The Clark dual and multiple periodic solutions of delay differential equations

  • Huafeng Xiao1Email author,

    Affiliated with

    • Jianshe Yu1 and

      Affiliated with

      • Zhiming Guo1

        Affiliated with

        Boundary Value Problems20112011:44

        DOI: 10.1186/1687-2770-2011-44

        Received: 18 May 2011

        Accepted: 11 November 2011

        Published: 11 November 2011

        Abstract

        We study the multiplicity of periodic solutions of a class of non-autonomous delay differential equations. By making full use of the Clark dual, the dual variational functional is considered. Some sufficient conditions are obtained to guarantee the existence of multiple periodic solutions.

        2000 Mathematics Subject Classification: 34K13; 34K18.

        Keywords

        periodic solution Clark dual Morse-Ekeland index delay differential equation asymptotic linearity

        1 Introduction

        The existence and multiplicity of periodic solutions of delay differential equations have been investigated since 1962. Various methods have been used to study such a problem [110]. Among those methods, critical point theory is a very important tool. By combining with Kaplan-Yorke method, it can be used indirectly to study the existence of periodic solutions of delay differential equation [1116]. By building the variational frame for some special systems, it can be used directly to study the existence of delay differential systems [17, 18]. However, the variational functionals in the above two cases are strongly indefinite. They are hard to be dealed with.

        In this article, we study multiple periodic solutions of the following non-autonomous delay differential equations
        x ( t ) = - f ( t , x ( t - π 2 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ1_HTML.gif
        (1)

        where x(t) ∈ ℝ n , fC (ℝ × ℝ n , ℝ n ). We assume that

        (f1) f(t, x) is odd with respect to x and π/2-periodic with respect to t, i.e.,
        f ( t , - x ) = - f ( t , x ) , f ( t + π 2 , x ) = f ( t , x ) , ( t , x ) × n ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equa_HTML.gif

        (f2) There exists a continuous differentiable function F(t, x), which is strictly convex with respect to x uniformly in t, such that f(t, x) is the gradient of F(t, x) with respect to x;

        (f3)
        f ( t , x ) = A 0 x + g ( t , x ) , g ( t , x ) = o ( x ) as x 0 uniformly in t ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equb_HTML.gif
        f ( t , x ) = A x + h ( t , x ) , h ( t , x ) = o ( x ) as x uniformly in t ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equc_HTML.gif

        where A0, A are positive definite constant matrices.

        By making use of the Clarke dual, we study the dual variational functional associated with (1), which is an indefinite functional. Since the dimension of its negative space is finite, we define it as the Morse index of the dual variational functional. This Morse index is significant. Then Z2-index theory can be used and some sufficient conditions are obtained to guarantee the existence of multiple periodic solutions of (1).

        The rest of this article is organized as follows: in Section 2, some preliminary results will be stated; in Section 3, linear system is discussed and the Morse index of the variational functional associated with linear system is defined; in Section 4, our main results will be stated and proved.

        2 Preliminaries

        Denote by ℕ, ℕ*, ℤ, ℝ the sets of all positive integers, nonnegative integers, integers, real numbers, respectively. We define S1 : = ℝ/(2π ℤ).

        For a matrix A, denote by σ(A) the set of eigenvalue of A. The identity matrix of order n is denoted by I n and for simplicity by I.

        Let x, yL2 (S1, ℝ n ). For every zC (S1, ℝ n ), if
        0 2 π ( x ( t ) , z ( t ) ) d t = - 0 2 π ( y ( t ) , z ( t ) ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equd_HTML.gif

        then y is called a weak derivative of x, denoted by http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq1_HTML.gif.

        The space H1 = W1,2(S1, ℝ n ) consists of 2π-periodic vector-valued functions with dimension n, which possess square integrable derivative of order 1. We can choose the usual norm and inner product in H1 as follows:
        x 2 = 0 2 π [ x ( t ) 2 + ( t ) 2 ] d t , < x , y > = 0 2 π [ ( x ( t ) , y ( t ) ) + ( ( t ) , ( t ) ) ] d t , x , y H 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Eque_HTML.gif

        where | · |, (·,·) denote the usual norm and inner product in ℝ n , respectively. Then H1 is a Hilbert space.

        Define the shift operator K : H1H1 by Kx(·) = x(· + π/2), for all xH1. Clearly, K is a bounded linear operator from H1 to H1. Set
        E = { x H 1 K 2 x = - x } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equf_HTML.gif
        Then E is a closed subspace of H1. If xE, its Fourier expansion is
        x ( t ) = k = 1 [ a k cos ( 2 k - 1 ) t + b k sin ( 2 k - 1 ) t ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ2_HTML.gif
        (2)

        where a k , b k ∈ ℝ n . In particular, E does not contain ℝ n as its subspace. In addition, 0 2 π x ( t ) d t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq2_HTML.gif for all xE.

        The dual variational functional corresponding to (1) defined on H1 is
        J ( y ) = 0 2 π [ 1 2 ( ( t + π 2 ) , y ( t ) ) + F * ( t , ( t ) ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ3_HTML.gif
        (3)
        By Hypothesis (f3), F(t, x)/|x| → +∞ as |x| → ∞ uniformly in t. Since F(t, ·) is strictly convex, Proposition 2.4 of [19] implies that F*(t, ·) ∈ C1(ℝ n , ℝ). Since f satisfies (f3), it follows the discussion in Chapter 7 of [19] that
        f * ( t , y ) = B y + o ( y ) as y uniformly in t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ4_HTML.gif
        (4)
        f * ( t , y ) = B 0 y + o ( y ) as y 0 uniformly in t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ5_HTML.gif
        (5)

        where B = A - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq3_HTML.gif and B 0 = A 0 - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq4_HTML.gif.

        Lemma 2.1. If yE is a critical point of J, then the function x defined by x ( t ) = f * ( t , ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq5_HTML.gif is a 2π-periodic solution of (1).

        Proof. Since f(·, x) is π/ 2-periodic, it follows that F(·, x) and then F * ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq6_HTML.gif are π/ 2-periodic. So is f * ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq7_HTML.gif. (4) implies that there exist positive constants a1, a2 such that |F*(t, y)| ≤ a1 + a2|y| s for some s > 2 and all t ∈ ℝ, yH1. We define φ by the formulas
        φ ( y ) = 0 2 π F * ( t , ( t ) ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equg_HTML.gif
        It follows Proposition B. 37 of [20] that φC1(H1, ℝ) and
        < φ ( y ) , z > = 0 2 π ( f * ( t , ( t ) ) , ż ( t ) ) d t , y , z H 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equh_HTML.gif

        We claim: φ'(y) ∈ E if yE.

        To prove the above claim, let zH1 and yE,
        < K 2 φ ( y ) , z > = < φ ( y ) , K - 2 z > = 0 2 π ( f * ( t , ( t ) ) , ż ( t - π ) ) d t = 0 2 π ( f * ( t + π , ( t + π ) ) , ż ( t ) ) d t = 0 2 π ( f * ( t , - ( t ) ) , ż ( t ) ) d t = 0 2 π ( - f * ( t , ( t ) ) , ż ( t ) ) d t = < - φ ( y ) , z > . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equi_HTML.gif

        The arbitrary of z implies that φ'(y) ∈ E.

        Since φC1(H1, ℝ), it is easy to check that JC1(H1, ℝ) and
        < J ( y ) , h > = 0 2 π ( 1 2 y ( t - π 2 ) - 1 2 y ( t + π 2 ) + f * ( t , ( t ) ) , ( t ) ) d t , h H 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equj_HTML.gif
        Assume that yE is a critical point of J. For any hH1, h = h1 + h2, where h1E, h2E. Then
        < J ( y ) , h > = < J ( y ) , h 1 > + < J ( y ) , h 2 > . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equk_HTML.gif
        Since φ'(y) ∈ E, it is easy to check that J'(y) ∈ E and < J'(y), h2 > = 0. Since y is a critical point of J on E, then < J'(y), h1 > = 0. Thus y is a critical point of J on H1. Applying the fundamental Lemma (cf. [19]), there exists c1 such that
        y t - π 2 + f * ( t , ( t ) ) = c 1 , a . e . on [ 0 , 2 π ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equl_HTML.gif
        Setting
        x ( t ) = f * ( t , ( t ) ) = - y t - π 2 + c 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equm_HTML.gif
        we obtain xH1, ( t - π 2 ) = ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq8_HTML.gif and by duality
        ( t ) = f ( t , x ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equn_HTML.gif
        Thus,
        t - π 2 = f ( t , x ( t ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equo_HTML.gif
        i.e.,
        ( t ) = - f t , x t - π 2 a . e . on [ 0 , 2 π ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equp_HTML.gif

        Moreover, x(0) = x(2π) since xH1. It follows a regular discussion that x ( t ) = f * ( t , ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq9_HTML.gif is a periodic solution of (1).   □

        Let X be a Hilbert space, Φ ∈ C1(X, ℝ), i.e., Φ is a continuously Fréchet differentiable functional defined on X. If X1 is a closed subspace of X, denote by X 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq10_HTML.gif the orthogonal complement of X1 in X. Fix a prime integer p > 1. Define a map μ : XX such that ||μx|| = ||x|| for any xX and μ p = id X , where id X is the identity map on X. Then μ is a linear isometric action of Z p on X, where Z p is the cyclic group with order p.

        A subset AX is called μ-invariant if μ(A) ⊂ A. A continuous map h : AE is called μ-equivariant if h(μx) = μh(x) for any xA. A continuous functional H : X → ℝ is called μ-invariant if H(μx) = H(x) for any xX.

        Φ is said to be satisfying (PS)-condition if any sequence {x j } ⊂ X for which {Φ(x j )} is bounded and Φ'(x j ) → 0 as j → ∞, possesses a convergent subsequence. A sequence {x j } is called (PS)-sequence if {Φ(x j )} is bounded and Φ'(x j ) → 0 as j → ∞.

        Lemma 2.2 [21]. Let Φ ∈ C1(X, ℝ) be a μ-invariant functional satisfying the (PS)-condition. Let Y and Z be closed μ-invariant subspaces of X with codimY and dimZ finite and
        c o d i m Y < d i m Z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equq_HTML.gif

        Assume that the following conditions are satisfied:

        (F1) Fix μ Y, ZFix μ = {0};

        (F2) infxYΦ(x) > -∞;

        (F3) there exist constants r > 0 and c < 0 such that Φ(x) ≤ c whenever xZ and ||x|| = r;

        (F4) if xFix μ and Φ'(x) = 0, then Φ(x) ≥ 0.

        Then there exists at least dimZ - codimY distinct Z p -orbits of critical points of Φ outside of Fix μ with critical value less than or equal to c.

        3 Morse index

        Let A be a positive definite constant matrix. We consider the periodic boundary value problem
        x ( t ) + A x ( t - π 2 ) = 0 x ( 0 ) = x ( 2 π ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ6_HTML.gif
        (6)
        Since F(t, x) = 1/2(Ax, x), it is easy to verify that its Legendre transform F*(t, y) is of the form F*(t, y) = 1/2(By, y), where B = A-1. The dual action of (6) is defined on H1 by
        χ A ( y ) = 0 2 π 1 2 [ ( ( t + π 2 ) , y ( t ) ) + ( B ( t ) , ( t ) ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ7_HTML.gif
        (7)
        A is positively definite, so is B. Thus there exists δ1 > 0 such that (By, y) ≥ δ1|y|2 for all y ∈ ℝ n . Wirtinger's inequality implies that the symmetric bilinear form given by
        ( ( y , z ) ) 1 = 0 2 π ( B ( t ) , ż ( t ) ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ8_HTML.gif
        (8)
        define an inner product on E. The corresponding norm || · ||1 is such that
        y 1 2 δ 1 L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ9_HTML.gif
        (9)

        Proposition 3.1. The norm || · ||1 is equivariant to the stand norm || · || of E.

        Proof. Since B is positive, Wirtinger's inequality and (9) imply that there exists a positive constant δ2 such that
        x 1 2 = 0 2 π ( B ( t ) , ( t ) ) d t δ 1 L 2 2 δ 1 δ 2 x 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equr_HTML.gif
        On the other hand, since B is a positive definite matrix, there exists a constant M > 0 such that
        x 2 L 2 2 1 M 0 2 π ( B ( t ) , ( t ) ) d t = 1 M x 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equs_HTML.gif

        Thus those two norms are equivariant to each other, which completes our proof.   □

        Let us define the linear operator L on E by setting
        ( ( L y , z ) ) 1 = 0 2 π y ( t + π 2 ) , ż ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ10_HTML.gif
        (10)
        One can easily check that L is a compact self-adjoint operator. (7) can be rewritten as
        2 χ A ( y ) = 0 2 π t + π 2 , y ( t ) + ( B ( t ) , ( t ) ) d t = ( ( ( I - L ) y , y ) ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ11_HTML.gif
        (11)

        It follows from the spectral theory that E can be decomposed as the orthogonal sum of Ker(I - L), E+ and E- with I - L positive definite (resp. negative definite) on E+ (resp. E-). Since L is compact, it has at most finite many eigenvalues (counting the multiplicity) greater than one. Thus the index dimE- < ∞.

        Definition 3.1. The index i(A) is the Morse index of χ A , i.e. the supermum of the dimension of the subspace of E on which χ A is negative definite.

        On the other hand, there exists a positive constant δ3 such that
        ( ( ( I - L ) y , y ) ) 1 δ 3 y 1 2 , y E + and ( ( ( I - L ) y , y ) ) 1 - δ 3 y 1 2 , y E - . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equt_HTML.gif
        Setting δ = δ1δ3 > 0, we reduce from (9) and (11) the estimates
        χ A ( y ) δ 2 L 2 2 , y E + and χ A ( y ) - δ 2 L 2 2 , y E - . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ12_HTML.gif
        (12)

        Proposition 3.2. The dimension of Ker(I - L) is equal to the number of linearly independent solutions of (6).

        Proof. If yE, by a Fourier argument, yKer(I - L) if and only if
        B ( t ) = y t + π 2 + c 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ13_HTML.gif
        (13)
        for some c2 ∈ ℝ n and a.e. t ∈ [0, 2π]. Since B = A-1 is invertible, y is continuous differentiable. Thus (13) hold for all t ∈ [0, 2π], and B E n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq11_HTML.gif. Set
        x ( t ) = ( φ y ) ( t ) = B ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ14_HTML.gif
        (14)
        In view of (13), (14) and the Clark dual, we have
        ( t ) = d d t ( B ( t ) ) = t + π 2 = A x t + π 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equu_HTML.gif

        Thus x is a solution of (6).    □

        Conversely, assume now that xE ⊕ ℝ n is a solution of (6). Then
        0 2 π A x t - π 2 d t = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equv_HTML.gif

        and hence there exists an unique yE such that x = φy, where φ is defined by (14).

        Thus
        - ( t ) = A x t - π 2 = t - π 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equw_HTML.gif
        for a.e. t ∈ [0, 2π]. Integrating the above equality, we have x(t) = y(t + π/2) + c3 for some c3 ∈ ℝ n and all t ∈ [0, 2π]. Consequently,
        A [ y t + π 2 + c 3 ] = ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equx_HTML.gif

        for a.e. t ∈ [0, 2π], which is equivariant to (13) and shows that yKer(I - L). Thus φ is an isomorphism between Ker(I - L) and the space of solutions of (6).

        Consider the following eigenvalue problem
        ( I - L ) y = λ y , y E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equy_HTML.gif
        For any zE, we have (((I - L)y, z))1 = >λ((y, z))1. Computing directly, we have
        0 2 π k = 1 [ ( 2 k - 1 ) B + ( - 1 ) k I ] [ - a k cos ( 2 k - 1 ) t + b k sin ( 2 k - 1 ) t ] , ż ( t ) d t = λ 0 2 π k = 1 ( 2 k - 1 ) B [ - a k cos ( 2 k - 1 ) t + b k sin ( 2 k - 1 ) t ] , ż ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equz_HTML.gif
        Denote by e i , i = 1, 2, ..., n the basic of ℝ n . Choosing
        z ( t ) = sin ( 2 k - 1 ) t cos ( 2 k - 1 ) t e i , k , i = 1 , 2 , , n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equaa_HTML.gif
        it follows that
        ( ( 2 k - 1 ) B + ( - 1 ) k I ) a k = λ ( 2 k - 1 ) B a k , ( ( 2 k - 1 ) B + ( - 1 ) k I ) b k = λ ( 2 k - 1 ) B b k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equab_HTML.gif
        Since B = A-1, then
        λ a k = ( ( - 1 ) k 2 k - 1 A + I ) a k , λ b k = ( ( - 1 ) k 2 k - 1 A + I ) b k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equac_HTML.gif
        Thus
        dim Ker ( I - L ) = 2 k = 1 dim Ker ( ( - 1 ) k 2 k - 1 A + I ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equad_HTML.gif

        Proposition 3.3. If σ(A) ∩ {4l + 1, l ∈ ℕ*} =, then I - L is invertible and codimE + = i(A).

        Proof. The above analysis implies that Ker (I - L) = {0} if σ(A) ∩ {4l + 1, l ∈ ℕ*} = ∅. Thus I - L is invertible. If we decompose E as E = E+E-, then codimE+ = i(A).    □

        4 Main results and their proofs

        Now we consider the multiplicity of periodic solutions of (1). The main result reads as follows:

        Theorem 4.1. Assume that (f1)-(f3) are satisfied. Moreover,

        (A1) σ (A) ∩ {4l + 1, l ∈ ℕ*} = ∅;

        (A2) i(A0) > i(A).

        Then (1) has at least i(A0) - i(A) pairs of nontrivial 2π-periodic solutions.

        By (f3), f (t, 0) = 0 uniformly in t. Since F(t, ·) is strictly convex, it follows that 0 is the unique equilibrium point of (1). Without loss of generality, we can assume that F (t, 0) = 0 uniformly in t. Since f (t, 0) = 0, then F*(t, 0) = 0 uniformly in t.

        Lemma 4.1. The functional J satisfies (PS)-condition.

        Proof. Denote by ((·, ·))1,∞ the inner product, defined by (8) replacing B by B. Let L be linear self-adjoint operator, defined by (10), under the inner product ((·, ·))1,∞. We define the operator N over E by setting
        ( ( N y , z ) ) 1 , = 0 2 π ( f * ( t , ) - B , ż ) d t , y , z E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equae_HTML.gif
        Let { y j } j = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq12_HTML.gif be a (PS)-sequence. Define f j : = y j - L y j + Ny j , j ∈ ℕ. Since
        < J ( y ) , z > = 0 2 π ( y t - π 2 + f * ( t , ) , ż ) d t = ( ( y - L y + N y , z ) ) 1 , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equaf_HTML.gif
        then f j → 0 in E as j → ∞. Then there exists R > 0 such that for all j ∈ ℕ, || f j || ≤ R. (A1) implies that M = I - L is invertible. Thus it follows from (4) that there exists a positive constant c4 such that, for all yE
        N y 1 , 1 2 M - 1 1 , - 1 y 1 , + c 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equag_HTML.gif
        where || · ||1,∞ denotes by the norm corresponding to ((·, ·))1,∞. Since || f j || ≤ R, we have
        y j 1 , M - 1 1 , ( N y j 1 , + f j 1 , ) 1 2 y j 1 , + M - 1 1 , ( c 4 + R ) , j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equah_HTML.gif

        The above inequality implies that {y j } is bounded. A standard argument as Lemma 4.5 in [19] shows that {y j } has a convergent sequence.    □

        Lemma 4.2. The functional J is bounded from below on a closed invariant subspace Y of E of codimension i(A).

        Proof. Consider the linear delay differential system
        x ( t ) = - A x t - π 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equai_HTML.gif
        Its dual variational functional is
        χ A ( y ) = 0 2 π 1 2 ( t + π 2 , y ( t ) ) + ( B ( t ) , ( t ) ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equaj_HTML.gif

        where ((·, ·))1,∞, L are defined as Lemma 4.1.

        Because of (A1) and Proposition 3.3, the operator I - L is invertible. We decompose E as E = E + E - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq13_HTML.gif, where E + ( resp.  E - ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq14_HTML.gif is the positive (resp. negative) definite eigenspace of I - L. Then codim E + = dim E - = i ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq15_HTML.gif. Let Y : = E + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq16_HTML.gif. Inequality (12) implies that there exists a positive constant δ4 such that, for each yY,
        χ A ( y ) = 0 2 π 1 2 t + π 2 , y ( t ) ) + ( B , ) d t δ 4 2 L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equak_HTML.gif
        It follows (4) that there exists a positive constant c5 such that
        f * ( t , y ) - B y δ 4 2 y + c 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equal_HTML.gif
        for each y ∈ ℝ n . Hence, by the mean value theorem,
        F * ( t , y ) - 1 2 ( B y , y ) 0 1 ( f * ( t , σ y ) - σ B y , y ) d σ 0 1 δ 4 2 σ y 2 + c 5 y d σ = δ 4 4 y 2 + c 5 y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equam_HTML.gif
        Consequently, we have, for yY,
        J ( y ) = χ A ( y ) + 0 2 π [ F * ( t , ) - 1 2 ( B , ) ] d t δ 4 2 L 2 2 - 0 2 π δ 4 4 2 + c 5 d t = δ 4 4 L 2 2 - c 5 L 1 δ 4 4 L 2 2 - c 5 2 π L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equan_HTML.gif

        and J is bounded from below on Y.

        Lemma 4.3. There exists an invariant subspace Z of E with dimension i(A0) and some r > 0 such that J(y) < 0 whenever yZ and ||y||1 = r.

        Proof. Consider the linear delay differential system
        x ( t ) = - A 0 x ( t - π 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equao_HTML.gif
        Its dual variational functional is
        χ A 0 ( y ) = 0 2 π 1 2 ( ( t + π 2 ) , y ( t ) ) + ( B 0 ( t ) , ( t ) ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equap_HTML.gif

        where ((·, ·))1,0, L0 are defined by (8) and (10) with B replaced by B0, respectively.

        We decompose E as E = E 0 + E 0 0 E 0 - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq17_HTML.gif, where E 0 0 = ker ( I - L 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq18_HTML.gif, E 0 + ( resp . E 0 - ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq19_HTML.gif is the positive (resp. negative) definite eigenspace of I - L0. Let Z : = E 0 - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq20_HTML.gif. Then Z is a finite dimensional space. Inequality (12) implies that there exists δ5 > 0 such that, for each yZ,
        χ A 0 ( y ) = 0 2 π 1 2 ( ( t + π 2 ) , y ( t ) ) + ( B 0 , ) d t - δ 5 2 L 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equaq_HTML.gif
        whenever yZ. By (5), there exists ρ > 0 such that
        f * ( t , y ) - B 0 y δ 5 2 y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equar_HTML.gif
        for y ∈ ℝ n with |y| ≤ ρ. Hence, by the mean value theorem, we have
        F * ( t , y ) - 1 2 ( B 0 y , y ) 0 1 ( f * ( t , σ y ) - σ B 0 y , y ) d σ 0 1 δ 5 2 σ y 2 d σ = δ 5 4 y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equ15_HTML.gif
        (15)
        whenever |y| ≤ ρ. Consequently, if yZ and 0 < |y| < ρ, we get
        J ( y ) = χ A 0 ( y ) + 0 2 π [ F * ( t , ) - 1 2 ( B 0 , ) ] d t - δ 5 2 L 2 2 + 0 2 π δ 5 4 2 d t = - δ 5 4 L 2 2 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equas_HTML.gif

        If J(y) = 0, then L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq21_HTML.gif, which implies that = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq22_HTML.gif for a.e. t ∈ [0, 2π]. Thus y ∈ ℝ n Z = {0}, which contradicts with 0 < |y| < ρ. Thus J(y) < 0.

        Proof of Theorem 4.1. Now, we apply Lemma 2.2 to prove Theorem 4.1.

        Define the action μ : EE by μx = -x. Then μ is a generator of group Z2 and Fix μ = {0}. Obviously, (F1) and (F4) hold. It is easy to check that J is μ-invariant. Lemma 4.1 implies that J satisfies (PS)-condition.

        Let Y, Z define as in Lemma 4.2 and 4.3. Then Y, Z are μ-invariant subspaces and (A2) implies that
        dim Z = i ( A 0 ) > i ( A ) = codim Y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equat_HTML.gif

        It follows from Lemmas 4.2 and 4.3 that (F2) and (F3) hold. Applying Lemma 2.2, J has at least i(A0) - i(A) pairs of distinct critical points with critical values less than or equal to c. Since J(0) = 0 and E ∩ ℝ n = {0}, if follows that all those i(A0) - i(A) pairs of distinct critical points are nonconstant.

        If y is a critical point of J, then x : = f * ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq23_HTML.gif is a 2π-periodic solution of (1). Clearly, y ≠ 0 implies that x ≠ 0. Let x1, x2 are two periodic solutions satisfying x1 = -x2. Setting 1 = f ( t , x 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq24_HTML.gif and 2 = f ( t , x 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_IEq25_HTML.gif, then
        1 = f ( t , x 1 ) = f ( t , - x 2 ) = - f ( t , x 2 ) = - 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-44/MediaObjects/13661_2011_Article_89_Equau_HTML.gif

        Integrating he above equality, we have y1 = - y2 + c6 for some c6 ∈ ℝ n . Since y1, y2E, it follows that c6 = 0. Then y1, y2 belongs the same Z2-orbit. Thus (1) has i(A0) - i(A) pairs of nontrivial periodic solutions.

        Corollary 4.1. Under the hypotheses of Theorem 4.1, (1) has at least 2[i(A0) - i(A)] nontrivial 2π-periodic solutions.

        Since every Z2-orbit has two elements and those two elements are different from each other, this corollary is obvious.

        Declarations

        Acknowledgements

        This project was supported by the National Natural Science Foundation of China (Nos. 11031002 and 10871053). The authors are very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript.

        Authors’ Affiliations

        (1)
        College of Mathematics and Information Sciences, Guangzhou University

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