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\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} \\
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  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]
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\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
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\section{Introduction}

We consider the superquadratic Hamiltonian system
\begin{equation}
\dot z = JH_z(t,z)     \label{1.1} 
\end{equation}
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 
\begin{equation}
\frac{H(t,z)}{|z|^{2}}\to  +\infty \text{ as }|z|\to  +\infty 
\text{ uniformly in } t.   \label{1.2}
\end{equation}
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] $
\begin{equation}
0<\mu H( t,z) \leq z\cdot H_{z}(t,z) .      \label{1.3}
\end{equation}
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
\begin{equation}
H(t,z)=|z|^{2}(\ln (1+|z|^{p}))^q , \quad  p, q>1.    \label{1.4}
\end{equation}

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
\begin{equation} \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} 
\end{equation}
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 
\begin{equation}
\frac{\int_0^1|z_m|^{\beta }dt}{\| \text{ }z_m\text{ }\| }
\to  0 \quad\text{as } m\to  \infty .   \label{2.3}
\end{equation}
Note that from (H3), $1<\lambda <1+\frac{\beta -1}{\beta }$. Let 
$\alpha =\frac{\beta -1}{\beta (\lambda -1)}$. Then 
\begin{equation}
\alpha >1, \quad \alpha \lambda -1=\alpha -\frac{1}{\beta }.   \label{2.4}
\end{equation}
By (H3), there exists $C_{4}>0$ such that  
\begin{equation}
|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}
\end{equation}
Denote $z_m=z_m^{+}+z_m^{-}+z_m^{0}\in E^{+}\oplus E^{-}\oplus E^{0}$. We have 
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
\frac{\| z_m^{+}\| }{\| z_m\| }\to  0 \quad\text{ as }m\to  \infty . \label{2.7}
\end{equation}
Similary, we have
\begin{equation}
\frac{\| z_m^{-}\| }{\| z_m\| }\to  0 \quad \text{ as } m\to \infty. \label{2.8}
\end{equation}
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}<Arz^{-},rz^{-}\rangle +\frac{1}{2}\langle
Arz^{+},rz^{+}\rangle -\int_0^1 H(t,z)dt   \\
&\leq -\frac{1}{2}\lambda ^{-}r^{2}\| z^{-}\|^2 
+\frac{1}{2}\| A\| r^{2}\| z^{+}\| ^{2}\leq 0 .
\end{align*}
%
ii) If $\| z^{-}\| \leq \gamma \| z^{+}+z^{0}\| $, we have
$$
1=\| z\| ^{2}=\| z^{-}\| ^{2}+\| z^{+}+z^{0}\|^2 \leq
(1+\gamma ^{2})\| z^{+}+z^{0}\| ^{2},
$$
i.e.,
\begin{equation}
\| z^{+}+z^{0}\| ^{2}\geq \frac{1}{1+\gamma ^{2}}>0 .   \label{2.9}
\end{equation}
Denote $\tilde K=\{z\in K: \| z^{-}\| \leq \gamma \| z^{+}+z^{0}\| \}$.

\noindent{\bf Claim: } There exists $\varepsilon _1>0$ such that, 
$\forall u\in \tilde K$, 
\begin{equation}
\mathop{\rm meas}\{t\in [0,1]: |u(t)|\geq \varepsilon _1\}\geq \varepsilon _1. 
   \label{2.10}
\end{equation}
For otherwise, $\forall k>0$, $\exists u_{k}\in \tilde K$ such that
\begin{equation}
\mathop{\rm meas}\{t\in [0,1]: |u_{k}(t)|\geq \frac{1}{k}\}<\frac{1}{k} .     \label{2.11}
\end{equation}
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
\begin{equation}
\| u_0^{0}+u_0^{+}\| ^2\geq \frac{1}{\gamma ^{2}+1}>0 .     \label{2.12}
\end{equation}
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., 
\begin{equation}
\int_0^1|u_{k}-u_0|^{2}dt\to  0 \quad \text{as } k\to  +\infty . \label{2.13}
\end{equation}
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
\begin{equation}
\mathop{\rm meas}\{t\in [0,1]: |u_0(t)|\geq \delta _1\}\geq \delta _2.   
\label{2.14}
\end{equation}
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
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
f(rz)\leq 0, \quad \text{for any }  z\in K \text{ and }  r\geq r_1. 
  \label{2.16}
\end{equation}
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
\begin{equation}
V(t,x)\leq d_2|x|^{\lambda +1}+d_3.     \label{3.1}
\end{equation}
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
\begin{equation}
\frac{\| z_m^{+}\| }{\| z_m\| }\to  0 \quad \text{ as } \quad 
m\to  +\infty .     \label{3.2} 
\end{equation}
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 
\begin{equation}
\frac{\| z_m^{0}\| }{\| z_m\| }\to  0 \quad \text{ as } 
m\to  +\infty .      \label{3.3}
\end{equation}
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 
\begin{equation}
\mathop{\rm meas}\{t\in [0,1]: |u(t)|\geq \varepsilon _1\}\geq \varepsilon _1, 
\quad \forall u\in K.         \label{3.4}
\end{equation}
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} 


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\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}