Dianchen Lu, Aly R.Seadawy and Asghar Ali

# Structure of traveling wave solutions for some nonlinear models via modified mathematical method

De Gruyter | Published online: December 31, 2018

# Abstract

We have employed the exp(-φ(ξ))-expansion method to derive traveling waves solutions of breaking solition (BS), Zakharov-Kuznetsov-Burgers (ZKB), Ablowitz-Kaup-Newell-Segur (AKNS) water wave, Unstable nonlinear Schrödinger (UNLS) and Dodd-Bullough-Mikhailov (DBM) equations. These models have valuable applications in mathematical physics. The results of the constructed model, along with some graphical representations provide the basic knowlegde about these models. The derived results have various applications in applied science.

## 1 Introduction

Partial differential equations (PDEs) have been measured with great significance due to its variety of applications in physics, applied mathematics and engineering. PDEs can be used to describe a wide variety of phenomena such as sound, heat, electrostatics, electrodynamics, fluids dynamics, elasticity and quantum mechanics. These seemingly distinct physical phenomena can be formalized similarly in terms of PDEs. Due to its broad/various applications and important mathematical properties, many methods have been presented to study in different aspects related with the solutions and physical phenomena of nonlinear wave equations. Hence, penetrating and constructing exact traveling wave solutions for nonlinear differential equations is a modern research area. Numerous effective methods were discussed to obtain solutions of nonlinear wave system of equations in different aspects [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].

Recently, many new powerful methods have been proposed for finding the exact traveling waves solution of nonlinear evolution equations such as: the inverse scattering transform method , the homogeneous balance method, modified simple equation method, modified extended direct algebraic method, the tanhsech method and the extended tanhcoth method, the soliton ansatz method [11, 12, 13, 14, 15, 16, 17, 18, 19, 20] and many more [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40]. In previous studies the authors [23, 24, 25, 26, 27, 28, 29] applied, auxiliary equation, extended mapping, modified simple equation, modified extended and G G expansion methods on breaking solition (BS), Zakharov-Kuznetsov-Burgers (ZKB), Ablowitz-Kaup-Newell-Segur (AKNS) water wave, unstable nonlinear Schrödinger (UNLS) and Dodd-Bullough-Mikhailov (DBM) equations, respectively. But here our aim is to investigate the novel exact and solitary wave solutions of these models by employing exp(-φ(ξ))-expansion method.

The description of method is given in Section 2. In section 3, we apply the present method on selective models. Results and discussion are presented in Section 4. Finally, the Conclusions are given in Section 5.

## 2 Description of the method

Consider PDE in the form

(1) G v , v t , v x , v y , v z , v x x , v y y , v z z , = 0 ,

where G is a polynomial function in v(x, y, z, t). Suppose,

(2) v ( x , y , z , t ) = V ( ξ ) , ξ = x + y + z ω t ,

Put (2) in (1),

(3) Q V , V , V , V , = 0 ,

where Q is a polynomial in V

Let (3) solution,

(4) V = A m ( exp ( φ ( ξ ) ) ) m + . . . , A m 0 ,

where φ(ξ) gratifies,

(5) φ ( ξ ) = exp ( φ ( ξ ) ) + μ 1 exp ( φ ( ξ ) ) + λ 1 ,

Casel 1. λ 1 2 4 μ 1 > 0 , μ 1 0 then (5) has solution,

(6) φ = ln λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ξ 0 ) λ 1 2 μ 1

Case 2. λ 1 2 4 μ 1 > 0 , μ 1 = 0 then (5) has solution,

(7) φ = ln λ 1 exp ( λ 1 ( ξ + ξ 0 ) ) 1

Case 3. λ 1 2 4 μ 1 = 0 , μ 1 0 , λ 1 0 , (5) has solution,

(8) φ = ln 2 ( λ 1 ( ξ + ξ 0 ) + 2 ) λ 1 2 ( ξ + ξ 0 )

Case 4. λ 1 2 4 μ 1 = 0 , μ 1 = 0 , λ 1 = 0 , (5) has solution,

(9) φ = ln ξ + ξ 0

Case 5. λ 1 2 4 μ 1 < 0 , (5) has the following solution

(10) φ ( ξ ) = ln 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ξ 0 ) λ 1 2 μ 1

Substituting 4) with 5) in 3), adjusting coefficients of exp(−(ξ)), m=0,1,2,3,... equal to zero, we achieve numerous equations that can be solved with use of Mathematica.

Putting all values of parameters with solution of (5) in (4), we obtain solution of (1).

## 3 Applications

### 3.1 (3+1)-dimensional BS equation

Consider general form of BS equation in 

(11) v x t + α 1 v x ( v x y + v x z ) + α 2 v x x ( v y + v z ) + α 3 ( v x x x y + v x x x z ) = 0 ,

Suppose the transformations,

(12) v ( x , y , z , t ) = V ( ξ ) , ξ = x + y + z ω t ,

Put (12) in (11), after integrating,

(13) ω V + ( α 1 + α 2 ) ( V ) 2 + 2 α 3 V = 0

Let (13) has solution,

(14) V = A 0 + A 1 exp ( φ ( ξ ) )

Substituting (14) with (5) in (13), we attained several equations

(15) A 0 = A 0 , A 1 = 12 α 3 ( α 1 + α 2 ) , ω = ( 2 λ 1 2 8 μ 1 ) α 3

Then (14) becomes,

(16) V = A 0 + 12 α 3 ( α 1 + α 2 ) exp ( φ ( ξ ) )

Case 1. λ 1 2 4 μ 1 > 0 , μ 1 0

(17) V 1 = A 0 + 24 α 3 μ 1 α 1 + α 2 λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ξ 0 ) λ 1

Case 2. λ 1 2 4 μ 1 > 0 , μ 1 = 0 ,

(18) V 2 = A 0 + 1 ( α 1 + α 2 ) 12 α 3 λ 1 exp ( λ 1 ( ξ + ξ 0 ) ) 1

Case 3. λ 1 2 4 μ 1 = 0 , μ 1 0 , λ 1 0 ,

(19) V 3 = A 0 1 ( α 1 + α 2 ) 6 α 3 λ 1 2 ( ξ + ξ 0 ) ( λ 1 ( ξ + ξ 0 ) + 2 )

Case 4. λ 1 2 4 μ 1 = 0 , μ 1 = 0 , λ 1 = 0 ,

(20) V 4 = A 0 + 12 α 3 ( α 1 + α 2 ) ( ξ + ξ 0 )

Case 5. If λ 1 2 4 μ 1 < 0 ,

(21) V 5 = A 0 + 24 α 3 μ 1 α 1 + α 2 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ξ 0 ) λ 1 ### Figure 1

Solitary waves of solutions (17), (21) on (a), (b) with: A0 = 1.5, λ1 = 2, μ1 = 0.7, α1 = α2 = 1.0, α3 = −1.0, ϵ = 0.5 and A0 = 1.6, λ1 = 2, μ1 = 2, α1 = α2 = 1.1, α3 = −3, ϵ = .6 respectively.

### 3.2 (3+1)-dimensional ZKB equation

The general form of three-dimensional Zakharov-Kuznetsov-Burgers equation [24, 25],

(22) v t + β 1 v v x + β 2 v x x x + β 3 ( v y y x + v z z x ) + β 4 v x x = 0 ,

Let the transformations,

(23) v ( x , t ) = V ( ξ ) , ξ = k x + l y + m z ω t ,

Put (23) in (22),

(24) ω V + β 1 k V V + β 4 k 2 V + ( β 2 k 3 + β 3 k l 2 + β 3 k m 2 ) V = 0.

Let (24) has solution form of (14). Substituting (14) with (5) into Eq.(24), after solving we have,

(25) A 0 = β 4 λ 1 k 2 + ω β 1 k A 1 = 2 β 4 k β 1 m = ± k 2 β 2 β 3 l 2 ,

Thus (14) can be written as:

(26) V = β 4 λ 1 k 2 + ω β 1 k + 2 β 4 k β 1 exp ( φ ( ξ ) )

Case 1. λ 1 2 4 μ 1 > 0 , μ 1 0

(27) V 6 = β 4 λ 1 k 2 + ω β 1 k + 4 k β 4 μ 1 β 1 λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ξ 0 ) λ 1 , k > l , β 3 > 0 , β 2 < 0.

Case 2. λ 1 2 4 μ 1 > 0 , μ 1 = 0 ,

(28) V 7 = β 4 λ 1 k 2 + ω β 1 k + 2 k β 4 λ 1 β 1 ( exp ( λ 1 ( ξ + ξ 0 ) ) 1 ) , k > l , β 3 > 0 , β 2 < 0.

Case 3. λ 1 2 4 μ 1 = 0 , μ 1 0 , λ 1 0 ,

(29) V 8 = β 4 λ 1 k 2 + ω β 1 k k β 4 λ 1 2 ( ξ + ξ 0 ) β 1 ( 2 λ 1 ( ξ + ξ 0 ) + 2 ) , k > l , β 3 > 0 , β 2 < 0.

Case 4. λ 1 2 4 μ 1 = 0 , μ 1 = 0 , λ 1 = 0 ,

(30) V 9 = β 4 λ 1 k 2 + ω β 1 k + 2 k β 4 β 1 ( ξ + ξ 0 ) , k > l , β 3 > 0 , β 2 < 0.

Case 5. λ 1 2 4 μ 1 < 0 ,

(31) V 10 = β 4 λ 1 k 2 + ω β 1 k + 4 k β 4 μ β 1 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ξ 0 ) λ 1 , k > l , β 3 > 0 , β 2 < 0. ### Figure 2

Exact solitary wave solutions (30) on (a), (31) at (b) with: β1 = 0.7, β2 = −1.0, β3 = 1.4, β4 = −3, k = 1.00, l = −0.5, ω = 0.6 and β1 = 4, β2 = −1, β3 = 3, β4 = 3, λ1 = −1, k = −5.1, μ1 = 2, l = 0.5, ω = −0.5, ϵ = 0.5 respectively.

### 3.3 (2+1)-dimensional AKNS equation

Let the generalized form in [26, 27]

(32) 4 v x t + v x x x t + 8 v x v x y + 4 v x x v y γ v x x = 0 ,

Consider,

(33) v ( x , y , t ) = V , ξ = x + y + k t ,

Putting (33) in (32), we obtaine

(34) ( 4 k γ ) V + 6 V 2 + k V = 0

Let (34) has solution form (14), after solving we have:

(35) A 0 = A 0 , A 1 = γ λ 1 2 4 μ 1 + 4 , k = γ λ 1 2 + 4 4 μ 1

Hence, (14) becomes as:

(36) V = A 0 + γ λ 1 2 4 μ 1 + 4 exp ( φ ( ξ ) )

Case 1. λ 1 2 4 μ 1 > 0 , μ 1 0

(37) V 11 = A 0 + 2 γ μ 1 λ 1 2 4 μ 1 + 4 λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ξ 0 ) λ 1

Case 2. λ 1 2 4 μ 1 > 0 , μ 1 = 0 ,

(38) V 12 = A 0 + 1 ( λ 1 2 4 μ 1 + 4 ) γ λ 1 exp ( λ 1 ( ξ + ξ 0 ) ) 1

Case 3. λ 1 2 4 μ 1 = 0 , μ 1 0 , λ 1 0 ,

(39) V 13 = A 0 1 ( λ 1 2 4 μ 1 + 4 ) γ λ 1 2 ( ξ + ξ 0 ) ( 2 λ 1 ( ξ + ξ 0 ) + 2 )

Case 4. λ 1 2 4 μ 1 = 0 , μ 1 = 0 , λ 1 = 0 ,

(40) V 14 = A 0 + γ ( λ 1 2 4 μ 1 + 4 ) ( ξ + ξ 0 )

Case 5. λ 1 2 4 μ 1 < 0 ,

(41) V 15 = A 0 + 2 γ μ 1 λ 1 2 4 μ 1 + 4 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ξ 0 ) λ 1

### 3.4 Unstable nonlinear Schrödinger equation

The general form of unstable Schrödinger equation,

(42) i u t + u x x + 2 η | u | 2 u 2 γ u = 0 ,

Consider,

(43) u ( x , t ) = V ( ξ ) e i δ , ξ = k x + ω t , δ = α x + β t

Put (43) in (42),

(44) k 2 V ( α 2 + β + 2 γ ) V + 2 η V 3 = 0 , ω = 2 α k

Let (44) has solution form:

(45) V = A 0 + A 1 exp ( φ ( ξ ) ) + A 2 ( exp ( φ ( ξ ) ) 2

a0, a1 and a2 are constants, which can be determined latter. Substituting (45) with (5) in (44), after solving we obtain:

(46) A 0 = α 2 + β + 2 γ λ 1 2 η ( λ 1 2 4 μ 1 ) , A 1 = 2 ( α 2 + β + 2 γ ) η ( λ 1 2 4 μ 1 ) , A 2 = 0 , ω = 2 α 2 ( α 2 + β + 2 γ ) ( λ 1 2 4 μ 1 )

we have demonstrated possible solutions regarding to (46).

Case I. λ 1 2 4 μ 1 > 0 , μ 1 0

(47) V 16 = α 2 + β + 2 γ λ 1 2 η ( λ 1 2 4 μ 1 ) 2 ( α 2 + β + 2 γ ) η ( λ 1 2 4 μ 1 ) 2 μ 1 λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ϵ 0 ) λ 1 e i δ

Case II. λ 1 2 4 μ 1 > 0 , μ 1 = 0 ,

(48) V 17 = α 2 + β + 2 γ λ 2 η ( λ 2 4 μ ) 2 ( α 2 + β + 2 γ ) η ( λ 2 4 μ ) λ ( exp ( λ ( ξ + ϵ 0 ) ) 1 ) e i δ

Case III. λ 1 2 4 μ 1 < 0 ,

(49) V 18 = α 2 + β + 2 γ λ 1 2 η ( λ 1 2 4 μ 1 ) 2 ( α 2 + β + 2 γ ) η ( λ 1 2 4 μ 1 ) 2 μ 1 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ϵ 0 ) λ 1 e i δ

### 3.5 DBM equation

General form in [29, 34],

(50) v x t + a e v + d e 2 v = 0 ,

Consider,

(51) v ( x , t ) = v ( ξ ) , ξ = k x + c t ,

Put (51) in (50),

(52) c V + a e v + d e 2 v = 0

Let V = ev substitute it and its derivatives in (52), we obtained:

(53) c k V V c k V 2 + a V 3 + d = 0

Suppose (53) has solution form of (45), after solving we have:

(54) A 0 = d 3 λ 1 2 + 2 μ 1 a 3 λ 1 2 4 μ 1 A 1 = 6 d 3 λ 1 a 3 λ 1 2 4 μ 1 A 2 = 6 d 3 a 3 λ 1 2 4 μ 1 , c = 3 a 2 / 3 d 3 k λ 1 2 4 μ

Case I. λ 1 2 4 μ 1 > 0 , μ 1 0

(55) V 19 = ( λ 1 2 + 2 μ 1 ) d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) 12 λ 1 μ 1 d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ϵ 0 ) λ 1 24 μ 2 d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) λ 1 2 4 μ 1 tanh λ 1 2 4 μ 1 2 ( ξ + ϵ 0 ) λ 1 2

Case II. λ 1 2 4 μ 1 > 0 , μ 1 = 0 ,

(56) V 20 = d 1 3 a 1 3 1 + 6 ( exp ( λ 1 ( ξ + ϵ 0 ) ) 1 ) + 6 ( exp ( λ 1 ( ξ + ϵ 0 ) ) 1 ) 2

Case III. λ 1 2 4 μ 1 < 0

(57) V 21 = ( λ 1 2 + 2 μ 1 ) d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) 12 λ 1 μ 1 d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ϵ 0 ) λ 1 24 μ 1 2 d 1 3 a 1 3 ( λ 1 2 4 μ 1 ) 4 μ 1 λ 1 2 tan 4 μ 1 λ 1 2 2 ( ξ + ϵ 0 ) λ 1 2

## 4 Discussion of the results

We attained that our result in (18) is likely similar to the Eqs. (3.14) and (3.24) in the . It is conversant that our result in (38) is approximately the same as the solution (13) and (19) in . Moreover, solution (47) is nearly equal to solution (17) in  and solution (10) in . Furthermore, our constructed solution (57) is likely the same as the solution (3.9) in  and solution (3.26b) in  respectively. our results are novel and have not been presented in any literature. ### Figure 3

Graph of (49) at (a), (56) on (b) with: ε = 0.5, μ1 = 4, λ1 = −2, β = −1, α = −1, γ = 0.5, η = 1 and ε = −0.5, μ1 = 0, λ1 = 1, a = −1, d = 1, k = 1 respectively.

## 5 Conclusion

We have successfully employed the exp(-φ(ξ))-expansion method to construct solutions of important selective waves models. The investigated results have numerous applications in applied sciences and play a fruitful rule in nonlinear sciences. Our technique is simple and straightforward, which is useful for solving different evolutions equations in mathematics and physics.

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Received: 2018-07-22
Accepted: 2018-10-19
Published Online: 2018-12-31

© 2018 D. Lu et al., published by De Gruyter

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