# A Comparative Study of Fractional-Order Diffusion Model within Atangana-Baleanu-Caputo Operator

Authors: Mohammad Alshammari; Naveed Iqbal; Davis Bundi Ntwiga
Journal: Journal of Function Spaces
 (2022)

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## Abstract

We provided two different approaches for solving fractional-order diffusion equations in this article. The fractional Atangana-Baleanu derivative operator in addition to the Laplace transform is used to generate several new approximate-analytical solutions to the time-fractional diffusion equations. The implementation of a sophisticated and straightforward approach to solving diffusion equations having a fractional-order derivative is the motivation and uniqueness behind the current work. The solutions to some illustrative problems are calculated to ensure that the actual and approximate solutions to the targeted problems are in close contact. The results we obtained have a higher rate of convergence and provide a closed-form solution, according to analysis. The proposed method’s key advantage is the small amount of calculations required. The suggested techniques can be applied to nonlinear fractional-order problems in a variety of applied science areas due to their simple and straightforward implementation. It can be used to overcome specific fractional-order physical problems in a variety of fields of applied sciences.


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## Body

## 1. Introduction

Fractional calculus (FC) is a branch of calculus that is an extension of ordinary calculus. In fact, the origins of FC may be traced back over 300 years. The idea of fractional calculus comes from Leibniz’s usage of the notationdny/dxn in his publication for the nth derivative. L’Hopital asked Leibniz about the derivative of order 1/2 in 1695 [1]. In nature, integer-order derivatives are local, whereas fractional-order derivatives are nonlocal. Several mathematicians, including Iqbal et al. [2], Alesemi et al. [3], Riemann [4], Alesemi et al. [5], and Podlubny [6], collaborated to establish the mathematical framework for fractional-order derivatives. Later on, a number of mathematicians focused their efforts on this topic. Because of its potential applications in several scientific fields such as biology, viscoelasticity, engineering, fluid mechanics, and other branches of science, fractional calculus has attracted a lot of attention [7–10].Mathematicians used the method proposed above to solve various fractional differential equations (FDEs) in general, as well as fractional partial differential equations (FPDEs). FPDEs are regarded to be one of the most basic mathematical tools for modelling many physical phenomena with greater accuracy than integer-order. Different phenomena in engineering and applied sciences are determined by nonlinear FPDEs. Nonlinear FPDEs are useful tools in a variety of fields, including material sciences, physics, fluid dynamics, chemistry, chemical kinetics, and a variety of other physical processes [11–15]. In literature, different methods are found for solving differential equations having fractional-order such as the modified Adomian decomposition method (MADM) [16], variational iteration method [17], optimal homotopy asymptotic method (OHAM) [18], differential transform method (DTM) [19], and homotopy perturbation method (HPM) [20–22].In this paper, we used the Atangana-Baleanu fractional derivative operator [23–26] and the Laplace transform to solve fractional-order diffusion equations using two analytical methodologies. The first methodology shows how the Laplace transform may be used to approximate the solution of a nonlinear differential using Adomian’s decomposition method. The Laplace decomposition method has been demonstrated to work with a wide range of physical problems and has been used for a large number of functional equations [27, 28]. The Laplace Variational Iteration Approach (LVIM), which combines the Laplace transform (LT) and the variational iteration method, is the second approach. Lagrange multiplies are chosen to solve nonlinear equations in this method based on the nature of the problem [29, 30]. He was the first to introduce VIM [31], a powerful method for solving a wide range of issues in applied sciences [32–34]. This method of solving linear and nonlinear equations has lately gained popularity among scholars. Unlike other methodologies, VIM does not require discretization, linearization, or perturbation.We use LTDM and VITM to solve diffusion equations of the form in this article.(1)
Fractional diffusion equation in one dimension as(1)∂ρμ∂κρ=∂2μ∂κ2−∂μ∂κ+μ∂2μ∂κ2−μ2+μ,0<ρ≤1,ϑ>0,with initial source
(2)μκ,0=eκ(2)
Fractional diffusion equation in one dimension as(3)∂ρμ∂κρ=∂2μ∂κ2+μ∂μ∂κ−μ2+μ,0<ρ≤1,ϑ>0,with initial source
(4)μκ,0=1+eκ(3)
Fractional diffusion equation in three dimensions as(5)∂ρμ∂κρ=∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2,0<ρ≤1,ϑ≥0,with initial source
(6)μκ,ℓ,ψ,0=sinκsinℓsinψ.The diffusion equation is the partial differential equation. The movement of atoms or molecules from a location of higher concentration to a region of lower concentration is referred to as diffusion. Adolf Fick, a physiologist, proposed Fick’s law of diffusion in 1885. Fick’s second law was later renamed the diffusion equation. Slow diffusion, diffusion-wave hybrid, classical diffusion, and the classical wave equation [35] were developed as generalisations of classical diffusion and wave equations. Diffusion equation applications include filtration, electromagnetism, electrochemistry, phase transitions, cosmology, biochemistry, acoustics, and biological group dynamics [36–38].


## 2. Preliminaries

In this part, we discussed some basic definitions of fractional calculus.Definition 1.
The fractional derivative in Caputo manner aswhere
(7)LCDϑρgϑ=1n−ρ∫0ϑϑ−kn−ρ−1gnkdk,n<ρ≤n+1.Definition 2.
The Laplace transform in connection with Caputo derivativeLCDϑρgϑis given as(8)LLCDϑρgϑν=1νn−ρνnLgκ,ϑν−νn−1gκ,0−⋯−gn−1κ,0.Definition 3.
The Atangana-Baleanu derivative in Caputo manner is as(9)ABCDϑρgϑ=Aρ1−ρ∫aϑg′kEρ−ρ1−ρ1−kρdk,where Aγ represent normalization function with A0=A1=1,g∈H1a,b,b>a,ρ∈0,1 and Eγ illustrates the Mittag-Leffler function.Definition 4.
The Atangana-Baleanu derivative in Riemann-Liouville manner is as(10)ABCDϑρgϑ=Aρ1−ρddϑ∫aϑgkEρ−γ1−ρ1−kρdk.Definition 5.
The Laplace transform connected with the Atangana-Baleanu operator is given as(11)ABDϑρgϑν=AγνρLgϑν−νρ−1g01−ρνρ+ρ/1−γ.Definition 6.
Consider0<ρ<1, andgis a function ofρ, then the fractional integral operator ofρis as(12)ABCIϑρgϑ=1−ρAρgϑ+ρAρΓρ∫aϑgkϑ−kρ−1dk.


## 3. Idea of LTDM

Here, we discuss the general methodology of LTDM for solving the fractional partial differential equation:(13)Dϑρμκ,ϑ+G¯1κ,ϑ+N1κ,ϑ=Fκ,ϑ,0<ρ≤1,with initial conditions
(14)μκ,0=ϕκ,∂∂ϑμκ,0=ζκ,where Dϑρ=∂ρ/∂ϑρ represents fractional AB operator of order ρ. G¯1,N1 are linear and nonlinear terms, and Fκ,ϑ demonstrates the source term.On applying Laplace transform to (13),
(15)LDϑρμκ,ϑ+G¯1κ,ϑ+N1κ,ϑ=LFκ,ϑ.By the Laplace differentiation property, we get(16)Lμκ,ϑ=Θκ,ν−νρ+ρ1−ρνρLG¯1κ,ϑ+N1κ,ϑ,where Θκ,ν=1/νρ+1νρg0κ+νρ−1g1κ+⋯+g1κ+νρ+ρ1−ρ/νρFκ,ϑ.On taking Laplace inverse transform,(17)μκ,ϑ=Θκ,ν−L−1νρ+ρ1−ρνρLG¯1κ,ϑ+N1κ,ϑ,where Θκ,ν represent the term coming from the source term. Let μκ,ϑ have series form solution as
(18)μκ,ϑ=∑m=0∞μmκ,ϑ,and the decomposition of nonlinear term N1 is as
(19)N1κ,ϑ=∑m=0∞Am,where Am represents Adomian polynomials as
(20)Am=1m!∂m∂ℓmN1∑k=0∞ℓkκk,∑k=0∞ℓkϑkℓ=0.By substituting Equations (18) and (20) into (17), we get
(21)∑m=0∞μmκ,ϑ=Θκ,ν−L−1νρ+ρ1−ρνρLG¯1∑m=0∞κm,∑m=0∞ϑm+∑m=0∞Am.The following terms are described:(22)μ0κ,ϑ=Θκ,ν,μ1κ,ϑ=L−1νρ+ρ1−ρνρLG¯1κ0,ϑ0+A0.Thus, all the terms form≥1 are determined as
(23)μm+1κ,ϑ=L−1νρ+ρ1−ρνρLG¯1κm,ϑm+Am.


## 4. VITM Formulation

Here, we discuss the general methodology of VITM for solving a fractional partial differential equation:(24)Dϑρμκ,ϑ+Mμκ,ϑ+Nμκ,ϑ−Pκ,ϑ=0,m−1<ρ≤m,with initial source
(25)μκ,0=g1κ,where Dϑρ=∂ρ/∂ϑρ represents fractional AB operator. M,N are linear and nonlinear terms, and P demonstrates the source term.On applying the Laplace transform to (24),
(26)LDϑρμκ,ϑ+LMμκ,ϑ+Nμκ,ϑ−Pκ,ϑ=0.By the Laplace differentiation property, we get(27)Lμκ,ϑ=νρνρ+ρ1−ρLMμκ,ϑ+Nμκ,ϑ−Pκ,ϑ.Thus, for Equation (27), the iteration technique is
(28)μm+1κ,ϑ=μmκ,ϑ+ρννρνρ+ρ1−ρLMμκ,ϑ+Nμκ,ϑ−Pκ,ϑ.ρν is Lagrange multiplier and
(29)ρν=−νρ+ρ1−ρνρ.On taking Laplace inverse transform, the series form solution for Equation (28) is
(30)μ0κ,ϑ=μ0+L−1ρsL−Pκ,ϑ,μ1κ,ϑ=L−1ρsLMμκ,ϑ+Nμκ,ϑ,⋮μn+1κ,ϑ=L−1ρsLMμ0κ,ϑ+μ1κ,ϑ+⋯,μnκ,ϑ+Nμ0κ,ϑ+μ1κ,ϑ,⋯,μnκ,ϑ.


## 5. Applications

The solutions to different time-fractional diffusion equations are derived using LTDM and VITM in this section.


### 5.1. Example

Consider fractional diffusion equation in one dimension as(31)∂ρμ∂κρ=∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ0<ρ≤1,ϑ>0,with initial source
(32)μκ,0=eκ.On applying the Laplace transform to (31),
(33)νρLμκ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ.On taking Laplace inverse transform,(34)μκ,ϑ=expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ.Letμκ,ϑ have series form solution as
(35)μκ,ϑ=∑m=0∞μmκ,ϑ,where μμκ=∑k=0∞Ak and μ2=∑k=0∞Bk illustrates Adomian polynomials that show nonlinear terms; thus, on putting certain terms in Equation (34), we get
(36)∑m=0∞μmκ,ϑ=expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2−∂μ∂κ+∑k=0∞Ak−∑k=0∞Bk+μ.The nonlinear term decomposition by means of Adomian polynomials is given in Equation (20) as
(37)A0=μ0μ0κ,A1=μ1μ0κ+μ0μ1κ,A2=μ2μ0κ+μ1μ1κ+μ0μ2κ,B0=μ02,B1=2μ0μ1,B2=2μ0μ2+μ22.By the comparison of Equation (36), both sides
(38)μ0κ,ϑ=eκ.Form=0,
(39)μ1κ,ϑ=eκρϑρΓρ+1+1−ρ.Form=1,
(40)μ2κ,ϑ=eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(41)μ3κ,ϑ=eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is(42)μκ,ϑ=∑m=0∞μmκ,ϑ=μ0κ,ϑ+μ1κ,ϑ+μ2κ,ϑ+μ3κ,ϑ+⋯,μκ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=eκ+ϑ.The analytical solution is by VITM.In the iterative formula for Equation (31), we get
(43)μm+1κ,ϑ=μmκ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2−∂μm∂κ+μm∂μm∂κ−μm2+μm,where
(44)μ0κ,ϑ=eκ.Form=0,1,2,⋯,
(45)μ1κ,ϑ=μ0κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2−∂μ0∂κ+μ0∂μ0∂κ−μ02+μ0,μ1κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ,μ2κ,ϑ=μ1κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2−∂μ1∂κ+μ1∂μ1∂κ−μ12+μ1,μ2κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ϑ=μ2κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2−∂μ2∂κ+μ2∂μ2∂κ−μ22+μ2,μ3κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ϑ=∑m=0∞μmκ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=eκ+ϑ.Atρ=1, the graphs in Figure 1 demonstrate the exact and approximate solutions in (AB fractional derivative) sense. For example 1, Figure 1 demonstrates our method’s solution for various fractional orders of ρ=1,0.8,0.6,0.4 and 0≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical illustration. Table 1 also shows a comparison of the exact solution and our approaches solution at various fractional orders using absolute error.Figure 1
The graphical layout for problem 1.Table 1
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 1.
ϑ=0.01Exact solutionOur method solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.801.0010005000000001.0010005000000000.0000000000E+007.3650000000E−067.6109000000E−050.11.1062766420000001.1062766420000000.0000000000E+008.1390000000E−068.4113000000E−050.21.2226247720000001.2226247710000001.0000000000E−098.9950000000E−069.2959000000E−050.31.3512093420000001.3512093420000000.0000000000E+009.9410000000E−061.0273600000E−040.41.4933172680000001.4933172690000001.0000000000E−091.0988000000E−051.1354200000E−040.51.6503708170000001.6503708170000000.0000000000E+001.2142000000E−051.2548200000E−040.61.8239418310000001.8239418300000001.0000000000E−091.3419000000E−051.3867800000E−040.72.0157674670000002.0157674670000000.0000000000E+001.4831000000E−051.5326400000E−040.82.2277675830000002.2277675820000001.0000000000E−091.6390000000E−051.6938200000E−040.92.4620639440000002.4620639440000000.0000000000E+001.8115000000E−051.8719800000E−041.02.7210014700000002.7210014690000001.0000000000E−092.0019000000E−052.0688500000E−04


### 5.2. Example

Consider fractional diffusion equation in one dimension as(46)∂ρμ∂κρ=∂2μ∂κ2+μ∂μ∂κ−μ2+μ0<ρ≤1,ϑ>0,with initial source
(47)μκ,0=1+eκ.On applying Laplace transform to (46),
(48)νρLμκ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2+μ∂μ∂κ−μ2+μ.Applying Laplace inverse transformation,(49)μκ,ϑ=1+expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+μ∂μ∂κ−μ2+μ.Letμκ,ϑ have a series form solution as
(50)μκ,ϑ=∑m=0∞μmκ,ϑ,where μμκ=∑k=0∞Ak and μ2=∑k=0∞Bk illustrates Adomian polynomials that show nonlinear terms; thus, on putting certain terms in Equation (49), we get
(51)∑m=0∞μmκ,ϑ=1+expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∑k=0∞Ak−∑k=0∞Bk+μ.The nonlinear term decomposition by means of Adomian polynomials is given in Equation (20) as
(52)A0=μ0μ0κ,A1=μ1μ0κ+μ0μ1κ,A2=μ2μ0κ+μ1μ1κ+μ0μ2κ,B0=μ02,B1=2μ0μ1,B2=2μ0μ2+μ22.By the comparison of Equation (51), both sides
(53)μ0κ,ϑ=1+eκ.Form=0,
(54)μ1κ,ϑ=1+eκρϑρΓρ+1+1−ρ.Form=1,
(55)μ2κ,ϑ=1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(56)μ3κ,ϑ=1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is as(57)μκ,ϑ=∑m=0∞μmκ,ϑ=μ0κ,ϑ+μ1κ,ϑ+μ2κ,ϑ+μ3κ,ϑ+⋯,μκ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=1+eκ+ϑ.The analytical solution is by VITM.In the iterative formula for Equation (46), we get
(58)μm+1κ,ϑ=μmκ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2+μm∂μm∂κ−μm2+μm,where
(59)μ0κ,ϑ=1+eκ.Form=0,1,2,⋯,
(60)μ1κ,ϑ=μ0κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2+μ0∂μ0∂κ−μ02+μ0,μ1κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ,μ2κ,ϑ=μ1κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2+μ1∂μ1∂κ−μ12+μ1,μ2κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ϑ=μ2κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2+μ2∂μ2∂κ−μ22+μ2,μ3κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ϑ=∑m=0∞μmκ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=1+eκ+ϑ.Figure2 depicts the accurate and approximate solutions in the (AB fractional derivative) sense for ρ=1 and at various fractional-orders inside the domain 0≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical depiction. Table 2 shows a comparison of the exact solution and the solution obtained using our approaches, demonstrating the correctness of the presented methods. Table 2 also includes a comparison of absolute errors at various fractional orders, indicating that results improve as fractional order moves to integer order.Figure 2
The graphical layout for problem 2.Table 2
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 2.
ϑ=0.01Exact solutionOur methods solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.802.0000001000000002.0000002000000001.0000000000E−071.0280000000E−061.2534000000E−050.12.1051710290000002.1051711290000001.0000000000E−071.0300000000E−061.2667000000E−050.22.2214028800000002.2214029800000001.0000000000E−071.0310000000E−061.2815000000E−050.32.3498589430000002.3498590430000001.0000000000E−071.0340000000E−061.2978000000E−050.42.4918248470000002.4918249470000001.0000000000E−071.0360000000E−061.3157000000E−050.52.6487214360000002.6487215360000001.0000000000E−071.0380000000E−061.3356000000E−050.62.8221189830000002.8221190820000009.9000000000E−081.0390000000E−061.3575000000E−050.73.0137529090000003.0137530080000009.9000000000E−081.0420000000E−061.3818000000E−050.83.2255411510000003.2255412510000001.0000000000E−071.0450000000E−061.4086000000E−050.93.4596033570000003.4596034570000001.0000000000E−071.0480000000E−061.4383000000E−051.03.7182821000000003.7182822000000001.0000000000E−071.0510000000E−061.4711000000E−05


### 5.3. Example

Consider the fractional diffusion equation in three dimensions as(61)∂ρμ∂κρ=∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2,0<ρ≤1,ϑ≥0,with initial source
(62)μκ,ℓ,ψ,0=sinκsinℓsinψ.On applying Laplace transform to (61),
(63)νρLμκ,ℓ,ψ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.On taking Laplace inverse transform,(64)μκ,ℓ,ψ,ϑ=sinκsinℓsinψ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.Letμκ,ℓ,ψ,ϑ have series form solution as
(65)μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ,(66)∑m=0∞μmκ,ℓ,ψ,ϑ=sinκsinℓsinψ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.By the comparison of Equation (66), both sides
(67)μ0κ,ℓ,ψ,ϑ=sinκsinℓsinψ.Form=0,
(68)μ1κ,ℓ,ψ,ϑ=−3sinκsinℓsinψρϑρΓρ+1+1−ρ.Form=1,
(69)μ2κ,ℓ,ψ,ϑ=−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(70)μ3κ,ℓ,ψ,ϑ=−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is as(71)μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ=μ0κ,ℓ,ψ,ϑ+μ1κ,ℓ,ψ,ϑ+μ2κ,ℓ,ψ,ϑ+μ3κ,ℓ,ψ,ϑ+⋯,μκ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ℓ,ψ,ϑ=exp−3ϑsinκsinℓsinψ.The analytical solution is by VITM.The iterative formula for Equation (61), we get
(72)μm+1κ,ℓ,ψ,ϑ=μmκ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2+∂2μm∂ℓ2+∂2μm∂ψ2,where
(73)μ0κ,ℓ,ψ,ϑ=sinκsinℓsinψ.Form=0,1,2,⋯,
(74)μ1κ,ℓ,ψ,ϑ=μ0κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2+∂2μ0∂ℓ2+∂2μ0∂ψ2,μ1κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ,μ2κ,ℓ,ψ,ϑ=μ1κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2+∂2μ1∂ℓ2+∂2μ1∂ψ2,μ2κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ℓ,ψ,ϑ=μ2κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2+∂2μ2∂ℓ2+∂2μ2∂ψ2,μ3κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get solution atρ=1 as μκ,ℓ,ψ,ϑ=exp−3ϑsinκsinℓsinψ.Figure3 depicts the accurate and approximate solutions in the (AB fractional derivative) sense for ρ=1 and at various fractional orders inside the domain −1≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical depiction. The LTDM and VITM solutions are closely connected to the exact solution, as seen in Figure 3 and Table 3.Figure 3
The graphical layout for problem 3.Table 3
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 3.
ϑ=0.01Exact solutionOur method solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.800.0000000000000000.0000000000000000.0000000000E+001.1470000000E−042.3021230000E−030.10.0402750524300000.0402750604900008.0612920340E−094.0267087740E−054.0274327760E−040.20.0801476898400000.0801477058800001.6038436970E−088.0131842970E−058.0146248110E−040.30.1192195181000000.1192195419000002.3813628870E−081.1919596380E−041.1921737920E−030.40.1571001442000000.1571001756000003.1386867730E−081.5706905820E−041.5709731480E−030.50.1934110777000000.1934111164000003.8707665300E−081.9337280920E−041.9340759560E−030.60.2277895116000000.2277895572000004.5607727380E−082.2774445860E−042.2778541300E−030.70.2598919480000000.2598920000000005.2002906860E−082.5984051810E−042.5988727020E−030.80.2893976300000000.2893976879000005.7893203750E−082.8934038720E−042.8939242370E−030.90.3160117465000000.3160118097000006.3194470960E−083.1594929850E−043.1600606410E−031.00.3394683781000000.3394684460000006.7906708470E−083.3940126310E−043.3946226980E−03


## 5.1. Example

Consider fractional diffusion equation in one dimension as(31)∂ρμ∂κρ=∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ0<ρ≤1,ϑ>0,with initial source
(32)μκ,0=eκ.On applying the Laplace transform to (31),
(33)νρLμκ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ.On taking Laplace inverse transform,(34)μκ,ϑ=expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2−∂μ∂κ+μ∂μ∂κ−μ2+μ.Letμκ,ϑ have series form solution as
(35)μκ,ϑ=∑m=0∞μmκ,ϑ,where μμκ=∑k=0∞Ak and μ2=∑k=0∞Bk illustrates Adomian polynomials that show nonlinear terms; thus, on putting certain terms in Equation (34), we get
(36)∑m=0∞μmκ,ϑ=expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2−∂μ∂κ+∑k=0∞Ak−∑k=0∞Bk+μ.The nonlinear term decomposition by means of Adomian polynomials is given in Equation (20) as
(37)A0=μ0μ0κ,A1=μ1μ0κ+μ0μ1κ,A2=μ2μ0κ+μ1μ1κ+μ0μ2κ,B0=μ02,B1=2μ0μ1,B2=2μ0μ2+μ22.By the comparison of Equation (36), both sides
(38)μ0κ,ϑ=eκ.Form=0,
(39)μ1κ,ϑ=eκρϑρΓρ+1+1−ρ.Form=1,
(40)μ2κ,ϑ=eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(41)μ3κ,ϑ=eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is(42)μκ,ϑ=∑m=0∞μmκ,ϑ=μ0κ,ϑ+μ1κ,ϑ+μ2κ,ϑ+μ3κ,ϑ+⋯,μκ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=eκ+ϑ.The analytical solution is by VITM.In the iterative formula for Equation (31), we get
(43)μm+1κ,ϑ=μmκ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2−∂μm∂κ+μm∂μm∂κ−μm2+μm,where
(44)μ0κ,ϑ=eκ.Form=0,1,2,⋯,
(45)μ1κ,ϑ=μ0κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2−∂μ0∂κ+μ0∂μ0∂κ−μ02+μ0,μ1κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ,μ2κ,ϑ=μ1κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2−∂μ1∂κ+μ1∂μ1∂κ−μ12+μ1,μ2κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ϑ=μ2κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2−∂μ2∂κ+μ2∂μ2∂κ−μ22+μ2,μ3κ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ϑ=∑m=0∞μmκ,ϑ=eκ+eκρϑρΓρ+1+1−ρ+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=eκ+ϑ.Atρ=1, the graphs in Figure 1 demonstrate the exact and approximate solutions in (AB fractional derivative) sense. For example 1, Figure 1 demonstrates our method’s solution for various fractional orders of ρ=1,0.8,0.6,0.4 and 0≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical illustration. Table 1 also shows a comparison of the exact solution and our approaches solution at various fractional orders using absolute error.Figure 1
The graphical layout for problem 1.Table 1
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 1.
ϑ=0.01Exact solutionOur method solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.801.0010005000000001.0010005000000000.0000000000E+007.3650000000E−067.6109000000E−050.11.1062766420000001.1062766420000000.0000000000E+008.1390000000E−068.4113000000E−050.21.2226247720000001.2226247710000001.0000000000E−098.9950000000E−069.2959000000E−050.31.3512093420000001.3512093420000000.0000000000E+009.9410000000E−061.0273600000E−040.41.4933172680000001.4933172690000001.0000000000E−091.0988000000E−051.1354200000E−040.51.6503708170000001.6503708170000000.0000000000E+001.2142000000E−051.2548200000E−040.61.8239418310000001.8239418300000001.0000000000E−091.3419000000E−051.3867800000E−040.72.0157674670000002.0157674670000000.0000000000E+001.4831000000E−051.5326400000E−040.82.2277675830000002.2277675820000001.0000000000E−091.6390000000E−051.6938200000E−040.92.4620639440000002.4620639440000000.0000000000E+001.8115000000E−051.8719800000E−041.02.7210014700000002.7210014690000001.0000000000E−092.0019000000E−052.0688500000E−04


## 5.2. Example

Consider fractional diffusion equation in one dimension as(46)∂ρμ∂κρ=∂2μ∂κ2+μ∂μ∂κ−μ2+μ0<ρ≤1,ϑ>0,with initial source
(47)μκ,0=1+eκ.On applying Laplace transform to (46),
(48)νρLμκ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2+μ∂μ∂κ−μ2+μ.Applying Laplace inverse transformation,(49)μκ,ϑ=1+expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+μ∂μ∂κ−μ2+μ.Letμκ,ϑ have a series form solution as
(50)μκ,ϑ=∑m=0∞μmκ,ϑ,where μμκ=∑k=0∞Ak and μ2=∑k=0∞Bk illustrates Adomian polynomials that show nonlinear terms; thus, on putting certain terms in Equation (49), we get
(51)∑m=0∞μmκ,ϑ=1+expκ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∑k=0∞Ak−∑k=0∞Bk+μ.The nonlinear term decomposition by means of Adomian polynomials is given in Equation (20) as
(52)A0=μ0μ0κ,A1=μ1μ0κ+μ0μ1κ,A2=μ2μ0κ+μ1μ1κ+μ0μ2κ,B0=μ02,B1=2μ0μ1,B2=2μ0μ2+μ22.By the comparison of Equation (51), both sides
(53)μ0κ,ϑ=1+eκ.Form=0,
(54)μ1κ,ϑ=1+eκρϑρΓρ+1+1−ρ.Form=1,
(55)μ2κ,ϑ=1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(56)μ3κ,ϑ=1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is as(57)μκ,ϑ=∑m=0∞μmκ,ϑ=μ0κ,ϑ+μ1κ,ϑ+μ2κ,ϑ+μ3κ,ϑ+⋯,μκ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=1+eκ+ϑ.The analytical solution is by VITM.In the iterative formula for Equation (46), we get
(58)μm+1κ,ϑ=μmκ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2+μm∂μm∂κ−μm2+μm,where
(59)μ0κ,ϑ=1+eκ.Form=0,1,2,⋯,
(60)μ1κ,ϑ=μ0κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2+μ0∂μ0∂κ−μ02+μ0,μ1κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ,μ2κ,ϑ=μ1κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2+μ1∂μ1∂κ−μ12+μ1,μ2κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ϑ=μ2κ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2+μ2∂μ2∂κ−μ22+μ2,μ3κ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ϑ=∑m=0∞μmκ,ϑ=1+eκ+1+eκρϑρΓρ+1+1−ρ+1+eκρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+1+eκρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ϑ=1+eκ+ϑ.Figure2 depicts the accurate and approximate solutions in the (AB fractional derivative) sense for ρ=1 and at various fractional-orders inside the domain 0≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical depiction. Table 2 shows a comparison of the exact solution and the solution obtained using our approaches, demonstrating the correctness of the presented methods. Table 2 also includes a comparison of absolute errors at various fractional orders, indicating that results improve as fractional order moves to integer order.Figure 2
The graphical layout for problem 2.Table 2
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 2.
ϑ=0.01Exact solutionOur methods solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.802.0000001000000002.0000002000000001.0000000000E−071.0280000000E−061.2534000000E−050.12.1051710290000002.1051711290000001.0000000000E−071.0300000000E−061.2667000000E−050.22.2214028800000002.2214029800000001.0000000000E−071.0310000000E−061.2815000000E−050.32.3498589430000002.3498590430000001.0000000000E−071.0340000000E−061.2978000000E−050.42.4918248470000002.4918249470000001.0000000000E−071.0360000000E−061.3157000000E−050.52.6487214360000002.6487215360000001.0000000000E−071.0380000000E−061.3356000000E−050.62.8221189830000002.8221190820000009.9000000000E−081.0390000000E−061.3575000000E−050.73.0137529090000003.0137530080000009.9000000000E−081.0420000000E−061.3818000000E−050.83.2255411510000003.2255412510000001.0000000000E−071.0450000000E−061.4086000000E−050.93.4596033570000003.4596034570000001.0000000000E−071.0480000000E−061.4383000000E−051.03.7182821000000003.7182822000000001.0000000000E−071.0510000000E−061.4711000000E−05


## 5.3. Example

Consider the fractional diffusion equation in three dimensions as(61)∂ρμ∂κρ=∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2,0<ρ≤1,ϑ≥0,with initial source
(62)μκ,ℓ,ψ,0=sinκsinℓsinψ.On applying Laplace transform to (61),
(63)νρLμκ,ℓ,ψ,ϑ−ν−1μκ,0νρ+ρ1−νρ=L∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.On taking Laplace inverse transform,(64)μκ,ℓ,ψ,ϑ=sinκsinℓsinψ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.Letμκ,ℓ,ψ,ϑ have series form solution as
(65)μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ,(66)∑m=0∞μmκ,ℓ,ψ,ϑ=sinκsinℓsinψ+L−1νρ+ρ1−νρνρL∂2μ∂κ2+∂2μ∂ℓ2+∂2μ∂ψ2.By the comparison of Equation (66), both sides
(67)μ0κ,ℓ,ψ,ϑ=sinκsinℓsinψ.Form=0,
(68)μ1κ,ℓ,ψ,ϑ=−3sinκsinℓsinψρϑρΓρ+1+1−ρ.Form=1,
(69)μ2κ,ℓ,ψ,ϑ=−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2.Form=2,
(70)μ3κ,ℓ,ψ,ϑ=−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3.The series form solution is as(71)μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ=μ0κ,ℓ,ψ,ϑ+μ1κ,ℓ,ψ,ϑ+μ2κ,ℓ,ψ,ϑ+μ3κ,ℓ,ψ,ϑ+⋯,μκ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get the solution atρ=1 as μκ,ℓ,ψ,ϑ=exp−3ϑsinκsinℓsinψ.The analytical solution is by VITM.The iterative formula for Equation (61), we get
(72)μm+1κ,ℓ,ψ,ϑ=μmκ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μm∂κ2+∂2μm∂ℓ2+∂2μm∂ψ2,where
(73)μ0κ,ℓ,ψ,ϑ=sinκsinℓsinψ.Form=0,1,2,⋯,
(74)μ1κ,ℓ,ψ,ϑ=μ0κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ0∂κ2+∂2μ0∂ℓ2+∂2μ0∂ψ2,μ1κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ,μ2κ,ℓ,ψ,ϑ=μ1κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ1∂κ2+∂2μ1∂ℓ2+∂2μ1∂ψ2,μ2κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2,μ3κ,ℓ,ψ,ϑ=μ2κ,ℓ,ψ,ϑ+L−1νρ+ρ1−νρνρLνρνρ+ρ1−νρ∂2μ2∂κ2+∂2μ2∂ℓ2+∂2μ2∂ψ2,μ3κ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3,μκ,ℓ,ψ,ϑ=∑m=0∞μmκ,ℓ,ψ,ϑ=sinκsinℓsinψ+−3sinκsinℓsinψρϑρΓρ+1+1−ρ+−32sinκsinℓsinψρ2ϑ2ρΓ2ρ+1+2ρ1−ρϑρΓρ+1+1−ρ2+−33sinκsinℓsinψρ3ϑ3ρΓ3ρ+1+3ρ21−ρϑ2ρΓ2ρ+1+3ρ1−ρ2ϑρΓρ+1+1−ρ3+⋯.Thus, we get solution atρ=1 as μκ,ℓ,ψ,ϑ=exp−3ϑsinκsinℓsinψ.Figure3 depicts the accurate and approximate solutions in the (AB fractional derivative) sense for ρ=1 and at various fractional orders inside the domain −1≤κ≥1. Our result is in good agreement with the actual solution, as shown by the graphical depiction. The LTDM and VITM solutions are closely connected to the exact solution, as seen in Figure 3 and Table 3.Figure 3
The graphical layout for problem 3.Table 3
Comparison of the exact and suggested method solution with the aid of absolute errors for problem 3.
ϑ=0.01Exact solutionOur method solutionAE of our methodsAE of our methodsAE of our methodsκρ=1ρ=1ρ=1ρ=0.9ρ=0.800.0000000000000000.0000000000000000.0000000000E+001.1470000000E−042.3021230000E−030.10.0402750524300000.0402750604900008.0612920340E−094.0267087740E−054.0274327760E−040.20.0801476898400000.0801477058800001.6038436970E−088.0131842970E−058.0146248110E−040.30.1192195181000000.1192195419000002.3813628870E−081.1919596380E−041.1921737920E−030.40.1571001442000000.1571001756000003.1386867730E−081.5706905820E−041.5709731480E−030.50.1934110777000000.1934111164000003.8707665300E−081.9337280920E−041.9340759560E−030.60.2277895116000000.2277895572000004.5607727380E−082.2774445860E−042.2778541300E−030.70.2598919480000000.2598920000000005.2002906860E−082.5984051810E−042.5988727020E−030.80.2893976300000000.2893976879000005.7893203750E−082.8934038720E−042.8939242370E−030.90.3160117465000000.3160118097000006.3194470960E−083.1594929850E−043.1600606410E−031.00.3394683781000000.3394684460000006.7906708470E−083.3940126310E−043.3946226980E−03


## 6. Conclusion

In this article, we employed the LTDM and VITM to investigate the importance of the well-studied AB derivative of fractional order. The proposed methods were implemented to solve fractional-order diffusion equations. Following our analysis, we discovered that the current strategies are the most effective tool for solving nonlinear fractional-order problems. The accurate and achieved solutions in closed contact have confirmed the technique’s reliability. Also, the graphical representation has confirmed the convergence of approximation results to exact solutions, as well as providing closed-form solutions to the targeted problems. The reliability of the proposed method has been confirmed by the convergence phenomenon. Furthermore, LTDM and VITM are often regarded as highly effective and efficient tools for handling a variety of real-world problems.



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