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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 20, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/20\hfil Boundary stabilization]
{Boundary stabilization of a coupled system of nondissipative
Schrodinger equations}
\author[H. Ilhem\hfil EJDE-2006/20\hfilneg]
{Hamchi Ilhem}  % in alphabetical order

\address{Hamchi Ilhem \hfil\break
Department of Mathematics, University of Batna,
05000, Algeria}
\email{hamchi\_ilhem@yahoo.fr}

\date{}
\thanks{Submitted June 20, 2005. Published February 9, 2006.}
\subjclass[2000]{93D15, 35Q40, 42B15}
\keywords{Boundary stabilization; Schrodinger equation; multiplier method}

\begin{abstract}
 We use the multiplier method and the approach in \cite{a1}
 to study the problem of exponential stabilization of a coupled
 system of two Schrodinger equations.
\end{abstract}

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

\section{Introduction}

Let $\Omega $ be a bounded domain of $\mathbb{R}^{n}$
($n\in \mathbb{N}^{\ast}$) with smooth boundary $\Gamma $,
 and let $(\Gamma _{0},\Gamma_{1})$ be a partition of $\Gamma $.
Consider the boundary feedback system:
\begin{gather}
iy_{t}+\Delta y+F_{1}(y,\nabla y)+P_{1}(z)=0\quad\text{in }Q,  \label{e1.1.a1} \\
iz_{t}+\Delta z+F_{2}(z,\nabla z)+P_{2}(y)=0\quad\text{in }Q,
\label{e1.1.a2} \\
y =z=0\quad\text{on }\Sigma _{0},  \label{e1.1.b1} \\
\frac{\partial y}{\partial \upsilon }+g(y_{t})=0,\quad
\frac{\partial z}{\partial \upsilon }+h(z_{t})=0\quad\text{on }
\Sigma _{1},  \label{e1.1.b2}\\
y(0)=y_{0},\quad z(0)=z_{0}\quad\text{in }\Omega .
\label{e1.1.c}
\end{gather}
where $T>0$, $Q=\Omega \times ]0,T[$,
$\Sigma _{l}=\Gamma_{l}\times ]0,T[$ $(l=0,1)$, $\upsilon $ is the
outward unit normal to $\Gamma $, $\frac{\partial }{\partial \upsilon }$
denotes the normal derivative, $y_{t}$ and $z_{t}$ the time derivative,
$\nabla $ and $\Delta $ are the Gradient and the Laplacian in the space
variables. Operators $F_{l}$, $P_{l}$ $(l=1,2)$ and the
functions $g$ and $h$ are defined in \S 2.

Our goal in this paper is to obtain, under a suitable geometrical conditions
on $(\Omega ,\Gamma _{0},\Gamma _{1})$, the exponential
stability of the system \eqref{e1.1.a1}--\eqref{e1.1.c}: For any initial data
$(y_{0},z_{0})$ in some space, the energy $E$ (see \eqref{e2.8})
of the solution tends to zero exponentially as
$t\to +\infty $.

In the case of one equation and where $F=0$ and $g$ is linear of the form
$g(s)=(h.\upsilon )s$, $h$ represents a vector
field satisfying some geometrical conditions, a lot of work have been
(see for example \cite{m1,m2}). Recently, in \cite{l1} they have obtain
a result of stabilization in the space
$L^{2}(\Omega )$ of one equation where $F=0$ and
linear feedback in the form $iy$. In these situations the system
is dissipative (the usual energy is decreasing). Notice, that this
property (the decrease of energy) specially $E(T)\leq E(0)$ has
a crucial role in studying the stability of
the solution. So, the approach in \cite{k1} can not be applied to the
non dissipative system.

To the best of our best knowledge, nonlinear boundary stabilization for the
coupled system of two Schrodinger equations with lower order terms
has  not been
considered in the literature. In this case the energy is not decreasing;
we do not have any information about the sign of the derivative of
the energy (see \eqref{e4.1}). This requires a careful treatment.

Our paper is divided in 3 sections. In \S 2, we give notation and
assumptions. In \S 3, we give the result of existence of solution. In \S 4
we give some formulas witch are needed for proving our result.
In \S 5 we use the multiplier method to show, that the energy of
the solution of system \eqref{e1.1.a1}--\eqref{e1.1.c}
 satisfies the inequalities:
\begin{equation}
\begin{gathered}
\int_{0}^{T}E(t)\leq \lambda _{1}(E(T)+E(0))
+\lambda _{2}(E(0)-E(T)),\quad\text{for all }T>0 \\
E'(t)\leq \lambda _{3}E(t),\quad \text{for all }t>0\,,
\end{gathered} \label{e1.2}
\end{equation}
where $\lambda _{1}\leq \lambda _{2}$.
$E'$ is the time derivative of the energy. As in \cite{a1}, we have
\begin{equation}
E(t)\leq cE(0)e^{-\omega t},\quad\text{for all }t>0.  \label{e1.3}
\end{equation}

\section{Notation and Assumptions}

Let $x_{0}$ be a fixed point in of $\mathbb{R}^{n}$.
Set $h=h(x)=x-x_{0}$ and
$R=\sup \{|h(x)|:x\in \overline{\Omega }\}$.
Assume that for some constant $h_{0}>0$, we have
\begin{equation}
\Gamma _{0}=\left\{ x\in \Gamma :h\upsilon \leq 0\right\},\quad
\Gamma_{1}=\left\{ x\in \Gamma :h\upsilon \geq h_{0}\right\} .  \label{e2.1}
\end{equation}
Assume that $F_{1}$, $F_{2}$ are linear differential operators of order one
in the space variables with $L_{\infty }(\overline{Q})$-coefficients
and $P_{1},P_{2}$ are linear operators of zero order with
$L_{\infty }(\overline{Q})$-coefficients:
\begin{gather}
F_{1}=F_{1}(y,\nabla y)=v_{1}(x).\nabla y-q_{1}(x)y,  \label{e2.2.a}
\\
F_{2}=F_{2}(z,\nabla z)=v_{2}(x).\nabla z-q_{2}(x)z, \label{e2.2.b}
\\
P_{1}=P_{1}(z)=\alpha _{1}(x)z,  \label{e2.2.c}
\\
P_{2}=P_{2}(y)=\alpha _{2}(x)y.  \label{e2.2.d}
\end{gather}
Where in \eqref{e2.2.c}--\eqref{e2.2.d}, $\alpha _{1}$ and $\alpha _{2}$ are
a bounded complex valued functions such that
\begin{gather}
\mathop{\rm Im}(\overline{\alpha _{1}\alpha _{2}})\geq 0\,,
\label{e2.3.a} \\
|\alpha _{1}|=|\alpha _{1}(x)|_{L^{\infty }(
\Omega )}\leq \frac{\min (1,\frac{n}{2})}{(
n+R)C+R}  \label{e2.3.b}
\end{gather}
where $q_{l}$ $(l=1,2)$ are a positives functions satisfying
\begin{equation}
\max_{l=1,2}(|q_{l}|)\leq \frac{\min
(1,\frac{n}{2})-|\alpha _{1}|((
n+R)C+R)}{2R(C+1)}.  \label{e2.4}
\end{equation}
Where $C$ is the positive constant satisfying
\begin{equation*}
\int_{\Omega }|y|^{2}\leq C\int_{\Omega }|\nabla y|^{2}.
\end{equation*}
We assume that for all $l=1,2$
$\ v_{l}=v_{l}(x)$ is a complex
$n$-vector field with $|v_{l}|\in L_{\infty }(\overline{Q})$,
the Hessian matrix $V_{l}$ of $v_{l}$ satisfies
$|V_{l}|\in L_{\infty }(\overline{Q})$ and the following
properties for $\mathop{\rm Im}v_{l}$ are achieved:
\begin{gather}
|\mathop{\rm Im}v_{l}.\mu |\leq
\mathop{\rm Im}v_{l}.\upsilon \quad\text{on }\Gamma _{1},  \label{e2.5.a} \\
\mathop{\rm Im}v_{l}.\upsilon = 0\quad\text{on }\Gamma _{0},  \label{e2.5.b}
\end{gather}
where $\mu $ represent the tangential unit vector on $\Gamma $.
Put
\begin{equation*}
\beta =\max_{l=1,2}\big\{|\alpha _{2}-\overline{\alpha _{1}
}|,|v_{l}|,|\nabla (\alpha _{2}-\overline{
\alpha _{1}})|,|V_{l}|,|\mathop{\rm div}(\mathop{\rm Im}
v_{l})|\big\}.
\end{equation*}
Assume that $g$ and $h$ are a complex valued functions such that there
exists a constants $g_{\ast },h_{\ast },g^{\ast },h^{\ast }>0$ such that for
all $s\in \mathbb{C}$ we have
\begin{gather}
g_{\ast }|s|\leq |g(s)|\leq g^{\ast }|
s|\quad\text{and}\quad g(s)\overline{s}\geq 0,  \label{e2.6}\\
h_{\ast }|s|\leq |h(s)|\leq h^{\ast }|
s|\quad\text{and}\quad h(s)\overline{s}\geq 0.  \label{e2.7}
\end{gather}
Put
\begin{equation*}
H_{\Gamma _{0}}^{1}(\Omega )=\big\{ u\in H^{1}(\Omega ):u=0\quad\text{on }
\Gamma _{0}\big\} .
\end{equation*}
We define the energy $E$ of \eqref{e1.1.a1}--\eqref{e1.1.c} by
\begin{equation}
E(t)=\frac 12\Big(\int_{\Omega }|\nabla y|
^{2}+\int_{\Omega }|\nabla z|^{2}+\int_{\Omega
}q_{1}|y|^{2}+\int_{\Omega }q_{2}|z|
^{2}dx\Big)-\mathop{\rm Re}\int_{\Omega }\overline{\alpha _{1}}y
\overline{z}.  \label{e2.8}
\end{equation}

\begin{remark} \label{rmk1} \rm
We note that, because \eqref{e2.3.b}, we have
\begin{equation*}
\zeta _{1}\Big(\int_{\Omega }|\nabla y|
^{2}+\int_{\Omega }|\nabla z|^{2}\Big)\leq \|
(y,z)\| _{H_{\Gamma _{0}}^{1}(\Omega )}^{2}\leq \zeta
_{2}\Big(\int_{\Omega }|\nabla y|
^{2}+\int_{\Omega }|\nabla z|^{2}\Big),
\end{equation*}
where
\begin{equation*}
\left\| (y,z)\right\| _{H_{\Gamma _{0}}^{1}(\Omega
)}^{2}=\int_{\Omega }|\nabla y|^{2}+\int_{\Omega
}|\nabla z|^{2}+\int_{\Omega }q_{1}|y|
^{2}+\int_{\Omega }q_{2}|z|^{2}dx-2\mathop{\rm Re}
\int_{\Omega }\alpha _{1}y\overline{z}.
\end{equation*}
\end{remark}

\section{Existence of solutions}

We define an operator
\begin{equation*}
A(u_{1},u_{2})=(i\Delta u_{1}+iF_{1}(u_{1},\nabla
u_{1})+iP_{1}(u_{2}),i\Delta u_{2}+iF_{2}(u_{2},\nabla
u_{2})+iP_{2}(u_{1}))
\end{equation*}
with domain
\begin{align*}
D(A)=\big\{&(u_{1},u_{2})\in V\times V:
\text{for }l=1,2\;\Delta u_{l}\in H_{\Gamma _{0}}^{1}(\Omega ), \\
&[\frac{\partial u_{1}}{\partial \upsilon }+g(i\Delta
u_{1}+iF_{1}+iP_{1})]_{\Gamma _{1}}=0, \\
&[\frac{\partial u_{2}}{\partial \upsilon }+h(
i\Delta u_{2}+iF_{2}+iP_{2})]_{\Gamma _{1}}=0 \big\}
\end{align*}
Where
$V=H^{2}(\Omega )\cap H_{\Gamma _{0}}^{1}(\Omega )$.
 Let $U=(u_{1},u_{2})$, we may rewrite system
\eqref{e1.1.a1}--\eqref{e1.1.c} as
\begin{equation}
\begin{gathered}
U_{t}=AU \\
U(0)=U_{0}.
\end{gathered}  \label{e3.1}
\end{equation}
then the solvability of \eqref{e1.1.a1}--\eqref{e1.1.c} is
equivalent to the one of \eqref{e3.1}.

We can prove that, if $g$ and $h$ are globally lipschitz, there exists a
positive constant $k$ such that the operator $A-kI$ is maximal dissipative
on $(H_{\Gamma _{0}}^{1}(\Omega ))^{2}$. Therefore, we have the following
theorem \cite{k1,l2}.

\begin{theorem} \label{thm2}
1. For every $(y_{0},z_{0})\in D(A)$ the system
\eqref{e1.1.a1}--\eqref{e1.1.c} has a unique solution
$(y,z)\in L^{\infty}([0,+\infty );D(A))\cap W^{1,\infty }([0,+\infty )
;(H_{\Gamma _{0}}^{1}(\Omega ))^{2})$.

\noindent 2. For every $(y_{0},z_{0})\in (H_{\Gamma
_{0}}^{1}(\Omega ))^{2}$ the system \eqref{e1.1.a1}--\eqref{e1.1.c}
has a unique solution
$(y,z)\in C([0,+\infty );(H_{\Gamma
_{0}}^{1}(\Omega ))^{2})$.
\end{theorem}

\section{Preliminary results}

We begin by establishing a formula concerning the derivative of the energy
of the system \eqref{e1.1.a1}--\eqref{e1.1.c} and  two estimates
concerning
$\int_{\Sigma _{1}}(g(y_{t})\overline{y_{t}}+h(
z_{t})\overline{z_{t}})$ and $\mathop{\rm Re}\int_{\Omega }(
v_{1}.\nabla y\overline{y_{t}}+(v_{2}.\nabla z+(\alpha _{2}-
\overline{\alpha _{1}})y)\overline{z_{t}})$.

\begin{lemma} \label{lem3}
For each solution $y$ of system \eqref{e1.1.a1}--\eqref{e1.1.c} we have
\begin{equation}
E'(t)=\mathop{\rm Re}\int_{\Omega }(v_{1}.\nabla y\overline{y_{t}
}+(v_{2}.\nabla z+(\alpha _{2}-\overline{\alpha _{1}})
y)\overline{z_{t}})-\int_{\Sigma _{1}}(g(
y_{t})\overline{y_{t}}+h(z_{t})\overline{z_{t}}).
\label{e4.1}
\end{equation}
\end{lemma}

\begin{proof}
We multiply both side \eqref{e1.1.a1} by $\overline{y}_{t}$ and use
Green's formula to obtain
\begin{equation*}
i\int_{\Omega }|y_{t}|^{2}+\int_{\Gamma }\frac{\partial y}{
\partial \upsilon }\overline{y_{t}}-\int_{\Omega }\nabla y.\nabla \overline{
y_{t}}+\int_{\Omega }(F_{1}+P_{1})\overline{y_{t}}=0.
\end{equation*}
Taking the real part, we obtain
\begin{equation*}
\mathop{\rm Re}\int_{\Omega }\nabla y.\nabla \overline{y_{t}}+\mathop{\rm Re}
\int_{\Omega }q_{1}y\overline{y_{t}}=\mathop{\rm Re}\int_{\Gamma }\frac{\partial y
}{\partial \upsilon }\overline{y_{t}}+\mathop{\rm Re}\int_{\Omega }(
v_{1}.\nabla y+\alpha _{1}z)\overline{y_{t}}.
\end{equation*}
But, by \eqref{e1.1.b1}--\eqref{e1.1.b2} and \eqref{e2.6} we find that
\begin{equation*}
\mathop{\rm Re}\int_{\Gamma }\frac{\partial y}{\partial \upsilon }\overline{y_{t}}
=-\mathop{\rm Re}\int_{\Gamma _{1}}g(y_{t})\overline{y_{t}}
=-\int_{\Gamma _{1}}g(y_{t})\overline{y_{t}}
\end{equation*}
and
\begin{equation}
\mathop{\rm Re}\int_{\Omega }\nabla y.\nabla \overline{y_{t}}+\mathop{\rm Re}
\int_{\Omega }q_{1}y\overline{y_{t}}=-\int_{\Gamma _{1}}g(y_{t})
\overline{y_{t}}+\mathop{\rm Re}\int_{\Omega }(v_{1}.\nabla y+\alpha
_{1}z)\overline{y_{t}}.  \label{e4.2.a}
\end{equation}
We obtain similar identity for $z$:
\begin{equation}
\mathop{\rm Re}\int_{\Omega }\nabla z.\nabla \overline{z_{t}}+\mathop{\rm Re}
\int_{\Omega }q_{2}z\overline{z_{t}}=-\int_{\Gamma _{1}}h(z_{t})
\overline{z_{t}}+\mathop{\rm Re}\int_{\Omega }(v_{2}.\nabla z+\alpha
_{2}y)\overline{z_{t}}. \label{e4.2.b}
\end{equation}
However,
\begin{equation*}
\mathop{\rm Re}\int_{\Omega }(\nabla y.\nabla \overline{y_{t}}+\nabla
z.\nabla \overline{z_{t}}+q_{1}y\overline{y_{t}}+q_{2}z\overline{z_{t}}-
\overline{\alpha _{1}}(\overline{z}y_{t}+\overline{z_{t}}y)
)=E'(t)\,.
\end{equation*}
Then by \eqref{e4.2.a}--\eqref{e4.2.b}, we find \eqref{e4.1}.
\end{proof}

\begin{lemma} \label{lem4}
There exists a constant $\overline{C}>0$ such that we have for all $t>0$,
\begin{equation}
\begin{aligned}
&\mathop{\rm Re}\int_{\Omega }(v_{1}.\nabla y\overline{y_{t}}
+(v_{2}.\nabla z+(\alpha _{2}-\overline{\alpha _{1}})y)
\overline{z_{t}}) \\
&\leq  2\beta \max (g^{\ast },h^{\ast })\Big(\int_{\Gamma
_{1}}g(y_{t})\overline{y_{t}}+\int_{\Gamma _{1}}h(
z_{t})\overline{z_{t}}\Big) \\
&\quad +\frac{\beta \overline{C}}{2}\Big(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{1}|
y|^{2}+\int_{\Omega }q_{2}|z|^{2}\Big).
\end{aligned}\label{e4.3}
\end{equation}
Moreover, if
\begin{equation}
\beta <\frac{1}{4\max (h^{\ast },g^{\ast })}  \label{e4.4}
\end{equation}
then, for all $T>0$, we have
\begin{equation}
\begin{aligned}
&\int_{\Sigma _{1}}g(y_{t})\overline{y_{t}}+\int_{\Sigma
_{1}}h(z_{t})\overline{z_{t}}\\
& \leq 2(E(0)-E(T)) +\frac{\beta \overline{C}}{2}
\Big(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{1}|
y|^{2}+\int_{\Omega }q_{2}|z|^{2}\Big).
\end{aligned}\label{e4.5}
\end{equation}
\end{lemma}

\begin{proof}
We start with the proof of \eqref{e4.3}.
Using \eqref{e1.1.a1}, we have
\begin{equation*}
\mathop{\rm Re}\int_{\Omega }v_{1}.\nabla y\overline{y_{t}}=-\mathop{\rm Re}\int_{\Omega
}iv_{1}.\nabla y(\Delta \overline{y}+\overline{v_{1}.\nabla y}-q
\overline{y}+\overline{\alpha _{1}z})\,.
\end{equation*}
Then
\begin{equation*}
\mathop{\rm Re}\int_{\Omega }v_{1}.\nabla y\overline{y_{t}}
=\mathop{\rm Im}\int_{\Omega
}v_{1}.\nabla y\Delta \overline{y}-\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla yq_{1}
\overline{y}+\mathop{\rm Im}\int_{\Omega }v_{1}.
\nabla y\overline{\alpha _{1}z}.
\end{equation*}
First, we consider the term
$\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla y\Delta \overline{y}$.
Applying Green's formula, we obtain:
\begin{equation*}
\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla y\Delta \overline{y}=\mathop{\rm Im}
\int_{\Gamma }\frac{\partial \overline{y}}{\partial \upsilon }v_{1}.\nabla
y-\mathop{\rm Im}\int_{\Omega }\nabla \overline{y}.\nabla (v_{1}.\nabla
y).
\end{equation*}
Indeed, with $s=v_{1}$, we recall the identity
\begin{equation}
\nabla (s.\nabla u)\nabla \overline{u}=S\nabla u.\nabla
\overline{u}+\frac{1}{2}s.\nabla (|\nabla u|^{2}),
\label{e*}
\end{equation}
where $S$ denotes the  Hessian of $s$.
Then
\[
\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla y\Delta \overline{y}
=\mathop{\rm Im} \int_{\Gamma _{1}}\frac{\partial \overline{y}}
{\partial \upsilon } v_{1}.\nabla y-\mathop{\rm Im}
\int_{\Omega }V_{1}\nabla y.\nabla \overline{y}
-\frac{1}{2}\int_{\Omega }\mathop{\rm Im}v_{1}\nabla (|\nabla y|)^{2}.
\]
We invoke the divergence identity:
\begin{equation}
\int_{\Omega }s.\nabla \psi =\int_{\Gamma }\psi s.\upsilon -\int_{\Omega
}\psi divs,  \label{e**}
\end{equation}
with $\psi =|\nabla y|^{2}$ and $s=v_{1}$. Then
\begin{align*}
&\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla y\Delta \overline{y} \\
&=\mathop{\rm Im} \int_{\Gamma _{1}}
 \frac{\partial \overline{y}}{\partial \upsilon }
 v_{1}.\nabla y-\mathop{\rm Im}\int_{\Omega }V_{1}\nabla y.\nabla
  \overline{y}
-\frac{1}{2}\int_{\Gamma }\mathop{\rm Im}v_{1}.\upsilon |\nabla y|
^{2}+\frac{1}{2}\int_{\Omega }\mathop{\rm Im}(divv_{1})|\nabla y|^{2}.
\end{align*}
Since, $s=(s.\upsilon ).\upsilon +(s.\mu ).\mu$ for all vector $s$
on $\Gamma $,
\begin{equation*}
|\nabla y|^{2}=|\frac{\partial y}{\partial \upsilon }
|^{2}+|\frac{\partial y}{\partial \mu }|^{2}.
\end{equation*}
Moreover,
\begin{equation*}
 \frac{\partial y}{\partial \mu }\big|_{\Gamma _{0}}= \nabla
y.\mu \big|_{\Gamma _{0}}=0,
\quad\mbox{or}\quad
|\nabla y|=|\frac{\partial y}{\partial \upsilon }|
\quad\text{on }\Gamma _{0,}.
\end{equation*}
Then
\begin{equation}
\mathop{\rm Re}\int_{\Omega }v_{1}.\nabla y\overline{y_{t}}=R_{\Gamma
_{0}}+R_{\Gamma _{1}}+R_{\Omega },  \label{e4.6}
\end{equation}
where
\begin{align*}
R_{\Gamma _{0}} &= \mathop{\rm Im}\int_{\Gamma _{0}}v_{1}.\upsilon |\nabla
y|^{2}-\frac{1}{2}\int_{\Gamma _{0}}(\mathop{\rm Im}v_{1}.\upsilon
)|\nabla y|^{2} \\
&= \frac{1}{2}\int_{\Gamma _{0}}(\mathop{\rm Im}v_{1}.\upsilon )
|\nabla y|^{2}\leq 0\quad (\text{by \eqref{e2.5.b}})
\end{align*}
and
\begin{align*}
R_{\Gamma _{1}}
&= \int_{\Gamma _{1}}\mathop{\rm Im}v_{1}.\upsilon |\frac{
\partial y}{\partial \upsilon }|^{2}+\mathop{\rm Im}\int_{\Gamma
_{1}}v_{1}.\mu \frac{\partial y}{\partial \mu }\frac{\partial \overline{y}
 }{\partial \upsilon }
-\frac{1}{2}\int_{\Gamma _{1}}(\mathop{\rm Im}v_{1}.\upsilon )
|\nabla y|^{2}
\\
&= \frac{1}{2}\int_{\Gamma _{1}}\mathop{\rm Im}v_{1}.\upsilon |\frac{
\partial y}{\partial \upsilon }|^{2}+\mathop{\rm Im}\int_{\Gamma
_{1}}v_{1}.\mu \frac{\partial y}{\partial \mu }\frac{\partial \overline{y}
}{\partial \upsilon }
 -\frac{1}{2}\int_{\Gamma _{1}}\mathop{\rm Im}v_{1}.\upsilon |\frac{
\partial y}{\partial \mu }|^{2}\,.
\end{align*}
Then
\begin{align*}
R_{\Gamma _{1}} &\leq \frac{1}{2}\int_{\Gamma _{1}}\mathop{\rm Im}v_{1}.\upsilon
|\frac{\partial y}{\partial \upsilon }|^{2}+\frac{1}{2}
\int_{\Gamma _{1}}|\mathop{\rm Im}v_{1}.\mu ||\frac{\partial y
}{\partial \upsilon }|^{2} \\
&\quad +\frac{1}{2}\int_{\Gamma _{1}}(|\mathop{\rm Im}v_{1}.\mu |-
\mathop{\rm Im}\upsilon _{1}.\upsilon )|\frac{\partial y}{\partial
\mu }|^{2}.
\end{align*}
By \eqref{e2.5.a},
$R_{\Gamma _{1}}\leq \beta \int_{\Gamma _{1}}|\frac{\partial y}{
\partial \upsilon }|^{2}$.
Then, by \eqref{e1.1.b2} and \eqref{e2.6}, we obtain
\begin{equation*}
R_{\Gamma _{1}}\leq \beta g^{\ast }\int_{\Gamma _{1}}g(y_{t})
\overline{y_{t}}
\end{equation*}
and (see \eqref{e4.6})
\begin{align*}
R_{\Omega } &= -\mathop{\rm Im}\int_{\Omega }V_{1}\nabla y.\nabla \overline{y}+
\frac{1}{2}\int_{\Omega }\mathop{\rm Im}(divv_{1})|\nabla
y|^{2} \\
&\quad-\mathop{\rm Im}\int_{\Omega }v_{1}.\nabla yq_{1}\overline{y}+\mathop{\rm Im}
\int_{\Omega }v_{1}.\nabla y\overline{\alpha _{1}z}.
\end{align*}
Then
\begin{equation*}
R_{\Omega }\leq \frac{5}{2}\beta \int_{\Omega }|\nabla y|
^{2}+\beta \frac{|q_{1}|}{2}\int_{\Omega }q_{1}|y|
^{2}+\beta |\alpha _{1}|^{2}\frac{C}{2}\int_{\Omega }|
\nabla z|^{2}.
\end{equation*}
We insert $R_{\Gamma _{l}}$ $(l=0,1)$ and $R_{\Omega }$ in
\eqref{e4.6} to obtain
\begin{equation}
\begin{aligned}
&\mathop{\rm Re}\int_{\Omega }v_{1}.\nabla y\overline{y_{t}} \\
&\leq \beta g^{\ast}\int_{\Gamma _{1}}g(y_{t})\overline{y_{t}}
+\frac{5}{2}\beta\int_{\Omega }|\nabla y|^{2}
 +\beta \frac{|q_{1}|}{2}\int_{\Omega }q_{1}|y|
^{2}+\beta |\alpha _{1}|^{2}\frac{C}{2}\int_{\Omega }|
\nabla z|^{2}.
\end{aligned}\label{e4.7.a}
\end{equation}
Using the same argument and by \eqref{e2.3.a},  we obtain
\begin{equation}
\begin{aligned}
&\mathop{\rm Re}\int_{\Omega }(v_{2}.\nabla z+(\alpha _{2}-\overline{
\alpha _{1}})y)\overline{z_{t}} \\
&\leq 2\beta h^{\ast}\int_{\Gamma _{1}}h(z_{t})\overline{z_{t}}
+4\beta \int_{\Omega }|\nabla z|^{2}
+|q_{2}|\beta \int_{\Omega }q_{2}|z|^{2}\\
&\quad +\frac{\beta }{2}\big((|\alpha _{2}|^{2}+2)
C+C'+1\big)\int_{\Omega }|\nabla y|^{2},
\end{aligned}  \label{e4.7.b}
\end{equation}
where we have put $\beta =\max (\beta ,\beta ^{2})$.
However, \eqref{e4.7.a}--\eqref{e4.7.b} gives
\begin{align*}
&\mathop{\rm Re}\int_{\Omega }(v_{1}.\nabla y\overline{y_{t}}+(
v_{2}.\nabla z+(\alpha _{2}-\overline{\alpha _{1}})y)
\overline{z_{t}})\\
&\leq 2\beta \max (g^{\ast },h^{\ast })(\int_{\Gamma
_{1}}g(y_{t})\overline{y_{t}}+\int_{\Gamma _{1}}h(
z_{t})\overline{z_{t}})\\
&\quad +\frac{\beta \overline{C}}{2}(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{2}|
z|^{2}+\int_{\Omega }q_{1}|y|^{2}),
\end{align*}
where $\overline{C}=\max \big\{|\alpha _{1}|^{2}C+8,(
|\alpha _{2}|^{2}+2)C+C'+6,|
q_{1}|,2|q_{2}|\big\}$.
Using \eqref{e4.1}, we have
\begin{align*}
&\int_{\Sigma _{1}}g(y_{t})\overline{y_{t}}+\int_{\Sigma
_{1}}h(z_{t})\overline{z_{t}} \\
&= E(0)-E(T)
 +\mathop{\rm Re}\int_{\Omega }(v_{1}.\nabla y\overline{y_{t}}+(
v_{2}.\nabla z+(\alpha _{2}-\overline{\alpha _{1}})y)
\overline{z_{t}}).
\end{align*}
 By \eqref{e4.3}, we find
\begin{align*}
&\int_{\Sigma _{1}}g(y_{t})\overline{y_{t}}+\int_{\Sigma
_{1}}h(z_{t})\overline{z_{t}} \\
&\leq E(0)-E(T)+2\beta \max
(g^{\ast },h^{\ast })\Big(\int_{\Gamma _{1}}g(
y_{t})\overline{y_{t}}+\int_{\Gamma _{1}}h(z_{t})
\overline{z_{t}}\Big)\\
&\quad+\frac{\beta \overline{C}}{2}\Big(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{2}|
z|^{2}+\int_{\Omega }q_{1}|y|^{2}\Big).
\end{align*}
Finally, by \eqref{e4.4}, we obtain \eqref{e4.5}.
\end{proof}

\section{Main result}

We begin by stating an identity given by the multiplier method.

\begin{lemma} \label{lem5}
Every solution $y$ of \eqref{e1.1.a1}--\eqref{e1.1.c} satisfies
\begin{equation}
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla
z|^{2}=I_{\Omega }+I_{\Sigma _{0}}+I_{\Sigma _{1}}+I_{Q},  \label{e5.1}
\end{equation}
where
\begin{gather*}
I_{\Omega }= -\frac{1}{2}\mathop{\rm Im}\int_{\Omega }(yh.\nabla
\overline{y}+zh.\nabla \overline{z})\big|_{0}^{T},
\\
I_{\Sigma _{0}}=\frac{1}{2}\int_{\Sigma _{0}}\Big(|\frac{\partial y
}{\partial \upsilon }|^{2}+|\frac{\partial z}{\partial
\upsilon }|^{2}\Big)h.\upsilon ,
\\
\begin{aligned}
I_{\Sigma _{1}}
&= \frac{1}{2}\mathop{\rm Im}\int_{\Sigma _{1}}h.\upsilon (y
\overline{y_{t}}+z\overline{z_{t}})+\frac{n}{2}\mathop{\rm Re}
\int_{\Sigma_{1}}\Big(\frac{\partial \overline{y}}{\partial \upsilon }y+\frac{
\partial \overline{z}}{\partial \upsilon }z\Big)\\
&\quad +\frac{1}{2}\int_{\Sigma _{1}}\Big(|\frac{\partial y}{\partial
\upsilon }|^{2}+|\frac{\partial z}{\partial \upsilon }|
^{2}\Big)h.\upsilon
+\mathop{\rm Re}\int_{\Sigma _{1}}h.\mu \Big(\frac{
\partial \overline{y}}{\partial \mu }\frac{\partial y}{\partial \upsilon }+
\frac{\partial \overline{z}}{\partial \mu }\frac{\partial z}{\partial
\upsilon }\Big)\\
&\quad -\frac{1}{2}\int_{\Sigma _{1}}h.\upsilon \Big(|\frac{\partial y}{
\partial \mu }|^{2}+|\frac{\partial z}{\partial \mu }|
^{2}\Big),
\end{aligned}
\\
\begin{aligned}
I_{Q} &= \mathop{\rm Re}\int_{Q}\big[(F_{1}+P_{1})h.\nabla
\overline{y}+(F_{2}+P_{2})h.\nabla \overline{z}\big]\\
&\quad +\frac{n}{2}\mathop{\rm Re}\int_{Q}[(F_{1}+P_{1})\overline{y}
+(F_{2}+P_{2})\overline{z}].
\end{aligned}
\end{gather*}
\end{lemma}

\begin{proof}
Multiply \eqref{e1.1.a1} by $h.\nabla \overline{y}$ and integrate over $Q$
to obtain
\begin{equation}
\begin{aligned}
0 &= \int_{Q}(iy_{t}+\Delta y+(F_{1}+P_{1}))h.\nabla \overline{y}\\
&=i\int_{Q}y_{t}h.\nabla \overline{y}
 +\int_{Q}\Delta yh.\nabla \overline{y}+\int_{Q}(F_{1}+P_{1})
h.\nabla \overline{y}.
\end{aligned} \label{e5.2}
\end{equation}
Taking the real part,
\begin{equation}
0=-\mathop{\rm Im}\int_{Q}y_{t}h.\nabla \overline{y}+\mathop{\rm Re}\int_{Q}\Delta
yh.\nabla \overline{y}+\mathop{\rm Re}\int_{Q}(F_{1}+P_{1})h.\nabla
\overline{y}.  \label{e5.3}
\end{equation}
Integrating by parts,
\begin{equation*}
\int_{Q}y_{t}h.\nabla \overline{y}=\int_{\Omega } yh.\nabla \overline{y
}\big|_{0}^{T}-\int_{Q}yh.\nabla \overline{y_{t}}\,.
\end{equation*}
Using the divergence identity \eqref{e**} with $\mathop{\rm div} h=n$, we
obtain
\begin{equation*}
\int_{Q}y_{t}h.\nabla \overline{y}=\int_{\Omega } yh.\nabla \overline{y
}\big|_{0}^{T}-\int_{\Sigma }y\overline{y_{t}}h.\nu +n\int_{Q}y\overline{
y_{t}}+\int_{Q}\overline{y_{t}}h.\nabla y.
\end{equation*}
Then
\begin{equation*}
2i\mathop{\rm Im}\int_{Q}y_{t}h.\nabla \overline{y}=\int_{\Omega }
 yh.\nabla \overline{y}\big|_{0}^{T}-\int_{\Sigma }y\overline{y_{t}}h.\nu +n\int_{Q}y
\overline{y_{t}},
\end{equation*}
so
\begin{equation}
2\mathop{\rm Im}\int_{Q}y_{t}h.\nabla \overline{y}
=\mathop{\rm Im}\int_{\Omega }
yh.\nabla \overline{y}\big|_{0}^{T}-\mathop{\rm Im}\int_{\Sigma }y\overline{
y_{t}}h.\nu +n\mathop{\rm Im}\int_{Q}y\overline{y_{t}}.  \label{e5.4}
\end{equation}
However,
\begin{align*}
\int_{Q}y\overline{y_{t}} &= i\int_{Q}y(-\Delta \overline{y}-\overline{
F_{1}+P_{1}})\\
&= -i\int_{\Sigma }\frac{\partial \overline{y}}{\partial \upsilon }
y+i\int_{Q}|\nabla y|^{2}-i\int_{Q}y\overline{F_{1}+P_{1}}.
\end{align*}
Then
\begin{equation*}
\mathop{\rm Im}\int_{Q}y\overline{y_{t}}=-\mathop{\rm Re}\int_{\Sigma }\frac{\partial
\overline{y}}{\partial \upsilon }y+\int_{Q}|\nabla y|^{2}-\mathop{\rm
Re}\int_{Q}y\overline{F_{1}+P_{1}}\,.
\end{equation*}
We substitute this expression in \eqref{e5.4} and we use \eqref{e1.1.b1} to
obtain
\begin{equation}
\begin{aligned}
-\mathop{\rm Im}\int_{Q}y_{t}h\nabla \overline{y}
 &= -\frac{1}{2}\mathop{\rm Im}
\int_{\Omega } yh\nabla \overline{y}\big|_{0}^{T}+\frac{1}{2}\mathop{\rm
Im}\int_{\Sigma _{1}}y\overline{y_{t}}h.\nu    \\
&\quad +\frac{n}{2}\mathop{\rm Re}\int_{\Sigma _{1}}
 \frac{\partial \overline{y}}{\partial \upsilon }y
 -\frac{n}{2}\int_{Q}|\nabla y|^{2}
 +\frac{n}{2}\mathop{\rm Re}\int_{Q}\overline{F_{1}+P_{1}}y.
\end{aligned}\label{e5.5}
\end{equation}
 Concerning the term $\int_{Q}\Delta yh\nabla \overline{y}$,
if we use Green's Formula, identity \eqref{e*}, we find
\begin{align*}
\int_{Q}\Delta yh\nabla \overline{y}
&= \int_{\Sigma }\frac{\partial y}{
\partial \upsilon }h\nabla \overline{y}-\int_{Q}\nabla y\nabla (
h\nabla \overline{y})\\
&= \int_{\Sigma }\frac{\partial y}{\partial \upsilon }h\nabla \overline{y}
-\int_{Q}|\nabla y|^{2}-\frac{1}{2}\int_{Q}h\nabla (
|\nabla y|^{2})\,.
\end{align*}
We use the divergence identity \eqref{e**} to find
\begin{equation}
\mathop{\rm Re}\int_{Q}\Delta yh\nabla \overline{y}=\mathop{\rm Re}\int_{\Sigma }\frac{
\partial y}{\partial \upsilon }h\nabla \overline{y}-\frac{1}{2}\int_{\Sigma
}|\nabla y|^{2}h.\upsilon +(\frac{n}{2}-1)
\int_{Q}|\nabla y|^{2}.  \label{e5.6}
\end{equation}
However, in $\Gamma$,
\begin{gather*}
|\nabla y|^{2}=|\frac{\partial y}{\partial \upsilon }
|^{2}+|\frac{\partial y}{\partial \mu }|^{2},
\\
\frac{\partial y}{\partial \upsilon }h\nabla \overline{y}=|\frac{
\partial y}{\partial \upsilon }|^{2}h.\upsilon +\frac{\partial y}{
\partial \upsilon }\frac{\partial \overline{y}}{\partial \mu }h.\mu
\end{gather*}
and in $\Gamma _{0}$,
$\frac{\partial y}{\partial \mu }=0$.
Then from \eqref{e5.6},
\begin{equation}
\begin{aligned}
\mathop{\rm Re}\int_{Q}\Delta yh\nabla \overline{y}
&= \frac{1}{2}\int_{\Sigma
_{0}}|\frac{\partial y}{\partial \upsilon }|^{2}h.\upsilon +
\frac{1}{2}\int_{\Sigma _{1}}|\frac{\partial y}{\partial \upsilon }
|^{2}h.\upsilon    \\
&\quad +\mathop{\rm Re}\int_{\Sigma _{1}}\frac{\partial y}{\partial \upsilon }\frac{
\partial \overline{y}}{\partial \mu }h.\mu -\frac{1}{2}\int_{\Sigma
_{1}}|\frac{\partial y}{\partial \mu }|^{2}h.\upsilon
+(\frac{n}{2}-1)\int_{Q}|\nabla y|^{2}.
\end{aligned} \label{e5.7}
\end{equation}
Finally, we insert \eqref{e5.5} and \eqref{e5.7} in \eqref{e5.3} to obtain
\begin{equation}
\begin{aligned}
\int_{Q}|\nabla y|^{2}
&= -\frac{1}{2}\mathop{\rm Im}\int_{\Omega
} yh\nabla \overline{y}\big|_{0}^{T}+\frac{1}{2}\int_{\Sigma
_{0}}|\frac{\partial y}{\partial \upsilon }|^{2}h.\upsilon +
\frac{1}{2}\mathop{\rm Im}\int_{\Sigma _{1}}y\overline{y_{t}}h.\nu
\\
&\quad +\frac{n}{2}\mathop{\rm Re}\int_{\Sigma _{1}}\frac{\partial \overline{y}}{
\partial \upsilon }y+\frac{1}{2}\int_{\Sigma _{1}}|\frac{\partial y}{
\partial \upsilon }|^{2}h.\upsilon +\mathop{\rm Re}\int_{\Sigma _{1}}
\frac{\partial y}{\partial \upsilon }\frac{\partial \overline{y}}{\partial
\mu }h.\mu    \\
&\quad -\frac{1}{2}\int_{\Sigma _{1}}|\frac{\partial y}{\partial \mu }
|^{2}h.\upsilon +\frac{n}{2}\mathop{\rm Re}\int_{Q}\overline{F_{1}+P_{1}}y+
\mathop{\rm Re}\int_{Q}\overline{F_{1}+P_{1}}h.\nabla \overline{y}.
\end{aligned}  \label{e5.8}
\end{equation}
We obtain a similar identity  for $z$. From such inequality and
 \eqref{e5.8}, we deduce \eqref{e5.1}.
\end{proof}

 Our main result is the following.

\begin{theorem} \label{thm6}
The solution of \eqref{e1.1.a1}--\eqref{e1.1.c} is exponentially stable.
\end{theorem}

\begin{proof}
 From \eqref{e4.1}, \eqref{e2.6}, \eqref{e2.7}, \eqref{e4.3},
 \eqref{e4.4} and Remark \ref{rmk1}, we have
\begin{equation*}
E'(t)\leq \lambda _{3}E(t),\quad\text{for all }t>0.
\end{equation*}
Now, we prove that there are two constants $\lambda _{1}$ and
$\lambda_{2}$, $(\lambda _{1}\leq \lambda _{2})$, such that
\begin{equation*}
\int_{0}^{T}E(t)\leq \lambda _{1}(E(T)+E(0))
+\lambda _{2}(E(0)-E(T)),\quad\text{for all }T>0.
\end{equation*}
We bound the terms: $I_{\Omega }$, $I_{\Sigma _{l}}(l=0,1)$
and $I_{Q}$ in \eqref{e5.1}.
For all $\varepsilon _{1}>0$,
\begin{equation*}
I_{\Omega }\leq R\big(\frac{C}{\varepsilon _{1}}+\varepsilon _{1}\big)
\frac{1}{\zeta _{1}}(E(T)+E(0)).
\end{equation*}
By Cauchy-Scwartz,
\begin{align*}
I_{Q} &\leq \frac{\beta }{2}(2R+\frac{n}{2}(C+1))
\Big(\int_{Q}|\nabla y|^{2}+\int_{Q}|
\nabla z|^{2}\Big)\\
&\quad+\frac{\max_{l=1,2}|q_{l}|}{2}
R(C+1)\Big(\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla z|^{2}\Big)\\
&\quad +\frac{n}{2}\mathop{\rm Re}\int_{Q}(\alpha _{1}z\overline{y}
+\alpha _{2}y\overline{z})
+\mathop{\rm Re}\int_{Q}(\alpha _{1}zh.\nabla \overline{y
}+\alpha _{2}yh.\nabla \overline{z})\\
&\quad -\frac{n}{2}\Big(\int_{Q}q_{1}|y|
^{2}+\int_{Q}q_{2}|z|^{2}\Big).
\end{align*}
However,
\begin{align*}
&\frac{n}{2}\mathop{\rm Re}\int_{Q}(\alpha _{1}z\overline{y}+\alpha _{2}y
\overline{z})\\
&= n\mathop{\rm Re}\int_{Q}\alpha _{1}z\overline{y}+\frac{n}{2
}\mathop{\rm Re}\int_{Q}(\alpha _{2}-\overline{\alpha _{1}})y
\overline{z} \\
&\leq n|\alpha _{1}|\frac{C}{2}\Big(\int_{Q}|
  \nabla y|^{2}+\int_{Q}|\nabla z|^{2}\Big)
+\frac{n}{2}\beta (C+1)\Big(\int_{Q}|\nabla
y|^{2}+\int_{Q}|\nabla z|^{2}\Big)
\end{align*}
and
\begin{align*}
&\mathop{\rm Re}\int_{Q}(\alpha _{1}zh.\nabla \overline{z}+\alpha
_{2}yh.\nabla \overline{y})\\
&= \mathop{\rm Re}\int_{Q}(\alpha
_{1}(zh.\nabla \overline{z}+\overline{y}h.\nabla y)+(
\alpha _{2}-\overline{\alpha _{1}})yh.\nabla \overline{y})\\
&\leq R|\alpha _{1}|\frac{(C+1)}{2}\Big(
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla z|^{2}\Big)
+\frac{\beta R(C+1)}{2}\int_{Q}|\nabla y|
^{2}\,.
\end{align*}
Therefore,
\begin{align*}
I_{Q} &\leq \frac{\beta }{2}\overline{\overline{C}}\Big(
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla z|^{2}\Big)+
\frac{\max_{l=1,2}(|q_{l}|)}{2}R(C+1)\Big(\int_{Q}|\nabla y|
^{2}+\int_{Q}|\nabla z|^{2}\Big)\\
&\quad +\frac{|\alpha _{1}|}{2}((n+R)C+R)
\Big(\int_{Q}|\nabla y|^{2}+\int_{Q}|
\nabla z|^{2}\Big)
  -\frac{n}{2}(\int_{Q}q_{1}|y|^{2}+\int_{Q}q_{2}|z|^{2}),
\end{align*}
where $\overline{\overline{C}}=2R+(\frac{3n}{2}+R)(C+1)$.
By \eqref{e2.4}, we find
\begin{align*}
I_{Q} &\leq \frac{\beta }{2}\overline{\overline{C}}\Big(
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla
z|^{2}\Big)\\
&\quad +\Big(\frac{\min (1,\frac{n}{2})}{2}-\frac{|\alpha
_{1}|}{2}((n+R)C+R)\Big)\Big(
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla
z|^{2}\Big)\\
&\quad +\frac{|\alpha _{1}|}{2}\big((n+R)C+R\big)
\Big(\int_{Q}|\nabla y|^{2}+\int_{Q}|
\nabla z|^{2}\Big)
  -\frac{n}{2}\Big(\int_{Q}q_{1}|y|
^{2}+\int_{Q}q_{2}|z|^{2}\Big)\,.
\end{align*}
So that
\begin{align*}
I_{Q} &\leq \Big(\frac{\min (1,\frac{n}{2})}{2}+\frac{\beta
}{2}\overline{\overline{C}}\Big)\Big(\int_{Q}|\nabla
y|^{2}+\int_{Q}|\nabla z|^{2}+\int_{\Omega
}q_{1}|y|^{2}+\int_{\Omega }q_{2}|z|^{2}\Big)\\
&\quad -\frac{n}{2}\Big(\int_{Q}q_{1}|y|
^{2}+\int_{Q}q_{2}|z|^{2}\Big).
\end{align*}
Concerning the term $I_{\Sigma _{l}}$, we have
$I_{\Sigma _{0}}\leq 0$.
Put
\begin{equation*}
I_{\Sigma _{1}}=I_{\Sigma _{1}}(y)+I_{\Sigma _{1}}(z),
\end{equation*}
where
\begin{align*}
I_{\Sigma _{1}}(y)
&= \frac{1}{2}\mathop{\rm Im}\int_{\Sigma
_{1}}h.\upsilon y\overline{y_{t}}+\frac{n}{2}\mathop{\rm Re}\int_{\Sigma _{1}}
\frac{\partial \overline{y}}{\partial \upsilon }y+\frac{1}{2}\int_{\Sigma
_{1}}|\frac{\partial y}{\partial \upsilon }|^{2}h.\upsilon  \\
&\quad +\mathop{\rm Re}\int_{\Sigma _{1}}h.\mu \frac{\partial \overline{y}}{\partial
\mu }\frac{\partial y}{\partial \upsilon }-\frac{1}{2}\int_{\Sigma
_{1}}h.\upsilon |\frac{\partial y}{\partial \mu }|^{2}
\end{align*}
and
\begin{align*}
I_{\Sigma _{1}}(z)
&= \frac{1}{2}\mathop{\rm Im}\int_{\Sigma
_{1}}h.\upsilon z\overline{z_{t}}+\frac{n}{2}\mathop{\rm Re}\int_{\Sigma _{1}}
\frac{\partial \overline{z}}{\partial \upsilon }z+\frac{1}{2}\int_{\Sigma
_{1}}|\frac{\partial z}{\partial \upsilon }|^{2}h.\upsilon  \\
&\quad +\mathop{\rm Re}\int_{\Sigma _{1}}h.\mu \frac{\partial \overline{z}}{\partial
\mu }\frac{\partial z}{\partial \upsilon }-\frac{1}{2}\int_{\Sigma
_{1}}h.\upsilon (|\frac{\partial y}{\partial \mu }|
^{2}+|\frac{\partial z}{\partial \mu }|^{2})\,.
\end{align*}
We start with $I_{\Sigma _{1}}(y)$.
Using Cauchy-Scwartz and \eqref{e2.1},  for
$\varepsilon_{2}>0$, we obtain
\begin{align*}
I_{\Sigma _{1}}(y)
&\leq RC'\varepsilon
_{2}\int_{Q}|\nabla y|^{2}+R\frac{1}{\varepsilon _{2}}
\int_{\Sigma _{1}}|y_{t}|^{2}+n\varepsilon _{2}C'\int_{Q}|\nabla y|^{2}+n\frac{1}{\varepsilon _{2}}
\int_{\Sigma _{1}}|\frac{\partial y}{\partial \upsilon }|^{2}
\\
&\quad +\frac{R}{2}\int_{\Sigma _{1}}|\frac{\partial y}{\partial \upsilon }
|^{2}+\frac{R}{2}\Big(\frac{1}{\tau }\int_{\Sigma _{1}}|
\frac{\partial y}{\partial \upsilon }|^{2}+\tau \int_{\Sigma
_{1}}|\frac{\partial y}{\partial \mu }|^{2}\Big)
-h_{0}\int_{\Sigma _{1}}|\frac{\partial y}{\partial \mu }|^{2}.
\end{align*}
Choosing $\tau =\frac{2h_{0}}{R}$ and using \eqref{e1.1.b2},
\eqref{e2.6} we obtain
\begin{equation*}
I_{\Sigma _{1}}(y)\leq C'\varepsilon _{2}(
R+n)\int_{Q}|\nabla y|^{2}+\Big(\frac{R}{g_{\ast
}\varepsilon _{2}}+\big(\frac{n}{\varepsilon _{2}}+R+\frac{R^{2}}{4h_{0}}
\big)g^{\ast }\Big)\int_{\Sigma _{1}}g(y_{t})\overline{y_{t}},
\end{equation*}
similarly for $I_{\Sigma _{1}}(z)$,
\begin{equation*}
I_{\Sigma _{1}}(z)\leq C'\varepsilon _{2}(
R+n)\int_{Q}|\nabla z|^{2}+\Big(\frac{R}{h_{\ast
}\varepsilon _{2}}+\big(\frac{n}{\varepsilon _{2}}+R+\frac{R^{2}}{4h_{0}}
\big)h^{\ast }\Big)\int_{\Sigma _{1}}h(z_{t})\overline{
z_{t}}.
\end{equation*}
Therefore,
\begin{align*}
I_{\Sigma } &\leq  C'\varepsilon _{2}(R+n)\Big(
\int_{Q}|\nabla y|^{2}+\int_{Q}|\nabla z|
^{2}\Big)\\
&\quad+\Big(\frac{R}{\min (g_{\ast },h_{\ast })\varepsilon _{2}}
+\big(\frac{n}{\varepsilon _{2}}+R+\frac{R^{2}}{4h_{0}}\big)
\max (g^{\ast },h^{\ast })\Big)\int_{\Sigma _{1}}(g(
y_{t})\overline{y_{t}}+h(z_{t})\overline{z_{t}})\,.
\end{align*}
Using \eqref{e4.5} we find
\begin{align*}
I_{\Sigma } &\leq \lambda_2 (E(0)-E(T))\\
&\quad +\big(C'\varepsilon _{2}(R+n)+\beta \overline{C}
\lambda_2 \big)
\Big(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{1}|
y|^{2}+\int_{\Omega }q_{2}|z|^{2}\Big),
\end{align*}
where
\begin{equation*}
\lambda_2 =2\Big(\frac{R}{\min (g_{\ast },h_{\ast })
\varepsilon _{2}}+\big(\frac{n}{\varepsilon _{2}}+R+\frac{R^{2}}{4h_{0}}
\big)\max (g^{\ast },h^{\ast })\Big).
\end{equation*}
We insert this in \eqref{e5.1} to obtain
\begin{align*}
&\frac{1}{2}\Big(\min (1,\frac{n}{2})-\beta (\overline{
C}+\lambda_2 \overline{\overline{C}})-2C'\varepsilon
_{2}(R+n)\Big)\\
&\times \Big(\int_{\Omega }|\nabla z|
^{2}+\int_{\Omega }|\nabla y|^{2}+\int_{\Omega }q_{2}|
z|^{2}+\int_{\Omega }q_{1}|y|^{2}\Big)\\
&\leq \lambda_1\big(E(T)+E(0)\big))+\lambda_2 \big(E(0)-E(T)\big),
\end{align*}
where, we have put
\begin{equation*}
\lambda_1=\frac{R}{\zeta _{1}}(\frac{C}{\varepsilon _{1}}
+\varepsilon _{1}).
\end{equation*}
We choose
\begin{equation*}
\beta \leq \frac{\min (1,\frac{n}{2})}{\overline{C}+\lambda_2 
\overline{\overline{C}}}
\end{equation*}
and
\begin{equation*}
\varepsilon _{2}<\min \Big(\frac{\min (1,\frac{n}{2})-\beta
(\overline{C}+\lambda_2 \overline{\overline{C}})}{2C'(R+n)},\sqrt{\frac{2\zeta _{1}^{2}n\max (g^{\ast
},h^{\ast })}{RC\min (g_{\ast },h_{\ast })}}\Big).
\end{equation*}
On the other hand, if we choose $\varepsilon _{1}$ such that
\begin{equation*}
\frac{\varepsilon _{2}CR}{\zeta _{1}\max (g^{\ast },h^{\ast })}
<\varepsilon _{1}<\frac{2\zeta _{1}}{\min (g_{\ast },h_{\ast })
\varepsilon _{2}},
\end{equation*}
then
$\lambda_1\leq \lambda_2$.
\end{proof}


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\end{document}