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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 99, pp. 1--10.\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-2006/99\hfil Stable time periodic solutions]
{Stable time periodic solutions for damped  Sine-Gordon equations}

\author[J. Zhu, Z. Li, Y. Liu \hfil EJDE-2006/99\hfilneg]
{Jianmin Zhu, Zhixiang Li, Yicheng Liu}  

\address{Jianmin Zhu \newline
Department of  Mathematics and System Science,
College of Science, National University of Defense Technology,
Changsha, 410073, China}
\email{zhujmmath@hotmail.com}

\address{Zhixiang Li \newline
Department of  Mathematics and System Science, College of Science,
National University of Defense Technology, Changsha, 410073,
China} 
\email{zhxli02@yahoo.com.cn}

\address{Yicheng Liu \newline
Department of  Mathematics and System Science, College of Science,
National University of Defense Technology, Changsha, 410073,
China} 
\email{liuyc2001@hotmail.com}


\date{}
\thanks{Submitted May 2, 2006. Published August 31, 2006.}
\subjclass[2000]{34K13, 35K50}
\keywords{Schaefer's fixed-point theorem;  time periodic solutions;
\hfill\break\indent
  Sine-Gordon equations; uniformly asymptotic stability}

\begin{abstract}
 We investigate existence and stability of time periodic solutions
 for damped Sine-Gordon equations with delay under reasonable
 assumptions. The key-step is constructing suitable Lyapunov
 functionals and establishing the priori bound for all possible
 periodic solutions.  We also use the Schaefer fixed-point
 theorem.
\end{abstract}

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

\section{Introduction}

In this paper, we are interested in obtaining existence and
stability of time periodic solutions for the following damped
Sine-Gordon equations with delay
\begin{gather*}
\begin{aligned}
 \frac{\partial^2}{\partial t^2}u(t,x)
&=\frac{\partial^2 }{\partial x^2}u(t,x)
-\delta\frac{\partial }{\partial t}u(t,x)
-\int_{t-\tau}^te^{-\alpha(t-s)}\frac{\partial }{\partial s}u(s,x)ds\\
&\quad +\beta\sin(u(t,x))+g(t,x),\quad 0< x< 1,\; t>0,
\end{aligned}\\
u(t,0)=u(t,1)=0, \quad t>0,
\end{gather*}
where $0<\tau< +\infty$, $\beta$ and $\delta$ are two constants.
The Sine-Gordon equation appears in a number of physical
applications, including the propagation of junction between two
superconductors, the motion of rigid pendula attached to a
stretched wire, and the dislocations in crystals.


The existence of  periodic solutions of nonlinear
partial differential equations with delay has been considered in several
works, see for example \cite{b1,f1,h1,h2,h5,j1,j2,j3,y1}
 and references listed therein.
Most of these results are established
by applying semigroup theory \cite{h2,h5}, Leray-Schauder continuation
theorem \cite{j2}, coincidence degree theory \cite{j3} and so on.
 Hino,  Naito et al. \cite{h5}
investigated the existence of (almost) periodic solutions for
damped wave equations by using the key assumption that there
exists a bounded solution. Burton and Zhang \cite{b1} obtained the
time periodic solution to some evolution equations with infinity
delay by means of Granas's fixed point theorem \cite{s2}. The
theory of partial differential equations with delay(s) has seen
considerable development, see the monographs of  Wu \cite{j1} and
Hale \cite{h1,h3}, where numerous properties of their solutions
are studied. For the more special work, we can read the references
therein \cite{y1,f1,r1,h4}.

The paper is organized as follows. In section 2 we will establish
existence of time periodic solution for abstract differential
equation in a certain Banach space. Some existence results for
time periodic solutions are given in section 3. At last section,
the uniformly asymptotic stable time periodic solution will be
shown.

 For the remainder of the introduction, we state the following
 lemma which will be used in the sequel.

\begin{lemma}[Schaefer \cite{s1}] \label{lem1.1}
 Let $(X,\|\cdot\|)$ be a normed linear space, $H$ a
continuous mapping of $X$ into $X$ which is compact on each
bounded subset $D$ of $X$. Then either
\begin{itemize}
\item[(i)] $x=\lambda Hx$ has a solution in $X$ for $\lambda=1$, or

\item[(ii)] the set of all such solutions, $0<\lambda<1$, is
  unbounded.
\end{itemize}
\end{lemma}

 \section{Time periodic solutions for abstract equations}

 In this section, we  obtain the existence results of
 periodic solutions for the  wave equation
\begin{equation} \label{e2.1}
\begin{gathered}
 \frac{\partial^2}{\partial t^2}u(t,x)+\delta\frac{\partial }{\partial
t}u(t,x)=\frac{\partial^2}{\partial x^2}u(t,x)+\lambda C(t,u),
\quad 0< x< 1,\; t>0,\\
u(t,0)=u(t,1)=0,\quad  t>0,
\end{gathered}
\end{equation}
where $C(t+T,u)=C(t,u)$ for some $T>0$, $\lambda\in (0,1)$. Let
$v(t,x)=\frac{\partial }{\partial t}u(t,x)$,
  we define a linear (unbounded) operator $A$ by
$$ A \begin{pmatrix}
 u\\
v \end{pmatrix}=\begin{pmatrix}
 v\\
\frac{\partial^2 u}{\partial x^2}-\delta v \end{pmatrix},\quad
\begin{pmatrix}
 u\\
v \end{pmatrix}
\in D(A)=(H^2\cap H_0^1)\times H_0^1,
$$
where $H_0^1=W_0^{1,2}(0,1)$, $H^2=W^{2,2}(0,1)$.
Set
$$
\tilde{C}(t,w)=\begin{pmatrix}
 0\\
C(t,u) \end{pmatrix}\quad\text{for }
w=\begin{pmatrix}
 u\\
v \end{pmatrix}\in L^2(0,1; R)\times L^2(0,1; R).
$$
We rewrite
\eqref{e2.1} as an abstract equation in $L^2(0,1; R)\times L^2(0,1; R)$.
That is,
\begin{equation} \label{e2.2}
 w'(t)=Aw(t)+\lambda \tilde{C}(t,w) .
\end{equation}
Let
\begin{equation} \label{e2.3}
\begin{gathered}
 X=\{w\in C(R,H_0^1\times H^0)|w(t+T)=w(t)\},\\
\|w\|_X=\sup\{(\int_0^1(u_x^2+v^2)dx)^{1/2}| 0\leq t\leq T\}.
\end{gathered}
\end{equation}
Then $(X,\|\cdot\|_X)$ is a Banach space.

\begin{theorem} \label{thm2.1}
Suppose the following conditions hold:
\begin{enumerate}
\item There exists $D>0$ such that if $w(t)$ is a
$T$-periodic solution of \eqref{e2.2} for some $\lambda\in (0,1)$, then
$\|w\|_X<D$;

\item  $C:[0,T]\times X\to H^0$ is continuous and
takes bounded sets into bounded sets. Then \eqref{e2.1} has a
$T$-periodic solution for $\lambda=1$.
\end{enumerate}
\end{theorem}

\begin{proof}
By the definition of the operator $A$, we see that $A$ generates
the strongly continuous semigroup $\{T(t)\}_{t\geq 0}$ and there
exist $c>0$, $\alpha>0$ such that $\alpha\in\rho(A)$ and
$\|T(t)\|\leq ce^{\alpha t}$ for $t\geq 0$. Take
$S(t)=T(t)e^{-\alpha t}$ for $t\geq 0$, then $\|S(t)\|\leq c$ for
$t\geq 0$. Therefore, $\{S(t)\}_{t\geq 0}$ is bounded strongly
continuous semigroup with the generator $A-\alpha I$. For the
constant $T>0$, from the equality
$\sigma(S(T))=\overline{e^{T\sigma(A-\alpha I)}}$
\cite[Theorem 1.8]{h5}, we conclude that $1\notin \sigma(S(T))$.
Thus there exists a positive number $N$ such that $\|(I-S(T))^{-1}\|\leq N$.
We will complete the proof with using the following two lemmas.

\begin{lemma} \label{lem2.2}
 If $L: X\to X$ is defined by
$$
Lw(t)=\int_{t-T}^tS(t-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))+\alpha w(s)]ds
$$
and if $w$ is a fixed point of  $L$ then $w$ satisfies \eqref{e2.2}.
\end{lemma}

\begin{proof}  For any $w\in X$, since
\begin{align*}
Lw(t+T)&=\int_{t}^{t+T}S(t+T-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))+\alpha w(s)]ds \\
&= \int_{t-T}^{t}S(t-r)[I-S(T)]^{-1}[\lambda\tilde{C}(r+T,w(r+T))+\alpha
w(r+T)]dr \\
&= \int_{t-T}^tS(t-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))+\alpha
w(s)]ds=Lw(t),
\end{align*}
we see that $L$ is well-defined. On the other hand, if $w(t)$ is a
fixed point of $L$, by the well-known equality
$\frac{d}{dt}S(t)w=(A-\alpha I)S(t)w$ for any $w\in X$, we have
\begin{align*}
\frac{d}{dt}w(t)
&=\frac{d}{dt}Lw(t)=[I-S(T)]^{-1}[\lambda\tilde{C}(t,w(t))+\alpha w(t)] \\
&\quad -S(T)[I-S(T)]^{-1}[\lambda\tilde{C}(t-T,w(t-T))+\alpha
w(t-T)] \\
&\quad +(A-\alpha
I)\int_{t-T}^tS(t-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))+\alpha
w(s)]ds \\
&= Aw(t)+\lambda\tilde{C}(t,w(t)).
\end{align*}
This completes the proof.
\end{proof}


\begin{lemma} \label{lem2.3}
The operator $L$ is a compact operator in $X$.
\end{lemma}

\begin{proof}
 Let $E(w(s))=\lambda\tilde{C}(s,w(s))+\alpha
w(s)$, by assumption (2) of Theorem \ref{thm2.1}, $E(\cdot)$ is continuous.
Then, for any $w_1,w_2\in X $,  we obtain
$$
Lw_1(t)-Lw_2(t)=\int_{t-T}^tS(t-s)[I-S(T)]^{-1}[E(w_1(s))-E(w_2(s))]ds.
$$
Thus
$$
\|Lw_1(t)-Lw_2(t)\|_{H^1_0\times H^0}\leq  cTN
\sup_{s\in[0, T]}\|E(w_1(s))-E(w_2(s))\|_{H^1_0\times H^0}.
$$
   Then the  continuity of $E(\cdot)$ implies that $L$ is continuous.

Next, we show $L$ maps the bounded sets into compact sets. Let $B$
be any bounded set in $X$ and
$M=\sup_{(t,w)\in [0,T]\times B}|C(t,u(t))|_{H^0}$.
Since $C$ takes $[0,T]\times B$ into a
bounded set, we see that
$$
\|Lw\|_X\leq  cTN[\sup_{(t,w)\in [0,T]\times
B}|C(t,u(t))|_{H^0}+\alpha K] \leq cTN[M+\alpha K],
$$
 where $K$ is the boundedness  of $B$. This implies that $L(B)$
is uniformly bounded.

On the other hand, for any $w\in B$, and $0\leq t_1<t_2\leq T$,
we have
\begin{align*}
&Lw(t_2)-Lw(t_1)\\
&=\int_{t_2-T}^{t_2}S(t_2-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))
   +\alpha w(s)]ds \\
&\quad -\int_{t_1-T}^{t_1}S(t_1-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))
   +\alpha w(s)]ds \\
&=\int_{t_1}^{t_2}S(t_2-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))
   +\alpha w(s)]ds \\
&\quad -\int_{t_1-T}^{t_2-T}S(t_1-s)[I-S(T)]^{-1}[\lambda\tilde{C}(s,w(s))
   +\alpha w(s)]ds\\
&\quad +[S(t_2-t_1)-I]\int_{t_2-T}^{t_1}S(t_1-s)[I-S(T)]^{-1}
[\lambda\tilde{C}(s,w(s))+\alpha w(s)]ds.
\end{align*}
Since $\lim_{t_2\to t_1}S(t_2-t_1)w=w$ for
$w\in X$, we conclude that the right-hand side of above equality
tends to zero as $t_2\to t_1$. This implies $L(B)$ is
 equicontinuous.
By Ascoli's lemma, $L(B)$ is a compact set. This completes the
proof.
\end{proof}

 From Assumption (1) in Theorem \ref{thm2.1}, we see that
the set $\{x\in X: x=\lambda Lx$, for some $\lambda\in(0,1)\}$ is
bounded. Then from Lemma \ref{lem2.2}, Lemma \ref{lem2.3} and
Lemma \ref{lem1.1}, we see that equation \eqref{e2.2} has a
$T$-periodic solution $w(t,x)$ for $\lambda=1$. That is,
\eqref{e2.1} has a $T$-periodic solution $u(t,x)$ for $\lambda=1$.
Then the proof of Theorem \ref{thm2.1} is complete.
\end{proof}


\section{Time periodic solutions for Sine-Gordon equations}

 In this section, we consider the existence of periodic solutions for
the  Sine-Gordon  equations with delay:
\begin{equation} \label{e3.1}
\begin{gathered}
\begin{aligned}
 \frac{\partial^2}{\partial t^2}u(t,x)
&=\frac{\partial^2 }{\partial x^2}u(t,x)
 -\delta\frac{\partial }{\partial t}u(t,x)
 -\int_{t-\tau}^te^{-\alpha(t-s)}\frac{\partial }{\partial s}u(s,x)ds\\
&\quad +\beta\sin(u(t,x))+g(t,x),\quad 0< x< 1,\; t>0,
\end{aligned}\\
u(t,0)=u(t,1)=0, \quad t>0,
\end{gathered}
\end{equation}
where $g$ is continuous.
We assume that there exist positive constants $K$ and $T$
satisfying
\begin{itemize}
\item[(A1)] $a=\int_{{-\tau}}^{^0}e^{\alpha s}ds<\delta$;

\item[(A2)] $\delta<\frac{\pi^2}{3} $ and
$0<|\beta|<\sqrt{(2\delta-\frac{(2+\delta)\delta}{\pi^2}-a)(2\delta
-\frac{\delta(5\delta+a)}{\pi^2})}$;

\item[(A3)] $g\neq 0, g(t+T,x)=g(t,x) $ and $|g(t,x)|\leq K$.
\end{itemize}

 First, we consider the homotopy equation of
 \eqref{e3.1},
\begin{equation} \label{e3.2}
 u_{tt}+\delta u_t=u_{xx}+\lambda[\beta\sin(u)+g(t,x)
-\int_{t-\tau}^te^{-\alpha(t-s)}u_s(s,x)ds],\quad 0<\lambda<1.
\end{equation}

\begin{lemma} \label{lem3.1}
Suppose that (A1), (A2) and (A3) hold. Then there is a constant $C_1>0$
such that all $T$-periodic solution $u(t)$ of \eqref{e3.2} satisfy
 $$
\sup_{0\leq t\leq T}\int_0^1{(u^2_x+u^2_t)dx}\leq C_1.
$$
\end{lemma}

\begin{proof}
Let $u(t,x)$ be a $T$-periodic solution of \eqref{e3.2} and define
 $$
V(t)=\int_0^1[v^2+2kuv+u_x^2]dx,
$$
where $v=u_t$, $k=\frac{\delta}{\pi^2}$. Since
$\int_0^1\pi^2u^2dx\leq\int_0^1u_x^2dx$ for $u\in H^1_0$, we obtain
\begin{equation} \label{e3.3}
(1-k)\int_0^1[u_{x}^2+v^2]dx \leq V(t)\leq
(1+k)\int_0^1[u_{x}^2+v^2]dx.
\end{equation}
Noting the assumptions (A2) and (A3), we see that
$|\beta\sin(u)+g(t,x)|\leq |\beta\|u|+K $. On the other hand, since
$$
0<|\beta|<\sqrt{(2\delta-\frac{(2+\delta)\delta}{\pi^2}-a)
(2\delta-\frac{\delta(5\delta+a)}{\pi^2})},
$$
then we choose positive constant $\varepsilon_1$ such that
$$
\frac{|\beta|}{2\delta-\frac{(2+\delta)\delta}{\pi^2}-a}
<\varepsilon_1<\frac{2\delta-\frac{\delta(5\delta+a)}{\pi^2}}{|\beta|}.
$$
Thus we have
$$
\frac{(2+\delta)\delta}{\pi^2}+a+|\beta|\varepsilon_1^{-1}
<2\delta
$$
and (since $|\beta|<2\delta$)
$$
\frac{\delta}{\pi^2}(\delta+2|\beta|+a)+|\beta|\varepsilon_1
<\frac{\delta(5\delta+a)}{\pi^2}+|\beta|\varepsilon_1<2\delta.
$$
Therefore, there exist constants $\varepsilon_2$ and
$\varepsilon_3$ such that
$$
\frac{(2+\delta)\delta}{\pi^2}+a+|\beta|\varepsilon_1^{-1}+\varepsilon_2
<2\delta
$$
and
$$
\frac{\delta}{\pi^2}(\delta+2|\beta|+a+\varepsilon_3)+|\beta|\varepsilon_1
<2\delta.
$$
Hence, by choosing $k=\delta/\pi^2$, we have
\begin{gather}
(2\delta-2k)-(k\delta+|\beta|\varepsilon_1^{-1}+\varepsilon_2+a)>0,
\label{e3.4}\\
 2k-\frac{1}{\pi^2}[k(\delta+2|\beta|+\varepsilon_3+a)+|\beta|\varepsilon_1]>0.
\label{e3.5}
\end{gather}
So that
\begin{align*}
&V'(t)\\
&=2\int_0^1[vv_t+u_xu_{xt}+k(u_tv+uv_t)]dx \quad
 (\mbox{since }\int_0^1u_xu_{xt}dx=-\int_0^1vu_{xx}dx) \\
&=2\int_0^1[vu_{tt}-vu_{xx}+k(u_{tt}u+v^2)]dx \\
&=2\int_0^1[-\delta v^2+\lambda v[\beta\sin(u)+g(t,x)-\int_{t-\tau}^te^{-\alpha(t-s)}v(s,x)ds]]dx \\
&\quad+2k\int_0^1[v^2+u_{xx}u-\delta vu+\lambda u[\beta\sin(u)+g(t,x)
  -\int_{t-\tau}^te^{-\alpha(t-s)}v(s,x)ds]]dx \\
&\leq 2\int_0^1[-(\delta-k)v^2-ku_{x}^2-\delta kuv]dx \\
&\quad +2\int_0^1(ku+v)(|\beta\|u|+K+\int_{t-\tau}^te^{-\alpha(t-s)}|v(s,x)|ds)dx \\
&\leq \int_0^1[-(2\delta-2k)v^2-2ku_{x}^2+k\delta(u^2+v^2)]dx \\
&\quad +\int_0^1(2|\beta| ku^2+\varepsilon_3ku^2 +k\frac{K^2}{\varepsilon_3}
 +ak u^2 +k\int_{t-\tau}^te^{-\alpha(t-s)}v^2(s,x)ds)dx \\
&\quad +\int_0^1(|\beta|\varepsilon_1u^2+\frac{|\beta|}{\varepsilon_1}v^2
+\varepsilon_2v^2+\frac{K^2}{\varepsilon_2}+a v^2
 +\int_{t-\tau}^te^{-\alpha(t-s)}v^2(s,x)ds)dx \\
&=\int_0^1[-(2\delta-2k)+k\delta+|\beta|\varepsilon_1^{-1}
 +\varepsilon_2+a]v^2dx \\
&\quad +\int_0^1-2ku_x^2+[k(\delta+2|\beta|+\varepsilon_3+a)
  +|\beta|\varepsilon_1]u^2dx \\
&\quad +\frac{K^2}{\varepsilon_2}+k\frac{K^2}{\varepsilon_3}
  +(1+k)\int_0^1\int_{t-\tau}^te^{-\alpha(t-s)}v^2(s,x)dsdx \\
& \quad (\mbox{since } \int_0^1\pi^2u^2dx\leq\int_0^1u_x^2dx) \\
&\leq -c_1\int_0^1v^2+u_x^2dx+c_2\int_0^1\int_{t-\tau}^t
e^{-\alpha(t-s)}v^2(s,x)dsdx+c_3,
\end{align*}
where $c_2=1+k$, $c_3=\frac{K^2}{\varepsilon_2}+k\frac{K^2}{\varepsilon_3}$
and (by \eqref{e3.4} and \eqref{e3.5})
\begin{align*}
c_1=\min\big\{&(2\delta-2k)-(k\delta+|\beta|\varepsilon_1^{-1}+\varepsilon_2+a), \\
&2k-\frac{1}{\pi^2}[k(\delta+2|\beta|+\varepsilon_3+a)+|\beta|\varepsilon_1]
\big\}>0.
\end{align*}
By \eqref{e3.3}, we see that there exist three positive constants
$r_1,r_2$ and $r_3$ such that
\begin{equation} \label{e3.6}
V'(t)\leq  -r_1V(t)
+r_2\int_0^1\int_{t-\tau}^te^{-\alpha(t-s)}v^2(s,x)ds\,dx+r_3.
\end{equation}
Since $V(t)$ and $v$ are $T$-periodic, we have
\begin{equation} \label{e3.7}
\begin{aligned}
\int_0^TV(t)dt
&\leq \frac{r_2}{r_1}\int_0^T\int_0^1\int_{t-\tau}^t
   e^{-\alpha(t-s)}v^2(s,x)dsdxdt+r_3T/r_1 \\
&=\frac{r_2a}{r_1}\int_0^T\int_0^1v^2(t,x)dx\,dt+r_3T/r_1.
\end{aligned}
\end{equation}
To obtain the boundedness of
$\int_{_0}^{^T}\int_{_0}^{^1}v^2dxdt$, we introduce the
function
$$
V_1(t)=\int_0^1[v^2+u_x^2+2\lambda\cos(u)]dx.
$$
Then
\begin{equation}
\begin{aligned}
V'_1(t)&=2\int_0^1\Big[-\delta v^2+\lambda v[\beta\sin(u)+g(t,x)\\
&\quad -\int_{t-\tau}^te^{-\alpha(t-s)}v(s,x)ds]-\lambda \beta\sin(u)v\Big]dx \\
&\leq \int_0^1\Big[-(2\delta-a-\varepsilon_4)v^2+
\int_{t-\tau}^te^{-\alpha(t-s)}v^2(s,x)ds\Big]dx
+\frac{K}{\varepsilon_4},
 \end{aligned}
\end{equation}
where $0<\varepsilon_4<2(\delta-a)$. Therefore,
$$
\int_0^T\int_0^1v^2(t,x)dxdt\leq
\frac{KT}{\varepsilon_4(2\delta-2a-\varepsilon_4)}:=K_1.
$$
By \eqref{e3.7}, we obtain
$$
\int_0^TV(t)dt\leq \frac{r_2a}{r_1}K_1+r_3T/r_1.
$$
 Since $v$ and $u_x$ are continuous,  there is a $t_0\in[0,T]$ with
$$
V(t_0)\leq \frac{\frac{r_2a}{r_1}K_1+r_3T/r_1}{T}:=K_2.
$$
Hence, if $t_0\leq t\leq t_0+T$, then by \eqref{e3.6}, we obtain
\begin{align*}
V(t)&= V(t_0)+\int_{t_0}^tV'(s)ds \\
&\leq K_2+r_2\int_{t_0}^{t_0+T}\int_0^1\int_{t-\tau}^t
  e^{-\alpha(t-s)}v^2(s,x)dsdxdt+r_3T  \\
&\leq K_2+r_2a K_1+r_3T.
\end{align*}
Note that \eqref{e3.3} yields
$$
\int_0^1{(u^2_x+v^2)dx}\leq \frac{K_2+r_2a K_1+r_3T}{1-\frac{\delta}{\pi^2}}
:=C_1.
$$
Thus
$$
\sup_{0\leq t\leq T}\int_0^1{(u^2_x+v^2)dx}\leq C_1.
$$
This completes the proof.
\end{proof}


\begin{theorem} \label{thm3.1}
Suppose that (A1)--(A3) hold. Then \eqref{e3.1} admits a nontrivial
$T$-periodic solution.
\end{theorem}

\begin{proof}
 Let $C(t,u)=\beta\sin(u)+g(t,x)-\int_{t-\tau}^te^{-\alpha(t-s)}u_s(s)ds$.
Since
\begin{align*}
\int_{t-\tau}^te^{-\alpha(t-s)}u_s(s)ds
&=\int_{-\tau}^0e^{\alpha s}u_s(t+s)ds \\
&=u(t)-u(t-\tau)e^{-\alpha\tau}-\alpha\int_{-\tau}^0e^{\alpha
s}u(t+s)ds,
\end{align*}  from the assumptions
(A1)--(A3), we see that $C$ is continuous and takes
bounded sets into bounded sets. By Lemma \ref{lem3.1} and Theorem \ref{thm2.1}, we
see that \eqref{e3.1} has a $T$-periodic solution. Since $g\neq 0$, we
see that the $T$-periodic solution is nontrivial.
This completes the proof.
\end{proof}

\section{Stable periodic solutions for Sine-Gordon equations}

In this section, we investigate the uniformly asymptotic stability of
time periodic solutions for Sine-Gordon equation
\begin{equation} \label{e4.1}
\begin{gathered}
 \frac{\partial^2 u(t,x)}{\partial
t^2}=\frac{\partial^2 u(t,x)}{\partial x^2}-\delta\frac{\partial
u(t,x)}{\partial t}
+\beta\sin(u(t,x))+g(t,x),\quad 0< x< 1,\; t>0,\\
u(t,0)=u(t,1)=0, \quad  t>0,
\end{gathered}
\end{equation}
where $g$ is continuous.
We assume  that
\begin{itemize}
\item[(H1)] $0<\delta<\frac{2\pi^2}{5} $ and
$0<|\beta|<\delta\sqrt{(2-\frac{(2+\delta)}{\pi^2})(2-\frac{5\delta}{\pi^2})}$.
\end{itemize}

\begin{theorem} \label{thm4.1}
Assume that (H1) and (A3) hold.
Then \eqref{e4.1} admits a nontrivial uniformly asymptotic stable
$T$-periodic solution.
\end{theorem}

\begin{proof} By applying Theorem \ref{thm3.1} and assumptions
(H1) and (A3),
  we conclude that  \eqref{e4.1} admits a nontrivial $T$-periodic solution
$\bar{u}(t)$.
Let $ u(t)$ be a solution for \eqref{e4.1}, and define
 $$
V_2(t)=\int_0^1[(u_t-\bar{u}_t)^2+2k(u-\bar{u})(u_t-\bar{u}_t)
+(u-\bar{u})_x^2]dx,
$$
where $k=\frac{\delta}{\pi^2}$. By the similar arguments
as in Lemma \ref{lem3.1}, we have
\begin{align*}
V_2'(t)
&=2\int_0^1[(u_t-\bar{u}_t)(u_t-\bar{u}_t)_t+(u-\bar{u})_x(u-\bar{u})_{xt}]dx \\
&\quad +2k\int_0^1[(u_t-\bar{u}_t)^2+(u-\bar{u})(u_t-\bar{u}_t)_t]dx \\
&= 2\int_0^1[-\delta(u_t-\bar{u}_t)^2+\beta[(u_t-\bar{u}_t)+k(u-\bar{u})](\sin(u)-\sin(\bar{u}))]dx \\
&\quad +2k\int_0^1[(u_t-\bar{u}_t)^2+(u-\bar{u})_{xx}(u-\bar{u})-\delta(u_t-\bar{u}_t)(u-\bar{u})]dx \\
&\leq 2\int_0^1[-(\delta-k)(u_t-\bar{u}_t)^2-k(u-\bar{u})_{x}^2-k\delta(u-\bar{u})(u_t-\bar{u}_t)]dx \\
&\quad +2|\beta|\int_0^1(k(u-\bar{u})+(u_t-\bar{u}_t))|u-\bar{u}|dx \\
&\leq \int_0^1[-(2\delta-2k)(u_t-\bar{u}_t)^2-2k(u-\bar{u})_{x}^2+k\delta((u-\bar{u})^2+(u_t-\bar{u}_t)^2)]dx \\
&\quad +\int_0^1(2|\beta| k(u-\bar{u})^2+|\beta|\varepsilon(u-\bar{u})^2
+\frac{|\beta|}{\varepsilon}(u_t-\bar{u}_t)^2dx \\
&=\int_0^1[-(2\delta-2k)+k\delta+|\beta|\varepsilon^{-1}](u_t-\bar{u}_t)^2dx \\
&\quad +\int_0^1-2k(u-\bar{u})_x^2+[k(\delta+2|\beta|)+|\beta|\varepsilon](u-\bar{u})^2dx \\
&\leq -\tilde{c}\int_0^1(u_t-\bar{u}_t)^2+(u-\bar{u})_x^2dx,
 \end{align*}
where
$$
\frac{|\beta|}{2\delta-\frac{2\delta+\delta^2}{\pi^2}}
<\varepsilon<\frac{2\delta-\frac{5\delta^2}{\pi^2}}{|\beta|}
$$
and
$$
\tilde{c}=\min\{2\delta-(\frac{2+|\beta|}{\pi^2}+|\beta|\varepsilon^{-1}),
\frac{1}{\pi^2}[2\delta-(\frac{\delta}{\pi^2}(\delta+2|\beta|)
+|\beta|\varepsilon)]\}.
$$
 Since
$$
\frac{1}{2}\int_0^1(u_t-\bar{u}_t)^2+(u-\bar{u})_x^2dx\leq V_2(t)\leq
 2\int_0^1(u_t-\bar{u}_t)^2+(u-\bar{u})_x^2dx,
$$
 we obtain that
$$
V'_2(t)\leq -\frac{\tilde{c}}{2}V_2(t).$$
This implies that for any $t>t_0$,
\begin{align*}
&\|u(t)-\bar{u}(t)\|_{H^1_0}+\|u_t(t)-\bar{u}_t(t)\|_{H^0}\\
&\leq 2e^{-\frac{\tilde{c}}{2}(t-t_0)}[\|u(t_0)-\bar{u}(t_0)\|_{H^1_0}
+\|u_t(t_0)-\bar{u}_t(t_0)\|_{H^0}].
\end{align*}
Thus the $T$-periodic solution $\bar{u}(t)$ is uniformly
asymptotic stable. Then the proof of Theorem \ref{thm4.1} is
complete.\end{proof}

Especially, in above arguments, let $u(t)$ be any $T$-periodic
solution of \eqref{e4.1}. Then, for any positive integer $n$, we have
\begin{align*}
&\|u(t)-\bar{u}(t)\|_{H^1_0}+\|u_t(t)-\bar{u}_t(t)\|_{H^0} \\
&=\|u(t+nT)-\bar{u}(t+nT)\|_{H^1_0}+\|u_t(t+nT)-\bar{u}_t(t+nT)\|_{H^0}\\
&\leq 2e^{-\frac{\tilde{c}}{2}(t+nT-t_0)}[\|u(t_0)-\bar{u}(t_0)\|_{H^1_0}
+\|u_t(t_0)-\bar{u}_t(t_0)\|_{H^0}].
\end{align*}
Let $n\to +\infty$, we obtain
$\|u(t)-\bar{u}(t)\|_{H^1_0}=0$ and
$\|u_t(t)-\bar{u}_t(t)\|_{H^0}=0$, which implies that
$u(t)=\bar{u}(t)$ a.e. in [0,T].
Thus we have the following result.

\begin{corollary}
Suppose that (H1) and (A3) hold. Then \eqref{e4.1} admits a unique
uniformly asymptotic stable  $T$-periodic solution.
\end{corollary}

Note that in Theorem \ref{thm4.1}, the assumption
$0<|\beta|<\delta\sqrt{(2-\frac{(2+\delta)}{\pi^2})(2-\frac{5\delta}{\pi^2})}$
is necessary. If this condition is omitted, the time periodic
solution maybe not uniformly asymptotic stable.
Now, we show an example to explain this case.

\noindent\textbf{Example}
Consider the equation
\begin{equation} \label{e4.2}
\begin{gathered}
 \frac{\partial^2 u(t,x)}{\partial t^2}+2\frac{\partial u(t,x)}{\partial
t}=\frac{\partial^2u(t,x)}{\partial x^2}+4\pi^2u(t,x),\quad 0< x<1,\;  t>0,\\
u(t,0)=u(t,1)=0, \quad t>0.
\end{gathered}
\end{equation}
It is easy to see that $u(t,x)=\sin(2\pi x)$ is a time periodic
 solution of \eqref{e4.2}. On the other hand, when
 $p=\sqrt{3\pi^2+1}-1$,
 $u(t,x)=\sin(2\pi x)+e^{pt}\sin(\pi x)$ is a solution of \eqref{e4.2} too,
but unbounded. Meanwhile $u(t,x)\equiv 0$ is time periodic
 solution of \eqref{e4.2}. Hence both  $u(t,x)\equiv 0$ and
$u(t,x)=\sin(2\pi x)$ are not uniformly asymptotic stable time periodic
solutions.


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\end{document}