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Optik 125 (2014) 5480–5489

Contents lists available at ScienceDirect

Optik

jo ur nal homepage: www.elsev ier .de / i j leo

opological solitons of resonant nonlinear Schödinger’sequation withual-power law nonlinearity by G′/G-expansion technique

. Mirzazadeha, M. Eslamib, Daniela Milovicc, Anjan Biswasd,e,∗

Department of Engineering Sciences, Faculty of Technology and Engineering East of Guilan, University of Guilan, P.C. 44891-63157, Rudsar-Vajargah, IranDepartment of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, IranFaculty of Electronic Engineering, Department of Telecommunications, University of Nis, Aleksandra Medvedeva 14, 18000 Nis, SerbiaDepartment of Mathematical Sciences, Delaware State University, Dover, DE 19901-2277, USAFaculty of Science, Department of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi Arabia

r t i c l e i n f o

rticle history:eceived 27 January 2014ccepted 5 March 2014

eywords:olitonsntegrability

a b s t r a c t

This paper considers the resonant nonlinear Schrödinger’s equation with dual-power law nonlinearity.The G′/G-expansion method is applied to integrate this equation. The soliton solutions are thus obtained.Both constant coefficients as well as time-dependent coefficients are considered. The results for paraboliclaw nonlinearity fall out as a special case.

© 2014 Elsevier GmbH. All rights reserved.

. Introduction

Optical solitons is one of the most active research areas in the field of Nonlinear Optics. There are several aspects to this area of researchhat makes it really exciting. The governing equation for the propagation of solitons through optical fibers is the well-known nonlinearchrödinger’s equation (NLSE). There are several forms of nonlinearity that are studied in this context. This paper is going to focus on one ofhe most general types of nonlinear media, namely the dual-power law nonlinearity. The three other popular forms of nonlinearity whichre Kerr law, power law and parabolic law fall out as special cases to dual-power law nonlinearity. Therefore, it suffices to consider thiseneralized form of nonlinear medium.

The integrability aspect will be the focus of the paper. While there are several forms of integration architecture available nowadays,his paper will address one of the most popular and powerful tools of integration. This is the G′/G-expansion scheme [2,3,6–20]. It must beoted that the problem of this paper is already studied earlier by the aid of simplest equation method and first integral method [2,3]. The

ntegration technique of this paper will extract topological soliton solutions to the NLSE with dual-power law nonlinearity. Subsequently,his NLSE will be studied, in the latter half of the paper, with time-dependent coefficients, thus making the study rounded. This is theesonant nonlinear Schrödinger’s equation (RNLSE) [1,4,5] which is studied in the context of nonlinear optics and Madelung fluids.

. Review of the G′/G-expansion method

In this section, we describe the G′/G-expansion method [2,3,6–20] for finding traveling wave solutions of NLEE and then subsequentlyt will be applied to solve the RNLSE with dual-power law nonlinearity [1,4,5].

Assuming that the NLEE takes the form

P(u, ut, ux, uxx, uxt, utt, . . .) = 0, (1)

here P is a polynomial. The essence of the G′/G-expansion method is presented in the following steps:

Step 1: In order to locate the traveling wave solutions of Eq. (1), we introduce the wave variable

u(x, t) = U(�), � = x − ct. (2)

∗ Corresponding author at: Department of Mathematical Sciences, Delaware State University, Dover, DE 19901-2277, USA. Tel.: +1 3026590169; fax: +1 3028577054.E-mail address: biswas.anjan@gmail.com (A. Biswas).

ttp://dx.doi.org/10.1016/j.ijleo.2014.03.042030-4026/© 2014 Elsevier GmbH. All rights reserved.

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M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489 5481

Substituting Eq. (2) into Eq. (1), we recover the ordinary differential equation (ODE)

Q(U, U′, U ′′, . . .

)= 0. (3)

Step 2: Eq. (3) is subsequently integrated provided all terms contain derivatives where integration constants are considered zeros.Step 3: Introduce the solution U(�) of Eq. (3) in a the form of finite series:

U(�) =N∑l=0

al

(G′(�)G(�)

)l, (4)

here al are real constants with aN /= 0 and N is a positive integer that needs to be determined. The function G(�) is the solution of theuxiliary linear ODE

G′′(�) + �G′(�) + �G(�) = 0, (5)

here � and � are real constants to be determined.Step 4: Determining N, can be accomplished by balancing the linear term of highest order derivatives with the highest order nonlinear

erm in Eq. (3).Step 5: Substituting the general solution of Eq. (5) together with Eq. (4) into Eq. (3) yields an algebraic equation involving powers of G′/G.

quating the coefficients of each power of G′/G to zero gives a system of algebraic equations for al, �, � and c . Then, we solve the systemith the aid of a computer algebra system, such as Maple, to determine these constants. Next, depending on the sign of the discriminant

= �2 − 4�, we get solutions of Eq. (3). So, we can obtain exact solutions of the given Eq. (1).

. Application of G′/G-expansion method to RNLSE

This section will apply the G′/G-expansion technique [7–9,19] to address the RNLSE with dual-power law nonlinearity. The study will beplit into two subsections. The first section will consider the equation with constant coefficients, while the following section will performhe integration of the NLSE with time-dependent coefficients.

.1. Constant coefficients

In this subsection, we shall apply the integration tool to establish the exact solution of the RNLSE with dual-power law nonlinearityith constant coefficients.

i t + a xx +(b| |2n + c| |4n

) + d

{(| |)xx

| |

} = 0. (6)

Using the wave transformation

(x, t) = U(�)ei(−�x+ωt+�), � = x + 2�at, (7)

e have

(a + d)U ′′ − (ω + a�2)U + bU2n+1 + cU4n+1 = 0. (8)

Balancing U′′ with U4n+1 we have N+ 2 = (4n + 1)N ; hence, N = 1/2n . We then assume that Eq. (8) has the following formal solutions7–9,19]:

U(�) = D(G′G

)1/2n

, D /= 0 (9)

here D is a constant to be determined later and G satisfies Eq. (5). Thus, we obtain

U′ = − 12nD(G′G

)(1/2n)+1

− 12nD�(G′G

)1/2n

− 12nD�(G′G

)(1/2n)−1

, (10)

U ′′ =(

14n2

+ 12n

)D(G′G

)(1/2n)+2

+(

12n2

+ 12n

)D�(G′G

)(1/2n)+1

+(

12n2

D� + 14n2

D�2)(

G′G

)1/2n

+(

12n2

− 12n

)D��

(G′G

)(1/2n)−1

+(

14n2

− 12n

)D�2

(G′G

)(1/2n)−2

. (11)

Substituting Eqs. (9)–(11) into Eq. (8) and collecting all terms with the same order of G′/G together, we convert the left-hand side ofq. (8) into a polynomial in G′/G. Setting each coefficient of each polynomial to zero, we derive a set of algebraic equations for �, ω, �, � and D :(

G′)(1/2n)+2

G coeff.:

a + d

4n2D + a + d

2nD + cD4n+1 = 0,

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482 M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489

(G′G

)(1/2n)+1coeff.:

a + d

2n2D� + a + d

2nD� + bD2n+1 = 0,

G′G

)1/2ncoeff.:

a + d

2n2D� + a + d

4n2D�2 −

(ω + a�2

)D = 0, (12)

(G′G

)(1/2n)−1coeff.:

a + d

2n2D�� − a + d

2nD�� = 0,

G′G

)(1/2n)−2coeff.:

a + d

4n2D�2 − a + d

2nD�2 = 0.

Solving that set of the above algebraic equations, we obtain

D =(

− (a + d)(1 + 2n)4n2c

)1/4n

, � = 0, � = −√

− n2b2(1 + 2n)

c(a + d)(1 + n)2,

ω = −(a�2 + b2(1 + 2n)

4c(1 + n)2

).

(13)

From Eqs. (5), (7), (9) and (13), we obtain the exact traveling wave solution of the R-NLSE with dual-power law nonlinearity as follows:

(x, t) =

⎧⎨⎩−b(1 + 2n)

2c(1 + n)c2e

√−(n2b2(1+2n))/(c(a+d)(1+n)2)(x+2a�t)

c1 + c2e

√−(n2b2(1+2n))/(c(a+d)(1+n)2)(x+2a�t)

⎫⎬⎭

1/2n

ei{−�x−(a�2+(b2(1+2n))/(4c(1+n)2))t+�}, (14)

here c1, c2, � and � are arbitrary constants.Eq. (14) is a new type of exact traveling wave solution to the R-NLSE with dual-power law nonlinearity. Especially, if we choose c1 = c2

n Eq. (14), we obtain the solitary wave solution of the R-NLSE with dual-power law nonlinearity, namely

(x, t) ={

−b(1 + 2n)4c(1 + n)

(1 + tanh

[√− n2b2(1 + 2n)

4c(a + d)(1 + n)2 (x + 2a�t)

])}1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}. (15)

Remark-1: In Eqs. (14) and (15), If we take n = 1, then we obtain the following exact traveling wave solutions

(x, t) ={

−3b4c

c2e√

−(3b2/4c(a+d))(x+2a�t)

c1 + c2e√

−(3n2b2/4c(a+d))(x+2a�t)

}1/2

ei{−�x−(a�2+(3b2/16c))t+�}, (16)

nd

(x, t) ={

−3b8c

(1 + tanh

[√− 3b2

16c(a + d)(x + 2a�t)

])}1/2

ei{−�x−(a�2+(3b2/16c))t+�}. (17)

It needs to be noted that (16) is a rational solution and is not a solitary wave. However, Eq. (17) is a topological 1-soliton solution of the-NLSE with parabolic law nonlinearity

i t + a xx +{b| |2 + c| |4

} + d

{(| |)xx

| |

} = 0. (18)

.2. Time-dependent coefficients

This subsection is applying the same integration tool, as in the previous case, to integrate RNLSE with dual-power law nonlinearity inresence of time-dependent coefficients [1,4,5]. The starting equation is therefore given by

i t + a(t) xx + {b(t)| |2n + c(t)| |4n} + d(t)

{(| |)xx

| |

} = 0, (19)

Eq. (19) will be converted to the ODE

(a(t) + d(t))U ′′ −(tdω(t)dt

+ ω(t) + �2a(t))U + b(t)U2n+1 + c(t)U4n+1 = 0, (20)

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M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489 5483

pon using the transformation

(x, t) = U(�)ei(−�x+ω(t)t+�), � = x + 2�

∫ t

0

a(t′) dt′. (21)

Following the balancing act procedure, we balance U′′ with U4n+m in Eq. (20) we have N = 1/2n. Using the same idea with that mentionedn Section 3.1, we may choose the solution of Eq. (20) in the form [7–9,19]

U(�) = E(G′G

)1/2n

, E /= 0, (22)

here E is a constant to be determined later and G satisfies Eq. (5).Using the method mentioned above, we obtain the following set of algebraic equations for �, ω(t), �, � and E:(G′G

)(1/2n)+2coeff.:

a(t) + d(t)4n2

E + a(t) + d(t)2n

E + c(t)E4n+1 = 0,

(G′G

)(1/2n)+1coeff.:

a(t) + d(t)2n2

E� + a(t) + d(t)2n

E� + b(t)E2n+1 = 0,

(G′G

)1/2ncoeff.:

a(t) + d(t)2n2

E� + a(t) + d(t)4n2

E�2 −(tdω(t)dt

+ ω(t) + �2a(t))E = 0, (23)

(G′G

)(1/2n)−1coeff.:

a(t) + d(t)2n2

E�� − a(t) + d(t)2n

E�� = 0,

(G′G

)(1/2n)−2coeff.:

a(t) + d(t)4n2

E�2 − a(t) + d(t)2n

E�2 = 0.

Solving system (23) in the Maple environment, we obtain the following solution:

E =(

− (a(t) + d(t))(1 + 2n)4n2c(t)

)1/4n

, � = 0, � = −√

− n2b2(t)(1 + 2n)

c(t)(a(t) + d(t))(1 + n)2,

ω = −1t

∫ t

0

{�2a(t′) + b2(t′)(1 + 2n)

4c(t′)(1 + n)2

}dt′.

(24)

From Eqs. (5), (21), (22) and (24), we obtain the exact traveling wave solution of the variable coefficients R-NLSE with dual-power lawonlinearity as follows:

(x, t) =

⎧⎨⎩−b(t)(1 + 2n)

2c(t)(1 + n)c2e

√−(n2b2(t)(1+2n)/c(t)(a(t)+d(t))(1+n)2)(x+2�

∫ t0a(t′) dt′)

c1 + c2e

√−(n2b2(t)(1+2n)/c(t)(a(t)+d(t))(1+n)2)(x+2�

∫ t0a(t′) dt′)

⎫⎬⎭

1/2n

× ei{−�x−

∫ t0

{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}(25)

here c1, c2, � and � are arbitrary constants.Eq. (25) is a new type of exact traveling wave solution to the variable coefficients R-NLSE with dual-power law nonlinearity. Especially, if

e choose c1 = c2 in Eq. (25), we obtain topological 1-soliton solution of the variable coefficients R-NLSE with dual-power law nonlinearity,amely

(x, t) ={

−b(t)(1 + 2n)4c(t)(1 + n)

(1 + tanh

[√− n2b2(t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2

(x + 2�

∫ t

0

a(t′)dt′)])}1/2n

i{−�x−∫ t

0{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}

× e . (26)

Remark-2: The exact solution (26) that we obtained for the variable coefficients R-NLSE with dual-power law nonlinearity is completelyame with the solution that was obtained by Triki et al [14] who used the solitary wave ansatz method.

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484 M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489

Remark-3: In Eqs. (25) and (26), If we take n = 1, then we obtain the following exact traveling wave solutions

(x, t) =

⎧⎪⎨⎪⎩−3b(t)

4c(t)c2e

√−(3b2(t)/4c(t)(a(t)+d(t)))

(x+2�

∫ t0a(t′) dt′

)

c1 + c2e

√−(3b2(t)/4c(t)(a(t)+d(t)))

(x+2�

∫ t0a(t′) dt′

)⎫⎪⎬⎪⎭

1/2

ei{−�x−

∫ t0

{�2a(t′)+(3b2(t′)/16c(t′))} dt′+�}(27)

nd

(x, t) ={

−3b(t)8c(t)

(1 + tanh

[√− 3b2(t)

16c(t)(a(t) + d(t))

(x + 2�

∫ t

0

a(t′)dt′)])}1/2

× ei{−�x−

∫ t0

{�2a(t′)+(3b2(t′)/16c(t′))} dt′+�}. (28)

Once again Eq. (27) is not a solitary wave while Eq. (28) is a topological 1-soliton solution of the variable coefficients R-NLSE witharabolic law nonlinearity

i t + a(t) xx +{b(t)| |2 + c(t)| |4

} + d(t)

{(| |)xx

| |

} = 0. (29)

. Application of improved G′/G-expansion method to RNLSE

This section will apply the G′/G-expansion method [2,3,6–18,20] to address the RNLSE with dual-power law nonlinearity. The studyill be split into two subsections. The first section will consider the equation with constant coefficients, while the following section willerform the integration of the NLSE with time-dependent coefficients.

.1. Constant coefficients

In order to obtain closed form solutions, we use the transformation

U(�) = V1/2n(�), (30)

hat will reduce Eq. (8) into the following ODE

(a + d)(1 − 2n)(V ′)2 + 2n(a + d)VV ′′ − 4n2(ω + a�2)V2 + 4n2bV3 + 4n2cV4 = 0. (31)

Suppose that the solutions of Eq. (31) can be expressed by a polynomial in G′/G as follows:

V(�) =N∑l=0

al

(G′G

)l(32)

here al are real constants with aN /= 0 and N is a positive integer which can be determined by balancing the highest order derivativeerm with the highest order nonlinear term after substituting ansatz (32) into Eq. (31), where G = G(�) satisfies Eq. (5).

Balancing the order of VV′′ and V4 in Eq. (31), we have N = 1. Therefore; Eq. (32) can be rewritten as

V(�) = a0 + a1

(G′G

). (33)

From Eqs. (7) and (33) we drive

V ′ = −a1

(G′G

)2

− a1�(G′G

)− a1�, (34)

V ′′ = 2a1

(G′G

)3

+ 3a1�(G′G

)2

+ (a1�2 + 2a1�)

(G′G

)+ a1��, (35)

V2 = a21

(G′G

)2

+ 2a0a1

(G′G

)+ a2

0, (36)

V3 = a31

(G′G

)3

+ 3a0a21

(G′G

)2

+ 3a20a1

(G′G

)+ a3

0, (37)

nd

V4 = a41

(G′G

)4

+ 4a0a31

(G′G

)3

+ 6a20a

21

(G′G

)2

+ 4a30a1

(G′G

)+ a4

0. (38)

Substituting Eqs. (33)–(38) into Eq. (31), collecting all terms with the same powers of G′/G and setting each coefficient to zero, we obtain

system of algebraic equations for a0, a1, ω, �, � and � as follows.(

G′G

)4coeff.:

4n(a + d)a21 + (a + d)(1 − 2n)a2

1 + 4n2ca41 = 0, (39)

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M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489 5485

(G′G

)3coeff.:

4n(a + d)a0a1 + 16n2ca0a31 + 6n(a + d)�a2

1 + 4n2ba31 + 2(a + d)(1 − 2n)�a2

1 = 0,

G′G

)2coeff.:

2n(a + d)a21�

2 + (a + d)(1 − 2n)a21�

2 + 6n(a + d)a0a1� + 12n2ba0a21 + 4n(a + d)a2

1� + 2(a + d)(1 − 2n)�a21

− 4n2(ω + a�2)a21 + 24n2ca2

0a21 = 0,

(G′G

)1coeff.:

2n(a + d)a0a1�2 + 2n(a + d)��a2

1 + 2(a + d)(1 − 2n)a21�� − 8n2(ω + a�2)a0a1 + 4n(a + d)a0a1� + 16n2ca3

0a1 + 12n2ba20a1 = 0,

G′G

)0coeff.:

(a + d)(1 − 2n)�2a21 + 2n(a + d)a0a1�� − 4n2(ω + a�2)a2

0 + 4n2ca40 + 4n2ba3

0 = 0.

Solving this system with the aid of Maple gives

a0 = −b(1 + 2n)4c(1 + n)

± �√

−c(a + d)(1 + 2n)4nc

,

a1 = ±√

−c(a + d)(1 + 2n)2nc

,

ω = −{a�2 + b2(1 + 2n)

4c(1 + n)2

},

� = n2b2(1 + 2n)

4c(a + d)(1 + n)2+ �2

4,

(40)

here � and � are arbitrary constants.Substituting the solution set (40) into Eq. (33), the solution formulae of Eq. (31) can be written as

V(�) = −b(1 + 2n)4c(1 + n)

±√

−c(a + d)(1 + 2n)2nc

(�

2+ G′G

). (41)

Substituting the general solutions of second order linear ODE into Eq. (41) gives three types of traveling wave solutions.Case-I: When � = �2 − 4� > 0, we obtain the hyperbolic function traveling wave solution

(G′G

)= −�

2+√�2 − 4�

2

{C1 sinh((

√�2 − 4�/2)�) + C2 cosh((

√�2 − 4�/2)�)

C1 cosh((√�2 − 4�/2)�) + C2 sinh((

√�2 − 4�/2)�)

}, (42)

(x, t) =

[− b(1 + 2n)

4c(1 + n)

{1 ∓

C1 sinh(√

−(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at)) + C2 cosh(√

−(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at))

C1 cosh(√

−(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at)) + C2 sinh(√

−(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at))

}]1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}, (43)

here C1 and C2 are arbitrary constants.On the other hand, assuming C1 /= 0 and C2 = 0 the traveling wave solution of Eq. (6) can be written as

(x, t) ={

−b(1 + 2n)4c(1 + n)

(1 ∓ tanh

[√− n2b2(1 + 2n)

4c(a + d)(1 + n)2 (x + 2a�t)

])}1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}. (44)

Next, assuming C1 = 0 and C2 /= 0, then we obtain

(x, t) ={

−b(1 + 2n)4c(1 + n)

(1 ∓ coth

[√− n2b2(1 + 2n)

4c(a + d)(1 + n)2 (x + 2a�t)

])}1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}. (45)

Eqs. (44) and (45), respectively, represent topological and singular 1-soliton solutions. The constraint condition for the existence ofhese soliton is

c(a + d) < 0.

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486 M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489

ase-II: When � = �2 − 4� < 0, we obtain the hyperbolic function traveling wave solution(G′G

)= −�

2+√

4� − �2

2

{−C1 sin((

√4� − �2/2)�) + C2 cos((

√4� − �2/2)�)

C1 cos((√

4� − �2/2)�) + C2 sin((√

4� − �2/2)�)

}, (46)

(x, t) =

[− b(1 + 2n)

4c(1 + n)

{1 ∓ i

−C1 sin(√

(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at)) + C2 cos(√

(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at))

C1 cos(√

(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at)) + C2 sin(√

(n2b2(1 + 2n)/4c(a + d)(1 + n)2)(x + 2�at))

}]1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}, (47)

here C1 and C2 are arbitrary constants.On the other hand, assuming C1 /= 0 and C2 = 0 the traveling wave solution of Eq. (6) can be written as

(x, t) ={

−b(1 + 2n)4c(1 + n)

(1 ± i tan

[√n2b2(1 + 2n)

4c(a + d)(1 + n)2 (x + 2a�t)

])}1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}. (48)

Next, assuming C1 = 0 and C2 /= 0, then we obtain

(x, t) ={

−b(1 + 2n)4c(1 + n)

(1 ∓ i cot

[√n2b2(1 + 2n)

4c(a + d)(1 + n)2 (x + 2a�t)

])}1/2n

× ei{−�x−(a�2+(b2(1+2n)/4c(1+n)2))t+�}. (49)

Eqs. (48) and (49) are complex-valued solutions to the RNLSE that are often refereed to as complexitons. The constraint for the existencef these solutions is:

c(a + d) > 0.

ase-III: When � = �2 − 4� = 0, we obtain the rational solutions(G′G

)=(

C2

C1 + C2�

)− �

2, (50)

(x, t) =[

±√

−c(a + d)(1 + 2n)2nc

C2

C1 + C2 (x + 2a�t)

]1/2n

ei{−�x−a�2t+�}. (51)

.2. Time-dependent coefficients

To obtain an analytic solution, we use the transformation

U(�) = V1/2n(�), (52)

hat will reduce Eq. (20) into the following ODE

(a(t) + d(t))(1 − 2n)(V ′)2 + 2n(a(t) + d(t))VV ′′ − 4n2(tdω(t)dt

+ ω(t) + �2a(t))V2 + 4n2b(t)V3 + 4n2c(t)V4 = 0. (53)

Balancing VV′′ with V3 in Eq. (53) we find that N = 1. Therefore, we can write the solution of Eq. (53) in the form of

V(�) = a0 + a1

(G′G

), a1 /= 0. (54)

By using (54) and the solution of (5), we can find (V′)2, VV′′, V2 and V3 as polynomials of G′/G. Performing step 5 of description of theethod yields to a set of simultaneous algebraic equations for a0, a1, ω, �, � and � . Solving this system by Maple gives

a0 = −b(t)(1 + 2n)4c(t)(1 + n)

± �√

−c(t)(a(t) + d(t))(1 + 2n)4nc(t)

,

a1 = ±√

−c(t)(a(t) + d(t))(1 + 2n)2nc(t)

,

ω = −1t

∫ t

0

{�2a(t′) + b2(t′)(1 + 2n)

4c(t′)(1 + n)2

}dt′,

� = n2b2(t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2+ �2

4,

(55)

where � and � are arbitrary constants.

Substituting the solution set (55) into Eq. (54), the solution formulae of Eq. (53) can be written as

V(�) = −b(t)(1 + 2n)4c(t)(1 + n)

±√

−c(t)(a(t) + d(t))(1 + 2n)2nc(t)

(�

2+ G′G

). (56)

ws

t

w

M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489 5487

Substituting the general solutions of second order linear ODE into Eq. (56) gives three types of traveling wave solutions.Case-I: When � = �2 − 4� > 0, we obtain the hyperbolic function traveling wave solution

(G′G

)= −�

2+√�2 − 4�

2

{C1 sinh((

√�2 − 4�/2)�) + C2 cosh((

√�2 − 4�/2)�)

C1 cosh((√�2 − 4�/2)�) + C2 sinh((

√�2 − 4�/2)�)

}, (57)

(x, t) =

[− b(t)(1 + 2n)

4c(t)(1 + n)

{1 ∓

C1 sinh(√

−(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�) + C2 cosh(√

−(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�)

C1 cosh(√

−(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�) + C2 sinh(√

−(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�)

}]1/2n

× ei{−�x−

∫ t

0{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}

, (58)

here C1 and C2 are arbitrary constants and � = x + 2�∫ t

0a(t′) dt′. On the other hand, assuming C1 /= 0 and C2 = 0 the traveling wave

olution of Eq. (19) can be written as

(x, t) ={

−b(t)(1 + 2n)4c(t)(1 + n)

(1 ∓ tanh

[√− n2b2(t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2

(x + 2�

∫ t

0

a(t′)dt′)])}1/2n

× ei{−�x−

∫ t0

{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}. (59)

Next, assuming C1 = 0 and C2 /= 0, then we obtain

(x, t) ={

−b(t)(1 + 2n)4c(t)(1 + n)

(1 ∓ coth

[√− n2b2(t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2

(x + 2�

∫ t

0

a(t′)dt′)])}1/2n

× ei{−�x−

∫ t0

{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}. (60)

Eqs. (59) and (60), respectively, represent topological and singular 1-soliton solutions. The constraint condition for the existence ofhese soliton is

c(t)(a(t) + d(t)) < 0.

Case-II: When � = �2 − 4� < 0, we obtain the hyperbolic function traveling wave solution

(G′G

)= −�

2+√

4� − �2

2

{−C1 sin((

√4� − �2/2)�) + C2 cos((

√4� − �2/2)�)

C1 cos((√

4� − �2/2)�) + C2 sin((√

4� − �2/2)�)

}, (61)

(x, t) =

[− b(t)(1 + 2n)

4c(t)(1 + n)

{1 ∓ i

−C1 sin(√

(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�) + C2 cos(√

(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�)

C1 cos(√

(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�) + C2 sin(√

(n2b2(t)(1 + 2n)/4c(t)(a(t) + d(t))(1 + n)2)�)

}]1/2n

× ei{−�x−

∫ t

0{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}

, (62)

here C1 and C2 are arbitrary constants and � = x + 2�∫ t

0a(t′)dt′.

On the other hand, assuming C1 /= 0 and C2 = 0 the traveling wave solution of Eq. (19) can be written as

(x, t) ={

−b(t)(1 + 2n)4c(t)(1 + n)

(1 ± i tan

[√n2b2(t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2

(x + 2�

∫ t

0

a(t′)dt′)])}1/2n

× ei{−�x−

∫ t0

{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}. (63)

Next, assuming C1 = 0 and C2 /= 0, then we obtain

{ ( [√2 2

( ∫ t )])}1/2n

(x, t) = −b(t)(1 + 2n)4c(t)(1 + n)

1 ∓ i cotn b (t)(1 + 2n)

4c(t)(a(t) + d(t))(1 + n)2x + 2�

0

a(t′)dt′

× ei{−�x−

∫ t0

{�2a(t′)+(b2(t′)(1+2n)/4c(t′)(1+n)2)} dt′+�}. (64)

5488 M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489

Fig. 1. Numerical simulation of topological soliton: a = 1, b = 1, c = −1, d = 1, c1 = 1, c2 = 0.1, n = 1/2.

wa

p

5

rnTitR

Fig. 2. Numerical simulation of topological soliton: a = 1, b = 1, c = −1, d = 1, c1 = 1, c2 = 0.1, n = 3/2.

Eqs. (63) and (64) are complexitons solutions of RNLSE. The constraint for the existence of these solutions is

c(t)(a(t) + d(t)) > 0.

Case-III: When � = �2 − 4� = 0, we obtain the rational solutions(G′G

)=(

C2

C1 + C2�

)− �

2, (65)

(x, t) =

⎡⎣±

√−c(t)(a(t) + d(t))(1 + 2n)

2nc(t)C2

C1 + C2

(x + 2�

∫ t0a(t′)dt′

)⎤⎦

1/2n

ei{−�x−�2

∫ t0a(t′) dt′+�}

. (66)

The soliton solutions of this paper are of two types. The first one is the topological soliton solution that are also referred to as shockaves or kinks. On the other hand, the singular solitons could possibly model optical rogons or rogue waves. The rational solutions are

lso a possible candidate to describe rogons, when the denominator approaches closer to zero.A couple of numerical simulations here illustrate the topological soliton solutions that are obtained. For both of these simulations,

arameter values selected are a = 1, b = 1, c = −1, d = 1, c1 = 1 and c2 = 0.1. Fig. 1 shows the profile of a topological soliton for n = 1/2.Fig. 2 is another profile of a topological soliton where n = 3/2.

. Conclusions

This paper addressed RNLSE, with dual-power law nonlinearity, by the aid of G′/G-expansion method. Topological soliton solutions areecovered. Both constant coefficients as well as time-dependent coefficients are taken into consideration. The results for parabolic lawonlinearity fell out as a special case to the dual-power law nonlinearity when the power-law nonlinearity parameter n collapses to unity.he results of this paper can be applied to several areas of applied science where RNLSE appears. The results of this paper will be applied

n future to address other aspects as well. When RNLSE will is considered with additional Hamiltonian perturbation terms, this integrationool can be applied to extract soliton solutions. Additionally, the case of vector RNLSE will be studied in future along with multi-dimensionalNLSE. Those results will be reported in future.

A

s

R

[

[

[[

[

[[

[

[

[[

M. Mirzazadeh et al. / Optik 125 (2014) 5480–5489 5489

cknowledgment

The authors would like to present their sincere note of thanks and appreciation to the referee for valuable and helpful comments withuggestions.

eferences

[1] A. Biswas, Soliton solutions of the perturbed resonant nonlinear Schrödinger’s equation with full nonlinearity by semi-inverse variational principle, Quantum Phys. Lett.1 (2) (2012) 79–84.

[2] M. Eslami, M. Mirzazadeh, A. Biswas, Soliton solutions of the resonant nonlinear Schrödinger’s equation in optical fibers with time-dependent coefficients by simplestequation approach, J. Mod. Opt. 60 (19) (2013) 1627–1636.

[3] M. Eslami, M. Mirzazadeh, A. Biswas, Optical solitons for the resonant nonlinear Schrödinger’s equation with time-dependent coefficients by the first integral method,Optik 125 (13) (2014) 3107–3116.

[4] H. Triki, T. Hayat, O.M. Aldossary, A. Biswas, Bright and dark solitons for the resonant nonlinear Schrödinger’s equation with time-dependent coefficients, Opt. LaserTechnol. 44 (7) (2012) 2223–2231.

[5] H. Triki, A. Yildirim, T. Hayat, O.M. Aldossary, A. Biswas, 1-Soliton solution of the generalized resonant nonlinear dispersive Schrödinger’s equation with time-dependentcoefficients, Adv. Sci. Lett. 16 (2012) 309–312.

[6] M.L. Wang, X.Z. Li, J.L. Zhang, The G′/G-expansion method and traveling wave solutions of nonlinear evolution equations in mathematical physics, Phys. Lett. A 372 (4)(2008) 417–423.

[7] E.M.E. Zayed, M.A.S. EL-Malky, The G′/G-expansion method for solving nonlinear Klein–Gordon equations, AIP Conf. Proc. ICNAAM Am. Inst. Phys. 1389 (2011) 2020–2024.[8] E.M.E. Zayed, M.A.S. EL-Malky, Exact solutions of nonlinear PDEs in mathematical physics using the G′/G-expansion method, Adv. Theor. Appl. Mech. 4 (2) (2011) 91–100.[9] E.M.E. Zayed, Equivalence of the G′/G-expansion method and the tanh-coth function method, AIP Conf. Proc. Am. Inst. Phys. 1281 (1) (2010) 2225–2228.10] E.M.E. Zayed, The G′/G-expansion method and its applications to some nonlinear evolution equations in the mathematical physics, J. Appl. Math. Comput. 30 (2009)

89–103.11] E.M.E. Zayed, K.A. Gepreel, The G′/G-expansion method for finding traveling wave solutions of nonlinear PDEs in mathematical physics, J. Math. Phys. 50 (1) (2009)

013502–013513.12] E.M.E. Zayed, K.A. Gepreel, Some applications of the G′/G-expansion method to nonlinear partial differential equations, Appl. Math. Comput. 212 (1) (2009) 1–13.13] E.M.E. Zayed, New traveling wave solutions for higher dimensional nonlinear evolution equations using a generalized G′/G-expansion method, J. Phys. A: Math. Theor.

42 (2009) 195202–195214.14] E.M.E. Zayed, K.A. Gepreel, Three types of traveling wave solutions for nonlinear evolution equations using the G′/G-expansion method, Int. J. Nonlinear Sci. 7 (4) (2009)

501–512.15] E.M.E. Zayed, Traveling wave solutions for higher dimensional nonlinear evolutions using the G′/G-expansion method, J. Appl. Math. Inform. 28 (1–2) (2010) 383–395.16] E.M.E. Zayed, S.A. Hoda Ibrahim, M.A.M. Abdelaziz, Traveling wave solutions for the Burgers equation and the KdV equation with variable coefficients using the generalized

G′/G-expansion method, Z. Naturforsch. A 65a (12) (2010) 1065–1070.17] E.M.E. Zayed, K.A. Gepreel, New applications of an improved G′/G-expansion method to construct the exact solutions of nonlinear PDEs, Int. J. Nonlinear Sci. Numer.

Simul. 11 (4) (2010) 273–283.18] E.M.E. Zayed, Applications of an extended G′/G-expansion method to find exact solutions of nonlinear PDEs in mathematical physics, Math. Probl. Eng. 2010 (2010),

Article ID 768573, http://dx.doi.org/10.1155/2010/768573, 19 pp.19] H. Zhang, New application of the G′/G-expansion method, Commun. Nonlinear Sci. Numer. Simul. 14 (2009) 3220–3225.20] S. Zhang, J.L. Tong, W. Wang, A generalized G′/G-expansion method for the mKdV equation with variable coefficients, Phys. Lett. A 372 (13) (2008) 2254–2257.