\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 14, 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  (login: ftp)}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/14\hfil Boundedness of solutions]
{Boundedness of solutions for a Li\'enard
 equation with multiple deviating arguments}

\author[Y. Yu, C. Zhao\hfil EJDE-2009/14\hfilneg]
{Yuehua Yu, Changhong Zhao}  % in alphabetical order

\address{Department of Mathematics, Hunan University of Arts
and Science, Changde, Hunan 415000, China} 
\email[Yuehua Yu]{jinli127@yahoo.com.cn} 
\email[Changhong Zhao]{hongchangzhao@yahoo.com.cn}

\thanks{Submitted December 15, 2008. Published January 13, 2009.}
\thanks{Supported by grants 06JJ2063 and 07JJ46001
from the Scientific Research Fund of \hfill\break\indent
Hunan Provincial Natural Science Foundation of China, and
08C616 from the Scientific \hfill\break\indent
Research Fund of Hunan Provincial Education Department of China}
\subjclass[2000]{34C25, 34K13, 34K25}
\keywords{Li\'enard equation; deviating argument; bounded solution}

\begin{abstract}
  We consider the  Li\'enard equation
  $$
  x''(t)+f_1 (x(t))  (x'(t))^{2}+f_2 (x(t))  x'(t)+g_0(x(t))
  +\sum_{j=1}^{m} g_{j}(x(t-\tau_{j}(t)))=p(t),
  $$
  where  $f_1$,   $f_2$,  $g_1 $ and  $g_2$ are continuous
  functions, the delays $\tau_j(t)\geq 0$  are bounded continuous,
  and $p(t)$ is a bounded continuous function.
  We obtain  sufficient conditions for all solutions and
  their derivatives to be bounded.
\end{abstract}

\maketitle 
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]

\section{Introduction}

 Consider the Li\'enard type equation with multiple deviating
  arguments
\begin{equation}
     x''(t)+f_1 (x(t))  (x'(t))^{2}+f_2 (x(t))  x'(t)+g_0(x(t))
      +\sum_{j=1}^{m} g_{j}(x(t-\tau_{j}(t)))=p(t),
\label{e1.1}
\end{equation}
 where  $f_1$,   $f_2$,  $g_1 $ and  $g_2$ are continuous
  functions on $R=(-\infty,+\infty)$, $\tau_j(t)\geq 0$,  $j= 1,2,\dots,m $
are bounded continuous  functions  on $R$, and $p(t)$ is a bounded
continuous function  on $R^+=[0,+\infty)$.
 Define
$$
a(x)=\exp\Big({\int_0^x} f_1(u)du\Big), \quad
  \varphi(x)={\int^x_0}a (u)[f_{ 2}(u)-a(u)]du, \quad
 y= a(x){\frac{dx}{dt}}+\varphi(x),
  %  \label{e1.2}
$$
 then we transform \eqref{e1.1} into the system
\begin{equation}
\begin{gathered}
  {\frac{dx(t)}{dt}}= \frac{1}{a(x(t))}[-\varphi(x(t))+y(t)], \\
  {\frac{dy(t)}{dt}}=a(x(t))\Big\{- y(t)-[g_{0}(x(t))
 - \varphi(x(t))]-\sum_{j=1}^{m} g_{j}(x(t-\tau_{j}(t)))+p(t)\Big\}.
\end{gathered}         \label{e1.3}
\end{equation}

In applied science some practical problems concerning physics,
mechanics and the engineering technique fields associated with
Li\'enard equation can be found in \cite{b1,h1,k2,y1}.
Hence, it has been the
object of intensive analysis by numerous authors. In particular,
there have been extensive results on boundedness of solutions of
Li\'enard equation with   delays in the literature. For example,
the authors in \cite{h2,k1,m1}
establish some sufficient conditions  to
ensure the boundedness for all solutions of \eqref{e1.1} without
delays; Zhang \cite{z1},  Liu and Huang \cite{l1} consider the boundedness
for all solutions of \eqref{e1.1} with constant delays;  We only
find that Liu and Huang \cite{l2} establish some sufficient conditions
to ensure the boundedness for all solutions of \eqref{e1.1} with a
deviating argument.  However, to the best of our knowledge, few
authors have considered boundedness of solutions of Li\'enard
equation with  multiple deviating arguments (See \cite{l2}). Thus, it is
worth while to continue to investigate the boundedness of
solutions of \eqref{e1.1} in this case.

A primary purpose of this paper is to study the boundedness of
solutions of \eqref{e1.3}. We will establish some sufficient
conditions  for all solutions of \eqref{e1.3} to be bounded. If
applying our results to \eqref{e1.1}, one will find that our
results are different from those in the references. An illustrative
example is given in the last section.


\section{Definitions and assumptions}

We assume that  $h=\max_{1\leq j\leq m}\{\sup_{t\in
\mathbb{R}}\tau_{j}(t)\}\geq 0$. Let $C([-h, 0], R)$ denote the
space of continuous functions $\phi:[-h,0]\to R$ with the supremum
norm $\|\cdot\|$. It is known in \cite{b1,h1,k2,y1} that for $g_1, g_2,
\varphi, \tau_{j}$ and $p$ continuous,
  given a continuous initial function $\phi\in C([-h, 0], R)$ and
  a number $y_0\in \mathbb{R}$, then there exists a solution of \eqref{e1.3} on an interval
  $[0, T)$ satisfying the initial condition and satisfying \eqref{e1.3} on
  $[0, T)$. If the solution remains bounded, then $T=+\infty$. We denote
  such a solution by $(x(t), y(t))=(x(t, \phi, y_0),  y(t, \phi, y_0))$ .

\noindent\textbf{Definition.}
Solutions of \eqref{e1.3} are uniformly bounded
  (UB) if for each $B_1>0$ there is a $B_2>0$ such that
$$
  (\phi, y_0)\in C([-h, 0], R)\times R \quad
  \text{and}\quad \|\phi\|+|y_0|\le B_1
$$
  imply that $|x(t, \phi, y_0)|+|y(t, \phi, y_0)|\le B_2$ for all
  $t\in \mathbb{R}^+$.

In this paper, we will  assume that the following  conditions:
\begin{itemize}
\item[(C1)]
 There exists a constant  $\underline{d}>1$  such that
$\underline{d} |u  |\leq \mathop{\rm sign}(u ) \varphi(u )$, for  all
$u \in \mathbb{R}$.
\item[(C2)] For $ j=0,  1,  2,  \dots, m$,
there exist nonnegative constants  $L_{j}$  and $q_{j}$ such that
for all $u  \in \mathbb{R}$,
\begin{gather*}
  \sum_{j=0}^{m}L_{j}<1, |(g_{0}(u)- \varphi(u)) |
   \leq L_{0} |u  |+q_{0}, \\ | g_1(u)  | \leq L_1 |u   | +q_1,  \dots,  | g_{m}(u)  | \leq L_{m} |u   |
   +q_{m}.
\end{gather*}
\end{itemize}

\section{Main result}

\begin{theorem} \label{thm3.1}
  Suppose that {\rm (C1), (C2)}  hold. Then solutions of \eqref{e1.3}
  are uniformly bounded.
\end{theorem}

\begin{proof}
 Let $(x(t), y(t))=(x(t, \phi, y_0),  y(t, \phi, y_0))$ be a
  solution of \eqref{e1.3} defined on $[0, T)$. We may assume that $T=+\infty$
  since the estimates which follow give an a priori bound on $(x(t),
  y(t))$.

Calculating the upper right derivative of $  |x (s)|$  and $  |y
(s)|$ along \eqref{e1.3}, in view of  (C1)  and (C2), we
have
\begin{equation}
D^+( |x (s)|)|_{s=t}
= \mathop{\rm sign} (x(t))\{\frac{1}{a(x(t))}[-\varphi(x(t))+y(t)]\}
\leq \frac{1}{a(x(t))}[ -\underline{d}  |x(t)|+|y(t)|] ,
\label{e3.1}
\end{equation}
and
\begin{equation}
\begin{aligned}
&D^+( |y (s)|)|_{s=t}\\
& =    \mathop{\rm sign} (y(t))a(x(t))\{-y(t)-[g_{0}(x(t))- \varphi(x(t))]
 -\sum_{j=1}^{m} g_{j}(x(t-\tau_{j}(t)))+p(t)\}\\
& \leq  a(x(t))\{-  |y(t)|+L_{0}|x(t) |+\sum_{j=1}^{m} L_{j}|x(t-\tau_{j}(t))|
+\sum_{j=0}^{m} q_{j}+|p(t)|\} .
\end{aligned}     \label{e3.2}
\end{equation}
 Let
\begin{equation}
M(t)=\max_{-h\leq s\leq t}\{\max\{|x (s)|, \ |y (s)| \}\},
\label{e3.3}
\end{equation}
 where $y(s)=y(0)$, for all $-h\leq s\leq 0$. It is
obvious that $\max\{|x (t)|, \ |y (t)| \}\leq M(t)$, and $M(t)$ is
non-decreasing, for $t\geq -h$.
Now, we consider two cases.

\noindent\textbf{Case (i):}
\begin{equation}
M(t)>  \max\{|x (t)|,\, |y (t)| \}\quad\text{for  all }t\geq 0\,.
\label{e3.4}
\end{equation}
We claim that
\begin{equation}
M(t)\equiv M(0), \quad\text{a   constant  for  all }
   t\geq 0. \label{e3.5}
\end{equation}
   Assume, by way of contradiction, that \eqref{e3.5} does not hold.
Then, there exists
$t_1>0$ such that $M(t_1)> M(0)$. Since
$$
\max\{|x (t)|,\, |y (t)| \}\leq M(0)
 \quad \text{ for  all  } -h\leq t\leq 0.
$$
There must   exist $\beta \in (0, \ t_1)$ such that
$$
\max\{|x (\beta)|,\, |y (\beta)| \}= M(t_1)\geq  M(\beta),
$$
which contradicts  $\eqref{e3.4}$. This contradiction implies
that \eqref{e3.5} holds. It follows that
\begin{equation}
\max\{|x (t)|, \ |y (t)| \} \leq M(t)= M(0)\quad
\text{for  all  } t\geq 0. \label{e3.6}
\end{equation}

\noindent \textbf{Case (ii):}
There is a point $t_{0}\geq 0$ such that
$M(t_{0})= \max\{|x (t_{0})|, \ |y (t_{0})| \}$.
Let
$$
\eta=\min\{\underline{d}-1,\,
1- \sum_{j=0}^{m}L_{j}\},\,
\theta= \sum_{j=0}^{m}q_{j} +\sup_{t \in \mathbb{R}^{+}}|p(t)|+1
$$
Then, if $M(t_{0})= \max\{|x (t_{0})|,\,
|y (t_{0})| \}=|x (t_{0})|$, in view of \eqref{e3.1}, we
have
\begin{equation}
\begin{aligned}
D^+( |x (s)|)|_{s=t_{0}}
&\leq  \frac{1}{a(x(t_{0}))}[-\underline{d}  |x(t_{0})|+|y(t_{0})| ] \\
&\leq  \frac{1}{a(x(t_{0}))} (-\underline{d}+1)M(t_{0})\\
&< \frac{1}{a(x(t_{0}))}[ -\eta M(t_{0})+\theta].
\end{aligned} \label{e3.7}
\end{equation}
If $M(t_{0})= \max\{|x (t_{0})|, \ |y (t_{0})| \}=|y (t_{0})|$, in
view of \eqref{e3.2}, we obtain
\begin{equation}
\begin{aligned}
&D^+( |y (s)|)|_{s=t_{0}}\\
&\leq  a(x(t_{0}))\{-  |y(t_{0})|+L_{0}|x(t_{0}) |+
  \sum_{j=1}^{m} L_{j}|x(t_{0}-\tau_{j}(t_{0}))|
 +\sum_{j=0}^{m} q_{j}+|p(t_{0})|\} \\
& <  a(x(t_{0}))[(-1+\sum_{j=0}^{m} L_{j})M(t_{0})+\theta ]\\
&\leq  a(x(t_{0}))[ -\eta M(t_{0})+\theta].
 \end{aligned} \label{e3.8}
\end{equation}
In addition, if $M(t_{0})\geq  \frac{\theta}{\eta}$,
it follows from \eqref{e3.7} and \eqref{e3.8} that  $M(t )$
is strictly decreasing in a small neighborhood
$(t_{0}, t_{0}+\delta_{0})$.  This contradicts that $M(t)$
 is non-decreasing.
   Hence,
\begin{equation}
\max\{|x (t_{0})|, \, |y (t_{0})| \}=M(t_{0})<  \frac{\theta}{\eta}.
\label{e3.9}
\end{equation}
 For $t>t_{0}$,  by the same approach
 used in the  proof of \eqref{e3.9}, we have
\begin{equation}
\max\{|x (t)|,\, |y (t)| \}<  \frac{\theta}{\eta} , \quad
 \text{if  }  M(t )=  \max\{|x (t)|,\, |y (t)| \}. \label{e3.10}
\end{equation}
On the other hand, if $M(t )>  \max\{|x (t)|,\, |y (t)| \}$, $t>t_{0}$.
We can choose $t_{0}\leq  t_2<t$ such that
for all $ s\in (t_2, t]$,
$$
M(t_2 )=  \max\{|x (t_2)|, \, |y (t_2)| \} < \frac{\theta}{\eta}, \quad
 M(s)>  \max\{|x (s)|, \, |y (s)| \}\,.
$$
Using  a similar argument as in the proof of Case (i),
we can show  that
$$
M(s)\equiv M(t_2) \quad \text{a  constant, for  all  }
   s\in (t_2, t],
$$
which implies
$$
\max\{|x (t)|, \, |y (t)| \} <  M(t)=
M(t_2)= \max\{|x (t_2)|, \, |y (t_2)| \}<\frac{\theta}{\eta}.
$$
In summary, the solutions of \eqref{e1.3} are uniformly bounded.
The proof  is  complete.
\end{proof}

\section{An example}

Consider the  Li\'enard equation with two deviating
arguments
\begin{equation}
\begin{aligned}
 &x''(t)+(x'(t))^{2}+[e^{-x(t)}(3x^{2}(t)+2)+e^{ x(t)}]  x'(t)
+ \frac{1}{2}\sin x(t)+x^{3}(t)+2x(t) \\
 &+\frac{1}{6} |x(t-|\sin t|)|
  +\frac{1}{6}\arctan x(t-\frac{1}{1+t^{2}})
=e^{\frac{1}{t^{2}+1}},
\end{aligned} \label{e4.1}
\end{equation}
All solutions and their derivatives are bounded.
Set
\begin{equation}
a(x)=e^{ x }, \quad
\varphi(x)={\int^x_0}(3u^{2}+2)du, \quad
y=  e^{ x }{\frac{dx}{dt}}+x^{3}+2x,
    \label{e4.2}
\end{equation}
then we can transform \eqref{e1.1} into the  system
\begin{equation}
\begin{gathered}
  {\frac{dx(t)}{dt}}= e^{-x(t)}[-(x^{3}(t)+2x(t))+y(t)], \\
\begin{aligned}
 {\frac{dy(t)}{dt}}
&=e^{ x(t)}\big[- y(t)-\frac{1}{2}\sin x(t)
  -\frac{1}{6} |x(t-|\sin t|)|\\
 &\quad -\frac{1}{6}\arctan x(t-\frac{1}{1+t^{2}})+e^{\frac{1}{t^{2}+1}}\big].
\end{aligned}
\end{gathered} \label{e4.3}
\end{equation}
  It is straight forward to check that all assumptions needed in
Theorem 3.1   are satisfied. Therefore, solutions of
system \eqref{e4.3} are uniformly bounded. This
implies that all solutions of \eqref{e4.1} and their derivatives are bounded.
\smallskip

\noindent\textbf{Remark.}   Equation \eqref{e4.1} is a very simple
Li\'enard  equation  with  two deviating   arguments.
Li\'enard equations with  constant delays have been studied in
\cite{h2,k1,l1,m1,v1,z1}, and with one deviating argument in \cite{l2}.
It is also clear that the results obtained in \cite{h2,k1,l1,l2,m1,v1}
can not be applied to  \eqref{e4.1}.
Since we proved  boundedness of
solutions to Li\'enard equation by a different method,
the results in this article are essentially new.

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\end{document}