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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 42, pp. 1--23.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/42\hfil Multi-dimensional Cahn-Hilliard equation]
{Periodic solutions of a multi-dimensional Cahn-Hilliard equation}

\author[J. Liu, Y. Wang, J. Zheng \hfil EJDE-2016/42\hfilneg]
{Ji Liu, Yifu Wang, Jiashan Zheng}

\address{Ji Liu (corresponding author) \newline
School of Mathematics and Statistics,
Beijing Institute of Technology,
Beijing 100081, China}
\email{cau\_lj@126.com, phone 18811789567}

\address{Yifu Wang \newline
School of Mathematics and Statistics,
Beijing Institute of Technology,
Beijing 100081, China}
\email{wangyifu@bit.edu.cn}

\address{Jiashan Zheng \newline
School of Mathematics and Statistics,
Beijing Institute of Technology,
Beijing 100081, China}
\email{zhengjiashan2008@163.com}

\thanks{Submitted August 7, 2014. Published January 29, 2016.}
\subjclass[2010]{35K25, 35B10, 35A01}
\keywords{Periodic solutions; Cahn-Hilliard equation; viscosity approach; 
\hfill\break\indent Schauder fixed point theorem}

\begin{abstract}
 This article concerns a multi-dimensional Cahn-Hilliard equation subject
 to Neumann boundary condition. We show existence of the periodic solutions
 by using the viscosity approach. By applying the Schauder fixed point theorem,
 we show existence of the solutions to the suitable approximate problem
 and then obtain the solutions of the considered periodic problem using
 a priori estimates. Our results extend those in \cite{20}.
\end{abstract}

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

\section{Introduction}

In 1958, Cahn and Hilliard \cite{3} derived the Cahn-Hilliard equation
\begin{equation}
u_{\tau}-\Delta(-\kappa\Delta{u}+g(u))=f,\label{e1.1}
\end{equation}
which is a model of phase separation in binary material.
Here $g(u)$ is the derivative of free energy $F(u)$.
If $F(u)$ is a smooth function, \eqref{e1.1} can be used to characterize the
spread of populations and the diffusion of an oil film over a solid surface,
see \cite{4, 16}. While $F(u)$ is not smooth, \eqref{e1.1} is used to describe
the phase separation with constraints, see for example \cite{2}.

Because of the applications of Cahn-Hilliard equation \eqref{e1.1} in physics,
there has been a great interest in studying the qualitative properties of
solutions to the Cahn-Hilliard equation.
For example, we can refer to \cite{6, 19} for existence, uniqueness and regularity
of the solutions, and \cite{7, 13} for asymptotic behavior of the solutions.
In addition, using the techniques of subdifferential operator,
 Kenmochi et al \cite{9} investigated the Cahn-Hilliard equation with constraints.
More recently, Kubo \cite{11} considered the strong solution and weak solution
to the Cahn-Hilliard equation with a time-dependent constraint and also
 discussed the relation between these solutions.

It is well known that one of the most interesting topics of the higher-order
parabolic equations, from a
theoretical and practical point of view, is existence of the periodic solutions,
which has been considered in several works \cite{12,14,18,20,22}.
Zhao et al \cite{22} studied existence and uniqueness of the time-periodic generalized
solutions to a fourth-order parabolic equation by the Galerkin method.
 Moreover, \cite{12, 14} are concerned with the existence, uniqueness and attractivity
of the time-periodic solutions to the Cahn-Hilliard equations with periodic
gradient-dependent potentials and sources. It should be remarked that
 \cite{12, 14, 22} are all in the case of one spatial dimension.
Also in one spatial dimension,  Yin et al \cite{20} used the qualitative theory
of parabolic equations to prove existence of the periodic solutions in the
classical sense to the following equation
$$
u_{\tau}+\kappa{u_{xxxx}=(A(\tau)u^3-B(\tau)u)_{xx}+f(x,\tau)},
$$
where $A(\tau)$ and $B(\tau)$ are positive, continuous and periodic
functions with the period $\omega>0$, and $f(\tau)$ is also a smooth
$\omega$-periodic function satisfying $\int^1_0f(x,\tau)dx=0$
for any $\tau\in[0,\omega]$. As for the case of higher dimensions,
Wang and Zheng \cite{18} recently showed the existence of periodic solutions
to the Cahn-Hillard equation with a constraint by applying the viscosity approach.

Motivated by the above works, the purpose of this paper is to show existence
of the periodic solutions to the problem
\begin{gather}
u_{\tau}(x,\tau)-\Delta(-\kappa\Delta{u(x,\tau)}+g(u(x,\tau)))
=f(x,\tau)\quad\text{in }Q_\omega:=\Omega\times(0,\omega),
\label{e1.2}\\
\frac{\partial{u}}{\partial{\nu}}(x,\tau)
=\frac{\partial}{\partial{\nu}}(-\kappa\Delta{u(x,\tau)}+g(u(x,\tau)))=0
\quad\text{on }\Sigma_{\omega}:=\partial\Omega\times(0,\omega),\label{e1.3} \\
u(x,0)=u(x,\omega)\quad\text{in }\Omega,\label{e1.4}
\end{gather}
where $\Omega$ is a bounded domain in $\mathbb{R}^N(1\leq{N}\leq3)$ with smooth
boundary,
${\frac{\partial}{\partial\nu}}$ stands for the outward normal derivative on
$\partial\Omega$, $f$ is a $\omega$-periodic function and
$g(u)=a_3u^3+a_2u^2+a_1u+a_0$ with constants $a_3>0$ and
$a_i\in\mathbb{R}~(i=0,1,2)$. In this case, the free energy
$ F(u)=\frac{a_3}{4}u^4+\frac{a_2}{3}u^3+\frac{a_1}{2}u^2+a_0u+C$,
where $C$ is a constant. Particularly, if $a_2=0$ and $a_1<0$, $F(u)$
is called double-well form potential. Since the principle part of \eqref{e1.2}
is a fourth-order operator, we take the viscosity approach in order to use the
standard theory of the second order parabolic equations.
More precisely, we study the approximate problem
\begin{equation}
\begin{gathered}
u_{\tau}(x,\tau)-\Delta(\varepsilon{u_{\tau}(x,\tau)}
-\kappa\Delta{u(x,\tau)}+g(u(x,\tau)))=f(x,\tau) \quad\text{in }Q_\omega,\\
{\frac{\partial{u}}{\partial{\nu}}(x,\tau)
=\frac{\partial}{\partial{\nu}}(-\kappa\Delta{u(x,\tau)}+g(u(x,\tau)))=0}
\quad\text{on }\Sigma_{\omega},\\
u(x,0)=u(x,\omega) \quad\text{in }\Omega,
\end{gathered} \label{e1.5}
\end{equation}
where $0<\varepsilon<1$. In order to apply the Schauder fixed point theorem
to show existence of the periodic solutions of \eqref{e1.5}, we need to
establish some a priori estimates on the solutions of \eqref{e1.5}
(cf. Lemma \ref{lem3.3} below).

The plan of this article is as follows.
In Section 2, we state some basic results in functional analysis
and give the main results.
In Section 3, we first establish some estimates of the solutions
for \eqref{e1.5}, and then obtain existence of the periodic solutions
for \eqref{e1.5} by the Schauder fixed point theorem.
In Section 4, based on the a priori estimates in Section 3,
we can take the limit as $\varepsilon\to 0$ and then obtain the
 periodic solutions of \eqref{e1.2}--\eqref{e1.4}.

\section{Preliminaries}

The notation and the basic results that we will use here are stated
as follows.

(1) We denote by $(\cdot,\cdot)$ and $|\cdot|_2$ the usual inner product
and the norm in $L^2(\Omega)$, respectively. Also, we denote the Hilbert
space $L^2(\Omega)$ by $H$.

(2) We denote $H^1(\Omega)$ by $V$ and its inner product by $(\cdot,\cdot)_V$,
where $(\eta_1,\eta_2)_V=(\eta_1,\eta_2)+(\nabla\eta_1,\nabla\eta_2)$
for any $\eta_1,\eta_2\in{H^1(\Omega)}$. As a result, the norm in
$H^1(\Omega)$ can be denoted by $|\eta|_V=(\eta,\eta)^{1/2}_V$.
$V^{*}$ denotes the dual space of $V$ and $\langle\cdot,\cdot\rangle_{V^{*},V}$
stands for the duality pairing between $V^{*}$ and $V$.

(3) We define $H_0:=\{\eta\in{H}|\int_{\Omega}\eta(x)dx=0\}$ which is the
closed subspace of $H$. We choose the notation $\pi_0$ to denote the projection
operator from $H$ onto $H_0$, that is,
$\pi_0[\eta](x)=\eta(x)-\frac{1}{|\Omega|}\int_{\Omega}\eta(y)dy$.
Also, we denote the inner product on $H_0$ by $(\cdot,\cdot)_0$.

(4) We denote by $V_0$ the space $V\cap{H_0}$ with the inner product
 $(\cdot,\cdot)_{V_0}$ and the norm $|\cdot|_{V_0}$, where
 $(\eta_1,\eta_2)_{V_0}=(\nabla\eta_1,\nabla\eta_2)$ for any $\eta_1,\eta_2\in{V_0}$.
 Furthermore, $F^{-1}_0$ and $\langle\cdot,\cdot\rangle_{V^{*}_0,V_0}$ denote
the duality mapping from $V^{*}_0$ onto $V_0$ and the duality pairing between
 $V^{*}_0$ and $V_0$, respectively. Thus, we see that $V^{*}_0$ is a Hilbert
space and its inner product can be defined as
\begin{equation}
(\eta_1,\eta_2)_{V^{*}_0}=\langle{\eta_1},F^{-1}_0\eta_2\rangle_{V^{*}_0,V_0}
=\langle{F^{-1}_0\eta_1,\eta_2}\rangle_{V_0,V^{*}_0}
\quad \text{for any }\eta_1,\eta_2\in{V^{*}_0}.\label{e2.1}
\end{equation}
It is observed that the Hilbert spaces stated above satisfy the following relations
$$
V\subset{H}\subset{V^{*}},\quad V_0\subset{H_0}\subset{V^{*}_0},
$$
where all the injections are compact and densely defined.
Throughout this article, we denote by $C_j>0(j=1,2,\ldots)$ the constants
induced by injection. Therefore, from the above injections, we have
\begin{equation}
\begin{gathered}
|\eta|_{V^{*}}\leq C_1|\eta|_2\quad \text{for any }\eta\in H,\\
|\eta|_2\leq C_2|\eta|_{V_0}\quad \text{for any } \eta\in V_0.
\end{gathered}\label{e2.2}
\end{equation}

(5) Let $\Delta_N$ be the Laplace operator with homogeneous Neumann boundary
condition in $H_0$ with its domain
\[
D(\Delta_N)=\big\{\eta\in{H^2(\Omega)\cap H_0:
{\frac{\partial\eta}{\partial{\nu}}=0} \text{ a.e. on } \partial\Omega}\big\}.
\]
Specially, $\Delta_N\eta=\Delta\eta$ a.e. on $\Omega$ for any
$\eta\in{D(\Delta_N)}$. We note that $-\Delta_N$ is invertible in $H_0$
and the inverse $(-\Delta_N)^{-1}$ is linear, continuous, positive and
selfadjoint in $H_0$ as well as its fractional power
$(-\Delta_N)^{1/2}$  \cite[Chapter 9, Section 11]{21}.
In addition, we have
\begin{equation}
|(-\Delta_N)^{1/2}\eta|_{H_0}
=|(-\Delta_N)^{-1}\eta|_{V_0}
=|\eta|_{V^{*}_0},\quad \forall{\eta}\in{H_0}. \label{e2.3}
\end{equation}
In this article, we always assume that the following condition holds
\begin{itemize}
\item[(H1)] $f\in{L^{\infty}(0,\omega;H)}$ is a $\omega-$periodic function and
satisfies $\int^{\omega}_0\int_{\Omega}f(x,\tau)\,dx\,d\tau=0$.
\end{itemize}
Now, we give the notion of the solution for \eqref{e1.2}--\eqref{e1.4}.

\begin{definition} \label{def2.1}\rm
 A function $u$ is called a solution of \eqref{e1.2}--\eqref{e1.4},
if the conditions below hold:
\begin{itemize}
\item[(H2)] $u\in{L^2(0,\omega;H^2(\Omega))\cap{L^{\infty}(0,\omega;V)}
\cap{W^{1,2}(0,\omega;V^{*})}}$,
${\frac{\partial{u}}{\partial{\nu}}=0}$ a.e. on $\Sigma_\omega$.

\item[(H3)] For all $\eta\in{H^2(\Omega)}$ with
${\frac{\partial\eta}{\partial\nu}\Big|_{\partial\Omega}=0}$,
\begin{align*}
&\int^{\omega}_0\langle{u_\tau(\tau),\eta}\rangle_{V^{*},V}d\tau
+\kappa\int^{\omega}_0(\Delta{u(\tau)},\Delta\eta)d\tau
-\int^{\omega}_0(g(u(\tau)),\Delta\eta)d\tau\\
&=\int^{\omega}_0(f(\tau),\eta)d\tau.
\end{align*}

\item[(H4)] $u(0)=u(\omega)\quad\text{in }H$.
\end{itemize}
\end{definition}

Now, we subtract $\frac{1}{|\Omega|}\int_{\Omega}f(x,\tau)dx$ from \eqref{e1.2}
 and obtain
\begin{equation}
\begin{aligned}
&\frac{d}{d\tau}\Big[u(x,\tau)-\frac{1}{|\Omega|}
\int^\tau_0\int_{\Omega}f(x,s)\,dx\,ds\big]
-\Delta(-\kappa\Delta{u(x,\tau)}+g(u(x,\tau)))\\
&=\pi_0[f(x,\tau)].
\end{aligned} \label{e2.4}
\end{equation}
Let
\[
 w(x,\tau)=u(x,\tau)-\frac{1}{|\Omega|}\int^\tau_0\int_{\Omega}f(x,s)\,dx\,ds.
\]
 Then \eqref{e2.4} can be rewritten as
\begin{equation}
\begin{aligned}
&w_\tau(x,\tau)-\Delta\Big[-\kappa\Delta{w(x,\tau)}+g\Big(w(x,\tau)
+\frac{1}{|\Omega|}\int^\tau_0\int_{\Omega}f(x,s)\,dx\,ds\Big)\Big]\\
&=\pi_0[f(x,\tau)].
\end{aligned}\label{e2.5}
\end{equation}
Therefore $\frac{1}{|\Omega|}\int_{\Omega}w(x,\tau)dx=m_0$ for some constant $m_0$.
Further, putting $v(x,\tau)=w(x,\tau)-m_0$, we can rewrite \eqref{e2.5} as
\begin{equation}
v_\tau(x,\tau)-\Delta_N(-\kappa\Delta_N{v(x,\tau)})
-\Delta_N\pi_0[g\left(v(x,\tau)+m(\tau)\right)]
=\pi_0[f(x,\tau)],\label{e2.6}
\end{equation}
with $\int_{\Omega} v(x,\tau)dx=0$ for all $\tau>0$, where
 $ m(\tau)=m_0+\frac{1}{|\Omega|}\int^\tau_0\int_{\Omega}f(x,s)\,dx\,ds$.

Now for any function $z\in H_0$, we can take $(-\Delta_N)^{-1}z$ as $\eta$
in $(H3)$. Hence by the arguments in\cite[Proposition 1.1]{5},
for any $z\in{H_0}$, it holds that
\begin{equation}
\begin{aligned}
&\int^{\omega}_0((-\Delta_N)^{-1}v_\tau(\tau),z)_{0}d\tau
+\kappa\int^{\omega}_0(-\Delta_N{v(\tau)},z)_0d\tau\\
&+\int^{\omega}_0(\pi_0[g(v(\tau)+m(\tau))],z)_0d\tau\\
&=\int^{\omega}_0((-\Delta_N)^{-1}\pi_0[f(\tau)],z)_0d\tau.
\end{aligned}\label{e2.7}
\end{equation}
From \eqref{e2.3}, \eqref{e2.7} and the definition of $F^{-1}_0$, we obtain
an equivalent form of \eqref{e1.2}, that is,
\begin{equation}
F^{-1}_0v_\tau(\tau)-\kappa\Delta_Nv(\tau)+\pi_0[g((v(\tau)+m(\tau)))]
=F^{-1}_0\pi_0[f(\tau)].\label{e2.8}
\end{equation}
Similarly, \eqref{e1.5} is equivalent to
\begin{equation}
\begin{gathered}
(F^{-1}_0+\varepsilon{I})v'_{\varepsilon}
(\tau)-\kappa\Delta_Nv_{\varepsilon}(\tau)
+\pi_0[g(v_{\varepsilon}(\tau)+m(\tau))]
=F^{-1}_0\pi_0[f(\tau)]\quad \text{in }Q_\omega,\\
{\frac{\partial{v_{\varepsilon}}}{\partial{\nu}}(x,\tau)=0}
\quad\text{on }\Sigma_{\omega},\\
v_{\varepsilon}(x,0)=v_{\varepsilon}(x,\omega)\quad\text{in }\Omega,
\end{gathered} \label{e2.9}
\end{equation}
where $\varepsilon\in(0,1)$,
$v'_{\varepsilon}(\tau)=\frac{d}{d\tau}v_{\varepsilon}(\tau)$
and $I$ is identity operator in $H_0$.

The main result of this article can be stated as follows.

\begin{theorem} \label{thm2.1}
Assume that {\rm (H1)} holds. Then for any constant $m_0$,
 \eqref{e1.2}--\eqref{e1.4} admits a solution $u(x,\tau)$ with
\[
\frac{1}{|\Omega|}\int_\Omega u(x,\tau)dx
=m_0+\frac{1}{|\Omega|}\int^\tau_0\int_{\Omega}f(x,s)\,dx\,ds.
\]
\end{theorem}

To prove this theorem, we use the viscosity approach. Therefore, we need to
investigate \eqref{e2.9} first. We have the following result which is proved
in next section.

\begin{theorem} \label{thm2.2}
Under the hypothesis of Theorem \ref{thm2.1}, \eqref{e2.9} admits a solution which has
the following properties:
\begin{itemize}
\item[(H2')] $v_{\varepsilon}\in{L^2(0,\omega;H^2(\Omega)\cap H_0)
\cap{L^{\infty}(0,\omega;V_0)}\cap{W^{1,2}(0,\omega;H_0)}}$,
$\frac{\partial{v_{\varepsilon}}}{\partial{\nu}}=0$ a.e. on
$\Sigma_{\omega}$.

\item[(H3')] For any $\eta\in{D(\Delta_N)}$ and $0<\tau<\omega$,
\begin{align*}
&\int^{\omega}_0((F^{-1}_0+\varepsilon{I})v'_{\varepsilon}(\tau)
 -\kappa\Delta_Nv_{\varepsilon}(\tau)
+\pi_0[g(v_{\varepsilon}(\tau)+m(\tau))],\eta)_0d\tau\\
&=\int^{\omega}_0(F^{-1}_0\pi_0[f(\tau)],\eta)_0d\tau
\quad\text{in }H_0.
\end{align*}


\item[(H4')] $v_{\varepsilon}(0)=v_{\varepsilon}(\omega)\quad\text{in }H_0$.
\end{itemize}
\end{theorem}

\section{Proof of Theorem \ref{thm2.2}}

For this purpose we use the Schauder fixed point theorem.
Firstly, we study the system
\begin{equation}
\begin{gathered}
(F^{-1}_{0}+\varepsilon{I})v'(\tau)-\kappa\Delta_Nv(\tau)
=\widehat{f}\quad\text{in }H_0,\\
v(0)=v(\omega)\quad\text{in }H_0,
\end{gathered}\label{e3.1}
\end{equation}
where $\hat{f}\in{L^{\infty}(0,\omega;H_0)}$.

\begin{theorem} \label{thm3.1}
Let $\hat{f}\in{L^{\infty}(0,\omega;H_0)}$. Then there exists a unique solution
$v(x,t)$ to problem \eqref{e3.1}.
\end{theorem}

We prove this theorem using Poincar\'e's mapping. Thus, we first introduce
the corresponding Cauchy problem
\begin{equation}
\begin{gathered}
(F^{-1}_{0}+\varepsilon{I})v'(\tau)-\kappa\Delta_Nv(\tau)=\widehat{f},
\quad 0<\tau<\omega,\\
v(0)=v_0\in{H_0}.
\end{gathered} \label{e3.2}
\end{equation}
With the help of the results in \cite{8,10}, we can see that \eqref{e3.2}
admits one and only one solution
$v\in{C([0,\omega];H_0)\cap{L^{\infty}_{loc}(0,\omega;V_0)}}$.
Consequently, with the unique solution $v(\tau)$, we can define a single-valued
mapping $P:v(0)\in{H_0}\to {v(\omega)\in{H_0}}$.

Define $\phi: H_0\to \mathbb{R}\bigcup\{+\infty\}$ by
$$
\phi(v)=\begin{cases}
\frac{\kappa}{2}|\nabla{v}|^2_2,& \text{if } v\in{V_0},\\
+\infty, &\text{otherwise}.
\end{cases}
$$
We see that $\phi$ is a proper, lower semicontinuous, and convex functional
on $H_0$.
Now, we give two lemmas which play an important role in the proof of 
Theorem \ref{thm3.1}.

\begin{lemma} \label{lem3.1}
There exists a constant $R>0$ such that $P$ is a self-mapping on the set
$$
B_R:=\{v\in{D(\phi);~\phi(v)\leq{R}}\},
$$
that is $P(B_R)\subset{B_R}$.
\end{lemma}

\begin{proof}
 Multiplying the equation in \eqref{e3.2} by $v'$, we have
\[
|v'|^2_{V^{*}_0}+\varepsilon|v'|^2_2+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v}|^2_2
= (\widehat{f},v')_0
\leq |\widehat{f}|_2|v'|_2
\leq \frac{1}{2\varepsilon}|\widehat{f}|^2_2
+\frac{\varepsilon}{2}|v'|^2_2,
\]
i.e.,
\begin{equation}
|v'|^2_{V^{*}_0}+\frac{\varepsilon}{2}|v'|^2_2
+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v}|^2_2
\leq\frac{1}{2\varepsilon}|\widehat{f}|^2_2.\label{e3.3}
\end{equation}
We also multiply the equation by $v$ and obtain
\begin{align*}
\kappa|\nabla{v}|^2_2
&= (\widehat{f},v)_0-(\varepsilon{v'},v)_0
 -\langle{F^{-1}_{0}{v'}},v\rangle_{V_0,V^{*}_0}\\
&\leq |\widehat{f}|_2|v|_2+\varepsilon|{v'}|_2|v|_2+|v'|_{V^{*}_0}|v|_{V^{*}_0}\\
&\leq \frac{2C^2_2}{\kappa}|\widehat{f}|^2_2+\frac{\kappa}{8}|\nabla{v}|^2_2
+\frac{2\varepsilon^2C^2_2}{\kappa}|v'|^2_2
+\frac{\kappa}{8}|\nabla v|^2_2+\frac{C^2_1C^2_2}{\kappa}|v'|^2_{V^{*}_0}
+\frac{\kappa}{4}|\nabla v|^2_2\\
&= \frac{2C^2_2}{\kappa}|\widehat{f}|^2_2
+\frac{2\varepsilon^2C^2_2}{\kappa}|v'|^2_2+\frac{C^2_1C^2_2}{\kappa}|v'|^2_{V^{*}_0}
+\frac{\kappa}{2}|\nabla{v}|^2_2,
\end{align*}
which implies
\begin{equation}
\frac{\kappa}{2}|\nabla{v}|^2_2
\leq\frac{2C^2_2}{\kappa}|\widehat{f}|^2_2
+\frac{2\varepsilon^2C^2_2}{\kappa}|v'|^2_2
+\frac{C^2_1C^2_2}{\kappa}|v'|^2_{V^{*}_0}\,.
\label{e3.4}
\end{equation}
Letting $\mu>0$ and performing $\eqref{e3.3}\times\mu+\eqref{e3.4}$, we obtain
\begin{align*}
& \mu\frac{d}{dt}(\frac{{\kappa}}{2}|\nabla{v}|^2_2)
 +\frac{\kappa}{2}|\nabla{v}|^2_2 \\
&\leq(\frac{\mu}{2\varepsilon}+\frac{2C^2_2}{\kappa})|\widehat{f}|^2_2
 +(\frac{C^2_1C^2_2}{\kappa}-\mu)|v'|^2_{V^{*}_0}
+(\frac{2\varepsilon^2C^2_2}{\kappa}-\frac{\mu\varepsilon}{2})|v'|^2_2.
\end{align*}
Choosing $\mu=\max\big\{\frac{C^2_1C^2_2}{\kappa},\frac{4C^2_2}{\kappa}\big\}$,
from $0<\varepsilon<1$ we have
\[
\frac{d}{dt}\phi(v)+\frac{1}{\mu}\phi(v)
\leq\big(\frac{1}{2\varepsilon}+\frac{2C^2_2}{\kappa\mu}\big)|\widehat{f}|^2_2.
\]
It follows from the Gronwall inequality that
\[
\phi(v(\omega))\leq{e^{-\frac{\omega}{\mu}}\phi(v(0))
+(1-e^{-\frac{\omega}{\mu}})
\big(\frac{\mu}{2\varepsilon}+\frac{2C^2_2}{\kappa}\big)
\|\widehat{f}\|^2_{L^{\infty}(0,\omega;H_0)}}.
\]
Set $R=(\frac{\mu}{2\varepsilon}+\frac{2C^2_2}{\kappa})
\|\widehat{f}\|^2_{L^{\infty}(0,\omega;H_0)}$. Then
 $\phi(v(\omega))\leq{R}$ provided that $\phi(v(0))\leq{R}$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem3.2}
The mapping $P$ is continuous in $H_0$.
\end{lemma}

\begin{proof}
Let $v_{0,n}\in{H_0}$ be such that $v_{0,n}\to {v_0}$ in $H_0$.
We denote the unique solution of \eqref{e3.2} by $v_n$ and
 $v$ corresponding to the initial data $v_{0,n}$ and $v_{0}$, respectively.
Then we have
\begin{equation}
F^{-1}_{0}(v'_{n}-v')+\varepsilon(v'_{n}-v')-\kappa\Delta_{N}(v_n-v)=0.\label{e3.5}
\end{equation}
Multiplying \eqref{e3.5} by $v_n-v$ and using integration by parts, we obtain
$$
\frac{1}{2}\frac{d}{dt}|v_n-v|^2_{V^{*}_0}
+\frac{\varepsilon}{2}\frac{d}{dt}|v_n-v|^2_2
+\kappa|\nabla(v_n-v)|^2_2=0.
$$
It can be easy to see that
$$
\frac{1}{2}\frac{d}{dt}|v_n-v|^2_{V^{*}_0}
+\frac{\varepsilon}{2}\frac{d}{dt}|v_n-v|^2_2\leq0.
$$
Therefore,
$$
\frac{1}{2}|v_n(\omega)-v(\omega)|^2_{V^{*}_0}
+\frac{\varepsilon}{2}|v_n(\omega)-v(\omega)|^2_2
\leq\frac{1}{2}|v_{n0}-v_0|^2_{V^{*}_0}+\frac{\varepsilon}{2}|v_{n0}-v_0|^2_2,
$$
which implies $v_n(w)\to {v(\omega)}$ in $H_0$ as $n\to \infty$.
Hence, $P$ is continuous in $H_0$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 On the one hand, it follows from the definition of $B_R$ and the convexity
of $\phi$ that $B_R$ is compact and convex in $H_0$. On the other hand,
Lemmas \ref{lem3.1} and \ref{lem3.2} ensure that $P$ maps $B_R$ to $B_R$ 
and is continuous in $H_0$. Thus, the Schauder fixed point theorem admits 
a fixed point $v^{*}_0\in{B_R}$ such that $Pv^{*}_0=v^{*}_0$, which implies that
the solution $v(x,t)$ of \eqref{e3.2} with $v_0=v^{*}_0$ is the desired
solution of \eqref{e3.1}.

Now, we prove that the solution for \eqref{e3.1} is unique. To this end,
let $v_1$ and $v_2$ be two solutions of \eqref{e3.1}. Then we have
\begin{equation}
F^{-1}_{0}(v'_{1}-v'_{2})+\varepsilon(v'_{1}-v'_{2})-\kappa\Delta_{N}(v_1-v_2)=0.
\label{e3.6}
\end{equation}
We multiply \eqref{e3.6} by $v_1-v_2$ and then get that
$$
\frac{1}{2}\frac{d}{dt}|v_1-v_2|^2_{V^{*}_0}
+\frac{\varepsilon}{2}\frac{d}{dt}|v_1-v_2|^2_2
+\kappa|\nabla(v_1-v_2)|^2_2=0.
$$
Integrating the equation over $(0,\omega)$ and by the periodic property,
 we obtain
$$
\int^{\omega}_0|\nabla(v_1(\tau)-v_2(\tau))|^2_2d\tau\leq0,
$$
which, together with \eqref{e2.2}, implies that
$$
\int^{\omega}_0\int_{\Omega}|v_1-v_2|^2\,dx\,d\tau\leq0.
$$
Hence, $v_1=v_2$ and the proof is complete.
\end{proof}

To apply the Schauder fixed point theorem to \eqref{e2.9}, we need
to establish a priori estimates for $v_{\varepsilon}$.

\begin{lemma} \label{lem3.3}
Let $v_{\varepsilon}$ be a solution of \eqref{e2.9}.
Then
\begin{gather}
\varepsilon\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_2d\tau
+\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
\leq\omega C^2_1\|f\|^2_{L^{\infty}(0,\omega;H)},\label{e3.7} \\
\sup_{\tau\in[0,\omega]}|v_{\varepsilon}(\tau)|^2_{V_0}
\leq\frac{2}{\kappa}\Big(3A_1+4A_2+\frac{3C^2_1+4\omega}{2}C^2_1
\|f\|^2_{L^{\infty}(0,\omega;H)}\Big),
\label{e3.8} \\
\begin{aligned}
&\|-\Delta_Nv_{\varepsilon}\|^2_{L^2(0,\omega;H_0)}\\
&\leq \frac{4\omega}{\kappa^2}\Big(\frac{a^2_2}{2a_3}+|a_1|\Big)(3A_1+4A_2)\\
&\quad +\frac{\omega}{\kappa^2}
 \Big[(3C^2_1+4\omega)C^2_1\Big(\frac{a^2_2}{a_3}+2|a_1|\Big)+C^4_1\Big]
\|f\|^2_{L^{\infty}(0,\omega;H)}
\end{aligned}\label{e3.9}
\\
\|v_{\varepsilon}(\tau)+m(\tau)\|^6_{C([0,\omega];L^6(\Omega))}\leq{A^3_3},
\label{e3.10}
\end{gather}
where $a_i$ $(i=0,1,2,3)$ are the coefficients of $g(\cdot)$,
\begin{gather*}
\begin{aligned}
A_1:&= |\Omega|\Big[\frac{9^3}{4}a_3
\Big(|m_0|+\frac{\omega}{|\Omega|^{1/2}}\|f\|_{L^{\infty}(0,\omega;H)}\Big)^4
+\Big(\frac{3a^2_1}{2a_3}+\frac{3a^2_2}{a_3}+1\Big) \\
&\quad \times \Big(|m_0|+\frac{\omega}{|\Omega|^{1/2}}
 \|f\|_{L^{\infty}(0,\omega;H)}\Big)^2 \\
&\quad +\frac{a^4_2}{4}\big(\frac{9}{a_3}\big)^3
 +\frac{a_3}{12}+\frac{3a^2_1+3}{a_3}\Big],
\end{aligned} \\
A_2:=6|\Omega|\Big[\frac{6a^4_2}{a^3_3}+\frac{a^2_1}{2a_3}
+\frac{a^{4/3}_0}{4}\big(\frac{6}{a_3}\big)^{1/3}\Big],
 \\
\begin{aligned}
A_3:&= C_3\Big\{\frac{2}{\kappa}(2\omega C^2_1C^2_2+2C^2_2+1)(3A_1+4A_2)
+4m^2_0|\Omega|(\omega C^2_1+1) \\
&\quad +\Big[4\omega^3C^2_1+4\omega^2
 \Big(\frac{2C^4_1C^2_2}{\kappa}+1\Big)
+2C^2_1\omega\Big(\frac{3C^4_1C^2_2+4C^2_2+2}{\kappa}+2\Big)\\
&\quad +\frac{3C^4_1(2C^2_2+1)}{\kappa}\Big]
\|f\|^2_{L^{\infty}(0,\omega;H)}\Big\}.
\end{aligned}
\end{gather*}
\end{lemma}

\begin{proof}
From \eqref{e2.2}, the definition of $\pi_0$ and $(H1)$,
 we know that
\begin{equation}
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}
\leq C^2_1\|\pi_0[f]\|^2_{L^2(0,\omega;H_0)}
\leq \omega C^2_1\|f\|^2_{L^{\infty}(0,\omega;H)}.\label{e3.11}
\end{equation}
It follows from the H\"{o}lder inequality and (H1) that for any $\tau\in[0,\omega]$
\begin{equation}
\begin{aligned}
|m(\tau)|&= \left|m_0+\frac{1}{|\Omega|}\int^{\tau}_0\int_{\Omega}f(x,s)\,dx\,ds\right|\\
&\leq |m_0|+\frac{\omega}{|\Omega|^{1/2}}\|f\|_{L^{\infty}(0,\omega;H)}.
\end{aligned}\label{e3.12}
\end{equation}

We multiply the equation in \eqref{e2.9} by $v'_{\varepsilon}$, and obtain
\begin{equation}
\begin{aligned}
&|v'_{\varepsilon}|^2_{V^{*}_0}+\varepsilon|v'_{\varepsilon}|^2_2
+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v_{\varepsilon}}|^2_2
+\int_{\Omega}v'_{\varepsilon}\pi_0[g(v_{\varepsilon}(\tau)+m(\tau))]dx\\
&= \langle{F^{-1}_0\pi_0[f]},v'_{\varepsilon}\rangle_{V_0,V^{*}_0}\\
&\leq |\pi_0[f]|_{V^{*}_0}|v'_{\varepsilon}|_{V^{*}_0}\\
&\leq \frac{1}{2}|\pi_0[f]|^2_{V^{*}_0}+\frac{1}{2}|v'_{\varepsilon}|^2_{V^{*}_0}.
\end{aligned}\label{e3.13}
\end{equation}
By the definition of $\pi_0$ and $g(\cdot)$, we have
\begin{equation}
\begin{aligned}
&\frac{1}{2}|v'_{\varepsilon}|^2_{V^{*}_0}+\varepsilon|v'_{\varepsilon}|^2_2
+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v_{\varepsilon}}|^2_2
+\frac{d}{dt}\int_{\Omega}\Big[\frac{a_3}{4}(v_{\varepsilon}(\tau)+m(\tau))^4\\
&+\frac{a_2}{3}(v_{\varepsilon}(\tau)+m(\tau))^3
 +\frac{a_1}{2}(v_{\varepsilon}(\tau)+m(\tau))^2
 +a_0(v_{\varepsilon}(\tau)+m(\tau))\Big]dx\\
&\leq \frac{1}{2}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned}\label{e3.14}
\end{equation}
From the periodic property, we integrate \eqref{e3.14} over $(0,\omega)$
and then obtain
\begin{equation}
2\varepsilon\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_2d\tau
+\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
\leq\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau.\label{e3.15}
\end{equation}
Combining this inequality with \eqref{e3.11}, we have
$$
\varepsilon\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_2d\tau
+\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
\leq\omega C^2_1\|f\|^2_{L^{\infty}(0,\omega;H)},
$$
which is \eqref{e3.7}.

Choose any $s,t\in[0,\omega]$ which satisfy $s<t$.
Integrating \eqref{e3.14} on $(s,t)$, we have
\begin{equation}
\begin{aligned}
&\frac{1}{2}\int^t_s|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
+\varepsilon\int^t_s|v'_{\varepsilon}(\tau)|^2_2d\tau
 +\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(t)|^2_2
+\int_{\Omega}\big[\frac{a_3}{4}(v_{\varepsilon}(t)+m(t))^4\\
&+\frac{a_2}{3}(v_{\varepsilon}(t)+m(t))^3
 +\frac{a_1}{2}(v_{\varepsilon}(t)+m(t))^2+a_0(v_{\varepsilon}(t)+m(t))\big]dx\\
&\leq \frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau+\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(s)|^2_2
+\int_{\Omega}\big[\frac{a_3}{4}(v_{\varepsilon}(s)+m(s))^4\\
&\quad +\frac{a_2}{3}(v_{\varepsilon}(s)+m(s))^3
 +\frac{a_1}{2}(v_{\varepsilon}(s)+m(s))^2+a_0(v_{\varepsilon}(s)+m(s))\big]dx.
\end{aligned}\label{e3.16}
\end{equation}
From Young's inequality, we obtain
\begin{gather}
\frac{a_2}{3}(v_{\varepsilon}(\tau)+m(\tau))^3
\leq\frac{a_3}{24}(v_{\varepsilon}(\tau)+m(\tau))^4+\frac{18a^4_2}{a^3_3},
\nonumber \\
\frac{a_1}{2}(v_{\varepsilon}(\tau)+m(\tau))^2
\leq\frac{a_3}{24}(v_{\varepsilon}(\tau)+m(\tau))^4+\frac{3a^2_1}{2a_3}
\nonumber \\
a_0(v_{\varepsilon}(\tau)+m(\tau))
\leq\frac{a_3}{24}(v_{\varepsilon}(\tau)+m(\tau))^4
+\frac{3a^{4/3}_0}{4}\Big(\frac{6}{a_3}\Big)^{1/3}.\label{e3.17}
\end{gather}
It follows from \eqref{e3.16} and \eqref{e3.17} that
\begin{equation}
\begin{aligned}
&\frac{1}{2}\int^t_s|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
+\varepsilon\int^t_s|v'_{\varepsilon}(\tau)|^2_2d\tau
 +\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(t)|^2_2\\
&+\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(t)+m(t))^4dx\\
&\leq \frac{\kappa}{2}|\nabla{v_{\varepsilon}}(s)|^2_2
+\frac{3a_3}{8}\int_{\Omega}(v_{\varepsilon}(s)+m(s))^4dx \\
&\quad +\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau\\
&\quad +A_2.
\end{aligned}\label{e3.18}
\end{equation}
Deleting the first two terms on the left-hand side of \eqref{e3.18}
and integrating  it on $(0,t)$ with respect to $s$, we have
\begin{align*}
&\frac{\kappa t}{2}|\nabla{v_{\varepsilon}}(t)|^2_2
 +\frac{a_3t}{8}\int_{\Omega}(v_{\varepsilon}(t)+m(t))^4dx \\
&\leq \frac{\kappa}{2}\int^{\omega}_0|\nabla{v_{\varepsilon}}(s)|^2_2ds
+\frac{3a_3}{8}\int^{\omega}_0\int_{\Omega}(v_{\varepsilon}(s)+m(s))^4\,dx\,ds
+\frac{\omega}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau\\
&\quad +A_2\omega.
\end{align*}
Letting $t=\omega$, one sees that
\begin{align*}
&\frac{\kappa\omega}{2}|\nabla{v_{\varepsilon}}(\omega)|^2_2
+\frac{a_3\omega}{8}\int_{\Omega}(v_{\varepsilon}(\omega)+m_0)^4dx\\
&\leq \frac{\kappa}{2}\int^{\omega}_0|\nabla{v_{\varepsilon}}(s)|^2_2ds
+\frac{3a_3}{8}\int^{\omega}_0\int_{\Omega}(v_{\varepsilon}(s)+m(s))^4\,dx\,ds
+\frac{\omega}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau\\
&\quad +A_2\omega,
\end{align*}
i.e.,
\begin{align*}
&\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(\omega)|^2_2
 +\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(\omega)+m_0)^4dx\\
&\leq \frac{\kappa}{2\omega}\int^{\omega}_0|\nabla{v_{\varepsilon}}(s)|^2_2ds
+\frac{3a_3}{8\omega}\int^{\omega}_0\int_{\Omega}(v_{\varepsilon}(s)+m(s))^4\,dx\,ds
+\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau\\
&\quad +A_2.
\end{align*}
From the periodic property, we have
\begin{equation}
\begin{aligned}
&\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(0)|^2_2
 +\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(0)+m_0)^4dx \\
&\leq \frac{\kappa}{2\omega}\int^{\omega}_0|\nabla{v_{\varepsilon}}(s)|^2_2ds
+\frac{3a_3}{8\omega}\int^{\omega}_0\int_{\Omega}(v_{\varepsilon}(s)+m(s))^4\,dx\,ds\\
&\quad +\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau+A_2.
\end{aligned} \label{e3.19}
\end{equation}

Multiplying the equation in \eqref{e2.9} by $v_{\varepsilon}$ and performing
a proper arrangement, we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v_{\varepsilon}|^2_{V^{*}_0}
+\frac{\varepsilon}{2}\frac{d}{dt}|v_{\varepsilon}|^2_2
+\kappa|\nabla{v_{\varepsilon}}|^2_2\\
&+a_3\int_{\Omega}
\left[(v_{\varepsilon}(\tau)+m(\tau))^4-m(\tau)(v_{\varepsilon}(\tau)
 +m(\tau))^3\right]dx\\
&+a_2\int_{\Omega}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^3dx
+\left[a_1-a_2m(\tau)\right]\int_{\Omega}\left[v_{\varepsilon}(\tau)
 +m(\tau)\right]^2dx \\
& -a_1m(\tau)\int_{\Omega}\left[v_{\varepsilon}(\tau)+m(\tau)\right]dx\\
&= \langle{F^{-1}_0\pi_0[f],v_{\varepsilon}}\rangle_{V_0,V^{*}_0}\\
&\leq |\pi_0[f]|_{V^{*}_0}|v_{\varepsilon}|_{V^{*}_0}\\
&\leq C_1|\pi_0[f]|_{V^{*}_0}|v_{\varepsilon}(\tau)+m(\tau)|_2
 +C_1|\pi_0[f]|_{V^{*}_0}|\Omega|^{1/2}|m(\tau)|.
\end{aligned}\label{e3.20}
\end{equation}
From Young's inequality, we obtain
\begin{gather*}
m(\tau)\left[v_{\varepsilon}(\tau)+m(\tau)\right]^3
\leq\frac{1}{12}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4
+\frac{9^3}{4}|m(\tau)|^4,
\\
a_2\left[v_{\varepsilon}(\tau)+m(\tau)\right]^3
\leq\frac{a_3}{12}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4
+\frac{a^4_2}{4}\big(\frac{9}{a_3}\big)^3,
\\
\begin{aligned}
\left[a_1-a_2m(\tau)\right]\left[v_{\varepsilon}(\tau)+m(\tau)\right]^2
&\leq \frac{a_3}{6}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4
 +\frac{3}{2a_3}\left[a_1-a_2m(\tau)\right]^2\\
&\leq \frac{a_3}{6}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4
+\frac{3a^2_1}{a_3}+\frac{3a^2_2}{a_3}|m(\tau)|^2,
\end{aligned}\\
\begin{aligned}
a_1m(\tau)\left[v_{\varepsilon}(\tau)+m(\tau)\right]
&\leq \frac{a_3}{6}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^2
 +\frac{3}{2a_3}a^2_1|m(\tau)|^2\\
&\leq \frac{a_3}{12}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4
 +\frac{a_3}{12}+\frac{3}{2a_3}a^2_1|m(\tau)|^2,
\end{aligned}\\
\begin{aligned}
|\pi_0[f]|_{V^{*}_0}|v_{\varepsilon}(\tau)+m(\tau)|_2
&\leq \frac{C^2_1}{4}|\pi_0[f]|^2_{V^{*}_0}+|v_{\varepsilon}(\tau)+m(\tau)|^2_2\\
&\leq \frac{C^2_1}{4}|\pi_0[f]|^2_{V^{*}_0}
+\frac{a_3}{12}\int_{\Omega}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4dx
+\frac{3}{a_3}|\Omega|
\end{aligned}
\end{gather*}
and
\begin{equation}
|\pi_0[f]|_{V^{*}_0}|\Omega|^{1/2}|m(\tau)|
\leq\frac{C^2_1}{4}|\pi_0[f]|^2_{V^{*}_0}+|\Omega||m(\tau)|^2.
\label{e3.21}
\end{equation}
In light of \eqref{e3.12}, \eqref{e3.20} and \eqref{e3.21}, we have
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v_{\varepsilon}|^2_{V^{*}_0}
+\frac{\varepsilon}{2}\frac{d}{dt}|v_{\varepsilon}|^2_2
+\kappa|\nabla{v_{\varepsilon}}|^2_2
+\frac{a_3}{2}\int_{\Omega}\left[v_{\varepsilon}(\tau)+m(\tau)\right]^4dx\\
&\leq\frac{C^2_1}{2}|\pi_0[f]|^2_{V^{*}_0}+A_1.
\end{aligned}\label{e3.22}
\end{equation}
From the periodic property, we integrate \eqref{e3.22} over $(0,\omega)$
and obtain
\begin{equation}
\begin{aligned}
&\kappa\int^{\omega}_0|\nabla{v_{\varepsilon}(\tau)}|^2_2d\tau
+\frac{a_3}{2}\int^{\omega}_0\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))^4\,dx\,d\tau\\
&\leq\frac{C^2_1}{2}\int^{\omega}_0|\pi_0[f]|^2_{V^{*}_0}d\tau+A_1\omega.
\end{aligned}\label{e3.23}
\end{equation}
Combining \eqref{e3.19} with \eqref{e3.23}, we have
\begin{equation}
\begin{aligned}
&\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(0)|^2_2
 +\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(0)+m_0)^4dx\\
&\leq \frac{C^2_1}{2\omega}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau
+A_1+\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau+A_2\\
&\leq A_1+A_2+\frac{\omega+C^2_1}{2\omega}\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}.
\end{aligned}\label{e3.24}
\end{equation}
Letting $s=0$ in \eqref{e3.18} and deleting the first two terms on the 
left-hand side, we obtain
\begin{equation}
\begin{aligned}
&\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(t)|^2_2
+\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(t)+m(t))^4dx\\
&\leq \frac{\kappa}{2}|\nabla{v_{\varepsilon}}(0)|^2_2
+\frac{3a_3}{8}\int_{\Omega}(v_{\varepsilon}(0)+m_0)^4dx
+\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau
+A_2\\
&\leq 3\Big[\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(0)|^2_2
+\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(0)+m_0)^4dx\Big]
+\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau
+A_2.\end{aligned}\label{e3.25}
\end{equation}
It follows from \eqref{e3.24} and \eqref{e3.25} that for any $t\in[0,\omega]$,
$$
\frac{\kappa}{2}|\nabla{v_{\varepsilon}}(t)|^2_2
+\frac{a_3}{8}\int_{\Omega}(v_{\varepsilon}(t)+m(t))^4dx
\leq 3A_1+4A_2+\frac{3C^2_1+4\omega}{2\omega}\int^{\omega}_0
|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau,
$$
which together with \eqref{e3.11} yields \eqref{e3.8}.

Multiplying the equation of \eqref{e2.9} by $-\Delta_N{v_{\varepsilon}}$ and
integrating it by parts, we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}|v_{\varepsilon}|^2_2
+\frac{\varepsilon}{2}\frac{d}{dt}|\nabla{v_{\varepsilon}}|^2_2
+\kappa|\Delta_Nv_{\varepsilon}|^2_2
+3a_3\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))^2|\nabla{v_{\varepsilon}}|^2dx\\
&+2a_2\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))|\nabla{v_{\varepsilon}}|^2dx
+a_1\int_{\Omega}|\nabla{v_{\varepsilon}}|^2dx\\
&= \langle{F^{-1}_0\pi_0[f],-\Delta_Nv_{\varepsilon}}\rangle_{V_0,V^{*}_0}\\
&\leq C_1|\pi_0[f]|_{V^{*}_0}|\Delta_Nv_{\varepsilon}|_2\\
&\leq \frac{C^2_1}{2\kappa}|\pi_0[f]|^2_{V^{*}_0}
+\frac{\kappa}{2}|\Delta_Nv_{\varepsilon}|^2_2.
\end{align*}
After a proper arrangement, we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v_{\varepsilon}|^2_2
 +\frac{\varepsilon}{2}\frac{d}{dt}|\nabla{v_{\varepsilon}}|^2_2
+\frac{\kappa}{2}|\Delta_Nv_{\varepsilon}|^2_2
+3a_3\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))^2|\nabla{v_{\varepsilon}}|^2dx\\
&+2a_2\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))|\nabla{v_{\varepsilon}}|^2dx\\
&\leq \frac{C^2_1}{2\kappa}|\pi_0[f]|^2_{V^{*}_0}
-a_1\int_{\Omega}|\nabla{v_{\varepsilon}}|^2dx.
\end{aligned}\label{e3.26}
\end{equation}
From Young's inequality, we have
\begin{equation}
2a_2(v_{\varepsilon}(\tau)+m(\tau))\leq2a_3(v_{\varepsilon}
(\tau)+m(\tau))^2+\frac{a^2_2}{2a_3}.\label{e3.27}
\end{equation}
It follows from \eqref{e3.26} and \eqref{e3.27} that
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v_{\varepsilon}|^2_2
+\frac{\varepsilon}{2}\frac{d}{dt}|\nabla{v_{\varepsilon}}|^2_2
+\frac{\kappa}{2}|\Delta_Nv_{\varepsilon}|^2_2
+a_3\int_{\Omega}(v_{\varepsilon}(\tau)+m(\tau))^2|\nabla{v_{\varepsilon}}|^2dx\\
&\leq \Big(\frac{a^2_2}{2a_3}+|a_1|\Big)
 \int_{\Omega}|\nabla{v_{\varepsilon}}|^2dx
+\frac{C^2_1}{2\kappa}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned}\label{e3.28}
\end{equation}
With the help of the periodic property, we integrate \eqref{e3.28} over
$(0,\omega)$ and then get that
$$
\int^{\omega}_0|\Delta_Nv_{\varepsilon}(\tau)|^2_2d\tau
\leq\frac{2}{\kappa}\Big(\frac{a^2_2}{2a_3}+|a_1|\Big)
\int^{\omega}_0\int_{\Omega}|\nabla{v_{\varepsilon}(\tau)}|^2\,dx\,d\tau
+\frac{C^2_1}{\kappa^2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau.
$$
Substituting \eqref{e3.8} and \eqref{e3.11} into the above inequality,
we obtain \eqref{e3.9}.

By \eqref{e3.7} and \eqref{e3.8}, we know that
$v_{\varepsilon}\in{W^{1,2}(0,\omega;V^{*}_0)}\cap{L^{\infty}(0,\omega;V_0)}$.
Therefore, $v_{\varepsilon}(\tau)+m(\tau)\in{W^{1,2}(0,\omega;V^{*})}
\cap{L^{\infty}(0,\omega;V)}$.
Since
$$
{W^{1,2}(0,\omega;V^{*})}\cap{L^{\infty}(0,\omega;V)}
\hookrightarrow{C([0,\omega];L^6(\Omega))},
$$
it is clear that there exists a positive constant $C_3$ such that
\begin{equation}
\begin{aligned}
&\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{C([0,\omega];L^6(\Omega))}\\
&\leq{C_3}(\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{W^{1,2}(0,\omega;V^{*})}
+\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{L^{\infty}(0,\omega;V)}).
\end{aligned}\label{e3.29}
\end{equation}

Now, we establish the estimates for
$\|v_{\varepsilon}(\tau)+m(\tau)\|_{W^{1,2}(0,\omega;V^{*})}$ and
$\|v_{\varepsilon}(\tau)+m(\tau)\|_{L^{\infty}(0,\omega;V)}$, respectively.
Since
\begin{align*}
\int^{\omega}_0|v_{\varepsilon}(\tau)+m(\tau)|^2_{V^{*}}d\tau
&\leq \int^{\omega}_0(|v_{\varepsilon}(\tau)|^2_{V^{*}_0}
 +2|v_{\varepsilon}(\tau)|_{V^{*}_0}|m(\tau)|_{V^{*}}+|m(\tau)|^2_{V^{*}})d\tau\\
&\leq 2(C^2_1C^2_2\int^{\omega}_0|v_{\varepsilon}(\tau)|^2_{V_0}d\tau
 +C^2_1\int^{\omega}_0|m(\tau)|^2_2d\tau)\\
&\leq 2\omega C^2_1(C^2_2\|v_{\varepsilon}(\tau)\|^2_{L^{\infty}
 (0,\omega;V_0)}+|\Omega||m(\tau)|^2),
\end{align*}
we obtain
\begin{equation} 
\begin{aligned}
\int^{\omega}_0\left|v_{\varepsilon}(\tau)+m(\tau)\right|^2_{V^{*}}d\tau
&\leq 2\omega C^2_1\Big[\frac{2C^2_2}{\kappa}(3A_1+4A_2)+2m^2_0|\Omega|\\
&\quad +\Big(\frac{3C^4_1C^2_2+4\omega C^2_1C^2_2}{\kappa}+2\omega^2\Big)
\|f\|^2_{L^{\infty}(0,\omega;H)}\Big].
\end{aligned}\label{e3.30}
\end{equation}
Similarly, we have
\begin{equation}
\begin{aligned}
&\int^{\omega}_0|v'_{\varepsilon}(\tau)+m'(\tau)|^2_{V^{*}}d\tau\\
&\leq \int^{\omega}_0\left(|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}
 +2|v'_{\varepsilon}(\tau)|_{V^{*}_0}|m'(\tau)|_{V^{*}}+|m'(\tau)|^2_{V^{*}}\right)d\tau\\
&\leq 2\Big(\int^{\omega}_0|v'_{\varepsilon}(\tau)|^2_{V^{*}_0}d\tau
+C^2_1\int^{\omega}_0|m'(\tau)|^2_2d\tau\Big).
\end{aligned}\label{e3.31}
\end{equation}
Moreover,
$$
\int^{\omega}_0|m'(\tau)|^2_2d\tau
= \int^{\omega}_0\Big|\frac{1}{|\Omega|}\int_{\Omega}f(x,\tau)dx\Big|^2_2d\tau
\leq \omega\|f\|^2_{L^{\infty}(0,\omega;H)}.
$$
Thus, together with \eqref{e3.7}, \eqref{e3.31} can be written as
\begin{equation}
\int^{\omega}_0\Big|\frac{d}{dt}(v_{\varepsilon}(\tau)+m(\tau))\Big|^2_{V^{*}}d\tau
\leq 4\omega C^2_1\|f\|^2_{L^{\infty}(0,\omega;H)}.\label{e3.32}
\end{equation}
It follows from \eqref{e3.30} and \eqref{e3.32} that
\begin{equation}
\begin{aligned}
&\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{W^{1,2}(0,\omega;V^{*})}\\
&\leq 2\omega C^2_1\Big[\frac{2C^2_2}{\kappa}(3A_1+4A_2)+2m^2_0|\Omega| \\
&\quad +\Big(\frac{3C^4_1C^2_2+4\omega C^2_1C^2_2}{\kappa}+2\omega^2+2\Big)
\|f\|^2_{L^{\infty}(0,\omega;H)}\Big].
\end{aligned} \label{e3.33}
\end{equation}
Also, since
\begin{align*}
&\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{L^{\infty}(0,\omega;V)}\\
&= \operatorname{ess\,sup}_{\tau\in[0,\omega]}
\Big[\int_{\Omega}|v_{\varepsilon}(\tau)+m(\tau)|^2dx
+\int_{\Omega}|\nabla(v_{\varepsilon}(\tau)+m(\tau))|^2dx\Big]\\
&\leq 2\operatorname{ess\,sup}_{\tau\in[0,\omega]}
\int_{\Omega}[v^2_{\varepsilon}(\tau)+|m(\tau)|^2]dx
+\operatorname{ess\,sup}_{\tau\in[0,\omega]}
\int_{\Omega}|\nabla{v_{\varepsilon}(\tau)}|^2dx\\
&\leq (2C^2_2+1)\|v_{\varepsilon}\|^2_{L^{\infty}(0,\omega;V_0)}
+2\operatorname{ess\,sup}_{\tau\in[0,\omega]}\int_{\Omega}|m(\tau)|^2dx,
\end{align*}
from \eqref{e3.8} and \eqref{e3.12} we have
\begin{equation}
\begin{aligned}
&\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{L^{\infty}(0,\omega;V)}\\
&\leq \frac{2}{\kappa}(2C^2_2+1)(3A_1+4A_2)+4{m^2_0}|\Omega|\\
&+\Big[\frac{(2C^2_2+1)(3C^2_1+4\omega)}{\kappa}C^2_1
+4\omega^2\Big]\|f\|^2_{L^{\infty}(0,\omega;H)}.
\end{aligned}\label{e3.34}
\end{equation}
Combining \eqref{e3.33} with \eqref{e3.34}, we obtain
\begin{equation}
\|v_{\varepsilon}(\tau)+m(\tau)\|^2_{C([0,\omega];L^6(\Omega))}\leq A_3.\\
\label{e3.35}
\end{equation}
Thus,
\begin{equation}
\|v_{\varepsilon}(\tau)+m(\tau)\|^6_{C([0,\omega];L^6(\Omega))}\leq{A^3_3},
\label{e3.36}
\end{equation}
which is \eqref{e3.10}. The proof is complete.
\end{proof}
Define a set
\begin{equation}
\begin{aligned}
Y_1:=\Big\{&\bar{v}\in{L^{\infty}(0,\omega;V_0)}\cap{W^{1,2}(0,\omega;H_0)}|
\bar{v}(0)=\bar{v}(\omega), \\
&\|\bar{v}(\tau)+m(\tau)\|^6_{C([0,\omega];L^6(\Omega))}\leq{A^3_3}\Big\}.
\end{aligned}\label{e3.37}
\end{equation}
Now, for any $\bar{v}\in{Y_1}$, we study the problem
\begin{equation}
\begin{gathered}
F^{-1}_0v'(\tau)+\varepsilon{v'(\tau)}-\kappa\Delta_Nv(\tau)
=-\pi_0[g(\bar{v}(\tau)+m(\tau))]+F^{-1}_0\pi_0[f(\tau)]\\
\text{in }H_0,\quad 0<\tau<\omega,\\
v(0)=v(\omega)\quad\text{in }H_0.
\end{gathered} \label{e3.38}
\end{equation}
For convenience, we denote the above system by $(E_{\varepsilon},\bar{v})$.

\begin{lemma} \label{lem3.4}
Let $v(t)$ be the solution of $(E_{\varepsilon},\bar{v})$.
Then the following estimates can be established
\begin{gather}
\begin{aligned}
&\int^{\omega}_{0}|v'(\tau)|^2_{V^{*}_0}d\tau
+\varepsilon\int^{\omega}_{0}|v'(\tau)|^2_2d\tau \\
&\leq\frac{\omega}{\varepsilon}
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]
+\omega C^2_1\|f\|^2_{L^{\infty}(0,\omega;H)},
\end{aligned}\label{e3.39}
\\
\begin{aligned}
&\int^{\omega}_0|v(\tau)|^{2}_2d\tau\\
&\leq\frac{2\omega C^4_2}{\kappa^2}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
+|\Omega|(a^2_2+2a^2_1)+C^4_1\|f\|^2_{L^{\infty}(0,\omega;H)}\right],
\end{aligned}\label{e3.40}
\\
\begin{aligned}
\kappa|\nabla{v}(t)|^2_2
&\leq2\Big(\frac{\omega}{\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right] \\
&\quad +2C^2_1\Big(\omega+\frac{C^2_1C^2_2}{\kappa}\Big)
 \|f\|^2_{L^{\infty}(0,\omega;H)}
\end{aligned}\label{e3.41}
\end{gather}
for $t\in[0,\omega]$ and
\begin{equation}
\begin{aligned}
&\int^{\omega}_0|\Delta_{N}v(\tau)|^2_2d\tau \\
&\leq\frac{2\omega}{\kappa^2}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
+|\Omega|(a^2_2+2a^2_1)+C^4_1\|f\|^2_{L^2(0,\omega;V^{*}_0)}\right],
\end{aligned} \label{e3.42}
\end{equation}
where $A_3$ is the same as that in Lemma \ref{lem3.3}.
\end{lemma}

\begin{proof}
For any $\tau\in[0,\omega]$, we have
\begin{equation}
\begin{aligned}
&\int_{\Omega}\big|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3]\big|^2dx\\
&= a^2_3\int_{\Omega}(\bar{v}(\tau)+m(\tau))^6dx-\frac{a^2_3}{|\Omega|}
\Big[\int_{\Omega}(\bar{v}(\tau)+m(\tau))^3dx\Big]^2 \\
&\leq a^2_3\int_{\Omega}(\bar{v}(\tau)+m(\tau))^6dx.
\end{aligned}\label{e3.43}
\end{equation}
Similarly, for any $\tau\in[0,\omega]$, we have
\begin{gather}
\int_{\Omega}\left|\pi_0[a_2(\bar{v}(\tau)+m(\tau))^2]\right|^2dx
\leq a^2_2\int_{\Omega}(\bar{v}(\tau)+m(\tau))^4dx\,,\label{e3.44}
\\
\int_{\Omega}\left|\pi_0[a_1(\bar{v}(\tau)+m(\tau))]\right|^2dx
\leq a^2_1\int_{\Omega}(\bar{v}(\tau)+m(\tau))^2dx.\label{e3.45}
\end{gather}
It follows from Young's inequality that
\begin{gather}
a^2_2\int_{\Omega}(\bar{v}(\tau)+m(\tau))^4dx
\leq a^2_2\int_{\Omega}(\bar{v}(\tau)+m(\tau))^6dx
+\frac{4a^2_2}{27}|\Omega|\,,\label{e3.46}
\\
\begin{aligned}
a^2_1\int_{\Omega}(\bar{v}(\tau)+m(\tau))^2dx
&\leq a^2_1\int_{\Omega}(\bar{v}(\tau)+m(\tau))^6dx
 +\frac{2a^2_1}{3^{\frac{3}{2}}}|\Omega|\\
&<a^2_1\int_{\Omega}(\bar{v}(\tau)+m(\tau))^6dx
+\frac{2a^2_1}{3}|\Omega|.
\end{aligned} \label{e3.47}
\end{gather}
By \eqref{e3.46}-\eqref{e3.47}, for any $\tau\in[0,\omega]$, we have
\begin{equation}
\begin{aligned}
&\int_{\Omega}\left|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)
+m(\tau))^2+a_1(\bar{v}(\tau)+m(\tau))]\right|^2dx\\
&\leq 3(a^2_3+a^2_2+a^2_1)\int_{\Omega}(\bar{v}(\tau)
 +m(\tau))^6dx+|\Omega|\Big(\frac{4a^2_2}{9}+2a^2_1\Big)\\
&\leq 3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1).
\end{aligned} \label{e3.48}
\end{equation}
Multiplying the equation in \eqref{e3.38} by $v'$ and integrating by parts, we have
\begin{align*}
&|v'|^2_{V^{*}_0}+\varepsilon|v'|^2_2+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v}|^2_2\\
&= (-\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2
 +a_1(\bar{v}(\tau)+m(\tau))+a_0],v')_0\\
&\quad +\langle{F^{-1}_0\pi_0[f]},v'\rangle_{V_0,V^{*}_0}\\
&\leq \left|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2+a_1(\bar{v}(\tau)+m(\tau))]\right|_2|v'|_2\\
&\quad +|\pi_0[f]|_{V^{*}_0}|v'|_{V^{*}_0}\\
&\leq \frac{\varepsilon}{2}|v'|^2_2+\frac{1}{2\varepsilon}\left|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3
+a_2(\bar{v}(\tau)+m(\tau))^2+a_1(\bar{v}(\tau)+m(\tau))]\right|^2_2\\
&\quad +\frac{1}{2}|v'|^2_{V^{*}_0}+\frac{1}{2}|\pi_0[f]|^2_{V^{*}_0},
\end{align*}
i.e.,
\begin{equation}
\begin{aligned}
&\frac{1}{2}|v'|^2_{V^{*}_0}+\frac{\varepsilon}{2}|v'|^2_2
+\frac{\kappa}{2}\frac{d}{dt}|\nabla{v}|^2_2\\
&\leq \frac{1}{2\varepsilon}\big|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3
+a_2(\bar{v}(\tau)+m(\tau))^2\\
&\quad +a_1(\bar{v}(\tau)+m(\tau))]\big|^2_2
+\frac{1}{2}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned}\label{e3.49}
\end{equation}
It follows from \eqref{e3.48} that
\begin{equation}
\begin{aligned}
&\frac{1}{2}|v'|^2_{V^{*}_0}+\frac{\varepsilon}{2}|v'|^2_2
 +\frac{\kappa}{2}\frac{d}{dt}|\nabla{v}|^2_2 \\
&\leq\frac{1}{2\varepsilon}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
 +|\Omega|(a^2_2+2a^2_1)\right]
+\frac{1}{2}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned}\label{e3.50}
\end{equation}
Integrating \eqref{e3.50} on $(0,\omega)$ and from \eqref{e3.11}, we obtain
\begin{align*}
&\frac{1}{2}\int^{\omega}_{0}|v'(\tau)|^2_{V^{*}_0}d\tau
+\frac{\varepsilon}{2}\int^{\omega}_{0}|v'(\tau)|^2_2d\tau\\
&\leq\frac{\omega}{2\varepsilon}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
 +|\Omega|(a^2_2+2a^2_1)\right]
+\frac{\omega C^2_1}{2}\|f\|^2_{L^{\infty}(0,\omega;H)},
\end{align*}
which is \eqref{e3.39}.

Multiplying the equation of \eqref{e3.38} by $v$ and integrating by parts, we have
\begin{equation} 
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v|^2_{V^{*}_0}+\frac{\varepsilon}{2}\frac{d}{dt}|v|^{2}_2
+\kappa|\nabla{v}|^{2}_2\\
&= (-\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2\\
&\quad +a_1(\bar{v}(\tau)+m(\tau))+a_0],v)_0
 +\langle{F^{-1}_0\pi_0[f]},v\rangle_{V_0,V^{*}_0}\\
&\leq \left|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2
 +a_1(\bar{v}(\tau)+m(\tau))]\right|_2|v|_2\\
&\quad +|\pi_0[f]|_{V^{*}_0}|v|_{V^{*}_0}\\
&\leq \frac{\kappa}{2}|\nabla{v}|^2_2+\frac{C^2_2}{\kappa}
 \Big|\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2\\
&\quad  +a_1(\bar{v}(\tau)+m(\tau))]\Big|^2_2
+\frac{C^2_1C^2_2}{\kappa}|\pi_0[f]|^2_{V^{*}_0},
\end{aligned}\label{e3.51}
\end{equation}
where the last two inequality signs follow from Young's inequality and \eqref{e2.2}.
Combining \eqref{e3.51} with \eqref{e3.48}, we obtain
\begin{equation} 
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v|^2_{V^{*}_0}+\frac{\varepsilon}{2}\frac{d}{dt}|v|^{2}_2
 +\frac{\kappa}{2}|\nabla{v}|^{2}_2 \\
&\leq\frac{C^2_2}{\kappa}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
 +|\Omega|(a^2_2+2a^2_1)\right]+\frac{C^2_1C^2_2}{\kappa}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned}\label{e3.52}
\end{equation}
Integrating \eqref{e3.52} on $(0,\omega)$, we obtain
\begin{equation} 
\begin{aligned}
&\frac{\kappa}{2}\int^{\omega}_0|\nabla{v}(\tau)|^{2}_2d\tau
\leq\frac{\omega C^2_2}{\kappa}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
+|\Omega|(a^2_2+2a^2_1)\right] \\
&\quad +\frac{C^2_1C^2_2}{\kappa}\|\pi_0[f]\|^2_{L^2(0,\omega;{V^{*}_0})}.
\end{aligned}\label{e3.53}
\end{equation}
Therefore, from \eqref{e2.2} and \eqref{e3.11}, we have
\begin{align*}
\frac{\kappa}{2}\int^{\omega}_0|{v}(\tau)|^{2}_2d\tau
&\leq\frac{\kappa C^2_2}{2}\int^{\omega}_0|\nabla{v}(\tau)|^{2}_2d\tau\\
&\leq\frac{\omega C^4_2}{\kappa}\left\{\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]
+C^4_1\|f\|^2_{L^{\infty}(0,\omega;H)}\right\},
\end{align*}
which is \eqref{e3.40}.

For any $s,t\in[0,\omega]$ satisfying $s<t$, we integrate \eqref{e3.50} on $(s,t)$
and obtain
\begin{equation} \begin{aligned}
&\frac{1}{2}\int^t_s|v'(\tau)|^2_{V^{*}_0}d\tau
 +\frac{\varepsilon}{2}\int^t_s|v'(\tau)|^2_2d\tau
+\frac{\kappa}{2}|\nabla{v}(t)|^2_2\\
&\leq \frac{\omega}{2\varepsilon}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
 +|\Omega|(a^2_2+2a^2_1)\right]
+\frac{1}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau
 +\frac{\kappa}{2}|\nabla{v}(s)|^2_2.
\end{aligned}\label{e3.54}
\end{equation}
Deleting the first two terms on the left-hand side of \eqref{e3.54} and integrating
it on $(0,t)$ with respect to $s$, we have
\begin{align*}
\frac{\kappa t}{2}|\nabla{v}(t)|^2_2
&\leq\frac{\omega^2}{2\varepsilon}\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]
+\frac{\omega}{2}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau \\
&\quad +\frac{\kappa}{2}\int^{\omega}_0|\nabla{v}(s)|^2_2ds.
\end{align*}
In view of \eqref{e3.53}, we obtain
\begin{equation} \begin{aligned}
\frac{\kappa t}{2}|\nabla{v}(t)|^2_2
&\leq\omega\Big(\frac{\omega}{2\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right] \\
&\quad +\Big(\frac{\omega}{2}+\frac{C^2_1C^2_2}{\kappa}\Big)
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}.
\end{aligned} \label{e3.55}
\end{equation}
Let $t=\omega$, then \eqref{e3.55} can be rewritten as
\begin{align*}
\frac{\kappa\omega}{2}|\nabla{v}(\omega)|^2_2
&\leq\omega\Big(\frac{\omega}{2\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right] \\
&\quad+\Big(\frac{\omega}{2}+\frac{C^2_1C^2_2}{\kappa}\Big)
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)},
\end{align*}
i.e.,
\begin{equation}
\begin{aligned}
\frac{\kappa}{2}|\nabla{v}(\omega)|^2_2
&\leq\Big(\frac{\omega}{2\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]\\
&\quad +\Big(\frac{1}{2}+\frac{C^2_1C^2_2}{\kappa\omega}\Big)
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}.
\end{aligned}\label{e3.56}
\end{equation}
It follows from the periodic property that
\begin{equation} \begin{aligned}
\frac{\kappa}{2}|\nabla{v}(0)|^2_2
&\leq\Big(\frac{\omega}{2\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]\\
&\quad +\Big(\frac{1}{2}+\frac{C^2_1C^2_2}{\kappa\omega}\Big)
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}.
\end{aligned}\label{e3.57}
\end{equation}
Choosing $s=0$ in \eqref{e3.54}, by \eqref{e3.57} we obtain
\begin{equation} \begin{aligned}
&\frac{1}{2}\int^t_0|v'(\tau)|^2_{V^{*}_0}d\tau+\frac{\varepsilon}{2}\int^t_0|v'(\tau)|^2_2d\tau
+\frac{\kappa}{2}|\nabla{v}(t)|^2_2\\
&\leq \Big(\frac{\omega}{\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]\\
&\quad +\Big(1+\frac{C^2_1C^2_2}{\kappa\omega}\Big)
\|\pi_0[f]\|^2_{L^2(0,\omega;V^{*}_0)}.
\end{aligned}\label{e3.58}
\end{equation}
Dropping the first two terms on the left-hand side of \eqref{e3.58}
and from \eqref{e3.11}, we obtain
\begin{align*}
\frac{\kappa}{2}|\nabla{v}(t)|^2_2
&\leq\Big(\frac{\omega}{\varepsilon}+\frac{C^2_2}{\kappa}\Big)
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]\\
&\quad +C^2_1\Big(\omega+\frac{C^2_1C^2_2}{\kappa}\Big)
\|f\|^2_{L^{\infty}(0,\omega;H)},
\end{align*}
which is \eqref{e3.41}.

Now, multiplying the equation of \eqref{e3.38} by $-\Delta_{N}v$ and
integrating by parts, we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}|v|^2_2
+\frac{\varepsilon}{2}\frac{d}{dt}|\nabla{v}|^2_2+\kappa|\Delta_{N}v|^2_2\\
&\leq \left|\pi_0\left[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)+m(\tau))^2
+a_1(\bar{v}(\tau)+m(\tau))\right]\right|_2|\Delta_{N}v|_2\\
&\quad +|\pi_0[f]|_{V^{*}_0}|\Delta_{N}v|_{V^{*}_0}\\
&\leq \frac{1}{\kappa}\left|\pi_0\left[a_3(\bar{v}(\tau)+m(\tau))^3
+a_2(\bar{v}(\tau)+m(\tau))^2+a_1(\bar{v}(\tau)+m(\tau))\right]\right|^2_2\\
&\quad +\frac{C^2_1}{\kappa}|\pi_0[f]|^2_{V^{*}_0}
 +\frac{\kappa}{2}|\Delta_{N}v|^2_2.
\end{align*}
After a proper arrangement and with the aid of \eqref{e3.48}, we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|v|^2_2
+\frac{\varepsilon}{2}\frac{d}{dt}|\nabla{v}|^2_2+\frac{\kappa}{2}|\Delta_{N}v|^2_2\\
&\leq\frac{1}{\kappa}\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right]
+\frac{C^2_1}{\kappa}|\pi_0[f]|^2_{V^{*}_0}.
\end{aligned} \label{e3.59}
\end{equation}
Integrating \eqref{e3.59} on $(0,\omega)$, we have
\begin{align*}
\frac{\kappa}{2}\int^{\omega}_0|\Delta_{N}v(\tau)|^2_2dx
&\leq\frac{\omega}{\kappa}\left[3A^3_3(a^2_3+a^2_2+a^2_1)
 +|\Omega|(a^2_2+2a^2_1)\right] \\
&\quad +\frac{C^2_1}{\kappa}\int^{\omega}_0|\pi_0[f(\tau)]|^2_{V^{*}_0}d\tau,
\end{align*}
i.e.,
$$
\int^{\omega}_0|\Delta_{N}v(\tau)|^2_2dx
\leq\frac{2\omega}{\kappa^2}\Big[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)
+C^4_1\|f\|^2_{L^{\infty}(0,\omega;H)}\Big].
$$
Therefore, we obtain \eqref{e3.42}. The proof is complete.
\end{proof}

Now, we prove Theorem \ref{thm2.2} by the Schauder fixed point theorem.

\begin{proof}[Proof of Theorem \ref{thm2.2}]
Based on Lemma \ref{lem3.4}, we define a set
$$
Y_2:=\{\bar{v}\in{Y_1}:\sup_{t\in[0,w]}\kappa|\nabla{\bar{v}}|^2_2
+\varepsilon|\bar{v}|^2_{W^{1,2}(0,\omega;H_0)}
+|\bar{v}|^2_{W^{1,2}(0,\omega;V^{*}_0)}\leq{M}\},
$$
where
\begin{align*}
M:&= \Big[3\omega C^2_1+\frac{2C^4_1C^2_2}{\kappa}
+\frac{2\omega C^4_1C^4_2}{\kappa^2}(\varepsilon+C^2_1)\Big]
\|f\|^2_{L^{\infty}(0,\omega;H)}\\
&\quad+\Big[\frac{3\omega}{\varepsilon}+\frac{2C^2_2}{\kappa}
+\frac{2\omega C^4_2}{\kappa^2}(\varepsilon+C^2_1)\Big]
\left[3A^3_3(a^2_3+a^2_2+a^2_1)+|\Omega|(a^2_2+2a^2_1)\right].
\end{align*}
We can see that $Y_2$ is a non-empty compact convex subset of $L^2(0,\omega;H_0)$.
Since for any $\tau\in[0,\omega]$, we have
$$
|\pi_0[f(\tau)]|^2_{H_0}
= \int_{\Omega}|f(x,\tau)|^2dx
-\frac{1}{|\Omega|}\Big(\int_{\Omega}f(x,\tau)dx\Big)^2
\leq \|f\|^2_{L^{\infty}(0,\omega;H)}.
$$
Thus, it follows from \eqref{e3.48} and (H1) that
\[
-\pi_0[a_3(\bar{v}(\tau)+m(\tau))^3+a_2(\bar{v}(\tau)
+m(\tau))^2+a_1(\bar{v}(\tau)+m(\tau))]
+\pi_0[f]\in{L^{\infty}(0,\omega;H_0)}.
\]
 As a result, by Theorem \ref{thm3.1}, there exists a unique solution $v_{\varepsilon}$
for each $\bar{v}\in{Y_2}$. From Lemma \ref{lem3.4}, it is clear that
$v_{\varepsilon}\in{Y_2}$. Hence, the mapping $S$ defined by
$S(\bar{v})=v_{\varepsilon}$ maps $Y_2$ into itself.

Next, we show that $v_{\varepsilon}$ is a solution of \eqref{e3.38}
and that $S$ is continuous in $Y_2$ with respect to the topology of
 $L^2(0,\omega;H_0)$. Let $\{\bar{v}_{n}\}$ be any convergent sequence in $Y_2$
with respect to the topology of $L^2(0,\omega;H_0)$.
 We denote the limit of $\{\bar{v}_{n}\}$ by $\bar{v}$.
Let $\{v_n\}$ be the sequence of solutions corresponding to
$\{\bar{v}_n\}$. It follows from lemma \ref{lem3.4} that $v_n(n=1,2,\ldots)$ satisfy
\eqref{e3.39} and \eqref{e3.40}. Thus, we can find a $v_{\varepsilon}$ and
a subsequence of $\{v_n\}$ which is denoted by $\{v_{n_k}\}$ such that
\begin{equation}
v_{n_k}\to {v_{\varepsilon}}\quad \text{weakly in }W^{1,2}(0,\omega;H_0).\label{e3.60}
\end{equation}
Therefore, by \eqref{e3.41},
\begin{gather}
v_{n_k}\to {v_{\varepsilon}}\quad\text{in }L^{2}(0,\omega;H_0), \label{e3.61}\\
v'_{n_k}\to {v'_{\varepsilon}}\quad \text{weakly in } L^{2}(0,\omega;H_0).
\label{e3.62}
\end{gather}
Since $W^{1,2}(0,\omega;H_0)\cap{L^{\infty}(0,\omega;V_0)}
\hookrightarrow{C([0,\omega];H_0)}$ \cite[Section 8, Corollary 5]{15}
 and the embedding is compact, it is clear that
\begin{equation}
v_{n_k}\to {v_{\varepsilon}}\quad\text{in } C([0,\omega];H_0).\label{e3.63}
\end{equation}
Thus,
$$
v_{n_k}(0)\to {v_{\varepsilon}(0)},\quad
v_{n_k}(\omega)\to {v_{\varepsilon}(\omega)}
\quad\text{in }H_0,
$$
which implies $v_{\varepsilon}$ satisfies the periodic condition
\begin{equation}
v_{\varepsilon}(0)=v_{\varepsilon}(\omega).\label{e3.64}
\end{equation}

Since $F^{-1}_0$ is linear and selfadjoint, it holds for any
$\eta\in{D(\Delta_N)}$ that
\begin{equation}
\begin{aligned}
\int^{\omega}_0(F^{-1}_0v'_{n_k}(\tau)-F^{-1}_0v'_{\varepsilon}(\tau),\eta)_0d\tau
&= \int^{\omega}_0(F^{-1}_0(v'_{n_k}(\tau)-v'_{\varepsilon}(\tau)),\eta)_0d\tau\\
&= \int^{\omega}_0(v'_{n_k}(\tau)-v'_{\varepsilon}(\tau),F^{-1}_0\eta)_0d\tau.
\end{aligned}\label{e3.65}
\end{equation}
Therefore, from \eqref{e3.62}, we know that
\begin{equation}
F^{-1}_0v'_{n_k}\to {F^{-1}_0v'_{\varepsilon}}
\quad \text{weakly in $L^2(0,\omega;H_0)$ as }k\to \infty.\label{e3.66}
\end{equation}
Also, for any $\eta\in{D(\Delta_N)}$, we have
\begin{align*}
\Big|\int^{\omega}_0(-\Delta_Nv_{n_k}-(-\Delta_Nv_{\varepsilon}),\eta)_0d\tau\Big|
&= \Big|\int^{\omega}_0(v_{n_k}-v_{\varepsilon},-\Delta_N\eta)_0d\tau\Big|\\
&\leq \|v_{n_k}-v_{\varepsilon}\|_{L^{2}(0,\omega;H_0)}
|\Delta_N\eta|_{2}\omega^{1/2}.
\end{align*}
In view of \eqref{e3.61}, we obtain
\begin{equation}
-\Delta_Nv_{n_k}\to -\Delta_Nv_{\varepsilon} \quad
\text{weakly in $L^2(0,\omega;H_0)$ as }k\to \infty.\label{e3.67}
\end{equation}
For any $v(\tau)\in{L^2(0,\omega;H_0)}$, let
$\psi(v(\tau))=\pi_0[a_3(v(\tau)+m(\tau))^3]$.
Then for any $v_1(\tau),v_2(\tau)\in{L^2(0,\omega;H_0)}$,
\begin{equation} \begin{aligned}
&\int^{\omega}_0(\psi(v_1(\tau))-\psi(v_2(\tau)),v_1(\tau)-v_2(\tau))_0d\tau\\
&= a_3\int^{\omega}_0\int_{\Omega}\left[v_1(\tau)-v_2(\tau)\right]^2
\Big\{\Big[(v_1(\tau)+m(\tau))+\frac{1}{2}(v_2(\tau)+m(\tau))\Big]^2\\
&\quad +\frac{3}{4}[v_2(\tau)+m(\tau)]^2\Big\}dx\,d\tau
\geq 0.
\end{aligned}\label{e3.68}
\end{equation}
Therefore, $\psi$ is a monotone operator in $L^2(0,\omega;H_0)$.
 Since $a_3(v(\tau)+m(\tau))^3$ is continuous with respect to
$v(\tau)$, it is  easy to prove that $\psi(\cdot)$ is hemicontinuous
\cite[Chapter II, Definition 1.3]{1} in $L^2(0,\omega;H_0)$.
In addition, it is clear to see that $\psi$ is everywhere defined in
$L^2(0,\omega;H_0)$. Thus, by \cite[Chapter II, Theorem 1.3]{1},
 $\psi(\cdot)$ is maximal monotone in $L^2(0,\omega;H_0)$.

It follows from the definition of $Y_2$ and \eqref{e3.43} that
$$
\|\pi_0[a_3(\bar{v}_{n_k}+m(\tau))^3]\|^2_{L^2(0,\omega;H_0)}\leq a^2_3A^3_3\omega.
$$
Thus, there exist a $\gamma(\tau)\in{L^2(0,\omega;H_0)}$ and a subsequence of
 $\{\bar{v}_{n_k}\}$ which is still denoted by $\{\bar{v}_{n_k}\}$ such that
\begin{equation}
\pi_0[a_3(\bar{v}_{n_k}(\tau)+m(\tau))^3]=\psi(\bar{v}_{n_k}(\tau))
\to {\gamma(\tau)} \quad
\text{weakly in }L^2(0,\omega;H_0)\label{e3.69}
\end{equation}
as $k\to \infty$.
It follows from \eqref{e3.69}, the maximal monotonicity of $\psi$ and the
\cite[Theorem A]{17}, we obtain
$\psi(\bar{v})=\gamma$, i.e.,
\begin{equation}
\pi_0[a_3(\bar{v}_{n_k}+m(\tau))^3]\to \pi_0[a_3(\bar{v}
+m(\tau))^3]\quad \text{weakly in }L^2(0,\omega;H_0).\label{e3.70}
\end{equation}
From \cite[Section 8, Corollary 5]{15},
we  know that
$W^{1,2}(0,\omega;H_0)\cap{L^{\infty}(0,\omega;V_0)}
\hookrightarrow{C([0,\omega];L^4(\Omega)\cap H_0)}$ and the embedding is compact.
Thus,
\begin{equation}
v_{n_k}\to {v_{\varepsilon}}\quad\text{in }C([0,\omega];L^4(\Omega)\cap H_0).
\label{e3.71}
\end{equation}
Since
\begin{align*}
&a^2_2\int^{\omega}_0\int_{\Omega}\left[(\bar{v}_{n_k}(\tau)+m(\tau))^2
-(\bar{v}(\tau)+m(\tau))^2\right]^2\,dx\,d\tau\\
&= a^2_2\int^{\omega}_0\int_{\Omega}(\bar{v}_{n_k}(\tau)-\bar{v}(\tau))^2(\bar{v}_{n_k}(\tau)+\bar{v}(\tau)+2m(\tau))^2\,dx\,d\tau\\
&\leq a^2_2\int^{\omega}_0\Big[\int_{\Omega}(\bar{v}_{n_k}(\tau)-\bar{v}(\tau))^4dx
\Big]^{1/2}
\Big[\int_{\Omega}(\bar{v}_{n_k}(\tau)+\bar{v}(\tau)+2m(\tau))^4dx\Big]^{1/2}d\tau\\
&\leq a^2_2\|\bar{v}_{n_k}(\tau)-\bar{v}(\tau))\|^2_{C([0,\omega];L^4(\Omega)\cap H_0)}
\int^{\omega}_0\Big[\int_{\Omega}(\bar{v}_{n_k}(\tau)+\bar{v}(\tau)
 +2m(\tau))^4dx\Big]^{1/2}d\tau,
\end{align*}
where the last second inequality sign follows from the H\"{o}lder inequality.
From \eqref{e3.71}, we see that
$$
a_2(\bar{v}_{n_k}(\tau)+m(\tau))^2\to {a_2(\bar{v}(\tau)+m(\tau))^2}\quad
\text{in $L^2(0,\omega;H_0)$ as }k\to \infty.
$$
Then, it is easy to prove that
\begin{equation}
\pi_0[a_2(\bar{v}_{n_k}(\tau)+m(\tau))^2]\to {\pi_0[a_2(\bar{v}(\tau)+m(\tau))^2]}
\quad \text{in }L^2(0,\omega;H_0) \label{e3.72}
\end{equation}
as $k\to \infty$.
Similarly, \eqref{e3.63} implies
\begin{equation}
\pi_0[a_1(\bar{v}_{n_k}(\tau)+m(\tau))]\to {\pi_0[a_1(\bar{v}(\tau)+m(\tau))]}
\quad \text{in }L^2(0,\omega;H_0)
\label{e3.73}
\end{equation}
as $k\to \infty$.

Consequently, with the help of \eqref{e3.62}, \eqref{e3.66}, \eqref{e3.67},
\eqref{e3.70}, \eqref{e3.72} and \eqref{e3.73}, for any $\eta\in{D(\Delta_N)}$,
we take $k\to \infty$ on both sides of the equality
\begin{align*}
&\int^{\omega}_0(F^{-1}_0v'_{n_k}(\tau),\eta)_0d\tau
+\int^{\omega}_0(\varepsilon{v'_{n_k}(\tau)},\eta)_0d\tau
+\int^{\omega}_0(-\kappa\Delta_Nv_{n_k}(\tau),\eta)_0d\tau\\
&= \int^{\omega}_0(\pi_0[g(\bar{v}_{n_k}(\tau)+m(\tau))],\eta)_0d\tau
+\int^{\omega}_0(F^{-1}_0f,\eta)_0d\tau,
\end{align*}
and then
\begin{align*}
&\int^{\omega}_0(F^{-1}_0v'_{\varepsilon}(\tau),\eta)_0d\tau
+\int^{\omega}_0(\varepsilon{v'_{\varepsilon}(\tau)},\eta)_0d\tau
+\int^{\omega}_0(-\kappa\Delta_Nv_{\varepsilon}(\tau),\eta)_0d\tau\\
&= \int^{\omega}_0(\pi_0[g(\bar{v}(\tau)+m(\tau))],\eta)_0d\tau
+\int^{\omega}_0(F^{-1}_0f,\eta)_0d\tau.
\end{align*}
This implies that $v_{\varepsilon}=S(\bar{v})$ is a unique solution
of \eqref{e2.9}. As a result, from \eqref{e3.61}, $S$ is continuous
in $Y_2$ with respect to the topology of $L^2(0,\omega;H_0)$.
By the Schauder fixed theorem, we can see that $S$ has at least one
fixed point in $Y_2$. The proof is complete.
\end{proof}

\section{Proof of main results}

In this section, based on Lemma \ref{lem3.3}, we prove Theorem \ref{thm2.1} 
by taking the limit as $\varepsilon\to 0$.

\begin{proof}[Proof of Theorem \ref{thm2.1}]
By Lemma \ref{lem3.3}, we can see that the constants on the right-hand side
of \eqref{e3.7}-\eqref{e3.10} are independent of $\varepsilon$.
Thus, it follows from \eqref{e3.7} and \eqref{e3.8} that
\begin{equation}
v_{\varepsilon}\in{W^{1,2}(0,\omega;V^{*}_0)\cap{L^{\infty}(0,\omega;V_0)}},
\label{e4.1}
\end{equation}
and there exists a $v\in{W^{1,2}(0,\omega;V^{*}_0)}\cap{L^{\infty}(0,\omega;V_0)}$
such that
\begin{gather}
v_{\varepsilon}\to {v}\quad \text{weakly in }W^{1,2}(0,\omega;V^{*}_0)\text{ as }
 \varepsilon\to 0,\label{e4.2}\\
v_{\varepsilon}\to {v}\quad \text{weakly star in }
L^{\infty}(0,\omega;V_0)\text{ as }\varepsilon\to 0,\label{e4.3}\\
v_{\varepsilon}\to {v}\quad\text{in }L^{2}(0,\omega;H_0)\text{ as }\varepsilon\to 0,
\label{e4.4}\\
v'_{\varepsilon}\to {v'}\quad \text{weakly in }L^{2}(0,\omega;V^{*}_0)\text{ as }
\varepsilon\to 0.\label{e4.5}
\end{gather}
Since $W^{1,2}(0,\omega;V^{*}_0)\cap{L^{\infty}(0,\omega;V_0)}
\hookrightarrow{C([0,\omega];H_0)}$ \cite[Section 8, Corollary 5]{15}
 and the embedding is compact, it follows from \eqref{e4.2} and \eqref{e4.3} that
$$
v_{\varepsilon}\to {v}\quad\text{in }C([0,\omega];H_0)\text{ as }\varepsilon\to 0.
$$
Thus,
$$
v_{\varepsilon}(0)\to {v(0)},\quad v_{\varepsilon}(\omega)\to {v(\omega)}
\quad\text{in }H_0,
$$
which implies
\begin{equation}
v(0)=v(\omega).\label{e4.6}
\end{equation}
Similarly as \eqref{e3.65}, for any $\eta\in{D(\Delta_N)}$, we have
\[
\int^{\omega}_0\langle{F^{-1}_0v'_{\varepsilon}(\tau)
 -F^{-1}_0v'(\tau),\eta}\rangle_{V_0,V^{*}_0}d\tau
=\int^{\omega}_0\langle{v'_{\varepsilon}(\tau)
 -v'(\tau),F^{-1}_0\eta}\rangle_{V^{*}_0,V_0}d\tau.
\]
By \eqref{e4.5}, we obtain
\begin{equation}
F^{-1}_0v'_{\varepsilon}\to {F^{-1}_0v'}\quad \text{weakly in }
L^{2}(0,\omega;V_0)\text{ as }\varepsilon\to 0.\label{e4.7}
\end{equation}
Furthermore, for any $\eta\in{D(\Delta_N)}$, we have
\begin{align*}
\int^{\omega}_0(-\Delta_Nv_{\varepsilon}(\tau)-(-\Delta_Nv(\tau)),\eta)_0d\tau
&= \int^{\omega}_0(v_{\varepsilon}(\tau)-v(\tau),-\Delta_N\eta)_0d\tau\\
&\leq \|v_{\varepsilon}-v\|_{L^{2}(0,\omega;H_0)}|\Delta_N\eta|_{2}\omega^{1/2}.
\end{align*}
It follows from \eqref{e4.4} that
\begin{equation}
-\Delta_Nv_{\varepsilon}\to -\Delta_Nv\quad \text{weakly in }
L^{2}(0,\omega;H_0)\text{ as }\varepsilon\to 0.\label{e4.8}
\end{equation}
From \eqref{e3.10} and the similar arguments as \eqref{e3.70}, we obtain
\begin{equation}
\pi_0[a_3(v_{\varepsilon}(\tau)+m(\tau))^3]
\to \pi_0[a_3(v(\tau)+m(\tau))^3]\quad \text{weakly in }
L^{2}(0,\omega;H_0)\label{e4.9}
\end{equation}
as $\varepsilon\to 0$.

By \cite[Section 8, Corollary 5]{15}, we see that both 
$W^{1,2}(0,\omega;V^{*}_0)\cap{L^{\infty}(0,\omega;V_0)}
\hookrightarrow C([0,\omega];L^4(\Omega)\cap H_0)$
and
$W^{1,2}(0,\omega;V^{*}_0)\cap{L^{\infty}
(0,\omega;V_0)}\hookrightarrow{C([0,\omega];H_0)}$ are compact.
Therefore, being similar as \eqref{e3.72} and \eqref{e3.73}, we have
\begin{equation}
\pi_0[a_2(v_{\varepsilon}(\tau)+m(\tau))^2]\to {\pi_0[a_2(v(\tau)+m(\tau))^2]}
\quad \text{in }L^2(0,\omega;H_0)\text{ as }k\to \infty
\label{e4.10}
\end{equation}
and
\begin{equation}
\pi_0[a_1(v_{\varepsilon}(\tau)+m(\tau))]\to {\pi_0[a_1(v(\tau)+m(\tau))]}
\quad \text{in } L^2(0,\omega;H_0)\text{ as } k\to \infty.\label{e4.11}
\end{equation}
With the help of \eqref{e4.7}-\eqref{e4.11}, for any $\eta\in{D(\Delta_N)}$,
we take $\varepsilon\to 0$ on both sides of the equation
\begin{align*}
&\varepsilon\int^{\omega}_0\langle{v'_{\varepsilon}(\tau),\eta}
\rangle_{V^{*}_0,V_0}d\tau\\
&= \int^{\omega}_0\Big(F^{-1}_0f(\tau)-\pi_0[g(v_{\varepsilon}(\tau)+m(\tau))]
+\kappa\Delta_Nv_{\varepsilon}(\tau),\eta\Big)_0d\tau \\
&\quad -\int^{\omega}_0\langle{F^{-1}_0v'_{\varepsilon}(\tau),\eta}
\rangle_{V_0,V^{*}_0}d\tau,
\end{align*}
and then get that for any $\eta\in{D(\Delta_N)}$,
\begin{align*}
0&= \int^{\omega}_0(F^{-1}_0f(\tau)-\pi_0[g(v(\tau)+m(\tau))]
+\kappa\Delta_Nv(\tau),\eta)_0d\tau\\
&\quad -\int^{\omega}_0\langle{F^{-1}_0v'(\tau),\eta}\rangle_{V_0,V^{*}_0}
d\tau\end{align*}
holds, which together with \eqref{e4.6} implies that $v$ is a solution of the
problem
\begin{gather*}
F^{-1}_0v'(\tau)-\kappa\Delta_Nv(\tau)+\pi_0[g(v(\tau)+m(\tau))]
=F^{-1}_0f(\tau),\quad 0<\tau<\omega,\\
v(0)=v(\omega).
\end{gather*}
As a result, from the equivalence of \eqref{e1.2} and \eqref{e2.8},
we know that $u(x,\tau)=v(x,\tau)+m(\tau)$ is the solution of
\eqref{e1.2}-\eqref{e1.4}.
\end{proof}

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\end{document}