# Analytical Treatment of the Generalized Hirota-Satsuma-Ito Equation Arising in Shallow Water Wave

Authors: Fan Yong-Yan; Jalil Manafian; Syed Maqsood Zia; Dinh Tran Ngoc Huy; Trung-Hieu Le
Journal: Advances in Mathematical Physics
 (2021)

---

## Abstract

In the current study, an analytical treatment is studied starting from the2+1-dimensional generalized Hirota-Satsuma-Ito (HSI) equation. Based on the equation, we first establish the evolution equation and obtain rational function solutions by means of the bilinear form with the help of the Hirota bilinear operator. Then, by the suggested method, the periodic, cross-kink wave solutions are also obtained. Also, the semi-inverse variational principle (SIVP) will be utilized for the generalized HSI equation. Two major cases were investigated from two different techniques. Moreover, the improved tanχξ method on the generalized Hirota-Satsuma-Ito equation is probed. The 3D, density, and contour graphs illustrating some instances of got solutions have been demonstrated from a selection of some suitable parameters. The existing conditions are handled to discuss the available got solutions. The current work is extensively utilized to report plenty of attractive physical phenomena in the areas of shallow water waves and so on.


---

## Body

## 1. Introduction

Nonlinear partial differential equations containing time variables are generally referred to as nonlinear evolution equations (NLEEs) which can describe the state or process changing along with times in physics, dynamics, and other nature sciences. For the past decades, a variety of methods have sprung up to obtain exact solutions of NLEEs such as the homogeneous balance method [1], the generalized auxiliary equation technique [2], the inverse scattering method [3], the homotopy perturbation method [4], the optimal Galerkin-homotopy asymptotic method [5, 6], the tanϕ/2-expansion method [7], new Kudryashov’s method [8], the function expansion method [9], the standard truncated Painlevé expansion method [10], Hirota’s bilinear method [11–15], He’s variational principle [16, 17], binary Darboux transformation [18], Lie group analysis [19, 20], Bäcklund transformation method [21], and the multiple Exp-function method [22]. Based on the above methods, plenty of exact solutions including soliton solution [23], lump solution [24], interaction solution [25], and rational solution [26] have been derived. In [27], some novel exact analytical solutions, such as soliton wave, periodic wave, and singular, and kink-singular wave solutions to the fractional Whitham-Broer-Kaup and generalized Hirota-Satsuma coupled KdV equations were investigated.In this paper, we consider the3+1-dimensional generalized Hirota-Satsuma-Ito (HSI) shallow water wave equation which will be read [28] as
(1)Ψxxxt+3ΨxΨtx+δ1Ψyt+δ2Ψxx+δ3Ψxy+δ4Ψxt+δ5Ψyy=0,by using the following bilinear transformation
(2)Ψ=2lnfx.Equation (1) is transformed to the bilinear form as below:
(3)Dx3Dt+δ1DyDt+δ2Dx2+δ3DxDy+δ4DxDt+δ5Dy2f.f=0,in which
(4)Dx2f.f=2ffxx−fx2,Dy2f.f=2ffyy−fy2,DxDyf.f=2ffxy−fxfy,DxDtf.f=2ffxt−fxft,Dx3Dtf.f=2ffxxxt−ftfxxx−3fxfxxt+3fxxfxt.Firstly, Hirota and Satsuma introduced the1+1-dimensional Hirota-Satsuma shallow water wave equation as a model equation describing the unidirectional propagation of shallow water waves [29], where we can write as
(5)Ψt−Ψxxt−3ΨΨt+3Ψx∫x∞Ψtdx+Ψx=0⇒DxDt−Dx3Dt+Dx2f.f=0,by applying the bilinear transformation Ψ=2lnfxx. Also, by introducing the Hirota bilinear method in integrability of nonlinear systems, the 2+1-dimensional Hirota-Satsuma shallow water wave equation [30] was studied as
(6)Ψxxxt+3ΨxΨtx+Ψyt+Ψxx=0⇒Dx3Dt+DyDt+Dx2f.f=0,by using the bilinear transformation Ψ=2lnfx.Chen and coauthors proposed the3+1-dimensional Hirota-Satsuma-Ito-like equation to describe the wave motion in fluid dynamics and shallow water [31]. Also, Liu et al. [32] investigated the N-soliton solution to construct the 2+1-dimensional generalized Hirota-Satsuma-Ito equation, from which some localized waves such as line solitons, lumps, periodic solitons, and their interactions. Kuo and Ma [33] studied on resonant multisoliton solutions to the 2+1-dimensional Hirota-Satsuma-Ito equations and the existence and nonexistence of solutions. Kaur and Wazwaz [34] used the bilinear form to the new reduced form of the 3+1-dimensional generalized BKP equation and obtained lump solutions with sufficient and necessary conditions. A variety of lump solutions, generated from quadratic functions, for the 3+1-dimensional BKP-Boussinesq equation have been obtained by using the Hirota bilinear form in [35]. Also, the same authors obtained the optical soliton solutions to the Schrödinger-Hirota equation [36]. The authors of [37] obtained the lump solutions by making use of Hirota bilinear form to the 3+1-dimensional generalized KP-Boussinesq equation. By using Hirota’s bilinear form and the extended Ansatz function method, Liu and Ye got the new exact periodic cross-kink wave solutions for the 2+1-dimensional KdV equation [38]. Liu and Xiong [39] obtained abundant multiwave, breather wave, and lump solutions by using the three wave method, the homoclinic breather approach, and the Hirota bilinear method for A variable-coefficient Boiti-Leon-Manna-Pempinelli (BLMP) equation. Also, Liu and He [40] utilized the Hirota bilinear form and concluded abundant lump solutions and lump-kink solutions of the new 3+1-dimensional generalized KP equation. Some three-wave solutions including kinky periodic solitary wave, periodic soliton, and kink solutions have been obtained to the 3+1-dimensional BLMP equation by the extended three-wave approach and the Hirota bilinear method in [41].The outline of this paper is organized as follows. In Section2, the new periodic solutions and multiple wave solutions of the 2+1-dimensional generalized HSI equation will be obtained by applying the Hirota bilinear method; in addition, the corresponding three-dimensional, contour, and density plots vividly show the physical structure of the periodic wave solutions. In Section 3, carrying the bilinear method to the cross-kink wave solutions will be obtained via choosing the specified function. In addition, we will plot several groups of maps to illustrate the cross-kink of the corresponding solutions by symbolic computation. We gave three cases of solitary solutions with the semi-inverse variational principle in Section 4. Finally, the improved tanχξ method and its application are given and investigated in Section 5. A few conclusions and outlook will be given in the final section.


## 2. New Periodic Wave Solutions for Generalized HSI Eq

By employing Hirota operator [42] for Equation (1), we have
(7)f=a1H1+a2H2+a3H3+a4H4,(8)H1=expΩ1x+Ω2y+Ω3t,H2=exp−Ω1x−Ω2y−Ω3t,H3=cosΩ4x+Ω5y+Ω6t,H4=coshΩ7x+Ω8y+Ω9t,(9)H5=sinΩ4x+Ω5y+Ω6t,H6=sinhΩ7x+Ω8y+Ω9t,(10)u=2∂∂xlnf=2a1Ω1H1−a2Ω1H2−a3H5Ω4+a4H6Ω7f,where Ωi,i=1,⋯,9,ai,j=1,⋯,4, are free values. By putting (10) into Equation (3), one gets
(11)2a424Ω73Ω9+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,2a3a4Ω43Ω6−3Ω42Ω7Ω9−3Ω4Ω6Ω72+Ω73Ω9−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,2a1a4Ω13Ω3+3Ω12Ω7Ω9+3Ω1Ω3Ω72+Ω73Ω9+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,−2a3a4Ω43Ω9+3Ω42Ω6Ω7−3Ω4Ω72Ω9−Ω6Ω73−2Ω4Ω7δ2−Ω4Ω8δ3−Ω4Ω9δ4−Ω5Ω7δ3−2Ω5Ω8δ5−Ω5Ω9δ1−Ω6Ω7δ4−Ω6Ω8δ1=0,−2a1a4Ω13Ω9+3Ω12Ω3Ω7+3Ω1Ω72Ω9+Ω3Ω73+2Ω1Ω7δ2+Ω1Ω8δ3+Ω1Ω9δ4+Ω2Ω7δ3+2Ω2Ω8δ5+Ω2Ω9δ1+Ω3Ω7δ4+Ω3Ω8δ1=0,2a324Ω43Ω6−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1=0,2a1a3Ω13Ω3−3Ω12Ω4Ω6−3Ω1Ω3Ω42+Ω43Ω6+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1=0,2a1a3Ω13Ω6+3Ω12Ω3Ω4−3Ω1Ω42Ω6−Ω3Ω43+2Ω1Ω4δ2+Ω1Ω5δ3+Ω1Ω6δ4+Ω2Ω4δ3+2Ω2Ω5δ5+Ω2Ω6δ1+Ω3Ω4δ4+Ω3Ω5δ1=0,8a1a24Ω13Ω3+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1=0.Solving the above equations, we get the following.Case 1.
(12)Ψ1=Ψ0−2a3sinΞ1Ω4a1e−Ω2δ5t/δ1+Ω2y+a2eΩ2δ5t/δ1−Ω2y+a3cosΞ1+a4coshΩ9t−Ω9δ1y/δ5,(13)Ξ1=Ω4x−Ω4Ω42δ5+δ1δ3−δ4δ5yδ1δ5,where a2,a3,a4,Ω2,Ω4,Ω9 are arbitrary values. Also, we need to satisfy the condition Ω7≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in parallel to the x axis in Figure 1.Case 2.
(14)Ψ2=Ψ0+2a1Ω1eΞ2−2a2Ω1e−Ξ2a1eΞ2+a2e−Ξ2+a3cos−Ω5δ5t/δ1+Ω5y+a4coshΩ9t−Ω9δ1y/δ5,(15)Ξ2=Ω1x+Ω1Ω12δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω1,Ω5,Ω9 are arbitrary values. Also, we need to satisfy the condition Ω1≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in the line of x−y axis in which they intersect at one point in Figure 2.Case 3.
(16)Ψ3=Ψ0+2a4sinhΞ3Ω7a1e−Ω2δ5t/δ1+Ω2y+a2eΩ2δ5t/δ1−Ω2y+a3cosΩ6t−Ω6δ1y/δ5+a4coshΞ3,(17)Ξ3=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω2,Ω6,Ω7 are arbitrary values. Also, we need to satisfy the condition Ω7≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in the line of x−y axis in which they intersect at one point in Figure 3.Case 4.
(18)Ψ4=Ψ0+2a1Ω7eΞ4−2a2Ω7e−Ξ4+2a4sinhΞ5Ω7a1eΞ4+a2e−Ξ4+a3cosΩ6t−Ω6δ1y/δ5+a4coshΞ5,Ξ4=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,Ξ5=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω5,Ω6,Ω7 are arbitrary values.Case 5.
(19)Ψ5=Ψ0+2a4sinhΩ7x+yΩ8Ω7a1eΩ3t+a2e−Ω3t+a3cosΩ6t+a4coshΩ7x+yΩ8,where a2,a3,a4,Ω3,Ω6,Ω7 are arbitrary values.Case 6.
(20)Ψ6=Ψ0+2a1Ω1eΩ1x+Ω2y−a2Ω1e−Ω1x−Ω2ya1eΩ1x+Ω2y+a2e−Ω1x−Ω2y+a3cosΩ6t+a4coshΩ9t,where a2,a3,a4,Ω1,Ω2,Ω6,Ω9 are arbitrary values.Case 7.
(21)Ψ7=Ψ0−2a3sinΩ4x+Ω5yΩ4a1eΩ3t+a2e−Ω3t+a3cosΩ4x+Ω5y+a4coshΩ9t,where a2,a3,a4,Ω3,Ω4,Ω5,Ω9 are arbitrary values.Case 8.
(22)Ψ8=Ψ0+2a1Ω7eΩ7x+yΩ8−a2Ω7e−Ω7x−yΩ8+a4sinhΩ7x+yΩ8Ω7a1eΩ7x+yΩ8+a2e−Ω7x−yΩ8+a3cosΩ6t+a4coshΩ7x+yΩ8,where a2,a3,a4,Ω6,Ω7,Ω8 are arbitrary values.Case 9.
(23)Ψ9=Ψ0+21/2a1−δ4eG1−1/2a2−δ4e−G1−a3sinG2Ω4+1/2a4sinhG1−δ4a1eG1+a2e−G1+a3cosG2+a4coshG1,(24)G1=83Ω4Ξ6Ω5−δ4tδ44Ω42−δ42+12−δ4x,G2=43Ω5Ξ6t4Ω42−δ42+Ω4x+Ω5y,Ξ6=4Ω42δ3+4Ω4Ω5δ5−δ3δ4,where a2,a3,a4,Ω6,Ω7,Ω8 are arbitrary values. Also, we need to satisfy condition 4Ω42−δ4≠0,δ4<0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in parallel to the y axis in Figure 4.Case 10.
(25)Ψ10=Ψ0+2a1Ω7eG1−a2Ω7e−G1−a3sinG2Ω4+a4sinhG1Ω7a1eG1+a2e−G1+a3cosG2+a4coshG1,Ξ7=2Ω42−2Ω72−δ4,G1=43δ32Ω42Ω7tδ5Ξ72+Ω7x+12Ω7δ34Ω72+δ4yδ5Ξ7,G2=−43Ω4Ω72δ32tΞ72δ5+Ω4x−12Ω4δ34Ω42−δ4yΞ7δ5,where a2,a3,a4,Ω4,Ω7 are arbitrary values. Also, we need to satisfy the condition 2Ω42−2Ω72−δ4δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in the line of x−y axis in which they intersect at one point in Figure 5.Case 11.
(26)Ψ11=Ψ0+2a1Ω1eG1−a2Ω1e−G1−a3sinG2Ω4+a4sinhG1Ω1a1eG1+a2e−G1+a3cosG2+a4coshG1,Ξ8=3Ω1δ3+2Ω2δ52Ω12−3Ω42+δ42,(27)G1=13Ω42Ξ8tΩ1δ5+Ω1x+Ω2y,G2=−13Ω4Ξ8tδ5+Ω4x+Ω43Ω13δ3+3Ω12Ω2δ5+3Ω1Ω42δ3−Ω2Ω42δ5+Ω2δ4δ5yΩ1δ5Ω12−3Ω42+δ4,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, we need to satisfy the condition Ω1δ5Ω12−3Ω42+δ4≠0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in the line of x−y axis in which they intersect at one point in Figure 6.Case 12.
(28)Ψ12=Ψ0+2a13Ω42−δ4eG1−a23Ω42−δ4e−G1−a3sinG2Ω4+a4sinhG13Ω42−δ4a1eG1+a2e−G1+a3cosG2+a4coshG1,(29)G1=3Ω42−δ4Ξ9tδ5+3Ω42−δ4x−32δ33Ω42−δ4yδ5,G2=−3Ω42−δ4Ξ9tΩ4δ5+Ω4x+Ω5y,(30)Ξ9=1123Ω4δ3+2Ω5δ5216Ω44−8Ω42δ4+δ42,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, we need to satisfy condition 3Ω42−δ4>0. By assigning particular values of the parameters, we can easily observe the wave motion as periodic waves in the line of x−y axis in which they intersect at one point in Figure 7.Case 13.
(31)Ψ13=Ψ0+2a1Ω1eG1−a2Ω1e−G1−a3sinG2Ω4+a4sinhG1Ω1a1eG1+a2e−G1+a3cosG2+a4coshG1,Ξ10=133Ω1δ3+2Ω2δ52Ω12−3Ω42+δ42,G1=Ω42Ξ10tΩ1δ5+Ω1x+Ω2y,G2=−Ω4Ξ10tδ5+Ω4x+Ω43Ω13δ3+3Ω12Ω2δ5+3Ω1Ω42δ3−Ω2Ω42δ5+Ω2δ4δ5yΩ1δ5Ω12−3Ω42+δ4,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, it needs to satisfy 3Ω42−δ4>0.Figure 1
Plot evolution of periodic waves (12) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω2=1.9,Ω4=.2,Ω9=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 2
Plot evolution of periodic waves (14) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω1=1.9,Ω5=.2,Ω9=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 3
Plot evolution of periodic waves (16) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω2=1.9,Ω6=.2,Ω7=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 4
Plot evolution of periodic waves (23) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω4=1.9,Ω5=.2,Ω7=2,δ1=1,δ2=1.3,δ3=2,δ4=−1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 5
Plot evolution of periodic waves (23) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω4=1.1,Ω7=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 6
Plot evolution of periodic waves (26) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω1=1.1,Ω2=1.5,Ω4=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 7
Plot evolution of periodic waves (28) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω4=1.5,Ω5=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)


## 3. New Cross-Kink Wave Solutions for Generalized HSI Eq

Based on the Hirota operator [42] for Equation (1), we get
(32)f=a1H1+a2H2+a3H3+a4H4,(33)H1=expΩ1x+Ω2y+Ω3t,H2=exp−Ω1x−Ω2y−Ω3t,H3=sinΩ4x+Ω5y+Ω6t,H4=sinhΩ7x+Ω8y+Ω9t,H5=cosΩ4x+Ω5y+Ω6t,(34)H6=coshΩ7x+Ω8y+Ω9t,(35)u=2∂∂xlnf=2a1Ω1H1−a2Ω1H2+a3H5Ω4+a4H6Ω7f,where Ωi,i=1,⋯,9,ai,j=1,⋯,4, are free parameters. By inserting (35) into Equation (3), one concludes
(36)2a424Ω73Ω9+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,2a3a4Ω43Ω6−3Ω42Ω7Ω9−3Ω4Ω6Ω72+Ω73Ω9−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,2a1a4Ω13Ω3+3Ω12Ω7Ω9+3Ω1Ω3Ω72+Ω73Ω9+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1+Ω72δ2+Ω7Ω8δ3+Ω7Ω9δ4+Ω82δ5+Ω8Ω9δ1=0,−2a3a4Ω43Ω9+3Ω42Ω6Ω7−3Ω4Ω72Ω9−Ω6Ω73−2Ω4Ω7δ2−Ω4Ω8δ3−Ω4Ω9δ4−Ω5Ω7δ3−2Ω5Ω8δ5−Ω5Ω9δ1−Ω6Ω7δ4−Ω6Ω8δ1=0−2a1a4Ω13Ω9+3Ω12Ω3Ω7+3Ω1Ω72Ω9+Ω3Ω73+2Ω1Ω7δ2+Ω1Ω8δ3+Ω1Ω9δ4+Ω2Ω7δ3+2Ω2Ω8δ5+Ω2Ω9δ1+Ω3Ω7δ4+Ω3Ω8δ1=0,2a324Ω43Ω6−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1=0,2a1a3Ω13Ω3−3Ω12Ω4Ω6−3Ω1Ω3Ω42+Ω43Ω6+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1−Ω42δ2−Ω4Ω5δ3−Ω4Ω6δ4−Ω52δ5−Ω5Ω6δ1=0,2a1a3Ω13Ω6+3Ω12Ω3Ω4−3Ω1Ω42Ω6−Ω3Ω43+2Ω1Ω4δ2+Ω1Ω5δ3+Ω1Ω6δ4+Ω2Ω4δ3+2Ω2Ω5δ5+Ω2Ω6δ1+Ω3Ω4δ4+Ω3Ω5δ1=0,8a1a24Ω13Ω3+Ω12δ2+Ω1Ω2δ3+Ω1Ω3δ4+Ω22δ5+Ω2Ω3δ1=0.Solving the above equations, we get the following.Case 1.
(37)Ψ1=Ψ0−2a3cosΞ1Ω4a1e−Ω2δ5t/δ1+Ω2y+a2eΩ2δ5t/δ1−Ω2y+a3sinΞ1+a4sinhΩ9t−Ω9δ1y/δ5,(38)Ξ1=Ω4x−Ω4Ω42δ5+δ1δ3−δ4δ5yδ1δ5,where a2,a3,a4,Ω2,Ω4,Ω9 are arbitrary values. Also, we need to satisfy the condition Ω7≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in parallel to the x axis in Figure 8.Case 2.
(39)Ψ2=Ψ0+2a1Ω1eΞ2−2a2Ω1e−Ξ2a1eΞ2+a2e−Ξ2+a3sin−Ω5δ5t/δ1+Ω5y+a4sinhΩ9t−Ω9δ1y/δ5,(40)Ξ2=Ω1x+Ω1Ω12δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω1,Ω5,Ω9 are arbitrary values. Also, we need to satisfy the condition Ω1≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 9.Case 3.
(41)Ψ3=Ψ0+2a4coshΞ3Ω7a1e−Ω2δ5t/δ1+Ω2y+a2eΩ2δ5t/δ1−Ω2y+a3sinΩ6t−Ω6δ1y/δ5+a4sinhΞ3,(42)Ξ3=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω2,Ω6,Ω7 are arbitrary values. Also, we need to satisfy the condition Ω7≠0,δ1δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 10.Case 4.
(43)Ψ4=Ψ0+2a1Ω7eΞ4−2a2Ω7e−Ξ4+2a4coshΞ5Ω7a1eΞ4+a2e−Ξ4+a3sinΩ6t−Ω6δ1y/δ5+a4sinhΞ5,Ξ4=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,Ξ5=Ω7x+Ω7Ω72δ5−δ1δ3+δ4δ5yδ1δ5,where a2,a3,a4,Ω5,Ω6,Ω7 are arbitrary values.Case 5.
(44)Ψ5=Ψ0+2a4coshΩ7x+yΩ8Ω7a1eΩ3t+a2e−Ω3t+a3sinΩ6t+a4sinhΩ7x+yΩ8,where a2,a3,a4,Ω3,Ω6,Ω7 are arbitrary values.Case 6.
(45)Ψ6=Ψ0+2a1Ω1eΩ1x+Ω2y−a2Ω1e−Ω1x−Ω2ya1eΩ1x+Ω2y+a2e−Ω1x−Ω2y+a3sinΩ6t+a4sinhΩ9t,where a2,a3,a4,Ω1,Ω2,Ω6,Ω9 are arbitrary values.Case 7.
(46)Ψ7=Ψ0−2a3cosΩ4x+Ω5yΩ4a1eΩ3t+a2e−Ω3t+a3sinΩ4x+Ω5y+a4sinhΩ9t,where a2,a3,a4,Ω3,Ω4,Ω5,Ω9 are arbitrary values.Case 8.
(47)Ψ8=Ψ0+2a1Ω7eΩ7x+yΩ8−a2Ω7e−Ω7x−yΩ8+a4coshΩ7x+yΩ8Ω7a1eΩ7x+yΩ8+a2e−Ω7x−yΩ8+a3sinΩ6t+a4sinhΩ7x+yΩ8,where a2,a3,a4,Ω6,Ω7,Ω8 are arbitrary values.Case 9.
(48)Ψ9=Ψ0+21/2a1−δ4eG1−1/2a2−δ4e−G1−a3cosG2Ω4+1/2a4coshG1−δ4a1eG1+a2e−G1+a3sinG2+a4sinhG1,(49)G1=83Ω4Ξ6Ω5−δ4tδ4+12−δ4x,G2=43Ω5Ξ6t1+Ω4x+Ω5y,Ξ6=4Ω42δ3+4Ω4Ω5δ5−δ3δ44Ω42−δ42,where a2,a3,a4,Ω6,Ω7,Ω8 are arbitrary values. Also, we need to satisfy the condition 4Ω42−δ4≠0,δ4<0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 11.Case 10.
(50)Ψ10=Ψ0+2a1Ω7eG1−a2Ω7e−G1−a3cosG2Ω4+a4coshG1Ω7a1eG1+a2e−G1+a3sinG2+a4sinhG1,Ξ7=2Ω42−2Ω72−δ4,G1=43δ32Ω42Ω7tδ5Ξ72+Ω7x+12Ω7δ34Ω72+δ4yδ5Ξ7,G2=−43Ω4Ω72δ32tΞ72δ5+Ω4x−12Ω4δ34Ω42−δ4yΞ7δ5,where a2,a3,a4,Ω4,Ω7 are arbitrary values. Also, we need to satisfy the condition 2Ω42−2Ω72−δ4δ5≠0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 12.Case 11.
(51)Ψ11=Ψ0+2a1Ω1eG1−a2Ω1e−G1−a3cosG2Ω4+a4coshG1Ω1a1eG1+a2e−G1+a3sinG2+a4sinhG1,Ξ8=133Ω1δ3+2Ω2δ52Ω12−3Ω42+δ42,(52)G1=Ω42Ξ8tΩ1δ5+Ω1x+Ω2y,G2=−Ω4Ξ8tδ5+Ω4x+Ω43Ω13δ3+3Ω12Ω2δ5+3Ω1Ω42δ3−Ω2Ω42δ5+Ω2δ4δ5yΩ1δ5Ω12−3Ω42+δ4,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, we need to satisfy the condition Ω1δ5Ω12−3Ω42+δ4≠0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 13.Case 12.
(53)Ψ12=Ψ0+2a13Ω42−δ4eG1−a23Ω42−δ4e−G1−a3cosG2Ω4+a4coshG13Ω42−δ4a1eG1+a2e−G1+a3sinG2+a4sinhG1,(54)G1=3Ω42−δ4Ξ9tδ5+3Ω42−δ4x−32δ33Ω42−δ4yδ5,G2=−3Ω42−δ4Ξ9tΩ4δ5+Ω4x+Ω5y,(55)Ξ9=1123Ω4δ3+2Ω5δ5216Ω44−8Ω42δ4+δ42,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, we need to satisfy the condition 3Ω42−δ4>0. By assigning particular values of the parameters, we can easily observe the wave motion as cross-kink waves in the line of x−y axis in which they intersect at one point in Figure 14.Case 13.
(56)Ψ13=Ψ0+2a1Ω1eG1−a2Ω1e−G1−a3cosG2Ω4+a4coshG1Ω1a1eG1+a2e−G1+a3sinG2+a4sinhG1,Ξ10=133Ω1δ3+2Ω2δ52Ω12−3Ω42+δ42,G1=Ω42Ξ10tΩ1δ5+Ω1x+Ω2y,G2=−Ω4Ξ10tδ5+Ω4x+Ω43Ω13δ3+3Ω12Ω2δ5+3Ω1Ω42δ3−Ω2Ω42δ5+Ω2δ4δ5yΩ1δ5Ω12−3Ω42+δ4,where a2,a3,a4,Ω1,Ω2,Ω4 are arbitrary values. Also, we need to satisfy the condition 3Ω42−δ4>0.Figure 8
Plot evolution of cross-kink waves (37) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω2=1.9,Ω4=.2,Ω9=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 9
Plot evolution of cross-kink waves (39) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω1=1.9,Ω5=.2,Ω9=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 10
Plot evolution of cross-kink waves (41) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω2=1.9,Ω6=.2,Ω7=2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 11
Plot evolution of cross-kink waves (48) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω4=1.9,Ω5=.2,Ω7=2,δ1=1,δ2=1.3,δ3=2,δ4=−1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 12
Plot evolution of cross-kink waves (48) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω4=1.1,Ω7=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 13
Plot evolution of cross-kink waves (51) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω1=1.1,Ω2=1.5,Ω4=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)Figure 14
Plot evolution of cross-kink waves (53) utilizing values a1=.5,a2=1.5,a3=2,a4=1,Ω1=1.1,Ω2=1.5,Ω4=1.2,δ1=1,δ2=1.3,δ3=2,δ4=1.3,δ5=2,Ψ0=1,t=10 and (a) 3D plot, (b) density plot, and (c) contour plot.
(a)(b)(c)


## 4. Application of SIVP for Equation (1)

By utilizingξ=kx+ay−ct in Equation (1), one becomes
(57)−ck2Ψ″″−6kcΨ′Ψ″+a2δ5−cδ4+aδ3+δ2−cδ1Ψ″=0.By points of Refs. [16, 17, 43] and by multiplying Equation (57) with Ψ′, we get
(58)J=∫−∞∞−2kcΨ′3+12ck2Ψ″2+12a2δ5−cδ4+aδ3+δ2−cδ1Ψ′2−ck2Ψ′Ψ′″dξ.Case 1.
With selection the below solution function(59)uξ=AsechBξ,then the stationary integral changes to
(60)J=−130A2B−21B2ck2−12kABc−5S,S=a2δ5−cδ4+aδ3+δ2−cδ1.
With the help of below(61)∂J∂A=0,∂J∂B=0,two nonlinear algebraic systems will be concluded as
(62)−115AB−21B2ck2−12kABc−5S+25A2B2kc=0,130A2−21B2ck2−12kABc−5S−130A2B−42Bck2−12Ack=0.
By solving the above cases, one gets(63)A=±S213cS,B=±21cS21ck.
The domain of definition is(64)ca2δ5−cδ4+aδ3+δ2−cδ1>0.
Lastly, the solitary wave solution gets(65)Ψx,y,t=±S213cSsech±21cS21cx+ay−ct,S=a2δ5−cδ4+aδ3+δ2−cδ1.Case 2.
By selecting the below solution function(66)uξ=Asech2Bξ,then the stationary integral changes to
(67)J=240B2ck2+70kABc+28SA2B105.
With the help of below,(68)∂J∂A=23A2B2kc+480B2ck2+140kABc+56SAB105=0,∂J∂B=480Bck2+70AckA2B105+240B2ck2+70kABc+28SA2105=0.
By solving the above cases, one gets(69)A=∓16S2135cS,B=±21cS30ck.
The domain of definition is(70)ca2δ5−cδ4+aδ3+δ2−cδ1>0.
Lastly, the bright wave solution one becomes(71)Ψx,y,t=∓16S213cSsech2±21cS30cx+ay−ct,S=a2δ5−cδ4+aδ3+δ2−cδ1.Case 3.
Suppose the dark soliton wave solution as below forms(72)uξ=Atanh2Bξ,then the stationary integral changes to
(73)J=−2A2B−120B2ck2+35kABc−14S105.
With the help of below,(74)∂J∂A=−4AB−120B2ck2+35kABc−14S105−23A2B2kc=0,∂J∂B=−2A2−120B2ck2+35kABc−14S105−2A2B−240Bck2+35Ack105=0.
By solving the above cases, one gets(75)A=∓16S2135cS,B=±21cS30ck.
The domain of definition is(76)ca2δ5−cδ4+aδ3+δ2−cδ1>0.
Then, the dark wave solution will be obtained as(77)Ψx,y,t=∓16S213cStanh2±21cS30cx+ay−ct,S=a2δ5−cδ4+aδ3+δ2−cδ1.


## 5. The Improved Rationaltanχξ/2-Expansion Method

The application of the improvedtanχξ/2-expansion technique will be studied where for the first time is given here. We first discuss the mathematical analysis of nonlinear partial differential equations (NPDEs). Hence, we consider the NPDEs in the following way.Step 1.
Assume a nonlinear partial differential equation is given in the general form as follows:(78)Lχ,χx,χy,χt,χxx,χyy,χtt,⋯=0.
After simple algebraic operations, this equation is transformed into an ordinary differential equation (ODE) with the below transformation:(79)χx,y,t=χξ,ξ=k1x+k2y−ct,as below nonlinear ODE
(80)Sχ,k1χ′,k2χ′,−cχ′,k12χ′′,k22χ′′,c2χ′′,⋯=0.Step 2.
Then, assume that the searched wave solutions of Equation (80) have the following representation:
(81)Ψξ=∑j=0σζjtanjχξ2+∑j=1σθjcotjχξ2,where ζj0≤j≤σ,θj0≤j≤σ are constants to be determined, such that ζσ,θσ≠0, and χ=χξ is the solution of the following first order differential equation:
(82)χ′=α1sinχ+α2cosχ+α3.The particular solutions of Equation (82) will be read as the following.Family 1.
Ifα12+α22−α32<0 and α2−α3≠0, afterwards Zξ/2=tan−1α1/α2−α3−α32−α12−α22/α2−α3tanα32−α12−α22/2ξ+ϕ0.Family 2.
Ifα12+α22−α32>0 and α2−α3≠0, afterwards Zξ/2=tan−1α1/α2−α3+α12+α22−α32/α2−α3tanhα12+α22−α32/2ξ+ϕ0.Family 3.
Ifα12+α22−α32>0, α2≠0 and α3=0, afterwards Zξ/2=tan−1α1/α2+α12+α22/α2tanhα12+α22/2ξ+ϕ0.Family 4.
Ifα12+α22−α32<0, α3≠0 and α2=0, afterwards Zξ/2=tan−1−α1/α3+α32−α12/α3tanα32−α12/2ξ+ϕ0.Family 5.
Ifα2−α3≠0 and α1=0, afterwards Zξ/2=tan−1α2+α3/α2−α3tanhα22−α32/2ξ+ϕ0.Family 6.
Ifα1=0 and α3=0, afterwards Zξ=tan−1e2α2ξ+ϕ0−1/e2α2ξ+ϕ0+1,2eα2ξ+ϕ0/e2α2ξ+ϕ0+1.Family 7.
Ifα2=0 and α3=0, afterwards Zξ=tan−12eα1ξ+ϕ0/e2α1ξ+ϕ0+1,e2α1ρ+ϕ0−1/e2α1ξ+ϕ0+1.Family 8.
Ifα12+α22=α32, afterwards Zξ/2=tan−1α1ξ+ϕ0+2/α2−α3ξ+ϕ0.Family 9.
Ifα1=α2=α3=kα1, afterwards Zξ/2=tan−1ekα1ξ+ϕ0−1.Family 10.
Ifα1=α3=kα1 and α2=−kα1, afterwards Zξ/2=−tan−1ekα1ξ+ϕ0/−1+ekα1ξ+ϕ0.Family 11.
Ifα3=α1, then Zξ/2=−tan−1α1+α2eα2ξ+ϕ0−1/α1−α2eα2ξ+ϕ0−1.Family 12.
Ifα1=α3, afterwards Zξ/2=tan−1α2+α3eα2ξ+ϕ0+1/α2−α3eα2ξ+ϕ0−1.Family 13.
Ifα3=−α1, afterwards Zξ/2=tan−1eα2ξ+ϕ0+α2−α1/eα2ξ+ϕ0−α2−α1.Family 14.
Ifα2=−α3, afterwards Zξ/2=tan−1α1eα1ξ+ϕ0/1−α3eα1ξ+ϕ0.Family 15.
Ifα2=0 and α1=α3, afterwards Zξ/2=−tan−1α3ξ+ϕ0+2/α3ξ+ϕ0.Family 16.
Ifα1=0 and α2=α3, afterwards Zξ/2=tan−1α3ξ+ϕ0.Family 17.
Ifα1=0 and α2=−α3, afterwards Zξ/2=−tan−11/α3ξ+ϕ0.Family 18.
Ifα1=0 and α2=0, afterwards Zξ=α3ξ+ϕ0+C.Family 19.
Ifα2=α3 afterwards Zξ/2=tan−1eα1ξ+ϕ0−α3/α1, where ϕ0 is an integration constant. Also, ζk,θkk=1,2,⋯,σ,α1,α2, and α3 are constants to be determined later.Step 3.
To determine the positive integerη, we usually balance linear terms of the highest order in the resulting equation with the highest order nonlinear terms appearing in Equation (80).Step 4.
We collect all the terms with the same order oftanχξ/2kk=0,1,2,⋯ together. Equating each coefficient of the polynomials of i to zero yields the set of algebraic equations for ζk,θkk=1,2,⋯,σ,α1,α2, and α3 with the aid of the Maple.Step 5.
Solving the algebraic equations in Step4, then substituting ζk,θkk=1,2,⋯,σ,α1,α2, and α3 in (81).


### 5.1. Application of Improvedtanχξ Method on the Generalized HSI Eq

By utilizing the following transformation,(83)ξ=k1x+k2y−ct.Based on the section before, the needed detail can be got in the below equation:(84)A1d4dξ4Ψξ+2A2ddξΨξd2dξ2Ψξ+A3d2dξ2Ψξ=0,with
(85)A1=−ck13,A2=−3ck12,A3=−cδ1k2−ck1δ4+δ2k12+δ3k1k2+δ5k22,where k1,k2,c are unspecified constants. The balance number will be obtained θ=1 by using the balance principle. Then, the exact solution is given as
(86)Ψξ=ζ0+ζ1tanχξ2+θ1cotχξ2.Firstly, we substitute the expressions ofχξ in (86) into (84) and collect all terms with the same order of tankχξ/2. Then, by equating the coefficient of each polynomial to zero containing a system of eleven nonlinear equations and by solving the nonlinear system, the specified coefficients will be got as the below cases.Case 1.
(87)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=0,θ1=k1α2+α3,k2=−k1α22k12−α32k12+δ4δ1,k1=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(88)Ψ1ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k1k12α22−k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(89)Ψ2ξ=ζ0±122δ5δ1δ3−2δ4δ5+δ12δ32−4δ12δ2δ5δ5cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k1k12α22+δ4δ1y−ct,k1=±122δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.Case 2.
(90)c=c,ζ0=ζ0,ζ1=0,α1=0,α2=−α3,α3=α3,k1=k1,k2=k2,θ1=θ1.
According to Family17, (86) becomes
(91)Ψ1ξ=ζ0−α3θ1k1x+k2y−ct+ϕ0.Case 3.
(92)c=δ2k12+δ3k1k2+δ5k22k13α12+α22−α32+δ1k2+δ4k1,ζ0=ζ0,ζ1=0,α1=α1,α2=α2,α3=α3,k1=k1,k2=k2,θ1=k1α2+α3.
According to Family1, (86) becomes
(93)Ψ1ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α1α2−α3−α32−α12−α22α2−α3tanα32−α12−α222ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α12+α22−α32+δ1k2+δ4k1t, α12+α22−α32<0, and α2−α3≠0. According to Family 2, (86) gets
(94)Ψ2ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α1α2−α3+α12+α22−α32α2−α3tanhα12+α22−α322ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α12+α22−α32+δ1k2+δ4k1t, α12+α22−α32>0, and α2−α3≠0. According to Family 6, (86) gets
(95)Ψ3ξ=ζ0+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α22+δ1k2+δ4k1t. According to Family 8, (86) gets
(96)Ψ4ξ==ζ0+k1α2+α3cotχξ2,tan12χξ=α1ξ+ϕ0+2α2−α3ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/δ1k2+δ4k1t and α12+α22−α32=0.Case 4.
(97)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=0,k2=−k1α22k12−α32k12+δ4δ1,k1=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(98)Ψ1ξ=ζ0+k1α2−α3tanχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k1k12α22−k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(99)Ψ2ξ=ζ0±122δ5δ1δ3−2δ4δ5+δ12δ32−4δ12δ2δ5δ5tanχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k1k12α22+δ4δ1y−ct,k1=±122δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.Case 5.
(100)c=c,ζ0=ζ0,ζ1=−2α3,α1=0,α2=−α3,α3=cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226ck12θ1,k1=k1,k2=k2,θ1=θ1.
According to Family17, (86) becomes
(101)Ψ1x,y,t=ζ0+2k1x+k2y−ct+ϕ0−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226ck12k1x+k2y−ct+ϕ0.Case 6.
(102)c=c,ζ0=ζ0,ζ1=−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226cα3k12,α1=0,α2=α3,α3=α3,k1=k1,k2=k2,θ1=2α3k1.
According to Family19, (86) becomes
(103)Ψ1x,y,t=ζ0−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226cα3k12eα1k1x+k2y−ct+ϕ0−α3α1+2α3k1α1eα1k1x+k2y−ct+ϕ0−α3.Case 7.
(104)c=δ2k12+δ3k1k2+δ5k224k13α22−α32+δ1k2+δ4k1,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=k1α2+α3.
According to Family5, (86) becomes
(105)Ψ1ξ=ζ0+k1α2−α3tanχξ2+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k224k13α22−α32+δ1k2+δ4k1t,so that 4k13α22−α32+δ1k2+δ4k1≠0.
According to Family6, (86) gets
(106)Ψ2ξ=ζ0+k1α2tanχξ2+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k224k13α22+δ1k2+δ4k1t.Case 8.
(107)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=k1α2+α3,k2=−k14α22k12−4α32k12+δ4δ1,k1=±142δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(108)Ψ1ξ=ζ0+k1α2−α3tanχξ2+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±142δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k14k12α22−4k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(109)Ψ2ξ=ζ0+k1α2tanχξ2+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k14k12α22+δ4δ1y−ct,k1=±142δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.


## 5.1. Application of Improvedtanχξ Method on the Generalized HSI Eq

By utilizing the following transformation,(83)ξ=k1x+k2y−ct.Based on the section before, the needed detail can be got in the below equation:(84)A1d4dξ4Ψξ+2A2ddξΨξd2dξ2Ψξ+A3d2dξ2Ψξ=0,with
(85)A1=−ck13,A2=−3ck12,A3=−cδ1k2−ck1δ4+δ2k12+δ3k1k2+δ5k22,where k1,k2,c are unspecified constants. The balance number will be obtained θ=1 by using the balance principle. Then, the exact solution is given as
(86)Ψξ=ζ0+ζ1tanχξ2+θ1cotχξ2.Firstly, we substitute the expressions ofχξ in (86) into (84) and collect all terms with the same order of tankχξ/2. Then, by equating the coefficient of each polynomial to zero containing a system of eleven nonlinear equations and by solving the nonlinear system, the specified coefficients will be got as the below cases.Case 1.
(87)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=0,θ1=k1α2+α3,k2=−k1α22k12−α32k12+δ4δ1,k1=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(88)Ψ1ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k1k12α22−k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(89)Ψ2ξ=ζ0±122δ5δ1δ3−2δ4δ5+δ12δ32−4δ12δ2δ5δ5cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k1k12α22+δ4δ1y−ct,k1=±122δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.Case 2.
(90)c=c,ζ0=ζ0,ζ1=0,α1=0,α2=−α3,α3=α3,k1=k1,k2=k2,θ1=θ1.
According to Family17, (86) becomes
(91)Ψ1ξ=ζ0−α3θ1k1x+k2y−ct+ϕ0.Case 3.
(92)c=δ2k12+δ3k1k2+δ5k22k13α12+α22−α32+δ1k2+δ4k1,ζ0=ζ0,ζ1=0,α1=α1,α2=α2,α3=α3,k1=k1,k2=k2,θ1=k1α2+α3.
According to Family1, (86) becomes
(93)Ψ1ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α1α2−α3−α32−α12−α22α2−α3tanα32−α12−α222ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α12+α22−α32+δ1k2+δ4k1t, α12+α22−α32<0, and α2−α3≠0. According to Family 2, (86) gets
(94)Ψ2ξ=ζ0+k1α2+α3cotχξ2,tan12χξ=α1α2−α3+α12+α22−α32α2−α3tanhα12+α22−α322ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α12+α22−α32+δ1k2+δ4k1t, α12+α22−α32>0, and α2−α3≠0. According to Family 6, (86) gets
(95)Ψ3ξ=ζ0+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/k13α22+δ1k2+δ4k1t. According to Family 8, (86) gets
(96)Ψ4ξ==ζ0+k1α2+α3cotχξ2,tan12χξ=α1ξ+ϕ0+2α2−α3ξ+ϕ0,so that ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k22/δ1k2+δ4k1t and α12+α22−α32=0.Case 4.
(97)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=0,k2=−k1α22k12−α32k12+δ4δ1,k1=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(98)Ψ1ξ=ζ0+k1α2−α3tanχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±122δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k1k12α22−k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(99)Ψ2ξ=ζ0±122δ5δ1δ3−2δ4δ5+δ12δ32−4δ12δ2δ5δ5tanχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k1k12α22+δ4δ1y−ct,k1=±122δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.Case 5.
(100)c=c,ζ0=ζ0,ζ1=−2α3,α1=0,α2=−α3,α3=cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226ck12θ1,k1=k1,k2=k2,θ1=θ1.
According to Family17, (86) becomes
(101)Ψ1x,y,t=ζ0+2k1x+k2y−ct+ϕ0−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226ck12k1x+k2y−ct+ϕ0.Case 6.
(102)c=c,ζ0=ζ0,ζ1=−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226cα3k12,α1=0,α2=α3,α3=α3,k1=k1,k2=k2,θ1=2α3k1.
According to Family19, (86) becomes
(103)Ψ1x,y,t=ζ0−cδ1k2+cδ4k1−δ2k12−δ3k1k2−δ5k226cα3k12eα1k1x+k2y−ct+ϕ0−α3α1+2α3k1α1eα1k1x+k2y−ct+ϕ0−α3.Case 7.
(104)c=δ2k12+δ3k1k2+δ5k224k13α22−α32+δ1k2+δ4k1,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=k1α2+α3.
According to Family5, (86) becomes
(105)Ψ1ξ=ζ0+k1α2−α3tanχξ2+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k224k13α22−α32+δ1k2+δ4k1t,so that 4k13α22−α32+δ1k2+δ4k1≠0.
According to Family6, (86) gets
(106)Ψ2ξ=ζ0+k1α2tanχξ2+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x+k2y−δ2k12+δ3k1k2+δ5k224k13α22+δ1k2+δ4k1t.Case 8.
(107)c=c,ζ0=ζ0,α1=0,α2=α2,α3=α3,ζ1=k1α2−α3,θ1=k1α2+α3,k2=−k14α22k12−4α32k12+δ4δ1,k1=±142δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32.
According to Family5, (86) becomes
(108)Ψ1ξ=ζ0+k1α2−α3tanχξ2+k1α2+α3cotχξ2,tan12χξ=α2+α3α2−α3tanhα22−α322ξ+ϕ0,ξ=±142δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α22−α32x−k14k12α22−4k12α32+δ4δ1y−ct,so that δ5α22−α32δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.
According to Family6, (86) gets
(109)Ψ2ξ=ζ0+k1α2tanχξ2+k1α2cotχξ2,tan12χξ=e2α2ξ+ϕ0−1e2α2ξ+ϕ0+1,2eα2ξ+ϕ0e2α2ξ+ϕ0+1,ξ=k1x−k14k12α22+δ4δ1y−ct,k1=±142δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32δ5α2,so that δ5δ1δ3−2δ4δ5+−4δ12δ2δ5+δ12δ32>0 and δ12δ32−4δ12δ2δ5>0.


## 6. Conclusion

In this study, the periodic, cross-kink, solitary, bright, and dark wave solutions of the2+1-dimensional generalized Hirota-Satsuma-Ito equation have been achieved. From the bilinear form of this equation, one test function or ansatz has been chosen. Through Maple, the evolution phenomenon of these waves is seen in Figures 1–14, respectively. Mainly, by choosing specific parameter constraints, all cases of 2D and 3D in solitons can be captured from the periodic and cross-kink wave solutions. Also, the improved tanχξ method on the generalized nonlinear wave equation studied and four sets of solutions were obtained. The obtained solutions are extended with numerical simulation to analyze graphically, which results into multiwave and cross-kink wave solutions. Moreover, we studied the solitary, bright, and dark soliton wave solutions of the generalized HSI equation by help of SIVP in the previous section. Finally, literature is full of nonlinear evolution that rich soliton structures are still to be constructed while applying these methods. Further investigations deserve to be made in order to ameliorate the improved tanχξ scheme, so that it may be possible to provide all the different solutions to a nonlinear system. These questions will constitute future works.



---

*Source: 1164838-2021-10-12.xml*