Exact solitary wave solutions for a system of some nonlinear space–time fractional differential equations

https://doi.org/10.1007/s12043-019-1864-6

Exact solitary wave solutions for a system of some nonlinear space–time fractional differential equations

MUHAMMAD HANIF^∗and M A HABIB

Department of Applied Mathematics, Noakhali Science and Technology University, Noakhali, Bangladesh

∗Corresponding author. E-mail: drhanifmuradnstu@yahoo.com

MS received 27 February 2019; revised 3 July 2019; accepted 21 August 2019

Abstract. We have enumerated new and exact general wave solutions, along with multiple exact soliton solutions of space–time nonlinear fractional differential equations (FDE), namely Zakharov–Kuznetsov–Benjamin–Bona–

Mahony (ZKBBM), foam drainage and symmetric regularised long-wave (SRLW) equations, by employing a relatively new technique called (G/G,1/G)-expansion method. Also, based on fractional complex transformation and the properties of the modified Riemann–Liouville fractional-order operator, the fractional partial differential equations transform into a form of ordinary differential equation (ODE). This method is a recollection of the commu- tation of the well-appointed (G/G)-expansion method introduced by Wanget al,Phys. Lett. A372, 417 (2008) In this paper, it is mentioned that the two-variable (G/G,1/G)-expansion method is more legitimate, modest, sturdy and effective in the sense of theoretical and pragmatical point of view. Lastly, the peculiarities of these analytic solutions are illustrated graphically by utilising the computer symbolic programming Wolfram Mathematica.

Keywords. (G/G,1/G)-expansion method; Zakharov–Kuznetsov–Benjamin–Bona–Mahony equation; foam drainage equation; symmetric regularised long-wave equation; fractional derivative; solitary wave solution.

PACS Nos 02.30.Jr; 02.30.Hq; 02.60.Lj

1. Introduction

The first concept of fractional calculus is found in 1695 [1], when Gottfried–Wilhelm–Leibniz suggested the possibility of fractional derivatives for the first time.

The foundations of the subject were laid by Liouville in a paper from 1832, and the fractional derivative of a power function obtained by Riemann in 1847 [2].

Fractional calculus is a powerful tool for modelling com- plex systems, especially for viscoelastic materials. The study of exact travelling and solitary wave solutions for nonlinear fractional differential equations (FDE) play an important role because of their prospective implementations in various scientific and technological fields, especially in solid-state physics, mathematical physics, mechanics, plasma physics, signal processing, bioengineering, optical fibres, geochemistry, stochas- tic dynamical systems, nonlinear optics, economics and business [3,4]. In the last few decades, many researchers and scholars have been attracted to solve nonlinear FDE by interposing several explicit, effective and powerful approaches. Researchers have introduced different methods such as the first integral method

[5,6], modified simple equation method [7–9], auxiliary equation method [10], fractional subequation method [11–13], fractional (G/G)-expansion method [14–16], ansatz method [17,18], fractional functional variable method [19], generalised (G/G)-expansion method [20], generalised Kudryashov method [21–25] and so on. Recently, based on the original (G/G)-expansion method, various modified techniques such as the gen- eralised (G/G)-expansion method, extended (G/G)- expansion method, (G/G²)-expansion method etc.

have been promoted. The main idea of the original single-variable (G/G)-expansion method is that, the exact travelling wave solution of nonlinear PDEs can be exposed by a polynomial in one-variable (G/G), in whichG =G(η)satisfies the second-order linear ordi- nary differential equation (LODE) G(η)+λG(η)+ μG(η) = 0, whereinλ andμ are non-zero constants and G is the derivative of G. This equation can also be used as an auxiliary equation. However, inspired and motivated by the ongoing research in this arena, we also introduce and make best use of (G/G,1/G)-expansion method, which can be envisaged as the modification of the original (G/G)-expansion method [26]. The

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key point of this two-variable (G/G,1/G)-expansion method is that the exact travelling wave solution of non- linear PDEs can be revealed by a polynomial in the two- variable (G/G) and (1/G) in whichG=G(η)gratifies the second-order LODEG(η)+λG(η)=μ, whereλ andμare constants. The degree of this polynomial can be discerned by introducing homogeneous balance prin- ciple between the highest order derivatives with highest order nonlinear term appearing in the conferred non- linear PDEs. Also, the coefficient of this polynomial can be ascertained by solving a set of algebraic equa- tions resulted by using this technique. More recently, some scholars like Yasar and Giresunlu [27] obtained exact solutions of space–time fractional Chan–Allen and Klein–Gordon equation by treating (G/G,1/G)- expansion method. Likewise, utilising similar tech- nique, Topsakalet al[28] have accomplished the exact and travelling wave solutions for space–time fractional modified Benjamin–Bona–Mahony (mBBM) and mod- ified nonlinear Kawahara equations. Hereafter, in this manuscript for the first time, we disclose the exact trav- elling and solitary wave solutions for the system of non- linear space–time FDE such as: Zakharov–Kuznetsov–

Benjamin–Bona–Mahony (ZKBBM), foam drainage and symmetric regularised long-wave (SRLW) equa- tions by applying two-variable (G/G,1/G)-expansion method. The principal motive for selecting this method is to give solutions in more general form. Moreover, the advantages of our proposed method over the original (G/G)-expansion method is that the solutions treat- ing the first method recapture the solutions treating the second one. So, we can say that the two-variable (G/G,1/G)-expansion method is an extension of the (G/G)-expansion method.

First, we describe the well-known space–time frac- tional ZKBBM equation, which arises as the gravity water wave phenomena in the unidirectional propa- gation of long waves in certain nonlinear dispersive systems. The form of the equation is represented by [29]

D_t^αv+D^α_xv−2avD^α_xv−b D^α_t D²_x^αv

=0,

t >0, 0< α≤1. (1)

In this equation,aandbare arbitrary constants.

Secondly, we consider the space–time fractional foam drainage equation [30] as

D_t^αw = 1

2wD^α_xD^α_xw−2w²D^α_xw+ D_x^αw2

,

0< α≤1. (2)

Finally, we designate the space–time fractional nonlin- ear SRLW equation, which emerges in various physical forms including ion sound waves in plasma. The equa- tion takes the following form [31]:

D_t^2αv+D^2α_x v+vD_t^α D_x^αv +D^α_t vD^α_xv+D_t^2α

D²_x^αv

=0, 0< α≤1. (3) This paper is organisd as follows: In §2, basic ideas of the modified Riemann–Liouville fractional-order derivative are given. In §3, we represent the structure of the renewed (G/G,1/G)-expansion method. In §4, we execute this method to seek new and further exact wave solutions of the space–time FDE written above. The nature of the solutions along with their graphical rep- resentation is provided in §5. In §6, we give the results and discussion and lastly, in §7, the concluding remarks of this present paper is given.

2. Basic ideas of the modified Riemann–Liouville fractional-order derivative

The Jumarie’s modified Riemann–Liouville derivative of order α with the continuous function f:R → R, x → f(x)is stated by the following expression [32]:

D_x^αf(x)

= 1 (−α)

x 0

(x−ξ)^−α−1(f(ξ)− f(0))dξ, where(x)is the gamma function which is defined as (x)=

_x

0

e^−tt^(x−1)dx.

Ifα <0 then D_x^αf(x)=

f^(α−1)(x) , D_x^αf(x)= 1

(1−α) d dx

× _x

0 (x −ξ)^−α(f(ξ)− f(0))dξ. (4) Also if 0< α <1,

D_x^αf(x)=

f⁽ⁿ⁾(x)_(α−n)

(5) ifn ≤α≤n+1 andn ≥1.

For fractional-order derivative, some essential postu- lates which we use further in this paper are selected as follows:

Postulate1.

D_x^αx^r = (1+r)

(1+r −α)x^r−α, (6) wherer is the real number.

Postulate2.

D_x^α(a f(x)+bg(x))=a D^α_x f(x)+b D^α_xg(x), (7)

wherea andbare arbitrary constants.

Postulate3.

D^α_x f(ξ)= df

dξ D^α_x(ξ), (8)

whereξ =g(x).

3. Formation of(G/G, 1/G)-expansion method In this section, we explain the major concept of (G/G,1/G)-expansion method for attaining exact solitary wave solutions of the nonlinear time FDE. As usual, we envisage the second-order LODE in G = G(η)as

G(η)+λG(η)=μ, (9)

and we assume two rational functionsφandψ as φ = G(η)

G(η), ψ = 1

G(η), (10)

whereλandμare two arbitrary constants andGis the derivative ofG.

From eqs (9) and (10) we obtain

φ= −φ²+μψ−λ, ψ = −φψ. (11) The solution of LODE (9), accomplish the following three distinct annotations:

Annotation1: Ifλ < 0, the general solution of LODE (9) gives

G(η)= A1 sinh √

−λ η

+A2 cosh √

−λ η +μ

λ, (12) where A1 and A2 are two arbitrary constants. There- fore, from eqs (10)–(12) the following relation can be obtained:

ψ² = −λ λ²ε+μ²

φ²−2μψ+λ , where

ε = A²₁−A²₂. (13)

Annotation2: Ifλ > 0, the general solution of LODE (9) is

G(η)= A1sin √

λ η

+A2cos √

λ η +μ

λ, (14)

and similarly, from eqs (10), (11) and (14) the corre- sponding relations are

ψ² = λ λ²ε−μ²

φ²−2μψ+λ , where

ε= A²₁+ A²₂. (15)

Annotation3: Finally, ifλ =0, the general solution of LODE (9) takes the form

G(η)= μ

2η²+A₁η+A₂. (16)

Consequently, we have

ψ² = 1

A²₁−2μA2

φ²−2μψ

. (17)

Now suppose we have a nonlinear FDE, say in two inde- pendent variablesxandtof the form

F

v,D_t^αv,D_x^αv,D_t^αD_t^αv,D_t^αD^α_xv,D^α_xD^α_xv,

. . .)=0, (18)

where F is a polynomial inv =v(x,t)and its various partial fractional derivatives. The main structure of the G/G,1/G

-expansion method are present as follows:

Step1: Take into account, the travelling wave variable v(x,t)=v(η), η= kx^α

(1+α) + ct^α

(1+α), (19) wherek andcare non-zero arbitrary constants. Substi- tuting eq. (19) and the values ofv = v(x,t)and their various fractional derivatives into eq. (18), it transforms into the following ordinary differential equation (ODE):

H

v, v, v, v, . . .

=0, (20)

whereindicates d/dηand so on. If needed, we integrate eq. (20) one or more times and assume the constant of integration to be zero.

Step2: In accordance with the(G/G,1/G)-expansion technique the exact solution of eq. (20) can be disclosed by a finite power series in two variables φ and ψ as follows:

v(η)= M i=0

aiφⁱ + M

j=1

bjφ^j−1ψ, (21) whereai(i=0,1,2, . . . ,M)and bj(j =1,2, . . . ,M) are constants, which will be determined later, as for instance, evaluating the values of positive integerMby homogeneous balance principle, balancing the highest- order derivative with the highest-order nonlinear term

appearing in eq. (20). But, balance numberMis not pos- itive all the times: sometimes it is fraction or negative.

In this case, we use the following transformation [33]:

(a) IfM=q/p, whereq/pis a fraction in the lowest terms, then

v(η)=u^q/p(η), (22)

then substitute eq. (22) into eq. (20) to attain an equation in the renewed functionu(η)with a pos- itive integer balance number.

(b) IfMis a negative number, then

v(η)=u^M(η), (23)

and set eq. (23) into eq. (20) to obtain an equation in the new functionu(η)with a positive integer balance number.

Step3. Substituting eq. (21) into eq. (20), also operat- ing eqs (11) and (13) will transform into a polynomial in φ and ψ, in which the degree of ψ is not longer than 1. Equalising all the coefficient of this polyno- mial to zero, yields a system of algebraic equations for ai (i=0,1,2, . . . ,M), bj(j =1,2, . . . ,M), k,c,μ, λ (λ <0), A1andA2.

Step 4. Solve the system which is attained in Step 3 with the aid of any computer symbolic program, like Mathematica and substituting the values ofai,bj,k,c, μ, λ (λ < 0), A1 and A2 into eq. (21) and finally, we come up with different types of exact wave solutions of eq. (18) represented by the hyperbolic functions.

Step5. Likewise, pursuing Steps 3 and 4, plugging eq.

(21) into eq. (20), treating eqs (11) and (15) (or eqs (11) and (17)), we come up with the solution of eq. (18) in the case of trigonometric functions (or by rational functions) as before.

4. Implementations of the method In this section, the

G/G,1/G

-expansion method has been assigned to visualise different exact wave solutions of the nonlinear space–time fractional ZKBBM equa- tion (1), foam drainage equation (2) and SRLW equation (3).

4.1 Space–time fractional ZKBBM equation

In this subsection, we shall make best use of the G/G,1/G

-expansion method to constitute the exact travelling and solitary wave solutions of the space–time fractional ZKBBM equation (1). Take into account, the following wave transformation

v(x,t)=v(ξ), ξ = kx^α

(1+α) − ct^α

(1+α), (24)

wherekandcare non-nzero constants. This wave trans- formation permits eq. (1) to be reduced to an ODE (k−c)v−2akvv+bck²v =0, (25) where prime denotes the derivative with respect toξ. Integrating eq. (25) once with respect toξ and setting the constant of integration to be zero, we get

(k−c)v−akv²+bck²v=0. (26) By using the homogeneous balance principle, balancing the highest-order derivative v with the highest-order nonlinear termv², balance number M =2 is obtained.

By putting the value of M = 2 in eq. (21), then the solution formula becomes

v(ξ)=a0+a1φ+a2φ²+b1ψ+b2φψ, (27) wherea0,a1,a2,b1, andb2are constants to be decided later. Therefore, the aforementioned three annotations are studied as follows:

Annotation1: Whenλ < 0 (Hyperbolic function solu- tions)

Plugging the value ofv(ξ)from eq. (27) into eq. (26), in addition with eqs. (11) and (13), the left-hand side of eq. (26) becomes a polynomial inφ andψ. Assuming each coefficient of the polynomial to be zero, we arrive at a system of algebraic equations as follows:

φ: −ca1+ka1+2bck²λa1−2aka0a1

+6bck²λ²μb2

μ²+λ²σ +2akλ²b1b2

μ²+λ²σ =0,

φ²: −aka²₁−ca2+ka2+8bck²λa2−2aka0a2

−2bck²λμ²a2

μ²+λ²σ +bck²λμb1

μ²+λ²σ + akλb₁² μ²+λ²σ + akλ²b²₂

μ²+λ²σ =0,

φ³:2bck²a1−2aka1a2+6bck²λμb2

μ²+λ²σ +2akλb1b2

μ²+λ²σ =0,

φ⁴:6bck²a2−aka₂²+ akλb²₂ μ²+λ²σ =0,

ψ: −4bck²λμa2−cb1+kb1+bck²λb1−2aka0b1

+4bck²λμ³a2

μ²+λ²σ −2bck²λμ²b1

μ²+λ²σ − 2akλμb₁² μ²+λ²σ =0,

φψ: −3bck²μa1−2aka1b1−cb2+kb2+5bck²λb2

−2aka0b2−12bck²λμ²b2

μ²+λ²σ −4akλμb1b2

μ²+λ²σ =0, φ²ψ: −10bck²μa₂+2bck²b₁

−2aka2b1−2aka1b2− 2akλμb²₂ μ²+λ²σ =0, φ³ψ:6bck²b₂−2aka₂b₂ =0,

Const.: −ca0+ka0−aka²₀+2bck²λ²a2

−2bck²λ²μ²a2

μ²+λ²σ =0.

On solving the above algebraic equations with the aid of a symbolic software program, like Mathematica, we come up with the following different sets of results:

Set1:

a₀ = 3

2a, a₁=0, a₂ = 3

2aλ, b₁ = −3μ 2aλ, b2 = ±3

−μ²−λ²σ 2aλ^3/2 , k = 1

√b√

λ, c = 1 2√

b√

λ. (28)

Setting the above values into eq. (27), we attain the exact wave solution of eq. (1) as

v(ξ)= 3

2a − 3μ

2aλ

A1 sinh ξ√

−λ

+ A2 cosh ξ√

−λ +^μ_λ

+3 −λ

−μ²−λ²σ A₁ cosh ξ√

−λ

+A₂ sinh ξ√

−λ 2aλ^3/2

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ + ^μ_λ2

− 3

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ2

2a

A₁ sinh ξ√

−λ

+A₂ cosh ξ√

−λ

+^μ_λ2 (29)

where σ = A²₁− A²₂ and

ξ = 1

√bλ

x^α

(1+α) − t^α 2(1+α)

.

Particularly, if we putA1 =0,μ=0 and A2 =0 in eq.

(29), the soliton solution is v(ξ)= 3

2a

1+√ σtanh

ξ√

−λ sech

ξ√

−λ

− tanh

ξ√

−λ2

. (30)

But if we take A2 = 0, μ = 0 and A1 = 0 then the solution becomes

v(ξ)= 3 2a

1+√

σcoth ξ√

−λ csch

ξ√

−λ

− coth

ξ√

−λ2

. (31)

Set2:

a0 = −1

a, a1=0, a2 = − 3

2aλ, b1 = 3μ 2aλ, b2 = ±3

−μ²−λ²σ 2aλ^3/2 , k = − i

√b√

λ, c= − i 2√

b√

λ. (32)

Substituting these values into eq. (27), the solution of eq. (1) takes the form

v(ξ)= −1

a + 3μ

2aλ(A1sinh(ξ√

−λ )+ A2cosh(ξ√

−λ )+ ^μ_λ)

+3 −λ

−μ²−λ²σ

A₁cosh ξ√

−λ

+A₂sinh ξ√

−λ 2aλ^3/2

A1sinh ξ√

−λ

+A2cosh ξ√

−λ +^μ_λ2

+ 3

A1cosh ξ√

−λ

+A2sinh ξ√

−λ2

2a

A1sinh ξ√

−λ

+A2cosh ξ√

−λ

+^μ_λ2 (33)

where σ = A²₁−A²₂ and

ξ = − i

√bλ

x^α

(1+α) − t^α 2(1+α)

.

For favourable values, setA₁=0,μ=0 andA₂=0 in eq. (33), and the solution yields

v(ξ)= 1 a

3√

λ³σ 2λ^3/2 tanh

ξ√

−λ sech

ξ√

−λ +3

2

tanh ξ√

−λ2

−1

. (34)

Also if we inputA2=0,μ=0 andA1=0, we obtain v(ξ)= 1

a

3√ λ³σ 2λ^3/2 coth

ξ√

−λ csch

ξ√

−λ +3

2

coth ξ√

−λ2

−1

. (35)

Set3.

a0 = 2bk²λ a

1−bk²λ, a1=0, a2 = 3bk² a

1−bk²λ,

b₁ = − 3bk²μ a

1−bk²λ, b₂ = 3i bk²

μ²+λ²σ

√λ

±a∓abk²λ and

c= k

1−bk²λ. (36)

By using these values into eq. (27), we derive the exact solution of eq. (1) as

v(ξ)= − 2bk²λ a

bk²λ−1−3bk² μ²+λ²σ A₁cosh ξ√

−λ

+A₂sinh ξ√

−λ a

1−bk²λ

A1sinh ξ√

−λ

+A2cosh ξ√

−λ

+^μ_λ2 + 3bk² a

bk²λ−1

×

⎛

⎝ μ A1sinh

ξ√

−λ

+A2cosh ξ√

−λ

+^μ_λ + λ

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ2

A1 sinh ξ√

−λ

+ A2 cosh ξ√

−λ +^μ_λ2

⎞

⎠, (37)

where σ = A²₁−A²₂ and

ξ =k

x^α

(1+α) − t^α 1−bk²λ

(1+α)

.

For special values, put A₁ =0,μ=0 and A₂ =0, v(ξ)= bk²λ

a

1−bk²λ

2−3√ σ tanh

ξ√

−λ

×sech ξ√

−λ

−3

tanh ξ√

−λ2 ,

(38) and if A2 =0,μ=0 and A1 =0 then

v(ξ)= bk²λ a

1−bk²λ

2−3√ σcoth

ξ√

−λ

×csch ξ√

−λ

−3

coth ξ√

−λ2 .(39) Annotation 2: When λ > 0 (Trigonometric function solutions)

In a similar manner, as mentioned in Annotation 1, putting the values of v(ξ) from eq. (27) into eq. (26) along with eqs. (11) and (15), the left-hand side of eq. (26) becomes a polynomial inφandψ. Tak- ing each coefficient of the polynomial equal to zero, we obtain a system of algebraic equations (for the sake of simplicity, the equations are not presented here) for a₀,a₁,a₂,b₁,b₂, μ, λ and σ. Solving this system of equations we obtain the following sets of results:

Set1.

a₀ = 3

2a, a₁ =0, a₂= 3 2aλ, b1 = −3μ

2aλ, b2 = ±3

−μ²+λ²σ 2aλ³² , k = 1

√b√

λ, c= 1 2√

b√

λ. (40)

Therefore, in this result the exact solution of eq. (1) transforms as

v(ξ)= 3

2a− 3μ

2aλ A1sin

ξ√ λ

+A2cos ξ√

λ +^μ_λ +3 −μ²+λ²σ

A1cos ξ√

λ

−A2sin ξ√

λ 2aλ

A1sin ξ√

λ

+A2cos ξ√

λ +^μ_λ2

+ 3

A1cos ξ√

λ

−A2sin ξ√

λ2

2a

A₁sin ξ√

λ

+A₂cos ξ√

λ

+^μ_λ2, (41) where

σ = A²₁+A²₂ and

ξ = 1

√bλ

x^α

(1+α)− t^α 2(1+α)

.

By setting A₁ =0,μ =0 and A₂ = 0, the periodic solution becomes

v(ξ)= 3 2a

tan

ξ√ λ2

−√ σtan

ξ√ λ

sec ξ√

λ +1

. (42) Besides, forA₂=0,μ=0 and A₁=0, we have v(ξ)= 3

2a

cot ξ√

λ2

+√ σcot

ξ√ λ

csc ξ√

λ +1

. (43) Set2.

a0 = − 2bk²λ a

bk²λ−1, a1=0, a2= − 3bk² a

bk²λ−1, b1 = 3bk²μ

a

bk²λ−1,b2= 3i bk²

μ²−λ²σ

√λ

±a∓abk²λ and

c= − k

bk²λ−1. (44)

Hence, the exact travelling wave solution of eq. (1) is v(ξ)= − 2bk²λ

a

bk²λ−1+ 3bk²μ

a

bk²λ−1 A1sin

ξ√ λ

+ A2cos ξ√

λ +^μ_λ

+3i bk² μ²−λ²σ A1cos

ξ√ λ

−A2sin ξ√

λ a

1−bk²λ A1sin

ξ√ λ

+A2cos ξ√

λ

+^μ_λ2 − 3bk²λ A1cos

ξ√ λ

−A2sin ξ√

λ2

a

bk²λ−1 A1sin

ξ√ λ

+A2cos ξ√

λ + ^μ_λ2

(45) where

σ = A²₁+ A²₂ and

ξ =k

x^α

(1+α) + t^α bk²λ−1

(1+α)

.

In particular, if we input A₁ =0, μ =0 and A₂ = 0, we achieve the solution of the following form:

v(ξ)= bk²λ a

1−bk²λ

3

tan ξ√

λ2

+3√ σtan

ξ√ λ

sec ξ√

λ +2

. (46) Also, for A2 =0,μ=0 and A1 =0,

v(ξ)= bk²λ a

1−bk²λ

3

cot ξ√

λ2

−3√ σcot

ξ√ λ

csc ξ√

λ +2

. (47) Annotation3. Whenλ=0 (Rational function solution) Using similar process, as mentioned in Annotation 1, substituting the values of v(ξ) from eq. (27), into eq. (26), along with eqs. (11) and (17), the left-hand side of eq. (26) becomes a polynomial inφandψ. Tak- ing each coefficient of the polynomial equal to zero, a system of algebraic equations is obtained (for the sake of simplicity, the equations are not represent here) for a0, a1, a2, b1, b2, μ, λand σ. Solving this system of equations, the following results are obtained:

Set1.

a0 =0, a1=0, a2 = 3bk²

a , b1 = −3bk²μ a , b2 = ∓3bk² A²₁−2μA₂

a and

c=k. (48)

Now, for the above values the new exact wave solution of eq. (1) is

v(ξ)= 6bk²

2A²₁+2A2

μξ − A²₁−2μA2

+μ

ξ

μξ −2 A²₁−2μA2

−2A2

a

μξ²+2ξA1+2A2

2 , (49)

where ξ =k

x^α

(1+α)− t^α (1+α)

.

Specifically, if we take A1 =0,μ= −1 andA2=0, the solution is as follows:

v(ξ)= 3bk² a²

1− ^ξ₂²2

ξ²+ξ√ 2−a

1−ξ²

2

. (50) On the other hand, A2 = 0,μ = 0 and A1 = 0, then the kink-type solution is

v(ξ)= −3bk²

aξ . (51)

4.2 Space–time fractional foam drainage equation The space–time fractional foam drainage equation is a model of the flow of liquid through the channels and nodes between the bubbles, driven by gravity and capil- larity. This equation is also stated as the model of waves on shallow water surface, ion-acoustic waves in plasma and the waves on foam may be illustrated in the field of a nonlinear PDE for the foam density as a function of time and vertical position [30]. For solving eq. (2), we envisage the travelling wave transformation

ξ=k

x− ct^α (1+α)

and η=mx− wt^α (1+α).

(52) By means of transformation (52), eq. (2) is converted into the following form:

2kw−m²ww+4mw²w−2m² w2

=0. (53) Again using the condition w = v⁻¹, then eq. (53) reduces to the following nonlinear ODE:

2kv²v+4m² v2

−m²vv+4mv =0. (54)

By the homogeneous balance principle, balancing between the highest-order derivative with highest-order nonlinear term in eq. (54), we achieve the value of inte- gerM =1. Hereafter, eq. (21) becomes

v(ξ)=a0+a1φ+b1ψ, (55) wherea₀,a₁andb₁are constants to be determined later.

Annotation1: Whenλ < 0 (Hyperbolic function solu- tions)

Substituting the value ofv(ξ)and its derivatives from eq. (53) into eq. (54), in cohesion with eqs (11) and (13), the left-hand side of eq. (54) turns out to be a polynomial inφandψ. Equating each coefficient of different powers to zero, we obtain a system of algebraic equations (for the sake of brevity, the equations are not given here) for a₀, a₁, b₁, μ, λ and σ. Solving these systems by any computer program like Mathematica, we have several types of results as follows:

a0 =0, a1= 2 mλ, b1 = ±2

−μ²−λ²σ

mλ^3/2 and k = m³λ

4 . (56)

Substituting these results into eq. (55), we get the exact solution of eq. (2) in the following form:

v(ξ)=

2

−(μ²+λ²σ) mλ^3/2

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ +^μ_λ

+2√

−λ

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ mλ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ

+^μ_λ . (57) According to the condition,w = v⁻¹ and simplifying solution (57), we get

w(ξ)= mλ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ + ^μ_λ 2

√

−λ

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ + √

−(μ√²+λ²σ) λ

, (58)

where σ = A²₁−A²₂ and

ξ = m³λ 4

x− ct^α (1+α)

.

When A1=0,μ=0 and A2 =0, the solitary wave solution is

w(ξ)= m√

λ 2i

tanh ξ√

−λ +√

σsech ξ√

−λ. (59) and when A2=0,μ=0 andA1=0, we obtain

w(ξ)= m√

λ 2i

coth ξ√

−λ +√

σcsch ξ√

−λ. (60) Annotation 2. When λ > 0 (Trigonometric function solutions)

In a similar procedure, putting the value ofv(ξ)and its derivatives from eq. (55) into eq. (54), along with eqs (11) and (15) and then solving the sets of algebraic equations, the following solutions will be obtained:

a0 =0, a1 = 2

mλ, b1= ∓2

−μ²+λ²σ mλ^3/2 and

k = m³λ

4 . (61)

Hence, the solution of eq. (2) is as follows:

v(ξ)= − 2

−μ²+λ²σ mλ^3/2

A1 sin ξ√

λ

+A2cos ξ√

λ +^μ_λ + 2

A1cos

ξ√ λ

−A2sin ξ√

λ m√

λ A₁sin

ξ√ λ

+A₂cos ξ√

λ +^μ_λ.

(62) Using the conditionw=v⁻¹, eq. (62) takes the follow- ing form:

w(ξ)= m√ λ

A1 sin ξ√

λ

+A2 cos ξ√

λ +^μ_λ 2

A1 cos

ξ√ λ

−A2 sin ξ√

λ

−√

−μ²+λ²σ λ

, (63)

where σ = A²₁+A²₂ and

ξ = m³λ 4

x− ct^α (1+α)

.

Particularly, if we selectA₁=0,μ=0 and A₂ =0, the periodic solution is

w(ξ)= − m√

λ 2

tan

ξ√ λ

+√ σsec

ξ√

λ. (64) ForA₂=0,μ=0 and A₁=0,

w(ξ)= m√

λ 2

cot

ξ√ λ

−√ σcsc

ξ√

λ. (65)

4.3 Space–time nonlinear fractional SRLW equation In this subsection, the space–time fractional SRLW equation will be solved by treating (G/G,1/G)- expansion method and various types of periodic and solitary wave solutions are acquired. For this purpose, we conceive the following transformation:

v(x,t)=v(ξ), ξ = kx^α

(1+α)+ ct^α

(1+α) +ξ0, (66) where k, c and ξ0 are constants along with k,c = 0.

Using the above transformation, the nonlinear ODE form of eq. (3) is

2k²c²v+2

k²+c²

v+kcv² =0. (67) By using the homogeneous balance principle, balancing the highest-order derivative v with the highest-order nonlinear termv², balance number M =2 is obtained.

Hence the solution formula is

v(ξ)=a0+a1φ+a2φ²+b1ψ+b2φψ, (68) wherea0,a1,a2,b1andb2are constants to be discussed later.

Annotation1. Whenλ < 0 (Hyperbolic function solu- tions)

Proceeding in a similar way as stated in §4.1(Anno- tation 1), the following results will be arrived:

Result1.

a₀ = 4k²λ −

1+k²λ

1+k²λ , a₁=0,

a₂ = − 6k²

−

1+k²λ, b₁ = 6k²μ

−

1+k²λ, b₂ = ±6

k⁴μ²+k⁴λ²σ

√λ+k²λ² ,

c= k

−

1+k²λ. (69)

Plugging these results into eq. (68), we achieve the exact travelling wave solution of eq. (3) as

v(ξ)= 4i k²λ

1+k²λ+ 6k²μ

−

1+k²λ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ +^μ_λ +6 −λ

k⁴μ²+k⁴λ²σ A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ λ

1+k²λ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ + ^μ_λ2

+ 6k²λ

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ2

−

1+k²λ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ

+^μ_λ2, (70)

where σ = A²₁−A²₂ and

ξ = kx^α

(1+α) + kt^α

−

1+k²λ

(1+α) +ξ0.

If we take A₁ = 0,μ=0 and A₂ =0, we come up with the periodic solution

v(ξ)= i k²λ 1+k²λ

4+6√

σtanh ξ√

−λ

×sech ξ√

−λ

−6

tanh ξ√

−λ2 .(71) Consequently, for A2 = 0, μ = 0 and A1 = 0, we obtain

v(ξ)= i k²λ 1+k²λ

4+6√

σcoth ξ√

−λ

×csch ξ√

−λ

−6

coth ξ√

−λ2 .(72) Result2.

a0 =2i√

2, a1=0, a2 = 3i√ 2 λ ,

b1 = −3i√ 2μ

λ , b2 = ±3√ 2

μ²+λ²σ λ^3/2 , k = − 1

√λ, c = i

√2√

λ. (73)

Therefore, the exact solution of eq. (3) reduces to

v(ξ)=2i√

2− 3i√

2μ λ

A1 sinh ξ√

−λ

+A2 cosh ξ√

−λ +^μ_λ +3√

2 −λ

μ²+λ²σ

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ λ³²

A1 sinh ξ√

−λ

+ A2 cosh ξ√

−λ +^μ_λ2

−3i√ 2

A1 cosh ξ√

−λ

+A2 sinh ξ√

−λ2

A₁ sinh ξ√

−λ

+A₂ cosh ξ√

−λ

+^μ_λ2, (74)

where σ = A²₁− A²₂ and

ξ = − x^α

√λ (1+α)+ i t^α

√2λ (1+α) +ξ0.

Specifically, for A1 =0,μ=0 and A2 =0, we obtain the solitary wave solution as

v(ξ)=√ 2i

2+3√ σtanh

ξ√

−λ sech

ξ√

−λ

−3

tanh ξ√

−λ2

. (75)

Likewise, if we set A2 = 0, μ = 0 and A1 = 0, we obtain

v(ξ)=√ 2i

2+3√ σcoth

ξ√

−λ csch

ξ√

−λ

−3

coth ξ√

−λ2

. (76)

Annotation 2. When λ > 0 (Trigonometric function solutions)

Following the uniform procedure which is described in §4.1 (Annotation 2), the following results will be obtained:

Result1.

a0 = 4k²λ

−

1+k²λ, a1 =0, a2= 6k²

−

1+k²λ, b₁ = − 6k²μ

−

1+k²λ, b₂ = ∓6

k⁴μ²−k⁴λ²σ

√λ+k²λ² ,

c= − k

−

1+k²λ. (77)

Hence, in this result the trigonometric function solution of eq. (3) is

v(ξ)= 4k²λ

−

1+k²λ

− 6k²μ

−

1+k²λ A₁sin

ξ√ λ

+A₂cos ξ√

λ +^μ_λ

−6 λ

k⁴μ²−k⁴λ²σ A₁cos

ξ√ λ

−A₂sin ξ√

λ λ

1+k²λ A1sin

ξ√ λ

+A2cos ξ√

λ +^μ_λ2

+ 6k²λ A1cos

ξ√ λ

−A2sinh ξ√

λ2

−

1+k²λ A1sin

ξ√ λ

+A2cos ξ√

λ +^μ_λ2,

(78) where

σ = A²₁+A²₂ and

ξ = kx^α

(1+α) − kt^α

−

1+k²λ

(1+α) +ξ0.

In particular, if we put A1 = 0,μ =0 and A2 = 0, the obtained periodic solution is

v(ξ)= k²λ

−

1+k²λ

4−6√ σtan

ξ√ λ

sec ξ√

λ

+6

tan ξ√

λ2

. (79)

If A2 =0,μ=0 and A1 =0 in eq. (78), v(ξ)= k²λ

−

1+k²λ

4+6√ σcot

ξ√ λ

×csc ξ√

λ +6

cot

ξ√ λ2

. (80) Result2.

a0 =3i√

2, a1=0, a2=3i√ 2

λ , b1= − 3i√ 2μ λ , b2 = ±3√

2

μ²−λ²σ

λ^3/2 , k = − i

√λ, c = 1

√2√ λ. (81) Now, the exact solution of eq. (3) takes the following form:

v(ξ)=3i√

2− 3i√

2μ λ

A1sin ξ√

λ

+A2cos ξ√

λ +^μ_λ +3√

2 μ²−λ²σ A1cos

ξ√ λ

−A2sin ξ√

λ λ

A1sin ξ√

λ

+A2cos ξ√

λ +^μ_λ2

+3i√ 2

A1cos

ξ√ λ

− A2sin ξ√

λ2

A1sin

ξ√ λ

+A2cos ξ√

λ

+ ^μ_λ2 (82) where

σ = A²₁+ A²₂ and

ξ = − i x^α

√λ(1+α)+ t^α

√2λ(1+α)+ξ0.

More preciously, for A1 = 0,μ = 0 and A2 = 0, the above solution transforms into the form

v(ξ)=i3√ 2

1−√

σtan ξ√

λ sec

ξ√ λ +

tan ξ√

λ2

. (83)

Also, if theA2=0,μ=0 and A1=0, we obtain

Figure 1. Shape of eq. (34) corresponding to the values (a)b=1.5,λ= −1 and its projection at (b)t =1.

Figure 2. Shape of eq. (39) corresponding to the values (a)b=1,k=2,a=1, λ= −1.2 and its projection at (b)t =2.

Figure 3. Shape of eq. (42) corresponding to the values (a)b=0.01,λ=1 and its projection at (b)t=1.

v(ξ)=i3√ 2

1+√

σcot ξ√

λ csc

ξ√ λ +

cot ξ√

λ2

. (84)

Annotation3. Whenλ=0 (Rational function solution) Taking the similar procedure described in §4.1(Anno- tation 3), the following results will be obtained:

a0 =0, a1 =0, b1 = −μa2,

b2 = ∓a2 A²₁−2μA2, k = ±(−1)^1/4√

a2

√6 ,

c= ±(−1)^3/4√ a₂

√6 and a2is a free parameter.(85) Substituting these results into eq. (68) and simplify- ing, we come up with the solution of eq. (3) as