\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 101, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/101\hfil Exact controllability for a wave equation] {Exact controllability for a wave equation with mixed boundary conditions in a non-cylindrical domain} \author[L. Cui, H. Gao \hfil EJDE-2014/101\hfilneg] {Lizhi Cui, Hang Gao} % in alphabetical order \address{Lizhi Cui \newline College of Applied Mathematics, Jilin University of Finance and Economics, Changchun 130117, China.\newline School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China} \email{cuilz924@126.com} \address{Hang Gao \newline School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China} \email{hangg@nenu.edu.cn} \thanks{Submitted October 15, 2013. Published April 11, 2014.} \subjclass[2000]{58J45, 35L05} \keywords{Exact controllability; wave equation; mixed boundary conditions; \hfill\break\indent non-cylindrical domain} \begin{abstract} In this article we study the exact controllability of a one-dimen\-sional wave equation with mixed boundary conditions in a non-cylindrical domain. The fixed endpoint has a Dirichlet-type boundary condition, while the moving end has a Neumann-type condition. When the speed of the moving endpoint is less than the characteristic speed, the exact controllability of this equation is established by Hilbert Uniqueness Method. Moreover, we shall give the explicit dependence of the controllability time on the speed of the moving endpoint. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction and statement of main results} Given $T>0$. For any $0T^{*}_k$, the equation \eqref{b1} is exactly controllable at the time $T$; i.e., for any initial value $(u^0,u^1)\in L^2(0,1)\times [V(0,1)]'$ and target $(u_d^0,u_d^1)\in L^2(0,\alpha_k(T))\times[V(0,\alpha_k(T))]'$, there exists a control $v\in [H^1(0,T)]'$ such that the corresponding solution $u$ of $\eqref {b1}$ satisfies $$\label{b2} u(T)=u_d^0\quad\text{and}\quad u_{t}(T)=u_d^1.$$ \end{theorem} \begin{remark} \label{rmk1.1}\rm It is easy to check that $$T^{*}_0:=\lim_{k\to0}T^{*}_k= \lim_{k\to0}\frac{{e}^\frac{2k(1+k)}{1-k}-1}{k}=2.$$ It is well known that the wave equation $\eqref{b1}$ in the cylindrical domain is exactly controllable at any time $T>T^*_0$. As we know, $T^*_0$ is sharp. However, we do not know whether the controllability time $T^*_k$ is sharp. \end{remark} To establish the exact controllability of \eqref{b1}, we first transform \eqref{b1} into an equivalent wave equation with variable coefficients in a cylindrical domain. To this aim, for any $(y, t)\in \widehat{Q}_T^k$, set $y=\alpha_k(t)x$ and $u(y,t)=u(\alpha_k(t)x,t)=w(x,t)$. Then it is easy to check that \eqref{b1} is transformed into the wave equation $$\label{b3} \begin{gathered} w_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}w_x]_x +[\frac{\gamma_k(x)}{\alpha_k(t)}]w_{tx}=0 \quad \text{in } Q,\\ w(0,t)=0,\quad w_x(1,t)=\overline{v}(t)\quad \text{on } (0,T),\\ w(x,0)=w^0(x), \quad w_{t}(x,0)=w^1(x)\quad \text{in } (0,1), \end{gathered}$$ where $$\label{b4} \begin{gathered} Q=(0, 1)\times(0, T),\quad \overline{v}(t)=\alpha_k(t)v(t),\quad \beta_k(x,t)=\frac{1-k^2x^2}{\alpha_k(t)},\\ \gamma_k(x)=-2kx, w^0=u^0 \quad \text{and } w^1=u^1+kxu^0_x. \end{gathered}$$ By a method similar to the one used in \cite{C1999}, it is easy to check that the equation \eqref{b3} has a unique solution $w$ by transposition $$w\in C([0,T];L^2(0,1))\cap C^1([0,T];[V(0,1)]').$$ Moreover, the exact controllability of \eqref{b1} (Theorem \ref{thm1}) is reduced to the following exact controllability result for \eqref{b3}. \begin{theorem} \label{thm2} Suppose that $T>T^{*}_k$. Then for any initial value $(w^0, w^1)\in L^2(0,1)\times [V(0,1)]'$ and target $(w_d^0,w_d^1)\in L^2(0,1)\times [V(0,1)]'$, there exists a control $\overline{v} \in [H^1(0,T)]'$ such that the corresponding solution $w$ of \eqref {b3} satisfies $$w(T)=w_d^0\quad\text{and}\quad w_{t}(T)=w_d^1.$$ \end{theorem} To prove Theorem \ref{thm2}, we adopt Hilbert Uniqueness Method. The key is to define a weighted energy function for a wave equation with variable coefficients in cylindrical domains. The rest of this paper is organized as follows. In Section 2, we derive an explicit energy equality for a wave equation with variable coefficients in cylindrical domains and further deduce two key inequalities for this equation. Section 3 is devoted to a proof of Theorem \ref{thm2}. \section{Two inequalities for the wave equation with variable coefficients} First we introduce some notation. Denote by $|\cdot|$ and $\|\cdot\|$ the norms of the spaces $L^2(0,1)$ and $V(0,1)$, respectively. Also, we use $L^2$, $V$ and $V'$ to represent the spaces $L^2(0,1), V(0,1)$ and $[V(0,1)]'$, respectively. Denote by $\langle\cdot,\cdot\rangle$ the duality product between the linear space $F$ and its dual space $F'$. Consider the wave equation with variable coefficients $$\label{2.10} \begin{gathered} \alpha_k(t) z_{tt}-[\beta_k(x,t)z_x]_x+\gamma_k(x)z_{tx}=0 \quad\text{in } Q,\\ z(0,t)=0,\quad \beta_k(1,t)z_x(1,t)-\gamma_k(1)z_{t}(1,t)=0 \quad\text{on } (0,T),\\ z(x,0)=z^0(x),\quad z_{t}(x,0)=z^1(x)\quad \text{in } (0,1), \end{gathered}$$ where $(z^0, z^1)\in V\times L^2$ is any given initial value, and $\alpha_k$, $\beta_k$ and $\gamma_k$ are the functions given in \eqref{b4}. By a similar method in \cite{C1999} and \cite{QR}, it is easy to check that \eqref{2.10} has a unique solution $z$ by transposition \begin{equation*} z\in C([0,T];V)\cap C^1([0,T];L^2). \end{equation*} Define the following energy function for \eqref{2.10}, \begin{equation*} E(t)=\frac{1}{2}\int_0^1[\alpha_k(t)|z_{t}(x,t)|^2 +\beta_k(x,t)|z_x(x,t)|^2]dx \quad\text{for } t\in [0, T], \end{equation*} where $z$ is the solution of \eqref{2.10}. It follows that \begin{equation*} E_0\triangleq E(0)=\frac{1}{2}\int_0^1[|z^1(x)|^2+\beta_k(x,0)|z^0_x(x)|^2]dx. \end{equation*} To prove Theorem \ref{thm2}, we need the following two key inequalities. \begin{theorem}\label{thm3} For any $T>0$, there exists a positive constant $C_1$ depending only only $T$, such that solutions $z$ of \eqref{2.10} satisfy $$\label{3.1} \int_0^{T} |z_{t}(1,t)|^2dt \leq C_1(\|z^0\|^2+|z^1|^2)\quad\text{for any } (z^0, z^1)\in V\times L^2.$$ \end{theorem} \begin{theorem}\label{thm4} Suppose that $T>T^{*}_k$. Then there exists a positive constant $C_2$ depending only on $T$, such that solutions $z$ of \eqref{2.10} satisfy $$\label{3.2} \int_0^{T} |z_{t}(1,t)|^2dt \geq C_2(\|z^0\|^2+|z^1|^2)\quad\text{for any } (z^0, z^1)\in V\times L^2.$$ \end{theorem} First, we prove two lemmas, which will be used in the proofs of these inequalities. The first lemma is related to an equivalent expression of the energy $E(t)$. \begin{lemma} \label{lemE} Suppose that $z$ is any solution of \eqref{2.10}. Then we have $$\label{E1} E(t)=\frac{1}{\alpha_k(t)}E_0-\frac{k}{\alpha_k(t)} \int_0^t\alpha_k(s)|z_{t}(1,s)|^2ds, \quad 0\leq t \leq T.$$ \end{lemma} \begin{proof} Multiplying both sides of the first equation of \eqref{2.10} by $z_{t}$ and integrating on $(0,1)\times(0,t)$, we obtain \begin{align*} 0&=\int_0^t\int_0^1 \big\{\alpha_k(s)z_{tt}(x,s)z_{t}(x,s) -[\beta_k(x,s)z_x(x,s)]_{x}z_{t}(x,s) \\ &\quad +\gamma_k(x)z_{tx}(x,s)z_{t}(x,s)\big\}dx\,ds\\ &\triangleq J_1+J_2+J_3. \end{align*} Next, we calculate the above three integrals. It is easy to check that $$\label{**} \begin{split} J_1&=\int_0^t\int_0^1 \frac{1}{2}\alpha_k(s)[|z_t(x,s)|^2]_t\,dx\,ds\\ &=\frac{1}{2}\int_0^1 \alpha_k(s)|z_t(x,s)|^2dx\big|_0^t -\frac{k}{2}\int_0^t\int_0^1 |z_t(x,s)|^2 \,dx\,ds. \end{split}$$ Further, by the second equation of \eqref{2.10}, it holds that \begin{align*} J_2&= -\int_0^t\beta_k(x,s)z_{x}(x,s)z_{t}(x,s)ds\big|_0^1 +\int_0^t\int_0^1\beta_k(x,s)z_{x}(x,s)z_{xt}(x,s)\,dx\,ds \\ &=- \int_0^t \beta_k(x,s)z_{x}(x,s)z_{t}(x,s)ds\big|_0^1 +\frac{1}{2}\int_0^1 \beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\ &\quad -\frac{1}{2}\int_0^t\int_0^1\beta_{k,t}(x,s)|z_{x}(x,s)|^2\,dx\,ds \\ &=-\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds +\frac{1}{2}\int_0^1 \beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\ &\quad -\frac{1}{2}\int_0^t\int_0^1\beta_{k,t}(x,s)|z_{x}(x,s)|^2\,dx\,ds. \end{align*} By \eqref{b4}, it is obvious that $$\beta_{k,t}(x,t)=-\frac{k(1-k^2x^2)}{(1+kt)^2}=-\frac{k}{(1+kt)}\beta_k(x,t).$$ This implies that $$\label{A263} \begin{split} J_2 &=-\int_0^t\int_0^1[\beta_k(x,s)z_x(x,s)]_{x}z_{t}(x,s)\,dx\,ds\\ &=-\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds +\frac{1}{2}\int_0^1 \beta_k(x,s)|z_{x}(x,s)|^2dx\big|_0^t\\ &\quad +\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1\beta_k(x,s)|z_{x}(x,s)|^2\,dx\,ds. \end{split}$$ Further, by the definition of $\gamma_k$, we find that \begin{align*} J_3 &= \frac{1}{2}\int_0^t \gamma_k(x)|z_{t}(x,s)|^2ds\big|_0^1 -\frac{1}{2} \int_0^t\int_0^1\gamma_{k,x}(x)|z_{t}(x,s)|^2\,dx\,ds\\ &=\frac{1}{2}\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds -\frac{1}{2} \int_0^t\int_0^1\gamma_{k,x}(x)|z_{t}(x,s)|^2\,dx\,ds. \end{align*} Since $\gamma_{k,x}(x)=-2k$, it follows that $$\label{A265} J_3=\frac{1}{2}\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds+ k\int_0^t\int_0^1|z_{t}(x,s)|^2\,dx\,ds.$$ By \eqref{**}-\eqref{A265} and the definition of $E(t)$, we see that \begin{align*} E(t)&=E_0+\frac{1}{2}\int_0^t \gamma_k(1)|z_{t}(1,s)|^2ds -\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1 \beta_k(x,s)|z_x(x,s)|^2 \,dx\,ds\\ &\quad -\frac{k}{2}\int_0^t\int_0^1 |z_{t}(x,s)|^2\,dx\,ds\\ &=E_0-\int_0^tk|z_{t}(1,s)|^2ds -\frac{1}{2}\int_0^t\frac{k}{(1+ks)}\int_0^1 \beta_k(x,s)|z_x(x,s)|^2 \,dx\,ds \\ &\quad -\frac{1}{2}\int_0^t\frac{k}{(1+ks)} \int_0^1 \alpha_k(x,s)|z_{t}(x,s)|^2\,dx\,ds\\ &=E_0-\int_0^t k|z_{t}(1,s)|^2ds-\int_0^t\frac{k}{(1+ks)}E(s)ds, \end{align*} which implies that $E_{t}(t)=-\frac{k}{1+kt}E(t)-k|z_{t}(1,t)|^2,\quad 0\leq t \leq T.$ It follows that \begin{equation*} [(1+kt)E(t)]_{t}=-k(1+kt)|z_{t}(1,t)|^2,\quad 0\leq t \leq T, \end{equation*} which completes the proof of Lemma \ref{lemE}. \end{proof} \begin{remark}\label{rmk2.1} \rm By \eqref{E1}, it is easy to check that $E(t)<\frac{1}{\alpha_k(t)}E_0T^{*}_k$, then $\frac{1}{k(1+k)}\ln(1+kT)-\frac{2}{1-k}>0$. This, together with \eqref{A29} indicates the desired estimate in Theorem \ref{thm4}. \end{proof} \section{Proof of Theorem \ref{thm2}} In this section we use the Hilbert Uniqueness Method. For Theorem \ref{thm2}, it suffices to show that for any given initial value $(w^0, w^1)\in L^2\times V'$ and target $(w_d^0,w_d^1)\in L^2\times V'$, one can find a control $\overline{v}=\overline{v}(t)\in [H^1(0,T)]'$ such that the corresponding solution $w$ of \eqref{b3} satisfies $$\label{2.8} w(T)=w^0_d\quad\text{and}\quad w_{t}(T)=w^1_d.$$ We divide the whole proof into three parts. \smallskip \noindent \textbf{Step 1.} First, we define a linear operator $\Lambda$ from $V\times L^2$ to $V'\times L^2$. Consider the wave equation $$\label{2.9} \begin{gathered} \xi_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}\xi_x]_x +[\frac{\gamma_k(x)}{\alpha_k(t)}]\xi_{tx}=0 \quad \text{in } Q,\\ \xi(0,t)=0,\quad \xi_x(1,t)=0 \quad \text{on } (0,T),\\ \xi(x,T)=w^0_d(x),\quad \xi_{t}(x,T)=w^1_d(x)\quad \text{in } (0,1). \end{gathered}$$ It is easy to check that \eqref{2.9} has a unique solution $$\xi\in C([0,T];L^2)\cap C^1([0,T];V')$$ and set $$(\xi^0,\xi^1)\triangleq(\xi(x,0),\xi_{t}(x,0))\in L^2\times V'.$$ Thus $$\label{2.0} (w^0-\xi^0,w^1-\xi^1)\in L^2\times V'.$$ On the other hand, for any $(z^0,z^1)\in V\times L^2$, we denote by $z$ the corresponding solution of \eqref{2.10}. Consider the wave equation $$\label{2.11} \begin{gathered} \eta_{tt}-[\frac{\beta_k(x,t)}{\alpha_k(t)}\eta_x]_x +[\frac{\gamma_k(x)}{\alpha_k(t)}]\eta_{tx}=0 \quad \text{in } Q,\\ \eta(0,t)=0,\quad \eta_x(1,t)=\frac{1}{\beta_k(1,t)} G_{z_{t}(1,t)}\quad \text{on } (0,T),\\ \eta(x,T)=\eta_{t}(x,T)=0\quad \text{in } (0,1). \end{gathered}$$ Notice that $G_{z_{t}(1,t)}\in (H^1(0,T))'$ is defined as $$\label{D} \begin{split} \langle G_{z_{t}(1,t)}, \phi\rangle_{(H^1(0,T))',H^1(0,T)} =-\int_0^{T}z_{t}(1,t)\phi_{t}(t)dt \quad \text{for any }\phi\in H^1(0,T). \end{split}$$ Now, we define the operator \begin{gather*} \Lambda: V\times L^2\to V'\times L^2, \\ (z^0,z^1)\to \big(\eta_{t}(x,0)+\gamma_k(x)\eta_x(x,0)-k\eta(x,0),-\eta(x,0)\big). \end{gather*} Therefore, $$\label{2.13} \begin{split} &\langle\Lambda(z^0,z^1),(z^0,z^1)\rangle\\ &=\int_0^1[\eta_{t}(x,0)z^0-k\eta(x,0)z^0 +\gamma_k(x)\eta_x(x,0)z^0-\eta(x,0)z^1]dx. \end{split}$$ For simplicity, we set $F=V\times L^2,F'=V'\times L^2$. \smallskip \noindent\textbf{Step 2.} We prove that $\Lambda$ is an isomorphism. To this aim, multiplying both sides of the first equation of \eqref{2.11} by $\alpha_k(t)z$ $(0\leq t \leq T)$ and integrating on $Q$, we obtain that $$\label{2.14} \begin{split} &-\int_0^T\beta_k(1,t)\eta_x(1,t) z(1,t)dt\\&= \int_0^1[\eta'(x,0)z^0-k\eta(x,0)z^0 -\eta(x,0)z^1+\gamma_k(x)\eta_x(x,0)z^0]dx. \end{split}$$ From \eqref{2.11}-\eqref{2.13} and \eqref{2.14}, we conclude that $$\label{2.16} \int_0^T|z_{t}(1,t)|^2dt =\langle \Lambda(z^0,z^1),(z^0,z^1)\rangle.$$ By Theorem \ref{thm3}, it holds that $\Lambda$ is a linear bounded operator. It remains to show that $\Lambda$ is onto. To this end, define the bilinear functional on $F \times F$: $A((\hat{z}^0,\hat{z}^1), (z^0,z^1)) =\langle \Lambda(\hat{z}^0,\hat{z}^1), (z^0,z^1)\rangle,$ where $(\hat{z}^0,\hat{z}^1)$, $(z^0,z^1)\in F\times F$. It is clear that $A$ is bounded. From \eqref{2.16} and Theorem \ref{thm4}, it follows that $A$ is coercive. Hence, applying Lax-Milgram Theorem, we derive that $\Lambda$ is onto. This completes the proof of Step 2. \smallskip \noindent \textbf{Step 3.} We prove that the exact controllability of \eqref{b3} is equivalent that $\Lambda$ is an isomorphism. Indeed, for any given $(w^0,w^1), (w_d^0,w_d^1)\in L^2\times V'$, we choose $$\overline{v}(\cdot)=\frac{1}{\beta_k(1,\cdot)}G_{z_{t}(1,\cdot)}\in (H^1(0,T))',$$ where $z$ is the solution of \eqref{2.10} associated to $(z^0, z^1)=\Lambda^{-1}((w^1-\xi^1)+\gamma_k(x)(w^0_x-\xi^0_x) -k(w^0-\xi^0),-(w^0-\xi^0))$ and $w$ is the solution of \eqref{b3}. From the definition of $\Lambda$, we conclude that $\Lambda(z^0,z^1)=(\eta'(x,0)+\gamma_k(x)\eta_x(x,0)-k\eta(x,0),-\eta(x,0))$, where $\eta$ is the solution of \eqref{2.11}. Then, $\eta$ satisfies $(\eta(x,0),\eta'(x,0))=(w^0-\xi^0,w^1-\xi^1)$. This implies that $w=\xi+\eta$ satisfies both \eqref{b3} and \eqref{2.8}. 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