# New Solutions for the Generalized BBM Equation in terms of Jacobi and Weierstrass Elliptic Functions

Authors: Alvaro H. Salas; Lorenzo J. Martinez H; David L. Ocampo R
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

The Jacobi elliptic function method is applied to solve the generalized Benjamin-Bona-Mahony equation (BBM). Periodic and soliton solutions are formally derived in a general form. Some particular cases are considered. A power series method is also applied in some particular cases. Some solutions are expressed in terms of the Weierstrass elliptic function.


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## Body

## 1. Introduction

The regularized long-wave (RLW) equation is a famous nonlinear wave equation which gives the phenomena of dispersion and weak nonlinearity, including magneto hydrodynamic wave in plasma, phonon packets in nonlinear crystals, and nonlinear transverse waves in shallow water or in ion acoustic. This equation is also called the BBM (Benjamin-Bona-Mahony) equation and reads(1)ut+uux+ux−μuxxt=0.It describes approximately the unidirectional propagation of long waves in certain nonlinear dispersive systems, and it was proposed by Benjamin et al. in 1972 [1] as a more satisfactory model than the KdV equation [2]:
(2)ut+uux+uxxx=0.It is easy to see that Equation (1) can be derived from the equal width equation [3]:
(3)ut+uux−μuxxt=0,by means of the change of variableu⟵u+1, that is, by replacing u with u+1. This last equation is considered an equally valid and accurate model for the same wave phenomena simulated by (1) and (2). On the other hand, some researches analyzed the generalized KdV equation with variable coefficients
(4)ut+σunux+μuxxt=0,because this model has important applications in several fields of science. Motivated by these facts, we will consider here the generalized equal width (RW) equation with constant coefficients(5)ut+σunux−μuxxt=0.Our aim is to present solutions different from those in [2, 4–6].


### 1.1. Trigonometric and Soliton Solutions

In order to obtain new solutions to Equation (5), we make the traveling wave transformation
(6)ux,t=uξ,whereξ=x−λt.Inserting (6) into (5) gives
(7)−λu′ξ+uξnσu′ξ+λμu′′′ξ=0.Integrating once with respect toξ to get
(8)−λuξ+σn+1un+1ξ+λμu′′ξ−c1=0,where c1 is the constant of integration. Now, we multiply Equation (8) by u′ξ, and we integrate the resulting equation with respect to ξ:
(9)−c122λ−c1uξ+c2+σuξn+2n+1n+2+12μu′ξ2−12λuξ2=0,where c2 is the constant of integration. If c1=c2=0, Equation (9) takes the form
(10)μ2u2ξ+σn+1n+2un+2ξ−λ2u′ξ2=0.Let(11)uξ=vrξ,rvξ≠0.Inserting Equation (11) into (10), we obtain
(12)v′ξ2=λvξ2μr2−2σvξnr+2μn+1n+2r2.From Equation (12), if r=1/n, then
(13)v′ξ2=λn2vξ2μ−2n2σvξ3μn+1n+2.A sech solution to Equation (13) is
(14)vξ=λn+1n+22σsech2n2λμξ+ξ0,ξ0=const.It is clear that function(15)ux,t=λn+1n+22σsech2n2λμx−λt+ξ0n,λ>0,is a solution to equation ut+σunux−μuxxt=0 for any choice of the parameters λ and ξ0.Finally, ifμ<0, then taking into account that
(16)sechn2λμx−λt+ξ0=secn2λ−μx−λt+ξ0,we obtain from (15) the trigonometric solution
(17)ux,t=λn+1n+22σsec2n2λ−μx−λt+ξ0n.


## 1.1. Trigonometric and Soliton Solutions

In order to obtain new solutions to Equation (5), we make the traveling wave transformation
(6)ux,t=uξ,whereξ=x−λt.Inserting (6) into (5) gives
(7)−λu′ξ+uξnσu′ξ+λμu′′′ξ=0.Integrating once with respect toξ to get
(8)−λuξ+σn+1un+1ξ+λμu′′ξ−c1=0,where c1 is the constant of integration. Now, we multiply Equation (8) by u′ξ, and we integrate the resulting equation with respect to ξ:
(9)−c122λ−c1uξ+c2+σuξn+2n+1n+2+12μu′ξ2−12λuξ2=0,where c2 is the constant of integration. If c1=c2=0, Equation (9) takes the form
(10)μ2u2ξ+σn+1n+2un+2ξ−λ2u′ξ2=0.Let(11)uξ=vrξ,rvξ≠0.Inserting Equation (11) into (10), we obtain
(12)v′ξ2=λvξ2μr2−2σvξnr+2μn+1n+2r2.From Equation (12), if r=1/n, then
(13)v′ξ2=λn2vξ2μ−2n2σvξ3μn+1n+2.A sech solution to Equation (13) is
(14)vξ=λn+1n+22σsech2n2λμξ+ξ0,ξ0=const.It is clear that function(15)ux,t=λn+1n+22σsech2n2λμx−λt+ξ0n,λ>0,is a solution to equation ut+σunux−μuxxt=0 for any choice of the parameters λ and ξ0.Finally, ifμ<0, then taking into account that
(16)sechn2λμx−λt+ξ0=secn2λ−μx−λt+ξ0,we obtain from (15) the trigonometric solution
(17)ux,t=λn+1n+22σsec2n2λ−μx−λt+ξ0n.


## 2. The Novel Solutions

In this section, we will give analytical solutions to Equation (5) for two special cases: n=1 and n=2. First of all, observe that Equation (5) is equivalent to Equation (9).


### 2.1. The Casen=1

Ifn=1, Equation (9) reads
(18)λ2u2ξ+σ6u3ξ−λμ2u′ξ2+c1uξ+c2=0.To solve this equation, we consider the reduced Duffing equation(19)y′′ξ=qy3ξ,q≠0,where the coefficient q does not depend on ξ. Observe that
(20)y′ξy′′ξ=y′ξqy3ξ,and integrating this equation w.r.t. ξ gives
(21)12y′ξ2=q4y4ξ+c3.Suppose thaty=yξ is a solution to Equation (19) and let
(22)uξ=Ay2ξ+λσ,whereAissomeconstant.We have(23)u′ξ=2Ayξy′ξ,(24)u′′ξ=2Ay′ξ2+2Ayξy′′ξ,(25)y′ξ2=q2yξ4+2c3.Taking into account Equations (18) and (23), we must have
(26)−A2qλμ+16A3σyξ6+Ac1−2A2c3λμ−Aλ22σyξ2+c2−λ33σ2+λc1σ=0.Equating to zero the coefficients ofy6, y2, and y0, we conclude that
(27)q=Aσ6λμ,c3=2c1σ−λ24Aλμσ.We have proved that the general solution to the nonlinear ODE(28)λ2u2ξ+σ6u3ξ−λμ2u′ξ2+c1uξ+c2=0may be written in the form uξ=Ay2ξ+λ/σ, where y=yξ is a solution to the Duffing equation
(29)y′′ξ=Aσ6λμy3ξ,withy′ξ2=Aσ12λμyξ4+2c1σ−λ22Aλμσ.According to [2], the general solution to Equation (29) may be expressed in terms of the Jacobian elliptic functions. More exactly, the general solution to Equation (19) is
(30)yξ=y0cnAσy026λμξ+ξ0,22+λσ,whereqy02≠0.The value ofξ0 is determined by solving equation y′0=y1. In principle, we may consider that ξ0 is any number as well as y0. Taking into account these facts, we have proved that the general solution to Equation (28) is
(31)uξ=y02cn2Aσy026λμξ+ξ0,22+λσ,y0≠0.Finally, an exact solution to the EW-equationut+σuux−μuxxt=0 is
(32)ux,t=Ay02cn2Aσy026λμx−λt+ξ0,22+λσ.Solution (32) involves four arbitrary constants: A, y0, λ, and ξ0 (these do not depend on ξ). We choose A so that Aσy02/6λμ>0.


### 2.2. The Casen=2

This case corresponds to the so-called modified BBM or modified equal width equation. Whenn=2, Equation (9) reads
(33)λ2u2ξ+σ12u4ξ−λμ2u′ξ2+c1uξ+c2=0.This equation is harder to solve. We seek solution to Equation (33) in the form
(34)uξ=p+q1+rcnwξ−ξ0,m.Observe that functiony=yξ=rcnwξ+ξ0,m satisfies the nonlinear differential equation
(35)y′′+w1−2my+2mwry3=0,for any constants γ and ξ0. This equation is Duffing equation y′′+αy+βy3=0 with coefficients α and β. The relationship between the frequency ω, the modulus m, and the coefficients is given by the system
(36)α=w1−2m,β=2mwr.Solving this system forω and m gives
(37)ω=α+rβ,m=rβ2α+rβ=rβ2w.To solve the modified BBM equation, we will make use of the power series method (PSM). This is a promising method that also may be applicable to obtain an approximate solution for those ODE’s that do not admit solution in closed form. In our case, we consider the function(38)Fξ=λ2u2ξ+σ12u4ξ−λμ2u′ξ2+c1uξ+c2,where
(39)uξ=p+q1+rcnwξ−ξ0,m.The values ofq and r are obtained from the initial conditions uξ0=0 and uξ0+Km/w=0, that is,
(40)p+q1+r=1,p+q1−r=0.Thus,(41)r=12p−1,(42)q=2pp−12p−1.Inserting the ansatz (39) into (38), we obtain an expression for Fξ in terms of the unknowns w, m, p, and λ. Observe that function uξ given by (39) is a solution to Equation (33), if and only if Fξ≡0. The series method consists in coefficients of Taylor series of Fξ around some point. Usually, this is the origin. Since the calculations are enormous, we cannot realize this work by hand. Instead, we may use Mathematica 12 or Maple 16. The Mathematica command has the form Series Fξ,ξ,ξ0,n. This gives us the nth degree Taylor polynomial of Fξ around the point ξ=ξ0. In our case, we will use the command Series Fξ,ξ,0,n for n≥3, since we need at least four equations to determine the five unknowns w, m, p, and λ in terms of the coefficients of Equation (33). Of course, the series method is applicable to solve the BBM when n=1. Our aim is to show two different methods. Let us proceed. We have that F2n+1ξ0=0 for all n=0,1,2,3,⋯, and
(43)Fξ=Fξ0+12F′′ξ0ξ−ξ02+124F′′′′ξ0ξ−ξ04+1720F′′′′′′ξ0ξ−ξ06+⋯,where
(44)Fξ0=11212c1+122+6λ+σ=0,(45)λ=−16σ+12c1+12c2,(46)F′′ξ0=−p−1w6c1p+3pwλμ+6pλ+2pσ−3wλμ12p2=0,(47)p=3wλμ6c1+6λ+3wλμ+2σ,(48)0=12p3F′′′′ξ0=p−1w2−18c1p+6c1p2+24c1mp2−27pλ+15p2λ+24mp2λ+36wλμ−48pwλμ−48mpwλμ+12p2wλμ+48mp2wλμ−15pσ+11p2σ+8mp2σ,(49)w=−p6c14mp+p−3+38mp+5p−9λ+8mp+11p−15σ12p−14mp+p−3λμ,(50)0=−12p4F′′′′′′ξ0=p−1w332p23pλ−24wλμ+24pwλμ+pσ+3pc1m2+4p−135pλ+111p2λ+450wλμ−708pwλμ+258p2wλμ−75pσ+67p2σ−90pc1+66p2c1m+270pλ−270p2λ+51p3λ−675wλμ+1125pwλμ−498p2wλμ+48p3wλμ+225pσ−300p2σ+92p3σ+135pc1,−90p2c1+6p3c1.Solving this last equation form and taking into account (45), (47), and (49), we obtain after long algebraic calculations following solution:
(51)λ=16−12c1−12c2−σ,(52)w=26Δμ12c1+12c2+σ,(53)p=−−6c1−24c2+6Δ12c2+σ,(54)m=216c1c2+432c22+6c1σ−σ2+126c2−24c2σ+36σΔ86c1+24c2−6Δ6c1+12c2−σ,(55)Δ=c1+4c26c1+12c2−σ,σ=σ,μ=μ.Direct calculations show that functionu=uξ defined by (39) with parameters given by (51)-(55) and (41) is a solution to Equation (41). Then, function u=ux−λt−ξ0 is a solution to the modified BBM equation
(56)ut+σu2ux−μuxxt=0.Observe thatcnx,1=sechx. It may be verified that the following function is a solution to Equation (56):
(57)ux,t=α1+β1+γcoshkx+λt+C,for the choice
(58)c1=−2α3γ4σ3γ2−23,c2=α4γ−2γ6γ+2σ12γ2−24,k=21−γ2γ2+2μ,λ=−α2γ2γ2+2σ3γ2−22,β=−2γ−1γ+1γ2−2.We obtain soliton solutions for1−γ2μ>0 and trigonometric solutions for 1−γ2μ<0.


### 2.3. Solution for a Generalized BBM Equation

Let us consider the following generalized BBM equation:(59)ut+aux+bupux−δuxxt=0.In order to obtain new solutions to Equation (59), we make the travelling wave transformation
(60)ux,t=v−2/pξ,whereξ=x−λt+ξ0.Inserting (60) into (59) gives
(61)pv′ξξpa−λvξ2+bp−3δλp+2vξv′′ξ+δλp2vξ2v3ξ+2δλp+1p+2v′ξξ3=0.Equation (61) is hard to integrate. We seek for an exact solution of the form
(62)vξ=A+B1+C℘ξ;g2,g3.Inserting ansatz (62) into (61) gives Rξ=0, where
(63)Rξ=c4℘ξ;g2,g34+c3℘ξ;g2,g33+c2℘ξ;g2,g32+c1℘ξ;g2,g3+c0,being
(64)c0=2aA2p2+4aABp2+2aB2p2−12A2C2δg3λp2+6A2Cδg2λp2−2A2λp2−12ABC2δg3λp2+24ABC2δg3λp+9ABCδg2λp2−6ABCδg2λp−4ABλp2+2bp2−8B2C2δg3λ−4B2C2δg3λp2+12B2C2δg3λp+3B2Cδg2λp2−6B2Cδg2λp−2B2λp2,c1=8aA2Cp2+12aABCp2+4aB2Cp2−24A2C3δg3λp2+6A2C2δg2λp2−8A2Cλp2+24A2δλp2−12ABC3δg3λp2+24ABC3δg3λp+6ABC2δg2λp2+12ABC2δg2λp−12ABCλp2+48ABδλp2+8bCp2−8B2C2δg2λ−B2C2δg2λp2+6B2C2δg2λp−4B2Cλp2+24B2δλp2,c2=Cp12aA2Cp+12aABCp+2aB2Cp−12A2C3δg3λp−6A2C2δg2λp−12A2Cλp+24A2δλp+18ABC2δg2λ−3ABC2δg2λp−12ABCλp+72ABδλ+36ABδλp+12bCp−2B2Cλp+72B2δλ+12B2δλp,c3=2C24aA2Cp2+2aABCp2−3A2C2δg2λp2−4A2Cλp2−12A2δλp2−2ABCλp2−12ABδλp2+24ABδλp+4bCp2+16B2δλ+2B2δλp2+12B2δλp,c4=2C3paA2Cp−A2Cλp−12A2δλp−12ABδλ−6ABδλp+bCp.Equating to zero the coefficients of℘jξ;g2,g3 gives a nonlinear algebraic system. Solving it with the aid of Wolfram Mathematica, we obtain
(65)B=−6A,C=−24δaA2p2+3aA2p+2aA2+2bbp2,λ=ap+2p+1A2+2bA2p+1p+2,g2=b2p448δ2ap+2p+1A2+2b2,g2=−b3p61728δ3ap+2p+1A2+2b3.Other solutions may be obtained choosingB from the condition
(66)12A2p230aA2+30b+15aA2p+bp−bp2−2Ap28aA2+28b−166aA2p−118bp−88aA2p2−4bp2+aA2p3+4bp3B−p−1p+2−24aA2−24b−38aA2p−2bp+aA2p2+bp2B2=0.Then, a solution to (61) is
(67)vξ=A−6A1−24δap+2p+1A2+2b/bp2℘ξ;g2,g3,where g2,g3, and λ are obtained from (65). The obtained solution is valid for
(68)Aδbap+2p+1A2+2bpp+1p+2≠0.We have solved the generalized BBM equation for anyp such that pp+1p+2≠0. The exact traveling wave solution is given by
(69)ux,t=A−6A1−24δaA2p2+3aA2p+2aA2+2b/bp2℘x−ap+2p+1A2+2b/A2p+1p+2t+ξ0;g2,g3−2/p.The numbersA and ξ0 are arbitrary. Particular cases are
(1)
p=1:(70)ux,t=A+6Ab48δ3aA2+b℘b+3A2a/3A2t−x−ξ0;b2/1923aA2+b2δ2,−b3/138243aA2+b3δ3−b−2(2)
p=2:(71)ux,t=A+6Ab12δ6aA2+b℘b+6A2a/6A2t−x−ξ0;b2/126aA2+b2δ2,−b3/2166aA2+b3δ3−b−1(3)
p=1/2:(72)ux,t=A+6Ab24δ15aA2+8b℘8b+15A2a/15A2t−x−ξ0;b2/4815aA2+8b2δ2,−b3/172815aA2+8b3δ3−b−4


### 2.4. Forbidden Values

Let us examine the cases whenpp+1p+2=0.Case 1 (p=0).
For this value, Equation (59) takes the form
(73)ut+aux+bux−δuxxt=0.
After doing a traveling wave transformation, we get the linear ode(74)a+b−λv′ξ+δv′′′ξ=0,whose general solution is given by
(75)vξ=δλλ−a−bc1eλ−a−b/δλξ−c2e−λ−a−b/δλξ+c3.Case 2 (p+1p+2=0).
Making the traveling wave transformationξ=x−λt+ξ0 and making use of the ansatz (62), we obtain that B=0, which is a trivial solution. So, in the cases p=−1 and p=−2, we cannot get nontrivial solutions to the generalized BBM equation.


## 2.1. The Casen=1

Ifn=1, Equation (9) reads
(18)λ2u2ξ+σ6u3ξ−λμ2u′ξ2+c1uξ+c2=0.To solve this equation, we consider the reduced Duffing equation(19)y′′ξ=qy3ξ,q≠0,where the coefficient q does not depend on ξ. Observe that
(20)y′ξy′′ξ=y′ξqy3ξ,and integrating this equation w.r.t. ξ gives
(21)12y′ξ2=q4y4ξ+c3.Suppose thaty=yξ is a solution to Equation (19) and let
(22)uξ=Ay2ξ+λσ,whereAissomeconstant.We have(23)u′ξ=2Ayξy′ξ,(24)u′′ξ=2Ay′ξ2+2Ayξy′′ξ,(25)y′ξ2=q2yξ4+2c3.Taking into account Equations (18) and (23), we must have
(26)−A2qλμ+16A3σyξ6+Ac1−2A2c3λμ−Aλ22σyξ2+c2−λ33σ2+λc1σ=0.Equating to zero the coefficients ofy6, y2, and y0, we conclude that
(27)q=Aσ6λμ,c3=2c1σ−λ24Aλμσ.We have proved that the general solution to the nonlinear ODE(28)λ2u2ξ+σ6u3ξ−λμ2u′ξ2+c1uξ+c2=0may be written in the form uξ=Ay2ξ+λ/σ, where y=yξ is a solution to the Duffing equation
(29)y′′ξ=Aσ6λμy3ξ,withy′ξ2=Aσ12λμyξ4+2c1σ−λ22Aλμσ.According to [2], the general solution to Equation (29) may be expressed in terms of the Jacobian elliptic functions. More exactly, the general solution to Equation (19) is
(30)yξ=y0cnAσy026λμξ+ξ0,22+λσ,whereqy02≠0.The value ofξ0 is determined by solving equation y′0=y1. In principle, we may consider that ξ0 is any number as well as y0. Taking into account these facts, we have proved that the general solution to Equation (28) is
(31)uξ=y02cn2Aσy026λμξ+ξ0,22+λσ,y0≠0.Finally, an exact solution to the EW-equationut+σuux−μuxxt=0 is
(32)ux,t=Ay02cn2Aσy026λμx−λt+ξ0,22+λσ.Solution (32) involves four arbitrary constants: A, y0, λ, and ξ0 (these do not depend on ξ). We choose A so that Aσy02/6λμ>0.


## 2.2. The Casen=2

This case corresponds to the so-called modified BBM or modified equal width equation. Whenn=2, Equation (9) reads
(33)λ2u2ξ+σ12u4ξ−λμ2u′ξ2+c1uξ+c2=0.This equation is harder to solve. We seek solution to Equation (33) in the form
(34)uξ=p+q1+rcnwξ−ξ0,m.Observe that functiony=yξ=rcnwξ+ξ0,m satisfies the nonlinear differential equation
(35)y′′+w1−2my+2mwry3=0,for any constants γ and ξ0. This equation is Duffing equation y′′+αy+βy3=0 with coefficients α and β. The relationship between the frequency ω, the modulus m, and the coefficients is given by the system
(36)α=w1−2m,β=2mwr.Solving this system forω and m gives
(37)ω=α+rβ,m=rβ2α+rβ=rβ2w.To solve the modified BBM equation, we will make use of the power series method (PSM). This is a promising method that also may be applicable to obtain an approximate solution for those ODE’s that do not admit solution in closed form. In our case, we consider the function(38)Fξ=λ2u2ξ+σ12u4ξ−λμ2u′ξ2+c1uξ+c2,where
(39)uξ=p+q1+rcnwξ−ξ0,m.The values ofq and r are obtained from the initial conditions uξ0=0 and uξ0+Km/w=0, that is,
(40)p+q1+r=1,p+q1−r=0.Thus,(41)r=12p−1,(42)q=2pp−12p−1.Inserting the ansatz (39) into (38), we obtain an expression for Fξ in terms of the unknowns w, m, p, and λ. Observe that function uξ given by (39) is a solution to Equation (33), if and only if Fξ≡0. The series method consists in coefficients of Taylor series of Fξ around some point. Usually, this is the origin. Since the calculations are enormous, we cannot realize this work by hand. Instead, we may use Mathematica 12 or Maple 16. The Mathematica command has the form Series Fξ,ξ,ξ0,n. This gives us the nth degree Taylor polynomial of Fξ around the point ξ=ξ0. In our case, we will use the command Series Fξ,ξ,0,n for n≥3, since we need at least four equations to determine the five unknowns w, m, p, and λ in terms of the coefficients of Equation (33). Of course, the series method is applicable to solve the BBM when n=1. Our aim is to show two different methods. Let us proceed. We have that F2n+1ξ0=0 for all n=0,1,2,3,⋯, and
(43)Fξ=Fξ0+12F′′ξ0ξ−ξ02+124F′′′′ξ0ξ−ξ04+1720F′′′′′′ξ0ξ−ξ06+⋯,where
(44)Fξ0=11212c1+122+6λ+σ=0,(45)λ=−16σ+12c1+12c2,(46)F′′ξ0=−p−1w6c1p+3pwλμ+6pλ+2pσ−3wλμ12p2=0,(47)p=3wλμ6c1+6λ+3wλμ+2σ,(48)0=12p3F′′′′ξ0=p−1w2−18c1p+6c1p2+24c1mp2−27pλ+15p2λ+24mp2λ+36wλμ−48pwλμ−48mpwλμ+12p2wλμ+48mp2wλμ−15pσ+11p2σ+8mp2σ,(49)w=−p6c14mp+p−3+38mp+5p−9λ+8mp+11p−15σ12p−14mp+p−3λμ,(50)0=−12p4F′′′′′′ξ0=p−1w332p23pλ−24wλμ+24pwλμ+pσ+3pc1m2+4p−135pλ+111p2λ+450wλμ−708pwλμ+258p2wλμ−75pσ+67p2σ−90pc1+66p2c1m+270pλ−270p2λ+51p3λ−675wλμ+1125pwλμ−498p2wλμ+48p3wλμ+225pσ−300p2σ+92p3σ+135pc1,−90p2c1+6p3c1.Solving this last equation form and taking into account (45), (47), and (49), we obtain after long algebraic calculations following solution:
(51)λ=16−12c1−12c2−σ,(52)w=26Δμ12c1+12c2+σ,(53)p=−−6c1−24c2+6Δ12c2+σ,(54)m=216c1c2+432c22+6c1σ−σ2+126c2−24c2σ+36σΔ86c1+24c2−6Δ6c1+12c2−σ,(55)Δ=c1+4c26c1+12c2−σ,σ=σ,μ=μ.Direct calculations show that functionu=uξ defined by (39) with parameters given by (51)-(55) and (41) is a solution to Equation (41). Then, function u=ux−λt−ξ0 is a solution to the modified BBM equation
(56)ut+σu2ux−μuxxt=0.Observe thatcnx,1=sechx. It may be verified that the following function is a solution to Equation (56):
(57)ux,t=α1+β1+γcoshkx+λt+C,for the choice
(58)c1=−2α3γ4σ3γ2−23,c2=α4γ−2γ6γ+2σ12γ2−24,k=21−γ2γ2+2μ,λ=−α2γ2γ2+2σ3γ2−22,β=−2γ−1γ+1γ2−2.We obtain soliton solutions for1−γ2μ>0 and trigonometric solutions for 1−γ2μ<0.


## 2.3. Solution for a Generalized BBM Equation

Let us consider the following generalized BBM equation:(59)ut+aux+bupux−δuxxt=0.In order to obtain new solutions to Equation (59), we make the travelling wave transformation
(60)ux,t=v−2/pξ,whereξ=x−λt+ξ0.Inserting (60) into (59) gives
(61)pv′ξξpa−λvξ2+bp−3δλp+2vξv′′ξ+δλp2vξ2v3ξ+2δλp+1p+2v′ξξ3=0.Equation (61) is hard to integrate. We seek for an exact solution of the form
(62)vξ=A+B1+C℘ξ;g2,g3.Inserting ansatz (62) into (61) gives Rξ=0, where
(63)Rξ=c4℘ξ;g2,g34+c3℘ξ;g2,g33+c2℘ξ;g2,g32+c1℘ξ;g2,g3+c0,being
(64)c0=2aA2p2+4aABp2+2aB2p2−12A2C2δg3λp2+6A2Cδg2λp2−2A2λp2−12ABC2δg3λp2+24ABC2δg3λp+9ABCδg2λp2−6ABCδg2λp−4ABλp2+2bp2−8B2C2δg3λ−4B2C2δg3λp2+12B2C2δg3λp+3B2Cδg2λp2−6B2Cδg2λp−2B2λp2,c1=8aA2Cp2+12aABCp2+4aB2Cp2−24A2C3δg3λp2+6A2C2δg2λp2−8A2Cλp2+24A2δλp2−12ABC3δg3λp2+24ABC3δg3λp+6ABC2δg2λp2+12ABC2δg2λp−12ABCλp2+48ABδλp2+8bCp2−8B2C2δg2λ−B2C2δg2λp2+6B2C2δg2λp−4B2Cλp2+24B2δλp2,c2=Cp12aA2Cp+12aABCp+2aB2Cp−12A2C3δg3λp−6A2C2δg2λp−12A2Cλp+24A2δλp+18ABC2δg2λ−3ABC2δg2λp−12ABCλp+72ABδλ+36ABδλp+12bCp−2B2Cλp+72B2δλ+12B2δλp,c3=2C24aA2Cp2+2aABCp2−3A2C2δg2λp2−4A2Cλp2−12A2δλp2−2ABCλp2−12ABδλp2+24ABδλp+4bCp2+16B2δλ+2B2δλp2+12B2δλp,c4=2C3paA2Cp−A2Cλp−12A2δλp−12ABδλ−6ABδλp+bCp.Equating to zero the coefficients of℘jξ;g2,g3 gives a nonlinear algebraic system. Solving it with the aid of Wolfram Mathematica, we obtain
(65)B=−6A,C=−24δaA2p2+3aA2p+2aA2+2bbp2,λ=ap+2p+1A2+2bA2p+1p+2,g2=b2p448δ2ap+2p+1A2+2b2,g2=−b3p61728δ3ap+2p+1A2+2b3.Other solutions may be obtained choosingB from the condition
(66)12A2p230aA2+30b+15aA2p+bp−bp2−2Ap28aA2+28b−166aA2p−118bp−88aA2p2−4bp2+aA2p3+4bp3B−p−1p+2−24aA2−24b−38aA2p−2bp+aA2p2+bp2B2=0.Then, a solution to (61) is
(67)vξ=A−6A1−24δap+2p+1A2+2b/bp2℘ξ;g2,g3,where g2,g3, and λ are obtained from (65). The obtained solution is valid for
(68)Aδbap+2p+1A2+2bpp+1p+2≠0.We have solved the generalized BBM equation for anyp such that pp+1p+2≠0. The exact traveling wave solution is given by
(69)ux,t=A−6A1−24δaA2p2+3aA2p+2aA2+2b/bp2℘x−ap+2p+1A2+2b/A2p+1p+2t+ξ0;g2,g3−2/p.The numbersA and ξ0 are arbitrary. Particular cases are
(1)
p=1:(70)ux,t=A+6Ab48δ3aA2+b℘b+3A2a/3A2t−x−ξ0;b2/1923aA2+b2δ2,−b3/138243aA2+b3δ3−b−2(2)
p=2:(71)ux,t=A+6Ab12δ6aA2+b℘b+6A2a/6A2t−x−ξ0;b2/126aA2+b2δ2,−b3/2166aA2+b3δ3−b−1(3)
p=1/2:(72)ux,t=A+6Ab24δ15aA2+8b℘8b+15A2a/15A2t−x−ξ0;b2/4815aA2+8b2δ2,−b3/172815aA2+8b3δ3−b−4


## 2.4. Forbidden Values

Let us examine the cases whenpp+1p+2=0.Case 1 (p=0).
For this value, Equation (59) takes the form
(73)ut+aux+bux−δuxxt=0.
After doing a traveling wave transformation, we get the linear ode(74)a+b−λv′ξ+δv′′′ξ=0,whose general solution is given by
(75)vξ=δλλ−a−bc1eλ−a−b/δλξ−c2e−λ−a−b/δλξ+c3.Case 2 (p+1p+2=0).
Making the traveling wave transformationξ=x−λt+ξ0 and making use of the ansatz (62), we obtain that B=0, which is a trivial solution. So, in the cases p=−1 and p=−2, we cannot get nontrivial solutions to the generalized BBM equation.


## 3. Analysis and Discussion

We obtained the traveling wave solutions to BBM equationut+σuux−μuxxt=0 and modified BBM equation ut+σu2ux−μuxxt=0. Other solutions were obtained in [7] using the extended Jacobi elliptic function expansion method. However, in our approach, we got nonzero integration constants. Most authors get zero integration constants to simplify matters, but this may cause loss of solutions. We also may consider the following combined BBM equation:
(76)ut+au+σu2ux−μuxxt=0.Making the traveling wave transformationu=vξ, ξ=x−λt+ξ0 gives
(77)−λv′ξ+avξ+σv2ξv′ξ+δλv′′′ξ=0.We now integrate once with respect toξ, and denoting the constant of integration by c1, we obtain the ode
(78)v′′ξ+c1δλ−1δvξ+a2δλvξ2+σ3δλvξ3=0.This last equation is a Duffing-Helmholtz equation. The general form of the undamped and unforced Duffing-Helmholtz equation reads(79)v′′ξ+n+pvξ+qv2ξ+rv3ξ=0.The general solution to Equation (79) may be expressed in the ansatz form
(80)vξ=A+B1+C℘ξ+D;g2,g3.Indeed, given the initial conditions(81)v0=v0,(82)v′0=v̇0,the solution to the initial value problem (80)-(81) is given by (80), where
(83)B=−6A3r+A2q+Ap+n3A2r+2Aq+p,C=123A2r+2Aq+p,g2=112−3A4r2−4A3qr−6A2pr−12Anr−4nq+p2,g3=12169A4pr2−3A4q2r+12A3pqr−4A3q3+18A2p2r−6A2pq2+36Anpr−12Anq2+27n2r−6npq+p3,D=±℘−1v0−A−BCA−v0;g2,g3.The numberA is a solution to the quartic
(84)3rA4+4qA3+6pA2+12nA−12nv0+6pv02+4qv03+3rv04+6v̇02=0.In soliton theory, we are interested in sech or tanh solutions to Equation (79). The soliton solutions may occur only when
(85)27nr2−9pqr+2q3=0.In this case, we will have the following soliton solution to Equation (79):
(86)vξ=−q3r+3Ar+q3rsech3Ar+q32rξ,for3Ar+q≠0.Let us examine condition (85) for (78). We must have
(87)a3+6aλσ+12c1σ2=0.Solving this equation forc1, we will have the following soliton solution to equation (18):
(88)ux,t=−a2σ+a+2Aσ2σsecha+2Aσδ4aAσ+4A2σ2−5a2x+1245a2σ−4aA−4A2σt+ξ0.In (88), the constants A and ξ0 are arbitrary. In the case when σ=0, Equation (18) converts into usual BBM equation ut+aux−μuxxt=0.In a more general fashion, we solved Equation (59). This equation was solved in [2, 5, 8] using the substitution u=v1/p. In the present paper, we solved it by using the substitution u=v−2/p. Let us examine the odes obtained by means of these different substitutions. Using u=v1/p gives
(89)n2vξ2a−λv′ξ+δλv3ξ+bn2vξ3v′ξ+δλn−12n−1v′ξ3−3δλn−1nvξv′ξv′′ξ=0.On the other hand, lettingu=v−2/p gives a different ode given by (61). So, our method gives other than that obtained in [2, 5, 8] solutions.The authors in [5] claim that there are no periodic solutions for the parameter values of different form p=1, p=2, or p=1/2. However, we obtained periodic solutions for p−1p−2p−1/2≠0.On the other hand, we may solve the ode (89) by means of the ansatz (62). That was the approach the author employed in [8] by means of the substitution u=v1/p. Indeed, let us consider the equation [8]:
(90)ut+ux+aupux+uxxx=0.Letu=v1/px−λt+ξ0. Then, the ode (90) converts into the ode
(91)−ap2vξ3v′ξ+p2vξ2λ−1v′ξ−v3ξ+3−2pp−1v′ξ3+3p−1pvξv′ξv′′ξ=0.In order to solve equation (91), we will assume the ansatz form (62). Plugging this ansatz into (91), we get
(92)d4℘ξ;g2,g34+d3℘ξ;g2,g33+d2℘ξ;g2,g32+d1℘ξ;g2,g3+d0=0,where
(93)d4=2AC3paA2Cp−ACλp+ACp−12Ap−6Bp+6B,d3=2C24aA3Cp2+3aA2BCp2−3A2C2g2p2−4A2Cλp2+4A2Cp2−12A2p2−2ABCλp2+2ABCp2−12ABp2−12ABp+2B2p2−6B2p+4B2,d2=Cp12aA3Cp+18aA2BCp+6aAB2Cp−12A2C3g3p−6A2C2g2p−12A2Cλp+12A2Cp+24A2p−3ABC2g2p−9ABC2g2−12ABCλp+12ABCp+36ABp−36AB−2B2Cλp+2B2Cp+12B2p−36B2,d1=8aA3Cp2+18aA2BCp2+12aAB2Cp2+2aB3Cp2−24A2C3g3p2+6A2C2g2p2−8A2Cλp2+8A2Cp2+24A2p2−12ABC3g3p2−12ABC3g3p+6ABC2g2p2−6ABC2g2p−12ABCλp2+12ABCp2+48ABp2−B2C2g2p2−3B2C2g2p−2B2C2g2−4B2Cλp2+4B2Cp2+24B2p2,d0=2aA3p2+6aA2Bp2+6aAB2p2+2aB3p2−12A2C2g3p2+6A2Cg2p2−2A2λp2+2A2p2−12ABC2g3p2−12ABC2g3p+9ABCg2p2+3ABCg2p−4ABλp2+4ABp2−4B2C2g3p2−6B2C2g3p−2B2C2g3+3B2Cg2p2+3B2Cg2p−2B2λp2+2B2p2.Solving the systemd0=d1=d2=d3=d4=0 gives the following solution:
(94)B=−3A2,C=3p+1p+2aAp2,g2=a2A2p43p+12p+22,g3=−a3A3p627p+13p+23,λ=2aA+p2+3p+2p+1p+2.Thus, a traveling wave solution to Equation (90) reads
(95)ux,t=A−3A/21+3p+1p+2/aAp2℘x−λt+ξ0;a2A2p4/3p+12p+22,−a3A3p6/27p+13p+23p.In a similar way, we may consider the following more general equation than that considered in [2]:
(96)ut+aux+bupux−δuxxt+γuxxt=0.In [9], the author solved the following generalization of the BBM equation:
(97)ut+αux+βun+γu2nux−δuxxt=0.Let us consider the following more general variant of BBM equation:(98)ut+αux+βun+γu2nux−δuxxt−κuxxx=0.We again may use the ansatz (62) to solve this equation. Making the traveling wave transformation u=v1/px−λt+ξ0 gives the ode
(99)n2vξ2v′ξα−λ+v′′′ξδλ−κ+βn2v′ξvξ3+γn2v′ξvξ4+3n−1nv′ξv′′ξvξκ−δλ+1−n2n−1v′ξ3κ−δλ=0.Plugging the ansatz (62) into the ode (99) and equating to zero the coefficients of ℘j will give a very large algebraic system. After some hard algebraic calculations, we get the following solution:
(100)B=−6AAγn+2+β+2βn5Aγn+2+β8n+4,C=12A2γδn+2+2Aβδ2n+1+αδ2n3+7n2+7n+2−κ2n3+7n2+7n+2An25Aγn+2+β8n+4,g2=A2n4Aγn+2+β4n+2212A2γδn+2+2Aβδ2n+1+αδ2n3+7n2+7n+2−κ2n3+7n2+7n+22,g3=−A3n6Aγn+2+β4n+23216A2γδn+2+2Aβδ2n+1+αδ2n3+7n2+7n+2−κ2n3+7n2+7n+23,λ=AAγn+2+β4n+2+α2n3+7n2+7n+2n+1n+22n+1.The forbidden values aren+1n+22n+1=0. The soliton solutions may be obtained under the condition
(101)2n−1n−2=0.Forn=2, we have the following soliton solution to Equation (98):
(102)ux,t=−5β8γ±5β8γsech25β2192αγδ−25β2δ−192γκx−192αγ−25β2192γt+ξ0.In the case whenn=1/2, we have the following solution to Equation (98):
(103)ux,t=−2β5γ±22β5γsech2β275αγδ−16β2δ−75γκx−75αγ−16β275γt+ξ02.Finally, more solutions may be obtained by means of the transformationu=v−2/n. We will not derive them here.


## 4. Conclusions

We successfully obtained exact solutions to regularized long-wave and generalized BBM equation by using different approaches. We showed the way to derive all traveling wave solutions to all known until now variants of BBM. We compared the known previously published solutions with the solutions obtained in this work. Other methods to find exact solutions to nonlinear differential equations may be found in [1–3, 7, 9–39].



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