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
C∂βf(s)∂sβ=s1−β∂f(s)∂s, | (1.1) |
where 0<β≤1. 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
C∂α∂tαu(x,t)=Dα∂2∂x2u(x,t), | (1.2) |
where 0<α≤1,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 Ω⊂RN (N≥1) 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
{C∂α∂tαu+(−Δ)βu(x,t)−mC∂α∂tαΔu=F(u(x,t)),x∈Ω,t∈(0,T),u(x,t)=0,x∈∂Ω,t∈(0,T),u(x,0)=u0(x)x∈Ω | (1.3) |
where α,β∈(0,1], m>0, the time fractional derivative C∂α∂tα 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
{C∂α∂tαu+(−Δ)βu(x,t)=F(u(x,t)),x∈Ω,t∈(0,T),u(x,t)=0,x∈∂Ω,t∈(0,T),u(x,0)=u0(x)x∈Ω. | (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 m→0+, 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 β→1−. 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 sα−1 or (tα−sα)−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 m→0+. In Section 6, we investigate the convergence of mild solutions when β→1−.
Definition 2.1. Conformable derivative model: Let B be a Banach space, and the function f:[0,∞)→B. Let C∂β∂tβ be the conformable derivative operator of order β∈(0,1], locally defined by
C∂βf(t)∂tβ:=limh→0f(t+ht1−β)−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),x∈Ω,en(x)=0,x∈∂Ω, |
admits the eigenvalues 0<λ1≤λ2≤⋯≤λn≤… with λn→∞ as n→∞. The corresponding eigenfunctions are en∈H10(Ω).
Definition 2.2. (Hilbert scale space). We recall the Hilbert scale space, which is given as follows
Hs(Ω)={f∈L2(Ω)|∞∑n=1λ2sn(∫Ωf(x)en(x)dx)2<∞} |
for any s≥0. It is well-known that Hs(Ω) is a Hilbert space corresponding to the norm
‖f‖Hs(Ω)=(∞∑n=1λ2sn(∫Ωf(x)en(x)dx)2)1/2,f∈Hs(Ω). |
Definition 2.3. Let Xa,q,α((0,T];B) denote the weighted space of all the functions ψ∈C((0,T];X) such that
‖ψ‖Xa,q,α((0,T];B):=supt∈(0,T]tae−qtα‖ψ(t,⋅)‖B<∞, |
where a,q>0 and 0<α≤1 (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=1⟨u(.,t),en⟩en(x). |
Taking the inner product of Problem (1.3) with en gives
{C∂α∂tα⟨u(.,t),en⟩+λβn⟨u(.,t),en⟩+mλnC∂α∂tα⟨u(.,t),en⟩=⟨F(.,t),en⟩,t∈(0,T),⟨u(.,0),en⟩=⟨u0,en⟩, | (3.1) |
where we repeat that
⟨Δu(.,t),en⟩=−λn⟨u(.,t),en⟩. |
The first equation of (3.1) is a differential equation with a conformable derivative as follows
C∂α∂tα⟨u(.,t),en⟩+λβn1+mλn⟨u(.,t),en⟩=11+mλn⟨F(.,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+mλn∫t0να−1exp(λβn1+mλnνα−tαα)⟨F(.,ν),en⟩dν. |
To simplify the solution formula, we will express the solution in operator equations. Let us set the following operators
Sm,α,β(t)f=∑n∈Nexp(−λβn1+mλntαα)⟨f,en⟩en, |
and
Pmf=∑n∈N(1+mλn)−1⟨f,en⟩en |
for any f∈L2(Ω) in the form f=∑n∈N⟨f,en⟩en. Then the inverse operator of Sm,α,β(t) is defined by
(Sm,α,β(t))−1f=∑n∈Nexp(λβn1+mλntαα)⟨f,en⟩en |
The mild solution is given by
u(t)=Sm,α,β(t)u0+∫t0να−1PmSm,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 v∈Hs+k−βk(Ω) for any k>0. Then we get
‖Sm,α,β(t)v‖Hs(Ω)≤Ckαkm−kt−αk‖v‖Hs+k−βk(Ω), | (3.3) |
and for 0≤ν≤t≤T,
‖Sm,α,β(t)(Sm,α,β(ν))−1v‖Hs(Ω)≤Ckαkm−k(tα−να)−k‖v‖Hs+k−βk(Ω). | (3.4) |
(b) If v∈Hs(Ω), then
‖Sm,α,β(t)v‖Hs(Ω)≤‖v‖Hs(Ω) | (3.5) |
and
‖Sm,α,β(t)(Sm,α,β(ν))−1v‖Hs(Ω)≤‖v‖Hs(Ω) | (3.6) |
for 0≤ν≤t≤T.
(c) If v∈Hs(Ω), then we have
‖Pmv‖Hs(Ω)≤‖v‖Hs(Ω) |
for any s.
Proof. For (a), in view of Parseval's equality, we get that
‖Sm,α,β(t)v‖2Hs(Ω)=∑n∈Nλ2snexp(−2λβn1+mλntαα)⟨v,en⟩2. | (3.7) |
Using the inequality e−y≤Cky−k, we find that
exp(−λβn1+mλntαα)≤Ck(λβn1+mλn)−kt−αkα−k=Ckαk(mk+λ−11)λ(1−β)knt−αk, | (3.8) |
where we use
(1+mλn)k≤Ck(1+mkλkn)≤Ck(mk+λ−11)λkn. |
It follows from (3.7) that
‖Sm,α,β(t)v‖2Hs(Ω)≤(Ckαk)2(mk+λ−11)2t−2αk∑n∈Nλ2s+2k−2βkn⟨f,en⟩2=(Ckαk)2(mk+λ−11)2t−2αk‖v‖2Hs+k−βk(Ω). |
By a similar explanation, we also get that for 0≤ν≤t≤T,
‖Sm,α,β(t)(Sm,α,β(ν))−1v‖2Hs(Ω)=∑n∈Nλ2snexp(λβn1+mλn2να−2tαα)⟨v,en⟩2≤(Ckαk)2(mk+λ−11)2(tα−να)−2k∑n∈Nλ2s+2k−2βkn⟨f,en⟩2=(Ckαk)2(mk+λ−11)2(tα−να)−2k‖v‖2Hs+k−βk(Ω), |
where we use the fact that
exp(λβn1+mλnνα−tαα)≤Ck(λβn1+mλn)−k(tα−να)−kα−k=Ckαk(mk+λ−11)λ(1−β)kn(tα−να)−k. |
Hence, (a) is proved.
For (b), in view of Parseval's equality, we get that
‖Sα,β(t)v‖2Hs(Ω)=∑n∈Nλ2snexp(−2λβn1+mλntαα)⟨v,en⟩2≤∑n∈Nλ2sn⟨v,en⟩2=‖v‖2Hs(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
‖Pmv‖2Hs(Ω)=∑n∈Nλ2sn(1+mλn)−2⟨v,en⟩2≤∑n∈Nλ2sn⟨v,en⟩2=‖v‖2Hs(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)≤Lg‖w1−w2‖Hr(Ω), | (4.1) |
for any w1,w2∈Hr(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 s≥r+k−βk for 0<k<12. Let the initial datum u0∈Hr+k−βk(Ω). Then Problem (1.3) has a unique solution um,α,β∈Lp(0,T;Hr(Ω)), where
1<p<1b,αk≤b<α−αk. |
In addition, we get
‖um,α,β(t)‖Hr(Ω)≤2Ckαk(mk+λ−11)eμ0TαTb−αkt−b‖u0‖Hr+k−βk(Ω), | (4.2) |
where Ck depends on k.
(ii) Let us assume that b<α2 and u0∈Hs+k−βk+β−1(Ω)∩Hr+k−βk(Ω). Then we have
‖∂∂tum,α,β(.,t)‖Hs(Ω)≲tα−αk−1‖u0‖Hs+k−βk+β−1(Ω)+(tα−1−b+t2α−b−kα−1)‖u0‖Hr+k−βk(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+∫t0να−1PmSm,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)u0‖Hr(Ω))≤Ckαk(mk+λ−11)t−αk‖u0‖Hr+k−βk(Ω). |
Hence, multiplying both sides of the above expression by tbe−μtα, we have that
tbe−μtα‖Bw0(t)‖Hr(Ω))≤Ckαk(mk+λ−11)tb−αk‖u0‖Hr+k−βk(Ω)≤Ckαk(mk+λ−11)Tb−αk‖u0‖Hr+k−βk(Ω), | (4.5) |
where we use b≥αk. This implies that Bw0∈Xb,μ,α((0,T];Hr(Ω)). Let any two functions w1,w2∈Xb,μ,α((0,T];Hr(Ω)). From (4.4) and (3.3), we obtain that
‖Bw1(t)−Bw2(t)‖Hr(Ω))=‖∫t0να−1PmSm,α,β(t)(Sm,α,β(ν))−1(G(w1(ν))−G(w2(ν)))dν‖Hr(Ω))≤Ckαkm−k∫t0να−1(tα−να)−k‖G(w1(ν))−G(w2(ν))‖Hr+k−βk(Ω))dν. | (4.6) |
Since the constraint s≥r+k−βk, we know that Sobolev embedding
Hs(Ω)↪Hr+k−βk(Ω). |
From some above observations and noting (4.1), we get that
tbe−μtα‖Bw1(t)−Bw2(t)‖Hr(Ω))≤Ckαk(mk+λ−11)tbe−μtα∫t0να−1(tα−να)−k‖G(w1(ν))−G(w2(ν))‖Hs(Ω))dν≤CkLgαk(mk+λ−11)tbe−μtα∫t0να−1(tα−να)−k‖w1(ν)−w2(ν)‖Hr(Ω))dν=CkLgαk(mk+λ−11)tb∫t0να−1−b(tα−να)−ke−μ(tα−να)νbe−μνα‖w1(ν)−w2(ν)‖Hr(Ω))dν. | (4.7) |
From the fact that
‖w1−w2‖Xb,μ,α((0,T];Hr(Ω))=sup0≤ν≤Tνbe−μνα‖w1(ν)−w2(ν)‖Hr(Ω)), |
we follows from (4.7) that
sup0≤t≤Ttbe−μtα‖Bw1(t)−Bw2(t)‖Hr(Ω))≤¯C‖w1−w2‖Xb,μ,α((0,T];Hr(Ω))sup0≤t≤T[tb∫t0να−1−b(tα−να)−ke−μ(tα−να)dν], | (4.8) |
where ¯C=CkLgαkm−k. Let us continue to treat the integral term as follows
J1,μ(t)=tb∫t0να−1−b(tα−να)−ke−μ(tα−να)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)=1αtα−αk∫10ξ−bα(1−ξ)−ke−μtα(1−ξ)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+a2≥−1, ρ>0 and t∈[0,T]. For h>0, the following limit holds
limρ→∞(supt∈[0,T]th∫10νa1(1−ν)a2e−ρt(1−ν)dν)=0. |
Since 0<b<α and 0<k<min(bα,1−bα), we easily to verify that the following conditions hold
{α−αk>0,−bα>−1, −k>−1,−bα−k≥−1. | (4.9) |
By Lemma 4.1 and (4.9), we have
limμ→+∞sup0≤t≤TJ1,μ(t)=0. |
This statement shows that there exists a μ0 such that
¯Csup0≤t≤TJ1,μ0(t)≤12. | (4.10) |
Combining (4.8) and (4.10), we obtain
‖Bw1−Bw2‖Xb,μ0((0,T];Hr(Ω))≤12‖w1−w2‖Xb,μ0((0,T];Hr(Ω)) | (4.11) |
for any w1,w2∈Xb,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,β∈Xb,μ0,α((0,T];Hr(Ω)). Hence, we can see that
um,α,β(t)=Sm,α,β(t)u0+∫t0να−1PmSm,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,β‖Xb,μ0,α((0,T];Hr(Ω))=‖Bum,α,β‖Xb,μ0,α((0,T];Hr(Ω))≤12‖um,α,β‖Xb,μ0((0,T];Hr(Ω))+‖B(v=0)‖Xb,μ0,α((0,T];Hr(Ω))=12‖um,α,β‖Xb,μ0,α((0,T];Hr(Ω))+‖Sm,α,β(t)u0‖Xb,μ0,α((0,T];Hr(Ω)), |
which combined with (4.5), we get
‖um,α,β‖Xb,μ0,α((0,T];Hr(Ω))≤2Ckαk(mk+λ−11)Tb−αk‖u0‖Hr+k−βk(Ω), |
which allows us to get that
‖um,α,β(t)‖Hr(Ω)≤˜C1t−b‖u0‖Hr+k−βk(Ω), | (4.13) |
where ˜C1 depends on k,μ0,b,α,m and
˜C1=2Ckαk(mk+λ−11)eμ0TαTb−αk, |
and we remind that Ck depends on k. Note that the improper integral ∫T0t−pbdt is convergent for 1<p<1b, we deduce that
um,α,β∈Lp(0,T;Hr(Ω)), |
and the following regularity holds
‖um,α,β‖Lp(0,T;Hr(Ω))≤˜C‖u0‖Hr+k−βk(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)=∫t0∂tK(t,s)ds+K(t,t)dt, |
we obtain the following equality
∂∂tum,α,β(.,t)=tα−1Qm,α,β(t)u0+tα−1G(um,α,β(x,t))+tα−1∫t0να−1PmQm,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=−∑n∈Nλβn1+mλnexp(−λβn1+mλntαα)⟨v,en⟩en |
for any v∈L2(Ω). In view of Parseval's equality, we get that
‖Qm,α,β(t)v‖2Hs(Ω)=∑n∈Nλ2sn(λβn1+mλn)2exp(−2λβn1+mλntαα)⟨v,en⟩2. |
Using (3.8) and noting that λβn1+mλn≤m−1λβ−1n, we get that
‖Qm,α,β(t)v‖2Hs(Ω)≤(Ckαkm−k−1)2t−2αk∑n∈Nλ2s+2k−2βk+2β−2n⟨v,en⟩2, |
which implies that
‖Qm,α,β(t)v‖Hs(Ω)≤Ckαkm−k−1t−αk‖v‖Hs+k−βk+β−1(Ω) | (4.15) |
for any v∈Hs+k−βk+β−1(Ω). In a similar technique as above, we also get that
‖Qm,α,β(t)(Sm,α,β)−1(ν)v‖Hs(Ω)≤Ckαkm−k−1(tα−να)−k‖v‖Hs+k−βk+β−1(Ω), | (4.16) |
for any 0≤ν≤t. 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
‖tα−1Qm,α,β(t)u0‖Hs(Ω)≤Ckαkm−k−1tα−αk−1‖u0‖Hs+k−βk+β−1(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
‖tα−1G(um,α,β(x,t))‖Hs(Ω)≤˜C1tα−1Lg‖um,α,β‖Hr(Ω)≤˜C1Lgtα−1−b‖u0‖Hr+k−βk(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
‖∫t0να−1PmQm,α,β(t)(Sm,α,β)−1(ν)G(um,α,β(ν))dν‖Hs(Ω)≤Ckαkm−k−1∫t0να−1(tα−να)−k‖G(um,α,β(ν))‖Hs+k−βk+β−1(Ω)dν. | (4.19) |
Since β≤1 and 0<k<1, we can easily verify that s+k−βk+β−1≤s, which implies the Sobolev embedding Hs(Ω)↪Hs+k−βk+β−1(Ω) is true. From these above observations and using (4.13), we derive that
∫t0να−1(tα−να)−k‖G(um,α,β(ν))‖Hs+k−βk+β−1(Ω)dν≤C(s,k,β)∫t0να−1(tα−να)−k‖G(um,α,β(ν))‖Hs(Ω)dν≤LgC(s,k,β)∫t0να−1(tα−να)−k‖um,α,β(ν)‖Hr(Ω)dν≤LgC(s,k,β)˜C1‖u0‖Hr+k−βk(Ω)(∫t0να−1−b(tα−να)−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
(∫t0να−1−b(tα−να)−kdν)2=(∫t0να−12−bνα−12(tα−να)−kdν)2≤(∫t0να−1−2bdν)(∫t0να−1(tα−να)−2kdν)=tα−2bα−2b(∫t0να−1(tα−να)−2kdν), | (4.21) |
where α>2b. By changing to a new variable z=να, we derive that dz=ανα−1dν. Hence, we infer that
∫t0να−1(tα−να)−2kdν=1α∫tα0(tα−z)−2kdz=tα(1−2k)α(1−2k), | (4.22) |
where we note that 0<k<12. Combining (4.21) and (4.22), we obtain that the following inequality
∫t0να−1−b(tα−να)−kdν≤tα−b−kα√α(1−2k)(α−2b). | (4.23) |
By (4.23) and following from (4.19) and (4.20), we have
‖∫t0να−1Qm,α,β(t)(Sm,α,β)−1(ν)G(um,α,β(ν))dν‖Hs(Ω)≤Ckαkm−k−1∫t0να−1(tα−να)−k‖G(um,α,β(ν))‖Hs+k−βk+β−1(Ω)dν≤Ckαkm−k−1LgC(s,k,β)˜C1√α(1−2k)(α−2b)tα−b−kα‖u0‖Hr+k−βk(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)≤‖tα−1Qm,α,β(t)u0‖Hs(Ω)+‖tα−1G(um,α,β(x,t))‖Hs(Ω)+tα−1‖∫t0να−1PmQm,α,β(t)(Sm,α,β)−1(ν)G(um,α,β(ν))dν‖Hs(Ω)≤Ckαkm−k−1tα−αk−1‖u0‖Hs+k−βk+β−1(Ω)+˜C1Lgtα−1−b‖u0‖Hr+k−βk(Ω)+Ckαkm−k−1LgC(s,k,β)˜C1√α(1−2k)(α−2b)t2α−b−kα−1‖u0‖Hr+k−βk(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 m→0+. 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 s≥r+k−βk for 0<k<12. Let the initial datum u0∈Hr+k−βk(Ω). Let um,α,β and u∗α,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)−u∗α,β(t)‖Hr(Ω)≲[m2γ−εγ+εl2+mk]T∗(α,γ,l,b,k)E1,α(Lgt), |
where αk≤b<α2, γ=k(1−β)+l(β−ε2)β+1−ε2, 0<l<k(1−β) and 1<ε<min(2,2k(1−β)+2lβl).
Remark 5.1. Note that m2γ−εγ+εl2+mk tends to zero when m→0+. Hence, we can deduce that ‖um,α,β(t)−u∗α,β(t)‖Hr(Ω)→0 when m→0+.
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
u∗α,β(t)=S∗α,β(t)u0+∫t0να−1S∗α,β(t)(S∗α,β(ν))−1G(u∗α,β(ν))dν, | (5.1) |
where we provide two operators that have the following Fourier series representation as follows
S∗α,β(t)f=∑n∈Nexp(−λβntαα)⟨f,en⟩en |
and
(S∗α,β(ν))−1f=∑n∈Nexp(λβnναα)⟨f,en⟩en. |
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)−u∗α,β(t)=(Sm,α,β(t)−S∗α,β(t))u0+∫t0να−1S∗α,β(t)(S∗α,β(ν))−1(G(um,α,β(ν))−G(u∗α,β(ν)))dν+∫t0να−1[PmSm,α,β(t)(Sm,α,β(ν))−1−PmS∗α,β(t)(S∗α,β(ν))−1]G(um,α,β(ν))dν+∫t0να−1(Pm−I)S∗α,β(t)(S∗α,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 |e−c−e−d|≤Cγmax(e−c,e−d)|c−d|γ with γ>0, let c=λβn1+mλntαα and d=λβntαα, we get the following inequality
|exp(−λβn1+mλntαα)−exp(−λβntαα)|≤Cγexp(−λβn1+mλntαα)(λβn−λβn1+mλn)γtαγα−γ≤CγCl(λβn1+mλn)−l(tαα)−l(λβn−λβn1+mλn)γtαγα−γ≤CγClαl−γλ−βln(mλ1+βn)γ(11+mλn)γ−ltαγ−αl | (5.3) |
by e−c>e−d 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)2ε≥1+z>z, which implies
(1+z)γ−l=(1+z)2εε(γ−l)2>zε(γ−l)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)γ−l>(mλn)ε(γ−l)2, which leads to the following inequality
(11+mλn)γ−l≤(mλn)ε(l−γ)2. | (5.5) |
Combining (5.3) and (5.5), we find that
|exp(−λβn1+mλntαα)−exp(−λβntαα)|≤C(α,γ,l)m2γ−εγ+εl2tαγ−αlλ−βl+βγ+γ+ε(l−γ)2n. | (5.6) |
Therefore, we have the following estimate
‖(Sm,α,β(t)−S∗α,β(t))u0‖2Hr(Ω))=∑n∈Nλ2rn|exp(−λβn1+mλntαα)−exp(−λβntαα)|2⟨u0,en⟩2≤|C(α,γ,l)|2m2γ−εγ+εlt2αγ−2αl∑n∈Nλ2r−2βl+2βγ+2γ+ε(l−γ)n⟨u0,en⟩2, |
which implies that
‖(Sm,α,β(t)−S∗α,β(t))u0‖Hr(Ω))≤C(α,γ,l)m2γ−εγ+εl2tαγ−αl‖u0‖Hr−βl+βγ+γ+ε(l−γ)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(1−β),1−ββk)=k(1−β), |
which implies that 2k(1−β)+2lβ>l, that is 2k(1−β)+2lβl>1. Then we can choose ε such that
1<ε<min(2,2k(1−β)+2lβl). |
Let us choose γ such that
γ=k(1−β)+l(β−ε2)β+1−ε2. |
It is easy to verify that γ>l+l(β−ε2)β+1−ε2=l and the following equality
r−βl+βγ+γ+ε(l−γ)2=r+k−βk. |
Then the estimate (5.7) becomes
‖M1‖Hr(Ω)=‖(Sm,α,β(t)−S∗α,β(t))u0‖Hr(Ω))≤C(α,γ,l)m2γ−εγ+εl2tαγ−αl‖u0‖Hr+k−βk(Ω). | (5.8) |
Step 2. Estimate of M2. Let ψ∈Hr(Ω). For 0≤ν≤t≤T, we can also get
‖S∗α,β(t)(S∗α,β(ν))−1ψ‖2Hr(Ω)=∑n∈Nλ2rnexp(2λβnνα−tαα)⟨ψ,en⟩2≤‖ψ‖2Hr(Ω), | (5.9) |
where we use that
exp(λβn1+mλnνα−tαα)≤1. |
By (5.9) and Sobolev embedding Hs(Ω))↪Hr(Ω), we find that
‖∫t0να−1S∗α,β(t)(S∗α,β(ν))−1(G(um,α,β(ν))−G(u∗α,β(ν)))dν‖Hr(Ω)≤∫t0να−1‖G(um,α,β(ν))−G(u∗α,β(ν))‖Hr(Ω)dν≤C(r,s)∫t0να−1‖G(um,α,β(ν))−G(u∗α,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
‖M2‖Hr(Ω)≤LgC(r,s)∫t0να−1‖um,α,β(ν)−u∗α,β(ν)‖Hr(Ω)dν. | (5.11) |
Step 3. Estimate of M3. Let f∈Hr+k−βk(Ω). Then using Parseval' s equality, we have the following identity
‖[PmSm,α,β(t)(Sm,α,β(ν))−1−PmS∗α,β(t)(S∗α,β(ν))−1]f‖2Hr(Ω))=∑n∈Nλ2rn(1+mλn)2|exp(−λβn1+mλntα−ναα)−exp(−λβntα−ναα)|2⟨f,en⟩2. | (5.12) |
By a similar explanation as in (5.6), we find that
|exp(−λβn1+mλntα−ναα)−exp(−λβntα−ναα)|≤C(α,γ,l)m2γ−εγ+εl2(tα−να)γ−lλ−βl+βγ+γ+ε(l−γ)2n. | (5.13) |
By (5.12) and (5.13), we get that
‖[PmSm,α,β(t)(Sm,α,β(ν))−1−PmS∗α,β(t)(S∗α,β(ν))−1]f‖Hr(Ω))≤C(α,γ,l)m2γ−εγ+εl2(tα−να)γ−l‖f‖Hr+k−βk(Ω). | (5.14) |
In view of (5.14) and s≥r+k−βk, we derive that
‖[PmSm,α,β(t)(Sm,α,β(ν))−1−PmS∗α,β(t)(S∗α,β(ν))−1]G(u∗α,β(ν))‖Hr(Ω))≤C(α,γ,l)m2γ−εγ+εl2(tα−να)γ−l‖G(um,α,β(ν))‖Hr+k−βk(Ω)≤C(α,γ,l,s)m2γ−εγ+εl2(tα−να)γ−l‖G(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)≤Lg‖um,α,β(ν)‖Hr(Ω)≤2CkLgαk(mk+λ−11)eμ0TαTb−αkν−b‖u0‖Hr+k−βk(Ω)≲(mk+λ−11)ν−b‖u0‖Hr+k−βk(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
‖M3‖Hr(Ω)≲m2γ−εγ+εl2(mk+λ−11)‖u0‖Hr+k−βk(Ω)∫t0να−1−b(tα−να)γ−ldν. | (5.17) |
Since γ>l and noting that b<α, we infer that
∫t0να−1−b(tα−να)γ−ldν≤tα(γ−l)∫t0να−1−bdν=tα(γ−l)+α−bα−b. | (5.18) |
By (5.17) and (5.18), we obtain
‖M3‖Hr(Ω)≲m2γ−εγ+εl2(mk+λ−11)tα(γ−l)+α−bα−b‖u0‖Hr+k−βk(Ω). | (5.19) |
Step 4. Estimate of M4. By Parseval's equality, we know that
‖(Pm−I)S∗α,β(t)(S∗α,β(ν))−1f‖2Hr(Ω))=∑n∈Nλ2rn(mλn1+mλn)2exp(−2λβntα−ναα)⟨f,en⟩2. | (5.20) |
Using the inequality e−z≤C(ε0)z−ε0, we get the following inequality
exp(−2λβntα−ναα)≤C(ε0,α)λ−2βε0n(tα−να)−2ε0. | (5.21) |
By the inequality (1+z)2>zε1 for 1<ε1<2, we find that the following inequality
(mλn1+mλn)2≤m2−ε1λ2−ε1n. | (5.22) |
By (5.20)–(5.22), we derive that
‖(Pm−I)S∗α,β(t)(S∗α,β(ν))−1f‖2Hr(Ω))≤C(ε0,α)m2−ε1(tα−να)−2ε0∑n∈Nλ2r+2−ε1−2βε0n⟨f,en⟩2. |
Hence, by using Parseval's equality, we obtain the following estimate
‖(Pm−I)S∗α,β(t)(S∗α,β(ν))−1f‖Hr(Ω))≤C(ε0,α)m1−ε12(tα−να)−ε0‖f‖Hr−βε0+1−ε12(Ω)). | (5.23) |
Next, we need to choose the appropriate parameters ε0,ε1. Let ε0=k∈(0,12) and ε1=2−2k, we can verify that 1<ε1<2. Hence, it follows from (5.23) that
‖(Pm−I)S∗α,β(t)(S∗α,β(ν))−1f‖Hr(Ω))≤C(k,α)mk(tα−να)−k‖f‖Hr−βk+k(Ω)). |
By (5.16) and Sobolev embedding Hs(Ω))↪Hr−βk+k(Ω), we derive that
‖(Pm−I)S∗α,β(t)(S∗α,β(ν))−1G(um,α,β(ν))‖Hr(Ω))≤C(k,α)mk(tα−να)−k‖G(um,α,β(ν))‖Hr−βk+k(Ω))≤C(k,α,s)mk(tα−να)−k‖G(um,α,β(ν))‖Hs(Ω))≲mk(tα−να)−k(mk+λ−11)ν−b‖u0‖Hr+k−βk(Ω), |
which implies that
‖M4‖Hr(Ω)≲mk(mk+λ−11)‖u0‖Hr+k−βk(Ω)∫t0να−1−b(tα−να)−kdν≲mk(mk+λ−11)tα−b−kα√α(1−2k)(α−2b)‖u0‖Hr+k−βk(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)−u∗α,β(t)‖Hr(Ω)≲‖M1‖Hr(Ω)+‖M2‖Hr(Ω)+‖M3‖Hr(Ω)+‖M4‖Hr(Ω)≤C(α,γ,l)m2γ−εγ+εl2tαγ−αl‖u0‖Hr+k−βk(Ω)+m2γ−εγ+εl2(mk+λ−11)tα(γ−l)+α−bα−b‖u0‖Hr+k−βk(Ω)+mk(mk+λ−11)tα−b−kα√α(1−2k)(α−2b)‖u0‖Hr+k−βk(Ω)+LgC(r,s)∫t0να−1‖um,α,β(ν)−u∗α,β(ν)‖Hr(Ω)dν. | (5.25) |
Here we note that 1−2k>0 and α−2b>0. Since the fact that γ>l and b<min(α,(1−k)α), it is obvious to see that
tαγ−αl≤Tαγ−αl, |
tα(γ−l)+α−bα−b≤Tα(γ−l)+α−bα−b, |
tα−b−kα√α(1−2k)(α−2b)≤Tα−b−kα√α(1−2k)(α−2b), |
which motivate us to put
T∗(α,γ,l,b,k)=max(Tαγ−αl,Tα(γ−l)+α−bα−b,Tα−b−kα√α(1−2k)(α−2b)). |
It follows from (5.25) that
‖um,α,β(t)−u∗α,β(t)‖Hr(Ω)≲[m2γ−εγ+εl2+mk]T∗(α,γ,l,b,k)+Lg∫t0να−1‖um,α,β(ν)−u∗α,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 v∈L1[0,T]. Consider some postive constant A,B,β′,γ′ such that β′+γ′>1 and
v(t)≤A+B∫t0(t−r)β′−1rγ′−1v(r)dr. |
Then for 0<t≤T, we get
v(t)≤AEβ′,γ′(B(Γ(β′))1β′+γ′−1t) |
Looking Lemma 5.1 and (5.26), we set
v(t)=‖um,α,β(t)−u∗α,β(t)‖Hr(Ω),A=[m2γ−εγ+εl2+m2γ−εγ+εl2+mk]T∗(α,γ,l,b,k), |
B=Lg,β′=1 and γ′=α. Then we deduce that
‖um,α,β(t)−u∗α,β(t)‖Hr(Ω)≲[m2γ−εγ+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+k−βk for 0<k<12. Let the initial datum u0∈Hr+k−βk+ε(Ω) for any ε>0. Then we get the following estimate
‖um,α,β−u∗∗m,α‖Xαk(0,T;Hr(Ω))≲Dβ(ε,k)‖u0‖Hr+k−βk+ε(Ω)+|Eβ(r,s,k)|‖u0‖Hr+k−βk(Ω), |
where the hidden constants depends on α,k,T,b. Here
Dβ(ε,k)=|1−λ1−β1|2k−βk+εβ+(1−β)ε,Eβ(r,s,k)=|1−λ1−β1|k+s−rβ+(1−β)s−r−k+βk |
for any ε>0.
Proof. For m>0, let um,α,β and u∗∗m,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+∫t0να−1PmSm,α,β(t)(Sm,α,β(ν))−1G(um,α,β(ν))dν. | (6.1) |
and
u∗∗m,α(t):=Sm,α,1(t)u0+∫t0να−1PmSm,α,1(t)(Sm,α,1(ν))−1G(u∗∗m,α(ν))dν. | (6.2) |
Subtracting (6.1) from (6.2) on each side, we derive that
um,α,β(t)−u∗∗m,α(t)=(Sm,α,β(t)u0−Sm,α,1(t)u0)+∫t0να−1[PmSm,α,β(t)(Sm,α,β(ν))−1−PmSm,α,1(t)(Sm,α,1(ν))−1]G(u∗∗m,α(ν))dν+∫t0να−1PmSm,α,β(t)(Sm,α,β(ν))−1[G(um,α,β(ν))−G(u∗∗m,α(ν))]dν=N1+N2+N3. | (6.3) |
Step 1. Estimate of N1. By Parseval' s equality, we have that the following equality
‖N1‖2Hr(Ω)=‖(Sm,α,β(t)u0−Sm,α,1(t)u0)‖2Hr(Ω)=∑n∈Nλ2rn|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|2⟨u0,en⟩2=∑λn>1λ2rn|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|2⟨u0,en⟩2+∑λn≤1λ2rn|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|2⟨u0,en⟩2=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
|e−c−e−d|≤Cγ′max(e−c,e−d)|c−d|γ′,γ′>0 |
with c=exp(−λβn1+mλntαα) and d=exp(−λn1+mλntαα), we derive that
|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|≤Cγ′exp(−λβn1+mλntαα)|λn−λβn1+mλn|γ′≤C(γ′,ε′)(λβn1+mλntαα)−ε′|λn−λβn1+mλn|γ′=C(γ′,ε′)α−ε′t−αε′(1+mλn)ε′−γ′λ−βε′n|λn−λβn|γ′. | (6.5) |
Since the fact that λn>1, we know that |λn−λβn|γ′=λγ′n(1−λβ−1n)γ′. Using the inequality 1−e−y≤C(μ)yμ for any μ>0, we find that
1−λβ−1n=1−exp[−(1−β)log(λn)]≤C(μ)(1−β)μlogμ(λn)≤C(μ)(1−β)μλμn, |
where we note that 0<log(y)≤y for any y>1, which implies that
|λn−λβn|γ′=λγ′n(1−λβ−1n)γ′≤C(μ,γ′)(1−β)μγ′λμγ′+γ′n. | (6.6) |
Combining (6.5) and (6.6), we derive that
|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|≤C(μ,γ′,ε′)α−ε′t−αε′(1+mλn)ε′−γ′(1−β)μγ′λ−βε′nλμγ′+γ′n. |
If we make the assumption ε′≤γ′, then (1+mλn)ε′−γ′≤1, which allows us to obtain that
N1,1≤|C(μ,γ′,ε′,α)|2t−2αε′(1−β)2μγ′∑λn>1λ2r+2μγ′+2γ′−2βε′n⟨u0,en⟩2≤|C(μ,γ′,ε′,α)|2t−2αε′(1−β)2μγ′‖u0‖2Hr+μγ′+γ′−βε′(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)|2t−2αk(1−β)2ε‖u0‖2Hr+k−βk+ε(Ω). | (6.7) |
Before mention to N1,2, we provide a set ¯N={n∈N:λn≤1}. Let us give the observation that if ¯N is an empty set, then N1,2=0. If ¯N is a non-empty set, then λ1≤1. For the term N1,2, we note that λβn1+mλn>λn1+mλn since λn≤1 and 0<β<1. Hence, we have that
exp(−λβn1+mλntαα)<exp(−λn1+mλntαα). |
By using the fact that
|e−c−e−d|≤Cγ1max(e−c,e−d)|c−d|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)|λn−λβn1+mλn|γ1≤C(γ1,ε1)(λn1+mλntαα)−ε1|λn−λβn1+mλn|γ1=C(γ1,ε1)α−ε1t−αε1(1+mλn)ε1−γ1λ−ε1n|λn−λβn|γ1. | (6.8) |
Since λ1≤λn≤1, it is obvious to see that
|λn−λβn|γ1=(λβn−λn)γ1=λβγ1n|1−λ1−βn|γ1≤λβγ1n|1−λ1−β1|γ1. | (6.9) |
Combining (6.8) and (6.9), we find that
|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|≤C(γ1,ε1,α)|1−λ1−β1|γ1t−αε1(1+mλn)ε1−γ1λ−ε1+βγ1n. | (6.10) |
Hence, it follows from (6.10) that
N1,2=∑λn≤1λ2rn|exp(−λβn1+mλntαα)−exp(−λn1+mλntαα)|2⟨u0,en⟩2≤|C(γ1,ε1,α)|2|1−λ1−β1|2γ1t−2αε1∞∑n=1(1+mλn)2ε1−2γ1λ2r−2ε1+2βγ1n⟨u0,en⟩2. |
Let ε1=k and γ1=2k+εβ−k. Since 0<β<1, we know that
ε1<γ1,2r−2ε1+2βγ1=2r+2k−2βk+2ε. |
Hence, in view of Parseval's equality, we get the following estimate
N1,2≤C(k,r,ε,α)|1−λ1−β1|4k−2βk+2εβt−2αk‖u0‖2Hr+k−βk+ε(Ω). | (6.11) |
Combining (6.4), (6.7) and (6.11), we obtain the following estimate
‖N1‖2Hr(Ω)=N1,1+N1,2≤C(k,r,ε,α)[|1−λ1−β1|4k−2βk+2εβ+(1−β)2ε]t−2αk‖u0‖2Hr+k−βk+ε(Ω). |
By taking the square root of both sides of the above expression and using the inequality √a+b≤√a+√b for any a,b≥0, we get
‖N1‖Hr(Ω)≤C(k,r,ε,α)Dβ(ε,k)t−αk‖u0‖Hr+k−βk+ε(Ω). | (6.12) |
Here we denote
Dβ(ε,k)=|1−λ1−β1|2k−βk+εβ+(1−β)ε,ε>0, |
where we observe that Dβ(ε,k)→0, β→1−.
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))−1−PmSm,α,1(t)(Sm,α,1(ν))−1]ψ‖Hr(Ω)≤C(k,r,ε0,α)[|1−λ1−β1|2k−βk+ε0β+(1−β)ε0](tα−να)−k‖ψ‖Hr+k−βk+ε0(Ω). |
Since s>r+k−βk, we know that ε0=s−r−k+βk>0, and using global Lipschitz property of G, we derive that
‖[PmSm,α,β(t)(Sm,α,β(ν))−1−PmSm,α,1(t)(Sm,α,1(ν))−1]G(u∗∗m,α(ν))‖Hr(Ω)≤¯C1Eβ(r,s,k)(tα−να)−k‖G(u∗∗m,α(ν))‖Hs(Ω)≤Kg¯C1Eβ(r,s,k)(tα−να)−k‖u∗∗m,α(ν)‖Hr(Ω), |
where ¯C1=C(k,r,s,β,α), and we have the following observation
Eβ(r,s,k)=|1−λ1−β1|k+s−rβ+(1−β)s−r−k+βk→0,β→1−. |
In view of (4.2), we obtain that the following upper bound
‖u∗∗m,α(ν)‖Hr(Ω)≤¯C2ν−b‖u0‖Hr+k−βk(Ω), |
where ¯C2=C(k,α,m,μ0,T,b). From two latter estimations as above, we infer that
‖N2‖Hr(Ω)≤∫t0να−1‖[PmSm,α,β(t)(Sm,α,β(ν))−1−PmSm,α,1(t)(Sm,α,1(ν))−1]G(u∗∗m,α(ν))‖Hr(Ω)dν≤¯C3Eβ(r,s,k)‖u0‖Hr+k−βk(Ω)∫t0να−1−b(tα−να)−kdν, | (6.13) |
where ¯C3=Kg¯C1¯C2. Using (4.23), we obtain that the following inequality
∫t0να−1−b(tα−να)−kdν≤tα−b−kα√α(1−2k)(α−2b). | (6.14) |
Combining (6.13) and (6.14), we obtain that
‖N2‖Hr(Ω)≤¯C3tα−b−kα√α(1−2k)(α−2b)Eβ(r,s,k)‖u0‖Hr+k−βk(Ω). | (6.15) |
Step 3. Estimate of N3. By a similar argument as in (4.6), we find that
‖∫t0να−1PmSm,α,β(t)(Sm,α,β(ν))−1[G(um,α,β(ν))−G(u∗∗m,α(ν))]dν‖Hr(Ω))≤Ckαkm−k∫t0να−1(tα−να)−k‖G(um,α,β(ν))−G(u∗∗m,α(ν))‖Hr+k−βk(Ω))dν. |
Since s>r+k−βk and using global Lipschitz property of G, we obtain that
‖G(um,α,β(ν))−G(u∗∗m,α(ν))‖Hr+k−βk(Ω))≤‖G(um,α,β(ν))−G(u∗∗m,α(ν))‖Hs(Ω))≤Kg‖um,α,β(ν)−u∗∗m,α(ν)‖Hr(Ω)). |
From the two above observations, we confirm the following statement
‖N3‖Hr(Ω)≤¯C3∫t0να−1(tα−να)−k‖um,α,β(ν)−u∗∗m,α(ν)‖Hr(Ω))dν, | (6.16) |
where ¯C3=Ckαkm−kKg. Combining (6.3), (6.12), (6.15) and (6.16), we deduce the following estimate
‖um,α,β(t)−u∗∗m,α(t)‖Hr(Ω)≤‖N1‖Hr(Ω)+‖N2‖Hr(Ω)+‖N3‖Hr(Ω)≤˜Ct−αkDβ(ε,k)‖u0‖Hr+k−βk+ε(Ω)+¯C3tα−b−kα√α(1−2k)(α−2b)Eβ(r,s,k)‖u0‖Hr+k−βk(Ω)+¯C3∫t0να−1(tα−να)−k‖um,α,β(ν)−u∗∗m,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
(∫t0να−1(tα−να)−k‖um,α,β(ν)−u∗∗m,α(ν)‖Hr(Ω))dν)2≤(∫t0να−1(tα−να)−2kdν)(∫t0να−1‖um,α,β(ν)−u∗∗m,α(ν)‖2Hr(Ω))dν)≤tα(1−2k)α(1−2k)(∫t0να−1‖um,α,β(ν)−u∗∗m,α(ν)‖2Hr(Ω))dν). | (6.18) |
Combining (6.17), (6.18) and the inequality (a+b+c)2≤3a2+3b2+3c2, we derive that
‖um,α,β(t)−u∗∗m,α(t)‖2Hr(Ω)≤3|˜C|2t−2αk|Dβ(ε,k)|2‖u0‖2Hr+k−βk+ε(Ω)+3|¯C3|2t2α−2b−2αkα(1−2k)(α−2b)|Eβ(r,s,k)|2‖u0‖2Hr+k−βk(Ω)+|¯C3|2tα(1−2k)α(1−2k)∫t0να−1‖um,α,β(ν)−u∗∗m,α(ν)‖2Hr(Ω))dν. | (6.19) |
Multiplying both sides of (6.19) by t2αk, we get the following estimate
t2αk‖um,α,β(t)−u∗∗m,α(t)‖2Hr(Ω)≤3|˜C|2|Dβ(ε,k)|2‖u0‖2Hr+k−βk+ε(Ω)+3|¯C3|2t2α−2bα(1−2k)(α−2b)|Eβ(r,s,k)|2‖u0‖2Hr+k−βk(Ω)+|¯C3|2tαα(1−2k)∫t0να−1‖um,α,β(ν)−u∗∗m,α(ν)‖2Hr(Ω))dν, |
which implies that
t2αk‖um,α,β(t)−u∗∗m,α(t)‖2Hr(Ω)≤3|˜C|2|Dβ(ε,k)|2‖u0‖2Hr+k−βk+ε(Ω)+3|¯C3|2T2α−2bα(1−2k)(α−2b)|Eβ(r,s,k)|2‖u0‖2Hr+k−βk(Ω)+|¯C3|2Tαα(1−2k)∫t0να−1−2αkν2αk‖um,α,β(ν)−u∗∗m,α(ν)‖2Hr(Ω))dν. | (6.20) |
Looking Lemma 5.1 and (6.20), we set v(t)=t2αk‖um,α,β(t)−u∗∗m,α(t)‖2Hr(Ω),
¯A=3|˜C|2|Dβ(ε,k)|2‖u0‖2Hr+k−βk+ε(Ω)+3|¯C3|2T2α−2bα(1−2k)(α−2b)|Eβ(r,s,k)|2‖u0‖2Hr+k−βk(Ω) |
and
¯B=|¯C3|2Tαα(1−2k),β′=1,γ′=α−2αk. |
By applying Lemma 5.1, we obtain that
t2αk‖um,α,β(t)−u∗∗m,α(t)‖2Hr(Ω)≤¯AE1,α−2αk(¯B(Γ(β′))1β′+γ′−1t)=¯AE1,α−2αk(¯Bt). | (6.21) |
In view of Lemma 3.1 as in [30], we obtain the following upper bound
E1,α−2αk(¯Bt)≤Cα, | (6.22) |
where Cα is a positive constant that depends on α. By (6.21) and (6.22), we get
tαk‖um,α,β(t)−u∗∗m,α(t)‖Hr(Ω)≤¯ACα. |
From Deinition 2.3, we find
‖um,α,β−u∗∗m,α‖Xαk(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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4. | Hassan Eltayeb, On the Multi-Dimensional Sumudu-Generalized Laplace Decomposition Method and Generalized Pseudo-Parabolic Equations, 2024, 13, 2075-1680, 91, 10.3390/axioms13020091 | |
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8. | Andreas Chatziafratis, Tohru Ozawa, New instability, blow-up and break-down effects for Sobolev-type evolution PDE: asymptotic analysis for a celebrated pseudo-parabolic model on the quarter-plane, 2024, 5, 2662-2963, 10.1007/s42985-024-00296-w | |
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