\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{gathered} y_{tt} - y_{xx} = 0 \quad 00, \\ 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} with the initial conditions y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad 00, \\ 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} with the initial conditions y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad 02$or by means of regular control$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$): $$\begin{gathered} \phi_{tt} - \phi_{xx} = 0 \quad 0 < x < 1, \ t>0, \\ \phi(0,t) =0 \quad t>0\\ \phi_x (1,t) + \xi(t) = 0 \quad t>0,\\ \xi_t(t) - \phi_t(1,t) = 0 \quad t>0 \label{e21} \end{gathered}$$ with the initial conditions \phi(x,0)=\phi_0(x), \quad \phi_t(x,0)=\phi_1(x),\quad 02$ 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 $$T \norm{\Phi_0}_\mathcal{H}^2 + 2\Big[\int_0^1 \phi_t x\phi_xdx\Big]_0^T = \int_0^T \Big (|\phi_{xt} (1,t)|^2 + 2|\phi_x (1,t)|^2\Big )dt. \label{e26}$$ 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 $$-2\| \Phi_0\|_\mathcal{H}^2\leq 2\Big[\int_0^1 \phi_t x\phi_xdx\Big]_0^T \leq 2\| \Phi_0\|_\mathcal{H}^2. \label{e27}$$ 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{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} 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 $$\| \Phi^n(t) \|_\mathcal{H}=\| \Phi^n_0 \|_\mathcal{H}=1, \quad \forall t\in \mathbb{R}\label{e211}$$ and $$\int_0^T \Big [\Big(\phi^n_x (1,t)-e^{-\alpha t}\phi^n_x (1,0)\Big)^2 + \Big(\alpha \phi^n_x(1,t) +\phi^n_{xt} (1,t)\Big)^2\Big ]dt\to 0 \label{e212}$$ where $\Phi^n(t)= (\phi^n(x,t),\phi_t^n(x,t),\xi^n(t))$ is the solution of the problem $$\Phi_t^n = A\Phi^n, \quad \Phi^n(0)=\Phi_0^n.\label{e213}$$ 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 $$\frac{T-2}{2}\norm{\Phi_0^n}^2_\mathcal{H}\leq \varepsilon_n+ \alpha |\phi_x^n(1,0)|^2.\label{e216}$$ 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 $$\Phi_t= A \Phi, \quad \Phi(0)=\Phi_0\label{e217}$$ 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 $$\xi(t)=-\phi_x(1,t)=-ce^{-\alpha t}=-\sum{a_n \cos(\mu_n) e^{i\mu_n t}}.\label{e221}$$ Noting that $\mu_n\in \mathbb{R}$, then using \eqref{e219} we deduce that $$\sum |a_n \cos(\mu_n)|^2 <\infty. \label{e222}$$ 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 $$\Psi_t = (\alpha I + A)\Psi, \quad \Psi(0)=\Psi_0=\Phi_0 \in \mathcal{H}. \label{e43}$$ Replacing $\phi$ by $e^{-\alpha t}\psi$ in \eqref{e28}-\eqref{e29} we obtain $$c_1 e^{-2\alpha T}\int_0^T \Big [\Big(\psi_x (1,t)-\psi_x(1,0)\Big)^2 + |\psi_{xt}(1,t)|^2\Big ]dt \leq \norm{\Psi_0}^2_{\mathcal{H}} \label{e44}$$ and $$\norm{\Psi_0}^2_{\mathcal{H}} \leq c_2 \int_0^T \Big [\Big(\psi_x (1,t)-\psi_x(1,0)\Big)^2+ |\psi_{xt}(1,t)|^2 \Big ]dt. \label{e45}$$ Defining $$\widetilde \Psi_0 =(\alpha I + A)^{-1}\Psi_0=(\alpha I + A)^{-1}\Phi_0.$$ Then $$\norm{\widetilde \Psi_0}_{\mathcal{H}}^2 =\norm{(\alpha I + A)^{-1}\Phi_0}_{\mathcal{H}}^2 =\norm{\Phi_0}_{D(\alpha I + A)'}^2.\label{e46}$$ Now, let $\widetilde \Psi$ the solution of the equation $$\widetilde \Psi_t=(\alpha I + A)\widetilde \Psi, \quad \widetilde \Psi(0)=(\alpha I + A)^{-1}\Phi_0.\label{e47}$$ Applying the inequalities \eqref{e44}-\eqref{e45} to $\widetilde \Psi$ we obtain $$c_1 e^{-2\alpha T}\int_0^T \Big[\Big(\widetilde \psi_x (1,t)-\widetilde \psi_x(1,0)\Big)^2+ |\widetilde \psi_{xt}(1,t) |^2\Big]dt \leq \norm{\Phi_0}_{D(\alpha I + A)'}^2 \label{e48}$$ and $$\norm{\Phi_0}_{D(\alpha I + A)'}^2 \leq c_2 \int_0^T \Big [\Big(\widetilde \psi_x (1,t)-\widetilde \psi_x(1,0)\Big)^2+ |\widetilde \psi_{xt}(1,t)|^2\Big ]dt. \label{e49}$$ Using \eqref{e47} we have $$\widetilde \Psi_t(0)=(\alpha I + A)\widetilde \Psi(0)=\Phi_0.$$ Then $\widetilde \Psi_t$ solve the equation $$\widetilde \Psi_{tt}=(\alpha I + A) \widetilde \Psi_t, \quad \widetilde \Psi_t(0)=\Phi_0 .\label{e410}$$ 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 $$c_1 e^{-2\alpha T}\int_0^T \Big [\Big(\int_0^t \psi_x (1,s)ds \Big)^2+ |\psi_{x}(1,t)|^2\Big ]dt \leq \norm{\Phi_0}_{D(\alpha I + A)'}^2 \label{e411}$$ and $$\norm{\Phi_0}_{D(\alpha I + A)'}^2 \leq c_2 \int_0^T \Big [\Big(\int_0^t \psi_x (1,s)ds\Big)^2+ |\psi_{x}(1,t)|^2\Big ]dt. \label{e412}$$ 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{gathered} y_{tt} - y_{xx} = 0 \quad 00, \\ 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} with the initial conditions y(x,0)=y_0(x), \quad y_t(x,0)=y_1(x), \quad 0 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 $$Y(x,t)\in C^0([0,T] ;\mathcal{H})\label{e38}$$ defined in the sense that the equation \eqref{e35} is satisfied for all$\Phi_0\in \mathcal{H}$and all$02$. 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 $$Y(T)=0. \label{e311}$$ \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 $$\| \Phi_0\|_{1}^2 = \int_0^T ( |\phi_x (1,t)|^2+ |\phi_{xt} (1,t)|^2)dt, \ \ \forall \Phi_0\in \mathcal{H}.\label{e312}$$ 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 $$v(t)=v_0(t)-\frac{d}{dt}v_1(t)=:-\phi_x(1,t)+\frac{d}{dt}\phi_{xt}(1,t) \label{e313}$$ where the derivative$\frac{d}{dt}$is defined in the sense of \eqref{e37}. Using the direct inequality \eqref{e24}, we have $$\| v_0(t)\|_{L^2(0,T)}+\| v_1(t)\|_{L^2(0,T)} \leq \sqrt{2(T+2)}\| \Phi_0\|_\mathcal{H}.\label{e314}$$ Now solve the backward problem $$\Psi_t=A\Psi+V, \quad \Psi(T)=0. \label{e315}$$ 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 $$\| \Psi \|_\mathcal{H} \leq \sqrt{2(T+2)}(\| v_0(t)\|_{L^2(0,T)}+\| v_1(t)\|_{L^2(0,T)}).\label{e316}$$ Next we define the operator$\Lambda$as $$\Lambda \Phi_0=-\Psi(0), \quad \forall \Phi_0\in \mathcal{H}. \label{e317}$$ 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 $$-(\Psi_0,\Phi_0)_\mathcal{H} = \int_0^T (|\phi_x(1,t)|^2+ | \phi_{xt}(1,t)|^2)dt.\label{e318}$$ This implies $$(\Lambda \Phi_0,\Phi_0)_\mathcal{H} = \| \Phi_0\|_1^2.\label{e319}$$ 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 $$\Lambda \Phi_0 =-Y_0.\label{e320}$$ 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)$$$v(t)=-e^{\alpha t}v_0(t) +e^{\alpha t}\int_T^t \Big( \int_0^s v_0(\tau)e^{\alpha \tau}d\tau\Big)ds, \quad v_0(t)\in L^2(0,T). \label{e413}$$ For the wellposedness of the equation \eqref{e32} with the control \eqref{e413} we first interpret \eqref{e35} into the following form $$L(\Phi_0)=\langle Y(x,t),S_A(t) \Phi_0\rangle_{D(A)\times D(A)'}, \quad \forall \Phi_0\in D(A)'\label{e414}$$ where the linear form$L$is defined by $$L(\Phi_0)=\langle Y_0,\Phi_0\rangle_{D(A)\times D(A)'} + \int_0^t v(s)\xi(s) ds, \quad \forall \Phi_0\in D(A)'. \label{e415}$$ \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 $$Y(x,t)\in C^0\Big([0,T] ;D(A)\Big)\label{e416}$$ defined in the sense that the equation \eqref{e414} is satisfied for all$\Phi_0\in D(A)'$and all$02$. 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 $$Y(T)=0. \label{e418}$$ \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 $$\| \Phi_0\|_{2} = \int_0^T [ ( \int_0^t \phi_x (1,s)e^{\alpha s}ds)^2+ e^{2\alpha t}|\phi_{x} (1,t)|^2 ]dt, \quad \forall \Phi_0\in {D(A)'}.\label{e419}$$ 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 $$v(t)=-e^{\alpha t}\phi_x(1,t)+e^{\alpha t} \int_T^t\Big(\int_0^s\phi_{x}(1,\tau)e^{\alpha \tau}d\tau\Big) ds \in L^2(0,T). \label{e420}$$ From the direct inequality \eqref{e41}, we have $$\| v_0\|_{L^2(0,T)} \leq c_1^{-1/2}\| \Phi_0 \|_{D(A)'}.\label{e421}$$ Next we solve the backward problem $$\Psi_t=A\Psi+V, \quad \Psi(T)=0. \label{e422}$$ 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 $$\| \Psi \|_{D(A)} \leq \sqrt{2}c_5c_3^{-1/2} \| v_0\|_{L^2(0,T)}. \label{e423}$$ Next we define the operator$\Lambda$as $$\Lambda \Phi_0=-\Psi(0), \quad \forall \Phi_0\in D(A)'. \label{e424}$$ 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 $$-\langle \Psi_0,\Phi_0\rangle_{D(A)\times D(A)'} = \int_0^T \Big[ \Big( \int_0^t \phi_x (1,s)e^{\alpha s}ds\Big)^2+ e^{2\alpha t}|\phi_{x} (1,t)|^2 \Big]dt.\label{e425}$$ This implies $$\langle \Lambda \Phi_0,\Phi_0\rangle_{D(A)\times D(A)'} = \| \Phi_0\|_2^2.\label{e426}$$ 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 $$\Lambda \Phi_0 =-Y_0.\label{e427}$$ 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\$. 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