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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

u ¨ (t)+Au(t)+F ( t , u ( t ) ) =0a.e. tR,
(1.1)

where A is an N×N symmetric matrix and F:R× R N R is T-periodic in t and satisfies the following assumption:

Assumption (A)′ F(t,x) is measurable in t for every x R N and continuously differentiable in x for a.e. t[0,T], and there exist aC( R + , R + ) and b: R + R + which is T-periodic and b L p (0,T; R + ) with p>1 such that

| F ( t , x ) | a ( | x | ) b(t), | F ( t , x ) | a ( | x | ) b(t)

for all x R N and a.e. t[0,T].

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

u ¨ (t)+F ( t , u ( t ) ) =0a.e. tR.
(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 μ>2 and L>0 such that

0<μF(t,x) ( F ( t , x ) , x ) ,|x|L,t[0,T].

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 θ>2 and μ>θ2 such that

lim sup | x | F ( t , x ) | x | θ <uniformly for a.e. t[0,T],
(1.3)
lim inf | x | ( F ( t , x ) , x ) 2 F ( t , x ) | x | μ >0uniformly for a.e. t[0,T].
(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(t,x) is measurable in t for every x R N and continuously differentiable in x for a.e. t[0,T], and there exist aC( R + , R + ) and b L 1 (0,T; R + ) such that

    | F ( t , x ) | a ( | x | ) b(t), | F ( t , x ) | a ( | x | ) b(t)

for all x R N and a.e. t[0,T]. Assume that (1.3), (1.4) and the following conditions hold:

F(t,x)0,(t,x)[0,T]× R N ,
(1.5)
lim | x | 0 F ( t , x ) | x | 2 =0 uniformly for a.e. t[0,T],
(1.6)
lim | x | F ( t , x ) | x | 2 > 2 π 2 T 2 uniformly for a.e. t[0,T].
(1.7)

Then system (1.2) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j.

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

( | u ( t ) | p 2 u ( t ) ) +F ( t , u ( t ) ) =0a.e. t[0,T],
(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:

lim | x | 0 F ( t , x ) | x | p =0< lim | x | F ( t , x ) | x | p uniformly for a.e. t[0,T].
(1.9)

Then system (1.8) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j.

When A= m 2 ω 2 I N , where ω=2π/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 ω 2 I N , F satisfies (A), (1.3), (1.4), (1.5), (1.6) and the following conditions:

lim | x | F ( t , x ) | x | 2 > 1 + 2 m 2 ω 2 uniformly for a.e. t[0,T].

Then system (1.1) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j.

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 ω 2 as its eigenvalues, where ω=2π/T, l i N, i=1,,r and 0rN. 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 ν[0,2) such that

    ( 2 + 1 ζ + η | x | ν ) F(t,x) ( F ( t , x ) , x ) ,x R N ,|x|>m a.e. t[0,T].

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 rN{0} and rN. Let λ i >0 (i{1,,r}) and λ i <0 (i{r+s+1,,N}) 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 ω 2 for all lN, where ω=2π/T, that is, λ i l 2 ω 2 (i=1,,r) for all lN.

The Assumption (A0) implies that one can find l i Z + :={0,1,2,} such that

l i 2 ω 2 < λ i < ( l i + 1 ) 2 ω 2 ,i=1,,r.
(1.10)

For the sake of convenience, we set

λ i + = max { λ i | i = 1 , , r } , λ i = min { λ i | i = 1 , , r } , λ i + = max { λ i | i = r + s + 1 , , N } , λ i = min { λ i | i = r + s + 1 , , N } .

Then

i + , i {1,,r}, i + , i {r+s+1,,N}.

Corresponding to (1.10), we know that there exist l i + , l i Z + such that

l i + 2 ω 2 < λ i + < ( l i + + 1 ) 2 ω 2 , l i 2 ω 2 < λ i < ( l i + 1 ) 2 ω 2 .

Moreover, set

h i = ( l i + 1 ) 2 ω 2 λ i ,i=1,,r,

and let h i 0 = min i { 1 , , r } { h i }. Then i 0 {1,,r}. Corresponding to (1.10), there exists l i 0 Z + such that

l i 0 2 ω 2 < λ i 0 < ( l i 0 + 1 ) 2 ω 2 .
(1.11)

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

(H1) For some kN, assume that k satisfies

( l i + 1 1 k ) 2 ω 2 λ i < ( l i + 1 ) 2 ω 2 for all i{1,,r}.
(1.12)

(H2) There exist positive constants m, ζ, η and ν[0,2) such that

( 2 + 1 ζ + η | x | ν ) F(t,x) ( F ( t , x ) , x ) ,x R N ,|x|>m, a.e. t[0,T].

(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 β k >min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 } such that

    F(t,x) β k | x | 2 ,x R N ,|x|> L k , a.e. t[0,T],
    (1.13)

where l i 0 and λ i 0 are defined by (1.11);

  1. (2)

    when r>0, s>0 and r+s<N, there exist L k >0 and β k >min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } such that (1.13) holds;

  2. (3)

    when r>0, s=0 and r+s<N, there exist L k >0 and β k >min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , λ i 2 } such that (1.13) holds;

  3. (4)

    when r>0, s=0 and r=N, there exist L k >0 and β k > ( l i 0 + 1 ) 2 ω 2 λ i 0 2 such that (1.13) holds;

  4. (5)

    when r=0, s>0 and s<N, there exist L k >0 and β k >min{ ω 2 2 k 2 , λ i 2 } such that (1.13) holds;

  5. (6)

    when r=0, s=0 and q=N, there exist L k >0 and β k > λ i 2 such that (1.13) holds;

(H4) there exist l k >0 and α k < σ k 2 such that

F(t,x) α k | x | 2 for all |x| l k and a.e. t[0,T],

where

σ k = min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 } if ( H 3 )  (1)  holds ; σ k = min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 , λ i 1 + λ i + } if ( H 3 )  (2)  holds ; σ k σ = min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , λ i 1 + λ i + } if ( H 3 )  (3)  holds ; σ k σ = min i { 1 , , N } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } if ( H 3 )  (4)  holds ; σ k = min { ω 2 ω 2 + k 2 , λ i 1 + λ i + } if ( H 3 )  (5)  holds ; σ k σ = λ i 1 + λ i + if ( H 3 )  (6)  holds ,

where σ implies that σ 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 kN satisfying (1.12) is finite. Let mK be the maximum integer satisfying (1.12), where

K= { k N k  satisfies  ( 1.12 ) } .

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

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

J u ˙ (t)+Au+H(t,u)=0a.e. t[0,T].
(1.14)

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

lim inf | x | ( x , H ( t , x ) ) 2 H ( t , x ) | x | μ a>0uniformly for a.e. t[0,2π],
(1.15)

and the so-called asymptotic noncrossing conditions

λ k 1 < lim inf | x | 2 H ( t , x ) | x | 2 lim sup | x | 2 H ( t , x ) | x | 2 λ k uniformly for a.e. t[0,2π],

where λ k 1 < λ k are consecutive eigenvalues of the operator L=Jd/dtA. Moreover, they also obtained system (1.14) has a nonzero T=2π periodic solution under (1.15) and the called crossing conditions

H ( t , u ) 1 2 λ k 1 | x | 2 for all  ( t , u ) [ 0 , 2 π ] × R 2 N , lim sup | x | 0 2 H ( t , x ) | x | 2 α < λ k < β lim inf | x | 2 H ( t , x ) | x | 2 uniformly for  t [ 0 , 2 π ] .

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], λ k 1 < λ k are consecutive eigenvalues of the operator L=Jd/dtA 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. λ i (i{1,,r}) 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<N, there exist L>0 and β> ω 2 2 such that

F(t,x)β | x | 2 ,x R N ,|x|>L, a.e. t[0,T];
(1.16)

(H4)′

lim | x | 0 F ( t , x ) | x | 2 =0 uniformly for a.e. t[0,T].

Then system (1.1) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j.

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 ν[0,p) such that

( p + 1 ζ + η | x | ν ) F(t,x) ( F ( t , x ) , x ) ,x R N ,|x|>m a.e. t[0,T];

(H6)

lim | x | 0 F ( t , x ) | x | p =0< lim | x | F ( t , x ) | x | p uniformly for a.e. t[0,T].

Then system (1.8) has a sequence of distinct nonconstant periodic solutions with period k j T satisfying k j N and k j as j.

2 Some preliminaries

Let

H k T 1 = { u : R R N | u  be absolutely continuous , u ( t ) = u ( t + k T )  and  u ˙ L 2 ( [ 0 , k T ] ) } .

Then H k T 1 is a Hilbert space with the inner product and the norm defined by

u,v= 0 k T ( u ( t ) , v ( t ) ) dt+ 0 k T ( u ˙ ( t ) , v ˙ ( t ) ) dt

and

u= [ 0 k T | u ( t ) | 2 d t + 0 k T | u ˙ ( t ) | 2 d t ] 1 / 2

for each u,v H k T 1 . Let

u ¯ = 1 k T 0 k T u(t)dtand u ˜ (t)=u(t) u ¯ .

Then one has

u ˜ 2 k T 12 0 k T | u ˙ ( t ) | 2 d t (Sobolev’s inequality) , u ˜ L 2 2 k 2 T 2 4 π 2 0 k T | u ˙ ( t ) | 2 d t (Wirtinger’s inequality)

(see Proposition 1.3 in [1]).

Lemma 2.1 If u H k T 1 , then

u 12 + k 2 T 2 12 k T u,

where u = max t [ 0 , k T ] |u(t)|.

Proof Fix t[0,kT]. For every τ[0,kT], we have

u(t)=u(τ)+ τ t u ˙ (s)ds.
(2.1)

Set

ϕ(s)={ s t + k T 2 , t k T / 2 s t , t + k T 2 s , t s t + k T / 2 .

Integrating (2.1) over [tkT/2,t+kT/2] and using the Hölder inequality, we obtain

k T | u ( t ) | = | t k T / 2 t + k T / 2 u ( τ ) d τ + t k T / 2 t + k T / 2 τ t u ˙ ( s ) d s d τ | t k T / 2 t + k T / 2 | u ( τ ) | d τ + t k T / 2 t τ t | u ˙ ( s ) | d s d τ + t t + k T / 2 t τ | u ˙ ( s ) | d s d τ = t k T / 2 t + k T / 2 | u ( τ ) | d τ + t k T / 2 t ( s t + k T 2 ) | u ˙ ( s ) | d s + t t + k T / 2 ( t + k T 2 s ) | u ˙ ( s ) | d s = t k T / 2 t + k T / 2 | u ( τ ) | d τ + t k T / 2 t + k T / 2 ϕ ( s ) | u ˙ ( s ) | d s ( k T ) 1 / 2 ( t k T / 2 t + k T / 2 | u ( τ ) | 2 d τ ) 1 / 2 + ( t k T / 2 t + k T / 2 [ ϕ ( s ) ] 2 d s ) 1 / 2 ( t k T / 2 t + k T / 2 | u ˙ ( s ) | 2 d s ) 1 / 2 = ( k T ) 1 / 2 ( t k T / 2 t + k T / 2 | u ( τ ) | 2 d τ ) 1 / 2 + ( k T ) 3 / 2 2 3 ( t k T / 2 t + k T / 2 | u ˙ ( s ) | 2 d s ) 1 / 2 ( k T + ( k T ) 3 12 ) 1 / 2 ( t k T / 2 t + k T / 2 | u ( τ ) | 2 d τ + t k T / 2 t + k T / 2 | u ˙ ( s ) | 2 d s ) 1 / 2 = ( k T + ( k T ) 3 12 ) 1 / 2 ( 0 k T | u ( τ ) | 2 d τ + 0 k T | u ˙ ( s ) | 2 d s ) 1 / 2 .

Hence, we have

u ( 1 k T + k T 12 ) 1 / 2 ( 0 k T | u ( s ) | 2 d s + 0 k T | u ˙ ( s ) | 2 d s ) 1 / 2 .

The proof is complete. □

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

Assume that F=F(t,x):R× R N R is T-periodic in t, F(t,x) is measurable in t for every x R N and continuously differentiable in x for a.e. t[0,T]. If there exist aC( R + , R + ) and b L p ([0,T], R + ) (p>1) such that

| F ( t , x ) | a ( | x | ) b(t),x R N , a.e. t[0,T],
(2.2)

then

c(u)= 0 k T F ( t , u ( t ) ) dt

is weakly continuous and uniformly differentiable on bounded subsets of H k T 1 .

Remark 2.1 In [[17], Lemma 2.2], F C 1 (R, R N ). 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 E 2 and E 2 = E 1 . Suppose that φ C 1 (E,R) satisfies (PS)-condition, and

(I1) φ(u)=1/2(Φu,u)+b(u), where Φu= Φ 1 P 1 u+ Φ 2 P 2 u and Φ i : E i E i bounded and self-adjoint, i=1,2;

(I2) b is compact, and

(I3) there exists a subspace E ˜ E and sets SE, Q E ˜ and constants α>β such that

  1. (i)

    S E 1 and φ | S α,

  2. (ii)

    Q is bounded and φ | Q β,

  3. (iii)

    S and ∂Q link.

Then φ possesses a critical value cα which can be characterized as

c= inf h Γ sup u Q φ ( h ( 1 , u ) ) ,

where

Γ { h C ( [ 0 , 1 ] × E , E ) | h satisfies the following ( Γ 1 ) - ( Γ 3 ) } ,

( Γ 1 ) h(0,u)=u,

( Γ 2 ) h(t,u)=u for uQ, and

( Γ 3 ) h(t,u)= e θ ( t , u ) Φ u+K(t,u), where θC([0,1]×E,R) 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 { u n } H T 1 , { u n } has a convergent subsequence if φ( u n ) is bounded and (1+ u n ) φ ( u n )0 as n.

3 Proofs of theorems

Proof of Theorem 1.1 It follows from Assumption (A)′ that the functional φ k on H k T 1 given by

φ k (u)= 1 2 0 k T | u ˙ ( t ) | 2 dt 1 2 0 k T ( Au ( t ) , u ( t ) ) dt 0 k T F ( t , u ( t ) ) dt

is continuously differentiable. Moreover, one has

φ k ( u ) , v = 0 k T [ ( u ˙ ( t ) , v ˙ ( t ) ) ( Au ( t ) , v ( t ) ) ( F ( t , u ( t ) ) , v ( t ) ) ] dt

for u,v H k T 1 and the solutions of system (1.1) correspond to the critical points of φ k (see [1]).

Obviously, there exists an orthogonal matrix Q such that

Q τ AQ=B=( λ 1 λ r 0 0 λ r + s + 1 λ N )
(3.1)

Let u=Qw. Then by (1.1),

Q w ¨ (t)+AQw(t)+F ( t , Q w ( t ) ) =0a.e. tR.

Furthermore

w ¨ (t)+ Q 1 AQw(t)+ Q 1 F ( t , Q w ( t ) ) =0a.e. tR,

that is,

w ¨ (t)+Bw(t)+ Q 1 F ( t , Q w ( t ) ) =0a.e. tR.
(3.2)

Let G(t,w)=F(t,Qw) and then G(t,w)= Q 1 F(t,Qw(t)). Let

ψ k (w)= 1 2 0 k T | w ˙ ( t ) | 2 dt 1 2 0 k T ( B w ( t ) , w ( t ) ) dt 0 k T G ( t , w ( t ) ) dt.

Then the critical points of ψ k correspond to solutions of system (3.2). It is easy to verify that φ k (u)= ψ k (w) 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 ψ k if and only if u=Qw is the critical point of φ 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 k T 1 . Let

I N =( 1 1 1 )=( e 1 , e 2 ,, e N ).

Note that

H k T 1 { i = 0 ( c i cos i k 1 ω t + d i sin i k 1 ω t ) | c i , d i R N , i = 0 , 1 , 2 } .

Define

H k T = { u H k T 1 | u = u ( t ) = i = 1 r e i j = 0 k l i ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } , H k T 0 = { u H k T 1 | u = u ( t ) = i = r + 1 r + s e i j = 0 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } , H k T + = { u H k T 1 | u = u ( t ) = i = 1 r e i j = k l i + 1 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) H k T + = + i = r + s + 1 N e i j = 0 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } .

Then H k T , H k T 0 and H k T + are closed subsets of H k T 1 and

  1. (1)
    H k T 1 = H k T H k T 0 H k T + ;
  2. (2)
    P k ( u , v ) = 0 , u H k T , v H k T 0 H k T + ,  or P k ( u , v ) = 0 , u H k T 0 , v H k T H k T + ,  or P k ( u , v ) = 0 , u H k T + , v H k T H k T 0 ,

where

P k (u,v)= 0 k T [ ( u ˙ ( t ) , v ˙ ( t ) ) ( Au ( t ) , v ( t ) ) ] dt,u,v H k T 1 .

Let

H k T 01 = { u H k T 0 | u = i = r + 1 r + s c i 0 e i , c i 0 R } , H k T 02 = { u H k T 0 | u = u ( t ) = i = r + 1 r + s e i j = 1 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } , H k T + 1 = { u H k T + | u = u ( t ) = i = 1 r e i j = k l i + 1 k l i + k 1 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } , H k T + 2 = { u H k T + | u = u ( t ) = i = 1 r e i j = k l i + k ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) H k T + 2 = + i = r + s + 1 N e i j = 0 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , c i j , d i j R } .

Then

H k T 0 = H k T 01 H k T 02 , H k T + = H k T + 1 H k T + 2 , H k T 1 = H k T H k T 01 H k T 02 H k T + 1 H k T + 2

and

P k (u,v)=0,u H k T + 1 ,v H k T + 2 .

Remark 3.1 When k=1, it is easy to see H T + 1 ={0}.

Step 2: Let

q k (u)= 1 2 0 k T [ | u ˙ ( t ) | 2 ( Au ( t ) , u ( t ) ) ] dt.

Next we consider the relationship between q k (u) and u 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 u H k T , since

    u=u(t)= i = 1 r e i j = 0 k l i ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) ,

then

q k ( u ) = 1 2 0 k T [ | u ˙ ( t ) | 2 ( Au ( t ) , u ( t ) ) ] d t = 1 2 0 k T [ ( i = 1 r e i j = 0 k l i j k 1 ω ( d i j cos j k 1 ω t c i j sin j k 1 ω t ) , i = 1 r e i j = 0 k l i j k 1 ω ( d i j cos j k 1 ω t c i j sin j k 1 ω t ) ) ( i = 1 r A e i j = 0 k l i ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , i = 1 r e i j = 0 k l i ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) ) ] d t = 1 2 i = 1 r 0 k T { [ j = 0 k l i j k 1 ω ( d i j cos j k 1 ω t c i j sin j k 1 ω t ) ] 2 λ i [ j = 0 k l i ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) ] 2 } d t = k T 4 i = 1 r j = 0 k l i [ ( j k 1 ω ) 2 λ i ] ( c i j 2 + d i j 2 )

and

u 2 = 0 k T ( | u ˙ ( t ) | 2 + | u ( t ) | 2 ) dt= k T 2 i = 1 r j = 0 k l i [ ( j k 1 ω ) 2 + 1 ] ( c i j 2 + d i j 2 ) .

Let

δ= min i { 1 , , r } { λ i ( l i ω ) 2 ( l i ω ) 2 + 1 } >0.

Then

q k (u) δ 2 u 2 ,u H k T .
(3.3)

Remark 3.2 Obviously, if one of (H3)(5) and (H3)(6) holds, then H k T ={0}. Hence,

q k (u)=0,u H k T .
  1. (b)

    For u H k T + 2 H k T 02 , let

    u=u(t)= u 1 (t)+ u 2 (t)+ u 3 (t),

where

u 1 ( t ) = i = 1 r e i j = k l i + k ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , u 2 ( t ) = i = r + s + 1 N e i j = 0 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , u 3 ( t ) = i = r + 1 r + s e i j = 1 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) .

Then

q k ( u ) = 1 2 0 k T [ | u ˙ ( t ) | 2 ( Au ( t ) , u ( t ) ) ] d t = 1 2 0 k T [ ( u ˙ 1 ( t ) + u ˙ 2 ( t ) + u ˙ 3 ( t ) , u ˙ 1 ( t ) + u ˙ 2 ( t ) + u ˙ 3 ( t ) ) ( A u 1 ( t ) + A u 2 ( t ) + A u 3 ( t ) , u 1 ( t ) + u 2 ( t ) + u 3 ( t ) ) ] d t = 1 2 0 k T [ ( u ˙ 1 ( t ) , u ˙ 1 ( t ) ) + ( u ˙ 2 ( t ) , u ˙ 2 ( t ) ) + ( u ˙ 3 ( t ) , u ˙ 3 ( t ) ) ( A u 1 ( t ) , u 1 ( t ) ) ( A u 2 ( t ) , u 2 ( t ) ) ( A u 3 ( t ) , u 3 ( t ) ) ] = k T 4 [ i = 1 r j = k l i + k ( j k 1 ω ) 2 ( c i j 2 + d i j 2 ) + i = r + s + 1 N j = 0 ( j k 1 ω ) 2 ( c i j 2 + d i j 2 ) + i = r + 1 r + s j = 1 ( j k 1 ω ) 2 ( c i j 2 + d i j 2 ) i = 1 r λ i j = k l i + k ( c i j 2 + d i j 2 ) + i = r + s + 1 N λ i j = 0 ( c i j 2 + d i j 2 ) ] = k T 4 { i = 1 r j = k l i + k [ ( j k 1 ω ) 2 λ i ] ( c i j 2 + d i j 2 ) + i = r + s + 1 N j = 0 [ ( j k 1 ω ) 2 + λ i ] ( c i j 2 + d i j 2 ) + i = r + 1 r + s j = 1 ( j k 1 ω ) 2 ( c i j 2 + d i j 2 ) }

and

u 2 = 0 k T ( | u ˙ ( t ) | 2 + | u ( t ) | 2 ) d t = k T 2 { i = 1 r j = k l i + k [ ( j k 1 ω ) 2 + 1 ] ( c i j 2 + d i j 2 ) + i = r + s + 1 N j = 0 [ ( j k 1 ω ) 2 + 1 ] ( c i j 2 + d i j 2 ) + i = r + 1 r + s j = 1 [ ( j k 1 ω ) 2 + 1 ] ( c i j 2 + d i j 2 ) } .

Since for fixed i{1,,r},

f(j)= ( j k 1 ω ) 2 λ i ( j k 1 ω ) 2 + 1 andg(j)= ( j k 1 ω ) 2 ( j k 1 ω ) 2 + 1

are strictly increasing on jN,

f(j)f(k l i +k)= ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 >0,jk l i +k

and

g(j)g(1)= ( k 1 ω ) 2 ( k 1 ω ) 2 + 1 = ω 2 ω 2 + k 2 >0.

Moreover, it is easy to verify that

( j k 1 ω ) 2 + λ i ( j k 1 ω ) 2 + 1 λ i 1 + λ i + ,jN{0},i=r+s+1,,N.

Let

σ k =min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 , λ i 1 + λ i + } .

Then

q k (u) σ k 2 u 2 ,u H k T + 2 H k T 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

    σ k =min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 } ;
  2. (ii)

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

    σ k =min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 , λ i 1 + λ i + } ;
  3. (iii)

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

    σ k σ=min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , λ i 1 + λ i + } ;
  4. (iv)

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

    σ k σ= min i { 1 , , N } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } ;
  5. (v)

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

    σ k =min { ω 2 ω 2 + k 2 , λ i 1 + λ i + } ;
  6. (vi)

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

    σ k σ= λ i 1 + λ i + .
  1. (c)

    For u H k T + 1 , since

    u = i = 1 r e i j = k l i + 1 k l i + k 1 ( c i j cos j k 1 ω t + d i j sin j k 1 ω t ) , q k ( u ) = k T 4 i = 1 r j = k l i + 1 k l i + k 1 [ ( j k 1 ω ) 2 λ i ] ( c i j 2 + d i j 2 )

and

u 2 = k T 2 i = 1 r j = k l i + 1 k l i + k 1 [ ( j k 1 ω ) 2 + 1 ] ( c i j 2 + d i j 2 ) .

Obviously, when k=1, u=0. So q 1 (u)=0. When k>1, it follows from

( l i + 1 1 k ) 2 ω 2 λ i < ( l i + 1 ) 2 ω 2 ,i{1,,r}

that

q k (u)0,u H k T + 1 .
(3.5)
  1. (d)

    Obviously, for u H k T 01 , we have

    q k (u)=0,u H k T 01 .
    (3.6)

Step 3: Assume that (H3)(2) holds. We prove that there exist ρ k >0 and b k >0 such that

φ k (u) b k >0,u ( H k T + 2 H k T 02 ) B ρ k .

Let

C k = 12 + k 2 T 2 12 k T .

Choosing ρ k =min{1, l k / C k }>0 and b k =( σ k 2 α k ) ρ k 2 >0, by Lemma 2.1, (H4) and (3.4), we have, for all u( H k T + 2 H k T 02 ) B ρ k ,

φ k ( u ) 1 2 0 k T | u ˙ ( t ) | 2 d t 1 2 0 k T ( Au ( t ) , u ( t ) ) d t 0 k T F ( t , u ( t ) ) d t σ k 2 u 2 α k 0 k T | u ( t ) | 2 d t ( σ k 2 α k ) u 2 = ( min { min i { 1 , , r } { ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 } , ω 2 ω 2 + k 2 , λ i 1 + λ i + } 2 α k ) ρ k 2 .

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

φ k (u) ( σ k 2 α k ) ρ k 2 >0,u ( H k T + 2 H k T 02 ) B ρ k .

Step 4: Let

Q k = { s h k | s [ 0 , s 1 ] } ( B s 2 ( H k T H k T 01 H k T + 1 ) ) ,

where h k H k T + 2 H k T 02 , s 1 and s 2 will be determined later. In this step, we prove φ k | Q k 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 { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } .

Case (i): if

d k :=d= ( l i 0 + 1 ) 2 ω 2 λ i 0 2 ,

then we choose

h k (t)=sin( l i 0 +1)ωt e i 0 ,tR.

Obviously, h k H k T + 2 and h ˙ k (t)=( l i 0 +1)ωcos( l i 0 +1)ωt e i 0 , tR. Then

h k L 2 2 = k T 2 , h ˙ k L 2 2 = k T ( l i 0 + 1 ) 2 ω 2 2 .

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

F(t,x) β k | x | 2 β k L ˆ k 2 =(d+ ε 0 k ) | x | 2 β k L ˆ k 2 ,x R N , a.e. t[0,kT],
(3.7)

where ε 0 k = β k d>0 and L ˆ k >max{1, L k }. Since H k T H k T 01 H k T + 1 is the finite dimensional space, there exists a constant K 1 k >0 such that

K 1 k u 2 u L 2 2 u 2 ,u H k T H k T 01 H k T + 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 H k T H k T 01 H k T + 1 ,

φ k ( s h k + u ) δ 2 u 2 + s 2 2 0 k T | h ˙ k ( t ) | 2 d t λ i 0 s 2 2 0 k T | h k ( t ) | 2 d t 0 k T F ( t , s h k ( t ) + u ( t ) ) d t δ 2 u 2 + s 2 2 k T ( l i 0 + 1 ) 2 ω 2 2 λ i 0 s 2 2 k T 2 ( d + ε 0 k ) 0 k T | s h k ( t ) + u ( t ) | 2 d t + β k L ˆ k 2 k T = δ 2 u 2 + s 2 2 k T ( l i 0 + 1 ) 2 ω 2 2 λ i 0 s 2 2 k T 2 ( d + ε 0 k ) ( s 2 h k L 2 2 + u L 2 2 ) + β k L ˆ k 2 k T δ 2 u 2 + ( k T ( l i 0 + 1 ) 2 ω 2 4 λ i 0 k T 4 d k T 2 k T ε 0 k 2 ) s 2 ( d + ε 0 k ) u L 2 2 + β k L ˆ k 2 k T k T ε 0 k 2 s 2 ε 0 k u L 2 2 + β k L ˆ k 2 k T k T ε 0 k 2 s 2 ε 0 k K 1 k u 2 + β k L ˆ k 2 k T .
(3.9)

Hence,

φ k (s h k +u)0,either s s 1  or u s 2 ,

where

s 1 = 2 β k L ˆ k 2 ε 0 k , s 2 = β k L ˆ k 2 k T ε 0 k K 1 k .

Case (ii): if d k = ω 2 /(2 k 2 ), then we choose

h k (t)=sin k 1 ωt e r + 1 H k T 02 ,tR.

Then

h ˙ k (t)= ω k cos k 1 ωt e r + 1 ,tR,

and

(A h k , h k )=0, h k L 2 2 = k T 2 , h ˙ k L 2 2 = T ω 2 2 k .
(3.10)

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

F(t,x) β k | x | 2 β k L ˆ k 2 = ( ω 2 2 k 2 + ε 0 k ) | x | 2 β k L ˆ k 2 ,x R N , a.e. t[0,T],
(3.11)

where L ˆ k >max{1, L k } and ε 0 k = β k ω 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 H k T H k T 01 H k T + 1 ,

φ k ( s h k + u ) δ 2 u 2 + s 2 2 0 k T | h ˙ k ( t ) | 2 d t 0 k T F ( t , s h k ( t ) + u ) d t δ 2 u 2 + s 2 2 T ω 2 2 k ( ω 2 2 k 2 + ε 0 k ) 0 k T | s h k ( t ) + u ( t ) | 2 d t + β k L ˆ k 2 k T = δ 2 u 2 + s 2 2 T ω 2 2 k ( ω 2 2 k 2 + ε 0 k ) ( s 2 h k L 2 2 + u L 2 2 ) + β k L ˆ k 2 k T = δ 2 u 2 + ( T ω 2 4 k T ω 2 4 k k T ε 0 k 2 ) s 2 ( ω 2 2 k 2 + ε 0 k ) u L 2 2 + β k L ˆ k 2 k T k T ε 0 k 2 s 2 ε 0 k u L 2 2 + β k L ˆ k 2 k T k T ε 0 k 2 s 2 ε 0 k K 1 k u 2 + β k L ˆ k 2 k T .

Hence,

φ k (s h k +u)0,either s s 1  or u s 2 ,

where

s 1 = 2 β k L ˆ k 2 ε 0 k , s 2 = β k L ˆ k 2 k T ε 0 k K 1 k .

Case (iii): if d k = λ i /2, then we choose

h k = 1 k T e i H k T + 2 .

Then

h ˙ k =0,(A h k , h k )= λ i ( h k , h k ), h k L 2 2 =1.

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

F(t,x) β k | x | 2 β k L ˆ k 2 = ( λ i 2 + ε 0 k ) | x | 2 β k L ˆ k 2 ,x R N , a.e. t[0,kT],
(3.12)

where L ˆ k >max{ 1 + 1 T , L k } and ε 0 k = β k λ 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 H k T H k T 01 H k T + 1 , we have

φ k ( s h k + u ) δ 2 u 2 + s 2 2 0 k T | h ˙ k ( t ) | 2 d t + λ i s 2 2 0 k T | h k ( t ) | 2 d t 0 k T F ( t , s h k ( t ) + u ( t ) ) d t δ 2 u 2 + λ i s 2 2 ( λ i 2 + ε 0 k ) 0 k T | s h k ( t ) + u ( t ) | 2 d t + β k L ˆ k 2 k T = δ 2 u 2 + λ i s 2 2 ( λ i 2 + ε 0 k ) ( s 2 h k L 2 2 + u L 2 2 ) + β k L ˆ k 2 k T = δ 2 u 2 + ( λ i 2 λ i 2 ε 0 k ) s 2 ( λ i 2 + ε 0 k ) u L 2 2 + β k L ˆ k 2 k T ε 0 k s 2 ε 0 k K 1 k u 2 + β k L ˆ k 2 k T .

Hence,

φ k (s e k +u)0,either s s 1  or u s 2 ,

where

s 1 = β k L ˆ k 2 k T ε 0 k , s 2 = β k L ˆ k 2 k T ε 0 k K 1 k .

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

s 1 = max { 2 β k L ˆ k 2 ε 0 k , 2 β k L ˆ k 2 ε 0 k , β k L ˆ k 2 k T ε 0 k } , s 2 = max { β k L ˆ k 2 k T ε 0 k K 1 k , β k L ˆ k 2 k T ε 0 k K 1 k , β k L ˆ k 2 ε 0 k K 1 k } ,

then

φ k (s h k +u)0,either s s 1  or u s 2 .
(3.13)

By (1.5), (3.3), (3.5) and (3.6), for all u H k T H k T 01 H k T + 1 , we have

φ k ( u ) = 1 2 0 k T | u ˙ ( t ) | 2 d t 1 2 0 k T ( Au ( t ) , u ( t ) ) d t 0 k T F ( t , u ( t ) ) d t δ 2 u 2 0 .
(3.14)

Thus, it follows from (3.13) and (3.14) that φ | Q k 0< b k .

Step 5: We prove that φ k satisfies (C)-condition in H k T 1 . The proof is similar to that in Theorem 1.1 in [17]. We omit it.

Step 6: We claim that φ k has a nontrivial critical point u k H k T 1 such that φ k ( u k ) 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

q k ( u ) = 1 2 u 2 1 2 0 k T ( ( A + I ) u ( t ) , u ( t ) ) d t = 1 2 ( I K ) u , u ) ,

where K: H k T 1 H k T 1 is the linear self-adjoint operator defined, using the Riesz representation theorem, by

0 k T ( ( A + I ) u ( t ) , v ( t ) ) dt= ( K u , v ) ,u,v H T 1 .

The compact imbedding of H k T 1 into C([0,kT]; R N ) implies that K is compact. In order to use Lemma 2.3, we let Φ=IK and define Φ i : E i E i , i=1,2 by

Φ i u,v= ( I K ) u , v ,u,v E i ,

where E 1 = H k T + 2 H k T 02 and E 2 = H k T H k T 01 H k T + 1 . Since K is a self-adjoint compact operator, it is easy to see that Φ i (i=1,2) are bounded and self-adjoint. Let

b(u)= 0 k T F ( t , u ( t ) ) dt.

Assumption (A)′ and Lemma 2.2 imply that b is weakly continuous and is uniformly differentiable on bounded subsets of E= H k T 1 . Furthermore, by standard theorems in [22], we conclude that b is compact. Let S k =( H k T + 2 H k T 02 ) B ρ k . Then S k and Q k link. Hence, by Step 1-Step 5, Lemma 2.3 and Remark 2.2, there exists a critical point u k H k T 1 such that φ k ( u k ) 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<N, (H1) holds for any kN. Moreover, since (H3)′ implies that (H3)(5) and (H4)′ implies that (H4), system (1.1) has kT-periodic solution for every kN.

Let d= ω 2 2 . Like the argument of case (ii) in the proof of Theorem 1.1, choose

e k (t)=sin k 1 ωt e r + 1 H k T 02 ,tR.

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

F(t,x)β | x | 2 β L 2 = ( ω 2 2 + ε 1 ) | x | 2 β L 2 ,x R N , a.e. t[0,kT],
(3.15)

where ε 1 =β ω 2 2 . In the proof of Theorem 1.1, if we replace (3.15) with (3.11), then we obtain

φ k (s e k +u)0,either s s 1  or u s 2 ,

where

s 1 = 2 β L 2 ε 1 = 2 β L 2 β ω 2 2 , s 2 = β L 2 k T ε 1 K 1 k .

Note that s 1 is independent of k. Hence, if u k is the critical point of φ 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

φ k ( u k ) sup u Q k φ k ( u ) sup s [ 0 , s 1 ] { s 2 2 0 k T | e ˙ k ( t ) | 2 d t s 2 2 0 k T ( A e k ( t ) , e k ( t ) ) d t } s 1 2 2 0 k T | e ˙ k ( t ) | 2 d t = β L 2 T ω 2 2 k ( β ω 2 2 ) β L 2 T ω 2 2 ( β ω 2 2 ) : = M .
(3.16)

Hence, φ k ( u k ) is bounded for any kN.

Obviously, we can find k 1 N/{1} such that k 1 > M b 1 , then we claim that u k is distinct from u 1 for all k k 1 . In fact, if u k = u 1 for some k k 1 , it is easy to check that

φ k ( u k )=k φ 1 ( u 1 )k b 1 .

Then by (3.16), we have k 1 k M b 1 , a contradiction. We also can find k 2 >max{ k 1 , k 1 M b k 1 } such that u k 1 k u k 1 for all k k 2 k 1 . Otherwise, if u k 1 k = u k 1 for some k k 1 , we have φ k 1 k ( u k 1 k )=k φ k 1 ( u k 1 )k b k 1 . Then by (3.16), we have k 2 k 1 k 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 N and k j as j. 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 { u n } is bounded if φ( u n ) is bounded and φ ( u n )(1+ u n )0 as n+. Other proofs are the same as in [15]. The proof of boundedness of { u n } is essentially the same as in Theorem 1.1 in [17] except that 2 is replaced by p, H k T 1 by

W k T 1 , p = { u : R R N | u  is absolutely continuous , u ( t ) = u ( t + T )  and  u ˙ L p ( [ 0 , T ] ) }

equipped with the norm

u= ( 0 k T | u ( t ) | p d t + 0 k T | u ˙ ( t ) | p d t ) 1 / p ,

and

F(t,x) β k | x | 2 ,x R N ,|x|>L

by

F(t,x)ε | x | p ,x R N ,|x|>L

for some ε>0. So, we omit the details. □

4 Examples

Example 4.1 Let T=2π and

A=( 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 ).

Then ω=1, r=2, λ 1 =7.5, λ 2 =7.4, λ 3 =0, λ 4 =3, λ 5 =4, λ i + =4 and λ i =3. Obviously, the matrix A satisfies Assumption (A0) and l 1 = l 2 =2 such that

l i 2 ω 2 < λ i < ( l i + 1 ) 2 ω 2 ,i=1,2.

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

F(t,x) 4 63 k 2 | x | 2 ( 11 | x | 3 / 2 1 + | x | 3 / 2 1 2 ) a.e. t[0,T].

Then F(t,x)0 for all x R N and a.e. t[0,T] and

lim | x | 0 F ( t , x ) | x | 2 = 2 63 k 2 uniformly for a.e. t[0,T],
(4.1)
lim | x | F ( t , x ) | x | 2 = 2 3 k 2 uniformly for a.e. t[0,T].
(4.2)

It is easy to verify that

( F ( t , x ) , x ) 2F(t,x)= 6 ln 11 63 k 2 | x | 2 11 | x | 3 / 2 1 + | x | 3 / 2 | x | 3 / 2 ( 1 + | x | 3 / 2 ) 2 .

Choose ξ=1, η=1 and ν=3/2. Moreover, obviously, there exists m>0 such that | x | 3 / 2 1 + | x | 3 / 2 > 2 3 . Then

( F ( t , x ) , x ) 2 F ( t , x ) F ( t , x ) ξ + η | x | ν = 3 2 ln 11 11 | x | 3 / 2 1 + | x | 3 / 2 | x | 3 / 2 1 + | x | 3 / 2 11 | x | 3 / 2 1 + | x | 3 / 2 1 2 >ln11>1.

Hence, (H2) holds.

When k=1,

min { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } = 1 2 and σ 1 =0.15.

By (4.2), we can find L 1 >0 such that

F(t,x) ( 2 3 1 10 ) | x | 2 = 17 30 | x | 2 ,|x|> L 1 , and a.e. t[0,T].

Let β 1 = 17 30 . Then (H3)(2) holds with k=1. Moreover, by (4.1), we can find l 1 >0 such that

F(t,x) ( 2 63 + 23 2520 ) | x | 2 0.0409 | x | 2 ,|x| l 1  and a.e. t[0,T].

Let α 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 { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } = 1 8 and σ 2 =0.15.

By (4.2), we can find L 2 >0 such that

F(t,x) ( 1 6 1 100 ) | x | 2 0.1567 | x | 2 ,|x|> L 2  and a.e. t[0,T].

Let β 2 =0.1567. Then (H3)(2) holds with k=2. Moreover, by (4.1), we can find l 2 >0 such that

F(t,x) ( 1 126 + 1 1000 ) | x | 2 0.00894 | x | 2 ,|x| l 2  and a.e. t[0,T].

Let α 2 =0.00894. Then (H4) holds. Note that 1 6 < 1 2 =min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 , λ i 2 }. 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 { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } = 1 18 and σ 3 =0.1.

By (4.2), we can find L 3 >0 such that

F(t,x) ( 2 27 1 100 ) | x | 2 0.0641 | x | 2 ,|x|> L 3  and a.e. t[0,T].

Let β 3 =0.0641. Then (H3)(2) holds with k=3. Moreover, by (4.1), we can find l 3 >0 such that

F(t,x) ( 2 567 + 1 1000 ) | x | 2 0.00453 | x | 2 ,|x| l 3  and a.e. t[0,T].

Let α 3 =0.00453. Then (H4) holds. Note that 2 27 < 1 8 =min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 × 2 2 , λ i 2 }< 1 2 =min{ ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 , λ i 2 }. So, when k=3, by Theorem 1.1, we cannot judge that system (1.1) has T-periodic solution and 2T-periodic solution. However, we can obtain that system (1.1) has a 3T-periodic solution. It is easy to verify that Example 4.1 does not satisfy the theorem in [19] even if k=1.

Example 4.2 Let

A=( 0 0 0 0 1 0 0 0 2 )

and

F(t,x) 2 π 2 T 2 | x | 2 ( e | x | 3 / 2 1 + | x | 3 / 2 1 ) a.e. t[0,T].

Then

lim | x | 0 F ( t , x ) | x | 2 = 0 uniformly for a.e.  t [ 0 , T ] , lim | x | F ( t , x ) | x | 2 = 2 π 2 T 2 ( e 1 ) uniformly for a.e.  t [ 0 , T ] .

Obviously, (A0), (A)′, (1.5), (H3)′ and (H4)′ hold. Let ξ=1, η=1 and ν= 3 2 . Similar to the argument in Example 4.1, we obtain (H2) also holds. Then by Theorem 1.2, system (1.1) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j.

Example 4.3 Let p=4 and

F(t,x) | x | p ( e | x | p 1 ) = | x | 4 ( e | x | 4 1 ) a.e. t[0,T].

Then (1.5) holds and

lim | x | 0 F ( t , x ) | x | 4 =0, lim | x | F ( t , x ) | x | 4 =+uniformly for a.e. t[0,T].

Let ξ=1, η=1 and ν=1/2. Then it is easy to obtain that there exists m>1 such that (H5) holds. By Theorem 1.3, system (1.8) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j. It is easy to see that Example 4.3 does not satisfy (1.3). Hence, Theorem 1.3 improved Theorem B.

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Acknowledgements

The authors would like to thank the anonymous referees for their valuable suggestions.

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XZ proposed the idea of the paper and finished the main proofs. XT provided some important techniques in the process of proofs.

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Zhang, X., Tang, X. Existence of subharmonic solutions for non-quadratic second-order Hamiltonian systems. Bound Value Probl 2013, 139 (2013). https://doi.org/10.1186/1687-2770-2013-139

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