# Improved Fractional Subequation Method and Exact Solutions to Fractional Partial Differential Equations

Authors: Jun Jiang; Yuqiang Feng; Shougui Li
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, the improved fractional subequation method is applied to establish the exact solutions for some nonlinear fractional partial differential equations. Solutions to the generalized time fractional biological population model, the generalized time fractional compound KdV-Burgers equation, the space-time fractional regularized long-wave equation, and the (3+1)-space-time fractional Zakharov-Kuznetsov equation are obtained, respectively.


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## Body

## 1. Introduction

Fractional differential equations are widely used to describe lots of important phenomena and dynamic processes in physics, engineering, electromagnetics, acoustics, viscoelasticity electrochemistry, material science, stochastic dynamical system, plasma physics, controlled thermonuclear fusion, nonlinear control theory, image processing, nonlinear biological systems and astrophysics, etc. [1–7]. In order to find the solutions of fractional differential equations, many powerful and efficient methods have been introduced and developed, such as Darboux transformations [8], the hyperbolic function method [9], the variational iteration method [10, 11], the autofinite element method [12, 13], the auxiliary equation method [14], the finite difference method [15, 16], the Adomian decomposition method [17, 18], the homogenous balance method [19], Hirotas bilinear method [20], the homotopy analysis method [21, 22], G′/G-expansion method [23, 24], the subequation method [25, 26], the first integral method [27, 28], the improved fractional subequation method [29, 30], the extended Jacobi elliptic function expansion method [31], the generalized Kudryashov method [32], the exponential rational function method [33], the exp-function method [34, 35], the multiple exp-function method [36], the extended simple equation method [37].In order to deal with nondifferentiable functions, Jumarie [38] has proposed a modification of the Riemann-Liouville definition which appears to provide a framework for a fractional calculus which is quite parallel to the classical calculus.Jumarie’s modified Riemann-Liouville derivative of orderα for a function f is defined as follows:
(1)Dxαfx=1Γ1−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.Some useful properties of modified Riemann-Liouville derivative are given below:(2)Dxαxγ=Γγ+1Γγ+1−αxγ−α,γ>0,(3)Dxαfxgx=gxDxαfx+fxDxαgx,(4)Dxαfgx=fg′gxDxαgx=Dgαfgxg′xα,which holds for nondifferentiable functions. Equations (2), (3), (4) which are important tools for fractional calculus. Based on these merits, the modified Riemann-Liouville derivative was successfully applied to the probability calculus, fractional Laplace problems, and fractional variational calculus.In this paper, we aim to find new exact solutions of some important partial fractional differential equations under Jumarie’s definition by improved fractional subequation method.In what follows, we introduce the aforementioned fractional partial differential equations. They are the generalized time fractional biological population model, the generalized time fractional compound KdV-Burgers equation, the space-time fractional regularized long-wave equation, and the (3+1)-space-time fractional Zakharov-Kuznetsov equation.Suppose timet>0, Dtαu is time modified Riemann-Liouville derivative of order α for a function u, 0<α≤1, the parameters k1,k2,⋯,k7 are any real constants.The generalized time fractional biological population model is given by(5)Dtαu+k1u2xx+k2u2yy+k3ux+k4uy+k5u2+k6u+k7=0,where u=ux,y,t is an unknown function.Whenk1=−1,k2=−1,k3=k4=k6=0,k5=−h≠0,k7=rh≠0, Equation (5) is the time fractional biological population model [39]:
(6)Dtαu=u2xx+u2yy+hu2−r,where u=ux,y,t denotes the population density and hu2−r represents the amount of population due to death and birth. Moreover, hu2−r leads to Verhulst law. Equation (5) has an important role to understand the dynamic process of population changes, and it is also an assistant to achieve precision about it.The generalized time fractional compound KdV-Burgers equation is given by(7)Dtαu+k1uux+k2u2ux+k3uxx+k4uxxx=0,where u=ux,t is an unknown function.Whenk1=k3=0, Equation (7) becomes the time fractional mKdV equation
(8)Dtαu+k2u2ux+k4uxxx=0;when k1=k3=0 Equation (7) becomes the time fractional KdV equation
(9)Dtαu+k2u2ux+k3uxx+k4uxxx=0;when k2=k4=0, Equation (7) becomes the time fractional Burgers equation [40]
(10)Dtαu+k1uux+k3uxx=0;when k1=0, Equation (7) becomes the time fractional mKdV-Burgers equation
(11)Dtαu+k2u2ux+k3uxx+k4uxxx=0;when k2=0, Equation (7) becomes the time fractional KdV-Burgers equation
(12)Dtαu+k1uux+k3uxx+k4uxxx=0.The space-time fractional regularized long-wave equation is given by [41]:
(13)Dtαu+k1Dxαu+k2uDxαu+k3DtαDx2αu=0,where u=ux,t is an unknown function, Dxαu is the modified Riemann-Liouville derivative of order α for a function u, and Dx2αu=DxαDxαu.The regularized long-wave equation, which describes approximately the unidirectional propagation of long waves in certain nonlinear dispersive systems, was proposed by Benjamin et al. in 1972. The regularized long-wave equation is considered an alternative to the KdV equation, which is modeled to govern a large number of physical phenomena such as shallow waters and plasma waves.The (3+1)-space-time fractional Zakharov-Kuznetsov equation given by [41].
(14)Dt?u+k1uDxαu+k2Dx3αu+k3DxαDy2αu+k4DxαDz2αu=0,where u=ux,y,z,t is an unknown function; Dxαu, Dyαu, and Dzαu are the modified Riemann-Liouville derivatives of the function u; Dx2αu=DxαDxαu; Dy2αu=DyαDyαu; Dz2αu=DzαDzαu; and Dx3αu=DxαDxαDxαu.The Zakharov-Kuznetsov equation was first derived for analysing weakly nonlinear ion acoustic waves in heavily magnetized lossless plasma and geophysical flows in two dimensions. The ZK equation is one of the two well-established canonical two-dimensional extensions of the KdV equation. The ZK equation governs the behavior of weakly nonlinear ionacoustic waves in a plasma comprising cold ions and hot isothermal electrons in the presence of a uniform magnetic field.Motivated by the above results, in this paper, we use the improved subequation method to find new exact solutions of the generalized time fractional biological population model, the generalized time fractional compound KdV-Burgers equation, the space-time fractional regularized long-wave equation, and the (3+1)-space-time fractional Zakharov-Kuznetsov equation, respectively.


## 2. A Brief Description of the Improved Fractional Subequation Method

In this section, basic steps of the improved subequation method [42] are presented.Consider the following nonlinear fractional differential equation,(15)Pu,Dtαu,Dx1αu,Dx1αu2,⋯,Dxmαu,Dx1αu2,Dx2αu2,⋯,Dxmαu2,⋯,Dx12αu,Dxm2αu,⋯=0,where t,x1,x2,⋯,xm are independent variables, ut,x1,x2,⋯,xm is an unknown function, and P is a polynomial of u,u2,⋯ and their partial fractional derivatives. Also, Dα⋅ symbolizes the modified Riemann-Liouville fractional derivative.Step 1. First of all, using a suitable fractional complex transform,(16)ut,x1,x2,⋯,xm=uξ,ξt,x1,x2,⋯∣,xm,Equation (15) converts into nonlinear ordinary differential equation given below:
(17)Qu,u′,u″,u′″,⋯=0,where u′,u″,u′″,⋯ denotes the derivations with respect to ξ.We used to utilize real transformation, but actually, complex transformations are more useful. They make some equations easier to simplify.Step 2. Suppose that the solution of ordinary differential equation (17) is
(18)u=∑i=n−1aiφi+a0+∑i=nnaiφi,where constants aii=−n,⋯,−1,1,⋯,n are going to be determined. Here, n is a positive integer, and it is obtained using the homogeneous balance of the highest order derivative and the nonlinear term seen in Equation (17).φ=φξ is the solution of the Riccati equation
(19)φ′=σ+φ2,where σ is a constant, and the solutions of Equation (19) are obtained by Zhang et al. [34] as follows:
20)φξ=−−σtanh−σξ,σ<0,−−σcoth−σξ,σ<0,σtanσξ,σ>0,−σcotσξ,σ>0,−Γ1+αξα+ω,σ=0,ω=const,In the previous literatures,u=a0+∑i=1naiφi was considered. In this paper, we assume u=∑i=−n−1aiφi+a0+∑i=1naiφi, and we can get a solution that has both hyperbolic tangent function and hyperbolic cotangent function or both tangent function and cotangent function.Step 3. Putting Equation (18) along with Equation (19) into Equation (17), we obtain a new polynomial in terms of φ. Then, all the coefficients of powers of φkk=0,1,2,⋯,−1,−2,⋯ are set equal to zero, we get a system of algebraic equations.Step 4. Finally, the system of algebraic equations is obtained in the previous step forai−n≤i≤n, and σ is solved by the Maple package. By substituting the newly obtained values into Equation (20), we get the exact solutions for the nonlinear fractional differential equation (15).Applying a suitable fractional complex transform of the improved fractional subequation method and the chain rule, nonlinear fractional differential equations with the modified Riemann-Liouville derivative can be converted into nonlinear ordinary differential equations. Then, using the solutions of a Riccati equation, we can find exact analytical solutions expressed by triangle functions, hyperbolic functions, or power functions.


## 3. Applications of the Improved Fractional Subequation Method

In this section, the improved fractional subequation method is utilized to solve some nonlinear fractional differential equations introduced in Section1.


### 3.1. The Generalized Time Fractional Biological Population Model

The generalized time fractional biological population model is given by(21)Dtαu+k1u2xx+k2u2yy+k3ux+k4uy+k5u2+k6u+k7=0,where u=ux,y,t is an unknown function.We considered two cases.Case 1.
Whenk1k2>0, let
(22)ux,y,t=uξ,ξ=ax+iak1k2y+btαΓ1+α,a≠0,then Equation (21) is reduced to the ordinary differential equation as follows:
(23)ak4k1k2u′=0,ak3+bu′+k5u2+k6u+k7=0.Whenk4≠0, Equation (23) has only constant solutions.Whenk4=0, Equation (23) has solutions in the form of (18). u′ is obtained from the homogeneous balance between the highest order derivative and the nonlinear term u2. We obtain the solution of Equation (21) as follows:
(24)uξ=a−1φ−1+a0+a1φ,Substituting Equation (24) together with its necessary derivatives into Equation (21), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(25)φ−2:−k3σa−1−bσa−1+k5a2−1,φ−1:2k5a−1a0+a−1k6,φ0:−ak3a−1−ba−1+ak3σa1+bσa1+k5a02+2k5a−1a1+k6a0+k7,φ1:2k5a0a1+k6a1,φ2:ak3a1+ba1+k5a12.Let the coefficients be zero. By solving the set of equations given above fora−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(26)a−1=0,a0=−k62k5,a1=a1,a=a,b=−ak3−a1k5,σ=λ4a12k52.Set 2(27)a−1=a−1,a0=−k62k5,a1=−λ16k52a−1,a=a,b=λ−16aa−1k3k516k5a−1,σ=16k52a−12λ.Set 3(28)a−1=a−1,a0=−k62k5,a1=0,a=a,b=λ−4aa−1k3k54k5a−1,σ=4k52a−12λ.where λ=4k5k7−k62.Case 2.
Whenk1k2<0, let
(29)ux,y,t=uξ,ξ=ax+a−k1k2y+btαΓ1+α,a≠0,then Equation (21) is reduced to the ordinary differential equation as follows:
(30)b+k3a+ak4−k1k2u′+k5u2+k6u+k7=0.The solution of Equation (30) is in the form of (18). n=1 is taken from the homogeneous balance between the highest order derivative u′ and the nonlinear term u2. We obtain the solution of Equation (30) as Equation (24). Substituting Equation (24) together with its necessary derivatives into Equation (30), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(31)φ−2:−a−k1k2k4σa−1−ak3σa−1−bσa−1+k5a−12,φ−1:2k5a−1a0+a−1k6,φ0:a−k1k2k4σa1+ak3σa1+bσa1−a−k1k2k4a−1,φ1:2k5a0a1+k6a1,φ2:a−k1k2k4a1+ak3a1+ba1+k5a12.Let the coefficients be zero. By solving the set of equations given above fora−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 4(32)a−1=0,a0=−k62k5,a1=a1,a=a,b=−ak3−ak4−k1k2−a1k5,σ=λ4a12k52.Set 5(33)a−1=a−1,a0=−k62k5,a1=−λ16k52a−1,a=a,b=λ−16μ16k5a−1,σ=16k52a−12λ.Set 6(34)a−1=a−1,a0=−k62k5,a1=0,a=a,b=λ−4μ4k5a−1,σ=4k52a−12λ⋅where μ=aa−1k3k5+aa−1k4k5−k1/k2.We find thata−1,a0,a1, and σ are equal in set 1 and set 4, set 2 and set 5, and set 3 and set 6, respectively. In this study, the solutions of differential equations are symbolized as ui,jx,y,t,i,j∈Z+, where i denotes obtained set number and j is the solution number of the Riccati equation, respectively. Thus, using set 1 to set 6, we obtain the solution of Equation (21) as ui,jx,y,t,i=1,2,⋯,6;j=1,2,⋯,5. ui,jx,y,t is the following:Whenk1k2>0, λ=4k5k7−k62<0, we have σ<0, then
(35)u1,1x,y,t=−k62k5−a1−λ2a1k5tanh−λ2a1k5ax+iak1k2y−ak3+a1k5tαΓ1+α,u1,2x,y,t=−k62k5−a1−λ2a1k5coth−λ2a1k5ax+iak1k2y−ak3+a1k5tαΓ1+α,u2,1x,y,t=−a−1−λ4k5a−1coth4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1tanh4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u2,2x,y,t=−a−1−λ4k5a−1tanh4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1coth4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u3,1x,y,t=a1−λ2a1k5a−1coth2a1k5a−1−λax+iak1k2y−λ−4aa−1k3k5tα4k5a−1Γ1+α−k62k5,u3,2x,y,t=a−1−λ2a1k5a−1tanh2a1k5a−1−λax+iak1k2y−λ−4aa−1k3k5tα4k5a−1Γ1+α−k62k5.Whenk1k2<0, λ=4k5k7−k62<0, we have σ<0, then
(36)u4,1x,y,t=−k62k5−a1−λ2a1k5tanh−λ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u4,2x,y,t=−k62k5−a1−λ2a1k5coth−λ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u5,1x,y,t=−a−1−λ4k5a−1coth4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1tanh4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α,u5,2x,y,t=−a−1−λ4k5a−1tanh4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1coth4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α,u6,1x,y,t=a−1−λ2a1k5a−1coth2a1k5a−1−λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5,u6,2x,y,t=a−1−λ2a1k5a−1tanh2a1k5a−1−λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5.Whenk1k2<0 and λ=4k5k7−k62<0, we have σ<0, then
(37)u1,3x,y,t=−k62k5+a1λ2a1k5tanλ2a1k5ax+iak1k2y+ak3+a1k5tαΓ1+α,u1,4x,y,t=−k62k5+a1λ2a1k5cotλ2a1k5ax+iak1k2y+ak3+a1k5tαΓ1+α,u2,3x,y,t=−a−1λ4k5a−1cot4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5+k5a−1λ4k52a−1tan4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u2,4x,y,t=−a−1λ4k5a−1tan4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1λ4k52a−1cot4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u3,3x,y,t=−a−1λ2a1k5a−1cot2a1k5a−1λax+iak1k2y+λ−4aa−1k3k5tα4k5a−11+α−k62k5,u3,4x,y,t=−a−1λ/2a1k5a−1tan2a1k5a−1/λax+iak1/k2y+λ−4aa−1k3k5tα/4k5a−11+α−k6/2k5,Whenk1k2<0 and λ=4k5k7−k62<0, we have σ<0, then
(38)u4,3x,y,t=−k62k5+a1λ2a1k5tanλ4a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u4,4x,y,t=−k62k5−a1λ2a1k5cotλ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u5,3x,y,t=−a−1λ4k5a−1cot4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5+k5a−1λ4k52a−1tan4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α,u5,4x,y,t=−a−1λ4k5a−1tan4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1λ4k52a−1cot4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α,u6,3x,y,t=a−1λ2k5a−1cot2k5a−1λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5,u6,4x,y,t=a−1λ2k5a−1tan2k5a−1λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5.Whenλ=4k5k7−k62=0, we have σ=0, then
(39)u1,5x,y,t=−k62k5−Γ1+αa1ax+iak1/k2y−ak3+a1k5tα/Γ1+αα+ω,u2,5x,y,t=−a−1ax+iak1/k2y−ak3tα/Γ1+αα+a−1ωΓ1+α−k62k5=u3,5x,y,t,u4,5x,y,t=−k62k5a1Γ1+αax+a−k1/k2y−ak3+ak3−k1/k2+a1k5tα/Γ1+αα+ω,u5,5x,y,t=−a−1Γ1+αaαx+−k1/k2y−k3+k4−k1/k2tαΓ1+αα+ω−k62k5.Solutionsui,1,ui,2i=1,2,⋯,6 describe the soliton. Solitons exist everywhere in the nature; they are special kinds of solitary waves. u2,1,u2,2,u5,1,andu5,2 describe the multiple soliton solutions. Solutions ui,3andui,4i=1,2,⋯,6 represent the exact periodic traveling wave solutions. Periodic solutions are traveling wave solutions.Figures1–6 present the solutions: u1,1,u2,1,u1,3,u2,3,u1,5,andu5,5 of the generalized time fractional biological population model with −10<x<10,0<t<10. Solution u1,1 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a1=1,a=2,andy=0; solution u2,1 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a−1=1,a=2,andy=0; solution u1,3 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=2,k7=1,a1=1,a=2,andy=0; solution u2,3 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a−1=1,a=2,andy=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=8,k7=1,a=2,a1=1,y=0,andω=0; solution u5,5 is presented for values α=0.5,k1=1,k2=−1,k3=0,k4=1,k5=2,k6=8,k7=1,a1=1,a=2,y=0,andω=0.Figure 1
u1,1x,0,t of Equation (21).Figure 2
u2,1x,0,t of Equation (21).Figure 3
u1,3x,0,t of Equation (21).Figure 4
u2,3x,0,t of Equation (21).Figure 5
u1,5x,0,t of Equation (21).Figure 6
u5,5x,0,t of Equation (21).Whenk1=−1,k2=−1,k3=k4=k6=0,k5=−h≠0,andk7=rh≠0, Equation (21) is the time fractional biological population model [39]
(40)Dtαu=u2xx+u2yy+hu2−r.We denote(41)ux,y,t=uξ,ξ=ax+iay+btαΓ1+α,a≠0,and find the solution of Equation (40) in the form of uξ=a−1φ−1+a0+a1φ. Then, by solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(42)a−1=0,a0=0,a1≠0,σ=−ra12.Leta1=c/h, then σ=−rh2/c2, we can obtain set 1 in [42].Set 2(43)a−1≠0,a0=0,a1=r4a−1,σ=−4a−12r.Leta−1=rh/4c, then σ=−rh2/4c2, we can obtain set 3 in [42].Set 3(44)a−1≠0,a0=0,a1=0,σ=−a−12r.Leta−1=rh/c, then σ=−rh2/c2, we can obtain set 2 in [42].Clearly, we get more solutions of the time fractional biological population model (40) than the literature [42].


### 3.2. The Generalized Compound KdV-Burgers Equation

The generalized compound KdV-Burgers equation is given by(45)Dtαu+k1uux+k2u2ux+k3uxx+k4uxxx=0,Let(46)ux,t=uξ,ξ=ax+btαΓ1+α,a≠0,then Equation (45) is reduced to ordinary differential equation as
(47)bu′+k1auu′+k2au2u′+k3a2u″+k4a3u′″=0.The solution of Equation (45) is in the form of (18), and here, n=1 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term u2u′. We obtain the solution of Equation (45) as follows:
(48)uξ=a−1φ−1+a0+a1φ.Substituting Equation (48) together with its necessary derivatives into Equation (47), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(49)φ−4:−6a3k4σ3a−1−ak2σa−13,φ−3:2a2k3σ2a−1−2ak2σa−12a0−ak1σa−12,φ−2:bσa−1−ak1σa−1a0−ak2a−13−ak2σa−1a02−ak2σa−12a1−8a3k4σ2a−1,φ−1:−ak1a−12−2ak2a−12a0+2a2k3σa−1,(50)φ0:ba−1−bσa1+ak1σa0a1−ak1a−1a0+ak2σa02a1+ak2σa−1a12−ak2a02a−1−ak2a−12a1+2a3k4σ2a1−2a3k4σa−1,φ1:ak1σa12+2ak2σa0a12+2a2k3σa1,φ2:−ba1+ak1a0a1+ak2a02a1+ak2a−1a12+ak2σa13+8a3k4σa1,φ3:2ak2a0a12+2a2k3a1+ak1a12,φ4:6a3k4a1+ak2a13.Let the coefficients ofφkξ to be zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Whenk2k4<0, it denotes μ=3ak12k4+2ak2k32−12bk2k4 and λ1=±−6k4/k2,λ2=−k1/2±∣k3∣−6k2k4/6k4, we haveSet 1(51)a−1=−λ1μ96a2k2k42,a0=−λ1k1+2k32λ1k2,a1=aλ1,a=a,b=b,σ=μ96a3k2k42.Set 2(52)a−1=0,a0=−λ1k1+2k32λ1k2,a1=aλ1,a=a,b=b,σ=μ24a3k2k42.Set 3(53)a−1=−μk312a2k2k422λ2+k1,a0=λ2k2,a1=0,a=a,b=b,σ=μ24a3k2k42.Thus, we obtain the solution of Equation (45) as ui,jx,t,i=1,2,3;j=1,2,3,4,5. Using set 1 to set 3, ui,jx,t is as follows:Whenk2k4<0 and aμk2<0, we have σ<0, then
(54)u1,1x,t=λ1μ∣a∣24a2k4−6aμk2coth14∣ak4∣−μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14∣ak4∣−μ6ak2tanh14∣ak4∣−μ6ak2ax+btαΓ1+αu1,2x,t=λ1μ∣a∣24a2k4−6aμk2tanh14∣ak4∣−μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14∣ak4∣−μ6ak2coth14∣ak4∣−μ6ak2ax+btαΓ1+αu2,1x,t=−λ1k1+2k32λ1k2−aλ12∣ak4∣−μ6ak2tanh12∣ak4∣−μ6ak2ax+btαΓ1+αu2,2x,t=−λ1k1+2k32λ1k2−aλ12ak4−μ6ak2coth,u3,1x,t=μk3a6a2k42λ2+k1−6aμk2coth+λ2k2,u3,2x,t=μk3a6a2k42λ2+k1−6aμk2tanh+λ2k2.Whenk2k4<0 and aμk2>0, we have σ>0, then
(55)u1,3x,t=−λ1μ∣a∣24a2k46aμk2cot14∣ak4∣μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2+aλ14∣ak4∣μ6ak2tan14ak4μ6ak2ax+btαΓ1+α,u1,4x,t=λ1μ∣a∣24a2k46aμk2tan14∣ak4∣μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14ak4μ6ak2cot14ak4μ6ak2ax+btαΓ1+α,u2,3x,t=−λ1k1+2k32λ1k2+aλ12ak4μ6ak2tan12ak4μ6ak2ax+btαΓ1+α,u2,4x,t=−λ1k1+2k32λ1k2−aλ12ak4μ6ak2coth12ak4μ6ak2ax+btαΓ1+α,u3,3x,t=−μk36a2k42λ2+k16a3μk2cot12k4μ6a3k2ax+btαΓ1+α+λ2k2,u3,4x,t=μk3a6a2k42λ2+k16aμk2tan12ak4μ6ak2ax+btαΓ1+α+λ2k2.Whenk2k4<0andμ=0, we have σ=0, then
(56)u1,5x,t=−λ1k1+2k32λ1k2+aλ1Γ1+αax+btα/Γ1+αα+ω=u2,5x,t.Solutionsui,1andui,2i=1,2,3 described the soliton. u1,1andu1,2 describe the multiple soliton solutions. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures7–10 present the solutions: u1,1,u2,1,u1,3,andu2,3 of the generalized compound KdV-Burgers equation with −10<x<10,0<t<1. Solutions u1,1 and u2,1 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,andy=0; solutions u1,3 and u2,3 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,andy=0.Figure 7
u1,1x,t of Equation (45).Figure 8
u2,1x,t of Equation (45).Figure 9
u1,3x,t of Equation (45).Figure 10
u2,3x,t of Equation (45).


### 3.3. The Space-Time Fractional Regularized Long-Wave Equation

Consider the space-time fractional regularized long-wave equation as follows:(57)Dtαu+k1Dxαu+k2uDxαu+k3DtαDx2αu=0.Let(58)ux,t=uξ,ξ=axαΓ1+α+btαΓ1+α,a≠0,then Equation (57) is reduced to the following ordinary differential equation:
(59)bu′+k1au′+k2auu′+k3a2bu′″=0.The solution of Equation (57) is in the form of (18), and here, n=2 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term uu′. We obtain the solution of Equation (57) as follows:
(60)uξ=a−2φ−2+a−1φ−1+a0+a1φ+a2φ2.Substituting Equation (60) together with its necessary derivatives into Equation (57), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(61)φ−5:−24k3a2ba−2σ3−2k2aa−22σ,φ−4:−6k3a2ba−1σ3−3k2aa−1a−2σ,φ−3:−2k1aa−2σ−2ba−2σ−2k2aa0a−2σ−k2aa−12σ−2k2aa−22−40k3a2ba−2σ2,φ−2:−k1aa−1σ−ba−1σ−k2aa1a−2σ−3k2aa−1a−2−k2aa0a−1σ−8k3a2ba−1σ2,φ−1:−2ba−2−2k1aa−2−k2aa−12−2k2aa0a−2−16k3a2ba−2σ,φ0:k1aa1σ−k1aa−1+a1bσ−a−1b−k2aa1a−2+k2aa−1a2σ+k2aa0a1σ−k2aa0a−1+2k3a2ba1σ2−2k3a2ba−1σ,(62)φ1:2k1aa2σ+2ba2σ+2k2aa0a2σ+k2aa12σ+16k3a2ba2σ2,φ2:k1aa1+ba1+k2aa2a−1+k2aa1a0+3k2aa1a2σ+8k3a2ba1σ,φ3:2k1aa2+2ba2+2k2aa22σ+2k2aa0a2+k2aa12+40k3a2ba2σ,φ4:6k3a2ba1+3k2aa1a2,φ5:24k3a2ba2+2k2aa22.Let the coefficients ofφkξ tbe zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(63)a−2=0,a−1=0,a0=a0,a1=0,a2=−12k3abk2,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Set 2(64)a−2=−3k1a+k2a0a+b216k2k3a3b,a−1=0,a0=a0,a1=0,a2=0,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Set 3(65)a−2=−3k1a+k2a0a+b216k2k3a3b,a−1=0,a0=a0,a1=0,a2=−12k3abk2,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Thus, we obtain the solution of Equation (57) as ui,jξ,i=1,2,3;j=1,2,3,4,5 and ξ=axα/Γ1+α+btα/Γ1+α; ui,jξ is as follows:Whenk1a+k2a0a+b/8k3a2b>0, we have σ<0, then
(66)u1,1ξ=a0−3k1a+k2a0a+b2k2atanh2k1a+k2a0a+b8k3a2bξ=u2,2ξ,u2,1ξ=−3k1a+k2a0a+b2k2acoth2k1a+k2a0a+b8k3a2bξ+a0=u1,2ξ,u3,1ξ=−3k1a+k2a0a+b2k2acoth2k1a+k2a0a+b8k3a2bξ+a0−3k1a+k2a0a+b2k2atanh2k1a+k2a0a+b8k3a2bξ=u3,2ξ.Whenk1a+k2a0a+b/8k3a2b<0, we have σ>0, then
(67)u1,3ξ=a0+3k1a+k2a0a+b2k2atan2−k1a+k2a0a+b8k3a2bξ=u2,4ξ,u2,3ξ=3k1a+k2a0a+b2k2acot2−k1a+k2a0a+b8k3a2bξ+a0=u1,4ξ,u3,3ξ=3k1a+k2a0a+b2k2acot2−k1a+k2a0a+b8k3a2bξ+a0+3k1a+k2a0a+b2k2atan2−k1a+k2a0a+b8k3a2bξ=u3,4ξ.Whenk1a+k2a0a+b/8k3a2b=0,a−2=0, we have σ=0, then
(68)u1,5ξ=a0−12k3abΓ21+αk2ξα+ω2=u3,5ξ.Solutionsui,1andui,2i=1,2,3 describe the multiple soliton. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures11–14 present the solutions: u1,1,u2,1,u2,3,andu1,5 of the generalized compound KdV-Burgers equation with 0<x<1,0<t<2. Solution u1,1 and solution u2,1 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,a0=0,andy=0; u2,3 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,a0=0,andy=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=−4,a0=0,ω=0,andy=0.Figure 11
u1,1x,t of Equation (57).Figure 12
u2,1x,t of Equation (57).Figure 13
u2,3x,t of Equation (57).Figure 14
u1,5x,t of Equation (57).


### 3.4. The (3+1)-Space-Time Fractional Zakharov-Kuznetsov Equation

The (3+1)-space-time fractional Zakharov-Kuznetsov Equation is given by
(69)Dtαu+k1uDxαu+k2Dx3αu+k3DxαDy2αu+k4DxαDz2αu=0,Let(70)ux,t=uξ,ξ=atαΓ1+α+bxαΓ1+α+cyαΓ1+α+dzαΓ1+α,a≠0,then Equation (69) is reduced to the ordinary differential equation as
(71)au′+k1buu′+k2b3u′″+k3bc2u′″+k4bd2u′″=0.The solution of Equation (71) is the form (18), and here, n=2 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term uu′. We obtain the solution of Equation (71) as
(72)uξ=a−2φ−2+a−1φ−1+a0+a1φ+a2φ2.Substituting Equation (72) together with its necessary derivatives into Equation (71), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(73)φ−5:−2k1ba−22σ+24λa−2σ3,φ−4:−3k1ba−1a−2σ+6λa−1σ3,φ−3:−2aa−2σ−2k1ba0a−2σ−k1ba−12σ−2k1ba−22+40λa−2σ2,φ−2:−aa−1σ−3k1ba−1a−2−k1ba0a−1σ−k1ba1a−2σ+8λa−1σ2,φ−1:−2aa−2−k1ba−12−2k1ba0a−2+16λa−2σ,φ0:aa1σ−aa−1−k1ba−2a1+k1ba−1a2σ+k1ba0a1σ−k1ba0a−1−2λa1σ2+2λa−1σ,φ1:2aa2σ+2k1ba0a2σ+k1ba12σ−16λa2σ2,φ2:aa1+k1ba−1a2+k1ba0a1+3k1ba1a2σ−8λa1σ,φ3:2aa2+2k1ba22σ+2k1ba0a2+k1ba12−40λa2σ,φ4:3k1ba1a2−6λa1,φ5:2k1ba22−24λa2.where λ=−k2b3−k3bc2−k4bd2.Let the coefficients ofφkξ be zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(74)a−2=0,a−1=0,a0=a0,a1=0,a2=12λk1,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Set 2(75)a−2=3a+k1a0b216k1b2λ,a−1=0,a0=a0,a0=a0,a1=0,a2=0,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Set 3(76)a−2=3a+k1a0b216k1b2λ,a−1=0,a0=a0,a1=0,a2=12λk1,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Thus, we obtain the solution of Equation (59) as ui,jξ,i=1,2,3;j=1,2,3,4,5 and ξ=atα/Γ1+α+bxα/Γ1+α+cyα/Γ1+α+dzα/Γ1+α; ui,jξ is tas follows:Whena+k1a0b/8bλ<0, we have σ<0, then
(77)u1,1ξ=a0−3a+k1a0b2k1btanh2−a+k1a0b8bλξ,u2,1ξ=−12λa+k1a0bk1coth2−a+k1a0b8bλξ+a0,u3,1ξ=−12λa+k1a0bk1coth2−a+k1a0b8bλξ+a0−3a+k1a0b2k1btanh2−a+k1a0b8bλξ,u1,2ξ=a0−3a+k1a0b2k1bcoth2−a+k1a0b8bλξ,u2,2ξ=−12λa+k1a0bk1tanh2−a+k1a0b8bλξ+a0,u3,2ξ=−12λa+k1a0bk1tanh2−a+k1a0b8bλξ+a0−3a+k1a0b2k1bcoth2−a+k1a0b8bλξ.Whena+k1a0b/8bλ>0, we have σ>0, then
(78)u1,3ξ=a0+3a+k1a0b2k1btan2a+k1a0b8bλξ,u2,3ξ=12λa+k1a0bk1cot2a+k1a0b8bλξ+a0,u3,3ξ=12λa+k1a0bk1cot2a+k1a0b8bλξ+a0+3a+k1a0b2k1btan2a+k1a0b8bλξ,u1,4ξ=a0+3a+k1a0b2k1bcot2a+k1a0b8bλξ,u2,4ξ=12λa+k1a0bk1tan2a+k1a0b8bλξ+a0,u3,4ξ=12λa+k1a0bk1tan2a+k1a0b8bλξ+a0+3a+k1a0b2k1bcot2a+k1a0b8bλξ.Whena+k1a0b/8bλ=0, we have σ=0, then
(79)u1,5ξ=a0+12λΓ21+αk1ξα+ω2=u3,5ξ.Solutionsui,1andui,2i=1,2,3 describe the multiple soliton. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures15–18 present the solutions: u1,1,u1,3,u2,3,andu1,5 of the generalized compound KdV-Burgers equation with 0<x<1,0<t<2. Solution u1,1 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,c=1,d=1,a0=0,andy=z=0; solution u1,3 and u2,3 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,c=1,d=1,a0=0,andy=z=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,c=1,d=1,a0=4,ω=0,andy=z=0.Figure 15
u1,1x,0,0,t of Equation (69).Figure 16
u1,3x,0,0,t of Equation (69).Figure 17
u2,3x,0,0,t of Equation (69).Figure 18
u1,5x,0,0,t of Equation (69).


## 3.1. The Generalized Time Fractional Biological Population Model

The generalized time fractional biological population model is given by(21)Dtαu+k1u2xx+k2u2yy+k3ux+k4uy+k5u2+k6u+k7=0,where u=ux,y,t is an unknown function.We considered two cases.Case 1.
Whenk1k2>0, let
(22)ux,y,t=uξ,ξ=ax+iak1k2y+btαΓ1+α,a≠0,then Equation (21) is reduced to the ordinary differential equation as follows:
(23)ak4k1k2u′=0,ak3+bu′+k5u2+k6u+k7=0.Whenk4≠0, Equation (23) has only constant solutions.Whenk4=0, Equation (23) has solutions in the form of (18). u′ is obtained from the homogeneous balance between the highest order derivative and the nonlinear term u2. We obtain the solution of Equation (21) as follows:
(24)uξ=a−1φ−1+a0+a1φ,Substituting Equation (24) together with its necessary derivatives into Equation (21), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(25)φ−2:−k3σa−1−bσa−1+k5a2−1,φ−1:2k5a−1a0+a−1k6,φ0:−ak3a−1−ba−1+ak3σa1+bσa1+k5a02+2k5a−1a1+k6a0+k7,φ1:2k5a0a1+k6a1,φ2:ak3a1+ba1+k5a12.Let the coefficients be zero. By solving the set of equations given above fora−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(26)a−1=0,a0=−k62k5,a1=a1,a=a,b=−ak3−a1k5,σ=λ4a12k52.Set 2(27)a−1=a−1,a0=−k62k5,a1=−λ16k52a−1,a=a,b=λ−16aa−1k3k516k5a−1,σ=16k52a−12λ.Set 3(28)a−1=a−1,a0=−k62k5,a1=0,a=a,b=λ−4aa−1k3k54k5a−1,σ=4k52a−12λ.where λ=4k5k7−k62.Case 2.
Whenk1k2<0, let
(29)ux,y,t=uξ,ξ=ax+a−k1k2y+btαΓ1+α,a≠0,then Equation (21) is reduced to the ordinary differential equation as follows:
(30)b+k3a+ak4−k1k2u′+k5u2+k6u+k7=0.The solution of Equation (30) is in the form of (18). n=1 is taken from the homogeneous balance between the highest order derivative u′ and the nonlinear term u2. We obtain the solution of Equation (30) as Equation (24). Substituting Equation (24) together with its necessary derivatives into Equation (30), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(31)φ−2:−a−k1k2k4σa−1−ak3σa−1−bσa−1+k5a−12,φ−1:2k5a−1a0+a−1k6,φ0:a−k1k2k4σa1+ak3σa1+bσa1−a−k1k2k4a−1,φ1:2k5a0a1+k6a1,φ2:a−k1k2k4a1+ak3a1+ba1+k5a12.Let the coefficients be zero. By solving the set of equations given above fora−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 4(32)a−1=0,a0=−k62k5,a1=a1,a=a,b=−ak3−ak4−k1k2−a1k5,σ=λ4a12k52.Set 5(33)a−1=a−1,a0=−k62k5,a1=−λ16k52a−1,a=a,b=λ−16μ16k5a−1,σ=16k52a−12λ.Set 6(34)a−1=a−1,a0=−k62k5,a1=0,a=a,b=λ−4μ4k5a−1,σ=4k52a−12λ⋅where μ=aa−1k3k5+aa−1k4k5−k1/k2.We find thata−1,a0,a1, and σ are equal in set 1 and set 4, set 2 and set 5, and set 3 and set 6, respectively. In this study, the solutions of differential equations are symbolized as ui,jx,y,t,i,j∈Z+, where i denotes obtained set number and j is the solution number of the Riccati equation, respectively. Thus, using set 1 to set 6, we obtain the solution of Equation (21) as ui,jx,y,t,i=1,2,⋯,6;j=1,2,⋯,5. ui,jx,y,t is the following:Whenk1k2>0, λ=4k5k7−k62<0, we have σ<0, then
(35)u1,1x,y,t=−k62k5−a1−λ2a1k5tanh−λ2a1k5ax+iak1k2y−ak3+a1k5tαΓ1+α,u1,2x,y,t=−k62k5−a1−λ2a1k5coth−λ2a1k5ax+iak1k2y−ak3+a1k5tαΓ1+α,u2,1x,y,t=−a−1−λ4k5a−1coth4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1tanh4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u2,2x,y,t=−a−1−λ4k5a−1tanh4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1coth4k5a−1−λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u3,1x,y,t=a1−λ2a1k5a−1coth2a1k5a−1−λax+iak1k2y−λ−4aa−1k3k5tα4k5a−1Γ1+α−k62k5,u3,2x,y,t=a−1−λ2a1k5a−1tanh2a1k5a−1−λax+iak1k2y−λ−4aa−1k3k5tα4k5a−1Γ1+α−k62k5.Whenk1k2<0, λ=4k5k7−k62<0, we have σ<0, then
(36)u4,1x,y,t=−k62k5−a1−λ2a1k5tanh−λ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u4,2x,y,t=−k62k5−a1−λ2a1k5coth−λ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u5,1x,y,t=−a−1−λ4k5a−1coth4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1tanh4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α,u5,2x,y,t=−a−1−λ4k5a−1tanh4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1−λ4k52a−1coth4k5a−1−λax−a−k1k2y+λ−16μtα16k5a−1Γ1+α,u6,1x,y,t=a−1−λ2a1k5a−1coth2a1k5a−1−λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5,u6,2x,y,t=a−1−λ2a1k5a−1tanh2a1k5a−1−λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5.Whenk1k2<0 and λ=4k5k7−k62<0, we have σ<0, then
(37)u1,3x,y,t=−k62k5+a1λ2a1k5tanλ2a1k5ax+iak1k2y+ak3+a1k5tαΓ1+α,u1,4x,y,t=−k62k5+a1λ2a1k5cotλ2a1k5ax+iak1k2y+ak3+a1k5tαΓ1+α,u2,3x,y,t=−a−1λ4k5a−1cot4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5+k5a−1λ4k52a−1tan4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u2,4x,y,t=−a−1λ4k5a−1tan4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α−k62k5−k5a−1λ4k52a−1cot4k5a−1λax−iak1k2y+λ−16aa−1k3k5tα16k5a−1Γ1+α,u3,3x,y,t=−a−1λ2a1k5a−1cot2a1k5a−1λax+iak1k2y+λ−4aa−1k3k5tα4k5a−11+α−k62k5,u3,4x,y,t=−a−1λ/2a1k5a−1tan2a1k5a−1/λax+iak1/k2y+λ−4aa−1k3k5tα/4k5a−11+α−k6/2k5,Whenk1k2<0 and λ=4k5k7−k62<0, we have σ<0, then
(38)u4,3x,y,t=−k62k5+a1λ2a1k5tanλ4a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u4,4x,y,t=−k62k5−a1λ2a1k5cotλ2a1k5ax+a−k1k2y−ak3+ak4−k1/k2+a1k5tαΓ1+α,u5,3x,y,t=−a−1λ4k5a−1cot4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5+k5a−1λ4k52a−1tan4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α,u5,4x,y,t=−a−1λ4k5a−1tan4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α−k62k5−k5a−1λ4k52a−1cot4k5a−1λax+a−k1k2y+λ−16μtα16k5a−1Γ1+α,u6,3x,y,t=a−1λ2k5a−1cot2k5a−1λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5,u6,4x,y,t=a−1λ2k5a−1tan2k5a−1λax+a−k1k2y+λ−4μtα4k5a−1Γ1+α−k62k5.Whenλ=4k5k7−k62=0, we have σ=0, then
(39)u1,5x,y,t=−k62k5−Γ1+αa1ax+iak1/k2y−ak3+a1k5tα/Γ1+αα+ω,u2,5x,y,t=−a−1ax+iak1/k2y−ak3tα/Γ1+αα+a−1ωΓ1+α−k62k5=u3,5x,y,t,u4,5x,y,t=−k62k5a1Γ1+αax+a−k1/k2y−ak3+ak3−k1/k2+a1k5tα/Γ1+αα+ω,u5,5x,y,t=−a−1Γ1+αaαx+−k1/k2y−k3+k4−k1/k2tαΓ1+αα+ω−k62k5.Solutionsui,1,ui,2i=1,2,⋯,6 describe the soliton. Solitons exist everywhere in the nature; they are special kinds of solitary waves. u2,1,u2,2,u5,1,andu5,2 describe the multiple soliton solutions. Solutions ui,3andui,4i=1,2,⋯,6 represent the exact periodic traveling wave solutions. Periodic solutions are traveling wave solutions.Figures1–6 present the solutions: u1,1,u2,1,u1,3,u2,3,u1,5,andu5,5 of the generalized time fractional biological population model with −10<x<10,0<t<10. Solution u1,1 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a1=1,a=2,andy=0; solution u2,1 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a−1=1,a=2,andy=0; solution u1,3 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=2,k7=1,a1=1,a=2,andy=0; solution u2,3 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=3,k7=1,a−1=1,a=2,andy=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=0,k5=2,k6=8,k7=1,a=2,a1=1,y=0,andω=0; solution u5,5 is presented for values α=0.5,k1=1,k2=−1,k3=0,k4=1,k5=2,k6=8,k7=1,a1=1,a=2,y=0,andω=0.Figure 1
u1,1x,0,t of Equation (21).Figure 2
u2,1x,0,t of Equation (21).Figure 3
u1,3x,0,t of Equation (21).Figure 4
u2,3x,0,t of Equation (21).Figure 5
u1,5x,0,t of Equation (21).Figure 6
u5,5x,0,t of Equation (21).Whenk1=−1,k2=−1,k3=k4=k6=0,k5=−h≠0,andk7=rh≠0, Equation (21) is the time fractional biological population model [39]
(40)Dtαu=u2xx+u2yy+hu2−r.We denote(41)ux,y,t=uξ,ξ=ax+iay+btαΓ1+α,a≠0,and find the solution of Equation (40) in the form of uξ=a−1φ−1+a0+a1φ. Then, by solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(42)a−1=0,a0=0,a1≠0,σ=−ra12.Leta1=c/h, then σ=−rh2/c2, we can obtain set 1 in [42].Set 2(43)a−1≠0,a0=0,a1=r4a−1,σ=−4a−12r.Leta−1=rh/4c, then σ=−rh2/4c2, we can obtain set 3 in [42].Set 3(44)a−1≠0,a0=0,a1=0,σ=−a−12r.Leta−1=rh/c, then σ=−rh2/c2, we can obtain set 2 in [42].Clearly, we get more solutions of the time fractional biological population model (40) than the literature [42].


## 3.2. The Generalized Compound KdV-Burgers Equation

The generalized compound KdV-Burgers equation is given by(45)Dtαu+k1uux+k2u2ux+k3uxx+k4uxxx=0,Let(46)ux,t=uξ,ξ=ax+btαΓ1+α,a≠0,then Equation (45) is reduced to ordinary differential equation as
(47)bu′+k1auu′+k2au2u′+k3a2u″+k4a3u′″=0.The solution of Equation (45) is in the form of (18), and here, n=1 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term u2u′. We obtain the solution of Equation (45) as follows:
(48)uξ=a−1φ−1+a0+a1φ.Substituting Equation (48) together with its necessary derivatives into Equation (47), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(49)φ−4:−6a3k4σ3a−1−ak2σa−13,φ−3:2a2k3σ2a−1−2ak2σa−12a0−ak1σa−12,φ−2:bσa−1−ak1σa−1a0−ak2a−13−ak2σa−1a02−ak2σa−12a1−8a3k4σ2a−1,φ−1:−ak1a−12−2ak2a−12a0+2a2k3σa−1,(50)φ0:ba−1−bσa1+ak1σa0a1−ak1a−1a0+ak2σa02a1+ak2σa−1a12−ak2a02a−1−ak2a−12a1+2a3k4σ2a1−2a3k4σa−1,φ1:ak1σa12+2ak2σa0a12+2a2k3σa1,φ2:−ba1+ak1a0a1+ak2a02a1+ak2a−1a12+ak2σa13+8a3k4σa1,φ3:2ak2a0a12+2a2k3a1+ak1a12,φ4:6a3k4a1+ak2a13.Let the coefficients ofφkξ to be zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Whenk2k4<0, it denotes μ=3ak12k4+2ak2k32−12bk2k4 and λ1=±−6k4/k2,λ2=−k1/2±∣k3∣−6k2k4/6k4, we haveSet 1(51)a−1=−λ1μ96a2k2k42,a0=−λ1k1+2k32λ1k2,a1=aλ1,a=a,b=b,σ=μ96a3k2k42.Set 2(52)a−1=0,a0=−λ1k1+2k32λ1k2,a1=aλ1,a=a,b=b,σ=μ24a3k2k42.Set 3(53)a−1=−μk312a2k2k422λ2+k1,a0=λ2k2,a1=0,a=a,b=b,σ=μ24a3k2k42.Thus, we obtain the solution of Equation (45) as ui,jx,t,i=1,2,3;j=1,2,3,4,5. Using set 1 to set 3, ui,jx,t is as follows:Whenk2k4<0 and aμk2<0, we have σ<0, then
(54)u1,1x,t=λ1μ∣a∣24a2k4−6aμk2coth14∣ak4∣−μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14∣ak4∣−μ6ak2tanh14∣ak4∣−μ6ak2ax+btαΓ1+αu1,2x,t=λ1μ∣a∣24a2k4−6aμk2tanh14∣ak4∣−μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14∣ak4∣−μ6ak2coth14∣ak4∣−μ6ak2ax+btαΓ1+αu2,1x,t=−λ1k1+2k32λ1k2−aλ12∣ak4∣−μ6ak2tanh12∣ak4∣−μ6ak2ax+btαΓ1+αu2,2x,t=−λ1k1+2k32λ1k2−aλ12ak4−μ6ak2coth,u3,1x,t=μk3a6a2k42λ2+k1−6aμk2coth+λ2k2,u3,2x,t=μk3a6a2k42λ2+k1−6aμk2tanh+λ2k2.Whenk2k4<0 and aμk2>0, we have σ>0, then
(55)u1,3x,t=−λ1μ∣a∣24a2k46aμk2cot14∣ak4∣μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2+aλ14∣ak4∣μ6ak2tan14ak4μ6ak2ax+btαΓ1+α,u1,4x,t=λ1μ∣a∣24a2k46aμk2tan14∣ak4∣μ6ak2ax+btαΓ1+α−λ1k1+2k32λ1k2−aλ14ak4μ6ak2cot14ak4μ6ak2ax+btαΓ1+α,u2,3x,t=−λ1k1+2k32λ1k2+aλ12ak4μ6ak2tan12ak4μ6ak2ax+btαΓ1+α,u2,4x,t=−λ1k1+2k32λ1k2−aλ12ak4μ6ak2coth12ak4μ6ak2ax+btαΓ1+α,u3,3x,t=−μk36a2k42λ2+k16a3μk2cot12k4μ6a3k2ax+btαΓ1+α+λ2k2,u3,4x,t=μk3a6a2k42λ2+k16aμk2tan12ak4μ6ak2ax+btαΓ1+α+λ2k2.Whenk2k4<0andμ=0, we have σ=0, then
(56)u1,5x,t=−λ1k1+2k32λ1k2+aλ1Γ1+αax+btα/Γ1+αα+ω=u2,5x,t.Solutionsui,1andui,2i=1,2,3 described the soliton. u1,1andu1,2 describe the multiple soliton solutions. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures7–10 present the solutions: u1,1,u2,1,u1,3,andu2,3 of the generalized compound KdV-Burgers equation with −10<x<10,0<t<1. Solutions u1,1 and u2,1 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,andy=0; solutions u1,3 and u2,3 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,andy=0.Figure 7
u1,1x,t of Equation (45).Figure 8
u2,1x,t of Equation (45).Figure 9
u1,3x,t of Equation (45).Figure 10
u2,3x,t of Equation (45).


## 3.3. The Space-Time Fractional Regularized Long-Wave Equation

Consider the space-time fractional regularized long-wave equation as follows:(57)Dtαu+k1Dxαu+k2uDxαu+k3DtαDx2αu=0.Let(58)ux,t=uξ,ξ=axαΓ1+α+btαΓ1+α,a≠0,then Equation (57) is reduced to the following ordinary differential equation:
(59)bu′+k1au′+k2auu′+k3a2bu′″=0.The solution of Equation (57) is in the form of (18), and here, n=2 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term uu′. We obtain the solution of Equation (57) as follows:
(60)uξ=a−2φ−2+a−1φ−1+a0+a1φ+a2φ2.Substituting Equation (60) together with its necessary derivatives into Equation (57), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(61)φ−5:−24k3a2ba−2σ3−2k2aa−22σ,φ−4:−6k3a2ba−1σ3−3k2aa−1a−2σ,φ−3:−2k1aa−2σ−2ba−2σ−2k2aa0a−2σ−k2aa−12σ−2k2aa−22−40k3a2ba−2σ2,φ−2:−k1aa−1σ−ba−1σ−k2aa1a−2σ−3k2aa−1a−2−k2aa0a−1σ−8k3a2ba−1σ2,φ−1:−2ba−2−2k1aa−2−k2aa−12−2k2aa0a−2−16k3a2ba−2σ,φ0:k1aa1σ−k1aa−1+a1bσ−a−1b−k2aa1a−2+k2aa−1a2σ+k2aa0a1σ−k2aa0a−1+2k3a2ba1σ2−2k3a2ba−1σ,(62)φ1:2k1aa2σ+2ba2σ+2k2aa0a2σ+k2aa12σ+16k3a2ba2σ2,φ2:k1aa1+ba1+k2aa2a−1+k2aa1a0+3k2aa1a2σ+8k3a2ba1σ,φ3:2k1aa2+2ba2+2k2aa22σ+2k2aa0a2+k2aa12+40k3a2ba2σ,φ4:6k3a2ba1+3k2aa1a2,φ5:24k3a2ba2+2k2aa22.Let the coefficients ofφkξ tbe zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(63)a−2=0,a−1=0,a0=a0,a1=0,a2=−12k3abk2,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Set 2(64)a−2=−3k1a+k2a0a+b216k2k3a3b,a−1=0,a0=a0,a1=0,a2=0,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Set 3(65)a−2=−3k1a+k2a0a+b216k2k3a3b,a−1=0,a0=a0,a1=0,a2=−12k3abk2,a=a,b=b,σ=−k1a+k2a0a+b8k3a2b.Thus, we obtain the solution of Equation (57) as ui,jξ,i=1,2,3;j=1,2,3,4,5 and ξ=axα/Γ1+α+btα/Γ1+α; ui,jξ is as follows:Whenk1a+k2a0a+b/8k3a2b>0, we have σ<0, then
(66)u1,1ξ=a0−3k1a+k2a0a+b2k2atanh2k1a+k2a0a+b8k3a2bξ=u2,2ξ,u2,1ξ=−3k1a+k2a0a+b2k2acoth2k1a+k2a0a+b8k3a2bξ+a0=u1,2ξ,u3,1ξ=−3k1a+k2a0a+b2k2acoth2k1a+k2a0a+b8k3a2bξ+a0−3k1a+k2a0a+b2k2atanh2k1a+k2a0a+b8k3a2bξ=u3,2ξ.Whenk1a+k2a0a+b/8k3a2b<0, we have σ>0, then
(67)u1,3ξ=a0+3k1a+k2a0a+b2k2atan2−k1a+k2a0a+b8k3a2bξ=u2,4ξ,u2,3ξ=3k1a+k2a0a+b2k2acot2−k1a+k2a0a+b8k3a2bξ+a0=u1,4ξ,u3,3ξ=3k1a+k2a0a+b2k2acot2−k1a+k2a0a+b8k3a2bξ+a0+3k1a+k2a0a+b2k2atan2−k1a+k2a0a+b8k3a2bξ=u3,4ξ.Whenk1a+k2a0a+b/8k3a2b=0,a−2=0, we have σ=0, then
(68)u1,5ξ=a0−12k3abΓ21+αk2ξα+ω2=u3,5ξ.Solutionsui,1andui,2i=1,2,3 describe the multiple soliton. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures11–14 present the solutions: u1,1,u2,1,u2,3,andu1,5 of the generalized compound KdV-Burgers equation with 0<x<1,0<t<2. Solution u1,1 and solution u2,1 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,a0=0,andy=0; u2,3 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,a0=0,andy=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=−4,a0=0,ω=0,andy=0.Figure 11
u1,1x,t of Equation (57).Figure 12
u2,1x,t of Equation (57).Figure 13
u2,3x,t of Equation (57).Figure 14
u1,5x,t of Equation (57).


## 3.4. The (3+1)-Space-Time Fractional Zakharov-Kuznetsov Equation

The (3+1)-space-time fractional Zakharov-Kuznetsov Equation is given by
(69)Dtαu+k1uDxαu+k2Dx3αu+k3DxαDy2αu+k4DxαDz2αu=0,Let(70)ux,t=uξ,ξ=atαΓ1+α+bxαΓ1+α+cyαΓ1+α+dzαΓ1+α,a≠0,then Equation (69) is reduced to the ordinary differential equation as
(71)au′+k1buu′+k2b3u′″+k3bc2u′″+k4bd2u′″=0.The solution of Equation (71) is the form (18), and here, n=2 is taken from the homogeneous balance between the highest order derivative u′″ and the nonlinear term uu′. We obtain the solution of Equation (71) as
(72)uξ=a−2φ−2+a−1φ−1+a0+a1φ+a2φ2.Substituting Equation (72) together with its necessary derivatives into Equation (71), the algebraic equation is arranged according to the powers of the function φkξ. Then, the following coefficients are obtained:
(73)φ−5:−2k1ba−22σ+24λa−2σ3,φ−4:−3k1ba−1a−2σ+6λa−1σ3,φ−3:−2aa−2σ−2k1ba0a−2σ−k1ba−12σ−2k1ba−22+40λa−2σ2,φ−2:−aa−1σ−3k1ba−1a−2−k1ba0a−1σ−k1ba1a−2σ+8λa−1σ2,φ−1:−2aa−2−k1ba−12−2k1ba0a−2+16λa−2σ,φ0:aa1σ−aa−1−k1ba−2a1+k1ba−1a2σ+k1ba0a1σ−k1ba0a−1−2λa1σ2+2λa−1σ,φ1:2aa2σ+2k1ba0a2σ+k1ba12σ−16λa2σ2,φ2:aa1+k1ba−1a2+k1ba0a1+3k1ba1a2σ−8λa1σ,φ3:2aa2+2k1ba22σ+2k1ba0a2+k1ba12−40λa2σ,φ4:3k1ba1a2−6λa1,φ5:2k1ba22−24λa2.where λ=−k2b3−k3bc2−k4bd2.Let the coefficients ofφkξ be zero. By solving the set of equations given above for a−1,a0,a1,a,b, and σ, we obtain solution sets as follows:Set 1(74)a−2=0,a−1=0,a0=a0,a1=0,a2=12λk1,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Set 2(75)a−2=3a+k1a0b216k1b2λ,a−1=0,a0=a0,a0=a0,a1=0,a2=0,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Set 3(76)a−2=3a+k1a0b216k1b2λ,a−1=0,a0=a0,a1=0,a2=12λk1,a=a,b=b,c=c,d=d,σ=a+k1a0b8bλ.Thus, we obtain the solution of Equation (59) as ui,jξ,i=1,2,3;j=1,2,3,4,5 and ξ=atα/Γ1+α+bxα/Γ1+α+cyα/Γ1+α+dzα/Γ1+α; ui,jξ is tas follows:Whena+k1a0b/8bλ<0, we have σ<0, then
(77)u1,1ξ=a0−3a+k1a0b2k1btanh2−a+k1a0b8bλξ,u2,1ξ=−12λa+k1a0bk1coth2−a+k1a0b8bλξ+a0,u3,1ξ=−12λa+k1a0bk1coth2−a+k1a0b8bλξ+a0−3a+k1a0b2k1btanh2−a+k1a0b8bλξ,u1,2ξ=a0−3a+k1a0b2k1bcoth2−a+k1a0b8bλξ,u2,2ξ=−12λa+k1a0bk1tanh2−a+k1a0b8bλξ+a0,u3,2ξ=−12λa+k1a0bk1tanh2−a+k1a0b8bλξ+a0−3a+k1a0b2k1bcoth2−a+k1a0b8bλξ.Whena+k1a0b/8bλ>0, we have σ>0, then
(78)u1,3ξ=a0+3a+k1a0b2k1btan2a+k1a0b8bλξ,u2,3ξ=12λa+k1a0bk1cot2a+k1a0b8bλξ+a0,u3,3ξ=12λa+k1a0bk1cot2a+k1a0b8bλξ+a0+3a+k1a0b2k1btan2a+k1a0b8bλξ,u1,4ξ=a0+3a+k1a0b2k1bcot2a+k1a0b8bλξ,u2,4ξ=12λa+k1a0bk1tan2a+k1a0b8bλξ+a0,u3,4ξ=12λa+k1a0bk1tan2a+k1a0b8bλξ+a0+3a+k1a0b2k1bcot2a+k1a0b8bλξ.Whena+k1a0b/8bλ=0, we have σ=0, then
(79)u1,5ξ=a0+12λΓ21+αk1ξα+ω2=u3,5ξ.Solutionsui,1andui,2i=1,2,3 describe the multiple soliton. Solutions ui,3andui,4i=1,2,3 represent the exact periodic traveling wave solutions.Figures15–18 present the solutions: u1,1,u1,3,u2,3,andu1,5 of the generalized compound KdV-Burgers equation with 0<x<1,0<t<2. Solution u1,1 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,c=1,d=1,a0=0,andy=z=0; solution u1,3 and u2,3 are presented for values α=0.5,k1=k2=1,k3=k4=−1,a=4,b=1,c=1,d=1,a0=0,andy=z=0; solution u1,5 is presented for values α=0.5,k1=k2=1,k3=k4=−1,a=−4,b=1,c=1,d=1,a0=4,ω=0,andy=z=0.Figure 15
u1,1x,0,0,t of Equation (69).Figure 16
u1,3x,0,0,t of Equation (69).Figure 17
u2,3x,0,0,t of Equation (69).Figure 18
u1,5x,0,0,t of Equation (69).


## 4. Results and Discussion

The generalized time fractional biological population model, the generalized time fractional compound KdV-Burgers equation, the space-time fractional regularized long-wave equation, and the (3+1)-space-time fractional Zakharov-Kuznetsov equation have gained the focus of many studies due to their frequent appearance in various applications.The improved fractional subequation method has several advantages according to other traditional methods. Applying a suitable fractional complex transformut,x1,x2,⋯,xm=uξ,ξ=ξt,x1,x2,⋯,xm and chain rule, the nonlinear fractional differential equations with the modified Riemann-Liouville derivative are converted into the nonlinear ordinary differential equations. This is a significant impact because neither Caputo definition nor Riemann-Liouville definition satisfies the chain rule. With the help of the Riccati equation, the method has been employed for finding the exact analytical solutions of these equations.These obtained solutions are traveling solutions. Furthermore, solutionsui,1ξ and ui,2ξ describe the solitons which are everywhere in nature. ui,3ξ and ui,4ξ represent the exact periodic traveling wave solutions.The improved fractional subequation method is reliable and effective for finding more solutions for some space-time fractional nonlinear differential equations. We can substituteφ′=r+pφ+qφ2 [39] for φ′=σ+φ2, then, we will obtain more solutions.



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