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Long-time behavior of a class of viscoelastic plate equations

  • This paper is concerned with the initial-boundary value problem for a class of viscoelastic plate equations on an arbitrary dimensional bounded domain. Under certain assumptions on the memory kernel and the source term, the global well-posedness of solutions and the existence of global attractors are obtained.

    Citation: Yang Liu. Long-time behavior of a class of viscoelastic plate equations[J]. Electronic Research Archive, 2020, 28(1): 311-326. doi: 10.3934/era.2020018

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  • This paper is concerned with the initial-boundary value problem for a class of viscoelastic plate equations on an arbitrary dimensional bounded domain. Under certain assumptions on the memory kernel and the source term, the global well-posedness of solutions and the existence of global attractors are obtained.



    In this paper, we study the following initial-boundary value problem for nonlinear viscoelastic plate equations

    utt+Δ2utg(tτ)Δ2u(τ)dτΔut=f(u)+h(x),  xΩ, t>0, (1)
    u(x,t)=u0(x,t), ut(x,0)=u1(x),  xΩ, t0, (2)
    u(x,t)=Δu(x,t)=0,  xΩ, tR, (3)

    where Ω is a bounded domain of RN with a smooth boundary Ω, the memory kernel g and the external forces f, h will be specified later.

    Problem 1-3 can be used to describe the vibrations of viscoelastic materials possessing a capacity of storage and dissipation of mechanical energy, see [17] for the details. And u(x,t) represents the displacement at time t of a particle having position x in a given reference configuration with the prescribed past history u0:Ω×(,0]R. In view of the main results of [5], we see that the viscoelastic term (namely the memory term) produces the effect of strong dissipation, which prevails the effect of weak damping term on the decay of solutions in time.

    There have been many works on the long-time behavior of viscoelastic plate equations, we refer the readers to [1,2,4,6,18,20,21,25] and the references therein. As for viscoelastic plate equations with past history, Pata [23] studied

    utt+αAu0g(τ)Au(tτ)dτ+μut=0,  t>0,
    u(t)=u0(t), ut(0)=u1,  t0,

    where A is a self-adjoint and strictly positive linear operator, α and μ are positive constants. Based on certain assumptions on g, he analyzed the exponential stability of the related semigroup. Guesmia and Messaoudi [13] investigated

    utt+Au0g(τ)Au(tτ)dτ=0,  t>0,
    u(t)=u0(t), ut(0)=u1,  t0.

    Under some assumptions on A and g, they established a general decay result which depends on the behavior of g. Jorge Silva and Ma [15] considered

    utt+αΔ2utg(tτ)Δ2u(τ)dτ  div(|u|p2u)Δut+f(u)=h(x),  xΩ, t>0,
    u(x,t)=u0(x,t), ut(x,t)=tu0(x,t),  xΩ, t0,
    u(x,t)=Δu(x,t)=0,  xΩ, tR,

    where hL2(Ω), and Ω is a bounded domain of RN (N1) with a smooth boundary Ω. Under some assumptions on f and g, they obtained the global well-posedness and regularity of weak solutions. Moreover, they proved the exponential decay of energy. Recently, Conti and Geredeli [9] studied

    utt+αΔ2utg(tτ)Δ2u(τ)dτ+f1(ut)+f2(u)=h(x),  xΩ, t>0,
    {u(x,0)=u0(x), ut(x,0)=u1(x),  xΩ,u(x,t)=ϕ0(x,t),  xΩ, t>0,
    u(x,t)=u(x,t)n=0,  xΩ, t>0,

    where hL2(Ω), Ω is a bounded domain of R3 with a smooth boundary Ω. Under some assumptions on f1, f2 and g, they obtained the existence and regularity of global attractors.

    In the works mentioned above, authors introduced a variable which reflects the relative displacement history so that the corresponding problem could be turned into an autonomous system. This scheme is so-called the past history approach [12] which suggests to consider some past history variables as additional components of the phase space corresponding to the equation under study.

    In the present paper, in order to study the long-time behaviour of solutions of problem 1-3, we employ the past history approach and the operator technique so that Eq. 1 can be transformed into an abstract system in the history phase space. And thus the operator technique combined with the energy estimates becomes a crucial tool for the proof of the existence of global attractors.

    This paper is organized as follows. In Section 2 some notations and assumptions on f and g are displayed. Moreover, 1-3 is transformed into a generalized problem, and the main results of this paper are stated. In Section 3 the global well-posedness of regular solutions is obtained. And the global well-posedness of weak solutions is established by the density arguments [6]. In Section 4 the existence of global attractors is derived by means of the existence of an absorbing set and the semigroup decomposition [14,16,26].

    Throughout the paper, in order to simplify the notations, we denote

    p:=Lp(Ω), :=L2(Ω).

    (,) denotes either the L2-inner product or a duality pairing between a space and its dual space. Moreover, |Ω| stands for the Lebesgue measure of Ω, Ci, i=1,2,3, denote some different positive constants, and Ci, i=1,2,3 represent the positive constants for inequalities

    uC1u, uC2Δu, uC3Δu.

    As in [3,10,27], we give the following assumptions on f in order to state the main results of this paper.

    (A1): f(0)=0, and there exists a constant b>0 such that

    |f(u)f(v)|b(|u|p2+|v|p2)|uv|,  u,vR,

    where

    2p< if N4, 2p2N4N4 if N>4.

    Moreover,

    lim sup|u|F(u)|u|20, (4)

    and

    lim sup|u|uf(u)ϱF(u)|u|20, (5)

    where 0<ϱ<1 and

    F(u)=u0f(s)ds.

    In addition, as in [7,19,22], we assume that g satisfies the following conditions.

    (A2): gC1(R+)L1(R+), g(t)0, g(t)0, t[0,), and

    κ:=10g(t)dt>0. (6)

    As in [2,3,8,10,27], we define the operator A:D(A)L2(Ω)L2(Ω)

    Au=Δ2u,  uD(A),

    where the dense domain

    D(A)={uH4(Ω)H10(Ω)|ΔuH2(Ω)H10(Ω)}.

    It is easy to verify that A is self-adjoint and strictly positive. Thus Aγ is also self-adjoint and strictly positive for any γ>0. Denote V4γ:=D(Aγ) and Vγ:=Vγ. Then, for any γR, Vγ and Lg,γ are Hilbert spaces equipped with inner products and norms

    (u,w)Vγ=(Aγ4u,Aγ4w), uVγ=Aγ4u,
    (v,w)g,γ=0g(τ)(v(τ),w(τ))Vγdτ, v2g,γ=0g(τ)v(τ)2Vγdτ,

    where

    Lg,γ:=L2g(R+;Vγ)={v:R+Vγ|0g(τ)v(τ)2Vγdτ<}.

    Thus

    V3:=H3(Ω)={uH3(Ω)H10(Ω)|ΔuH10(Ω)},
     V2:=H2(Ω)H10(Ω), V1:=H10(Ω), V0:=L2(Ω).

    In this way, problem 1-3 can be seen as

    utt+Autg(tτ)Au(τ)dτ+A12ut=f(u)+h,  t>0, (7)
    u(t)=u0(t), ut(0)=u1,  t0.

    Now we are in a position to define the auxiliary function

    vt(τ)=u(t)u(tτ),  τ>0, t0.

    Thus the viscoelastic dissipation in 7 can be rewritten as

    tg(tτ)Au(τ)dτ=0g(τ)Au(tτ)dτ=(1κ)Au+0g(τ)Avt(τ)dτ.

    Therefore, problem 1-3 is transformed into the following system

    {utt+κAu+0g(τ)Avt(τ)dτ+A12ut=f(u)+h,  t>0,vtt(τ)=ut(t)vtτ(τ),  τ>0, t>0, (8)

    with

    {u(0)=u0, ut(0)=u1,v0(τ)=v0(τ), (9)

    where

    u0=u0(0),v0(τ)=u0(0)u0(τ),  τ>0.

    Definition 2.1. (u(t),ut(t),vt) is called a weak solution of problem 8, 9 if uC([0,T];V2), utC([0,T];V0), vtC([0,T];Lg,2), u(0)=u0 in V2, ut(0)=u1 in V0, v0=v0 in Lg,2, and

    {(ut,w1)+κt0(A12u,A12w1)dτ+t0(vs,w1)g,2ds  +(A14u,A14w1)=t0(f(u)+h,w1)dτ+(u1,w1)+(A14u0,A14w1),(vt,w2)g,2=(u,w2)g,2(u0,w2)g,2t0(vsτ,w2)g,2ds+(v0,w2)g,2.

    for any w1V2, w2Lg,2 and a.e. t(0,T]. }

    Thus, in order to deal with problem 1-3, we study the modified problem 8, 9. In fact, for a solution (u,ut,vt) of problem 8, 9, we have

    (utt,w1)+κ(A12u,A12w1)+(vt,w1)g,2+(A14ut,A14w1)=(f(u)+h,w1).

    In view of [17,Chapter 2,Section 4], we see that g(t):=G(t), where G(t) is the viscoelastic flexural rigidity. From

    κA12u=(1+0G(τ)dτ)A12u=A12u+(limτG(τ)G(0))A12u,

    it follows that

    (utt,w1)+(A12u,A12w1)+(limτG(τ)G(0))(A12u,A12w1)  +(vt,w1)g,2+(A14ut,A14w1)=(f(u)+h,w1). (10)

    Since

    (limτG(τ)G(0))(A12u,A12w1)+(vt,w1)g,2=(limτG(τ)G(0))(A12u,A12w1)+t0g(τ)(A12vt(τ),A12w1)dτ+tg(τ)(A12vt(τ),A12w1)dτ,

    and

    vt(τ)={u(t)u0(tτ),τt,u(t)u(tτ),τ<t,

    we deduce that

    (limτG(τ)G(0))(A12u,A12w1)+(vt,w1)g,2=(limτG(τ)G(0))(A12u,A12w1)t0G(τ)(A12u(t),A12w1)dτ+t0G(τ)(A12u(tτ),A12w1)dτtG(τ)(A12u(t),A12w1)dτ+tG(τ)(A12u0(tτ),A12w1)dτ=t0G(τ)(A12u(tτ),A12w1)dτ+tG(τ)(A12u0(tτ),A12w1)dτ.

    Substituting this into 10, we obtain

    (utt,w1)+(A12u,A12w1)+t0G(τ)(A12u(tτ),A12w1)dτ  +tG(τ)(A12u0(tτ),A12w1)dτ+(A14ut,A14w1)=(f(u)+h,w1).

    Due to

    t0G(τ)(A12u(tτ),A12w1)dτ+tG(τ)(A12u0(tτ),A12w1)dτ=tg(tτ)(A12u(τ),A12w1)dτ,

    we conclude that

    (utt,w1)+(A12u,A12w1)tg(tτ)(A12u(τ),A12w1)dτ  +(A14ut,A14w1)=(f(u)+h,w1),

    which shows that (u,ut) is a solution of problem 1-3.

    The main results of this paper are stated as follows.

    Theorem 2.2. Let (A1) and (A2) be fulfilled. Assume that hV0 and (u0,u1,v0)Z:=V2×V0×Lg,2. Then problem 8, 9 admits a unique solution (u,ut,vt)C([0,);Z) depending continuously on initial data.

    Define the mapping S(t):ZZ by

    S(t)(u0,u1,v0)=(u(t),ut(t),vt).

    Then it is easy to see from Theorem 2.2 that {S(t)}t0 is a C0-semigroup generated by problem 8, 9.

    Theorem 2.3. Let (A1) and (A2) be fulfilled. And there exists a constant ρ>0 such that g(t)+ρg(t)0 for all t[0,). Assume that hV0 and (u0,u1,v0)Z. Then S(t) possesses a global attractor in Z.

    Theorem 3.1. Let (A1) and (A2) be fulfilled. Assume that hV0, u0V3, u1V1, v0Lg,3. Then problem 8, 9 admits a unique solution uL(0,;V3), utL(0,;V1)L2(0,;V2), vtL(0,;Lg,3), which depends continuously on initial data.

    Proof. Let {ωj}j=1 be an orthogonal basis of V2 and an orthonormal basis of V0 given by eigenfunctions of A. As in [11,15], we select {ej}j=1 of the form {lk˜ωj}k,j=1, where {lk}k=1 is an orthonormal basis of L2g(R+)C0(R+) and ˜ωj=ωjωjV2. Then {ej}j=1 is an orthonormal basis of Lg,2.

    We construct the approximate solutions of problem 8, 9

    un(t)=nj=1ξjn(t)ωj, vtn(τ)=nj=1ζjn(t)ej(τ),  n=1,2,,

    which satisfy

    {(untt,ωj)+κ(A12un,A12ωj)+(vtn,ωj)g,2  +(A14unt,A14ωj)=(f(un),ωj)+(h,ωj),(vtnt,ej)g,2=(unt,ej)g,2(vtnτ,ej)g,2,  j=1,2,,n, (11)

    with

    {un(0)=nj=1ξjn(0)ωju0 in V3,unt(0)=nj=1ξjn(0)ωju1 in V1,v0n=nj=1ζjn(0)ejv0 in Lg,3. (12)

    The approximate problem 11, 12 can be reduced to an ordinary differential system in the variables ξjn(t) and ζjn(t). In terms of standard theory for ODEs, there exists a solution (un(t),unt(t),vtn) on some interval [0,Tn) with TnT. The following estimates will allow us to extend the local solutions to [0,T] with any T>0.

    Replacing ωj in 111 with unt and ej in 112 with vtn, summing for j and adding the two results, we obtain

    En(t)+A14unt2=(vtnτ,vtn)g,2, (13)

    where

    En(t)=12unt2+κ2A12un2+12vtn2g,2ΩF(un)dx(h,un). (14)

    Since limτ0vtn(τ)=0, we deduce from (A2) that

    (vtnτ,vtn)g,2=120τ(g(τ)A12vtn(τ)2)dτ120g(τ)A12vtn(τ)2dτ0.

    Hence, by integrating 13 with respect to t from 0 to t, we get

    En(t)+t0A14unτ2dτEn(0). (15)

    It follows from 4 in (A1) that, for any η>0, there exists a constant Cη>0 such that

    ΩF(un)dxηun2+Cη|Ω|.

    By virtue of Cauchy's inequality with ϵ>0, we get

    (h,un)hunϵC22A12un2+14ϵh2.

    Consequently, taking sufficiently small η and ϵ such that

    C1:=κ2ηC22ϵC22>0,

    we deduce from 14 that

    En(t)12unt2+C1A12un2+12vtn2g,2C2(h2+|Ω|). (16)

    Hence, from 15, 16 and 12, it follows that

    unt2+A12un2+vtn2g,2+t0A14unτ2dτC3. (17)

    Replacing ωj in 111 with A12unt and ej in 112 with A12vtn, summing for j and adding the two results, we obtain

    12ddt(A14unt2+κA34un2+vtn2g,3)+A12unt2=(f(un),A12unt)+(h,A12unt)(vtnτ,vtn)g,3.

    Noting that

    (vtnτ,vtn)g,30,
    (f(un),A12unt)C4A12un2p2+14A12unt2,

    and

    (h,A12unt)h2+14A12unt2,

    we conclude from 17 that

    A14unt2+A34un2+vtn2g,3+t0A12unτ2dτC5. (18)

    Therefore, there exist u, vt and subsequences of {un}, {vtn}, still represented by the same notations and we shall not repeat, such that, as n,

    unu weakly star in L(0,T;V3), (19)
    untut weakly star in L(0,T;V1) and weakly in L2(0,T;V2), (20)
    vtnvt weakly star in L(0,T;Lg,3),

    for any T>0. According to the Aubin-Lions lemma, we have

    unu in L2(0,T;V2).

    Moreover, from 18-20, it follows that

    unu in C([0,T];V2). (21)

    We now claim that for any t[0,T] and fixed j,

    t0(f(un),ωj)dτt0(f(u),ωj)dτ, (22)

    as n.

    Indeed, for any wV2, we have

    (f(un)f(u),w)b((|un|p2+|u|p2)|unu|,w).

    If p>2, then when N4,

    (f(un)f(u),w)b(unp24(p2)+up24(p2))unu4w,

    when N>4,

    (f(un)f(u),w)b(unp2N(p2)3+up2N(p2)3)unu2NN2w2NN4.

    Hence

    (f(un)f(u),w)C6(unp2V2+up2V2)unuV1wV2C7unuV1wV2. (23)

    If p=2, then it is clear that 23 remains valid. Therefore,

    |t0(f(un)f(u),wj)dτ|C8t0unuV1dτ.

    Thus assertion 22 follows from 21.

    For fixed j,

    (vtnτ,ej)g,2=0g(τ)(A12vtn(τ),A12ej(τ))dτ0g(τ)(A12vtn(τ),A12ejτ(τ))dτ.

    Hence

    limn(vtnτ,ej)g,2=(vtτ,ej)g,2.

    Consequently, for fixed j, integrating 11 with respect to t and taking n, we get

    {(ut,ωj)+κt0(A12u,A12ωj)dτ+t0(vs,ωj)g,2ds  +(A14u,A14ωj)=t0(f(u)+h,ωj)dτ+(u1,ωj)+(A14u0,A14ωj),(vt,ej)g,2=(u,ej)g,2(u0,ej)g,2t0(vsτ,ej)g,2ds+(v0,ej)g,2.

    Moreover, it is easy to see from 12 that u(0)=u0 in V3, ut(0)=u1 in V1, v0=v0 in Lg,3. Therefore, (u,ut,vt) is a solution of problem 8, 9.

    Next we prove continuous dependence of (u(t),ut(t),vt) on (u0,u1,v0). Suppose that (u,ut,vt) and (ˉu,ˉut,ˉvt) are two regular solutions of problem 8, 9 with initial data (u0,u1,v0) and (ˉu0,ˉu1,ˉv0), respectively. Set ˜u=ˉuu and ˜vt=ˉvtvt. Then

    {˜utt+κA˜u+0g(τ)A˜vt(τ)dτ+A12˜ut=f(ˉu)f(u),˜vtt=˜ut˜vtτ, (24)
    {˜u(0)=˜u0=ˉu0u0, ˜ut(0)=˜u1=ˉu1u1,˜v0(τ)=˜v0=ˉv0v0.

    By the arguments similar to [7,Lemma 4.9], we obtain

    12ddt(˜ut2+κA12˜u2+˜vt2g,2)+A14˜ut2=(f(ˉu)f(u),˜ut)(˜vtτ,˜vt)g,2. (25)

    By the arguments similar to the proof of 23, we have

    (f(ˉu)f(u),˜ut)C9A12˜uA14˜utC94ϵA12˜u2+C9ϵA14˜ut2.

    Note that (˜vtτ,˜vt)g,20. Hence, by taking ϵ=1C9, we deduce from 25 that

    12ddt(˜ut2+κA12˜u2+˜vt2g,2)C294A12˜u2.

    As a consequence, by Gronwall's inequality, we obtain

    ˜ut2+A12˜u2+˜vt2g,2C10(˜u12+A12˜u02+˜v02g,2). (26)

    In particular, by taking (u0,u1,v0)=(ˉu0,ˉu1,ˉv0), it is obvious that (u,ut,vt) is the unique solution of problem 8, 9.

    Proof of Theorem 2.2. For u0V2, u1V0, v0Lg,2, there exist {u0m}V3, {u1m}V1, {v0m}Lg,3 such that

    u0mu0 in V2, u1mu1 in V0, v0mv0 in Lg,2.

    According to Theorem 3.1, for any mN+, problem 8, 9 admits a unique regular solution (um,umt,vtm) satisfying

    {umtt+κAum+0g(τ)Avtm(τ)dτ+A12umt  =f(um)+h,  t(0,),vtmt(τ)=umt(t)vtmτ(τ),  τ(0,), t(0,),
    {um(0)=u0m, umt(0)=u1m,v0m(τ)=v0m(τ).

    Hence umC([0,T];V2), umtC([0,T];V0). Moreover, according to [24,Theorem 3.2], we have vtmC([0,T];Lg,2).

    Set ym=um2um1 and zm=vtm2vtm1. Then, by the arguments similar to the proof of 26, we get

    ymt2+A12ym2+zm2g,2C11(ymt(0)2+A12ym(0)2+zm(0)2g,2). (27)

    By 27 and the arguments similar to the proof of 17, we have

    umu in C([0,T];V2), (28)
    umtut in C([0,T];V0), (29)
    vtmvt in C([0,T];Lg,2). (30)

    Thus (u,ut,vt) is a global weak solution of problem 8, 9.

    Suppose that (u,ut,vt) and (ˉu,ˉut,ˉvt) are two solutions of problem 8, 9 with initial data (u0,u1,v0), (ˉu0,ˉu1,ˉv0), respectively. Then there exist

    (u0m,u1m,v0m)V3×V1×Lg,3, (ˉu0m,ˉu1m,ˉv0m)V3×V1×Lg,3,

    such that

    (u0m,u1m,v0m)(u0,u1,v0) in V2×V0×Lg,2, (31)
    (ˉu0m,ˉu1m,ˉv0m)(ˉu0,ˉu1,ˉv0) in V2×V0×Lg,2. (32)

    Set ˜um=ˉumum and ˜vtm=ˉvtmvtm. Then, on account of 26, we obtain

    ˜umt2+A12˜um2+˜vtm2g,2C12(˜u1m2+A12˜u0m2+˜v0m2g,2).

    Therefore, in terms of 28-32, the conclusions of Theorem 2.2 are derived immediately.

    In this section, for the sake of convenience, we denote

    S(t)(u0,u1,v0)2Z:=u2V2+ut2V0+vt2g,2.

    Lemma 4.1. Under the conditions of Theorem 2.3, S(t) possesses an absorbing set in Z.

    Proof. Let U=ut+εu, t[0,), where ε is a positive constant to be determined later. Then 81 becomes

    Ut+A12UεUεA12u+ε2u+κAu+0g(τ)Avt(τ)dτ=f(u)+h. (33)

    Note that

    (vtτ,vt)g,2ρ2vt2g,2.

    Multiplying 33 by U in V0 and 82 by vt in Lg,2, integrating over Ω and adding the two results, we obtain

    E1(t)+E2(t)0, (34)

    where

    E1(t)=12(U2+κA12u2+vt2g,2+ε2u2εA14u22ΩF(u)dx2(h,u)),

    and

    E2(t)=A14U2εU2ε2A14u2+ε3u2+εκA12u2+ε0g(τ)(A12vt(τ),A12u(t))dτε(f(u),u)ε(h,u)+ρ2vt2g,2.

    Hence

    E2(t)εϱE1(t)=εκ(2ϱ)2A12u2ε2(1ϱ2)A14u2+ε3(1ϱ2)u2+4i=1Λi,

    where

    Λ1=A14U2ε(1+ϱ2)U2,
    Λ2=ρ2vt2g,2+ε0g(τ)(A12vt(τ),A12u(t))dτϱε2vt2g,2,
    Λ3=ε(ϱΩF(u)dx(f(u),u)),
    Λ4=ε(1ϱ)(h,u).

    Applying Cauchy's inequality with ϵ1>0, we get

    Λ2ρ2vt2g,2ϵ1ε(1κ)A12u2ε4ϵ1vt2g,2ϱε2vt2g,2.

    It follows from 5 that, for any η>0, there exists a constant Cη>0 such that

    Λ3ε(ηu2+Cη|Ω|)εηC22A12u2εCη|Ω|.

    Moreover,

    Λ11C21U2ε(1+ϱ2)U2,

    and

    Λ4ε(1ϱ)(ϵ2C22A12u2+14ϵ2h2).

    Consequently, by taking sufficiently small ϵ1, η and ϵ2 such that

    C13:=κ(2ϱ)2ϵ1(1κ)ηC22ϵ2(1ϱ)C22>0,

    we deduce that

    E2(t)εϱE1(t)εC13A12u2ε2(1ϱ2)A14u2+[1C21ε(1+ϱ2)]U2+(ρ2ε4ϵ1ϱε2)vt2g,2εCη|Ω|ε(1ϱ)4ϵ2h2.

    Choosing

    εmin{2C13(2ϱ)C23, 2(2+ϱ)C21, 2ϵ1ρ1+2ϵ1ϱ},

    we obtain

    E2(t)εϱE1(t)C14(h2+|Ω|) (35)

    and

    κA12u2εA14u2C15A12u2. (36)

    Since

    U2ut2ε2u2,

    we conclude from 36 and the arguments similar to the proof of 16 that

    E1(t)C16(ut2+A12u2+vt2g,2)C17(h2+|Ω|). (37)

    It follows from 34 and 35 that

    E1(t)+εϱE1(t)C14(h2+|Ω|),

    which yields

    E1(t)E1(0)eεϱt+C14εϱ(h2+|Ω|).

    This, together with 37, gives

    S(t)(u0,u1,v0)2ZE1(0)C16eεϱt+C14+εϱC17εϱC16(h2+|Ω|).

    Hence S(t) possesses an absorbing set with the radius R>C14+εϱC17εϱC16(h2+|Ω|).

    Proof. We decompose u=ˆu+ˇu and vt=ˆvt+ˇvt satisfying

    {ˆutt+κAˆu+A12ˆut+0g(τ)Aˆvt(τ)dτ=0,ˆvtt=ˆutˆvtτ,ˆu(0)=u0, ˆut(0)=u1, ˇv0(τ)=v0(τ), (38)
    {ˇutt+κAˇu+A12ˇut+0g(τ)Aˇvt(τ)dτ=Φ,ˇvtt=ˇutˇvtτ,ˇu(0)=0, ˇut(0)=0, ˇv0(τ)=0, (39)

    where

    Φ=f(u)+h.

    Let ψ=Aδˇu, φt=Aδˇvt, 0<δ14. Then it follows from 39 that

    {ψtt+κAψ+A12ψt+0g(τ)Aφt(τ)dτ=AδΦ,φtt=ψtφtτ,ψ(0)=0, ψt(0)=0, φ0(τ)=0. (40)

    Let Ψ=ψt+εψ, where ε is a positive constant to be determined later. Then 401 can be written in the form

    Ψt+A12ΨεΨεA12ψ+ε2ψ+κAψ+0g(τ)Aφt(τ)dτ=AδΦ. (41)

    Multiplying 41 by Ψ in V0 and 402 by φt in Lg,2, integrating over Ω and adding the two results, we obtain

    E3(t)+2A14Ψ22εΨ22ε2A14ψ2+2ε3ψ2+2εκA12ψ2  +2ε(φt(τ),ψ(t))g,2=2(AδΦ,Ψ)2(φtτ,φt)g,2,

    where

    E3(t)=Ψ2+κA12ψ2εA14ψ2+ε2ψ2+φt2g,2.

    Hence

    E3(t)+2i=1Λi2ε2A14ψ2+2ε3ψ2+2εκA12ψ22(AδΦ,Ψ), (42)

    where

    Λ1=2A14Ψ22εΨ2,

    and

    Λ2=2ε(φt(τ),ψ(t))g,2+ρφt2g,2.

    Note that

    Λ1A14Ψ2+(1C212ε)Ψ2,

    and

    Λ2(ρε2ϵ)φt2g,22εϵ(1κ)A12ψ2.

    Consequently, taking

    ϵκ(2σ1)2(1κ),

    with some constant 0<σ1<2, we deduce from 42 that

    E3(t)+A14Ψ2+(1C212ε)Ψ22ε2A14ψ2+2ε3ψ2  +σ1εκA12ψ2+(ρε2ϵ)φt2g,22(AδΦ,Ψ).

    We further choose

    ε<min{1(σ2+2)C21, 2ϵρ1+2ϵσ2, (σ1σ2)κ2C23, κC23},

    with some constant 0<σ2<σ1. Thus

    E3(t)+σ2εE3(t)+A14Ψ22(AδΦ,Ψ) (43)

    and

    E3(t)ψt2+(κεC23)A12ψ2+φt2g,2. (44)

    Applying Hölder's inequality and Cauchy's inequality to the right side of 43, we get

    E3(t)+σ2εE3(t)Φ2V4δ1. (45)

    For any wV14δ, we have

    (f(u),w)b|u|p12NN+2(14δ)w2NN2(14δ)C18up1V2wV14δ.

    Since

    ΦV4δ1f(u)V4δ1+hV4δ1,

    we deduce from 45 that

    E3(t)E3(0)eσ2εt+C19.

    Combining this with 403 and 44, we obtain

    ˇu2V2+4δ+ˇut2V4δ+ˇvt2g,2+4δC20,  t[0,). (46)

    Taking into account

    ˇvt(τ)={ˇu(t),τt,ˇu(t)ˇu(tτ),0<τ<t,

    we get

    ˇvtτ(τ)={0,τt,ˇut(tτ),0<τ<t.

    Let Υ=t0ˇvt. Then Υ is bounded in Lg,2+4δH1g(R+,V4δ) due to 46, where H1g(R+,V4δ) is a Hilbert space of V4δ-valued functions v on R+ such that both v and vτ belong to Lg,4δ. Note that supvΥv(τ)2V2L1g(R+). Hence, in view of [24,Lemma 5.5], we see that Υ is relatively compact in Lg,2. Since V2+4δ×V4δ↪↪V2×V0, we conclude from 46 that there exists a t0=t0(B) such that tt0S2(t)B is relatively compact. Thus the operators S2(t) are uniformly compact for t large. Furthermore, as for problem 38, it is easy to check that

    ˆu2V2+ˆut2V0+ˆvt2g,2C21eσ2εt,  t[0,).

    This means that S1(t) is a continuous mapping from Z into itself such that

    sup(u0,u1,v0)BS1(t)(u0,u1,v0)Z0,

    as t. Therefore, by virtue of [26,Chapter I,Theorem 1.1] and Lemma 4.1, the proof of Theorem 2.3 is complete.

    The author would like to thank the editors and the referees for the valuable comments and suggestions.



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