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The developable surfaces with pointwise 1-type Gauss map of Frenet type framed base curves in Euclidean 3-space

  • Received: 03 September 2022 Revised: 10 October 2022 Accepted: 21 October 2022 Published: 31 October 2022
  • MSC : 53A04, 53A05, 58K05

  • In this study, the ruled developable surfaces with pointwise 1-type Gauss map of Frenet-type framed base (Ftfb) curve are introduced in Euclidean 3-space. The tangent developable surfaces, focal developable surfaces, and rectifying developable surfaces with singular points are considered. Then the conditions for the Gauss map of these surfaces to be pointwise 1-type are obtained separately. In order to form a basis for the study, first, the basic concepts related to the Ftfb curve and the Gauss map of a surface are recalled. Later, the necessary and sufficient conditions are found for these surfaces to be of the pointwise 1-type of the Gauss map. Finally, examples for each type of these surfaces are given, and their graphics are illustrated.

    Citation: Yanlin Li, Kemal Eren, Kebire Hilal Ayvacı, Soley Ersoy. The developable surfaces with pointwise 1-type Gauss map of Frenet type framed base curves in Euclidean 3-space[J]. AIMS Mathematics, 2023, 8(1): 2226-2239. doi: 10.3934/math.2023115

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  • In this study, the ruled developable surfaces with pointwise 1-type Gauss map of Frenet-type framed base (Ftfb) curve are introduced in Euclidean 3-space. The tangent developable surfaces, focal developable surfaces, and rectifying developable surfaces with singular points are considered. Then the conditions for the Gauss map of these surfaces to be pointwise 1-type are obtained separately. In order to form a basis for the study, first, the basic concepts related to the Ftfb curve and the Gauss map of a surface are recalled. Later, the necessary and sufficient conditions are found for these surfaces to be of the pointwise 1-type of the Gauss map. Finally, examples for each type of these surfaces are given, and their graphics are illustrated.



    In general, microscopic particles spread out in a random manner. This occurs in a large number of phenomena in engineering, biology, ecology and medicine (see [4,5,7,14,19,22,23,26,31,34,36]). They have been studied by many researchers. Several formulations of diffusion exist in the market, and there are several types of treatments as well. Anomalous diffusion (unlike the classical one) cannot be elucidated by the standard ways, as the corresponding mean square displacement does not adhere to the linear growth law assumed in the classical problems. It is found that fractional derivatives are better suited for these disordered phenomena. They outperform by far the nonlinear models which suffer from high complexity and high computational cost.

    In this paper, we are interested in structures described by Timoshenko models. This type of model, in the integer-order case (second order time derivative), has been investigated by a fairly large number of researchers, and several results have appeared in the literature. The basic model is described as follows:

    {ρ1φttk(φx+ψ)x=0,0<x<1,t>0,ρ2ψttbψxx+k(φx+ψ)=0,0<x<1,t>0,

    where φ stands for the transverse movement and ψ stands for the angle of rotation of the beam. The constant ρ1 stands for the mass density, ρ2 represents the moment of mass inertia, k is the shear modulus of elasticity and b is the rigidity coefficient of the cross-section.

    The system is conservative. Several kinds of damping devices have been utilized to stabilize these structures. First, two frictional dampings (based on velocity feedback) have been added, one in each equation [32]:

    {ρ1φttk(φx+ψ)x+φt=0,0<x<1,t>0,ρ2ψttbψxx+k(φx+ψ)+ψt=0,0<x<1,t>0.

    This is the most favorable situation after the case of structural dampings (based on the Laplacian of the velocity feedback). Of course, these are stronger than the mere frictional dampings. Quickly, it has been realized that only one damping process (in either the first or second equation) is enough to stabilize the system in an exponential manner. Undoubtedly, from the application point of view, this is more convenient because it is less costly and easier to implement. However, from the point of view mathematics, the problem becomes more challenging. Indeed, in this case we lost one nice term, which is responsible for the dissipation of that component. That is to say, one needs to show that only one damping in one component may not only force that component to go to zero exponentially, but also the other (undamped) one (in the energy norm). Researchers will then need to come up with new arguments to overcome this loss. For instance, the frictional damping term φt has been dropped and the problem given by

    {ρ1φttk(φx+ψ)x=0,0<x<1,t>0,ρ2ψttbψxx+k(φx+ψ)+ψt=0,0<x<1,t>0

    is discussed in [25]. A much weaker damping namely, a viscoelastic damping in the rotational direction was considered first in [3]:

    {ρ1φttk(φx+ψ)x+φt=0,0<x<1,t>0,ρ2ψttbψxx+t0ω(ts)ψxx(s)ds+k(φx+ψ)=0,0<x<1,t>0.

    The authors proved that the system is exponentially (polynomially) stable depending on whether the involved relaxation function ω is exponentially (polynomially) decaying to zero. Subsequently, many extensions, improvements and generalizations, in several directions, have appeared in the literature (for example, [10,15,20,21,24,35,37,38,39] and the references therein). In particular, boundary dampings, nonlinear dampings and coupling with thermal equations have been considered. Moreover, the class of admissible relaxation functions has been extensively enlarged.

    As mentioned above, when the structure operates in a complex medium, integer-order derivatives are no longer appropriate to describe the phenomena. The mean square displacement is not linear. It is nonlinear and proportional to "t" to some power γ0. The system is then classified according to the ranges or values of γ whether they are equal to 0, between 0 and 1, between 1 and 2, equal to 2, above 2, etc. The combination of the balance law and the constitutive relationship between the stress and the strain will then involve a nonlocal term in time containing a singular kernel. This suggests fractional calculus as an adequate platform to study such phenomena.

    We shall consider a viscoelastic damping which may be embodied in the material of the structure or added as a controller. It will act in the rotational direction to slow down the vibrations or absorb shocks that can reduce the lifetime of the structure, damage it or completely destroy it. Similar dampers are called "rotary" dampers in the market. We will address the fractional problem:

    {ρ1CDβφk(φx+ψ)x=0,0<x<1,t>0,1<β<2,ρ2CDβψbψxx+t0ω(ts)ψxx(s)ds+k(φx+ψ)=0,0<x<1,t>0, (1.1)

    with the following Dirichlet boundary conditions and initial values:

    {φ(0,t)=φ(1,t)=ψ(0,t)=ψ(1,t)=0,t0,φ(.,0)=φ0,φt(.,0)=φ1,ψ(.,0)=ψ0,ψt(.,0)=ψ1, (1.2)

    where CDβ denotes the Caputo fractional derivative operator (defined below). To the best of our knowledge, the present analysis of this problem, has not been discussed so far.

    Let U=(φ,ψ)T and U0=(φ0,ψ0)T; our system may be recast into the form

    CDβU=MUt0N(ts)U(s)ds,

    where

    M=(k2x/ρ1kx/ρ1kx/ρ2b2x/ρ2kId/ρ2),

    and

    N(t)=(000ω(t)2x/ρ2).

    We consider the space

    H=H10(0,1)×H10(0,1),

    and the domain

    D(M)={U=(φ,ψ)TH:φ,ψH2(0,1)H10(0,1)}.

    In [9], the authors proved the existence and uniqueness for a similar abstract problem. We report here briefly this result. Let (X,.) be a Banach space and P, (S(t))t0 be closed linear operators defined on the domains D(P) and D(S(t))D(P) dense in X, respectively. We denote by .G the graph norm in D(P), R(ν,P):=(νIP)1 and ρ(P) is the resolvent of P. The problem

    CDβw=Pwt0S(ts)w(s)ds,1<β<2,

    with the initial data w(0)=w0 and w(0)=0 in D(P) (see also [1, 30]) admits the classical solution

    wC((0,);D(P))C1((0,);X)

    such that t1βΓ(2β)(ww0)C2((0,);X), provided that the following conditions hold:

    (A1) For some 0<θ0π/2 and every θ<θ0, there is a constant U>0 such that

    Σ0,βω:={νC:ν0,|arg(ν)|<βω}ρ(P),

    and R(ν,P)<U/|ν|, νΣ0,βω with ω=θ+π/2.

    (A2) For each wD(P), S(.)w is strongly measurable on (0,). There exists a locally integrable function f(t) with Laplace transform ˆf(ν), Re(ν)>0 and S(t)wf(t)wG, t>0, wD(P). In addition, S:Σ0,π/2L(D(P);X) has an analytic extension ˜S to Σ0,ω, verifying the relations ˜S(ν)w˜S(ν)wG, wD(P) and ˜S(ν)=O(1|ν|) as |ν|.

    (A3) There exists a subspace F that is dense in (D(P),.G) and C>0 such that P(F)D(P), S(ν)(F)D(P), PS(ν)wCw, wF and νΣ0,ω.

    Assuming that U0D(M), it follows from the above result that we have a classical solution in

    UC(R+,D(M))C1(R+,H)

    such that t1βΓ(2β)(φφ0),t1βΓ(2β)(ψψ0)C2((0,);L2(0,1)).

    In fractional calculus, it is known that CDα(CDαφ(t,x))CD2αφ(t,x). Consequently, rewriting our system using CDα(CDαφ) and CDα(CDαψ) will result in additional terms. Actually, the relationship between both forms is given by

    CDα(CDαφ(t,x))=CD2αφ(t,x)+φ(0,x)t12αΓ(22α).

    We shall consider, rather the simplified system given by

    {ρ1CDα(CDαφ)k(φx+ψ)x=0,1/2<α<1,ρ2CDα(CDαψ)bψxx+t0ω(ts)ψxx(s)ds+k(φx+ψ)=0 (1.3)

    for 0<x<1,t>0 with the initial and boundary conditions (1.2). Clearly, the two fractional derivatives coincide when φ(0,x)=0. The existence and uniqueness are then guaranteed by [9] (see also Section 3 below). Regarding the stability, it will not be affected, as 12α<0.

    For this problem (1.2) and (1.3), we shall first finalize the discussion on the existence and uniqueness of a solution by using the notion of resolvents (instead of semi-groups). Then, we will highlight the difficulty of treating the stability for this problem. The utilization of a fractional chain rule version gives rise to a problematic term in the fractional derivative of the "energy" functional. Even though we found a way to control it, a smallness condition on the involved kernel is still imposed. We obtain a Mittag-Leffler- type stability provided that the kernel itself decays somehow in a similar manner.

    Various other fractional systems modeling Timoshenko beams have been derived under different conditions and in different circumstances. The governing equations often involve either Caputo or Riemann-Liouville fractional derivatives. They have been tackled mostly from the engineering (response determination, resonance, free vibrations, influences on mechanical responses, parametric analyses, steady-state response, transient response, etc.) or numerical analysis perspective. A number of analytical and semi-analytical approaches have been employed (see [6,27,28] and the references therein). This involves, for instance, Galerkin schemes, central difference schemes, direct numerical integration schemes, averaging methods and various kinds of approximations of fractional derivatives. The utilized methods were compared to the traditional ones (extended actually to the fractional case), such as the Laplace transform, Fourier integral transform and Mellin transform, which have been very instrumental. Error estimates and numerical experiments have been conducted to justify the usefulness, effectiveness and accuracy of fractional models in describing the dynamics of beams. On the other hand, integer-order (with second-order time-derivatives of the states as leading terms) Timoshenko models with fractional damping have also been investigated in many research papers (see [16,18] and the references therein). The fractional damping may be weak (described by a fractional derivative of the state) or strong (described by a fractional derivative of the Laplacian of the state). In the latter case, it may be looked at as a viscoelastic damping process with a singular kernel. More precisely, it takes the form of the convolution of a singular kernel (power type) with the time-derivative of the Laplacian of the state.

    In the next section, we prepare some preliminaries. Section 3 contains the existence and uniqueness results. It is followed by a section in which we prove two useful identities and define our "energy" functional. Section 5 is devoted to the introduction of several functionals. In Section 6, we treat the stability issue. Section 6 is concerned with some numerical simulations. A conclusion is provided in the last section.

    In this section, we present the definition of the Riemann-Liouville fractional derivative, the Caputo fractional derivative and the Mittag-Leffler functions (see also [13,17,29,33] for more on fractional calculus). Some useful formulas are also given.

    Definition 1. The Riemann-Liouville fractional integral of order γ>0 is given by

    Iγχ(t)=1Γ(γ)t0(ts)γ1χ(s)ds,γ>0

    for any measurable function χ provided that the right-hand side exists. Here, Γ(γ) is the usual gamma function.

    Definition 2. The fractional derivative of order γ in the sense of Caputo is given by

    CDγχ(t)=1Γ(nγ)t0(ts)nγ1χ(n)(s)ds,n1<γ<n,

    whereas the fractional derivative of order γ in the sense of Riemann-Liouville is defined by

    RLDγχ(t)=1Γ(nγ)dndtnt0(ts)nγ1χ(s)ds,n1<γ<n,

    provided that the right-hand sides exist.

    The passage to and from the Riemann-Liouville fractional derivative and the Caputo fractional derivative is given by

    RLDγχ(t)=χ(0)tγΓ(1γ)+CDγχ(t),0<γ<1,t>0.

    We recall the definitions of the one-parametric and two-parametric Mittag-Leffler functions, respectively:

    Eα(z):=n=0znΓ(αn+1),Re(α)>0,

    and

    Eα,β(z):=n=0znΓ(αn+β),Re(α)>0,Re(β)>0.

    Clearly, Eα,1(z)Eα(z).

    Proposition 1. [40] If χ(t) is a differentiable function satisfying

    CDαχ(t)γχ(t),0<α<1

    for some γ>0, then χ(t)χ(0)Eα(γtα), t0. When the derivative is of the Riemann-Liouville type, we obtain tα1Eα,α(γtα) rather than Eα(γtα).

    Proposition 2. [13] For α,β>0, the identity

    λtαEα,α+β(λtα)=Eα,β(λtα)1Γ(β)

    holds.

    Proposition 3. [13, p. 61] For μ,α,β>0, we have the relation

    1Γ(μ)t0(ts)μ1Eα,β(λsα)sβ1ds=tμ+β1Eα,μ+β(λtα),t>0.

    Proposition 4. [29, p. 99] If I1γχ(t)C1([0,)), 0<γ<1 and f(t) is a continuous function, then it holds that

    RLDγt0χ(ts)f(s)ds=t0f(ts)RLDγχ(s)ds+f(t)limt0+I1γχ(t),t>0.

    We shall utilize repeatedly the following fractional product rule.

    Proposition 5. [2] Let f(t) and g(t) be absolutely continuous functions on [0,T], T>0. Then, for t[0,T] and 0<α<1, we have

    f(t)CDαg(t)+g(t)CDαf(t)=CDα(fg(t))+αΓ(1α)t0dξ(tξ)1αξ0f(η)dη(tη)αξ0g(s)ds(ts)α.

    Taking f(t)g(t) will result in the well-known inequality (instead of the ordinary chain rule)

    CDαf2(t)2f(t)CDαf(t).

    The vector version also holds.

    Here, we delineate the assumptions on the initial data and the kernel. Moreover, the existence and uniqueness issue is finalized. We shall assume that our initial data satisfy U0D(M) (this is defined below and involves the knowledge of fractional derivatives of the states at zero). The kernel ω is assumed to satisfy the following condition:

    (A) The kernel ω:(0,)(0,) is a nonnegative continuous function satisfying I1αω(t)C1([0,)),

    RLDαω(t)Cωω(t),t>0,

    and

    ˉω:=0ω(s)ds<b

    for some positive constant Cω.

    Proposition 6. Assuming (A), we have

    ω(t)˜ωtα1Eα,α(Cωtα),t0,

    ω(t) is summable and

    t0ω(s)ds˜ωt0sα1Eα,α(Cωsα)ds=˜ωtαEα,α+1(Cωtα)˜ω/Cω,t>0

    for some ˜ω>0.

    Proof. The proofs follow by a direct application of Propositions 1–3.

    Our problem (1.2) and (1.3) may be rewritten as a system of fractional order α, 1/2<α<1. As a matter of fact, letting U=(φ,˜φ,ψ,˜ψ)T, ˜φ=CDαφ, ˜ψ=CDαψ and U0=(φ0,˜φ0,ψ0,˜ψ0)T, it is evident that

    CDαU=MUt0N(ts)U(s)ds, (3.1)

    where

    M=(0Id00k2x/ρ10kx/ρ10000Idkx/ρ20b2x/ρ2kId/ρ20),

    and

    N(t)=(00000000000000ω(t)2x/ρ20).

    We set

    H=H10(0,1)×L2(0,1)×H10(0,1)×L2(0,1),

    and

    D(M)={U=(φ,˜φ,ψ,˜ψ)TH:φ,ψH2(0,1)H10(0,1),˜φ,˜ψH10(0,1)}.

    Assuming that U0D(M), it follows from the theorem in [30] that we have the classical solution

    UC(R+,D(M))C1(R+,H)

    satisfying tαφtC1((0,),L2(0,1)) and tαψtC1((0,),L2(0,1)).

    It is proved in [8] that the abstract problem given by

    {CDγU(t,x)=PU(t,x)+t0S(ts)U(s,x)ds+f(t,U(t)),0<γ<1,U(0,x)=U0(x)X, (3.2)

    possesses a unique solution under the above assumptions (A2), (A3) and

    (A1)'For some π/2<ϕ<π, there exists a constant C=C(ϕ)>0 such that

    Σ0,ϕ:={νC:|arg(ν)|<ϕ}ρ(P),

    and R(ν,P)<C/|ν|, νΣ0,ϕ.

    The function f:(0,)×XX was assumed to be continuous and locally Lipschitz in the second variable uniformly, with respect to the first variable.

    The following resolvent operator notion was employed.

    Definition 3. The family of bounded linear operators (Rγ(t))t0 determines a γ-resolvent for (3.2) if the following is true

    (a) The mapping Rγ(t):[0,)L(X):=L(X;X) is strongly continuous and Rγ(0)=I.

    (b) wD(P), Rγ(.)wC([0,);D(P))Cγ((0,);X) (Cγ((0,);X) is the space of continuous functions w for which CDγw exists and is continuous), and

    CDγRγ(t)w=PRγ(t)w+t0S(ts)Rγ(s)wds=Rγ(t)Pw+t0Rγ(ts)S(s)wds,t0.

    It is shown that the family

    Rγ(t):=12πiγ1Γϰeϰt(ϰγIPSˆ(ϰ))1dϰ,t0

    is a γ-resolvent for (3.2). Here, Γ:={teiθ:tr}{reiζ:θζθ}{teiθ:tr} is oriented counterclockwise, where π/2<θ<ϕ and r>r1 (a positive number determined in the proofs).

    Theorem 1. For U0D(M) and f(t,U(t))0, the function

    U(t):=Rγ(t)U0C([0,);D(M))Cγ((0,);X)

    is a solution of (3.2).

    Bearing in mind that, in our case, the Laplacian is the infinitesimal generator of an analytic semigroup, the conditions (A1), (A2) and (A3) are fulfilled. We deduce the existence of a unique (strong) solution in the space

    Hα:={uC([0,),D(A))C1([0,),H10(Ω)):t12α(uu0)C2([0,),L2(Ω))},

    where

    D(A):=H2(Ω)H10(Ω).

    Before tackling the stability issue, we need to prove two identities, define the "energy" functional and compute its fractional derivative. The superscript C in CDα will be dropped for notation convenience. Moreover, we will denote

    Tχ(t):=α2Γ(1α)10t0dξ(tξ)1α(ξ0χ(η)dη(tη)α)2dx,t0,
    (ωχ)(t):=10t0ω(ts)|χ(t)χ(s)|2dsdx,t0,

    and . the L2-norm.

    The analysis below is valid on arbitrary intervals [0,T], T>0. Since the evaluations are independent of time, they will be valid on all [0,+).

    Proposition 7. Assume that I1αω(t)C1([0,)), 0<α<1 and χH10(0,1) is a differentiable function. Then, it holds that

    10Dαχxt0ω(ts)χx(s)dsdx=12(RLDαωχx)(t)+12(t0ω(ts)ds)Dαχx2α2Γ(1α)t0dξ(tξ)1αξ0ω(η)dη(tη)αξ0[χx(s)2]ds(ts)α12Dα(ωχx)(t)+αΓ(1α)×10t0dξ(tξ)1αξ0(χx)(η)dη(tη)α(ξ0(ts)α(s0ω(sτ)χx(τ)dτ)ds)dx,t0.

    Proof. Using Proposition 5 and the identity

    10t0ω(ts)|χx(t)χx(s)|2dsdx=χx2t0ω(ts)ds+t0ω(ts)χx(s)2ds210χxt0ω(ts)χx(s)dsdx,t0,

    we find that

    Dα(ωχx)(t)=(Dαt0ω(ts)ds)χx2+(t0ω(ts)ds)Dαχx2αΓ(1α)t0dξ(tξ)1α×ξ0ω(η)dη(tη)αξ0[χx(s)2]ds(ts)α+Dαt0ω(ts)χx(s)2ds210Dαχxt0ω(ts)χx(s)dsdx210χxDαt0ω(ts)χx(s)dsdx+2αΓ(1α)10t0dξ(tξ)1αξ0[(χx)(η)]dη(tη)α+2αΓ(1α)10t0dξ(tξ)1αξ0[(χx)(η)]dη(tη)α(ξ0(ts)α(s0ω(sτ)χx(τ)dτ)ds)dx,t0.

    By virtue of Proposition 4 and the summability of ω, we see that

    Dαt0ω(ts)ds=t0RLDαω(ts)ds+limt0+I1αω(t),t0,
    Dαt0ω(ts)χx(s)2ds=RLDαt0ω(ts)χx(s)2dstαΓ(1α)(t0ω(ts)χx(s)2ds)|t=0=t0RLDαω(ts)χx(s)2ds+χx(t)2limt0+I1αω(t),t>0,

    and

    10χxDαt0ω(ts)χx(s)dsdx=10χxt0RLDαω(ts)χx(s)dsdx+χx(t)2limt0+I1αω(t),t>0.

    Therefore,

    Dα(ωχx)(t)=χx2t0RLDαω(ts)ds+(t0ω(ts)ds)Dαχx2αΓ(1α)t0dξ(tξ)1αξ0ω(η)dη(tη)αξ0[χx(s)2]ds(ts)α+t0RLDαω(ts)χx(s)2ds210Dαχxt0ω(ts)χx(s)dsdx210χxt0RLDαω(ts)χx(s)dsdx+2αΓ(1α)10t0dξ(tξ)1αξ0(χx)(η)dη(tη)α(ξ0(ts)α(s0ω(sτ)χx(τ)dτ)ds)dx.

    Having in mind that

    (RLDαωχx)(t)=χx2t0RLDαω(ts)ds+t0RLDαω(ts)χx(s)2ds210χxt0RLDαω(ts)χx(s)dsdx,t0,

    we infer that

    Dα(ωχx)(t)=(RLDαωχx)(t)+(t0ω(s)ds)Dαχx2αΓ(1α)t0dξ(tξ)1αξ0ω(η)dη(tη)αξ0[χx(s)2]ds(ts)α210Dαχxt0ω(ts)χx(s)dsdx+2αΓ(1α)10t0dξ(tξ)1αξ0(χx)(η)dη(tη)α(ξ0(ts)α(s0ω(sτ)χx(τ)dτ)ds)dx (4.1)

    for t0. This finishes the proof.

    Proposition 8. For absolutely continuous functions χ and ω such that I1αω, 0<α<1 is absolutely continuous, we have

    Dαt0ω(ts)[χ(t)χ(s)]ds=t0RLDαω(ts)(χ(t)χ(s))ds+(t0ω(s)ds)Dαχ(t)αΓ(1α)t0dξ(tξ)1α(ξ0ω(η)dη(tη)α)(ξ0χ(s)ds(ts)α),t0.

    Proof. Clearly, from Propositions 4 and 5, we can write

    Dαt0ω(ts)[χ(t)χ(s)]ds=Dα(χ(t)t0ω(ts)ds)Dαt0ω(ts)χ(s)ds=(t0ω(s)ds)Dαχ(t)+χ(t)[t0RLDαω(ts)ds+I1αω(0)]αΓ(1α)t0dξ(tξ)1α(ξ0ω(η)dη(tη)α)(ξ0χ(s)ds(ts)α)t0RLDαω(ts)χ(s)dsχ(t)I1αω(0),t0.

    The conclusion follows.

    The following energy functional is dictated by Proposition 7:

    E(t)=12[ρ1Dαφ2+ρ2Dαψ2+(bt0ω(s)ds)ψ2+kφx+ψ2]+12(ωψx)(t),t0.

    Employing (4.1), we have the following along the solutions of (1.3) and (1.2)

    DαE(t)=ρ110DαφDα(Dαφ)dxTDαφ(t)+ρ210DαψDα(Dαψ)dxTDαψ(t)12(Dαt0ω(s)ds)ψx2+12(bt0ω(s)ds)Dαψx2+α2Γ(1α)10t0dξ(tξ)1αξ0ω(η)dη(tη)αξ0[(ψx)2(s)]ds(ts)αdx+k10(φx+ψ)Dα(φx+ψ)dxkTφx+ψ(t)+12(RLDαωψx)(t)+12(t0ω(ts)ds)Dαψx2α2Γ(1α)10t0dξ(tξ)1αξ0ω(η)dη(tη)αξ0[(ψx)2(s)]ds(ts)αdx10Dαψxt0ω(ts)ψx(s)dsdx+αΓ(1α)10t0dξ(tξ)1α(ξ0[ψx(η)]dη(tη)α)(ξ0(ts)α(s0ω(sτ)ψx(τ)dτ)ds)dx.

    On the other hand, from the equations of the system (1.3), we have

    ρ110DαφDα(Dαφ)dx+ρ210DαψDα(Dαψ)dx=k10(φx+ψ)Dα(φx+ψ)dxb10ψxDαψxdx+10Dαψxt0ω(ts)ψx(s)dsdx.

    Therefore,

    DαE(t)=12(Dαt0ω(s)ds)ψx2+12(RLDαωψx)(t)TDαφ(t)TDαψ(t)bTψx(t)kTφx+ψ(t)+αΓ(1α)10t0dξ(tξ)1α(ξ0[ψx(η)]dη(tη)α)(ξ0(ts)α(s0ω(sτ)ψx(τ)dτ)ds)dx,t>0. (4.2)

    As the last term in (4.2) may be estimated by

    10t0dξ(tξ)1α(ξ0[ψx(η)]dη(tη)α)ξ0(ts)α(s0ω(sτ)ψx(τ)dτ)dsdxbΓ(1α)2αTψx(t)+1b10t0dξ(tξ)1α(ξ0(ts)α(s0ω(sτ)ψx(τ)dτ)ds)2dx,

    it is clear that, for t>0,

    DαE(t)I1αω(t)2ψx2Cω2(ωψx)(t)TDαφ(t)TDαψ(t)kTφx+ψ(t)b2Tψx(t)+αbΓ(1α)10t0dξ(tξ)1α(ξ0(ts)α(s0ω(sτ)ψx(τ)dτ)ds)2dx, (4.3)

    where we have used the Assumption (A). This is a delicate situation because we do not know the sign of the right-hand side of (4.3). Actually, even if we know that the sign is negative, we cannot deduce dissipativity.

    Several functionals, with which we will modify our energy functional, will be introduced in this section, and their fractional derivatives will be evaluated.

    Lemma 1. The \alpha -fractional derivative of the functional

    \begin{equation*} U_{1}(t): = \left\Vert \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right\Vert ^{2}+\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds, \;t\geq 0 \end{equation*}

    satisfies

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{1}(t)\leq -\frac{C_{\omega }}{2}\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds+\frac{4\tilde{\omega}^{2}}{ C_{\omega }}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\left( \frac{4\bar{\omega}\left[ \tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}}{C_{\omega }}+\tilde{\omega}\right) \left\Vert \psi _{x}\right\Vert ^{2} \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }d\eta \right) ^{2}dx, \;t\geq 0. \end{array} \end{equation*}

    Proof. Proposition 5 allows us to write

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{1}(t) = 2\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) D^{\alpha }\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }d\eta \right) ^{2}dx+D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds \end{array} \end{equation*}

    for t\geq 0, and the relationship between the Riemann-Liouville fractional derivative and the Caputo fractional derivative gives

    \begin{equation*} D^{\alpha }\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) = ^{RL}D^{\alpha }\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx-\frac{t^{-\alpha }}{\Gamma (1-\alpha )}\left. \left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) \right\vert _{t = 0}. \end{equation*}

    Moreover, the summability of \omega and Proposition 4 lead to

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{1}(t) = 2\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) \left[ \int\nolimits_{0}^{t}\, ^{RL}D^{\alpha }\omega (t-s)\psi _{x}(s)ds+\psi _{x}(t)\lim\limits_{t\rightarrow 0^{+}}I^{1-\alpha }\omega (t)\right] dx \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }d\eta \right) ^{2}dx \\ +\int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. \omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds+\left\Vert \psi _{x}(t)\right\Vert ^{2}\lim\limits_{t\rightarrow 0}I^{1-\alpha }\omega (t), \end{array} \end{equation*}

    and by the definition of the Riemann-Liouville fractional derivative, it follows that

    \begin{equation} \begin{array}{c} D^{\alpha }U_{1}(t)\leq \frac{2\bar{\omega}}{C_{\omega }}\int \nolimits_{0}^{1}\left[ \int\nolimits_{0}^{t}\, ^{RL}D^{\alpha }\omega (t-s) \left[ \psi _{x}(s)-\psi _{x}(t)\right] ds+\psi _{x}(t)I^{1-\alpha }\omega (t)\right] ^{2}dx \\ +\frac{C_{\omega }}{2}\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds-\frac{\alpha }{\Gamma (1-\alpha )} \int\nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha } }\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }d\eta \right) ^{2}dx \\ -C_{\omega }\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds+\left\Vert \psi _{x}(t)\right\Vert ^{2}\lim\limits_{t\rightarrow 0}I^{1-\alpha }\omega (t), \;t\geq 0. \end{array} \end{equation} (5.1)

    From Proposition 3, we have

    \begin{equation} I^{1-\alpha }\omega (t)\leq \frac{\tilde{\omega}}{\Gamma (1-\alpha )} \int\nolimits_{0}^{t}(t-s)^{\alpha -1}E_{\alpha , \alpha }(-C_{\omega }(t-s)^{\alpha })s^{-\alpha }ds = \tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\leq \tilde{\omega}, \end{equation} (5.2)

    and using

    \begin{equation*} \psi _{x}^{2}(s)\leq 2\left[ \psi _{x}(s)-\psi _{x}(t)\right] ^{2}+2\psi _{x}^{2}(t), \end{equation*}

    the relation (5.1) leads to

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{1}(t)\leq \frac{-C_{\omega }}{2}\int\nolimits_{0}^{t}\omega (t-s)\left\Vert \psi _{x}(s)\right\Vert ^{2}ds+\left( \frac{2\bar{\omega} \left[ \tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}}{ C_{\omega }}+\tilde{\omega}\right) \left\Vert \psi _{x}\right\Vert ^{2} \\ +\frac{4\bar{\omega}}{C_{\omega }}\left( \int\nolimits_{0}^{t}\, \left\vert ^{RL}D^{\alpha }\omega (t-s)\right\vert ds\right) \int\nolimits_{0}^{1}\int\nolimits_{0}^{t}\, \left\vert ^{RL}D^{\alpha }\omega (t-s)\right\vert \left[ \psi _{x}(s)-\psi _{x}(t)\right] ^{2}dsdx \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }d\eta \right) ^{2}dx, \;t\geq 0. \end{array} \end{equation*}

    The announced estimation is obtained by noticing that

    \begin{equation} 0\leq \int\nolimits_{0}^{t}\, \left\vert ^{RL}D^{\alpha }\omega (s)\right\vert ds = I^{1-\alpha }\omega (0)-I^{1-\alpha }\omega (t)\leq \tilde{ \omega}. \end{equation} (5.3)

    The proof is complete.

    Our second functional is given in the next lemma.

    Lemma 2. For the functional

    \begin{equation*} U_{2}(t): = -\int\nolimits_{0}^{1}(\rho _{1}\varphi D^{\alpha }\varphi +\rho _{2}\psi D^{\alpha }\psi )dx, \;t > 0, \end{equation*}

    and \delta _{2} > 0, it holds that

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{2}(t)\leq -\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\left( \delta _{2}+b-\int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert \psi _{x}\right\Vert ^{2} \\ +\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t)+2\rho _{1} \mathcal{T}_{\varphi _{x}+\psi }(t)+\left( 2\rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t), \;t\geq 0. \end{array} \end{equation*}

    Proof. Applying the operator D^{\alpha } to the functional U_{2}(t), we obtain

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{2}(t) = -\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}+ \frac{\alpha \rho _{1}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\varphi ^{\prime }(\eta )d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx+k\left\Vert \varphi _{x}+\psi \right\Vert ^{2} \\ -\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert \psi _{x}\right\Vert ^{2}+\int\nolimits_{0}^{1}\psi _{x}\int\nolimits_{0}^{t}\omega (t-s)\left[ \psi _{x}(t)-\psi _{x}(s)\right] dsdx \\ +\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(\eta )d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx, \;t\geq 0. \end{array} \end{equation*}

    Next, the estimations

    \begin{equation*} \int\nolimits_{0}^{1}\psi _{x}\int\nolimits_{0}^{t}\omega (t-s)\left[ \psi _{x}(t)-\psi _{x}(s)\right] dsdx\leq \delta _{2}\left\Vert \psi _{x}\right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t), \;\delta _{2} > 0, \end{equation*}
    \begin{equation*} \frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\varphi ^{\prime }(\eta )d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx\leq \mathcal{T} _{\varphi }(t)+\mathcal{T}_{D^{\alpha }\varphi }(t), \end{equation*}

    and

    \begin{equation*} \frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(\eta )d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx\leq \mathcal{T} _{\psi }(t)+\mathcal{T}_{D^{\alpha }\psi }(t) \end{equation*}

    imply that

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{2}(t)\leq -\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert \psi _{x}\right\Vert ^{2} \\ +\delta _{2}\left\Vert \psi _{x}\right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t)+\rho _{1}\left[ \mathcal{T}_{\varphi }(t)+ \mathcal{T}_{D^{\alpha }\varphi }(t)\right] +\rho _{2}\left[ \mathcal{T} _{\psi }(t)+\mathcal{T}_{D^{\alpha }\psi }(t)\right] , \;t\geq 0. \end{array} \end{equation*}

    Finally, by virtue of the relation

    \begin{equation} \mathcal{T}_{\varphi }(t)\leq \mathcal{T}_{\varphi _{x}}(t)\leq 2\mathcal{T} _{\varphi _{x}+\psi }(t)+2\mathcal{T}_{\psi _{x}}(t), \end{equation} (5.4)

    we get

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{2}(t)\leq -\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\left( \delta _{2}+b-\int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert \psi _{x}\right\Vert ^{2} \\ +\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t)+2\rho _{1} \mathcal{T}_{\varphi _{x}+\psi }(t)+\left( 2\rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t), \;t\geq 0. \end{array} \end{equation*}

    Our third functional is defined below.

    Lemma 3. If

    \begin{equation*} U_{3}(t): = -\rho _{2}\int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx, \;t\geq 0, \end{equation*}

    then, for t > t_{0} > 0,

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{3}(t)\leq \rho _{2}\left( \delta _{2}+\frac{2\alpha \delta _{1} }{\Gamma (1-\alpha )}-\omega _{0}\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2}+b\delta _{2}\left\Vert \psi _{x}\right\Vert ^{2}+k\delta _{2}\left\Vert \varphi _{x}+\psi \right\Vert ^{2} \\ +\bar{\omega}\left( 1+\frac{k+b}{4\delta _{2}}\right) (\omega \square \psi _{x})(t)-\frac{\tilde{\omega}\rho _{2}}{4\delta _{2}}(^{RL}D^{\alpha }\omega \square \psi _{x})(t)+\frac{\rho _{2}C_{2}E_{\alpha }(-C_{\omega }t^{\alpha })}{\delta _{1}}\mathcal{T}_{\psi }(t)+\frac{\rho _{2}}{\delta _{3}}\mathcal{ T}_{D^{\alpha }\psi }(t) \\ +\frac{\alpha \rho _{2}\delta _{3}}{2\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx, \end{array} \end{equation*}

    where \omega _{0} = \int\nolimits_{0}^{t_{0}}\omega (s)ds and \delta _{i} > 0, i = 1, 2, 3.

    Proof. Differentiating U_{3}(t) and using Problem (1.2) and (1.3), we find that

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{3}(t) = -\rho _{2}\int\nolimits_{0}^{1}D^{\alpha }\psi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx-b\int\nolimits_{0}^{1}\psi _{xx}\int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx \\ +\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\psi _{xx}(s)ds\right) \int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx \\ +k\int\nolimits_{0}^{1}(\varphi _{x}+\psi )\int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx \\ +\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation*}

    Next, we integrate by parts and use Proposition 8 to get

    \begin{equation} \begin{array}{c} D^{\alpha }U_{3}(t) = -\rho _{2}\int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\, ^{RL}D^{\alpha }\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx-\rho _{2}\left( \int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2} \\ +\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{ (t-s)^{\alpha }}\right) dx+\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) ds\right) ^{2}dx \\ +\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \int\nolimits_{0}^{1}\psi _{x}\int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) dsdx \\ +k\int\nolimits_{0}^{1}(\varphi _{x}+\psi )\int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx \\ +\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation} (5.5)

    By assumption (A), relation (5.3) and the Young inequality, we infer that

    \begin{equation} \begin{array}{c} \int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\, ^{RL}D^{\alpha }\omega (t-s)\left[ \psi (t)-\psi (s)\right] dsdx\leq \delta _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\tilde{\omega}}{4\delta _{2}} (\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t) \\ \leq \delta _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}-\frac{\tilde{ \omega}}{4\delta _{2}}(^{RL}D^{\alpha }\omega \square \psi _{x})(t), \;\delta _{2} > 0, \end{array} \end{equation} (5.6)
    \begin{equation} \int\nolimits_{0}^{1}\psi _{x}\int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) dsdx\leq \delta _{2}\left\Vert \psi _{x}\right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t), \;\delta _{2} > 0, \end{equation} (5.7)
    \begin{equation} \int\nolimits_{0}^{1}\left[ \int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) ds\right] ^{2}dx\leq \bar{\omega}(\omega \square \psi _{x})(t), \end{equation} (5.8)
    \begin{equation*} \begin{array}{c} \int\nolimits_{0}^{1}(\varphi _{x}+\psi )\int\nolimits_{0}^{t}\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx\leq \delta _{2}\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t), \;\delta _{2} > 0,\\ \frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{ (t-s)^{\alpha }}\right) dx \\ \leq \frac{\alpha \delta _{1}}{\Gamma (1-\alpha )}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\alpha }{4\delta _{1}\Gamma (1-\alpha )} \int\nolimits_{0}^{1}\left[ \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{ (t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{(t-s)^{\alpha }}\right) \right] ^{2}dx \end{array} \end{equation*} (5.9)

    for \delta _{1} > 0, and

    \begin{equation*} \begin{array}{l} \frac{\alpha }{2\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\left[ \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{ (t-s)^{\alpha }}\right) \right] ^{2}dx \\ \leq \frac{\alpha }{2\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\left[ \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) ^{2}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{(t-s)^{\alpha }}\right) ^{2}\right] dx \\ \leq \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) ^{2}\mathcal{T}_{\psi }(t), \;t > 0. \end{array} \end{equation*}

    Next, observing that E_{\alpha }(-C_{\omega }t^{\alpha })\leq C_{1}t^{-\alpha } (for some positive constant C_{1} ) away from zero, we have

    \begin{equation*} \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) ^{2}\leq \left[ \Gamma (1-\alpha )E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\leq C_{2}E_{\alpha }(-C_{\omega }t^{\alpha }), \;t > t_{0} > 0, \end{equation*}

    where C_{2}: = C_{1}\Gamma ^{2}(1-\alpha)/\alpha. Therefore,

    \begin{equation} \begin{array}{c} \frac{\alpha }{2\Gamma (1-\alpha )}\int\nolimits_{0}^{1}D^{\alpha }\psi \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{ (t-s)^{\alpha }}\right) dx \\ \leq \frac{\alpha \delta _{1}}{\Gamma (1-\alpha )}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{C_{2}E_{\alpha }(-C_{\omega }t^{\alpha })}{2\delta _{1}}\mathcal{T}_{\psi }(t), \;t > t_{0} > 0. \end{array} \end{equation} (5.10)

    Our last evaluation is as follows:

    \begin{equation} \begin{array}{c} \frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ \leq \frac{1}{\delta _{3}}\mathcal{T}_{D^{\alpha }\psi }(t)+\frac{\alpha \delta _{3}}{2\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int\nolimits_{0}^{t} \frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx \end{array} \end{equation} (5.11)

    for \delta _{3} > 0. The relations (5.6)–(5.11), when inserted in (5.5), yield

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{3}(t)\leq \rho _{2}\left( \delta _{2}+\frac{2\alpha \delta _{1} }{\Gamma (1-\alpha )}-\int\nolimits_{0}^{t_{0}}\omega (s)ds\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2}+b\delta _{2}\left\Vert \psi _{x}\right\Vert ^{2}+k\delta _{2}\left\Vert \varphi _{x}+\psi \right\Vert ^{2} \\ +\bar{\omega}\left( 1+\frac{k+b}{4\delta _{2}}\right) (\omega \square \psi _{x})(t)-\frac{\tilde{\omega}\rho _{2}}{4\delta _{2}}(^{RL}D^{\alpha }\omega \square \psi _{x})(t)+\frac{\rho _{2}C_{2}E_{\alpha }(-C_{\omega }t^{\alpha })}{\delta _{1}}\mathcal{T}_{\psi }(t)+\frac{\rho _{2}}{\delta _{3}}\mathcal{ T}_{D^{\alpha }\psi }(t) \\ +\frac{\alpha \rho _{2}\delta _{3}}{2\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx \end{array} \end{equation*}

    for t > t_{0} > 0. This ends the proof.

    The role of our fourth functional is to control the last term in the evaluation of D^{\alpha }U_{3}(t). We set

    \begin{equation*} U_{4}(t): = \int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s) \left[ \psi (t)-\psi (s)\right] ds\right) ^{2}dx, \;t\geq 0. \end{equation*}

    Lemma 4. The Caputo fractional derivative of the above functional U_{4}(t) fulfills the following for \delta _{1}, \delta _{2} > 0 and t > t_{0} > 0:

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{4}(t)\leq \bar{\omega}\delta _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\bar{\omega}\left( 1+\frac{\bar{\omega}}{\delta _{2}}+\frac{ 2\alpha \delta _{1}}{\Gamma (1-\alpha )}\right) \left( \omega \square \psi _{x}\right) (t)+\tilde{\omega}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t) \\ +\frac{C_{2}E_{\alpha }(-C_{\omega }t^{\alpha })}{\delta _{1}}\mathcal{T} _{\psi _{x}}(t)-\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1} \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx. \end{array} \end{equation*}

    Proof. A simple fractional differentiation of order \alpha shows that

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{4}(t) = 2\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\left[ \psi (t)-\psi (s)\right] ds\right) D^{\alpha }\left( \int\nolimits_{0}^{t}\omega (t-s)\left[ \psi (t)-\psi (s) \right] ds\right) dx \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx, \end{array} \end{equation*}

    and in view of Proposition 8, we see that

    \begin{equation} \begin{array}{c} D^{\alpha }U_{4}(t) = 2\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\left[ \psi (t)-\psi (s)\right] ds\right) \int\nolimits_{0}^{t}\, ^{RL}D^{\alpha }\omega (t-s)\left( \psi (t)-\psi (s)\right) dsdx \\ +2\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\left[ \psi (t)-\psi (s)\right] ds\right) dx\left( \int\nolimits_{0}^{t}\omega (s)ds\right) \int\nolimits_{0}^{1}D^{\alpha }\psi (t)dx \\ -\frac{2\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)\left[ \psi (t)-\psi (s)\right] ds\right) \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\psi ^{\prime }(s)ds}{ (t-s)^{\alpha }}\right) dx \\ -\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx. \end{array} \end{equation} (5.12)

    This identity (5.12) can be evaluated by Young's inequality and a similar argument as for relation (5.10) above, as follows:

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{4}(t)\leq \bar{\omega}\delta _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\bar{\omega}\left( 1+\frac{\bar{\omega}}{\delta _{2}}+\frac{ 2\alpha \delta _{1}}{\Gamma (1-\alpha )}\right) \left( \omega \square \psi _{x}\right) (t)+\tilde{\omega}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t) \\ +\frac{C_{2}E_{\alpha }(-C_{\omega }t^{\alpha })}{\delta _{1}}\mathcal{T} _{\psi _{x}}(t)-\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1} \int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left( \int\nolimits_{0}^{\eta }\omega (\eta -s)\left( \psi (\eta )-\psi (s)\right) ds\right) ^{\prime }d\eta \right) ^{2}dx\; \end{array} \end{equation*}

    for \delta _{1}, \delta _{2} > 0 and t > t_{0} > 0. The proof is completed.

    The fractional derivative of our fifth functional gives rise to the useful term -\left\Vert \varphi _{x}+\psi \right\Vert ^{2}.

    Lemma 5. Let

    \begin{equation*} U_{5}(t) = \rho _{2}\int\nolimits_{0}^{1}D^{\alpha }\psi (\varphi _{x}+\psi )dx+\frac{b\rho _{1}}{k}\int\nolimits_{0}^{1}\psi _{x}D^{\alpha }\varphi dx- \frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\varphi \int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx, \;t\geq 0. \end{equation*}

    Then, for \delta _{2}, \delta _{3} > 0 ,

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{5}(t)\leq \frac{k}{12}\left[ \varphi _{x}^{2}(1)+\varphi _{x}^{2}(0)\right] -k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{ \omega}}{4k^{2}}E_{\alpha }(-C_{\omega }t^{\alpha })\left\Vert \psi _{x}\right\Vert ^{2} \\ +\left[ \delta _{2}+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{ \omega}}{4\delta _{2}k^{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx \\ +\frac{\rho _{1}}{k}\left( b+\frac{1}{\delta _{3}}\right) \mathcal{T} _{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t)+\rho _{2}\mathcal{T}_{\varphi _{x}+\psi }(t)+\frac{b\rho _{1}}{k}\mathcal{T} _{\psi _{x}}(t) \\ +\frac{3}{k}\left[ \left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)+\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0)\right] \\ +\frac{\alpha \rho _{1}\delta _{3}}{2k\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) ^{2}dx, \;t\geq 0. \end{array} \end{equation*}

    Proof. Clearly, from Proposition 5 and the equations of the system, we have

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{5}(t) = \int\nolimits_{0}^{1}(\varphi _{x}+\psi )\left[ b\psi _{xx}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{xx}(s)ds-k(\varphi _{x}+\psi ) \right] dx \\ +\rho _{2}\int\nolimits_{0}^{1}D^{\alpha }\psi D^{\alpha }(\varphi _{x}+\psi )dx-\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{(\varphi _{x}+\psi )^{\prime }(s)ds}{(t-s)^{\alpha }}\right) dx \\ +\frac{b\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx+\frac{b}{k}\int\nolimits_{0}^{1}\psi _{x}k(\varphi _{x}+\psi )_{x}dx -\frac{\alpha b\rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ \psi _{x}(\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{ \left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ -\frac{1}{k}\int\nolimits_{0}^{1}k(\varphi _{x}+\psi )_{x}\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx-\frac{\rho _{1}}{k} \int\nolimits_{0}^{1}D^{\alpha }\varphi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx \\ +\frac{\alpha \rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx, \end{array} \end{equation*}

    and after integration by parts,

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{5}(t) = \left[ \varphi _{x}\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) \right] _{0}^{1}-k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2} \\ +\left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx-\frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\varphi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx \\ -\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{(\varphi _{x}+\psi )^{\prime }(s)ds}{(t-s)^{\alpha }}\right) dx \\ -\frac{\alpha b\rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ \psi _{x}(\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{ \left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ +\frac{\alpha \rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation*}

    Repeated use of Young's inequality leads to

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{5}(t)\leq \frac{k}{12}\left[ \varphi _{x}^{2}(1)+\varphi _{x}^{2}(0)\right] -k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2} \\ +\left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx+\rho _{2} \mathcal{T}_{D^{\alpha }\psi }(t)+\rho _{2}\mathcal{T}_{\varphi _{x}+\psi }(t) \\ +\frac{3}{k}\left[ \left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)+\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0)\right] \\ +\frac{b\rho _{1}}{k}\mathcal{T}_{\psi _{x}}(t)+\frac{b\rho _{1}}{k}\mathcal{ T}_{D^{\alpha }\varphi }(t)-\frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{ \alpha }\varphi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx \\ +\frac{\alpha \rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation*}

    Observe that

    \begin{equation*} \begin{array}{l} \frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\varphi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx \\ = \frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\varphi \left[ \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. \omega (t-s)\psi _{x}(s)ds+\psi _{x}(t)\lim\limits_{t\rightarrow 0^{+}}I^{1-\alpha }\omega (t)\right] dx \\ \leq \frac{\rho _{1}}{k}\int\nolimits_{0}^{1}D^{\alpha }\varphi \left[ \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. \omega (t-s)\left( \psi _{x}(s)-\psi _{x}(t)\right) ds+\psi _{x}(t)I^{1-\alpha }\omega (t)\right] dx \\ \leq \delta _{2}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{\omega}}{4\delta _{2}k^{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{\omega}}{4k^{2}}E_{\alpha }(-C_{\omega }t^{\alpha })\left\Vert \psi _{x}\right\Vert ^{2}, \end{array} \end{equation*}

    and

    \begin{equation*} \begin{array}{c} \frac{\alpha \rho _{1}}{k\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ \leq \frac{\rho _{1}}{k\delta _{3}}\mathcal{T}_{D^{\alpha }\varphi }(t)+ \frac{\alpha \rho _{1}\delta _{3}}{2k\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) ^{2}dx. \end{array} \end{equation*}

    Therefore,

    \begin{equation} \begin{array}{l} D^{\alpha }U_{5}(t)\leq \frac{k}{12}\left[ \varphi _{x}^{2}(1)+\varphi _{x}^{2}(0)\right] -k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{ \omega}}{4k^{2}}E_{\alpha }(-C_{\omega }t^{\alpha })\left\Vert \psi _{x}\right\Vert ^{2} \\ +\left[ \delta _{2}+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\rho _{1}^{2}\tilde{ \omega}}{4\delta _{2}k^{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx \\ +\frac{\rho _{1}}{k}\left( b+\frac{1}{\delta _{3}}\right) \mathcal{T} _{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t)+\rho _{2}\mathcal{T}_{\varphi _{x}+\psi }(t)+\frac{b\rho _{1}}{k}\mathcal{T} _{\psi _{x}}(t) \\ +\frac{3}{k}\left[ \left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)+\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0)\right] \\ +\frac{\alpha \rho _{1}\delta _{3}}{2k\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-\eta )^{-\alpha }\left[ \int\nolimits_{0}^{\eta }\omega (\eta -s)\psi _{x}(s)ds\right] ^{\prime }d\eta \right) ^{2}dx \end{array} \end{equation} (5.13)

    for \delta _{2}, \delta _{3} > 0. The proof is complete.

    To deal with some of the boundary terms in (5.13), we need the functional

    \begin{equation*} U_{6}(t) = \rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi \left( b\psi _{x}(t)-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx, \;t\geq 0, \end{equation*}

    where m(x) = 2-4x, so that m(0) = -m(1) = 2.

    Lemma 6. The functional U_{6}(t) verifies along the solutions of (1.2) and (1.3)

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{6}(t)\leq -\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)-\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0) \\ +\left[ 4b^{2}+\frac{kb^{2}}{2\delta _{4}}+\rho _{2}^{2}\tilde{\omega} E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert \psi _{x}\right\Vert ^{2}+2k\left( \delta _{4}+\delta _{2}\right) \left\Vert \varphi _{x}+\psi \right\Vert ^{2} \\ +\left[ 2b\rho _{2}+\delta _{2}+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\bar{\omega} \left( 4+\frac{k}{2\delta _{2}}\right) (\omega \square \psi _{x})(t) \\ +\frac{\tilde{\omega}\rho _{2}^{2}}{\delta _{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\rho _{2}b^{2}\mathcal{T}_{\psi _{x}}(t)+4\rho _{2}\left( 1+\frac{1}{\delta _{3}}\right) \mathcal{T} _{D^{\alpha }\psi }(t) \\ +\frac{\alpha \rho _{2}\delta _{3}}{2\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) ^{2}dx \end{array} \end{equation*}

    for t\geq 0 and \delta _{i} > 0, \; i = 2, 3, 4.

    Proof. Clearly, a direct application of Proposition 5 gives

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{6}(t) = \rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\left( D^{\alpha }\psi \right) \left( b\psi _{x}(t)-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ +\rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi \left( bD^{\alpha }\psi _{x}(t)-D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ -\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}m(x)\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \times \left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( b\psi _{x}(s)-\int\nolimits_{0}^{s}\omega (s-\sigma )\psi _{x}(\sigma )d\sigma \right) ^{\prime }ds\right) dx, \end{array} \end{equation*}

    and using the second equation of our system gives

    \begin{equation} \begin{array}{l} D^{\alpha }U_{6}(t)\leq \int\nolimits_{0}^{1}m(x)\left( b\psi _{xx}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{xx}(s)ds\right) \left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ -k\int\nolimits_{0}^{1}m(x)(\varphi _{x}+\psi )\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx+b\rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi D^{\alpha }\psi _{x}(t)dx \\ -\rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx+\rho _{2}b^{2}\mathcal{T} _{\psi _{x}}(t)+4\rho _{2}\left( 1+\frac{1}{\delta _{3}}\right) \mathcal{T} _{D^{\alpha }\psi }(t) \\ +\frac{\alpha \rho _{2}\delta _{3}}{2\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) ^{2}dx, \;\delta _{3} > 0, \;t\geq 0. \end{array} \end{equation} (5.14)

    Notice that, employing the relations (5.2) and (5.3), integrations by parts and Young's inequality gives

    \begin{equation} \begin{array}{l} \rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi D^{\alpha }\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)dsdx \\ = \rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi \left[ \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. \omega (t-s)\psi _{x}(s)ds+\psi _{x}(t)\lim\limits_{t\rightarrow 0^{+}}I^{1-\alpha }\omega (t)\right] dx \\ = \rho _{2}\int\nolimits_{0}^{1}m(x)D^{\alpha }\psi \left[ \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. \omega (t-s)\left( \psi _{x}(s)-\psi _{x}(t)\right) ds+\psi _{x}(t)I^{1-\alpha }\omega (t)\right] dx \\ \leq \delta _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\tilde{ \omega}\rho _{2}^{2}}{\delta _{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\left( \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\rho _{2}^{2}\left\Vert \psi _{x}\right\Vert ^{2}\right) , \end{array} \end{equation} (5.15)
    \begin{equation} \begin{array}{c} \int\nolimits_{0}^{1}m(x)(\varphi _{x}+\psi )\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ \leq 2\left( \delta _{4}+\delta _{2}\right) \left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\frac{b^{2}}{2\delta _{4}}\left\Vert \psi _{x}\right\Vert ^{2}+\frac{\bar{\omega}}{2\delta _{2}}(\omega \square \psi _{x})(t), \end{array} \end{equation} (5.16)
    \begin{equation} \begin{array}{l} \int\nolimits_{0}^{1}m(x)\left( b\psi _{xx}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{xx}(s)ds\right) \left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) dx \\ = -\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)-\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0) \\ -\frac{1}{2}\int\nolimits_{0}^{1}m^{\prime }(x)\left[ \left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \psi _{x}+\int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) ds\right] ^{2}dx \\ \leq -\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)-\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0) \\ +4b^{2}\left\Vert \psi _{x}\right\Vert ^{2}+4\bar{\omega}(\omega \square \psi _{x})(t), \end{array} \end{equation} (5.17)

    and

    \begin{equation} \begin{array}{c} \int\nolimits_{0}^{1}m(x)D^{\alpha }\psi D^{\alpha }\psi _{x}(t)dx = \frac{1}{ 2}\int\nolimits_{0}^{1}m(x)\frac{d}{dx}\left( D^{\alpha }\psi \right) ^{2}dx \\ = \frac{1}{2}\left[ m(x)\left( D^{\alpha }\psi \right) ^{2}\right] _{0}^{1}- \frac{1}{2}\int\nolimits_{0}^{1}m^{\prime }(x)\left( D^{\alpha }\psi \right) ^{2}dx\leq 2\left\Vert D^{\alpha }\psi \right\Vert ^{2}. \end{array} \end{equation} (5.18)

    It suffices now to apply the previous estimations (5.15)–(5.18) in (5.14) to obtain the following for \delta _{i} > 0 , i = 2, 3, 4 :

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{6}(t)\leq -\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(1)-\left( b\psi _{x}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{x}(s)ds\right) ^{2}(0) \\ +\left[ 4b^{2}+\frac{kb^{2}}{2\delta _{4}}+\rho _{2}^{2}\tilde{\omega} E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert \psi _{x}\right\Vert ^{2}+2k\left( \delta _{4}+\delta _{2}\right) \left\Vert \varphi _{x}+\psi \right\Vert ^{2}+\bar{\omega}\left( 4+\frac{k}{2\delta _{2}}\right) (\omega \square \psi _{x})(t) \\ +\left[ 2b\rho _{2}+\delta _{2}+\tilde{\omega}E_{\alpha }(-C_{\omega }t^{\alpha })\right] \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{ \tilde{\omega}\rho _{2}^{2}}{\delta _{2}}(\left\vert ^{RL}D^{\alpha }\omega \right\vert \square \psi _{x})(t)+\rho _{2}b^{2}\mathcal{T}_{\psi _{x}}(t)+4\rho _{2}\left( 1+\frac{1}{\delta _{3}}\right) \mathcal{T} _{D^{\alpha }\psi }(t) \\ +\frac{\alpha \rho _{2}\delta _{3}}{2\Gamma (1-\alpha )}\int \nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }} \left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) ^{2}dx, \;t\geq 0. \end{array} \end{equation*}

    The proof is over.

    The derivative of the next functional provides us with the boundary terms needed to control the other remaining corresponding boundary terms in Lemma 5. Let

    \begin{equation*} U_{7}(t): = \rho _{1}\int\nolimits_{0}^{1}m(x)\varphi _{x}D^{\alpha }\varphi dx, \;t\geq 0. \end{equation*}

    Lemma 7. The above functional U_{7}(t) satisfies

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{7}(t)\leq 2\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-k\left[ \varphi _{x}^{2}(1)+\varphi _{x}^{2}(0)\right] +6k\left\Vert \varphi _{x}+\psi \right\Vert ^{2} \\ +7k\left\Vert \psi _{x}\right\Vert ^{2}+8\rho _{1}\mathcal{T}_{\varphi _{x}+\psi }(t)+8\rho _{1}\mathcal{T}_{\psi _{x}}(t)+4\rho _{1}\mathcal{T} _{D^{\alpha }\varphi }(t), \;t\geq 0. \end{array} \end{equation*}

    Proof. Proceeding similarly to the previous lemmas, it appears that

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{7}(t) = \rho _{1}\int\nolimits_{0}^{1}m(x)D^{\alpha }\varphi _{x}D^{\alpha }\varphi dx+\rho _{1}\int\nolimits_{0}^{1}m(x)\varphi _{x}D^{\alpha }\left( D^{\alpha }\varphi \right) dx \\ -\frac{\alpha \rho _{1}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}m(x)\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ \varphi _{x}(\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{ \left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ \leq -\frac{\rho _{1}}{2}\int\nolimits_{0}^{1}m^{\prime }(x)\left( D^{\alpha }\varphi \right) ^{2}dx+k\int\nolimits_{0}^{1}m(x)\varphi _{x}(\varphi _{x}+\psi )_{x}dx+4\rho _{1}\mathcal{T}_{\varphi _{x}}(t)+4\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t) \end{array} \end{equation*}

    or

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{7}(t)\leq 2\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{k}{2}\left[ m(x)\varphi _{x}^{2}\right] _{0}^{1}-\frac{k}{2} \int\nolimits_{0}^{1}m^{\prime }(x)\varphi _{x}^{2}dx+k\left\Vert \varphi _{x}\right\Vert ^{2} \\ +k\left\Vert \psi _{x}\right\Vert ^{2}+4\rho _{1}\mathcal{T}_{\varphi _{x}}(t)+4\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t), \;t\geq 0. \end{array} \end{equation*}

    Next, we employ the inequality \left\Vert \varphi _{x}\right\Vert ^{2}\leq 2\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+2\left\Vert \psi _{x}\right\Vert ^{2} to reach

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{7}(t)\leq 2\rho _{1}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-k\left[ \varphi _{x}^{2}(1)+\varphi _{x}^{2}(0)\right] +6k\left\Vert \varphi _{x}+\psi \right\Vert ^{2}+7k\left\Vert \psi _{x}\right\Vert ^{2} \\ +8\rho _{1}\mathcal{T}_{\varphi _{x}+\psi }(t)+8\rho _{1}\mathcal{T}_{\psi _{x}}(t)+4\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t), \;t\geq 0. \end{array} \end{equation*}

    The proof is complete.

    Consider the problem

    \begin{equation*} \left\{ \begin{array}{l} -w_{xx} = \psi _{x}, \;x\in (0, 1), \\ w(0) = w(1) = 0. \end{array} \right. \end{equation*}

    The last functional is

    \begin{equation*} U_{8}(t): = \int\nolimits_{0}^{1}(\rho _{1}wD^{\alpha }\varphi +\rho _{2}\psi D^{\alpha }\psi )dx. \end{equation*}

    Lemma 8. The above functional U_{8}(t) satisfies the following for \delta _{2}, \delta _{5} > 0 and t\geq 0:

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{8}(t)\leq \left( \rho _{1}\delta _{5}+\rho _{2}\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}}{4\delta _{5}} \left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t) \\ +\left[ \delta _{2}-\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \right] \left\Vert \psi _{x}\right\Vert ^{2}+\left( \rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t). \end{array} \end{equation*}

    Proof. Proposition 5 implies that

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{8}(t) = \rho _{1}\int\nolimits_{0}^{1}D^{\alpha }wD^{\alpha }\varphi dx+\rho _{1}\int\nolimits_{0}^{1}wD^{\alpha }\left( D^{\alpha }\varphi \right) dx+\rho _{2}\int\nolimits_{0}^{1}\psi D^{\alpha }(D^{\alpha }\psi )dx \\ +\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}-\frac{\alpha \rho _{1} }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ w(\eta ) \right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\varphi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx \\ -\frac{\alpha \rho _{2}}{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\left[ \psi (\eta )\right] ^{\prime }d\eta }{ (t-\eta )^{\alpha }}\right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ D^{\alpha }\psi (s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation*}

    Considering the solutions of (1.3), this relation may be estimated by

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{8}(t)\leq \rho _{1}\delta _{5}\left\Vert D^{\alpha }w\right\Vert ^{2}+\frac{\rho _{1}}{4\delta _{5}}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}+k\int\nolimits_{0}^{1}w(\varphi _{x}+\psi )_{x}dx \\ +\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\int\nolimits_{0}^{1}\psi \left[ b\psi _{xx}-\int\nolimits_{0}^{t}\omega (t-s)\psi _{xx}(s)ds-k(\varphi _{x}+\psi ) \right] dx \\ +\rho _{1}\mathcal{T}_{w}(t)+\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{\psi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t), \end{array} \end{equation*}

    and because \left\Vert D^{\alpha }w\right\Vert ^{2}\leq \left\Vert D^{\alpha }\psi \right\Vert ^{2} (by applying D^{\alpha } to both sides, multiplying by D^{\alpha }w and integrating by parts), we find that

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{8}(t)\leq \rho _{1}\delta _{5}\left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}}{4\delta _{5}}\left\Vert D^{\alpha }\varphi \right\Vert ^{2}-k\int\nolimits_{0}^{1}w_{x}(\varphi _{x}+\psi )dx+\rho _{2}\left\Vert D^{\alpha }\psi \right\Vert ^{2} \\ +\int\nolimits_{0}^{1}\psi \left[ \left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \psi _{xx}+\int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{xx}(t)-\psi _{xx}(s)\right) ds-k(\varphi _{x}+\psi )\right] dx \\ +\left( \rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1} \mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t). \end{array} \end{equation*}

    Therefore,

    \begin{equation*} \begin{array}{l} D^{\alpha }U_{8}(t)\leq \left( \rho _{1}\delta _{5}+\rho _{2}\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}}{4\delta _{5}} \left\Vert D^{\alpha }\varphi \right\Vert ^{2} \\ -\int\nolimits_{0}^{1}\psi _{x}\left[ \left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \psi _{x}+\int\nolimits_{0}^{t}\omega (t-s)\left( \psi _{x}(t)-\psi _{x}(s)\right) ds\right] dx \\ +\left( \rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1} \mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t) \end{array} \end{equation*}

    or

    \begin{equation*} \begin{array}{c} D^{\alpha }U_{8}(t)\leq \left( \rho _{1}\delta _{5}+\rho _{2}\right) \left\Vert D^{\alpha }\psi \right\Vert ^{2}+\frac{\rho _{1}}{4\delta _{5}} \left\Vert D^{\alpha }\varphi \right\Vert ^{2}+\frac{\bar{\omega}}{4\delta _{2}}(\omega \square \psi _{x})(t) \\ +\left[ \delta _{2}-\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \right] \left\Vert \psi _{x}\right\Vert ^{2}+\left( \rho _{1}+\rho _{2}\right) \mathcal{T}_{\psi _{x}}(t)+\rho _{1}\mathcal{T}_{D^{\alpha }\varphi }(t)+\rho _{2}\mathcal{T}_{D^{\alpha }\psi }(t). \end{array} \end{equation*}

    The proof is complete.

    We are now ready to state and prove our first theorem. Let

    \begin{equation*} U(t): = NE(t)+\sum\nolimits_{i = 1}^{8}M_{i}U_{i}(t), \;t\geq 0, \end{equation*}

    where N and M_{i} are positive constants to be determined inside the proof.

    In this subsection, we consider the case when the wave speeds of propagation are equal.

    Theorem 2. Assume that the initial data satisfy \mathcal{U} _{0}\in D(M) and \omega satisfies assumption (A). If \frac{ \rho _{1}}{k} = \frac{\rho _{2}}{b}, then the solution of (1.2) and (1.3) goes to rest in a Mittag-leffler manner provided that \tilde{\omega} is small enough, i.e., there exist two positive constants \mu and L (depending on E(0) ) such that

    \begin{equation*} E(t)\leq LE_{\alpha }(-\mu t^{\alpha }), \;t\geq 0. \end{equation*}

    Proof. With the help of all previous lemmas above, we compute D^{\alpha }U(t). The idea of the proof is to reach a fractional differential equation of the form

    \begin{equation} D^{\alpha }U(t)\leq -C_{3}U(t), \;t > t_{0} > 0. \end{equation} (6.1)

    Proposition 1 then allows one to conclude the estimation in the theorem but only on (t_{0}, \infty). Employing a continuity argument, we obtain a similar one on [0, t_{0}].

    Observe first that we can make the terms in E_{\alpha }(-C_{\omega }t^{\alpha }) as small as we wish by increasing the time t_{0} . Consequently, the parameter \delta _{1} may be ignored for the time being. Second, take M_{1} = \frac{2N}{b}, M_{6} = \frac{3M_{5}}{k}, \delta _{4} = \frac{k}{26} and M_{7} = \frac{M_{5}}{12} . For small \tilde{\omega} (and therefore small \delta _{2}), we are left with two sets of conditions:

    \begin{equation} \left\{ \begin{array}{l} \left( \frac{7\rho _{1}}{6}+\rho _{2}\right) M_{5} < kN, \\ \left[ \frac{7}{12}+\frac{1}{k}\left( b+\frac{1}{\delta _{3}}\right) \right] \rho _{1}M_{5}+\rho _{1}M_{8} < N, \\ \delta _{3}\rho _{2}M_{3}+\left[ \frac{5}{4}+\frac{12}{k}\left( 1+\frac{1}{ \delta _{3}}\right) \right] \rho _{2}M_{5}+\rho _{2}M_{8} < N, \\ \left( \frac{b\rho _{1}+3\rho _{2}b^{2}}{k}+\frac{14\rho _{1}+3\rho _{2}}{12} \right) M_{5}+M_{8}\left( \rho _{1}+\rho _{2}\right) < \frac{bN}{2}, \\ \delta _{3}M_{5}\left( \rho _{1}+3\rho _{2}\right) < \frac{2kN}{b}, \end{array} \right. \end{equation} (6.2)

    and

    \begin{equation*} \left\{ \begin{array}{l} \frac{M_{8}}{\delta _{5}} < \frac{M_{5}}{3}, \\ 3\left( \frac{1}{4}+\frac{2b}{k}\right) M_{5}+M_{8}\left( \frac{\rho _{1}}{ \rho _{2}}\delta _{5}+1\right) < M_{3}\omega _{0}, \\ \left[ \frac{b-\omega _{0}}{4}+\frac{7k}{12}+\frac{51b^{2}}{k}\right] M_{5} < bM_{8}, \\ \frac{\rho _{2}M_{3}}{2\delta _{3}} < M_{4}. \end{array} \right. \end{equation*}

    Take N large enough ( \delta _{3} may also be large) so that the first set of conditions of (6.2) holds. For a large value of \delta _{3}, there remains

    \begin{equation*} \left\{ \begin{array}{l} \frac{M_{8}}{\delta _{5}} < \frac{M_{5}}{3}, \\ 3\left( \frac{1}{4}+\frac{2b}{k}\right) M_{5}+M_{8}\left( \frac{\rho _{1}}{ \rho _{2}}\delta _{5}+1\right) < M_{3}\omega _{0}, \\ \left[ \frac{b-\omega _{0}}{4}+\frac{7k}{12}+\frac{51b^{2}}{k}\right] M_{5} < bM_{8}. \end{array} \right. \end{equation*}

    Next, choosing M_{3} large so that the second relation is verified (with \delta _{5} large), we are left only with

    \begin{equation*} \left[ \frac{b-\omega _{0}}{4}+\frac{7k}{12}+\frac{51b^{2}}{k}\right] M_{5} < bM_{8}. \end{equation*}

    Proceeding backward, we can select the ignored terms so as to verify all of the conditions. Thus, we get a relation of the form (6.1). The proof is complete.

    In practical problems, the speeds of propagation are not necessarily equal. We will need the following result for the basic fractional differential equation:

    \begin{equation*} D^{\alpha }x(t) = f(t, x(t)) \end{equation*}

    with 0 as the equilibrium. Let \mathcal{K} be the class of strictly increasing continuous functions h:[0, +\infty)\rightarrow \lbrack 0, +\infty) satisfying h(0) = 0.

    Proposition 9. [11] Assume that f is a nonnegative bounded function such that I^{\alpha }f, 0 < \alpha < 1 is also bounded. Then, \lim \inf_{t\rightarrow \infty }f(t) = 0.

    Proposition 10. [12] If there exist a Lyapunov function Z(t, x(t)) and two functions \vartheta _{1}(.) and \vartheta _{2}(.) in \mathcal{K} such that, for all x\neq 0, \vartheta _{1}(\left\Vert x(t)\right\Vert)\leq Z(t, x(t))\leq \vartheta _{2}(\left\Vert x(t)\right\Vert) and D^{\alpha }Z(x(t), t)\leq 0, 0 < \alpha \leq 1, then the equilibrium is Lyapunov uniformly stable.

    Theorem 3. Assume that \mathcal{U}_{0}\in D(M) and the speeds of propagation are not necessarily equal. Then, we have that \lim \inf_{t\rightarrow \infty }E(t) = 0. If \psi _{x}(0) = 0, then the system is Lyapunov uniformly stable for small values of \tilde{\omega} .

    We need to come up with a Lyapunov function whose fractional derivative is non-positive. We recall, from the previous section, that after putting back the non-zero term

    \begin{equation*} \left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx \end{equation*}

    in U_{5}(t), as follows

    \begin{equation*} D^{\alpha }U(t)\leq -C_{4}E(t)+M_{5}\left( \frac{b\rho _{1}}{k}-\rho _{2}\right) \int\nolimits_{0}^{1}D^{\alpha }\psi _{x}D^{\alpha }\varphi dx, \;t > t_{0} > 0, \end{equation*}

    we pass to the higher-order energy. Rewriting E(t) in the form E(t) = E(t, \varphi, \psi) to account for the dependence on \varphi and \psi, we define the ` 2\alpha -order' energy by

    \begin{equation*} E_{s}(t): = E(t, D^{\alpha }\varphi , D^{\alpha }\psi ), \;t\geq 0, \end{equation*}

    that is,

    \begin{equation*} \begin{array}{c} E_{s}(t): = \frac{1}{2}\left[ \rho _{1}\left\Vert D^{\alpha }\left( D^{\alpha }\varphi \right) \right\Vert ^{2}+\rho _{2}\left\Vert D^{\alpha }\left( D^{\alpha }\psi \right) \right\Vert ^{2}+\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \left\Vert D^{\alpha }\psi _{x}\right\Vert ^{2}\right. \\ \left. +k\left\Vert D^{\alpha }\left( \varphi _{x}+\psi \right) \right\Vert ^{2}+(\omega \square D^{\alpha }\psi _{x})(t)\right] , \;t\geq 0. \end{array} \end{equation*}

    We find that

    \begin{equation*} \begin{array}{l} D^{\alpha }E_{s}(t) = \rho _{1}D^{\alpha }\left( D^{\alpha }\varphi \right) D^{\alpha }\left[ D^{\alpha }\left( D^{\alpha }\varphi \right) \right] +\rho _{2}D^{\alpha }\left( D^{\alpha }\psi \right) D^{\alpha }\left[ D^{\alpha }\left( D^{\alpha }\psi \right) \right] -\frac{1}{2}I^{1-\alpha }\omega (t)\left\Vert D^{\alpha }\psi _{x}\right\Vert ^{2} \\ +\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) D^{\alpha }\psi _{x}D^{\alpha }\left( D^{\alpha }\psi _{x}\right) +kD^{\alpha }\left( \varphi _{x}+\psi \right) D^{\alpha }\left[ D^{\alpha }\left( \varphi _{x}+\psi \right) \right] \\ +\frac{1}{2}D^{\alpha }(\omega \square D^{\alpha }\psi _{x})(t)-\left( b-\int\nolimits_{0}^{t}\omega (s)ds\right) \mathcal{T}_{D^{\alpha }\left( \psi _{x}\right) }-\rho _{1}\mathcal{T}_{D^{\alpha }\left( D^{\alpha }\varphi \right) }-\rho _{2}\mathcal{T}_{D^{\alpha }\left( D^{\alpha }\psi \right) }-k\mathcal{T}_{D^{\alpha }\left( \varphi _{x}+\psi \right) } \\ +\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{(t-\eta )^{\alpha }} \right) \left( \int\nolimits_{0}^{\xi }\frac{\left[ \left( D^{\alpha }\psi _{x}\right) ^{2}(s)\right] ^{\prime }ds}{(t-s)^{\alpha }}\right) dx. \end{array} \end{equation*}

    From Proposition 7, we have the following for t\geq 0 :

    \begin{equation*} \begin{array}{l} \frac{1}{2}D^{\alpha }(\omega \square D^{\alpha }\psi _{x}) = \frac{1}{2} (^{RL}D^{\alpha }\omega \square D^{\alpha }\psi _{x})(t) \\ -\int\nolimits_{0}^{1}\, D^{\alpha }\left( D^{\alpha }\psi _{x}\right) \int\nolimits_{0}^{t}\omega (t-s)D^{\alpha }\psi _{x}(s)dsdx+\frac{1}{2} \left( \int\nolimits_{0}^{t}\omega (t-s)ds\right) D^{\alpha }\left\Vert D^{\alpha }\psi _{x}\right\Vert ^{2} \\ -\frac{\alpha }{2\Gamma (1-\alpha )}\int\nolimits_{0}^{t}\frac{d\xi }{ (t-\xi )^{1-\alpha }}\int\nolimits_{0}^{\xi }\frac{\omega (\eta )d\eta }{ (t-\eta )^{\alpha }}\int\nolimits_{0}^{\xi }\frac{\left[ \left\Vert D^{\alpha }\psi _{x}(s)\right\Vert ^{2}\right] ^{\prime }ds}{(t-s)^{\alpha }} \\ +\frac{\alpha }{\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\int\nolimits_{0}^{\xi } \frac{\left[ \left( D^{\alpha }\psi _{x}\right) (\eta )\right] ^{\prime }d\eta }{(t-\eta )^{\alpha }}\left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )D^{\alpha }\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) dx. \end{array} \end{equation*}

    Therefore, for \delta _{6} > 0, it follows that

    \begin{equation} \begin{array}{l} D^{\alpha }E_{s}(t)\leq \frac{1}{2}(^{RL}D^{\alpha }\omega \square D^{\alpha }\psi _{x})(t)-\frac{1}{2}I^{1-\alpha }\omega (t)\left\Vert D^{\alpha }\psi _{x}\right\Vert ^{2}+\left( \delta _{6}-\omega _{0}\right) \mathcal{T} _{D^{\alpha }\psi _{x}} \\ -\rho _{1}\mathcal{T}_{D^{\alpha }\left( D^{\alpha }\varphi \right) }-\rho _{2}\mathcal{T}_{D^{\alpha }\left( D^{\alpha }\psi \right) }-k\mathcal{T} _{D^{\alpha }\left( \varphi _{x}+\psi \right) } \\ +\frac{\alpha }{2\delta _{6}\Gamma (1-\alpha )}\int\nolimits_{0}^{1}\int \nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha }}\left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )D^{\alpha }\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) ^{2}dx. \end{array} \end{equation} (6.3)

    Proposition 11. Assume that k, f are two continuous functions on (0, +\infty) such that

    \begin{equation*} I^{1-\alpha }k(t), I^{1-\alpha }f(t)\in C^{1}([0, \infty )). \end{equation*}

    Then,

    \begin{equation*} \begin{array}{c} \left( \int\nolimits_{0}^{t}k(s)ds\right) D^{\alpha }f(t) = \int\nolimits_{0}^{t}k(t-s)\left[ D^{\alpha }f(t)-D^{\alpha }f(s) \right] ds-\int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }k\right. (t-s)\left[ f(t)-f(s)\right] ds \\ +I^{1-\alpha }k(t)f(t)-f(0)I^{1-\alpha }k(t)-k(t)I^{1-\alpha }f(0), \;t\geq 0. \end{array} \end{equation*}

    Proof. Clearly, for t\geq 0 ,

    \begin{equation*} \begin{array}{c} \int\nolimits_{0}^{t}k(t-s)\left[ D^{\alpha }f(t)-D^{\alpha }f(s)\right] ds-\int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }k\right. (t-s)\left[ f(t)-f(s)\right] ds \\ = \left( \int\nolimits_{0}^{t}k(s)ds\right) D^{\alpha }f(t)-\int\nolimits_{0}^{t}k(t-s)D^{\alpha }f(s)ds -\left[ I^{1-\alpha }k(t)-I^{1-\alpha }k(0)\right] f(t)+\int \nolimits_{0}^{t}\left. ^{RL}D^{\alpha }k\right. (t-s)f(s)ds. \end{array} \end{equation*}

    On one hand, by Proposition 4, we find

    \begin{equation*} \left. ^{RL}D^{\alpha }\right. \int\nolimits_{0}^{t}k(t-s)f(s)ds = \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. k(t-s)f(s)ds+f(t)I^{1-\alpha }k(0), \end{equation*}

    and on the other hand, the relationship between both derivatives gives

    \begin{equation*} \begin{array}{l} \left. ^{RL}D^{\alpha }\right. \int\nolimits_{0}^{t}k(s)f(t-s)ds = \int\nolimits_{0}^{t}k(s)\left. ^{RL}D^{\alpha }\right. f(t-s)ds+k(t)I^{1-\alpha }f(0) \\ = \int\nolimits_{0}^{t}k(t-s)\left[ D^{\alpha }f(s)+\frac{s^{-\alpha }f(0)}{ \Gamma (1-\alpha )}\right] ds+k(t)I^{1-\alpha }f(0) \\ = \int\nolimits_{0}^{t}k(t-s)D^{\alpha }f(s)ds+\frac{f(0)}{\Gamma (1-\alpha ) }\int\nolimits_{0}^{t}k(s)(t-s)^{-\alpha }ds+k(t)I^{1-\alpha }f(0) \\ = \int\nolimits_{0}^{t}k(t-s)D^{\alpha }f(s)ds+f(0)I^{1-\alpha }k(t)+k(t)I^{1-\alpha }f(0). \end{array} \end{equation*}

    Therefore,

    \begin{equation*} \begin{array}{c} \int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }\right. k(t-s)f(s)ds+f(t)I^{1-\alpha }k(0) = \int\nolimits_{0}^{t}k(t-s)D^{\alpha }f(s)ds+f(0)I^{1-\alpha }k(t)+k(t)I^{1-\alpha }f(0), \end{array} \end{equation*}

    and

    \begin{equation} \begin{array}{c} \int\nolimits_{0}^{t}k(t-s)\left[ D^{\alpha }f(t)-D^{\alpha }f(s)\right] ds-\int\nolimits_{0}^{t}\left. ^{RL}D^{\alpha }k\right. (t-s)\left[ f(t)-f(s)\right] ds \\ = \left( \int\nolimits_{0}^{t}k(s)ds\right) D^{\alpha }f-I^{1-\alpha }k(t)f(t)+f(0)I^{1-\alpha }k(t)+k(t)I^{1-\alpha }f(0), \;t\geq 0. \end{array} \end{equation} (6.4)

    The proof of the proposition is complete.

    Multiplying the relation (6.4) by g(t), then integrating over (0, 1) and applying Young's inequality, we obtain the following:

    Corollary 1. Assume that k, f are two continuous functions on (0, +\infty) such that

    \begin{equation*} I^{1-\alpha }k(t), f(t)\in C^{1}(0, \infty ), \end{equation*}

    and f, D^{\alpha }f, g\in L^{2}(0, 1). Then,

    \begin{equation*} \begin{array}{c} \left( \int\nolimits_{0}^{t}k(s)ds\right) \int\nolimits_{0}^{1}g(t)\left( D^{\alpha }f\right) dx\leq 2\varepsilon \left\Vert g(t)\right\Vert ^{2}+ \frac{1}{\varepsilon }\left( \int\nolimits_{0}^{t}k(s)ds\right) (k\square D^{\alpha }f)(t)+\frac{1}{\varepsilon }\left( \int\nolimits_{0}^{t}\left\vert ^{RL}D^{\alpha }k\right\vert ds\right) \\ \times (\left\vert ^{RL}D^{\alpha }k\right\vert \square f)(t)+\frac{1}{ \varepsilon }\left[ I^{1-\alpha }k(t)\right] ^{2}\left\Vert f\right\Vert ^{2}+\frac{1}{\varepsilon }\left[ I^{1-\alpha }k(t)\right] ^{2}\left\Vert f(0)\right\Vert ^{2}+\frac{1}{4\varepsilon }k^{2}(t)\left\Vert I^{1-\alpha }f(0)\right\Vert ^{2}, \;t\geq 0. \end{array} \end{equation*}

    In the case that g = D^{\alpha }f , this corollary gives the following:

    Corollary 2. Assume that k, f are two continuous functions on (0, +\infty) such that I^{1-\alpha }k(t), f(t)\in C^{1}(0, \infty). Then,

    \begin{equation*} \begin{array}{c} \left( \int\nolimits_{0}^{t}k(s)ds-2\varepsilon \right) \left\Vert D^{\alpha }f\right\Vert ^{2}\leq \frac{1}{\varepsilon }\left( \int\nolimits_{0}^{t}k(s)ds\right) (k\square D^{\alpha }f)(t)+\frac{1}{ \varepsilon }\left( \int\nolimits_{0}^{t}\left\vert ^{RL}D^{\alpha }k\right\vert ds\right) (\left\vert ^{RL}D^{\alpha }k\right\vert \square f)(t) \\ +\frac{1}{\varepsilon }\left[ I^{1-\alpha }k(t)\right] ^{2}\left\Vert f\right\Vert ^{2}+\frac{1}{\varepsilon }\left[ I^{1-\alpha }k(t)\right] ^{2}\left\Vert f(0)\right\Vert ^{2}+\frac{1}{4\varepsilon } k^{2}(t)\left\Vert I^{1-\alpha }f(0)\right\Vert ^{2}, \;t\geq 0. \end{array} \end{equation*}

    Proof of Theorem 3. (Sketch) As in the previous section, we suggest using the new functional

    \begin{equation*} U_{9}(t): = \int\nolimits_{0}^{1}\left( \int\nolimits_{0}^{t}\omega (t-s)D^{\alpha }\psi _{x}(s)ds\right) ^{2}dx, \;t\geq 0 \end{equation*}

    to deal with the problematic term

    \begin{equation*} \int\nolimits_{0}^{1}\int\nolimits_{0}^{t}\frac{d\xi }{(t-\xi )^{1-\alpha } }\left( \int\nolimits_{0}^{\xi }(t-s)^{-\alpha }\left( \int\nolimits_{0}^{s}\omega (s-\tau )D^{\alpha }\psi _{x}(\tau )d\tau \right) ^{\prime }ds\right) ^{2}dx, \end{equation*}

    and proceed similarly. Combining (6.3), D^{\alpha }U and the above corollaries, for

    \begin{equation*} V(t): = N\left( E(t)+E_{s}(t)\right) +\sum\nolimits_{i = 1}^{9}M_{i}U_{i}(t), \end{equation*}

    we arrive at

    \begin{equation*} D^{\alpha }V(t)\leq -C_{5}E(t)+C_{6}\left[ E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}\left\Vert \psi _{x}(0)\right\Vert ^{2}, \;t > t_{0} > 0 \end{equation*}

    for small \tilde{\omega} and large t_{0} for some C_{5} , C_{6} > 0. Applying I^{\alpha } to both sides gives

    \begin{equation*} \begin{array}{c} V(t)-V(0) = \frac{1}{\Gamma (\alpha )}\int\nolimits_{0}^{t_{0}}(t-s)^{\alpha -1}D^{\alpha }V(s)ds+\frac{1}{\Gamma (\alpha )}\int \nolimits_{t_{0}}^{t}(t-s)^{\alpha -1}D^{\alpha }V(s)ds \\ \leq \frac{1}{\Gamma (\alpha )}\int\nolimits_{0}^{t_{0}}(t-s)^{\alpha -1}D^{\alpha }V(s)ds+\frac{1}{\Gamma (\alpha )}\int \nolimits_{t_{0}}^{t}(t-s)^{\alpha -1}\left\{ -C_{5}E(t)+C_{6}\left[ E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}\left\Vert \psi _{x}(0)\right\Vert ^{2}\right\} ds. \end{array} \end{equation*}

    Notice that, by the continuity of D^{\alpha }V(s) (below, all constants C_{i}, i = 7, ..., 12 are positive),

    \begin{equation*} \int\nolimits_{0}^{t_{0}}(t-s)^{\alpha -1}D^{\alpha }V(s)ds\leq \int\nolimits_{0}^{t_{0}}(t_{0}-s)^{\alpha -1}\left\vert D^{\alpha }V(s)\right\vert ds\leq C_{7}t_{0}{}^{\alpha }. \end{equation*}

    Thus,

    \begin{equation*} C_{5}I_{t_{0}}^{\alpha }E(t)\leq V(0)+\frac{C_{7}t_{0}{}^{\alpha }}{\Gamma (\alpha )}+C_{6}I^{\alpha }\left[ E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}\left\Vert \psi _{x}(0)\right\Vert ^{2}, \end{equation*}

    and by Proposition 3, for t\geq t_{0} , we have

    \begin{equation*} I^{\alpha }\left[ E_{\alpha }(-C_{\omega }t^{\alpha })\right] ^{2}\leq I^{\alpha }\left[ C_{8}t^{-\alpha }E_{\alpha }(-C_{\omega }t^{\alpha }) \right] \leq C_{9}E_{\alpha }(-C_{\omega }t^{\alpha })\leq C_{10}. \end{equation*}

    Hence,

    \begin{equation*} C_{11}E(t)+I_{t_{0}}^{\alpha }E(t)\leq C_{12}. \end{equation*}

    The first conclusion follows from Proposition 9 and the second one from Proposition 10.

    In this section, we shall present some numerical examples to validate the results obtained for both the equal speed of propagation and non-equal speed of propagation cases.

    Example 1. Consider the system (1.2) and (1.3) with the following selected functions and parameters: \varphi _{0}(x) = e^{- \frac{x}{4}}\cos (\pi x), \psi _{0}(x) = 2e^{-\frac{x}{4}}\sin (2\pi x), \varphi _{1}(x) = \psi _{1}(x) = 0, \omega (t) = e^{-2t}, \beta = 1.96, k = b = 2, \rho _{1} = \rho _{2} = 2, x\in \lbrack 0, 10], t\in \lbrack 0, 2].

    The condition (A) and the equality \frac{k}{\rho _{1}} = \frac{b}{\rho _{2}} are satisfied. By means of Theorem 2, the solutions \varphi (x, t) and \psi (x, t) go toward zero in a Mittag-Leffler manner. This finding is illustrated in Figures 1 and 2, depicting the variation of \varphi (x, t) and \psi (x, t) as functions of spatial and temporal variables in the range [0, 10]\times \lbrack 0, 2]. Figures 3 and 4 show the dissipativity of the vibrations relative to the x -axis as the time variable rises.

    Figure 1.  The solution \varphi (x, t) of the problem (1.2) and (1.3).
    Figure 2.  The solution \psi (x, t) of the problem (1.2) and (1.3).
    Figure 3.  The solution \varphi (x, t) of the problem (1.2) and (1.3) for different values of t.
    Figure 4.  The solution \psi (x, t) of the problem (1.2) and (1.3) for different values of t.

    Example 2. Consider the system (1.2) and (1.3) involving the functions and parameters that are taken as \varphi _{0}(x) = 0.5\sin (\frac{7}{4}\pi x), \psi _{0}(x) = 2x\cos (\pi x), \varphi _{1}(x) = \psi _{1}(x) = 0, \omega (t) = e^{-4t}, \beta = 1.96, k = 2, b = 1.5, \rho _{1} = \frac{2}{3}, \rho _{2} = 1, x\in \lbrack 0, 10], t\in \lbrack 0, 2].

    The conditions (A) and \frac{k}{\rho _{1}}\neq \frac{b}{\rho _{2}} are met. Thus, Theorem 3 can be applied, which means that the solutions \varphi (x, t) and \psi (x, t) go toward zero as a Mittag-Leffler function. This may be visualized in Figures 5 and 6, representing the behavior of solutions \varphi (x, t) and \psi (x, t) in [0, 10]\times \lbrack 0, 2]. The oscillations of \varphi (x, t) and \psi (x, t) with regard to the spatial variable, decrease as the temporal variable increases, as illustrated in Figures 7 and 8.

    Figure 5.  The solution \varphi (x, t) of the problem (1.2) and (1.3).
    Figure 6.  The solution \psi (x, t) of the problem (1.2) and (1.3).
    Figure 7.  The solution \varphi (x, t) of the problem (1.2) and (1.3) for different values of t.
    Figure 8.  The solution \psi (x, t) of the problem (1.2) and (1.3) for different values of t.

    In this work, we have examined the dynamics of a Timoshenko beam operating in an anomalous media. It is derived from the classical Timoshenko model by replacing the integer-order derivatives with fractional ones between one and two. Actually, sequential fractional derivatives are better suited for the multiplier technique. It is shown that a viscoelastic damping process acting on the rotational component is capable of driving the structure to equilibrium with a Mittag-Leffler rate in the case of equal speed of propagation and a relaxation function satisfying a fractional inequality whose solutions are bounded above by Mittag-leffler functions. This generalizes a similar case of exponential stability which happens in the classical (integer-order) case. When the speeds of propagation are not equal, the situation is much more complicated and differs from the classical situation. We have obtained only a uniform Lyapunov stability without a specific rate.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are very grateful for the financial support and the facilities provided by King Fahd University of Petroleum and Minerals (Interdisciplinary Research Center for Intelligent Manufacturing & Robotics) through project number INMR2300. They are also very grateful to the anonymous referees, whose comments and suggestions helped to improve the original version of this work.

    The authors declare no conflict of interest.



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    沈阳化工大学材料科学与工程学院 沈阳 110142

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