\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 98, pp. 1--7.\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-2015/98\hfil Exact controllability for a string equation]
{Exact controllability for a string equation in domains with moving
boundary in one dimension}

\author[H. Sun, H. Li, L. Lu \hfil EJDE-2015/98\hfilneg]
{Haicong Sun, Huifen Li, Liqing Lu}

\address{Haicong Sun \newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{18835110799@163.com}

\address{Huifen Li \newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{lhf1024896246@163.com}

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

\thanks{Submitted December 14, 2014. Published April 14, 2015.}
\makeatletter
\@namedef{subjclassname@2010}{\textup{2010} Mathematics Subject Classification}
\makeatother
\subjclass[2010]{93B05}
\keywords{Exact controllability; string equation;  moving boundary; 
\hfill\break\indent Hilbert uniqueness method; multiplier method}

\begin{abstract}
 We consider a string equation in a domain with moving boundary.
 By using the multiplier method in non-cylindrical domains,
 we establish the exact boundary controllability in
 domains with moving boundary, and obtain
 a weaker condition on the time controllability.
\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}

 Let $T>0$. For any given $ k\in(0,1)$, set $ \alpha_k(t)=1+kt$ for $t\in[0,T]$.
 Denote by $\hat {Q}^k_T$ the  non-cylindrical domain in $R^2$,
 $$
\hat{Q}^k_T=\{(x,t)\in R^2;0<x<\alpha_k(t),t\in(0,T)\}.
$$
Consider the  controlled string equation in $\hat{Q}^k_T$:
\begin{equation}
 \begin{gathered}
 u_{tt}-u_{xx}=0 \quad\text{in } \hat{Q}^k_T,\\
 u(0,t)=0,\quad u(\alpha_k(t),t)=v  \quad\text{on } (0,T),\\
 u(0)=u^0,\quad  u_t(0)=u^1 \quad\text{in }(0,1),
   \end{gathered} \label{e1.1}
\end{equation}
where $u$ is the state variable, $v$ is the control variable and
$(u^0,u^1)\in L^2(0,1)\times H^{-1}(0,1)$ is any given
initial data. \eqref{e1.1} may describe the motion of a string with a
fixed endpoint and a moving one.
By Milla Miranda \cite{m1},
 for $0<k<1$, any $(u^0,u^1)\in L^2(0,1)\times H^{-1}(0,1)$
and $v\in L^2(0,T)$, \eqref{e1.1} admits a unique solution in the sense
of transposition.

The goal of this article is to study the exact controllability of
 \eqref{e1.1} in the following sense.

\begin{definition} \label{def1.1}\rm
Problem \eqref{e1.1} is said to be exactly controllable at time $T$, if for
 any initial state $(u^0,u^1)\in L^2(0,1)\times H^{-1}(0,1)$  and any 
preassigned state $(u^0_T,u^1_T)\in L^2(0,\alpha_k(T))\times H^{-1}(0,\alpha_k(T))$, 
there is a control  $v\in L^2(0,T)$ such that the corresponding solution 
of \eqref{e1.1} in the sense of transposition satisfies
$$
u(T)= u^0_T,\quad u_t(T)=u^1_T.
$$
\end{definition}

For $k\in(0,1)$, set
\begin{equation}
\bar{T}_k=\frac{2}{1-k}\,.\label{e1.2}
\end{equation}
Our main result  is stated as follows.

\begin{theorem} \label{thm1.1} 
Suppose that $k\in(0,1)$. Then for any given $T>\bar{T}_k$, Problem \eqref{e1.1} 
is exactly controllable at time $T$.
\end{theorem}

\begin{remark} \label{rmk1.1}\rm
 It is easy to check that
$$
\bar{T}\triangleq\lim_{k\to 0}\bar{T}_k=\lim_{k\to 0}\frac{2}{1-k}=2.
$$
And we can easily find that  $\bar{T}_k<T^*_k$, where 
$T_{k}^{*}=\frac {e^{\frac {2k(1+k)}{1-k}}-1}{k}$ is the controllability
 time in \cite{c5}.
Indeed, we have 
$$
T_{k}^{*}=\frac {e^{\frac {2k(1+k)}{1-k}}-1}{k}>
\frac{2(1+k)}{1-k}>\bar{T}_k\,,
$$
for $0<k<1$.
\end{remark}


Similar to \cite[Theorem 4.1]{k1}, we use the Hilbert Uniqueness Method (HUM)
to seek a control $v$ in the special
 form $ v(t)=w_x(\alpha_k(t),t)$, where $ w $ is the solution of the 
homogeneous problem
 \begin{equation}
 \begin{gathered}
 w_{tt}-w_{xx}=0 \quad\text{in } \hat{Q}^k_T,\\
 w(0,t)=0 ,\quad w(\alpha_k(t),t)=0  \quad\text{on } (0,T),\\
 w(0)=w^0 ,\quad w_t(0)=w^1 \quad \text{in } (0,1),
   \end{gathered} \label{e1.3}
\end{equation}
where $k\in(0,1),(w^0,w^1)\in H^1_0(0,1)\times L^2(0,1)$
is any given initial value, $\alpha_k(t)=1+kt$ is the function given
in \eqref{e1.1}.
According to the existence theorem in \cite{c3}, Problem \eqref{e1.3} has a
unique solution
  $$
w\in C([0,T];H_0^1(0,\alpha_k(t)))\cap C^1([0,T];L^2(0,\alpha_k(t))).
$$
 We define the energy of the above system as
\begin{equation}
E(t)=\frac{1}{2}\int_0^{\alpha_k(t)}(w^2_x+w^2_t)dx\quad\text{for }
  t\geq0\,. \label{e1.4}
\end{equation}
Next, we deduce a growth estimate of the energy of problem \eqref{e1.3}.

  \begin{theorem} \label{thm1.2} 
Let $w(x,t)$ be a solution of the problem \eqref{e1.3} in $\hat{Q}^k_T$.
Then $E(t)$ is nonincreasing for $t\geq0$ and
\begin{equation}
\frac{1-k}{(1+k)(1+kt)}E(0)\leq E(t)\leq \frac{1+k}{(1-k)(1+kt)}E(0)\quad
\text{for } t\geq0.\label{e1.5}
\end{equation}
\end{theorem}

\begin{remark} \label{rmk1.3}\rm
 In the case of $k=1$, some results have been obtained in \cite{c5}. 
However, we do not  extend the approach developed in this paper to the case $k=1$.
\end{remark}

\begin{remark} \label{rmk1.4} \rm 
We can obtain the same results as in this article for a  more general
function $\alpha_k(t)$, as long as it meets the 
condition $0<\alpha'_k(t)<1$. 
\end{remark}

  \begin{remark} \label{rmk1.5}\rm
It would be quite interesting to the study Dirichlet control for multi-dimensional
wave equations in non-cylindrical domains by the same approach as this paper.
We shall consider Neumann control for wave equations in domains with moving boundary
in the forthcoming papers.
\end{remark}

Control and stabilization for wave  equations in domains with
moving boundary has been widely studied and many results are given, see
\cite{c2,c3,c4,l1,m1,m2} and the references therein. 
Under a restrictive assumption on the domain
with moving boundary, Bardos et al \cite{b1} considered the exact controllability 
and stabilization of \eqref{e1.1} for multi-dimensional
  wave equations in non-cylindrical domains. Using a suitable change of variables,
Cavalcanti et al \cite{c1} studied the existence and asymptotic behavior of global 
regular solutions of the mixed problem for the Kirchhoff nonlinear model.
By using the same transformation, Cui et al \cite{c5} considered the exact boundary
controllability for a one-dimensional wave equation and obtained the 
controllability time.
A damped Klein-Gordon equation in a non-cylindrical domain was studied in \cite{g1}, 
there the authors obtained the existence
of global solutions and the exponential decay of the energy.
The stabilization and controllability
for the wave equation with variable coefficients in domains with moving boundary
was investigated in \cite{l2}.
Under some appropriate geometric conditions, the energy decay estimates
was established and the exact controllability was also obtained by the Rimannian
geometry method. It is well known that
the Riemannian geometry method was first introduced in \cite{y1} for the
controllability of the wave equation with variable coefficients.

Motivated by \cite{b1,c5,l2}, we study the
exact boundary controllability of  \eqref{e1.1}. Instead of transforming
the problem from a non-cylindrical domain into a cylindrical domain,
 we study the problem directly in non-cylindrical domains. 
By using a modified multiplier method,
we obtain a controllability time which is smaller than that in \cite{c5}.

The rest of this article is organized as follows. 
In section 2, we prove three lemmas which will be needed
in the sequel. 
Section 3 we  prove our main results.

\section{Preliminaries}

In this section, we establish three key lemmas which are needed in proving 
our main results. The first lemma gives a equality on the energy of the 
solution to problem \eqref{e1.3}. 

\begin{lemma} \label{lem2.1} 
Let $w(x,t)$ be a solution of  \eqref{e1.3} in $\hat{Q}^k_T$. Then
\begin{equation}
E(T)-E(0)=\frac{k(k^2-1)}{2\sqrt{1+k^2} }\int_0^T w_x^2(\alpha_k(t),t)dt.
\label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}
 Multiply\eqref{e1.3} by  $w_t$ and integrate on $\hat {Q}^k_T$, we
obtain 
\begin{align*}
0&=\int_0^T\int_0^{\alpha_k(t)}{(w_{tt}-w_{xx})w_t}\,dx\,dt \\
&=\int_0^T\int_0^{\alpha_k(t)}[(\frac{1}{2}w_t^2)_t
 +(\frac{1}{2}w_x^2)_t-(w_xw_t)_x]\,dx\,dt
\end{align*}
Let us denote by $\Sigma$ the boundary of $\hat {Q}^k_T$ and by
 $\nu=(\nu_x,\nu_t)$ the unit outward normal at $(x,t)$ on $\Sigma$.
First, Gauss-Green formula implies
\begin{equation}
0=\int_\Sigma[\frac{1}{2}(w_t^2+w_x^2)\nu_t-(w_xw_t)\nu_x]d\Sigma.\label{e2.2}
\end{equation}
 Since $w$ is a solution to \eqref{e1.3}, $w(0,t)=0$ and $ w(\alpha_k(t),t)=0$,
then we have
$kw_x(\alpha_k(t),t)+w_t(\alpha_k(t),t)=0$ and $w_t(0,t)=0$.
 Substituting these equalities in \eqref{e2.2},
\eqref{e2.1} is proved.
\end{proof}

\begin{remark} \label{rmk2.1} \rm
 Without loss of generality, let $0<t_1<t_2<T$, replacing $T$ by $t_2$ and 
$0$ by $t_1$, we have 
$$
E(t_2)-E(t_1)=\frac{k(k^2-1)}{2\sqrt{1+k^2} }\int_{t_1}^{t_2} w_x^2(\alpha_k(t),t)dt. 
$$ 
Moreover it is easy to check that the energy of \eqref{e1.3} is decreasing for 
$k\in(0,1)$ and is conserved for $k=1$.
\end{remark}

\begin{lemma} \label{lem2.2} 
Suppose that $w$ is the solution of \eqref{e1.3}. Let $T>0$ be given, 
we have the  equality
\begin{equation}
\begin{aligned}
&\frac{(1-k^2)}{\sqrt{1+k^2} }\int_{0}^{T} w_x^2(\alpha_k(t),t)dt\\
&=2TE(T)-2\int_0^1xw_t(0)w_x(0)dx+2\int_0^{1+kT}xw_t(T)w_x(T)dx.
\end{aligned} \label{e2.3}
\end{equation}
\end{lemma}

\begin{proof} 
(i) Multiplying \eqref{e1.3} by  $2xw_x$ and integrating by parts
on $\hat {Q}^k_T$,
we have
\begin{align*}
0&=\int_0^T\int_0^{\alpha_k(t)}(w_{tt}-w_{xx})2xw_x \,dx\,dt\\
&=\int_0^T\int_0^{\alpha_k(t)}[(2xw_xw_t)_t-(xw_t^2+xw_x^2)_x+w_t^2+w_x^2]\,dx\,dt\\
&=\int_\Sigma[(2xw_tw_x)\nu_t-(xw_t^2+xw_x^2)\nu_x]d\Sigma+2\int_0^TE(t)dt\\
&=\int_0^{1+kT}2xw_t(T)w_x(T)dx-\int_0^12xw^1w_x(x,0)\\
&\quad +\frac{k^2-1}{\sqrt{1+k^2}}
\int_0^T(1+kt)w_x^2(\alpha_k(t),t)dt+2\int_0^TE(t)dt.
\end{align*}
This implies 
\begin{equation}
\begin{aligned}
&\frac{1-k^2}{\sqrt{1+k^2}}\int_{0}^{T}(1+kt) w_x^2(\alpha_k(t),t)dt\\
&= 2\int_0^TE(t)dt-\int_0^12xw_t(0)w_x(0,x)dx+
\int_0^{1+kT}2xw_t(T)w_x(T)dx.
\end{aligned}\label{e2.4}
\end{equation}
(ii) Multiplying \eqref{e1.3} by $2tw_t$ and integrating by parts on $\hat{Q}^k_T$
yields
\begin{equation}
\frac{1-k^2}{\sqrt{1+k^2}}\int_0^T kt w_x^2(\alpha_k(t),t)dt
=2\int_0^T E(t)dt-2 T E(T).\label{e2.5}
\end{equation}
Equality \eqref{e2.3} follows easily from \eqref{e2.4} and \eqref{e2.5}.
\end{proof}

Using Cauchy's inequality, we can obtain easily the following result.

\begin{lemma} \label{lem2.3}
 Denote by $w$ the solution of \eqref{e1.3}. For $t\in(0,T)$ and $k\in(0,1)$, 
we have the  estimate
\begin{equation}
|\int_0^{1+kt}2xw_tw_xdx|\leq2(1+kt)E(t). \label{e2.6}
\end{equation}
\end{lemma}

\section{Proof of main results}

In this section, we prove the exact controllability for the string equation 
\eqref{e1.1} in the non-cylindrical domain
 $\hat {Q}^k_T$ (Theorem \ref{thm1.1}) for $0<k<1$ by the Hilbert Uniqueness Method
(HUM).  We first need to prove Theorem  \ref{thm1.2}.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 From \eqref{e2.1} and \eqref{e2.3}, it is easy to check that
\begin{equation}
\begin{aligned}
&\frac{2}{k}(E(0)-E(T))\\
&=2TE(T)-2\int_0^1xw_t(0)w_x(0)dx+\int_0^{1+kT}2xw_t(T)w_x(T)dx.
\end{aligned} \label{e3.1}
\end{equation}
 Rearranging this equality, we have
 $$
E(0)+k\int_0^1xw^1w_x(0,x)dx=(1+kT)E(T)+k\int_0^{1+kT}xw_tw_x\,dx.
$$
 Using \eqref{e2.6}, one can easily obtain
\begin{gather*}
(1-k)E(0)\leq(1+k)(1+kT)E(T),\\
(1+k)E(0)\geq(1-k)(1+kT)E(T),
\end{gather*}
for $T>0$. Then we have the estimates:
 $$
\frac{1-k}{(1+k)(1+kT)}E(0)\leq E(T)
\leq\frac{1+k}{(1-k)(1+kT)}E(0).
$$
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
We apply HUM as in \cite[chapter 4]{k1}. 
For any $(w^0,w^1)\in H_0^1(0,1)\times L^2(0,1)$,
let $w$ be the solution of \eqref{e1.3}. Consider
the  problem
\begin{equation}
 \begin{gathered}
  u_{tt}-u_{xx}=0 \quad\text{in } \hat{Q}^k_T,\\
  u(0,t)=0 ,\quad u(\alpha_k(t),t)=w_x(\alpha_k(t),t)  \quad\text{on }  (0,T),\\
  u(T)=0 ,\quad  u_t(T)=0 \quad\text{in } (0,1).
\end{gathered} \label{e3.2}
\end{equation}
It is well known that \eqref{e3.2} admits a unique solution such that
$$
(u(0),u_t(0))\in L^2(0,1)\times H^{-1}(0,1).
$$
Then, we introduce a map
$\Lambda:H_0^1(0,1)\times L^2(0,1)\to H^{-1}(0,1)\times L^2(0,1) $
defined by
$$
\Lambda(w^0,w^1)=(u^1,-u^0),
$$
where $u^0=u(0),\ u^1=u_t(0)$.

Then  the map $\Lambda$ is an isomorphism of $H_0^1(0,1)\times L^2(0,1)$
onto $H^{-1}(0,1)\times L^2(0,1)$.
 To simplify our analysis, we introduce the following notation:
$$
F:=H_0^1(0,1)\times L^2(0,1) \quad  F':=H^{-1}(0,1)\times L^2(0,1).
$$
In fact, multiplying equation  \eqref{e3.2} by $w$ and integrating on ${Q}^k_T$, 
we obtain that
\begin{equation}
\int_0^1(w^0u^1-w^1u^0)dx=\frac{1-k^2}{\sqrt{1+k^2}}
\int_0^{T}w^2_x(\alpha_k(t),t)dt.\label{e3.4}
\end{equation}
Hence we have
\begin{equation}
\langle\Lambda(w^0,w^1),(w^0,w^1)\rangle_{F',F}
=\frac{1-k^2}{\sqrt{1+k^2}}\int_0^{T}w^2_x
(\alpha_k(t),t)dt,\label{e3.5}
\end{equation}
for every $(w^0,w^1)\in F$.

Recalling  estimate \eqref{e1.5}  and equality \eqref{e2.1} , we  have
\begin{equation}
\frac{(1-k)T-2}{(1-k)(1+kT)}E(0)
\leq\langle\Lambda(w^0,w^1),(w^0,w^1)\rangle_{F',F}
\leq\frac{(1+k)T+2}{(1+k)(1+kT)}E(0).\label{e3.6}
\end{equation}
From these  inequalities, we conclude that $\Lambda$ is a coercive linear map
for $T>\bar{T}_k$ and  is bounded. Therefore, $\Lambda$ is a
surjection by Lax-Milgram Theorem. It follows that $\Lambda$ is an isomorphism.

Since $\Lambda$ is  an isomorphism, for any  initial value 
$(u^0,u^1)\in L^2(0,1)\times H^{-1}(0,1) $,
there exists $(w^0,w^1)\in H_0^1(0,1)\times L^2(0,1)$ such that
$$
\Lambda(w^0,w^1)=(u^1,-u^0).
$$
Then $u$ is the solution of \eqref{e1.1} with $v=w_x(\alpha_k(t),t)$. 
Furthermore, $(u(0),u_t(0))=(u^0,u^1)$ and
$(u(T),u_t(T))=(0,0)$. This completes the proof.
\end{proof}

\subsection*{Acknowledgments}
This work is supported by the National Science
Foundation of China (Nos. 11401351, 61104129, 61174082, 61374089), and by
the Shanxi Scholarship Council of China (2013-013).


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\end{document}