# Some Methods about Finding the Exact Solutions of Nonlinear Modified BBM Equation

Authors: Fan Niu; Jianming Qi; Zhiyong Zhou
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

Finding exact solutions of nonlinear equations plays an important role in nonlinear science, especially in engineering and mathematical physics. In this paper, we employed the complex method to get eight exact solutions of the modified BBM equation for the first time, including two elliptic function solutions, two simply periodic solutions, and four rational function solutions. We used the exp−ϕz-expansion methods to get fourteen forms of solutions of the modified BBM equation. We also used the sine-cosine method to obtain eight styles’ exact solutions of the modified BBM equation. Only the complex method can obtain elliptic function solutions. We believe that the complex method presented in this paper can be more effectively applied to seek solutions of other nonlinear evolution equations.


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## Body

## 1. Introduction

It is well known that modern natural science is undergoing profound changes. Nonlinear science runs through the fields of mathematical science, life science, space science, and electrical engineering [1] and has become an important frontier of contemporary scientific research. In the field of electrical engineering, the security of distribution network systems reliability and economy involve many nonlinear sciences [2–4]. The development of nonlinear sciences provides more accurate reference information for power system operation and scheduling [5] and also promotes the development of nonlinear sensitive electronic devices, which establish the foundation for the extensive use of nonlinear electrical equipment on the load side and grid side of the power system [6, 7], such as the nonlinear controller in the differential model of the integrated control system of the generator, which improves the standard and transient stability of the generator excitation system in the power system [8]. Therefore, constructing a more accurate nonlinear mathematical model in the power system and solving the exact solution of the equations in the model are of great significance to the analysis of the power system [9–11].The exact solutions of nonlinear partial differential equations have always attracted much attention; many effective methods for solving nonlinear partial differential equations have been continuously proposed and improved [12, 13]. Gardner et al. [14] proposed the nonlinear Fourier transform method. Lou and Tang [15] established and improved the multilinear separation variable method to solve a large class of 2+1−dimensional nonlinear systems. The projection Riccati equation expansion method was proposed by Musette and Conte [16]. He and Wu [17] proposed the exponential function method. Khater et al. [18–20] also had made the outstanding and huge contribution in related fields. Many scholars later proposed other methods, such as direct algebra method [21], tanh function expansion method [22], and Weierstrass elliptic function expansion method [23]. There are also methods based on symbolic calculations for solitary wave solutions of nonlinear partial differential equations [24, 25].In 1972, Benjamin-Bona-Mahony equation was introduced which is a nonlinear partial differential equation with small amplitude and long wave in the simulation of fluid mechanics by Benjamin, Bona, and Mahony [26]. One of the important models of long-wave one-way propagation in weakly nonlinear dispersive media is the BBM equation [27], later developed into mBBM. In 1991, Zhang [28] used the undetermined coefficient method to construct bell-shaped and twisted solitary wave solutions. In 1998, Shang [29] used direct and hypothetical methods to find the twisted and anti-twisted solitary wave solution clocks, shaped solitary wave solutions, trigonometric function waves, and singular traveling wave solutions for BBM and modified BBM equations. In 2003, Liu [30] constructed the exact solution of the BBM equation by using the Jacobi elliptic function expansion method. In 2004, Tautusan [31] constructed an exact solitary wave solution of the BBM equation and the mBBM equation by combining auxiliary equations and hyperbolic function assumptions. In 2013, Liu [32] used Painlevé direct truncation method to find the exact solution of mBBM equation. In 2017, Yang and Feng [33] applied the improved simple equation method to find the exact solution of the mBBM equation. In 2020, Gupta [34] generalised Kudryashov technique which has been implemented to construct new solutions of modified BBM equation. In this paper, we used the complex method [35, 36] and the exp−ϕz-expansion method [37] and sine-cosine method [38] to find the exact solutions of modified Benjamin-Bona-Mahony equation.Modified BBM equation (see [39, 40]) is given as(1)ut+ux+u2ux+βuxxt=0.In (1), β is constant. By putting(2)ux,t=wz,z=kx−λt,into (1), and integrating it deduces(3)βλk2w″−13w3+λ−1w−b=0.In order to get the exact solution of mBBM equation, we employ complex method to seek for the exact solution of (3), and we also use the exp−ϕz-expansion method to get some exact solutions and use sine-cosine method to find the exact solutions of modified Benjamin-Bona-Mahony equation.


## 2. Introduction of Complex Method and Some Lemmas and Main Result

In order to give our complex method, we have to know some concepts and symbols. First, we setm∈ℕ≔1,2,3,…,, rj∈ℕ0=ℕ∪0, r=r0,r1,…,rm, j=0,1,…,m. Then, we can get a differential monomial by(4)Mrwz≔wzr0w′zr1w″zr2,…,wmzrm,pr≔r0+2r1+⋯+m+1rm, and degM are regarded as the weight and degree of Mrw, separately.The differential polynomialPw,w′,…,wm can be defined as follows:(5)Pw,w′,…,wm≔∑r∈IarMrw.In equation (5), ar are constants, and I is a finite index set.The total weight and degree ofPw,w′,…,wm are marked as WP≔maxr∈Ipr and degP≔maxr∈IdegMr, separately.Considering the complex ordinary differential equations,(6)Pw,w′,…,wm=bwn+c.In equation (6), b≠0, c are constants, n∈ℕ.We takep,q∈ℕ, and we consider that the meromorphic solutions w of equation (6) have one or more poles. We can get that equation (6) satisfies the p,q condition, where p indicates that the equation has p distinct meromorphic solutions and q indicates that their multiplicity of the pole at z=0 is q.We have a hard time finding thep,q condition of equation (6), so we need a way to find the weak p,q condition shown as follows.To find out the weakp,q condition of equation (6), we can substitute Laurent series(7)wz=∑k=−q∞ckzk,q>0,c−q≠0,into equation (6); then, we can find out the p distinct Laurent singular parts as follows:(8)∑k=−q−1ckzk.Given two complex numbersω1, ω2, and Imω1/ω2>0, L=L2ω1,2ω2 are discrete subset L2ω1,2ω2=ω|ω=2nω1+2mω2,n,m∈ℤ, which is isomorphic to ℤ×ℤ. Let the discriminant Δ=Δc1,c2≔c13−27c22 and(9)sn=snL≔∑ω∈L∖01ωn.A meromorphic functionwz means that wz is holomorphic in the complex plane ℂ except for poles. ℘z,g2,g3 is the Weierstrass elliptic function with invariants g2 and g3.Iff is an elliptic function, or a rational function of eαz,α∈ℂ, or a rational function of z, the meromorphic function f belongs to the class W.Weierstrass elliptic function℘z≔℘z,g2,g3 is a meromorphic function with double periods ω1,ω2 defined as(10)℘z;w1,w2≔1z2+∑μ,ν∈Z,μ2+ν2≠01z+μw1+νw22−1μw1+νw22,which satisfies the following equation:(11)℘′z2=4℘z3−g2℘z−g3,where g2=60s4,g3=140s6, and Δg2,g3≠0.Or alternating equation (11) to the form(12)℘′z2=4℘z−e1℘z−e2℘z−e2,where e1=℘ω1,e2=℘ω2,e3=℘ω1+ω2.Contrarily, given two complex numbersg2 and g3 and Δg2,g3≠0, then there will be double periods ω1,ω2 Weierstrass elliptic function ℘z which the solutions will possess.In 2009, A. E. Eremenko et al. [41] investigated the m-order Briot-Bouquet equation (BBEq) as follows:(13)Fw,wm=∑j=0mFjwwmj=0.In equation (13), Fjw are constant coefficients polynomials, m∈ℕ. There are the following lemmas of the m order BBEq.Recently, Yuan et al. [35, 36, 39, 42] summarized the work of Eremenko et al. and introduced the complex method to find the exact solutions of nonlinear evolution equations in mathematical physics for the first time. Lemmas 1 and 2 play an important role in the complex method.Lemma 1 (see [43–45]).
Letp,l,m,n∈ℕ, degPw,wm<n. Consider that an m-order Briot–Bouquet equation(14)Pwm,w=bwn+csatisfies weak p,q condition; then, all the meromorphic solutions w will belong to the class W. For some values of parameters, if the solution w exists, then other meromorphic solutions form a one-parametric family wz−z0,z0∈ℂ. And then each solution can be written as the following forms with pole at z=0:(15)wz=∑i=1l−1∑j=2qi−1jc−ijj−1!dj−2dzj−214℘′z+Bi℘z−Ai2−℘z+∑i=1l−1c−i12℘′z+Bi℘z−Ai+∑j=2ql−1jc−ljj−1!dj−2dzj−2℘z+c0.
In equation (15), c−ij are given by equation (5), and Bi2=4Ai3−g2Ai−g3, ∑i=1lc−i1=0.
Each rational function solutionw≔Rz can be shown as follows:(16)Rz=∑i=1l∑j=1qcijz−zij+c0,with l≤p distinct poles of multiplicity q.
Every simply periodic solution is a rational functionRξ of ξ=eαzα∈ℂ. Rξ has l≤p distinct poles of multiplicity q and can be shown as follows:(17)Rξ=∑i=1l∑j=1qcijξ−ξij+c0.Lemma 2 (see [45, 46]).
The Weierstrass elliptic functions℘z≔℘z,g2,g3 have two successive degeneracies and addition formula:(i)
Degeneracy to simply periodic functions (i.e., rational functions of one exponentialekz) according to(18)℘z,3d2,−d3=2d−3d2coth23d2z,if one root ej is double Δg2,g3=0.(ii)
Degeneracy to rational functions ofz according to(19)℘z,0,0=1z2,if one root ej is triple (g2=g3=0).(ii)
Addition formula(20)℘z−z0=−℘z−℘z0+14℘′z+℘′z0℘z−℘z02.Through the above lemma and results, we can introduce complex method to find exact solutions of some PDEs. Here are the five steps:(1)
Substitute the transformT:ux,t⟶wz,x,t⟶z into a given PDE to produce a nonlinear ODE (6) or (8)(2)
Put equation (7) or (8) into equation (6) to find out weak p,q condition(3)
By determinant relation equations (15)–(17), we, respectively, find the elliptic, rational, and simply periodic solutions wz of equation (6) or (8) with pole at z=0(4)
By Lemmas1 and 2, we can get meromorphic solutions and the addition formula(5)
Put the inverse transformT−1 into the meromorphic solutions wz−z0, we can obtain all exact solutions ux,t of the original PDETheorem 1.
When we employ complex method, we suppose−1/3βλk2≠0; then, all meromorphic solutions w of (3) belong to the class W. And we found that (3) has the following three forms of solutions:(a)
The elliptic function solutions:(21)wdz=±126βλk2−℘+c4℘c2+4℘2c+2℘′d−℘g2−cg212c2−g2℘+4c3−3cg2℘′+4℘3+12c℘2−3g2℘−cg2d,where,g3=0,d2=4c3−g2c,g2 and c are arbitrary constants.(b)
The simply periodic solutions:(22)ws,1z=α32βλk2cothα2z−z0−cothα2z−z0−z1−cothα2z1.(i)
In order to better describews,1z, we make a graph of the function solution under specific parameters, as shown in Figure 1 in Section 7 and(23)ws,2z=α32βλk2tanhα2z−z0,wherez0∈ℂ,λ−1=βλk2α21/2+3/2sinh2α/2z1,b=−3/2βλk2tanhα/2z1/sinh2α/2z1,z1≠0 in the former formula, or λ−1=βλk2α2/2,b=0.(c)
The rational function solutions:(24)wr,1z=±6βλk21z−z0.(i)
In order to better describewr,1z, we make a graph of the function solution under specific parameters, as shown in Figure 2 in Section 7 and(25)wr,2z=±6βλk21z12z1z−z0−z1z−z0−z1−1,wherez0∈ℂ,λ=1,b=0 in the former formula, or given z1≠0,λ−1=6βλk2/z12,−b=∓2/36βλk2/z123/2.Figure 1
The solution of mBBM equation corresponding tows,1z, take α=2, β=1/31+3csch12, λ=3, k=1,z0=2,z1=1.Figure 2
The solution of mBBM equation corresponding towr,1z, take β=1/6, λ=2, k=1,z0=2.


## 3. Introduction of Exp−ϕz-Expansion Method and Main Result

Consider the following form of a nonlinear partial differential equation (PDE):(26)Pμ,μx,μt,μxx,μtt,…,=0,where P is a polynomial with an unknown function μx,y,t and its derivatives in which nonlinear terms and highest order derivatives are involved. And it can be processed as follows:Step 1. Inserting the traveling wave transformμx,t=wz,z=kx−λt into equation (26), alternating it to the following ordinary differential equation (ODE):(27)Kw,w′,w″,w‴,…,=0,whereK is a polynomial of wz and its derivatives.Step 2. Regarding that equation (27) has the following traveling wave solution:(28)wz=∑j=0nCjexp−ϕzj,whereCj0≤j≤n are constants and will be determined later, and Cj≠0 and ϕ=ϕz satisfies the ODE as follows:(29)ϕ′z=exp−ϕz+μexpϕz+δ.(i)
Equation (29) has different style solutions as follows:Ifδ2−4μ>0,μ≠0, then(30)ϕz=ln−δ2−4μtanhδ2−4μ/2z+c−δ2μ,(31)ϕz=ln−δ2−4μcothδ2−4μ/2z+c−δ2μ.Ifδ2−4μ<0,μ≠0, then(32)ϕz=ln−4μ−δ2tan4μ−δ2/2z+c−δ2μ,(33)ϕz=ln−4μ−δ2cot4μ−δ2/2z+c−δ2μ.Ifδ2−4μ<0,μ=0,δ≠0, then(34)ϕz=−lnδexpδz+c−1.Ifδ2−4μ=0,μ≠0,δ≠0, then(35)ϕz=ln−2δz+c+2δ2z+c.Ifδ2−4μ=0,μ=0,δ=0, then(36)ϕz=lnz+c.In the above equation,Cn≠0, δ, and μ are constants and will be determined later and c is an arbitrary constant. We consider the homogeneous balance between nonlinear terms and highest order derivatives of equation (27), and then we can get the positive integer n.Step 3. Substitute equation (28) into (27) and accounting for the function exp−ϕz, we obtain a polynomial of exp−ϕz. We calculate all the coefficients of the same power of exp−ϕz to zero and then we get a set of algebraic equations. By solving the algebraic equations, we obtain the values of Cn≠0, δ, μ; by putting these into equation (16) along with equations (30)–(36), we can get the determination of the solutions of equation (26).Theorem 2.
By employing the expexp−ϕz-expansion method, we found that there will be three forms of solutions of (3).
Ifδ2−4μ>0, μ≠0,(37)w11=6βλδk2−26βλkaμδ2−4μtanhz+cδ2−4μ/2−δ.
In order to better describew11z, we make a graph of the function solution under specific parameters, as shown in Figure 3 in Section 7. Also,(38)w12=6βλδk2−26βλkμδ2−4μcothz+cδ2−4μ/2−δ,w21=−6βλδk2+26βλkμδ2−4μtanhz+cδ2−4μ/2−δ,w22=−6βλδk2+26βλkμδ2−4μcothz+cδ2−4μ/2−δ.Ifδ2−4μ<0, μ≠0,(39)w13=6βλδk2−26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,In order to better describew13z, we make a graph of the function solution under specific parameters, as shown in Figure 4 in Section 7. Also,(40)w14=6βλδk2−26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ,w23=−6βλδk2+26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,w24=−6βλδk2+26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ.Ifδ2−4μ>0, μ=0,δ≠0,(41)w15=6βλδk2+6βλkδeδz+c−1,w25=−6βλδk2−6βλkδeδz+c−1.Ifδ2−4μ=0, μ≠0,δ≠0,(42)w16=6βλδk2−6βλkδ2z+c2δz+c+4,w26=−6βλδk2+6βλkδ2z+c2δz+c+4.Ifδ2−4μ=0, μ=0,δ=0,(43)w17=6βλkz+c,w27=−6βλkz+c.Figure 3
The solution of mBBM equation corresponding tow11z, take β=1,λ=1,δ=4,μ=3,k=1,a=1,c=2.Figure 4
The solution of mBBM equation corresponding tow12z, take β=1,λ=1,δ=4,μ=3,k=1,a=1,c=2.


## 4. Introduction of Sine-Cosine Method and Main Result

On the other hand, the sine-cosine method was proved to be effective in handing problems where compactions are generated. In what follows, we simply describe this method because details can be found in [38].The sine-cosine method admits the use of the solution in the form(44)ux,t=λ1cosβ1μz,z<π2μ,or in the form(45)ux,t=λ1cosβ1μz,z<πμ,and zero otherwise. The parameters λ1, μ, and β1 will be determined.In equation (26), P is a polynomial with an unknown function μx,t and its derivatives in which nonlinear terms and highest order derivatives are involved. And it can be processed as follows:Step 1. Insert the traveling wave transformμx,t=wz,z=kx−λt into equation (26) alternating it to the following ordinary differential equation (27). In equation (27), K is a polynomial of wz and its derivatives.Step 2. Regarding that equation (27) has the following traveling wave solution:(46)wz=λ1cosβ1μz,z<π2μ,(i)
or in the form(47)wz=λ1cosβ1μz,z<πμ,and zero otherwise in equation (27). The parameters λ1, μ, and β1 will be determined.Step 3. Calculate all the coefficients of the same power of sine or cosine to zero and then we get a set of algebraic equations. By solving the algebraic equations, we obtain the values ofλ1, μ, and β1; we can get the determination of the solutions of equation (26).Theorem 3.
If we employ sine-cosine method, then all meromorphic solutionsw of equation (3) have the following forms of solutions:(1)
By using the sine method, we obtain the solution of (3) as follows:(48)wT1z=±i6−6λcscλ−1kβλz.In order to better describewT1z, we make a graph of the function solution under specific parameters, as shown in Figure 5 in Section 7 or(49)wT2z=±i6−6λcsc−λ−1kβλz.(2)
By using the cosine method, we obtain the solution of (3) as follows:(50)wT3z=±i6−6λsecλ−1kβλz.In order to better describewT3z, we make a graph of the function solution under specific parameters, as shown in Figure 6 in Section 7 or(51)wT4z=±i6−6λsec−λ−1kβλz.Figure 5
The solution of mBBM equation corresponding towT1z, take λ=2,β=3,k=1.Figure 6
The solution of mBBM equation corresponding towT3z, take λ=2,β=3,k=1.


## 5. Proof of Theorems

### 5.1. Proof of Theorem1

Noting that our hypothesisβλk2/3≠0 and putting (7) into (3), we have q=1, p=2, c−1=±6βλk2, c0=0, c1=∓λ−1/6βλk2, c2=−b/βλk2+4, c4=1−λb/24β2λ2k4, where c3 is arbitrary.Hence, (3) satisfies weak 1,2 condition and is a 2-order Briot-Bouquet differential equation. Obviously, (3) satisfies the dominant condition. By Lemma 1, we know that all meromorphic solutions of (3) belong to W, and then we will give the forms of all meromorphic solutions of (3).By (16), we infer the indeterminant rational solutions of (3) with pole at z=0 that(52)R1z=c11z+c12z−z1+c10.SubstitutingR1z into (3), we obtain two classes; one is as follows:(53)R1,1z=±6βλk21z,where λ=1andb=0. The other is(54)R1,2z=±−6βλk21z12z1z−z1z−z1−1,where λ−1=6βλk2/z12andb=±2/36βλk2/z123/2.So, all rational solutions of (3) are as follows:(55)wr,1z=±6βλk21z−z0,(56)wr,2z=±6βλk2z12z1z−z0−z1z−z0−z1−1,where z0∈ℂ,λ=1,b=0 in the former formula, or given z1≠0,λ−1=6βλk2/z12,b=±2/36βλk2/z123/2.In order to have simply periodic solutions, setξ=expαz , put w=Rξ into (3); then,(57)Aα2ξR′+ξ2R″+BR+CR3+D=0.Putting(58)R2ξ=c21ξ−1+c22ξ−ξ1+c20into (57), we get that(59)R2,1ξ=α1−ξ132βλk2,R2,2ξ=α32βλk22ξ−1+1.Here,(60)λ−1=12+6ξ11−ξ12βλk2,−b=8Cξ11+ξ1α31−ξ1332βλk23/2,in the former formula, or λ−1=βλk2α2/2,b=0.Substitutingξ=eαz into the above relation, we obtain simply periodic solutions of (3) with pole at z=0(61)ws0,1z=α32βλk2cothα2z−cothα2z−z1−cothα2z1,ws0,2z=α32βλk2tanhα2z.So, all simply periodic solutions of (3) are obtained by(62)ws,1z=α32βλk2cothα2z−z0−cothα2z−z0−z1−cothα2z1,ws,2z=α32βλk2tanhα2z−z0.Here,(63)z0∈ℂ,λ−1=βλk2α212+32sinh2α/2z1,b=−32βλk2tanhα/2z1sinh2α/2z1,z1≠0,in the former formula, or λ−1=βλk2α2/2,b=0.From (15) of Lemma 1, we have indeterminant relations of elliptic solutions of (3) with pole at z=0:(64)wd0z=c−12℘′z+F℘z−E+c30,where F2=4E3−g2E−g2. Applying the conclusion II of Lemma 2 to wd0z, and noting the results of rational solutions obtained above, we deduce that c30=0, E=F=0,g3=0. Then, we obtain that(65)wd0z=±126βλk2℘′z℘z,where g3=0. Therefore, all elliptic solutions of (3) are as follows:(66)wdz=±126βλk2℘′zz−z0℘zz−z0,where z0∈ℂ, g3=0. By using of III of Lemma 2, we rewrite it to the form(67)wdz=±126βλk2−℘+c4℘c2+4℘2c+2℘′d−℘g2−cg212c2−g2℘+4c3−3cg2℘′+4℘3+12c℘2−3g2℘−cg2d′,where g3=0, d2=4c3−g2c, g2, and c are arbitrary.


### 5.2. Proof of Theorem2

Taking the homogeneous balance betweenw″ and w3 in (3), we obtain(68)wz=C0+C1exp−ϕz,where C1≠0, C0 are constants which need to be determined, and ϕz satisfies equation ϕ′z=exp−ϕz+μexpϕz+δ; here δ and μ are arbitrary constants.From equation (68), we insert w, w3, w′′ into (3) and sort out the coefficients of exp−ϕz to zero, so we get(69)C1=±6βλk,C0=±6βλδk2,where δ=±2βλ2βλk2μ+λ−1/βλk and b=0.By substituting equation (69) into (68), we obtain(70)w1z=6βλδk2+6βλke−ϕz,or(71)w2z=−6βλδk2−6βλke−ϕz.We apply equations (30)–(36) into (70) and (71), respectively, to obtain traveling wave solutions of the mBBM equation as follows.Ifδ2−4μ>0, μ≠0,(72)w11=6βλδk2−26βλkμδ2−4μtanhz+cδ2−4μ/2−δ,w12=6βλδk2−26βλkμδ2−4μcothz+cδ2−4μ/2−δ,w21=−6βλδk2+26βλkμδ2−4μtanhz+cδ2−4μ/2−δ,w22=−6βλδk2+26βλkμδ2−4μcothz+cδ2−4μ/2−δ.Ifδ2−4μ<0, μ≠0,(73)w13=6βλδk2−26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,w14=6βλδk2−26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ,w23=−6βλδk2+26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,w24=−6βλδk2+26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ.Ifδ2−4μ>0, μ=0,δ≠0,(74)w15=6βλδk2+6βλkδeδz+c−1,w25=−6βλδk2−6βλkδeδz+c−1.Ifδ2−4μ=0, μ≠0,δ≠0,(75)w16=6βλδk2−6βλkδ2z+c2δz+c+4,w26=−6βλδk2+6βλkδ2z+c2δz+c+4.Ifδ2−4μ=0, μ=0,δ=0,(76)w17=6βλkz+c,(77)w27=−6βλkz+c.


### 5.3. Proof of Theorem3

The sine-cosine method admits the use of the solution in the form(78)wz=λ1cosβ1μz,z<π2μ,or in the form(79)wz=λ1sinβ1μz,z<πμ,and zero otherwise. The parameters λ1,μ,β1 will be determined. We substitute(80)wz=λ1cosβ1μzinto (3) and obtain(81)λβk2β12λ1μ2−λβk2β1λ1μ2cosβ1−2μz+λλ1−λ1−λβk2β12λ1μ2cosβ1μz−13λ13cos3β1μz−b=0.Collecting the coefficients of each pair of cosine functions of the same exponent and setting it equal to zero, we obtain the following system of algebraic equations:(82)β1−1≠0,β1−2=3β1,k2ββ12λλ1μ2−k2ββ1λλ1μ2=13λ13,λλ1−λ1=k2ββ12λλ1μ2.Solving this system gives(83)β1=−1,μ=±λ−1kβλ,λ1=±i6−6λ.At last, we get(84)wc1z=±i6−6λsecλ−1kβλz,or(85)wc2z=±i6−6λsec−λ−1kβλz.We substitute(86)wz=λ1sinβ1μzinto (3) and obtain(87)λβk2β12λ1μ2−λβk2β1λ1μ2sinβ1−2μz+λλ1−λ1−λβk2β12λ1μ2sinβ1μz−13λ13sin3β1μz−b=0.Collecting the coefficients of each pair of cosine functions of the same exponent and setting it equal to zero, we obtain the following system of algebraic equations:(88)β1−1≠0,β1−2=3β1,k2ββ12λλ1μ2−k2ββ1λλ1μ2=13λ13,λλ1−λ1=k2ββ12λλ1μ2.Solving this system gives(89)β1=−1,μ=±λ−1kβλ,λ1=±i6−6λ.At last, we obtain(90)wc1z=±i6−6λcscλ−1kβλz,or(91)wc2z=±i6−6λcsc−λ−1kβλz.


## 5.1. Proof of Theorem1

Noting that our hypothesisβλk2/3≠0 and putting (7) into (3), we have q=1, p=2, c−1=±6βλk2, c0=0, c1=∓λ−1/6βλk2, c2=−b/βλk2+4, c4=1−λb/24β2λ2k4, where c3 is arbitrary.Hence, (3) satisfies weak 1,2 condition and is a 2-order Briot-Bouquet differential equation. Obviously, (3) satisfies the dominant condition. By Lemma 1, we know that all meromorphic solutions of (3) belong to W, and then we will give the forms of all meromorphic solutions of (3).By (16), we infer the indeterminant rational solutions of (3) with pole at z=0 that(52)R1z=c11z+c12z−z1+c10.SubstitutingR1z into (3), we obtain two classes; one is as follows:(53)R1,1z=±6βλk21z,where λ=1andb=0. The other is(54)R1,2z=±−6βλk21z12z1z−z1z−z1−1,where λ−1=6βλk2/z12andb=±2/36βλk2/z123/2.So, all rational solutions of (3) are as follows:(55)wr,1z=±6βλk21z−z0,(56)wr,2z=±6βλk2z12z1z−z0−z1z−z0−z1−1,where z0∈ℂ,λ=1,b=0 in the former formula, or given z1≠0,λ−1=6βλk2/z12,b=±2/36βλk2/z123/2.In order to have simply periodic solutions, setξ=expαz , put w=Rξ into (3); then,(57)Aα2ξR′+ξ2R″+BR+CR3+D=0.Putting(58)R2ξ=c21ξ−1+c22ξ−ξ1+c20into (57), we get that(59)R2,1ξ=α1−ξ132βλk2,R2,2ξ=α32βλk22ξ−1+1.Here,(60)λ−1=12+6ξ11−ξ12βλk2,−b=8Cξ11+ξ1α31−ξ1332βλk23/2,in the former formula, or λ−1=βλk2α2/2,b=0.Substitutingξ=eαz into the above relation, we obtain simply periodic solutions of (3) with pole at z=0(61)ws0,1z=α32βλk2cothα2z−cothα2z−z1−cothα2z1,ws0,2z=α32βλk2tanhα2z.So, all simply periodic solutions of (3) are obtained by(62)ws,1z=α32βλk2cothα2z−z0−cothα2z−z0−z1−cothα2z1,ws,2z=α32βλk2tanhα2z−z0.Here,(63)z0∈ℂ,λ−1=βλk2α212+32sinh2α/2z1,b=−32βλk2tanhα/2z1sinh2α/2z1,z1≠0,in the former formula, or λ−1=βλk2α2/2,b=0.From (15) of Lemma 1, we have indeterminant relations of elliptic solutions of (3) with pole at z=0:(64)wd0z=c−12℘′z+F℘z−E+c30,where F2=4E3−g2E−g2. Applying the conclusion II of Lemma 2 to wd0z, and noting the results of rational solutions obtained above, we deduce that c30=0, E=F=0,g3=0. Then, we obtain that(65)wd0z=±126βλk2℘′z℘z,where g3=0. Therefore, all elliptic solutions of (3) are as follows:(66)wdz=±126βλk2℘′zz−z0℘zz−z0,where z0∈ℂ, g3=0. By using of III of Lemma 2, we rewrite it to the form(67)wdz=±126βλk2−℘+c4℘c2+4℘2c+2℘′d−℘g2−cg212c2−g2℘+4c3−3cg2℘′+4℘3+12c℘2−3g2℘−cg2d′,where g3=0, d2=4c3−g2c, g2, and c are arbitrary.


## 5.2. Proof of Theorem2

Taking the homogeneous balance betweenw″ and w3 in (3), we obtain(68)wz=C0+C1exp−ϕz,where C1≠0, C0 are constants which need to be determined, and ϕz satisfies equation ϕ′z=exp−ϕz+μexpϕz+δ; here δ and μ are arbitrary constants.From equation (68), we insert w, w3, w′′ into (3) and sort out the coefficients of exp−ϕz to zero, so we get(69)C1=±6βλk,C0=±6βλδk2,where δ=±2βλ2βλk2μ+λ−1/βλk and b=0.By substituting equation (69) into (68), we obtain(70)w1z=6βλδk2+6βλke−ϕz,or(71)w2z=−6βλδk2−6βλke−ϕz.We apply equations (30)–(36) into (70) and (71), respectively, to obtain traveling wave solutions of the mBBM equation as follows.Ifδ2−4μ>0, μ≠0,(72)w11=6βλδk2−26βλkμδ2−4μtanhz+cδ2−4μ/2−δ,w12=6βλδk2−26βλkμδ2−4μcothz+cδ2−4μ/2−δ,w21=−6βλδk2+26βλkμδ2−4μtanhz+cδ2−4μ/2−δ,w22=−6βλδk2+26βλkμδ2−4μcothz+cδ2−4μ/2−δ.Ifδ2−4μ<0, μ≠0,(73)w13=6βλδk2−26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,w14=6βλδk2−26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ,w23=−6βλδk2+26βλkμ4μ−δ2tanz+c4μ−δ2/2−δ,w24=−6βλδk2+26βλkμ4μ−δ2cotz+c4μ−δ2/2−δ.Ifδ2−4μ>0, μ=0,δ≠0,(74)w15=6βλδk2+6βλkδeδz+c−1,w25=−6βλδk2−6βλkδeδz+c−1.Ifδ2−4μ=0, μ≠0,δ≠0,(75)w16=6βλδk2−6βλkδ2z+c2δz+c+4,w26=−6βλδk2+6βλkδ2z+c2δz+c+4.Ifδ2−4μ=0, μ=0,δ=0,(76)w17=6βλkz+c,(77)w27=−6βλkz+c.


## 5.3. Proof of Theorem3

The sine-cosine method admits the use of the solution in the form(78)wz=λ1cosβ1μz,z<π2μ,or in the form(79)wz=λ1sinβ1μz,z<πμ,and zero otherwise. The parameters λ1,μ,β1 will be determined. We substitute(80)wz=λ1cosβ1μzinto (3) and obtain(81)λβk2β12λ1μ2−λβk2β1λ1μ2cosβ1−2μz+λλ1−λ1−λβk2β12λ1μ2cosβ1μz−13λ13cos3β1μz−b=0.Collecting the coefficients of each pair of cosine functions of the same exponent and setting it equal to zero, we obtain the following system of algebraic equations:(82)β1−1≠0,β1−2=3β1,k2ββ12λλ1μ2−k2ββ1λλ1μ2=13λ13,λλ1−λ1=k2ββ12λλ1μ2.Solving this system gives(83)β1=−1,μ=±λ−1kβλ,λ1=±i6−6λ.At last, we get(84)wc1z=±i6−6λsecλ−1kβλz,or(85)wc2z=±i6−6λsec−λ−1kβλz.We substitute(86)wz=λ1sinβ1μzinto (3) and obtain(87)λβk2β12λ1μ2−λβk2β1λ1μ2sinβ1−2μz+λλ1−λ1−λβk2β12λ1μ2sinβ1μz−13λ13sin3β1μz−b=0.Collecting the coefficients of each pair of cosine functions of the same exponent and setting it equal to zero, we obtain the following system of algebraic equations:(88)β1−1≠0,β1−2=3β1,k2ββ12λλ1μ2−k2ββ1λλ1μ2=13λ13,λλ1−λ1=k2ββ12λλ1μ2.Solving this system gives(89)β1=−1,μ=±λ−1kβλ,λ1=±i6−6λ.At last, we obtain(90)wc1z=±i6−6λcscλ−1kβλz,or(91)wc2z=±i6−6λcsc−λ−1kβλz.


## 6. Comparison

Implementing the exp−ϕz-expansion method, we found that there are fourteen forms of solutions of mBBM equation. Using the complex method, we found eight solutions of mBBM equation. In the above two methods, the solutions of equations (76), (77), and (55) are the same. Finally, using sine-cosine method, we found eight styles’ solutions of mBBM equation. From the above results, we can find more solutions by the exp−ϕz-expansion method, whereas we can obtain elliptic function solutions just by the complex method. Also, we know that, by using the exp−ϕz-expansion method and complex method, we find the hyperbolic functions solutions for the mBBM equation. By using the sine-cosine method, we only find the trigonometric function solutions for mBBM equation. Each of three methods has its own characteristics. We firmly believe that the complex method presented in this paper can be more effectively applied to seek solutions of other nonlinear evolution equations now and in the future. These three methods are very useful tools for finding the exact solutions of nonlinear evolution equations.


## 7. Computer Simulations

In this section, we illustrate some results through computer simulations. We carry out further analysis to the properties of simply periodic solutionsws,1z,w11z,wT1z,wr,1z,w12z,wT3z of the mBBM equation in the figures.From Figures1 and 2, we can obtain the dynamic behavior of solving ws,1z,w11z. From Figure 3, the lump solution wT1z has one global maximum point.From Figure4, wr,1z represents the singular soliton for the special parameters. From Figure 5, we can get the continuous dynamic behavior of solving w12z. From Figure 6, wT3z has many distinct generation poles.


## 8. Conclusions

Complex method is very effective tool for seeking the exact solutions of nonlinear evolution equations, and equation (3) is one of most important auxiliary equations because many nonlinear evolution equations can be transformed into it. In this paper, we employed the complex method, exp−ϕz-expansion method, and sine-cosine method to get the exact solutions of modified BBM equation. In this way, we can reduce the dimension of the nonlinear evolution equations related to engineering and mathematical physics. We obtain twenty-eight forms of solutions of mBBM equation. It shows these methods are very efficient and powerful in solving the exact solutions of nonlinear evolution equations now and in the future. We can apply the idea of this research to other nonlinear evolution equations. Our work shows that there exist some classes of rational solutions wr,1z, wr,2z and simple periodic solutions ws,1z, ws,2z which are not degenerated successively by the elliptic function solutions and are new.



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*Source: 5564162-2021-04-07.xml*