Open Access

Existence of periodic solutions for a class of second order discrete Hamiltonian systems

Advances in Difference Equations20162016:68

https://doi.org/10.1186/s13662-016-0787-5

Received: 17 January 2016

Accepted: 21 February 2016

Published: 8 March 2016

Abstract

By using the variational minimizing method and the saddle point theorem, the periodic solutions for non-autonomous second-order discrete Hamiltonian systems are considered. The results obtained in this paper complete and extend previous results.

Keywords

periodic solutions second-order discrete Hamiltonian systems saddle point theorem least action principle

1 Introduction and main results

Consider the second-order discrete Hamiltonian system
$$ \Delta^{2}u(n-1)=\nabla F\bigl(n,u(n)\bigr), $$
(1.1)
where \(\Delta^{2}u(n)=\Delta(\Delta u(n))\) and \(\nabla F(n,x)\) denotes the gradient of F with respect to the second variable. F satisfies the following assumption:
  1. (A)

    \(F(n,x)\in C^{1}(\mathbb{R}^{N},\mathbb{R})\) for any \(n\in \mathbb{Z}\), \(F(n+T,x)=F(n,x)\) for \((n,x)\in\mathbb{Z}\times\mathbb {R}^{N}\), T is a positive integer.

     

Since Guo and Yu developed a new method to study the existence and multiplicity of periodic solutions of difference equations by using critical point theory (see [14]), the existence and multiplicity of periodic solutions for system (1.1) have been extensively studied and lots of interesting results have been worked out; see [516] and the references therein. System (1.1) is a discrete form of classical second-order Hamiltonian systems, which has been paid much attention to by many mathematicians in the past 30 years; see [1724] for example.

In particular, when the nonlinearity \(\nabla F(n,x)\) is bounded, Guo and Yu [3] obtained one periodic solution to system (1.1). When the gradient of the potential energy does not exceed sublinear growth, i.e. there exist \(M_{1}>0\), \(M_{2}>0\), and \(\alpha\in [0,1)\), such that
$$ \bigl|\nabla F(n,x)\bigr|\leq M_{1}|x|^{\alpha}+M_{2}, \quad\forall (n,x)\in \mathbb{Z}[1,T]\times\mathbb{R}^{N}, $$
(1.2)
where \(\mathbb{Z}[a,b]:=\mathbb{Z}\cap[a,b]\) for every \(a,b\in\mathbb {Z}\) with \(a\leq b\), Xue and Tang [12, 13] considered the periodic solutions of system (1.1), which completed and extended the results in [3] under the condition where
$$ \lim_{|x|\rightarrow\infty}|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x)=+ \infty, $$
(1.3)
or
$$ \lim_{|x|\rightarrow\infty}|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x)=- \infty. $$
(1.4)
Under weaker conditions on \(\nabla F(n,x)\), i.e.,
$$ \lim_{|x|\rightarrow\infty}|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x)< + \infty, $$
(1.5)
or
$$ \lim_{|x|\rightarrow\infty}|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x)>- \infty, $$
(1.6)
Tang and Zhang [11] considered the periodic solutions of system (1.1), which completed and extended the results in [12, 13].

In this paper, we will further investigate periodic solutions to the system (1.1) under the conditions of (1.5) or (1.6). Our main results are the following theorems.

Theorem 1.1

Suppose that \(F(n,x)=F_{1}(n,x)+F_{2}(x)\), where \(F_{1}\) and \(F_{2}\) satisfy (A) and the following conditions:
  1. (1)
    there exist \(f,g:\mathbb{Z}[1,T]\rightarrow\mathbb{R^{+}}\) and \(\alpha\in[0,1)\) such that
    $$ \bigl|\nabla F_{1}(n,x)\bigr|\leq f(n)|x|^{\alpha}+g(n),\quad \textit{for all } (n,x)\in\mathbb{Z}[1,T]\times\mathbb{R}^{N}; $$
     
  2. (2)
    there exist constants \(r>0\) and \(\gamma\in[0,2)\) such that
    $$ \bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y\bigr) \geq-r|x-y|^{\gamma}, \quad\textit{for all } x,y\in\mathbb{R}^{N}; $$
     
  3. (3)
    $$ \liminf_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac {1}{8\sin^{2}\frac{\pi}{T}}\sum _{n=1}^{T}f^{2}(n). $$
     
Then system (1.1) has at least one T-periodic solution.

Theorem 1.2

Suppose that \(F(n,x)=F_{1}(n,x)+F_{2}(x)\), where \(F_{1}\) and \(F_{2}\) satisfy (A), (1), (2), and the following conditions:
  1. (4)
    there exist \(\delta\in[0,2)\) and \(C>0\) such that
    $$ \bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y\bigr)\leq C|x-y|^{\delta}, \quad\textit{for all } x,y\in\mathbb{R}^{N}; $$
     
  2. (5)
    $$ \limsup_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)< -\frac {3}{8\sin^{2}\frac{\pi}{T}}\sum _{n=1}^{T}f^{2}(n). $$
     
Then system (1.1) has at least one T-periodic solution.

Theorem 1.3

Suppose that \(F(n,x)=F_{1}(n,x)+F_{2}(x)\), where \(F_{1}\) and \(F_{2}\) satisfy (A), (1), and the following conditions:
  1. (6)
    there exists a constant \(0< r<\frac{6}{T^{2}-1}\), such that
    $$ \bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y\bigr) \geq-r|x-y|^{2},\quad \textit{for all } x,y\in\mathbb{R}^{N}; $$
     
  2. (7)
    $$ \liminf_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac {3}{ (24-4(T^{2}-1)r )\sin^{2}\frac{\pi}{T}}\sum _{n=1}^{T}f^{2}(n). $$
     
Then system (1.1) has at least one T-periodic solution.

Theorem 1.4

Suppose that \(F=F_{1}+F_{2}\), where \(F_{1}\) and \(F_{2}\) satisfy (A), (1), and the following conditions:
  1. (8)
    there exist \(h:\mathbb{Z}[1,T]\rightarrow\mathbb{R}^{+}\) and \((\lambda,u)\)-subconvex potential \(G:\mathbb{R}^{N}\rightarrow\mathbb {R}\) with λ>1/2 and \(1/2<\mu<2\lambda^{2}\), such that
    $$ \bigl(\nabla F_{2}(n,x),y\bigr)\geq-h(n)G(x-y), \quad\textit{for all } x,y\in\mathbb {R}^{N} \textit{ and } n\in\mathbb{Z}[1,T]; $$
     
  2. (9)
    $$\begin{aligned}& \limsup_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F_{1}(n,x)< -\frac{3}{8\sin^{2}\frac{\pi}{T}}\sum _{n=1}^{T}f^{2}(n), \\& \limsup_{|x|\rightarrow+\infty}|x|^{-\beta}\sum _{n=1}^{T}F_{2}(n,x)< -8\mu\max _{|s|\leq1}G(s)\sum_{n=1}^{T}h(n), \end{aligned}$$
    where \(\beta=\log_{2\lambda}(2\mu)\).
     

Then system (1.1) has at least one T-periodic solution.

Remark 1.5

Theorems 1.1-1.3 extend some existing results. On the one hand, we decomposed the potential F into \(F_{1}\) and \(F_{2}\). On the other hand, if \(F_{2}=0\), the theorems in [11], Theorems 1 and 2, are special cases of Theorem 1.1 and Theorem 1.2, respectively. Some examples of F are given in Section 4, which are not covered in the references. Moreover, our Theorem 1.4 is a new result.

2 Some important lemmas

\(H_{T}\) can be equipped with the inner product
$$ \langle u,v\rangle=\sum_{n=1}^{T} \bigl[\bigl(\Delta u(n),\Delta v(n)\bigr)+\bigl(u(n),v(n)\bigr)\bigr],\quad \forall u,v \in H_{T}, $$
by which the norm \(\|\cdot\|\) can be induced by
$$ \|u\|= \Biggl(\sum_{n=1}^{T} \bigl[\bigl|\Delta u(n)\bigr|^{2}+\bigl|u(n)\bigr|^{2}\bigr] \Biggr)^{\frac {1}{2}},\quad \forall u\in H_{T}. $$
Define
$$\Phi(u)=\frac{1}{2}\sum_{t=1}^{T}\bigl| \Delta u(t)\bigr|^{2}-\sum_{t=1}^{T}F \bigl(t,u(t)\bigr) $$
and
$$\bigl\langle \Phi'(u), v\bigr\rangle =\sum _{t=1}^{T}\bigl(\Delta u(t),\Delta v(t)\bigr)-\sum _{t=1}^{T}\bigl(\nabla F\bigl(t,u(t) \bigr),v(t)\bigr), $$
for \(u,v\in H_{T}\).

By (A), it is easy to see that Φ is continuously differentiable, and the critical points of Φ are the T-periodic solutions of system (1.1).

The following lemma is a discrete form of Wirtinger’s inequality and Sobolev’s inequality (see [19]).

Lemma 2.1

[11]

If \(u\in H_{T}\) and \(\sum_{t=1}^{T}u(t)=0\), then
$$\begin{aligned}& \sum_{t=1}^{T}\bigl|u(t)\bigr|^{2}\leq \frac{1}{4\sin^{2}\frac{\pi}{T}}\sum_{t=1}^{T}\bigl|\Delta u(t)\bigr|^{2},\\& \|u\|^{2}_{\infty}:= \Bigl(\max_{t\in\mathbb{Z}[1,T]}\bigl|u(t)\bigr| \Bigr)^{2}\leq\frac{T^{2}-1}{6T}\sum_{t=1}^{T}\bigl| \Delta u(t)\bigr|^{2}. \end{aligned}$$

Lemma 2.2

[25]

Let \(E=V\oplus X\), where E is a real Banach space and \(V\neq\{0\}\) and is finite dimensional. Suppose \(I\in C^{1}(E, \mathbb{R})\), it satisfies (PS), and
  1. (i)

    there is a constant α and a bounded neighborhood D of 0 in V such that \(I\mid_{\partial D}\leq\gamma\), and

     
  2. (ii)

    there is a constant \(\beta>\gamma\) such that \(I\mid_{X}\geq\beta\).

     
Then I possesses a critical value \(c\geq\beta\). Moreover, c can be characterized as
$$c=\inf_{h\in\Gamma}\max_{s\in\overline{D}}I\bigl(h(s)\bigr), $$
where
$$\Gamma=\bigl\{ h\in C(\overline{D},E)\mid h(s)=s, s\in\partial D\bigr\} . $$

3 Proof of theorems

For convenience, we denote
$$ R_{1}= \Biggl(\sum_{n=1}^{T}f^{2}(n) \Biggr)^{1/2},\qquad R_{2}=\sum_{n=1}^{T}f(n), \quad\mbox{and}\quad R_{3}=\sum_{n=1}^{T}g(n). $$

Proof of Theorem 1.1

According to (3), there exists \(a_{1}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$\liminf_{x\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac{a_{1}}{2}R_{1}^{2}. $$
From (1) and Lemma 2.1, for any \(u\in H_{T}\), one has
$$\begin{aligned} &\Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr]\Biggr\vert \\ &\quad=\Biggl\vert \sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{1}\bigl(n,\bar {u}+s\tilde{u}(n)\bigr),\tilde{u}(n)\bigr)\,ds\Biggr\vert \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}f(n)\bigl|\bar{u}+s\tilde {u}(n)\bigr|^{\alpha}\bigl|\tilde{u}(n)\bigr|\,ds+ \sum_{n=1}^{T} \int_{0}^{1}g(n)\bigl|\tilde{u}(n)\bigr|\,ds \\ &\quad\leq\sum_{n=1}^{T}f(n) \bigl(|\bar{u}|+\bigl| \tilde{u}(n)\bigr|\bigr)^{\alpha }\bigl|\tilde{u}(n)\bigr|+\sum _{n=1}^{T}g(n)\bigl|\tilde{u}(n)\bigr| \\ &\quad\leq\sum_{n=1}^{T}f(n)| \bar{u}|^{\alpha}\bigl|\tilde {u}(n)\bigr|+\sum_{n=1}^{T}f(n)\bigl| \tilde{u}(n)\bigr|^{\alpha+1}+\sum_{n=1}^{T}g(n)\bigl| \tilde{u}(n)\bigr| \\ &\quad\leq|\bar{u}|^{\alpha} \Biggl(\sum_{n=1}^{T}f^{2}(n) \Biggr)^{1/2} \Biggl(\sum_{n=1}^{T}\bigl| \tilde{u}(n)\bigr|^{2} \Biggr)^{1/2}+\|\tilde{u}\| _{\infty}^{\alpha+1}\sum_{n=1}^{T}f(n) +\|\tilde{u}\|_{\infty}\sum_{n=1}^{T}g(n) \\ &\quad\leq\frac{1}{2a_{1}}\sum_{n=1}^{T}\bigl| \tilde{u}(n)\bigr|^{2}+\frac {a_{1}}{2}R_{1}^{2}| \bar{u}|^{2\alpha}+R_{2}\|\tilde{u}\|_{\infty }^{\alpha+1}+R_{3} \|\tilde{u}\|_{\infty} \\ &\quad\leq\frac{1}{8a_{1}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\frac{a_{1}}{2}R_{1}^{2}| \bar{u}|^{2\alpha }+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &\qquad{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
(3.1)
From (2) and Lemma 2.1, for any \(u\in H_{T}\), we have
$$\begin{aligned} &\sum_{n=1}^{T} \bigl[F_{2}\bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr] \\ &\quad=\sum _{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar{u}+s\tilde {u}(n)\bigr)-\nabla F_{2}( \bar{u}),s\tilde{u}(n)\bigr)\,ds \\ &\quad\geq-\sum_{n=1}^{T} \int_{0}^{1}rs^{\gamma-1}\bigl|\tilde {u}(n)\bigr|^{\gamma}\,ds \\ &\quad\geq-\frac{rT}{\gamma}\|\tilde{u}\|_{\infty}^{\gamma} \\ &\quad\geq-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl( \sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
(3.2)
Combining (3.1) with (3.2), for all \(u\in H_{T}^{1}\) one has
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr] \\ &{}+\sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr]+\sum_{n=1}^{T}F(n, \bar{u}) \\ \geq{}& \biggl(\frac{1}{2}-\frac{1}{8a_{1}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}- \biggl( \frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}-\frac{rT}{\gamma} \biggl(\frac {T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2} \\ &{}+|\bar{u}|^{2\alpha} \Biggl(|\bar{u}|^{-2\alpha}\sum _{n=1}^{T}F(n,\bar{u})-\frac{a_{1}}{2}R_{1}^{2} \Biggr). \end{aligned}$$
Hence, \(\varphi(u)\rightarrow\infty\) as \(\|u\|\rightarrow\infty\). From this result, if \(\{u_{k}\}\subset H_{T}\) is a minimizing sequence for φ, i.e., \(\varphi(u_{k})\rightarrow\inf\varphi \), \(k\rightarrow\infty\), then \(\{u_{k}\}\) is bounded. Since \(H_{T}\) is finite dimensional, going if necessary to a subsequence, we can assume that \(\{u_{k}\}\) converges to some \(u_{0}\in H_{T}\). Because of φ is continuously differentiable on \(H_{T}\), one has
$$ \varphi(u_{0})=\inf\varphi \quad\mbox{and}\quad \varphi'(u_{0}). $$

Obviously, \(u_{0}\in H_{T}\) is a T-periodic solution of system (1.1). □

Proof of Theorem 1.2

Step 1. To prove φ satisfies the (PS) condition. Suppose that \({u_{k}}\) is a (PS) sequence, that is, \(\varphi'(u_{k})\rightarrow0\) as \(k\rightarrow\infty \) and \({\varphi(u_{k})}\) is bounded. According to (5), there exists \(a_{2}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$\limsup_{x\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)< - \biggl(a_{2}+ \frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2}. $$
In the same way as (3.1), for any \(u\in H_{T}\), one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl(\nabla F_{1}\bigl(n,u_{k}(n)\bigr), \tilde{u}_{k}(n)\bigr)\Biggr\vert \leq{}&\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}+\frac{a_{2}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \end{aligned}$$
(3.3)
and
$$\begin{aligned} \sum_{n=1}^{T}\bigl(\nabla F_{2} \bigl(u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr)\geq- \frac {rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \|\tilde{u}_{k}\| \geq{}&\bigl\langle \varphi'(u_{k}),\tilde {u}_{k}\bigr\rangle \\ ={}&\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl(\nabla F\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr) \\ \geq{}& \biggl(1-\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}} \biggr)\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}-\frac {a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2} \end{aligned}$$
(3.4)
for all large k.
By Lemma 2.1, one has
$$ \|\tilde{u}_{k}\|\leq\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} . $$
(3.5)
By (3.4) and (3.5), for all \(u\in H_{T}^{1}\) one has
$$\begin{aligned} \frac{a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \geq{}& \biggl(1-\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}} \biggr) \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr] \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2} \\ \geq{}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+C_{1}, \end{aligned}$$
(3.6)
where
$$\begin{aligned} C_{1} ={}&\min_{s\in[0,+\infty)}\biggl\{ {\frac{4a_{2}\sin^{2}\frac {\pi}{T}-1}{8a_{2}\sin^{2}\frac{\pi}{T}}s^{2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}s^{\alpha+1}} -\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2}s^{\gamma} \\ &{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr]s\biggr\} . \end{aligned}$$
By the choice of \(a_{2}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\), \(-\infty < C_{1}<0\). Hence
$$ \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \leq a_{2}R_{1}^{2}|\bar {u}_{k}|^{2\alpha}-2C_{1}, $$
(3.7)
and then
$$ \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\leq\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2}, $$
(3.8)
where \(0< C_{2}<+\infty\).
From Theorem 1.1, one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u_{k}(n)\bigr)-F_{1}(n, \bar{u}_{k})\bigr]\Biggr\vert \leq{}&\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}+\frac{a_{2}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
(3.9)
By (4), we obtain
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{2} \bigl(u_{k}(n)\bigr)-F_{2}(\bar{u}_{k})\bigr] \\ &\quad=\sum_{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar {u}_{k}+s\tilde{u}_{k}(n)\bigr)- \nabla F_{2}(\bar{u}_{k}),s\tilde {u}_{k}(n) \bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}Cs^{\delta-1}\bigl|\tilde {u}_{k}(t)\bigr|^{\delta}\,ds\leq\frac{CT}{\delta}\| \tilde{u}_{k}\|_{\infty }^{\delta} \\ &\quad\leq\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta/2} \Biggl( \sum_{=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\delta/2}. \end{aligned}$$
Combining the boundedness of \(\{\varphi(u_{k})\}\) and (3.7)-(3.9), one has
$$\begin{aligned} C_{3} \leq{}&\varphi(u_{k}) \\ ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u_{k}(n)\bigr)-F_{1}(n, \bar{u}_{k})\bigr] \\ &{}+\sum_{n=1}^{T}\bigl[F_{2} \bigl(u_{k}(n)\bigr)-F_{2}(\bar{u}_{k})\bigr]+\sum _{n=1}^{T}F(n,\bar {u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{2}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+ \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}+\frac {a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \\ &{}+\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\delta/2}+\sum_{n=1}^{T}F(n, \bar{u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{2}\sin^{2}\frac{\pi }{T}} \biggr) \bigl(a_{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}-2C_{1} \bigr) + \frac{a_{2}}{2}R_{1}^{2}|\bar{u}_{k}|^{2\alpha}+ \sum_{n=1}^{T}F(n,\bar {u}_{k}) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}\bigl(\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2} \bigr)^{\alpha+1} \\ &{}+\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta/2}\bigl(\sqrt {a_{2}}R_{1}| \bar{u}_{k}|^{\alpha}+C_{2}\bigr)^{\delta} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3}\bigl(\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2} \bigr) \\ \leq{}&|\bar{u}_{k}|^{2\alpha} \Biggl[|\bar{u}_{k}|^{-2\alpha } \sum_{n=1}^{T}F(n,\bar{u}_{k})+ \biggl(a_{2}+\frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2} \Biggr] \\ &{}+C_{4}|\bar{u}_{k}|^{\alpha(\alpha+1)}+C_{5}|\bar {u}_{k}|^{\alpha}+C_{6}|\bar{u}_{k}|^{\alpha\delta}+C_{7} \end{aligned}$$
for large k. By the choice of \(a_{2}\), \(\{\bar{u}_{k}\}\) is bounded. From (3.7), \(\{u_{k}\}\) is bounded. In view of \(H_{T}\) is finite dimensional Hilbert space, φ satisfies the (PS) condition.
Step 2. Let \(\tilde{H}_{T}=\{u\in H_{T}:\bar{u}=0 \}\). We show that, for \(u\in\tilde{H}_{T}\),
$$ \varphi(u)\rightarrow+\infty,\quad \|u\|\rightarrow\infty. $$
(3.10)
From (1) and Lemma 2.1, one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F(n,0)\bigr]\Biggr\vert ={}&\Biggl\vert \sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{1}\bigl(n,s u(n)\bigr),u(n)\bigr)\,ds\Biggr\vert \\ \leq{}&\sum_{n=1}^{T}f(n)\bigl|u(n)\bigr|^{\alpha+1}+ \sum_{n=1}^{T}g(n)\bigl|u(n)\bigr| \\ \leq{}& R_{2}\|u\|_{\infty}^{\alpha+1}+R_{3}\|u \|_{\infty} \\ \leq{}& \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha +1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2} \end{aligned}$$
for all \(u\in\tilde{H}_{T}\). It follows from (2) that
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(0)\bigr] \\ &\quad=\sum_{n=1}^{T} \int _{0}^{1}\bigl(\nabla F_{2}\bigl(su(n) \bigr)-\nabla F_{2}(0),u(n)\bigr)\,ds \\ &\quad\geq-\sum_{n=1}^{T} \int_{0}^{1}rs^{\gamma-1}\bigl|u(n)\bigr|^{\gamma }\,ds \geq-\frac{rT}{\gamma}\|u\|_{\infty}^{\gamma} \\ &\quad\geq-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl( \sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u(n)\bigr)-F(n,0)\bigr]+\sum_{n=1}^{T}F(n,0) \\ \geq{}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}+\sum_{n=1}^{T}F(n,0) \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$

In view of Lemma 2.1, \(\|u\|\rightarrow+\infty\) in \(\tilde{H}_{T}\) if and only if \((\sum_{n=1}^{T}|\Delta u(n)|^{2})^{1/2}\rightarrow\infty\). Hence (3.10) is satisfied.

Step 3. By (5), for all \(u\in(\tilde{H}_{T})^{\bot }=\mathbb{R}^{N}\), one has
$$\varphi(u)=- \sum_{n=1}^{T}F\bigl(n,u(n) \bigr)\rightarrow-\infty, \quad\| u\| \rightarrow\infty. $$

Above all, all conditions of Lemma 2.2 are satisfied. So, by Lemma 2.2, system (1.1) has at least one T-periodic solution. □

Proof of Theorem 1.3

By (7), there exists \(a_{3}>\frac {3}{(12-2(T^{2}-1)r)\sin^{2}\frac{\pi}{T}}\) satisfying
$$\liminf_{|x|\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac{a_{3}}{2}R_{1}^{2}. $$
Similar to (3.1), we have
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{1} \bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr] \\ &\quad\geq-\frac{1}{8a_{3}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}-\frac{a_{3}}{2}R_{1}^{2}| \bar{u}|^{2\alpha }- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &\qquad{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
By (6) and Lemma 2.1, one has
$$\begin{aligned} \sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr] &=\sum _{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar {u}+s\tilde{u}(n)\bigr)-\nabla F_{2}( \bar{u}),s\tilde{u}(n)\bigr)\,ds \\ &\geq-\sum_{n=1}^{T} \int_{0}^{1}rs\bigl|\tilde{u}(n)\bigr|^{2}\,ds\geq - \frac{(T^{2}-1)r}{12}\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}. \end{aligned}$$
So, for any \(u\in H_{T}\), we have
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u(n)\bigr)-F(n,\bar{u})\bigr]+\sum_{n=1}^{T}F(n, \bar{u}) \\ \geq{}& \biggl(\frac{1}{2}-\frac{1}{8a_{3}\sin^{2}\frac{\pi }{T}}-\frac{(T^{2}-1)r}{12} \biggr)\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2} \\ &{}+|\bar{u}|^{2\alpha} \Biggl(|\bar{u}|^{-2\alpha}\sum _{n=1}^{T}F(n,\bar{u})-\frac{a_{3}}{2}R_{1}^{2} \Biggr). \end{aligned}$$

Therefore, \(\varphi(u)\rightarrow+\infty\) as \(\|u\|\rightarrow+\infty\) due to the choice of \(a_{3}\) and \(r<\frac{6}{T^{2}-1}\). The rest is similar to the proof of Theorem 1.1. □

Proof of Theorem 1.4

First, we prove that φ satisfies the (PS) condition. Suppose that \(\{u_{k}\}\subset H_{T}\) is a (PS) sequence of φ, that is, \(\varphi'(u_{k})\rightarrow0\) as \(k\rightarrow\infty\) and \(\{\varphi(u_{k})\}\) is bounded. By (9), there exists \(a_{4}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$ \limsup_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)< - \biggl(a_{4}+ \frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2}. $$
(3.11)
By the \((\lambda,\mu)\)-subconvexity of \(G(x)\), we have
$$ G(x)\leq \bigl(2\mu|x|^{\beta}+1 \bigr)G_{0} $$
(3.12)
for all \(x\in\mathbb{R}^{N}\), where \(G_{0}=\max_{|s|\leq1}G(s)\), \(\beta =\log_{2\lambda}(2\mu)<2\).
Then
$$\begin{aligned} \sum_{n=1}^{T}\bigl(\nabla F_{2}\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n) \bigr) &\geq-\sum_{n=1}^{T}h(n)G( \bar{u}_{k}) \\ &\geq-\sum_{n=1}^{T}h(n) \bigl(2\mu| \bar{u}_{k}|^{\beta }+1\bigr)G_{0} \\ &=-2\mu R_{4}|\bar{u}_{k}|^{\beta}-R_{4}, \end{aligned}$$
(3.13)
where \(R_{4}=G_{0}\sum_{n=1}^{T}h(n)\). For large k, according to (3.3) and (3.13) we have
$$\begin{aligned} \|\tilde{u}_{k}\| \geq{}&\bigl\langle \varphi'(u_{k}),\tilde {u}_{k}\bigr\rangle \\ ={}&\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl(\nabla F\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr) \\ \geq{}& \biggl(1-\frac{1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}-\frac{a_{4}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}-2\mu R_{4}|\bar {u}_{k}|^{\beta}-R_{4}. \end{aligned}$$
(3.14)
By (3.5) and (3.14), one has
$$\begin{aligned} &\frac{a_{4}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+2\mu R_{4}|\bar {u}_{k}|^{\beta} \\ &\quad\geq \biggl(1-\frac{1}{8a_{4}\sin^{2}\frac {\pi}{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \\ &\qquad{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2}-R_{4} \\ &\qquad{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr] \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \\ &\quad\geq \frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+C_{8}, \end{aligned}$$
(3.15)
where
$$\begin{aligned} C_{8} ={}&\min_{s\in[0,+\infty)}\biggl\{ \biggl( \frac{1}{2}-\frac {1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)s^{2} - \biggl( \frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}s^{\alpha+1} \\ &{}-R_{4}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin \frac{\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr]s\biggr\} . \end{aligned}$$
By the choice of \(a_{4}\), \(-\infty< C_{8}<0\). By (3.15), we have
$$ \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \leq a_{4}R_{1}^{2}|\bar {u}_{k}|^{2\alpha}+4 \mu R_{4}|\bar{u}_{k}|^{\beta}-2C_{8}, $$
(3.16)
and then
$$ \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\leq\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9}, $$
(3.17)
where \(C_{9}>0\). By (8) and (3.12), for any \(u\in H_{T}\), we get
$$\begin{aligned} &\sum_{n=1}^{T} \bigl[F_{2}\bigl(n,u(n)\bigr)-F_{2}(n,\bar{u})\bigr] \\ &\quad=-\sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{2}\bigl(n,\bar {u}_{k}+s\tilde{u}_{k}(n)\bigr),\tilde{u}_{k}(n) \bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}h(n)G\bigl(\bar {u}_{k}+(s+1) \tilde{u}_{k}(n)\bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}h(n) \bigl(2\mu\bigl|\bar {u}_{k}+(s+1)\tilde{u}_{k}(n)\bigr|^{\beta}+1 \bigr)G_{0}\,ds \\ &\quad\leq4\mu\sum_{n=1}^{T}h(n) \bigl(| \bar{u}_{k}|^{\beta}+2^{\beta }\bigl|\tilde{u}_{k}(n)\bigr|^{\beta} \bigr)G_{0}+R_{4} \\ &\quad\leq2^{\beta+2}\mu R_{4}\|\tilde{u}_{k} \|_{\infty}^{\beta }+4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4} \\ &\quad\leq \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta +2}\mu R_{4} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{\beta /2}+4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4}. \end{aligned}$$
(3.18)
Combining the boundedness of \(\{\varphi(u_{k})\}\) and (3.16)-(3.18), one has
$$\begin{aligned} C_{10} \leq{}&\varphi(u_{k}) \\ ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u_{k}(n)\bigr)-F(n,\bar{u}_{k})\bigr]+\sum _{n=1}^{T}F(n,\bar {u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{4}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+ \frac {a_{4}}{2}R_{1}^{2}|\bar{u}_{k}|^{2}\\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\\ &{}+ \biggl(\frac {T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2} \mu R_{4} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{\beta/2} +4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4}+ \sum_{n=1}^{T}F(n,\bar{u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{4}\sin^{2}\frac{\pi }{T}} \biggr) \bigl(a_{4}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+4\mu R_{4}|\bar {u}_{k}|^{\beta}-2C_{8} \bigr)+\frac{a_{4}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \bigl( \sqrt{a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2 \sqrt{\mu R_{4}}|\bar {u}_{k}|^{\beta/2}+C_{9} \bigr)^{\alpha+1} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \bigl(\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2}\mu R_{4} \bigl(\sqrt{a_{4}}R_{1}| \bar{u}_{k}|^{\alpha}+2\sqrt{\mu R_{4}}| \bar{u}_{k}|^{\beta/2}+C_{9} \bigr)^{\beta}+4\mu R_{4}|\bar {u}_{k}|^{\beta}\\ &{}+R_{4}+\sum _{n=1}^{T}F(n,\bar{u}_{k}) \\ \leq{}& \biggl(1+\frac{1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)a_{4}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+ \biggl(6+\frac{1}{2a_{4}\sin ^{2}\frac{\pi}{T}} \biggr)\mu R_{4}|\bar{u}_{k}|^{\beta}\\ &{}- \biggl(1+ \frac {1}{4a_{4}\sin^{2}\frac{\pi}{T}} \biggr)C_{8} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \bigl(2^{\alpha}a_{4}^{\frac{\alpha+1}{2}}R_{1}^{\alpha+1}| \bar {u}_{k}|^{\alpha(\alpha+1)} +2^{3\alpha+1}\mu^{\frac{\alpha+1}{2}}R_{4}^{\frac{\alpha+1}{2}}| \bar {u}_{k}|^{\frac{\beta(\alpha+1)}{2}} +2^{2\alpha}C_{9}^{\alpha+1} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \bigl(\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2}\mu R_{4} \bigl(2^{\beta-1}a_{4}^{\frac{\beta}{2}}R_{1}^{\beta}| \bar {u}_{k}|^{\alpha\beta}+2^{3\beta-2}\mu^{\frac{\beta}{2}}R_{4}^{\frac {\beta}{2}}| \bar{u}_{k}|^{\frac{\beta^{2}}{2}} +2^{2(\beta-1)}C_{9}^{\beta} \bigr)\\ &{}+R_{4}+\sum_{n=1}^{T}F(n, \bar {u}_{k}) \\ ={}&|\bar{u}_{k}|^{2\alpha}\Biggl[|\bar{u}_{k}|^{-2\alpha} \sum_{n=1}^{T}F_{1}(n, \bar{u}_{k})+ \biggl(a_{4}+\frac{1}{8\sin^{2}\frac{\pi }{T}} \biggr)R_{1}^{2} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}\sqrt{a_{4}}R_{1}R_{3}|\bar {u}_{k}|^{-\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{\alpha }a_{4}^{\frac{\alpha+1}{2}}R_{1}^{\alpha+1}| \bar{u}_{k}|^{\alpha(\alpha-1)} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{2\beta+1}\mu a_{4}^{\frac {\beta}{2}}R_{1}^{\beta}R_{4}| \bar{u}_{k}|^{\alpha(\beta-2)}\Biggr] \\ &{}+|\bar{u}_{k}|^{\beta}\Biggl[|\bar{u}_{k}|^{-\beta} \sum_{n=1}^{T}F_{2}(n, \bar{u}_{k})+ \biggl(6+\frac{1}{2a_{4}\sin^{2}\frac{\pi }{T}} \biggr)\mu R_{4}\\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{4\beta } \mu^{\frac{\beta+2}{2}}R_{4}^{\frac{\beta+2}{2}}|\bar{u}_{k}|^{\frac {1}{2}\beta^{2}-\beta} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{3\alpha +1} \mu^{\frac{\alpha+2}{2}}R_{2}R_{4}^{\frac{\alpha+1}{2}}|\bar {u}_{k}|^{\frac{\beta(\alpha-1)}{2}} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}2R_{3}\sqrt{\mu R_{4}}|\bar {u}_{k}|^{-\beta/2}\Biggr] \\ &{}- \biggl(1+\frac{1}{4a_{4}\sin^{2}\frac{\pi}{T}} \biggr)C_{8}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{2\alpha }R_{2}C_{9}^{\alpha+1}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3}C_{9} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{3\beta}\mu R_{4}C_{9}^{\beta}+R_{4}. \end{aligned}$$

Combining (3.11) and the above inequality, we see that \(\{|\bar {u}|\}\) is bounded. By (3.16), \(\{u_{k}\}\) is bounded. Since \(H_{T}\) is a finite dimensional Hilbert space, φ satisfies the (PS) condition.

Similar to the proof of Theorem 1.2, all conditions of Lemma 2.2 are satisfied. So, the proof of Theorem 1.4 is completed. □

4 Examples

In this section, we give some examples to illustrate our results.

Example 4.1

Let \(F=F_{1}+F_{2}\), with
$$\begin{aligned} &F_{1}(n,x)= \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4}+(2T-n)|x|^{3/2}+ \bigl(k(n),x\bigr),\\ &F_{2}(x)=C(x)-\frac{3r}{4}|x|^{4/3}, \end{aligned}$$
where \(k:\mathbb{Z}[1,T]\longrightarrow\mathbb{R}\) and \(k(n+T)=k(n)\), for all \(n\in\mathbb{Z}\), \(r>0\), \(C(x)=\frac{3r}{4} (|x_{1}|^{4/3}+|x_{2}|^{4/3}+\cdots+|x_{N}|^{4/3})\). It is easy to see that
$$\begin{aligned} \bigl|\nabla F_{1}(n,x)\bigr| &\leq\frac{7}{8}|T+1-2n||x|^{3/4}+ \frac {3}{2}|2T-n||x|^{1/2}+\bigl|k(n)\bigr| \\ &\leq\frac{7}{8}\bigl(|T+1-2n|+\varepsilon\bigr)|x|^{3/4}+ \frac {9T^{2}}{\varepsilon^{2}}+\bigl|k(n)\bigr|. \end{aligned}$$
For all \((n,x)\in\mathbb{Z}[1,T]\times\mathbb{R}^{N}\), where \(\varepsilon>0\),
$$ \bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y\bigr) \geq-r|x-y|^{4/3}. $$
Thus, (1), (2) hold with \(\alpha=3/4\), \(\gamma=4/3\), and
$$ f(n)=\frac{7}{8}\bigl(|T+1-2n|+\varepsilon\bigr),\qquad g(n)= \frac {9T^{2}}{\varepsilon^{2}}+\bigl|k(n)\bigr|. $$
So, we have
$$\begin{aligned} &|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x) \\ &\quad=|x|^{-3/2}\sum_{n=1}^{T} \biggl[ \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4}+(2T-n)|x|^{3/2} +C(x)-\frac{3r}{4}|x|^{4/3}+\bigl(k(n),x\bigr) \biggr] \\ &\quad=\frac{T(3T-1)}{2}+\frac{T(C(x)-\frac {3r}{4}|x|^{4/3})}{|x|^{3/2}}+ \Biggl(\sum _{n=1}^{T}k(n),|x|^{-3/2}x \Biggr). \end{aligned}$$
On the other hand, one has
$$ \frac{1}{8\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}f^{2}(n)= \frac{1}{8\sin ^{2}\frac{\pi}{T}}\sum_{n=1}^{T} \biggl[ \frac{7}{8} \bigl(|T+1-2n|+\varepsilon\bigr) \biggr]^{2}\leq \frac{49 [T(T^{2}-1+6\varepsilon T +2\varepsilon^{2}) ]}{1{,}536\sin^{2}\frac {\pi}{T}}. $$
If \(T\in\{2,3,4,5,6,7\}\), we can choose \(\varepsilon>0\) such that
$$ \liminf_{|x|\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)=\frac{T(3T-1)}{2}>\frac{1}{8\sin^{2}\frac{\pi}{T}} \sum_{n=1}^{T}f^{2}(n). $$
So, (3) holds. By Theorem 1.1, system (1.1) has at least one T-periodic solution.

Example 4.2

Let \(F=F_{1}+F_{2}\), with
$$\begin{aligned}& F_{1}(n,x)= \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4}-(2T-n)|x|^{3/2}+ \bigl(k(n),x\bigr),\\& F_{2}(x)=-\frac{4r}{5}|x|^{5/4}, \end{aligned}$$
where \(k:\mathbb{Z}[1,T]\longrightarrow\mathbb{R}^{N}\) and \(k(n+T)=k(n)\) for all \(n\in\mathbb{Z}\), \(r>0\).
In a way similar to Example 4.1, it is easy to see that condition (1) and (4) are satisfied with \(\alpha=3/4\). So,
$$\begin{aligned} &|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x) \\ &\quad=|x|^{-3/2}\sum_{n=1}^{T} \biggl[ \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4} -(2T-n)|x|^{3/2}- \frac{4r}{5}|x|^{5/4}+\bigl(k(n),x\bigr) \biggr] \\ &\quad=-\frac{T(3T-1)}{2}-\frac{4r}{5}|x|^{-1/4}+ \Biggl(\sum _{n=1}^{T}k(n),|x|^{-3/2}x \Biggr). \end{aligned}$$
If \(T\in\{2,3,4,5\}\), we can choose \(\varepsilon>0\) small enough such that
$$\begin{aligned} \limsup_{|x|\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x) =-\frac{T(3T-1)}{2}< - \frac{3}{8\sin ^{2}\frac{\pi}{T}}\sum_{n=1}^{T}f^{2}(n), \end{aligned}$$
which implies that (5) holds. By Theorem 1.2, system (1.1) has at least one T-periodic solution.

Example 4.3

Let \(F=F_{1}+F_{2}\), with
$$\begin{aligned}& F_{1}(n,x)= \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4}- \biggl(\frac {T-3n}{2} \biggr)|x|^{3/2}+\bigl(k(n),x\bigr),\\& F_{2}(x)=C(x)-\frac{r}{2}|x|^{2}, \end{aligned}$$
where \(k:\mathbb{Z}[1,T]\longrightarrow\mathbb{R}\) and \(k(n+T)=k(n)\) for all \(n\in\mathbb{Z}\), \(r>0\), \(C(x)=\frac{r}{2} (|x_{1}|^{4}+|x_{2}|^{2}+\cdots+|x_{N}|^{2})\), \(0< r<\frac {6}{T^{2}-1}\).
In a way similar to Example 4.1, it is easy to see that conditions (1) and (6) are satisfied with \(\alpha=3/4\). So
$$\begin{aligned} &|x|^{-2\alpha}\sum_{n=1}^{T}F(n,x) \\ &\quad= |x|^{-3/2}\sum_{n=1}^{T} \biggl[ \biggl(\frac{T+1}{2}-n \biggr)|x|^{7/4}- \biggl(\frac{T-3n}{2} \biggr)|x|^{3/2}+C(x)-\frac{r}{2}|x|^{2}+ \bigl(k(n),x \bigr) \biggr] \\ &\quad=\frac{T(T+3)}{4}+\frac{T(C(x)-\frac {r}{2}|x|^{2})}{|x|^{3/2}}+ \Biggl(\sum _{n=1}^{T}k(n),|x|^{-3/2}x \Biggr) \\ &\quad=\frac{T(T+3)}{4}+\frac {rT(|x|_{1}^{4}-|x|_{1}^{2})}{2|x|^{3/2}}+ \Biggl(\sum _{n=1}^{T}k(n),|x|^{-3/2}x \Biggr). \end{aligned}$$
If \(T\in\{2,3\}\), we choose \(\varepsilon>0\), such that
$$ \liminf_{|x|\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)=\frac{T(T+3)}{4}>\frac{3}{ (24-4(T^{2}-1)r )\sin^{2}\frac{\pi}{T}} \sum_{n=1}^{T}f^{2}(n), $$
which implies that (7) holds. By Theorem 1.3, system (1.1) has at least one T-periodic solution.

Declarations

Acknowledgements

Research was supported by the Postdoctoral fund in China (Grant No. 2013M531717) and NSFC (11561043).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Electrical and Information Engineering, Lanzhou University of Technology
(2)
Department of Applied Mathematics, Lanzhou University of Technology

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