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Open Physics

formerly Central European Journal of Physics

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Volume 16, Issue 1

Issues

Volume 13 (2015)

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

Dianchen Lu / Aly R.Seadawy
  • Corresponding author
  • Mathematics Department, Faculty of Science, Taibah University, Al-Madinah Al-Munawarah, Saudi Arabia
  • Mathematics Department, Faculty of Science, Beni-Suef University, Beni-Suef, Egypt
  • Email
  • Other articles by this author:
  • De Gruyter OnlineGoogle Scholar
/ Asghar Ali
Published Online: 2018-12-31 | DOI: https://doi.org/10.1515/phys-2018-0107

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.

Keywords: Breaking soliton equation; Three-dimensional ZKB equation; Ablowitz-Kaup-Newell-Segur equation; Unstable nonlinear NLS; Dodd-Bullough-Mikhailov equation (DBME); Mathematical method

PACS: 02.30.Jr; 04.30.Nk; 05.45.Yv

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 GG 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

Gv,vt,vx,vy,vz,vxx,vyy,vzz,=0,(1)

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

v(x,y,z,t)=V(ξ),ξ=x+y+zωt,(2)

Put (2) in (1),

QV,V,V,V,=0,(3)

where Q is a polynomial in V

Let (3) solution,

V=Am(exp(φ(ξ)))m+...,Am0,(4)

where φ(ξ) gratifies,

φ(ξ)=exp(φ(ξ))+μ1exp(φ(ξ))+λ1,(5)

Casel 1. λ124μ1>0,μ10 then (5) has solution,

φ=lnλ124μ1tanhλ124μ12(ξ+ξ0)λ12μ1(6)

Case 2. λ124μ1>0,μ1=0then (5) has solution,

φ=lnλ1exp(λ1(ξ+ξ0))1(7)

Case 3. λ124μ1=0,μ10,λ10, (5) has solution,

φ=ln2(λ1(ξ+ξ0)+2)λ12(ξ+ξ0)(8)

Case 4. λ124μ1=0,μ1=0,λ1=0, (5) has solution,

φ=lnξ+ξ0(9)

Case 5. λ124μ1<0, (5) has the following solution

φ(ξ)=ln4μ1λ12tan4μ1λ122(ξ+ξ0)λ12μ1(10)

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 [23]

vxt+α1vx(vxy+vxz)+α2vxx(vy+vz)+α3(vxxxy+vxxxz)=0,(11)

Suppose the transformations,

v(x,y,z,t)=V(ξ),ξ=x+y+zωt,(12)

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

ωV+(α1+α2)(V)2+2α3V=0(13)

Let (13) has solution,

V=A0+A1exp(φ(ξ))(14)

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

A0=A0,A1=12α3(α1+α2),ω=(2λ128μ1)α3(15)

Then (14) becomes,

V=A0+12α3(α1+α2)exp(φ(ξ))(16)

Case 1. λ124μ1>0,μ10

V1=A0+24α3μ1α1+α2λ124μ1tanhλ124μ12(ξ+ξ0)λ1(17)

Case 2. λ124μ1>0,μ1=0,

V2=A0+1(α1+α2)12α3λ1exp(λ1(ξ+ξ0))1(18)

Case 3. λ124μ1=0,μ10,λ10,

V3=A01(α1+α2)6α3λ12(ξ+ξ0)(λ1(ξ+ξ0)+2)(19)

Case 4. λ124μ1=0,μ1=0,λ1=0,

V4=A0+12α3(α1+α2)(ξ+ξ0)(20)

Case 5. If λ124μ1<0,

V5=A0+24α3μ1α1+α24μ1λ12tan4μ1λ122(ξ+ξ0)λ1(21)
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.
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],

vt+β1vvx+β2vxxx+β3(vyyx+vzzx)+β4vxx=0,(22)

Let the transformations,

v(x,t)=V(ξ),ξ=kx+ly+mzωt,(23)

Put (23) in (22),

ωV+β1kVV+β4k2V+(β2k3+β3kl2+β3km2)V=0.(24)

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

A0=β4λ1k2+ωβ1kA1=2β4kβ1m=±k2β2β3l2,(25)

Thus (14) can be written as:

V=β4λ1k2+ωβ1k+2β4kβ1exp(φ(ξ))(26)

Case 1. λ124μ1>0,μ10

V6=β4λ1k2+ωβ1k+4kβ4μ1β1λ124μ1tanhλ124μ12(ξ+ξ0)λ1,k>l,β3>0,β2<0.(27)

Case 2. λ124μ1>0,μ1=0,

V7=β4λ1k2+ωβ1k+2kβ4λ1β1(exp(λ1(ξ+ξ0))1),k>l,β3>0,β2<0.(28)

Case 3. λ124μ1=0,μ10,λ10,

V8=β4λ1k2+ωβ1kkβ4λ12(ξ+ξ0)β1(2λ1(ξ+ξ0)+2),k>l,β3>0,β2<0.(29)

Case 4. λ124μ1=0,μ1=0,λ1=0,

V9=β4λ1k2+ωβ1k+2kβ4β1(ξ+ξ0),k>l,β3>0,β2<0.(30)

Case 5. λ124μ1<0,

V10=β4λ1k2+ωβ1k+4kβ4μβ14μ1λ12tan4μ1λ122(ξ+ξ0)λ1,k>l,β3>0,β2<0.(31)
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.
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]

4vxt+vxxxt+8vxvxy+4vxxvyγvxx=0,(32)

Consider,

v(x,y,t)=V,ξ=x+y+kt,(33)

Putting (33) in (32), we obtaine

(4kγ)V+6V2+kV=0(34)

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

A0=A0,A1=γλ124μ1+4,k=γλ12+44μ1(35)

Hence, (14) becomes as:

V=A0+γλ124μ1+4exp(φ(ξ))(36)

Case 1. λ124μ1>0,μ10

V11=A0+2γμ1λ124μ1+4λ124μ1tanhλ124μ12(ξ+ξ0)λ1(37)

Case 2. λ124μ1>0,μ1=0,

V12=A0+1(λ124μ1+4)γλ1exp(λ1(ξ+ξ0))1(38)

Case 3. λ124μ1=0,μ10,λ10,

V13=A01(λ124μ1+4)γλ12(ξ+ξ0)(2λ1(ξ+ξ0)+2)(39)

Case 4. λ124μ1=0,μ1=0,λ1=0,

V14=A0+γ(λ124μ1+4)(ξ+ξ0)(40)

Case 5. λ124μ1<0,

V15=A0+2γμ1λ124μ1+44μ1λ12tan4μ1λ122(ξ+ξ0)λ1(41)

3.4 Unstable nonlinear Schrödinger equation

The general form of unstable Schrödinger equation[28],

iut+uxx+2η|u|2u2γu=0,(42)

Consider,

u(x,t)=V(ξ)eiδ,ξ=kx+ωt,δ=αx+βt(43)

Put (43) in (42),

k2V(α2+β+2γ)V+2ηV3=0,ω=2αk(44)

Let (44) has solution form:

V=A0+A1exp(φ(ξ))+A2(exp(φ(ξ))2(45)

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

A0=α2+β+2γλ12η(λ124μ1),A1=2(α2+β+2γ)η(λ124μ1),A2=0,ω=2α2(α2+β+2γ)(λ124μ1)(46)

we have demonstrated possible solutions regarding to (46).

Case I. λ124μ1>0,μ10

V16=α2+β+2γλ12η(λ124μ1)2(α2+β+2γ)η(λ124μ1)2μ1λ124μ1tanhλ124μ12(ξ+ϵ0)λ1eiδ(47)

Case II. λ124μ1>0,μ1=0,

V17=α2+β+2γλ2η(λ24μ)2(α2+β+2γ)η(λ24μ)λ(exp(λ(ξ+ϵ0))1)eiδ(48)

Case III. λ124μ1<0,

V18=α2+β+2γλ12η(λ124μ1)2(α2+β+2γ)η(λ124μ1)2μ14μ1λ12tan4μ1λ122(ξ+ϵ0)λ1eiδ(49)

3.5 DBM equation

General form in [29, 34],

vxt+aev+de2v=0,(50)

Consider,

v(x,t)=v(ξ),ξ=kx+ct,(51)

Put (51) in (50),

cV+aev+de2v=0(52)

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

ckVVckV2+aV3+d=0(53)

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

A0=d3λ12+2μ1a3λ124μ1A1=6d3λ1a3λ124μ1A2=6d3a3λ124μ1,c=3a2/3d3kλ124μ(54)

Case I. λ124μ1>0,μ10

V19=(λ12+2μ1)d13a13(λ124μ1)12λ1μ1d13a13(λ124μ1)λ124μ1tanhλ124μ12(ξ+ϵ0)λ124μ2d13a13(λ124μ1)λ124μ1tanhλ124μ12(ξ+ϵ0)λ12(55)

Case II. λ124μ1>0,μ1=0,

V20=d13a131+6(exp(λ1(ξ+ϵ0))1)+6(exp(λ1(ξ+ϵ0))1)2(56)

Case III. λ124μ1<0

V21=(λ12+2μ1)d13a13(λ124μ1)12λ1μ1d13a13(λ124μ1)4μ1λ12tan4μ1λ122(ξ+ϵ0)λ124μ12d13a13(λ124μ1)4μ1λ12tan4μ1λ122(ξ+ϵ0)λ12(57)

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 [23]. It is conversant that our result in (38) is approximately the same as the solution (13) and (19) in [27]. Moreover, solution (47) is nearly equal to solution (17) in [28] and solution (10) in [33]. Furthermore, our constructed solution (57) is likely the same as the solution (3.9) in [34] and solution (3.26b) in [35] respectively. our results are novel and have not been presented in any literature.

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.
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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About the article

Received: 2018-07-22

Accepted: 2018-10-19

Published Online: 2018-12-31


Citation Information: Open Physics, Volume 16, Issue 1, Pages 854–860, ISSN (Online) 2391-5471, DOI: https://doi.org/10.1515/phys-2018-0107.

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© 2018 D. Lu et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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