# Abundant Soliton Structures to the (2+1)-Dimensional Heisenberg Ferromagnetic Spin Chain Dynamical Model

Authors: Kang-Jia Wang; Feng Shi; Guo-Dong Wang
Journal: Advances in Mathematical Physics
 (2023)

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## Abstract

In this paper, we aim to investigate the (2+1)-dimensional Heisenberg ferromagnetic spin chain equation that is used to describe the nonlinear dynamics of magnets. Two recent effective technologies, namely, the variational method and subequation method, are employed to construct the abundant soliton solutions. By these two methods, diverse solutions such as the bright soliton, dark soliton, bright-dark soliton, perfect periodic soliton, and singular periodic soliton are successfully extracted. The numerical results are illustrated in the form of 3-D plots and 2-D curves by choosing proper parametric values to interpret the dynamics of wave profiles. Finally, the physical interpretation of the acquired results is elaborated in detail. The results obtained in this study are helpful to explain some physical meanings of some nonlinear physical models in electromagnetic waves.


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## Body

## 1. Introduction

Many complex problems arising in nature can be described by nonlinear partial differential equations (NLPDEs). The exact solutions of the NLPDEs have been a hot topic for mathematicians and physicists. The soliton travels in a nonlinear media while maintaining a constant size, shape, and energy. The study of the soliton solutions can provide a theoretical reference for the research of nonlinear physical models, soliton control, optical switching equipment, and so on. Up to now, many effective and powerful methods have been obtained for constructing the soliton solutions of the NLPDEs: the extended trial equation method [1, 2], generalized exponential rational function method [3], sine-Gordon expansion method [4–6], generalized (G′/G) expansion method [7], extended rational sine-cosine and sinh-cosh method [8, 9], F-expansion method [10–12], Cole-Hopf transformation [13, 14], exp-function method [15–17], extended tanh-function method [18–20], Bäcklund transformation [21], Sardar subequation method [22, 23], and so on [24–31]. In this paper, we aim to study the (2+1)-dimensional Heisenberg ferromagnetic spin chain equation (HFSCE) that is given by [32]:
(1)iχt+γ1χxx+γ2χyy+γ3χxy−γ4χ2χ=0,where γ1=λ4v+v2, γ2=λ4v1+v2, γ3=3λv2, and γ4=3λ4A. λ is the lattice parameter, v and v1 are the coefficients of bilinear exchange interactions along the x- and y-directions, respectively. v2 refers to the neighboring interaction along the diagonal, and A represents the uniaxial crystal field anisotropy parameter.The (2+1)-dimensional HFSCE is used to describe the nonlinear dynamics of magnets. In [33], the two methods, improved auxiliary equation and generalized Riccati mapping methods, are used to seek for the diverse solitons. In [34], the Jacobi elliptic expansion method is adopted and bright and dark soliton solutions are obtained. In [35], the theory of dynamical systems is introduced for constructing the bounded traveling wave solutions. In [36], the modified Khater method is carried out to deal with Eq. (1). The generalized tanh methods is employed to solve Eq. (1) in [37], and dark, dark-bright, or combined optical and singular soliton solutions are attained. In [38], the authors use two approaches, namely, the modified F-expansion and projective Riccati equation methods to investigate (Eq. (1)). Inspired by these research results, here in this work, we mainly seek its abundant soliton solutions by using two effective methods, the variational method (VM) and subequation method (SEM). The VM which is based on the variational theory can provide the optimized solutions in only one step via the stationary condition. In addition, the SEM that is straightforward and effective can construct abundant solutions via the symbolic calculation. The structure of this paper is organized as follows. In Section 2, we introduce the algorithms of the two methods. In Section 3, the two methods are applied to find the soliton solutions. In Section 3, the behaviors of the solutions and their physical interpretations are presented. Finally, a conclusion is reached in Section 5.


## 2. The Two Methods

This section is to give a brief introduction to the VM and SEM. Considering a partial differential equation (PDE) with the form as(2)Fu,ux,uxx,uxxx,..⋯,ut,..⋯=0..Applying the following transformation,(3)ux,t=Uξ,ξ=kx+rt,where k and r are arbitrary nonzero constants. Substituting Eq. (3) into Eq. (2), we can convert Eq. (3) into an ordinary differential equation (ODE):
(4)FU,Uξ,Uξξ,Uξξξ,.⋯,..⋯=0.


### 2.1. The VM

Aided by the semi-inverse method [39–45], the variational principle of Eq. (4) can be developed as
(5)JU=∫LU,U′,U″,⋯dξ.Assuming that the solutions of Eq. (4) take the following different forms.Hypothesis one:Uξ=H1sechξ.Hypothesis two:Uξ=H2sech2ξ.Hypothesis three:Uξ=H3sinhξ/coshξ3/2.Hypothesis four:Uξ=H4cosϖξ.Then, taking the above-assumed expressions into Eq. (5) and applying the stationary conditions asResult one:dJ/dH1=0.Result two:dJ/dH2=0.Result three:dJ/dH3=0.Result four:dJ/dH4=0.Solving the above different results, the expressions ofH1, H2, H3, and H4 can be easily determined.


### 2.2. The SEM

By the SEM [46–49], the solution of Eq. (4) can be assumed as
(6)Uξ=∑i=0cσiφiξ,where σii=0,1,2.⋯,c are the constants to be determined later. There is
(7)φ′ξ=Ξ+φ2ξ.Here,Ξ is a constant. Substituting Eq. (6) together with Eq. (7) into Eq. (4), we can determine the value of c by the balancing theory. With the determined c, we can get an overdetermined system of nonlinear algebraic equations by setting the coefficients of each power of φγ to zero. Solving it, we can determine the expression of Eq. (6).The solutions of Eq.(7) for the different conditions are
(8)φγ=−−Ξtanh−Ξξ,Ξ<0,−−Ξcoth−Ξξ,Ξ<0,ΞtanΞξ,Ξ>0,−ΞcotΞξ,Ξ>0,−1ξ+μ,μisaconstant,Ξ=0.


## 2.1. The VM

Aided by the semi-inverse method [39–45], the variational principle of Eq. (4) can be developed as
(5)JU=∫LU,U′,U″,⋯dξ.Assuming that the solutions of Eq. (4) take the following different forms.Hypothesis one:Uξ=H1sechξ.Hypothesis two:Uξ=H2sech2ξ.Hypothesis three:Uξ=H3sinhξ/coshξ3/2.Hypothesis four:Uξ=H4cosϖξ.Then, taking the above-assumed expressions into Eq. (5) and applying the stationary conditions asResult one:dJ/dH1=0.Result two:dJ/dH2=0.Result three:dJ/dH3=0.Result four:dJ/dH4=0.Solving the above different results, the expressions ofH1, H2, H3, and H4 can be easily determined.


## 2.2. The SEM

By the SEM [46–49], the solution of Eq. (4) can be assumed as
(6)Uξ=∑i=0cσiφiξ,where σii=0,1,2.⋯,c are the constants to be determined later. There is
(7)φ′ξ=Ξ+φ2ξ.Here,Ξ is a constant. Substituting Eq. (6) together with Eq. (7) into Eq. (4), we can determine the value of c by the balancing theory. With the determined c, we can get an overdetermined system of nonlinear algebraic equations by setting the coefficients of each power of φγ to zero. Solving it, we can determine the expression of Eq. (6).The solutions of Eq.(7) for the different conditions are
(8)φγ=−−Ξtanh−Ξξ,Ξ<0,−−Ξcoth−Ξξ,Ξ<0,ΞtanΞξ,Ξ>0,−ΞcotΞξ,Ξ>0,−1ξ+μ,μisaconstant,Ξ=0.


## 3. Applications

The following complex transformation is introduced:(9)χ=ψξeiϕ,ξ=ax+by−ηt,ϕ=mx+ny−kt+ϕ0.Taking Eq. (9) into Eq. (1) produces the following the imaginary and real parts, respectively, as
(10)η=2aγ1m+2bγ2n+γ3bm+an,(11)k−γ1m2−nγ2n+γ3mψ−γ4ψ3+γ1a2+γ2b2+γ3abψ″=0,where ψ″=d2ψ/dξ2.


### 3.1. Application of the VM

With help of the semi-inverse method, we can get the Euler-Lagrange equation of Eq. (11) as
(12)ℓ=12k−γ1m2−nγ2n+γ3mψ2−34γ4ψ4−12γ1a2+γ2b2+γ3abψ′2.So, the variational formulation of Eq. (11) can be established as
(13)Jψ=∫12k−γ1m2−nγ2n+γ3mψ2−34γ4ψ4−12γ1a2+γ2b2+γ3abψ′2dξ.For simplicity, we rewrite Eq. (13) as
(14)Jψ=∫μ1ψ2−μ2ψ4−μ3ψ′2dξ,where μ1=1/2k−γ1m2−nγ2n+γ3m, μ2=3/4γ4, and μ3=1/2γ1a2+γ2b2+γ3ab.


#### 3.1.1. Bright Soliton

We assume the solution of Eq. (11) has the following form:
(15)ψξ=Asechξ.Bringing the above equation into Eq. (14) yields
(16)JA=∫0∞μ1A2sechξ2−μ2A4sechξ4−μ3A2sechξ2tanhξ2dξ=−A23μ3−3μ1+2A2μ2.Taking the stationary value condition for Eq. (16) with respect to A, we attain
(17)−23Aμ3−3μ1+4A2μ2=0,which leads to
(18)A=±3μ1−μ34μ2.With this, the solution of Eq. (11) can be obtained as
(19)ψξ=±3μ1−μ34μ2sechξ.Thus, the bright soliton solution of Eq. (1) is attained as
(20)χ1±x,y,t=±3μ1−μ34μ2sechax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


#### 3.1.2. Bright-Like Soliton

We can also suppose the solution of Eq. (11) as
(21)ψξ=Asech2ξ.In the same way, substituting Eq. (21) into Eq. (14) results in
(22)JA=−4105A28μ3−35μ1+48A2μ2.Taking the stationary value condition of Eq. (22) yields
(23)−2105A228μ3−35μ1+24A2μ2=0.Solving Eq. (23), we have
(24)A=±75μ1−4μ348μ2.Thus, we can get the solution of Eq. (4) as
(25)ψξ=±75μ1−4μ348μ2sech2ξ.Then, the bright-like soliton solution of Eq. (1) is found as
(26)χ2±x,y,t=±75μ1−4μ348μ2sech2ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


#### 3.1.3. Bright-Dark Soliton

The solution of Eq. (11) can be assumed as
(27)ψζ=Asinhξcoshξ3/2.Taking the above equation into Eq. (14) yields
(28)Jψ=∫0∞μ1Asinhξcoshξ3/22−μ2Asinhξcoshξ3/24−μ3Acoshξ−32Asinh2ξcoshξ5/22dξ=−1320A264A2μ2+55μ3π−80μ1π.Computing the stationary condition of the above equation as(29)∂J∂A=0,which leads to
(30)−4A3μ25+132A16μ1−11μ3π=0.So, we can determine the expression ofA as
(31)A=±18516μ1−11μ3π2μ2.Then, Eq. (27) can be written as
(32)ψζ=±18516μ1−11μ3π2μ2sinhξcoshξ3/2.Thus, we gain the bright-dark soliton solution of Eq. (1) as
(33)χ3±x,y,t=±18516μ1−11μ3π2μ2sinhax+by−2aγ1m+2bγ2n+γ3bm+antcoshax+by−2aγ1m+2bγ2n+γ3bm+ant3/2eimx+ny−kt+ϕ0.


#### 3.1.4. Periodic Soliton

The periodic solution of Eq. (11) is assumed as
(34)ψξ=Θcosϖξ,ϖ>0,,where ϖ is the frequency to be further determined. Injecting Eq. (34) into Eq. (7), there is
(35)Jψ=∫0T/4μ1Θcosϖξ2−μ2Θcosϖξ4−μ3−Θϖsinϖξ2dξ=−Θ2164ϖ2μ3−4μ1+3Θ2nπ.Taking its stationary value condition with respect toΘ yields
(36)−Θ42ϖ2μ3−2μ1+3Θ2nπ=0.From which, we can determine the frequency as(37)ϖ=2μ1−3Θ2μ22μ3>0.Therefore, we attain the periodic solution of Eq. (4) as
(38)ψξ=Θcos2μ1−3Θ2μ22μ3ξ.So, the periodic soliton solution of Eq. (1) is obtained as
(39)χ4±x,y,t=Θcos2μ1−3Θ2μ22μ3ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


### 3.2. Application of the SEM

To use the SEM, we first rewrite Eq. (11) as
(40)ψ″+δ1ψ−δ2ψ3=0,where δ1=k−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab and δ2=γ4/γ1a2+γ2b2+γ3ab.Based on the SEM, the solution of Eq. (40) is considered as
(41)ψξ=∑i=0cσiφiξ.Balancingψ″ and ψ3, it gives
(42)c=1,so Eq. (41) becomes
(43)ψξ=σ0+σ1φξ.And there is(44)φ′ξ=Ξ+φ2ξ.Taking Eq. (43) along with Eq. (44) into Eq. (40) results intoφ0: δ0σ1−δ03σ2=0,φ1: 2Ξσ1+σ1δ1−3σ02σ1δ2=0,φ2: −3σ0σ12δ2=0,φ3:2σ1−σ13δ2=0.Solving the above equations, we have(45)σ0=0,σ1=±2δ2,Ξ=δ12.Then, we get the solutions as follows:Case 1.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab<0, we have
(46)χ5±x,y,t=±−k−γ1m2−nγ2n+γ3mγ4tanh−k−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0,(47)χ6±x,y,t=±−k−γ1m2−nγ2n+γ3mγ4coth−k−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0.Case 2.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab>0, we have
(48)χ7±x,y,t=±k−γ1m2−nγ2n+γ3mγ4tank−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0,χ8±x,y,t=±k−γ1m2−nγ2n+γ3mγ4cotk−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0.Case 3.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab=0, we have
(49)χ9±x,y,t=±2γ1a2+γ2b2+γ3abγ41k−γ1m2−nγ2n+γ3m/2γ1a2+γ2b2+γ3ab−1/2ax+by−ηt+μ.
For Eqs. (46)~(49), there is η=2aγ1m+2bγ2n+γ3bm+an.


## 3.1. Application of the VM

With help of the semi-inverse method, we can get the Euler-Lagrange equation of Eq. (11) as
(12)ℓ=12k−γ1m2−nγ2n+γ3mψ2−34γ4ψ4−12γ1a2+γ2b2+γ3abψ′2.So, the variational formulation of Eq. (11) can be established as
(13)Jψ=∫12k−γ1m2−nγ2n+γ3mψ2−34γ4ψ4−12γ1a2+γ2b2+γ3abψ′2dξ.For simplicity, we rewrite Eq. (13) as
(14)Jψ=∫μ1ψ2−μ2ψ4−μ3ψ′2dξ,where μ1=1/2k−γ1m2−nγ2n+γ3m, μ2=3/4γ4, and μ3=1/2γ1a2+γ2b2+γ3ab.


### 3.1.1. Bright Soliton

We assume the solution of Eq. (11) has the following form:
(15)ψξ=Asechξ.Bringing the above equation into Eq. (14) yields
(16)JA=∫0∞μ1A2sechξ2−μ2A4sechξ4−μ3A2sechξ2tanhξ2dξ=−A23μ3−3μ1+2A2μ2.Taking the stationary value condition for Eq. (16) with respect to A, we attain
(17)−23Aμ3−3μ1+4A2μ2=0,which leads to
(18)A=±3μ1−μ34μ2.With this, the solution of Eq. (11) can be obtained as
(19)ψξ=±3μ1−μ34μ2sechξ.Thus, the bright soliton solution of Eq. (1) is attained as
(20)χ1±x,y,t=±3μ1−μ34μ2sechax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


### 3.1.2. Bright-Like Soliton

We can also suppose the solution of Eq. (11) as
(21)ψξ=Asech2ξ.In the same way, substituting Eq. (21) into Eq. (14) results in
(22)JA=−4105A28μ3−35μ1+48A2μ2.Taking the stationary value condition of Eq. (22) yields
(23)−2105A228μ3−35μ1+24A2μ2=0.Solving Eq. (23), we have
(24)A=±75μ1−4μ348μ2.Thus, we can get the solution of Eq. (4) as
(25)ψξ=±75μ1−4μ348μ2sech2ξ.Then, the bright-like soliton solution of Eq. (1) is found as
(26)χ2±x,y,t=±75μ1−4μ348μ2sech2ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


### 3.1.3. Bright-Dark Soliton

The solution of Eq. (11) can be assumed as
(27)ψζ=Asinhξcoshξ3/2.Taking the above equation into Eq. (14) yields
(28)Jψ=∫0∞μ1Asinhξcoshξ3/22−μ2Asinhξcoshξ3/24−μ3Acoshξ−32Asinh2ξcoshξ5/22dξ=−1320A264A2μ2+55μ3π−80μ1π.Computing the stationary condition of the above equation as(29)∂J∂A=0,which leads to
(30)−4A3μ25+132A16μ1−11μ3π=0.So, we can determine the expression ofA as
(31)A=±18516μ1−11μ3π2μ2.Then, Eq. (27) can be written as
(32)ψζ=±18516μ1−11μ3π2μ2sinhξcoshξ3/2.Thus, we gain the bright-dark soliton solution of Eq. (1) as
(33)χ3±x,y,t=±18516μ1−11μ3π2μ2sinhax+by−2aγ1m+2bγ2n+γ3bm+antcoshax+by−2aγ1m+2bγ2n+γ3bm+ant3/2eimx+ny−kt+ϕ0.


### 3.1.4. Periodic Soliton

The periodic solution of Eq. (11) is assumed as
(34)ψξ=Θcosϖξ,ϖ>0,,where ϖ is the frequency to be further determined. Injecting Eq. (34) into Eq. (7), there is
(35)Jψ=∫0T/4μ1Θcosϖξ2−μ2Θcosϖξ4−μ3−Θϖsinϖξ2dξ=−Θ2164ϖ2μ3−4μ1+3Θ2nπ.Taking its stationary value condition with respect toΘ yields
(36)−Θ42ϖ2μ3−2μ1+3Θ2nπ=0.From which, we can determine the frequency as(37)ϖ=2μ1−3Θ2μ22μ3>0.Therefore, we attain the periodic solution of Eq. (4) as
(38)ψξ=Θcos2μ1−3Θ2μ22μ3ξ.So, the periodic soliton solution of Eq. (1) is obtained as
(39)χ4±x,y,t=Θcos2μ1−3Θ2μ22μ3ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


## 3.1.1. Bright Soliton

We assume the solution of Eq. (11) has the following form:
(15)ψξ=Asechξ.Bringing the above equation into Eq. (14) yields
(16)JA=∫0∞μ1A2sechξ2−μ2A4sechξ4−μ3A2sechξ2tanhξ2dξ=−A23μ3−3μ1+2A2μ2.Taking the stationary value condition for Eq. (16) with respect to A, we attain
(17)−23Aμ3−3μ1+4A2μ2=0,which leads to
(18)A=±3μ1−μ34μ2.With this, the solution of Eq. (11) can be obtained as
(19)ψξ=±3μ1−μ34μ2sechξ.Thus, the bright soliton solution of Eq. (1) is attained as
(20)χ1±x,y,t=±3μ1−μ34μ2sechax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


## 3.1.2. Bright-Like Soliton

We can also suppose the solution of Eq. (11) as
(21)ψξ=Asech2ξ.In the same way, substituting Eq. (21) into Eq. (14) results in
(22)JA=−4105A28μ3−35μ1+48A2μ2.Taking the stationary value condition of Eq. (22) yields
(23)−2105A228μ3−35μ1+24A2μ2=0.Solving Eq. (23), we have
(24)A=±75μ1−4μ348μ2.Thus, we can get the solution of Eq. (4) as
(25)ψξ=±75μ1−4μ348μ2sech2ξ.Then, the bright-like soliton solution of Eq. (1) is found as
(26)χ2±x,y,t=±75μ1−4μ348μ2sech2ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


## 3.1.3. Bright-Dark Soliton

The solution of Eq. (11) can be assumed as
(27)ψζ=Asinhξcoshξ3/2.Taking the above equation into Eq. (14) yields
(28)Jψ=∫0∞μ1Asinhξcoshξ3/22−μ2Asinhξcoshξ3/24−μ3Acoshξ−32Asinh2ξcoshξ5/22dξ=−1320A264A2μ2+55μ3π−80μ1π.Computing the stationary condition of the above equation as(29)∂J∂A=0,which leads to
(30)−4A3μ25+132A16μ1−11μ3π=0.So, we can determine the expression ofA as
(31)A=±18516μ1−11μ3π2μ2.Then, Eq. (27) can be written as
(32)ψζ=±18516μ1−11μ3π2μ2sinhξcoshξ3/2.Thus, we gain the bright-dark soliton solution of Eq. (1) as
(33)χ3±x,y,t=±18516μ1−11μ3π2μ2sinhax+by−2aγ1m+2bγ2n+γ3bm+antcoshax+by−2aγ1m+2bγ2n+γ3bm+ant3/2eimx+ny−kt+ϕ0.


## 3.1.4. Periodic Soliton

The periodic solution of Eq. (11) is assumed as
(34)ψξ=Θcosϖξ,ϖ>0,,where ϖ is the frequency to be further determined. Injecting Eq. (34) into Eq. (7), there is
(35)Jψ=∫0T/4μ1Θcosϖξ2−μ2Θcosϖξ4−μ3−Θϖsinϖξ2dξ=−Θ2164ϖ2μ3−4μ1+3Θ2nπ.Taking its stationary value condition with respect toΘ yields
(36)−Θ42ϖ2μ3−2μ1+3Θ2nπ=0.From which, we can determine the frequency as(37)ϖ=2μ1−3Θ2μ22μ3>0.Therefore, we attain the periodic solution of Eq. (4) as
(38)ψξ=Θcos2μ1−3Θ2μ22μ3ξ.So, the periodic soliton solution of Eq. (1) is obtained as
(39)χ4±x,y,t=Θcos2μ1−3Θ2μ22μ3ax+by−2aγ1m+2bγ2n+γ3bm+anteimx+ny−kt+ϕ0.


## 3.2. Application of the SEM

To use the SEM, we first rewrite Eq. (11) as
(40)ψ″+δ1ψ−δ2ψ3=0,where δ1=k−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab and δ2=γ4/γ1a2+γ2b2+γ3ab.Based on the SEM, the solution of Eq. (40) is considered as
(41)ψξ=∑i=0cσiφiξ.Balancingψ″ and ψ3, it gives
(42)c=1,so Eq. (41) becomes
(43)ψξ=σ0+σ1φξ.And there is(44)φ′ξ=Ξ+φ2ξ.Taking Eq. (43) along with Eq. (44) into Eq. (40) results intoφ0: δ0σ1−δ03σ2=0,φ1: 2Ξσ1+σ1δ1−3σ02σ1δ2=0,φ2: −3σ0σ12δ2=0,φ3:2σ1−σ13δ2=0.Solving the above equations, we have(45)σ0=0,σ1=±2δ2,Ξ=δ12.Then, we get the solutions as follows:Case 1.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab<0, we have
(46)χ5±x,y,t=±−k−γ1m2−nγ2n+γ3mγ4tanh−k−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0,(47)χ6±x,y,t=±−k−γ1m2−nγ2n+γ3mγ4coth−k−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0.Case 2.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab>0, we have
(48)χ7±x,y,t=±k−γ1m2−nγ2n+γ3mγ4tank−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0,χ8±x,y,t=±k−γ1m2−nγ2n+γ3mγ4cotk−γ1m2−nγ2n+γ3m2γ1a2+γ2b2+γ3abax+by−ηteimx+ny−kt+ϕ0.Case 3.
Fork−γ1m2−nγ2n+γ3m/γ1a2+γ2b2+γ3ab=0, we have
(49)χ9±x,y,t=±2γ1a2+γ2b2+γ3abγ41k−γ1m2−nγ2n+γ3m/2γ1a2+γ2b2+γ3ab−1/2ax+by−ηt+μ.
For Eqs. (46)~(49), there is η=2aγ1m+2bγ2n+γ3bm+an.


## 4. Numerical Results and Physical Interpretation

In this section, we mainly illustrate the results gained in Section3 through the 3-D plot and 2-D curve to interpret the dynamics of wave profiles by selecting the proper parameters.We plot the 3-D plots of absolute value of the solutionsχ1+x,y,t, χ3+x,y,t, χ4+x,y,t, χ5+x,y,t, and χ7+x,y,t in Figures 1(a), 2(a), 3(a), 4(a), and 5(a), respectively, for t=0. Besides, we also draw the 2-D curves of the absolute value of the solutions χ1+x,y,t, χ3+x,y,t, χ4+x,y,t, χ5+x,y,t, and χ7+x,y,t in Figures 1(b), 2(b), 3(b), 4(b), and 5(b), respectively, for y=0 and t=0. It can be observed that the absolute values of χ1+x,y,t and χ5+x,y,t are the bright soliton wave and dark soliton, the absolute values of χ3+x,y,t is the bright-dark soliton, and the absolute values of χ4+x,y,t and χ7+x,y,t are perfect periodic soliton and singular periodic soliton.Figure 1
The behavior ofχ1+x,y,t with the parameters λ=1, v=1, v1=1, v2=−2, A=1, a=1, b=1, m=1, n=1, k=1, and ϕ0=0. (a) The 3-D plot for t=0. (b) The 2-D curve for y=0 and t=0.
(a)(b)Figure 2
The behavior ofχ3+x,y,t with the parameters λ=1, v=1, v1=1, v2=−2, A=1, a=1, b=1, m=1, n=1, k=1, and ϕ0=0. (a) The 3-D plot for t=0. (b) The 2-D curve for y=0 and t=0.
(a)(b)Figure 3
The behavior ofχ4+x,y,t with the parameters Θ=2, λ=1, v=1, v1=1, v2=−2, A=1, a=1, b=1, m=1, n=1, k=1, and ϕ0=0. (a) The 3-D plot for t=0. (b) The 2-D curve for y=0 and t=0.
(a)(b)Figure 4
The behavior ofχ5+x,y,t with the parameters λ=1, v=1, v1=1, v2=1, A=1, a=1, b=1, m=1, n=1, k=4, and ϕ0=0. (a) The 3-D plot. (b) The 2-D curve for y=0 and t=0.
(a)(b)Figure 5
The behavior ofχ5+x,y,t with the parameters λ=1, v=1, v1=1, v2=1, A=1, a=1, b=1, m=1, n=1, k=8, and ϕ0=0. (a) The 3-D plot for t=0. (b) The 2-D curve for y=0 and t=0.
(a)(b)


## 5. Conclusion and Future Recommendation

In this work, abundant soliton solutions of the (2+1)-dimensional HFSCE such as the bright soliton, dark soliton, bright-dark soliton, perfect periodic soliton, and singular periodic soliton are successfully constructed by means of the VM and SEM. The absolute parts of the solutions are presented through the 3-D plot and 2-D curve by assigning the proper parameters. In addition, we give the physical explanation of the solutions in detail. Through the calculation process, it can be found that the VM can reduce the order of the differential equation to simplify the equation via the variational theory. In addition, the SEM is effective and direct and can construct the different types of solitons. The methods adopted in this study can be used to investigate the other PDEs arising in the physics.In recent years, the fractal and fractional derivatives [50–61] have attracted extensive attention in different fields. How to apply the proposed methods to study the fractal and fractional PDEs will be the focus of our future research.



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