\documentclass[twoside]{article} \usepackage{amssymb,amsmath} \pagestyle{myheadings} \markboth{\hfil Periodic solutions of superquadratic systems \hfil EJDE--2002/08} {EJDE--2002/08\hfil Guihua Fei \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 08, pp. 1--12. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % On periodic solutions of superquadratic Hamiltonian systems % \thanks{ {\em Mathematics Subject Classifications:} 58E05, 58F05, 34C25. \hfil\break\indent {\em Key words:} periodic solution, Hamiltonian system, linking theorem. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted September 20, 2001. Published Janaury 15, 2002.} } \date{} % \author{Guihua Fei} \maketitle \begin{abstract} We study the existence of periodic solutions for some Hamiltonian systems $\dot z=JH_{z}(t,z)$ under new superquadratic conditions which cover the case $H(t,z)=|z|^{2}(\ln (1+|z|^{p}))^q$ with $p, q>1$. By using the linking theorem, we obtain some new results. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \section{Introduction} We consider the superquadratic Hamiltonian system $$\dot z = JH_z(t,z) \label{1.1}$$ where $H\in C^1([0,1]\times \mathbb{R}^{2N},\mathbb{R})$ is a 1-periodic function in $t$, $J=\begin{pmatrix} 0 & -I_N \\ I_N & 0 \end{pmatrix}$ is the standard $2N\times 2N$ symplectic matrix, and $$\frac{H(t,z)}{|z|^{2}}\to +\infty \text{ as }|z|\to +\infty \text{ uniformly in } t. \label{1.2}$$ We assume $H$ satisfies the following conditions. \begin{enumerate} \item[(H1)] $H(t,z)\geq 0$, for all $(t,z)\in [0,1]\times \mathbb{R}^{2N}$. \item[(H2)] $H(t,z)=o(|z|^{2})$ as $|z|\to 0$ uniformly in $t$. \end{enumerate} In \cite{r1}, Rabinowitz established the existence of periodic solutions for (\ref{1.1}) under the following superquadratic condition: there exist $\mu >0$ and $r_1>0$ such that for all $|z|\geq r_1$ and $t\in \left[ 0,1\right]$ $$0<\mu H( t,z) \leq z\cdot H_{z}(t,z) . \label{1.3}$$ Since then, the condition (\ref{1.3}) has been used extensively in the literature; see [1-14] and the references therein. It is easy to see that (\ref{1.3}) does not include some superquadratic nonlinearity like $$H(t,z)=|z|^{2}(\ln (1+|z|^{p}))^q , \quad p, q>1. \label{1.4}$$ In this paper, we shall study the periodic solutions of (\ref{1.1}) under some superquadratic conditons which cover the cases like (\ref{1.4}). We assume $H$ satisfies the following condition. \begin{enumerate} \item[(H3)] There exist constants $\beta >1$, $1<\lambda <1+\frac{\beta -1}{\beta }$, $c_1, c_2>0$ and $L>0$ such that \begin{gather*} z\cdot H_{z}(t,z)-2H(t,z) \geq c_1|z|^{\beta}, \quad \forall |z|\geq L, \ \forall t\in [0,1]; \\ |H_{z}(t,z)| \leq c_2|z|^{\lambda}, \quad \forall |z|\geq L, \ \forall t\in [0,1]. \end{gather*} \end{enumerate} \begin{theorem} \label{thm1.1} Suppose $H\in C^1([0,1]\times \mathbb{R}^{2N},\mathbb{R})$ is 1-periodic in $t$ and satisfies (\ref{1.2}), (H1)--(H3). Then (\ref{1.1}) possesses a nonconstant 1-periodic solution. \end{theorem} A straightforward computation shows that if $H$ satisfies (\ref{1.4}), for any $T>0$, the system (\ref{1.1}) has a nonconstant T-periodic solution with minimal period $T$. One can see Remark 2.2 and Corollary 2.3 for more examples. For the second order Hamiltonian system $$\label{1.5} \begin{gathered} \ddot{u}(t)+V'(t,u(t))=0, \\ u(0)-u(1)=\dot{u}(0)-\dot{u}(1)=0 \end{gathered}$$ we have a similar result. \begin{theorem} \label{thm1.2} Suppose $V\in C^1([0,1]\times \mathbb{R}^N,\mathbb{R})$ is 1-periodic in $t$ and satisfies \begin{enumerate} \item[(V1)] $V(t,x)\geq 0$, for all $(t,x)\in [0,1]\times \mathbb{R}^N$ \item[(V2)] $V(t,x)=o(|x|^{2})$ as $|x|\to 0$ uniformly in $t$ \item[(V3)] $V(t,x)/|x|^{2}\to +\infty$ as $|x|\to +\infty$ uniformly in $t$ \item[(V4)] There exist constants $1<\lambda \leq \beta$, $d_1, d_2>0$ and $L>0$ such that \begin{gather} x\cdot V'(t,x) -2V(t,x)\geq d_1|x|^{\beta}, \quad \forall |x|\geq L, \ \forall t\in [0,1]; \nonumber\\ |V'(t,x)| \leq d_2|x|^{\lambda}, \quad \forall |x|\geq L, \ \forall t\in [0,1]. \label{1.6} \\ (\text{ or } V(t,x) \leq d_2|x|^{\lambda +1}, \quad \forall |x|\geq L, \ \forall t\in [0,1] ). \label{1.7} \end{gather} \end{enumerate} Then (\ref{1.5}) possesses a nonconstant 1-periodic solution. \end{theorem} We shall use the linking theorem \cite[Theorem 5.29]{r2} to prove our results. The idea comes from \cite{m1,r1,r2}. Theorem 1.1 is proved in Section 2 while the proof of Theorem 1.2 is carried out in Section 3. \section{First order Hamiltonian system } Let $S^1 =\mathbb{R}/(2\pi \mathbb{Z})$ and $E=W^{1/2,2}(S^1 ,\mathbb{R}^{2N} )$. Then $E$ is a Hilbert space with norm $\| \cdot \|$ and inner product $\langle\cdot,\cdot\rangle$. We define \begin{gather} \langle Ax, y\rangle = \int^1_0 (-J\dot x , y)\,dt, \quad \forall x,y\in E; \label{2.1} \\ f(z)=\frac 12 \langle Az,z \rangle -\int^1_0 H(t,z)\,dt , \quad \forall z\in E. \label{2.2} \end{gather} Then $A$ is a bounded selfadjoint operator and $\ker A=\mathbb{R}^{2N}$. (H1)--(H3) imply that $$|H(t,z)|\leq a_1+a_2|z|^{\lambda +1}, \quad \forall z\in \mathbb{R}^{2N}.$$ This implies that $f\in C^1(E,\mathbb{R})$ and looking for the solutions of (\ref{1.1}) is equivalent to looking for the critical points of $f$ \cite{r1,r2}. Let $E^{0}=\ker (A)$, $E^{+}=$ positive definite subspace of $A$, and $E^{-}=$ negative definite subspace of $A$. Then $E=E^{0}\oplus E^{-}\oplus E^{+}$. \begin{lemma} \label{lm2.1} Under the conditions of Theorem 1.1, $f$ satisfies the (PS) condition. \end{lemma} \paragraph{Proof.} Let $\{z_m\}$ be a (PS)-sequence, i.e., $$|f(z_m)|\leq M; \quad f'(z_m)\to 0 \quad \text{ as } \ m\to \infty .$$ We want to show that $\{z_m\}$ is bounded. Then by a standard argument, $\{z_m\}$ has a convergent subsequence \cite{r2}. Suppose $\{z_m\}$ is not bounded, then passing to a subsequence if necessary, $\| z_m\| \to +\infty$ as $m\to +\infty$. By (H3), there exists $C_3>0$ such that for all $z\in \mathbb{R}^{2N}$, $t\in [0,1]$ $$z\cdot H_{z}(t,z)-2H(t,z)\geq C_1|z|^{\beta }-C_3.$$ Therefore, we have \begin{align*} 2f(z_m)-\langle f'(z_m),z_m\rangle& =\int_0^1 [z_m\cdot H_{z}(t,z_m)-2H(t,z_m)]dt \\ &\geq \int_0^1[C_1|z_m|^{\beta }-C_3]dt =C_1\int_0^1|z_m|^{\beta }dt-C_3. \end{align*} This implies $$\frac{\int_0^1|z_m|^{\beta }dt}{\| \text{ }z_m\text{ }\| } \to 0 \quad\text{as } m\to \infty . \label{2.3}$$ Note that from (H3), $1<\lambda <1+\frac{\beta -1}{\beta }$. Let $\alpha =\frac{\beta -1}{\beta (\lambda -1)}$. Then $$\alpha >1, \quad \alpha \lambda -1=\alpha -\frac{1}{\beta }. \label{2.4}$$ By (H3), there exists $C_{4}>0$ such that $$|H_{z}(t,z)|^{\alpha }\leq C_2^{\alpha }|z|^{\lambda \alpha }+C_{4}, \quad \forall (t,z)\in [0,1]\times \mathbb{R}^{2N}. \label{2.5}$$ Denote $z_m=z_m^{+}+z_m^{-}+z_m^{0}\in E^{+}\oplus E^{-}\oplus E^{0}$. We have \begin{aligned} \langle f'(z_m),z_m^{+}\rangle &=\langle Az_m^{+}, z_m^{+}\rangle -\int_0^1[H_{z}(t,z_m)\cdot z_m^{+}]dt \\ &\geq \langle Az_m^{+}, z_m^{+}\rangle -\int_0^1|H_{z}(t,z_m)||z_m^{+}|dt \\ &\geq \langle Az_m^{+}, z_m^{+}\rangle -(\int_0^1|H_{z}(t,z_m)|^{\alpha })^{\frac{1}{\alpha }}\cdot C_{\alpha }\| z_m^{+}\| , \end{aligned} \label{2.6} where $C_{\alpha }>0$ is a constant independent of $m$. By (\ref{2.5}), \begin{align*} \int_0^1|H_{z}(t,z_m)|^{\alpha }dt&\leq \int_0^1 (C_2^{\alpha }|z_m|^{\lambda \alpha }+C_{4})dt \\ &\leq C_{5}(\int_0^1|z_m|^{\beta }dt)^{1/\beta} (\int_0^1|z_m|^{(\alpha \lambda -1)\cdot \frac{\beta }{\beta -1}}dt)^ {1-\frac{1}{\beta }}+C_{4} \\ &\leq C_{6}(\int_0^1|z_m|^{\beta })^{1/\beta}\| z_m\| ^{(\alpha \lambda -1)}+C_{4} . \end{align*} Combining this inequality with (\ref{2.3}) and (\ref{2.4}) yields that $$\frac{(\int_0^1|H_{z}(t,z_m)|^{\alpha }dt)^{\frac{1}{\alpha }}}{\| z_m\| } \leq [\frac{C_{6}(\int_0^1|z_m|^{\beta }dt)^{1/\beta}} {\| z_m\| ^{1/\beta}} \cdot \frac{\| z_m\| ^{(\alpha \lambda -1)}} {\| z_m\| ^{\alpha -\frac{1}{\beta }}} +\frac{C_{4}}{\| z_m\| ^{\alpha }}]^{\frac{1}{\alpha }}\to 0$$ as $m\to \infty$. By (\ref{2.6}) we have $$\frac{\langle Az_m^{+},z_m^{+}\rangle }{\| z_m\| \text{ }\| z_m^{+}\| } \leq \frac{\| f'(z_m)\| \text{ }\| z_m^{+}\| } {\| z_m\| \text{ }\| z_m^{+}\| } +\frac{(\int_0^1|H_{z}(t,z_m)|^{\alpha }dt)^{\frac{1}{\alpha }}}{\| z_m\| } \cdot \frac{C_{\alpha }\| z_m^{+}\| }{\| z_m^{+}\| }\to 0$$ as $m\to \infty$. This implies $$\frac{\| z_m^{+}\| }{\| z_m\| }\to 0 \quad\text{ as }m\to \infty . \label{2.7}$$ Similary, we have $$\frac{\| z_m^{-}\| }{\| z_m\| }\to 0 \quad \text{ as } m\to \infty. \label{2.8}$$ By (H3) there exist $C_7,C_8>0$ such that $$z\cdot H_{z}(t,z_m)-2H(t,z)\geq C_7|z|-C_8, \quad \forall (t,z)\in [0,1]\times \mathbb{R}^{2N}.$$ This implies \begin{align*} 2f(z_m)-\langle f'(z_m),z_m\rangle &=\int_0^1[z_m\cdot H_{z}(t,z_m)-2H(t,z_m)]dt\geq \int_0^1[C_7|z_m|-C_8]dt \\ &\geq \int_0^1[C_7|z_m^{0}|-C_7|z_m^{+}|-C_7|z_m^{-}|-C_8]dt \\ &\geq C_9\| z_m^{0}\| -C_{10}(\| z_m^{+}\| +\| z_m^{-}\| +1) . \end{align*} Therefore, by (\ref{2.7}) and (\ref{2.8}) $$\frac{\| z_m^{0}\| }{\| z_m\| }\to 0 \quad \text{ as } m\to \infty .$$ Combine this with (\ref{2.7}) and (\ref{2.8}), we get $$1=\frac{\| z_m\| }{\| z_m\| }\leq \frac{\| z_m^{+}\| +\| z_m^{-}\| +\| z_m^{0}\| }{\| z_m\| }\to 0 \quad \text{ as } m\to \infty ,$$ a contradiction. Therefore, $\{z_m\}$ must be bounded. \hfill$\Box$ \paragraph{Proof of Theorem 1.1} We prove that $f$ satisfies the conditions of Theorem 5.29 in \cite{r2}. \noindent{\bf Step 1: } By (H1)--(H3), we have $$H(t,z)\leq a_1+a_2|z|^{\lambda +1}, \quad \forall (t,z)\in [0,1]\times \mathbb{R}^{2N} .$$ By (H2), for any $\varepsilon >0$, there exists $\delta >0$ such that $$H(t,z)\leq \varepsilon |z|^{2}, \quad \forall t\in [0, 1], \ |z|\leq \delta .$$ Therefore, there exists $M=M(\varepsilon )>0$ such that $$H(t,z)\leq \varepsilon |z|^{2}+M|z|^{\lambda +1}, \quad \forall (t,z)\in [0,1]\times \mathbb{R}^{2N} .$$ Note that $\lambda +1>2$. By the same arguements as in \cite[Lemma 6.16]{r2}, there exist $\rho >0$ and $\tilde{a}>0$, such that for $z\in E_1=E^{+}$ $$f(z)\geq \tilde{a} \quad \text{ if } \| z\| =\rho ,$$ i.e., $f$ satisfies $(I_7)(i)$ in \cite[Theorem 5.29]{r2} with $S=\partial B_{\rho}\cap E_1$. \noindent{\bf Step 2: } Let $e\in E^{+}$ with $\| e\| =1$ and $\tilde{E}=E^{-}\oplus E^{0}\oplus span\{e\}$. We denote $$K=\big\{z\in \tilde{E}: \| z\| =1\big\}, \quad \lambda ^- =\inf_{z\in E^{-},\| z^{-}\| =1}|\langle Az^{-},z^{-}\rangle |, \quad \gamma =(\frac{\| A\| }{\lambda ^{-}})^{1/2}.$$ For $z\in K$, we write $z=z^{-}+z^{0}+z^{+}\in \tilde{E}$. \\ i) If $\| z^{-}\| >\gamma \| z^{+}+z^{0}\|$, by (H1) we have, for any $r >0$, \begin{align*} f(rz)&=\frac{1}{2}0 . \label{2.9} Denote $\tilde K=\{z\in K: \| z^{-}\| \leq \gamma \| z^{+}+z^{0}\| \}$. \noindent{\bf Claim: } There exists $\varepsilon _1>0$ such that, $\forall u\in \tilde K$, $$\mathop{\rm meas}\{t\in [0,1]: |u(t)|\geq \varepsilon _1\}\geq \varepsilon _1. \label{2.10}$$ For otherwise, $\forall k>0$, $\exists u_{k}\in \tilde K$ such that $$\mathop{\rm meas}\{t\in [0,1]: |u_{k}(t)|\geq \frac{1}{k}\}<\frac{1}{k} . \label{2.11}$$ Write $u_{k}=u_{k}^{-}+u_{k}^{0}+u_{k}^{+}\in \tilde {E}$. Notice that $\dim (E^0\oplus span\{e\})<+\infty$ and $\| u_{k}^{0}+u_{k}^{+}\| \leq 1$. In the sense of subsequence, we have $$u_{k}^{0}+u_{k}^{+}\to u_0^{0}+u_0^{+}\in E^{0}\oplus span\{e\} \quad \text{as } k\to +\infty .$$ Then (\ref{2.9}) implies that $$\| u_0^{0}+u_0^{+}\| ^2\geq \frac{1}{\gamma ^{2}+1}>0 . \label{2.12}$$ Note that $\| u_{k}^{-}\| \leq 1$, in the sense of subsequence $u_{k}^{-}\rightharpoonup u_0^{-}\in E^{-}$ as $k\to +\infty$. Thus in the sense of subsequences, $$u_{k}\rightharpoonup u_0=u_0^{-}+u_0^{0}+u_0^{+} \quad \text{as } k\to +\infty .$$ This means that $u_{k}\to u_0$ in $L^{2}$, i.e., $$\int_0^1|u_{k}-u_0|^{2}dt\to 0 \quad \text{as } k\to +\infty . \label{2.13}$$ By (\ref{2.12}) we know that $\| u_0\| >0$. Therefore, $\int_0^1|u_0|^{2}dt>0$. Then there exist $\delta _1>0$, $\delta _2>0$ such that $$\mathop{\rm meas}\{t\in [0,1]: |u_0(t)|\geq \delta _1\}\geq \delta _2. \label{2.14}$$ Otherwise, for all $n>0$, we must have \begin{gather*} \mathop{\rm meas}\{t\in [0,1]: |u_0(t)|\geq \frac{1}{n}\}=0, \quad\text{i.e., } \mathop{\rm meas}\{t\in [0,1]: |u_0(t)|<\frac{1}{n}\}=1; \\ 0<\int_0^1|u_0|^{2}dt< \frac{1}{n^{2}}\cdot 1\to 0 \quad \text{as } n\to +\infty . \end{gather*} We get a contradiction. Thus (\ref{2.14}) holds. Let $\Omega _0=\{t\in [0,1]: |u_0(t)|\geq \delta _1\}$, $\Omega _{k}=\{t\in [0,1]: |u_{k}(t)|<1/k\}$, and $\Omega ^{\bot }_k=[0,1]\backslash \Omega _{k}$. By (\ref{2.11}), we have $$\mathop{\rm meas}(\Omega _{k}\cap \Omega _0)= \mathop{\rm meas}(\Omega _0-\Omega _0\cap \Omega ^{\bot }_k) \geq \mathop{\rm meas}(\Omega _0)-\mathop{\rm meas}(\Omega _0\cap \Omega ^{\bot }_k) \geq \delta _2 - \frac{1}{k} . \label{2.15}$$ Let $k$ be large enough such that $\delta _2-\frac{1}{k} \geq \frac{1}{2}\delta _2$ and $\delta _1-\frac{1}{k} \geq \frac{1}{2}\delta _1$. Then we have $$|u_k(t)-u_0(t)|^2\geq (\delta _1-\frac{1}{k} )^2\geq (\frac{1}{2}\delta _1)^2, \quad \forall t\in \Omega _{k}\cap \Omega _0.$$ This implies that \begin{align*} \int_0^1|u_{k}-u_0|^{2}dt&\geq \int_{\Omega _{k}\cap \Omega _0}|u_{u}-u_0|^{2} dt \geq (\frac{1}{2}\delta _1)^{2}\cdot \mathop{\rm meas}(\Omega _{k}\cap \Omega _0) \\ &\geq (\frac{1}{2}\delta _1)^{2}\cdot (\delta _2-\frac{1}{k}) \geq (\frac{1}{2}\delta _1)^{2}(\frac{1}{2}\delta _2)>0 . \end{align*} This is a contradiction to (\ref{2.13}). Therefore the claim is true and (\ref{2.10}) holds. For $z=z^{-}+z^{0}+z^{+}\in \tilde K$, let $\Omega _{z}=\{t\in [0,1]: |z(t)|\geq \varepsilon _1\}$. By (\ref{1.2}), for $M=\frac{\| A\| }{\varepsilon _1^{3}}>0$, there exists $L_1>0$ such that $$H(t,x)\geq M|x|^{2}, \quad \forall |x|\geq L_1, \text{ uniformly in } \ t.$$ Choose $r_1\geq L_1/ \varepsilon _1$. For $r\geq r_1$, $$H(t,rz(t))\geq M|rz(t)|^{2}\geq Mr^{2}\varepsilon _1^{2} , \quad \forall t\in \Omega _{z}.$$ By (H1), for $r\geq r_1$ \begin{align*} f(rz)&=\frac{1}{2}r^{2}\langle Az^{+},z^{+}\rangle + \frac{1}{2}r^{2} \langle Az^{-},z^{-}\rangle -\int_0^1H(t,rz)dt \\ &\leq \frac{1}{2}\| A\| r^{2}-\int_{\Omega _{z}}H(t,rz)dt \leq \frac{1}{2}\| A\| r^{2}-Mr^{2}\varepsilon _1^{2}\cdot \mathop{\rm meas}(\Omega _{z}) \\ &\leq \frac{1}{2}\| A\| r^{2}-M\varepsilon _1^{3}r^{2} =-\frac{1}{2}\| A\| r^{2}<0. \end{align*} Therefore, we have proved that $$f(rz)\leq 0, \quad \text{for any } z\in K \text{ and } r\geq r_1. \label{2.16}$$ Let $E_2=E^{-}\oplus E^{0}$, $Q=\{re: 0\leq r\leq 2r_1\}\oplus \{z\in E_2: \| z\| \leq 2r_1\}$. By (H1) and (2.16) we have $f|_{\partial Q}\leq 0$, i.e., $f$ satisfies $(I_7)(ii)$ in \cite[Theorem 5.29]{r2}. \noindent{\bf Step 3: } By Lemma 2.1, $f$ satisfies the (PS) condition. Similar to the proof of \cite[Theorem 6.10]{r2}, by the linking theorem \cite[Theorem 5.29]{r2}, there exists a critical point $z^{*}\in E$ of $f$ such that $f(z^{*})\geq \tilde{a}>0$. Moreover, $z^{*}$ is a classical solution of (\ref{1.1}) and $z^{*}$ is nonconstant by (H1). \hfill$\Box$ \begin{remark} \label{rmk2.2} \rm i) Suppose $H(t,z)=\frac{1}{2}\langle B(t)z,z\rangle +\tilde H(t,z)$ with $B(t)$ being a $2N\times 2N$ matrix, continuous and 1-periodic in $t$ and $\tilde H(t,z)$ satisfies (\ref{1.2}) and (H1)-(H3). We have the same conclusion as Theorem 1.1. The proof is similar and we omit it. ii) Suppose $H(t,z)=H(z)$ is independent on $t$, i.e., (\ref{1.1}) is an autonomous Hamiltonian system. Then under similar conditions as (\ref{1.2}) and (H1)-(H3), for any $T>0$, the system (\ref{1.1}) has a nonconstant $T$-periodic solution. Moreover, if $H(z)\in C^{2}(\mathbb{R}^{2N},\mathbb{R})$ and satisfies some strictly convex conditions such as $H''(x)$ is positive defininte for $x\neq 0$, then for any $T>0$, (\ref{1.1}) has a nonconstant $T$-periodic solution with minimal period $T$. We omit the proof which is similar to the one in \cite{f1,f2}. iii) Suppose (\ref{1.4}) holds, i.e., $$H(t,z)=H(z)=|z|^{2}(\ln (1+|z|^{p}))^q, \quad \forall (t,z)\in [0,1]\times \mathbb{R}^{2N},$$ where $p>1$ and $q>1$. Obviously, (\ref{1.2}), (H1) and (H2) hold. Note that \begin{gather*} z\cdot H_{z}(z)-2H(z)=|z|^{2}q(\ln (1+|z|^q))^{q-1} \frac{p|z|^{p}}{1+|z|^{p}}\geq |z|^{2}\frac{pq(\ln 2)^{q-1}}{2}, \quad \forall |z|\geq 1. \\ |H_{z}(z)|\leq 2(\ln (1+|z|^{p}))^q|z|+\frac{p|z|^{p}}{1+|z|^{p}}q(\ln (1+|z|^p))^{q-1}|z| \leq 2|z|^{\frac{5}{4}}, \ \forall |z|\geq L, \end{gather*} for $L$ being large enough. This implies (H3). By directly computation, $H''(z)$ is positive definite for $z\neq 0$. Therefore, for any $T>0$, (\ref{1.1}) possesses a $T$-periodic solution with minimal period $T$. iv) There are many examples which satisfy (H1)-(H3) and (\ref{1.2}) but do not satisfy (\ref{1.3}). For example $$H(t,z)=|z|^{2}\ln (1+|z|^{2})\ln (1+2|z|^{3}).$$ \end{remark} \begin{corollary} \label{coro2.3} Suppose $H(t,z)=|z|^{2}h(t,z)$ with $h\in C^1([0,1]\times \mathbb{R}^{2N}, \mathbb{R})$ being 1-periodic in $t$ and satisfies \begin{enumerate} \item[(H$1'$)] $h(t,z)\geq 0$, for all $(t,z)\in [0,1]\times \mathbb{R}^{2N}$. \item[(H$2'$)] $h(t,z)\to 0$ as $|z|\to 0$; $h(t,z)\to +\infty$ as $|z|\to +\infty$. \item[(H$3'$)] There exist $0\leq \delta <1$, $L>0$, $\varepsilon _0>0$ and $M>0$ such that \begin{gather*} |z|^{\delta }h_{z}(t,z)\cdot z\geq \varepsilon _0, \quad |z||h_{z}(t,z)|\leq Mh, \quad \forall |z|\geq L; \\ \frac{h(t,z)}{|z|^{\gamma }}\to 0 \quad \text{ as } |z|\to \infty \text{ for any } \gamma >0 . \end{gather*} \end{enumerate} Then system (\ref{1.1}) possesses a nonconstant 1-periodic solution. \end{corollary} \noindent{\bf Proof } Obviously, $(H1')-(H3')$ imply (\ref{1.2}), (H1) and $(H2)$. \begin{gather*} z\cdot H_{z}(t,z)-2H(t,z)=|z|^{2}|h_{z}(t,z)\cdot z\geq \varepsilon _0|z|^{2-\delta}, \quad \forall |z|\geq L; \\ \begin{aligned} |H_{z}(t,z)|&\leq |2h(t,z)||z|+|z|^{2}|h_{z}(t,z)| \\ &\leq (2+M)|z|h(t,z)\leq (2+M)|z|^{1+\gamma}, \ \ \forall |z|\geq L'. \end{aligned} \end{gather*} Let $\beta =2-\delta$ and $\lambda =1+\gamma$ with $0<\gamma < (1-\delta )/(2-\delta)$. Then (H3) holds. By Theorem 1.1 we get the conclusion. \hfill$\Box$ \section{Second order Hamiltonian System } Let $E=W^{1,2}(S^1,\mathbb{R}^N)$ with the norm $\|\cdot \|$ and inner product $\langle \cdot , \cdot \rangle$. Then $E\subset C(S^1,\mathbb{R}^N)$ and $\| u\| ^{2}=\int_0^1(|\dot u|^{2}+|u|^{2})dt$. Define \begin{gather*} \langle Kx,y\rangle =\int_0^1x\cdot ydt, \quad \forall x, y\in E; \\ f(z)=\frac{1}{2} \langle (id-K)z,z \rangle -\int_0^1V(t,z)dt, \quad \forall z\in E. \end{gather*} Then $K$ is compact, $\ker (id-K)=\mathbb{R}^N$, and the negative definite subspace of $\mathop{\rm id}-K$, $M^{-}(\mathop{\rm id}-K)=\{0\}$, i.e., $E=E^{0}\oplus E^{+}$ where $E^{0}=\ker (id-K)$ and $E^{+}$ is the positive definite subspace of $id-K$. Note that (V1)--(V4) imply $$V(t,x)\leq d_2|x|^{\lambda +1}+d_3. \label{3.1}$$ This implies that $f\in C^1(E,\mathbb{R})$ and critical points of $f$ are 1-periodic solutions of (\ref{1.5}) \cite{m1}. \begin{lemma} \label{lm3.1} Suppose (V1)--(V4) hold. Then $f$ satisfies the (PS) condition. \end{lemma} \noindent{Proof} Let $\{z_m\}$ be a (PS) sequence. Suppose $\{z_m\}$ is not bounded. Passing to a subsequence if necessary, $\|z_m\|\to +\infty$ as $m\to \infty$. Then by (V4) $$2f(z_m)-\langle f'(z_m),z_m\rangle =\int_0^1[z_m\cdot V'(t,z_m)-2V(t,z_m)]dt \geq d_1\int_0^1|z_m|^{\beta }dt-d_{4}.$$ This implies $$\frac{\int_0^1|z_m|^{\beta }dt}{\| z_m\| } \to 0 \quad \text{ as } m\to +\infty .$$ If (\ref{1.6}) holds, we have \begin{align*} \langle f'(z_m),z_m^{+}\rangle &=\langle (id-K)z_m^{+},z_m^{+}\rangle -\int_0^1V'(t,z_m)\cdot z_m^{+}dt \\ &\geq \langle (id-K)z_m^{+},z_m^{+}\rangle -\| z_m^{+}\| _{\infty }\int_0^1|V'(t,z_m)|dt \\ &\geq \langle (id-K)z_m^{+},z_m^{+}\rangle -d_5\| z_m^{+}\| (\int_0^1|z_m|^{\lambda }dt +d_6). \end{align*} Since $\lambda \leq \beta$, we have $$\frac{\| z_m^{+}\| }{\| z_m\| }\to 0 \quad \text{ as } \quad m\to +\infty . \label{3.2}$$ If (\ref{1.7}) holds, we have \begin{align*} f(z_m)&=\frac{1}{2}\langle (id-K)z_m^{+},z_m^{+}\rangle -\int_0^1V(t,z_m)dt \\ &\geq \frac{1}{2}\langle (id-K)z_m^{+},z_m^{+}\rangle -d_{5}\int_0^1|z_m|^{1+\lambda }dt-d_7 \\ &\geq \langle (id-K)z_m^{+},z_m^{+}\rangle -d_8\| z_m\| \int_0^1|z_m|^{\lambda }dt-d_7. \end{align*} Since $\lambda \leq \beta$, we obtain (\ref{3.2}). On the other hand, (V1)--(V4) imply \begin{align*} x\cdot V'(t,x)-2V(t,x)&\geq d_9|x|-d_{10}, \quad \forall t\in S^1\times \mathbb{R}^N. \\ 2f(z_m)-\langle f'(z_m),z_m\rangle &=\int_0^1[z_m\cdot V'(t,z_m)-2V(t,z_m)]dt \\ &\geq d_9\int_0^1|z_m|dt-d_{10} \\ &\geq d_9\int_0^1|z_m^{0}|dt-d_9\int_0^1|z_m^{+}|dt-d_{10} \\ &\geq d_9\| z_m^{0}\| -d_{11}\| z_m^{+}\| -d_{10}. \end{align*} This implies $$\frac{\| z_m^{0}\| }{\| z_m\| }\to 0 \quad \text{ as } m\to +\infty . \label{3.3}$$ By (\ref{3.2}) and (\ref{3.3}), we get a contradiction. Therefore $\{z_m\}$ is bounded. By a standard argument, $\{z_m\}$ has a convergent subsequence \cite{m1}. \hfill$\Box$ \paragraph{Proof of Theorem 1.2} As in Step 1 of the proof of Theorem 1.1, by (V2) and (\ref{3.1}), there exist $\tilde{a}>0$, $\rho >0$ such that $$f(z)\geq \tilde{a}, \quad \forall z\in E^{+} \quad \text{with } \| z\| =\rho .$$ Choose $e\in E^{+}$ with $\| e\| =1$. Let $\tilde {E}=\mathop{\rm span} \{e\}\oplus E^{0}$ and $K=\{u\in \tilde{E}: \ \| u\| =1\}$. Note that $\dim \tilde{E}<+\infty$. By using similar arguments as in the proof of (\ref{2.10}), there exists $\varepsilon _1>0$ such that $$\mathop{\rm meas}\{t\in [0,1]: |u(t)|\geq \varepsilon _1\}\geq \varepsilon _1, \quad \forall u\in K. \label{3.4}$$ By (V1), (V3) and similar arguments as in the proof of Theorem 1.1, there exists $r_1>0$ such that $$f|_{\partial Q}\leq 0, \quad \text{where} \quad Q=\{re: 0\leq r\leq 2r_1\}\oplus \{z\in E^{0}: \| z\| \leq 2r_1\}.$$ Now by Lemma 3.1, \cite[Theorem 5.29]{r2}, and (V1), $f$ has a nonconstant critical point $z^{*}$ such that $f(z^{*})\geq \tilde{a}>0$. $z^{*}$ is 1-periodic solution of (\ref{1.5}). \hfill$\Box$ \begin{remark} \label{rmk3.2} \rm (i) Suppose $V(t,x)=V(x)$ is independent on $t$ and $V(x)$ satisfies (V1)--(V4). Then for any $T>0$, (\ref{1.5}) possesses a nonconstant $T$-periodic solution. (ii) There are many examples which satisfy (V1)--(V4) but do not satisfy a condition similar to (\ref{1.3}). For example, \begin{align*} &V(t,x)=[1+(\sin 2\pi t)^{2}]\cdot |x|^{2}\ln (1+2|x|^{2}); \quad \text{or} \\ &V(t,x)=|x|^{2}\ln (1+|x|^{2})\ln (1+2|x|^{4}). \end{align*} \end{remark} By using similar arguments as in the proof of Theorem 1.2, we can prove the following corollary. Details are omited. \begin{corollary} \label{coro3.3} Suppose $V(t,x)=|x|^{2}h(t,x)$ with $h\in C^1(S^1\times \mathbb{R}^N,\mathbb{R})$ satisfies \begin{enumerate} \item[(V$1'$)] $h(t,x)\geq 0, \quad \forall (t, x)\in S^1 \times \mathbb{R}^N$. \item[(V$2'$)] $h(t,x)\to 0$ as $|x|\to 0$; $\quad h(t,x)\to +\infty$ as $|x|\to +\infty$. \item[(V$3'$)] There exist $L>0$, $\lambda >0$, $C_1, C_2>0$ such that for $t\in S^1$ $$C_1|x|(h'(t,x)\cdot x)\geq h(t,x), \quad h(t,x)\leq C_2|x|^{\lambda}, \quad \forall |x|\geq L.$$ \end{enumerate} Then (\ref{1.5}) possesses a nonconstant 1-periodic solution. \end{corollary} \begin{thebibliography}{00} \frenchspacing \bibitem{c1} K. C. Chang, {\it Infinite dimensional Morse theory and multiple solution problems}, Progress in nonlinear differential equations and their applications, V.6 (1993). \bibitem{e1} I. Ekeland, {\it Convexity Method in Hamiltonian Mechanics}, Springer-Verlag. Berlin. (1990). \bibitem{e2} I. Ekeland \& H. 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Applications to nonlinear partial differential equations and Hamiltonian systems}, Springer-Verlag, Berlin, 1990. \end{thebibliography} \noindent\textsc{Guihua Fei}\\ Department of Mathematics and statistics\\ University of Minnesota \\ Duluth, MN 55812, USA \\ e-mail: gfei@d.umn.edu \end{document}