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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 242, 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 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/242\hfil Existence of positive periodic solutions]
{Existence of positive periodic solutions for  neutral Li\'enard differential
equations with a singularity}

\author[F. Kong, S. Lu, Z. Liang  \hfil EJDE-2015/242\hfilneg]
{Fanchao Kong, Shiping Lu, Zaitao Liang}

\address{Fanchao Kong \newline
College of Mathematics and Computer Science,
Anhui Normal University,
Wuhu 241000, Anhui,  China}
\email{fanchaokong88@yahoo.com}

\address{Shiping Lu \newline
College of Mathematics and Statistics, Nanjing University of Information
Science and Technology, Nanjing 210044, China}
\email{lushiping26@sohu.com}

\address{Zaitao Liang \newline
Department of Mathematics,
College of Science, Hohai University, Nanjing 210098, China}
\email{liangzaitao@sina.cn}

\thanks{Submitted May 6, 2015. Published September 21, 2015.}
\subjclass[2010]{34B16, 34B20, 34B24}
\keywords{Positive periodic solution; neutral differential equation;
\hfill\break\indent  deviating argument; singular; Mawhin's continuation}

\begin{abstract}
 By applying Mawhin's continuation theorem, we study the existence of
 positive periodic solutions for a second-order neutral functional differential
 equation
 $$
 ((x(t)-cx(t-\sigma)))''+f(x(t)) x'(t)+g(t,x(t-\delta))=e(t),
 $$
 where $g$ has a strong singularity at $x=0$ and satisfies a small force
 condition at $ x=\infty$, which is different from the corresponding ones 
 known in the literature.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

In recent years, the existence of periodic solutions for the second order 
differential equations with a singularity have been studied in many literature. 
See \cite{1993-Fonda-p1294--1311}-\cite{2014-Wang-p227--234} and the 
references therein.

Wang \cite{2014-Wang-p227--234}  studied the  Li\'enard equation with a 
singularity and a deviating argument
\begin{equation}
x''(t)+f(x(t))x'(t)+g(t,x(t-\sigma))=0, \label{eSP}
\end{equation}
where $0\leq\sigma<T$ is a constant, 
$f:  \mathbb{R}\to  \mathbb{R}$, 
$g:  \mathbb{R}\times (0,+\infty)\to \mathbb{R}$ is an 
$L^2$-Carath\'eodory function, $g(t,x)$ is a $T$-periodic function in
the first argument and can be singular at $x=0$, i. e., $g(t,x)$ can
be unbounded as $x\to0^{+}$.

Let \eqref{eSP} be of repulsive type and set
\[
\overline{g}(x)=\frac{1}{T}\int_{0}^{T} g(t,x)dt,~x>0.
\]
Assume that
\[
\varphi(t)=\lim_{x\to+\infty}\sup\frac{g(t,x)}{x},
\]
exists uniformly for a. e. $t\in[0,T]$, i.e., for any $\varepsilon>0$,
there is $g_\varepsilon\in L^2(0,T)$ such that 
\[
g(t,x)\leq(\varphi(t)+\varepsilon)x+g_\varepsilon,
\]
for all $x>0$ and a. e. $t\in[0,T]$. Assume that
$\varphi\in C( {\mathbb{R},\mathbb{R}})$ and $\varphi(t+T)=\varphi(t)$, 
$t\in {\mathbb{R}}$.

Wang established the following theorem.

\begin{theorem} \label{thm1.1}
 Assume that the following conditions are satisfied:
\begin{itemize}
\item[(H1)] (Balance) There exist constants $0<D_1<D_2$ such that 
if $x$ is a positive continuous $T$-periodic function satisfying
\[
\int_{0}^{T} g(t,x(t))dt=0,
\]
then
\[
 D_1\leq x(\tau)\leq D_2,~{\rm for ~some} ~\tau\in [0,T].
\]

\item[(H2)] (Degree) $\overline{g}(x)<0$ for all $x\in (0,D_1)$, 
and $\overline{g}(x)>0$ for all $x>D_2$.

\item[(H3)] (Decomposition) $g(t,x)=g_0(x)+g_1(t,x)$, where 
$g_0\in C((0,+\infty), \mathbb{R})$ and 
$g_1:[0,T]\times[0,+\infty)\to \mathbb{R}$ is an $L^2$-Carath\'eodory function,
i. e., $g_1$ is measurable with respect to the first variable, continuous
 with respect to the second one, and for any $b>0$ there is 
$h_b\in L^2((0,T);[0,+\infty))$ such that $|g_1(t,x)|\leq h_b(t)$ for a.e.
 $t\in [0,T]$ and all $x\in [0,b]$.

\item[(H4)] (Strong force  at $x=0$) $\int_{0}^{1} g_0(x)dx=-\infty$.

\item[(H5)] (Small force  at $x=\infty$)
$$\| \varphi\|_\infty<(\frac{\sqrt{\pi}}{T})^2.
$$
\end{itemize}
Then \eqref{eSP} has at least one positive $T$-periodic solution.
\end{theorem}

Meanwhile,  the problem of the existence of periodic solutions to the  neutral 
functional differential equation was studied in many papers, 
see \cite{1995-Zhang-p378--392}-\cite{2004-Lu-p433--448} and the references therein. 
For example, in \cite{2006-Liu-p121--132}, Liu and Huang studied the following 
neutral functional differential equation
\[
(u(t)+Bu(t-\tau))'=g_1(t,u(t))+g_2(u(t-\tau_1))+p(t).
\]
And in  in \cite{2004-Lu-p433--448}, Lu and Ge studied the existence of periodic 
solutions for a kind of second-order neutral functional differential equation
of the form
\begin{align*}
&\frac{d^2}{dt^2}\Big(u(t)-\sum^{n}_{j=1} c_{j}u(t-r_{j})\Big)\\
&=f(u(t))u'(t)+\alpha(t)g(u(t))
 +\sum^{n}_{j=1} \beta_{j}g(u(t-\gamma_{j}))+p(t),
\end{align*}
where $f$, $g\in  C(\mathbb{R}; \mathbb{R})$, $a(t)$, $p(t)$, $\beta_j(t)$, 
$\gamma_j(t)$ $(j= 1, 2, \dots , n)$ are continuous periodic functions defined 
on $\mathbb{R}$ with period $T > 0$, $c_j$, $r_j\in \mathbb{R}$ are constants 
with $r_j > 0$ $(j = 1, 2, \dots , n)$. By using the continuation theorem of 
coincidence degree theory and some new analysis techniques, the authors 
obtained some new results on the existence of periodic solution.

However, to the  best of our knowledge, the studying of positive periodic 
solutions for the neutral  functional differential equation with a 
singularity is relatively infrequent. As we know, in order to establish 
the existence of positive periodic solutions, a key condition is that 
the greatest lower bound must be estimated because of the singularity.
 However, it is difficult to verify the  greatest lower bound, especially 
for the neutral  functional differential equations with a singularity. 
Besides, because of the singularity, the third  condition of Mawhin's 
continuation theorem is not easy to verify.

 Inspired by the above facts, in this paper, we consider the following  
neutral Li\'enard  differential equation with a singularity and a deviating
argument
\begin{equation}
 ((x(t)-cx(t-\sigma)))''+f(x(t)) x'(t)+g(t,x(t-\delta))=e(t),   \label{1-2}
\end{equation}
where   $c$ is a constant with $|c|<1$,  $0\leq\sigma,~\delta<T$, 
$f:\mathbb{R}\to \mathbb{R}$ is continuous,  
$g:[0,T] \times (0,+\infty)\to  \mathbb{R}$ is a continuous function  
and can be singular at $u=0$, i. e., $g(t,u)$ can be unbounded as 
$u\to0^{+}$. $e(t)$ is $T$-periodic with $\int_{0}^{T} e(t)dt=0$. 
And we can easily see that when $c=0$, the \eqref{1-2} transforms into \eqref{eSP}.     To sum up, our results are essentially new.

The  rest of this paper is  organized as follows. In  Section 2, we state 
some necessary definitions and lemmas. In Section 3, we prove the main result. 
At last,  we will give an example of an application in section 4.

\section{Preliminaries}

To prove the announced result, we  state the following necessary definitions 
and lemmas.
Denote the operator $A$ by
\[
A:C_T\to C_T,\quad (Ax)(t)=x(t)-cx(t-\sigma),
\]
where $C_T=\{\varphi\in C(\mathbb{R},\mathbb{R}),\;\varphi(t+T)=\varphi(t)\}$,
 with  norm $||\varphi||_0=\max_{t\in[0,T]}|\varphi(t)|$. 
Clearly, $C_T$ is a Banach space.
Define the operator
\[
L:D(L)\subset X\to Y,\quad Lx=(Ax)' ,
\]
where $D(L)=\{x\in C^1( \mathbb{R}, \mathbb{R} ),\;x(t)=x(t+T)\}$.
Define
\[
N:C_T\to C_T,\quad (Nx)(t)=-f(x(t)) x'(t)-g(t,x(t-\delta))+e(t).
\]
Then \eqref{1-2} can be rewritten by $Lx=Nx$.

\begin{lemma}[\cite{2003-Lu-p195--209}]\label{lem2.1} 
  If $|c|<1$ then $A$ has continuous inverse on $C_T$ and
\begin{itemize}
\item[(1)] $\| A^{-1} x\|\leq\frac{\| x\|_0}{|1-|c||}$ for all $x\in C_T$;

\item[(2)] $\int_{0}^{T}|(A^{-1}f)(t)|dt
\leq\frac{1}{|1-|c||}\int_{0}^{T}|f(t)|dt$ for all $f\in C_T$;

\item[(3)] $\int_{0}^{T}|A^{-1}f|^2(t)dt
\leq\frac{1}{(1-|c|)^2}\int_{0}^{T}f^2(t)dt$ for all $f\in C_T$.
\end{itemize}
\end{lemma}

From Hale's terminology \cite{1977-Hale-p}, a solution of the \eqref{1-2} 
is $x \in C(\mathbb{R},\mathbb{R} )$ such that $Ax \in C^1(\mathbb{R},\mathbb{R} )$ 
and  \eqref{1-2} is satisfied on $\mathbb{R}$. In general, $x$ is not from 
$C^1(\mathbb{R},\mathbb{R} )$. Nevertheless, it is easy to see that $(Au)'=Au'$.
 Thus, a $T$-periodic solution $x$ of the \eqref{1-2} must be from 
$C^1(\mathbb{R},\mathbb{R} )$. According to Lemma  \ref{lem2.1}, we can easily 
obtain that $  \ker L =\mathbb{R} $,
$  \operatorname{Im}L =\{x:x\in X,\int_{0}^{T}x(s)ds=0\}$. 
Thus $L$ is a Fredholm operator with index zero.

Let the projections $P$ and $Q$ be
\begin{gather*}
P:C_T\to \ker L ,Px=\frac{1}{T}\int_0^{T}x(s)ds;\\
Q:C_T\to C_T\setminus  \operatorname{Im}L,Qy=\frac{1}{T}\int_0^{T}x(s)ds.
\end{gather*}
Let $  L_p=L|_{D(L)\cap   \ker P }:C_T\cap   \ker P\to \operatorname{Im}L $.  
Then $L_P$ has continuous inverse $L^{-1}_p$ on $  \operatorname{Im}L $ defined by
\[
(L^{-1}_py)(t)= A^{-1}\Big( \int_{0}^{T}G(t,s)y(s)ds\Big),
\]
where
\[
G_k(t)=\begin{cases}
 \frac{s-T}{T}, &0\leq t\leq s ;\\
\frac{s}{T},  &s\leq t\leq T.
\end{cases}
\]

\begin{lemma}[\cite{1977-Gaines-p}]\label{lem2.2}   
Let $X$ and $Y$ be two real Banach spaces, and
$\Omega$ is an open and bounded set of $X$, and 
$L: D(L)\subset X \to Y$ is a Fredholm operator of index zero and
the operator $N:\bar{\Omega}\subset X\to Y$ is said to be 
$L$-compact in $\bar{\Omega}$. In addition, if the following conditions hold:
\begin{itemize}
\item[(1)] $Lx\neq\lambda Nx$ for all $(x,\lambda)\in\partial\Omega\times(0,1)$;

\item[(2)] $QN x\neq0$ for all $ x\in \ker L\cap\partial\Omega$;

\item[(3)] $\deg \{ JQN ,\Omega\cap  \ker L ,0\}\neq0$,
where $ J: ImQ \to  \ker L $ is a homeomorphism.
\end{itemize}
Then $ Lx=Nx $ has at least one solution in $D(L)\cap\bar{\Omega}$.
\end{lemma}

 For the sake of convenience, we list the following assumptions:
\begin{itemize}
\item[(H1)] There exist positive constants $D_1$ and $D_2$ with $D_1<D_2$ such that
\begin{itemize}
\item[(1)] for each  positive continuous $T$-periodic function $x(t)$ satisfying \\
$\int_{0}^{T} g(t, x(t))dt=0$,  there exists a positive point
 $\tau\in [0,T]$ such that 
$$ 
D_1\leq x(\tau)\leq D_2;
$$

\item[(2)]  $\overline{g}(x)<0$ for all $x\in(0,D_1)$ and 
$ \overline{g}(x)>0$ for all $x>D_2$, where
$\overline{g}(x)=\frac{1}{T}\int_{0}^{T} g(t,x)dt$, $x>0$.
\end{itemize}

\item[(H2)] $g(t,x )=g_1(t,x )+g_0(x )$, where 
$g_1: [0,T]\times(0,+\infty)\to  \mathbb{R}$ is a
continuous function and
\begin{itemize}
\item[(1)] there exist positive constants $m_0$ and $m_1$ such that
\[
g(t,x)\leq m_0x +m_1,\quad\text{for all }(t,x)\in [0,T]\times(0,+\infty);
\]

\item[(2)] $\int_{0}^{1} g_0(x)dx=-\infty$.
\end{itemize}
\end{itemize}

\section{Main results}

\begin{theorem}\label{thm3.1}  
Suppose that the conditions (H1)-(H2) hold, $| c|<1$ and
\[
\frac{ | c| (1+| c|)+m_0T^2}{(1-| c|)^2}<1,
\]
 then the \eqref{1-2} has at least one positive $T$-periodic solution.
\end{theorem}

\begin{proof} 
Consider the  operator equation
\[
Lx=\lambda Nx,\quad \lambda\in(0,1).
\]
Let $\Omega_1=\{x\in\overline{\Omega}, Lx=\lambda Nx ,\lambda\in(0,1)\}$.
 If $x\in \Omega_1$, then  $x$ must satisfy
\begin{equation}
 ((Au)'(t))'+\lambda f(u(t))u'(t)+\lambda g(t,u(t-\delta))=\lambda e(t).  \label{3-1}
\end{equation}
Integrating  \eqref{3-1}  on the interval $[0,T]$, we have
\begin{equation}
\int_{0}^{T}g(t, u(t-\delta ))dt=0.   \label{3-2}
\end{equation}
It follows from (H1)(1) that there exist positive constants $D_1$, $D_2$ 
and $\tau\in[0,T]$ such that
\begin{equation}
 D_1\leq u(\tau)\leq D_2. \label{3-3}
\end{equation}
Then, we obtain
\begin{equation}\label{3-4}
\| u\|_0 
=\max_{t\in [0,T]}|u(t)|
\leq\max_{t\in [0,T]}|u(\tau)+\int_{\tau}^{t}u'(s)ds|
\leq D_2+ \int_{0}^{T}|u'(s)| ds.
\end{equation}
Multiplying the both sides of \eqref{3-1} by $u(t)$ and integrating on 
the interval $[0,T]$, we obtain
\begin{equation}
\begin{aligned}
&\int_{0}^{T}( (Au)'(t) )'u(t)dt\\
&=- \lambda \int_{0}^{T}f(u(t))u'(t)u(t)dt
 -\lambda \int_{0}^{T}g(t,u(t-\delta))u(t)dt
 +\lambda \int_{0}^{T}e(t)u(t)dt\\
&=-\lambda \int_{0}^{T}g(t,u(t-\delta))u(t)dt
 +\lambda \int_{0}^{T}e(t)u(t)dt. 
\end{aligned} \label{3-5}
\end{equation}
Furthermore,
\begin{equation}
\begin{aligned}
\int_{0}^{T}( (Au)'(t) )'u(t)dt 
&=-\int_{0}^{T} (Au)'(t) u'(t)dt\\
&=- \int_{0}^{T} (Au)'(t) [u'(t)-c u'(t-\sigma)+c u'(t-\sigma)]dt\\
&=- \int_{0}^{T}  (Au)'(t) [(Au')(t)+c u'(t-\sigma)]dt\\
&=-\int_{0}^{T}|(Au')(t)|^2dt-\int_{0}^{T}c u'(t-\sigma) (Au)'(t) dt. 
\end{aligned} \label{3-6}
\end{equation}
Substituting  \eqref{3-6} in \eqref{3-5}, we obtain
\begin{align*}
&\int_{0}^{T}|(Au')(t)|^2dt\\
&=-\int_{0}^{T}c u'(t-\sigma) (Au)'(t) dt
 +\lambda \int_{0}^{T}g(t,u(t-\delta))u(t)dt 
 -\lambda \int_{0}^{T}e(t)u(t)dt \\
&\leq | c| \int_{0}^{T}|u'(t-\sigma)|| (Au)'(t) |dt
 +\int_{0}^{T}|g(t,u(t-\delta))||u(t)|dt 
 + \int_{0}^{T}|e(t)||u(t)|dt.
\end{align*}
It follows from (H2)(1) that
\begin{equation}
\begin{aligned}
\int_{0}^{T}|(Au')(t)|^2dt
&\leq | c| \int_{0}^{T}|u'(t-\sigma)|| (Au)'(t) |dt
+m_0\int_{0}^{T}|u(t)|^{2}dt\\
&\quad +m_1\int_{0}^{T} |u(t)|dt+ \int_{0}^{T}|e(t)||u(t)|dt.
\end{aligned}  \label{3-7}
\end{equation}
Moreover, by applying H\"{o}lder inequality and Minkowski inequality, we can have
\begin{align}
 &\int_{0}^{T}|u'(t-\sigma)|| (Au)'(t) |dt \nonumber \\
&\leq \Big(\int_{0}^{T} | (Au)'(t) |^2dt\Big)^{1/2}
\Big(\int_{0}^{T} |u'(t-\sigma)|^2dt\Big)^{1/2} \nonumber\\
&=\Big(\int_{0}^{T} |(Au)'(t)|^2dt\Big)^{1/2}
\Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2} \nonumber\\
&=\Big[\Big(\int_{0}^{T} |u'(t)-cu'(t-\sigma)|^2dt\Big)^{1/2}\Big]
 \Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2} \nonumber\\
&\leq \Big[\Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2}
 +\Big(\int_{0}^{T}|c u'(t-\sigma)|^2dt\Big)^{1/2}\Big]
 \Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2} \nonumber\\
&\leq\Big[\Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2}
 +| c|\Big(\int_{0}^{T}| u'(t)|^2dt\Big)^{1/2}\Big] 
 \Big(\int_{0}^{T} |u'(t)|^2dt\Big)^{1/2} \nonumber\\
&=(1+| c|) \int_{0}^{T}|u'(t)|^2 dt.  \label{3-8}
\end{align}  
Substituting  \eqref{3-8} into  \eqref{3-7} and by \eqref{3-4}, we  can obtain
\begin{equation}
\begin{aligned}
\int_{0}^{T}|(Au')(t)|^2dt
& \leq | c| (1+| c|) \int_{0}^{T}|u'(t)|^2dt +m_0T\| u\|_0^{2} \\
&\quad +(m_1+\| e\|_0)T\| u\|_0\\
&\leq  | c| (1+| c|) \int_{0}^{T}|u'(t)|^2dt +m_0T(D_2+\int_{0}^{T}|u'(s)|ds)^{2} \\
&\quad +(m_1+\| e\|_0)T(D_2+\int_{0}^{T}|u'(s)|ds). 
\end{aligned} \label{3-9}
\end{equation}
By applying the third part of Lemma  \eqref{lem2.1}, we have
\begin{equation}
\int_{0}^{T}|u'(t)|^2dt=\int_{0}^{T}|(A^{-1}A)u'(t)|^2dt
\leq\frac{1}{(1-| c|)^2}\int_{0}^{T}|(Au')(t)|^2dt.   \label{3-10}
\end{equation}
Substituting  \eqref{3-10} into \eqref{3-9}  and by applying H\"{o}lder inequality, 
 we obtain
 \begin{align*}
&\int_{0}^{T}|(Au')(t)|^2dt\\
&\leq [ | c| (1+| c|)+m_0T^2] \int_{0}^{T}|u'(t)|^2dt\\
&\quad + [2m_0D_2+m_1+\| e\|_0]T\sqrt{T}\Big( \int_{0}^{T}|u'(t)|^2 dt\Big)^{1/2}
+m_0TD^2_2+(m_1+\| e\|_0)TD_2\\
&\leq \frac{ | c| (1+| c|)+m_0T^2}{(1-| c|)^2}\int_{0}^{T}|(Au')(t)|^2dt\\
&\quad +\frac{[2m_0D_2+m_1+\| e\|_0]T\sqrt{T}}{1-| c|}
\Big(\int_{0}^{T}|(Au')(t)|^2dt\Big)^{1/2}
+m_0TD^2_2 \\
&\quad +(m_1+\| e\|_0)TD_2.
\end{align*}
It follows from $\frac{ | c| (1+| c|)+m_0T^2}{(1-| c|)^2}<1$ that there 
exist a positive constant $M$ such that
\[
\int_{0}^{T}|(Au')(t)|^2dt\leq M,
\]
which combining with  \eqref{3-10} gives
\begin{equation}
\int_{0}^{T}|u'(t)|^2dt \leq\frac{M}{(1-| c|)^2} . \label{3-11}
\end{equation}
Then by  \eqref{3-4}, we obtain
\begin{equation}
 \| u\|_0\leq D_2+ \frac{\sqrt{TM}}{ 1-| c|  }:=M_1 .\label{3-12}
\end{equation}
Since $[Au](t)$ is $T$-periodic, there exists $t_0\in [0,T]$ 
such that $[Au'](t_0)=0$. Hence, we have that, for $t\in[0,T]$,
\begin{equation}
\begin{aligned}
 |[Au'](t)|
&=|[Au'](t_0)+\int_{t_0}^{t}([Au'](s))'ds|\\
&\leq \lambda \int_{0}^{T}|f(u(t))||u'(t)|dt
 +\lambda \int_{0}^{T}|g(t,u(t-\delta))|dt.
\end{aligned} \label{3-13}
\end{equation}
Set $F_{M_1}=\max_{|u|\leq M_1}|f(u)|$, then by  \eqref{3-11} we obtain
\begin{equation}
\int_{0}^{T}|f(u(t))||u'(t)|dt
\leq F_{M_1}\int_{0}^{T}|u'(t)|dt
\leq \frac{F_{M_1} \sqrt{TM}}{ 1-| c| }.\label{3-14}
\end{equation}
Write 
$$
I_{+}=\{t\in[0,T]:g(t, u(t-\delta ))\geq0\};\quad
I_{-}=\{t\in[0,T]:g(t, u(t-\delta ))\leq0\}. 
$$
Then it follows from  \eqref{3-2} and (H2)(1) that
\begin{equation}
\begin{aligned}
\int_{0}^{T}| g(t,u(t-\delta))|dt&=\int_{I_{+}}g(t,u(t-\delta))dt-\int_{I_{-}}g(t,u(t-\delta))dt\\&=2\int_{I_{+}}g(t,u(t-\delta))dt\\
&\leq2m_0\int_{0}^{T} u (t-\delta)dt+2 \int_{0}^{T}m_1dt\\
&\leq 2m_0T \| u\| _0+2Tm_1.  
\end{aligned} \label{3-15}
\end{equation}
According to  \eqref{3-14} and  \eqref{3-15}, we  have
\begin{align*}
 \| Au'\|_0 
&\leq \lambda \int_{0}^{T}|f(u(t))||u'(t)|dt
 +\lambda \int_{0}^{T}|g(t,u(t-\delta))|dt\\
&\leq \lambda\Big(\frac{F_{M_1} \sqrt{TM}}{ 1-| c| }+2m_0T M_1+2Tm_1\Big),
\end{align*}
which combining with  the first  part of Lemma  \ref{lem2.1},  we  see that
\begin{align*}
| u'(t)|&=|[A^{-1}Au'](t)|\leq\frac{\| Au'\|_0}{|1-|c||}\\
 &\leq \frac{ \frac{F_{M_1} \sqrt{TM}}{ 1-| c| }+2m_0T M_1+2Tm_1}{|1-|c||},
\end{align*}
i. e.,
\begin{equation}
 \| u'\|_0\leq \frac{ \frac{F_{M_1} \sqrt{TM}}{ 1-| c| }+2m_0T M_1+2Tm_1}{|1-|c||}
:=A_3. \label{3-16}
\end{equation}
On the other hand, it follows from  \eqref{3-1} and (H2) that
\begin{equation}
\begin{aligned}
((Au)'(t+\delta))'
&=-\lambda f(u(t+\delta))u'(t+\delta)-\lambda [g_1(t+\delta, u(t))+g_0(u(t))]\\
&\quad +\lambda e(t+\delta). 
\end{aligned}  \label{3-17}
\end{equation}
Multiplying both sides of \eqref{3-17} by $u'(t)$, we have
\begin{equation}
\begin{aligned}
((Au)'(t+\delta))'u'(t)
&=-\lambda f(u(t+\delta ))u'(t+\delta )u'(t)\\
&\quad -\lambda [g_1(t+\delta, u(t))+g_0(u(t))] u'(t) 
+\lambda e(t+\delta )u'(t). 
\end{aligned} \label{3-18}
\end{equation}
Let $\tau\in [0,T]$ be as in \eqref{3-3}. For any $t\in[\tau,T]$, 
integrating \eqref{3-18} on the interval $[\tau,T]$, we have
\begin{align*}
\lambda\int_{u(\tau)}^{u(t)}g_0(u)du&=\lambda\int_{\tau}^{t}g_0(u(t))u'(t)dt\\
&=-\int_{\tau}^{t}((Au)'(t+\delta))'u'(t)dt
-\lambda \int_{\tau}^{t}f(u(t+\delta ))u'(t+\delta )u'(t)dt\\
&\quad -\lambda\int_{\tau}^{t}g_1(t+\delta,u(t))u'(t)dt
 +\lambda\int_{\tau}^{t} e(t +\delta)u'(t)dt.
\end{align*}
Set $G_{M_1}=\max_{|u|\leq M_1}|g_1(t,u)|$, then from the inequality above, 
we obtain
\begin{align*}
&\lambda|\int_{u(\tau)}^{u(t)}g_0(u)du|\\
&=\lambda|\int_{\tau}^{t}g_0(u(t))u'(t)dt|\\
&\leq  \int_{0}^{T}|((Au)'(t+\delta))'u'(t)|dt 
 +\lambda\int_{0}^{T}|f(u(t+\delta ))||u'(t+\delta )||u'(t)|dt \\
&\quad +\lambda\int_{0}^{T}|g_1(t+\delta,u(t))||u'(t)|dt
 +\lambda\int_{0}^{T} |e(t +\delta )||u'(t)|dt\\
&\leq \| u'\|_0 \int_{0}^{T}|((Au)'(t+\delta))' |dt 
 +\lambda F_{M_1}  \| u'\|^2_0T
 +\lambda G_{M_1}\| u'\|_0T
 +\lambda \| e\|_0\| u'\|_0T,
\end{align*}
i. e.,
\begin{equation}
\begin{aligned}
\lambda |\int_{u(\tau)}^{u(t)}g_0(u)du|
&\leq \| u'\|_0 \int_{0}^{T}|((Au)'(t+\delta))' |dt +\lambda F_{M_1}  \| u'\|^2_0T \\
&\quad +\lambda G_{M_1}\| u'\|_0T +\lambda \| e\|_0\| u'\|_0T.
\end{aligned} \label{3-19}
\end{equation}
Moreover,
\begin{align*}
\int_{0}^{T}|((Au)'(t+\delta))' |dt
&= \int_{0}^{T}|((Au)'(t ))' |dt\\
& \leq \lambda \Big( \int_{0}^{T}|f(u(t))||u'(t)|dt
 +  \int_{0}^{T}|g(t,u(t-\delta))|dt\Big),
\end{align*}
which combining with  \eqref{3-14} and \eqref{3-15} yields
\begin{equation}
\begin{aligned}
\int_{0}^{T}|((Au)'(t ))' |dt
& \leq \lambda \Big( \int_{0}^{T}|f(u(t))||u'(t)|dt
 +  \int_{0}^{T}|g(t,u(t-\delta))|dt\Big)\\
&\leq\lambda \Big(\frac{F_{M_1} \sqrt{TM}}{ 1-| c| }+2m_0T M_1+2Tm_1\Big).
\end{aligned}  \label{3-20}
\end{equation}
Substituting  \eqref{3-20}  into  \eqref{3-19} and combining with 
 \eqref{3-16}, obtain
\begin{align*}
 |\int_{u(\tau)}^{u(t)}g_0(u)du|
&\leq A_3 \Big(\frac{F_{M_1} \sqrt{TM}}{ 1-| c| }+2m_0T M_1+2Tm_1\Big)  
+  F_{M_1} A_3^2T \\
&\quad +  G_{M_1}A_3T +\| e\|_0A_3T<+\infty .
\end{align*}
According to (H2)(2), we can see that there exists a constant $M_2>0$ 
such that, for $t\in[\tau,T]$,
\begin{equation}
 u(t)\geq M_2.  \label{3-21}
\end{equation}
For the case $t\in[0,\tau]$, we can handle similarly.

Let us define
\begin{gather*}
0<A_1=\min\{D_1,M_2\}, \\
 A_2=\max\{D_2,M_1\}.
\end{gather*}
Then by  \eqref{3-3}, \eqref{3-12} and  \eqref{3-21}, we obtain
\begin{equation}
  A_1\leq u(t)\leq A_2.  \label{3-22}
\end{equation}
Set
\[
\Omega=\{x=(u,v)^\top\in X:\frac{A_1}{2}< u(t)< A_2+1,||u'||_0<A_3+1 \}.
\]
Then  condition (1) of Lemma  \ref{lem2.2} is satisfied. 

Assume that there exists $x\in \partial\Omega\cap \ker L $ such that 
$ QN x=\frac{1}{T}\int_{0}^{T} N x(s)ds=0$, i. e.,
\begin{equation}
\frac{1}{T}\int_{0}^{T}[-f(u(t))u'(t)-g(t, u(t-\delta ))
+ e(t)]dt=0,  \label{3-23}
\end{equation}
then we have
\[
\frac{1}{T}\int_{0}^{T}g(t,u(t-\delta ))dt=0.
\]
It follows from the (H1)(1) we can see that
\[
\frac{A_1}{2}<D_1 \leq u(t)\leq D_2<A_2+1,
\]
which contradicts the assumption $x\in \partial\Omega$.
So  for all $x\in  \ker L \cap\partial\Omega$, we have $ QN x\neq0$. 
Therefore,  condition (2) of Lemma  \ref{lem2.2} is satisfied.

Finally, we prove that  condition (3) of Lemma \ref{lem2.1} is also satisfied.
Let
\[
z=Kx=x-\frac{A_1+A_2}{2}  ,
\]
then, we have
\[
x=z+ \frac{A_1+A_2}{2}  .
\]
Define $ J:ImQ\to \ker L $ is a linear isomorphism with
$ J (u )= u $,
and define
\[
H(\mu,x)=\mu K x+(1-\mu)  JQN x,~~\forall(x,\mu)\in\Omega\times[0,1].
\]
Then,
\begin{equation}
H(\mu,x)= \mu x-\frac{\mu(A_1+A_2)}{2}
 +\frac{1-\mu}{T} \int_{0}^{T}g(t, x) dt . \label{3-24}
\end{equation}
Now we claim that $H(\mu,x)$ is a homotopic mapping.
 Assume, by way of contradiction, i. e., there exists 
$\mu_0\in [0,1]$ and $x_0\in \partial\Omega$ such that $H(\mu_0,x_0)=0$.

Substituting $\mu_0$ and $x_0$ into  \eqref{3-24}, we have
\begin{equation}
 H(\mu_0,x_0)=  \mu_0 x_0-\frac{\mu_0(A_1+A_2)}{2} 
 +(1-\mu_0) \overline{g}(x_0).  \label{3-25}
\end{equation}
It follows $H(\mu_0,x_0)=0$  that   $x_0=A_1$ or $A_2$. Furthermore,
If $x_0=A_1$, it follows from  (H1)(2) that $ \overline{g} (x_0)<0$, then  we have
\begin{equation}
 \mu_0 x_0-\frac{\mu_0(A_1+A_2)}{2} 
+(1-\mu_0)\overline{ g} (x_0)<\mu_0(x_0-\frac{A_1+A_2}{2})<0 .\label{3-26}
\end{equation}
If $x_0=A_2$, it follows from  (H1)(2) that $ \overline{g} (x_0)>0$, then  we have
\begin{equation}
 \mu_0 x_0-\frac{\mu_0(A_1+A_2)}{2} +(1-\mu_0) \overline{g} (x_0)
>\mu_0(x_0-\frac{A_1+A_2}{2})>0. \label{3-27}
\end{equation}
Combining with  \eqref{3-26} and  \eqref{3-27}, we can see that 
$H(\mu_0,x_0)\neq0$, which contradicts the assumption. 
Therefore $H(\mu,x)$ is a homotopic mapping and $x^\top H(\mu,x)\neq0$, 
for all $(x,\mu)\in (\partial\Omega\cap   \ker L )\times[0,1]$, then
\begin{align*}
\deg({\rm JQN},\Omega\cap {\rm \ker L},0)
&=\deg(H(0,x),\Omega\cap{\rm  \ker L},0)\\
&=\deg(H(1,x),\Omega\cap{\rm  \ker L},0)\\
&=\deg(Kx,\Omega\cap{\rm  \ker L},0)\\
&=\sum_{x\in K^{-1}(0)} {\rm sgn}|K'(x)|\\
&=1\neq0.
\end{align*}
Thus,  condition (3) of Lemma \ref{lem2.2} is also satisfied.

Therefore, by applying Lemma \ref{lem2.1}, we can conclude  
that \eqref{1-2} has at least one positive $T$-periodic solution.
\end{proof}


\section{Example}

In this section, we provide an example to illustrate results from the
 previous sections.

\begin{example} \label{examp4.1} \rm
Consider the  neutral Li\'enard differential  equation with a singularity
and a deviating argument,
\begin{equation}
\begin{aligned}
&( (u(t)-0.1u(t-\pi))'))'+\Big(\frac{u^2(t)}{3+u(t)}+9\Big)u'(t)\\
&+\frac{1}{2}(1+\frac{1}{2}\sin8t)u(t-\delta) 
 -\frac{1}{u (t-\delta)}= \sin 8t.  
 \end{aligned} \label{4-1}
 \end{equation}
Corresponding to Theorem \ref{thm3.1} and  \eqref{1-2}, we have
\begin{gather*}
f(u(t))=\frac{u^2(t)}{3+u (t)}+9,\quad e(t)= \sin8t,\\
g(t,u(t-\delta))= \frac{1}{2}\Big(1+\frac{1}{2}\sin8t \Big)
u  (t-\delta)-\frac{1}{u (t-\delta)}.
\end{gather*}
Then, we choose 
$$
\sigma=\pi,\quad c =0.1,\quad T=\frac{\pi}{4}, \quad
m_0=\frac{3}{4},\quad D_1=2,\quad D_2=3.
$$
Thus, $| c|<1$ and the conditions (H1) and (H2) are satisfied. 
Meanwhile, we have
\[
\frac{ | c| (1+| c|)+m_0T^2}{(1-| c|)^2}\approx 0.706 <1.
\]
Hence, by applying Theorem \ref{thm3.1}, we can see that \eqref{4-1} 
has at least one positive $\frac{\pi}{4}$-periodic solution.
\end{example}

\begin{remark} \rm
 Since only a few papers consider positive periodic solutions for the  
neutral Li\'enard equation. One can easily see that all the results in
 \cite{1993-Fonda-p1294--1311}-\cite{1977-Hale-p} and the references therein 
are not applicable to \eqref{4-1} for obtaining positive periodic 
solutions with period $\frac{\pi}{4}$. 
This implies that the results in this paper are essentially new.
\end{remark}


\subsection*{Acknowledgments}
The research was supported by the the National Natural Science Foundation 
of China (Grant No.11271197). The authors would like to express their 
gratitude to Professor Giovanni Monica Bisci for his valuable guidance, 
and also to the reviewers for their
valuable comments and constructive suggestions, which help  us  improve
this article.

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\end{document}