# Breather Wave Solutions and Interaction Solutions for Two Mixed Calogero-Bogoyavlenskii-Schiff and Bogoyavlensky-Konopelchenko Equations

Authors: Hongcai Ma; Caoyin Zhang; Aiping Deng
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

In this paper, based on a bilinear differential equation, we study the breather wave solutions by employing the extended homoclinic test method. By constructing the different forms, we also consider the interaction solutions. Furthermore, it is natural to analyse dynamic behaviors of three-dimensional plots.


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## Body

## 1. Introduction

Recently, great attention has been paid to the study about exact solutions of nonlinear partial differential equations. So, it becomes more important to seek exact solutions of nonlinear partial differential equations (NLPDEs), which occur in many fields, such as chemistry, biology, optics, classical mechanics, acoustics, engineering, and social sciences. At present, many mathematicians have proposed a large number of methods to seek exact solutions, such as Bäcklund transformation [1], Hirota bilinear methods [2], homoclinic breather limit approach [3, 4], and Darboux transformation [5–12]. Among these methods, the Hirota bilinear method is one of the most critical and powerful methods. Recently, some new exact solutions of nonlinear partial differential equations have been constructed [13–22] by means of bilinear operator theories, so it has become an important research direction to study the dynamic properties of these new equations. In this article, the breather wave solutions will be discussed. On the basis of lump solution [23], the interaction solutions will be obtained.The two mixed Calogero-Bogoyavlenskii-Schiff (CBS) and Bogoyavlensky-Konopelchenko (BK) equations [23] are usually written as
(1)ut+u2x,y+3uxuy+δ1uy+δ2w2y+δ3ux+δ43ux2+u3x+δ53w2y2+w4y+δ63uyw2y+u3y=0,where ux=w and δi, i=1,⋯,6, are arbitrary constants. When the constants satisfy δ3=δ4=δ5=δ6=0, and δ5=δ6=0, the Calogero-Bogoyavlenskii-Schiff (CBS) and Bogoyavlensky-Konopelchenko (CBS-BK) equations will become a generalized Calogero-Bogoyavlenskill-Schiff (gCBS) equation [24] and a generalized Calagero-Bogoyavlenskii Konopelchenko equation [25], respectively. The CBS equation was first constructed by Bogoyavlenskii and Schiff in different ways [26, 27]. Namely, Bogoyavlenskii used the modified Lax formalism, whereas Schiff derived the same equation by reducing the self-dual Yang-Mills equation. In 2019, a class explicit lump solutions of the CBS-BK equation are constructed by using the Hirota bilinear approaches by Ren et al. [23]. The (2+1)-dimensional CBS equation also can be derived from the Korteweg-de Vries equation [28, 29]. Moreover, the BK equation is used as the interaction of a Rieman wave propagation [30], so we called the (2+1)-dimensional nonlinear partial differential equation (1) as gCBS-BK equation. These two equations have been widely studied in different ways [29, 31–40].


## 2. The Bilinear Equation for gCBS-BK Equation

If we take(2)u=2∂xlnf,where fx,y,t is an unknown real function, the bilinear equation of Equation (1) can be presented
(3)DtDx+Dx3Dy+δ1DxDy+δ2Dy2+δ3Dx2+δ4Dx4+δ5Dy4+δ6DxDy3f·f=0,where Dt,Dx are all bilinear derivative operators and D-operator [2] is defined by
(4)DxmDtnax,t·bx,t=∂t−∂t′n∂x−∂x′max,tbx′,t′x′=x,t′=t,where m and n are the positive integers, ax,t is the function of x and t, and bx,t is the function of the formal variables x′ and t′.


## 3. Breather Wave Solutions of CBS-BK Equation

In this section, we will use the extended homoclinic text method [41, 42] to get the breather wave solutions of Equation (1). To start with,
(5)fx,y,t=k1expξ1+exp−ξ1+k2cosξ2+a9,where ξ1 and ξ2 are defined by
(6)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,where ai, i=1,⋯,8, k1, and k2 are all real numbers. Substituting Equation (5) into Equation (3), we can get the following.where a1, a5, and a7 are some free real numbers.Case 1.
(7)a5=0,a7=−a63a1a22δ6−a1a62δ6+4a23δ5−4a2a62δ5+a13+a1δ1+2a2δ2a1,a9=0.
Substituting Equation (7) into Equation (5), through the transformation (2), we have
(8)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2,where ξ1 and ξ2 are given by
(9)ξ1=a1x+a2y+a3t+a4,ξ2=a6y−a63a1a22δ6−a1a62δ6+4a23δ5−4a2a62δ5+a13+a1δ1+2a2δ2a1t+a8,where ai, i=1,⋯,8, δ1, δ2, δ5, and δ6 are real numbers. Figure 1 described the evolution of solution (8).Case 2.
(10)a2=0,a3=a1a7a5,a6=0,a9=0,Figure 1
Spatiotemporal structure of solution (8) with the parameter selections a1=1, a2=1, a3=1, a6=1, k1=1, δ1=1, δ2=1, δ5=1, and δ6=1.Substituting Equation (10) into Equation (5), through the transformation (2), we have
(11)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2,where ξ1 and ξ2 are determined by
(12)ξ1=a1x+a1a7a5t+a4,ξ2=a5x+a7t+a8,where a1, a4, a5, a7, and a8 are real numbers. Therefore, the dynamic behavior can be performed in Figure 2.Case 3.
(13)a2=0,a3=a12a12a5δ4+2a53δ4+a7a5,a6=0,a9=0,k1=−k22a524k12.Figure 2
Spatiotemporal structure of solution (11) with the parameter selections a1=1, a5=1, and a7=1.Substituting Equation (13) into Equation (5), through the transformation (2), we have
(14)ux,y,t=−2k22a52/4k12a1expξ1+2k22a52/4k12a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2,where ξ1 and ξ2 are given by
(15)ξ1=a1x+a12a12a5δ4+2a53δ4+a7a5t+a4,ξ2=a5x+a7t+a8,where a1, a4, a5, a7, a8, k1, k2, and δ4 are free real numbers. Figure 3 described the evolution of solution (14).Case 4.
Substitutinga1, k1, and a7 into Equation (5), through the transformation (2), we have
(16)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2.Figure 3
Spatiotemporal structure of solution (14) with the parameter selections a1=1, a5=1, a7=1, k1=1, k2=1, and δ4=1.The evolution of solution (16) is described in Figure 4.Figure 4
Spatiotemporal structure of solution (16) with the parameter selections a2=1, a5=1, a6=1, k2=1, δ1=1, δ2=1, δ3=1, and δ5=1.ξ1 and ξ2 are given by
(17)ξ1=−a2a5a6x+a2y+a3t+a4,ξ2=a5x+a6y+8a22a64δ5−8a66δ5+4a22a53a6+4a53a63−3a22a52δ3−3a23a53a6δ1−3a22a62δ2−3a52a62δ3−3a5a63δ1+a64δ23a5a22+a62t+a8,where a2, a3, a4, a5, a6, a8, δ1, and δ2 are real numbers.Case 5.
(18)a2=0,a3=a1a12a6−2a5δ32a5,a7=a63δ6+a12a62+a52a6−a5δ3−a6δ1.Substituting Equation (18) into Equation (5), through the transformation (2), we have
(19)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2.The evolution of solution (19) is described in Figure 5.Figure 5
Spatiotemporal structure of solution (19) a1=1, a5=1, a6=1, k1=1, k2=1, δ3=1, and δ6=1.ξ1 and ξ2 are given by
(20)ξ1=a1x+a1a12a6−2a5δ32a5t+a4,ξ2=a5x+a6y+a63δ6+a12a62+a52a6−a5δ3−a6δ1t+a8,where a1, a4, a5, a6, a8, k1, k2, δ3, and δ6 are free real numbers. The three-dimensional dynamic figure can be drawn as Figure 5.Case 6.
(21)a1=0,a3=−a2a22δ6−a52+δ1,a6=0.Substituting Equation (21) into Equation (5), through the transformation (2), we have
(22)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2,where ξ1 and ξ2 are defined by
(23)ξ1=a2y−a2a22δ6−a52+δ1t+a4,ξ2=a5x+a7t+a8,where a2, a4, a5, a7, a8, k1, k2, δ1, δ5, and δ6 are some free real numbers. The figure is given as Figure 6.Case 7.
(24)a2=−a6,a3=4a13a6−3a12δ3+3a1a6δ1−a62δ23a1,a7=4a13a6−3a12δ3−3a1a6δ1−a62δ13a1,k1=k224,where a1, a6, a7, k2, δ1, δ2, δ3, and δ5 are free real numbers. Substituting Equation (24) into Equation (5), through the transformation (2), we have
(25)ux,y,t=2k22/4a1expξ1−2k22/4a1exp−ξ1−2k2a5sinξ2k22/4expξ1+exp−ξ1+k2cosξ2.Figure 6
Spatiotemporal structure of solution (22) with the parameter selections a2=1, a5=1, a7=1, k1=1, k2=1, δ1=1, δ5=1, and δ6=1.The figure is given as Figure7.Figure 7
Spatiotemporal structure of solution (25) with the parameter selections a1=1, a6=1, a7=1, k2=1, δ1=1, δ2=1, δ3=1, and δ5=1.ξ1 and ξ2 are followed by
(26)ξ1=a1x+a2y−4a13a6+3a12δ3+3a1a6δ1+a62δ23a62δ1t+a4,ξ2=−a1x+a6y+4a13a6+3a12δ3−3a1a6δ1+a62δ13a1t+a8,where a1, a4, a5, a6, a7, a8, k2, δ1, δ2, δ3, and δ5 are free real numbers.Case 8.
(27)a3=−4a13a6+3a12δ3+3a1a6δ1+a62δ23a62δ1,a5=−a1,a7=4a13a6+3a12δ3−3a1a6δ1+a62δ13a1,where a1, a6, a7, k2, δ1, δ2, δ3, and δ5 are free real numbers. Substituting Equation (27) into Equation (5), through the transformation (2), we have
(28)ux,y,t=2k1a1expξ1−2k1a1exp−ξ1−2k2a5sinξ2k1expξ1+exp−ξ1+k2cosξ2.The figure is drawn as Figure8.Figure 8
Spatiotemporal structure of solution (28) with the parameter selections a1=1, a6=1, a7=1, k2=1, δ1=1, δ2=1, δ3=1, and δ5=1.ξ1 and ξ2 are defined by
(29)ξ1=a1x+a2y−4a13a6+3a12δ3+3a1a6δ1+a62δ23a62δ1t+a4,ξ2=−a1x+a6y+4a13a6+3a12δ3−3a1a6δ1+a62δ13a1t+a8,where a1, a2, a4, a6, a7, a8, k2, δ1, δ2, δ3, and δ5 are free real numbers.


## 4. Interaction Solutions of CBS-BK System

### 4.1. Interaction between a Lump and One-Kink Soliton

With the help of Maple, we will discuss the interaction between a lump and one-kink soliton by takingfx,y,t as a combination of positive quadratic function and one exponential function, that is,
(30)fx,y,t=ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(31)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. In order to get the interaction solutions of Equation (1), substituting Equation (30) into Equation (2),
(32)ux,y,t=4a1ξ1+4a2ξ2+2k1expξ3ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(33)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. Substituting Equation (30) into Equation (3), through complex analysis and calculations, we can have the following.Case 1.
(34)a1=0,a6=0,k1=0,δ2=0,δ4=0,δ5=0,q1=p1a2δ6p12−a3a2,δ1=−a7a5,where a2, a3, a5, a7, p1, and δ6 are free real numbers. Substituting Equation (34) into Equation (32), we have
(35)ux,y,t=2a5a5x+a6y+a7t+a8a9+a2y+a3t+a42+a5x+a7t+a82+b1expp1y+p1a2δ6p12−a3/a2t+r1.Case 2.
(36)a1=0,a5=0,δ2=0,δ5=0,δ6=0,a3=a7a2a6,δ1=a6k12+a7a6,q1=−a6k13δ4+a6k1δ3−a7p1a6,where a2, a6, a7, k1, p1, δ3, and δ4 are some free real numbers. Substituting Equation (36) into Equation (32), we have
(37)ux,y,t=4a1ξ1+4a2ξ2+expξ3ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(38)ξ1=a2y+a7a2a6t+a4,ξ2=a6y+a7t+a8,ξ3=k1x+p1y−a6k13δ4+a6k1δ3−a7p1a6t+r1,where a2, a4, a6, a7, a8, k1, p1, r1, δ3, and δ4 are some free real numbers.In order to obtain the dynamic feature, we choose Case2 to analyse. The three-dimensional dynamic graphs are drawn as Figure 9. We can find that the lump waves and the exponential function waves interact with each other and keep moving in the opposite direction.Figure 9
Spatiotemporal structure of solution (37) with the parameter selections: (a) t=−10, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1; (b) t=0, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1; (c) t=−10, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1.
(a)(b)(c)


### 4.2. Interaction between a Lump and Periodic Waves

In order to get interaction solutions between a lump and periodic waves, we will takef as the combination of positive function and hyperbolic cosine function. Therefore, f can be determined by
(39)fx,y,t=ξ12+ξ22+b1cosξ3+a9,where variables are defined by
(40)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. Substituting Equation (39) into Equation (2), we can get the interaction solutions of Equation (1):
(41)ux,y,t=4a1ξ1+4a2ξ2−2b1k1sinξ3ξ12+ξ22+b1cosξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(42)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are real numbers. Through long and tedious calculations, we can get the following relations between the parameters.where a5, a7, b1, p1, δ1, δ2, and δ6 are free real numbers.where a5, a6, a7, p1, b1, δ1, and δ6 are free real numbers.where a2, a3, a5, a7, p1, and δ6 are some free real numbers.where a5, p1, δ1, and δ6 are free real numbers.Case 1.
(43)a1=0,a2=0,a3=0,a6=0,q1=p1P12δ6−δ1,δ3=−a7a5,δ4=−b12p12δ2a54,δ5=δ2p12,Case 2.
(44)a1=0,a2=0,a3=0,k1=0,q1=−p1δ1,δ2=−a5p12δ6a6,δ3=a6p12δ6−2a6δ1−2a7a5,Case 3.
(45)a1=0,a6=0,k1=0,q1=p1a2p12δ6+a3a2,δ1=−a3a2,δ2=0,δ3=−a7a5,δ4=0,δ5=0,Case 4.
(46)a2=0,a3=−a1δ3,k1=0,q1=p13δ6−p1δ1,δ2=0,δ4=0,δ5=0,When we change the coefficients of the equation, the value of Equation (47) will be different accordingly. In order to obtain the dynamic feature, we choose Case 2 to analyse. Taking Equation (44) into Equation (41), we can get
(47)ux,y,t=2a5a5+a7t+a8a9+a42+a5x+a6y+a7t+a82+b1cosp1y−p1δ1t+r1.With the help of Maple, the three-dimensional dynamic graphs are drawn as Figure10. We can find that lump waves and periodic waves interact with each other and keep moving in the opposite direction.Figure 10
Spatiotemporal structure of solution (47) with the parameter selections: (a) t=0, a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1; (b) t=10, a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1; (c) a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1.
(a)(b)(c)


## 4.1. Interaction between a Lump and One-Kink Soliton

With the help of Maple, we will discuss the interaction between a lump and one-kink soliton by takingfx,y,t as a combination of positive quadratic function and one exponential function, that is,
(30)fx,y,t=ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(31)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. In order to get the interaction solutions of Equation (1), substituting Equation (30) into Equation (2),
(32)ux,y,t=4a1ξ1+4a2ξ2+2k1expξ3ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(33)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. Substituting Equation (30) into Equation (3), through complex analysis and calculations, we can have the following.Case 1.
(34)a1=0,a6=0,k1=0,δ2=0,δ4=0,δ5=0,q1=p1a2δ6p12−a3a2,δ1=−a7a5,where a2, a3, a5, a7, p1, and δ6 are free real numbers. Substituting Equation (34) into Equation (32), we have
(35)ux,y,t=2a5a5x+a6y+a7t+a8a9+a2y+a3t+a42+a5x+a7t+a82+b1expp1y+p1a2δ6p12−a3/a2t+r1.Case 2.
(36)a1=0,a5=0,δ2=0,δ5=0,δ6=0,a3=a7a2a6,δ1=a6k12+a7a6,q1=−a6k13δ4+a6k1δ3−a7p1a6,where a2, a6, a7, k1, p1, δ3, and δ4 are some free real numbers. Substituting Equation (36) into Equation (32), we have
(37)ux,y,t=4a1ξ1+4a2ξ2+expξ3ξ12+ξ22+expξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(38)ξ1=a2y+a7a2a6t+a4,ξ2=a6y+a7t+a8,ξ3=k1x+p1y−a6k13δ4+a6k1δ3−a7p1a6t+r1,where a2, a4, a6, a7, a8, k1, p1, r1, δ3, and δ4 are some free real numbers.In order to obtain the dynamic feature, we choose Case2 to analyse. The three-dimensional dynamic graphs are drawn as Figure 9. We can find that the lump waves and the exponential function waves interact with each other and keep moving in the opposite direction.Figure 9
Spatiotemporal structure of solution (37) with the parameter selections: (a) t=−10, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1; (b) t=0, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1; (c) t=−10, a2=1, p1=1, a3=1, a7=1, a5=1, and δ6=1.
(a)(b)(c)


## 4.2. Interaction between a Lump and Periodic Waves

In order to get interaction solutions between a lump and periodic waves, we will takef as the combination of positive function and hyperbolic cosine function. Therefore, f can be determined by
(39)fx,y,t=ξ12+ξ22+b1cosξ3+a9,where variables are defined by
(40)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are all real numbers. Substituting Equation (39) into Equation (2), we can get the interaction solutions of Equation (1):
(41)ux,y,t=4a1ξ1+4a2ξ2−2b1k1sinξ3ξ12+ξ22+b1cosξ3+a9,where ξ1, ξ2, and ξ3 are defined by
(42)ξ1=a1x+a2y+a3t+a4,ξ2=a5x+a6y+a7t+a8,ξ3=k1x+p1y+q1t+r1,where ai, i=1,⋯,9, p1, k1, r1, and q1 are real numbers. Through long and tedious calculations, we can get the following relations between the parameters.where a5, a7, b1, p1, δ1, δ2, and δ6 are free real numbers.where a5, a6, a7, p1, b1, δ1, and δ6 are free real numbers.where a2, a3, a5, a7, p1, and δ6 are some free real numbers.where a5, p1, δ1, and δ6 are free real numbers.Case 1.
(43)a1=0,a2=0,a3=0,a6=0,q1=p1P12δ6−δ1,δ3=−a7a5,δ4=−b12p12δ2a54,δ5=δ2p12,Case 2.
(44)a1=0,a2=0,a3=0,k1=0,q1=−p1δ1,δ2=−a5p12δ6a6,δ3=a6p12δ6−2a6δ1−2a7a5,Case 3.
(45)a1=0,a6=0,k1=0,q1=p1a2p12δ6+a3a2,δ1=−a3a2,δ2=0,δ3=−a7a5,δ4=0,δ5=0,Case 4.
(46)a2=0,a3=−a1δ3,k1=0,q1=p13δ6−p1δ1,δ2=0,δ4=0,δ5=0,When we change the coefficients of the equation, the value of Equation (47) will be different accordingly. In order to obtain the dynamic feature, we choose Case 2 to analyse. Taking Equation (44) into Equation (41), we can get
(47)ux,y,t=2a5a5+a7t+a8a9+a42+a5x+a6y+a7t+a82+b1cosp1y−p1δ1t+r1.With the help of Maple, the three-dimensional dynamic graphs are drawn as Figure10. We can find that lump waves and periodic waves interact with each other and keep moving in the opposite direction.Figure 10
Spatiotemporal structure of solution (47) with the parameter selections: (a) t=0, a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1; (b) t=10, a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1; (c) a5=1, a6=1, a7=1, p1=1, b1=1, δ1=1, and δ6=1.
(a)(b)(c)


## 5. Conclusions

In this paper, based on a bilinear differential equation, we study the breather wave solutions and the interaction solutions of the mixed Calogero-Bogoyavlenskii-Schiff and Bogoyavlensky-Konopelchenko equations. Compared with the existing results in the literature, our results are new. It will be extensively used to report many attractive physical phenomena in the fields of acoustics, heat transfer, fluid dynamics, classical mechanics, and so on. It is demonstrated that the Hirota operators are very simple and powerful in constructing new nonlinear differential equations, which possess nice math properties. It is interesting to study the interaction solutions between soliton solutions and period solution by makingf as a combination of exponential function and trigonometry function. However, this method can be applied to those equations which have Hirota bilinear forms. Furthermore, we also can study the quardrilinear forms and even polylinearity forms of this equation in the future. These questions may also be interesting and worth studying.



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