# A Novel Analysis of Generalized Perturbed Zakharov–Kuznetsov Equation of Fractional-Order Arising in Dusty Plasma by Natural Transform Decomposition Method

Authors: Sharifah E. Alhazmi; Shaimaa A. M. Abdelmohsen; Maryam Ahmed Alyami; Aatif Ali; Joshua Kiddy K. Asamoah
Journal: Journal of Nanomaterials
 (2022)

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## Abstract

The natural transform decomposition method (NTDM) is a relatively new transformation method for finding an approximate differential equation solution. In the current study, the NTDM has been used for obtaining an approximate solution of the fractional-order generalized perturbed Zakharov–Kuznetsov (GPZK) equation. The method has been tested for three nonlinear cases of the fractional-order GPZK equation. The absolute errors are analyzed by the proposed method and the q-homotopy analysis transform method (q-HATM). 3D and 2D graphs have shown the proposed method’s accuracy and effectiveness. NTDM gives a much-closed solution after a few terms.


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## Body

## 1. Introduction

The power of applying fractional calculus to physical problems is that, when dealing with the integer order of derivative, which depends on the function’s behavior, the fractional derivative produces the whole story of this function. For this reason, studying the behavior of the function fractionally is sometimes called the memory effect. This effect leads to many applications of differential equations. The importance of fractional differential equations (FDEs) cannot be denied in the recent advancement of the real world. There are models which provide a better way of managing the systems through fractional differential equations. These equations may arise in electronic circuits, physics, engineering [1, 2], bioscience, etc. [3, 4]. Finance also deals with the fractional-order differential equations to handle clients in more suitable and affordable packages for dealing with financial crises [5, 6]. Other differential equation (DE) applications are image processing and signal processing [7, 8]. These equations may be linear or nonlinear, depending upon the geometry of the problem. Simple problems can be demonstrated by ordinary differential equations (ODEs), while complex problems can be demonstrated through partial differential equations (PDEs). Most linear problems have the exact solution, but it is hard to find the exact solution to complex nonlinear problems. To handle these problems, the researchers used numerical, analytical, and some homotopy-based methods to approximate such a nonlinear problem. Some famous methods which handle the DEs of fractional and integer order are the residual power series method [9, 10], Laplace decomposition method (LDM) [11], q-homotopy analysis method (q-HAM) [12], Adomian decomposition method (ADM) [13], reduced differential transform method (RDTM) [14], variational iteration method (VIM) [15], optimal homotopy asymptotic method (OHAM) [16, 17], homotopy perturbation method (HPM) [18], homotopy analysis method (HAM) [19], etc. Besides these methods, many researchers have introduced many numerical methods to handle differential equations [20]. Transformations also help for the solution approximation of DEs. In the present study, we apply a relatively new method named the natural transform decomposition method (NTDM). We decompose the nonlinear terms with the help of Adomian polynomials and then apply the natural transformation to obtain the solution to the problem. Many researchers have applied NTDM to handle DEs of fractional order [21, 22]. The time-fractional GPZK equation (3+1) dimension with the following form is taken to analyze NTDM [23].
(1)Dtαq+β1qλqχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,0<α≤1,t>0,where α is the fractional-order of the Caputo’s derivative, q is the electrostatic potential, ζ represents the smallness parameter, λ is a positive number, and β1, β2, and β3 are constants. Equation (1) describes the nonlinear dust-ion-acoustic waves in the magnetized plasmas [23]. The study of ion-acoustic waves and structures in dense quantum plasmas has sparked much interest in recent years.The remaining paper is organized as follows: The preliminary definitions are given in Section2 contains. Section 3 introduces the core concept of NTDM. In Section 4, NTDM is applied to three fractional-order ZK equations. There is a conclusion to the method applications, and in the end, the bibliography is given.


## 2. Preliminaries

This section introduces some basic definition including the fractional Caputo’s definition, the fractional order Riemann-Liouville (R-L) integral and some basic definitions of transformation regarding derivative and integration.Definition 1.
The R-L fractional integralΙtα of order α≥0, ft is defined as follows:
(2)Ιtαft=1Γα∫0tt−ηα−1fηdη,α>0,t>0,Ιt0ft=ft.Definition 2.
Caputo’s fractional derivative of orderα>0 is defined as follows:
(3)Dtαqx,t=∂αq℘,t∂tα=1Γn−α∫0tt−ηn−α−1∂nq℘,η∂ηn,ifn−1<α<n,∂nq℘,t∂tn,ifα=n∈N.Definition 3.
Natural transform ofqt is given as [17]
(4)ℕ+qt=ℝs,v=1v∫0∞e−st/vqtdt;s,v>0,
s and v are the transform variables.Definition 4.
Inverse natural transform ofℝs,v is written as follows:
(5)ℕ−ℝs,v=qt=12πi∫c−i∞c+i∞est/vℝs,vds,the integral is along the complex plane.Definition 5.
Ifqnt is the nth derivative of the function qt, then its natural transform is as follows:
(6)ℕ+qnt=ℝns,v=snvnℝs,v−∑k=0n−1sn−k+1vn−kqn0,n≥1.Theorem 6.
If natural transform ofgt and ktare, respectively, Gs,v and Ks,v and are defined in set A, then
(7)ℕg∗k=vGs,vKs,v,where g∗k is a convolution of g and k.


## 3. Basic Idea of NTDM

Consider the fractional order PDE of the form(8)Dtαq℘,t=f℘,t+Lq℘,t+ℵq℘,t,0<t≤1,m−1<α<m,m∈N,whereα is fractional derivative, L represents linear, ℵ shows the nonlinear operator, and f℘,t is known function. The initial condition is given as follows:
(9)u℘,0=g℘.When we use the natural transform on Equation (8), we get
(10)ℕ+Dtαq℘,t=ℕ+f℘,t+ℕ+Lq℘,t+ℵq℘,t.Using the natural transform’s differentiation characteristic to Equation (10), we have
(11)sαvαℕ+q℘,t−sα−1vαq℘,0=ℕ+f℘,t+ℕ+Lq℘,t+ℵq℘¯,t.Equation (11), after rearranging, is
(12)ℕ+q℘,t=g℘s+vαsαℕ+f℘,t+vαsαℕ+Lq℘,t+ℵq℘,t.For the NTDM solution, q℘,t expands as the infinite series
(13)q℘,t=∑i=0∞qi℘,t.The infinite series defines the nonlinear terms as follows:
(14)ℵq℘,t=∑i=0∞Ai,(15)Ai=1i!didλiℵ∑i=0∞λiqiλ=0,i=0,1,2⋯The Adomian polynomials are represented byAi. Equation (12) is modified by substituting Equations (13) and (14).
(16)ℕ+∑i=0∞q℘,t=g℘s+vαsαℕ+f℘,t+vαsαℕ+L∑i=0∞q℘,t+∑i=0∞Ai,using the natural transform’s linearity as follows:
(17)ℕ+q0℘,t=q℘,0s+vαsαℕ+f℘,t,ℕ+q1℘,t=vαsαℕ+Lq0℘,t+A0,⋮ℕ+qi+1℘,t=vαsαℕ+Lqi℘,t+Ai,i≥0.The solution components of Equation (28) are obtained by taking the inverse natural transform as follows:
(18)q0℘,t=g℘,t,qi+1℘,t=ℕ−vαsαℕ+Lqi℘,t+Ai,where the termg℘,t is derived from the specified source term and initial condition.The approximatem-term solution of Equations (8) and (9) is
(19)q℘,t=q0℘,t+q1℘,t+q2℘,t+⋯+qm−1℘,t.


## 4. Convergence Analysis of NTDM

Theorem 7.
LetH be the Hilbert space defined by H=L2α,βX0,T the set of applications
(20)q=α,βX0,T⟶Rwith∫α,βX0,Tq2x,sdsdϑ<+∞.
In light of the above assumptions, we now consider the GPZK equation of fractional order and denote(21)Lq=∂α∂tαq.The GPZK equation is then written in operator form
(22)Lq=−β1qkux−β2qxxx−β3qxyy−β3qxzz−ζqxxxxx.If the following hypotheses are true, the NTDM is convergent:
(23)H1Lq−Lη,q−η≥kq−η2;k>0,∀q,η∈H.
H2 whatever maybe M>0, there exists a constant CM>0, such that for q,η∈H with q≤M,η≤M, we have for every Lq−Lη,q−η≤CMq−ηw for every w∈H.


## 5. Applications of NTDM

### 5.1. Problem 1

Consider the following3+1 dimension GPZK equation [23]
(24)Dtαq+β1qqχχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,where q=qχ,y,z,t together with initial conditions
(25)qχ,y,z,0=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4.The exact solution is given as [23]
(26)qχ,y,z,t=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy−β1e0ρt+ϕ4.The (24) is rearranged as follows:
(27)Dtαq=−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Apply natural transform to Equation (27), we have
(28)ℕ+Dtαq=ℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Use the natural transform’s differentiation characteristic to Equation (28), we have
(29)sαvαqχ,y,z,t−vα−1sαqχ,y,z,0=ℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.After the application of inverse transform, we have
(30)qχ,y,z,t=qχ,y,z,0s+ℕ−vαsαℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Using the recursive relation and replacing the nonlinear termqqx by Adomian polynomials, Equation (30) yields
(31)qχ,y,z,t=qχ,y,z,0s+ℕ−vαsαℕ+−β1Ai−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ,i=0,1,2,⋯We got the solution components as follows by using the NTDM concept(32)q0χ,y,z,t=ℕ−qχ,y,z,0sq1χ,y,z,t=ℕ−vαsαℕ+−β1A0−β2q0χχχ−β3q0χyy−β3q0χzz−ζq0χχχχχq2χ,y,z,t=ℕ−vαsαℕ+−β1A1−β2q1χχχ−β3q1χyy−β3q1χzz−ζq1χχχχχq3χ,y,z,t=ℕ−vαsαℕ+−β1A2−β2q2χχχ−β3q2χyy−β3q2χzz−ζq2χχχχχ⋮,whereA0,, A1, and A2 are the Adomian polynomials and can be calculated using Equation (15). By solving, we get the solution as follows:
(33)q0χ,y,z,t=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4q1χ,y,z,t=−6720e0ζρ5tαΓα+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ5q2χ,y,z,t=−33600β1e02ζρ6t2αΓ2α+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ6q3χ,y,z,t=201600β1e02ζp7Γ2α+1t3α−β1e0−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4/Γ2α+1+1120ζρ4/Γα+12−2240ζρ4/Γ2α+1Γ3α+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ11.Adding the components, we obtain the 3rd order solution as follows:
(34)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t+q3χ,y,z,t−33600β1e02ζp6t2αΓ2α+1−z−β2p2/β3−κ2+pχ+κy+ϕ6−6720e0ζp5tα/Γα+1−z−β2p2/β3−κ2+pχ+κy+ϕ5+201600β1e02ζp7Γ2α+1t3α−β1e0−z−β2p2/β3−κ2+pχ+κy+ϕ4/Γ2α+1+1120ζp4/Γα+12−2240ζp4/Γ2α+1Γ3α+1−z−β2p2/β3−κ2+pχ+κy+ϕ11+e0−1680ζp4/β1−z−β2p2/β3−κ2+pχ+κy+ϕ4.


### 5.2. Problem 2

Consider the following3+1 dimension GPZK equation [23]
(35)Dtαq+β1q2qχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,0<α≤1,t>0,and initial condition as follows:
(36)qχ,y,z,0=e0+610iζρ2β1−10−iβ1e0ζp2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2.The exact solution is given as [24]
(37)qχ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z−β1e02ρt+ρχ+κy+ϕ2.By using basics concepts of NTDM, the solution components obtained as follows:(38)q0χ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2q1χ,y,z,t=12i10β1e02ζρ3tαΓα+1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ3q2χ,y,z,t=36i10β13/2e04ζρ4t2αΓ2α+1−z10−iβ1e0ζρ2−β2ρ2−β3κ2/β3+ρχ+κy+ϕ4.Adding the components, we get the solution as follows:
(39)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2+12i10β1e02ζρ3tαΓα+1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ3+36i10β13/2e04ζρ4t2αΓ2α+1−z10−iβ1e0ζρ2−β2ρ2−β3κ2/β3+ρχ+κy+ϕ4.


### 5.3. Problem 3

Consider the following3+1 dimension GPZK equation [23]
(40)Dtαq+β1q4qχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,and initial condition as follows:
(41)qχ,y,z,0=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z.The exact solution is given as [18]
(42)qχ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z+24ρ5ζt.Using NTDM concept, the solution components are obtained as follows:
(43)q0χ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z,q1χ,y,z,t=24+24i304ζ5/4p6tαsec2−z−20ζρ4−β2ρ2−β3κ2/β3+ρχ+κyβ14Γα+1q2χ,y,z,t=1152+1152i304ζ9/4ρ11t2αβ14Γ2α+1tan−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy×sec2−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy.Adding the components, we get the solution as follows:
(44)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z++24+24i304ζ5/4ρ6tαsec2−z−20ζρ4−β2ρ2−β3κ2/β3+ρχ+κyβ14Γα+1+1152+1152i304ζ9/4ρ11t2αβ14Γ2α+1tan−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy×sec2−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy.


## 5.1. Problem 1

Consider the following3+1 dimension GPZK equation [23]
(24)Dtαq+β1qqχχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,where q=qχ,y,z,t together with initial conditions
(25)qχ,y,z,0=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4.The exact solution is given as [23]
(26)qχ,y,z,t=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy−β1e0ρt+ϕ4.The (24) is rearranged as follows:
(27)Dtαq=−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Apply natural transform to Equation (27), we have
(28)ℕ+Dtαq=ℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Use the natural transform’s differentiation characteristic to Equation (28), we have
(29)sαvαqχ,y,z,t−vα−1sαqχ,y,z,0=ℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.After the application of inverse transform, we have
(30)qχ,y,z,t=qχ,y,z,0s+ℕ−vαsαℕ+−β1qqχ−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ.Using the recursive relation and replacing the nonlinear termqqx by Adomian polynomials, Equation (30) yields
(31)qχ,y,z,t=qχ,y,z,0s+ℕ−vαsαℕ+−β1Ai−β2qχχχ−β3qχyy−β3qχzz−ζqχχχχχ,i=0,1,2,⋯We got the solution components as follows by using the NTDM concept(32)q0χ,y,z,t=ℕ−qχ,y,z,0sq1χ,y,z,t=ℕ−vαsαℕ+−β1A0−β2q0χχχ−β3q0χyy−β3q0χzz−ζq0χχχχχq2χ,y,z,t=ℕ−vαsαℕ+−β1A1−β2q1χχχ−β3q1χyy−β3q1χzz−ζq1χχχχχq3χ,y,z,t=ℕ−vαsαℕ+−β1A2−β2q2χχχ−β3q2χyy−β3q2χzz−ζq2χχχχχ⋮,whereA0,, A1, and A2 are the Adomian polynomials and can be calculated using Equation (15). By solving, we get the solution as follows:
(33)q0χ,y,z,t=e0−1680ζρ4β1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4q1χ,y,z,t=−6720e0ζρ5tαΓα+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ5q2χ,y,z,t=−33600β1e02ζρ6t2αΓ2α+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ6q3χ,y,z,t=201600β1e02ζp7Γ2α+1t3α−β1e0−z−β2ρ2/β3−κ2+ρχ+κy+ϕ4/Γ2α+1+1120ζρ4/Γα+12−2240ζρ4/Γ2α+1Γ3α+1−z−β2ρ2/β3−κ2+ρχ+κy+ϕ11.Adding the components, we obtain the 3rd order solution as follows:
(34)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t+q3χ,y,z,t−33600β1e02ζp6t2αΓ2α+1−z−β2p2/β3−κ2+pχ+κy+ϕ6−6720e0ζp5tα/Γα+1−z−β2p2/β3−κ2+pχ+κy+ϕ5+201600β1e02ζp7Γ2α+1t3α−β1e0−z−β2p2/β3−κ2+pχ+κy+ϕ4/Γ2α+1+1120ζp4/Γα+12−2240ζp4/Γ2α+1Γ3α+1−z−β2p2/β3−κ2+pχ+κy+ϕ11+e0−1680ζp4/β1−z−β2p2/β3−κ2+pχ+κy+ϕ4.


## 5.2. Problem 2

Consider the following3+1 dimension GPZK equation [23]
(35)Dtαq+β1q2qχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,0<α≤1,t>0,and initial condition as follows:
(36)qχ,y,z,0=e0+610iζρ2β1−10−iβ1e0ζp2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2.The exact solution is given as [24]
(37)qχ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z−β1e02ρt+ρχ+κy+ϕ2.By using basics concepts of NTDM, the solution components obtained as follows:(38)q0χ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2q1χ,y,z,t=12i10β1e02ζρ3tαΓα+1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ3q2χ,y,z,t=36i10β13/2e04ζρ4t2αΓ2α+1−z10−iβ1e0ζρ2−β2ρ2−β3κ2/β3+ρχ+κy+ϕ4.Adding the components, we get the solution as follows:
(39)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t=e0+610iζρ2β1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ2+12i10β1e02ζρ3tαΓα+1−10−iβ1e0ζρ2−β2ρ2−β3κ2/β3z+ρχ+κy+ϕ3+36i10β13/2e04ζρ4t2αΓ2α+1−z10−iβ1e0ζρ2−β2ρ2−β3κ2/β3+ρχ+κy+ϕ4.


## 5.3. Problem 3

Consider the following3+1 dimension GPZK equation [23]
(40)Dtαq+β1q4qχ+β2qχχχ+β3qχyy+β3qχzz+ζqχχχχχ=0,and initial condition as follows:
(41)qχ,y,z,0=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z.The exact solution is given as [18]
(42)qχ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z+24ρ5ζt.Using NTDM concept, the solution components are obtained as follows:
(43)q0χ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z,q1χ,y,z,t=24+24i304ζ5/4p6tαsec2−z−20ζρ4−β2ρ2−β3κ2/β3+ρχ+κyβ14Γα+1q2χ,y,z,t=1152+1152i304ζ9/4ρ11t2αβ14Γ2α+1tan−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy×sec2−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy.Adding the components, we get the solution as follows:
(44)qχ,y,z,t=q0χ,y,z,t+q1χ,y,z,t+q2χ,y,z,t=23/4−154ζ4ρβ14tanρχ+κy−−20ζρ4−β2ρ2−β3κ2β3z++24+24i304ζ5/4ρ6tαsec2−z−20ζρ4−β2ρ2−β3κ2/β3+ρχ+κyβ14Γα+1+1152+1152i304ζ9/4ρ11t2αβ14Γ2α+1tan−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy×sec2−z−20ζρ4−β2ρ2−β3κ2β3+ρχ+κy.


## 6. Results and Discussion

This section presents the application of NTDM for solving fractional-order nonlinear GPZK equations. The values for the arbitrary constants have been taken asβ1=1, β2=2, β3=0.1, y=2, z=2, ζ=0.1, eo=3, ρ=0.5, қ=0.5, and ϕ=1 for problem 1-3. Problem 1 is approximated up to 3rd order. Problem 2 and problem 3 are approximated up to 2nd order by the NTDM algorithm. The approximate and exact real part of the solution is shown by Figures 1 and 2, respectively, by 3D plots for problem 1. The 3rd order approximate and exact solution of the imaginary part of problem 1 has been displayed by Figures 3 and 4, respectively. The solution obtained by NTDM has been compared by plots in Figures 5 and 6 for fractional values of α. Figures 7 and 8 display the 3D plots of the 2nd order real part solution and exact solution, respectively, for problem 2. Figures 9 and 10 show the 2nd order imaginary part solution and exact solution, respectively, for problem 2. The solution gained by NTDM has been compared by 2D plots in Figures 11 and 12 for fractional values of α.Figure 1
NTDM solution of the real part of problem 1.Figure 2
Exact solution of the real part of problem 1.Figure 3
NTDM solution of the imaginary part of problem 1.Figure 4
Exact solution of the imaginary part of problem 1.Figure 5
Real part solution comparison for fractional values ofα for problem 1.Figure 6
Imaginary part solution comparison for fractional values ofα for problem 1.Figure 7
NTDM solution of the real part of problem 2.Figure 8
Exact solution of the real part of problem 2.Figure 9
NTDM solution of the imaginary part of problem 2.Figure 10
Exact solution of the imaginary part of problem 2.Figure 11
Real part solution comparison for fractional values ofα for problem 2.Figure 12
Imaginary part solution comparison for fractional values ofα for problem 2.Similarly, Figures13 and 14 display the real part solution, and Figures 15 and 16 display the imaginary part solution for problem 3. In Tables 1 and 2, the absolute error is compared with the q-HATM solution, which shows the convergence of NTDM. Figure 17 shows the real part solution comparison for fractional values of α for problem 3. Figure 18 shows the imaginary part solution comparison for different fractional values of α for problem 3. In Tables 1 and 2, the absolute error is compared with the q-HATM solution, which shows the convergence of NTDM. Similarly, Tables 3 and 4 are compared for problem 2, while Tables 5 and 6 show the comparison for problem 3.Figure 13
NTDM solution of the real part of problem 3.Figure 14
Exact solution of the real part of problem 3.Figure 15
NTDM solution of the imaginary part of problem 3.Figure 16
Exact solution of the imaginary part of problem 3.Table 1
Real part solution comparison for fractional values ofα for problem 1 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error NTDMAbs. error [27]-153.00343.00363.003693.003713.003712.007727×10−122.007727×10−12-103.008123.008053.007983.007963.007961.639577×10−121.639577×10−12-52.981072.979652.979172.979072.979075.8479×10−115.8479×10−1102.999023.000453.000973.001083.001084.721068×10−114.721068×10−1153.00663.006343.006233.006213.006216.424195×10−126.424195×10−12103.001643.001523.001483.001473.001475.728751×10−145.728751×10−14153.000233.00023.00023.000193.000191.092459×10−131.092459×10−13Table 2
Imaginary part solution comparison for fractional values ofα for problem 1 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error NTDMAbs. error [38]-150.001515710.001463440.001436880.001430110.001430111.789912×10−121.789912×10−12-10-0.00722543-0.00801857-0.00835063-0.0084304-0.00843042.285489×10−112.285489×10−11-5-0.0123648-0.0108878-0.0102153-0.0100505-0.01005056.996747×10−116.996747×10−1100.01758860.01714010.0169050.01684580.01684586.948674×10−126.948674×10−1250.0007766440.0004196390.0002955250.0002672660.0002672664.853387×10−134.853387×10−1310-0.00160753-0.00158678-0.0015772-0.00157485-0.001574857.746282×10−137.746282×10−1315-0.000870554-0.00084329-0.000832837-0.000830384-0.0008303845.348467×10−145.348467×10−14Figure 17
Real part solution comparison for fractional values ofα for problem 3.Figure 18
Imaginary part solution comparison for different fractional values ofα for problem 3.Table 3
Real part solution comparison for fractional values ofα for problem 2 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error-152.98742.986572.986212.986122.986121.55851×10−9-102.978382.976962.976382.976242.976249.833575×10−9-52.967712.966592.966372.966332.966335.097527×10−802.977782.982042.9842.984482.984488.531288×10−853.020443.025083.026523.026833.026834.077737×10−9103.032913.032043.031613.03153.03153.67115×10−8153.022143.02073.020163.020033.020032.767871×10−9Table 4
Imaginary part solution comparison for fractional values ofα for problem 2 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error-150.006408350.006372690.006328980.00631650.00631655.566582×10−9-100.003295880.002524350.002111050.002005350.002005361.470755×10−8-5-0.0117248-0.0148465-0.0162419-0.0165841-0.01658412.044629×10−80-0.0429444-0.0469602-0.0481431-0.0483942-0.04839437.870128×10−85-0.0448139-0.0411819-0.0396028-0.0392184-0.03921839.6700550×10−810-0.0128468-0.0095154-0.00836997-0.00811094-0.008110956.485631×10−9150.003094160.004003630.004293290.004356890.004356881.14261×10−8Table 5
Real part solution comparison for fractional values ofα for problem 3 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error-150.6579720.65797200.65797200.65797200.65797201.354472×10−14-100.6579950.65799500.6579960.65799600.6579963.119727×10−14-50.6580780.6580780.6580780.6580790.6580793.108624×10−1400.6581020.6581020.6581020.6581020.6581021.354472×10−1450.6580330.6580330.6580320.6580320.6580323.885781×10−14100.657970.6579690.6579690.6579690.6579698.437695×10−15150.6580020.6580030.6580030.6580030.6580033.397282×10−14Table 6
Imaginary part solution comparison for fractional values ofα for problem 3 at t=0.01.
xα=0.5α=0.7α=0.9α=1.0ExactAbs. error-15-0.658012-0.6580130-0.658013-0.658013-0.6580133.6526340×10−14-10-0.658092-0.6580920-0.658093-0.658093-0.6580932.331468×10−14-5-0.6580930-0.6580930-0.6580930-0.6580930-0.6580932.3425710×10−140-0.658014-0.658013-0.658013-0.658013-0.65801303.663736×10−145-0.657968-0.657968-0.6579680-0.657968-0.6579682.775558×10−1510-0.6580210-0.6580220-0.658022-0.6580220-0.6580223.808065×10−1415-0.658097-0.658097-0.658098-0.658098-0.6580981.876277×10−14The above conversation summarizes that NTDM is appropriate for solving DEs of fractional order.


## 7. Conclusion

The GPZK equation of fractional order has been solved by NTDM in the current study. Three nonlinear cases have been shown convergent by comparing the results with the exact solution and q-HATM solution. The effectiveness of NTDM has been shown by showing numerical results. The plots for different numerical values ofα confirm the convergence of NTDM as α tend 1 the NTDM solution overlaps the exact solution. This discussion summarizes that the NTDM is suitable for approximating the complex nonlinear PDEs and ODEs of integer and fractional order.



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