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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 140, 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 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/140\hfil Periodic solutions]
{Periodic solutions for $p$-Laplacian functional differential
 equations with two deviating arguments}

\author[C. Song, X. Gao \hfil EJDE-2011/140\hfilneg]
{Changxiu Song, Xuejun Gao} 

\address{Changxiu Song \newline
School of Applied Mathematics,
Guangdong University of Technology, Guangzhou 510006, China}
\email{scx168@sohu.com}

\address{Xuejun Gao \newline
School of Applied Mathematics,
Guangdong University of Technology, Guangzhou 510006, China}
\email{gaoxxj@163.com}


\thanks{Submitted March 5, 2011. Published October 27, 2011.}
\thanks{Supported by grants 10871052 and 109010600 
 NNSF of China,  and by grant \hfill\break\indent
 10151009001000032 from NSF of Guangdong}
\subjclass[2000]{34B15}
\keywords{$p$-Laplacian operator; periodic solutions;
 coincidence degree; \hfill\break\indent deviating arguments}

\begin{abstract}
 Using the theory of coincidence degree, we prove the existence
 of periodic solutions for the $p$-Laplacian functional differential
 equations with deviating arguments.
\end{abstract}

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

\section{Introduction}

 In recent years, the existence of periodic solutions for
the Duffing equation, Rayleigh equation and Li\'enard equation
has received a lot of attention; see \cite{c1,d1,h1,l1,l2,l3,m1}.
For example, Liu \cite{l1} studied periodic solutions for
the $p$-Laplacian Li\'enard equation with a deviating argument.
Using Mawhin's continuation theorem, some results on the existence
of periodic solution are obtained.  But the  $p$-Laplacian Li\'enard
equation with two deviating arguments has been studied far less
often.

 In this article, we study the existence of periodic solutions for
the following Li\'enard equation with two deviating arguments:
\begin{equation}
(\phi_p(x'(t)))'+f(x(t))x'(t)+g_1(t,x(t-\tau_1(t)))
+g_2(t,x(t-\tau_2(t)))=e(t),\label{e1.1}
\end{equation}
where $f, \tau_1, \tau_2, e\in C(\mathbb{R},\mathbb{R})$;
$g_1, g_2\in C(\mathbb{R}^2,\mathbb{R})$;
$\tau_1(t), \tau_2(t), g_1(t,x), g_2(t,x), e(t)$ are periodic
functions with period $T$;
$\phi_p(\cdot )$ is the $p$-Laplacian operator,
$1<p<\infty$. By using the theory of coincidence degree, we obtain
some results to guarantee the existence of periodic solutions.
Even for $p=2$, the results in this paper are also new.

In what follows, the $L^p-$norm in $L^p([0,T],\mathbb{R})$ is
defined by
\[
 \|x\|_p=(\int_0^T |x(t)|^p dt)^{1/p},
\]
and the $L^\infty-$norm in $L^\infty([0,T],\mathbb{R})$ is
$ \|x\|_\infty=\max_{t\in [0,T]}|x(t)|$.
Let the Sobolev space $W^{1,p}([0,T],\mathbb{R}]$ be denoted by $W$.


\begin{lemma}[\cite{z1}] \label{lem1.1}
 Suppose $u\in W$ and $u(0)=u(T)=0$. Then
\[
\|u\|_\infty \leq (T/2)^{1/q}\|u'\|_p.
\]
\end{lemma}

The following Mawhin's continuous theorem is useful in obtaining
the existence of $T$-periodic solutions of \eqref{e1.1}.

\begin{lemma}[\cite{g1}] \label{lem1.2}
Let $X$ and $Y$ be two Banach spaces. Suppose that $L:D(L)\in X\to
Y$ 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 all the following
conditions are satisfied:
\begin{itemize}
\item[(1)] $Lx\neq \lambda Nx$, for all  $x\in \partial\Omega\cap D(L)$,
and all $\lambda\in (0,1)$;

\item[(2)] $Nx \notin   \operatorname{Im}L$, for all
$ x\in \partial\Omega\cap \ker  L$;

\item[(3)] $\deg \{JQN,\Omega \cap \ker  L,0\}\neq 0$, where
$J:\operatorname{Im} Q\to \ker   L$ is an isomorphism,
\end{itemize}
then equation $Lx=Nx$ has a solution on $\Omega\cap D(L)$.
\end{lemma}

\section{Main results}

To use coincidence degree theory in the study
of $T$-periodic solutions for \eqref{e1.1}, we
rewrite \eqref{e1.1} in the  form
\begin{equation}
\begin{gathered}
 x'(t)=\phi_q(y(t))\\
y'(t)=-f(x(t))x'(t)-g_1(t,x(t-\tau_1(t)))-g_2(t,x(t-\tau_2(t)))+e(t).
\end{gathered}\label{e2.1}
\end{equation}

If $z(t)=(x(t),y(t))^T$ is a $T$-periodic solution of \eqref{e2.1},
then $x(t)$ must be a $T$-periodic solution of \eqref{e1.1}.
Thus, the problem of finding a $T$-periodic solution for \eqref{e1.1}
 reduces to finding one for \eqref{e2.1}.

We set the following notation: $T>0$ is a constant,
$C_T=\{x\in  C(\mathbb{R},\mathbb{R}): x(t+T)\equiv x(t)\}$
 with the norm
  $\|x\|_\infty=\max_{t\in [0,T]}|x(t)|, X=Z=\{z=(x,y)\in C(\mathbb{R},\mathbb{R}^2): z(t)\equiv
  z(x+T)\}$ with the norm
$\|z\|=\max\{\|x\|_\infty,\|y\|_\infty\}$. Clearly, $X$
and $Z$ are Banach spaces. Also let
$L:\operatorname{Dom} L\subset X\to Z$ be defined by
\[
 (Lz)(t)=z'(t)=\begin{pmatrix}x'(t)\\
  y'(t)\end{pmatrix},
\]
and $N:X\to Z$ defined by
\[
 (Nz)(t)=\begin{pmatrix}
  \phi_q(y(t))\\
  -f(x(t))x'(t)-g_1(t,x(t-\tau_1(t)))-g_2(t,x(t-\tau_2(t)))+e(t)
\end{pmatrix}
\]
It is easy to see that $\ker  L=\mathbb{R}^2$,
$ \operatorname{Im}L =\{z\in  Z:\int_0^Tz(s)ds=0\}$.
So $L$ is a Fredholm operator with index   zero.
Let $P:X\to \ker  L$ and $Q:Z\to \operatorname{Im} Q$ be defined by
\[
Pu=\frac{1}{T}\int_0^Tu(s)ds,\quad  u\in X;\quad
Qv=\frac{1}{T}\int_0^Tv(s)ds,\quad  v\in Z,
\]
  and let $K_p$ denote the inverse of
$L|_{\ker P\cap \operatorname{Dom}L}$.
Obviously, $\ker  L=\operatorname{Im}Q=\mathbb{R}^2$ and
\begin{equation}
  (K_pz)(t)=\int_0^tz(s)ds-\frac{1}{T}\int_0^T\int_0^tz(s)\,ds\,dt.
 \label{e2.2}
\end{equation}
From this equality, one can easily see that $N$ is $L$-compact on
$\overline{\Omega}$, where $\Omega$ is an open bounded subset of
$X$.

\begin{theorem} \label{thm2.1}
Suppose that there exist  constants $d>0$ $r_1\geq 0$ and
$r_2\geq 0$ such that
\begin{itemize}
\item[(H1)] $g_1(t,u)+g_2(t,v)-e(t)>0$  for all $t\in \mathbb{R}$,
$|\max\{u,v\}|>d$;

\item[(H2)]  $\lim_{x\to -\infty}\sup_{t\in
[0,T]}\frac{|g_1(t,x)|}{|x|^{p-1}}\leq r_1$;
$\lim_{x\to -\infty}\sup_{t\in [0,T]}\frac{|g_2(t,x))|}{|x|^{p-1}}\leq
r_2$.
\end{itemize}
Then \eqref{e1.1} has at least one $T$-periodic solution,
if $4(r_1+r_2)T (T/2)^{p/q}<1$.
\end{theorem}

\begin{proof}
 Consider the parametric equation
\begin{equation}
(Lz)(t)=\lambda (Nz)(t),\quad   \lambda\in (0,1). \label{e2.3}
\end{equation}
Let $z(t)=\begin{pmatrix}x(t)\\
  y(t)\end{pmatrix}$ be a possible $T$-periodic solution of
\eqref{e2.3} for some $\lambda\in (0,1)$. One can see $x=x(t)$ is a
$T$-periodic solution of the  equation
\begin{equation}
(\phi_p(x'(t)))'+\lambda^{p-1}f(x(t))x'(t)+\lambda^pg_1
(t,x(t-\tau_1(t)))+\lambda^pg_2(t,x(t-\tau_2(t)))
=\lambda^pe(t).\label{e2.4}
\end{equation}
Integrating both sides of \eqref{e2.4} over $[0,T]$, we have
\begin{equation}
\int_0^T [g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt=0,
\label{e2.5}
\end{equation}
which implies that there exists $\eta\in [0,T]$ such that
\[
g_1(\eta,x(\eta-\tau_1(\eta)))+g_2(\eta,x(\eta-\tau_2(\eta)))
-e(\eta)=0.
\]
From assumption (H1), we know that there exists
$\xi\in \mathbb{R}$ such that $|x(\xi)|\leq d$.

Let $\xi=kT+t_0$, where $t_0\in [0,T]$ and $k$ is an integer.
Let $\chi (t)=x(t+t_0)-x(t_0)$.
Then $\chi (0)=\chi (T)=0$ and $\chi\in
W^{1,p}([0,T],\mathbb{R})$. By Lemma \ref{lem1.1}, we have
\[
\|x\|_\infty\leq \|\chi\|\infty+d\leq
(\frac{T}{2})^{1/q}\|\chi'\|_p+d=(\frac{T}{2})^{1/q}\|x'\|_p+d.
\]
On the other hand, multiplying the two sides of \eqref{e2.4} by $x(t)$
and integrating them over $[0,T]$, we obtain
\[
-\|x'\|_p^p=-\lambda^p\int_o^Tx(t)[g_1(t,x(t-\tau_1(t)))
+g_2(t,x(t-\tau_2(t)))-e(t)]dt;
\]
i.e.,
\begin{equation}
 \|x'\|_p^p\leq \|x\|_\infty
\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt.
\label{e2.6}
\end{equation}
From assumption (H2),  there exists a constant $\rho>0$
such that
\begin{equation}
|g_1(t,x)|\leq r_1|x|^{p-1},\quad
|g_2(t,x)|\leq r_2|x|^{p-1},\quad
\forall  t\in \mathbb{R},\; x<-\rho. \label{e2.7}
\end{equation}
Let
\begin{gather*}
E_1=\{t\in[0,T]:\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}<-\rho\},\\
E_2=\{t\in[0,T]:\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}>\rho\},\\
E_3=\{t\in [0,T]:|\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}|\leq\rho\},\\
E_4=\{t\in [0,T]:-\rho\leq x(t-\tau_1(t))\leq \rho,
 -\rho\leq x(t-\tau_2(t))\leq \rho\},\\
E_5=\{t\in[0,T]:x(t-\tau_1(t))<- \rho,-\rho\leq x(t-\tau_2(t))
  \leq \rho\}, \\
E_6=\{t\in [0,T]:-\rho\leq x(t-\tau_1(t))\leq \rho,
x(t-\tau_2(t))<- \rho\}.
\end{gather*}
By \eqref{e2.4} it is easy to see that
\[
\int_0^T[g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt=0.
\]
Hence
\begin{align*}
&\int_{E_2}|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\
&=\int_{E_2}[g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt\\
&= -\Big(\int_{E_1}+\int_{E_3}\Big)[g_1(t,x(t-\tau_1(t)))
+g_2(t,x(t-\tau_2(t)))-e(t)]dt\\
&\leq \Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t)))
 +g_2(t,x(t-\tau_2(t)))-e(t)|dt.
\end{align*} % {e2.8}
From the above inequality and \eqref{e2.7} we obtain
\begin{align*}
&\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\
&\leq 2\Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t)))
 +g_2(t,x(t-\tau_2(t)))-e(t)|dt\\
&\leq 2\Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t)))|dt
+2\Big(\int_{E_1}+\int_{E_3}\Big)|g_2(t,x(t-\tau_2(t)))|dt\\
&\quad +2\int_0^T|e(t)|dt \\
&\leq 2r_1\int_{E_1}|x(t-\tau_1(t)|^{p-1}dt
 +2r_2\int_{E_1}|x(t-\tau_2(t)|^{p-1}dt\\
&\quad +2\int_{E_3}(|g_1(t,x(t-\tau_1(t)))|
 +|g_2(t,x(t-\tau_2(t)))|)dt+2\int_0^T|e(t)|dt\\
&\leq 2(r_1+r_2)T\|x\|_\infty^{p-1}
 +2\Big(\int_{E_4}+\int_{E_5}+\int_{E_6}\Big)
 \Big(|g_1(t,x(t-\tau_1(t)))|\\
&\quad +|g_2(t,x(t-\tau_2(t)))|\Big)dt +2\int_0^T|e(t)|dt\\
&\leq 2(r_1+r_2)T\|x\|_\infty^{p-1}+2T(g_{1\rho}+g_{2\rho})
 +2r_1\int_{E_5}|x(t-\tau_1(t)|^{p-1}dt\\
&\quad+2Tg_{2\rho}+2r_2\int_{E_6}|x(t-\tau_2(t)|^{p-1}dt
 +2Tg_{1\rho}+2\int_0^T|e(t)|dt\\
&\leq 4(r_1+r_2)T\|x\|_\infty^{p-1}
 +4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt,
\end{align*} % {e2.9}
where
\[
g_{1\rho}=\max_{t\in [0,T], |x|\leq
\rho}|g_1(t,x(t-\tau_1(t)))|, \quad
g_{2\rho}=\max_{t\in [0,T],|x|\leq \rho}|g_2(t,x(t-\tau_2(t)))|.
\]
From \eqref{e2.6} and the above inequality, we have
\begin{align*}
\|x'\|_p^p
&\leq \|x\|_\infty \int_0^T|g_1(t,x(t-\tau_1(t)))
 +g_2(t,x(t-\tau_2(t)))-e(t)|dt\\
&\leq \|x\|_\infty[4(r_1+r_2)T\|x\|_\infty^{p-1}
 +4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt]\\
&=4(r_1+r_2)T
\Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big)^p  +[4T(g_{1\rho}+g_{2\rho})\\
&\quad +2\int_0^T|e(t)|dt]\Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big).
\end{align*}

Case (1): $\|x'(t)\|=0$, from \eqref{e2.6} we see $\|x\|_\infty\leq
d$.

Case (2): $\|x'(t)\|>0$, then we know that
\[
[(\frac{T}{2})^{1/q}\|x'\|_p+d]^p
=(\frac{T}{2})^{p/q}\|x'\|^p_p[1+(\frac{T}{2})^{-1/q}
 \frac{d}{\|x'(t)\|_p}]^p.
\]
From mathematical analysis, there is a constant
$\delta>0$ such that
\begin{equation}
(1+x)^p<1+(1+p)x, \quad \forall  x\in [0,\delta].\label{e2.10}
\end{equation}
If $(\frac{T}{2})^{-1/q}\frac{d}{\|x'(t)\|_p}>\delta$, then
we have  $\|x'\|_p<(\frac{T}{2})^{-1/q}\frac{d}{\delta}$.

If $(\frac{T}{2})^{-1/q}\frac{d}{\|x'(t)\|_p}\leq\delta$, by
\eqref{e2.10} we know that
\begin{equation}
[(\frac{T}{2})^{1/q}\|x'\|_p+d]^p\leq
(\frac{T}{2})^{p/q}\|x'(t)\|_p^p+(p+1)(\frac{T}{2})
^{p-1)/q}d\|x'(t)\|^{p-1}_p.\label{e2.11}
\end{equation}
By \eqref{e2.11}, we obtain
\begin{align*}
\|x'\|_p^p
&\leq 4(r_1+r_2)T
(\frac{T}{2})^{p/q}\|x'(t)\|_p^p+(p+1)(\frac{T}{2})^{p-1)/q}
d\|x'(t)\|^{p-1}_p\\
&\quad +(4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt)
\Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big).
\end{align*}
As $p>1$, $4(r_1+r_2)T (\frac{T}{2})^{p/q}<1$, there exists a
constant $R_2>0$ such that $\|x'\|_p\leq R_2$.

Let $R_1=\max\{(\frac{T}{2})^{-1/q}\frac{d}{\delta},R_2\}$.
Then we have
$\|x\|_\infty\leq (\frac{T}{2})^{1/q}R_1:=R_0$.
By the second equation of \eqref{e2.1} we obtain
\[
y'(t)=-f(x(t))x'(t)-\lambda g_1(t,x(t-\tau_1(t)))-\lambda
g_2(t,x(t-\tau_2(t)))+\lambda e(t).
\]
Hence
\begin{align*}
\int_0^T |y'(t)|dt
&\leq f_{R_0}\int_0^T|x'(t)|dt+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt\\
&\leq f_{R_0}T^{1/q}\|x'\|_p+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt\\
&\leq f_{R_0}T^{1/q}R_1+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt:=R_3,
\end{align*}
where
\[
f_{R_0}=\max_{|s|\leq R_0}|f(s)|,\quad
g_{1R_0}=\max_{t\in [0,T],s\leq R_0}|g_1(t,s)|,\quad
g_{2R_0}=\max_{t\in [0,T],s\leq R_0}|g_2(t,s)|.
\]
By the first equation of \eqref{e2.1} we have
$\int_0^T\phi_q(y(t))dt=0$,
which implies there exists a constant $t_1\in [0,T]$ such that
$y(t_1)=0$. So
\[
|y(t)|=|\int_{t_1}^ty'(s)ds|\leq \int_0^T|y'(s)|ds\leq R_3,
\]
and $\|y\|_\infty\leq R_3$.

Let $R_4> \max\{R_0,R_3\}$, $\Omega=\{z\in Z: \|z\|<R_4\}$, then
$Lz\neq \lambda NZ$, for all
$z\in \operatorname{Dom} L\cap \partial\Omega$,  $\lambda\in(0,1)$.
Since
\[
QNz=\frac{1}{T}\int_0^T\begin{pmatrix}
  \phi_q(y(t))\\
  -f(x(t))x'(t)-g_1(t,x(t-\tau_1(t)))
-g_2(t,x(t-\tau_2(t)))+e(t)\end{pmatrix},
\]
for any $z\in \ker  L\cap \partial\Omega$, if $QNz=0$, we
obtain $y=0$, $|x|=R_4>d$. But when $|x|=R_4$, we know
that  $-g_1(t,x)-g_2(t,x)+e(t)<0$, which yields a contradiction.
So conditions   (1) and (2) of Lemma \ref{lem1.2} is satisfied.

Define the isomorphism $J:\operatorname{Im} Q\to \ker  L$ as follows:
\[
J(x,y)^T=(-y,x)^T.
\]
Let $H(\mu, z)=\mu x+(1-\mu)JQNz, (\mu,z)\in [0,1]\times
  \Omega$, then we have
\[
  H(\mu,z)=\begin{pmatrix}
\mu x+(1-\mu)\frac{1}{T}\int_0^T[g_1(t,x)+g_2(t,x)-e(t)]dt\\
  \mu y+(1-\mu)\phi_q (y)\end{pmatrix},
\]
where $ (\mu,z)\in  [0,1]\times (\ker L\cap \partial\Omega)$.
It is obvious that $ H(\mu,z)=\mu x+(1-\mu) JQNz\neq 0$
for $(\mu,z)\in  [0,1]\times (\ker  L\cap \partial\Omega)$. Hence
\[
 \deg \{JQN,\Omega\cap \ker  L,0\}=\deg \{I,\Omega\cap \ker  L,0\}
=1\neq   0.
\]
So the condition (3) of Lemma \ref{lem1.2} is satisfied. By applying
  Lemma \ref{lem1.2}, we conclude that equation $Lz=Nz$ has a solution
  $z(t)=(x(t),y(t))^T$; i.e., \eqref{e1.1} has a $T$-periodic solution
  $x(t)$.
\end{proof}

\begin{theorem} \label{thm2.2}
Suppose that there exist constants $d>0$, $r_1\geq 0$,
and $r_2\geq 0$ such that {\rm (H1)} holds and
\begin{itemize}
\item[(H2*)] $\lim_{x\to +\infty}\sup_{t\in
[0,T]}\frac{|g_1(t,x)|}{|x|^{p-1}}\leq r_1$;
$\lim_{x\to +\infty}\sup_{t\in [0,T]}\frac{|g_2(t,x))|}{|x|^{p-1}}\leq
r_2$.
\end{itemize}
Then \eqref{e1.1} has at least one $T$-periodic solution,
if $4(r_1+r_2)T (\frac{T}{2})^{p/q}<1$.
\end{theorem}

\begin{theorem} \label{thm2.3}
Suppose that $p> 2$ and there exist constants $d>0$,
$b_1\geq 0$, $b_2\geq 0$ such that {\rm (H1)} holds and
\begin{itemize}

\item[(H3)] $|g_i(t,u)-g_i(t,v)|\leq b_i|u-v|$ for all
$t, u, v\in \mathbb{R}$, $i=1,2$.
\end{itemize}
Then \eqref{e1.1} has at least one $T$-periodic solution.
\end{theorem}

\begin{proof}
 By the proof of Theorem \ref{thm2.1}, we have
\[
\|x\|_\infty\leq \|\chi\|_\infty+d\leq
(\frac{T}{2})^{1/q}\|\chi'\|_p+d=(\frac{T}{2})^{1/q}\|x'\|_p+d,
\]
and
\[
 \|x'\|_p^p\leq \|x\|_\infty
\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt.
\]
From assumption (H3), we have
\begin{align*}
&\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\
&\leq \int_0^T|g_1(t,x(t-\tau_1(t)))-g_1(t,0)|dt
 +\int_0^T|g_1(t,0)|dt\\
&\quad +|g_2(t,x(t-\tau_2(t)))-g_1(t,0)|dt+\int_0^T|g_2(t,0)|dt+\int_0^T|e(t)|dt\\
&\leq b_1\int_0^T|x(t-\tau_1(t))|dt+\int_0^T|g_1(t,0)|dt+b_2\int_0^T|x(t-\tau_2(t))|dt\\
&\quad +\int_0^T|g_2(t,0)|dt+\int_0^T|e(t)|dt\\
&\leq  T(b_1+b_2)\|x\|_\infty+Tb
\end{align*}
where $b=\max\{|g_1(t,0)|+|g_2(t,0)|+|e(t)|\}$.
Thus,
\begin{align*}
 \|x'\|_p^p
&\leq \|x\|_\infty[T(b_1+b_2)\|x\|_\infty+Tb]\\
& \leq T(b_1+b_2)[(\frac{T}{2})^{1/q}\|x'\|_p+d]^2
 +Tb(\frac{T}{2})^{1/q}\|x'\|_p+Tbd.
\end{align*}
As $p>2$, there exists a constant $R_2>0$ such that
$\|x'\|_p\leq R_2$.
The rest of  the proof is same to Theorem \ref{thm2.1} and is omitted.
\end{proof}

\begin{corollary} \label{coro2.1}
Suppose that $p=2$ and  conditions {\rm (H1), (H3)} hold.
Then \eqref{e1.1} has at least one $T$-periodic solution,
if $T(b_1+b_2)(\frac{T}{2})^\frac{2}{q}<1$.
\end{corollary}

\begin{remark} \label{rmk2.1} \rm
If  condition (H1) is replaced by
\begin{itemize}
\item[(H1*)] $g_1(t,u)+g_2(t,v)-e(t)<0$  for all
$t\in \mathbb{R}$, $|\max\{u,v\}|>d$,
\end{itemize}
then the results in this article still hold.
\end{remark}

\begin{example} \label{exmp2.1} \rm
Consider the equation
\begin{equation}
(\phi_3(x'(t)))'+e^{x(t)}x'(t)+g_1(t,x(t-\sin t))+g_2(t,x(t-\cos
t))=\frac{1}{\pi}\sin t, \label{e2.12}
\end{equation}
where $p=3$, $T=2\pi$,  $\tau_1(t)=\sin t$, $\tau_2(t)=\cos t$,
\begin{gather*}
g_1(t,x)=\begin{cases}
e^{\sin^2t}x^3+\frac{1}{\pi}\sin t, & x\geq 0,\\[3PT]
\frac{x^2}{18e\pi^3}e^{\sin^2t}+\frac{1}{\pi}\sin t,& x<0,
\end{cases}
\\
g_2(t,x)=\begin{cases}
e^{\cos^2t}x^3, & x\geq 0,\\[3PT]
\frac{x^2}{18e\pi^3}e^{\cos^2t},& x<0.
\end{cases}
\end{gather*}
By \eqref{e2.12}, we can get $d=1/10$
(Actually, $d$ can be an arbitrarily small positive),
$r_1=r_2=1/(18\pi^3)$, $4(r_1+r_2)T (T/2)^{p/q}<1$ and
check that  (H1)--(H2) hold.
Thus, according to Theorem \ref{thm2.1},  equation \eqref{e2.12} has at
least one $2\pi$-periodic solution.
\end{example}

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\end{document}