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On an initial boundary value problem for fractional pseudo-parabolic equation with conformable derivative

  • In this paper, we study the initial boundary value problem of the pseudo-parabolic equation with a conformable derivative. We focus on investigating the existence of the global solution and examining the derivative's regularity. In addition, we contributed two interesting results. Firstly, we proved the convergence of the mild solution of the pseudo-parabolic equation to the solution of the parabolic equation. Secondly, we examine the convergence of solution when the order of the derivative of the fractional operator approaches 1. Our main techniques used in this paper are Banach fixed point theorem and Sobolev embedding. We also apply different techniques to evaluate the convergence of generalized integrals encountered.

    Citation: Huy Tuan Nguyen, Nguyen Van Tien, Chao Yang. On an initial boundary value problem for fractional pseudo-parabolic equation with conformable derivative[J]. Mathematical Biosciences and Engineering, 2022, 19(11): 11232-11259. doi: 10.3934/mbe.2022524

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  • In this paper, we study the initial boundary value problem of the pseudo-parabolic equation with a conformable derivative. We focus on investigating the existence of the global solution and examining the derivative's regularity. In addition, we contributed two interesting results. Firstly, we proved the convergence of the mild solution of the pseudo-parabolic equation to the solution of the parabolic equation. Secondly, we examine the convergence of solution when the order of the derivative of the fractional operator approaches 1. Our main techniques used in this paper are Banach fixed point theorem and Sobolev embedding. We also apply different techniques to evaluate the convergence of generalized integrals encountered.



    Fractional calculus is one of today's most popular mathematical tools to model real-world problems. More specifically, it has been applied to model evolutionary systems involving memory effects on dynamical systems. Partial differential equations (PDEs) with fractional operators are important in describing phenomena in many fields such as physics, biology and chemistry [1,2]. Based on the generalizations of fractional derivatives by famous mathematicians such as Euler, Lagrange, Laplace, Fourier, Abel and Liouville, today's mathematicians have explored and introduced many more types of fractional derivatives such as Riemann-Liouville, Caputo, Liouville, Weyl, Riesz and Hifler [3,4,5,6].

    PDEs with conformable derivatives attract interested mathematicians using different approaches because of their wide range of applications, such as electrical circuits[7] and chaotic systems in dynamics[8]. We recognize that the conformable and classical derivatives have a close relationship. There is an interesting observation that: If f is a real function and s>0, then f has a conformable fractional derivative of order β at s if and only if it is (classically) differentiable at s, and

    Cβf(s)sβ=s1βf(s)s, (1.1)

    where 0<β1. Another surprising observation is that Eq (1.1) will not hold if f is defined in a general Banach space. We can better understand why the ODEs with the conformable derivative on R have been studied so much. In addition, the relevant research in infinite-dimensional spaces, such as Banach or Hilbert space, is still limited, which motivates us to investigate some types of PDEs with conformable derivatives in Hilbert or Sobolev spaces.

    Besides, in some phenomena, the conformable derivative is better simulated than the classical derivative. In [9], the authors considered the conformable diffusion equation

     Cαtαu(x,t)=Dα2x2u(x,t), (1.2)

    where 0<α1,x>0,t>0, u(x,t) is the concentration, and Dα represents the generalized diffusion coefficient, which was applied in the description of a subdiffusion process. In particular, the conformable diffusion equation (1.2) reduces to the normal diffusion equation if α=1. A natural and fundamental question is, "Does the conformable diffusion model predict better than the normal diffusion model?". The results in [9] show that the conformable derivative model agrees better with the experimental data than the normal diffusion equation.

    Let ΩRN (N1) be a bounded domain with smooth boundary Ω, and T>0 is a given positive number. In this paper, we investigate the Sobolev equation with a conformable derivative as follows

    {Cαtαu+(Δ)βu(x,t)mCαtαΔu=F(u(x,t)),xΩ,t(0,T),u(x,t)=0,xΩ,t(0,T),u(x,0)=u0(x)xΩ (1.3)

    where α,β(0,1], m>0, the time fractional derivative Cαtα is the conformable derivative of order α, defined in Definition 1. The function F represents the external forces or the advection term of a diffusion phenomenon, etc., and the function u0 is the initial condition specified later. The operator (Δ)β is the fractional Laplacian operator, which is well-defined in [10] (see page 3). In the sense of distribution, the study of weak solutions to Problem (1.3) is still limited compared to the classical problem. Thus, the fundamental knowledge in the distributive sense for Problem (1.3) is still open and challenging, which is the main reason and motivation for us to study the problem from the perspective of the semigroup.

    Next, we mention some results related to Problem (1.3). There are two interesting observations regarding Problem (1.3).

    If we take m=0 in Problem (1.3), then we obtain an initial boundary value problem of the parabolic equation with a conformable operator as follows

    {Cαtαu+(Δ)βu(x,t)=F(u(x,t)),xΩ,t(0,T),u(x,t)=0,xΩ,t(0,T),u(x,0)=u0(x)xΩ. (1.4)

    The latest results on the well-posedness of solutions to Problem (1.4) are shown in more detail in [11], and the authors used the Hilbert scales space technique to prove the local existence of the mild solution to Problem (1.4).

    If α=1, the main equation of Problem (1.3) becomes the classical equation

    ut+(Δ)βu(x,t)mΔut=F(u(x,t)). (1.5)

    Equation (1.5) is familiar to mathematicians about PDEs, called the pseudo-parabolic equation, also known as the Sobolev equation. The pseudo-parabolic equation describes a series of important physical processes, such as the permeation of a homogeneous liquid through fractured rock, population aggregation and one-way propagation of the nonlinear dispersion length wave [12,13]. Equation (1.5) has been studied extensively; for details, see [12,13,14] and references given there. Concerning the study of the existence and blowup of solutions to pseudo-parabolic equations, we refer the reader to [15,16,17].

    For the convenience of readers, we next list some interesting results related to pseudo-parabolic with fractional derivative. Luc et al. in [18] considered fractional pseudo-parabolic equation with Caputo derivative

    {Dαt(u+kAu)+Aβu=F(t,x,u),in(0,T]×Ω,u(t,x)=0,on(0,T]×Ω,u(0,x)=u0(x),inΩ, (1.6)

    where 0<α<1, A=Δ, Dα is Caputo fractional derivative operator of order α. They studied the local and global existence of solutions to Problem (1.6) when the nonlinear term F is the global Lipschitz. Based on the work in [18], there are many related results in the spirit of a semigroup representation of the form of the Fourier series. In [12], the authors studied nonlinear time-fractional pseudo-parabolic equations with Caputo derivative on both bounded and unbounded domains by different methods and techniques from [18]. In [19], the authors studied the nonlocal in time problem for a pseudo-parabolic equation with fractional time and space in the linear case. Tuan et al. [20] derived the nonlinear pseudo-parabolic equation with a nonlocal type of integral condition. In [21], the authors considered the time-space pseudo-parabolic equation with the Riemann-Liouville time-fractional derivative, and they applied the Galerkin method to show the global and local existence of solutions.

    As far as we know, there has not been any work that considers the initial boundary value problem (1.3) with a conformable derivative. The main results and methods of the present paper are described in detail as follows

    ● Firstly, we prove the existence of the global solution to Problem (1.3). The main idea is to use Banach fixed point theorem with the new weighted norm used in [22]. In order to prove the regularity and the derivative of the mild solution, we need to apply some complicated techniques on Hilbert scales for nonlinearity terms.Compared with [11], our method has very different characteristics. It is important to emphasize that proving the existence of the global solution is difficult, which is demonstrated in our current paper, but not in the paper [11].

    ● Secondly, we investigate the convergence of solution to Problem (1.3) when m0+, which does not appear in the works related to fractional pseudo-parabolic equation. This result allows us to get the relationship between the solution of Sobolev equation and parabolic diffusion equation. To overcome the difficulty, we need to control the improper integrals and control the parameters. This pioneering work can open up some new research directions for finding the relationship between the solutions of the pseudo-parabolic equation and the parabolic equation.

    ● Finally, we prove the convergence of the solution when the order of derivative β1. This direction of research was motivated by the recent paper [23]. Since the current model has a nonlinear source function, the processing technique for the proof in this paper seems to be more complicated than that of [23].

    The greatest difficulty in solving this problem is the study of many integrals containing singular terms, such as sα1 or (tαsα)m. To overcome these difficulties, we need to use ingenious calculations and techniques to control the convergence of several generalized integrals.

    This paper is organized as follows. Section 2 provides some definitions. In Section 3, we give the definition of the mild solution and some important lemmas for the proof of the main results. Section 4 shows the global existence of the solution to Problem (1.3). In addition, we present the regularity result for the derivative of the mild solution. Section 5 shows the convergence of the solution to Problem (1.3) when m0+. In Section 6, we investigate the convergence of mild solutions when β1.

    Definition 2.1. Conformable derivative model: Let B be a Banach space, and the function f:[0,)B. Let Cβtβ be the conformable derivative operator of order β(0,1], locally defined by

    Cβf(t)tβ:=limh0f(t+ht1β)f(t)hinB

    for each t>0. (For more details on the above definition, we refer the reader to [24,25,26,27].)

    In this section, we introduce the notation and the functional setting used in our paper. Recall the spectral problem

    {Δen(x)=λnen(x),xΩ,en(x)=0,xΩ,

    admits the eigenvalues 0<λ1λ2λn with λn as n. The corresponding eigenfunctions are enH10(Ω).

    Definition 2.2. (Hilbert scale space). We recall the Hilbert scale space, which is given as follows

    Hs(Ω)={fL2(Ω)|n=1λ2sn(Ωf(x)en(x)dx)2<}

    for any s0. It is well-known that Hs(Ω) is a Hilbert space corresponding to the norm

    fHs(Ω)=(n=1λ2sn(Ωf(x)en(x)dx)2)1/2,fHs(Ω).

    Definition 2.3. Let Xa,q,α((0,T];B) denote the weighted space of all the functions ψC((0,T];X) such that

    ψXa,q,α((0,T];B):=supt(0,T]taeqtαψ(t,)B<,

    where a,q>0 and 0<α1 (see [22]). If q=0, we denote Xa,q((0,T];B) by Xa((0,T];B).

    In order to find a precise formulation for solutions, we consider the mild solution in terms of the Fourier series

    u(x,t)=n=1u(.,t),enen(x).

    Taking the inner product of Problem (1.3) with en gives

    {Cαtαu(.,t),en+λβnu(.,t),en+mλnCαtαu(.,t),en=F(.,t),en,t(0,T),u(.,0),en=u0,en, (3.1)

    where we repeat that

    Δu(.,t),en=λnu(.,t),en.

    The first equation of (3.1) is a differential equation with a conformable derivative as follows

    Cαtαu(.,t),en+λβn1+mλnu(.,t),en=11+mλnF(.,t),en.

    In view of the result in Theorem 5, [26] and Theorem 3.3, [28] the solution to Problem (1.3) is

    u(.,t),en=exp(λβn1+mλntαα)u0,en+11+mλnt0να1exp(λβn1+mλnναtαα)F(.,ν),endν.

    To simplify the solution formula, we will express the solution in operator equations. Let us set the following operators

    Sm,α,β(t)f=nNexp(λβn1+mλntαα)f,enen,

    and

    Pmf=nN(1+mλn)1f,enen

    for any fL2(Ω) in the form f=nNf,enen. Then the inverse operator of Sm,α,β(t) is defined by

    (Sm,α,β(t))1f=nNexp(λβn1+mλntαα)f,enen

    The mild solution is given by

    u(t)=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1F(ν)dν. (3.2)

    For a qualitative analysis of the solution to (3.2), we need the bounded result for the operators in Hilbert scales space.

    Lemma 3.1. (a) Let vHs+kβk(Ω) for any k>0. Then we get

    Sm,α,β(t)vHs(Ω)CkαkmktαkvHs+kβk(Ω), (3.3)

    and for 0νtT,

    Sm,α,β(t)(Sm,α,β(ν))1vHs(Ω)Ckαkmk(tανα)kvHs+kβk(Ω). (3.4)

    (b) If vHs(Ω), then

    Sm,α,β(t)vHs(Ω)vHs(Ω) (3.5)

    and

    Sm,α,β(t)(Sm,α,β(ν))1vHs(Ω)vHs(Ω) (3.6)

    for 0νtT.

    (c) If vHs(Ω), then we have

    PmvHs(Ω)vHs(Ω)

    for any s.

    Proof. For (a), in view of Parseval's equality, we get that

    Sm,α,β(t)v2Hs(Ω)=nNλ2snexp(2λβn1+mλntαα)v,en2. (3.7)

    Using the inequality eyCkyk, we find that

    exp(λβn1+mλntαα)Ck(λβn1+mλn)ktαkαk=Ckαk(mk+λ11)λ(1β)kntαk, (3.8)

    where we use

    (1+mλn)kCk(1+mkλkn)Ck(mk+λ11)λkn.

    It follows from (3.7) that

    Sm,α,β(t)v2Hs(Ω)(Ckαk)2(mk+λ11)2t2αknNλ2s+2k2βknf,en2=(Ckαk)2(mk+λ11)2t2αkv2Hs+kβk(Ω).

    By a similar explanation, we also get that for 0νtT,

    Sm,α,β(t)(Sm,α,β(ν))1v2Hs(Ω)=nNλ2snexp(λβn1+mλn2να2tαα)v,en2(Ckαk)2(mk+λ11)2(tανα)2knNλ2s+2k2βknf,en2=(Ckαk)2(mk+λ11)2(tανα)2kv2Hs+kβk(Ω),

    where we use the fact that

    exp(λβn1+mλnναtαα)Ck(λβn1+mλn)k(tανα)kαk=Ckαk(mk+λ11)λ(1β)kn(tανα)k.

    Hence, (a) is proved.

    For (b), in view of Parseval's equality, we get that

    Sα,β(t)v2Hs(Ω)=nNλ2snexp(2λβn1+mλntαα)v,en2nNλ2snv,en2=v2Hs(Ω),

    which allows us to conclude the proof of (3.5). The proof of (3.6) is similar to (3.5), and we omit it here. For (c), noting that (1+mλn)1<1, we can claim it as follows

    Pmv2Hs(Ω)=nNλ2sn(1+mλn)2v,en2nNλ2snv,en2=v2Hs(Ω).

    The proof of Lemma 3.1 is completed.

    Let G:Hr(Ω)Hs(Ω) such that G(0)=0 and

    G(w1)G(w2)Hs(Ω)Lgw1w2Hr(Ω), (4.1)

    for any w1,w2Hr(Ω) and Lg is a postive constant.

    Theorem 4.1. (i) Let G:Hr(Ω)Hs(Ω) such that (4.1) holds. Here r,s satisfy that sr+kβk for 0<k<12. Let the initial datum u0Hr+kβk(Ω). Then Problem (1.3) has a unique solution um,α,βLp(0,T;Hr(Ω)), where

    1<p<1b,αkb<ααk.

    In addition, we get

    um,α,β(t)Hr(Ω)2Ckαk(mk+λ11)eμ0TαTbαktbu0Hr+kβk(Ω), (4.2)

    where Ck depends on k.

    (ii) Let us assume that b<α2 and u0Hs+kβk+β1(Ω)Hr+kβk(Ω). Then we have

    tum,α,β(.,t)Hs(Ω)tααk1u0Hs+kβk+β1(Ω)+(tα1b+t2αbkα1)u0Hr+kβk(Ω). (4.3)

    Here the hidden constant depends on k,b,α,m,p,β,Lg (Lg is defined in (4.1)).

    Proof. Let us define B:Xb,μ,α((0,T];Hr(Ω))Xb,μ,α((0,T];Hr(Ω)), μ>0 by

    Bw(t):=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(w(ν))dν. (4.4)

    Let the zero function w0(t)=0. From the fact G(0)=0, we know that

    Bw0(t)=Sm,α,β(t)u0.

    In view of (3.4) as in Lemma 3.3, we obtain the following estimate

    Bw0(t)Hr(Ω))=Sm,α,β(t)u0Hr(Ω))Ckαk(mk+λ11)tαku0Hr+kβk(Ω).

    Hence, multiplying both sides of the above expression by tbeμtα, we have that

    tbeμtαBw0(t)Hr(Ω))Ckαk(mk+λ11)tbαku0Hr+kβk(Ω)Ckαk(mk+λ11)Tbαku0Hr+kβk(Ω), (4.5)

    where we use bαk. This implies that Bw0Xb,μ,α((0,T];Hr(Ω)). Let any two functions w1,w2Xb,μ,α((0,T];Hr(Ω)). From (4.4) and (3.3), we obtain that

    Bw1(t)Bw2(t)Hr(Ω))=t0να1PmSm,α,β(t)(Sm,α,β(ν))1(G(w1(ν))G(w2(ν)))dνHr(Ω))Ckαkmkt0να1(tανα)kG(w1(ν))G(w2(ν))Hr+kβk(Ω))dν. (4.6)

    Since the constraint sr+kβk, we know that Sobolev embedding

    Hs(Ω)Hr+kβk(Ω).

    From some above observations and noting (4.1), we get that

    tbeμtαBw1(t)Bw2(t)Hr(Ω))Ckαk(mk+λ11)tbeμtαt0να1(tανα)kG(w1(ν))G(w2(ν))Hs(Ω))dνCkLgαk(mk+λ11)tbeμtαt0να1(tανα)kw1(ν)w2(ν)Hr(Ω))dν=CkLgαk(mk+λ11)tbt0να1b(tανα)keμ(tανα)νbeμναw1(ν)w2(ν)Hr(Ω))dν. (4.7)

    From the fact that

    w1w2Xb,μ,α((0,T];Hr(Ω))=sup0νTνbeμναw1(ν)w2(ν)Hr(Ω)),

    we follows from (4.7) that

    sup0tTtbeμtαBw1(t)Bw2(t)Hr(Ω))¯Cw1w2Xb,μ,α((0,T];Hr(Ω))sup0tT[tbt0να1b(tανα)keμ(tανα)dν], (4.8)

    where ¯C=CkLgαkmk. Let us continue to treat the integral term as follows

    J1,μ(t)=tbt0να1b(tανα)keμ(tανα)dν.

    In order to control the above integral, we need to change the variable ν=tξ1α. Then we get the following statement

    J1,μ(t)=1αtααk10ξbα(1ξ)keμtα(1ξ)dξ.

    Next, we provide the following lemma which can be found in [22], Lemma 8, page 9.

    Lemma 4.1. Let a1>1, a2>1 such that a1+a21, ρ>0 and t[0,T]. For h>0, the following limit holds

    limρ(supt[0,T]th10νa1(1ν)a2eρt(1ν)dν)=0.

    Since 0<b<α and 0<k<min(bα,1bα), we easily to verify that the following conditions hold

    {ααk>0,bα>1, k>1,bαk1. (4.9)

    By Lemma 4.1 and (4.9), we have

    limμ+sup0tTJ1,μ(t)=0.

    This statement shows that there exists a μ0 such that

    ¯Csup0tTJ1,μ0(t)12. (4.10)

    Combining (4.8) and (4.10), we obtain

    Bw1Bw2Xb,μ0((0,T];Hr(Ω))12w1w2Xb,μ0((0,T];Hr(Ω)) (4.11)

    for any w1,w2Xb,μ,α((0,T];Hr(Ω)). This statement tells us that B is the mapping from Xb,μ,α((0,T];Hr(Ω)) to itself. By applying Banach fixed point theorem, we deduce that B has a fixed point um,α,βXb,μ0,α((0,T];Hr(Ω)). Hence, we can see that

    um,α,β(t)=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(um,α,β(ν))dν. (4.12)

    Let us show the regularity property of the mild solution um,α,β. Indeed, using the triangle inequality and (4.11) and noting that B(v=0)=Sm,α,β(t)u0, we obtain

    um,α,βXb,μ0,α((0,T];Hr(Ω))=Bum,α,βXb,μ0,α((0,T];Hr(Ω))12um,α,βXb,μ0((0,T];Hr(Ω))+B(v=0)Xb,μ0,α((0,T];Hr(Ω))=12um,α,βXb,μ0,α((0,T];Hr(Ω))+Sm,α,β(t)u0Xb,μ0,α((0,T];Hr(Ω)),

    which combined with (4.5), we get

    um,α,βXb,μ0,α((0,T];Hr(Ω))2Ckαk(mk+λ11)Tbαku0Hr+kβk(Ω),

    which allows us to get that

    um,α,β(t)Hr(Ω)˜C1tbu0Hr+kβk(Ω), (4.13)

    where ˜C1 depends on k,μ0,b,α,m and

    ˜C1=2Ckαk(mk+λ11)eμ0TαTbαk,

    and we remind that Ck depends on k. Note that the improper integral T0tpbdt is convergent for 1<p<1b, we deduce that

    um,α,βLp(0,T;Hr(Ω)),

    and the following regularity holds

    um,α,βLp(0,T;Hr(Ω))˜Cu0Hr+kβk(Ω),

    where ˜C depends on k,μ0,b,α,m,p. Our next aim is to claim the derivative of the mild solution um,α,β. Applying the following formula

    d(t0K(t,s)ds)=t0tK(t,s)ds+K(t,t)dt,

    we obtain the following equality

    tum,α,β(.,t)=tα1Qm,α,β(t)u0+tα1G(um,α,β(x,t))+tα1t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dν, (4.14)

    where the operator Qm,α,β(t) is defined by

    Qm,α,β(t)v=nNλβn1+mλnexp(λβn1+mλntαα)v,enen

    for any vL2(Ω). In view of Parseval's equality, we get that

    Qm,α,β(t)v2Hs(Ω)=nNλ2sn(λβn1+mλn)2exp(2λβn1+mλntαα)v,en2.

    Using (3.8) and noting that λβn1+mλnm1λβ1n, we get that

    Qm,α,β(t)v2Hs(Ω)(Ckαkmk1)2t2αknNλ2s+2k2βk+2β2nv,en2,

    which implies that

    Qm,α,β(t)vHs(Ω)Ckαkmk1tαkvHs+kβk+β1(Ω) (4.15)

    for any vHs+kβk+β1(Ω). In a similar technique as above, we also get that

    Qm,α,β(t)(Sm,α,β)1(ν)vHs(Ω)Ckαkmk1(tανα)kvHs+kβk+β1(Ω), (4.16)

    for any 0νt. Let us go back to the right hand side of (4.14). By (4.15), we evaluate the first term on the right hand side of (4.14) as follows

    tα1Qm,α,β(t)u0Hs(Ω)Ckαkmk1tααk1u0Hs+kβk+β1(Ω). (4.17)

    Using global Lipschitz of G as in (4.1) and the fact that G(0)=0, the second term on the right hand side of (4.14) is estimated as follows

    tα1G(um,α,β(x,t))Hs(Ω)˜C1tα1Lgum,α,βHr(Ω)˜C1Lgtα1bu0Hr+kβk(Ω), (4.18)

    where we use (4.13). Let us now to treat the third integral term in (4.14). By (4.16), we obtain that

    t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dν. (4.19)

    Since β1 and 0<k<1, we can easily verify that s+kβk+β1s, which implies the Sobolev embedding Hs(Ω)Hs+kβk+β1(Ω) is true. From these above observations and using (4.13), we derive that

    t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dνC(s,k,β)t0να1(tανα)kG(um,α,β(ν))Hs(Ω)dνLgC(s,k,β)t0να1(tανα)kum,α,β(ν)Hr(Ω)dνLgC(s,k,β)˜C1u0Hr+kβk(Ω)(t0να1b(tανα)kdν). (4.20)

    Let us now treat the integral term on the right hand side of (4.20). Controlling it is really not that simple task. By applying Hölder inequality, we find that

    (t0να1b(tανα)kdν)2=(t0να12bνα12(tανα)kdν)2(t0να12bdν)(t0να1(tανα)2kdν)=tα2bα2b(t0να1(tανα)2kdν), (4.21)

    where α>2b. By changing to a new variable z=να, we derive that dz=ανα1dν. Hence, we infer that

    t0να1(tανα)2kdν=1αtα0(tαz)2kdz=tα(12k)α(12k), (4.22)

    where we note that 0<k<12. Combining (4.21) and (4.22), we obtain that the following inequality

    t0να1b(tανα)kdνtαbkαα(12k)(α2b). (4.23)

    By (4.23) and following from (4.19) and (4.20), we have

    t0να1Qm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dνCkαkmk1LgC(s,k,β)˜C1α(12k)(α2b)tαbkαu0Hr+kβk(Ω). (4.24)

    Summarizing the above results (4.14), (4.17), (4.18), (4.24) and using the triangle inequality, we obtain the following assertion

    tum,α,β(.,t)Hs(Ω)tα1Qm,α,β(t)u0Hs(Ω)+tα1G(um,α,β(x,t))Hs(Ω)+tα1t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1tααk1u0Hs+kβk+β1(Ω)+˜C1Lgtα1bu0Hr+kβk(Ω)+Ckαkmk1LgC(s,k,β)˜C1α(12k)(α2b)t2αbkα1u0Hr+kβk(Ω)

    which shows (4.3). The proof is completed.

    The main purpose of this section is to investigate the convergence of mild solutions to Problem (1.3) when m0+. Our result gives us an interesting connection between the solution of the Sobolev equation and the parabolic equation.

    Theorem 5.1. Let G:Hr(Ω)Hs(Ω) such that G(0)=0 and (4.1) holds. Here r,s satisfy that sr+kβk for 0<k<12. Let the initial datum u0Hr+kβk(Ω). Let um,α,β and uα,β be the mild solutions to Problem (1.3) with m>0 and m=0 respectively. Then we get the following estimate

    um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)E1,α(Lgt),

    where αkb<α2, γ=k(1β)+l(βε2)β+1ε2, 0<l<k(1β) and 1<ε<min(2,2k(1β)+2lβl).

    Remark 5.1. Note that m2γεγ+εl2+mk tends to zero when m0+. Hence, we can deduce that um,α,β(t)uα,β(t)Hr(Ω)0 when m0+.

    Proof. In the case m=0, thanks for the results on [11], the mild solution to Problem (1.3) is given by the following operator equation

    uα,β(t)=Sα,β(t)u0+t0να1Sα,β(t)(Sα,β(ν))1G(uα,β(ν))dν, (5.1)

    where we provide two operators that have the following Fourier series representation as follows

    Sα,β(t)f=nNexp(λβntαα)f,enen

    and

    (Sα,β(ν))1f=nNexp(λβnναα)f,enen.

    It's worth emphasizing that the existence of the solution to Equation (5.1) has been demonstrated in [11]. Subtracting (5.1) from (4.12), we get the following equality by some simple calculations

    um,α,β(t)uα,β(t)=(Sm,α,β(t)Sα,β(t))u0+t0να1Sα,β(t)(Sα,β(ν))1(G(um,α,β(ν))G(uα,β(ν)))dν+t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]G(um,α,β(ν))dν+t0να1(PmI)Sα,β(t)(Sα,β(ν))1G(um,α,β(ν))dν=M1+M2+M3+M4. (5.2)

    Next, we estimate the four terms on the right hand of (5.2) in Hr(Ω) space. We divide this process into four steps as below.

    Step 1. Estimate of M1. In view of the inequality |eced|Cγmax(ec,ed)|cd|γ with γ>0, let c=λβn1+mλntαα and d=λβntαα, we get the following inequality

    |exp(λβn1+mλntαα)exp(λβntαα)|Cγexp(λβn1+mλntαα)(λβnλβn1+mλn)γtαγαγCγCl(λβn1+mλn)l(tαα)l(λβnλβn1+mλn)γtαγαγCγClαlγλβln(mλ1+βn)γ(11+mλn)γltαγαl (5.3)

    by ec>ed and (3.8).

    Here γ and l are two positive constants that are later chosen. For 1<ε<2 and any z>0, we easily verify that (1+z)2ε1+z>z, which implies

    (1+z)γl=(1+z)2εε(γl)2>zε(γl)2 (5.4)

    for any γ>l>0.

    In (5.4), by choosing z=1+mλn and after some simple calculation, we have (1+mλn)γl>(mλn)ε(γl)2, which leads to the following inequality

    (11+mλn)γl(mλn)ε(lγ)2. (5.5)

    Combining (5.3) and (5.5), we find that

    |exp(λβn1+mλntαα)exp(λβntαα)|C(α,γ,l)m2γεγ+εl2tαγαlλβl+βγ+γ+ε(lγ)2n. (5.6)

    Therefore, we have the following estimate

    (Sm,α,β(t)Sα,β(t))u02Hr(Ω))=nNλ2rn|exp(λβn1+mλntαα)exp(λβntαα)|2u0,en2|C(α,γ,l)|2m2γεγ+εlt2αγ2αlnNλ2r2βl+2βγ+2γ+ε(lγ)nu0,en2,

    which implies that

    (Sm,α,β(t)Sα,β(t))u0Hr(Ω))C(α,γ,l)m2γεγ+εl2tαγαlu0Hrβl+βγ+γ+ε(lγ)2(Ω). (5.7)

    Next, we explain how to choose the parameters l,ε,γ. Since β<1, we can choose l such that

    0<l<min(k(1β),1ββk)=k(1β),

    which implies that 2k(1β)+2lβ>l, that is 2k(1β)+2lβl>1. Then we can choose ε such that

    1<ε<min(2,2k(1β)+2lβl).

    Let us choose γ such that

    γ=k(1β)+l(βε2)β+1ε2.

    It is easy to verify that γ>l+l(βε2)β+1ε2=l and the following equality

    rβl+βγ+γ+ε(lγ)2=r+kβk.

    Then the estimate (5.7) becomes

    M1Hr(Ω)=(Sm,α,β(t)Sα,β(t))u0Hr(Ω))C(α,γ,l)m2γεγ+εl2tαγαlu0Hr+kβk(Ω). (5.8)

    Step 2. Estimate of M2. Let ψHr(Ω). For 0νtT, we can also get

    Sα,β(t)(Sα,β(ν))1ψ2Hr(Ω)=nNλ2rnexp(2λβnναtαα)ψ,en2ψ2Hr(Ω), (5.9)

    where we use that

    exp(λβn1+mλnναtαα)1.

    By (5.9) and Sobolev embedding Hs(Ω))Hr(Ω), we find that

    t0να1Sα,β(t)(Sα,β(ν))1(G(um,α,β(ν))G(uα,β(ν)))dνHr(Ω)t0να1G(um,α,β(ν))G(uα,β(ν))Hr(Ω)dνC(r,s)t0να1G(um,α,β(ν))G(uα,β(ν))Hs(Ω)dν. (5.10)

    By the global Lipschitz property of G as in (4.1) and noting (5.10), it follows that

    M2Hr(Ω)LgC(r,s)t0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. (5.11)

    Step 3. Estimate of M3. Let fHr+kβk(Ω). Then using Parseval' s equality, we have the following identity

    [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]f2Hr(Ω))=nNλ2rn(1+mλn)2|exp(λβn1+mλntαναα)exp(λβntαναα)|2f,en2. (5.12)

    By a similar explanation as in (5.6), we find that

    |exp(λβn1+mλntαναα)exp(λβntαναα)|C(α,γ,l)m2γεγ+εl2(tανα)γlλβl+βγ+γ+ε(lγ)2n. (5.13)

    By (5.12) and (5.13), we get that

    [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]fHr(Ω))C(α,γ,l)m2γεγ+εl2(tανα)γlfHr+kβk(Ω). (5.14)

    In view of (5.14) and sr+kβk, we derive that

    [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]G(uα,β(ν))Hr(Ω))C(α,γ,l)m2γεγ+εl2(tανα)γlG(um,α,β(ν))Hr+kβk(Ω)C(α,γ,l,s)m2γεγ+εl2(tανα)γlG(um,α,β(ν))Hs(Ω). (5.15)

    By using global Lipschitz property of G as in (4.1) and noting (4.2), we infer that

    G(um,α,β(ν))Hs(Ω)Lgum,α,β(ν)Hr(Ω)2CkLgαk(mk+λ11)eμ0TαTbαkνbu0Hr+kβk(Ω)(mk+λ11)νbu0Hr+kβk(Ω), (5.16)

    where the hidden constant depends on k,α,Lg,μ0,b. Combining (5.15) and (5.16), we get the following estimate

    M3Hr(Ω)m2γεγ+εl2(mk+λ11)u0Hr+kβk(Ω)t0να1b(tανα)γldν. (5.17)

    Since γ>l and noting that b<α, we infer that

    t0να1b(tανα)γldνtα(γl)t0να1bdν=tα(γl)+αbαb. (5.18)

    By (5.17) and (5.18), we obtain

    M3Hr(Ω)m2γεγ+εl2(mk+λ11)tα(γl)+αbαbu0Hr+kβk(Ω). (5.19)

    Step 4. Estimate of M4. By Parseval's equality, we know that

    (PmI)Sα,β(t)(Sα,β(ν))1f2Hr(Ω))=nNλ2rn(mλn1+mλn)2exp(2λβntαναα)f,en2. (5.20)

    Using the inequality ezC(ε0)zε0, we get the following inequality

    exp(2λβntαναα)C(ε0,α)λ2βε0n(tανα)2ε0. (5.21)

    By the inequality (1+z)2>zε1 for 1<ε1<2, we find that the following inequality

    (mλn1+mλn)2m2ε1λ2ε1n. (5.22)

    By (5.20)–(5.22), we derive that

    (PmI)Sα,β(t)(Sα,β(ν))1f2Hr(Ω))C(ε0,α)m2ε1(tανα)2ε0nNλ2r+2ε12βε0nf,en2.

    Hence, by using Parseval's equality, we obtain the following estimate

    (PmI)Sα,β(t)(Sα,β(ν))1fHr(Ω))C(ε0,α)m1ε12(tανα)ε0fHrβε0+1ε12(Ω)). (5.23)

    Next, we need to choose the appropriate parameters ε0,ε1. Let ε0=k(0,12) and ε1=22k, we can verify that 1<ε1<2. Hence, it follows from (5.23) that

    (PmI)Sα,β(t)(Sα,β(ν))1fHr(Ω))C(k,α)mk(tανα)kfHrβk+k(Ω)).

    By (5.16) and Sobolev embedding Hs(Ω))Hrβk+k(Ω), we derive that

    (PmI)Sα,β(t)(Sα,β(ν))1G(um,α,β(ν))Hr(Ω))C(k,α)mk(tανα)kG(um,α,β(ν))Hrβk+k(Ω))C(k,α,s)mk(tανα)kG(um,α,β(ν))Hs(Ω))mk(tανα)k(mk+λ11)νbu0Hr+kβk(Ω),

    which implies that

    M4Hr(Ω)mk(mk+λ11)u0Hr+kβk(Ω)t0να1b(tανα)kdνmk(mk+λ11)tαbkαα(12k)(α2b)u0Hr+kβk(Ω), (5.24)

    where we use (4.23). Combining (5.8), (5.11), (5.19) and (5.24), we obtain that

    um,α,β(t)uα,β(t)Hr(Ω)M1Hr(Ω)+M2Hr(Ω)+M3Hr(Ω)+M4Hr(Ω)C(α,γ,l)m2γεγ+εl2tαγαlu0Hr+kβk(Ω)+m2γεγ+εl2(mk+λ11)tα(γl)+αbαbu0Hr+kβk(Ω)+mk(mk+λ11)tαbkαα(12k)(α2b)u0Hr+kβk(Ω)+LgC(r,s)t0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. (5.25)

    Here we note that 12k>0 and α2b>0. Since the fact that γ>l and b<min(α,(1k)α), it is obvious to see that

    tαγαlTαγαl,
    tα(γl)+αbαbTα(γl)+αbαb,
    tαbkαα(12k)(α2b)Tαbkαα(12k)(α2b),

    which motivate us to put

    T(α,γ,l,b,k)=max(Tαγαl,Tα(γl)+αbαb,Tαbkαα(12k)(α2b)).

    It follows from (5.25) that

    um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)+Lgt0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. (5.26)

    To continue to go further in the proof, we now need to recall the following Lemma introduced in [29].

    Lemma 5.1. Let vL1[0,T]. Consider some postive constant A,B,β,γ such that β+γ>1 and

    v(t)A+Bt0(tr)β1rγ1v(r)dr.

    Then for 0<tT, we get

    v(t)AEβ,γ(B(Γ(β))1β+γ1t)

    Looking Lemma 5.1 and (5.26), we set

    v(t)=um,α,β(t)uα,β(t)Hr(Ω),A=[m2γεγ+εl2+m2γεγ+εl2+mk]T(α,γ,l,b,k),

    B=Lg,β=1 and γ=α. Then we deduce that

    um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)E1,α(Lgt).

    The proof of Theorem 5.1 is completed.

    Theorem 6.1. Let G:Hr(Ω)Hs(Ω) such that (4.1) holds. Here r,s satisfy that s>r+kβk for 0<k<12. Let the initial datum u0Hr+kβk+ε(Ω) for any ε>0. Then we get the following estimate

    um,α,βum,αXαk(0,T;Hr(Ω))Dβ(ε,k)u0Hr+kβk+ε(Ω)+|Eβ(r,s,k)|u0Hr+kβk(Ω),

    where the hidden constants depends on α,k,T,b. Here

    Dβ(ε,k)=|1λ1β1|2kβk+εβ+(1β)ε,Eβ(r,s,k)=|1λ1β1|k+srβ+(1β)srk+βk

    for any ε>0.

    Proof. For m>0, let um,α,β and um,α be the mild solutions to Problem (1.3) with 0<β<1 and β=1 respectively. Let us recall the formula of these two solutions

    um,α,β(t):=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(um,α,β(ν))dν. (6.1)

    and

    um,α(t):=Sm,α,1(t)u0+t0να1PmSm,α,1(t)(Sm,α,1(ν))1G(um,α(ν))dν. (6.2)

    Subtracting (6.1) from (6.2) on each side, we derive that

    um,α,β(t)um,α(t)=(Sm,α,β(t)u0Sm,α,1(t)u0)+t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))dν+t0να1PmSm,α,β(t)(Sm,α,β(ν))1[G(um,α,β(ν))G(um,α(ν))]dν=N1+N2+N3. (6.3)

    Step 1. Estimate of N1. By Parseval' s equality, we have that the following equality

    N12Hr(Ω)=(Sm,α,β(t)u0Sm,α,1(t)u0)2Hr(Ω)=nNλ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2=λn>1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2+λn1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2=N1,1+N1,2. (6.4)

    For the term N1,1, since λn>1 and 0<β<1, we note that λβn1+mλn<λn1+mλn. Hence, we have the following inequality

    exp(λβn1+mλntαα)>exp(λn1+mλntαα).

    In view of the above inequality

    |eced|Cγmax(ec,ed)|cd|γ,γ>0

    with c=exp(λβn1+mλntαα) and d=exp(λn1+mλntαα), we derive that

    |exp(λβn1+mλntαα)exp(λn1+mλntαα)|Cγexp(λβn1+mλntαα)|λnλβn1+mλn|γC(γ,ε)(λβn1+mλntαα)ε|λnλβn1+mλn|γ=C(γ,ε)αεtαε(1+mλn)εγλβεn|λnλβn|γ. (6.5)

    Since the fact that λn>1, we know that |λnλβn|γ=λγn(1λβ1n)γ. Using the inequality 1eyC(μ)yμ for any μ>0, we find that

    1λβ1n=1exp[(1β)log(λn)]C(μ)(1β)μlogμ(λn)C(μ)(1β)μλμn,

    where we note that 0<log(y)y for any y>1, which implies that

    |λnλβn|γ=λγn(1λβ1n)γC(μ,γ)(1β)μγλμγ+γn. (6.6)

    Combining (6.5) and (6.6), we derive that

    |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(μ,γ,ε)αεtαε(1+mλn)εγ(1β)μγλβεnλμγ+γn.

    If we make the assumption εγ, then (1+mλn)εγ1, which allows us to obtain that

    N1,1|C(μ,γ,ε,α)|2t2αε(1β)2μγλn>1λ2r+2μγ+2γ2βεnu0,en2|C(μ,γ,ε,α)|2t2αε(1β)2μγu02Hr+μγ+γβε(Ω).

    Let us choose μ=εk and γ=ε=k for any ε>0. Then we get the following estimate

    N1,1|C(ε,k,α)|2t2αk(1β)2εu02Hr+kβk+ε(Ω). (6.7)

    Before mention to N1,2, we provide a set ¯N={nN:λn1}. Let us give the observation that if ¯N is an empty set, then N1,2=0. If ¯N is a non-empty set, then λ11. For the term N1,2, we note that λβn1+mλn>λn1+mλn since λn1 and 0<β<1. Hence, we have that

    exp(λβn1+mλntαα)<exp(λn1+mλntαα).

    By using the fact that

    |eced|Cγ1max(ec,ed)|cd|γ1,γ1>0

    with c=exp(λβn1+mλntαα) and d=exp(λn1+mλntαα), we derive that

    |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(γ1)exp(λn1+mλntαα)|λnλβn1+mλn|γ1C(γ1,ε1)(λn1+mλntαα)ε1|λnλβn1+mλn|γ1=C(γ1,ε1)αε1tαε1(1+mλn)ε1γ1λε1n|λnλβn|γ1. (6.8)

    Since λ1λn1, it is obvious to see that

    |λnλβn|γ1=(λβnλn)γ1=λβγ1n|1λ1βn|γ1λβγ1n|1λ1β1|γ1. (6.9)

    Combining (6.8) and (6.9), we find that

    |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(γ1,ε1,α)|1λ1β1|γ1tαε1(1+mλn)ε1γ1λε1+βγ1n. (6.10)

    Hence, it follows from (6.10) that

    N1,2=λn1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2|C(γ1,ε1,α)|2|1λ1β1|2γ1t2αε1n=1(1+mλn)2ε12γ1λ2r2ε1+2βγ1nu0,en2.

    Let ε1=k and γ1=2k+εβk. Since 0<β<1, we know that

    ε1<γ1,2r2ε1+2βγ1=2r+2k2βk+2ε.

    Hence, in view of Parseval's equality, we get the following estimate

    N1,2C(k,r,ε,α)|1λ1β1|4k2βk+2εβt2αku02Hr+kβk+ε(Ω). (6.11)

    Combining (6.4), (6.7) and (6.11), we obtain the following estimate

    N12Hr(Ω)=N1,1+N1,2C(k,r,ε,α)[|1λ1β1|4k2βk+2εβ+(1β)2ε]t2αku02Hr+kβk+ε(Ω).

    By taking the square root of both sides of the above expression and using the inequality a+ba+b for any a,b0, we get

    N1Hr(Ω)C(k,r,ε,α)Dβ(ε,k)tαku0Hr+kβk+ε(Ω). (6.12)

    Here we denote

    Dβ(ε,k)=|1λ1β1|2kβk+εβ+(1β)ε,ε>0,

    where we observe that Dβ(ε,k)0, β1.

    Step 2. Estimate of N2. We confirm the following result for any ε0>0 using the method similar to Step 1,

    [PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]ψHr(Ω)C(k,r,ε0,α)[|1λ1β1|2kβk+ε0β+(1β)ε0](tανα)kψHr+kβk+ε0(Ω).

    Since s>r+kβk, we know that ε0=srk+βk>0, and using global Lipschitz property of G, we derive that

    [PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))Hr(Ω)¯C1Eβ(r,s,k)(tανα)kG(um,α(ν))Hs(Ω)Kg¯C1Eβ(r,s,k)(tανα)kum,α(ν)Hr(Ω),

    where ¯C1=C(k,r,s,β,α), and we have the following observation

    Eβ(r,s,k)=|1λ1β1|k+srβ+(1β)srk+βk0,β1.

    In view of (4.2), we obtain that the following upper bound

    um,α(ν)Hr(Ω)¯C2νbu0Hr+kβk(Ω),

    where ¯C2=C(k,α,m,μ0,T,b). From two latter estimations as above, we infer that

    N2Hr(Ω)t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))Hr(Ω)dν¯C3Eβ(r,s,k)u0Hr+kβk(Ω)t0να1b(tανα)kdν, (6.13)

    where ¯C3=Kg¯C1¯C2. Using (4.23), we obtain that the following inequality

    t0να1b(tανα)kdνtαbkαα(12k)(α2b). (6.14)

    Combining (6.13) and (6.14), we obtain that

    N2Hr(Ω)¯C3tαbkαα(12k)(α2b)Eβ(r,s,k)u0Hr+kβk(Ω). (6.15)

    Step 3. Estimate of N3. By a similar argument as in (4.6), we find that

    t0να1PmSm,α,β(t)(Sm,α,β(ν))1[G(um,α,β(ν))G(um,α(ν))]dνHr(Ω))Ckαkmkt0να1(tανα)kG(um,α,β(ν))G(um,α(ν))Hr+kβk(Ω))dν.

    Since s>r+kβk and using global Lipschitz property of G, we obtain that

    G(um,α,β(ν))G(um,α(ν))Hr+kβk(Ω))G(um,α,β(ν))G(um,α(ν))Hs(Ω))Kgum,α,β(ν)um,α(ν)Hr(Ω)).

    From the two above observations, we confirm the following statement

    N3Hr(Ω)¯C3t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν, (6.16)

    where ¯C3=CkαkmkKg. Combining (6.3), (6.12), (6.15) and (6.16), we deduce the following estimate

    um,α,β(t)um,α(t)Hr(Ω)N1Hr(Ω)+N2Hr(Ω)+N3Hr(Ω)˜CtαkDβ(ε,k)u0Hr+kβk+ε(Ω)+¯C3tαbkαα(12k)(α2b)Eβ(r,s,k)u0Hr+kβk(Ω)+¯C3t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν, (6.17)

    where ˜C=C(k,r,ε,α). By the Hölder inequality and in combination with (4.22), we get the estimate of the third term on the right hand side of (6.17) as follows

    (t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν)2(t0να1(tανα)2kdν)(t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν)tα(12k)α(12k)(t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν). (6.18)

    Combining (6.17), (6.18) and the inequality (a+b+c)23a2+3b2+3c2, we derive that

    um,α,β(t)um,α(t)2Hr(Ω)3|˜C|2t2αk|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2t2α2b2αkα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2tα(12k)α(12k)t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν. (6.19)

    Multiplying both sides of (6.19) by t2αk, we get the following estimate

    t2αkum,α,β(t)um,α(t)2Hr(Ω)3|˜C|2|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2t2α2bα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2tαα(12k)t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν,

    which implies that

    t2αkum,α,β(t)um,α(t)2Hr(Ω)3|˜C|2|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2T2α2bα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2Tαα(12k)t0να12αkν2αkum,α,β(ν)um,α(ν)2Hr(Ω))dν. (6.20)

    Looking Lemma 5.1 and (6.20), we set v(t)=t2αkum,α,β(t)um,α(t)2Hr(Ω),

    ¯A=3|˜C|2|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2T2α2bα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)

    and

    ¯B=|¯C3|2Tαα(12k),β=1,γ=α2αk.

    By applying Lemma 5.1, we obtain that

    t2αkum,α,β(t)um,α(t)2Hr(Ω)¯AE1,α2αk(¯B(Γ(β))1β+γ1t)=¯AE1,α2αk(¯Bt). (6.21)

    In view of Lemma 3.1 as in [30], we obtain the following upper bound

    E1,α2αk(¯Bt)Cα, (6.22)

    where Cα is a positive constant that depends on α. By (6.21) and (6.22), we get

    tαkum,α,β(t)um,α(t)Hr(Ω)¯ACα.

    From Deinition 2.3, we find

    um,α,βum,αXαk(0,T;Hr(Ω))¯ACα.

    The proof is completed.

    Huy Tuan Nguyen is supported by the Van Lang University. Chao Yang is supported by the Ph.D. Student Research and Innovation Fund of the Fundamental Research Funds for the Central Universities (3072022GIP2403).

    The authors declare there is no conflict of interest.



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