Research

# 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

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

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 ,

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