Improved Fractional Subequation Method and Exact Solutions...

Research ArticleImproved Fractional Subequation Method and Exact Solutions toFractional Partial Differential Equations

Jun Jiang,1 Yuqiang Feng ,1,2 and Shougui Li2

1School of Science, Wuhan University of Science and Technology, Wuhan, 430065 Hubei, China2Hubei Province Key Laboratory of Systems Science in Metallurgical Process, Wuhan, 430065 Hubei, China

Correspondence should be addressed to Yuqiang Feng; [email protected]

Received 2 February 2020; Accepted 13 April 2020; Published 17 May 2020

Academic Editor: Xinguang Zhang

Copyright © 2020 Jun Jiang et al. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, the improved fractional subequation method is applied to establish the exact solutions for some nonlinear fractionalpartial differential equations. Solutions to the generalized time fractional biological population model, the generalized timefractional 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.

1. Introduction

Fractional differential equations are widely used to describelots of important phenomena and dynamic processes in phys-ics, engineering, electromagnetics, acoustics, viscoelasticityelectrochemistry, material science, stochastic dynamical sys-tem, plasma physics, controlled thermonuclear fusion, nonlin-ear control theory, image processing, nonlinear biologicalsystems and astrophysics, etc. [1–7]. In order to find the solu-tions of fractional differential equations, many powerful andefficient methods have been introduced and developed, suchas Darboux transformations [8], the hyperbolic functionmethod [9], the variational iteration method [10, 11], theautofinite element method [12, 13], the auxiliary equationmethod [14], the finite difference method [15, 16], theAdomian decomposition method [17, 18], the homogenous

balance method [19], Hirotas bilinear method [20], thehomotopy analysis method [21, 22], ðG′/GÞ-expansionmethod [23, 24], the subequation method [25, 26], the firstintegral method [27, 28], the improved fractional subequa-tion method [29, 30], the extended Jacobi elliptic functionexpansion 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-Liouvilledefinition which appears to provide a framework for a frac-tional 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:

Dαx f xð Þ =

1Γ 1 − αð Þ

ðx0x − ξð Þ−α−1 f ξð Þ − f 0ð Þ½ �dξ, α < 0,

1Γ 1 − αð Þ

ddx

ðx0x − ξð Þ−α f ξð Þ − f 0ð Þ½ �dξ, 0 < α < 1,

f nð Þ xð Þh i α−nð Þ

, n ≤ α < n + 1, n ≥ 1:

8>>>>>>><>>>>>>>:

ð1Þ

HindawiJournal of Function SpacesVolume 2020, Article ID 5840920, 18 pageshttps://doi.org/10.1155/2020/5840920

https://orcid.org/0000-0002-1208-0509

https://creativecommons.org/licenses/by/4.0/

https://creativecommons.org/licenses/by/4.0/

https://doi.org/10.1155/2020/5840920

Some useful properties of modified Riemann-Liouvillederivative are given below:

Dαxx

γ = Γ γ + 1ð ÞΓ γ + 1 − αð Þ x

γ−α, γ > 0, ð2Þ

Dαx f xð Þg xð Þ½ � = g xð ÞDα

x f xð Þ + f xð ÞDαxg xð Þ, ð3Þ

Dαx f g xð Þ½ � = f g′ g xð Þ½ �Dα

xg xð Þ =Dαg f g xð Þ½ � g′ xð Þ

h iα,

ð4Þwhich holds for nondifferentiable functions. Equations (2),(3), (4) which are important tools for fractional calculus. Basedon these merits, the modified Riemann-Liouville derivativewas successfully applied to the probability calculus, fractionalLaplace problems, and fractional variational calculus.

In this paper, we aim to find new exact solutions of someimportant partial fractional differential equations under Jum-arie’s definition by improved fractional subequation method.

In what follows, we introduce the aforementioned frac-tional partial differential equations. They are the generalizedtime fractional biological population model, the generalizedtime fractional compound KdV-Burgers equation, the space-time fractional regularized long-wave equation, and the(3 + 1)-space-time fractional Zakharov-Kuznetsov equation.

Suppose time t > 0, Dαt u 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 populationmodel is given by

Dαt u + k1 u2

� �xx+ k2 u2

� �yy+ k3ux + k4uy + k5u

2 + k6u + k7 = 0,

ð5Þ

where u = uðx, y, tÞ is an unknown function.When k1 = −1, k2 = −1, k3 = k4 = k6 = 0, k5 = −h ≠ 0, k7 =

rh ≠ 0, Equation (5) is the time fractional biological popula-tion model [39]:

Dαt u = u2

� �xx+ u2� �

yy+ h u2 − r� �

, ð6Þ

where u = uðx, y, tÞ denotes the population density and hðu2 − rÞ represents the amount of population due to deathand birth. Moreover, hðu2 − rÞ leads to Verhulst law. Equa-tion (5) has an important role to understand the dynamicprocess of population changes, and it is also an assistant toachieve precision about it.

The generalized time fractional compound KdV-Burgersequation is given by

Dαt u + k1uux + k2u

2ux + k3uxx + k4uxxx = 0, ð7Þ

where u = uðx, tÞ is an unknown function.When k1 = k3 = 0, Equation (7) becomes the time frac-

tional mKdV equation

Dαt u + k2u

2ux + k4uxxx = 0 ; ð8Þ

when k1 = k3 = 0 Equation (7) becomes the time fractionalKdV equation

Dαt u + k2u

2ux + k3uxx + k4uxxx = 0 ; ð9Þ

when k2 = k4 = 0, Equation (7) becomes the time fractionalBurgers equation [40]

Dαt u + k1uux + k3uxx = 0 ; ð10Þ

when k1 = 0, Equation (7) becomes the time fractionalmKdV-Burgers equation

Dαt u + k2u

2ux + k3uxx + k4uxxx = 0 ; ð11Þ

when k2 = 0, Equation (7) becomes the time fractional KdV-Burgers equation

Dαt u + k1uux + k3uxx + k4uxxx = 0: ð12Þ

The space-time fractional regularized long-wave equa-tion is given by [41]:

Dαt u + k1D

αxu + k2uD

αxu + k3D

αt D

2xαu = 0, ð13Þ

where u = uðx, tÞ is an unknown function, Dαxu is the modi-

fied Riemann-Liouville derivative of order α for a functionu, and D2α

x u =DαxðDα

xuÞ.The regularized long-wave equation, which describes

approximately the unidirectional propagation of long wavesin certain nonlinear dispersive systems, was proposed byBenjamin et al. in 1972. The regularized long-wave equationis considered an alternative to the KdV equation, which ismodeled to govern a large number of physical phenomenasuch as shallow waters and plasma waves.

The (3 + 1)-space-time fractional Zakharov-Kuznetsovequation given by [41].

D?t u + k1uD

αxu + k2D

3αx u + k3D

αxD

2αy u + k4D

αxD

2αz u = 0, ð14Þ

where u = uðx, y, z, tÞ is an unknown function; Dαxu,D

αyu, and

Dαz u are the modified Riemann-Liouville derivatives of the

function u; D2αx u =Dα

xðDαxuÞ; D2α

y u =DαyðDα

yuÞ; D2αz u =Dα

z

ðDαz uÞ; and D3α

x u =DαxðDα

xðDαxuÞÞ.

The Zakharov-Kuznetsov equation was first derived foranalysing weakly nonlinear ion acoustic waves in heavilymagnetized lossless plasma and geophysical flows in twodimensions. The ZK equation is one of the two well-established canonical two-dimensional extensions of theKdV equation. The ZK equation governs the behavior ofweakly nonlinear ionacoustic waves in a plasma comprisingcold ions and hot isothermal electrons in the presence of auniform magnetic field.

Motivated by the above results, in this paper, we use theimproved subequation method to find new exact solutions ofthe generalized time fractional biological population model,the generalized time fractional compound KdV-Burgers equa-tion, the space-time fractional regularized long-wave equation,

2 Journal of Function Spaces

and the (3 + 1)-space-time fractional Zakharov-Kuznetsovequation, respectively.

2. A Brief Description of the ImprovedFractional Subequation Method

In this section, basic steps of the improved subequationmethod [42] are presented.

Consider the following nonlinear fractional differentialequation,

P�u,Dα

t u,Dαx1u,Dα

x1u2,⋯,Dα

xmu,Dα

x1u2,

Dαx2u2,⋯,Dα

xmu2,⋯,D2α

x1u,D2α

xmu,⋯

�= 0,

ð15Þ

where t, x1, x2,⋯, xm are independent variables, uðt, x1, x2,⋯, xmÞ is an unknown function, and P is a polynomial of u,u2,⋯ and their partial fractional derivatives. Also, Dαð⋅Þ sym-bolizes the modified Riemann-Liouville fractional derivative.

Step 1. First of all, using a suitable fractional complextransform,

u t, x1, x2,⋯, xmð Þ = u ξð Þ, ξ t, x1, x2,⋯∣,xmð Þ, ð16Þ

Equation (15) converts into nonlinear ordinary differen-tial equation given below:

Q u, u′, u″, u′″,⋯� �

= 0, ð17Þ

where u′, u″, u′″,⋯ denotes the derivations with respectto ξ.

We used to utilize real transformation, but actually, com-plex transformations are more useful. They make some equa-tions easier to simplify.

Step 2. Suppose that the solution of ordinary differentialequation (17) is

u = 〠−1

i=naiφ

i + a0 + 〠n

i=naiφ

i, ð18Þ

where constants aiði = −n,⋯, −1, 1,⋯, nÞ are going to bedetermined. Here, n is a positive integer, and it is obtainedusing the homogeneous balance of the highest order deriva-tive and the nonlinear term seen in Equation (17).

φ = φðξÞ is the solution of the Riccati equation

φ′ = σ + φ2, ð19Þ

where σ is a constant, and the solutions of Equation (19) areobtained by Zhang et al. [34] as follows:

φ ξð Þ =

−ffiffiffiffiffiffi−σ

ptanh j−σξ,

�σ < 0,

−ffiffiffiffiffiffi−σ

pcoth j−σξ,

�σ < 0,

ffiffiffiσ

ptan jσξ� �

, σ > 0,

−ffiffiffiσ

pcot jσξ� �

, σ > 0,

−Γ 1 + αð Þξα + ω

, σ = 0, ω = const,

8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:

20Þ

In the previous literatures, u = a0 +∑ni=1aiφ

i was consid-ered. In this paper, we assume u =∑−1

i=−naiφi + a0 +∑n

i=1aiφi,

and we can get a solution that has both hyperbolic tangentfunction and hyperbolic cotangent function or both tangentfunction and cotangent function.

Step 3. Putting Equation (18) along with Equation (19)into Equation (17), we obtain a new polynomial in termsof φ. Then, all the coefficients of powers of φkðk = 0, 1, 2,⋯, −1, −2,⋯Þ are set equal to zero, we get a system of alge-braic equations.

Step 4. Finally, the system of algebraic equations isobtained in the previous step for aið−n ≤ i ≤ nÞ, and σ issolved by the Maple package. By substituting the newlyobtained values into Equation (20), we get the exact solutionsfor the nonlinear fractional differential equation (15).

Applying a suitable fractional complex transform of theimproved fractional subequation method and the chain rule,nonlinear fractional differential equations with the modifiedRiemann-Liouville derivative can be converted into nonlin-ear ordinary differential equations. Then, using the solutionsof a Riccati equation, we can find exact analytical solutionsexpressed by triangle functions, hyperbolic functions, orpower functions.

3. Applications of the Improved FractionalSubequation Method

In this section, the improved fractional subequation methodis utilized to solve some nonlinear fractional differentialequations introduced in Section 1.

3.1. The Generalized Time Fractional Biological PopulationModel. The generalized time fractional biological populationmodel is given by

Dαt u + k1 u2

� �xx+ k2 u2

� �yy+ k3ux + k4uy

+ k5u2 + k6u + k7 = 0,

ð21Þ

where u = uðx, y, tÞ is an unknown function.We considered two cases.

Case 1. When k1k2 > 0, let

u x, y, tð Þ = u ξð Þ, ξ = ax + ia

ffiffiffiffiffik1k2

sy + btα

Γ 1 + αð Þ , a ≠ 0, ð22Þ

3Journal of Function Spaces

then Equation (21) is reduced to the ordinary differentialequation as follows:

ak4

ffiffiffiffiffik1k2

su′ = 0,

ak3 + bð Þu′ + k5u2 + k6u + k7 = 0:

8>><>>: ð23Þ

When k4 ≠ 0, Equation (23) has only constant solutions.When k4 = 0, Equation (23) has solutions in the form of

(18). u′ is obtained from the homogeneous balance betweenthe highest order derivative and the nonlinear term u2. Weobtain the solution of Equation (21) as follows:

u ξð Þ = a − 1φ−1 + a0 + a1φ, ð24Þ

Substituting Equation (24) together with its necessaryderivatives into Equation (21), the algebraic equation isarranged according to the powers of the function φkðξÞ.Then, the following coefficients are obtained:

φ−2 : −k3σa−1 − bσa−1 + k5a2−1,

φ−1 : 2k5a−1a0 + a−1k6,φ0 : −ak3a−1 − ba−1 + ak3σa1 + bσa1

+ k5a20 + 2k5a−1a1 + k6a0 + k7,

φ1 : 2k5a0a1 + k6a1,φ2 : ak3a1 + ba1 + k5a

21:

ð25Þ

Let the coefficients be zero. By solving the set of equationsgiven above for a−1, a0, a1, a, b, and σ, we obtain solution setsas follows:

Set 1

a−1 = 0,

a0 = −k62k5

,

a1 = a1,a = a,b = −ak3 − a1k5,

σ = λ

4a21k25:

ð26Þ

Set 2

a−1 = a−1,

a0 = −k62k5

,

a1 =−λ

16k25a−1,

a = a,

b = λ − 16aa−1k3k516k5a−1

,

σ = 16k25a2−1λ

:

ð27Þ

Set 3

a−1 = a−1,

a0 = −k62k5

,

a1 = 0,a = a,

b = λ − 4aa−1k3k54k5a−1

,

σ = 4k25a2−1λ

:

ð28Þ

where λ = 4k5k7 − k26.

Case 2. When k1k2 < 0, let

u x, y, tð Þ = u ξð Þ, ξ = ax + a

ffiffiffiffiffiffiffiffi−k1k2

sy + btα

Γ 1 + αð Þ , a ≠ 0, ð29Þ

then Equation (21) is reduced to the ordinary differentialequation as follows:

b + k3a + ak4


s !u′ + k5u

2 + k6u + k7 = 0: ð30Þ

The solution of Equation (30) is in the form of (18). n = 1is taken from the homogeneous balance between the highestorder derivative u′ and the nonlinear term u2. We obtain thesolution of Equation (30) as Equation (24). SubstitutingEquation (24) together with its necessary derivatives intoEquation (30), the algebraic equation is arranged accordingto the powers of the function φkðξÞ. Then, the following coef-ficients are obtained:

φ−2 : − a


sk4σa−1 − ak3σa−1 − bσa−1 + k5a

2−1,

φ−1 : 2k5a−1a0 + a−1k6,

φ0 : a


sk4σa1 + ak3σa1 + bσa1 − a


sk4a−1,

φ1 : 2k5a0a1 + k6a1,


φ2 : a


sk4a1 + ak3a1 + ba1 + k5a

21:

ð31Þ

Let the coefficients be zero. By solving the set of equationsgiven above for a−1, a0, a1, a, b, and σ, we obtain solution setsas follows:

Set 4

a−1 = 0,

a0 = −k62k5

,

a1 = a1,a = a,

b = −ak3 − ak4


s− a1k5,

σ = λ

4a21k25:

ð32Þ

Set 5

a−1 = a−1,

a0 = −k62k5

,

a1 =−λ

16k25a−1,

a = a,

b = λ − 16μ16k5a−1

,

σ = 16k25a2−1λ

:

ð33Þ

Set 6

a−1 = a−1,

a0 = −k62k5

,

a1 = 0,a = a,

b = λ − 4μ4k5a−1

,

σ = 4k25a2−1λ

⋅

ð34Þ

where μ = aa−1k3k5 + aa−1k4k5ffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2

p.

We find that a−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

uði,jÞðx,y,tÞ, ði, j ∈ Z+Þ, where i denotes obtained set numberand j is the solution number of the Riccati equation, respec-tively. Thus, using set 1 to set 6, we obtain the solution ofEquation (21) as ui,jðx, y, tÞ, ði = 1, 2,⋯, 6 ; j = 1, 2,⋯, 5Þ.ui,jðx, y, tÞ is the following:

When k1k2 > 0, λ = 4k5k7 − k26 < 0, we have σ < 0, then

u1,1 x, y, tð Þ = −k62k5

−a1

ffiffiffiffiffiffi−λ

p

2 a1k5j j tanh" ffiffiffiffiffiffi

−λp

2 a1k5j j

ax

+ ia

ffiffiffiffiffik1k2

sy −

ak3 + a1k5ð ÞtαΓ 1 + αð Þ

!#,

u1,2 x, y, tð Þ = −k62k5

−a1


p

2 a1k5j j coth" ffiffiffiffiffiffi

−λp

2 a1k5j j

ax

+ ia

ffiffiffiffiffik1k2

sy −

ak3 + a1k5ð ÞtαΓ 1 + αð Þ

!#,

u2,1 x, y, tð Þ = −a−1


p

4 k5a−1j j coth"4 k5a−1j jffiffiffiffiffiffi

−λp

ax − ia

ffiffiffiffiffik1k2

sy

+ λ − 16aa−1k3k5ð Þtα16k5a−1Γ 1 + αð Þ

!#−

k62k5

−k5a−1j j


p

4k25a−1tanh

"4 k5a−1j jffiffiffiffiffiffi

−λp

ax

− ia

ffiffiffiffiffik1k2

sy + λ − 16aa−1k3k5ð Þtα

16k5a−1Γ 1 + αð Þ

!#,

u2,2 x, y, tð Þ = −a−1


p

4 k5a−1j j tanh"4 k5a−1j jffiffiffiffiffiffi

−λp

ax − ia

ffiffiffiffiffik1k2

sy


!#−

k62k5

−k5a−1j j


p

4k25a−1coth


−λp

ax

− ia

ffiffiffiffiffik1k2


16k5a−1Γ 1 + αð Þ

!#,

u3,1 x, y, tð Þ = a1ffiffiffiffiffiffi−λ

p

2 a1k5a−1j j coth"2 a1k5a−1j jffiffiffiffiffiffi

−λp

ax

+ ia

ffiffiffiffiffik1k2

sy −

λ − 4aa−1k3k5ð Þtα4k5a−1Γ 1 + αð Þ

!#−

k62k5

,


u3,2 x, y, tð Þ = a−1ffiffiffiffiffiffi−λ

p

2 a1k5a−1j j tanh"2 a1k5a−1j jffiffiffiffiffiffi

−λp

ax

+ ia

ffiffiffiffiffik1k2

sy −

λ − 4aa−1k3k5ð Þtα4k5a−1Γ 1 + αð Þ

!#−

k62k5

:

ð35Þ

When k1k2 < 0, λ = 4k5k7 − k26 < 0, we have σ < 0, then

u4,1 x, y, tð Þ = −k62k5

−a1


p

2 a1k5j j tanh" ffiffiffiffiffiffi

−λp

2 a1k5j j

ax + a


sy

−ak3 + ak4

ffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2

p+ a1k5

� �tα

Γ 1 + αð Þ

!#,

u4,2 x, y, tð Þ = −k62k5

−a1


p

2 a1k5j j coth" ffiffiffiffiffiffi

−λp

2 a1k5j j

ax + a


sy

−ak3 + ak4


p+ a1k5

� �tα

Γ 1 + αð Þ

!#,

u5,1 x, y, tð Þ = −a−1


p

4 k5a−1j j coth"4 k5a−1j jffiffiffiffiffiffi

−λp

ax − a


sy

+ λ − 16μð Þtα16k5a−1Γ 1 + αð Þ

!#−

k62k5

−k5a−1j j


p

4k25a−1tanh


−λp

ax

− a


sy + λ − 16μð Þtα

16k5a−1Γ 1 + αð Þ

!#,

u5,2 x, y, tð Þ = −a−1


p

4 k5a−1j j tanh"4 k5a−1j jffiffiffiffiffiffi

−λp

ax − a


sy

+ λ − 16μð Þtα16k5a−1Γ 1 + αð Þ

!#−

k62k5

−k5a−1j j


p

4k25a−1coth


−λp

ax

− a



16k5a−1Γ 1 + αð Þ

!#,


p

2 a1k5a−1j j coth"2 a1k5a−1j jffiffiffiffiffiffi

−λp

ax + a


sy

+ λ − 4μð Þtα4k5a−1Γ 1 + αð Þ

!#−

k62k5

,


p

2 a1k5a−1j j tanh"2 a1k5a−1j jffiffiffiffiffiffi

−λp

ax + a


sy


!#−

k62k5

:

ð36Þ

When k1k2 < 0 and λ = 4k5k7 − k26 < 0, we have σ < 0, then

u1,3 x, y, tð Þ = −k62k5

+ a1ffiffiffiλ

p

2 a1k5j j tan" ffiffiffi

λp

2 a1k5j j

ax

+ ia

ffiffiffiffiffik1k2

sy + ak3 + a1k5ð Þtα

Γ 1 + αð Þ

!#,

u1,4 x, y, tð Þ = −k62k5

+ a1ffiffiffiλ

p

2 a1k5j j cot" ffiffiffi

λp

2 a1k5j j

ax

+ ia

ffiffiffiffiffik1k2

sy + ak3 + a1k5ð Þtα

Γ 1 + αð Þ

!#,

u2,3 x, y, tð Þ = −a−1

ffiffiffiλ

p

4 k5a−1j j cot"4 k5a−1j jffiffiffi

λp

ax − ia

ffiffiffiffiffik1k2

sy


!#−

k62k5

+ k5a−1j jffiffiffiλ

p

4k25a−1tan"4 k5a−1j jffiffiffi

λp

ax

− ia

ffiffiffiffiffik1k2


16k5a−1Γ 1 + αð Þ

!#,

u2,4 x, y, tð Þ = −a−1

ffiffiffiλ

p

4 k5a−1j j tan"4 k5a−1j jffiffiffi

λp

ax − ia

ffiffiffiffiffik1k2

sy


!#−

k62k5

−k5a−1j j

ffiffiffiλ

p

4k25a−1cot"4 k5a−1j jffiffiffi

λp

ax

− ia

ffiffiffiffiffik1k2


16k5a−1Γ 1 + αð Þ

!#,

u3,3 x, y, tð Þ = −a−1

ffiffiffiλ

p

2 a1k5a−1j j cot"2 a1k5a−1j jffiffiffi

λp

ax

+ ia

ffiffiffiffiffik1k2


4k5a−1 1 + αð Þ

!#−

k62k5

,

ð37Þ

u3,4ðx, y, tÞ = −ða−1ffiffiffiλ

p/2 ja1k5a−1jÞ tan ½ð2ja1k5a−1j/

ffiffiffiλ

p Þðax + ia

ffiffiffiffiffiffiffiffiffiffik1/k2

py + ððλ − 4aa−1k3k5Þtα/4k5a−1ð1 + αÞÞÞ� − k6/

2k5,


When k1k2 < 0 and λ = 4k5k7 − k26 < 0, we have σ < 0, then

u4,3 x, y, tð Þ = −k62k5

+ a1ffiffiffiλ

p

2 a1k5j j tan" ffiffiffi

λp

4 a1k5j j

ax + a


sy

−ak3 + ak4


p+ a1k5

� �tα

Γ 1 + αð Þ

!#,

u4,4 x, y, tð Þ = −k62k5

−a1

ffiffiffiλ

p

2 a1k5j j cot" ffiffiffi

λp

2 a1k5j j

ax + a


sy

−ak3 + ak4


p+ a1k5

� �tα

Γ 1 + αð Þ

!#,

u5,3 x, y, tð Þ = −a−1

ffiffiffiλ

p


λp

ax + a


sy

+ λ − 16μð Þtα16k5a−1Γ 1 + αð Þ

!#−

k62k5

+ k5a−1j jffiffiffiλ

p

4k25a−1tan"4 k5a−1j jffiffiffi

λp

� ax + a



16k5a−1Γ 1 + αð Þ

!#,

u5,4 x, y, tð Þ = −a−1

ffiffiffiλ

p


λp

ax + a


sy

+ λ − 16μð Þtα16k5a−1Γ 1 + αð Þ

!#−

k62k5

−k5a−1j j

ffiffiffiλ

p

4k25a−1cot"4 k5a−1j jffiffiffi

λp

� ax + a



16k5a−1Γ 1 + αð Þ

!#,

u6,3 x, y, tð Þ = a−1ffiffiffiλ

p


λp

ax + a


sy


!#−

k62k5

,

u6,4 x, y, tð Þ = a−1ffiffiffiλ

p


λp

ax + a


sy


!#−

k62k5

:

ð38Þ

When λ = 4k5k7 − k26 = 0, we have σ = 0, then

Solutions ui,1, ui,2ði = 1, 2,⋯, 6Þ describe the soliton. Sol-itons exist everywhere in the nature; they are special kinds ofsolitary waves. u2,1, u2,2, u5,1, and u5,2 describe the multiplesoliton solutions. Solutions ui,3 and ui,4 ði = 1, 2,⋯, 6Þ repre-sent the exact periodic traveling wave solutions. Periodicsolutions are traveling wave solutions.

Figures 1–6 present the solutions: u1,1, u2,1, u1,3, u2,3,u1,5, and u5,5 of the generalized time fractional biological

population model with −10 < x < 10, 0 < t < 10. Solutionu1,1 is presented for values α = 0:5, k1 = k2 = 1, k3 = k4 = 0,k5 = 2, k6 = 3, k7 = 1, a1 = 1, a = 2, and y = 0; solution u2,1 ispresented for values α = 0:5, k1 = k2 = 1, k3 = k4 = 0, k5 = 2,k6 = 3, k7 = 1, a−1 = 1, a = 2, and y = 0; solution u1,3 is pre-sented for values α = 0:5, k1 = k2 = 1, k3 = k4 = 0, k5 = 2, k6= 2, k7 = 1, a1 = 1, a = 2, and y = 0; solution u2,3 is presentedfor values α = 0:5, k1 = k2 = 1, k3 = k4 = 0, k5 = 2, k6 = 3, k7 =

u1,5 x, y, tð Þ = −k62k5

−Γ 1 + αð Þa1

ax + iaffiffiffiffiffiffiffiffiffiffiffiffiffiffik1/k2ð Þp

y − ak3 + a1k5ð Þtα/Γ 1 + αð Þð Þh iα

+ ω,

u2,5 x, y, tð Þ = −a−1 ax + ia

ffiffiffiffiffiffiffiffiffiffiffiffiffiffik1/k2ð Þp

y − ak3tα/Γ 1 + αð Þð Þ

h iα+ a−1ω

Γ 1 + αð Þ −k62k5

= u3,5 x, y, tð Þ,

u4,5 x, y, tð Þ = −k62k5

a1Γ 1 + αð Þax + a

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2ð Þp

y − ak3 + ak3ffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2

p+ a1k5

� �tα/Γ 1 + αð Þ

� �h iα+ ω

,

u5,5 x, y, tð Þ = −a−1Γ 1 + αð Þ aα x +

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2ð Þ

py −

k3 + k4ffiffiffiffiffiffiffiffiffiffiffiffiffi−k1/k2

p� �tα

Γ 1 + αð Þ

0@

1A

α

+ ω

24

35 −

k62k5

:

ð39Þ


1, a−1 = 1, a = 2, and y = 0; solution u1,5 is presented for valuesα = 0:5, k1 = k2 = 1, k3 = k4 = 0, k5 = 2, k6 =

ffiffiffi8

p, k7 = 1, a = 2,

a1 = 1, y = 0, andω = 0; solution u5,5 is presented for valuesα = 0:5, k1 = 1, k2 = −1, k3 = 0, k4 = 1, k5 = 2, k6 =

ffiffiffi8

p, k7 = 1,

a1 = 1, a = 2, y = 0, andω = 0.

When k1 = −1, k2 = −1, k3 = k4 = k6 = 0, k5 = −h ≠ 0, andk7 = rh ≠ 0, Equation (21) is the time fractional biologicalpopulation model [39]

Dαt u = u2

� �xx+ u2� �

yy+ h u2 − r� �

: ð40Þ

–10–5051086420

–1

–0.9

–0.8

–0.7

–0.6

–0.5

xt

Figure 1: u1,1ðx, 0, tÞ of Equation (21).

–10–5051086420

–1

–0.5

0.5

0

x

t


–10–5051086420

0

1000

2000

3000

x

t


–10–5051086420

–400

–300

–200

–100

0

100

200

x

t


–10–50 51086420

10

–8

–6

–4

–2

x

t


–10–5051086420

–5

–4

–3

–2

–1

x

t



We denote

u x, y, tð Þ = u ξð Þ, ξ = ax + iay + btα

Γ 1 + αð Þ , a ≠ 0, ð41Þ

and find the solution of Equation (40) in the form of uðξÞ = a−1φ

−1 + a0 + a1φ. Then, by solving the set of equationsgiven above for a−1, a0, a1, a, b, and σ, we obtain solution setsas follows:

Set 1

a−1 = 0,a0 = 0,a1 ≠ 0,

σ = −ra21

:

ð42Þ

Let a1 = c/h, then σ = −rh2/c2, we can obtain set 1 in [42].Set 2

a−1 ≠ 0,a0 = 0,

a1 =r

4a−1,

σ = −4a2−1r

:

ð43Þ

Let a−1 = rh/4c, then σ = −rh2/4c2, we can obtain set 3 in[42].

Set 3

a−1 ≠ 0,a0 = 0,a1 = 0,

σ = −a2−1r

:

ð44Þ

Let a−1 = rh/c, then σ = −rh2/c2, we can obtain set 2 in[42].

Clearly, we get more solutions of the time fractional bio-logical population model (40) than the literature [42].

3.2. The Generalized Compound KdV-Burgers Equation. Thegeneralized compound KdV-Burgers equation is given by

Dαt u + k1uux + k2u

2ux + k3uxx + k4uxxx = 0, ð45Þ

Let

u x, tð Þ = u ξð Þ, ξ = ax + btα

Γ 1 + αð Þ , a ≠ 0, ð46Þ

then Equation (45) is reduced to ordinary differential equa-tion as

bu′ + k1auu′ + k2au2u′ + k3a

2u″ + k4a3u′″ = 0: ð47Þ

The solution of Equation (45) is in the form of (18),and here, n = 1 is taken from the homogeneous balancebetween the highest order derivative u′″ and the nonlinearterm u2u′. We obtain the solution of Equation (45) as follows:

u ξð Þ = a−1φ−1 + a0 + a1φ: ð48Þ

Substituting Equation (48) together with its necessaryderivatives into Equation (47), the algebraic equation isarranged according to the powers of the function φkðξÞ. Then,the following coefficients are obtained:

φ−4 : −6a3k4σ3a−1 − ak2σa3−1,

φ−3 : 2a2k3σ2a−1 − 2ak2σa2−1a0 − ak1σa2−1,

φ−2 : bσa−1 − ak1σa−1a0 − ak2a3−1 − ak2σa−1a

20

− ak2σa2−1a1 − 8a3k4σ2a−1,

φ−1 : −ak1a2−1 − 2ak2a2−1a0 + 2a2k3σa−1,

ð49Þ

φ0 : ba−1 − bσa1 + ak1σa0a1 − ak1a−1a0+ ak2σa

20a1 + ak2σa−1a

21 − ak2a

20a−1

− ak2a2−1a1 + 2a3k4σ2a1 − 2a3k4σa−1,

φ1 : ak1σa21 + 2ak2σa0a21 + 2a2k3σa1,

φ2 : −ba1 + ak1a0a1 + ak2a20a1 + ak2a−1a

21

+ ak2σa31 + 8a3k4σa1,

φ3 : 2ak2a0a21 + 2a2k3a1 + ak1a21,

φ4 : 6a3k4a1 + ak2a31:

ð50Þ

Let the coefficients of φkðξÞ to be zero. By solving the set ofequations given above for a−1, a0, a1, a, b, and σ, we obtainsolution sets as follows:

When k2k4 < 0, it denotes μ = 3ak21k4 + 2ak2k23 − 12bk2k4 and λ1 = ±

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−6k4/k2

p, λ2 = −k1/2 ± ðð∣k3 ∣

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi−6k2k4

p Þ/6k4Þ,we have

Set 1

a−1 = −λ1μ

96a2k2k24,

a0 = −λ1k1 + 2k32λ1k2

,

a1 = aλ1,a = a,b = b,

σ = μ

96a3k2k24:

ð51Þ


Set 2

a−1 = 0,

a0 = −λ1k1 + 2k32λ1k2

,

a1 = aλ1,a = a,b = b,

σ = μ

24a3k2k24:

ð52Þ

Set 3

a−1 = −μk3

12a2k2k24 2λ2 + k1ð Þ,

a0 =λ2k2

,

a1 = 0,a = a,b = b,

σ = μ

24a3k2k24:

ð53Þ

Thus, we obtain the solution of Equation (45) as ui,jðx, tÞ, ði = 1, 2, 3 ; j = 1, 2, 3, 4, 5Þ. Using set 1 to set 3, ui,jðx, tÞ is as follows:

When k2k4 < 0 and aμk2 < 0, we have σ < 0, then

u1,1 x, tð Þ = λ1μ ∣ a ∣24a2k4

ffiffiffiffiffiffiffiffi−6aμk2

scoth

� 14 ∣ ak4 ∣

ffiffiffiffiffiffiffiffiffi−μ6ak2

r

� ax + btα

Γ 1 + αð Þ�

−λ1k1 + 2k32λ1k2

−aλ1

4 ∣ ak4 ∣


rtanh

� 14 ∣ ak4 ∣

�ffiffiffiffiffiffiffiffiffi−μ6ak2

rax + btα

Γ 1 + αð Þ�

u1,2 x, tð Þ = λ1μ ∣ a ∣24a2k4


stanh

� 14 ∣ ak4 ∣


r

� ax + btα

Γ 1 + αð Þ�

−λ1k1 + 2k32λ1k2

−aλ1

4 ∣ ak4 ∣


rcoth

� 14 ∣ ak4 ∣

�ffiffiffiffiffiffiffiffiffi−μ6ak2

rax + btα

Γ 1 + αð Þ�

u2,1 x, tð Þ = −λ1k1 + 2k32λ1k2

−aλ1

2 ∣ ak4 ∣


rtanh

� 12 ∣ ak4 ∣


rax + btα

Γ 1 + αð Þ� �

u2,2 x, tð Þ = −λ1k1 + 2k32λ1k2

−aλ12 ak4j j


rcoth,

u3,1 x, tð Þ = μk3 aj j6a2k4 2λ2 + k1ð Þ


scoth + λ2

k2,



stanh + λ2

k2:

ð54Þ

When k2k4 < 0 and aμk2 > 0, we have σ > 0, then

u1,3 x, tð Þ = −λ1μ ∣ a ∣24a2k4

ffiffiffiffiffiffiffi6aμk2

scot� 14 ∣ ak4 ∣

ffiffiffiffiffiffiffiffiffiμ

6ak2

r

� ax + btα

Γ 1 + αð Þ�

−λ1k1 + 2k32λ1k2

+ aλ14 ∣ ak4 ∣


6ak2

rtan� 14 ak4j j


6ak2

r

� ax + btα

Γ 1 + αð Þ�

,

u1,4 x, tð Þ = λ1μ ∣ a ∣24a2k4


stan� 14 ∣ ak4 ∣


6ak2

r

� ax + btα

Γ 1 + αð Þ�

−λ1k1 + 2k32λ1k2

−aλ14 ak4j j


6ak2

rcot� 14 ak4j j


6ak2

r

� ax + btα

Γ 1 + αð Þ�

,

u2,3 x, tð Þ = −λ1k1 + 2k32λ1k2

+ aλ12 ak4j j


6ak2

rtan

� 12 ak4j j


6ak2

rax + btα

Γ 1 + αð Þ� �

,

u2,4 x, tð Þ = −λ1k1 + 2k32λ1k2

−aλ12 ak4j j


6ak2

rcoth

� 12 ak4j j


6ak2

rax + btα

Γ 1 + αð Þ� �

,

u3,3 x, tð Þ = −μk3

6a2k4 2λ2 + k1ð Þ

ffiffiffiffiffiffiffi6a3μk2

scot

� 12 k4j j

ffiffiffiffiffiffiffiffiffiffiffiμ

6a3k2

rax + btα

Γ 1 + αð Þ� �

+ λ2k2

,




stan

� 12 ak4j j


6ak2

rax + btα

Γ 1 + αð Þ� �

+ λ2k2

:

ð55Þ

When k2k4 < 0 and μ = 0, we have σ = 0, then

u1,5 x, tð Þ = −λ1k1 + 2k32λ1k2

+ aλ1Γ 1 + αð Þax + btα/Γ 1 + αð Þð Þð Þα + ω

= u2,5 x, tð Þ:ð56Þ

Solutions ui,1 and ui,2 ði = 1, 2, 3Þ described the soliton.u1,1 and u1,2 describe the multiple soliton solutions. Solutionsui,3 and ui,4 ði = 1, 2, 3Þ represent the exact periodic travelingwave solutions.

Figures 7–10 present the solutions: u1,1, u2,1, u1,3, and u2,3of the generalized compound KdV-Burgers equation with −10 < x < 10, 0 < t < 1. Solutions u1,1 and u2,1 are presentedfor 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, and y = 0.

3.3. The Space-Time Fractional Regularized Long-WaveEquation. Consider the space-time fractional regularizedlong-wave equation as follows:

Dαt u + k1D

αxu + k2uD

αxu + k3D

αt D

2αx u = 0: ð57Þ

Let

u x, tð Þ = u ξð Þ, ξ = axα

Γ 1 + αð Þ + btα

Γ 1 + αð Þ , a ≠ 0, ð58Þ

then Equation (57) is reduced to the following ordinary dif-ferential equation:

bu′ + k1au′ + k2auu′ + k3a2bu′″ = 0: ð59Þ

The solution of Equation (57) is in the form of (18), andhere, n = 2 is taken from the homogeneous balance betweenthe highest order derivative u′″ and the nonlinear term uu′.We obtain the solution of Equation (57) as follows:

u ξð Þ = a−2φ−2 + a−1φ

−1 + a0 + a1φ + a2φ2: ð60Þ

Substituting Equation (60) together with its necessaryderivatives into Equation (57), the algebraic equation isarranged according to the powers of the function φkðξÞ. Then,the following coefficients are obtained:

φ−5 : −24k3a2ba−2σ3 − 2k2aa2−2σ,φ−4 : −6k3a2ba−1σ3 − 3k2aa−1a−2σ,φ−3 : −2k1aa−2σ − 2ba−2σ − 2k2aa0a−2σ

− k2aa2−1σ − 2k2aa2−2 − 40k3a2ba−2σ2,

φ−2 : −k1aa−1σ − ba−1σ − k2aa1a−2σ

− 3k2aa−1a−2 − k2aa0a−1σ − 8k3a2ba−1σ2,

φ−1 : −2ba−2 − 2k1aa−2 − k2aa2−1 − 2k2aa0a−2

− 16k3a2ba−2σ,

φ0 : k1aa1σ − k1aa−1 + a1bσ − a−1b − k2aa1a−2+ k2aa−1a2σ + k2aa0a1σ − k2aa0a−1+ 2k3a2ba1σ2 − 2k3a2ba−1σ,

ð61Þ

φ1 : 2k1aa2σ + 2ba2σ + 2k2aa0a2σ+ k2aa

21σ + 16k3a2ba2σ2,

φ2 : k1aa1 + ba1 + k2aa2a−1 + k2aa1a0+ 3k2aa1a2σ + 8k3a2ba1σ,

φ3 : 2k1aa2 + 2ba2 + 2k2aa22σ + 2k2aa0a2+ k2aa

21 + 40k3a2ba2σ,

φ4 : 6k3a2ba1 + 3k2aa1a2,φ5 : 24k3a2ba2 + 2k2aa22:

ð62Þ

Let the coefficients of φkðξÞ tbe zero. By solving the set ofequations given above for a−1, a0, a1, a, b, and σ, we obtainsolution sets as follows:

Set 1

a−2 = 0,a−1 = 0,a0 = a0,a1 = 0,

a2 = −12k3abk2

,

a = a,b = b,

σ = −k1a + k2a0a + b

8k3a2b:

ð63Þ

Set 2

a−2 = −3 k1a + k2a0a + bð Þ2

16k2k3a3b,

a−1 = 0,

a0 = a0,


a1 = 0,a2 = 0,

a = a,

b = b,

σ = −k1a + k2a0a + b

8k3a2b:

ð64Þ

Set 3

a−2 = −3 k1a + k2a0a + bð Þ2

16k2k3a3b,

a−1 = 0,a0 = a0,a1 = 0,

a2 = −12k3abk2

,

a = a,b = b,

σ = −k1a + k2a0a + b

8k3a2b:

ð65Þ

Thus, we obtain the solution of Equation (57) as ui,jðξÞ,ði = 1, 2, 3 ; j = 1, 2, 3, 4, 5Þ and ξ = ðaxα/Γð1 + αÞÞ + ðbtα/Γð1 + αÞÞ; ui,jðξÞ is as follows:

When ðk1a + k2a0a + bÞ/ð8k3a2bÞ > 0, we have σ < 0,then

u1,1 ξð Þ = a0 −3 k1a + k2a0a + bð Þ

2k2atanh2

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik1a + k2a0a + b

8k3a2b

sξ

!= u2,2 ξð Þ,

u2,1 ξð Þ = −3 k1a + k2a0a + bð Þ

2k2acoth2


8k3a2b

sξ

!+ a0 = u1,2 ξð Þ,

u3,1 ξð Þ = −3 k1a + k2a0a + bð Þ

2k2acoth2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik1a + k2a0a + b

8k3a2b

sξ

!

+ a0 −3 k1a + k2a0a + bð Þ

2k2atanh2


8k3a2b

sξ

!= u3,2 ξð Þ:

ð66Þ

–10–5051086420

–1–0.8–0.6–0.4–0.2

00.20.40.60.8

x

t

Figure 8: u2,1ðx, tÞ of Equation (45).

–10–505100.80.60.40.20

–60

–50

–40

–30

–20

–10

0

10

xt


–10–505100.80.60.40.20

–10

0

10

20

30

40

50

60

x

t



When ðk1a + k2a0a + bÞ/ð8k3a2bÞ < 0, we have σ > 0,then

u1,3 ξð Þ = a0 +3 k1a + k2a0a + bð Þ

2k2atan2

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1a + k2a0a + b

8k3a2b

sξ

!= u2,4 ξð Þ,

u2,3 ξð Þ = 3 k1a + k2a0a + bð Þ2k2a

cot2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1a + k2a0a + b

8k3a2b

sξ

!

+ a0 = u1,4 ξð Þ,

u3,3 ξð Þ = 3 k1a + k2a0a + bð Þ2k2a

cot2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1a + k2a0a + b

8k3a2b

sξ

!

+ a0 +3 k1a + k2a0a + bð Þ

2k2atan2

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−k1a + k2a0a + b

8k3a2b

sξ

!= u3,4 ξð Þ:

ð67Þ

When ðk1a + k2a0a + bÞ/ð8k3a2bÞ = 0, a−2 = 0, we haveσ = 0, then

u1,5 ξð Þ = a0 −12k3abΓ2 1 + αð Þk2 ξα + ω� �2 = u3,5 ξð Þ: ð68Þ

Solutions ui,1 and ui,2ði = 1, 2, 3Þ describe the multiplesoliton. Solutions ui,3 and ui,4ði = 1, 2, 3Þ represent the exactperiodic traveling wave solutions.

Figures 11–14 present the solutions: u1,1, u2,1, u2,3, andu1,5 of the generalized compound KdV-Burgers equationwith 0 < x < 1, 0 < t < 2. Solution u1,1 and solution u2,1 arepresented for values α = 0:5, k1 = k2 = 1, k3 = k4 = −1, a = −4,b = 1, a0 = 0, and y = 0; u2,3 is presented for values α = 0:5,k1 = k2 = 1, k3 = k4 = −1, a = 4, b = 1, a0 = 0, and y = 0; solu-

tion u1,5 is presented for values α = 0:5, k1 = k2 = 1, k3 = k4 =−1, a = 4, b = −4, a0 = 0, ω = 0, and y = 0.

3.4. The (3 + 1)-Space-Time Fractional Zakharov-KuznetsovEquation. The (3 + 1)-space-time fractional Zakharov-Kuznetsov Equation is given by

Dαt u + k1uD

αxu + k2D

3αx u + k3D

αxD

2αy u + k4D

αxD

2αz u = 0, ð69Þ

Let

u x, tð Þ = u ξð Þ, ξ = atα

Γ 1 + αð Þ + bxα

Γ 1 + αð Þ + cyα

Γ 1 + αð Þ+ dzα

Γ 1 + αð Þ , a ≠ 0,ð70Þ

then Equation (69) is reduced to the ordinary differentialequation as

au′ + k1buu′ + k2b3u′″ + k3bc

2u′″ + k4bd2u′″ = 0: ð71Þ

–10–505100.80.60.40.20–150

–100

–50

0

50

x

t


10.8 0.80.6 0.6

0.4 0.40.2 0.2 00

–0.4

–0.3

–0.2

–0.1

0

x

t


10.8 0.80.6 0.6

0.4 0.40.2 0.2 00

1032

1032

1032

1032

1032

1032

1032

0

x

t



The solution of Equation (71) is the form (18), and here,n = 2 is taken from the homogeneous balance between thehighest order derivative u′″ and the nonlinear term uu′.We obtain the solution of Equation (71) as

u ξð Þ = a−2φ−2 + a−1φ

−1 + a0 + a1φ + a2φ2: ð72Þ

Substituting Equation (72) together with its necessaryderivatives into Equation (71), the algebraic equation isarranged according to the powers of the function φkðξÞ.Then, the following coefficients are obtained:

φ−5 : −2k1ba2−2σ + 24λa−2σ3,φ−4 : −3k1ba−1a−2σ + 6λa−1σ3,φ−3 : −2aa−2σ − 2k1ba0a−2σ − k1ba

2−1σ

− 2k1ba2−2 + 40λa−2σ2,

φ−2 : −aa−1σ − 3k1ba−1a−2 − k1ba0a−1σ

− k1ba1a−2σ + 8λa−1σ2,

φ−1 : −2aa−2 − k1ba2−1 − 2k1ba0a−2 + 16λa−2σ,

φ0 : aa1σ − aa−1 − k1ba−2a1 + k1ba−1a2σ

+ k1ba0a1σ − k1ba0a−1 − 2λa1σ2 + 2λa−1σ,

φ1 : 2aa2σ + 2k1ba0a2σ + k1ba21σ − 16λa2σ2,

φ2 : aa1 + k1ba−1a2 + k1ba0a1 + 3k1ba1a2σ − 8λa1σ,φ3 : 2aa2 + 2k1ba22σ + 2k1ba0a2 + k1ba

21 − 40λa2σ,

φ4 : 3k1ba1a2 − 6λa1,φ5 : 2k1ba22 − 24λa2:

ð73Þ

where λ = −k2b3 − k3bc

2 − k4bd2.

Let the coefficients of φkðξÞ be zero. By solving the set ofequations given above for a−1, a0, a1, a, b, and σ, we obtainsolution sets as follows:

Set 1

a−2 = 0,a−1 = 0,a0 = a0,a1 = 0,

a2 =12λk1

,

a = a,b = b,c = c,d = d,

σ = a + k1a0b8bλ :

ð74Þ

Set 2

a−2 =3 a + k1a0bð Þ2

16k1b2λ,

a−1 = 0,a0 = a0,a0 = a0,a1 = 0,a2 = 0,a = a,b = b,c = c,d = d,

σ = a + k1a0b8bλ :

ð75Þ

10.8 0.80.6 0.60.4 0.4

0.2 0.2 00

200400600800

10001200140016001800

x

t


10.8 0.80.6 0.6

0.4 0.40.2 0.2 00

3000

2000

1000

0

1000

2000

3000

x

t



Set 3

a−2 =3 a + k1a0bð Þ2

16k1b2λ,

a−1 = 0,a0 = a0,a1 = 0,

a2 =12λk1

,

a = a,b = b,c = c,d = d,

σ = a + k1a0b8bλ :

ð76Þ

Thus, we obtain the solution of Equation (59) as ui,jðξÞ,ði = 1, 2, 3 ; j = 1, 2, 3, 4, 5Þ and ξ = ðatα/Γð1 + αÞÞ + ðbxα/Γð1+ αÞÞ + ðcyα/Γð1 + αÞÞ + ðdzα/Γð1 + αÞÞ; ui,jðξÞ is tas follows:

When ða + k1a0bÞ/ð8bλÞ < 0, we have σ < 0, then

u1,1 ξð Þ = a0 −3 a + k1a0bð Þ

2k1btanh2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi−a + k1a0b

8bλ

rξ

!,

u2,1 ξð Þ = −12λ a + k1a0bð Þ

k1coth2


8bλ

rξ

!+ a0,

u3,1 ξð Þ = −12λ a + k1a0bð Þ

k1coth2


8bλ

rξ

!

+ a0 −3 a + k1a0bð Þ

2k1btanh2


8bλ

rξ

!,

u1,2 ξð Þ = a0 −3 a + k1a0bð Þ

2k1bcoth2


8bλ

rξ

!,

u2,2 ξð Þ = −12λ a + k1a0bð Þ

k1tanh2


8bλ

rξ

!+ a0,

u3,2 ξð Þ = −12λ a + k1a0bð Þ

k1tanh2


8bλ

rξ

!

+ a0 −3 a + k1a0bð Þ

2k1bcoth2


8bλ

rξ

!:

ð77Þ

When ða + k1a0bÞ/ð8bλÞ > 0, we have σ > 0, then

u1,3 ξð Þ = a0 +3 a + k1a0bð Þ

2k1btan2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia + k1a0b

8bλ

rξ

!,

u2,3 ξð Þ = 12λ a + k1a0bð Þk1

cot2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia + k1a0b

8bλ

rξ

!+ a0,


cot2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia + k1a0b

8bλ

rξ

!

+ a0 +3 a + k1a0bð Þ

2k1btan2


8bλ

rξ

!,

u1,4 ξð Þ = a0 +3 a + k1a0bð Þ

2k1bcot2


8bλ

rξ

!,


tan2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia + k1a0b

8bλ

rξ

!+ a0,


tan2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia + k1a0b

8bλ

rξ

!

+ a0 +3 a + k1a0bð Þ

2k1bcot2


8bλ

rξ

!:

ð78Þ

When ða + k1a0bÞ/ð8bλÞ = 0, we have σ = 0, then

u1,5 ξð Þ = a0 +12λΓ2 1 + αð Þk1 ξα + ω� �2 = u3,5 ξð Þ: ð79Þ

Solutions ui,1 and ui,2ði = 1, 2, 3Þ describe the multiplesoliton. Solutions ui,3 and ui,4 ði = 1, 2, 3Þ represent the exactperiodic traveling wave solutions.

Figures 15–18 present the solutions: u1,1, u1,3, u2,3, andu1,5 of the generalized compound KdV-Burgers equationwith 0 < x < 1, 0 < t < 2. Solution u1,1 is presented forvalues α = 0:5, k1 = k2 = 1, k3 = k4 = −1, a = −4, b = 1, c = 1,d = 1, a0 = 0, and y = z = 0; solution u1,3 and u2,3 are pre-sented for values α = 0:5, k1 = k2 = 1, k3 = k4 = −1, a = 4, b= 1, c = 1, d = 1, a0 = 0, and y = z = 0; solution u1,5 is pre-sented for values α = 0:5, k1 = k2 = 1, k3 = k4 = −1, a = −4,b = 1, c = 1, d = 1, a0 = 4, ω = 0, and y = z = 0.

4. Results and Discussion

The generalized time fractional biological population model,the generalized time fractional compound KdV-Burgersequation, the space-time fractional regularized long-waveequation, and the (3 + 1)-space-time fractional Zakharov-Kuznetsov equation have gained the focus of many studiesdue to their frequent appearance in various applications.

The improved fractional subequation method has severaladvantages according to other traditional methods. Applyinga suitable fractional complex transform uðt, x1, x2,⋯,xmÞ =uðξÞ, ξ = ξðt, x1, x2,⋯,xmÞ and chain rule, the nonlinear frac-tional differential equations with the modified Riemann-Liouville derivative are converted into the nonlinear ordinarydifferential equations. This is a significant impact because


neither Caputo definition nor Riemann-Liouville definitionsatisfies the chain rule. With the help of the Riccati equation,the method has been employed for finding the exact analyti-cal solutions of these equations.

These obtained solutions are traveling solutions. Further-more, solutions ui,1ðξÞ and ui,2ðξÞ describe the solitons whichare everywhere in nature. ui,3ðξÞ and ui,4ðξÞ represent theexact periodic traveling wave solutions.

The improved fractional subequation method is reliableand effective for finding more solutions for some space-time fractional nonlinear differential equations. We can sub-stitute φ′ = r + pφ + qφ2 [39] for φ′ = σ + φ2, then, we willobtain more solutions.

Data Availability

All data included in this study are available upon request bycontact with the corresponding author.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research is supported by the Open Project of StateKey Laboratory of Environment-Friendly Energy Materials(19kfhg08), the Natural Science Foundation (61473338), andHubei Province Key Laboratory of Systems Science in Metal-lurgical Process (Y201705).

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50.8 40.6 3

0.4 20.2 1 00

0

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Figure 15: u1,1ðx, 0, 0, tÞ of Equation (69).

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