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Recent coating materials for textile-based solar cells

  • Electronic textiles may be used to monitor a person’s vital function, to offer information or navigation, and to support people in daily life. One crucial problem, however, is the energy supply needed for such electronic functions. One of the possibilities discussed for this purpose is the integration of solar cells into garments. While silicon-based solar cells are mostly rigid or can in the meantime also be prepared in the shape of relatively flexible foils, the approach to integrate solar cell functionality into a textile fabric goes one step further. In this case, dye-sensitized solar cells (DSSCs), polymer or organic solar cells are advantageous. During the last years, promising advances are reported in the literature, discussing new materials for the different layers of such flexible solar cells. This review gives an overview of the recent materials used for creating flexible solar cells on textile fabrics or fibers.

    Citation: Andrea Ehrmann, Tomasz Blachowicz. Recent coating materials for textile-based solar cells[J]. AIMS Materials Science, 2019, 6(2): 234-251. doi: 10.3934/matersci.2019.2.234

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  • Electronic textiles may be used to monitor a person’s vital function, to offer information or navigation, and to support people in daily life. One crucial problem, however, is the energy supply needed for such electronic functions. One of the possibilities discussed for this purpose is the integration of solar cells into garments. While silicon-based solar cells are mostly rigid or can in the meantime also be prepared in the shape of relatively flexible foils, the approach to integrate solar cell functionality into a textile fabric goes one step further. In this case, dye-sensitized solar cells (DSSCs), polymer or organic solar cells are advantageous. During the last years, promising advances are reported in the literature, discussing new materials for the different layers of such flexible solar cells. This review gives an overview of the recent materials used for creating flexible solar cells on textile fabrics or fibers.


    The Moore-Gibson-Thompson (MGT) equation is one of the equations of nonlinear acoustics describing acoustic wave propagation in gases and liquids [13,15,30] and arising from modeling high frequency ultrasound waves [9,18] accounting for viscosity and heat conductivity as well as effect of the radiation of heat on the propagation of sound. This research field is highly active due to a wide range of applications such as the medical and industrial use of high intensity ultrasound in lithotripsy, thermotherapy, ultraound cleaning, etc. The classical nonlinear acoustics models include the Kuznetson's equation, the Westervelt equation and the Kokhlov-Zabolotskaya-Kuznetsov equation.

    In order to gain a better understanding of the nonlinear MGT equation, we shall begin with the linearized model. In [15], Kaltenbacher, Lasiecka and Marchand investigated the following linearized MGT equation

    τuttt+αutt+c2Au+bAut=0. (1.1)

    For equation (1.1), they disclosed a critical parameter γ=αc2τb and showed that when γ>0, namely in the subcritical case, the problem is well-posed and its solution is exponentially stable; while γ=0, the energy is conserved. Since its appearance, an increasing interest has been developed to study the MGT equation, see [4,5,8,10,14]. Caixeta, Lasiecka and Cavalcanti [4] considered the following nonlinear equation

    τuttt+αutt+c2Au+bAut=f(u,ut,utt). (1.2)

    They proved that the underlying PDE generates a well-posed dynamical system which admits a global and finite dimensional attractor. They also overcomed the difficulty of lacking the Lyapunov function and the lack of compactness of the trajectory.

    Now, we concentrate on the stabilization of MGT equation with memory which has received a considerable attention recently. For instance, Lasiecka and Wang [17] studied the following equation:

    τuttt+αutt+bAut+c2Aut0g(ts)Aw(s)ds=0, (1.3)

    where αc2τb0 and the form of w classifies the memory into three types. By imposing the assumption on the relaxation function g, for a positive constant c0, as

    g(t)c0g(t), (1.4)

    they discussed the effect of memory described by three types on decay rates of the energy when αc2τb>0. Moreover, in the critical case αc2τb=0, they proved an exponential rate of decay for the solution of (1.3) under "the right mixture" of memory. Lasiecka and Wang [18] showed the general decay result of the equation (1.3) when w=u, and established their result under weaker condition on g. In [9], Filippo et al. investigated the critical case of equation (1.3) (that is αbc2τ=0) for w=u and g satisfies (1.4), and obtained an exponential decay result if and only if A is a bounded operator. When t0 is replaced by 0, (1.3) turns to

    τuttt+αutt+bAut+c2Au0g(s)Aw(ts)ds=0. (1.5)

    Alves et al. [1] investigated the uniform stability of equation (1.5) encompassing three different types of memory in a history space set by the linear semigroup theory. Moreover, we refer the reader to [3,6,7,12,24,25,26,28] for other works of the equation(s) with memory.

    More recently, Filippo and Vittorino [11] considered the fourth-order MGT equation

    utttt+αuttt+βutt+γAutt+δAut+ϱAu=0. (1.6)

    They investigated the stability properties of the related solution semigroup. And, according to the values of certain stability numbers depending on the strictly positive parameters α, β, γ, δ, ϱ, they established the necessary and sufficient condition for exponential stability. For other related results on the higher-order equations, please see [20,27,34,35,36,37] and the references therein.

    Motivated by the above results, we intend to study the following abstract version of the fourth-order Moore-Gibson-Thompson (MGT) equation with a memory term

    utttt+αuttt+βutt+γAutt+δAut+ϱAut0g(ts)Au(s)ds=0, (1.7)

    where α, β, γ, δ, ϱ are strictly positive constants, A is a strictly positive self-adjoint linear operator defined in a real Hilbert space H where the (dense) embedding D(A)H need not to be compact. And we consider the following initial conditions

    u(0)=u0,ut(0)=u1,utt(0)=u2,uttt(0)=u3. (1.8)

    A natural question that arised in dealing with the general decay of fourth-order MGT equation with memory:

    ● Can we get a general decay result for a class of relaxation functions satisfying g(t)ξ(t)M(g(t)) for M to be increasing and convex function near the origin and ξ(t) to be a nonincreasing function?

    Mustafa answered this question for viscoelastic wave equations in [31,32]. Messaoudi and Hassan [29] considered the similar question for memory-type Timoshenko system in the cases of equal and non-equal speeds of wave propagation. Moreover, they extended the range of polynomial decay rate optimality from p[1,32) to p[1,2) when g satisfies g(t)ξ(t)gp(t). We refer to [19] for the non-equal wave speeds case. And, Liu et al. [22,23] also concerned with the similar question for third-order MGT equations with memory term.

    The aim of this paper is to establish the well-posedness and answer the above mention question for fourth-order MGT equation with memory (1.7). We first use the Faedo-Galerkin method to prove the well-posedness result. We then use the idea developed by Mustafa in [31,32], taking into consideration the nature of fourth-order MGT equation, to prove new general decay results for the case γδα>0 and βαϱδ>0, based on the perturbed energy method and on some properties of convex functions. Our result substantially improves and generalizes the earlier related results in previous literature.

    The rest of our paper is organized as follows. In Section 2, we give some assumptions and state our main results. In Section 3, we give the proof of well-posedness. In Section 4, we state and prove some technical lemmas that are relevant in the entire work. In Section 5, we prove the general decay result.

    In this section, we consider the following assumptions and state our main results. We use c>0 to denote a positive constant which does not depend on the initial data.

    First, we consider the following assumptions as in [11] for (A1), in [18] for (A3), (A5) and in [31] for (A2), (A4):

    (A1) γδα>0 and βαϱδ>0.

    (A2) g:R+R+  is a non-increasing differentiable function such that

    0<g(0)<2αϱδ(αγδ),    ϱ+0g(s)ds=l>0.

    (A3) g(t)0 almost everywhere.

    (A4) There exists a non-increasing differentiable function ξ : R+R+ and a C1 function M: [0,)[0,) which is either linear or strictly increasing and strictly convex C2 function on (0,r], rg(0), with M(0)=M(0)=0, such that

    g(t)ξ(t)M(g(t)), t0. (2.1)

    (A5) There exists λ0>0 such that A satisfies u2λ0A12u2 for all uH.

    Remark 1. ([31,Remark 2.8]) (1) From assumption (A2), we deduce that

    g(t)0ast+andg(t)ϱlt, t>0.

    Furthermore, from the assumption (A4), we obtain that there exists t00 large enough such that

    g(t0)=randg(t)r, tt0.

    The non-increasing property of g(t) and ξ(t) gives

    0<g(t0)g(t)g(0)and0<ξ(t0)ξ(t)ξ(0), t[0,t0].

    A combination of these with the continuity of H, for two constants a,d>0, yields

    aξ(t)M(g(t))d, t[0,t0].

    Consequently, for any t[0,t0], we get

    g(t)ξ(t)M(g(t))a=ag(0)g(0)ag(0)g(t)

    and, hence,

    g(t)g(0)ag(t), t[0,t0]. (2.2)

    (2) If M is a strictly increasing and strictly convex C2 function on (0,r], with M(0)=M(0)=0, then it has an extension ¯M, which is strictly increasing and strictly convex C2 function on (0,). For example, if we set M(r)=A, M(r)=B, M(r)=C, we can define ¯M, for any t>r, by

    ¯M=C2t2+(BCr)t+(A+C2r2Br).

    Then, inspired by the notations in [11], we define the Hilbert spaces

    Hr:=D(Ar2),rR.

    In order to simplify the notation, we denote the usual space H0 by H. The phase space of our problem is

    H=D(A12)×D(A12)×D(A12)×H.

    Moreover, we will denote the inner product of H by (,) and its norm by .

    After that, we introduce the following energy functional

    E(t)=12[uttt+αutt+αϱδut2+δα(ϱG(t)ϱ)A12utt+αA12ut+αϱδA12u2+δαϱG(t)A12utt+αA12ut2+(γδα)A12utt2+(γδα)αϱδA12ut2+2t0g(ts)(A12u(t)A12u(s),A12utt+αA12ut)ds+αϱδ(gA12u)(t)α(gA12u)(t)+αg(t)A12u2+(βαϱδ)utt2+αϱδ(βαϱδ)ut2],

    where G(t)=t0g(s)ds and for any vL2loc(R+;L2(Ω)),

    (gv)(t):=Ωt0g(ts)(v(t)v(s))2dsdx.

    As in [31], we set, for any 0<ν<1,

    Cν=0g2(s)νg(s)g(s)dsandh(t)=νg(t)g(t).

    The following lemmas play an important role in the proof of our main results.

    Lemma 2.1. ([31] ) Assume that condition (A2) holds. Then for any uL2loc(R+; L2(Ω)), we have

    Ω(t0g(ts)(A12u(s)A12u(t))ds)2dxCν(hA12u)(t), t0.

    Lemma 2.2. (Jensen's inequality) Let P: [b,c]R be a convex function. Assume that functions f:Ω[b,c] and h:ΩR are integrable such that h(x)0, for any xΩ and Ωh(x)dx=k>0. Then

    P(1kΩf(x)h(x)dx)1kΩP(f(x))h(x)dx.

    Lemma 2.3. ([2])(The generalized Young inequality) If f is a convex function defined on a real vector space X and ite convex conjugate is denoted by f, then

    ABf(A)+f(B), (2.3)

    where

    f(s)=s(f)1(s)f[(f)1(s)]. (2.4)

    We are now in a position to state the well-posedness and the general decay result for problem (1.7)-(1.8).

    Theorem 2.4. (Well-posedness) Assume that (A1)(A5) hold. Then, for given (u0,u1,u2,u3)H and T>0, there exists a unique weak solution u of problem (1.7)-(1.8) such that

    uC([0,T];D(A12))C1([0,T];H).

    Theorem 2.5. (General decay) Let (u0,u1,u2,u3)H. Assume that (A1)-(A5) hold. Then there exist positive constants k1 and k2 such that, along the solution of problem (1.7)-(1.8), the energy functional satisfies

    E(t)k2M11(k1tg1(r)ξ(s)ds), tg1(t), (2.5)

    where M1(t)=rt1sM(s)ds and M1 is strictly decreasing and convex on (0,r], with limt0M1(t)=+.

    Remark 2. Assume that M(s)=sp, 1p<2 in (A4), then by simple calculations, we see that the decay rate of E(t) is given by, for constants ¯k, ˜k and C,

    E(t){Cexp(˜kt0ξ(s)ds),ifp=1,¯k(1+t0ξ(s)ds)1p1,if1<p<2. (2.6)

    In this section, we will prove the global existence and uniqueness of the solution of problem (1.7)-(1.8). Firstly, we give the following lemmas.

    Lemma 3.1. If 0<g(0)<2αϱδ(αγδ), then there is σ>0 such that α(γδα)δg(0)2(ασ)ϱ>0.

    Proof. Since g(0)<2αϱδ(αγδ), all we need is to show

    2αϱδ(ασ)(γδα)2αϱδ(αγδ)asσ0,

    which is trivially true.

    Lemma 3.2. Assume that (A1)-(A5) hold. Then, the energy functional E(t) satisfies, for all t0,

    12[uttt+αutt+αϱδut2+δα(ϱG(t)ϱ)A12utt+αA12ut+αϱδA12u2+(γδα)A12utt2+(γδα)αϱδA12ut2α(gA12u)(t)+αg(t)A12u2+(βαϱδ)utt2+αϱδ(βαϱδ)ut2]E(t)12[uttt+αutt+αϱδut2+δα(ϱG(t)ϱ)A12utt+αA12ut+αϱδA12u2+(γδα)A12utt2+(γδα)αϱδA12ut2α(gA12u)(t)+2αϱδ(gA12u)(t)+αg(t)A12u2+(βαϱδ)utt2+αϱδ(βαϱδ)ut2+2δαϱG(t)A12utt+αA12ut2].

    Proof. From the definition of E(t), we have

    E(t)=12[uttt+αutt+αϱδut2+δα(ϱG(t)ϱ)A12utt+αA12ut+αϱδA12u2+δαϱG(t)A12utt+αA12ut2+(γδα)A12utt2+(γδα)αϱδA12ut2+2t0g(ts)(A12u(t)A12u(s),A12utt+αA12ut)ds+αϱδ(gA12u)(t)α(gA12u)(t)+αg(t)A12u2+(βαϱδ)utt2+αϱδ(βαϱδ)ut2].

    Then, we estimate the sixth term of the above equality

    2|t0g(ts)(A12u(t)A12u(s),A12utt+αA12ut)ds|t0g(ts)[αϱδA12u(t)A12u(s)2+δαϱA12utt+αA12ut2]ds=αϱδ(gA12u)(t)+δαϱG(t)A12utt+αA12ut2.

    A combination of the above results, we complete the proof of lemma.

    Now, we prove the well-posedness result of problem (1.7)-(1.8).

    Proof of Theorem 2.1. The proof is given by Faedo-Galerkin method and combines arguments from [16,39,38]. We present only the main steps.

    Step 1. Approximate problem

    We construct approximations of the solution u by the Faedo-Galerkin method as follows. For every m1, let Wm=span{w1,,wm} be a Hilbertian basis of the space H10(Ω). We choose four sequences (um0), (um1), (um2) and (um3) in Wm such that um0u0 strongly in D(A12), um1u1 strongly in D(A12), um2u2 strongly in D(A12) and um3u3 strongly in H. We define now the approximations:

    um(t)=mj=1amj(t)wj(x), (3.1)

    where um(t) are solutions to the finite dimensional Cauchy problem (written in normal form):

    Ωumtttt(t)wjdx+αΩumttt(t)wjdx+βΩumtt(t)wjdx+γΩA12umtt(t)A12wjdx+δΩA12umt(t)A12wjdx+ϱΩA12um(t)A12wjdxt0g(ts)ΩA12um(t)A12wjdxds=0 (3.2)

    with initial conditions

    (um(0),umt(0),umtt(0),umttt(0))=(um0,um1,um2,um3). (3.3)

    According to the standard theory of ordinary differential equation, the finite dimensional problem (3.2)-(3.3) has a local solution (um(t),umt(t),umtt(t),umttt(t)) in some interval [0,Tm) with 0<TmT, for every mN. Next, we present some estimates that allow us to extend the local solutions to the interval [0,T], for any given T>0.

    Step 2. Weak solutions

    Multiplying equation (3.2) by amjttt+αamjtt+αϱδamjt and integrating over Ω, we have

    ddtEm(t)+α(βαϱδ)umtt2αg(t)2A12um2+δ2αϱg(t)A12umtt+αϱδA12um2+α2(gA12um)(t)=[α(γδα)δg(t)2αϱ]A12umtt2+t0g(ts)(A12um(t)A12um(s),A12umtt)ds+αϱ2δ(gA12um)(t),

    where

    Em(t)=12[umttt+αumtt+αϱδumt2+δα(ϱG(t)ϱ)A12umtt+αA12umt+αϱδA12um2+δαϱG(t)A12umtt+αA12umt2+(γδα)A12umtt2+(γδα)αϱδA12umt2+2t0g(ts)(A12um(t)A12um(s),A12umtt+αA12umt)ds+αϱδ(gA12um)(t)α(gA12um)(t)+αg(t)A12um2+(βαϱδ)umtt2+αϱδ(βαϱδ)umt2]. (3.4)

    From assumptions (A1)(A3) and Lemma 3.1, we get, for ε(0,α),

    t0g(ts)(A12um(t)A12um(s),A12umtt)ds(αε)ϱ2δ(gA12um)(t)δ2(αε)ϱA12umtt2t0g(ts)ds=(αε)ϱ2δ(gA12um)(t)+δ(g(0)g(t))2(αε)ϱA12umtt2

    and so

    [α(γδα)δg(t)2αϱ]A12umtt2+t0g(ts)(A12um(t)A12um(s),A12umtt)ds+αϱ2δ(gA12um)(t)[α(γδα)δg(t)2αϱ]A12umtt2(αε)ϱ2δ(gA12um)(t)+δ(g(0)g(t))2(αε)ϱA12umtt2+αϱ2δ(gA12um)(t)=[α(γδα)δg(0)2(αε)ϱ]A12umtt2+εϱ2δ(gA12um)(t)[δg(t)2(αε)ϱδg(t)2αϱ]A12umtt20. (3.5)

    Therefore, we have

    ddtEm(t)+α(βαϱδ)umtt2αg(t)2A12um2+δ2αϱg(t)A12umtt+αϱδA12um2+α2(gA12um)(t)0. (3.6)

    Integrating (3.6) from 0 to tTm, one has

    Em(t)+t0[α(βαϱδ)umtt2αg(τ)2A12um2+δ2αϱg(τ)A12umtt+αϱδA12um2+α2(gA12um)(τ)]dτEm(0). (3.7)

    Now, since the sequences (um0)mN, (um1)mN, (um2)mN and (um3)mN converge and using (A1)(A3), we can find a positive constant C independent of m such that

    Em(t)C. (3.8)

    Therefore, using the fact ϱt0g(s)dsl, the last estimate (3.8) together with (3.4) give us, for all mN,Tm=T, we deduce that

    (um)mNisboundedinL(0,T;D(A12))(umt)mNisboundedinL(0,T;D(A12))(umtt)mNisboundedinL(0,T;D(A12))(umttt)mNisboundedinL(0,T;H). (3.9)

    Consequently, we may conclude that

    umuweakinL(0,T;D(A12))umtutweakinL(0,T;D(A12))umttuttweakinL(0,T;D(A12))umtttutttweakinL(0,T;H).

    From (3.9), we get that (um)mN is bounded in L(0,T;D(A12)). Then, (um)mN is bounded in L2(0,T;D(A12)). Since (umt)mN is bounded in L(0,T;D(A12)), (umt)mN is bounded in L2(0,T;D(A12)). Consequently, (umtt)mN is bounded in L2(0,T;D(A12)) and (umttt)mN is bounded in L2(0,T;H). Moreover, (um)mN is bounded in H3(0,T;H1(Ω)).

    Since the embedding H3(0,T;H1(Ω))L2(0,T;H(Ω)) is compact, using Aubin-Lions theorem [21], we can extract a subsequence (un)nN of (um)mN such that

    unustronglyinL2(0,T;H(Ω)).

    Therefore,

    unustronglyanda.e.on(0,T)×Ω.

    The proof now can be completed arguing as in [21].

    Step 3. Uniqueness

    It is sufficient to show that the only weak solution of (1.7)-(1.8) with u0=u1=u2=u3=0 is

    u0. (3.10)

    According to the energy estimate (3.8) and noting that E(u(0))=0, we obtain

    E(u(t))=0,t[0,T].

    So, we have

    uttt+αutt+αϱδut2=A12utt+αA12ut+αϱδA12u2=A12utt2=A12ut2=A12u2=utt2=0,t[0,T].

    And this implies (3.10). Thus, we conclude that problem (1.7)-(1.8) has at most one solution.

    In this section, we state and prove some lemmas needed to establish our general decay result.

    Lemma 4.1. Let (u,ut,utt,uttt) be the solution of (1.7). Assume that (A1)-(A3) hold. Then, we have

    ddtE(t)α(βαϱδ)utt2[α(γδα)δg(0)2(αε)ϱ]A12utt2+αg(t)2A12u2[δg(t)2(αε)ϱδg(t)2αϱ]A12utt2δ2αϱg(t)A12utt+αϱδA12u2α2(gA12u)(t)+εϱ2δ(gA12u)0.

    Proof. Multiplying (1.7) by uttt+αutt+αϱδut and integrating over Ω yield

    ddtE(t)=α(βαϱδ)utt2[α(γδα)δg(t)2αϱ]A12utt2+αg(t)2A12u2δ2αϱg(t)A12utt+αϱδA12u2+t0g(ts)(A12u(t)A12u(s),A12utt)dsα2(gA12u)(t)+αϱ2δ(gA12u)(t). (4.1)

    We proceed to show that, for a constant ε(0,α),

    |t0g(ts)(A12u(t)A12u(s),A12utt)ds|(αε)ϱ2δ(gA12u)(t)+δ(g(0)g(t))2(αε)ϱA12utt2. (4.2)

    Then, combining (4.1) and (4.2), we can obtain

    ddtE(t)α(βαϱδ)utt2[α(γδα)δg(0)2(αε)ϱ]A12utt2+αg(t)2A12u2[δg(t)2(αε)ϱδg(t)2αϱ]A12utt2δ2αϱg(t)A12utt+αϱδA12u2α2(gA12u)(t)+εϱ2δ(gA12u).

    According to (A1)-(A3) and Lemma 3.1, we complete the proof of lemma.

    Lemma 4.2. Assume that (A1)-(A5) hold. Then, the functional F1(t) defined by

    F1(t)=Ω(utt+αut+αϱδu)(uttt+αutt+αϱδut)dx

    satisfies the estimate

    F1(t)δ2αA12utt+αA12ut+αϱδA12u2+2αλ0δ(βαϱδ)2utt2+2αδ(γδα)2A12utt2+uttt+αutt+αϱδut2+2α(ϱl)2δA12u2+2αδCν(hA12u)(t). (4.3)

    Proof. Taking the derivative of F1(t) with respect to t, exploiting (1.7) and integrating by parts, we get

    F1(t)=Ω[(βαϱδ)utt](utt+αut+αϱδu)dxδαA12utt+αA12ut+αϱδA12u2+uttt+αutt+αϱδut2Ω(γδα)A12utt(A12utt+αA12ut+αϱδA12u)dx+Ω(t0g(ts)A12u(s)ds)(A12utt+αA12ut+αϱδA12u)dx.

    Using Young's inequality, Lemma 2.1, (A5) and the fact γδα>0 and βαϱδ>0, we have

    Ω[(βαϱδ)utt](utt+αut+αϱδu)dx2αλ0δ(βαϱδ)2utt2+δ8αλ0utt+αut+αϱδu22αλ0δ(βαϱδ)2utt2+δ8αA12utt+αA12ut+αϱδA12u2

    and

    Ω(t0g(ts)A12u(s)ds)(A12utt+αA12ut+αϱδA12u)dx=Ω(t0g(ts)(A12u(s)A12u(t))ds)(A12utt+αA12ut+αϱδA12u)dx+Ω(t0g(ts)A12u(t)ds)(A12utt+αA12ut+αϱδA12u)dx2αδCν(hA12u)(t)+δ4αA12utt+αA12ut+αϱδA12u2+2α(ϱl)2δA12u2.

    Also, we have

    Ω(γδα)A12utt(A12utt+αA12ut+αϱδA12u)dx2αδ(γδα)2A12utt2+δ8αA12utt+αA12ut+αϱδA12u2.

    Then, combining the above inequalities, we complete the proof of (4.3).

    Lemma 4.3. Assume that (A1)-(A5) hold. Then the functional F2(t) defined by

    F2(t)=Ω(uttt+αutt+αϱδut)t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdx

    satisfies the estimate

    F2(t)G(t)4uttt+αutt+αϱδut2+[(ϱl)2α22ε1+4λ0g2(0)α2G(t)]A12ut2+[λ0(ϱl)22+(δ2α2+ϱ2)+3(ϱl)2]ε1A12utt+αA12ut+αϱδA12u2+[α2ϱ2(ϱl)2λ0ε12δ2+(ϱl)22ε1+3(ϱl)2(αϱδ)2ε1+(ϱl)22(αϱδ)2]A12u2+[(ϱl)22ε1+12ε1(γδα)2+12(αϱδ)2(γδα)2]A12utt2+[(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)Cν+2α2ϱ2λ0G(t)δ2](hA12u)(t)+[2(βαϱδ)2ε1+4g2(0)G(t)]utt2, (4.4)

    where 0<ε1<1.

    Proof. By differentiating F2(t) with respect to t, using (1.7) and integrating by parts, we obtain

    F2(t)=Ω[βutt+γAutt+δAut+ϱAut0g(ts)Au(s)dsαϱδutt]×t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdxg(0)Ω(uttt+αutt+αϱδut)(utt+αut)dxΩ(uttt+αutt+αϱδut)t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdxt0g(s)dsuttt+αutt+αϱδut2=Ω[(βαϱδ)utt+δα(Autt+αAut+αϱδAu)+(γδα)Auttt0g(s)dsAu(t)]t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdxt0g(s)dsuttt+αutt+αϱδut2Ω(uttt+αutt+αϱδut)t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdx+(t0g(s)ds)Ωt0g(ts)(Au(t)Au(s))ds(utt+αut)dx+αϱδΩ(t0g(ts)(A12u(t)A12u(s))ds)2dxg(0)Ω(uttt+αutt+αϱδut)(utt+αut)dx.

    Now, we estimate the terms in the right-hand side of the above identity.

    Using Young's inequality, we obtain, for 0<ε1<1,

    Ω[(βαϱδ)utt+δα(Autt+αAut+αϱδAu)+(γδα)Auttt0g(s)dsAu(t)]t0g(ts)[(utt+αut+αϱδu)αϱδu(s)]dsdx[λ0(ϱl)22+(δ2α2+ϱ2)+2(ϱl)2]ε1A12utt+αA12ut+αϱδA12u2+2(βαϱδ)2ε1utt2+[α2ϱ2(ϱl)2λ0ε12δ2+(ϱl)22ε1+2(ϱl)2(αϱδ)2ε1+(ϱl)22(αϱδ)2]A12u2+[(ϱl)22ε1+12ε1(γδα)2+12(αϱδ)2(γδα)2]A12utt2+(ϱl)2α22ε1A12ut2+(ε1α2ϱ2λ04δ2+14ε1+1)Cν(hA12u)(t)

    and

    (t0g(s)ds)Ωt0g(ts)(Au(t)Au(s))ds(utt+αut)dx=(t0g(s)ds)Ωt0g(ts)(A12u(t)A12u(s))ds(A12utt+αA12ut)dx12ε1Cν(hA12u)(t)+(ϱl)22ε1A12utt+αA12ut212ε1Cν(hA12u)(t)+(ϱl)2ε1A12utt+αA12ut+αϱδA12u2+(ϱl)2(αϱδ)2ε1A12u2.

    Also, we have

    αϱδΩ(t0g(ts)(A12u(t)A12u(s))ds)2dxαϱδCν(hA12u)(t)

    and

    g(0)Ω(uttt+αutt+αϱδut)(utt+αut)dxG(t)4uttt+αutt+αϱδut2+g2(0)G(t)utt+αut2G(t)4uttt+αutt+αϱδut2+2g2(0)G(t)utt2+2λ0g2(0)α2G(t)A12ut2.

    Exploiting Young's inequality and (A5), we get

    Ω(uttt+αutt+αϱδut)t0g(ts)[(utt+αut+αϱδu)(t)αϱδu(s)]dsdx=Ω(uttt+αutt+αϱδut)t0g(ts)(utt+αut)(t)dsdxαϱδΩ(uttt+αutt+αϱδut)t0g(ts)(u(t)u(s))dsdxG(t)2uttt+αutt+αϱδut2+2g2(0)G(t)utt2+2λ0α2g2(0)G(t)A12ut2+2α2ϱ2λ0G(t)δ2(α2Cν+1)(hA12u)(t).

    A combination of all the above estimates gives the desired result.

    As in [11], we introduce the following auxiliary functional

    F3(t)=Ω(uttt+αutt)utdx+ϱ2A12u2.

    Lemma 4.4. Assume that (A1)-(A5) hold. Then the functional F3(t) satisfies the estimate

    F3(t)(3δ8ε2δ4)A12ut2+ε2δ38α2ϱ2λ0uttt+αutt+αϱδut2+2γ2δA12utt2+(ε2δ34ϱ2λ0+4α2ϱ2λ0ε2δ3+2β2λ0δ)utt2+1δCν(hA12u)(t)+2(ϱl)2δA12u2, (4.5)

    where 0<ε2<1.

    Proof. Using the equation (1.7), a direct computation leads to the following identity

    F3(t)=Ω(uttt+αutt)uttdx+Ω(utttt+αuttt)utdx+ϱ(A12u,A12ut)=(uttt,utt)+αutt2β(utt,ut)γ(A12utt,A12ut)δA12ut2+(t0g(ts)A12u(s)ds,A12ut). (4.6)

    Now, the first and third terms in the right-hand side of (4.6) can be estimated as follows:

    (uttt,utt)ε2δ316α2ϱ2λ0uttt2+4α2ϱ2λ0ε2δ3utt2ε2δ38α2ϱ2λ0uttt+αutt+αϱδut2+ε2δ38α2ϱ2λ0αutt+αϱδut2+4α2ϱ2λ0ε2δ3utt2ε2δ38α2ϱ2λ0uttt+αutt+αϱδut2+α2(ε2δ34α2ϱ2λ0+4ϱ2λ0ε2δ3)utt2+ε2δ4A12ut2

    and

    β(utt,ut)2β2λ0δutt2+δ8λ0ut22β2λ0δutt2+δ8A12ut2,

    where 0<ε2<1.

    Using Young's inequality and Lemma 2.1, we get

    γ(A12utt,A12ut)2γ2δA12utt2+δ8A12ut2

    and

    (t0g(ts)A12u(s)ds,A12ut)=(t0g(ts)(A12u(s)A12u(t)+A12u(t))ds,A12ut)1δΩ(t0g(ts)(A12u(t)A12u(s))ds)2dx+δ4A12ut2+2(ϱl)2δA12u2+δ8A12ut21δCν(hA12u)(t)+3δ8A12ut2+2(ϱl)2δA12u2.

    Then, combining the above inequalities, we obtain the desired result.

    Lemma 4.5. Assume that (A1)-(A5) hold. Then the functional F4(t) defined by

    F4(t)=Ωt0f(ts)|A12u(s)|2dsdx

    satisfies the estimate

    F4(t)12(gA12u)(t)+3(ϱl)A12u2, (4.7)

    where f(t)=tg(s)ds.

    Proof. Noting that f(t)=g(t), we see that

    F4(t)=f(0)A12u2Ωt0g(ts)|A12u(s)|2dsdx=f(0)A12u2Ωt0g(ts)|A12u(s)A12u(t)|2dsdx2ΩA12ut0g(ts)(A12u(s)A12u(t))dsdx(t0g(s)ds)A12u2=(gA12u)(t)2ΩA12ut0g(ts)(A12u(s)A12u(t))dsdx+f(t)A12u2.

    Exploiting Young's inequality and the fact t0g(s)dsϱl, we obtain

    2ΩA12ut0g(ts)(A12u(s)A12u(t))dsdx2(ϱl)A12u2+12(ϱl)(t0g(ts)ds)Ωt0g(ts)(A12u(s)A12u(t))2dsdx2(ϱl)A12u2+12(gA12u)(t).

    Moreover, taking account of f(t)f(0)=ϱl, we have

    f(t)A12u2(ϱl)A12u2.

    Combining the above estimates, we arrive at the desired result.

    Lemma 4.6. Assume that (A1)(A5) hold. The functional L(t) defined by

    L(t)=NE(t)+F1(t)+N2F2(t)+N3F3(t)

    satisfies, for a suitable choice of N,N2,N3,

    L(t)E(t)

    and the estimate, for all tt0,

    L(t)c[utt2+A12utt2+uttt+αutt+αϱδut2+A12utt+αA12ut+αϱδu2]4(ϱl)A12u2+18(gA12u)(t), (4.8)

    where t0 has been introduced in Remark 2.1.

    Proof. Combining Lemmas 4.1-4.4 and recalling that g=νgh, we obtain, for all tt0,

    L(t)[α(βαϱδ)N2αλ0δ(βαϱδ)2(2(βαϱδ)2ε1+4g2(0)G(t))N2(ε2δ34ϱ2λ0+4α2ϱ2λ0ε2δ3+2β2λ0δ)N3]utt2[(α(γδα)δg(0)2(αε)ϱ)N+(δg(t)2(αε)ϱδg(t)2αϱ)N2αδ(γδα)2((ϱl)22ε1+12ε1(γδα)2+12(αϱδ)2(γδα)2)N22γ2δN3]A12utt2α2N(gA12u)(t)[(3δ8ε2δ4)N3((ϱl)2α22ε1+4λ0g2(0)α2G(t))N2]A12ut2(G(t)4N21ε2δ3N38α2ϱ2λ0)uttt+αutt+αϱδut2+εϱν2δN(gA12u)(t)[δ2α(λ0(ϱl)22+(δ2α2+ϱ2)+3(ϱl)2)ε1N2]×A12utt+αA12ut+αϱδA12u2[αg(t)2N2α(ϱl)2δ(α2ϱ2(ϱl)2λ0ε12δ2+(ϱl)22ε1+3(ϱl)2(αϱδ)2ε1+(ϱl)22(αϱδ)2)N22(ϱl)2δN3]A12u2[εϱ2δN2αδCν((ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)Cν+2α2ϱ2λ0G(t)δ2)N2N3δCν](hA12u)(t).

    At this point, we need to choose our constants very carefully. First, we choose

    ε1=αδ2N2[λ0α2(ϱl)2+2(δ2+α2ϱ2)+6α2(ϱl)2]andε2=1N3.

    The above choice yields

    L(t)[α(βαϱδ)N2αλ0δ(βαϱδ)2(2(βαϱδ)2ε1+4g2(0)G(t))N2δ34ϱ2λ04α2ϱ2λ0δ3N232β2λ0δN3]utt2[(α(γδα)δg(0)2(αε)ϱ)N+(δg(t)2(αε)ϱδg(t)2αϱ)N2αδ(γδα)2((ϱl)22ε1+12ε1(γδα)2+12(αϱδ)2(γδα)2)N22γ2δN3]A12utt2α2N(gA12u)(t)[3δ8N3δ4((ϱl)2α22ε1+4λ0g2(0)α2G(t))N2]A12ut2+εϱν2δN(gA12u)(t)(G(t)4N21δ38α2ϱ2λ0)uttt+αutt+αϱδut2δ4αA12utt+αA12ut+αϱδA12u2[αg(t)2N2α(ϱl)2δ(α2ϱ2(ϱl)2λ0ε12δ2+(ϱl)22ε1+3(ϱl)2(αϱδ)2ε1+(ϱl)22(αϱδ)2)N22(ϱl)2δN3]A12u2[εϱ2δN2αδCν((ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)Cν+2α2ϱ2λ0G(t)δ2)N2N3δCν](hA12u)(t).

    Then, we choose N2 large enough so that

    G(t)4N21δ38α2ϱ2λ0>0.

    Next, we choose N3 large enough so that

    3δ8N3δ4((ϱl)2α22ε1+4λ0g2(0)α2G(t))N2>0.

    Now, as ν2g(s)νg(s)g(s)<g(s), it is easy to show, using the Lebesgue dominated convergence theorem, that

    νCν=0ν2g(s)νg(s)g(s)ds0,asν0.

    Hence, there is 0<ν0<1 such that if ν<ν0, then

    νCν<116(2αδ+(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)N2+N3δ).

    Now, let us choose N large enough and choose ν satisfying

    εϱ4δN2α2ϱ2λ0G(t)δ2N2>0andν=δ4εϱN<ν0,

    which means

    εϱ2δN2α2ϱ2λ0G(t)δ2N2Cν(2αδ+(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)N2+N3δ)>εϱ2δN2α2ϱ2λ0G(t)δ2N2116ν=εϱ4δN2α2ϱ2λ0G(t)δ2N2>0

    and

    α(βαϱδ)N2αλ0δ(βαϱδ)2(2(βαϱδ)2ε1+4g2(0)G(t))N2δ34ϱ2λ04α2ϱ2λ0δ3N232β2λ0δN3>0,(α(γδα)δg(0)2(αε)ϱ)N+(δg(t)2(αε)ϱδg(t)2αϱ)N2αδ(γδα)2((ϱl)22ε1+12ε1(γδα)2+12(αϱδ)2(γδα)2)N22γ2δN3>0,αg(t)2N2α(ϱl)2δ(α2ϱ2(ϱl)2λ0ε12δ2+(ϱl)22ε1+3(ϱl)2(αϱδ)2ε1+(ϱl)22(αϱδ)2)N22(ϱl)2δN3>4(ϱl).

    So we arrive at, for positive constant c,

    L(t)c[utt2+A12utt2+uttt+αutt+αϱδut2+A12utt+αA12ut+αϱδu2]4(ϱl)A12u2+18(gA12u)(t).

    On the other hand, from Lemma 3.2, we find that

    |L(t)NE(t)|Ω|utt+αut+αϱδu||uttt+αutt+αϱδut|dx+N2Ω|uttt+αutt+αϱδut|t0g(ts)×|(utt+αut+αϱδu)(t)αϱδu(s)|dsdx+N3Ω|uttt+αutt||ut|dx+N3ϱ2A12u2cE(t).

    Therefore, we can choose N even large (if needed) so that (4.8) is satisfied.

    In this section, we will give an estimate to the decay rate for the problem (1.7)-(1.8).

    Proof of Theorem 2.2. Our proof starts with the observation that, for any tt0,

    t00g(s)Ω|A12u(t)A12u(ts)|2dxdsg(0)at00g(s)Ω|A12u(t)A12u(ts)|2dxdsg(0)at0g(s)Ω|A12u(t)A12u(ts)|2dxdscE(t),

    which are derived from (2.2) and Lemma 4.1 and can be used in (4.8).

    Taking F(t)=L(t)+cE(t), which is obviously equivalent to E(t), we get, for all tt0,

    L(t)c[utt2+A12utt2+uttt+αutt+αϱδut2+A12utt+αA12ut+αϱδu2]4(ϱl)A12u2+18(gA12u)(t)mE(t)+c(gA12u)(t)mE(t)cE(t)+ctt0g(s)Ω|A12u(t)A12u(ts)|2dxds,

    where m is a positive constant. Then, we obtain that

    F(t)=L(t)+cE(t)mE(t)+ctt0g(s)Ω|A12u(t)A12u(ts)|2dxds. (5.1)

    We consider the following two cases relying on the ideas presented in [31].

    (ⅰ) M(t) is linear.

    We multiply (5.1) by ξ(t), then on account of (A1)-(A4) and Lemma 4.1, we obtain, for all tt0,

    ξ(t)F(t)mξ(t)E(t)+cξ(t)tt0g(s)Ω|A12u(t)A12u(ts)|2dxdsmξ(t)E(t)+ctt0ξ(s)g(s)Ω|A12u(t)A12u(ts)|2dxdsmξ(t)E(t)ctt0g(s)Ω|A12u(t)A12u(ts)|2dxdsmξ(t)E(t)ct0g(s)Ω|A12u(t)A12u(ts)|2dxdsmξ(t)E(t)cE(t).

    Therefore,

    ξ(t)F(t)+cE(t)mξ(t)E(t).

    As ξ(t) is non-increasing and F(t)E(t), we have

    ξ(t)F(t)+cE(t)E(t)

    and

    (ξF+cE)(t)mξ(t)E(t), tt0.

    It follows immediately that

    E(t)mξ(t)E(t), tt0.

    We may now integrate over (t0,t) to conclude that, for two positive constants k1 and k2

    E(t)k2exp(k1tt0ξ(s)ds), tt0.

    By the continuity of E(t), we have

    E(t)k2exp(k1t0ξ(s)ds), t>0.

    (ⅱ) M is nonlinear.

    First, we define the functional

    L(t)=L(t)+F4(t).

    Obviously, L(t) is nonnegative. And, by Lemma 4.5 and Lemma 4.6, there exists b>0 such that

    L(t)=L(t)+F4(t)c[utt2+A12utt2+uttt+αutt+αϱδut2+A12utt+αA12ut+αϱδu2](ϱl)A12u238(gA12u)(t)bE(t).

    Therefore, integrating the above inequality over (t0,t), we see at once that

    L(t0)L(t)L(t0)btt0E(s)ds.

    It is sufficient to show that

    0E(s)ds< (5.2)

    and

    E(t)ctt0,t>t0.

    Now, we define a functional λ(t) by

    λ(t):=tt0g(s)A12u(t)A12u(ts)2ds.

    Clearly, we have

    λ(t)t0g(s)A12u(t)A12u(ts)2dscE(t), tt0. (5.3)

    After that, we define another functional I(t) by

    I(t):=qtt0A12u(t)A12u(ts)2ds.

    Now, the following inequality holds under Lemma 4.1 and (5.2) that

    tt0A12u(t)A12u(ts)2ds2tt0(A12u(t)2+A12u(ts)2)ds4tt0(E(t)+E(ts))ds8tt0E(0)ds<. (5.4)

    Then (5.4) allows for a constant 0<q<1 chosen so that, for all tt0,

    0<I(t)<1; (5.5)

    otherwise we get an exponential decay from (5.1). Moreover, recalling that M is strict convex on (0,r] and M(0)=0, then

    M(θx)θM(x),for0θ1andx(0,r].

    From assumptions (A2) and (A4), (5.5) and Lemma 2.2, it follows that

    λ(t)=tt0g(s)A12u(t)A12u(ts)2ds=1qI(t)tt0I(t)(g(s))qA12u(t)A12u(ts)2ds1qI(t)tt0I(t)ξ(s)M(g(s))qA12u(t)A12u(ts)2dsξ(t)qI(t)tt0M(I(t)g(s))qA12u(t)A12u(ts)2dsξ(t)qM(1I(t)tt0I(t)g(s)qA12u(t)A12u(ts)2ds)=ξ(t)qM(qtt0g(s)A12u(t)A12u(ts)2ds).

    According to ¯M is an extension of M (see Remark 2.1(2)), we also have

    λ(t)ξ(t)q¯M(qtt0g(s)A12u(t)A12u(ts)2ds).

    In this way,

    tt0g(s)A12u(t)A12u(ts)2ds1q¯M1(qλ(t)ξ(t))

    and (5.1) becomes

    F(t)mE(t)+ctt0g(s)Ω|A12u(t)A12u(ts)|2dxdsmE(t)+c¯M1(qλ(t)ξ(t)), tt0. (5.6)

    Let 0<ε0<r, we define the functional F1(t) by

    F1(t):=¯M(ε0E(t)E(0))F(t)+E(t), t0.

    Then, recalling that E(t)0, ¯M>0 and ¯M>0 as well as making use of estimate (5.6), we deduce that F1(t)E(t) and also, for any tt0, we have

    F1(t)mE(t)¯M(ε0E(t)E(0))+c¯M(ε0E(t)E(0))¯M1(qλ(t)ξ(t))+E(t). (5.7)

    Taking account of Lemma 2.3, we obtain

    ¯M(ε0E(t)E(0))¯M1(qλ(t)ξ(t))¯M(¯H(ε0E(t)E(0)))+¯M(¯M1(qλ(t)ξ(t)))=¯M(¯M(ε0E(t)E(0)))+qλ(t)ξ(t) (5.8)

    where

    ¯M(¯M(ε0E(t)E(0)))=¯M(ε0E(t)E(0))(¯M)1(¯M(ε0E(t)E(0)))¯M[(¯M)1(¯M(ε0E(t)E(0)))]=ε0E(t)E(0)¯M(ε0E(t)E(0))¯M(ε0E(t)E(0))ε0E(t)E(0)¯M(ε0E(t)E(0)). (5.9)

    So, combining (5.7), (5.8) and (5.9), we obtain

    F1(t)(mE(0)cε0)E(t)E(0)¯M(ε0E(t)E(0))+cqλ(t)ξ(t)+E(t).

    From this, we multiply the above inequality by ξ(t) to get

    ξ(t)F1(t)(mE(0)cε0)ξ(t)E(t)E(0)¯M(ε0E(t)E(0))+cqλ(t)+ξ(t)E(t).

    Then, using the fact that, as ε0E(t)E(0)<r, ¯M(ε0E(t)E(0))=M(ε0E(t)E(0)) and (5.3), we get

    ξ(t)F1(t)(mE(0)cε0)ξ(t)E(t)E(0)M(ε0E(t)E(0))cE(t).

    Consequently, defining F2(t)=ξ(t)F1(t)+cE(t), then since F1(t)E(t), we arrive at

    F2(t)E(t), (5.10)

    and with a suitable choice of ε0, we get, for some positive constant k and for any tt0,

    F2(t)kξ(t)(E(t)E(0))M(ε0E(t)E(0)). (5.11)

    Define

    R(t)=λ1F2(t)E(0),λ1>0andM2(t)=tM(ε0t).

    Moreover, it suffices to show that M2(t),M2(t)>0 on (0,1] by the strict convexity of M on (0,r]. And, it is easily seen that

    F2(t)kξ(t)M2(E(t)E(0)). (5.12)

    According to (5.10) and (5.12), there exist λ2,λ3>0 such that

    λ2R(t)E(t)λ3R(t). (5.13)

    Then, it follows that there exists k1>0 such that

    k1ξ(t)R(t)M2(R(t)), tt0. (5.14)

    Next, we define

    M1(t):=rt1sM(s)ds.

    And based on the properties of M, we know that M1 is strictly decreasing function on (0,r] and limt0M1(t)=+.

    Now, we integrate (5.14) over (t0,t) to obtain

    tt0R(s)M2(R(s))dsk1tt0ξ(s)ds

    so

    k1tt0ξ(s)dsM1(ε0R(t))M1(ε0R(t0)),

    which implies that

    M1(ε0R(t))k1tt0ξ(s)ds.

    It is easy to obtain that

    R(t)1ε0M11(k1tt0ξ(s)ds), tt0. (5.15)

    A combining of (5.13) and (5.15) gives the proof.

    The authors are grateful to the anonymous referees and the editor for their useful remarks and comments.



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