# Numerical and Analytical Simulations of Nonlinear Time Fractional Advection and Burger’s Equations

Authors: Hassan Khan; Qasim Khan; Fairouz Tchier; Ibrarullah; Evren Hincal; Gurpreet Singh; F. M. O. Tawfiq; Shahbaz Khan
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, the higher nonlinear problems of fractional advection-diffusion equations and systems of nonlinear fractional Burger’s equations are solved by using two sophisticated procedures, namely, the q-homotopy analysis transform method and the residual power series method. The proposed methods are implemented with the Caputo operator. The present techniques are utilised in a very comprehensive and effective manner to obtain the solutions to the suggested fractional-order problems. The nonlinearity of the problem was controlled tactfully. The numerical results of a few examples are calculated and analyzed. The tables and graphs are constructed to understand the higher accuracy and applicability of the current method. The obtained results that are in good contact with the actual dynamics of the given problem, which is verified by the graphs and tables. The present techniques require fewer calculations and are associated with a higher degree of accuracy, and therefore can be extended to solve other high nonlinear fractional problems.


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## Body

## 1. Introduction

The most powerful tool for researchers to simulate various physical phenomena in applied sciences and nature is known as fractional partial differential equations (FPDEs). The following physical phenomena have been accurately modelled by FPDEs: optics [1], economics [2], fluid traffic [3], electrodynamics [4], hepatitis B virus [5], tuberculosis [6], air foil [7], modelling of earth quack nonlinear oscillation [8], propagation of spherical waves [9], Chaos theory [10], fractional COVID-19 model [11], finance [12], pine wilt disease [13], Zener [14], cancer chemotherapy [15], traffic flow model [16], Poisson-Nernst-Planck diffusion [17], diabetes [18], biomedical and biological [19], and many other numerous applications in various branches of applied mathematics (see [20–22]). Due to these numerous applications, FPDEs and factional calculus have gained more attention from researchers as compared to ordinary calculus.To obtain the approximate solutions to the above models, researchers use and develop a variety of analytical approaches. The frequently used methods are optimal homotopy asymptotic method (OHAM) [23], Iterative Laplace transform method [24], extended direct algebraic method (EDAM) [25], Adomian decomposition method (ADM) [26], the Finite difference method (FDM) [27], the homotopy perturbation transform technique along with transformation (HPTM) [28], the (G/G′)-expansion method [29], the Haar wavelet method (HWM) [30], standard reductive perturbation method [31], the variational iteration procedure with transformation (VITM) [32], and the differential transform method (DTM) [33]. In this context, Hassan et al. have presented the solutions of some nonlinear FPDEs which can be seen in [34–36]. Some useful methods can be seen in [37–39].Obtaining the analytical solutions of FPDEs and their systems has been a difficult task for researchers in recent years. In this circumstance, Prakash and Kaur use q-HATM [40] to solve the time-fractional Navier Stokes equation. Similarly, q-HATM was implemented by Kumar et al. and used to obtain the solutions of the fractional long-wave equations in the regularised form [41]. The Schrodinger generalized form of fractional order was solved by Veeresha et al. using q-HATM [42]. The q-HATM convergence was done by El-Tawil and Huseen [43]. The RPSM technique was used by Alquran to solve the drainage equation in [44] and the fractional-order Phi-4 equation in [45]. The Whitham-Broer-Kaup equation of fractional order was analysed by Wang and Chen by using RPSM [46]. RPSM has been used in [47] to find the solution of the fractional Biswas-Milovic equation of multidimensions. Komashynska et al. used the RPSM [48] to solve a system of multipantograph delay differential equations. In [49], RPSM was used to find an approximation solution for the fractional-order Sharma-Tasso-Olever equation. The solutions to the fractional order Schrodinger equations were determined using RPSM in [50]. The RPSM has been used to solve a variety of problems, including the gas dynamic equation [51], the Emden-Fowler equation, Berger-Fisher equation, and the Benney-Lin equation in its fractional format, solved by RPSM in [52]. Similarly, RPSM was used by Al-Smadi [53] to overcome the solutions to initial value problems.In this paper, the solutions of systems of fractional Berger’s equations and nonlinear advection equations were examined by combining two analytical techniques, q-HAM and RPSM. The results are compared to each other as well as the exact solution to the given problems. The current methods are used to compare the solutions methodologies for the analysis of the higher nonlinear problems of fractional advection-diffusion equations and systems of nonlinear fractional Burger’s equations in the Caputo sense. The obtained approximate series solutions are fast converging towards the exact solutions of the targeted problems, according to quantitative analysis. It is found that the techniques under discussion are simple and effective for the solutions of FPDE systems. The graphs and tables for the solutions to the targeted problems via RPSM and q-HATM. It is confirmed that the obtained solutions are in good contact with the exact solution to the problems. The fractional order solutions are convergent towards the integer order solutions, which shows the reliability of the fractional solutions. The presented techniques have a wide range of applications and could be used to examine approximate analytical solutions of other nonlinear FPDEs and multidimensional systems of FPDEs in the future.This paper will be formatted as follows: in the second section, some basic definitions are discussed. In Section3, the methodology is described, and in Section 4, two separate techniques are used to compare certain numerical results. The conclusion and references are found in the fifth and final section of the paper.


## 2. Basic Definitions

In this section we will discuss some important definitions


### 2.1. Definition

The Caputo operator in [54, 55] is given as.
(1)Dtδft=dnftdtn,δ=n∈N,1Γn−δ∫0tt−xn−δ−1fnxdx,n<δ<n+1,n∈N.


### 2.2. Definition

An expansion of power series (PS) at pointt=t0 is known as fractional PS and is given by [56].
(2)∑n=0∞ant−t0nδ=a0+a1t−t0δ+a2t−t02δ+⋯,&∑n=0∞fnxt−t0nδ=f0x+f1xt−t0δ+f2xt−t02δ+⋯,n−1<δ≤n,t≥t0,note FPS can be expanded at point t0 is
(3)υx,t=∑n=0∞Dtnδυx,t0Γnδ+1t−t0nδ,0≤n−1<δ≤m,x∈I,t0≤t<t0+R,which is the Taylor’s series expansion form.


### 2.3. Laplace Transform (LT)

The LT for continuous functiongt is defined as [57]
(4)Gs=Lgt=∫0∞e−stgtdt,where Gs is the LT for the function gt.


### 2.4. Definition

The LTLυx,t of Caputo fractional derivative is given by [57]
(5)LDtnδυx,t=snδLυx,t−∑k=0n−1snδ−k−1υkx,0,n−1<nδ≤n.


### 2.5. Definition

The LT of two functionkt×gt is defined by [57]
(6)Lk×g=Lkt×Lgt,Ask×g, represent the product between k and g,
(7)k×gt=∫0tktgt−xdt.


### 2.6. Definition

The LT of fractional derivative is defined as [57]
(8)LDtδgx=sδGs−∑k=0n−1sδ−1−kgk0+,n<δ≤n+1.where the LT of gx is denoted by Gs.


### 2.7. Theorem

Assume that the series solution∑k=0∞ukx,t=∑k=0∞ukx,t1/nk is convergent to the solution u for a prescribed value of h. If the truncated series
(9)∑k=0mukx,t=∑k=0mukx,t1nis used as an approximation to the solution ux,t of problem, then an upper bound for the error, Em is estimated as
(10)Em≤r/nm+11−r/nuox,tProof.
See [43]


### 2.8. Theorem

suppose thatut∈Ct0,t0+R and Dtkαut∈Ct0,t0+Rfork=0,1,2,⋯,n+1,where0<α≤1. Then u could be represented by:
(11)ut=∑k=0nDt0kαut0Γkα+1t−t0kα+Jt0n+1αDt0n+1αut,t0≤t≤t0+R.Proof.
See [58].


### 2.9. Theorem

IfDt0n+1αut≤M on t0≤t≤d, where 0<α≤1, then the reminder Rnt of the generalized Taylor’s series will satisfies the inequality:
(12)Rnt≤MΓn+1α+1t−t0n+1α,t0≤t≤D.Proof.
See [58].


### 2.10. Theorem

Suppose thatu has a Fractional power series representation at t0 of the form
(13)ut=∑n=0∞cnt−t0nα,0≤m−1<α≤m,t0≤t<t0+R,where R is the radius of convergence. Then u is analytic in t0,t0+R.Proof.
See [58].


## 2.1. Definition

The Caputo operator in [54, 55] is given as.
(1)Dtδft=dnftdtn,δ=n∈N,1Γn−δ∫0tt−xn−δ−1fnxdx,n<δ<n+1,n∈N.


## 2.2. Definition

An expansion of power series (PS) at pointt=t0 is known as fractional PS and is given by [56].
(2)∑n=0∞ant−t0nδ=a0+a1t−t0δ+a2t−t02δ+⋯,&∑n=0∞fnxt−t0nδ=f0x+f1xt−t0δ+f2xt−t02δ+⋯,n−1<δ≤n,t≥t0,note FPS can be expanded at point t0 is
(3)υx,t=∑n=0∞Dtnδυx,t0Γnδ+1t−t0nδ,0≤n−1<δ≤m,x∈I,t0≤t<t0+R,which is the Taylor’s series expansion form.


## 2.3. Laplace Transform (LT)

The LT for continuous functiongt is defined as [57]
(4)Gs=Lgt=∫0∞e−stgtdt,where Gs is the LT for the function gt.


## 2.4. Definition

The LTLυx,t of Caputo fractional derivative is given by [57]
(5)LDtnδυx,t=snδLυx,t−∑k=0n−1snδ−k−1υkx,0,n−1<nδ≤n.


## 2.5. Definition

The LT of two functionkt×gt is defined by [57]
(6)Lk×g=Lkt×Lgt,Ask×g, represent the product between k and g,
(7)k×gt=∫0tktgt−xdt.


## 2.6. Definition

The LT of fractional derivative is defined as [57]
(8)LDtδgx=sδGs−∑k=0n−1sδ−1−kgk0+,n<δ≤n+1.where the LT of gx is denoted by Gs.


## 2.7. Theorem

Assume that the series solution∑k=0∞ukx,t=∑k=0∞ukx,t1/nk is convergent to the solution u for a prescribed value of h. If the truncated series
(9)∑k=0mukx,t=∑k=0mukx,t1nis used as an approximation to the solution ux,t of problem, then an upper bound for the error, Em is estimated as
(10)Em≤r/nm+11−r/nuox,tProof.
See [43]


## 2.8. Theorem

suppose thatut∈Ct0,t0+R and Dtkαut∈Ct0,t0+Rfork=0,1,2,⋯,n+1,where0<α≤1. Then u could be represented by:
(11)ut=∑k=0nDt0kαut0Γkα+1t−t0kα+Jt0n+1αDt0n+1αut,t0≤t≤t0+R.Proof.
See [58].


## 2.9. Theorem

IfDt0n+1αut≤M on t0≤t≤d, where 0<α≤1, then the reminder Rnt of the generalized Taylor’s series will satisfies the inequality:
(12)Rnt≤MΓn+1α+1t−t0n+1α,t0≤t≤D.Proof.
See [58].


## 2.10. Theorem

Suppose thatu has a Fractional power series representation at t0 of the form
(13)ut=∑n=0∞cnt−t0nα,0≤m−1<α≤m,t0≤t<t0+R,where R is the radius of convergence. Then u is analytic in t0,t0+R.Proof.
See [58].


## 3. RPSM and Q-HATM Procedures

To understand main concept of RPSM and q-HATM [59] for system of FPDEs, we consider the following FPDEs system,
(14)Dtδυx,y,z,t=Nυx,y,z,t+Rυx,y,z,t,0<δ≤1,t>0,Dtδϑx,y,z,t=Nϑx,y,z,t+Rϑx,y,z,t,0<δ≤1,t>0,Dtδϖx,y,z,t=Nϖx,y,z,t+Rϖx,y,z,t,0<δ≤1,t>0,with initial condition,
(15)υx,y,z,0=fx,y,z,ϑx,y,z,0=gx,y,z,ϖx,y,z,0=hx,y,z,where Dtδ the Caputo fractional derivative, R is linear and N is nonlinear operator, respectively, in Eq. (14).


### 3.1. RPSM Procedure for FPDEs System [60]

Eq. (14) can be simplified as
(16)υx,y,z,t=∑n=0kfnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,ϑx,y,z,t=∑n=0kgnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,ϖx,y,z,t=∑n=0khnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,where υx,y,z,t, ϑx,y,z,t, and ϖx,y,z,t is the kth truncated series of form
(17)υkx,y,z,t=∑n=0kfnx,y,ztnδΓ1+nδ,ϑkx,y,z,t=∑n=0kgnx,y,ztnδΓ1+nδ,ϖkx,y,z,t=∑n=0khnx,y,ztnδΓ1+nδ.In RPSM thezeroth approximate solution of υx,y,z,t, ϑx,y,z,t, and ϖx,y,z,t is given by
(18)υ0x,y,z,t=υx,0=fx,y,z,ϑ0x,y,z,t=ϑx,0=gx,y,z,ϖ0x,y,z,t=ϖx,0=hx,y,z.Eq. (17), implies that,
(19)υkx,y,z,t=fx,y,z+∑n=1kfnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϑkx,y,z,t=gx,y,z+∑n=1kgnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϖkx,y,z,t=hx,y,z+∑n=1khnx,y,ztnδΓ1+nδ,k=1,2,⋯,for Eq. (14), the residual function is given by
(20)Resυx,y,z,t=Dtδυx,y,z,t−Nυx,y,z,t−Rυx,y,z,t,Resϑx,y,z,t=Dtδϑx,y,z,t−Nϑx,y,z,t−Rϑx,y,z,t,Resϖx,y,z,t=Dtδϖx,y,z,t−Nϖx,y,z,t−Rϖx,y,z,t,so, the kth residual function becomes
(21)Resυ,kx,y,z,t=Dtδυkx,y,z,t−Nυkx,y,z,t−Rυkx,y,z,t,Resϑ,kx,y,z,t=Dtδϑkx,y,z,t−Nϑkx,y,z,t−Rϑkx,y,z,t,Resϖ,kx,y,z,t=Dtδϖkx,y,z,t−Nϖkx,y,z,t−Rϖkx,y,z,t.As in [61, 62], it is clear that Resx,y,z,t=0 and limnto∞Reskx,y,z,t=Resx,y,z,t. Therefore, DtnδResυx,y,z,t=0, DtnδResϑx,y,z,t=0, and DtnδResϖx,y,z,t=0. In the Caputo definition, the constant has zero derivative therefore DtnδResx,y,z,0=DtnδReskx,y,z,0=0, k=0,1,⋯,n which mean that Dtnδ of Resυx,y,z,t, Resϑx,y,z,t, Resϖx,y,z,t, and Reskx,y,z,t are t=0 matching at n=0,1,⋯,k;.To determinef1x,y,z,f2x,y,z,f3x,y,z,⋯, g1x,y,z,g2x,y,z,g3x,y,z,⋯, and h1x,y,z,h2x,y,z,h3x,y,z,⋯, we substitute k=0,1,⋯, in Eq. (17), and then the obtained results are put in Eq. (19). In the final step, we apply Dtkδ on both sides we obtained the following
(22)DtkδResυ,kx,y,z,0=0,k=0,1,⋯,DtkδResϑ,kx,y,z,0=0,k=0,1,⋯,DtkδResϖ,kx,y,z,0=0,k=0,1,⋯.


### 3.2. Q-HATM Procedure for FPDEs System [63]

Using LT, Eq. (14) can be simplified as
(23)sδLυx,y,z,t−∑k=0n−1sδ−k−1υkx,y,z,0+LRυx,y,z,t+Nυx,y,z,t=Lfx,y,z,t,sδLϑx,y,z,t−∑k=0n−1sδ−k−1ϑkx,y,z,0+LRϑx,y,z,t+Nϑx,y,z,t=Lgx,y,z,t,sδLϖx,y,z,t−∑k=0n−1sδ−k−1ϖkx,y,z,0+LRϖx,y,z,t+Nϖx,y,z,t=Lhx,y,z,t,so Eq. (23), implies that
(24)Lυx,y,z,t−1sδ∑k=0n−1sδ−k−1υkx,y,z,0+1sδLRυx,y,z,t+Nυx,y,z,t−fx,y,z,t=0,Lϑx,y,z,t−1sδ∑k=0n−1sδ−k−1ϑkx,y,z,0+1sδLRϑx,y,z,t+Nϑx,y,z,t−gx,y,z,t=0,Lϖx,y,z,t−1sδ∑k=0n−1sδ−k−1ϖkx,y,z,0+1sδLRϖx,y,z,t+Nϖx,y,z,t−hx,y,z,t=0.We define the nonlinear operator is(25)Nθx,y,z,t;q=Lθx,y,z,t;q−1sδ∑k=on−1sδ−k−1θkx,y,z,t;q0++1sδLRυx,y,z,t+Nυx,y,z,t−fx,y,z,t,Nθx,y,z,t;q=Lθx,y,z,t;q−1sδ∑k=on−1sδ−k−1θkx,t;q0++1sδLRϑx,y,z,t+Nϑx,y,z,t−gx,y,z,t,Nθx,y,z,t;q=Lθx,t;q−1sδ∑k=on−1sδ−k−1θkx,y,z,t;q0++1sδLRϖx,y,z,t+Nϖx,y,z,t−hx,y,z,t,where q∈0,1/n, θx,y,z,t;q is real function of x,y,z, t, and q.Homotopy can be constructed as(26)1−nqLθx,y,z,t;q−υ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q,1−nqLθx,y,z,t;q−ϑ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q,1−nqLθx,y,z,t;q−ϖ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q.In Eq. (26), the auxiliary parameter, nonzero auxiliary function and embedding parameter are h≠0, Hx,y,z,t and n≥1, q∈0,1/n, respectively. L denotes Laplacian operator and υ0, ϑ0, and ϖ0 are the initial conditions. The following results are obtained at q=0 and q=1/n,
(27)θx,y,z,t;0=υ0x,y,z,tandθx,y,z,t;1n=υx,y,z,t,θx,y,z,t;0=ϑ0x,y,z,tandθx,y,z,t;1n=ϑx,y,z,t,θx,y,z,t;0=ϖ0x,y,z,tandθx,y,z,t;1n=ϖx,y,z,ts.By using Taylor theoremθx,y,z,t;q can be expressed as
(28)θx,y,z,t;q=υ0x,y,z,t+∑m=1∞υmx,y,z,tqm,θx,y,z,t;q=ϑ0x,y,z,t+∑m=1∞ϑmx,y,z,tqm,θx,y,z,t;q=ϖ0x,y,z,t+∑m=1∞ϖmx,y,z,tqm,where
(29)υmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0,ϑmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0,ϖmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0.After simplification, we have(30)υmx,y,z,t=υ0x,y,z,t+∑m=1∞υmx,y,z,t1nm,ϑmx,y,z,t=ϑ0x,y,z,t+∑m=1∞ϑmx,y,z,t1nm,ϖmx,y,z,t=ϖ0x,y,z,t+∑m=1∞ϖmx,y,z,t1nm,Taking them-times derivatives of Eq. (26), put q=0, implies the zeroth order solution
(31)Lυmx,y,z,t−kmυm−1x,y,z,t=ℏHx,y,z,tRmυ⟵m−1,Lϑmx,y,z,t−kmϑm−1x,y,z,t=ℏHx,y,z,tRmϑ⟵m−1,Lϖmx,y,z,t−kmϖm−1x,y,z,t=ℏHx,y,z,tRmϖ⟵m−1.where the vectors are defined as
(32)υ⟵m=υ0x,y,z,t,υ1x,y,z,t,⋯,υmx,y,z,t,ϑ⟵m=ϑ0x,y,z,t,ϑ1x,y,z,t,⋯,ϑmx,y,z,t,ϖ⟵m=ϖ0x,y,z,t,ϖ1x,y,z,t,⋯,ϖmx,y,z,t.The following recursive formula is obtained by taking inverse LT of Eq. (31),
(33)υmx,y,z,t=kmx,y,z,tυm−1x,y,z,t+ℏL−1Hx,y,z,tRmυ⟵m−1,ϑmx,y,z,t=kmx,y,z,tϑm−1x,y,z,t+ℏL−1Hx,y,z,tRmϑ⟵m−1,ϖmx,y,z,t=kmx,y,z,tϖm−1x,y,z,t+ℏL−1Hx,y,z,tRmϖ⟵m−1.Where(34)Rmυ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,Rmϑ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,Rmϖ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,(35)km=0,m≤1,n,m>1.Eq. (33) and Eq. (35) are known q-HATM series solutions for the given system.


## 3.1. RPSM Procedure for FPDEs System [60]

Eq. (14) can be simplified as
(16)υx,y,z,t=∑n=0kfnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,ϑx,y,z,t=∑n=0kgnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,ϖx,y,z,t=∑n=0khnxtnδΓ1+nδ,0<δ≤1,−∞<x,y,z<∞,0≤t<R,where υx,y,z,t, ϑx,y,z,t, and ϖx,y,z,t is the kth truncated series of form
(17)υkx,y,z,t=∑n=0kfnx,y,ztnδΓ1+nδ,ϑkx,y,z,t=∑n=0kgnx,y,ztnδΓ1+nδ,ϖkx,y,z,t=∑n=0khnx,y,ztnδΓ1+nδ.In RPSM thezeroth approximate solution of υx,y,z,t, ϑx,y,z,t, and ϖx,y,z,t is given by
(18)υ0x,y,z,t=υx,0=fx,y,z,ϑ0x,y,z,t=ϑx,0=gx,y,z,ϖ0x,y,z,t=ϖx,0=hx,y,z.Eq. (17), implies that,
(19)υkx,y,z,t=fx,y,z+∑n=1kfnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϑkx,y,z,t=gx,y,z+∑n=1kgnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϖkx,y,z,t=hx,y,z+∑n=1khnx,y,ztnδΓ1+nδ,k=1,2,⋯,for Eq. (14), the residual function is given by
(20)Resυx,y,z,t=Dtδυx,y,z,t−Nυx,y,z,t−Rυx,y,z,t,Resϑx,y,z,t=Dtδϑx,y,z,t−Nϑx,y,z,t−Rϑx,y,z,t,Resϖx,y,z,t=Dtδϖx,y,z,t−Nϖx,y,z,t−Rϖx,y,z,t,so, the kth residual function becomes
(21)Resυ,kx,y,z,t=Dtδυkx,y,z,t−Nυkx,y,z,t−Rυkx,y,z,t,Resϑ,kx,y,z,t=Dtδϑkx,y,z,t−Nϑkx,y,z,t−Rϑkx,y,z,t,Resϖ,kx,y,z,t=Dtδϖkx,y,z,t−Nϖkx,y,z,t−Rϖkx,y,z,t.As in [61, 62], it is clear that Resx,y,z,t=0 and limnto∞Reskx,y,z,t=Resx,y,z,t. Therefore, DtnδResυx,y,z,t=0, DtnδResϑx,y,z,t=0, and DtnδResϖx,y,z,t=0. In the Caputo definition, the constant has zero derivative therefore DtnδResx,y,z,0=DtnδReskx,y,z,0=0, k=0,1,⋯,n which mean that Dtnδ of Resυx,y,z,t, Resϑx,y,z,t, Resϖx,y,z,t, and Reskx,y,z,t are t=0 matching at n=0,1,⋯,k;.To determinef1x,y,z,f2x,y,z,f3x,y,z,⋯, g1x,y,z,g2x,y,z,g3x,y,z,⋯, and h1x,y,z,h2x,y,z,h3x,y,z,⋯, we substitute k=0,1,⋯, in Eq. (17), and then the obtained results are put in Eq. (19). In the final step, we apply Dtkδ on both sides we obtained the following
(22)DtkδResυ,kx,y,z,0=0,k=0,1,⋯,DtkδResϑ,kx,y,z,0=0,k=0,1,⋯,DtkδResϖ,kx,y,z,0=0,k=0,1,⋯.


## 3.2. Q-HATM Procedure for FPDEs System [63]

Using LT, Eq. (14) can be simplified as
(23)sδLυx,y,z,t−∑k=0n−1sδ−k−1υkx,y,z,0+LRυx,y,z,t+Nυx,y,z,t=Lfx,y,z,t,sδLϑx,y,z,t−∑k=0n−1sδ−k−1ϑkx,y,z,0+LRϑx,y,z,t+Nϑx,y,z,t=Lgx,y,z,t,sδLϖx,y,z,t−∑k=0n−1sδ−k−1ϖkx,y,z,0+LRϖx,y,z,t+Nϖx,y,z,t=Lhx,y,z,t,so Eq. (23), implies that
(24)Lυx,y,z,t−1sδ∑k=0n−1sδ−k−1υkx,y,z,0+1sδLRυx,y,z,t+Nυx,y,z,t−fx,y,z,t=0,Lϑx,y,z,t−1sδ∑k=0n−1sδ−k−1ϑkx,y,z,0+1sδLRϑx,y,z,t+Nϑx,y,z,t−gx,y,z,t=0,Lϖx,y,z,t−1sδ∑k=0n−1sδ−k−1ϖkx,y,z,0+1sδLRϖx,y,z,t+Nϖx,y,z,t−hx,y,z,t=0.We define the nonlinear operator is(25)Nθx,y,z,t;q=Lθx,y,z,t;q−1sδ∑k=on−1sδ−k−1θkx,y,z,t;q0++1sδLRυx,y,z,t+Nυx,y,z,t−fx,y,z,t,Nθx,y,z,t;q=Lθx,y,z,t;q−1sδ∑k=on−1sδ−k−1θkx,t;q0++1sδLRϑx,y,z,t+Nϑx,y,z,t−gx,y,z,t,Nθx,y,z,t;q=Lθx,t;q−1sδ∑k=on−1sδ−k−1θkx,y,z,t;q0++1sδLRϖx,y,z,t+Nϖx,y,z,t−hx,y,z,t,where q∈0,1/n, θx,y,z,t;q is real function of x,y,z, t, and q.Homotopy can be constructed as(26)1−nqLθx,y,z,t;q−υ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q,1−nqLθx,y,z,t;q−ϑ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q,1−nqLθx,y,z,t;q−ϖ0x,y,z,t=ℏqHx,y,z,tNθx,y,z,t;q.In Eq. (26), the auxiliary parameter, nonzero auxiliary function and embedding parameter are h≠0, Hx,y,z,t and n≥1, q∈0,1/n, respectively. L denotes Laplacian operator and υ0, ϑ0, and ϖ0 are the initial conditions. The following results are obtained at q=0 and q=1/n,
(27)θx,y,z,t;0=υ0x,y,z,tandθx,y,z,t;1n=υx,y,z,t,θx,y,z,t;0=ϑ0x,y,z,tandθx,y,z,t;1n=ϑx,y,z,t,θx,y,z,t;0=ϖ0x,y,z,tandθx,y,z,t;1n=ϖx,y,z,ts.By using Taylor theoremθx,y,z,t;q can be expressed as
(28)θx,y,z,t;q=υ0x,y,z,t+∑m=1∞υmx,y,z,tqm,θx,y,z,t;q=ϑ0x,y,z,t+∑m=1∞ϑmx,y,z,tqm,θx,y,z,t;q=ϖ0x,y,z,t+∑m=1∞ϖmx,y,z,tqm,where
(29)υmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0,ϑmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0,ϖmx,y,z,t=1m!∂mθx,y,z,t;q∂qmq=0.After simplification, we have(30)υmx,y,z,t=υ0x,y,z,t+∑m=1∞υmx,y,z,t1nm,ϑmx,y,z,t=ϑ0x,y,z,t+∑m=1∞ϑmx,y,z,t1nm,ϖmx,y,z,t=ϖ0x,y,z,t+∑m=1∞ϖmx,y,z,t1nm,Taking them-times derivatives of Eq. (26), put q=0, implies the zeroth order solution
(31)Lυmx,y,z,t−kmυm−1x,y,z,t=ℏHx,y,z,tRmυ⟵m−1,Lϑmx,y,z,t−kmϑm−1x,y,z,t=ℏHx,y,z,tRmϑ⟵m−1,Lϖmx,y,z,t−kmϖm−1x,y,z,t=ℏHx,y,z,tRmϖ⟵m−1.where the vectors are defined as
(32)υ⟵m=υ0x,y,z,t,υ1x,y,z,t,⋯,υmx,y,z,t,ϑ⟵m=ϑ0x,y,z,t,ϑ1x,y,z,t,⋯,ϑmx,y,z,t,ϖ⟵m=ϖ0x,y,z,t,ϖ1x,y,z,t,⋯,ϖmx,y,z,t.The following recursive formula is obtained by taking inverse LT of Eq. (31),
(33)υmx,y,z,t=kmx,y,z,tυm−1x,y,z,t+ℏL−1Hx,y,z,tRmυ⟵m−1,ϑmx,y,z,t=kmx,y,z,tϑm−1x,y,z,t+ℏL−1Hx,y,z,tRmϑ⟵m−1,ϖmx,y,z,t=kmx,y,z,tϖm−1x,y,z,t+ℏL−1Hx,y,z,tRmϖ⟵m−1.Where(34)Rmυ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,Rmϑ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,Rmϖ⟵m−1=1m−1!∂m−1Nθx,y,z,t;q∂qm−1q=0,(35)km=0,m≤1,n,m>1.Eq. (33) and Eq. (35) are known q-HATM series solutions for the given system.


## 4. Numerical Results

### 4.1. Example

Consider nonlinear advection-diffusion equation of the form [64]
(36)∂δυ∂tδ=−υ∂υ∂x+υ−υ2,0<δ≤1,with ICs
(37)υx,0=e−x,and exact solution at δ=1(38)υx,t=et−x.


#### 4.1.1. RPSM-Solution

(1) 1st Iteration. Using RPSM, the kth truncated series of Eq. (36), can obtain
(39)υkx,t=∑n=0kfnxtnδΓ1+nδ,the 1st approximate is given as
(40)υ0x,t=υx,0=fx,Eq. (39) should be written as
(41)υkx,t=fx+∑n=1kfnxtnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (41), we get
(42)υ1x,t=fx+f1xtδΓ1+δ,where υx,0=fx=e−x,
(43)υ1x,t=e−x+f1xtδΓ1+δ.The residual function of Eq. (36), is given by
(44)Resυx,t=∂δυ∂tδ+υ∂υ∂x−υ+υ2.Thekth residual function Resυx,t is given by,
(45)Resυkx,t=∂δυk∂tδ+υk∂υk∂x−υk+υk2,put k=1 in the Eq. (45), we obtain
(46)Resυ1x,t=f1x+e−x+f1xtδΓ1+δ−e−x+f1′xtδΓ1+δ−e−x+f1xtδΓ1+δ+e−2x+f1xe−xtδΓ1+δ+2e−xf1xe−xtδΓ1+δ,we have
(47)Resυ1x,0=0.Eq. (46), will become
(48)f1x=e−x.(2) 2nd Iteration. for k=2 Eq. (41), can be written as
(49)υ2x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ,where fx=e−x, and f1x=e−x,
(50)υ2x,t=e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ,put k=2 in Eq. (45), we obtain
(51)Resυ2x,t=∂δυ2∂tδ+υ2∂υ2∂x−υ2+υ12,(52)Resυ2x,t=e−x+f2xtδΓ1+δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ−e−x−e−xtδΓ1+δ+f2′xt2δΓ1+2δ−e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ2,as we know that
(53)Dtk−1δResυkx,t=0,for k=2 Eq. (53), become as
(54)DtδResυ2x,t=0,using Dtδ on Eq. (52), we get
(55)DtδResυ2x,t=f2x+e−x+f2xtδΓ1+δ−e−x+f2′xtδΓ1+δ−e−x+f2xtδΓ1+δ+e−x+f2xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (55), we obtain
(56)f2x=e−x.(3) 3rd Iteration. Put k=3 in Eq. (41), we obtain
(57)υ3x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ,where fx=e−x, f1x=e−x, and f2x=e−x,
(58)υ3x,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ,put k=3 in Eq. (45) we obtain
(59)Resυ3x,t=∂δυ3∂tδ+υ3∂υ3∂x−υ3+υ32,(60)Resυ3x,t=e−x+e−xtδΓ1+δ+f3xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ−e−x−e−xtδΓ1+δ−e−xt2δΓ1+2δ+f3′xt3δΓ1+3δ−e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ2,put k=3 in Eq. (53), we obtain
(61)Dt2δResυ3x,t=0,applying Dt2δ on both sides of the Eq. (60), we get
(62)Dt2δResυ2x,t=f3x+e−x+f3xtδΓ1+δ−e−x+f3′xtδΓ1+δ−e−x+f3xtδΓ1+δ+e−x+f3xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (62), we obtain
(63)f2x=e−x.The RPSM solution of Eq. (36), is given as
(64)υx,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ+⋯,υx,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+e−xt3δΓ1+3δ+⋯.


#### 4.1.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (36), and simplifying
(65)sδLυx,t−∑k=0n−1sδ−k−1υkx,0+Lυ∂υ∂x−υ+υ2=0,Lυx,t−1sδsδ−1υ0x,0+1sδLυ∂υ∂x−υ+υ2=0,Lυx,t−e−xs+1sδLυ∂υ∂x−υ+υ2=0.The nonlinear termN is defined as
(66)Nθx,t;q=Lθx,t;q−e−xs+1sδLυ∂υ∂x−υ+υ2.Using the procedure of q-HATM(67)υmx,t=kmυm−1x,t+hL−1Rmυm−1,put m=1 in the Eq. (67), we obtain
(68)υ1x,t=k1υ0x,t+hL−1R1υ0,(69)Rmυm−1=Lυm−1−1−kmne−xs+1sδL∑i=0m−1υi∂υm−1−i∂x−υm−1+∑i=0m−1υiυm−1−i,put m=1 in Eq. (69), we obtain
(70)R1υ0=Lυ0−1−k1ne−xs+1sδLυ0∂υ0∂x−υ0+υ0υ0,R1υ0=Le−x−1−k1ne−xs+1sδLe−x∂∂xe−x−e−x+e−xe−x,R1υ0=e−xs−e−xs+1sδL−e−2x−e−x+e−2x,R1υ0=−e−xsδ+1,put in Eq. (68), we obtain
(71)υ1x,t=hL−1−e−xsδ+1,υ1x,t=−e−xhtδΓ1+δ.(2) 2nd Iteration. Put m=2 in the Eq. (67), we obtain
(72)υ2x,t=k2υ1x,t+hL−1R2υ1,put m=2 in Eq. (69), we obtain
(73)R2υ1=Lυ1−1−k2ne−xs+1sδLυ0∂υ1∂x+υ1∂υ0∂x−υ1+υ1υ0+υ0υ1,=−e−xhsδ+1+e−xhts2δ+1,put in Eq. (72), we obtain
(74)υ2x,t=−e−xnhtδΓ1+δ+hL−1−e−xhsδ+1+e−xhts2δ+1,υ2x,t=−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1.(3) 3rd Iteration. Put m=3 in the Eq. (67), we obtain
(75)υ3x,t=k3υ2x,t+hL−1R3υ2,put m=3 in Eq. (69), we obtain
(76)R3υ2=Lυ2−e−xs+1sδLυ0∂υ2∂x+υ1∂υ1∂x+υ2∂υ0∂x−υ2+υ0υ2+υ1υ1+υ2υ0,=−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,put in Eq. (75), we obtain
(77)υ3x,t=n−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1+hL−1−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,υ3x,t=−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1.The q-HATM solution of Eq. (36), is given as
(78)υx,t=υ0x,t+υ1x,t+υ2x,t+υ3x,t+⋯,υx,t=e−x−e−xhtδΓ1+δ−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1+⋯.


### 4.2. Example

The system of 3D Burgers’ equation are [65, 66]
(79)υtδ+υυx+ϑυy+ϖυz−υxx−υyy−υzz=0,ϑtδ+υϑx+ϑϑy+ϖϑz−ϑxx−ϑyy−ϑzz=0,ϖtδ+υϖx+ϑϖy+ϖϖz−ϖxx−ϖyy−ϖzz=0,the initial condition of Eq. (79), are
(80)υx,y,z,0=−0.5x+y+z,ϑx,y,z,0=x−0.5y+z,ϖx,y,z,0=x+y−0.5z.The exact solution of the Eq. (79) at δ=1, are
(81)υx,y,z,t=−0.5x+y+z−2.25xt1−2.25t2,ϑx,y,z,t=x−0.5y+z−2.25yt1−2.25t2,ϖx,y,z,t=x+y−0.5z−2.25zt1−2.25t2.


#### 4.2.1. RPSM-Solution

(1) 1st Iteration. The kth truncated series of the solution of Eq. (79), using RPSM, we obtain
(82)υkx,y,z,t=∑n=0kfnx,y,ztnδΓ1+nδ,ϑkx,y,z,t=∑n=0kgnx,y,ztnδΓ1+nδ,ϖkx,y,z,t=∑n=0khnx,y,ztnδΓ1+nδ,the zeroth RPSM approximate solution of Eq. (79), is given by
(83)υ0x,y,z,t=υx,y,z,0=fx,y,z,ϑ0x,y,z,t=ϑx,y,z,0=gx,y,z,ϖ0x,y,z,t=ϖx,y,z,0=hx,y,z,the Eq. (82), should be written as
(84)υkx,y,z,t=fx,y,z+∑n=1kfnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϑkx,y,z,t=gx,y,z+∑n=1kgnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϖkx,y,z,t=hx,y,z+∑n=1khnx,y,ztnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (84), we get
(85)υ1x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,where
(86)υx,y,z,0=fx,y,z=−0.5x+y+z,ϑx,y,z,0=gx,y,z=x−0.5y+z,ϖx,y,z,0=hx,y,z=x+y−0.5z,υ1x,y,z,t=−0.5x+y+z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=x−0.5y+z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=x+y−0.5z+h1x,y,ztδΓ1+δ.The residual function of Eq. (79), is given by
(87)Resυx,y,z,t=∂δυ∂tδ+υ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Resϑx,y,z,t=∂δϑ∂tδ+υ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Resϖx,y,z,t=∂δϖ∂tδ+υ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2,where the kth residual function of Resυx,y,t, Resϑx,y,t, and Resϖx,y,t is given by
(88)Resυkx,y,z,t=∂δυk∂tδ+υk∂υk∂x+ϑk∂υk∂y+ϖk∂υk∂z−∂2υk∂x2−∂2υk∂y2−∂2υk∂z2,Resϑkx,y,z,t=∂δϑk∂tδ+υk∂ϑk∂x+ϑk∂ϑk∂y+ϖk∂ϑk∂z−∂2ϑk∂x2−∂2ϑk∂y2−∂2ϑk∂z2,Resϖkx,y,z,t=∂δϖk∂tδ+υk∂ϖk∂x+ϑk∂ϖk∂y+ϖk∂ϖk∂z−∂2ϖk∂x2−∂2ϖk∂y2−∂2ϖk∂z2,put k=1 in the Eq. (88), we obtain
(89)Resυ1x,y,z,t=∂δυ1∂tδ+υ1∂υ1∂x+ϑ1∂υ1∂y+ϖ1∂υ1∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,Resϑ1x,y,z,t=∂δϑ1∂tδ+υ1∂ϑ1∂x+ϑ1∂ϑ1∂y+ϖ1∂ϑ1∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,Resϖ1x,y,z,t=∂δϖ1∂tδ+υ1∂ϖ1∂x+ϑ1∂ϖ1∂y+ϖ1∂ϖ1∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,(90)=f1x,y,z+−0.5x+y+z+f1x,y,ztδΓ1+δ−0.5+f1′x,y,ztδΓ1+δ+x−0.5y+z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ+x+y−0.5z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ,=g1x,y,z+−0.5x+y+z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ+x−0.5y+z+g1x,y,ztδΓ1+δ−0.5+g1′x,y,ztδΓ1+δ+x+y−0.5z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ,=h1x,y,z+−0.5x+y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x−0.5y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x+y−0.5z+h1x,y,ztδΓ1+δ−0.5+h1′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ,we know that
(91)Resυ1x,y,z,0=0,Resϑ1x,y,z,0=0,Resϖ1x,y,z,0=0,put in Eq. (90), we obtain
(92)f1x,y,z=−2.25x,g1x,y,z=−2.25y,h1x,y,z=−2.25z.(2) 2nd Iteration. Put k=2 in Eq. (84), we obtain
(93)υ2x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,where
(94)υx,y,z,0=fx,y,z=−0.5x+y+z,f1x,y,z=−2.25xϑx,y,z,0=gx,y,z=x−0.5y+z,g1x,y,z=−2.25y,ϖx,y,z,0=hx,y,z=x+y−0.5z,h1x,y,z=−2.25z,υ2x,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,put k=2 in the Eq. (88), we obtain
(95)Resυ2x,y,z,t=∂δυ2∂tδ+υ2∂υ2∂x+ϑ2∂υ2∂y+ϖ2∂υ2∂z−∂2υ2∂x2−∂2υ2∂y2−∂2υ2∂z2,Resϑ2x,y,z,t=∂δϑ2∂tδ+υ2∂ϑ2∂x+ϑ2∂ϑ2∂y+ϖ2∂ϑ2∂z−∂2ϑ2∂x2−∂2ϑ2∂y2−∂2ϑ2∂z2,Resϖ2x,y,z,t=∂δϖ2∂tδ+υ2∂ϖ2∂x+ϑ2∂ϖ2∂y+ϖ2∂ϖ2∂z−∂2ϖ2∂x2−∂2ϖ2∂y2−∂2ϖ2∂z2,(96)Resυ2x,y,z,t=−2.25x+f2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ,Resϑ2x,y,z,t=−2.25y+g2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ,Resϖ2x,y,z,t=−2.25z+h2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ,we know that
(97)Dtk−1δResυkx,y,z,t=0,Dtk−1δResϑkx,y,z,t=0,Dtk−1δResϖkx,y,z,t=0,put k=2 in Eq. (97), we obtain
(98)DtδResυ2x,y,z,t=0,DtδResϑ2x,y,z,t=0,DtδResϖ2x,y,z,t=0,applying Dtδ, on both sides of the Eq. (96), we get
(99)DtδResυ2x,y,z,t=f2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−2.25+f2′x,y,ztδΓ1+δ+−2.25x+f2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25z+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ,DtδResϑ2x,y,z,t=g2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25x+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−2.25+g2′x,y,ztδΓ1+δ+−2.25y+g2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25z+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−g2′′x,y,ztδΓ1+δ,DtδResϖ2x,y,z,t=h2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25x+h2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−2.25+h2′x,y,ztδΓ1+δ+−2.25z+h2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ,we know that
(100)DtδResυ2x,y,z,0=0,DtδResϑ2x,y,z,0=0,DtδResϖ2x,y,z,0=0,put in Eq. (99), we obtain
(101)f2x,y,z=−2.25x+4.5y+4.5z,g2x,y,z=4.5x−2.25y+4.5z,h2x,y,z=4.5x+4.5y−2.25z.The solution of Eq. (79), in term of RPSM is given by
(102)υx,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ+⋯,ϑx,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ+⋯,ϖx,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ+⋯,υx,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+−2.25x+4.5y+4.5zt2δΓ1+2δ+⋯,ϑx,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+4.5x−2.25y+4.5zt2δΓ1+2δ+⋯,ϖx,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+4.5x+4.5y−2.25zt2δΓ1+2δ+⋯.


#### 4.2.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (79), and simplifying we obtain
(103)Lυx,y,z,t−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2=0,Lϑx,y,z,t−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2=0,Lϖx,y,z,t−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2=0.The nonlinear termN is defined as
(104)Nθx,y,z,t;q=Lθx,y,z,t;q−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2.Use the procedure of q-HATM(105)υmx,y,z,t=kmυm−1x,y,z,t+hL−1Rmυm−1,ϑmx,y,z,t=kmϑm−1x,y,z,t+hL−1Rmϑm−1,ϖmx,y,z,t=kmϖm−1x,y,z,t+hL−1Rmϖm−1,put m=1 in the Eq. (105), we obtain
(106)υ1x,y,z,t=k1υ0x,y,z,t+hL−1R1υ0,ϑ1x,y,z,t=k1ϑ0x,y,z,t+hL−1R1ϑ0,ϖ1x,y,z,t=k1ϖ0x,y,z,t+hL−1R1ϖ0,(107)Rmυm−1=Lυm−1−1−kmn−0.5x+y+zs+1sδL∑i=0m−1υi∂υm−1−i∂x+∑i=0m−1ϑi∂υm−1−i∂y+∑i=0m−1ϖi∂υm−1−i∂z−∂2υm−1∂x2−∂2υm−1∂y2−∂2υm−1∂z2,Rmϑm−1=Lϑm−1−1−kmnx−0.5y+zs+1sδL∑i=0m−1υi∂ϑm−1−i∂x+∑i=0m−1ϑi∂ϑm−1−i∂y+∑i=0m−1ϖi∂ϑm−1−i∂z−∂2ϑm−1∂x2−∂2ϑm−1∂y2−∂2ϑm−1∂z2,Rmϖm−1=Lϖm−1−1−kmnx+y−0.5zs+1sδL∑i=0m−1υi∂ϖm−1−i∂x+∑i=0m−1ϑi∂ϖm−1−i∂y+∑i=0m−1ϖi∂ϖm−1−i∂z−∂2ϖm−1∂x2−∂2ϖm−1∂y2−∂2ϖm−1∂z2,put m=1 in Eq. (107), we get
(108)R1υ0=Lυ0−1−k1n−0.5x+y+zs+1sδLυ0∂υ0∂x+ϑ0∂υ0∂y+ϖ0∂υ0∂z−∂2υ0∂x2−∂2υ0∂y2−∂2υ0∂z2,R1ϑ0=Lϑ0−1−k1nx−0.5y+zs+1sδLυ0∂ϑ0∂x+ϑ0∂ϑ0∂y+ϖ0∂ϑ0∂z−∂2ϑ0∂x2−∂2ϑ0∂y2−∂2ϑ0∂z2,R1ϖ0=Lϖ0−1−k1nx+y−0.5zs+1sδLυ0∂ϖ0∂x+ϑ0∂ϖ0∂y+ϖ0∂ϖ0∂z−∂2ϖ0∂x2−∂2ϖ0∂y2−∂2ϖ0∂z2,=1sδL2.25x=2.25xsδ+1,=1sδL2.25y=2.25ysδ+1,=1sδL2.25z=2.25zsδ+1.Put in Eq. (106), we obtain
(109)υ1x,y,z,t=2.25hxtδΓδ+1,ϑ1x,y,z,t=2.25hytδΓδ+1,ϖ1x,y,z,t=2.25hztδΓδ+1.(2) 2nd Iteration. Put m=2 in the Eq. (105), we obtain
(110)υ2x,y,z,t=k2υ1x,y,z,t+hL−1R2υ1,ϑ2x,y,z,t=k2ϑ1x,y,z,t+hL−1R2ϑ1,ϖ2x,y,z,t=k2ϖ1x,y,z,t+hL−1R2ϖ1,put m=2 in Eq. (107), we obtain
(111)R2υ1=Lυ1−1−k2n−0.5x+y+zs+1sδLυ0∂υ1∂x+υ1∂υ0∂x+ϑ0∂υ1∂y+ϑ1∂υ0∂y+ϖ0∂υ1∂z+ϖ1∂υ0∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,R2ϑ1=Lϑ1−1−k2nx−0.5y+zs+1sδLυ0∂ϑ1∂x+υ1∂ϑ0∂x+ϑ0∂ϑ1∂y+ϑ1∂ϑ0∂y+ϖ0∂ϑ1∂z+ϖ1∂ϑ0∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,R2ϖ1=Lϖ1−1−k2nx+y−0.5zs+1sδLυ0∂ϖ1∂x+υ1∂ϖ0∂x+ϑ0∂ϖ1∂y+ϑ1∂ϖ0∂y+ϖ0∂ϖ1∂z+ϖ1∂ϖ0∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,after simplification we obtain
(112)R2υ1=4.5hysδ+1+4.5hzsδ+1,R2ϑ1=4.5hxsδ+1+4.5hzsδ+1,R2ϖ1=4.5hxsδ+1+4.5hysδ+1.Put Eq. (112) in Eq. (110), we obtain
(113)υ2x,y,z,t=2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑ2x,y,z,t=2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖ2x,y,z,t=2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.The solution of Eq. (79), in term of q-HATM is given by
(114)υx,y,z,t=υ0x,y,z,t+υ1x,y,z,t+υ2x,y,z,t,ϑx,y,z,t=ϑ0x,y,z,t+ϑ1x,y,z,t+ϑ2x,y,z,t,ϖx,y,z,t=ϖ0x,y,z,t+ϖ1x,y,z,t+ϖ2x,y,z,t,υx,y,z,t=−0.5x+y+z+2.25xtδΓ1+δ+2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑx,y,z,t=x−0.5y+z+2.25ytδΓ1+δ+2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖx,y,z,t=x+y−0.5z+2.25ztδΓ1+δ+2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.


## 4.1. Example

Consider nonlinear advection-diffusion equation of the form [64]
(36)∂δυ∂tδ=−υ∂υ∂x+υ−υ2,0<δ≤1,with ICs
(37)υx,0=e−x,and exact solution at δ=1(38)υx,t=et−x.


### 4.1.1. RPSM-Solution

(1) 1st Iteration. Using RPSM, the kth truncated series of Eq. (36), can obtain
(39)υkx,t=∑n=0kfnxtnδΓ1+nδ,the 1st approximate is given as
(40)υ0x,t=υx,0=fx,Eq. (39) should be written as
(41)υkx,t=fx+∑n=1kfnxtnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (41), we get
(42)υ1x,t=fx+f1xtδΓ1+δ,where υx,0=fx=e−x,
(43)υ1x,t=e−x+f1xtδΓ1+δ.The residual function of Eq. (36), is given by
(44)Resυx,t=∂δυ∂tδ+υ∂υ∂x−υ+υ2.Thekth residual function Resυx,t is given by,
(45)Resυkx,t=∂δυk∂tδ+υk∂υk∂x−υk+υk2,put k=1 in the Eq. (45), we obtain
(46)Resυ1x,t=f1x+e−x+f1xtδΓ1+δ−e−x+f1′xtδΓ1+δ−e−x+f1xtδΓ1+δ+e−2x+f1xe−xtδΓ1+δ+2e−xf1xe−xtδΓ1+δ,we have
(47)Resυ1x,0=0.Eq. (46), will become
(48)f1x=e−x.(2) 2nd Iteration. for k=2 Eq. (41), can be written as
(49)υ2x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ,where fx=e−x, and f1x=e−x,
(50)υ2x,t=e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ,put k=2 in Eq. (45), we obtain
(51)Resυ2x,t=∂δυ2∂tδ+υ2∂υ2∂x−υ2+υ12,(52)Resυ2x,t=e−x+f2xtδΓ1+δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ−e−x−e−xtδΓ1+δ+f2′xt2δΓ1+2δ−e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ2,as we know that
(53)Dtk−1δResυkx,t=0,for k=2 Eq. (53), become as
(54)DtδResυ2x,t=0,using Dtδ on Eq. (52), we get
(55)DtδResυ2x,t=f2x+e−x+f2xtδΓ1+δ−e−x+f2′xtδΓ1+δ−e−x+f2xtδΓ1+δ+e−x+f2xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (55), we obtain
(56)f2x=e−x.(3) 3rd Iteration. Put k=3 in Eq. (41), we obtain
(57)υ3x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ,where fx=e−x, f1x=e−x, and f2x=e−x,
(58)υ3x,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ,put k=3 in Eq. (45) we obtain
(59)Resυ3x,t=∂δυ3∂tδ+υ3∂υ3∂x−υ3+υ32,(60)Resυ3x,t=e−x+e−xtδΓ1+δ+f3xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ−e−x−e−xtδΓ1+δ−e−xt2δΓ1+2δ+f3′xt3δΓ1+3δ−e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ2,put k=3 in Eq. (53), we obtain
(61)Dt2δResυ3x,t=0,applying Dt2δ on both sides of the Eq. (60), we get
(62)Dt2δResυ2x,t=f3x+e−x+f3xtδΓ1+δ−e−x+f3′xtδΓ1+δ−e−x+f3xtδΓ1+δ+e−x+f3xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (62), we obtain
(63)f2x=e−x.The RPSM solution of Eq. (36), is given as
(64)υx,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ+⋯,υx,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+e−xt3δΓ1+3δ+⋯.


### 4.1.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (36), and simplifying
(65)sδLυx,t−∑k=0n−1sδ−k−1υkx,0+Lυ∂υ∂x−υ+υ2=0,Lυx,t−1sδsδ−1υ0x,0+1sδLυ∂υ∂x−υ+υ2=0,Lυx,t−e−xs+1sδLυ∂υ∂x−υ+υ2=0.The nonlinear termN is defined as
(66)Nθx,t;q=Lθx,t;q−e−xs+1sδLυ∂υ∂x−υ+υ2.Using the procedure of q-HATM(67)υmx,t=kmυm−1x,t+hL−1Rmυm−1,put m=1 in the Eq. (67), we obtain
(68)υ1x,t=k1υ0x,t+hL−1R1υ0,(69)Rmυm−1=Lυm−1−1−kmne−xs+1sδL∑i=0m−1υi∂υm−1−i∂x−υm−1+∑i=0m−1υiυm−1−i,put m=1 in Eq. (69), we obtain
(70)R1υ0=Lυ0−1−k1ne−xs+1sδLυ0∂υ0∂x−υ0+υ0υ0,R1υ0=Le−x−1−k1ne−xs+1sδLe−x∂∂xe−x−e−x+e−xe−x,R1υ0=e−xs−e−xs+1sδL−e−2x−e−x+e−2x,R1υ0=−e−xsδ+1,put in Eq. (68), we obtain
(71)υ1x,t=hL−1−e−xsδ+1,υ1x,t=−e−xhtδΓ1+δ.(2) 2nd Iteration. Put m=2 in the Eq. (67), we obtain
(72)υ2x,t=k2υ1x,t+hL−1R2υ1,put m=2 in Eq. (69), we obtain
(73)R2υ1=Lυ1−1−k2ne−xs+1sδLυ0∂υ1∂x+υ1∂υ0∂x−υ1+υ1υ0+υ0υ1,=−e−xhsδ+1+e−xhts2δ+1,put in Eq. (72), we obtain
(74)υ2x,t=−e−xnhtδΓ1+δ+hL−1−e−xhsδ+1+e−xhts2δ+1,υ2x,t=−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1.(3) 3rd Iteration. Put m=3 in the Eq. (67), we obtain
(75)υ3x,t=k3υ2x,t+hL−1R3υ2,put m=3 in Eq. (69), we obtain
(76)R3υ2=Lυ2−e−xs+1sδLυ0∂υ2∂x+υ1∂υ1∂x+υ2∂υ0∂x−υ2+υ0υ2+υ1υ1+υ2υ0,=−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,put in Eq. (75), we obtain
(77)υ3x,t=n−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1+hL−1−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,υ3x,t=−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1.The q-HATM solution of Eq. (36), is given as
(78)υx,t=υ0x,t+υ1x,t+υ2x,t+υ3x,t+⋯,υx,t=e−x−e−xhtδΓ1+δ−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1+⋯.


## 4.1.1. RPSM-Solution

(1) 1st Iteration. Using RPSM, the kth truncated series of Eq. (36), can obtain
(39)υkx,t=∑n=0kfnxtnδΓ1+nδ,the 1st approximate is given as
(40)υ0x,t=υx,0=fx,Eq. (39) should be written as
(41)υkx,t=fx+∑n=1kfnxtnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (41), we get
(42)υ1x,t=fx+f1xtδΓ1+δ,where υx,0=fx=e−x,
(43)υ1x,t=e−x+f1xtδΓ1+δ.The residual function of Eq. (36), is given by
(44)Resυx,t=∂δυ∂tδ+υ∂υ∂x−υ+υ2.Thekth residual function Resυx,t is given by,
(45)Resυkx,t=∂δυk∂tδ+υk∂υk∂x−υk+υk2,put k=1 in the Eq. (45), we obtain
(46)Resυ1x,t=f1x+e−x+f1xtδΓ1+δ−e−x+f1′xtδΓ1+δ−e−x+f1xtδΓ1+δ+e−2x+f1xe−xtδΓ1+δ+2e−xf1xe−xtδΓ1+δ,we have
(47)Resυ1x,0=0.Eq. (46), will become
(48)f1x=e−x.(2) 2nd Iteration. for k=2 Eq. (41), can be written as
(49)υ2x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ,where fx=e−x, and f1x=e−x,
(50)υ2x,t=e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ,put k=2 in Eq. (45), we obtain
(51)Resυ2x,t=∂δυ2∂tδ+υ2∂υ2∂x−υ2+υ12,(52)Resυ2x,t=e−x+f2xtδΓ1+δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ−e−x−e−xtδΓ1+δ+f2′xt2δΓ1+2δ−e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+f2xt2δΓ1+2δ2,as we know that
(53)Dtk−1δResυkx,t=0,for k=2 Eq. (53), become as
(54)DtδResυ2x,t=0,using Dtδ on Eq. (52), we get
(55)DtδResυ2x,t=f2x+e−x+f2xtδΓ1+δ−e−x+f2′xtδΓ1+δ−e−x+f2xtδΓ1+δ+e−x+f2xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (55), we obtain
(56)f2x=e−x.(3) 3rd Iteration. Put k=3 in Eq. (41), we obtain
(57)υ3x,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ,where fx=e−x, f1x=e−x, and f2x=e−x,
(58)υ3x,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ,put k=3 in Eq. (45) we obtain
(59)Resυ3x,t=∂δυ3∂tδ+υ3∂υ3∂x−υ3+υ32,(60)Resυ3x,t=e−x+e−xtδΓ1+δ+f3xt2δΓ1+2δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ−e−x−e−xtδΓ1+δ−e−xt2δΓ1+2δ+f3′xt3δΓ1+3δ−e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ+e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+f3xt3δΓ1+3δ2,put k=3 in Eq. (53), we obtain
(61)Dt2δResυ3x,t=0,applying Dt2δ on both sides of the Eq. (60), we get
(62)Dt2δResυ2x,t=f3x+e−x+f3xtδΓ1+δ−e−x+f3′xtδΓ1+δ−e−x+f3xtδΓ1+δ+e−x+f3xtδΓ1+δ2,put DtδResυ2x,0=0 in Eq. (62), we obtain
(63)f2x=e−x.The RPSM solution of Eq. (36), is given as
(64)υx,t=fx+f1xtδΓ1+δ+f2xt2δΓ1+2δ+f3xt3δΓ1+3δ+⋯,υx,t=e−x+e−xtδΓ1+δ+e−xt2δΓ1+2δ+e−xt3δΓ1+3δ+⋯.


## 4.1.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (36), and simplifying
(65)sδLυx,t−∑k=0n−1sδ−k−1υkx,0+Lυ∂υ∂x−υ+υ2=0,Lυx,t−1sδsδ−1υ0x,0+1sδLυ∂υ∂x−υ+υ2=0,Lυx,t−e−xs+1sδLυ∂υ∂x−υ+υ2=0.The nonlinear termN is defined as
(66)Nθx,t;q=Lθx,t;q−e−xs+1sδLυ∂υ∂x−υ+υ2.Using the procedure of q-HATM(67)υmx,t=kmυm−1x,t+hL−1Rmυm−1,put m=1 in the Eq. (67), we obtain
(68)υ1x,t=k1υ0x,t+hL−1R1υ0,(69)Rmυm−1=Lυm−1−1−kmne−xs+1sδL∑i=0m−1υi∂υm−1−i∂x−υm−1+∑i=0m−1υiυm−1−i,put m=1 in Eq. (69), we obtain
(70)R1υ0=Lυ0−1−k1ne−xs+1sδLυ0∂υ0∂x−υ0+υ0υ0,R1υ0=Le−x−1−k1ne−xs+1sδLe−x∂∂xe−x−e−x+e−xe−x,R1υ0=e−xs−e−xs+1sδL−e−2x−e−x+e−2x,R1υ0=−e−xsδ+1,put in Eq. (68), we obtain
(71)υ1x,t=hL−1−e−xsδ+1,υ1x,t=−e−xhtδΓ1+δ.(2) 2nd Iteration. Put m=2 in the Eq. (67), we obtain
(72)υ2x,t=k2υ1x,t+hL−1R2υ1,put m=2 in Eq. (69), we obtain
(73)R2υ1=Lυ1−1−k2ne−xs+1sδLυ0∂υ1∂x+υ1∂υ0∂x−υ1+υ1υ0+υ0υ1,=−e−xhsδ+1+e−xhts2δ+1,put in Eq. (72), we obtain
(74)υ2x,t=−e−xnhtδΓ1+δ+hL−1−e−xhsδ+1+e−xhts2δ+1,υ2x,t=−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1.(3) 3rd Iteration. Put m=3 in the Eq. (67), we obtain
(75)υ3x,t=k3υ2x,t+hL−1R3υ2,put m=3 in Eq. (69), we obtain
(76)R3υ2=Lυ2−e−xs+1sδLυ0∂υ2∂x+υ1∂υ1∂x+υ2∂υ0∂x−υ2+υ0υ2+υ1υ1+υ2υ0,=−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,put in Eq. (75), we obtain
(77)υ3x,t=n−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1+hL−1−e−xnhsδ+1−e−xh2sδ+1+2e−xh2s2δ+1+e−xnhs2δ+1−e−xh2s3δ+1,υ3x,t=−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1.The q-HATM solution of Eq. (36), is given as
(78)υx,t=υ0x,t+υ1x,t+υ2x,t+υ3x,t+⋯,υx,t=e−x−e−xhtδΓ1+δ−e−xnhtδΓ1+δ−e−xh2tδΓδ+1+e−xh2t2δΓ2δ+1−e−xn2htδΓ1+δ−e−xnh2tδΓδ+1+e−xnh2t2δΓ2δ+1−e−xnh2tδΓδ+1−e−xh3tδΓδ+1+2e−xh3t2δΓ2δ+1+e−xnh2t2δΓ2δ+1−e−xh3t3δΓ3δ+1+⋯.


## 4.2. Example

The system of 3D Burgers’ equation are [65, 66]
(79)υtδ+υυx+ϑυy+ϖυz−υxx−υyy−υzz=0,ϑtδ+υϑx+ϑϑy+ϖϑz−ϑxx−ϑyy−ϑzz=0,ϖtδ+υϖx+ϑϖy+ϖϖz−ϖxx−ϖyy−ϖzz=0,the initial condition of Eq. (79), are
(80)υx,y,z,0=−0.5x+y+z,ϑx,y,z,0=x−0.5y+z,ϖx,y,z,0=x+y−0.5z.The exact solution of the Eq. (79) at δ=1, are
(81)υx,y,z,t=−0.5x+y+z−2.25xt1−2.25t2,ϑx,y,z,t=x−0.5y+z−2.25yt1−2.25t2,ϖx,y,z,t=x+y−0.5z−2.25zt1−2.25t2.


### 4.2.1. RPSM-Solution

(1) 1st Iteration. The kth truncated series of the solution of Eq. (79), using RPSM, we obtain
(82)υkx,y,z,t=∑n=0kfnx,y,ztnδΓ1+nδ,ϑkx,y,z,t=∑n=0kgnx,y,ztnδΓ1+nδ,ϖkx,y,z,t=∑n=0khnx,y,ztnδΓ1+nδ,the zeroth RPSM approximate solution of Eq. (79), is given by
(83)υ0x,y,z,t=υx,y,z,0=fx,y,z,ϑ0x,y,z,t=ϑx,y,z,0=gx,y,z,ϖ0x,y,z,t=ϖx,y,z,0=hx,y,z,the Eq. (82), should be written as
(84)υkx,y,z,t=fx,y,z+∑n=1kfnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϑkx,y,z,t=gx,y,z+∑n=1kgnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϖkx,y,z,t=hx,y,z+∑n=1khnx,y,ztnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (84), we get
(85)υ1x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,where
(86)υx,y,z,0=fx,y,z=−0.5x+y+z,ϑx,y,z,0=gx,y,z=x−0.5y+z,ϖx,y,z,0=hx,y,z=x+y−0.5z,υ1x,y,z,t=−0.5x+y+z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=x−0.5y+z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=x+y−0.5z+h1x,y,ztδΓ1+δ.The residual function of Eq. (79), is given by
(87)Resυx,y,z,t=∂δυ∂tδ+υ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Resϑx,y,z,t=∂δϑ∂tδ+υ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Resϖx,y,z,t=∂δϖ∂tδ+υ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2,where the kth residual function of Resυx,y,t, Resϑx,y,t, and Resϖx,y,t is given by
(88)Resυkx,y,z,t=∂δυk∂tδ+υk∂υk∂x+ϑk∂υk∂y+ϖk∂υk∂z−∂2υk∂x2−∂2υk∂y2−∂2υk∂z2,Resϑkx,y,z,t=∂δϑk∂tδ+υk∂ϑk∂x+ϑk∂ϑk∂y+ϖk∂ϑk∂z−∂2ϑk∂x2−∂2ϑk∂y2−∂2ϑk∂z2,Resϖkx,y,z,t=∂δϖk∂tδ+υk∂ϖk∂x+ϑk∂ϖk∂y+ϖk∂ϖk∂z−∂2ϖk∂x2−∂2ϖk∂y2−∂2ϖk∂z2,put k=1 in the Eq. (88), we obtain
(89)Resυ1x,y,z,t=∂δυ1∂tδ+υ1∂υ1∂x+ϑ1∂υ1∂y+ϖ1∂υ1∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,Resϑ1x,y,z,t=∂δϑ1∂tδ+υ1∂ϑ1∂x+ϑ1∂ϑ1∂y+ϖ1∂ϑ1∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,Resϖ1x,y,z,t=∂δϖ1∂tδ+υ1∂ϖ1∂x+ϑ1∂ϖ1∂y+ϖ1∂ϖ1∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,(90)=f1x,y,z+−0.5x+y+z+f1x,y,ztδΓ1+δ−0.5+f1′x,y,ztδΓ1+δ+x−0.5y+z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ+x+y−0.5z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ,=g1x,y,z+−0.5x+y+z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ+x−0.5y+z+g1x,y,ztδΓ1+δ−0.5+g1′x,y,ztδΓ1+δ+x+y−0.5z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ,=h1x,y,z+−0.5x+y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x−0.5y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x+y−0.5z+h1x,y,ztδΓ1+δ−0.5+h1′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ,we know that
(91)Resυ1x,y,z,0=0,Resϑ1x,y,z,0=0,Resϖ1x,y,z,0=0,put in Eq. (90), we obtain
(92)f1x,y,z=−2.25x,g1x,y,z=−2.25y,h1x,y,z=−2.25z.(2) 2nd Iteration. Put k=2 in Eq. (84), we obtain
(93)υ2x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,where
(94)υx,y,z,0=fx,y,z=−0.5x+y+z,f1x,y,z=−2.25xϑx,y,z,0=gx,y,z=x−0.5y+z,g1x,y,z=−2.25y,ϖx,y,z,0=hx,y,z=x+y−0.5z,h1x,y,z=−2.25z,υ2x,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,put k=2 in the Eq. (88), we obtain
(95)Resυ2x,y,z,t=∂δυ2∂tδ+υ2∂υ2∂x+ϑ2∂υ2∂y+ϖ2∂υ2∂z−∂2υ2∂x2−∂2υ2∂y2−∂2υ2∂z2,Resϑ2x,y,z,t=∂δϑ2∂tδ+υ2∂ϑ2∂x+ϑ2∂ϑ2∂y+ϖ2∂ϑ2∂z−∂2ϑ2∂x2−∂2ϑ2∂y2−∂2ϑ2∂z2,Resϖ2x,y,z,t=∂δϖ2∂tδ+υ2∂ϖ2∂x+ϑ2∂ϖ2∂y+ϖ2∂ϖ2∂z−∂2ϖ2∂x2−∂2ϖ2∂y2−∂2ϖ2∂z2,(96)Resυ2x,y,z,t=−2.25x+f2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ,Resϑ2x,y,z,t=−2.25y+g2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ,Resϖ2x,y,z,t=−2.25z+h2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ,we know that
(97)Dtk−1δResυkx,y,z,t=0,Dtk−1δResϑkx,y,z,t=0,Dtk−1δResϖkx,y,z,t=0,put k=2 in Eq. (97), we obtain
(98)DtδResυ2x,y,z,t=0,DtδResϑ2x,y,z,t=0,DtδResϖ2x,y,z,t=0,applying Dtδ, on both sides of the Eq. (96), we get
(99)DtδResυ2x,y,z,t=f2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−2.25+f2′x,y,ztδΓ1+δ+−2.25x+f2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25z+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ,DtδResϑ2x,y,z,t=g2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25x+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−2.25+g2′x,y,ztδΓ1+δ+−2.25y+g2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25z+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−g2′′x,y,ztδΓ1+δ,DtδResϖ2x,y,z,t=h2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25x+h2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−2.25+h2′x,y,ztδΓ1+δ+−2.25z+h2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ,we know that
(100)DtδResυ2x,y,z,0=0,DtδResϑ2x,y,z,0=0,DtδResϖ2x,y,z,0=0,put in Eq. (99), we obtain
(101)f2x,y,z=−2.25x+4.5y+4.5z,g2x,y,z=4.5x−2.25y+4.5z,h2x,y,z=4.5x+4.5y−2.25z.The solution of Eq. (79), in term of RPSM is given by
(102)υx,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ+⋯,ϑx,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ+⋯,ϖx,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ+⋯,υx,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+−2.25x+4.5y+4.5zt2δΓ1+2δ+⋯,ϑx,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+4.5x−2.25y+4.5zt2δΓ1+2δ+⋯,ϖx,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+4.5x+4.5y−2.25zt2δΓ1+2δ+⋯.


### 4.2.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (79), and simplifying we obtain
(103)Lυx,y,z,t−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2=0,Lϑx,y,z,t−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2=0,Lϖx,y,z,t−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2=0.The nonlinear termN is defined as
(104)Nθx,y,z,t;q=Lθx,y,z,t;q−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2.Use the procedure of q-HATM(105)υmx,y,z,t=kmυm−1x,y,z,t+hL−1Rmυm−1,ϑmx,y,z,t=kmϑm−1x,y,z,t+hL−1Rmϑm−1,ϖmx,y,z,t=kmϖm−1x,y,z,t+hL−1Rmϖm−1,put m=1 in the Eq. (105), we obtain
(106)υ1x,y,z,t=k1υ0x,y,z,t+hL−1R1υ0,ϑ1x,y,z,t=k1ϑ0x,y,z,t+hL−1R1ϑ0,ϖ1x,y,z,t=k1ϖ0x,y,z,t+hL−1R1ϖ0,(107)Rmυm−1=Lυm−1−1−kmn−0.5x+y+zs+1sδL∑i=0m−1υi∂υm−1−i∂x+∑i=0m−1ϑi∂υm−1−i∂y+∑i=0m−1ϖi∂υm−1−i∂z−∂2υm−1∂x2−∂2υm−1∂y2−∂2υm−1∂z2,Rmϑm−1=Lϑm−1−1−kmnx−0.5y+zs+1sδL∑i=0m−1υi∂ϑm−1−i∂x+∑i=0m−1ϑi∂ϑm−1−i∂y+∑i=0m−1ϖi∂ϑm−1−i∂z−∂2ϑm−1∂x2−∂2ϑm−1∂y2−∂2ϑm−1∂z2,Rmϖm−1=Lϖm−1−1−kmnx+y−0.5zs+1sδL∑i=0m−1υi∂ϖm−1−i∂x+∑i=0m−1ϑi∂ϖm−1−i∂y+∑i=0m−1ϖi∂ϖm−1−i∂z−∂2ϖm−1∂x2−∂2ϖm−1∂y2−∂2ϖm−1∂z2,put m=1 in Eq. (107), we get
(108)R1υ0=Lυ0−1−k1n−0.5x+y+zs+1sδLυ0∂υ0∂x+ϑ0∂υ0∂y+ϖ0∂υ0∂z−∂2υ0∂x2−∂2υ0∂y2−∂2υ0∂z2,R1ϑ0=Lϑ0−1−k1nx−0.5y+zs+1sδLυ0∂ϑ0∂x+ϑ0∂ϑ0∂y+ϖ0∂ϑ0∂z−∂2ϑ0∂x2−∂2ϑ0∂y2−∂2ϑ0∂z2,R1ϖ0=Lϖ0−1−k1nx+y−0.5zs+1sδLυ0∂ϖ0∂x+ϑ0∂ϖ0∂y+ϖ0∂ϖ0∂z−∂2ϖ0∂x2−∂2ϖ0∂y2−∂2ϖ0∂z2,=1sδL2.25x=2.25xsδ+1,=1sδL2.25y=2.25ysδ+1,=1sδL2.25z=2.25zsδ+1.Put in Eq. (106), we obtain
(109)υ1x,y,z,t=2.25hxtδΓδ+1,ϑ1x,y,z,t=2.25hytδΓδ+1,ϖ1x,y,z,t=2.25hztδΓδ+1.(2) 2nd Iteration. Put m=2 in the Eq. (105), we obtain
(110)υ2x,y,z,t=k2υ1x,y,z,t+hL−1R2υ1,ϑ2x,y,z,t=k2ϑ1x,y,z,t+hL−1R2ϑ1,ϖ2x,y,z,t=k2ϖ1x,y,z,t+hL−1R2ϖ1,put m=2 in Eq. (107), we obtain
(111)R2υ1=Lυ1−1−k2n−0.5x+y+zs+1sδLυ0∂υ1∂x+υ1∂υ0∂x+ϑ0∂υ1∂y+ϑ1∂υ0∂y+ϖ0∂υ1∂z+ϖ1∂υ0∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,R2ϑ1=Lϑ1−1−k2nx−0.5y+zs+1sδLυ0∂ϑ1∂x+υ1∂ϑ0∂x+ϑ0∂ϑ1∂y+ϑ1∂ϑ0∂y+ϖ0∂ϑ1∂z+ϖ1∂ϑ0∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,R2ϖ1=Lϖ1−1−k2nx+y−0.5zs+1sδLυ0∂ϖ1∂x+υ1∂ϖ0∂x+ϑ0∂ϖ1∂y+ϑ1∂ϖ0∂y+ϖ0∂ϖ1∂z+ϖ1∂ϖ0∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,after simplification we obtain
(112)R2υ1=4.5hysδ+1+4.5hzsδ+1,R2ϑ1=4.5hxsδ+1+4.5hzsδ+1,R2ϖ1=4.5hxsδ+1+4.5hysδ+1.Put Eq. (112) in Eq. (110), we obtain
(113)υ2x,y,z,t=2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑ2x,y,z,t=2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖ2x,y,z,t=2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.The solution of Eq. (79), in term of q-HATM is given by
(114)υx,y,z,t=υ0x,y,z,t+υ1x,y,z,t+υ2x,y,z,t,ϑx,y,z,t=ϑ0x,y,z,t+ϑ1x,y,z,t+ϑ2x,y,z,t,ϖx,y,z,t=ϖ0x,y,z,t+ϖ1x,y,z,t+ϖ2x,y,z,t,υx,y,z,t=−0.5x+y+z+2.25xtδΓ1+δ+2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑx,y,z,t=x−0.5y+z+2.25ytδΓ1+δ+2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖx,y,z,t=x+y−0.5z+2.25ztδΓ1+δ+2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.


## 4.2.1. RPSM-Solution

(1) 1st Iteration. The kth truncated series of the solution of Eq. (79), using RPSM, we obtain
(82)υkx,y,z,t=∑n=0kfnx,y,ztnδΓ1+nδ,ϑkx,y,z,t=∑n=0kgnx,y,ztnδΓ1+nδ,ϖkx,y,z,t=∑n=0khnx,y,ztnδΓ1+nδ,the zeroth RPSM approximate solution of Eq. (79), is given by
(83)υ0x,y,z,t=υx,y,z,0=fx,y,z,ϑ0x,y,z,t=ϑx,y,z,0=gx,y,z,ϖ0x,y,z,t=ϖx,y,z,0=hx,y,z,the Eq. (82), should be written as
(84)υkx,y,z,t=fx,y,z+∑n=1kfnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϑkx,y,z,t=gx,y,z+∑n=1kgnx,y,ztnδΓ1+nδ,k=1,2,⋯,ϖkx,y,z,t=hx,y,z+∑n=1khnx,y,ztnδΓ1+nδ,k=1,2,⋯,put k=1 in Eq. (84), we get
(85)υ1x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ,where
(86)υx,y,z,0=fx,y,z=−0.5x+y+z,ϑx,y,z,0=gx,y,z=x−0.5y+z,ϖx,y,z,0=hx,y,z=x+y−0.5z,υ1x,y,z,t=−0.5x+y+z+f1x,y,ztδΓ1+δ,ϑ1x,y,z,t=x−0.5y+z+g1x,y,ztδΓ1+δ,ϖ1x,y,z,t=x+y−0.5z+h1x,y,ztδΓ1+δ.The residual function of Eq. (79), is given by
(87)Resυx,y,z,t=∂δυ∂tδ+υ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Resϑx,y,z,t=∂δϑ∂tδ+υ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Resϖx,y,z,t=∂δϖ∂tδ+υ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2,where the kth residual function of Resυx,y,t, Resϑx,y,t, and Resϖx,y,t is given by
(88)Resυkx,y,z,t=∂δυk∂tδ+υk∂υk∂x+ϑk∂υk∂y+ϖk∂υk∂z−∂2υk∂x2−∂2υk∂y2−∂2υk∂z2,Resϑkx,y,z,t=∂δϑk∂tδ+υk∂ϑk∂x+ϑk∂ϑk∂y+ϖk∂ϑk∂z−∂2ϑk∂x2−∂2ϑk∂y2−∂2ϑk∂z2,Resϖkx,y,z,t=∂δϖk∂tδ+υk∂ϖk∂x+ϑk∂ϖk∂y+ϖk∂ϖk∂z−∂2ϖk∂x2−∂2ϖk∂y2−∂2ϖk∂z2,put k=1 in the Eq. (88), we obtain
(89)Resυ1x,y,z,t=∂δυ1∂tδ+υ1∂υ1∂x+ϑ1∂υ1∂y+ϖ1∂υ1∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,Resϑ1x,y,z,t=∂δϑ1∂tδ+υ1∂ϑ1∂x+ϑ1∂ϑ1∂y+ϖ1∂ϑ1∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,Resϖ1x,y,z,t=∂δϖ1∂tδ+υ1∂ϖ1∂x+ϑ1∂ϖ1∂y+ϖ1∂ϖ1∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,(90)=f1x,y,z+−0.5x+y+z+f1x,y,ztδΓ1+δ−0.5+f1′x,y,ztδΓ1+δ+x−0.5y+z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ+x+y−0.5z+f1x,y,ztδΓ1+δ1+f1′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ−f1′′x,y,ztδΓ1+δ,=g1x,y,z+−0.5x+y+z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ+x−0.5y+z+g1x,y,ztδΓ1+δ−0.5+g1′x,y,ztδΓ1+δ+x+y−0.5z+g1x,y,ztδΓ1+δ1+g1′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ−g1′′x,y,ztδΓ1+δ,=h1x,y,z+−0.5x+y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x−0.5y+z+h1x,y,ztδΓ1+δ1+h1′x,y,ztδΓ1+δ+x+y−0.5z+h1x,y,ztδΓ1+δ−0.5+h1′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ−h1′′x,y,ztδΓ1+δ,we know that
(91)Resυ1x,y,z,0=0,Resϑ1x,y,z,0=0,Resϖ1x,y,z,0=0,put in Eq. (90), we obtain
(92)f1x,y,z=−2.25x,g1x,y,z=−2.25y,h1x,y,z=−2.25z.(2) 2nd Iteration. Put k=2 in Eq. (84), we obtain
(93)υ2x,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,where
(94)υx,y,z,0=fx,y,z=−0.5x+y+z,f1x,y,z=−2.25xϑx,y,z,0=gx,y,z=x−0.5y+z,g1x,y,z=−2.25y,ϖx,y,z,0=hx,y,z=x+y−0.5z,h1x,y,z=−2.25z,υ2x,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ,ϑ2x,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ,ϖ2x,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ,put k=2 in the Eq. (88), we obtain
(95)Resυ2x,y,z,t=∂δυ2∂tδ+υ2∂υ2∂x+ϑ2∂υ2∂y+ϖ2∂υ2∂z−∂2υ2∂x2−∂2υ2∂y2−∂2υ2∂z2,Resϑ2x,y,z,t=∂δϑ2∂tδ+υ2∂ϑ2∂x+ϑ2∂ϑ2∂y+ϖ2∂ϑ2∂z−∂2ϑ2∂x2−∂2ϑ2∂y2−∂2ϑ2∂z2,Resϖ2x,y,z,t=∂δϖ2∂tδ+υ2∂ϖ2∂x+ϑ2∂ϖ2∂y+ϖ2∂ϖ2∂z−∂2ϖ2∂x2−∂2ϖ2∂y2−∂2ϖ2∂z2,(96)Resυ2x,y,z,t=−2.25x+f2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ−f2′′x,y,zt2δΓ1+2δ,Resϑ2x,y,z,t=−2.25y+g2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ−g2′′x,y,zt2δΓ1+2δ,Resϖ2x,y,z,t=−2.25z+h2x,y,ztδΓ1+δ+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ−h2′′x,y,zt2δΓ1+2δ,we know that
(97)Dtk−1δResυkx,y,z,t=0,Dtk−1δResϑkx,y,z,t=0,Dtk−1δResϖkx,y,z,t=0,put k=2 in Eq. (97), we obtain
(98)DtδResυ2x,y,z,t=0,DtδResϑ2x,y,z,t=0,DtδResϖ2x,y,z,t=0,applying Dtδ, on both sides of the Eq. (96), we get
(99)DtδResυ2x,y,z,t=f2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+f2x,y,zt2δΓ1+2δ−2.25+f2′x,y,ztδΓ1+δ+−2.25x+f2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+f2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+f2x,y,zt2δΓ1+2δf2′x,y,ztδΓ1+δ+−2.25z+f2x,y,ztδΓ1+δ1+f2′x,y,zt2δΓ1+2δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ,DtδResϑ2x,y,z,t=g2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25x+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+g2x,y,zt2δΓ1+2δ−2.25+g2′x,y,ztδΓ1+δ+−2.25y+g2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+g2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+g2x,y,zt2δΓ1+2δg2′x,y,ztδΓ1+δ+−2.25z+g2x,y,ztδΓ1+δ1+g2′x,y,zt2δΓ1+2δ−g2′′x,y,ztδΓ1+δ−f2′′x,y,ztδΓ1+δ−g2′′x,y,ztδΓ1+δ,DtδResϖ2x,y,z,t=h2x,y,z+−0.5x+y+z−2.25xtδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25x+h2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x−0.5y+z−2.25ytδΓ1+δ+h2x,y,zt2δΓ1+2δh2′x,y,ztδΓ1+δ+−2.25y+f2x,y,ztδΓ1+δ1+h2′x,y,zt2δΓ1+2δ+x+y−0.5z−2.25ztδΓ1+δ+h2x,y,zt2δΓ1+2δ−2.25+h2′x,y,ztδΓ1+δ+−2.25z+h2x,y,ztδΓ1+δ−0.5−2.25tδΓ1+δ+h2′x,y,zt2δΓ1+2δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ−h2′′x,y,ztδΓ1+δ,we know that
(100)DtδResυ2x,y,z,0=0,DtδResϑ2x,y,z,0=0,DtδResϖ2x,y,z,0=0,put in Eq. (99), we obtain
(101)f2x,y,z=−2.25x+4.5y+4.5z,g2x,y,z=4.5x−2.25y+4.5z,h2x,y,z=4.5x+4.5y−2.25z.The solution of Eq. (79), in term of RPSM is given by
(102)υx,y,z,t=fx,y,z+f1x,y,ztδΓ1+δ+f2x,y,zt2δΓ1+2δ+⋯,ϑx,y,z,t=gx,y,z+g1x,y,ztδΓ1+δ+g2x,y,zt2δΓ1+2δ+⋯,ϖx,y,z,t=hx,y,z+h1x,y,ztδΓ1+δ+h2x,y,zt2δΓ1+2δ+⋯,υx,y,z,t=−0.5x+y+z−2.25xtδΓ1+δ+−2.25x+4.5y+4.5zt2δΓ1+2δ+⋯,ϑx,y,z,t=x−0.5y+z−2.25ytδΓ1+δ+4.5x−2.25y+4.5zt2δΓ1+2δ+⋯,ϖx,y,z,t=x+y−0.5z−2.25ztδΓ1+δ+4.5x+4.5y−2.25zt2δΓ1+2δ+⋯.


## 4.2.2. Q-HATM Solution

(1) 1st Iteration. Taking LT of Eq. (79), and simplifying we obtain
(103)Lυx,y,z,t−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2=0,Lϑx,y,z,t−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2=0,Lϖx,y,z,t−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2=0.The nonlinear termN is defined as
(104)Nθx,y,z,t;q=Lθx,y,z,t;q−−0.5x+y+zs+1sδLυ∂υ∂x+ϑ∂υ∂y+ϖ∂υ∂z−∂2υ∂x2−∂2υ∂y2−∂2υ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x−0.5y+zs+1sδLυ∂ϑ∂x+ϑ∂ϑ∂y+ϖ∂ϑ∂z−∂2ϑ∂x2−∂2ϑ∂y2−∂2ϑ∂z2,Nθx,y,z,t;q=Lθx,y,z,t;q−x+y−0.5zs+1sδLυ∂ϖ∂x+ϑ∂ϖ∂y+ϖ∂ϖ∂z−∂2ϖ∂x2−∂2ϖ∂y2−∂2ϖ∂z2.Use the procedure of q-HATM(105)υmx,y,z,t=kmυm−1x,y,z,t+hL−1Rmυm−1,ϑmx,y,z,t=kmϑm−1x,y,z,t+hL−1Rmϑm−1,ϖmx,y,z,t=kmϖm−1x,y,z,t+hL−1Rmϖm−1,put m=1 in the Eq. (105), we obtain
(106)υ1x,y,z,t=k1υ0x,y,z,t+hL−1R1υ0,ϑ1x,y,z,t=k1ϑ0x,y,z,t+hL−1R1ϑ0,ϖ1x,y,z,t=k1ϖ0x,y,z,t+hL−1R1ϖ0,(107)Rmυm−1=Lυm−1−1−kmn−0.5x+y+zs+1sδL∑i=0m−1υi∂υm−1−i∂x+∑i=0m−1ϑi∂υm−1−i∂y+∑i=0m−1ϖi∂υm−1−i∂z−∂2υm−1∂x2−∂2υm−1∂y2−∂2υm−1∂z2,Rmϑm−1=Lϑm−1−1−kmnx−0.5y+zs+1sδL∑i=0m−1υi∂ϑm−1−i∂x+∑i=0m−1ϑi∂ϑm−1−i∂y+∑i=0m−1ϖi∂ϑm−1−i∂z−∂2ϑm−1∂x2−∂2ϑm−1∂y2−∂2ϑm−1∂z2,Rmϖm−1=Lϖm−1−1−kmnx+y−0.5zs+1sδL∑i=0m−1υi∂ϖm−1−i∂x+∑i=0m−1ϑi∂ϖm−1−i∂y+∑i=0m−1ϖi∂ϖm−1−i∂z−∂2ϖm−1∂x2−∂2ϖm−1∂y2−∂2ϖm−1∂z2,put m=1 in Eq. (107), we get
(108)R1υ0=Lυ0−1−k1n−0.5x+y+zs+1sδLυ0∂υ0∂x+ϑ0∂υ0∂y+ϖ0∂υ0∂z−∂2υ0∂x2−∂2υ0∂y2−∂2υ0∂z2,R1ϑ0=Lϑ0−1−k1nx−0.5y+zs+1sδLυ0∂ϑ0∂x+ϑ0∂ϑ0∂y+ϖ0∂ϑ0∂z−∂2ϑ0∂x2−∂2ϑ0∂y2−∂2ϑ0∂z2,R1ϖ0=Lϖ0−1−k1nx+y−0.5zs+1sδLυ0∂ϖ0∂x+ϑ0∂ϖ0∂y+ϖ0∂ϖ0∂z−∂2ϖ0∂x2−∂2ϖ0∂y2−∂2ϖ0∂z2,=1sδL2.25x=2.25xsδ+1,=1sδL2.25y=2.25ysδ+1,=1sδL2.25z=2.25zsδ+1.Put in Eq. (106), we obtain
(109)υ1x,y,z,t=2.25hxtδΓδ+1,ϑ1x,y,z,t=2.25hytδΓδ+1,ϖ1x,y,z,t=2.25hztδΓδ+1.(2) 2nd Iteration. Put m=2 in the Eq. (105), we obtain
(110)υ2x,y,z,t=k2υ1x,y,z,t+hL−1R2υ1,ϑ2x,y,z,t=k2ϑ1x,y,z,t+hL−1R2ϑ1,ϖ2x,y,z,t=k2ϖ1x,y,z,t+hL−1R2ϖ1,put m=2 in Eq. (107), we obtain
(111)R2υ1=Lυ1−1−k2n−0.5x+y+zs+1sδLυ0∂υ1∂x+υ1∂υ0∂x+ϑ0∂υ1∂y+ϑ1∂υ0∂y+ϖ0∂υ1∂z+ϖ1∂υ0∂z−∂2υ1∂x2−∂2υ1∂y2−∂2υ1∂z2,R2ϑ1=Lϑ1−1−k2nx−0.5y+zs+1sδLυ0∂ϑ1∂x+υ1∂ϑ0∂x+ϑ0∂ϑ1∂y+ϑ1∂ϑ0∂y+ϖ0∂ϑ1∂z+ϖ1∂ϑ0∂z−∂2ϑ1∂x2−∂2ϑ1∂y2−∂2ϑ1∂z2,R2ϖ1=Lϖ1−1−k2nx+y−0.5zs+1sδLυ0∂ϖ1∂x+υ1∂ϖ0∂x+ϑ0∂ϖ1∂y+ϑ1∂ϖ0∂y+ϖ0∂ϖ1∂z+ϖ1∂ϖ0∂z−∂2ϖ1∂x2−∂2ϖ1∂y2−∂2ϖ1∂z2,after simplification we obtain
(112)R2υ1=4.5hysδ+1+4.5hzsδ+1,R2ϑ1=4.5hxsδ+1+4.5hzsδ+1,R2ϖ1=4.5hxsδ+1+4.5hysδ+1.Put Eq. (112) in Eq. (110), we obtain
(113)υ2x,y,z,t=2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑ2x,y,z,t=2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖ2x,y,z,t=2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.The solution of Eq. (79), in term of q-HATM is given by
(114)υx,y,z,t=υ0x,y,z,t+υ1x,y,z,t+υ2x,y,z,t,ϑx,y,z,t=ϑ0x,y,z,t+ϑ1x,y,z,t+ϑ2x,y,z,t,ϖx,y,z,t=ϖ0x,y,z,t+ϖ1x,y,z,t+ϖ2x,y,z,t,υx,y,z,t=−0.5x+y+z+2.25xtδΓ1+δ+2.25nhxtδΓδ+1+4.5h2ytδΓδ+1+4.5h2ztδΓδ+1,ϑx,y,z,t=x−0.5y+z+2.25ytδΓ1+δ+2.25nhytδΓδ+1+4.5h2xtδΓδ+1+4.5h2ztδΓδ+1,ϖx,y,z,t=x+y−0.5z+2.25ztδΓ1+δ+2.25nhztδΓδ+1+4.5h2xtδΓδ+1+4.5h2ytδΓδ+1.


## 5. Results and Discussion

Here, we will discuss the numerical solutions. In Figure1, 3D plots of (a) RPSM, (b) Exact, and (c) q-HATM ψ−solutions at δ=1 of Example 4.1 are presented. Which are in closed contact with the exact solution of Example 4.1. In Figure 2, 2D plots of (a) RPSM, (b) Exact, and (c) q-HATM ψ−solutions at δ=1 of Example 4.1 are presented, from which the validity of the proposed methods are confirmed. While in Figure 3, 3D plots of Example 4.1 for fractional order δ=1,0.9 are plotted. Numerical values of Example 4.1 are presented in Table 1. In Figure 4, 2D-solutions plots and Figure 5 of (a) RPSM, (b) Exact, and (c) q-HATM, at δ=1 of Example 4.2 are plotted. The RPSM and q-HATM solutions are very closed to the exact solution. The RPSM and q-HATM solutions graphs at different fractional order are plotted in Figure 6.Figure 1
3D plots of (a) RPSM, (b) Exact, and (c) q-HATMυ−solutions at δ=1 of Example 4.1.
(a)(b)(c)Figure 2
2D plots of (a) RPSM, (b) Exact, and (c) q-HATMυ−solutions at δ=1 of Example 4.1.
(a)(b)(c)Figure 3
The 3D plots of example 4.1 for fractional orderδ=1,0.9.
(a)(b)Table 1
Solution comparison of RPSM, q-HATM, and Exact with different time and spaces of Example 4.1.
xtRPSMRPSMq-HATMq-HATMExactδ=0.9δ=1δ=0.9δ=1δ=10.250.0010.78041830.77957990.78041830.77957990.77957990.500.60779030.60713740.60779030.60713740.60713740.750.47334760.47283910.47334760.47283910.472839110.36864350.36824750.36864350.36824750.36824750.250.0050.78571180.78270450.78571180.78270450.78270450.500.61191300.60957090.61191300.60957090.60957090.750.47655830.47473420.47655830.47473420.474734210.37114400.36972340.37114400.36972340.3697234Figure 4
2D-solutions plots of (a) RPSM, (b) Exact, and (c) q-HATM, atδ=1 of Example 4.2.
(a)(b)(c)Figure 5
3D-solutions plots of (a) RPSM, (b) Exact, and (c) q-HATM, atδ=1 of Example 4.2.
(a)(b)(c)Figure 6
The 3D plots at fractional orderδ of Example 4.2.
(a)(b)


## 6. Conclusion

In this paper, the solutions of nonlinear systems of fractional Burger’s equations and advection diffusion equation are calculated by using q-HATM and RPSM. The proposed methods provide the results with higher degree of accuracy and using very few terms of their series solutions. The solutions comparison of both techniques are compared with the actual solutions of each problem. The comparison has shown the best compromise between the solutions of the suggested techniques. The fractional-order solutions are calculated successfully and shown to be convergent towards the integer-order solutions. The novelty of this paper is given in the nonlinear fractional solutions which provide the more accurate results as compared to other studies in literature. In the future, the suggested techniques can be utilised easily for the solutions of higher dimensional and nonlinear FPDEs and their systems because of their simple and straightforward implementation.



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*Source: 3666348-2022-10-06.xml*