Open Access

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

Boundary Value Problems20132013:139

DOI: 10.1186/1687-2770-2013-139

Received: 23 October 2012

Accepted: 10 May 2013

Published: 30 May 2013

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.

Keywords

second-order Hamiltonian systems subharmonic solution critical point linking theorem

1 Introduction and main results

Consider the second-order Hamiltonian system
u ¨ ( t ) + Au ( t ) + F ( t , u ( t ) ) = 0 a.e.  t R ,
(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 a C ( 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 ) ) = 0 a.e.  t R .
(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 < μ F ( 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 > θ 2 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 > 0 uniformly 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 a C ( 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 ) ) = 0 a.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 = 2 π / 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 = 2 π / T , l i N , i = 1 , , r and 0 r 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 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 r N { 0 } and r N . 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 l N , where ω = 2 π / T , that is, λ i l 2 ω 2 ( i = 1 , , r ) for all l N .

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 k N , 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 k N satisfying (1.12) is finite. Let m K 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 k N . Hence, Theorem 1.1 implies that system (1.11) has nonzero kT-periodic solutions for every k N .

Remark 1.2 In [18], Costa and Magalhães studied the first-order Hamiltonian system
J u ˙ ( t ) + Au + H ( t , u ) = 0 a.e.  t [ 0 , T ] .
(1.14)
They obtained that system (1.14) has a T = 2 π periodic solution under the following non-quadraticity conditions:
lim inf | x | ( x , H ( t , x ) ) 2 H ( t , x ) | x | μ a > 0 uniformly for a.e.  t [ 0 , 2 π ] ,
(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 , 2 π ] ,
where λ k 1 < λ k are consecutive eigenvalues of the operator L = J d / d t A . Moreover, they also obtained system (1.14) has a nonzero T = 2 π 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 = J d / d t 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. 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 ) ) d t + 0 k T ( u ˙ ( t ) , v ˙ ( t ) ) d t
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 ) d t and 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 , k T ] . For every τ [ 0 , k T ] , we have
u ( t ) = u ( τ ) + τ t u ˙ ( s ) d s .
(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 [ t k T / 2 , t + k T / 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 a C ( 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 ) ) d t

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 ( Φ u , u ) + b ( u ) , where Φ u = 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 S E , 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 c α 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 u Q , and

( Γ 3 ) h ( t , u ) = e θ ( t , u ) Φ u + K ( t , u ) , where θ C ( [ 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 d t 1 2 0 k T ( Au ( t ) , u ( t ) ) d t 0 k T F ( t , u ( t ) ) d t
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 ) ) ] d t

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 A Q = B = ( λ 1 λ r 0 0 λ r + s + 1 λ N )
(3.1)
Let u = Q w . Then by (1.1),
Q w ¨ ( t ) + A Q w ( t ) + F ( t , Q w ( t ) ) = 0 a.e.  t R .
Furthermore
w ¨ ( t ) + Q 1 A Q w ( t ) + Q 1 F ( t , Q w ( t ) ) = 0 a.e.  t R ,
that is,
w ¨ ( t ) + B w ( t ) + Q 1 F ( t , Q w ( t ) ) = 0 a.e.  t R .
(3.2)
Let G ( t , w ) = F ( t , Q w ) and then G ( t , w ) = Q 1 F ( t , Q w ( t ) ) . Let
ψ k ( w ) = 1 2 0 k T | w ˙ ( t ) | 2 d t 1 2 0 k T ( B w ( t ) , w ( t ) ) d t 0 k T G ( t , w ( t ) ) d t .

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 = Q w 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 ) ) ] d t , 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 ) ) ] d t .
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 ) d t = 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 and g ( j ) = ( j k 1 ω ) 2 ( j k 1 ω ) 2 + 1
are strictly increasing on j N ,
f ( j ) f ( k l i + k ) = ( l i + 1 ) 2 ω 2 λ i ( l i + 1 ) 2 ω 2 + 1 > 0 , j k 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 + , j N { 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 ) ω t e i 0 , t R .
Obviously, h k H k T + 2 and h ˙ k ( t ) = ( l i 0 + 1 ) ω cos ( l i 0 + 1 ) ω t e i 0 , t R . 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 , k T ] ,
(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 ω t e r + 1 H k T 02 , t R .
Then
h ˙ k ( t ) = ω k cos k 1 ω t e r + 1 , t R ,
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 , k T ] ,
(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 ) ) d t = ( K u , v ) , u , v H T 1 .
The compact imbedding of H k T 1 into C ( [ 0 , k T ] ; R N ) implies that K is compact. In order to use Lemma 2.3, we let Φ = I K 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 ) ) d t .

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 k N . Moreover, since (H3)′ implies that (H3)(5) and (H4)′ implies that (H4), system (1.1) has kT-periodic solution for every k N .

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 ω t e r + 1 H k T 02 , t R .
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 , k T ] ,
(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 k N .

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 = 2 π 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 ) 2 F ( 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 > ln 11 > 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.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Kunming University of Science and Technology
(2)
School of Mathematical Sciences and Computing Technology, Central South University

References

  1. Mawhin J, Willem M: Critical Point Theory and Hamiltonian Systems. Springer, New York; 1989.View Article
  2. Tang CL: Periodic solutions of nonautonomous second order systems with sublinear nonlinearity. Proc. Amer. Math. Soc. 1998, 126: 3263–3270. 10.1090/S0002-9939-98-04706-6MathSciNetView Article
  3. Ailva EAB: Subharmonic solutions for subquadratic Hamiltonian systems. J. Differ. Equ. 1995, 115: 120–145. 10.1006/jdeq.1995.1007View Article
  4. Schechter M: Periodic non-autonomous second-order dynamical systems. J. Differ. Equ. 2006, 223: 290–302. 10.1016/j.jde.2005.02.022MathSciNetView Article
  5. Rabinowitz PH CBMS Regional Conf. Ser. in Math. 65. In Minimax Methods in Critical Point Theory with Applications to Differential Equations. Am. Math. Soc., Providence; 1986.
  6. Rabinowitz PH: Periodic solutions of Hamiltonian systems. Commun. Pure Appl. Math. 1978, 31: 157–184. 10.1002/cpa.3160310203MathSciNetView Article
  7. Tao ZL, Yan S, Wu SL: Periodic solutions for a class of superquadratic Hamiltonian systems. J. Math. Anal. Appl. 2007, 331: 152–158. 10.1016/j.jmaa.2006.08.041MathSciNetView Article
  8. Chang KC Progress in Nonlinear Differential Equations and Their Applications 6. Infinite Dimensional Morse Theory and Multiple Solution Problems 1993.View Article
  9. Ekeland I: Convexity Method in Hamiltonian Mechanics. Springer, Berlin; 1990.View Article
  10. Ekeland I, Hofer H: Periodic solutions with prescribed period for convex autonomous Hamiltonian systems. Invent. Math. 1985, 81: 155–188. 10.1007/BF01388776MathSciNetView Article
  11. Fei G, Qiu Q: Minimal periodic solutions of nonlinear Hamiltonian systems. Nonlinear Anal. 1996, 27: 821–839. 10.1016/0362-546X(95)00077-9MathSciNetView Article
  12. Fei G, Kim S, Wang T: Minimal period estimates of periodic solutions for superquadratic Hamiltonian systems. J. Math. Anal. Appl. 1999, 238: 216–233. 10.1006/jmaa.1999.6527MathSciNetView Article
  13. Fei G: On periodic solutions of superquadratic Hamiltonian systems. Electron. J. Differ. Equ. 2002, 2002: 1–12.
  14. Tao ZL, Tang CL: Periodic and subharmonic solutions of second order Hamiltonian systems. J. Math. Anal. Appl. 2004, 293: 435–445. 10.1016/j.jmaa.2003.11.007MathSciNetView Article
  15. Ma S, Zhang Y: Existence of infinitely many periodic solutions for ordinary p -Laplacian systems. J. Math. Anal. Appl. 2009, 351: 469–479. 10.1016/j.jmaa.2008.10.027MathSciNetView Article
  16. Ye YW, Tang CL: Periodic and subharmonic solutions for a class of superquadratic second order Hamiltonian systems. Nonlinear Anal. 2009, 71: 2298–2307. 10.1016/j.na.2009.01.064MathSciNetView Article
  17. Zhang X, Tang X: Subharmonic solutions for a class of non-quadratic second order Hamiltonian systems. Nonlinear Anal., Real World Appl. 2012, 13: 113–130. 10.1016/j.nonrwa.2011.07.013MathSciNetView Article
  18. Costa DG, Magalhães CA: A unified approach to a class of strongly indefinite functions. J. Differ. Equ. 1996, 125: 521–547. 10.1006/jdeq.1996.0039View Article
  19. Kyristi ST, Papageorgiou NS: On superquadratic periodic systems with indefinite linear part. Nonlinear Anal. 2010, 72: 946–954. 10.1016/j.na.2009.07.035MathSciNetView Article
  20. Benci V, Rabinowitz PH: Critical point theorems for indefinite functions. Invent. Math. 1979, 52: 241–273. 10.1007/BF01389883MathSciNetView Article
  21. Bartolo P, Benci V, Fortunato D: Abstract critical point theorems and applications to some nonlinear problems with strong resonance at infinity. Nonlinear Anal. 1983, 7: 241–273.MathSciNetView Article
  22. Krosnoselski MA: Topological Methods in the Theory of Nonlinear Integral Equations. Macmillan Co., New York; 1964.

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