# Numerical Analysis of Fractional-Order Parabolic Equations via Elzaki Transform

Authors: Muhammad Naeem; Omar Fouad Azhar; Ahmed M. Zidan; Kamsing Nonlaopon; Rasool Shah
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This research article is dedicated to solving fractional-order parabolic equations, using an innovative analytical technique. The Adomian decomposition method is well supported by Elzaki transformation to establish closed-form solutions for targeted problems. The procedure is simple, attractive, and preferred over other methods because it provides a closed-form solution for the given problems. The solution graphs are plotted for both integer and fractional-order, which shows that the obtained results are in good contact with problems’ exact solution. It is also observed that the solution of fractional-order problems is convergent to the integer-order problem. Moreover, the validity of the proposed method is analyzed by considering some numerical examples. The theory of the suggested approach is fully supported by the obtained results for the given problems. In conclusion, the present method is a straightforward and accurate analytical technique that can solve other fractional-order partial differential equations.


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## Body

## 1. Introduction

The present research work is dedicated to studying the analytical solution of fractional-order parabolic equations. The literature is well recognized that a broad range of physics, engineering, nuclear physics, and mathematics problems can be defined as unique boundary and initial value problems. Homogeneous beam’s transverse vibrations are controlled by fractional single fourth-order parabolic partial differential equations (PDEs). Such problem types occur in viscoelastic and inelastic flow mathematical modeling, layer deflection theories, and beam deformation [1–12]. Analyses of these problems have taken several physicist’s and mathematician’s attention [13–15].The time fractional parabolic PDEs with variable coefficient:(1)∂βμ∂τβ+κϕ,φ,ψ∂4μ∂ϕ4+1φμϕ,φ,ψ∂4μ∂φ4+1ψρϕ,φ,ψ∂4μ∂ψ4=gϕ,φ,ψ,τ,1<β≤2,τ≥0,where κϕ,φ,ψ, μϕ,φ,ψ, and ρϕ,φ,ψ are positive. With initial conditions,
(2)μϕ,φ,ψ,τ=f0ϕ,φ,ψ,μτϕ,φ,ψ,τ=k0ϕ,φ,ψ,with boundary conditions
(3)μa,φ,ψ,τ=h0φ,ψ,τ,μb,φ,ψ,τ=h1φ,ψ,τ,μϕ,a,ψ,τ=g0φ,ψ,τ,μϕ,b,ψ,τ=g1φ,ψ,τ,μϕ,φ,a,τ=k0φ,ψ,τ,μϕ,φ,b,τ=k1φ,ψ,τ,μϕϕa,φ,ψ,τ=h0¯φ,ψ,τ,μϕϕb,φ,ψ,τ=h1¯φ,ψ,τ,μφφϕ,a,ψ,τ=g0¯φ,ψ,τ,μφφϕ,b,ψ,τ=g1¯φ,ψ,τ,μψψϕ,φ,a,τ=k0¯φ,ψ,τ,μψψϕ,φ,b,τ=k1¯φ,ψ,τ.For which,hℓ, gℓ, kℓ, hℓ, gℓ, and kℓ are continuous variables, and ℓ differs between 0 and 1 and beam’s flexural stiffness ratio [1] in its volume per unit mass, like, and its mentions [1, 3, 4, 6, 8, 10, 11]. Many researchers [10, 16, 17] have attempted to study the analytical solutions of the parabolic equation of the fourth-order. Different techniques have been suggested recently, such as the B-spline method [18], decomposition method [19], the implicit scheme [20], the explicit scheme [11], and the spline method [21] to analyze the solution of the partial differential fourth-order parabolic equation. Biazar and Ghazvini [22] have used He’s iterative technique for the solution of parabolic PDE’s. The modified version of this method was introduced in [23] to solve singular fourth-order parabolic PDEs. The fourth-order parabolic PDE analytical solution was examined in [24]. The modified Laplace discussed variational iteration technique [25] to solve singular fourth-order parabolic PDEs.G. Adomian is an American scientist who has developed the Adomian decomposition method. It focuses on the search for a set of solutions and the decomposition of the nonlinear operator into a sequence in which Adomian polynomials [26] are recurrently computed to use the terms. This method is improved with the aid of Elzaki transformation such that the improved method is known as the Elzaki decomposition method (EDM). Elzaki Transform (ET) is a modern integral transform introduced by Tarig Elzaki in 2010. ET is a modified transform of Sumudu and Laplace transforms. It is important to note that there are many differential equations with variable coefficients that Sumudu and Laplace cannot accomplish transforms but can be conveniently done by using ET [27–30]. Many mathematicians have been solving differential equations with the aid of ET, such as Navier-Stokes equations [30], heat-like equations [31] and Burgers–Huxley equation [32].


## 2. Preliminaries

### 2.1. Definition

The Abel-Riemann of fractional operatorDβ of order β is given as [27–30]
(4)Dβνζ=djdζjνζ,β=j,1Γj−βddζj∫0ζνζζ−ψβ−j+1dψ,j−1<β<j,where j∈Z+, β∈R+, and
(5)D−βνζ=1Γβ∫0ζζ−ψβ−1νψdψ,0<β≤1.


### 2.2. Definition

The fractional-order Abel-Riemann integration operatorJβ is defined as [27–30]
(6)Jβνζ=1Γβ∫0ζζ−ψβ−1νζdζ,ζ>0,β>0.The operator of basic properties:(7)Jβζj=Γj+1Γj+β+1ζj+ψ,Dβζj=Γj+1Γj−β+1ζj−ψ.


### 2.3. Definition

The Caputo fractional operatorDβ of β is defined as [27–30]
(8)CDβνζ=1Γj−β∫0ζνjψζ−ψβ−j+1dψ,j−1<β<j,djdζjνζ,j=β.


## 2.1. Definition

The Abel-Riemann of fractional operatorDβ of order β is given as [27–30]
(4)Dβνζ=djdζjνζ,β=j,1Γj−βddζj∫0ζνζζ−ψβ−j+1dψ,j−1<β<j,where j∈Z+, β∈R+, and
(5)D−βνζ=1Γβ∫0ζζ−ψβ−1νψdψ,0<β≤1.


## 2.2. Definition

The fractional-order Abel-Riemann integration operatorJβ is defined as [27–30]
(6)Jβνζ=1Γβ∫0ζζ−ψβ−1νζdζ,ζ>0,β>0.The operator of basic properties:(7)Jβζj=Γj+1Γj+β+1ζj+ψ,Dβζj=Γj+1Γj−β+1ζj−ψ.


## 2.3. Definition

The Caputo fractional operatorDβ of β is defined as [27–30]
(8)CDβνζ=1Γj−β∫0ζνjψζ−ψβ−j+1dψ,j−1<β<j,djdζjνζ,j=β.


## 3. Idea of NDM

The general fractional-order PDE is given as(9)Dβμϕ,τ+Lμϕ,τ+Nμϕ,τ=qϕ,τ,ϕ,τ≥0,1<β≤1,In Equation (9), we represent the linear part of the equation with L and the nonlinear part with N, and Dβ=∂β/∂τβ denotes the Caputo fractional derivatives.With initial condition,(10)μϕ,0=kϕ,We have applied the Elzaki transformation to Equation (9)
(11)EDβμϕ,τ+ELμϕ,τ+Nμϕ,τ=Eqϕ,τ,and using Elzaki Transform’s differentiation property, we get
(12)1sβEμϕ,τ−s2−βμϕ,0=Eqϕ,τ−ELμϕ,τ+Nμϕ,τ,Eμϕ,τ=s2μϕ,0+sβEqϕ,τ−sβELμϕ,τ+Nμϕ,τ.Now,μϕ,0=kϕ.
(13)Eμϕ,τ=s2kϕ+sβEqϕ,τ−sβELμϕ,τ+Nμϕ,τ.The following infinite series represent the EDM solutionμϕ,τ.
(14)μϕ,τ=∑j=0∞μjϕ,τ,and Adomian polynomials as
(15)Nμϕ,τ=∑j=0∞Aj,(16)Aj=1j!djdλjN∑j=0∞λjμjλ=0,j=0,1,2,⋯.We get replacement Equation (14) and Equation (15) in Equation (13).
(17)E∑j=0∞μjϕ,τ=s2kϕ+sβEqϕ,τ−sβEL∑j=0∞μjϕ,τ+∑j=0∞Aj.Applying the Elzaki transformation’s linearity,(18)Eμ0ϕ,τ=s2kϕ+sβEqϕ,τ,(19)Eμ1ϕ,τ=−sβELμ0ϕ,τ+A0.We can generally write(20)Eμj+1ϕ,τ=−sβELμjϕ,τ+Aj,j≥1.Equation (18) and Equation (20) implement the inverse Elzaki transformation
(21)μ0ϕ,τ=kϕ+E−1sβEqϕ,τ,μj+1ϕ,τ=−E−1sβELμjϕ,τ+Aj.


## 4. Numerical Implementation

### 4.1. Problem

Consider fractional-order one-dimensional parabolic equation:(22)∂βμ∂τβ+1ϕ+ϕ4120∂4μ∂ϕ4=0,1<β≤2,τ≥0,with initial conditions
(23)μϕ,0=0,μτϕ,0=1+ϕ5120,with boundary conditions
(24)μ12,τ=1+1/25120sinτ,μ1,τ=121120sinτ,∂2μ∂ϕ212,τ=16123sinτ,∂2μ∂ϕ21,τ=16sinτ.The Elzaki transform of Equation (22):
(25)1sβμϕ,s,u−s2−βμϕ,0−s3−βμτϕ,0=−E1ϕ+ϕ4120∂4μ∂ϕ4.Simplify and replace Equation (23) condition.
(26)μϕ,s,u=s20+s31+ϕ5120−E1ϕ+ϕ4120∂4μ∂ϕ4.Use of inverse Elzaki transformation(27)μϕ,τ=E−1s31+ϕ5120−sβE1ϕ+ϕ4120∂4μ∂ϕ4,(28)μϕ,τ=1+ϕ5120τ−E−1sβE1ϕ+ϕ4120∂4μ∂ϕ4.Equation (28) correction function is provided by
(29)∑ℓ=0∞μℓ+1ϕ,τ=1+ϕ5120τ−E−1sβE1ϕ+ϕ4120∑ℓ=0∞∂4μℓ∂ϕ4,the first term
(30)μ0ϕ,τ=1+ϕ5120τ,then we got
(31)μℓ+1ϕ,τ=−E−1sβE1ϕ+ϕ4120∑ℓ=0∞∂4μℓ∂ϕ4,for j=0,
(32)μ1ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ0∂ϕ4,μ1ϕ,τ=−E−11+ϕ5/120uβsβ+2=−1+ϕ5120τβ+1Γβ+2.The following terms are(33)μ2ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ1∂ϕ4=1+ϕ5120τ2β+1Γ2β+2,μ3ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ2∂ϕ4=−1+ϕ5120τ3β+1Γ3β+2,⋮The series form of problem (1) such as:(34)μϕ,τ=μ0ϕ,τ+μ1ϕ,τ+μ2ϕ,τ+μ3ϕ,τ+μ4ϕ,τ⋯,μϕ,τ=1+ϕ5120τ−τβ+1Γβ+2+τ2β+1Γ2β+2−τ3β+1Γ3β+2+τ4β+1Γ4β+2⋯,when β=2, then integer EDM solution is
(35)μϕ,τ=1+ϕ5120τ−τ33!+τ55!−τ77!+τ99!⋯.The exact result is given as(36)μϕ,τ=1+ϕ5120sinτ.In Figure1, the exact and the EDM solutions of problem 1 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 2, the EDM solutions of problem 1 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 1
The graphs of exact and EDM result forβ=2 of problem 1.Figure 2
The fractional-order graphs ofβ=0.8 and 0.6 of problem 1.


### 4.2. Problem

Consider fractional-order two-dimensional parabolic equation:(37)∂βμ∂τβ+21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4=0,1<β≤2,τ≥0,with initial conditions
(38)μϕ,φ,0=0,μτϕ,φ,0=2+ϕ66!+φ66!,with boundary conditions
(39)μ12,φ,τ=2+1/266!+φ66!sinτ,μ12,φ,τ=2+166!+φ66!sinτ,μϕϕ12,φ,τ=1/244!sinτ,μϕϕ12,φ,τ=124sinτ,μφφϕ,12,τ=1/244!sinτ,μφφϕ,12,τ=124sinτ.In the Elzaki transformation of Equation (37), we get
(40)1sβμϕ,φ,s,u−s2−βμϕ,0−s3−βμτϕ,0=−E21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,Simplify and replace Equation (38) condition.
(41)μϕ,φ,s,u=s20+s32+ϕ66!+φ66!−sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,using inverse Elzaki transformation
(42)μϕ,φ,τ=E−1s32+ϕ66!+φ66!−sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,(43)μϕ,φ,τ=2+ϕ66!+φ66!τ−E−1sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,Equation (43) correction function is provided by
(44)∑ℓ=0∞μℓ+1ϕ,φ,τ=2+ϕ66!+φ66!τ−E−1sβE21ϕ2+ϕ46!∑ℓ=0∞∂4μℓ∂ϕ4+21φ2+φ46!∑ℓ=0∞∂4μℓ∂φ4,the first term
(45)μ0ϕ,φ,τ=2+ϕ66!+φ66!τ,then we get
(46)μℓ+1ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∑ℓ=0∞∂4μℓ∂ϕ4+21φ2+φ46!∑ℓ=0∞∂4μℓ∂φ4,for j=0,
(47)μ1ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ0∂ϕ4+21φ2+φ46!∂4μ0∂φ4,μ1ϕ,τ=−E−12+ϕ66!+φ66!uβsβ+2=−2+ϕ66!+φ66!τβ+1Γβ+2.The following terms are(48)μ2ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ1∂ϕ4+21φ2+φ46!∂4μ1∂φ4,μ2ϕ,φ,τ=2+ϕ66!+φ66!τ2β+1Γ2β+2,μ3ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ2∂ϕ4+21φ2+φ46!∂4μ2∂φ4,μ3ϕ,φ,τ=−2+ϕ66!+φ66!τ3β+1Γ3β+2,..In the series form of problem (2), we get(49)μϕ,φ,τ=μ0ϕ,φ,τ+μ1ϕ,φ,τ+μ2ϕ,φ,τ+μ3ϕ,φ,τ+μ4ϕ,φ,τ⋯,μϕ,φ,τ=2+ϕ66!+φ66!τ−τβ+1Γβ+2+τ2β+1Γ2β+2−τ3β+1Γ3β+2+τ4β+1Γ4β+2⋯.Then,β=2, the integer EDM result as
(50)μϕ,φ,τ=2+ϕ66!+φ66!τ−τ33!+τ55!−τ77!+τ99!⋯.The exact solution is(51)μϕ,φ,τ=2+ϕ66!+φ66!sinτ.In Figure3, the exact and the EDM solutions of problem 2 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 4, the EDM solutions of problem 2 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 3
The exact and EDM solution forβ=2 of problem 2.Figure 4
The fractional-order graphs ofβ=0.8 and 0.6 of problem 2.


### 4.3. Problem

Consider fractional-order three-dimensional parabolic equation:(52)∂βμ∂τβ+2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4=0,(53)1<β≤2,τ≥0,with initial conditions
(54)μϕ,φ,ψ,0=ϕ+φ+ψ−cosϕ+cosφ+cosψ,(55)μτϕ,φ,ψ,0=cosϕ+cosφ+cosψ−ϕ+φ+ψ,with boundary conditions
(56)μ0,φ,ψ,τ=−1+φ+ψ−cosφ−cosψe−τ,μπ3,φ,ψ,τ=2π−36+φ+ψ−cosφ−cosψe−τ,μϕ,0,ψ,τ=−1+ϕ+ψ−cosϕ−cosψe−τ,μϕ,π3,ψ,τ=2π−36+ϕ+ψ−cosϕ−cosψe−τ,μϕ,φ,0,τ=−1+ϕ+φ−cosϕ−cosφe−τ,μϕ,φ,π3,τ=2π−36+ϕ+φ−cosϕ−cosφe−τ,μϕ0,φ,ψ,τ=μφϕ,0,ψ,τ=μψϕ,φ,0,τ=e−τ,μϕπ3,φ,ψ,τ=μφϕ,π3,ψ,τ=μψϕ,φ,π3,τ=3+22e−τ.In the Elzaki transformation of Equation (52), we get
(57)1sβμϕ,φ,ψ,s,u−s2−βμϕ,φ,ψ,0−s3−βμτϕ,φ,ψ,0=−E2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,Simplify and replace Equation (54) condition.
(58)μϕ,φ,ψ,s,u=s2ϕ+φ+ψ−cosϕ+cosφ+cosψ+s3cosϕ+cosφ+cosψ−ϕ+φ+ψ−sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,using the inverse Elzaki transform
(59)μϕ,φ,ψ,τ=E−1s2ϕ+φ+ψ−cosϕ+cosφ+cosψ+s3cosϕ+cosφ+cosψ−ϕ+φ+ψ−E−1sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,(60)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ−E−1sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4.Equation (59) correction function is provided by
(61)∑ℓ=0∞μℓ+1ϕ,φ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ−E−1sβE2φ+ψ2cosϕ−1∑ℓ=0∞∂4μℓ∂ϕ4+2ϕ+ψ2cosφ−1∑ℓ=0∞∂4μℓ∂φ4+2φ+ϕ2cosψ−1∑ℓ=0∞∂4μℓ∂ψ4,the first term
(62)μ0ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ,then we get
(63)μℓ+1ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∑ℓ=0∞∂4μℓ∂ϕ4+2ϕ+ψ2cosφ−1∑ℓ=0∞∂4μℓ∂φ4+2φ+ϕ2cosψ−1∑ℓ=0∞∂4μℓ∂ψ4,for j=0,
(64)μ1ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∂4μ0∂ϕ4+2ϕ+ψ2cosφ−1∂4μ0∂φ4+2φ+ϕ2cosψ−1∂4μ0∂ψ4,μ1ϕ,φ,ψ,τ=E−1ϕ+φ+ψ−cosϕ+cosφ+cosψuβsβ+1−uβsβ+2,μ1ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτβΓβ+1−τβ+1Γβ+2.The following terms are(65)μ2ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∂4μ1∂ϕ4+2ϕ+ψ2cosφ−1∂4μ1∂φ4+2φ+ϕ2cosψ−1∂4μ1∂ψ4,μ2ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτ2βΓ2β+1−τ2β+1Γ2β+2,μ3ϕ,φ,ψ,τ=E−1sβE2φ+ψ2cosϕ−1∂4μ2∂ϕ4+2ϕ+ψ2cosφ−1∂4μ2∂φ4+2φ+ϕ2cosψ−1∂4μ2∂ψ4,μ3ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτ3βΓ3β+1−τ3β+1Γ3β+2,⋮The series form of problem (3) such as(66)μϕ,φ,ψ,τ=μ0ϕ,φ,ψ,τ+μ1ϕ,φ,ψ,τ+μ2ϕ,φ,ψ,τ+μ3ϕ,φ,ψ,τ+μ4ϕ,φ,ψ,τ⋯,μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ+τβΓβ+1−τβ+1Γβ+2+τ2βΓ2β+1−τ2β+1Γ2β+2+τ3βΓ3β+1−τ3β+1Γ3β+2⋯.In the integer-order solution of EDM of Equation (52) at β=2, we get
(67)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ+τ22!−τ33!+τ44!−τ55!+τ66!−77!⋯.The exact solution is given as(68)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψe−τ.In Figure5, the exact and the EDM solutions of problem 3 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 6, the EDM solutions of problem 3 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 5
The exact and EDM solution forβ=2 of problem 3.Figure 6
For fractional-order graphs ofβ=0.8 and 0.6 of problem 3.


## 4.1. Problem

Consider fractional-order one-dimensional parabolic equation:(22)∂βμ∂τβ+1ϕ+ϕ4120∂4μ∂ϕ4=0,1<β≤2,τ≥0,with initial conditions
(23)μϕ,0=0,μτϕ,0=1+ϕ5120,with boundary conditions
(24)μ12,τ=1+1/25120sinτ,μ1,τ=121120sinτ,∂2μ∂ϕ212,τ=16123sinτ,∂2μ∂ϕ21,τ=16sinτ.The Elzaki transform of Equation (22):
(25)1sβμϕ,s,u−s2−βμϕ,0−s3−βμτϕ,0=−E1ϕ+ϕ4120∂4μ∂ϕ4.Simplify and replace Equation (23) condition.
(26)μϕ,s,u=s20+s31+ϕ5120−E1ϕ+ϕ4120∂4μ∂ϕ4.Use of inverse Elzaki transformation(27)μϕ,τ=E−1s31+ϕ5120−sβE1ϕ+ϕ4120∂4μ∂ϕ4,(28)μϕ,τ=1+ϕ5120τ−E−1sβE1ϕ+ϕ4120∂4μ∂ϕ4.Equation (28) correction function is provided by
(29)∑ℓ=0∞μℓ+1ϕ,τ=1+ϕ5120τ−E−1sβE1ϕ+ϕ4120∑ℓ=0∞∂4μℓ∂ϕ4,the first term
(30)μ0ϕ,τ=1+ϕ5120τ,then we got
(31)μℓ+1ϕ,τ=−E−1sβE1ϕ+ϕ4120∑ℓ=0∞∂4μℓ∂ϕ4,for j=0,
(32)μ1ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ0∂ϕ4,μ1ϕ,τ=−E−11+ϕ5/120uβsβ+2=−1+ϕ5120τβ+1Γβ+2.The following terms are(33)μ2ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ1∂ϕ4=1+ϕ5120τ2β+1Γ2β+2,μ3ϕ,τ=−E−1sβE1ϕ+ϕ4120∂4μ2∂ϕ4=−1+ϕ5120τ3β+1Γ3β+2,⋮The series form of problem (1) such as:(34)μϕ,τ=μ0ϕ,τ+μ1ϕ,τ+μ2ϕ,τ+μ3ϕ,τ+μ4ϕ,τ⋯,μϕ,τ=1+ϕ5120τ−τβ+1Γβ+2+τ2β+1Γ2β+2−τ3β+1Γ3β+2+τ4β+1Γ4β+2⋯,when β=2, then integer EDM solution is
(35)μϕ,τ=1+ϕ5120τ−τ33!+τ55!−τ77!+τ99!⋯.The exact result is given as(36)μϕ,τ=1+ϕ5120sinτ.In Figure1, the exact and the EDM solutions of problem 1 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 2, the EDM solutions of problem 1 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 1
The graphs of exact and EDM result forβ=2 of problem 1.Figure 2
The fractional-order graphs ofβ=0.8 and 0.6 of problem 1.


## 4.2. Problem

Consider fractional-order two-dimensional parabolic equation:(37)∂βμ∂τβ+21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4=0,1<β≤2,τ≥0,with initial conditions
(38)μϕ,φ,0=0,μτϕ,φ,0=2+ϕ66!+φ66!,with boundary conditions
(39)μ12,φ,τ=2+1/266!+φ66!sinτ,μ12,φ,τ=2+166!+φ66!sinτ,μϕϕ12,φ,τ=1/244!sinτ,μϕϕ12,φ,τ=124sinτ,μφφϕ,12,τ=1/244!sinτ,μφφϕ,12,τ=124sinτ.In the Elzaki transformation of Equation (37), we get
(40)1sβμϕ,φ,s,u−s2−βμϕ,0−s3−βμτϕ,0=−E21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,Simplify and replace Equation (38) condition.
(41)μϕ,φ,s,u=s20+s32+ϕ66!+φ66!−sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,using inverse Elzaki transformation
(42)μϕ,φ,τ=E−1s32+ϕ66!+φ66!−sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,(43)μϕ,φ,τ=2+ϕ66!+φ66!τ−E−1sβE21ϕ2+ϕ46!∂4μ∂ϕ4+21φ2+φ46!∂4μ∂φ4,Equation (43) correction function is provided by
(44)∑ℓ=0∞μℓ+1ϕ,φ,τ=2+ϕ66!+φ66!τ−E−1sβE21ϕ2+ϕ46!∑ℓ=0∞∂4μℓ∂ϕ4+21φ2+φ46!∑ℓ=0∞∂4μℓ∂φ4,the first term
(45)μ0ϕ,φ,τ=2+ϕ66!+φ66!τ,then we get
(46)μℓ+1ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∑ℓ=0∞∂4μℓ∂ϕ4+21φ2+φ46!∑ℓ=0∞∂4μℓ∂φ4,for j=0,
(47)μ1ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ0∂ϕ4+21φ2+φ46!∂4μ0∂φ4,μ1ϕ,τ=−E−12+ϕ66!+φ66!uβsβ+2=−2+ϕ66!+φ66!τβ+1Γβ+2.The following terms are(48)μ2ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ1∂ϕ4+21φ2+φ46!∂4μ1∂φ4,μ2ϕ,φ,τ=2+ϕ66!+φ66!τ2β+1Γ2β+2,μ3ϕ,φ,τ=−E−1sβE21ϕ2+ϕ46!∂4μ2∂ϕ4+21φ2+φ46!∂4μ2∂φ4,μ3ϕ,φ,τ=−2+ϕ66!+φ66!τ3β+1Γ3β+2,..In the series form of problem (2), we get(49)μϕ,φ,τ=μ0ϕ,φ,τ+μ1ϕ,φ,τ+μ2ϕ,φ,τ+μ3ϕ,φ,τ+μ4ϕ,φ,τ⋯,μϕ,φ,τ=2+ϕ66!+φ66!τ−τβ+1Γβ+2+τ2β+1Γ2β+2−τ3β+1Γ3β+2+τ4β+1Γ4β+2⋯.Then,β=2, the integer EDM result as
(50)μϕ,φ,τ=2+ϕ66!+φ66!τ−τ33!+τ55!−τ77!+τ99!⋯.The exact solution is(51)μϕ,φ,τ=2+ϕ66!+φ66!sinτ.In Figure3, the exact and the EDM solutions of problem 2 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 4, the EDM solutions of problem 2 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 3
The exact and EDM solution forβ=2 of problem 2.Figure 4
The fractional-order graphs ofβ=0.8 and 0.6 of problem 2.


## 4.3. Problem

Consider fractional-order three-dimensional parabolic equation:(52)∂βμ∂τβ+2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4=0,(53)1<β≤2,τ≥0,with initial conditions
(54)μϕ,φ,ψ,0=ϕ+φ+ψ−cosϕ+cosφ+cosψ,(55)μτϕ,φ,ψ,0=cosϕ+cosφ+cosψ−ϕ+φ+ψ,with boundary conditions
(56)μ0,φ,ψ,τ=−1+φ+ψ−cosφ−cosψe−τ,μπ3,φ,ψ,τ=2π−36+φ+ψ−cosφ−cosψe−τ,μϕ,0,ψ,τ=−1+ϕ+ψ−cosϕ−cosψe−τ,μϕ,π3,ψ,τ=2π−36+ϕ+ψ−cosϕ−cosψe−τ,μϕ,φ,0,τ=−1+ϕ+φ−cosϕ−cosφe−τ,μϕ,φ,π3,τ=2π−36+ϕ+φ−cosϕ−cosφe−τ,μϕ0,φ,ψ,τ=μφϕ,0,ψ,τ=μψϕ,φ,0,τ=e−τ,μϕπ3,φ,ψ,τ=μφϕ,π3,ψ,τ=μψϕ,φ,π3,τ=3+22e−τ.In the Elzaki transformation of Equation (52), we get
(57)1sβμϕ,φ,ψ,s,u−s2−βμϕ,φ,ψ,0−s3−βμτϕ,φ,ψ,0=−E2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,Simplify and replace Equation (54) condition.
(58)μϕ,φ,ψ,s,u=s2ϕ+φ+ψ−cosϕ+cosφ+cosψ+s3cosϕ+cosφ+cosψ−ϕ+φ+ψ−sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,using the inverse Elzaki transform
(59)μϕ,φ,ψ,τ=E−1s2ϕ+φ+ψ−cosϕ+cosφ+cosψ+s3cosϕ+cosφ+cosψ−ϕ+φ+ψ−E−1sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4,(60)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ−E−1sβE2φ+ψ2cosϕ−1∂4μ∂ϕ4+2ϕ+ψ2cosφ−1∂4μ∂φ4+2φ+ϕ2cosψ−1∂4μ∂ψ4.Equation (59) correction function is provided by
(61)∑ℓ=0∞μℓ+1ϕ,φ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ−E−1sβE2φ+ψ2cosϕ−1∑ℓ=0∞∂4μℓ∂ϕ4+2ϕ+ψ2cosφ−1∑ℓ=0∞∂4μℓ∂φ4+2φ+ϕ2cosψ−1∑ℓ=0∞∂4μℓ∂ψ4,the first term
(62)μ0ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ,then we get
(63)μℓ+1ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∑ℓ=0∞∂4μℓ∂ϕ4+2ϕ+ψ2cosφ−1∑ℓ=0∞∂4μℓ∂φ4+2φ+ϕ2cosψ−1∑ℓ=0∞∂4μℓ∂ψ4,for j=0,
(64)μ1ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∂4μ0∂ϕ4+2ϕ+ψ2cosφ−1∂4μ0∂φ4+2φ+ϕ2cosψ−1∂4μ0∂ψ4,μ1ϕ,φ,ψ,τ=E−1ϕ+φ+ψ−cosϕ+cosφ+cosψuβsβ+1−uβsβ+2,μ1ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτβΓβ+1−τβ+1Γβ+2.The following terms are(65)μ2ϕ,φ,ψ,τ=−E−1sβE2φ+ψ2cosϕ−1∂4μ1∂ϕ4+2ϕ+ψ2cosφ−1∂4μ1∂φ4+2φ+ϕ2cosψ−1∂4μ1∂ψ4,μ2ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτ2βΓ2β+1−τ2β+1Γ2β+2,μ3ϕ,φ,ψ,τ=E−1sβE2φ+ψ2cosϕ−1∂4μ2∂ϕ4+2ϕ+ψ2cosφ−1∂4μ2∂φ4+2φ+ϕ2cosψ−1∂4μ2∂ψ4,μ3ϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψτ3βΓ3β+1−τ3β+1Γ3β+2,⋮The series form of problem (3) such as(66)μϕ,φ,ψ,τ=μ0ϕ,φ,ψ,τ+μ1ϕ,φ,ψ,τ+μ2ϕ,φ,ψ,τ+μ3ϕ,φ,ψ,τ+μ4ϕ,φ,ψ,τ⋯,μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ+τβΓβ+1−τβ+1Γβ+2+τ2βΓ2β+1−τ2β+1Γ2β+2+τ3βΓ3β+1−τ3β+1Γ3β+2⋯.In the integer-order solution of EDM of Equation (52) at β=2, we get
(67)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψ1−τ+τ22!−τ33!+τ44!−τ55!+τ66!−77!⋯.The exact solution is given as(68)μϕ,φ,ψ,τ=ϕ+φ+ψ−cosϕ+cosφ+cosψe−τ.In Figure5, the exact and the EDM solutions of problem 3 at β=1 are shown by subgraphs, respectively. From the given figure, it can be seen that both the EDM and exact results are in close contact with each other. In Figure 6, the EDM solutions of problem 3 are investigated at different fractional order β=0.8 and 0.6. It is analyzed that time-fractional problem results are convergent to an integer-order effect as time-fractional analysis to integer order.Figure 5
The exact and EDM solution forβ=2 of problem 3.Figure 6
For fractional-order graphs ofβ=0.8 and 0.6 of problem 3.


## 5. Conclusion

In the present article, an efficient analytical technique is used to solve fractional-order parabolic equations. The present method is the combination of two well-known methods, namely, Elzaki transform and Adomian decomposition method. The Elzaki transform is applied to the given problem, which makes it easier. After this, we implemented Adomian decomposition method and then inverse Elzaki transform to get closed form analytical solutions for the given problems. The proposed method required small number of calculation to attain closed form solutions and is therefore considered to be one of the best analytical method to solve fractional-order partial differential equations.



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