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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 178, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/178\hfil Subharmonic solutions]
{Subharmonic solutions for nonautonomous
 second-order Hamiltonian systems}

\author[M. Timoumi\hfil EJDE-2012/178\hfilneg]
{Mohsen Timoumi}  % in alphabetical order

\address{Mohsen Timoumi \newline
 Department of Mathematics,
 Faculty of Sciences, 5000 Monastir, Tunisia}
\email{m\_timoumi@yahoo.com}

\thanks{Submitted March 9, 2012. Published October 12, 2012.}

\subjclass[2000]{34C25}
\keywords{Hamiltonian systems; subharmonics; minimal periods;
\hfill\break\indent  saddle point theorem}

\begin{abstract}
 In this article, we prove the existence of subharmonic solutions for
 the non-autonomous second-order Hamiltonian system 
 $\ddot{u}(t)+V'(t,u(t))=0$.
 Also we study the minimality of their periods, when the nonlinearity
 $V'(t,x)$ grows faster than   $|x|^{\alpha}$, $\alpha\in[0,1[$ 
 at infinity.  The proof is based on the Least Action Principle 
 and the Saddle Point Theorem.
\end{abstract}

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


\section{Introduction}

 Consider the non-autonomous second-order Hamiltonian system
\begin{equation}
\ddot{u}(t)+V'(t,u(t))=0, \label{eS}
\end{equation}
where $V:\mathbb{R}\times\mathbb{R}^{N}\to\mathbb{R}$, $(t,x)\to V(t,x)$ is a continuous function,
$T$-periodic $(T > 0)$ in the first variable and differentiable with respect
to the second variable such that the gradient
$V'(t,x)=\frac{\partial V}{\partial x}(t,x)$ is continuous on
$\mathbb{R}\times\mathbb{R}^{N}$.  In this work, we are interested in the existence
of subharmonic solutions  of \eqref{eS}. Assuming that $T>0$ is the minimal
 period of the time dependence of $V(t,x)$, by subharmonic solution of
\eqref{eS} we mean a $kT$-periodic solution, where $k$ is any integer;
 when moreover the solution is not $T$-periodic we call it a true subharmonic.

Using variational methods, there have been various types of results concerning
the existence of subharmonic solutions to system \eqref{eS}.
 Many solvability  conditions are given, such as a convexity condition 
\cite{f3,w1},
a super-quadratic condition \cite{r1,t3}, a subquadratic condition \cite{f3,j1},
a periodic condition \cite{s1}, a bounded nonlinearity condition [1,2,5], and a
 sublinear condition \cite{t2}. In particular, under the assumptions that
 there exists a constant $M > 0$ such that
\begin{gather}
|V'(x)|\leq M,\quad \forall x\in\mathbb{R}^{N}, \label{e1.1}\\
\lim_{|x|\to\infty}(V'(x)-\bar{e})x=+\infty, \label{e1.2}
\end{gather}
where $e:\mathbb{R}\to\mathbb{R}^{N}$ is a continuous periodic function having minimal
 period $T > 0$, and $\bar{e}$ is the mean value of $e$, A. Fonda and
 Lazer in \cite{f1}  showed that the system
\begin{equation}
\ddot{u}(t)+V'(u(t))=e(t) \label{e1.3}
\end{equation}
admitted periodic solutions with minimal period $kT$, for any sufficiently
large prime number $k$. After that,  Tang and  Wu in \cite{t2}
generalized these results without studying the minimality of periods.
Precisely, it was assumed that the nonlinearity satisfied the following
 restrictions:
\begin{gather}
|V'(t,x)|\leq f(t)|x|^{\alpha}+g(t),\ \forall x\in\mathbb{R}^{N},\quad
\text{a. e. } t\in[0,T], \label{e1.4}\\
\frac{1}{|x|^{2\alpha}}\int^{T}_0V(t,x)dt\to+\infty
\quad\text{as } |x|\to +\infty, \label{e1.5}
\end{gather}
here $f,g\in L^{1}(0,T;\mathbb{R}^{+})$ are $T$-periodic and $\alpha\in[0,1[$.

In \cite{f1,t2}, the nonlinearity is required to grow at infinity at most like
$|x|^{\alpha}$ with $\alpha\in [0,1[$. In this article, we will firstly,
establish the existence of subharmonic solutions for the system \eqref{eS}
when the nonlinearity $V'(t,x)$ is required to have a sublinear growth at
 infinity faster than $|x|^{\alpha},\ \alpha\in[0,1[$. Our first main result is
as follows.


\begin{theorem} \label{thm1.1}
 Let $\omega\in C([0,\infty[,\mathbb{R}^{+})$ be a nonincreasing positive function
with the properties:
\begin{gather}
\liminf_{s\to\infty}\frac{\omega(s)}{\omega(s^{1/2})} > 0, \label{ea}\\
\omega(s)\to 0,\quad \omega(s)s\to\infty \quad \text{as }s\to\infty. \label{eb}
\end{gather}
Assume that $V$ satisfies:
 There exist two $T$-periodic functions $f\in L^{2}(0,T;\mathbb{R}^{+})$
 and $g\in L^{1}(0,T;\mathbb{R}^{+})$ such that
\begin{equation}
|V'(t,x)|\leq f(t)\omega(|x|)|x|+g(t),\quad \forall x\in\mathbb{R}^{N},\text{a.e. }
 t\in[0,T]; \label{V1}
\end{equation}
\begin{equation}
\frac{1}{[\omega(|x|)|x|]^{2}}\int^{T}_0V(t,x)dt \to+\infty\quad
\text{as } |x|\to\infty; \label{V2}
\end{equation}
 There is a subset $C$ of $[0,T]$ with $\operatorname{meas}(C)>0$ and
$h\in L^{1}(0,T;\mathbb{R})$ such that
\begin{gather}
\lim_{|x|\to\infty}V(t,x)=+\infty,\quad\text{a.e. }t\in C, \label{V3i}\\
V(t,x)\geq h(t)\quad\text{for all $x\in\mathbb{R}^{N}$, a.e. $t\in [0,T]$}. \label{V3ii}
\end{gather}
Then for all positive integer $k$, the system \eqref{eS} has at least one
$kT$-periodic solution $u_{k}$ satisfying
$$
\lim_{k\to\infty}\|u_{k}\|_{\infty}=+\infty,
$$
where $\|u\|_{\infty}=\sup_{t\in\mathbb{R}}|u(t)|$.
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
Let
$$
V(t,x)=\gamma(t)\frac{|x|^{2}}{ln(2+|x|^{2})},\quad \forall x\in\mathbb{R}^{N},\;
\forall t\in\mathbb{R},
$$
where
$$\gamma(t)=\begin{cases}
\sin(2\pi t/T), & t\in [0,T/2]\\
0, & t\in [ T/2,T].
\end{cases}
$$
Taking $\omega(s)=\frac{1}{ln(2+s^{2})}$, $C=[0,\frac{T}{2}]$.
By a simple computation, we prove that $V(t,x)$ satisfies \eqref{V1}--(V3)
and does not satisfy the conditions \eqref{e1.1}, \eqref{e1.2} nor
\eqref{e1.4}, \eqref{e1.5}.
\end{remark}

\begin{corollary} \label{coro1.1}
 Assume that {\rm \eqref{V1}} holds and there exists a subset $C$ of $[0,T]$
with $\operatorname{meas}(C)>0$ and $h\in L^{1}(0,T;\mathbb{R})$ such that
\begin{gather}
\lim_{|x|\to\infty}{{V(t,x)}\over{[\omega(|x|)|x|]^{2}}}=+\infty,
\quad\text{a.e. }t\in C, \label{V4i}
\\
V(t,x)\geq h(t),\quad\text{for $all x\in\mathbb{R}^{N}$, a.e. $t\in [0,T]$.}
\label{V4ii}
\end{gather}
Then the conclusion of Theorem \ref{thm1.1} holds.
\end{corollary}

There are a few results studying the minimality of periods 
of the subharmonics, see \cite{w1} for the case of convexity, and \cite{f1} for the 
case of bounded gradient. We study this problem  and obtain the following result.

\begin{theorem} \label{thm1.2}
 Assume that $V$ satisfies \eqref{V1} and 
\begin{equation}
\frac{V'(t,x)x}{[\omega(|x|)|x|]^{2}}\to+\infty\quad
\text{as $|x|\to\infty$, uniformly for $t\in [0,T]$}. \label{V5}
\end{equation}
Then, for all integer $k\geq 1$, Equation \eqref{eS} possesses a $kT$-periodic
 solution $u_{k}$ such that $\lim_{k\to\infty}\|u_{k}\|_{\infty}=+\infty$.
If moreover $V$ satisfies the assumption:
\begin{equation} \label{V6}
\parbox{9cm}{
If $u(t)$ is a periodic function with minimal period $rT$
with $r$ rational, and $V'(t,u(t))$  is a periodic function with
minimal period $rT$, then  $r$ is necessarily an integer.}
\end{equation}
Then, for any sufficiently large prime number $k$, $kT$ is the minimal period
of $u_{k}$.
\end{theorem}

As an example of a function $V$ we have 
$$
V(t,x)=(2+\cos(\frac{2\pi}{T}t))\frac{|x|^{2}}{\ln(2+|x|^{2})},
\quad \omega(s)=\frac{1}{\ln(2+s^{2})}.
$$
Then $V(t,x)$ satisfies \eqref{V1} \eqref{V5} and \eqref{V6},
Our main tools, for proving our results, are the Least Action Principle
 and the Saddle Point Theorem.

\section{Proof of theorems} 

Let us fix a positive integer $k$ and consider the continuously differentiable 
function
$$
\varphi_{k}(u)=\int^{kT}_0[{1\over 2}|\dot{u}(t)|^{2}-V(t,u(t))]dt
$$
defined on the space $H^{1}_{kT}$ of $kT$-periodic absolutely continuous 
vector functions whose derivatives have square-integrable norm. 
This set is a Hilbert space with the norm
$$
\|u\|_{k}=\Big[\int^{kT}_0|u(t)|^{2}dt + \int^{kT}_0|\dot{u}(t)|^{2}dt
\Big]^{1/2},\quad  u\in H^{1}_{kT}
$$
and the associated inner product
$$
\langle u,v \rangle_{k}=\int^{kT}_0[u(t)v(t) + \dot{u}(t)\dot{v}(t)]dt,\quad
 u,v\in H^{1}_{kT}.
$$
For $u\in H^{1}_{kT}$, let $\bar{u}={1\over{kT}}\int^{kT}_0u(t)dt$ and
$\tilde{u}(t)=u(t)-\bar{u}$, then we have 
Sobolev's inequality,
\begin{equation}
\|\tilde{u}\|^{2}_{\infty}
\leq {{kT}\over{12}}\int^{kT}_0|\dot{u}(t)|^{2}dt,
  \label{e2.1}
\end{equation}
and Wirtinger's\ inequality,
\begin{equation}
\int^{kT}_0|\tilde{u}(t)|^{2}dt\leq\frac{k^{2}T^{2}}{4\pi^{2}}
 \int^{kT}_0|\dot{u}(t)|^{2}dt\,.\label{e2.2}
\end{equation}
It is easy to see that the norm $\|u\|_{k}$ is equivalently to the norm
$$
\|u\|=\Big[\int^{kT}_0|\dot{u}(t)|^{2}dt+|\bar{u}|^{2}\Big]^{1/2}.
$$
In the following, we will use this last norm.
It is well known that $\varphi_{k}$ is continuously differentiable with
$$
\varphi'_{k}(u)v=\int^{kT}_0[\dot{u}(t)\dot{v}(t)-V'(t,u(t))v(t)]dt,\quad
 \forall u,v\in H^{1}_{kT}
$$
and its critical points correspond to the $kT$-periodic solutions of the
system \eqref{eS}.

\subsection*{Proof of  Theorem \ref{thm1.1}}
Here, we will show that, for every positive integer $k$, one can find a
 $kT$-periodic solution $u_{k}$ of \eqref{eS} in such a way that the 
sequence $(u_{k})$ satisfies
\begin{equation}
\lim_{k\to\infty}\frac{1}{k}\varphi_{k}(u_{k})=-\infty. \label{e2.3}
\end{equation}
This will be done by using some estimates on the critical levels of
$\varphi_{k}$ given by the Saddle Point Theorem.
The following lemma will be needed for the study of the geometry of
the functionals $\varphi_{k}$.

\begin{lemma} \label{lem2.1} 
Assume that \eqref{V1}, \eqref{V2} hold. Then there exist a nonincreasing 
positive function $\theta\in C(]0,\infty[,\mathbb{R}^{+})$ and a positive 
constant $c_0$ satisfying the following conditions:
\begin{itemize}
\item[(i)] $\theta(s)\to 0$, $\theta(s)s\to +\infty$ as $s\to\infty$,

\item[(ii)] $\|V'(t,u)\|_{L^{1}}\leq c_0[\theta(\|u\|)\|u\|+1]$, for all
$u\in H^{1}_{kT}$, 

\item[(iii)]
\[
\frac{1}{[\theta(|x|)|x|]^{2}}\int^{kT}_0V(t,x)dt \to+\infty\quad
\text{as } |x|\to +\infty.
\]
\end{itemize}
\end{lemma}

\begin{proof}
For $u\in E$, let
$A=\{t\in[0,kT]/|u(t)|\geq\|u\|^{1/2}\}$.
By \eqref{V1}, we have
\begin{align*}
&\|V'(t,u)\|_{L^{1}} \\
&\leq \int^{T}_0\big[f(t)\omega(|u(t)|)|u(t)|+g(t)\big]dt\\
&\leq \|f\|_{L^{2}}\big(\int^{T}_0[\omega(|u(t)|)|u(t)|]^{2}dt\big)^{1/2}
 +\|g\|_{L^{1}}\\
&\leq \|f\|_{L^{2}}\Big(\int_{A}[\omega^{2}(\|u\|^{1/2})|u(t)|^{2}dt
 + \int_{[0,kT]-A}\sup_{s\geq 0}\omega^{2}(s)\|u\|dt\Big)^{1/2}+ \|g\|_{L^{1}}\\
&\leq \|f\|_{L^{2}}\big[\omega^{2}(\|u\|^{1/2}) \|u\|_{L^{2}}^{2}
 + kT\sup_{s\geq 0}\omega^{2}(s)\|u\|\big]^{1/2}+ \|g\|_{L^{1}}.
\end{align*}
So there exists a positive constant $c_0$ such that
$$
\|V'(t,u)\|_{L^{1}}\leq c_0\big(\big[\omega^{2}(\|u\|^{1/2})
 \|u\|^{2}+\|u\|\big]^{1/2} +1\big).
$$
Take
\begin{equation}
\theta(s)=\big[\omega^{2}(s^{1/2})+\frac{1}{s}\big]^{1/2},\ s > 0,
\label{e2.4}
\end{equation}
then $\theta$ satisfies (ii) and it is easy to see that $\theta$ satisfies (i).

Now, by (a), we have 
$\rho=\liminf_{s\to\infty}\frac{\omega^{2}(s)}{\omega^{2}(s^{1/2})} >0$.
By \eqref{V2}, for any $\gamma > 0$, there exists a positive constant
 $c_1$ such that
\begin{equation}
\int^{kT}_0V(t,x)dt \geq \gamma[\omega(|x|)|x|]^{2} - c_1. \label{e2.5}
\end{equation}
Combining \eqref{e2.4} and \eqref{e2.5} yields
\begin{equation}
\frac{\int^{kT}_0V(t,x)dt}{[\theta(|x|)|x|]^{2}}
\geq \frac{\gamma[\omega(|x|)|x|]^{2}-c_1}{\omega^{2} |x|^{1/2})|x|^{2}+|x|}.
 \label{e2.6}
\end{equation}
By the definition of $\rho$, there exists $R > 0$ such that for all $s\geq R$
\begin{equation}
\frac{\omega^{2}(s)s^{2}}{\omega^{2}(s^{1/2})s^{2}+s}\geq \frac{\rho}{2}, \label{e2.7}
\end{equation}
and
\begin{equation}
\frac{c_1}{\omega^{2}(s)s^{2}+s}\leq\frac{\gamma\rho}{4}, \label{e2.8}
\end{equation}
Therefore, by \eqref{e2.6}-\eqref{e2.8}, we have
$$
\frac{\int^{kT}_0V(t,x)dt}{[\theta(|x|)|x|]^{2}}\geq \frac{\gamma\rho}{4},\quad
 \forall |x|\geq R.
$$
Since $\gamma$ is arbitrary chosen, condition (iii) holds.
The proof of Lemma \ref{lem2.1} is complete.
\end{proof}

Now, we need to show that, for every positive integer $k$, one can 
find a critical point $u_{k}$ of the functional $\varphi_{k}$ in such a
 way that \eqref{e2.3} holds. To this aim, we will apply the Saddle Point
 Theorem to each of the $\varphi_{k}$' s. Let us fix $k$ and write 
$H^{1}_{kT}=\mathbb{R}^{N}\oplus \tilde{H}^{1}_{kT}$ where $\mathbb{R}^{N}$ 
is identified with the set of constant functions and $\tilde{H}^{1}_{kT}$ 
consists of functions $u$ in $H^{1}_{kT}$ such that $\int^{kT}_0u(t)dt=0$.
First, we prove the Palais-Smale condition. 

\begin{lemma} \label{lem2.2} 
Assume that \eqref{V1} and \eqref{V2} hold. Then $\varphi_{k}$ satisfies 
the Palais-Smale condition.
\end{lemma}

\begin{proof}
 Let $(u_{n})$ be a sequence in $H^{1}_{kT}$ such that $(\varphi_{k}(u_{n}))$
 is bounded and $\varphi'_{k}(u_{n})\to 0$ as $n\to\infty$. 
In particular, for a positive constant $c_2$ we will have
\begin{equation}
\varphi'_{k}(u_{n})\tilde{u}_{n}
=\int^{kT}_0[|\dot{u_{n}}(t)|^{2}-V'(t,u_{n}(t))\tilde{u}_{n}(t)]dt
\leq c_2\|\tilde{u}_{n}\|. \label{e2.9}
\end{equation}
Since $\theta$ is non increasing and $\|u\|\geq max(|\bar{u}|,\|\tilde{u}\|)$,
we obtain
\begin{equation}
\theta(\|u\|)\leq \min(\theta(|\bar{u}|),\theta(\|\tilde{u}\|)). \label{e2.10}
\end{equation}
Combining Sobolev's inequality, Lemma \ref{lem2.1} (ii) and \eqref{e2.10},
 we can find a positive constant $c_3$ such that
\begin{equation}
\begin{split}
|\int^{kT}_0V'(t,u_{n})\tilde{u}_{n}dt|
&\leq c_0\|\tilde{u}_{n}\|_{\infty}[\theta(\|u_{n}\|)\|u_{n}\|+1]\\
&\leq c_0\|\tilde{u}_{n}\|_{\infty}\big[\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\|
+  \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1\big]\\
&\leq c_3\|\tilde{u}_{n}\|\big[\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\|+
 \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1\big]
\end{split} \label{e2.11}
\end{equation}
for all $n\in\mathbb{N}$.

From Wirtinger's inequality, there exists a constant $c_4>0$ such that
\begin{equation}
\|\dot{u}_{n}\|_{L^{2}}\leq\|\tilde{u}_{n}\|\leq c^{-1/2}_4\|\dot{u}_{n}\|_{L^{2}}.
\label{e2.12}
\end{equation}
Therefore, by \eqref{e2.9} and \eqref{e2.11}, we obtain
\begin{equation}
\begin{split}
c_2\|\tilde{u}_{n}\|
&\geq \varphi'_{k}(u_{n}).\tilde{u}_{n} =\int^{kT}_0|\dot{u_{n}}|^{2}dt
 -\int^{kT}_0V'(t,u_{n})\tilde{u}_{n}dt\\
&\geq c_4\|\tilde{u}_{n}\|^{2}-c_3\|\tilde{u}_{n}\|
 [\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\|
 +  \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1].
\end{split} \label{e2.13}
\end{equation}
Assume that $(\|\tilde{u}_{n}\|)$ is unbounded, by going to a subsequence
if necessary, we can assume that $\|\tilde{u}_{n}\|\to\infty$ as $n\to\infty$.
 Since $\theta(s)\to 0$ as $s\to \infty$, we deduce from \eqref{e2.13}
that there exists a positive constant $c_5$ such that
\begin{equation}
\|\tilde{u}_{n}\|\leq c_5\theta(|\bar{u}_{n}|)|\bar{u}_{n}|
=c_5\big[\omega^{2}(|\bar{u}_{n}|^{1/2})|\bar{u}_{n}|^{2}+ |\bar{u}_{n}|\big]^{1/2}
\label{e2.14}
\end{equation}
for $n$ large enough. Since $\omega$ is bounded, it follows that
$|\bar{u}_{n}|\to\infty$ as $n\to\infty$.

Now, by the Mean Value Theorem and Lemma \ref{lem2.1} (ii), we obtain
\begin{equation}
\begin{split}
|\int^{kT}_0(V(t,u_{n})-V(t,\bar{u}_{n}))dt|
&=|\int^{kT}_0\int^{1}_0V'(t,\bar{u}_{n})+s\tilde{u}_{n})\tilde{u}_{n}\,ds\,dt|
\\
&\leq\|\tilde{u}_{n}\|_{\infty}\int^{1}_0\int^{kT}_0|V'(t,\bar{u}_{n}+s\tilde{u}_{n})|
 \,ds\,dt
\\
&\leq c_0\|\tilde{u}_{n}\|_{\infty}\int^{1}_0\big[\theta(\|\bar{u}_{n}
+s\tilde{u}_{n}\|)\|\bar{u}_{n}+s\tilde{u}_{n}\| +1\big]ds.
\end{split} \label{e2.15}
\end{equation}
Since $\|\bar{u}_{n}+s\tilde{u}_{n}\|\geq \|\bar{u}_{n}\|$ for all
 $s\in [0,1]$, we deduce from \eqref{e2.1}, \eqref{e2.14} and \eqref{e2.15}
that there exists a positive constant $c_6$ such that
\begin{equation}
\begin{split}
&|\int^{kT}_0(V(t,u_{n})-V(t,\bar{u}_{n}))dt|\\
&\leq c_6\Big([\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}
+ \theta(|\bar{u}_{n}|)[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}
+\theta(|\bar{u}_{n}|)|\bar{u}_{n}|\Big). 
\end{split} \label{e2.16}
\end{equation}
Thus, by \eqref{e2.14} and \eqref{e2.16}, we obtain for a positive constant
$c_7$,
\begin{align*}
&\varphi_{k}(u_{n})\\
&=\frac{1}{2}\|\dot{u}_{n}\|^{2}_{L^{2}}-\int^{kT}_0(V(t,u_{n})-V(t,\bar{u}_{n}))dt
 - \int^{kT}_0V(t,\bar{u}_{n})dt
\\
&\leq c_7\Big([\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}
 +  \theta(|\bar{u}_{n}|)[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}
+\theta(|\bar{u}_{n}|)|\bar{u}_{n}|\Big)-\int^{kT}_0V(t,\bar{u}_{n})dt
\\
&=c_7[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}\Big[1+\theta(|\bar{u}_{n}|)
+{1\over{\theta(|\bar{u}_{n}|)|\bar{u}_{n}|}}
-\frac{1}{c_7[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}}
\int^{kT}_0V(t,\bar{u}_{n})dt\Big]
\end{align*}
so $\varphi_{k}(u_{n})\to -\infty\ as\ n\to\infty$. This contradicts the boundedness
 of $(\varphi_{k}(u_{n}))$. Therefore $(\|\tilde{u}_{n}\|)$ is bounded.

It remains to prove that $(|\bar{u}_{n}|)$ is bounded. Assume the contrary. 
By taking a subsequence, if necessary, we can assume that 
$\|\bar{u}_{n}\|\to\infty$ as $n\to \infty$. By the preceding calculus, 
we obtain for some positive constants $c_8, c_9$ such that
\begin{align*}
&\varphi_{k}(u_{n})\\
&\leq c_8\Big[\|\tilde{u}_{n}\|^{2} +\|\tilde{u}_{n}\|\theta(|\bar{u}_{n}|)
 |\bar{u}_{n}|+\theta(|\bar{u}_{n}|)\|\tilde{u}_{n}\|+1\Big]
 -\int^{kT}_0V(t,\bar{u}_{n})dt
\\
&\leq c_9\big[1+\theta(|\bar{u}_{n}|)|\bar{u}_{n}|+ \theta(|\bar{u}_{n}|)\big]
 -\int^{kT}_0V(t,\bar{u}_{n})dt
\\
&\leq c_9[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}
 \Big[\frac{1+\theta(|\bar{u}_{n}|)}
{[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}}
+\frac{1}{\theta(|\bar{u}_{n}|)|\bar{u}_{n}|} 
-\frac{1}{c_9[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2}}
 \int^{kT}_0V(t,\bar{u}_{n})dt\Big]
\end{align*}
so $\varphi_{k}(u_{n})\to-\infty$ as $n\to\infty$, which also contradicts
 the boundedness of $(\varphi_{k}(u_{n}))$. So $(|\bar{u}_{n}|)$ is bounded
 and then $(\|u_{n}\|)$ is also bounded. By a standard argument,
we conclude that $(u_{n})$ possesses a convergent subsequence and the 
proof is complete. 
\end{proof}

Now, it is easy to show that \eqref{V2} yields
\begin{equation}
\varphi_{k}(u)=-\int^{kT}_0V(t,u)dt\to-\infty\quad\text{as }
 |u|\to\infty\text{ in } \mathbb{R}^{N}. \label{e2.17}
\end{equation}
On the other hand, by the Mean Value Theorem, \eqref{V1} and H\"older's inequality,
for all $u\in \tilde{H}^{1}_{kT}$ and $a\in\mathbb{R}^{N}$ $|a| > 0$, we have
\begin{align*}
&|\int^{kT}_0(V(t,u)-V(t,a))dt|\\
&=|\int^{kT}_0\int^{1}_0V'(t,a+s(u-a))(u-a)\,ds\,dt|\\
&\leq\|u-a\|_{\infty}\int^{1}_0\int^{kT}_0|V'(t,a+s(u-a))|\,dt\,ds\\
&\leq\|u-a\|_{\infty}\int^{1}_0\int^{kT}_0\big[f(t)\omega(|a+s(u-a)|)|a+s(u-a)|
  +g(t)\big]dt \\
&\leq\|u-a\|_{\infty}\Big(\|f\|_{L^{2}}\int^{1}_0
 \big[\int^{kT}_0(\omega(|a+s(u-a)|)|a+s(u-a)|)^{2}dt\big]^{1\over 2}ds
 + \|g\|_{L^{1}}\Big).
\end{align*}
For $s\in[0,1]$, take
$$
A(s)=\{t\in [0,kT]/|a+s(u(t)-a)|\geq |a|\}.
$$
By a classical calculation as in the proof of Lemma \ref{lem2.1}, we obtain some
positive constants $c_{10}$ and $c(a)$ such that
$$
|\int^{kT}_0(V(t,u)-V(t,a))dt|\leq c_{10}\omega(a)\|u\|^{2}+c(a)(\|u\|+1).
$$
Since $\omega(|a|)\to 0$ as $|a|\to\infty$, there exists $|a|>0$ such that
 $c_{10}\omega(|a|)\leq\frac{1}{4}c^{2}_4$ and then we obtain
$$
|\int^{kT}_0(V(t,u)-V(t,a))dt|\leq \frac{1}{4}c^{2}_4\|u\|^{2}+c(a)(\|u\|+1)
$$
which implies that
\begin{equation}
\varphi_{k}(u)\geq\frac{1}{4}c^{2}_4\|u\|^{2}-c(a)(\|u\|+1)-\int^{kT}_0V(t,a)dt
\to\infty
\quad\text{as }\|u\|\to\infty. \label{e2.18}
\end{equation}
We deduce from Lemma \ref{lem2.2} and \eqref{e2.17}, \eqref{e2.18} that all the
Saddle Point Theorem's assumptions are satisfied. Therefore the functional
$\varphi_{k}$ possesses at least a critical point $u_{k}$ satisfying
\begin{equation}
-\infty<\inf_{\tilde{H}^{1}_{kT}}\varphi_{k}\leq \varphi_{k}(u_{k})
\leq\sup_{\mathbb{R}^{N}+e_{k}}\varphi_{k} \label{e2.19}
\end{equation}
where $e_{k}(t)=k \cos(\frac{\sigma t}{k})x_0$ for $t\in\mathbb{R}$,
$\sigma={{2\pi}\over T}$ and some $x_0\in\mathbb{R}^{N}$ with $|x_0|=1$.
The first part of Theorem \ref{thm1.1} is proved.

Next, we will prove that the sequence $(u_{k})_{k\geq 1}$ obtained above 
satisfies \eqref{e2.3}. For this aim, the following two lemmas will be needed.


\begin{lemma}[\cite{t1}] \label{lem2.3}
 If \eqref{V3i} holds, then for every $\delta > 0$ there exists a measurable 
subset $C_{\delta}$ of $C$ with $\operatorname{meas}(C-C_{\delta})<\delta$ 
such that
$$ 
V(t,x)\to +\infty\quad\text{as 
$|x|\to \infty$, uniformly in }t\in C_{\delta}.
$$
\end{lemma}

\begin{lemma} \label{lem2.4} Suppose that $V$ satisfies \eqref{V3i} - \eqref{V3ii}, 
then
\begin{equation}
\limsup_{k\to\infty} \sup_{x \in\mathbb{R}^{N}}\frac{1}{k}\varphi_{k}(x+e_{k})=-\infty.
 \label{e2.20}
\end{equation}
\end{lemma}

\begin{proof}
Let $x\in \mathbb{R}^{N}$, we have
$$
\varphi_{k}(x+e_{k})=\frac{1}{4}kT\sigma^{2}-\int^{kT}_0V(t,x+e_{k}(t))dt.
$$
By \eqref{V3i} and Lemma \ref{lem2.3}, for $\delta=\frac{1}{2}\operatorname{meas}(C)$ 
and all $\gamma > 0$, there exist a measurable subset $C_{\delta}\subset C$ with 
$\operatorname{meas}(C-C_{\delta})<\delta$ and $r > 0$ such that
\begin{equation}
V(t,x)\geq\gamma,\quad \forall |x|\geq r\; \forall t\in C_{\delta}. \label{e2.21}
\end{equation}
Let
$$
B_{k}=\{t\in [0,kT]: |x+e_{k}(t)|\leq r\}.
$$
Then we have
\begin{equation}
\operatorname{meas}(B_{k})\leq {{k\delta}\over 2}. \label{e2.22}
\end{equation}
In fact, if $\operatorname{meas}(B_{k}) > k\delta/ 2$,
there exists $t_0\in B_{k}$ such that
\begin{gather}
\frac{k\delta}{8}\leq t_0\leq {{k T}\over 2}-\frac{k\delta}{8}, \label{e2.23}\\
\frac{k T}{2}+\frac{k\delta}{8}\leq t_0\leq kT-\frac{k\delta}{8}, \label{e2.24}
\end{gather}
and there exists $t_1\in B_{k}$ such that
\begin{gather}
|t_1-t_0|\geq {{k\delta}\over 8}, \label{e2.25}\\
|t_1-(kT-t_0)|\geq {{k\delta}\over 8}. \label{e2.26}
\end{gather}
It follows from \eqref{e2.26} that
\begin{equation}
|{{t_0+t_1}\over {2k}}-{T\over 2}|\geq \frac{\delta}{16}. \label{e2.27}
\end{equation}
By \eqref{e2.23} and \eqref{e2.24}, one has
\begin{equation}
\frac{\delta}{16}\leq\frac{t_0+t_1}{2k}\leq T-\frac{\delta}{16}. \label{e2.28}
\end{equation}
Combining \eqref{e2.27} and \eqref{e2.28}, yields
\begin{equation}
|\sin(\frac{t_0+t_1}{2k}\sigma)|\geq \sin(\frac{\sigma\delta}{16}). \label{e2.29}
\end{equation}
On the other hand, by \eqref{e2.25} we have
$$
|\cos(\frac{\sigma t_0}{k})-\cos(\frac{\sigma t_1}{k})|
=2|\sin(\frac{t_0+t_1} {2k}\sigma)||\sin(\frac{t_0-t_1}{2k}\sigma)|\geq 2\sin^{2}
(\frac{\sigma\delta}{16}),
$$
and
$$
|\cos(\frac{\sigma t_0}{2k})-\cos(\frac{\sigma t_1}{2k})|
=\frac{1}{k} |(x+e_{k}(t_1))-(x+e_{k}(t_0))|\leq\frac{2r}{k},
$$
which is impossible for large $k$. Hence \eqref{e2.22} holds.

Now, let
$C_{k}=\cup^{k-1}_{j=0}(jT+C_{\delta})$.
It follows from \eqref{e2.22} that for all $k$,
$$
\operatorname{meas}(C_{k}-B_{k})\geq\frac{k\delta}{2}.
$$
By \eqref{e2.21}, we have
\begin{align*}
k^{-1}\varphi_{k}(x+e_{k})
&=\frac{1}{4}T\sigma^{2}-k^{-1}\int^{kT}_0V(t,x+e_{k}(t))dt\\
&\leq \frac{1}{4}T\sigma^{2}-k^{-1}\int_{[0,kT]-(C_{k}-B_{k})}h(t)dt
 -k^{-1}\gamma \operatorname{meas}(C_{k}-B_{k})\\
&\leq \frac{1}{4}T\sigma^{2}-\int^{T}_0|h(t)|dt-\frac{\delta\gamma}{2}
\end{align*}
for all large $k$, which implies
$$
\limsup_{k\to\infty}\sup_{x\in\mathbb{R}^{N}}k^{-1}\varphi_{k}(x+e_{k})
\leq\frac{1}{4} T\sigma^{2}+\int^{T}_0|h(t)|dt-\frac{\delta\gamma}{2}.
$$
By the arbitrariness of $\gamma$, we obtain
$$
\limsup_{k\to\infty}\sup_{x\in\mathbb{R}^{N}}k^{-1}\varphi_{k}(x+e_{k})=-\infty.
$$
The proof of Lemma \ref{lem2.4} is complete. 
\end{proof}

It remains to prove that the sequence $(\|u_{k}\|_{\infty})$ of solutions
 of \eqref{eS} obtained above, is unbounded. Arguing by contradiction, 
assume that $(\|u_{k}\|_{\infty})$ is bounded, then there exists $R > 0$ 
such that $(\|u_{k}\|_{\infty})\leq R$ for all $k \geq 1$. We have
\begin{equation}
k^{-1}\varphi_{k}(u_{k})\geq-k^{-1}\int^{kT}_0V(t,u_{k})dt. \label{e2.30}
\end{equation}
Since $V$ is $T$-periodic in $t$ and continuous, then there exists a constant
 $\rho > 0$ such that
$$
|V(t,x)|\leq \rho,\quad \forall x\in\mathbb{R}^{N},\; |x|\leq R,\text{ a.e. } t\in\mathbb{R}.
$$
Therefore,
\begin{equation}
k^{-1}\varphi_{k}(u_{k})\geq -\rho T \label{e2.31}
\end{equation}
which contradicts \eqref{e2.19} with \eqref{e2.20}.
The proof of Theorem \ref{thm1.1} is complete.


\subsection*{Proof of Theorem \ref{thm1.2}} 
The following lemma will be needed.

\begin{lemma} \label{lem2.5} 
Let \eqref{V1}, \eqref{V5} hold. Then for all $\rho>0$, there exists 
a constant $c_{\rho}\geq 0$ such that for all $x\in\mathbb{R}^{N}$, $|x|>1$ 
and for a.e. $t\in[0,T]$,
\begin{equation}
V(t,x)\geq V(t,0)+\frac{\rho}{2}[\omega(|x|)|x|]^{2}(1-\frac{1}{|x|^{2}})
-c_{\rho}ln(|x|)-\frac{1}{2}a\sup_{r\geq 0}\omega(r)-g(t). \label{e2.32}
\end{equation}
\end{lemma}

\begin{proof}
For $x\in \mathbb{R}^{N}$, $|x | > 1$, we have
\begin{equation}
\begin{split}
V(t,x)&=V(t,0)+\int^{1}_0V'(t,sx)x\,ds\\
&=V(t,0)+\int^{\frac{1}{|x|}}_0V'(t,sx)x\,ds
+\int^{1}_{\frac{1}{|x|}}V'(t,sx)x\,ds.
\end{split} \label{e2.33}
\end{equation}
By \eqref{V1}, we have
\begin{equation}
\begin{split}
|\int^{\frac{1}{|x|}}_0V'(t,sx)x\,ds|
&\leq|x| \int^{\frac{1} {|x|}}_0 [f(t)\omega(|sx|)|sx|+ g(t)]dt\\
&\leq|x|[f(t)\sup_{r\geq 0}\omega(r)|x|\int^{\frac{1}{|x|}}_0s\,ds
 + g(t){\frac{1}{|x|}}]\\
&\leq\frac{1}{2}f(t)\sup_{r\geq 0}\omega(r)+g(t).
\end{split} \label{e2.34}
\end{equation}
Let $\rho>0$, then by \eqref{V5}, there exists a positive constant
 $c_{\rho}$ such that
\begin{equation}
V'(t,x)x\geq \rho[\omega(|x|)|x|]^{2}-c_{\rho}. \label{e2.35}
\end{equation}
Therefore,
\begin{equation}
\begin{split}
\int^{1}_{\frac{1}{|x|}}V'(t,sx)x\,ds
&=\int^{1}_{\frac{1}{|x|}}V'(t,sx)sx \frac{ds}{s}\\
&\geq \int^{1}_{\frac{1}{|x|}}(\rho[\omega(|sx|)|sx|]^{2}-c_{\rho})\frac{ds}{s}\\
&\geq \frac{\rho}{2}[\omega(|x|)|x|]^{2}(1-\frac{1}{|x|^{2}})
-c_{\rho} b(t)ln(|x|).
\end{split} \label{e2.36}
\end{equation}
Combining \eqref{e2.33}, \eqref{e2.34} and \eqref{e2.36}, we obtain \eqref{e2.32}
and Lemma \ref{lem2.5} is proved.
\end{proof}

Now, since $a_0=\liminf_{s\to\infty}\frac{\omega(s)}{\omega(s^{1\over 2})}>0$, 
 for $s$ large enough, we have
\begin{equation}
\frac{1}{\omega(s)}\leq \frac{1}{a_0\omega(s^{1/2})} \label{e2.37}
\end{equation}
which implies that for $|x|$ large enough
\begin{equation}
\frac{ln(|x|)}{[\omega(|x|)|x|]^{2}}\leq \frac{ln(|x|)}{|x|}
\frac{1}{a^{2}_0[\omega(|x|^{1/2})|x|^{1/2}]^{2}}\to 0\quad\text{as }
 |x|\to\infty. \label{e2.38}
\end{equation}
Combining \eqref{e2.32}, \eqref{e2.38} we obtain $(V_4)$.
By applying Corollary \ref{coro1.1}, we obtain a sequence $(u_{k})$ of $kT$-periodic
solutions of \eqref{eS} such that $\lim_{k\to\infty}\|u_{k}\|_{\infty} =+\infty$.

It remains to analyst the minimal periods of the subharmonic solutions 
found with the previous results. For this, we will split the problem into two parts. 
Firstly, we claim that for a sufficiently large integer $k$, the subharmonic 
solution $u_{k}$ is not $T$-periodic. 
In fact, let $S_{T}$ be the set of $T$-periodic solutions of \eqref{eS}, 
we will show that $S_{T}$ is bounded in $H^{1}_{T}$. 
Assume by contradiction that there exists a sequence $(u_{n})$ in $S_{T}$ 
such that $\|u_{n}\|_1\to\infty$ as $n\to\infty$. 
Let us write $u_{n}(t)=\bar{u}_{n}+\tilde{u}_{n}(t)$, 
where $\bar{u}_{n}$ is the mean value of $u_{n}$. 
Multiplying both sides of the identity
\begin{equation}
\ddot{u}_{n}(t)+V'(t,u_{n}(t))=0 \label{e2.39}
\end{equation}
by $\tilde{u}_{n}(t)$ and integrating, we obtain by \eqref{V1} and H\"older's
inequality
\begin{equation}
\begin{split}
\int^{T}_0|\dot{u}_{n}|^{2}dt
&=-\int^{T}_0\ddot{u}_{n}\tilde{u}_{n}dt \\
&=\int^{T}_0V'(t,u_{n})\tilde{u}_{n}dt \\
&\leq\|\tilde{u}_{n}\|_{\infty}\int^{T}_0\big[f(t)\omega(|u_{n}(t)|)|u_{n}(t)|
+ g(t)\big]dt \\
&\leq\|\tilde{u}_{n}\|_{\infty}\Big[\|f\|_{L^{2}}
 \big(\int^{T}_0 [\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt\big)^{1/2}
+ \|g\|_{L^{1}}\Big].
\end{split} \label{e2.40}
\end{equation}
By \eqref{e2.1}, \eqref{e2.2} and \eqref{e2.40}, there exists a positive
constant $c_{11}$ such that
\begin{equation}
\|\tilde{u}_{n}\|_1\leq c_{11}\big[\big(\int^{T}_0[\omega(|u_{n}(t)|)
|u_{n}(t)|]^{2}dt\big)^{1/2}+1\big]. \label{e2.41}
\end{equation}
Let $\rho>0$ and let $c_{\rho}$ be a constant satisfying \eqref{e2.35}.
 Multiplying both sides of the identity \eqref{e2.39} by $u_{n}(t)$
and integrating
\begin{equation}
\begin{split}
\int^{T}_0|\dot{u}_{n}|^{2}dt
&= -\int^{T}_0\ddot{u}_{n}u_{n}dt \\
&=\int^{T}_0V'(t,u_{n})u_{n}dt \\
&\geq \rho\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt-c_{\rho}T.
\end{split}\label{e2.42}
\end{equation}
We deduce from \eqref{e2.42} and Wirtinger inequality that there exists
a positive constant $c_{12}$ such that
\begin{equation}
\|\tilde{u}_{n}\|^{2}_1\geq c_{12}\big[\rho\int^{T}_0[\omega(|u_{n}(t)|)
|u_{n}(t)|]^{2}dt-c_{\rho}T\big]. \label{e2.43}
\end{equation}
Combining \eqref{e2.41} with \eqref{e2.43}, we can find a positive
 constant $c_{13}$ such that
\begin{equation}
\rho\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt-c_{\rho}T\leq
c_{13} \big[\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt+1\big].
\label{e2.44}
\end{equation}
Since $\rho$ is arbitrary chosen,
\begin{equation}
(\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt)\quad \text{is bounded.}
 \label{e2.45}
\end{equation}
Combining \eqref{e2.41} and \eqref{e2.45} yields $(\tilde{u}_{n})$ is bounded
in $H^{1}_{T}$ and then $|\bar{u}_{n}|\to\infty$ as $n\to\infty$.
Since the embedding $H^{1}_{T}\to L^{2}(0,T;\mathbb{R}^{N})$, $u\to u$ is compact,
then we can assume, by going to a subsequence if necessary, that
$\tilde{u}_{n}(t)\to\tilde{u}(t)$ as $n\to\infty$, a.e. $t\in [0,T]$. We deduce that
\begin{equation}
|u_{n}(t)|\to\infty\quad \text{as $n\to\infty$, a.e. $t\in [0,T]$.}
\label{e2.46}
\end{equation}
Fatou's lemma and \eqref{e2.46} imply
\begin{equation}
\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt\to \infty\quad
\text{as }n\to\infty \label{e2.47}
\end{equation}
which contradicts \eqref{e2.45}. Therefore $S_{T}$ is bounded in $H^{1}_{T}$.
As a consequence, $\varphi_1(S_{T})$ is bounded, and since for any $u\in S_{T}$
one has $\varphi_{k}(u)=k\varphi_1(u)$, then there exists a positive constant
$c_{14}$ such that
\begin{equation}
{1\over k}|\varphi_{k}(u)| \leq c_{14},\quad \forall u\in S_{T},\; \forall k\geq 1.
 \label{e2.48}
\end{equation}
Consequently by \eqref{e2.3}, for $k$ large enough, $u_{k}\notin S_{T}$.
Finally, assumption \eqref{V6} requires that the minimal period of each solution
$u_{k}$ of \eqref{eS} is an integer multiple of $T$. So if $k$ is chosen
 to be a prime number, the minimal period of $u_{k}$ has to be $kT$.
The proof of Theorem \ref{thm1.2} is complete.

\subsection*{Acknowledgments} 
I wish to thank the anonymous referee for his/her suggestions and interesting 
remarks.

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\end{document}