\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 $$\ddot{u}(t)+V'(t,u(t))=0, \label{eS}$$ 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 $$\ddot{u}(t)+V'(u(t))=e(t) \label{e1.3}$$ 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 $$|V'(t,x)|\leq f(t)\omega(|x|)|x|+g(t),\quad \forall x\in\mathbb{R}^{N},\text{a.e. } t\in[0,T]; \label{V1}$$ $$\frac{1}{[\omega(|x|)|x|]^{2}}\int^{T}_0V(t,x)dt \to+\infty\quad \text{as } |x|\to\infty; \label{V2}$$ 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 $$\frac{V'(t,x)x}{[\omega(|x|)|x|]^{2}}\to+\infty\quad \text{as |x|\to\infty, uniformly for t\in [0,T]}. \label{V5}$$ 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: $$\label{V6} \parbox{9cm}{ If u(t) is a periodic function with minimal period rT with r rational, and V'(t,u(t)) is a periodic function with minimal period rT, then r is necessarily an integer.}$$ 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, $$\|\tilde{u}\|^{2}_{\infty} \leq {{kT}\over{12}}\int^{kT}_0|\dot{u}(t)|^{2}dt, \label{e2.1}$$ and Wirtinger's\ inequality, $$\int^{kT}_0|\tilde{u}(t)|^{2}dt\leq\frac{k^{2}T^{2}}{4\pi^{2}} \int^{kT}_0|\dot{u}(t)|^{2}dt\,.\label{e2.2}$$ 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 $$\lim_{k\to\infty}\frac{1}{k}\varphi_{k}(u_{k})=-\infty. \label{e2.3}$$ 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 $$\theta(s)=\big[\omega^{2}(s^{1/2})+\frac{1}{s}\big]^{1/2},\ s > 0, \label{e2.4}$$ 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 $$\int^{kT}_0V(t,x)dt \geq \gamma[\omega(|x|)|x|]^{2} - c_1. \label{e2.5}$$ Combining \eqref{e2.4} and \eqref{e2.5} yields $$\frac{\int^{kT}_0V(t,x)dt}{[\theta(|x|)|x|]^{2}} \geq \frac{\gamma[\omega(|x|)|x|]^{2}-c_1}{\omega^{2} |x|^{1/2})|x|^{2}+|x|}. \label{e2.6}$$ By the definition of $\rho$, there exists $R > 0$ such that for all $s\geq R$ $$\frac{\omega^{2}(s)s^{2}}{\omega^{2}(s^{1/2})s^{2}+s}\geq \frac{\rho}{2}, \label{e2.7}$$ and $$\frac{c_1}{\omega^{2}(s)s^{2}+s}\leq\frac{\gamma\rho}{4}, \label{e2.8}$$ 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 $$\varphi'_{k}(u_{n})\tilde{u}_{n} =\int^{kT}_0[|\dot{u_{n}}(t)|^{2}-V'(t,u_{n}(t))\tilde{u}_{n}(t)]dt \leq c_2\|\tilde{u}_{n}\|. \label{e2.9}$$ Since $\theta$ is non increasing and $\|u\|\geq max(|\bar{u}|,\|\tilde{u}\|)$, we obtain $$\theta(\|u\|)\leq \min(\theta(|\bar{u}|),\theta(\|\tilde{u}\|)). \label{e2.10}$$ Combining Sobolev's inequality, Lemma \ref{lem2.1} (ii) and \eqref{e2.10}, we can find a positive constant $c_3$ such that $$\begin{split} |\int^{kT}_0V'(t,u_{n})\tilde{u}_{n}dt| &\leq c_0\|\tilde{u}_{n}\|_{\infty}[\theta(\|u_{n}\|)\|u_{n}\|+1]\\ &\leq c_0\|\tilde{u}_{n}\|_{\infty}\big[\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\| + \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1\big]\\ &\leq c_3\|\tilde{u}_{n}\|\big[\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\|+ \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1\big] \end{split} \label{e2.11}$$ for all $n\in\mathbb{N}$. From Wirtinger's inequality, there exists a constant $c_4>0$ such that $$\|\dot{u}_{n}\|_{L^{2}}\leq\|\tilde{u}_{n}\|\leq c^{-1/2}_4\|\dot{u}_{n}\|_{L^{2}}. \label{e2.12}$$ Therefore, by \eqref{e2.9} and \eqref{e2.11}, we obtain $$\begin{split} c_2\|\tilde{u}_{n}\| &\geq \varphi'_{k}(u_{n}).\tilde{u}_{n} =\int^{kT}_0|\dot{u_{n}}|^{2}dt -\int^{kT}_0V'(t,u_{n})\tilde{u}_{n}dt\\ &\geq c_4\|\tilde{u}_{n}\|^{2}-c_3\|\tilde{u}_{n}\| [\theta(\|\tilde{u}_{n}\|)\|\tilde{u}_{n}\| + \theta(|\bar{u}_{n}|)|\bar{u}_{n}|+1]. \end{split} \label{e2.13}$$ 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 $$\|\tilde{u}_{n}\|\leq c_5\theta(|\bar{u}_{n}|)|\bar{u}_{n}| =c_5\big[\omega^{2}(|\bar{u}_{n}|^{1/2})|\bar{u}_{n}|^{2}+ |\bar{u}_{n}|\big]^{1/2} \label{e2.14}$$ 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 $$\begin{split} |\int^{kT}_0(V(t,u_{n})-V(t,\bar{u}_{n}))dt| &=|\int^{kT}_0\int^{1}_0V'(t,\bar{u}_{n})+s\tilde{u}_{n})\tilde{u}_{n}\,ds\,dt| \\ &\leq\|\tilde{u}_{n}\|_{\infty}\int^{1}_0\int^{kT}_0|V'(t,\bar{u}_{n}+s\tilde{u}_{n})| \,ds\,dt \\ &\leq c_0\|\tilde{u}_{n}\|_{\infty}\int^{1}_0\big[\theta(\|\bar{u}_{n} +s\tilde{u}_{n}\|)\|\bar{u}_{n}+s\tilde{u}_{n}\| +1\big]ds. \end{split} \label{e2.15}$$ 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 $$\begin{split} &|\int^{kT}_0(V(t,u_{n})-V(t,\bar{u}_{n}))dt|\\ &\leq c_6\Big([\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2} + \theta(|\bar{u}_{n}|)[\theta(|\bar{u}_{n}|)|\bar{u}_{n}|]^{2} +\theta(|\bar{u}_{n}|)|\bar{u}_{n}|\Big). \end{split} \label{e2.16}$$ 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 $$\varphi_{k}(u)=-\int^{kT}_0V(t,u)dt\to-\infty\quad\text{as } |u|\to\infty\text{ in } \mathbb{R}^{N}. \label{e2.17}$$ 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 $$\varphi_{k}(u)\geq\frac{1}{4}c^{2}_4\|u\|^{2}-c(a)(\|u\|+1)-\int^{kT}_0V(t,a)dt \to\infty \quad\text{as }\|u\|\to\infty. \label{e2.18}$$ 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 $$-\infty<\inf_{\tilde{H}^{1}_{kT}}\varphi_{k}\leq \varphi_{k}(u_{k}) \leq\sup_{\mathbb{R}^{N}+e_{k}}\varphi_{k} \label{e2.19}$$ 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 $$\limsup_{k\to\infty} \sup_{x \in\mathbb{R}^{N}}\frac{1}{k}\varphi_{k}(x+e_{k})=-\infty. \label{e2.20}$$ \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 $$V(t,x)\geq\gamma,\quad \forall |x|\geq r\; \forall t\in C_{\delta}. \label{e2.21}$$ Let $$B_{k}=\{t\in [0,kT]: |x+e_{k}(t)|\leq r\}.$$ Then we have $$\operatorname{meas}(B_{k})\leq {{k\delta}\over 2}. \label{e2.22}$$ 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 $$|{{t_0+t_1}\over {2k}}-{T\over 2}|\geq \frac{\delta}{16}. \label{e2.27}$$ By \eqref{e2.23} and \eqref{e2.24}, one has $$\frac{\delta}{16}\leq\frac{t_0+t_1}{2k}\leq T-\frac{\delta}{16}. \label{e2.28}$$ Combining \eqref{e2.27} and \eqref{e2.28}, yields $$|\sin(\frac{t_0+t_1}{2k}\sigma)|\geq \sin(\frac{\sigma\delta}{16}). \label{e2.29}$$ 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 $$k^{-1}\varphi_{k}(u_{k})\geq-k^{-1}\int^{kT}_0V(t,u_{k})dt. \label{e2.30}$$ 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, $$k^{-1}\varphi_{k}(u_{k})\geq -\rho T \label{e2.31}$$ 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. $t\in[0,T]$, $$V(t,x)\geq V(t,0)+\frac{\rho}{2}[\omega(|x|)|x|]^{2}(1-\frac{1}{|x|^{2}}) -c_{\rho}ln(|x|)-\frac{1}{2}a\sup_{r\geq 0}\omega(r)-g(t). \label{e2.32}$$ \end{lemma} \begin{proof} For $x\in \mathbb{R}^{N}$, $|x | > 1$, we have $$\begin{split} V(t,x)&=V(t,0)+\int^{1}_0V'(t,sx)x\,ds\\ &=V(t,0)+\int^{\frac{1}{|x|}}_0V'(t,sx)x\,ds +\int^{1}_{\frac{1}{|x|}}V'(t,sx)x\,ds. \end{split} \label{e2.33}$$ By \eqref{V1}, we have $$\begin{split} |\int^{\frac{1}{|x|}}_0V'(t,sx)x\,ds| &\leq|x| \int^{\frac{1} {|x|}}_0 [f(t)\omega(|sx|)|sx|+ g(t)]dt\\ &\leq|x|[f(t)\sup_{r\geq 0}\omega(r)|x|\int^{\frac{1}{|x|}}_0s\,ds + g(t){\frac{1}{|x|}}]\\ &\leq\frac{1}{2}f(t)\sup_{r\geq 0}\omega(r)+g(t). \end{split} \label{e2.34}$$ Let $\rho>0$, then by \eqref{V5}, there exists a positive constant $c_{\rho}$ such that $$V'(t,x)x\geq \rho[\omega(|x|)|x|]^{2}-c_{\rho}. \label{e2.35}$$ Therefore, $$\begin{split} \int^{1}_{\frac{1}{|x|}}V'(t,sx)x\,ds &=\int^{1}_{\frac{1}{|x|}}V'(t,sx)sx \frac{ds}{s}\\ &\geq \int^{1}_{\frac{1}{|x|}}(\rho[\omega(|sx|)|sx|]^{2}-c_{\rho})\frac{ds}{s}\\ &\geq \frac{\rho}{2}[\omega(|x|)|x|]^{2}(1-\frac{1}{|x|^{2}}) -c_{\rho} b(t)ln(|x|). \end{split} \label{e2.36}$$ 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 $$\frac{1}{\omega(s)}\leq \frac{1}{a_0\omega(s^{1/2})} \label{e2.37}$$ which implies that for $|x|$ large enough $$\frac{ln(|x|)}{[\omega(|x|)|x|]^{2}}\leq \frac{ln(|x|)}{|x|} \frac{1}{a^{2}_0[\omega(|x|^{1/2})|x|^{1/2}]^{2}}\to 0\quad\text{as } |x|\to\infty. \label{e2.38}$$ 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 $$\ddot{u}_{n}(t)+V'(t,u_{n}(t))=0 \label{e2.39}$$ by $\tilde{u}_{n}(t)$ and integrating, we obtain by \eqref{V1} and H\"older's inequality $$\begin{split} \int^{T}_0|\dot{u}_{n}|^{2}dt &=-\int^{T}_0\ddot{u}_{n}\tilde{u}_{n}dt \\ &=\int^{T}_0V'(t,u_{n})\tilde{u}_{n}dt \\ &\leq\|\tilde{u}_{n}\|_{\infty}\int^{T}_0\big[f(t)\omega(|u_{n}(t)|)|u_{n}(t)| + g(t)\big]dt \\ &\leq\|\tilde{u}_{n}\|_{\infty}\Big[\|f\|_{L^{2}} \big(\int^{T}_0 [\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt\big)^{1/2} + \|g\|_{L^{1}}\Big]. \end{split} \label{e2.40}$$ By \eqref{e2.1}, \eqref{e2.2} and \eqref{e2.40}, there exists a positive constant $c_{11}$ such that $$\|\tilde{u}_{n}\|_1\leq c_{11}\big[\big(\int^{T}_0[\omega(|u_{n}(t)|) |u_{n}(t)|]^{2}dt\big)^{1/2}+1\big]. \label{e2.41}$$ 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 $$\begin{split} \int^{T}_0|\dot{u}_{n}|^{2}dt &= -\int^{T}_0\ddot{u}_{n}u_{n}dt \\ &=\int^{T}_0V'(t,u_{n})u_{n}dt \\ &\geq \rho\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt-c_{\rho}T. \end{split}\label{e2.42}$$ We deduce from \eqref{e2.42} and Wirtinger inequality that there exists a positive constant $c_{12}$ such that $$\|\tilde{u}_{n}\|^{2}_1\geq c_{12}\big[\rho\int^{T}_0[\omega(|u_{n}(t)|) |u_{n}(t)|]^{2}dt-c_{\rho}T\big]. \label{e2.43}$$ Combining \eqref{e2.41} with \eqref{e2.43}, we can find a positive constant $c_{13}$ such that $$\rho\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt-c_{\rho}T\leq c_{13} \big[\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt+1\big]. \label{e2.44}$$ Since $\rho$ is arbitrary chosen, $$(\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt)\quad \text{is bounded.} \label{e2.45}$$ 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 $$|u_{n}(t)|\to\infty\quad \text{as n\to\infty, a.e. t\in [0,T].} \label{e2.46}$$ Fatou's lemma and \eqref{e2.46} imply $$\int^{T}_0[\omega(|u_{n}(t)|)|u_{n}(t)|]^{2}dt\to \infty\quad \text{as }n\to\infty \label{e2.47}$$ 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 $${1\over k}|\varphi_{k}(u)| \leq c_{14},\quad \forall u\in S_{T},\; \forall k\geq 1. \label{e2.48}$$ Consequently by \eqref{e2.3}, for $k$ large enough, $u_{k}\notin S_{T}$. 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