\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small 2004-Fez conference on Differential Equations and Mechanics \newline {\em Electronic Journal of Differential Equations}, Conference 11, 2004, pp. 175--185.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \setcounter{page}{175} \begin{document} \title[\hfilneg EJDE/Conf/11 \hfil Decay estimates for solutions] {Decay estimates for solutions of some systems for elasticity with nonlinear boundary feedback} \author[Naji Yebari\hfil EJDE/Conf/11 \hfilneg] {Naji Yebari} \address{Naji Yebari \hfill\break D\'epartement de Math\'ematiques\\ Universit\'e Abdelmalek Essaadi\\ B. P. 2121, Tetouan, Maroc} \email{nyebari@hotmail.com} \date{} \thanks{Published October 15, 2004.} \subjclass[2000]{74KB05, 74K10, 74K20, 93D15} \keywords{Beams; linear elasticity; plates; stabilization of systems by feedback} \begin{abstract} We study the energy decay rate for the Euler-Bernoulli and Kirchhoff plate equations. Under suitable growth assumptions on the nonlinear dissipative boundary feedback functions, we obtain some new results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction and main results} We consider the following Euler-Bernoulli beam equation with nonlinear boundary feedback controls: $$\begin{gathered} y_{tt}+y_{xxxx}=0 \quad \text{in } (0,1), \; t>0,\\ y(0,t)=y_x(0,t)=0 \quad \text{(clamped at x=0)}, \; t>0, \\ -y_{xx}(1,t)=h(y_{xt}(1,t))\quad \text{(moment),} \; t>0, \\ y_{xxx}(1,t)=g(y_t(1,t))\quad \text{(force),} \; t>0, \\ y(.,0)=y_0 \in H_E^2,\quad y_t(.,0)=y_1 \in L^2(0,1), \end{gathered} \label{e1.1}$$ where $H_E^2=\{ u\in H^2(0,1);\;u(0)=u'(0)=0\}$, $x$ stands for the position and $t$ the time. The flexural rigidity of the beam and the mass density are assumed to be equal to one. One end is controlled by a point force and point bending moment which are assumed to be nonlinear functions of the observation. By observation we mean velocity and angular velocity of the transversal deflexion at the end. Throughout this paper, $g$ and $h$ are assumed to satisfy the following hypothesis \begin{gather} g\in C^0(\mathbb{R}),\text{ nondecreasing, }g(0)=0,\; g(s)s>0\;\forall s\neq 0\,, \label{e1.2} \\ h\in C^0(\mathbb{R}),\text{ nondecreasing, }h(0)=0,\; h(s)s>0\;\forall s\neq 0\,. \label{e1.3} \end{gather} The total energy of system \eqref{e1.1} is $$\widetilde{E}(t)=\frac 12\int_0^1 (y_t^2(x,t)+y_{xx}^2(x,t) )dx. \label{e1.4}$$ Formally, $$\frac d{dt}\widetilde{E}(t)=-y_t(1,t)\;g(y_t(1,t))-y_{xt}(1,t)% \;h(y_{xt}(1,t)) \label{e1.5}$$ Assumptions \eqref{e1.2} and \eqref{e1.3} imply that the energy $\widetilde{E}(t)$ is non-increasing and is a Lyapounov function. It is well-known that for any initial data $(y_0,y_1)$ $\in \mathcal{H}=H_E^2\times L^2(0,1)$, Problem \eqref{e1.1} admits a unique weak solution $y$ such that $(y(t),y_t(t))\in \mathcal{H}$ and $y\in C^0(\mathbb{R}^{+};H^4(0,1))\cap C^1(\mathbb{R}^{+};L^2(0,1))$ (see Conrad and Pierre \cite{c2}). The study of the strong asymptotic stability of the solution of \eqref{e1.1} in $\mathcal{H}$, using the invariance principle of Lasalle has been proved by Conrad and Pierre \cite{c2} in the case where $g$ and $h$ are multivalued maximal monotone graphs. Our aim in this study is to estimate the rate of decay of the energy $\widetilde{E}(t)$ when the nonlinear feedback functions $g$ and $h$ satisfy suitable growth conditions. However, the extension of this method to the case where $g$ and $h$ are maximal monotone graphs seems difficult. Our two main results are as follows. \begin{theorem} \label{thm1} Assume \eqref{e1.2} and \eqref{e1.3} hold. Then for every solution $y$ of system \eqref{e1.1}, we have \noindent (i) If there exist positive constants $C_1,C_2,C_3,C_4$ such that for all $x\in \mathbb{R}$, $$\begin{gathered} C_1| x| \leq | g(x)| \leq C_2| x| , \\ C_3| x| \leq | h(x)| \leq C_4| x| , \end{gathered} \label{e1.6}$$ then given any $M>1$, there exists a constant $\omega >0$ such that $\widetilde{E}(t)\leq M\widetilde{E}(0)e^{-\omega t}\quad \forall t>0.$ (ii) If there exist positive constants $C_1, C_2, C_3, C_4$, and $p$, $q$ in $[1,+\infty [$ such that $\max (p,q)=p\vee q>1$ such that for all $x\in \mathbb{R}$, $$\begin{gathered} C_1\min ( | x| ,| x| ^p) \leq |g(x)| \leq C_2| x| , \\ C_3\min ( | x| ,| x| ^q) \leq | h(x)| \leq C_4| x| , \end{gathered} \label{e1.7}$$ then, given any $M>1$, there exists a constant $\omega >0$ depending continuously on $\widetilde{E}(0)$ such that $\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t)^{-\frac 2{( p\vee q) -1}} \quad \forall t\geq 0.$ \end{theorem} \begin{theorem} \label{thm2} Assume \eqref{e1.2} and \eqref{e1.3} hold. Then for every solution $y$ of system \eqref{e1.1} we have \noindent (i)If there exist positive constants $C_1,C_2, C_3, C_4$ and $p,q \in ]0,1]$ with $\min (p,q)=p\wedge q<1$ such that for all $x\in \mathbb{R}$, $$\begin{gathered} C_1| x| \leq | g(x)| \leq C_2\max ( | x| ,| x| ^p) , \\ C_3| x| \leq | h(x)| \leq C_4\max ( | x| ,| x| ^q) , \end{gathered} \label{e1.8}$$ then, given any $M>1$, there exists a constant $\omega >0$ depending continuously on $\widetilde{E}(0)$ such that $\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t) ^{-\frac{2(p\wedge q)}{1-(p\wedge q)}},\quad \forall t\geq 0.$ (ii) If there exists positive constants $C_1, C_2, C_3, C_4$ and $(p,q)\in ]0,1]\times [1,+\infty [$ with $\frac 1p\vee q>1$ such that for all $x\in \mathbb{R}$, $$\begin{gathered} C_1 | x| \leq | g(x)| \leq C_2\max ( | x| ,| x| ^p) , \\ C_3\min ( | x| ,| x| ^q) \leq | h(x)| \leq C_4| x| , \end{gathered} \label{e1.9}$$ then given any $M>1$, there exists a constant $\omega >0$ depending continuously on $\widetilde{E}(0)$ such that $\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t) ^{-\frac 2{( \frac 1p\vee q) -1}}\quad \forall t\geq 0.$ \end{theorem} Theorems \ref{thm1} and \ref{thm2} will also be valid for the following model of a Kirchhoff plate equation (see Ciarlet \cite{c1} and Lagnese \cite{l1}), in star-shaped domain by nonlinear boundary feedback: $$\begin{gathered} y_{tt}+\Delta ^2y=0 \quad \text{in } \Omega \times ] 0,+\infty [ , \\ \Delta y+(1-\mu )B_1y=v_1 \quad \text{on } \Gamma \times ] 0,+\infty[ , \\ \frac \partial {\partial \nu }\Delta y+(1-\mu )\frac \partial {\partial \tau }B_2y=v_2 \quad \text{on } \Gamma \times ] 0,+\infty [ , \\ y(0)=y_0 \in H^2(\Omega ),\quad y_t(0)=y_1 \in L^2(\Omega ). \end{gathered} \label{e1.10}$$ We assume that $\Omega$ is a bounded strongly star-shaped domain of $\mathbb{R}^2$ with respect to $x_0\in \Omega$ and having smooth boundary $\Gamma =\partial \Omega$ of class $C^2$, which means there exists a positive constant $\delta$ such that $$m(x).\nu (x)\geq \delta ^{-1}\quad \forall x\in \Gamma, \label{e1.11}$$ where $\nu (x)=(\nu _1(x),\nu _2(x))$ is the unit outer normal vector to $\Gamma$, $m(x)=x-x_0$, and the dot .'' denotes the scalar product in $\mathbb{R}^2$. $\tau (x)=(-\nu _2(x),\nu _1(x))$ is a unit tangent vector. We denote by $\frac \partial {\partial \nu }$ $($resp. $\frac \partial {\partial \tau }$, ) the normal derivative (resp. tangent derivative$)$. The constant $0<\mu <1/2$ is the Poisson coefficient and the boundary operators $B_1,B_2$ are defined by $B_1y=2\nu _1\nu _2\frac{\partial ^2y}{\partial x_1\partial x_2} -\nu _1^2\frac{\partial ^2y}{\partial x_2^2}-\nu _2^2\frac{\partial ^2y}{\partial x_1^2},\quad B_2y=(\nu _1^2-\nu _2^2)\frac{\partial ^2y}{\partial x_1\partial x_2}+\nu _1\nu _2(\frac{\partial ^2y}{\partial x_1^2} -\frac{\partial ^2y}{\partial x_2^2}),$ where $y=y(x_1,x_2,t)$ is the vertical displacement of the point $x=(x_{1,}x_2)\in \Omega$ at the time $t$ of the plate. The aim of this paper is to study the uniform energy decay rate of the system \eqref{e1.10} by the nonlinear feedback laws $v_1$ and $v_2$ given as follows \begin{gather} v_1(t)=-\beta \frac{\partial y}{\partial \nu }-h(\frac{\partial y_t}{% \partial \nu }),\quad v_2(t)=\alpha y+g(y_t)\quad \text{on }\Gamma \times ] 0,+\infty [ , \label{e1.12}\\ (\alpha,\beta) \in (L^\infty (\Gamma ))^{2},\quad 0<\alpha _0\leq \alpha (x)\leq \alpha _1,\quad 0<\beta _0\leq \beta (x)\leq \beta _1\quad \forall x\in \Gamma . \label{e1.13} \end{gather} It is well-known that for any initial data $(y_0,y_1)\in V=H^2(\Omega )\times L^2(\Omega )$, the system \eqref{e1.10} has a unique weak solution $y$ such that $(y(t),y_t(t))\in V$ and $y\in C^0(\mathbb{R}^{+},H^2(\Omega ))\cap C^1(\mathbb{R} ^{+},L^2(\Omega ))$ (see Rao \cite{r1}). We introduce the energy associated with the system \eqref{e1.10} as follows $$E(t)=\frac 12( \int_\Omega | y_t| ^2dx+a(y,y)+\int_\Gamma (\alpha | y| ^2+\beta | \frac{\partial y}{\partial \nu }| ^2)d\Gamma ), \label{e1.14}$$ where $$a(y,y)=\int_\Omega (( \frac{\partial ^2y}{\partial x_1^2})^{2} +(\frac{\partial ^2y}{\partial x_2^2})^{2} +2\mu \frac{\partial ^2y}{\partial x_1^2 }\frac{\partial ^2y}{\partial x_2^2}+2(1-\mu )(\frac{\partial ^2y}{\partial x_1\partial x_{2}})^{2}) dx.$$ Using Green's formula, the derivative of the energy $E(t)$ is $$\frac{dE}{dt}=-\int_\Gamma (g(y_t)y_t+h(\frac{\partial y_t}{\partial \nu }) \frac{\partial y_t}{\partial \nu })d\Gamma . \label{e1.15}$$ Assumptions \eqref{e1.2} and \eqref{e1.3} imply that the energy is non-increasing and is a Lyapounov function. It is easy to prove the strong stabilization by applying Holmgren's theorem (see Lions \cite{l2}) and Lasalle's invariance principle (see \cite{s1}). The problem of estimating the rate of decay of the energy $E(t)$ has been studied extensively by many authors, among which we can mention Lagnese \cite{l1}, Komornik - Zuazua \cite{k1} and Zuazua \cite{z1}. In the nonlinear cases, under suitable growth conditions on the functions $g$ and $h$, Rao \cite{r1} has established exponential or rational rates of decay of the energy for any positive functions $\alpha ,\beta$ given by \eqref{e1.13}. We obtain here an improvement of these results in the sense that we prove the estimates when the nonlinear feedback functions $g$ and $h$ satisfy more general growth conditions. Now, we state our main results. \begin{theorem} \label{thm3} Assume \eqref{e1.2}, \eqref{e1.3}, \eqref{e1.11}, \eqref{e1.13}. Then for every solution $y$ of system \eqref{e1.10} with feedback laws \eqref{e1.12}, we have \noindent (i) In addition, if we assume that $g$ and $h$ such that \eqref{e1.7} holds. Then, given any $M>1$, there exists a constant $\omega >0$ depending continuously on $E(0)$ such that $E(t)\leq ME(0)(1+\omega t)^{-\frac 2{( p\vee q)-1}},\quad \forall t\geq 0.$ (ii) In addition, if we assume that $g$ and $h$ such that \eqref{e1.8} holds. Then, given any $M>1$, there exists a constant $\omega >0$ depending continuously on $E(0)$ such that $E(t)\leq ME(0)(1+\omega t)^{-\frac{2(p\wedge q)}{1-(p\wedge q)}}, \quad \forall t\geq 0.$ (iii) In addition, if we assume that $g$ and $h$ such that \eqref{e1.9} holds. Then, given any $M>1$, there exists a constant $\omega>0$ depending continuously on $E(0)$ such that $E(t)\leq ME(0)(1+\omega t)^{-\frac 2{( \frac 1p\vee q)-1}}\quad \forall t\geq 0.$ \end{theorem} \section{Proof of Theorem \ref{thm1}} Let $y$ be a smooth solution of \eqref{e1.1}. We define the functional $\rho (t)=\rho _1(t)+\rho _2(t)+\rho _3(t),$ where \begin{gather*} \rho _1=4\int_0^1xy_ty_xdx,\quad \rho _2=\frac 14y(1,t)\int_0^1x^2(3-2x)y_tdx ,\\ \rho _3=\frac 14y_x(1,t)\int_0^1x^2(x-1)y_tdx. \end{gather*} We can show that there exist positive constants $K_0,K_1$ and $K_2$ such that for any $t\geq0$, the following estimates hold \begin{gather} | \rho (t)| \leq K_0\widetilde{E}(t), \label{e2.1} \\ \frac{d\rho }{dt}(t)\leq -\widetilde{E}(t)+K_1(y_{xt}^2(1,t)+y_t^2(1,t)) +K_2( g^2(y_t(1,t))+h^2(y_{xt}(1,t))). \label{e2.2} \end{gather} Given $\varepsilon >0$, we introduce the perturbed energy by $$\widetilde{E}_\varepsilon (t)=\widetilde{E}(t)+\varepsilon \rho (t)\big[ \widetilde{E}(t)\big] ^{\frac{( p\vee q) -1}2}. \label{e2.3}$$ This together with the non-increasing of the energy $\widetilde{E}% (t)$ implies that for any $M>1$ $$M^{-\frac 12}\big[ \widetilde{E}_\varepsilon (t)\big] ^{\frac{( p\vee q) +1 }2} \leq \big[ \widetilde{E}(t)\big] ^{\frac{( p\vee q) +1}2} \leq M^{1/2}\big[ \widetilde{E}_\varepsilon (t)\big]^{\frac{( p\vee q) +1}2} \label{e2.4}$$ with $$\varepsilon \leq K_0^{-1}\big[ \widetilde{E}(0)\big]^{\frac{1-(p\vee q) }2} (1-M^{\frac{-1}{( p\vee q) +1}}). \label{e2.5}$$ Now, we calculate the derivative of the perturbed energy $\widetilde{E}_\varepsilon (t)$ $$\widetilde{E}_\varepsilon '(t)=\widetilde{E}'(t)+\varepsilon \frac{(p\vee q)-1}2\rho (t)\widetilde{E}'(t)[ \widetilde{E}(t)] ^{\frac{( p\vee q) -3}% 2}+\varepsilon \rho '(t)[ \widetilde{E}(t)] ^{\frac{( p\vee q) -1}2}, \label{e2.6}$$ on the other hand, from \eqref{e1.6}, \eqref{e1.7} and \eqref{e2.2} one obtains $$\rho '(t)\leq -\widetilde{E}(t)+K_3y_t^2(1,t)+K_4y_{xt}^2(1,t) \label{e2.7}$$ where $K_3=K_1+K_2C_2^2$ and $K_4=K_1+K_2C_4^2$. Plugging \eqref{e2.1} and \eqref{e2.7} into equation \eqref{e2.6}, one obtains $$\widetilde{E}_\varepsilon '(t) \leq (-1+\varepsilon \frac{(p\vee q)-1}2K_0[ \widetilde{E}(0)] ^{\frac{( p\vee q)-1}2})(-\widetilde{E}'(t))+F_{1}+F_{2} -\varepsilon [ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2} \label{e2.8}$$ where $F_{1}=\varepsilon K_3[ \widetilde{E}(t)] ^{\frac{( p\vee q) -1}2}y_t^2(1,t)$ and $F_{2}=\varepsilon K_4[\widetilde{E}(t)] ^{\frac{( p\vee q) -1}2} y_{xt}^2(1,t))$. Now we distinguish the case $p\vee q=1$ and $p\vee q>1$. \noindent (i) Case $p\vee q=1$.\; In this case \eqref{e2.8} yields $\widetilde{E}_\epsilon '(t)\leq (-1+\varepsilon K_3/C_1)y_t(1,t)g(y_t(1,t)+(-1+\varepsilon K_4/C_3)y_{xt}(1,t)h(y_{xt}(1,t))-\varepsilon \widetilde{E}(t).$ If we choose $\varepsilon \leq \min (C_1/K_3,C_3/K_4,K_0^{-1}[ \widetilde{E}(0)] ^{^{\frac{1-( p\vee q) }2}}(1-M^{\frac{-1}{( p\vee q) +1}}))$, this implies $\widetilde{E}_\varepsilon '(t)\leq -\varepsilon \widetilde{E}% (t)\leq -\varepsilon M^{\frac{-1}2}\widetilde{E}_\varepsilon (t),$ so we obtain $\widetilde{E}(t)\leq M\widetilde{E}(0)e^{-\varepsilon M^{\frac{-1}2} t},\quad \forall t>0.$ (ii) Case $p\vee q>1$.\; If $y_{xt}^2(1,t)>1$, it follows from hypothesis \eqref{e1.2}, \eqref{e1.3} and \eqref{e1.7} that $$F_{2}\leq \varepsilon (K_4/C_3)[ \widetilde{E}(0)] ^{\frac{(p\vee q) -1}2}y_{xt}(1,t)h(y_{xt}(1,t)). \label{e2.9}$$ However, while $y_{xt}^2(1,t)\leq 1$, we have min ($| y_{xt}(1,t)| ,| y_{xt}(1,t)|^q)=| y_{xt}(1,t)| ^q$ and $|y_{xt}(1,t)| ^{( p\vee q) +1}\leq |y_{xt}(1,t)|^{q+1}\leq(1/C_3)y_{xt}(1,t)h(y_{xt}(1,t)),$ by Young's inequality, we have for any $\delta >0$, \begin{aligned} F_{2} &\leq \varepsilon \frac{( p\vee q) -1}{( p\vee q) +1}\delta ^{-\frac{( p\vee q) +1}{( p\vee q) -1}}[ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2}\\ &\quad + \varepsilon \frac 2{C_3(p\vee q+1)}(K_4\delta )^{\frac{( p\vee q) +1}2}y_{xt}(1,t)h(y_{xt}(1,t)). \end{aligned}\label{e2.10} Combining \eqref{e2.9} and \eqref{e2.10}, one has $$F_{2}\leq \varepsilon \frac{( p\vee q) -1}{( p\vee q) +1}\delta ^{-\frac{( p\vee q) +1}{( p\vee q) -1}}[ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2}+ \varepsilon K_5y_{xt}(1,t)h(y_{xt}(1,t)), \label{e2.11}$$ where $K_5=\frac 2{C_3(p\vee q+1)}(K_4\delta )^{\frac{( p\vee q) +1}2}+(K_4/C_3)[ \widetilde{E}(0)] ^{\frac{( p\vee q) -1}2}.$ Similarly, we can show that $$F_{1} \leq \varepsilon \frac{( p\vee q) -1}{( p\vee q) +1}\delta ^{-\frac{( p\vee q) +1}{( p\vee q) -1}}[ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2}+ \varepsilon K_6y_t(1,t)g(y_t(1,t)), \label{e2.12}$$ with $K_6=\frac 2{C_1(p\vee q+1)}(K_4\delta )^{\frac{( p\vee q) +1}2}+(K_3/C_1)[ \widetilde{E}(0)] ^{\frac{( p\vee q) -1}2}.$ Inserting \eqref{e1.5},\eqref{e2.11} and \eqref{e2.12} into \eqref{e2.8}, we obtain \begin{align*} \widetilde{E}_\varepsilon '(t) &\leq \varepsilon (-1+2\frac{( p\vee q) -1}{( p\vee q) +1}\delta ^{-\frac{( p\vee q) +1}{( p\vee q) -1}})[ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2}+(-1+\varepsilon \lambda_0)y_t(1,t)g(y_t(1,t)) \\ &\quad +(-1+\varepsilon \lambda _1)y_{xt}(1,t)h(y_{xt}(1,t)), \end{align*} where $\lambda _i=K_{6-i}+K_0\frac{(p\vee q)-1}2[ \widetilde{E}(0)] ^{\frac{( p\vee q) -1}2}$ for $i=0,1$. This implies that $$\widetilde{E}_\varepsilon '(t)\leq -\mu \varepsilon [ \widetilde{E}(t)] ^{\frac{( p\vee q) +1}2}, \label{e2.13}$$ provided $\delta$ is chosen such that for some $\mu >0$, $2\frac{( p\vee q) -1}{( p\vee q) +1}\delta ^{-\frac{( p\vee q) +1}{( p\vee q) -1}}\leq 1-\mu$ and $\varepsilon$ is chosen as follows $-1+\varepsilon (K_{6-i}+K_0\frac{(p\vee q)-1}2[\widetilde{E}(0)] ^{\frac{( p\vee q) -1}2})\leq 0$ for $i=0,1$. Combining \eqref{e2.4} and \eqref{e2.13}, we get $$\widetilde{E}_\varepsilon '(t)\leq -\mu \varepsilon M^{\frac{-1}2} [ \widetilde{E}_\varepsilon (t)] ^{\frac{( p\vee q)+1}2}. \label{e2.14}$$ Finally, solving the differential inequality \eqref{e2.14} and using \eqref{e2.4} we obtain $\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t)^{-\frac 2{( p\vee q) -1}},$ with $\omega =\frac{( p\vee q) -1}2\mu \varepsilon M ^{\frac{-( p\vee q) }{( p\vee q) +1}}[ \widetilde{E} (0)] ^{\frac{( p\vee q) -1}2}$. This completes the proof of theorem \ref{thm1}. \section{Proof of Theorem \ref{thm2}} (i) First, by the conditions \eqref{e1.8} and \eqref{e2.2}, we can deduce the following estimate $$\rho '(t)\leq -\widetilde{E}(t)+K ( g^2(y_t(1,t))+h^2(y_{xt}(1,t))) , \label{e3.1}$$ with $K=K_2+K_1(1/C_1^2+1/C_2^2)$. Next, given $\varepsilon >0$, we introduce the perturbed energy by $$\widetilde{E}_\varepsilon (t)=\widetilde{E}(t)+\varepsilon \rho (t)[ \widetilde{E}(t)] ^{\frac{1-(p\wedge q)}{2(p\wedge q)}}. \label{e3.2}$$ Then, for any $M>1$, we have \eqref{e2.4} and \eqref{e2.5} with $1/p$ instead of $p$. We have also \eqref{e2.6} with $1/p$ instead of $p$. Plugging \eqref{e1.5}, \eqref{e2.1} and \eqref{e3.1} into equation \eqref{e2.6} one obtains $$\widetilde{E}_\varepsilon '(t) \leq (-1+\varepsilon K_0% \frac{1-(p\wedge q)}{2(p\wedge q)}[ \widetilde{E}(0)] ^{\frac{% 1-(p\wedge q)}{2(p\wedge q)}})(-\widetilde{E}'(t) ) +D_{1}+D_{2}- \varepsilon [ \widetilde{E}(t)] ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}. \label{e3.3}$$ where $D_{1}= \varepsilon K (\widetilde{E}(t)) ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}g^{2}(y_{t}(1,t)$ and $D_{2}= \varepsilon K( \widetilde{E}(t)) ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}h^{2}(y_{xt}(1,t)$. If $y_{xt}^2(1,t)\geq 1$, it follows from hypothesis \eqref{e1.2}, \eqref{e1.3} and \eqref{e1.8}, we have $$D_{2}\leq \varepsilon C_4K[ \widetilde{E}(0)] ^{\frac{1-(p\wedge q)}{2(p\wedge q)}}h(y_{xt}(1,t))y_{xt}(1,t). \label{e3.4}$$ However, while $y_{xt}^2(1,t)<1$, we have $\max(|y_{xt}(1,t)| ,| y_{xt}(1,t)| ^q)=|y_{xt}(1,t)| ^q$, so $| h(y_{xt}(1,t))| ^{\frac{1+(p\wedge q)}{(p\wedge q)} }\leq C_4y_{xt}(1,t)h(y_{xt}(1,t),$ by Young's inequality we obtain $$D_{2}\leq (\varepsilon /4)[ \widetilde{E}(t)] ^{\frac{% 1+(p\wedge q)}{2(p\wedge q)}}+\varepsilon C_4(4K)^{\frac{1+(p\wedge q)}{% 2(p\wedge q)} }y_{xt}(1,t)h(y_{xt}(1,t)). \label{e3.5}$$ Setting $\theta _i=C_{4-i}((4K)^{\frac{1+(p\wedge q)}{2(p\wedge q)}% }+K$ $(\widetilde{E}(0)) ^{\frac{1-(p\wedge q)}{2(p\wedge q)}})$ for $i=0$ and $i=2$ and combining \eqref{e3.4} and \eqref{e3.5}, one obtains $$D_{2}\leq (\varepsilon /4)[ \widetilde{E}(t)] ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}+\varepsilon \theta _0y_{xt}(1,t)h(y_{xt}(1,t)). \label{e3.6}$$ Similarly we can show that $$D_{1}\leq (\varepsilon /4)[ \widetilde{E}(t)] ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}+\varepsilon \theta _2y_t(1,t)g(y_t(1,t)) \label{e3.7}$$ Inserting \eqref{e1.5}, \eqref{e3.6} and \eqref{e3.7} into \eqref{e3.3}, it follows $\widetilde{E}_\varepsilon '(t)\leq -\frac \varepsilon 2[ \widetilde{E}(t)] ^{\frac{1+(p\wedge q)}{2(p\wedge q)}}+C(\varepsilon )(y_t(1,t)g(y_t(1,t))+y_{xt}(1,t)h(y_{xt}(1,t))),$ where $C(\varepsilon )=\varepsilon (\theta _0+\theta _2)-1+\varepsilon K_0\frac{1-(p\wedge q)}{2(p\wedge q)} (\widetilde{E}% (0)) ^{\frac{1-(p\wedge q)}{2(p\wedge q)}}$. Using \eqref{e2.4}--\eqref{e2.5} with $1/p$ instead of $p$, we can show that $$\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t)^{-\frac{2(p\wedge q)}{ 1-(p\wedge q)}}, \label{e3.8}$$ where $\omega =\varepsilon \frac{1-(p\wedge q)}{4(p\wedge q)} M^{-\frac 1{1+(p\wedge q)}} (\widetilde{E}(0)) ^{\frac{1-(p\wedge q)}{2(p\wedge q)}}$ and $\varepsilon$ is chosen such that \begin{align*} \varepsilon &\leq \min\Big(((\theta _0+\theta _2)+K_0 \frac{1-(p\wedge q)}{2(p\wedge q)}(\widetilde{E}(0)) ^{\frac{1-(p\wedge q)}{2(p\wedge q)}})^{-1}, \\ &\quad K_0^{-1}( \widetilde{E}(0)) ^{\frac{(p\wedge q)-1}{2(p\wedge q)}}(1-M^{-\frac{(p\wedge q)}{(p\wedge q+1)}})\Big). \end{align*} (ii) First, by the conditions \eqref{e1.9} and \eqref{e2.2}, we have $$\rho '(t)\leq -\widetilde{E}(t)+Ay_{xt}^2(1,t)+Bg^2(y_t(1,t)), \label{e3.9}$$ with $A=K_1+K_2C_4^2$ and $B=(K_1/C_1^2)+K_2$. Next, given $\varepsilon >0$, we introduce the perturbed energy by $$\widetilde{E}_\varepsilon (t)=\widetilde{E}(t)+\varepsilon \rho (t)[ \widetilde{E}(t)] ^{\frac{( \frac 1p\vee q) -1}2}. \label{e3.10}$$ Then, for any $M>1$, we have \eqref{e2.4} and \eqref{e2.5} with $1/p$ instead of $p$. We have also \eqref{e2.6} with $1/p$ instead of $p$. Plugging \eqref{e2.1} and \eqref{e3.9} into \eqref{e2.4} one obtains $$\widetilde{E}_\varepsilon '(t) \leq (-1 +\varepsilon K_0\frac{( \frac 1p\vee q) -1} 2[ \widetilde{E}(0)] ^{^{\frac{( \frac 1p\vee q) -1}2}})(-\widetilde{E}'(t)) +E_1+E_2 - \varepsilon [ \widetilde{E}(t)] ^{^{\frac{( \frac 1p\vee q) +1}2}} \label{e3.11}$$ where $E_1 = \varepsilon A( \widetilde{E}(t)) ^{\frac{( \frac 1p\vee q) -1}2}y_{xt}^2(1,t),\quad E_2 = \varepsilon B (\widetilde{E}(t)) ^{\frac{( \frac 1p\vee q) -1} 2}g^2(y_t(1,t))).$ If $y_{xt}^2(1,t)>1$, it follows from hypothesis \eqref{e1.2}, \eqref{e1.3} and \eqref{e1.9}, that $$E_1 \leq (A/C_3)\varepsilon [ \widetilde{E}(0)] ^{\frac{( \frac 1p\vee q) -1}2}y_{xt}(1,t)h(y_{xt}(1,t)) \label{e3.12}$$ However, while $y_{xt}^2(1,t)\leq 1$, we have $min(| y_{xt}(1,t)| ,| y_{xt}(1,t)|^q)=| y_{xt}(1,t)| ^q$, $| y_{xt}(1,t)| ^{1+( \frac 1p\vee q) }\leq | y_{xt}(1,t)| ^{q+1} \leq (1/C_3)y_{xt}(1,t)h(y_{xt}(1,t)),$ and by Young's inequality, for any $\delta >0$, we have $$E_1 \leq \varepsilon /\beta 'C_3)(A\delta )^{\beta '}y_{xt}(1,t)h(y_{xt}(1,t))+(\varepsilon /\beta )\delta ^{-\beta }[ \widetilde{E}(t)] ^{^{\frac{( \frac 1p\vee q) +1}2}}. \label{e3.13}$$ Combining \eqref{e3.11} and \eqref{e3.12}, one has $$E_1 \leq (\varepsilon /\beta )\delta ^{-\beta }[ \widetilde{E} (t)] ^{^{\frac{( \frac 1p\vee q) +1}2}}+ \varepsilon A(\delta )\;y_{xt}(1,t)h(y_{xt}(1,t)), \label{e3.14}$$ with $A(\delta )=(A/C_3)[ \widetilde{E}(0)] ^{\frac{ ( \frac 1p\vee q) -1}2}+(1/\beta 'C_3)(A\delta )^{\beta '}$. If $y_t^2(1,t)\geq 1$, it follows from hypothesis \eqref{e1.2}, \eqref{e1.3} and \eqref{e1.9}, that $$E_2 \leq \varepsilon BC_2[ \widetilde{E}(0)] ^{^{\frac{( \frac 1p\vee q) -1}2}}y_t(1,t)g(y_t(1,t)). \label{e3.15}$$ However, while $y_t^2(1,t)<1$, we have max($| y_t(1,t)| ,| y_t(1,t)| ^p)=| y_{t}(1,t)| ^p$, $| g(y_t(1,t))| ^{1+( \frac 1p\vee q) }\leq (C_2)^{( \frac 1p\vee q) }y_t(1,t)g(y_t(1,t))$ and by Young's inequality, we can deduce that $$E_2 \leq \varepsilon (4B)^{\frac{( \frac 1p\vee q) -1} 2}(C_2)^{\frac 1p\vee q}y_t(1,t)g(y_t(1,t)) +\varepsilon 4^{-\beta }[ \widetilde{E}(t)] ^{\frac{( \frac 1p\vee q) +1}2}. \label{e3.16}$$ Combining \eqref{e3.14} and \eqref{e3.15}, we obtain $$E_2 \leq \varepsilon (C_2B[ \widetilde{E} (0)] ^{\frac{( \frac 1p\vee q) -1}2}+ (4B)^{\frac{( \frac 1p\vee q) -1} 2}(C_2)^{\frac 1p\vee q})y_t(1,t)g(y_t(1,t)))+\frac \varepsilon 4[ \widetilde{E}(t)] ^{\frac{^{(\frac 1p\vee q)+1}}2}. \label{e3.17}$$ Inserting \eqref{e3.13} and \eqref{e3.16} into \eqref{e2.6}, one obtains \begin{aligned} \widetilde{E}_\varepsilon '(t) &\leq -\varepsilon (1-\frac14-(\delta ^{-\beta }/\beta )) [ \widetilde{E}(t)] ^{\frac{^{(\frac 1p\vee q)+1}}2} +(-1+\varepsilon \sigma_1)y_t(1,t)g(y_t(1,t))\\ &\quad +(-1+\varepsilon \sigma _2)y_{xt}(1,t)h(y_{xt}(1,t)) \end{aligned}\label{e3.18} where $\sigma _1=(C_2B+K_0\frac{( \frac 1p\vee q) -1} 2)[ \widetilde{E}(0)] ^{\frac{^{(\frac 1p\vee q)-1}} 2}+(4B)^{\beta '}(C_2)^{\frac 1p\vee q})$ and $\sigma _2=A(\delta ) + K_0\frac{( \frac 1p\vee q) -1}2[ \widetilde{E}(0)] ^{\frac{^{(\frac 1p\vee q)-1}}2}.$ By choosing $\delta$ so that $\frac 14+(\delta ^{-\beta }/\beta )=\frac 12$, and $\varepsilon \leq \min (\sigma _1^{-1},\sigma _2^{-1}, K_0^{-1}[ \widetilde{E}% (0)] ^{^{\frac{1-(\frac 1p\vee q) }2}}(1-M^{\frac{-1}{(\frac 1p\vee q)+1}})).$ This with \eqref{e3.17} and using \eqref{e2.4} with $1/p$ instead of $p$, we obtain \begin{equation*} \widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega t)^{-\frac{^2}{(\frac 1p\vee q)-1}}, %\label{e3.19} \end{equation*} with $\omega =\frac \varepsilon 4((\frac 1p\vee q)-1)M^{-\frac{% ^{\frac 1p\vee q}}{^{(\frac 1p\vee q)+1}}}[ \widetilde{E}(0)] ^{% \frac{^{(\frac 1p\vee q)-1}}2}$. The proof of theorem \ref{thm2} is thus complete. \section{Proof of Theorem \ref{thm3}} (i) It follows from \eqref{e1.7} that \eqref{e1.7} is also true when $p$ is replaced by $p\vee q$ and $q$ is replaced by $p\vee q$. So we get the desired estimate, by applying Rao's theorem 1.1 (ii) (see \cite{r1}). \noindent (ii) \eqref{e1.8} implies that \eqref{e1.8} is also true when $p$ is replaced by $p\wedge q$ and $q$ is replaced by $p\wedge q$. This is exactly the condition (1.28)-(1.29) with exponent $p \wedge q <1$ of Rao's theorem 1.2 (see \cite{r1}), consequently, the proof of this part is thus complete. \noindent (iii) We adopt the method used by Rao \cite{r1} but in our case we have two nonlinear feedback functions $g$ and $h$ satisfying more general growth conditions. We introduce the same functional defined in \cite{r1}, $$\rho (t)=\int_\Omega y_t(m.\nabla y)dx+C_0\int_\Omega y_t\varphi dx, \label{e4.1}$$ where $C_0$ is a positive constant and $\varphi$ is the solution of the problem \begin{gather*} \Delta ^2\varphi =0 \quad \text{in } \Omega , \\ \varphi =y,\quad \frac{\partial \varphi }{\partial \nu } =\frac{\partial y}{\partial \nu } \quad \text{on } \Gamma , \end{gather*} we verify that the following estimates hold $\int_\Omega \varphi ^2dx\leq \gamma ^2\int_\Gamma \big((y)^2+( \frac{\partial y}{\partial \nu })^2\big)d\Gamma ,\quad a(\varphi ,y)=a(\varphi ,\varphi )\geq 0,$ where $\gamma$ is a constant depending only on the domain $\Omega$. We can show that there exist positive constants $K_0, K_1, K_2$ are such that for any $t\geq 0$, \begin{gather} | \rho (t)| \leq K_0E(t),\quad \forall t\geq 0, \label{e4.2} \\ \rho '(t)\leq -E(t)+K_1\int_\Gamma ((y_t)^2+(\frac{ \partial y_t}{\partial \nu })^2)d\Gamma +K_2\int_\Gamma (g^2(y_t)+h^2(\frac{\partial y_t}{\partial \nu }))d\Gamma . \label{e4.3} \end{gather} This with \eqref{e1.9} implies $$\rho '(t)\leq -E(t)+A\int_\Gamma | \frac{\partial y_t% }{\partial \nu }| ^2d\Gamma +B\int_\Gamma g^2(y_t)d\Gamma , \label{e4.4}$$ where $A=K_1+KC_4^2$ and $B=(K_1/C_1^2)+K_2$. % Next, given $\varepsilon >0$, we denote \eqref{e4.5} the % perturbed energy \eqref{e3.10} without the sign. Then, for any $M>1$, % we denote \eqref{e4.6}, \eqref{e4.7} and \eqref{e4.8} the relationships % \eqref{e2.4}, \eqref{e2.5} and \eqref{e2.6} % with $p$ replaced by $1/p$, but % without the sign in the respective formulas. Plugging \eqref{e4.2} and \eqref{e4.4} into \eqref{e2.6} with $p$ replaced by $1/p$, one obtains $$E_\varepsilon '(t)\leq (-1+\varepsilon K_0\frac{(\frac 1p\vee q)-1}2[ E(0)] ^{\frac{( \frac 1p\vee q) -1}2})(-E '(t))-\varepsilon [ E(t)] ^{\frac{( \frac 1p\vee q) +1}2}+ L, \label{e4.9}$$ where $L = \varepsilon [ E(t)] ^{\frac{( \frac 1p\vee q) -1} 2}(A\int_\Gamma | \frac{\partial y_t}{\partial \nu }| ^2 d\Gamma +B\int_\Gamma g^2(y_t)d\Gamma ).$ >From \eqref{e1.2}, \eqref{e1.3} and \eqref{e1.9} we have \begin{gather} (g(s))^2 \leq C_2g(s)s\;\;\forall | s| \geq 1; \quad |g(s)| ^{( \frac 1p\vee q) +1}\leq (C_2)^{\frac 1p\vee q}g(s)s\;\;\forall | s| \leq 1; \label{e4.10} \\ | s| ^2\leq (1/C_3)h(s)s\;\;\forall | s| \geq 1;\quad | s| ^{( \frac 1p\vee q) +1}\leq (1/C_3)h(s)s\;\;\forall | s| \leq 1. \label{e4.11} \end{gather} Using \eqref{e4.10} and \eqref{e4.11}, we have $$L \leq Q+\varepsilon C[ E(0)] ^{\frac{( \frac 1p\vee q)-1} 2} \Big(\int_{\{ |\frac{\partial y_t}{\partial \nu }| \geq 1\} } h(\frac{\partial y_t}{\partial \nu }) \frac{\partial y_t}{\partial \nu }d\Gamma +\int_{\{ |y_t| \geq 1\} }g(y_t)y_td\Gamma \Big), \label{e4.12}$$ where $Q=\varepsilon [ E(t)] ^{\frac{(\frac1p\vee q) -1}2} \Big(A\int_{\{ | \frac{\partial y_t% }{\partial \nu }| \leq 1\} }(\frac{\partial y_t}{\partial \nu } )^2d\Gamma +B\int_{\{ | y_t| \leq 1\} }g^2(y_t)d\Gamma\Big)$ and $C=(A+B)(C_2+1/C_3)$, With the exponents $\beta =\frac{( \frac 1p\vee q) +1}{( \frac 1p\vee q) -1}$, $\beta '=\frac{( \frac 1p\vee q) +1}2$, applying Young's inequality to $Q$, it follows from H\^{o}lder's inequality and \eqref{e4.10}, \eqref{e4.11} that for any parameter $\delta >0$, \begin{aligned} Q &\leq 2\varepsilon \delta ^{-\beta }/\beta [ E(t)] ^{\frac{( \frac 1p\vee q) +1}2} +(\varepsilon /\beta ')| \Gamma | ^{\beta '/\beta }\\ &\times\Big[ (A\delta )^{\beta '}/C_3\int_{\{ | \frac{ \partial y_t}{\partial \nu }| \leq 1\} }h(\frac{\partial y_t}{ \partial \nu })\frac{\partial y_t}{\partial \nu }d\Gamma +(B\delta )^{\beta '}C_2{}^{( \frac 1p\vee q) }\int_{\{| y_t| \leq 1\} }g(y_t)y_td\Gamma \Big]. \end{aligned}\label{e4.13} Inserting \eqref{e4.10} into \eqref{e4.12} gives $$L \leq \varepsilon \eta (-E '(t)) + 2\varepsilon \delta ^{-\beta }/\beta [ E(t)] ^{\frac{( \frac 1p\vee q) +1}2} \label{e4.14}$$ where $\eta =C( E(0)) ^{\frac{( \frac 1p\vee q) -1}2}+(1/\beta ')| \Gamma | ^{\beta '/\beta }((A\delta )^{\beta '}/C_3+(B\delta )^{\beta '}(C_2)^{( \frac 1p\vee q) })$. Inserting \eqref{e4.14} into \eqref{e4.9}, it follows \begin{aligned} E_\varepsilon '(t) &\leq \Big(-1+\varepsilon (\eta +K_0\frac{(\frac 1p\vee q)-1}2(E(0)) ^{\frac{(\frac 1p\vee q)-1}2})\Big)(-E '(t))\\ &\quad -\varepsilon (1-2\delta ^{-\beta }/\beta )(E(t)) ^{\frac{( \frac 1p\vee q) +1} 2}. \end{aligned} \label{e4.15} By choosing $\delta$ so that $\delta =(4/\beta )^{(1/\beta )}$, and $\varepsilon$ as $\varepsilon \leq \min((\eta +K_0\frac{ (\frac 1p\vee q)-1}2( E(0)) ^{\frac{( \frac 1p\vee q) -1}2})^{-1},K_0^{-1}({ E}(0)) ^{^{\frac{1-( \frac 1p\vee q) }2}}(1-M^{\frac{-1}{ ( (\frac 1p\vee q)+1)}}) \; ).$ This with \eqref{e4.15} and using \eqref{e2.4} with $p$ replaced by $1/p$, we obtain $E(t)\leq ME(0)(1+\omega t)^{-\frac 2{( \frac 1p\vee q) -1}},$ with $$\omega =\frac \varepsilon 4((\frac 1p\vee q)-1)M^{-\frac{ \frac 1p\vee q}{( \frac 1p\vee q) +1}}(E(0)) ^{\frac{( \frac 1p\vee q) -1}2}.$$ The proof of theorem \ref{thm3} is thus complete. \begin{thebibliography}{9} \bibitem{c1} P. G. Ciarlet and P. Destuynder, \emph{A justification of a nonlinear model in plate theory}, Comp. Method appl. Mech. Engng. 17/18, (1979) pp. 227-258. \bibitem{c2} F. Conrad and M. Pierre, \emph{Stabilization of Euler-Bernoulli beam by nonlinear boundary feedback}, Rapport de Recherches, INRIA, (1990). \bibitem{c3} F. Conrad and B. Rao, \emph{Decay of solutions of the wave equation in a star-shaped domain with nonlinear boundary feedback} - Asymptotic Analysis 7, (1993) pp. 159-177. \bibitem{k1} V. Komornik and E. Zuazua, \emph{A direct method for the boundary stabilization of the wave equation}, J. Math. pures app. 69, (1990) pp. 33-54. \bibitem{l1} J. E. Lagnese, \emph{Boundary stabilization of thin Plates}, Siam Publications Philadelphia, Pennsylvania, (1989). \bibitem{l2} J. L. Lions, \emph{Contr\^{o}labilit\'{e} exacte, perturbations et stabilisation de syst\`{e}mes distribu\'{e}s}, Vol. 1, Masson, Paris, (1988). \bibitem{r1} B. Rao, \emph{Stabilisation of Kirchhoff Plate equation in a star-shaped domain by nonlinear boundary feedback}, Nonlinear Analysis, Theory Methods and Applications, Vol. 20, 6, (1993) pp. 605-626. \bibitem{s1} M. Slemrod, \emph{Feedback stabilization of a linear system in Hilbert space with an a priori bounded control}, Math. Control Signals Systems, 2, (1989) pp. 265-285. \bibitem{z1} E. Zuazua, \emph{Uniform stabilization of the wave equation by nonlinear boundary feedback}, Siam J. Control and Optimization, 28, (1989) pp. 265-268. \end{thebibliography} \end{document}