# The Analysis of Fractional-Order Nonlinear Systems of Third Order KdV and Burgers Equations via a Novel Transform

Authors: A. A. Alderremy; Shaban Aly; Rabia Fayyaz; Adnan Khan; Rasool Shah; Noorolhuda Wyal
Journal: Complexity
 (2022)

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## Abstract

In this article, we solve nonlinear systems of third order KdV Equations and the systems of coupled Burgers equations in one and two dimensions with the help of two different methods. The suggested techniques in addition with Laplace transform and Atangana–Baleanu fractional derivative operator are implemented to solve four systems. The obtained results by implementing the proposed methods are compared with exact solution. The convergence of the method is successfully presented and mathematically proved. The results we get are compared with exact solution through graphs and tables which confirms the effectiveness of the suggested techniques. In addition, the results obtained by employing the proposed approaches at different fractional orders are compared, confirming that as the value goes from fractional order to integer order, the result gets closer to the exact solution. Moreover, suggested techniques are interesting, easy, and highly accurate which confirm that these methods are suitable methods for solving any partial differential equations or systems of partial differential equations as well.


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## Body

## 1. Introduction

In recent years, fractional calculus has surpassed ordinary calculus in popularity. The standard calculus has reached its pinnacle in terms of discovery. Mathematicians and engineers require fractional calculus as a solution. This permits a more accurate description of real-world phenomena than the traditional “integer” order. Numerous mathematicians, including Fourier, Laplace, Riesz, and others, were engaged and made substantial contribution to the subject. Modern definitions of fractional order derivatives and integrals, such as the Atangana–Baleanu fractional integral [1], the Caputo fractional derivative [2], and the Caputo–Fabrizio fractional derivative [3], have ushered in a new era in the literature of fractional derivatives. Models based on fractional calculus can accurately depict numerous engineering, physics, and chemistry processes, among others [4]. In addition, fractional calculus is used to simulate the frequency-dependent damping behavior of a variety of viscoelastic materials [5], the dynamics of interfaces between nanoparticles and substrates [6], economics [7], and numerous other applications [8–11].Finding the actual or approximate solutions of FDEs is crucial in all of these areas of study, but because we lack a method for obtaining the precise solution of these sorts of FDEs, we must focus on approximating the exact solution. Determining the exact answer to such FDEs and other scientific applications is a difficult task in mathematics. Unlike the approximate answer [12], the exact solution enables us to comprehend the problem’s mechanism and complexity. Obtaining exact analytical expressions to FDEs is exceedingly difficult, if not impossible, due to the complexity of computation involved in these equations. As a result, it is necessary to seek out some useful approximations and numerical techniques, such as the homotopy perturbation method [13], variation iteration method [14], residual power series method [15], approximate-analytical method [16], Elzaki transform decomposition method [17], Iterative Laplace transform method [18], Adomian decomposition method [19], reduced differential transform method, and others [20–23].In this paper, we provided two analytic approaches in conjunction with the Laplace transformation and fractional derivatives in Antagana–Baleanu solution to satisfy fractional-order problems [24]. The first method is the mixing of Laplace transform (LT) and variational iteration method known as variational iteration transform method (VITM) which was first developed by He [25] and is an effective solution for a broad variety of problems in scientific fields [26, 27]. The second significant methodology for solving nonlinear functional equations is the combination of the Adomian decomposition method and Laplace transform, which was first developed by George Adomian (1923–1996) in the 1980s. The technique depends on the decomposition of a nonlinear equation result into a series of functions. A polynomial produced by a power series expansion of an analytic function returns each series term. This method for solving several nonlinear fractional-order differential equations is interesting, straightforward, and accurate.Harry Bateman introduced the Burgers equation in 1915 [28], and it was subsequently dubbed by the Burgers equation. The Burgers equation has many applications in science and engineering, especially when dealing with nonlinear problems. Burgers equation applications have grown in prominence and attention among mathematical scientists and researchers. This equation is acknowledged to represent a range of phenomena, such as dynamic modeling, heat conduction, acoustic waves, and turbulence [29–31]. In 1895, Korteweg and Vries initially derived the Korteweg-De Vries (KDV) equation. The KDV equation is used to predict long waves, tides, solitary waves, and wave propagation in a shallow canal. The KDV equation is utilized in several disciplines, including fluid mechanics, signal processing, hydrology, viscoelasticity, and fractional kinetics.


## 2. Preleminaries

In this section, we presented some basic definitions of fractional calculus related to our present work.Definition 1.
The derivative by Caputo having order fraction is defined as(1)Dℑσchℑ=1n−σ∫0ℑℑ−kn−σ−1hnkdk,wheren<σ≤n+1.Definition 2.
The derivative by Caputo having order fraction with the aid of Laplace transformDℑσLChℑ is defined as(2)LDℑσLChℑω=1ωn−σωnLhξ,ℑω−ωn−1hξ,0−⋯−hn−1ξ,0.Definition 3.
The fractional Atangana–Baleanu derivative in terms of Caputo manner is defined as(3)DℑσABChℑ=Bσ1−σ∫aℑh′kEσ−σ1−σ1−kσdk.where normalization function is denoted by Bσ with B0=B1=1,h∈H1a,b,b>a,σ∈0,1 and Eσ is the Mittag–Leffler function.Definition 4.
In terms of Riemann–Liouville, the Atangana–Baleanu fractional derivative is defined as(4)DℑσABChℑ=Bσ1−σddℑ∫aℑhkEσ−γ1−σ1−kσdk.Definition 5.
The Atangana–Baleanu operator in connection with Laplace transform is given by(5)DℑσABhℑω=BσωσLhℑω−ωσ−1h01−σωσ+σ/1−γ.Definition 6.
Consider0<σ<1, and h is a function of order σ, then the fractional integral operator for σ is defined as(6)IℑσABChℑ=1−σBσhℑ+σBσΓσ∫aℑhkℑ−kσ−1dk.


## 3. Idea of LTDM

Here, we discuss the methodology of LTDM for solving fractional-order partial differential equations.(7)DℑσABCφξ,ℑ+G¯1ξ,ℑ+N1ξ,ℑ=ℱξ,ℑ,0<σ≤1,having initial terms(8)φξ,0=ξξ,∂∂ℑφξ,0=ζξ.Here, the fractional-order AB operator is indicated fromDℑσABC=∂σ/∂ℑσ having order σ, G¯1,N1 are linear and nonlinear operator, and ℱξ,ℑ represent the source term.On taking the Laplace transform of (7), we get(9)LDℑσABCφξ,ℑ+G¯1ξ,ℑ+N1ξ,ℑ=Lℱξ,ℑ.Using the differentiation property of LT, we obtain(10)Lφξ,ℑ=Θξ,ω−ωσ+σ1−σωσLG¯1ξ,ℑ+N1ξ,ℑ,where Θξ,ω=1/ωσ+1ωσh0ξ+ωσ−1h1ξ+⋯+g1ξ+ωσ+σ1−σ/ωσℱξ,ℑ.Now, using the inverse Laplace transform, we get(11)φξ,ℑ=Θξ,ω−L−1ωσ+σ1−σωσLG¯1ξ,ℑ+N1ξ,ℑ,where Θξ,ω shows the term that come from the source term. LTDM generates the result of the infinite series of φξ,ℑ(12)φξ,ℑ=∑m=0∞φmξ,ℑ,and decomposing the nonlinear operator N1 as(13)N1ξ,ℑ=∑m=0∞Am,where the Adomian polynomials are represented by Am(14)Am=1m!∂m∂ℓmN1∑k=0∞ℓkξk,∑k=0∞ℓkℑkℓ=0,and putting equations (12) and (14) into (11), we get(15)∑m=0∞φmξ,ℑ=Θξ,ω−L−1ωσ+σ1−σωσLG¯1∑m=0∞ξm,∑m=0∞ℑm+∑m=0∞Am.The terms listed below are defined as(16)φ0ξ,ℑ=Θξ,ω,φ1ξ,ℑ=L−1ωσ+σ1−σωσLG¯1ξ0,ℑ0+A0.As a result, all of the components form≥1 are determined as(17)φm+1ξ,ℑ=L−1ωσ+σ1−σωσLG¯1ξm,ℑm+Am.


## 4. VITM Formulation

Here, we discuss the methodology of VITM for solving fractional-order partial differential equations.(18)DℑσABCφξ,ℑ+ℳφξ,ℑ+Nφξ,ℑ−Fξ,ℑ=0,m−1<σ≤m,having initial term(19)φξ,0=g1ξ.Here, the fractional-order AB operator is indicated fromDℑσABC=∂σ/∂ℑσ. ℳ,N are linear and non,linear operator and F represent the source term.On taking the Laplace transform of (18), we get(20)LDℑσφξ,ℑ+Lℳφξ,ℑ+Nφξ,ℑ−Fξ,ℑ=0.Using the differentiation property of LT, we obtain(21)Lφξ,ℑ=ωσωσ+σ1−σLℳφξ,ℑ+Nφξ,ℑ−Fξ,ℑ.The method of iteration for the equation (21)(22)φm+1ξ,ℑ=φmξ,ℑ+σsωσωσ+σ1−σLℳφξ,ℑ+Nφξ,ℑ−Fξ,ℑ.σs is Lagrange multiplier and(23)σs=−ωσ+σ1−σωσ,Equation (22) series form solution is obtained by using the inverse Laplace transform.(24)φ0ξ,ℑ=φ0+L−1σsL−Fξ,ℑ,φ1ξ,ℑ=L−1σsLℳφξ,ℑ+Nφξ,ℑ,⋮φn+1ξ,ℑ=L−1σsLℳφ0ξ,ℑ+φ1ξ,ℑ+⋯,φnξ,ℑ+Nφ0ξ,ℑ+φ1ξ,ℑ,…,φnξ,ℑ.


## 5. Applications

To show the validity and capability of the suggested techniques, we implemented proposed methods for solving four nonlinear systems.


### 5.1. Problem 1

Consider system of homogeneous KdV equation having order three(25)∂σφ∂ℑσ=∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ,∂σν∂ℑσ=−2∂3ν∂ξ3+φ∂ν∂ξ0<σ≤1,with initial source(26)φξ,0=3−6tanh2ξ2,νξ,0=−3l2tanh2ξ2.On taking the Laplace transform of (25), we get(27)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ.ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L−2∂3ν∂ξ3+φ∂ν∂ξ.We obtain when we use the Laplace inverse transform(28)φξ,ℑ=3−6tanh2ξ2+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ.νξ,ℑ=−3l2tanh2ξ2+L−1ωσ+σ1−ωσωσL−2∂3ν∂ξ3+φ∂ν∂ξ.Assume that the solution,φξ,ℑ and νξ,ℑ in series form as(29)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,where φφξ=∑m=0∞Am, ννξ=∑m=0∞ℬm, and ννξ=∑m=0∞ℬm are Adomian polynomials that characterize the nonlinear terms, and so equation (28) is rewritten as(30)∑m=0∞φmξ,ℑ=3−6tanh2ξ2+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3+∑m=0∞Am+∑m=0∞ℬm.∑m=0∞νmξ,ℑ=−3l2tanh2ξ2+L−1ωσ+σ1−ωσωσL−2∂3ν∂ξ3+∑m=0∞Cm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(31)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=ν0ν0ξ,ℬ1=ν1ν0ξ+ν0ν1ξ,ℬ2=ν2ν0ξ+ν1ν1ξ+ν0ν2ξ,C0=φ0ν0ξ,C1=φ1ν0ξ+φ0ν1ξ,C2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ,As a result, when comparing the two sides of equation (30)(32)φ0ξ,ℑ=3−6tanh2ξ2,ν0ξ,ℑ=−3l2tanh2ξ2.Form=0,(33)φ1ξ,ℑ=−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ,ν1ξ,ℑ=3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ.Form=1,(34)φ2ξ,ℑ=322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,The approximate solution to the series is written as(35)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯φξ,ℑ=3−6tanh2ξ2−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ+322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=−3l2tanh2ξ2+3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ+3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(36)φξ,ℑ=3−6tanh2ℑ+ξ2,νξ,ℑ=−3l2tanh2ℑ+ξ2.


#### 5.1.1. VITM Analytical Results

For Equation (25), we have the iteration formula:(37)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φm∂ξ3+φm∂φm∂ξ+νm∂νm∂ξ,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3νm∂ξ3φm∂νm∂ξ,where(38)φ0ξ,ℑ=3−6tanh2ξ2,ν0ξ,ℑ=−3l2tanh2ξ2.Form=0,1,2,⋯,(39)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ0∂ξ3+φ0∂φ0∂ξ+ν0∂ν0∂ξ,φ1ξ,ℑ=−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν0∂ξ3φ0∂ν0∂ξ,ν1ξ,ℑ=3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ1∂ξ3+φ1∂φ1∂ξ+ν1∂ν1∂ξ,φ2ξ,ℑ=322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν1∂ξ3φ1∂ν1∂ξ,ν1ξ,ℑ=3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=∑m=0∞φmξ,ℑ=3−6tanh2ξ2−6sech2ξ2tanhξ2σℑσΓσ+1=+1−σ+322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=−3l2tanh2ξ2+3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ+3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(40)φξ,ℑ=3−6tanh2ℑ+ξ2,νξ,ℑ=−3l2tanh2ℑ+ξ2.The analytical solution and exact solution is shown in Figures1(a) and 1(b) at σ=1 and −5≤ψ≥5. Figure 1(c) shows the absolute error, and Figure 1(d) gives the solution at various fractional-order graph for φξ,ℑ. The behavior of the exact solution and analytical solution for νξ,ℑ is seen in Figures 2(a) and 2(b). Tables 1 and 2 show the comparison of the exact and our methods solution in addition with the absolute error at different fractional-order. From the figure and tables, it is clear that our methods solution is in good agreement with the exact solution.Figure 1
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ of problem 1.
(a)(b)(c)(d)Figure 2
Exact solution and analytical solution forνξ,ℑ problem 1.
(a)(b)Table 1
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.802.9999999850000003.0000000000000001.5000000000E-081.5000000000E-081.5000000000E-080.12.9849950500000002.9849950650000001.5000000000E-084.8177000000E-051.7264100000E-040.22.9403385240000002.9403385380000001.4000000000E-089.5412000000E-053.4187000000E-040.32.8669121160000002.8669121300000001.4000000000E-081.4077300000E-045.0438500000E-040.42.7661440730000002.7661440860000001.3000000000E-081.8342200000E-046.5718200000E-040.52.6399509450000002.6399509570000001.2000000000E-082.2262900000E-047.9764400000E-040.62.4906618060000002.4906618160000001.0000000000E-082.5779500000E-049.2363200000E-040.72.3209299610000002.3209299700000009.0000000000E-092.8847600000E-041.0335470000E-030.82.1336376490000002.1336376570000008.0000000000E-093.1437800000E-041.1263450000E-030.91.9317992900000001.9317992950000005.0000000000E-093.3536800000E-041.2015390000E-031.01.7184683350000001.7184683380000003.0000000000E-093.5145100000E-041.2591560000E-03Table 2
Comparison of the proposed method and exact results with Absolute Error (AE) at various fractional order forνξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.0000000000106070.0000000000000001.0606601700E-111.0606601700E-111.0606601700E-110.1−0.000010610102200−0.0000105678064204.2295783020E-081.6438202410E-079.6316691610E-070.2−0.000042187034300−0.0000421032931208.3741185290E-083.2548962340E-071.9071994720E-060.3−0.000094107345200−0.0000939838027601.2354244410E-074.8020666600E-072.8137865210E-060.4−0.000165361111500−0.0001652001485001.6096295920E-076.2567014160E-073.6661525080E-060.5−0.000254593128400−0.0002543977659001.9536257270E-077.5938997350E-074.4497048860E-060.6−0.000360156490500−0.0003599302736002.2621689430E-078.7933098140E-075.1525210840E-060.7−0.000480175029200−0.0004799218942002.5313503520E-079.8396949140E-075.7656675160E-060.8−0.000612610692900−0.0006123348322002.7586074000E-071.0723124430E-066.2833302080E-060.9−0.000755331965700−0.0007550376911002.9427450770E-071.1438942760E-066.7027805770E-061.0−0.000906179730200−0.0009058713463003.0838397500E-071.1987439840E-067.0241846510E-06


### 5.2. Problem 2

Consider the generalized coupled Hirota Satsuma KdV system(41)∂σφ∂ℑσ=12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ,∂σν∂ℑσ=3φ∂ν∂ξ−∂3ν∂ξ3,∂σℓ∂ℑσ=3φ∂ℓ∂ξ−∂3ℓ∂ξ30<σ≤1,with initial source(42)φξ,0=−13+2tanh3ξ,νξ,0=tanhξ,ℓξ,0=83tanhξ.On taking the Laplace transform of (41), we get(43)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ,ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L3φ∂ν∂ξ−∂3ν∂ξ3,ωσLℓξ,ℑ−ω−1ℓξ,0ωσ+σ1−ωσ=L3φ∂ℓ∂ξ−∂3ℓ∂ξ3.We obtain when we use the Laplace inverse transform(44)φξ,ℑ=−13+2tanh3ξ+L−1ωσ+σ1−ωσωσL12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ.νξ,ℑ=tanhξ+L−1ωσ+σ1−ωσωσL3φ∂ν∂ξ−∂3ν∂ξ3,ℓξ,ℑ=83tanhξ+L−1ωσ+σ1−ωσωσL3φ∂ℓ∂ξ−∂3ℓ∂ξ3.Assume that the solution,φξ,ℑ, νξ,ℑ, and ℓξ,ℑ in series form as(45)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,ℓξ,ℑ=∑m=0∞ℓmξ,ℑ,where φφξ=∑m=0∞Am, νℓξ=∑m=0∞ℬm, φνξ=∑m=0∞Cm, and φℓξ=∑m=0∞Dm are Adomian polynomials that characterize the nonlinear terms, and so equation (30) is rewritten as(46)∑m=0∞φmξ,ℑ=−13+2tanh3ξ+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3−3∑m=0∞Am+3∑m=0∞ℬm.∑m=0∞νmξ,ℑ=tanhξ+L−1ωσ+σ1−ωσωσL3∑m=0∞Cm−∂3ν∂ξ3,∑m=0∞ℓmξ,ℑ=83tanhξ+L−1ωσ+σ1−ωσωσL3∑m=0∞Dm−∂3ℓ∂ξ3.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (7),(47)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=ν0ℓ0ξ+ℓ0ν0ξ,ℬ1=ν0ℓ0ξ+ν1ℓ0ξ+ℓ1ν0ξ+ℓ0ν1ξ,ℬ2=ν2ℓ0ξ+ν1ℓ1ξ+ν0ℓ2ξ+ℓ2ν0ξ+ℓ1ν1ξ+ℓ0ν2ξ,C0=φ0ν0ξ,C1=φ1ν0ξ+φ0ν1ξ,C2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ,D0=φ0ℓ0ξ,D1=φ1ℓ0ξ+φ0ℓ1ξ,D2=φ2ℓ0ξ+φ1ℓ1ξ+φ0ℓ2ξ,As a result, when comparing the two sides of (46),(48)φ0ξ,ℑ=−13+2tanh2ξ,ν0ξ,ℑ=tanhξ,ℓ0ξ,ℑ=83tanhξ.Form=0,(49)φ1ξ,ℑ=4sech2ξtanhξσℑσΓσ+1+1−σ,ν1ξ,ℑ=sech2ξσℑσΓσ+1+1−σ,ℓ1ξ,ℑ=83sech2ξσℑσΓσ+1+1−σ.Form=1,(50)φ2ξ,ℑ=4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ℓ2ξ,ℑ=−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(51)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯ℓξ,ℑ=∑m=0∞ℓmξ,ℑ=ℓ0ξ,ℑ+ℓ1ξ,ℑ+ℓ2ξ,ℑ+⋯φξ,ℑ=−13+2tanh2ξ+4sech2ξtanhξσℑσΓσ+1+1−σ+4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=tanhξ+sech2ξσℑσΓσ+1+1−σ−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯ℓξ,ℑ=83tanhξ+83sech2ξσℑσΓσ+1+1−σ−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1,(52)φξ,ℑ=−13+2tanh2ℑ+ξ,νξ,ℑ=tanhℑ+ξ,ℓξ,ℑ=83tanhℑ+ξ.


#### 5.2.1. VITM Analytical Results

For equation (41), we have the iteration formula(53)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φm∂ξ3−3φm∂φm∂ξ+3∂∂ξνmℓm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂νm∂ξ−∂3νm∂ξ3,ℓm+1ξ,ℑ=ℓmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ℓm∂ξ−∂3ℓm∂ξ3,where(54)φ0ξ,ℑ=−13+2tanh2ξ,ν0ξ,ℑ=tanhξ,ℓ0ξ,ℑ=83tanhξ.Form=0,1,2,⋯,(55)φ1ξ,ℑ==φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ0∂ξ3−3φ0∂φ0∂ξ+3∂∂ξν0ℓ0,φ1ξ,ℑ=4sech2ξtanhξσℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ν0∂ξ−∂3ν0∂ξ3,ν1ξ,ℑ=sech2ξσℑσΓσ+1+1−σ,ℓ1ξ,ℑ=ℓ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ0∂ν0∂ξ−∂3ν0∂ξ3,ℓ1ξ,ℑ=83sech2ξσℑσΓσ+1+1−σ,φ2ξ,ℑ==φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ1∂ξ3−3φ1∂φ1∂ξ+3∂∂ξν1ℓ1,φ2ξ,ℑ=4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ν2ξ,ℑ=−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ℓ2ξ,ℑ=ℓ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ℓ2ξ,ℑ=−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=−13+2tanh2ξ+4sech2ξtanhξσℑσΓσ+1+1−σ+4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=tanhξ+sech2ξσℑσΓσ+1+1−σ−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯ℓξ,ℑ=83tanhξ+83sech2ξσℑσΓσ+1+1−σ−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(56)φξ,ℑ=−13+2tanh2ℑ+ξ,νξ,ℑ=tanhℑ+ξℓξ,ℑ=83tanhℑ+ξThe analytical solution and exact solution forφξ,ℑ of example 2 at σ=1 and −5≤ψ≥5 are shown in Figures 3(a) and 3(b), whereas Figures 3(c) and 3(d) show the absolute error and the solution at various fractional-order. The graphical behavior of exact solution and analytical solution for νξ,ℑ are shown in Figures 4(a) and 4(b), while Figures 4(c) and 4(d) show the absolute error and the solution at different fractional-order. Figures 5(a) and 5(b) give the solution graph for ℓξ,ℑ. Tables 3–5 show the comparison of the exact and suggested methods solution in addition with the absolute error at various fractional order. From the results of the figures and Tables, it is confirmed that our method solution converges quickly towards exact solution.Figure 3
Exact solution, analytical solution, absolute error, and various fractional order solutions forφξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 4
Exact solution, analytical solution, absolute error, and various fractional order solutions forνξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 5
Exact solution and analytical solution forℓξ,ℑ problem 2.
(a)(b)Table 3
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.333333313300000−0.3333333333000002.0000000000E-082.0000000000E-082.0000000000E-080.1−0.313426424600000−0.3134264438000001.9200000000E-086.3598200000E-056.4898550000E-040.2−0.255343407800000−0.2553434248000001.7000000000E-081.2227320000E-041.2475490000E-030.3−0.163500606900000−0.1635006205000001.3600000000E-081.7185690000E-041.7533551000E-030.4−0.044480856100000−0.0444808656700009.5700000000E-092.0958106000E-042.1381671200E-030.50.0939165789000000.0939165733000005.6000000000E-092.3429753000E-042.3902801500E-030.60.2436640045000000.2436640025000002.0000000000E-092.4637090000E-042.5134153000E-030.70.3973409319000000.3973409332000001.3000000000E-092.4731770000E-042.5230449000E-030.80.5487048209000000.5487048246000003.7000000000E-092.3933690000E-042.4416045000E-030.90.6929714497000000.6929714558000006.1000000000E-092.2486240000E-042.2939183000E-031.00.8268459177000000.8268459237000006.0000000000E-092.0621260000E-042.1036596000E-03Table 4
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forνξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of techniques methodsAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0000999999996700.0001000000000003.3000000000E-131.6117443060E-041.6442500900E-030.10.0997670002700000.0997670012500009.8000000000E-101.5957435000E-041.6279175300E-030.20.1974714226000000.1974714245000001.9000000000E-091.5489750000E-041.5801969000E-030.30.2914041235000000.2914041262000002.7000000000E-091.4749940000E-041.5047167000E-030.40.3800345229000000.3800345262000003.3000000000E-091.3791040000E-041.4068874000E-030.50.4621957984000000.4621958021000003.7000000000E-091.2675890000E-041.2931204000E-030.60.5371207210000000.5371207248000003.8000000000E-091.1469190000E-041.1700156000E-030.70.6044312472000000.6044312511000003.9000000000E-091.0230770000E-041.0436745000E-030.80.6640926721000000.6640926758000003.7000000000E-099.0109100000E-059.1923020000E-040.90.7163465584000000.7163465619000003.5000000000E-097.8482200000E-058.0061750000E-041.00.7616361502000000.7616361534000003.2000000000E-096.7692400000E-056.9054610000E-04Table 5
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forℓξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0002666666658000.0002666666667009.0000000000E-134.2979848170E-041.5398420700E-030.10.2660453341000000.2660453367000002.6000000000E-094.2553160000E-041.5245483000E-030.20.5265904603000000.5265904653000005.0000000000E-094.1305980000E-041.4798594000E-030.30.7770776627000000.7770776698000007.1000000000E-093.9333160000E-041.4091735000E-030.41.0134253940000001.0134254030000009.0000000000E-093.6776200000E-041.3175580000E-030.51.2325221290000001.2325221380000009.0000000000E-093.3802300000E-041.2110150000E-030.61.4323219230000001.4323219330000001.0000000000E-083.0584500000E-041.0957270000E-030.71.6118166590000001.6118166700000001.1000000000E-082.7282100000E-049.7741000000E-040.81.7709137920000001.7709138020000001.0000000000E-082.4029100000E-048.6086700000E-040.91.9102574890000001.9102574990000001.0000000000E-082.0928600000E-047.4978500000E-041.02.0310297340000002.0310297420000008.0000000000E-091.8051200000E-046.4670200000E-04


### 5.3. Problem 3

Consider the coupled Burgers equation in one dimension(57)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1,with initial source(58)φξ,0=cosξ,νξ,0=cosξ.On taking the Laplace transform of (57), we get(59)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν.ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν.We obtain when we use the Laplace inverse transform(60)φξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν.νξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν.Assume that the solution,φξ,ℑ and νξ,ℑ, in series form as(61)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,where φφξ=∑m=0∞Am, φνξ=∑m=0∞ℬm, and ννξ=∑m=0∞Cm are Adomian polynomials that characterize the nonlinear terms, and so (60) is rewritten as(62)∑m=0∞φmξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+2∑m=0∞Am−∑m=0∞ℬm.∑m=0∞νmξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∑m=0∞Cm−∑m=0∞ℬm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(63)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1.As a result, when comparing the two sides of (62),(64)φξ,0=cosξ,νξ,0=cosξ.Form=0,(65)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1.Form=1,(66)φ2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(67)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯φξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1,(68)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.In closed form,φξ,ℑ=cosξexp−ℑ and νξ,ℑ=cosξexp−ℑ.


#### 5.3.1. VITM Analytical Results

For (57), we have the iteration formula:(69)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+2νm∂ν∂ξ−∂∂ξφmνm,where(70)φξ,0=cosξ,νξ,0=cosξ.Form=0,1,2,⋯,(71)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ℑ=−cosξσℑσΓ1−σ+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+2ν0∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=−cosξσℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+2ν1∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯νξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯We achieve the exact solution by puttingσ=1,(72)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.In closed form,φξ,ℑ=cosξexp−ℑ and νξ,ℑ=cosξexp−ℑ.In Figures6(a) and 6(b), we display the solution graph at σ=1 for φξ,ℑ,ϕξ,ℑ in the domain −5≤ψ,ϕ≥5, and Figures 6(c) and 6(d) show solution graph of absolute error and various fractional order solution. Also Table 6 demonstrates the error comparison of the exact and suggested methods solution at various orders. It is verified from the figures and table that our solution is closely related with the exact solution.Figure 6
Exact solution, analytical solution, absolute error. and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 3.
(a)(b)(c)(d)Table 6
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and νξ,ℑ of problem 3.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.9999000050000000.9999000000000005.0000000000E-099.5960000000E-055.7744580000E-040.10.9949046699000000.9949046649000005.0000000000E-099.5480600000E-055.7456100000E-040.20.9799685760000000.9799685711000004.9000000000E-099.4047100000E-055.6593530000E-040.30.9552409602000000.9552409555000004.7000000000E-099.1674100000E-055.5165500000E-040.40.9209688925000000.9209688879000004.6000000000E-098.8385000000E-055.3186280000E-040.50.8774948080000000.8774948036000004.4000000000E-098.4212800000E-055.0675630000E-040.60.8252530855000000.8252530813000004.2000000000E-097.9199200000E-054.7658660000E-040.70.7647657069000000.7647657031000003.8000000000E-097.3394300000E-054.4165490000E-040.80.6966370421000000.6966370386000003.5000000000E-096.6856000000E-054.0231040000E-040.90.6215478104000000.6215478073000003.1000000000E-095.9649700000E-053.5894610000E-041.00.5402482784000000.5402482757000002.7000000000E-095.1847400000E-053.1199530000E-04


### 5.4. Problem 4

Consider the coupled Burgers equation two dimension(73)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1,with initial source(74)φξ,0=cosξ,νξ,0=cosξ.On taking the Laplace transform of (73), we get(75)ωσLφξ,ϖ,ℑ−ω−1φξ,ϖ,0ωσ+σ1−ωσ=L∂2φ∂ξ2+∂2φ∂ϖ2+2φ∂φ∂ξ−∂∂ξφν.ωσLνξ,ϖ,ℑ−ω−1νξ,ϖ,0ωσ+σ1−ωσ=L∂2ν∂ξ2+∂2ν∂ϖ2+2ν∂ν∂ξ−∂∂ξφν.We obtain when we use the Laplace inverse transform(76)φξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+∂2φ∂ϖ2+2φ∂φ∂ξ−∂∂ξφν.νξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∂2ν∂ϖ2+2ν∂ν∂ξ−∂∂ξφν.Assume that the solution,φξ,ϖ,ℑ and νξ,ϖ,ℑ in series form as(77)φξ,ϖ,ℑ=∑m=0∞φmξ,ϖ,ℑ,νξ,ϖ,ℑ=∑m=0∞νmξ,ϖ,ℑ,where φφξ=∑m=0∞Am, φνξ=∑m=0∞ℬm, and ννξ=∑m=0∞Cm are Adomian polynomials that characterize the nonlinear terms, and so (76) is rewritten as(78)∑m=0∞φmξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+∂2φ∂ϖ2+2∑m=0∞Am−∑m=0∞ℬm.∑m=0∞νmξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∂2φ∂ϖ2+∑m=0∞Cm−∑m=0∞ℬm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(79)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=φ0ν0ξ+ν0φ0ξ,ℬ1=φ0ν0ξ+φ1ν0ξ+ν1φ0ξ+ν0φ1ξ,ℬ2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ+ν2φ0ξ+ν1φ1ξ+ν0φ2ξ,C0=ν0ν0ξ,C1=ν1ν0ξ+ν0ν1ξ,C2=ν2ν0ξ+ν1ν1ξ+ν0ν2ξ.As a result, when comparing the two sides of (78),(80)φξ,0=cosξ,νξ,0=cosξ.Form=0,(81)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.Form=1,(82)φ2ξ,ϖ,ℑ=2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ϖ,ℑ=2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(83)φξ,ϖ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ+ℑ+φ1ξ+ℑ+φ2ξ+ℑ+⋯νξ,ϖ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ+ℑ+ν1ξ+ℑ+ν2ξ+ℑ+⋯φξ,ϖ,ℑ=cosξ+ϖ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ−1+1−σ2+⋯νξ,ϖ,ℑ=cosξ+ϖ−2cosξ+ϖσℑσΓσ+1+σ+1+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+σ+12+⋯We achieve the exact solution by puttingσ=1,(84)φξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯,νξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯.In closed form,φξ,ϖ,ℑ=cosξ+ϖexp−2ℑ and νξ,ϖ,ℑ=cosξ+ϖexp−2ℑ.


#### 5.4.1. VITM Analytical Results

For (73), we have the iteration formula:(85)φm+1ξ,ϖ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+∂2φm∂ϖ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ϖ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+∂2νm∂ϖ2+2νm∂νm∂ξ−∂∂ξφmνm,where(86)φ0ξ,ϖ,ℑ=cosξ+ϖ,ν0ξ,ϖ,ℑ=cosξ+ϖ.Form=0,1,2,⋯,(87)φ1ξ,ϖ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+∂2φ0∂ϖ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ϖ,ℑ=−2cosξ+ϖσℑσΓσ+1+1−σ,ν1ξ,ϖ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+∂2ν0∂ϖ2+2ν0∂ν0∂ξ−∂∂ξφ0ν0,ν1ξ,ϖ,ℑ=−2cosξ,ℑσℑσΓσ+1+1−σ,φ2ξ,ϖ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+∂2φ1∂ϖ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ϖ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+∂2ν1∂ϖ2+2ν1∂ν1∂ξ−∂∂ξφ0ν0,ν2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,νξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2We achieve the exact solution by puttingσ=1,(88)φξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯,νξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯.In closed form,φξ,ϖ,ℑ=cosξ+ϖexp−2ℑ, and νξ,ϖ,ℑ=cosξ+ϖexp−2ℑ.Figures7(a) and 7(b) show the behavior of the exact and analytical solutions, respectively, whereas Figures 7(c) and 7(d) show the graphical perspective for various fractional orders. Table 7 also shows the behaviour of the exact and suggested method solutions in addition with absolute error at various orders of σ. Finally, the figures and tables show that the suggested techniques have higher degree of accuracy and rapid convergence towards the exact results.Figure 7
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 4.
(a)(b)(c)(d)Table 7
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and vξ,ℑ of problem 4.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.8774070629000000.8774070454000001.7500000000E-082.8290530000E-041.0135215000E-030.10.8251705643000000.8251705478000001.6500000000E-082.6606260000E-049.5318140000E-040.20.7646892342000000.7646892189000001.5300000000E-082.4656140000E-048.8331750000E-040.30.6965673819000000.6965673680000001.3900000000E-082.2459660000E-048.0462770000E-040.40.6214856587000000.6214856463000001.2400000000E-082.0038770000E-047.1789830000E-040.50.5401942562000000.5401942454000001.0800000000E-081.7417660000E-046.2399600000E-040.60.4535054112000000.4535054022000009.0000000000E-091.4622520000E-045.2385880000E-040.70.3622852902000000.3622852829000007.3000000000E-091.1681290000E-044.1848760000E-040.80.2674453342000000.2674453288000005.4000000000E-098.6233300000E-053.0893480000E-040.90.1699331529000000.1699331495000003.4000000000E-095.4792200000E-051.9629540000E-041.00.0707230556400000.0707230542300001.4100000000E-092.2803470000E-058.1694500000E-05


## 5.1. Problem 1

Consider system of homogeneous KdV equation having order three(25)∂σφ∂ℑσ=∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ,∂σν∂ℑσ=−2∂3ν∂ξ3+φ∂ν∂ξ0<σ≤1,with initial source(26)φξ,0=3−6tanh2ξ2,νξ,0=−3l2tanh2ξ2.On taking the Laplace transform of (25), we get(27)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ.ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L−2∂3ν∂ξ3+φ∂ν∂ξ.We obtain when we use the Laplace inverse transform(28)φξ,ℑ=3−6tanh2ξ2+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3+φ∂φ∂ξ+ν∂ν∂ξ.νξ,ℑ=−3l2tanh2ξ2+L−1ωσ+σ1−ωσωσL−2∂3ν∂ξ3+φ∂ν∂ξ.Assume that the solution,φξ,ℑ and νξ,ℑ in series form as(29)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,where φφξ=∑m=0∞Am, ννξ=∑m=0∞ℬm, and ννξ=∑m=0∞ℬm are Adomian polynomials that characterize the nonlinear terms, and so equation (28) is rewritten as(30)∑m=0∞φmξ,ℑ=3−6tanh2ξ2+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3+∑m=0∞Am+∑m=0∞ℬm.∑m=0∞νmξ,ℑ=−3l2tanh2ξ2+L−1ωσ+σ1−ωσωσL−2∂3ν∂ξ3+∑m=0∞Cm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(31)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=ν0ν0ξ,ℬ1=ν1ν0ξ+ν0ν1ξ,ℬ2=ν2ν0ξ+ν1ν1ξ+ν0ν2ξ,C0=φ0ν0ξ,C1=φ1ν0ξ+φ0ν1ξ,C2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ,As a result, when comparing the two sides of equation (30)(32)φ0ξ,ℑ=3−6tanh2ξ2,ν0ξ,ℑ=−3l2tanh2ξ2.Form=0,(33)φ1ξ,ℑ=−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ,ν1ξ,ℑ=3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ.Form=1,(34)φ2ξ,ℑ=322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,The approximate solution to the series is written as(35)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯φξ,ℑ=3−6tanh2ξ2−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ+322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=−3l2tanh2ξ2+3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ+3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(36)φξ,ℑ=3−6tanh2ℑ+ξ2,νξ,ℑ=−3l2tanh2ℑ+ξ2.


### 5.1.1. VITM Analytical Results

For Equation (25), we have the iteration formula:(37)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φm∂ξ3+φm∂φm∂ξ+νm∂νm∂ξ,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3νm∂ξ3φm∂νm∂ξ,where(38)φ0ξ,ℑ=3−6tanh2ξ2,ν0ξ,ℑ=−3l2tanh2ξ2.Form=0,1,2,⋯,(39)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ0∂ξ3+φ0∂φ0∂ξ+ν0∂ν0∂ξ,φ1ξ,ℑ=−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν0∂ξ3φ0∂ν0∂ξ,ν1ξ,ℑ=3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ1∂ξ3+φ1∂φ1∂ξ+ν1∂ν1∂ξ,φ2ξ,ℑ=322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν1∂ξ3φ1∂ν1∂ξ,ν1ξ,ℑ=3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=∑m=0∞φmξ,ℑ=3−6tanh2ξ2−6sech2ξ2tanhξ2σℑσΓσ+1=+1−σ+322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=−3l2tanh2ξ2+3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ+3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(40)φξ,ℑ=3−6tanh2ℑ+ξ2,νξ,ℑ=−3l2tanh2ℑ+ξ2.The analytical solution and exact solution is shown in Figures1(a) and 1(b) at σ=1 and −5≤ψ≥5. Figure 1(c) shows the absolute error, and Figure 1(d) gives the solution at various fractional-order graph for φξ,ℑ. The behavior of the exact solution and analytical solution for νξ,ℑ is seen in Figures 2(a) and 2(b). Tables 1 and 2 show the comparison of the exact and our methods solution in addition with the absolute error at different fractional-order. From the figure and tables, it is clear that our methods solution is in good agreement with the exact solution.Figure 1
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ of problem 1.
(a)(b)(c)(d)Figure 2
Exact solution and analytical solution forνξ,ℑ problem 1.
(a)(b)Table 1
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.802.9999999850000003.0000000000000001.5000000000E-081.5000000000E-081.5000000000E-080.12.9849950500000002.9849950650000001.5000000000E-084.8177000000E-051.7264100000E-040.22.9403385240000002.9403385380000001.4000000000E-089.5412000000E-053.4187000000E-040.32.8669121160000002.8669121300000001.4000000000E-081.4077300000E-045.0438500000E-040.42.7661440730000002.7661440860000001.3000000000E-081.8342200000E-046.5718200000E-040.52.6399509450000002.6399509570000001.2000000000E-082.2262900000E-047.9764400000E-040.62.4906618060000002.4906618160000001.0000000000E-082.5779500000E-049.2363200000E-040.72.3209299610000002.3209299700000009.0000000000E-092.8847600000E-041.0335470000E-030.82.1336376490000002.1336376570000008.0000000000E-093.1437800000E-041.1263450000E-030.91.9317992900000001.9317992950000005.0000000000E-093.3536800000E-041.2015390000E-031.01.7184683350000001.7184683380000003.0000000000E-093.5145100000E-041.2591560000E-03Table 2
Comparison of the proposed method and exact results with Absolute Error (AE) at various fractional order forνξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.0000000000106070.0000000000000001.0606601700E-111.0606601700E-111.0606601700E-110.1−0.000010610102200−0.0000105678064204.2295783020E-081.6438202410E-079.6316691610E-070.2−0.000042187034300−0.0000421032931208.3741185290E-083.2548962340E-071.9071994720E-060.3−0.000094107345200−0.0000939838027601.2354244410E-074.8020666600E-072.8137865210E-060.4−0.000165361111500−0.0001652001485001.6096295920E-076.2567014160E-073.6661525080E-060.5−0.000254593128400−0.0002543977659001.9536257270E-077.5938997350E-074.4497048860E-060.6−0.000360156490500−0.0003599302736002.2621689430E-078.7933098140E-075.1525210840E-060.7−0.000480175029200−0.0004799218942002.5313503520E-079.8396949140E-075.7656675160E-060.8−0.000612610692900−0.0006123348322002.7586074000E-071.0723124430E-066.2833302080E-060.9−0.000755331965700−0.0007550376911002.9427450770E-071.1438942760E-066.7027805770E-061.0−0.000906179730200−0.0009058713463003.0838397500E-071.1987439840E-067.0241846510E-06


## 5.1.1. VITM Analytical Results

For Equation (25), we have the iteration formula:(37)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φm∂ξ3+φm∂φm∂ξ+νm∂νm∂ξ,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3νm∂ξ3φm∂νm∂ξ,where(38)φ0ξ,ℑ=3−6tanh2ξ2,ν0ξ,ℑ=−3l2tanh2ξ2.Form=0,1,2,⋯,(39)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ0∂ξ3+φ0∂φ0∂ξ+ν0∂ν0∂ξ,φ1ξ,ℑ=−6sech2ξ2tanhξ2σℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν0∂ξ3φ0∂ν0∂ξ,ν1ξ,ℑ=3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂3φ1∂ξ3+φ1∂φ1∂ξ+ν1∂ν1∂ξ,φ2ξ,ℑ=322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ−2∂3ν1∂ξ3φ1∂ν1∂ξ,ν1ξ,ℑ=3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=∑m=0∞φmξ,ℑ=3−6tanh2ξ2−6sech2ξ2tanhξ2σℑσΓσ+1=+1−σ+322sech2ξ2+7sech4ξ2−15sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=−3l2tanh2ξ2+3l2sech2ξ2tanhξ2σℑσΓσ+1+1−σ+3l242sech2ξ2+21sech4ξ2−24sech6ξ2σ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(40)φξ,ℑ=3−6tanh2ℑ+ξ2,νξ,ℑ=−3l2tanh2ℑ+ξ2.The analytical solution and exact solution is shown in Figures1(a) and 1(b) at σ=1 and −5≤ψ≥5. Figure 1(c) shows the absolute error, and Figure 1(d) gives the solution at various fractional-order graph for φξ,ℑ. The behavior of the exact solution and analytical solution for νξ,ℑ is seen in Figures 2(a) and 2(b). Tables 1 and 2 show the comparison of the exact and our methods solution in addition with the absolute error at different fractional-order. From the figure and tables, it is clear that our methods solution is in good agreement with the exact solution.Figure 1
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ of problem 1.
(a)(b)(c)(d)Figure 2
Exact solution and analytical solution forνξ,ℑ problem 1.
(a)(b)Table 1
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.802.9999999850000003.0000000000000001.5000000000E-081.5000000000E-081.5000000000E-080.12.9849950500000002.9849950650000001.5000000000E-084.8177000000E-051.7264100000E-040.22.9403385240000002.9403385380000001.4000000000E-089.5412000000E-053.4187000000E-040.32.8669121160000002.8669121300000001.4000000000E-081.4077300000E-045.0438500000E-040.42.7661440730000002.7661440860000001.3000000000E-081.8342200000E-046.5718200000E-040.52.6399509450000002.6399509570000001.2000000000E-082.2262900000E-047.9764400000E-040.62.4906618060000002.4906618160000001.0000000000E-082.5779500000E-049.2363200000E-040.72.3209299610000002.3209299700000009.0000000000E-092.8847600000E-041.0335470000E-030.82.1336376490000002.1336376570000008.0000000000E-093.1437800000E-041.1263450000E-030.91.9317992900000001.9317992950000005.0000000000E-093.3536800000E-041.2015390000E-031.01.7184683350000001.7184683380000003.0000000000E-093.5145100000E-041.2591560000E-03Table 2
Comparison of the proposed method and exact results with Absolute Error (AE) at various fractional order forνξ,ℑ of problem 1.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.0000000000106070.0000000000000001.0606601700E-111.0606601700E-111.0606601700E-110.1−0.000010610102200−0.0000105678064204.2295783020E-081.6438202410E-079.6316691610E-070.2−0.000042187034300−0.0000421032931208.3741185290E-083.2548962340E-071.9071994720E-060.3−0.000094107345200−0.0000939838027601.2354244410E-074.8020666600E-072.8137865210E-060.4−0.000165361111500−0.0001652001485001.6096295920E-076.2567014160E-073.6661525080E-060.5−0.000254593128400−0.0002543977659001.9536257270E-077.5938997350E-074.4497048860E-060.6−0.000360156490500−0.0003599302736002.2621689430E-078.7933098140E-075.1525210840E-060.7−0.000480175029200−0.0004799218942002.5313503520E-079.8396949140E-075.7656675160E-060.8−0.000612610692900−0.0006123348322002.7586074000E-071.0723124430E-066.2833302080E-060.9−0.000755331965700−0.0007550376911002.9427450770E-071.1438942760E-066.7027805770E-061.0−0.000906179730200−0.0009058713463003.0838397500E-071.1987439840E-067.0241846510E-06


## 5.2. Problem 2

Consider the generalized coupled Hirota Satsuma KdV system(41)∂σφ∂ℑσ=12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ,∂σν∂ℑσ=3φ∂ν∂ξ−∂3ν∂ξ3,∂σℓ∂ℑσ=3φ∂ℓ∂ξ−∂3ℓ∂ξ30<σ≤1,with initial source(42)φξ,0=−13+2tanh3ξ,νξ,0=tanhξ,ℓξ,0=83tanhξ.On taking the Laplace transform of (41), we get(43)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ,ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L3φ∂ν∂ξ−∂3ν∂ξ3,ωσLℓξ,ℑ−ω−1ℓξ,0ωσ+σ1−ωσ=L3φ∂ℓ∂ξ−∂3ℓ∂ξ3.We obtain when we use the Laplace inverse transform(44)φξ,ℑ=−13+2tanh3ξ+L−1ωσ+σ1−ωσωσL12∂3φ∂ξ3−3φ∂φ∂ξ+3∂∂ξνℓ.νξ,ℑ=tanhξ+L−1ωσ+σ1−ωσωσL3φ∂ν∂ξ−∂3ν∂ξ3,ℓξ,ℑ=83tanhξ+L−1ωσ+σ1−ωσωσL3φ∂ℓ∂ξ−∂3ℓ∂ξ3.Assume that the solution,φξ,ℑ, νξ,ℑ, and ℓξ,ℑ in series form as(45)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,ℓξ,ℑ=∑m=0∞ℓmξ,ℑ,where φφξ=∑m=0∞Am, νℓξ=∑m=0∞ℬm, φνξ=∑m=0∞Cm, and φℓξ=∑m=0∞Dm are Adomian polynomials that characterize the nonlinear terms, and so equation (30) is rewritten as(46)∑m=0∞φmξ,ℑ=−13+2tanh3ξ+L−1ωσ+σ1−ωσωσL∂3φ∂ξ3−3∑m=0∞Am+3∑m=0∞ℬm.∑m=0∞νmξ,ℑ=tanhξ+L−1ωσ+σ1−ωσωσL3∑m=0∞Cm−∂3ν∂ξ3,∑m=0∞ℓmξ,ℑ=83tanhξ+L−1ωσ+σ1−ωσωσL3∑m=0∞Dm−∂3ℓ∂ξ3.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (7),(47)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=ν0ℓ0ξ+ℓ0ν0ξ,ℬ1=ν0ℓ0ξ+ν1ℓ0ξ+ℓ1ν0ξ+ℓ0ν1ξ,ℬ2=ν2ℓ0ξ+ν1ℓ1ξ+ν0ℓ2ξ+ℓ2ν0ξ+ℓ1ν1ξ+ℓ0ν2ξ,C0=φ0ν0ξ,C1=φ1ν0ξ+φ0ν1ξ,C2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ,D0=φ0ℓ0ξ,D1=φ1ℓ0ξ+φ0ℓ1ξ,D2=φ2ℓ0ξ+φ1ℓ1ξ+φ0ℓ2ξ,As a result, when comparing the two sides of (46),(48)φ0ξ,ℑ=−13+2tanh2ξ,ν0ξ,ℑ=tanhξ,ℓ0ξ,ℑ=83tanhξ.Form=0,(49)φ1ξ,ℑ=4sech2ξtanhξσℑσΓσ+1+1−σ,ν1ξ,ℑ=sech2ξσℑσΓσ+1+1−σ,ℓ1ξ,ℑ=83sech2ξσℑσΓσ+1+1−σ.Form=1,(50)φ2ξ,ℑ=4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ℓ2ξ,ℑ=−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(51)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯ℓξ,ℑ=∑m=0∞ℓmξ,ℑ=ℓ0ξ,ℑ+ℓ1ξ,ℑ+ℓ2ξ,ℑ+⋯φξ,ℑ=−13+2tanh2ξ+4sech2ξtanhξσℑσΓσ+1+1−σ+4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=tanhξ+sech2ξσℑσΓσ+1+1−σ−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯ℓξ,ℑ=83tanhξ+83sech2ξσℑσΓσ+1+1−σ−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1,(52)φξ,ℑ=−13+2tanh2ℑ+ξ,νξ,ℑ=tanhℑ+ξ,ℓξ,ℑ=83tanhℑ+ξ.


### 5.2.1. VITM Analytical Results

For equation (41), we have the iteration formula(53)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φm∂ξ3−3φm∂φm∂ξ+3∂∂ξνmℓm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂νm∂ξ−∂3νm∂ξ3,ℓm+1ξ,ℑ=ℓmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ℓm∂ξ−∂3ℓm∂ξ3,where(54)φ0ξ,ℑ=−13+2tanh2ξ,ν0ξ,ℑ=tanhξ,ℓ0ξ,ℑ=83tanhξ.Form=0,1,2,⋯,(55)φ1ξ,ℑ==φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ0∂ξ3−3φ0∂φ0∂ξ+3∂∂ξν0ℓ0,φ1ξ,ℑ=4sech2ξtanhξσℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ν0∂ξ−∂3ν0∂ξ3,ν1ξ,ℑ=sech2ξσℑσΓσ+1+1−σ,ℓ1ξ,ℑ=ℓ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ0∂ν0∂ξ−∂3ν0∂ξ3,ℓ1ξ,ℑ=83sech2ξσℑσΓσ+1+1−σ,φ2ξ,ℑ==φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ1∂ξ3−3φ1∂φ1∂ξ+3∂∂ξν1ℓ1,φ2ξ,ℑ=4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ν2ξ,ℑ=−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ℓ2ξ,ℑ=ℓ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ℓ2ξ,ℑ=−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=−13+2tanh2ξ+4sech2ξtanhξσℑσΓσ+1+1−σ+4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=tanhξ+sech2ξσℑσΓσ+1+1−σ−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯ℓξ,ℑ=83tanhξ+83sech2ξσℑσΓσ+1+1−σ−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(56)φξ,ℑ=−13+2tanh2ℑ+ξ,νξ,ℑ=tanhℑ+ξℓξ,ℑ=83tanhℑ+ξThe analytical solution and exact solution forφξ,ℑ of example 2 at σ=1 and −5≤ψ≥5 are shown in Figures 3(a) and 3(b), whereas Figures 3(c) and 3(d) show the absolute error and the solution at various fractional-order. The graphical behavior of exact solution and analytical solution for νξ,ℑ are shown in Figures 4(a) and 4(b), while Figures 4(c) and 4(d) show the absolute error and the solution at different fractional-order. Figures 5(a) and 5(b) give the solution graph for ℓξ,ℑ. Tables 3–5 show the comparison of the exact and suggested methods solution in addition with the absolute error at various fractional order. From the results of the figures and Tables, it is confirmed that our method solution converges quickly towards exact solution.Figure 3
Exact solution, analytical solution, absolute error, and various fractional order solutions forφξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 4
Exact solution, analytical solution, absolute error, and various fractional order solutions forνξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 5
Exact solution and analytical solution forℓξ,ℑ problem 2.
(a)(b)Table 3
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.333333313300000−0.3333333333000002.0000000000E-082.0000000000E-082.0000000000E-080.1−0.313426424600000−0.3134264438000001.9200000000E-086.3598200000E-056.4898550000E-040.2−0.255343407800000−0.2553434248000001.7000000000E-081.2227320000E-041.2475490000E-030.3−0.163500606900000−0.1635006205000001.3600000000E-081.7185690000E-041.7533551000E-030.4−0.044480856100000−0.0444808656700009.5700000000E-092.0958106000E-042.1381671200E-030.50.0939165789000000.0939165733000005.6000000000E-092.3429753000E-042.3902801500E-030.60.2436640045000000.2436640025000002.0000000000E-092.4637090000E-042.5134153000E-030.70.3973409319000000.3973409332000001.3000000000E-092.4731770000E-042.5230449000E-030.80.5487048209000000.5487048246000003.7000000000E-092.3933690000E-042.4416045000E-030.90.6929714497000000.6929714558000006.1000000000E-092.2486240000E-042.2939183000E-031.00.8268459177000000.8268459237000006.0000000000E-092.0621260000E-042.1036596000E-03Table 4
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forνξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of techniques methodsAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0000999999996700.0001000000000003.3000000000E-131.6117443060E-041.6442500900E-030.10.0997670002700000.0997670012500009.8000000000E-101.5957435000E-041.6279175300E-030.20.1974714226000000.1974714245000001.9000000000E-091.5489750000E-041.5801969000E-030.30.2914041235000000.2914041262000002.7000000000E-091.4749940000E-041.5047167000E-030.40.3800345229000000.3800345262000003.3000000000E-091.3791040000E-041.4068874000E-030.50.4621957984000000.4621958021000003.7000000000E-091.2675890000E-041.2931204000E-030.60.5371207210000000.5371207248000003.8000000000E-091.1469190000E-041.1700156000E-030.70.6044312472000000.6044312511000003.9000000000E-091.0230770000E-041.0436745000E-030.80.6640926721000000.6640926758000003.7000000000E-099.0109100000E-059.1923020000E-040.90.7163465584000000.7163465619000003.5000000000E-097.8482200000E-058.0061750000E-041.00.7616361502000000.7616361534000003.2000000000E-096.7692400000E-056.9054610000E-04Table 5
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forℓξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0002666666658000.0002666666667009.0000000000E-134.2979848170E-041.5398420700E-030.10.2660453341000000.2660453367000002.6000000000E-094.2553160000E-041.5245483000E-030.20.5265904603000000.5265904653000005.0000000000E-094.1305980000E-041.4798594000E-030.30.7770776627000000.7770776698000007.1000000000E-093.9333160000E-041.4091735000E-030.41.0134253940000001.0134254030000009.0000000000E-093.6776200000E-041.3175580000E-030.51.2325221290000001.2325221380000009.0000000000E-093.3802300000E-041.2110150000E-030.61.4323219230000001.4323219330000001.0000000000E-083.0584500000E-041.0957270000E-030.71.6118166590000001.6118166700000001.1000000000E-082.7282100000E-049.7741000000E-040.81.7709137920000001.7709138020000001.0000000000E-082.4029100000E-048.6086700000E-040.91.9102574890000001.9102574990000001.0000000000E-082.0928600000E-047.4978500000E-041.02.0310297340000002.0310297420000008.0000000000E-091.8051200000E-046.4670200000E-04


## 5.2.1. VITM Analytical Results

For equation (41), we have the iteration formula(53)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φm∂ξ3−3φm∂φm∂ξ+3∂∂ξνmℓm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂νm∂ξ−∂3νm∂ξ3,ℓm+1ξ,ℑ=ℓmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ℓm∂ξ−∂3ℓm∂ξ3,where(54)φ0ξ,ℑ=−13+2tanh2ξ,ν0ξ,ℑ=tanhξ,ℓ0ξ,ℑ=83tanhξ.Form=0,1,2,⋯,(55)φ1ξ,ℑ==φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ0∂ξ3−3φ0∂φ0∂ξ+3∂∂ξν0ℓ0,φ1ξ,ℑ=4sech2ξtanhξσℑσΓσ+1+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φm∂ν0∂ξ−∂3ν0∂ξ3,ν1ξ,ℑ=sech2ξσℑσΓσ+1+1−σ,ℓ1ξ,ℑ=ℓ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ0∂ν0∂ξ−∂3ν0∂ξ3,ℓ1ξ,ℑ=83sech2ξσℑσΓσ+1+1−σ,φ2ξ,ℑ==φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ12∂3φ1∂ξ3−3φ1∂φ1∂ξ+3∂∂ξν1ℓ1,φ2ξ,ℑ=4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ν2ξ,ℑ=−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ℓ2ξ,ℑ=ℓ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ3φ1∂ν1∂ξ−∂3ν1∂ξ3ℓ2ξ,ℑ=−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=−13+2tanh2ξ+4sech2ξtanhξσℑσΓσ+1+1−σ+4sech2ξ1−3tanh2ξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=tanhξ+sech2ξσℑσΓσ+1+1−σ−sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯ℓξ,ℑ=83tanhξ+83sech2ξσℑσΓσ+1+1−σ−83sech2ξtanhξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1(56)φξ,ℑ=−13+2tanh2ℑ+ξ,νξ,ℑ=tanhℑ+ξℓξ,ℑ=83tanhℑ+ξThe analytical solution and exact solution forφξ,ℑ of example 2 at σ=1 and −5≤ψ≥5 are shown in Figures 3(a) and 3(b), whereas Figures 3(c) and 3(d) show the absolute error and the solution at various fractional-order. The graphical behavior of exact solution and analytical solution for νξ,ℑ are shown in Figures 4(a) and 4(b), while Figures 4(c) and 4(d) show the absolute error and the solution at different fractional-order. Figures 5(a) and 5(b) give the solution graph for ℓξ,ℑ. Tables 3–5 show the comparison of the exact and suggested methods solution in addition with the absolute error at various fractional order. From the results of the figures and Tables, it is confirmed that our method solution converges quickly towards exact solution.Figure 3
Exact solution, analytical solution, absolute error, and various fractional order solutions forφξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 4
Exact solution, analytical solution, absolute error, and various fractional order solutions forνξ,ℑ of problem 2.
(a)(b)(c)(d)Figure 5
Exact solution and analytical solution forℓξ,ℑ problem 2.
(a)(b)Table 3
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.80−0.333333313300000−0.3333333333000002.0000000000E-082.0000000000E-082.0000000000E-080.1−0.313426424600000−0.3134264438000001.9200000000E-086.3598200000E-056.4898550000E-040.2−0.255343407800000−0.2553434248000001.7000000000E-081.2227320000E-041.2475490000E-030.3−0.163500606900000−0.1635006205000001.3600000000E-081.7185690000E-041.7533551000E-030.4−0.044480856100000−0.0444808656700009.5700000000E-092.0958106000E-042.1381671200E-030.50.0939165789000000.0939165733000005.6000000000E-092.3429753000E-042.3902801500E-030.60.2436640045000000.2436640025000002.0000000000E-092.4637090000E-042.5134153000E-030.70.3973409319000000.3973409332000001.3000000000E-092.4731770000E-042.5230449000E-030.80.5487048209000000.5487048246000003.7000000000E-092.3933690000E-042.4416045000E-030.90.6929714497000000.6929714558000006.1000000000E-092.2486240000E-042.2939183000E-031.00.8268459177000000.8268459237000006.0000000000E-092.0621260000E-042.1036596000E-03Table 4
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forνξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of techniques methodsAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0000999999996700.0001000000000003.3000000000E-131.6117443060E-041.6442500900E-030.10.0997670002700000.0997670012500009.8000000000E-101.5957435000E-041.6279175300E-030.20.1974714226000000.1974714245000001.9000000000E-091.5489750000E-041.5801969000E-030.30.2914041235000000.2914041262000002.7000000000E-091.4749940000E-041.5047167000E-030.40.3800345229000000.3800345262000003.3000000000E-091.3791040000E-041.4068874000E-030.50.4621957984000000.4621958021000003.7000000000E-091.2675890000E-041.2931204000E-030.60.5371207210000000.5371207248000003.8000000000E-091.1469190000E-041.1700156000E-030.70.6044312472000000.6044312511000003.9000000000E-091.0230770000E-041.0436745000E-030.80.6640926721000000.6640926758000003.7000000000E-099.0109100000E-059.1923020000E-040.90.7163465584000000.7163465619000003.5000000000E-097.8482200000E-058.0061750000E-041.00.7616361502000000.7616361534000003.2000000000E-096.7692400000E-056.9054610000E-04Table 5
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forℓξ,ℑ of problem 2.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.0002666666658000.0002666666667009.0000000000E-134.2979848170E-041.5398420700E-030.10.2660453341000000.2660453367000002.6000000000E-094.2553160000E-041.5245483000E-030.20.5265904603000000.5265904653000005.0000000000E-094.1305980000E-041.4798594000E-030.30.7770776627000000.7770776698000007.1000000000E-093.9333160000E-041.4091735000E-030.41.0134253940000001.0134254030000009.0000000000E-093.6776200000E-041.3175580000E-030.51.2325221290000001.2325221380000009.0000000000E-093.3802300000E-041.2110150000E-030.61.4323219230000001.4323219330000001.0000000000E-083.0584500000E-041.0957270000E-030.71.6118166590000001.6118166700000001.1000000000E-082.7282100000E-049.7741000000E-040.81.7709137920000001.7709138020000001.0000000000E-082.4029100000E-048.6086700000E-040.91.9102574890000001.9102574990000001.0000000000E-082.0928600000E-047.4978500000E-041.02.0310297340000002.0310297420000008.0000000000E-091.8051200000E-046.4670200000E-04


## 5.3. Problem 3

Consider the coupled Burgers equation in one dimension(57)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1,with initial source(58)φξ,0=cosξ,νξ,0=cosξ.On taking the Laplace transform of (57), we get(59)ωσLφξ,ℑ−ω−1φξ,0ωσ+σ1−ωσ=L∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν.ωσLνξ,ℑ−ω−1νξ,0ωσ+σ1−ωσ=L∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν.We obtain when we use the Laplace inverse transform(60)φξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν.νξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν.Assume that the solution,φξ,ℑ and νξ,ℑ, in series form as(61)φξ,ℑ=∑m=0∞φmξ,ℑ,νξ,ℑ=∑m=0∞νmξ,ℑ,where φφξ=∑m=0∞Am, φνξ=∑m=0∞ℬm, and ννξ=∑m=0∞Cm are Adomian polynomials that characterize the nonlinear terms, and so (60) is rewritten as(62)∑m=0∞φmξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+2∑m=0∞Am−∑m=0∞ℬm.∑m=0∞νmξ,ℑ=cosξ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∑m=0∞Cm−∑m=0∞ℬm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(63)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1.As a result, when comparing the two sides of (62),(64)φξ,0=cosξ,νξ,0=cosξ.Form=0,(65)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1.Form=1,(66)φ2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(67)φξ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ,ℑ+φ1ξ,ℑ+φ2ξ,ℑ+⋯νξ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ,ℑ+ν1ξ,ℑ+ν2ξ,ℑ+⋯φξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯νξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2+⋯We achieve the exact solution by puttingσ=1,(68)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.In closed form,φξ,ℑ=cosξexp−ℑ and νξ,ℑ=cosξexp−ℑ.


### 5.3.1. VITM Analytical Results

For (57), we have the iteration formula:(69)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+2νm∂ν∂ξ−∂∂ξφmνm,where(70)φξ,0=cosξ,νξ,0=cosξ.Form=0,1,2,⋯,(71)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ℑ=−cosξσℑσΓ1−σ+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+2ν0∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=−cosξσℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+2ν1∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯νξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯We achieve the exact solution by puttingσ=1,(72)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.In closed form,φξ,ℑ=cosξexp−ℑ and νξ,ℑ=cosξexp−ℑ.In Figures6(a) and 6(b), we display the solution graph at σ=1 for φξ,ℑ,ϕξ,ℑ in the domain −5≤ψ,ϕ≥5, and Figures 6(c) and 6(d) show solution graph of absolute error and various fractional order solution. Also Table 6 demonstrates the error comparison of the exact and suggested methods solution at various orders. It is verified from the figures and table that our solution is closely related with the exact solution.Figure 6
Exact solution, analytical solution, absolute error. and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 3.
(a)(b)(c)(d)Table 6
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and νξ,ℑ of problem 3.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.9999000050000000.9999000000000005.0000000000E-099.5960000000E-055.7744580000E-040.10.9949046699000000.9949046649000005.0000000000E-099.5480600000E-055.7456100000E-040.20.9799685760000000.9799685711000004.9000000000E-099.4047100000E-055.6593530000E-040.30.9552409602000000.9552409555000004.7000000000E-099.1674100000E-055.5165500000E-040.40.9209688925000000.9209688879000004.6000000000E-098.8385000000E-055.3186280000E-040.50.8774948080000000.8774948036000004.4000000000E-098.4212800000E-055.0675630000E-040.60.8252530855000000.8252530813000004.2000000000E-097.9199200000E-054.7658660000E-040.70.7647657069000000.7647657031000003.8000000000E-097.3394300000E-054.4165490000E-040.80.6966370421000000.6966370386000003.5000000000E-096.6856000000E-054.0231040000E-040.90.6215478104000000.6215478073000003.1000000000E-095.9649700000E-053.5894610000E-041.00.5402482784000000.5402482757000002.7000000000E-095.1847400000E-053.1199530000E-04


## 5.3.1. VITM Analytical Results

For (57), we have the iteration formula:(69)φm+1ξ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+2νm∂ν∂ξ−∂∂ξφmνm,where(70)φξ,0=cosξ,νξ,0=cosξ.Form=0,1,2,⋯,(71)φ1ξ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ℑ=−cosξσℑσΓ1−σ+1−σ,ν1ξ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+2ν0∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=−cosξσℑσΓσ+1+1−σ,φ2ξ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+2ν1∂ν∂ξ−∂∂ξφ0ν0,ν1ξ,ℑ=cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯νξ,ℑ=cosξ−cosξσℑσΓσ+1+1−σ+cosξσ2ℑ2σΓ2σ+1+2σ1−σℑσΓ1−σ+1−σ2+⋯We achieve the exact solution by puttingσ=1,(72)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.In closed form,φξ,ℑ=cosξexp−ℑ and νξ,ℑ=cosξexp−ℑ.In Figures6(a) and 6(b), we display the solution graph at σ=1 for φξ,ℑ,ϕξ,ℑ in the domain −5≤ψ,ϕ≥5, and Figures 6(c) and 6(d) show solution graph of absolute error and various fractional order solution. Also Table 6 demonstrates the error comparison of the exact and suggested methods solution at various orders. It is verified from the figures and table that our solution is closely related with the exact solution.Figure 6
Exact solution, analytical solution, absolute error. and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 3.
(a)(b)(c)(d)Table 6
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and νξ,ℑ of problem 3.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.9999000050000000.9999000000000005.0000000000E-099.5960000000E-055.7744580000E-040.10.9949046699000000.9949046649000005.0000000000E-099.5480600000E-055.7456100000E-040.20.9799685760000000.9799685711000004.9000000000E-099.4047100000E-055.6593530000E-040.30.9552409602000000.9552409555000004.7000000000E-099.1674100000E-055.5165500000E-040.40.9209688925000000.9209688879000004.6000000000E-098.8385000000E-055.3186280000E-040.50.8774948080000000.8774948036000004.4000000000E-098.4212800000E-055.0675630000E-040.60.8252530855000000.8252530813000004.2000000000E-097.9199200000E-054.7658660000E-040.70.7647657069000000.7647657031000003.8000000000E-097.3394300000E-054.4165490000E-040.80.6966370421000000.6966370386000003.5000000000E-096.6856000000E-054.0231040000E-040.90.6215478104000000.6215478073000003.1000000000E-095.9649700000E-053.5894610000E-041.00.5402482784000000.5402482757000002.7000000000E-095.1847400000E-053.1199530000E-04


## 5.4. Problem 4

Consider the coupled Burgers equation two dimension(73)∂σφ∂ℑσ=∂2φ∂ξ2+2φ∂φ∂ξ−∂∂ξφν,∂σν∂ℑσ=∂2ν∂ξ2+2ν∂ν∂ξ−∂∂ξφν0<σ≤1,with initial source(74)φξ,0=cosξ,νξ,0=cosξ.On taking the Laplace transform of (73), we get(75)ωσLφξ,ϖ,ℑ−ω−1φξ,ϖ,0ωσ+σ1−ωσ=L∂2φ∂ξ2+∂2φ∂ϖ2+2φ∂φ∂ξ−∂∂ξφν.ωσLνξ,ϖ,ℑ−ω−1νξ,ϖ,0ωσ+σ1−ωσ=L∂2ν∂ξ2+∂2ν∂ϖ2+2ν∂ν∂ξ−∂∂ξφν.We obtain when we use the Laplace inverse transform(76)φξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+∂2φ∂ϖ2+2φ∂φ∂ξ−∂∂ξφν.νξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∂2ν∂ϖ2+2ν∂ν∂ξ−∂∂ξφν.Assume that the solution,φξ,ϖ,ℑ and νξ,ϖ,ℑ in series form as(77)φξ,ϖ,ℑ=∑m=0∞φmξ,ϖ,ℑ,νξ,ϖ,ℑ=∑m=0∞νmξ,ϖ,ℑ,where φφξ=∑m=0∞Am, φνξ=∑m=0∞ℬm, and ννξ=∑m=0∞Cm are Adomian polynomials that characterize the nonlinear terms, and so (76) is rewritten as(78)∑m=0∞φmξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2φ∂ξ2+∂2φ∂ϖ2+2∑m=0∞Am−∑m=0∞ℬm.∑m=0∞νmξ,ϖ,ℑ=cosξ+ϖ+L−1ωσ+σ1−ωσωσL∂2ν∂ξ2+∂2φ∂ϖ2+∑m=0∞Cm−∑m=0∞ℬm.The decomposition of nonlinear terms by Adomian polynomials is defined as in equation (14),(79)A0=φ0φ0ξ,A1=φ1φ0ξ+φ0φ1ξ,A2=φ2φ0ξ+φ1φ1ξ+φ0φ2ξ,ℬ0=φ0ν0ξ+ν0φ0ξ,ℬ1=φ0ν0ξ+φ1ν0ξ+ν1φ0ξ+ν0φ1ξ,ℬ2=φ2ν0ξ+φ1ν1ξ+φ0ν2ξ+ν2φ0ξ+ν1φ1ξ+ν0φ2ξ,C0=ν0ν0ξ,C1=ν1ν0ξ+ν0ν1ξ,C2=ν2ν0ξ+ν1ν1ξ+ν0ν2ξ.As a result, when comparing the two sides of (78),(80)φξ,0=cosξ,νξ,0=cosξ.Form=0,(81)φξ,ℑ=cosξ1−ℑ+ℑ22−⋯,νξ,ℑ=cosξ1−ℑ+ℑ22−⋯.Form=1,(82)φ2ξ,ϖ,ℑ=2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ϖ,ℑ=2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2.The approximate solution to the series is written as(83)φξ,ϖ,ℑ=∑m=0∞φmξ,ℑ=φ0ξ+ℑ+φ1ξ+ℑ+φ2ξ+ℑ+⋯νξ,ϖ,ℑ=∑m=0∞νmξ,ℑ=ν0ξ+ℑ+ν1ξ+ℑ+ν2ξ+ℑ+⋯φξ,ϖ,ℑ=cosξ+ϖ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ−1+1−σ2+⋯νξ,ϖ,ℑ=cosξ+ϖ−2cosξ+ϖσℑσΓσ+1+σ+1+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+σ+12+⋯We achieve the exact solution by puttingσ=1,(84)φξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯,νξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯.In closed form,φξ,ϖ,ℑ=cosξ+ϖexp−2ℑ and νξ,ϖ,ℑ=cosξ+ϖexp−2ℑ.


### 5.4.1. VITM Analytical Results

For (73), we have the iteration formula:(85)φm+1ξ,ϖ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+∂2φm∂ϖ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ϖ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+∂2νm∂ϖ2+2νm∂νm∂ξ−∂∂ξφmνm,where(86)φ0ξ,ϖ,ℑ=cosξ+ϖ,ν0ξ,ϖ,ℑ=cosξ+ϖ.Form=0,1,2,⋯,(87)φ1ξ,ϖ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+∂2φ0∂ϖ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ϖ,ℑ=−2cosξ+ϖσℑσΓσ+1+1−σ,ν1ξ,ϖ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+∂2ν0∂ϖ2+2ν0∂ν0∂ξ−∂∂ξφ0ν0,ν1ξ,ϖ,ℑ=−2cosξ,ℑσℑσΓσ+1+1−σ,φ2ξ,ϖ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+∂2φ1∂ϖ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ϖ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+∂2ν1∂ϖ2+2ν1∂ν1∂ξ−∂∂ξφ0ν0,ν2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,νξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2We achieve the exact solution by puttingσ=1,(88)φξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯,νξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯.In closed form,φξ,ϖ,ℑ=cosξ+ϖexp−2ℑ, and νξ,ϖ,ℑ=cosξ+ϖexp−2ℑ.Figures7(a) and 7(b) show the behavior of the exact and analytical solutions, respectively, whereas Figures 7(c) and 7(d) show the graphical perspective for various fractional orders. Table 7 also shows the behaviour of the exact and suggested method solutions in addition with absolute error at various orders of σ. Finally, the figures and tables show that the suggested techniques have higher degree of accuracy and rapid convergence towards the exact results.Figure 7
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 4.
(a)(b)(c)(d)Table 7
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and vξ,ℑ of problem 4.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.8774070629000000.8774070454000001.7500000000E-082.8290530000E-041.0135215000E-030.10.8251705643000000.8251705478000001.6500000000E-082.6606260000E-049.5318140000E-040.20.7646892342000000.7646892189000001.5300000000E-082.4656140000E-048.8331750000E-040.30.6965673819000000.6965673680000001.3900000000E-082.2459660000E-048.0462770000E-040.40.6214856587000000.6214856463000001.2400000000E-082.0038770000E-047.1789830000E-040.50.5401942562000000.5401942454000001.0800000000E-081.7417660000E-046.2399600000E-040.60.4535054112000000.4535054022000009.0000000000E-091.4622520000E-045.2385880000E-040.70.3622852902000000.3622852829000007.3000000000E-091.1681290000E-044.1848760000E-040.80.2674453342000000.2674453288000005.4000000000E-098.6233300000E-053.0893480000E-040.90.1699331529000000.1699331495000003.4000000000E-095.4792200000E-051.9629540000E-041.00.0707230556400000.0707230542300001.4100000000E-092.2803470000E-058.1694500000E-05


## 5.4.1. VITM Analytical Results

For (73), we have the iteration formula:(85)φm+1ξ,ϖ,ℑ=φmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φm∂ξ2+∂2φm∂ϖ2+2φm∂φm∂ξ−∂∂ξφmνm,νm+1ξ,ϖ,ℑ=νmξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2νm∂ξ2+∂2νm∂ϖ2+2νm∂νm∂ξ−∂∂ξφmνm,where(86)φ0ξ,ϖ,ℑ=cosξ+ϖ,ν0ξ,ϖ,ℑ=cosξ+ϖ.Form=0,1,2,⋯,(87)φ1ξ,ϖ,ℑ=φ0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ0∂ξ2+∂2φ0∂ϖ2+2φ0∂φ0∂ξ−∂∂ξφ0ν0,φ1ξ,ϖ,ℑ=−2cosξ+ϖσℑσΓσ+1+1−σ,ν1ξ,ϖ,ℑ=ν0ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν0∂ξ2+∂2ν0∂ϖ2+2ν0∂ν0∂ξ−∂∂ξφ0ν0,ν1ξ,ϖ,ℑ=−2cosξ,ℑσℑσΓσ+1+1−σ,φ2ξ,ϖ,ℑ=φ1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2φ1∂ξ2+∂2φ1∂ϖ2+2φ1∂φ1∂ξ−∂∂ξφ0ν0,φ2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,ν2ξ,ϖ,ℑ=ν1ξ,ℑ−L−1ωσ+σ1−ωσωσLωσωσ+σ1−ωσ∂2ν1∂ξ2+∂2ν1∂ϖ2+2ν1∂ν1∂ξ−∂∂ξφ0ν0,ν2ξ,ϖ,ℑ=2cosξ,ℑσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,φξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2,νξ,ϖ,ℑ=cosξ,ℑ−2cosξ+ϖσℑσΓσ+1+1−σ+2cosξ+ϖσ2ℑ2σΓ2σ+1+2σ1−σℑσΓσ+1+1−σ2We achieve the exact solution by puttingσ=1,(88)φξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯,νξ,ϖ,ℑ=cosξ+ϖ1−2ℑ+4ℑ22!−⋯.In closed form,φξ,ϖ,ℑ=cosξ+ϖexp−2ℑ, and νξ,ϖ,ℑ=cosξ+ϖexp−2ℑ.Figures7(a) and 7(b) show the behavior of the exact and analytical solutions, respectively, whereas Figures 7(c) and 7(d) show the graphical perspective for various fractional orders. Table 7 also shows the behaviour of the exact and suggested method solutions in addition with absolute error at various orders of σ. Finally, the figures and tables show that the suggested techniques have higher degree of accuracy and rapid convergence towards the exact results.Figure 7
Exact solution, analytical solution, absolute error, and various fractional order solution forφξ,ℑ and νξ,ℑ of problem 4.
(a)(b)(c)(d)Table 7
Comparison of the proposed method and exact results with absolute error (AE) at various fractional order forφξ,ℑ and vξ,ℑ of problem 4.
ℑ=0.0001Exact solutionProposed techniques solutionAE of proposed techniquesAE of proposed techniquesAE of proposed techniquesΥσ=1σ=1σ=1σ=0.9σ=0.800.8774070629000000.8774070454000001.7500000000E-082.8290530000E-041.0135215000E-030.10.8251705643000000.8251705478000001.6500000000E-082.6606260000E-049.5318140000E-040.20.7646892342000000.7646892189000001.5300000000E-082.4656140000E-048.8331750000E-040.30.6965673819000000.6965673680000001.3900000000E-082.2459660000E-048.0462770000E-040.40.6214856587000000.6214856463000001.2400000000E-082.0038770000E-047.1789830000E-040.50.5401942562000000.5401942454000001.0800000000E-081.7417660000E-046.2399600000E-040.60.4535054112000000.4535054022000009.0000000000E-091.4622520000E-045.2385880000E-040.70.3622852902000000.3622852829000007.3000000000E-091.1681290000E-044.1848760000E-040.80.2674453342000000.2674453288000005.4000000000E-098.6233300000E-053.0893480000E-040.90.1699331529000000.1699331495000003.4000000000E-095.4792200000E-051.9629540000E-041.00.0707230556400000.0707230542300001.4100000000E-092.2803470000E-058.1694500000E-05


## 6. Conclusion

The LTDM and VITM were used for solving of coupled nonlinear partial differential equations. On comparing the results of these methods with the exact solution, it is observed that proposed methods are extremely simple and easy to handle the nonlinear terms. The obtained results converge quickly in the form of series towards the exact solution. Four nonlinear systems are solved which shows that the suggested techniques solution are in strong agreement with the exact solution. It is confirmed that the proposed methods need much less computational work which shows fast convergence. Furthermore, LTDM and VITM are very effective and efficient for finding out the approximate analytic solutions for a wide range of real world problems arising in engineering and science.



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*Source: 4935809-2022-06-22.xml*