\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 60, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/60\hfil 2D Zakharov-Kuznetsov-Burgers equations]
{2D Zakharov-Kuznetsov-Burgers equations with variable dissipation on a strip}

\author[N. A. Larkin \hfil EJDE-2015/60\hfilneg]
{Nikolai A. Larkin}

\address{Nikolai A. Larkin \newline
Departamento de Matem\'atica, Universidade Estadual
de Maring\'a, Av. Colombo 5790: Ag\^encia UEM, 87020-900,
Maring\'a, PR, Brazil}
\email{nlarkine@uem.br}

\thanks{Submitted November 12, 2014. Published March 10, 2015.}
\subjclass[2000]{35Q53, 35B35}
\keywords{KdV-Burgers equation; dispersive equations; exponential decay}

\begin{abstract}
 An initial-boundary value problem for a  2D Zakharov-Kuznetsov-Burgers
 type equation with dissipation located in a neighborhood of $x=-\infty$
 and posed  on a channel-type strip was considered. The existence and
 uniqueness results for regular and weak  solutions in weighted spaces as
 well as exponential decay of small solutions without restrictions on the
 width of a strip were proven both for regular solutions in an elevated norm
 and for weak solutions in  the $L^2$-norm.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}\label{introduction}

We are concerned with an initial-boundary value problem (IBVP) for
the two-dimensional Zakharov-Kuznetsov-Burgers (ZKB) equation
with a variable dissipativity located in a neighborhood of $x=-\infty$
\begin{equation}\label{kdvb}
u_t+u_x-(a(x)u_x)_x+uu_x +u_{xxx}+u_{xyy}=0\
\end{equation}
posed on a strip modeling an infinite channel
$\{(x,y)\in\mathbb{R}^2:\ x\in \mathbb{R},\,y \in (0,B), \, B>0\}$.
Here $a(x): \mathbb{R}\to \mathbb{R} $ is a sufficiently smooth nonnegative
function such that
\begin{equation} \label{a}
a(x) \geq 0 \text{ in } \mathbb{R}, \;
\sup_{\mathbb{R}}|\partial_x^i a(x)|\leq C(i);\; a(x)\geq a_0>0\;
\forall x< -r,
\end{equation}
where $a_0,r$ are arbitrary positive constants, $i=0,1,2$.
This equation is a two-dimensional analog of the well-known
Korteweg-de Vries-Burgers  equation
\begin{equation}\label{kdv}
u_t+u_x-u_{xx}+uu_x+u_{xxx}=0
\end{equation}
 which includes dissipation due to viscosity of a medium and dispersion and has been
studied by various researchers due to its applications in Mechanics
and Physics \cite{bona1,bona2,marcelo}.
 One can find  extensive bibliography and sharp results on
 decay rates of solutions to the Cauchy problem (IVP) for (1.3) in
\cite{bona1}. Exponential decay of solutions to the initial problem
for \eqref{kdvb} with additional damping has been  established in
\cite{marcelo}. Equations \eqref{kdvb} and \eqref{kdv} are typical
examples of so-called dispersive equations which attract
considerable attention of both pure and applied mathematicians in
the past decades.

 Quite recently, the interest on
dispersive equations became to be extended to  multi-dimensional
models such as Kadomtsev-Petviashvili (KP) and Zakharov-Kuznetsov
(ZK) equations \cite{zk}. As far as the ZK equation and its
generalizations are concerned, the results on IVPs
 can be found in
\cite{fam1,farah,pastor1,pastor2,sautlin,ribaud,temam2} and IBVPs
were studied in \cite{dorlar1,fam2,familark,lar14,lar1,lartron,temam2,wang}. In
\cite{lar1,lartron} it was shown that IBVP for the ZK equation posed on
a half-strip unbounded in $x$ direction with the Dirichlet
conditions on the boundaries  possesses regular solutions which
decay exponentially as $t\to \infty$ provided initial data are
sufficiently small and the width of a half-strip is not too large.
This means that the ZK equation may create an internal dissipative
mechanism for some types of IBVPs.

 The goal of our note is to prove
that the ZKB equation posed on a strip also may create a dissipative
effect without adding any artificial damping even when a variable
dissipativity $(-a(x)u_x)_x$ is acting only for $x<-r$. We must mention that
IBVP for the ZK equation on a strip $(x\in(0,1),\,y\in\mathbb{R})$
has been studied in \cite{dorlar1,temam1} and IBVPs on a strip
$(y\in(0,L),\,x\in \mathbb{R})$  for the ZK equation were considered
in \cite{fambayk} and for the ZK equation with some internal variable
damping  $[-(a_1(x,y)u_x)_x-(a_2(x,y)u_y)_y]$
in \cite{fam3}. In the domain $(y\in(0,B),\,x\in\mathbb{R},\,t>0 )$,
the term $u_x$ in \eqref{kdvb} can be  scaled out by a simple change
of variables. Nevertheless, it can not be safely ignored for
problems posed  both on finite and semi-infinite intervals as well
as on infinite in $y$ direction bands without changes in the
original domain \cite{dorlar1,rozan}.

The main results of our paper are the existence  and uniqueness of
regular and weak  global-in-time solutions for  \eqref{kdvb} posed
on a strip with the Dirichlet boundary conditions  and the
exponential decay rate of these solutions as well as continuous
dependence on initial data.
We must say that exploiting of an exponential weight function is
crucial for obtaining necessary global estimates as well as for
definition of regular and weak solutions.
This fact yearlier has been observed in \cite{kato} while studying the
Cauchy problem for the 1D KdV equation.

This article has the following structure. Section 1 is Introduction.
Section \ref{problem} contains formulation of the problem. In
Section \ref{regexist}, we prove  global existence and uniqueness
theorems for regular solutions in some weighted spaces and
continuous dependence on initial data. In Section \ref{regdecay}, we
prove exponential decay of small regular solutions in an elevated
norm corresponding to the $H^1(\mathcal{S})$-norm.
In Section \ref{weak},  we prove the existence, uniqueness and continuous
dependence on initial data for weak solutions as well
as the exponential decay rate of the $L^2(\mathcal{S})$-norm for small solutions
without limitations on the width of the strip.

\section{Problem and preliminaries}\label{problem}

Let $B,T,r$ be finite positive numbers. Define
$\mathcal{S}=\{(x,y)\in\mathbb{R}^2:\ x\in\mathbb{R},\ y\in(0,B)\};$
$\mathcal{S}_r=\{(x,y)\in\mathbb{R}^2:\ x\in (-r,+\infty),\,
y\in(0,B)\}$ and $\mathcal{S}_T=\mathcal{S}\times (0,T)$.

Hereafter subscripts $u_x,\ u_{xy}$, etc. denote the partial
derivatives, as well as $\partial_x$ or $\partial_{xy}^2$ when it is
convenient. Operators $\nabla$ and $\Delta$ are the gradient and
Laplacian acting over $\mathcal{S}$. By $(\cdot,\cdot)$ and
$\|\cdot\|$ we denote the inner product and the norm in
$L^2(\mathcal{S})$, and $\|\cdot\|_{H^k}$ stands for norms in the
$L^2$-based Sobolev spaces. We will use also the spaces $H^s\cap
L^2_b$, where $L^2_b=L^2(e^{2bx}dx)$, see \cite{kato}.

Consider the  IBVP
\begin{gather}
Lu\equiv u_t-(a(x)u_{x})_x+uu_x+u_{xxx}+u_{xyy}=0\ \ \text{in}\
\mathcal{S}_T; \label{e1}
\\
u(x,0,t)= u(x,B,t)=0,\; x\in \mathbb{R},\ t>0; \label{e2}
\\
u(x,y,0)=u_0(x,y),\quad (x,y)\in\mathcal{S}. \label{e3}
\end{gather}

\section{Existence of regular solutions}\label{regexist}

\subsection{Regularized problem}

First, for  fixed $h\in(0,1)$ sufficently small and $m\in\mathbb{N}$
sufficiently large consider the  regularized problem
\begin{gather}
L_h u_h\equiv u_{ht}-(a(x)u_{hx})_x+u_h u_{hx}+u_{hxxx}+u_{hxyy}
+h\partial_x^4 u_h=0,\quad \text{in }\mathcal{S}_T; \label{eh1}
\\
u_h(x,0,t)= u_h(x,B,t)=0,\quad x\in \mathbb{R},\; t>0; \label{eh2}
\\
u_h(x,y,0)=u_{0m}(x,y),\quad  (x,y)\in\mathcal{S}, \label{eh3}
\end{gather}
where $u_{0m}$ is an independent of $h$ approximation of $u_0$
such that for all $m\in\mathbb{N}$
\begin{gather}
\begin{aligned}
 J_m
&=\int_{\mathcal{S}}\{u_{0m}^2+e^{2bx}(u_{0m}^2+\big[u_{0m}u_{0mx}
 +\Delta u_{0mx}\big]^2+|\nabla u_{0m}|^2 \\
&\quad +|\nabla u_{0mx}|^2+|\partial_x^4 u_{0m}|^2)\}\,dx\,dy<\infty,
\end{aligned}\label{eJ_m}\\
\begin{aligned}
J_{hm}&=\int_{\mathcal{S}}\{u_{0m}^2+e^{2bx}(u_{0m}^2+\big[u_{0m}u_{0mx}
 +\Delta u_{0mx}\big]^2+|\nabla u_{0m}|^2 \\
&\quad +|\nabla u_{0mx}|^2+h|\partial_x^4 u_{0m}|^2)\}\,dx\,dy<\infty,
\end{aligned}\label{eJ_h}\\
\begin{aligned}
J_{0m}&=\int_{\mathcal{S}}\{u_{0m}^2+e^{2bx}(u_{0m}^2+\big[u_{0m}u_{0mx}
 +\Delta u_{0mx}\big]^2+|\nabla u_{0m}|^2\\
&\quad +|\nabla u_{0mx}|^2)\}\,dx\,dy <\infty,
\end{aligned} \label{eJ_{0m}}
\\
J_{0}=\int_{\mathcal{S}}\{u_{0}^2+e^{2bx}(u_{0}^2+\big[u_{0}u_{0x}
 +\Delta u_{0x}\big]^2+|\nabla u_{0}|^2
 +|\nabla u_{0x}|^2)\}\,dx\,dy<\infty \label{eJ_{0}}
\end{gather}
and $ \lim_{m\to \infty}J_{0m}=J_0$.

Obviously, for $m\geq m^*$ sufficiently large,
$$
J_{0m}\leq 2J_0,\quad J_{hm}\leq J_m,\quad \|u_{0m}\|^2\leq 2\|u_0\|^2,\quad
(e^{2bx},u_{0m}^2)\leq 2(e^{2bx},u_0^2).
$$


\subsection*{Approximate solutions}
 We will construct  solutions to
\eqref{eh1}-\eqref{eh3} by the Faedo-Galerkin method: let ${w_j(y)}$
be orthonormal in $L^2(\mathcal{S})$ eigenfunctions of the
Dirichlet problem
\begin{gather}
w_{jyy}+\lambda_j w_j=0, \quad y\in (0,B); \label{2.4} \\
w_j(0)=w_j(B)=0. \label{2.5}
\end{gather}
Define approximate solutions of \eqref{eh1}-\eqref{eh3} as follows:
\begin{equation}
 u^N_h(x,y,t)=\sum^N_{j=1} w_j(y)g_{hj}(x,t), \label{UN}
\end{equation}
where $g_{hj}(x,t)$ are solutions to the following Cauchy problem for
the system of $N$ nonlinear parabolic equations:
\begin{gather}
\begin{aligned}
&\frac{\partial}{\partial t}g_{hj}(x,t)+\frac{\partial^3}{\partial
x^3}g_{hj}(x,t)-\frac{\partial}{\partial x}(a(x)g_{hjx}(x,t))
 -\lambda_j\frac{\partial}{\partial x} g_{hj}(x,t)  \\
&+\int^B_0 u^N_h(x,y,t)u^N_{hx}(x,y,t)w_j(y)\, dy
+h\frac{\partial^4}{\partial x^4} g_{hj}(x,t)=0,
\end{aligned}\label{2.6}\\
g_{hj}(x,0)=\int^B_0 w_j(y)u_{0m}(x,y)\,dy,\quad j=1,\dots,N. \label{2.7}
\end{gather}
It is known that for $g_{hj}(x,0)$ sufficiently smooth the Cauchy problem
for the parabolic system
\eqref{2.6}-\eqref{2.7} has a unique regular solution
(at least local in $t$) \cite{friedman,kato1,pazy}.
 To prove the existence of  global regular solutions for
 \eqref{eh1}-\eqref{eh3}, we need uniform in $N$  global in $t$
 estimates of approximate solutions $u^N_h(x,y,t)$.

\subsection*{Estimate I}
 Multiply the j-th equation of \eqref{2.6} by
 $g_{hj}$, sum up over $j=1,\dots,N$ and integrate the result with respect
 to $x$ over $\mathbb{R}$ to obtain
$$
\frac{d}{dt}\|u^N_h\|^2(t)+2(a(x),|u^N_{hx}|^2)(t)+2h\|u^N_{hxx}\|^2(t)= 0 .
$$
Since $u^N_{0m}=\sum^N_{j=1} w_j(y)g_{hj}(x,0)$ is an approximation of $u_0$,
then for $N,m$ sufficiently large $\|u^N_{0m}\|^2\leq 2\|u_{0m}\|^2\|\leq 4\|u_0\|^2$.
 Hence it follows for  $N,m$ sufficiently large  that for all $t\in(0,T)$
\begin{gather}
\|u^N_h\|^2(t) +2h\int_0^t\|u^N_{hxx}\|^2(s)\,ds
 \leq\|u^N_{0m}\|^2\leq 4\|u_{0}\|^2 \label{E1},\\
\int_0^t\int_{\mathcal{S}-\mathcal{S}_r}|u^N_{hx}(x,y,t)|^2(s)\,dxy\,ds
\leq \frac{\|u^N_{0m}\|^2}{2a_0}\leq  2\frac{\|u_{0}\|^2}{a_0}\label{E1r}.
\end{gather}
In our calculations we  drop the indices $h,N,m$ where it is not
ambiguous.

\subsection*{Estimate II}
For some positive $b$, multiply the j-th
equation of \eqref{2.6} by $e^{2bx}g_j$ , sum up over $j=1,\dots,N$
and integrate the result with respect  to $x$ over $\mathbb{R}$.
Dropping the indices $N,h$, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u^2)(t)+(e^{2bx},[2a(x)
 +6b-16hb^2]u^2_x)(t)+2b(e^{2bx},u^2_y)(t) \\
&+2h(e^{2bx},u_{xx}^2)(t)-\frac{4b}{3}(e^{2bx},u^3)(t)
 -(e^{2bx},A(b,a,h)u^2)(t)=0,
\end{aligned}\label{2}
\end{equation}
where $A(b,a,h)$ is a continuous function depending on $b,h$, $a(x)$
and derivatives of $a(x)$.
In our calculations, we will frequently use the following
multiplicative inequalities \cite{lady}.

\begin{proposition} \label{GN}
(i) For all $u \in H^1(\mathbb{R}^2)$,
\begin{equation}
{\|u\|}_{L^4(\mathbb{R}^2)}^2
\leq 2 {\|u\|}_{L^2(\mathbb{R}^2)}{\|\nabla u\|}_{L^2(\mathbb{R}^2)}.
 \label{p1}
\end{equation}
(ii) For all $u \in H^1(D)$,
\begin{equation}
{\|u\|}_{L^4(D)}^2 \leq C_D {\|u\|}_{L^2(D)}{\|u\|}_{H^1(D)}, \label{p2}
\end{equation}
where the constant $C_D$ depends on a way of continuation of
$u \in  H^1(D)$ as $ \tilde{u}(\mathbb{R}^2)$ such that $\tilde{u}(D)=u(D)$.
\end{proposition}

Extending $u^N_h(x,y,t)$ for a fixed $t$ into exterior of
$\mathcal{S}$ by 0 and exploiting  inequality \eqref{p1}, we find
 \begin{equation}
\frac{4b}{3}(e^{2bx}u^3)(t)\leq
b(e^{2bx},u^2_y)(t) \\+2b(e^{2bx},u^2_x)(t)+2(b^3
+\frac{8b}{9}\|u_{0m}^N\|^2)(e^{2bx},u^2)(t).\label{nl}
\end{equation}
Substituting this into \eqref{2} and taking into account \eqref{a},
 for $h$ sufficiently small and $N,m$ sufficiently large we arrive to
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u^2)(t)+4b (e^{2bx},u^2_x)(t)+b(e^{2bx},u^2_y)(t)
+h(e^{2bx},u_{xx}^2)(t)\\
&\leq C(b,a)(1+\|u_0\|^2)(e^{2bx},u^2)(t). \label{2.9}
\end{aligned}
\end{equation}
By  Gronwall's lemma,
$$
(e^{2bx},u^2)(t)\leq C(b,a,T,\|u_0\|)(e^{2bx},u^2_0),
$$
where $C(b,a,T,\|u_0\|)$ depends on $a(x)$ and its derivatives.
Returning to \eqref{2.9},
\begin{equation}
\begin{aligned}
&(e^{2bx},|u^N_h|^2)(t)+\int_0^t (e^{2bx},|\nabla u^N_h|^2+h|u^N_{hxx}|^2)(\tau )d
\tau \\
&\leq C(b,a,T,\|u_0\|)(e^{2bx},u^2_0)\quad \forall t \in (0,T),
\end{aligned}\label{E2}
\end{equation}
whence
$$
e^{-2br}\int_0^t\|u_x\|_{L^2(\mathcal{S}_r)}(\tau)\,d\tau
\leq C(b,a,T,\|u_0\|)(e^{2bx},u^2_0).
$$
Adding \eqref{E1r}, we obtain
\begin{equation}
\int_0^t\|u^N_{hx}\|_{L^2(\mathcal{S})}(\tau)\,d\tau
\leq C(b,a,r,T,\|u_0\|)(e^{2bx},u^2_0) \label{ux}
\end{equation}
and the constants in \eqref{E2}, \eqref{ux} do not depend on $N,h,m$.


\subsection*{Estimate III}
Multiplying the j-th equation of \eqref{2.6} by
$-(e^{2bx} g_{jx})_x$, and dropping the index $N$, we arrive to 
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u_x^2)(t)+(e^{2bx},[2a(x)+6b-16hb^2]u^2_{xx})(t)+2b(e^{2bx},u_{xy}^2)(t) \\
&+2h(e^{2bx},u_{xxx}^2)(t)+(e^{2bx},A(b,a,h)u_x^2)(t)
 +(e^{2bx},u_x^3)(t)-2b(e^{2bx} u,u_x^2)(t)=0.
\end{aligned}\label{e3b}
\end{equation}
Using Proposition \ref{GN}, we estimate
\begin{align*}
I_1=(e^{2bx},u_x^3)(t)
&\leq \|u_x\|(t)\|e^{bx}u_x\|^2(t)_{L^4(\mathcal{S})} \\
&\leq 2\|u_x\|(t)\|e^{bx}u_x\|(t)\|\nabla(e^{bx} u_x)\|(t) \\
&\leq \delta(e^{2bx},2u_{xx}^2+u_{xy}^2)(t)+2\big[\delta
b^2+\frac{\|u_x\|^2(t)}{2\delta}\big](e^{2bx},u_x^2)(t).
\end{align*}
Similarly,
\begin{align*}
I_2
&=2b(e^{2bx},uu_x^2)(t)\\
&\leq \delta(e^{2bx},2u_{xx}^2+u_{xy}^2)(t)
 +\big[2b^2\delta+\frac{4b^2}{\delta}\|u^N_{0m}\|^2(t)\big](e^{2bx},u_x^2)(t).
\end{align*}
Substituting $I_1,I_2$ into \eqref{e3}, taking $\delta$ and a fixed
$h>0$ suffuciently small and making use of \eqref{E1}, \eqref{ux}, we
obtain for $\forall t\in(0,T):$
\begin{equation}
\begin{aligned}
&(e^{2bx},|u^N_{hx}|^2)(t)+\int_0^t(e^{2bx},|\nabla u^N_{hx}|^2+h|u^N_{hxxx}|^2 )(s)\,ds \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},u_{0mx}^2).
\end{aligned}\label{E3}
\end{equation}

\subsection*{Estimate IV}
Multiplying the j-th equation of \eqref{2.6} by
$-2(e^{2bx}\lambda g_{j})$, and dropping the index $N$, we arrive to 
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u_y^2)(t)+(e^{2bx},[2a(x)+6b-16hb^2]u^2_{xy})(t)+2b(e^{2bx},u_{yy}^2)(t) \\
&+2(1-b)(e^{2bx},u_x u_y^2)(t)+2h(e^{2bx},u_{xxy}^2)(t)
+(e^{2bx},A(b,a,h)u_y^2)(t)=0.
\end{aligned} \label{4}
\end{equation}
Using Proposition \ref{GN}, we estimate
\begin{align*}
I&=2(1-b)(e^{2bx},u_x u_y^2)(t)\\
&\leq 2C_D(1+b)\|u_x\|(t)\|e^{bx}u_y\|(t)\|
e^{bx}u_y\|_{H^1(\mathcal{S})}(t) \\
&\leq\delta(e^{2bx}, 2u_{xy}^2+u_{yy}^2)(t)+\big[2\delta
(1+b^2) +\frac{C_D^2(1+b)^2\|u_x\|^2(t)}{\delta}\big](e^{2bx},u_y^2)(t).
\end{align*}
For $\delta,h$ sufficiently small, we transform \eqref{4} into the inequality
\begin{align*}
&\frac{d}{dt}(e^{2bx},
u_y^2)(t)+4b(e^{2bx},u_{xy}^2)(t)+b(e^{2bx},u_{yy}^2)(t) 
+2h(e^{2bx},u_{xxy}^2)(t)\\
&\leq C(b,a)[1+\|u_x\|(t)^2](e^{2bx},u_y^2)(t).
\end{align*}
Using \eqref{ux} and the Gronwall lemma, we obtain for all
$t\in(0,T)$,
\begin{equation}
\begin{aligned}
&(e^{2bx},|u^N_{hy}|^2)(t)+\int_0^t(e^{2bx},|u^N_{hyy}|^2+h|u^N_{hxxy}|^2)(s)\,ds \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},u_{0my}^2). \label{E4}
\end{aligned}
\end{equation}
This and \eqref{E3} imply that for all finite $r>0$ and all
$t\in(0,T)$,
\begin{equation}
\|u^N\|^2(t)
\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},|\nabla u_{0m}|^2). \label{Ert}
\end{equation}

\subsection*{Estimate V} Multiplying the j-th equation of \eqref{2.6} by
$(e^{2bx} g_{jxx})_{xx}$, and dropping the indices $N,h$, we arrive to 
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},
u_{xx}^2)(t)+(e^{2bx},[2a(x)+6b-16hb^2]u_{xxx}^2)(t)+2b(e^{2bx},u_{xxy}^2)(t) \\
&+2h(e^{2bx},u_{xxxx}^2)(t)+\sum_{i=1}^2(e^{2bx},A_i(b,a,h)|\partial^i_x u|^2)(t) \\
&-2b(e^{2bx},u u_{xx}^2)(t)+5(e^{2bx} u_x,u_{xx}^2)(t)=0,
\end{aligned}\label{5}
\end{equation}
where $A_i(b,a,h)$ are constants. Using \eqref{p1}, we find
\begin{align*}
I&=-2b(e^{2bx},u u_{xx}^2)(t)+5(e^{2bx} u_x,u_{xx}^2)(t) \\
&\leq 2\delta(e^{2bx},2u_{xxx}^2+u_{xxy}^2)(t)
 +\big[4b^2\delta+\frac{25}{\delta}\|u_x\|(t)^2 \\
&+\frac{4b^2}{\delta}\|u\|^2(t)\big](e^{2bx},u_{xx}^2)(t).
\end{align*}
Substituting $I$ in \eqref{5} for $\delta,h$ sufficiently small, we obtain
\begin{align*}
&\frac{d}{dt}(e^{2bx},u_{xx}^2)(t)+4b(e^{2bx},u_{xxx}^2)(t)+ b(e^{2bx},u_{xxy}^2)(t)
+2h(e^{2bx},u_{xxxx}^2)(t)\\
&\leq C(b,a)[1+\|u_x\|^2(t)+\|u\|(t)^2](e^{2bx},u_{xx}^2)(t).
\end{align*}
Taking into account \eqref{E1}, \eqref{ux}, we find
\begin{equation}
\begin{aligned}
&(e^{2bx},|u^N_{hxx}|^2)(t)+\int_0^t(e^{2bx},|\nabla
u^N_{hxx}|^2+h|u^N_{hxxxx}|^2)(s)\,ds \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},u_{0mxx}^2)\quad
\forall t\in(0,T).
\end{aligned} \label{E5}
\end{equation}

\subsection*{Estimate VI} Differentiate \eqref{2.6} by $t$ and multiply
the result by $e^{2bx} g_{jt}$ to obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u_t^2)(t)+(e^{2bx},[2a(x)+6b-16hb^2]u_{xt}^2)(t)+2b(e^{2bx},u_{ty}^2)(t) \\
&+2h(e^{2bx},u_{txx}^2)(t)-(e^{2bx},A(b,a,h)u_{t}^2)(t)
+(2-2b)(e^{2bx} u_x, u_{t}^2)(t)=0.
\end{aligned}\label{6}
\end{equation}
 Using \eqref{p1}, we estimate
 \begin{align*}
 I&=(2-2b)(e^{2bx} u_x, u_{t}^2)(t)
\leq  2(2+2b)\|u_x\|(t)\|e^{bx}u_t\|(t)\|\nabla(e^{bx}  u_t)\|(t) \\
 &\leq \delta(e^{2bx},2u_{xt}^2+u_{ty}^2)(t)
 +\big[2b^2\delta+\frac{(2+2b)^2\|u_x\|(t)^2}{\delta}\big](e^{2bx},u_t^2)(t).
\end{align*}
Taking $\delta,h$ sufficiently small and substituting $I$ into \eqref{6},
 we obtain
\begin{align*}
&\frac{d}{dt}(e^{2bx},u_t^2)(t)+4b(e^{2bx},u_{xt}^2)(t)+b(e^{2bx},u_{ty}^2)(t) \\
&\leq C(b,a)[1+\|u_x\|(t)^2](e^{2bx} , u_{t}^2)(t).
\end{align*}
Using \eqref{eh1} and \eqref{UN}, we calculate
$$
(e^{2bx},|u_{ht}^N|^2)(0)\leq CJ_{hm}.
$$
This implies that for all $t\in 0,T)$,
\begin{equation}
\begin{aligned}
&(e^{2bx},|u^N_{ht}|^2)(t)+\int_0^t(e^{2bx},|\nabla
u^N_{hs}|^2+h|u^N_{hsxx}|^2)(s)\,ds \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},u_{ht}^2)(0) \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)\|)J_{hm} \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)\|)J_{m}.
\end{aligned} \label{E6}
\end{equation}

\subsection*{Estimate VII} Multiplying the j-th equation of \eqref{2.6} by
$-e^{2bx} g_{jx}$,  dropping indices $h,N$, we arrive, to 
\begin{equation}
\begin{aligned}
&(e^{2bx},[u_{xy}^2+(1-3hb)u_{xx}^2])(t)\\
&=(e^{2bx}[u_{t},u_x)(t)+(e^{2bx},uu_x^2)(t) -(e^{2bx},[4hb^3-2b^2
+ba-\frac{a_x}{2}]u^2_x)(t).
\end{aligned} \label{e7}
\end{equation}
Using \eqref{p1}, we estimate
\[
I=(e^{2bx},uu_x^2)(t)\leq \delta(e^{2bx},2u_{xx}^2+u_{xy}^2)(t)+
\big[2b^2\delta+\frac{\|u^N_{0m}\|^2}{\delta}\big](e^{2bx},u_x^2)(t).
\]
Taking $\delta,h$ sufficiently small and $N,m$ sufficiently large,
using \eqref{E3}, (3.27), \eqref{E6} and substituting $I$, into \eqref{e7}, we obtain
\begin{equation}
\begin{aligned}
(e^{2bx}|u^N_{hxx}|^2+|u^N_{hxy}|^2)(t)
&\leq
C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)\|)J_{hm} \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{m} \quad
\forall t\in(0,T).
\end{aligned}\label{E7}
\end{equation}

\subsection*{Estimate VIII} We shall need the following lemma.

\begin{lemma} \label{supr}
Let $u(x,y): \mathcal{S}\to \mathbb{R}$ be such that
$$ 
\int_{\mathcal{S}}e^{2bx} [u^2(x,y)+|\nabla
u(x,y)|^2+u_{xy}^2(x,y)]\,dx\,dy < \infty
$$ 
and for all $x\in\mathbb{R}$ there is some $y_0\in [0,B]$ such that
$u(x,y_0)=0$. Then
\begin{equation}
\begin{aligned}
\sup_{\mathcal{S}}|e^{bx}u(x,y,t)|^2
&\leq \delta(1+2b^2)(e^{2bx},u_y^2)(t)+
2\delta(e^{2bx},u_{xy}^2)(t) \\
&\quad +\frac{2\delta_1}{\delta}(e^{2bx},
u_x^2)(t)+\frac{1}{\delta}\big[\frac{1}{\delta_1}+2\delta_1
b^2\big](e^{2bx},u^2)(t),
\end{aligned}\label{esup}
\end{equation}
where $\delta, \delta_1$ are arbitrary positive numbers.
\end{lemma}

\begin{proof} 
Denote $v=e^{bx}u$. Then simple calculations give
\[
\sup_{\mathcal{S}} v^2(x,y,t)\leq
\delta[\|v_y\|^2(t)+\|v_{xy}\|^2(t)] +\frac{1}{\delta}[\|v_x\|^2(t)+\|v\|^2(t)].
\]
Returning to the function $u(x,y,t)$, we complete the proof.
\end{proof}

Multiplying the j-th equation of \eqref{2.6} by $e^{2bx} g_{jxxx}$, 
 dropping indices $h,N$, we arrive to
\begin{equation}
\begin{aligned}
(e^{2bx},u_{xxy}^2+(1-hb)u_{xxx}^2)(t)
&=-(e^{2bx}[u_t-(a(x)u_{x})_x],u_{xxx})(t)\\
&\quad -(e^{2bx} uu_x,u_{xxx})(t)+2b^2(e^{2bx},u_{xy}^2)(t).
\end{aligned}\label{8}
\end{equation}
Using Lemma \ref{supr} and \eqref{E1}, we estimate
\begin{align*}
I&=(e^{2bx} uu_x,u_{xxx})(t)
 \leq \|u\|(t)\|e^{bx}u_{xxx}\|(t)\sup_{\mathcal{S}}|e^{bx}u_x(x,y,t)| \\
&\leq \epsilon\|u_0\|^2(e^{2bx},u_{xxx}^2)(t)+\frac{1}{4\epsilon}
\big[\frac{1}{\delta}(1+2b^2)(e^{2bx},u_x^2)(t) \\
&\quad +\frac{2}{\delta}(e^{2bx}, u_{xx}^2)(t)+\delta(1+2b^2)(e^{2bx},u_{xy}^2)(t)
 +2\delta(e^{2bx},u_{xxy}^2)(t)\big].
\end{align*}
Taking $\epsilon$ and $\delta$ sufficiently small, positive and
substituting $I$ into \eqref{8}, we find
\begin{equation}
\begin{aligned}
(e^{2bx}, |\nabla u^N_{hxx}|^2)(t)
&\leq C(b,r,a,T,\|u_0\|)J_{hm} \\
& \leq C(b,r,a,T,\|u_0\|)J_m   \quad \forall t\in(0,T).
\end{aligned}\label{E81}
\end{equation}
Consequently,  from the equalities
$$ -(e^{2bx} [u^N_{ht}-(a(x)u^N_{hx})_x+u^N_{hxxx}+u^N_{hxyy}+u^N_h
u^N_{hx}+hu^N_{hxxxx}],u^N_{hyy})(t)=0
$$
and
$$(e^{2bx} [u^N_{ht}-(a(x)u^N_{hx})_x+u^N_{hxxx}+u^N_{hxyy}+u^N_h
u^N_{hx}+hu^N_{hxxxx}],u^N_{hxyy})(t)=0
$$
it follows that
\begin{equation}
(e^{2bx}, |u^N_{hyy}|^2+|u^N_{hxyy}|^2)(t)\leq C(b,r,a,T,\|u_0\|)J_{hm} \quad
\forall t\in(0,T). \label{E82}
\end{equation}

Jointly, estimates \eqref{E3},\eqref{E4}, \eqref{E5},
\eqref{E7},\eqref{E81}, \eqref{E82} read
\begin{equation}
\begin{aligned}
&\|u^N_{h}\|^2(t)+(e^{2bx}, |u^N_h|^2+|\nabla u^N_h|^2 +|\nabla u^N_{hx}|^2+|\nabla
u^N_{hy}|^2+|\nabla u^N_{hxx}|^2)(t) \\
&\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}\quad \forall t\in(0,T).
\end{aligned} \label{E83}
\end{equation}
This and \eqref{E6} imply
\begin{equation} \label{E85}
h^2(e^{2bx},u_{hxxxx}^2)(t) \leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}.
\end{equation}
In other words,
\begin{equation}
e^{bx}u^N_h, e^{bx}u^N_{hx}\in L^{\infty}(0,T;H^2(\mathcal{S}))
\label{E84}
\end{equation}
and these inclusions are uniform in $N,h>0$ while $m$ is fixed and
sufficiently large.

\subsection*{Estimate IX} 
Differentiating the j-th equation of \eqref{2.6}
with respect to $x$ and multiplying the result by  $e^{2bx}
\partial_x^4 g_j$,  dropping indices $h,N$, we arrive to
\begin{equation}
\begin{aligned}
&(e^{2bx},u_{xxxy}^2+(1-hb)u_{xxxx}^2)(t)\\
&=2b^2(e^{2bx},u_{xxy}^2)(t)-(e^{2bx} u_{xt},u_{xxxx})(t)\\
&\quad +(e^{2bx}[a_{xx}u_x+2a_xu_{xx}+au_{xxx}],u_{xxxx})(t) 
 -(e^{2bx}[u_x^2+uu_{xx}],\partial_x^4u)(t).
\end{aligned} \label{e9}
\end{equation}

Using Lemma \ref{supr} and \eqref{E83}, we estimate
\begin{align*}
I_1&= (e^{2bx},u_x^2,\partial_x^4 u)(t)\leq
\|u_x\|(t)\|e^{bx}\partial_x^4
u\|(t)\sup_{\mathcal{S}}|e^{bx}u_x(x,y,t)| \\
&\leq \frac{\epsilon_1}{2}(e^{2bx},|\partial_x^4
u|^2)(t)+\frac{1}{2\epsilon_1}\|u_x\|^2(t)\big[(1+2b^2)(e^{2bx},u_x^2)(t) \\
&\quad +2(e^{2bx},u_{xx}^2)(t)+(1+2b^2)(e^{2bx},u_{xy}^2)(t)+2(e^{2bx},u_{xxy}^2)(t)\big] \\
&\leq \frac{\epsilon_1}{2}(e^{2bx},|\partial_x^4
u|^2)(t)+\frac{1}{2\epsilon_1}C(b,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm},
\end{align*}
\begin{align*}
I_2&=(e^{2bx} u,u_{xx}\partial_x^4 u)(t)\leq\|e^{bx}\partial_x^4
u\|(t)\|u\|(t)\sup_{\mathcal{S}}|e^{bx}u_{xx}(x,y,t| \\
&\leq\frac{\epsilon_1}{2}\|u_0\|^2(t)(e^{2bx},|\partial_x^4
u|^2)(t)+\frac{1}{2\epsilon_1}\{2\delta(e^{2bx},u_{xxxy}^2)(t) \\
&\quad +\delta(1+2b^2)(e^{2bx},u_{xxy}^2)(t)+\frac{2}{\delta}(e^{2bx},u_{xxx}^2)(t) \\
&\quad +\frac{1}{\delta}(1+2b^2)(e^{2bx},u_{xx}^2)(t)\}.
\end{align*}

Applying Young's inequality, taking $h,\epsilon_1,\delta$
sufficiently small positive, substituting $I_1,I_2$ into \eqref{e9}
and integrating the result, we arrive to 
\begin{equation}
\int_0^t(e^{2bx}, |u^N_{hxxxy}|^2+|u^N_{hxxxx}|^2)(s)\,ds
\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}  \label{E9}
\end{equation}
for all $t\in(0,T)$.

\subsection*{Estimate X} Multiplying the j-th equation of \eqref{2.6} by
$-e^{2bx}\lambda^2 g_{jx}$,  dropping indices $h,N$, we arrive to
\begin{equation}
\begin{aligned}
&(e^{2bx},(1-3hb) u_{xxyy}^2+u_{xyyy}^2)(t)\\
&=-(e^{2bx} u_{ty},u_{xyyy})(t) +(2b^2-4hb^3)(e^{2bx},u_{xyy}^2)(t)\\
&\quad +(e^{2bx}[a_x u_{xy}+au_{xyy}],u_{xyyy})(t) \\
&\quad -(e^{2bx} u_y u_x,u_{xyyy})(t)+(e^{2bx} uu_{xy},u_{xyyy})(t).
\end{aligned}\label{e10}
\end{equation}
We estimate
\begin{gather*}
I_1=-(e^{2bx} u_{ty},u_{xyyy})(t)\leq
\frac{\epsilon}{2}(e^{2bx},u_{xyyy}^2)(t)+\frac{1}{2\epsilon}(e^{2bx},u_{yt}^2)(t), \\
\begin{aligned}
I_2&=(e^{2bx} u_y u_x,u_{xyyy})(t)\leq\|u_x\|(t)\|e^{bx}u_{xyyy}\|(t)
\sup_{\mathcal{S}}|e^{bx}u_y(x,y,t| \\
&\leq  \frac{\epsilon}{2}(e^{2bx},u_{xyyy}^2)(t)
 +\frac{\|u_x\|(t)^2}{2\epsilon}\big[(1+2b^2)(e^{2bx},u_y^2)(t) \\
&\quad +2(e^{2bx},u_{xy}^2)(t)+(1+2b^2)(e^{2bx},u_{yy}^2)(t)
+2(e^{2bx},u_{xyy}^2)(t)\big],
\end{aligned}\\
\begin{aligned}
I_3&=(e^{2bx} uu_{xy},u_{xyyy})(t)\leq\|u\|(t)\|e^{bx}u_{xyyy})\|(t)
 \sup_{\mathcal{S}}|e^{bx}u_{xy}(x,y,t| \\
&\leq \frac{\|u_0\|^2\epsilon_1}{2}(e^{2bx},u_{xyyy}^2)(t)
 +\frac{1}{2\epsilon_1}\big[ 2\delta(e^{2bx}, u_{xxyy}^2)(t) \\
&\quad +\frac{2}{\delta}(e^{2bx},u_{xxy}^2)(t)+\delta(1+2b^2)(e^{2bx},u_{xyy}^2)(t)
+\frac{1}{\delta}(1+2b^2)(e^{2bx},u_{xy}^2)(t)\big].
\end{aligned}
\end{gather*}
Substituting $I_1-I_3$ in \eqref{e10} and choosing
$h,\epsilon,\epsilon_1, \delta$ sufficiently small,
positive, after integration, we transform \eqref{e10}  into the  inequality
\begin{equation}
\int_0^T(e^{2bx},[|u^N_{hxxyy}|^2+|u^N_{hxyyy}|^2])(t)\,dt
\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}. \label{E101}
\end{equation}
Similarly, we obtain from the scalar product
$$
(e^{2bx}\big[u^N_{ht}-(au^N_{hx})_x+u^N_{hxxx}
+u^N_{hxyy}+u^N_h u^N_{hx}+h\partial^4_x u^N_h\big],u^N_{hyyyy})(t)=0
$$
the estimate
\begin{equation}
\int_0^T(e^{2bx},|u^N_{hyyy}|^2)(t)\,dt
\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}.\label{E102}
\end{equation}

 Estimates \eqref{E83}, \eqref{E84}, \eqref{E9}, \eqref{E101},
\eqref{E102} guarantee that
\begin{equation}
\begin{aligned}
&\|e^{bx}u^N_h\|_{L^{\infty}(0,T;H^2(\mathcal{S}))\cap
L^2(0,T;H^3(\mathcal{S}))}+ \|e^{bx}u^N_{hx}\|_{L^{\infty}(0,T;H^2(\mathcal{S}))\cap
L^2(0,T;H^3(\mathcal{S}))} \\
&+ \|e^{bx}u^N_{ht}\|_{L^{\infty}(0,T;L^2(\mathcal{S}))\cap
L^2(0,T;H^1(\mathcal{S}))}\\
&\leq C(b,r,T,\|u_0\|,\|e^{bx}u_0\|)J_{hm}
\end{aligned}\label{EN}
\end{equation}
and since by \eqref{eJ_m}, \eqref{eJ_h}, $J_{hm}\leq J_m$,
these estimates do not depend on $N,h$. Independence of Estimates
\eqref{E1},\eqref{EN} of $N$ allow us first to pass to the limit as
$N\to \infty$ in \eqref{2.6} and to prove the following result:

\begin{theorem} \label{u_h}
Let $r,B, T$ be finite positive numbers, $h\in(0,1)$; a smooth nonnegative
 function $a(x)$ be defined by \eqref{a} and a given function   
$u_{0m}(x,y):\mathbb{R}^2\to \mathbb{R}$ be such that
$u_{0m}(x,0)=u_{0m}(x,B)=0$ and for some $b>0$
\begin{align*}
J_{hm}&= \int_{\mathcal{S}}\{u_{0m}^2+e^{2bx}[u_{0m}^2+|\nabla u_{0m}|^2+|\nabla
u_{0mx}|^2+u_{0m}^2 u_{0mx}^2 \\
&\quad  + |\Delta u_{0mx}|^2+h|\partial_x^4 u_{0m}|^2\big]\}\,dx\,dy < \infty.
\end{align*}
 Then for a fixed $h>0$ sufficiently small and for a fixed $m$ sufficiently 
large there exists a regular solution $u_h(x,y,t)$ to \eqref{eh1}-\eqref{eh3}:
 \begin{gather*}
u_{hm}\in L^{\infty}(0,T;L^2(\mathcal{S})),\quad 
u_{hmx}\in  L^2(0,T;L^2(\mathcal{S})); \\
e^{bx}u_{hm},\;e^{bx}u_{hmx} \in L^{\infty}(0,T;H^2(\mathcal{S}))\cap
 L^2(0,T;H^3(\mathcal{S})); \\
e^{bx}u_{hmt} \in L^{\infty}(0,T;(L^2(\mathcal{S}))\cap
 L^2(0,T;H^1(\mathcal{S}))
 \end{gather*}
which (dropping the index $m$) for a.e. $ t\in(0,T)$ satisfies the equality
\begin{equation}
 \begin{aligned}
&(e^{bx}\big[u_{ht}-(a(x))u_{hx})_x+u_{hxxx}+u_hu_{hx}+u_{hxyy}\big]\phi(x,y,t))(t)\\
&+h(e^{bx}u_{hxxxx},\phi(x,y,t))(t)=0,
\end{aligned} \label{defu_h}
\end{equation}
where $\phi(x,y,t)$ is an arbitrary function from
 $L^{\infty}(0,T;L^2_b(\mathcal{S}))$.
\end{theorem}

\begin{proof}
Dropping the indices $h,m$, rewrite \eqref{2.6} in the form
 \begin{align}
&(e^{bx}\big[u^N_{t}-(a(x))u^N_{x})_x+u^N_{xxx}+u^N  u^N_{x}
+u^N_{xyy}\big],\Phi^N(y)\Psi(x,t))(t) \\
&+h(e^{bx}u^N_{xxxx},\Phi^N(y)\Psi(x,t))(t)=0,
\end{align}
where $\Phi^N(y)$ is an arbitrary function from the linear
combinations $\sum_{i=1}^N \alpha_i w_i(y)$ and $\Psi(x,t)$ is an
arbitrary function from $L^{\infty}(0,T;L^2_b(\mathcal{S}))$. 
Taking into account estimates \eqref{E1}, \eqref{EN} and fixing $\Phi^N$,
 we can easily pass to the limit as $N\to\infty$ in linear terms. 
To pass to the limit in the nonlinear term, we must use \eqref{Ert} and
repeat arguments of \cite{kato}. Since linear combinations
$[\sum_{i=1}^N \alpha_i w_i(y)]\Psi(x,t)$ are dense in
$L^2(0,T;L^2(\mathcal{S})\cap L^2_b(\mathcal{S}))$, we arrive to \eqref{defu_h}. 
This proves the existence of regular solutions to \eqref{eh1}-\eqref{eh3}.
\end{proof}

Moreover, estimates \eqref{E1}, \eqref{EN} do not depend on $N,h$, 
hence they are valid also for the limit solution $u_h$. 
This allow us to pass to the limit as $h\to 0$ in \eqref{defu_h} and 
to prove the following result.

\begin{theorem} \label{regsol}
Let $r, B, T$ be finite positive numbers; a smooth nonnegative function $a(x)$ 
be defined by \eqref{a} and given function   $u_0(x,y):\mathbb{R}^2\to \mathbb{R}$ 
be such that $u_0(x,0)=u_0(x,B)=0$ and for some $b>0$,
$$
J_0= \int_{\mathcal{S}}\{u_0^2+e^{2bx}[u_0^2+|\nabla u_0|^2
+|\nabla u_{0x}|^2+u_0^2 u_{0x}^2 + |\Delta u_{0x}|^2]\}\,dx\,dy < \infty.
$$
 Then there exists a regular solution $u(x,y,t)$ to \eqref{e1}--\eqref{e3}
such that
\begin{gather*}
 u\in L^{\infty}(0,T;L^2(\mathcal{S})),\quad 
u_x\in  L^2(0,T;L^2(\mathcal{S})); \\
e^{bx}u,\;e^{bx}u_x \in L^{\infty}(0,T;H^2(\mathcal{S}))\cap
 L^2(0,T;H^3(\mathcal{S})); \\
e^{bx}u_t \in L^{\infty}(0,T;L^2(\mathcal{S}))\cap
 L^2(0,T;H^1(\mathcal{S}))
 \end{gather*}
which for a.e. $t\in(0,T)$ satisfies the identity
\begin{equation}
(e^{bx}\big[u_t-(a(x)u_{x})_x+u_{xxx}+uu_x+u_{xyy}\big],\phi(x,y,t))(t)=0,
\label{regularsol}
\end{equation}
where $\phi(x,y,t)$ is an arbitrary function from 
$L^{\infty}(0,T;L^2_b(\mathcal{S}))$.
\end{theorem}

\begin{proof}
First, dropping the index $m$, we assume that 
  $\phi(x,y,t) \in L^{\infty}(0,T;L^2_b(\mathcal{S})\cap H^1(\mathcal{S}))$
in \eqref{defu_h}, 
and rewrite it as 
\begin{align*}
&(e^{bx}\big[u_{ht}-(a(x)u_{hx})_x+u_{hxxx}+u_hu_{hx}+u_{hxyy}\big],
\phi(x,y,t))(t)\\
&- h(e^{bx} u_{hxxx},\phi_x(x,y,t))(t) + b\phi(x,y,t))(t)=0.
\end{align*}
 Due to estimate \eqref{E81},
 $$
\lim_{h\to 0} h(e^{bx} u_{hxxx},\phi_x(x,y,t))(t)+ b\phi(x,y,t))(t)=0. 
$$
Hence, passing to limit as $h\to 0$, we obtain the equality
\begin{equation}
(e^{bx}\big[u_{mt}-(a(x)u_{mx})_x+u_{mxxx}+u_m u_{mx}
+u_{mxyy}\big],\phi(x,y,t))(t)=0,
\end{equation}
where $\phi(x,y,t) \in L^{\infty}(0,T;L^2_b(\mathcal{S}))$.
Obviously, for $u_{0m}(x,y)$ sufficiently smooth and $m$ fixed,
 $\lim_{h \to 0}J_{hm}=J_{0m}$. Taking into account that 
$\lim_{m\to \infty} J_{0m}=J_0$, we have for $m$ sufficiently large that 
$J_{0m}\leq 2J_0$ and consequently, estimates \eqref{E1}, \eqref{EN}
 do not depend on $m$, we can pass to the limit as $m\to\infty$ in the 
last equality and come to \eqref{regularsol}, where
$\phi(x,y,t) \in L^{\infty}(0,T;L^2_b(\mathcal{S}))$.
This proves the
existence of regular solutions to \eqref{e1}--\eqref{e3}.
\end{proof}

\begin{remark} \label{limit estimates} \rm
Estimates \eqref{E1}, \eqref{E2}, \eqref{ux} are valid also for  the limit
function $u(x,y,t)$ in the form
\begin{gather}
 \|u\|(t)\leq \|u_0\|^2,\label{Esharp}\\
\begin{aligned}
&\|u\|^2(t)+(e^{2bx},u^2)(t)+\int_0^t\{ \|u_x\|^2(s)+(e^{2bx},|\nabla u|^2)(s)\}\,ds \\
&\leq C(r,b,a,T,\|u_0\|,\|e^{bx}u_0\|)\big[\|u_0\|^2+(e^{2bx},u_0^2)\big] \quad \forall
t\in(0,T).
\end{aligned}\label{E11}
\end{gather}
\end{remark}


subsection*{Uniqueness of a regular solution}

\begin{theorem} \label{uniq} 
A regular solution from Theorem \ref{regsol} is uniquely defined.
\end{theorem}

\begin{proof}
Let $u_1,\,u_2$ be two distinct regular solutions of
\eqref{e1}--\eqref{e3}, then $z=u_1-u_2$ satisfies the 
initial-boundary value problem
\begin{gather}
z_t-(a(x)z_{x})_x+z_{xxx}+z_{xyy} +\frac{1}{2}(u_1^2-u_2^2)_x=0
\quad \text{in } \mathcal{S}_T, \label{u1}\\
z(x,0,t)=z(x,B,t)=0,\quad x\in \mathbb{R},\quad t>0,\label{u2}\\
z(x,y,0)=0 \quad (x,y)\in \mathcal{S}. \label{u3}
\end{gather}
Multiplying \eqref{u1} by $2e^{2bx}z$, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},z^2)(t)+(e^{2bx}[2a(x)+6b],z_x^2)(t) -(e^{2bx}[4b^2a+2ba_x+8b^3],z^2)(t) \\
&+2b(e^{2bx}, z_y^2)(t)+(e^{2bx} [u_{1x}+u_{2x}],z^2)(t)
-b(e^{2bx}(u_1+u_2),z^2)(t)=0.
\end{aligned}\label{u4}
\end{equation}
We estimate
\begin{align*}
I_1&=(e^{2bx}(u_{1x}+u_{2x}),z^2)(t)\leq
\|u_{1x}+u_{2x}\|(t)\|e^{bx}z\|(t)^2_{L^4(\mathcal{S})} \\
&\leq 2\|u_{1x}+u_{2x}\|(t)\|e^{bx}z\|(t)\|\nabla(e^{bx}z)\|(t) \\
&\leq \delta(e^{2bx},[2{z_x}^2+{z_y}^2])(t)+[2{b^2}\delta
 +\frac{2}{\delta}(\|u_{1x}\|^2(t)
 +\|u_{2x}\|^2(t))](e^{2bx},z^2)(t), \\
I_2&=b(e^{2bx}(u_{1}+u_{2}),z^2)(t)
 \leq b\|u_{1}+u_{2}\|(t)\|e^{bx}z\|^2_{L^4(\mathcal{S})} \\
&\leq 2b\|u_{1}+u_{2}\|(t)\|e^{bx}z\|(t)\|\nabla(e^{bx}z)\|(t) \\
&\leq \delta(e^{2bx},2z_x^2+z_y^2)(t)+[2b^2\delta+\frac{2b^2}{\delta}(\|u_{1}\|^2(t)
+\|u_{2}\|^2(t))](e^{2bx},z^2)(t).
\end{align*}
Substituting $I_1,I_2$ into \eqref{u4} and taking $\delta>0$
sufficiently small, we find
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},z^2)(t)+2b(e^{2bx},z_x^2)(t)+b(e^{2bx},z_y^2)(t)\\
&\leq C(b,a)\big[1+\|u_1\|(t)^2
 +\|u_2\|(t)^2+\|u_{1x}\|(t)^2+\|u_{2x}\|(t)^2\big](e^{2bx},z^2)(t).
\end{aligned}\label{u5}
\end{equation}
By \eqref{Esharp} and \eqref{E11},
$$
u_i\in L^{\infty}(0,T;L^2(\mathcal{S})),\quad
u_{ix}\in L^2(0,T;L^2(\mathcal{S}))\quad i=1,2,
$$
hence by  Gronwall's lemma,
$$
(e^{2bx}, z^2)(t)=0 \quad\forall \:t\in(0,T)
$$
and $u_1=u_2$ a.e. in $\mathcal{S}_T$.
\end{proof}

\begin{remark} \label{rmk2} \rm
Changing initial condition \eqref{u3} for $z(x,y,0)=z_0(x,y)\ne 0$,
and repeating the proof of Theorem \ref{regsol}, from \eqref{u5} we obtain 
$$
(e^{2bx},z^2)(t)\leq C(b,r,a,T,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},z^2_0)\quad \forall t\in(0,T).
$$
This means continuous dependence of regular solutions on the initial
data.
\end{remark}

\section{Decay of regular solutions}\label{regdecay}

 In this section we prove exponential decay of regular
 solutions in an elevated weighted norm corresponding to the
 $H^1(\mathcal{S})$ norm. We start with  Theorem  \ref{decay1} which
 is crucial for the main result.

\begin{theorem} \label{decay1}
Let  $b\in(0, b_0),\;a_x(x)\leq 0,\,
\|u_0\|\leq \frac{3\pi}{8B}$ and $u(x,y,t)$ be a regular solution to
\eqref{e1}--\eqref{e3}. Then for all finite $B>0$,
\begin{gather}
 \|e^{bx}u\|^2(t)\leq e^{-\chi t}\|e^{bx}u_0\|^2(0),\label{dec1}\\
\int_0^t e^{\chi s}(e^{2bx},|\nabla u|^2)(s)\,ds\leq 
C(b,\|u_0\|)(1+t)(e^{2bx},u_0^2), \label{dec2}
\end{gather}
 where
 \begin{equation}
 b_0=\frac{\pi^2}{4B^2}\big[\frac{1}{\sup_{\mathbb{R}}[a(x)+\sqrt{a^2(x)
+\frac{5\pi^2}{8B^2}}]}\big],\quad \chi=b_0\frac{\pi^2}{2B^2},
 \end{equation}
\end{theorem}

\begin{proof}
Multiplying \eqref{e1} by $2e^{2bx} u$, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx},u^2)(t)+(e^{2bx}[2a(x)+6¨b],u^2_x)(t)+2b(e^{2bx},u^2_y)(t) \\
&-\frac{4b}{3}(e^{2bx},u^3)(t) -(e^{2bx}\{2b[2a(x)b+a_x(x)]+8b^3\},u^2)(t)=0.
\end{aligned}\label{d1}
\end{equation}
The following proposition is necessary for our proof.

\begin{proposition}\label{propcr}
\begin{equation}
\int_{\mathbb{R}}\int_0^B e^{2bx}u^2(x,y,t)\,dy\,dx\leq
\frac{B^2}{\pi^2}\int_{\mathbb{R}}\int_0^B
e^{2bx}u^2_y(x,y,t)\,dy\,dx. \label{mineq}
\end{equation}
\end{proposition}

\begin{proof}
Since $u(x,0,t)=u(x,B,t)=0$, fixing $(x,t)$, we can use with respect
to $y$ the following Steklov inequality: if $f(y)\in H^1_0(0,\pi)$,
then
$$
\int_0^{\pi}f^2(y)\,dy\leq \int_0^{\pi} |f_y(y)|^2\,dy.
$$
After a corresponding process of scaling we complete the proof.
\end{proof}

Taking into account \eqref{Esharp}, we estimate
\[
I=\frac{4b}{3}(e^{2bx},u^3)(t)
\leq b(e^{2bx},u_y^2+2u_x^2)(t) 
+\big [2b^3+\frac{16b}{9}\|u_0\|^2\big](e^{2bx},u^2)(t).\label{eI}
\]
Using \eqref{mineq} and substituting $I$ in \eqref{d1},
 we arrive to 
\begin{align*}
&\frac{d}{dt}(e^{2bx},u^2)(t)+4b(e^{2bx},u^2_x)(t) 
+b\big[\frac{\pi^2}{B^2}-2a_x(x)\\
&-\{4a(x)b+ 10b^2+\frac{16}{9}\|u_0\|^2\}\big](e^{2bx},u^2)(t)\leq 0
\end{align*}
which can be rewritten as
\begin{equation} 
\frac{d}{dt}(e^{2bx},u^2)(t) +\chi (e^{2bx},u^2)(t)\leq 0, \label{d2}
\end{equation}
where
$$
\chi=b\big[\frac{\pi^2}{B^2}-2a_x(x)-4ba(x)-10b^2-\frac{16\|u_0\|^2}{9}\big].
$$
Since we need $\chi>0$, define
\begin{equation}
0<4ba(x)+10b^2\leq\frac{\pi^2}{4B^2},\quad
\frac{16\|u_0\|^2}{9}\leq\frac{\pi^2}{4B^2}. \label{algebra}
\end{equation}
Solving \eqref{algebra}, we find
\begin{align}
 b_0=\frac{\pi^2}{4B^2}\big[\frac{1}{\sup_{\mathbb{R}}[a(x)+\sqrt{a^2(x)
+\frac{5\pi^2}{8B^2}}]}\big],\quad \chi=b_0\frac{\pi^2}{2B^2}\,.
 \end{align}
From \eqref{d2} we obtain
$$
 (e^{2bx},u^2)(t)\leq e^{-\chi t}(e^{2bx},|u_0|^2).
$$
This inequality  implies \eqref{dec1}.

To prove \eqref{dec2}, we return to \eqref{d1} and multiply it by 
$e^{\chi t}$ to obtain
\begin{equation}
\begin{aligned}
   &\frac{d}{dt}[e^{\chi t}(e^{2bx},u^2)(t)]+e^{\chi t}
\big[(e^{2bx},[2a(x)+6¨b]u_x^2)(t)+2b(e^{2bx},u_y^2)(t)\big] \\
&  =\frac{4b e^{\chi t}}{3}(e^{2bx},u^3)(t)+e^{\chi t}(e^{2bx},A(\chi,b,a)u^2)(t),
\end{aligned}\label{ed2}
\end{equation}
where $ A(\chi, b,a)=2ba_x(x)+4b^2a(x)+8b^3+\chi$.
    Substituting \eqref{eI} in \eqref{ed2}, we obtain
\begin{equation}
\frac{d}{dt}[e^{\chi t}(e^{2bx},u^2)(t)]+e^{\chi t}(e^{2bx},|\nabla u|^2)(t)
\leq C(b,\chi, a,\|u_0\|)e^{\chi t}(e^{2bx},u^2)(t).
\end{equation}
Integrating and using \eqref{dec1} imply
\begin{equation}
e^{\chi t}(e^{2bx},u^2)(t)+\int_0^t e^{\chi s}(e^{2bx},|\nabla u|^2)(s)\,ds \\
\leq C(b,\chi, a,\|u_0\|)(1+t)(e^{2bx},u_0^2). \label{ed3}
\end{equation}
The proof is complete.
 \end{proof}

 Observe that differently from \cite{lar1,lartron}, we do not have
 any restrictions on the width of a strip $B$.
The main result of this section is the following assertion.

\begin{theorem} \label{decay2}
Let l the conditions of Theorem \ref{decay1} be fulfilled. Then
regular solutions of \eqref{e1}--\eqref{e3} satisfy 
\begin{equation}
(e^{2bx},u^2+|\nabla u|^2)(t)\leq C(b,\chi,\|u_0\|)(1+t)e^{-\chi t}(e^{2bx},
\big[u_0^2
+|u_0|^3 +|\nabla u_0|^2\big]) \label{decaymain}
\end{equation}
or
$$
\|e^{bx}u\|^2_{H^1(\mathcal{S})}(t)\leq C(b,\chi,\|u_0\|)(1+t)e^{-\chi
t}(e^{2bx},u_0^2+|u_0|^3 +|\nabla u_0|^2).
$$
\end{theorem}

\begin{proof} 
We start with the following lemma.

\begin{lemma} \label{lem1}
Regular solutions of \eqref{e1}--\eqref{e3} satisfy 
\begin{equation}
\begin{aligned}
&e^{\chi t}(e^{2bx}, |\nabla u|^2)(t)+\int_0^t
e^{\chi s}\{(e^{2bx}[2a(x)+6b],u_{xx}^2)(s)+\frac{b}{2}(e^{2bx},u^4)(s) \\
&+(e^{2bx}[2a(x)+8b], u_{xy}^2)(s)+2b(e^{2bx},u_{yy}^2)(s)\}\,ds\\
&=\frac{e^{\chi t}}{3}(e^{2bx},u^3)(t)
+\int_0^t \{e^{\chi s}(e^{2bx}[\chi+2ba_x(x)+4b^2a(x)+8b^3],u^2_y)(s) \\
&\quad +4b(e^{2bx},uu_y^2)(s)-(e^{2bx}\big[\frac{4a(x)b^2+2ba_x(x)+8b^3}{3}
 -\chi\big],u^3)(s)\}\,ds \\
&\quad +\int_0^t e^{\chi s}(e^{2bx}[\chi+a_{xx}(x)+4b^2a(x)+8b^3],u^2_x)(s)\,ds \\
&\quad +2\int_0^t e^{\chi s}(e^{2bx}[a(x)+4b] u,u_x^2)(s)\,ds
 +(e^{2bx},|\nabla u_0|^2-\frac{u_0^3}{3}).
\end{aligned} \label{e1l1}
\end{equation}
\end{lemma}

\begin{proof}
First we transform the scalar product
\begin{equation}
-(e^{bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big], \\
\big[2(e^{bx}u_x)_x+2e^{bx}u_{yy}+e^{bx}u^2\big])(t)=0 \label{e2l1}
\end{equation}
into the  equality
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(e^{2bx}, |\nabla
u|^2-\frac{u^3}{3})(t)+(e^{2bx}[2a(x)+6b],u_{xx}^2)(t) \\
&+2b(e^{2bx}, u_{yy}^2)(t)+(e^{2bx}[2a(x)+8b],u_{xy}^2)(t)+\frac{b}{2}(e^{2bx},u^4)(t) \\
&=(e^{2bx}[a_{xx}(x)+4a(x)b^2+8b^3],u_x^2)(t)
-(e^{2bx}\big[\frac{4a(x)b^2+2a_x(x)b+8b^3}{3}\big],u^3)(t) \\
&\quad +(e^{2bx}[4a(x)b^2+2ba_x(x)+8b^3],u_y^2)(t)
+4b(e^{2bx},uu_y^2)(t) \\
&\quad +(e^{2bx}[2a(x)+8b],uu_x^2)(t).
\end{aligned}\label{e3l1}
\end{equation}
To prove \eqref{e3l1}, we estimate separate terms in \eqref{e2l1} as
follows:
\begin{align*}
&I_1=-2(e^{bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big],(e^{bx}u_x)_x)(t) \\
&=2(e^{2bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big]_x,u_x)(t) \\
&=\frac{d}{dt}(e^{2bx},u_x^2)(t)+(e^{2bx}[2a(x)+6b],u_{xx}^2)(t)+2b(e^{2bx},u_{xy}^2)(t) \\
&\quad -(e^{2bx}[a_{xx}(x)+4a(x)b^2+8b^3],u_x^2)(t)+(e^{2bx}
u^2,u_{xxx})(t) \\
&\quad -8b(e^{2bx},uu_x^2)(t)+\frac{8b^3}{3}(e^{2bx},u^3)(t);
\end{align*}
\begin{align*}
&I_2=-2(e^{bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big],e^{bx}u_{yy})(t) \\
&=2(e^{bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big]_y,e^{bx}u_y)(t) \\
&=\frac{d}{dt}(e^{2bx},u_y^2)(t)+(e^{2bx}[2a(x)+6¨b],u_{xy}^2)(t)+2b(e^{2bx},u_{yy}^2)(t) \\
&\quad -(e^{2bx}[2ba_x(x)+4b^2a(x)+8b^3],u_y^2)(t)+(e^{2bx}
u,u_{xyy})(t)-4b(e^{2bx},uu_y^2)(t);
\end{align*}
\begin{align*}
&I_3=-(e^{bx}\big[u_t-[a(x)u_{x}]_x+u_{xxx}+u_{xyy}+uu_x\big],e^{bx}u^2)(t) \\
&=-\frac{d}{dt}(e^{2bx},\frac{u^3}{3})(t)+\frac{2b}{3}(e^{2bx}[2ba(x)+a_x(x)],u^3)(t)
 +\frac{b}{2}(e^{2bx},u^4)(t) \\
&\quad -2(e^{2bx} a(x),uu_x^2)(t)-(e^{2bx},u_{xxx}+u_{xyy})(t).
\end{align*}

Summing $I_1+I_2+I_3$, we obtain \eqref{e3l1}. In turn, multiplying
it by $e^{\chi t}$ and integrating the result over $(0,t)$, we arrive 
to \eqref{e1l1}. The proof of Lemma \ref{lem1} is complete.
\end{proof}

Using  \eqref{p1}, we estimate
\begin{align*}
I_4&=\frac{e^{\chi t}}{3}(e^{2bx},u^3)(t)
\leq \frac{2e^{\chi t}}{3}\|u_0\|\|e^{bx}u\|(t)\|\nabla(e^{bx}u)\|(t) \\
&\leq \frac{e^{\chi t}}{2}\{(e^{2bx},|\nabla u|^2)(t)+[\frac{b^2}{2}
+\frac{4\|u_0\|^2}{9}](e^{2bx},u^2)(t)\}.
\end{align*}
Substituting $I_4$ in \eqref{e1l1}, we obtain
\begin{equation}
\begin{aligned}
&e^{\chi t}(e^{2bx}, |\nabla u|^2)(t)+2\int_0^t
e^{\chi s}\{(e^{2bx}[2a(x)+6b],u_{xx}^2)(s)+\frac{b}{2}(e^{2bx},u^4)(s) \\
&+(e^{2bx}[2a(x)+8b], u_{xy}^2)(s)+2b(e^{2bx},u_{yy}^2)(s)\}\,ds\\
&\leq 2\int_0^t \{e^{\chi s}(e^{2bx}[\chi+2ba_x(x)+4b^2a(x)+8b^3],u^2_y)(s) \\
&\quad +4b(e^{2bx},uu_y^2)(s)-(e^{2bx}\big[\frac{4a(x)b^2+2ba_x(x)+8b^3}{3}-\chi\big],
 u^3)(s)\}\,ds \\
&\quad +2\int_0^t e^{\chi s}(e^{2bx}[\chi+a_{xx}(x)+4b^2a(x)+8b^3],u^2_x)(s)\,ds \\
&\quad +4\int_0^t e^{\chi s}(e^{2bx}[a(x)+4b] u,u_x^2)(s)\,ds \\
&\quad +[b^2+\frac{8\|u_0\|^2}{9}](e^{2bx},u^2)(t)+2(e^{2bx},|\nabla u_0|^2-\frac{u_0^3}{3}).
\end{aligned} \label{e4l1}
\end{equation}
By Proposition \ref{GN},
\begin{align*}
I_1&=-(e^{2bx}\big[\frac{4a(x)b^2+2ba_x(x)+8b^3}{3}-\chi\big],u^3)(s)\\
&\leq (e^{2bx},|\nabla u|^2)(s)+C(\chi,b,a,\|u_0\|)(e^{2bx},u^2)(s).
\end{align*}
Similarly,
\begin{align*}
I_2&=4(e^{2bx}[a(x)+4b]u,u_x^2)(s)\\
&\leq \delta(e^{2bx},2u_{xx}^2+u_{xy}^2)(s) 
+\frac{1}{\delta}C(\chi,a,b,\|u_0\|)(e^{2bx},u_x^2)(s)
\end{align*}
and
\begin{align*}
I_3&=8b(e^{2bx},uu_y^2)(s)\leq
8bC_D\|u_0\|\|e^{bx}u_y\|(s)\|e^{bx}u_y\|_{H^1(\mathcal{S})}(s) \\
&\leq\delta(e^{2bx},2u_{xy}^2+u_{yy}^2)(s)+\big[(2b^2+1)\delta+\frac{16b^2\|u_0\|^2
C_D^2}{\delta}\big](e^{2bx}, u_y^2)(s).
\end{align*}
Taking $\delta=2b$ and using \eqref{dec1}, \eqref{dec2},  from
\eqref{e4l1} we obtain
$$
e^{\chi t}(e^{2bx},|\nabla u|^2)(t)\leq C(b,\chi,a,\|u_0\|)(1+t)(e^{2bx},
u_0^2+|u_0|^3+|\nabla u_0|^2).
$$
Adding \eqref{dec1}, we complete
the proof of Theorem \ref{decay2}.
\end{proof}

\section{Weak solutions}\label{weak}

Here we prove the existence, uniqueness and continuous
dependence on initial data  as well as exponential decay
results for weak solutions of \eqref{e1}--\eqref{e3} with
$u_0\in L^2(\mathcal{S})\cap L^2_b(\mathcal{S})$.

\begin{theorem}\label{weakexist}
Let $u_0 \in L^2 (\mathcal{S})\cap L^2_b(\mathcal{S})$. Then for all
finite positive $T$ and $B$ there exists at least one function $u(x,y,t)$  such that
\begin{gather*}
u \in L^{\infty}(0,T;L^2(\mathcal{S})),\quad
u_x\in L^2(0,T;L^2(\mathcal{S})),\\
e^{bx}u \in L^{\infty}(0,T;L^2(\mathcal{S}))\cap L^2(0,T;H^1(\mathcal{S})).
\end{gather*}
Moreover,
\begin{gather*}
u^m\rightharpoonup  u\quad \text{*-weakly in } 
 L^{\infty}(0,T;L^2(\mathcal{S})\cap L^2_b(\mathcal{S})),\\
e^{bx}u^m\rightharpoonup e^{bx}u \quad \text{weakly in } L^{2}(0,T;H^1(\mathcal{S})),
\end{gather*}
where $u^m$ are regular solutions to \eqref{e1}--\eqref{e3} provided by 
Theorem \ref{regsol}.  For a.e. $t\in(0,T)$
the following integral identity takes a place
\begin{equation}
\begin{aligned}
&(e^{bx} u,v)(t)+\int_0^t\{-(e^{bx} u,v_s)(s)+(e^{bx}
u_x,\big[v_{xx}+(a(x)+2b)v_x \\
&+(a(x)b+b^2)v\big])(s)-\frac{1}{2}(e^{2bx} u^2,bv+v_x)(s)
 +(e^{bx} u_y,bv_y+v_{xy})(s)\}\,ds\\
&=(e^{bx} u_0,v(x,y,0)),
\end{aligned}\label{weakdefin}
\end{equation}
where
$$ v \in L^{\infty}(0,T;L^2_b(\mathcal{S}))
 \cap L^2(0,T;H^2(\mathcal{S}))\quad v_t\in L^2(0,T;L^2_b(\mathcal{S}))
$$
is an arbitrary function.
\end{theorem}

\begin{proof}
To justify our calculations, we must operate with
sufficiently smooth solutions $u^m(x,y,t)$. With this purpose, we
consider first initial functions $u_{0m}(x,y)$,  which satisfy
conditions of Theorem \ref{regsol}, and obtain estimates 
\eqref{Esharp}, \eqref{E11} for functions $u^m(x,y,t)$. 
By Theorem \ref{regsol}, we can write for a.e. $t\in (0,T)$
\begin{equation}
(e^{bx}\big[u^m_t-(a(x)u^m_{x})_x+u^m_{xxx}+u^m u^m_x+u^m_{xyy}\big],
\phi(x,y,t))(t)=0,
\label{regularsolm}
\end{equation}
where $\phi(x,y,t)$ is an arbitrary function from 
$$
L^{\infty}(0,T;L^2_b(\mathcal{S}))\cap L^2(0,T;H^2(\mathcal{S})),
$$
and the inner product at the left-hand side of \eqref{regularsolm} 
is an integrable function on $(0,T)$. Integrating \eqref{regularsolm} 
over (0,t),  after standard calculations we arrive to the integral equality
\begin{equation}
\begin{aligned}
&(e^{bx} u^m,v)(t)+\int_0^t\{-(e^{bx} u^m,v_s)(s)+(e^{bx}
u^m_x,\big[v_{xx}+(a(x)+2b)v_x \\
&+(a(x)b+b^2)v\big])(s) -\frac{1}{2}(e^{2bx} |u^m|^2,bv+v_x)(t)
+(e^{bx} u^m_y,bv_y+v_{xy})(s)\}\,ds\\
&=(e^{bx} u_{0m},v(x,y,0)).
\end{aligned} \label{e5.3}
\end{equation}
Using estimates \eqref{Esharp}, \eqref{E11}, we pass to the limit as
$m\to\infty$ and come to \eqref{weakdefin}.
\end{proof}

\begin{remark} \label{rmk3} \rm
There is an alternate manner to define a weak solution as a distribution,
 using estimates \eqref{Esharp}, \eqref{E11}, and passing to the limit 
directly in  \eqref{regularsolm},
$$
(e^{bx}\big[u^m_t-(a(x)u^m_{x})_x+u^m_{xxx}+u^m u^m_x+u^m_{xyy}\big],\phi(x,y,t))(t)
=0.
$$
Here we can estimate $e^{bx}u^m_t\in L^2(0,T;H^{-2}(\mathcal{S}))$ which 
we will need to pass to the limit in the nonlinear term.
\end{remark}

\begin{remark} \label{rmk4} \rm
It is easy to verify that regular solutions from Theorem \ref{regsol}, 
$u^m$, satisfy  \eqref{e1}--\eqref{e3}), hence \eqref{regularsolm} 
and \eqref{e5.3}. And vice versa, if regular solutions $u^m$  satisfy 
\eqref{e5.3}, then for a.e. $t\in(0,T)$ they also satisfy  \eqref{regularsolm}, 
and consequently, \eqref{e1}--\eqref{e3}.
\end{remark}

\subsection*{Uniqueness of a weak solution}

\begin{theorem} \label{weakuniq}
A weak solution of Theorem \ref{weakexist} is uniquely defined.
\end{theorem}

\begin{proof}
Actually, this proof is provided by Theorem \ref{uniq}. It is
sufficient to approximate an initial function $u_0\in
L^2(\mathcal{S})\cap L^2_b(\mathcal{S})$ by regular functions $u_{0m}$ in the form:
$$
\lim_{m\to\infty}\|u_{0m}-u_0\|_{L^2(\mathcal{S})\cap L^2_b(\mathcal{S})}=0,
$$
where $u_{0m}$ satisfies the conditions of Theorem \ref{regsol}.
This guarantees the existence of the unique regular solution to
\eqref{e1}--\eqref{e3}, $u^m$, and allows us, due to Remark \ref{rmk4}, 
to repeat all the calculations which have
been done during the proof of Theorem \ref{uniq} and arrive to 
\begin{align*}
&\frac{d}{dt}(e^{2bx},z_m^2)(t)+2b(e^{2bx},z_{mx}^2)(t)+b(e^{2bx},z_{my}^2)(t) \\
&\leq C(b,a)\big[1+\|u_{1m}\|(t)^2+\|u_{2m}\|(t)^2+\|u_{1xm}\|(t)^2
 +\|u_{2xm}\|(t)^2\big](e^{2bx},z_m^2)(t).
\end{align*}
By the generalized Gronwall`s lemma,
\begin{align*}
(e^{2bx},z_m^2)(t)
&\leq \exp\{\int_0^t C(b,a)\big[1+\|u_{1m}\|^2(s)+\|u_{2m}\|^2(s)
+\|u_{1xm}\|^2(s) \\
&\quad +\|u_{2xm}\|^2(s)\big]\,ds\}(e^{2bx},z_{0m}^2)(t).
\end{align*}
Functions $u_{1m}$ and $u_{2m}$ for $m$ sufficiently large satisfy
the estimate \eqref{E11},
\begin{align*}
&\|u_{im}\|(t)^2+\int_0^t \|u_{imx}\|(s)^2\,ds\\
&\leq C(r,T,\|u_{0m}\|,\|e^{bx}u_{0m}\|)[\|u_{0m}\|^2  +(e^{2bx},u_{0m}^2)]\\
&\leq C(r,T,\|u_0\|,\|e^{bx}u_0\|)[\|u_0\|^2+(e^{2bx},u_0^2)],\quad i=1,2.
\end{align*}
Hence,
\begin{align*}
&\exp\{\int_0^t
C\big[1+\|u_{1m}\|(s)^2+\|u_{2m}\|(s)^2+\|u_{1xm}\|(s)^2 
+\|u_{2xm}\|(s)^2\big]\,ds\}\\
&\leq C(b,a,T,r,\|u_0\|,\|e^{bx}u_{0}\|).
\end{align*}
Since $e^{bx}z(x,y,t)$ is a weak limit of regular
solutions $\{e^{bx}z_m(x,y,t)\}$, then 
$$
(e^{2bx},z^2)(t)\leq (e^{2bx}, z_m^2)(t)= 0.
$$ 
This implies $u_1\equiv u_2$ a.e. in
$\mathcal{S}_T$. The proof of Theorem \ref{weakuniq} is complete.
\end{proof}

\begin{remark} \label{rmk5} \rm
Changing initial condition $z(x,y,0)\equiv 0$  for
$z(x,y,0)=z_0(x,y)\ne 0$, and repeating the proof of Theorem
\ref{weakuniq}, we obtain  that
$$
(e^{2bx},z^2)(t)\leq C(b,a,T,r,\|u_0\|,\|e^{bx}u_0\|)(e^{2bx},z^2_0)\quad \forall t\in(0,T).
$$
This means continuous dependence of weak solutions on initial data.
\end{remark}


\subsection*{Decay of weak solutions}

\begin{theorem} \label{weakdecay}
Let  $b\in(0, b_0)$, $a_x(x)\leq 0$,
$\|u_0\|\leq 3\pi/(16B)$ and $u(x,y,t)$ be a weak solution of
\eqref{e1}--\eqref{e3}. Then for all finite $B>0$,
\begin{equation}
\|e^{bx}u\|^2(t)\leq e^{-\chi t}\|e^{bx}u_0\|^2(0),\label{dec1w}
\end{equation}
 where
\begin{equation}
 b_0=\frac{\pi^2}{4B^2}
\big[\frac{1}{\sup_{\mathbb{R}}[a(x)+\sqrt{a^2(x)+\frac{5\pi^2}{8B^2}}]}\big],
\quad \chi=b_0\frac{\pi^2}{2B^2},
\end{equation}
\end{theorem}

\begin{proof}
Similarly to the proof of the uniqueness result for a weak solution,
we approximate $u_0 \in L^2(\mathcal{S})\cap L^2_b(\mathcal{S})$ 
by sufficiently smooth functions $u_{0m}$, satisfying the conditions 
of Theorem \ref{regsol}, in order to work with regular solutions. Acting
in the same manner as by the proof of Theorem \ref{decay1}, we arrive to 
\begin{equation}
 \|e^{bx}u_m\|^2(t)\leq e^{-\chi t}\|e^{bx}u_0\|^2(0),
\label{wdecay}
\end{equation}
 where
 $$
\chi=b_0\frac{\pi^2}{2B^2}.
$$
Since $u(x,y,t)$ is a weak limit of regular solutions $\{u_m(x,y,t)\}$,
$$
(e^{2bx}, u^2)(t)\leq (e^{2bx}, u_m^2)(t)\leq e^{-\chi t}(e^{2bx},u_0^2).
$$
The proof of Theorem \ref{weakdecay} is complete.
\end{proof}

In this Theorem we have a more strict condition for the decay of weak solutions 
$\|u_0\|\leq \frac{3\pi}{8\sqrt{2}B}$ instead of $\|u_0\|\leq \frac{3\pi}{8B}$ in the
case of decay for regular solution which follows from \eqref{algebra}. 
In the case of weak solutions, we use in \eqref{algebra} instead of $u_0$ 
its approximation $u_{0m}$.
This implies for $m$ sufficiently large
$$ 
64B^2\|u_{0m}\|^2\leq 64B^22\|u_0\|^2\leq 9\pi^2
$$
and consequently, $\|u_0\|\leq 3\pi/(\sqrt{2}8B)$.


\subsection*{Acknowledgments} 
The author appreciate very much the helpful comments of the anonymous referee.

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\end{document}