# Analytical Investigation of Noyes–Field Model for Time-Fractional Belousov–Zhabotinsky Reaction

Authors: Mohammed Kbiri Alaoui; Rabia Fayyaz; Adnan Khan; Rasool Shah; Mohammed S. Abdo
Journal: Complexity
 (2021)

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## Abstract

In this article, we find the solution of time-fractional Belousov–Zhabotinskii reaction by implementing two well-known analytical techniques. The proposed methods are the modified form of the Adomian decomposition method and homotopy perturbation method with Yang transform. In Caputo manner, the fractional derivative is used. The solution we obtained is in the form of series which helps in investigating the analytical solution of the time-fractional Belousov–Zhabotinskii (B-Z) system. To verify the accuracy of the proposed methods, an illustrative example is taken, and through graphs, the solution is shown. Also, the fractional-order and integer-order solutions are compared with the help of graphs which are easy to understand. It has been verified that the solution obtained by using the given approaches has the desired rate of convergence to the exact solution. The proposed technique’s principal benefit is the low amount of calculations required. It can also be used to solve fractional-order physical problems in a variety of domains.


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## Body

## 1. Introduction

Fractional calculus is concerned with fractional-order derivatives and integrals [1]. Although fractional calculus has a similar background to classical calculus, it has received little attention for a long time. Fractional calculus and fractional differential equations, on the other hand, have grown in popularity in recent decades as a result of their significant prospective applications in physics and engineering [2–6]. For various physical phenomena, a considerable number of new differential equations (models) including fractional derivatives and integration have been constructed. These models have proven to be effective. Many phenomena in physics, chemistry, engineering, and other sciences can be better understood using fractional calculus methods, such as the theory of fractional noninteger order derivatives and integrals. These mathematical phenomena enable a more accurate description of an actual object than using “integer” methods [7–10].Fractional partial differential equations (FPDEs) are a contemporary tool in calculus that may be used to simulate a variety of phenomena in applied sciences and engineering. Because it proved difficult to simulate nonlinear real-world processes in ordinary calculus, the subject of fractional calculus became popular among researchers [11–17]. In this context, FPDEs approximation and analytical solutions are critical for accurately describing the dynamics of important physical processes. In view of the above statement, mathematicians have devised and applied a variety of approximate and analytical procedures to identify the solutions to a number of important mathematical models that represent real-world issues. Despite the fact that calculating the analytical and even approximate solutions of certain nonlinear FPDEs and systems of FPDEs is extremely challenging, mathematicians continue to do their best in this respect. Around the years, a great deal of study has been conducted in the fields of science and engineering all over the world, and various methodologies have been developed to provide the best obtainable solutions. This is an unstoppable process, and new techniques are being developed on a daily basis since the world is confronted with new, insoluble, and challenging difficulties and problems [18–22]. Different methods for solving FDEs have been proposed in the literature, including the fractional differential transform method (FDTM) [23], modified homotopy perturbation method (MHPM) [24], Adomian decomposition method (ADM) [25], Sumudu transform method (STM) [26], differential transform method (DTM) [27], homotopy analysis method (HAM) [28], variational iteration method (VIM) [29], and fractional Fourier transform method (FFTM) [30].The Yang transform decomposition method (YTDM) and the homotopy perturbation Yang transform method (HPYTM) are expanded for time-fractional Belousov–Zhabotinskii (B-Z) system solution in this article. Xiao-Jun Yang proposed the Yang transform, which may be used to solve a variety of differential equations with constant coefficients. On the other hand, the Adomian decomposition method [31, 32] is a well-known technique for solving linear and nonlinear and homogeneous and nonhomogeneous differential and partial differential equations, and integrodifferential and FDEs that provides exact solutions in the form of a convergent series. He introduced HPM in 1998 [33, 34]. The result is considered to be an in series solution with a high number of terms that quickly converge to the actual derived solution in this technique. The method is capable of adequately solving nonlinear PDEs. When the HPTM results were compared with the real solution to the problems, a higher level of accuracy was proven. Nonlinear wave equations [35], nonlinear problem bifurcation [36], and boundary value problems [37] have all been solved using this technique. In this paper, we used a new approximation analytical method called HPYTM. The hybrid form of Yang transform and HPM is the newly designed technique. The current methods are shown to be quite effective in obtaining the analytical solution of the time-fractional Belousov–Zhabotinskii (B-Z) system. The results of the recommended procedures are convincing, providing exact solutions to the targeted problems. The fractional problem findings obtained utilising the given approaches are also used to analyse the problems from a fractional perspective. The existing techniques can be adjusted to solve various fractional PDEs and related systems, which has been confirmed.Our goal in this paper is to use two analytical techniques for solving a fractional-order nonlinear oscillatory system known as Belousov–Zhabotinskii (B-Z). The B-Z chemical reaction family is fascinating because it can show both temporal and spatial travelling concentration waves, as well as dramatic colour changes [38]. For the B-Z reaction, the simplified fractional Noyes–Field model is as follows [39]:(1)Dταξ=Ψ1ξφφ+βζ+ξ−ξ2−ξζ,0<α≤1,Dταζ=Ψ2ζφφ+γζ−δξζ,where Ψ1 and Ψ2 are the diffusing constants for ξ and ζ concentration, respectively, γ and β are given constants, δ≠1 and ϱ are positive parameters, and α is the fractional order.


## 2. Preliminaries

In this section, we presented some basic definitions of fractional calculus along with Yang transform theory properties.


### 2.1. Definition

The fractional-order derivative in Caputo manner is given as(2)Dταξφ,τ=1Γk−α∫0ττ−ρk−α−1ξkφ,ρdρ,k−1<α≤k,k∈N.


### 2.2. Definition

The Yang–Laplace transform was introduced by Xiao-Jun Yang in 2018. The Yang transform for a functionξτ is determined by Yξτ or M(u) and is given as(3)Yξτ=Mu=∫0∞e−τ/uξτdτ,τ>0,u∈−τ1,τ2.The inverse transform (inverse symbol) is defined as(4)Y−1Mu=ξτ,where Y−1 is inverse Yang operator.


### 2.3. Definition

The Yang transform for nth derivatives is defined as(5)Yξnτ=Muun−∑k=0n−1ξk0un−k−1,∀n=1,2,3,….


### 2.4. Definition

The Yang transform for fractional-order derivatives is defined as(6)Yξατ=Muuγ−∑k=0n−1ξk0uα−k+1,0<α≤n.


## 2.1. Definition

The fractional-order derivative in Caputo manner is given as(2)Dταξφ,τ=1Γk−α∫0ττ−ρk−α−1ξkφ,ρdρ,k−1<α≤k,k∈N.


## 2.2. Definition

The Yang–Laplace transform was introduced by Xiao-Jun Yang in 2018. The Yang transform for a functionξτ is determined by Yξτ or M(u) and is given as(3)Yξτ=Mu=∫0∞e−τ/uξτdτ,τ>0,u∈−τ1,τ2.The inverse transform (inverse symbol) is defined as(4)Y−1Mu=ξτ,where Y−1 is inverse Yang operator.


## 2.3. Definition

The Yang transform for nth derivatives is defined as(5)Yξnτ=Muun−∑k=0n−1ξk0un−k−1,∀n=1,2,3,….


## 2.4. Definition

The Yang transform for fractional-order derivatives is defined as(6)Yξατ=Muuγ−∑k=0n−1ξk0uα−k+1,0<α≤n.


## 3. Homotopy Perturbation Yang Transform Method

Consider a general nonlinear homogeneous partial differential equation with initial conditions of the type to demonstrate the basic idea of this method:(7)Dταξφ,τ+P1ξφ,τ+Q1ξφ,τ=0,0<α≤1,with some initial sources such as(8)ξφ,0=hφ,where Dτα is the fractional differential operator with respect to τ. P1 and Q1 are the linear and nonlinear differential operator with respect to φ, respectively.Using Yang transformation to (7), we have(9)YDταξφ,τ+YP1ξφ,τ+Q1ξφ,τ=0,(10)1uαMu−uξ0+YP1ξφ,τ+Q1ξφ,τ=0.Equation (9) implies that(11)Mξ=uhφ−uαYP1ξφ,τ+Q1ξφ,τ.Now, by taking inverse Yang transform, we get(12)ξφ,τ=Hφ−Y−1uαYP1ξφ,τ+Q1ξφ,τ,where Hφ represents the term arising from the source term and the prescribed initial conditions. Now, we apply the homotopy perturbation method:(13)ξφ,τ=∑k=0∞pkξkφ,τ,and the nonlinear term can be decomposed as(14)Q1ξφ,τ=∑k=0∞pkHkξ,for a set of He’s polynomials Hk defined by(15)Hkξ0,ξ1,…,ξk=1k!∂k∂pkQ1∑i=0∞piξip=0,k=0,1,2,3…Substituting equations (14) and (15) in (12), we get(16)∑k=0∞pkξkφ,τ=Hφ−p×Y−1uαYP1∑k=0∞pkξkφ,τ+∑k=0∞pkHkξ.The Yang transform and the homotopy perturbation approach using He’s polynomials are coupled in this method. The following approximations are obtained by comparing the coefficients of like powers ofp.(17)p0:ξ0φ,τ=Hφ,p1:ξ1φ,τ=−Y−1uαYP1ξ0φ,τ+H0ξ,p2:ξ2φ,τ=−Y−1uαYP1ξ1φ,τ+H1ξ,⋅⋅⋅pk:ξkφ,τ=−Y−1uαYP1ξk−1φ,τ+Hk−1ξ,k>0,k∈N.Thus, we can calculate easily componentξkφ,τ, which rapidly leads us to the convergent series. By taking p⟶1, we get(18)ξφ,τ=limM⟶∞∑k=1Mξkφ,τ.The result we get is in form of series and quickly converges to the problem exact solution.


## 4. Idea of YTDM

The solution by YTDM for partial differential equations having fractional order is described in this section.(19)Dταξφ,τ=Pφ,τ1+Q1φ,τ+ℛ1φ,τ,0<α≤1,with some initial sources such as(20)ξφ,0=ξφ,where Dτα=∂α/∂τα is the fractional derivative in Caputo manner of order α, P1 and Q1 are linear and nonlinear functions, and ℛ1 is the source term, respectively.By applying Yang transform on both sides of equation (19), we obtain(21)YDταξφ,τ=YP1φ,τ+Q1φ,τ+ℛ1φ,τ.By differentiation property of Yang transform, we obtain(22)1uαMu−uξ0=YP1φ,τ+Q1φ,τ+ℛ1φ,τ.Equation (20) implies that(23)Mξ=uξ0+uαYP1φ,τ+Q1φ,τ+ℛ1φ,τ,and taking the inverse Yang transform of equation (23), we get(24)ξφ,τ=ξ0+Y−1uαYP1φ,τ+Q1φ,τ+ℛ1φ,τ.YTDM defines the infinite sequence solution ofξφ,τ as(25)ξφ,τ=∑m=0∞ξmφ,τ.The Adomian polynomialQ1 decomposition of nonlinear terms is given as(26)Q1φ,τ=∑m=0∞Am.All types of nonlinearity are represented by Adomian polynomials as(27)Am=1m!∂m∂δmQ1∑k=0∞δkφk,∑k=0∞δkτkδ=0.Putting equations (23) and (25) into (24) gives(28)∑m=0∞ξmφ,τ=ξ0+Y−1uαYℛ1φ,τ+Y−1uαYP1∑m=0∞φm,∑m=0∞τm+∑m=0∞Am.The following terms are described:(29)ξ0φ,τ=ξ0+Y−1uαYℛ1φ,τ,ξ1φ,τ=Y−1uαYP1φ0,τ0+A0.In general, form≥1, it is calculated as(30)ξm+1φ,τ=Y−1uαYP1φm,τm+Am.


## 5. Applications

The solutions to the time-fractional Belousov–Zhabotinskii (B-Z) system are derived using YTDM and HPYTM in this section.


### 5.1. Example

Consider the time-fractional B-Z system havingγ=β=0, then equation (1) is reduced to(31)∂αξ∂τα=∂2ξ∂φ2+ξ−ξ2−ξζ,0<α≤1,∂αζ∂τα=∂2ζ∂φ2−δξζ,having initial conditions as(32)φ,0=1eδ/6φ+12,ζφ,0=1−δeδ/6φeδ/6φ+2eδ/6φ+12.The exact solution of equation (28) when α=1 is(33)ξφ,τ=e5δ/3τeδ/6φ+e5δ/6τ2,ζφ,τ=1−δeδ/6φeδ/6φ+2e5δ/6τeδ/6φ+e5δ/6τ2,where ϱ and δ≠1 are positive parameters.Remark 1.
. The exact solution can take the following form as well:(34)e5δ/3τeδ/6φ+e5δ/6τ2=14tanh2δ24φ−5δ12τ−12,1−δeδ/6φeδ/6φ+2e5δ/6τeδ/6φ+e5δ/6τ2=δ−14tanh2δ24φ−5δ12τ−2tanhδ24φ−5δ12τ−3.
Taking Yang transformation of equation (28), we get(35)Y∂αξ∂τα=Y∂2ξ∂φ2+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(36)1uαMu−uξ0=Y∂2ξ∂φ2+ξ−ξ2−ξζ,1uαMu−uζ0=Y∂2ζ∂φ2−δξζ,Mu=uξ0+uαY∂2ξ∂φ2+ξ−ξ2−ξζ,Mu=uζ0+uαY∂2ζ∂φ2−δξζ.
The inverse Yang transform implies that(37)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2−δξζ,φ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δξζ.
Now, we apply the homotopy perturbation method as follows:(38)ξφ,τ=ξ0+ξ1p+ξ2p2+⋯,ζφ,τ=ζ0+ζ1p+ζ2p2+⋯,∑k=0∞pkξkφ,τ=1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+∑k=0∞pkξkφ,τ−∑k=0∞pkHk1φ,τ,∑k=0∞pkζkφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+pY−1uαY∑k=0∞pkζkφ,τφφ−∑k=0∞pkHk2φ,τ,where the nonlinear terms are represented by He’s polynomials Hkφ. For example, the first few components of He’s polynomials are given by(39)H01φ=2ξ0ξ1+ξ0ζ1+ξ1ζ0,H11φ=2ξ0ξ2+ξ12+ξ0ζ2+ξ1ζ1+ξ2ζ0,H21φ=2ξ0ξ3+2ξ1ξ2+ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0,⋮H02φ=ξ0ζ1+ξ1ζ0,H12φ=ξ0ζ2+ξ1ζ1+ξ2ζ0,H22φ=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
When the coefficients of like powers ofp are compared, we get(40)p0:ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ1−δeδ/6φeδ/6φ+2eδ/6φ+12,p1:ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13,p2:ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,p3:ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,p4:ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18,⋮
Now, takingp⟶1 gives the approximate solution as follows:(41)ξφ,τ=ξ0+ξ1+ξ2+ξ3+ξ4+⋯=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯.
YTDM Solution
Taking Yang transformation of equation (28), we get(42)Y∂αξ∂τα=Y∂2ξ∂φ2+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(43)1uαMu−uξ0=Y∂2ξ∂φ2+ξ−ξ2−ξζ,1uαMu−uζ0=Y∂2ζ∂φ2−δξζ..
The inverse Yang transform implies that(44)ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2−δξζ,ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δξζ.
Assume that the unknownξφ,τ and ζφ,τ functions, in infinite series form, have the following solution:(45)ξφ,τ=∑m=0∞ξmφ,τ,ζφ,τ=∑m=0∞ξmφ,τ,where the Adomian polynomials ξ2=∑m=0∞Am and ξζ=∑m=0∞ℬm and the nonlinear terms have been characterised. Using certain terms, equation (44) can be rewritten in the form(46)∑m=0∞ξmφ,τ=ξφ,0+Y−1uαY∂2ξ∂φ2+ξ−∑m=0∞Am−∑m=0∞ℬm,∑m=0∞ζmφ,τ=ζφ,0+Y−1uαY∂2ζ∂φ2−δ∑m=0∞ℬm,∑m=0∞ξmφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−∑m=0∞Am−∑m=0∞ℬm,∑m=0∞ζmφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δ∑m=0∞ℬm.
All forms of nonlinearity can be represented by the Adomian polynomials as, according to equation (27),(47)A0=ξ02,A1=2ξ0ξ1,A2=2ξ0ξ2+ξ12,A3=2ξ0ξ3+2ξ1ξ2,ℬ0=ξ0w0,ℬ1=ξ1ζ0+ξ0ζ1,ℬ2=ξ0ζ2+ξ1ζ1+ξ2ζ0,ℬ3=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
Thus, on comparing both sides of equation (46),(48)ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12.
Form=0,(49)ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13.
Form=1,(50)ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14.
Form=2,(51)ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16.
Form=3,(52)ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648ϱΓ4α+1eδ/6φ+18.
The remaining YTDM solution elementsxim and ζm for m≥3 are similarly simple to get. As a result, we define the series of possibilities as follows:(53)ξφ,τ=∑m=0∞ξmφ,τ=ξ0φ,τ+ξ1φ,τ+ξ2φ,τ+ξ3φ,τ+ξ4φ,τ+⋯,ζφ,τ=∑m=0∞ζmφ,τ=ζ0φ,τ+ζ1φ,τ+ζφ,τ+ζ3φ,τ+ζ4φ,τ+⋯,ξφ,τ=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+5δδ−1eδ/6φ3Γα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯.
In Figure1, the exact and analytical solution graph of ξφ,τ is shown. In Figure 2, the different fractional graph of α=1,0.7,0.5, and 0.3 of ξφ,τ is shown. Similarly, in Figure 3, the exact and analytical solution graph of ζφ,τ is shown. In Figure 4, the different fractional graph of α=1,0.7,0.5, and 0.3 of ζφ,τ is shown.Figure 1
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 2
The different fractional-order solution graph ofα for ξφ,τ.Figure 3
The exact solution and analytical solution graph atα=1 for ζφ,τ.Figure 4
The different fractional-order solution graph ofα for ζφ,τ.


### 5.2. Example

Consider the time-fractional B-Z system havingγ=δ and β=1, then equation (1) is reduced to(54)∂αξ∂τα=∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,0<α≤1,∂αζ∂τα=∂2ζ∂φ2+δζ−δξζ,having initial conditions(55)ξφ,0=1eδ/6φ+12,ζφ,0=δ−1eδ/6φ+12,The exact solution of equation (54) when α=1 is(56)ξφ,τ=e5δ/3τeδ/6φ+e5δ/6τ2,ζφ,τ=δ−1e5δ/3ττeδ/6φ+e5δ/6τ2,where δ≠1 is a positive parameter.Remark 2.
. The exact solution can take the following form as well:(57)e5δ/3τeδ/6φ+e5δ/6τ2=14tanhδ24φ−5δ12τ−12,δ−1e5δ/3τeδ/6φ+e5δ/6τ2=δ−14tanhδ24φ−5δ12τ−12.
Taking Yang transformation of equation (54), we get(58)Y∂αξ∂τα=Y∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2+δζ−δξζ.
Applying the differential property of the Yang transform, we get(59)1uαMu−uξ0=Y∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,1uαMu−uζ≤0=Y∂2ζ∂φ2+δζ−δξζ,Mu=uξ0+uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,Mu=uζ0+uαY∂2ζ∂φ2+δζ−δξζ.
The inverse Yang transform implies that(60)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2+δζ−δξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=δ−1eδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δξζ.
Now, we apply the homotopy perturbation method as follows:(61)ξφ,τ=ξ0+ξ1p+ξ2p2+ξ3p3+ξ4p4+⋯,ζφ,τ=ζ0+ζ1p+ζ2p2+ζ3p3+ζ4p4+⋯,∑k=0∞pkξkφ,τ=1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+∑k=0∞pkξkφ,τ+∑k=0∞pkξkφ,τ−∑k=0∞pkHk1φ,τ,∑k=0∞pkζkφ,τ=δ−1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+δ∑k=0∞pkξkφ,τ−∑k=0∞pkHk2φ,τ,where the nonlinear terms are represented by He’s polynomials Hkφ. For example, the first few components of He’s polynomials are given by(62)H01φ=2ξ0ξ1+ξ0ζ1+ξ1ζ0,H11φ=2ξ0ξ2+ξ12+ξ0ζ2+ξ1ζ1+ξ2ζ0,H21φ=2ξ0ξ3+2ξ1ξ2+ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0,⋮H02φ=ξ0ζ1+ξ1ζ0,H12φ=ξ0ζ2+ξ1ζ1+ξ2ζ0,H22φ=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
When the coefficients of like powers ofp are compared, we get(63)p0:ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=δ−1ϱeδ/6φ+12,p1:ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3ϱΓα+1eδ/6φ+13,p2:ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118ϱΓ2α+1eδ/6φ+14,p3:ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16,p4:ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18⋮.
Now, takingp⟶1 gives the approximate solution as follows:(64)ξφ,τ=ξ0+ξ1+ξ2+ξ3+ξ4+⋯=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯,ζφ,τ=ζ0+ζ1+ζ2+ζ3+ζ4+⋯=δ−1eδ/6φ+12+5δδ−1eδ/6φ3Γα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯.
YTDM Solution
Taking Yang transformation of equation (41), we get(65)Y∂αξ∂τα=Y∂2ξ∂φ2+ϱζ+ξ−ξ2−ϱξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(66)1uαMu−uξ0=Y∂2ξ∂φ2+ϱζ+ξ−ξ2−ϱξζ,1uαMu−uζ0=Y∂2ζ∂φ2+δζ−δξζ.
The inverse Yang transform implies that(67)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2+δζ−δξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=δ−1eδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δξζ.
Assume that the unknownξφ,τ and ζφ,τ functions, in infinite series form, have the following solution:(68)ξφ,τ=∑m=0∞ξmφ,τ,ζφ,τ=∑m=0∞ζmφ,τ,where the Adomian polynomials ξ2=∑m=0∞Am and ξζ=∑m=0∞ℬm and the nonlinear terms have been characterised. Using certain terms, equation (67) can be rewritten in the form as follows:(69)∑m=0∞ξmφ,τ=ξφ,0+Y−1uαY∂2ξ∂φ2+ϱζ+ξ−∑m=0∞Am−ϱ∑m=0∞ℬm,∑m=0∞ζmφ,τ=ζφ,0+Y−1uαY∂2ζ∂φ2+δζ−δ∑m=0∞ℬm,∑m=0∞ξmφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ϱζ+ξ−∑m=0∞Am−ϱ∑m=0∞ℬm,∑m=0∞ζmφ,τ=δ−1ϱeδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δ∑m=0∞ℬm.
All forms of nonlinearity can be represented by the Adomian polynomials, according to equation (25), as(70)A0=ξ02,A1=2ξ0ξ1,A2=2ξ0ξ2+ξ12,A3=2ξ0ξ3+2ξ1ξ2,ℬ0=ξ0w0,ℬ1=ξ1ζ0+ξ0ζ1,ℬ2=ξ0ζ2+ξ1ζ1+ξ2ζ0,ℬ3=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
Thus, on comparing both side of equation (69),(71)ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=δ−1eδ/6φ+12.
Form=0,(72)ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13.
Form=1,(73)ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14.
Form=2,(74)ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16.
Form=3,(75)ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648ϱΓ4α+1eδ/6φ+18.
The remaining YTDM solution elementsxim and ζm for m≥3 are similarly simple to get. As a result, we define the series of possibilities as follows:(76)ξφ,τ=∑m=0∞ξmφ,τ=ξ0φ,τ+ξ1φ,τ+ξ2φ,τ+ξ3φ,τ+ξ4φ,τ+⋯,ζφ,τ=∑m=0∞ζmφ,τ=ζ0φ,τ+ζ1φ,τ+ζφ,τ+ζ3φ,τ+ζ4φ,τ+⋯,ξφ,τ=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯,ζφ,τ=δ−1ϱeδ/6φ+12+5δδ−1eδ/6φ3ϱΓα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118ϱΓ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18⋯.
In Figure5, the exact and analytical solution graph of ξφ,τ is shown. In Figure 6, the different fractional graph of α=1,0.7,0.5, and 0.3 of ξφ,τ is shown. Similarly, in Figure 7, the exact and analytical solution graph of ζφ,τ is shown. In Figure 8, the different fractional graph of α=1,0.7,0.5, and 0.3 of ζφ,τ is shown.Figure 5
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 6
The different fractional-order solution graph ofα for ξφ,τ.Figure 7
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 8
The different fractional-order solution graph ofα for ζφ,τ.


## 5.1. Example

Consider the time-fractional B-Z system havingγ=β=0, then equation (1) is reduced to(31)∂αξ∂τα=∂2ξ∂φ2+ξ−ξ2−ξζ,0<α≤1,∂αζ∂τα=∂2ζ∂φ2−δξζ,having initial conditions as(32)φ,0=1eδ/6φ+12,ζφ,0=1−δeδ/6φeδ/6φ+2eδ/6φ+12.The exact solution of equation (28) when α=1 is(33)ξφ,τ=e5δ/3τeδ/6φ+e5δ/6τ2,ζφ,τ=1−δeδ/6φeδ/6φ+2e5δ/6τeδ/6φ+e5δ/6τ2,where ϱ and δ≠1 are positive parameters.Remark 1.
. The exact solution can take the following form as well:(34)e5δ/3τeδ/6φ+e5δ/6τ2=14tanh2δ24φ−5δ12τ−12,1−δeδ/6φeδ/6φ+2e5δ/6τeδ/6φ+e5δ/6τ2=δ−14tanh2δ24φ−5δ12τ−2tanhδ24φ−5δ12τ−3.
Taking Yang transformation of equation (28), we get(35)Y∂αξ∂τα=Y∂2ξ∂φ2+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(36)1uαMu−uξ0=Y∂2ξ∂φ2+ξ−ξ2−ξζ,1uαMu−uζ0=Y∂2ζ∂φ2−δξζ,Mu=uξ0+uαY∂2ξ∂φ2+ξ−ξ2−ξζ,Mu=uζ0+uαY∂2ζ∂φ2−δξζ.
The inverse Yang transform implies that(37)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2−δξζ,φ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δξζ.
Now, we apply the homotopy perturbation method as follows:(38)ξφ,τ=ξ0+ξ1p+ξ2p2+⋯,ζφ,τ=ζ0+ζ1p+ζ2p2+⋯,∑k=0∞pkξkφ,τ=1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+∑k=0∞pkξkφ,τ−∑k=0∞pkHk1φ,τ,∑k=0∞pkζkφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+pY−1uαY∑k=0∞pkζkφ,τφφ−∑k=0∞pkHk2φ,τ,where the nonlinear terms are represented by He’s polynomials Hkφ. For example, the first few components of He’s polynomials are given by(39)H01φ=2ξ0ξ1+ξ0ζ1+ξ1ζ0,H11φ=2ξ0ξ2+ξ12+ξ0ζ2+ξ1ζ1+ξ2ζ0,H21φ=2ξ0ξ3+2ξ1ξ2+ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0,⋮H02φ=ξ0ζ1+ξ1ζ0,H12φ=ξ0ζ2+ξ1ζ1+ξ2ζ0,H22φ=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
When the coefficients of like powers ofp are compared, we get(40)p0:ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ1−δeδ/6φeδ/6φ+2eδ/6φ+12,p1:ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13,p2:ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,p3:ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,p4:ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18,⋮
Now, takingp⟶1 gives the approximate solution as follows:(41)ξφ,τ=ξ0+ξ1+ξ2+ξ3+ξ4+⋯=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯.
YTDM Solution
Taking Yang transformation of equation (28), we get(42)Y∂αξ∂τα=Y∂2ξ∂φ2+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(43)1uαMu−uξ0=Y∂2ξ∂φ2+ξ−ξ2−ξζ,1uαMu−uζ0=Y∂2ζ∂φ2−δξζ..
The inverse Yang transform implies that(44)ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2−δξζ,ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−ξ2−ξζ,ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δξζ.
Assume that the unknownξφ,τ and ζφ,τ functions, in infinite series form, have the following solution:(45)ξφ,τ=∑m=0∞ξmφ,τ,ζφ,τ=∑m=0∞ξmφ,τ,where the Adomian polynomials ξ2=∑m=0∞Am and ξζ=∑m=0∞ℬm and the nonlinear terms have been characterised. Using certain terms, equation (44) can be rewritten in the form(46)∑m=0∞ξmφ,τ=ξφ,0+Y−1uαY∂2ξ∂φ2+ξ−∑m=0∞Am−∑m=0∞ℬm,∑m=0∞ζmφ,τ=ζφ,0+Y−1uαY∂2ζ∂φ2−δ∑m=0∞ℬm,∑m=0∞ξmφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ξ−∑m=0∞Am−∑m=0∞ℬm,∑m=0∞ζmφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+Y−1uαY∂2ζ∂φ2−δ∑m=0∞ℬm.
All forms of nonlinearity can be represented by the Adomian polynomials as, according to equation (27),(47)A0=ξ02,A1=2ξ0ξ1,A2=2ξ0ξ2+ξ12,A3=2ξ0ξ3+2ξ1ξ2,ℬ0=ξ0w0,ℬ1=ξ1ζ0+ξ0ζ1,ℬ2=ξ0ζ2+ξ1ζ1+ξ2ζ0,ℬ3=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
Thus, on comparing both sides of equation (46),(48)ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12.
Form=0,(49)ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13.
Form=1,(50)ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14.
Form=2,(51)ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16.
Form=3,(52)ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648ϱΓ4α+1eδ/6φ+18.
The remaining YTDM solution elementsxim and ζm for m≥3 are similarly simple to get. As a result, we define the series of possibilities as follows:(53)ξφ,τ=∑m=0∞ξmφ,τ=ξ0φ,τ+ξ1φ,τ+ξ2φ,τ+ξ3φ,τ+ξ4φ,τ+⋯,ζφ,τ=∑m=0∞ζmφ,τ=ζ0φ,τ+ζ1φ,τ+ζφ,τ+ζ3φ,τ+ζ4φ,τ+⋯,ξφ,τ=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯ζφ,τ=1−δeδ/6φeδ/6φ+2eδ/6φ+12+5δδ−1eδ/6φ3Γα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯.
In Figure1, the exact and analytical solution graph of ξφ,τ is shown. In Figure 2, the different fractional graph of α=1,0.7,0.5, and 0.3 of ξφ,τ is shown. Similarly, in Figure 3, the exact and analytical solution graph of ζφ,τ is shown. In Figure 4, the different fractional graph of α=1,0.7,0.5, and 0.3 of ζφ,τ is shown.Figure 1
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 2
The different fractional-order solution graph ofα for ξφ,τ.Figure 3
The exact solution and analytical solution graph atα=1 for ζφ,τ.Figure 4
The different fractional-order solution graph ofα for ζφ,τ.


## 5.2. Example

Consider the time-fractional B-Z system havingγ=δ and β=1, then equation (1) is reduced to(54)∂αξ∂τα=∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,0<α≤1,∂αζ∂τα=∂2ζ∂φ2+δζ−δξζ,having initial conditions(55)ξφ,0=1eδ/6φ+12,ζφ,0=δ−1eδ/6φ+12,The exact solution of equation (54) when α=1 is(56)ξφ,τ=e5δ/3τeδ/6φ+e5δ/6τ2,ζφ,τ=δ−1e5δ/3ττeδ/6φ+e5δ/6τ2,where δ≠1 is a positive parameter.Remark 2.
. The exact solution can take the following form as well:(57)e5δ/3τeδ/6φ+e5δ/6τ2=14tanhδ24φ−5δ12τ−12,δ−1e5δ/3τeδ/6φ+e5δ/6τ2=δ−14tanhδ24φ−5δ12τ−12.
Taking Yang transformation of equation (54), we get(58)Y∂αξ∂τα=Y∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,Y∂αζ∂τα=Y∂2ζ∂φ2+δζ−δξζ.
Applying the differential property of the Yang transform, we get(59)1uαMu−uξ0=Y∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,1uαMu−uζ≤0=Y∂2ζ∂φ2+δζ−δξζ,Mu=uξ0+uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,Mu=uζ0+uαY∂2ζ∂φ2+δζ−δξζ.
The inverse Yang transform implies that(60)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2+δζ−δξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=δ−1eδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δξζ.
Now, we apply the homotopy perturbation method as follows:(61)ξφ,τ=ξ0+ξ1p+ξ2p2+ξ3p3+ξ4p4+⋯,ζφ,τ=ζ0+ζ1p+ζ2p2+ζ3p3+ζ4p4+⋯,∑k=0∞pkξkφ,τ=1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+∑k=0∞pkξkφ,τ+∑k=0∞pkξkφ,τ−∑k=0∞pkHk1φ,τ,∑k=0∞pkζkφ,τ=δ−1eδ/6φ+12+pY−1uαY∑k=0∞pkξkφ,τφφ+δ∑k=0∞pkξkφ,τ−∑k=0∞pkHk2φ,τ,where the nonlinear terms are represented by He’s polynomials Hkφ. For example, the first few components of He’s polynomials are given by(62)H01φ=2ξ0ξ1+ξ0ζ1+ξ1ζ0,H11φ=2ξ0ξ2+ξ12+ξ0ζ2+ξ1ζ1+ξ2ζ0,H21φ=2ξ0ξ3+2ξ1ξ2+ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0,⋮H02φ=ξ0ζ1+ξ1ζ0,H12φ=ξ0ζ2+ξ1ζ1+ξ2ζ0,H22φ=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
When the coefficients of like powers ofp are compared, we get(63)p0:ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=δ−1ϱeδ/6φ+12,p1:ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3ϱΓα+1eδ/6φ+13,p2:ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ=25δ2δ−1eδ/6φ2eδ/6φ−118ϱΓ2α+1eδ/6φ+14,p3:ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16,p4:ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18⋮.
Now, takingp⟶1 gives the approximate solution as follows:(64)ξφ,τ=ξ0+ξ1+ξ2+ξ3+ξ4+⋯=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯,ζφ,τ=ζ0+ζ1+ζ2+ζ3+ζ4+⋯=δ−1eδ/6φ+12+5δδ−1eδ/6φ3Γα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18⋯.
YTDM Solution
Taking Yang transformation of equation (41), we get(65)Y∂αξ∂τα=Y∂2ξ∂φ2+ϱζ+ξ−ξ2−ϱξζ,Y∂αζ∂τα=Y∂2ζ∂φ2−δξζ.
Applying the differential property of the Yang transform, we get(66)1uαMu−uξ0=Y∂2ξ∂φ2+ϱζ+ξ−ξ2−ϱξζ,1uαMu−uζ0=Y∂2ζ∂φ2+δζ−δξζ.
The inverse Yang transform implies that(67)ξφ,τ=ξ0+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=ζ0+Y−1uαY∂2ζ∂φ2+δζ−δξζ,ξφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ζ+ξ−ξ2−ξζ,ζφ,τ=δ−1eδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δξζ.
Assume that the unknownξφ,τ and ζφ,τ functions, in infinite series form, have the following solution:(68)ξφ,τ=∑m=0∞ξmφ,τ,ζφ,τ=∑m=0∞ζmφ,τ,where the Adomian polynomials ξ2=∑m=0∞Am and ξζ=∑m=0∞ℬm and the nonlinear terms have been characterised. Using certain terms, equation (67) can be rewritten in the form as follows:(69)∑m=0∞ξmφ,τ=ξφ,0+Y−1uαY∂2ξ∂φ2+ϱζ+ξ−∑m=0∞Am−ϱ∑m=0∞ℬm,∑m=0∞ζmφ,τ=ζφ,0+Y−1uαY∂2ζ∂φ2+δζ−δ∑m=0∞ℬm,∑m=0∞ξmφ,τ=1eδ/6φ+12+Y−1uαY∂2ξ∂φ2+ϱζ+ξ−∑m=0∞Am−ϱ∑m=0∞ℬm,∑m=0∞ζmφ,τ=δ−1ϱeδ/6φ+12+Y−1uαY∂2ζ∂φ2+δζ−δ∑m=0∞ℬm.
All forms of nonlinearity can be represented by the Adomian polynomials, according to equation (25), as(70)A0=ξ02,A1=2ξ0ξ1,A2=2ξ0ξ2+ξ12,A3=2ξ0ξ3+2ξ1ξ2,ℬ0=ξ0w0,ℬ1=ξ1ζ0+ξ0ζ1,ℬ2=ξ0ζ2+ξ1ζ1+ξ2ζ0,ℬ3=ξ0ζ3+ξ1ζ2+ξ2ζ1+ξ3ζ0.
Thus, on comparing both side of equation (69),(71)ξ0φ,τ=1eδ/6φ+12,ζ0φ,τ=δ−1eδ/6φ+12.
Form=0,(72)ξ1φ,τ=5δeδ/6φ3Γα+1eδ/6φ+13,ζ1φ,τ=5δδ−1eδ/6φ3Γα+1eδ/6φ+13.
Form=1,(73)ξ2φ,τ=25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14,ζ2φ,τ25δ2δ−1eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14.
Form=2,(74)ξ3φ,τ=−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16,ζ3φ,τ=−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16.
Form=3,(75)ξ4φ,τ=−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e22δ/3φ+1648Γ4α+1eδ/6φ+18,ζ4φ,τ=−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e22δ/3φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e22δ/3φ+1648ϱΓ4α+1eδ/6φ+18.
The remaining YTDM solution elementsxim and ζm for m≥3 are similarly simple to get. As a result, we define the series of possibilities as follows:(76)ξφ,τ=∑m=0∞ξmφ,τ=ξ0φ,τ+ξ1φ,τ+ξ2φ,τ+ξ3φ,τ+ξ4φ,τ+⋯,ζφ,τ=∑m=0∞ζmφ,τ=ζ0φ,τ+ζ1φ,τ+ζφ,τ+ζ3φ,τ+ζ4φ,τ+⋯,ξφ,τ=1eδ/6φ+12+5δeδ/6φ3Γα+1eδ/6φ+13+25δ2eδ/6φ2eδ/6φ−118Γ2α+1eδ/6φ+14−25δ3Γ2α+1e2δ/3φ9Γα+12Γ3α+1eδ/6φ+16−25δ3eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108Γ3α+1eδ/6φ+16−25δ4Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127Γα+12Γ4α+1eδ/6φ+18−125δ4Γ3α+1e2δ/3φ2eδ/6φ−127Γα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324Γ4α+1eδ/6φ+18−625δ4eδ/6φ17e2δ/32φ+1648Γ4α+1eδ/6φ+18⋯,ζφ,τ=δ−1ϱeδ/6φ+12+5δδ−1eδ/6φ3ϱΓα+1eδ/6φφ+13+25δ2δ−1eδ/6φ2eδ/6φ−118ϱΓ2α+1eδ/6φ+14−25δ3δ−1Γ2α+1e2δ/3φ9ϱΓα+12Γ3α+1eδ/6φ+16−25δ3δ−1eδ/6φ15e2δ/3φ−20e3δ/2φ+6eδ/6φ−5108ϱΓ3α+1eδ/6φ+16−25δ4δ−1Γ2α+1e2δ/3φ11e2δ/3φ−5eδ/6φ−127ϱΓα+12Γ4α+1eδ/6φ+18−125δ4δ−1Γ3α+1e2δ/3φ2eδ/6φ−127ϱΓα+1Γ2α+1Γ4α+1eδ/6φ+17+25δ4δ−1e2δ/3φ124e2δ/3φ+100e2δ/32φ+85eδ/6φ−4324ϱΓ4α+1eδ/6φ+18−625δ4δ−1eδ/6φ17e2δ/32φ+1648ϱΓ4α+1eδ/6φ+18⋯.
In Figure5, the exact and analytical solution graph of ξφ,τ is shown. In Figure 6, the different fractional graph of α=1,0.7,0.5, and 0.3 of ξφ,τ is shown. Similarly, in Figure 7, the exact and analytical solution graph of ζφ,τ is shown. In Figure 8, the different fractional graph of α=1,0.7,0.5, and 0.3 of ζφ,τ is shown.Figure 5
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 6
The different fractional-order solution graph ofα for ξφ,τ.Figure 7
The exact solution and analytical solution graph atα=1 for ξφ,τ.Figure 8
The different fractional-order solution graph ofα for ζφ,τ.


## 6. Conclusion

Finding the analytical solution to fractional partial differential equations is a tough task in most circumstances. This article makes a successful attempt to solve time-fractional Belousov–Zhabotinskii reaction for this purpose. It is confirmed that the proposed methods are the best tool for solving FPDEs. The displayed graphs confirm the strong relationship between the exact and analytical results. The plotted graphs confirmed the accuracy of the suggested techniques. In order to understand the behavior of the given problems, solutions at different fractional order is taken and is shown with the help of graphs. The convergence phenomenon has confirmed the reliability of the suggested method.



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