# Solution of Space-Time Fractional Differential Equations Using Aboodh Transform Iterative Method

Authors: Michael A. Awuya; Gbenga O. Ojo; Nazim I. Mahmudov
Journal: Journal of Mathematics
 (2022)

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## Abstract

A relatively new and efficient approach based on a new iterative method and the Aboodh transform called the Aboodh transform iterative method is proposed to solve space-time fractional differential equations, the fractional order is considered in the Caputo sense. This method is a combination of the Aboodh transform and the new iterative method and gives the solution in series form with easily computable components. The nonlinear term is easily handled by the new iterative method, to affirm the simplicity and performance of the proposed method, five examples were considered, and the solution plots were presented to show the effect of the fractional order. The outcome reveals that the approach is accurate and easy to implement.


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## Body

## 1. Introduction

Fractional Calculus can be described as the field of mathematics that consists of ordinary and partial derivatives of positive noninteger order. It is the generalization of classical integral and differential equations [1, 2]. One major attractive property of fractional calculus is the nonlocal property.Recently, various problems in Biology and Physics has been modeled with fractional order derivative, an analytical solution of the Fornberg–Whithan equation was presented in [3], fractional model of the Rosenau–Hyman equation which is a KdV-like equation was considered in [4], for application of fractional derivative to Biology population model see [5], the numerical study of HIV-1 infection of CD4+ T-cell was presented in [6], Caputo–Fabrizio fractional model of photocatalytic degradation of dyes was studied in [7], a wavelet based numerical scheme for fractional order SEIR epidemic of measles by using Genocchi polynomials was presented in [8], and the investigation of fractional order susceptible-infected-recovered epidemic model of childhood disease was presented in [9]. Therefore, it is extremely important to find an effective method of solving fractional differential equations, as only the solutions can give a better comprehension of the underlying problems. Many researchers have presented different methods for solving fractional differential equations such as reproducing kernel discretization method [10], Chebyshev wavelet collocation method, [11] Tichonov regularization method [12], Chebyshev collocation method, [13] q-homotopy analysis Shehu transform method [14], Fractional differential transform, [15] Fractional variational iterational method [16], and iterative Laplace transform method [17].In 2016, the new iterative method was presented by Daftardar–Gejji and Jafari to solve functional equations [18], but now the iterative method has been used to solve many integral and fractional order differential equations. [5, 19, 20] But most of these methods considered a single term time-fractional order differential equations.In this paper, the main objective is to extend the Aboodh transform iterative method to solve space-time fractional differential equations with more than a single term fractional derivative. The fractional derivative is considered in Caputo sense both for time and space, whenα=β=1, the space-time fractional differential equations becomes the classical differential equations. The rest of this paper is arranged as follows: in Section 2, we gave some definitions and a preliminary concept of Aboodh transform. In Section 3, we described briefly the Aboodh transform iterative method for space-time fractional derivative while in Section 4, a few examples were considered to describe the efficiency of the method. Finally, we concluded in Section 5.


## 2. Definitions and Preliminaries

In this section, we give some definitions and notions about Aboodh transform.Definition 1.
Caputo time-fractional derivative of orderα>0 for the function Qx,t is defined as follows [1, 2]:(1)DtαQx,t=1Γn−α∫0tt−τn−α−1Qnx,τdτ,n−1<α≤n.
Similarly, the Caputo space fractional derivative of orderβ>0 for the function Qx,t is defined as follows:(2)DxβQx,t=1Γn−β∫0xx−tn−β−1Qnx,tdt,n−1<β≤n.Remark 1.
DtαQx,t=DxβQx,t=0, whenever Qx,t is a constant.Remark 2.
Dtαtb=Γb+1/Γb−α+1tb−α,ifn−1<α≤n,b>α−1,0,n−1<α≤n,b≤α−1.Definition 2.
One parameter Mittag-Leffler function is given as follows [5]:(3)Eαz=∑k=0∞zkΓ1+kα,α,z∈CReα≥0.Definition 3.
The Aboodh transform ofQt is defined as follows [5]:(4)AQt=1v∫0∞Qte−vtdt=Av,t≥0.
The inverse Aboodh transform of functionQt if AQt=Av is defined as follows:(5)Qt=A−1Av.Remark 3.
The Aboodh transform of the functionQt satisfy the linearity property [5].Definition 4.
The Aboodh transform for Caputo time-fractional derivative of orderβ is given as follows [5]:(6)ADtβQx,t;v=vβAQx,t−∑k=0m−1Qkx,0v2−β+k,m−1<β≤m.


## 3. Basic Idea of the Proposed Method

Consider the space-time fractional partial differential equation of the form.(7)DtαQx,t=ΦQx,t,DxβQx,t,Dx2βQx,t,Dx3βQx,t,0<α,β≤1,with the initial conditions(8)Qkx,0=hk,k=0,1,…,m−1,Qx,t is the unknown function to be determine and ΦQx,t,DxβQx,t,Dx2βQx,t,Dx3βQx,t can be linear or nonlinear operator of Qx,t, DxβQx,t, Dx2βQx,t, and Dx3βQx,t For convenience we represent Qx,t with Q, so by applying the Aboodh transform to both sides of equation (7) we have the following equation:(9)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+AΦQ,DxβQ,Dx2βQ,Dx3βQ,taking the inverse Aboodh transform, we get the following equation:(10)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+AΦQ,DxβQ,Dx2βQ,Dx3βQ.The Aboodh transform iterative method gives the solution in form of an infinite series.(11)Qx,t=∑i=0∞Qi.SinceΦQ,DxβQ,Dx2βQ,Dx3βQ is either a linear or nonlinear operator which can be decomposed as follows:(12)ΦQ,DxβQ,Dx2βQ,Dx3βQ=ΦQ0,DxβQ0,Dx2βQ0,Dx3βQ0+∑i=0∞Φ∑k=0iQk,DxβQk,Dx2βQk,Dx3βQ−Φ∑k=1i−1Qk,DxβQk,Dx2βQk,Dx3βQ.Substituting equations (12) and (11) into equation (10) we obtain the following equation:(13)∑i=0∞Qix,t=A−11vα∑k=0m−1Qkx,0v2−α+k+AΦQ0,DxβQ0,Dx2βQ0,Dx3βQ0+A−11vαA∑i=0∞Φ∑k=0iQk,DxβQk,Dx2βQk,Dx3βQk−A−11vαAΦ∑k=1i−1Qk,DxβQk,Dx2βQk,Dx3βQk.Now, recursively, we compute the terms.(14)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k,Q1x,t=A−11vαAΦQ0,DxβQ0,Dx2βQ0,Dx3βQ0,⋮Qm+1x,t=A−11vαA∑i=0∞Φ∑k=0iQk,DxβQk,Dx2βQk,Dx3βQk−A−11vαA∑i=0∞Φ∑k=1i−1Qk,DxβQk,Dx2βQk,Dx3βQk,m=1,2,….The series converges rapidly, for convergence see [18, 21]. So the m-term analytically approximate solution of equation (7) is given by the following equation:(15)Qx,t≈∑i=0m−1Qi.


## 4. Application

Here, the Aboodh transform iterative method is applied to solve five distinct space-time fractional differential equations with suitable initial conditions.Example 1.
Consider the fractional Airy’s-like equation with an additional term [22].(16)DtαQx,t=DxβQ+Q,0<α,β≤1.
With the initial condition,(17)Qx,0=x3.
Applying the Aboodh transform on both sides of equation (16), we obtain the following equation:(18)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+ADxβQ+Q,taking the inverse Aboodh transform on equation (18), we have the following equation:(19)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+ADxβQ+Q.
Using the Aboodh transform iterative procedure, we obtain the following equation:(20)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k=A−1Qx,0v2=x3,Q1x,t=A−11vαADxβQ0+Q0=A−1Γ4x3−βv2+αΓ4−β+x3v2+α=Γ4x3−βtαΓα+1Γ4−β+x3tαΓα+1,Q2x,t=A−11vαADxβQ0+Q1+Q0+Q1−A−11vαADxβQ0+Q0=A−1Γ4x3−βv2+αΓ4−β+Γ4x3−2βv2+2αΓ4−2β+2Γ4x3−βv2+2αΓ4−β+x3v2+α+x3v2+2α−A−1Γ4x3−βv2+αΓ4−β+x3v2+α=Γ4x3−2βt2αΓ2α+1Γ4−2β+2Γ4x3−βt2αΓ4−βΓ2α+1+x3t2αΓ2α+1,⋮and so on. The series solution is given by the following equation:(21)Qx,t=Q0+Q1+Q2+⋯.
Figure1 represent the solution plots of equation (16) when α=β=.02,.04,.06,.08,2,.4,.6,.8 at x=1 and t = 1, respectively. While the remaining are the surface plots.Figure 1
Comparison of the solution at various values of alpha and beta.Example 2.
Consider the nonlinear space-time fractional Fokker–Planck equation [23].(22)DtαQx,t=DxβxQ3−4xQ2x+Q2xx,0<α,β≤1.
With the initial condition,(23)Qx,0=x2.
Applying the Aboodh transform on equation (22), we obtain the following equation:(24)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+ADxβx3Q−4xQ2x+Q2xx,taking the inverse Aboodh transform, we have the following equation:(25)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+ADxβx3Q−4xQ2x+Q2xx.
Using the Aboodh transform iterative method procedure, we obtain the following equation:(26)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k=A−1Qx,0v2=x2,Q1x,t=A−11vαADxβx3Q0−4xQ02x+Q02xx=A−12x3−βΓ4−βv2+α=2x3−βtαΓ4−βΓα+1,Q2x,t=A−11vαADxβx3Q0+Q1−4xQ0+Q12x+Q0+Q1xx2−A−11vαADxβx3Q0−4xQ02x+Q02xx=A−12x3−βΓ4−βv2+α+2Γ3−βx4−2β3Γ4−βΓ3−2βv2+2α−4−β16x3−βΓ4−βv2+2α−A−15−2β16x4−2βΓ4−βΓα+12v2+3α+45−β4−βx3−βΓ4−βv2+2α+A−146−2β5−2βΓ2α+1x4−2βΓ4−βΓα+12v2+3α−2x3−βΓ4−βv2+α=2Γ5−βx4−2βt2α3Γ4−βΓ5−2βΓ2α+1+4−4β4−βx3−βt2αΓ4−βΓ2α+1+8−8β5−2βΓ2α+1x4−2βt3αΓ4−βΓα+12Γ3α+1,⋮and so on. The series solution is obtained as follows:(27)Qx,t=Q0+Q1+Q2+⋯=x2+2x3−βtαΓ4−βΓα+1+2Γ5−βx4−2βt2α3Γ4−βΓ5−2βΓ2α+1+4−4β4−βx3−βt2αΓ4−βΓ2α+1+8−8β5−2βΓ2α+1x4−2βt3αΓ4−βΓα+12Γ3α+1+⋯.
Settingβ=1 in equation (27), we obtain the following equation:(28)Qx,t=x21+tαΓα+1+t2αΓ2α+1+⋯=x2∑i=0∞tiαΓiα+1.
The solution obtained in equation (28) converges to the exact solution in a closed form as i⟶∞,(29)Qx,t=x2limi⟶∞∑i=0∞tiαΓiα+1=x2Eαtα.
So, by settingα=1, we obtain the following equation:(30)Qx,t=x2et.
Which is the same solution obtained in [23]. Figure 2 represents the solution plots of equation (22) when α=β=.02,.04,.06,.08,.2,.6,.8 at x=1 and t = 1 respectively. While the remaining are the surface plots.Figure 2
Comparison of the solution at various values of alpha and beta.Example 3.
Consider the one-dimensional space-time diffusion equation [24].(31)Dtα=Dx2βQ+DxβxQ,0<α,β≤1.
With the initial condition,(32)Qx,0=1.
Applying the Aboodh transform on equation (31), we obtain the following equation:(33)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+ADx2βQ+DxβxQ.
Taking the Aboodh transform inverse of equation (33), we obtain the following equation:(34)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+ADx2βQ+DxβxQ.
Using the Aboodh transform iterative procedure, we obtain the following equation:(35)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k=A−1Qx,0v2=1,Q1x,t=A−11vαADx2βQ0+DxβxQ0=A−1x1−βv2+αΓ2−β=x1−βtαΓα+1Γ2−β,Q2x,t=A−11vαADx2βQ0+Q1+DxβxQ0+Q1−A−11vαADx2βQ0+DxβxQ0=A−1x1−βΓ2−βv2+α+Γ3−βx2−2βΓ3−2βΓ2−βv2+2α−A−1x1−βΓ2−βv2+α=2−βx2−2βt2αΓ3−2βΓ2α+1.
The series solution is obtained as follows:(36)Qx,t=Q0+Q1+Q2+⋯=1+x1−βtαΓ2−βΓα+1+2−βx2−2βtαΓ3−2βΓ2α+1+⋯.
Settingβ=1 in equation (36), we obtain the following equation:(37)Qx,t=1+tαΓα+1+t2αΓ2α+1+⋯=∑i=0∞tiαΓiα+1.
The solution obtained in equation (37) converges to the exact solution in a closed form as i⟶∞,(38)Qx,t=limi⟶∞∑i=0∞tiαΓiα+1=Eαtα.
So, by settingα=1/2, we obtain the following equation:(39)Qx,t=E1/2t1/2.
Which is the solution obtained in [24] using the natural transform method. Figure 3 represents the solution plots of equation (31) when α=β=.02,.04,.06,.08,.2,.4,.6,.8 at x=1 and t = 1, respectively. While the remaining are the surface plots.Figure 3
Comparison of the solution at various values of alpha and beta.Example 4.
Consider the space-time fractional Airy’s partial differential equations [22].(40)DtαQx,t=Dx3βQ,0<α,β≤1.
With the initial condition,(41)Qx,0=16x3.
Applying the Aboodh transform on equation (40), we get the following equation:(42)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+ADx3βQ,taking the inverse Aboodh transform of equation (42), we get the following equation:(43)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+ADx3βQ.
Using the Aboodh transform iterative method procedure, we obtain the following equation:(44)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k=A−1Qx,0v2=16x3,Q1x,t=A−11vαADx3βQ0=A−1x3−3βΓ4−3βv2+α=x3−3βtαΓ4−3βΓα+1,Q2x,t=A−11vαADx3βQ1+Q0−A−11vαADx3βQ0=A−1Γ4x3−3β6Γ4−3βv2+α−A−1x3−3βΓ4−3βv2+α=0,⋮and so on. The series solution is obtained as follows:(45)Qx,t=Q0+Q1+Q2+⋯=16x3+x3−3βtαΓ4−3βΓα+1+0+⋯,for all i>1, Qix,t=0. Setting β=1 in equation (45), we obtain the following equation:(46)Qx,t=16x3+tαΓα+1+0+0+⋯=16x3+tαΓα+1.
We obtain the exact solution whenα=1,(47)Qx,t=16x3+t,which is the solution obtained in [22]. Figure 4 represents the solution plots of equation (40) when α=β=.02,.04,.06,.08,.2,.4,.6,.8 at x=1 and t = 1 respectively. While Figure 5 is the surface plots.Figure 4
Comparison of the solution at various values of alpha and beta.Figure 5
Comparison of the solution at various values of alpha and beta. (a) Early strength. (b) Early strength loss rate.Example 5.
Consider the nonlinear space-time fractional Fokker–Planck equation which consists of a single term time-fractional order and three terms of space fractional order [23].(48)DtαQx,t=DxβxQ3−Dxβ4Q2x+Dx2βQ2,0<α,β≤1.
Subject to the initial condition,(49)Qx,0=x2.
Applying the Aboodh transform on equation (48), we obtain the following equation:(50)AQx,t=1vα∑k=0m−1Qkx,0v2−α+k+ADxβxQ3−Dxβ4Q2x+Dx2βQ2,taking the inverse Aboodh transform, we get the following equation:(51)Qx,t=A−11vα∑k=0m−1Qkx,0v2−α+k+ADxβxQ3−Dxβ4Q2x+Dx2βQ2.
Using Aboodh transform iterative procedure, we get the following equation:(52)Q0x,t=A−11vα∑k=0m−1Qkx,0v2−α+k=A−1Qx,0v2=x2,Q1x,t=A−11vαADxβxQ03−Dxβ4Q02x+Dx2βQ02=A−124x4−2βΓ5−2βv2+α−22x3−βΓ4−βv2+α=24x4−2βtαΓ5−2βΓα+1−22x3−βtαΓ4−βΓα+1,Q2x,t=A−11vαADxβxQ0+Q13−Dxβ4Q0+Q12x+Dx2βQ0+Q12−A−11vαADxβxQ03−Dxβ4Q02x+Dx2βQ02=A−18Γ6−2βx5−3βΓ5−2βΓ6−3βv2+2α−22Γ5−βx4−2β3Γ4−βΓ5−2βv2+2α+Γ4x3−β3Γ4−βv2+α+A−1−1152Γ8−4βΓ2α+1x7−5βΓ5−2βΓα+12Γ8−5βv2+3α−192Γ6−2βx5−3βΓ5−2βΓ6−3βv2+2α+A−1−1936Γ6−2βΓ2α+1x5−3βΓ4−βΓα+12Γ6−3βv2+3α+4,224Γ7−3βΓ2α+1x6−4βΓ5−2βΓα+12Γ4−βΓ7−4βv2+3α+A−1176Γ5−βx4−2βΓ4−βΓ5−2βv2+2α−4Γ4x3−βΓ4−βv2+α+576Γ9−4βΓ2α+1x8−6βΓ5−2βΓα+12Γ9−6βv2+3α+A−1−1056Γ8−3βΓ2α+1x7−5βΓ5−2βΓα+12Γ4−βΓ8−5βv2+3α+48Γ7−2βx6−4βΓ5−2βΓ7−4βv2+2α+A−1484Γ7−2βΓ2α+1x6−4βΓ4−βΓα+12Γ7−4βv2+3α+44Γ6−βx5−3βΓ4−βΓ6−3βv2+2α+24x4−2βΓ5−2βv2+α−A−124x4−2βΓ5−2βv2+α−22x3−βΓ4−βv2+α=8Γ6−2βx5−3βt2αΓ5−2βΓ6−3βΓ2α+1−1152Γ8−4βΓ2α+1x7−5βt3αΓ5−2βΓα+12Γ8−5βΓ3α+1−22Γ5−βx4−2βt2α3Γ4−βΓ5−2βΓ2α+1+4224Γ7−3βΓ2α+1x6−4βt3αΓ5−2βΓα+12Γ4−βΓ7−4βΓ3α+1−1436Γ6−2βΓ2α+1x5−3βt3αΓ4−βΓα+12Γ6−3βΓ3α+1−192Γ6−2βx5−3βt2αΓ5−2βΓ6−3βΓ2α+1+176Γ5−βx4−2βx4−2βt2αΓ4−βΓ5−2βΓ2α+1+576Γ9−4βΓ2α+1x8−6βt3αΓ5−2βΓα+12Γ9−6βΓ3α+1−1056Γ8−3βΓ2α+1x7−5βt3αΓ5−2βΓα+12Γ4−βΓ8−5βΓ3α+1+48Γ7−2βx6−4βt2αΓ5−2βΓ7−4βΓ2α+1+484Γ7−2βΓ2α+1x6−4βt3αΓ4−βΓα+12Γ7−4βΓ3α+1+44Γ6−βx5−3βt2αΓ4−βΓ6−3βΓ2α+1,⋮and so on. The series solution is obtained as follows:(53)Qx,t=Q0+Q1+Q2+⋯.
Settingα=β=1 we get the following equation:(54)Qx,t=x2+x2t+x2t22+⋯=x21+t+t22+⋯.
Hence,(55)Qx,t=x2∑i=0∞tii!=x2et,which agrees with the exact solution obtained in [23], also it is similar to the solution obtained in Example 2. The reason being that in Example 2, only one space fractional derivative term was considered while here three terms of space fractional derivative was considered.


## 5. Conclusion and Future Work

We proposed the Aboodh transform iterative method for the solution of space-time fractional differential equation with fractional order derivative in more than one term. The proposed method is efficient and effective, the method combined the Aboodh transform which is a modification of the Laplace transform with the new iterative method. To the best of our knowledge, no attempt has been recorded regarding the approximate analytical solution of space-time fractional differential equations using the Aboodh transform iterative method which is the novelty of this study.The new iterative method decomposes the linear and nonlinear term. Some examples were considered, ifα=β=1 the fractional differential equations becomes the classical differential equations. Aboodh transforms iterative method yields closed form solutions in this study and exact solutions in some cases. Also, the effect of the fractional orders α and β are displayed in Figures 1 to 5, this is left for the readers in different fields of study to transcribe for different applications.In the future, we hope to extend the Aboodh transform iterative method to solve boundary value problems with consideration for other fractional order differential equations which till date have not been solved either analytically or numerically.



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