Periodic solutions for nonautonomous second order Hamiltonian systems with sublinear nonlinearity

  • Zhiyong Wang1Email author and

    Affiliated with

    • Jihui Zhang2

      Affiliated with

      Boundary Value Problems20112011:23

      DOI: 10.1186/1687-2770-2011-23

      Received: 14 May 2011

      Accepted: 13 September 2011

      Published: 13 September 2011

      Abstract

      Some existence and multiplicity of periodic solutions are obtained for nonautonomous second order Hamiltonian systems with sublinear nonlinearity by using the least action principle and minimax methods in critical point theory.

      Mathematics Subject Classification (2000): 34C25, 37J45, 58E50.

      Keywords

      Control function Periodic solutions The least action principle Minimax methods

      1 Introduction and main results

      Consider the second order systems
      { u ¨ ( t ) = F ( t , u ( t ) ) a .e t [ 0 , T ] , u ( 0 ) u ( T ) = u ˙ ( 0 ) u ˙ ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ1_HTML.gif
      (1.1)
      where T > 0 and F : [0, T] × ℝ → ℝ satisfies the following assumption:
      1. (A)
        F (t, x) is measurable in t for every x ∈ ℝ and continuously differentiable in x for a.e. t ∈ [0, T], and there exist aC(ℝ+, ℝ+), bL1(0, T ; ℝ+) 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%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equa_HTML.gif
         

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

      The existence of periodic solutions for problem (1.1) has been studied extensively, a lot of existence and multiplicity results have been obtained, we refer the readers to [113] and the reference therein. In particular, under the assumptions that the nonlinearity ∇F (t, x) is bounded, that is, there exists p(t) ∈ L1(0, T ; ℝ+) such that
      | F ( t , x ) | p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ2_HTML.gif
      (1.2)
      for all x ∈ ℝ and a.e. t ∈ [0, T], and that
      0 T F ( t , x ) d t a s | x | + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ3_HTML.gif
      (1.3)
      Mawhin and Willem in [3] have proved that problem (1.1) admitted a periodic solution. After that, when the nonlinearity ∇F (t, x) is sublinear, that is, there exists f(t), g(t) ∈ L1(0, T ; ℝ+) and α ∈ [0, 1) such that
      | F ( t , x ) | f ( t ) | x | α + g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ4_HTML.gif
      (1.4)
      for all x ∈ ℝ and a.e. t ∈ [0, T], Tang in [7] have generalized the above results under the hypotheses
      1 | x | 2 α 0 T F ( t , x ) d t a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ5_HTML.gif
      (1.5)
      Subsequently, Meng and Tang in [13] further improved condition (1.5) with α ∈ (0, 1) by using the following assumptions
      liminf | x | + 1 | x | 2 α 0 T F ( t , x ) d t > T 2 4 0 T f ( t ) d t 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ6_HTML.gif
      (1.6)
      limsup | x | + 1 | x | 2 α 0 T F ( t , x ) d t < - T 8 0 T f ( t ) d t 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ7_HTML.gif
      (1.7)
      Recently, authors in [14] investigated the existence of periodic solutions for the second order nonautonomous Hamiltonian systems with p-Laplacian, here p > 1, it is assumed that the nonlinearity ∇F (t, x) may grow slightly slower than |x|p-1, a typical example with p = 2 is
      F ( t , x ) = t | x | ln ( 1 0 0 + | x | 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ8_HTML.gif
      (1.8)
      solutions are found as saddle points to the corresponding action functional. Furthermore, authors in [12] have extended the ideas of [14], replacing in assumptions (1.4) and (1.5) the term |x| with a more general function h(|x|), which generalized the results of [3, 7, 10, 11]. Concretely speaking, it is assumed that there exist f(t), g(t) ∈ L1(0, T; ℝ+) and a nonnegative function hC([0, +∞), [0, +∞)) such that
      | F ( t , x ) | f ( t ) h ( | x | ) + g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equb_HTML.gif
      for all x ∈ ℝ and a.e. t ∈ [0, T], and that
      1 h 2 ( | x | ) 0 T F ( t , x ) d t a s | x | + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equc_HTML.gif
      where h be a control function with the properties:
      ( a ) h ( s ) h ( t ) ( b ) h ( s + t ) C * ( h ( s ) + h ( t ) ) ( c ) 0 h ( t ) K 1 t α + K 2 ( d ) h ( t ) + s t , s , t [ 0 , + ) , s , t [ 0 , + ) , t [ 0 , + ) , a s t + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equd_HTML.gif

      if α = 0, h(t) only need to satisfy conditions (a)-(c), here C*, K1 and K2 are positive constants. Moreover, α ∈ [0, 1) is posed. Under these assumptions, periodic solutions of problem (1.1) are obtained. In addition, if the nonlinearity ∇F (t, x) grows more faster at infinity with the rate like | x | ln ( 1 0 0 + | x | 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq1_HTML.gif, f(t) satisfies some certain restrictions and α is required in a more wider range, say, α ∈ [0,1], periodic solutions have also been established in [12] by minimax methods.

      An interesting question naturally arises: Is it possible to handle both the case such as (1.8) and some cases like (1.4), (1.5), in which only f(t) ∈ L1(0, T ; ℝ+) and α ∈ [0, 1)? In this paper, we will focus on this problem.

      We now state our main results.

      Theorem 1.1. Suppose that F satisfies assumption (A) and the following conditions:

      (S1) There exist constants C ≥ 0, C* > 0 and a positive function hC(ℝ+, ℝ+) with the properties:
      ( i ) h ( s ) h ( t ) + C ( i i ) h ( s + t ) C * ( h ( s ) + h ( t ) ) ( i i i ) t h ( t ) - 2 H ( t ) - ( i v ) H ( t ) t 2 0 s t , s , t + , s , t + , a s t + , a s t + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Eque_HTML.gif
      where H ( t ) : = 0 t h ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq2_HTML.gif. Moreover, there exist fL1(0, T; ℝ+) and gL1(0, T; ℝ+) such that
      | F ( t , x ) | f ( t ) h ( | x | ) + g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equf_HTML.gif

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

      (S2) There exists a positive function hC(ℝ+, ℝ+) which satisfies the conditions (i)-(iv) and
      liminf | x | + 1 H ( | x | ) 0 T F ( t , x ) d t > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equg_HTML.gif
      Then, problem (1.1) has at least one solution which minimizes the functional φ given by
      φ ( u ) : = 1 2 0 T | u ( t ) | 2 d t + 0 T [ F ( t , u ( t ) ) - F ( t , 0 ) ] d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equh_HTML.gif
      on the Hilbert space H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif defined by
      H T 1 : = u : [ 0 , T ] | u  is absolutely  continuous  , u ( 0 ) = u ( T ) , u L 2 ( 0 , T ; ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equi_HTML.gif
      with the norm
      | | u | | : = 0 T | u ( t ) | 2 d t + 0 T | u ( t ) | 2 d t 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equj_HTML.gif
      Theorem 1.2. Suppose that (S1) and assumption (A) hold. Assume that
      ( S 3 ) limsup | x | + 1 H ( | x | ) 0 T F ( t , x ) d t < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equk_HTML.gif

      Then, problem (1.1) has at least one solution in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Theorem 1.3. Suppose that (S1), (S3) and assumption (A) hold. Assume that there exist δ > 0, ε > 0 and an integer k > 0 such that
      - 1 2 ( k + 1 ) 2 ω 2 | x | 2 F ( t , x ) - F ( t , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ9_HTML.gif
      (1.9)
      for all x ∈ ℝ and a.e. t ∈ [0, T], and
      F ( t , x ) - F ( t , 0 ) - 1 2 k 2 ω 2 ( 1 + ε ) | x | 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ10_HTML.gif
      (1.10)

      for all |x| ≤ δ and a.e. t ∈ [0, T], where ω = 2 π T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq4_HTML.gif. Then, problem(1.1) has at least two distinct solutions in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Theorem 1.4. Suppose that (S1), (S2) and assumption (A) hold. Assume that there exist δ > 0, ε > 0 and an integer k ≥ 0 such that
      - 1 2 ( k + 1 ) 2 ω 2 | x | 2 F ( t , x ) - F ( t , 0 ) - 1 2 k 2 ω 2 | x | 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ11_HTML.gif
      (1.11)

      for all |x| ≤ δ and a.e. t ∈ [0, T]. Then, problem (1.1) has at least three distinct solutions in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Remark 1.1.
      1. (i)

        Let α ∈ [0, 1), in Theorems 1.1-1.4, ∇F(t, x) does not need to be controlled by |x|2αat infinity; in particular, we can not only deal with the case in which ∇F(t, x) grows slightly faster than |x|2αat infinity, such as the example (1.8), but also we can treat the cases like (1.4), (1.5).

         
      2. (ii)

        Compared with [12], we remove the restriction on the function f(t) as well as the restriction on the range of α ∈ [0, 1] when we are concerned with the cases like (1.8).

         
      3. (iii)

        Here, we point out that introducing the control function h(t) has also been used in [12, 14], however, these control functions are different from ours because of the distinct characters of h(t).

         

      Remark 1.2. From (i) of (S1), we see that, nonincreasing control functions h(t) can be permitted. With respect to the detailed example on this assertion, one can see Example 4.3 of Section 4.

      Remark 1.3. There are functions F(t, x) satisfying our theorems and not satisfying the results in [114]. For example, consider function
      F ( t , x ) = f ( t ) | x | 2 ln ( 1 0 0 + | x | 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equl_HTML.gif
      where f(t) ∈ L1(0, T; ℝ+) and f(t) > 0 for a.e. t ∈ [0, T]. It is apparent that
      | F ( t , x ) | 4 f ( t ) | x | ln ( 1 0 0 + | x | 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ12_HTML.gif
      (1.12)

      for all x ∈ ℝ and t ∈ [0, T]. (1.12) shows that (1.4) does not hold for any α ∈ [0, 1), moreover, note f(t) only belongs to L1(0, T; ℝ+) and no further requirements on the upper bound of 0 T f ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq5_HTML.gif are posed, then the approach of [12] cannot be repeated. This example cannot be solved by earlier results, such as [113].

      On the other hand, take h ( t ) = t ln ( 1 0 0 + t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq6_HTML.gif, H ( t ) = 0 t s ln ( 1 0 0 + s 2 ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq7_HTML.gif, C = 0, C* = 1, then by simple computation, one has
      ( i ) h ( s ) h ( t ) s t , s , t + , ( i i ) h ( s + t ) = s + t ln ( 1 0 0 + ( s + t ) 2 ) h ( s ) + h ( t ) s , t + , ( i i i ) t h ( t ) - 2 H ( t ) = t 2 ln ( 1 0 0 + t 2 ) - 2 0 t 1 ln ( 1 0 0 + s 2 ) d 1 2 s 2 = - 0 t 2 s 3 ( 1 0 0 + s 2 ) ln 2 ( 1 0 0 + s 2 ) d s - a s t + , ( i v ) H ( t ) t 2 = 0 t s ln ( 1 0 0 + s 2 ) d s t 2 0 a s t + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equm_HTML.gif
      and
      1 H ( | x | ) 0 T F ( t , x ) d t = 2 0 T f ( t ) d t > 0 a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equn_HTML.gif

      Hence, (S1) and (S2) are hold, by Theorem 1.1, problem (1.1) has at least one solution which minimizes the functional φ in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      What's more, Theorem 1.1 can also deal with some cases which satisfy the conditions (1.4) and (1.5). For instance, consider function
      F ( t , x ) = ( 0 . 6 T - t ) | x | 3 2 + ( q ( t ) , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equo_HTML.gif
      where q(t) ∈ L1(0, T; ℝ). It is not difficult to see that
      | F ( t , x ) | 3 2 | 0 . 6 T - t | | x | 1 2 + | q ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equp_HTML.gif

      for all x ∈ ℝ and a.e. t ∈ [0, T]. Choose h ( t ) = t 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq8_HTML.gif, H ( t ) = 2 3 t 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq9_HTML.gif, C = 0, C* = 1, f ( t ) = 3 2 | 0 . 6 T - t | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq10_HTML.gif and g(t) = |q(t)|, then (S1) and (S2) hold, by Theorem 1.1, problem (1.1) has at least one solution which minimizes the functional φ in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif. However, we can find that the results of [14] cannot cover this case. More examples are drawn in Section 4.

      Our paper is organized as follows. In Section 2, we collect some notations and give a result regrading properties of control function h(t). In Section 3, we are devote to the proofs of main theorems. Finally, we will give some examples to illustrate our results in Section 4.

      2 Preliminaries

      For u H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq11_HTML.gif, let ū : = 1 T 0 T u ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq12_HTML.gif and ũ ( t ) : = u ( t ) - ū http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq13_HTML.gif, then one has
      | | ũ | | 2 T 1 2 0 T | u ( t ) | 2 d t ( S o b o l e v s i n e q u a l i t y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equq_HTML.gif
      and
      0 T | ũ ( t ) | 2 d t T 2 4 π 2 0 T | u ( t ) | 2 d t ( W i r t i n g e r s i n e q u a l i t y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equr_HTML.gif

      where | | ũ | | : = max 0 t T | ũ ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq14_HTML.gif.

      It follows from assumption (A) that the corresponding function φ on H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif given by
      φ ( u ) : = 1 2 0 T | u ( t ) | 2 d t + 0 T [ F ( t , u ( t ) ) - F ( t , 0 ) ] d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equs_HTML.gif
      is continuously differentiable and weakly lower semi-continuous on H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif (see[2]). Moreover, one has
      ( φ ( u ) , v ) = 0 T ( u ( t ) , v ( t ) ) d t + 0 T ( F ( t , u ( t ) ) , v ( t ) ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equt_HTML.gif

      for all u , v H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq15_HTML.gif. It is well known that the solutions of problem (1.1) correspond to the critical point of φ.

      In order to prove our main theorems, we prepare the following auxiliary result, which will be used frequently later on.

      Lemma 2.1. Suppose that there exists a positive function h which satisfies the conditions (i), (iii), (iv) of (S1), then we have the following estimates:
      ( 1 ) 0 < h ( t ) ε t + C 0 ( 2 ) h 2 ( t ) H ( t ) 0 ( 3 ) H ( t ) + f o r a n y ε > 0 , C 0 > 0 , t + , a s t + , a s t + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equu_HTML.gif
      Proof. It follows from (iv) of (S1) that, for any ε > 0, there exists M1 > 0 such that
      H ( t ) ε t 2 t M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ13_HTML.gif
      (2.1)
      By (iii) of (S1), there exists M2 > 0 such that
      t h ( t ) - 2 H ( t ) 0 t M 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ14_HTML.gif
      (2.2)
      which implies that
      h ( t ) 2 H ( t ) t ε t t M , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ15_HTML.gif
      (2.3)
      where M := max{M1, M2}. Hence, we obtain
      h ( t ) ε t + h ( M ) + C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ16_HTML.gif
      (2.4)

      for all t > 0 by (i) of (S1). Obviously, h(t) satisfies (1) due to the definition of h(t) and (2.4).

      Next, we come to check condition (2). Recalling the property (iv) of (S1) and (2.2), we get
      0 < h 2 ( t ) H ( t ) = h 2 ( t ) H 2 ( t ) H ( t ) 2 t 2 H ( t ) = 4 H ( t ) t 2 0 a s t + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equv_HTML.gif

      Therefore, condition (2) holds.

      Finally, we show that (3) is also true. By (iii) of (S1), one arrives at, for every β > 0, there exists M3 > 0 such that
      t h ( t ) - 2 H ( t ) - 2 β t M 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ17_HTML.gif
      (2.5)
      Let θ ≥ 1, using (2.5) and integrating the relation
      d d θ H ( θ t ) θ 2 = θ t h ( θ t ) - 2 H ( θ t ) θ 3 - 2 β θ 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equw_HTML.gif
      over an interval [1, S] ⊂ [1, +∞), we obtain
      H ( S t ) S 2 - H ( t ) β 1 S 2 - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equx_HTML.gif
      Thus, since lim S + H ( S t ) S 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq16_HTML.gif by (iv) of (S1), one has
      H ( t ) β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equy_HTML.gif
      for all tM3. That is,
      H ( t ) + a s t + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equz_HTML.gif

      which completes the proof. □

      3 Proof of main results

      For the sake of convenience, we will denote various positive constants as C i , i = 1, 2, 3,.... Now, we are ready to proof our main results.

      Proof of Theorem 1.1. For u H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq17_HTML.gif, it follows from (S1), Lemma 2.1 and Sobolev's inequality that
      0 T [ F ( t , u ( t ) ) - F ( t , ū ) ] d t = 0 T 0 1 ( F ( t , ū + s ũ ( t ) ) , ũ ( t ) ) d s d t 0 T 0 1 f ( t ) h ( | ū + s ũ ( t ) | ) | ũ ( t ) | d s d t + 0 T 0 1 g ( t ) | ũ ( t ) | d s d t 0 T 0 1 f ( t ) h ( | ū | + | ũ ( t ) | ) + C | ũ ( t ) | d s d t + | | ũ | | 0 T g ( t ) d t 0 T 0 1 f ( t ) C * h ( | ū | ) + h ( | ũ ( t ) | ) + C | ũ ( t ) | d s d t + | | ũ | | 0 T g ( t ) d t C * [ h ( | ū | ) + h ( | ũ ( t ) | ) ] | | ũ | | 0 T f ( t ) d t + C | | ũ | | 0 T f ( t ) d t + | | ũ | | 0 T g ( t ) d t C * 3 C * T | | ũ | | 2 + C * T 3 h 2 ( | ū | ) 0 T f ( t ) d t 2 + | | ũ | | 0 T g ( t ) d t + C * [ h ( | | ũ | | ) + C ] | | ũ | | 0 T f ( t ) d t + C | | ũ | | 0 T f ( t ) d t 1 4 0 T | u ( t ) | 2 d t + C 1 h 2 ( | ū | ) + C * ε | | ũ | | + C 0 + C | | ũ | | 0 T f ( t ) d t + C | | ũ | | 0 T f ( t ) d t + | | ũ | | 0 T g ( t ) d t 1 4 + ε C 2 0 T | u ( t ) | 2 d t + C 1 h 2 ( | ū | ) + C 3 0 T | u ( t ) | 2 d t 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ18_HTML.gif
      (3.1)
      which implies that
      φ ( u ) = 1 2 0 T | u ( t ) | 2 d t + 0 T [ F ( t , u ( t ) ) - F ( t , ū ) ] d t + 0 T F ( t , ū ) d t - 0 T F ( t , 0 ) d t 1 4 - ε C 2 0 T | u ( t ) | 2 d t - C 3 0 T | u ( t ) | 2 d t 1 2 - 0 T F ( t , 0 ) d t + H ( | ū | ) 1 H ( | ū | ) 0 T F ( t , ū ) d t - C 1 h 2 ( | ū | ) H ( | ū | ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ19_HTML.gif
      (3.2)
      Taking into account Lemma 2.1 and (S2), one has
      H ( | ū | ) 1 H ( | ū | ) 0 T F ( t , ū ) d t - C 1 h 2 ( | ū | ) H ( | ū | ) + a s | ū | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ20_HTML.gif
      (3.3)
      As ||u|| → +∞ if and only if | ū | 2 + 0 T | u ( t ) | 2 d t 1 2 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq18_HTML.gif, for ε small enough, (3.2) and (3.3) deduce that
      φ ( u ) + a s | | u | | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaa_HTML.gif

      Hence, by the least action principle, problem (1.1) has at least one solution which minimizes the function φ in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif. □

      Proof of Theorem 1.2. First, we prove that φ satisfies the (PS) condition. Suppose that { u n } H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq19_HTML.gif is a (PS) sequence of φ, that is, φ'(u n ) → 0 as n → +∞ and {φ(u n )} is bounded. In a way similar to the proof of Theorem 1.1, we have
      0 T ( F ( t , u n ( t ) ) , ũ n ( t ) ) d t 1 4 + ε C 2 0 T | u n ( t ) | 2 d t + C 1 h 2 ( | ū n | ) + C 3 0 T | u n ( t ) | 2 d t 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equab_HTML.gif
      for all n. Hence, we get
      | | ũ n | | ( φ ( u n ) , ũ n ) = 0 T | u n ( t ) | 2 d t + 0 T ( F ( t , u n ( t ) ) , ũ n ( t ) ) d t 3 4 - ε C 2 0 T | u n ( t ) | 2 d t - C 1 h 2 ( | ū n | ) - C 3 0 T | u n ( t ) | 2 d t 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ21_HTML.gif
      (3.4)
      for large n. On the other hand, it follows from Wirtinger's inequality that
      | | ũ n | | T 2 4 π 2 + 1 1 2 0 T | u n ( t ) | 2 d t 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ22_HTML.gif
      (3.5)
      for all n. Combining (3.4) with (3.5), we obtain
      C 4 h ( | ū n | ) 0 T | u n ( t ) | 2 d t 1 2 - C 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ23_HTML.gif
      (3.6)
      for all large n. By (3.1), (3.6), Lemma 2.1 and (S3), one has
      φ ( u n ) = 1 2 0 T | u n ( t ) | 2 d t + 0 T [ F ( t , u n ( t ) ) - F ( t , ū n ) ] d t + 0 T F ( t , ū n ) d t - 0 T F ( t , 0 ) d t 3 4 + ε C 2 0 T | u n ( t ) | 2 d t + C 1 h 2 ( | ū n | ) + C 3 0 T | u ( t ) | 2 d t 1 2 + 0 T F ( t , ū n ) d t - 0 T F ( t , 0 ) d t C 6 [ C 4 h ( | ū n | ) + C 5 ] 2 + C 1 h 2 ( | ū n | ) + C 3 [ C 4 h ( | ū n | ) + C 5 ] + 0 T F ( t , ū n ) d t - 0 T F ( t , 0 ) d t C 7 h 2 ( | ū n | ) + C 8 h ( | ū n | ) + C 9 + 0 T F ( t , ū n ) d t - 0 T F ( t , 0 ) d t H ( | ū n | ) C 7 h 2 ( | ū n | ) H ( | ū n | ) + C 8 h ( | ū n | ) H ( | ū n | ) + 1 H ( | ū n | ) 0 T F ( t , ū n ) d t + C 9 - 0 T F ( t , 0 ) d t - a s | ū n | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equac_HTML.gif

      This contradicts the boundedness of {φ(u n )}. So, { ū n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq20_HTML.gif is bounded. Notice (3.6) and (1) of Lemma 2.1, hence {u n } is bounded. Arguing then as in Proposition 4.1 in [3], we conclude that the (PS) condition is satisfied.

      In order to apply the saddle point theorem in [2, 3], we only need to verify the following conditions:

      (φ 1) φ(u) → +∞ as ||u|| → +∞in H ̃ T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq21_HTML.gif, where H ̃ T 1 : = u H T 1 | ū = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq22_HTML.gif,

      (φ 2) φ(u) → -∞ as |u(t)| → +∞.

      In fact, for all u H ̃ T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq23_HTML.gif, by (S1), Sobolev's inequality and Lemma 2.1, we have
      0 T [ F ( t , u ( t ) ) - F ( t , 0 ) ] d t = 0 T 0 1 ( F ( t , ū + s u ( t ) ) , u ( t ) ) d s d t 0 T f ( t ) h ( | s u ( t ) | ) | u ( t ) | d t + 0 T g ( t ) | u ( t ) | d t 0 T f ( t ) [ h ( | u ( t ) | ) + C ] | u ( t ) | d t + | | u | | 0 T g ( t ) d t ε | | u | | 2 0 T f ( t ) d t + ( C 0 + C ) | | u | | 0 T f ( t ) d t + | | u | | 0 T g ( t ) d t ε C 1 0 0 T | u ( t ) | 2 d t + C 1 1 0 T | u ( t ) | 2 d t 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equad_HTML.gif
      which implies that
      φ ( u ) = 1 2 0 T | u ( t ) | 2 d t + 0 T [ F ( t , u ( t ) ) - F ( t , 0 ) ] d t 1 2 - ε C 1 0 0 T | u ( t ) | 2 d t - C 1 1 0 T | u ( t ) | 2 d t 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equ24_HTML.gif
      (3.7)
      By Wirtinger's inequality, one has
      | | u | | + 0 T | u ( t ) | 2 d t 1 2 + o n H ̃ T 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equae_HTML.gif

      Hence, for ε small enough, (φ1) follows from (3.7).

      On the other hand, by (S3) and Lemma 2.1, we get
      0 T F ( t , u ( t ) ) d t - as  | u ( t ) | + in  , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaf_HTML.gif
      which implies that
      φ ( u ) = 0 T F ( t , u ( t ) ) d t - 0 T F ( t , 0 ) d t - as  | u ( t ) | + in  . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equag_HTML.gif

      Thus, (φ2) is verified. The proof of Theorem 1.2 is completed. □

      Proof of Theorem 1.3. Let E = H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq24_HTML.gif,
      H k : = j = 1 k ( a j cos j ω t + b j sin ω t ) | a j , b j , j = 1 , 2 , , k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equah_HTML.gif
      and ψ = -φ. Then, ψC1(E, ℝ) satisfies the (PS) condition by the proof of Theorem 1.2. In view of Theorem 5.29 and Example 5.26 in [2], we only need to prove that
      ( ψ 1 ) lim inf | | u | | - 2 ψ ( u ) > 0 ( ψ 2 ) ψ ( u ) 0 ( ψ 3 ) ψ ( u ) - a s u 0 i n H k , f o r a l l u i n H k , a n d a s | | u | | i n H k - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equai_HTML.gif
      We see that
      F ( t , x ) - F ( t , 0 ) = 0 1 ( F ( t , s x ) , x ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaj_HTML.gif
      for all x ∈ ℝ and a.e. t ∈ [0, T]. By (S1) and Lemma 2.1, one has
      F ( t , x ) - F ( t , 0 ) 0 1 ( f ( t ) h ( | s x | ) + g ( t ) , x ) d s f ( t ) [ h ( | x | ) + C ] | x | + g ( t ) | x | f ( t ) [ ε | x | + C 0 + C ] | x | + g ( t ) | x | = ε f ( t ) | x | 2 + [ f ( t ) ( C 0 + C ) + g ( t ) ] | x | Q ( t ) | x | 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equak_HTML.gif
      for all |x| ≥ δ, a.e. t ∈ [0, T] and some Q(t) ∈ L1(0, T; ℝ+) given by
      Q ( t ) : = ε f ( t ) δ - 1 + [ f ( t ) ( C 0 + C ) + g ( t ) ] δ - 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equal_HTML.gif
      Now, it follows from (1.10) that
      F ( t , x ) - F ( t , 0 ) - 1 2 k 2 ω 2 ( 1 + ε ) | x | 2 + Q ( t ) | x | 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equam_HTML.gif
      for all x ∈ ℝ and a.e. t ∈ [0, T]. Hence, we obtain
      ψ ( u ) = - 1 2 0 T | u ( t ) | 2 d t - 0 T [ F ( t , u ( t ) ) - F ( t , 0 ) ] d t - 1 2 0 T | u ( t ) | 2 d t + 1 2 k 2 ω 2 ( 1 + ε ) 0 T | u ( t ) | 2 d t - 0 T Q ( t ) | u ( t ) | 3 d t 1 2 ε 0 T | u ( t ) | 2 d t + 1 2 k 2 ω 2 ( 1 + ε ) | ū | 2 T - | | u | | 3 0 T Q ( t ) d t C 1 2 | | u | | 2 - C 1 3 | | u | | 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equan_HTML.gif

      for all uH k . Then, (ψ1) follows from the above inequality.

      For u H k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq25_HTML.gif, by (1.9), one has
      ψ ( u ) - 1 2 0 T | u ( t ) | 2 d t + 1 2 ( k + 1 ) 2 ω 2 0 T | u ( t ) | 2 d t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equao_HTML.gif

      So, (ψ2) is obtained. At last, (ψ3) follows from (φ1) which are appeared in the proof of Theorem 1.2. Then the proof of Theorem 1.3 is completed. □

      Proof of Theorem 1.4. From the proof of Theorem 1.1, we know that φ is coercive which implies that φ satisfies the (PS) condition. With the similar manner to [4, 7], we can get the multiplicity results, here we omit the details. □

      4 Examples

      In this section, we give some examples to illustrate our results.

      Example 4.1. Consider the function
      F ( t , x ) = 1 3 T - t | x | 2 ln ( 1 0 0 + | x | 2 ) + ( d ( t ) , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equap_HTML.gif

      where d(t) ∈ L1(0, T; ℝ). Let h ( t ) = t ln ( 1 0 0 + t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq6_HTML.gif, then H ( t ) = 0 t s ln ( 1 0 0 + s 2 ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq7_HTML.gif, by a direct computation, (S1) and (S3) hold. Then, by Theorem 1.2, we conclude that problem (1.1) has one solution in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif. However, as the reason of Remark 1.3, the results in [113] cannot be applied.

      Example 4.2. Consider the function
      F ( t , x ) = 2 3 T - t | x | 2 ln ( 1 0 0 + | x | 2 ) + A ( t ) | x | + B ( t ) , | x | > 1 , - 1 4 ω 2 | x | 2 + 1 2 ω 2 + 3 2 T - 9 4 t | x | 4 - 1 4 ω 2 + 5 6 T - 5 4 t | x | 6 , | x | 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaq_HTML.gif

      where A(t), B(t) are suitable functions which insure assumption (A) hold. Also, put h ( t ) = t ln ( 1 0 0 + t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq6_HTML.gif, H ( t ) = 0 t s ln ( 1 0 0 + s 2 ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq7_HTML.gif, we see that (S1), (S2) and (1.11) hold. By virtue of Theorem 1.4, problem (1.1) has at least three distinct solutions in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Example 4.3. Consider the function
      F ( t , x ) = 2 3 T - t ln ( 1 0 0 + | x | 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equar_HTML.gif
      We observe that
      | F ( t , x ) | 2 3 T - t 2 | x | 1 0 0 + | x | 2 2 2 3 T - t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equas_HTML.gif
      which means ∇F(t, x) is bounded, moreover, one has
      0 T F ( t , x ) d t + a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equat_HTML.gif

      Then, by the results in [3, 7, 12], problem (1.1) has one solution which minimizes the functional φ in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      In fact, our Theorem 1.1 can also handle this case. In this situation, let h ( t ) = t 1 0 0 + t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq26_HTML.gif, H ( t ) = 0 t s 1 0 0 + s 2 d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq27_HTML.gif, and choose C = 2, C* = 1 f ( t ) = 2 2 3 T - t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq28_HTML.gif, g(t) ≡ 0, we infer
      ( i ) h ( s ) h ( t ) + C s t , s , t + , ( i i ) h ( s + t ) = s + t 1 0 0 + ( s + t ) 2 h ( s ) + h ( t ) s , t + , ( i i i ) t h ( t ) - 2 H ( t ) = t 2 1 0 0 + t 2 - 2 1 2 ln ( 1 0 0 + t 2 ) - 1 2 ln 1 0 0 a s t + , - ( i v ) H ( t ) t 2 = 0 t s 1 0 0 + s 2 d s t 2 0 a s t + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equau_HTML.gif
      and
      1 H ( | x | ) 0 T F ( t , x ) d x > 0 a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equav_HTML.gif

      So, by Theorem 1.1, problem (1.1) has one solution which minimizes the functional φ in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Remark 4.1. Unlike the control functions in [12], where h(t) is nondecreasing, here control function h ( t ) = t 1 0 0 + t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq26_HTML.gif is bounded but not increasing.

      Example 4.4. Consider the function
      F ( t , x ) = 1 3 T - t | x | 4 3 + ( k ( t ) , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaw_HTML.gif
      where k(t) ∈ L1(0, T; ℝ). It is easy to check that
      | F ( t , x ) | 4 3 1 3 T - t | x | 1 3 + | k ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equax_HTML.gif
      The above inequality leads to (1.4) hold with
      f ( t ) = 4 3 1 3 T - t , g ( t ) = | k ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equay_HTML.gif
      Take α = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq29_HTML.gif, then
      1 | x | 2 α 0 T F ( t , x ) d t - a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equaz_HTML.gif

      So, by the theorems in [3, 7, 12, 13], problem (1.1) has at least one solution in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif.

      Indeed, our Theorem 1.2 can also deal with this case. Let h ( t ) = t 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq30_HTML.gif, H ( t ) = 3 4 t 4 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq31_HTML.gif, and choose C = 0, C* = 1, f ( t ) = 4 3 4 3 T - t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq32_HTML.gif, g(t) = |k(t)|, we know
      ( i ) h ( s ) h ( t ) s t , s , t + , ( i i ) h ( s + t ) = ( s + t ) 1 3 8 ( h ( s ) + h ( t ) ) s , t + , ( i i i ) t h ( t ) - 2 H ( t ) = - 1 2 t 4 3 - a s t + , ( i v ) H ( t ) t 2 = 3 4 t 2 3 0 a s t + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equba_HTML.gif
      Furthermore, one has
      1 H ( | x | ) 0 T F ( t , x ) d x < 0 a s | x | + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_Equbb_HTML.gif

      Hence, (S1) and (S3) are true, by Theorem 1.2, problem (1.1) has at least one solution in H T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-23/MediaObjects/13661_2011_Article_21_IEq3_HTML.gif. However, we can find that the results in [14] cannot deal with this case.

      Declarations

      Acknowledgements

      The authors would like to thank Professor Huicheng Yin for his help and many valuable discussions, and the first author takes the opportunity to thank Professor Xiangsheng Xu and the members at Department of Mathematics and Statistics at Mississippi State University for their warm hospitality and kindness. This Project is Supported by National Natural Science Foundation of China (Grant No. 11026213, 10871096), Natural Science Foundation of the Jiangsu Higher Education Institutions (Grant No. 10KJB110006) and Foundation of Nanjing University of Information Science and Technology (Grant No. 20080280).

      Authors’ Affiliations

      (1)
      Department of Mathematics, Nanjing University of Information Science and Technology
      (2)
      Jiangsu Key Laboratory for NSLSCS, School of Mathematics Sciences, Nanjing Normal University

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      © Wang and Zhang; licensee Springer. 2011

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