\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{equation}
\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}
\end{equation}
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
\begin{equation}
\widetilde{E}(t)=\frac 12\int_0^1 (y_t^2(x,t)+y_{xx}^2(x,t)
)dx. \label{e1.4}
\end{equation}
Formally,
\begin{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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
\begin{equation}
m(x).\nu (x)\geq \delta ^{-1}\quad \forall x\in \Gamma, \label{e1.11}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
\widetilde{E}_\varepsilon (t)=\widetilde{E}(t)+\varepsilon \rho
(t)\big[ \widetilde{E}(t)\big] ^{\frac{( p\vee q) -1}2}.
\label{e2.3}
\end{equation}
This together with the non-increasing of the energy $\widetilde{E}%
(t)$ implies that for any $M>1$
\begin{equation}
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}
\end{equation}
with
\begin{equation}
\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}
\end{equation}
Now, we calculate the derivative of the perturbed energy
$\widetilde{E}_\varepsilon (t)$
\begin{equation}
\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}
\end{equation}
on the other hand, from \eqref{e1.6}, \eqref{e1.7} and \eqref{e2.2} one
obtains
\begin{equation}
\rho '(t)\leq -\widetilde{E}(t)+K_3y_t^2(1,t)+K_4y_{xt}^2(1,t)
\label{e2.7}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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{equation}
\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}
\end{equation}
Combining \eqref{e2.9} and \eqref{e2.10}, one has
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
\widetilde{E}_\varepsilon '(t)\leq -\mu \varepsilon [
\widetilde{E}(t)] ^{\frac{( p\vee q) +1}2},
\label{e2.13}
\end{equation}
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
\begin{equation}
\widetilde{E}_\varepsilon '(t)\leq -\mu \varepsilon M^{\frac{-1}2}
[ \widetilde{E}_\varepsilon (t)] ^{\frac{( p\vee q)+1}2}.  \label{e2.14}
\end{equation}
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
\begin{equation}
\rho '(t)\leq -\widetilde{E}(t)+K (
g^2(y_t(1,t))+h^2(y_{xt}(1,t))) ,  \label{e3.1}
\end{equation}
with $K=K_2+K_1(1/C_1^2+1/C_2^2)$. Next, given
$\varepsilon >0$, we introduce the perturbed energy by
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
Similarly we can show that
\begin{equation}
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}
\end{equation}
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
\begin{equation}
\widetilde{E}(t)\leq M\widetilde{E}(0)(1+\omega
t)^{-\frac{2(p\wedge q)}{ 1-(p\wedge q)}},  \label{e3.8}
\end{equation}
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
\begin{equation}
\rho '(t)\leq -\widetilde{E}(t)+Ay_{xt}^2(1,t)+Bg^2(y_t(1,t)),
\label{e3.9}
\end{equation}
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
\begin{equation}
\widetilde{E}_\varepsilon (t)=\widetilde{E}(t)+\varepsilon \rho
(t)[ \widetilde{E}(t)] ^{\frac{( \frac 1p\vee q) -1}2}.
\label{e3.10}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
Combining \eqref{e3.11} and \eqref{e3.12}, one has
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
Combining \eqref{e3.14} and \eqref{e3.15}, we obtain
\begin{equation}
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}
\end{equation}
Inserting \eqref{e3.13} and \eqref{e3.16} into \eqref{e2.6}, one obtains
\begin{equation}
\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}
\end{equation}
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},
\begin{equation}
\rho (t)=\int_\Omega y_t(m.\nabla
y)dx+C_0\int_\Omega y_t\varphi dx,  \label{e4.1}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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{equation}
\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}
\end{equation}
Inserting \eqref{e4.10} into \eqref{e4.12} gives
\begin{equation}
L \leq \varepsilon \eta (-E '(t)) + 2\varepsilon \delta
^{-\beta }/\beta [ E(t)] ^{\frac{( \frac 1p\vee q) +1}2}  \label{e4.14}
\end{equation}
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{equation}
\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}
\end{equation}
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}