\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 100, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2003 Texas State University-San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/100\hfil Global attractor for an equation] {Global attractor for an equation modelling a thermostat} \author[R. de C. D. S. Broche, L. A. F. de Oliveira, \& A. L. Pereira\hfil EJDE--2003/100\hfilneg] {Rita de C\'assia D. S. Broche, L. Augusto F. de Oliveira,\\ \& Ant\^onio L. Pereira} % in alphabetical order \address{Departamento de Matem\'atica, Instituto de Matem\'atica e Estat\'\i tica, Universidade de S\~ao Paulo, Brazil} \email[Rita de C\'assia D. S. Broche]{ritac@ime.usp.br} \email[L. Augusto F. de Oliveira]{luizaug@ime.usp.br} \email[Ant\^onio L. Pereira]{alpereir@ime.usp.br} \date{} \thanks{Submitted August 1, 2003. Published September 26, 2003.} \thanks{R. B. was partially supported by CAPES - Brazil.\hfill\break\indent L. de O. was partially supported by grant MECD 023/01 from CAPES Brazil} \subjclass[2000]{35B40, 35K60} \keywords{Nonlocal boundary conditions, parabolic equations, attractors} \begin{abstract} In this work we show that the system considered by Guidotti and Merino in \cite{GM} as a model for a thermostat has a global attractor and, assuming that the parameter is small enough, the origin is globally asymptotically stable. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem} [section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{rem}[theorem]{Remark} \section{Introduction} The purpose of this note is to answer a question proposed by Guidotti and Merino \cite{GM}, concerning the global stability for the trivial solution of the nonlinear and nonlocal boundary-value problem \begin{equation} \label{pro} \begin{gathered} u_t= u_{xx}, \quad x\in (0, \pi) \; t>0\\ u_x(0,t)= \tanh(\beta u(\pi ,t)),\quad t>0 \; \beta > 0\\ u_x(\pi ,t)= 0, \quad t>0\\ u(x,0)=u_0(x),\quad x \in(0,\pi). \end{gathered} \end{equation} This problem was proposed in \cite{GM} as a rudimentary model for a thermostat. To achieve this goal, we first show the existence of a global compact attractor $\mathcal{A}_\beta $ for (\ref{pro}), for any positive value of the parameter $\beta$. We then prove that $\mathcal{A}_\beta = \{0\}$ if $ 0 < \beta< 1/\pi$, thus showing that the trivial solution is globally asymptotically stable in the phase space, for these values of $\beta$. \section{Global semi-flux in a fractional power space $X^\alpha$} As in \cite{GM} we adopt here the following weak formulation for (\ref{pro}): $u$ is a solution of (\ref{pro}) if \begin{equation} \label{pro2} \begin{gathered} \int_0^\pi u_t \varphi dx + \int_0^\pi u_x \varphi_x dx = -\tanh(\beta u(\pi))\varphi (0),\ t>0,\\ u(0)=u_0, \\ \end{gathered} \end{equation} for all $\varphi \in H^1(0, \pi )=H^1$. Consider the linear operator $A\in \mathcal{L}(H^{1},(H^{1})')$ induced by the continuous bilinear form $a(\cdot ,\cdot ):H^1 \times H^1 \to \mathbb{R}$ given by $a(u,v)= ((u,v))_{H^1}$, that is, $$ {\langle Au,v\rangle_{(H^1)' \times H^1} = a(u,v) = ((u,v))_{H^1}, \forall u,v \in H^1}. $$ We may interpret $A$ as the unbounded closed nonnegative self-adjoint operator $A:D(A)\subset L^2(0,\pi )\to L^2(0,\pi)=L^2$ defined by $$ Au(x) = - u''(x) + u(x), x \in (0, \pi ), $$ for any $u \in D(A)=\{u \in H^2(0,\pi ): u'(0) = u'(\pi) = 0\}$. Let $\{\lambda_n\}$ and $\{e_n\}$ denote the eigenvalues and eigenfuctions of $A$, respectively. As it is easy to see, $A$ is a sectorial operator in $L^2(0, \pi )$ and, therefore, its fractional powers are well defined (cf. Henry \cite{Henry}). Let $X^\alpha = D(A^\alpha )$, $\alpha \geq 0$, be the domain of $A^\alpha$. It is well known that $X^\alpha$ endowed with the inner product $$ (u,v)_\alpha = (A^\alpha u, A^\alpha v)_{L^2} = \sum_{n=0}^{\infty}|\lambda_n|^{2\alpha}(u,e_n)_{L^2}(v,e_n)_{L^2} $$ is a Hilbert space. In particular, we have $X^0 = L^2$, $X^1 = D(A)$ and $ X^{1/2} = H^1$. Following Amann \cite{Amann} or Teman \cite{Tem} we have, for any $\theta \in [0,1]$ $$ X^{\frac{1-\theta}{2}} = [ H^1, L^2 ]_\theta , $$ where $[\cdot, \cdot]_\theta$ denotes the complex interpolation functor. On the other hand, for any $s\in [0,1]$, $$ H^s(0, \pi ) = [ H^1,L^2 ]_{1-s}. $$ Letting $\theta = 1-s$, we obtain $X^\alpha = H^{2\alpha}$, for any $\alpha \in [0, 1/2]$. Denoting $X^{-1/2}= \left(X^{1/2}\right)' = (H^1)'$ and considering the linear operator $A \in \mathcal{L}(H^1, (H^1)')$ as a unbounded operator in $(H^1)' = X^{-1/2}$ given by $D(A) = X^{1/2}$ and $$ \langle Au, \varphi \rangle_{-1/2,1/2} = (u,\varphi)_{1/2} = ((u,\varphi))_{H^1}, $$ for any $u,\varphi \in H^1 = X^{1/2}$, we rewrite equation (\ref{pro2}) as an evolution equation \begin{equation} \label{pro3} \begin{gathered} u_t = - Au + F(u)\quad\mbox{in } X^{-1/2} \; t>0,\\ u(0)= u_0 \end{gathered} \end{equation} where $F: X^\alpha \to X^{-1/2}$ is defined by $$ \langle F(u),\varphi \rangle_{-1/2,1/2} = -\tanh(\beta u(\pi))\varphi (0) + \int_0^\pi u \varphi dx, $$ for $u \in X^\alpha$ and $\varphi \in X^{1/2}$ , that is, $ F(u) = - \gamma_0^\ast \tanh(\beta \gamma_\pi (u)) + u$ in $X^{-1/2}$, where $\gamma_\pi \in \mathcal{L}(X^\alpha , \mathbb{R})$ is given by $\gamma_\pi(u)=u(\pi)$ and $ \gamma_0^\ast \in \mathcal{L}(\mathbb{R}, X^{-1/2})$ is the adjoint operator of $\gamma_0 \in \mathcal{L} (X^{1/2},\mathbb{R})$ given by $\gamma_0(u)=u(0)$. To have a well-posed problem in $X^\alpha$, we make some restrictions on $\alpha$. We impose first that $X^\alpha \hookrightarrow \mathcal{C}([0,\pi])$, which is accomplished by requiring that $\alpha>1/4$. Now, according to \cite{Amann,Henry}, $-A$ is the infinitesimal generator of an analytic semigroup $\{e^{- At}; t \geq 0\}$ in $\mathcal{L}(X^{-1/2})$; since $F$ maps $X^\alpha$ into $X^{-1/2}$, we impose also that $0\leq \alpha-(-\frac{1}{2})<1$. It turns out that the condition $\frac{1}{4}<\alpha<\frac{1}{2}$ implies that (\ref{pro3}) has an unique global solution $u:[0,\infty)\to X^\alpha$, for any $u_0\in X^\alpha$. This follows immediately from Theorem 3.3.3 in \cite{Henry} and from the fact that $F$ is globally Lipschitz continuous: \begin{align*} &|\langle F(u)-F(v),\varphi \rangle_{-1/2,1/2}|\\ & \leq |\tanh (\beta \gamma_\pi (u)) - \tanh(\beta \gamma_\pi (v))| |\varphi(0)| + |(u-v,\varphi)_{L^2}|\\ & \leq \beta \|\gamma_\pi\|_{\mathcal{L}(X^\alpha ,\mathbb{R})} \|\gamma_0\|_{\mathcal{L}(X^{1/2},\mathbb{R})} \|u - v\|_\alpha \|\varphi \|_{1/2} + \|u - v\|_{L^2}\|\varphi\|_{L^2}\\ & \leq \big( \beta \|\gamma_\pi\|_{\mathcal{L}(X^\alpha,\mathbb{R})} \|\gamma_0\|_{\mathcal{L}(X^{1/2},\mathbb{R})} + k \big) \|u-v\|_\alpha \|\varphi \|_{1/2}, \end{align*} for all $\varphi$ in $X^{1/2}$ and any $u$, $v$ in $X^\alpha$, which implies $$ \|F(u)-F(v)\|_{-1/2} \leq K\|u-v\|_\alpha, $$ for all $u,v \in X^\alpha$, where $K= \left(\beta \|\gamma_\pi\|_{\mathcal{L}(X^\alpha,\mathbb{R})} \|\gamma_0\|_{\mathcal{L}(X^{1/2},\mathbb{R})}\;\;+ k \right)$ and $k$ is the embedding constant of $X^\alpha$ in $L^2$. Since $F$ maps bounded sets of $X^\alpha$ into bounded sets of $X^{-1/2}$, it follows by \cite[Theorem 3.3.4]{Henry} that the flow defined by (\ref{pro3}) is global. \section{Main Results} We denote by $\{T(t);\; t\geq 0\} \subset \mathcal{L}(X^{-1/2})$ the semigroup generated by (\ref{pro3}). Since the spectrum of $ A : X^{1/2} \subset X^{-1/2}\to X^{-1/2}$ is given by $\sigma (A)= \{n^2 + 1; n=0,1,...\}$, for any $0<\delta < 1$, we have, by \cite[Theorem 1.4.3]{Henry}, \begin{equation} \label{est} \|e^{-At}\|_{\mathcal{L}(X^{-1/2})}\leq Ce^{-\delta t},\quad \|A^\alpha e^{-At}\|_{\mathcal{L}(X^{-1/2})} \leq C_\alpha t^{-\alpha}e^{-\delta t}, \end{equation} for $t>0$. Since \begin{align*} |\langle F(u),\varphi \rangle_{-1/2,1/2}| &\leq |\tanh (\beta u(\pi))||\varphi(0)|\;+\; |(u,\varphi)_{L^2}|\\ &\leq |\varphi(0) | + \|u\|_{L^2}\|\varphi\|_{L^2}\\ &\leq \sqrt{2\pi}\|\varphi\|_{1/2} + \| u \|_{L^2} \|\varphi \|_{1/2}, \end{align*} for all $\varphi \in X^{1/2}$, we have that for all $u \in X^\alpha$, \begin{equation} \label{est1} \|F(u)\|_{-1/2} \leq \sqrt{2\pi} + \| u \|_{L^2}\,. \end{equation} \begin{lemma} \label{absorbing} % lm3.1 Let $\beta \in (0,\infty )$, $\alpha \in (1/4,1/2)$. Denote by $B_{\varepsilon}$ the ball with center $0$ and radius $\pi(\sqrt {\pi} + \varepsilon )$ in $ L^2$. Then we have \begin{enumerate} \item For any $ u_0 \in X^{\alpha}$ there exists $t^{*} = t^{*}(u_0)$, depending only on the $L^2$-norm of $u_0$, such that the positive semiorbit $T(t)u_0$ is in $B_{\varepsilon} $ for $t \geq t^{*}(u_0)$; \item While $T(t) u_0$ is outside $B_{\varepsilon} $ its $L^2$-norm is decreasing. \end{enumerate} \end{lemma} \begin{proof} Let $u_0\in X^\alpha$, $\epsilon>0$ and, for simplicity, denote by $ u(\cdot,t)= T(t)u_0$ the solution of (\ref{pro}) through $u_0$. Then, we have \begin{equation} \label{derivative} \begin{aligned} \frac{d}{dt}\frac{1}{2}\int_0^{\pi}u(x,t)^2dx &=\int_0^{\pi}u(x,t)u_t(x,t)dx\\ &=-\tanh (\beta u(\pi ,t))u(0,t) - \int_0^{\pi} u_x(x,t)^2dx,\: t>0. \end{aligned} \end{equation} To obtain estimates for this derivative we consider the subsets \begin{gather*} S_1(u_0)=\{ t \in (0,\infty) : u(0,t)u(\pi ,t)\geq 0\},\\ S_2(u_0)=\{ t \in (0,\infty): u(0,t)u(\pi,t)<0\} = (0,\infty) \setminus S_1(u_0). \end{gather*} If $t\in S_2(u_0)$, there exists $y(t) \in (0,\pi )$ such that $u(y(t),t)=0$ and then \begin{eqnarray*} |u(x,t)|&\leq &|u(y(t),t)| +\int_0^\pi |u_x(x,t)|dx\: \leq \: \sqrt{\pi}\|u_x(\cdot,t)\|_{L^2}, \end{eqnarray*} for all $x\in [0,\pi]$. Therefore, \begin{eqnarray*} \|u(\cdot,t)\|^2_{L^2}&\leq&\pi^2\|u_x(\cdot,t)\|^2_{L^2}, \end{eqnarray*} for any $t \in S_2(u_0)$. Hence, for all $t \in S_2(u_0)$, \begin{equation} \label{decayS2} \begin{aligned} \frac{d}{dt}\frac{1}{2}\|u(\cdot,t)\|^2_{L^2} &= |\tanh (\beta u(\pi ,t))||u(0,t)| - \|u_x(\cdot,t)\|^2_{L^2} \\ &\leq \sqrt{\pi}\|u_x(\cdot,t)\|_{L^2} - \|u_x(\cdot,t)\|^2_{L^2}\,. \end{aligned} \end{equation} If $\|u(\cdot,t)\|_{L^2}>\pi(\sqrt{\pi}+\epsilon)$, then \begin{equation} \label{decayS2hip} \frac{d}{dt}\frac{1}{2}\|u(\cdot,t)\|^2_{L^2} \leq -\epsilon(\sqrt{\pi}+\epsilon). \end{equation} To compute the derivative when $t\in S_1(u_0)$, we need to estimate $\|u_x(\cdot,t)\|_{L^2}$. Let $m:(0,\infty)\to \mathbb{R}^+$ be the continuous function $m(t)=\min \left\{ |u(0,t)|,|u(\pi ,t)| \right\}$ and \[ J(u_0) = \big\{ t\in S_1(u_0), \; m(t) \leq \frac{1}{2\pi}\|u(\cdot,t)\|_{L^2}\big\}. \] From $$ |u(x,t)|\leq \min \left\{ |u(0,t)|,|u(\pi ,t)|\right\} + \int_0^\pi |u_x(x,t)|dx $$ for $x\in [0,\pi]$ and $t>0$, we have $$ \|u(\cdot,t)\|^2_{L^2}\leq \pi\left( m(t) + \sqrt{\pi}\|u_x(\cdot,t)\|_{L^2}\right)^2 \leq 2\pi^2 \left(m(t)^2 + \|u_x(\cdot,t)\|^2_{L^2}\right)\,. $$ Therefore, $$ \|u_x(\cdot,t)\|^2_{L^2}\geq \frac{1}{2\pi^2}\|u(\cdot,t)\|^2_{L^2} - m(t)^2. $$ Thus, if $t\in J(u_0)$, then $$ \|u_x(\cdot,t)\|^2_{L^2}\geq \frac{1}{2\pi^2}\|u(\cdot,t)\|^2_{L^2} - \frac{1}{4\pi^2}\|u(\cdot,t)\|^2_{L^2} = \frac{1}{4\pi^2}\|u(\cdot,t)\|^2_{L^2} . $$ Therefore, for all $t\in J(u_0)$, \begin{equation} \label{decayJ} \begin{aligned} \frac{d}{dt}\frac{1}{2} \|u(\cdot,t)\|^2_{L^2} & = - \tanh (\beta u(\pi ,t)) u(0,t) - \|u_x(\cdot,t)\|^2_{L^2} \\ & \leq - \|u_x(\cdot,t)\|^2_{L^2} \\ & \leq - \frac{1}{4\pi^2} \|u(\cdot,t)\|^2_{L^2}. \end{aligned} \end{equation} If $\|u(\cdot,t)\|_{L^2}>\pi(\sqrt{\pi}+\epsilon)$, we obtain \begin{equation} \label{decayJhip} \frac{d}{dt}\frac{1}{2}\|u(\cdot,t)\|^2_{L^2} \leq - \frac{1}{4}(\sqrt{\pi}+\varepsilon)^2 \end{equation} On the other hand, if $t\in S_1(u_0)\setminus J(u_0)$, then \begin{equation} \label{decayJc} \begin{aligned} \frac{d}{dt} \frac{1}{2}\|u(\cdot,t)\|^2_{L^2} &=-\tanh(\beta u(\pi,t))u(0,t) - \|u_x(\cdot,t)\|^2_{L^2} \\ &\leq -\tanh(\beta u(\pi ,t))u(0,t) \\ &= -\tanh(\beta |u(\pi ,t)|)|u(0,t)| \\ &\leq -\tanh\big(\frac{\beta \|u(\cdot, t)\|_{L^2} }{2 \pi}\big) \frac{ \|u(\cdot, t)\|_{L^2} }{2 \pi} \end{aligned} \end{equation} If $\|u(\cdot,t)\|_{L^2}>\pi(\sqrt{\pi}+\epsilon)$, we obtain \begin{equation} \label{decayJchip} \frac{d}{dt} \frac{1}{2}\|u(\cdot,t)\|^2_{L^2} \leq -\tanh\big(\frac{\beta (\sqrt{\pi}+\epsilon) }{2 }\big) \frac{(\sqrt{\pi}+\epsilon) }{2 } \end{equation} Letting $\varepsilon_1 = \min\big\{\varepsilon(\sqrt{\pi}+\varepsilon), \frac{1}{4}(\sqrt{\pi}+\varepsilon)^2, \tanh\big(\frac{\beta (\sqrt{\pi}+\epsilon) }{2}\big) \frac{ (\sqrt{\pi}+\epsilon)}{2} \big\} $, we conclude using (\ref{decayS2hip}), (\ref{decayJhip}) and (\ref{decayJchip}), that \begin{equation} \label{decay} \frac{d}{dt}\|u(\cdot,t)\|_{L^2}^2 \leq - 2 \varepsilon_1 \end{equation} This proves our second assertion. Suppose $u(t,u_0)$ is outside $B_{\varepsilon}$ for $0 \leq t \leq \bar{t}$. Then $\|u(\cdot,\bar{t})\|^2_{L^2} \leq \|u_0\|^2_{L^2} - 2 \varepsilon_1 \bar{t}$. Therefore, there must exist a $ t^{*} = t^*(u_0) \leq \frac{1}{2\varepsilon_1}\left( \|u_0\|^2_{L^2} - \pi^2( \sqrt{\pi} + \varepsilon)^2\right) $ such that $u(\cdot , t^{*})$ belongs to $B_{\varepsilon}$. We claim that $\|u(\cdot,t)\|_{L^2} \leq \pi (\sqrt{\pi}+ \varepsilon )$ for all $t\geq t^*$. Otherwise, there would exist $t_1\geq t^*$ and $\delta>0$ such that $\| u(\cdot,t_1)\|_{L^2} = \pi(\sqrt{\pi}+\varepsilon)$ and $\| u(\cdot,t)\|_{L^2} > \pi(\sqrt{\pi}+\varepsilon)$ for $t\in (t_1,t_1+\delta)$, which is a contradiction with the fact that $t\mapsto \|u(\cdot,t)\|_{L^2}$ is non increasing. This proves our first assertion. \end{proof} \begin{theorem} \label{attractor} If $\beta \in (0, \infty )$ and $\alpha \in (1/4, 1/2)$, then $\{T(t); t\geq 0 \}$ has a global attractor $\mathcal{A}_\beta$. \end{theorem} \begin{proof}. Let $u_0 \in X^{\alpha}$ and $u(\cdot,t) = T(t) u_0$. By the variation of constant formula and estimates (\ref{est}), (\ref{est1}), we have \begin{equation} \label{varconst} % \label{1.1} \begin{aligned} \|u(\cdot,t)\|_{\alpha} &\leq Ce^{-\delta t}\|u_{0}\|_{\alpha} +C_\alpha { \int_{0}^{t}e^{-\delta(t-s)}(t-s)^{-\alpha}\|F(u(\cdot,s))\|_{-1/2}ds}, \\ &\leq Ce^{-\delta t}\|u_{0}\|_{\alpha} + C_\alpha {\int_{0}^{t}e^{-\delta (t-s)}(t-s)^{-\alpha}(\sqrt{2\pi} +\|u(\cdot,s)\|_{L^2})ds}. \end{aligned} \end{equation} If $t^{*}(u_0) $ is as given by Lemma \ref{absorbing}, for $t>t^*$ we have \begin{equation} \label{estimt*} \begin{aligned} \|u(\cdot,t)\|_{\alpha} &\leq Ce^{-\delta t}\|u_{0}\|_{\alpha} +C_\alpha { \int_{0}^{t^*}e^{-\delta (t-s)}(t-s)^{-\alpha}(\sqrt{2\pi} + \|u(\cdot,s)\|_{L^2})ds} \\ &\quad + C_\alpha {\int_{t^*}^{t}e^{-\delta(t-s)}(t-s)^{-\alpha }(\sqrt{2\pi} + \|u(\cdot,s)\|_{L^2})ds} \\ &\leq Ce^{-\delta t}\|u_0\|_{\alpha} + C_\alpha { \int_{0}^{t^*}e^{-\delta(t-s)}(t-s)^{-\alpha}(\sqrt{2\pi } + \|u(\cdot,s)\|_{L^2})ds} \\ &\quad + C_\alpha(\sqrt{2\pi} + \pi(\sqrt{\pi}+ \varepsilon )){ \int_{0}^{\infty}e^{-\delta (t-s)}(t-s)^{-\alpha}ds} \\ &\leq Ce^{-\delta t}\|u_{0}\|_{\alpha} + C_\alpha e^{-\delta t}(\sqrt{2\pi } + \|u_0\|_{L^2}){ \int_{0}^{t^*}e^{ \delta s}(t-s)^{-\alpha}ds} + M_1 \\ &\leq e^{-\delta t}\left( C\|u_{0}\|_{\alpha} + C_\alpha(\sqrt{2\pi } + \|u_0\|_{L^2})e^{\delta t^*}(t^*)^{1-\alpha}(1-\alpha)^{-1}\right) + M_1, \end{aligned} \end{equation} where $M_1 = C_\alpha(\sqrt{2\pi} + \pi(\sqrt{\pi}+ \varepsilon )){ \int_{0}^{\infty}e^{-\delta (t-s)}(t-s)^{-\alpha}ds} $. From this formula, and the continuous inclusion of $ X_{\alpha}$ in $L^2$, it is easy to see that one can choose $t_1>0$, depending only on the norm of $u_0$ in $X_{\alpha}$, so that $$ \|u(\cdot,t)\|_{\alpha} \leq 2M_1, $$ for all $t \geq t_1$ and, therefore, the semigroup $\{T(t); t\geq 0\}$ is bounded dissipative. If $t < t^{*}$ the same estimate (without the last term and with $t$ in the place of $t^{*}$) shows that \begin{equation} \label{estimt} \|u(\cdot,t)\|_{\alpha} \leq e^{-\delta t} C\|u_{0}\|_{\alpha} + C_\alpha(\sqrt{2\pi } + \|u_0\|_{L^2}) t^{1-\alpha}(1-\alpha)^{-1} \end{equation} From \ref{estimt*} and \ref{estimt} it follows that orbits of bounded sets are bounded. Since $A$ has compact resolvent and $F$ maps bounded sets in $X^\alpha$ into bounded sets in $X^{-1/2}$, it follows from \cite[Theorem 4.2.2]{Hale} that $T(t)$ is compact for all $t>0$. The result follows then from \cite[Theorem 3.4.6]{Hale}. \end{proof} \begin{rem} \label{rmk3.2} \rm We observe that \ref{estimt*} above also gives an estimate for the size of the attractor. \end{rem} \begin{theorem} \label{thm3.2} If $\beta \in (0,1/\pi)$ and $\alpha \in (1/4, 1/2)$, then $\mathcal{A}_\beta = \{0\}$. \end{theorem} \begin{proof} Let $\varepsilon>0$ be given. We will use the estimates obtained in Lemma \ref{attractor} for the decay of the $L^2$-norm of a solution $u(\cdot,t)$ when $t\in S_1(u_0)$. If $t\in S_2(u_0)$, we have \begin{equation} \label{decayS2improve} \begin{aligned} \frac{d}{dt}\frac{1}{2}\|u(\cdot,t)\|^2_{L^2} &=|\tanh(\beta u(\pi,t))||u(0,t)| - \|u_x(\cdot,t)\|_{L^2}^2\\ &\leq \beta |u(\pi,t))||u(0,t)| -\|u_x(\cdot,t)\|_{L^2}^2\\ &\leq \beta \left( \sqrt{\pi}\|u_x(\cdot,t)\|_{L^2}\right)^2 - \|u_x(\cdot,t)\|^2_{L^2}\\ &\leq (\beta \pi - 1)\|u_x(\cdot,t)\|^2_{L^2}\\ &\leq - \frac{1 - \beta \pi}{\pi^2}\|u(t)\|^2_{L^2} \end{aligned} \end{equation} If $\|u(\cdot,t)\|_{L^2} \geq \varepsilon$ and $\varepsilon_2 =\min\big\{ \frac{1- \beta \pi}{\pi^2} \varepsilon^2, \frac{\varepsilon^2}{ 4 \pi^2}, \tanh\big(\frac{\beta \varepsilon }{2 \pi}\big) \left(\frac{\varepsilon }{2 \pi}\right) \big\} $, we obtain using (\ref{decayS2improve}), (\ref{decayJ}) and (\ref{decayJc}), that \begin{equation} \label{decayimprove} \frac{d}{dt}\|u(\cdot,t)\|_{L^2}^2 \leq - 2 \varepsilon_2. \end{equation} Suppose $ \|u(\cdot,t)\|_{L^2} \geq \varepsilon$ for $0 \leq t \leq \bar{t}$. Then $\|u(\cdot,\bar{t})\|^2_{L^2} \leq \|u_0\|^2_{L^2} - 2 \varepsilon_2 \bar{t}$. Therefore, there must exist a $ t^{*} = t^*(u_0) \leq \frac{1}{2\varepsilon_2}\left( \|u_0\|^2_{L^2} - \varepsilon^2\right) $ such that $ \|u(\cdot,t)\|_{L^2} \leq \varepsilon$ for $t\geq t^*$. Since the attractor $\mathcal{A}_{\beta}$ is a bounded subset of $L^2$, there exists $t^*(\varepsilon)$ such that $ \mathcal{A}_{\beta} = T(t^*)\mathcal{A}_{\beta} \subset V_{\varepsilon}$, where $V_{\varepsilon}$ is the ball of radius $\varepsilon $ in $L^2$. Since $\varepsilon$ is arbitrary, we conclude that $ A_{\beta} = \{ 0 \}$ as claimed. \end{proof} \begin{thebibliography}{00} \bibitem{Amann} H. Amann, {\it Parabolic Evolution Equations and Nonlinear Boundary Conditions}, Journal of Differential Equations, {\bf 72} (1988), 201-269. \bibitem{GM} P. Guidotti and S. 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