# On the Exact Solutions of Nonlinear Potential Yu–Toda–Sasa–Fukuyama Equation by Different Methods

Authors: Qinghao Zhu; Jianming Qi
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this article, the exact solutions to the potential Yu–Toda–Sasa–Fukuyama equation are successfully examined by the extended complex method andG′/G-expansion method. Consequently, we find solutions for three models of Weierstrass elliptic functions, simply periodic functions, and rational function solutions. The obtained results will play an important role in understanding and studying potential Yu–Toda–Sasa–Fukuyama equation. It is observed that the extended complex method and G′/G-expansion method are reliable and will be used extensively to seek for exact solutions of any other nonlinear partial differential equations (NPDEs).


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## Body

## 1. Introduction

As is known to all, the development of modern natural science is not only linear. Nonlinear science is considered as the most essential forefront of the fundamental understanding of nature. Even nonlinear systems look like counterintuitive, chaotic, or unbelievable, but it is more in line with reality, among which the behavior analysis of energy dynamic system solutions is one of the hot spots in recent research. Especially in biologists science, physicists science, information technology, space science and so on [1], the convenience and stability of society are greatly enhanced.In recent years, the application of nonlinear science in power systems has become more and more extensive. It provides a large number of reliable references for the operation and planning of the power system. The application of nonlinear equations promoted the development of nonlinear sensitive electronic devices on the load side and grid side of the power system. The stable operation of power system at each level can be effectually protected by exploring the nonlinear phenomena in the case of ferromagnetic resonance overvoltage situation. The harmonic linearization method is proposed to solve the asymmetric nonlinear oscillation problem in the parallel operation of power stations, which will help to select the grid structure reasonably and effectively improve the stability of the grid structure [2]. The high-frequency disturbances that may be encountered can be predicted by constructing a nonlinear prediction model with random disturbances [3]. Nonlinear equations can not only simulate the actual operating conditions of the power system but also simulate possible hazards and play an extremely important role in the stable operation of the power grid [4].The study and development of nonlinear dynamic systems have been conducted for more than a century. Since the 1970s, the bifurcation theory and chaotic phenomenon of nonlinear dynamics have become a hot topic of nonlinear differential equations. Nonlinear system of equations is used to describe and model the complex nonlinear systems. Recently, in order to obtain various solitary and periodic solutions of nonlinear equations, scientists have come up with many innovative and effective methods.Much attention has been paid to the exact solutions of nonlinear partial differential equations. Mathematicians and scientists have continuously proposed and improved methods for solving nonlinear differential equations [5, 6]. Julier et al. [7] used the new mean and covariance parameterization, which can be directly transformed by the system equation to predict the transformed mean and covariance. In addition, there are many effective and magical methods to obtain exact solutions of nonlinear differential equations, such as Bäcklund transformation method [8], continuation method [9], Painlevé truncation extension method [10, 11], Hirota bilinear method [12, 13], Exp−φz-Expansion method [14, 15], and so on.The3+1−dimensional potential Yu–Toda–Sasa–Fukuyama equation is a nonlinear partial differential equation which is often used to solve problems in fluid mechanics. In 2014, Hu et al. [16] used three-wave method to obtain some new kink multisoliton solutions about potential Yu–Toda–Sasa–Fukuyama equation. In 2017, for further study of potential Yu–Toda–Sasa–Fukuyama equation, Roshid [17] constructed the exact solution of the equation by using the lump solution method. Zhang and Zong [18] obtained the exact solution by using the F-expansion method. In 2018, Foroutan et al. [19] studied the 3+1−dimensional potential-YTSF equation by implementing the Hirota bilinear method. In 2019, Zhao and He [20] employed the bilinear method to study the 3+1−dimensional potential-YTSF equation in fluid dynamics. In this article, we used the extended complex method [21–24] and G′/G-expansion method to find the exact solutions of modified 3+1−dimensional potential-YTSF equation.Modified3+1−dimensional potential-YTSF equation (see [25]) is considered as(1)−4uxt+uxxxz+4uxuxz+2uxxuz+3uyy=0.Taking the transformationξ=kx+ly+mz+wt, where k,l,m, and w are arbitrary constants, equation (1) turns into(2)k3mu″′+3k2mu′2+3l2−4kwu′+d=0.In (2), d is the integration constant. By putting(3)u′=v,into (2) and integrating it, we get(4)k3mv″+3k2mv2+3l2−4kwv+d=0.In this paper, the complex method andG′/G−expansion method are employed by us to seek for the exact solution of (4). These two methods will play a significant role in constructing exact solutions for the nonlinear partial differential equations via the 3+1−dimensional potential-YTSF equation.


## 2. Introduction of the Extended Complex Method and Main Result

Because the extended complex method involves the Weierstrass elliptic function, first we give the brief introduction of Weierstrass elliptic function:℘ξ≔℘ξ,g2,g3 is a meromorphic function with double periods ω1,ω2 [26–28] and is defined as(5)℘ξ;w1,w2≔1ξ2+∑μ,ν∈Z,μ2+ν2≠01ξ+μw1+νw22−1μw1+νw22,which satisfies the following equation:(6)℘′ξ2=4℘ξ3−g2℘ξ−g3,where g2=60s4,g3=140s6, and Δg2,g3≠0, and has the another formula(7)℘ξ−ξ0=−℘ξ−℘ξ0+14℘′ξ+℘′ξ0℘ξ−℘ξ02.Next, we will employ the recently established extended complex method [21–24] to seek for traveling wave solutions to the equations mentioned above.Step 1. Insert the traveling wave transform T: ux,y,z,t=uξ,ξ=kx+ly+mz+wt into a given PDE and alternate it to the following integer-order ordinary differential equation (ODE):(8)Ku,u′,u″,u‴,…=0,whereK is a polynomial of uξ and its derivatives.Step 2. Next, we will find out weak p,q condition. To find out the weak p,q condition of equation (8), we can substitute Laurent series(9)uξ=∑k=−q∞ckξk,q>0,c−q≠0,into equation (8), and then we can find out the p distinct Laurent singular parts as follows:(10)∑k=−q−1ckξk,wherep indicates that the equation has p distinct meromorphic solutions and q indicates that their multiplicity of the pole at ξ=0 is q.Step 3. Take the indeterminate forms(11)uξ=∑i=1l−1∑j=2qi−1jc−ijj−1!dj−2dzj−214℘′ξ+Bi℘ξ−Ai2−℘ξ+∑i=1l−1c−i12℘′ξ+Bi℘ξ−Ai+∑j=2ql−1jc−ljj−1!dj−2dzj−2℘ξ+c0,(12)uξ=∑i=1l∑j=1qcijξ−ξij+c0,(13)uξ=∑i=1l∑j=1qcijξ−ξij+c0,whereζ=eαξα∈ℂinto equation (8), respectively, and produce a set of algebraic equations; we get Weierstrass elliptic function solutions, simply periodic solutions, and rational function solutions with a pole at z=0, in which c−ij are given by equation (9), Bi2=4Ai3−g2Ai−g3, Σi=1lc−i1=0, and equations (12) and (13) have l≤p distinct poles of multiplicity q.Step 4. On the basis of Step 3, we can easily obtain meromorphic solutions at arbitrary pole. Putting the inverse transform T−1 into the solutions uξ, we can easily get all exact solutions ux,y,z,t in the original FPDE.Theorem 1.
If we employ the extended complex method and supposekm≠0, then there are three types solutions w of equation (4) listed below:(a)
All elliptic solutions are as follows:(14)veξ=−2k−℘ξ+14℘′ξ+F℘ξ−E2+2kE−3l2−4kw6k2m,where12dk2m=−12k3mg2+3l2−4kw2. Noting ℜξ=∫℘ξdξ and ξ0∈ℂ,(15)ueξ0ξ=−2k−ℜξ+14ℜ′ξ+Fℜξ−E2+2kE−3l2−4kw6k2mξ+c1,where12k2md=−12k3mg2+3l2−4kw2, F2=4E3−g2E−g3, here g3 and E=ℜξ0 are arbitrary constants, and c1 is an integral constant.(b)
The simply periodic solutions are obtained by(16)vsξ=−2kα2coth2α2ξ−ξ0−k6α2−3l2−4kw6k2m,where12k2md=−k6m2α4+3l2−4kw2, α∈ℂ, and ξ0∈ℂ. So,(17)usξ0ξ,=4kαcothαξ−ξ02+2kαlncothαξ−ξ02−1,−2kαlncothαξ−ξ02+1−k6α2ξ−3l2−4kw6k2mξ+c2.Herec2 is an integral constant.(c)
The rational function solutions:(18)v1,2ξ=−2kξ−ξ02−3l2−4kw6k2m,where12k2md=3l2−4kw2,ξ0∈ℂ. So,(19)u1,2ξ=2kξ−ξ0−3l2−4kw6k2mξ+c3.Herec3 is an integral constant.


## 3. Introduction ofG′/G-Expansion Method and Main Result

Next, we employ the recently establishedG′/G-expansion method [29, 30] to seek for traveling wave solutions to the equations mentioned above.Consider the following form of a nonlinear partial differential equation (PDE):(20)Pux,uy,uz,ut,uxy,utt,uxz…=0,where P is a polynomial with an unknown function μx,y,z,t and its derivatives in which nonlinear terms and highest order derivatives are involved. Next the detail steps can be processed as follows:Step 1. Inserting the traveling wave transform ux,y,z,t=uξ,ξ=kx+ly+mz+wt into equation (20), the following ordinary differential equation (ODE) is obtained:(21)Ku,u′,u″,u‴,…=0,whereK is a polynomial of uz and its derivatives.Step 2. Regarding that, the solution of equation (21) can be expressed by a polynomial in G′/G in the following form:(22)uξ=αmG′Gm+αm−1G′Gm−1+…,whereG=Gξ satisfies the second-order linear ordinary differential equation (LODE) as follows:(23)G′′+λG′+μG=0.These unknown constantsαm, αm−1, … are about to be determined. The omitted parts in (22) are also a polynomial in G′/G, the degree of which is equal to or less than m−1. The positive integer m can be determined by considering the uniform equilibrium between the highest derivative and the nonlinear term in ODE (21).Step 3. Substituting equation (22) into equation (21) and using (23), the left-hand side of equation (21) is converted into another polynomial in G′/G. Calculating all the coefficients to zero of the polynomial yields a set of algebraic equations for αm, …, α1.Step 4. The constants αm, …, α1 can be solved by the system of algebraic equations obtained in Step 3. Because the general solutions of second-order equation (23) are well known, depending on the sign of the discriminant Δ=λ2−4μ, the exact solutions of the given equation (20) are thus obtained. Here(24)G′G=−λ2+λ2−4μ2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/2,λ2−4μ>0,−λ2+λ2−4μ2−C1sinλ2−4μξ/2+C2cosλ2−4μξ/2C1cosλ2−4μξ/2+C2sinλ2−4μξ/2,λ2−4μ<0,−λ2+2C2−C1λ−C2λξ2C1+C2ξ,λ2−4μ=0,whereC1 and C2 are arbitrary constants.Theorem 2.
If we employ theG′/G-expansion method, then we obtain the following traveling wave solutions of equation (4):Case 1. If Δ=λ2−4μ>0,(25)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m.Here(26)G′G=−λ2+λ2−4μ2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants. Now, we give three forms of uξ:(1)
IfC1C2≠0, then(27)G′G=−λ2+λ2−4μ2C1sinhλ2−4μ2ξ+C2coshλ2−4μ2ξ/C1coshλ2−4μ2ξ+C2sinhλ2−4μ2ξv1ξ=2k3λ2m−8k3mμ+4km−3l26k2m,−kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22.(28)u1ξ,=∫v1ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ,−∫kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22dξ.Because of the rational hyperbolic functions forms, we cannot accumulate the exact expressions ofu1ξ.(2)
IfC1=0,C2≠0, then G′/Gξ=−λ/2+λ2−4μ/2cothλ2−4μ/2ξ. Hence,(29)v2ξ,=2k3λ2m−8k3mμ+4km−3l26k2m,−2kλλ2−4μcothλ2−4μ/2ξ−kλ2−4μ/2coth2λ2−4μ/2ξ,(30)u2ξ,=∫v2ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλln|sinhλ2−4μ/2ξ|,+kλ2−4μcothλ2−4μ/2ξ+lncothλ2−4μ/2ξ−12−lncothλ2−4μ/2ξ+12+c4.(3)
IfC1≠0,C2=0, then G′/Gξ=−λ/2+λ2−4μ/2tanhλ2−4μ/2ξ. Hence,(31)v3ξ,=2k3λ2m−8k3mμ+4km−3l26k2m,−2kλλ2−4μtanhλ2−4μ/2ξ−kλ2−4μ2λ2−4μ/2tanh2λ2−4μ/2ξ,(32)u3ξ,=∫v3ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλlncoshλ2−4μ/2ξ,+kλ2−4μtanhλ2−4μ/2ξ+lntanhλ2−4μ/2ξ−12−lntanhλ2−4μ/2ξ+12+c5,wherec4,c5 are integral numbers.Case 2. If Δ=λ2−4μ<0,(33)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m.Here(34)G′G=−λ2+4μ−λ22−C1sinλ2−4μξ/2+C2cosλ2−4μξ/2C1cosλ2−4μξ/2+C2sinλ2−4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.Almost the same as case 1, it is just a little different form ofuξ. If C1C2≠0 and C1=0,C2≠0, forms of uξ are almost the same as case 1.IfC1≠0,C2=0, then G′/Gξ=−λ/2−λ2−4μ/2tanhλ2−4μξ/2.Hence,(35)v4ξ,=2k3λ2m−8k3mμ+4km−3l26k2m,+2kλλ2−4μtanhλ2−4μ/2ξ−kλ2−4μ/2tanh2λ2−4μ/2ξ,(36)u4ξ,=∫v4ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ+4kλlncoshλ2−4μ/2ξ,+kλ2−4μtanhλ2−4μ/2ξ+lntanhλ2−4μ/2ξ−12−lntanhλ2−4μ/2ξ+12+c6,wherec6 is an integral number.Case 3. If Δ=λ2−4μ=0,(37)vξ=−2k2C2−C1λ−C2λξ2C1+C2ξ2−2kλ2C2−C1λ−C2λξ2C1+C2ξ−k3λ2m+8k3mμ−4km+3l26k2m,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.(38)u5ξ,=∫v5ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ+2kC2C2ξ+C1+c7,wherec7 is an integral number.


## 4. Proof of Theorems

### 4.1. Proof of Theorem1

Putting (9) into (4), we have q=2, p=1, A=k3m, B=3l2−4kw, C=3k2m, D=d, c−2=−2k, c−1=0, c0=4kw−3l2/6k2m, c1=0, c2=6dk2m−24k2m−3l2−34kw2/120k5m2, c3=0, where c4 is arbitrary. Hence, (4) satisfies weak 1,2 condition.By (12), we have the indeterminant rational style solutions of (4) with pole at ξ=0 that(39)v1ξ=c11ξ2+c12ξ+c.Substitutingv1ξ into (4), we obtain the following result:(40)v1ξ=−k3ξ2−3l2−4kw6k2m,where 12k2md=3l2−4kw2. So,(41)u1ξ=∫v1ξdξ=k3ξ−3l2−4kw6k2mξ+c8.The rational solutions of (4) are as follows:(42)v1,2ξ=−2kξ−ξ02−3l2−4kw6k2m,where 12k2md=3l2−4kw2,ξ0∈ℂ. So,(43)u1,2ξ=∫v1,2ξdξ=2kξ−ξ0−3l2−4kw6k2mξ+c9.Next we will seek for simply periodic solutions, settingζ=expαξ and putting v=Rζ into (4); then,(44)Aα2ζR′+ζ2R′′+BR+CR2+D=0,and putting(45)v2ζ=c2ζ−12+c1ζ−ζ1+c10into (44), we get that(46)v2ζ=−2kα2ζ−12+−2kα2ζ−1−k6α2−3l2−4kw6k2m,where 12k2md=−k6m2α4+3l2−4kw2. Substituting ζ=eαξ into above relation, we obtain simple periodic solutions (4), and the pole point is located at ξ=0.(47)vsξ,=−2kα2eαξ−12−2kα2eαξ−1−k6α2−3l2−4kw6k2m,=−2kα2eαξeαξ−12−k6α2−3l2−4kw6k2m,=−2kα2coth2α2ξ−k6α2−3l2−4kw6k2m.So,(48)usξ,=∫vsξdξ,=4kαcothαξ2+2kαlncothαξ2−1,−2kαlncothαξ2+1−k6α2−3l2−4kw6k2m+c11.So, all simply periodic solutions of (4) are obtained by(49)vsξ0ξ=−2kα2coth2α2ξ−ξ0−k6α2−3l2−4kw6k2m,where 12k2md=−k6m2α4+3l2−4kw2, ξ0∈ℂ.So,(50)usξ0ξ,=∫vsξ0ξdξ,=4kαcothαξ−ξ02+2kαlncothαξ−ξ02−1,−2kαlncothαξ−ξ02+1−k6α2ξ−3l2−4kw6k2mξ+c12,where c8,c9,c10,c11,c12 are constants.From (11), we have indeterminant relations of elliptic solutions of (4) and the pole point is located at ξ=0.(51)veξ=c−2℘ξ+c0.Substitutingveξ into the (4), we obtain that(52)veξ=−2k℘ξ−3l2−4kw6k2m.Here12dk2m=−12k3mg2+3l2−4kw2. So,(53)ueξ=∫veξdξ=−2kℜξ−3l2−4kw6k2mξ+c13,where ℜξ=∫℘ξdξ is also an elliptic function.Thus, all the elliptic solutions for (4) are as follows:(54)veξ0ξ=−2k℘ξ−ξ0−3l2−4kw6k2m.So,(55)ueξ0ξ=∫veξ0ξdξ=−2kℜξ−ξ0−3l2−4kw6k2mξ+c14,where ξ0∈ℂ. By using (7), we rewrite it to the form(56)veξ0ξ=−2k−℘ξ+14℘′ξ+F℘ξ−E2+2kE−3l2−4kw6k2m,where 12dk2m=−12k3mg2+3l2−4kw2. Then, from (55) and (7), we deduce that(57)ueξ0ξ=∫veξ0ξdξ=−2k−ℜξ+14ℜ′ξ+Fℜξ−E2+2kE−3l2−4kw6k2mξ+c15,where c13,c14,c15 are integral numbers, and here 12k2md=−12k3mg2+3l2−4kw2, F2=4E3−g2E−g3, where g3 and E=ℜξ0 are arbitrary constants..The proof is over.


### 4.2. Proof of Theorem2

Considering the homogeneous equilibrium betweenv″ and v2 in (4), we get m=2. So, we can set the solution of (4) as follows:(58)v=α2G′G2+α1G′G+α0;α2≠0.α0,α1,α2 here are constants to be determined later.Next, we will use (58) and (23) together and collect all terms of the same power as G′/G.First, from (58), we get(59)v′ξ=2α2G′ξG′′ξGξ2−2α2G′ξ3Gξ3+α1G′′ξGξ−α1G′ξ2Gξ2.Substituting (23) into (59), we obtain(60)v′ξ=−2α2G′ξ2λGξ2−2α2G′ξμGξ−2α2G′ξ3Gξ3−α1λG′ξGξ−α1μ−α1G′ξ2Gξ2,and then we deduce(61)v′′ξ,=6α2G′ξλμGξ+2α2μ2+8α2G′ξ2μGξ2+4α2G′ξ2λ2Gξ2,+10α2G′ξ3λGξ3+6α2G′ξ4Gξ4+3α1G′ξ2λGξ2+α1G′ξλ2Gξ,+α1λμ+2α1G′ξμGξ+2α1G′ξ3Gξ3.Now, we put (58), (60), and (61) into (4) and collect all terms with the same power of G′ξ/Gξ together:(62)6k3mα2+3k2mα22G′ξGξ4+10k3λmα2+2k3mα1+6k2mα1α2G′ξGξ3,+4k3mα2λ2+3k3mα1λ+8k3mμα2+6k2mα0α2+3k2mα12−4kwα2+3l2α2G′ξGξ2,+k3mα1λ2+6k3λmμα2+2k3mα1μ+6k2mα0α1−4kwα1+3l2α1G′ξGξ,+k3mα1λμ+2k3mα2μ2+3k2mα02−4α0kw+3α0l2+d=0.For theG′ξ/Gξ function with the same homomorphism power, its undetermined coefficients are extracted and set to zero to obtain the following equations:(63)6k3mα2+3k2mα22=0,10k2λmα2+2k3mα1+6k3mα1α2=0,k3mα1λ2+6k3λmμα2+2k3mα1μ+6k2mα0α1−4kmα1+3l2α1=0,k3mα1λμ+2k3mα2μ2+3k2mα02−4α0kw+3α0l2+d=0.Solving the above simultaneous equations, we get(64)d=−k6λ4m2−8k6λ2m2μ+16k6μ2m2+8k4λ2mw+64k4mμw−48k2w2+48kl2w−9l412k2m,α0=−k3λ2m+8k3mμ−4kw+3l26k2m,α1=−2kλ,α2=−2k.By using (64), (58) is written as(65)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m.Replacing the general solution for (23) and (65) with (4), we obtain the following exact solution for (4):Case 1. If Δ=λ2−4μ>0,(66)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m,where(67)G′G=−λ2+λ2−4μ2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.Now, we give three forms ofuξ.(1)
IfC1C2≠0, then G′/G=−λ/2+λ2−4μ/2C1sinhλ2−4μ/2ξ+C2coshλ2−4μ/2ξ/C1coshλ2−4μ/2ξ+C2sinhλ2−4μ/2ξ.(68)v1ξ=2k3λ2m−8k3mμ+4km−3l26k2m−kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22,(69)u1ξ=∫v1ξdξ=2k3λ2m−8k3mμ+4km−3l26k2mξ−∫kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22dξ.Because of the rational hyperbolic functions forms, we cannot accumulate the exact expressions ofu1ξ.(2)
IfC1=0,C2≠0, then G′/Gξ=−λ/2+λ2−4μ/2cothλ2−4μ/2ξ. Hence,(70)v2ξ,=2k3λ2m−8k3mμ+4km−3l26k2m−2kλλ2−4μcothλ2−4μ2ξ−kλ2−4μ2coth2λ2−4μ2ξ,(71)u2ξ,=∫v2ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλln|sinhλ2−4μ2ξ|,+kλ2−4μcothλ2−4μ2ξ+lncothλ2−4μ2ξ−1−lncothλ2−4μ2ξ+1+c16.(3)
IfC1≠0,C2=0, then G′/Gξ=−λ/2+λ2−4μ/2tanhλ2−4μ/2ξ. Hence,(72)v3ξ,=2k3λ2m−8k3mμ+4km−3l26k2m−2kλλ2−4μtanhλ2−4μ2ξ−kλ2−4μ2tanh2λ2−4μ2ξ,(73)u3ξ,=∫v3ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλlncoshλ2−4μ2ξ+kλ2−4μtanhλ2−4μ2ξ+lntanhλ2−4μ/2ξ−12−lntanhλ2−4μ/2ξ+12+c17,wherec16,c17 are integral numbers.Case 2. If Δ=λ2−4μ<0,(74)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m,where(75)G′G=−λ2+4μ−λ22−C1sin−λ2+4μξ/2+C2cos−λ2+4μξ/2C1cos−λ2+4μξ/2+C2sin−λ2+4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.Almost the same as case 1, it is just a little different form ofuξ. If C1C2≠0 and C1=0,C2≠0, forms of uξ are the same as case 1.IfC1≠0,C2=0, then G′/Gξ=−λ/2+λ2−4μ/2tanhλ2−4μ/2ξ.Hence,(76)v4ξ,=2k3λ2m−8k3mμ+4km−3l26k2m+2kλλ2−4μtanhλ2−4μ2ξ−kλ2−4μ2tanh2λ2−4μ2ξ,(77)u4ξ,=∫v4ξdξ=2k3λ2m−8k3mμ+4km−3l26k2mξ+4kλlncoshλ2−4μ2ξ,+kλ2−4μtanhλ2−4μ2ξ+lntanhλ2−4μ2ξ−1−lntanhλ2−4μ2ξ+1+c18,wherec18 is an integral number.Case 3. If Δ=λ2−4μ=0,(78)v5ξ=−2k2C2−C1λ−C2λξ2C1+C2ξ2−2kλ2C2−C1λ−C2λξ2C1+C2ξ−k3λ2m+8k3mμ−4km+3l26k2m,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.(79)u5ξ,=∫vξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ+2kC2C2ξ+C1+c19,wherec19 is an integral number.


## 4.1. Proof of Theorem1

Putting (9) into (4), we have q=2, p=1, A=k3m, B=3l2−4kw, C=3k2m, D=d, c−2=−2k, c−1=0, c0=4kw−3l2/6k2m, c1=0, c2=6dk2m−24k2m−3l2−34kw2/120k5m2, c3=0, where c4 is arbitrary. Hence, (4) satisfies weak 1,2 condition.By (12), we have the indeterminant rational style solutions of (4) with pole at ξ=0 that(39)v1ξ=c11ξ2+c12ξ+c.Substitutingv1ξ into (4), we obtain the following result:(40)v1ξ=−k3ξ2−3l2−4kw6k2m,where 12k2md=3l2−4kw2. So,(41)u1ξ=∫v1ξdξ=k3ξ−3l2−4kw6k2mξ+c8.The rational solutions of (4) are as follows:(42)v1,2ξ=−2kξ−ξ02−3l2−4kw6k2m,where 12k2md=3l2−4kw2,ξ0∈ℂ. So,(43)u1,2ξ=∫v1,2ξdξ=2kξ−ξ0−3l2−4kw6k2mξ+c9.Next we will seek for simply periodic solutions, settingζ=expαξ and putting v=Rζ into (4); then,(44)Aα2ζR′+ζ2R′′+BR+CR2+D=0,and putting(45)v2ζ=c2ζ−12+c1ζ−ζ1+c10into (44), we get that(46)v2ζ=−2kα2ζ−12+−2kα2ζ−1−k6α2−3l2−4kw6k2m,where 12k2md=−k6m2α4+3l2−4kw2. Substituting ζ=eαξ into above relation, we obtain simple periodic solutions (4), and the pole point is located at ξ=0.(47)vsξ,=−2kα2eαξ−12−2kα2eαξ−1−k6α2−3l2−4kw6k2m,=−2kα2eαξeαξ−12−k6α2−3l2−4kw6k2m,=−2kα2coth2α2ξ−k6α2−3l2−4kw6k2m.So,(48)usξ,=∫vsξdξ,=4kαcothαξ2+2kαlncothαξ2−1,−2kαlncothαξ2+1−k6α2−3l2−4kw6k2m+c11.So, all simply periodic solutions of (4) are obtained by(49)vsξ0ξ=−2kα2coth2α2ξ−ξ0−k6α2−3l2−4kw6k2m,where 12k2md=−k6m2α4+3l2−4kw2, ξ0∈ℂ.So,(50)usξ0ξ,=∫vsξ0ξdξ,=4kαcothαξ−ξ02+2kαlncothαξ−ξ02−1,−2kαlncothαξ−ξ02+1−k6α2ξ−3l2−4kw6k2mξ+c12,where c8,c9,c10,c11,c12 are constants.From (11), we have indeterminant relations of elliptic solutions of (4) and the pole point is located at ξ=0.(51)veξ=c−2℘ξ+c0.Substitutingveξ into the (4), we obtain that(52)veξ=−2k℘ξ−3l2−4kw6k2m.Here12dk2m=−12k3mg2+3l2−4kw2. So,(53)ueξ=∫veξdξ=−2kℜξ−3l2−4kw6k2mξ+c13,where ℜξ=∫℘ξdξ is also an elliptic function.Thus, all the elliptic solutions for (4) are as follows:(54)veξ0ξ=−2k℘ξ−ξ0−3l2−4kw6k2m.So,(55)ueξ0ξ=∫veξ0ξdξ=−2kℜξ−ξ0−3l2−4kw6k2mξ+c14,where ξ0∈ℂ. By using (7), we rewrite it to the form(56)veξ0ξ=−2k−℘ξ+14℘′ξ+F℘ξ−E2+2kE−3l2−4kw6k2m,where 12dk2m=−12k3mg2+3l2−4kw2. Then, from (55) and (7), we deduce that(57)ueξ0ξ=∫veξ0ξdξ=−2k−ℜξ+14ℜ′ξ+Fℜξ−E2+2kE−3l2−4kw6k2mξ+c15,where c13,c14,c15 are integral numbers, and here 12k2md=−12k3mg2+3l2−4kw2, F2=4E3−g2E−g3, where g3 and E=ℜξ0 are arbitrary constants..The proof is over.


## 4.2. Proof of Theorem2

Considering the homogeneous equilibrium betweenv″ and v2 in (4), we get m=2. So, we can set the solution of (4) as follows:(58)v=α2G′G2+α1G′G+α0;α2≠0.α0,α1,α2 here are constants to be determined later.Next, we will use (58) and (23) together and collect all terms of the same power as G′/G.First, from (58), we get(59)v′ξ=2α2G′ξG′′ξGξ2−2α2G′ξ3Gξ3+α1G′′ξGξ−α1G′ξ2Gξ2.Substituting (23) into (59), we obtain(60)v′ξ=−2α2G′ξ2λGξ2−2α2G′ξμGξ−2α2G′ξ3Gξ3−α1λG′ξGξ−α1μ−α1G′ξ2Gξ2,and then we deduce(61)v′′ξ,=6α2G′ξλμGξ+2α2μ2+8α2G′ξ2μGξ2+4α2G′ξ2λ2Gξ2,+10α2G′ξ3λGξ3+6α2G′ξ4Gξ4+3α1G′ξ2λGξ2+α1G′ξλ2Gξ,+α1λμ+2α1G′ξμGξ+2α1G′ξ3Gξ3.Now, we put (58), (60), and (61) into (4) and collect all terms with the same power of G′ξ/Gξ together:(62)6k3mα2+3k2mα22G′ξGξ4+10k3λmα2+2k3mα1+6k2mα1α2G′ξGξ3,+4k3mα2λ2+3k3mα1λ+8k3mμα2+6k2mα0α2+3k2mα12−4kwα2+3l2α2G′ξGξ2,+k3mα1λ2+6k3λmμα2+2k3mα1μ+6k2mα0α1−4kwα1+3l2α1G′ξGξ,+k3mα1λμ+2k3mα2μ2+3k2mα02−4α0kw+3α0l2+d=0.For theG′ξ/Gξ function with the same homomorphism power, its undetermined coefficients are extracted and set to zero to obtain the following equations:(63)6k3mα2+3k2mα22=0,10k2λmα2+2k3mα1+6k3mα1α2=0,k3mα1λ2+6k3λmμα2+2k3mα1μ+6k2mα0α1−4kmα1+3l2α1=0,k3mα1λμ+2k3mα2μ2+3k2mα02−4α0kw+3α0l2+d=0.Solving the above simultaneous equations, we get(64)d=−k6λ4m2−8k6λ2m2μ+16k6μ2m2+8k4λ2mw+64k4mμw−48k2w2+48kl2w−9l412k2m,α0=−k3λ2m+8k3mμ−4kw+3l26k2m,α1=−2kλ,α2=−2k.By using (64), (58) is written as(65)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m.Replacing the general solution for (23) and (65) with (4), we obtain the following exact solution for (4):Case 1. If Δ=λ2−4μ>0,(66)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m,where(67)G′G=−λ2+λ2−4μ2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.Now, we give three forms ofuξ.(1)
IfC1C2≠0, then G′/G=−λ/2+λ2−4μ/2C1sinhλ2−4μ/2ξ+C2coshλ2−4μ/2ξ/C1coshλ2−4μ/2ξ+C2sinhλ2−4μ/2ξ.(68)v1ξ=2k3λ2m−8k3mμ+4km−3l26k2m−kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22,(69)u1ξ=∫v1ξdξ=2k3λ2m−8k3mμ+4km−3l26k2mξ−∫kλ2−4μk2C1sinhλ2−4μξ/2+C2coshλ2−4μξ/2C1coshλ2−4μξ/2+C2sinhλ2−4μξ/22dξ.Because of the rational hyperbolic functions forms, we cannot accumulate the exact expressions ofu1ξ.(2)
IfC1=0,C2≠0, then G′/Gξ=−λ/2+λ2−4μ/2cothλ2−4μ/2ξ. Hence,(70)v2ξ,=2k3λ2m−8k3mμ+4km−3l26k2m−2kλλ2−4μcothλ2−4μ2ξ−kλ2−4μ2coth2λ2−4μ2ξ,(71)u2ξ,=∫v2ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλln|sinhλ2−4μ2ξ|,+kλ2−4μcothλ2−4μ2ξ+lncothλ2−4μ2ξ−1−lncothλ2−4μ2ξ+1+c16.(3)
IfC1≠0,C2=0, then G′/Gξ=−λ/2+λ2−4μ/2tanhλ2−4μ/2ξ. Hence,(72)v3ξ,=2k3λ2m−8k3mμ+4km−3l26k2m−2kλλ2−4μtanhλ2−4μ2ξ−kλ2−4μ2tanh2λ2−4μ2ξ,(73)u3ξ,=∫v3ξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ−4kλlncoshλ2−4μ2ξ+kλ2−4μtanhλ2−4μ2ξ+lntanhλ2−4μ/2ξ−12−lntanhλ2−4μ/2ξ+12+c17,wherec16,c17 are integral numbers.Case 2. If Δ=λ2−4μ<0,(74)vξ=−2kG′G2−2kλG′G−k3λ2m+8k3mμ−4km+3l26k2m,where(75)G′G=−λ2+4μ−λ22−C1sin−λ2+4μξ/2+C2cos−λ2+4μξ/2C1cos−λ2+4μξ/2+C2sin−λ2+4μξ/2,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.Almost the same as case 1, it is just a little different form ofuξ. If C1C2≠0 and C1=0,C2≠0, forms of uξ are the same as case 1.IfC1≠0,C2=0, then G′/Gξ=−λ/2+λ2−4μ/2tanhλ2−4μ/2ξ.Hence,(76)v4ξ,=2k3λ2m−8k3mμ+4km−3l26k2m+2kλλ2−4μtanhλ2−4μ2ξ−kλ2−4μ2tanh2λ2−4μ2ξ,(77)u4ξ,=∫v4ξdξ=2k3λ2m−8k3mμ+4km−3l26k2mξ+4kλlncoshλ2−4μ2ξ,+kλ2−4μtanhλ2−4μ2ξ+lntanhλ2−4μ2ξ−1−lntanhλ2−4μ2ξ+1+c18,wherec18 is an integral number.Case 3. If Δ=λ2−4μ=0,(78)v5ξ=−2k2C2−C1λ−C2λξ2C1+C2ξ2−2kλ2C2−C1λ−C2λξ2C1+C2ξ−k3λ2m+8k3mμ−4km+3l26k2m,andξ=kx+ly+mz+wt. C1 and C2 are arbitrary constants.(79)u5ξ,=∫vξdξ,=2k3λ2m−8k3mμ+4km−3l26k2mξ+2kC2C2ξ+C1+c19,wherec19 is an integral number.


## 5. Comparison

In this paper, about equation (4), we consider the case where integral constant d is not zero. This point is different from reference [31]. Obviously, we consider a much more complex scenario. By the G′/G-expansion method and the extended complex method, we found various forms for solutions of potential Yu–Toda–Sasa–Fukuyama equation. Using the extended complex method, we found elliptic function solutions, hyperbolic function solutions, and rational solutions. We also can find more solutions by the G′/G-expansion method including hyperbolic function solutions and rational solutions. However, only by applying extended complex methods, we can obtain elliptic function solutions. If the appropriate parameters are selected, both methods obtain the same rational solution. From the perspective of the calculation process, the extended complex method is more concise and clear. The G′/G-expansion method is relatively complex, but more formal solutions can be obtained. Both methods have their own characteristics.We firmly believe that the extended complex andG′/G-expansion methods proposed in this paper can be more efficiently applied to find solutions to other nonlinear evolution equations. Both methods are effective tools to obtain exact solutions of the nonlinear evolution equations.


## 6. Computer Simulations

In this section, we attempt to interpret the results obtained by two different methods by computer simulated images and further analyze the properties ofusξ0ξ, u2ξ, and u4ξ in the 3+1−dimensional potential-YTSF equation.Figure1 shows the 3D images of usξ0ξ by considering the values α=3, ξ0=5, k=2, l=4, m=3, w=5, z=2, and d=−2912/2; t takes three different values: t=−2, t=0, and t=2. Three graphs demonstrate poles of solution usξ0ξ on the domain.Figure 1
The three-dimensional (3D) images ofusξ0ξ in three different situations.
(a)(b)(c)Figure2 shows the 3D images of u2ξ by considering the values k=1, λ=4, m=2, μ=3, l=3, w=5, and z=3; t takes the following three different values: t=−2, t=0, and t=2. However, the images change dramatically accompanied with different t.Figure 2
The 3D images ofu2ξ in three different situations.
(a)(b)(c)Figure3 shows the 3D images of u4ξ by considering the values k=1, λ=2, m=2, μ=2, l=3, w=5, and z=3; t takes the following three different values: t=−2, t=0, and t=2. Three graphs show the singularities on the domain.Figure 3
The 3D images ofu4ξ in three different situations.
(a)(b)(c)Presented profiles of the above 3D graphs will help to understand wave function fluctuations dynamics to a greater extent.Figure4 shows the 2D images of usξ0z by taking the values α=3, ξ0=5, k=2, l=4, m=3, w=5, y=0, z=2, and d=−2912/2; t takes the following three different values: t=−2, t=0, and t=2.Figure 4
The two-dimensional (2D) images ofus0ξ in three different situations.
(a)(b)(c)Figure5 shows the 2D images of u2ξ by taking the values k=1, λ=4, m=2, μ=3, l=3, w=5, y=0,andz=3; t takes the following three different values: t=−2, t=0, and t=2.Figure 5
The 2D images ofu2ξ in three situations.
(a)(b)(c)Figure6 shows the 2D images of u4ξ by taking the values k=1, λ=2, m=2, μ=2, l=3, w=5, y=0, and z=3; t takes the following three different values: t=−2, t=0, and t=2.Figure 6
The 2D images ofu4ξ in three situations.
(a)(b)(c)The above 2D images all intuitively and clearly reflect the high frequencies and amplitude of function fluctuations.


## 7. Conclusions and Future Outlook of the Study

By traveling wave transformation, many of the PDEs can be converted into nonlinear differential equations similar to equation (4). In this paper, we employed the extended complex method and the G′/G-expansion method to study the exact solutions about the potential Yu–Toda–Sasa–Fukuyama equation. By traveling wave transformation, we can reduce the dimensions of the PDEs to nonlinear evolution differential equations related to mathematical physics and engineering.The results in the full text strongly demonstrate that the above methods are very effective and powerful in solving the exact solutions of the nonlinear evolution equations. We can apply these ideas and methods to other nonlinear evolution equations. Our work shows that there is a simple periodic solution,usξ0ξ, which does not continuously degenerate successively to the elliptic function solutions.In the future, we may consider the unifiedF-expansion method [32] and generalized direct algebraic method [33] to further investigate the potential Yu–Toda–Sasa–Fukuyama equation. In 2021, Behera et al. [34] used the modified G′/G2-expansion method for investigating the exact solutions of the traveling waves about Drinfeld–Sokolov–Wilson (DSW) equation and Benjamin–Ono (BO) equation. This is also a new research direction that we will focus on and have great interest in the future.



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*Source: 2179375-2022-02-21.xml*