Global solutions to a class of nonlinear damped wave operator equations

  • Zhigang Pan1Email author,

    Affiliated with

    • Zhilin Pu2 and

      Affiliated with

      • Tian Ma1

        Affiliated with

        Boundary Value Problems20122012:42

        DOI: 10.1186/1687-2770-2012-42

        Received: 6 October 2011

        Accepted: 13 April 2012

        Published: 13 April 2012

        Abstract

        This study investigates the existence of global solutions to a class of nonlinear damped wave operator equations. Dividing the differential operator into two parts, variational and non-variational structure, we obtain the existence, uniformly bounded and regularity of solutions.

        Mathematics Subject Classification 2000: 35L05; 35A01; 35L35.

        Keywords

        nonlinear damped wave operator equations global solutions uniformly bounded regularity

        1 Introduction

        In recent years, there have been extensive studies on well-posedness of the following nonlinear variational wave equation with general data:
        t 2 u - c ( u ) x c ( u ) x u = 0 in ( 0 , ) × R , u | t = 0 = u 0 on R , t u | t = 0 = u 1 on R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ1_HTML.gif
        (1.1)
        where c(·) is given smooth, bounded, and positive function with c'(·) ≥ 0 and c'(u0) > 0,u0H1(R),u1(x) ∈ L2(R). Equation (1.1) appears naturally in the study for liquid crystals [14]. In addition, Chang et al. [5], Su [6] and Kian [7] discussed globally Lipschitz continuous solutions to a class one dimension quasilinear wave equations
        u t t - p ρ ( x ) , u x x = ρ ( x ) h ρ ( x ) , u , u x , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = ω 0 ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ2_HTML.gif
        (1.2)
        where (x,t) ∈ R × R+, u0(x),ω0(x) ∈ R. Furthermore, Nishihara [8] and Hayashi [9] obtained the global solution to one dimension semilinear damped wave equation
        u t t + u t - u x x = f ( u ) , ( t , x ) R + × R + ( u , u t ) ( 0 , x ) = ( u 0 , u 1 ) ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ3_HTML.gif
        (1.3)

        Ikehata [10] and Vitillaro [11] proved global existence of solutions for semilinear damped wave equations in R N with noncompactly supported initial data or in the energy space, in where the nonlinear term f(u) = |u| p or f(u) = 0 is too special; some authors [1214] discussed the regularity of invariant sets in semilinear wave equation, but they didn't refer to any the initial value condition of it. Unfortunately, it is difficulty to classify a class wave operator equations, since the differential operator structure is too complex to identify whether have variational property. Our aim is to classify a class of nonlinear damped wave operator equations in order to research them more extensively and go beyond the results of [12].

        In this article, we are interested in the existence of global solutions of the following nonlinear damped wave operator equations:
        d 2 u d t 2 + k d u d t = G ( u ) , k > 0 u ( x , 0 ) = φ ( x ) , u t ( x , 0 ) = ψ ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ4_HTML.gif
        (1.4)

        where G : X 2 × R + X 1 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq1_HTML.gif is a mapping, X2X1, X1, X2 are Banach spaces and X 1 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq2_HTML.gif is the dual spaces of X1, R+ = [0, ∞), u = u(x,t). If k > 0, (1.4) is called damped wave equation. We obtain the existence, uniformly bounded and regularity of solutions by dividing the differential operator G(u) into two parts, variational and non-variational structure.

        2 Preliminaries

        First we introduce a sequence of function spaces:
        X H 2 X 2 X 1 H , X 2 H 1 H , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ5_HTML.gif
        (2.1)
        where H, H1, H2 are Hilbert spaces, X is a linear space, X1, X2 are Banach spaces and all inclusions are dense embeddings. Suppose that
        L : X X 1 is one to one dense linear operator , L u , v H = u , v H , u , v X . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ6_HTML.gif
        (2.2)
        In addition, the operator L has an eigenvalue sequence
        L e k = λ k e k , ( k = 1 , 2 , . . . ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ7_HTML.gif
        (2.3)
        such that {e k } ⊂ X is the common orthogonal basis of H and H2. We investigate the existence of global solutions of the Equation (1.4), so we need define its solution. Firstly, in Banach space X, introduce
        L p ( ( 0 , T ) , X ) = u : ( 0 , T ) X | 0 T u p d t < , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equa_HTML.gif
        where p = (p1, p2,..., p m ),p i ≥ 1(1 ≤ im),
        u p = k = 1 m u k p k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equb_HTML.gif
        where | · | k is semi-norm in X, and X = i = 1 m i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq3_HTML.gif. Similarily, we can define
        W 1 , p ( ( 0 , T ) , X ) = u : ( 0 , T ) X | u , u L p ( ( 0 , T ) , X ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equc_HTML.gif

        Let L loc p ( ( 0 , ) , X ) = u ( t ) X | u L p ( ( 0 , T ) , X ) , T > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq4_HTML.gif.

        Definition 2.1. Set (φ, ψ) ∈ X2 × H1, u W loc 1 , 0 , , H 1 L loc ( ( 0 , ) , X 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq5_HTML.gif is called a globally weak solution of (1.4), if for ∀vX1, it has
        u t , v H + k u , v H = 0 t G u , v d t + k φ , v H + ψ , v H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ8_HTML.gif
        (2.4)
        Definition 2.2. Let Y1,Y2 be Banach spaces, the solution u(t, φ, ψ) of (1.4) is called uniformly bounded in Y1 × Y2, if for any bounded domain Ω1 × Ω2Y1 × Y2, there exists a constant C which only depends the domain Ω1 × Ω2, such that
        u Y 1 + u t Y 2 C , ( φ , ψ ) Ω 1 × Ω 2 and t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equd_HTML.gif
        Definition 2.3. A mapping G : X 2 X 1 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq6_HTML.gif is called weakly continuous, if for any sequence {u n } ⊂ X2, u n u0 in X2,
        lim n G ( u n ) , v = G ( u 0 ) , v , v X 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Eque_HTML.gif

        Lemma 2.1. [15]Let H2, H be Hilbert spaces, and H2H be a continuous embedding. Then there exists a orthonormal basis {e k } of H, and also is one orthogonal basis of H2.

        Proof. Let I : H2H be imbedded. According to assume I is a linear compact operator, we define the mapping A : H2H as follows
        A u , v H 2 = I u , v H = u , v H , v H 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equf_HTML.gif
        obviously, A : H2H2 is linear symmetrical compact operator and positive definite. Therefore, A has a complete eigenvalue sequence {λ k } and eigenvector sequence k H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq7_HTML.gif such that
        A k = λ k k , k = 1 , 2 , . . . , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equg_HTML.gif
        and k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq8_HTML.gif is orthogonal basis of H2. Hence
        i , j H = A i , j H 2 = λ i i , j H 2 = 0 , if i j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equh_HTML.gif

        it implies i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq9_HTML.gif is also orthogonal sequence of H. Since H2H is dense, i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq9_HTML.gif is also orthogonal sequence of H, so { e i } = i / i H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq10_HTML.gif is norm orthogonal basis of H. The proof is completed.

        Now, we introduce an important inequality

        Lemma 2.2. [16] (Gronwall inequality) Let x(t), y(t), z(t) be real function on [a, b], where x(t) ≥ 0,∀atb, z(t) ∈ C[a, b], y(t) is differentiable on [a, b]. If the inequality as follows is hold
        z ( t ) y ( t ) + a t x ( τ ) z ( τ ) d τ , a t b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ9_HTML.gif
        (2.5)
        then
        z ( t ) y ( a ) e a t x ( s ) d s + a t e a t x ( τ ) d y d s d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ10_HTML.gif
        (2.6)

        3 Main results

        Suppose that G = A + B : X 2 × R + X 1 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq11_HTML.gif. Throughout of this article, we assume that
        1. (i)
          There exists a function FC 1 : X 2R 1 such that
          A u , L v = - D F ( u ) , v , u , v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ11_HTML.gif
          (3.1)
           
        2. (ii)
          Function F is coercive, if
          F ( u ) u X 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ12_HTML.gif
          (3.2)
           
        3. (iii)
          B as follows
          B u , L v C 1 F ( u ) + C 2 v H 1 2 , u , v X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ13_HTML.gif
          (3.3)
           

        for some g L loc 1 ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq12_HTML.gif.

        Theorem 3.1. Set G : X 2 × R + X 1 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq1_HTML.gifis weakly continuous, (φ, ψ) ∈ X2 × H1, then we obtain the results as follows:

        (1) If G = A satisfies the assumption (i) and (ii), then there exists a globally weak solution of (1.4)
        u W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equi_HTML.gif

        and u is uniformly bounded in X2 × H1;

        (2) If G = A + B satisfies the assumption (i), (ii) and (iii), then there exists a globally weak solution of (1.4)
        u W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equj_HTML.gif
        (3) Furthermore, if G = A + B satisfies
        G u , v 1 2 v H 2 + C F ( u ) + g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ14_HTML.gif
        (3.4)

        for some g L loc 1 ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq12_HTML.gif, then u W loc 2 , 2 ( 0 , ) , H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq13_HTML.gif.

        Proof. Let {e k } ⊂ X be the public orthogonal basis of H and H2, satisfies (2.3).

        Note
        X n = i = 1 n α i e i | α i R 1 , X ̃ n = j = 1 n β j ( t ) e j | β j C 2 0 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ15_HTML.gif
        (3.5)
        From the assumption, we know L X n = X n , L X n ̃ = X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq14_HTML.gif, apply the Galerkin method to make truncate in X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq15_HTML.gif:
        d 2 u i d t 2 + k d u i d t = G ( u n ) , e i , 1 i n u i ( x , 0 ) = φ , e i H , u i ( x , 0 ) = ψ , e i H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ16_HTML.gif
        (3.6)
        there exists u n = i = 1 n u i ( t ) e i C 2 ( 0 , ) , X n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq16_HTML.gif for any v X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq17_HTML.gif satisfies
        0 t d 2 u n d t 2 + k d u n d t , v H d t = 0 t G u n , v d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ17_HTML.gif
        (3.7)
        for any vX n , it yields that
        d u n d t , v H + k u n , v H = 0 t G u n , v d t + k φ , v H + ψ , v H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ18_HTML.gif
        (3.8)
        1. (1)
          If G = A , u n X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq18_HTML.gif substitute v = d d t L u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq19_HTML.gif into (3.7), we get
          0 t d 2 u n d t 2 + k d u n d t , d d t L u n H 1 d t = 0 t G u n , d d t L u n d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equk_HTML.gif
           
        combine condition (2.2) with (3.1), we get
        0 t Ω d 2 u n d t 2 d u n d t d x d t + 0 t Ω k d u n d t d u n d t d x d t + 0 t D F ( u n ) d u n d t d x d t = 0 0 t 1 2 d d t d u n d t H 1 2 d t + k 0 t d u n d t H 1 2 d t + 0 t d d t F ( u n ) d t = 0 1 2 d u n d t H 1 2 - 1 2 ψ n H 1 2 + k 0 t d u n d t H 1 2 d t + F ( u n ) - F ( φ n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equl_HTML.gif
        consequently, we get
        F ( u n ) + 1 2 u n H 1 2 + k 0 t u n H 1 2 d t = F ( u n ) + 1 2 ψ n H 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ19_HTML.gif
        (3.9)
        Assume φH2, combine(2.2)with(2.3), we know {e n } is also the orthogonal basis of H1, then φ n φ in H2, ψ n ψ in H1, owing to H2X2 is embedded, so
        φ n φ i n X 2 ψ n ψ i n X 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ20_HTML.gif
        (3.10)
        due to the condition (3.6), from (3.9)and (3.10) we easily know
        { u n } W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) i s b o u n d e d . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equm_HTML.gif
        consequently, assume that
        u n u 0 i n W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) a . e . t > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equn_HTML.gif
        i.e. u n u0in X2a.e. t > 0, and G is weakly continuous, so
        lim n G u n , v = G u 0 , v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equo_HTML.gif
        By (3.8), we have
        lim n d u n d t H + k u n , v H = lim n 0 t G u n , v d t + k φ , v H + ψ , v H d u 0 d t , v H + k u 0 , v H = 0 t G u 0 , v d t + k φ , v H + ψ , v H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equp_HTML.gif
        it indicates for any v n = 1 X n X 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq20_HTML.gif, it holds. Hence, for any vX2, we have
        d u 0 d t , v H + k u 0 , v H = 0 t G u 0 , v d t + k φ , v H + ψ , v H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ21_HTML.gif
        (3.11)

        Consequently, u0 is a globally weak solution of (1.4).

        Furthermore, by (3.9) and (3.10), for any R > 0, there exists a constant C such that if
        φ X 2 + ψ H 1 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ22_HTML.gif
        (3.12)
        then the weak solution u(t, φ, ψ) of (1.4) satisfies
        u ( t , φ , ψ ) X 2 + u t ( t , φ , ψ ) H 1 C . t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ23_HTML.gif
        (3.13)
        Assume (φ,ψ) ∈ X2 × H1 satisfies (3.12), by H2X2 is dense. May fix φ n H2 such that
        φ n X 2 + ψ H 1 R , lim n φ n = φ in X 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equq_HTML.gif

        by (3.13), the solution {u(t, φ n , ψ)} of (1.4) is bounded in W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq21_HTML.gif a.e. t > 0.

        Therefore, assume u(t, φ n , ψ) ⇀ u in W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq21_HTML.gif then u(t) is a weak solution of (1.4), it satisfies uniformly bounded of (3.13). So the conclusion (1) is proved.
        1. (2)
          If G = A + B , u n X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq22_HTML.gif, substitute v = d d t L u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq19_HTML.gif into (3.7), we get
          0 t d 2 u n d t 2 , d d t L u n H 1 + k d u n d t , d d t L u n 1 H 1 d t = 0 t A u n , d u n d t + B u n , d u n d t d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equr_HTML.gif
           
        combine the condition (2.2) and (3.1), we have
        0 t Ω d 2 u n d t 2 d u n d t d x d t + k 0 t Ω d u n d t 2 d u n d t d x d t + 0 t D F ( u n ) d u n d t d t = 0 t B u n , d u n d t d t 0 t 1 2 d d t d u n d t H 1 2 d t + k 0 t d u n d t H 1 2 d t + 0 t d d t F ( u n ) d t = 0 t B u n , d u n d t d t 1 2 u n H 1 2 - 1 2 ψ n H 1 2 + k 0 t u n H 1 2 d t + F ( u n ) + F ( φ n ) = 0 t B u n , d u n d t d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equs_HTML.gif
        consequently, we have
        F ( u n ) + 1 2 u n H 1 2 + k 0 t u n H 1 2 d t = 0 t B u n , d u n d t d t + F ( φ n ) + 1 2 ψ n H 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ24_HTML.gif
        (3.14)
        by the condition (3.3),(3.14)implies
        F ( u n ) + 1 2 u n H 1 2 C 0 t F ( u n ) + 1 2 u n H 1 2 d t + f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ25_HTML.gif
        (3.15)

        where f ( t ) = 0 t g ( τ ) d t + 1 2 ψ H 1 2 + sup n F ( φ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq23_HTML.gif.

        by Gronwall inequality [Lemma(2.2)], from (3.15) we easily know:
        F ( u n ) + 1 2 u n H 1 2 f ( 0 ) e C t + 0 t f ( τ ) e C ( t - τ ) d τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equ26_HTML.gif
        (3.16)
        it implies that, for any 0 < T < ∞
        { u n } W 1 , ( 0 , T ) , X 2 L 0 , T , X 2 is bounded . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equt_HTML.gif
        now, use the same way as (1), we can obtain the result (2).
        1. (3)
          If the condition (3.4) is hold, u n X n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq24_HTML.gif, substitute v = d 2 u d t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_IEq25_HTML.gif into (3.7), we can get
          0 t d 2 u n d t 2 , d 2 u n d t 2 H + k d u n d t , d 2 u n d t 2 H d t = 0 t G u n , d 2 u n d t 2 d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equu_HTML.gif
           
        then
        0 t d 2 u n d t 2 , d 2 u n d t 2 H d t + k 2 0 t d d t u n ( t ) H 2 d t 0 t 1 2 u n ( t ) H 2 + C F ( u n ) + g ( t ) d t 0 t d 2 u n d t 2 , d 2 u n d t 2 H d t + k 2 u n H 2 k 2 ψ n H 2 + 0 t 1 2 d 2 u n d t 2 H 2 + C F ( u n ) + g ( τ ) d τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equv_HTML.gif
        by (3.16), it implies that
        0 t d 2 u n d t 2 H 2 d τ C , ( C > 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equw_HTML.gif
        consequently, for any 0 < T < ∞
        { u n } W 2 , 2 ( ( 0 , T ) , H ) is bounded . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-42/MediaObjects/13661_2011_Article_137_Equx_HTML.gif

        it implies that uW2,2((0,T), H), the main theorem (3.1) has been proved.

        Declarations

        Acknowledgements

        The author is very grateful to the anonymous referees whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. Foundation item: the National Natural Science Foundation of China (No. 10971148).

        Authors’ Affiliations

        (1)
        Yangtze Center of Mathematics, Sichuan University
        (2)
        College of Mathematics and Software Science, Sichuan Normal University

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