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

The fractional problem for partial di ﬀ erential 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 di ﬀ usion 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.


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][2][3][4][5][6][7][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 [23][24][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, for α, β ∈ ð0, 1Þ, we are interested to study the following problem: with the initial condition or the terminal condition 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 initialboundary 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.
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 Section 2, we focus premilinaries with some background on the definition and evaluations of Mittag-Leffler functions.

Premilinaries
Let us consider the Mittag-Leffler function, which is defined by ðz ∈ ℂÞ, for α > 0 and β ∈ ℝ. When β = 1, it is abbreviated as E α ðzÞ = E α,1 ðzÞ. Lemma 2.1. The following equality holds (See [33]): where the Wright function Φ α ðθÞ is defined by In addition, Φ α ðθÞ is a probability density function, that is, Lemma 2.2. For α ∈ ð0, 1Þ and b > −1, the following properties hold (See [33]): Let a given positive number σ ≥ 0. Let us also define the Hilbert scale space as follows: with the following norm Here we give the following lemma, which will help our proofs later: Then we get the following: Proof. Let us now to study the difference jE α, Since the definition of Wright function as in Lemma 2.1, we get that Since j ≥ 1 and 0 < β ≤ 1, we know easily that exp ð−j 2β t α θÞ > exp ð−j 2 t α θÞ. Hence, we find that Using the inequality 1 − e −z ≤ C ε z ε for any ε > 0, we find that Combining Problems (13) and (15), we derive that For any ε ′ > 0 and noting that log ðjÞ ≤ j for any j ≥ 1, it is obvious to see that This implies that Journal of Function Spaces From some above observations, we get that By a similar argument as above, we also obtain the desired result, Problem (12).

Initial Value Problem
In this section, we focus the following initial value problem under the linear case: where v 0 and source function H are defined later.
for any 0 < s < p.
Proof. The mild solution to Problem (20) and the mild solution to Problem (20) By subtracting both sides of the two expressions above, we get the following difference: Let us first consider the term M 1 . By applying Parseval's equality and Lemma 2.3, we find that where any delta > 0. Hence, we know that the upper bound Let us now treat the second term M 2 . By using Parseval's equality, we get that In view of the second estimate of Lemma 2.3, we derive that 3

Journal of Function Spaces
Combining Problems (27) and (28), we derive that It is obvious to see that the integral term Ð t 0 ðt − rÞ ðα−1+2αεÞ dr is convergent. Hence, we obtain that the following estimate: Combining Problems (24), (25), and (30), we find that Since p > s, we can choose Then we get
Proof. The mild solution to terminal value Problem (1) for 0 < β < 1 is given by where The mild solution to terminal value Problem (1) for β = 1 is given by Taking the difference of Problems (35) and (37) on both sides, we get the following bound: Journal of Function Spaces Step 1. Estimation of the Term J 1 . In order to evaluate J 1 , we need to control the component It is obvious to compute the above term as follows: Since the fact that we know that By a similar explanation as above, we find that where the hidden constant depends on α, T, ε, ε′. From two above observation, we find that where the hidden constant depends on α, T, ε. Hence, we obtain that It implies that the following bound Step 2. Estimation of the Term J 3 . By using Parseval's equality and noting that Problem (44), we find that where we have used the fact that E α,α ð−j 2 ðT − rÞ α Þ ≤ C α . Hence, we find that Step 3. Estimation of the Term J 2 . By using Parseval's equality, we derive that