\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 239, pp. 1--19.\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{8mm}}

\begin{document}
\title[\hfilneg EJDE-2015/239\hfil Boundary controllability]
{Boundary controllability for a nonlinear \\ beam equation}

\author[X.-M. Cao \hfil EJDE-2015/239\hfilneg]
{Xiao-Min Cao}

\address{Xiao-Min Cao \newline
School of Mathematical Sciences,
Shanxi University, Taiyuan 030006, China}
\email{caoxm@sxu.edu.cn}

\thanks{Submitted May 4, 2015. Published September 17, 2015.}
\subjclass[2010]{93B05, 35L75, 93C10, 93C20}
\keywords{Nonlinear beam equation; locally exact controllability;
\hfill\break\indent  equilibrium; smooth control}

\begin{abstract}
 This article concerns a nonlinear system modeling the bending vibrations
 of a nonlinear beam of length $L>0$. First, we derive the existence of
 long time solutions near an equilibrium. Then we prove that the nonlinear
 beam is locally exact controllable around the equilibrium in $H^4(0,L)$
 and with control functions in $H^2(0,T)$. The approach we used are open
 mapping theorem, local controllability established by linearization,
 and the induction.

\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction and statement of main results}

We consider a controllability problem for a system
modeling the bending vibrations of a nonlinear beam of length $L >0$.
Let $\phi$ denote the deflection of the beam, the left end of
the beam is fixed, and an appropriate shear force $u$ exerted on the
right end of the beam, then the equations of motion describing beam
bending are
\begin{equation} \label{e1.1}
\begin{gathered}
\phi_{tt}+(a(x,\phi)\phi'')''=b(x,\phi,\phi',\phi''), \quad
x\in(0,L),\; t\in (0,T),\\
\phi(x,0)=\phi_0(x), \quad
\phi_{t}(x,0)=\phi_1(x), \quad x\in(0,L),\\
\phi(0,t)=0,\quad \phi'(0,t)=0,\quad t\in(0,T),\\
\phi(L,t)=0,\quad \phi'(L,t)=u(t),\quad t\in(0,T),
\end{gathered}
\end{equation}
 where $a(x, y)$, $b(x, y_1,y_2,y_3)$ are smooth functions on
$[0,L]\times \mathbb{R}$ and $[0,L]\times \mathbb{R}^3$, respectively, such that
 \begin{gather}
 a(x,y)>0, \quad \forall (x,y)\in [0,L]\times \mathbb{R}\\
 b(x,0,0,0)=0, \quad \forall x\in [0,L]
\end{gather}

Let $\phi_0,\phi_1,\widehat{\phi}_0,\widehat{\phi}_1$ be given
functions and $T > 0$ be given. If there is boundary function
$u$ on $(0, T)$ such that the solution of  \eqref{e1.1}
satisfies $\phi(T) = \widehat{\phi}_0, \phi_t(T) = \widehat{\phi}_1$
on $[0,L]$, we say the system \eqref{e1.1} is exactly controllable from
$(\phi_0, \phi_1)$ to $(\widehat{\phi}_0, \widehat{\phi}_1)$ at time
T by controlled moment on the right end.

Boundary exact controllability on linear beam problems has been studied
for many years, see
\cite{Krabs,lagnese,Lagnese-L-L,L-T,Lions,Rao,Russell}, and many others.

To the best of our knowledge, there is little about the
boundary control of nonlinear beam equation in the existing related
papers. Recently, Yao and G. Weiss \cite{yao2} consider a dynamical
system with boundary input and output
describing the bending vibrations of a quasi-linear beam, where the
nonlinearity comes from Hooke's law. they show that the structure of the
boundary input and output forces the system to admit global solutions
at least when the initial data and the boundary input are small in a
certain sense. And they
prove that the norm of the state of the system decays exponentially if
the input becomes zero after
a finite time. Cindea and Tucsnak \cite{Cindea} study the exact
controllability of a nonlinear plate equation by the
means of a control which acts on an internal region of the plate.
For rectangular domains, they obtain that the Berger system is locally
exactly controllable in arbitrarily small time
and for every open and nonempty control region.

Recently, using moment theory and Nash-Moser theorem,
 Beauchard \cite{Beauchard} prove that the linear beam
equation with clamped ends is locally controllable in a
$H^{5+\varepsilon}\times H^{3+\varepsilon} ((0, 1),R)$ neighborhood
of a particular trajectory of the free system, with $\varepsilon >
0$ and with control functions in $H^1_0 ((0, T),R)$.

 In the present work we will study boundary controllability for the nonlinear
beam equation \eqref{e1.1} by using some ideas of \cite{schmidt} for
the nonlinear system of wave equations. First, we will derive the existence
of long time solutions near an equilibrium.
Then, the locally exact controllability of the system
around the equilibrium will be established.

Let us choose some Sobolev spaces to formulate the problems.
To get a smooth control, we assume initial data
$\phi_0\in H^4(0,L), \phi_1\in H^2(0,L)$ to study the controllability of the system
around the equilibrium in $H^4(0,L)$ at time $T$ via a boundary control
$u\in H^2(0,T)$. Obviously, control function we obtained is more smooth
than the previous results \cite{Beauchard,lagnese}.

 We say $\omega\in H^4(0,L)$ is an equilibrium of the system \eqref{e1.1}
if
\begin{equation} (a(x,\omega)\omega'')''=b(x,\omega,\omega',\omega'').
\end{equation}

Let $\phi_0\in H^4(0,L)$, $\phi_1\in H^2(0,L)$, $u\in H^2(0,T)$. We say
these functions satisfy compatibility conditions if
\begin{gather*}
 \phi_0(0)=0, \phi'_0(0)=0,\phi_1(0)=0,
\phi'_1(0)=0 ,\\ \phi_0(L)=0,\phi'_0(L)=u(0),
\phi_1(L)=0,\phi'_1(L)=\dot{u}(0).
\end{gather*}
Set
\begin{equation} V(0,L)=\{\varphi|\varphi\in
H^2(0,L),\varphi(0)=\varphi(L)=\varphi'(0)=0.\}
\end{equation}

The next result shows that near one equilibrium, the system
has solutions of long time.

\begin{theorem} \label{thm1.1}
Let $w \in H^4(0,L) \cap V(0,L)$ be an equilibrium
of \eqref{e1.1}. Let $T > 0$ be arbitrary given. Then there is
$\varepsilon_T>0$, which depends on the time $T$, such that if
$\phi_0\in H^4(0,L)\cap V(0,L), \phi_1\in V(0,L)$ satisfy
$$
\|\phi_0-\omega\|_4<\varepsilon_T, \|\phi_1\|_2<\varepsilon_T,
$$
and $u\in H^2(0,T)$ satisfies the compatibility conditions
with $(\phi_0,\phi_1)$ and $\|u\|_2<\varepsilon_T$, then 
system \eqref{e1.1} has a solution
$$\phi\in C([0,T],H^4(0,L))\cap
C^1([0,T],H^2(0,L)) \cap C^2([0,T],L^2(0,L)).
$$
\end{theorem}

Near an equilibrium, we have the following locally exact controllability
results.

\begin{theorem} \label{thm1.2}
Let $\omega \in H^4(0,L) \cap V(0,L)$ be an
equilibrium of  \eqref{e1.1}. Then, for $T>0$ given, there is
$\varepsilon_T>0$ such that for any
$(\phi^i_0,\phi_1^i)\in (H^4(0,L) \cap V(0,L)) \times V(0,L) (i=1,2)$ with
$$
\|\phi^i_0-\omega\|_4<\varepsilon_T, \|\phi^i_1\|_2<\varepsilon_T,
$$
we can find $u\in H^2(0,T)$
which is compatible with $(\phi^1_0,\phi^1_1)$ such that the
solution of the system \eqref{e1.1} with the initial data
$(\phi^1_0,\phi^1_1)$ satisfies
$$
\phi(T)=\phi_0^2, \quad\phi_t(T)=\phi_1^2.
$$
\end{theorem}

\section{Existence of long time solutions near equilibria}

The existence of short time solutions for the nonlinear beam
equation can be proved using standard arguments such as the
nonlinear semigroups theory or the Galerkin method and fixed point
arguments \cite{Ackleh,dafer,Lions-M,Slemrod}.
 We only study some energy estimates of the short time solutions to
have long time solutions when initial data are close to an
equilibrium here.

We suppose that the equilibrium is the zero in this section. If an
equilibrium $\omega \in H^4(0,L) \cap V(0,L)$ is not zero, we can
make a transform by $\phi=\omega+\psi$, and consider the problem
\begin{gather*}
\psi_{tt}+(a(x,\omega+\psi)(\psi''+\omega''))''
=b(x,\omega+\psi,\omega'+\psi',\omega''+\psi''), \\
 x\in(0,L),t\in (0,T),\\
\psi(x,0)=\phi_0-\omega, \quad
\psi_{t}(x,0)=\phi_1(x), \quad x\in(0,L),\\
\psi(0,t)=0,\psi'(0,t)=0,\quad t\in(0,T),\\
\psi(L,t)=0,\quad \psi'(L,t)=u(t)-\omega'(L),\quad t\in(0,T),
\end{gather*}

 Let $\phi\in C([0,T],H^4(0,L))\cap C^1([0,T],H^2(0,L)) \cap C^2([0,T],L^2(0,L))$
be a solution of  \eqref{e1.1} for some $T>0$. Suppose that
$u\in H^2(0,T)$.
We introduce
\begin{gather*}
\mathcal{E}(t)=\|\phi\|_4^2+\|\phi_t\|_2^2+\|\phi_{tt}\|^2,\\
\mathcal{E}_\Gamma(t)=u^2(t)+\dot{u}^2(t)+\ddot{u}^2(t),\\
\mathcal{Q}(t)=\|\phi_t\|^2+\|\phi''\|^2+\|\phi'_t\|^2
+\|\phi'''\|^2+\|\phi_{tt}\|^2+\|\phi''_t\|^2.
\end{gather*}
For solutions of \eqref{e1.1} near the zero equilibrium, we have
the following result.

\begin{theorem} \label{thm2.1}
Let $\gamma>0$ be given and $\phi$ be a solution
of \eqref{e1.1} on the interval $[0,T]$ for some $T>0$ such that
the condition
 \begin{equation}\label{c21}
\sup_{0\le t \le T}\|\phi(t)\|_4 \leq \gamma.
\end{equation}
 holds. Then there is $c_\gamma > 0$, which depends
on the $\gamma$ but is independent of initial data $(\phi_0, \phi_1)$
and boundary functions $u$, such that
\begin{equation} \label{Q}
\mathcal{Q}(t)\le c_\gamma\mathcal{Q}(0)+c_\gamma \int_0^t
\Big[(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)+\mathcal{E}^\frac{3}{2}(\tau))
\mathcal{E}(\tau)+\mathcal{E}_\Gamma(\tau)\Big]d\tau.
\end{equation}
 and
\begin{equation}\label{Q1}
\mathcal{Q}(t)\le \mathcal{E}(t)\le c_\gamma\mathcal{Q}(t)+c_\gamma\mathcal{E}_\Gamma(t),
\end{equation}
for $t\in [0,T]$.
\end{theorem}

Here we list a few basic properties of Sobolev spaces to be invoked
in the sequel.

(i) Let $s_1>s_2\ge 0$. For any $\varepsilon>0$ there is
$c_\varepsilon>0$ such that 
\[ 
\|\varphi\|_{s_2}^2\le \varepsilon\|\varphi\|_{s_1}^2
+c_\varepsilon\|\varphi\|^2, \quad \forall
\varphi \in H^{s_1}(0,L).
\]

(ii) If $m>\frac{1}{2}$, then for each $k=0,1,\cdots$, we have $
H^{m+k}(0,L)\subset C^k([0,L])$ with continuous inclusion.

(iii) If $r:=\min\{s_1,s_2,s_1+s_2-1\}\ge 0$, then there is a
constant $c>0$ such that
\begin{equation} \label{e2.6}
\|fg\|_r\le c\|f\|_{s_1}\|g\|_{s_2},
\forall f\in H^{s_1}(0,L),g\in H^{s_2}(0,L).
\end{equation}


\begin{lemma} \label{lem2.1}
 (i) Let $f(x,y)$ be a smooth function on
$[0,L]\times \mathbb{R} $. Set $F(x)=f(x,\phi)$, for $0\le k\le 4$, then
there is $c=c(\sup_{x\in [0,L]}|\phi|)>0$, such that
\begin{equation} \label{e2.7}
\|F\|_k\le c\sum_{j=0}^{k}(1+\|\phi\|_4)^j.
\end{equation}

(ii) Let $f(x,y_1,y_2,y_3)$ be a smooth function on $[0,L]\times \mathbb{R}^3$.
Set 
\[
G(x)=f(x,\phi,\phi',\phi''),
\]
then there is $c=c\big(\sup_{x\in [0,L]}|\phi|,\sup_{x\in
[0,L]}|\phi'|,\sup_{x\in [0,L]}|\phi''|\big)>0$, such that
\begin{equation} \label{e2.8}
\|G\|\le c.
\end{equation}
\end{lemma}

\begin{proof}
 (i) Inequality \eqref{e2.7} is clearly true for $k=0$.
By using  \eqref{e2.6} and the induction of the order $k$,
inequality \eqref{e2.7} follows.

(ii) A standard method as to the linearly elliptic problem can give 
inequality \eqref{e2.8},
for example see Taylor \cite{Taylor}.
\end{proof}

 Observing the partial differential of the function $a(x,\phi)$ and
$b(x,\phi,\phi',\phi'')$, using the formula \eqref{e2.6} and Lemma \ref{lem2.1},
 we have the following lemma.

 \begin{lemma} \label{lem2.2}
Let $\gamma>0$ be given and $\phi$
be a solution of  \eqref{e1.1} on the interval [0,T] for some
$T>0$ such that the condition \eqref{c21} holds true.
Then there is $c_\gamma>0$, which
depends on the $\gamma$, such that
\begin{gather*}
\|a_t\|_2\le c_\gamma\mathcal{E}^{1/2}(t),\quad
\|a'_t\|_1\le c_\gamma \big(\mathcal{E}^{1/2}(t)+\mathcal{E}(t)\big),\\
\|a''_t\|\le c_\gamma \big(\mathcal{E}^{1/2}(t)+\mathcal{E}(t)+\mathcal{E}^{\frac{3}{2}}(t)\big),\quad
\|b_t\|\le c_\gamma\mathcal{E}^{1/2}(t).
\end{gather*}
\end{lemma}

 \begin{lemma} \label{lem2.3}
Let $\gamma>0$ be given and $\phi$
be a solution of  \eqref{e1.1} on the interval [0,T] for some
$T>0$ such that the condition \eqref{c21} holds true. Set
\[
\Upsilon_1(t)=\|\phi_t\|^2+\|\phi''\|^2, \quad
\Upsilon_2(t)=\|\phi'_t\|^2+\|\phi'''\|^2, \quad
\Upsilon_3(t)=\|\phi_{tt}\|^2+\|\phi_t''\|^2.
\]
 Then there is $c_\gamma>0$, which
depends on the $\gamma$, such that
\begin{gather} \label{upsi1}
\Upsilon_1(t)\le c_\gamma\Upsilon_1(0)+c_\gamma \int_0^t \big[(1+
\mathcal{E}^\frac{1}{2}(\tau))\mathcal{E}(\tau)+\mathcal{E}_{\Gamma}(\tau)\big]d\tau,
 \\
\label{upsi230}
\begin{aligned}
&\Upsilon_2(t)+\Upsilon_3(t)\\
&\le c_\gamma\big(\Upsilon_2(0)+\Upsilon_3(0)\big) 
 +c_\gamma \int_0^t \Big[\Big(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)
+\mathcal{E}^\frac{3}{2}(\tau)\Big) \mathcal{E}(\tau)+\mathcal{E}_{\Gamma}(\tau)
\Big]d\tau
\end{aligned}
\end{gather}
for $0\le t\le T$.
\end{lemma}

\begin{proof}
Let
\[
 P_1(t)=\|\phi_t\|^2+(a\phi'',\phi'').
\]
 We obtain
\begin{align*}
 \dot{P}_1(t)
&= 2(\phi_t,\phi_{tt})+2(a\phi'',\phi''_t)+(a_t\phi'',\phi'')\\
&= 2(\phi_t,b-(a\phi'')'')+2(a\phi'',\phi''_t)+(a_t\phi'',\phi'')\\
&= 2(\phi_t,b)+2\dot{u}(t)(a\phi'')(L,t)+(a_t\phi'',\phi'').
\end{align*}
It follows that
\begin{align*}
 \Upsilon_1(t)
&\leq c_\gamma P_1(t) \\
&\leq c_\gamma P_1(0)+c_\gamma \int_0^t (\|\phi_t\|\|b\|
 + \|a_t\|_2\|\phi''\|^2)dt+ c_\gamma \int_0^t
 (\dot{u}^2(t)+{\phi''}^2(L,t)) dt\\
&\leq c_\gamma\Upsilon_1(0)+c_\gamma \int_0^t \Big[(1+
\mathcal{E}^\frac{1}{2}(\tau))\mathcal{E}(\tau)+\mathcal{E}_{\Gamma}(\tau)\Big]d\tau .
\end{align*}

To obtain inequality \eqref{upsi230}, first we differentiate  equation
\eqref{e1.1} with respect to $x$, then we have
\begin{equation}
\phi'_{tt}+(a\phi'')'''=b'.
\end{equation}
Let
\begin{equation}
P_2(t)=\|\phi'_{t}\|^2+(a\phi''',\phi''').
\end{equation}
 We obtain \begin{align*}
 \dot{P}_2(t)
&= 2(\phi'_{t},\phi'_{tt})+2(a\phi''',\phi_{t}''')+(a_t\phi''',\phi''')\\
&= 2(\phi'_{t},b'-(a\phi'')''')+2(a\phi''',\phi_{t}''')+(a_t\phi''',\phi''')\\
&=2(\phi'_{t},b')-2\dot{u}(t)b(L,t)+2\phi''_{t}(L,t)(a\phi''')(L,t)
 -2\phi''_{t}(0,t)(a\phi''')(0,t)\\
&\quad -2(\phi''_{t},a'\phi''')
 +2(\phi''_{t},2a'\phi'''+a''\phi'')+(a_t\phi''',\phi''').
\end{align*}
So
\begin{equation} \label{upsi2}
\begin{aligned}
\Upsilon_2(t)
& \le c_\gamma P_2(t) \\
&\leq c_\gamma P_2(0)+c_\gamma \int_0^t \Big[(1+\mathcal{E}^{1/2}(\tau))
\mathcal{E}(\tau)+\dot{u}^2(\tau)\Big]d\tau \\
&\quad + \varepsilon \int_0^t\big(
 {\phi''_{t}}^2(0,t)+{\phi''_{t}}^2(L,t)\big)dt
 +c_{\gamma,\varepsilon}\int_0^t\big({\phi'''}^2(0,t)+{\phi'''}^2(L,t)\big)dt.
\end{aligned}
\end{equation}
Furthermore, we differentiate
\eqref{e1.1} with respect to t, then we have
\begin{equation} \label{difft}
\phi_{ttt}+(a\phi''_t)''+(a_t\phi'')''=b_t.
\end{equation}
 Let
\begin{equation}
P_3(t)=\|\phi_{tt}\|^2+(a\phi_t'',\phi_t'').
\end{equation}
We deduce that
\begin{align*}
 \dot{P}_3(t)
&= 2(\phi_{tt},\phi_{ttt})+2(a\phi_t'',\phi_{tt}'')+(a_t\phi_t'',\phi_t'')\\
&= 2(\phi_{tt},b_t-(a_t\phi'')''-(a\phi''_t)'')+2(a\phi_t'',\phi_{tt}'')
 +(a_t\phi_t'',\phi_t'')\\
&=2(\phi_{tt},b_t)-2(\phi_{tt},a_t\phi^{(4)})
 -4(\phi_{tt},a'_t\phi''')-2(\phi_{tt},a''_t\phi'')\\
&\quad +2\ddot{u}(t)(a\phi''_t)(L,t)+(a_t\phi_t'',\phi_t'').
\end{align*}
So
\begin{equation}\label{upsi31}
\begin{aligned}
 \Upsilon_3(t)
 &\leq c_\gamma P_3(t) \\
 &\leq c_\gamma P_3(0)+c_\gamma \int_0^t \big[(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)
 +\mathcal{E}^\frac{3}{2}(\tau)) \mathcal{E}(\tau)\big]d\tau \\
 &\quad +\varepsilon \int_0^t {\phi''_t}^2(L,t)dt
 +c_{\gamma,\varepsilon}\int_0^t\ddot{u}^2(t)dt.
\end{aligned}
\end{equation}
From \eqref{upsi2} and \eqref{upsi31}, we conclude that
 \begin{equation}\label{upsi23}
\begin{aligned}
 \Upsilon_2(t)+ \Upsilon_3(t)
&\leq c_\gamma (P_2(0)+P_3(0))
+\varepsilon \int_0^t \big({\phi''_t}^2(0,t)+{\phi''_t}^2(L,t)\big)dt\\
&\quad +c_\gamma \int_0^t \big[(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)+\mathcal{E}^\frac{3}{2}(\tau))
\mathcal{E}(\tau)+\dot{u}^2(t)\big]d\tau \\
&\quad +c_{\gamma,\varepsilon}\int_0^t
\big({\phi'''}^2(0,t)+{\phi'''}^2(L,t)+\ddot{u}^2(t)\big)dt.
\end{aligned}
\end{equation}
Next, we estimate the terms $\int_0^t {\phi''_{t}}^2(0,t)dt$ and $\int_0^t
{\phi''_{t}}^2(L,t)dt$.

Differentiating twice \eqref{e1.1} with respect to $x$, multiplying the
two sides of the equation by $(L-x)\phi'''$ and integrating from 0 to
$L$ by parts, we obtain
\begin{equation} \label{diffxx}
\phi''_{tt}+(a\phi'')^{(4)}=b'',
\end{equation}
\begin{gather*}
\int_0^L \phi''_{tt}(L-x)\phi'''dx
=\frac{\partial}{\partial t}(\phi''_{t},(L-x)\phi''')
 +\frac{1}{2}L {\phi''_{t}}^2(0,t)
-\frac{1}{2}\int_0^L {\phi''_{t}}^2 dx,
\\
\begin{aligned}
&\int_0^L (b''-(a\phi'')^{(4)})(L-x)\phi'''dx\\
&=(b'',(L-x)\phi''')-b'(0,t)L\phi'''(0,t)+\ddot{u}_1(t)L\phi'''(0,t)
 -\frac{1}{2}L({a\phi^{(4)}}^2)(0,t)\\
&\quad -\frac{1}{2}\int_0^L (a'(L-x)-a){\phi^{(4)}}^2dx
 +\int_0^L (3a'\phi^{(4)}+3a''\phi'''
 +a'''\phi'')(L-x)\phi^{(4)}dx\\
&\quad -b(L,t)\phi'''(L,t)+b(0,t)\phi'''(0,t)+\int_0^L
(a\phi'')''\phi^{(4)}dx.
\end{aligned}
\end{gather*}
Whence \begin{equation} \label{ident1}
\begin{aligned}
&\frac{1}{2}L {\phi''_{t}}^2(0,t)\\
&= -\frac{\partial}{\partial t}(\phi''_{t},(L-x)\phi''')
 +\frac{1}{2}\int_0^L {\phi''_{t}}^2 dx
 +(b'',(L-x)\phi''')\\
&\quad -b'(0,t)L\phi'''(0,t)-\frac{1}{2}L({a\phi^{(4)}}^2)(0,t)
-\frac{1}{2}\int_0^L (a'(L-x)-a){\phi^{(4)}}^2dx\\
&\quad +\int_0^L (3a'\phi^{(4)}+3a''\phi'''+a'''\phi'')(L-x)\phi^{(4)}dx\\
&\quad -b(L,t)\phi'''(L,t)+b(0,t)\phi'''(0,t)
 +\int_0^L (a\phi'')''\phi^{(4)}dx.
\end{aligned}
\end{equation}

It follows that
\begin{equation}\label{phi0}
\begin{aligned}
\int_0^t {\phi''_{t}}^2(0,t) dt
&\leq c_\gamma \Big(\Upsilon_2(0)+\Upsilon_3(0)+\Upsilon_2(t)+\Upsilon_3(t)\Big)\\
&\quad +c_\gamma \int_0^t \Big[\Big(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)
 +\mathcal{E}^\frac{3}{2}(\tau)\Big) \mathcal{E}(\tau)\Big]d\tau.
\end{aligned}
\end{equation}

Similarly, with respect to the term
$\int_0^t {\phi''_{t}}^2(L,t)dt$, multiplying the two sides of \eqref{diffxx}
by $x\phi'''$ and integrating from 0 to $L$ by parts, we obtain
\begin{equation} \label{e2.22}
\int_0^L \phi''_{tt}x\phi'''dx=\frac{\partial}{\partial
t}(\phi''_{t},x\phi''')+\frac{1}{2}L {\phi''_{t}}^2(L,t)
-\frac{1}{2}\int_0^L {\phi''_{t}}^2 dx,
\end{equation}
and
\begin{equation} \label{e2.23}
\begin{aligned}
&\int_0^L (b''-(a\phi'')^{(4)})x\phi'''dx\\
&=(b'',x\phi''')-b'(L,t)L\phi'''(L,t)+\ddot{u}(t)L\phi'''(L,t)
 +\frac{1}{2}L({a\phi^{(4)}}^2)(L,t)\\
&\quad -\frac{1}{2}\int_0^L (a'x+a){\phi^{(4)}}^2dx
 +\int_0^L (3a'\phi^{(4)}+3a''\phi'''+a'''\phi'')x\phi^{(4)}dx\\
&\quad +b(L,t)\phi'''(L,t)-b(0,t)\phi'''(0,t)-\int_0^L (a\phi'')''\phi^{(4)}dx.
\end{aligned}
\end{equation}
Using \eqref{e2.22} and \eqref{e2.23} we have
\begin{equation}\label{ident2}
\begin{aligned}
&\frac{1}{2}L {\phi''_{t}}^2(L,t)\\
&= -\frac{\partial}{\partial t}(\phi''_{t},x\phi''')
 +\frac{1}{2}\int_0^L {\phi''_{t}}^2 dx
 +(b'',x\phi''')-b'(L,t)L\phi'''(L,t)\\
&\quad +\ddot{u}(t)L\phi'''(L,t)+\frac{1}{2}L({a\phi^{(4)}}^2)(L,t)
 -\frac{1}{2}\int_0^L (a'x+a){\phi^{(4)}}^2dx\\
&\quad +\int_0^L (3a'\phi^{(4)}+3a''\phi'''+a'''\phi'')x\phi^{(4)}dx\\
&\quad +b(L,t)\phi'''(L,t)-b(0,t)\phi'''(0,t)
 -\int_0^L (a\phi'')''\phi^{(4)}dx.
\end{aligned}
\end{equation}
We deduce the inequality
\begin{equation}\label{phiL}
\begin{aligned}
\int_0^t {\phi''_{t}}^2(L,t) dt
&\leq c_\gamma \Big(\Upsilon_2(0)+\Upsilon_3(0)+\Upsilon_2(t)
 +\Upsilon_3(t)\Big)\\
&\quad +c_\gamma \int_0^t \Big[\Big(1+\mathcal{E}^{1/2}(\tau)+\mathcal{E}(\tau)
 +\mathcal{E}^\frac{3}{2}(\tau)\Big) \mathcal{E}(\tau)+\ddot{u}^2(t)\Big]d\tau.
\end{aligned}
\end{equation}
Finally, Substituting  inequality \eqref{phi0} and \eqref{phiL}
in \eqref{upsi23}, choosing $\varepsilon>0$ small enough such that the term
$\varepsilon c_\gamma [\Upsilon_2(t)+\Upsilon_3(t)]$ can be moved to
the left hand side of the inequality, then  \eqref{upsi230} is
obtained.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
 Lemma \ref{lem2.3} gives  inequality \eqref{Q}.
To prove inequality \eqref{Q1}, we notice that
$a\phi^{(4)}= b-\phi_{tt}-2a'\phi'''-a''\phi'' $, then there is
 $c_\gamma>0$, such that
 \begin{align*} %\label{phi4}
\|\phi^{(4)}\|^2
&\le c_\gamma(a\phi^{(4)},\phi^{(4)})\\
&= c_\gamma(b-\phi_{tt}-2a'\phi'''-a''\phi'',\phi^{(4)})\\
&\leq c_{\gamma,\varepsilon}
(\|b\|^2+\|\phi_{tt}\|^2+\|\phi'''\|^2+\|\phi''\|^2)
 +\varepsilon\|\phi^{(4)}\|^2.
\end{align*}
Choosing $\varepsilon$ sufficiently small, then there is $c_\gamma>0$, such that
\[
\|\phi^{(4)}\|^2 \leq c_\gamma
 (\|b\|^2+\|\phi_{tt}\|^2+\|\phi'''\|^2+\|\phi''\|^2)
\leq c_\gamma \mathcal{Q}(t).
\]
From the definition of $\mathcal{Q}(t)$ and $\mathcal{E}(t)$\ and using the
Poincar\'{e} inequality,  inequality \eqref{Q1} holds.
\end{proof}

Using the standard method as for the global solution of partial differential equation,
from Theorem \ref{thm2.1} we have the following proof.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 Clearly, it will suffice to prove
Theorem \ref{thm1.1} for the zero equilibrium $\omega=0$.

Let $T_1>0$ be arbitrary given. We take $\gamma=1$. Let
\begin{equation} \label{c1}
c_1=c_\gamma\ge 1
\end{equation}
be fixed such that the corresponding
inequalities \eqref{Q} and \eqref{Q1} of Theorem \ref{thm2.1} hold for $t$ in the
existence interval of the solution $\phi$.

We shall prove that, if the initial data $(\phi_0,\phi_1)$ and boundary
value $u$ are compatible and satisfy
\begin{equation}\label{pe0}
\mathcal{E}(0)+\max_{0\le t\le T_1}\mathcal{E}_\Gamma(t)+\int_0^{T_1}\mathcal{E}_\Gamma(t)dt
\le \frac{1}{16c_1^3}e^{-4c_1^2T_1},
\end{equation}
 then the solution of problem \eqref{e1.1} exists at least on the interval
$[0, T_1]$.

We set
\begin{equation} \label{eta1}
\eta=\frac{1}{4c_1}< \frac{1}{2}.
\end{equation}
Since $\mathcal{E}(0)\le \frac{\eta}{4}$, the solution of short time must
satisfy
\begin{equation} \label{Ed}
\mathcal{E}(t)\le \eta\le \frac{1}{2}
\end{equation}
 for some interval $[0, \delta]$.
 Let $\delta_0$ be the largest number such that \eqref{Ed} is true for
$t \in [0, \delta_0)$. We shall prove
$\delta_0 \ge T_1$ by contradiction.

Suppose that $\delta_0 < T_1$. In this interval
$t \in [0, \delta_0]$, the condition \eqref{c21} is true.
We apply Theorem \ref{thm2.1} and the inequalities \eqref{Q} and \eqref{Q1},
via \eqref{c1}--\eqref{Ed}. Then we conclude that
\begin{equation}\label{pe1}
\mathcal{E}(t)\le 2c_1^2[\mathcal{E}(0)
 +\max_{0\le t\le T_1}\mathcal{E}_\Gamma(t)+\int_0^{T_1}\mathcal{E}_\Gamma(t)dt]
+4c_1^2\int_0^t\mathcal{E}(t)dt,
\end{equation}
for $t \in [0, \delta_0]$. By \eqref{pe0} and \eqref{pe1},
the Gronwall inequality yields
$$
\mathcal{E}(\delta_0)\le \eta/2< \eta.
$$
This is a contradiction.
\end{proof}


\section{Locally exact controllability}

Let $T>0$ given.
The first step of the proof for the local exact controllability
depends on the following fact: Let $X$ and $Y$ be Banach spaces and
$\Phi: U\to Y$, where $U$ is an open subset of $X$, be
Frech\'et differentiable. If $\Phi'(x_0): X\to Y$ is
surjective, then there is an open neighborhood of $y_0=\Phi(x_0)$
contained in the image $\Phi(U)$.

Given an equilibrium $\omega\in H^4(0,L)\cap V(0,L)$, We invoke
Theorem \ref{thm1.1} to define the map for $u\in H^2(0,T)$ by setting
$$
F(u)=(\psi(\cdot,T),\psi_t(\cdot,T)),
$$
where $\psi$ is the solution to
\begin{equation} \label{psi0}
\begin{gathered}
\psi_{tt}+(a(x,\psi)\psi'')''=b(x,\psi,\psi',\psi''),
\quad x\in(0,L),t\in (0,T),\\
\psi(x,0)=\omega(x), \quad \psi_{t}(x,0)=0,\quad x\in(0,L),\\
\psi(0,t)=0,\quad \psi'(0,t)=0,\quad t\in(0,T),\\
\psi(L,t)=0,\quad \psi'(L,t)=u+\omega'(L),\quad t\in(0,T),
\end{gathered}
\end{equation}
Let $\varepsilon_T>0$ be given by Theorem \ref{thm1.1}. Then
\begin{equation}
F: B(0,\varepsilon_T)\to (H^4(0,L)\cap V(0,L))\times V(0,L),
\end{equation}
where $B(0,\varepsilon_T)\subset H^2(0,T)$ is the
ball with the radius $\varepsilon_T$ centered at 0. We note that
$F(0)=(\omega(x),0)$.

We need to evaluate
$F'(0)(u)=D_\lambda F(\lambda u)|_{\lambda=0}$.
It is easy to check that $F'(0)(u)=(\psi(\cdot,T),\psi_t(\cdot,T))$,
where $\psi(x,t)$ is the solution of the linear system
\begin{equation} \label{psi}
\begin{gathered}
\psi_{tt}+(p(x)\psi'')''= b_1(x)\psi +b_2(x)\psi'+b_{3}(x)\psi'',
\quad x\in(0,L),t\in (0,T),\\
\psi(x,0)=0, \quad \psi_{t}(x,0)=0, \quad x \in(0,L),\\
\psi(0,t)=0,\quad \psi'(0,t)=0,\quad t\in (0,T),\\
\psi(L,t)=0,\quad \psi'(L,t)=u(t),\quad t\in (0,T),
\end{gathered}
\end{equation}
and
\begin{gather*}
p(x)=a(x,\omega),\\
b_1(x)=b_{y_1}(x,\omega,\omega',\omega'')-(a_y(x,\omega)\omega'')'',\\
b_2(x)=b_{y_2}(x,\omega,\omega',\omega'')-2(a_y(x,\omega)\omega'')',\\
b_3(x)=b_{y_3}(x,\omega,\omega',\omega'')-a_y(x,\omega)\omega''.
\end{gather*}
We now verify that $F'(0)$ is
surjective. In the language of control theory surjectivity is just
exact controllability, which for a reversible system such as \eqref{psi} is
equivalent to null controllability.

Explicitly one has to show that, for $T> 0$ given, given
$\psi_0\in H^4(0,L)\cap V(0,L),\psi_1\in V(0,L)$, one can find
$u\in H^2(0,T)$ such that the solution to
\begin{equation} \label{control}
\begin{gathered}
\psi_{tt}+(p(x)\psi'')''= b_1(x)\psi +b_2(x)\psi'+b_{3}(x)\psi'',
\quad x\in(0,L),t\in (0,T),\\
\psi(x,0)=\psi_0(x),\quad \psi_{t}(x,0)=\psi_1(x), \quad x\in (0,L),\\
\psi(0,t)=0,\quad \psi'(0,t)=0,\quad t\in (0,T),\\
\psi(L,t)=0,\quad \psi'(L,t)=u(t),\quad t\in (0,T),
\end{gathered}
\end{equation} satisfies $\psi(\cdot,T)=0$ and
$\psi_t(\cdot,T)=0$.

\begin{theorem} \label{thm3.1}
Given an equilibrium $\omega\in H^4(0,L)\cap V(0,L)$.
Let $T>0$ given. Then for any
$$
(\psi^0,\psi^1)\in (H^4(0,L)\cap V(0,L))\times V(0,L),
$$
there is a control
$u\in H^2(0,T)$ such that the solution
$$
\psi \in C([0,T],H^4(0,L))\cap C^1([0,T],H^2(0,L)) \cap C^2([0,T],L^2(0,L))
$$
of problem \eqref{control} satisfies $\psi(\cdot,T)=0$ and
$\psi_t(\cdot,T)=0$.
\end{theorem}

 Unfortunately, our proof does not give any information on the time $T$
needed for the controlled motion.


\subsection{Distributed control}

 As to the exact controllability of linear
systems by distributed control, there is a long history and a lot of
results where many approaches are involved. Here the distributed
control means that solutions $(\psi(t),\psi_t(t)) $ of the
controlled system \eqref{control} are only in the space
$L^2(0,L) \times H^{-2}(0,L)$ for $t \in [0,T ]$.
One of the useful approaches is the
multiplier method, which introduced by Ho \cite{Ho} where Lions \cite{Lions}
provided a key technique of multipliers for observability estimates, to control
the linear system by its duality system. In the case of constant coefficients,
 Lagnese \cite{lagnese,Lagnese-L-L}, study the controllability of linear
beam by using the multiplier method. In this subsection, we will consider the
controllability of linear beam with variable coefficients.

As noted in \cite{lagnese}, everything is therefore going to rely on a prior
inequality
\begin{equation}\label{pI}
\|(\phi_0,\phi_1)\|^2_{H_0^2\times L^2}\le
C_T\int_0^T {\phi''}^2(L,t)dt,\quad
\forall(\phi_0,\phi_1)\in H_0^2(0,L)\times L^2(0,L)
\end{equation}
where $\phi$ is the solution of the system
\begin{equation} \label{dual}
\begin{gathered}
\phi_{tt}+(p(x)\phi'')''= b_1(x)\phi -(b_2(x)\phi)'+(b_{3}(x)\phi)'',
\quad x\in(0,L),t\in (0,T),\\
\phi(x,0)=\phi_0(x), \quad \phi_{t}(x,0)=\phi_1(x),\quad x\in (0,L),\\
 \phi(0,t)=0,\phi'(0,t)=0,\quad t\in(0,T),\\
\phi(L,t)=0,\phi'(L,t)=0, \quad t\in(0,T),
\end{gathered}
\end{equation}

Given $\phi_0\in H_0^2(0,L),\phi_1\in L^2(0,L)$. We define $\phi$ as
the solution to \eqref{dual} and we define $\eta$ as the solution of
\begin{equation} \label{eta}
\begin{gathered}
\eta_{tt}+(p(x)\eta'')''=b_1(x)\eta+b_2(x)\eta'+b_3(x)\eta''£¬
\quad x\in(0,L),t\in (0,T), \\
\eta(x,T)=0, \quad \eta_{t}(x,T)=0,\quad x\in (0,L),\\
\eta(0,t)=0,\quad \eta'(0,t)=0,\quad t\in(0,T),\\
\eta(L,t)=0,\quad \eta'(L,t)=\phi''(L,t), \quad t\in(0,T),
\end{gathered}
\end{equation}
 We define the operator
$\Lambda : H_0^2(0,L)\times L^2(0,L) \to H^{-2}(0,L)\times L^2(0,L)$ by
 \begin{equation} \label{Lambda}
\lambda (\phi_0,\phi_1)= (\eta_t(0),-\eta(0))
\end{equation}
The solution $\eta$ of \eqref{eta} is a weak solution defined by transposition,
in such a way that Green's formula makes sense. Therefore
\begin{equation}\label{oper}
\langle \Lambda(\phi_0,\phi_1),(\phi_0,\phi_1)\rangle
=\int_0^T p(L){\phi''}^2(L,t)dt.
\end{equation}

\begin{lemma} \label{lem3.1}
Let $\phi$ be a solution of \eqref{dual}.
Then there exist constants $c,c_0>0$ such that for $T\ge t>0$:
 \begin{equation}\label{EN}
e^{-ct}E(0)-N(T)\le E(t)\le [E(0)+N(T)]e^{ct},
\end{equation}
 where we introduced the energy
$$
E(t)=\int_0^L\Big(\phi_t^2+p(x){\phi''}^2\Big)dx
$$
and set $N(T)=c_0\int_0^T\|\phi\|^2_1 dt$.
\end{lemma}

\begin{proof}
 Multiplying equation \eqref{dual} by $\phi_t$ and integrating over
$(s,t)\times (0,L)$ by parts in $t$ on the left-hand side we obtain that
for all $s,t$:
\begin{equation}
E(t)=E(s)+2\int_s^t\int_0^L
\phi_t\Big(b_1(x)\phi -(b_2(x)\phi)'+(b_{3}(x)\phi)''\Big)\,dx\,dt.
\end{equation}
By Schwartz inequality, we obtain for $t\ge s\ge 0$:
$$
2\int_s^t\int_0^L
\phi_t\Big(b_1(x)\phi -(b_2(x)\phi)'+(b_{3}(x)\phi)''\Big)\,dx\,dt
\le N(T)+c\int_s^t E(\tau)d\tau.
$$
So
\begin{gather} \label{EN1}
E(t)\le [E(s)+N(T)]+c\int_s^t E(\tau)d\tau,\\
\label{EN2}
E(s)\le [E(t)+N(T)]+c\int_s^t E(\tau)d\tau.
\end{gather}
We apply the classical argument of the Gronwall's inequality to
 \eqref{EN1} and \eqref{EN2}, where we note that the terms into
the square brackets are
 independent of $s$ in \eqref{EN2}, we thus obtain for $t\ge s\ge 0$:
\begin{equation}\label{EN3}
E(t)\le [E(s)+N(T)]e^{c(t-s)};\quad E(s)\le [E(t)+N(T)]e^{c(t-s)}.
\end{equation}
Setting $s=0$ and thus $t>0$ in \eqref{EN3} yields \eqref{EN}.
\end{proof}

The observability inequality for the system \eqref{dual} is covered in
the following lemma.

 \begin{lemma} \label{lem3.2}
Let an equilibrium $\omega \in H^4(0,L)\cap V(0,L)$ be given.
Then for $T>0$, $\Lambda$ is an isomorphism from $H_0^2(0,L) \times L^2(0,L)$
onto $H^{-2}(0,L) \times L^2(0,L)$. In particular, there are constants
$C_1> 0$ and $C_2 > 0$ such that the inequality
\begin{equation} \label{observe0}
C_1\|(\phi_0,\phi_1)\|^2_{H_0^2\times L^2}
\le \int_0^T p(L){\phi''}^2(L,t)dt
\le C_2\|(\phi_0,\phi_1)\|^2_{H_0^2\times L^2}
\end{equation}
holds true for $(\phi_0,\phi_1)\in {H_0^2(0,L)\times L^2(0,L)}$.
\end{lemma}

\begin{proof}
Assume that $\phi_0\in H_0^2(0,L),\phi_1\in L^2(0,L)$.
Then \eqref{dual} admits a unique solution
$\phi\in C([0,T];H_0^2(0,L))\cap C^1([0,T];L^2(0,L))$ and the energy $E(t)$
 of the system \eqref{dual} satisfies
$$
E(0)=E(\phi_0,\phi_1)=\int_0^L(\phi_1^2+p(x){\phi_0''}^2)dx.
$$
That is, we are going to prove that there exist two constants
$C_1,C_2>0$ such that:
\begin{equation}\label{observability inequality}
C_1E(0)\le \int_0^T p(L){\phi''}^2(L,t)dt\le C_2E(0).
\end{equation}

We always have the right side of \eqref{observability inequality},
so we just need to prove that
$$
\int_0^T p(L){\phi''}^2(L,t)dt \ge C_1E(0).
$$
We use multiplier $h(x)\phi'$, where $h(x)$ is a function on the
interval $[0,L]$, and we obtain
\begin{gather*}
\int_0^L \int_0^T \phi_{tt}h(x)\phi'\,dx\,dt
=(h(x)\phi',\phi_t)|_0^T +\frac{1}{2}\int_0^T \int_0^L h'(x)\phi^2_t\,dx\,dt.
\\
\begin{aligned}
&\int_0^T \int_0^L(p(x)\phi'')''h(x)\phi'\,dx\,dt\\
&= -\frac{1}{2}\int_0^T p(x)h(x){\phi''}^2|_0^Ldt+\frac{3}{2}\int_0^T
\int_0^Lp(x)h'(x){\phi''}^2\,dx\,dt\\
&\quad -\frac{1}{2}\int_0^T \int_0^L p'(x)h(x){\phi''}^2\,dx\,dt
-\frac{1}{2}\int_0^T \int_0^L (p(x)h''(x))'{\phi'}^2\,dx\,dt.
\end{aligned} \\
\begin{aligned}
&\int_0^T \int_0^L(b_1(x)\phi -(b_2(x)\phi)'+(b_{3}(x)\phi)'')h\phi'\,dx\,dt\\
&= \int_0^T\int_0^L
\Big(m_1(x)\phi^2+m_2(x){\phi'}^2\Big)\,dx\,dt.
\end{aligned}
\end{gather*}
These equalities give
\begin{align*}
&(h(x)\phi',\phi_t)|_0^T +\frac{1}{2}\int_0^T \int_0^L
h'(x)\phi_t^2\,dx\,dt+\frac{3}{2}\int_0^T
\int_0^Lp(x)h'(x){\phi''}^2\,dx\,dt\\
& -\frac{1}{2}\int_0^T \int_0^L
p'(x)h(x){\phi''}^2\,dx\,dt
-\frac{1}{2}\int_0^T \int_0^L (p(x)h''(x))'{\phi'}^2\,dx\,dt\\
& -\frac{1}{2}\int_0^T p(x)h(x){\phi''}^2|_0^Ldt\\
&= \int_0^T\int_0^L
\Big(m_1(x)\phi^2+m_2(x){\phi'}^2\Big)\,dx\,dt.
\end{align*}

Next we use another multiplier $h'(x)\phi$. Integrating by part on
$[0,T]\times [0,L]$, we obtain
\begin{gather*}
\int_0^L \int_0^T \phi_{tt}h'(x)\phi \,dx\,dt=(h'(x)\phi,\phi_t)|_0^T
-\int_0^T \int_0^L h'(x)\phi^2_t\,dx\,dt.
\\
\begin{aligned}
&\int_0^L \int_0^T (p(x)\phi'')''h'(x)\phi \,dx\,dt\\
&= \int_0^T \int_0^Lp(x)h'(x)(\phi'')^2\,dx\,dt+\frac{1}{2}\int_0^T
\int_0^L (p(x)h'''(x))''\phi^2\,dx\,dt \\
&\quad -\int_0^T \int_0^L
\Big(p(x)h'''(x)+(p(x)h''(x))'\Big){\phi' }^2\,dx\,dt\,,
\end{aligned} \\
\int_0^T \int_0^L\Big(b_1(x)\phi -(b_2(x)\phi)'
 +(b_{3}(x)\phi)''\Big)h'(x)\phi \,dx\,dt\\
= \int_0^T\int_0^L \Big(m_3(x)\phi^2+m_4(x){\phi'}^2\Big)\,dx\,dt,
\end{gather*}
where functions $m_i(x)(i=1,2,3,4)$ are the function of
$b_1(x)$, $b_2(x)$, $b_3(x)$, $h(x)$ and their derivatives.
Considering the fact that functions $m_i(x)(i=1,2,3,4)$ are all lower order term,
we omit specific functions form here.

By using the above equalities, we have
\begin{equation} \label{e3.17}
\begin{aligned}
&\frac{1}{2}\int_0^T p(x)h(x){\phi''}^2|_0^Ldt\\
&= \Big(h(x)\phi'-\frac{1}{2}h'(x)\phi,\phi_t\Big)\Big|_0^T 
 +\int_0^T \int_0^L h'(x)\phi^2_t\,dx\,dt\\
&\quad +\frac{1}{2}\int_0^T \int_0^L
p(x)h'(x){\phi''}^2\,dx\,dt \\
&\quad +\frac{1}{2}\int_0^T \int_0^L
(p(x)h'(x)-p'(x)h(x)){\phi''}^2\,dx\,dt\\
&\quad -\int_0^T \int_0^L [d_1(x){\phi}^2+d_2(x){\phi'}^2] \,dx\,dt,
\end{aligned}
\end{equation}
Where functions $d_1(x),\,d_2(x)$ are the function of
$b_1(x)$, $b_2(x)$, $b_3(x)$, $h(x)$ and their derivatives.

Let $h(x)$ be the solution of the problem
\begin{equation} \label{h}
\begin{gathered}
 h_x=\frac{b}{p}h+1,\\
h(0)=0.
\end{gathered}
\end{equation}
where $b(x) = \max(p_x, 0)$, e.g.,
 $$
h(x) =e^{\int_0^x\frac{b}{p}dx} \int_0^x e^{-\int_0^s\frac{b}{p}d\tau} ds,\quad
 0<x<L.
$$
 Therefore, if we introduce
$$
Y=\Big(h(x)\phi'-\frac{1}{2}h'(x)\phi,\phi_t\Big)\Big|_0^T\,.
$$
From \eqref{e3.17}, we obtain
\begin{equation} \label{e3.19}
\frac{1}{2}\int_0^T p(L)h(L){\phi''}^2(L,t)dt
\ge \int_0^T E(t)dt-|Y|-C\int_0^T \int_0^L
({\phi}^2+{\phi'}^2)\,dx\,dt.
\end{equation}
For $|Y|$, we have
\begin{align*}
|Y| &\leq \varepsilon\int_0^L{\phi_t}^2dx|_0^T
 +C_\varepsilon\int_0^L({\phi}^2+{\phi'}^2)dx|_0^T\\
&\leq \varepsilon\Big( E(T)+E(0)\Big)
 +C_\varepsilon\Big(\|\phi(T)\|^2_{H^1}+\|\phi(0)\|^2_{H^1}\Big).
\end{align*}
Using the inequality in Lemma \ref{lem3.1}, we compute
\begin{equation} \label{e3.21}
\int_0^T E(t)dt
\ge \Big(\int_0^T e^{-ct}dt\Big) E(0)-TN(T)
\ge k_T[E(0)+E(T)]-\frac{N(T)}{2},
\end{equation}
where $k_T=(\int_0^T e^{-ct}dt) \frac{e^{-cT}}{2}$ and
$E(T)\ge e^{-cT}E(0)-N(T)$.

 Using inequalities \eqref{e3.19}-\eqref{e3.21} and
\eqref{observe0}, we obtain
\begin{align*}
&\frac{1}{2}\int_0^T p(L)h(L){\phi''}^2(L,t)dt\\
&\ge k_T[E(0)+E(T)]-\frac{N(T)}{2}-C\int_0^T \int_0^L ({\phi}^2+{\phi'}^2)\,dx\,dt\\
&\quad -\varepsilon\Big\{\Big(E(T)+E(0)\Big)+C_\varepsilon
 \Big(\|\phi(T)\|^2_{H^1}+\|\phi(0)\|^2_{H^1}\Big)\Big\}\\
&\ge (k_T-\varepsilon)[E(0)+E(T)]-C_T\|\phi\|^2_{L^\infty([0,T];H^1(0,L))}\\
&\ge (k_T-\varepsilon)[E(0)+e^{-cT}E(0)-N(T)]
 -C_T\|\phi\|^2_{L^\infty([0,T];H^1(0,L))}.
\end{align*}
Choosing $\varepsilon>0$ small enough, such that
$k_T-\varepsilon>0$, we obtain
\begin{equation}\label{aE0}
\alpha E(0)\le \int_0^T p(L)h(L){\phi''}^2(L)dt
 +C_T\|\phi\|^2_{L^\infty([0,T];H^1(0,L))},
\end{equation}
 where $\alpha=2(k_T-\varepsilon)(1+e^{-CT})$.

By using a compactness and uniqueness argument similar to Lagnese
\cite{Lagnese-L-L}, we can easily prove that there exist constant $C>0$,such that
$$
\|\phi\|_{L^\infty(0,T;H^1(0,L))}^2\le C\int_0^T{\phi''}^2(L,t)dt.
$$
So the proof is complete.
\end{proof}

Therefore, $\Lambda$ defines an isomorphism from
$H_0^2(0,L)\times L^2(0,L)$ onto $H^{-2}(0,L)\times L^2(0,L)$.
For $\psi_0\in L^2(0,L),\psi_1\in H^{-2}(0,L) $ given , we solve
$$
\Lambda(\overline{\phi}_0,\overline{\phi}_1)=(\psi_1,-\psi_0).
$$
Then we define $\overline{\phi}$ as the corresponding solution of
\eqref{dual} and we take
$$
u(t)=\overline{\phi''}(L,t)\in L^2(0,T),
$$
which is the control driving the system to rest at time $t=T$.

Let us now introduce a new norm
$$
\|(\phi_0,\phi_1)\|_{\rm new}^2=\int_0^T p(L){\phi''}^2(L,t)dt.
$$
It follows from \eqref{observe0} that for $T>0, \|(\phi_0,\phi_1)\|_{\hbox{new}}$
is a norm which is equivalent to the $H_0^2(0,L)\times L^2(0,L)$ norm.

However, the above control strategy only gives distributed control
functions because solutions ($\eta,\eta_t$) of the controlled system
\eqref{eta} are only in $L^2\times H^{-2}$ no matter $(\phi_0,\phi_1) $
are smooth or not.

\subsection{Smooth control}

Smooth control has been considered by
Lasiecka and Triggiani \cite{L-T2,L-T3}, Tataru \cite{Tataru1}.
Here we shall modify the above control strategy to obtain smooth control
to meet the need of Theorem \ref{thm3.1} by induction on the order of the space.

Firstly, we define operator $B$ by
\begin{equation} \label{B}
Bu=-(p(x)u'')''+b_1(x)u -(b_2(x)u)'+(b_{3}(x)u)'',\quad\forall u\in H^4(0,L).
\end{equation}
Let $T>0$ be given. We assume that
$z \in C^\infty(-\infty,\infty)$ is such that $0\le z(t) \le 1 $ with
\begin{equation}\label{z}
z(t) = \begin{cases}
 0, & t \ge T \\
 1, & t \le 0 \\
\end{cases}
\end{equation}
For $(\phi_0,\phi_1)\in (H^4(0,L)\cap H_0^2(0,L))\times H_0^2(0,L)$
given, we solve  \eqref{dual} and then, instead of \eqref{eta}, we
solve the problem
\begin{equation} \label{eta1b}
\begin{gathered}
\eta_{tt}+(p(x)\eta'')''=b_1\eta+b_2\eta'+b_3\eta'',
\quad
x\in(0,L),t\in (0,T),\\
\eta(x,T)=0, \quad \eta_{t}(x,T)=0,\quad x\in(0,L),\\
\eta(0,t)=0, \quad \eta'(0,t)=0,\quad t\in(0,T),\\
\eta(L,t)=0, \quad \eta'(L,t)=z(t)\phi''(L,t), \quad t\in(0,T),
\end{gathered}
\end{equation}

Let $\Lambda$ be given by \eqref{Lambda} where $\eta$ is the
solution of \eqref{eta1b} this time.
It is easy to check that, for any
$(\phi_0, \phi_1), (\varphi_0,\varphi_1) \in H_0^2(0,L )\times L^2(0,L)$,
 \begin{equation} \label{observe}
\langle \Lambda(\phi_0,\phi_1),(\varphi_0,\varphi_1)\rangle _{L^2\times
L^2}
=\int_0^T z(t)p(L){\phi''(L,t)}\varphi''(L,t)dt,
\end{equation}
where $\phi$ and $\varphi$ are solutions of  \eqref{dual} with
initial data $(\phi_0,\phi_1)$ and $(\varphi_0,\varphi_1) $,
respectively.

We shall show that problem \eqref{eta1b} provides smooth controls to
Theorem \ref{thm3.1} by the following lemma.

\begin{lemma} \label{lem3.3}
Let $\Lambda$ be given by \eqref{Lambda} where $\eta$ is the
solution of  \eqref{eta1b}. Then there are constants $c_1 > 0$
and $c_2 > 0$ such that
\begin{equation} \label{3.1}
\begin{aligned}
 c_1\|(\phi_0, \phi_1)\|_{H^4(0,L)\times H^2(0,L)}
&\le \|\Lambda(\phi_0, \phi_1)\|_{L^2(0,L)\times H^2(0,L)}\\
&\le c_2\|(\phi_0, \phi_1)\|_{H^4(0,L)\times H^2(0,L)},
\end{aligned}
\end{equation}
for all $(\phi_0, \phi_1)\in (H^4(0,L)\cap H_0^2(0,L))\times H_0^2(0,L)$,
\end{lemma}

\begin{proof} From the previous argument,
\begin{equation}
\begin{aligned}
 c_1\|(\phi_0,\phi_1)\|_{H^2(0,L)\times L^2(0,L)}
&\le \|\Lambda(\phi_0, \phi_1)\|_{H^{-2}(0,L)\times L^2(0,L)}\\
&\le c_2\|(\phi_0, \phi_1)\|_{H^2(0,L)\times L^2(0,L)}
\end{aligned}
\end{equation}
holds.
Let $(\phi_0, \phi_1)\in (H^4(0,L)\cap H_0^2(0,L))\times H_0^2(0,L)$
be given, Suppose that $\phi$ is the solution of problem \eqref{dual}
corresponding to the initial data $(\phi_0, \phi_1)$.

For $(\varphi_0, \varphi_1)\in (H^6(0,L)\cap H_0^2(0,L))\times
(H^4(0,L)\cap H_0^2(0,L))$, let $\varphi$ is the solution of \eqref{dual}
 corresponding to the initial data $(\varphi_0,
\varphi_1)$. Then $\varphi_t$ and $\varphi_{tt}$ are the solutions
of \eqref{dual} corresponding to the initial data
$(\varphi_1,B\varphi_0)$ and $(B\varphi_0,B\varphi_1)$,
respectively.

Using the initial data $(\phi_0, \phi_1) $ and
$(B\varphi_0,B\varphi_1)$ in the formula \eqref{observe}, we obtain
 \begin{equation}
(\eta(0),B\varphi_1)-(\eta_t(0),B\varphi_0)
=-\int_0^T z(t)p(L){\phi''(L,t)}\varphi_{tt}''(L,t)dt.
\end{equation}
On the one hand, integrating by parts with respect to the
variable $t$ on $[0, T]$, we obtain
 \begin{align*}
&- \int_0^T z(t)p(L){\phi''(L,t)}\varphi_{tt}''(L,t)dt \\
&= p(L)\phi_0''(L)\varphi_1''(L)
 +p(L)\int_0^T z_t(t)\phi''(L,t)\varphi_{t}''(L,t)dt \\
&\quad +p(L)\int_0^T z(t)\phi_t''(L,t)\varphi_{t}''(L,t)dt
 \end{align*}
On the other hand, using  \eqref{B},
\begin{align*}
&(\eta(0),B\varphi_1)\\
&= -\int_0^L  \eta(0)(p(x)\varphi_1'')''dx+\int_0^L\eta(0)\Big(b_1(x)\varphi_1
-(b_2(x)\varphi_1)'+(b_{3}(x)\varphi_1)''\Big)dx\\
&=  p(L)\phi_0''(L)\varphi_1''(L)
-\int_0^L(p(x)\eta''(0))''\varphi_1dx\\
&\quad +\int_0^L\Big(b_1(x)\eta(0)+b_2(x)\eta'(0)
+b_{3}(x)\eta''(0)\Big)\varphi_1dx
\end{align*}
then
\begin{align*}
&(\eta(0),B\varphi_1)-(\eta_t(0),B\varphi_0)\\
&= p(L)\phi_0''(L)\varphi_1''(L)
-\int_0^L(p(x)\eta''(0))''\varphi_1dx\\
&\quad +\int_0^L\Big(b_1(x)\eta(0)+b_2(x)\eta'(0)
+b_{3}(x)\eta''(0)\Big)\varphi_1dx-(\eta_t(0),B\varphi_0).
\end{align*}
So we have the  identity
\begin{equation} \label{I}
 I(\phi,\varphi) =(B^\ast\eta(0),\varphi_1)-(\eta_t(0),B\varphi_0),
\end{equation}
where
\begin{gather*}
I(\phi,\varphi)=p(L)\int_0^T z_t(t)\phi''(L,t)\varphi_{t}''(L,t)dt
+p(L)\int_0^T z(t)\phi_t''(L,t)\varphi_{t}''(L,t)dt,
\\
B^\ast\eta(0)=-(p(x)\eta''(0))''+b_1(x)\eta(0)+b_2(x)\eta'(0)
+b_{3}(x)\eta''(0).
\end{gather*}

Since $(H^6(0,L)\cap H_0^2(0,L))\times (H^4(0,L)\cap H_0^2(0,L))$ is
dense in $(H^4(0,L)\cap H_0^2(0,L))\times H_0^2(0,L)$, the
identity \eqref{I} is true for all 
$(\varphi_0,\varphi_1)\in (H^4(0,L)\cap H_0^2(0,L))\times H_0^2(0,L)$.
Letting $\varphi_0=0$ in \eqref{I}, we obtain
\begin{equation} \label{I1}
I(\phi,\varphi)=(B^\ast\eta(0),\varphi_1),
\end{equation}
for $ \varphi_1\in H_0^2(0,L )$
where $\varphi$ is the solution of \eqref{dual} for the initial
data $(0,\varphi_1)$.

Moreover, by inequality \eqref{observe0}, we have the
estimate
\begin{equation} \label{I2}
\begin{aligned}
&I(\phi,\varphi)|\\ 
&= p(L)\int_0^T z_t(t)\phi''(L,t)\varphi_{t}''(L,t)dt
 +p(L)\int_0^T z(t)\phi_t''(L,t)\varphi_{t}''(L,t)dt \\
&\le c\Big[\int_0^T p(L)(\phi''(L,t))^2dt+\int_0^T
 p(L)(\phi_t''(L,t))^2dt\Big]^{1/2} \\
&\quad\times \Big(\int_0^T p(L)(\varphi_t''(L,t))^2dt\Big)^{1/2} \\
&\le  c\Big(E(\phi_0,\phi_1)+E(\phi_1,B\phi_0)\Big)^{1/2}
  \Big(\int_0^T p(L)(\varphi_t''(L,t))^2dt\Big)^{1/2} \\
&\le c\Big(\|\phi_0\|^2_4+\|\phi_1\|^2_2\Big)^{1/2}\|\varphi_1\|_2.
\end{aligned}
\end{equation}
In terms of \eqref{I1}-\eqref{I2}, we obtain
\begin{equation} \label{B*}
\|B^\ast\eta(0)\|_{-2}
\le \sup_{\|\varphi_1\|_2=1}\Big(B^\ast\eta(0),\varphi_1\Big) \\
\le  c(\|\phi_0\|^2_4+\|\phi_1\|^2_2)^{1/2}.
\end{equation}
Now using the ellipticity of the operator $B^\ast$ and from \eqref{B*},
we have
\begin{align*}
\|\eta(0)\|_2
&\le  c\Big(\|\eta(0)\|_0+\|B^\ast\eta(0)\|_{-2}\Big) \\
&\le c\Big(\|\eta(0)\|_0+(\|\phi_0\|^2_4+\|\phi_1\|^2_2)^{1/2}\Big) \\
&\le c(\|\phi_0\|^2_4+\|\phi_1\|^2_2)^{1/2}.
\end{align*}
where $\|\eta(0)\|_0\le  c(\|\phi_0\|^2_2+\|\phi_1\|^2_0)^{1/2}$ is used.
A similar argument yields
\begin{equation}
 \|\eta_t(0)\|_{0}\le  c(\|\phi_0\|^2_4+\|\phi_1\|^2_2)^{1/2},
\end{equation}
after we let $\varphi_0\in (H^4(0,L)\cap H_0^2(0,L)),\varphi_1=0$ in
\eqref{I}.

Next, let us prove the left hand side of inequality \eqref{3.1}. We
set $\phi_0 =\varphi_0$ and $\phi_1 = \varphi_1$ in \eqref{I}, 
\begin{equation} \label{I3}
\begin{aligned}
I(\phi,\phi)
&\ge  \int_0^{T_1} p(L)(\phi''_t(L,t))^2dt
  -c_\varepsilon\int_0^T p(L)(\phi''(L,t))^2dt\\
&\quad -\varepsilon\int_0^T p(L)(\phi_t''(L,t))^2]dt \\
&\ge  \int_0^{T_1}
p(L)(\phi_t''(L,t))^2dt-c_\varepsilon
E(\phi_0,\phi_1)-\varepsilon E(\phi_1,B\phi_0) \\
&\ge C_1E(\phi_1,B\phi_0)-c_\varepsilon(\|\phi_0\|^2_2+\|\phi_1\|^2_0).
\end{aligned}
\end{equation}
Since
\begin{align*}
&(B^\ast\eta(0),\phi_1)\\
&=-\int_0^L
p(x)\eta''(0)\phi''_1dx+\int_0^L\Big(b_1(x)\eta(0)+b_2(x)\eta'(0)
+b_{3}(x)\eta''(0)\Big)\phi_1dx,
\end{align*}
we have
\begin{equation} \label{B1}
\begin{aligned}
& (B^\ast\eta(0),\phi_1)-(\eta_t(0),B\phi_0) \\
&\le  c(\|\eta(0)\|_2\|\phi_1\|_2+\|\eta_t(0)\|_0\|B\phi_0\|_{0}) \\
&\le c\Big(\|\eta(0)\|^2_2+\|\eta_t(0)\|^2_{0}\Big)^{1/2}
\Big(\|\phi_1\|^2_2+\|B\phi_0\|^2_{0}\Big)^{1/2}.
\end{aligned}
\end{equation}
Using \eqref{I3}-\eqref{B1} and the induction assumption
$$
c(\|\phi_0\|^2_2+\|\phi_1\|^2_{0})\le
\|\eta(0)\|^2_{0}+\|\eta_t(0)\|^2_{-2},
$$
it follows that
\begin{align*}
\|\phi_0\|^2_4+\|\phi_1\|^2_2
&\leq c E(\phi_1,B\phi_0)+c\Big( \|\phi_0\|^2_2+\|\phi_1\|^2_0\Big) \\
&\leq c\Big(\|\eta(0)\|^2_2+\|\eta_t(0)\|^2_0\Big)
+c\Big(\|\phi_0\|^2_2+\|\phi_1\|^2_{0}\Big) \\
&\leq c\Big(\|\eta(0)\|^2_2+\|\eta_t(0)\|^2_0\Big).
\end{align*}
The proof  is complete.
\end{proof}

A similar argument can used for establishing the inequality
\begin{equation}\label{smooth1}
\begin{aligned}
c_1\|(\phi_0, \phi_1)\|_{H^6(0,L)\times H^4(0,L)}
&\le \|\Lambda(\phi_0, \phi_1)\|_{H^2(0,L)\times H^4(0,L)}\\
&\le c_2\|(\phi_0,\phi_1)\|_{H^6(0,L)\times H^4(0,L)}.
\end{aligned}
\end{equation}
for all $(\phi_0, \phi_1)\in (H^6(0,L)\cap H_0^2(0,L))
\times (H^4(0,L)\cap H_0^2(0,L))$.

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 It follows that  the operator
$$
\Lambda:(H^6(0,L)\cap H_0^2(0,L))\times (H^4(0,L)\cap H_0^2(0,L))\to
{V(0,L)\times (H^4(0,L)\cap V(0,L))}
$$ 
is surjective.
Let $(\psi_0,\psi_1)\in (H^4(0,L)\cap V(0,L))\times V(0,L)$ be
given, then there is $(\phi_0, \phi_1)\in (H^6(0,L)\cap
H_0^2(0,L))\times (H^4(0,L)\cap H_0^2(0,L))$ such that the control
$$
u(t)=z(t)\phi''(L,t)
$$ 
which drives system \eqref{control} to rest at the time $T$, 
where $\phi$ is the solution of \eqref{dual} with the initial 
data $(\phi_0, \phi_1)$.

Since $\phi_t$, $\phi_{tt}$ is the solution of \eqref{dual} with the 
initial data $(\phi_1,B\phi_0)$ and $(B\phi_0,B\phi_1)$ respectively,
we conclude that $\phi''_t(L,t)$,$\phi''_{tt}(L,t)\in L^2(0,T)$ 
from Lemma \ref{lem3.2}. Then $u\in H^2(0,T)$. 
\end{proof}

If we change the boundary control condition into $\phi(0,t)=0$, $\phi''(0,t)=0$,
$\phi(L,t)=0$, $\phi''(L,t)=u(t)$ or others, the methods in this 
article still be applicable.

\subsection*{Acknowledgements}
This research is supported by the
National Science Foundation of China, grants no. 61104129, no. 11201272
and no. 11171195.
I want to thank Professor Peng-Fei Yao for his valuable guidance. 
The author would like to thank the editor and the reviewer 
for their valuable comments and constructive suggestions
which help to improve the presentation of this article.


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\end{document}
