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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 37, 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 2011 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2011/37\hfil Existence of periodic solutions]
{Existence of periodic solutions for second order delay
 differential equations with impulses}

\author[L. Pan\hfil EJDE-2011/37\hfilneg]
{Lijun Pan}

\address{Lijun Pan \newline
School of Mathematics, Jia Ying University,
Meizhou Guangdong, 514015, China}
\email{plj1977@126.com}

\thanks{Submitted January 28, 2011. Published March 3, 2011.}
\thanks{Supported by grant 9151008002000012 from
the Natural Science Foundation of \hfill\break\indent
Guangdong Province, China}
\subjclass[2000]{34K13, 34K45}
\keywords{Second-order delay differential equations;
 impulses; \hfill\break\indent periodic solution; coincidence degree.}

\begin{abstract}
 Using the coincidence degree theory by Mawhin,
 we prove the existence of periodic solutions for the
 second-order delay  differential equations with impulses
 \begin{gather*}
  x''(t)+f(t,x'(t))+g(x(t-\tau(t))=p(t),\quad t\geq0,\; t\neq t_k,\\
  \Delta x(t_k)=I_k(x(t_k),x'(t_k)),\\
  \Delta x'(t_k)=J_k(x(t_k),x'(t_k)).
 \end{gather*}
 We obtain new existence results and illustrated them by an example.
\end{abstract}

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

\section{Introduction}

 This article  concerns the existence of periodic
solutions for the second-order delay differential equations
with impulses
\begin{equation}
\begin{gathered}
 x''(t)+f(t,x'(t))+g(x(t-\tau(t))=p(t),t\geq0,t\neq t_k,\\
 \Delta x(t_k)=I_k(x(t_k),x'(t_k)),\\
 \Delta x'(t_k)=J_k(x(t_k),x'(t_k))
\end{gathered} \label{e1.1}
\end{equation}
where $\Delta x(t_k)=x(t^{+}_k)-x(t^{-}_k)$,
$x(t^{+}_k)=\lim_{t\to t^{+}_k}x(t)$,
$x(t^{-}_k)=\lim_{t\to t^{-}_k}x(t)$ and
$x(t^{-}_k)=x(t_k)$;
also
$\Delta x'(t_k)=x'(t^{+}_k)-x'(t^{-}_k)$,
$x'(t^{+}_k)=\lim_{t\to t^{+}_k}x'(t)$,
$x'(t^{-}_k)=\lim_{t\to t^{-}_k}x'(t)$ and
$x'(t^{-}_k)=x'(t_k)$.

We assume that the following conditions:
\begin{itemize}

\item[(H1)] $f\in C(\mathbb{R}^2,\mathbb{R})$ and
$f(t+T,x)=f(t,x)$, $g\in C(\mathbb{R},\mathbb{R})$,
$p, \tau\in C(\mathbb{R},\mathbb{R})$ with
$\tau(t+T)=\tau(t)$, $p(t+T)=p(t)$;

\item[(H2)]  $\{t_k\}$ satisfies $t_k<t_{k+1}$ and
$\lim_{k\to\pm\infty}t_k=\pm\infty$, $k\in Z$,
$I_k(x,y),J_k(x,y)\in C(\mathbb{R}^2,\mathbb{R})$,
and there is a positive $n$ such that
$\{t_k\}\cap[0,T]=\{t_1,t_2,\dots,t_{n}\}$,
 $t_{k+n}=t_k+T,I_{k+n}(x,y)=I_k(x,y),J_{k+n}(x,y)=J_k(x,y)$.
\end{itemize}

 Impulsive differential equations are mathematical apparatus  for
simulations of process and phenomena observed in control
theory,physics,chemistry, population dynamics, biotechnologies,
industrial robotics, economics, etc. So there have been quite a
few results on properities of their solutions in recent years
\cite{b1,c1,g2,l1,s1}.
 In paiticular, the existence of periodic solutions for
first order differential equations with impulses has been studied
 in \cite{q1,y1}.
Li and Shen \cite{s1} have studied the existence of periodic
solutions for duffing equations with delays and impulses.
 In present paper, by using Mawhin's continuation theorem,
we will establish some theorems on the existence of periodic
solutions of  \eqref{e1.1}.
 The results is related to not only $f(t,x)$ and $g(y)$ but also the
impulses  $I_k(x,y)$ and $J_k(x,y)$ and the delay $\tau(t)$.
In addition, we give an example to illustrate our new results.

 For background material on periodic solutions of first or second
order differential equations without impulses, the references
\cite{f1,k1,l3,l4, m1,n1,o1,w1} may be consulted.

\section{Preliminaries}

 We establish the theorems of existence of periodic solution based
on the following Mawhin's continuation theorem.

Let $PC(\mathbb{R},\mathbb{R})=\{x:\mathbb{R}\to\mathbb{R},x(t)$
be continuous everywhere except for some $t_k$ at which
$x(t^{+}_k)$ and $x(t^{-}_k)$ exist and $x(t^{-}_k)=x(t_k)\}$,
$PC^{1}(\mathbb{R},\mathbb{R})=\{x:\mathbb{R}\to\mathbb{R},x(t)$
is continuous everywhere except for some $t_k$ at which
$x'(t^{+}_k)$ and $x'(t^{-}_k)$ exist and
$x'(t^{-}_k)=x'(t_k)\}$.
Let $X=\{x(t)\in PC^{1}(\mathbb{R},\mathbb{R}),x(t+T)=x(t)\}$ with
norm $\|x\|=\max\{|x|_{\infty},|x'|_{\infty}\}$, where
$|x|_{\infty}=\sup_{t\in[0,T]}|x(t)|$,
$Y=PC(\mathbb{R},\mathbb{R})\times \mathbb{R}^{n}\times \mathbb{R}^{n}$, with norm
$\|y\|=\max\{|u|_{\infty},|c|\}$, where
$u\in PC(\mathbb{R},\mathbb{R}),c=(c_1, \dots c_{2n})\in R^{n}\times \mathbb{R}^{n}$, $|c|=\max_{1\leq k\leq2n}\{|c_k|\}$. Then $X$ and $Y$
are Banach spaces. $L:D(L)\subset X\to Y$ is a Fredholm operator
of index zero, where $D(L)$ denotes the domain of $L$.
$P:X\to X,Q:Y\to Y$ are projectors such that
$$
\operatorname{Im}P=\ker L,\quad
\ker Q=\operatorname{Im}L,\quad
X=\ker L\oplus \ker P,\quad
Y=\operatorname{Im}L\oplus \operatorname{Im}Q.
$$
It follows that
$$
L|_{D(L)\cap \ker P}:D(L)\cap \ker P\to \operatorname{Im}L
$$
is invertible and we define the inverse of that map by $K_{p}$.
Let $\Omega$ be an open bounded subset of $X$,
$D(L)\cap\overline{\Omega}\neq\emptyset$,
the map $N:X\to Y$ will be called
 $L$-compact in $\overline{\Omega}$,
if $QN(\overline{\Omega})$ is bounded and
 $K_{p}(I-Q)N:\overline{\Omega}\to X$ is compact.

\begin{lemma}[\cite{g2}] \label{lem2.1}
 Let $L$ be a Fredholm operator of index zero and let $N$ be
$L$-compact on $\overline{\Omega}$. Assume that the
following conditions are satisfied:
\begin{itemize}
\item[(i)] $ Lx\neq\lambda Nx,\forall x\in\partial\Omega\cap D(L),
\lambda\in(0,1)$;
\item[(ii)] $QNx\neq 0$, for all $x\in\partial\Omega\cap \ker L$;

\item[(iii)]  $\deg\{JQNx,\Omega\bigcap \ker L,0\}\neq0$,
where $J:\operatorname{Im}Q\to \ker L$ is
an isomorphism.
\end{itemize}
Then the equation $Lx=Nx$ has at least one solution in
$\overline{\Omega}\bigcap D(L)$.
\end{lemma}

We define the operators $L:D(L)\subset X\to Y$ by
\begin{equation}
Lx=(x'',\Delta x(t_1),\dots,\Delta x(t_{n}),
\Delta x'(t_1),\dots,\Delta x'(t_{n})),
\label{e2.1}
\end{equation}
and $N:X\to Y$ by
\begin{equation}
\begin{split}
Nx&=(-f(t,x'(t))-g(x(t-\tau(t)))+ p(t),\\
&\quad I_1(x(t_1)),\dots, I_{n}(x(t_{n})),
J_1(x'(t_1)),\dots,J_{n}(x'(t_{n}))).
\end{split} \label{e2.2}
\end{equation}
It is easy to see that  \eqref{e1.1} can be converted
into the abstract equation $Lx=Nx$.

\begin{lemma}[\cite{l2}] \label{lem2.2}
$L$ is a Fredholm operator of index zero with
\begin{equation}
\ker L=\{x(t)=c,t\in R\}, \label{e2.3}
\end{equation}
and
\begin{equation}
\begin{split}
\operatorname{Im}L
&=\{(y,a_1,\dots,a_{n},b_1,\dots,b_{n})\in Y:\\
&\quad
\int^T_0y(s)ds+\sum^{n}_{k=1}b_k(T-t_k)+
\sum^{n}_{k=1}a_k+x'(0)T=0\}.
\end{split} \label{e2.4}
\end{equation}
Furthermore, let the linear continuous projector operator
$P:X\to X$ and  $Q:Y\to Y$ be defined by
\begin{equation}
Px=x(0), \label{e2.5}
\end{equation}
and
\begin{equation}
\begin{split}
&Q(y,a_1,\dots,a_{n},b_1,\dots,b_{n})\\
&=\frac{2}{T^2}[\int^T_0(T-s)y(s)ds+
\sum^{n}_{k=1}b_k(T-t_k)+\sum^{n}_{k=1}a_k+x'(0)T],0,\dots,0).
\end{split} \label{e2.6}
\end{equation}
Then the linear operator
$K_{p}:\operatorname{Im}L\to D(L)\cap \ker P$ can be written as
\begin{equation}
\begin{split}
&K_{p}(y,a_1,\dots,a_{n},b_1,\dots,b_{n})\\
&= \int^T_0(T-s)y(s)ds+
\sum^{n}_{k=1}b_k(T-t_k)+\sum^{n}_{k=1}a_k+x'(0)T.
\end{split} \label{e2.7}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.3}
Suppose $\Omega\subset X$ is bounded open set,then $N$
is $L$-compact in $\overline{\Omega}$.
\end{lemma}

\begin{proof}
It is easy to see that $QN(\overline{\Omega})$ is bound. By using the
Ascoli-Arzela theorem, we can prove that $K_{p}(I-Q)Nx$ is compact.
Thus $N$ is $L$-compact in $\overline{\Omega}$.
\end{proof}

\begin{lemma}[\cite{l4}] \label{lem2.4}
 Suppose $\alpha>0$, $x(t)\in PC^{1}(\mathbb{R},\mathbb{R})$
with $x(t+T)=x(t)$, Then
\begin{equation}
\int^T_0 \int^{t}_{t-\alpha}|x'(s)|^2\,ds\,dt
=\alpha\int^T_0|x'(t)|^2dt \label{e2.8}
\end{equation}
and
\begin{equation}
\int^T_0 \int^{t+\alpha}_{t}|x'(s)|^2\,ds\,dt
=\alpha\int^T_0|x'(t)|^2dt. \label{e2.9}
\end{equation}
\end{lemma}

Let
\begin{gather*}
A_1(t,\alpha)=\sum_{t-\alpha\leq t_k\leq t}a_k,\quad
A_2(t,\alpha)=\sum_{t\leq t_k\leq t+\alpha}a_k, \\
B_1(t,\alpha)=\sum_{t-\alpha\leq t_k\leq t}a'_k,\quad
B_2(t,\alpha)=\sum_{t\leq t_k\leq t+\alpha}a'_k,\\
A(\alpha)=\Big(\int^T_0A^2_1(t,\alpha)dt\Big)^{1/2}
 +\Big(\int^T_0A^2_2(t,\alpha)dt\Big)^{1/2}, \\
B(\alpha)=\Big(\int^T_0B^2_1(t,\alpha)dt\Big)^{1/2}
+\Big(\int^T_0B^2_2(t,\alpha)dt\Big)^{1/2},\\
C(\alpha)=\int^T_0A^2_1(t,\alpha)dt
 +\int^T_0A^2_2(t,\alpha)dt,\\
D(\alpha)=\int^T_0A_1(t,\alpha)B_1(t)dt
 +\int^T_0A_2(t,\alpha)B_2(t)dt,\\
E(\alpha)=\int^T_0B^2_1(t,\alpha)dt
 +\int^T_0B^2_2(t,\alpha)dt
\end{gather*}
 The following Lemma is crucial for us to establish theorems
related to the delay $\tau(t)$  and $I_k(x,y)$.

\begin{lemma} \label{lem2.5}
Suppose $\tau(t)\in C(\mathbb{R},\mathbb{R})$ with
$\tau(t+T)=\tau(t)$ and $\tau(t)\in[-\alpha,\alpha]$ for all
$t\in[0,T]$,
$x(t)\in PC^{1}(\mathbb{R},\mathbb{R})$ with $x(t+T)=x(t)$
and there is a positive $n$ such that
$\{t_k\}\cap[0,T]=\{t_1,t_2,\dots,t_{n}\}$,
$\Delta x(t_k)= \lambda I_k(x(t_k),x'(t_k))$
for all $\lambda\in(0,1)$ and
$t_{k+n}=t_k+T,I_{k+n}(x,y)=I_k(x,y)$.
Furthermore there exist nonnegative constants
$a_k,a_k$ such that $|I_k(x,y)|\leq a_k|x|+a'_k$. Then
\begin{equation}
\begin{split}
&\int^T_0|x(t)-x(t-\tau(t))|^2dt\\
&\leq2\alpha^2\int^T_0|x'(t)|^2dt+
2\alpha A(\alpha)|x(t)|_{\infty}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}\\
&\quad + 2\alpha B(\alpha) \Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}
+C(\alpha)|x(t)|^2_{\infty}+D(\alpha)|x(t)|_{\infty}
+E(\alpha).
\end{split}\label{e2.10}
\end{equation}
\end{lemma}

\begin{proof}
 If $\tau(t)\in[0,\alpha]$, then for all $t\in[0,T]$,
using Schwarz inequality, we obtain
 \begin{align*}
&|x(t)-x(t-\tau(t))|^2\\
&=|\int^{t}_{t-\tau(t)}x'(s)ds+\lambda
\sum_{t-\tau(t)\leq t_k<t}I_k(x(t_k)|^2\\
&\leq\Big(\int^{t}_{t-\alpha}|x'(s)|ds\Big)^2
 +2\lambda\Big(\int^{t}_{t-\alpha}|x'(s)|ds\Big)
\sum_{t-\alpha\leq t_k<t}|I_k(x(t_k)|\\
&\quad + \Big(\lambda\sum_{t-\alpha\leq t_k<t}|I_k(x(t_k))
 |\Big)^2\\
&\leq\alpha\int^{t}_{t-\alpha}|x'(s)|^2ds
+ 2\int^{t}_{t-\alpha}|x'(s)|ds
\sum_{t-\alpha\leq t_k<t}
[a_k|x(t)|_{\infty}+a'_k]\\
&\quad+ \Big[\sum_{t-\alpha\leq t_k<t}(a_k|x(t)|_{\infty}
+a'_k)\Big]^2.
\end{align*}%\label{e2.11}
By the Schwarz inequality and Lemma \ref{lem2.4}, we obtain
\begin{align*}
&\int^T_0|x(t)-x(t-\tau(t))|^2dt\\
&\leq\alpha\int^T_0\int^{t}_{t-\alpha}|x'(s)|^2\,ds\,dt\\
&\quad +2|x(t)|_{\infty}\int^T_0A_1(t,\alpha)
 \int^{t}_{t-\alpha}|x'(s)|\,ds\,dt+
 2\int^T_0B_1(t,\alpha)
 \int^{t}_{t-\alpha}|x'(s)|\,ds\,dt\\
&\quad +|x(t)|^2_{\infty}\int^T_0A^2_1(t,\alpha)dt+
 |x(t)|_{\infty}\int^T_0A_1(t,\alpha)B_1(t,\alpha)dt
 +\int^T_0B^2_1(t,\alpha)dt\\
&\leq\alpha\int^T_0\int^{t}_{t-\alpha}|x'(s)|^2\,ds\,dt+
2|x(t)|_{\infty}\\
&\quad\times \Big(\int^T_0A^2_1(t,\alpha)dt\Big)^{1/2}
\Big(\int^T_0\Big(\int^{t}_{t-\alpha}|x'(s)|ds\Big)^2dt\Big)^{1/2}\\
&\quad +2\Big(\int^T_0B^2_1(t,\alpha)dt\Big)^{1/2}
\Big(\int^T_0\Big(\int^{t}_{t-\alpha}|x'(s)|ds\Big)^2dt\Big)^{1/2}\\
&\quad +|x(t)|^2_{\infty}\int^T_0A^2_1(t,\alpha)dt+
 |x(t)|_{\infty}\int^T_0[A_1(t,\alpha) B_1(t,\alpha)]dt
 +\int^T_0B^2_1(t,\alpha)dt\\
&\leq\alpha^2\int^T_0|x'(t)|^2dt+
2\alpha|x(t)|_{\infty}\Big(\int^T_0A^2_1(t,\alpha)dt\Big)^{1/2}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}\\
&\quad + 2\alpha\Big(\int^T_0B^2_1(t,\alpha)dt\Big)^{1/2}
(\int^T_0|x'(t)|^2dt)^{1/2}\\
&\quad + |x(t)|^2_{\infty}\int^T_0A^2_1(t,\alpha)dt+
|x(t)|_{\infty}\int^T_0A_1(t,\alpha)B_1(t,\alpha)dt
+\int^T_0B^2_1(t,\alpha)dt.
\end{align*} %\label{e2.12}
If $\tau(t)\in[-\alpha,0]$, then for all $t\in[0,T]$, similarly,
we obtain
\begin{align*}
&\int^T_0|x(t)-x(t-\tau(t))|^2dt\\
&\leq\alpha^2\int^T_0|x'(t)|^2dt+
2\alpha|x(t)|_{\infty}\Big(\int^T_0A^2_2(t,\alpha)dt\Big)^{1/2}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}\\
&\quad +2\alpha(\int^T_0B^2_2(t,\alpha)dt)^{1/2}
(\int^T_0|x'(t)|^2dt)^{1/2}\\
&\quad +|x(t)|^2_{\infty}\int^T_0A^2_2(t,\alpha)dt+
|x(t)|_{\infty}\int^T_0A_2(t,\alpha)
B_2(t,\alpha)dt
+\int^T_0B^2_2(t,\alpha)dt.
\end{align*} %\label{e2.13}}
Let $\Delta_1=\{t:t\in[0,T],\tau(t)\geq0\}$,
$\Delta_2=\{t:t\in[0,T],\tau(t)<0\}$.
Then for for all $t\in[0,T]$,
\begin{align*}
&\int^T_0|x(t)-x(t-\tau(t))|^2dt\\
&=\int_{\Delta_1}|x(t)-x(t-\tau(t))|^2dt+
\int_{\Delta_2}|x(t)-x(t-\tau(t))|^2dt\\
&\leq2\alpha^2\int^T_0|x'(t)|^2dt+
2\alpha A(\alpha)|x(t)|_{\infty}
(\int^T_0|x'(t)|^2dt)^{1/2}\\
&\quad +2\alpha B(\alpha)
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}
+C(\alpha)|x(t)|^2_{\infty}+ D(\alpha)|x(t)|_{\infty}
+E(\alpha).
\end{align*} %\label{e2.14}
\end{proof}

\section{Main results}

For the next theorem we use the following conditions:
\begin{itemize}
\item[(H3)] There are constants $\sigma,\beta\geq 0$ such that
\begin{gather}
|f(t,x)|\leq\sigma|x|,\quad \forall (t,x)\in[0,T]\times \mathbb{R},
 \label{e3.1}\\
xf(t,x)\geq \beta|x|^2,\quad \forall (t,x)\in[0,T]\times \mathbb{R};
 \label{e3.2}
\end{gather}

\item[(H4)] there are constants $\beta_i\geq0$ ($i=1,2,3$)
 such that
\begin{gather}
 |g(x)|\geq\beta_1+\beta_2|x|,  \label{e3.3}\\
|g(x)-g(y)|\leq\beta_3|x-y|; \label{e3.4}
\end{gather}

\item[(H5)]  there are constants $\gamma_i>0$ ($i=1,2,3$), such that
 $|\int^{x+\lambda I_k(x,y)}_{x}g(s)ds|
 \leq |I_k(x,y)|(\gamma_1+\gamma_2|x|+\gamma_3|I_k(x,y)|),\quad
\forall\lambda\in(0,1)$;

\item[(H6)] there are constants $a_k,a'_k\geq0$
such that $|I_k(x,y)|\leq a_k|x|+a'_k$;

\item[(H7)] $yJ_k(x,y)\leq 0$ and
 there are constants $b_k\geq0$ such that
$|J_k(x,y)|\leq b_k$.
\end{itemize}

\begin{theorem} \label{thm3.1}
Suppose {\rm (H1)--(H7)} hold.
Then  \eqref{e1.1} has at least one $T$-periodic solution
provided the following two conditions hold
\begin{gather}
\sum^{n}_{k=1}a_k<1, \label{e3.5}\\
\begin{split}
&\Big[\gamma_2(\sum^{n}_{k=1}a_k)+\gamma_3
(\sum^{n}_{k=1}a^2_k)\Big]M^2
+\beta_3\Big[2|\tau(t)|_{\infty}^2\\
&+ 2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})M
+C(|\tau(t)|_{\infty})M^2\Big]^{1/2}<\beta,
\end{split}
 \label{e3.6}
\end{gather}
 where
$$
M=\frac{1}{1-\sum^{n}_{k=1}a_k}
(\frac{\sigma}{\beta_2T^{1/2}}+
T^{1/2}).
$$
\end{theorem}

\begin{proof}
Consider the equation
$Lx=\lambda Nx$, with $\lambda\in(0,1)$,
where $L$ and $N$ are defined by \eqref{e2.1} and \eqref{e2.2}.
Let
$$
\Omega_1=\{x\in D(L): \ker L,Lx=\lambda Nx \text{ for some }
 \lambda\in(0,1)\}\,.
$$
For $x\in \Omega_1$, we have
\begin{equation}
 \begin{gathered}
  x''(t)+\lambda f(t,x'(t))+\lambda g(t,x(t-\tau(t))
 =\lambda p(t),\quad  t\neq t_k,\\
  \Delta x(t_k)=\lambda I_k(x(t_k),x'(t_k)),\\
 \Delta x'(t_k)=\lambda J_k(x(t_k),x'(t_k)).
\end{gathered} \label{e3.7}
\end{equation}
Integrating them on $[0,T]$, using Schwarz inequality, we have
\begin{align*}
&|\int^T_0g(x(t-\tau(t))dt|\\
&= |\int^T_0p(t)dt-\int^T_0f(t,x'(t))dt+\sum^{n}_{k=1}
 J_k(x(t_k),x'(t_k))|\\
&\leq T|p(t)|_{\infty}+\sigma\int^T_0|x'(t)|dt+ \sum^{n}_{k=1}b_k\\
&\leq \sigma T^{1/2}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}
+T|p(t)|_{\infty}+\sum^{n}_{k=1}b_k.
\end{align*} % \label{e3.8}
 From the above formula, there is a $t_0\in[0,T]$ such that
$$
|g(x(t_0-\tau(t_0))|
\leq\frac{\sigma}{T^{1/2}}
(\int^T_0|x'(t)|^2dt)^{1/2}+|p(t)|_{\infty}+
\frac{1}{T}\sum^{n}_{k=1}b_k.
$$ % \label{e3.9}
It follows from \eqref{e3.3} that
$$
\beta_1+\beta_2|x(t_0-\tau(t_0))|
\leq  \frac{\sigma}{T^{1/2}}
(\int^T_0|x'(t)|^2dt)^{1/2}+|p(t)|_{\infty}+
\frac{1}{T}\sum^{n}_{k=1}b_k.
$$ % \label{e3.10}
Thus
\[
|x(t_0-\tau(t_0))|
\leq\frac{\sigma}{\beta_2T^{1/2}}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}+d,
\] %\label{e3.11}
where
$d=\big(||p(t)|_{\infty}+\frac{1}{T}\sum^{n}_{k=1}
b_k- \beta_1|\big)/\beta_2$.
So there
must be an integer $m$ and a point $t_1\in[0,T]$ such that
$t_0-\tau(t_0)=mT+t_1$. Hence
$$
|x(t_1)|=|x(t_0-\tau(t_0))|\leq
\frac{\sigma}{\beta_2T^{1/2}} \Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}+d,
$$
which implies
\[
x(t)=x(t_1)+\int^{t}_{t_1}x'(s)ds+\sum_{t_1\leq t_k<t}
I_k(x(t_k),x'(t_k)).
\] % \label{e3.12}
This  yields
\begin{align*}
|x(t)|_{\infty}
&\leq |x(t_1)|+\int^{t}_{t_1}|x'(s)|ds+
\sum_{t_1\leq t_k<t}|I_k(x(t_k))|\\
&\leq \frac{\sigma}{\beta_2T^{1/2}}
(\int^T_0|x'(t)|^2dt)^{1/2}+d
+ \int^T_0|x'(t)|dt+\sum^{n}_{k=1}a_k|x|_{\infty}
 +\sum^{n}_{k=1}a'_k\\
 &\leq |x|_{\infty}\sum^{n}_{k=1}a_k+(\frac{\sigma}{\beta_2T^{1/2}}+T^{1/2})
 \Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}+
 d +\sum^{n}_{k=1}a'_k.
\end{align*}% \label{e3.13}
It follows that
\begin{equation}
\begin{split}
|x(t)|_{\infty}
&\leq\frac{d+\sum^{n}_{k=1}a'_k}{1-\sum^{n}_{k=1}a_k}
+\frac{1}{1-\sum^{n}_{k=1}a_k}
(\frac{\sigma}{\beta_2T^{1/2}}+
T^{1/2})(\int^T_0|x'(t)|^2dt)^{1/2}\\
&=u_1 +M(\int^T_0|x'(t)|^2dt)^{1/2},
\end{split}  \label{e3.14}
\end{equation}
where $u_1$ is a positive constant.
On the other hand, multiplying both side of \eqref{e3.7}
 by $x'(t)$, we have
\begin{align*}
&\int^T_0x''(t)x'(t)dt+\lambda\int^T_0f(t,x'(t))x'(t)dt&+
\lambda\int^T_0g(t,x(t-\tau(t))x'(t)dt\\
&=\lambda\int^T_0p(t)x'(t)dt.
\end{align*} % \label{e3.15}
Since
\[
\int^T_0x''(t)x'(t)dt=-\frac{1}{2}
\sum^{n}_{i=1}[(x'(t^{+}_k))^2-(x'(t_k))^2],
\]
it follows from assumption (H7) that
\begin{align*}
&(x'(t^{+}_k))^2-(x'(t_k))^2\\
&= (x'(t^{+}_k)+x'(t_k))(x'(t^{+}_k)-(x'(t_k))\\
&=\Delta x'(t_k)(2x'(t_k)+\Delta x'(t_k))\\
&=\lambda J_k(x(t_k),x'(t_k))(2x'(t_k)
 +\lambda J_k(x(t_k),x'(t_k))\\
&=2\lambda J_k(x(t_k),x'(t_k))x'(t_k)
 + [\lambda J_k(x(t_k),x'(t_k))]^2
\leq b^2_k.
\end{align*} % \label{e3.16}
In view of \eqref{e3.2}, by Schwarz inequality, we obtain
\begin{equation}
\begin{split}
&\beta\int^T_0|x'(t)|^2dt\\
&\leq-\int^T_0g(x(t-\tau(t))x'(t)dt+
\int^T_0p(t)x'(t)dt+\frac{1}{2}
\sum^{n}_{k=1}b^2_k\\
&=\int^T_0[g(x(t)-g(x(t-\tau(t))]x'(t)dt-\int^T_0g(x(t))x'(t)dt\\
&\quad + \int^T_0p(t)x'(t)dt+\frac{1}{2} \sum^{n}_{i=1}b^2_k\\
&\leq\int^T_0|g(x(t))-g(x(t-\tau(t))||x'(t)|dt+
|p(t)|_{\infty}\int^T_0|x'(t)|dt\\
&\quad +|\int^T_0g(x(t))x'(t)dt|+\frac{1}{2}\sum^{n}_{i=1}b^2_k\\
&\leq \Big[\Big(\int^T_0|g(x(t))-g(x(t-\tau(t)))|^2dt\Big)^{1/2}
+ |p(t)|_{\infty}T^{1/2}\Big]
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}\\
&\quad +|\int^T_0g(x(t))x'(t)dt|+\frac{1}{2}
\sum^{n}_{i=1}b^2_k.
\end{split} \label{e3.17}
\end{equation}
 From (H5) and (H6), we have
\begin{align*}
&|\int^T_0g(x(t))x'(t)dt|\\
&=|\int^{x(t_1)}_{x(0)}g(s)ds+ \int^{x(t_2)}_{x(t^{+}_1)}g(s)ds
 +\dots+\int^{x(T)}_{x(t^{+}_{n})}g(s)ds|\\
&=|\int^{x(T)}_{x(0)}g(s)ds-\sum^{n}_{k=1}\int^{x(t^{+}_k)}_{x(t_k)}g(s)ds|\\
&\leq\sum^{n}_{k=1}|\int^{x(t_k)+\lambda I_k(x(t_k),x'(t_k))}_{x(t_k)}g(s)ds|\\
&\leq\sum^{n}_{k=1}[|I_k(x(t_k),x'(t_k))|(\gamma_1+\gamma_2|x(t_k)|
+\gamma_3|I_k(x(t_k),x'(t_k))|)]\\
&\leq[\gamma_2(\sum^{n}_{k=1}a_k)+\gamma_3
(\sum^{n}_{k=1}a^2_k)]|x(t)|^2_{\infty}+
u_2|x(t)|_{\infty}+u_3,\\
\end{align*}% \label{e3.18}
where $u_2,u_3$ are positive constants. From \eqref{e3.14}, we have
\begin{equation}
\begin{split}
&|\int^T_0g(x(t))x'(t)dt|\\
&\leq[\gamma_2(\sum^{n}_{k=1}a_k)+\gamma_3
(\sum^{n}_{k=1}a^2_k)]M^2\int^T_0|x'(t)|^2dt+
u_4(\int^T_0|x'(t)|^2dt)^{1/2}+u_5,
\end{split} \label{e3.19}
\end{equation}
where $u_4, u_5$ are positive constants.
Applying Lemma \ref{lem2.5}, we obtain
\begin{align*}
&\int^T_0|g(x(t)-g(x(t-\tau(t)))|^2dt\\
&\leq\beta^2_3\int^T_0|x(t)-x(t-\tau(t))|^2dt\\
&\leq \beta^2_3[2|\tau(t)|_{\infty}^2\int^T_0|x'(t)|^2dt+
2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})|x(t)|_{\infty}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2} \\
&\quad +2|\tau(t)|_{\infty} B(|\tau(t)|_{\infty})
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}
+ C(|\tau(t)|_{\infty})|x(t)|^2_{\infty}\\
&\quad +D(|\tau(t)|_{\infty})|x(t)|_{\infty}
+E(|\tau(t)|_{\infty})].
\end{align*}% \label{e3.20}
Substituting \eqref{e3.14} into the above inequality, we have
\begin{align*}
&\int^T_0|g(x(t)-g(x(t-\tau(t)))|^2dt\\
&\leq\beta^2_3[2|\tau(t)|_{\infty}^2+2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})M\\
&\quad +C(|\tau(t)|_{\infty})M^2]
\int^T_0|x'(t)|^2dt+u_6\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}+u_7,
\end{align*} % \label{e3.21}
where $u_6,u_7$ are positive constants.
 Using the inequality
 \begin{equation}
(a+b)^{1/2}\leq a^{1/2}+b^{1/2} \ \ \ for\ \ \ a\geq0,b\geq0,
 \label{e3.22}
\end{equation}
we have
\begin{align*}
&\Big(\int^T_0|g(x(t))-g(x(t-\tau(t)))|^2dt\Big)^{1/2}\\
&\leq \beta_3[2|\tau(t)|_{\infty}^2
 +2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})M\\
&\quad +C(|\tau(t)|_{\infty})M^2]^{1/2}
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/2}+u^{1/2}_6
\Big(\int^T_0|x'(t)|^2dt\Big)^{1/4}+u^{1/2}_7.
\end{align*}%  \label{e3.23}
Substituting the above formula and \eqref{e3.19} in
 \eqref{e3.17}, we obtain
\begin{align*}
&\big\{\beta-[\gamma_2(\sum^{n}_{k=1}a_k)+\gamma_3
(\sum^{n}_{k=1}a^2_k)]M^2-\beta_3[2|\tau(t)|_{\infty}^2\\
&+ 2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})M
 +C(|\tau(t)|_{\infty})M^2]^{1/2}\big\}
\int^T_0|x'(t)|^2dt\\
&\leq u_8 (\int^T_0|x'(t)|^2dt)^{\frac{3}{4}}+
u_9(\int^T_0|x'(t)|^2dt)^{1/2}+u_{10},
\end{align*}%  \label{e3.24}
where $u_8,u_9,u_{10}$ are positive constants.
Then there is a constant $M_1>0$ such that
\begin{equation}
\int^T_0|x'(t)|^2dt\leq M_1.  \label{e3.25}
\end{equation}
 From \eqref{e3.14}, we have
$$
|x(t)|_{\infty}\leq d+M(\int^T_0|x'(t)|^2dt)^{1/2}\leq d+M
(M_1)^{1/2}. %  \label{e3.26}
$$
Then there is a constant $M_2>0$ such that
$|x(t)|_{\infty}\leq M_2$. %\label{e3.27}
Furthermore, integrating \eqref{e3.7} on $[0,T]$,
using Schwarz inequality, we obtain
\begin{align*}
\int^T_0|x''(t)|dt
&=\int^T_0|-f(t,x(t))-g(x(t-\tau(t)))+p(t)|dt\\
&\leq\int^T_0|f(t,x'(t))|dt+
\int^T_0|g(x(t-\tau(t)))|dt+
\int^T_0|p(t)|dt\\
&\leq \sigma\int^T_0|x'(t)|dt+g_{\delta}T
 +T|p(t)|_{\infty}\\
&\leq\sigma T^{1/2}(\int^T_0|x'(t)|^2dt)^{1/2}+g_{\delta}T+
 T|p(t)|_{\infty}\\
&\leq \sigma T^{1/2}(M_1)^{1/2}+g_{\delta}T+
 T|p(t)|_{\infty},
\end{align*} % \label{e3.28}
where $h_{\delta}=\max_{|x|\leq\delta}|g(x)|$. That is to say
that there is a constant $M_3>0$ such that
\begin{equation}
\int^T_0|x''(t)|dt\leq M_3. \label{e3.29}
\end{equation}
 From \eqref{e3.25}, it is easy to see that there are
$t_2\in[0,T]$ and $u_{11}>0$ such that
$|x'(t_2)|\leq u_{11}$, then for $t\in[0,T]$
\begin{equation}
|x'(t)|_{\infty} \leq|x'(t_2)|+\int^T_0|x''(t)|dt+
\sum^{n}_{k=1}b_k. \label{e3.30}
\end{equation}
Hence there is a constant $M_4>0$ such that
\begin{equation}
|x'(t)|_{\infty}\leq M_4.  \label{e3.31}
\end{equation}
It follows that there is a constant $B>\max\{M_2,M_4\}$
such that $\|x\|\leq B$,
Thus $\Omega_1$ is bounded.

 Let $\Omega_2=\{ x\in \ker L,QNx=0\}$. Suppose
$x\in \Omega_2$, then $x(t)=c\in R$
and  satisfies
\begin{equation}
QN(x,0)=(-\frac{2}{T^2}\int^T_0[f(t,0)+g(c)-p(t)]dt,0,\dots,0)=0.
 \label{e3.32}
\end{equation}
Then
\begin{equation}
\int^T_0[f(t,0)+g(c)-p(t)]dt=0. \label{e3.33}
\end{equation}
It follows from \eqref{e3.33} that there must be a
 $t_0\in[0,T]$ such that
\begin{equation}
g(c)=-f(t_0,0)+p(t_0). \label{e3.34}
\end{equation}
From \eqref{e3.34} and assumption (H3), (H4), we have
\begin{equation}
\beta_1+\beta_2|c|\leq|g(c)|\leq|f(t_0,0)|+|p(t_0)|
\leq\sigma\times0+|p(t)|_{\infty}.
\label{e3.35}
\end{equation}
Thus
\begin{equation}
|c|\leq\frac{||p(t)|_{\infty}-\beta_1|}{\beta_2} \label{e3.36}
\end{equation}
which implies $\Omega_2$ is bounded.
Let $\Omega$ be a non-empty open bounded subset of $X$ such that
$\Omega\supset\overline{\Omega_1}\cup\overline{\Omega_2}
\cup\overline{\Omega_3}$,
where
$\Omega_3=\{x\in X: |x| <||p(t)|_{\infty}-\beta_1|/\beta_2+1\}$.
By Lemmas \ref{lem2.2} and \ref{lem2.3}, we can see that $L$
is a Fredholm operator of index zero and $N$ is $L$-compact
on $\overline{\Omega}$. Then
by the above argument,
\begin{itemize}
\item[(i)]  $ Lx\neq\lambda Nx$ for all
$x\in\partial\Omega\cap D(L),\lambda\in(0,1)$;

\item[(ii)] $QNx\neq0$ for all $x\in\partial\Omega\cap \ker L$.

\end{itemize}
At last we prove that (iii) of Lemma \ref{lem2.1} is satisfied. We take
$H(x,\mu):\Omega\times[0,1]\to X$,
$$
H(x,\mu)=\mu x+\frac{2(1-\mu)}{T^2}\int^T_0[-f(t,x'(t))
+g(x(t-\tau(t))+p(t)]dt.
$$ %\label{e3.37}
From assumptions (H3) and (H4), we can easily obtain
$H(x,\mu)\neq0$, for all
$(x,\mu)\in\partial\Omega\cap \ker L\times[0,1]$, which results in
\begin{align*}
\deg\{JQNx,\Omega\cap \ker L,0\}
&=\deg\{H(x,0),\Omega\cap \ker L,0\}\\
&=\deg\{H(x,1),\Omega\cap \ker L,0\}\neq0,
\end{align*} % \label{e3.38}
where $J(x,0,\dots,0)=x$.
Therefore, by Lemma \ref{lem2.1}, Equation \eqref{e1.1} has at least one
$T$-periodic solution.
\end{proof}

\begin{theorem} \label{thm3.2}
Suppose {\rm (H1)-(H2), (H4)-(H6)} hold and the following two
conditions hold:
\begin{itemize}
\item[(H8)]  there is an constant $\sigma\geq0$ such that
\begin{gather*}
|f(t,x)|\leq\sigma|x|,\quad \forall (t,x)\in[0,T]\times \mathbb{R},
\\ % \label{e3.39}
xf(t,x)\leq-\beta|x|^2,\forall (t,x)\in[0,T]\times \mathbb{R},
% \label{e3.40}
\end{gather*}

\item[(H9)] $yJ_k(x,y)\geq 0$ and
 there are constants $b_k\geq0$ such that
$|J_k(x,y)|\leq b_k$.
\end{itemize}
Then  \eqref{e1.1} has at least one $T$-periodic solution
provided \eqref{e3.5} and \eqref{e3.6} hold.
\end{theorem}

 The proof of the above theorem is similar to that of
Theorem \ref{thm3.1}, so we omit it.

\subsection*{Example}
Consider the  equation
\begin{equation} \begin{gathered}
 x''(t)+\frac{1}{3}x'(t)+\frac{1}{15}x(t-\frac{1}{10}\cos t)
 =\sin t,\quad t\neq k,\\
 \Delta x(k)=\frac{\sin(k\pi/3)}{120}x(k)+
            \frac{x'(t_k)}{1+x^{'2}(t_k)},\\
 \Delta x'(k)=-\frac{2x^2(t_k)x'(t_k)}{1+x^{4}(t_k)x^{'2}(t_k)},
\end{gathered} \label{e3.41}
\end{equation}
where $t_k=k$, $f(t,x)=\frac{1}{3}x$,
$g(y)=\frac{1}{15}y$, $p(t)=\sin t$,
$\tau(t)=\frac{1}{10}\cos t$,
$I_k(x,y)= \frac{\sin\frac{k\pi}{3}}{120}x+\frac{y}{1+y^2}$,
$J_k(x,y)=-\frac{2x^2y}{1+x^{4}y^2}$,
it is easy to see that $|\tau(t)|_{\infty}=\frac{1}{10}$,
$T=2\pi,\{k\}\cap[0,2\pi]=\{1,2,3,4,5,6\}$,
$\sigma=\beta=\frac{1}{3}$,
$\beta_1=0$, $\beta_2=\beta_3=\frac{1}{15}$.
Since $|I_k(x,y)|\leq\frac{1}{120}|x|+\frac{1}{2},
|J_k(x,y)|\leq1$,$|\int^{x+I_k(x,y)}_{x}g(s)ds|\leq|I_k(x,y)|
(\frac{1}{15}|x|+\frac{1}{30}|I_k(x,y)|)$, then we take
$a_k=\frac{1}{120}$, $a'_k=\frac{1}{2}$,
$b'_k=1$ ($k=1,2,3,4,5,6$),
$\gamma_1=0$, $\gamma_2=1/15$, $\gamma_3=1/30$.
Thus assumption (H1)--(H7) hold and
\begin{gather*}
\sum^{6}_{k=1}a_k=\frac{1}{20}<1,\\ %\label{e3.42}
M=\frac{1}{1-\sum^{n}_{k=1}a_k}
(\frac{\sigma}{\beta_2T^{1/2}}+
T^{1/2})=
\frac{1}{1-\frac{1}{20}}(\frac{\frac{1}{3}}{\frac{1}{15}(2\pi)^{1/2}}
+(2\pi)^{1/2})<6. % \label{e3.43}}
\end{gather*}
Thus
\begin{align*}
&\Big[\gamma_2(\sum^{n}_{k=1}a_k)+\gamma_3
(\sum^{n}_{k=1}a^2_k)\Big]M^2
+\beta_3[2|\tau(t)|_{\infty}^2\\
&+ 2|\tau(t)|_{\infty} A(|\tau(t)|_{\infty})M
+C(|\tau(t)|_{\infty})M^2]^{1/2}<\beta.
\end{align*} % \label{e3.44}
By Theorem \ref{thm3.1}, Equation \eqref{e3.41} has at least one
$2\pi$-periodic solution.

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\end{document}