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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 101, 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}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/101\hfil Exact controllability for a wave equation]
{Exact controllability for a wave equation with mixed boundary conditions 
 in a non-cylindrical domain}

\author[L. Cui, H. Gao \hfil EJDE-2014/101\hfilneg]
{Lizhi Cui, Hang Gao}  % in alphabetical order

\address{Lizhi Cui \newline
College of Applied Mathematics, Jilin University of
Finance and Economics, Changchun 130117, China.\newline
School of Mathematics and Statistics, Northeast Normal
University, Changchun 130024, China}
\email{cuilz924@126.com}

\address{Hang Gao \newline
School of Mathematics and Statistics, Northeast Normal University, 
Changchun 130024, China}
\email{hangg@nenu.edu.cn}

\thanks{Submitted October 15, 2013. Published April 11, 2014.}
\subjclass[2000]{58J45, 35L05}
\keywords{Exact controllability; wave equation; mixed boundary conditions;
\hfill\break\indent non-cylindrical domain}

\begin{abstract}
 In this article we study the exact controllability of  a one-dimen\-sional
 wave equation with mixed boundary conditions in a non-cylindrical domain.
 The fixed endpoint has a Dirichlet-type boundary condition, while the moving
 end has a Neumann-type condition. When the speed of the moving endpoint
 is less than the characteristic speed, the exact controllability of
 this equation is established by Hilbert Uniqueness Method.
 Moreover, we shall give the explicit dependence of the controllability time
 on the speed of the moving endpoint.
\end{abstract}

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

\section{Introduction and statement of main results}

Given $T>0$. For any $0<k<1$, set 
\begin{equation} \label{H1}
\alpha_k(t)=1+kt\quad\text{for } t\in [0, T].
\end{equation}
Also, define the non-cylindrical domain
$$
\widehat{Q}_T^k=\{(y,t)\in \mathbb{R}^2;\ 0< y< \alpha_k(t),\; t\in [0,T]\}
$$
and write 
$$
V(0,\alpha_k(t))=\{\varphi\in H^1(0,\alpha_k(t));\varphi(0)=0\}\quad \text{for } 
t \in [0, T],
$$
which is a subspace of $H^1(0,\alpha_k(t))$ and we denote
by $[V(0,\alpha_k(t))]'$ its conjugate space.

Consider the  wave equation
\begin{equation}\label{b1}
\begin{gathered}
 u_{tt}-u_{yy}=0 \quad \text{in } \widehat{Q}_T^k,\\
u(0,t)=0,\quad u_y(\alpha_k(t),t)=v(t) \quad \text{on } (0,T), \\
u(y,0)=u^0(y),\quad u_t(y,0)=u^1(y)\quad \text{in } (0,1),
\end{gathered}
\end{equation}
where  $v$ is the control variable, $u$ is the state variable and 
$(u^0,u^1)\in L^2(0,1)\times [V(0,1)]'$ is
any given initial value. By  \cite{C1999} and \cite{MMM1996}, 
it is easy to check that
the equation \eqref{b1} has a unique solution $u$ by transposition
$$
u\in C([0,T];L^2(0,\alpha_k(t)))\cap C^1([0,T];[V(0,\alpha_k(t))]').
$$


The main purpose  of this article is  to study the exact controllability 
of \eqref{b1}. There are numerous publications on the 
controllability problems of wave equations in a cylindrical domain.
However, there are only a few works on the exact controllability for
wave equations defined in non-cylindrical domains. We refer to
\cite{CG1981,FGL2004,MMM1996,c2013,c2014} for some known results in this respect.
In \cite{CG1981},  the exact
controllability of a multi-dimensional wave equation with constant coefficients in
a non-cylindrical domain was established, while a control entered the system through
the whole non-cylindrical domain.
In \cite{FGL2004,MMM1996,c2013,c2014},   some  controllability results for
 the wave equations with Dirichlet boundary conditions in suitable non-cylindrical 
domains were  investigated, respectively.

In \cite{FGL2004} and \cite{MMM1996},  exact
controllability of a wave equation  in certain non-cylindrical domain was studied.
 But in the one-dimensional case, some  conditions on the moving boundary 
were required, e.g.
\begin{equation} \label{b333}
\int_0^{\infty}|\alpha_k'(t)|dt<\infty.
\end{equation}
In \cite{c2013} and \cite{c2014}, the exact Dirichlet   
boundary controllability of the following systems
were discussed,
\begin{gather*}%\label{b12}
 u_{tt}-u_{yy}=0 \quad\text{in }\widehat{Q}_T^k,\\
u(0,t)=0,\quad u(\alpha_k(t),t)=v(t)\quad \text{on } (0,T), \\
u(y,0)=u^0,\quad u_t(y,0)=u^1 \quad\text{in } (0,1),
\end{gather*}
and
\begin{gather*} %\label{b11}
 u_{tt}-u_{yy}=0 \quad \text{in } \widehat{Q}_T^k,\\
u(0,t)=v(t),\quad u(\alpha_k(t),t)=0 \quad \text{on } (0,T), \\
u(y,0)=u^0,\quad u_t(y,0)=u^1 \quad\text{in } (0,1),
\end{gather*}
In \cite{c2013, c2014} and in this research,
we deal with the different case. It is easy to check that the condition
\begin {equation*}
\label{b33}
\int_0^{\infty}|\alpha_k'(t)|dt=\infty
\end{equation*}
is satisfied on the moving boundary

To overcome these difficulties, in this article, 
we transform \eqref{b1} into an equivalent wave equation with variable
coefficients in the cylindrical domain and establish the exact controllability 
of this equation by  Hilbert Uniqueness Method.
In \cite{C1999}, the Neumann boundary controllability problem for
a multi-dimensional wave equation with variable  coefficients in a 
cylindrical domain was studied.
However in \cite{C1999}, in the one-dimensional case,
the condition \eqref{b333} was required.
In this paper, the key point is to construct a different adjoint equation.
Then we define weighted energy function  for
this adjoint equation
and characterize  the energy explicitly (see \eqref{E1}).

Throughout this article, we  set
\begin{equation} \label{*}
T_k^{*}=\frac{{e}^\frac{2k(1+k)}{1-k}-1}{k}.
\end{equation}
The main result of this paper is stated as follows.

\begin{theorem} \label{thm1}
For any given  $T>T^{*}_k$, the equation \eqref{b1} is exactly controllable 
at the time $T$; i.e., for any initial value 
$(u^0,u^1)\in L^2(0,1)\times [V(0,1)]'$ and target
$(u_d^0,u_d^1)\in L^2(0,\alpha_k(T))\times[V(0,\alpha_k(T))]'$,
there exists a control
$v\in [H^1(0,T)]'$ such that the corresponding solution $u$
of  $\eqref {b1}$ satisfies
\begin{equation} \label{b2}
u(T)=u_d^0\quad\text{and}\quad   u_{t}(T)=u_d^1.
\end{equation}
\end{theorem}

\begin{remark} \label{rmk1.1}\rm
It is easy to check that
$$
T^{*}_0:=\lim_{k\to0}T^{*}_k=
\lim_{k\to0}\frac{{e}^\frac{2k(1+k)}{1-k}-1}{k}=2.
$$
It is well known that the wave equation $\eqref{b1}$ in the cylindrical domain  
is exactly controllable at any time $T>T^*_0$. As we know, $T^*_0$ is sharp. 
However, we do not know whether the controllability time $T^*_k$ is sharp.
\end{remark}

To establish the exact controllability of \eqref{b1},
we first transform \eqref{b1} into  an equivalent wave equation with variable
coefficients in a cylindrical domain.
To this aim,  for any $(y, t)\in \widehat{Q}_T^k$,
set $y=\alpha_k(t)x$ and $u(y,t)=u(\alpha_k(t)x,t)=w(x,t)$.
Then it is easy to check that \eqref{b1} is transformed into the  wave equation
\begin{equation} \label{b3}
\begin{gathered}
w_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}w_x]_x
+[\frac{\gamma_k(x)}{\alpha_k(t)}]w_{tx}=0 \quad \text{in } Q,\\
w(0,t)=0,\quad w_x(1,t)=\overline{v}(t)\quad \text{on } (0,T),\\
w(x,0)=w^0(x), \quad w_{t}(x,0)=w^1(x)\quad \text{in } (0,1),
\end{gathered}
\end{equation}
where
\begin{equation}\label{b4}
\begin{gathered}
Q=(0, 1)\times(0, T),\quad
 \overline{v}(t)=\alpha_k(t)v(t),\quad
 \beta_k(x,t)=\frac{1-k^2x^2}{\alpha_k(t)},\\
\gamma_k(x)=-2kx, w^0=u^0 \quad \text{and } w^1=u^1+kxu^0_x.
\end{gathered}
\end{equation}
By a method similar to the one used in \cite{C1999}, 
it is easy to check that the equation \eqref{b3}
has a unique solution $w$ by transposition
$$ 
w\in C([0,T];L^2(0,1))\cap C^1([0,T];[V(0,1)]').
$$
Moreover, the exact controllability
of  \eqref{b1} (Theorem \ref{thm1}) is reduced
to the following exact controllability result for \eqref{b3}.

\begin{theorem} \label{thm2}
Suppose that $T>T^{*}_k$.
Then for any initial value $(w^0, w^1)\in L^2(0,1)\times [V(0,1)]'$
and  target $(w_d^0,w_d^1)\in L^2(0,1)\times [V(0,1)]'$,
there exists a control
$\overline{v} \in [H^1(0,T)]'$ such that the corresponding solution $w$
of \eqref {b3} satisfies 
$$
w(T)=w_d^0\quad\text{and}\quad w_{t}(T)=w_d^1.
$$
\end{theorem}

To prove Theorem \ref{thm2},
we adopt  Hilbert Uniqueness Method. The key is  to define a weighted 
 energy function for  a wave equation
with variable coefficients in cylindrical domains.

The rest of this paper is organized as follows.
In Section 2, we  derive an explicit energy equality for a wave equation
 with variable coefficients in cylindrical domains and further deduce
two key inequalities for this equation.  Section 3 is devoted to a proof 
of  Theorem \ref{thm2}.


\section{Two inequalities for the wave equation with variable coefficients}

First we introduce some notation.  
Denote by $|\cdot|$ and $\|\cdot\|$ the norms of the spaces $L^2(0,1)$
and  $V(0,1)$, respectively.  Also, we use $L^2$, $V$ and $V'$ 
to represent the spaces $L^2(0,1), V(0,1)$ and $[V(0,1)]'$, respectively.  
Denote by  $\langle\cdot,\cdot\rangle$ the duality product between
the  linear space $F$ and its dual space $F'$.

Consider the  wave equation with variable coefficients
\begin{equation} \label{2.10}
\begin{gathered}
\alpha_k(t) z_{tt}-[\beta_k(x,t)z_x]_x+\gamma_k(x)z_{tx}=0
\quad\text{in } Q,\\
z(0,t)=0,\quad \beta_k(1,t)z_x(1,t)-\gamma_k(1)z_{t}(1,t)=0 \quad\text{on } (0,T),\\
z(x,0)=z^0(x),\quad z_{t}(x,0)=z^1(x)\quad \text{in } (0,1),
\end{gathered}
\end{equation}
where  $(z^0, z^1)\in V\times L^2$ is any given initial value, 
and $\alpha_k$, $\beta_k$ and
$\gamma_k$ are the functions given in \eqref{b4}. By a similar method 
in  \cite{C1999} and \cite{QR},
it is easy to check that \eqref{2.10}
has a unique solution $z$ by transposition
\begin{equation*}
z\in C([0,T];V)\cap C^1([0,T];L^2).
\end{equation*}
Define the following  energy function for  \eqref{2.10},
\begin{equation*}
E(t)=\frac{1}{2}\int_0^1[\alpha_k(t)|z_{t}(x,t)|^2
+\beta_k(x,t)|z_x(x,t)|^2]dx \quad\text{for } t\in [0, T],
\end{equation*}
where $z$ is the solution of \eqref{2.10}. It follows that
\begin{equation*}
E_0\triangleq E(0)=\frac{1}{2}\int_0^1[|z^1(x)|^2+\beta_k(x,0)|z^0_x(x)|^2]dx.
\end{equation*}

To prove Theorem \ref{thm2}, we need the following two key inequalities.

\begin{theorem}\label{thm3}
For any $T>0$, there exists a positive constant $C_1$ depending only only $T$, 
such that solutions $z$ of  \eqref{2.10} satisfy
\begin{equation} \label{3.1}
\int_0^{T} |z_{t}(1,t)|^2dt
\leq C_1(\|z^0\|^2+|z^1|^2)\quad\text{for any } (z^0, z^1)\in V\times L^2.
\end{equation}
\end{theorem}

\begin{theorem}\label{thm4}
Suppose that $T>T^{*}_k$.
Then there exists a positive constant $C_2$ depending only on $T$,
such that solutions $z$ of  \eqref{2.10} satisfy
\begin{equation}\label{3.2}
\int_0^{T} |z_{t}(1,t)|^2dt
\geq C_2(\|z^0\|^2+|z^1|^2)\quad\text{for any } (z^0, z^1)\in V\times L^2.
\end{equation}
\end{theorem}

First, we prove two lemmas, which will be used in the proofs of these inequalities.
The first lemma is related to an equivalent expression of the energy $E(t)$.

\begin{lemma} \label{lemE}
Suppose that $z$ is any solution of \eqref{2.10}. Then we have
\begin{equation} \label{E1}
E(t)=\frac{1}{\alpha_k(t)}E_0-\frac{k}{\alpha_k(t)}
\int_0^t\alpha_k(s)|z_{t}(1,s)|^2ds, \quad
0\leq t \leq T.
\end{equation}
\end{lemma}


\begin{proof}
Multiplying both sides of the first equation of \eqref{2.10} by $z_{t}$ 
and integrating on $(0,1)\times(0,t)$, we obtain
\begin{align*}
0&=\int_0^t\int_0^1 \big\{\alpha_k(s)z_{tt}(x,s)z_{t}(x,s)
-[\beta_k(x,s)z_x(x,s)]_{x}z_{t}(x,s) \\
&\quad +\gamma_k(x)z_{tx}(x,s)z_{t}(x,s)\big\}dx\,ds\\
&\triangleq  J_1+J_2+J_3.
\end{align*}

Next, we  calculate the above three integrals.
It is easy to check that
\begin{equation}\label{**}
\begin{split}
J_1&=\int_0^t\int_0^1 \frac{1}{2}\alpha_k(s)[|z_t(x,s)|^2]_t\,dx\,ds\\
&=\frac{1}{2}\int_0^1 \alpha_k(s)|z_t(x,s)|^2dx\big|_0^t
-\frac{k}{2}\int_0^t\int_0^1
|z_t(x,s)|^2 \,dx\,ds.
\end{split}
\end{equation}
Further, by the second equation of \eqref{2.10}, it holds that
\begin{align*}
J_2&= -\int_0^t\beta_k(x,s)z_{x}(x,s)z_{t}(x,s)ds\big|_0^1
+\int_0^t\int_0^1\beta_k(x,s)z_{x}(x,s)z_{xt}(x,s)\,dx\,ds
\\
&=- \int_0^t \beta_k(x,s)z_{x}(x,s)z_{t}(x,s)ds\big|_0^1
 +\frac{1}{2}\int_0^1 \beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\
&\quad -\frac{1}{2}\int_0^t\int_0^1\beta_{k,t}(x,s)|z_{x}(x,s)|^2\,dx\,ds
\\
&=-\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds +\frac{1}{2}\int_0^1
\beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\
&\quad -\frac{1}{2}\int_0^t\int_0^1\beta_{k,t}(x,s)|z_{x}(x,s)|^2\,dx\,ds.
\end{align*}
By \eqref{b4}, it is obvious that
$$
\beta_{k,t}(x,t)=-\frac{k(1-k^2x^2)}{(1+kt)^2}=-\frac{k}{(1+kt)}\beta_k(x,t).
$$
This implies that
\begin{equation}\label{A263}
\begin{split}
J_2
&=-\int_0^t\int_0^1[\beta_k(x,s)z_x(x,s)]_{x}z_{t}(x,s)\,dx\,ds\\
&=-\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds
+\frac{1}{2}\int_0^1 \beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\
&\quad +\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1\beta_k(x,s)|z_{x}(x,s)|^2\,dx\,ds.
\end{split}
\end{equation}
Further, by the definition of $\gamma_k$, we find that
\begin{align*}
J_3
&= \frac{1}{2}\int_0^t \gamma_k(x)|z_{t}(x,s)|^2ds\big|_0^1
-\frac{1}{2} \int_0^t\int_0^1\gamma_{k,x}(x)|z_{t}(x,s)|^2\,dx\,ds\\
&=\frac{1}{2}\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds
-\frac{1}{2} \int_0^t\int_0^1\gamma_{k,x}(x)|z_{t}(x,s)|^2\,dx\,ds.
\end{align*}
Since $\gamma_{k,x}(x)=-2k$, it follows that
\begin{equation}\label{A265}
J_3=\frac{1}{2}\int_0^t
\gamma_k(1)|z_{t}(1,s)|^2ds+
k\int_0^t\int_0^1|z_{t}(x,s)|^2\,dx\,ds.
\end{equation}
By \eqref{**}-\eqref{A265} and the definition of $E(t)$, we see that
\begin{align*}
E(t)&=E_0+\frac{1}{2}\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds
-\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1 \beta_k(x,s)|z_x(x,s)|^2 \,dx\,ds\\
&\quad -\frac{k}{2}\int_0^t\int_0^1 |z_{t}(x,s)|^2\,dx\,ds\\
&=E_0-\int_0^tk|z_{t}(1,s)|^2ds
-\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1 \beta_k(x,s)|z_x(x,s)|^2 \,dx\,ds
\\
&\quad -\frac{1}{2}\int_0^t\frac{k}{(1+ks)}
 \int_0^1 \alpha_k(x,s)|z_{t}(x,s)|^2\,dx\,ds\\
&=E_0-\int_0^t
k|z_{t}(1,s)|^2ds-\int_0^t\frac{k}{(1+ks)}E(s)ds,
\end{align*}
which implies that
\[
E_{t}(t)=-\frac{k}{1+kt}E(t)-k|z_{t}(1,t)|^2,\quad 0\leq t \leq T.
\]
It follows that
\begin{equation*}
[(1+kt)E(t)]_{t}=-k(1+kt)|z_{t}(1,t)|^2,\quad 0\leq t \leq T,
\end{equation*}
which completes the proof of Lemma \ref{lemE}.
\end{proof}

\begin{remark}\label{rmk2.1} \rm
By \eqref{E1}, it is easy to check that
$E(t)<\frac{1}{\alpha_k(t)}E_0<E_0.$
\end{remark}

By the multiplier method, we have the following estimate for any  solution
of \eqref{2.10}.


\begin{lemma}\label{lemF}
Let $q \in C^1([0,1])$. Then any
solution $z$ of \eqref{2.10} satisfies
\begin{equation} \label{A31}
\begin{split}
&[\frac{1}{2}\int_0^{T} \beta_k(x,t)q(x)|z_x(x,t)|^2dt]\big|_0^1
+\frac{1}{2}\int_0^{T}\alpha_k(t)q(1)|z_{t}(1,t)|^2dt\\
&= \frac{1}{2} \int_0^T\!\int_0^1
q_{x}(x)[\alpha_k(t)|z_{t}(x,t)|^2+\beta_k(x,t)|z_x(x,t)|^2]\,dx\,dt\\
&\quad-\int_0^T\!\int_0^1
\alpha_{k,t}(t)q(x)z_{t}(x,t)z_x(x,t)\,dx\,dt\\
&\quad-\frac{1}{2}\int_0^T\!\int_0^1
\beta_{k,x}(x,t)q(x)|z_x(x,t)|^2\,dx\,dt\\
&\quad+\Big\{\int_0^1 [\alpha_k(t)q(x)z_{t}(x,t)z_x(x,t)+
\frac{1}{2}\gamma_k(x)q(x)|z_x(x,t)|^2]dx\Big\}\Big|_0^T.
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
Multiplying  the first equation of  \eqref{2.10} by $qz_{x}$
and integrating on $Q$, we obtain
\begin{align*}
0&= \int_0^T\!\int_0^1 \alpha_k(t)z_{tt}(x,t)q(x)z_{x}(x,t)\,dx\,dt\\
&\quad -\int_0^T\!\int_0^1[\beta_k(x,t)z_x(x,t)]_{x}q(x)z_{x}(x,t)\,dx\,dt
\\
&\quad +\int_0^T\!\int_0^1\gamma_k(x)z_{tx}(x,t)q(x)z_{x}(x,t)\,dx\,dt
\\
&\triangleq L_1+L_2+L_3.
\end{align*}
Now, we  calculate $L_1, L_2$ and $L_3$. 
First, it is easy to check that
\begin{equation}\label{A32}
\begin{split}
L_1&= \int_0^1
\alpha_k(t)q(x)z_{t}(x,t)z_x(x,t)dx\big|_0^{T}-\int_0^T\!\int_0^1
\alpha_{k,t}(t)q(x)z_{t}(x,t)z_x(x,t)\,dx\,dt\\
&\quad-\frac{1}{2}\int_0^{T}\alpha_k(t)q(x)|z_{t}(x,t)|^2dt\big|_0^1
+\frac{1}{2}\int_0^T\!\int_0^1
\alpha_k(t)q_{x}(x)|z_{t}(x,t)|^2\,dx\,dt
\\
&= \int_0^1 \alpha_k(t)q(x)z_{t}(x,t)z_x(x,t)dx\big|_0^{T}
-\int_0^T\!\int_0^1 \alpha_{k,t}(t)q(x)z_{t}(x,t)z_x(x,t)\,dx\,dt\\
&
\quad-\frac{1}{2}\int_0^{T}\alpha_k(t)q(1)|z_{t}(1,t)|^2dt
+\frac{1}{2}\int_0^T\!\int_0^1
\alpha_k(t)q_{x}(x)|z_{t}(x,t)|^2\,dx\,dt.
\end{split}
\end{equation}
Further,
\begin{align*}
L_2&=-\int_0^T\!\int_0^1[\beta_k(x,t)z_x(x,t)]_{x} q(x)z_{x}(x,t)\,dx\,dt\\
&=-\int_0^{T} \beta_k(x,t)q(x)|z_{x}(x,t)|^2dt\big|_0^1\\
&\quad +\int_0^T\!\int_0^1
[\beta_k(x,t)q_{x}(x)|z_{x}(x,t)|^2
+\beta_k(x,t)z_{x}(x,t)q(x)z_{xx}(x,t)]\,dx\,dt
\\
&= -\int_0^{T} \beta_k(x,t)q(x)|z_{x}(x,t)|^2dt\big|_0^1
+\int_0^T\!\int_0^1 \beta_k(x,t)q_{x}(x)|z_{x}(x,t)|^2\,dx\,dt\\
&\quad +\frac{1}{2} \int_0^{T}\beta_k(x,t)q(x)|z_{x}(x,t)|^2dt\big|_0^1
-\frac{1}{2}\int_0^T\!\int_0^1[\beta_k(x,t)q(x)]_x|z_{x}(x,t)|^2\,dx\,dt.
\end{align*}
It follows that
\begin{equation} \label{A33}
\begin{split}
L_2&=-\frac{1}{2}\int_0^{T} \beta_k(x,t)q(x)|z_{x}(x,t)|^2dt\big|_0^1\\
&\quad +\frac{1}{2}\int_0^T\!\int_0^1
[\beta_k(x,t)q_{x}(x)|z_{x}(x,t)|^2
-\beta_{k,x}(x,t)q(x)|z_{x}(x,t)|^2]\,dx\,dt.
\end{split}
\end{equation}
Further,
\begin{equation} \label{A34}
L_3= \frac{1}{2}\int_0^1 \gamma_k(x)q(x)|z_{x}(x,t)|^2dx\big|_0^{T}.
\end{equation}
By \eqref{A32}-\eqref{A34}, we get the desired result in Lemma \ref{lemF}.
\end{proof}

Next, we  prove Theorems \ref{thm3} and \ref{thm4}.


\begin{proof}[Proof of Theorem \ref{thm3}]
Choose $q(x)=x$. Notice that $\alpha_{k,t}(t)=k$,
$\beta_{k,x}(x,t)=\frac{-2k^2x}{1+kt}$ and
$\gamma_k(x)=-2kx$. By \eqref{A31}, it follows that
\begin{equation}\label{A211}
\begin{split}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}\big)
\int_0^{T}\alpha_k(t)|z_{t}(1,t)|^2dt\\
&=\int_0^{T}E(t)dt
-\int_0^T\!\int_0^1kxz_{t}(x,t)z_x(x,t)\,dx\,dt+\int_0^T\!\int_0^1
\frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt\\
&\quad+\Big\{\int_0^1\big[\alpha_k(t)xz_{t}(x,t)z_x(x,t)
-kx^2|z_x(x,t)|^2]dx
\Big\}\Big|_0^T.
\end{split}
\end{equation}

Now, we  estimate the terms in the right-hand side of \eqref{A211}.
Using the Young inequality, we obtain 
\begin{equation}\label{A12} \begin{split}
&\int_0^{T}E(t)dt
+\int_0^T\!\int_0^1 \frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt
-\int_0^T\!\int_0^1kxz_{t}(x,t)z_x(x,t)\,dx\,dt\\
&\leq \int_0^{T}E(t)dt+
\int_0^T\!\int_0^1\frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt\\
&\quad +\frac{1}{2}\int_0^T\!\int_0^1\frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt
+\frac{1}{2}\int_0^T\!\int_0^1\alpha_k(t) |z_{t}(x,t)|^2\,dx\,dt\\
&= \int_0^{T}E(t)dt+ \frac{3}{2}\int_0^T\!\int_0^1
\frac{k^2x^2}{1-k^2x^2}\beta_k(x,t)|z_x(x,t)|^2\,dx\,dt
\\
&\quad +\frac{1}{2}\int_0^T\!\int_0^1\alpha_k(t) |z_{t}(x,t)|^2\,dx\,dt\\
&\leq \int_0^{T}E(t)dt+ \frac{3}{2}\frac{k^2}{1-k^2}\int_0^T\!\int_0^1
\beta_k(x,t)|z_x(x,t)|^2\,dx\,dt
\\
&\quad +\frac{1}{2}\int_0^T\!\int_0^1\alpha_k(t) |z_{t}(x,t)|^2\,dx\,dt\\
&\leq \int_0^{T}E(t)dt+\Big(\frac{3k^2}{1-k^2}+1\Big)
\int_0^{T}E(t)dt\\
&=\frac{2+k^2}{1-k^2} \int_0^{T}E(t)dt.
\end{split}
\end{equation}
Further, for any $t\in[0,T]$ and $0<\varepsilon<1$, by the Young inequality,
it holds that
\begin{align*}
&\big|\int_0^1 [\alpha_k(t)xz_{t}(x,t)z_x(x,t) -kx^2|z_x(x,t)|^2]dx\big|\\
&\leq \sqrt{1+kt}\Big[\frac{1}{2\varepsilon}\int_0^1\alpha_k(t)|z_{t}(x,t)|^2dx
+\frac{\varepsilon}{2}\int_0^1x^2|z_x(x,t)|^2dx\Big]\\
&\quad +k\int_0^1x^2|z_x(x,t)|^2dx\\
&\leq \frac{\sqrt{1+kt}}{2\varepsilon}\int_0^1\alpha_k(t)|z_{t}(x,t)|^2dx+
\Big(\frac{\sqrt{1+kt}}{2}\varepsilon+k\Big)\int_0^1x^2|z_x(x,t)|^2dx\\
&\leq \frac{\sqrt{1+kt}}{\varepsilon}\frac{1}{2}\int_0^1\alpha_k(t)|z_{t}(x,t)|^2dx\\
&\quad + \frac{2\big(\frac{\sqrt{1+kt}}{2}\varepsilon+k\big)(1+kt)}{1-k^2}
\frac{1}{2}\int_0^1\beta_k(x,t)|z_x(x,t)|^2dx.
\end{align*}
Choose 
$$
\varepsilon=\frac{1-k}{\sqrt{1+kt}}.
$$
Then it is easy to check that $\varepsilon\in(0, 1)$ and
\[
\frac{\sqrt{1+kt}}{\varepsilon}=
\frac{2\left(\frac{\sqrt{1+kt}}{2}\varepsilon+k\right)(1+kt)}{1-k^2}
=\frac{1+kt}{1-k}.
\]
By \eqref{E1}, it follows that
\begin{align*}
&\big|\int_0^1 [\alpha_k(t)xz_{t}(x,t)z_x(x,t)-kx^2|z_x(x,t)|^2]dx\big|\\
&=\frac{1+kt}{1-k}E(t)\\
&\leq\frac{1+kt}{1-k}\frac{1}{1+kt}E_0=\frac{1}{1-k}E_0.
\end{align*}
This implies that
\begin{equation}\label{A251}
\Big|\Big\{\int_0^1[\alpha_k(t)xz_{t}(x,t)z_x(x,t)
-kx^2|z_x(x,t)|^2]dx
\Big\}\Big|_0^T\Big|\leq\frac{2}{1-k}E_0.
\end{equation}
By \eqref{A211}, \eqref{A12}, \eqref{A251} and Remark \ref{rmk2.1}, we find that
\begin{equation} \label{A13}
\begin{split}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}\big)
\int_0^{T}\alpha_k(t)|z_{t}(1,t)|^2dt\\
&\leq \frac{2+k^2}{1-k^2}
\int_0^{T}E_0dt+\frac{2}{1-k}E_0=
\big(\frac{2+k^2}{1-k^2}T+\frac{2}{1-k}\big)E_0.
\end{split}
\end{equation}
By \eqref{b4}, it follows that
$$\beta_k(x,0)=1-k^2x^2.$$
It is obvious that
\begin{equation}\label{A14}
E_0=\frac{1}{2}\int_0^1[|z^1|^2
+\beta_k(x,0)|z^0_x|^2]dx\leq\frac{1}{2}(\|z^0\|^2+|z^1|^2).
\end{equation}
By \eqref{A13} and \eqref{A14}, noting that 
$1\leq\alpha_k(t)\leq(1+kT)$ for $0\leq t \leq T$, one can find a positive constant
$C_1=\frac{1}{2}(\frac{1}{2}+\frac{2k^2}{1-k^2})^{-1}
(\frac{2+k^2}{1-k^2}T+\frac{2}{1-k})$ such that
$$
\int_0^{T}|z_{t}(1,t)|^2dt \leq C_1(\|z^0\|^2+|z^1|^2),
$$
which completes the proof.
\end{proof}



\begin{proof}[Proof of Theorem \ref{thm4}]
By the  Young inequality, for any  $\varepsilon\in (0,\frac{1}{2})$,
it is easy to check that
\begin{align*} %\label{A22}
&\int_0^{T}E(t)dt +\int_0^T\!\int_0^1
\frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt
-\int_0^T\!\int_0^1kxz_{t}(x,t)z_x(x,t)\,dx\,dt\\
&\geq \int_0^T\!\int_0^1\big\{ \frac{1-\varepsilon}{2}\alpha_k(t) |z_{t}(x,t)|^2
+\big[\frac{1}{2}\beta_k(x,t)
+\big(1-\frac{1}{2\varepsilon}\big)
\frac{k^2x^2}{1+kt}\big]|z_x(x,t)|^2\big\}\,dx\,dt\\
&= \int_0^T\!\int_0^1\big\{
(1-\varepsilon)\frac{\alpha_k(t)}{2} |z_{t}(x,t)|^2
+[1+\big(2-\frac{1}{\varepsilon}\big)
\frac{k^2x^2}{1-k^2x^2}]\frac{\beta_k(x,t)}{2}|z_x(x,t)|^2\big\}\,dx\,dt\\
&\geq \int_0^T\!\int_0^1\big\{(1-\varepsilon)\frac{\alpha_k(t)}{2} |z_{t}(x,t)|^2
+[1+\big(2-\frac{1}{\varepsilon}\big)
\frac{k^2}{1-k^2}]\frac{\beta_k(x,t)}{2}|z_x(x,t)|^2\big\}\,dx\,dt.
\end{align*}
Take $\varepsilon=\frac{k}{1+k}\in(0,\frac{1}{2})$, then we find that
$$
1-\varepsilon=1+\big(2-\frac{1}{\varepsilon}\big)\frac{k^2}{1-k^2}.
$$
It follows that
\begin{equation}\label{A23}
\begin{split}
&\int_0^{T}E(t)dt +\int_0^T\!\int_0^1
\frac{k^2x^2}{1+kt}|z_x(x,t)|^2\,dx\,dt
-\int_0^T\!\int_0^1kxz_{t}(x,t)z_x(x,t)\,dx\,dt
\\
&\geq \big(1-\frac{k}{1+k}\big)\int_0^{T}E(t)dt
= \frac{1}{1+k}\int_0^{T}E(t)dt.
\end{split}
\end{equation}
Therefore, substituting  \eqref{E1}, \eqref{A251} and  \eqref{A23}
into \eqref{A211} indicates that
\begin{align*}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}\big)
\int_0^{T}\alpha_k(t)|z_{t}(1,t)|^2dt\\
&\geq\frac{1}{1+k}\int_0^{T}E(t)dt-\frac{2}{1-k}E_0\\
&=\frac{1}{1+k}\int_0^{T}[\frac{1}{1+kt}E_0
-\frac{k}{\alpha_k(t)}\int_0^t\alpha_k(s)|z_{t}(1,s)|^2ds]
dt-\frac{2}{1-k}E_0,
\end{align*}
which implies that
\begin{align*}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}\big)
\int_0^{T}\alpha_k(t)|z_{t}(1,t)|^2dt
+\frac{1}{k+1}\int_0^{T}[\frac{k}{\alpha_k(t)}
\int_0^t\alpha_k(s)|z_{t}(1,s)|^2ds]dt\\
&\geq\frac{1}{1+k}\int_0^{T}\frac{1}{1+kt}E_0dt-\frac{2}{1-k}E_0.
\end{align*}
It follows that
\begin{equation}\label{xin}
\begin{split}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}+\frac{kT}{1+k}\big)
\int_0^{T}\alpha_k(t)|z_{t}(1,t)|^2dt \\
& \geq\frac{1}{1+k}\int_0^{T}\frac{1}{1+kt}E_0dt-\frac{2}{1-k}E_0
=[\frac{1}{k(1+k)}\ln(1+kT)-\frac{2}{1-k}]E_0.
\end{split}
\end{equation}
From \eqref{xin} and  \eqref{A14}, it holds that
\begin{equation}\label{A29}
\begin{split}
&\big(\frac{1}{2}+\frac{2k^2}{1-k^2}+\frac{kT}{1+k}\big)
(1+kT)\int_0^{T}|z_{t}(1,t)|^2dt\\
&\geq \frac{1-k^2}{2}[\frac{1}{k(1+k)}\ln(1+kT)-\frac{2}{1-k}]
(\|z^0\|^2+|z^1|^2).
\end{split}
\end{equation}
Notice that if $T>T^{*}_k$, then
$\frac{1}{k(1+k)}\ln(1+kT)-\frac{2}{1-k}>0$.
This, together with \eqref{A29} indicates
the desired estimate in Theorem \ref{thm4}.
\end{proof}

\section{Proof of Theorem \ref{thm2}}

In this section we use the Hilbert Uniqueness Method. For Theorem
\ref{thm2}, it suffices to show that for any given initial value
$(w^0, w^1)\in L^2\times V'$
and target $(w_d^0,w_d^1)\in L^2\times V'$,
one can find a control $\overline{v}=\overline{v}(t)\in [H^1(0,T)]'$
such that the corresponding solution $w$ of  \eqref{b3}
satisfies
\begin{equation}
\label{2.8}
w(T)=w^0_d\quad\text{and}\quad w_{t}(T)=w^1_d.
\end{equation}

We divide the whole proof into  three parts.
\smallskip

\noindent \textbf{Step 1.} First, we define a linear operator 
$\Lambda$ from $V\times L^2$ to $V'\times L^2 $.
Consider  the  wave equation
\begin{equation} \label{2.9}
\begin{gathered}
\xi_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}\xi_x]_x
+[\frac{\gamma_k(x)}{\alpha_k(t)}]\xi_{tx}=0
\quad \text{in }  Q,\\
\xi(0,t)=0,\quad \xi_x(1,t)=0 \quad \text{on } (0,T),\\
\xi(x,T)=w^0_d(x),\quad \xi_{t}(x,T)=w^1_d(x)\quad \text{in } (0,1).
\end{gathered}
\end{equation}
It is easy to check that \eqref{2.9} has a unique solution
$$
\xi\in C([0,T];L^2)\cap C^1([0,T];V')
$$
and set
$$
(\xi^0,\xi^1)\triangleq(\xi(x,0),\xi_{t}(x,0))\in L^2\times V'.
$$
Thus
\begin{equation} \label{2.0}
(w^0-\xi^0,w^1-\xi^1)\in L^2\times V'.
\end{equation}

On the other hand, for any $(z^0,z^1)\in V\times L^2$, we denote by $z$ 
the corresponding solution of \eqref{2.10}.
Consider  the  wave equation
\begin{equation}\label{2.11}
\begin{gathered}
\eta_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}\eta_x]_x
+[\frac{\gamma_k(x)}{\alpha_k(t)}]\eta_{tx}=0
\quad \text{in }  Q,\\
\eta(0,t)=0,\quad \eta_x(1,t)=\frac{1}{\beta_k(1,t)} G_{z_{t}(1,t)}\quad
\text{on } (0,T),\\
\eta(x,T)=\eta_{t}(x,T)=0\quad \text{in } (0,1).
\end{gathered}
\end{equation}
Notice that $G_{z_{t}(1,t)}\in (H^1(0,T))'$ is defined as 
\begin{equation} \label{D}
\begin{split}
\langle G_{z_{t}(1,t)}, \phi\rangle_{(H^1(0,T))',H^1(0,T)}
=-\int_0^{T}z_{t}(1,t)\phi_{t}(t)dt \quad
\text{for any }\phi\in H^1(0,T).
\end{split}
\end{equation}
Now, we define the operator
\begin{gather*}
\Lambda: V\times L^2\to  V'\times L^2, \\
(z^0,z^1)\to \big(\eta_{t}(x,0)+\gamma_k(x)\eta_x(x,0)-k\eta(x,0),-\eta(x,0)\big).
\end{gather*}
Therefore,
\begin{equation}\label{2.13}
\begin{split}
&\langle\Lambda(z^0,z^1),(z^0,z^1)\rangle\\
&=\int_0^1[\eta_{t}(x,0)z^0-k\eta(x,0)z^0
+\gamma_k(x)\eta_x(x,0)z^0-\eta(x,0)z^1]dx.
\end{split}
\end{equation}
For simplicity, we set
$F=V\times L^2,F'=V'\times L^2$.
\smallskip


\noindent\textbf{Step 2.} We  prove that
$\Lambda$ is an isomorphism.
To this aim, multiplying  both sides of the first equation of \eqref{2.11}
 by $\alpha_k(t)z$ $(0\leq t \leq T)$
and integrating on $Q$, we obtain that
\begin{equation}\label{2.14}
\begin{split}
&-\int_0^T\beta_k(1,t)\eta_x(1,t) z(1,t)dt\\&=
\int_0^1[\eta'(x,0)z^0-k\eta(x,0)z^0
-\eta(x,0)z^1+\gamma_k(x)\eta_x(x,0)z^0]dx.
\end{split}
\end{equation}
From \eqref{2.11}-\eqref{2.13} and \eqref{2.14},
we conclude that
\begin{equation} \label{2.16}
\int_0^T|z_{t}(1,t)|^2dt
=\langle \Lambda(z^0,z^1),(z^0,z^1)\rangle.
\end{equation}
By Theorem \ref{thm3}, it holds that $\Lambda$
is a linear bounded operator.

It remains to show that $\Lambda$ is onto.
To this end, define the  bilinear functional on $F \times F$:
\[
A((\hat{z}^0,\hat{z}^1), (z^0,z^1))
=\langle \Lambda(\hat{z}^0,\hat{z}^1), (z^0,z^1)\rangle,
\]
where $(\hat{z}^0,\hat{z}^1)$, $(z^0,z^1)\in F\times F$.
It is clear that  $A$ is bounded.
From \eqref{2.16} and Theorem \ref{thm4}, it follows that
$A$ is coercive. Hence, applying Lax-Milgram Theorem,
we derive that $\Lambda$ is onto.
This completes the proof of Step 2.
\smallskip


\noindent \textbf{Step 3.} 
We  prove that the exact controllability of \eqref{b3} is equivalent that
 $\Lambda$ is an isomorphism.
Indeed, for any given $(w^0,w^1), (w_d^0,w_d^1)\in L^2\times V'$,
we choose  
$$
\overline{v}(\cdot)=\frac{1}{\beta_k(1,\cdot)}G_{z_{t}(1,\cdot)}\in (H^1(0,T))',
$$
where $z$ is the solution of  \eqref{2.10} associated to
$(z^0, z^1)=\Lambda^{-1}((w^1-\xi^1)+\gamma_k(x)(w^0_x-\xi^0_x)
-k(w^0-\xi^0),-(w^0-\xi^0))$ and $w$ is the solution of \eqref{b3}.
From  the definition of  $\Lambda$,
we conclude that 
$\Lambda(z^0,z^1)=(\eta'(x,0)+\gamma_k(x)\eta_x(x,0)-k\eta(x,0),-\eta(x,0))$,
where $\eta$ is the solution of \eqref{2.11}.
Then,  $\eta$ satisfies $(\eta(x,0),\eta'(x,0))=(w^0-\xi^0,w^1-\xi^1)$.
This implies that $w=\xi+\eta$ satisfies both \eqref{b3} and \eqref{2.8}. 
This completes the proof of  Theorem \ref{thm2}.

\subsection*{Acknowledgments}
This work is partially supported by the NSF of China under
grants  11171060 and 11371084. 

\begin{thebibliography}{0}

\bibitem{CG1981} C. Bardos, G. Chen;
 \emph{Control and stabilization for the wave equation, part~III:
domain with moving boundary}, SIAM J. Control Optim., 19 (1981), 123--138.

\bibitem{FGL2004} F. D. Araruna, G. O. Antunes, L. A. Medeiros, 
\emph{Exact controllability for the semilinear string equation
in the non cylindrical domains}, Control Cybernet., 33 (2004), 237--257.

\bibitem{C1999} M. M. Cavalcanti;
\emph{Exact controllability of the wave equation with
mixed boundary condition and time coefficients}, 
Arch.  Math. (BRNO), 35 (1999), 29--57.

\bibitem{JLL1988} J. L. Lions;
 \emph{Exact controllability, stabilizability and
perturbation for distributed systems}, SIAM Rev., 30 (1988), 1--68.

\bibitem{MMM1996} M. Milla Miranda;
\emph{Exact controllability for the wave equation in domains with variable boundary},
 Rev. Mat. Univ. Complut. Madrid, 9 (1996), 435--457.

\bibitem{c2013} Lizhi Cui, Xu Liu, Hang Gao;
\emph{Exact controllability for a one-dimensional wave equation in
non-cylindrical domains}, Journal of Mathematical Analysis and Applications, 
402 (2013), 612--625.

\bibitem{c2014} Lizhi Cui, Libo Song;
\emph{Controllability for a  Wave Equation with Moving Boundary}, 
Journal of Applied Mathematics,   doi:10.1155/2014/827698 (2014).

\bibitem{QR} J. P. Quinn, D. L. Russell;
\emph{Asymptotic stability and energy decay rates for
solutions of hyperbolic equations with boundary damping},
 Proc. Soc. Edinburgh Sect. A 77 (1977), 97--127.

\end{thebibliography}

\end{document}