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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 31, pp. 1--13.\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-2015/31\hfil Exact controllability of a wave equation]
{Exact controllability problem of a wave equation in non-cylindrical domains}

\author[H. Wang, Y. He, S. Li \hfil EJDE-2015/31\hfilneg]
{Hua Wang, Yijun He, Shengjia Li}

\address{Hua Wang \newline
School of Mathematical Sciences,
Shanxi University, Taiyuan 030006, China}
\email{197wang@163.com}

\address{Yijun He \newline
School of Mathematical Sciences, 
Shanxi University, 
Taiyuan 030006, China}
\email{heyijun@sxu.edu.cn}

\address{Shengjia Li (corresponding author)\newline
School of Mathematical Sciences, Shanxi University,
Taiyuan 030006, China}
\email{shjiali@sxu.edu.cn}

\thanks{Submitted December 6, 2014. Published January 30, 2015.}
\subjclass[2000]{35L05, 93B05}
\keywords{Exact controllability;  non-cylindrical domain;\hfill\break\indent 
 Hilbert uniqueness method}

\begin{abstract}
 Let $\alpha: [0,  \infty)\to(0,  \infty)$ be a twice continuous differentiable
 function which satisfies that $\alpha(0)=1$, $\alpha'$ is monotone and
 $0<c_1\le \alpha'(t)\le c_2<1$ for some constants $c_1, c_2$.
 The exact controllability  of a one-dimensional wave equation in a
 non-cylindrical domain is proved.  This equation characterizes small
 vibrations of a string with one of its endpoint fixed and the other moving
 with speed $\alpha'(t)$.  By using the Hilbert Uniqueness Method,
 we obtain the exact controllability results of this equation with
 Dirichlet boundary control on one endpoint.  We also give an estimate
 on the controllability time that depends only on $c_1$ and $c_2$.
\end{abstract}

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

\section{Introduction and main results}

Suppose $\alpha: [0,  \infty)\to(0,  \infty)$ is a twice continuous
differentiable function satisfying the following assumptions:
\begin{itemize}
\item[(A1)] $0<c_1\le \alpha'(t)\le c_2<1$ for all $0\le t<\infty$;

\item[(A2)] $\alpha'$ is monotone;

\item[(A3)] $\alpha(0)=1$.
\end{itemize}
Let $T>0$.  We define the non-cylindrical domain $\widehat{Q}_{T}^{\alpha}$ by
\[
  \widehat{Q}_{T}^{\alpha}=\{(y,  t)\in\mathbb{R}^{2}: 0<y<\alpha(t),
   t \in (0,  T)\}.
\]
 This article concerns the exact controllability of the one-dimensional
 wave equation
 \begin{equation}\label{11}
   \begin{gathered}
  u_{tt}(y, t)-u_{yy}(y, t)=0,   \quad (y, t)\in \widehat{Q}_{T}^{\alpha}, \\
   u(0, t)=0, \quad  u(\alpha(t), t)=v(t), \quad t\in  (0,  T),\\
  u(y, 0)=u^{0}(y), \quad  u_{t}(y, 0)=u^{1}(y),  \quad y\in (0, 1),
  \end{gathered}
 \end{equation}
where $v\in L^2(0, T)$ and $(u^{0},  u^{1})\in L^{2}(0,  1)\times H^{-1}(0,  1)$. 
Since $\sup_{t\in(0, T)}|\alpha'(t)|<1$, by \cite{MM1}, 
the system of \eqref{11} admits a unique solution in the sense of transposition. 
Here, as in \cite{MM2},
$u\in L^{\infty}(0, T;L^2(0, \alpha(t))$ is called a solution by 
transposition of problem \eqref{11} if $u$ verifies
\begin{equation}
\begin{aligned}
&\int_0^T\int_0^{\alpha(t)}u(y, t)\hat{h}(y, t)\,dy\,dt \\
& =  \int_0^1[u^1(y)\theta(y, 0)-u^0(y)\theta_t(y, 0)]dy
- \int_0^T v(t)\theta_y(\alpha(t), t)dt,
\end{aligned}
\end{equation}
for all $\hat{h}\in L^1(0, T; L^2(0, \alpha(t))$,
where $\theta$ is the weak solution of the  problem
\begin{equation}\label{25}
\begin{gathered}
\theta_{tt}(y, t)-\theta_{yy}(y, t) = \hat{h},   \quad
  (y, t)\in \widehat{Q}_{T}^{\alpha}, \\
\theta(0, t)= \theta(\alpha(t), 0 )=0,  \quad t\in (0, T), \\
\theta(T)=\theta'(T)=0,  \quad x \in (0, 1).
  \end{gathered}
\end{equation}

The exact controllability problem of system \eqref{11} is stated as follows.

\begin{definition}\label{def1.1} \rm
  We say  system  \eqref{11} is exactly controllable at time $T$,  
if for any $(u^0, u^1)\in L^2(0, 1)\times H^{-1}(0, 1)$, 
$(u^0_d,  u^1_d)\in L^2(0,  \alpha(T))\times H^{-1}(0,  \alpha(T))$,  
there exists $v\in L^2(0, T)$  such that the solution by transposition $u$ 
of \eqref{11}  satisfies
$u(T)=u^0_d$ and $u_{t}(T)=u^1_d$.
\end{definition}

For a function $\alpha$ satisfying conditions (A1)--(A3), 
 we define 
  \begin{gather}\label{T*}
T^*=\frac{1}{c_2}\big\{\exp\big(\frac{2c_2^2(1-c_1)(1+c_2)}{c_1(1-c_2)^2}\big)
-1\big\}, \\
\label{T^*1}
  T_1^*=\frac{1}{c_2}\big\{\exp\big(\frac{2c_2^2(1-c_1)}{c_1(1-c_2)^3}\big)-1\big\}.
\end{gather}
One of the main results of this article as follows.

\begin{theorem}\label{thm1}
  For any given $T>T^*$,  \eqref{11} is exactly controllable at time $T$.
\end{theorem}

  Similarly, for the exact controllability problem, when the control is 
acting on the fixed endpoint,
\begin{equation}
 \begin{gathered}
  u_{tt}(y, t)-u_{yy}(y, t)=0,  \quad (y, t)\in \widehat{Q}_{T}^{\alpha}, \\
  u(0,  t)=v(t), \ u(\alpha(t),  t)=0,  \quad t\in (0,  T),\\
  u(y, 0)=u^{0}(y), \quad  u_{t}(y, 0)=u^{1}(y),  \quad y\in (0, 1),
  \end{gathered} \label{11'}
\end{equation}
 we have the following result.

\begin{theorem}\label{thm2}
  For any given $T>T_1^*$,  \eqref{11'} is exactly controllable at time $T$.
\end{theorem}

\begin{remark} \rm
When $\alpha(t)=1+kt$ for some constant $k\in(0, 1)$,  $T^*$ is 
reduced to $T_k^*$ defined in \cite{CLG},  and Theorem \ref{thm1} 
is reduced to \cite[Theorem 1.1]{CLG}.
\end{remark}

\begin{remark} \rm
 Theorem \ref{thm2} extends the results in \cite{CS1} and \cite{CS2}.
 In fact, when $\alpha(t)=1+kt$, an exact controllability result of 
system \eqref{11'} has been proved for $0<k<1-\frac{1}{\sqrt{e}}$
in \cite{CS1} and for $0<k<1-\frac{2}{1+e^2}$ in \cite{CS2}. 
We also note that the controllability time $T_1^*$ given here is better 
than the constants $T_k^*$ in \cite{CS1} and \cite{CS2} in this case.
\end{remark}

\begin{remark} \rm
  We note that there are many functions $\alpha(t)$ satisfying conditions 
(A1)--(A3) but are not the form $1+kt$, for example $\alpha(t)=1+(t+\arctan t)/c$
 where $c$ is any constant that is greater than 2.
\end{remark}

Recently, several works on the  controllability problems of wave equations
 in non-cylindrical domains have been published.  The existence 
of solutions of the initial boundary value problem for the nonlinear 
wave equation in non-cylindrical domains has been studied  
in \cite{BCo,M}. The controllability problem for a multi-dimensional
 wave equation in a non-cylindrical domain has been investigated in 
\cite{BCh,MM1,MM2}.  About the one-dimension cases, there have been
 extensive study of the controllability problem in a non-cylindrical domain.
 We refer the reader to \cite{AAM,CLG,CS1,CS2}.

When $\alpha(t)=1+kt$ for some constant $0<k<1$,  in \cite{CLG},  the exact 
controllability of the system \eqref{11} has been acquired. 
 When $\alpha(t)=1+kt$, Cui and Song obtained that the system \eqref{11'}
 is exactly controllable for $0<k<1-\frac{1}{\sqrt{e}}$ in \cite{CS1} and 
is exactly controllable for $0<k<1-\frac{2}{1+e^2}$ in \cite{CS2}.

There are also other results on the exact controllability problem for wave
 equations of variable coefficients in cylindrical domains, 
see \cite{FYZ,MM2,Yao,Zhang} and the references therein. 
So, our first aim is to transform \eqref{11} and \eqref{11'} into 
wave equations with variable coefficients in a cylindrical domain.

Let $x=\frac{y}{\alpha(t)}$ and $w(x, t)=u(y,  t)=u(\alpha(t)x, t)$  
for $(y, t)\in\widehat{Q}^{\alpha}_{T}$. Then, it is straightforward 
to show that $(x,  t)$ varies in $Q_T:=(0, 1)\times(0, T)$ and \eqref{11} 
is transformed into the wave equation with variable coefficients,
\begin{equation}\label{12}
  \begin{gathered}
    w_{tt}-\big[\frac{\beta(x, t)}{\alpha(t)}w_{x}\big]_{x}
+\frac{\gamma(x, t)}{\alpha(t)}w_{tx}
    +\frac{\tau(x, t)}{\alpha(t)}w_{x}=0,   \quad \text{in }  Q_T, \\
    w(0, t)=0, \quad w(1, t)=v(t)  \quad  t\in (0,  T),\\
    w(x, 0)=w^{0}(x), \quad  w_{t}(x, 0)=w^{1}(x),  \quad x\in (0,  1),
  \end{gathered}
  \end{equation}
where $\beta(x, t)=\frac{1-\alpha'^2(t)x^{2}}{\alpha(t)}$, 
 $\gamma(x, t)=-2\alpha'(t)x$,  $\tau(x,  t)=-\alpha''(t)x$,  
$w^{0}=u^{0}$,   $w^{1}=u^{1}+\alpha'(0)xu^{0}_{x}$.

From \cite{MM2},  we know that for $(u^{0}, u^{1})\in L^{2}(0, 1)\times H^{-1}(0, 1) $
 and $v\in L^{2}(0, T)$, \eqref{12} admits a unique solution 
$w\in C([0, T]; L^{2}(0, 1))\cap C^{1}([0, T]; H^{-1}(0, 1))$ in the sense of 
transposition, where $w$ is called a solution by transposition of
 problem \eqref{12} if
\begin{align*}
&\int_0^T\int_0^1 wh \,dx\,dt \\
& =  \int_0^1[-w^0(x)z_t(x, 0)+\alpha'(0)w^0(x)z(x, 0)+w'(x)z(x, 0)]dx \\
&\quad - \int_0^T\beta(1, t)z_x(1, t)v(t)dt 
 +\int_0^1[\gamma_x(x, 0)w^0(x)z(x, 0)+\gamma(x, 0)w^0_x(x)z(x, 0)]dx,
  \end{align*}
for every $h\in L^1(0, T; L^2(0, 1))$ and $z$ is the weak solution of 
the  problem
\begin{equation}\label{22}
  \begin{gathered}
   L^*z = h,  \quad \text{in}\  Q_T, \\
    z(0, t)=z(1, t)=0, \quad t\in  (0, T), \\
    z(x, T)=z_{t}(x, T)=0,  \quad x\in (0, 1),
  \end{gathered}
\end{equation}
where the formal adjoint $L^*$ of $L$ is defined by
\begin{equation}
 L^*z = \alpha(t)z_{tt}-[\beta(x, t)z_{x}]_{x}+\gamma(x, t)z_{xt}+\tau(x, t)z_{x}.
\end{equation}

Thus, Theorem \ref{thm1} can be restated as the following exact controllability
 result for  equation \eqref{12}.

\begin{theorem}\label{t2}
  For any $T>T^*$ where $T^*$ is given by \eqref{T*}, any 
$(w^0, w^1)\in L^2(0, 1)\times H^{-1}(0, 1)$ and 
$(w^0_d, w^1_d)\in L^2(0, 1)\times H^{-1}(0, 1)$,  we can always find 
a control $v\in L^2(0, T)$ such that the corresponding solution by 
transposition $w$ of \eqref{12} satisfies
  $w(T)=w^0_d$,  $w_t(T)=w^1_d$.
\end{theorem}

Similarly, \eqref{11'} can be transformed into the
wave equation with  variable coefficients,
\begin{equation}
\begin{gathered}
    w_{tt}-\big[\frac{\beta(x, t)}{\alpha(t)}w_{x}\big]_{x}
+\frac{\gamma(x, t)}{\alpha(t)}w_{tx}
    +\frac{\tau(x, t)}{\alpha(t)}w_{x}=0,   \quad \text{in }  Q_T, \\
    w(0, t)=v(t), \quad w(1, t)=0,  \quad t\in (0,  T),\\
    w(x, 0)=w^{0}(x), \quad w_{t}(x, 0)=w^{1}(x),   \quad x\in (0,  1),
  \end{gathered} \label{12'}
\end{equation}
and Theorem \ref{thm2} can be restated as the following exact controllability
 result for  equation\eqref{12'}.

\begin{theorem}\label{t3}
  For any $T>T^*_1$ where $T^*_1$ is given by \eqref{T^*1}, any 
$(w^0, w^1)\in L^2(0, 1)\times H^{-1}(0, 1)$ and 
$(w^0_d, w^1_d)\in L^2(0, 1)\times H^{-1}(0, 1)$,  we can always find a 
control $v\in L^2(0, T)$ such that the corresponding solution by transposition
 $w$ of \eqref{12'} satisfies
  $w(T)=w^0_d$,  $w_t(T)=w^1_d$.
\end{theorem}

\section{Description of the Hilbert uniqueness method}

In this section, we describe the Hilbert uniqueness method which is 
used in the proof of  Theorems \ref{t2} and \ref{t3}.
 Next, we consider Theorem \ref{t2} in detail.

Firstly, for any $(w^0_d,  w^1_d)\in L^2(0,  1)\times H^{-1}(0,  1)$,   
 the  system
\begin{equation}
  \begin{gathered}
    \alpha(t)\xi_{tt}-\big[\beta(x,  t)\xi_{x}\big]_{x}+\gamma(x, t)\xi_{tx}
    +\tau(x,  t)\xi_{x}=0,  \quad \text{in } Q_T, \\
    \xi(0, t)=0,  \quad  \xi(1, t)=0, \quad t\in  (0,  T), \\
    \xi(x, T)=w^0_d(x),  \quad \xi_{t}(x, T)=w^1_d(x),  \quad x\in (0,  1)
  \end{gathered}
\end{equation}
has a unique solution 
$\xi \in C([0,  T]; L^2(0,  1))\cap C^1([0,  T]; H^{-1}(0,  1))$ 
in the sense of transportation.

Secondly, for any $(z^0,  z^1)\in H_0^1(0, 1)\times L^2(0, 1)$, we solve
\begin{equation}\label{13}
  \begin{gathered}
   \alpha(t)z_{tt}-[\beta(x, t)z_{x}]_{x}+\gamma(x, t)z_{xt}+\tau(x, t)z_{x}=0,  
\quad \text{in }  Q_T, \\
    z(0, t)=z(1, t)=0, \quad t\in  (0, T), \\
    z(x, 0)=z^{0}(x), \quad z_{t}(x, 0)=z^{1}(x),  \quad x\in (0, 1),
  \end{gathered}
\end{equation}
 and 
\begin{equation}\label{115}
  \begin{gathered}
    \alpha(t)\eta_{tt}-\big[\beta(x, t)\eta_{x}\big]_{x}+\gamma(x, t)\eta_{tx}
    +\tau(x,  t)\eta_{x}=0,  \quad \text{in}\ Q_T, \\
    \eta(0, t)=0, \quad  \eta(1, t)=z_{x}(1, t),  \quad t\in  (0, T), \\
    \eta(x, T)=0, \quad \eta_{t}(x, T)=0,   \quad x\in(0, 1).
  \end{gathered}
\end{equation}
Then we define a linear operator 
$\Lambda: H^{1}_0(0, 1)\times L^{2}(0, 1)\to H^{-1}(0, 1)\times L^{2}(0, 1)$,
by
$$
(z^{0},  z^{1})\mapsto (\eta_{t}(\cdot, 0)+\gamma(\cdot, 0)\eta_{x}(\cdot, 0)
-\alpha'(0)\eta(\cdot,  0),  -\eta(\cdot, 0)),
$$
Lastly, the problem is reduced to prove  the existence of some 
 $(z^0,  z^1)\in H^1_0(0,  1)\times L^2(0,  1)$ such that
 \begin{equation}\label{116}
   \Lambda (z^0,  z^1)=([w^1-\xi_t(0)]-\alpha'(0)[w^0-\xi(0)]
+\gamma(0)[w^0_x-\xi_x(0)] -[w^0-\xi(0)]).
 \end{equation}

To solve \eqref{116}, we observe that
 \begin{equation}\label{4.3}
\int_0^1\beta(1,  t)|z_x(1, t)|^2dt=\langle\Lambda(z^0,  z^1), 
 (z^0,  z^1)\rangle_{H^{-1}(0, 1)\times L^2(0,  1), H_0^1(0,  1)\times L^2(0,  1)}.
\end{equation}

 In section 3, we  prove the following observability inequality for
 system \eqref{13}:   there exists a constant $C>0$ such that
  \begin{equation}\label{21}
\int_0^T\beta(1, t)|z_x(1, t)|^2dt
\ge C\big(\|z^0\|^2_{H_0^1(0, 1)}+\|z^1\|^2_{L^2(0, 1)}\big).
\end{equation}
Also, we prove that $\Lambda$ is a bounded linear operator; i.e., 
there exists a constant $C>0$ such that
\begin{equation}\label{117}
    \int^{T}_0\beta(1,  t)|z_{x}(1,  t)|^{2}dt
\leq C (||z^{0}||^{2}_{H^{1}_0(0,  1)}+||z^{1}||^{2}_{L^{2}(0,  1)}).
  \end{equation}
Combining \eqref{21}, \eqref{117} and the Lax-Milgram Theorem, we can 
show that $\Lambda$ is an isomorphism.

Then, the equation\eqref{116} has a unique solution  
$(z^0,  z^1)\in H^1_0(0,  1)\times L^2(0,  1)$, and the function $z_x(1,t)$
is the desired control such that the solution $w$ of \eqref{12}  satisfies 
 $w(T)=w^0_d$,  $w_t(T)=w^1_d$.

For the proof of Theorem \ref{t3}, the steps are similar to those of  
Theorem \ref{t2}.
In this case, instead of \eqref{115}, we consider the following homogeneous 
wave equation
\begin{equation}
  \begin{gathered}
    \alpha(t)\eta_{tt}-\big[\beta(x, t)\eta_{x}\big]_{x}+\gamma(x, t)\eta_{tx}
    +\tau(x,  t)\eta_{x}=0,  \quad \text{in}\ Q_T, \\
    \eta(0, t)=z_x(0, t), \quad  \eta(1, t)=0,  \quad t\in  (0, T), \\
    \eta(x, T)=0, \quad \eta_{t}(x, T)=0,  \quad x\in (0, 1),
  \end{gathered}
\end{equation}
and  define a linear operator $\Lambda$ just same as \eqref{116}, then we 
observe that
\begin{equation}
\int_0^1\beta(0,  t)|z_x(0, t)|^2dt
=-\langle\Lambda(z^0,  z^1),  (z^0,  z^1)\rangle_{H^{-1}(0, 1)\times L^2(0,  1), H_0^1(0,  1)\times L^2(0,  1)}.
\end{equation}
 We omit the details of the proof here.

\section{Observability estimates}

The main purpose of this section is to prove the observability inequalities 
for system \eqref{13}.
To prove those estimates, we need some technical lemmas.

From \cite{MM2}, we know that: for any $(z^0,  z^1)\in H_0^1(0, 1)\times L^2(0, 1)$, 
the equation \eqref{13} has a unique weak solution 
$z\in C([0, T]; H_0^1(0, 1))\cap C^1([0, T]; L^2(0, 1))$ in the sense of 
transportation.

The energy for \eqref{13} is defined as
\begin{equation}
 E(t)=\frac{1}{2}\int^{1}_0[\alpha(t)|z_{t}(x, t)|^{2}
 +\beta(x, t)|z_{x}(x, t)|^{2}]dx,  \quad \text{for }   t\geq0,
\end{equation}
where $z$ is the solution of \eqref{13}.
Since $\alpha(0)=1$, we have
\begin{equation}
E_0:=E(0)=\frac12\int_0^1\big[|z^1(x)|^2+\beta(x, 0)|z^0_x(x)|^2\big]dx.
\end{equation}

First, we prove a lemma which is related to the decay rate of the energy 
$E(t)$.

\begin{lemma} \label{lm3.1}
If $\alpha(0)=1$,  $0<c_1\le \alpha'(t)\le c_2<1$ and $\alpha'$ is monotone,  
then
\begin{equation}\label{14}
\frac{c_3E_0}{\alpha(t)}\le E(t)\le\frac{c_4E_0}{\alpha(t)},
\end{equation}
where
\begin{equation}\label{c3c4}
(c_3,  c_4)=\begin{cases}
\big(\frac{1-c_2}{1-c_1},  \frac{c_2}{c_1}\big), 
  & \text{if $ \alpha'$ is increasing}, \\[4pt]
\big(\frac{c_1}{c_2},  \frac{1-c_1}{1-c_2}\big),  & \text{if $\alpha'$ is decreasing}.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof}
For any $0<t\leq T$, through multiplying the first equation of 
\eqref{13} by $z_{t}$ and integrating the result on $(0, 1)\times (0, t)$,  
we conclude that
\begin{align*}
  0 
&=\int_0^{t}\int_0^{1}\big\{\alpha(s)z_{tt}(x, s)z_{t}(x, s)
-[\beta(x, s)z_{x}(x, s)]_{x}z_{t}(x, s)\\
 &\quad +\gamma(x, s)z_{xt}(x, s)z_{t}(x, s)
 +\tau(x, s)z_{x}(x, s)z_{t}(x, s)\big\}\,dx\,ds\\
&:=I_1+I_2+I_3+I_4,
\end{align*}
where
\begin{gather*}
I_{1}=\frac{1}{2}\int^{1}_0\alpha(s)|z_{t}(x, s)|^{2}dx\big|^{t}_0
-\frac{1}{2}\int^{t}_0\int^{1}_0\alpha'(s)|z_{t}(x, s)|^{2}\,dx\,ds,\\
\begin{aligned}
I_2 
&= \frac12\int_0^1\beta(x,  s)|z_x(x, s)|^2dx\big|^t_0
 -\frac12\int_0^t\int_0^1\beta_t(x, s)|z_x(x,  s)|^2\,dx\,ds\\
&= \frac12\int_0^1\beta(x,  s)|z_x(x, s)|^2dx\big|^t_0
+\frac12\int_0^t\int_0^1\frac{\alpha'(s)}{\alpha(s)}\beta(x, s)|z_x(x,  s)|^2\,dx\,ds\\
&\quad +\int_0^t\int_0^1\frac{\alpha'(s)\alpha''(s)}{\alpha(s)}x^2
 |z_x(x,  s)|^2\,dx\,ds,
\end{aligned}
\\
  I_{3}=\int^{t}_0\int^{1}_0\alpha'(s)|z_{t}(x, s)|^{2}\,dx\,ds, \\
I_{4}=-\int^{t}_0\int^{1}_0x\alpha''(s)z_{x}(x, s)z_{t}(x, s)\,dx\,ds.
\end{gather*} 
We thereby obtain:
\begin{align*}
E(t)&= E_0-\int_0^t\frac{\alpha'(s)}{\alpha(s)}E(s)ds
-\int_0^t\int_0^1\frac{\alpha'(s)\alpha''(s)}{\alpha(s)}x^2|z_x(x,  s)|^2\,dx\,ds \\
&\quad +\int_0^t\int_0^1\alpha''(s)xz_x(x, s)z_t(x,  s)\,dx\,ds.
\end{align*}
\begin{equation}
E'(t)=-\frac{\alpha'(t)}{\alpha(t)}E(t)
-\int_0^1\frac{\alpha'(t)\alpha''(t)}{\alpha(t)}x^2|z_x(x,  t)|^2dx
+\int_0^1\alpha''(t)xz_x(x, t)z_t(x,  t)dx.
\end{equation}
We subdivide the proof into two cases:
\smallskip

(1) $\alpha'$ is increasing; that is, $\alpha''(t)\ge0$. 
 By using the inequalities
\begin{align*}
&-\frac{\alpha'(t)\alpha''(t)}{2\epsilon(t)\alpha(t)}x^2|z_x(x, t)|^2
-\frac{\epsilon(t)\alpha(t)\alpha''(t)}{2\alpha'(t)}|z_t(x, t)|^2\\
&\le \alpha''(t)xz_x(x, t)z_t(x, t)\\
&\le \frac{\alpha'(t)\alpha''(t)}{2\alpha(t)}x^2|z_x(x, t)|^2
+\frac{\alpha(t)\alpha''(t)}{2\alpha'(t)}|z_t(x, t)|^2,
\end{align*}
where $\epsilon(t)=\frac{\alpha'(t)}{1-\alpha'(t)}$,  we easily obtain
\begin{equation}
-(\frac{\alpha'(t)}{\alpha(t)}+\frac{\alpha''(t)}{1-\alpha'(t)})E(t)
\le E'(t)\le -(\frac{\alpha'(t)}{\alpha(t)}-\frac{\alpha''(t)}{\alpha'(t)})E(t),
\end{equation}
so
\begin{equation}
\frac{(1-\alpha'(t))E_0}{(1-\alpha'(0))\alpha(t)}
\le E(t)
\le\frac{\alpha'(t)E_0}{\alpha'(0)\alpha(t)}.
\end{equation}
Using $0<c_1\le \alpha'(t)\le c_2<1$,  we conclude that
\begin{equation}
\frac{c_3E_0}{\alpha(t)}\le E(t)\le\frac{c_4E_0}{\alpha(t)},
\end{equation}
where $c_3=\frac{1-c_2}{1-c_1}$,  $c_4=\frac{c_2}{c_1}$.
\smallskip

(2) $\alpha'$ is decreasing; that is, $\alpha''(t)\le0$. By using the inequalities
\begin{align*}
&\frac{\alpha'(t)\alpha''(t)}{2\alpha(t)}x^2|z_x(x, t)|^2
+\frac{\alpha(t)\alpha''(t)}{2\alpha'(t)}|z_t(x, t)|^2\\
&\le \alpha''(t)xz_x(x, t)z_t(x, t)\\
&\le -\frac{\alpha'(t)\alpha''(t)}{2\epsilon(t)\alpha(t)}x^2|z_x(x, t)|^2
-\frac{\epsilon(t)\alpha(t)\alpha''(t)}{2\alpha'(t)}|z_t(x, t)|^2,
\end{align*}
where $\epsilon(t)=\frac{\alpha'(t)}{1-\alpha'(t)}$,  we easily get
\begin{equation}
-(\frac{\alpha'(t)}{\alpha(t)}-\frac{\alpha''(t)}{\alpha'(t)})E(t)
\le E'(t)\le -(\frac{\alpha'(t)}{\alpha(t)}+\frac{\alpha''(t)}{1-\alpha'(t)})E(t),
\end{equation}
so
\begin{equation}
\frac{\alpha'(t)E_0}{\alpha'(0)\alpha(t)}\le E(t)
\le\frac{(1-\alpha'(t))E_0}{(1-\alpha'(0))\alpha(t)}.
\end{equation}
Using $0<c_1\le \alpha'(t)\le c_2<1$,  we conclude that
\begin{equation}
\frac{c_3E_0}{\alpha(t)}\le E(t)\le\frac{c_4E_0}{\alpha(t)},
\end{equation}
where $c_3=\frac{c_1}{c_2}$,  $c_4=\frac{1-c_1}{1-c_2}$.
\end{proof}

\begin{remark} \rm
When $\alpha''(t)\equiv0$,  that is,  $\alpha(t)=1+kt$ for some 
constant $k\in(0, 1)$,  then $c_3=c_4=1$,
Lemma \ref{lm3.1} is reduced to Lemma 3.1 in \cite{CLG}.
\end{remark}

Next, similar to the proof of \cite[Lemma 3.2]{CLG}, we can get  
the following estimate for each weak solution $z$ of \eqref{13} by the 
multiplier method.

\begin{lemma}
  For any function $q\in C^{1}([0, 1])$,  the solution $z$ of \eqref{13} 
satisfies the  estimate
\begin{equation} \label{17}
\begin{aligned}
  &\frac{1}{2}\int^{T}_0\beta(x, t)q(x)|z_{x}(x, t)|^{2}dt\Big|^{1}_0 \\
  &=\frac{1}{2}\int^{T}_0\int^{1}_0q'(x)[\alpha(t)|z_t(x, t)|^2
 +\beta(x, t)|z_x(x, t)|^2]\,dx\,dt \\
  &\quad -\int^{T}_0\int^{1}_0\alpha'(t)q(x) z_{x}(x, t)z_{t}(x, t)\,dx\, dt
-\frac{1}{2}\int^{T}_0\int^{1}_0\beta_{x}(x, t)q(x)|z_{x}(x, t)|^{2}\,dx\,dt  \\
  &\quad +\int^{1}_0[\alpha(t) q(x) z_{x}(x, t)z_{t}(x, t)
-x\alpha'(t)q(x)|z_{x}(x, t)|^{2}]dx\Big|^{T}_0.
\end{aligned} %\label{multi}
 \end{equation}
\end{lemma}

Finally, we derive the continuity estimate.

\begin{theorem}\label{t1}
  Assume $T>0$,  for any $(z^{0}, z^{1})\in H^{1}_0(0, 1)\times L^{2}(0, 1)$,  
there exists a constant $C>0$ such that the solution of \eqref{13} satisfies 
the following two estimates:
  \begin{gather}\label{119}
    \int^{T}_0\beta(1,  t)|z_{x}(1,  t)|^{2}dt
\leq C (||z^{0}||^{2}_{H^{1}_0(0,  1)}+||z^{1}||^{2}_{L^{2}(0,  1)}), \\
\label{118}
  \int_0^T\beta(0,  t)|z_x(0,  t)|^2dt
\leq C (||z^{0}||^{2}_{H^{1}_0(0,  1)}+||z^{1}||^{2}_{L^{2}(0,  1)});
\end{gather}
  so $z_x(0, \cdot)\in L^2(0, T)$ and $z_x(1, \cdot)\in L^2(0, T)$.
\end{theorem}

 \begin{proof}
First, we prove inequality \eqref{119}.
Let $q(x)=x$ for $x\in [0, 1]$ in \eqref{17} and noticing that 
$\beta_x(x, t)=-\frac{2\alpha'^2(t)x}{\alpha(t)}$,
$\gamma(x, t)=-2\alpha'(t)x$,  it follows that
\begin{equation} \label{112}
\begin{aligned}
& \frac12\int_0^T\beta(1, t)|z_x(1, t)|^2dt \\
&=\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)xz_t(x, t)z_x(x, t)\,dx\,dt \\
&\quad +\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x^2|z_x(x, t)|^2\,dx\,dt \\
&\quad +\int_0^1[\alpha(t)xz_t(x, t)z_x(x, t)-\alpha'(t)x^2|z_x(x, t)|^2]dx\Big|_0^T.
\end{aligned} %\label{beta1t}
\end{equation}

 We estimate every terms on the right side of \eqref{112}. 
 By the assumption for $\alpha$,  we have $1\le\alpha(t)\le 1+c_2T$ 
and $0<\frac{1-c_2^2}{1+c_2T}\le\beta(x, t)\le1$ for any $(x, t)\in Q_T$, 
 these inequalities together with \eqref{14} and the boundedness 
of $\alpha'(t)$ imply
\begin{equation}
\begin{aligned}
&\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)xz_t(x, t)z_x(x, t)\,dx\,dt \\
&+\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x^2|z_x(x, t)|^2\,dx\,dt \\
&\le \int_0^TE(t)dt+C\int_0^T\int_0^1[|z_t(x, t)|^2+|z_x(x, t)|^2]\,dx\,dt \\
&\le \int_0^TE(t)dt+C\int_0^T\int_0^1[\alpha(t)|z_t(x, t)|^2+\beta(x, t)|z_x(x, t)|^2]\,dx\,dt \\
&\le CE_0.
\end{aligned} \label{CE0}
\end{equation}
For each $t\in[0,  T]$ and $\epsilon(t)>0$,  it holds that
\begin{align*}
&\Big|\int_0^1[\alpha(t)xz_t(x,  t)z_x(x,  t)-\alpha'(t)x^2|z_x(x,  t)|^2]dx\Big|\\
&\leq  \int_0^1[\alpha(t)|z_t(x,  t)||z_x(x,  t)|+\alpha'(t)|z_x(x, t)|^2]dx\\
&\leq  \frac{1}{2\epsilon(t)}\int^1_0\alpha^2(t)|z_t(x,  t)|^2dx
 +\frac{\epsilon(t)}{2}\int^1_0|z_x(x,  t)|^2dx
 +\int^1_0\alpha'(t)|z_x(x,  t)|^2dx\\
&\leq \frac{\alpha(t)}{2\epsilon(t)}\int^1_0\alpha(t)|z_t(x,  t)|^2dx
 + [\frac{\epsilon(t)}{2}+\alpha'(t)]
\frac{\alpha(t)}{1-\alpha'^2(t)}\int^1_0\beta(x,  t)|z_x(x,  t)|^2dx.
\end{align*}
 Choosing $\epsilon(t)=1-\alpha'(t)$,  then it is easy to see
$$
\epsilon(t)>0\ \text{and}\  \frac{\alpha(t)}{\epsilon}
=[\frac{\epsilon}{2}+\alpha'(t)]\frac{2\alpha(t)}{1-\alpha'^2(t)}
  =\frac{\alpha(t)}{1-\alpha'(t)}.
$$
This implies that
$$
\Big|\int_0^1[\alpha(t)xz_t(x, t)z_x(x, t)-\alpha'(t)x^2|z_x(x, t)|^2]dx\Big|
\le \frac{\alpha(t)}{1-\alpha'(t)}E(t)\le\frac{\alpha(t)}{1-c_2}E(t).
 $$
Then,  using \eqref{14},  it follows that
\begin{equation}\label{c5}
\Big|\int_0^1[\alpha(t)xz_t(x, t)z_x(x, t)-\alpha'(t)x^2|z_x(x, t)|^2]dx\Big|_0^T
\Big| \le c_5E_0,
\end{equation}
where $c_5=\frac{2c_4}{1-c_2}$.
Therefore,  combining \eqref{112},  \eqref{CE0} and \eqref{c5},  it follows that
$$
\int_0^T\beta(1,  t)|z_x(1,  t)|^2dt\le CE_0
\le C\big(\|z^0\|^2_{H_0^1(0,  1)}+\|z^1\|^2_{L^2(0,  1)}\big).
$$

Next, we prove the inequality \eqref{118}.
  Let $q(x)=x-1$ for $x\in [0, 1]$ in \eqref{17} and noticing that 
$\beta_x(x, t)=-\frac{2\alpha'^2(t)x}{\alpha(t)}$,
$\gamma(x, t)=-2\alpha'(t)x$,  it follows that
\begin{equation} \label{112*}
\begin{aligned}
&\frac12\int_0^T\beta(0, t)|z_x(0, t)|^2dt \\
&=\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)(x-1)z_t(x, t)z_x(x, t)\,dx\,dt \\
&\quad +\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x(x-1)|z_x(x, t)|^2\,dx\,dt \\
&\quad +\int_0^1[\alpha(t)(x-1)z_t(x, t)z_x(x, t)-\alpha'(t)x(x-1)|z_x(x, t)|^2]dx
\Big|_0^T. 
\end{aligned}%\label{beta2t}
\end{equation}

 Through estimating every terms on the right side of \eqref{112*},  
similar to the derive of \eqref{CE0},  it follows that
\begin{equation} \label{327}
\begin{aligned}
&\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)(x-1)z_t(x, t)z_x(x, t)\,dx\,dt \\
&+\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x(x-1)|z_x(x, t)|^2\,dx\,dt
\leq CE_0.
\end{aligned}
\end{equation}
Since
\begin{align*}
 & \Big|\int_0^1[\alpha(t)(x-1)z_t(x,  t)z_x(x,  t)-\alpha'(t)x(x-1)|z_x(x,  t)|^2]dx
\Big|\\
 & \leq  \int_0^1[\alpha(t)|z_t(x,  t)||z_x(x,  t)|+\alpha'(t)|z_x(x,  t)|^2]dx,
\end{align*}
 similar to the derive of \eqref{c5},  it follows that
\begin{equation} \label{325}
  \Big|\int_0^1[\alpha(t)(x-1)z_t(x, t)z_x(x, t)
-\alpha'(t)x(x-1)|z_x(x, t)|^2]dx\big|^T_0\Big|\leq c_5E(0),
\end{equation}
where $c_5=\frac{2c_4}{1-c_2}$.

From \eqref{112*}, \eqref{327} and \eqref{325},  it follows that
\[
  \int^T_0\beta(0,  t)|z_x(0,  t)|^2dt\leq CE_0
\leq(||z^0||^2_{H^1_0(0,  1)}+||z^1||^2_{L^2(0,  1)}).
\]
\end{proof}

Now, we give the proof of the observability inequalities.

\begin{theorem}\label{t4}
  For  $T>T^*$ where $T^*$ satisfies \eqref{T*} and any 
$(z^0,  z^1)\in H_0^1(0,  1)\times L^2(0,  1)$,  there exists a constant
 $C>0$ such that the corresponding solution of \eqref{13} satisfies
  \begin{equation}
\int_0^T\beta(1, t)|z_x(1, t)|^2dt\ge C\big(\|z^0\|^2_{H_0^1(0, 1)}
+\|z^1\|^2_{L^2(0, 1)}\big).
\end{equation}
\end{theorem}

\begin{proof}
 If we choose $\epsilon(t)=\frac{\alpha'(t)}{1+\alpha'(t)}$,  
then it is obvious that
 $$
0<\epsilon(t)<\frac12, \quad \text{and}\quad
 1-\epsilon(t)=1+\Big(2-\frac{1}{\epsilon(t)}\Big)
\frac{\alpha'^2(t)}{1-\alpha'^2(t)}
 =\frac{1}{1+\alpha'(t)}. 
$$
Thus,  using 
$$
x^2=\frac{\alpha(t)x^2}{1-\alpha'^2(t)x^2}\beta(x, t)
\le\frac{\alpha(t)}{1-\alpha'^2(t)}\beta(x, t)
$$
and \eqref{14},  it follows that
\begin{equation}
\begin{aligned}
&\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)xz_t(x, t)z_x(x, t)\,dx\,dt \\
&+\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x^2|z_x(x, t)|^2\,dx\,dt \\
&\ge \int_0^T\int_0^1\frac{1-\epsilon(t)}{2}\alpha(t)|z_t(x, t)|^2\,dx\,dt \\
&\quad +\int_0^T\int_0^1\big\{\frac12\beta(x, t)
+(1-\frac{1}{2\epsilon(t)})\frac{\alpha'^2(t)}{\alpha(t)}x^2\big\}
 |z_x(x, t)|^2\,dx\,dt \\
&\ge \int_0^T\int_0^1\frac{1-\epsilon(t)}{2}\alpha(t)|z_t(x, t)|^2\,dx\,dt \\
&\quad +\int_0^T\int_0^1\big[1+\frac{(2-\frac{1}{\epsilon(t)})
 \alpha'^2(t)}{1-\alpha'^2(t)}\big]\frac12\beta(x, t)
|z_x(x, t)|^2\,dx\,dt \\
&=\int_0^T\frac{1}{1+\alpha'(t)}E(t)dt\\
&\ge c_6E_0\int_0^T\frac{1}{\alpha(t)}dt,
\end{aligned}  \label{c6}
\end{equation}
where $c_6=c_3/(1+c_2)$.
By \eqref{112},  \eqref{c5} and \eqref{c6},  we obtain
\begin{align*}
\frac12\int_0^T\beta(1, t)|z_x(1, t)|^2dt 
&\ge c_6E_0\int_0^T\frac{1}{1+c_2t}dt-c_5E_0\\
&= \big(\frac{c_6}{c_2}\log(1+c_2T)-c_5\big)E_0.
\end{align*}
If we choose $T^*$ as in \eqref{T*},  then it is easy to see that
\begin{equation}
  T^*=\frac{1}{c_2}\big\{\exp\big(\frac{c_2c_5}{c_6}\big)-1\big\},
\end{equation}
 so for $T>T^*$,
$$
\int_0^T\beta(1, t)|z_x(1, t)|^2dt \ge C\big(\|z^0\|^2_{H_0^1(0, 1)}
+\|z^1\|^2_{L^2(0, 1)}\big),
$$
holds for  $C=\frac{2c_6}{c_2}\log(1+c_2T)-2c_5>0$.
\end{proof}

\begin{theorem}\label{t6}
  For $T>T^*_1$ where $T^*_1$ satisfies \eqref{T^*1} and any 
$(z^0,  z^1)\in H_0^1(0,  1)\times L^2(0,  1)$,  there exists a constant 
$C>0$ such that the corresponding solution of \eqref{13} satisfies
  \begin{equation}\label{58}
\int_0^T\beta(0,  t)|z_x(0,  t)|^2dt
\geq C (||z^{0}||^{2}_{H^{1}_0(0,  1)}+||z^{1}||^{2}_{L^{2}(0,  1)}).
\end{equation}
\end{theorem}

\begin{proof}
Choosing $\epsilon(x,  t)=\frac{\alpha'(t)(1-\alpha'(t)x)}{\alpha(t)}$,  
it is easy to see that
\begin{align*}
 &   |\alpha'(t)z_x(x,  t)z_t(x,  t)|\\
 & \leq   \frac{\alpha'^2(t)}{2\epsilon(x,  t)}|z_t(x,  t)|^2
 +\frac{\epsilon(x,  t)}{2}|z_x(x,  t)|^2\\
 &=  \frac{\alpha'(t)}{1-\alpha'(t)x}\frac{\alpha(t)}{2}|z_t(x,  t)|^2+
           \frac{\alpha'(t)}{1+\alpha'(t)x}\frac{\beta(x,  t)}{2}|z_x(x,  t)|^2.
\end{align*}
 Since $x-1\leq 0$ for $x\in[0, 1]$, we have
 \begin{equation} \label{330}
\begin{aligned}
&\int_0^TE(t)dt-\int_0^T\int_0^1\alpha'(t)(x-1)z_t(x,  t)z_x(x,  t)\,dx\,dt\\
&\quad +\int_0^T\int_0^1\frac{\alpha'^2(t)}{\alpha(t)}x(x-1)|z_x(x,  t)|^2\,dx\,dt \\
&\geq \int_0^TE(t)dt+\int_0^T\int_0^1\frac{\alpha'(t)(x-1)}{1-\alpha'(t)x}
 \frac{\alpha(t)}{2}|z_t(x,  t)|^2\,dx\,dt\\
&\quad +\frac{\alpha'(t)(x-1)}{1+\alpha'(t)x}\frac{\beta(x,  t)}{2}
 |z_x(x,  t)|^2\,dx\,dt \\
&\quad +\int_0^T\int_0^1\frac{2\alpha'^2(t)x(x-1)}{1-\alpha'^2(t)x^2}
 \frac{\beta(x,  t)}{2}|z_x(x,  t)|^2\,dx\,dt \\
&= \frac12\int^T_0\int^1_0\frac{1-\alpha'(t)}{1-\alpha'(t)x}
 [\alpha(t)|z_t(x,  t)|^2+\beta(x,  t)|z_x(x,  t)|^2]\,dx\,dt \\
&  \geq  \int^T_0(1-\alpha'(t))E(t)dt \\
&\geq c_6^*E_0\int^T_0\frac{1}{\alpha(t)}dt,
 \end{aligned}
\end{equation}
 where $c_6^*=(1-c_2)c_3$.

 From \eqref{112*},  \eqref{325} and \eqref{330},  we arrive at
 \begin{align*}
   &  \frac12\int^T_0\beta(0,  t)|z_x(0,  t)|^2dt  \geq  c_6^*E_0\int^T_0\frac{1}{\alpha(t)}dt-c_5E_0\\
   & \geq  c_6^*E_0\int^T_0\frac{1}{1+c_2t}dt-c_5E_0=(\frac{c_6^*}{c_2}\log(1+c_2T)-c_5)E_0.
 \end{align*}
 If we choose $T^*_1$ as in \eqref{T^*1}, then it is easy to see that
 $$
T^*_1=\frac{1}{c_2}\big\{\exp(\frac{c_2c_5}{c^*_6})-1\big\};
$$
thus,  when $T>T^*_1$,
 \[
   \int^T_0\beta(0,  t)|z_x(0,  t)|^2dt  
\ge C\big(\|z^0\|^2_{H_0^1(0,  1)}+\|z^1\|^2_{L^2(0,  1)}\big).
\]
 holds for  $C=\frac{2c_6^*}{c_2}\log(1+c_2T)-2c_5>0$.
\end{proof}

\subsection*{Acknowledgments}
The first author was partially supported by the grant No. 61403239 of the NSFC.
The second author was partially supported by the grant No. 11401351 of the NSFC
and by the grant No. 2014011005-2 of the Science Foundation of Shanxi Province, 
China.
The third author was partially supported by the grant No. 61174082 of the NSFC,
and by the grant No. 2011-006 of Shanxi Scholarship Council of China.

 \begin{thebibliography}{99}

\bibitem{AAM} F.  D.  Araruna,  G.  O.  Antunes, L.  A.  Medeiros;
  \emph{Exact controllability for the semilinear string equation in the 
non cylindrical domains},  Control cybernet,  \textbf{33}(2004),  237-257.

\bibitem{BCh} C.  Bardos, G.  Chen;
\emph{Control and stabilization for the wave equation,  part III: 
domain with moving boundary},  SIAM J. Control Optim.,  \textbf{19}(1981),  123-138.

\bibitem{BCo} C.  Bardos, J.  Cooper;
\emph{A nonlinear wave equation in a time dependent domain},  
J.  Math.  Anal.  Appl.,  \textbf{42}(1973),  29-60.

\bibitem{CLG}L. Cui,  X. Liu, H. Gao;
\emph{Exact controllability for a one-dimensional wave equation in 
non-cylindrical domains},  J.  Math.  Anal.  Appl.,  \textbf{402} (2013),  612-625.

\bibitem{CS1} L. Cui, L. Song;
\emph{Controllability for a wave equation with moving boundary},  
J.  Appl.  Math.,  vol.  2014,  Article ID 827698,  6 pages,  2014.  
doi:10. 1155/2014/827698

\bibitem{CS2} L. Cui, L. Song;
\emph{Exact controllability for a wave equation with fixed boundary control}, 
Boundary Value Problems 2014, 2014:47.

\bibitem{FYZ} X.  Fu,  J.  Yong,  X.  Zhang;
 \emph{Exact controllability for multidimensional semilinear hyperbolic equations}, 
 SIAM J.  Control Optim.,  \textbf{46}(2007),  1578-1614.

\bibitem{M} L.  A.  Medeiros;
 \emph{Nonlinear wave equations in domains with variable boundary}, 
 Arch.  Rat.  mech.  Anal.,  \textbf{47}(1972),  47-58.

\bibitem{MM1} M.  Milla Miranda;
\emph{Exact controllability for the wave equation in domains with variable boundary},
  Rev.  Mat.  Univ.,  \textbf{9} (1996),  435-457.

\bibitem{MM2} M.  Milla Miranda;
\emph{HUM and the wave equation with variable coefficients},  
Asymptot.  Anal.,  \textbf{11} (1995),  317-341.

\bibitem{Yao} P.  Yao;
\emph{On the observability inequalities for exact controllability of wave 
equations with variable coefficients},  SIAM J.  Control Optim., 
 \textbf{37}(1999),  1568-1599.

\bibitem{Zhang} X.  Zhang;
\emph{A unified controllability/observability theory for some stochastic 
and deterministic partial differential equations},  in: 
Proceedings of the International Congress of Mathematicians,  Vol.  IV, 
 Hindustan Book Agency,  New Delhi,  2010,  pp.  3008-3034.

\end{thebibliography}

\end{document}