Theory and Modern Applications

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

## 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.

## 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.

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## Acknowledgements

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

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Correspondence to Wen Guan.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The main idea of this paper was proposed by KY and WG, WG prepared the manuscript initially and KY performed a part of the steps of the proofs in this research. All authors read and approved the final manuscript.

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Guan, W., Yang, K. Existence of periodic solutions for a class of second order discrete Hamiltonian systems. Adv Differ Equ 2016, 68 (2016). https://doi.org/10.1186/s13662-016-0787-5