# Further Results about Seeking for the Exact Solutions of the Nonlinear2+1-Dimensional Jaulent-Miodek Equation

Authors: Jingsen Hu; Jianming Qi
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

Nonlinear science is a great revolution of modern natural science. As a result of its rise, the various branches of subjects characterized by nonlinearity have been developed vigorously. In particular, more attention to acquiring the exact solutions of a wide variety of nonlinear equations has been paid by people. In this paper, three methods for solving the exact solutions of the nonlinear2+1-dimensional Jaulent-Miodek equation are introduced in detail. First of all, the exact solutions of this nonlinear equation are obtained by using the exp−ϕz-expansion method, tanh method, and sine-cosine method. Secondly, the relevant results are verified and simulated by using Maple software. Finally, the advantages and disadvantages of the above three methods listed in the paper are analyzed, and the conclusion was drawn by us. These methods are straightforward and concise in very easier ways.


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## Body

## 1. Introduction

Since the 20th century, with the integration of various disciplines, people have gradually discovered nonlinear phenomena which exist in different disciplines but have common characteristics. The resulting nonlinear science is quickly and widely used in medical diagnosis, information technology, electrical engineering, and other fields, which greatly promotes the progress and development of human society [1].Nonlinear science has laid a solid theoretical foundation for the safety, reliability, and stability of power systems [2–4]. By studying the nonlinear differential equations that appear in the ferromagnetic resonance overvoltage phenomenon, the safe operation of power grids at all levels can be effectively protected [5]. Aimed at the problem of asymmetric nonlinear oscillation in parallel operation of power stations, the harmonic linearization method is put forward, which is helpful to reasonably select the power grid structure and effectively improve the stability of power grid structure [6]. By constructing the nonlinear system model with random disturbance, the high-frequency faults of the power grid caused by disturbance can be greatly reduced [4, 7, 8]. According to the nonlinear characteristics of the actual power system, a reliable and stable nonlinear power grid structure model is constructed, and the exact solution of the equation in the structure model is obtained, which plays a decisive role in the stable operation of the electrified wire netting [9, 10].Study on the exact solutions of a wide variety of nonlinear equations can not only help people discover new nonlinear phenomena and their laws but also help ensure the reliability and stability of nonlinear calculations. Therefore, more and more scholars have proposed different methods for studying the exact solutions of a wide variety of nonlinear equations. Gardner et al. [11] used the backscattering argument of the Schrödinger equation to propose the inverse scattering transformation method. Wang and Li [12] used the principle of homogeneous balance to derive the nonlinear transformation of a great deal of nonlinear PDE and obtained the exact solution of the equation. Yu and Sun [13] presented a new 3+1-dimensional KP-like nonlinear partial differential equation and construct the lump solutions, rationally localized in all directions in the space, to its two-dimensionally reduced cases. The method is based on a generalized bilinear differential equation, so we can build the lump solutions to the presented KP-like equation from special polynomial solutions to the aforementioned generalized bilinear equation which is helpful to improve the research efficiency. Yu and Sun [14] introduced the 3+1-dimensional Kadomtsev-Petviashvili-Boussinesq-like equation. Based on this constructed bilinear Bäcklund transformation, some classes of exponential and rational traveling wave solutions with arbitrary wave numbers are presented. It provides a new idea for the follow-up research. Matinfar and Hosseini [15] formally studied the dynamic behavior of the solution obtained in the presence of linear and nonlinear effects. The research process is clear and interesting. Akinyemi et al. [16] studied singular solitons of the 2+1-dimensional improved nonlinear Schrödinger equation with spatiotemporal dispersions, group velocity dispersions, and power law nonlinearity by using the Kudryashov method, a complex transformation as well as symbolic computations. The results show that the suggested approach is incredibly accurate and powerful and can be used to explore solitons and other forms of solutions of nonlinear wave models in applied sciences and engineering. In addition, there are many effective methods for solving exact solutions of nonlinear differential equations, for instance, Bäcklund transformation method [17], Painlevé truncation extension method [18, 19], continuation method [20], trial equation method [21], and Hirota bilinear method [22–25].In 1976, Jaulent and Miodek [26] proposed the coupled Jaulent-Miodek equation when studying the relationship between the nonlinear evolution equation and the energy-dependent Schrödinger potential. In 2007, Feng and Li [27] used the theory of the plane dynamic system to study the existence of solitary waves and periodic waves of the coupled Jaulent-Miodek equation and obtained all possible explicit expressions. In 2009, Wazwaz [28] obtained multiple kink solutions and multiple singular kink solutions by studying the 2+1-dimensional nonlinear evolution equation generated by the Jaulent-Miodek hierarchy. In 2013, Cai et al. [29] used the CK direct method to obtain new exact multiple kink and singular kink solutions through the symmetric transformation of the 2+1-dimensional Jaulent-Miodek equation. In 2014, Ma et al. [30] used the auxiliary equation method and the Clarkson-Kruskal direct method to study 2+1-dimensional nonlinear model systems and obtained some new exact solutions. In 2015, Matinfar et al. [31] studied some 2+1-dimensional integrable models generated by the Jaulent-Miodek hierarchy and deduced the quasiperiodic solution of the 2+1-dimensional integrable model. In 2016, Li et al. [32] obtained the classical Lie symmetry of the extended 2+1-dimensional Jaulent-Miodek equation through the direct symmetry method and obtained a large number of new exact solutions by solving the reduced equation. In 2017, Wei et al. [33] studied the 2+1-dimensional Jaulent-Miodek equation using the homogeneous balance method and obtained four new solitary wave solutions of the equation. In 2018, Motsepa et al. [34] derived the conservation law and obtained the traveling wave solutions of the 2+1-dimensional Jaulent-Miodek equation. In 2018, Gu et al. [35] used the complex method to seek for the exact solutions of the 2+1-dimensional Jaulent-Miodek equation. In 2019, Shehata and Zahran [36] used the Jaulent-Miodek equation to construct the exact traveling wave solutions of parametric equations with different methods. In 2020, Kaewta et al. [37] transformed the 2+1-dimensional Jaulent-Miodek equation into a fourth-order partial differential equation and then obtained the exact solution which is simple and reliable.The2+1-dimensional Jaulent-Miodek equation (see [35, 38, 39]) is considered:
(1)a1uxt+a2ux2uxx−uxxxx−a3uxxuy−a4uxuxy+a5uyy=0.In equation (1), ai is constant (i=1,2,⋯,5.)Substituting traveling wave transform(2)ux,y,t=vz,z=x+ly+λtinto equation (1) and integrating it deduce
(3)v‴−a1λ+a5l2v′+lb2v′2−a23v′3−δ=0.In equation (3), b=a3+a4, where l, b, and λ are constants and δ is the integral constant. By setting w=v′, equation (3) becomes
(4)w″−a1λ+a5l2w+lb2w2−a23w3−δ=0.For the nonlinear2+1-dimensional Jaulent-Miodek equation, this paper adopts the exp−ϕz-expansion method, tanh method, and sine-cosine method to obtain the exact traveling wave solutions in turn.


## 2. Introduction of theexp−ϕz-Expansion Method and Main Result

Discuss a nonlinear partial differential equation (PDE) of the following form:(5)Pμ,μx,μt,μxx,μxy,μyy,μyt,μtt⋯=0.In equation (5), for the polynomial P containing the nonlinear term, the highest order polynomial, and the unknown function μx,y,t, the following steps can be followed.Step 1.
By substituting traveling wave transformμx,y,t=wz,z=x+ly+λt and putting them into equation (5), one can get the new equation, which is an ordinary differential equation (ODE):
(6)Kw,w′,w″,w‴,⋯=0.
K in equation (6) is both a polynomial of wz and a polynomial of the derivatives.Step 2.
The traveling wave solutions of equation (6) are shown as indicated below:
(7)wz=∑j=0ncjexp−ϕzj.
Equation (7) contains a constant term cj0≤j≤n, and it will be discussed later; at the same time, cj≠0 and ϕ=ϕz satisfy the ODE which will get
(8)ϕ′z=exp−ϕz+μexpϕz+δ1.Next, different forms of equation (8) are shown under different conditions.Ifδ12−4μ>0,μ≠0, then
(9)ϕz=ln−δ12−4μtanhδ12−4μ/2z+c−δ12μ,(10)ϕz=ln−δ12−4μcothδ12−4μ/2z+c−δ12μ.Ifδ12−4μ<0,μ≠0, then
(11)ϕz=ln−4μ−δ12tan4μ−δ12/2z+c−δ12μ,(12)ϕz=ln−4μ−δ12cot4μ−δ12/2z+c−δ12μ.Ifδ12−4μ<0,μ=0,δ1≠0, then
(13)ϕz=−lnδ1expδ1z+c−1.Ifδ12−4μ=0,μ≠0,δ1≠0, then
(14)ϕz=ln−2δ1z+c+2δ12z+c.Ifδ12−4μ=0,μ=0,δ1=0, then
(15)ϕz=lnz+c.In the above equations,cn≠0, δ1, and μ are undetermined constants, and the constant c can take any value. We consider the homogeneous balance between nonlinear terms and highest order derivatives of equation (6), and then, we can get the positive integer n.Step 3.
Through simultaneous equations (6) and (7), the polynomial about exp−ϕz can be obtained. Calculating all the coefficients of the same power of exp−ϕz to zero, we can get a set of simultaneous equations.Theorem 1.
Based on theexp−ϕz-expansion method, we can get fourteen kinds of solutions to equation (3).
Ifδ12−4μ>0, μ≠0,
(16)v11=lb+6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1−1tanh2δ12−4μz+c/2−δ1+d1,(17)v12=lb+6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1−1coth2δ12−4μz+c/2−δ1+d2,(18)v21=lb−6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1tanh2δ12−4μz+c/2−δ1−1+d3,(19)v22=lb−6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1coth2δ12−4μz+c/2−δ1−1+d4,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d1, d2, d3, and d4 are integral constants.Ifδ12−4μ<0, μ≠0,
(20)v13=lb+6a2δ12a2z+26/a2μ4μ−δ12lntan4μ−δ12z+c/2−δ12+1tan24μ−δ12z+c/2−δ1+d5,(21)v14=lb+6a2δ12a2z+26/a2μ4μ−δ12lncot24μ−δ12z+c/2−δ1cot4μ−δ12z+c/2−δ12+1+d6,(22)v23=lb−6a2δ12a2z+26/a2μ4μ−δ12lntan24μ−δ12z+c/2−δ1tan4μ−δ12z+c/2−δ12+1+d7,(23)v24=lb−6a2δ12a2z+26/a2μ4μ−δ12lncot4μ−δ12z+c/2−δ12+1cot24μ−δ12z+c/2−δ1+d8,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d5, d6, d7, and d8 are integral constants.Ifδ12−4μ<0, μ=0,δ1≠0,
(24)v15=lb+6a2δ12a2z−6/a2eδ1z+c−1+d9,(25)v25=lb−6a2δ12a2z+6/a2eδ1z+c−1+d10,where μ=0, δ=−1/2l3b3∓36a23/2δ13/12a22, and d9 and d10 are integral constants.Ifδ12−4μ=0, μ≠0,δ1≠0,
(26)v16=lb2a2z+6a2lnδ1z+c+2+d11,(27)v26=lb2a2z−6a2lnδ1z+c+2+d12,where μ=1/4δ12, δ=−l3b3/24a22, and d11 and d12 are integral constants.Ifδ12−4μ=0, μ=0,δ1=0,
(28)v17=lb2a2z+6a2lnz+c+d13,(29)v27=lb2a2z−6a2lnz+c+d14,where μ=0, δ=−l3b3/24a22, and d13 and d14 are integral constants.


## 3. Introduction of thetanh Method and Main Result

The canonicaltanh method proposed by [40, 41] includes the following steps.Step 1.
Thetanh method takes the form of introducing a new independent variable:
(30)Y=tanhμz.The derivatives of the independent variable change as follows:(31)ddz=μ1−Y2ddY,(32)d2dz2=μ21−Y2−2YddY+1−Y2d2dY2,where other derivatives of this kind can be obtained in a similar way.Step 2.
The premise of using thetanh method is to assume that the solution wz has the form
(33)wz=∑m=0MbmYm,where M is a positive integer and its value needs to be determined here. Then, by substituting equations (31) and (32) into the simplified ODE, we can get equations about the Y powers.Step 3.
Through balancing the highest order of linear and nonlinear terms, we can get parameterM. By equating the power coefficients of Y in the obtained equations, a set of algebraic equations about bm, μ, and c can be obtained. After the above parameters are determined, an analytical solution can be obtained by using equation (33) under closed condition.Theorem 2.
By employing thetanh method, we found that there will be two forms of solutions of (3).
(34)vT1=lb2a2z−32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d15,(35)vT2=lb2a2z+32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d16,where d15 and d16 are integral constants.


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

It is proven that the method of sine-cosine is effective in dealing with compaction problems. Next, the report describes this method briefly below, as the details have been shown in [42].The sine-cosine method applies to the following form:(36)ux,t=λ1cosβ1μz,∣z∣<π2μ,or another form
(37)ux,t=λ1sinβ1μz,∣z∣<πμ,otherwise, it is zero. The values of λ1, μ, and β1 parameters were measured. In equation (5), for the polynomial P containing the nonlinear term, the highest order polynomial, and the unknown function μx,y,t, the following steps can be followed.Step 1.
By substituting traveling wave transformμx,y,t=wz,z=x+ly+λt into equation (5), it is transformed into ordinary differential equation (6). K in equation (6) is both a polynomial of wz and a polynomial of the derivatives.Step 2.
The traveling wave solutions of equation (6) are shown as indicated below:
(38)wz=λ1cosβ1μz,∣z∣<π2μ,or another form
(39)wz=λ1sinβ1μz,∣z∣<πμ,otherwise, it is zero. The values of λ1, μ, and β1 parameters were measured.Step 3.
For the cosine or sine functions of the same homomorphic power, extract its undetermined coefficients and set them to zero, so we have simultaneous equations. The determination of the solutions of equation (5) can be obtained by obtaining the values of λ1, μ, and β1.Theorem 3.
Based on the sine-cosine method, the four forms of all meromorphic solutionsv of equation (3) are shown as indicated below.
(1)
The cosine method is used to solve equation (3) as follows:
(40)vc1z=6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d17,vc2z=−6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d18,where d17 and d18 are integral constants.(2)
The sine method is used to solve equation (3) as follows:
(41)vs1z=−6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d19,vs2z=6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d20,where d19 and d20 are integral constants.


## 5. Proof of Theorems

### 5.1. Proof of Theorem1

By homogeneous equilibrium ofw″ and w3 in equation (4), we can get
(42)wz=c0+c1exp−ϕz.In the above formula,c1≠0 and c0 are undetermined constants. Moreover, ϕz satisfies equation (8), where the constants δ1 and μ can be arbitrarily taken.From equation (42), we insert w, w3, and w″ into equation (4). By calculating all the coefficients of the same power of exp−ϕz to zero, the following equations can be obtained:
(43)e−3ϕzcoefficient:2c1−a2c133=0,e−2ϕzcoefficient:3c1δ1+lbc122−a2c0c12=0,e−1ϕzcoefficient:2c1μ+c1δ12−a1c1λ−a5l2c1+lbc0c1−a2c02c1=0,e−0ϕzcoefficient:c1δ1μ−a1c0λ−a5c0l2+lbc022−a2c033−δ=0.Simultaneously, the above equations can be obtained:(44)c0=lb+6a2δ12a2,c1=6a2,orc0=lb−6a2δ12a2,c1=−6a2.Substitute the result of the above formula into (42) to get
(45)w1z=lb+6a2δ12a2+6a2e−ϕz,(46)orw2z=lb−6a2δ12a2−6a2e−ϕz.We apply equations (9)–(15) into equations (45) and (46), respectively; the traveling wave solutions of the 2+1-dimensional Jaulent-Miodek equations are shown as indicated below.Ifδ12−4μ>0, μ≠0,
(47)w11=lb+6a2δ12a2−26/a2μδ12−4μtanhz+cδ12−4μ/2−δ1,v11=∫w11dz=lb+6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1−1tanh2δ12−4μz+c/2−δ1+d1,(48)w12=lb+6a2δ12a2−26/a2μδ12−4μcothz+cδ12−4μ/2−δ1,v12=∫w12dz=lb+6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1−1coth2δ12−4μz+c/2−δ1+d2,(49)w21=lb−6a2δ12a2+26/a2μδ12−4μtanhz+cδ12−4μ/2−δ1,v21=∫w21dz=lb−6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1tanh2δ12−4μz+c/2−δ1−1+d3,(50)w22=lb−6a2δ12a2+26/a2μδ12−4μcothz+cδ12−4μ/2−δ1,v22=∫w22dz=lb−6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1coth2δ12−4μz+c/2−δ1−1+d4,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d1, d2, d3, and d4 are integral constants.Ifδ12−4μ<0, μ≠0,
(51)w13=lb+6a2δ12a2−26/a2μ4μ−δ12tanz+c4μ−δ12/2−δ1,v13=∫w13dz=lb+6a2δ12a2z+26/a2μ4μ−δ12lntan4μ−δ12z+c/2−δ12+1tan24μ−δ12z+c/2−δ1+d5,(52)w14=lb+6a2δ12a2−26/a2μ4μ−δ12cotz+c4μ−δ12/2−δ1,v14=∫w14dz=lb+6a2δ12a2z+26/a2μ4μ−δ12lncot24μ−δ12z+c/2−δ1cot4μ−δ12z+c/2−δ12+1+d6,(53)w23=lb−6a2δ12a2+26/a2μ4μ−δ12tanz+c4μ−δ12/2−δ1,v23=∫w23dz=lb−6a2δ12a2z+26/a2μ4μ−δ12lntan24μ−δ12z+c/2−δ1tan4μ−δ12z+c/2−δ12+1+d7,(54)w24=lb−6a2δ12a2+26/a2μ4μ−δ12cotz+c4μ−δ12/2−δ1,v24=∫w24dz=lb−6a2δ12a2z+26/a2μ4μ−δ12lncot4μ−δ12z+c/2−δ12+1cot24μ−δ12z+c/2−δ1+d8,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d5, d6, d7, and d8 are integral constants.Ifδ12−4μ<0, μ=0,δ1≠0,
(55)w15=lb+6a2δ12a2+6/a2δ1eδ1z+c−1,v15=∫w15dz=lb+6a2δ12a2z−6/a2eδ1z+c−1+d9,(56)w25=lb−6a2δ12a2−6/a2δ1eδ1z+c−1,v25=∫w25dz=lb−6a2δ12a2z+6/a2eδ1z+c−1+d10,where μ=0, δ=−1/2l3b3∓36a23/2δ13/12a22, and d9 and d10 are integral constants.Ifδ12−4μ=0, μ≠0,δ1≠0,
(57)w16=lb+6a2δ12a2−6/a2δ12z+c2δ1z+c+4,v16=∫w16dz=lb2a2z+6a2lnδ1z+c+2+d11,(58)w26=lb−6a2δ12a2+6/a2δ12z+c2δ1z+c+4,v26=∫w26dz=lb2a2z−6a2lnδ1z+c+2+d12,where μ=1/4δ12, δ=−l3b3/24a22, and d11 and d12 are integral constants.Ifδ12−4μ=0, μ=0,δ1=0,
(59)w17=lb2a2+6/a2z+c,v17=∫w17dz=lb2a2z+6a2lnz+c+d13,(60)w27=lb2a2−6/a2z+c,v27=∫w27dz=lb2a2z−6a2lnz+c+d14,where μ=0, δ=−l3b3/24a22, and d13 and d14 are integral constants.


### 5.2. Proof of Theorem2

Thetanh method takes the form of introducing a new independent variable:
(61)Y=tanhμz.The derivatives of the independent variable change as follows:(62)ddz=μ1−Y2ddY,(63)d2dz2=μ21−Y2−2YddY+1−Y2d2dY2.For the nonlinear2+1-dimensional Jaulent-Miodek equation,
(64)w″−a1λ+a5l2w+lb2w2−a23w3−δ=0,w=∑m=0MbmYm.Balancing the highest order of linear and nonlinear terms, we get(65)3M=M+2.It is derived that M=1; therefore,
(66)wz=b0+b1Y=b0+b1tanhμz.In the above formula,(67)dwdY=b1,d2wdY2=0,where b0, b1, and μ appearing in the above are all undetermined constants; substituting (62) and (63) into (64), we get
(68)μ21−Y2−2Yw′+1−Y2w″−a1λ+a5l2w+lb2w2−a23w3−δ=0.Substituting (66) and (67) into (68), we get
(69)μ21−Y2−2Yb1−a1λ+a5l2b0+b1Y+lb2b0+b1Y2−a23b0+b1Y3−δ=0.In equation (69), a1, a5, l, b, and λ are all constants and δ is the integral constant. Extract the coefficients before all the powers of Y, and set each coefficient to zero, so as to obtain the simultaneous equations about the undetermined coefficients b0, b1, and μ:
(70)Y3coefficient:2b1μ2−a2b133=0,Y2coefficient:lbb122−a2b0b12=0,Y1coefficient:lbb0b1−a5b1l2−a1b1λ−2b1μ2−a2b02b1=0,Y0coefficient:lbb022−a1b0λ−a5b0l2−a2b033−δ=0.Solving the above simultaneous equations can get(71)b0=lb2a2,b1=±3lb2−12a2a1λ+a5l24a22,μ=±lb2−4a2a1λ+a5l28a2,δ=lb3−6a2lba1λ+a5l212a22.Substitutingb0, b1, and μ into equation (68), we can get solutions for two cases:
(72)wT1=lb2a2+3lb2−12a2a1λ+a5l24a22tanhlb2−4a2a1λ+a5l28a2z,vT1=lb2a2z−32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d15,(73)wT2=lb2a2−3lb2−12a2a1λ+a5l24a22tanh−lb2−4a2a1λ+a5l28a2z,vT2=lb2a2z+32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d16,where d15 and d16 are integral constants.


### 5.3. Proof of Theorem3

The sine-cosine method applies to the following form:(74)wz=λ1cosβ1μz,∣z∣<π2μ,or another form
(75)wz=λ1sinβ1μz,∣z∣<πμ,otherwise, it is zero; the values of λ1, μ, and β1 parameters were measured.We substitute(76)wz=λ1cosβ1μzinto equation (4) and obtain
(77)λ1β1μ2β1−1cosβ1−2μz−λ1β12μ2−a1λ+a5l2λ1cosβ1μz+lb2λ12cos2β1μz−a23λ13cos3β1μz−δ=0.For the cosine functions of the same homomorphic power, extract its undetermined coefficients and set them to zero, and the following equations can be obtained:(78)β1−1≠0,β1−2=3β1,λ1β1μ2β1−1=a23λ13,a1λ+a5l2λ1=−λ1β12μ2,lb2λ12=0.Solving the above simultaneous equations can get(79)β1=−1,μ=±−a1λ−a5l2,λ1=±−6a1λ+a5l2a2.In the end, we get the result(80)wc1z=−6a5l2−6a1λa2sec−a5l2−a1λz,vc1z=∫wc1zdz=6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d17,wc2z=−−6a5l2−6a1λa2sec−a5l2−a1λz,vc2z=∫wc2zdz=6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d18,where d17 and d18 are integral constants.We substitute(81)wz=λ1sinβ1μzinto equation (4) and obtain
(82)λ1β1μ2β1−1sinβ1−2μz−λ1β12μ2−a1λ+a5l2λ1sinβ1μz+lb2λ12sin2β1μz−a23λ13sin3β1μz−δ=0.For the sine functions of the same homomorphic power, extract its undetermined coefficients and set them to zero, and the following equations can be obtained:(83)β1−1≠0,β1−2=3β1,λ1β1μ2β1−1=a23λ13,a1λ+a5l2λ1=−λ1β12μ2,lb2λ12=0.Solving the above simultaneous equations can get(84)β1=−1,μ=±−a1λ−a5l2,λ1=±−6a1λ+a5l2a2.In the end, we get the result:(85)ws1z=−6a5l2−6a1λa2csc−a5l2−a1λz,vs1z=∫ws1zdz=−6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d19,ws2z=−−6a5l2−6a1λa2csc−a5l2−a1λz,vs2z=∫ws2zdz=6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d20,where d19 and d20 are integral constants.


## 5.1. Proof of Theorem1

By homogeneous equilibrium ofw″ and w3 in equation (4), we can get
(42)wz=c0+c1exp−ϕz.In the above formula,c1≠0 and c0 are undetermined constants. Moreover, ϕz satisfies equation (8), where the constants δ1 and μ can be arbitrarily taken.From equation (42), we insert w, w3, and w″ into equation (4). By calculating all the coefficients of the same power of exp−ϕz to zero, the following equations can be obtained:
(43)e−3ϕzcoefficient:2c1−a2c133=0,e−2ϕzcoefficient:3c1δ1+lbc122−a2c0c12=0,e−1ϕzcoefficient:2c1μ+c1δ12−a1c1λ−a5l2c1+lbc0c1−a2c02c1=0,e−0ϕzcoefficient:c1δ1μ−a1c0λ−a5c0l2+lbc022−a2c033−δ=0.Simultaneously, the above equations can be obtained:(44)c0=lb+6a2δ12a2,c1=6a2,orc0=lb−6a2δ12a2,c1=−6a2.Substitute the result of the above formula into (42) to get
(45)w1z=lb+6a2δ12a2+6a2e−ϕz,(46)orw2z=lb−6a2δ12a2−6a2e−ϕz.We apply equations (9)–(15) into equations (45) and (46), respectively; the traveling wave solutions of the 2+1-dimensional Jaulent-Miodek equations are shown as indicated below.Ifδ12−4μ>0, μ≠0,
(47)w11=lb+6a2δ12a2−26/a2μδ12−4μtanhz+cδ12−4μ/2−δ1,v11=∫w11dz=lb+6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1−1tanh2δ12−4μz+c/2−δ1+d1,(48)w12=lb+6a2δ12a2−26/a2μδ12−4μcothz+cδ12−4μ/2−δ1,v12=∫w12dz=lb+6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1−1coth2δ12−4μz+c/2−δ1+d2,(49)w21=lb−6a2δ12a2+26/a2μδ12−4μtanhz+cδ12−4μ/2−δ1,v21=∫w21dz=lb−6a2δ12a2z+26/a2μδ12−4μlntanh2δ12−4μz+c/2−δ1tanh2δ12−4μz+c/2−δ1−1+d3,(50)w22=lb−6a2δ12a2+26/a2μδ12−4μcothz+cδ12−4μ/2−δ1,v22=∫w22dz=lb−6a2δ12a2z+26/a2μδ12−4μlncoth2δ12−4μz+c/2−δ1coth2δ12−4μz+c/2−δ1−1+d4,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d1, d2, d3, and d4 are integral constants.Ifδ12−4μ<0, μ≠0,
(51)w13=lb+6a2δ12a2−26/a2μ4μ−δ12tanz+c4μ−δ12/2−δ1,v13=∫w13dz=lb+6a2δ12a2z+26/a2μ4μ−δ12lntan4μ−δ12z+c/2−δ12+1tan24μ−δ12z+c/2−δ1+d5,(52)w14=lb+6a2δ12a2−26/a2μ4μ−δ12cotz+c4μ−δ12/2−δ1,v14=∫w14dz=lb+6a2δ12a2z+26/a2μ4μ−δ12lncot24μ−δ12z+c/2−δ1cot4μ−δ12z+c/2−δ12+1+d6,(53)w23=lb−6a2δ12a2+26/a2μ4μ−δ12tanz+c4μ−δ12/2−δ1,v23=∫w23dz=lb−6a2δ12a2z+26/a2μ4μ−δ12lntan24μ−δ12z+c/2−δ1tan4μ−δ12z+c/2−δ12+1+d7,(54)w24=lb−6a2δ12a2+26/a2μ4μ−δ12cotz+c4μ−δ12/2−δ1,v24=∫w24dz=lb−6a2δ12a2z+26/a2μ4μ−δ12lncot4μ−δ12z+c/2−δ12+1cot24μ−δ12z+c/2−δ1+d8,where μ=4a5l2+4a1λ+2δ12a2−l2b2/8a2, δ=l3b3−6a2lba5l2+a1λ/12a22, and d5, d6, d7, and d8 are integral constants.Ifδ12−4μ<0, μ=0,δ1≠0,
(55)w15=lb+6a2δ12a2+6/a2δ1eδ1z+c−1,v15=∫w15dz=lb+6a2δ12a2z−6/a2eδ1z+c−1+d9,(56)w25=lb−6a2δ12a2−6/a2δ1eδ1z+c−1,v25=∫w25dz=lb−6a2δ12a2z+6/a2eδ1z+c−1+d10,where μ=0, δ=−1/2l3b3∓36a23/2δ13/12a22, and d9 and d10 are integral constants.Ifδ12−4μ=0, μ≠0,δ1≠0,
(57)w16=lb+6a2δ12a2−6/a2δ12z+c2δ1z+c+4,v16=∫w16dz=lb2a2z+6a2lnδ1z+c+2+d11,(58)w26=lb−6a2δ12a2+6/a2δ12z+c2δ1z+c+4,v26=∫w26dz=lb2a2z−6a2lnδ1z+c+2+d12,where μ=1/4δ12, δ=−l3b3/24a22, and d11 and d12 are integral constants.Ifδ12−4μ=0, μ=0,δ1=0,
(59)w17=lb2a2+6/a2z+c,v17=∫w17dz=lb2a2z+6a2lnz+c+d13,(60)w27=lb2a2−6/a2z+c,v27=∫w27dz=lb2a2z−6a2lnz+c+d14,where μ=0, δ=−l3b3/24a22, and d13 and d14 are integral constants.


## 5.2. Proof of Theorem2

Thetanh method takes the form of introducing a new independent variable:
(61)Y=tanhμz.The derivatives of the independent variable change as follows:(62)ddz=μ1−Y2ddY,(63)d2dz2=μ21−Y2−2YddY+1−Y2d2dY2.For the nonlinear2+1-dimensional Jaulent-Miodek equation,
(64)w″−a1λ+a5l2w+lb2w2−a23w3−δ=0,w=∑m=0MbmYm.Balancing the highest order of linear and nonlinear terms, we get(65)3M=M+2.It is derived that M=1; therefore,
(66)wz=b0+b1Y=b0+b1tanhμz.In the above formula,(67)dwdY=b1,d2wdY2=0,where b0, b1, and μ appearing in the above are all undetermined constants; substituting (62) and (63) into (64), we get
(68)μ21−Y2−2Yw′+1−Y2w″−a1λ+a5l2w+lb2w2−a23w3−δ=0.Substituting (66) and (67) into (68), we get
(69)μ21−Y2−2Yb1−a1λ+a5l2b0+b1Y+lb2b0+b1Y2−a23b0+b1Y3−δ=0.In equation (69), a1, a5, l, b, and λ are all constants and δ is the integral constant. Extract the coefficients before all the powers of Y, and set each coefficient to zero, so as to obtain the simultaneous equations about the undetermined coefficients b0, b1, and μ:
(70)Y3coefficient:2b1μ2−a2b133=0,Y2coefficient:lbb122−a2b0b12=0,Y1coefficient:lbb0b1−a5b1l2−a1b1λ−2b1μ2−a2b02b1=0,Y0coefficient:lbb022−a1b0λ−a5b0l2−a2b033−δ=0.Solving the above simultaneous equations can get(71)b0=lb2a2,b1=±3lb2−12a2a1λ+a5l24a22,μ=±lb2−4a2a1λ+a5l28a2,δ=lb3−6a2lba1λ+a5l212a22.Substitutingb0, b1, and μ into equation (68), we can get solutions for two cases:
(72)wT1=lb2a2+3lb2−12a2a1λ+a5l24a22tanhlb2−4a2a1λ+a5l28a2z,vT1=lb2a2z−32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d15,(73)wT2=lb2a2−3lb2−12a2a1λ+a5l24a22tanh−lb2−4a2a1λ+a5l28a2z,vT2=lb2a2z+32a2ln−sech12l2b2−4a2a5l2+a1λ2a2z2+d16,where d15 and d16 are integral constants.


## 5.3. Proof of Theorem3

The sine-cosine method applies to the following form:(74)wz=λ1cosβ1μz,∣z∣<π2μ,or another form
(75)wz=λ1sinβ1μz,∣z∣<πμ,otherwise, it is zero; the values of λ1, μ, and β1 parameters were measured.We substitute(76)wz=λ1cosβ1μzinto equation (4) and obtain
(77)λ1β1μ2β1−1cosβ1−2μz−λ1β12μ2−a1λ+a5l2λ1cosβ1μz+lb2λ12cos2β1μz−a23λ13cos3β1μz−δ=0.For the cosine functions of the same homomorphic power, extract its undetermined coefficients and set them to zero, and the following equations can be obtained:(78)β1−1≠0,β1−2=3β1,λ1β1μ2β1−1=a23λ13,a1λ+a5l2λ1=−λ1β12μ2,lb2λ12=0.Solving the above simultaneous equations can get(79)β1=−1,μ=±−a1λ−a5l2,λ1=±−6a1λ+a5l2a2.In the end, we get the result(80)wc1z=−6a5l2−6a1λa2sec−a5l2−a1λz,vc1z=∫wc1zdz=6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d17,wc2z=−−6a5l2−6a1λa2sec−a5l2−a1λz,vc2z=∫wc2zdz=6a2lnsec−a5l2−a1λz+tan−a5l2−a1λz+d18,where d17 and d18 are integral constants.We substitute(81)wz=λ1sinβ1μzinto equation (4) and obtain
(82)λ1β1μ2β1−1sinβ1−2μz−λ1β12μ2−a1λ+a5l2λ1sinβ1μz+lb2λ12sin2β1μz−a23λ13sin3β1μz−δ=0.For the sine functions of the same homomorphic power, extract its undetermined coefficients and set them to zero, and the following equations can be obtained:(83)β1−1≠0,β1−2=3β1,λ1β1μ2β1−1=a23λ13,a1λ+a5l2λ1=−λ1β12μ2,lb2λ12=0.Solving the above simultaneous equations can get(84)β1=−1,μ=±−a1λ−a5l2,λ1=±−6a1λ+a5l2a2.In the end, we get the result:(85)ws1z=−6a5l2−6a1λa2csc−a5l2−a1λz,vs1z=∫ws1zdz=−6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d19,ws2z=−−6a5l2−6a1λa2csc−a5l2−a1λz,vs2z=∫ws2zdz=6a2lncsc−a5l2−a1λz+cot−a5l2−a1λz+d20,where d19 and d20 are integral constants.


## 6. Computer Simulations

In this section, we use computer simulation images to explain the results obtained according to three different methods and further analyze the nature of the simple periodic solutionsv11z, vT1z, and vc1z in the 2+1-dimensional Jaulent-Miodek equation.Figure1 shows the three-dimensional images of v11z by considering the values a1=2, a2=2, a5=2, l=2, b=5, λ=2, c=2, δ1=3, μ=2, and d1=7; from Figures 1(a)–1(c), t takes the following three different values: t=−2, t=0, and t=2. Figures 1(a)–1(c) demonstrate discontinuities of solution v11z on the domain.Figure 1
The three-dimensional images ofv11z in three situations.
(a)(b)(c)Figure2 shows the three-dimensional images of vT1z by considering the values a1=2, a2=2, a5=2, l=2, b=5, λ=2, and d15=7; from Figures 2(a)–2(c), t takes the following three different values: t=−2, t=0, and t=2. Figures 2(a)–2(c) describe the kink wave and parabola solution to the nonlinear evolution equation (3).Figure 2
The three-dimensional images ofvT1z in three situations.
(a)(b)(c)Figure3 shows the three-dimensional images of vc1z by considering the values a1=3, a2=2, a5=2, l=2, b=5, λ=−3, and d19=7; from Figures 3(a)–3(c), t takes the following three different values: t=−2, t=0, and t=2. Figures 3(a)–3(c) show the singularities on the domain. However, when t>0, the image changes dramatically.Figure 3
The three-dimensional images ofvc1z in three situations.
(a)(b)(c)Figure4 shows the two-dimensional images of v11z by considering the values a1=2, a2=2, a5=2, l=2, b=5, λ=2, c=2, δ1=3, μ=2, and d1=7; from Figures 4(a)–4(c), t takes the following three different values: t=−2, t=0, and t=2.Figure 4
The two-dimensional images ofv11z in three situations.
(a)(b)(c)Figure5 shows the two-dimensional images of vT1z by considering the values a1=2, a2=2, a5=2, l=2, b=5, λ=2, and d15=7; from Figures 5(a)–5(c), t takes the following three different values: t=−2, t=0, and t=2.Figure 5
The two-dimensional images ofvT1z in three situations.
(a)(b)(c)Figure6 shows the two-dimensional images of vc1z by considering the values a1=3, a2=2, a5=2, l=2, b=5, λ=−3, and d19=7; from Figures 6(a)–6(c), t takes the following three different values: t=−2, t=0, and t=2.Figure 6
The two-dimensional images ofvc1z in three situations.
(a)(b)(c)The above two-dimensional images all intuitively and accurately reflect the periodicity of function fluctuations.


## 7. Discussions and Conclusions

By the exp−ϕz-expansion method, we have obtained fourteen forms of solutions to the Jaulent-Miodek equation. Utilizing the tanh method, we obtained four forms of solutions of the Jaulent-Miodek equation. In the end, utilizing the sine-cosine method, we obtained two forms of solutions to the Jaulent-Miodek equation. Combining the existing results found in this paper, more results are easily obtained by the exp−ϕz-expansion method.In [35], the authors adopt this complex method to seek for analytic solution of equation (3); the elliptic function solution is obtained by the complex method. From our paper and [35], it is easy to see that by utilizing the exp−ϕz-expansion method and complex method, both can obtain the hyperbolic function solutions for the Jaulent-Miodek equation. Unfortunately, utilizing the sine-cosine approach, we find only the trigonometric function solutions for the Jaulent-Miodek equation.In [37], the authors used both methods, namely, the G′/G2-expansion method and the Jacobi elliptic equation method to calculate and analyze the explicit exact traveling wave solutions of the 2+1-dimensional Jaulent-Miodek equation as given by equation (1). By applying the G′/G2 expansion method to the equation, the solutions were obtained by the author including the triangular hyperbolic function solutions and rational function solutions. Implemented by the Jacobi elliptic equation method (analytical equation (1)), three sets of parameter values were reached. Every set has produced fifteen exact terms for the solution of Jacobi elliptic function solutions. By comparing the methods to [37], the methods in our article seem more concise and straightforward.All three methods in this paper have their own characteristics. We trust that the above method presented here and the methods in [35, 37] together can be applied more effectively to find out solutions for nonlinear evolution equations currently and henceforth. The above-mentioned methods are very helpful tools for unearthing exact solutions to the nonlinear evolution equations.



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