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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 77, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/77\hfil Periodic solutions of Hamiltonian systems]
{Existence of periodic solutions for non-autonomous second-order \\
 Hamiltonian systems}

\author[Y. Wu, T. An  \hfil EJDE-2013/77\hfilneg]
{Yue Wu, Tianqing An}  

\address{Yue Wu \newline
College of Science, Hohai University,
Nanjing 210098, China}
\email{wyue007@126.com}

\address{Tianqing An \newline
College of Science, Hohai University,
Nanjing 210098,  China}
\email{antq@hhu.edu.cn}

\thanks{Submitted January 11, 2013. Published March 19, 2013.}
\subjclass[2000]{34C25, 58F05}
\keywords{Periodic solution; Hamiltonian systems; critical point;
\hfill\break\indent variational method}

\begin{abstract}
 The purpose of this paper is to study the existence of periodic
 solutions for a class of non-autonomous second order Hamiltonian systems.
 New results are obtained by using the least action principle and
 the minimax methods, without the so-called Ahmad-Lazer-Paul type condition.
\end{abstract}

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

\section{Introduction and main results}

Consider the second-order Hamiltonian system
\begin{equation}\label{eq}
\begin{gathered}
         \ddot u(t)=\nabla F\big(t,u(t)\big),\\
      u(T)-u(0)=\dot{u}(T)-\dot{u}(0)=0,
  \end{gathered}
   \end{equation}
where $T>0$ and $ F:[0,T]\times\mathbb{R}^{N}\to\mathbb{R}$
satisfies the following assumption:
\begin{itemize}
\item[(A)] $F(t,x)$ is measurable in $t$ for every
$x\in\mathbb{R}^{N}$, continuously differentiable in $x$ for a.e.
$t\in[0,T]$, and there exist $a\in C(\mathbb{R}^{+},\mathbb{R}^{+})$,
$b\in \mathscr{L}^1 (0,T;\mathbb{R}^{+}) $ such that
$$
|F(t,x)|\leq a(|x|)b(t),\quad |\nabla F(t,x)|\leq a(|x|)b(t)
$$
for all $x \in \mathbb{R}^{N}$ and a.e. $t \in [0,T]$.
\end{itemize}
The corresponding functional $\varphi : H^1_T \to \mathbb{R}$,
$$
\varphi (u)=\frac{1}{2}\int^T_0|\dot{u}(t)|^2dt
+    \int_0^T   F \big( t,u(t) \big)dt
$$
is continuously differentiable and weakly
lower semi-continuous on $H^1_T$ (see \cite{MW89}), where
$H_T^1$ is the usual Sobolev space with the norm
$$
\|u\|=\Big[ \int^T_0|u(t)|^2dt+   \int_0^T    |\dot{u}(t)|^2dt
\Big]^{1/2}.
$$
It is well know that the solutions of problem \eqref{eq}
correspond to the critical points of $\varphi$.

Problem \eqref{eq} has been extensively studied in the past
thirty years; see for example the references in this article.
 Under some suitable
solvability conditions, such as the coercivity condition (cf.
\cite{Berg77}), the periodicity condition (cf. \cite{Mawhin81}), the
convexity condition (cf. \cite{Mawhin87}), the subadditive condition
(cf. \cite{Tang95}), the existence and multiplicity results are
obtained. We note that in many contributions (for example, see
\cite{An11,MaTang02,Tang98A, WuXP99,YangRG08,ZhangXY08,ZhaoFK04}),
the following condition was assumed:
\begin{equation}\label{eq2}
  \lim_{|x|\to \infty}|x|^{-2\alpha}\int^T_0   F(t,x)dt=\infty
\quad \text{or}\quad -\infty,
\end{equation}
where $\alpha $ is a constant. In this article, instead of
\eqref{eq2}, we discuss the existence of periodic solutions of
\eqref{eq} under a weak condition that
$\liminf_{|x| \to \infty }|x|^{-2\alpha}\int^T_0   F(t,x)dt$ or
$\limsup_{|x| \to \infty }|x|^{-2\alpha}\int^T_0   F(t,x)dt$
 has appropriate lower or upper bound.

 Our main results are as follows:


\begin{theorem}\label{thm1}
Suppose that $F(t,x)=F_1(t,x)+F_2(x)$, where $F_1$ and $F_2$
satisfy assumption {\rm{(A)}} and the
 following conditions:
\begin{itemize}
\item[(F1)] there exist $f,g \in \mathscr{L}^1(0,T;\mathbb{R}^{+})$ and $\gamma \in [0,1)$
 such that
    $$
|\nabla F_1(t,x)|\leq f(t)|x|^{\gamma}+g(t),
$$
    for all $x\in \mathbb{R}^{N}$ and a.e. $t \in [0,T]$;

\item[(F2)] there exist constants $r >0$ and $\alpha \in [0,2)$ such that
    $$
(\nabla F_2(x)- \nabla F_2(y),x-y )\geq -r|x-y|^{\alpha},
$$
    for all $x,y \in \mathbb{R}^{N}$;

item[(F3)]
$$
\liminf_{|x|\to \infty}|x|^{-2\gamma}\int_0^T F(t,x)\,dt
\geq \frac{T^2}{8\pi^2} \int_0^T f^2(t)\,dt.
$$
\end{itemize}
Then problem \eqref{eq} has at least one periodic solution which
minimizes $\varphi$ on $H_T^1$.
\end{theorem}

\begin{theorem}\label{thm2}                                                              % Th 1.2
Suppose that $F(t,x)=F_1(t,x)+F_2(x)$, where $F_1$ and $F_2$
satisfy assumptions {\rm (A), (F1), (F2)} and the following
conditions:
\begin{itemize}
\item[(F4)] there exist $\delta \in [0,2)$ and $C>0$ such that
         \begin{equation*}
              \left( \nabla F_2(x)-\nabla F_2(y),x-y \right ) \leq C|x-y|^{\delta},
         \end{equation*}
   for all $x,y \in \mathbb{R}^N$;

\item[(F5)]
   $$
\limsup_{|x|\to \infty}|x|^{-2\gamma}\int_0^T F(t,x)\,dt
\leq -\frac{3T^2}{8\pi^2}\int_0^T f^2(t)\,dt.
$$
\end{itemize}
Then problem \eqref{eq} has at least one periodic solution which
minimizes $\varphi$ on $H_T^1$.
\end{theorem}

\begin{theorem}\label{thm3}                                                    % Th 1.3
Suppose that $F(t,x)=F_1(t,x)+F_2(x)$, where $F_1$ and $F_2$
satisfy assumptions {\rm (A), (F1)}, and the following
conditions:
\begin{itemize}
\item[(F2')]   there exists a constant $0<r<4\pi^2/T^2$, such that
         \begin{equation*}
               \left (\nabla F_2(x)- \nabla F_2(y),x-y \right )\geq -r|x-y|^2;
         \end{equation*}
    for all $x,y \in \mathbb{R}^N$;

\item[(F3')]
    $$\liminf_{|x|\to \infty}|x|^{-2\gamma}\int_0^T F(t,x)\,dt
\geq \frac{T^2}{2(4\pi^2-rT^2)}\int_0^T f^2(t)\,dt.
$$
\end{itemize}
Then problem \eqref{eq} has at least one periodic solution which
minimizes $\varphi$ on $H_T^1$.
\end{theorem}

\begin{theorem}\label{thm4}                                                              % Th 1.4
Suppose that $F=F_1+F_2$, where $F_1$ and $F_2$ satisfy
assumptions {\rm (A), (F1)} and the following conditions:
\begin{itemize}
\item[(F6)] there exist $k \in \mathscr{L}^1 (0,T;\mathbb{R}^+)$ and
$(\lambda,\mu)$-subconvex
        potential $G:\mathbb{R}^N \to \mathbb{R}$ with $\lambda>1/2$ and
$0<\mu < 2\lambda^2$, such that
        \begin{equation*}
           \left( \nabla F_2(t,x),y \right ) \geq -k(t)G(x-y),
        \end{equation*}
    for all $x,y \in \mathbb{R}^N$ and a.e. $t\in [0,T]$;

\item[(F7)]
\begin{gather*}
\limsup_{|x|\to \infty}|x|^{-2\gamma}\int_0^T F_1(t,x)\,dt
\leq -\frac{3T^2}{8\pi^2}\int_0^T f^2(t)\,dt,
\\
\limsup_{|x|\to\infty}|x|^{-\beta}\int_0^TF_2(t,x)\,dt
\leq -8\mu \max_{|s|\leq 1}G(s)\int_0^T k(t)\,dt ,
\end{gather*}
where $\beta=\log_{2\lambda}(2\mu)$.
\end{itemize}
Then problem \eqref{eq} has at least one periodic solution which
minimizes $\varphi$ on $H_T^1$.
\end{theorem}

\begin{remark} \label{rmk1.5}\rm
Theorems \ref{thm1}--\ref{thm3}   extend some existing
results. On the one hand, we decomposed the potential $F$ into $F_1$
and $F_2$. On the other hand, we weaken the so-called
Ahmad-Lazer-Paul type condition \eqref{eq2} as conditions
(F3), (F5) and (F3').
Note that \cite[Theorem 2]{YangRG08} and
\cite[Theorem 1]{MaTang02} are the direct corollaries of
Theorem \ref{thm1} and Theorem \ref{thm3} respectively.
If $F_2=0$, \cite[Theorems 1 and  2]{TangXH10} are special cases of
 Theorem \ref{thm1} and Theorem \ref{thm2} respectively.
Some examples of $F$ are given in section 3, which are
not covered in the references. Moreover, our Theorem \ref{thm4} is a new result.
\end{remark}

\section{Proof of Theorems}

For $u \in H_T^1$, let
\[
\bar{u}=\frac{1}{T}\int_0^T u(t)\,dt ,\quad \tilde{u}(t)=u(t)-\bar{u}.
\]
The following inequalities are well known (cf. \cite{MW89}):
\begin{gather*}
\|\tilde{u}\|_{\infty}^2\leq \frac{T}{12}\|\dot u\|^2_{L^2}
\quad\text{(Sobolev's inequality)},\\
\|\tilde{u} \|_{L^2}^2\leq
\frac{T^2}{4\pi^2}\|\dot u\|^2_{L^2}\quad \text{(Wirtinger's inequality)}
\end{gather*}
For convenience, we denote
$$
M_1=\Big( \int_0^T f^2(t)\,dt  \Big)^{1/2},\quad
M_2=\int_0^T f(t)\,dt,\quad
M_3=\int_0^T g(t)\,dt.
$$
Now we give the proofs of the main results.

\begin{proof}[Proof of Theorem \ref{thm1}]
  By (F3), we can choose an $a_1>T^2/(4\pi^2)$ such that
  \begin{equation}\label{3.1}
    \liminf_{|x| \to \infty}|x|^{-2\gamma}\int_0^T F(t,x)\,dt>\frac{a_1}{2}M_1^2.
  \end{equation}
By (F1) and the Sobolev's inequality, for any $u \in H_T^1$,
\begin{equation} \label{3.2}
\begin{aligned}
&\big|\int_0^T[F_1(t,u(t))-F_1(t,\bar{u})]\,dt\big|\\
&=\big| \int_0^T{\int_0^1   \left(\nabla F_1 \big(t,\bar{u}+s\tilde{u}(t)
\big),\tilde{u}(t)\right) \,ds}\,dt \big|  \\
&\leq \int_0^T{\int_0^1    f(t) | \bar u + s \tilde {u}(t)
|^{\gamma} |\tilde{u}(t)| \,ds}\,dt +
\int_0^T{\int_0^1    g(t) |\tilde{u}(t)| \,ds}\,dt
\\
&\leq  |\bar u|^{\gamma} \Big(\int_0^T f^2(t)\,dt \Big)^{1/2}
  \Big(\int_0^T |\tilde u(t)|^2\,dt \Big)^{1/2}\\
&\quad   + \|\tilde u\|^{\gamma +1}_{\infty}   \int_0^T f(t)\,dt
 + \|\tilde u\|_{\infty}   \int_0^T  g(t)\,dt \\
&\leq \frac{1}{2{{a}_1}}\|\tilde{u}\|_{L^2}^2+
 \frac{{{a}_1}}{2}M_1^2{{|\bar{u}|}^{2\gamma }}
 +{{M}_2}\|\tilde{u}\|_{\infty }^{\gamma +1}+{{M}_3}
 {{\|\tilde{u}\|}_{\infty }} \\
 &\leq  \frac{{{T}^2}}{8{{\pi }^2}{{a}_1}}
\|\dot{u}\|_{L^2}^2 + \frac{{{a}_1}}{2}M_1^2{{
|\bar{u}|}^{2\gamma }}
+ {{\big(\frac{T}{12}\big) }^{  \frac{\gamma +1}{2}}}    {{M}_2}
 \|\dot{u}\|_{L^2}^{\gamma +1}
+ {{\big(\frac{T}{12}\big) }^{1/2}}
   {{M}_3}{{\|\dot{u}\|}_{L^2}}
\end{aligned}
\end{equation}
Similarly, by (F2) and the Sobolev's inequality, for any $u \in
H_T^1$,
\begin{equation} \label{3.3}
\begin{aligned}
 \int_0^T   {[ {{F}_2}(u(t))-{{F}_2}(\bar{u})]dt}
 &=   \int_0^T   {\int_0^1  {\frac{1}{s}\left( \nabla {{F}_2}
 (\bar{u}+s\tilde{u}(t))-\nabla {{F}_2}(\bar{u}),s\tilde{u}(t) \right)}\,ds\,dt} \\
 &\ge - \int_0^T   {\int_0^1   rs^{\alpha -1}}
 {{\left| \tilde{u}(t) \right|}^{\alpha }}\,ds\,dt\\
 &\ge -\frac{rT}{\alpha }
 \|\tilde{u}\|_{\infty }^{\alpha }\\
 &\ge -\frac{rT}{\alpha }{{\big(\frac{T}{12}\big)}^{{\alpha }/{2}}}
 \|\dot{u}\|_{L^2}^{\alpha }
\end{aligned}
\end{equation}
It follows from \eqref{3.2} and \eqref{3.3} that
\begin{align*}
  \varphi (u)
&=\frac{1}{2}\|\dot{u}\|_{L^2}^2 +   \int_0^T   {\left[ {{F}_1}
 (t,u(t))-{{F}_1}(t,\bar{u}) \right]dt}\\
&\quad +   \int_0^T   {\left[ {{F}_2}(u(t))-{{F}_2}
 (\bar{u}) \right]dt}+   \int_0^T   {F(t,\bar{u})}dt \\
 &\ge \Big( \frac{1}{2}-\frac{{{T}^2}}{8{{\pi }^2}{{a}_1}} \Big)
 \|\dot{u}\|_{L^2}^2-{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}
 {{M}_2}\|\dot{u}\|_{L^2}^{\gamma +1}-{{\big(\frac{T}{12}\big)}^{1/2}}
 {{M}_3}{{\|\dot{u}\|}_{L^2}}\\
 &\quad -\frac{rT}{\alpha }{{\big(\frac{T}{12}\big)}^{\alpha/2}}
 \|\dot{u}\|_{L^2}^{\alpha } +{{|\bar{u}|}^{2\gamma }}
\Big( {{|\bar{u}|}^{-2\gamma }}   \int_0^T
{F(t,\bar{u})dt}-\frac{{{a}_1}}{2}M_1^2 \Big)
\end{align*}
for all $u \in H^1_T$, which implies that $\varphi (u) \to \infty $
as $\|u\| \to \infty $,   due to \eqref{3.1} and $\gamma <1$.

By the least action principle
(see \cite[Theorem 1.1 and Corollary 1.1]{MW89}), the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
\emph{Step 1.} We firstly show that $\varphi$ satisfies the (PS)
condition. Suppose that $\{u_n\}$ is a (PS) sequence, that is,
${\varphi }'({{u}_n})\to 0 $ as $n\to 0$ and $\left\{ \varphi
\left( {{u}_n} \right) \right\}$ is bounded. By (F5), we can
choose an ${{a}_2}>T^2/(4\pi^2)$ such that
\begin{equation}\label{3.4}
  \underset{|x|\to \infty }{\mathop{\lim \sup }}\,{{|x|}
^{-2\gamma }}\int_0^T{F\left( t,x \right)dt}
<-\Big(\frac{{{a}_2}}{2}+ \frac{\sqrt{{{a}_2}}T}{2\pi }
\Big)M_1^2.
\end{equation}
In a way similar to the proof of Theorem \ref{thm1}, one has
\begin{equation}
\begin{aligned}
&\int_0^T {\left( \nabla{{F}_1}(t,{{u}_n}(t)),{{{\tilde{u}}}_n}(t) \right)
\,dt}\\
&\le \frac{{{T}^2}}{8{{\pi }^2}{{a}_2}}
  \left\| {{{\dot{u}}}_n} \right\|_{L^2}^2+\frac{{{a}_2}}
  {2}M_1^2{{|\bar{u}_n|}^{2\gamma}}
 +{{\big(\frac{T}{12}\big)}^{\frac{( \gamma +1)}{2}}}
  {{M}_2}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma+1}
  +{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{\left\| {{{\dot{u}}}_n}
 \right\|}_{L^2}}
\end{aligned}\label{3.5}
\end{equation}
and
\[
\int_0^T   {\left( \nabla
{{F}_2}({{u}_n}(t)),{{{\tilde{u}}}_n}(t) \right)dt}
  \ge-\frac{rT}{\alpha}{{\big(\frac{T}{12}\big)}^{{\alpha }/{2}}}  \left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\alpha }
  \int_0^T{r(t)dt}
\]
 for all $n$. Hence one has
\begin{equation}
\begin{aligned}
\left\| {{{\tilde{u}}}_n} \right\|&\ge \left( {\varphi
}'({{u}_n}),{{{\tilde{u}}}_n} \right)\\
&=\left\| {{{\dot{u}}}_n}\right\|_{L^2}^2+\int_0^T
 {\left( \nabla F(t,{{u}_n}(t)),{{{\tilde{u}}}_n}(t) \right)}dt \\
 & \ge \Big( 1-\frac{{{T}^2}}{8{{\pi }^2}{{a}_2}} \Big)
 \left\| {{{\dot{u}}}_n}\right\|_{L^2}^2-\frac{{{a}_2}}{2}M_1^2
 {{\left| {{{\bar{u}}}_n}\right|}^{2\gamma }}
 -{{\big(\frac{T}{12}\big)}^{\frac{\left(\gamma +1 \right)}{2}}}    {{M}_2}
 \left\|{{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1}\\
 &\quad -{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}
 -\frac{rT}{\alpha}{{\big(\frac{T}{12}\big)}^{{\alpha}/{2}}}   \left\|{{{\dot{u}}}_n}\right\|_{L^2}^{\alpha
 }
\end{aligned}\label{3.6}
\end{equation}
for large $n$. It follows from Wirtinger's inequality that
\begin{equation}\label{3.7}
\left\| {{{\tilde{u}}}_n} \right\|\le
 \frac{{{\left( {{T}^2}+4{{\pi }^2} \right)}^{{1}/{2}\;}}}{2\pi }{{\left\| {{{\dot{u}}}_n}
 \right\|}_{L^2}}.
\end{equation}
By \eqref{3.6} and \eqref{3.7},
\begin{equation}
\begin{aligned}
 \frac{{{a}_2}}{2}M_1^2{{|\bar{u}_n|}^{2\gamma }}
  &\ge \Big( 1-\frac{{{T}^2}}{8{{\pi }^2}{{a}_2}} \Big)
  \left\| {{{\dot{u}}}_n} \right\|_{L^2}^2-{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1 }{2}\;}}
      {{M}_2}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1}-{{\big(\frac{T}{12}\big)}^{\frac{1}{2}\;}}
     {{M}_3}{{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}  \\
 &\quad-\frac{rT}{\alpha}{{\big(\frac{T}{12}\big)}^{\frac{\alpha }{2}\;}}   \left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\alpha }
 -\frac{{{\left( {{T}^2}+4{{\pi }^2} \right)}^{{1}/{2}\;}}}{2\pi }
 {{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}\\
 &\ge \frac{1}{2}\left\|{{{\dot{u}}}_n}\right\|_{L^2}^2+{{C}_1},
\end{aligned} \label{3.8}
\end{equation}
where
\begin{align*}
{{C}_1}=\min_{{s\in [0,+\infty)}}\Big\{
&\frac{4{{\pi}^2}{{a}_2}-{{T}^2}} {8{{\pi}^2}{{a}_2}}{{s}^2}
-{{\big(\frac{T}{12}\big)}^{\frac{\gamma
+1}{2}}}{{M}_2}{{s}^{{\gamma +1}}}
-\big[\frac{rT}{\alpha}{{\big(\frac{T}{12}\big)}^{\alpha/2}}\big]
{{s}^{\alpha}} \\
&\quad -\big[{{\big(\frac{T}{12}\big)}^{1/2}{{M}_3}
+\frac{{{\big({{T}^2}+4{{\pi }^2} \big)}^{1/2}}}{2\pi }}
\big]s\Big\}.
\end{align*}
Note that $a_2>T^2/(4\pi^2)$ implies
$-\infty<C_1<0$. Hence, it follows from \eqref{3.8} that
\begin{equation}\label{3.9}
  \left\|{{{\dot{u}}}_n} \right\|_{L^2}^2
  \le {{a}_2}M_1^2{{|\bar{u}_n|}^{2\gamma
  }}-2{{C}_1},
\end{equation}
and then
\begin{equation}\label{3.10}
  {{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}\le \sqrt{{{a}_2}}{{M}_1}
  {{|\bar{u}_n|}^{\gamma }}+{{C}_2},
\end{equation}
where $0<C_2<+\infty$. In a way similar to the proof of
Theorem \ref{thm1}, we have
\begin{equation}
\begin{aligned}
&\big|\int_0^T {[{{F}_1}(t,u(t))-{{F}_1}(t,\bar{u}) ]dt} \big|\\
&\le {{M}_1}{{|\bar{u}|}^{\gamma }}{{\left\|{\tilde{u}} \right\|}_{L^2}} +{{M}_2}\|\tilde{u}\|_{\infty }^{\gamma +1}+{{M}_3}{{\|\tilde{u}\|}_{\infty }} \\
&\le \frac{\pi }{\sqrt{{{a}_2}}T}\left\| {{{\tilde{u}}}_n}
\right\|_{L^2}^2
  + \frac{\sqrt{{{a}_2}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }}
  + {{M}_2} \left\| {{{\tilde{u}}}_n} \right\|_{\infty }^{\gamma +1}
 +{{M}_3}{{\left\| {{{\tilde{u}}}_n} \right\|}_{\infty }} \\
&\le \frac{T}{4\pi \sqrt{{{a}_2}}} \left\| {{{\dot{u}}}_n}
\right\|_{L^2}^2 + \frac{\sqrt{{{a}_2}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }}
 +  {\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}
 M_2 \left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1} \\
&\quad +   {{\big(\frac{T}{12}\big)}^{1/2}} {{M}_3}{{\left\| {{{\dot{u}}}_n}
\right\|}_{L^2}}.
\end{aligned}  \label{3.11}
\end{equation}
By (F4), we obtain
\begin{align*}
&\int_0^T  {\left[{{F}_2}({{u}_n}(t))-{{F}_2}({{{\bar{u}}}_n}) \right]dt}\\
&=\int_0^T{\int_0^1{\frac{1}{s} \left( \nabla {{F}_2}({{{\bar{u}}}_n} +s{{{\tilde{u}}}_n}(t))-\nabla {{F}_2}({{{\bar{u}}}_n}),s{{{\tilde{u}}}_n}(t) \right)}\,ds\,dt} \\
 &\le \int_0^T{\int_0^1{   C{{s}^{\delta -1}}{{\left| {{{\tilde{u}}}_n}(t) \right|}^{\delta }}\,ds\,dt}}\le \frac{CT}{\delta }\left\| {{{\tilde{u}}}_n} \right\|_{\infty }^{\delta }\\
 &\le \frac{CT}{\delta }{{\big(\frac{T}{12}\big)}^{\delta/2}}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\delta }.
\end{align*}
It follows from the boundedness of $\{\varphi (u_n)\}$ and
\eqref{3.9}-\ref{3.11} that
\begin{align*}
 {{C}_3}
 &\le \varphi ({{u}_n}) \\
 & =\frac{1}{2}\left\|{{{\dot{u}}}_n} \right\|_{L^2}^2
 +\int_0^T{\left[ {{F}_1}\left( t,{{u}_n}(t) \right)
 -{{F}_1}\left( t,{{{\bar{u}}}_n} \right) \right]dt}
 +\int_0^T{\left[ {{F}_2}\left( {{u}_n}(t) \right)
 -{{F}_2}\left( {{{\bar{u}}}_n} \right) \right]dt}\\
 &\quad+\int_0^T{F(t,{{{\bar{u}}}_n})dt} \\
 &\le \Big( \frac{1}{2}+\frac{T}{4\pi \sqrt{{{a}_2}}} \Big)\left\| {{{\dot{u}}}_n} \right\|_{L^2}^2+\frac{\sqrt{{{a}_2}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }}+{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}{{M}_2}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1}\\
 &\quad+{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}+\frac{CT}{\delta }{{\big(\frac{T}{12}\big)}^{\delta/2}}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\delta }+\int_0^T{F(t,{{{\bar{u}}}_n})dt} \\
 &\le\Big( \frac{1}{2}+\frac{T}{4\pi \sqrt{{{a}_2}}} \Big)\left( {{a}_2}M_1^2{{|\bar{u}_n|}^{2\gamma }}-2{{C}_1} \right)+{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}   {{M}_2}{{\left( \sqrt{{{a}_2}}{{M}_1}{{|\bar{u}_n|}^{\gamma }}  + {{C}_2} \right)}^{\gamma +1}}\\
 &\quad +{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}
 \left( \sqrt{{{a}_2}}{{M}_1}{{|\bar{u}_n|}^{\gamma }}+{{C}_2} \right)+\frac{CT}{\delta }{{\big(\frac{T}{12}\big)}^{\delta/2}}{{\left( \sqrt{{{a}_2}}{{M}_1}{{|\bar{u}_n|}^{\gamma }}+{{C}_2} \right)}^{\delta}}\\
 &\quad+\frac{\sqrt{{{a}_2}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }}
+\int_0^T{F(t,{{{\bar{u}}}_n})dt}
\\
 & \le \Big( \frac{{{a}_2}}{2}+\frac{\sqrt{{{a}_2}}T}{2\pi } \Big)M_1^2
 {{|\bar{u}_n|}^{2\gamma }}+{{\big(\frac{T}{12}\big)}
 ^{\frac{\gamma +1}{2}}}{{M}_2}
 \Big( {{2}^{\gamma }}{{\left( \sqrt{{{a}_2}}{{M}_1} \right)}^{\gamma +1}}
 {{|\bar{u}_n|}^{\gamma (\gamma +1)}}+{{2}^{\gamma }}C_2^{\gamma +1} \Big) 
\\
 & \quad +\frac{CT}{\delta }{{\big(\frac{T}{12}\big)}^{\delta/2}}
\left( {{2}^{\delta -1}}
 {{\left( \sqrt{{{a}_2}}{{M}_1} \right)}^{\delta }}{{|\bar{u}_n|}^{\gamma \delta }}
 +{{2}^{\delta-1}}C_2^{\delta } \right)\\
&\quad +{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}
 \left( \sqrt{{{a}_2}}{{M}_1}{{|\bar{u}_n|}^{\gamma }}+{{C}_2} \right)
 -\Big( 1+\frac{T}{2\pi \sqrt{{{a}_2}}} \Big){{C}_1}
  +   \int_0^T   {F\left( t,{{{\bar{u}}}_n} \right)dt} 
\\
 &= {{|\bar{u}_n|}^{2\gamma }}\Big[ {{|\bar{u}_n|}^{-2\gamma }}
    \int_0^T   {F( t,{{{\bar{u}}}_n} )dt} 
  + \Big( \frac{{{a}_2}}{2} + \frac{\sqrt{{{a}_2}}T}{2\pi } \Big)M_1^2\\
&\quad + {{\Big(\frac{{{a}_2}T}{12}\Big) }^{\frac{\gamma +1}{2}}}
   {{2}^{\gamma }}M_1^{\gamma +1}{{M}_2}{{|\bar{u}_n|}
   ^{\gamma \left( \gamma -1 \right)}} 
  +{{\Big(\frac{{{a}_2}T}{12}\Big)}^{1/2}}{{M}_1}
 {{M}_3}{{|\bar{u}_n|}^{-\gamma }}\\
&\quad +\frac{CT}{\delta }
 {{\Big(\frac{{{a}_2}T}{12}\Big)}^{\delta/2}}{{2}^{\delta -1}}M_1^{\delta }
 {{|\bar{u}_n|}^{\gamma \left( \delta -2 \right)}} \Big]
 +{{\big(\frac{T}{12}\big)   }^{\frac{\gamma +1}{2}}}{{2}^{\gamma }}{{M}_2}C_2^{\,\gamma +1} \\
 & \quad +\frac{CT}{\delta }{{\big(\frac{T}{12}\big)}^{\delta/2}}{{2}^{\delta -1}}C_2^{\delta }
 +{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{C}_2}
-\Big( 1+\frac{T}{2\pi \sqrt{{{a}_2}}} \Big){{C}_1}
\end{align*}
for large $n$. The above inequality and \eqref{3.4} imply that
$\{|\bar u|\}$ is bounded. Hence $\{ u_n\}$ is bounded by
\eqref{3.9}. Arguing as in the proof of Proposition 4.1 of \cite{MW89}, we
conclude that (PS) condition is satisfied.

\emph{Step 2. }\quad Let  $\widetilde H_T^1=\{u \in
H_T^1 : \bar u =0 \}$. We show that for  $u\in \widetilde
H_T^1$,
\begin{equation}\label{3.12}
 \varphi (u)\to +\infty \quad (\|u\|\to \infty).
\end{equation}
In fact, by (F1) and Sobolev's inequality, one has
\begin{align*}
 \big|\int_0^T{\left[ {{F}_1}\left( t,u(t) \right)-{{F}_1}
\left( t,0 \right)\right]}dt\big|
 &=\big| \int_0^T{\int_0^1{\left( \nabla {{F}_1}(t,su(t)),u(t) \right)}}\,ds\,dt
\big| \\
 & \le \int_0^T{f(t){{\left| u(t) \right|}^{\gamma +1}}dt}
 +\int_0^T{g(t)\left| u(t) \right|dt} \\
 & \le {{\big(\frac{T}{12}\big)}^{\frac{\alpha +1}{2}}}
 {{M}_2}\|\dot{u}\|_{L^2}^{\alpha +1}
 +{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{\|\dot{u}\|}_{L^2}}
\end{align*}
for all $u \in \widetilde H_T^1$. It follows from (F2) that
\begin{align*}
\int_0^T{\left[ {{F}_2}\left(u(t)
\right)-{{F}_2}\left( 0 \right) \right]}dt
&=\int_0^T{\int_0^1{\left( \nabla
{{F}_2}(su(t))-\nabla {{F}_2}\left( 0 \right),u(t) \right)}}\,ds\,dt \\
 & \ge -\int_0^T{\int_0^1{r{{s}^{\alpha -1}}{{\left| u(t) \right|}^{\alpha }}
\,ds\,dt}}\\
&\ge -\frac{rT}{\alpha }\left\| u \right\|_{\infty }^{\alpha }\\
 &\ge -\frac{rT}{\alpha }{{\left( \frac{T}{12}\right)} ^{\alpha/2}}
\left\| u \right\|_{L^2}^{\alpha}.
\end{align*}
Hence, we have
\begin{align*}
\varphi (u)
&=\frac{1}{2}\|\dot{u}\|_{L^2}^2+\int_0^T{\left[ F\left( t,u(t) \right)
-F\left( t,0 \right) \right]}dt+\int_0^T{F(t,0)dt} \\
 & \ge \frac{1}{2}\|\dot{u}\|_{L^2}^2-{{\big(\frac{T}{12}\big)}^{\frac{\alpha +1 }{2}}}{{M}_2}\|\dot{u}\|_{L^2}^{\alpha +1}-{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}{{\|\dot{u}\|}_{L^2}}\\
  &\quad-\frac{rT}{\alpha }{{\big(\frac{T}{12}\big)}^{\alpha/2}}\left\|u\right\|_{L^2}^{\alpha}+\int_0^T{F(t,0)dt}.
\end{align*}
By Wirtinger's inequality, $\|u\| \to \infty$ if and only if
$\|\dot{u}\|_{L^2} \to \infty$ in $\widetilde H_T^1$. Hence
\eqref{3.12} is satisfied.

\emph{Step 3.} By (F5), we can easily see that
$\int_0^T    F(t,x)dt \to -\infty$ as $|x|\to \infty$ for all
$x\in \mathbb{R}^N$. Thus, for all $u\in(\widetilde H_T^1)^\bot=\mathbb{R}^N$,
$$
\varphi (u)=\int_0^T    F(t,u)dt \to -\infty\quad \text{as } |u|\to \infty.
$$
Now, the proof is completed by saddle point theorem
(cf. \cite[Theorem 4.6]{Rab86})
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
By (F3'), we can
choose an $a_3>\frac{T^2}{4\pi^2-rT^2}$ such that
\begin{equation}\label{3.13}
\liminf_{|x| \to
\infty}|x|^{-2\gamma}\int_0^T F(t,x)\,dt>\frac{a_3}{2}M_1^2.
\end{equation}
The condition (F2') and the Sobolev's inequality imply that
\begin{align*}
\int_0^T{\left[{{F}_2}(u(t)-{{F}_2}(\bar{u}))\right]dt}
 &=\int_0^T{\int_0^1{\frac{1}{s}\left(\nabla{{F}_2}
 (\bar{u}+s\tilde{u}(t))-\nabla {{F}_2}(\bar{u}),s\tilde{u}(t) \right)}\,ds\,dt} \\
 &\ge-\int_0^T{\int_0^1{r}}s{{\left|\tilde{u}(t) \right|}^2}\,ds\,dt
 -\frac{{r{T}^2}}{8{{\pi }^2}}\|\dot{u}\|_{L^2}^2.
\end{align*}
It follows immediately from the similar method of the proof
of Theorem \ref{thm1} that
\begin{align*}
  \varphi (u)
& =\frac{1}{2}\|\dot{u}\|_{L^2}^2+\int_0^T   {F(t,u(t))}dt\\
 & \ge \left( \frac{1}{2}-\frac{{{T}^2}}{8{{\pi }^2}{{a}_3}}-\frac{rT^2}{8\pi^2} \right)
 \|\dot{u}\|_{L^2}^2-{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}
 {{M}_2}\|\dot{u}\|_{L^2}^{\gamma +1}\\
 &{\quad} -{{\big(\frac{T}{12}\big)}^{1/2}}
 {{M}_3}{{\|\dot{u}\|}_{L^2}}
 +{{|\bar{u}|}^{2\gamma }}
\Big({{|\bar{u}|}^{-2\gamma}}\int_0^T{F(t,\bar{u})dt}-\frac{{{a}_3}}{2}M_1^2
 \Big)
\end{align*}
for all $u \in H^1_T$, which implies that $\varphi (u) \to \infty $
as $\|u\| \to \infty $ by \eqref{3.13}, due to the facts that $\gamma <1$,
$r< \frac{4\pi^2}{T^2}$ and $ \| u\| \to \infty $ if and only if
\[
{{\big( {{|\bar{u}|}^2}+\|\dot{u}\|_{L^2}^2 \big)}^{1/2}}\to \infty.
\]
By the least action principle, Theorem \ref{thm3} holds.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm4}]
We firstly show that $\varphi$ satisfies the (PS) condition. Suppose
that $\{u_n\}$ satisfies ${\varphi }'({{u}_n})\to 0 $ as $n\to 0$
and $\left\{ \varphi \left( {{u}_n} \right) \right\}$ is bounded.
By (F7), we can choose an $a_4>T^2/(4\pi^2$ such that
\begin{equation}\label{3.14}
\limsup_ {|x|\to \infty }{{|x|}^{-2\gamma
}} \int_0^T   {{{F}_1}\left( t,x \right)dt}
<-\big(\frac{{{a}_4}}{2}+\frac{\sqrt{{{a}_4}}T}{2\pi }
\big)M_1^2.
\end{equation}
By the ($\lambda$,$\mu$)-subconvexity of $G(x)$, we have
\begin{equation}\label{3.15}
G(x)\le \left( 2\mu {{|x|}^{\beta }}+1 \right){{G}_0}
\end{equation}
for all $x \in \mathbb{R}^N$, and a.e. $t \in [0,T]$, where
${{G}_0}=\max_{|s|\le 1}  G(s)$,
 $\beta ={{\log }_{2\lambda }}( 2\mu)<2$.
Then
\begin{equation}
\begin{aligned}
 \int_0^T{\left( \nabla {{F}_2}\left( t,{{u}_n}(t)
\right),{{{\tilde{u}}}_n}(t) \right)dt}
&\ge-\int_0^T{k(t)G({{{\bar{u}}}_n})dt}\\
&\ge -\int_0^T{k(t)\left( 2\mu {{|\bar{u}_n|}^{\beta }}+1 \right){{G}_0}dt}\\
&=-2\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}-{{M}_4},
\end{aligned}\label{3.16}
\end{equation}
where ${{M}_4}={{G}_0}\int_0^T{k(t)dt}$. It follows from
\eqref{3.5} and \eqref{3.16} that for large $n$,
\begin{equation}
\begin{aligned}
 \left\| {{{\tilde{u}}}_n} \right\|
 &\ge \left( \varphi ({{u}_n}),{{{\tilde{u}}}_n} \right)\\
 &=\left\| {{{\dot{u}}}_n} \right\|_{L^2}^2+\int_0^T{\left( \nabla F(t,{{u}_n}(t)),{{{\tilde{u}}}_n}(t) \right)}dt\\
 & \ge \Big( 1-\frac{{{T}^2}}{8{{\pi }^2}
 {{a}_4}} \Big)\left\| {{{\dot{u}}}_n} \right\|_{L^2}^2-\frac{{{a}_4}}{2}M_1^2{{|\bar{u}_n|}^{2\gamma }}-{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}\;}}{{M}_2}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1} \\
 & \quad -{{\big(\frac{T}{12}\big)}^{{1}/{2}\;}}{{M}_3}{{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}
 -2\mu {{M}_4}{{|\bar{u}_n|}^{\beta
 }}-{{M}_4}.
\end{aligned}\label{3.17}
\end{equation}
Then \eqref{3.7} and \eqref{3.17} imply that
\begin{equation}
\begin{aligned}
\frac{{{a}_4}}{2}M_1^2{{|\bar{u}_n|}^{2\gamma }}
+2\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}
&\ge \Big( 1-\frac{{{T}^2}}{8{{\pi}^2}{{a}_4}} \Big) \left\| {{{\dot{u}}}_n}
\right\|_{L^2}^2-{{\big(\frac{T}{12}\big)}^{\frac{\gamma
+1}{2}}}   {{M}_2}\left\|{{{\dot{u}}}_n}\right\|_{L^2}^{\gamma
+1} \\
 & \quad -\Big( {{\big(\frac{T}{12}\big)}^{1/2}}  {{M}_3}+\frac{{{\left( {{T}^2}
 +4{{\pi }^2} \right)}^{{1}/{2}\;}}}{2\pi } \Big){{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}-{{M}_4}\\
 & \ge \frac{1}{2}\left\| {{{\dot{u}}}_n}
 \right\|_{L^2}^2+{{C}_4},
\end{aligned} \label{3.18}
\end{equation}
where
\begin{align*}
{{C}_4}&=\min_{s\in[0,+\infty)}\Big\{ \frac{8{{\pi
}^2}{{a}_4}-{{T}^2}}{8{{\pi }^2}{{a}_4}}{{s}^2}
-{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}{{M}_2}{{s}^{\gamma +1}}-{M}_4\\
&\quad -\Big[{{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}
+\frac{{{\left( {{T}^2}+4{{\pi}^2} \right)}^{1/2}}}{2\pi } \Big]s\Big\}.
\end{align*}
Note that $-\infty < C_4<0$ due to $ a_4>\frac{T^2}{4\pi^2}$. By
\eqref{3.18}, one has
\begin{equation}\label{3.19}
\left\| {{{\dot{u}}}_n} \right\|_{L^2}^2\le {{a}_4}M_1^2
{{|\bar{u}_n|}^{2\gamma }}+4\mu
{{M}_4}{{|\bar{u}_n|}^{\beta }}-2{{C}_4},
\end{equation}
and then
\begin{equation}\label{3.20}
  {{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}}
  \le \frac{\sqrt{2{{a}_4}}}{2}{{M}_1}{{|\bar{u}_n|}^{\gamma }}
  +\sqrt{2\mu {{M}_4}}{{|\bar{u}_n|}^{\beta/2}}+{{C}_{5}},
\end{equation}
where $C_5>0$. It follows from (F6) and \eqref{3.15} that
\begin{equation}
\begin{aligned}
&\int_0^T{\left[ {{F}_2}\left( t,u(t)
\right)-{{F}_2}\left( t,\bar{u} \right) \right]dt}\\
&=-\int_0^T{\int_0^1{\left( \nabla
{{F}_2}\left( t,{{{\bar{u}}}_n} +s{{{\tilde{u}}}_n}(t)
\right),-{{{\tilde{u}}}_n}(t) \right)}\,ds\,dt} \\
 & \le \int_0^T{\int_0^1{k(t)G\left( {{{\bar{u}}}_n}+(s+1){{{\tilde{u}}}_n}(t) 
 \right)\,ds}\,dt} \\
 & \le \int_0^T{\int_0^1{k(t)\left( 2\mu {{\left| {{{\bar{u}}}_n}
 +(s+1){{{\tilde{u}}}_n}(t) \right|}^{\beta }}+1 \right){{G}_0}\,ds}\,dt}\\
 & \le 4\mu \int_0^T{k(t)\left( {{|\bar{u}_n|}^{\beta }}+{{2}^{\beta }}
 {{\left| {{{\tilde{u}}}_n}(t) \right|}^{\beta }} \right){{G}_0}dt}
 +{{G}_0}\int_0^T   {k(t)dt} \\
 & \le {{2}^{\beta +2}}\mu {{M}_4}\left\| {{{\tilde{u}}}_n} 
 \right\|_{\infty }^{\beta }
 +4\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}+{{M}_4} \\
 & \le {{\big(\frac{T}{12}\big)}^{\beta/2}}
 {{2}^{\beta +2}}\mu {{M}_4}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\beta }
 +4\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}+{{M}_4}
\end{aligned} \label{3.21}
\end{equation}
for all $u\in H_T^1$. By the boundedness of $\{\varphi (u_n)\}$ and
the inequalities \eqref{3.11}, \eqref{3.19}-\eqref{3.21}, one has
\begin{align*}
{{C}_{6}}
&\le \varphi ({{u}_n}) \\
 & =\frac{1}{2}\left\|{{{\dot{u}}}_n}\right\|_{L^2}^2
+\int_0^T{\left[ {{F}_1}\left( t,{{u}_n}(t) \right)-{{F}_1}
\left( t,{{{\bar{u}}}_n} \right) \right]dt}\\
&\quad +\int_0^T{\left[ {{F}_2}\left( t,{{u}_n}(t) \right)
-{{F}_2}\left( t,{{{\bar{u}}}_n} \right) \right]dt}
 +\int_0^T{F(t,{{{\bar{u}}}_n})dt} \\
 & \le \Big( \frac{1}{2}+\frac{T}{4\pi \sqrt{{{a}_4}}} \Big)
 \left\| {{{\dot{u}}}_n} \right\|_{L^2}^2
 +\frac{\sqrt{{{a}_4}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }}
+{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}{{M}_2}
\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\gamma +1}
\\
 &\quad  + {{\big(\frac{T}{12}\big)}^{1/2}}{{M}_3}
 {{\left\| {{{\dot{u}}}_n} \right\|}_{L^2}} 
 + {{\big(\frac{T}{12}\big)}^{\beta/2}}{{2}^{\beta+2}}
 \mu{{M}_4}\left\| {{{\dot{u}}}_n} \right\|_{L^2}^{\beta} 
+ 4\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}  + {{M}_4} \\
&\quad  +  \int_0^T    {F(t,{{{\bar{u}}}_n})dt}
\\
&\le \Big( \frac{1}{2}+\frac{T}{4\pi \sqrt{{{a}_4}}} \Big)
 \left( {{a}_4}M_1^2 {{|\bar{u}_n|}^{2\gamma }}
 +4\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}-2{{C}_4} \right)
 +\frac{\sqrt{{{a}_4}}T}{4\pi }M_1^2{{|\bar{u}_n|}^{2\gamma }} 
\\
&\quad +{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}} 
 {{M}_2}{{\left( \sqrt{{a}_4}  {{M}_1}{{|\bar{u}_n|}^{\gamma }}
 +2\sqrt{\mu {{M}_4}}{{|\bar{u}_n|}^{\beta/2}}+{{C}_{5}}
  \right)}^{\gamma +1}}   \\
&\quad + {{\big(\frac{T}{12}\big)}^{1/2}} p{{M}_3}\Big( \sqrt{{a}_4}{{M}_1}
 {{|\bar{u}_n|}^{\gamma }}
 +2\sqrt{\mu {{M}_4}}{{|\bar{u}_n|}^{\beta/2}}+{{C}_{5}} \Big)\\
&\quad +{{\big(\frac{T}{12}\big)}^{\beta/2}}{{2}^{\beta +2}}\mu {{M}_4} 
 {{\left(\sqrt{{a}_4}{{M}_1}{{|\bar{u}_n|}^{\gamma }}
 +2\sqrt{\mu {{M}_4}}{{\left|{{{\bar{u}}}_n}\right|}^{\beta/2}}+{{C}_{5}}
 \right)}^{\beta}}\\
&\quad
+\mu {{M}_4}{{|\bar{u}_n|}^{\beta}}+{{M}_4}
+\int_0^T   {F(t,{{{\bar{u}}}_n})dt} 
\\
 & \le \Big(\frac{{{a}_4}}{2}+\frac{\sqrt{{{a}_4}}T}{2\pi } \Big)
 M_1^2{{|\bar{u}_n|}^{2\gamma }}
 +\Big( 6+\frac{T}{\pi \sqrt{{{a}_4}}} \Big)\mu {{M}_4}{{|\bar{u}_n|}^{\beta }}
 -\Big( 1+\frac{T}{2\pi \sqrt{{{a}_4}}} \Big){{C}_4} \\
 &\quad
 +{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}{{M}_2}
\Big(
 {{2}^{\gamma}}{{a}_4}^{\frac{\gamma+1}{2}}
 {{M}_1}^{\gamma+1}{{|\bar{u}_n|}^{\gamma (\gamma +1)}}
 +{{2}^{3\gamma+1}}\mu^{\frac{\gamma+1}{2}} 
{M}_4^{\frac{\gamma+1}{2}}{{|\bar{u}_n|}^{\frac{\beta (\gamma +1)}{2}}}\\
 &\quad +{{2}^{2\gamma }}C_{5}^{\gamma +1} \Big) 
  +{{\big(\frac{T}{12}\big)}^{ \frac{\beta }{2}}}{{2}^{\beta +2}}\mu {{M}_4}
 \Big({{2}^{\beta -1}}{{{a}_4}^{\frac{\beta}{2}}}{{M}_1}^{\beta}{{|\bar{u}_n|}
 ^{\gamma \beta }}
 +{{2}^{3\beta -2}}\mu^{\frac{\beta}{2}} {{M}_4}
 ^{\frac{\beta}{2}}{{|\bar{u}_n|}^{\frac{{{\beta }^2}}{2}}}\\
&\quad +{{2}^{2\left( \beta -1 \right)}}C_{5}^{\beta } \Big)
+{{\left( \frac{T}{12}\right)}^{1/2}}{{M}_3}\left(
{\sqrt{{{a}_4}}}{{M}_1}{{|\bar{u}_n|}^{\gamma }}
 +2\sqrt{\mu {{M}_4}}{{|\bar{u}_n|}^{\beta/2}}+{{C}_{5}} \right)\\
&\quad +{{M}_4} +\int_0^T   {F(t,{{{\bar{u}}}_n})dt} 
\\
& ={{|\bar{u}_n|}^{2\gamma }}\Big[ {{|\bar{u}_n|}^{-2\gamma }}
 \int_0^T   {{{F}_1}(t,{{{\bar{u}}}_n})dt} + \Big( \frac{{{a}_4}}{2}
  + \frac{\sqrt{{{a}_4}}T}{2\pi } \Big)M_1^2\\
&\quad +{{\big(\frac{T}{12}\big)}^{1/2}}{\sqrt{{{a}_4}}}{{M}_1}{{M}_3}
 {{|\bar{u}_n|}^{-\gamma }}
 + {{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}
 {{2}^{\gamma -\frac{1}{2}}}\sqrt{{{a}_4}}{{M}_1}{{M}_2}
 {{|\bar{u}_n|}^{\gamma (\gamma -1)}} \\
&\quad +{{\big(\frac{T}{12}\big)} ^{\beta/2}}{{2}^{2\beta  +1}}
 \mu {{{a}_4}^{\frac{\beta}{2}}} {{M}_1}^{\beta}{{M}_4}{{|\bar{u}_n|}
^{\gamma (\beta -2)}} \Big] \\
&\quad +{{|\bar{u}_n|}^{\beta }}
  \Big[ {{|\bar{u}_n|}^{-\beta }}   \int_0^T   {{{F}_2}(t,{{{\bar{u}}}_n})dt}
 +  \Big( 6 +\frac{T}{\pi \sqrt{{{a}_4}}} \Big)\mu {{M}_4}\\
&\quad  + {{\big(\frac{T}{12}\big)}^{\beta/2}}{{2}^{4\beta
 }}\mu^{\frac{\beta+2}{2}}{{M}_4}^{  \frac{\beta+2}{2}}
 {{|\bar{u}_n|}^{\frac{1}{2}{{\beta }^2}-2}}  \\
&\quad
  +{{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}{{M}_2}{{2}^{3\gamma+1
 }}\mu^{\frac{\gamma+1}{2}}{M_2}{M_4}^{\frac{\gamma+1}{2}}
 {{|\bar{u}_n|}^{\frac{\beta (\gamma -1)}{2}}}
 +{{\big(\frac{T}{12}\big)}^{1/2}}2{{M}_3}\sqrt{\mu {{M}_4}}
 {{|\bar{u}_n|}^{-\beta/2}} \Big] \\
 &\quad
 - \Big( 1 + \frac{T}{2\pi \sqrt{{{a}_4}}} \Big){{C}_4}
  + {{\big(\frac{T}{12}\big)}^{\frac{\gamma +1}{2}}}   {{2}^{2\gamma }}
 {{M}_2}C_{5}^{\gamma +1}  +  {{\big(\frac{T}{12}\big)}^{1/2}} {{M}_3}{{C}_{5}}\\
&\quad  + {{\big(\frac{T}{12}\big)}^{\beta/2}}  {{2}^{3\beta }}\mu {{M}_4}C_{5}^{\beta }  + {{M}_4}
 \end{align*}
for large $n$. The above inequality and \eqref{3.14} imply that
$\{|\bar u|\}$ is bounded. Hence $\{u_n\}$ is bounded by
\eqref{3.19}. By using the usual method, the (PS) condition holds.

Similar to the proof of Theorem \ref{thm2}, we can verify that functional
satisfies the other conditions of the saddle point theorem. We omit
the details.
\end{proof}


\section{Examples}

In this section, we give some examples of $F$ to illustrate that our
results are new.

\begin{example} \label{examp3.1}\rm
Let $F=F_1+F_2$, with
\begin{gather*}
 F_1(t,x)=\sin\big( \frac{2\pi t}{T}\big)|x|^{7/4}+(0.6T-t)|x|^{3/2}
+\left(h(t),x \right),\\
F_2(x)=C(x)-\frac{3r}{4}|x|^{4/3},
\end{gather*}
where $h\in \mathscr{L}^1(0,T;\mathbb{R}^N)$, $r>0$,
$C(x)=\frac{3r}{4}(|x_1|^4+|x_2|^{4/3}+\cdots +|x_N|^{4/3})$.
\end{example}

 By Young's inequality, it is easy to see that
\begin{align*}
  |\nabla F_1(t,x)|
&\le \frac{7}{4}\Big| \sin\Big( \frac{2\pi t}{T}\Big)
  \Big||x|^{3/4}+\frac{3}{2}|0.6T-t||x|^{1/2}+|h(t)|\\
  &\le \frac{7}{4}\Big(\Big| \sin\Big( \frac{2\pi t}{T}\Big)
  \Big|+\varepsilon \Big) |x|^{3/4}+\frac{T^3}{\varepsilon^2}+|h(t)|
\end{align*}
for all $x \in \mathbb{R}^N$ and a.e. $t \in [0,T]$, where $\varepsilon >0$. And
\[
  \left( \nabla F_2(x)-\nabla F_2(y),x-y \right)\ge -r|x-y|^{4/3}
\]
for all $x,y \in \mathbb{R}^N$. Thus, (F1), (F2) hold with
$\gamma=3/4$, $\alpha=4/3$ and
$$
f(t)=\frac{7}{4}( | \sin( \frac{2\pi t}{T}) | + \varepsilon ),
\quad g(t)= \frac{T^3}{\varepsilon^2}+|h(t)|.
$$
However, $F$ does not satisfy \eqref{eq2}. In fact
\begin{align*}
  & |x|^{-2\gamma}\int_0^T F(t,x)dt\\
  &=|x|^{-3/2} \int_0^T  \big[\sin\big( \frac{2\pi t}{T}  \big) |x|^{7/4}
+(0.6T-t)|x|^{3/2}+\big(C(x)-\frac{3r}{4}|x|^{4/3}\big)
+  (h(t),x)\big]dt\\
   &=0.1T^2+\frac{T(C(x)-\frac{3r}{4}|x|^{4/3})}{|x|^{3/2}}
+\Big( \int_0^T h(t)dt, |x|^{-3/2}x \Big)
\end{align*}
On the other hand, we have
$$
\frac{T^2}{8\pi^2}\int_0^T    f^2(t)dt= \frac{49T^3}{128\pi^2}
\Big( \frac{1}{2}+\frac{4\varepsilon}{\pi}+\varepsilon^2 \Big)
$$
If $T<\frac{128\pi^2}{245} $, we choose $\varepsilon>0$ sufficient
small such that
$$
\liminf_{|x|\to \infty}|x|^{-2\gamma}\int_0^T   F(t,x)dt=0.1T^2
>\frac{T^2}{8\pi^2}\int_0^T    f^2(t)dt
$$
which implies that (F3) holds. Then $F=F_1+F_2$ is not convex,
not $\gamma$-subadditive, not periodic, not a.e. uniformly
coercive, and $\nabla F$ is not sublinear. Thus, $F$ is not covered
by results in the references.

\begin{example} \label{examp3.2}\rm
Let $F=F_1+F_2$, with
\begin{gather*}
F_1(t,x)=(0.5T-t)|x|^{7/4}+(0.4T-t)|x|^{3/2}+\left(h(t),x\right),\\
F_2(x)=-\frac{4r}{5}|x|^{5/4},
\end{gather*}
where $h\in \mathscr{L}^1(0,T;\mathbb{R}^N)$, $r>0$.
\end{example}

Similar to Example \ref{examp3.1}, we can
see that all conditions of Theorem \ref{thm2} hold but $F$ is not
covered by results in the references.


\begin{example} \label{examp3.3}\rm
Let $F=F_1+F_2$, with
\begin{gather*}
F_1(t,x)=(0.5T-t)|x|^{7/4}+(0.6T-t)|x|^{3/2}+\left(h(t),x\right),\\
F_2(x)=C(x)-\frac{r}{2}|x|^2,
\end{gather*}
where $h\in \mathscr{L}^1(0,T;\mathbb{R}^N)$,
$C(x)=\frac{r}{2}(|x_1|^4+|x_2|^2+\cdots+|x_N|^2)$,
$0 < r<\frac{4\pi^2}{T^2}$.
\end{example}

In a way similar to Example \ref{examp3.1}, it is easy
to see that condition (F1) and (F2') are satisfied with
$\gamma=3/4$. However, $F$ does not satisfies \eqref{eq2}.
In fact,
\begin{align*}
  &|x|^{-2\gamma}\int_0^T F(t,x)dt\\
  &=|x|^{-2/3}   \int_0^T  \big[(0.5T-t)|x|^{7/4}
  +(0.6T -t)|x|^{3/2}
  + \left(C(x) - \frac{r}{2}|x|^2 \right)+ (h(t),x)
  \big]dt\\
  &=0.1T^2+\frac{T\left( C(x)-\frac{r}{2}|x|^2\right)}{|x|^{3/2}}
 +\Big(\int_0^T   h(t)dt,x|x|^{-3/2}\Big)\\
  &=0.1T^2+\frac{rT(|x_1|^4-|x_1|^2)}{2|x|^{3/2}}
+\Big(\int_0^T   h(t)dt,x|x|^{-3/2}\Big).
\end{align*}
We can choose $\varepsilon>0$  small enough and some suitable $T$
such that
\begin{align*}
\liminf_{|x|\to\infty}|x|^{-2\gamma}\int_0^T   F(t,x)dt=0.1T^2
>\frac{T^2}{2(4\pi^2-rT^2)}\int_0^T   f^2(t,x)dt,
\end{align*}
which implies that (F3') holds. $F$ is also not covered by
results in the references.

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\end{document}