\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 77, 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 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/77\hfil Existence of periodic solutions]
{Existence of periodic solutions for Rayleigh equations with
state-dependent delay}

\author[J. O. Alzabut,  C. Tun\c{c} \hfil EJDE-2012/77\hfilneg]
{Jehad  O. Alzabut, Cemil Tun\c{c}}  % in alphabetical order

\address{Jehad  O. Alzabut \newline
Department of Mathematics and Physical Sciences, 
Prince Sultan University \\
P.O. Box 66833, Riyadh 11586, Saudi Arabia}
\email{jalzabut@psu.edu.sa}

\address{Cemil Tun\c{c} \newline
Department of Mathematics,
Faculty of Arts and Science, Y\"uz\"unc\"u Yil University\\
65080 Van, Turkey}
\email{cemtunc@yahoo.com}

\thanks{Submitted December 22, 2011. Published May 14, 2012.}
\subjclass[2000]{34K13}
\keywords{Rayleigh Equation; continuation theorem; periodic solutions}

\begin{abstract}
 We establish sufficient conditions for the existence of
 periodic solutions for a  Rayleigh-type equation with
 state-dependent delay. Our approach is based on the continuation
 theorem in degree theory, and some analysis techniques. An 
 example illustrates that our approach to this problem is new.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

 Lord Rayleigh  (John William Strutt: 1842--1919) \cite{s1}
introduced the equation
\begin{equation}\label{eq10}
    x''(t)+f(x'(t))+ax(t)=0
\end{equation}
to model the oscillations of a clarinet reed. This
equation is used for  studying problems arising in acoustics, and is
referred in the literature as Rayleigh equation. Later on, the
Rayleigh equation of the form
\begin{equation}\label{eq11}
    x''(t)+f(x'(t))+g(t,x(t))=0
\end{equation}
was studied in the  monographs \cite{d1,g1,g4}. In
many circumstances, however,  it is known that the forces
intervening in the system depend depend not only at the current
time considered, but also on previous times. 
Thus, the forced Rayleigh equation with delay 
\begin{equation}\label{eq12}
    x''(t)+f(t,x'(t))+g(t,x(t-\tau))=p(t)
\end{equation}
has been taken into consideration, see \cite{w1,w2,w3}. Recently,
it has been recognized that \eqref{eq12} has widespread
applications in many applied sciences such as physics, mechanics
and engineering techniques fields. In such applications, it is
crucial to know the periodic behavior of solutions for Rayleigh
equation. This justifies the intensive interest among researchers
in investigating the existence of periodic solutions for this
equation in the last decade. Publications
\cite{h2,g3,l1,l2,l3,l5,l6,l7,p1,w4,z1,z2,z3,z4} are
devoted to various generalizations of equation \eqref{eq12}.
Nevertheless, one can realize that all the results obtained in the
above mentioned papers have been proved under the assumptions that
$\tau$ is a constant, $g$ is bounded and
$\int_0^{2\pi}p(t)\,{\rm d} t=0$.
However, it is known that the delay may not be only related
to time $t$ but also it relates to the current state $x$. Thus, it
is worth while to consider a type of Rayleigh equation with
state-dependent delay. In this paper, particularly, we consider
Rayleigh equation of the form
\begin{equation}\label{eq1}
    x''(t)+f(t,x(t))+g(x(t-\tau(t,x(t))))=p(t).
\end{equation}
We shall utilize the continuation theorem of degree theory to
obtain sufficient conditions for the existence of periodic
solutions of  \eqref{eq1}. The main result is proved by
bypassing the boundedness of $g$ and the integral condition on
$p$. To the best of authors' observations, there exists no paper
establishing sufficient conditions for the existence of periodic
solutions for \eqref{eq1}.  Thus,  our result presents a new approach.

\section{Preliminaries}

Let
$$
C_{2\pi}=\big\{x:x\in C(\mathbb{R},\mathbb{R}), x(t+2\pi)\equiv x(t),\;
\forall\;t \in \mathbb{R}\big\}
$$
 with the norm $\| x\|_0=\max_{t \in [0,2\pi]} | x(t) |$, for 
$x \in C_{2\pi}$ and
$$
C_{2\pi}^{1}=\big\{x:x\in
C^{1}(\mathbb{R},\mathbb{R}), x(t+2\pi)\equiv x(t),\; \forall\;t
\in \mathbb{R}\big\}
$$
 with the norm $\| x \|_1=\max_{t
\in [0,2\pi]} \{\| x(t) \|_0,\| x'(t)\|_0\}$, for $x \in C_{2\pi}^{1}$.
We shall consider  \eqref{eq1} under the assumptions that
$ f \in C_{2\pi}(\mathbb{R}^2,\mathbb{R})$ with
 $f(\cdot,0)=0$, $g \in C_{2\pi}(\mathbb{R},\mathbb{R})$,
$\tau \in C_{2\pi}(\mathbb{R}^2, \mathbb{R}^{+})$ and
 $p \in C_{2\pi}(\mathbb{R}, \mathbb{R})$.

Let $0 \leq \tau(t) \in C_{2\pi}$, then there must exist two
integers $k \ge 0$ and $m \ge 1$ such that
\begin{equation}\label{tau}
    \tau(t) \in [2\pi k,2\pi (k+m)],\quad
\tau(t) \notin (0,2\pi k)\cup    (2\pi (k+m),\infty).
\end{equation}
Denote $\Delta_{i}=\big\{t:t \in [0,2\pi],\;\tau(t) \in [2\pi
(k+i),2\pi (k+i+1)]\big\}$, $i=0,1,2,\dots,m-1$,
\begin{displaymath}
\tau_0(t)= \begin{cases}
\tau(t), & t \in \Delta_0,\\
2\pi (k+1), & t \in [0,2\pi]\backslash \Delta_0,
\end{cases}
\end{displaymath}
and
\begin{displaymath}
\tau_j(t)= \begin{cases}
\tau(t), & t \in \Delta_j,\\
2\pi (k+j), & t \in [0,2\pi]\backslash \Delta_j.
\end{cases}
\end{displaymath}
Then, it is clear that $\cup_{i=0}^{m-1} \Delta_{i}=[0,2\pi]$;
$2\pi (k+1)-\tau_0(t) \in [0,2\pi]$
 and  $\tau_j(t)-2\pi (k+j) \in [0,2\pi]$
for all $ t \in [0,2\pi]$, $j=1,2,\dots,m-1$.

Let $\delta_0=\sup_{t \in [0,2\pi]}[2\pi(k+1)-\tau_0(t)]$,
$\delta_j=\sup_{t \in [0,2\pi]}[\tau_j(t)-2\pi (k+j)]$, then we have $\delta_0,
\delta_{m-1}\in [0,2\pi]$, $\delta_j=2\pi,\;j=1,2,\dots,m-2$.

The following  lemma plays a key role in proving the main result.

\begin{lemma}[\cite{d2}] \label{lem2}
Let $\tau(t,x(t)) \in C_{2\pi}$ satisfying \eqref{tau} and
$x \in C_{2\pi}^{1}$, then
$$
\int _0^{2 \pi} | x(t-\tau(t,x(t)))-x(t) |^2\,{\rm d} t
 \leq \Big(\beta_0+\beta_{m-1}+\sum_{j=1}^{m-2}\beta_j \Big)
\int_0^{2 \pi}| x'(t)|^2\,{\rm d} t, \;2<m<\infty
$$
and
$$
\int _0^{2 \pi} | x(t-\tau(t,x(t)))-x(t)|^2\,{\rm d} t 
\leq \big(\beta_0+\beta_{m-1} \big)\int_0^{2 \pi}| x'(t)|^2\,{\rm d} t, \;1
 \leq m \leq 2,
$$
where
$$
\beta_0=\max_{\sigma \in [0,2\pi-\delta_0]}
\int_{\sigma}^{\sigma+\delta_0}\tau(t,x(t))\,{\rm d} t,\;\beta_{m-1}
=\max_{\sigma \in [0,2\pi-\delta_{m-1}]}
\int_{\sigma}^{\sigma+\delta_{m-1}}\tau(t,x(t))\,{\rm d}t
$$
and
$$
\beta_j=\int_0^{2\pi}\tau(t,x(t))\,{\rm d}t,\;\;j=1,2,\dots,m-2.
$$
\end{lemma}

\begin{lemma}[\cite{l4}] \label{lem3}
Let $x \in C_{2\pi}^{1}$ and there exists a constant
$\xi \in \mathbb{R}$ such that $x(\xi)=0$. Then we have
$$
\int_0^{2\pi}| x(t) |^2\,{\rm d}t
 \leq 4 \int_0^{2\pi}| x'(t)|^2\,{\rm d}t.
$$
\end{lemma}

Degree theory has been used to prove the existence of solutions of
a wide variety of differential, integral, functional and
difference equations. Of particular interest is its use in the
investigation of periodic solutions. We shall use a result by
Mawhin \cite{m1} to prove the existence of a $2\pi$-periodic
solution of equation \eqref{eq1}. We refer the reader to
\cite{g2} for more information.  Here are some basic concepts
in the framework of this theory.

Define a linear operator
$$
L:D(L)\subset C_{2\pi}^{1} \to C_{2\pi}, \quad Lx=x'',
$$
where $D(L)=\{x:x \in C^2(\mathbb{R},\mathbb{R}),
x(t+2\pi)\equiv x(t)\}$ and a nonlinear operator
$$
N: C_{2 \pi}^{1}\to C_{2 \pi},\quad
Nx=-f(t,x(t))-g(x(t-\tau(t,x(t))))+p(t).
$$
It is easy to see that
$$
\ker (L)=\{a,\;a \in \mathbb{R}\},\quad \operatorname{Im}
L=\Big\{y:y \in C_{2 \pi},\; \int_0^{2\pi}y(s)\,{\rm d}s=0\Big\}.
$$
Therefore, $\operatorname{Im} L$ is closed in $C_{2\pi}$ and
$\dim  \ker  L= \operatorname{codim} \operatorname{Im} L =1$. It
follows that the operator $L$ is a Fredholm operator with index
zero.

Define the continuous projectors
\begin{gather*}
P: C_{2\pi} \to \ker  L,\quad Px=x(0), \\
Q: C_{2 \pi} \to C_{2\pi}/\operatorname{Im} L,\quad
Qy=\frac{1}{2 \pi} \int_0^{2 \pi}y(s)\,{\rm d}s.
\end{gather*}
It is easy to see that
$\operatorname{Im} P=\ker  L\quad\mbox{and}\quad
\ker Q=\operatorname{Im} L$. Set the operators
$$
L_{p}=L \big|_{D(L)\cap \ker  P}: D(L)\cap \ker  P \to \operatorname{Im} L.
$$
Then, $L_{p}$ has a unique continuous inverse operator
$L_{p}^{-1}$ on $\operatorname{Im} L$ defined by
$$
(L_{p}^{-1}y)(t)=\int_0^{2 \pi}G(t,s)y(s)ds,
$$
where
\begin{displaymath}
G(t,s)= \begin{cases}
\frac{s(2\pi-t)}{2\pi}, & 0 \leq s <t \\
\frac{t(2\pi-s)}{2\pi}, & t \leq s \leq 2\pi.
\end{cases}
\end{displaymath}

\begin{lemma}[\cite{g1}] \label{degree}
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 $\Omega \subset X$ is an
open bounded set. If the following conditions hold:
\begin{itemize}
  \item[(i)] $Lx \ne \lambda Nx$, for all
$ x \in \partial \Omega \cap D(L)$,  for all $\lambda \in  (0,1)$;

\item[(ii)] $Nx \notin \operatorname{Im} L$, for all
$x \in \partial \Omega \cap \ker  L$;

\item[(iii)] The Brouwer degree $\deg \{QN,\Omega \cap \ker  L, 0 \}\ne 0$.
\end{itemize}
Then equation $Lx=Nx$ has at least one solution in
$\overline{\Omega}$.
\end{lemma}

\section{Existence result}

\begin{theorem}\label{maintheorem}
Assume that there exist constants $K >0$,  $d>0$ and $L \ge 0$
such that
\begin{itemize}
  \item[(C1)] $| f(t,x)| \leq K$ for all $(t,x)\in \mathbb{R}^2$;
  \item[(C2)] $xg(x)<0$ and $| g(x) | \leq  \| p \|_0$ implies $| x | \leq  d$;
  \item[(C3)] $| g(x_1)-g(x_2) | \leq L| x_1-x_2|$, for all
 $x_1,x_2 \in \mathbb{R}$.
\end{itemize}
If
\begin{equation}\label{condition}
   2L \Big(\beta_0+\beta_{m-1}+\sum_{j=1}^{m-2}\beta_j
\Big)^{1/2}<1.
\end{equation}
Then  \eqref{eq1} has at least one $2\pi$-periodic solution.
\end{theorem}

\begin{proof}
 Consider the auxiliary equation
\begin{equation}\label{main313}
x''(t)+\lambda f(t,x(t))+\lambda g(x(t-\tau(t,x(t))))=\lambda
    p(t),\quad \lambda \in (0,1).
\end{equation}
To complete the proof  of this theorem, one can see that  it
suffices to show that all possible $2 \pi$-periodic solutions of
\eqref{main313} are bounded. In other words, we shall prove that
there exist positive constants $M_2$ and  $M_4$ independent of
$\lambda$ and $x$ such that if $x(t)$ is a $2\pi$-periodic
solution of equation \eqref{main313} then
 $    \| x \|_0<M_2$ and $    \| x'\|_0<M_4$.

Let $x(t)$ be any $2 \pi$-periodic solution of
\eqref{main313}. Then there exist $t_1, t_2 \in [0,2\pi]$
such that
\begin{equation}\label{optimal}
 x(t_1)=\min_{t \in [0,2\pi]} x(t),\;\; x(t_2)
=\max_{t \in [0,2\pi]}x(t).
\end{equation}
We claim that there exists $t^{*} \in [0,2\pi]$ such that
\begin{equation}\label{mainy}
| x(t^{*}) | \leq d.
\end{equation}
From \eqref{optimal}, it follows that $x'(t_1)=0$ and thus
$x''(t_1)\ge 0$. Therefore, we have
\begin{equation}\label{12ee}
    g(x(t_1-\tau(t_1,x(t_1)))) \leq p(t_1)-f(t_1,x(t_1)).
\end{equation}
In a similar manner, we deduce that $x'(t_2)=0$ and thus
$x''(t_2)\leq 0$. Hence,
\begin{equation}\label{13ee}
    g(x(t_2-\tau(t_2,x(t_2)))) \ge
    p(t_2)-f(t_2,x(t_2)).
\end{equation}
In view of \eqref{12ee} and \eqref{13ee}, we may write
\begin{gather}\label{14e}
g(x(t_1-\tau(t_1,x(t_1)))) \leq     p(t_1)\leq \| p \|_0, \\
\label{15ee}
g(x(t_2-\tau(t_2,x(t_2)))) \ge     p(t_2)-K \ge  - \| p \|_0.
\end{gather}
Combining \eqref{14e} and \eqref{15ee},  we can find a point
$\xi \in [0,2\pi]$ such that
$$
| g(x(\xi-\tau(\xi,x(\xi))))| \leq \| p \|_0.
$$
By  (C2), the above inequality implies
$$
\big |  x(\xi-\tau(\xi,x(\xi))) \big | \leq d.
$$
Since $x(t)$ is a $2\pi$-periodic function then there exists
$t^{*} \in [0,2\pi]$ such that $\xi-\tau(\xi,x(\xi))=2\pi
k+t^{*}$. Therefore, one can see that \eqref{mainy} holds.
It follows that
\begin{equation}\label{main 9}
    \| x \|_0 \leq | x(t^{*}) | + \int_0^{2 \pi} | x'(s) | \,{\rm d}s \leq d
+ \int_0^{2 \pi} | x'(s) |    \,{\rm d}s.
\end{equation}
Let
$$
E_1=\{t \in [0,2\pi]:\;| x(t) | > d\},\quad
E_2=\{t \in [0,2\pi]:\;| x(t) | \leq d\}.
$$
Multiplying both sides of \eqref{main313} by $x(t)$ and
integrating over $[0,2\pi]$, we have
\begin{align*}
-\int_0^{2\pi}\Big( x'(t) \Big)^2\,{\rm d}t
&= -\lambda \int_0^{2\pi}g\Big(x(t-\tau(t,x(t)))\Big)x(t)\,{\rm d}t
-\lambda \int_0^{2\pi}f(t,x(t))x(t)\,{\rm d}t\\
&\quad +\lambda \int_0^{2\pi}p(t)x(t)\,{\rm d}t
\end{align*}
or
\begin{align*}
\int_0^{2\pi}| x'(t) |^2\,{\rm d}t
&= \lambda \int_0^{2\pi}g\Big(x(t-\tau(t,x(t)))\Big)x(t)\,{\rm d}t
+\lambda \int_0^{2\pi}f(t,x(t))x(t)\,{\rm d}t\\
&\quad -\lambda \int_0^{2\pi}p(t)x(t)\,{\rm d}t.
\end{align*}
It follows that
\begin{align*}
&\int_0^{2\pi}| x'(t) |^2\,{\rm d}t\\
&=  \lambda \int_0^{2\pi}\Big[g(x(t-\tau(t,x(t))))-g(x(t))\Big]x(t)\,{\rm d}t
  +\lambda \int_0^{2\pi}g(x(t))x(t)\,{\rm d}t\\
&\quad +\lambda \int_0^{2\pi}f(t,x(t))x(t)\,{\rm d}t-\lambda
  \int_0^{2\pi}p(t)x(t)\,{\rm d}t
\end{align*}
or
\begin{align*}
&\int_0^{2\pi}| x'(t) |^2\,{\rm d}t\\
 &=  \lambda \int_0^{2\pi}\Big[g(x(t-\tau(t,x(t))))-g(x(t))\Big]x(t)\,{\rm d}t
+\lambda \int_{E_1}g(x(t))x(t)\,{\rm d}t\\
&\quad +\lambda \int_{E_2}g(x(t))x(t)\,{\rm d}t
 +\lambda \int_0^{2\pi}f(t,x(t))x(t)\,{\rm d}t-\lambda
  \int_0^{2\pi}p(t)x(t)\,{\rm d}t.
\end{align*}
This implies
\begin{align*}
  \int_0^{2\pi}| x'(t) |^2\,{\rm d}t
&\leq  \int_0^{2\pi}\Big | g(x(t-\tau(t,x(t))))-g(x(t))\Big | | x(t) | \,{\rm d}t
 +g_{d}\int_0^{2\pi}| x(t) | \,{\rm d}t  \\
&\quad + K \int_0^{2\pi}| x(t) | \,{\rm d}t+\int_0^{2\pi}| p(t) | | x(t)|  \,{\rm d}t,
\end{align*}
where $g_{d}=\max_{t \in E_2} | g(x(t)) |$. Furthermore,
\begin{align*}
&\int_0^{2\pi}| x'(t) |^2\,{\rm d}t \\
&\leq  L \int_0^{2\pi} \Big | x(t-\tau(t,x(t)))-x(t)   \Big| | x(t) | \,{\rm d}t+g_{d}
(2\pi)^{1/2}\| x \|_2 + K\| x \|_2+\| p\|_2 \| x  \|_2,
\end{align*}
where $\|  x \|_2=\Big( \int_0^{2 \pi} | x(s)
|^2 \,{\rm d}s \Big)^{1/2}$.
 It follows that
\begin{align*}
\int_0^{2\pi}| x'(t) |^2\,{\rm d}t
&\leq   L \left(\int_0^{2\pi} \Big| x(t-\tau(t,x(t)))-x(t) \Big|^2
\,{\rm d}t \right)^{1/2}\| x \|_2 +g_{d}
(2\pi)^{1/2}\| x \|_2\\ &\quad + K\| x
\|_2+\| p \|_2 \| x
  \|_2.
\end{align*}
By the consequence of Lemma \ref{lem2}, we obtain
\begin{equation} \label{es1}
\begin{aligned} 
\int_0^{2\pi}| x'(t) |^2\,{\rm d}t
&\leq  L \Big(\beta_0+\beta_{m-1}+\sum_{j=1}^{m-2}\beta_j
\Big)^{1/2} \Big( \int_0^{2\pi}| x'(t) |^2
\,{\rm d}t \Big)^{1/2} \| x \|_2  \\
&\quad +g_{d} (2\pi)^{1/2}\| x \|_2
 + K\| x \|_2+\| p \|_2 \| x  \|_2,
\end{aligned}
\end{equation}
Denote $u(t)=x(t)-x(t^{*})$ where $t^{*}$ is defined as in
\eqref{main 9}. Then we have
$$
| x(t) | \leq | x(t^{*}) | + | x(t)-x(t^{*}) | \leq d +| u(t) |.
$$
Using the Minkowski inequality \cite{h1}, we obtain
\begin{equation}\label{ewq}
    \| x \|_2=\Big( \int_0^{2\pi}| x(t) |^2 \,{\rm d}t \Big)^{1/2}
    \leq (2\pi)^{1/2}d+ \Big( \int_0^{2\pi}| u(t) |^2 \,{\rm d}t
    \Big)^{1/2}.
\end{equation}
However, since $u(t^{*})=0,\; u(t+2\pi)=u(t)$ and $u'(t)=x'(t)$
then by the consequence of Lemma \ref{lem3}, we have
$$\Big( \int_0^{2\pi}| u(t) |^2 \,{\rm d}t
    \Big)^{1/2} \leq 2\Big( \int_0^{2\pi}| u'(t) |^2 \,{\rm d}t
    \Big)^{1/2} = 2 \Big( \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t
    \Big)^{1/2}.
$$
Substituting back in \eqref{ewq}, we obtain
\begin{equation}\label{ews}
    \Big( \int_0^{2\pi}| x(t) |^2 \,{\rm d}t \Big)^{1/2}
    \leq (2\pi)^{1/2}d+ 2 \Big( \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t
    \Big)^{1/2}.
\end{equation}
It follows from \eqref{es1} and \eqref{ews} that
\begin{align*}
\int_0^{2\pi}| x'(t) |^2\,{\rm d}t
&\leq    L \Big( \beta_0+\beta_{m-1}+\sum_{j=1}^{m-2}\beta_j \Big)^{1/2}
\Big( \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t \Big)^{1/2}\\
&\quad\times  \Big( (2\pi)^{1/2}d+2\Big( \int_0^{2\pi} | x'(t) |^2 \,{\rm d}t \Big)^{1/2}\Big)\\
&\quad +\left(g_{d}(2\pi)^{1/2}+K+\| p
\|_2\right)\Big( (2\pi)^{1/2}d+2\Big(
\int_0^{2\pi}| x'(t) |^2 \,{\rm d}t  \Big)^{1/2}\Big)
\\
 &= 2L\Big( \beta_0+\beta_{m-1}+\sum_{j=1}^{m-2}\beta_j \Big)^{1/2}
   \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t\\
 &\quad + (2\pi)^{1/2}d L \Big( \beta_0+\beta_{m-1}
   +\sum_{j=1}^{m-2}\beta_j \Big)^{1/2}\Big( \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t
    \Big)^{1/2}\\
 &\quad +2 \left(g_{d}(2\pi)^{1/2}+K+\| p \|_2\right)
    \Big( \int_0^{2\pi}| x'(t) |^2 \,{\rm d}t  \Big)^{1/2}\\
 &\quad   +(2\pi)^{1/2}d \left(g_{d}(2\pi)^{1/2}+K+\| p    \|_2\right).
\end{align*}
By  \eqref{condition}, we deduce that there exists a
constant $M_1>0$ such that
$$
\int_0^{2\pi}| x'(t) |^2\,{\rm d}t \leq M_1.
$$
From \eqref{main 9}, we end up with
$$
\| x \|_0 \leq d+(2\pi)^{1/2}M_1^{1/2}:=M_2.
$$
In view of equation \eqref{eq1}, one can obtain
\begin{equation}\label{son}
    \| x''(t) \| \leq g_{M_2}+K+\| p \|_0:=M_{3},
\end{equation}
where $g_{M_2}=\max_{| x | \leq M_2}| g(x) |$.
However, since $x(0)=x(2\pi)$ then there exists a constant
$\eta \in [0,2\pi]$ such that $x'(\eta)=0$. Therefore, by \eqref{son} we
have
$$
\| x' \|_0 \leq | x'(\eta)| + \int_0^{2\pi} | x''(s) | \,{\rm d}s \leq 2\pi M_{3}:=M_4.
$$
Clearly, $M_2$ and $M_4$ are independent of $\lambda$ and $x$.
Take $\Omega=\{x: x \in X, \;\| x \|_0<M_2,\; \| x'\|_0<M_4\}$ and
$\Omega_1=\{x:x \in \ker L,\;Nx \in \operatorname{Im} L\}$.
 Clearly for all $x \in \Omega_1,\; x \equiv c$ is a constant and
 $f(t,c)+g(c)=p(t)$ thus by assumption
(C2) we have $| c | \leq d$ and hence $\Omega_1 \subset
\Omega$. This tells that conditions (i)--(ii) of Lemma
\ref{degree} are satisfied. Let
$$
 H(x,\mu)=\mu x -\frac{1-\mu}{2 \pi}\int_0^{2 \pi}
\Big( f(t,x)+g(x)-p(t) \Big)\,{\rm d}t.
$$
Then, one can easily realize that $H(x,\mu)\ne 0$ for all
 $(x,\mu) \in (\partial \Omega \cap \ker  L)\times [0,1]$. Hence
\begin{align*}
  \deg\{QN,\Omega\cap \ker  L,0\}
&= \deg\{H(x,0),\Omega\cap \ker  L,0\}
 = \deg\{H(x,1),\Omega\cap \ker  L,0\} \\
&= \deg\{1,\Omega\cap \ker  L,0\}\ne 0.
\end{align*}
Therefore, condition (iii) of Lemma \ref{degree} holds. This shows
that equation \eqref{eq1} has at least one $2\pi$-periodic
solution. The proof is complete.
\end{proof}

\begin{example} \rm
Consider the equation
\begin{equation}\label{seer1}
    x''(t)+\frac{1}{2}\sin x(t)-\frac{x^3
\Big(t-\frac{1}{40}\tau(t,x(t))\Big)}{1+x^2
\Big(t-\frac{1}{20}\tau(t,x(t))\Big)}=e^{\cos^2 t},
\end{equation}
where $f(t,x)=\frac{1}{2}\sin x(t)$,
$g(x)=-x^3/(1+x^2)$,
 $\tau(t,x(t))=\frac{1}{80}| \cos (t+10x(t)) |$
and $p(t)=e^{\cos^2 t}$.
Clearly, $K=1/2$, $L=1$ and $\tau(t,x(t)) \in [0,4\pi]$
for $t \in [0,2 \pi]$. Therefore,
$k=0$, $m=2$, $\delta_0=\delta_1=2\pi$,
$\beta_0=\beta_1=\pi/40$. Thus, it is
straightforward to realize that conditions (C1)--(C3) and
\eqref{condition} hold. By  Theorem
\ref{maintheorem}, equation \eqref{seer1} has at least one
$2\pi$-periodic solution.
\end{example}

We remark that the results obtained in
\cite{d2,h2,g3,l1,l2,l3,l5,l6,l7,p1,w4,z1,z2,z3,z4}
can not be applied to \eqref{seer1}. This tells that the
result in this paper is essentially new.

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 \end{document}

