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This paper constructs exact solutions for the (2 + 1)-dimensional KdV-Calogero-Bogoyavlenkskii-Schiff equation with the help of symbolic computation. By means of the truncated Painlev expansion, the (2 + 1)-dimensional KdV-Calogero-Bogoyavlenkskii-Schiff equation can be written as a trilinear equation, through the trilinear-linear equation, we can obtain the explicit representation of exact solutions for the (2 + 1)-dimensional KdV-Calogero-Bogoyavlenkskii-Schiff equation. We have depicted the profiles of the exact solutions by presenting their three-dimensional plots and the corresponding density plots.

Over the past few decades, the research of the nonlinear evolution equations (NLEEs) is flourishing because of the rich findings of these equations. NLEEs display many interesting nonlinear dynamic behaviors, such as plasma physics, optical systems, ocean, superfluids, hydrodynamics and other nonlinear fields [

In this paper, we focus on the following (2 + 1)-dimensional KdV-Calogero-Bogoyavlenkskii-Schiff (KdV-CBS) equation [

4 u t − α ( 4 u u y + 2 u x ∂ x − 1 u y + 2 u x x y ) − β ( 6 u u x + u x x x ) = 0 , (1)

which with two arbitrary constant α and β , Equation (1) can be reduced to some other equations with physical meanings. If we setting α ≠ 0 , β = 0 , Equation (1) is reduced to Calogero-Bogoyavlenskii-Schiff equation [

4 u t − α ( 4 u u y + 2 u x ∂ x − 1 u y + u x x y ) = 0. (2)

If setting { α = 0 , β ≠ 0 } , Equation (1) is famous KdV equation

4 u t − β ( u x x x + 6 u u x ) = 0. (3)

Let v x = u y , Equation (1) is transformed into the following system

u t − α ( 4 u u y + 2 u x v + u x x y ) − β ( 6 u u x + u x x x ) = 0 , (4a)

v x − u y = 0. (4b)

By applying Painlevé truncated extension, some types of explicit solutions (1) were studied in [

The outline of this paper is as follows. In Section 2, we convert the original KdV-CBS equation to a trilinear equation by using the truncated Painlevé expansion. In Section 3, we derived exact solutions for the KdV-CBS equation by taking function f as a positive quadratic function; another kind of exact solutions were got by taking function f as a positive quadratic and an exponential function; by taking function f as two exponential terms and the quadratic function, we obtained three kinds of exact solutions respectively via different parameters. A short conclusion is included in the last section.

Based on the Painlevé analysis proposed in Ref. [

u = u 0 + u 1 f + u 2 f 2 , v = v 0 + v 1 f + v 2 f 2 , (5)

with u 0 , u 1 , v 0 , v 1 , v 2 , f being the function of x , y and t. We have

v 1 = f x y , v 2 = − 2 f x f y , u 2 = − 2 f x 2 , u 0 = λ − f x x x 2 f x + f x x 2 4 f x 2 , u 1 = 2 f x x , v 0 = − 3 β λ α + 2 f t α f x − 2 λ f y f x + 3 β 4 α f x x 2 f x 2 + f x x f x y f x 2 − f x x y f x − β 2 α f x x x f x , (6)

the Equation (6) is from Ref. [

α f x x x y f x 2 + 4 α λ f x 2 f x y + β f x x x x f x 2 − α f f x f x x x f x y − 4 β f x x x f x f x x − 3 α f x f x x f x y − 4 λ α f x x f x f y + 3 α f x y f x x 2 + 3 β f x x 3 − 4 f x 2 f x t + 4 α 4 f x 2 f x f x = 0. (7)

By solving the Equation (5), we can get

u = λ − 1 2 f x x x f x + f x x 2 4 f x 2 , (8a)

v = − 3 β λ α + 2 f t α f x − 2 λ f y f x + 3 β 4 α f x x 2 f x 2 + f x x f x y f x 2 − f x x y f x − β 2 α f x x x f x . (8b)

Based on the idea in Ref. [

f = ( a 1 x + a 2 y + a 3 t + a 4 ) 2 + ( a 5 x + a 6 y + a 7 t + a 8 ) 2 + a 9 + m exp ( k 1 x + k 2 y + k 3 t + k 4 ) + n exp ( − ( k 1 x + k 2 y + k 3 t + k 4 ) ) , (9)

where a i , k , k j ,1 ≤ i ≤ 9,1 ≤ j ≤ 4 are real parameters to be determined. Substituting function (9) into trilinear Equation (7), we give five kinds of exact solutions of Equation (4a) and Equation (4b) via function (9).

Case I

Substitution function (9) into trilinear Equation (7), a set of constraining equations for the parameters can be obtained as following

a 1 = 0 , a 2 = a 2 , a 3 = a 2 λ α , a 4 = a 4 , a 5 = a 5 , a 6 = − a 5 β λ , a 7 = a 7 , a 8 = a 8 , a 9 = a 9 , m = 0 , n = 0 , k 1 = k 1 , k 2 = k 2 , k 3 = k 3 , k 4 = k 4 , β = β , λ = λ (10)

where

λ ≠ 0 , a 9 > 0 (11)

to guarantee that the corresponding f is positive. By substituting Equation (10) into function (9), then we get the function f reads

f = ( a 2 y + a 2 λ α t + a 4 ) 2 + ( a 5 x − a 5 β λ y + a 7 t + a 8 ) 2 + a 9 . (12)

By means of Equation (8a) and Equation (8b), the solutions of (4a) and (4b) can be obtained

u = λ + a 5 2 2 A 1 2 , (13a)

v = − 3 β λ α + 2 B 1 A 1 − 2 λ C 1 A 1 − β a 5 2 α A 1 2 , (13b)

where

f = g 2 + h 2 + a 9 , g = a 2 y + a 2 λ α t + a 4 , h = a 5 x − a 5 β λ y + a 7 t + a 8 , A 1 = 2 a 5 f , B 1 = 2 a 2 α λ f + 2 a 7 g , C 1 = 2 a 2 f − 2 a 5 β λ g . (14)

The solutions of u via Equation (13a) which can be seen in

Case II

a 1 = a 1 , a 2 = − a 1 β α , a 3 = a 3 , a 4 = a 4 , a 5 = 0 , a 6 = a 6 , a 7 = a 6 α λ , a 8 = a 8 , a 9 = a 9 , m = m , k 1 = 0 , k 2 = k 2 , k 3 = k 2 α λ , k 4 = k 4 , (15)

which should satisfy the constraint conditions

α ≠ 0 , a 9 > 0 , m > 0 , (16)

to ensure that the corresponding f is positive and well defined. By substituting Equation (15) into function (9), then the function f reads

f = ( a 1 x − a 1 β α y + a 3 t + a 4 ) 2 + ( a 6 y + a 6 α λ t + a 8 ) 2 + m exp ( k 2 y + k 2 α λ t + k 4 ) + a 9 . (17)

Using Equation (8a) and Equation (8b), the solutions for (4a) and (4b) can be obtained

u = λ + a 1 2 4 g 2 , (18a)

v = − 3 β λ α + A 2 α λ B 2 α a 1 g + 3 β a 1 2 16 λ g 2 − a 1 3 β α g 2 , (18b)

where

f = g 2 + h 2 + m exp ( ξ ) + a 9 , g = a 1 x − a 1 β α y + a 3 t + a 4 , h = a 6 y + a 6 α λ t + a 8 , A 2 = 2 a 3 g + 2 a 6 α λ h + k 2 m α λ exp ( ξ ) , B 2 = − 2 a 1 β g α + 2 a 6 h + k 2 m exp ( ξ ) , ξ = k 2 y + k 2 α λ t + k 4 , (19)

The solutions of u via Equation (18a) can be seen in

Case III

a 1 = a 1 , a 2 = a 2 , a 3 = a 3 , a 4 = a 4 , a 5 = 0 , a 6 = a 6 , a 7 = a 6 α λ , a 8 = a 8 , a 9 = a 9 , α = α , β = a 2 α a 1 , k 1 = 0 , k 2 = k 2 , k 3 = k 2 α λ , k 4 = k 4 , λ = λ , m = m , n = n , (20)

where

a 1 ≠ 0 , a 9 > 0 , m > 0 , n > 0. (21)

to ensure that the corresponding f is positive and well defined. By substituting Equation (20) into function (9),

f = ( a 1 x + a 2 y + a 3 t + a 4 ) 2 + ( a 6 x + a 6 α λ t + a 8 ) 2 + a 9 + m exp ( k 2 y + k 2 α λ t + k 4 ) + n exp ( − ( k 2 y + k 2 α λ t + k 4 ) ) , (22)

where a i , k j ,1 ≤ i ≤ 9,1 ≤ j ≤ 4 are all real parameters to be determined. Then we get the solutions of the Equation (4a) and Equation (4b)

u = λ + a 1 2 4 g 2 , (23a)

v = − 3 a 2 λ a 1 + 2 B 3 − λ α C 3 α A 3 + a 1 3 a 2 A 3 2 , (23b)

with

f = h 2 + g 2 + a 9 + m exp ( ξ ) + n exp ( − ξ ) , g = a 1 x + a 2 y + a 3 t + a 4 , h = a 6 x + a 6 α λ t + a 8 , ξ = k 2 y + k 2 α λ t + k 4 , A 3 = 2 a 1 g , C 3 = 2 a 2 g + 2 a 6 h + m k 2 exp ( ξ ) − n k 2 exp ( − ξ ) B 3 = 2 a 3 g + 2 a 6 α λ h + m k 2 α λ exp ( ξ ) − n k 2 α λ exp ( ξ ) (24)

The exact solutions for Equation (23a) and Equation (23b) can be seen in

Case IV

a 1 = a 1 , a 2 = − a 1 β α , a 3 = − a 1 a 7 α a 6 , a 4 = − a 1 a 8 a 6 α , a 5 = − α a 6 β , a 6 = a 6 , a 7 = a 7 , a 8 = a 8 , a 9 = a 9 , m = m , n = n , α = α , β = β , k 1 = 0 , k 2 = k 2 , k 3 = k 2 α λ , k 4 = k 4 , λ = λ (25)

where

α β a 6 ≠ 0 , a 9 > 0 , m > 0 , n > 0. (26)

to guarantee that the corresponding f is positive and well defined. We substitute Equation (25) into Equation (9), hence, reinstall function f as the following formula:

f = ( a 1 x − a 1 β α y − a 1 a 7 α a 6 t − a 1 a 8 a 6 α ) 2 + ( a 5 x + a 6 y + a 7 t + a 8 ) 2 + a 9 + m exp ( k 2 y + k 2 α λ t + k 4 ) + n exp ( − ( k 2 y + k 2 α λ t + k 4 ) ) , (27)

where a i , k j ,1 ≤ i ≤ 9,1 ≤ j ≤ 4 are all real parameters to be determined. Then, we get

u = λ + D 4 2 A 4 2 , (28a)

v = − β λ α + 2 C 4 α A 4 + 2 λ B 4 A 4 + D 4 ( 3 β D 4 + 4 α E 4 ) 4 α A 4 2 , (28b)

where

f = h 2 + g 2 + α exp ( ξ ) + β exp ( − ξ ) + a 9 , h = a 1 x − a 1 β α y − a 1 a 7 α a 6 t − a 1 a 8 a 6 α , g = a 5 x + a 6 y + a 7 t + a 8 , ξ = k 2 y + k 2 α λ t + k 4 ,

A 4 = 2 ( a 1 h + a 5 g ) , B 4 = − 2 a 1 β α h + 2 a 6 g + k 2 ( m exp ( ξ ) − n exp ( − ξ ) ) , C 4 = − 2 a 1 a 7 α a 6 h + 2 a 7 g + k 2 α λ ( m exp ( ξ ) − n exp ( − ξ ) ) , D 4 = 2 ( a 1 2 + a 5 2 ) , E 4 = 2 ( a 5 a 6 − a 1 2 β α ) . (29)

The exact solution for Equation (28b) can be seen in

Case V.

a 1 = 0 , a 2 = a 2 , a 3 = α λ a 2 , a 4 = a 4 , a 5 = a 6 k 1 k 2 a 6 = a 6 , a 7 = a 7 , a 8 = a 8 , a 9 = a 9 , α = − β k 1 k 2 , β = β , k 1 = k 1 , k 2 = k 1 , k 3 = a 7 k 2 a 6 , k 4 = k 4 , m = m , n = n . (30)

with

k 2 a 6 ≠ 0 , a 9 > 0 , m > 0 , n > 0. (31)

to guarantee that the corresponding f is positive and well defined. We substitute Equation (30) into Equation (9), then function f reads

f = ( a 2 y + α λ a 2 t + a 4 ) 2 + ( a 6 k 1 k 2 x + a 6 y + a 7 t + a 8 ) 2 + a 9 + m exp ( k 1 x + k 2 y + a 7 k 2 a 6 t + k 4 ) + n exp ( − ( k 1 x + k 2 y + a 7 k 2 a 6 t + k 4 ) ) , (32)

where a i , k j ,1 ≤ i ≤ 9,1 ≤ j ≤ 4 are all real parameters to be determined. Then, the rational solution of system (4a) and (4a) can be got again.

u = λ − k 1 3 ( m exp ( ξ ) − n exp ( − ξ ) ) 2 A 5 + D 5 4 A 5 2 , v = − 3 β λ α + 2 B 5 A 5 + 2 λ C 5 A 5 , + 3 β D 5 2 4 α A 5 2 + D 5 E 5 A 5 2 − F 5 A 5 − β G 5 2 α A 5 , (33)

and where

f = g 2 + h 2 + m exp ( ξ ) + n exp ( − ξ ) + a 9 , g = a 2 y + α λ a 2 t + a 4 , h = a 6 k 1 k 2 x + a 6 y + a 7 t + a 8 , ξ = k 1 x + k 2 y + a 7 k 2 a 6 t + k 4 , A 5 = 2 a 6 k 1 k 2 + k 1 h + m k 1 exp ( ξ ) − n k 1 exp ( − ξ ) ,

B 5 = 2 α λ a 2 g + 2 a 7 h + a 7 k 2 a 6 ( m exp ( ξ ) + n exp ( − ξ ) ) , C 5 = 2 a 2 g + 2 a 6 h + k 2 ( m exp ( ξ ) + n exp ( − ξ ) ) , D 5 = 2 a 6 2 k 1 2 k 2 2 + k 1 2 ( m exp ( ξ ) + n exp ( − ξ ) ) , E 5 = k 1 k 2 ( m exp ( ξ ) + n exp ( − ξ ) ) + 2 a 6 a 6 k 1 k 2 , F 5 = k 1 2 k 2 ( m exp ( ξ ) + n exp ( − ξ ) ) . (34)

The exact solutions for Equation (33) can be seen in

In this paper, by using the truncated Painlev expansion, the (2 + 1)-dimensional KdV-CBS equation can be changed to a trilinear Equation (7), and by means of symbolic computations, we presented five types of exact solutions ((13a), (13b), (18a), (18b), (23a), (23b), (28a), (28b), (33)) to the (2 + 1)-dimensional KdV-CBS equation. Three-dimensional plots and the corresponding density plots of these exact solutions are given respectively in

This work is supported by the National Natural Science Foundation of China under grant number 11571008.

The authors declare no conflicts of interest regarding the publication of this paper.

Li, Y. and Chaolu, T. (2020) Exact Solutions for (2 + 1)-Dimensional KdV-Calogero-Bogoyavlenkskii-Schiff Equation via Symbolic Computation. Journal of Applied Mathematics and Physics, 8, 197-209. https://doi.org/10.4236/jamp.2020.82015