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

Xingyong Zhang1* and Xianhua Tang2

Author Affiliations

1 Department of Mathematics, Faculty of Science, Kunming University of Science and Technology, Kunming, Yunnan, 650500, P.R. China

2 School of Mathematical Sciences and Computing Technology, Central South University, Changsha, Hunan, 410083, P.R. China

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Boundary Value Problems 2013, 2013:139  doi:10.1186/1687-2770-2013-139


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/139


Received:23 October 2012
Accepted:10 May 2013
Published:30 May 2013

© 2013 Zhang and Tang; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 intfor every x R N and continuously differentiable inxfor a.e. t [ 0 , T ] , and there exist a C ( R + , R + ) and b : R + R + which isT-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 [1-7] 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 [8-12] 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 thatFsatisfies

(A) F ( t , x ) is measurable intfor every x R N and continuously differentiable inxfor 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 thatFsatisfies (A), (1.3) and (1.4) with 2 replaced byp, (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 , Fsatisfies (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):

(H) there exist positive constantsm,ζ,ηand 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)Ahas 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 1.1Assume that (A0) holds andFsatisfies (A)′, (1.5) and the following conditions.

(H1) For some k N , assume thatksatisfies

( 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 constantsm, ζ, ηand 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) 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);

(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;

(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;

(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;

(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;

(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 ,

whereσimplies that σ k is independent ofk. Then system (1.1) has a nonzerokT-periodic solution. Especially, for cases (H3)(1) and (H3)(4), system (1.1) has a nonconstantkT-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 1.2Suppose that (A0) holds andFsatisfies (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 1.3Suppose thatFsatisfies (A), (1.5) and the following conditions:

(H5) there exist positive constantsm, ζ, ηand 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 2.1If 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 isT-periodic in t, F ( t , x ) is measurable intfor every x R N and continuously differentiable inxfor 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])

LetEbe 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

(i) S E 1 and φ | S α ,

(ii) Qis bounded and φ | Q β ,

(iii) Sand∂Qlink.

Thenφpossesses 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 ) andKis 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)

H k T 1 = H k T H k T 0 H k T + ;

(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.

(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 .

(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:

(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 } ;

(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 + } ;

(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 + } ;

(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 } ;

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

σ k = min { ω 2 ω 2 + k 2 , λ i 1 + λ i + } ;

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

σ k σ = λ i 1 + λ i + .

(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)

(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