# Analytical Solutions for Nonlinear Dispersive Physical Model

Authors: Wen-Xiu Ma; Mohamed R. Ali; R. Sadat
Journal: Complexity
 (2020)

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## Abstract

Nonlinear evolution equations widely describe phenomena in various fields of science, such as plasma, nuclear physics, chemical reactions, optics, shallow water waves, fluid dynamics, signal processing, and image processing. In the present work, the derivation and analysis of Lie symmetries are presented for the time-fractional Benjamin–Bona–Mahony equation (FBBM) with the Riemann–Liouville derivatives. The time FBBM equation is reduced to a nonlinear fractional ordinary differential equation (NLFODE) using its Lie symmetries. These symmetries are derivations using the prolongation theorem. Applying the subequation method, we then use the integrating factor property to solve the NLFODE to obtain a few travelling wave solutions to the time FBBM.


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## Body

## 1. Introduction

Partial differential equations running into the thinking of most of the researchers as it represented the importance in several topics of scientific fields as mechanics, optical fibers, medical sciences (as breast cancer), biological science, turbulent bursts, and oceans waves [1–11].The differential model has a broad application in many phenomena as in [12–14]. Recently, nonlinear fractional differential equations (NLFDEs) show significantly in engineering and applications of other sciences, for example, electrochemistry, physics, electromagnetics, and signal data processing [15–22].Getting exact solutions for these forms of equations became an important issue; then, most researchers try to achieve this target. The most effective method for obtaining exact solutions for NLFPDEs is the Lie symmetry reduction method. There are many papers for using Lie’s method to obtain explicit solutions for NLFPDEs [23–26].In our paper, we drive the symmetry vectors for the time FBBM equation and present new closed-form solutions for it. The FBBM equation has many forms [27–30], and we choose to work on the following form:(1)Dtαψ=−ψx−ψψx+ψxxt,where ψxxt is the dissipative term.The manuscript is prearranged as follows. In Section2, Lie’s group method for FPDEs is exposed. In Section 3, we apply the Lie group reduction method to obtain Lie point symmetry for the time FBBM equation (1). At the end of Section 4, we use these similarity variables to get the reduced equation. In Section 4, we use two methods for solving the resulting ordinary differential equation, the first method is the subequation method and the second method is the integrating factors method to get new solutions that have the properties and form the travelling wave form for the FBBME. In the end, conclusions are written in Section 5.


## 2. Notations and Introductory

### 2.1. Fractional Riemann–Liouville Derivative

In this section, we show some definitions for RL fractional derivative [31], which can be considered as follows:(2)Dtαf=dnftdtn,ifn=α,n∈ℕ,dndtnIn−αft,if0≤n−1<α<n,n∈ℕ,(3)In−αft=1Γn−α∫0tt−sfsds,n>0,where Dtα is the total differentiation of integer number of orders α,α>0, the Gamma function is Γn−α, and Inft is the (RL) fractional integral of an order of n.Definition 1.
The partial derivative of orderα for Riemann–Liouville definition is presented by(4)∂tα=∂nf∂tn,n=α,∂n∂tn1Γn−α∫0tt−sn−α−1fs,xds,0≤n−1<α<n.


### 2.2. Notations for Lie Symmetry Reduction Method for the Time FPDEs

In this section, we show in detail the main notations and definitions that will be used for obtaining the symmetries of NLFPDEs.Here, we will consider timing NLFPDEs of the form [31–33](5)∂tαψ=Ft,x,ψ,ψx,ψxx,0<α≤1.Assume, equation (2) has a Lie vector X in the form(6)X=ξ1x,t,u∂∂x+ξ2x,t,u∂∂t+ηx,t,u∂∂u,where ξ1, ξ2, and η can be called as the infinitesimals of the transformations the independent and the dependent variables x,t,ψ, respectively. Let a one-parameter Lie algebra of infinitesimal transformations be of the following form:(7)t¯=t+εξ2t,x,ψ+Oε2,x¯=x+εξ1t,x,ψ+Oε2,ψ¯=ψ+εηt,x,ψ+Oε2,∂αψ¯∂t¯=∂αψ∂tα+εηα0t,x,ψ+Oε2,∂ψ¯∂x¯=∂ψ∂x+εηxt,x,ψ+Oε2,∂3ψ¯∂x¯2∂t¯=∂3ψ∂x2∂t+εηxxtt,x,ψ+Oε2,where ε « 1 can be defined as a group parameter, in most cases we take it equal one. The explicit expressions of ηx,ηxx, and ηxxt, which can be called the prolongation of the infinitesimals and are given by(8)ηx=Dxη−ψxDxξ1−ψtDtξ2,ηxx=Dxηx−ψxtDxξ1−ψxxDtξ2,ηxxt=Dxηxt−ψxxxDxξ1−ψxxtDxξ2,where Di is the total differentiation operator [34] with respect to the independent variables xi (i = 1, 2, then x1=x,x2=t):(9)Di=∂∂xi+∑α=1qψiα∂∂ψα+∑j=1p∑α=1qψjiα∂∂ψjα+⋯+∑j1=1p⋯∑jn−1p∑α=1qψj1,…,jn−1iα∂∂ψ1,…,jn−1iα.Theorem 1.
Equation (1) concedes a one-parameter group of infinitesimal transformations in equation (2) with the Lie Vector X if and just if the accompanying infinitesimal conditions hold:(10)Prα,3XΔΔ=0=0,where Δ=Dtαu−Ft,x,u,ux,uxx,…… and Pr is the 3rd prolongation of the infinitesimal generator X.Definition 2.
Prolonged vector is given by [31](11)PrnX=X+∑i=1p∑α=1qξiα∂∂uiα+⋯+∑j1=1p⋯∑jn=1p∑α=1qξj1,….,jnα∂∂uj1,….,jnα,where q is the numbers of dependent variables, p is the numbers of independent variables, ∂/∂uj1α=∂/∂uxα, and PDE involve derivatives up to order n. Also, the invariance condition [35] gives(12)ξ2t,x,ut=0=0.
Theαthextended infinitesimal, which deals with fractional derivatives, has the following form [36–38]:(13)ηα0=∂αη∂tα+ηu−αDtξ2∂αu∂tα−u∂αηu∂tα+μ−∑n=1∞αnDtnξ1Dtα−nux+∑n=1∞αn∂nηu∂tn−αn+1Dtn+1ξ2Dtα−nu,where(14)μ=∑n=2∞∑m=2n∑k=2m∑r=2k−1αnnmkr1k!tn−αΓn+1−α−ur∂m∂tmuk−r∂n−m+k∂tn−m∂uk.
Remember that(15)αn=−1n−1αΓn−αΓn−αΓn+1.
Due to linearization of the infinitesimalη in u and the presence of ∂kη/∂ψk, μ will vanish, where k≥2 in equation (14).Lemma 1.
The functionψ=θx,t can be defined as an invariant solution of (2) if and only if(i)ψ=θx,t is an invariant surface of equation (2)(ii)ξ2x,t,θθt+ξ1x,t,θθx=ηx,t,θ


## 2.1. Fractional Riemann–Liouville Derivative

In this section, we show some definitions for RL fractional derivative [31], which can be considered as follows:(2)Dtαf=dnftdtn,ifn=α,n∈ℕ,dndtnIn−αft,if0≤n−1<α<n,n∈ℕ,(3)In−αft=1Γn−α∫0tt−sfsds,n>0,where Dtα is the total differentiation of integer number of orders α,α>0, the Gamma function is Γn−α, and Inft is the (RL) fractional integral of an order of n.Definition 1.
The partial derivative of orderα for Riemann–Liouville definition is presented by(4)∂tα=∂nf∂tn,n=α,∂n∂tn1Γn−α∫0tt−sn−α−1fs,xds,0≤n−1<α<n.


## 2.2. Notations for Lie Symmetry Reduction Method for the Time FPDEs

In this section, we show in detail the main notations and definitions that will be used for obtaining the symmetries of NLFPDEs.Here, we will consider timing NLFPDEs of the form [31–33](5)∂tαψ=Ft,x,ψ,ψx,ψxx,0<α≤1.Assume, equation (2) has a Lie vector X in the form(6)X=ξ1x,t,u∂∂x+ξ2x,t,u∂∂t+ηx,t,u∂∂u,where ξ1, ξ2, and η can be called as the infinitesimals of the transformations the independent and the dependent variables x,t,ψ, respectively. Let a one-parameter Lie algebra of infinitesimal transformations be of the following form:(7)t¯=t+εξ2t,x,ψ+Oε2,x¯=x+εξ1t,x,ψ+Oε2,ψ¯=ψ+εηt,x,ψ+Oε2,∂αψ¯∂t¯=∂αψ∂tα+εηα0t,x,ψ+Oε2,∂ψ¯∂x¯=∂ψ∂x+εηxt,x,ψ+Oε2,∂3ψ¯∂x¯2∂t¯=∂3ψ∂x2∂t+εηxxtt,x,ψ+Oε2,where ε « 1 can be defined as a group parameter, in most cases we take it equal one. The explicit expressions of ηx,ηxx, and ηxxt, which can be called the prolongation of the infinitesimals and are given by(8)ηx=Dxη−ψxDxξ1−ψtDtξ2,ηxx=Dxηx−ψxtDxξ1−ψxxDtξ2,ηxxt=Dxηxt−ψxxxDxξ1−ψxxtDxξ2,where Di is the total differentiation operator [34] with respect to the independent variables xi (i = 1, 2, then x1=x,x2=t):(9)Di=∂∂xi+∑α=1qψiα∂∂ψα+∑j=1p∑α=1qψjiα∂∂ψjα+⋯+∑j1=1p⋯∑jn−1p∑α=1qψj1,…,jn−1iα∂∂ψ1,…,jn−1iα.Theorem 1.
Equation (1) concedes a one-parameter group of infinitesimal transformations in equation (2) with the Lie Vector X if and just if the accompanying infinitesimal conditions hold:(10)Prα,3XΔΔ=0=0,where Δ=Dtαu−Ft,x,u,ux,uxx,…… and Pr is the 3rd prolongation of the infinitesimal generator X.Definition 2.
Prolonged vector is given by [31](11)PrnX=X+∑i=1p∑α=1qξiα∂∂uiα+⋯+∑j1=1p⋯∑jn=1p∑α=1qξj1,….,jnα∂∂uj1,….,jnα,where q is the numbers of dependent variables, p is the numbers of independent variables, ∂/∂uj1α=∂/∂uxα, and PDE involve derivatives up to order n. Also, the invariance condition [35] gives(12)ξ2t,x,ut=0=0.
Theαthextended infinitesimal, which deals with fractional derivatives, has the following form [36–38]:(13)ηα0=∂αη∂tα+ηu−αDtξ2∂αu∂tα−u∂αηu∂tα+μ−∑n=1∞αnDtnξ1Dtα−nux+∑n=1∞αn∂nηu∂tn−αn+1Dtn+1ξ2Dtα−nu,where(14)μ=∑n=2∞∑m=2n∑k=2m∑r=2k−1αnnmkr1k!tn−αΓn+1−α−ur∂m∂tmuk−r∂n−m+k∂tn−m∂uk.
Remember that(15)αn=−1n−1αΓn−αΓn−αΓn+1.
Due to linearization of the infinitesimalη in u and the presence of ∂kη/∂ψk, μ will vanish, where k≥2 in equation (14).Lemma 1.
The functionψ=θx,t can be defined as an invariant solution of (2) if and only if(i)ψ=θx,t is an invariant surface of equation (2)(ii)ξ2x,t,θθt+ξ1x,t,θθx=ηx,t,θ


## 3. Lie Symmetry and Reduction of FBBM Equation

In this partition, the Lie symmetry reduction method was applied to find the similarity variables for a one-dimensional time (FBBM) equation. Suppose that (1) is an invariant under (2); we have that(16)ψ¯tα+ψ¯ψ¯x+ψ¯x−ψ¯xxt=0.Thus,ψx,t satisfies equation (1). Applying the third prolongation to (1), we have the accompanying deciding condition, which is given as(17)ηα0+ψxη+ψ+1ηx−ηxxt=0.Substituting (7) and (8) into (16) and equaling coefficients in derivatives for x and power of u to zero, the system of equations is obtained:(18)αn∂tnηψ−αn+1Dtn+1ξ2=0,n=1,2,3….,ξψ2=ξx2=ξψ1=ξt1=ηψψ=0,1−αξt2+2ξx1=0,ψηx−ηxxt+∂tαη−ψ∂tαηψ+ηx=0,2ξxt1−ηtψ=0.By solving the obtained equations in (18), we get the following infinitesimal:(19)ξ1=c1+α−1c2x,ξ2=2c2t,η=−α+1c2ψ+1,where c1andc2 are constants. By the previous infinitesimal, equation (1) has two vector fields in the form(20a)X1=∂∂x,(20b)X2=α−1x∂∂x+2t∂∂t−α+1ψ+1∂∂ψ.Case 1.
For the infinitesimal generator in (20a), we have a characteristic equation in the following form:(21)dx1=dt0=dψ0.
By solving the previous equation, we get the variablest and ψ.Putting ψ=ft into (1), we obtain the following fractional ODE:(22)Dtαft=0.
By solving the above equation, we obtain(23)ψ=a1tα−1,where a1 is constant of integration.Case 2.
ForX2 in equation (20b), the similarity variables for the infinitesimal generator X2 can be obtained from the equation:(24)dxα−1x=dt2t=−dψα+1ψ+1.
The previous equation is called the characteristic equation; by solving it, we have the similarity variable as a result in the form:(25)ξ=xt−α−1/2.
The group invariant solution(26)ψx,t=t−α+1/2fξ−1=gξ−1,where fξ is a new arbitrary function of ξ and gξ=t−α+1/2fξ.By using equation (26), equation (1) is transformed into FODE.Theorem 3.
The transformation in (25) and (26), which is obtained from the similarity group method, reduces equation (1) to NLFODE as below:(27)P2/α−11/2−3α/2,αfξ−tα+1/2Γ1−α+−32α+12fξξ−12α−12fξξξ=0.Using the operator EK fractional differential operator [32, 34],(28)Pβξ2,αfξ=∏j=0n−1ξ2+j−1βddξKβξ2+α,n−αfξ,(29)n=α+1,α≠N,α,α∈N,where(30)Kβξ2,αfξ=1Γα∫1∞ψ−1α−1ψ−ξ2+αfξψ1/βdψ,α>0,fξ,α=0.Proof.
Letn−1<α≤1,n=1,2,3,4,… Depending on the Riemann–Liouville (RL) derivatives, definitions, and similarity variables in (25) and (26), we obtain(31)Dtαgξ=∂n∂tn1Γn−α∫0tt−sn−α−1s−α+1/2fxs1−α/2ds.
Letv=t/s and ds=−t/v2dv. Thus, (31) becomes(32)Dtαgξ=∂n∂tntn−3α+1/21Γn−α∫1∞v−1n−α−1v−n+−3α+1/2fξvα−1/2dv.
Substitute the EK fractional operator in (30) into (32), we have(33)Dtαgξ=∂n∂tntn−3α+1/2K2/α−11−α/2,n−αfξ.
For simplicity, letξ=xt−α−1/2 and ϕ∈0,∞; we acquire.
t∂/∂tϕξ=tx−α−1/2t−α−1/2−1ϕ′ξ=−α−1/2ξ∂/∂ξϕξ. Hence, equation (33) will be rewritten as(34)∂n∂tntn−3α+1/2K2/α−11−α/2,n−αfξ=∂n−1∂tn−1∂∂ttn−3α+1/2K2/α−11−α/2,n−αfξ=∂n−1∂tn−1tn−3α+1/2−1n−3α+12−α−12ξ∂∂ξK2/α−11−α/2,n−αfξ.Repeating n−1 times, we have(35)∂n∂tntn−3α+1/2K2/α−11−α/2,n−αfξ=∂n−1∂tn−1∂∂ttn−3α+1/2K2/α−11−α/2,n−αfξ=∂n−1∂tn−1tn−3α+1/2−1n−3α+12−α−12ξ∂∂ξK2/α−11−α/2,n−αfξ⋮=t−3α+1/2∏j=0n−1j−3α−12−α−12ξ∂∂ξK2/α−11−α/2,n−αfξ.
Using the definition of EK fractional differential operator in (28) to rewrite (35), we obtain(36)Dtαgξ=t−3α+1/2P2/α−1−3α+1/2,αfξ.Remark 1.
The fractional derivative must achieve the linearization property [37, 39]:(37)Dtαht+kt=Dtαht+Dtαkt,(38)Dtαtγ=Γγ+1tγ−αΓγ−α+1,γ>α−1.
Using the invariant group solution in (26), (37), and (38), we obtain(39)Dtαψx,t=Dtαgξ−Dtα1.
Hence,(40)Dtαψx,t=t−3α+1/2P2/α−1−3α+1/2,αfξ−t−αΓ1−α.
Thus, (1) can be reduced to(41)P2/α−11/2−3α/2,αfξ−tα+1/2Γ1−α+−32α+12fξξ−12α−12ξfξξξ=0,and the theorem is totally proofed.


## 4. Explicit Solutions for FBBM Equation

### 4.1. Clarifications for the Subequation Method

The subequation method [39] is presented in this section. Consider the NLFPDE in the form(42)Pψ,ψt,ψx,Dtαψ,Dxαψ,…,,0<α≤1,where ψ is a dependent variable, P is a series of ψ and its fractional derivatives, and DtαψandDxαψ are the Riemann–Liouville (RL) derivatives of ψ w.r.t tandx. Here, we present the principles for the subequation technique. By using the d’Alembert transformation,(43)ψx,t=ψζ,ζ=x+ct,where c is constant that will be determined later, and we can rewrite (41) as NLFODE:(44)Pψ,cψ′,ψ″,cαDζαψ,Dζαψ,…,,0<α≤1.According to the subequation procedure, assume that the wave solution will be written in the following form:(45)ψζ=a0+∑i−1naiϕζi,where ai,i=1,…,n are constants, which will be determined later, n belongs to integers numbers, which are determined by equaling the highest order derivatives and nonlinear terms in (44) together, and the functionϕζ achieves the Riccati equation of fractional order(46)Dζαϕζ=σ+ϕζ2,where σ is a constant. Some trigonometric solutions of the fractional Riccati equation (46) are(47)ϕζ=−−σtanhα−σζ,σ<0,−−σcothα−σζ,σ<0,σtanασζ,σ>0,−σcotασζ,σ>0,−Γ1+αζα+w,w=constant,σ=0.By substituting forms (45) into (44) and setting the coefficients of ϕζ to be zero, we obtain an algebraic system in ai,i=1,…,nandc. By solving the determinate system, we obtain the constants ai,i=1,…,nandc. Substituting these constants and the solutions of (47) into (45), we obtain the closed form solutions of (42).


### 4.2. Applying the Subequation Method to the Time FBBM Equation

We now implement a subequation method to (1). We will use the transformation(48)ψx,t=ψζ,ζ=x+ct,where c is a constant, and this will transform (1) into an NLFODE:(49)cαDζαψ=−ψζ−ψψζ+Cψζζζ.We now assume that (49) has the solution in the form(50)ψζ=a0+∑i−1naiϕζi,where ai (i = 1, ..., n) are constants, which will be determined, and ϕζ achieves equation (46).Balancing the highest order derivative terms with nonlinear terms in equation (49), we obtain n = 2. Hence,(51)ψζ=a0+a1ϕζ+a2ϕζ2.We then substitute (51) along with (46) into (49), then collect the coefficients of ϕζ, and set them to equal zero. A set of algebraic equations are obtained in knowns c, a0,a1,anda2. Solving these algebraic equations with the help of the software program (Maple), we get the following values.Thus, from (47), we obtain five forms of explicit travelling wave solutions of (1), namely,(52)c=112σ,σ=σ,a0=−cα−1+23σ2,a1=0,a2=σ.Thus, from (47), we obtain five forms of explicit travelling wave solutions of (1), namely,(53)ψ1x,t=a0+σ−−σtanhα−σx+ct2,σ<0,(54)ψ2x,t=a0+σ−−σcothα−σx+ct2,σ<0,(55)ψ3x,t=a0+σσtanασx+ct2,σ>0,(56)ψ4x,t=a0+σ−σcotασx+ct2,σ>0,(57)ψ5x,t=a0+σ−Γ1+αx+ctα+w2,σ=0,where a0is arbitrary constant. We plot the result in equation (57) in the three dimensions, contour plot, and density plot, as shown in Figures 1–3, respectively.Figure 1
3D plot for (57) with σ=10andα=0.75.Figure 2
Contour plot of (57) at σ=10andα=0.75.Figure 3
Density plot of (57) at σ=10andα=0.75.


### 4.3. Applying Simple Transformation

We solve the conformable FBBM equation using simple transformation to change the fraction order in partial derivative to nonsolvable ODE. For the reduction of (1) to ODE, we use the following transformation:(58)ψx,t=ψζ,whereζ=vx−ktαα,where v and k are arbitrary constants; we can rewrite (1) as NLODE:(59)kv2ψζζζ=k−vψζ−vψψζ.This equation has no implicit solution but possesses two integrating factors. We apply the integrating factor technique to obtain an analytical solution for (59).Equation (59) has two integrating factors (IF) as follows:(60)μ1=ψζ,μ2=1.Using these integrating factors by the same steps in [39] and neglecting the constants of integration, equation (59) will be reduced to(61)3kv2ψζ2=3k−vψ2−vψ3.By solving this equation, we obtain travelling wave solution for (1):(62)ψζ=1v3−vtan12−k2+kv−ζ+ckv2+k−v+ktan12−k2+kv−ζ+ckv2.Replacingζ=vx−ktα/α,(63)ψx,t=1v3−vtan12−k2+kv−vx+ktα/α+ckv2+k−v+ktan12−k2+kv−vx+ktα/α+ckv2.In other manner, equation (59) have two Lie vectors. The first one of them reduces it to(64)frr=−fr2f−−k+v+krkv2f,where,f=ψζ,r=ψ.Equation (64) has closed form solution, but, in the back substitution step, we are unable to get ψx,t even if we neglect the values of constants. So, from here, we can say the integrating factor method for reducing and solve ODEs, occasionally, more effectiveness than the Lie reduction method. Result obtained in (63) is plotted in Figure 4 at different values of α. We observe that, by decreasing the value of α, the top of the wave has a parabolic shape.Figure 4
ψx,t at c=1,v=−1,k=1,andα=0.4,0.9.Comparing our result in (63) with results in [5], specially equation (17), we find that the two solutions are travelling wave solutions, but the amplitude and direction of flow are different.


## 4.1. Clarifications for the Subequation Method

The subequation method [39] is presented in this section. Consider the NLFPDE in the form(42)Pψ,ψt,ψx,Dtαψ,Dxαψ,…,,0<α≤1,where ψ is a dependent variable, P is a series of ψ and its fractional derivatives, and DtαψandDxαψ are the Riemann–Liouville (RL) derivatives of ψ w.r.t tandx. Here, we present the principles for the subequation technique. By using the d’Alembert transformation,(43)ψx,t=ψζ,ζ=x+ct,where c is constant that will be determined later, and we can rewrite (41) as NLFODE:(44)Pψ,cψ′,ψ″,cαDζαψ,Dζαψ,…,,0<α≤1.According to the subequation procedure, assume that the wave solution will be written in the following form:(45)ψζ=a0+∑i−1naiϕζi,where ai,i=1,…,n are constants, which will be determined later, n belongs to integers numbers, which are determined by equaling the highest order derivatives and nonlinear terms in (44) together, and the functionϕζ achieves the Riccati equation of fractional order(46)Dζαϕζ=σ+ϕζ2,where σ is a constant. Some trigonometric solutions of the fractional Riccati equation (46) are(47)ϕζ=−−σtanhα−σζ,σ<0,−−σcothα−σζ,σ<0,σtanασζ,σ>0,−σcotασζ,σ>0,−Γ1+αζα+w,w=constant,σ=0.By substituting forms (45) into (44) and setting the coefficients of ϕζ to be zero, we obtain an algebraic system in ai,i=1,…,nandc. By solving the determinate system, we obtain the constants ai,i=1,…,nandc. Substituting these constants and the solutions of (47) into (45), we obtain the closed form solutions of (42).


## 4.2. Applying the Subequation Method to the Time FBBM Equation

We now implement a subequation method to (1). We will use the transformation(48)ψx,t=ψζ,ζ=x+ct,where c is a constant, and this will transform (1) into an NLFODE:(49)cαDζαψ=−ψζ−ψψζ+Cψζζζ.We now assume that (49) has the solution in the form(50)ψζ=a0+∑i−1naiϕζi,where ai (i = 1, ..., n) are constants, which will be determined, and ϕζ achieves equation (46).Balancing the highest order derivative terms with nonlinear terms in equation (49), we obtain n = 2. Hence,(51)ψζ=a0+a1ϕζ+a2ϕζ2.We then substitute (51) along with (46) into (49), then collect the coefficients of ϕζ, and set them to equal zero. A set of algebraic equations are obtained in knowns c, a0,a1,anda2. Solving these algebraic equations with the help of the software program (Maple), we get the following values.Thus, from (47), we obtain five forms of explicit travelling wave solutions of (1), namely,(52)c=112σ,σ=σ,a0=−cα−1+23σ2,a1=0,a2=σ.Thus, from (47), we obtain five forms of explicit travelling wave solutions of (1), namely,(53)ψ1x,t=a0+σ−−σtanhα−σx+ct2,σ<0,(54)ψ2x,t=a0+σ−−σcothα−σx+ct2,σ<0,(55)ψ3x,t=a0+σσtanασx+ct2,σ>0,(56)ψ4x,t=a0+σ−σcotασx+ct2,σ>0,(57)ψ5x,t=a0+σ−Γ1+αx+ctα+w2,σ=0,where a0is arbitrary constant. We plot the result in equation (57) in the three dimensions, contour plot, and density plot, as shown in Figures 1–3, respectively.Figure 1
3D plot for (57) with σ=10andα=0.75.Figure 2
Contour plot of (57) at σ=10andα=0.75.Figure 3
Density plot of (57) at σ=10andα=0.75.


## 4.3. Applying Simple Transformation

We solve the conformable FBBM equation using simple transformation to change the fraction order in partial derivative to nonsolvable ODE. For the reduction of (1) to ODE, we use the following transformation:(58)ψx,t=ψζ,whereζ=vx−ktαα,where v and k are arbitrary constants; we can rewrite (1) as NLODE:(59)kv2ψζζζ=k−vψζ−vψψζ.This equation has no implicit solution but possesses two integrating factors. We apply the integrating factor technique to obtain an analytical solution for (59).Equation (59) has two integrating factors (IF) as follows:(60)μ1=ψζ,μ2=1.Using these integrating factors by the same steps in [39] and neglecting the constants of integration, equation (59) will be reduced to(61)3kv2ψζ2=3k−vψ2−vψ3.By solving this equation, we obtain travelling wave solution for (1):(62)ψζ=1v3−vtan12−k2+kv−ζ+ckv2+k−v+ktan12−k2+kv−ζ+ckv2.Replacingζ=vx−ktα/α,(63)ψx,t=1v3−vtan12−k2+kv−vx+ktα/α+ckv2+k−v+ktan12−k2+kv−vx+ktα/α+ckv2.In other manner, equation (59) have two Lie vectors. The first one of them reduces it to(64)frr=−fr2f−−k+v+krkv2f,where,f=ψζ,r=ψ.Equation (64) has closed form solution, but, in the back substitution step, we are unable to get ψx,t even if we neglect the values of constants. So, from here, we can say the integrating factor method for reducing and solve ODEs, occasionally, more effectiveness than the Lie reduction method. Result obtained in (63) is plotted in Figure 4 at different values of α. We observe that, by decreasing the value of α, the top of the wave has a parabolic shape.Figure 4
ψx,t at c=1,v=−1,k=1,andα=0.4,0.9.Comparing our result in (63) with results in [5], specially equation (17), we find that the two solutions are travelling wave solutions, but the amplitude and direction of flow are different.


## 5. Conclusions

In this paper, we show the importance and the effective of the Lie symmetry reduction method on the FBBM equations. We obtain time FBBM equation’s Lie symmetry generators and then reduce the equation to FODE using these symmetry vectors. The projected analysis is extremely effective and dependable for getting similarity solutions for fractional differential equations. New travelling solutions were derived for the FBBM equation using the subequation method.



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