\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 06, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/06\hfil Exact boundary controllability] {Exact boundary controllability for a series of membranes elastically connected} \author[W. D. Bastos, R. S. O. Nunes, J. M. S. Pitot \hfil EJDE-2017/06\hfilneg] {Waldemar D. Bastos, Ruikson S. O. Nunes, Jo\~{a}o Manoel Soriano Pitot} \address{Waldemar D. Bastos \newline UNESP - S\~{a}o Paulo State University, IBILCE, Department of Mathematics, Rua Crist\'{o}v\~{a}o Colombo, 2265, Jd. Nazareth, CEP: 15054-000, S\~{a}o Jos\'{e} do Rio Preto, SP, Brazil} \email{waldemar@ibilce.unesp.br} \address{Ruikson S. O. Nunes \newline UFMT- Federal University of Mato Grosso, ICET, Department of Mathematics, Avenida Fernando Corr\^ea, 2367, Boa Esperan\c{c}a, CEP: 78060-900, Cuiab\'a, MT, Brazil} \email{ruiksonsillas@hotmail.com} \address{Jo\~{a}o Manoel Soriano Pitot \newline UNESP - S\~{a}o Paulo State University, IBILCE, Department of Mathematics, Rua Crist\'{o}v\~{a}o Colombo, 2265, Jd. Nazareth, CEP: 15054-000, S\~{a}o Jos\'e do Rio Preto, SP, Brazil} \email{john.pitot@gmail.com} \thanks{Submitted September 28, 2016. Published January 5, 2017.} \subjclass[2010]{35L52, 35L53, 35B40, 35B45, 93B05, 49J20} \keywords{Boundary control; exact controllability; coupled wave equations; \hfill\break\indent Neumann control} \begin{abstract} In this article we study the exact controllability with Neumann boundary controls for a system of linear wave equations coupled in parallel by lower order terms on piecewise smooth domains of the plane. We obtain square integrable controls for initial state with finite energy and time of controllability near the optimal value. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} In this article we study a system of $m\geq 2$ coupled wave equations \begin{equation}\label{1.1} \begin{gathered} U_{tt}- \Delta U + AU= 0 \quad \text{in } \Omega \times ]0,T[,\\ U(\cdot,0)=U_0\quad \text{in } \Omega,\\ U_{t}(\cdot,0)=U_1\quad \text{in } \Omega,\\ \frac{\partial U}{\partial \eta}=f \quad \text{on } \partial\Omega \times ]0,T[, \end{gathered} \end{equation} where $\Omega $ is a bounded simply connected domain of the plane with piecewise smooth boundary $\partial \Omega $. The exterior unit normal vector, defined almost everywhere in $\partial \Omega $ is denoted by $\eta $ and by $\frac{\partial U}{\partial \eta }$ we denote the normal derivative of $U=(u^1,\dots,u^{m})^T$ where ${}^T$ stands for transpose. As usual we write $U_{tt}=(u_{tt}^1,\dots,u_{tt}^{m})^T$, $\Delta U=(\Delta u^1,\dots,\Delta u^{m})^T$ where the subscript ${}_{tt}$ denotes the second derivative with respect to $t$ and $\Delta $ the Laplacian with respect to space variables. $A=[a_{ij}]_{m\times m}$ is assumed to be a real diagonalizable matrix with nonnegative eigenvalues. The system \eqref{1.1} includes those used to model physical phenomena in which several elastic bodies are attached together by elastic layers. For instance, following \cite{r9, r17} we see that a system composed by $m$ identical membranes connected in parallel by elastic layers having the transverse displacement given by $U=(u^1,\dots,u^{m})^T$ must satisfy the relations \begin{equation}\label{1.2} \begin{gathered} u_{tt}^{i}-\Delta u^{i}+k_{i-1}(u^{i}-u^{i-1})+k_i(u^{i}-u^{i+1})=0 \quad i=1,\dots,m, \\ u^{0}\equiv u^{m+1}\equiv 0, \quad k_i \text{ real constants } i=0,1,\dots,m. \end{gathered} \end{equation} In this case the coupling matrix is the tridiagonal symmetric matrix $$A= \begin{bmatrix} k_0+k_1 & -k_1 & & & & & \\ -k_1 & k_1+k_2 & -k_2 & & & & \\ & -k_2 & k_2+k_3 & -k_3 & & & \\ & & -k_3 & \ddots & & & \\ & & & & & k_{m-2}+k_{m-1} & -k_{m-1} \\ & & & & & -k_{m-1} & k_{m-1}+k_{m} \end{bmatrix}. $$ In this work we are interested in studying exact boundary controllability for system \eqref{1.1}. Before announcing the results of this work we establish some notation. For a general bounded domain $\Xi \subset\mathbb{R}^{N}$ where $N$ is a positive integer, we denote $(L^2(\Xi ),\|\cdot \| _{L^2(\Xi )})$ and $(H^1(\Xi ),\|\cdot\|_{H^1(\Xi )})$ the usual Lebesgue space of square integrable functions on $\Xi $ and the Sobolev space of functions of $L^2(\Xi )$ whose first order distributional derivatives are also in $L^2(\Xi )$. These spaces are endowed with their usual norms \cite{r7}. We also introduce the product space $\mathcal{H}^1(\Xi )=H^1(\Xi )\times \dots\times H^1(\Xi )=[H^1 (\Xi)]^{m}$ with norm given by $$ \| U\| _{\mathcal{H}^1(\Xi )} =\big[\|u^1\| _{H^1(\Xi )}^2+\dots+\| u^{m}\| _{H^1(\Xi )}^2 \big]^{1/2},\quad U=(u^1,\dots,u^{m})^T\in \mathcal{H}^1(\Xi ) $$ and the space $\mathcal{L}^2(\Xi )=L^2(\Xi )\times \dots \times L^2(\Xi )=[L^2(\Xi )]^{m}$ with norm $$ \| U\| _{\mathcal{L}^2(\Xi )} =\big[ \| u^1\| _{L^2(\Xi )}^2+\dots+\| u^{m}\| _{L^2(\Xi )}^2\big]^{1/2}, \quad U=(u^1,\dots,u^{m})^T\in \mathcal{L}^2(\Xi ). $$ The product space $\mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )$ is endowed with the natural norm $$ \| (U,V)\| _{\mathcal{H} ^1(\Xi )\times \mathcal{L}^2(\Xi )}^2=\| U\| _{ \mathcal{H}^1(\Xi )}^2+\| V\| _{\mathcal{L}^2(\Xi)}^2. $$ The spaces $\mathcal{H}_{\rm loc}^1(\Xi )$, $\mathcal{L}_{\rm loc}^2(\Xi )$ and $\mathcal{H}_0^1(\Xi )$ have their usual meaning. The theorem bellow is the main result of this article. \begin{theorem} \label{thm1.1} Let $A=[a_{ij}]_{m\times m}$ be a real diagonalizable matrix with nonnegative eigenvalues and $\Omega $ be a bounded simply connected domain of the plane with piecewise smooth boundary $\partial \Omega $. It is assumed that $\partial \Omega $ has no cuspid point and that $\Omega $ lays in one side of $\partial \Omega $. Then, for any $T_{\ast }>\operatorname{diam}(\Omega )$ and any initial state $U_0\in \mathcal{H}^1(\Omega )$, $U_1\in \mathcal{L}^2(\Omega )$ there exist a control $f\in \mathcal{L}^2(\partial \Omega \times ]0,T_{\ast }[)$ so that the solution $U\in \mathcal{H}^1(\Omega \times ]0,T_{\ast }[)$ of \eqref{1.1} satisfies the final condition $ U(\cdot,T_{\ast })=U_{t}(\cdot,T_{\ast })=0$ on $\Omega $. \end{theorem} This theorem extends to systems of $m>2$ equations and time of control near optimal value the results presented in \cite{r6} for a system with two equations. We employ the extension method introduced by Russell \cite{r18} as improved by Lagnese \cite{r12} for the wave equation with smooth initial state. We explore properties of the solutions of the Cauchy problem for $U_{tt}-\Delta U+AU=0$ such as local energy decay, regularity on noncharacteristic surfaces and time analyticity. Following the evolution of the method to be employed and also the didactic aspect we firstly prove controllability for a large enough time interval following Russell's original procedure \cite{r18}. After, following the lines of \cite{r16} we obtain time analyticity for the solution operator associated to the Cauchy problem for $U_{tt}-\Delta U+AU=0$ and proceed as in \cite{r12} to prove Theorem \ref{thm1.1}. The literature on control and stabilization for coupled systems has increased enormously in the last two decades. Just to mention, energy decay and boundary stabilization for the system \eqref{1.2} with $m=2$, $k_0=k_2=0$ has been treated by several authors. Indeed, in \cite{r1,r2,r11} the authors estimate the energy decay rate inside a bounded smooth domain $\Omega\subset \mathbb{R}^{N}$ under nonlinear boundary feedbacks. Similar questions for two waves with different speed has been studied in \cite{r14} for smooth domains of $\mathbb{R}^{N}$, $(N\leq 3)$. Controllability for a system like \eqref{1.1} with $m=2$ and Dirichlet type control has been studied in \cite{r3} by employing the Two Level Energy Method. Concerning controllability results for coupled parabolic problems we mention the survey \cite{r10}. A variety of problems for systems highly coupled of partial differential equations, possibly of different type, has been considered in the treatise \cite{r13}. This article is organized as follows. In Section 2 we obtain a local energy decay for solutions to the Cauchy problem for the system $U_{tt}-\Delta U+AU=0$ and obtain controllability in large enough time interval. Section 3 is dedicated to prove time analyticity of solutions to the system $U_{tt}-\Delta U+AU=0$. Finally, in Section 4 we prove Theorem \ref{thm1.1} and illustrate how the method works for the case in which the control acts only in a part of the membrane's boundary. \section{Energy decay and controllability in large time} In this section we prove the local energy decay of solutions to the Cauchy problem \begin{equation}\label{2.1} \begin{gathered} U_{tt}-\Delta U+AU=0 \quad \text{in }\mathbb{R}^{2+1}, \\ U(\cdot,0)=U_0,\quad U_{t}(\cdot,0)=U_1 \quad \text{in }\mathbb{R}^2, \end{gathered} \end{equation} and use it to obtain controllability for the system \eqref{1.1} in an interval of time sufficiently large. Let $0\leq \lambda _1\leq \dots \leq \lambda _{m}$ be non negative eigenvalues of the real diagonalizable matrix $A=[a_{ij}]_{m\times m}$. \begin{theorem} \label{thm2.1} Let $\Xi \subset\mathbb{R}^2$, be a bounded domain. There exists constants $T_0>\operatorname{diam}(\Xi )$ and $K=K(\Xi ,A,T_0)>0$ such that the solution $U\in \mathcal{H}_{\rm loc}^1(\mathbb{R}^{2+1})$ of the Cauchy problem \eqref{2.1} satisfies \begin{equation}\label{2.2} \| U(\cdot,t)\| _{\mathcal{H}^1(\Xi)}^2 +\| U_{t}(\cdot,t)\| _{\mathcal{L}^2(\Xi )}^2 \leq \frac{K}{t^2}\big\{\| U_0\| _{\mathcal{H}^1(\Xi )}^2+\| U_1\| _{\mathcal{L}^2(\Xi )}^2\big\} \end{equation} for every $t>T_0$ whenever the initial data $(U_0,U_1)\in \mathcal{H}^1(\mathbb{R}^2)\times \mathcal{L}^2(\mathbb{R}^2)$ has compact support in $\Xi $. \end{theorem} \begin{proof} First observe that the transposed matrix $A^T$ is also diagonalizable and has the same eigenvalues of $A$. Let $\{v^{i}=(\alpha _1^{i},\dots,\alpha _{m}^{i}): i=1,\dots,m\}$ be a basis for $\mathbb{R}^{m}$ whose elements are eigenvectors of $A^T$ with $v^{i}$ associated to $\lambda _i$. We have for each $i=1,\dots,m$ the set of relations \begin{equation}\label{2.3} a_{1j}\alpha _1^{i} + a_{2j}\alpha _2^{i} + \dots + a_{mj}\alpha _{m}^{i} = \lambda _i\alpha _j^{i}, \quad j=1,\dots,m. \end{equation} Let $B$ be the real matrix $[\alpha _j^{i}]_{m\times m}$ and let $W=(w^1,\dots,w^{m})^T$ be given by $W=BU$ where $U=(u^1,\dots,u^{m})^T$ is the solution of the Cauchy problem \eqref{2.1}. Hence \begin{equation}\label{2.4} w^{i}=\alpha _1^{i}u^1+\alpha _2^{i}u^2+\dots+\alpha _{m}^{i}u^{m}, \quad i=1,\dots,m. \end{equation} For each $i=1,\dots,m$, an easy computation leads us to $$ w_{tt}^{i}=\Delta (\alpha _1^{i}u^1 +\alpha_2^{i}u^2+\dots+\alpha _{m}^{i}u^{m}) -\sum_{j=1}^{m}(a_{1j}\alpha_1^{i}+a_{2j}\alpha _2^{i} +\dots+a_{mj}\alpha _{m}^{i})u^{j}. $$ Now, by using the relations \eqref{2.3} and \eqref{2.4} we obtain $$ w_{tt}^{i}=\Delta w^{i}-\lambda _i(\alpha _1^{i}u^1+\alpha _2^{i}u^2+\dots+\alpha _{m}^{i}u^{m}) =\Delta w^{i}-\lambda _iw^{i}, \quad i=1,\dots,m. $$ Putting $U_0=(u_0^1,u_0^2,\dots,u_0^{m})^T$, $U_1=(u_1^1,u_1^2,\dots,u_1^{m})^T$ and noting that $W$ must satisfy the initial conditions $W(\cdot,0)=BU_0$ and $W_{t}(\cdot,0)=BU_1$ we see that each entry $w^{i}$ of the new independent variable $W$ satisfies the Cauchy problem \begin{equation}\label{2.5} \begin{gathered} w_{tt}^{i}-\Delta w^{i}+\mu_i^2w^{i}=0 \quad \text{in } \mathbb{R}^{2+1}, \\ w^{i}(\cdot,0)=\alpha _1^{i}u_0^1+\alpha _2^{i}u_0^2+\dots+\alpha_{m}^{i}u_0^{m} \quad \text{in } \mathbb{R}^2, \\ w_{t}^{i}(\cdot,0)=\alpha _1^{i}u_1^1+\alpha _2^{i}u_1^2+\dots+\alpha _{m}^{i}u_1^{m} \quad \text{in } \mathbb{R}^2, \end{gathered} \end{equation} where $\mu _i^2=\lambda _i$ for $i=1,\dots,m$. Since, $w^{i}(\cdot,0)$ and $w_{t}^{i}(\cdot,0)$ have compact support in $\Xi $, using \cite[Lemma 2.1]{r6} we obtain the estimate $$ \| w^{i}(\cdot,t)\| _{H^1(\Xi )}^2+\| w_{t}^{i}(\cdot,t)\| _{L^2(\Xi )}^2 \leq \frac{K_i}{t^2}\big\{\| w^{i}(\cdot,0)\| _{H^1(\Xi )}^2+\| w_{t}^{i}(\cdot,0)\| _{L^2(\Xi )}^2\big\} $$ for every $t>T_0^{i}$, for some $T_0^{i}>\operatorname{diam}(\Xi )$ and some constant $K_i=K_i(\Xi,T_0^{i},\lambda _i)$ $>0$. The constant $T_0^{i}$ depends on $\lambda _i$ and $\operatorname{diam}(\Xi )$. From the above inequality it follows \begin{equation}\label{2.6} \begin{aligned} & \| w^{i}(\cdot,t)\| _{H^1(\Xi)} ^2+\| w_{t}^{i}(\cdot,t)\| _{L^2(\Xi )}^2\\ & \leq \frac{K_i}{t^2}(| \alpha _1^{i}| ^2+| \alpha _2^{i}| ^2+\dots+| \alpha _{m}^{i}|^2)\big\{\| U_0\| _{\mathcal{H}^1(\Xi )}^2+\| U_1\| _{\mathcal{L}^2(\Xi )}^2\big\} \end{aligned} \end{equation} for every $t>T_0^{i}$, $i=1,\dots,m$. Now we set $\| B\| ^2=\sum | \alpha_j^{i}| ^2$, $M=\max \{K_i;i=1,\dots,m\}$ and $T_0=\max \{T_0^{i};i=1,\dots,m\}$, and by summing up inequalities \eqref{2.6} we obtain \begin{equation}\label{2.7} \| W(\cdot,t)\| _{\mathcal{H}^1(\Xi )}^2+\| W_{t}(\cdot,t)\| _{\mathcal{L}^2(\Xi )}^2\leq \frac{M}{t^2}\| B\| ^2\big\{\| U_0\| _{ \mathcal{H}^1(\Xi )}^2+\| U_1\| _{\mathcal{L}^2(\Xi )}^2\big\} \end{equation} for every $t>T_0$. By setting $B^{-1}=[\beta _j^{i}]_{m\times m}$ and observing that $u^{i}(\cdot,t)=\beta _1^{i}w^1(\cdot,t)+\dots+\beta _{m}^{i}w^{m}(\cdot,t)$, $i=1,\dots,m $, we obtain \begin{equation}\label{2.8} \begin{aligned} & \| U(\cdot,t)\| _{\mathcal{H}^1(\Xi )}^2+\| U_{t}(\cdot,t)\| _{\mathcal{L}^2(\Xi )}^2\\ &\leq\| B^{-1}\| ^2\big\{\| W(\cdot,t)\| _{\mathcal{H} ^1(\Xi )}^2+\| W_{t}(\cdot,t)\| _{\mathcal{L}^2(\Xi )}^2\big\}. \end{aligned} \end{equation} Now we put $K=M\| B\| ^2\| B^{-1}\| ^2$ and combine \eqref{2.7} and \eqref{2.8} to obtain \eqref{2.2}. \end{proof} We can now prove the first result on exact controllability for the system \eqref{1.1}. \begin{theorem} \label{thm2.2} Let $A=[a_{ij}]_{m\times m}$\ be a real diagonalizable matrix with nonnegative eigenvalues and $\Omega \subset \mathbb{R}^2$, be a bounded simply connected domain with piecewise smooth boundary $\partial \Omega $. It is assumed that $\partial \Omega $\ has no cuspid point and that $\Omega $\ lays in one side of $\partial \Omega $.\ Then, there exists a large enough $T>0$\ such that for each initial state $ U_0\in \mathcal{H}^1(\Omega )$, $U_1\in \mathcal{L}^2(\Omega )$ there exist a control $f\in \mathcal{L}^2(\partial \Omega \times ]0,T[)$ so that the solution $U\in \mathcal{H}^1(\Omega \times ]0,T[)$ of the system \eqref{1.1} satisfies the final condition $ U(\cdot,T)=U_{t}(\cdot,T)=0$ on $\Omega$. \end{theorem} \begin{proof} For a fixed $\delta >0$ we set $\Omega _{\delta }=\{x+y:x\in \Omega ,| y| <\delta \}$. Associated with the Cauchy problem \eqref{2.1} we define, for $t>0$, the bounded linear operator $\mathbf{S}_{t}:\mathcal{H}_0^1(\Omega _{\delta })\times \mathcal{L} ^2(\Omega _{\delta })\to \mathcal{H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })$ by setting \begin{equation}\label{2.9} \mathbf{S}_{t}(V_0,V_1)(x)=(V(x,t),V_{t}(x,t)), \quad x\in \Omega _{\delta } \end{equation} where $V$ is the solution to the Cauchy problem \eqref{2.1} with initial data $(V_0,V_1)\in \mathcal{H}_0^1(\Omega _{\delta })\times \mathcal{ L}^2(\Omega _{\delta })$ extended by zero outside $\Omega _{\delta}$. Now, let $T_0$ be given by Theorem \ref{thm2.1} with $\Xi =\Omega _{\delta }$. It follows from the Theorem \ref{thm2.1} the estimate \begin{equation}\label{2.10} \| \mathbf{S}_{T}(V_0,V_1)\| _{\mathcal{ H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })}^2\leq \frac{K}{T^2}\big\{\| V_0\| _{\mathcal{H}^1(\Omega _{\delta })}^2+\| V_1\| _{\mathcal{L}^2(\Omega _{\delta })}^2\big\}, \end{equation} for every $T>T_0$, $(V_0,V_1)\in \mathcal{H}_0^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })$ and a constant $K$ independent of $(V_0,V_1)$. For arbitrary $T>0$ consider the Cauchy problem \begin{equation}\label{2.11} \begin{gathered} Z_{tt}-\Delta Z+AZ=0 \quad \text{in } \mathbb{R}^{2+1}, \\ Z(\cdot,T)=Z_0, \quad Z_{t}(\cdot,T)=Z_1 \quad \text{in } \mathbb{R}^2, \end{gathered} \end{equation} with initial state $(Z_0,Z_1)\in \mathcal{H}_0^1(\Omega_{\delta })\times \mathcal{L}^2(\Omega _{\delta })$ extended by zero outside $\Omega_{\delta}$, at initial time $T$. Associated to the Cauchy problem \eqref{2.11} we define the bounded linear operator $\widehat{\mathbf{S}}_{T}:\mathcal{H}_0^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega_{\delta })\to \mathcal{H}^1(\Omega _{\delta }) \times \mathcal{L}^2(\Omega _{\delta })$ by setting \begin{equation}\label{2.12} \widehat{\mathbf{S}}_{T}(Z_0,Z_1)(x)=(Z(x,0),Z_{t}(x,0)), \quad x\in \Omega_{\delta}. \end{equation} Note that the function $Z(\cdot,T-\tau )$ satisfies $Z_{\tau \tau }-\Delta Z+AZ=0 $ with data $(Z_0,Z_1)$ at time $\tau =0$. Now, assuming that $T>T_0$ and applying the Theorem \ref{thm2.1} to the solution of \eqref{2.11} we obtain $$ \| Z(\cdot,0)\| _{\mathcal{H}^1(\Omega _{\delta })}^2+\| Z_{\tau }(\cdot,0)\| _{\mathcal{L} ^2(\Omega _{\delta })}^2\leq \frac{K}{T^2}\big\{\| Z_0\| _{\mathcal{H}^1(\Omega _{\delta })}^2+\| Z_1\| _{\mathcal{L}^2(\Omega _{\delta })}^2\big\}. $$ Hence, \begin{equation}\label{2.13} \| \widehat{\mathbf{S}}_{T}(Z_0,Z_1)\| _{\mathcal{H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta})}^2 \leq \frac{K}{T^2}\big\{\| Z_0\| _{\mathcal{H} ^1(\Omega _{\delta })}^2+\| Z_1\| _{\mathcal{L} ^2(\Omega _{\delta })}^2\big\}, \end{equation} for every $T>T_0$ and $(Z_0,Z_1)\in \mathcal{H} _0^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })$. Let $E$ be an extension operator taking an arbitrary initial state $(V_0,V_1)\in \mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )$ into its extension $(\widetilde{V}_0,\widetilde{V}_1)\in \mathcal{H}^1 (\mathbb{R}^2)\times \mathcal{L}^2(\mathbb{R}^2)$ with compact support in $\Omega _{\delta }$. Let $\widetilde{V}\in\mathcal{H}_{\rm loc}^1(\mathbb{R} ^{2+1})$ be the solution of \eqref{2.1} with initial state $(\widetilde{V}_0, \widetilde{V}_1)$. Let $\varphi \in C_0^{\infty}(\mathbb{R}^2)$ be a cut off function such that $\varphi =1$ on $\Omega _{\delta /2}$ and $\varphi =0$ on $\mathbb{R}^2\backslash \Omega_{\delta}$. Let $\widetilde{Z}\in \mathcal{H}_{\rm loc}^1(\mathbb{R}^{2+1})$ be the solution of the problem \eqref{2.11} with $(Z_0,Z_1)=\varphi ( \widetilde{V}(\cdot,T),\widetilde{V}_{t}(\cdot,T))$. Notice that the state $(\widetilde{Z}(\cdot,0),\widetilde{Z}_{t}(\cdot,0))$ is expressed in terms of the operators $\mathbf{S}_{T}$, $\widehat{\mathbf{S}}_{T}$ and the extension $E$ as $$ (\widetilde{Z}(\cdot,0),\widetilde{Z}_{t}(\cdot,0))=(\widehat{\mathbf{S}} _{T}\varphi \mathbf{S}_{T}E)(V_0,V_1) $$ for every $T>T_0$ and arbitrary initial data $(V_0,V_1)\in \mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )$. Here $\varphi $ stands for the operator multiplication by $\varphi $. Now let us define the function $U^{\circ }=\widetilde{V}-\widetilde{Z}$. Observe that $U^{\circ }$ satisfies \begin{equation}\label{2.14} U_{tt}^{\circ }-\Delta U^{\circ }+AU^{\circ }=0 \quad \text{in } \mathbb{R}^{2+1}. \end{equation} Since $\widetilde{Z}(\cdot,T)=\varphi \widetilde{V}(\cdot,T)$ and $\widetilde{Z}_{t}(\cdot,T)=\varphi \widetilde{V}_{t}(\cdot,T)$ and $\varphi =1$ on $\Omega_{\delta /2}$ it follows that $U^{\circ }(\cdot,T)=U_{t}^{\circ }(\cdot,T)=0$ on $\Omega $. Consider $U_0\in \mathcal{H}^1(\Omega )$, $U_1\in \mathcal{L}^2(\Omega )$ given in the statement of the Theorem \ref{thm2.2}. To have $U^{\circ }(\cdot,0)=U_0$ and $U_{t}^{\circ }(\cdot,0)=U_1$ in $\Omega $ we might solve for the variable $(V_0,V_1)\in \mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )$ the equation $(V_0,V_1)-(\widetilde{Z}(\cdot,0),\widetilde{Z}_{t}(\cdot,0))=(U_0,U_1)$ in $\Omega $. More precisely we must solve the equation \begin{equation}\label{2.15} (V_0,V_1)-(R\widehat{\mathbf{S}}_{T}\varphi \mathbf{S}_{T}E)(V_0,V_1)=(U_0,U_1) \end{equation} where $R$ denotes the operator restriction to $\Omega $. Denoting $\mathbf{K}_{T}=R\widehat{\mathbf{S}}_{T}\varphi \mathbf{S}_{T}E$, we prove that $Id-\mathbf{K}_{T}$ is invertible. We use the energy decay to prove that $\mathbf{K}_{T}$ is a contraction in $\mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )$ if the parameter $T$ is taken sufficiently large. Indeed, by using \eqref{2.13} we have \begin{align*} \| R\widehat{\mathbf{S}}_{T}\varphi \mathbf{S}_{T}E(V_0,V_1) \|^2_{\mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega)} &=\| (\widetilde{Z}(\cdot,0),\widetilde{Z}_{t}(\cdot,0))\| _{ \mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )}^2\\ & \leq \| (\widetilde{Z}(\cdot,0),\widetilde{Z}_{t}(\cdot,0))\| _{ \mathcal{H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta})}^2\\ &=\| \widehat{\mathbf{S}}_{T}\varphi (\widetilde{V}(\cdot,T), \widetilde{V}_{t}(\cdot,T))\| _{\mathcal{H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })}^2\\ &\leq \frac{K}{T^2}\big\{\| \varphi \widetilde{V}(\cdot,T)\| _{ \mathcal{H}^1(\Omega _{\delta })}^2+\| \varphi \widetilde{V} _{t}(\cdot,T)\| _{\mathcal{L}^2(\Omega _{\delta })}^2\big\}\\ &\leq \frac{\widetilde{K}}{T^2}\big\{\| \widetilde{V}(\cdot,T)\| _{\mathcal{H} ^1(\Omega _{\delta })}^2+\| \widetilde{V}_{t}(\cdot,T)\| _{ \mathcal{L}^2(\Omega _{\delta })}^2\big\}\\ &=\frac{\widetilde{K}}{T^2}\| \mathbf{S}_{T}E(V_0,V_1)\|_{\mathcal{H}^1 (\Omega _{\delta })\times \mathcal{L}^2(\Omega_{\delta })}^2 \end{align*} where $\widetilde{K}$ depends on $K$ and $\varphi $. Now, using \eqref{2.10} we obtain \begin{align*} \frac{\widetilde{K}}{T^2}\big\{\| \widetilde{V}(\cdot,T)\| _{ \mathcal{H}^1(\Omega _{\delta })}^2 + \| \widetilde{V} _{t}(\cdot,T)\| _{\mathcal{L}^2(\Omega _{\delta })}^2\big\} &\leq \frac{\widetilde{K}}{T^2}\frac{K}{T^2}\big\{\| \widetilde{V} _0\| _{\mathcal{H}^1(\Omega _{\delta })}^2 +\|\widetilde{V}_1\| _{\mathcal{L}^2(\Omega _{\delta })}^2\big\}\\ &=\frac{\widetilde{K}K}{T^{4}}\| E(V_0,V_1)\| _{ \mathcal{H}^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })}^2 \\ &\leq \frac{\rm Const.}{T^{4}}\| (V_0,V_1)\| _{ \mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )}^2 \end{align*} where Const. represents a convenient constant. Hence $$ \| \mathbf{K}_{T}(V_0,V_1)\| _{\mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )}^2\leq \frac{\rm Const.}{T^{4}}\| (V_0,V_1)\| _{\mathcal{H}^1(\Omega )\times \mathcal{L} ^2(\Omega )}^2, $$ for all $(V_0,V_1)\in \mathcal{H}^1(\Omega )\times \mathcal{L }^2(\Omega )$ and $T>T_0$. At this point we choose and fix a value for $T>T_0$ such that $\frac{\rm Const.}{T^{4}}<1$. For such $T$ the operator $\mathbf{K}_{T}$ is a contraction on $\mathcal{H}^1(\Omega )\times \mathcal{L}^2(\Omega )$. Let $(\mathcal{V}_0,\mathcal{V}_1)\in \mathcal{H}^1(\Omega )\times \mathcal{L} ^2(\Omega )$ be the unique solution of \eqref{2.15}. From the construction we see that $E(\mathcal{V}_0,\mathcal{V}_1)-(\widehat{\mathbf{S}} _{T}\varphi \mathbf{S}_{T}E)(\mathcal{V}_0,\mathcal{V}_1)$ is an extension of $(U_0,U_1)$ to the entire space $\mathbb{R}^2$. Let us define $(U_0^{\circ },U_1^{\circ })=:E(\mathcal{V}_0, \mathcal{V}_1)-(\widehat{\mathbf{S}}_{T}\varphi \mathbf{S}_{T}E)(\mathcal{V }_0,\mathcal{V}_1)$. Since all the Cauchy problems involved in the construction of $(U_0^{\circ },U_1^{\circ })$ have the property of finite velocity of propagation and all the initial states considered have compact support we conclude that $(U_0^{\circ },U_1^{\circ })$ also have compact support. Once we have the appropriate extension to the initial data we return to the beginning of the proof and start solving the problem \eqref{2.1} with initial data $(U_0^{\circ },U_1^{\circ })$, then we localize its state at the time $T$ to obtain the correct initial state to the backward problem \eqref{2.11}. We then built the solution $U^{\circ }$ to the system \eqref{2.14} satisfying \begin{equation}\label{2.16} U^{\circ }(\cdot,0)=U_0, \quad U_{t}^{\circ }(\cdot,0)=U_1, \quad U^{\circ }(\cdot,T)=0, \quad U_{t}^{\circ }(\cdot,T)=0 \quad \text{on } \Omega . \end{equation} Observe that $U_{tt}^{\circ }-\Delta U^{\circ } \in \mathcal{L}_{\rm loc}^2(\mathbb{R}^2\times \mathbb{R})$. From \cite[Theorem 2]{r19} it follows that trace of the normal derivative of $U^{\circ }$ along the surface $\partial \Omega \times ]0,T[$ is well defined and is locally square integrable. Hence $\frac{\partial U^{\circ }}{ \partial \eta }\in \mathcal{L}^2(\partial \Omega \times ]0,T[)$. To finish the proof we define $U=:U^{\circ }| _{\Omega \times ]0,T[}$; the restriction of $U$ to the domain $\Omega \times ]0,T[$ and $f=:\frac{\partial U^{\circ }}{\partial \eta }$ and observe that they all meet the conditions of the theorem. \end{proof} \section{Time analyticity of solutions to the system $U_{tt}-\Delta U+AU=0$} Let $\Xi \subset\mathbb{R}^2$ be a bounded domain, $\mu >0$ and $v_0$, $v_1\in C_0^{\infty } (\mathbb{R}^2)$ be functions such that $\operatorname{supp}v_0$, $\operatorname{supp}v_1\subset \Xi $. If $v$ is the solution of the Cauchy problem \begin{equation}\label{3.1} \begin{gathered} v_{tt}-\Delta v+\mu ^2v=0 \quad \text{in } \mathbb{R}^{2+1}, \\ v(\cdot,0)=v_0,\quad v_{t}(\cdot,0)=v_1 \quad \text{in } \mathbb{R}^{2+1}, \end{gathered} \end{equation} then $v$ is given, for every $t>0$, by the formula \begin{equation}\label{3.2} \begin{aligned} v(x,t)&=\frac{1}{2\pi }\frac{\partial }{\partial t} \int_{| y-x| \operatorname{diam}(\Xi )$ and observe that for $\ t\geq T_0$ we have $\Xi \subset \{y:| y-x| 2$, see \cite{r15}). For each $t>0$ we define the bounded linear operator $S_{t}:H_0^1(\Xi )\times L^2(\Xi )\to H^1(\Xi )\times L^2(\Xi )$ by setting \begin{equation}\label{3.4} S_{t}(v_0,v_1)(x)=(v(x,t),v_{t}(x,t)), \quad x\in \Xi \end{equation} where $v$ is the solution to the Cauchy problem \eqref{3.1} with initial data $(v_0,v_1)\in H_0^1(\Xi )\times L^2(\Xi )$ extended by zero outside $\Xi $. Our goal now is to extend the family $\{S_{t}:t>T_0\}$ to the complex parameter $t=\zeta \in \Sigma $ where $\Sigma $ is an appropriate domain in the complex plane. To do so we start by exploiting the explicit formula for $v(x,t)$ and $v_{t}(x,t))$, $x\in \overline{\Xi }$ and $t\geq T_0>\operatorname{diam}(\Xi )$. To handle properly the formula \eqref{3.3} we introduce the functions \begin{equation}\label{3.5} \gamma _k(x,y,t)=\frac{\cos (\mu \sqrt{t^2-| y-x| ^2})}{(t^2-|y-x| ^2)^{k/2}}, \quad \theta _k(x,y,t)=\frac{\sin (\mu \sqrt{t^2-| y-x| ^2})}{(t^2-| y-x| ^2)^{k/2}} \end{equation} $x,y\in \overline{\Xi }$, $t\geq T_0$ and $k=1,2,\dots$. Observe that \eqref{3.3} reduces to $$ v(x,t)=\frac{1}{2\pi }\frac{\partial }{\partial t}\int_{\Xi }\gamma _1(x,y,t)v_0(y)dy+\frac{1}{2\pi }\int_{\Xi }\gamma _1(x,y,t)v_1(y)dy. $$ Firstly we claim that the functions \eqref{3.5} satisfy \begin{equation}\label{3.6} | \gamma _k(x,y,t)| ,\,\, | \theta _k(x,y,t)| \leq \frac{\varrho ^{k}}{T_0^{k}},\quad k=1,2,\dots \end{equation} for some $\varrho >0$ and every $x,y\in \overline{\Xi }$ and $t\geq T_0$. Indeed, if we introduce the function $\chi (s)=\frac{1}{\sqrt{1-s^2}}$, $-1\operatorname{diam}(\overline{\Xi })$ we have $0\leq | \frac{y-x}{t}|<1$, we then rewrite formulas \eqref{3.5} as \begin{gather} \label{3.7} \gamma _k(x,y,t)=\frac{1}{t^{k}}\cos (\mu \sqrt{t^2-| y-x| ^2})\chi (| \frac{y-x}{t}| )^{k},\quad k=1,2,\dots, \\ \label{3.8} \theta _k(x,y,t)=\frac{1}{t^{k}} \sin (\mu \sqrt{t^2-| y-x| ^2}) \chi (| \frac{y-x}{t}| )^{k},\quad k=1,2,\dots \end{gather} Now we choose and fix $\kappa $ such that $\frac{\operatorname{diam}(\Xi )}{T_0}<\kappa<1 $ and let $\varrho $ be the maximum of the function $\chi $ in $[-\kappa,\kappa ]$. For $x,y\in \overline{\Xi }$ and $t\geq T_0$ we have $| \frac{y-x}{t}| \leq \frac{\operatorname{diam}(\Xi )}{T_0}<\kappa $ which implies $\chi (| \frac{y-x}{t}| )\leq \varrho $. Hence $| \gamma _k(x,y,t)|$, $| \theta_k(x,y,t)| \leq \frac{\varrho ^{k}}{t^{k}}\leq \frac{\varrho ^{k} }{T_0^{k}}$\ for every $x,y\in \overline{\Xi }$ and $t\geq T_0$. This shows our claim. Now observe that the derivatives of the functions \eqref{3.7} and \eqref{3.8} are expressed in terms of themselves. Indeed, some computation led us \begin{gather} \label{3.9} \frac{\partial }{\partial t} \gamma _k=-\mu t\theta _{k+1}-kt\gamma _{k+2}, \\ \label{3.10} \frac{\partial }{\partial t}\theta_k=\mu t\gamma _{k+1}-kt\theta _{k+2}, \\ \label{3.11} \frac{\partial }{\partial x_i}\gamma _k =-\mu (y_i-x_i)\theta _{k+1}-k(y_i-x_i)\gamma _{k+2},\quad i=1,2,\\ \label{3.12} \frac{\partial }{\partial x_i}\theta_k=\mu (y_i-x_i)\gamma _{k+1}-k(y_i-x_i) \theta _{k+2},\quad i=1,2. \end{gather} By using \eqref{3.9}-\eqref{3.12} to obtain higher order derivatives of $\gamma _1$ and using the estimates \eqref{3.6} we conclude that $v\in C^{\infty }(\overline{\Xi }\times \lbrack T_0,\infty ))$ and differentiations in \eqref{3.3} may be carried out beneath the integral. For instance, observing that \begin{gather}\label{3.13} \begin{aligned} \frac{\partial ^2}{\partial t^2}\gamma _k &=-\mu \theta _{k+1}-\mu ^2t^2\gamma _{k+2}+\mu t^2(k+1)\theta _{k+3}\\ &\quad -k\gamma _{k+2}+\mu kt^2\theta_{k+3}+k(k+2)t^2\gamma _{k+4}, \end{aligned}\\ \label{3.14} \begin{aligned} \frac{\partial ^2}{\partial t^2}\theta _k &=\mu \gamma _{k+1}-\mu ^2t^2\theta _{k+2}-\mu t^2(k+1)\gamma _{k+3}\\ &\quad -k\theta _{k+2}-\mu kt^2\gamma_{k+3}+k(k+2)t^2\theta _{k+4} \end{aligned} \end{gather} we obtain \begin{equation}\label{3.15} \begin{aligned} v(x,t)&=\frac{1}{2\pi }\Big[ -\mu t\int_{\Xi }\theta _2(x,y,t)v_0(y)dy\\ &\quad -t\int_{\Xi }\gamma _3(x,y,t)v_0(y)dy+\int_{\Xi}\gamma _1(x,y,t)v_1(y)dy\Big] \end{aligned} \end{equation} and \begin{equation}\label{3.16} \begin{aligned} v_{t}(x,t) &=\frac{1}{2\pi }\Big[-\mu \int_{\Xi }\theta_2(x,y,t)v_0(y)dy -\mu ^2t^2\int_{\Xi }\gamma _3(x,y,t)v_0(y)dy\\ &\quad +3\mu t^2\int_{\Xi }\theta_{4}(x,y,t)v_0(y)dy -\int_{\Xi }\gamma_3(x,y,t)v_0(y)dy \\ &\quad +3t^2\int_{\Xi }\gamma _{5}(x,y,t)v_0(y)dy -\mu t\int_{\Xi }\theta_2(x,y,t)v_1(y)dy \\ &\quad -t\int_{\Xi }\gamma_3(x,y,t)v_1(y)dy\Big] \end{aligned} \end{equation} as long as $t\geq T_0$ and $x\in \overline{\Xi }$. From the discussion above we see that the bounded linear operator $S_{t}$ defined in \eqref{3.4} is compact, for each $t\geq T_0$. Let $\Sigma _0$ be the region of the complex plane given by $$ \Sigma _0=\big\{\zeta :\zeta =T_0+z,| \arg (z)| \leq \frac{\pi }{4}\big\}. $$ Let $B(H_0^1(\Xi )\times L^2(\Xi ),H^1(\Xi )\times L^2(\Xi ))$ be the space of all bounded linear operators from $H_0^1(\Xi )\times L^2(\Xi )$ into $H^1(\Xi )\times L^2(\Xi )$. We claim that the map $$ [T_0,\infty )\ni t\to S_{t}\in B(H_0^1(\Xi )\times L^2(\Xi ),H^1(\Xi )\times L^2(\Xi )) $$ extends to $\Sigma _0$ by setting \begin{equation}\label{3.17} S_{\zeta }(v_0,v_1)(x)=(v(x,\zeta ),v_{t}(x,\zeta )), \quad x\in \Xi \end{equation} where $v(x,\zeta )$ and $v_{t}(x,\zeta )$ are obtained from \eqref{3.15} and \eqref{3.16} by changing $t$ by $\zeta $. Moreover, $S_{\zeta }$ is compact, for every $\zeta \in \Sigma _0$. Observe that formulas \eqref{3.15} and \eqref{3.16} show that $v(x,t)$ and $v_{t}(x,t)$ are expressed as linear combinations of terms of the form \begin{equation}\label{3.18} t^{p}\int_{\Xi }\gamma _{\nu }(x,y,t)f(y)dy\quad \text{and}\quad t^{q}\int_{\Xi }\theta_k(x,y,t)f(y)dy \end{equation} where $p,q\in \{0,1,2\}$, $\nu\in \{1,3,5\}$, $k \in \{2,4\}$, $f\in \{v_0,v_1\}$ and $x\in \overline{\Xi }$, $t\geq T_0$.To prove our claim it suffices to extend the integrals in \eqref{3.18} to the parameter $t=\zeta \in \Sigma _0$ and certify that the resulting integrals have the regularity to make sure that $S_{\zeta }$ take values in $H^1(\Xi )\times L^2(\Xi )$. To begin, let $\kappa >0$ be such that $\frac{\operatorname{diam}(\Xi )}{T_0}<\kappa <1$. For each $\zeta =T_0+z\in \Sigma _0$ and all $x,y\in \overline{\Xi }$ we have $$ \operatorname{Re}(\zeta ^2-| x-y| ^2)=T_0^2 +2T_0\operatorname{Re}(z)-\operatorname{Im}(z)^2 +\operatorname{Re}(z)^2-| x-y| ^2\geq T_0^2-| x-y| ^2 $$ since $| \arg (z)| \leq \pi/4$. It follows \begin{equation}\label{3.19} \operatorname{Re}(\zeta ^2-| x-y| ^2)\geq T_0^2\Big(1-| \frac{x-y}{T_0}|^2\Big)>T_0^2(1-\kappa ^2)>0 \end{equation} and $| \arg (\zeta ^2-| x-y|^2)| \leq \pi/2$. By choosing that value of $(\zeta ^2-| x-y| ^2)^{1/2}$ which have positive real part, one obtains, for fixed $x,y\in \overline{\Xi }$, an analytic function of $\zeta $ whose values lie in the sector $$ \Sigma _{\kappa }=\big\{\zeta :\zeta =\sqrt{1-\kappa ^2}T_0+z,| \arg (z)| \leq \frac{\pi }{4}\big\} $$ for all $x,y\in \overline{\Xi }$. Now, by using \eqref{3.19} we obtain $$ | \gamma _k(x,y,\zeta)| \leq \frac{| \cos (\mu \sqrt{\zeta ^2-| y-x| ^2})| }{T_0^{k}(1-\kappa ^2)^{k/2}}, \quad | \theta _k(x,y,\zeta )| \leq \frac{| \sin (\mu \sqrt{\zeta ^2-| y-x| ^2})| }{ T_0^{k}(1-\kappa ^2)^{k/2}}, $$ for every $x,y\in \overline{\Xi }$, $\zeta \in \Sigma _0$ and $k=1,2,\dots$. Since $(\zeta ^2-| x-y| ^2)^{1/2}$ takes values in $\Sigma_{\kappa }$ for every $x,y\in \overline{\Xi }$, it follows that $\cos (\mu \sqrt{\zeta ^2-| y-x| ^2})$ and $\sin(\mu \sqrt{\zeta ^2-| y-x| ^2})$ are locally bounded as functions of $\zeta $, for every $x,y\in \overline{\Xi }$, i.e., for each open ball $V_{\zeta _0}=\{\zeta \in \mathbb{C}:| \zeta -\zeta _0| 0$ such that $$ | \cos (\mu \sqrt{ \zeta ^2-| y-x| ^2})| ,| \sin (\mu\sqrt{\zeta ^2-| y-x| ^2})| \leq C_{V_{\zeta _0}} $$ for every $\zeta \in V_{\zeta _0}$ and $x,y\in \overline{\Xi }$. Hence, \begin{equation}\label{3.20} | \gamma _k(x,y,\zeta )|, | \theta _k(x,y,\zeta )| \leq \frac{C_{V_{\zeta _0}}}{T_0^{k}(1-\kappa ^2)^{k/2}}, \quad k=1,2,\dots \end{equation} for every $x,y\in \overline{\Xi }$, and $\zeta \in $ $V_{\zeta_0}$. Hence, formulas \eqref{3.9}-\eqref{3.12} together with \eqref{3.20} shows that integrals \eqref{3.18} are well defined for $t=\zeta \in \Sigma _0$ as well as all differentiations in \eqref{3.18} may be carried out beneath the integral. Hence, the map defined in \eqref{3.17} is well defined in $\Sigma _0$ and it takes values in $B(H_0^1(\Xi )\times L^2(\Xi ),H^1(\Xi )\times L^2(\Xi)) $. Moreover, the maps $x\to v(x,\zeta )$, $x\to v_{t}(x,\zeta )$ are in $C^{\infty }(\overline{\Xi })$ for every $\zeta \in \Sigma _0$. It follows from this discussion that $S_{\zeta }$ is compact, for every $\zeta \in \Sigma _0$. Even though we obtain the analyticity of the maps $\zeta \to v(x,\zeta )$, $\zeta \to v_{t}(x,\zeta )$ for each $x\in \overline{\Xi }$ the regularity with respect to $\zeta $ is even better as the next theorem shows. \begin{theorem} \label{thm3.1} Let $\Xi \subset\mathbb{R}^2$ be a bounded domain, $T_0>\operatorname{diam}(\Xi )$ a real number and $\Sigma _0=\{\zeta :\zeta =T_0+z,| \arg (z)| \leq\frac{\pi }{4}\}$. The map $$ \Sigma _0\ni \zeta \to S_{\zeta }\in B(H_0^1(\Xi )\times L^2(\Xi ),H^1(\Xi )\times L^2(\Xi )) $$ defined by $$ S_{\zeta }(v_0,v_1)(x)=(v(x,\zeta ),v_{t}(x,\zeta )), \quad x\in \Xi $$ where $v(x,\zeta )$ and $v_{t}(x,\zeta )$ are given by \eqref{3.15} and \eqref{3.16} respectively ($t=\zeta $) is analytic inside the sector $\Sigma _0$. \end{theorem} \begin{proof} Let $P_0:H^1(\Xi )\times L^2(\Xi)\to H^1(\Xi )$, $P_1:H^1(\Xi )\times L^2(\Xi )\to L^2(\Xi )$ be the projections of $H^1(\Xi )\times L^2(\Xi )$ onto $H^1(\Xi )$ and $L^2(\Xi )$ respectively. It suffices to prove analyticity of the maps \begin{gather*} \Sigma _0\ni \zeta \to P_0S_{\zeta }\in B(H_0^1(\Xi )\times L^2(\Xi ),H^1(\Xi )), \\ \Sigma _0\ni \zeta \to P_1S_{\zeta }\in B(H_0^1(\Xi )\times L^2(\Xi ),L^2(\Xi )), \end{gather*} respectively. Let $\langle \cdot,\cdot\rangle _{H^1(\Xi)} $ and $\langle \cdot,\cdot\rangle_{L^2(\Xi )}$ denote the inner product in $H^1(\Xi )$ and $L^2(\Xi )$ respectively. As weak analyticity implies strong analyticity (see \cite[theorem 3.12, p. 152]{r8}) it is enough to show that the complex functions \begin{gather*} \Sigma _0\ni \zeta \to F_0(\zeta ) =:\langle P_0S_{\zeta }(v_0,v_1),w\rangle _{H^1(\Xi )}\in\mathbb{C}, \quad w\in H^1(\Xi ), \\ \Sigma _0\ni \zeta \to F_1(\zeta ) =:\langle P_1S_{\zeta }(v_0,v_1),w\rangle _{L^2(\Xi )}\in\mathbb{C},\quad w\in L^2(\Xi ), \end{gather*} for $(v_0,v_1)\in H_0^1(\Xi )\times L^2(\Xi )$ are analytic in the interior of $\Sigma _0$. It suffices to prove it only for $F_0$, to $F_1$ the procedure is analogous. Remember that $P_0S_{\zeta }(v_0,v_1)=\ v(\cdot,\zeta )$ where $v$ is given in \eqref{3.15}. Hence, for each $(v_0,v_1)\in H_0^1(\Xi )\times L^2(\Xi )$ and $w\in H^1(\Xi )$ we have $$ \langle P_0S_{\zeta }(v_0,v_1),w\rangle _{H^1(\Xi )}=\int_{\Xi }v(x,\zeta )\overline{w(x)}dx+\sum_{i=1}^2\int_{\Xi }\frac{ \partial v}{\partial x_i}(x,\zeta )\overline{\frac{\partial w}{\partial x_i}(x)}dx. $$ Now, considering that $v(x,\zeta )$ is a linear combination of terms of the form \eqref{3.18} (with $t=\zeta $) and taking into account the formulas \eqref{3.11} and \eqref{3.12} for the derivatives $\frac{\partial }{\partial x_i}\gamma _k$ and $\frac{\partial }{\partial x_i}\theta _k$ respectively, we see that $v(x,\zeta )$ and $\frac{\partial v}{\partial x_i}(x,\zeta )$ are linear combinations of terms of the form $$ \zeta ^{p}\int_{\Xi }\gamma _{\nu }(x,y,\zeta )(y_i-x_i)^{s}f(y)dy \,\,\, \text{and} \,\,\,\zeta ^{q}\int_{\Xi }\theta _k(x,y,\zeta )(y_i-x_i)^{s}f(y)dy $$ where $s\in \{0,1\}$, $f\in \{v_0,v_1\}$ and $p,q,k$ and $\nu $ are nonnegative integers. It follows that $\langle P_0S_{\zeta }(v_0,v_1),w\rangle $ is a linear combination of terms of the form \begin{gather}\label{3.21} \int_{\Xi }\zeta ^{p}\int_{\Xi }\gamma _{\nu }(x,y,\zeta )(y_i-x_i)^{s}f(y)\overline{\big(\frac{\partial }{\partial x_i}\big)^{l}w(x)}\,dy\,dx, \\ \label{3.22} \int_{\Xi }\zeta ^{q}\int_{\Xi }\theta _k(x,y,\zeta )(y_i-x_i)^{s}f(y)\overline{\big(\frac{\partial }{\partial x_i}\big)^{l}w(x)}\,dy\,dx, \end{gather} where $s,l\in \{0,1\}$, $f\in \{v_0,v_1\}$ and $p,q,k$ and $\nu $ are nonnegative integers. Now, by setting $\psi (x,y)=(y_i-x_i)^{s}f(y)\overline{(\frac{\partial }{\partial x_i})^{l}w(x)}$ in \eqref{3.21} and \eqref{3.22} we conclude that to prove that $F_0$ is analytic inside $\Sigma _0$ it suffices to prove analyticity of \begin{gather} \label{3.23} \Sigma _0\ni \zeta \to \zeta ^{p}\int_{\Xi \times \Xi }\psi (x,y) \gamma _{\nu }(x,y,\zeta )\,dx\,dy \\ \label{3.24} \Sigma _0\ni \zeta \to \zeta ^{p}\int_{\Xi \times \Xi }\psi (x,y) \theta _k(x,y,\zeta )\,dx\,dy \end{gather} inside the sector $\Sigma _0$. Let $\zeta _0$ be an arbitrary point in the interior of $\Sigma _0$ and $V_{\zeta _0}$ some open ball centered in $\zeta _0$ and contained in the interior of $\Sigma _0$. Now, observing that \begin{gather*} \frac{\partial \gamma _{\nu }}{\partial\zeta }(x,y,\zeta ) = -\mu \zeta \theta_{\nu +1}(x,y,\zeta )-\nu \zeta \gamma _{\nu +2}(x,y,\zeta ),\quad \nu =1,2,\dots \\ \frac{\partial \theta _k}{\partial \zeta }(x,y,\zeta ) = \mu \zeta \gamma _{k+1}(x,y,\zeta )-k\zeta \theta _{k+2}(x,y,\zeta ),\quad k=1,2,\dots, \end{gather*} for every $x,y\in \Xi $ and $\zeta \in \Sigma _0$, and using the estimate \eqref{3.20} we conclude that $| \frac{\partial \gamma _{\nu}}{\partial \zeta }(x,y,\zeta )| $ and $| \frac{\partial \theta _k}{\partial \zeta }(x,y,\zeta )| $ are bounded in $\overline{\Xi }\times \overline{\Xi }\times V_{\zeta _0}$. Since $\psi $ is integrable in the bounded set $\Xi \times \Xi $ we have \begin{gather*} \int_{\Xi \times \Xi }| \psi (x,y)\gamma _{\nu }(x,y,\zeta)| \,dx\,dy \leq \mathrm{Const.} \int_{\Xi \times \Xi }|\psi (x,y)| \,dx\,dy,\\ \int_{\Xi \times \Xi }| \psi (x,y)\theta _k(x,y,\zeta )| \,dx\,dy\leq\mathrm{Const.}\int_{\Xi \times \Xi }| \psi (x,y)| \,dx\,dy, \end{gather*} for every $x,y\in \Xi $, $\zeta \in V_{\zeta _0}$. Hence, we may differentiate \eqref{3.23} and \eqref{3.24} with respect to $\zeta \in V_{\zeta _0}$ under the integrals. From this, it follows that $F_0$ is analytic inside $\Sigma _0$. \end{proof} Now, for each $i=1,\dots,m$ and $t>0$ we define the bounded linear operator $S_{t}^{i}:H_0^1(\Xi )\times L^2(\Xi )\to H^1(\Xi )\times L^2(\Xi )$ by setting \begin{equation}\label{3.25} S_{t}^{i}(w_0^{i},w_1^{i})(x)=(w^{i}(x,t),w_{t}^{i}(x,t)), \quad x\in \Xi \end{equation} where $w^{i}$ is the solution to the Cauchy problem \begin{equation}\label{3.26} \begin{gathered} w_{tt}^{i}-\Delta w^{i}+\mu _i^2w^{i}=0 \quad \text{in } \mathbb{R}^{2+1}, \\ w^{i}(\cdot,0)=w_0^{i} \quad \text{in } \mathbb{R}^2, \\ w_{t}^{i}(\cdot,0)=w_1^{i} \quad \text{in } \mathbb{R}^2, \end{gathered} \end{equation} where $\mu _i^2=\lambda _i$ and initial data $(w_0^{i},w_1^{i})\in H_0^1(\Xi )\times L^2(\Xi )$ extended by zero outside $\Xi $. Now fix $T_0>\operatorname{diam}(\Xi )$. Then applying Theorem \ref{thm3.1} to the family of operators $\{S_{t}^{i}:t>T_0\}$ shows that it admits an extension to the sector $\Sigma _0$, analytic in its interior. As in Theorem \ref{thm3.1}, for each $\zeta \in \Sigma _0$, $S_{\zeta }^{i}$ is given by \begin{equation}\label{3.27} S_{\zeta}^{i}(w_0^{i},w_1^{i})(x)=(w^{i}(x,\zeta ),w_{t}^{i}(x,\zeta )),\quad x\in \Xi \end{equation} where $w^{i}(x,\zeta )$ and $w_{t}^{i}(x,\zeta )$ are given by \eqref{3.15} and \eqref{3.16} respectively after appropriate adjustments. Now we define \begin{equation}\label{3.28} \mathbf{S}_{\zeta }^{\ast }:\mathcal{H} _0^1(\Xi )\times \mathcal{L}^2(\Xi )\to \mathcal{H} ^1(\Xi )\times \mathcal{L}^2(\Xi ) \end{equation} by setting \begin{equation}\label{3.29} \begin{aligned} & \mathbf{S}_{\zeta }^{\ast}((w_0^1,\dots,w_0^{m}),(w_1^1,\dots,w_1^{m}))\\ &=((P_0S_{\zeta }^1(w_0^1,w_1^1),\dots,P_0S_{\zeta }^{m}(w_0^{m},w_1^{m})), (P_1S_{\zeta}^1(w_0^1,w_1^1),\dots,\\ &\quad P_1S_{\zeta }^{m}(w_0^{m},w_1^{m}))) \\ &=((w^1(\cdot,\zeta ),\dots,w^{m}(\cdot,\zeta )),(w_{t}^1(\cdot,\zeta ),\dots,w_{t}^{m}(\cdot,\zeta ))) \end{aligned} \end{equation} where $P_0$ and $P_1$ are the projections introduced in the proof of Theorem \ref{thm3.1}. It follows from that theorem that the map \begin{equation}\label{3.30} \Sigma _0\ni \zeta \to \mathbf{ S}_{\zeta }^{\ast }\in B(\mathcal{H}_0^1(\Xi )\times \mathcal{L}^2(\Xi ),\mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )) \end{equation} is analytic inside $\Sigma _0$. Now consider the Cauchy problem \begin{equation}\label{3.31} \begin{gathered} U_{tt}-\Delta U+AU=0 \quad \text{in } \mathbb{R}^{2+1}, \\ U(\cdot,0)=U_0,\quad U_{t}(\cdot,0)=U_1 \quad \text{in } \mathbb{R}^2, \end{gathered} \end{equation} with initial state $(U_0,U_1)\in \mathcal{H}_0^1(\Xi)\times \mathcal{L}^2(\Xi )$. For $t>0$ we define the bounded linear operator $\mathbf{S}_{t}:\mathcal{H}_0^1(\Xi )\times \mathcal{L}^2(\Xi)\to \mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )$ by setting \begin{equation}\label{3.32} \mathbf{S}_{t}(U_0,U_1)(x)=(U(x,t),U_{t}(x,t)), \quad x\in \Xi \end{equation} where $U$ is the solution to the Cauchy problem \eqref{3.31}. \begin{theorem} \label{thm3.2} The family of operators $\{\textbf{S}_{t}; t>T_0 \}$ defined by \eqref{3.32} extends to the complex parameter $\xi \in \Sigma_0$ as a family of compact operators that is analytic in the interior of the sector $\Sigma_0$. \end{theorem} \begin{proof} As in section 2, we denote $ U_0=(u_0^1,u_0^2,\dots,u_0^{m})^T $, $U_{1}=(u_1^1,u_1^2,\dots,u_1^{m})^T$, $W=(w^1,\dots,w^{m})^T$ where $ W=BU$. Note that $w^{i}$ satisfy \eqref{3.26} with initial conditions $w_0^{i}=\alpha _1^{i}u_0^1+\alpha _2^{i}u_0^2+\dots+\alpha_{m}^{i}u_0^{m}$ and $w_1^{i}=\alpha _1^{i}u_1^1+\alpha_2^{i}u_1^2+\dots+\alpha _{m}^{i}u_1^{m}$, $i=1,\dots,m$. From \eqref{3.29} we have \begin{equation}\label{3.33} \begin{aligned} &\mathbf{S}_{t}^{\ast}((w_0^1,\dots,w_0^{m}),(w_1^1,\dots,w_1^{m}))\\ &=((w^1(\cdot,t),\dots,w^{m}(\cdot,t)),(w_{t}^1(\cdot,t),\dots, w_{t}^{m}(\cdot,t)))\\ &=(W(\cdot,t),W_{t}(\cdot,t))=(BU(\cdot,t),BU_{t}(\cdot,t)). \end{aligned} \end{equation} Let us introduce the projections $\mathbf{P}_0:\mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )\to \mathcal{H}^1(\Xi )$ and $\mathbf{P}_1:\mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )\to \mathcal{L}^2(\Xi )$ of $\mathcal{H}^1(\Xi )\times \mathcal{L}^2(\Xi )$ onto $\mathcal{H}^1(\Xi )$ and $\mathcal{L}^2(\Xi )$ respectively. From \eqref{3.33} it follows \begin{gather*} BU(\cdot,t)=\mathbf{P}_0\mathbf{S}_{t}^{\ast }(~(w_0^1,\dots,w_0^{m}),(w_1^1,\dots,w_1^{m})), \\ BU_{t}(\cdot,t)=\mathbf{P}_1\mathbf{S}_{t}^{\ast }(~(w_0^1,\dots,w_0^{m}),(w_1^1,\dots,w_1^{m})). \end{gather*} Now, observing that $ (w_0^1,\dots,w_0^{m})^T=BU_0$ and $ (w_1^1,\dots,w_1^{m})^T=BU_1$ we obtain $$U(\cdot,t)=[B^{-1}\mathbf{P}_0\mathbf{S}_{t}^{\ast }B]U_0\,\,\,\text{and} \,\,\, U_{t}(\cdot,t)=[B^{-1}\mathbf{P}_1\mathbf{S}_{t}^{\ast }B]U_1.$$ Therefore \begin{equation}\label{3.34} \mathbf{S}_{t}(U_0,U_1)=([B^{-1} \mathbf{P}_0\mathbf{S}_{t}^{\ast }B]U_0,[B^{-1}\mathbf{P}_1\mathbf{S} _{t}^{\ast }B]U_1) \end{equation} for all $(U_0,U_1)\in \mathcal{H}_0^1(\Xi )\times \mathcal{L}^2(\Xi )$ and $t\geq T_0$. From the relationship \eqref{3.34} among the operators $\mathbf{S}_{t}$ and $\mathbf{S}_{t}^{\ast }$ and the analyticity of the map \eqref{3.30} the result follows. \end{proof} \section{Control near optimal time} Now we go over the proof of Theorem \ref{thm1.1}. Let $\Omega \subset\mathbb{R}^2$ be a bounded simply connected domain with piecewise smooth boundary $\partial \Omega $. Given any $T_{\ast }>\operatorname{diam}(\Omega )$ we choose $\delta >0$ and $T_0>0$ such that $$ \operatorname{diam}(\Omega )<\operatorname{diam}(\Omega _{\delta })T_0\}$ to the sector $\Sigma _0$, analytically in its interior. Note that operators $\mathbf{S}_{T}$ and $\widehat{\mathbf{S}}_{T}$ obey the following relations $$ \mathbf{P}_0\widehat{\mathbf{S}}_{T}(V_0,V_1)=\mathbf{P}_0\mathbf{S} _{T}(V_0,-V_1),~\ \ \mathbf{P}_1\widehat{\mathbf{S}}_{T}(V_0,V_1)= \mathbf{P}_1\mathbf{S}_{T}(-V_0,V_1) $$ for $(V_0,V_1)\in \mathcal{H}_0^1(\Omega _{\delta })\times \mathcal{L}^2(\Omega _{\delta })$. Since, by Theorem \ref{thm3.2} $\{\mathbf{S}_{T}:T>T_0\}$ admits such an extension, the same applies to $\{\widehat{\mathbf{S}}_{T}:T>T_0\}$ and hence to $\{\mathbf{K}_{T}:T>T_0\}$. Now we apply to the family of compact operators $\{\mathbf{K}_{\zeta }:\zeta \in \Sigma _0\}$ Atkinson's result as in \cite[Theorem 1.9, page 370]{r8}. This theorem states that either $1$ is eigenvalue of $\mathbf{K}_{\zeta }$ for every $\zeta \in \Sigma _0$ or $Id-\mathbf{K}_{\zeta }$ is invertible for all but at most a finite number of values of $\zeta $ in each compact subset of $\Sigma _0$. From the proof of Theorem \ref{thm2.2} we know that $\mathbf{K}_{\zeta }$ is a contraction if $\zeta \in \mathbb{R}\cap \Sigma _0$ is sufficiently large. This excludes the first alternative. It follows that for a convenient $\varepsilon >0$, the compact set $[T_0+\varepsilon ,T_{\ast }-\varepsilon ]\subset \Sigma _0$ includes some $T$ for which $Id-\mathbf{K}_{T}$ is invertible. As observed before, this suffices to conclude the proof of Theorem \ref{thm1.1}. $\hfill$ $\square$ We end up this section showing how to apply the procedure above to treat the case in which part of membranes boundary is hold fixed. Let \begin{gather}\label{4.1} \Omega=\{(r\cos\theta ,r\sin\theta );02r_2$ and initial state $U_0\in \mathcal{H}^1(\Omega )$, $U_1\in \mathcal{L}^2(\Omega )$, $U_0=0$ on $\Gamma_0$, there exist a control $f\in \mathcal{L}^2(\partial \Omega /\Gamma_0\times ]0,T_{\ast }[)$ so that the solution $U\in \mathcal{H}^1(\Omega \times ]0,T_{\ast }[)$ of the system \begin{equation}\label{4.3} \begin{gathered} U_{tt}-\Delta U+AU=0 \quad \text{in } \Omega \times ]0,T[, \\ U(\cdot,0)=U_0 \quad \text{in } \Omega, \\ U_{t}(\cdot,0)=U_1 \quad \text{in } \Omega, \\ U=0 \quad \text{on } \Gamma_0\times ]0,T[, \\ \frac{\partial U}{\partial \eta }=f \quad \text{on } \partial\Omega/ \Gamma_0 \times ]0,T[, \end{gathered} \end{equation} satisfies the final condition $U(\cdot,T_{\ast })=U_{t}(\cdot,T_{\ast})=0 $ on $\Omega $. The proof of this result follows the same lines of the proof of Theorem \ref{thm1.1} with additional care on the extension operator $E$ used in the beginning. Consider the angular sector $\Omega _{\infty }=\{(r\cos\theta ,r\sin\theta);r>0,\;0<\theta <\pi /n\}$. Fix $T_0$ and $\delta >0$ such that $\deltar_2+\delta /2$, and more: $V_0=0$ on the edges $\theta =0$ and $\theta =\pi /n$ of $\Omega _{\infty }$. Next we extend each entry of $(V_0,V_1)$ to the plane as odd functions with respect to each entire line determined by the angles $\theta =i(\pi /n)$, $i=1,\dots,n$. We denote $(\widetilde{V}_0,\widetilde{V}_1)$ the resulting extensions and define $E(V_0,V_1):=(\widetilde{V}_0,\widetilde{V}_1)$. Observe that $E(V_0,V_1)$ has compact support in the set defined by $\Omega _{\delta }=:\{x\in\mathbb{R}^2:r_1-\delta <|x|0$. By using a localizing function $\varphi \in C_0^{\infty }(\mathbb{R}^2)$ of radial type satisfying $\varphi =1$ on $\Omega _{\delta /2}$ and vanishing outside $\Omega _{\delta }$ we see that the solution to the backward problem \eqref{2.11} also vanish on the lines $\theta =i(\pi /n)$, $i=1,\dots,n$. From this we conclude that if we use $\Omega _{\delta }=:\{x\in \mathbb{R}^2:r_1-\delta <|x|