On the Inverse Problem of the Fractional Heat-Like Partial Differential Equations: Determination of the Source Function

Abstract

The study in this paper mainly concerns the inverse problem of determining an unknown source function in the linear fractional differential equation with variable coefficient using Adomian decomposition method (ADM). We apply ADM to determine the continuous right hand side functions 𝑓(𝑥) and 𝑓(𝑡) in the heat-like diffusion equations 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑥) and 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑡), respectively. The results reveal that ADM is very effective and simple for the inverse problem of determining the source function.

1. Introduction

Fractional differential equations (FDEs) are obtained by generalizing differential equations to an arbitrary order. They are used to model physical systems with memory. Since FDEs have memory, nonlocal relations in space and time complex phenomena can be modeled by using these equations. Due to this fact, materials with memory and hereditary effects, fluid flow, rheology, diffusive transport, electrical networks, electromagnetic theory and probability, signal processing, and many other physical processes are diverse applications of FDEs. Since FDEs are used to model complex phenomena, they play a crucial role in engineering, physics, and applied mathematics. Therefore, they are generating an increasing interest from engineers and scientist in the recent years. As a result, FDEs are quite frequently encountered in different research areas and engineering applications [1].

The book written by Oldham and Spanier [2] played an outstanding role in the development of the fractional calculus. Also, it was the first book that was entirely devoted to a systematic presentation of the ideas, methods, and applications of the fractional calculus. Afterwards, several fundamental works on various aspects of the fractional calculus include extensive survey on fractional differential equations by Miller and Ross [3], Podlubny [4], and others. Further, several references to the books by Oldham and Spanier [2], Miller and Ross [3], and Podlubny [4] show that applied scientists need first of all an easy introduction to the theory of fractional derivatives and fractional differential equations, which could help them in their initial steps in adopting the fractional calculus as a method of research [5].

In general, FDEs do not have exact analytical solutions; hence, the approximate and numerical solutions of these equations are studied [6–8]. Analytical approximations for linear and nonlinear FDEs are obtained by variational iteration method, Adomian decomposition method, homotopy perturbation method, Lagrange multiplier method, BPs operational matrices method, and so forth. An effective and easy-to-use method for solving such equations is needed. Large classes of linear and nonlinear differential equations, both ordinary and partial, can be solved by the Adomian decomposition method [9–12].

Solving an equation with certain data in a specified region is called direct problem. On the other hand, determining an unknown input by using output is called an inverse problem. This unknown input could be some coefficients, or it could be a source function in equation. Based on this unknown input the inverse problem is called inverse problem of coefficient identification or inverse problem of source identification, respectively. Generally inverse problems are ill-posed problems; that is, they are very sensitive to errors in measured input. In order to deal with this ill-posedness, regularization methods have been developed. Inverse problems have many practical applications such as geophysics, optics, quantum mechanics, astronomy, medical imaging and materials testing, X-ray tomography, and photoelasticity. Theoretical and applied aspects of inverse problems have been under intense study lately, especially for the fractional equation [13–16].

In this paper, we investigate inverse problems of the linear heat-like differential equations of fractional orders 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑥) and 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑡) where the function 𝑢(𝑥,𝑡) is assumed to be a causal function of time and space. Time fractional derivative operator 𝐷^𝛼_𝑡 is considered as in Caputo sense [17]. We use the Adomian decomposition method [9, 10] to obtain source functions 𝑓(𝑥) and 𝑓(𝑡) under the initial and mixed boundary conditions. By this method, we determine the source functions 𝑓(𝑥) and 𝑓(𝑡) in a rapidly converging series form when they exist. Compared with previous researches [10, 12, 18], the method we use in this paper is more effective and accurate.

The structure of this paper is given as follows. First, we give some basic definitions of fractional calculus. Inverse problem of finding the source function in one-dimensional fractional heat-like equations with mixed boundary conditions is given in Section 2. After that, we give some illustrative examples of this method for all cases in Section 3. Finally, the conclusion is given in Section 4.

1.1. Fractional Calculus

In this section, we give basic definitions and properties of the fractional calculus [17, 18].

Definition 1. A real function 𝑓(𝑥), 𝑥>0, is said to be in the space 𝐶_𝜇, 𝜇∈ℝ if there exists a real number 𝑝>𝜇 such that 𝑓(𝑥)=𝑥^𝑝𝑓₁(𝑥), where 𝑓₁(𝑥)∈𝐶[0,∞), and it is said to be in the space 𝐶^𝑚_𝜇 if 𝑓^(𝑚)∈𝐶_𝜇, 𝑚∈ℕ.

Definition 2. The Riemann-Liouville fractional integral operator of order 𝛼≥0, of a function 𝑓∈𝐶_𝜇, 𝜇≥−1 is defined as 𝐽^𝛼𝑓(𝑥)=1Γ(𝛼)^𝑥₀(𝑥−𝑡)^𝛼−1𝑓(𝑡)𝐽𝑑𝑡,𝛼>0,𝑥>0₀𝑓(𝑥)=𝑓(𝑥). (1) Some of the basic properties of this operator are given as follows.

For 𝑓∈𝐶_𝜇, 𝜇≥−1, 𝛼,𝛽≥0 and 𝛾>−1: (1)𝐽^𝛼𝐽^𝛽𝑓(𝑥)=𝐽^𝛼+𝛽𝑓(𝑥),(2)𝐽^𝛼𝐽^𝛽𝑓(𝑥)=𝐽^𝛽𝐽^𝛼𝑓(𝑥),(3)𝐽^𝛼𝑥^𝛾=Γ𝛾+1Γ𝑥𝛼+𝛾+1^𝛼+𝛾. (2) The other properties can be found in [17].

Definition 3. The fractional derivative of 𝑓(𝑥) in the Caputo sense is defined as 𝐷^𝛼𝑓(𝑥)=𝐽^𝑚−𝛼𝐷^𝑚𝑓(𝑥)=1Γ(𝑚−𝛼)^𝑥₀(𝑥−𝑡)^{𝑚−𝛼−1}𝑓^(𝑚)(𝑡)𝑑𝑡, (3) where 𝑚−1<𝛼≤𝑚, 𝑚∈ℕ, 𝑥>0, 𝑓∈𝐶^𝑚₋₁.

Useful properties of 𝐷^𝛼 are given as follows.

Lemma 4. If 𝑚−1<𝛼≤𝑚, 𝑚∈ℕ, and 𝑓∈𝐶^𝑚_𝜇, 𝜇≥−1, then 𝐷^𝛼𝐽^𝛼𝑓(𝑥)(𝑥),𝐽=𝑓_𝛼𝐷^𝛼𝑓(𝑥)=𝑓(𝑥)−^𝑚−1_𝑘=0𝑓^(𝑘)0⁺𝑥^𝑘𝑘!,𝑥>0. (4) Since traditional initial and boundary conditions are allowed in problems including Caputo fractional derivatives, it is considered here. In this paper, we deal with the fractional heat-like equations where the unknown function 𝑢=𝑢(𝑥,𝑡) is an arbitrary function of time and space.

Definition 5. The Caputo time fractional derivative operator of order 𝛼>0 is defined as follows where 𝑚 is the smallest integer that exceeds 𝛼: 𝐷^𝛼_𝑡𝑢()=𝜕𝑥,𝑡_𝛼𝑢(𝑥,𝑡)𝜕𝑡^𝛼=1Γ(𝑚−𝛼)×∫^𝑡₀(𝑡−𝜏)^{𝑚−𝛼−1}𝜕^𝑚𝑢(𝑥,𝜏)𝜕𝑡^𝑚𝜕𝑑𝜏,for𝑚−1<𝛼<𝑚_𝑚𝑢(𝑥,𝑡)𝜕𝑡^𝑚,for𝛼=𝑚∈IN. (5) For more details about Caputo fractional differential operator, we refer to [17].

Definition 6. The Mittag-Leffler function with two-parameters is defined by the series expansion as shown below, where the real part of 𝛼 is strictly positive [19] 𝐸_𝛼,𝛽(𝑧)=^∞_𝑛=0𝑧^𝑛Γ.𝛼𝑛+𝛽 (6)

2. Inverse Problem of Determining Source Function

In this section, we deal with inverse problem of finding the source function, in one-dimensional fractional heat-like equations with mixed boundary conditions. To determine the unknown source function we have developed new methods through ADM as in the following subsections.

2.1. Determination of Unknown Source Functions Depending on 𝑥

We consider the following inverse problem of determining the source function 𝑓(𝑥): 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥()(𝑥),𝑢()𝑥,𝑡+𝑓𝑥>0,𝑡>0,0<𝛼≤1,𝑥,0=𝑓₁(𝑥),𝑢(0,𝑡)=ℎ₁(𝑡),𝑢_𝑥(0,𝑡)=ℎ₂(𝑡), (7) where the functions ℎ₁(𝑡), ℎ₂(𝑡)∈𝐶^∞[0,∞) and ℎ(𝑥), 𝑓(𝑥), 𝑓₁(𝑥)∈𝐶^∞[0,∞). In order to determine the source function for this kind of inverse problems, we apply ADM. First, we apply the time-dependent Riemann-Liouville fractional integral operator 𝐽^𝛼_𝑡 to both sides of (7) to get rid of fractional derivative 𝐷^𝛼_𝑡 as shown below: 𝐽^𝛼_𝑡𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡)+𝐽^𝛼_𝑡𝑓(𝑥). (8) Then we get 𝑢(𝑥,𝑡)=𝑢(𝑥,0)+𝐽^𝛼_𝑡𝑓(𝑥)+𝐽^𝛼_𝑡ℎ(𝑥)𝑢_𝑥𝑥(𝑥,𝑡). (9) In ADM the solution 𝑢(𝑥,𝑡) is written in the following series form [9]: 𝑢(𝑥,𝑡)=^∞_𝑛=0𝑢_𝑛(𝑥,𝑡), (10) where 𝑢 and 𝑢_𝑛, 𝑛∈ℕ, are defined in 𝐶^∞[0,∞)×𝐶¹_𝜇[0,∞). After substituting the decomposition (10) into (9) and setting the recurrence scheme as follows: 𝑢₀(𝑥,𝑡)=𝑢(𝑥,0)+𝐽^𝛼_𝑡𝑓(𝑥),𝑢_𝑛+1(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑢_𝑛𝑥𝑥(𝑥,𝑡),𝑛=0,1,…, (11) we get ADM polynomials below 𝑢₀(𝑥,𝑡)=𝑓₁(𝑥)(𝑥)𝑡+𝑓^𝛼Γ(),𝑢𝛼+1₁(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑢_0𝑥𝑥(𝑥)=ℎ(𝑥)𝑓₁^′′(𝑥)𝑡^𝛼Γ()(𝑥)𝑓𝛼+1+ℎ^′′(𝑥)𝑡^2𝛼Γ(),𝑢2𝛼+1₂(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑢_1𝑥𝑥(𝑥)=ℎ(𝑥)ℎ^′′(𝑥)𝑓^′₁(𝑥)+ℎ^′(𝑥)𝑓₁^′′(𝑥)+ℎ^′(𝑥)𝑓₁^′′′(𝑥)+ℎ(𝑥)𝑓₁^(𝑖𝑣)(𝑥𝑡)^2𝛼Γ()+ℎ2𝛼+1_′′(𝑥)𝑓^′′(𝑥)+2ℎ^′(𝑥)𝑓^′′′(𝑥)+ℎ(𝑥)𝑓^(𝑖𝑣)(𝑥𝑡)^3𝛼Γ(),⋮3𝛼+1 (12) After writing these polynomials in (10), the solution 𝑢(𝑥,𝑡) is given by 𝑢(𝑥,𝑡)=𝑓₁(𝑥)(𝑥)𝑡+𝑓^𝛼Γ()(𝑥)𝑓𝛼+1+ℎ₁^′′(𝑥)𝑡^𝛼Γ()(𝑥)𝑓𝛼+1+ℎ^′′(𝑥)𝑡^2𝛼Γ()(𝑥ℎ2𝛼+1+ℎ)^′′(𝑥)𝑓^′₁(𝑥)+ℎ^′(𝑥)𝑓₁^′′(𝑥)+ℎ^′(𝑥)𝑓₁^′′′(𝑥)+ℎ(𝑥)𝑓₁^(𝑖𝑣)(𝑥𝑡)^2𝛼Γ()+ℎ2𝛼+1_′′(𝑥)𝑓^′′(𝑥)+2ℎ^′(𝑥)𝑓^′′′(𝑥)+ℎ(𝑥)𝑓^(𝑖𝑣)(𝑥𝑡)^3𝛼Γ(3𝛼+1)+⋯. (13) If we arrange it with respect to like powers of 𝑡, then we get 𝑢(𝑥,𝑡)=𝑓₁(𝑥)+𝑓(𝑥)+ℎ(𝑥)𝑓₁^′′(𝑥𝑡)^𝛼Γ()(𝑥𝑓𝛼+1+ℎ)^′′(𝑥)+ℎ^′′(𝑥)𝑓^′₁(𝑥)+ℎ^′(𝑥)𝑓₁^′′(𝑥)+ℎ^′(𝑥)𝑓₁^′′′(𝑥)+ℎ(𝑥)𝑓₁^(𝑖𝑣)(𝑥𝑡)^2𝛼Γ()+ℎ2𝛼+1_′′(𝑥)𝑓^′′(𝑥)+2ℎ^′(𝑥)𝑓^′′′(𝑥)+ℎ(𝑥)𝑓^(𝑖𝑣)(𝑥𝑡)^3𝛼Γ(3𝛼+1)+⋯. (14) To determine the unknown source function, first we expand the boundary conditions 𝑢(0,𝑡)=ℎ₁(𝑡) and 𝑢_𝑥(0,𝑡)=ℎ₂(𝑡) into the following series for the space whose bases are {𝑡^𝑛𝛼/Γ(𝑛𝛼+1)}^∞_𝑛=0, 0<𝛼≤1: ℎ₁(𝑡)=ℎ₁(0)+ℎ^′₁(0)𝑡^𝛼Γ(𝛼+1)+ℎ₁^′′(0)𝑡^2𝛼Γ()ℎ2𝛼+1+⋯,₂(𝑡)=ℎ₂(0)+ℎ^′₂(0)𝑡^𝛼Γ(𝛼+1)+ℎ₂^′′(0)𝑡^2𝛼Γ(2𝛼+1)+⋯. (15)(16) On the other hand, if we rewrite the boundary conditions 𝑢(0,𝑡) and 𝑢_𝑥(0,𝑡) from (14), then we have ℎ₁(𝑡)=𝑓₁(0)+𝑓(0)+ℎ(0)𝑓₁^′′(0𝑡)^𝛼Γ()(0𝑓𝛼+1+ℎ)^′′(0)+ℎ^′′(0)𝑓^′₁(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ(0)𝑓₁^(𝑖𝑣)(0𝑡)^2𝛼Γ()+ℎ2𝛼+1_′′(0)𝑓^′′(0)+2ℎ^′(0)𝑓^′′′(0)+ℎ(0)𝑓^(𝑖𝑣)(0𝑡)^3𝛼Γ()ℎ3𝛼+1+⋯,₂(𝑡)=𝑓^′₁(0)+𝑓^′(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ(0)𝑓₁^′′′(0𝑡)^𝛼Γ()+ℎ𝛼+1_′(0)𝑓^′′(0)+ℎ^′′(0)𝑓^′₁(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ(0)𝑓₁^(𝑖𝑣)(0(0𝑓)+ℎ)^′′′(0)+ℎ^′′′(0)𝑓^′₁(0)+2ℎ^′′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ^′′(0)𝑓₁^′′′(0)+2ℎ^′(0)𝑓₁^(𝑖𝑣)(0)+ℎ(0)𝑓₁^(𝑣)(0𝑡)^2𝛼Γ()+ℎ2𝛼+1_′(0)ℎ^′′(0)𝑓^′′(0)+2ℎ^′(0)𝑓^′′′(0)+ℎ(0)𝑓^(𝑖𝑣)(0(0ℎ)+ℎ)^′′′(0)𝑓^′′(0)+3ℎ^′′(0)𝑓^′′′(0)+3ℎ^′(0)𝑓^(𝑖𝑣)(0)+ℎ(0)𝑓^(𝑣)(0𝑡)^3𝛼Γ(3𝛼+1)+⋯. (17)(18) Equating (15) and (17) yields the following: ℎ₁(0)=𝑓₁(0),ℎ^′₁(0)=𝑓(0)+ℎ(0)𝑓₁^′′(0),ℎ₁^′′(0)=ℎ(0)𝑓^′′(0)+ℎ^′′(0)𝑓^′₁(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ(0)𝑓₁^(𝑖𝑣)(0,⋮) (19) and equating (16) and (18) yields the following: ℎ₂(0)=𝑓^′₁(0),ℎ^′₂(0)=𝑓^′(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ(0)𝑓₁^′′′(0),ℎ₂^′′(0)=ℎ^′(0)𝑓^′′(0)+ℎ^′′(0)𝑓^′₁(0)+ℎ^′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ(0)𝑓₁^(𝑖𝑣)(0(0𝑓)+ℎ)^′′′(0)+ℎ^′′′(0)𝑓^′₁(0)+2ℎ^′′(0)𝑓₁^′′(0)+ℎ^′(0)𝑓₁^′′′(0)+ℎ^′′(0)𝑓₁^′′′(0)+2ℎ^′(0)𝑓₁^(𝑖𝑣)(0)+ℎ(0)𝑓₁^(𝑣)(0,⋮) (20)(21) Using the above data in the following Taylor series expansion of unknown function 𝑓(𝑥) we get 𝑓(𝑥)=𝑓(0)+𝑓^′(0)𝑥+𝑓^′′(0)𝑥²2!+𝑓^′′′(0)𝑥³3!+⋯. (22) Consequently, we determine 𝑓(𝑥) as follows: 𝑓(𝑥)=ℎ^′₁(0)−ℎ(0)𝑓₁^′′(0+ℎ_′2)(0)−ℎ^′(0)𝑓₁^′′(0)−ℎ(0)𝑓₁^′′′(0𝑥+ℎ_1′′)(0)ℎ(0)−ℎ^′′(0)𝑓^′₁(0)−ℎ^′(0)𝑓₁^′′(0)−ℎ^′(0)𝑓₁^′′′(0)−ℎ(0)𝑓₁^(𝑖𝑣)(0𝑥)²2!+⋯, (23) where ℎ(0)≠0.

2.2. Determination of Unknown Source Functions Depending on 𝑡

We consider the following inverse problem of determining the source function 𝑓(𝑡): 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=ℎ(𝑥)𝑢_𝑥𝑥()(𝑡),𝑢()𝑥,𝑡+𝑓𝑥>0,𝑡>0,0<𝛼≤1,𝑥,0=𝑓₁(𝑥),𝑢(0,𝑡)=ℎ₁(𝑡),𝑢_𝑥(0,𝑡)=ℎ₂(𝑡), (24) where ℎ₁(𝑡), ℎ₂(𝑡)∈𝐶^∞[0,∞), ℎ(𝑥), 𝑓₁(𝑥)∈𝐶^∞[0,∞), and 𝑓(𝑡)∈𝐶¹_𝜇[0,∞), 𝜇≥−1. As in the previous case, we apply ADM to determine the unknown function 𝑓(𝑡).

First, to reduce the problem, we define new functions in the following form: 𝑤(𝑡)=𝐽^𝛼_𝑡𝑓(𝑡)𝑢(𝑥,𝑡)=𝑣(𝑥,𝑡)+𝑤(𝑡). (25) Then the reduced problem is given as follows: 𝐷^𝛼_𝑡𝑣(𝑥,𝑡)=ℎ(𝑥)𝑣_𝑥𝑥(𝑥,𝑡), (26) with the following initial and mixed boundary conditions 𝑣(𝑥,0)=𝑓₁(𝑥)(0),𝑣()−𝑤0,𝑡=ℎ₁(𝑡)(𝑡),𝑣−𝑤_𝑥(0,𝑡)=ℎ₂(𝑡). (27) By using ADM as in the previous section, we determine the function 𝑤(𝑡) which leads to the source function 𝑓(𝑡). Let us apply 𝐽^𝛼_𝑡 to both sides of (26) as shown below 𝐽^𝛼_𝑡𝐷^𝛼_𝑡𝑣(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑣_𝑥𝑥(𝑥,𝑡). (28) Then we get 𝑣(𝑥,𝑡)=𝑣(𝑥,0)+𝐽^𝛼_𝑡ℎ(𝑥)𝑣_𝑥𝑥(𝑥,𝑡). (29) Now we define the solution 𝑣(𝑥,𝑡) by the following decomposition series according to ADM 𝑣(𝑥,𝑡)=^∞_𝑛=0𝑣_𝑛(𝑥,𝑡). (30) Substituting (30) into (29), we obtain 𝑣₀()(),𝑣𝑥,𝑡=𝑣𝑥,0_𝑛+1(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑣_𝑛𝑥𝑥(𝑥,𝑡),𝑛=0,1,…. (31) Hence, the recurrence scheme is obtained as follows: 𝑣₀(𝑥,𝑡)=𝑣(𝑥,0)=𝑓₁(𝑥)(0),𝑣−𝑤₁(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑣_0𝑥𝑥(𝑥)=ℎ(𝑥)𝑓₁^′′(𝑥)𝑡^𝛼Γ(),𝑣𝛼+1₂(𝑥,𝑡)=𝐽^𝛼_𝑡ℎ(𝑥)𝑣_1𝑥𝑥(𝑥)=ℎ²(𝑥)𝑓₁^(𝑖𝑣)(𝑥)𝑡^2𝛼Γ(),⋮2𝛼+1 (32) Consequently, from (30), the solution 𝑣(𝑥,𝑡) is given as shown below 𝑣(𝑥,𝑡)=𝑣₀(𝑥,𝑡)+𝑣₁(𝑥,𝑡)+𝑣₂(𝑥,𝑡)+⋯=𝑓₁(𝑥)−𝑤(0)+ℎ(𝑥)𝑓₁^′′(𝑥)𝑡^𝛼Γ(𝛼+1)+ℎ²(𝑥)𝑓₁^(𝑖𝑣)(𝑥)𝑡^2𝛼Γ(2𝛼+1)+⋯. (33) By using the boundary condition 𝑣(0,𝑡)=ℎ₁(𝑡)+𝑤(𝑡) and 𝑤(0)=0, we have 𝑤(𝑡)=𝑓₁(0)−ℎ₁(𝑡)+ℎ(0)𝑓₁^′′(0)𝑡^𝛼Γ(𝛼+1)+ℎ²(0)𝑓₁^(𝑖𝑣)(0)𝑡^2𝛼Γ(2𝛼+1)+⋯, (34) which implies the following: 𝐽^𝛼_𝑡𝑓(𝑡)=𝑓₁(0)−ℎ₁(𝑡)+ℎ(0)𝑓₁^′′(0)𝑡^𝛼Γ(𝛼+1)+ℎ²(0)𝑓₁^(𝑖𝑣)(0)𝑡^2𝛼Γ(2𝛼+1)+⋯. (35) Since 𝐷^𝛼_𝑡𝑤(𝑥,𝑡)=𝐷^𝛼_𝑡𝐽^𝛼_𝑡𝑓(𝑡)=𝑓(𝑡), we obtain the source function 𝑓(𝑡) as follows: 𝑓(𝑡)=𝐷^𝛼_𝑡𝑓₁(𝑥)−ℎ₁(𝑡)+ℎ(𝑥)𝑓₁^′′(𝑥)𝑡^𝛼Γ(𝛼+1)+ℎ²(𝑥)𝑓₁^(𝑖𝑣)(𝑥)𝑡^2𝛼Γ().2𝛼+1+⋯ (36)

3. Examples

Example 1. We consider the inverse problem of determining source function 𝑓(𝑥) in the following one-dimensional fractional heat-like PDE: 𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=2𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑥),𝑥>0,0<𝛼≤1,𝑡>0, (37) subject to the following initial and nonhomogeneous mixed boundary conditions: 𝑢(𝑥,0)=𝑒^𝑥𝑢()+sin𝑥,0,𝑡=𝑒^2𝑡,𝑢_𝑥(0,𝑡)=𝑒^2𝑡+1. (38) Now, let us apply the time-dependent Riemann Liouville fractional integral operator 𝐽^𝛼_𝑡 to both sides of (37) 𝐽^𝛼_𝑡𝐷^𝛼_𝑡𝑢(𝑥,𝑡)=2𝐽^𝛼_𝑡𝑢_𝑥𝑥(𝑥,𝑡)+𝐽^𝛼_𝑡𝑓(𝑥) (39) which implies 𝑢(𝑥,𝑡)−𝑢(𝑥,0)=2𝐽^𝛼_𝑡𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑥)𝐽^𝛼_𝑡(1). (40) Then, from the initial condition we get 𝑢(𝑥,𝑡)=𝑒^𝑥(𝑥)𝑡+sin𝑥+𝑓^𝛼Γ(𝛼+1)+2𝐽^𝛼_𝑡𝑢_𝑥𝑥(𝑥,𝑡). (41) Now, we apply ADM to the problem. In (41), the sum of the first three terms is identified as 𝑢₀. So 𝑢₀=𝑒^𝑥(𝑥)𝑡+sin𝑥+𝑓^𝛼Γ(),𝑢𝛼+1_𝑘+1=2𝐽^𝛼_𝑡𝑢_𝑘𝑥𝑥(𝑥,𝑡),𝑘≥0. (42) For 𝑘=0, we have 𝑢₁=2𝐽^𝛼_𝑡𝑢_0𝑥𝑥(𝑥,𝑡)=2𝑒^𝑥𝑡^𝛼Γ()𝑡𝛼+1−2sin𝑥^𝛼Γ(𝛼+1)+2𝑓^′′(𝑥)𝑡^2𝛼Γ(2𝛼+1)+⋯, (43) similarly, for 𝑘=1, we have 𝑢₂=2𝐽^𝛼_𝑡𝑢_1𝑥𝑥(𝑥,𝑡)=4𝑒^𝑥𝑡^2𝛼Γ()𝑡2𝛼+1+4sin𝑥^2𝛼Γ(2𝛼+1)+4𝑓^(𝑖𝑣)(𝑥)𝑡^3𝛼Γ(3𝛼+1)+⋯, (44) and for 𝑘=2, we have 𝑢₃=2𝐽^𝛼_𝑡𝑢_2𝑥𝑥(𝑥,𝑡)=8𝑒^𝑥𝑡^3𝛼Γ()𝑡3𝛼+1−8sin𝑥^3𝛼Γ(3𝛼+1)+8𝑓^(𝑣𝑖)(𝑥)𝑡^4𝛼Γ()⋮4𝛼+1+⋯ (45) Then using ADM polynomials, we get the solution 𝑢(𝑥,𝑡) as follows: 𝑢(𝑥,𝑡)=𝑢₀+𝑢₁+𝑢₂+𝑢₃+⋯=𝑒^𝑥(𝑥)𝑡+sin𝑥+𝑓^𝛼Γ(𝛼+1)+2𝑒^𝑥𝑡^𝛼Γ()𝑡𝛼+1−2sin𝑥^𝛼Γ(𝛼+1)+2𝑓^′′(𝑥)𝑡^2𝛼Γ(2𝛼+1)+4𝑒^𝑥𝑡^2𝛼Γ()𝑡2𝛼+1+4sin𝑥^2𝛼Γ(2𝛼+1)+4𝑓^(𝑖𝑣)(𝑥)𝑡^3𝛼Γ(3𝛼+1)+8𝑒^𝑥𝑡^3𝛼Γ()𝑡3𝛼+1−8sin𝑥^3𝛼Γ(3𝛼+1)+8𝑓^(𝑣𝑖)(𝑥)𝑡^4𝛼Γ(4𝛼+1)+⋯. (46) After arranging it according to like powers of 𝑡, we have 𝑢(𝑥,𝑡)=𝑒^𝑥𝑡+sin𝑥+^𝛼Γ()𝑓(𝑥)𝛼+1+2𝑒^𝑥+𝑡−2sin𝑥_2𝛼Γ(2𝛼+1)2𝑓^′′(𝑥)+4𝑒^𝑥+𝑡+4sin𝑥_3𝛼Γ(3𝛼+1)4𝑓^(𝑖𝑣)(𝑥)+8𝑒^𝑥+𝑡−8sin𝑥_4𝛼Γ(4𝛼+1)8𝑓^(𝑣𝑖)(𝑥)+16𝑒^𝑥+16sin𝑥+⋯. (47) Now, by applying the boundary condition given in (38), we obtain 𝑢()𝑡0,𝑡=1+^𝛼Γ()𝑓(0)+𝑡𝛼+1+2^2𝛼Γ(2𝛼+1)2𝑓^′′(0)+𝑡+4_3𝛼Γ(3𝛼+1)4𝑓^(𝑖𝑣)(0)+𝑡+8_4𝛼Γ(4𝛼+1)8𝑓^(𝑣𝑖)(0)+16+⋯. (48) From (15), it must be equal to the following Taylor series expansion of 𝑒^2𝑡 in the space whose bases are {𝑡^𝑛𝛼/Γ(𝑛𝛼+1)}^∞_𝑛=0, 0<𝛼≤1: 𝑒^2𝑡𝑡=1+2^𝛼Γ()𝑡𝛼+1+4^2𝛼Γ()𝑡2𝛼+1+8^3𝛼Γ()𝑡3𝛼+1+16^4𝛼Γ(4𝛼+1)+⋯. (49) Hence, from the equality of the coefficients of corresponding terms, we get 𝑓(0)=𝑓^′′(0)=𝑓^(𝑖𝑣)(0)=𝑓^(𝑣𝑖)(0)=⋯=0. (50) From (47), we have 𝑢_𝑥(𝑥,𝑡)=𝑒^𝑥𝑡+cos𝑥+^𝛼Γ()𝑓𝛼+1^′(𝑥)+2𝑒^𝑥+𝑡−2cos𝑥_2𝛼Γ(2𝛼+1)2𝑓^′′′(𝑥)+4𝑒^𝑥+𝑡+4cos𝑥_3𝛼Γ(3𝛼+1)4𝑓^(𝑣)(𝑥)+8𝑒^𝑥+𝑡−8cos𝑥_4𝛼Γ(4𝛼+1)8𝑓^{(𝑣𝑖𝑖)}(𝑥)+16𝑒^𝑥+16cos𝑥+⋯. (51) So, 𝑢_𝑥()𝑡0,𝑡=2+^𝛼Γ()𝑓𝛼+1^′(0)+𝑡^2𝛼Γ(2𝛼+1)2𝑓^′′′(0)+𝑡+8_3𝛼Γ(3𝛼+1)4𝑓^(𝑣)(0+𝑡_4𝛼)Γ(4𝛼+1)8𝑓^{(𝑣𝑖𝑖)}(0)+32+⋯. (52) From the derivative boundary condition given in (38), it must be equal to the following series expansion of 𝑒^2𝑡+1 in the space whose bases are {𝑡^𝑛𝛼/Γ(𝑛𝛼+1)}^∞_𝑛=0, 0<𝛼≤1: 𝑒^2𝑡𝑡+1=2+2^𝛼Γ()𝑡𝛼+1+4^2𝛼Γ()𝑡2𝛼+1+8^3𝛼Γ()𝑡3𝛼+1+16^4𝛼Γ(4𝛼+1)+⋯. (53) Then, we find the following data: 𝑓^′(0)=2,𝑓^′′′(0)=−2,𝑓^(𝑣)(0)𝑓=2,_{(𝑣𝑖𝑖)}(0)=−2,…. (54) Next, using (50) and (54), we have the Taylor series expansion of 𝑓(𝑥) as follows: 𝑓(𝑥)=𝑓(0)+𝑓^′(0)𝑥+𝑓^′′(0)𝑥²2!+𝑓^′′′(0)𝑥³3!+𝑓^(𝑖𝑣)(0)𝑥⁴𝑥4!+⋯=2𝑥−2³𝑥3!+2⁵𝑥5!+2⁷7!+⋯. (55) That is, 𝑓(𝑥)𝑥=2𝑥−³+𝑥3!⁵+𝑥5!7⁷!+⋯ (56) which is the series expansion of the function 2sin𝑥. Consequently, we determine the source function 𝑓(𝑥) as 𝑓(𝑥)=2sin𝑥. (57)

Example 2. We consider the inverse problem of determining source function 𝑓(𝑡) in the following one-dimensional fractional heat-like diffusion equation: 𝐷^𝛼_𝑡𝑢()=1𝑥,𝑡2𝑥²𝑢_𝑥𝑥(𝑥,𝑡)+𝑓(𝑡),𝑥>0,0<𝛼≤1,𝑡>0, (58) subject to following initial and mixed boundary conditions 𝑢(𝑥,0)=𝑥²+12()=𝑒,𝑢0,𝑡^2𝑡2,𝑢_𝑥(0,𝑡)=0. (59) Now let us determine the source function 𝑓(𝑡). To reduce the problem, we define new functions as follows: 𝑤(𝑡)=𝐽^𝛼_𝑡𝑓(𝑡),𝑢(𝑥,𝑡)=𝑣(𝑥,𝑡)+𝑤(𝑡). (60) Then, our reduced problem is given as follows: 𝐷^𝛼_𝑡𝑣()=1𝑥,𝑡2𝑥²𝑣_𝑥𝑥()𝑣()()(0)𝑥,𝑡,0<𝛼≤1,𝑡>0𝑥,0=𝑢𝑥,0−𝑤=𝑥²+12,𝑣()()(𝑡)=𝑒0,𝑡=𝑢0,𝑡−𝑤^2𝑡2(𝑡),𝑣−𝑤_𝑥(0,𝑡)=0. (61) Applying 𝐽^𝛼_𝑡 to both sides of (61), then we get 𝑣()()=1𝑥,𝑡−𝑣𝑥,02𝑥²𝐽^𝛼_𝑡𝑣_𝑥𝑥(𝑥,𝑡) (62) which implies 𝑣(𝑥,𝑡)=𝑥²+12+12𝑥²𝐽^𝛼_𝑡𝑣_𝑥𝑥(𝑥,𝑡). (63) By using ADM for (63), we obtain 𝑣₀=𝑥²+12,𝑣_𝑘+1=12𝑥²𝐽^𝛼_𝑡𝑣_𝑘𝑥𝑥(𝑥,𝑡),𝑘≥0. (64) Then, for 𝑘=0, we get 𝑣₁=12𝑥²𝐽^𝛼_𝑡𝑣_0𝑥𝑥(𝑥,𝑡)=𝑥²𝑡^𝛼Γ(),𝛼+1 (65) similarly, for 𝑘=1, we get 𝑣₂=12𝑥²𝐽^𝛼_𝑡𝑣_1𝑥𝑥(𝑥,𝑡)=𝑥²𝑡^2𝛼Γ(),2𝛼+1 (66) and for 𝑘=2, we get 𝑣₃=12𝑥²𝐽^𝛼_𝑡𝑣_2𝑥𝑥(𝑥,𝑡)=𝑥²𝑡^3𝛼Γ(),⋮3𝛼+1 (67) As a result, we get the solution 𝑣 as follows: 𝑣(𝑥,𝑡)=𝑣₀+𝑣₁+𝑣₂+𝑣₃+⋯=𝑥²+12+𝑥²𝑡^𝛼Γ(𝛼+1)+𝑥²𝑡^2𝛼Γ(2𝛼+1)+𝑥²𝑡^3𝛼Γ()=13𝛼+1+⋯2+𝑥²𝑡1+^𝛼Γ()+𝑡𝛼+1Γ(^2𝛼)+𝑡2𝛼+1_3𝛼Γ().3𝛼+1+⋯ (68) Therefore, from the boundary condition we have 𝑤(𝑡)=𝑒^2𝑡2−12. (69) Using (69) in the definition 𝐷^𝛼_𝑡𝑤(𝑡)=𝐷^𝛼_𝑡𝐽^𝛼_𝑡𝑓(𝑡)=𝑓(𝑡), finally we obtain the source function 𝑓(𝑡) as 𝑓(𝑡)=𝐷^𝛼_𝑡((𝑒^2𝑡/2)−(1/2)); that is, 𝑓(𝑡)=12𝐷^𝛼_𝑡𝑒^2𝑡. (70) Here, 𝐷^𝛼_𝑡𝑒^2𝑡=𝑡^−𝛼𝐸_1,1−𝛼(2𝑡), (71) where 𝐸_1,1−𝛼 is Mittag-Leffler function with two parameters given as; (6).

4. Conclusion

The best part of this method is that one can easily apply ADM to the fractional partial differential equations like applying ADM to ordinary differential equations.