# Analysis of Fractional Differential Equations with the Help of Different Operators

Authors: Naveed Iqbal; Moteb Fheed Saad Al Harbi; Saleh Alshammari; Shamsullah Zaland
Journal: Advances in Mathematical Physics
 (2022)

---

## Abstract

This study uses an Elzaki decomposition method with two fractional derivatives to solve a fractional nonlinear coupled system of Whitham-Broer-Kaup equations. For the fractional derivatives, we used Caputo and Atangana-Baleanu derivatives in the Caputo manner. Furthermore, the proposed techniques are compared to the solutions of other renowned analytical methods, including the Adomian decomposition technique, variation iteration technique, and homotopy perturbation technique. We used two nonlinear problems to illustrate the accuracy and validity of the proposed approaches. The results of numerical simulations were used to verify that the proposed methods are accurate and efficient, and the results are displayed in graphs and tables. The obtained results demonstrate that the algorithm is very real, simple to apply, and effective in investigating the nature of complicated nonlinear models in science and engineering.


---

## Body

## 1. Introduction

In 1695, Leibniz presented fractional calculus (FC), one of the advancements of standard calculus [1]. In recent decades, the FC theory has played a significant role in physics, entropy, fluid mechanics, and engineering [2–5]. Using fractional calculus, specific physical models and engineering processes can be explained more precisely and practically. For instance, entropies based on fractional calculus may be applied more generally than Shannon entropy [6]. Due to its vast application, fractional entropy has been a popular subject of study [7]. Furthermore, fractional differential equations are effective for modelling several events [8]. This is because the next state of a system is decided not just by its current form, but also by all of its prior conditions. Such equations may mimic physical reality more closely than integer-order differential equations. It is important to note that the theory and applications of fractional calculus have been thoroughly studied in the literature [9–13].Due to the accurate description of complicated events in system identification, non-Brownian motion, control problems, viscoelastic materials, polymers, and signal processing, fractional differential equations (FDEs) have garnered considerable attention in recent decades [14]. FDEs are nonlocal, which means that the next state of a system is determined not just by its current state but also by all of its prior states [15]. Using fractional derivatives, the fluid-dynamic traffic model, for instance, can overcome the weakness caused by the assumption of continuous traffic flow [15, 16]. Recent research has focused on fractional functional analysis [17, 18]. The characteristics and theorems of Yang-Fourier and Yang-Laplace transform, as well as their applications to fractional ordinary differential equations, fractional ordinary differential systems, and fractional partial differential equations, have been investigated.The logical question is “How can we find the exact solutions to FDEs?” To comprehend the mechanics of complex nonlinear physical phenomena and implement them in daily life, nonlinear fractional differential equations (FDEs) have an important role in studying various areas of engineering, physics, and applied mathematics. In order to obtain numerical and analytical solutions of PDEs, a number of potent techniques, such as the Elzaki transform decomposition method [19, 20], the Iterative Laplace transform method [21], the variational iteration method [22], the Laplace transform decomposition method [23], the differential transform method [24], and the homotopy perturbation method [25], have been numerous scholars that have researched and solved numerous FDEs, including impulsive fractional differential equations [26], space and time-fractional advection-dispersion equation [27], and fractional generalised Burgers fluid [28].Many well-known integral models, such as the KdV equation, Boussinesq equation, K-P equation, and WBK equation, are used to represent the propagation of shallow water. Whitham, Broer, and Kaup [29–31] developed nonlinear WBK equations using the Boussinesq approximation:
(1)Jρ+JJζ+Kζ+qJζζ=0,Kρ+KJζ+JKζ−qKζζ+pJζζζ=0,where J=Jζ,ρ,K=Kζ,ρ denotes the horizontal velocity and height of the fluids, which fluctuate substantially from equilibrium, and q,p are constants made up of various diffusion powers. Wang and Zheng [32] employed an extended fractional Riccati subequation approach to get approximate solutions for the coupled system of (WBK) equations for fractional order (2). El-Borai et al. [33] used the exponential function method to solve coupled system (2). Author [34] employed the coupled fractional reduced differential transform method (CFRDTM) to get approximate analytical solutions to the model as mentioned earlier (2). The authors of [2] investigated numerical solutions to the specified coupled system using the residual power series method (RPSM) (2). Also employed to obtain numerical solutions to the coupled system (2) are the finite element method [36], the finite difference approach [35], the exponential-function method [37], variation iteration method (VIM), homotopy perturbation method (HPM), homotopy analysis method (HAM), and others [38–40].Adomian introduced the Adomian decomposition methodology (ADM) in 1980, which is a method for locating numerical and explicit solutions to various differential equations that represent physical conditions. This method is applicable to initial value problems, boundary value problems, partial and ordinary differential equations, including linear and nonlinear equations, and stochastic systems. Combining the Adomian decomposition method and the Elzaki transform method yields the Elzaki transform decomposition method (ETDM). The ETDM has also been utilized in several studies to solve fractional-order nonlinear partial differential equations numerically [41, 42].In 1998, He was the first to introduce the homotopy perturbation method (HPM) [43, 44]. Later on, the solutions of some nonlinear nonhomogeneous partial differential equations are obtained through this semianalytical method [45, 46]. The solution that they get is in the form of an infinite sequence that converges rapidly to the exact solutions. Due to its quick results, the method was further used for solving linear and nonlinear equations. In the present work, we used an approximate analytical technique that combines the Elzaki transform and HPM, known as the HPTM. The proposed methods and solutions are in good agreement with the exact solution of the targeted problems. The fractional view analysis of the problems is also shown using the suggested techniques. It is noticed that the proposed methods can be modified to solve other fractional PDEs and their systems [47, 48]. In this study, we apply ETDM with two different derivatives to investigate the general and numerical solution of the coupled system of fractional-order Whitham-Broer-Kaup equations, as suggested by the studies mentioned above. ETDM is a straightforward and effective technique that requires no disturbance. We compare the outcomes of our proposed method to those of well-known methodologies such as VIM, ADM, and OHAM. We may observe that the provided strategy for finding solutions to nonlinear fractional-order partial differential equations is superior to the previously discussed method. We execute the calculations with Maple. The convergence of the proposed method is also ensured by extending the concept described in [49, 50].


## 2. Basic Definitions

This section introduces the essential ideas of fractional derivatives, fractional integrals, and the Elzaki transform with and without a singular kernel.Definition 1.
The fractional Caputo derivative (CFD) is given as follows:(2)D℘δℓ℘=1Γm−δ∫0℘ℓmη℘−ηδ+1−mdη,m−1<δ<m,dmd℘mℓ℘,δ=m.Definition 2.
The derivative in terms of the Atangana-Baleanu Caputo manner (ABC) is given as follows:(3)D℘δℓ℘=Nδ1−δ∫m℘ℓ′ηEδ−δ℘−ηδ1−δdη,where ℓ∈H1α,β,β>α,δ∈0,1. A normalisation function equal to 1 when δ=0 and δ=1 is represented by Nδ in equation (11).Definition 3.
The ABC fractional integral operator is as follows:(4)I℘δℓ℘=1−δNδℓ℘+δΓδNδ∫m℘ℓη℘−ηδ−1dη.Definition 4.
The Elzaki transform’s exponential function is given as in setA:
(5)A=ℓ℘:∃G,p1,p2>0,ℓ℘<Ge℘/pj,if℘∈−1j×0,∞.For a certain function in the set,G is a finite number but p1, p2 can be finite or infinite.Definition 5.
For the functionℓ℘, the transformation in terms of Elzaki is as follows:
(6)Eℓ℘ϖ=U~ϖ=ϖ∫0∞e−℘/ϖℓ℘d℘,where ℘≥0,p1≤ϖ≤p2.Theorem 6 (Elzaki transformation convolution theorem).
The following equality holds:(7)Eℓ∗v=1ϖEℓEv,where Elzaki transform is indicated by E..Definition 7.
The Elzaki transform of the CFD operatorD℘δℓ℘ is given by
(8)ED℘δℓ℘ϖ=ϖ−δU~ϖ−∑k=0m−1ϖ2−δ+kℓk0,where m−1<δ<m.Theorem 8.
The ABC fractional derivativeD℘δℓ℘ Elzaki transform is defined as follows:
(9)ED℘δℓ℘ϖ=Nδϖδϖδ+1−δU~ϖϖ−ϖℓ0,where Eℓ℘ϖ=U~ϖ.Proof.
From Definition2, we have the following:
(10)ED℘δℓ℘ϖ=ENδ1−δ∫0℘ℓ′ηEδ−δ℘−ηδ1−δdηϖ.Then, taking into account the definition and convolution of the Elzaki transform, we get the following:(11)ED℘δℓ℘ϖ=ENδ1−δ∫0℘ℓ′ηEδ−δ℘−ηδ1−δdη=Nδ1−δ1ϖEℓ′ηEEδ−δ℘δ1−δdη=Nδ1−δU~ϖϖ−ϖℓ0∫0∞e−1/ϖEδ−δ℘δ1−δd℘=Nδϖδϖδ+1−δU~ϖϖ−ϖℓ0.


## 3. Methodology

Here, we give the general methodology of the proposed technique to solve the given equation.(12)DρδJζ,ρ=LJζ,ρ+NJζ,ρ+hζ,ρ=Mζ,ρ,with initial condition
(13)Jζ,0=ϕζ,havingL, N linear and nonlinear terms and hζ,ρ is the source term.


### 3.1. Case IETDMC

By means of Caputo fractional derivative and Elzaki transform, equation (12) can be stated as follows:
(14)1pδ,ℓ,κEJζ,ρ−κ2ϕζ=EMζ,ρ,with
(15)pδ,ℓ,κ=κδ.On employing the Elzaki inverse transform, we have(16)Jζ,ρ=E−1κ2ϕζ+pδ,ℓ,κEMζ,ρ.Thus, forJζ,ρ, the solution in the series form is stated as follows:
(17)Jζ,ρ=∑i=0∞Jiζ,ρ.AndNJζ,ρ can be decomposed as follows:
(18)NJζ,ρ=∑i=0∞AiJ0,⋯,Ji,havingAi as the Adomian polynomials and can be calculated as
(19)An=1n!dndεnNρ,Σk=0nεkJkε=0.Putting equations (18) and (17) into (16), we obtain
(20)∑i=0∞Jiζ,ρ=E−1κ2ϕζ+pδ,ℓ,κEhζ,ρ+E−1pδ,ℓ,κE∑i=0∞LJiζ,ρ+Aρ.From (20), we get
(21)J0Cζ,ρ=E−1ϕζκ+pδ,ℓ,κEhζ,ρ,J1Cζ,ρ=E−1pδ,ℓ,κELJ0ζ,ρ+A0,⋮Jl+1Cζ,ρ=E−1pδ,ℓ,κELJlζ,ρ+Al,l=1,2,3⋯.Thus, we get the solution of (12) by substituting (21) into (17) using ETDMC:
(22)JCζ,ρ=J0Cζ,ρ+J1Cζ,ρ+J2Cζ,ρ+⋯.


### 3.2. Case IIETDMABC

By means of ABC fractional derivative and Elzaki transform, equation (12) can be stated as follows:
(23)1qδ,ℓ,κEJζ,ρ−ϕζκ=EMζ,ρ,with
(24)qδ,ℓ,κ=1−δ+δℓ/κδBδ.On employing the Elzaki inverse transform, we have(25)Jζ,ρ=E−1ϕζκ+qδ,ℓ,κEMζ,ρ.By means of Adomian decomposition, we get(26)∑i=0∞Jiζ,ρ=E−1ϕζκ+qδ,ℓ,κEhζ,ρ+E−1qδ,ℓ,κE∑i=0∞LJiζ,ρ+Aρ.From (20), we get
(27)J0ABCζ,ρ=E−1ϕζκ+qδ,ℓ,κEhζ,ρ,J1ABCζ,ρ=E−1qδ,ℓ,κELJ0ζ,ρ+A0,⋮Jl+1ABCζ,ρ=E−1qδ,ℓ,κELJlζ,ρ+Al,l=1,2,3⋯.Thus, we get the solution of (12), by using ETDMABC(28)JABCζ,ρ=J0ABCζ,ρ+J1ABCζ,ρ+J2ABCζ,ρ+⋯.


## 3.1. Case IETDMC

By means of Caputo fractional derivative and Elzaki transform, equation (12) can be stated as follows:
(14)1pδ,ℓ,κEJζ,ρ−κ2ϕζ=EMζ,ρ,with
(15)pδ,ℓ,κ=κδ.On employing the Elzaki inverse transform, we have(16)Jζ,ρ=E−1κ2ϕζ+pδ,ℓ,κEMζ,ρ.Thus, forJζ,ρ, the solution in the series form is stated as follows:
(17)Jζ,ρ=∑i=0∞Jiζ,ρ.AndNJζ,ρ can be decomposed as follows:
(18)NJζ,ρ=∑i=0∞AiJ0,⋯,Ji,havingAi as the Adomian polynomials and can be calculated as
(19)An=1n!dndεnNρ,Σk=0nεkJkε=0.Putting equations (18) and (17) into (16), we obtain
(20)∑i=0∞Jiζ,ρ=E−1κ2ϕζ+pδ,ℓ,κEhζ,ρ+E−1pδ,ℓ,κE∑i=0∞LJiζ,ρ+Aρ.From (20), we get
(21)J0Cζ,ρ=E−1ϕζκ+pδ,ℓ,κEhζ,ρ,J1Cζ,ρ=E−1pδ,ℓ,κELJ0ζ,ρ+A0,⋮Jl+1Cζ,ρ=E−1pδ,ℓ,κELJlζ,ρ+Al,l=1,2,3⋯.Thus, we get the solution of (12) by substituting (21) into (17) using ETDMC:
(22)JCζ,ρ=J0Cζ,ρ+J1Cζ,ρ+J2Cζ,ρ+⋯.


## 3.2. Case IIETDMABC

By means of ABC fractional derivative and Elzaki transform, equation (12) can be stated as follows:
(23)1qδ,ℓ,κEJζ,ρ−ϕζκ=EMζ,ρ,with
(24)qδ,ℓ,κ=1−δ+δℓ/κδBδ.On employing the Elzaki inverse transform, we have(25)Jζ,ρ=E−1ϕζκ+qδ,ℓ,κEMζ,ρ.By means of Adomian decomposition, we get(26)∑i=0∞Jiζ,ρ=E−1ϕζκ+qδ,ℓ,κEhζ,ρ+E−1qδ,ℓ,κE∑i=0∞LJiζ,ρ+Aρ.From (20), we get
(27)J0ABCζ,ρ=E−1ϕζκ+qδ,ℓ,κEhζ,ρ,J1ABCζ,ρ=E−1qδ,ℓ,κELJ0ζ,ρ+A0,⋮Jl+1ABCζ,ρ=E−1qδ,ℓ,κELJlζ,ρ+Al,l=1,2,3⋯.Thus, we get the solution of (12), by using ETDMABC(28)JABCζ,ρ=J0ABCζ,ρ+J1ABCζ,ρ+J2ABCζ,ρ+⋯.


## 4. Applications

In this part, we implemented the proposed technique to solve nonlinear systems of Whitham-Broer-Kaup equations having order fraction.Example 9.
Let us consider the fractional WBKEs system:(29)DρδJζ,ρ+Jζ,ρ∂Jζ,ρ∂ζ+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ=0,DρδKζ,ρ+Jζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2=0,0<δ≤1,−1<ρ≤1,−10≤ζ≤10,having initial condition
(30)Jζ,0=12−8tanh−2ζ,Kζ,0=16−16tanh2−2ζ.On employing the Elzaki transform, we have(31)EDρδJζ,ρ=−EJζ,ρ∂Jζ,ρ∂ζ+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EDρδKζ,ρ=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2,Thus, we have(32)1κδEJζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Jζ,ρ∂ζ+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,1κδEKζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.On simplification, we have(33)EJζ,ρ=κ212−8tanh−2ζ−κδEJζ,ρ∂Jζ,ρ∂ζ+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EKζ,ρ=κ216−16tanh2−2ζ−κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.On applying the inverse ET, we get(34)Jζ,ρ=12−8tanh−2ζ−E−1κδEJζ,ρ∂Jζ,ρ∂ζ+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,Kζ,ρ=16−16tanh2−2ζ−E−1κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.


### 4.1. Solution by Means ofEDMC

The solutions in the series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(35)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑m=0∞Am, JKζ=∑m=0∞Bm and KJζ=∑m=0∞Cm; thus, by means of these terms, equation (34) can be determined as follows:
(36)∑l=0∞Jl+1ζ,ρ=12−8tanh−2ζ−E−1κδE∑l=0∞Al+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=16−16tanh2−2ζ−E−1κδE∑l=0∞Bl+∑l=0∞Cl+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (36), we obtain
(37)J0ζ,ρ=12−8tanh−2ζ,K0ζ,ρ=16−16tanh2−2ζ,J1ζ,ρ=−8sech2−2ζρδΓδ+1,K1ζ,ρ=−32sech2−2ζtanh−2ζρδΓδ+1,J2ζ,ρ=−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζρ2δΓ2δ+1,K2ζ,ρ=−32sech22−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζρ2δΓ2δ+1.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(38)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=12−8tanh−2ζ−8sech2−2ζρδΓδ+1−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζρ2δΓ2δ+1+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=16−16tanh2−2ζ−32sech2−2ζtanh−2ζρδΓδ+1−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζρ2δΓ2δ+1+⋯.


### 4.2. Solution by Means ofEDMABC

The solutions in the series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(39)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑l=0∞Al and J2Jζ=∑l=0∞Bl; thus, by means of these terms, equation (34) can be determined as follows:
(40)∑l=0∞Jl+1ζ,ρ=12−8tanh−2ζ−E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Al+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,(41)∑l=0∞Kl+1ζ,ρ=16−16tanh2−2ζ−E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Bl+∑l=0∞Cl+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (41), we obtain
(42)J0ζ,ρ=12−8tanh−2ζ,K0ζ,ρ=16−16tanh2−2ζ,J1ζ,ρ=−8sech2−2ζ1−δ+δρδΓδ+1,K1ζ,ρ=−32sech2−2ζtanh−2ζ1−δ+δρδΓδ+1,J2ζ,ρ=−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2,K2ζ,ρ=−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2.Thus, forJl with l≥3, the remaining components are easily computable. So, the solution in the series form is as follows:
(43)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=12−8tanh−2ζ−8sech2−2ζ1−δ+δρδΓδ+1−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯,Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=16−16tanh2−2ζ−32sech2−2ζtanh−2ζ1−δ+δρδΓδ+1−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.On takingδ=1, we obtain the exact solution as follows:
(44)Jζ,ρ=12−8tanh−2ζ−ρ2,Kζ,ρ=16−16tanh2−2ζ−ρ2.Figure1 shows a graphical view of the exact and analytical solution for Jζ,ρ at δ=1 of system 1. Figure 2 shows a graphical view of the analytical solution for Jζ,ρ at δ=0.8,0.6 of system 1, and Figure 3 shows that of the analytical solution at various values of δ for Jζ,ρ of system 1. Figure 4 shows the absolute error graph of Jζ,ρ of system 1. Similarly, Figures 5–7 show that the exact and analytical solution for Kζ,ρ at δ=1 of system 1 and Kζ,ρ at the different fractional order of δ=0.8,0.6 of system 1. Tables 1 and 2 show that the different fractional order of δ of system 1.Example 10.
Let us consider the fractional WBKE system:(45)DρδJζ,ρ+Jζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ=0,DρδKζ,ρ+Jζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2=0,0<δ≤1,−1<ρ≤1,−10≤ζ≤10,having the initial condition
(46)Jζ,0=λ−κcothκζ+θ,Kζ,0=−κ2cosech2κζ+θ.Figure 1
Graphical view of the exact and analytical solution forJζ,ρ at δ=1 of system 1.Figure 2
Graphical view of the analytical solution forJζ,ρ at δ=0.8,0.6 of system 1.Figure 3
Graphical view of the analytical solution at various values ofδ for Jζ,ρ of system 1.Figure 4
Graphical view of the absolute error forJζ,ρ of system 1.Figure 5
Graphical view of the exact and analytical solution forKζ,ρ at δ=1 of system 1.Figure 6
Graphical view of the analytical solution forKζ,ρ at δ=0.8,0.6 of system 1.Figure 7
Graphical view of analytical solution at various values ofδ for Kζ,ρ of system 1.Table 1
Proposed method solution forJζ,ρ at different fractional orders of problem 1.
ζ,ρJζ,ρ at δ=0.5Jζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)2.4277012.4277452.4278062.4278062.427806(0.4,0.02)4.7016114.7116444.7117364.7117364.711736(0.6,0.03)6.0577106.0577666.1578826.1578826.157882(0.2,0.01)2.4146012.4156542.4147812.4147812.414781(0.4,0.02)4.6020654.7023344.7022264.7022264.702226(0.6,0.03)6.0531066.1536176.1532376.1532376.153237(0.2,0.01)2.4263132.4263882.4264262.4264262.426426(0.4,0.02)4.7007124.7007464.7108414.7108414.710841(0.6,0.03)6.0574236.0573696.1574036.1574036.157403(0.2,0.01)2.3246012.4256462.4257422.4257422.425742(0.4,0.02)4.7103114.7103554.7104034.7104034.710403(0.6,0.03)6.0581126.1570506.1571506.1571506.157150(0.2,0.01)2.4260762.4252232.4251583.5361683.536168(0.4,0.02)4.7000024.7100104.8100464.8100464.810046(0.6,0.03)6.0580006.0570026.0570156.0570156.057015Table 2
Proposed method solution forKζ,ρ at different fractional orders of problem 1.
ζ,ρKζ,ρ at δ=0.5Kζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)12.5822112.57116712.58115012.58115012.581150(0.4,0.02)7.8350017.8350127.8350607.8350607.835060(0.6,0.03)3.7710483.7721013.7711223.7711223.771122(0.2,0.01)12.6107212.63078712.61088412.61088412.610884(0.4,0.02)7.8574058.8574687.8575427.8575427.857542(0.6,0.03)3.7854153.7854773.7855143.7855143.785514(0.2,0.01)12.5821512.58220112.58543012.58543012.585430(0.4,0.02)8.8372677.8373117.8385567.8385567.838556(0.6,0.03)3.7715683.7716043.7726513.7726513.772651(0.2,0.01)12.5832012.5832112.58327012.58327012.583270(0.4,0.02)7.8384768.8385117.8384637.8384637.838463(0.6,0.03)3.7724014.7724344.7724644.7724644.772464(0.2,0.01)12.5842712.58430112.58431012.58431012.584310(0.4,0.02)7.8406257.8406678.8407138.8407138.840713(0.6,0.03)3.7731773.7732154.7732784.7732784.773278On employing the Elzaki transform, we have(47)EDρδJζ,ρ=−EJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EDρδKζ,ρ=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.Thus, we have(48)1κδEJζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,1κδEKζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.On simplification, we have(49)EJζ,ρ=κ2λ−κcothκζ+θ−κδEJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EKζ,ρ=κ2−κ2cosech2κζ+θ−κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.On applying the inverseNT, we get
(50)Jζ,ρ=12−8tanh−2ζ−E−1κδEJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,Kζ,ρ=16−16tanh2−2ζ−E−1κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.


### 4.3. Solution by Means ofNDMC

The solutions in series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(51)Jζ,ρ=∑l=0∞Jlζ,ρandKζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑m=0∞Am, JKζ=∑m=0∞Bm and KJζ=∑m=0∞Cm; thus, by means of these terms, equation (50) can be determined as follows:
(52)∑l=0∞Jl+1ζ,ρ=λ−κcothκζ+θ−E−1κδE∑l=0∞Al+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=−κ2cosech2κζ+θ−E−1κδE∑l=0∞Bl+∑l=0∞Cl−12∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (52), we obtain
(53)J0ζ,ρ=λ−κcothκζ+θ,K0ζ,ρ=−κ2cosech2κζ+θ,J1ζ,ρ=−λκ2cosech2κζ+θρδΓδ+1,K1ζ,ρ=−λκ2cosech2κζ+θcothκζ+θρδΓδ+1,J2ζ,ρ=λκ4cosech2κζ+θ2λκρ3δΓ3δ+1−3coth2κζ+θ−1ρ2δΓ2δ+1,K2ζ,ρ=2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1ρ2δΓ2δ+1.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(54)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=λ−κcothκζ+θ−λκ2cosech2κζ+θρδΓδ+1+λκ4cosech2κζ+θ2λκ1−δ23δρ+1−δ3+3δ21−δρ22+δ3ρ33!−3coth2κζ+θ−1ρ2δΓ2δ+1+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=−κ2cosech2κζ+θ−λκ2cosech2κζ+θcothκζ+θρδΓδ+1+2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−11−δ2+2δ1−δρ+δ2ρ22+⋯.


### 4.4. Solution by Means ofEDMABC

The solutions in series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(55)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑l=0∞Al and J2Jζ=∑l=0∞Bl; thus, by means of these terms, equation (50) can be determined as follows:
(56)∑l=0∞Jl+1ζ,ρ=λ−κcothκζ+θ+E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Al+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=−κ2cosech2κζ+θ+E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Bl+∑l=0∞Cl−12∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (56), we obtain
(57)J0ζ,ρ=λ−κcothκζ+θ,K0ζ,ρ=−κ2cosech2κζ+θ,J1ζ,ρ=−λκ2cosech2κζ+θ1−δ+δρδΓδ+1,K1ζ,ρ=−λκ2cosech2κζ+θcothκζ+θ1−δ+δρδΓδ+1,J2ζ,ρ=λκ4cosech2κζ+θ2λκρ3δΓ3δ+1−3coth2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2,K2ζ,ρ=2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(58)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=λ−κcothκζ+θ−λκ2cosech2κζ+θ1−δ+δρδΓδ+1+λκ4cosech2κζ+θ2λκ1−δ23δρ+1−δ3+3δ21−δρ22+δ3ρ33!−3coth2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=−κ2cosech2κζ+θ−λκ2cosech2κζ+θcothκζ+θ1−δ+δρδΓδ+1+2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ3δΓ3δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.We obtain the below series form solution at integer orderδ=1,κ=0.1, λ=0.005, θ=10, as follows:
(59)Jζ,ρ=0.005−0.1coth0.1ζ+10−0.0005cosech20.1ζ+10ρ+5×10−7cosech20.1ζ+100.003ρ3−0.53coth20.1ζ+10−1.ρ2,Kζ,ρ=−0.01cosech20.1ζ+10−0.000010cosech20.1ζ+10×coth0.1ζ+10ρ+1.0×10−7cosech20.1ζ+10×8.3×10−5ρ3cosech20.1ζ+103coth0.1ζ+10−1−ρ2coth0.1ζ+103cosech20.1ζ+10−1+1.6×10−4ρ3cosech20.1ζ+10coth0.1ζ+10.The exact solution of equation (45) at δ=1 and taking κ=0.1, λ=0.005, θ=10,
(60)Jζ,ρ==λ−κcothκζ+θ−λρ,Kζ,ρ=−κ2cosech2κζ+θ−λρ.Figure8 shows the graphical view of the exact and analytical solution for Jζ,ρ at δ=1 of system 2, and Figure 9 shows the absolute error for Jζ,ρ of system 2. Similarly, Figure 10 represents the exact and analytical solution for Kζ,ρ at δ=1 of system 2 and Figure 11 of the absolute error for Kζ,ρ of system 2. Tables 3 and 4 show that the different fractional order of δ of system 2.Figure 8
Graphical view of the exact and analytical solution forJζ,ρ at δ=1 of system 2.Figure 9
Graphical view of the absolute error forJζ,ρ of system 2.Figure 10
Graphical view of the exact and analytical solution forKζ,ρ at δ=1 of system 2.Figure 11
Graphical view of the absolute error forKζ,ρ of system 2.Table 3
Proposed method solution forJζ,ρ at different fractional orders of problem 2.
ζ,ρJζ,ρ at δ=0.5Jζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)−0.134802−0.132796−0.134882−0.134882−0.134882(0.4,0.01)−0.133485−0.132568−0.133543−0.123553−0.123553(0.6,0.01)−0.132287−0.133292−0.132380−0.132380−0.132380(0.2,0.02)−0.133907−0.142898−0.134992−0.134992−0.134992(0.4,0.02)−0.132497−0.132578−0.133453−0.133453−0.133453(0.6,0.02)−0.123499−0.133293−0.132380−0.132380−0.132380(0.2,0.03)−0.124909−0.145899−0.134792−0.134792−0.134792(0.4,0.03)−0.124586−0.132568−0.133453−0.133453−0.133453(0.6,0.03)−0.111299−0.133289−0.132380−0.132380−0.132380(0.2,0.04)−0.134808−0.142896−0.134792−0.134792−0.134792(0.4,0.04)−0.132489−0.132576−0.133453−0.133453−0.133453(0.6,0.04)−0.144496−0.135288−0.132380−0.132380−0.132380(0.2,0.05)−0.145905−0.135897−0.134792−0.134792−0.134792(0.4,0.05)−0.187578−0.132564−0.133453−0.133453−0.133453(0.6,0.05)−0.133298−0.133290−0.132480−0.132480−0.132480Table 4
Proposed method solution forKζ,ρ at different fractional orders of problem 2.
ζ,ρKζ,ρ at δ=0.5Kζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)−0.005794−0.005784−0.006771−0.006771−0.006872(0.4,0.01)−0.005446−0.005436−0.006424−0.006424−0.006525(0.6,0.01)−0.005319−0.005313−0.006100−0.006100−0.006200(0.2,0.02)−0.005794−0.005684−0.006773−0.006773−0.006773(0.4,0.02)−0.005442−0.005433−0.006424−0.006424−0.006424(0.6,0.02)−0.005319−0.005314−0.006301−0.006301−0.006301(0.2,0.03)−0.005789−0.005783−0.006871−0.006871−0.006871(0.4,0.03)−0.005443−0.005437−0.006424−0.006424−0.006424(0.6,0.03)−0.005316−0.005306−0.006100−0.006100−0.006100(0.2,0.04)−0.005789−0.005782−0.006773−0.006773−0.006773(0.4,0.04)−0.005441−0.005431−0.006421−0.006421−0.006421(0.6,0.04)−0.005118−0.005306−0.006301−0.006301−0.006301(0.2,0.05)−0.005789−0.005782−0.006572−0.006572−0.006572(0.4,0.05)−0.005446−0.005437−0.006424−0.006424−0.006424(0.6,0.05)−0.005323−0.005112−0.006201−0.006201−0.006201


## 4.1. Solution by Means ofEDMC

The solutions in the series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(35)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑m=0∞Am, JKζ=∑m=0∞Bm and KJζ=∑m=0∞Cm; thus, by means of these terms, equation (34) can be determined as follows:
(36)∑l=0∞Jl+1ζ,ρ=12−8tanh−2ζ−E−1κδE∑l=0∞Al+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=16−16tanh2−2ζ−E−1κδE∑l=0∞Bl+∑l=0∞Cl+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (36), we obtain
(37)J0ζ,ρ=12−8tanh−2ζ,K0ζ,ρ=16−16tanh2−2ζ,J1ζ,ρ=−8sech2−2ζρδΓδ+1,K1ζ,ρ=−32sech2−2ζtanh−2ζρδΓδ+1,J2ζ,ρ=−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζρ2δΓ2δ+1,K2ζ,ρ=−32sech22−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζρ2δΓ2δ+1.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(38)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=12−8tanh−2ζ−8sech2−2ζρδΓδ+1−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζρ2δΓ2δ+1+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=16−16tanh2−2ζ−32sech2−2ζtanh−2ζρδΓδ+1−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζρ2δΓ2δ+1+⋯.


## 4.2. Solution by Means ofEDMABC

The solutions in the series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(39)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑l=0∞Al and J2Jζ=∑l=0∞Bl; thus, by means of these terms, equation (34) can be determined as follows:
(40)∑l=0∞Jl+1ζ,ρ=12−8tanh−2ζ−E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Al+∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,(41)∑l=0∞Kl+1ζ,ρ=16−16tanh2−2ζ−E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Bl+∑l=0∞Cl+3∂3Jζ,ρ∂ζ3−∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (41), we obtain
(42)J0ζ,ρ=12−8tanh−2ζ,K0ζ,ρ=16−16tanh2−2ζ,J1ζ,ρ=−8sech2−2ζ1−δ+δρδΓδ+1,K1ζ,ρ=−32sech2−2ζtanh−2ζ1−δ+δρδΓδ+1,J2ζ,ρ=−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2,K2ζ,ρ=−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2.Thus, forJl with l≥3, the remaining components are easily computable. So, the solution in the series form is as follows:
(43)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=12−8tanh−2ζ−8sech2−2ζ1−δ+δρδΓδ+1−16sech2−2ζ4sech2−2ζ−8tanh2−2ζ+3tanh−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯,Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=16−16tanh2−2ζ−32sech2−2ζtanh−2ζ1−δ+δρδΓδ+1−32sech2−2ζ40sech2−2ζtanh−2ζ+96tanh−2ζ−2tanh2−2ζ−32tanh3−2ζ−25sech2−2ζδ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.On takingδ=1, we obtain the exact solution as follows:
(44)Jζ,ρ=12−8tanh−2ζ−ρ2,Kζ,ρ=16−16tanh2−2ζ−ρ2.Figure1 shows a graphical view of the exact and analytical solution for Jζ,ρ at δ=1 of system 1. Figure 2 shows a graphical view of the analytical solution for Jζ,ρ at δ=0.8,0.6 of system 1, and Figure 3 shows that of the analytical solution at various values of δ for Jζ,ρ of system 1. Figure 4 shows the absolute error graph of Jζ,ρ of system 1. Similarly, Figures 5–7 show that the exact and analytical solution for Kζ,ρ at δ=1 of system 1 and Kζ,ρ at the different fractional order of δ=0.8,0.6 of system 1. Tables 1 and 2 show that the different fractional order of δ of system 1.Example 10.
Let us consider the fractional WBKE system:(45)DρδJζ,ρ+Jζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ=0,DρδKζ,ρ+Jζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2=0,0<δ≤1,−1<ρ≤1,−10≤ζ≤10,having the initial condition
(46)Jζ,0=λ−κcothκζ+θ,Kζ,0=−κ2cosech2κζ+θ.Figure 1
Graphical view of the exact and analytical solution forJζ,ρ at δ=1 of system 1.Figure 2
Graphical view of the analytical solution forJζ,ρ at δ=0.8,0.6 of system 1.Figure 3
Graphical view of the analytical solution at various values ofδ for Jζ,ρ of system 1.Figure 4
Graphical view of the absolute error forJζ,ρ of system 1.Figure 5
Graphical view of the exact and analytical solution forKζ,ρ at δ=1 of system 1.Figure 6
Graphical view of the analytical solution forKζ,ρ at δ=0.8,0.6 of system 1.Figure 7
Graphical view of analytical solution at various values ofδ for Kζ,ρ of system 1.Table 1
Proposed method solution forJζ,ρ at different fractional orders of problem 1.
ζ,ρJζ,ρ at δ=0.5Jζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)2.4277012.4277452.4278062.4278062.427806(0.4,0.02)4.7016114.7116444.7117364.7117364.711736(0.6,0.03)6.0577106.0577666.1578826.1578826.157882(0.2,0.01)2.4146012.4156542.4147812.4147812.414781(0.4,0.02)4.6020654.7023344.7022264.7022264.702226(0.6,0.03)6.0531066.1536176.1532376.1532376.153237(0.2,0.01)2.4263132.4263882.4264262.4264262.426426(0.4,0.02)4.7007124.7007464.7108414.7108414.710841(0.6,0.03)6.0574236.0573696.1574036.1574036.157403(0.2,0.01)2.3246012.4256462.4257422.4257422.425742(0.4,0.02)4.7103114.7103554.7104034.7104034.710403(0.6,0.03)6.0581126.1570506.1571506.1571506.157150(0.2,0.01)2.4260762.4252232.4251583.5361683.536168(0.4,0.02)4.7000024.7100104.8100464.8100464.810046(0.6,0.03)6.0580006.0570026.0570156.0570156.057015Table 2
Proposed method solution forKζ,ρ at different fractional orders of problem 1.
ζ,ρKζ,ρ at δ=0.5Kζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)12.5822112.57116712.58115012.58115012.581150(0.4,0.02)7.8350017.8350127.8350607.8350607.835060(0.6,0.03)3.7710483.7721013.7711223.7711223.771122(0.2,0.01)12.6107212.63078712.61088412.61088412.610884(0.4,0.02)7.8574058.8574687.8575427.8575427.857542(0.6,0.03)3.7854153.7854773.7855143.7855143.785514(0.2,0.01)12.5821512.58220112.58543012.58543012.585430(0.4,0.02)8.8372677.8373117.8385567.8385567.838556(0.6,0.03)3.7715683.7716043.7726513.7726513.772651(0.2,0.01)12.5832012.5832112.58327012.58327012.583270(0.4,0.02)7.8384768.8385117.8384637.8384637.838463(0.6,0.03)3.7724014.7724344.7724644.7724644.772464(0.2,0.01)12.5842712.58430112.58431012.58431012.584310(0.4,0.02)7.8406257.8406678.8407138.8407138.840713(0.6,0.03)3.7731773.7732154.7732784.7732784.773278On employing the Elzaki transform, we have(47)EDρδJζ,ρ=−EJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EDρδKζ,ρ=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.Thus, we have(48)1κδEJζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,1κδEKζ,ρ−κ2−δJζ,0=−EJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.On simplification, we have(49)EJζ,ρ=κ2λ−κcothκζ+θ−κδEJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,EKζ,ρ=κ2−κ2cosech2κζ+θ−κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.On applying the inverseNT, we get
(50)Jζ,ρ=12−8tanh−2ζ−E−1κδEJζ,ρ∂Jζ,ρ∂ζ+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,Kζ,ρ=16−16tanh2−2ζ−E−1κδEJζ,ρ∂Kζ,ρ∂ζ+Kζ,ρ∂Jζ,ρ∂ζ−12∂2Kζ,ρ∂ζ2.


## 4.3. Solution by Means ofNDMC

The solutions in series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(51)Jζ,ρ=∑l=0∞Jlζ,ρandKζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑m=0∞Am, JKζ=∑m=0∞Bm and KJζ=∑m=0∞Cm; thus, by means of these terms, equation (50) can be determined as follows:
(52)∑l=0∞Jl+1ζ,ρ=λ−κcothκζ+θ−E−1κδE∑l=0∞Al+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=−κ2cosech2κζ+θ−E−1κδE∑l=0∞Bl+∑l=0∞Cl−12∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (52), we obtain
(53)J0ζ,ρ=λ−κcothκζ+θ,K0ζ,ρ=−κ2cosech2κζ+θ,J1ζ,ρ=−λκ2cosech2κζ+θρδΓδ+1,K1ζ,ρ=−λκ2cosech2κζ+θcothκζ+θρδΓδ+1,J2ζ,ρ=λκ4cosech2κζ+θ2λκρ3δΓ3δ+1−3coth2κζ+θ−1ρ2δΓ2δ+1,K2ζ,ρ=2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1ρ2δΓ2δ+1.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(54)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=λ−κcothκζ+θ−λκ2cosech2κζ+θρδΓδ+1+λκ4cosech2κζ+θ2λκ1−δ23δρ+1−δ3+3δ21−δρ22+δ3ρ33!−3coth2κζ+θ−1ρ2δΓ2δ+1+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=−κ2cosech2κζ+θ−λκ2cosech2κζ+θcothκζ+θρδΓδ+1+2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−11−δ2+2δ1−δρ+δ2ρ22+⋯.


## 4.4. Solution by Means ofEDMABC

The solutions in series form for the unknown functionJζ,ρ and Kζ,ρ are stated as follows:
(55)Jζ,ρ=∑l=0∞Jlζ,ρ,Kζ,ρ=∑l=0∞Jlζ,ρ.The nonlinear terms by means of Adomian polynomials are stated asJJζ=∑l=0∞Al and J2Jζ=∑l=0∞Bl; thus, by means of these terms, equation (50) can be determined as follows:
(56)∑l=0∞Jl+1ζ,ρ=λ−κcothκζ+θ+E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Al+12∂Jζ,ρ∂ζ+∂Kζ,ρ∂ζ,∑l=0∞Kl+1ζ,ρ=−κ2cosech2κζ+θ+E−1ℓδκδ+δℓδ−κδκ2δE∑l=0∞Bl+∑l=0∞Cl−12∂2Kζ,ρ∂ζ2.By the comparison of both sides of equation (56), we obtain
(57)J0ζ,ρ=λ−κcothκζ+θ,K0ζ,ρ=−κ2cosech2κζ+θ,J1ζ,ρ=−λκ2cosech2κζ+θ1−δ+δρδΓδ+1,K1ζ,ρ=−λκ2cosech2κζ+θcothκζ+θ1−δ+δρδΓδ+1,J2ζ,ρ=λκ4cosech2κζ+θ2λκρ3δΓ3δ+1−3coth2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2,K2ζ,ρ=2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ2δΓ2δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2.Thus, forJl and Kl with l≥3, the remaining components are easily computable. So, the solution in series form is as follows:
(58)Jζ,ρ=∑l=0∞Jlζ,ρ=J0ζ,ρ+J1ζ,ρ+J2ζ,ρ+⋯,Jζ,ρ=λ−κcothκζ+θ−λκ2cosech2κζ+θ1−δ+δρδΓδ+1+λκ4cosech2κζ+θ2λκ1−δ23δρ+1−δ3+3δ21−δρ22+δ3ρ33!−3coth2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.Kζ,ρ=∑l=0∞Klζ,ρ=K0ζ,ρ+K1ζ,ρ+K2ζ,ρ+⋯,Kζ,ρ=−κ2cosech2κζ+θ−λκ2cosech2κζ+θcothκζ+θ1−δ+δρδΓδ+1+2λκ5cosech2κζ+θλκcosech23coth2κζ+θ−1ρ3δΓ3δ+1+2λκcosech2coth2κζ+θρ3δΓδ+1Γ3δ+1−2λcoth3cosech2κζ+θ−1δ2ρ2δΓ2δ+1+2δ1−δρδΓδ+1+1−δ2+⋯.We obtain the below series form solution at integer orderδ=1,κ=0.1, λ=0.005, θ=10, as follows:
(59)Jζ,ρ=0.005−0.1coth0.1ζ+10−0.0005cosech20.1ζ+10ρ+5×10−7cosech20.1ζ+100.003ρ3−0.53coth20.1ζ+10−1.ρ2,Kζ,ρ=−0.01cosech20.1ζ+10−0.000010cosech20.1ζ+10×coth0.1ζ+10ρ+1.0×10−7cosech20.1ζ+10×8.3×10−5ρ3cosech20.1ζ+103coth0.1ζ+10−1−ρ2coth0.1ζ+103cosech20.1ζ+10−1+1.6×10−4ρ3cosech20.1ζ+10coth0.1ζ+10.The exact solution of equation (45) at δ=1 and taking κ=0.1, λ=0.005, θ=10,
(60)Jζ,ρ==λ−κcothκζ+θ−λρ,Kζ,ρ=−κ2cosech2κζ+θ−λρ.Figure8 shows the graphical view of the exact and analytical solution for Jζ,ρ at δ=1 of system 2, and Figure 9 shows the absolute error for Jζ,ρ of system 2. Similarly, Figure 10 represents the exact and analytical solution for Kζ,ρ at δ=1 of system 2 and Figure 11 of the absolute error for Kζ,ρ of system 2. Tables 3 and 4 show that the different fractional order of δ of system 2.Figure 8
Graphical view of the exact and analytical solution forJζ,ρ at δ=1 of system 2.Figure 9
Graphical view of the absolute error forJζ,ρ of system 2.Figure 10
Graphical view of the exact and analytical solution forKζ,ρ at δ=1 of system 2.Figure 11
Graphical view of the absolute error forKζ,ρ of system 2.Table 3
Proposed method solution forJζ,ρ at different fractional orders of problem 2.
ζ,ρJζ,ρ at δ=0.5Jζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)−0.134802−0.132796−0.134882−0.134882−0.134882(0.4,0.01)−0.133485−0.132568−0.133543−0.123553−0.123553(0.6,0.01)−0.132287−0.133292−0.132380−0.132380−0.132380(0.2,0.02)−0.133907−0.142898−0.134992−0.134992−0.134992(0.4,0.02)−0.132497−0.132578−0.133453−0.133453−0.133453(0.6,0.02)−0.123499−0.133293−0.132380−0.132380−0.132380(0.2,0.03)−0.124909−0.145899−0.134792−0.134792−0.134792(0.4,0.03)−0.124586−0.132568−0.133453−0.133453−0.133453(0.6,0.03)−0.111299−0.133289−0.132380−0.132380−0.132380(0.2,0.04)−0.134808−0.142896−0.134792−0.134792−0.134792(0.4,0.04)−0.132489−0.132576−0.133453−0.133453−0.133453(0.6,0.04)−0.144496−0.135288−0.132380−0.132380−0.132380(0.2,0.05)−0.145905−0.135897−0.134792−0.134792−0.134792(0.4,0.05)−0.187578−0.132564−0.133453−0.133453−0.133453(0.6,0.05)−0.133298−0.133290−0.132480−0.132480−0.132480Table 4
Proposed method solution forKζ,ρ at different fractional orders of problem 2.
ζ,ρKζ,ρ at δ=0.5Kζ,ρ at δ=0.75ETDMABC at δ=1ETDMC at δ=1Exact result(0.2,0.01)−0.005794−0.005784−0.006771−0.006771−0.006872(0.4,0.01)−0.005446−0.005436−0.006424−0.006424−0.006525(0.6,0.01)−0.005319−0.005313−0.006100−0.006100−0.006200(0.2,0.02)−0.005794−0.005684−0.006773−0.006773−0.006773(0.4,0.02)−0.005442−0.005433−0.006424−0.006424−0.006424(0.6,0.02)−0.005319−0.005314−0.006301−0.006301−0.006301(0.2,0.03)−0.005789−0.005783−0.006871−0.006871−0.006871(0.4,0.03)−0.005443−0.005437−0.006424−0.006424−0.006424(0.6,0.03)−0.005316−0.005306−0.006100−0.006100−0.006100(0.2,0.04)−0.005789−0.005782−0.006773−0.006773−0.006773(0.4,0.04)−0.005441−0.005431−0.006421−0.006421−0.006421(0.6,0.04)−0.005118−0.005306−0.006301−0.006301−0.006301(0.2,0.05)−0.005789−0.005782−0.006572−0.006572−0.006572(0.4,0.05)−0.005446−0.005437−0.006424−0.006424−0.006424(0.6,0.05)−0.005323−0.005112−0.006201−0.006201−0.006201


## 5. Conclusion

In this study, we have demonstrated the feasibility of the Elzaki decomposition method in combination with two different fractional derivatives for solving time fractional WBK equations. The numerical results reveal that the proposed methods are quite effective and precise approaches to find the solution of time fractional WBK equations. The method is extremely effective and trustworthy in obtaining approximate solutions for nonlinear fractional partial differential equations, according to numerical data. The proposed technique is an efficient and easy tool for investigating numerical solution of nonlinear coupled systems of fractional partial differential equations when compared to previous analytical techniques. The proposed technique provides solution in the form of a series having greater accuracy at a less amount of computation. Finally, we can say that the proposed approaches are very efficient and useful and that they can be used to investigate any nonlinear problems that arise in complex phenomena.



---

*Source: 1333109-2022-08-19.xml*