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

  • Xingyong Zhang1Email author and

    Affiliated with

    • Xianhua Tang2

      Affiliated with

      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ1_HTML.gif
      (1.1)

      where A is an N × N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq1_HTML.gif symmetric matrix and F : R × R N R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq2_HTML.gif is T-periodic in t and satisfies the following assumption:

      Assumption (A)′ F ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq3_HTML.gif is measurable in t for every x R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq4_HTML.gif and continuously differentiable in x for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq5_HTML.gif, and there exist a C ( R + , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq6_HTML.gif and b : R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq7_HTML.gif which is T-periodic and b L p ( 0 , T ; R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq8_HTML.gif with p > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq9_HTML.gif such that
      | F ( t , x ) | a ( | x | ) b ( t ) , | F ( t , x ) | a ( | x | ) b ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equa_HTML.gif

      for all x R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq10_HTML.gif and a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq5_HTML.gif.

      When A = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq11_HTML.gif, system (1.1) reduces to the second-order Hamiltonian system
      u ¨ ( t ) + F ( t , u ( t ) ) = 0 a.e.  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ2_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq12_HTML.gif and L > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq13_HTML.gif such that
      0 < μ F ( t , x ) ( F ( t , x ) , x ) , | x | L , t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equb_HTML.gif
      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq14_HTML.gif and μ > θ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq15_HTML.gif such that
      lim sup | x | F ( t , x ) | x | θ < uniformly for a.e.  t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ3_HTML.gif
      (1.3)
      lim inf | x | ( F ( t , x ) , x ) 2 F ( t , x ) | x | μ > 0 uniformly for a.e.  t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ4_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq3_HTML.gif is measurable in t for every x R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq4_HTML.gif and continuously differentiable in x for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq5_HTML.gif, and there exist a C ( R + , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq6_HTML.gif and b L 1 ( 0 , T ; R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq16_HTML.gif such that
        | F ( t , x ) | a ( | x | ) b ( t ) , | F ( t , x ) | a ( | x | ) b ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equc_HTML.gif
         
      for all x R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq10_HTML.gif and a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq5_HTML.gif. Assume that (1.3), (1.4) and the following conditions hold:
      F ( t , x ) 0 , ( t , x ) [ 0 , T ] × R N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ5_HTML.gif
      (1.5)
      lim | x | 0 F ( t , x ) | x | 2 = 0 uniformly for a.e. t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ6_HTML.gif
      (1.6)
      lim | x | F ( t , x ) | x | 2 > 2 π 2 T 2 uniformly for a.e. t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ7_HTML.gif
      (1.7)

      Then system (1.2) has a sequence of distinct periodic solutions with period k j T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq17_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq18_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif.

      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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ8_HTML.gif
      (1.8)

      where p > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq9_HTML.gif. 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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ9_HTML.gif
      (1.9)

      Then system (1.8) has a sequence of distinct periodic solutions with period k j T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq17_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq18_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif.

      When A = m 2 ω 2 I N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq21_HTML.gif, where ω = 2 π / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq22_HTML.gif and I N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq23_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq24_HTML.gif, 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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equd_HTML.gif

      Then system (1.1) has a sequence of distinct periodic solutions with period k j T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq17_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq18_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif.

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq25_HTML.gif as its eigenvalues, where ω = 2 π / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq22_HTML.gif, l i N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq26_HTML.gif, i = 1 , , r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq27_HTML.gif and 0 r N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq28_HTML.gif. 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq29_HTML.gif 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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Eque_HTML.gif
         

      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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq30_HTML.gif and r N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq31_HTML.gif. Let λ i > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq32_HTML.gif ( i { 1 , , r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq33_HTML.gif) and λ i < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq34_HTML.gif ( i { r + s + 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq35_HTML.gif) 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq36_HTML.gif for all l N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq37_HTML.gif, where ω = 2 π / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq22_HTML.gif, that is, λ i l 2 ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq38_HTML.gif ( i = 1 , , r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq39_HTML.gif) for all l N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq40_HTML.gif.

      The Assumption (A0) implies that one can find l i Z + : = { 0 , 1 , 2 , } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq41_HTML.gif such that
      l i 2 ω 2 < λ i < ( l i + 1 ) 2 ω 2 , i = 1 , , r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ10_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equf_HTML.gif
      Then
      i + , i { 1 , , r } , i + , i { r + s + 1 , , N } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equg_HTML.gif
      Corresponding to (1.10), we know that there exist l i + , l i Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq42_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equh_HTML.gif
      Moreover, set
      h i = ( l i + 1 ) 2 ω 2 λ i , i = 1 , , r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equi_HTML.gif
      and let h i 0 = min i { 1 , , r } { h i } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq43_HTML.gif. Then i 0 { 1 , , r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq44_HTML.gif. Corresponding to (1.10), there exists l i 0 Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq45_HTML.gif such that
      l i 0 2 ω 2 < λ i 0 < ( l i 0 + 1 ) 2 ω 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ11_HTML.gif
      (1.11)

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

      (H1) For some k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif, 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ12_HTML.gif
      (1.12)
      (H2) There exist positive constants m, ζ, η and ν [ 0 , 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq47_HTML.gif 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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equj_HTML.gif
      (H3) Assume that one of the following cases holds:
      1. (1)
        when r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq48_HTML.gif, s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and r + s = N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq50_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > min { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq52_HTML.gif such that
        F ( t , x ) β k | x | 2 , x R N , | x | > L k , a.e. t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ13_HTML.gif
        (1.13)
         
      where l i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq53_HTML.gif and λ i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq54_HTML.gif are defined by (1.11);
      1. (2)

        when r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq48_HTML.gif, s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and r + s < N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq55_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > min { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , ω 2 2 k 2 , λ i 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq56_HTML.gif such that (1.13) holds;

         
      2. (3)

        when r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq48_HTML.gif, s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq57_HTML.gif and r + s < N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq55_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > min { ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , λ i 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq58_HTML.gif such that (1.13) holds;

         
      3. (4)

        when r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq48_HTML.gif, s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq57_HTML.gif and r = N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq59_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > ( l i 0 + 1 ) 2 ω 2 λ i 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq60_HTML.gif such that (1.13) holds;

         
      4. (5)

        when r = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq61_HTML.gif, s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and s < N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq62_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > min { ω 2 2 k 2 , λ i 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq63_HTML.gif such that (1.13) holds;

         
      5. (6)

        when r = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq64_HTML.gif, s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq65_HTML.gif and q = N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq66_HTML.gif, there exist L k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq51_HTML.gif and β k > λ i 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq67_HTML.gif such that (1.13) holds;

         
      (H4) there exist l k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq68_HTML.gif and α k < σ k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq69_HTML.gif such that
      F ( t , x ) α k | x | 2 for all | x | l k and a.e. t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equk_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equl_HTML.gif

      where σ implies that σ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq70_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif satisfying (1.12) is finite. Let m K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq71_HTML.gif be the maximum integer satisfying (1.12), where
      K = { k N k  satisfies  ( 1.12 ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equm_HTML.gif

      Then K = { 1 , 2 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq72_HTML.gif. Hence, Theorem 1.1 implies that system (1.1) has nonzero kT-periodic solutions ( k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq73_HTML.gif). For cases (H3)(5) and (H3)(6), since r = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq64_HTML.gif, (1.12) holds for every k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif. Hence, Theorem 1.1 implies that system (1.11) has nonzero kT-periodic solutions for every k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif.

      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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ14_HTML.gif
      (1.14)
      They obtained that system (1.14) has a T = 2 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq74_HTML.gif 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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ15_HTML.gif
      (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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equn_HTML.gif
      where λ k 1 < λ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq75_HTML.gif are consecutive eigenvalues of the operator L = J d / d t A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq76_HTML.gif. Moreover, they also obtained system (1.14) has a nonzero T = 2 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq74_HTML.gif 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 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equo_HTML.gif

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq75_HTML.gif are consecutive eigenvalues of the operator L = J d / d t A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq76_HTML.gif or d 2 / d t 2 + A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq77_HTML.gif. 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq78_HTML.gif ( i { 1 , , r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq79_HTML.gif) 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq64_HTML.gif, s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and s < N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq62_HTML.gif, there exist L > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq13_HTML.gif and β > ω 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq80_HTML.gif such that
      F ( t , x ) β | x | 2 , x R N , | x | > L , a.e. t [ 0 , T ] ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ16_HTML.gif
      (1.16)
      (H4)′
      lim | x | 0 F ( t , x ) | x | 2 = 0 uniformly for a.e. t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equp_HTML.gif

      Then system (1.1) has a sequence of distinct periodic solutions with period k j T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq81_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq18_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif.

      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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq82_HTML.gif 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 ] ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equq_HTML.gif
      (H6)
      lim | x | 0 F ( t , x ) | x | p = 0 < lim | x | F ( t , x ) | x | p uniformly for a.e. t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equr_HTML.gif

      Then system (1.8) has a sequence of distinct nonconstant periodic solutions with period k j T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq17_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq18_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif.

      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 ] ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equs_HTML.gif
      Then H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equt_HTML.gif
      and
      u = [ 0 k T | u ( t ) | 2 d t + 0 k T | u ˙ ( t ) | 2 d t ] 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equu_HTML.gif
      for each u , v H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq84_HTML.gif. Let
      u ¯ = 1 k T 0 k T u ( t ) d t and u ˜ ( t ) = u ( t ) u ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equv_HTML.gif
      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) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equw_HTML.gif

      (see Proposition 1.3 in [1]).

      Lemma 2.1 If u H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq85_HTML.gif, then
      u 12 + k 2 T 2 12 k T u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equx_HTML.gif

      where u = max t [ 0 , k T ] | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq86_HTML.gif.

      Proof Fix t [ 0 , k T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq87_HTML.gif. For every τ [ 0 , k T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq88_HTML.gif, we have
      u ( t ) = u ( τ ) + τ t u ˙ ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ17_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equy_HTML.gif
      Integrating (2.1) over [ t k T / 2 , t + k T / 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq89_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equz_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaa_HTML.gif

      The proof is complete. □

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

      Assume that F = F ( t , x ) : R × R N R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq90_HTML.gif is T-periodic in t, F ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq3_HTML.gif is measurable in t for every x R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq4_HTML.gif and continuously differentiable in x for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq5_HTML.gif. If there exist a C ( R + , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq91_HTML.gif and b L p ( [ 0 , T ] , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq92_HTML.gif ( p > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq9_HTML.gif) such that
      | F ( t , x ) | a ( | x | ) b ( t ) , x R N , a.e. t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ18_HTML.gif
      (2.2)
      then
      c ( u ) = 0 k T F ( t , u ( t ) ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equab_HTML.gif

      is weakly continuous and uniformly differentiable on bounded subsets of H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif.

      Remark 2.1 In [[17], Lemma 2.2], F C 1 ( R , R N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq93_HTML.gif. 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq94_HTML.gif and E 2 = E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq95_HTML.gif. Suppose that φ C 1 ( E , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq96_HTML.gif satisfies (PS)-condition, and

      (I1) φ ( u ) = 1 / 2 ( Φ u , u ) + b ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq97_HTML.gif, where Φ u = Φ 1 P 1 u + Φ 2 P 2 u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq98_HTML.gif and Φ i : E i E i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq99_HTML.gif bounded and self-adjoint, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq100_HTML.gif;

      (I2) b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq101_HTML.gif is compact, and

      (I3) there exists a subspace E ˜ E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq102_HTML.gif and sets S E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq103_HTML.gif, Q E ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq104_HTML.gif and constants α > β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq105_HTML.gif such that
      1. (i)

        S E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq106_HTML.gif and φ | S α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq107_HTML.gif,

         
      2. (ii)

        Q is bounded and φ | Q β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq108_HTML.gif,

         
      3. (iii)

        S and ∂Q link.

         
      Then φ possesses a critical value c α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq109_HTML.gif which can be characterized as
      c = inf h Γ sup u Q φ ( h ( 1 , u ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equac_HTML.gif
      where
      Γ { h C ( [ 0 , 1 ] × E , E ) | h satisfies the following ( Γ 1 ) - ( Γ 3 ) } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equad_HTML.gif

      ( Γ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq110_HTML.gif h ( 0 , u ) = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq111_HTML.gif,

      ( Γ 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq112_HTML.gif h ( t , u ) = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq113_HTML.gif for u Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq114_HTML.gif, and

      ( Γ 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq115_HTML.gif h ( t , u ) = e θ ( t , u ) Φ u + K ( t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq116_HTML.gif, where θ C ( [ 0 , 1 ] × E , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq117_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq118_HTML.gif, { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq119_HTML.gif has a convergent subsequence if φ ( u n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq120_HTML.gif is bounded and ( 1 + u n ) φ ( u n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq121_HTML.gif as n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq122_HTML.gif.

      3 Proofs of theorems

      Proof of Theorem 1.1 It follows from Assumption (A)′ that the functional φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif on H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equae_HTML.gif
      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaf_HTML.gif

      for u , v H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq84_HTML.gif and the solutions of system (1.1) correspond to the critical points of φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ19_HTML.gif
      (3.1)
      Let u = Q w http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq124_HTML.gif. Then by (1.1),
      Q w ¨ ( t ) + A Q w ( t ) + F ( t , Q w ( t ) ) = 0 a.e.  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equag_HTML.gif
      Furthermore
      w ¨ ( t ) + Q 1 A Q w ( t ) + Q 1 F ( t , Q w ( t ) ) = 0 a.e.  t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equah_HTML.gif
      that is,
      w ¨ ( t ) + B w ( t ) + Q 1 F ( t , Q w ( t ) ) = 0 a.e.  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ20_HTML.gif
      (3.2)
      Let G ( t , w ) = F ( t , Q w ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq125_HTML.gif and then G ( t , w ) = Q 1 F ( t , Q w ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq126_HTML.gif. 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equai_HTML.gif

      Then the critical points of ψ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq127_HTML.gif correspond to solutions of system (3.2). It is easy to verify that φ k ( u ) = ψ k ( w ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq128_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq129_HTML.gif if and only if u = Q w http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq124_HTML.gif is the critical point of φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif. Therefore, we only need to consider the special case that A = B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq130_HTML.gif is the diagonal matrix defined by (3.1). We divide the proof into six steps.

      Step 1: Decompose the space H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif. Let
      I N = ( 1 1 1 ) = ( e 1 , e 2 , , e N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaj_HTML.gif
      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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equak_HTML.gif
      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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equal_HTML.gif
      Then H k T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq131_HTML.gif, H k T 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq132_HTML.gif and H k T + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq133_HTML.gif are closed subsets of H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif and
      1. (1)
        H k T 1 = H k T H k T 0 H k T + ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equam_HTML.gif
         
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equan_HTML.gif
         
      where
      P k ( u , v ) = 0 k T [ ( u ˙ ( t ) , v ˙ ( t ) ) ( Au ( t ) , v ( t ) ) ] d t , u , v H k T 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equao_HTML.gif
      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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equap_HTML.gif
      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaq_HTML.gif
      and
      P k ( u , v ) = 0 , u H k T + 1 , v H k T + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equar_HTML.gif

      Remark 3.1 When k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq134_HTML.gif, it is easy to see H T + 1 = { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq135_HTML.gif.

      Step 2: Let
      q k ( u ) = 1 2 0 k T [ | u ˙ ( t ) | 2 ( Au ( t ) , u ( t ) ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equas_HTML.gif
      Next we consider the relationship between q k ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq136_HTML.gif and u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq137_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq138_HTML.gif, 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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equat_HTML.gif
         
      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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equau_HTML.gif
      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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equav_HTML.gif
      Let
      δ = min i { 1 , , r } { λ i ( l i ω ) 2 ( l i ω ) 2 + 1 } > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaw_HTML.gif
      Then
      q k ( u ) δ 2 u 2 , u H k T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ21_HTML.gif
      (3.3)
      Remark 3.2 Obviously, if one of (H3)(5) and (H3)(6) holds, then H k T = { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq139_HTML.gif. Hence,
      q k ( u ) = 0 , u H k T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equax_HTML.gif
      1. (b)
        For u H k T + 2 H k T 02 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq140_HTML.gif, let
        u = u ( t ) = u 1 ( t ) + u 2 ( t ) + u 3 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equay_HTML.gif
         
      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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equaz_HTML.gif
      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 ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equba_HTML.gif
      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 ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbb_HTML.gif
      Since for fixed i { 1 , , r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq33_HTML.gif,
      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbc_HTML.gif
      are strictly increasing on j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq141_HTML.gif,
      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbd_HTML.gif
      and
      g ( j ) g ( 1 ) = ( k 1 ω ) 2 ( k 1 ω ) 2 + 1 = ω 2 ω 2 + k 2 > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Eqube_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbf_HTML.gif
      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 + } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbg_HTML.gif
      Then
      q k ( u ) σ k 2 u 2 , u H k T + 2 H k T 02 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ22_HTML.gif
      (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 } ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbh_HTML.gif
         
      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 + } ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbi_HTML.gif
         
      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 + } ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbj_HTML.gif
         
      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 } ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbk_HTML.gif
         
      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 + } ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbl_HTML.gif
         
      6. (vi)
        if (H3)(6) holds, then (3.4) holds with
        σ k σ = λ i 1 + λ i + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbm_HTML.gif
         
      1. (c)
        For u H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq142_HTML.gif, 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbn_HTML.gif
         
      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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbo_HTML.gif
      Obviously, when k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq134_HTML.gif, u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq143_HTML.gif. So q 1 ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq144_HTML.gif. When k > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq145_HTML.gif, it follows from
      ( l i + 1 1 k ) 2 ω 2 λ i < ( l i + 1 ) 2 ω 2 , i { 1 , , r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbp_HTML.gif
      that
      q k ( u ) 0 , u H k T + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ23_HTML.gif
      (3.5)
      1. (d)
        Obviously, for u H k T 01 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq146_HTML.gif, we have
        q k ( u ) = 0 , u H k T 01 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ24_HTML.gif
        (3.6)
         
      Step 3: Assume that (H3)(2) holds. We prove that there exist ρ k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq147_HTML.gif and b k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq148_HTML.gif such that
      φ k ( u ) b k > 0 , u ( H k T + 2 H k T 02 ) B ρ k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbq_HTML.gif
      Let
      C k = 12 + k 2 T 2 12 k T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbr_HTML.gif
      Choosing ρ k = min { 1 , l k / C k } > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq149_HTML.gif and b k = ( σ k 2 α k ) ρ k 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq150_HTML.gif, by Lemma 2.1, (H4) and (3.4), we have, for all u ( H k T + 2 H k T 02 ) B ρ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq151_HTML.gif,
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbs_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbt_HTML.gif
      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 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbu_HTML.gif

      where h k H k T + 2 H k T 02 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq152_HTML.gif, s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq153_HTML.gif and s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq154_HTML.gif will be determined later. In this step, we prove φ k | Q k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq155_HTML.gif. 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbv_HTML.gif
      Case (i): if
      d k : = d = ( l i 0 + 1 ) 2 ω 2 λ i 0 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbw_HTML.gif
      then we choose
      h k ( t ) = sin ( l i 0 + 1 ) ω t e i 0 , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbx_HTML.gif
      Obviously, h k H k T + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq156_HTML.gif and h ˙ k ( t ) = ( l i 0 + 1 ) ω cos ( l i 0 + 1 ) ω t e i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq157_HTML.gif, t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq158_HTML.gif. Then
      h k L 2 2 = k T 2 , h ˙ k L 2 2 = k T ( l i 0 + 1 ) 2 ω 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equby_HTML.gif
      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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ25_HTML.gif
      (3.7)
      where ε 0 k = β k d > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq159_HTML.gif and L ˆ k > max { 1 , L k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq160_HTML.gif. Since H k T H k T 01 H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq161_HTML.gif is the finite dimensional space, there exists a constant K 1 k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq162_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ26_HTML.gif
      (3.8)
      By (3.3), (3.5), (3.6), (3.7) and (3.8), we know that for all s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and u = u + u 01 + u + 1 H k T H k T 01 H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq163_HTML.gif,
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ27_HTML.gif
      (3.9)
      Hence,
      φ k ( s h k + u ) 0 , either  s s 1  or  u s 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equbz_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equca_HTML.gif
      Case (ii): if d k = ω 2 / ( 2 k 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq164_HTML.gif, then we choose
      h k ( t ) = sin k 1 ω t e r + 1 H k T 02 , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcb_HTML.gif
      Then
      h ˙ k ( t ) = ω k cos k 1 ω t e r + 1 , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcc_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ28_HTML.gif
      (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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ29_HTML.gif
      (3.11)
      where L ˆ k > max { 1 , L k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq160_HTML.gif and ε 0 k = β k ω 2 2 k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq165_HTML.gif. By (3.3), (3.5), (3.6), (3.8) and (3.11), we know that for all s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and u = u + u 01 + u + 1 H k T H k T 01 H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq163_HTML.gif,
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcd_HTML.gif
      Hence,
      φ k ( s h k + u ) 0 , either  s s 1  or  u s 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equce_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcf_HTML.gif
      Case (iii): if d k = λ i / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq166_HTML.gif, then we choose
      h k = 1 k T e i H k T + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcg_HTML.gif
      Then
      h ˙ k = 0 , ( A h k , h k ) = λ i ( h k , h k ) , h k L 2 2 = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equch_HTML.gif
      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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ30_HTML.gif
      (3.12)
      where L ˆ k > max { 1 + 1 T , L k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq167_HTML.gif and ε 0 k = β k λ i / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq168_HTML.gif. By (3.3), (3.5), (3.6), (3.8) and (3.12), for all s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and u = u + u 01 + u + 1 H k T H k T 01 H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq163_HTML.gif, 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equci_HTML.gif
      Hence,
      φ k ( s e k + u ) 0 , either  s s 1  or  u s 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcj_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equck_HTML.gif
      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 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcl_HTML.gif
      then
      φ k ( s h k + u ) 0 , either  s s 1  or  u s 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ31_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq169_HTML.gif, 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ32_HTML.gif
      (3.14)

      Thus, it follows from (3.13) and (3.14) that φ | Q k 0 < b k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq170_HTML.gif.

      Step 5: We prove that φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif satisfies (C)-condition in H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif. The proof is similar to that in Theorem 1.1 in [17]. We omit it.

      Step 6: We claim that φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif has a nontrivial critical point u k H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq171_HTML.gif such that φ k ( u k ) b k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq172_HTML.gif. Especially, we claim that, for cases (H3)(1) and (H3)(4), since A is a positive semidefinite matrix, (1.5) implies that u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq173_HTML.gif 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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcm_HTML.gif
      where K : H k T 1 H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq174_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcn_HTML.gif
      The compact imbedding of H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif into C ( [ 0 , k T ] ; R N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq175_HTML.gif implies that K is compact. In order to use Lemma 2.3, we let Φ = I K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq176_HTML.gif and define Φ i : E i E i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq99_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq100_HTML.gif by
      Φ i u , v = ( I K ) u , v , u , v E i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equco_HTML.gif
      where E 1 = H k T + 2 H k T 02 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq177_HTML.gif and E 2 = H k T H k T 01 H k T + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq178_HTML.gif. Since K is a self-adjoint compact operator, it is easy to see that Φ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq179_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq100_HTML.gif) are bounded and self-adjoint. Let
      b ( u ) = 0 k T F ( t , u ( t ) ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcp_HTML.gif

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq180_HTML.gif. Furthermore, by standard theorems in [22], we conclude that b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq101_HTML.gif is compact. Let S k = ( H k T + 2 H k T 02 ) B ρ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq181_HTML.gif. Then S k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq182_HTML.gif and Q k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq183_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq171_HTML.gif such that φ k ( u k ) b k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq172_HTML.gif, which implies that u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq173_HTML.gif is nonzero. For cases (H3)(1) and (H3)(4), since A is a positive semidefinite matrix, it follows from (1.5) that u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq173_HTML.gif is nonconstant. The proof is complete.  □

      Proof of Theorem 1.2 Obviously, when r = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq64_HTML.gif, s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq49_HTML.gif and s < N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq62_HTML.gif, (H1) holds for any k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif. Moreover, since (H3)′ implies that (H3)(5) and (H4)′ implies that (H4), system (1.1) has kT-periodic solution for every k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif.

      Let d = ω 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq184_HTML.gif. 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcq_HTML.gif
      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 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ33_HTML.gif
      (3.15)
      where ε 1 = β ω 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq185_HTML.gif. 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcr_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcs_HTML.gif
      Note that s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq153_HTML.gif is independent of k. Hence, if u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq173_HTML.gif is the critical point of φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq123_HTML.gif, then it follows from (3.3), (3.5), (3.6), the definitions of critical value c in Lemma 2.3 and Q k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq186_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equ34_HTML.gif
      (3.16)

      Hence, φ k ( u k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq187_HTML.gif is bounded for any k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq46_HTML.gif.

      Obviously, we can find k 1 N / { 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq188_HTML.gif such that k 1 > M b 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq189_HTML.gif, then we claim that u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq173_HTML.gif is distinct from u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq190_HTML.gif for all k k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq191_HTML.gif. In fact, if u k = u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq192_HTML.gif for some k k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq191_HTML.gif, it is easy to check that
      φ k ( u k ) = k φ 1 ( u 1 ) k b 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equct_HTML.gif

      Then by (3.16), we have k 1 k M b 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq193_HTML.gif, a contradiction. We also can find k 2 > max { k 1 , k 1 M b k 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq194_HTML.gif such that u k 1 k u k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq195_HTML.gif for all k k 2 k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq196_HTML.gif. Otherwise, if u k 1 k = u k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq197_HTML.gif for some k k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq191_HTML.gif, we have φ k 1 k ( u k 1 k ) = k φ k 1 ( u k 1 ) k b k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq198_HTML.gif. Then by (3.16), we have k 2 k 1 k M b k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq199_HTML.gif, 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq81_HTML.gif satisfying k j N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq200_HTML.gif and k j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq19_HTML.gif as j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq20_HTML.gif. 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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq201_HTML.gif is bounded if φ ( u n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq120_HTML.gif is bounded and φ ( u n ) ( 1 + u n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq202_HTML.gif as n + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq203_HTML.gif. Other proofs are the same as in [15]. The proof of boundedness of { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq201_HTML.gif is essentially the same as in Theorem 1.1 in [17] except that 2 is replaced by p, H k T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq83_HTML.gif 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 ] ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcu_HTML.gif
      equipped with the norm
      u = ( 0 k T | u ( t ) | p d t + 0 k T | u ˙ ( t ) | p d t ) 1 / p , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcv_HTML.gif
      and
      F ( t , x ) β k | x | 2 , x R N , | x | > L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcw_HTML.gif
      by
      F ( t , x ) ε | x | p , x R N , | x | > L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_Equcx_HTML.gif

      for some ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-139/MediaObjects/13661_2012_Article_393_IEq204_HTML.gif. So, we omit the details. □

      4 Examples

      Example 4.1 Let T = 2 π