\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 97, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/97\hfil Uniform boundedness of solutions]
{Uniform boundedness of solutions for a class of
Li\'enard equations}

\author[G.-R. Ye, H.-S. Ding, X.-L. Wu\hfil EJDE-2009/97\hfilneg]
{Guo-Rong Ye, Hui-Sheng Ding, Xi-Lang Wu}  

\address{College of Mathematics and Information Science,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email[G.-R. Ye]{yeguorong2006@sina.com}
\email[H.-S. Ding]{dinghs@mail.ustc.edu.cn}
\email[X.-L. Wu]{wuxilang99@sina.com}


\thanks{Submitted May 15, 2009. Published August 11, 2009.}
\subjclass[2000]{34K25}
\keywords{Li\'enard equation; boundedness of solutions}

\begin{abstract}
 In this article, we study a class of  Li\'{e}nard equations
 $$
 x''(t)+f(x(t))x'(t)+g_1(x(t))+g_2(x(t-\tau(t))=e(t).
 $$
 Under some suitable conditions, we ensure that all solutions
 of the above Li\'{e}nard equations are uniformly bounded.
 Our assumptions are less restrictive than those in \cite{Liu03};
 thus we extend some previous results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

As it is we all know, Li\'{e}nard equations appears in a number of
physical models and is important in describing fluid mechanical and
nonlinear elastic mechanical phenomena. Thus, there has been
great interest for many mathematicians to study the dynamical
behavior of all kinds of Li\'{e}nard equations (cf.
\cite{Bereanu,cac,Fonda,Huang,Liu03,Liu,toni,Villari,Cieutat04,Cieutat05}
and references therein). Especially, several authors have
contributed to the study on boundedness of solutions to Li\'{e}nard
equations (cf. \cite{Fonda,Huang,Liu,Liu03,Villari} and references
therein). For example, in 1998, the authors in \cite{Fonda}
discussed the bounded solutions of the Li\'{e}nard equation
\begin{equation*}
x''(t)+f(x)x'+g(x) =e(t).
\end{equation*}
Recently, the authors in \cite{Liu} studied the boundedness of
solutions to the following Li\'{e}nard equation with a deviating
argument:
\begin{equation}\label{eq}
x''(t)+f(x(t))x'(t)+g_1(x(t))+ g_2(x(t-\tau(t)))=e(t),
\end{equation}
where $f$, $g_1$ and $g_2$ are
continuous functions on $\mathbb{R}$, $\tau(t) \geq 0 $ is a bounded
continuous function on $\mathbb{R}$, and $e(t)$ is a bounded
continuous function on $\mathbb{R}^+ =[0,+\infty)$.

The authors in \cite{Liu} established a theorem which ensure that
all solutions of \eqref{eq} are uniformly bounded,
under the following two assumptions:
\begin{itemize}
\item[(C1)] There exists a constant $d>1$ such that
$d|u|\leq \mathop{\rm sgn}(u)\varphi(u)$ for all $u\in\mathbb{R}$, where
$$
\varphi(u)=\int^u_0[f(x)-1]dx.
$$

\item[(C2)] There exist nonnegative constants $L_1,L_2,q_1,q_2$ such
that $L_1+L_2<1$ and
$$
|g_1(u)-\varphi(u)|\leq L_1|u|+q_1,\quad |g_2(u)|\leq L_2|u|
+q_2,\quad \forall u\in\mathbb{R}.
$$
\end{itemize}

In this article, we will make further study on this problem. As one
will see, under weaker assumptions than (C1) and (C2), we also get
the same conclusion to \cite{Liu}. Next, let us recall some
notations and basic results.


Throughout this paper, we denote
\begin{equation*}
\varphi(x)=\int_0^x [f(u)-1]du,  \quad  y=\frac{dx}{dt} +\varphi(x).
\end{equation*}
Then  \eqref{eq} is transformed into the system
\begin{equation}\label{eq2}
 \begin{gathered}
 \frac{dx(t)}{dt} = -\varphi(x(t))+y(t), \\
 \frac{dy(t)}{dt} = -y(t)-[g_1(x(t))-\varphi (x(t))]-
g_2(x(t-\tau(t))+e(t).
\end{gathered}
 \end{equation}

Let $h = \sup_{t \in \mathbb{R}} \tau(t) \geq 0 $.
$C([-h,0],\mathbb{R})$ denotes the space of continuous functions
$\phi:[-h,0]\rightarrow \mathbb{R}$ with the supremum norm
$\|\cdot\|$. It is well known (cf. \cite{Burton,Hale}) that for any
given continuous initial function $\phi \in C([-h,0],\mathbb{R})$
and a number $y_0$, there exists a solution of \eqref{eq2} on an
interval $[0,T)$ satisfying the initial conditions and \eqref{eq2}
on $[0,T)$. If the solution remains bounded, then $T =+\infty$. We
denote such a solution by $x(t) = x(t,\phi, y_0)$,
$y(t)=y(t,\phi,y_0)$.

\begin{definition}[\cite{Liu}] \rm
 Solutions of \eqref{eq2} are called uniformly bounded if for each
$B_1 >0$ there is a $B_2 >0$ such that
$(\phi, y_0) \in C([-h,0], \mathbb{R})\times \mathbb{R}$ and
$\|\phi\|+|y_0| \leq B_1$ implies that
$|x(t,\phi,y_0)|+ |y(t,\phi,y_0)| \leq B_2$ for
all $t \in \mathbb{R}^+$.
\end{definition}

\section{Main results}

For our convenience, we list the following assumptions:
\begin{itemize}
\item[(A1)]
$|u|< \mathop{\rm sgn}(u)\varphi(u)$ for all $u\in\mathbb{R}$.

\item[(A2)]
There exist two nondecreasing functions $G,\Phi$ defined on
$\mathbb{R}^+$ such that
\begin{gather*}
|g_1(u)-\varphi(u)|\leq \Phi(|u|),\quad
|g_2(u)|\leq G(|u|),\quad \forall u\in\mathbb{R},
\\
\limsup_{x\to +\infty}[\Phi(x)+G(x)-x+\overline{e}]<0,\quad
\overline{e}=\sup_{t\in\mathbb{R}^+}|e(t)|.
\end{gather*}
\end{itemize}

\begin{theorem}\label{thm1}
Suppose that {\rm (A1), (A2)} hold. Then solutions of \eqref{eq2}
are uniformly bounded.
\end{theorem}

\begin{proof}
 Let $x(t)=x(t,\phi,y_0)$, $y(t)=y(t,\phi,y_0)$ be a solution of
\eqref{eq2}. Calculating the upper right derivatives of $ |x(s)|$
and $ |y(s)|$, in view of  (A1) and (A2), we have
\begin{align*}
\ D^+(|x(s)|)|_{s=t}
&=  \mathop{\rm sgn}(x(t))\{ -\varphi (x(t))+y(t)\} \\
&<  -|x(t)|+|y(t)|,\\
D^+(|y(s)|)|_{s=t}&=  \mathop{\rm sgn}(y(t)) \{-y(t)-[g_1(x(t))-\varphi
(x(t))]-g_2(x(t-\tau(t))+e(t)\} \\
&\leq -|y(t)|+\Phi(|x(t)|)+G(|x(t-\tau(t))|)+\overline{e}.
\end{align*}
Let
$$
M(t)=\max_{ -h \leq s\leq t}\{\max \{|x(s)|,|y(s)|\}\},
\quad   t \geq 0.
$$
By (A2), there is a constant $M>0$ such that
\begin{equation}\label{m}
\Phi(x)+G(x)-x+\overline{e}<0,\quad x\geq M.
\end{equation}
For any given $ t_0\geq 0$, we consider five cases.

Case (i):  $M(t_0)> \max \{|x(t_0)|,|y(t_0)|\}$.
It follows from the continuity of $x(t)$ and $y(t)$ that
there exists $\delta_1> 0$ such that
$$
\max \{|x(t)|,|y(t)|\} < M(t_0), \quad  \forall t \in (t_0,t_0+\delta_1).
$$
Thus, one can conclude
$M(t)=M(t_0)$, for all $t \in (t_0,t_0+\delta_1)$.

Case (ii): $M(t_0)=\max \{|x(t_0)|, |y(t_0)|\}<M $.
Also, by the continuity of $x(t)$ and $y(t)$, there exists
$\delta_2> 0$ such that
$$
\max \{|x(t)|,|y(t)|\} <M, \quad
 \forall t \in (t_0,t_0+\delta_2).
$$
Therefore,
$M(t)<M$, for all $t \in (t_0,t_0+\delta_2)$.

Case (iii): $M(t_0)=\max \{|x(t_0)|,|y(t_0)|\}=|x(t_0)| \geq M$, and
$|x(t_0)|>|y(t_0)|$.
Since
$$
\ D^+(|x(s)|)|_{s=t_0} < -|x(t_0)|+|y(t_0)|<0,
$$
there exists $\delta_3> 0$ such that
$$
|x(t)| < |x(t_0)| =M(t_0)\quad  \forall t \in (t_0,t_0+\delta_3).
$$
On the other hand, by the continuity of $y(t)$, without loss, one
can assume that
$$
|y(t)|< |x(t_0)|=M(t_0), \quad \forall t \in (t_0,t_0+\delta_3).
$$
So
$$
\max \{|x(t)|,|y(t)|\} < M(t_0), \quad
\forall t \in (t_0,t_0+\delta_3),
$$
which implies $M(t)=M(t_0)$, for all $t \in (t_0,t_0+\delta_3)$.

Case (iv): $M(t_0)=\max \{|x(t_0)|,|y(t_0)|\}=|y(t_0)| \geq M$, and
$|x(t_0)|<|y(t_0)|$.
By \eqref{m}, we have
\begin{align*}
D^+(|y(s)|)|_{s=t_0}
&\leq -|y(t_0)|+\Phi(|x(t_0)|)+G(|x(t_0-\tau(t_0))|)+\overline{e}\\
&\leq -M(t_0)+\Phi(M(t_0))+G(M(t_0))+\overline{e}<0,
\end{align*}
which yields that there exists
$\delta_4 > 0$ such that
$$
|y(t)| < |y(t_0)| =M(t_0),\quad
 \forall t \in (t_0,t_0+\delta_4).
$$
On the other hand, without loss of generality, one can assume that
$$
|x(t)|< |y(t_0)|=M(t_0), \quad \forall t \in (t_0,t_0+\delta_4).
$$
So one can conclude
$$
\max \{|x(t)|,|y(t)|\} < M(t_0), \quad  \forall t \in
(t_0,t_0+\delta_4).
$$
Thus $M(t)=M(t_0)$ for all $t \in (t_0,t_0+\delta_4)$.

Case (v): $M(t_0)=\max \{|x(t_0)|,|y(t_0)|\}=|x(t_0)| =|y(t_0)|\geq M$.
We have
$$
D^+(|x(s)|)|_{s=t_0} < -|x(t_0)|+|y(t_0)|=0.
$$
Also, similar to the proof of Case (iv), one can show that
$$
D^+(|y(s)|)|_{s=t_0} <0.
$$
Thus, there exists $\delta_5> 0$ such that
$$
|x(t)| < |x(t_0)| =M(t_0),\quad |y(t)|< |y(t_0) |
=M(t_0) \quad \forall t \in (t_0,t_0+\delta_5).
$$
Therefore, $M(t)=M(t_0)$ for all $t \in (t_0,t_0+\delta_5)$.
In summary, for each $t_0\geq 0$, there exists
$\delta > 0$ such that
$$
M(t)\leq \max \{M(t_0), M\},\quad \forall
t \in (t_0,t_0+\delta).
$$
Let
$$
\alpha=\begin{cases}
\inf\{t\geq 0:M(t)>\max\{ M(0),M \}\} &\\
\quad\text{if } \{t\geq 0:M(t)>\max\{ M(0),\ M \}\}\neq\emptyset,
\\
+\infty \\
\quad \text{if } \{t\geq 0:M(t)>\max\{ M(0),M \}\}=\emptyset.
\end{cases}
$$
We claim that $\alpha=+\infty $. If $\alpha<+\infty$, then
\begin{equation}\label{7}
M(t)\leq \max\{ M(0), M \},\quad \forall t\in [0,\alpha].
\end{equation}
It follows from the above proof that there
is a constant $\delta'>0$ such that
\begin{equation}\label{8}
M(t)\leq \max\{ M(\alpha), M \},\quad \forall t\in (\alpha,\alpha+\delta').
\end{equation}
Combing \eqref{7} and \eqref{8}, we have
$$
M(t)\leq \max\{ M(0), M \},\quad \forall t\in [0,\alpha+\delta'),
$$
which yields $\alpha\geq \alpha+\delta'$.
This is a contradiction. Thus, $\alpha=+\infty $, which implies
$$
M(t) \leq \max\{ M(0), M \},\quad \forall t\geq 0.
$$
Then, we have
$$
|x(t)| \leq \max\{ M(0), M \},\quad
|y(t)| \leq \max\{ M(0), M \},\quad \forall t\geq 0.
$$
Therefore, solutions of \eqref{eq2} are uniformly bounded.
\end{proof}

\begin{remark}\label{rmk1}\rm
One can easily conclude (A1) and (A2) from the assumptions (C1) and
(C2).
So Theorem \ref{thm1} is a generalization of
\cite[Theorem 3.1]{Liu}. In addition, our assumptions are
weaker than (C1) and (C2) in essence (see Remark \ref{rmk2}).
\end{remark}

Next, we give an example to illustrate our results.

\begin{example}\label{exam1}\rm
Consider the following Li\'{e}nard equation:
\begin{equation}\label{1}
x''(t)+f(x(t))x'(t)+g_1(x(t))+ g_2(x(t-\tau(t))=e(t),
\end{equation}
where
\begin{gather*}
f(x)=\frac{e^{-x}-xe^{-x}}{2}+2,\quad
g_1(x)=\frac{xe^{-x}+3x+x^{1/3}}{2},\\
g_2(x)=x^{1/3},\quad \tau(t)=\cos^2 t,\quad e(t)=\sin t.
\end{gather*}
Then
$$
\varphi(x)=\int_0^x[f(u)-1]du=\frac{1}{2}xe^{-x}+x,
$$
and
$$
\mathop{\rm sgn}(x)\varphi(x)=\big(\frac{1}{2}e^{-x}+1\big)|x|>|x|,\quad
\forall x\in\mathbb{R}.
$$
So (A1) holds. In addition, let
$$
\Phi(x)=\frac{x+x^{1/3}}{2},\quad G(x)=x^{1/3}.
$$
Then
$$
|g_1(u)-\varphi(u)|=\big|\frac{u+u^{1/3}}{2}\big|
\leq \Phi(|u|),\quad
|g_2(u)|=G(|u|),\quad \forall u\in\mathbb{R},
$$
and
\begin{gather*}
\limsup_{x\to +\infty}[\Phi(x)+G(x)-x+\overline{e}]=\limsup_{x\to
+\infty}\big[\frac{x+x^{1/3}}{2}+x^{1/3}-x+1\big]<0,
\\
\overline{e}=\sup_{t\in\mathbb{R}^+}|e(t)|=1.
\end{gather*}
So (A2) holds. Then Theorem \ref{thm1} shows that solutions
of  \eqref{1} are uniformly bounded.
\end{example}


\begin{remark}\label{rmk2}\rm
In the above example, there is no a constant $d>1$ such
that
$$
\mathop{\rm sgn}(x)\varphi(x)\geq d |x|,\quad \forall x\in\mathbb{R}.
$$
So (C1) does not hold. Thus, \cite[Theorem 3.1]{Liu} can not be
applied.
\end{remark}

\subsection*{Acknowledgments}
Hui-Sheng Ding acknowledges support from the NSF of China
(10826066), the NSF of Jiangxi Province of China (2008GQS0057),
the Youth Foundation of Jiangxi Provincial Education Department
(GJJ09456), and the Youth Foundation of Jiangxi Normal University.
Guo-Rong Ye and Xi-Lang Wu acknowledge support from the Graduate
Innovation Foundation of Jiangxi Normal University
(JXSD-Y-09045).


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\end{document}