# Remarks on the Initial and Terminal Value Problem for Time and Space Fractional Diffusion Equation

Authors: Hoang Luc Nguyen
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The fractional problem for partial differential equation has many applications in science and technology. The main objective of the paper is to investigate the convergence of the mild solution of the diffusion equation with time and space fractional. We consider the problem in two cases which are forward problem and inverse problem. We use new techniques to overcome some of the complex assessments.


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## Body

## 1. Introduction

Fractional calculation has been shown to provide many important applications in natural sciences, such as in biological systems, signal processing, fluid mechanics, electrical networks, optical, and viscosity [1–8]. With the development of mathematics, there are now many different definitions of fractional derivatives, for example, Riemann-Liouville, Caputo, Hadamard, and Riesz. Let us refer many various papers on fractional differential equation, for example, Manimaran et al., Tuan et al., Long et al., Long L.D. et al., and Ngoc et al. [9–14]; Adiguzel et al., Li et al., Afshari et al., Alqahtani et al., Karapınar et al., Salim et al., Karapinar et al., and Abdeljawad et al. [15–22]; and Bachir et al., Salim et al., and Baitichea et al. [23–25]. Although most of them have been extensively studied, most mathematicians are interested and studied two derivatives which are Caputo and Riemann-Liouville derivatives.In this paper, forα,β∈0,1, we are interested to study the following problem:
(1)∂tαux,t+−Δβux,t=Hx,t,x,t∈0,π×0,T,u0,t=uπ,t=0,t∈0,T,with the initial condition
(2)ux,0=u0x,0<x<π,or the terminal condition
(3)ux,T=fx,0<x<π.There are many results related to the Problem (1) in both aspects: theoretical analysis and numerical analysis. The existence and well-posedness of Problems (1)–(2) and (1)–(3) has been studied in [26]. Jin et al. [27] applied two semidiscrete schemes of Galerkin FEM method in order to approximate the solution of Problems (1) and (2). In [28], the authors investigated a reaction-diffusion equation with a Caputo fractional derivative in time. In [29], the authors established the existence and uniqueness of the weak solution and the regularity of the solution for coupled fractional diffusion system. Mu et al. [30] investigated some initial-boundary value problems for time-fractional diffusion equations. Let us now mention some previous works on terminal value problem Problems (1)–(3). The main current applications of the terminal value problem are hydrodynamic inversion and spoil the image. In [31], the authors used variable total variation to approximate the backward problem for a time-space fractional diffusion equation. Under the interesting paper [32], Ngoc et al. considered the terminal value problem for nonlinear model.
(4)D0+αu−uxx=Fu.Our main purpose of this paper is to study the convergence of Problem (1) when β⟶1−. This result gives us the relationship between the solutions of the two Problem (1) with the case 0<β<1 and β=1. To the best of our knowledge, the research direction on this convergence topic is still limited. The main techniques to solve the our problem is to use Mittag-Leffler evaluations with the combination of the Wright function.This paper is organized as follows. In Section2, we focus premilinaries with some background on the definition and evaluations of Mittag-Leffler functions.


## 2. Premilinaries

Let us consider the Mittag-Leffler function, which is defined by(5)Eα,βz=∑n=0∞znΓnα+β.z∈ℂ, for α>0 and β∈ℝ. When β=1, it is abbreviated as Eαz=Eα,1z.Lemma 2.1.
The following equality holds (See [33]):
(6)Eα,1−z=∫0∞Φαθe−zθdθ,forz∈ℂ,where the Wright function Φαθ is defined by
(7)Φαθ≔∑j=0∞θjj!Γ−αj+1−α,0<α<1.
In addition,Φαθ is a probability density function, that is,
(8)Φαθ≥0,forθ>and∫0∞Φαθ=1.Lemma 2.2.
Forα∈0,1 and b>−1, the following properties hold (See [33]):
(9)∫0∞θbΦαθdθ=Γb+1Γbα+1.Let a given positive numberσ≥0. Let us also define the Hilbert scale space as follows:
(10)ℍσΩ=ψ∈L2Ω:∑j=1∞j2σψ,φj2<+∞,with the following norm ψℍσΩ=∑j=1∞j2σψ,φj21/2.Here we give the following lemma, which will help our proofs later:Lemma 2.3.
Letε,ε′>0. Then we get the following:
(11)Eα,1−j2βtα−Eα,1−j2tα≤C1α,εtαε1−βεε′j2ε+εε′.(12)Eα,α−j2βtα−Eα,α−j2tα≤C2α,εtαε1−βεε′j2ε+εε′.Proof.
Let us now to study the differenceEα,1−j2βtα−Eα,1−j2tα for 0<β<1. Since the definition of Wright function as in Lemma 2.1, we get that
(13)Eα,1−j2βtα−Eα,1−j2tα=∫0∞Φαθexp−j2βtαθdθ−∫0∞Φαθexp−j2tαθdθ.
Sincej≥1 and 0<β≤1, we know easily that exp−j2βtαθ>exp−j2tαθ. Hence, we find that
(14)exp−j2βtαθ−exp−j2tαθ=exp−j2βtαθ1−exp−j2−j2βtαθ.
Using the inequality1−e−z≤Cεzε for any ε>0, we find that
(15)exp−j2βtαθ−exp−j2tαθ≤Cεj2−j2βεtαεθε.
Combining Problems (13) and (15), we derive that
(16)Eα,1−j2βtα−Eα,1−j2tα≤Cεj2−j2βεtαε∫0∞θεΦαθdθ=CεΓε+1αε+1j2−j2βεtαε.
For anyε′>0 and noting that logj≤j for any j≥1, it is obvious to see that
(17)j2−j2β=j21−exp−2−2βlogj≤j22−2βε′logjε′≲1−βε′j2+ε′.
This implies that(18)j2−j2βε≲1−βεε′j2ε+εε′.
From some above observations, we get that(19)Eα,1−j2βtα−Eα,1−j2tα≤C1α,εtαε1−βεε′j2ε+εε′.
By a similar argument as above, we also obtain the desired result, Problem (12).


## 3. Initial Value Problem

In this section, we focus the following initial value problem under the linear case:(20)∂tαv=−−Δβvx,t+Hx,t,x,t∈0,π×0,T,v0,t=vπ,t=0,t∈0,T,vx,0=v0x,x∈0,π,where v0 and source function H are defined later.Theorem 3.1.
Letv0∈ℍpΩ and H∈L∞0,T;ℍpΩ for any p>0. Then we get
(21)vβ.,t−v∗.,tℍsΩ≲1−βp−s/2v0ℍpΩ+HL∞0,T;ℍpΩfor any 0<s<p.Proof.
The mild solution to Problem (20) with 0<β<1 is defined by
(22)vβx,t=∑j=1∞Eα,1−j2βtα∫0πv0xφjxdxφjx+∑j=1∞∫0tt−rα−1Eα,α−j2βt−rαHjrdrφjx,and the mild solution to Problem (20) with β=1 is defined by
(23)v∗x,t=∑j=1∞Eα,1−j2tα∫0πv0xφjxdxφjx+∑j=1∞∫0tt−rα−1Eα,α−j2t−rαHjrdrφjx.
By subtracting both sides of the two expressions above, we get the following difference:(24)vβx,t−v∗x,t=∑j=1∞Eα,1−j2βtα−Eα,1−j2tα∫0πv0xφjxdxφjx+∑j=1∞∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2t−rαHjrdrφjx=M1x,t+M2x,t.
Let us first consider the termM1. By applying Parseval’s equality and Lemma 2.3, we find that
(25)M1.,tℍsΩ2=∑j=1∞j2sEα,1−j2βtα−Eα,1−j2tα2∫0πv0xφjxdx2≤C1α,ε,δ2t2αε1−β2εδ∑j=1∞j2s+4ε+2εδ∫0πv0xφjxdx2,where any delta>0. Hence, we know that the upper bound
(26)M1.,tℍsΩ≲1−βεδv0ℍs+2ε+εδΩ.
Let us now treat the second termM2. By using Parseval’s equality, we get that
(27)M2.,tℍsΩ2=∑j=1∞j2s∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2rt−αHjrdr2∑j=1∞j2s∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2t−rα2Hjr2dr.
In view of the second estimate of Lemma2.3, we derive that
(28)Eα,α−j2βt−rα−Eα,α−j2t−rα2≤C2α,ε,ε′2t−r2αε1−β2εε′j4ε+2εε′
Combining Problems (27) and (28), we derive that
(29)M2.,tℍsΩ2≤C2α,ε,ε′21−β2εε′∫0tt−rα−1+2αε∑j=1∞j2s+4ε+2εε′Hjr2dr=C2α,ε,ε′21−β2εε′∫0tt−rα−1+2αεHrℍs+2ε+εε′Ωdr≤C2α,ε,ε′21−β2εε′∫0tt−rα−1+2αεdrHL∞0,T;ℍs+2ε+εε′Ω2.
It is obvious to see that the integral term∫0tt−rα−1+2αεdr is convergent. Hence, we obtain that the following estimate:
(30)M2.,tℍsΩ≤C2α,ε,ε′1−βεε′HL∞0,T;ℍs+2ε+εε′Ω.
Combining Problems (24), (25), and (30), we find that
(31)vβ.,t−v∗.,tℍsΩ≤M1.,tℍsΩ+M2.,tℍsΩ≲1−βεδv0ℍs+2ε+εδΩ+1−βεε′HL∞0,T;ℍs+2ε+εε′Ω.
Sincep>s, we can choose
(32)ε=p−s4,δ=ε′=2.
This implies that(33)vβ.,t−v∗.,tℍsΩ≲1−βp−s/2v0ℍpΩ+HL∞0,T;ℍpΩ.


## 4. Terminal Value Problem

Theorem 4.1.
Letf∈ℍbΩ and H∈L∞0,T;ℍbΩ. Then we get
(34)uβ.,t−u∗.,tLm0,T;ℍsΩ≲1−βb−s−2β−2/2fℍbΩ+HL∞0,T;ℍbΩ+1−βb−s+2β+2/2HL∞0,T;ℍbΩ,for 1<m<1/α and b>s+2β+2.Proof.
The mild solution to terminal value Problem (1) for 0<β<1 is given by
(35)uβx,t=∑j=1∞Eα,1−j2βtαEα,1−j2βTα∫0πfxφjxdxφjx−∑j=1∞Eα,1−j2βtαEα,1−j2βTα∫0TT−rα−1Eα,α−j2βT−rαHjrdrφjx+∑j=1∞∫0tt−rα−1Eα,α−j2βt−rαHjrdrφjx,where
(36)Hjr=∫0πHx,rφjxdx.
The mild solution to terminal value Problem (1) for β=1 is given by
(37)u∗x,t=∑j=1∞Eα,1−j2tαEα,1−j2Tα∫0πfxφjxdxφjx−∑j=1∞Eα,1−j2tαEα,1−j2Tα∫0TT−rα−1Eα,α−j2βT−rαHjrdrφjx+∑j=1∞∫0tt−rα−1Eα,α−j2t−rαHjrdrφjx.
Taking the difference of Problems (35) and (37) on both sides, we get the following bound:
(38)uβx,t−u∗x,t=∑j=1∞Eα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα∫0πfxφjxdxφjx−∑j=1∞Eα,1−j2βtαEα,1−j2βTα∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rαHjrdrφjx+∑j=1∞Eα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα∫0TT−rα−1Eα,α−j2T−rαHjrdrφjx+∑j=1∞∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2t−rαHjrdrφjx=J1x,t+J2x,t+J3x,t+J4x,t.
Step 1. Estimation of the TermJ1.
In order to evaluateJ1, we need to control the component
(39)M1=Eα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα.
It is obvious to compute the above term as follows:(40)M1=Eα,1−j2βtα−Eα,1−j2tαEα,1−j2βTα−Eα,1−j2βTα−Eα,1−j2TαEα,1−j2βTαEα,1−j2Tα.
Since the fact that(41)Eα,1−j2βTα≥Cα−1+j2βTα≤Cαj2βTα+1,we know that
(42)Eα,1−j2βtα−Eα,1−j2tαEα,1−j2βTα≤C1α,εTα+1Cαtαε1−βεε′j2ε+εε′+2β.
By a similar explanation as above, we find that(43)Eα,1−j2βTα−Eα,1−j2TαEα,1−j2βTαEα,1−j2Tα≲Tαε1−βεε′j2ε+εε′+2β+2,where the hidden constant depends on α,T,ε,ε′. From two above observation, we find that
(44)M1=Eα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα≲1−βεε′j2ε+εε′+2β+2,where the hidden constant depends on α,T,ε. Hence, we obtain that
(45)J1ℍsΩ2=∑j=1∞j2sEα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα2∫0πfxφjxdx2≲1−β2εε′∑j=1∞j2s+4ε+2εε′+4β+4∫0πfxφjxdx2.
It implies that the following bound(46)J1ℍsΩ≲1−βεε′fℍs+2ε+εε′+2β+2Ω.
Step 2. Estimation of the TermJ3.
By using Parseval’s equality and noting that Problem (44), we find that
(47)J3ℍsΩ2=∑j=1∞j2sEα,1−j2βtαEα,1−j2βTα−Eα,1−j2tαEα,1−j2Tα2∫0TT−rα−1Eα,α−j2T−rαHjrdr2≲1−β2εε′∑j=1∞j2s+4ε+2εε′+4β+4∫0TT−rα−1dr∫0TT−rα−1Hjr2dr2,where we have used the fact that Eα,α−j2T−rα≤Cα. Hence, we find that
(48)J3ℍsΩ2≲Tαα1−β2εε′∫0TT−rα−1Hrℍs+2ε+εε′+2β+2Ωdr≲1−β2εε′HL∞0,T;ℍs+2ε+εε′+2β+2Ω2.
Step 3. Estimation of the TermJ2.
By using Parseval’s equality, we derive that(49)J2ℍsΩ2=∑j=1∞j2sEα,1−j2βtαEα,1−j2βTα2∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rαHjrdr2.
It is easy to verify that(50)Eα,1−j2βtαEα,1−j2βTα≲1+Tαj2β1+tαj2β≤Tαt−α.
Using Hölder’s inequality, we derive that(51)∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rαHjrdr2≤∫0TT−rα−1dr∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rα2Hjr2dr≲∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rα2Hjr2dr.
By a similar explanation, we can get that the following bound:(52)Eα,α−j2βtα−Eα,α−j2tα≤C2α,ε,γtαε1−βεγj2ε+εγ,for any γ>0. This implies that
(53)Eα,α−j2βT−rα−Eα,α−j2T−rα2≲T−r2αε1−β2εγj4ε+2εγ.
Hence, we get that the following bound:(54)∫0TT−rα−1Eα,α−j2βT−rα−Eα,α−j2T−rα2Hjr2dr≲1−β2εγ∫0TT−rα+2αε−1j4ε+2εγHjr2dr.
Combining Problems (49), (50), and (54), we derive that
(55)J2ℍsΩ2≲t−2α1−β2εγ∫0TT−rα+2αε−1∑j=1∞j2s+4ε+2εγHjr2dr=t−2α1−β2εγ∫0TT−rα+2αε−1Hrℍs+2ε+εγΩ2dr≤t−2α1−β2εγ∫0TT−rα+2αε−1drHL∞0,T;ℍs+2ε+εγΩ2.
It is obvious to see that(56)∫0TT−rα+2αε−1dr=Tα+2αεα+2αε.
So, we obtain that the following confirmation(57)J2ℍsΩ≲1−βεγt−αHL∞0,T;ℍs+2ε+εγΩ.
Step 4. Estimation of the TermJ4.
By using Parseval’s equality and Hölder’s inequality, we get that(58)J4ℍsΩ2=∑j=1∞j2s∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2t−rαHjrdr2≤∫0tt−rα−1dr∑j=1∞j2s∫0tt−rα−1Eα,α−j2βT−rα−Eα,α−j2t−rα2Hjr2dr.
By a similar techniques as in Probelem (54), we derive that
(59)∫0tt−rα−1Eα,α−j2βt−rα−Eα,α−j2t−rα2Hjr2dr≲1−β2εγ∫0tt−rα+2αε−1j4ε+2εγHjr2dr.
By review two latter observations, we can deduce that(60)J4ℍsΩ2≲1−β2εγ∫0tt−rα+2αε−1∑j=1∞j2s+4ε+2εγHjr2dr=1−β2εγ∫0tt−rα+2αε−1Hrℍs+2ε+εγΩ2dr≤1−β2εγ∫0tt−rα+2αε−1drHL∞0,T;ℍs+2ε+εγΩ2.
The above inequality implies that the following estimate:(61)J4ℍsΩ≲1−βεγ∫0tt−rα+2αε−1drHL∞0,T;ℍs+2ε+εγΩ
By similar computation as above, we deduce that(62)J4ℍsΩ≲1−βεγHL∞0,T;ℍs+2ε+εγΩ.
Combining four steps as above, we deduce that(63)uβ.,t−u∗.,tℍsΩ≤J1ℍsΩ+J2ℍsΩ+J3ℍsΩ+J4ℍsΩ≲1−βεε′fℍs+2ε+εε′+2β+2Ω+HL∞0,T;ℍs+2ε+εε′+2β+2Ω+1−βεγt−α+1HL∞0,T;ℍs+2ε+εγΩ.
Let us choose(64)ε=b−s−2β−24,ε′=2,γ=2b−s+2β+2b−s−2β−2.
Then from some above observations, we deduce that the following estimate:(65)uβ.,t−u∗.,tℍsΩ≲1−βb−s−2β−2/2fℍbΩ+HL∞0,T;ℍbΩ+1−βb−s+2β+2/2t−α+1HL∞0,T;ℍbΩ.
This estimate implies that the desired result, Problem (34).


## 5. Conclusion

In this work, we consider the fractional problem for partial differential equation. We investigate the convergence of the mild solution of the diffusion equation with time and space fractional. Moreover, we consider the problem in two cases which are forward problem and inverse problem by using new techniques to overcome some of the complex assessments.



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