# Computational Technique to Study Analytical Solutions to the Fractional Modified KDV-Zakharov-Kuznetsov Equation

Authors: Mohamed A. Abdoon; Faeza Lafta Hasan; Nidal E. Taha
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

In this article, we study and investigate the analytical solutions of the space-time nonlinear fractional modified KDV-Zakharov-Kuznetsov (mKDV-ZK) equation. We have got new exact solutions of the fractional mKDV-ZK equation by using first integral method; we found new types of hyperbolic solutions and trigonometric solutions by symbolic computation.


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## Body

## 1. Introduction

Partial differential equations have their both effective and active roles in describing natural phenomena and complicated phenomena in theoretical and applied physics. It has a vital role in engineering sciences, biology, viscoelasticity, fluid mechanics, population technology, and describing flowing motion since population growth passing the human behaviorism up to astronomy, so many phenomena can be explained by using PDEs [1–3].Guy [4] has proposed modified Riemann-Liouville derivative; these derivatives have their wider area applications, particularly fractal calculus in both sides on genetic and nongenetic; the last side has its own application that most of them rotate on diseases and pandemics spreading; this gives a proper description in this area [5, 6]. It is pointed that the complicated physical systems are described by using comfortable fractal derivative. Fractal derivative can be changed to differential equation by using variable transformation; several methods have been proposed to obtain exact and approximate solution of fractional differential equation [7–18], such as the homotopy perturbation method [19], the homotopy analysis method [20], fractional subequation method [21], the Lagrange characteristic method [22], and so on [23, 24]. In this article, we study exact and approximate solution of some nonlinear time-fractional partial differential equations using the first integral method.


## 2. Introductions and Basic Definitions

Guy’s modified Riemann-Liouville derivative, of orderα, can be defined by the following expression:
(1)Dsαfs=1Γ−α∫0xx−ξ−α−1fξ−f0dξ,α<0,1Γ1−αddx∫0xx−ξ−αfξ−f0dξ,0<α<1,fnxα−n,n≤α<n+1,n≥1.Moreover, some properties for the modified Riemann-Liouville derivative can be given as follows:(2)Dsαsr=Γ1+rΓ1+r−α,DsαfsgS=fsDsαgs+gsDsαts,Dsαfgs=fg′gsDsαgs=Dsαfgsg′tα.Then, the time-fractional differential equation with independent variablesX=x1,x2,⋯,xm,tand a dependent variableuis given as
(3)Wu,Dtαu,ux1,ux2,ux3,Dt2αu,ux1x1,ux2x2,ux3x3,⋯=0,using the variable transformation
(4)ux1,x2,⋯,xm,t=uζ,ζ=x1+l1x2+⋯+lm−1xm+λtαΓ1+α,where m,li, and λ≠0 are constants. equation (3) is reduced to a nonlinear ordinary differential equation
(5)F=u,uζ,uζζ,uζζ,⋯=0,We suppose that equation (5) has a solution in the form
(6)uζ=fζ,which introduces a new independent variableYζ=X′ζ; we get a new system of ODEs:
(7)X′ζ=Yζ,(8)Y′ζ=FXζ,Yζ,by using the division theorem for two variables in complex domain CX,Y which is based on the Hilbert-Nullstellensatz theorem [25]. An exact solution to equation (3) is when we obtain a first integral to equation (7) which can applied to equation (5) to obtain a first-order ordinary differential equation. Now, we wish to quickly recall the division theory.Theorem 1 (the division theorem).
Assume thatPx,y and Qx,y are polynomials of two variables x and y in Cx,y and Px,y is irreducible in Cx,y. If Qx,y vanishes at all points of Px,y, then there exists a polynomial GX,Y in Cx,y such that
(9)QX,Y=PX,YGX,Y.


## 3. The First Integral Method: Formula (1)

We need to apply the FIM to the following form:(10)u″ζ−Tuζ,u′ζu′ζ−Ruζ=0,where Tu,u′ is a polynomial in u and u′ and Ru is a polynomial with real coefficients.Now, chooseTu,u′=0 andRu=Au+Bu3, so (10) changes. We get
(11)u″ζ−Auζ−Bu3ζ=0.Using (2.9) and (4), (10) is equivalent to the two-dimensional autonomous system:
(12)X′ζ=Yζ,(13)Y′ζ=AXζ+BX3ζ.Now, we are applying the division theorem to find the first integral to (12), supposing that X=Xζ and Y=Yζ are the nontrivial solutions to (12) and qx,y=∑i=0NaiXYi=0, which is an irreducible polynomial in the complex domain CX,Y; thus,
(14)qXζ,Yζ=∑i=0NaiXζYiζ=0,where aiXi=0,1,2.⋯⋯⋯,N are polynomials and aNX≠0; equation (14) is called the first integral method. There exists a polynomial δX+θXY in the complex domain Cx,y such that
(15)dqdζ=dqdX.dXdζ+dqdY.dYdζ=δX+θxY∑i=0NaiXYi.Case 1.
Suppose thatN=1. Then, (15) becomes
(16)∑i=01a•iXYiY+∑i=01iaixyi−1Ax+Bx3=δx+θxy∑i=01aiyi.Setting all coefficients ofYii=0,1, we take
(17)a′x=Ax.ax.Also,(18)Ax+Bx3,−δax=0,where
(19)ax=a1x,a0xt,(20)Ax=θx0δxθx.Then,aixi=0,1 are polynomials; from (14), we deduce that a1x is a constant, θx=0, and taking a1x=1; then, from (17), we have
(21)ax=1∫δxdx.Compare the degrees ofgX,a1X, anda0X; we conclude thatdegδx=1; suppose thatδx=A1x+B0; then, from (17), we finda0x, as follows:
(22)a0X=A1X22+B0X+B1,whereA1,B0, and B1 are constants. Substituting δx,a1x, and a0x into (18) and comparing all coefficient of x to be zero, we obtain
(23)A1=22B,B0=0,B1=±A2B,B2=A22B.It follows from equation (14) with equation (23), then
(24)y=±B2x2+AB.Combining (24) with (12), we obtain the following exact solution of equation (11):
(25a)IfAB>0,uζ=±ABtanA2ζ+ζ0,IfAB<0,uζ=∓ABtanhA2ζ+ζ0,where ζ0 is arbitrary constant.Case 2.
Suppose thatN=1; equation (15), equating the coefficients, gives
(26)∑i=02ai′XYiY+∑i=02iaiXYi−1AX+BX3=δX+θXy∑i=02aiXYi.We have(27)a′X=AX·aX,(28)0,AX+BX3,−δaX=0,where
(29)ax=a2X,a1X,a0Xt,Ax=θX00δXθX02AX+2BX3δXθX.Then,aiXi=0,1,2are polynomials; from (27), also takinga2Xis a constant,θX=0, anda2X=1; then, from (27), we have
(30)ax=1∫δXdX∫2AX+2BX3+δXa1XdX.Remark 2.
Ifdegδx=k>0, then deuce dega1x=k+1and dega0x=2k+1 from (30), supposing that δx=A1x+B0 and a1x≠0. Then, we get
(31)a1x=12A1X2+B0X+B1,a0x=18A21−4BX4+12A1B0xX3+12A1B20−2AA1+B1A12X2+B1B0x+B2.Substitutinga1x and a0x into (28) and setting all coefficients xii=5,4,3,2,1 to zero, then yield, and c0=0, we get
(32)12BA1=18A1A12−4B,BB0=12A12B0A12−4B,12AA1+BB1=12A1A1B02−2AA1+B1A12+12A1B02,AB0=A1B0B1+12B0AB02−2AA1+B1A12,B1A=A1B2+B1B02,B0B2=0.Then, we deduce(33)A1=22B,B0=0,B1=±A2B,B2=A22B.It follows from (26) and (33) that
(34)y=±B2x2+AB.Combining (34) with (12), we obtain the following exact solution of equation (11):
(35)IfAB>0,uζ=±ABtanA2ζ+ζ0,IfAB<0,uζ=∓ABtanhA2ζ+ζ0.


## 4. Application

### 4.1. The (3+1)-Dimensional Space-Time-Fractional mKDV-ZK Equation

In the present subsection, we apply the current method to solve the (3+1)-dimensional mKDV-ZK space-time-fractional equation of the form
(36)Dtαu+βu2ux+euxxx+ruxyy+suxzz=0t>0,0<α<1,where β,e,r, and s are nonzero constants, α is a parameter representing the order of the time-space-fractional derivative. When α=1, equation (36) is defined the modified KDV-ZK equation as
(37)ut+βu2ux+euxxx+ruxyy+suxzz=0,when r=0, s=0, and β,e≠0; equation (36) is defined the modified KDV fractional equation as
(38)Dtαu+βu2ux+euxxx=00<α<1,t>0.The solutions of mKDV-ZK equation have been introduced by several papers for the importance of applying this equation in various physics fields (see references [26, 27]. Then, we obtain the soliton solutions to equation (36). Therefore, by (4), we utilize the transformations as follows:
(39)ux,y,z,t=Fξ,(40)ξ=x+y+z−λtαΓ1+α,where λ≠0Equation (36) transforms to ordinary differential equation by substituting (39); we get
(41)−λF′+βF2F′+eF′″+rF′″+sF′″=0,where F′=dF/dξ when integrating one time and equalizing the constant of integration to zero; we have
(42)−λF+β3F3+e+r+sF″=0,then
(43)F″=λe+r+sF−β3e+r+sF3.We apply (7) to obtain the soliton solutions for equation (36); then equation (43) becomes
(44)X′ξ=YξY′ξ=λe+r+sXξ−β3e+r+sX3ξ.Now, to get the first integral to equation (44); suppose that X=Xξ, and Y=Yξ are the nontrivial solutions to equation (44) and qXξ,Yξ=∑i=0Naixyi=0, an irreducible polynomial in the complex domain ℂX,Y, such that aiXi=0,1,2,⋯N and aNX≠0. According to the concept of division theorem and applying equation (15), there exist two polynomials δX+θXY in ℂX,Ysuch that
(45)dqdξ=∂q∂XdXdξ+∂q∂YdYdξ=δX+θXY∑i=0NaiXYi.In our study, we will discuss three cases, which are explained as follows:Case 1.
We start our study by assumingN=1; equation (45) becomes
(46)∑i=01ai′XYi+1+∑i=01iaiXYi−1AX2+BX=δX+θXY∑i=01aiXYi.Equating the coefficients ofYii=1,0, we have system of ODE,
(47a)a1′X=a1Xθx,(47b)a0′X=a1Xδx+a0Xθx,(47c)a1Xλe+r+sX−β3e+r+sX3=a0Xδx.According to equations (17), (18), and (19), we can write system (47a), (47b), and (47c) as follows:
(48a)a′x=Ax.ax,and
(48b)λe+r+sX−β3e+r+sX3,−δax=0,where
(49)ax=a1x,a0xt,Ax=θx0δxθx.Then,aixi=0,1are polynomials; from (21), we deduce thata1xis a constant,θx=0, and takinga1x=1; then from equation (48a) as equation (21), we have
(50)ax=1∫δxdx.Compare the degrees ofδx,a1X, and a0X; we conclude that degδx=1; suppose that δx=A1x+B0, then from (23), we find a0x, as follows:
(51)a0X=AX22+BX+C0,where A,B, and C0 are arbitrary integration constants. Then, when solving a system of nonlinear algebraic equations which is obtained by substituting a1X,a0X, and gX in equation (48b) and equating all coefficients of powers X from both sides of equation (48b), we get
(52)A⟶−2/3β−e−r−s,B⟶0,C0⟶3/2λ−e−r−sβ,A⟶2/3β−e−r−s,B⟶0,C0⟶−3/2λ−e−r−sβ.We obtain two solutions ofYby substituting first and second sets of solutions inqXξ,Yξ=∑i=0Naixyi=0, then
(53a)Y=X2β−3λ6−e−r−sβ,(53b)Y=−X2β−3λ6−e−r−sβ.By combining equations (53a) and (53b) with (43), respectively, we get the exact solutions of equation (44) as follows:
(54)f1ζ=3λtanh1/22λ−e−r−sζ/e+r+s−23λβC1β,where C1 and C2 are an arbitrary integration constants; then, the solution of equation (36) is as follows:
(55)u1x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λta/г1+a/e+r+s−23λβC1β.Case 2.
When takingN=2 in (45) and qX,Y=0, this implies dq/dξ=0,(56)∑i=02ai′XYi+1++∑i=02iaiXYi−1λe+r+sXξ−β3e+r+sX3ξ=δX+θXY∑i=02aiXYi,on both sides of (56) by equating the coefficients of Yii=2,1,0, and according to equations (27) and (28), we have
(57a)a′X=AX·aX,(57b)0,λe+r+sXξ−β3e+r+sX3ξ,−δaX=0,where
(58)ax=a2x,a1x,a0xt,(59)Ax=θx00δxθx02.λe+r+sXξ−β3e+r+sX3ξδxθx,(60)a1Xλe+r+sXξ−β3e+r+sX3ξ=a0XgX.Sinceaixi=0,1,2 are polynomials of X, we deduce that a2X is a constant from (57a), and θx=0. For simplicity, take a2X=1 as remark and balance the degrees of δX,a1X, and a0X if degδx=k>0; then, deuce dega1x=k+1 and dega0x=2k+1; therefore, we obtaindegδX=1only, and suppose that δX=AX+B and A≠0; after that, we calculate a1X, and a0X. From (57b), we have
(61)ax=1∫δXdX∫2.λe+r+sXξ−β3e+r+sX3ξ+δXa1XdX.Then, we get(62a)a1X=A2X2+BX+C0,(62b)a0X=d+12ABX3+X43A2e+3A2r+3A2s+4β24e+r+s+X2B2e+B2r+B2s−2λ2e+r+s+BXC0+12AX2C0,where C0 and d are constants. By substituting a2X,a1X, a0X, and δX in the last equation in (60), then system of nonlinear algebraic equations obtained by putting all the coefficients of power X to be zero, and by solving this system, we get sets of values as follows:
(63)A=0,B=0,C0=0,A=−22/3β−e−r−s,B=0,C0=6λ−e−r−sβ,d=−3λ22e+r+sβ,A=22/3β−e−r−s,B=0,C0=−6λ−e−r−sβ,d=−3λ22e+r+sβ.The first set represents the trivial solution therefore neglected; by the second and third sets, we get two solutionsY1 and Y2, respectively, as follows:
(64a)Y1=122/3X2β−6λ−e−r−sβ−6λ−2/3X2β−e−r−sβ2−4X4β+2e+r+sβ/−e−r−s6e+r+s−2X2λ−e−r−s−2βλX2+3λ22e+r+sβ,(64b)Y2=122/3X2β−6λ−e−r−sβ+6λ−2/3X2β−e−r−sβ2−4X4β+2e+r+sβ/−e−r−s6e+r+s−2X2λ−e−r−s−2βλX2+3λ22e+r+sβ.And combining equations (64a) and (64b) with (43), respectively, we will get two similar solutions of (44) as
(65)fζ=3λtanh1/22−e−r−sζλ/e+r+s−23λβC1β,where C1 is an arbitrary integration constant; then, the solution of (36) is
(66)u2x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λtα/Γ1+α/e+r+s−23λβC1βCase 3.
When takingN=3 in (45), and qX,Y=0, this implies dq/dξ=0,(67)∑i=03ai′XYi+1++∑i=03iaiXYi−1λe+r+sXξ−β3e+r+sX3ξ=δX+θXY∑i=03aiXYi.The same as the steps in Case 1 and Case 2, we get two solutionsY1,Y2, andY3as follows:
(68a)Y1=−6eX2β−6rX2β−6sX2β+36λe+r+s6−e−r−s3/2β,(68b)Y2=ⅈX2β−3λ6βe+r+s,(68c)Y3=−ⅈX2β−3λ6βe+r+s,And, combining equations (68a), (70b), and (68c) with (43), respectively, we will get three similar solutions of (44) as
(69a)f1ζ=3λtanh1/22−e−r−sζλ/e+r+s−23βλC1β,f2ζ=3λtanh1/2−ⅈ2λζ/e+r+s−23βλC2β,f3ζ=3λTanh1/2ⅈ2λζ/e+r+s−23βλC3β,whereC1,C2, andC3are an arbitrary integration constants; then, the solution of (36) is as follows:
(70a)u3x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λtα/Γ1+α/e+r+s−23βλC1β,(70b)u4x,y,z,t=3λtanh1/2−i2λx+y+z−λtα/Γ1+α/e+r+s−23βλC2β(70c)u5x,y,z,t=3λtanh1/2i2λx+y+z−λtα/Γ1+α/e+r+s−23βλC3β.


## 4.1. The (3+1)-Dimensional Space-Time-Fractional mKDV-ZK Equation

In the present subsection, we apply the current method to solve the (3+1)-dimensional mKDV-ZK space-time-fractional equation of the form
(36)Dtαu+βu2ux+euxxx+ruxyy+suxzz=0t>0,0<α<1,where β,e,r, and s are nonzero constants, α is a parameter representing the order of the time-space-fractional derivative. When α=1, equation (36) is defined the modified KDV-ZK equation as
(37)ut+βu2ux+euxxx+ruxyy+suxzz=0,when r=0, s=0, and β,e≠0; equation (36) is defined the modified KDV fractional equation as
(38)Dtαu+βu2ux+euxxx=00<α<1,t>0.The solutions of mKDV-ZK equation have been introduced by several papers for the importance of applying this equation in various physics fields (see references [26, 27]. Then, we obtain the soliton solutions to equation (36). Therefore, by (4), we utilize the transformations as follows:
(39)ux,y,z,t=Fξ,(40)ξ=x+y+z−λtαΓ1+α,where λ≠0Equation (36) transforms to ordinary differential equation by substituting (39); we get
(41)−λF′+βF2F′+eF′″+rF′″+sF′″=0,where F′=dF/dξ when integrating one time and equalizing the constant of integration to zero; we have
(42)−λF+β3F3+e+r+sF″=0,then
(43)F″=λe+r+sF−β3e+r+sF3.We apply (7) to obtain the soliton solutions for equation (36); then equation (43) becomes
(44)X′ξ=YξY′ξ=λe+r+sXξ−β3e+r+sX3ξ.Now, to get the first integral to equation (44); suppose that X=Xξ, and Y=Yξ are the nontrivial solutions to equation (44) and qXξ,Yξ=∑i=0Naixyi=0, an irreducible polynomial in the complex domain ℂX,Y, such that aiXi=0,1,2,⋯N and aNX≠0. According to the concept of division theorem and applying equation (15), there exist two polynomials δX+θXY in ℂX,Ysuch that
(45)dqdξ=∂q∂XdXdξ+∂q∂YdYdξ=δX+θXY∑i=0NaiXYi.In our study, we will discuss three cases, which are explained as follows:Case 1.
We start our study by assumingN=1; equation (45) becomes
(46)∑i=01ai′XYi+1+∑i=01iaiXYi−1AX2+BX=δX+θXY∑i=01aiXYi.Equating the coefficients ofYii=1,0, we have system of ODE,
(47a)a1′X=a1Xθx,(47b)a0′X=a1Xδx+a0Xθx,(47c)a1Xλe+r+sX−β3e+r+sX3=a0Xδx.According to equations (17), (18), and (19), we can write system (47a), (47b), and (47c) as follows:
(48a)a′x=Ax.ax,and
(48b)λe+r+sX−β3e+r+sX3,−δax=0,where
(49)ax=a1x,a0xt,Ax=θx0δxθx.Then,aixi=0,1are polynomials; from (21), we deduce thata1xis a constant,θx=0, and takinga1x=1; then from equation (48a) as equation (21), we have
(50)ax=1∫δxdx.Compare the degrees ofδx,a1X, and a0X; we conclude that degδx=1; suppose that δx=A1x+B0, then from (23), we find a0x, as follows:
(51)a0X=AX22+BX+C0,where A,B, and C0 are arbitrary integration constants. Then, when solving a system of nonlinear algebraic equations which is obtained by substituting a1X,a0X, and gX in equation (48b) and equating all coefficients of powers X from both sides of equation (48b), we get
(52)A⟶−2/3β−e−r−s,B⟶0,C0⟶3/2λ−e−r−sβ,A⟶2/3β−e−r−s,B⟶0,C0⟶−3/2λ−e−r−sβ.We obtain two solutions ofYby substituting first and second sets of solutions inqXξ,Yξ=∑i=0Naixyi=0, then
(53a)Y=X2β−3λ6−e−r−sβ,(53b)Y=−X2β−3λ6−e−r−sβ.By combining equations (53a) and (53b) with (43), respectively, we get the exact solutions of equation (44) as follows:
(54)f1ζ=3λtanh1/22λ−e−r−sζ/e+r+s−23λβC1β,where C1 and C2 are an arbitrary integration constants; then, the solution of equation (36) is as follows:
(55)u1x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λta/г1+a/e+r+s−23λβC1β.Case 2.
When takingN=2 in (45) and qX,Y=0, this implies dq/dξ=0,(56)∑i=02ai′XYi+1++∑i=02iaiXYi−1λe+r+sXξ−β3e+r+sX3ξ=δX+θXY∑i=02aiXYi,on both sides of (56) by equating the coefficients of Yii=2,1,0, and according to equations (27) and (28), we have
(57a)a′X=AX·aX,(57b)0,λe+r+sXξ−β3e+r+sX3ξ,−δaX=0,where
(58)ax=a2x,a1x,a0xt,(59)Ax=θx00δxθx02.λe+r+sXξ−β3e+r+sX3ξδxθx,(60)a1Xλe+r+sXξ−β3e+r+sX3ξ=a0XgX.Sinceaixi=0,1,2 are polynomials of X, we deduce that a2X is a constant from (57a), and θx=0. For simplicity, take a2X=1 as remark and balance the degrees of δX,a1X, and a0X if degδx=k>0; then, deuce dega1x=k+1 and dega0x=2k+1; therefore, we obtaindegδX=1only, and suppose that δX=AX+B and A≠0; after that, we calculate a1X, and a0X. From (57b), we have
(61)ax=1∫δXdX∫2.λe+r+sXξ−β3e+r+sX3ξ+δXa1XdX.Then, we get(62a)a1X=A2X2+BX+C0,(62b)a0X=d+12ABX3+X43A2e+3A2r+3A2s+4β24e+r+s+X2B2e+B2r+B2s−2λ2e+r+s+BXC0+12AX2C0,where C0 and d are constants. By substituting a2X,a1X, a0X, and δX in the last equation in (60), then system of nonlinear algebraic equations obtained by putting all the coefficients of power X to be zero, and by solving this system, we get sets of values as follows:
(63)A=0,B=0,C0=0,A=−22/3β−e−r−s,B=0,C0=6λ−e−r−sβ,d=−3λ22e+r+sβ,A=22/3β−e−r−s,B=0,C0=−6λ−e−r−sβ,d=−3λ22e+r+sβ.The first set represents the trivial solution therefore neglected; by the second and third sets, we get two solutionsY1 and Y2, respectively, as follows:
(64a)Y1=122/3X2β−6λ−e−r−sβ−6λ−2/3X2β−e−r−sβ2−4X4β+2e+r+sβ/−e−r−s6e+r+s−2X2λ−e−r−s−2βλX2+3λ22e+r+sβ,(64b)Y2=122/3X2β−6λ−e−r−sβ+6λ−2/3X2β−e−r−sβ2−4X4β+2e+r+sβ/−e−r−s6e+r+s−2X2λ−e−r−s−2βλX2+3λ22e+r+sβ.And combining equations (64a) and (64b) with (43), respectively, we will get two similar solutions of (44) as
(65)fζ=3λtanh1/22−e−r−sζλ/e+r+s−23λβC1β,where C1 is an arbitrary integration constant; then, the solution of (36) is
(66)u2x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λtα/Γ1+α/e+r+s−23λβC1βCase 3.
When takingN=3 in (45), and qX,Y=0, this implies dq/dξ=0,(67)∑i=03ai′XYi+1++∑i=03iaiXYi−1λe+r+sXξ−β3e+r+sX3ξ=δX+θXY∑i=03aiXYi.The same as the steps in Case 1 and Case 2, we get two solutionsY1,Y2, andY3as follows:
(68a)Y1=−6eX2β−6rX2β−6sX2β+36λe+r+s6−e−r−s3/2β,(68b)Y2=ⅈX2β−3λ6βe+r+s,(68c)Y3=−ⅈX2β−3λ6βe+r+s,And, combining equations (68a), (70b), and (68c) with (43), respectively, we will get three similar solutions of (44) as
(69a)f1ζ=3λtanh1/22−e−r−sζλ/e+r+s−23βλC1β,f2ζ=3λtanh1/2−ⅈ2λζ/e+r+s−23βλC2β,f3ζ=3λTanh1/2ⅈ2λζ/e+r+s−23βλC3β,whereC1,C2, andC3are an arbitrary integration constants; then, the solution of (36) is as follows:
(70a)u3x,y,z,t=3λtanh1/22λ−e−r−sx+y+z−λtα/Γ1+α/e+r+s−23βλC1β,(70b)u4x,y,z,t=3λtanh1/2−i2λx+y+z−λtα/Γ1+α/e+r+s−23βλC2β(70c)u5x,y,z,t=3λtanh1/2i2λx+y+z−λtα/Γ1+α/e+r+s−23βλC3β.


## 5. Conclusion

In this study, the equation of the space-time nonlinear fractional modified KDV-Zakharov-Kuznetsov (mKDV-ZK) equation has been discussed by using the first integral method, accuracy of finding the mentioned solutions, adding, verifying, and checking them by using symbolic computation. For the illustration of solutions, thus drawing diagrams of solutions have been created. It has been noticed that all solutions in all situationsN=1,2,3 are fully similar and repeating; however, we have found that we can gain the same solution as considered to be the only solution to this equation whenever N increases. This method is effective, direct, and much more accurate; it can be applied upon other dynamic and engineering models (Figure 1).Figure 1
(a–c) Illustrations of the solitary wave solutions in different dimensions; in the physical description, we give more complete illustrations of (55), (66), and (70a) by using maple plots for the solitary wave solutions u1x,y,z,t,u2x,y,t, and u3x,y,t, of equation (36), with N=1,2, and 3, respectively, when e=0.5, r=0.5, s=−2, C1=0.01, λ=0.2, β=0.01, and α=0.5.
(a)(b)(c)



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*Source: 2162356-2022-06-20.xml*