new exact solutions of coupled boussinesq-burgers

Available online at www.sciencedirect.com

Journal of Ocean Engineering and Science 2 (2017) 34–46 www.elsevier.com/locate/joes

New exact solutions of coupled Boussinesq–Burgers equations by

Exp-function method

L.K. Ravi, S. Saha Ray

∗, S. Sahoo

National Institute of Technology, Department of Mathematics, Rourkela 769008, India

Received 3 August 2016; received in revised form 8 September 2016; accepted 18 September 2016 Available online 22 September 2016

Abstract

In the present paper, we build the new analytical exact solutions of a nonlinear differential equation, specifically, coupled Boussinesq–Burgers equations by means of Exp-function method. Then, we analyze the results by plotting the three dimensional soliton graphs for each case, which exhibit the simplicity and effectiveness of the proposed method. The primary purpose of this paper is to employ a new approach, which allows us victorious and efficient derivation of the new analytical exact solutions for the coupled Boussinesq–Burgers equations. © 2016 Shanghai Jiaotong University. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license. ( http://creativecommons.org/licenses/by-nc-nd/4.0/ )

Keywords: Exp-function method; Solitary wave solutions; Non-linear evolution equations; Coupled Boussinesq–Burgers equations.

a

t

q

a

a

w

v

b

a

t

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1. Introduction

A well-known model is the coupled Boussinesq–Burgers(BB) equations [1]

u t = −2u u x +

1

2

v x , (1.1)

v t =

1

2

u x x x − 2 (u v) x , (1.2)

where x and t represent the normalized space and time re-spectively. Here u ( x , t ) represents the horizontal velocity andat the leading order it is the depth averaged horizontal field,while v ( x , t ) denotes the height of the water surface abovethe horizontal level at the bottom.

The Boussinesq–Burgers equations emerge in the investi-gation of fluid flow and represent the proliferation of shallowwater waves [1] . Kaup [2] has been investigated IST integra-bility of Boussinesq–Burgers equation. The multi-phase peri-odic solutions of BB equations have been has studied in [3,4] .The soliton and multi-soliton of Eqs. (1.1) and ( 1.2 ) has beendiscussed in [5–7] .The exact travelling solutions of BB equa-tions have been discussed in [8,9] . El et al. [10,11] removed

∗ Corresponding author. E-mail address: [email protected] (S. Saha Ray).

E

t

e

http://dx.doi.org/10.1016/j.joes.2016.09.001 2468-0133/© 2016 Shanghai Jiaotong University. Published by Elsevier B.V. This( http://creativecommons.org/licenses/by-nc-nd/4.0/ )

n initial discontinuity of BB equations by applying Whithamheory of modulations. Kamchatnov et al. [12] described auasi-classical soliton trains soliton trains of BB equationsrising from a large initial pulse. By using Lie symmetrynalysis Mhlanga and Khalique [13] describe the travellingave solution of generalized coupled BB equation. The en-elope solition and periodic wave solutions have been studiedy Ebadi et al. [14] .

Recently a variety of powerful methods [15–18] have beenpplied in numerous physical models [19–21] and analyzedo obtain the exact solutions to nonlinear evolution equations22,23] .

In the present paper, we implement Exp-function methodo find the new exact Soliton Solitary wave solutions of Eqs.1.1) and ( 1.2 ). Exp-function method introduced by He and

u [24] in 2006, is successfully applied to many kind ofonlinear differential equations.

This method has been successfully implemented on Burg-rs equations, KdVs–Burgers–Kuramoto equation, two di-ensional Bratu-type equation, Generalized Fisher equation,

igher order boundary value problem, modified Zakharov–uznetsov equation; see [25–28] . This clearly indicates thatxp-function method is easy, concise and an effective method

o implement to nonlinear evolution equations arising in math-matical physics. The main advantage of this method over

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L.K. Ravi et al. / Journal of Ocean Engineering and Science 2 (2017) 34–46 35

t

s

2

i

c

o

4

2

N

uη

e

N

t

f

u

w

k

3B

(

w

w

η

g

w

w

l

u

v

w

a

i

u

v

w

w

A

B

a

ξ

δ

ξ

a

o

e

a

u

he other methods is that it gives more general solutions withome free parameters.

The rest of the paper is organized as follow: in Section , we proposed the algorithm of Exp-function method. Thenn Section 3 , we implement the proposed method to solveoupled Boussinesq–Burger equations and also we analyzedbtained the solutions by three dimensional graphs in Section . Some conclusions are presented in Section 5 .

. Basic idea of Exp-function method

Consider a given nonlinear wave equation

(u, u t , u x , u xx , u tt , u xt , . . . ) = 0, (2.1)

We seek its wave solutions

= u(η) ,

= kx + wt . (2.2)

Consequently, ( 2.1 ) is reduced to the ordinary differentialquation:

(u, wu

′ , ku

′ , k 2 u

′′ , w

2 u

′′ , kwu

′′ , . . . ) = 0 (2.3)

The Exp-function method is based on the assumption thathe travelling wave solution can be expressed in the followingorm:

(η) =

∑ c n = −d a n exp (nη) ∑ p

m= −q b m

exp (mη) (2.4)

here c , d , p and q are positive integers, a n and b m

are un-nown constants.

. Application of Exp-function to coupled

oussinesq–Burgers equations

Using the transformation u = u ( η), v = v ( η), η= kx + wt Eqs.1.1) and ( 1.2 ) become the ordinary differential equations,

u

′ + 2kuu

′ − 1

2

kv ′ = 0, (3.1)

v ′ − 1

2

k 3 u

′′′ + 2k(uv) ′ = 0, (3.2)

On integrating Eqs. (3.1) and ( 3.2 ) once with reference toand assuming that the constant of integration to be zero, weet

u + k u

2 − 1

2

kv = 0 (3.3)

v − 1

2

k 3 u

′′ + 2kuv = 0 (3.4)

According to Exp-function method, we assume that the so-ution of Eqs. (3.1) and ( 3.2 ) can be expressed in the form,

(η) =

∑ h n= −g a n exp (nη) ∑ p

m= −q b m

exp (mη)

=

a −g exp (−gη) + · · · + a h exp (hη)

b −q exp (−qη) + · · · + b p exp (pη) , (3.5)

(η) =

∑ j n= −i d n exp (nη) ∑ t

m= −s c m

exp (mη)

=

d −i exp (−iη) + · · · + d j exp ( jη)

c −s exp (−sη) + · · · + c t exp (tη) . (3.6)

here h , g , p , q , j , i , t and s are positive integers a n , b m

, d n

nd c m

are unknown constants. For simplicity, we set h = p = 1, g = q = 1 and j = t = 1,

= s = 1, then Eqs. (3.5) and ( 3.6 ) reduced to

(η) =

a −1 exp (−η) + a 0 + a 1 exp (η)

b −1 exp ( −η) + b 0 + exp (η) , (3.7)

(η) =

d −1 exp (−η) + d 0 + d 1 exp (η)

c −1 exp ( −η) + c 0 + exp (η) . (3.8)

Substituting Eqs. (3.7) and ( 3.8 ) into Eqs. (3.3) and ( 3.4 ),e have

1

A

{ δ−3 exp (−3 η) + δ−2 exp (−2η) + δ−1 exp (−η)

+ δ0 + δ1 exp (η) + δ2 exp (2η) + δ3 exp (3 η) } = 0, (3.9)

1

B

{ ξ−4 exp (−4η) + ξ−3 exp (−3 η) + ξ−2 exp (−2η)

+ ξ−1 exp (−η) + ξ0 + ξ1 exp (η)

+ ξ2 exp (2η) + ξ3 exp (3 η) + ξ4 exp (4η) } = 0. (3.10)

here

= ( b −1 exp (−η) + b 0 + exp (η) ) 2

( c −1 exp (−η) + c 0 + exp (η) ) ,

= ( b −1 exp (−η) + b 0 + exp (η) ) 3

( c −1 exp (−η) + c 0 + exp (η) ) ,

nd δ−3 , δ−2 , δ−1 , δ0 , δ1 , δ2 , δ3 , ξ −4 , ξ −3 , ξ −2 , ξ −1 , ξ 0 ,1 , ξ 2 , ξ 4 are defined in Appendix 1.

By setting,

−3 = δ−2 = δ−1 = δ0 = δ1 = δ2 = δ3 = 0, and

−4 = ξ−3 = ξ−2 = ξ−1 = ξ0 = ξ1 = ξ2 = ξ3 = ξ4 = 0

,

nd solving these system of algebraic equations simultane-usly with the aid of symbolic computation system of math-matical software, we obtain the following results,

Case I.

−1 = 0, a 0 = −√

w b 0 √

2

, a 1 = 0, b −1 = 0, d −1 = 0,

d 0 = −w b 0 , d 1 = 0, c −1 = b

2 0 ,

c 0 = 2 b 0 , k =

√

2w ,

Substituting these values into Eqs. (3.7) and ( 3.8 ), we get:

(x, t ) = −√

w b 0 √

2 [ b 0 + cos h(kx + wt ) + sin h(kx + wt )] ,

36 L.K. Ravi et al. / Journal of Ocean Engineering and Science 2 (2017) 34–46

(kx + wt ) ) + sin h(kx + wt ) .

, d 1 = 0, c −1 = b

2 0 , c 0 = 2 b 0 , k = −

√

2w

t:

(kx + wt ) ) + sin h(kx + wt ) .

:

(kx + wt ) ) + sin h(kx + wt ) ,

:

(kx + wt ) ) + sin h(kx + wt ) .

v(x, t ) = − w b 0

2 b 0 + cos h(kx + wt ) + b

2 0 ( cos h(kx + wt ) − sin h

Case II.

a −1 = 0, a 0 =

√

w b 0 √

2

, a 1 = 0, b −1 = 0, d −1 = 0, d 0 = −w b 0

Substituting these values into Eqs. (3.9) and ( 3.10 ), we ge

u(x, t ) =

√

w b 0 √

2 ( b 0 + cos h(kx + wt ) + sin h(kx + wt ) ) ,

v(x, t ) = − w b 0

2 b 0 + cos h(kx + wt ) + b


Case III.

a −1 = 0, a 0 = 0, a 1 = −√

w √

2

, b −1 = 0, d −1 = 0,

d 0 = −w b 0 , d 1 = 0, c −1 = b

2 0 ,

c 0 = 2 b 0 , k =

√

2w ,

Substituting these values into Eqs. (3.7) and ( 3.8 ), we get

u(x, t ) = −√

w ( cos h(kx + wt ) + sin h(kx + wt ) ) √

2 ( b 0 + cos h(kx + wt ) + sin h(kx + wt ) ) ,

v(x, t ) = − w b 0

2 b 0 + cos h(kx + wt ) + b


Case IV.

a −1 = 0, a 0 = 0, a 1 =

√

w √

2

, b −1 = 0, d −1 = 0,

d 0 = −w b 0 , d 1 = 0, c −1 = b

2 0 ,

c 0 = 2 b 0 , k = −√

2w ,


u(x, t ) =

√

w ( cos h(kx + wt ) + sin h(kx + wt ) ) √

2 ( b 0 + cos h( kx + wt ) + sin h(kx + wt ) ) ,

v(x, t ) = − w b 0

2 b 0 + cos h(kx + wt ) + b


Case V.

a −1 =

√

w b −1 √

2

, a 0 =

1

4

[ √

2w b 0 −√

2 M 1 ] , a 1 = 0, d −1 = 0,


d

c

c

:

u wt ) + cos h(kx + wt ) )

]

os h(2 kx + 2 wt ) + sin h(2 kx + 2 wt ) ) ,

v t ) + sin h(kx + wt ) ) ]

w b −1 + 2

√

w b

2 0 −

w

a ,

d

c

c

:

uwt ) + cos h(kx + wt ) )

]

os h(2 kx + 2 wt ) + sin h(2 kx + 2 wt ) ] ,

v t ) + sin h(kx + wt ) ) ]

√

w b −1 + 2

√

w b

2 0 +

w

a

0 =

−w

2 b 0 + w

3 / 2 M 1

2w

, d 1 = 0,

−1 =

[3 w

5 / 2 b −1 b 0 − w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

−w

5 / 2 b 0 + w

2 M 1 ,

0 =

2[2 w

5 / 2 b −1 − w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

−w

5 / 2 b 0 + w

2 M 1 , k = −

√

2w ,


(x, t ) =

[2

√

w b −1 + ( √

w b 0 − M 1 ) ( sin h(kx +2

√

2 ( b −1 + b 0 ( cos h(kx + wt ) + sin h(kx + wt ) ) + c

(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 − 2

√

w b 0 M 1 )( cos h(kx + w

× 1 (2

(−3

√

w b −1 b 0 +

√

w b

3 0 + ( b −1 − b

2 0 ) M 1 +

(−4

√

2 b 0 M 1 ) ( cos h(kx + wt ) + sin h(kx + wt ) ) +

(√

w b 0 −M 1 ) ( cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ) ) ) ,

here M 1 =

√

−4w b −1 + wb

2 0 .

Case VI.

−1 = −√

w b −1 √

2

, a 0 =

1

4

[ −√

2w b 0 −√

2 M 1 ] , a 1 = 0, d −1 = 0

0 =

−w

2 b 0 − w

3 / 2 M 1

2w

, d 1 = 0,

−1 =

[−3 w

5 / 2 b −1 b 0 + w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

]

w

5 / 2 b 0 + w

2 M 1 ,

0 =

2

[−2 w

5 / 2 b −1 + w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

[w

5 / 2 b 0 + w

2 M 1 ] , k =

√

2w ,


(x, t ) =

[−2

√

w b −1 − ( √

w b 0 + M 1 ) ( sin h(kx +2

√

2

[b −1 + b 0 ( cos h(kx + wt ) + sin h(kx + wt ) ) + c

(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 + 2

√


× 1

[2

(−3

√

w b −1 b 0 +

√

w b

3 0 + (−b −1 + b

2 0 ) M 1 +

(−4


(√

w b 0 +M 1 ) ( cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ) ) ] ,

here M 1 =

√

−4w b −1 + wb

2 0 .

Case VII.

−1 =

√

w b −1 √

2

, a 0 =

1

4

[ √

2w b 0 +

√

2 M 1 ] , a 1 = 0, d −1 = 0,


:

wt ) + cos h(kx + wt ) ) ]

os h(2 kx + 2 wt ) + sin h(2 kx + 2 wt ) ] ,

) + sinh(kx + wt ) ) ]

√

w b −1 + 2

√

w b

2 0 +

,

:

wt ) + cos h(kx + wt ) ) ]

cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ]

x + wt ) + sin h(kx + wt ) ) ]

w b −1 + 2

√

w b

2 0 −

− M 1 ) ( cos h(2kx + 2wt ) + sin h(2kx + 2wt )) ) ]

d 0 =

−w

2 b 0 − w

3 / 2 M 1

2w

, d 1 = 0,

c −1 =

[−3 w

5 / 2 b −1 b 0 + w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

]

w

5 / 2 b 0 + w

2 M 1 ,

c 0 =

2

[−2 w

5 / 2 b −1 + w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

[w

5 / 2 b 0 + w

2 M 1 ] , k = −

√

2w ,


u(x, t ) =

[2

√

w b −1 +

(√

w b 0 + M 1 )( sin h(kx +

2

√

2

[b −1 + b 0 ( cos h(kx + wt ) + sin h(kx + wt ) ) + c

v(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 + 2

√

w b 0 M 1 )( cosh(kx + wt

× 1 [2

(−3

√

w b −1 b 0 +

√

w b

3 0 + (−b −1 + b

2 0 ) M 1 + (−4

2 b 0 M 1 )( cos h(kx + wt ) + sin h(kx + wt )) + ( √

w b 0

M 1 )( cos h(2kx + 2wt ) + sin h(2kx + 2wt )) ) ] ,

where M 1 =

√

−4w b −1 + wb

2 0 .

Case VIII.

a −1 = −√

w b −1 √

2

, a 0 =

1

4

[ −√

2w b 0 +

√

2 M 1 ] , a 1 = 0, d −1 = 0

d 0 =

−w

2 b 0 + w

3 / 2 M 1

2w

, d 1 = 0,

c −1 =

[3 w

5 / 2 b −1 b 0 − w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

]

−w

5 / 2 b 0 + w

2 M 1 ,

c 0 =

2

[2 w

5 / 2 b −1 − w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

[−w

5 / 2 b 0 + w

2 M 1 ] , k =

√

2w ,


u(x, t ) =

[−2

√

w b −1 −(√

w b 0 − M 1 )( sin h(kx +

2

√

2

[b −1 + b 0 ( cos h(kx + wt ) + sin h(kx + wt ) ) +

v(x, t ) = −[√

w × (−4w b −1 + 2wb

2 0 − 2

√

w b 0 M 1 ) × ( cos h(k

× 1 [2

(−3

√

w b −1 b 0 +

√

w b

3 0 + ( b −1 − b

2 0 ) M 1 + (−4

√

2 b 0 M 1 )( cos h(kx + wt ) + sin h(kx + wt )) + ( √

w b 0

where M 1 =

√

−4w b −1 + wb

2 0 .

Case IX.

a −1 = 0, a 0 =

1

4

( √

2w b 0 + M 1 ) , a 1 =

√

w √

2

, d −1 = 0,

d 0 =

−w

2 b 0 + w

3 / 2 M 1

2w

, d 1 = 0,


c

c

:

u 2

√

w ( cos h(kx + wt )

) + sin h(kx + wt ) ) +

v wt ) + sin h(kx + wt ) ) ]

w b −1 + 2

√

w b

2 0 −

− M 1 )

w

a

d

c

c

:

u 2

√

w ( cos h(kx + wt )

t ) + sin h(kx + wt ) ) +

v t ) + sin h(kx + wt ) ) ]

w b −1 + 2

√

w b

2 0 −

− M 1 )

w

−1 =

(3 w

5 / 2 b −1 b 0 − w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1 )

−w

5 / 2 b 0 + w

2 √

w(−4 b −1 + b

2 0 )

,

0 =

2(2 w

5 / 2 b −1 − w

5 / 2 b

2 0 + w

2 b 0 M 1 )

(−w

5 / 2 b 0 + w

2 M 1 ) , k = −

√

2w ,


(x, t ) =

[( cos h(kx + wt ) + sin h(kx + wt ) )

(√

w b 0 + M 1 ) +

+ sin h(kx + wt ) ) ] × 1 (2

√

2 [ b −1 + b 0 ( cos h(kx + wt

cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ] )

(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 − 2

√

w b 0 M 1 ) × ( cos h(kx +

× 1 [2

(−3

√

w b −1 b 0 +

√

w b

3 0 + ( b −1 − b

2 0 ) M 1 +

(−4

√


(√

w b 0

( cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ) ) ] ,

here M 1 =

√

−4w b −1 + wb

2 0 .

Case X.

−1 = 0, a 0 =

1

4

(−√

2w b 0 −√

2 M 1 ) , a 1 = −√

w √

2

, d −1 = 0,

0 =

−w

2 b 0 + w

3 / 2 M 1

2w

, d 1 = 0,

−1 =

(3 w

5 / 2 b −1 b 0 − w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1 )

−w

5 / 2 b 0 + w

2 M 1 ,

0 =

2(2 w

5 / 2 b −1 − w

5 / 2 b

2 0 + w

2 b 0 M 1 )

(−w

5 / 2 b 0 + w

2 M 1 ) , k =

√

2w ,


(x, t ) = −[( cos h(kx + wt ) + sin h(kx + wt ) )

(√

w b 0 + M 1 +sin h(kx + wt ) ) ) ] + × 1 (

2

√

2 [ b −1 + b 0 ( cos h(kx + w

cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ] )

(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 − 2

√


× 1 [2

(−3

√

w b −1 b 0 +

√

w b

3 0 + ( b −1 − b

2 0 ) M 1 +

(−4

√


(√

w b 0


here M 1 =

√

−4w b −1 + wb

2 0 .


:

2

√

w ( cos h(kx + wt )

t ) + sin h(kx + wt ) )

t ) + sin h(kx + wt ) ) ]

4

√

w b −1 + 2

√

w b

2 0 +

+ M 1 )

:

2

√

w ( cos h(kx + wt )

t ) + sin h(kx + wt ) ) +

t ) + sin h(kx + wt ) ) ]

√

w b −1 + 2

√

w b

2 0 +

+ M 1 )

Case XI.

a −1 = 0, a 0 =

1

4

( √

2w b 0 −√

2 M 1 ) , a 1 =

√

w √

2

, d −1 = 0,

d 0 =

−w

2 b 0 − w

3 / 2 M 1

2w

, d 1 = 0,

c −1 =

[−3 w

5 / 2 b −1 b 0 + w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

]

w

5 / 2 b 0 + w

2 M 1 ,

c 0 =

2

[−2 w

5 / 2 b −1 + w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

[w

5 / 2 b 0 + w

2 M 1 ] , k = −

√

2w ,


u(x, t ) =

[( cos h(kx + wt ) + sin h(kx + wt ) )

(√

w b 0 − M 1 ) +

+ sin h(kx + wt ) ) ] × 1 [ 2

√

2 ( b −1 + b 0 ( cos h(kx + w

+ cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ) ]

v(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 + 2

√


× 1 [(2

(−3

√

w b −1 b 0 +

√

w b

3 0 + (−b −1 + b

2 0 ) M 1 +

(−2 b 0 M 1 ) ( cos h(kx + wt ) + sin h(kx + wt ) ) +

(√

w b 0


where M 1 =

√

−4w b −1 + wb

2 0 .

Case XII.

a −1 = 0, a 0 =

1

4

(−√

2w b 0 +

√

2 M 1 , a 1 = −√

w √

2

, d −1 = 0,

d 0 =

−w

2 b 0 − w

3 / 2 M 1

2w

, d 1 = 0,

c −1 =

[−3 w

5 / 2 b −1 b 0 + w

5 / 2 b

3 0 − w

2 ( b −1 − b

2 0 ) M 1

]

w

5 / 2 b 0 + w

2 M 1 ,

c 0 =

2

[−2 w

5 / 2 b −1 + w

5 / 2 b

2 0 + w

2 b 0 M 1 ]

( w

5 / 2 b 0 + w

2 M 1 ) , k =

√

2w ,


u(x, t ) = [ ( cos h(kx + wt ) + sin h(kx + wt ) ) (−√

w b 0 + M 1 −+ sin h(kx + wt ) ) ) ] × 1 (

2

√

2 [ b −1 + b 0 ( cos h(kx + w

( cos h(2kx + 2wt ) + sin h(2kx + 2wt ) ) ] )

v(x, t ) = −[√

w

(−4w b −1 + 2wb

2 0 + 2

√


× 1 [2

(−3

√

w b −1 b 0 +

√

w b

3 0 + (−b −1 + b

2 0 ) M 1 +

(−4


(√

w b 0


where M 1 =

√

−4w b −1 + wb

2 0 .


Fig. 1. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case I, when b 0 = w = 1.

Fig. 2. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case II, when b 0 = w = 1.

Fig. 3. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case III, when b 0 = w = 1.

4 obtained by new proposed method

Eqs. (1.1) and ( 1.2 ) have been used to draw the graphs as shown

ior BB have been presented for visualizing the underlying mech-

a pact on concerning the exchange of amplitude and nature of the s t here, the 3D graphs describe the behavior of u ( x, t ) and v ( x, t ) i e and shape for each obtained solitary wave solutions.

. The numerical simulations for solutions of BB equation

In this present numerical experiment, the exact solutions ofn Figs. 1 –12.

In the present numerical simulation, the solutions graphs fnism of the governing equations. Figs. 1 –12 give a great imolitary waves because of the variant of the parameters. Righn space x at time t , which represents the change of amplitud


Fig. 4. 3 -D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case IV, when b 0 = w = 1.

Fig. 5. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case V, when b 0 =2, b −1 =1, w=1.

Fig. 6. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case VI, when b 0 =2, b −1 =1, w=1.

ction method to derive the exact solitary wave solutions of the ave been graphically demonstrated in order to justify the accuracy

te that Exp-function method is straightforward and concise math- nlinear evolution equations. Therefore, we hope that this method

5. Conclusion

In this paper, we have successfully implemented Exp-funcoupled Boussinesq–Burgers equations. The obtained results hand efficiency of the proposed method. The results demonstraematical tool to establish the exact analytical solutions of no


Fig. 7. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case VII, when b 0 =0, b −1 = −1, w=1.

Fig. 8. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case VIII, when b 0 =2, b −1 =1, w=1.

Fig. 9. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case IX, when b 0 =2, b −1 =1, w=1.

c evolution equations which frequently take place in engineering, a

A

d gratitude to the anonymous learnt reviewers for their valuable c

an be more effectively used to investigate other nonlinear pplied mathematics and physical sciences.

cknowledgments

The authors would like to express their sincere thanks anomments and suggestions for the improvement of the paper.


Fig. 10. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case X, when b 0 =2, b −1 =1, w=1.

Fig. 11. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case XI, when b 0 =0, b −1 = −1, w=1.

Fig. 12. 3-D solitary wave solutions graphs of Eqs. (3.7) and ( 3.8 ) respectively, in case XII, when b 0 =0, b −1 = −1, w=1.


A

δ

δ

δ

δ

δ

δ

δ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ppendix 1

−3 = k a

2 −1 c −1 + w a −1 b −1 c −1 − 1

2

k b

2 −1 d −1 ,

−2 = 2k a −1 a 0 c −1 + w a 0 b −1 c −1 + w a −1 b 0 c −1

+ ka

2 −1 c 0 + w a −1 b −1 c 0 − k b −1 b 0 d −1 − 1

2

kb

2 −1 d 0 ,

−1 = ka

2 −1 + w a −1 b −1 + w a −1 c −1 + ka

2 0 c −1 + 2k a −1 a 1 c −1

+ w a 1 b −1 c −1 + w a 0 b 0 c −1 + 2k a −1 a 0 c 0

+ w a 0 b −1 c 0 + w a −1 b 0 c 0 − k b −1 d −1 − 1

2

kb

2 0 d −1

−k b −1 b 0 d 0 − 1

2

kb

2 −1 d 1 ,

0 =

1

2

(4k a −1 a 0 + 2w a 0 b −1 + 2w a −1 b 0 + 2w a 0 c −1

+4k a 0 a 1 c −1 + 2w a 1 b 0 c −1 + 2w a −1 c 0 +2k a

2 0 c 0 + 4k a −1 a 1 c 0 + 2w a 1 b −1 c 0 + 2w a 0 b 0 c 0

−2k b 0 d −1 − 2k b −1 d 0 − kb

2 0 d 0

−2k b −1 b 0 d 1 ) ,

1 = w a −1 + ka

2 0 + 2k a −1 a 1 + w a 1 b −1 + w a 0 b 0

+ w a 1 c −1 + ka

2 1 c −1 + w a 0 c 0 + 2k a 0 a 1 c 0

+ w a 1 b 0 c 0 − 1

2

k d −1 − k b 0 d 0 − k b −1 d 1 − 1

2

kb

2 0 d 1 ,

2 = w a 0 + 2k a 0 a 1 + w a 1 b 0 + w a 1 c 0 + ka

2 1 c 0

−1

2

k d 0 − k b 0 d 1 ,

3 = w a 1 + k a

2 1 −

1

2

k d 1 ,

−4 = 2k a −1 b

2 −1 d −1 + wb

3 −1 d −1 ,

−3 = −1

2

k 3 a 0 b

2 −1 c −1 +

1

2

k 3 a −1 b −1 b 0 c −1 + 2k a 0 b

2 −1 d −1

+4k a −1 b −1 b 0 d −1 + 3 wb

2 −1 b 0 d −1

+2k a −1 b

2 −1 d 0 + wb

3 −1 d 0

−2 = 2 k 3 a −1 b −1 c −1 − 2 k 3 a 1 b

2 −1 c −1 +

1

2

k 3 a 0 b −1 b 0 c −1

−1

2

k 3 a −1 b

2 0 c −1 − 1

2

k 3 a 0 b

2 −1 c 0

+

1

2

k 3 a −1 b −1 b 0 c 0 + 4k a −1 b −1 d −1 + 3 wb

2 −1 d −1

+2k a 1 b

2 −1 d −1 + 4k a 0 b −1 b 0 d −1

+2k a −1 b

2 0 d −1 + 3 w b −1 b

2 0 d −1 + 2k a 0 b

2 −1 d 0

+4k a −1 b −1 b 0 d 0 + 3 wb

2 −1 b 0 d 0

2k a −1 b

2 −1 d 1 + wb

3 −1 d 1 ,

−1 = −1

2

k 3 a 0 b

2 −1 +

1

2

k 3 a −1 b −1 b 0 + 3 k 3 a 0 b −1 c −1

−3

2

k 3 a −1 b 0 c −1 − 3

2

k 3 a 1 b −1 b 0 c −1

+2 k 3 a −1 b −1 c 0 − 2 k 3 a 1 b

2 −1 c 0 +

1

2

k 3 a 0 b −1 b 0 c 0

−1

2

k 3 a −1 b

2 0 c 0 + 4k a 0 b −1 d −1

+4k a −1 b 0 d −1 + 6 w b −1 b 0 d −1 + 4k a 1 b −1 b 0 d −1

+2k a 0 b

2 0 d −1 + wb

3 0 d −1 + 4k a −1 b −1 d 0

+3 wb

2 −1 d 0 + 2k a 1 b

2 −1 d 0 + 4k a 0 b −1 b 0 d 0 + 2k a −1 b

2 0 d 0

+3 w b −1 b

2 0 d 0 + 2k a 0 b

2 −1 d 1

+4k a −1 b −1 b 0 d 1 + 3 wb

2 −1 b 0 d 1 ,

0 =

1

2

(4 k 3 a −1 b −1 − 4 k 3 a 1 b

2 −1 + k 3 a 0 b −1 b 0 − k 3 a −1 b

2 0

−4 k 3 a −1 c −1 + 4 k 3 a 1 b −1 c −1

+ k 3 a 0 b 0 c −1 − k 3 a 1 b

2 0 c −1 + 6 k 3 a 0 b −1 c 0 − 3 k 3 a −1 b 0 c 0

−3 k 3 a 1 b −1 b 0 c 0 + 4k a −1 d −1

+6 w b −1 d −1 + 8 k a 1 b −1 d −1 + 8 k a 0 b 0 d −1 + 6 wb

2 0 d −1

+4k a 1 b

2 0 d −1 + 8 k a 0 b −1 d 0

+8 k a −1 b 0 d 0 + 12w b −1 b 0 d 0 + 8 k a 1 b −1 b 0 d 0 + 4k a 0 b

2 0 d 0

+2wb

3 0 d 0 + 8 k a −1 b −1 d 1

+6 wb

2 −1 d 1 + 4k a 1 b

2 −1 d 1 + 8 k a 0 b −1 b 0 d 1 + 4k a −1 b

2 0 d 1

+6 w b −1 b

2 0 d 1 ) ,

1 = 3 k 3 a 0 b −1 − 3

2

k 3 a −1 b 0 − 3

2

k 3 a 1 b −1 b 0

−1

2

k 3 a 0 c −1 +

1

2

k 3 a 1 b 0 c −1 − 2 k 3 a −1 c 0

+2 k 3 a 1 b −1 c 0 +

1

2

k 3 a 0 b 0 c 0 − 1

2

k 3 a 1 b

2 0 c 0 + 2k a 0 d −1

+3 w b 0 d −1 + 4k a 1 b 0 d −1

+2k a −1 d 0 + 3 w b −1 d 0 + 4k a 1 b −1 d 0 + 4k a 0 b 0 d 0

+3 wb

3 0 d 0 + 2k a 1 b

2 0 d 0 + 4k a 0 b −1 d 1

+4k a −1 b 0 d 1 + 6 w b −1 b 0 d 1 + 4k a 1 b −1 b 0 d 1

+2k a 0 b

2 0 d 1 + wb

3 0 d 1 ,

2 = −2 k 3 a −1 + 2 k 3 a 1 b −1 +

1

2

k 3 a 0 b 0 − 1

2

k 3 a 1 b

2 0

−1

2

k 3 a 0 c 0 +

1

2

k 3 a 1 b 0 c 0 + w d −1

+2k a 1 d −1 + 2k a 0 d 0 + 3 w b 0 d 0 + 4k a 1 b 0 d 0 + 2k a −1 d 1

+3 w b −1 d 1 + 4k a 1 b −1 d 1

+4k a 0 b 0 d 1 + 3 wb

2 0 d 1 + 2k a 1 b

2 0 d 1 ,

3 = −1

2

k 3 a 0 +

1

2

k 3 a 1 b 0 + w d 0 + 2k a 1 d 0

+2k a 0 d 1 + 3 w b 0 d 1 + 4k a 1 b 0 d 1 ,

4 = w d 1 + 2k a 1 d 1 .


[[[[

[

[[

[

[

[[

[

[

[

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new exact solutions of coupled boussinesq-burgers

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