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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 140, pp. 1--12.\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-2009/140\hfil Periodic solutions]
{Periodic solutions for a Li\'enard equation
 with two deviating arguments}

\author[Y. Wang, J. Tian  \hfil EJDE-2009/140\hfilneg]
{Yong Wang, Junkang Tian}  

\address{Yong Wang \newline
School of Sciences, Southwest Petroleum University, Chengdu, Sichuan
610500, China} 
\email{odinswang@yahoo.com.cn, wangyong@swpu.edu.cn}

\address{Junkang Tian \newline
School of Sciences, Southwest Petroleum University, Chengdu, Sichuan
610500, China} 
\email{tianjunkang1980@163.com}

\thanks{Submitted July 7, 2009. Published October 30, 2009.}
\thanks{Supported by the SWPU Science and Technology Fund of China 
 and by the Open \hfill\break\indent
 Fund of State Key Laboratory of Oil and Gas 
 Researvoir Geology and Exploitation  \hfill\break\indent
 (Southwest Petroleum University)}
\subjclass[2000]{34C25, 34D40} 
\keywords{Periodic solution; Li\'enard equation; deviating argument}

\begin{abstract}
 In this work, we prove the existence and uniqueness of periodic
 solutions for a Li\'enard equation with two deviating arguments.
 Our main tools are the Mawhin's continuation theorem and
 the Schwarz inequality. We obtain our results under weaker
 conditions than those in  \cite{z07jcam},  as shown by an example in
 the last section of this artticle.
\end{abstract}

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

\section{Introduction}

The Li\'enard equation can be derived from many fields, such
as physics, mechanics and engineering technique fields. An
important question is whether this equation can support periodic
solutions. In the past several years, the existence of periodic
solutions to Li\'enard equation has been widely discussed, notably
by  Li\'enard \cite{l28rge} and by Levinson and Smith \cite{l42jdm}.
Recently, Zhou and Long \cite{z07jcam} studied the existence and
uniqueness of periodic solutions of the following Li\'enard equation
with two deviating arguments
\begin{equation}\label{a1}
x''(t)+f(x(t))x'(t)+g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{2}(t)))=e(t),
\end{equation}
where $f,\tau_{1},\tau_{2},e \in C(\mathbb{R},\mathbb{R})$,
$g_{1},g_{2} \in C(\mathbb{R}^{2},\mathbb{R})$,
$\tau_{1}(t),\tau_{2}(t),g_{1}(t,x),g_{2}(t,x),e(t)$ are
periodic functions with period $T$, with respect to $t$,
$T$-periodic for short.

In recent years, there have been many publications on the existence of
periodic solutions of the Li\'enard equation of the type \eqref{a1}; see
for example
\cite{c04na,l07jcam,l07cam,c01na,l06na,l03amc,l04jmaa,w02na,wy}.
However, as far as we know, there are fewer results on the
existence and uniqueness of periodic solutions to \eqref{a1}.
Applying Mawhin's continuation theorem and some analysis techniques,
Zhou and Long \cite{z07jcam} provided a sufficient condition for
the existence and uniqueness of periodic solutions to
\eqref{a1}, but their results can be improved.

The main purpose of this paper is to provide a new
sufficient condition for guaranteeing the existence and uniqueness
of $T$-periodic solutions to \eqref{a1}, by using Mawhin's
continuation theorem and Schwarz inequality. Our results hold under
weaker conditions than those in \cite{z07jcam},
and that are verifiable as shown by an example in the last section.

\section{Preliminaries}

For convenience, we define
\begin{gather*}
|x|_{\infty}=\max_{t\in [0,T]}|x(t)| , \quad
|x'|_{\infty}=\max_{t\in [0,T]}|x'(t)|,\\
|x|_{k}=\Big(\int_{0}^{T}|x(t)|^{k}dt \Big)^{1/k}, \quad
\bar{e}=\frac{1}{T}\int_{0}^{T}e(t)dt
\end{gather*}
Let
\[
C_{T}^{1}:=\{x\in C^{1}(\mathbb{R}  ,\mathbb{R} ): x \textrm{ is
$T$-periodic}\},\quad
C_{T}:=\{x\in C(\mathbb{R}  ,\mathbb{R} ): x \textrm{ is
$T$-periodic}\},
\]
which are  Banach spaces with the norms
$$
\|x\|_{C_{T}^{1}}=\max\{|x|_{\infty},|x'|_{\infty}\},\quad
\|x\|_{C_{T}}=|x|_{\infty}.
$$
The following conditions will be used in this paper:
\begin{itemize}
\item[(H0)] There exist $C_1\geq 0$, $C_2\geq
0$, $b_1\geq 0$ and $b_2\geq 0$ such that
$|f(x_1)-f(x_2)|\leq C_1|x_1-x_2|$,
$|f(x)|\leq C_2$ and $|g_i(t,u)-g_i(t,v)|\leq b_i|u-v|$,
for all $x_1,x_2,x,t,u,v\in\mathbb{R}$, $i=1,2$.
\end{itemize}

The following Mawhin's continuation theorem is useful in obtaining
the existence of $T$-periodic solutions of \eqref{a1}.

\begin{lemma}[{\cite[p. 40]{gre}}] \label{lem1}
 Let $X$ and $Y$ be two Banach spaces. Suppose that
$L: D(L) \subset X \to Y$ is a Fredholm
operator with index zero and $N: X \to Y$ is L-compact on
$\overline{\Omega}$, where $\Omega$ is an open bounded subset of $X$.
Moreover, assume that all the following conditions are satisfied:
\begin{itemize}
\item[(i)] $Lx\neq \lambda Nx$, for all $x\in\partial{\Omega}\cap
 D(L), \lambda\in(0,1)$;

\item[(ii)]  $Nx \notin \mathop{\rm Im}L$, for all
$x\in\partial{\Omega}\cap \ker L$;

\item[(iii)]  the Brouwer degree $\deg \{JQN, \Omega\cap \ker L,
0\}\neq 0$,
 where $J: \mathop{\rm Im}Q \to \ker L$ is an isomorphism.
\end{itemize}
 Then equation $Lx=Nx$ has at least one solution on
$\overline{\Omega}\cap D(L)$.
\end{lemma}

\begin{lemma} \label{lem2}
 If $x\in C^{2}(\mathbb{R},\mathbb{R})$ with
$x(t+T)=x(t)$, then
$$
|x'|_{2}^{2}\leq \big(\frac{T}{2\pi}\big)^{2}|x''|_{2}^{2}.
$$
\end{lemma}

The proof of the above lemma  is a direct consequence of the Wirtinger
inequality; see for example \cite{hgh}.
Consider the homotopic equation of \eqref{a1}, for $\lambda\in(0,1)$,
\begin{equation}\label{b1}
x''(t)+\lambda f(x(t))x'(t)+\lambda
g_{1}(t,x(t-\tau_{1}(t)))+\lambda g_{2}(t,x(t-\tau_{2}(t)))=\lambda
e(t).
\end{equation}
We have the following lemma.

\begin{lemma} \label{lem3}  Suppose that the following conditions are
satisfied:
\begin{itemize}
\item[(H1)] one of the following conditions holds:
\begin{enumerate}
\item $(g_{i}(t,u)-g_{i}(t,v))(u-v)>0$  for
all $t,u,v\in\mathbb{R}$, $u\neq v$, $i=1,2$,
\item $(g_{i}(t,u)-g_{i}(t,v))(u-v)<0$ for
all $t,u,v\in\mathbb{R}$, $u\neq v$, $i=1,2$;
\end{enumerate}

\item[(H2)] there exists $d\geq 0$ such that one of the following
conditions holds:
\begin{enumerate}
\item $x(g_{1}(t,x)+g_{2}(t,x)-\bar{e})>0$,
for all $t\in\mathbb{R}$, $|x|> d$,
\item $x(g_{1}(t,x)+g_{2}(t,x)-\bar{e})<0$,
for all $t\in\mathbb{R}$, $|x|> d$;
\end{enumerate}
\end{itemize}
If $x(t)$ is a $T$-periodic solution of \eqref{b1}, then
\begin{equation}\label{b2}
|x|_{\infty}\leq d+\frac{\sqrt{T}}{2}|x'|_{2}.
\end{equation}
\end{lemma}

\begin{proof}
Let $x(t)$ be an arbitrary $T$-periodic solution of
\eqref{b1}. Then, integrating \eqref{b1} from 0 to $T$, we have
\begin{equation}\label{b3}
\int_{0}^{T}[g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{2}(t)))-e(t)]dt=0,
\end{equation}
which implies that there exists $t_{1}\in\mathbb{R}$ such that
\begin{equation}\label{b6}
g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))+g_{2}(t_{1},
x(t_{1}-\tau_{2}(t_{1})))-\bar{e}=0.
\end{equation}
Now we show the following statement.

\noindent{\bf Claim.}  If $x(t)$ is a $T$-periodic solution of
\eqref{b1}, then there exists $t_{2}\in\mathbb{R}$ such that
\begin{equation}\label{b7}
|x(t_{2})|\leq d.
\end{equation}

Assume, by way of contradiction, that \eqref{b7} does not hold. Then
\begin{equation}\label{b8}
|x(t)|> d\quad \textrm{for all}\,\,t\in\mathbb{R},
\end{equation}
which, together with (H1), (H2) and \eqref{b6},
imply that one of the following four relations holds:
\begin{gather}
x(t_1-\tau_1(t_1))>x(t_1-\tau_2(t_1))>d, \label{e2.7}\\
x(t_1-\tau_2(t_1))>x(t_1-\tau_1(t_1))>d, \label{e2.8} \\
x(t_1-\tau_1(t_1))<x(t_1-\tau_2(t_1))<-d, \label{e2.9} \\
x(t_1-\tau_2(t_1))<x(t_1-\tau_1(t_1))<-d. \label{e2.10}
\end{gather}
Suppose that \eqref{e2.7} holds, in view of (H1)(1),
(H1)(2), (H2)(1) and (H2)(2), we  consider
fours cases as follows:

{\sl Case} (i): If (H1)(1) and (H2)(1) hold,
according to \eqref{e2.7}, we have
\begin{align*}
0&< g_{1}(t_{1},x(t_{1}-\tau_{2}(t_{1})))
   +g_{2}(t_{1},x(t_{1} -\tau_{2}(t_{1})))-\bar{e}\\
&< g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
  +g_{2}(t_{1},x(t_{1} -\tau_{2}(t_{1})))-\bar{e},
\end{align*}
which contradicts  \eqref{b6}. Thus \eqref{b7} is true.

{\sl Case} (ii): If (H1)(2) and (H2)(1) hold,
according to \eqref{e2.7}, we have
\begin{align*}
0&< g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
   +g_{2}(t_{1},x(t_{1} -\tau_{1}(t_{1})))-\bar{e}\\
&< g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
   +g_{2}(t_{1},x(t_{1} -\tau_{2}(t_{1})))-\bar{e},
\end{align*}
which contradicts  \eqref{b6}. Thus
\eqref{b7} is true.

{\sl Case} (iii): If (H1)(1) and (H2)(2) hold,
according to \eqref{e2.7}, we have
\begin{align*}
0&> g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
   +g_{2}(t_{1}, x(t_{1}-\tau_{1}(t_{1})))-\bar{e}\\
&> g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
  +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-\bar{e},
\end{align*}
which contradicts  \eqref{b6}. Thus
\eqref{b7} is true.

{\sl Case} (iv): If (H1)(2) and (H2)(2) hold,
according to \eqref{e2.7}, we have
\begin{align*}
0&> g_{1}(t_{1},x(t_{1}-\tau_{2}(t_{1})))
   +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-\bar{e}\\
&> g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
   +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-\bar{e},
\end{align*}
which contradicts \eqref{b6}. Thus \eqref{b7} is true.

Suppose that \eqref{e2.8}(or \eqref{e2.9}, or \eqref{e2.10}) holds;
using methods similar to those used in Case (i)--(iv), we can show
that \eqref{b7} is true.
This completes the proof of the above claim.
\smallskip

Let $t_2=kT+\widetilde{t}_2$, where $\widetilde{t}_2\in[0,T]$ and
$k$ is an integer. Then noticing $x(t)=x(t+T)$ and \eqref{b7}, for
any $t\in[\widetilde{t}_2,\widetilde{t}_2+T]$, we obtain
$$
|x(t)|=\Big|x(\widetilde{t}_2)+\int_{\widetilde{t}_2}^{t}x'(s)ds\Big|
\leq d+\int_{\widetilde{t}_2}^{t}|x'(s)|ds
$$
and
$$
|x(t)|=\Big|x(\widetilde{t}_2+T)+\int_{\widetilde{t}_2+T}^{t}x'(s)ds\Big|
\leq d+\Big|-\int^{\widetilde{t}_2+T}_{t}x'(s)ds\Big|\leq
d+\int^{\widetilde{t}_2+T}_{t}|x'(s)|ds.
$$
Combining  the two inequalities above, we obtain
$$
|x(t)|\leq d+\frac{1}{2}\int_{0}^{T}|x'(s)|ds.
$$
Using Schwarz inequality yields
\begin{equation}\label{b13}
|x|_{\infty}=\max_{t\in[\widetilde{t}_2,\widetilde{t}_2+T]}|x(t)|\leq
d+\frac{1}{2}\int_{0}^{T}|x'(s)|ds\leq
d+\frac{1}{2}|1|_{2}|x'|_{2}=d+\frac{1}{2}\sqrt{T}|x'|_{2}.
\end{equation}
This completes the proof.
\end{proof}

\begin{lemma} \label{lem4}
 Suppose {\rm (H0)--(H2)} hold. Also
suppose the following condition holds
\begin{itemize}
\item[(H3)] $C_2\frac{T}{2\pi}+(b_1+b_2)\frac{T^{2}}{4\pi}<1$.
\end{itemize}
If $x(t)$ is a $T$-periodic solution of \eqref{a1}, then
$|x'|_{\infty}\leq D$, where
\[
D= \frac{[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|:
0\leq t\leq T\}+|e|_{\infty}]T}{2
\big(1-C_2\frac{T}{2\pi}-b_1\frac{T^{2}}{4\pi}
-b_2\frac{T^{2}}{4\pi}\big)}.
\]
\end{lemma}

\begin{proof} Let $x(t)$ be a $T$-periodic solution of \eqref{a1}.
 From (H1) and (H2), we can easily show that
\eqref{b2} also holds. Multiplying $x''(t)$ and \eqref{a1} and then
integrating it from 0 to $T$, by Lemma 2, (H0), \eqref{b2}
and Schwarz inequality, we have
\begin{align*}
&|x''|_{2}^{2}\\
&= -\int_{0}^{T}f(x(t))x'(t)x''(t)dt
  -\int_{0}^{T}g_1(t,x(t-\tau_1(t)))x''(t)dt \\
&\quad -\int_{0}^{T}g_2(t,x(t-\tau_2(t)))x''(t)dt
 +\int_{0}^{T}e(t)x''(t)dt \\
&\leq \int_{0}^{T}|f(x(t))\|x'(t)\|x''(t)|dt
 +\int_{0}^{T}|g_1(t,x(t-\tau_1(t)))\|x''(t)|dt \\
&\quad +\int_{0}^{T}|g_2(t,x(t-\tau_2(t)))\|x''(t)|dt
 +\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2\int_{0}^{T}|x'(t)\|x''(t)|dt
 +\int_{0}^{T}[|g_1(t,x(t-\tau_1(t)))-g_1(t,0)|+|g_1(t,0)|]|x''(t)|dt \\
&\quad +\int_{0}^{T}[|g_2(t,x(t-\tau_2(t)))-g_2(t,0)|+|g_2(t,0)|]|x''(t)|dt
 +\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2|x'|_2|x''|_{2}+b_1\int_{0}^{T}|x(t-\tau_1(t))\|x''(t)|dt
 +\int_{0}^{T}|g_1(t,0)\|x''(t)|dt \\
&\quad +b_2\int_{0}^{T}|x(t-\tau_2(t))\|x''(t)|dt
 +\int_{0}^{T}|g_2(t,0)\|x''(t)|dt+\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2\frac{T}{2\pi}|x''|_{2}^{2}+(b_1+b_2)\sqrt{T}|x|_{\infty}|x''|_{2} \\
&\quad +[\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
 T\}+|e|_{\infty}]\sqrt{T}|x''|_2 \\
&\leq \big[C_2\frac{T}{2\pi}+(b_1+b_2)\frac{T^2}{4\pi}\big]|x''|_{2}^{2} \\
&\quad +\big[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}\big]\sqrt{T}|x''|_2,
\end{align*}
which, together with  (H3), implies
\begin{equation}\label{b14}
|x''|_2\leq\frac{[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}]\sqrt{T}}{1-C_2\frac{T}{2\pi}-b_1\frac{T^{2}}{4\pi}-b_2\frac{T^{2}}{4\pi}}.
\end{equation}
Since $x(0)=x(T)$, there exists $t_0\in[0,T]$ such that $x'(t_0)=0$,
for any $t\in[t_0,t_0+T]$, we obtain
\begin{gather*}
|x'(t)|=\Big|x'(t_0)+\int_{t_0}^{t}x''(s)ds\Big|\leq
\int_{t_0}^{t}|x''(s)|ds, \\
|x'(t)|=\Big|x'(t_0+T)+\int_{t_0+T}^{t}x''(s)ds\Big|
\leq \Big|-\int^{t_0+T}_{t}x''(s)ds\Big|\leq
\int^{t_0+T}_{t}|x''(s)|ds.
\end{gather*}
Combining these two inequalities, we obtain
$$
|x'(t)|\leq \frac{1}{2}\int_{0}^{T}|x''(s)|ds.
$$
Using Schwarz inequality yields
\begin{equation}\label{b15}
|x'|_{\infty}=\max_{t\in[t_0,t_0+T]}|x'(t)|\leq
\frac{1}{2}\int_{0}^{T}|x''(s)|ds\leq
\frac{1}{2}|1|_{2}|x''|_{2}=\frac{1}{2}\sqrt{T}|x''|_{2}.
\end{equation}
By (\ref{b14}) and (\ref{b15}), we obtain
$$
|x'|_{\infty}\leq \frac{[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|:
0\leq t\leq
T\}+|e|_{\infty}]T}{2\left(1-C_2\frac{T}{2\pi}-b_1\frac{T^{2}}{4\pi}-b_2
\frac{T^{2}}{4\pi}\right)}+:D.
$$
This completes the proof.
\end{proof}

\begin{lemma} \label{lem5}
 Suppose {\rm (H0)--(H3)} hold. Also assume
 the  condition
\begin{itemize}
\item[(H4)]
$C_1D\frac{T^{2}}{4\pi}+C_2\frac{T}{2\pi}+(b_1+b_2)
 \frac{T^{2}}{4\pi}<1$.
\end{itemize}
Then \eqref{a1} has at most one $T$-periodic solution.
\end{lemma}

\begin{proof}
Suppose that $x_{1}(t)$ and $x_{2}(t)$ are two
$T$-periodic solutions of \eqref{a1}. Set $Z(t)=x_{1}(t)-x_{2}(t)$.
Then, we have
\begin{equation}\label{b16}
\begin{aligned}
&Z''(t)+[f(x_1(t))x_1'(t)-f(x_2(t))x_2'(t)]
+[g_{1}(t,x_{1}(t-\tau_{1}(t)))-g_{1}(t,x_{2}(t-\tau_{1}(t)))] \\
&+[g_{2}(t,x_{1}(t-\tau_{2}(t)))-g_{2}(t,x_{2}(t-\tau_{2}(t)))]=0.
\end{aligned}
\end{equation}
Since $x_{1}(t)$ and $x_{2}(t)$ are two $T$-periodic solutions of
\eqref{a1}, integrating (\ref{b16}) from 0 to $T$, we obtain
\begin{align*}
&\int_{0}^{T}\big[g_{1}(t,x_{1}(t-\tau_{1}(t)))
 -g_{1}(t,x_{2}(t-\tau_{1}(t)))\\
&+g_{2}(t,x_{1}(t-\tau_{2}(t)))
 -g_{2}(t,x_{2}(t-\tau_{2}(t)))\big]dt
=0.
\end{align*}
Thus, in view of Mean Value Theorem for integrals, it follows that
there exists $\tilde{t}\in\mathbb{R}$ such that
\begin{equation}\label{b17}
g_{1}(\tilde{t},x_{1}(\tilde{t}-\tau_{1}(\tilde{t})))
-g_{1}(\tilde{t},x_{2}(\tilde{t}-\tau_{1}(\tilde{t})))
+g_{2}(\tilde{t},x_{1}(\tilde{t}-\tau_{2}(\tilde{t})))
-g_{2}(\tilde{t},x_{2}(\tilde{t}-\tau_{2}(\tilde{t})))=0.
\end{equation}
By (H1), (\ref{b17}) implies
$$
Z(\tilde{t}-\tau_{1}(\tilde{t}))Z(\tilde{t}-\tau_{2}(\tilde{t}))
=(x_{1}(\tilde{t}-\tau_{1}(\tilde{t}))
 -x_{2}(\tilde{t}-\tau_{1}(\tilde{t})))
 (x_{1}(\tilde{t}-\tau_{2}(\tilde{t}))
 -x_{2}(\tilde{t}-\tau_{2}(\tilde{t})))\leq0.
$$
Since $Z(t)=x_1(t)-x_2(t)$ is a continuous function in $\mathbb{R}$,
it follows that there exists $\hat{t}\in\mathbb{R}$ such that
\begin{equation}\label{b18}
Z(\hat{t})=0.
\end{equation}
Set $\hat{t}=nT+\bar{t}$, where $\bar{t}\in[0,T]$ and $n$ is an
integer. Noticing $Z(t+T)=Z(t)$, we get
\begin{equation}\label{b19}
Z(\bar{t})=Z(nT+\bar{t})=Z(\hat{t})=0.
\end{equation}
Hence, for any $t\in[\bar{t},\bar{t}+T]$, we obtain
$$
|Z(t)|=\Big|Z(\bar{t})+\int_{\bar{t}}^{t}Z'(s)ds\Big|
 \leq\int_{\bar{t}}^{t}|Z'(s)|ds
$$
and
$$
|Z(t)|=\Big|Z(\bar{t}+T)+\int_{\bar{t}+T}^{t}Z'(s)ds\big|
=\big|-\int^{\bar{t}+T}_{t}Z'(s)ds\big|
\leq\int^{\bar{t}+T}_{t}|Z'(s)|ds.
$$
Combining these two inequalities, we obtain
$$
|Z(t)|\leq\frac{1}{2}\int_{0}^{T}|Z'(s)|ds.
$$
Using Schwarz inequality yields
\begin{equation}\label{b20}
|Z|_{\infty}=\max_{t\in[\bar{t},\bar{t}+T]}|Z(t)|
\leq\frac{1}{2}\int_{0}^{T}|Z'(s)|ds\leq\frac{1}{2}|1|_{2}|Z'|_{2}
=\frac{1}{2}\sqrt{T}|Z'|_{2}.
\end{equation}

Multiplying $Z''(t)$ and (\ref{b16}) and then integrating it from
$0$ to $T$, by Lemma 2, Lemma 4, (H0), (\ref{b20}) and
Schwarz inequality, we have
%\label{b21}
\begin{align*}
|Z''|_{2}^{2}
&= -\int_{0}^{T}[f(x_1(t))x_1'(t)-f(x_2(t))x_2'(t)]Z''(t)dt \\
&\quad -\int_{0}^{T}[g_{1}(t,x_{1}(t-\tau_{1}(t)))-g_{1}(t,x_{2}(t-\tau_{1}(t)))]Z''(t)dt \\
&\quad -\int_{0}^{T}[g_{2}(t,x_{1}(t-\tau_{2}(t)))-g_{2}(t,x_{2}(t-\tau_{2}(t)))]Z''(t)dt \\
&\leq \int_{0}^{T}|f(x_1(t))\|x'_{1}(t)-x'_{2}(t)\|Z''(t)|dt \\
&\quad +\int_{0}^{T}|f(x_{1}(t))-f(x_{2}(t))\|x'_{2}(t)\|Z''(t)|dt \\
&\quad +b_1\int_{0}^{T}|x_{1}(t-\tau_{1}(t))-x_{2}(t-\tau_{1}(t))\|Z''(t)|dt \\
&\quad +b_2\int_{0}^{T}|x_{1}(t-\tau_{2}(t))-x_{2}(t-\tau_{2}(t))\|Z''(t)|dt \\
&\leq \int_{0}^{T}C_2|Z'(t)\|Z''(t)|dt+\int_{0}^{T}C_1D|Z(t)\|Z''(t)|dt \\
&\quad +b_1\int_{0}^{T}|Z(t-\tau_1(t))\|Z''(t)|dt+b_2\int_{0}^{T}|Z(t-\tau_2(t))\|Z''(t)|dt \\
&\leq
C_2|Z'|_2|Z''|_2+C_1D\sqrt{T}|Z|_{\infty}|Z''|_2+(b_1+b_2)\sqrt{T}|Z|_{\infty}|Z''|_2 \\
&\leq
\big[C_1D\frac{T^{2}}{4\pi}+C_2\frac{T}{2\pi}+(b_1+b_2)
\frac{T^{2}}{4\pi}\big]|Z''|_{2}^{2}.
\end{align*}
Since $Z(t), Z'(t)$, $Z''(t)$ are continuous $T$-periodic
functions, by (H4), (\ref{b20}) and the above inequality, we obtain
$$
Z(t)=Z'(t)=Z''(t)=0 \quad \textrm{for all } t\in\mathbb{R}.
$$
Thus, $x_{1}(t)\equiv x_{2}(t)$, for all $t\in\mathbb{R}$. Hence,
(\ref{a1}) has at most one $T$-periodic solution. This completes
the proof.
\end{proof}

\begin{lemma} \label{lem6}
 Suppose {\rm (H0)--(H3)} hold. Then the
set of $T$-periodic solutions of  \eqref{b1} are bounded in
$C_{T}^{1}$.
\end{lemma}

\begin{proof} Let $S\subset C_T^1$ be the set of $T$-periodic
solutions of \eqref{b1}. If $S=\emptyset$, the proof is complete.
Suppose $S\neq \emptyset$, and let $x\in S$. Multiplying $x''(t)$
and \eqref{b1} and then integrating it from 0 to $T$, by Lemma 2,
(H0), \eqref{b2} and Schwarz inequality, we have
\begin{align*}
&|x''|_{2}^{2}\\
&= -\lambda\int_{0}^{T}f(x(t))x'(t)x''(t)dt
 -\lambda\int_{0}^{T}g_1(t,x(t-\tau_1(t)))x''(t)dt \\
&\quad -\lambda\int_{0}^{T}g_2(t,x(t-\tau_2(t)))x''(t)dt
 +\lambda\int_{0}^{T}e(t)x''(t)dt \\
&\leq \int_{0}^{T}|f(x(t))\|x'(t)\|x''(t)|dt
 +\int_{0}^{T}|g_1(t,x(t-\tau_1(t)))\|x''(t)|dt \\
&\quad +\int_{0}^{T}|g_2(t,x(t-\tau_2(t)))\|x''(t)|dt
 +\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2\int_{0}^{T}|x'(t)\|x''(t)|dt
 +\int_{0}^{T}[|g_1(t,x(t-\tau_1(t)))-g_1(t,0)|+|g_1(t,0)|]|x''(t)|dt \\
&\quad +\int_{0}^{T}[|g_2(t,x(t-\tau_2(t)))-g_2(t,0)|
 +|g_2(t,0)|]|x''(t)|dt+\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2|x'|_2|x''|_{2}+b_1\int_{0}^{T}|x(t-\tau_1(t))\|x''(t)|dt
 +\int_{0}^{T}|g_1(t,0)\|x''(t)|dt \\
&\quad +b_2\int_{0}^{T}|x(t-\tau_2(t))\|x''(t)|dt
 +\int_{0}^{T}|g_2(t,0)\|x''(t)|dt+\int_{0}^{T}|e(t)\|x''(t)|dt \\
&\leq C_2\frac{T}{2\pi}|x''|_{2}^{2}+(b_1+b_2)
 \sqrt{T}|x|_{\infty}|x''|_{2} \\
&\quad +[\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}]\sqrt{T}|x''|_2 \\
&\leq \big[C_2\frac{T}{2\pi}+(b_1+b_2)\frac{T^2}{4\pi}\big]|x''|_{2}^{2} \\
&\quad +\big[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}\big]\sqrt{T}|x''|_2,
\end{align*}
which, together with (H3), implies that there exists
$M_0>0$ such that
\begin{equation}\label{b22}
|x''|_2<M_0.
\end{equation}
This, together with Lemma 2 and Lemma 3, leads to
\begin{equation}\label{b23}
|x|_{\infty}<d+\frac{\sqrt{T^{3}}}{4\pi}M_0.
\end{equation}
On the other hand, since $x(0)=x(T)$, there exists
$\bar{t}_0\in[0,T]$ such that $x'(\bar{t}_0)=0$. For any
$t\in[\bar{t}_0,\bar{t}_0+T]$, from (\ref{b22}), we obtain
$$
|x'(t)|=\big|x'(\bar{t}_0)+\int_{\bar{t}_0}^{t}x''(s)ds\big|
\leq\int_{0}^{T}|x''(s)|ds\leq|1|_{2}|x''|_{2}<\sqrt{T}M_0,
$$
which implies
\begin{equation}\label{b24}
|x'|_{\infty}=\max_{t\in[\bar{t}_0,\bar{t}_0+T]}|x'(t)|<\sqrt{T}M_0.
\end{equation}
Let
$M=\max\big\{d+\frac{\sqrt{T^{3}}}{4\pi}M_0,\sqrt{T}M_0\big\}$,
by (\ref{b23}) and (\ref{b24}), we have $\|x\|<M$. This completes
the proof.
\end{proof}


\section{Main results}

Now we are in the position to give our main results.

\begin{theorem} \label{thm1.}
 Suppose {\rm (H0)--(H2), (H4)} hold. Then \eqref{a1} has a
unique $T$-periodic solution.
\end{theorem}

\begin{proof}
 Lemma 5 states that \eqref{a1} has at most one
$T$-periodic solution. Thus, to prove Theorem 1, it suffices to show
that \eqref{a1} has at least one $T$-periodic solution. To do this,
we apply Lemma 1.

By Lemma 6, there exists $M>d$ such that, for any $T$-periodic
solution $x(t)$ of \eqref{b1}
\begin{equation}\label{c1}
\|x\|<M.
\end{equation}
Set
\begin{equation}\label{c2}
\Omega=\{x: x\in C_{T}^{1},\|x\|<M\}.
\end{equation}
Define a linear operator $L: D(L)\subset C_{T}^{1}\to C_{T}$
by setting
$D(L)=\{x: x\in C_{T}^{1}, x''\in C(\mathbb{R,\mathbb{R}})$,
for $x\in D(L)$, and
\begin{equation}\label{c3}
Lx=x''.
\end{equation}
We also define a nonlinear operator $N: C_{T}^{1}\to C_{T}$,
by
\begin{equation}\label{c4}
Nx=-f(x(t))x'(t)-g_1(t,x(t-\tau_1(t)))-g_2(t,x(t-\tau_2(t)))+e(t).
\end{equation}
Then \eqref{b1} is equivalent to the operator equation
\begin{equation}\label{c5}
Lx=\lambda Nx,\quad \lambda\in(0,1).
\end{equation}
It is easy to see that
$$
\ker L=\mathbb{R}\quad,\textrm{and}\quad
\mathop{\rm Im}L=\{x: x\in C_{T}, \int_{0}^{T}x(s)ds=0\},
$$
then, $L$ is a Fredholm operator with index zero.
Also let projectors $P: C_{T}^{1}\to \ker L$ and
$Q: C_{T}\to C_{T}/\mathop{\rm Im}L$ defined by
\begin{gather*}
Px=x(0)\quad\textrm{where } x\in C_{T}^{1},\\
Qx=\frac{1}{T}\int_{0}^{T}x(s)ds\quad\textrm{where } x\in C_{T},
\end{gather*}
hence, $\mathop{\rm Im} P =\mathop{\rm Im} Q =\ker  L =\mathbb{R}$
and
$\ker  Q =\mathop{\rm Im} L $.
Define the isomorphism as follows
\begin{equation}\label{c6}
J: \mathop{\rm Im}Q\to \ker L, \quad J(x)=x.
\end{equation}
Let
$$
L_{P}:=L_{D(L)\cap \ker \,P}: D(L)\cap \ker \,P\to \mathop{\rm Im}L,
$$
then we know, by \cite[p. 41--42]{gre}, that $L_{P}$ has a continuous
inverse $L_{P}^{-1}$ on $\mathop{\rm Im} L$ defined by
\begin{equation}\label{c7}
(L^{-1}_{P}y)(t)=\int_{0}^{T}G(s,t)y(s)ds,
\end{equation}
where
$$
G(s,t)=
\begin{cases}
-\frac{s}{T}(T-t),& 0\leq s\leq t;\\
-\frac{t}{T}(T-s),& t\leq s\leq T.
\end{cases}
$$
Using Ascoli-Arzela theorem we have, from (\ref{c2}) and (\ref{c7}),
that $L^{-1}_{P}(I-Q)N(\overline{\Omega})$ is compact. On the other
hand, $QN(\overline{\Omega})$ is bounded by the continuity of
function $QN$. Thus $N$ is $L$-compact on $\overline{\Omega}$. By
(\ref{c2}) and (\ref{c5}), condition (i) of Lemma 1 is
satisfied.\\
In view of (H2)(1) and (H2)(2), we will consider
two cases:

{\sl Case}(i): If (H2)(1) holds. Since
\begin{align*}
QNx
&= -\frac{1}{T}\int_{0}^{T}[f(x(t))x'(t)+g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt\\
&= -\frac{1}{T}\int_{0}^{T}[f(x(t))x'(t)+g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-\bar{e}]dt;
\end{align*}
for any $x\in\partial{\Omega}\cap \ker L$, $x=M$ or $x=-M$, $x'=0$, we
obtain
\begin{gather}\label{c8}
QN(M)=-\frac{1}{T}\int_{0}^{T}[g_1(t,M)+g_2(t,M)-\bar{e}]dt< 0,\\
\label{c9}
QN(-M)=-\frac{1}{T}\int_{0}^{T}[g_1(t,-M)+g_2(t,-M)-\bar{e}]dt> 0
\end{gather}
which implies the condition (ii) of Lemma 1 is satisfied.
Define
\begin{align*}
H(x,\mu)&= -\mu x+(1-\mu)QNx\\
&=-\mu x-(1-\mu)\frac{1}{T}\int_{0}^{T}\big[f(x(t))x'(t)
 +g_1(t,x(t-\tau_1(t)))\\
&\quad +g_2(t,x(t-\tau_2(t)))-e(t)\big]dt\\
&=  -\mu x-(1-\mu)\frac{1}{T}\int_{0}^{T}\big[f(x(t))x'(t)
 +g_1(t,x(t-\tau_1(t)))\\
&\quad +g_2(t,x(t-\tau_2(t)))-\bar{e}\big]dt
\end{align*}
in view of (\ref{c8}) and (\ref{c9}), we get
$xH(x,\mu)<0$, for all
$x\in\partial{\Omega}\cap \ker L$ and $\mu\in[0,1]$.
Hence, $H(x,\mu)$ is a homotopic transformation, together with
(\ref{c6}) and by using homotopic invariance theorem, we have
$$
\deg \left\{JQN, \Omega\cap \ker L, 0\right\}=\deg \{QN, \Omega\cap \ker L,
0\}=\deg \{-x, \Omega\cap \ker L, 0\}\neq 0,
$$
so condition (iii) of Lemma 1 is satisfied.

{\sl Case}(ii): If (H2)(2) holds. Since
\begin{align*}
QNx
&= -\frac{1}{T}\int_{0}^{T}[f(x(t))x'(t)+g_1(t,x(t-\tau_1(t)))
 +g_2(t,x(t-\tau_2(t)))-e(t)]dt\\
&= -\frac{1}{T}\int_{0}^{T}[f(x(t))x'(t)+g_1(t,x(t-\tau_1(t)))
 +g_2(t,x(t-\tau_2(t)))-\bar{e}]dt;
\end{align*}
for any $x\in\partial{\Omega}\cap \ker L$, $x=M$ or $x=-M$, $x'=0$, we
obtain
\begin{gather}\label{c10}
QN(M)=-\frac{1}{T}\int_{0}^{T}[g_1(t,M)+g_2(t,M)-\bar{e}]dt>0,\\
\label{c11}
QN(-M)=-\frac{1}{T}\int_{0}^{T}[g_1(t,-M)+g_2(t,-M)-\bar{e}]dt< 0
\end{gather}
which implies the condition (ii) of Lemma 1 is satisfied.
Define
\begin{align*}
H(x,\mu)&= \mu x+(1-\mu)QNx\\
&= \mu x-(1-\mu)\frac{1}{T}\int_{0}^{T}\big[f(x(t))x'(t)
 +g_1(t,x(t-\tau_1(t)))\\
&\quad +g_2(t,x(t-\tau_2(t)))-e(t)\big]dt\\
&=  \mu x-(1-\mu)\frac{1}{T}\int_{0}^{T}\big[f(x(t))x'(t)
 +g_1(t,x(t-\tau_1(t)))\\
&\quad +g_2(t,x(t-\tau_2(t)))-\bar{e}\big]dt,
\end{align*}
in view of (\ref{c10}) and (\ref{c11}), we get
$xH(x,\mu)>0$, for all $x\in\partial{\Omega}\cap \ker L$
and $\mu\in[0,1]$.
Hence, $H(x,\mu)$ is a homotopic transformation, together with
(\ref{c6}) and by using homotopic invariance theorem, we have
$$
\deg \{JQN, \Omega\cap \ker L, 0\}=\deg \{QN, \Omega\cap \ker L, 0\}=\deg \{x,
\Omega\cap \ker L, 0\}\neq 0,
$$
so condition (iii) of Lemma 1 is satisfied. Therefore, it follows
from Lemma 1 that \eqref{a1} has at least one $T$-periodic solution.
This completes the proof.
\end{proof}

In \cite{z07jcam}, Zhou and Long studied \eqref{a1}
and obtained the got the following results.

\begin{theorem} \label{thm1}
Assume {\rm (H0), (H1)}, and that the following conditions hold:
\begin{itemize}

\item[(A2)] there exists $d\geq 0$ such that one of the following
conditions holds:
\begin{enumerate}
\item $x(g_{1}(t,x)+g_{2}(t,x)-e(t))>0$, for
all $t\in\mathbb{R}$, $|x|> d$,
\item $x(g_{1}(t,x)+g_{2}(t,x)-e(t))<0$, for
all $t\in\mathbb{R}$, $|x|> d$;
\end{enumerate}

\item[(A4)] $C_1D_1\frac{T^{2}}{2\pi}+C_2\frac{T}{2\pi}+(b_1+b_2)
\frac{T^{2}}{2\pi}<1$,
where
\[
D_1=\frac{[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}]T}{1-C_2\frac{T}{2\pi}-b_1
\frac{T^{2}}{2\pi}-b_2\frac{T^{2}}{2\pi}}.
\]
\end{itemize}
 Then \eqref{a1} has a unique $T$-periodic solution.
\end{theorem}
If $e(t)\neq $ constant,  it is easy to verify that the condition
(H2) is weaker than the condition (A2) since
$\min_{t\in\mathbb{R}}e(t)< \bar{e}< \max_{t\in\mathbb{R}}e(t)$. On
the other hand, noticing $\frac{1}{4\pi}<\frac{1}{2\pi}$ and
$D<\frac{1}{2}D_1$, we can see that the condition {\rm (H$_4$)} is
also weaker than the condition {\rm (A$_4$)}. Therefore, our results
improve those in \cite{z07jcam}.

\section{Example and remark}

In this section, we apply the main results obtained in previous
sections to an example.

Consider the existence and uniqueness of a
$2\pi$-periodic solution to the Li\'enard equation
\begin{equation}\label{d1}
x''(t)+\frac{1}{10}\cos t x'(t)+g_1(t,x(t-\cos t))+g_2(t,x(t-\sin
t))=e(t),
\end{equation}
where $T=2\pi$, $\tau_1(t)=\cos t$, $\tau_2(t)=\sin t$,
$g_1(t,x)=\frac{1}{80\pi(1+\cos^{2} t)}\arctan x$,
$g_2(t,x)=\frac{1}{60\pi}(1+\sin^{2} t)\arctan x$ and
$e(t)=\frac{1}{\pi}\sin t$.


 It is obvious that the conditions (A0) and
(A1) in \cite[Theorem 1]{z07jcam} hold. However, we can
easily check that (A2) does not hold, which implies that
(A4)  does not hold. Hence,  \cite[Theorem 1]{z07jcam} can
not be applied. Meanwhile, Theorem 1 in this
paper remains applicable, as we show now.

By (\ref{d1}), we can get $b_1=\frac{1}{80\pi}, b_2=\frac{1}{30\pi}$
and $C_1=C_2=\frac{1}{10}$. Noticing
$\bar{e}=\frac{1}{T}\int_{0}^{T}e(t)dt=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{\pi}\sin
tdt=0$, we can get $d=\frac{1}{10}$ (Actually, $d$ can be an
arbitrarily small positive constant.) and check that (H0)--(H2) hold.
On the other hand, noticing that
\begin{align*}
D&= \frac{[(b_1+b_2)d+\max\{|g_1(t,0)|+|g_2(t,0)|: 0\leq t\leq
T\}+|e|_{\infty}]T}{2\big(1-C_2\frac{T}{2\pi}-b_1
\frac{T^{2}}{4\pi}-b_2\frac{T^{2}}{4\pi}\big)}\\
&=\frac{[\frac{1}{10}(\frac{1}{80\pi}+\frac{1}{30\pi})
+\frac{1}{\pi}]2\pi}{2(1-\frac{1}{10}-\frac{1}{80}-\frac{1}{30})}
\approx
1.176,
\end{align*}
it is easy to verify that (H4) holds since
$C_1D\frac{T^{2}}{4\pi}+C_2\frac{T}{2\pi}+(b_1+b_2)
\frac{T^{2}}{4\pi}=\frac{1}{10}\times
1.176\times \pi+\frac{1}{10}+\frac{1}{80}+\frac{1}{30}\approx
0.515<1 $. Thus, Theorem 1 in this study shows that \eqref{a1} has a
unique $2\pi$-periodic solution. Hence our results
improve those in \cite{z07jcam}.

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 \end{document}