\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 257, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2016/257\hfil 
 Exact controllability of the Euler-Bernoulli plate]
{Exact controllability of the Euler-Bernoulli plate with variable
coefficients and simply supported boundary condition}

\author[F. Yang \hfil EJDE-2016/257\hfilneg]
{Fengyan Yang}

\address{Fengyan Yang \newline
Key Laboratory of Systems and Control,
Institute of Systems Science,
Academy of Mathematics and Systems Science,
Chinese Academy of Sciences, Beijing 100190, China}
\email{yangfengyan12@mails.ucas.ac.cn}

\thanks{Submitted August 2, 2016. Published September 22, 2016.}
\subjclass[2010]{93B05, 93B27, 93C20, 35G16}
\keywords{Exact controllability; Euler-Bernoulli plate; variable coefficients;
\hfill\break\indent  Riemannian geometry; multiplier method}

\begin{abstract}
 This article studies the exact controllability of an Euler-Bernoulli
 plate equation with variable coefficients, subject to the simply
 supported boundary condition. By the Riemannian geometry approach,
 the duality method, the multiplier technique, and the compactness-uniqueness
 argument, we establish the corresponding observability inequality and
 obtain the exact controllability results.
\end{abstract}

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

\section{Introduction}

Let $A(x)=(a_{ij}(x))$ be a symmetric, positive matrix for each $x\in\mathbb{R}^n$,
where $a_{ij}(x)$ are $C^\infty$ functions in $\mathbb{R}^n$ , such that
 $$
\sum^n_{i,j=1}a_{ij}(x)\xi_i\xi_j>0,\quad \forall x\in\mathbb{R}^n,
\; 0\neq\xi=(\xi_1,\ldots,\xi_n)^T\in\mathbb{R}^n.
$$
We introduce
$$
g=A^{-1}(x)\quad \text{for }x\in\mathbb{R}^n,
$$
as a Riemannian metric on $\mathbb{R}^n$ and consider the couple
$(\mathbb{R}^n, g)$ as a Riemannian manifold. We denote by
$g=\langle\cdot,\cdot\rangle_g$ the inner product. Then
$$
\langle X,Y\rangle_g=\langle A^{-1}(x)X,Y\rangle\quad \text{for }
X,Y\in \mathbb{R}^n_x , x\in \mathbb{R}^n,
$$
where $\langle\cdot,\cdot\rangle $ is the Euclidean product of $\mathbb{R}^n$.

Let $\Omega\subset \mathbb{R}^n $ be an open, bounded set with a sufficient
smooth boundary $\Gamma=\overline{\Gamma_0\cup \Gamma_1}$ and
$\Gamma_0\cap \Gamma_1=\emptyset$, where $\Gamma_1$ is nonempty.
We consider the following Euler-Bernoulli plate model
\begin{equation}\label{1}
 \begin{gathered}
 u_{tt}+\mathscr{A}^2u=0\quad \text{in } Q=(0,T)\times \Omega,\\
 u=0\quad \text{on }\Sigma_0=(0,T)\times \Gamma_0, \\
 u=\varphi \quad \text{on }\Sigma_1=(0,T)\times \Gamma_1, \\
 \mathscr{A}u+a(x)Bu=0\quad \text{on }\Sigma_0, \\
 \mathscr{A}u+a(x)Bu=\psi \quad \text{on }\Sigma_1, \\
 u(0)=u_0, u_t(0)=u_1\quad \text{on }\Omega.
 \end{gathered}
\end{equation}
with two controls $\varphi$ and $\psi$, where $u_{tt}$ stands for
$\partial^2u/\partial t^2$,
\[
\mathscr{A}u =\sum^n_{i,j=1}\frac{\partial}{\partial x_i}\Big(a_{ij}(x)
\frac{\partial u}{\partial x_j}\Big)
\]
 and $B$ is a boundary operator, defined by
$$
 Bu=-\sum^n_{i=2}e_i\langle e_i,\nabla_{\Gamma_g}u\rangle_g+ku_{\nu_\mathscr{A}}.
$$
Here $\nu $ is the outside normal along $\Gamma $, $\nu_\mathscr{A} = A(x)\nu$,
and $u_{\nu_\mathscr{A}}=\langle \nabla_g u,\nu\rangle
=\langle A(x)\nabla u,\nu\rangle $. For $2\leq i\leq n$, $e_i $ is the
tangential vector fields on $\Gamma$ such that
$e_1=\nu_\mathscr{A}/{|\nu_\mathscr{A}|_g}, e_2,\ldots,e_n$ form a unit
orthogonal basis of $(\mathbb{R}^n_x ,g(x))$ for each $x\in\Gamma$,
$\nabla_{\Gamma_g}$ is the gradient of Riemannian manifold $(\Gamma,g)$.
 $k$ and $a(x)$ are bounded positive functions on $\Gamma$ and $\Omega$
respectively, which are related to the material. The boundary condition
we consider here is known as the simply supported boundary condition of
the plate (see \cite{Eller 2015,Lagnese 1988}), which arises from the
physical models and includes moments of inertia realistically present
in the system.

In the case of constant coefficients where $ A(x)$ is the unit matrix
and $n=2$, exact controllability results of problem \eqref{1}
have been obtained by Horn \cite{Horn 1992}. The objective of this paper
is to generalize the exact controllability results to the case where $A(x)$
is a non-constant, symmetric, positive n-order matrix and represents some
property of the materials, for example, the mass of the plate is not uniformly
distributed with respect to spatial position. The problem is of practical
and theoretical importance. From the physical point of view,
the variable-coefficient model is more realistic. Meanwhile, this together
with the simply supported boundary condition also introduces additional
non-trivial complications for the mathematical analysis.

The high-dimensional Euler-Bernoulli equations ($n\geq2$), as a kind
of classical partial differential equation, are used to describe the
vibration of elastic thin plates. Stimulated by the extensive applications in
the architectural structures, automobile and aerospace industries, etc.
(see \cite{Szilard 2004,Ventsel 2001}), there have been a great amount
of research on the control problems of Euler Bernoulli plates.
 We shall only cite the literature closely related to this paper,
the exact controllability of the Euler Bernoulli plates with different
choices of controls active in the varying boundary conditions.
For the constant coefficient case, we refer the reader to
\cite{Horn 1992, Komornik 1994, Lagnese 1988, 
Lasiecka and Triggiani 1989,Lasiecka and Triggiani 1990,
Lions 1988, Zuazua 1987}, and the references therein.
Particularly, in \cite{Lions 1988}, Lions considered the exact controllability
of the Euler-Bernoulli model with one control acting through Neumann boundary
condition. Later, Lasiecka and Triggiani \cite{Lasiecka and Triggiani 1989}
studied the situation where control acts only on the Dirichlet boundary condition,
in which they also managed to get rid of some geometrical conditions by
further adding a Neumann control. And in \cite{Lasiecka and Triggiani 1990},
they discussed the exact controllability problem with boundary controls for
displacement $u$ and moment $\Delta u$, which act in the Dirichlet boundary
conditions. Horn \cite{Horn 1992} derived the exact controllability of
the Euler-Bernoulli plate with a simply supported boundary condition only
via bending moments on the space of optimal regularity.
For the variable coefficient case, Yao \cite{Yao 2000} used the Riemannian
geometry approach to give checkable conditions for the exact controllability
of two Euler-Bernoulli models with clamped and hinged boundary conditions
respectively, which has been extended by many others like
\cite{Chai 2009, Guo 2007, Guo 2006, Li 2016,Li 2014}.
In particular, Guo and Zhang \cite{Guo 2007} showed that the exact
controllability of an Euler-Bernoulli plate with variable coefficients
and partial boundary Neumann control is equivalent to the exponential
stability of its closed-loop system under proportional output feedback.

The Riemannian geometry is a useful tool for the controllability of variable
-- coefficient systems mainly due to its two virtues: The Bochner technique
can be used to simplify computation to obtain the multiplier identities,
and the curvature theory provides the global information on the existence
of an escape vector field which guarantees the exact controllability.
 Given this, we shall use the Riemannian geometry approach to study our problem.

Since the dynamics of system \eqref{1} are time-reversible and it is well
known that exact controllability is equivalent to null controllability in that case,
we attempt to prove the following property: Given any
$(u_0,u_1)\in H^1_0(\Omega)\times H^{-1}(\Omega)$, there exist some $T>0$
and controls $(\varphi,\psi)\in H^1_0(0,T;L^2(\Gamma_1))\times L^2(\Sigma_1)$
such that the corresponding solution of problem \eqref{1} satisfies
$$
 u(T)\equiv u_t(T)\equiv0.
$$

\begin{remark} \label{rmk1.1} \rm
The above corresponding regularity results for problem \eqref{1} can be
obtained by the cosine operator theory in a similar argument as in the
case of constant coefficients (see \cite{Lasiecka and Triggiani}),
during which, however, some computations on Riemannian manifold are needed
to deal with the variable coefficients. Besides, it is worth noting that
the recent work by Wen et al.\ \cite{Wen 2014} gave the well-posedness
and regularity of two types of Euler--Bernoulli equations with variable
coefficients and Dirichlet boundary control, in which semigroup theory
and the multiplier technique with Riemannian geometry are utilized.
This method can also apply to the same question for our problem \eqref{1},
because the operator $\mathbf{A}$ we define below is quite similar to the operator
$A $ which is fundamentally used in \cite{Wen 2014}.
\end{remark}

This article is organized as follows:
In Section 2, we will introduce the escape vector field and state our
primary results. In Section 3, we use the duality method to find the
observability inequality. The proofs of the results are given in
the last section.

\section{Main results}

We denote the Levi-Civita connection in the metric $g$ by $D$.
Let $X$ be a vector field on $(\mathbb{R}^n,g).$ The covariant differential
$DX$ of $X$ determines a bilinear form on $\mathbb{R}^n_x\times\mathbb{R}^n_x$
for each $x\in\mathbb{R}^n$ by
 $$
DX(Y,Z)=\langle D_ZX,Y\rangle_g, \forall Y,Z\in\mathbb{R}^n_x,
$$
 where $D_ZX$ is the covariant derivative of $X$ with respect to $Z$.

\begin{definition} \label{def2.1} \rm
A vector field $H$ is said to be an escape vector field for the metric
$g$ on $\overline{\Omega}$ if there exists a constant $\rho_0>0$ such that
\begin{equation}
 DH(x)\geq\rho_0 g(x) \quad\text{for all } x\in \overline{\Omega}.
\end{equation}
\end{definition}

\begin{remark} \label{rmk} \rm
Escape vector field was introduced by Yao \cite{Yao 1999} as a checkable
assumption for the exact controllability of the wave equation with variable
coefficients. Actually, the existence of such a vector field can also
guarantee the exact controllability of an Euler-Bernoulli plate equation
with variable coefficients and the simply supported boundary condition
(see our results below).
\end{remark}

If $h$ is a strictly convex function in the metric $g$ on $\overline{\Omega}$,
then $H=Dh$ is such an escape vector field owing to $D^2h$, i.e.,
the Hessian of $h$, is positive. It is well known that the square of the
 distance function initiating from a given point $x_0\in \Omega$ in the
metric $g$ is strictly convex in a neighborhood of $x_0$ (see, e.g.,\cite{Wu 1989}),
then the escape vector field certainly exists locally.
Fortunately, the sectional curvature of the Riemannian metric $g$ can provide
the global information on its existence. Here are some relevant results
from \cite{Yao 1999} and \cite{Yao 2011}:

\begin{proposition} \label{prop2.1}
Let $x_0\in \mathbb{R}^n$ be given. For any $x\in \mathbb{R}^n$,
$\kappa(x,\Pi) $ denotes the sectional curvature of a two-dimensional subspace
$\Pi\subset\mathbb{R}^n_x$ in the metric g, set
$$
\kappa(\Omega)=\sup_{x\in\Omega,\, \Pi\subset\mathbb{R}^n_x} \kappa(x,\Pi).
$$
Let $B_g(x_0,\gamma)$ be a geodesic ball in $(\mathbb{R}^n, g)$
centered at $x_0$ with radius $\gamma$. Denote by $\rho(x)=d_g(x,x_0)$
the distance function of the metric g from $x$ to $x_0$.
If $\gamma > 0$ satisfies $4\gamma^2\kappa(\Omega)<\pi^2$ and
 $ \overline{\Omega}\subset B_g(x_0,\gamma)$, then $H=\rho D\rho$
is an escape vector field for the metric $g$ on $\overline{\Omega}$ .
\end{proposition}

\begin{proposition} \label{prop2.2}
 Suppose $(\mathbb{R}^n, g)$ is a Riemannian manifold, then

(a) If $(\mathbb{R}^n, g)$ has non-positive sectional curvature,
then there exists an escape vector field for the metric $g$ on the whole
space $\mathbb{R}^n$.

(b) If $(\mathbb{R}^n, g)$ is noncompact, complete, and its sectional curvature
is positive everywhere on $\mathbb{R}^n$, then there exists an escape vector
field in the metric $g$ on the whole space $\mathbb{R}^n$.
\end{proposition}

Now we present the main results.

\begin{theorem} \label{thm2.1}
Let $H$ be an escape vector field for the metric $g$ on $\overline{\Omega}$
and let $T>0$ be given. Let $\|k\|^2_{L^\infty(\Gamma)}< k_0$, which will
be given concretely in Section 4.
Then system \eqref{1} is exactly controllable on the space
$H^1_0(\Omega)\times H^{-1}(\Omega)$ with controls
$(\varphi,\psi)\in H^1_0( 0,T;L^2(\Gamma_1))\times L^2(\Sigma_1)$, where
$$
\Gamma_1=\{x | \langle H,\nu\rangle>0, x\in\Gamma\}.
$$
\end{theorem}

\section{Observability inequality}

The dual problem of system \eqref{1} can be readily derived as follows
\begin{equation}\label{2}
\begin{gathered}
w_{tt}+\mathscr{A}^2w=0 \quad \text{in } Q,\\
w=0 \quad \text{on }\Sigma, \\
\mathscr{A}w+a(x)Bw=0 \quad \text{on }\Sigma, \\
w(0)=w_0, w_t(0)=w_1 \quad \text{on }\Omega.
\end{gathered}
\end{equation}
Let $ \mathbf{A}:L^2(\Omega)\to L^2(\Omega) $ be a linear operator defined by
$$
\mathbf{A} f=\mathscr{A}^2f, D(\mathbf{A})=\{f\in H^4(\Omega):f|_\Gamma=0,
\mathscr{A}f+a(x)Bf|_\Gamma=0\}.
$$
It is easy to check that $\mathbf{A}$ is a positive, self-adjoint operator.
According to the interpolation results in \cite{Lions 1972}, we have
the following space identifications:
\begin{equation}\label{11}
\begin{gathered}
D(\mathbf{A}^\theta)=H^{4\theta}(\Omega), \quad 0<\theta<\frac{1}{8} ,\\
D(\mathbf{A}^\theta)=\{f\in H^{4\theta}(\Omega):
f|_\Gamma=0\}, \frac{1}{8}<\theta<\frac{5}{8}.
\end{gathered}
\end{equation}
In particular, $\mathbf{A}^{1/2}f=-\mathscr{A}f $ and
$D(\mathbf{A}^{1/2})=H^2(\Omega)\cap H^1_0(\Omega)$.

We introduce the energy of system \eqref{2} by
$$
2E(t)=\int_{\Omega}[(\mathbf{A}^{1/4}w)^2+(\mathbf{A}^{-1/4}w_t)^2]{\rm d}x.
$$
Differentiating the above identity with respect to $t$, we have
\begin{align*}
E'(t)&=(A^{1/4}w_t,A^{1/4}w)+(A^{-1/4}w_{tt},A^{-1/4}w_t) \\
&= (A^{1/4}w_t,A^{1/4}w)-(A^{3/4}w,A^{-1/4}w_t)=0,
\end{align*}
then $ E(t)\equiv E(0)$ for all $t>0$.

For $(w_0,w_1)\in H^1_0(\Omega)\times H^{-1}(\Omega)$, we solve problem
\eqref{2} to obtain the solution $w$. Then we solve the terminal value problem
\begin{equation}\label{3}
\begin{gathered}
u_{tt}+\mathscr{A}^2u=0 \quad \text{in } Q,\\
u(T)=u_t(T)=0 \quad \text{on }\Omega, \\
u|_{\Sigma_0}=0, u|_{\Sigma_1}=-(\mathscr{A}w)_{\nu_\mathscr{A}},\\
\mathscr{A}u+a(x)Bu=0 \quad \text{on }\Sigma_0, \\
\mathscr{A}u+a(x)Bu=-a(x)\sum^n_{i=2}e_i\langle e_i,\nabla_{\Gamma_g}
u\rangle_g-w_{\nu_\mathscr{A}}\quad \text{on }\Sigma_1.
 \end{gathered}
\end{equation}
Further, we define an operator
 $\Lambda:H^1_0(\Omega)\times H^{-1}(\Omega) \to
 H^{-1}(\Omega)\times H^1_0(\Omega) $ by
$$
\Lambda(w_0,w_1)=(u_t(0),-u(0)) \quad \text{on } \Omega.
$$
Using  equations \eqref{2} and \eqref{3}, we obtain
\begin{align*}
&(\Lambda(w_0,w_1),(w_0,w_1))_{L^2(\Omega)\times L^2(\Omega)} \\
&=(u_t(0),w_0)-(u(0),w_1)=[(u,w_t)-(u_t,w)]|^T_0 \\
&=\int_Q(w_{tt}u-u_{tt}w) {\rm d}Q
 =\int_Q(w\mathscr{A}^2u-u\mathscr{A}^2w) {\rm d}Q \\
&=\int_\Sigma[w(\mathscr{A}u)_{\nu_\mathscr{A}}
 -w_{\nu_\mathscr{A}}\mathscr{A}u-u(\mathscr{A}w)_{\nu_\mathscr{A}}
 +u_{\nu_\mathscr{A}}\mathscr{A}w]{\rm d}\Sigma \\
&=\int_{\Sigma_1}[w_{\nu_\mathscr{A}}^2+(\mathscr{A}w)_{\nu_\mathscr{A}}^2]{\rm d}
\Sigma.
\end{align*}

By the duality method given by  Lions \cite{Lions 1988}, the exact controllability
 of problem \eqref{1} on the space $H^1_0(\Omega)\times H^{-1}(\Omega)$ is
 equivalent to the following statement:

There is a $C_T>0$ such that
\begin{equation}\label{4}
\int_{\Sigma_1}[w_{\nu_\mathscr{A}}^2+(\mathscr{A}w)_{\nu_\mathscr{A}}^2]{\rm d}
\Sigma\geq C_T\|(w_0,w_1)\|^2_{H^1_0(\Omega)\times H^{-1}(\Omega)}.
\end{equation}
Using a result in \cite{Yao 2011}, the norm
$$
\|(w_0,w_1)\|^2_\star= \||\nabla_g(\mathscr{A}(\mathscr{A}^{-1}w_0))
|_g\|^2_{L^2(\Omega)}+\||\nabla_g(\mathscr{A}^{-1}w_1)|_g\|^2_{L^2(\Omega)}
$$
is equivalent norm on $H^1_0(\Omega)\times H^{-1}(\Omega)$.
Then  inequality \eqref{4} becomes
$$
\int_{\Sigma_1}[w_{\nu_\mathscr{A}}^2+(\mathscr{A}w)_{\nu_\mathscr{A}}^2]
{\rm d}\Sigma\geq C_TE(0).
$$
Let $z=\mathbf{A}^{-1/2}w$ and define
\begin{equation}\label{50}
 \mathbf{D} \xi = \zeta \quad 
\text{if  $\mathscr{A}\zeta = 0$  in $\Omega$, and
 $\zeta|_\Gamma=\xi$}.
 \end{equation}
Elliptic regularity theory (see \cite{Lions 1972}) gives
\begin{equation}\label{12}
\mathbf{D}\in\mathscr{L}(L^2(\Gamma)\to H^{1/2}(\Omega)).
\end{equation}
Clearly, $z$ satisfies the boundary conditions
$$
z|_\Gamma=\mathscr{A}z|_\Gamma=0.
$$
Moreover, we find that
\begin{align*}
z_{tt}&=\mathbf{A}^{-1/2}w_{tt}=-\mathbf{A}^{-1/2}\mathscr{A}^2w \\
&=-\mathbf{A}^{-1/2}\mathbf{A}^{1/2}(\mathscr{A}^2z-\mathbf{D}(\mathscr{A}^2z|_\Gamma)) \\
&=-\mathscr{A}^2z+\mathbf{D}(\mathscr{A}^2z|_\Gamma).
\end{align*}
Since $w|_\Gamma=0$, we obtain
\begin{equation}
 \mathscr{A}^2z|_\Gamma
=-\mathscr{A}w|_\Gamma=a(x)Bw
=-ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}.
\end{equation}
Consequently, $z$ satisfies the  equation
\begin{equation}\label{5}
 \begin{gathered}
z_{tt}+\mathscr{A}^2z=-\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}),\\
z|_\Gamma=\mathscr{A}z|_\Gamma=0, \\
z(0)=z_0,z_t(0)=z_1.
 \end{gathered}
\end{equation}
Then the observability inequality becomes
\begin{equation}\label{46}
\int_{\Sigma_1}[(\mathscr{A}z)_{\nu_\mathscr{A}}^2+(\mathscr{A}^2z)
_{\nu_\mathscr{A}}^2]{\rm d}\Sigma \geq C_TE(0),
\end{equation}
where the energy is now represented as
$$
 2E(t)=\int_{\Omega}[(\mathbf{A}^{3/4}z)^2+(\mathbf{A}^{1/4}z_t)^2]{\rm d}x.
$$

\section{Proofs of the results}

We consider $u$ as a regular solution to the problem
\begin{equation}\label{6}
 u_{tt}+\mathscr{A}^2u=f \quad \text{in } (0,\infty)\times\Omega,
\end{equation}
where $f$ is a given function.

The following lemma from \cite{Yao 2011} will play an important role in
establishing our multiplier identities.

\begin{lemma} \label{lem4.1}
Let $f$, $h$ be functions on $\mathbb{R}^n$ and let $H$ be a vector field
on $\mathbb{R}^n$. Then
\begin{align*}
&\langle\nabla_gf,\nabla_g(H(h))\rangle_g+\langle\nabla_gh,\nabla_g(H(f))\rangle_g\\
&=\operatorname{div}(\langle\nabla_gf,\nabla_gh\rangle_gH)
-\langle\nabla_gf,\nabla_gh\rangle_g\operatorname{div} H
+DH(\nabla_gh,\nabla_gf)+DH(\nabla_gf,\nabla_gh),
\end{align*}
where $\operatorname{div} H$ is the divergence of the vector field $H$
in the Euclidean metric.
\end{lemma}

Next are our main geometric multiplier identities.

\begin{lemma} \label{lem4.2}
Let $H$ be a vector field on $\overline{\Omega}$ and let $p$ be a function
on $\overline{\Omega}$, set $q=\operatorname{div} H$. Suppose that $u$ is a solution to
problem \eqref{6}. Then (1)
\begin{equation} \label{7}
\begin{aligned}
&\int_\Sigma\{2[qu_t+H(u_t)](u_t)_{\nu_\mathscr{A}}
+2H(\mathscr{A}u)(\mathscr{A}u)_{\nu_\mathscr{A}}
 -u^2_t\langle\nabla_g q,\nu\rangle \\
 &-(2u_t\mathscr{A}u_t+|\nabla_g u_t|^2_g+|\nabla_g (\mathscr{A}u)|^2_g)
\langle H,\nu\rangle\}{\rm d}\Sigma \\
&=\int_Q\{2DH(\nabla_g u_t,\nabla_g u_t)+2DH(\nabla_g (\mathscr{A}u),
\nabla_g (\mathscr{A}u)) \\
&\quad +[|\nabla_g u_t|^2_g-|\nabla_g (\mathscr{A}u)|^2_g]q
-u^2_t\mathscr{A}q+2fH(\mathscr{A}u)\}{\rm d}Q-2(u_t,H(\mathscr{A}u))|^T_0.
 \end{aligned}
\end{equation}
and (2)
\begin{equation} \label{8}
\begin{aligned}
&\int_\Sigma\big\{2p[u_t(u_t)_{\nu_\mathscr{A}}-\mathscr{A}
u(\mathscr{A}u)_{\nu_\mathscr{A}}]+[(\mathscr{A}u)^2-u_t^2]
p_{\nu_\mathscr{A}}\big\}{\rm d}\Sigma \\
&=2(u_t,p\mathscr{A}u)\big|^T_0
 +\int_Q\big\{\mathscr{A}p[(\mathscr{A}u)^2-u_t^2]
 +2p[|\nabla_g u_t|^2_g \\
&\quad -|\nabla_g (\mathscr{A}u)|^2_g  -f\mathscr{A}u]\big\}{\rm d}Q.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We multiply  equation \eqref{6} by $2H(\mathscr{A}u )$ and $2p\mathscr{A}u $,
 respectively. Then integrating over $Q$ by parts with Lemma \ref{lem4.1} yields
 these identities.
\end{proof}

Using these multiplier identities, we can derive the following estimates.

\begin{lemma} \label{lem4.3}
Let $T>0$ be given and let $H$ be an escape vector field for the metric
$g$ on $\overline{\Omega}$. Assume $z$ is the solution to \eqref{5}.
Then there is a $C_{T,1}>0$ such that
\begin{equation}\label{15}
\| (z_t)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}
+\| (\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}\geq C_{T,1}E(0).
\end{equation}
\end{lemma}

\begin{lemma} \label{lem4.4}
Let $z$ be the solution to  \eqref{5}. Then there is a $C_{T,2}>0$ such that
\begin{equation}\label{26}
\| (z_t)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
+\| (\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}\leq C_{T,2}E(0).
\end{equation}
\end{lemma}

\begin{proof}[Proof of Lemma \ref{lem4.3}]
Since $H$ is escaping on $\overline{\Omega}$, there is $\rho_0>0 $ such that
\begin{equation}\label{9}
 DH(X,X)\geq\rho_0|X|^2_g \quad\text{for }X\in \mathbb{R}^n_x , x\in\overline{\Omega}.
\end{equation}
By the boundary conditions, $z=\mathscr{A}z=0$ on $\Gamma$, we have
$$
\nabla_gz_t=\sum^n_{i=1}\langle\nabla_gz_t,e_i\rangle_ge_i
=\langle\nabla_gz_t,\frac{\nu_\mathscr{A}}{|\nu_\mathscr{A}|_g}\rangle_g
\frac{\nu_\mathscr{A}}{|\nu_\mathscr{A}|_g}
=\frac{(z_t)_{\nu_\mathscr{A}}}{|\nu_\mathscr{A}|^2_g}\nu_\mathscr{A}.
$$
Similarly, $\nabla_g(\mathscr{A}z)
=\frac{(\mathscr{A}z)_{\nu_\mathscr{A}}}{|\nu_\mathscr{A}|^2_g}\nu_\mathscr{A}$.
Thus,
\begin{gather}\label{60}
|\nabla_gz_t|^2_g=\frac{(z_t)_{\nu_\mathscr{A}}^2}{|\nu_\mathscr{A}|^2_g}, \quad
H(z_t) =\frac{\langle H,\nu\rangle}{|\nu_\mathscr{A}|^2_g}(z_t)_{\nu_\mathscr{A}},
\\
\label{61}
|\nabla_g(\mathscr{A}z)|^2_g
=\frac{(\mathscr{A}z)_{\nu_\mathscr{A}}^2}{|\nu_\mathscr{A}|^2_g}, \quad
 H(\mathscr{A}z)=\frac{\langle H,\nu\rangle}{|\nu_\mathscr{A}|^2_g}
(\mathscr{A}z)_{\nu_\mathscr{A}}.
\end{gather}
Using the boundary conditions of problem \eqref{5}, the relations \eqref{60}
and \eqref{61} in identity \eqref{7} with
$f=-\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})$, we obtain
\begin{equation} \label{10}
\begin{aligned}
&\int_\Sigma[(z_t)^2_{\nu_{\mathscr{A}}}+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]
 \langle H,\nu\rangle/|\nu_{\mathscr{A}}|^2_g{\rm d}\Sigma \\
&=\int_Q \Big\{[2DH(\nabla_g z_t,\nabla_g z_t)+2DH(\nabla_g (\mathscr{A}z),
\nabla_g (\mathscr{A}z))] \\
&\quad +[|\nabla_g z_t|^2_g-|\nabla_g (\mathscr{A}z)|^2_g]\operatorname{div}
H-z^2_t\mathscr{A}q 
-2\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})H(\mathscr{A}z)\Big\}{\rm d}Q \\
&\quad -2(z_t,H(\mathscr{A}z))|^T_0.
\end{aligned}
\end{equation}
Firstly,
\begin{equation} \label{13}
\int_\Sigma[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]\langle H,\nu\rangle/
|\nu_{\mathscr{A}}|^2_g{\rm d}\Sigma
\leq C\int_{\Sigma_1}[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma.
\end{equation}

Next, we shall estimate all terms on the right-hand side of \eqref{10}.
For the first term, by means of \eqref{9}, we obtain
\begin{equation}
\begin{aligned}
& \int_Q[2DH(\nabla_g z_t,\nabla_g z_t)+2DH(\nabla_g (\mathscr{A}z),
\nabla_g (\mathscr{A}z))]{\rm d}Q \\
&\geq 2\rho_0\int_Q[|\nabla_g z_t|^2_g+|\nabla_g (\mathscr{A}z)|^2_g]{\rm d}Q
=4\rho_0TE(0).
\end{aligned}
\end{equation}
For the  second term, using the boundary conditions of problem \eqref{5}
in identity \eqref{8} with $p={\operatorname{div} H}/2$ and
$f=-\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})$, we obtain
\begin{equation} \label{24}
\begin{aligned}
 \big|\int_Q[|\nabla_gz_t|^2_g-|\nabla_g(\mathscr{A}z)|^2_g]
\operatorname{div} H{\rm d}Q\big|
\leq \varepsilon \|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
+C_\varepsilon L(z),
\end{aligned}
\end{equation}
where
\begin{align*}
 L(z)&= \|z(0)\|^2_{L^2(\Omega)}+\|z(T)\|^2_{L^2(\Omega)}+\|z\|^2_{L^2(Q)}
+\|z_t(0)\|^2_{L^2(\Omega)}+\|z_t(T)\|^2_{L^2(\Omega)}\\
&\quad +\|z_t\|^2_{L^2(Q)}
 +\||D^2z|_g(0)\|^2_{L^2(\Omega)}+\||D^2z|_g(T)\|^2_{L^2(\Omega)}
 +\||D^2z|_g\|^2_{L^2(Q)},
\end{align*}
are the lower terms relative to the energy $E(t)$.

For the third term, we have
\begin{equation}
 \int_Q-z^2_t\mathscr{A}q{\rm d}Q\geq-\sup_{x\in\Omega}|\mathscr{A}q|
 \|z_t\|^2_{L^2(Q)}.
\end{equation}

For the fourth term, by \eqref{11} and \eqref{12}, 
$\mathbf{A}^\theta\mathbf{D}\in \mathcal {L}(L^2(\Gamma)\to L^2(\Omega))$
for $\theta<{1/8}$,  we have
\begin{equation}
\begin{aligned}
&|(\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}),H(\mathscr{A}z))_{L^2(Q)}|\\
&=|(\mathbf{A}^{-\theta}\mathbf{A}^\theta \mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}),
 H(\mathscr{A}z))_{L^2(Q)}| \\
&\leq\epsilon\|\mathbf{A}^\theta\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})\|^2_{L^2(Q)}
 +C_\epsilon\|(\mathbf{A}^{-\theta} H(\mathscr{A}z)\|^2_{L^2(Q)} \\
&\leq\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
+C_\epsilon T\|\mathbf{A}^{-\theta}H(\mathscr{A}z)\|^2_{C[0,T;L^2(\Omega)]}.
\end{aligned}
\end{equation}
Applying Lemma \ref{lem4.4}, we obtain
\begin{equation}
\begin{aligned}
\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
&=\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}
 +\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_0)} \\
&\leq\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}
+\epsilon C_{T,2}E(0).
\end{aligned}
\end{equation}

For the last term, we have
\begin{equation}
\begin{aligned}
 |(z_t,H(\mathscr{A}z))|
&\leq \sup_{x\in\Omega}|H|_g\int_\Omega|z_t| |\nabla_g(\mathscr{A}z)|_g{\rm d}x \\
 &\leq\epsilon\int_\Omega|\nabla_g(\mathscr{A}z)|^2_g{\rm d}x
+C_\epsilon\int_\Omega z^2_t{\rm d}x \\
&\leq 2\epsilon E(0)+C_\epsilon\int_\Omega z^2_t{\rm d}x.
\end{aligned}
\end{equation}
Thus
\begin{equation}\label{14}
-2(z_t,H(\mathscr{A}z))|^T_0
\geq-8\epsilon E(0)-2C_\epsilon(\|z_t(0)\|^2_{L^2(\Omega)}
+\|z_t(T)\|^2_{L^2(\Omega)}).
\end{equation}

Combining \eqref{10}--\eqref{14}, we have
\begin{equation}
\begin{aligned}
&C\int_{\Sigma_1}[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma \\
&\geq 4(\rho_0T-2\epsilon-\frac{\epsilon}{2} C_{T,2})E(0)
 -2C_\epsilon L(z)-\varepsilon \|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)} \\
&-2\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}
 -2C_\epsilon T\|\mathbf{A}^{-\theta} H(\mathscr{A}z)\|^2_{C[0,T;L^2(\Omega)]}.
\end{aligned}
\end{equation}
Then for $\epsilon$ small enough, there are constants $C_i>0$ for
$1\leq i\leq3$ such that
\begin{equation}\label{16}
C_1\int_{\Sigma_1}[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma+C_2L(z)\geq C_3E(0),
\end{equation}
for all solutions $z$ to  \eqref{5}. Then inequality \eqref{15} follows
 by Lemma \ref{lem4.5} below.
\end{proof}

\begin{lemma} \label{lem4.5}
Let inequality \eqref{16} hold for all solutions $z$ of \eqref{5}. 
Then there is a $C>0$ such that
\begin{equation}\label{19}
\int_{\Sigma_1}[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma\geq CE(0).
\end{equation}
\end{lemma}

To prove this lemma, we  need the following uniqueness result from 
\cite{Pederson 1958}.

\begin{proposition} \label{prop4.1}
Let $\hat{\Gamma}$ be a relatively open subset of $\Gamma$. 
If $w$ solves the problem
\begin{equation}
 \begin{gathered}
\mathscr{A}^2w=F(w,Dw,D^2w,D^3w) \quad \text{on }\Omega,\\
w=w_{\nu_\mathscr{A}}=\mathscr{A}w=(\mathscr{A}w)_{\nu_\mathscr{A}}=0
\quad \text{on }\hat{\Gamma}, \\
 \end{gathered}
\end{equation}
then $w=0$ on $\Omega$ .
\end{proposition}

\begin{proof}[Proof of Lemma \ref{lem4.5}]
\textbf{Step 1.} 
Let $Y=\{z\in H^3(Q) : z$ is a solution to problem \eqref{5} satisfying 
$(z_t)_{\nu_\mathscr{A}}|_{\Sigma_1}=(\mathscr{A}z)_{\nu_\mathscr{A}}|_{\Sigma_1}=0\}$.
Then
\begin{equation}\label{18}
 Y = \{0\}.
\end{equation}
Indeed, from  inequality \eqref{16}, we have
$$
C_2L(z)\geq C_3E(0) \quad \text{for all } z\in Y ,
$$
which implies that any bounded closed set in $Y\cap H^3(Q)$ 
is compact in $H^3(Q)$. Then $Y$ is a finite-dimensional linear space.
For any $z\in Y$, we can readily obtain that $z_t\in Y$.
Then $\partial_t:Y\to Y $ is a linear operator. Let $Y\neq \{0\}$, 
then $\partial_t$ has at least one eigenvalue $ \lambda$. 
Assume that $v \neq 0$ is one of its eigenfunctions, then $v_t=\lambda v$. 
Further, $v$ is a nonzero solution to the problem
\begin{equation}\label{17}
\begin{gathered}
\mathscr{A}^2v=-\lambda^2v-\mathbf{D}(ka(x)(\mathscr{A}v)_{\nu_{\mathscr{A}}}) 
\quad\text{on }\Omega,\\
v=(v_t)_{\nu_\mathscr{A}}
=\mathscr{A}v=(\mathscr{A}v)_{\nu_\mathscr{A}}=0 \quad\text{on }\Gamma_1. 
 \end{gathered}
\end{equation}
However, by Proposition \ref{prop4.1}, problem \eqref{17} only has zero solution, 
this contradiction shows that \eqref{18} holds.
\smallskip

\noindent\textbf{Step 2.} 
Suppose that the estimate \eqref{19} is not true. 
Then there are $(z^k_0,z^k_1)\in H^3_0(\Omega)\times H^1_0(\Omega)$,
 whose solutions are denoted by $z^k$, such that
\begin{equation}\label{20}
 E(z^k,0)=1, \quad \int_{\Sigma_1}[(z^k_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z^k)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma
\leq \frac{1}{k} \quad \text{for } k\geq1.
\end{equation}
Then $\|z^k\|^2_{H^3(Q)}=2T$ for all $k\geq1$. Thus there is a subsequence, 
still denoted by ${z^k}$, such that
\begin{gather}\label{21}
 {z^k}\text{ converges in $H^2(\Omega)$ for each $t\in [0,T]$,  and} \\
\label{22}
 {z^k}\text{ converges in } H^2(Q).
\end{gather}
It follows from relations \eqref{16}, \eqref{20}, \eqref{21} and \eqref{22} 
that ${z^k}$ converges in $H^3(Q)$. Then there exists a solution $z^0$ to 
problem \eqref{5} such that
$$
z^k\to z^0\quad \text{as }k\to \infty \text{ in } H^3(Q).
$$
Then
$$
E(z^0,0)=1, \int_{\Sigma_1}[(z^0_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z^0)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma=0.
$$
Thus 
$$ 
(z^0_t)_{\nu_\mathscr{A}}|_{\Sigma_1}
=(\mathscr{A}z^0)_{\nu_\mathscr{A}}|_{\Sigma_1}=0.
$$
Then $0\neq z^0\in Y$, it contradicts the relation \eqref{18}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem4.4}]
We choose a vector field $H$ on $\overline{\Omega}$ such that 
$$
H=A(x)\nu \quad \text{for } x\in \Gamma,
$$
and let $f=-\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})$.
 Then using the boundary conditions of problem \eqref{5}, relations \eqref{60} 
and \eqref{61} in identity \eqref{7}, it gives
\begin{equation} \label{23}
\begin{aligned}
&\int_\Sigma[(z_t)^2_{\nu_{\mathscr{A}}}
+(\mathscr{A}z)^2_{\nu_{\mathscr{A}}}]{\rm d}\Sigma \\
&=\int_Q\{2DH(\nabla_g z_t,\nabla_g z_t)
 +2DH(\nabla_g (\mathscr{A}z),\nabla_g (\mathscr{A}z)) \\
&\quad +(|\nabla_g z_t|^2_g-|\nabla_g (\mathscr{A}z)|^2_g)
 \operatorname{div} H-z^2_t\mathscr{A}q \\
&\quad -2\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})H(\mathscr{A}z)\}{\rm d}Q
-2(z_t,H(\mathscr{A}z))|^T_0.
\end{aligned}
\end{equation}
We shall estimate all terms on the right-hand side of \eqref{23} separately.
For the first term, we have
\begin{equation}
\begin{aligned}
& \int_Q[2DH(\nabla_g z_t,\nabla_g z_t)
 +2DH(\nabla_g (\mathscr{A}z),\nabla_g (\mathscr{A}z))]{\rm d}Q \\
&\leq C\int_Q[|\nabla_g z_t|^2_g+|\nabla_g (\mathscr{A}z)|^2_g]{\rm d}Q=2CTE(0).
\end{aligned}
\end{equation}

We have already estimated the second term in the  proof of Lemma \ref{lem4.3}.

For the third term, we have
\begin{equation}
 -\int_Qz^2_t\mathscr{A}q{\rm d}Q
\leq T\sup_{x\in\Omega}|\mathscr{A}q| \|z_t\|^2_{C[0,T;L^2(\Omega)]}.
\end{equation}

For the fourth term, we have
\begin{equation}
\begin{aligned}
&\big|\int_Q\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})H(\mathscr{A}z){\rm d}Q\big|\\
&\leq\epsilon\|\mathbf{D}(ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}})\|^2_{L^2(Q)}
 +\sup_{x\in\Omega}|H|^2_g C_\epsilon\||\nabla_g (\mathscr{A}z)|_g\|^2_{L^2(Q)} \\
&\leq\epsilon\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
+2T\sup_{x\in\Omega}|H|^2_gC_\epsilon E(0).
\end{aligned}
\end{equation}

For the last term, we have
\begin{align*}
&-2(z_t,H(\mathscr{A}z))|^T_0 \\
&\leq2\int_\Omega[|z_t(0)| |H(\mathscr{A}z)(0)|+|z_t(T)|
|H(\mathscr{A}z)(T)|]{\rm d}x \\
&\leq\|z_t(0)\|^2_{L^2(\Omega)}+\|z_t(T)\|^2_{L^2(\Omega)}
 +\sup_{x\in\Omega}|H|^2_g\int_\Omega[|\nabla_g(\mathscr{A}z)(0)|^2_g
 +|\nabla_g(\mathscr{A}z)(T)|^2_g]{\rm d}x \\
&\leq2\|z_t\|^2_{C[0,T;L^2(\Omega)]}+4\sup_{x\in\Omega}|H|^2_gE(0).
\end{align*}
Since $z_t=\mathscr{A}z=0$ on $\Gamma$, according to the Poincare's
inequality, we have
\begin{equation}\label{25}
\|z_t\|^2\leq C\||\nabla_gz_t|_g\|^2_{L^2(\Omega)},
\|\mathscr{A}z\|^2\leq C\||\nabla_g(\mathscr{A}z)|_g\|^2_{L^2(\Omega)}.
\end{equation}

Combining \eqref{24}, \eqref{23}--\eqref{25}, we obtain the desired 
estimate \eqref{26}.
\end{proof}

Using  some ideas from \cite{Horn 1992}, we can eliminate the term
 $\|(z_t)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)}$ from the inequality \eqref{15}.
Firstly, we have the following lemma.

\begin{lemma} \label{lem4.6}
Let $\alpha>0$ be a given constant and define 
$\Sigma^\alpha= [-\alpha,T+\alpha]\times\Gamma$.
 Assume $z$ satisfies problem \eqref{5}. Then for any $\epsilon>0$, 
there is a $C_{T,3}>0$ such that
\begin{equation} \label{33}
\begin{aligned}
&\|(z_t)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)} \\
&\leq\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}
 +\epsilon C_{T,3}E(0)+(\frac{8}{\epsilon}+2)
 C_{K,D,T}\|ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha)} \\
&\quad +C(\|z_0\|^2_{L^2(\Omega)}+\|z_1\|^2_{L^2(\Omega)}).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 We shall take four steps to prove it.
\smallskip

\noindent\textbf{Step 1.} 
Let $z$ be a complex solution to problem \eqref{5}. By using the cosine 
operator theory (see \cite{Fattorini 1985}), we obtain
\begin{equation}
z(t)=e^{i\mathscr{A}t}\tilde{z}_0+e^{-i\mathscr{A}t}\tilde{z}_1
+\mathscr{A}^{-1}\int^t_0\frac{1}{2i}(e^{i\mathscr{A}(t-\tau)}
-e^{-i\mathscr{A}(t-\tau)})\mathbf{D} f(\tau){\rm d}\tau,
\end{equation}
where
$$
f=-ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}, \quad
\tilde{z}_0=\frac{z_0}{2}-\frac{i}{2}\mathscr{A}^{-1}z_1, \quad
\tilde{z}_1=\frac{z_0}{2}+\frac{i}{2}\mathscr{A}^{-1}z_1.
$$
To simplify notation, we define
\begin{equation}
\begin{gathered}
A_1=(\mathscr{A}e^{i\mathscr{A}t}\tilde{z}_0)_{\nu_\mathscr{A}}, \quad
B_1=\frac{1}{2}(\int^t_0e^{-i\mathscr{A}(t-\tau)}
 \mathbf{D} f(\tau){\rm d}\tau)_{\nu_\mathscr{A}},\\
A_2=(\mathscr{A}e^{-i\mathscr{A}t}\tilde{z}_1)_{\nu_\mathscr{A}}, \quad
B_2=\frac{1}{2}(\int^t_0e^{i\mathscr{A}(t-\tau)}
 \mathbf{D} f(\tau){\rm d}\tau)_{\nu_\mathscr{A}}.
\end{gathered}
\end{equation}
Using these definitions, we have
\begin{gather}
(z_t)_{\nu_\mathscr{A}}=i(A_1-A_2)+(B_1+B_2), \\
(\mathscr{A}z)_{\nu_\mathscr{A}}=(A_1+A_2)+i(B_1-B_2).
\end{gather}
Thus, we obtain
\begin{align}\label{27}
|(z_t)_{\nu_\mathscr{A}}|^2-|(\mathscr{A}z)_{\nu_\mathscr{A}}|^2
=4Re(-A_1\bar{A}_2+iA_1\bar{B}_1-iA_2\bar{B}_2+\bar{B}_1B_2).
\end{align}
\smallskip

\noindent\textbf{Step 2.}
Let $\phi(t)\in C^\infty_0(\mathbb{R})$ be such that 
$0\leq\phi(t)\leq1,\phi(t)\equiv 1$ on $[0,T]$, and $\phi(t)\equiv 0 $ 
on $(-\infty,-\alpha)\cup(T+\alpha,\infty)$.
From \eqref{27}, we obtain
\begin{equation} \label{30}
\begin{aligned}
&\|(z_t)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma_1)} \\
&\leq \|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}
+4\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}A_1\bar{A}_2{\rm d}x{\rm d}t\big| \\
&\quad +4\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}A_1\bar{B}_1{\rm d}
 x{\rm d}t\big|
 +4\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}A_2\bar{B}_2{\rm d}
 x{\rm d}t\big| \\
&\quad +4\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}\bar{B}_1B_2{\rm d}x{\rm d}t\big|.
\end{aligned}
\end{equation}
\smallskip

\noindent\textbf{Step 3.}
\begin{equation}  \label{28}
\begin{aligned}
A_1\bar{A}_2
&=(\mathscr{A}e^{i\mathscr{A}t}\tilde{z}_0)_{\nu_\mathscr{A}}
\times(\mathscr{A}e^{i\mathscr{A}t}\bar{\tilde{z}}_1)_{\nu_\mathscr{A}} \\
&=(\sum^\infty_{n=1}\lambda_ne^{-i\lambda_nt}(\tilde{z}_0,\phi_n)
(\phi_n)_{\nu_\mathscr{A}})\times(\sum^\infty_{m=1}
\lambda_me^{-i\lambda_mt}(\bar{\tilde{z}}_1,\phi_m)(\phi_m)_{\nu_\mathscr{A}}),
\end{aligned}
\end{equation}
where $\lambda_i$ and $\phi_i$ denote the eigenvalues and eigenfunctions
corresponding to the operator $-\mathscr{A}$ with $|\phi_i|=1$.
Since
$$|(\phi_n)_{\nu_\mathscr{A}}|\leq C|\mathscr{A}\phi_n|
\leq C\lambda_n|\phi_n|=C\lambda_n,
$$
we have
\begin{equation}\label{29}
\int_{\Gamma_1}|(\phi_n)_{\nu_\mathscr{A}}| |(\phi_m)_{\nu_\mathscr{A}}
|{\rm d}\Gamma\leq C\lambda_n\lambda_m.
\end{equation}
Combining \eqref{28} and \eqref{29}, we find
\begin{align*}
&\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}A_1\bar{A}_2{\rm d}
x{\rm d}t\big| \\
&\leq C\sum^\infty_{n=1}\sum^\infty_{m=1}\lambda^2_n\lambda^2_m|(\tilde{z}_0,\phi_n)|
 |(\bar{\tilde{z}}_1,\phi_m)|\big|\int^\infty_{-\infty}\phi(t)
e^{-i(\lambda_n+\lambda_m)t}{\rm d}t\big|.
\end{align*}
Since $\phi(t)\in C^\infty_0(\mathbb{R})$,  for any $N$, we have
\begin{equation}
\big|\int^\infty_{-\infty}\phi(t)e^{-i(\lambda_n+\lambda_m)t}{\rm d}t\big|
\leq\frac{c_\phi}{|\lambda_n+\lambda_m|^N} .
\end{equation}
Thus,
\begin{equation} \label{31}
\begin{aligned}
\big|\int^\infty_{-\infty}\phi(t)\int_{\Gamma_1}A_1\bar{A}_2{\rm d}x{\rm d}t
\big|
&\leq C_\phi(\sum^\infty_{n=1}\frac{|(\tilde{z}_0,\phi_n)|^2}{\lambda^{N-4}_n}
 +\sum^\infty_{m=1}\frac{|(\bar{\tilde{z}}_1,\phi_m)|^2}{\lambda^{N-4}_m}) \\
&=C_\phi(\|\mathscr{A}^{2-\frac{N}{2}}\tilde{z}_0\|^2_{L^2(\Omega)}
 +\|\mathscr{A}^{2-\frac{N}{2}}\bar{\tilde{z}}_1\|^2_{L^2(\Omega)}) \\
&\leq C(\|z_0\|^2_{H^{4-N}(\Omega)}+\|z_1\|^2_{H^{2-N}(\Omega)}).
\end{aligned}
\end{equation}
\smallskip

\noindent\textbf{Step 4.}
Before we complete the proof of Lemma \ref{lem4.6}, we shall need the following 
result to estimate the remaining three terms on the right-hand side of 
inequality \eqref{30}.

\begin{proposition} \label{prop4.2}
Let $y$ be a solution of the  problem
\begin{equation}\label{34}
\begin{gathered}
 y_t=i\mathbf{A}^{1/2}y+\mathbf{D}\xi,\\
 y(0)=y_0\in \mathbf{D}(\mathbf{A}^{1/4}).
 \end{gathered}
\end{equation}
Then
\begin{equation}
\|y_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
\leq C_{T,3}\|\mathscr{A}^{1/2}y_0\|^2_{L^2(\Omega)}
+C_{K,D,T} \|\xi\|^2_{L^2(\Sigma)}.
\end{equation}
\end{proposition}

The above proposition will be proven later.
Applying the result of Proposition \ref{prop4.2} with $\xi=0$, we obtain
\begin{equation}
\begin{aligned}
 \|A_1\|^2_{L^2(\Sigma^\alpha)}+\|A_2\|^2_{L^2(\Sigma^\alpha)}
&\leq C_{T,3}(\|\mathscr{A}^{3/2}\tilde{z}_0\|^2_{L^2(\Omega)}
+\|\mathscr{A}^{3/2}\tilde{z}_1\|^2_{L^2(\Omega)}) \\
&\leq C_{T,3}(\|\mathscr{A}^{3/2}z_0\|^2_{L^2(\Omega)}
+\|\mathscr{A}^{1/2}z_1\|^2_{L^2(\Omega)}),
\end{aligned}
\end{equation}
with $y_0=0$, we obtain
\[
\|B_1\|^2_{L^2(\Sigma^\alpha)}+\|B_2\|^2_{L^2(\Sigma^\alpha)}
\leq C_{K,D,T} \|f\|^2_{L^2(\Sigma^\alpha)}= C_{K,D,T} \|ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha)}.
\]
Then
\begin{equation} \label{32}
\begin{aligned}
&\int^{T+\alpha}_{-\alpha}\int_{\Gamma_1}(|A_1\bar{B}_1|+|A_2\bar{B}_2|
+|\bar{B}_1B_2|){\rm d}x{\rm d}t \\
&\leq \int_{\Sigma^\alpha_1}[\epsilon(|A_1|^2+|A_2|^2)
 +(C_\epsilon+\frac{1}{2})(|B_1|^2+|B_2|^2)]{\rm d}\Sigma \\
&\leq\epsilon C_{T,3}(\|\mathscr{A}^{3/2}z_0\|^2_{L^2(\Omega)}
 +\|\mathscr{A}^{1/2}z_1\|^2_{L^2(\Omega)}) \\
&\quad  +(C_\epsilon+\frac{1}{2})C_{K,D,T}\|ka(x)
(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha)}.
\end{aligned}
\end{equation}
Combining \eqref{30}, \eqref{31} and \eqref{32}, we obtain the desired
inequality \eqref{33}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop4.2}]
Let
\begin{equation}
y(t)=y_1+y_2=e^{-i\mathscr{A}t}y_0+\int^t_0e^{-i\mathscr{A}(t-\tau)}
\mathbf{D}\xi(\tau){\rm d}\tau.
\end{equation}
Clearly, $y$ satisfies \eqref{34}.
We shall do the proof by several steps.
\smallskip

\noindent\textbf{Step 1.}
We firstly prove the  estimate
\begin{equation}\label{40}
\|y_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
\leq C_{T,4}(\|\mathbf{D}\xi\|^2_{L^1[0,T;L^2(\Omega)]}
+\|\mathscr{A}^{1/2}y\|^2_{L^\infty[0,T;L^2(\Omega)]}).
\end{equation}

\begin{proof}
 We multiply \eqref{34} by $h(\bar{y})$, where $h|_\Gamma=\nu_\mathscr{A}$, 
and integrate over $Q$ by parts, with Lemma \ref{lem4.1} to obtain
\begin{equation} \label{a}
\begin{aligned}
&\operatorname{Im} \int_Qy_th(\bar{y}){\rm d}Q \\
&=\operatorname{Im}\Big(-\int_Qi\mathscr{A}yh(\bar{y})+\mathbf{D}\xi h(\bar{y}){\rm d}Q\Big) \\
&=\int_Q\operatorname{Re}[-\operatorname{div} h(\bar{y})A(x)\nabla y
 +\langle\nabla_gh(\bar{y}),\nabla_gy\rangle]{\rm d}Q
+\operatorname{Im}\int_Q\mathbf{D}\xi h(\bar{y}){\rm d}Q \\
&=\int_Q\{\operatorname{Re}[-\operatorname{div}
 h(\bar{y})A(x)\nabla y+Dh(\nabla_gy,\nabla_g\bar{y})] \\
&\quad +\frac{1}{2}\operatorname{div}(\langle\nabla_gy,
\nabla_g\bar{y}\rangle_gh)-\frac{1}{2}\langle\nabla_gy,
\nabla_g\bar{y}\rangle_g\operatorname{div} h\}{\rm d}Q+\operatorname{Im}\int_Q\mathbf{D}\xi h(\bar{y}){\rm d}Q \\
&=\operatorname{Im}\int_Q\mathbf{D}\xi h(\bar{y}){\rm d}Q
-\frac{1}{2}\int_\Sigma|y_{\nu_\mathscr{A}}|^2{\rm d}\Sigma \\
& +\int_Q[\operatorname{Re}Dh(\nabla_gy,\nabla_g\bar{y})
-\frac{1}{2}|\nabla_gy|^2_g\operatorname{div} h]{\rm d}Q,
\end{aligned}
\end{equation}
where the notation ``$\operatorname{Im}$" and ``$\operatorname{Re}$"
denote the imaginary part and the real part of a complex number, respectively.

On the other hand, using the divergence theorem, we find
\begin{equation}
\operatorname{div}\bar{y}y_t h=\bar{y}y_t \operatorname{div} h+y_th(\bar{y})
+[\bar{y}h(y)]_t-\bar{y}_th(y).
\end{equation}
Thus,
\begin{equation} \label{36}
\begin{aligned}
\operatorname{Im}\int_Qy_th(\bar{y}) {\rm d}Q
&= -\frac{i}{2}\int_\Sigma\bar{y}y_t|\nu_\mathscr{A}|^2_g{\rm d}\Sigma
 +\frac{i}{2}\int_Q\bar{y}y_t\operatorname{div} h {\rm d}Q \\
&\quad +\frac{i}{2}\int_\Omega[\bar{y}(T)h(y)(T)-\bar{y}(0)h(y)(0)]{\rm d}\Omega.
\end{aligned}
\end{equation}
Combining \eqref{a} and \eqref{36}, we obtain
\begin{equation} \label{37}
\begin{aligned}
&\frac{1}{2}\int_\Sigma|y_{\nu_\mathscr{A}}|^2{\rm d}\Sigma \\
&= \operatorname{Im}\int_Q\mathbf{D}\xi h(\bar{y}) {\rm d}Q
 -\frac{i}{2}\int_Q\bar{y}y_t\operatorname{div} h{\rm d}Q \\
&\quad +\frac{i}{2}\int_\Omega[\bar{y}(0)h(y)(0)-\bar{y}(T)h(y)(T)]{\rm d}\Omega \\
&\quad +\int_Q[\operatorname{Re}Dh(\nabla_gy,\nabla_g\bar{y})
 -\frac{1}{2}|\nabla_gy |^2_g\operatorname{div} h]{\rm d}Q.
\end{aligned}
\end{equation}

Next, we multiply \eqref{34} by $\bar{y}$ and integrate over $Q$ by parts to obtain
\[
\int_Qy_t\bar{y}{\rm d}Q
=-i\int_Q\bar{y}\mathscr{A}y{\rm d}Q+\int_Q\bar{y} \mathbf{D}\xi {\rm d}Q
=i\int_Q\langle\nabla_g\bar{y},\nabla_gy\rangle_g{\rm d}Q
+\int_Q\bar{y} \mathbf{D}\xi {\rm d}Q.
\]
Thus,
\begin{equation} \label{38}
\begin{aligned}
 \big|\int_Qy_t\bar{y} {\rm d}Q\big|
&\leq\int_Q|\nabla_gy|^2_g{\rm d}Q+\int_Q|\bar{y} \mathbf{D}\xi |{\rm d}Q \\
 &\leq\int_Q|\nabla_gy|^2_g{\rm d}Q+\frac{1}{2}\int_Q(|\bar{y}|^2
+|\mathbf{D}\xi|^2){\rm d}Q \\
 &\leq C\int_Q|\nabla_gy|^2_g{\rm d}Q+\frac{1}{2}T
 \|\mathbf{D}\xi\|^2_{L^\infty[0,T;L^2(\Omega)]} \\
 &\leq C_T(\||\nabla_gy|_g\|^2_{L^\infty[0,T;L^2(\Omega)]}
+\|\mathbf{D}\xi\|^2_{L^1[0,T;L^2(\Omega)]}).
\end{aligned}
\end{equation}
Furthermore, we can bound all terms of the right-hand side of \eqref{37} as follows:
\begin{gather}
\begin{aligned}
& \operatorname{Im}\int_Q\mathbf{D} \xi h(\bar{y}){\rm d}Q\\
&\leq\big|\int_Q\mathbf{D} \xi h(\bar{y}){\rm d}Q\big| \\
&\leq \frac{1}{2}T\|\mathbf{D}\xi\|^2_{L^\infty[0,T;L^2(\Omega)]}
 +\frac{1}{2}\sup_{x\in\Omega}|h|^2_g\int^T_0\||\nabla_g\bar{y}
 |_g\|^2_{L^2(\Omega)}{\rm d}t \\
&\leq \frac{1}{2}T\|\mathbf{D}\xi\|^2_{L^1[0,T;L^2(\Omega)]}
+\frac{1}{2}\sup_{x\in\Omega}|h|^2_gT\||\nabla_g\bar{y}
|_g\|^2_{L^\infty[0,T;L^2(\Omega)]};
\end{aligned}
\\
\big|-\frac{i}{2}\int_Q\bar{y}y_t\operatorname{div} h{\rm d}Q\big|
\leq \frac{1}{2}\sup_{x\in\Omega}|\operatorname{div} h|
\big|\int_Qy_t\bar{y}{\rm d}Q\big|;
\\
\begin{aligned}
& \big|\frac{i}{2}\int_\Omega[\bar{y}(0)h(y)(0)-\bar{y}(T)h(y)(T)]{\rm d}x\big| \\
&\leq\frac{1}{2}\int_\Omega[|\bar{y}(0)\langle h,\nabla_gy\rangle_g(0)|
 +|\bar{y}(T)\langle h,\nabla_gy\rangle_g(T)|]{\rm d}x \\
&\leq\frac{1}{4}\sup_{x\in\Omega}|h|_g\int_\Omega[|\nabla_gy(0)|^2_g
 +|\bar{y}(0)|^2+|\nabla_gy(T)|^2_g+|\bar{y}(T)|^2]{\rm d}x \\
&\leq\frac{1}{2}\sup_{x\in\Omega}|h|_g(\||\nabla_gy|_g
 \|^2_{L^\infty[0,T;L^2(\Omega)]}+\|\bar{y}\|^2_{L^\infty[0,T;L^2(\Omega)]}) \\
&\leq C\sup_{x\in\Omega}|h|_g\||\nabla_gy|_g\|^2_{L^\infty[0,T;L^2(\Omega)]};
\end{aligned}
\\
\int_Q\operatorname{Re}Dh(\nabla_gy,\nabla_g\bar{y}){\rm d}Q
\leq C_T\||\nabla_gy|_g\|^2_{L^\infty[0,T;L^2(\Omega)]};
\\
\label{39}
-\int_Q\frac{1}{2}|\nabla_gy|^2_g\operatorname{div} h{\rm d}Q
\leq C_T\sup_{x\in\Omega}|\operatorname{div} h| \||\nabla_gy|_g
\|^2_{L^\infty[0,T;L^2(\Omega)]}.
\end{gather}
Finally, by combining \eqref{37}, \eqref{38} - \eqref{39},
 we obtain the desired inequality \eqref{40}.
\end{proof}
\smallskip

\noindent\textbf{Step 2.} Estimates for $y_1$
\[
\|\mathscr{A}^{1/2}y_1(t)\|^2_{L^2(\Omega)}
=\|\mathscr{A}^{1/2}e^{-i\mathscr{A}t}y_0\|^2_{L^2(\Omega)}
=\|\mathscr{A}^{1/2}y_0\|^2_{L^2(\Omega)}=\mathrm{constant}.
\]
Therefore,
\begin{equation}\label{44}
\|\mathscr{A}^{1/2}y_1\|^2_{L^\infty[0,T;L^2(\Omega)]}=
\|\mathscr{A}^{1/2}y_0\|^2_{L^2(\Omega)}.
\end{equation}
\smallskip

\noindent\textbf{Step 3.}
Estimates for $y_2$. We shall prove
\begin{equation}\label{43}
\|y_2\|^2_{L^\infty[0,T;H^1_0(\Omega)]}\leq C_K\|\xi\|^2_{L^2(\Sigma)}.
\end{equation}

\begin{proof}
We define a closed and dense operator $L:L^2(\Sigma)\to L^2(Q)$ by
\begin{equation}
(Lf)(t)=\mathscr{A}\int^t_0e^{-i\mathscr{A}(t-\tau)}\mathbf{D} f(\tau){\rm d}\tau.
\end{equation}
Then we can obtain
\begin{equation}
(L^\ast\Phi)(t)=\mathbf{D}^\ast\mathscr{A}\int^t_0e^{-i\mathscr{A}(t-\tau)}
\Phi(\tau){\rm d}\tau,
\end{equation}
where $\Phi=\mathbf{D} f$ .

Let $\eta=\int^t_0e^{-i\mathscr{A}(t-\tau)}\Phi(\tau){\rm d}\tau$,
 then $\eta$ satisfies the equation
\begin{equation}
\begin{gathered}
 \eta_t=i\mathbf{A}^{1/2}\eta+\Phi,\\
 \eta(0)=0.
 \end{gathered}
\end{equation}
As in the proof of Step 1, we can show that
\begin{equation}\label{41}
\|\eta_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
\leq C_{T,5}(\|\Phi\|^2_{L^1[0,T;L^2(\Omega)]}
+\|\mathscr{A}^{1/2}\eta\|^2_{L^\infty[0,T;L^2(\Omega)]}).
\end{equation}
Moreover.
\begin{align}\label{42}
\|\mathscr{A}^{1/2}\eta\|_{L^2(\Omega)}\leq\int^t_0\|\mathscr{A}^{1/2}\Phi(\tau)\|_{L^2(\Omega)}{\rm d}\tau
\leq\|\mathscr{A}^{1/2}\Phi\|_{L^1[0,T;L^2(\Omega)]}.
\end{align}
Combining \eqref{41} and \eqref{42} yields
\begin{equation}
\|\eta_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma)}
\leq C_{T,5}\|\Phi\|^2_{L^1[0,T;H^1_0(\Omega)]}.
\end{equation}
In addition, 
\begin{equation}
\begin{aligned}
 (\mathbf{D}^*\mathscr{A}\eta,f)_{L^2(\Gamma)}
&=(\mathscr{A}\eta,\mathbf{D} f)_{L^2(\Omega)} \\
&=\int_\Gamma[\eta_{\nu_\mathscr{A}}\mathbf{D} f-\eta(\mathbf{D} f)_{\nu_\mathscr{A}}]{\rm d}x
 +\int_\Omega\eta\mathscr{A}(\mathbf{D} f){\rm d}x \\
&=(\eta_{\nu_\mathscr{A}},f)_{L^2(\Gamma)},
 \end{aligned}
\end{equation}
the last equality  holds because of the definition of the operator $\mathbf{D}$
and $\eta\in D(\mathbf{A}^{1/2})$.
Therefore, $\eta_{\nu_\mathscr{A}}=\mathbf{D}^\ast\mathscr{A}\eta=L^\ast\Phi$.
It tell us that
\begin{equation}
L^\ast\in\mathscr{L}(L^1[0,T;H^1_0(\Omega)]\to L^2(\Sigma)).
\end{equation}
And then, we have
\begin{equation}
L\in\mathscr{L}(L^2(\Sigma)\to L^\infty[0,T;H^{-1}(\Omega)]).
\end{equation}

Let $K$ be defined by
\begin{equation}
Kf=\mathscr{A}^{-1}Lf;
\end{equation}
then $K\in\mathscr{L}(L^2(\Sigma)\to L^\infty[0,T;H^1_0(\Omega)])$.
Since $K\xi=y_2$, we obtain
\begin{equation}\label{45}
\|y_2\|^2_{L^\infty[0,T;H^1_0(\Omega)]}\leq C_K\|\xi\|^2_{L^2(\Sigma)}.
\end{equation}
Thus, inequality \eqref{43} holds.
\end{proof}
\smallskip

\noindent\textbf{Step 4.} 
Combining \eqref{44} and \eqref{45}, we find
\begin{equation} \label{54}
\begin{aligned}
\|\mathscr{A}^{1/2}y \|^2_{L^\infty[0,T;L^2(\Omega)]}
&=\|\mathscr{A}^{1/2}y_1+\mathscr{A}^{1/2}y_2\|^2_{L^\infty[0,T;L^2(\Omega)]} \\
&\leq 2\|\mathscr{A}^{1/2}y_1\|^2_{L^\infty[0,T;L^2(\Omega)]}
 +2\|y_2\|^2_{L^\infty[0,T;H^1_0(\Omega)]} \\
&\leq 2\|\mathscr{A}^{1/2}y_0\|^2_{L^2(\Omega)}+2C_K\|\xi\|^2_{L^2(\Sigma)}.
\end{aligned}
\end{equation}
By substituting inequality \eqref{54} into \eqref{40} and recalling that
$\mathbf{D}\in\mathscr{L}(L^2(\Gamma)\to L^2(\Omega))$, the desired result of
Proposition \ref{prop4.2} is found.

Now, we are ready to complete the proof of the observability inequality 
\eqref{46}.
Combining the results of Lemmas \ref{lem4.3} and \ref{lem4.6} with 
$\epsilon=\frac{C_{T,1}}{2C_{T,3}}$, we obtain
\begin{equation} \label{47}
\begin{aligned}
 C_{T,1}E(0)&\leq 4\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}
+2C(\|z_0\|^2_{L^2(\Omega)}+\|z_1\|^2_{L^2(\Omega)}) \\
&\quad +4(\frac{8C_{T,3}}{C_{T,1}}+1)C_{K,D,T}\|ka(x)
(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha)}.
\end{aligned}
\end{equation}
From Lemma \ref{lem4.4}, we find
\begin{equation} \label{48}
\begin{aligned}
 \int^{T+\alpha}_{-\alpha}\|ka(x)(\mathscr{A}z)_{\nu_\mathscr{A}}
\|^2_{L^2(\Gamma)}{\rm d}t
&\leq a\|k\|^2_{L^\infty(\Gamma)}
\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha)} \\
&\leq a\|k\|^2_{L^\infty(\Gamma)}C_{T,2}E(0).
\end{aligned}
\end{equation}
Thus, if
\[
\|k\|^2_{L^\infty(\Gamma)}< k_0=\frac{C^2_{T,1}}
{4aC_{T,2}C_{K,D,T}(8C_{T,3}+C_{T,1})},
\]
 where $a=\sup_{x\in\Gamma} {|a(x)|^2}$, by combining \eqref{47}
and \eqref{48}, we obtain
\begin{equation} \label{49}
\begin{aligned}
 C_TE(0)&\leq\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}
+C\big(\|z_0\|^2_{L^2(\Omega)}+\|z_1\|^2_{L^2(\Omega)}\big) \\
 &\leq\|(\mathscr{A}z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}
+\|(\mathscr{A}^2z)_{\nu_\mathscr{A}}\|^2_{L^2(\Sigma^\alpha_1)}+CL(z).
\end{aligned}
\end{equation}
Again, the lower terms in the right-hand side of inequality \eqref{49}
can be absorbed by using the following compactness-uniqueness argument:
\smallskip

\noindent\textbf{Step 1.} 
Let $U=\{z\in H^3(Q):  z$ is a solution to problem \eqref{5} satisfying 
$(\mathscr{A}z)_{\nu_\mathscr{A}}|_{\Sigma^\alpha_1}
=(\mathscr{A}^2z)_{\nu_\mathscr{A}}|_{\Sigma^\alpha_1}=0\}$. Then
\begin{equation}\label{52}
 U = {0}.
\end{equation}
Indeed, from inequality \eqref{49}, we have
$$
CL(z)\geq C_TE(0) \quad \text{for all } z\in U ,
$$
which implies that any bounded closed set in $U\cap H^3(Q)$ is compact 
in $H^3(Q)$. Then $U$ is a finite-dimensional linear space.

For any $ z\in U$, we have $z_t\in U$.
 Then $\partial_t:U\to U $ is a linear operator. Let $U\neq {0}$, then
 $\partial_t$ has at least one eigenvalue $\lambda$. Assume that $v \neq 0$ 
is one of its eigenfunctions, then $v_t=\lambda v$. 
Further, $v$ is a nonzero solution to the  problem
\begin{equation}\label{51}
\begin{gathered}
 \mathscr{A}^2v=-\lambda^2v-\mathbf{D}(ka(x)(\mathscr{A}v)_{\nu_{\mathscr{A}}})
\quad\text{on }\Omega,\\
 v=\mathscr{A}v=(\mathscr{A}v)_{\nu_\mathscr{A}}=(\mathscr{A}^2v)_{\nu_\mathscr{A}}=0
\quad \text{on }\Gamma_1.
 \end{gathered}
\end{equation}
Let $\Psi=\mathbf{A}^{1/2}v$, using relations \eqref{50} and \eqref{51}, 
it is easy to find that $\Psi$ satisfies the  problem
\begin{equation}
\begin{gathered}
 \mathscr{A}^2\Psi=-\lambda^2\Psi \quad \text{on }\Omega,\\
 \Psi=\Psi_{\nu_\mathscr{A}}=\mathscr{A}\Psi=(\mathscr{A}\Psi)_{\nu_\mathscr{A}}=0
\quad \text{on }\Gamma_1. 
 \end{gathered}
\end{equation}
By Proposition \ref{prop4.1}, the above problem only has zero solution, 
thus $\Psi\equiv0$ in $\Omega\cup\Gamma_1$, i.e., 
$\mathscr{A}v\equiv0$ in $\Omega\cup\Gamma_1$. Moreover, 
$v|_\Gamma=0$, therefore we can obtain $v\equiv0$, this contradiction 
shows that \eqref{52} holds.

Since the subsequent proof is basically the same as that in 
Step 2 of Lemma \ref{lem4.5}, 
we omit it.

Finally, we obtain
\begin{equation}\label{53}
 \int_{\Sigma^\alpha_1}[(\mathscr{A}z)_{\nu_\mathscr{A}}^2
+(\mathscr{A}^2z)_{\nu_\mathscr{A}}^2]{\rm d}\Sigma\geq C_TE(0).
\end{equation}
Introducing the new variable $\tilde{z}=z(t-\alpha)$ into \eqref{53} yields
\begin{equation}
\int^{T+2\alpha}_0\int_{\Gamma_1}[(\mathscr{A}\tilde{z})_{\nu_\mathscr{A}}^2
+(\mathscr{A}^2\tilde{z})_{\nu_\mathscr{A}}^2]{\rm d}\Sigma\geq C_TE(0).
\end{equation}
Since both $\tilde{z}$ and $z$ are solutions to the same problem \eqref{5}, 
the inequality \eqref{46} holds with $T$ replaced by $T+2\alpha$.
\end{proof}

\subsection*{Acknowledgements}
The author would like to thank the anonymous referee for careful reading
of the manuscript, and for the constructive comments.
 
This work is supported by National Natural Science Foundation (NNSF) 
of China under Grant nos 61573342 and 61473126.

This article is a modified and extended version of an Invited
session paper at the 34th Chinese Control Conference held in 2015.

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\end{document}