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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 107, pp. 1--11.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/107\hfil Periodic solutions]
{Periodic solutions for a  kind of rayleigh
 equation  with  two deviating arguments}

\author[Y. Wu, B. Xiao, H.Zhang\hfil EJDE-2006/107\hfilneg]
{Yuanheng Wu, Bing Xiao, Hong Zhang} % in alphabetical order

\address{Yuanheng Wu \newline
College  of  Continuing  Education,
Guangdong  University of  Foreign Studies,
Guangzhou  510420, China}
\email{wyhcd2006@yahoo.com.cn}

\address{Bing Xiao \newline
Department of Mathematics,  Hunan University of Arts and Science,
Changde, Hunan 415000, China}
\email{changde1218@yahoo.com.cn}

\address{Hong Zhang \newline
Department of Mathematics,  Hunan University of Arts and Science,
Changde, Hunan 415000, China}
\email{hongzhang320@yahoo.com.cn}

\date{}
\thanks{Submitted January 27, 2006. Published September 8, 2006.}
\thanks{Supported by the NNSF of China and by project 05JJ40009 from
the Hunan Provincial \hfill\break\indent
Natural Science Foundation of China}
\subjclass[2000]{34C25, 34D40}
\keywords{Rayleigh equation; deviating argument; periodic solution;
\hfill\break\indent coincidence degree}

\begin{abstract}
  In this paper, we use the coincidence degree theory to
  establish new results  on the existence of  $T$-periodic
  solutions for the Rayleigh equation  with two deviating
  arguments of the form
$$
    x''+f(x(t), x'(t))+g_{1}(t,x(t-\tau_{1}(t)))
    +g_{2}(t,x(t-\tau_{2}(t)))=p(t).
$$
\end{abstract}

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

\section{Introduction}

 Consider  the Rayleigh  equation  with two deviating arguments
\begin{equation}
    x''+f(x(t),x'(t))
    +g_{1}(t,x(t-\tau_{1}(t)))
    +g_{2}(t,x(t-\tau_{2}(t)))=p(t),
\label{e1.1}
\end{equation}
where   $\tau_{1}$, $\tau_{2}$, $p:\mathbb{R}\to \mathbb{R}$
   and $f$,  $g_{1}$, $g_{2}:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ are continuous
    functions, $f(x, 0)=0$, $\tau_{1}$, $\tau_{2}$ and $p$ are $T$-periodic,
    $g_{1}$ and $g_{2}$ are $T$-periodic in the first argument,  and $T>0$. In recent years,
    the problem of the existence of periodic solutions of  \eqref{e1.1}  has
been extensively studied in the literature. We refer the reader to
\cite{g2,s1,s2,s3,x1} and the references cited therein. Moreover,
    in the above-mentioned literature, we find the following
assumptions:
\begin{itemize}
\item[(H0)]  $g_{1}(t,x)    +g_{2}(t,x)=g(x)$,
$g(x)\in C(\mathbb{R}, \mathbb{R})$ and there exist constants $k_{1}\geq 0$
  and $k_{2}\geq 0$ such that one of the following conditions
  holds:\\
 (1)  $ xg(x)>0$,  for  all $|x|>k_{1}$,
 and $g(x)\geq - k_{2}$,  for  all $x\leq -k_{1}$,
 \\
(2) $ xg(x)>0$, for  all $|x|>k_{1}$,
 and  $g(x)\leq k_{2}$, for  all $x\geq k_{1}$;

\item[(H1)]  $g_{1}(t,x)+g_{2}(t,x)=g(x)$, $g(x)\in C^{1}(\mathbb{R}, \mathbb{R})$ and there exists a constant
$K\geq 0$ such that
$$
|g'(x)|\leq K, \forall x\in \mathbb{R};
$$

\item[(H2)]  $f(x, y)=f(y)$,  and there exist constants $r\geq 0$
and $K>0$ such that
  $$
|f(y)|\leq r|y|+K,\forall y\in \mathbb{R}  ;
  $$

\item[(H3)] $f(x, y)=f(y)$,  and there exists constants $n\geq 1$ and
$\sigma>0$ such that
  $$
yf(y)\geq \sigma |y|^{n+1},\quad \forall y\in \mathbb{R}
\quad\mbox{or}\quad
yf(y)\leq -\sigma |y|^{n+1},\quad \forall y\in \mathbb{R}.
  $$

\end{itemize}
These conditions have been considered  for the existence of periodic
solutions of \eqref{e1.1}. However, to the best of our knowledge,
few authors have considered  \eqref{e1.1}
    without the  assumptions (H0)--(H3). Thus, it is worth
    while to continue to investigate the existence of periodic solutions of
      \eqref{e1.1} in this case.

    The main purpose of this paper is to establish sufficient conditions for
    the existence of $T$-periodic solutions of   \eqref{e1.1}. The results
    of this paper are new and they complement previously  known
    results. In particular, we do not use  assumptions (H0)--(H3),
and we illustrate our results with  examples in Section 4.

     For ease of exposition, throughout this paper we will adopt the following
     notation:
$$
    |x|_k=\Big(\int^T_0|x(t)|^kdt\Big)^{1/k}, \quad
    |x|_\infty=\max_{t\in [0,T]}|x(t)|.
$$
Let
\begin{gather*}
X=\{x|x\in C^{1}(\mathbb{R}, \mathbb{R}): x(t+T)=x(t), \mbox{ for  all }
 t\in \mathbb{R}\},
\\
Y=\{x|x\in C(\mathbb{R}, \mathbb{R}), x(t+T)=x(t),  \mbox{ for  all }
 t\in \mathbb{R}\}
\end{gather*}
be two Banach spaces with the norms
$$
 \|x\|_{X}=\max\{|x|_\infty, |x'|_\infty\}, \quad\mbox{and} \quad
 \|x\|_{Y}=|x|_\infty.
$$
Define a linear operator  $L: D(L)\subset X\to Y$, with
$D(L)=\{x|x\in X:x''\in C(\mathbb{R}, \mathbb{R})\}$
and for $x \in D(L)$,
\begin{equation}
  Lx=x''.  \label{e1.2}
\end{equation}
We also define the nonlinear operator $N: X\to Y$  by
\begin{equation}
Nx=-f(x(t), x'(t)) -g_{1}(t,x(t-\tau_{1}(t)))
    -g_{2}(t,x(t-\tau_{2}(t)))+p(t).\label{e1.3}
\end{equation}
It is easy to see that
$$
\ker L=\mathbb{R},\quad \mbox{and}\quad
\mathop{\rm Im}L=\{x : x\in Y , \int^{T}_{0}x(s)ds=0 \}.
$$
Thus,  the operator $L$ is a Fredholm operator with index zero.
Define the continuous projectors   $P: X \to \ker L$ \
and  $ Q:Y \to Y$ by setting
$$
 Px(t)=x(0)=x(T), \quad
 Qx(t)=\frac{1}{T}\int^{T}_{0}x(s)ds.
$$
and let
$$
L_{P}=L|_{D(L)\cap \ker P}:D(L)\cap \ker P\to
\mathop{\rm Im}L
$$
Then, according to \cite{s1}, we have that $L_{P}$ has continuous
inverse  $L^{-1}_{P}$ on $\mathop{\rm Im}L$ defined  by
$$
L^{-1}_{P}y(t)=-\frac{t}{T}\int^{T}_{0}(t-s)y(s)ds+\int^{t}_{0}(t-s)y(s)ds.
\label{e1.4}
$$


\section{Preliminary Results}
\
In view of  \eqref{e1.2} and \eqref{e1.3}, the operator equation
$ Lx=\lambda Nx $ is equivalent to the  equation
\begin{equation}
    x''+\lambda [f(x(t),x'(t))+g_{1}(t,x(t-\tau_{1}(t)))
    +g_{2}(t,x(t-\tau_{2}(t)))]=\lambda p(t),
    \label{e2.1}
\end{equation}
where $\lambda\in (0,  1)$.

For convenience of use, we introduce the Continuation Theorem;
see \cite{g1}.

\begin{lemma} \label{lem2.1}
   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 the following conditions  are
     satisfied:
\begin{enumerate}
\item  $Lx\neq \lambda Nx$, for all
$x\in \partial \Omega\cap D(L) , \lambda\in (0,1)$;

\item $Nx \not \in \mathop{\rm Im}L$, for all $x\in \partial \Omega\cap
    \ker L$;

\item  The Brouwer  degree,
$\deg\{QN, \Omega\cap \ker L , 0 \}$, is not equal to zero.
\end{enumerate}
 Then the equation $Lx= Nx$ has at  least one $T$-periodic
    solution in $\overline{\Omega}$.
\end{lemma}

The following lemma  will be useful for proving our main results in
Section 3.

\begin{lemma} \label{lem2.2}
Assume that  the following conditions  are satisfied:
\begin{itemize}
\item[(A1)] One of the following conditions holds:
 \begin{enumerate}
 \item  $(g_{i}(t,u_{1})-g_{i}(t,u_{2}))(u_{1}-u_{2})>0$,
    for $i=1,2, u_{i}\in \mathbb{R}$, for all $t\in \mathbb{R}$ and
    $u_{1}\neq u_{2}$,
 \item  $(g_{i}(t,u_{1})-g_{i}(t,u_{2}))(u_{1}-u_{2})<0$,
   for $i=1,2, u_{i}\in \mathbb{R}$, for all $t\in \mathbb{R}$ and
   $u_{1}\neq u_{2}$;
  \end{enumerate}

\item[(A2)] There exists a constant $d>0$
    such that one of the following conditions holds:
 \begin{enumerate}
 \item $x(g_{1}(t,x)+g_{2}(t,x)-p(t))>0$, for all $t\in \mathbb{R}$,
 $|x|\ge d$,
 \item $x(g_{1}(t,x)+g_{2}(t,x)-p(t))<0$, for all $t\in \mathbb{R}$,
 $|x|\ge d$.
 \end{enumerate}
\end{itemize}
    If $x(t)$ is a $T$-periodic solution of $\eqref{e2.1}$, then
\begin{equation}
    |x|_\infty\le d+\sqrt{T}|x'|_2.
    \label{e2.2}
\end{equation}
\end{lemma}

\begin{proof}  Let $x(t)$ be a $T$-periodic solution of \eqref{e2.1}.
    Set
$$
    x(t_{\max})=\max_{t\in \mathbb{R}} x(t), \quad
    x(t_{\min})=\min_{t\in \mathbb{R}} x(t),
$$
 where $t_{\max}, t_{\min} \in \mathbb{R}$.
Then
\begin{equation}
    x'(t_{\max})=0, \quad x''(t_{\max})\le 0, \quad\mbox{and} \quad
    x'(t_{\min})=0, \quad x''(t_{\min})\ge 0. \label{e2.3}
\end{equation}
    In view of $f(x, 0)=0$ and \eqref{e2.1}, Equation \eqref{e2.3} implies
\begin{gather}
\begin{aligned}
 &g_{1}(t_{\max},x(t_{\max}-\tau_{1}(t_{\max})))
    +g_{2}(t_{\max},x(t_{\max}-\tau_{2}(t_{\max})))-p(t_{\max})\\
 &=-\frac{x''(t_{\max})} {\lambda}\ge 0,
\end{aligned} \label{e2.4}\\
\begin{aligned}
& g_{1}(t_{\min},x(t_{\min}-\tau_{1}(t_{\min})))
    +g_{2}(t_{\min},x(t_{\min}-\tau_{2}(t_{\min})))-p(t_{\min})\\
&=-\frac{x''(t_{\min})}{\lambda}\le 0.
\end{aligned}\label{e2.5}
\end{gather}
Since $g_{1}(t,x(t-\tau_{1}(t)))    +g_{2}(t,x(t-\tau_{2}(t)))-p(t)$
is a continuous function on $\mathbb{R}$, it follows from \eqref{e2.4}
     and \eqref{e2.5} that there
    exists a constant  $t_{1}\in \mathbb{R}$ such that
\begin{equation}
g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
    +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-p(t_{1})=0.
\label{e2.6}
\end{equation}
Next we show that the following claim is true.
\smallskip

\noindent\textbf{Claim:}
If $x(t)$ is a $T$-periodic solution of \eqref{e2.1}, then
    there exists a constant $t_{2}\in \mathbb{R}$ such that
\begin{equation}
    |x(t_{2})|\leq d .
    \label{e2.7}
\end{equation}

\begin{proof}
 Assume, by way of contradiction, that \eqref{e2.7} does not hold. Then
\begin{equation}
|x(t)|> d, \quad \mbox{for   all } t\in \mathbb{R},
 \label{e2.8}
\end{equation}
which, together with  (A2) and \eqref{e2.6},  implies that
one of the following relations holds:
\begin{gather}
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}\\
x(t_{1}-\tau_{1}(t_{1}))< x(t_{1}-\tau_{2}(t_{1}))<- d;\label{e2.11}\\
x(t_{1}-\tau_{2}(t_{1}))< x(t_{1}-\tau_{1}(t_{1}))<- d.\label{e2.12}
\end{gather}
Suppose that \eqref{e2.9} holds, in view of (A1)(1), (A1)(2),  (A2)(1) and
(A2)(2), we consider following four cases:

\noindent Case (i).  If (A2)(1) and (A1)(1) hold,  according to
    \eqref{e2.9}, we obtain
\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})))-p(t_{1})
\\
& <  g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
    +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-p(t_{1}),
\end{align*}
which contradicts \eqref{e2.6}. This contradiction implies
\eqref{e2.7}.

\noindent    Case (ii).  If (A2)(1) and (A1)(2) hold,  according to
    \eqref{e2.9}, we obtain
\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})))-p(t_{1})
\\
& <  g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
    +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-p(t_{1}),
\end{align*}
which contradicts  \eqref{e2.6}. This contradiction implies
\eqref{e2.7} .

\noindent  Case (iii).  If (A2)(2) and (A1)(1) hold,  according to
    \eqref{e2.9}, we obtain
\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})))-p(t_{1})
\\
& >  g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
    +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-p(t_{1}),
\end{align*}
which contradicts  \eqref{e2.6}. This contradiction implies \eqref{e2.7}.

\noindent Case (iv).  If (A2)(2) and (A1)(2) hold,  according to
    \eqref{e2.9}, we obtain
\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})))-p(t_{1})
\\
& >  g_{1}(t_{1},x(t_{1}-\tau_{1}(t_{1})))
    +g_{2}(t_{1},x(t_{1}-\tau_{2}(t_{1})))-p(t_{1}),
\end{align*}
which contradicts  \eqref{e2.6}. This
contradiction implies \eqref{e2.7}.

Suppose that \eqref{e2.10} (or \eqref{e2.11}, or \eqref{e2.12}) holds,
using the methods similarly to those used in Cases (i)--(iv), we can show that
\eqref{e2.7} is true. This completes the proof of the above claim.
\end{proof}

 Let $t_{2}=m T+t_0$,  where  $ t_0\in [0,T]$ and $m$ is an integer.
Then,
    using the Schwarz inequality and the relation
$$
    |x(t)|=|x(t_0)+\int^t_{t_0}x'(s)ds|\le d+\int^T_0|x'(s)|ds,
    t\in [0, T],
$$
we obtain
$$
    |x|_\infty=\max_{t\in[0,T]}|x(t)|\le d+\sqrt{T}
    |x'|_2.
$$
This completes the proof.
\end{proof}

\section{Main Results}

\begin{theorem} \label{thm3.1}
Suppose that (A1)(1)  and (A2)(1) hold,
    and there exist nonnegative constants $m_{1}$, $m_{2}$ ,  $m_{3}$
and $m_{4}$ such that $2m_{1}+4m_{3}<\frac{1}{2T^{2}}$,
    and one of the following conditions holds:
\begin{enumerate}
\item $f(x,y)\leq 0 $  for all  $x\in \mathbb{R}$, $y\in \mathbb{R}$,
$|g_{2}(t,x)|\leq m_{3}|x|+m_{4}$  for all  $t\in \mathbb{R}$,
   $x\in \mathbb{R}$, and
$$
g_{1}(t,x)+g_{2}(t,x)-p(t)     \leq m_{1}x+m_{2},\quad \forall
t\in \mathbb{R},\; x\geq d;
$$
\item  $f(x, y)\ge 0$  for all  $x\in \mathbb{R}, y\in \mathbb{R}$,
$|g_{2}(t,x)|\leq m_{3}|x|+m_{4}$  for all  $t\in \mathbb{R}$,
  $x\in \mathbb{R}$,
and
$$
g_{1}(t,x)+g_{2}(t,x)-p(t)
    \ge m_{1}x-m_{2},\quad \forall  t\in \mathbb{R},\; x\leq - d.
$$
\end{enumerate}
Then  \eqref{e1.1} has at least one $T$-periodic
    solution.
\end{theorem}

\begin{proof}
 We shall seek to apply Lemma \ref{lem2.1}.  To do this,
 it suffices to prove that the set of all possible $T$-periodic solutions of
 \eqref{e2.1}  are bounded.
Let $x(t)$ be a $T$-periodic solution of \eqref{e2.1}.
Integrating \eqref{e2.1} from $0$ to $T$,  we have
\begin{equation}
\int^T_0 f(x(t), x'(t))dt+\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{2}(t)))-p(t)] dt=0.
\label{e3.1}
\end{equation}
 Set
\begin{gather*}
[x(t-\tau_{1}(t))<-d]=\{t|t\in [0, T], x(t-\tau_{1}(t))<-d \},\\
[x(t-\tau_{1}(t))\geq -d]=\{t|t\in [0, T], x(t-\tau_{1}(t))\geq -d \},\\
[x(t-\tau_{1}(t))>d]=\{t|t\in [0, T], x(t-\tau_{1}(t))>d \},\\
[x(t-\tau_{1}(t))\leq d]=\{t|t\in [0, T], x(t-\tau_{1}(t))\leq d \}.
\end{gather*}
Then, in view of (A2)(1),  \eqref{e3.1} implies
\begin{align}
&\int_{[x(t-\tau_{1}(t))<-d]} |g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt \notag \\
&=-\int_{[x(t-\tau_{1}(t))<-d]}
    [g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt \notag \\
& =\int_{[x(t-\tau_{1}(t))\geq -d]}[g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt  \notag\\
&\quad -\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt   \notag\\
& =\int_{[x(t-\tau_{1}(t))\geq -d]}[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt  \label{e3.2}\\
&\quad -\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{2}(t)))-p(t)]dt \notag\\
&\quad -\int^T_0g_{2}(t,x(t-\tau_{1}(t)))dt
+\int^T_0 g_{2}(t,x(t-\tau_{2}(t)))dt \notag\\
&=\int_{[x(t-\tau_{1}(t))\geq -d]}[g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt \notag \\
&\quad -\int^T_0g_{2}(t,x(t-\tau_{1}(t)))dt
 +\int^T_0g_{2}(t,x(t-\tau_{2}(t)))dt+\int^T_0f(x(t), x'(t))dt, \notag
\end{align}
and
\begin{equation}
\begin{aligned}
&\int_{[x(t-\tau_{1}(t))>d]} |g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt \\
& =\int_{[x(t-\tau_{1}(t))>d]}
    [g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt \\
& =-\int_{[x(t-\tau_{1}(t))\leq d]}[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt\\
&\quad +\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt\\
&=-\int_{[x(t-\tau_{1}(t))\leq d]}[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt \\
&\quad+\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{2}(t)))-p(t)]dt\\
&\quad +\int^T_0g_{2}(t,x(t-\tau_{1}(t)))dt
-\int^T_0 g_{2}(t,x(t-\tau_{2}(t)))dt \\
&=-\int_{[x(t-\tau_{1}(t))\leq d]}[g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)]dt \\
&\quad +\int^T_0g_{2}(t,x(t-\tau_{1}(t)))dt
  -\int^T_0g_{2}(t,x(t-\tau_{2}(t)))dt-\int^T_0f(x(t),x'(t))dt.
\end{aligned} \label{e3.3}
\end{equation}
Now suppose that  (1) (or  (2)) holds. We shall
consider two cases as follows.

\noindent\textbf{Case 1:}  If (1) holds, it follows from  \eqref{e2.2}
and \eqref{e3.2} that
\begin{align}
&\int_{[x(t-\tau_{1}(t))<-d]}|g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt   \notag\\
&\le \int_{[x(t-\tau_{1}(t))\geq -d]}
  |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt   \notag \\
&\quad+\int^T_0|g_{2}(t,x(t-\tau_{1}(t)))|dt
  +\int^T_0|g_{2}(t,x(t-\tau_{2}(t)))|dt  \notag\\
&\le \int_{\{t|t\in [0, T], |x(t-\tau_{1}(t))|\le d \}}
    |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt  \notag\\
&\quad +\int_{[x(t-\tau_{1}(t))>d]}
    |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt  \label{e3.4}\\
&\quad +\int_{0}^{T}(m_{3}|x(t-\tau_{1}(t))|+m_{4})dt
  +\int_{0}^{T}(m_{3}|x(t-\tau_{2}(t))| +m_{4})dt   \notag\\
&\le T(\max\{|g_{1}(t,x)+g_{2}(t,x)-p(t)|:t\in \mathbb{R}, |x|\le d\})   \notag\\
&\quad  +\int_{0}^{T}(m_{1} |x(t-\tau_{1}(t))|+m_{2})dt
 +2T(m_{3}|x|_{\infty}+m_{4})  \notag\\
&\le T(\max\{|g_{1}(t,x)+g_{2}(t,x)-p(t)|:t\in \mathbb{R}, |x|\le d\}
+m_{2}+2m_{4})   \notag\\
&\quad +T(m_{1}+2m_{3})|x|_{\infty}   \notag\\
&\le T(\theta_{1}+m_{2}+2m_{4})+T(m_{1}+2m_{3})(\sqrt{T}
|x'|_{2}+d), \notag
\end{align}
where $\theta_{1}=\max\{|g_{1}(t,x)+g_{2}(t,x)-p(t)|:t\in \mathbb{R},
|x|\le d\}$. Then, \eqref{e3.4} implies
\begin{equation}
\begin{aligned}
&\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt\\
&= \int_{[x(t-\tau_{1}(t))<-d]}|g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt     \\
&\quad +\int_{[x(t-\tau_{1}(t))\geq -d]}
     |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt\\
&\le 2T(\theta_{1}+m_{2}+m_{4})+2T(m_{1}+m_{3})(\sqrt{T}
|x'|_{2}+d),
\end{aligned} \label{e3.5}
\end{equation}
and
\begin{equation}
\begin{aligned}
\int^T_0|f(x(t), x'(t))|dt
&= -\int^T_0f(x(t), x'(t))dt \\
&= \int^T_0[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{2}(t)))-p(t)] dt \\
&= \int^T_0[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)] dt\\
&\quad -\int^T_0g_{2}(t,x(t-\tau_{1}(t))) dt
  +\int^T_0g_{2}(t,x(t-\tau_{2}(t))) dt \\
&\le \int^T_0|g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt\\
&\quad +\int^T_0|g_{2}(t,x(t-\tau_{1}(t))) |dt
 +\int^T_0|g_{2}(t,x(t-\tau_{2}(t))) |dt \\
&\le 2T(\theta_{1}+m_{2}+2m_{4})+2T(m_{1}+2m_{3})(\sqrt{T}
|x'|_{2}+d).
\end{aligned}\label{e3.6}
\end{equation}

\noindent\textbf{Case 2:}   If (2) holds, it follows from
\eqref{e2.2} and \eqref{e3.3} that
\begin{equation}
\begin{aligned}
&\int_{[x(t-\tau_{1}(t))>d]}|g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt \\
&\le \int_{[x(t-\tau_{1}(t))\leq d]}
    |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt
    \\
&\quad +\int^T_0|g_{2}(t,x(t-\tau_{1}(t)))|dt
  +\int^T_0|g_{2}(t,x(t-\tau_{2}(t)))|dt\\
&\le \int_{\{t|t\in [0, T],\ |x(t-\tau_{1}(t))|\le d \}}
    |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt\\
&\quad +\int_{[x(t-\tau_{1}(t))<-d]}
    |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt\\
&\quad +\int_{0}^{T}(m_{3}|x(t-\tau_{1}(t))|+m_{4})dt
   +\int_{0}^{T}(m_{3}|x(t-\tau_{2}(t))| +m_{4})dt\\
&\le T(\max\{|g_{1}(t,x)+g_{2}(t,x)-p(t)|:t\in \mathbb{R}, |x|\le d\})\\
&\quad+\int_{0}^{T}(m_{1} |x(t-\tau_{1}(t))|+m_{2})dt
 +2T(m_{3}|x|_{\infty}+m_{4}) \\
&\le T(\theta_{1}+m_{2}+2m_{4})+T(m_{1}+2m_{3})(\sqrt{T}
|x'|_{2}+d),
\end{aligned} \label{e3.7}
\end{equation}
which implies
\begin{equation}
\begin{aligned}
&\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt \\
&=\int_{[x(t-\tau_{1}(t))>d]}|g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt     \\
&\quad +\int_{[x(t-\tau_{1}(t))\leq d]}
     |g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt\\
&\le 2T(\theta_{1}+m_{2}+m_{4})+2T(m_{1}+m_{3})(\sqrt{T}
|x'|_{2}+d),
\end{aligned} \label{e3.8}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\int^T_0|f(x(t), x'(t))|dt\\
&= \int^T_0f(x(t), x'(t))dt \\
&= -\int^T_0[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{2}(t)))-p(t)] dt \\
&\le \int^T_0|g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)| dt \\
&\quad +\int^T_0|g_{2}(t,x(t-\tau_{1}(t))) |dt
 +\int^T_0|g_{2}(t,x(t-\tau_{2}(t))) |dt \\
&\le 2T(\theta_{1}+m_{2}+2m_{4})+2T(m_{1}+2m_{3})(\sqrt{T}|x'|_{2}+d).
\end{aligned}\label{e3.9}
\end{equation}
   Multiplying \eqref{e2.1} by $x(t)$ and then integrating from
    $0$ to $T$,  by \eqref{e2.3}, \eqref{e3.5} ,  \eqref{e3.6}, \eqref{e3.8}
and  \eqref{e3.9}, we have
\begin{align}
&|x'|_2^2  \notag\\
&= \lambda\int^T_0\{f(x(t), x'(t))
+[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{2}(t)))
-p(t)] \}x(t)dt  \notag\\
&= \lambda\int^T_0\{f(x(t), x'(t))
+[g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)  \notag\\
&\quad -g_{2}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{2}(t)))] \}x(t)dt  \notag\\
&\le \int^T_0|f(x(t), x'(t))||x(t)|dt  \notag\\
&\quad +\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)||x(t)| dt  \notag\\
&\quad +\int^T_0|g_{2}(t,x(t-\tau_{1}(t))) ||x(t)|dt
+\int^T_0|g_{2}(t,x(t-\tau_{2}(t))) ||x(t)|dt \label{e3.10}\\
&\le |x|_{\infty} \{\int^T_0|f(x(t),x'(t))|dt  \notag\\
&\quad +\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))
 +g_{2}(t,x(t-\tau_{1}(t)))-p(t)|  dt
 +2T(m_{3}|x|_{\infty}+m_{4})\}  \notag\\
&\le 2T[(2\theta_{1}+2m_{2}+4m_{4})+(2m_{1}+4m_{3})(\sqrt{T}
|x'|_{2}+d)](\sqrt{T} |x'|_{2}+d)  \notag\\
&=2(2m_{1}+4m_{3})T^{2}|x'|^{2}_{2}+2T[(2\theta_{1}+2m_{2}+4m_{4})
+2(2m_{1}+4m_{3})d]\sqrt{T} |x'|_{2}  \notag\\
&\quad+2Td[(2\theta_{1}+2m_{2}+4m_{4})+(2m_{1}+4m_{3})d]. \notag
\end{align}
    Since $0\leq 2m_{1}+4m_{3}<\frac{1}{2T^{2}}$, \eqref{e3.10} implies that
    there exists a positive constant  $D_1$ such that
\begin{equation}
    |x'|_2 \le D_1 \mbox{ \ and \ } |x|_{\infty}\le \sqrt{T}
|x'|_{2}+d\leq D_1. \label{e3.11}
\end{equation}
In view of \eqref{e3.5},  \eqref{e3.6}, \eqref{e3.8} and  \eqref{e3.9},
it follows from \eqref{e2.1} that
\begin{equation}
\begin{aligned}
&\int^T_0|x''(t)|dt \\
&\le  \int^T_0|f(x(t), x'(t))|dt
 +\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{1}(t)))-p(t)\\
&\quad -g_{2}(t,x(t-\tau_{1}(t)))+g_{2}(t,x(t-\tau_{2}(t)))|dt\\
&\le \int^T_0|f (x(t), x'(t)) |dt  +\int^T_0|g_{1}(t,x(t-\tau_{1}(t)))
+g_{2}(t,x(t-\tau_{1}(t)))-p(t)|dt\\
&\quad +\int^T_0|g_{2}(t,x(t-\tau_{1}(t)))|dt
  +\int^T_0|g_{2}(t,x(t-\tau_{2}(t)))|dt\\
&\le  2T[(2\theta_{1}+2m_{2}+4m_{4})+(2m_{1}+4m_{3})(\sqrt{T}
|x'|_{2}+d)]\\
&\le  2T[(2\theta_{1}+2m_{2}+4m_{4})+(2m_{1}+4m_{3})(\sqrt{T}
D_{1}+d)]
:=  D_{2}.
\end{aligned} \label{e3.12}
\end{equation}
 Since $x(0)=x(T)$, it follows that there exists a constant
$\zeta\in [0,T]$ such that
$ x'(\zeta)=0$ and
$$
    |x'(t)|=|x'(\zeta)+\int^t_{\zeta}x''(s)ds|
\leq \int^T_0|x''(t)|dt\le  D_2, \quad \forall  t\in [0, T],
$$
which, together with \eqref{e3.11}, implies
$$
    \|x\|_{X}\leq |x|_\infty+|x'|_\infty<D_1+ D_2+1:=M_1.
$$
If $x\in \Omega_{1}=\{x|x\in \ker L \cap X $  and
$Nx\in \mathop{\rm Im}L\}$, then there exists a constant $M_{2}$
such that
$$
x(t)\equiv M_{2} \  \mbox{ and} \ \int^{T}_{0}[g_{1}(t,
M_{2})+g_{2}(t, M_{2})-p(t)]dt=0. \ \ \ \label{e3.13}
$$
Thus,
\begin{equation}
|x(t)|\equiv |M_{2}|<d, \quad \mbox{for  all }  x(t)\in \Omega_{1}.
\label{e3.14}
\end{equation}
Let $M=M_{1}+d+1$. Set
$$
\Omega=\{x|x\in X, |x|_\infty<M, \ |x'|_\infty<M\}.
$$
It is easy to see from \eqref{e1.3}  and \eqref{e1.4} that $N$ is
$L$-compact on $\overline{\Omega}$. We have from \eqref{e3.13},
\eqref{e3.14} and the fact
$M>\max\{M_{1} , d \}$ that  the conditions (1) and (2)
in Lemma \ref{lem2.1} hold.

 Furthermore, we define a continuous
    function $H(x,\mu)$  by setting
$$
H(x,\mu)=-(1-\mu)x-\mu\cdot
   \frac{1}{T}\int^{T}_{0}[g_{1}(t,x)+g_{2}(t,x)-p(t)]dt; \quad
 \mu\in [0,1].
$$
In view of  (A2)(1), we have
$$
  xH(x,\mu)\not= 0  \quad  \mbox{for  all }  x\in \partial \Omega\cap \ker L.
$$
Hence, using the  homotopy invariance  theorem, we obtain
\begin{align*}
  \deg\{QN, \Omega\cap \ker L , 0 \}
&=\deg\{-\frac{1}{T}\int^{T}_{0}[
    g_{1}(t,x)+g_{2}(t,x)-p(t)]dt, \Omega\cap \ker L , 0\}\\
&=deg\{-x, \Omega\cap \ker L , 0   \}\neq 0.
\end{align*}
    In view of  the discussions above, from Lemma \ref{lem2.1}  we complete
the proof of Theorem  \ref{thm3.1}.
\end{proof}

    A similar argument leads to the following result.

\begin{theorem} \label{thm3.2}
 Suppose that (A1)(2)  and (A2)(2) holds,
 and there exist nonnegative constants $m_{1}$, $m_{2}$ ,  $m_{3}$
and $m_{4}$ such that  $2m_{1}+4m_{3}<\frac{1}{2T^{2}}$,
    and one of the following two conditions holds:
\begin{enumerate}
\item  $f(x, y)\geq 0$  for all  $x\in \mathbb{R}, y\in \mathbb{R}$,
    $|g_{2}(t,x)|\leq m_{3}|x|+m_{4}$  for all
$t\in \mathbb{R}$, $ x\in \mathbb{R}$,
and $g_{1}(t,x)+g_{2}(t,x)-p(t) \geq - m_{1}x-m_{2}$,
for all $ t\in \mathbb{R}$, $x\geq d$;

 \item $f(x, y)\leq 0$  for all  $x \in \mathbb{R}$, $y\in \mathbb{R}$,
 $|g_{2}(t,x)|\leq m_{3}|x|+m_{4}$  for all  $t\in \mathbb{R}$,
   $x\in \mathbb{R}$,
and
$$
g_{1}(t,x)+g_{2}(t,x)-p(t)
    \leq -m_{1}x+m_{2},\quad \text{for all } t\in \mathbb{R},\; x\leq - d.
$$
\end{enumerate}
Then   \eqref{e1.1} has at least one $T$-periodic   solution.
\end{theorem}

\section{Examples and Remarks}

\begin{example} \label{exa4.1} \rm
Let $g(t,x)=x^{13}+\frac{1}{72\pi^{2}} x$ for  $t\in \mathbb{R}$,
$x\leq 0$, and
    $g(t,x)=\frac{1}{36\pi^{2}} x$ for  $t\in \mathbb{R}$, $x> 0$.
Then the Rayleigh equation
\begin{equation}
    x''- (x')^{4}+g(t,x(t-\sin(t))) =e^{\cos^2 t},
    \label{e4.1}
\end{equation}
has at least one $2\pi$-periodic solution.
\end{example}

\begin{proof}
Let $g_{2}(t,x)=\frac{1}{72\pi^{2}} x$ for $t\in \mathbb{R}$,
$x\in \mathbb{R}$,
 $g_{1}(t,x)=x^{13}$ for  $t\in \mathbb{R}$, $x\leq 0$, and
    $g_{1}(t,x)=\frac{1}{72\pi^{2}} x$ for
 $t\in \mathbb{R}$, $x> 0$. Then \eqref{e4.1} is equivalent
to the equation
\begin{equation}
    x''- (x')^{4}+g_{1}(t,x(t-\sin(t)))+g_{2}(t,x(t-\sin(t))) =e^{\cos^2 t}.
    \label{e4.2}
\end{equation}
   From \eqref{e4.2}, we have
$f (x, y)=- y^{4}\leq 0$,  $\tau_{1}(t)=\tau_{2}(t)    =\sin t$,
$p(t)=e^{\cos^2 t}$ and
$g_{1}(t,x)+g_{2}(t,x)-p(t) =\frac{1}{36\pi^{2}} x -e^{\cos^2 t}
\leq \frac{1}{36\pi^{2}} x+e$, for all $ t\in \mathbb{R}$, $x>0$.
It is straightforward to check that all the conditions needed
in Theorem \ref{thm3.1} are satisfied. Therefore,  \eqref{e4.2} has at least one
$2\pi$-periodic solution. This implies that \eqref{e4.1} has at least one
$2\pi$-periodic solution.
\end{proof}

\begin{remark} \label{rmk4.1} \rm
 Equation \eqref{e4.1} is a very simple version of  Rayleigh equation.
    Obviously, the conditions (H0)--(H3)  are  not
satisfied. Therefore, the results in \cite{g2,s1,s2,s3,x1} and the
    references cited therein cannot be applied to \eqref{e4.1}. This
    implies that the results of this paper are essentially new.
\end{remark}

\begin{example} \label{exa4.2} \rm
  Let $g_{1}(t,x)=-\frac{1}{72\pi^{2}} x$ for
 $t\in \mathbb{R}$, $x\in \mathbb{R}$,
     $g_{2}(t,x)=-x^{13}$ for  $t\in \mathbb{R}$, $x\leq 0$, and
    $g_{2}(t,x)=-\frac{1}{72\pi^{2}} x$ for  $t\in \mathbb{R}$, $x> 0$.
Then, the Rayleigh equation
\begin{equation}
    x''+x^{4}(x')^{6}+g_{1}(t,x(t-\cos(t)))+g_{2}(t,x(t-\sin(t)))
=\frac{1}{4} \cos^2 t.     \label{e4.3}
\end{equation}
    has at least one $2\pi$-periodic solution.
\end{example}

\begin{proof}  From \eqref{e4.3}, we can obtain
$f (x, y)=x^{4}y^{6}$, $\tau_{1}(t)=\cos(t)$,
$\tau_{2}(t) =\sin (t)$,  $p(t)=\frac{1}{4}cos^2 t$
and $g_{1}(t,x)+g_{2}(t,x)-p(t) =-\frac{1}{36\pi^{2}} x
-\frac{1}{4}cos^2 t\geq -\frac{1}{36\pi^{2}} x-\frac{1}{4}$,
for $ t\in \mathbb{R}$, $x>0$.  It is obvious that all
     the conditions needed in Theorem \ref{thm3.2} are
satisfied.  Hence, by Theorem \ref{thm3.2}, equation \eqref{e4.3} has at least
one     $2\pi$-periodic solution.
\end{proof}

\begin{remark} \label{rmk4.2} \rm
  In view of \eqref{e4.3},  it is clear that (H0)--(H3), do not hold
   for \eqref{e4.3},     and so the results obtained in
\cite{g2,s1,s2,s3,x1} and the
    references cited therein cannot be applied to  \eqref{e4.3}.
\end{remark}

\begin{remark} \label{rmk4.3} \rm
Using the methods similarly to those used  for \eqref{e1.1},
we can study the Rayleigh equation with multiple
    deviating arguments
\begin{equation}
    x''+f(x(t), x'(t)) +\sum_{i=1}^{n}g_{i}(t,x(t-\tau_{i}(t)))
    =p(t),
\label{e4.4}
\end{equation}
where  $\tau_{i}(i=1,2,\dots, n)$, $p:\mathbb{R}\to \mathbb{R}$
   and  $f$,
$g_{i}:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$
are continuous  functions, $f(x, 0)=0$, $\tau_{i}$ and $p$ are $T$-periodic,
$g_{i}$ are $T$-periodic in the first argument, and $T>0$ ($i=1,2,\dots, n$).
One may also establish the results similarly to those in
Theorems \ref{thm3.1} and \ref{thm3.2} under some minor additional
assumptions on $g_{i}(t,x)$ ($i=1,2,\dots, n$).
\end{remark}

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\end{thebibliography}

\end{document}