partial^{\varphi}_{i} {\bf n} \)-\varepsilon
\kappa\Delta^\varphi (\chi\Pi v).
\end{align}
Note that the pressure estimates in \eqref{qpressurem} indicate that at most, one can estimate only $\norm{S_{{\bf n}}}_{{\mathbb X}^{m-2}}$ and hence
$\norm{\partial_z v}_{{\mathbb X}^{m-2}}$ at this stage. However, we shall prove a control of $\norm{\partial_z v}_{{\mathbb X}^{m-1}}$ based on the vorticity equation in
the next
subsection.

\begin{prop}\label{propdzvm-2}
Any smooth solution of \eqref{NSv} satisfies the estimate
 \begin{align} \label{estSnm-2}
& \norm{S_{{\bf n}}}_{{\mathbb X}^{m-2}}^2    + \varepsilon \int_{0}^t \norm{ \nabla S_{{\bf n}}}_{{\mathbb X}^{m-2}}^2  \\
&\nonumber\le \Lambda_0\norm{S_{{\bf n}}(0)}_{{\mathbb X}^{m-2}}^2+\int_0^t \Lambda_\infty \varepsilon^2\(\norm{\nabla v}_{{\mathbb X}^{m-2,2}}^2+\abs{h}_{{\mathbb X}^{m-2,\frac{5}{2}}}^2\)
\\\nonumber&\quad  +\int_0^t \Lambda_\infty \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2+\sigma^2 \abs{h}_{{\mathbb X}^{m-2,3}}^2+\norm{v}_{{\mathbb X}^{m-1,1}}^2  + \norm{\partial_z
v}_{{\mathbb X}^{m-2,1}}^2 +\norm{ S_{{\bf n}} }_{{\mathbb X}^{m-2}}^2  .
\end{align}
\end{prop}
\begin{proof}
We start with the estimates of $F_S$. It follows from \eqref{gues} and Lemma \ref{propeta} that
\begin{align}
\label{FS1es}&
\norm{ F_{S}^1}_{{\mathbb X}^{m-2}} \le\Lambda_\infty\( \norm{\nabla  v }_{{\mathbb X}^{m-2}}  + \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \norm{v}_{{\mathbb X}^{m-2}} \),\\
\label{FS2es}&
\norm{ F_{S}^2 }_{{\mathbb X}^{m-2}} \le \Lambda_\infty \norm{\nabla q}_{{\mathbb X}^{m-2,1}},\\
\label{FS3es}&
\norm{ F_{S}^3 }_{{\mathbb X}^{m-2}} \le\Lambda_\infty \varepsilon   \(\norm{\nabla v}_{{\mathbb X}^{m-2}}  + \abs{h}_{{\mathbb X}^{m-2, \frac{5}{2} }}\),\\
\label{FS4es}&
\norm{ F_{S}^4 }_{{\mathbb X}^{m-2}}
\le \Lambda_\infty \varepsilon\(\abs{h}_{{\mathbb Y}^{\frac{m-2}{2}+2}}\norm{\nabla^2 v}_{{\mathbb X}^{m-2}}+\abs{h}_{{\mathbb X}^{m-2,\frac{3}{2}}}\norm{\nabla^2 v}_{{\mathbb Y}^{\frac{m-2}{2}}}\)
\\&\nonumber \  \le \Lambda_\infty \( \varepsilon \norm{\nabla v}_{{\mathbb X}^{m-2,1}} +\norm{\partial_zv}_{{\mathbb X}^{m-2}}+\norm{v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,-\frac{1}{2}}}+
\norm{\nabla q}_{{\mathbb X}^{m-2}}+\abs{h}_{{\mathbb X}^{m-2,\frac{3}{2}}}\).
\end{align}
Here in the second inequality of \eqref{FS4es}, one has used \eqref{dzzves11} and \eqref{dzzves221} with $k=m-2$. Hence,
\begin{equation}\label{FSes}
\norm{ F_{S}}_{{\mathbb X}^{m-2}} \le\Lambda_\infty\(\varepsilon\(\norm{\nabla
v}_{{\mathbb X}^{m-2,1}}+\abs{h}_{{\mathbb X}^{m-2,\frac{5}{2}}}\)+\norm{\partial_zv}_{{\mathbb X}^{m-2}}+\norm{v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \norm{\nabla q}_{{\mathbb X}^{m-2,1}}
\).
\end{equation}

It follows from applying $\frac{1}{\partial_z\varphi}Z^\alpha(\partial_z\varphi\cdot) $ for $| \alpha | \leq m-2 $ to \eqref{eqPiS}, \eqref{transportW} and
\eqref{deltaphi} that
\begin{equation}\label{SNalpha}
\partial^{\varphi}_{t}  Z^\alpha S_{{\bf n}}+ \(v \cdot \nabla^\varphi\)Z^\alpha S_{{\bf n}}  -\varepsilon \Delta^\varphi Z^\alpha  S_{{\bf n}} = Z^\alpha F_{S} + \mathcal{C}_{S},
\end{equation}
where the commutator is given by
\begin{equation}\label{CS}
\mathcal{C}_{S}= \mathcal  C_{S}^1 + \mathcal C_{S}^2
\end{equation}
with
\begin{align}
\mathcal{C}_{S}^1&=\frac{1}{\partial_z\varphi}\[Z^\alpha,\partial_z\varphi\]\(\partial_t+v_y\cdot\nabla_y\)S_{\bf n}+ \[Z^\alpha, v_{y}\]\cdot  \nabla_{y} S_{{\bf n}} +
\[Z^\alpha, v_{z}\] \partial_{z} S_{{\bf n}}
\\\nonumber&:= \mathcal{C}_{S1}^1 + \mathcal{C}_{S2}^1 + \mathcal{C}_{S3}^1
\end{align}
and
\begin{align}
\mathcal{C}_{S}^2& =- \varepsilon {1 \over \partial_{z} \varphi} [Z^\alpha, \nabla] \cdot \big( E \nabla S_{{\bf n}}\big)-\varepsilon {1 \over \partial_{z}\varphi}
\nabla\cdot  \big( \[Z^\alpha, E \]\nabla S_{{\bf n}}\big)-\varepsilon {1 \over \partial_{z}\varphi} \nabla\cdot  \big( E [Z^\alpha,  \nabla \big]
S_{{\bf n}}\big)
\\\nonumber&:= \mathcal{C}_{S1}^2 + \mathcal{C}_{S2}^2 + \mathcal{C}_{S3}^2.
\end{align}
Since $Z^\alpha S_{{\bf n}}=0$ on the boundary, one can obtain
\begin{equation}\label{m-21}
\frac{1}{2} \frac{d}{dt}\int_{\Omega}  \abs{Z^\alpha S_{{\bf n}}}^2 \, d\mathcal{V}_{t} + \varepsilon \int_{\Omega} \abs{\nabla^\varphi Z^\alpha S_{{\bf n}}}^2 \, d\mathcal{V}_{t}
=\int_{\Omega}  \(Z^\alpha F_{S} + \mathcal{C}_{S} \) \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}.
\end{equation}
The right hand side of \eqref{m-21} can be estimated as follows. \eqref{FSes} implies immediately that
\begin{align}\label{FSmes}
\int_{\Omega}  Z^\alpha F_{S} \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}
\le \Lambda_\infty&\(\varepsilon\(\norm{\partial_z v}_{{\mathbb X}^{m-2,1}}+\abs{h}_{{\mathbb X}^{m-2,\frac{5}{2}}}\)+\norm{\partial_zv}_{{\mathbb X}^{m-2}}\right.
\\\nonumber&\ \left.+\norm{v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \norm{\nabla q}_{{\mathbb X}^{m-2,1}} \)\norm{ S_{{\bf n}} }_{{\mathbb X}^{m-2}}.
\end{align}
Next, we estimate the part involving $C_S^1$. \eqref{com} yields
\begin{equation}\label{CSy}
\norm{\mathcal  C_{S1}^1}+\norm{\mathcal  C_{S1}^2}\le\Lambda_\infty\(  \norm{v}_{{\mathbb X}^{m-2}}+ \norm{S_{{\bf n}}}_{{\mathbb X}^{m-2 }} + \abs{h}_{{\mathbb X}^{m-2,\frac{1}{2} }}  \).
\end{equation}
Additional care is needed to estimate $\norm{\mathcal  C_{S3}^1}$, since $\partial_{z}S_{{\bf n}}$ can not be
controlled. By expanding the commutator and using \eqref{idcom}, one sees clearly that it suffices to estimate terms of the form
$$\norm{ Z^\beta v_{z} \partial_{z} Z^\gamma S_{{\bf n}} }$$
with $|\beta |+ | \gamma |\leq m-2$, $ | \gamma |  \leq m-3$. Since
$$ Z^\beta v_{z} \partial_{z} Z^\gamma S_{{\bf n}}= {1 \over z(z+b) } Z^\beta v_{z} Z_{3} Z^\gamma S_{{\bf n}},$$
which can be further rewritten as a sum of terms of the form
\begin{equation}\label{CSz0}
c_{\tilde \beta }Z^{\tilde \beta }\( {1 \over z(z+b) } v_{z} \)   \, Z_{3} Z^\gamma S_{{\bf n}},
\end{equation}
where $c_{\tilde \beta}$ are harmless bounded functions and $|\tilde \beta | \leq | \beta |$.  Indeed, this comes from the fact that $ Z_{3}\( {1
\over z(z+b) }\)= -{2z+b \over z(z+b) } $. If $\tilde \beta = 0$,  since \eqref{Wdef} implies that $v_{z}=0$ on the boundary, then
$$
\norm{ c_{\tilde \beta }Z^{\tilde \beta }\(  {1 \over z(z+b) } v_{z} \)   \, Z_{3} Z^\gamma S_{{\bf n}}}\lesssim  \norm{\partial_z v_z}_{L^\infty}   \norm{
S_{{\bf n}}}_{{\mathbb X}^{m-2}}.
$$
If $\tilde \beta \neq 0$, one can use \eqref{gues} to obtain that
\begin{align*}
&\norm{ c_{\tilde \beta }Z^{\tilde \beta }\(  {1 \over z(z+b) } v_{z} \)   \, Z_{3} Z^\gamma S_{{\bf n}}}
\\&\quad\lesssim \norm{Z \(  {1 \over z(z+b) } v_{z} \)}_{{\mathbb Y}^{\frac{m-3}{2}}} \norm{Z_{3} S_{{\bf n}} }_{{\mathbb X}^{m-3}}
+ \norm{Z \(  {1 \over z(z+b) } v_{z} \)}_{{\mathbb X}^{m-3}} \norm{Z_{3} S_{{\bf n}} }_{{\mathbb Y}^{\frac{m-3}{2}}}.
\end{align*}
Observe that by again that $v_{z}=0$ on the boundary,
$$ \norm{Z \(  {1 \over z(z+b) } v_{z} \)}_{{\mathbb Y}^{\frac{m-3}{2}}}  \lesssim \norm{ \partial_z v_{z} }_{{\mathbb Y}^{\frac{m }{2}}} .$$
On the other hand, since
$$\norm{Z \(  {1 \over z(z+b) } v_{z} \)}_{{\mathbb X}^{m-3}} \lesssim  \norm{   {1 \over z(z+b) } Z v_{z}  }_{{\mathbb X}^{m-3}}  +  \norm{   {1 \over z(z+b) } v_{z}
}_{{\mathbb X}^{m-3}},
$$
it suffices to estimate
$$ \norm{   {1 \over z(z+b) } Z^\beta Z v_{z}  },\quad \norm{   {1 \over z(z+b) } Z^\beta v_{z}  }, \quad |\beta| \leq m-3.$$
Indeed, since $v_{z}=0$ on the boundary, it follows from the Hardy inequality that
$$ \norm{   {1 \over z(z+b) } Z^\beta Z v_{z}  }\lesssim \norm{\partial_z Z^\beta Z v_{z}  }\text{ and } \norm{   {1 \over z(z+b) } Z^\beta v_{z}  }\lesssim
\norm{\partial_z Z^\beta v_{z}  }.$$
We have thus proven that
$$ \norm{\mathcal{C}_{S3}^1} \le  \norm{ \partial_z v_{z} }_{{\mathbb Y}^{\frac{m }{2}}}  \norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}} + \norm{ S_{{\bf n}}
}_{{\mathbb Y}^{\frac{m}{2}}}\norm{ \partial_z v_{z} }_{{\mathbb X}^{m-2}}.
$$
By Lemma \ref{propeta}, it holds that
\begin{align*}
\norm{ \partial_z v_{z} }_{{\mathbb Y}^{\frac{m }{2}}}\le \Lambda_\infty\text{ and }\norm{ \partial_z v_{z}}_{{\mathbb X}^{m-2}} \le  \Lambda_\infty\(\norm{v}_{{\mathbb X}^{m-2}} +\norm{\partial_z
v}_{{\mathbb X}^{m-2}} +\abs{h }_{{\mathbb X}^{m-1 ,\frac{1}{2}}} \).
\end{align*}
Hence,
\begin{equation}\label{CSz}
\norm{\mathcal{C}_{S3}^1}  \le \Lambda_\infty \(\norm{v}_{{\mathbb X}^{m-2}} +\norm{\partial_z v}_{{\mathbb X}^{m-2}} +\abs{h }_{{\mathbb X}^{m-1 ,\frac{1}{2}}}+\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}\).
\end{equation}
This together with \eqref{CSy} yields
\begin{equation}\label{CS1}
\int_{\Omega}   \mathcal{C}_{S}^1  \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t} \le \Lambda_\infty \(\norm{v}_{{\mathbb X}^{m-2}} +\norm{\partial_z v}_{{\mathbb X}^{m-2}} +\abs{h }_{{\mathbb X}^{m-1
,\frac{1}{2}}}+\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}\)\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}.
\end{equation}

Next, we shall estimate the term involving $\mathcal{C}_{S}^2$. As mentioned in the beginning of this section, we need to employ a different
argument from Proposition 8.3 in \cite{MasRou}. Due to \eqref{idcom} to handle the term involving $\mathcal{C}_{S1}^2$, it suffices to
estimate
$$ \varepsilon \int_{\Omega } Z^\beta\partial_{z}\big( E\nabla S_{{\bf n}}\big) \cdot Z^\alpha S_{{\bf n}} \,dy dz$$
with $|\beta | \leq m-3$, which can be reduced to the estimate of
$$ \varepsilon \int_{\Omega } \(Z^{\beta'}\partial_{z}  E Z^{\beta''}\nabla S_{{\bf n}}+Z^{\beta'} E Z^{\beta''} \partial_{z} \nabla S_{{\bf n}}  \) \cdot
Z^\alpha S_{{\bf n}} \,dy dz$$
with $\beta'+\beta''=\beta$.
It follows from \eqref{gues}, \eqref{dzzves11} and \eqref{dzzves221} with $k=m-3$ that
\begin{align}\label{keyiss11}
&\varepsilon\norm{Z^{\beta'}\partial_{z}  E Z^{\beta''}\nabla S_{{\bf n}}}
 \lesssim  \norm{\partial_{z}  E}_{{\mathbb Y}^{\frac{m-3}{2}}}\varepsilon\norm{\nabla S_{{\bf n}}}_{{\mathbb X}^{m-3}}
+\norm{\partial_{z}  E}_{{\mathbb X}^{m-3}}\varepsilon\norm{\nabla S_{{\bf n}}}_{{\mathbb Y}^{\frac{m-3}{2}}}
\\&\nonumber \quad\le \Lambda_\infty \(\varepsilon \norm{\partial_z   v}_{{\mathbb X}^{m-3,1}}   +\norm{\partial_zv}_{{\mathbb X}^{m-3}}+\norm{v}_{
{{\mathbb X}^{m-2}}}+\abs{h}_{{\mathbb X}^{m-2,-\frac{1}{2}}}+\norm{\nabla  q}_{{\mathbb X}^{m-3}}+ \abs{h}_{{\mathbb X}^{m-3,\frac{3}{2}}}\),
\end{align}
and \eqref{dzzves22'} with $k=m-2$ implies that
\begin{align}\label{keyiss12}
&\varepsilon\norm{Z^{\beta'}  E Z^{\beta''}\partial_{z}\nabla S_{{\bf n}}}
\lesssim  \norm{   E}_{{\mathbb Y}^{\frac{m-3}{2}}}\varepsilon\norm{\partial_{z} \nabla S_{{\bf n}}}_{{\mathbb X}^{m-3}}
+\norm{   E}_{{\mathbb X}^{m-3}}\varepsilon\norm{\partial_{z} \nabla S_{{\bf n}}}_{{\mathbb Y}^{\frac{m-3}{2}}}
\\&\nonumber\quad \le \Lambda_\infty \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{ h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} \right.
\\&\nonumber\qquad\qquad\   +\norm{\partial_zv}_{{\mathbb X}^{m-2}} +\norm{v}_{ {{\mathbb X}^{m-2,1}}} +\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla
q}_{{\mathbb X}^{m-3}}\Big).
\end{align}
Thus,
\begin{align}\label{CS22}
 \int_{\Omega} \mathcal{C}_{S1}^2 \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}
\le& \Lambda_\infty \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{ h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} \right.
\\&\nonumber\qquad     +\norm{\partial_zv}_{{\mathbb X}^{m-2}} +\norm{v}_{ {{\mathbb X}^{m-2,1}}} +\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{m-3}}\Big).
\end{align}
By expanding the second commutator $\mathcal{C}_{S2}^2$, one sees that it suffices to estimate terms of the form
$$\varepsilon \int_{\Omega }  \nabla \cdot \(  Z^{\beta} E Z^{ \gamma} \nabla S_{{\bf n}}\) \, \cdot Z^\alpha S_{{\bf n}} \, dydz$$
with $\beta +   \gamma = \alpha$, $\beta \neq 0$. If $\gamma=0$ and hence $\beta=\alpha$, since $Z^\alpha S_{{\bf n}}=0$ on the boundary, one can then
integrate by parts to get
\begin{align}
&\abs{\varepsilon \int_{\Omega }  \nabla \cdot \(  Z^{\alpha} E   \nabla S_{{\bf n}}\) \, \cdot Z^\alpha S_{{\bf n}} \, dydz}
=
\abs{\varepsilon \int_{\Omega }       Z^{\alpha} E   \nabla S_{{\bf n}}  \cdot \nabla Z^\alpha S_{{\bf n}} \, dydz}
\\&\nonumber\quad\le \varepsilon \norm{Z^{\alpha} E}\norm{\nabla S_{{\bf n}}}_{L^\infty}\norm{\nabla Z^\alpha S_{{\bf n}}}\le \Lambda_\infty
\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}}\varepsilon^\frac{1}{2} \norm{\nabla Z^\alpha S_{{\bf n}}},
\end{align}
where in the last inequality one has used the fact that $\Lambda_{\infty}$ involves $\sqrt{\varepsilon}  \norm{\partial_{zz}v}_{L^\infty}$.
If $\gamma\neq 0$ and hence $1\le |\beta|\le m-3$, then one can expand $\nabla\cdot$, by \eqref{gues} and Lemma \ref{dzzlem}, to estimate
$$\varepsilon \int_{\Omega }   \(       Z^{\beta} \nabla E Z^{ \gamma} \nabla S_{{\bf n}}+  Z^{\beta} E   Z^{ \gamma}  \nabla^2 S_{{\bf n}}  \)\, \cdot Z^\alpha
S_{{\bf n}} \, dydz.$$
It follows from \eqref{gues}, \eqref{dzzves11} and \eqref{dzzves221} with $k=m-3$ that
\begin{align}\label{keyiss112}
&\varepsilon\norm{ Z^{\beta}\nabla E Z^{ \gamma} \nabla S_{{\bf n}}}
 \lesssim  \norm{Z\nabla  E}_{{\mathbb Y}^{\frac{m-4}{2}}}\varepsilon\norm{Z\nabla S_{{\bf n}}}_{{\mathbb X}^{m-4}}
+\norm{Z\nabla  E}_{{\mathbb X}^{m-4}}\varepsilon\norm{Z\nabla S_{{\bf n}}}_{{\mathbb Y}^{\frac{m-4}{2}}}
\\&\nonumber\quad \le \Lambda_\infty \(\varepsilon \norm{\partial_z   v}_{{\mathbb X}^{m-3,1}}   +\norm{\partial_zv}_{{\mathbb X}^{m-3}}+\norm{v}_{
{{\mathbb X}^{m-2}}}+\abs{h}_{{\mathbb X}^{m-2,-\frac{1}{2}}}+\norm{\nabla  q}_{{\mathbb X}^{m-3}}+ \abs{h}_{{\mathbb X}^{m-3,\frac{3}{2}}}\),
\end{align}
and \eqref{dzzves22'} with $k=m-2$ that
\begin{align}
&\varepsilon\norm{Z^{\beta}  E   Z^{\gamma} \nabla^2 S_{{\bf n}}} \lesssim  \norm{Z   E}_{{\mathbb Y}^{\frac{m-4}{2}}}\varepsilon\norm{Z\nabla^2 S_{{\bf n}}}_{{\mathbb X}^{m-4}}
+\norm{ Z  E}_{{\mathbb X}^{m-4}}\varepsilon\norm{Z\nabla^2 S_{{\bf n}}}_{{\mathbb Y}^{\frac{m-4}{2}}}\\&\nonumber\quad \le \Lambda_\infty \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{
h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} \right.
\\&\nonumber\qquad\qquad\   +\norm{\partial_zv}_{{\mathbb X}^{m-2}} +\norm{v}_{ {{\mathbb X}^{m-2,1}}} +\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla
q}_{{\mathbb X}^{m-3}}\Big).
\end{align}
Hence,
\begin{align}\label{CS222}
  \nonumber\int_{\Omega} \mathcal{C}_{S2}^2 \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}
&\le \Lambda_\infty  \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{ h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} +\norm{\partial_zv}_{{\mathbb X}^{m-2}} +\norm{v}_{
{{\mathbb X}^{m-2,1}}}\right.
\\&\qquad  +\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{m-3}}\Big)\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}+\Lambda_\infty  \abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}}\varepsilon^\frac{1}{2}
\norm{\nabla Z^\alpha S_{{\bf n}}}  .
\end{align}
To handle the term involving $\mathcal{C}_{S3}^2$, due to \eqref{idcom}, one needs to estimate
$$ \varepsilon \int_{\Omega }\nabla\cdot\( E Z^\beta\partial_{z}  S_{{\bf n}} \)\cdot Z^\alpha S_{{\bf n}} \,dy dz$$
with $|\beta | \leq m-3$, which can be bounded easily by using \eqref{dzzves221} with $k=m-3$ and \eqref{dzzves22'}
with $k=m-2$ so that
\begin{align}\label{CS2223}
  \int_{\Omega} \mathcal{C}_{S3}^2 \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}
  &\le \Lambda_\infty  \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{ h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} +\norm{\partial_zv}_{{\mathbb X}^{m-2}}\right.
\\&\nonumber\qquad\quad  +\norm{v}_{ {{\mathbb X}^{m-2,1}}}+\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{m-3}}\Big)\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}
.
\end{align}
In view of \eqref{CS22}, \eqref{CS222}, \eqref{CS2223}, one has actually proven that
\begin{align} \label{CS2}
 \nonumber \int_{\Omega} \mathcal{C}_{S}^2 \cdot Z^\alpha S_{{\bf n}}\, d\mathcal{V}_{t}
&\le \Lambda_\infty  \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-3,2}}+\abs{ h}_{{\mathbb X}^{m-3,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}} +\norm{\partial_zv}_{{\mathbb X}^{m-2}} +\norm{v}_{
{{\mathbb X}^{m-2,1}}}\right.
\\&\qquad\quad  +\norm{\nabla  q}_{{\mathbb X}^{m-3,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{m-3}}\Big)\norm{ S_{{\bf n}}}_{{\mathbb X}^{m-2}}+\Lambda_\infty
\abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}}\varepsilon^\frac{1}{2} \norm{\nabla Z^\alpha S_{{\bf n}}}  .
\end{align}

Consequently, by \eqref{FSmes}, \eqref{CS1} and \eqref{CS2}, one deduces from \eqref{m-21} that
 \begin{align}
& \frac{1}{2} \frac{d}{dt} \int_{\Omega} \abs{Z^\alpha S_{{\bf n}}}^2 \, d\mathcal{V}_{t}    +\varepsilon \int_{\Omega}\abs{\nabla^\varphi Z^\alpha S_{{\bf n}}}^2 \, d\mathcal{V}_{t} \\
&\nonumber\le \Lambda_\infty\(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-2,1}}+\abs{h}_{{\mathbb X}^{m-2,\frac{5}{2}}}\)+\norm{ S_{{\bf n}}
}_{{\mathbb X}^{m-2}}+\norm{\partial_zv}_{{\mathbb X}^{m-2}}+\norm{v}_{{\mathbb X}^{m-1}} \right.
\\\nonumber&\quad \left.+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \norm{\nabla q}_{{\mathbb X}^{m-2,1}} +\norm{\partial_z\nabla  q}_{{\mathbb X}^{m-3}}\)\norm{ S_{{\bf n}}
}_{{\mathbb X}^{m-2}}+\Lambda_\infty  \abs{h}_{{\mathbb X}^{m-2,\frac{1}{2}}}\varepsilon^\frac{1}{2} \norm{\nabla Z^\alpha S_{{\bf n}}} .
\end{align}
To conclude, one can use Lemma \ref{mingrad} to replace $ \norm{\nabla^\varphi S_{{\bf n}}}_{{\mathbb X}^{m-2}}$ by $\norm{\nabla  S_{{\bf n}}}_{{\mathbb X}^{m-2}}$ in the
left hand side and then use Cauchy's inequality to absorb the last term. On the other hand, one can follow the derivation of \eqref{qpressurem}
to get that
  \begin{align}\label{qpressurem'}
  \nonumber&    \norm{   q }_{{\mathbb X}^{m-2,1}}+\norm{ \nabla q }_{{\mathbb X}^{m-2,1}}+\norm{ \partial_{zz} q}_{{\mathbb X}^{m-2}}
      \\\nonumber&  \quad\leq \Lambda_\infty\(\abs{h}_{{\mathbb X}^{m-1,{1 \over 2 }}}+\sigma \abs{h}_{{\mathbb X}^{m-2,3}}+\norm{v}_{{\mathbb X}^{m-1,1}}  + \norm{\nabla
      v}_{{\mathbb X}^{m-2,1}} +  \varepsilon\norm{ \nabla v }_{{{\mathbb X}^{m-2, 2}}}+\varepsilon\abs{h}_{{{\mathbb X}^{m-2,\frac{5}{2}}}}\),
  \end{align}
which reduces the order of time derivatives to the spatial derivatives.
Finally, we integrate the resulting inequality in time to obtain \eqref{estSnm-2}.
\end{proof}

\subsection{Estimate of  $\norm{\partial_z v}_{{\mathbb X}^{m-1}}$}\label{sectionnorm2}

Note that Proposition \ref{propdzvm-2} only provides the control of $\norm{\partial_z v}_{{\mathbb X}^{m-2}}$, and this is due to the appearance of
$(\nabla^\varphi)^2 q$ in the source term in the equation of $S_{{\bf n}}$. To get around this, a natural way is to use the vorticity instead of
$S_{{\bf n}}$.

Set $\omega = \nabla^\varphi \times v$. Then
\begin{equation}\label{omega1}
\omega \times {\bf n}= \frac{1}{2} \( \nabla^\varphi v\, {\bf n} - (\nabla^\varphi v)^t{\bf n}\)= S^\varphi v\, {\bf n} -   (\nabla^\varphi v)^t{\bf n},
\end{equation}
and hence by \eqref{Su},
$$ \omega \times {\bf n} =   \frac{1}{2}  \partial_{{\bf n}} u  - g^{ij}\big( \partial_{j}v \cdot {\bf n} \big) \partial_{y^i} .$$
Consequently, as in Lemma \ref{lemdzS} one can get that
\begin{equation}\label{dzvm-1O}
\norm{\partial_{z} v}_{{\mathbb X}^{m-1}} \le \Lambda_\infty\(\norm{v}_{{\mathbb X}^{m-1,1}}+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}} } + \norm{\omega}_{{\mathbb X}^{m-1}}\).
\end{equation}
To estimate $\norm{\omega}_{{\mathbb X}^{m-1}}$, one notes that the first equation in \eqref{NSv} implies
\begin{equation}\label{eqomega}
\partial^{\varphi}_{t} \omega + v \cdot \nabla^\varphi \omega - \omega \cdot \nabla^\varphi v = \varepsilon \Delta^\varphi \omega.
\end{equation}
As in the previous subsection, applying $ \frac{1}{\partial_z\varphi}Z^\alpha(\partial_z\varphi\cdot)$ for $|\alpha|\le m-1$ to
\eqref{eqomega} yields
\begin{equation}\label{eqomegaalpha}
\partial^{\varphi}_{t} Z^\alpha  \omega + (v \cdot \nabla^\varphi) \, Z^\alpha  \omega - \varepsilon \Delta Z^\alpha \omega= F.\end{equation}
Here $F$ is given by
   \begin{equation}
   \label{Fomega}
    F= Z^\alpha\big(\omega \cdot \nabla^\varphi v)  + \mathcal{C}_{S},
   \end{equation}
   where $\mathcal{C}_{S}$  is given, as in \eqref{CS}, by
    \begin{equation}
       \label{CSomega}
       \mathcal{C}_{S}= \mathcal  C_{S}^1 + \mathcal C_{S}^2\end{equation}
with
\begin{align}
\mathcal{C}_{S}^1&=\frac{1}{\partial_z\varphi}\[Z^\alpha,\partial_z\varphi\]\(\partial_t+v_y\cdot\nabla_y\)\omega+ \[Z^\alpha, v_{y}\]\cdot  \nabla_{y} \omega +
\[Z^\alpha, v_{z}\] \partial_{z}\omega
\\\nonumber&:= \mathcal{C}_{S1}^1 + \mathcal{C}_{S2}^1 + \mathcal{C}_{S3}^1
\end{align}
and
\begin{align}
\mathcal{C}_{S}^2& =- \varepsilon {1 \over \partial_{z} \varphi} [Z^\alpha, \nabla] \cdot \big( E \nabla \omega\big)-\varepsilon {1 \over \partial_{z}\varphi}
\nabla\cdot  \big( \[Z^\alpha, E \]\nabla \omega\big)-\varepsilon {1 \over \partial_{z}\varphi} \nabla\cdot  \big( E [Z^\alpha,  \nabla \big]
\omega\big)
\\\nonumber&:= \mathcal{C}_{S1}^2 + \mathcal{C}_{S2}^2 + \mathcal{C}_{S3}^2.
\end{align}

The main difficulty lies in that the vorticity does not vanish on the boundary. Thus, set
\begin{equation}\label{splitomega}
Z^\alpha \omega= \omega^\alpha_{nh}+ \omega^{\alpha}_{h}
\end{equation}
where $\omega^\alpha_{nh} $ sloves
\begin{equation}\label{eqomeganh}
\begin{cases}
\partial^{\varphi}_{t}  \omega^\alpha_{nh} + (v \cdot \nabla^\varphi)   \omega^\alpha_{nh} - \varepsilon \Delta^\varphi  \omega^\alpha_{nh}= F &\text{in }\Omega
\\ \omega^\alpha_{nh} = 0 &\text{on }\{z=0,-b\}
\\  \omega^\alpha_{nh}|_{t=0}= \omega(0)&
\end{cases}
\end{equation}
and $ \omega^{\alpha}_{h}$ solves
\begin{equation}\label{eqomegah}
\begin{cases}
\partial^{\varphi}_{t}  \omega^\alpha_{h} + (v \cdot \nabla^\varphi)   \omega^\alpha_{h} - \varepsilon \Delta^\varphi  \omega^\alpha_{h}= 0 &\text{in }\Omega
\\ \omega^\alpha_{h} = Z^\alpha \omega &\text{on }\{z=0,-b\}
\\  \omega^\alpha_{h}|_{t=0}= 0.&
\end{cases}
\end{equation}

The estimate of $\omega^\alpha_{nh}$ is  given as follows.
\begin{prop}
For $|\alpha|\le m-1$, the solution $\omega_{nh}^\alpha$ of \eqref{eqomeganh} satisfies the estimate
\begin{align}\label{propomeganh}
\norm{\omega^\alpha_{nh}(t)}^2 + \varepsilon \int_{0}^t \norm{\nabla \omega^\alpha_{nh} }^2 \le& \Lambda_0\norm{ \omega(0)}_{{\mathbb X}^{m-1}}^2+\int_0^t \Lambda_\infty
\varepsilon^2\(\norm{\nabla v}_{{\mathbb X}^{m-2,2}}^2+\abs{ h}_{{\mathbb X}^{m-2,\frac{5}{2}}}^2\)
\\&\nonumber +\int_0^t \Lambda_\infty \(\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}^2+\norm{\partial_zv}_{{\mathbb X}^{m-1}}^2 +\norm{v}_{ {{\mathbb X}^{m-1,1}}}^2+\norm{ \omega^\alpha_{nh}}^2 \)
.
\end{align}
\end{prop}
\begin{proof}
Since $\omega^\alpha_{nh}=0$ on the boundary, it follows from \eqref{eqomeganh} that
\begin{equation}\label{omeganh1}
\frac{1}{2} \frac{d}{dt} \int_{\Omega}  \abs{\omega^\alpha_{nh}}^2 \, d\mathcal{V}_{t}
+ \varepsilon \int_{\Omega}  \abs{ \nabla^\varphi \omega^\alpha_{nh}}^2 \, d\mathcal{V}_{t}
=\int_{\Omega}F \cdot \omega^\alpha_{nh}\, d\mathcal{V}_{t}.
\end{equation}
We now estimate the right hand side of \eqref{omeganh1}. \eqref{gues} implies that
\begin{equation}\label{cssss}
\norm{Z^\alpha \(\omega \cdot \nabla^\varphi v \)}
\le\Lambda_\infty \( \norm{\omega }_{{\mathbb X}^{m-1}}+ \norm{ \nabla^\varphi v}_{{\mathbb X}^{m-1}}\)
\le \Lambda_\infty\( \norm{\omega }_{{\mathbb X}^{m-1}}+ \norm{ \nabla  v}_{{\mathbb X}^{m-1}}+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\).
\end{equation}
To estimate the part involving $\mathcal{C}_{S}$, one first can change $S_{{\bf n}}$ into $\omega$ and $m$ into $m+1$ in \eqref{CS2} to get
\begin{align}\label{CS2omega}
\int_{\Omega} \mathcal{C}_{S}^2 \cdot \omega^\alpha_{nh}\, d\mathcal{V}_{t}
&\le \Lambda_\infty  \(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-2,2}}+\abs{ h}_{{\mathbb X}^{m-2,\frac{5}{2}}}\)+\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} \right.
\\&\nonumber\qquad\quad  +\norm{\partial_zv}_{{\mathbb X}^{m-1}} +\norm{v}_{ {{\mathbb X}^{m-1,1}}} \Big)\norm{ \omega^\alpha_{nh}} +\Lambda_\infty
\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\varepsilon^\frac{1}{2} \norm{\nabla \omega^\alpha_{nh}}  .
\end{align}
For the part involving $\mathcal{C}_{S}^1$, one can change $S_{{\bf n}}$ into $\omega$ and $m$ into $m+1$ in \eqref{CSy} to obtain
\begin{equation}\label{CSyomega}
\norm{\mathcal  C_{S1}^1}+\norm{\mathcal  C_{S1}^2}\le\Lambda_\infty\(  \norm{v}_{{\mathbb X}^{m-1}}+ \norm{\omega}_{{\mathbb X}^{m-1 }} + \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2} }}  \).
\end{equation}
The commutator $\mathcal{C}_{S3}^1$ requires much more care. Indeed, one can not change $m$ into $m+1$ in \eqref{CSz} since it would involve
$|h|_{{\mathbb X}^{m,\frac{1}{2}} }$. As in \eqref{CSz0}, this commutator can be expanded into a sum of terms of the form
$$ c_{  \beta }Z^{  \beta }\(  {1  \over {z (z+b)}} v_{z} \)   \, Z_{3} Z^\gamma \omega$$
such that  $|  \beta| + |  \gamma| \leq m-1$, $| \gamma | \leq m-2$. As the arguments in previous sections, by \eqref{gues} and $v_{z}=0$ on the
boundary that, one deduces
\begin{align}\label{tqqq0}
\norm{Z^{  \beta }\(  {1  \over {z (z+b)}} v_{z} \)   \, Z_{3} Z^\gamma \omega}
&\le\Lambda_\infty\( \norm{ \omega }_{{\mathbb X}^{m-2,1}}+  \norm{ Z\( {1  \over {z (z+b)}} v_{z}\)}_{{\mathbb X}^{m-2}}\) \\
&\le\Lambda_\infty\( \norm{ \omega }_{{\mathbb X}^{m-2,1}}+ \norm{{1  \over {z (z+b)}}  v_{z}}_{{\mathbb X}^{m-1}} \).\nonumber
\end{align}
It follows from $v_z=0$ on the boundary and the Hardy inequality that, since $  v_{z}  =  v\cdot {\bf N} - \partial_{t} \eta  $,
\begin{align}\label{tqqq00}
&\norm{{1  \over {z (z+b)}}  v_{z}}_{{\mathbb X}^{m-1}} \lesssim  \norm{\partial_{z}  v_{z}}_{{\mathbb X}^{m-1}}
 \\\nonumber&\quad\le \Lambda_\infty \(\norm{\partial_z v}_{{\mathbb X}^{m-1}} +\norm{  v}_{{\mathbb X}^{m-1}} + \abs{  h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\)+ \sum_{| \alpha |= m- 1} \norm{ v
 \cdot \partial_{z}Z^\alpha {\bf N} - \partial_{z} Z^\alpha \partial_{t} \eta },
\end{align}
where the last term requires furthermore analysis. Due to  \eqref{eqeta}, it holds that
$$ |Z_{3} \hat \eta| \lesssim  |\tilde \chi(|\xi| z ) \hat{h}|$$
where $\tilde \chi$ has a slightly bigger support than $\chi$. Hence $Z_{3}$ acts as a zero order operator:
\begin{equation}\label{Z3mieux}
\norm{\nabla Z_{3} \eta } \lesssim \abs{h}_{1\over 2}.
\end{equation}
This yields that if $\alpha_{3} \neq 0$, one gains at least one derivative, and thus,
\begin{equation}\label{tqqq1} \norm{ v \cdot \partial_{z}Z^\alpha {\bf N} - \partial_{z} Z^\alpha \partial_{t} \eta }  \le\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-2,\frac{3}{2}}} +
\abs{\partial_{t}
 h}_{{\mathbb X}^{m-2,\frac{1}{2}}} \) \le  \Lambda_\infty \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}.
 \end{equation}
Hence it suffices to estimate this term for the case with $| \alpha |=m-1$ and $\alpha_{3}= 0$. Note that \eqref{eqeta} implies that
\begin{equation}\label{etaconv}
\eta=\(1+\frac{z}{b}\){1 \over z^2} \psi\( {\cdot \over z}\)\star_{y} h:= \(1+\frac{z}{b}\)\psi_{z} \star_{y} h,
\end{equation}
where $\star_{y}$ stands for the convolution in the $y$ variable and $\psi$ is  in  $L^1(\mathbb{R}^2)$. Then
\begin{align}
 v \cdot \partial_{z}Z^\alpha {\bf N} - \partial_{z} Z^\alpha \partial_{t} \eta=&-
 \frac{1}{b}\( v_{y}\cdot \psi_z\star_{y} \nabla_{y} Z^{\alpha} h
+\psi_{z} \star_{y} \partial_{t} Z^{\alpha} h\)
\\& -\(1+\frac{z}{b}\)\partial_z\( v_{y}\cdot \psi_z\star_{y} \nabla_{y} Z^{\alpha} h
+\psi_{z} \star_{y} \partial_{t} Z^{\alpha} h\) := \mathcal{T}_{\alpha}=\mathcal{T}_{\alpha1}+\mathcal{T}_{\alpha2}.\nonumber
\end{align}
It is clear that
\begin{equation} \label{Talpha0}
 \norm{ \mathcal{T}_{\alpha1} }  \le\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \abs{\partial_{t} h}_{{\mathbb X}^{m-1,-\frac{1}{2}}} \) \le  \Lambda_\infty \abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}.
 \end{equation}
For the second term, one can separate it into two cases. For $-b< z\le -\frac{b}{2}$, then
$$
\mathcal{T}_{\alpha2}=- \frac{1}{bz} Z_3\( v_{y}\cdot \psi_z\star_{y} \nabla_{y} Z^{\alpha} h
+\psi_{z} \star_{y} \partial_{t} Z^{\alpha} h\).
$$
It then follows from \eqref{Z3mieux} that
\begin{equation}\label{tqqq} \norm{ \mathcal{T}_{\alpha2} }_{L^2\( \{-b\le z\le -\frac{b}{2}\}\)}   \le\Lambda_\infty\( \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} + \abs{\partial_{t}
 h}_{{\mathbb X}^{m-1,-\frac{1}{2}}} \) \le  \Lambda_\infty \abs{h}_{{\mathbb X}^{m,-\frac{1}{2}}}.
 \end{equation}
For $-\frac{b}{2}\le z<0$, $\mathcal{T}_{\alpha2}$ can be written as
\begin{equation}\label{Talphadecfin}
\mathcal{T}_{\alpha2}= -\(1+\frac{z}{b}\)\(v_y(t,y,0)\cdot\partial_{z} \(  \psi_{z} \star_{y} \nabla_{y} Z^{\alpha} h \)
+\partial_{z}\(  \psi_{z} \star_{y} \partial_{t} Z^{\alpha} h \)\) +  \mathcal{R},
\end{equation}
 where
 $$ \abs{\mathcal{R}} \leq \Lambda_\infty|z(z+b)| \abs{ \partial_{z} \( \psi_{z} \star  Z^\alpha \nabla h\)}
 \leq \Lambda_\infty\abs{ Z_{3} \( \psi_{z} \star  Z^\alpha \nabla h \)}.$$
By using \eqref{Z3mieux} again, one can get that
\begin{equation}\label{Talphadecfin1}
\norm{\mathcal{R}}_{L^2\({\Omega}  \cap \{  -\frac{b}{2}\le z<0\}\)}
\leq \Lambda_\infty \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}} }.\end{equation}
To estimate the first term in \eqref{Talphadecfin}, one notes that
\begin{align*}  &  v_y(t,y,0)\cdot \partial_{z} \(  \psi_{z} \star_{y} \nabla_{y} Z^{\alpha} h \)
+\partial_{z}\(  \psi_{z} \star_{y} \partial_{t} Z^{\alpha} h \)  = \partial_{z} \( \psi_{z} \star_{y}\( v_y(t,y,0)\cdot \nabla_{y} Z^\alpha h
+ \partial_{t} Z^\alpha h \) \)\\
& \qquad\qquad\qquad\quad +\partial_{z} \(\int_{\mathbb{R}^2}\(  v_{y}(t,y,0) - v_{y}(t,y',0) \)\cdot \psi_{z}(y-y') \nabla_{y} Z^\alpha h(t,y')
\).
\end{align*}
For the second term in the right hand side of the above, one can employ the Taylor formula for $v_{y}$ to get  that
$$ \abs{ (v_{y}(t,y,0) - v_{y}(t,y', 0))\partial_{z} \psi_{z}(y-y') } \leq \Lambda_\infty  {1 \over z^2} \tilde{\psi}\({y-y' \over z}\), $$
where $\tilde{\psi} $ is still an $L^1$ function. This yields that
\begin{align}\label{Talphadecfin2}
&\sup_{z \in (- \frac{b}{2}, 0)}\abs{ \partial_{z} \(\int_{\mathbb{R}^2}\(  v_{y}(t,y,0) - v_{y}(t,y',0) \)\cdot \psi_{z}(y-y') \nabla_{y}
Z^\alpha h(t,y') \)}_{0}\leq \Lambda_\infty \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}.
\end{align}
For the first term, one shall use \eqref{bordh} to get
$$v_y\cdot\nabla_{y} Z^\alpha h  + \partial_{t} Z^\alpha h  = -\partial_z^\varphi v\cdot{\bf N} Z^\alpha h+ \mathcal{C}^\alpha(h)+  V^\alpha \cdot {\bf N},
$$
which implies,  since $|\alpha|= m-1$,
$$\abs{ v_y\cdot\nabla_{y} Z^\alpha h + \partial_{t} Z^\alpha h  }_{ \frac{1}{2} }
       \leq \Lambda_\infty\( \abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}}+ \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\).$$
It thus holds that
\begin{align}\label{Talphadecfin3}
\norm{  \partial_{z} \( \psi_{z} \star_{y}\( v_y(t,y,0)\cdot \nabla_{y} Z^\alpha h  - \partial_{t} Z^\alpha h \) \)}   & \lesssim  \abs{
v_y\cdot\nabla_{y} Z^\alpha h  - \partial_{t} Z^\alpha h  }_{ \frac{1}{2} } \\
&\leq  \Lambda_\infty \(\abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}}+ \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\),\nonumber
\end{align}
Collecting all the estimates \eqref{Talphadecfin1}--\eqref{Talphadecfin3}, one deduces from \eqref{Talphadecfin} that
\begin{equation}\label{tqqq2} \norm{ \mathcal{T}_{\alpha}}_{L^2\({\Omega}  \cap \{z\ge -\frac{b}{2}\}\)} \leq \Lambda_\infty \(\abs{v}_{{\mathbb X}^{m-1,\frac{1}{2}}}+
\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\).
\end{equation}
In view of the estimates \eqref{tqqq0}, \eqref{tqqq00}, \eqref{tqqq1}, \eqref{Talpha0}, \eqref{tqqq} and \eqref{tqqq2}, by the trace estimates,
one finally gets
\begin{equation}\label{CSzomega}
\norm{\mathcal{C}_{S3}^1} \le \Lambda_\infty \(\norm{\partial_z v}_{{\mathbb X}^{m-1}} +\norm{  v}_{{\mathbb X}^{m-1,1}} + \abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}\).
\end{equation}

As a consequence of \eqref{cssss}, \eqref{CS2omega}, \eqref{CSyomega}, \eqref{CSzomega} and
Lemma \ref{mingrad}, we deduce from \eqref{omeganh1} that
\begin{align}
&\norm{\omega^\alpha_{h}(t)}^2 + \varepsilon \int_{0}^t \norm{\nabla \omega^\alpha_{h} }^2  \le \Lambda_0\norm{ \omega(0)}_{{\mathbb X}^{m-1}}^2+\int_0^t \Lambda_\infty
\(\varepsilon\(\norm{\nabla v}_{{\mathbb X}^{m-2,2}}+\abs{ h}_{{\mathbb X}^{m-2,\frac{5}{2}}}\) \right.
\\&\nonumber\qquad \qquad  +\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}}+\norm{\partial_zv}_{{\mathbb X}^{m-1}} +\norm{v}_{ {{\mathbb X}^{m-1,1}}} \Big)\norm{ \omega^\alpha_{nh}} +\Lambda_\infty
\varepsilon^\frac{1}{2}\abs{h}_{{\mathbb X}^{m-1,\frac{1}{2}}} \norm{\nabla \omega^\alpha_{h} }  .
\end{align}
We then conlude the proposition by using Cauchy's inequality.
\end{proof}

It remains to estimate $\omega^\alpha_{h}$ of \eqref{eqomegah}. Note that Lemma \ref{lembord} and the trace estimate imply that
\begin{align}
\label{omegab2}
\sqrt{\varepsilon}  \int_{0}^t\abs{Z^\alpha \omega}_0^2 &\leq \sqrt{\varepsilon} \int_{0}^t\Lambda_\infty\(\abs{v}_{{\mathbb X}^{m-1,1}}^2 + \abs{h}_{{\mathbb X}^{m-1,1}}^2  \)
 \\\nonumber& \leq   \sqrt{\varepsilon}\int_{0}^t \Lambda_\infty  \(\norm{\nabla v}_{{\mathbb X}^{m-1,1}}\norm{  v}_{{\mathbb X}^{m-1,1}}  + \norm{
 v}_{{\mathbb X}^{m-1,1}}^2+\abs{h}_{{\mathbb X}^{m-1,1}}^2\)  .
\end{align}
Thus the main difficulty will be to handle the non-homogeneous boundary value problem, \eqref{eqomegah}, whose boundary value is at a low level
of regularity (it is $L^2_{t,y}$ and no more) due to \eqref{omegab2}. This creates two
difficulties: the first is that one cannot lift the boundary condition easily and  perform a standard energy estimate; the second one  is
that due to the lower boundary estimate, one cannot expect an estimate of  $\norm{\omega^\alpha_{h}}$ in $L^\infty(0,T)$ independent of
$\varepsilon$, as was well explained in Section 10.2 of \cite{MasRou}: in using the model of the heat equation, one may expect a
control in $H^\frac{1}{4}(0,T)\subset L^4(0,T)$.

\begin{prop}\label{nor2}
Assume that $\Lambda_\infty(t)+\int_0^t\norm{\partial_{zz}v}_{1,\infty}\le M$ for $M>0$. Then for $|\alpha|\le m-1$, the solution $\omega_{h}^\alpha$ of
\eqref{eqomegah} satisfies the estimate
\begin{equation}\label{nor2estimate}
\norm{ \omega^\alpha_{h} }_{L^4_T L^2}^2 \leq  \Lambda(M) \int_0^T \( \norm{  v}_{{\mathbb X}^{m-1,1}}^2+\abs{h}_{{\mathbb X}^{m-1,1}}^2\)+\varepsilon\int_0^T\norm{\nabla
v}_{{\mathbb X}^{m-1,1}}^2.
\end{equation}
\end{prop}
\begin{proof}
Since the situation here is similar to Section 10.2 in \cite{MasRou}, thus this proposition is a restatement of Proposition 10.4 in
\cite{MasRou}. However, for completeness, we will sketch the main idea and some steps of the proof and state our estimates with a slight
modification.

First, it is convenient to eliminate the convection term in \eqref{eqomegah} by considering Lagrangian coordinates. Define a
parametrization of $\Omega(t)$, $X(t,\cdot):\Omega\mapsto \Omega(t)$, by
\begin{equation}\label{lagrange1}
\begin{cases}
\partial_{t} X(t,x)= u(t, X(t,x))= v(t,\Phi(t, \cdot)^{-1}\circ X)
\\ X(0, x)= \Phi(0,x),
\end{cases}
\end{equation}
where the map $\Phi(t, \cdot)$ is defined by \eqref{diff}. Denote $J(t,x)= | \mbox{det }\nabla X(t,x)|$ for the Jacobian of the change of
coordinates. Then $J(t,x)= J_{0}(x)$ by the divergence free condition. Define
\begin{equation}\label{Omegaalphadef}
\Omega^\alpha=e^{-\gamma t} \omega_{h}^\alpha (t, \Phi^{-1}\circ X),
\end{equation}
where $\gamma >0$ is a large parameter to be chosen. Then  $\Omega^\alpha$ solves
\begin{equation}\label{omegalagrange}
a_{0}\big( \partial_{t}\Omega^\alpha+ \gamma \Omega^\alpha\big) - \varepsilon  \partial_{i}\( a_{ij} \partial_{j}\Omega^\alpha \)= 0\quad\text{in
}\Omega.
\end{equation}
Here $ a_{0}= |J_{0}|^{1\over 2}$ and the matrix $ (a_{ij})$ is defined by
$$ ( a_{ij} ) =  |J_{0}|^{1 \over 2} P^{-1}\text{ with } P_{ij}=  \partial_{i}X \cdot \partial_{j} X.$$

Due to the assumption $\Lambda_\infty(t)+\int_0^t\norm{\partial_{zz}v}_{1,\infty}\le M$, Lemma 10.5 in \cite{MasRou} holds. Then the
following estimates hold:
\begin{eqnarray}
\label{J0}   & &\norm{J}_{W^{1, \infty}}+\norm{J^{-1}}_{W^{1, \infty}} \leq  \Lambda_{0}, \\
\label{nablaX} & & \norm{ \nabla X}_{L^\infty}  + \norm{ \partial_{t} \nabla X }_{L^\infty} \leq \Lambda(M) e^{ t\Lambda(M)},  \\
\label{dtnablaX} & & \norm{ \nabla X}_{1, \infty} +  \norm{ \partial_{t} \nabla X}_{1, \infty}  \leq \Lambda(M)e^{ t\Lambda(M)}, \\
\label{ZnablaX}  & &\sqrt{\varepsilon}\norm{ \nabla^2 X }_{1, \infty}+\sqrt{\varepsilon}\norm{ \partial_{t} \nabla^2 X}_{L^\infty} \leq   \Lambda(M)(1+t) e^{t
\Lambda(M)}.
\end{eqnarray}
Indeed, the estimate \eqref{J0} follows directly by $J(t,x)= J_{0}(x)$. Next, \eqref{lagrange1} implies that
\begin{equation}\label{eqDX}
\partial_{t} \nabla X= \nabla v \nabla \Phi^{-1} \nabla X
\end{equation}
and hence
$$\norm{\nabla X}_{L^\infty} \leq  \norm{\nabla X(0)}_{L^\infty}+ \Lambda_{0} \int_{0}^t \norm{\nabla v}_{L^\infty} \norm{ \nabla X}_{L^\infty}
\leq \Lambda_{0} + \Lambda(M) \int_{0}^t \norm{ \nabla X}_{L^\infty}.$$
This yields the first part of  \eqref{nablaX} by the Gronwall inequality.
Next, applying one spatial conormal derivative to \eqref{eqDX}, one can get
$$\norm{\nabla X}_{1, \infty} \leq  \norm{\nabla X(0)}_{1, \infty}+ \Lambda_{0} \int_{0}^t \norm{\nabla v}_{1, \infty} \norm{ \nabla X}_{1,
\infty}
\leq \Lambda_{0} + \Lambda(M) \int_{0}^t \norm{ \nabla X}_{1, \infty}.$$
and hence the first part of \eqref{dtnablaX} follows from the Gronwall inequality.
The estimates hold also for the time derivative in \eqref{nablaX} and \eqref{dtnablaX} by using again the equation \eqref{eqDX}. To prove
\eqref{ZnablaX}, one applies $\sqrt{\varepsilon}\partial_{x_i}$ to \eqref{eqDX} to find that
\begin{equation}\label{eqD2X}
\partial_{t} \sqrt{\varepsilon}\partial_{x_i} \nabla X= \sqrt{\varepsilon}  \(\partial_{x_i}\nabla v \nabla \Phi^{-1}\nabla X+\nabla v \partial_{x_i}\nabla \Phi^{-1}\nabla
X+\nabla v \nabla \Phi^{-1} \partial_{x_i}\nabla X\).
\end{equation}
It follows from \eqref{dtnablaX} that
$$ \sqrt{\varepsilon }\norm{ \nabla^2 X}_{1, \infty}  \leq \Lambda(M)(1+t) + \Lambda(M) \int_{0}^t \sqrt{\varepsilon} \norm{ \nabla^2 X}_{1, \infty}
+ \Lambda(M) e^{\Lambda(M) t} \int_{0}^t   \sqrt{\varepsilon} \norm{ \nabla^2 v}_{1, \infty}$$
and hence, by using the assumption $\int_0^t\norm{\partial_{zz}v}_{1,\infty}\le M$,
$$  \sqrt{\varepsilon }\norm{\nabla^2 X}_{1, \infty}  \leq \Lambda(M) ( 1+t) e^{\Lambda(M) t} + \Lambda(M) \int_{0}^t \sqrt{\varepsilon} \| \nabla^2 X\|_{1,
\infty}$$
and so the  first part of \eqref{ZnablaX} follows from the Gronwall inequality. For the second part of \eqref{ZnablaX},
it follows by using again \eqref{eqD2X} and the fact that
$\Lambda_\infty$ involves the control of $ \sqrt{\varepsilon}\norm{ \nabla ^2 v}_{L^\infty}$.

As consequences of the estimates \eqref{J0}--\eqref{ZnablaX}, one gets
\begin{equation}\label{compact}
a_{0}\geq m ,\ a_{3,3} \geq m,  \     (a_{ij}) \geq c_{0} \mbox{Id}
\end{equation}
by a suitable choice of $m$ and $c_0$ depending on $M$ for $t\in [0,1]$ and that
\begin{align}\label{compact2}
&\norm{( a_{0}, a_{ij}) }_{L^\infty} + \sqrt{\varepsilon}\norm{\partial_{z} ( a_{0}, a_{ij})  }_{L^\infty}  \leq \Lambda(M),
 \\&\norm{\partial_{t,y}( a_{0}, a_{ij}) }_{L^\infty} + \sqrt{\varepsilon}  \norm{\partial_{t,y}\nabla a_{ij}}_{L^\infty} \leq \Lambda(M).
\end{align}
Note that these coefficients are lack of uniform regularity with respect to normal variables.
To get the estimates for solutions of \eqref{omegalagrange}, the authors in  \cite{MasRou} use the paradifferential calculous to prove Theorem
10.6 in \cite{MasRou}, which yields that there exists $\gamma_{0}$ depending only on $M$ such that for $\gamma \geq \gamma_{0}$, the solution of
\eqref{omegalagrange} satisfies the estimate
\begin{equation}
\norm{\Omega^{\alpha}}_{H_T^{1 \over 4} L^2 }^2 \leq
\sqrt{\varepsilon} \Lambda(M)\int_{0}^T  \abs{\Omega^{\alpha}}_0^2.
\end{equation}
By the Minkowski inequality and the one-dimensional Sobolev embedding $H^{1 \over 4} \subset L^4$,
$$\norm{\Omega^\alpha}_{L^4_T L^2 }^2
      \lesssim \norm{\Omega^\alpha}_{ L^2 L^4_T}^2
       \lesssim\norm{\Omega^{\alpha}}_{L^2 H_T^{1 \over 4} }^2=\norm{\Omega^{\alpha}}_{H_T^{1 \over 4} L^2 }^2,$$
one then has
\begin{equation}
\norm{\Omega^\alpha}_{L^4_T L^2 }^2
      \le
\sqrt{\varepsilon} \Lambda(M)\int_{0}^T  \abs{\Omega^{\alpha}}_0^2 .
\end{equation}
This and \eqref{Omegaalphadef} show that
$$ \norm{\omega^\alpha_{h}}_{L^4_T L^2 }^2   \leq  \Lambda(M) \sqrt{\varepsilon} \int_{0}^T\abs{Z^\alpha \omega}_0^2.$$
Thus \eqref{nor2estimate} follows from \eqref{omegab2} and Cauchy's inequality.
\end{proof}

\section{$L^\infty$ estimates}\label{secinfty}

In order to close the a priori estimates, we shall now estimate the $L^\infty$ norms contained in $\Lambda_\infty$, $\norm{\partial_z v
}_{{\mathbb Y}^{\frac{m}{2}+2}}$ and $\sqrt{\varepsilon} \norm{\partial_{zz}v}_{L^\infty}$, and $\int_0^t\sqrt{\varepsilon}\norm{\partial_{zz}v}_{1,\infty}$ that was
used in Proposition \ref{nor2}. We will prove that they can be bounded in terms of the quantities in the left hand side of the estimates of
Propositions \ref{propdzvm-2} and \ref{conormv1}, which will then be shown to be in $L^\infty$ in time.

The key point is to use again the quantity $S_{\bf n}$, and one has the following:
\begin{lem}\label{propLinfty1}
For any $k\in \mathbb{N}$:
\begin{equation}\label{dzv1infty1}
\norm{\partial_z v }_{{\mathbb Y}^k}\le\Lambda\(\frac{1}{c_0},\abs{h}_{{\mathbb Y}^{k+1}}\) \( \norm{S_{\bf n} }_{{\mathbb Y}^{k}}+\norm{v }_{{\mathbb Y}^{k+1}}\).
\end{equation}
Also,
\begin{equation}\label{dzv1infty11}
 \sqrt{\varepsilon}\norm{\partial_{zz}v}_{L^\infty}\leq  \Lambda_0\(\sqrt{\varepsilon} \norm{\partial_{z}S_{{\bf n}}}_{L^\infty} + \norm{S_{{\bf n}}}_{1,
 \infty}  + \norm{v}_{2, \infty}\)
 \end{equation}
 and
 \begin{equation}\label{dzv1infty111}
 \sqrt{\varepsilon}\norm{\partial_{zz}v}_{1,\infty}\leq  \Lambda_0\(\sqrt{\varepsilon} \norm{\partial_{z}S_{{\bf n}}}_{1,\infty} + \norm{S_{{\bf n}}}_{2,
 \infty}  + \norm{v}_{3, \infty}\).
 \end{equation}
\end{lem}
\begin{proof}
This follows again from \eqref{eqdzvnk} and \eqref{eqdzvPin}.
\end{proof}

As a consequence of  Lemma \ref{propLinfty1}, it suffices to estimate $ S_{\bf n} $. Before proceeding further, one has the following
observation.
\begin{lem}\label{iii}
For any smooth cut-off function $\chi$ such that $\chi= 0$ in a vicinity of $\partial\Omega =\{z=0,-b\}$, we have for $m > k+3/2$:
\begin{equation}\label{inftyint}
\norm{\chi f}_{ W^{k, \infty}}\lesssim\norm{f}_{m}.
\end{equation}
\end{lem}
\begin{proof}
It follows directly by the Sobolev embedding $H^k(\Omega)\subset L^\infty(\Omega)$ for $k>3/2$ and the fact that away from the boundary
$\partial\Omega$ the conormal Sobolev norm $\norm{\cdot}_{k}$ is equivalent to the $H^k$ norm.
\end{proof}

Hence, one needs only to estimate $ \chi S_{\bf n}  $ for $\chi(z)$ compactly supported and such that $\chi = 1$ in a vicinity of
$\partial\Omega$. In this step, as in \cite{MasRou12,MasRou}, it is convenient to use a coordinate system where  the Laplacian has the simplest
expression. We shall thus use a normal geodesic coordinate system in the vicinity of $\partial\Omega$. Note that this coordinate
system has not been used before since it requires more regularity for the boundary: to get an $H^m$ (or $\mathcal{C}^m$) coordinate system, the
boundary needs to be $H^{m+1}$ (or $\mathcal{C}^{m+1}$). Nevertheless, at this step, this is not a problem since one needs only to estimate a low
number of derivatives
of the velocity, say $\frac{m}{2}+2$, while the boundary is $H^m$ and $m$ can be as large as needed. We shall choose the cut-off
function $\chi$ in order to get a well defined coordinate system in the vicinity of $\partial\Omega$. For sake of brevity, we consider only the
estimates near $\{z=0\}$, and the estimates near $\{z=-b\}$ may follow in the same way and a bit simpler.

Define a different parametrization of the vicinity of $\{z=0\}$ by
\begin{align} \label{diffgeo}
\begin{split}
\Psi(t, \cdot) : \,   \Omega= \mathbb{R}^2 \times (-\infty, 0) & \rightarrow   \Omega(t) \\
   (y,z) &\mapsto    \left( \begin{array}{ll}  y \\ h(t,y)   \end{array} \right) + z {\bf n}^b(t,y),
\end{split}
\end{align}
where ${\bf n}^b= {\bf N}^b/ |{\bf N}^b|$ is the unit exterior normal with ${\bf N}^b=(-\partial_{1}h, - \partial_{2} h, 1)$. Note that $\Psi(t, \cdot) $ is a
diffeomorphism from $\mathbb{R}^2\times( - \delta, 0)$ to a vicinity of $\partial \Omega(t)$  for some  $\delta$ which depends only on $c_{0}$,
and  for every $t \in [0, T^{\varepsilon,\sigma}]$ thanks to \eqref{apriori}. By this parametrization, the induced Riemannian metric has the block structure
\begin{equation}\label{blockg}
g(t,y,z)= \( \begin{array}{cc} \tilde{g}(t,y,z) & 0  \\ 0  & 1 \end{array} \).
\end{equation}
Hence, the Laplacian in this coordinate system reads:
\begin{equation}\label{laplacien}
\Delta_{g}f= \partial_{zz}f +\frac{1}{2} \partial_{z} \big( \ln |g| \big) \partial_{z} f  + \Delta_{\tilde g} f,
\end{equation}
where $|g|$ denotes the determinant of the matrix $g$ and $\Delta_{\tilde{g}}$ is defined by
\begin{equation}
\Delta_{\tilde{g}} f=  { 1 \over |\tilde{g}|^{1 \over 2} }  \sum_{1 \leq i, \, j \leq 2} \partial_{y^i}\big(\tilde{g}^{ij} |\tilde{g}|^{1 \over
2} \partial_{y^j} f \big).
\end{equation}

To use this coordinate system, one shall first localize the equation for $S^\varphi v$ in a vicinity of  $\{z=0\}$. Set
\begin{equation}\label{Schidef}
S^\chi= \chi(z) S^\varphi v,
\end{equation}
where $\chi(z) \in [0, 1]$ is smooth compactly supported near $\{z=0\}$ and takes the value $1$ in a vicinity of $\{z=0\}$. \eqref{eqS} yields
that
\begin{equation}\label{Schieq}
\partial_{t}^\varphi S^\chi + (v \cdot \nabla^\varphi)S^\chi - \varepsilon \Delta^\varphi S^\chi=  F_{S^\chi}\quad \text{in }\Omega,
\end{equation}
where
\begin{equation}\label{Fchidef}
F_{S^\chi}=  F^\chi + F_{v}
\end{equation}
with
\begin{eqnarray}
& & F^\chi= \big(V_{z} \partial_{z} \chi\big) S^\varphi v  -  \varepsilon \nabla^\varphi \chi \cdot \nabla^\varphi S^\varphi v - \varepsilon \Delta^\varphi
\chi \, S^\varphi v ,\label{fcex} \\
& & F_{v} = - \chi \big( D^\varphi)^2 q -{\chi \over 2}\big( (\nabla ^\varphi v)^2 + ((\nabla^\varphi v)^t)^2\big).\label{fvex}
\end{eqnarray}
Next, define implicitly $\tilde{S}$ in $\Omega(t)$ by $\tilde S (t,  \Phi(t,y,z))= S^\chi(t,y,z)$. Then \eqref{Schieq} yields
\begin{equation}
\partial_{t}   \tilde S  + u \cdot \nabla \tilde S  - \varepsilon \Delta  \tilde S = F_{S^\chi}(t, \Phi(t, \cdot)^{-1})\quad\text{in }\Omega(t).
\end{equation}
Finally, define $S^\Psi$ in $ \Omega $ by  $S^\Psi(t,y,z)= \tilde S (t,\Psi(t,y,z))\equiv  S^\chi(t, \Phi(t, \cdot)^{-1} \circ \Psi)$. It then
follows from (8.14) and \eqref{laplacien} that
\begin{equation}\label{eqSPsi}
\partial_{t}  S^\Psi + w \cdot \nabla  S^\Psi - \varepsilon\big(  \partial_{zz}   S^\Psi  + {1 \over 2 } \partial_{z} \big( \ln |g| \big) \partial_{z}
S^\Psi  + \Delta_{\tilde g}   S^\Psi \big)= F_{S^\chi} (t, \Phi^{-1} \circ \Psi(t,\cdot)) \quad \text{in }\Omega,
\end{equation}
where the vector field $w$ is given by
\begin{equation}\label{defW}
w=\overline{\chi} (\nabla \Psi)^{-1}\big( u(t,  \Psi ) - \partial_{t} \Psi \big)\equiv\overline{\chi} (\nabla \Psi)^{-1}\big( v(t, \Phi^{-1}
\circ \Psi ) - \partial_{t} \Psi \big).
\end{equation}
Note that $S^\Psi$ is compactly supported in $z$ in a vicinity of $\{z=0\}$. The function $\overline{\chi}(z)$ is a function with a slightly
larger support such that $\overline \chi S^\Psi= S^\Psi$. The introduction of this function allows to have $w$ also supported in a vicinity of
$\{z=0\}$. Note that on $\{z=0\}$, $w_3=0$. Indeed, since
\begin{equation}\label{wbord1}
\nabla \Psi(t,y,0)= \left( \begin{array}{ccc}
        1  & 0 & {\bf n}_{1}^b \\ 0 & 1 & {\bf n}_{2}^b   \\ \partial_{1} h & \partial_{2} h  & {\bf n}_{3}^b \end{array} \right),
\end{equation}
thus for any $Y\in \mathbb{R}^3$,
\begin{equation}\label{cdv} ((\nabla\Psi(t,y,0))^{-1} Y)_{3}= Y \cdot {\bf n}^b.
\end{equation}
Hence, \eqref{defW} implies that on $\{z=0\}$,
\begin{equation}\label{w3bord}
w_{3}= v \cdot {\bf n}^b - \partial_{t} h {\bf n}_{3}^b= {1 \over |{\bf N}^b|} \big( v \cdot {\bf N}^b - \partial_{t}h)= 0
\end{equation}
thanks to the kinematic boundary condition.

Now, set
\begin{equation}\label{etadef}
S_{{\bf n}}^\Psi(t,y,z)=\Pi^b(t,y) S^\Psi(t,y,z) {\bf n}^b(t,y)
\end{equation}
with $\Pi^b= \mbox{Id}- {\bf n}^b \otimes {\bf n}^b$. Note that $\Pi^b$ and ${\bf n}^b$  are independent of $z$. Moreover, since the equation
\eqref{eqSPsi} is compactly supported in $z$ in a vicinity of $\{z=0\}$, this yields that $S_{{\bf n}}^\Psi$ solves
\begin{equation}\label{eqSnPsi}
\partial_{t} S_{{\bf n}}^\Psi + w \cdot \nabla  S_{{\bf n}}^\Psi - \varepsilon\big(  \partial_{zz}  + \frac{1}{2} \partial_{z} \big( \ln |g| \big) \partial_{z} \big)
S_{{\bf n}}^\Psi  = F_{{\bf n}}^\Psi\quad \text{in }\{z<0\},
\end{equation}
where
\begin{equation}\label{Fetadef}
F_{{\bf n}}^\Psi=  \Pi^b F_{S^\chi} {\bf n}^b + F_{{\bf n}}^{\Psi, 1}  + F_{{\bf n}}^{\Psi, 2}
\end{equation}
with
\begin{eqnarray}
& &   \label{Feta1}
F_{{\bf n}}^{\Psi, 1}=  \big(( \partial_{t} + w_{y} \cdot \nabla_{y}) \Pi^b\big)S^\Psi {\bf n}^b + \Pi^b S^\Psi\big( \partial_{t}+ w_{y}\cdot \nabla_{y})
n^b, \\& & \label{Feta2} F_{{\bf n}}^{\Psi, 2}=  - \varepsilon \Pi^b\big( \Delta_{\tilde{g}} S^\Psi\big) {\bf n}^b.
\end{eqnarray}
On $\{z=0\}$, $S_{{\bf n}}^\Psi = S_{{\bf n}}=0$. Furthermore, it is convenient to eliminate the term $\varepsilon \partial_{z} \ln |g| \partial_{z}$ in
the equation \eqref{eqSnPsi}.
Set
\begin{equation}\label{etadefbis}
\rho = |g|^{1 \over 4} S_{{\bf n}}^\Psi,
\end{equation}
then
\begin{equation}\label{eqrhobis}
\partial_{t}  \rho + w \cdot \nabla   \rho - \varepsilon  \partial_{zz}   \rho= \mathcal{H}\quad \text{in }\{z<0\},
\end{equation}
where
\begin{equation}\label{Fg}
\mathcal{H}= |g|^{ 1 \over 4 } \(F_{{\bf n}}^\Psi + F_{g}\)\text{
with }
F_{g}=  \rho|g|^{-\frac{1}{2}} \( \partial_{t} + w \cdot \nabla - \varepsilon \partial_{zz} \) |g|^{1 \over 4}.
\end{equation}
Trivially, on $\{z=0\}$, $\rho= 0$.

\subsection{Estimate of $\norm{\partial_z v}_{{\mathbb Y}^{\frac{m}{2}+2}}$}\label{sec linf1}

We now establish the first $L^\infty$ estimate. Note that it is equivalent to estimate $S_{{\bf n}}$ or $S_{{\bf n}}^\Psi$ or  $\rho$.  Indeed,  by the
definition \eqref{etadef},
using the chain rule and the fact that $Z$ is tangent to $\{z=0\}$, one has
$$ \norm{S_{{\bf n}}^\Psi}_{{\mathbb Y}^{k}}  \leq \Lambda \(\frac{1}{c_0},\abs{h}_{{\mathbb Y}^{k+1}}\) \norm{ \Pi^b S^\varphi v \, {\bf n}^b}_{{\mathbb Y}^{k}}.$$
Since $|\Pi- \Pi^b| + | {\bf n}- {\bf n}^b|= \mathcal{O}(z)$ in the vicinity of $\{z=0\}$, thus
\begin{equation}\label{etaSn}
\norm{S_{{\bf n}}^\Psi}_{{\mathbb Y}^{k}}  \leq \Lambda \(\frac{1}{c_0},\abs{h}_{{\mathbb Y}^{k+1}}\) \(\norm{ S_{\bf n}}_{{\mathbb Y}^{k}}+\norm{ v}_{{\mathbb Y}^{k+1}}\).
\end{equation}
Similar arguments show that
\begin{equation}\label{Sneta}
\norm{S_{{\bf n}}}_{{\mathbb Y}^{k}}  \leq \Lambda \(\frac{1}{c_0},\abs{h}_{{\mathbb Y}^{k+1}}\) \(\norm{ S_{\bf n}^\Psi}_{{\mathbb Y}^{k}}+\norm{ v}_{{\mathbb Y}^{k+1}}\).
\end{equation}
On the other hand, it is easy to see that it is equivalent to estimate $\rho$ or $S_{{\bf n}}^\Psi$. By \eqref{dzv1infty1}, it thus suffices to
estimate $\rho$.

Set
\begin{equation}
\widetilde{\Lambda}_{\infty}(t)=\Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb X}^{\frac{m}{2}+7}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{m}{2}+2}} +
\norm{v}_{{\mathbb X}^{\frac{m }{2}+7}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{m }{2}+6}}   \).
\end{equation}

\begin{prop}\label{propdzvinfty}
For $m\ge 14$, it holds that
\begin{equation}
\norm{ \rho(t) }_{{\mathbb Y}^{\frac{m}{2}+2}} \lesssim\norm{ \rho(0)}_{{\mathbb Y}^{\frac{m}{2}+2}}
+ \int_{0}^t \widetilde{\Lambda}_{\infty}\(1+\varepsilon\norm{\nabla S_{\bf n} }_{{\mathbb X}^{m-2}}\)  .
\end{equation}
\end{prop}
\begin{proof}
Apply $Z^\alpha$, for $\alpha\in \mathbb{N}^{1+3}$ with $|\alpha|=k\le \frac{m}{2}+2$, to \eqref{eqrhobis} to obtain
\begin{equation}\label{eqetaFP0}
\partial_{t}Z^\alpha{\rho}+w\cdot\nabla Z^\alpha{\rho}-\varepsilon\partial_{zz}Z^\alpha{\rho}=Z^\alpha {\mathcal{H}}+\mathcal{C}_S\quad \text{in
}\{z<0\},
\end{equation}
where
\begin{equation}\label{Ht}
\mathcal{C}_S=\mathcal{C}_S^1+\mathcal{C}_S^2
\end{equation}
with
\begin{equation}
\mathcal{C}_S^1=\[Z^\alpha,w_y\]\cdot  \nabla_y \rho+\[Z^\alpha,w_3\]\cdot  \partial_z \rho:=\mathcal{C}_{Sy}+\mathcal{C}_{Sz}\text{ and
}\mathcal{C}_S^2=-\varepsilon\[Z^\alpha,\partial_{zz}\] \rho.
\end{equation}
The maximum principle on \eqref{eqetaFP0} yields that, since $Z^\alpha{\rho}=0$ on $\{z=0\}$,
\begin{align}\label{linfty1}
\norm{Z^\alpha{\rho}}_{L^\infty}\le \norm{Z^\alpha{\rho}(0)}_{L^\infty}+\int_0^t\(\norm{Z^\alpha {\mathcal{H}}}_{L^\infty}+\norm{
\mathcal{C}_S}_{L^\infty}\).
\end{align}
The right hand side of \eqref{linfty1} can be estimated as follows. For the commutator $\mathcal{C}_S^1$, the direct estimates yield
\begin{equation}\label{eeqq10}
\norm{\mathcal{C}_{Sy}}_{L^\infty}\lesssim \norm{w_y}_{{\mathbb Y}^k}\norm{\rho}_{{\mathbb Y}^k}.
\end{equation}
To estimate $\mathcal{C}_{Sz}$, by expanding the commutator and using \eqref{idcom}, one needs to estimate terms of the form
$$\norm{ Z^\beta  w_{3}  \partial_{z} Z^\gamma  {\rho} }_{L^\infty}$$
with $\beta +  \gamma \le \alpha$ and $|\gamma|\le |\alpha|-1$. Since $w_3=0$ on $\{z=0\}$, so
\begin{equation}\label{eeqq}
\norm{ Z^\beta  w_{3}  \partial_{z} Z^\gamma  {\rho} }_{L^\infty}=\norm{\frac{ Z^\beta  w_{3}}{z(z+b)}  Z_3 Z^\gamma  {\rho}
}_{L^\infty}\lesssim\norm{\partial_{z}w_3}_{{\mathbb Y}^k} \norm{ \rho}_{{\mathbb Y}^k}.
\end{equation}
For the commutator $\mathcal{C}_S^2$, using \eqref{idcom} repeatedly leads to
\begin{align}\label{idcom10}
- \[Z^\alpha,\partial_{zz}\] \rho&=\partial_z\(\[Z^\alpha, \partial_{z}\]{\rho}\) +\[Z^\alpha,\partial_z\] \partial_{z}{\rho}
\\\nonumber& =\sum_{| \beta | \leq k-1} \partial_z\( c_{\beta } \partial_{z} ( Z^\beta  \rho)\) +\sum_{| \beta | \leq k-1} c_{\beta } \partial_{z} (
Z^\beta \partial_z\rho)
\\\nonumber& =\sum_{| \beta' | \leq k-1}  c_{\beta' } Z^{\beta'} \partial_{z} \rho   +\sum_{| \beta' | \leq k-1} \tilde c_{\beta'}  Z^{\beta'}
\partial_{zz}\rho.
\end{align}
Hence,
\begin{equation}\norm{ \mathcal{C}_S^2 }_{L^\infty}\lesssim \varepsilon \norm{ \partial_z  {\rho} }_{{\mathbb Y}^{k-1}}+\varepsilon \norm{ \partial_{zz}  {\rho} }_{{\mathbb Y}^{k-1}}.
\end{equation}
Note that using the equation \eqref{eqrhobis} implies
\begin{equation}
\varepsilon \norm{ \partial_{zz}  {\rho} }_{{\mathbb Y}^{k-1}} \le \norm{\partial_{t}  \rho + w \cdot \nabla   \rho-\mathcal{H}}_{{\mathbb Y}^{k-1}}
\le \norm{ \rho}_{{\mathbb Y}^{k}} +\norm{  w \cdot \nabla   \rho}_{{\mathbb Y}^{k-1}}+\norm{ \mathcal{H}}_{{\mathbb Y}^{k-1}}.
\end{equation}
Recall from \eqref{dzzves11} that
\begin{equation}
\varepsilon \norm{ \partial_z  {\rho} }_{{\mathbb Y}^{k-1}}\le \varepsilon \Lambda\(\frac{1}{c_0},\abs{h}_{{\mathbb Y}^{k+1}}\)\norm{ \nabla^2  v }_{{\mathbb Y}^{k-1}}\le \Lambda_\infty.
\end{equation}
As \eqref{eeqq10} and \eqref{eeqq},
\begin{equation}\label{eeqq333}\norm{  w \cdot \nabla   \rho}_{{\mathbb Y}^{k-1}}\lesssim \(\norm{ w_y}_{{\mathbb Y}^k}+\norm{\partial_{z}w_3}_{{\mathbb Y}^k}\) \norm{ \rho}_{{\mathbb Y}^k}.
\end{equation}
Thus,
\begin{equation}\label{eeqq111}\norm{ \mathcal{C}_S^2 }_{L^\infty}\lesssim \(1+\norm{ w_y}_{{\mathbb Y}^k}+\norm{\partial_{z}w_3}_{{\mathbb Y}^k}\) \norm{ \rho}_{{\mathbb Y}^k}+\norm{
\mathcal{H}}_{{\mathbb Y}^{k-1}}+\Lambda_\infty.\end{equation}
Consequently, in light of the estimates \eqref{eeqq10}, \eqref{eeqq} and \eqref{eeqq111} into \eqref{linfty1}, one obtains
\begin{equation}\label{rhoes1}
\norm{  \rho }_{{\mathbb Y}^k} \lesssim\norm{  \rho(0) }_{{\mathbb Y}^k}
+ \int_{0}^t  \norm{ \mathcal{H}}_{{\mathbb Y}^{k}}+\(1+\norm{ w_y}_{{\mathbb Y}^k}+\norm{\partial_{z}w_3}_{{\mathbb Y}^k}\) \norm{ \rho}_{{\mathbb Y}^k}+\Lambda_\infty.
\end{equation}

Now we estimate $\mathcal{H}$. Note first that
$$
\norm{ |g|^{\frac{1}{4}}F_{g} }_{{\mathbb Y}^{k}}\le \Lambda\(\frac{1}{c_0},\norm{\rho}_{{\mathbb Y}^{k}}+\norm{w}_{{\mathbb Y}^{k}}+\abs{h}_{{\mathbb Y}^{k+3}}\).
$$
Next, it follows from \eqref{Feta1} and \eqref{Feta2} that
$$ \norm{ |g|^{\frac{1}{4}}F_{\bf n}^{\Psi, 1} }_{{\mathbb Y}^{k}} \le \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+2}}
     + \norm{w }_{{\mathbb Y}^{k}}+ \norm{\nabla v }_{{\mathbb Y}^{k}}\)$$
     and
     $$  \norm{ |g|^{\frac{1}{4}}F_{\bf n}^{\Psi,2} }_{{\mathbb Y}^{k}} \le \varepsilon \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+3}}
    \)\norm{\nabla v }_{{\mathbb Y}^{k+2}}.$$
Using \eqref{fcex} and \eqref{fvex}, the fact that $F^\chi$  is supported away from
$\{z=0\}$ and Lemma \ref{iii}, one gets
$$  \norm{ |g|^{\frac{1}{4}}\Pi^b F ^\chi {\bf n}^b }_{{\mathbb Y}^{k}}\le \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+2}}
+\norm{ v }_{{\mathbb X}^{k+4}}\)$$
and
$$  \norm{ |g|^{\frac{1}{4}}\Pi^b F_v {\bf n}^b }_{{\mathbb Y}^{k}}\le \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+2}}+\norm{\nabla v }_{{\mathbb Y}^{k}}
     \)\(1+ \norm{\nabla^2 q }_{{\mathbb Y}^{k}}\).$$
Recalling (8.27), (8.22) and (8.11), and collecting these estimates, one arrives at
\begin{align}\label{hhes}
\norm{\mathcal{H}}_{{\mathbb Y}^{k}}\le \Lambda\( {1 \over c_{0}}, \abs{h}_{{\mathbb Y}^{k+3}}+\norm{ v }_{{\mathbb X}^{k+4}}+\norm{\nabla v }_{{\mathbb Y}^{k}}
     \)\(1+\varepsilon\norm{\nabla v }_{{\mathbb Y}^{k+2}}+ \norm{\nabla^2 q }_{{\mathbb Y}^{k}}\).
\end{align}
Recall from \eqref{qinfty} that
\begin{align} \label{qinfty1}
\norm{\nabla^2 q }_{{\mathbb Y}^{\frac{m}{2}+2}}
 &\nonumber\leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{m+6}{4}+4}}  +\abs{h}_{{\mathbb X}^{\frac{m+6}{2}+4}}+\norm{\nabla
 v}_{{\mathbb Y}^{\frac{m+6}{4}+3}} +  \norm{v}_{{\mathbb X}^{\frac{m+6}{2}+4}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{m+6}{2}+3}}   \)
\\& \leq \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb Y}^{\frac{m}{2}+5}}  +\abs{h}_{{\mathbb X}^{\frac{m}{2}+7}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{m}{2}+2}} +
\norm{v}_{{\mathbb X}^{\frac{m }{2}+7}}  + \norm{\nabla v}_{{\mathbb X}^{\frac{m }{2}+6}}   \)
\end{align}
if $m\ge 10$. Hence, \eqref{rhoes1} implies
\begin{align}
\norm{  \rho }_{{\mathbb Y}^{\frac{m}{2}+2}} \lesssim\norm{  \rho_0 }_{{\mathbb Y}^{\frac{m}{2}+2}}
+ \int_{0}^t  &  \Lambda\( {1 \over c_{0}},   \abs{h}_{{\mathbb X}^{\frac{m}{2}+7}}+\norm{\nabla v}_{{\mathbb Y}^{\frac{m}{2}+2}} +  \norm{v}_{{\mathbb X}^{\frac{m }{2}+7}}
+ \norm{\nabla v}_{{\mathbb X}^{\frac{m }{2}+6}}   \)
\\&\nonumber\qquad\qquad\times\(1+\varepsilon\norm{\nabla v }_{{\mathbb Y}^{\frac{m}{2}+4}}\)  .
\end{align}
Then the desired estimates (8.29) follows for $m\ge 14$ so that $\frac{m}{2}+4+1\le m-2$.
\end{proof}

\subsection{Estimate of $\sqrt{\varepsilon}\norm{\partial_{zz} v}_{L^\infty}$}\label{sec linf2}

The next $L^\infty$ estimate is the only place where one needs to use the compatibility condition on the initial data.  As in the previous
subsection, by  \eqref{dzv1infty11}, one can reduce
the problem to the estimate of $\sqrt{\varepsilon} \norm{\partial_{z} \rho }_{L^\infty}.$
\begin{prop}\label{dzzvLinfty}
Assume that the initial data satisfies the compatibility condition $S_{\bf n}(0)=0$ on $\{z=0,-b\}$. Then it holds that for $m \geq 6$,
\begin{equation}
\sqrt{\varepsilon}\norm{\partial_{z} \rho (t)}_{L^\infty}\lesssim  \sqrt{\varepsilon} \norm{\partial_{z} \rho_{0}}_{L^\infty} + \int_{0}^t  {\Lambda_\infty \over \sqrt{t-
\tau}}.
\end{equation}
\end{prop}
\begin{proof}
The proof follows the spirit of the proof of Proposition 9.8 in \cite{MasRou}. Recall that $ \rho $ satisfies \eqref{eqrhobis} in $\{z<0\}$ with
$\rho=0$ on $\{z=0\}$. Note that one can not apply $\sqrt{\varepsilon}\partial_{z}$ to \eqref{eqrhobis}
and then use the maximum principle due to boundary condition. We shall use a precise description of the
solution of \eqref{eqrhobis}. Indeed, one can use the one-dimensional heat kernel of  $\{z<0\}$:
\begin{equation}
G(t,z,z')= {1 \over \sqrt{4\pi \varepsilon t}}\( e^{- { (z-z')^2 \over 4\varepsilon t }} -  e^{- { (z+z')^2 \over 4\varepsilon t }} \)
\end{equation}
to write that
\begin{align}\label{duhamel}
\sqrt{\varepsilon}\partial_{z} \rho(t,y,z)=&\int_{0}^{+ \infty} \sqrt{\varepsilon}\partial_{z}G(t,z, z') \rho_{0}(y, z') dz'
\\&\nonumber+ \int_{0}^t \sqrt{\varepsilon} \partial_{z} G(t-\tau,z,z')\( \mathcal{H}(\tau,y,z') - w \cdot \nabla \rho\) dz' d\tau.
\end{align}
Since $\rho_{0}=0$ on $\{z=0\}$, thanks to the compatibility condition, one can integrate by parts the first term to obtain
\begin{equation}\label{dzzeta1}
\sqrt{\varepsilon}\norm{\partial_{z} \rho (t)}_{L^\infty}\lesssim  \sqrt{\varepsilon} \norm{\partial_{z} \rho_{0}}_{L^\infty}  + \int_{0}^t  {1 \over \sqrt{t-
\tau}} \(\norm{\mathcal{H}}_{L^\infty}+ \norm{ w \cdot \nabla \rho }_{L^\infty}\).
\end{equation}
Next, it follows from \eqref{hhes} with $k=0$ that
\begin{equation}\label{eeqq444}
\norm{ \mathcal{H}}_{L^\infty} \le \Lambda_\infty\( 1+ \norm{\nabla^2 q}_{L^\infty}\)  \le \Lambda_\infty.
\end{equation}
On the other hand, as \eqref{eeqq333},
\begin{equation}\label{eeqq555}
\norm{w \cdot \nabla \rho }_{L^\infty}   \le \(\norm{w_y}_{L^\infty} +\norm{\partial_{z}w_3}_{L^\infty} \) \norm{ \rho}_{1,\infty}  \le \Lambda_\infty.
\end{equation}
Consequently, plugging \eqref{eeqq444} and \eqref{eeqq555} into \eqref{dzzeta1} yields that
\begin{equation}
\sqrt{\varepsilon}\norm{\partial_{z} \rho (t)}_{L^\infty}\lesssim  \sqrt{\varepsilon} \norm{\partial_{z} \rho_{0}}_{L^\infty} + \int_{0}^t  {\Lambda_\infty \over \sqrt{t-
\tau}},
\end{equation}
which completes the proof.
\end{proof}

\subsection{Estimate of $\sqrt{\varepsilon}\norm{\partial_{zz} v}_{1,\infty}$}

The last $L^\infty$ estimate is the one that was used in Proposition \ref{nor2}. By \eqref{dzv1infty111}, one can again reduce the problem to the
estimate of $\sqrt{\varepsilon
}\norm{ \partial_{z}  \rho}_{1,\infty}$.
\begin{prop}\label{propro1}
For $m \geq 6$, it holds that:
\begin{equation}\label{Linfty0t1}
\int_0^t\sqrt{\varepsilon }\norm{ \partial_{z}  \rho}_{1,\infty} \lesssim   \sqrt{t}  \norm{ \rho _{0}}_{1, \infty}+ t \int_{0}^t {\Lambda_\infty\over \sqrt{t -
\tau}}.
\end{equation}
\end{prop}
\begin{proof}
We will use a different argument from the proof of Lemma 9.9 in \cite{MasRou}.  A direct use of  the Duhamel formula \eqref{duhamel} yields
\begin{equation}
\sqrt{\varepsilon }\norm{ \partial_{z}  \rho}_{1,\infty} \lesssim {1 \over \sqrt{t}} \norm{ \rho _{0}}_{1, \infty}+ \int_{0}^t{1 \over \sqrt{t- \tau}}\(
\norm{ \mathcal{H}}_{1, \infty} + \norm{ w \cdot \nabla \rho }_{1, \infty} \).
\end{equation}
Next, it follows as the previous arguments that
\begin{equation}
\norm{ \mathcal{H}}_{1,\infty} \le \Lambda_\infty\( 1+ \norm{\nabla^2 q}_{1,\infty}\)  \le \Lambda_\infty
\end{equation}
and
\begin{equation}
\norm{w \cdot \nabla \rho }_{1,\infty}   \le \Lambda_\infty \(\norm{w_y}_{2,\infty}+ \norm{\partial_{z}w_3}_{2,\infty} \)\norm{ \rho }_{2, \infty} \le \Lambda_\infty.
\end{equation}
Consequently,
\begin{equation}
\sqrt{\varepsilon }\norm{ \partial_{z}  \rho}_{1,\infty} \lesssim {1 \over \sqrt{t}} \norm{ \rho _{0}}_{1, \infty}+ \int_{0}^t {\Lambda_\infty\over \sqrt{t - \tau}}.
\end{equation}
Integration in time yields
\begin{equation}
\int_0^t\sqrt{\varepsilon }\norm{ \partial_{z}  \rho}_{1,\infty} \lesssim   \sqrt{t}  \norm{ \rho _{0}}_{1, \infty}+ t \int_{0}^t {\Lambda_\infty\over \sqrt{t -
\tau}}.
\end{equation}
which completes the proof of the proposition.
\end{proof}

\section{Proof of Theorem \ref{main}}\label{finalsec}

Collecting the estimates  obtained in Sections \ref{secconormal}--\ref{secinfty}, one can prove Theorem \ref{main} in the similar way as that for
Theorem
1.1 of \cite{MasRou} with slight modifications.

Recall $\mathcal{N}(T)$ and $\mathcal{Q}(T)$. For two parameters $R$ and $c_{0}$ to be chosen $1/c_{0}\ll R$, define
\begin{align}\label{TTTTTT}
        T^{\varepsilon, \delta}_{*}= \sup_{T>0}\left\{T\in [0, 1] \mid  \mathcal{N}(t) \leq R, \ \abs{h(t)}_{2, \infty} \leq {1 \over c_{0}},\
        \partial_{z} \varphi(t) \geq \frac{{c_{0}}}{2}\right.
        \\ \left.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ and }  g  -   \partial_{z}^\varphi q (t)  \geq  {c_{0}\over
2}\text{ on } \{z=0\}, \ \forall t \in [0, T].
        \right\}.\nonumber
\end{align}
Proposition \ref{propro1} yields
\begin{equation} \int_{0}^T\norm{\partial_{zz}v}_{1,\infty} \leq \Lambda(R).
\end{equation}
This allows one to use Proposition \ref{nor2}, which together with Propositions \ref{conormv1}, \ref{conormvm}, \ref{propdzvm-2},
\ref{propdzvinfty} and \ref{dzzvLinfty} implies that, by a suitable linear combination,
\begin{equation}\label{Em1}
\mathcal{N}(t)
       \leq \Lambda\( {1 \over c_{0}}, \mathcal{N}(0)\) + \Lambda(R)\( T^{1 \over 2} +\mathcal{W}(T)+\mathcal{W}(T)^\frac{1}{2}\),
\end{equation}
   where
$$\mathcal{W}(T):=\int_0^T  \abs{\partial_t^mh}_{0}^2 + \sigma     \abs{\partial_t^mh}_{1}^2
+ \norm{\partial_t^m v} ^2 + \norm{\partial_z v}_{{\mathbb X}^{m-1}}^2. $$
It follows from the Cauchy-Schwarz inequality that
\begin{equation}
\mathcal{W}(T)\le \mathcal{N}(T)T^\frac{1}{2} \le R T^\frac{1}{2}. \end{equation}
Hence, one deduces from \eqref{Em1} that
       \begin{equation}
       \label{Em2}
        \mathcal{N}(t)
       \leq \Lambda\( {1 \over c_{0}}, \mathcal{N}(0)\) + \Lambda(R) T^{1 \over 4}.
 \end{equation}
On the other hand, since $\mathcal{N}(T)$ involves time derivatives, one gets easily that
\begin{equation}\label{choixc01}
\abs{h(t)}_{2, \infty} \leq \abs{h(0)}_{2, \infty}+ \Lambda(R) T,\end{equation}
\begin{equation}\label{choixc02}
\partial_{z} \varphi(t) \geq  \partial_{z} \varphi(0)- \Lambda(R) T
\end{equation}
and
\begin{equation}\label{taylorcon}
g  -   \partial_{z}^\varphi q (t) \geq  g  -   \partial_{z}^\varphi q (0) - \Lambda(R) T.
\end{equation}
Consequently, one can choose $c_0$ so that $\abs{h(0)}_{2, \infty}\le \frac{1}{2c_0}$ and then $R= 2 \Lambda\( {1 \over c_{0}},
\mathcal{N}(0)\)$, then there exists $T_{*}$ which depends only on
$ R$ so that for $T \leq \mbox{Min }(T_{*}, T^{\varepsilon, \delta}_{*})$,
$$  \mathcal{N}(t) \leq \frac{3R}{4}, \quad \partial_{z} \varphi(t) \geq \frac{{3c_{0}}}{4}, \ \abs{h(t)}_{2, \infty} \leq {3 \over 4c_{0}}\text{
and }  g  -   \partial_{z}^\varphi q (t)  \geq  {3c_{0}\over
4}\text{ on } \{z=0\}, \ \forall t \in [0, T].$$
This yields  $T^{\varepsilon, \delta }_{*} \geq T_{*}$ by the definition \eqref{TTTTTT} and also the estimate \eqref{mainborne1}. The proof of Theorem
\ref{main} is thus completed. \hfill$\Box$

redpajama_set_name: RedPajamaArXiv