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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 114, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/114\hfil Observability and controllability]
{Observability and controllability for a vibrating string with
dynamical boundary control}

\author[A. Wehbe\hfil EJDE-2010/114\hfilneg]
{Ali Wehbe}

\address{Ali Wehbe \newline
Lebanese University\\
Faculty of sciences I\\
Hadath-Beirut, Lebanon}
\email{ali.wehbe@ul.edu.lb}

\thanks{Submitted June 21, 2010. Published August 17, 2010.}
\subjclass[2000]{35B37, 35D05, 93C20, 73K50}
\keywords{Wave equation; dynamical control; exact controllability;
\hfill\break\indent HUM method}

\begin{abstract}
 We consider the exact controllability of a wave equation by means
 of dynamical boundary control.  Unlike the classical control,
 a difficulty is due to the presence  of the dynamical type.
 First, we establish a new weak observability  results. Next,
 by the HUM method, we prove that the system is  exactly controllable
 by means of regular dynamical boundary control.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newcommand{\norm}[1]{\|#1\|}

\section{Introduction and statement of the main result}

The aim of this paper is to investigate the observability and  the
exact controllability of the one-dimensional system
\begin{equation}
\begin{gathered}
y_{tt} - y_{xx} = 0  \quad 0<x<1,\; t>0, \\
y(0,t) =0  \quad t>0,\\
y_x (1,t) + \eta(t) = 0 \quad t>0,  \\
\eta_t(t) - y_t(1,t) = v(t) \quad t>0
\end{gathered} \label{e11}
\end{equation}
with the initial conditions
\begin{equation}
y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad 0<x<1, \quad
\eta(0)=\eta_0\in \mathbb{R}\label{e12}
\end{equation}
where $v(t)$ denotes the dynamical boundary control.

In a previous paper \cite{w}, we have considered the energy decay
rate of the following one-dimensional wave equation with dynamical
boundary control
\begin{equation}
\begin{gathered}
y_{tt} - y_{xx} = 0  \quad 0<x<1,\; t>0, \\
y(0,t) =0  \quad t>0,\\
y_x (1,t) + \eta(t) = 0 \quad t>0,  \\
\eta_t(t) - y_t(1,t) = -\eta(t) \quad t>0
\end{gathered} \label{e11p}
\end{equation}
with the  initial conditions
\begin{equation}
y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad 0<x<1, \quad
\eta(0)=\eta_0\in \mathbb{R}\label{e12p}
\end{equation}
where $\eta(t)$ denotes the dynamical boundary control. We proved that
the uniform decay rate of the system \eqref{e11p}-\eqref{e12p} is not
true in the energy space. In addition, using a spectral approach,
we established the optimal energy decay rate $1/t$ for smooth
initial data.

A physical implementation of the dynamic control
may be used in pressurized gas tanks with servo controlled actors,
as well as in standard mass-spring dampers (see \cite{m} and the
references herein). We mention that the dynamical controls form a
part of indirect mechanisms proposed by Russell (see \cite{ru1}
and the references herein), see also \cite{we} and \cite{rw}.

Now let $y$ be a smooth solution of the system \eqref{e11}.
We define the associated energy
\begin{equation}
E(t)=\frac{1}{2}\Big\{\int_0^1(y_x^2+y_t^2)dx+\eta^2\Big\}.
\end{equation}
Denoting by $Y(x,t)=(y(x,t),y_t(x,t),\eta(t))$ the state of the 
system \eqref{e11} and by
$V=(0,0,v)$ the control. we can formulate the system \eqref{e11}-\eqref{e12} 
as an abstract problem
\begin{equation}
Y_t=AY+V,\quad Y(0)=Y_0\in \mathcal{H} \label{e14}
\end{equation}
where  $A$ is an m-dissipative operator on an appropriate Hilbert
space $\mathcal{H}$. We obtain thus a weak formulation of the
original problem \eqref{e11}.

In this paper, our aim is to study the exact controllability of the
system \eqref{e11}.
For this aim, we will adapt the
Hilbert Uniqueness Method  \cite{k,lions,lions2} to the abstract
problem \eqref{e14}.

First, by a multiplier method, we establish an inverse observability
inequality with the usual norm for initial data in $\mathcal{H}$ and
consequently, by the HUM method, we prove that the problem
\eqref{e14} is exactly controllable by means of singular control
$v\in H^1(0,T)'$.

Next, to prove the exact controllability of \eqref{e14} by means
of regular control $v\in L^2(0,T)$, we have to establish
observability results with a weaker norm (see \cite{lions}).
Here lies the main difficulty in this paper. In fact,
the operator $A$ of the problem \eqref{e14} is not invertible in the
energy space, so the method used by Rao in \cite{r2} can not be adapted
in this case. Indeed, the observability inequalities
obtained with the usual norm can not be extended, directly using
$A^{-1}$, to initial data in $D(A)'$. To overcome this difficulty,
we establish new intermediate observability results with the usual
norm and then, by a suitable change of variable, we extend these
results to initial data in $D(A)'$.

In the case of static feedback, the two conditions $y_x (1,t) + \eta(t) = 0$ 
and $\eta_t(t) - y_t(1,t) = v(t)$ are
replaced by the condition $y_x(1,t)+g(y(1,t))=v(t)$, the exact 
controllability of the system \eqref{e11}-\eqref{e12}
was well studied by different approaches (see \cite{k,lions,lions2,hz} 
and the references herein).

The paper is organized as follows: In section 2, we consider the
homogeneous problem associated to \eqref{e11}. Using a
multiplier method, we first establish  direct and inverse observability
results with the usual norm; i.e, for initial data in $\mathcal{H}$. Next,
we establish new intermediate observability results which leads,
 by a suitable change, to extend these observability inequalities to
initial data in $D(A)'$. In section 3, using the HUM method, we
prove that the abstract problem \eqref{e14} is exactly
controllable by means of either a singular control $v\in
H(0,T)'$ for usual initial data $Y_0\in\mathcal{H}$ and $T>2$ or
by means of regular control $v\in L^{2}(0,T)$ for smooth initial
data $Y_0\in D(A)$ and $T>2$.

\section{Observability results}\label{s2}

In this section, our aim to establish all observability results
necessary to the controllability of the system \eqref{e11} by
singular and regular control $v$. For this purpose we consider the
following homogeneous system ($v=0$):
\begin{equation}
\begin{gathered}
\phi_{tt} - \phi_{xx} = 0 \quad 0 < x < 1, \ t>0, \\
\phi(0,t) =0  \quad t>0\\
\phi_x (1,t) + \xi(t) = 0 \quad t>0,\\
\xi_t(t) - \phi_t(1,t) = 0 \quad t>0 \label{e21}
\end{gathered}
\end{equation}
with the initial conditions
\begin{equation}
\phi(x,0)=\phi_0(x), \quad \phi_t(x,0)=\phi_1(x),\quad 0<x<1,\quad
\xi(0)=\xi_0\in \mathbb{R}. \label{e22}
\end{equation}
First, we will study the well-posedness of the problem \eqref{e21}.

\subsection{Well-posedness of the problem}

To write formally the system \eqref{e21}-\eqref{e22}, we, first,
introduce
$$
V =\big \{ \phi \in H^1(0,1) :\phi(0)= 0  \big \}
$$
and define the energy space $\mathcal{H} = V \times L^2 (0,1)
\times\mathbb{R}$, endowed with the inner product
$$
(\Phi,\widetilde \Phi )_\mathcal{H} = \int_0^1 \phi_x \widetilde
\phi_x dx +\int_0^1 \psi\widetilde \psi dx + \xi \widetilde \xi,\quad
\Phi=(\phi,\psi,\xi),\quad
\widetilde \Phi = (\widetilde \phi, \widetilde \psi,\widetilde \xi)
\in \mathcal{H}.
$$
Next we define the linear unbounded operator $A$ on
\begin{gather*}
D( A) = \big \{ \Phi=(\phi,\psi,\xi )\in
\mathcal{H} :\phi \in H^2(0,1), \quad \psi \in V \  \text{and } \
\phi_x (1)+\xi =0 \big \},\\
A\Phi =(\psi,\phi_{xx},\psi(1)),\quad   \forall
 \Phi=(\phi,\psi,\xi) \in D(A).
\end{gather*}
Then setting $\Phi(x,t)=(\phi(x,t),\phi_t (x,t) ,\xi (t) ) \in D(A)$,
the state of the system \eqref{e21}, we formally
transform the problem \eqref{e21}-\eqref{e22} into an evolutionary
equation:
\begin{equation}
\Phi_t = A\Phi, \quad \Phi(0)=\Phi_0 \in \mathcal{H}. \label{e23}
\end{equation}
It is easy to check that $A$ is skew adjoint and $m$-dissipative on
$\mathcal{H}$ and therefore generates a strongly continuous group
of isometries $S_A (t)$ on the energy space $\mathcal{H}$ (see
\cite{p,b}). So, we have the following existence and
uniqueness result.

\begin{proposition} \label{prop2.1}
(a-) Assume that $\Phi_0 \in \mathcal{H}$. The system
\eqref{e23} admits a unique weak solution $\Phi(t)$ satisfying
$$
\Phi(t) \in C^0 ( {\mathbb{R}}^+ ;\mathcal{H}).
$$

(b-) Assume that $\Phi_0 \in D(A)$. The system \eqref{e23}
admits a unique strong solution $\Phi(t)$ satisfying
$$
\Phi(t) \in C^0 ( {\mathbb{R}}^+ ;D(A))\cap C^1 ({\mathbb{R}}^+
;\mathcal{H} )
$$
and we have
$$
\norm{\Phi (t)}_\mathcal{H}=\norm{\Phi_0}_\mathcal{H}, \quad
\forall t\in {\mathbb{R}}^+.
$$
\end{proposition}

Then we will establish two observability results for usual initial
data.

\subsection{Observability results for initial data in $\mathcal{H}$}

In this part, by a multiplier method, we establish the following
observability results.

\begin{theorem} \label{thm2.2}
Let $T>2$ be arbitrarily. Then for every $\Phi_0\in \mathcal{H}$
the solution $\Phi$ of the system \eqref{e23} satisfies the following
inequalities
\begin{gather}
\frac{1}{T+2}\int_0^T \big[|\phi_x (1,t)|^2+ |
\phi_{xt} (1,t)|^2\big]dt\leq \norm{\Phi_0}^2_\mathcal{H},
\label{e24}
\\
\norm{\Phi_0}^2_\mathcal{H} \leq \frac{2}{T-2}\int_0^T
\big[|\phi_x (1,t)|^2+ | \phi_{xt} (1,t)|^2\big]dt. \label{e25}
\end{gather}
\end{theorem}

\begin{proof}
Assume that $\Phi_0\in D(A^2)$. Multiplying the equation \eqref{e21}
by $2x\phi_x$ and integrating by parts, we obtain
$$
\int_0^1 \int_0^T \Big(\phi_t^2 + \phi_x^2 \Big ) \,dx\,dt
=-2\Big[\int_0^1 \phi_t x\phi_xdx\Big]_0^T
+\int_0^T \Big(|\phi_{xt} (1,t)
|^2 + |\phi_x (1,t)|^2\Big )dt.
$$
This implies
\begin{equation}
T \norm{\Phi_0}_\mathcal{H}^2 + 2\Big[\int_0^1 \phi_t
x\phi_xdx\Big]_0^T  = \int_0^T \Big (|\phi_{xt}
(1,t)|^2 + 2|\phi_x (1,t)|^2\Big )dt.
\label{e26}
\end{equation}
On the other hand, using Cauchy-Schwartz inequality, we deduce that
$$
|2\int_0^1 \phi_t x\phi_xdx|\leq \norm{\Phi_0}_\mathcal{H}^2, \quad
 \forall t\in \mathbb{R}^+.
$$
Finally we have
\begin{equation}
-2\| \Phi_0\|_\mathcal{H}^2\leq 2\Big[\int_0^1 \phi_t
x\phi_xdx\Big]_0^T  \leq 2\| \Phi_0\|_\mathcal{H}^2.
\label{e27}
\end{equation}
Inserting \eqref{e27} in \eqref{e26} we obtain \eqref{e24} and
\eqref{e25} for every $\Phi_0\in D(A^2)$.
By a density argument we prove \eqref{e24} and \eqref{e25} for
every $\Phi_0\in \mathcal{H}$. The proof is thus complete.
\end{proof}

\begin{remark} \label{rmk2.3}\rm
 (i) There exists no constant $c>0$ such that
$$
\norm{\Phi_0}^2_\mathcal{H} \leq c\int_0^T|\phi_{xt} (1,t)|^2dt.
$$
In fact, it easy to see that the operator $A$ has $0$ as an eigenvalue,
with an associated eigenfunction
$\Psi_0=(x,0,-1)$. Let $\Phi_0=\Psi_0$ then $\Phi=\Phi_0$ is the
solution of the problem \eqref{e23} and we have
$$
\norm{\Phi_0}^2_\mathcal{H}=2, \quad\text{and} \quad
\int_0^T|\phi_{xt} (1,t)|^2dt=0.
$$
 (ii) There exists no constant $c>0$ such that
$$
\norm{\Phi_0}^2_\mathcal{H} \leq c\int_0^T|\phi_x (1,t)|^2dt.
$$
In fact, the skew operator $A$ has $i\mu_n\in i\mathbb{R}$ ,
$n\in \mathbb{Z}$, isolated eigenvalues with algebraic multiplicity one
and $|\mu_n|$ goes to infinity as $n$ goes to infinity.
Moreover, $\mu_n$ has the following asymptotic expansion
(see \cite{w})
$$
\mu_n=n\pi +\frac{\pi}{2}+\frac{1}{n\pi}-\frac{1}{2n^2\pi}
+ O(\frac{1}{n^3}), \quad \text{as } n\to\infty.
$$
The associated eigenvectors:
$$
\Psi_0=(x,0,-1),\quad
\Psi_n=(\frac{1}{\mu_n}\sin(\mu_nx),i\sin(\mu_nx),-\cos(\mu_n)), \quad
\forall n\in \mathbb{Z}^*.
$$
Let $\Phi_0^n=\Psi_n$, $n\in\mathbb{Z}^*$, then $\Phi^n=e^{i\mu_n t}\Psi_n$
is the solution of the problem \eqref{e23} and we have
$$
\norm{\Phi_0^n}^2_\mathcal{H}=1+|\cos(\mu_n)|^2\to 1, \quad
  \text{and}\quad \int_0^T|\phi_x (1,t)|^2dt
=T|\cos(\mu_n)|^2\to 0.
$$
We conclude that the usual inverse observability inequalities
obtained for the classical wave equation (see \cite{lions,k})
does not hold in this case.

(iii) The observability inequality \eqref{e25} leads, by the HUM
 method, to the exact controllability of \eqref{e11} by means of
singular control $v\in H^1(0,1)'$.
\end{remark}


\subsection{Observability results for initial data in $D(A)'$}


To prove that the system \eqref{e11} is exactly controllable by
means  of regular control $v\in L^2(0,T)$, we have to establish an
inverse observability inequality with a weaker norm
 in \cite[pp. 122-127]{lions}.
For this aim, we will extend the inverse observability inequality,
obtained for usual initial data, to initial data in $D(A)'$. Since
the operator $A$ is not invertible, the classical methods based on
using $A^{-1}$ to obtain the extension (see \cite{r2}) can not be
adapted for this system. To overcome this difficulty, we first
establish two intermediate observability inequalities based on the
following theorem.

\begin{theorem} \label{thm2.4}
Let $T>2$ and $\alpha>1$ be a real number. Then there exist constants
$c_1(T)>0$ and $c_2(T)>0$ such that for every $\Phi_0\in\mathcal{H}$
the solution $\Phi$ of the system \eqref{e23} satisfies the following
inequalities
\begin{gather}
c_1\int_0^T \Big [\Big(\phi_x (1,t)-e^{-\alpha t}\phi_x
(1,0)\Big)^2+ \Big(\alpha \phi_x(1,t)+\phi_{xt} (1,t)
\Big)^2\Big ]dt\leq \norm{\Phi_0}^2_\mathcal{H},   \label{e28}
\\
\norm{\Phi_0}^2_\mathcal{H} \leq c_2 \int_0^T \Big [\Big(\phi_x
(1,t)-e^{-\alpha t}\phi_x (1,0)\Big)^2+ \Big(\alpha
\phi_x(1,t)+\phi_{xt} (1,t) \Big)^2\Big ]dt. \label{e29}
\end{gather}
\end{theorem}

\begin{proof}
It is sufficient to prove the estimates \eqref{e28} and \eqref{e29}
for $\Phi_0\in D(A)$ the case of $\Phi_0\in \mathcal{H}$, then
follows by a density argument.
First, a direct computation gives
\begin{equation}
\begin{aligned}
&\int_0^T \Big [\Big(\phi_x (1,t)-e^{-\alpha t}\phi_x (1,0)\Big)^2
+ \Big(\alpha \phi_x(1,t)+\phi_{xt} (1,t) \Big)^2\Big ]dt \\
&\leq 2(1+\alpha^2)\int_0^T \Big [|\phi_x
(1,t)|^2+|\phi_{xt} (1,t)|^2\Big ]dt
+\Big(\frac{1}{\alpha}-\frac{1}{\alpha}e^{-2\alpha
T}\Big)\phi_x^2(1,0).
\end{aligned}\label{e210}
\end{equation}
On the other hand, using the definition of the norm we have
$ \phi_x^2(1,0)=\xi_0^2\leq \norm{\Phi_0}_{\mathcal{H}}^2$.
Then inserting \eqref{e24} into \eqref{e210}, we obtain the
direct inequality \eqref{e28}, and we have
$$
c_1^{-1}=2(T+2)(1+\alpha^2)+(\frac{1}{\alpha}
 -\frac{1}{\alpha}e^{-2\alpha T}).
$$
Next, we  verify the inverse inequality \eqref{e29} by contradiction.
Assume that \eqref{e29} fails, then there exists
a sequence $(\Phi^n)_{n\in \mathbb{N}}$ such that
\begin{equation}
\| \Phi^n(t) \|_\mathcal{H}=\| \Phi^n_0 \|_\mathcal{H}=1,
\quad \forall t\in \mathbb{R}\label{e211}
\end{equation}
and
\begin{equation}
\int_0^T \Big [\Big(\phi^n_x (1,t)-e^{-\alpha t}\phi^n_x (1,0)\Big)^2
+ \Big(\alpha \phi^n_x(1,t)
+\phi^n_{xt} (1,t)\Big)^2\Big ]dt\to 0 \label{e212}
\end{equation}
where $\Phi^n(t)= (\phi^n(x,t),\phi_t^n(x,t),\xi^n(t))$ is
the solution of the  problem
\begin{equation}
\Phi_t^n = A\Phi^n, \quad \Phi^n(0)=\Phi_0^n.\label{e213}
\end{equation}
Since $|\phi_x^n(1,0)|^2\leq \| \Phi_0^n\|_\mathcal{H}^2=1$
then there exists a subsequence $\phi_x^n(1,0)$, still indexed by
$n$ for convenience, that converges to a constant $-1\leq c\leq 1$
as $n\to +\infty$. From \eqref{e212} we deduce that
\begin{gather}
\phi_x^n(1,t) \to ce^{-\alpha t}, \quad \text{in } L^2(0,T)\label{e214}
\\
\varepsilon_n=\int_0^T \Big(\alpha \phi^n_x(1,t)+\phi^n_{xt} (1,t)
\Big)^2dt\to 0.\label{e215}
\end{gather}
Since $\alpha >1$, then using the inequality \eqref{e25} we have
\begin{equation}
\frac{T-2}{2}\norm{\Phi_0^n}^2_\mathcal{H}\leq \varepsilon_n+
\alpha |\phi_x^n(1,0)|^2.\label{e216}
\end{equation}
Using the linearity of the problem, \eqref{e214}-\eqref{e216} and
the trace theorem, we conclude that, for any $\varepsilon>0$ there
exists $n_0\in\mathbb{N}$ such that for all $n,m\geq n_0$,
$$
\frac{T-2}{2}\| \Phi^n(t)- \Phi^m(t)\|_\mathcal{H}^2
\leq 2(\varepsilon_{n}+\varepsilon_m) + \alpha |\phi_{x}^n
(1,0)- \phi_{x}^m (1,0)|^2\leq \varepsilon.
$$
Then $(\Phi^n(t))$ is a cauchy sequence in $\mathcal{H}$. This implies
that
$$
\Phi^n(t)\to \Phi(t), \quad \text{strongly in } \mathcal{H}.
$$
Using \eqref{e25}, \eqref{e211}, \eqref{e213} and \eqref{e214}
we deduce that $\Phi(t)=( \phi(x,t),\phi_t(x,t), \xi(t))$ solves
the problem
\begin{equation}
\Phi_t= A \Phi, \quad
\Phi(0)=\Phi_0\label{e217}
\end{equation}
and the supplementary conditions
\begin{gather}
\phi_x(1,t)=ce^{-\alpha t}, \quad t>0,\label{e218}\\
\norm{\Phi(t)}_{\mathcal{H}}=1,\quad t>0.\label{e219}
\end{gather}
Now, let $\Phi=(\phi,\phi_t,\xi)$ be the solution of
\eqref{e217}-\eqref{e218} then, using Remark \ref{rmk2.3} (ii),
we have
$$
\Phi(x,t)=\sum{a_n \Psi_n(x) e^{i\mu_n t}},\quad a_n\in\mathbb{C}
$$
where the sequence $\mu_n$ satisfies $\sin \mu_n=-\mu_n\cos \mu_n$.
This implies
\begin{equation}
\xi(t)=-\phi_x(1,t)=-ce^{-\alpha t}=-\sum{a_n \cos(\mu_n) e^{i\mu_n t}}.\label{e221}
\end{equation}
Noting that $\mu_n\in \mathbb{R}$, then using \eqref{e219} we
deduce that
\begin{equation}
\sum |a_n \cos(\mu_n)|^2 <\infty. \label{e222}
\end{equation}
Using \eqref{e221} and \eqref{e222} we conclude, from Riesz-Fisher
theorem \cite[pp. 110]{be}, that the function $ce^{-\alpha t}$
is a $\mathcal{B}^2$ almost periodic function. Then the Parseval
equation is true for $ce^{-\alpha t}$ \cite[pp. 109]{be}; i.e.,
$$
\sum |a_n \cos(\mu_n)|^2 =M\{c^2e^{-2\alpha t}\}
$$
where the mean value $M\{c^2e^{-2\alpha t}\}$ is given by
$$
M\{c^2e^{-2\alpha t}\}=\lim_{X\to+\infty}\frac{1}{X}\int_0^{X}
c^2e^{-2\alpha t}dt=0.
$$
This, together the fact that $\cos \mu_n\not = 0$, implies
that $a_n=0$ for all $n\in \mathbb{Z}^\star$ and $c=0$.
Applying Holmgren's theorem  \cite{LM}, the system
\eqref{e217}-\eqref{e218} admits the unique trivial solution
$\Phi=0$, this contradicts \eqref{e219}.
The proof is thus complete.
\end{proof}

Next, by a suitable change of variable, we will establish a direct
and inverse observability inequality for initial data in $D(A)'$.

\begin{theorem} \label{thm2.5}
Let $T>2$ and $\alpha>1$ be an arbitrarily real number. Then there
exist constants $c_3(T)>0$ and $c_4(T)>0$ such that
the solution of the system \eqref{e23} satisfies the following
inequalities
\begin{gather}
c_3\int_0^T \Big [\Big(\int_0^t \phi_x (1,s)e^{\alpha s}ds\Big)^2
+ e^{2\alpha t}|\phi_x(1,t)|^2\Big ]dt
\leq \norm{\Phi_0}^2_{D(A)'},   \label{e41}
\\
\norm{\Phi_0}^2_{D(A)'} \leq c_4 \int_0^T \Big
[\Big(\int_0^t \phi_x (1,s)e^{\alpha s}ds\Big)^2+ e^{2\alpha t}
|\phi_x(1,t)|^2\Big ]dt. \label{e42}
\end{gather}
\end{theorem}

\begin{proof}
It is sufficient to prove \eqref{e41} and \eqref{e42} for
$\Phi_0\in D(A)$ the general case follows by a density argument.
Let $\Phi_0\in D(A)$ then the problem \eqref{e23} has a unique
solution $\Phi \in D(A)$. We define a new function
$\Psi(x,t)$ by
$$
\Psi(x,t)=e^{\alpha t}\Phi(x,t).
$$
It easy to see that $\Psi$ solve the equation
\begin{equation}
\Psi_t = (\alpha I + A)\Psi, \quad
\Psi(0)=\Psi_0=\Phi_0 \in \mathcal{H}. \label{e43}
\end{equation}
Replacing $\phi$ by $e^{-\alpha t}\psi$ in \eqref{e28}-\eqref{e29}
we obtain
\begin{equation}
c_1 e^{-2\alpha T}\int_0^T \Big [\Big(\psi_x (1,t)-\psi_x(1,0)\Big)^2
+ |\psi_{xt}(1,t)|^2\Big ]dt
\leq \norm{\Psi_0}^2_{\mathcal{H}}   \label{e44}
\end{equation}
and
\begin{equation}
\norm{\Psi_0}^2_{\mathcal{H}} \leq c_2 \int_0^T
\Big [\Big(\psi_x (1,t)-\psi_x(1,0)\Big)^2+ |\psi_{xt}(1,t)|^2
\Big ]dt. \label{e45}
\end{equation}
Defining
$$
\widetilde \Psi_0 =(\alpha I + A)^{-1}\Psi_0=(\alpha I + A)^{-1}\Phi_0.
$$
Then
\begin{equation}
\norm{\widetilde \Psi_0}_{\mathcal{H}}^2
=\norm{(\alpha I + A)^{-1}\Phi_0}_{\mathcal{H}}^2
=\norm{\Phi_0}_{D(\alpha I + A)'}^2.\label{e46}
\end{equation}
Now, let $\widetilde \Psi$ the solution of the equation
\begin{equation}
\widetilde \Psi_t=(\alpha I + A)\widetilde \Psi, \quad
\widetilde \Psi(0)=(\alpha I + A)^{-1}\Phi_0.\label{e47}
\end{equation}
Applying the inequalities \eqref{e44}-\eqref{e45} to $\widetilde \Psi$
we obtain
\begin{equation}
c_1 e^{-2\alpha T}\int_0^T 
\Big[\Big(\widetilde \psi_x (1,t)-\widetilde \psi_x(1,0)\Big)^2+
|\widetilde \psi_{xt}(1,t) |^2\Big]dt 
\leq \norm{\Phi_0}_{D(\alpha I + A)'}^2  \label{e48}
\end{equation}
and
\begin{equation}
\norm{\Phi_0}_{D(\alpha I + A)'}^2 \leq c_2
\int_0^T \Big [\Big(\widetilde \psi_x (1,t)-\widetilde
\psi_x(1,0)\Big)^2+ |\widetilde \psi_{xt}(1,t)|^2\Big ]dt. \label{e49}
\end{equation}
Using \eqref{e47} we have
$$
\widetilde \Psi_t(0)=(\alpha I + A)\widetilde \Psi(0)=\Phi_0.
$$
Then $\widetilde \Psi_t$ solve the  equation
\begin{equation}
\widetilde \Psi_{tt}=(\alpha I + A) \widetilde \Psi_t, \quad
\widetilde \Psi_t(0)=\Phi_0 .\label{e410}
\end{equation}
This implies that $\widetilde \Psi_t=\Psi$ and
$$
\widetilde \psi_x(1,t)-\widetilde \psi_x(1,0)=\int_0^t \psi_x(1,s)ds,
\quad \widetilde \psi_{xt}(1,t)=\psi_x(1,t).
$$
Using \eqref{e48} and \eqref{e49} we obtain
\begin{equation}
c_1 e^{-2\alpha T}\int_0^T
\Big [\Big(\int_0^t \psi_x (1,s)ds \Big)^2+ |\psi_{x}(1,t)|^2\Big ]dt
\leq \norm{\Phi_0}_{D(\alpha I + A)'}^2  \label{e411}
\end{equation}
and
\begin{equation}
\norm{\Phi_0}_{D(\alpha I + A)'}^2 \leq c_2 \int_0^T
\Big [\Big(\int_0^t \psi_x (1,s)ds\Big)^2+
 |\psi_{x}(1,t)|^2\Big ]dt. \label{e412}
\end{equation}
On the other hand, we have
$$
\norm{\Phi_0}_{D(A)}^2\leq\norm{\Phi_0}_{D(\alpha I+A)}^2
\leq(1+\alpha)\norm{\Phi_0}_{D(A)}^2.
$$
This implies that $\norm{\cdot}_{D(\alpha I+A)'}$ and
$\norm{\cdot}_{D(A)'}$ are equivalent.
Replacing $\psi_x(1,t)$ by $e^{\alpha t}\phi_x(1,t)$, 
we obtain \eqref{e41} and \eqref{e42} with
$$
c_3= c_1 e^{-2\alpha T},\quad  c_4=c_2.
$$
The proof is complete.
\end{proof}

\section{Exact controllability of the system}

In this section we study the exact controllability result
in $\mathcal{H}$ the controlled system
\begin{equation}
\begin{gathered}
y_{tt} - y_{xx} = 0  \quad  0<x<1,\ \ t>0, \\
y(0,t) =0  \quad  t>0,\\
y_x (1,t) + \eta(t) = 0 \quad  t>0,  \\
\eta_t(t) - y_t(1,t) = v(t) \quad  t>0
\end{gathered}\label{e31}
\end{equation}
with the  initial conditions
\begin{equation}
y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad
 0<x<1, \quad \eta(0)=\eta_0\in \mathbb{R}. \label{ei}
\end{equation}
Setting $Y(x,t)=(y(x,t),y_t(x,t),\eta(t))$ the state of the system
\eqref{e31}-\eqref{ei} we formally transform the problem into
an evolutionary problem
\begin{equation}
Y_t=AY+V,\quad Y(0)=Y_0\in \mathcal{H} \label{e32}
\end{equation}
where $V=(0,0,v)$.

\subsection{Exact controllability for initial data in $\mathcal{H}$}

The observability inequalities for usual initial data obtained  in
the subsection 2.2 leads, by the HUM method, to the exact
controllability of the system \eqref{e32} by means of singular
control $v\in H^1(0,T)'$. Now, let $\Phi=(\phi,\phi_t,\xi)$ be a
solution of the homogeneous problem \eqref{e23}. Multiplying the
equation \eqref{e32} by $\Phi$ and integrating by parts so that we
obtain formally
\begin{equation}
(Y_0,\Phi_0)_\mathcal{H} + \int_0^t v(s)\xi(s) ds= (Y(x,t),
\Phi(x,t))_\mathcal{H}.\label{e33}
\end{equation}
Identify the Hilbert space $\mathcal{H}$ with its dual and define
the linear form $L$ by setting
\begin{equation}
L(\Phi_0)=  (Y_0,\Phi_0)_\mathcal{H}  + \int_0^t v(s)\xi(s) ds, \quad
\forall \Phi_0\in \mathcal{H}
 \label{e34}
\end{equation}
 we obtain a weak formulation of the problem \eqref{e32}.
\begin{equation}
L(\Phi_0)=(Y(x,t), \Phi(x,t))_\mathcal{H}=(Y(x,t),S_A(t)
\Phi_0)_\mathcal{H}, \quad \forall \Phi_0\in
\mathcal{H}\label{e35}
\end{equation}
where $S_A(t)$ the group of isometries associated to the
homogeneous problem \eqref{e23}.

Next, we consider the exact controllability of the equation
\eqref{e32} for usual initial data $Y_0\in \mathcal{H}$. We choose
the control
\begin{equation}
v(t)=v_0(t) - \frac{d}{dt}v_1(t),\quad
v_0\in L^2(0,T),\quad
\frac{d}{dt}v_1(t)\in H^1(0,T)' \label{e36}
\end{equation}
where the derivative $\frac{d}{dt}$ is defined in
the sense of $H^1(0,T)'$
\begin{equation}
- \int_0^T\frac{d}{dt}v_1(t)\mu(t)dt
=\int_0^T v_1(t)\frac{d}{dt}\mu(t)dt ,\quad
\forall \mu\in H^1(0,T).\label{e37}
\end{equation}

\begin{theorem} \label{thm3.1}
Let $T > 0$ and $v$ be chosen in \eqref{e36}. For every
$Y_0\in \mathcal{H}$,  the controlled system \eqref{e32}
admits a unique weak solution $Y(x,t)$ such that
\begin{equation}
Y(x,t)\in C^0([0,T] ;\mathcal{H})\label{e38}
\end{equation}
defined in the sense that the equation \eqref{e35} is satisfied
for all $\Phi_0\in \mathcal{H}$ and all $0<t<T$. Moreover the
linear mapping
\begin{equation}
(Y_0,v_0,v_1) \to Y\label{e39}
\end{equation}
is continuous form $\mathcal{H}\times L^2(0,T)\times L^2(0,T)$
into $\mathcal{H}$.
\end{theorem}

\begin{proof}
Let $\Phi_0\in\mathcal{H}$ and $\Phi=(\phi,\phi_t,\xi)$ be the
solution of the system \eqref{e23}. We have
\begin{equation}
\begin{aligned}
&|\int_0^t v(s)\xi(s) ds|\\
&= |\int_0^t (v_0-\frac{d}{ds}v_1(s))\phi_x(1,s) ds|\\
&=|\int_0^t v_0\phi_x(1,s) ds+\int_0^t v_1(s)\phi_{xs}(1,s) ds|\\
&\leq \| v_0\|_{L^2(0,T)}  \| \phi_x (1, .)\|_{L^2(0,T)}+
\| v_1\|_{L^2(0,T)}  \|\phi_{xt} (1, .)\|_{L^2(0,T)}\\
&\leq (\| v_0\|_{L^2(0,T)}+\| v_1\|_{L^2(0,T)})(
\|\phi_x (1, .)\|_{L^2(0,T)}+\|\phi_{xt} (1, .)\|_{L^2(0,T)}).
\end{aligned}\label{e310}
\end{equation}
Using \eqref{e24}, \eqref{e34} and \eqref{e310} we obtain
$$
| L(\Phi_0)| \leq \Big[\sqrt{2(T+2)}(\| v_0\|_{L^2(0,T)}+\| v_1
\|_{L^2(0,T)})+\| Y_0\|_\mathcal{H}\Big] \|
\Phi_0\|_\mathcal{H}
$$
for all $\Phi_0\in \mathcal {H}$. This
implies that the linear form $L$ is continuous in the space
$\mathcal{H}$. And we have
$$
\| L\|_{\mathcal{L}(\mathcal{H},\mathbb{R})}
\leq \sqrt{2(T+2)}(\| v_0\|_{L^2(0,T)}+\| v_1
\|_{L^2(0,T)})+\| Y_0\|_\mathcal{H}.
$$
From Riesz's representation theorem, there exist a unique
$Z(x,t)\in \mathcal{H}$ solution of the following problem
$$
L(\Phi_0)= (Z(x,t),\Phi_0)_\mathcal{H}, \quad
\forall \Phi_0\in \mathcal{H}.
$$
Finally, we define $Y(x,t)$ by $S_A(t)Y(x,t)=-Z(x,t)$ and we
deduce that $Y(x,t)$ is the unique solution of the problem
\eqref{e35}. And we have
$$
\| Y(x,t)\|_\mathcal{H}\leq \sqrt{2(T+2)}(\| v_0\|_{L^2(0,T)}+\| v_1
\|_{L^2(0,T)})+\| Y_0\|_\mathcal{H}, \quad \forall t\in
[0,T].
$$
This implies that the linear application \eqref{e39} is
continuous from $\mathcal{H}\times L^2(0,T)\times L^2(0,T)$ into
$\mathcal{H}$. The proof is thus complete.
\end{proof}

\begin{theorem} \label{thm3.2}
Let $T>2$. For all $Y_0\in\mathcal{H}$, there exists a control
$v(t)=v_0(t)-\frac{d}{dt}v_1(t)$, $v_0,v_1\in L^2(0,T)$ such that
the weak solution $Y(x,t)$ of the controlled problem \eqref{e32}
satisfies the final condition
\begin{equation}
Y(T)=0. \label{e311}
\end{equation}
\end{theorem}

\begin{proof}
Let $\Phi$ be the solution of the homogeneous system \eqref{e23}
with initial data $\Phi_0\in \mathcal{H}$. We define the semi-norm
\begin{equation}
\| \Phi_0\|_{1}^2 = \int_0^T ( |\phi_x (1,t)|^2+
|\phi_{xt} (1,t)|^2)dt, \ \ \forall \Phi_0\in
\mathcal{H}.\label{e312}
\end{equation}
Thanks to inequalities \eqref{e24} and \eqref{e25}, we know that
\eqref{e312} defines an equivalent norm in the energy space
$\mathcal{H}$. Now, choosing the controller $v(t)$ as
\begin{equation}
v(t)=v_0(t)-\frac{d}{dt}v_1(t)=:-\phi_x(1,t)+\frac{d}{dt}\phi_{xt}(1,t)
\label{e313}
\end{equation}
where the derivative $\frac{d}{dt}$ is defined in the sense of
\eqref{e37}. Using the direct inequality \eqref{e24}, we have
\begin{equation}
\| v_0(t)\|_{L^2(0,T)}+\| v_1(t)\|_{L^2(0,T)} \leq
\sqrt{2(T+2)}\| \Phi_0\|_\mathcal{H}.\label{e314}
\end{equation}
Now solve the backward problem
\begin{equation}
\Psi_t=A\Psi+V, \quad \Psi(T)=0. \label{e315}
\end{equation}
Using Theorem \ref{thm2.4} the problem \eqref{e315} admits a unique weak
solution $\Psi(x,t)\in C^0([0,T] ;\mathcal{H})$, and we have
 \begin{equation}
\| \Psi \|_\mathcal{H} \leq \sqrt{2(T+2)}(\|
v_0(t)\|_{L^2(0,T)}+\| v_1(t)\|_{L^2(0,T)}).\label{e316}
\end{equation}
Next we define the operator $\Lambda$ as
\begin{equation}
\Lambda \Phi_0=-\Psi(0), \quad \forall \Phi_0\in \mathcal{H}.
\label{e317}
\end{equation}
By virtue of inequalities \eqref{e314} and \eqref{e316} we obtain
$$
\| \Lambda \Phi_0\|_\mathcal{H}\leq \sqrt{2(T+2)}
(\| v_0(t)\|_{L^2(0,T)}+ \| v_1(t)\|_{L^2(0,T)} )
\leq 2(T+2)\|\Phi_0\|_\mathcal{H}.
$$
This implies that $\Lambda$ is a linear continuous operator from
$\mathcal{H}$ into $\mathcal{H}$. Multiplying the backward
problem \eqref{e315} by $\Phi$ and integrating by parts we
obtain
\begin{equation}
-(\Psi_0,\Phi_0)_\mathcal{H} = \int_0^T
(|\phi_x(1,t)|^2+  |
\phi_{xt}(1,t)|^2)dt.\label{e318}
\end{equation}
This implies
\begin{equation}
(\Lambda \Phi_0,\Phi_0)_\mathcal{H} = \|
\Phi_0\|_1^2.\label{e319}
\end{equation}
Thanks to the Lax-Milgram theorem, we deduce that $\Lambda$ is an
isomorphism from $\mathcal{H}$ onto $\mathcal{H}$. In particular,
given any $-Y_0\in \mathcal{H} $, there exists a unique $\Phi_0\in
\mathcal{H}$ such that
\begin{equation}
\Lambda \Phi_0 =-Y_0.\label{e320}
\end{equation}
This equality implies that the weak solution $Y(x,t)$ of
backward problem \eqref{e315}, with $v$ given by \eqref{e313}
satisfy the initial value condition $Y(x,0)=Y_0$ and
that final condition $Y(x,T)=0$. The proof is  complete.
\end{proof}

\subsection{Exact controllability for initial data in $D(A)$}

Now we consider the exact controllability of the equation \eqref{e32}
 by means of a regular control $v\in L^2(0,T)$
\begin{equation}
v(t)=-e^{\alpha t}v_0(t) +e^{\alpha t}\int_T^t
\Big( \int_0^s v_0(\tau)e^{\alpha \tau}d\tau\Big)ds, \quad
 v_0(t)\in L^2(0,T). \label{e413}
\end{equation}

For the wellposedness of the equation \eqref{e32} with the control
\eqref{e413} we first interpret \eqref{e35} into the
following form
\begin{equation}
L(\Phi_0)=\langle Y(x,t),S_A(t) \Phi_0\rangle_{D(A)\times D(A)'}, \quad
\forall \Phi_0\in D(A)'\label{e414}
\end{equation}
where the linear form $L$ is defined by
\begin{equation}
L(\Phi_0)=\langle Y_0,\Phi_0\rangle_{D(A)\times D(A)'}
 + \int_0^t v(s)\xi(s) ds, \quad \forall \Phi_0\in D(A)'.
 \label{e415}
\end{equation}

\begin{theorem} \label{thm3.3}
Let $T > 0$ and $v\in L^2(0,T)$ defined by \eqref{e413}.
For every $Y_0\in D(A)$  the controlled
system \eqref{e32} admits a unique weak solution satisfying
\begin{equation}
Y(x,t)\in C^0\Big([0,T] ;D(A)\Big)\label{e416}
\end{equation}
defined in the sense that the equation \eqref{e414} is satisfied for
all $\Phi_0\in D(A)'$ and all $0<t<T$.
Moreover the linear application
\begin{equation}
(Y_0,v_0) \to Y\label{e417}
\end{equation}
is continuous form $D(A)\times L^2(0,T)$ into $D(A)$.
\end{theorem}

\begin{proof}
Let $\Phi_0\in D(A)'$ and $\Phi=(\phi,\phi_t,\xi)$ be the solution
of the system \eqref{e23}.
It easy to see that
\begin{align*}
&|\int_0^t v(s)\xi(s) ds|\\
&=  |\int_0^t e^{\alpha s}v_0(s)\phi_x(1,s)ds
+\int_0^t\Big(\int_0^s e^{\alpha \tau}
\phi_x(1,\tau) d\tau\Big)\Big(\int_0^s e^{\alpha \tau}
v_0(\tau )d\tau\Big)ds|\\
&\leq e^{\alpha T}\| v_0\|_{L^2(0,T)}  \| \phi_x (1, .)\|_{L^2(0,T)}+
\norm{\tilde v_0}_{L^2(0,T)}\| \tilde \phi_x (1, .)\|_{L^2(0,T)}
\end{align*}
where
$$
\tilde \phi_x(1,t)=\int_0^te^{\alpha s}\phi_x(1,s)ds, \quad
\tilde v_0(t)=\int_0^t e^{\alpha s}v_0 (s)ds.
$$
We deduce that
$$
\big|\int_0^t v(s)\xi(s) ds\big|
\leq c_5 \| v_0\|_{L^2(0,T)} (\| \phi_x (1, .)\|_{L^2(0,T)}+
\| \tilde \phi_x (1, .)\|_{L^2(0,T)})
$$
where $c_5$ is a constant given by
$$
c_5=1+\sqrt{\frac{T(e^{2\alpha T}-1)}{2\alpha}}.
$$
Using \eqref{e41} and \eqref{e415} we obtain
$$
| L(\Phi_0)| \leq (\sqrt{2}c_5c_3^{-1/2}\| v_0\|_{L^2(0,T)}
+\| Y_0\|_{D(A)}) \| \Phi_0\|_{D(A)'}, \quad \forall \Phi_0\in D(A)'.
$$
This implies that the linear form $L$ is continuous in the space
${D(A)'}$, and we have
$$
\| L\|_{\mathcal{L}(D(A)',\mathbb{R})}
\leq \sqrt{2}c_5c_3^{-1/2}\| v_0\|_{L^2(0,T)}+\| Y_0\|_{D(A)}.
$$
 From the Riesz representation theorem, there exist a unique
$Z(x,t)\in {D(A)}$ solution of the following problem
$$
L(\Phi_0)=<Z(x,t),\Phi_0\rangle_{D(A)\times D(A)'}, \quad
 \forall \Phi_0\in {D(A)'}.
$$
Finally, we define $Y(x,t)$ by $S_A(t)Y(x,t)=-Z(x,t)$ and we deduce
that $Y(x,t)$ is the unique solution of the problem
\eqref{e414}, and we have
$$
\| Y(x,t)\|_{D(A)}\leq \sqrt{2}c_5c_3^{-1/2}\| v_0\|_{L^2(0,T)}
+\| Y_0\|_{D(A)}, \quad  \forall t\in [0,T].
$$
This implies that the linear application \eqref{e417} is continuous
from ${D(A)}\times L^2(0,T)$ into ${D(A)}$.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.4}
Let $T>2$. For all $Y_0\in{D(A)}$, there exists a control
$v(t)\in L^2(0,T)$ such that the weak solution
$Y(x,t)$ of the controlled problem \eqref{e32} satisfies the
final condition
\begin{equation}
Y(T)=0. \label{e418}
\end{equation}
\end{theorem}

\begin{proof}
Let $\Phi_0\in {D(A)'}$ and $\Phi$ be the solution of the
homogeneous system \eqref{e23}.
We define the semi-norm
\begin{equation}
\| \Phi_0\|_{2} = \int_0^T [ ( \int_0^t \phi_x (1,s)e^{\alpha s}ds)^2+
e^{2\alpha t}|\phi_{x} (1,t)|^2 ]dt, \quad
 \forall \Phi_0\in {D(A)'}.\label{e419}
\end{equation}
Thanks to inequalities \eqref{e41} and \eqref{e42}, we know
that \eqref{e419} defines an equivalent
norm in the energy space ${D(A)'}$. Now, choosing the controller
$v(t)$ by
\begin{equation}
v(t)=-e^{\alpha t}\phi_x(1,t)+e^{\alpha t}
\int_T^t\Big(\int_0^s\phi_{x}(1,\tau)e^{\alpha \tau}d\tau\Big) ds
\in L^2(0,T).
\label{e420}
\end{equation}
 From the direct inequality \eqref{e41}, we have
\begin{equation}
\| v_0\|_{L^2(0,T)} \leq c_1^{-1/2}\| \Phi_0 \|_{D(A)'}.\label{e421}
\end{equation}
Next we solve the backward problem
\begin{equation}
\Psi_t=A\Psi+V, \quad \Psi(T)=0. \label{e422}
\end{equation}
Using Theorem \ref{thm2.4} the problem \eqref{e422} admits a unique weak
solution
$\Psi(x,t)\in C^0([0,T] ;D(A))$. And we have
 \begin{equation}
\| \Psi \|_{D(A)} \leq \sqrt{2}c_5c_3^{-1/2} \| v_0\|_{L^2(0,T)}.
\label{e423}
\end{equation}
Next we define the operator $\Lambda$ as
\begin{equation}
\Lambda \Phi_0=-\Psi(0), \quad \forall \Phi_0\in D(A)'. \label{e424}
\end{equation}
By virtue of inequalities \eqref{e421} and \eqref{e423} we obtain
$$
\| \Lambda \Phi_0\|_{D(A)}\leq \sqrt{2}c_5c_1^{-1}
e^{\alpha T} \| \Phi_0 \|_{D(A)'}.
$$
This implies that $\Lambda$ is a linear continuous operator
from $D(A)'$ into $D(A)$.
Now multiplying the backward problem \eqref{e422} by $\Phi$
and integrating by parts we obtain
\begin{equation}
-\langle \Psi_0,\Phi_0\rangle_{D(A)\times D(A)'}
= \int_0^T \Big[ \Big( \int_0^t \phi_x (1,s)e^{\alpha s}ds\Big)^2+
e^{2\alpha t}|\phi_{x} (1,t)|^2 \Big]dt.\label{e425}
\end{equation}
This implies
\begin{equation}
\langle \Lambda \Phi_0,\Phi_0\rangle_{D(A)\times D(A)'}
= \| \Phi_0\|_2^2.\label{e426}
\end{equation}
Thanks to the the Lax-Milgram theorem, we deduce that $\Lambda$
is an isomorphism from $D(A)'$ into $D(A)$.
In particular, given any $-Y_0\in D(A) $, there exists a unique
$\Phi_0\in D(A)'$  such that
\begin{equation}
\Lambda \Phi_0 =-Y_0.\label{e427}
\end{equation}
This equality implies that the weak solution $Y(x,t)$ of backward
problem \eqref{e424}, with $v$ given by \eqref{e420}
satisfy the initial value condition $Y(x,0)=Y_0$ and that final
condition $Y(x,T)=0$. The proof is thus complete.
\end{proof}

\subsection*{Acknowledgments}
The author would like to thank professor Bopeng Rao
for his very helpful discussions and suggestions.

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\end{document}