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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 177, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/177\hfil Li\'enard type equation]
{Li\'enard type p-Laplacian neutral Rayleigh
equation with a deviating argument}

\author[A. Anane, O. Chakrone, L. Moutaouekkil\hfil EJDE-2010/177\hfilneg]
{Aomar Anane, Omar Chakrone, Loubna Moutaouekkil}  % in alphabetical order

\address{Universit\'e Mohamed I, Facult\'e des Sciences,
D\'epartement de Math\'ematiques et Informatique,
Oujda, Maroc}
\email[Aomar Anane]{anane@sciences.univ-oujda.ac.ma}
\email[Omar Chakrone]{chakrone@yahoo.fr}
\email[Loubna Moutaouekkil]{loubna\_anits@yahoo.fr}

\thanks{Submitted September 15, 2010. Published December 22, 2010.}
\subjclass[2000]{34C25, 34B15}
\keywords{Periodic solution; neutral Rayleigh equation;
 Li\'enard equation; \hfill\break\indent
 Deviating argument; p-Laplacian;
 Man\'asevich-Mawhin continuation}

\begin{abstract}
 Based on Man\'asevich-Mawhin continuation theorem, we prove
 the existence of  periodic solutions for Li\'enard type
 $p$-Laplacian neutral Rayleigh equations with a deviating
 argument,
 $$
 (\phi_p(x(t)-c x(t-\sigma))')'+f(x(t))x'(t)+
 g(t,x(t-\tau(t)))=e(t).
 $$
 An example is provided to illustrate our results.
\end{abstract}

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


\section{Introduction}

The existence of  periodic solutions for Li\'enard type
$p$-Laplacian equation with a deviating argument
\begin{equation} \label{e1.1}
(\phi_p(x'(t)))'+f(x(t))x'(t)+ g(t,x(t-\tau(t)))=e(t)
\end{equation}
has been studied using the coincidence degree theory
\cite{l1}.
Zhu and Lu \cite{z2}, studied the existence  of  periodic solution for
$p$-Laplacian neutral functional differential equation with a
deviating argument when $p>2$
\begin{equation} \label{e1.2}
(\phi_p(x(t)-c x(t-\sigma))')'+g(t,x(t-\tau(t)))=e(t).
\end{equation}
They obtained some results by transforming \eqref{e1.2} into a
two-dimensional system to which Mawhin's continuation theorem was
applied.

Peng \cite{p1} discussed the existence  of  periodic solution for
$p$-Laplacian neutral Rayleigh  equation with a deviating argument
\begin{equation} \label{e1.3}
(\phi_p(x(t)-c x(t-\sigma))')'+f(x'(t))+g(t,x(t-\tau(t)))=e(t)
\end{equation}
and obtained the existence of periodic solutions
under the assumption  $f(0)=0$ and $\int_0^T e(t)dt=0$.

Throughout this paper, $2<p<\infty$ is a fixed real number. The
conjugate exponent of $p $ is denoted by $q$; i.e.,
$\frac{1}{ p}+\frac{1}{ q}=1$. Let
$\phi_p:\mathbb{R}\to \mathbb{R}$ be defined by
$\phi_p(s)=|s|^{p-2}s$ for $s\neq0$, and
$\phi_p(0)=0$.
In this article, we will investigate the existence of
periodic solution to the Li\'enard type p-Laplacian neutral Rayleigh
equation
\begin{equation} \label{e1.4}
(\phi_p(x(t)-c x(t-\sigma))')'+f(x(t))x'(t)+g(t,x(t-\tau(t)))=e(t)
\end{equation}
where  $f$, $e$ and $\tau$ are real continuous functions on
$\mathbb{R}$. $\tau$ and $e$ are periodic with period $T$, $T>0$ is
fixed. $g$ is continuous function defined on $\mathbb{R}^2$ and
$T$-periodic in the first argument, $c$ and $\sigma$ are constants
such that $|c|\neq1$.

\section{Preliminaries}

Let $\mathcal{C}_T=\{x\in\mathcal{C}(\mathbb{R},\mathbb{R}):
x(t+T)=x(t)\}$ and
$\mathcal{C}_T^{1}=\{x\in\mathcal{C}^{1}(\mathbb{R},\mathbb{R}):
x(t+T)=x(t)\}$.
$\mathcal{C}_T$ is a Banach space endowed with the norm
$\|x\|_{\infty}=\max|x(t)|_{t\in[0,T]}$. $\mathcal{C}_T^{1}$ is a
Banach space endowed with the norm
$\|x\|=\max\{\|x\|_{\infty},\|x'\|_{\infty}\}$. In what follows, we
will use $\|.\|_p$ to denote the $L^{P}$-norm. We also define a
linear operator $A:\mathcal{C}_T\to \mathcal{C}_T$,
$$
(Ax)(t)=x(t)-cx(t-\sigma).
$$

\begin{lemma}[\cite{l2,z1}] \label{lem2.1}
If $|c|\neq1$, then $A$ has continuous bounded inverse on
$\mathcal{C}_T$, and
\begin{itemize}
\item[(1)] $\|A^{-1}x\|_\infty\leq\frac{ \|x\|_\infty}{ |1-|c||}$,
 for all $x\in\mathcal{C}_T$;

\item[(2)]
\[
(A^{-1}x)(t)=   \begin{cases}
 \sum_{j\geq0}c^{j}x(t-j\sigma),   &|c|<1\\
 -\sum_{j\geq1}c^{-j}x(t+j\sigma), &|c|>1.
\end{cases} %E4
\]

\item[(3)]
$$
\int_0^T|(A^{-1}x)(t)|dt\leq\frac{1}{|1-|c||}\int_0^T|x(t)|dt,
\quad \forall x\in\mathcal{C}_T.
$$
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{p1}]\label{lem2.2}
If $|c|\neq1$ and $p>1$, then
\begin{equation} \label{e2.1}
\int_0^T|(A^{-1}x)(t)|^pdt\leq\frac{
1}{ |1-|c||^p}\int_0^T|x(t)|^pdt,\quad \forall
x\in\mathcal{C}_T.
\end{equation}
\end{lemma}

For the $T$-periodic boundary value problem
\begin{equation} \label{e2.2}
(\phi_p(x'(t)))'=\widetilde{f}(t,x,x'),\quad
x(0)=x(T),\quad x'(0)=x'(T),
\end{equation}
where $\widetilde{f}\in\mathcal{C}(\mathbb{R}^{3},\mathbb{R})$, we
have the following result.

\begin{lemma}[\cite{m1}] \label{lem2.3}
Let $\Omega$ be an open bounded set in $\mathcal{C}_T^{1}$, and
let the following conditions hold:
\begin{itemize}
\item [(i)] For each $\lambda\in(0,1)$, the problem
 $$
    (\phi_p(x'(t)))'=\lambda\widetilde{f}(t,x,x'),\quad
 x(0)=x(T),\quad x'(0)=x'(T)
 $$ has no solution on $\partial\Omega$.

\item [(ii)] The equation
 $$
 F(a)=\frac{1}{T}\int^T_0\widetilde{f}(t,a,0)dt=0
 $$
 has no solution on  $\partial\Omega\cap\mathbb{R}$.

\item [(iii)] The Brouwer degree of
    $F$, $\deg(F,\Omega\cap\mathbb{R},0)\neq0$.

\end{itemize}
    Then the $T$-periodic boundary value problem \eqref{e2.2}
has at least one periodic solution on $\overline{\Omega}$.
\end{lemma}

\section{Main results}

\begin{theorem}\label{thm3.1}
 Suppose that $p>2$ and there exist constants $r_1\geq0$,
 $r_2\geq 0$, $d>0$ and $k>0$ such that
\begin{itemize}
\item [(A1)] $|f(x)|\leq  k+r_1|x|^{p-2}$ for $x\in\mathbb{R}$;
\item [(A2)] $x[g(t,x)-e(t)]<0$ for $|x|>d $ and $ t\in\mathbb{R}$;
\item [(A3)] $\lim_{x\to -\infty}\frac{ |g(t,x)-e(t)|}{ |x|^{p-1}}=r_2$.
\end{itemize}
Then \eqref{e1.4} has at least one $T$-periodic solution if
$$
\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p.
$$
\end{theorem}

\begin{proof}
Consider the homotopic equation of \eqref{e1.4} as
follows:
\begin{equation} \label{e3.1}
(\phi_p(x(t)-c x(t-\sigma))')'+\lambda f(x(t))x'(t)+\lambda
g(t,x(t-\tau(t)))=\lambda e(t),\quad \lambda\in(0,1).
\end{equation}
We claim that the set of all possible periodic solution of
\eqref{e3.1} are bounded in $\mathcal{C}_T^{1}$.

Let $x(t)\in \mathcal{C}_T^{1}$ be an arbitrary solution of
\eqref{e3.1} with period $T$. By integrating two sides of
\eqref{e3.1} over $[0,T]$, and noticing that $x'(0)=x'(T)$, we have
\begin{equation} \label{e3.2}
\int _0^T[g(t,x(t-\tau(t)))- e(t)]dt=0.
\end{equation}
By the integral mean value theorem, there is a constant
$\xi\in[0,T]$ such that
 $g(\xi,x(\xi-\tau(\xi)))- e(\xi)=0$. So
from assumption (A2), we can get $|x(\xi-\tau(\xi))|\leq d$.
Let $\xi-\tau(\xi)=mT+\overline{\xi}$, where
$\overline{\xi}\in[0,T]$, and $m$ is an integer. Then, we have
$$
|x(t)|=|x(\overline{\xi})+\int_{\overline{\xi}}^{t}x'(s)ds|\leq
d+\int_{\overline{\xi}}^{t}|x'(s)|ds,\quad
t\in[\overline{\xi},\overline{\xi}+T],
$$
and
$$
|x(t)|=|x(t-T)|=|x(\overline{\xi})-\int_{t-T}^{\overline{\xi}}x'(s)ds|\leq
d+\int_{t-T}^{\overline{\xi}}|x'(s)|ds,\quad
t\in[\overline{\xi},\overline{\xi}+T].
$$
Combining the above two inequalities, we obtain
\begin{equation} \label{e3.3}
\begin{aligned}
\|x\|_{\infty}
&=\max_{t\in[0,T]}|x(t)|
=\max_{t\in[\overline{\xi},\overline{\xi}+T]}|x(t)|\\
&\leq\max_{t\in[\overline{\xi},\overline{\xi}+T]}
\Big\{d+\frac{1}{2}\Big(\int_{\overline{\xi}}^{t}
|x'(s)|ds+\int_{t-T}^{\overline{\xi}}|x'(s)|ds\Big)\Big\}\\
&\leq d+\frac{1}{2}\int_0^T|x'(s)|ds.
\end{aligned}
\end{equation}
In view of $\frac{1}{
2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p$, there
exist a constant $\varepsilon>0$ such that
$$
\frac{1}{ 2^{p-1}}(1+|c|)T^{p-1}(r_1+T(r_2+\varepsilon))<|1-|c||^p.
$$
  From assumption (A3),  there exist a constant
$\rho>d$ such that
\begin{equation} \label{e3.4}
|g(t,x(t-\tau(t)))-
e(t)|dt\leq(r_2+\varepsilon)|x|^{p-1}\;\;\;for\;
t\in\mathbb{R}\;\;and\; \;x<-\rho.
\end{equation}
Denote $E_1=\{t\in[0,T],x(t-\tau(t))\leq-\rho\}$,
$E_2=\{t\in[0,T],|x(t-\tau(t))|<\rho\}$,
$E_3=\{t\in[0,T],x(t-\tau(t))\geq\rho\}$.
By  \eqref{e3.2}, it is easy to see that
\begin{equation} \label{e3.5}
\Big(\int_{E_1}+\int_{E_2}+\int_{E_3}\Big)[g(t,x(t-\tau(t)))-
e(t)]dt=0.
\end{equation}
Hence
\begin{equation} \label{e3.6}
\begin{aligned}
\int_{E_3}|g(t,x(t-\tau(t)))-
e(t)|dt
&=-\int_{E_3}[g(t,x(t-\tau(t)))- e(t)]dt\\&
=\Big(\int_{E_1}+\int_{E_2}\Big)[g(t,x(t-\tau(t)))-e(t)]dt\\
& \leq \Big(\int_{E_1}+\int_{E_2}\Big)|g(t,x(t-\tau(t)))- e(t)|dt.
\end{aligned}
\end{equation}
Therefore, by \eqref{e3.4} and \eqref{e3.6}, we obtain
\begin{equation} \label{e3.7}
\begin{aligned}
\int_0^T|g(t,x(t-\tau(t)))-e(t)|dt
&=\Big(\int_{E_1}+\int_{E_2}+\int_{E_3}\Big)|g(t,x(t-\tau(t)))-
e(t)|dt\\
&\leq 2\Big(\int_{E_1}+\int_{E_2}\Big)|g(t,x(t-\tau(t)))- e(t)|dt\\
&\leq 2 \int_{E_1}(r_2+\varepsilon)|x(t-\tau(t))|^{p-1}dt
 +2\widetilde{g}_{\rho}T\\
& \leq 2 (r_2+\varepsilon)T \|x\|_{\infty}^{p-1}+2\widetilde{g}_{\rho}T.
\end{aligned}
\end{equation}
Where $\widetilde{g}_{\rho}=\max_{t\in E_2}|g(t,x(t-\tau(t)))-
e(t)|$. Multiplying both sides of \eqref{e3.1} by
$(Ax)(t)=x(t)-cx(t-\sigma)$ and integrating them over $[0,T]$, we
have
\begin{equation} \label{e3.8}
\begin{aligned}
\|Ax'\|_p^p
& =\lambda\int_0^T (Ax)(t)\left[f(x(t))x'(t)+
g(t,x(t-\tau(t)))- e(t)\right]dt\\
& \leq (1+|c|)\|x\|_{\infty}\int_0^T\left [|f(x(t))x'(t)| +
|g(t,x(t-\tau(t)))- e(t)|\right]dt.
\end{aligned}
\end{equation}
  From assumption (A1), we obtain.
\begin{equation} \label{e3.9}
\int_0^T|f(x(t))x'(t)|dt \leq k\int_0^T|x'(t)|dt+ r_1
\int_0^T|x'(t)||x(t)|^{p-2}dt.
\end{equation}
Using H\"older inequality, and substituting \eqref{e3.3}
into \eqref{e3.9}, we obtain
\begin{equation} \label{e3.10}
\int_0^T|f(x(t))x'(t)|dt \leq
kT^{1/q}\|x'\|_p+r_1T^{1/q}\|x'\|_p
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-2}.
\end{equation}
  From \eqref{e3.3} and \eqref{e3.7}, we have
\begin{equation} \label{e3.11}
\int_0^T|g(t,x(t-\tau(t)))- e(t)|dt \leq
2\;\widetilde{g}_{\rho}\; T + 2(r_2+\varepsilon)T
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}.
\end{equation}
 Substituting \eqref{e3.10}, \eqref{e3.11}
and \eqref{e3.3} into \eqref{e3.8}, we obtain
\begin{equation} \label{e3.12}
\begin{aligned}
& \|Ax'\|_p^p\\
&\leq (1+|c|)\Big[ kT^{1/q}\|x'\|_p
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)\\
&\quad +\Big(d+\frac{1}{2} \int_0^T|x'(t)|dt\Big)^{p-1}r_1T^{1/q}\|x'\|_p\\
&\quad +2(r_2+\varepsilon)T
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^p
+2\widetilde{g}_{\rho} T
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)\Big].
\end{aligned}
\end{equation}

Case(1). If $\int_0^T|x'(t)|dt=0$, from \eqref{e3.3}, we have
$\|x\|_{\infty}<d$.

Case(2). If $\int_0^T|x'(t)|dt>0$, then
\begin{equation} \label{e3.13}
\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}
=\Big(\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}
\Big(1+\frac{2d}{\int_0^T|x'(t)|dt}\Big)^{p-1}.
\end{equation}
By elementary analysis, there is a constant $\delta>0$ such that
\begin{equation} \label{e3.14}
(1+u)^{p-1}\leq1+pu,\quad \forall u \in [0,\delta].
\end{equation}

If $ 2d / \int_0^T|x'(t)|dt>\delta$, then
$\int_0^T|x'(t)|dt<2d/\delta$, so from \eqref{e3.3}, we have
$\|x\|_{\infty}<d+(d/\delta)$.

If $2d/ \int_0^T|x'(t)|dt\leq\delta$, by \eqref{e3.13} and
\eqref{e3.14},
\begin{equation} \label{e3.15}
\begin{aligned}
&\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}\\
&\leq\Big(\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}
\Big(1+\frac{2pd}{\int_0^T|x'(t)|dt}\Big)\\
&\leq \big(\frac{1}{2}\big)^{p-1}
\Big(\int_0^T|x'(t)|dt\Big)^{p-1}+\big(\frac{1}{2}\big)^{p-2}pd
\Big(\int_0^T|x'(t)|dt\Big)^{p-2}\\
&\leq \big(\frac{1}{2}\big)^{p-1}
T^{\frac{p-1}{q}}\|x'\|_p^{p-1}+\big(\frac{1}{2}\big)^{p-2}p
d T^{\frac{p-2}{q}}\|x'\|_p^{p-2}.
\end{aligned}
\end{equation}
Similarly, from\eqref{e3.14}, there is a constant $\delta'>0$
such that
\begin{equation} \label{e3.16}
(1+u)^p\leq1+(1+p)u,\quad \forall u \in [0,\delta']
\end{equation}

If $ 2d/ \int_0^T|x'(t)|dt>\delta'$, then
$\int_0^T|x'(t)|dt<2d/ \delta'$, so from \eqref{e3.3}, we have
$\|x\|_{\infty}<d+(d/\delta')$.

If $2d/ \int_0^T|x'(t)|dt\leq\delta'$, by \eqref{e3.16}, we
have
\begin{equation} \label{e3.17}
\begin{aligned}
&\Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^p\\
&\leq\Big(\frac{1}{2}\int_0^T|x'(t)|dt\Big)^p
\Big(1+\frac{2(p+1)d}{\int_0^T|x'(t)|dt}\Big)\\
&\leq \big(\frac{1}{2}\big)^p
\Big(\int_0^T|x'(t)|dt\Big)^p+\big(\frac{1}{2}\big)^{p-1}(p+1)d
\Big(\int_0^T|x'(t)|dt\Big)^{p-1}\\
&\leq \big(\frac{1}{2}\big)^pT^{\frac{p}{q}}\|x'\|_p^p
+\big(\frac{1}{2}\big)^{p-1}(p+1)d T^{\frac{p-1}{q}}\|x'\|_p^{p-1}.
\end{aligned}
\end{equation}
Substituting \eqref{e3.15} and \eqref{e3.17} into \eqref{e3.12}
and using H\"older inequality, we obtain
\begin{equation} \label{e3.18}
\begin{aligned}
\|Ax'\|_p^p
&\leq (1+|c|)[(\frac{1}{2^{p-1}}(r_2+\varepsilon)T^p
+\frac{1}{2^{p-1}}r_1T^{\frac{p}{q}})\|x'\|_p^p+a_0\|x'\|_p^{p-1}\\
&\quad +a_1\|x'\|_p^2+a_2\|x'\|_p+ 2\widetilde{g}_{\rho}Td],
\end{aligned}
\end{equation}
where $a_0,a_1$ and $a_2$ are constants depending on
$T,r_1,k,r_2,d,p$ and $c$.  Then from Lemma\eqref{e2.2}, we have
$$
|1-|c||^p\|x'\|_p^p=|1-|c||^p\|A^{-1}Ax'\|_p^p\leq\|Ax'\|_p^p\,.
$$
So it follows from \eqref{e3.18} that
\begin{equation} \label{e3.19}
\begin{aligned}
 |1-|c||^p\|x'\|_p^p
&\leq (1+|c|)[(\frac{1}{2^{p-1}}
 T^{p-1}(r_1+T(r_2+\varepsilon)))\|x'\|_p^p+a_0\|x'\|_p^{p-1}\\
&\quad +a_1\|x'\|_p^2+a_2\|x'\|_p+2\widetilde{g}_{\rho}Td ].
\end{aligned}
\end{equation}
As $p>2$ and $\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p$, there exists a
constant $R_3>0$ such that
\begin{equation} \label{e3.20}
\|x'\|_p\leq R_3.
\end{equation}
Which together with \eqref{e3.3} implies that there is a
positive number $R_4$ such that
\begin{equation} \label{e3.21}
\|x\|_{\infty}\leq R_4.
\end{equation}
  From \eqref{e3.1}, we have
\begin{equation} \label{e3.22}
\begin{aligned}
&\int_0^T|(\phi_p(A x')(t))'|dt\\
&\leq  \int_0^T[|f(x(t))x'(t)|+|g(t,x(t-\tau(t)))+|e(t)|]dt\\&\leq k T
 ^{1/q}\|x'\|_p+\int_0^Tr_1|x|^{p-2}|x'(t)|+Tg_{R_4}+\int_0^T|e(t)|dt\\&\leq k T
 ^{1/q}\|x'\|_p+r_1\|x\|_{\infty}^{p-2}T
 ^{1/q}\|x'\|_p+Tg_{R_4}+\int_0^T|e(t)|dt\\&\leq kT
 ^{1/q}R_3+r_1R_4^{p-2}T
 ^{1/q}R_3+Tg_{R_4}+\int_0^T|e(t)|dt=R_5,
\end{aligned}
\end{equation}
where $g_{R_4}=\max_{|x|\leq R_4,
t\in[0,T]}|g(t,x(t-\tau(t)))|$. As $(Ax)(0)=(Ax)(T)$, there exists
$t_0\in ]0,T[$ such that $(Ax')(t_0)=0$, while $\phi_p(0)=0$
we see $\phi_p(A x')(t_0)=0$. Thus, for any $t\in [0,T]$, we
have
$$
|\phi_p(A x')(t))|=|\int_{t_0}^{t}\phi_p(A
x')(s))ds|\leq \int_0^T|(\phi_p(A x')(s))'|dt\leq R_5.
$$
  From which, it follows that
\begin{equation} \label{e3.23}
\|Ax'\|_{\infty}\leq R_5^{q-1}.
\end{equation}
  From Lemma \ref{lem2.1}, we derive
\begin{equation} \label{e3.24}
\|x'\|_{\infty}=\|A^{-1}Ax'\|_{\infty}
\leq\frac{\|Ax'\|_{\infty}}{|1-|c||}
\leq\frac{R_5^{q-1}}{|1-|c||}=R_6.
\end{equation}
Now, let $y(t)=(Ax)(t)$, we can see that \eqref{e3.1} is
equivalent to the equation
\begin{equation} \label{e3.25}
(\phi_p(y'(t)))'+\lambda f((A^{-1}y)(t))(A^{-1}y')(t)+\lambda
g(t,(A^{-1}y)(t-\tau(t)))=\lambda e(t).
\end{equation}
So, if $y$ is an periodic solution of \eqref{e3.25},
then $x=A^{-1}y$ is $T$-periodic solution of \eqref{e3.1}.

Let $R_7=2(1+|c|)\max\{R_4,R_6,d\}$,
$\Omega=\{y\in \mathcal{C}_T^{1}:\|y\|<R_7\}$, we can see
that \eqref{e3.25} has no solution on $\partial\Omega $ for
$ \lambda\in(0,1)$. In fact, if $y=Ax $ is a solution
\eqref{e3.25} on $\partial\Omega$, then
$\|y\|=R_7,\|y\|_{\infty}=R_7$ or $\|y'\|_{\infty}=R_7$. If
$\|y\|_{\infty}=R_7$, then
$\|x\|_{\infty}\geq\frac{\|y\|_{\infty}}{1+|c|}=2\max\{R_4,R_6,d\}>R_4$,
from \eqref{e3.21} which is a contradiction. Similarly,
$\|y'\|_{\infty}=R_7$ is also impossible. If
$y\in\partial\Omega\cap\mathbb{R}$, then $y$ is a constant and
$|y|=R_7$, $x=A^{-1}y=\frac{y}{1-c}$,
$|x|\geq2\max\{R_4,R_6,d\}$. Let
$$
F(y)=\frac{1}{T}\int_0^T[e(t)-f((A^{-1}y)(t))(A^{-1}y')(t)
-g(t,(A^{-1}y)(t-\tau(t)))].
$$
Then
$F(y)=\frac{1}{T}\int_0^T[e(t)-g(t,\frac{y}{1-c})]dt$ for
$y\in\partial\Omega\cap\mathbb{R} $.  From (A2), we know that
$F(y)\neq0$ on $\partial\Omega\cap\mathbb{R}$, so condition (ii)
in Lemma \ref{lem2.3} is satisfied.
Define
$$
H(y,\mu)=\mu(A^{-1}y)+(1-\mu)F(y),$$
$y\in\partial\Omega\cap\mathbb{R}$, $\mu\in[0,1]$.
Then
$$
(-A^{-1}y)H(y,\mu)=-\mu(A^{-1}y)^2-(1-\mu)(A^{-1}y)
\frac{1}{T}\int_0^T[e(t)-g(t,(A^{-1}y)(t-\tau(t)))]dt.
$$
 From (A2) we obtain $(A^{-1}y)H(y,\mu)>0$.
Thus $H(y,\mu)$ is a homotopic transformation  and
$\deg[F,\Omega\cap\mathbb{R},0]=\deg[A^{-1}y,\Omega\cap\mathbb{R},0]
\neq 0$. So, for \eqref{e3.25}, all of conditions of
Lemma \ref{lem2.3} are satisfied.
Applying Lemma \ref{lem2.3}, we conclude that
\begin{equation} \label{e3.26}
(\phi_p(y'(t)))'+ f((A^{-1}y)(t))(A^{-1}y')(t)+
g(t,(A^{-1}y)(t-\tau(t)))= e(t)
\end{equation}
has at least one $T$-periodic solution $\overline{y}$. Therefore,
$\overline{x}=A^{-1}\overline{y}$ is an $T$-periodic solution of
\eqref{e1.4}.
\end{proof}

Similarly, we can prove the following Theorem.

\begin{theorem}\label{thm3.2}
 Suppose that $p>2$ and that there exist constants $r_1\geq0$,
 $r_2\geq0$, $d>0$ and $k>0$ such that

\begin{itemize}
\item [(A1)] $|f(x)|\leq  k+r_1|x|^{p-2}$ for $x\in\mathbb{R}$;
\item [(A2)] $x[g(t,x)-e(t)]<0$ for $|x|>d $ and
    $t\in\mathbb{R}$;
\item [(A3)] $\lim_{x\to +\infty}
 \frac{|g(t,x)-e(t)|}{|x|^{p-1}}=r_2$.
\end{itemize}
then \eqref{e1.4} has at least one $T$-periodic solution if
$$
\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p.
$$
\end{theorem}

\section{Example}

In this section, we  illustrate  Theorem \ref{thm3.1} with
the following example.
 Consider the  equation
\begin{equation} \label{e4.1}
(\phi_3(x(t)-5 x(t-\pi))')'+f(x(t))x'(t)+
g(t,x(t-\sin(t)))=e^{\cos^2t},
\end{equation}
where
$p=3$, $c=5$, $\sigma=4$, $\mathrm{T}=2\pi$,
$\tau(t)=\sin t$, $e(t)=e^{\cos^2t}$,
$f(x)=2+\frac{\sqrt{|x|}}{\pi^2}$,
$$
g(t,x) = \begin{cases}
 -xe^{\sin^2t},  & x\geq 0 \\
 \frac{ x^2}{18\pi^2}, & x<0 .
\end{cases}
$$
Let
$d=3\pi\sqrt{2e}$, $r_1=\frac{1}{\pi^2}$,
$r_2=\frac{1}{18\pi^2}$,
$k=4+\frac{\max_{|x|\leq1}|f(x)|}{\pi^2}$.
We can easily check the condition (A1), (A2) and (A3) of
Theorem \ref{thm3.1} hold. Furthermore,
$$
\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)
=6+\frac{2\pi}{3}<|1-|c||^p=64.
$$
 By Theorem \ref{thm3.1}, \eqref{e4.1} has at least one
$2\pi$-periodic solution.

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\end{document}