# Final Value Problem for Parabolic Equation with Fractional Laplacian and Kirchhoff’s Term

Authors: Nguyen Hoang Luc; Devendra Kumar; Le Dinh Long; Ho Thi Kim Van
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study a diffusion equation of the Kirchhoff type with a conformable fractional derivative. The global existence and uniqueness of mild solutions are established. Some regularity results for the mild solution are also derived. The main tools for analysis in this paper are the Banach fixed point theory and Sobolev embeddings. In addition, to investigate the regularity, we also further study the nonwell-posed and give the regularized methods to get the correct approximate solution. With reasonable and appropriate input conditions, we can prove that the error between the regularized solution and the search solution is towards zero whenδ tends to zero.


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## Body

## 1. Introduction

The aim of this study is to investigate the final value for the space fractional diffusion equation(1)∂Cα∂tαvx,t+∇vL2−Δβvx,t=Fx,t,x∈Ω,t∈0,T,vx,t=0,x∈∂Ω,t∈0,T,vx,T=fx,x∈Ω,where the symbol ∂Cαvt/∂tα is called the conformable derivative which is defined clearly in Section 2. Here, Ω⊂ℝdd≥1 is a bounded domain with the smooth boundary ∂Ω, and T>0 is a given positive number. The function F represents the external forces or the advection term of a diffusion phenomenon, etc., and the function f is the final datum which will be specified later.The applications of the conformable derivative are interested in various models such as the harmonic oscillator, the damped oscillator, and the forced oscillator (see, e.g., [1]), electrical circuits (see, e.g., [2]), chaotic systems in dynamics (see, e.g., [3]), and quantum mechanics (see, e.g., [4]). From the paper, see, e.g., [5], we must confirm that the study of the ODE problem with the conformable derivative is very different from the study of the PDE problem with a conformable derivative. Results and research methods of the well-posedness for the ODE and PDE model are not the same and are completely different. The following two remarks confirm what we have just pointed out.Remark 1.
Let us first discuss conformable ODEs. Letv be the functions whose domain of its value is ℝ. If α=1, ∂Cα/∂tα becomes the classical derivative. If 0<α<1, by the paper of [6], we know that the relation between the conformable derivative and the classical derivative by the following lemma.Lemma 2.
Ifv:0,T⟶ℝ, then a conformable derivative of order α at s>0 of v exists if and only if it is differentiable at s, and the following equality is true:(2)∂Cαvs∂sα=s1−α∂vs∂s.Remark 3.
In the following, we mention the PDEs with conformable derivative whereD is a Sobolev space, such as L2Ω, Wγ,pΩ, and DAγ. When we study the PDE model, we often do with a multivariable function v:0,T⟶D, where D is a Sobolev space. This means that, for each t, vt can take on values in many classes of spaces with D1↪D↪D2⋯. Some illustrated examples given in [5] say that (2) may be not true on Sobolev spaces.Let us mention some recent works on diffusion equations with a conformable derivative, for example, [2, 5, 7–16]. Some interesting papers on fractional diffusion equations can be found in [17–24] and the references therein.Whenα=1, the main equation of Problem (1) appears in many population dynamics. By the work of Chipot and Lovat [25], we know that the diffusion coefficient B is dependent on the entire population in the domain instead of local density; that is, the moves are guided by considering the global state of the vehicle. The function u is a descriptive population density (e.g., bacteria) spread. According to article [26], we find that model (1) is a type of Kirchhoff equation, arising in vibration theory; see, for example, [27].(i)
This paper is the first study on the final value problem for a diffusion equation with a Kirchhoff-type equation and conformable derivative. Since our models are nonlinear, in order to establish the existence and uniqueness of solutions, we have to use the Banach contracting mapping theorem combined with some techniques to evaluate inequality and some Sobolev embeddings. One of the most difficult points is finding the appropriate functional spaces for the solution(ii)
The second result is to investigating the regularized solution for our problem. We show the ill-posedness of the problem and give Fourier regularization. The most difficult thing that we have to overcome is finding the appropriate space, to prove that the regularized solution converges with the exact solutionIt can be said that our article is one of the first results, giving a general and comprehensive picture, considering both the frequency and the inaccuracy of Kirchhoff’s diffusion equation with fractional time and space derivative. Using complex and interoperable assessment techniques, we find the right keys and tools to achieve both of our goals.This paper is organized as follows. In Section3, we present the existence of the backward Problem (1) with the simple case F=0. In the appropriate terms of the terminal data f, we show that the mild solution of (1) in the case β<1 converges with the mild solution of the same problem in the case β=1 when β⟶1−. Finally, in Section 4, we consider a backward problem with an inhomogeneous source term. The first part of this section discusses the existence of a mild solution under the appropriate conditions of the source function F. Furthermore, we also give an example, which shows that the problem is not stable, and then look for the approximate solution. Using the Fourier truncation method, we involve the regularized solution. Convergence error between the regularized solution and the correct solution has also been established, with some suitable conditions of input value data.


## 2. Preliminaries

### 2.1. Conformable Derivative Model

Let the functionv:0,∞⟶D, where D is a Banach space.If for eacht>0, the limitation(3)∂Cαvt∂tα≔limε⟶0vt+εt1−α−vtεinD,finitely exists, then it is called the conformable derivative of order α∈0,1 of v. We can refer the reader to [6, 8, 14, 28, 29].We introduce fractional powers ofA as follows:(4)DAv=g∈L2Ω:∑j=1∞g,wj2λj2v<∞.The spaceDAν is a Banach space in the following with the corresponding norm:(5)gDAv≔∑j=1∞g,wj2λj2v1/2,g∈DAv.The information for negative fractional powerA−ν can be provided by [30]. For any θ>0, we introduce the following Hölder continuous space of exponent θ(6)Cθ0,T;B=v∈C0,T;B:sup0≤s<t≤Tv.,t−v.,sBt−sθ<∞,corresponding to the following norm:(7)vCθ0,T:B=sup0≤s<t≤Tv.,t−v.,sBt−sθ.For any0<θ<1, let us introduce the following space:(8)Cθ0,T;B=v∈C0,T;L2Ω:sup0<t≤Ttθv.,tB<∞,corresponding to the norm vCθ0,T;B≔sup0<t≤Ttθv.,tB.Let us define the space as follows:(9)Xβ,αΩ=g∈L2Ω,∑j=1∞λj2+2βexp2TαM1λjβαg,wj2<∞.


## 2.1. Conformable Derivative Model

Let the functionv:0,∞⟶D, where D is a Banach space.If for eacht>0, the limitation(3)∂Cαvt∂tα≔limε⟶0vt+εt1−α−vtεinD,finitely exists, then it is called the conformable derivative of order α∈0,1 of v. We can refer the reader to [6, 8, 14, 28, 29].We introduce fractional powers ofA as follows:(4)DAv=g∈L2Ω:∑j=1∞g,wj2λj2v<∞.The spaceDAν is a Banach space in the following with the corresponding norm:(5)gDAv≔∑j=1∞g,wj2λj2v1/2,g∈DAv.The information for negative fractional powerA−ν can be provided by [30]. For any θ>0, we introduce the following Hölder continuous space of exponent θ(6)Cθ0,T;B=v∈C0,T;B:sup0≤s<t≤Tv.,t−v.,sBt−sθ<∞,corresponding to the following norm:(7)vCθ0,T:B=sup0≤s<t≤Tv.,t−v.,sBt−sθ.For any0<θ<1, let us introduce the following space:(8)Cθ0,T;B=v∈C0,T;L2Ω:sup0<t≤Ttθv.,tB<∞,corresponding to the norm vCθ0,T;B≔sup0<t≤Ttθv.,tB.Let us define the space as follows:(9)Xβ,αΩ=g∈L2Ω,∑j=1∞λj2+2βexp2TαM1λjβαg,wj2<∞.


## 3. Backward Problem for Homogeneous Case

In this section, we consider the final value problem for the homogeneous equation with a space fractional derivative as follows:(10)∂Cα∂tαvx,t+B∇vL2−Δβvx,t=0,x∈Ω,t∈0,T,vx,t=0,x∈∂Ω,t∈0,T,vx,T=fx,x∈Ω,where 0<M0≤Bξ≤M1 and ξ∈0,T. The following theorem states the existence and uniqueness of the solution of Problem (10).Theorem 4.
Letf∈Xβ,αΩ. Then, Problem (10) has a unique mild solution v∈C0,T;H1Ω which satisfies that(11)vx,t=∑j=1∞expλjβ∫tα/αTα/αB∇v.,sL2dsf,wjwjx.Furthermore, this solution is not stable in theL2 norm.Proof.
We express a mild solution of (10) by Fourier series as follows:(12)vx,t=∑j=1∞v.,t,wjwjx.
It follows from Problem (10) and the equality −Δβv.,t,wj=λjβv.,t,wj that(13)∂ℓα∂tαv.,t,wj+λjβB∇vL2v.,t,wj=0,t∈0,T,u.,0,wj=u0,wj.
Note that this formula(14)∂ℓα∂tαu.,t,wj=t1−α∂∂tu.,t,wj,is correct; we get that(15)∂∂tv.,t,wj+λjβtα−1B∇vL2v.,t,wj=0.
Multiply both sides of equation (15) by the quantity exp∫0tα/αλjβB∇v.,sL2ds, we reach the following assertion:(16)∂∂tv.,t,wjexp∫0tα/αλjβBv.,sL2ds=0,where we have used the fact that(17)∂∂texp∫0tα/αλjβB∇v.,sL2ds=exp∫0tα/αλjβB∇v.,sL2dstα−1λjβB∇v.,tL2.
Integrating the two sides of the latter equation 0 tot, we obtain the following confirmation:(18)v.,t,wjexp∫0tα/αλjβB∇v.,sL2ds=v.,0,wj.
It yields that(19)v.,t,wj=exp−∫0tα/αλjβB∇v.,sL2dsv.,0,wj.
Therefore, we find that(20)vx,t=∑j=1∞expλjβ∫tα/αTα/αB∇v.,sL2dsf,wjwjx.
Forv∈L∞0,T;H1Ω, we consider the following function:(21)Qvx,t=∑j=1∞expλjβ∫tα/αTα/αB∇v.,sL2dsf,wjwjx.
We shall prove by induction ifw1,w2∈L∞0,T;H1Ω, then(22)Qkw1−Qkw2L2Ω≤fXβ,αΩ2Kb2Tα−tα/α1/2k!w1−w2L∞0,T;L2Ω∀q≤1.
Form=1, using the inequality ea−eb≤a−bmaxea,eb for any a,b∈ℝ, we have(23)Qw1−Qw2H1Ω2=∑j=1∞λj2+2βexpλjβ∫tα/αTα/αB∇w1.,sL2ds−expλjβ∫tα/αTα/αB∇w2.,sL2ds2f,wj2≤∑j=1∞λj2−2β∫tα/αTα/αB∇w1.,sL2ds−∫tα/αTα/αB∇w2.,sL2ds2exp2Tα−2tααM1λjf,wj2≤Kb2fXβ,αΩ2∫tα/αTα/α∇w1−w2.,sL2Ω2ds≤Kb2fXβ,αΩ2Tα−tααw1−w2L∞0,T;H1Ω2.
Assume that (22) holds for m=k. We show that (22) holds for m=k+1. Indeed, we have(24)Qk+1w1−Qk+1w2H1Ω2=∑j=1∞λj2expλjβ∫tα/αTα/αB∇Qkw1.,sL2ds−expλjβ∫tα/αTα/αB∇Qkw2.,sL2ds2f,wj2≤fXβ,αΩ2Kb2∫tα/αTα/αQkw1.,s−Qkw2.,sH1Ω2ds≤fXβ,αΩ2Kb2∫tα/αTα/αfXβ,αΩ2Kb2Tα−tα/αkk!ds≤fXβ,αΩ2Kb2Tα−tα/αk+1k+1!w1−w2L∞0,T;H1Ω2.
By the theory of the induction principle, (22) holds for all w1,w2∈L∞0,T;H1Ω. Since the fact that(25)limk⟶+∞fXβ,αΩ2Kb2Tα−tα/αkk!1/2=0,there exists a positive integer number k0 such that Qk0 is a contraction. It follows that the equation Qk0v=v has a unique solution v∈L∞0,T;H1Ω. It is easy to see that v is also a fixed point of Q.☐Theorem 5.
Assume thatf∈Xβ+γ,αΩ for any γ>β. Let us choose ν0 such that(26)2Kb2fXβ,αΩ2Tα1−δCδ2ν0−2δα1−δ1−δ<1.
Let any0<ε<γ−β. Then, there exists Cε>0 such that(27)vα,β−wαLν0∞0,T;H1Ω2≤Cε1−βεfXβ+γ,αΩ1−2Kb2fXβ,αΩTα1−δCδ2ν0−2δ/α1−δ1−δ,where(28)Lν0∞0,T;H1Ω=v∈L∞0,T;H1Ω,vLν0∞0,T;H1Ω=sup0≤t≤Teν0t−TvtH1Ω<∞.Proof.
Letvα,β be the solution of Problem (11). Let wα be the solution to Problem (11) with β=1. Then, we get(29)vα,βx,t=∑j=1∞expλjβ∫tα/αTα/αB∇vα,β.,sL2dsf,wjwjx,wαx,t=∑j=1∞expλj∫tα/αTα/αB∇wα.,sL2dsf,wjwjx.
We have(30)vα,βx,t−wαx,t=∑j=1∞expλjβ∫tα/αTα/αB∇vα,β.,sL2dsf,wjwjx−∑j=1∞expλjβ∫tα/αTα/αB∇wα.,sL2dsf,wjwjx+∑j=1∞expλjβ∫tα/αTα/αB∇wα.,sL2dsf,wjwjx−∑j=1∞expλj∫tα/αTα/αB∇wα.,sL2dsf,wjwjx=D1+D2,whereby(31)D1=∑j=1∞expλjβ∫tα/αTα/αB∇wα.,sL2dsf,wjwjx,D2=∑j=1∞expλj∫tα/αTα/αB∇wα.,sL2dsf,wjwjx.
The termD1 is bounded by(32)D1H1Ω2≤∑j=1∞λj2β+2exp2TαM1λjβα∫tα/αTα/αB∇vα,β.,sL2ds−∫tα/αTα/αB∇wα.,sL2ds2f,wj2≤Kb2fXβ,αΩ2∫tα/αTα/α∇vα,β−wα·,sL2Ω2ds.
The termD2 is estimated as follows:(33)D2H1Ω2≤∑j=1∞λj2β+2exp2TαM1λjβαλjβ−λj2∫tα/αTα/αB∇wα.,sL2ds2f,wj2≤M12Tα−tαα2∑j=1∞λj2β+2exp2TαM1λjβαλjβ−λj2f,wj2.
Consider the following subset:(34)A1=j∈ℕ,λj≤1,A2=j∈ℕ,λj>1.
Ifj∈A1, then using the inequality 1−e−z≤Cεzε, we get(35)λjβ−λj=λjβ1−λj1−β=λjβCε1−βελj−ε=Cε1−βελjβ−ε,which allows us to obtain(36)∑j∈A1λj2βexp2TαM1λjβαλjβ−λj2f,wj2≤Cε21−β2ε∑j∈A1λj4β+2−2εexp2TαM1λjβαf,wj2.
Ifj∈A2, then using the inequality 1−e−z≤Cεzε, we find(37)λjβ−λj=λj1−λjβ−1=λjCε1−βελjε=Cε1−βελjβ+ε.
Hence, we obtain(38)∑j∈A2λj2β+2exp2TαM1λjβαλjβ−λj2f,wj2≤Cε21−β2ε∑j∈A1λj4β+2+2εexp2TαM1λjβαf,wj2.
Combining (36) and (38), we find that(39)λj2βexp2TαM1λjβαλjβ−λj2f,wj2≤CCε21−β2ε∑j∈A1λj4β+2ε+2exp2TαM1λjβαf,wj2.
Let us choose0<ε<γ−β. Then, we follow from (33) and the latter equality that(40)D22≤CM12Tα−tαα2Cε21−β2ε∑j∈A1λj4β+2ε+2exp2TαM1λjβαf,wj2≤CM12Tα−tαα2Cε21−β2ε∑j∈A1λj2β+2γ2exp2TαM1λjβαf,wj2=CM12Tα−tαα2Cε21−β2εfXβ+γ,αΩ2.
This above inequality together with (32) and (3) yields that(41)e2νt−Tvα,β.,t−wα.,tH1Ω2≤2e2νt−TD1H1Ω2+2e2νt−TD2H1Ω2≤2Kb2fXβ,αΩ2e2νt−T∫tα/αTα/αvα,β.,s−wα.,sH1Ω2ds+2CM12e2νt−TTα−tαα2Cε21−β2εfXβ+γ,αΩ2.
It is easy to get that(42)e2νt−T∫tα/αTα/αvα,β.,s−wα.,sH1Ω2ds=∫tα/αTα/αe2νt−se2νs−Tvα,β.,s−wα.,sH1Ω2ds≤∫tα/αTα/αe2νt−sdsvα,β−wαLν∞0,T;H1Ω2.It follows from the inequality(43)e−z≤Cδz−δ,0<δ<1,we get(44)∫tα/αTα/αe2νt−sds≤Cδ2ν−2δ∫tα/αTα/αs−t−δds.
The inequalitya+bα≤aα+bα leads to(45)∫tα/αTα/αs−t−δds=Tα/α−t1−δ−tα/α−t1−δ1−δ≤Tα/α−tα/α1−δ1−δ≤Tα1−δα1−δ1−δ,which allows us to get immediately that(46)∫tα/αTα/αe2νt−sds≤Tα1−δCδ2ν−2δα1−δ1−δ·
It follows from (41) and (42) that for any t∈0,T(47)e2νt−Tvα,β.,t−wα.,tH1Ω2≤2Kb2fXβ,αΩ2Tα1−δCδ2ν−2δα1−δ1−δvα,β−wαLν∞0,T;H1Ω2+2CM12Tαα2Cε21−β2εfXβ+γ,αΩ2.
Since the right-hand side of (47) is independent of t, we deduce that(48)vα,β−wαLν∞0,T;H1Ω2≤2Kb2fXβ,αΩ2Tα1−δCδ2ν−2δα1−δ1−δvα,β−wαLν∞0,T;H1Ω2+2CM12Tαα2Cε21−β2εfXβ+γ,αΩ2.
Then, we find that(49)vα,β−wαLν0∞0,T;H1Ω2≤2CM12Tα/α2Cε21−β2εfXβ+γ,αΩ21−2Kb2fXβ,αΩ2Tα1−δCδ2ν0−2δ/α1−δ1−δ.
☐


## 4. Backward Problem for Inhomogeneous Case

In this section, we consider the final value problem for homogeneous equation as follows:(50)∂ℓα∂tαux,t+B∇uL2−Δβux,t=Fx,t,x∈Ω,t∈0,T,ux,t=0,x∈∂Ω,t∈0,T,ux,T=0,x∈Ω,where F is defined later.


### 4.1. Existence and Uniqueness of the Mild Solution

In this subsection, we state the existence and uniqueness of the mild solution. In order to give the main results, we require the conditionF which belongs to the space L2p0,T;Xβ,αΩ.Theorem 6.
Let0<β≤1 and F be the source function that belongs to L2p0,T;Xβ,αΩ for any 1<p<1/1−α. Let B be the functions which satisfy M0≤Bz≤M1, z∈0,T and(51)Bz1−Bz2≤Kbz1−z2,z1,z2∈ℝ.
Then, Problem (50) has a unique mild solution u∈Lμ0∞0,T;H1Ω, where μ0 is small enough. The function u satisfies that(52)ux,t=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjdswj.
Furthermore, this solution is not stable in theL2 norm.Proof.
By a simple calculation, we get the following equality:(53)v.,t,wj=exp−∫0tα/αλjβB∇v.,sL2dsu.,0,wj+∫0tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds.
By lettingt=T and noting that vx,T=0, we find that(54)exp−∫0Tα/αλjβB∇v.,sL2dsu.,0,wj+∫0Tsα−1exp−∫sα/αTα/αλjβB∇v.,rL2drF.,s,wjds=0.
Therefore, we obtain(55)u.,0,wj=−∫0Tsα−1exp∫0sα/αλjβB∇v.,rL2drF.,s,wjds.
Combining (53) and (55), we deduce that(56)v.,t,wj=∫0tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds−∫0Tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds=−∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drF.,s,wjds.
Let us denote byLμ∞0,T;V the functional subspace of L∞0,T;V corresponding to the norm(57)gLμ∞0,T;V≔max0≤t≤Texpμt−Tg.,tV,∀g∈Lμ∞0,T;V,where(58)V=μ∈ℝ,g∈L2Ω,∑j=1∞expμt−Tg,wj2<∞·
Set the following function:(59)Pvt=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drF.,s,wjdswj,and we let(60)Ms,t,j,w=exp∫tα/αsα/αλjβB∇w.,rL2dr.
So, using Parseval’s equality, we get that(61)Pw1−Pw2H1Ω2=∑j=1∞λj∫tTsα−1Ms,t,j,w1−Ms,t,j,w2F.,s,wjds2≤∑j=1∞λj∫tTsα−1ds∫0tsα−1Ms,t,j,w1−Ms,t,j,w22F.,s,wj2ds.
Using the inequalityea−eb≤a−bmaxea,eb, we continue to treat the term Ms,t,j,w1−Ms,t,j,w2 as follows:(62)Ms,t,j,w1−Ms,t,j,w2=exp∫tα/αsα/αλjβB∇w1.,rL2dr−exp∫tα/αsα/αλjβB∇w2.,rL2dr≤expTαM1λjβαKbλjβ∫tα/αsα/αu.,r−v.,rH1dr.
Therefore, applying the Hölder inequality, we get(63)Ms,t,j,v1−Ms,t,j,v22≤exp2TαM1λjβαKb2λj2βT2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds.
Inserting (61) and (63) yields the following inequality:(64)exp2μt−TPv1−Pv2H1Ω2≤Kb2T3αα3∫tα/αsα/αe2μt−se2μs−Tv1.,s−v2.,sL2ds×∫tα/αsα/αsα−1∑j=1∞λj2β+2exp2TαM1λjβαF.,s,wj2ds≤Kb2T3α3α∫tα/αsα/αexp2μt−sds×max0≤s≤Texp2μs−Tv1.,s−v2.,sH1Ω2∫0tsα−1F.,sXβ,αΩ2ds.
Take anyδ∈0,1. By a similar explanation as (46), we find that(65)∫tα/αsα/αexp2μt−sds≤Tα1−δCδ2μ−2δα1−δ1−δ.
By applying the Hölder inequality, we also obtain that(66)∫0tsα−1F.,sXβ,αΩ2ds≤∫0tsα−1p∗ds1/p∗∫0tF.,sXβ,αΩ2pds1/p≤p−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2.
From some observations as above, we deduce that(67)exp2μt−TPv1−Pv2H1Ω2≤Kb2T3α3αTα1−δCδ2μ−2δα1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2v1−v2Lμ∞0,T;H1Ω2.
Since the right-hand side of the latter estimate is independent oft, we find that(68)Pv1−Pv2Lμ∞0,T;H1Ω2≤Kb2T3α3αTα1−δCδ2μ−2δα1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2v1−v2Lμ∞0,T;H1Ω2.
Let us chooseμ0 such that(69)μ02δ>Kb2T3α3αTα1−δCδ2α1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2.
Then, we can conclude thatP is a contraction mapping in the space Lμ0∞0,T;H1Ω. Next, we continue to show that if v∈Lμ0∞0,T;H1Ω, then Pv∈Lμ0∞0,T;H1Ω. If v1=0, then(70)Pv1t=∑j=1∞∫tTsα−1F.,s,wjdswj.
Hence, from Parseval’s equality, we find that(71)Pv1H1Ω2=∑j=1∞λj2∫tTsα−1F.,s,wjds2≤∫tTsα−1ds∑j=1∞∫tTsα−1λj2F.,s,wj2ds≤Tαα∫tTsα−1F.,sH1Ω2ds≤Tαα∫0tsα−1p∗ds1/p∗∫0tF.,sH1Ω2pds1/p≤p−1αp−1Tαp−1/p−1FL2p0,T;H1Ω2.
This says thatPv1 belongs to the space Lμ0∞0,T;H1Ω. Using (68), we arrive at the confirmation that Pv belongs to Lμ0∞0,T;H1Ω if v∈Lμ0∞0,T;H1Ω. For any m∈ℕ, let um be the function that satisfies the following integral equation:(72)umx,t=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇um.,rL2drF.,s,wjdswjx.
Let us assume that(73)Fmx,t=1λm∑j=1∞wjx.
It is not difficult to verify thatFm∈L∞0,T;Xβ,αΩ, so we get that Fm∈L2p0,T;Xβ,αΩ.
Using Theorem6, we conclude that equation (72) has a unique solution um∈L∞0,T;H1Ω. By the fact that Bz≥M0∀z∈ℝ, we obtain the following estimate:(74)um·,tL2Ω2=1λm2∫tTsα−1exp∫tα/αsα/αλjβB∇um.,rL2dr2≥∫tTsα−1expM0λjβsα−tα/αds2λm2.
The estimate is true for allt∈0,T, so it is easy to see that(75)umL∞0,T;L2Ω≥∫0Tsα−1expM0λjβsα/αdsλm≥expλmβM0Tα/αM0λmβ+1.
Whenm tends to +∞, we can check that fmL2Ω=1/λm go to zero when m⟶+∞ and(76)limm⟶+∞umC0,T;L2Ω=limm⟶+∞expλmβM0Tα/αM0λmβ+1=+∞.
This shows that Problem (50) is ill-posed in the sense of Hadamard in the L2-norm.☐


### 4.2. Fourier Truncation Method

In this section, we will provide a regularized solution and solve the problem by the Fourier truncation method as follows:(77)uN,δx,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇uN,δ.,rL2drFδ.,s,wjdswjx.Here,N≔Nδ goes to infinity as δ tends to zero which is called a parameter regularization. The function F is disturbed by the observed data Fδ∈L∞0,T;L2Ω provided by(78)Fδ−FL∞0,T;L2Ω≤δ.The main results of this subsection are given by the theorem below.Theorem 7.
Letν>0 such that F belongs to the space L∞0,T;Xβ+ν,αΩ. Let Fδ be as above. Let us assume that Problem (50) has a unique mild solution u∈L∞0,T;DAν+θ for θ>0. Let us choose N such that(79)limδ⟶0λNνexpTαM1λNβαδ=0,limδ⟶0λN=+∞.
Hereν≥1/2. Then, there exists a positive μ¯ large enough such that Problem has a unique solution vN,δ∈Lμ¯∞0,T;DAν. Moreover, we have the following estimate:(80)vN,δ−uLμ¯∞0,T;DAν2≤6T2αα2λN2νexp2TαM1λNβαδ2+6λN−2θuL∞0,T;DAν+θ2.Remark 8.
SinceλN~N2/d, we can choose a natural number N such that(81)λN=α1−blog1/bTαM11/β.Proof.
Part 1: prove that the nonlinear integral equation (77) has a unique mild solution.
Let anyv∈C0,T;H1Ω, we denote by the following function(82)Gvx,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drFδ.,s,wjdswjx.
By applying Parseval’s equality, we follow from (82) that(83)Gv1.,t−Gv2.,tDAν2=∑j=1Nλj2ν∫tTsα−1Ms,t,j,v1−Ms,t,j,v2F.,s,wjds2≤TααλN2ν∑j=1N∫0tsα−1Ms,t,j,w1−Ms,t,j,w22Fδ.,s,wj2ds.
If1≤j≤N, then we have in view of (63) that(84)Ms,t,j,v1−Ms,t,j,v22≤exp2TαM1λjβαKb2λj2βT2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds≤exp2TαM1λNβαλN2βKb2T2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds.
The above two observations (65) lead to(85)exp2μt−TGv1.,t−Gv2.,tDAν2≤Kb2T3αα3exp2TαM1λNβαλN2β+2ν∫tα/αsα/αe2μt−se2μs−Tv1.,s−v2.,sH1ds∫tα/αsα/αsα−1∑j=1∞Fδ.,s,wj2ds≤Kb2T3αα3exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δmax0≤s≤Texp2μs−Tv1.,s−v2.,sH1Ω2∫0tsα−1Fδ.,sL2Ω2ds≤Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δFδL∞0,T;L2Ω2v1−v2L∞0,T;H1Ω2.
Because the right-hand side of the latter estimate is independent oft and noting the Sobolev embedding DAν↪H1Ω, we arrive at(86)Gv1−Gv2Lμ∞0,T;DAν2≤Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δFδL∞0,T;L2Ω2v1−v2Lμ∞0,T;DAν2.
Let us chooseμ1 such that(87)Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2α1−δ1−δFδL∞0,T;L2Ω2<μ12δ.
It is easy to see thatG is a contracting mapping on the space Lμ1∞0,T;DAν. Therefore, we can conclude that there exists a uniqueness solution vN,δ for Problem (77).
Next, we continue to give the upper bound of the termvN,δ·,t−u·,tDAν2. First, we have(88)vN,δx,t−ux,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wjdswjx+∑j=1N∫tTsα−1Ms,t,j,vN,δ−Ms,t,j,uF.,s,wjdswjx+∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjdswj.
The above equality and Parseval’s equality allow us to get that(89)vN,δ·,t−u·,tDAν2≤3∑j=1Nλj2ν∫tTsα−1exp∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wjds2⏟J1+3∑j=1Nλj2ν∫tTsα−1Ms,t,j,vN,δ−Ms,t,j,uF.,s,wjds2⏟J2+3∑j=N+1∞λj2ν∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjds2⏟J3.
Since the conditionBz≤M1∀z∈ℝ and applying Hölder inequality, the quantity J1 is bounded by(90)J1≤3λN2ν∫tTsα−1∑j=1N∫tTsα−1exp2∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wj2ds≤3TααλN2νexp2TαM1λNβα∫0tsα−1Fδ.,s−FL2Ω2ds≤3TααλN2νexp2TαM1λNβα∫tTsα−1dsFδ−FL∞0,T;L2Ω2≤3T2αα2λN2νexp2TαM1λNβαδ2,where we have used the fact that Fδ−FL∞0,T;L2Ω≤δ.
The quantityJ2 is estimated as follows:(91)exp2μt−TJ2≤3exp2μt−T∫tTsα−1ds∑j=1N∫tTsα−1λj2νMs,t,j,vN,δ−Ms,t,j,u2F.,s,wj2ds.
We have in view of (63) that(92)Ms,t,j,vN,δ−Ms,t,j,u2≤exp2TαM1λjβαλj2βKb2T2αα2∫tα/αsα/αvN,δ.,r−u.,rDAν2dr.
This leads to the following estimate:(93)exp2μt-TJ2≤3Kb2T3αα3∫tα/αsα/αe2μt-s⋅e2μs-TvN,δ.,r-u.,rDAν2⋅dr∫0Tsα-1∑j=1∞λj2β+2νexp2TαM1λjβαF.,s,wj2ds≤3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1-δCδ2μ-2δα1-δ1-δ⋅vN,δ-uLμ∞0,T;DAν.
The termJ3 is estimated as follows:(94)J3=3∑j=N+1∞λj2νu.,t,wj2=3∑j=N+1∞λj−2θλj2ν+2θu.,t,wj2≤3λN−2θuL∞0,T;DAν+θ2.
Combining (89), (90), (93), and (94), we find that(95)exp2μt−TvN,δ·,t−u·,tDAν2≤exp2μt−TJ1+J2+J3≤3T2αα2λN2νexp2TαM1λNβαδ2+3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δα1−δ1−δvN,δ−uLμ∞0,T;DAν+3λN−2θuL∞0,T;DAν+θ2.
We chooseμ¯ such that both the following inequalities are satisfied:(96)3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2α1−δ1−δ≤12μ¯2δ,Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2α1−δ1−δFδL∞0,T;L2Ω2<μ¯2δ.
Some observations above give us the following confirmation:(97)vN,δ−uLμ¯∞0,T;DAν2≤3T2α/α2λN2νexp2TαM1λNβ/αδ2+3λN−2θuL∞0,T;DAν+θ21−3Kb2T4α/α4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δ/α1−δ1−δ.
Since the fact that(98)1−3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δα1−δ1−δ≥12,
We easily obtain the desired result (80).☐


## 4.1. Existence and Uniqueness of the Mild Solution

In this subsection, we state the existence and uniqueness of the mild solution. In order to give the main results, we require the conditionF which belongs to the space L2p0,T;Xβ,αΩ.Theorem 6.
Let0<β≤1 and F be the source function that belongs to L2p0,T;Xβ,αΩ for any 1<p<1/1−α. Let B be the functions which satisfy M0≤Bz≤M1, z∈0,T and(51)Bz1−Bz2≤Kbz1−z2,z1,z2∈ℝ.
Then, Problem (50) has a unique mild solution u∈Lμ0∞0,T;H1Ω, where μ0 is small enough. The function u satisfies that(52)ux,t=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjdswj.
Furthermore, this solution is not stable in theL2 norm.Proof.
By a simple calculation, we get the following equality:(53)v.,t,wj=exp−∫0tα/αλjβB∇v.,sL2dsu.,0,wj+∫0tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds.
By lettingt=T and noting that vx,T=0, we find that(54)exp−∫0Tα/αλjβB∇v.,sL2dsu.,0,wj+∫0Tsα−1exp−∫sα/αTα/αλjβB∇v.,rL2drF.,s,wjds=0.
Therefore, we obtain(55)u.,0,wj=−∫0Tsα−1exp∫0sα/αλjβB∇v.,rL2drF.,s,wjds.
Combining (53) and (55), we deduce that(56)v.,t,wj=∫0tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds−∫0Tsα−1exp−∫sα/αtα/αλjβB∇v.,rL2drF.,s,wjds=−∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drF.,s,wjds.
Let us denote byLμ∞0,T;V the functional subspace of L∞0,T;V corresponding to the norm(57)gLμ∞0,T;V≔max0≤t≤Texpμt−Tg.,tV,∀g∈Lμ∞0,T;V,where(58)V=μ∈ℝ,g∈L2Ω,∑j=1∞expμt−Tg,wj2<∞·
Set the following function:(59)Pvt=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drF.,s,wjdswj,and we let(60)Ms,t,j,w=exp∫tα/αsα/αλjβB∇w.,rL2dr.
So, using Parseval’s equality, we get that(61)Pw1−Pw2H1Ω2=∑j=1∞λj∫tTsα−1Ms,t,j,w1−Ms,t,j,w2F.,s,wjds2≤∑j=1∞λj∫tTsα−1ds∫0tsα−1Ms,t,j,w1−Ms,t,j,w22F.,s,wj2ds.
Using the inequalityea−eb≤a−bmaxea,eb, we continue to treat the term Ms,t,j,w1−Ms,t,j,w2 as follows:(62)Ms,t,j,w1−Ms,t,j,w2=exp∫tα/αsα/αλjβB∇w1.,rL2dr−exp∫tα/αsα/αλjβB∇w2.,rL2dr≤expTαM1λjβαKbλjβ∫tα/αsα/αu.,r−v.,rH1dr.
Therefore, applying the Hölder inequality, we get(63)Ms,t,j,v1−Ms,t,j,v22≤exp2TαM1λjβαKb2λj2βT2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds.
Inserting (61) and (63) yields the following inequality:(64)exp2μt−TPv1−Pv2H1Ω2≤Kb2T3αα3∫tα/αsα/αe2μt−se2μs−Tv1.,s−v2.,sL2ds×∫tα/αsα/αsα−1∑j=1∞λj2β+2exp2TαM1λjβαF.,s,wj2ds≤Kb2T3α3α∫tα/αsα/αexp2μt−sds×max0≤s≤Texp2μs−Tv1.,s−v2.,sH1Ω2∫0tsα−1F.,sXβ,αΩ2ds.
Take anyδ∈0,1. By a similar explanation as (46), we find that(65)∫tα/αsα/αexp2μt−sds≤Tα1−δCδ2μ−2δα1−δ1−δ.
By applying the Hölder inequality, we also obtain that(66)∫0tsα−1F.,sXβ,αΩ2ds≤∫0tsα−1p∗ds1/p∗∫0tF.,sXβ,αΩ2pds1/p≤p−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2.
From some observations as above, we deduce that(67)exp2μt−TPv1−Pv2H1Ω2≤Kb2T3α3αTα1−δCδ2μ−2δα1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2v1−v2Lμ∞0,T;H1Ω2.
Since the right-hand side of the latter estimate is independent oft, we find that(68)Pv1−Pv2Lμ∞0,T;H1Ω2≤Kb2T3α3αTα1−δCδ2μ−2δα1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2v1−v2Lμ∞0,T;H1Ω2.
Let us chooseμ0 such that(69)μ02δ>Kb2T3α3αTα1−δCδ2α1−δ1−δp−1αp−1Tαp−1/p−1FL2p0,T;Xβ,αΩ2.
Then, we can conclude thatP is a contraction mapping in the space Lμ0∞0,T;H1Ω. Next, we continue to show that if v∈Lμ0∞0,T;H1Ω, then Pv∈Lμ0∞0,T;H1Ω. If v1=0, then(70)Pv1t=∑j=1∞∫tTsα−1F.,s,wjdswj.
Hence, from Parseval’s equality, we find that(71)Pv1H1Ω2=∑j=1∞λj2∫tTsα−1F.,s,wjds2≤∫tTsα−1ds∑j=1∞∫tTsα−1λj2F.,s,wj2ds≤Tαα∫tTsα−1F.,sH1Ω2ds≤Tαα∫0tsα−1p∗ds1/p∗∫0tF.,sH1Ω2pds1/p≤p−1αp−1Tαp−1/p−1FL2p0,T;H1Ω2.
This says thatPv1 belongs to the space Lμ0∞0,T;H1Ω. Using (68), we arrive at the confirmation that Pv belongs to Lμ0∞0,T;H1Ω if v∈Lμ0∞0,T;H1Ω. For any m∈ℕ, let um be the function that satisfies the following integral equation:(72)umx,t=−∑j=1∞∫tTsα−1exp∫tα/αsα/αλjβB∇um.,rL2drF.,s,wjdswjx.
Let us assume that(73)Fmx,t=1λm∑j=1∞wjx.
It is not difficult to verify thatFm∈L∞0,T;Xβ,αΩ, so we get that Fm∈L2p0,T;Xβ,αΩ.
Using Theorem6, we conclude that equation (72) has a unique solution um∈L∞0,T;H1Ω. By the fact that Bz≥M0∀z∈ℝ, we obtain the following estimate:(74)um·,tL2Ω2=1λm2∫tTsα−1exp∫tα/αsα/αλjβB∇um.,rL2dr2≥∫tTsα−1expM0λjβsα−tα/αds2λm2.
The estimate is true for allt∈0,T, so it is easy to see that(75)umL∞0,T;L2Ω≥∫0Tsα−1expM0λjβsα/αdsλm≥expλmβM0Tα/αM0λmβ+1.
Whenm tends to +∞, we can check that fmL2Ω=1/λm go to zero when m⟶+∞ and(76)limm⟶+∞umC0,T;L2Ω=limm⟶+∞expλmβM0Tα/αM0λmβ+1=+∞.
This shows that Problem (50) is ill-posed in the sense of Hadamard in the L2-norm.☐


## 4.2. Fourier Truncation Method

In this section, we will provide a regularized solution and solve the problem by the Fourier truncation method as follows:(77)uN,δx,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇uN,δ.,rL2drFδ.,s,wjdswjx.Here,N≔Nδ goes to infinity as δ tends to zero which is called a parameter regularization. The function F is disturbed by the observed data Fδ∈L∞0,T;L2Ω provided by(78)Fδ−FL∞0,T;L2Ω≤δ.The main results of this subsection are given by the theorem below.Theorem 7.
Letν>0 such that F belongs to the space L∞0,T;Xβ+ν,αΩ. Let Fδ be as above. Let us assume that Problem (50) has a unique mild solution u∈L∞0,T;DAν+θ for θ>0. Let us choose N such that(79)limδ⟶0λNνexpTαM1λNβαδ=0,limδ⟶0λN=+∞.
Hereν≥1/2. Then, there exists a positive μ¯ large enough such that Problem has a unique solution vN,δ∈Lμ¯∞0,T;DAν. Moreover, we have the following estimate:(80)vN,δ−uLμ¯∞0,T;DAν2≤6T2αα2λN2νexp2TαM1λNβαδ2+6λN−2θuL∞0,T;DAν+θ2.Remark 8.
SinceλN~N2/d, we can choose a natural number N such that(81)λN=α1−blog1/bTαM11/β.Proof.
Part 1: prove that the nonlinear integral equation (77) has a unique mild solution.
Let anyv∈C0,T;H1Ω, we denote by the following function(82)Gvx,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇v.,rL2drFδ.,s,wjdswjx.
By applying Parseval’s equality, we follow from (82) that(83)Gv1.,t−Gv2.,tDAν2=∑j=1Nλj2ν∫tTsα−1Ms,t,j,v1−Ms,t,j,v2F.,s,wjds2≤TααλN2ν∑j=1N∫0tsα−1Ms,t,j,w1−Ms,t,j,w22Fδ.,s,wj2ds.
If1≤j≤N, then we have in view of (63) that(84)Ms,t,j,v1−Ms,t,j,v22≤exp2TαM1λjβαKb2λj2βT2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds≤exp2TαM1λNβαλN2βKb2T2αα2∫tα/αsα/αv1.,r−v2.,rH1Ω2ds.
The above two observations (65) lead to(85)exp2μt−TGv1.,t−Gv2.,tDAν2≤Kb2T3αα3exp2TαM1λNβαλN2β+2ν∫tα/αsα/αe2μt−se2μs−Tv1.,s−v2.,sH1ds∫tα/αsα/αsα−1∑j=1∞Fδ.,s,wj2ds≤Kb2T3αα3exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δmax0≤s≤Texp2μs−Tv1.,s−v2.,sH1Ω2∫0tsα−1Fδ.,sL2Ω2ds≤Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δFδL∞0,T;L2Ω2v1−v2L∞0,T;H1Ω2.
Because the right-hand side of the latter estimate is independent oft and noting the Sobolev embedding DAν↪H1Ω, we arrive at(86)Gv1−Gv2Lμ∞0,T;DAν2≤Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2μ−2δα1−δ1−δFδL∞0,T;L2Ω2v1−v2Lμ∞0,T;DAν2.
Let us chooseμ1 such that(87)Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2α1−δ1−δFδL∞0,T;L2Ω2<μ12δ.
It is easy to see thatG is a contracting mapping on the space Lμ1∞0,T;DAν. Therefore, we can conclude that there exists a uniqueness solution vN,δ for Problem (77).
Next, we continue to give the upper bound of the termvN,δ·,t−u·,tDAν2. First, we have(88)vN,δx,t−ux,t=−∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wjdswjx+∑j=1N∫tTsα−1Ms,t,j,vN,δ−Ms,t,j,uF.,s,wjdswjx+∑j=1N∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjdswj.
The above equality and Parseval’s equality allow us to get that(89)vN,δ·,t−u·,tDAν2≤3∑j=1Nλj2ν∫tTsα−1exp∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wjds2⏟J1+3∑j=1Nλj2ν∫tTsα−1Ms,t,j,vN,δ−Ms,t,j,uF.,s,wjds2⏟J2+3∑j=N+1∞λj2ν∫tTsα−1exp∫tα/αsα/αλjβB∇u.,rL2drF.,s,wjds2⏟J3.
Since the conditionBz≤M1∀z∈ℝ and applying Hölder inequality, the quantity J1 is bounded by(90)J1≤3λN2ν∫tTsα−1∑j=1N∫tTsα−1exp2∫tα/αsα/αλjβB∇vN,δ.,rL2drFδ.,s−F.,s,wj2ds≤3TααλN2νexp2TαM1λNβα∫0tsα−1Fδ.,s−FL2Ω2ds≤3TααλN2νexp2TαM1λNβα∫tTsα−1dsFδ−FL∞0,T;L2Ω2≤3T2αα2λN2νexp2TαM1λNβαδ2,where we have used the fact that Fδ−FL∞0,T;L2Ω≤δ.
The quantityJ2 is estimated as follows:(91)exp2μt−TJ2≤3exp2μt−T∫tTsα−1ds∑j=1N∫tTsα−1λj2νMs,t,j,vN,δ−Ms,t,j,u2F.,s,wj2ds.
We have in view of (63) that(92)Ms,t,j,vN,δ−Ms,t,j,u2≤exp2TαM1λjβαλj2βKb2T2αα2∫tα/αsα/αvN,δ.,r−u.,rDAν2dr.
This leads to the following estimate:(93)exp2μt-TJ2≤3Kb2T3αα3∫tα/αsα/αe2μt-s⋅e2μs-TvN,δ.,r-u.,rDAν2⋅dr∫0Tsα-1∑j=1∞λj2β+2νexp2TαM1λjβαF.,s,wj2ds≤3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1-δCδ2μ-2δα1-δ1-δ⋅vN,δ-uLμ∞0,T;DAν.
The termJ3 is estimated as follows:(94)J3=3∑j=N+1∞λj2νu.,t,wj2=3∑j=N+1∞λj−2θλj2ν+2θu.,t,wj2≤3λN−2θuL∞0,T;DAν+θ2.
Combining (89), (90), (93), and (94), we find that(95)exp2μt−TvN,δ·,t−u·,tDAν2≤exp2μt−TJ1+J2+J3≤3T2αα2λN2νexp2TαM1λNβαδ2+3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δα1−δ1−δvN,δ−uLμ∞0,T;DAν+3λN−2θuL∞0,T;DAν+θ2.
We chooseμ¯ such that both the following inequalities are satisfied:(96)3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2α1−δ1−δ≤12μ¯2δ,Kb2T4αα4exp2TαM1λNβαλN2β+2νTα1−δCδ2α1−δ1−δFδL∞0,T;L2Ω2<μ¯2δ.
Some observations above give us the following confirmation:(97)vN,δ−uLμ¯∞0,T;DAν2≤3T2α/α2λN2νexp2TαM1λNβ/αδ2+3λN−2θuL∞0,T;DAν+θ21−3Kb2T4α/α4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δ/α1−δ1−δ.
Since the fact that(98)1−3Kb2T4αα4FL∞0,T;Xβ+ν,αΩ2Tα1−δCδ2μ−2δα1−δ1−δ≥12,
We easily obtain the desired result (80).☐



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*Source: 7238678-2021-07-03.xml*