# Existence of subharmonic solutions for non-quadratic second-order Hamiltonian systems

## Abstract

In this paper, some existence theorems are obtained for subharmonic solutions of second-order Hamiltonian systems with linear part under non-quadratic conditions. The approach is the minimax principle. We consider some new cases and obtain some new existence results.

MSC:34C25, 58E50, 70H05.

## 1 Introduction and main results

Consider the second-order Hamiltonian system

(1.1)

where A is an $N×N$ symmetric matrix and $F:\mathbb{R}×{\mathbb{R}}^{N}\to \mathbb{R}$ is T-periodic in t and satisfies the following assumption:

Assumption (A)′ $F\left(t,x\right)$ is measurable in t for every $x\in {\mathbb{R}}^{N}$ and continuously differentiable in x for a.e. $t\in \left[0,T\right]$, and there exist $a\in C\left({\mathbb{R}}^{+},{\mathbb{R}}^{+}\right)$ and $b:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ which is T-periodic and $b\in {L}^{p}\left(0,T;{\mathbb{R}}^{+}\right)$ with $p>1$ such that

$|F\left(t,x\right)|\le a\left(|x|\right)b\left(t\right),\phantom{\rule{2em}{0ex}}|\mathrm{\nabla }F\left(t,x\right)|\le a\left(|x|\right)b\left(t\right)$

for all $x\in {\mathbb{R}}^{N}$ and a.e. $t\in \left[0,T\right]$.

When $A=0$, system (1.1) reduces to the second-order Hamiltonian system

(1.2)

There have been many existence results for system (1.2) (for example, see [17] and references therein). In 1978, Rabinowitz [6] obtained the nonconstant periodic solutions for system (1.2) under the following AR-condition: there exist $\mu >2$ and $L>0$ such that

$0<\mu F\left(t,x\right)\le \left(\mathrm{\nabla }F\left(t,x\right),x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }|x|\ge L,t\in \left[0,T\right].$

From then on, the condition has been used extensively in the literature; see [812] and the references therein. In [13], Fei also obtained the existence of nonconstant solutions for system (1.2) under a kind of new superquadratic condition. Subsequently, Tao and Tang [14] gave the following more general one than Fei’s: there exist $\theta >2$ and $\mu >\theta -2$ such that

(1.3)
(1.4)

They also considered the existence of subharmonic solutions and obtained the following result.

Theorem A (See [14], Theorem 2)

Suppose that F satisfies

1. (A)

$F\left(t,x\right)$ is measurable in t for every $x\in {\mathbb{R}}^{N}$ and continuously differentiable in x for a.e. $t\in \left[0,T\right]$, and there exist $a\in C\left({\mathbb{R}}^{+},{\mathbb{R}}^{+}\right)$ and $b\in {L}^{1}\left(0,T;{\mathbb{R}}^{+}\right)$ such that

$|F\left(t,x\right)|\le a\left(|x|\right)b\left(t\right),\phantom{\rule{2em}{0ex}}|\mathrm{\nabla }F\left(t,x\right)|\le a\left(|x|\right)b\left(t\right)$

for all $x\in {\mathbb{R}}^{N}$ and a.e. $t\in \left[0,T\right]$. Assume that (1.3), (1.4) and the following conditions hold:

$F\left(t,x\right)\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }\left(t,x\right)\in \left[0,T\right]×{\mathbb{R}}^{N},$
(1.5)
$\underset{|x|\to 0}{lim}\frac{F\left(t,x\right)}{{|x|}^{2}}=0\phantom{\rule{1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right],$
(1.6)
$\underset{|x|\to \mathrm{\infty }}{lim}\frac{F\left(t,x\right)}{{|x|}^{2}}>\frac{2{\pi }^{2}}{{T}^{2}}\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$
(1.7)

Then system (1.2) has a sequence of distinct periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$.

Recently, Ma and Zhang [15] considered the following p-Laplacian system:

(1.8)

where $p>1$. By using some techniques, they obtained the following more general result than Theorem A.

Theorem B (See [15], Theorem 1)

Suppose that F satisfies (A), (1.3) and (1.4) with 2 replaced by p, (1.5) and the following condition:

$\underset{|x|\to 0}{lim}\frac{F\left(t,x\right)}{{|x|}^{p}}=0<\underset{|x|\to \mathrm{\infty }}{lim}\frac{F\left(t,x\right)}{{|x|}^{p}}\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$
(1.9)

Then system (1.8) has a sequence of distinct periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$.

When $A={m}^{2}{\omega }^{2}{I}_{N}$, where $\omega =2\pi /T$ and ${I}_{N}$ is the unit matrix of order N. Ye and Tang [16] obtained the following result.

Theorem C (See [16], Theorem 2)

Suppose that $A={m}^{2}{\omega }^{2}{I}_{N}$, F satisfies (A), (1.3), (1.4), (1.5), (1.6) and the following conditions:

$\underset{|x|\to \mathrm{\infty }}{lim}\frac{F\left(t,x\right)}{{|x|}^{2}}>\frac{1+2m}{2}{\omega }^{2}\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$

Then system (1.1) has a sequence of distinct periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$.

Recently, in [17], we considered a more general case than that in [16]. We considered the case that A only has 0 or ${l}_{i}^{2}{\omega }^{2}$ as its eigenvalues, where $\omega =2\pi /T$, ${l}_{i}\in \mathbb{N}$, $i=1,\dots ,r$ and $0\le r\le N$. In [17], we used the following condition which presents some advantages over (1.3) and (1.4):

1. (H)

there exist positive constants m, ζ, η and $\nu \in \left[0,2\right)$ such that

In this paper, we consider some new cases which can be seen as a continuance of our work in [17].

Next, we state our main results. Assume that $r\in \mathbb{N}\cup \left\{0\right\}$ and $r\le N$. Let ${\lambda }_{i}>0$ ($i\in \left\{1,\dots ,r\right\}$) and $-{\lambda }_{i}<0$ ($i\in \left\{r+s+1,\dots ,N\right\}$) be the positive and negative eigenvalues of A, respectively, where r and s denote the number of positive eigenvalues and zero eigenvalues of A (counted by multiplicity), respectively. Moreover, we denote by q the number of negative eigenvalues of A (counted by multiplicity). We make the following assumption:

Assumption (A0) A has at least one nonzero eigenvalue and all positive eigenvalues are not equal to ${l}^{2}{\omega }^{2}$ for all $l\in \mathbb{N}$, where $\omega =2\pi /T$, that is, ${\lambda }_{i}\ne {l}^{2}{\omega }^{2}$ ($i=1,\dots ,r$) for all $l\in \mathbb{N}$.

The Assumption (A0) implies that one can find ${l}_{i}\in {\mathbb{Z}}^{+}:=\left\{0,1,2,\dots \right\}$ such that

${l}_{i}^{2}{\omega }^{2}<{\lambda }_{i}<{\left({l}_{i}+1\right)}^{2}{\omega }^{2},\phantom{\rule{1em}{0ex}}i=1,\dots ,r.$
(1.10)

For the sake of convenience, we set

$\begin{array}{r}{\lambda }_{{i}^{+}}=max\left\{{\lambda }_{i}|i=1,\dots ,r\right\},\phantom{\rule{2em}{0ex}}{\lambda }_{{i}^{-}}=min\left\{{\lambda }_{i}|i=1,\dots ,r\right\},\\ {\lambda }_{{i}_{+}}=max\left\{{\lambda }_{i}|i=r+s+1,\dots ,N\right\},\phantom{\rule{2em}{0ex}}{\lambda }_{{i}_{-}}=min\left\{{\lambda }_{i}|i=r+s+1,\dots ,N\right\}.\end{array}$

Then

${i}^{+},{i}^{-}\in \left\{1,\dots ,r\right\},\phantom{\rule{2em}{0ex}}{i}_{+},{i}_{-}\in \left\{r+s+1,\dots ,N\right\}.$

Corresponding to (1.10), we know that there exist ${l}_{{i}^{+}},{l}_{{i}^{-}}\in {\mathbb{Z}}^{+}$ such that

${l}_{{i}^{+}}^{2}{\omega }^{2}<{\lambda }_{{i}^{+}}<{\left({l}_{{i}^{+}}+1\right)}^{2}{\omega }^{2},\phantom{\rule{2em}{0ex}}{l}_{{i}^{-}}^{2}{\omega }^{2}<{\lambda }_{{i}^{-}}<{\left({l}_{{i}^{-}}+1\right)}^{2}{\omega }^{2}.$

Moreover, set

${h}_{i}={\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i},\phantom{\rule{1em}{0ex}}i=1,\dots ,r,$

and let ${h}_{{i}_{0}}={min}_{i\in \left\{1,\dots ,r\right\}}\left\{{h}_{i}\right\}$. Then ${i}_{0}\in \left\{1,\dots ,r\right\}$. Corresponding to (1.10), there exists ${l}_{{i}_{0}}\in {\mathbb{Z}}^{+}$ such that

${l}_{{i}_{0}}^{2}{\omega }^{2}<{\lambda }_{{i}_{0}}<{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}.$
(1.11)

Theorem 1.1 Assume that (A0) holds and F satisfies (A)′, (1.5) and the following conditions.

(H1) For some $k\in \mathbb{N}$, assume that k satisfies

${\left({l}_{i}+1-\frac{1}{k}\right)}^{2}{\omega }^{2}\le {\lambda }_{i}<{\left({l}_{i}+1\right)}^{2}{\omega }^{2}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}i\in \left\{1,\dots ,r\right\}.$
(1.12)

(H2) There exist positive constants m, ζ, η and $\nu \in \left[0,2\right)$ such that

$\left(2+\frac{1}{\zeta +\eta {|x|}^{\nu }}\right)F\left(t,x\right)\le \left(\mathrm{\nabla }F\left(t,x\right),x\right),\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{N},|x|>m,\phantom{\rule{0.1em}{0ex}}\mathit{\text{a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$

(H3) Assume that one of the following cases holds:

1. (1)

when $r>0$, $s>0$ and $r+s=N$, there exist ${L}_{k}>0$ and ${\beta }_{k}>min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}}\right\}$ such that

$F\left(t,x\right)\ge {\beta }_{k}{|x|}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathbb{R}}^{N},|x|>{L}_{k},\phantom{\rule{0.1em}{0ex}}\mathit{\text{a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right],$
(1.13)

where ${l}_{{i}_{0}}$ and ${\lambda }_{{i}_{0}}$ are defined by (1.11);

1. (2)

when $r>0$, $s>0$ and $r+s, there exist ${L}_{k}>0$ and ${\beta }_{k}>min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}$ such that (1.13) holds;

2. (3)

when $r>0$, $s=0$ and $r+s, there exist ${L}_{k}>0$ and ${\beta }_{k}>min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\lambda }_{{i}_{-}}}{2}\right\}$ such that (1.13) holds;

3. (4)

when $r>0$, $s=0$ and $r=N$, there exist ${L}_{k}>0$ and ${\beta }_{k}>\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2}$ such that (1.13) holds;

4. (5)

when $r=0$, $s>0$ and $s, there exist ${L}_{k}>0$ and ${\beta }_{k}>min\left\{\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}$ such that (1.13) holds;

5. (6)

when $r=0$, $s=0$ and $q=N$, there exist ${L}_{k}>0$ and ${\beta }_{k}>\frac{{\lambda }_{{i}_{-}}}{2}$ such that (1.13) holds;

(H4) there exist ${l}_{k}>0$ and ${\alpha }_{k}<\frac{{\sigma }_{k}}{2}$ such that

$F\left(t,x\right)\le {\alpha }_{k}{|x|}^{2}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}|x|\le {l}_{k}\phantom{\rule{0.1em}{0ex}}\mathit{\text{and a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right],$

where

where σ implies that ${\sigma }_{k}$ is independent of k. Then system (1.1) has a nonzero kT-periodic solution. Especially, for cases (H3)(1) and (H3)(4), system (1.1) has a nonconstant kT-periodic solution.

Remark 1.1 For cases (H3)(1)-(H3)(4), from (1.10) and (1.12), it is easy to see that the number of $k\in \mathbb{N}$ satisfying (1.12) is finite. Let $m\in K$ be the maximum integer satisfying (1.12), where

Then $K=\left\{1,2,\dots ,m\right\}$. Hence, Theorem 1.1 implies that system (1.1) has nonzero kT-periodic solutions ($k=1,2,\dots ,m$). For cases (H3)(5) and (H3)(6), since $r=0$, (1.12) holds for every $k\in \mathbb{N}$. Hence, Theorem 1.1 implies that system (1.11) has nonzero kT-periodic solutions for every $k\in \mathbb{N}$.

Remark 1.2 In [18], Costa and Magalhães studied the first-order Hamiltonian system

(1.14)

They obtained that system (1.14) has a $T=2\pi$ periodic solution under the following non-quadraticity conditions:

(1.15)

and the so-called asymptotic noncrossing conditions

where ${\lambda }_{k-1}<{\lambda }_{k}$ are consecutive eigenvalues of the operator $L=-Jd/dt-A$. Moreover, they also obtained system (1.14) has a nonzero $T=2\pi$ periodic solution under (1.15) and the called crossing conditions

One can also establish the similar results for the second-order Hamiltonian system (1.1). Some related contents can be seen in [19]. It is worth noting that in [18] and [19], ${\lambda }_{k-1}<{\lambda }_{k}$ are consecutive eigenvalues of the operator $L=-Jd/dt-A$ or $-{d}^{2}/d{t}^{2}+A$. In our Theorem 1.1 and Theorem 1.2, we study the existence of subharmonic solutions for system (1.1) from a different perspective. ${\lambda }_{i}$ ($i\in \left\{1,\dots ,r\right\}$) in our theorems are the eigenvalues of the matrix A. Obviously, it is much easier to seek the eigenvalue of a matrix. In Section 4, we present an interesting example satisfying our Theorem 1.1 but not satisfying the theorem in [19].

Theorem 1.2 Suppose that (A0) holds and F satisfies (A)′, (1.5), (H2) and the following conditions:

(H3)′ when $r=0$, $s>0$ and $s, there exist $L>0$ and $\beta >\frac{{\omega }^{2}}{2}$ such that

$F\left(t,x\right)\ge \beta {|x|}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathbb{R}}^{N},|x|>L,\phantom{\rule{0.1em}{0ex}}\mathit{\text{a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right];$
(1.16)

(H4)′

$\underset{|x|\to 0}{lim}\frac{F\left(t,x\right)}{{|x|}^{2}}=0\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$

Then system (1.1) has a sequence of distinct periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$.

In the final theorem, we present a result about the existence of subharmonic solutions for system (1.8). Using a condition like (H2) and similar to the argument of Remark 1.1 in [17], we can improve Theorem B.

Theorem 1.3 Suppose that F satisfies (A), (1.5) and the following conditions:

(H5) there exist positive constants m, ζ, η and $\nu \in \left[0,p\right)$ such that

$\left(p+\frac{1}{\zeta +\eta {|x|}^{\nu }}\right)F\left(t,x\right)\le \left(\mathrm{\nabla }F\left(t,x\right),x\right),\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{N},|x|>m\phantom{\rule{0.1em}{0ex}}\mathit{\text{a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right];$

(H6)

$\underset{|x|\to 0}{lim}\frac{F\left(t,x\right)}{{|x|}^{p}}=0<\underset{|x|\to \mathrm{\infty }}{lim}\frac{F\left(t,x\right)}{{|x|}^{p}}\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly for a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right].$

Then system (1.8) has a sequence of distinct nonconstant periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$.

## 2 Some preliminaries

Let

Then ${H}_{kT}^{1}$ is a Hilbert space with the inner product and the norm defined by

$〈u,v〉={\int }_{0}^{kT}\left(u\left(t\right),v\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{kT}\left(\stackrel{˙}{u}\left(t\right),\stackrel{˙}{v}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt$

and

$\parallel u\parallel ={\left[{\int }_{0}^{kT}{|u\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\right]}^{1/2}$

for each $u,v\in {H}_{kT}^{1}$. Let

$\overline{u}=\frac{1}{kT}{\int }_{0}^{kT}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{˜}{u}\left(t\right)=u\left(t\right)-\overline{u}.$

Then one has

$\begin{array}{r}{\parallel \stackrel{˜}{u}\parallel }_{\mathrm{\infty }}^{2}\le \frac{kT}{12}{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{(Sobolev’s inequality)},\\ {\parallel \stackrel{˜}{u}\parallel }_{{L}^{2}}^{2}\le \frac{{k}^{2}{T}^{2}}{4{\pi }^{2}}{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{(Wirtinger’s inequality)}\end{array}$

(see Proposition 1.3 in [1]).

Lemma 2.1 If $u\in {H}_{kT}^{1}$, then

${\parallel u\parallel }_{\mathrm{\infty }}\le \sqrt{\frac{12+{k}^{2}{T}^{2}}{12kT}}\parallel u\parallel ,$

where ${\parallel u\parallel }_{\mathrm{\infty }}={max}_{t\in \left[0,kT\right]}|u\left(t\right)|$.

Proof Fix $t\in \left[0,kT\right]$. For every $\tau \in \left[0,kT\right]$, we have

$u\left(t\right)=u\left(\tau \right)+{\int }_{\tau }^{t}\stackrel{˙}{u}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds.$
(2.1)

Set

$\varphi \left(s\right)=\left\{\begin{array}{cc}s-t+\frac{kT}{2},\hfill & t-kT/2\le s\le t,\hfill \\ t+\frac{kT}{2}-s,\hfill & t\le s\le t+kT/2.\hfill \end{array}$

Integrating (2.1) over $\left[t-kT/2,t+kT/2\right]$ and using the Hölder inequality, we obtain

$\begin{array}{rcl}kT|u\left(t\right)|& =& |{\int }_{t-kT/2}^{t+kT/2}u\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t-kT/2}^{t+kT/2}{\int }_{\tau }^{t}\stackrel{˙}{u}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau |\\ \le & {\int }_{t-kT/2}^{t+kT/2}|u\left(\tau \right)|\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t-kT/2}^{t}{\int }_{\tau }^{t}|\stackrel{˙}{u}\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t}^{t+kT/2}{\int }_{t}^{\tau }|\stackrel{˙}{u}\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \\ =& {\int }_{t-kT/2}^{t+kT/2}|u\left(\tau \right)|\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t-kT/2}^{t}\left(s-t+\frac{kT}{2}\right)|\stackrel{˙}{u}\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds\\ +{\int }_{t}^{t+kT/2}\left(t+\frac{kT}{2}-s\right)|\stackrel{˙}{u}\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int }_{t-kT/2}^{t+kT/2}|u\left(\tau \right)|\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t-kT/2}^{t+kT/2}\varphi \left(s\right)|\stackrel{˙}{u}\left(s\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\left(kT\right)}^{1/2}{\left({\int }_{t-kT/2}^{t+kT/2}{|u\left(\tau \right)|}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{1/2}\\ +{\left({\int }_{t-kT/2}^{t+kT/2}{\left[\varphi \left(s\right)\right]}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}{\left({\int }_{t-kT/2}^{t+kT/2}{|\stackrel{˙}{u}\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}\\ =& {\left(kT\right)}^{1/2}{\left({\int }_{t-kT/2}^{t+kT/2}{|u\left(\tau \right)|}^{2}\phantom{\rule{0.2em}{0ex}}d\tau \right)}^{1/2}+\frac{{\left(kT\right)}^{3/2}}{2\sqrt{3}}{\left({\int }_{t-kT/2}^{t+kT/2}{|\stackrel{˙}{u}\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}\\ \le & {\left(kT+\frac{{\left(kT\right)}^{3}}{12}\right)}^{1/2}{\left({\int }_{t-kT/2}^{t+kT/2}{|u\left(\tau \right)|}^{2}\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{t-kT/2}^{t+kT/2}{|\stackrel{˙}{u}\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}\\ =& {\left(kT+\frac{{\left(kT\right)}^{3}}{12}\right)}^{1/2}{\left({\int }_{0}^{kT}{|u\left(\tau \right)|}^{2}\phantom{\rule{0.2em}{0ex}}d\tau +{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}.\end{array}$

Hence, we have

${\parallel u\parallel }_{\mathrm{\infty }}\le {\left(\frac{1}{kT}+\frac{kT}{12}\right)}^{1/2}{\left({\int }_{0}^{kT}{|u\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(s\right)|}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/2}.$

The proof is complete. □

Lemma 2.2 (see [[17], Lemma 2.2])

Assume that $F=F\left(t,x\right):\mathbb{R}×{\mathbb{R}}^{N}\to \mathbb{R}$ is T-periodic in t, $F\left(t,x\right)$ is measurable in t for every $x\in {\mathbb{R}}^{N}$ and continuously differentiable in x for a.e. $t\in \left[0,T\right]$. If there exist $a\in C\left({\mathbb{R}}^{+},{\mathbb{R}}^{+}\right)$ and $b\in {L}^{p}\left(\left[0,T\right],{\mathbb{R}}^{+}\right)$ ($p>1$) such that

$|\mathrm{\nabla }F\left(t,x\right)|\le a\left(|x|\right)b\left(t\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathbb{R}}^{N},\phantom{\rule{0.1em}{0ex}}\mathit{\text{a.e.}}\phantom{\rule{0.1em}{0ex}}t\in \left[0,T\right],$
(2.2)

then

$c\left(u\right)={\int }_{0}^{kT}F\left(t,u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt$

is weakly continuous and uniformly differentiable on bounded subsets of ${H}_{kT}^{1}$.

Remark 2.1 In [[17], Lemma 2.2], $F\in {C}^{1}\left(\mathbb{R},{\mathbb{R}}^{N}\right)$. In fact, in its proof, it is not essential that F is continuously differentiable in t.

We use Lemma 2.3 below due to Benci and Rabinowitz [20] to prove our results.

Lemma 2.3 (see [20] or [[5], Theorem 5.29])

Let E be a real Hilbert space with $E={E}_{1}\oplus {E}_{2}$ and ${E}_{2}={E}_{1}^{\perp }$. Suppose that $\phi \in {C}^{1}\left(E,\mathbb{R}\right)$ satisfies (PS)-condition, and

(I1) $\phi \left(u\right)=1/2\left(\mathrm{\Phi }u,u\right)+b\left(u\right)$, where $\mathrm{\Phi }u={\mathrm{\Phi }}_{1}{P}_{1}u+{\mathrm{\Phi }}_{2}{P}_{2}u$ and ${\mathrm{\Phi }}_{i}:{E}_{i}\to {E}_{i}$ bounded and self-adjoint, $i=1,2$;

(I2) ${b}^{\prime }$ is compact, and

(I3) there exists a subspace $\stackrel{˜}{E}\subset E$ and sets $S\subset E$, $Q\subset \stackrel{˜}{E}$ and constants $\alpha >\beta$ such that

1. (i)

$S\subset {E}_{1}$ and $\phi {|}_{S}\ge \alpha$,

2. (ii)

Q is bounded and $\phi {|}_{\partial Q}\le \beta$,

3. (iii)

Then φ possesses a critical value $c\ge \alpha$ which can be characterized as

$c=\underset{h\in \mathrm{\Gamma }}{inf}\underset{u\in Q}{sup}\phi \left(h\left(1,u\right)\right),$

where

$\mathrm{\Gamma }\equiv \left\{h\in C\left(\left[0,1\right]×E,E\right)|h\phantom{\rule{0.1em}{0ex}}\mathit{\text{satisfies the following}}\phantom{\rule{0.1em}{0ex}}\left({\mathrm{\Gamma }}_{1}\right)\text{-}\left({\mathrm{\Gamma }}_{3}\right)\right\},$

$\left({\mathrm{\Gamma }}_{1}\right)$ $h\left(0,u\right)=u$,

$\left({\mathrm{\Gamma }}_{2}\right)$ $h\left(t,u\right)=u$ for $u\in \partial Q$, and

$\left({\mathrm{\Gamma }}_{3}\right)$ $h\left(t,u\right)={e}^{\theta \left(t,u\right)\mathrm{\Phi }}u+K\left(t,u\right)$, where $\theta \in C\left(\left[0,1\right]×E,\mathbb{R}\right)$ and K is compact.

Remark 2.2 As shown in [21], a deformation lemma can be proved with replacing the usual (PS)-condition with condition (C), and it turns out that Lemma 2.3 holds true under condition (C). We say φ satisfies condition (C), i.e., for every sequence $\left\{{u}_{n}\right\}\subset {H}_{T}^{1}$, $\left\{{u}_{n}\right\}$ has a convergent subsequence if $\phi \left({u}_{n}\right)$ is bounded and $\left(1+\parallel {u}_{n}\parallel \right)\parallel {\phi }^{\prime }\left({u}_{n}\right)\parallel \to 0$ as $n\to \mathrm{\infty }$.

## 3 Proofs of theorems

Proof of Theorem 1.1 It follows from Assumption (A)′ that the functional ${\phi }_{k}$ on ${H}_{kT}^{1}$ given by

${\phi }_{k}\left(u\right)=\frac{1}{2}{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{1}{2}{\int }_{0}^{kT}\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-{\int }_{0}^{kT}F\left(t,u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt$

is continuously differentiable. Moreover, one has

$〈{\phi }_{k}^{\prime }\left(u\right),v〉={\int }_{0}^{kT}\left[\left(\stackrel{˙}{u}\left(t\right),\stackrel{˙}{v}\left(t\right)\right)-\left(\mathit{Au}\left(t\right),v\left(t\right)\right)-\left(\mathrm{\nabla }F\left(t,u\left(t\right)\right),v\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt$

for $u,v\in {H}_{kT}^{1}$ and the solutions of system (1.1) correspond to the critical points of ${\phi }_{k}$ (see [1]).

Obviously, there exists an orthogonal matrix Q such that

${Q}^{\tau }AQ=B=\left(\begin{array}{c}{\lambda }_{1}\\ \ddots \\ {\lambda }_{r}\hfill \\ 0\hfill \\ \ddots \hfill \\ 0\hfill \\ -{\lambda }_{r+s+1}\hfill \\ \ddots \hfill \\ -{\lambda }_{N}\hfill \end{array}\right)$
(3.1)

Let $u=Qw$. Then by (1.1),

Furthermore

that is,

(3.2)

Let $G\left(t,w\right)=F\left(t,Qw\right)$ and then $\mathrm{\nabla }G\left(t,w\right)={Q}^{-1}\mathrm{\nabla }F\left(t,Qw\left(t\right)\right)$. Let

${\psi }_{k}\left(w\right)=\frac{1}{2}{\int }_{0}^{kT}{|\stackrel{˙}{w}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{1}{2}{\int }_{0}^{kT}\left(Bw\left(t\right),w\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-{\int }_{0}^{kT}G\left(t,w\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt.$

Then the critical points of ${\psi }_{k}$ correspond to solutions of system (3.2). It is easy to verify that ${\phi }_{k}\left(u\right)={\psi }_{k}\left(w\right)$ and G satisfies all the conditions of Theorem 1.1 and Theorem 1.2 if F satisfies them. Hence, w is the critical point of ${\psi }_{k}$ if and only if $u=Qw$ is the critical point of ${\phi }_{k}$. Therefore, we only need to consider the special case that $A=B$ is the diagonal matrix defined by (3.1). We divide the proof into six steps.

Step 1: Decompose the space ${H}_{kT}^{1}$. Let

${I}_{N}=\left(\begin{array}{c}1\\ 1\\ \ddots \hfill \\ 1\hfill \end{array}\right)=\left({e}_{1},{e}_{2},\dots ,{e}_{N}\right).$

Note that

${H}_{kT}^{1}\subset \left\{\sum _{i=0}^{\mathrm{\infty }}\left({c}_{i}cosi{k}^{-1}\omega t+{d}_{i}sini{k}^{-1}\omega t\right)|{c}_{i},{d}_{i}\in {\mathbb{R}}^{N},i=0,1,2\cdots \right\}.$

Define

$\begin{array}{c}{H}_{kT}^{-}=\left\{u\in {H}_{kT}^{1}|u=u\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=0}^{k{l}_{i}}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\},\hfill \\ {H}_{kT}^{0}=\left\{u\in {H}_{kT}^{1}|u=u\left(t\right)=\sum _{i=r+1}^{r+s}{e}_{i}\sum _{j=0}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\},\hfill \\ {H}_{kT}^{+}=\left\{u\in {H}_{kT}^{1}|u=u\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=k{l}_{i}+1}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right)\hfill \\ \phantom{{H}_{kT}^{+}=}+\sum _{i=r+s+1}^{N}{e}_{i}\sum _{j=0}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\}.\hfill \end{array}$

Then ${H}_{kT}^{-}$, ${H}_{kT}^{0}$ and ${H}_{kT}^{+}$ are closed subsets of ${H}_{kT}^{1}$ and

1. (1)
${H}_{kT}^{1}={H}_{kT}^{-}\oplus {H}_{kT}^{0}\oplus {H}_{kT}^{+};$
2. (2)

where

${P}_{k}\left(u,v\right)={\int }_{0}^{kT}\left[\left(\stackrel{˙}{u}\left(t\right),\stackrel{˙}{v}\left(t\right)\right)-\left(\mathit{Au}\left(t\right),v\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u,v\in {H}_{kT}^{1}.$

Let

$\begin{array}{c}{H}_{kT}^{01}=\left\{u\in {H}_{kT}^{0}|u=\sum _{i=r+1}^{r+s}{c}_{i0}{e}_{i},{c}_{i0}\in \mathbb{R}\right\},\hfill \\ {H}_{kT}^{02}=\left\{u\in {H}_{kT}^{0}|u=u\left(t\right)=\sum _{i=r+1}^{r+s}{e}_{i}\sum _{j=1}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\},\hfill \\ {H}_{kT}^{+1}=\left\{u\in {H}_{kT}^{+}|u=u\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=k{l}_{i}+1}^{k{l}_{i}+k-1}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\},\hfill \\ {H}_{kT}^{+2}=\left\{u\in {H}_{kT}^{+}|u=u\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right)\hfill \\ \phantom{{H}_{kT}^{+2}=}+\sum _{i=r+s+1}^{N}{e}_{i}\sum _{j=0}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),{c}_{ij},{d}_{ij}\in \mathbb{R}\right\}.\hfill \end{array}$

Then

${H}_{kT}^{0}={H}_{kT}^{01}\oplus {H}_{kT}^{02},\phantom{\rule{2em}{0ex}}{H}_{kT}^{+}={H}_{kT}^{+1}\oplus {H}_{kT}^{+2},\phantom{\rule{2em}{0ex}}{H}_{kT}^{1}={H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{02}\oplus {H}_{kT}^{+1}\oplus {H}_{kT}^{+2}$

and

${P}_{k}\left(u,v\right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{+1},\mathrm{\forall }v\in {H}_{kT}^{+2}.$

Remark 3.1 When $k=1$, it is easy to see ${H}_{T}^{+1}=\left\{0\right\}$.

Step 2: Let

${q}_{k}\left(u\right)=\frac{1}{2}{\int }_{0}^{kT}\left[{|\stackrel{˙}{u}\left(t\right)|}^{2}-\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt.$

Next we consider the relationship between ${q}_{k}\left(u\right)$ and $\parallel u\parallel$ on those subspaces defined above. We only consider the case that (H3)(2) holds. For others, the conclusions are easy to be seen from the argument of this case.

1. (a)

For $\mathrm{\forall }u\in {H}_{kT}^{-}$, since

$u=u\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=0}^{k{l}_{i}}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),$

then

$\begin{array}{rcl}{q}_{k}\left(u\right)& =& \frac{1}{2}{\int }_{0}^{kT}\left[{|\stackrel{˙}{u}\left(t\right)|}^{2}-\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{2}{\int }_{0}^{kT}\left[\left(\sum _{i=1}^{r}{e}_{i}\sum _{j=0}^{k{l}_{i}}j{k}^{-1}\omega \left({d}_{ij}cosj{k}^{-1}\omega t-{c}_{ij}sinj{k}^{-1}\omega t\right),\\ \sum _{i=1}^{r}{e}_{i}\sum _{j=0}^{k{l}_{i}}j{k}^{-1}\omega \left({d}_{ij}cosj{k}^{-1}\omega t-{c}_{ij}sinj{k}^{-1}\omega t\right)\right)\\ -\left(\sum _{i=1}^{r}A{e}_{i}\sum _{j=0}^{k{l}_{i}}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),\\ \sum _{i=1}^{r}{e}_{i}\sum _{j=0}^{k{l}_{i}}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{2}\sum _{i=1}^{r}{\int }_{0}^{kT}\left\{{\left[\sum _{j=0}^{k{l}_{i}}j{k}^{-1}\omega \left({d}_{ij}cosj{k}^{-1}\omega t-{c}_{ij}sinj{k}^{-1}\omega t\right)\right]}^{2}\\ -{\lambda }_{i}{\left[\sum _{j=0}^{k{l}_{i}}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right)\right]}^{2}\right\}\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{kT}{4}\sum _{i=1}^{r}\sum _{j=0}^{k{l}_{i}}\left[{\left(j{k}^{-1}\omega \right)}^{2}-{\lambda }_{i}\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\end{array}$

and

${\parallel u\parallel }^{2}={\int }_{0}^{kT}\left({|\stackrel{˙}{u}\left(t\right)|}^{2}+{|u\left(t\right)|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dt=\frac{kT}{2}\sum _{i=1}^{r}\sum _{j=0}^{k{l}_{i}}\left[{\left(j{k}^{-1}\omega \right)}^{2}+1\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right).$

Let

$\delta =\underset{i\in \left\{1,\dots ,r\right\}}{min}\left\{\frac{{\lambda }_{i}-{\left({l}_{i}\omega \right)}^{2}}{{\left({l}_{i}\omega \right)}^{2}+1}\right\}>0.$

Then

${q}_{k}\left(u\right)\le -\frac{\delta }{2}{\parallel u\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{-}.$
(3.3)

Remark 3.2 Obviously, if one of (H3)(5) and (H3)(6) holds, then ${H}_{kT}^{-}=\left\{0\right\}$. Hence,

${q}_{k}\left(u\right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{-}.$
1. (b)

For $\mathrm{\forall }u\in {H}_{kT}^{+2}\oplus {H}_{kT}^{02}$, let

$u=u\left(t\right)={u}_{1}\left(t\right)+{u}_{2}\left(t\right)+{u}_{3}\left(t\right),$

where

$\begin{array}{c}{u}_{1}\left(t\right)=\sum _{i=1}^{r}{e}_{i}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),\hfill \\ {u}_{2}\left(t\right)=\sum _{i=r+s+1}^{N}{e}_{i}\sum _{j=0}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),\hfill \\ {u}_{3}\left(t\right)=\sum _{i=r+1}^{r+s}{e}_{i}\sum _{j=1}^{\mathrm{\infty }}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right).\hfill \end{array}$

Then

$\begin{array}{rcl}{q}_{k}\left(u\right)& =& \frac{1}{2}{\int }_{0}^{kT}\left[{|\stackrel{˙}{u}\left(t\right)|}^{2}-\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{2}{\int }_{0}^{kT}\left[\left({\stackrel{˙}{u}}_{1}\left(t\right)+{\stackrel{˙}{u}}_{2}\left(t\right)+{\stackrel{˙}{u}}_{3}\left(t\right),{\stackrel{˙}{u}}_{1}\left(t\right)+{\stackrel{˙}{u}}_{2}\left(t\right)+{\stackrel{˙}{u}}_{3}\left(t\right)\right)\\ -\left(A{u}_{1}\left(t\right)+A{u}_{2}\left(t\right)+A{u}_{3}\left(t\right),{u}_{1}\left(t\right)+{u}_{2}\left(t\right)+{u}_{3}\left(t\right)\right)\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{2}{\int }_{0}^{kT}\left[\left({\stackrel{˙}{u}}_{1}\left(t\right),{\stackrel{˙}{u}}_{1}\left(t\right)\right)+\left({\stackrel{˙}{u}}_{2}\left(t\right),{\stackrel{˙}{u}}_{2}\left(t\right)\right)+\left({\stackrel{˙}{u}}_{3}\left(t\right),{\stackrel{˙}{u}}_{3}\left(t\right)\right)\\ -\left(A{u}_{1}\left(t\right),{u}_{1}\left(t\right)\right)-\left(A{u}_{2}\left(t\right),{u}_{2}\left(t\right)\right)-\left(A{u}_{3}\left(t\right),{u}_{3}\left(t\right)\right)\right]\\ =& \frac{kT}{4}\left[\sum _{i=1}^{r}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}{\left(j{k}^{-1}\omega \right)}^{2}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)+\sum _{i=r+s+1}^{N}\sum _{j=0}^{\mathrm{\infty }}{\left(j{k}^{-1}\omega \right)}^{2}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\\ +\sum _{i=r+1}^{r+s}\sum _{j=1}^{\mathrm{\infty }}{\left(j{k}^{-1}\omega \right)}^{2}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\\ -\sum _{i=1}^{r}{\lambda }_{i}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)+\sum _{i=r+s+1}^{N}{\lambda }_{i}\sum _{j=0}^{\mathrm{\infty }}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\right]\\ =& \frac{kT}{4}\left\{\sum _{i=1}^{r}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}\left[{\left(j{k}^{-1}\omega \right)}^{2}-{\lambda }_{i}\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)+\sum _{i=r+s+1}^{N}\sum _{j=0}^{\mathrm{\infty }}\left[{\left(j{k}^{-1}\omega \right)}^{2}+{\lambda }_{i}\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\\ +\sum _{i=r+1}^{r+s}\sum _{j=1}^{\mathrm{\infty }}{\left(j{k}^{-1}\omega \right)}^{2}\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\right\}\end{array}$

and

$\begin{array}{rcl}{\parallel u\parallel }^{2}& =& {\int }_{0}^{kT}\left({|\stackrel{˙}{u}\left(t\right)|}^{2}+{|u\left(t\right)|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{kT}{2}\left\{\sum _{i=1}^{r}\sum _{j=k{l}_{i}+k}^{\mathrm{\infty }}\left[{\left(j{k}^{-1}\omega \right)}^{2}+1\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\\ +\sum _{i=r+s+1}^{N}\sum _{j=0}^{\mathrm{\infty }}\left[{\left(j{k}^{-1}\omega \right)}^{2}+1\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\\ +\sum _{i=r+1}^{r+s}\sum _{j=1}^{\mathrm{\infty }}\left[{\left(j{k}^{-1}\omega \right)}^{2}+1\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\right\}.\end{array}$

Since for fixed $i\in \left\{1,\dots ,r\right\}$,

$f\left(j\right)=\frac{{\left(j{k}^{-1}\omega \right)}^{2}-{\lambda }_{i}}{{\left(j{k}^{-1}\omega \right)}^{2}+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(j\right)=\frac{{\left(j{k}^{-1}\omega \right)}^{2}}{{\left(j{k}^{-1}\omega \right)}^{2}+1}$

are strictly increasing on $j\in \mathbb{N}$,

$f\left(j\right)\ge f\left(k{l}_{i}+k\right)=\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }j\ge k{l}_{i}+k$

and

$g\left(j\right)\ge g\left(1\right)=\frac{{\left({k}^{-1}\omega \right)}^{2}}{{\left({k}^{-1}\omega \right)}^{2}+1}=\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}}>0.$

Moreover, it is easy to verify that

$\frac{{\left(j{k}^{-1}\omega \right)}^{2}+{\lambda }_{i}}{{\left(j{k}^{-1}\omega \right)}^{2}+1}\ge \frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }j\in \mathbb{N}\cup \left\{0\right\},i=r+s+1,\dots ,N.$

Let

${\sigma }_{k}=min\left\{\underset{i\in \left\{1,\dots ,r\right\}}{min}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\},\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}\right\}.$

Then

${q}_{k}\left(u\right)\ge \frac{{\sigma }_{k}}{2}{\parallel u\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{+2}\oplus {H}_{kT}^{02}.$
(3.4)

Remark 3.3 From the above discussion, it is easy to see the following conclusions:

1. (i)

if (H3)(1) holds, then (3.4) holds with

${\sigma }_{k}=min\left\{\underset{i\in \left\{1,\dots ,r\right\}}{min}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\},\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}}\right\};$
2. (ii)

if (H3)(2) holds, then (3.4) holds with

${\sigma }_{k}=min\left\{\underset{i\in \left\{1,\dots ,r\right\}}{min}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\},\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}\right\};$
3. (iii)

if (H3)(3) holds, then (3.4) holds with

${\sigma }_{k}\equiv \sigma =min\left\{\underset{i\in \left\{1,\dots ,r\right\}}{min}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\},\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}\right\};$
4. (iv)

if (H3)(4) holds, then (3.4) holds with

${\sigma }_{k}\equiv \sigma =\underset{i\in \left\{1,\dots ,N\right\}}{min}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\};$
5. (v)

if (H3)(5) holds, then (3.4) holds with

${\sigma }_{k}=min\left\{\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}\right\};$
6. (vi)

if (H3)(6) holds, then (3.4) holds with

${\sigma }_{k}\equiv \sigma =\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}.$
1. (c)

For $\mathrm{\forall }u\in {H}_{kT}^{+1}$, since

$\begin{array}{c}u=\sum _{i=1}^{r}{e}_{i}\sum _{j=k{l}_{i}+1}^{k{l}_{i}+k-1}\left({c}_{ij}cosj{k}^{-1}\omega t+{d}_{ij}sinj{k}^{-1}\omega t\right),\hfill \\ {q}_{k}\left(u\right)=\frac{kT}{4}\sum _{i=1}^{r}\sum _{j=k{l}_{i}+1}^{k{l}_{i}+k-1}\left[{\left(j{k}^{-1}\omega \right)}^{2}-{\lambda }_{i}\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right)\hfill \end{array}$

and

${\parallel u\parallel }^{2}=\frac{kT}{2}\sum _{i=1}^{r}\sum _{j=k{l}_{i}+1}^{k{l}_{i}+k-1}\left[{\left(j{k}^{-1}\omega \right)}^{2}+1\right]\left({c}_{ij}^{2}+{d}_{ij}^{2}\right).$

Obviously, when $k=1$, $u=0$. So ${q}_{1}\left(u\right)=0$. When $k>1$, it follows from

${\left({l}_{i}+1-\frac{1}{k}\right)}^{2}{\omega }^{2}\le {\lambda }_{i}<{\left({l}_{i}+1\right)}^{2}{\omega }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }i\in \left\{1,\dots ,r\right\}$

that

${q}_{k}\left(u\right)\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{+1}.$
(3.5)
1. (d)

Obviously, for $\mathrm{\forall }u\in {H}_{kT}^{01}$, we have

${q}_{k}\left(u\right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{01}.$
(3.6)

Step 3: Assume that (H3)(2) holds. We prove that there exist ${\rho }_{k}>0$ and ${b}_{k}>0$ such that

${\phi }_{k}\left(u\right)\ge {b}_{k}>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in \left({H}_{kT}^{+2}\oplus {H}_{kT}^{02}\right)\cap \partial {B}_{{\rho }_{k}}.$

Let

${C}_{k}=\sqrt{\frac{12+{k}^{2}{T}^{2}}{12kT}}.$

Choosing ${\rho }_{k}=min\left\{1,{l}_{k}/{C}_{k}\right\}>0$ and ${b}_{k}=\left(\frac{{\sigma }_{k}}{2}-{\alpha }_{k}\right){\rho }_{k}^{2}>0$, by Lemma 2.1, (H4) and (3.4), we have, for all $u\in \left({H}_{kT}^{+2}\oplus {H}_{kT}^{02}\right)\cap \partial {B}_{{\rho }_{k}}$,

$\begin{array}{rcl}{\phi }_{k}\left(u\right)& \ge & \frac{1}{2}{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{1}{2}{\int }_{0}^{kT}\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-{\int }_{0}^{kT}F\left(t,u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\\ \ge & \frac{{\sigma }_{k}}{2}{\parallel u\parallel }^{2}-{\alpha }_{k}{\int }_{0}^{kT}{|u\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ \ge & \left(\frac{{\sigma }_{k}}{2}-{\alpha }_{k}\right){\parallel u\parallel }^{2}\\ =& \left(\frac{min\left\{{min}_{i\in \left\{1,\dots ,r\right\}}\left\{\frac{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}-{\lambda }_{i}}{{\left({l}_{i}+1\right)}^{2}{\omega }^{2}+1}\right\},\frac{{\omega }^{2}}{{\omega }^{2}+{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{1+{\lambda }_{{i}_{+}}}\right\}}{2}-{\alpha }_{k}\right){\rho }_{k}^{2}.\end{array}$

For cases (H3)(1) and (H3)(3)-(H3)(6), correspondingly, by (H4) and Remark 3.3, similar to the above argument, we can also obtain that

${\phi }_{k}\left(u\right)\ge \left(\frac{{\sigma }_{k}}{2}-{\alpha }_{k}\right){\rho }_{k}^{2}>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in \left({H}_{kT}^{+2}\oplus {H}_{kT}^{02}\right)\cap \partial {B}_{{\rho }_{k}}.$

Step 4: Let

${Q}_{k}=\left\{s{h}_{k}|s\in \left[0,{s}_{1}\right]\right\}\oplus \left({B}_{{s}_{2}}\cap \left({H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}\right)\right),$

where ${h}_{k}\in {H}_{kT}^{+2}\oplus {H}_{kT}^{02}$, ${s}_{1}$ and ${s}_{2}$ will be determined later. In this step, we prove ${\phi }_{k}{|}_{\partial {Q}_{k}}\le 0$. We only consider the case that F satisfies (H3)(2). For other cases, the results can be seen easily from the argument of case (H3)(2).

Assume that F satisfies (H3)(2). Let

${d}_{k}=min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}.$

Case (i): if

${d}_{k}:=d=\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},$

then we choose

${h}_{k}\left(t\right)=sin\left({l}_{{i}_{0}}+1\right)\omega t\cdot {e}_{{i}_{0}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \mathbb{R}.$

Obviously, ${h}_{k}\in {H}_{kT}^{+2}$ and ${\stackrel{˙}{h}}_{k}\left(t\right)=\left({l}_{{i}_{0}}+1\right)\omega cos\left({l}_{{i}_{0}}+1\right)\omega t\cdot {e}_{{i}_{0}}$, $\mathrm{\forall }t\in \mathbb{R}$. Then

${\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}=\frac{kT}{2},\phantom{\rule{2em}{0ex}}{\parallel {\stackrel{˙}{h}}_{k}\parallel }_{{L}^{2}}^{2}=\frac{kT{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}}{2}.$

By (H3)(2), (1.5) and the periodicity of F, we have

(3.7)

where ${\epsilon }_{0k}={\beta }_{k}-d>0$ and ${\stackrel{ˆ}{L}}_{k}>max\left\{1,{L}_{k}\right\}$. Since ${H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$ is the finite dimensional space, there exists a constant ${K}_{1k}>0$ such that

${K}_{1k}{\parallel u\parallel }^{2}\le {\parallel u\parallel }_{{L}^{2}}^{2}\le {\parallel u\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in {H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}.$
(3.8)

By (3.3), (3.5), (3.6), (3.7) and (3.8), we know that for all $s>0$ and $u={u}^{-}+{u}^{01}+{u}^{+1}\in {H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$,

$\begin{array}{rcl}{\phi }_{k}\left(s{h}_{k}+u\right)& \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}{\int }_{0}^{kT}{|{\stackrel{˙}{h}}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{{\lambda }_{{i}_{0}}{s}^{2}}{2}{\int }_{0}^{kT}{|{h}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ -{\int }_{0}^{kT}F\left(t,s{h}_{k}\left(t\right)+u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\\ \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}\cdot \frac{kT{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}}{2}-\frac{{\lambda }_{{i}_{0}}{s}^{2}}{2}\cdot \frac{kT}{2}\\ -\left(d+{\epsilon }_{0k}\right){\int }_{0}^{kT}{|s{h}_{k}\left(t\right)+u\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ =& -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}\cdot \frac{kT{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}}{2}-\frac{{\lambda }_{{i}_{0}}{s}^{2}}{2}\cdot \frac{kT}{2}\\ -\left(d+{\epsilon }_{0k}\right)\left({s}^{2}{\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}+{\parallel u\parallel }_{{L}^{2}}^{2}\right)+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\left(\frac{kT{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}}{4}-\frac{{\lambda }_{{i}_{0}}kT}{4}-\frac{dkT}{2}-\frac{kT{\epsilon }_{0k}}{2}\right){s}^{2}\\ -\left(d+{\epsilon }_{0k}\right){\parallel u\parallel }_{{L}^{2}}^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -\frac{kT{\epsilon }_{0k}}{2}{s}^{2}-{\epsilon }_{0k}{\parallel u\parallel }_{{L}^{2}}^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -\frac{kT{\epsilon }_{0k}}{2}{s}^{2}-{\epsilon }_{0k}{K}_{1k}{\parallel u\parallel }^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT.\end{array}$
(3.9)

Hence,

where

${s}_{1}=\sqrt{\frac{2{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}}{{\epsilon }_{0k}}},\phantom{\rule{2em}{0ex}}{s}_{2}=\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}{K}_{1k}}}.$

Case (ii): if ${d}_{k}={\omega }^{2}/\left(2{k}^{2}\right)$, then we choose

${h}_{k}\left(t\right)=sin{k}^{-1}\omega t\cdot {e}_{r+1}\in {H}_{kT}^{02},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \mathbb{R}.$

Then

${\stackrel{˙}{h}}_{k}\left(t\right)=\frac{\omega }{k}cos{k}^{-1}\omega t\cdot {e}_{r+1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \mathbb{R},$

and

$\left(A{h}_{k},{h}_{k}\right)=0,\phantom{\rule{2em}{0ex}}{\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}=\frac{kT}{2},\phantom{\rule{2em}{0ex}}{\parallel {\stackrel{˙}{h}}_{k}\parallel }_{{L}^{2}}^{2}=\frac{T{\omega }^{2}}{2k}.$
(3.10)

By (H3)(2), (1.5) and the periodicity of F, we have

(3.11)

where ${\stackrel{ˆ}{L}}_{k}>max\left\{1,{L}_{k}\right\}$ and ${\epsilon }_{0k}^{\prime }={\beta }_{k}-\frac{{\omega }^{2}}{2{k}^{2}}$. By (3.3), (3.5), (3.6), (3.8) and (3.11), we know that for all $s>0$ and $u={u}^{-}+{u}^{01}+{u}^{+1}\in {H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$,

$\begin{array}{rcl}{\phi }_{k}\left(s{h}_{k}+u\right)& \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}{\int }_{0}^{kT}{|{\stackrel{˙}{h}}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-{\int }_{0}^{kT}F\left(t,s{h}_{k}\left(t\right)+u\right)\phantom{\rule{0.2em}{0ex}}dt\\ \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}\cdot \frac{T{\omega }^{2}}{2k}-\left(\frac{{\omega }^{2}}{2{k}^{2}}+{\epsilon }_{0k}^{\prime }\right){\int }_{0}^{kT}{|s{h}_{k}\left(t\right)+u\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ +{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ =& -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}\cdot \frac{T{\omega }^{2}}{2k}-\left(\frac{{\omega }^{2}}{2{k}^{2}}+{\epsilon }_{0k}^{\prime }\right)\left({s}^{2}{\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}+{\parallel u\parallel }_{{L}^{2}}^{2}\right)\\ +{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ =& -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\left(\frac{T{\omega }^{2}}{4k}-\frac{T{\omega }^{2}}{4k}-\frac{kT{\epsilon }_{0k}^{\prime }}{2}\right){s}^{2}-\left(\frac{{\omega }^{2}}{2{k}^{2}}+{\epsilon }_{0k}^{\prime }\right){\parallel u\parallel }_{{L}^{2}}^{2}\\ +{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -\frac{kT{\epsilon }_{0k}^{\prime }}{2}{s}^{2}-{\epsilon }_{0k}^{\prime }{\parallel u\parallel }_{{L}^{2}}^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -\frac{kT{\epsilon }_{0k}^{\prime }}{2}{s}^{2}-{\epsilon }_{0k}^{\prime }{K}_{1k}{\parallel u\parallel }^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT.\end{array}$

Hence,

where

${s}_{1}=\sqrt{\frac{2{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}}{{\epsilon }_{0k}^{\prime }}},\phantom{\rule{2em}{0ex}}{s}_{2}=\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}^{\prime }{K}_{1k}}}.$

Case (iii): if ${d}_{k}={\lambda }_{{i}_{-}}/2$, then we choose

${h}_{k}=\frac{1}{\sqrt{kT}}\cdot {e}_{{i}_{-}}\in {H}_{kT}^{+2}.$

Then

${\stackrel{˙}{h}}_{k}=0,\phantom{\rule{2em}{0ex}}\left(A{h}_{k},{h}_{k}\right)=-{\lambda }_{{i}_{-}}\left({h}_{k},{h}_{k}\right),\phantom{\rule{2em}{0ex}}{\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}=1.$

By (H3)(2), (1.5) and the periodicity of F, we have

(3.12)

where ${\stackrel{ˆ}{L}}_{k}>max\left\{\sqrt{1+\frac{1}{T}},{L}_{k}\right\}$ and ${\epsilon }_{0k}^{″}={\beta }_{k}-{\lambda }_{{i}_{-}}/2$. By (3.3), (3.5), (3.6), (3.8) and (3.12), for all $s>0$ and $u={u}^{-}+{u}^{01}+{u}^{+1}\in {H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$, we have

$\begin{array}{rcl}{\phi }_{k}\left(s{h}_{k}+u\right)& \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{s}^{2}}{2}{\int }_{0}^{kT}{|{\stackrel{˙}{h}}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt+\frac{{\lambda }_{{i}_{-}}{s}^{2}}{2}{\int }_{0}^{kT}{|{h}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ -{\int }_{0}^{kT}F\left(t,s{h}_{k}\left(t\right)+u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\\ \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{\lambda }_{{i}_{-}}{s}^{2}}{2}-\left(\frac{{\lambda }_{{i}_{-}}}{2}+{\epsilon }_{0k}^{″}\right){\int }_{0}^{kT}{|s{h}_{k}\left(t\right)+u\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ =& -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\frac{{\lambda }_{{i}_{-}}{s}^{2}}{2}-\left(\frac{{\lambda }_{{i}_{-}}}{2}+{\epsilon }_{0k}^{″}\right)\left({s}^{2}{\parallel {h}_{k}\parallel }_{{L}^{2}}^{2}+{\parallel u\parallel }_{{L}^{2}}^{2}\right)+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ =& -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}+\left(\frac{{\lambda }_{{i}_{-}}}{2}-\frac{{\lambda }_{{i}_{-}}}{2}-{\epsilon }_{0k}^{″}\right){s}^{2}-\left(\frac{{\lambda }_{{i}_{-}}}{2}+{\epsilon }_{0k}^{″}\right){\parallel u\parallel }_{{L}^{2}}^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT\\ \le & -{\epsilon }_{0k}^{″}{s}^{2}-{\epsilon }_{0k}^{″}{K}_{1k}{\parallel u\parallel }^{2}+{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT.\end{array}$

Hence,

where

${s}_{1}=\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}^{″}}},\phantom{\rule{2em}{0ex}}{s}_{2}=\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}^{″}{K}_{1k}}}.$

Combining cases (i), (ii) and (iii), if we let

$\begin{array}{c}{s}_{1}=max\left\{\sqrt{\frac{2{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}}{{\epsilon }_{0k}}},\sqrt{\frac{2{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}}{{\epsilon }_{0k}^{\prime }}},\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}^{″}}}\right\},\hfill \\ {s}_{2}=max\left\{\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}{K}_{1k}}},\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}kT}{{\epsilon }_{0k}^{\prime }{K}_{1k}}},\sqrt{\frac{{\beta }_{k}{\stackrel{ˆ}{L}}_{k}^{2}}{{\epsilon }_{0k}^{″}{K}_{1k}}}\right\},\hfill \end{array}$

then

(3.13)

By (1.5), (3.3), (3.5) and (3.6), for all $u\in {H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$, we have

$\begin{array}{rcl}{\phi }_{k}\left(u\right)& =& \frac{1}{2}{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{1}{2}{\int }_{0}^{kT}\left(\mathit{Au}\left(t\right),u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-{\int }_{0}^{kT}F\left(t,u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\\ \le & -\frac{\delta }{2}{\parallel {u}^{-}\parallel }^{2}\\ \le & 0.\end{array}$
(3.14)

Thus, it follows from (3.13) and (3.14) that $\phi {|}_{\partial {Q}_{k}}\le 0<{b}_{k}$.

Step 5: We prove that ${\phi }_{k}$ satisfies (C)-condition in ${H}_{kT}^{1}$. The proof is similar to that in Theorem 1.1 in [17]. We omit it.

Step 6: We claim that ${\phi }_{k}$ has a nontrivial critical point ${u}_{k}\in {H}_{kT}^{1}$ such that ${\phi }_{k}\left({u}_{k}\right)\ge {b}_{k}>0$. Especially, we claim that, for cases (H3)(1) and (H3)(4), since A is a positive semidefinite matrix, (1.5) implies that ${u}_{k}$ is nonconstant.

In fact, it is easy to see that

$\begin{array}{rcl}{q}_{k}\left(u\right)& =& \frac{1}{2}{\parallel u\parallel }^{2}-\frac{1}{2}{\int }_{0}^{kT}\left(\left(A+I\right)u\left(t\right),u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{2}〈\left(I-K\right)u,u\right)〉,\end{array}$

where $K:{H}_{kT}^{1}\to {H}_{kT}^{1}$ is the linear self-adjoint operator defined, using the Riesz representation theorem, by

${\int }_{0}^{kT}\left(\left(A+I\right)u\left(t\right),v\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt=〈\left(Ku,v\right)〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u,v\in {H}_{T}^{1}.$

The compact imbedding of ${H}_{kT}^{1}$ into $C\left(\left[0,kT\right];{\mathbb{R}}^{N}\right)$ implies that K is compact. In order to use Lemma 2.3, we let $\mathrm{\Phi }=I-K$ and define ${\mathrm{\Phi }}_{i}:{E}_{i}\to {E}_{i}$, $i=1,2$ by

$〈{\mathrm{\Phi }}_{i}u,v〉=〈\left(I-K\right)u,v〉,\phantom{\rule{1em}{0ex}}u,v\in {E}_{i},$

where ${E}_{1}={H}_{kT}^{+2}\oplus {H}_{kT}^{02}$ and ${E}_{2}={H}_{kT}^{-}\oplus {H}_{kT}^{01}\oplus {H}_{kT}^{+1}$. Since K is a self-adjoint compact operator, it is easy to see that ${\mathrm{\Phi }}_{i}$ ($i=1,2$) are bounded and self-adjoint. Let

$b\left(u\right)=-{\int }_{0}^{kT}F\left(t,u\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt.$

Assumption (A)′ and Lemma 2.2 imply that b is weakly continuous and is uniformly differentiable on bounded subsets of $E={H}_{kT}^{1}$. Furthermore, by standard theorems in [22], we conclude that ${b}^{\prime }$ is compact. Let ${S}_{k}=\left({H}_{kT}^{+2}\oplus {H}_{kT}^{02}\right)\cap \partial {B}_{{\rho }_{k}}$. Then ${S}_{k}$ and $\partial {Q}_{k}$ link. Hence, by Step 1-Step 5, Lemma 2.3 and Remark 2.2, there exists a critical point ${u}_{k}\in {H}_{kT}^{1}$ such that ${\phi }_{k}\left({u}_{k}\right)\ge {b}_{k}>0$, which implies that ${u}_{k}$ is nonzero. For cases (H3)(1) and (H3)(4), since A is a positive semidefinite matrix, it follows from (1.5) that ${u}_{k}$ is nonconstant. The proof is complete.  □

Proof of Theorem 1.2 Obviously, when $r=0$, $s>0$ and $s, (H1) holds for any $k\in \mathbb{N}$. Moreover, since (H3)′ implies that (H3)(5) and (H4)′ implies that (H4), system (1.1) has kT-periodic solution for every $k\in \mathbb{N}$.

Let $d=\frac{{\omega }^{2}}{2}$. Like the argument of case (ii) in the proof of Theorem 1.1, choose

${e}_{k}\left(t\right)=sin{k}^{-1}\omega t{e}_{r+1}\in {H}_{kT}^{02},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \mathbb{R}.$

By (H3)′, (1.5) and the T-periodicity of F, we have

(3.15)

where ${\epsilon }_{1}=\beta -\frac{{\omega }^{2}}{2}$. In the proof of Theorem 1.1, if we replace (3.15) with (3.11), then we obtain

where

${s}_{1}=\sqrt{\frac{2\beta {L}^{2}}{{\epsilon }_{1}}}=\sqrt{\frac{2\beta {L}^{2}}{\beta -\frac{{\omega }^{2}}{2}}},\phantom{\rule{2em}{0ex}}{s}_{2}=\sqrt{\frac{\beta {L}^{2}kT}{{\epsilon }_{1}{K}_{1k}}}.$

Note that ${s}_{1}$ is independent of k. Hence, if ${u}_{k}$ is the critical point of ${\phi }_{k}$, then it follows from (3.3), (3.5), (3.6), the definitions of critical value c in Lemma 2.3 and ${Q}_{k}$ that

$\begin{array}{rcl}{\phi }_{k}\left({u}_{k}\right)& \le & \underset{u\in {Q}_{k}}{sup}{\phi }_{k}\left(u\right)\\ \le & \underset{s\in \left[0,{s}_{1}\right]}{sup}\left\{\frac{{s}^{2}}{2}{\int }_{0}^{kT}{|{\stackrel{˙}{e}}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt-\frac{{s}^{2}}{2}{\int }_{0}^{kT}\left(A{e}_{k}\left(t\right),{e}_{k}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt\right\}\\ \le & \frac{{s}_{1}^{2}}{2}{\int }_{0}^{kT}{|{\stackrel{˙}{e}}_{k}\left(t\right)|}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{\beta {L}^{2}T{\omega }^{2}}{2k\left(\beta -\frac{{\omega }^{2}}{2}\right)}\\ \le & \frac{\beta {L}^{2}T{\omega }^{2}}{2\left(\beta -\frac{{\omega }^{2}}{2}\right)}:=M.\end{array}$
(3.16)

Hence, ${\phi }_{k}\left({u}_{k}\right)$ is bounded for any $k\in \mathbb{N}$.

Obviously, we can find ${k}_{1}\in \mathbb{N}/\left\{1\right\}$ such that ${k}_{1}>\frac{M}{{b}_{1}}$, then we claim that ${u}_{k}$ is distinct from ${u}_{1}$ for all $k\ge {k}_{1}$. In fact, if ${u}_{k}={u}_{1}$ for some $k\ge {k}_{1}$, it is easy to check that

${\phi }_{k}\left({u}_{k}\right)=k{\phi }_{1}\left({u}_{1}\right)\ge k{b}_{1}.$

Then by (3.16), we have ${k}_{1}\le k\le \frac{M}{{b}_{1}}$, a contradiction. We also can find ${k}_{2}>max\left\{{k}_{1},\frac{{k}_{1}M}{{b}_{{k}_{1}}}\right\}$ such that ${u}_{{k}_{1}k}\ne {u}_{{k}_{1}}$ for all $k\ge \frac{{k}_{2}}{{k}_{1}}$. Otherwise, if ${u}_{{k}_{1}k}={u}_{{k}_{1}}$ for some $k\ge {k}_{1}$, we have ${\phi }_{{k}_{1}k}\left({u}_{{k}_{1}k}\right)=k{\phi }_{{k}_{1}}\left({u}_{{k}_{1}}\right)\ge k{b}_{{k}_{1}}$. Then by (3.16), we have $\frac{{k}_{2}}{{k}_{1}}\le k\le \frac{M}{{b}_{{k}_{1}}}$, a contradiction. In the same way, we can obtain that system (1.1) has a sequence of distinct periodic solutions with period ${k}_{j}T$ satisfying ${k}_{j}\in \mathbb{N}$ and ${k}_{j}\to \mathrm{\infty }$ as $j\to \mathrm{\infty }$. The proof is complete. □

Proof of Theorem 1.3 Except for verifying (C) condition, the proof is the same as in Theorem B (that is Theorem 1 in [15]). To verify (C) condition, we only need to prove the sequence $\left\{{u}_{n}\right\}$ is bounded if $\phi \left({u}_{n}\right)$ is bounded and $\parallel {\phi }^{\prime }\left({u}_{n}\right)\parallel \left(\parallel 1+\parallel {u}_{n}\parallel \right)\to 0$ as $n\to +\mathrm{\infty }$. Other proofs are the same as in [15]. The proof of boundedness of $\left\{{u}_{n}\right\}$ is essentially the same as in Theorem 1.1 in [17] except that 2 is replaced by p, ${H}_{kT}^{1}$ by

equipped with the norm

$\parallel u\parallel ={\left({\int }_{0}^{kT}{|u\left(t\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{kT}{|\stackrel{˙}{u}\left(t\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/p},$

and

$F\left(t,x\right)\ge {\beta }_{k}{|x|}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathbb{R}}^{N},|x|>L$

by

$F\left(t,x\right)\ge \epsilon {|x|}^{p},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathbb{R}}^{N},|x|>L$

for some $\epsilon >0$. So, we omit the details. □

## 4 Examples

Example 4.1 Let $T=2\pi$ and

$A=\left(\begin{array}{ccccc}7.5& 0& 0& 0& 0\\ 0& 7.4& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& -3& 0\\ 0& 0& 0& 0& -4\end{array}\right).$

Then $\omega =1$, $r=2$, ${\lambda }_{1}=7.5$, ${\lambda }_{2}=7.4$, ${\lambda }_{3}=0$, ${\lambda }_{4}=-3$, ${\lambda }_{5}=-4$, ${\lambda }_{{i}_{+}}=4$ and ${\lambda }_{{i}_{-}}=3$. Obviously, the matrix A satisfies Assumption (A0) and ${l}_{1}={l}_{2}=2$ such that

${l}_{i}^{2}{\omega }^{2}<{\lambda }_{i}<{\left({l}_{i}+1\right)}^{2}{\omega }^{2},\phantom{\rule{1em}{0ex}}i=1,2.$

It is easy to verify that (H1) holds with $k=1,2,3$. Let

Then $F\left(t,x\right)\ge 0$ for all $x\in {\mathbb{R}}^{N}$ and a.e. $t\in \left[0,T\right]$ and

(4.1)
(4.2)

It is easy to verify that

$\left(\mathrm{\nabla }F\left(t,x\right),x\right)-2F\left(t,x\right)=\frac{6ln11}{63{k}^{2}}{|x|}^{2}\cdot {11}^{\frac{{|x|}^{3/2}}{1+{|x|}^{3/2}}}\cdot \frac{{|x|}^{3/2}}{{\left(1+{|x|}^{3/2}\right)}^{2}}.$

Choose $\xi =1$, $\eta =1$ and $\nu =3/2$. Moreover, obviously, there exists $m>0$ such that $\frac{{|x|}^{3/2}}{1+{|x|}^{3/2}}>\frac{2}{3}$. Then

$\frac{\left(\mathrm{\nabla }F\left(t,x\right),x\right)-2F\left(t,x\right)}{\frac{F\left(t,x\right)}{\xi +\eta {|x|}^{\nu }}}=\frac{\frac{3}{2}ln11\cdot {11}^{\frac{{|x|}^{3/2}}{1+{|x|}^{3/2}}}\cdot \frac{{|x|}^{3/2}}{1+{|x|}^{3/2}}}{{11}^{\frac{{|x|}^{3/2}}{1+{|x|}^{3/2}}}-\frac{1}{2}}>ln11>1.$

Hence, (H2) holds.

When $k=1$,

$min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}=\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\sigma }_{1}=0.15.$

By (4.2), we can find ${L}_{1}>0$ such that

Let ${\beta }_{1}=\frac{17}{30}$. Then (H3)(2) holds with $k=1$. Moreover, by (4.1), we can find ${l}_{1}>0$ such that

Let ${\alpha }_{1}=0.0409$. Then (H4) holds. By Theorem 1.1, we obtain that system (1.1) has a T-periodic solution.

When $k=2$,

$min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}=\frac{1}{8}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\sigma }_{2}=0.15.$

By (4.2), we can find ${L}_{2}>0$ such that

Let ${\beta }_{2}=0.1567$. Then (H3)(2) holds with $k=2$. Moreover, by (4.1), we can find ${l}_{2}>0$ such that

Let ${\alpha }_{2}=0.00894$. Then (H4) holds. Note that $\frac{1}{6}<\frac{1}{2}=min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2},\frac{{\lambda }_{{i}_{-}}}{2}\right\}$. So, when $k=2$, by Theorem 1.1, we cannot judge that system (1.1) has a T-periodic solution. However, we can obtain that system (1.1) has a 2T-periodic solution.

When $k=3$,

$min\left\{\frac{{\left({l}_{{i}_{0}}+1\right)}^{2}{\omega }^{2}-{\lambda }_{{i}_{0}}}{2},\frac{{\omega }^{2}}{2{k}^{2}},\frac{{\lambda }_{{i}_{-}}}{2}\right\}=\frac{1}{18}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\sigma }_{3}=0.1.$

By (4.2), we can find ${L}_{3}>0$ such that

Let ${\beta }_{3}=0.0641$. Then (H3)(2) holds with $k=3$. Moreover, by (4.1), we can find ${l}_{3}>0$ such that

$F\left(t,x\right)\le \left(\frac{2}{567}+\frac{1}{1000}\right){}^{}$