# Three Different Methods for New Soliton Solutions of the Generalized NLS Equation

Authors: Anwar Ja’afar Mohamad Jawad
Journal: Abstract and Applied Analysis
 (2017)

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## Abstract

Three different methods are applied to construct new types of solutions of nonlinear evolution equations. First, the Csch method is used to carry out the solutions; then the Extended Tanh-Coth method and the modified simple equation method are used to obtain the soliton solutions. The effectiveness of these methods is demonstrated by applications to the RKL model, the generalized derivative NLS equation. The solitary wave solutions and trigonometric function solutions are obtained. The obtained solutions are very useful in the nonlinear pulse propagation through optical fibers.


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## Body

## 1. Introduction

Partial differential equations describe various nonlinear phenomena in natural and applied sciences such as fluid dynamics, plasma physics, solid state physics, optical fibers, acoustics, biology, and mathematical finance. Partial differential equations which arise in real-world physical problems are often too complicated to be solved exactly. It is of significant importance to solve nonlinear partial differential equations (NLPDEs) from both theoretical and practical points of view. The analysis of some physical phenomena is investigated by the exact solutions of nonlinear evolution equations (NLEEs) [1–9].In this paper, the third-order generalized NLS equation is studied, which is proposed by Radhakrishnan, Kundu, and Lakshmanan (RKL) [10]. The normalized RKL model can be written as (1)iqt+qxx+2q2q+iαqxxx+iβq2qx+iγq4qx+δq4q=0.Equation (1) describes the propagation of femtosecond optical pulses, q(x,t) represents normalized complex slowly varying amplitude of the pulse envelope, and α,β,γ, and δ are real constants. Some solitary wave solutions and combined Jacobian elliptic function solution were constructed by different methods [3, 4].The Csch method is used to carry out the solutions. Then, the Extended Tanh-Coth method and the modified simple equation method are used to obtain the soliton solutions of this equation.


## 2. Traveling Wave Solution

Consider the nonlinear partial differential equation in the form(2)Fu,ut,ux,utt,uxx,uxxx,…=0,where u(x,t) is a traveling wave solution of nonlinear partial differential equation (2). We use the transformations,(3)ux,t=fξ,where ξ=kx-λt. This enables us to use the following changes:(4)∂∂t·=-λddξ·,∂∂x·=kddξ·.Using (4) to transfer the nonlinear partial differential equation (2) to nonlinear ordinary differential equation,(5)Qf,f′,f′′,f′′′,…=0.The ordinary differential equation (5) is then integrated as long as all terms contain derivatives, where we neglect the integration constants.


## 3. The Generalized NLS Equation (RKL)

In this section, the generalized third-order NLS equation (RKL) (1) is chosen to illustrate the effectiveness of three methods.The solution of (1) may be supposed as(6)qx,t=eiθuξ,whereθ=k1x+k2t,ξ=kx-λt.Substituting (6) into (1) and by defining the derivatives,(7)qt=-λu′ξ+ik2uξeiθ,qx=ku′ξ+ik1uξeiθ,qxx=k2u′′+2ikk1u′-k12ueiθ,qxxx=k3u′′′+2ik2k1u′′-kk12u′+ik1k2u′′-2kk12u′-ik13ueiθ;then decomposing (1) into real and imaginary parts yields a pair of relations which represented nonlinear ordinary differential equations. The real part is(8)k21-3αk1u′′+αk13-k12-k2u+2-3k1βu3+δ-5k1γu5=0,while the imaginary part is(9)αk3u′′′+2kk1-λ-3αkk12u′+3βku2u′+5kγu4u′=0.Integrating (9) once and setting the integration constant to zero, we obtain(10)αk3u′′+2kk1-λ-3αkk12u+βku3+kγu5=0.Equations (8) and (10) will be equivalent, provided that(11)k21-3αk1αk3=αk13-k12-k22kk1-λ-3αkk12=2-3βk1βk=δ-5γk1kγ,from which we get the parametric constraints(12)δ=1α+2k1γ,β=2α,α=-kk2-8kk12+3λk1∓kk2-8kk12+3λk12+32kk13λ-2kk116kk13;multiplying both sides of (10) by u′ and integrating with respect to ξ with zero constant, we get(13)u′2+βαk13-k12-k2αk2u2+β2αk2u4+γ3αk2u6=0;assume that(14)c2=βαk13-k12-k2αk2,c4=β2αk2,c6=γ3αk2.Then(15)u′2+c2u2+c4u4+c6u6=0.


## 4. Methodology

In this section we will apply three different methods to solve (15). These methods are Csch method, Extended Tanh-Coth method, and the modified simple equation method (MSEM).


### 4.1. Csch Function Method

The solution of many nonlinear equations can be expressed in the form [11](16)uξ=Acschτμξand their derivative(17)u′ξ=-Aτμcschτμξ·coth⁡μξ,u′′ξ=Aτμ2τ+1cschτ+2μξ+τcschτμξ,where A, μ, and τ are parameters to be determined and μ and λ are the wave number and the wave speed, respectively. We substitute (16)-(17) into the reduced equation (15); we get(18)A2τ2μ2csch2τμξ+A2τ2μ2csch2τ+2μξ+c2A2csch2τμξ+c4A4csch4τμξ+c6A6csch6τμξ=0.Balance the terms of the Csch functions to find τ(19)2τ+2=6τ,Thenτ=12.We next collect all terms in (18) with the same power in cschkμξ and set their coefficients to zero to get a system of algebraic equations among the unknowns A, μ, and τ and solve the subsequent system(20)A214μ2+c2A2=0,A214μ2+c6A6=0.Solving the system of equations in (20), we get(21)μ=2ic2=2ikβααk13-k12-k2,A=∓c2c64=∓3βγαk13-k12-k24;then(22)uξ=∓3βγαk13-k12-k24cschμξtherefore(23)qx,t=∓3βγαk13-k12-k24eik1x+k2tcsch⁡2ikβααk13-k12-k2kx-λt;Figure 1 represents the solitary wave in (23) for k=k1=k2=1,λ=-1,γ=-1, δ=-10/3, β=3/2, α=3/4, and then qx,t=1.54csch⁡3.1622(x+t).Figure 1
The solitary wave in (23) for 0≤t≤1,0≤x≤10.


### 4.2. Tanh-Coth Method

The key step is to introduce the ansatz, the new independent variable [12, 13] (24)Y=tanh⁡ξthat leads to the change of variables:(25)dUdξ=1-Y2dUdY,(26)d2Udξ2=-2Y1-Y2dUdY+1-Y22d2UdY2,(27)d3Udξ3=21-Y23Y2-1dUdY-6Y1-Y22d2UdY2+1-Y23d3UdY3.Assume(28)Vξ=u2ξ.Equation (15) can be written as(29)V′2+4c2V2+4c4V3+4c6V4=0.The next step is that the solution of (29) is expressed in the form(30)Vξ=∑i=0maiYi+∑i=1mbiY-i,where the parameter m can be found by balancing the highest-order linear term with the nonlinear terms in (29).We balanceV4 with dV/dY2, to obtain 4m=m+12; then m=1. The Tanh-Coth method admits the use of the finite expansion for(31)V=a0+a1Y+b1Y-1,V′=a1-b1Y-2.λ,k,a0,a1,b1 are to be determined. Substituting (31) into (25) and then into (29) will yield a set of algebraic equations because all coefficients of Y have to vanish(32)1-Y22a1-b1Y-22+4c2a0+a1Y+b1Y-12+4c4a0+a1Y+b1Y-13+4c6a0+a1Y+b1Y-14=0.Equation (32) can be written as(33)a121-2Y2+Y4-2a1b1Y-2-2+Y2+b12Y-4-2Y-2+1+4c2a02+2a1b1+2a0a1Y+a12Y2+2a0b1Y-1+b12Y-2+4c4a03+3a02a1Y+3a0a12Y2+a13Y3+3a02Y-1+2a0a1+a12Yb1+3a0Y-2+a1Y-1b12+b13Y-3+4c6a04+4a03a1Y+6a02a12Y2+4a0a13Y3+a14Y4+4a03Y-1+3a02a1+3a0a12Y+a13Y2b1+6a02Y-2+2a0a1Y-1+a12b12+4a0Y-3+a1Y-2b13+b14Y-4=0.Equating expressions at Yi,i=-4,-3,-2,-1,0,1,2,3,4 to zero, we have the following system of equations:(34)CoefficientsofY-4:1+4c6b12b12=0CoefficientsofY-3:c4+4c6a0b13=0CoefficientsofY-2:-2a1b1-2b12+4c2b12+12c4a0b12+4c66a02+4a1b1b12=0CoefficientsofY-18c2a0+12c4a02+a1b1+4c64a03+12a0a1b1b1=0CoefficientsofY0:a12+4a1b1+b12+4c2a02+4c4a03+6a0a1b1+4c6a04+12a02a1b1+6a12b12=0CoefficientsofY:8c2a0a1+12c4a02+a1b1a1+16c6a0a1a02+3a1b1=0CoefficientsofY2:-2a12-2a1b1+4c2a12+12c4a0a12+4c6a126a02+4a1b1=0CoefficientsofY3:c4+4c6a0a13=0CoefficientsofY4:1+4c6a12a12=0.Solving the system of equations (34), we get(35)b1=∓i2c6,a0=-12c6,a1=0,c2=1,c4=2c6.Substitute for c6 from (14), and then(36)b1=∓ik2,a0=-k2,a1=0,c4=1k1k,k=2αk13-k12-k2,γ=3α4k2,δ=31+2αk14k2;therefore(37)Vξ=k2-1∓icoth⁡ξ,uξ=k-1∓icoth⁡ξ.Then(38)qx,t=eik1x+k2tk-1∓icoth⁡kx-λtfor k=1/2, k1=k2=1,λ=3k, α=1/4, and then (39)qx,t=12-1+icoth⁡12x-3t.Figure 2 represents the solitary wave in (39).Figure 2
The solitary wave in (39) for 0≤t≤1,0≤x≤10.


### 4.3. The Modified Simple Equation Method

We look for solutions of (29) in the form [14](40)V=A0+A1ψξψ,Vξ=A1ψξξψ-ψξ2ψ2.Then (29) can be written as(41)A12ψξξψ-ψξ2ψ22+4c2A0+A1ψξψ2+4c4A0+A1ψξψ3+4c6A0+A1ψξψ4=0.Then (41) can be written as(42)A12ψξξ2ψ2-2ψξξψξ2ψ3+ψξ4ψ4+4c2A02+2A0A1ψξψ+A12ψξ2ψ2+4c4A03+3A02A1ψξψ+3A0A12ψξ2ψ2+A13ψξ3ψ3+4c6A04+4A03A1ψξψ+6A02A12ψξ2ψ2+4A0A13ψξ3ψ3+A14ψξ4ψ4=0.Equating expressions in (42) at ψ-1,ψ-2,ψ-3, and ψ-4 to zero, we have the following system of equations:(43)1+4c6A12A12=0,ψξξ-2A1c4+4c6A0ψξA12=0,ψξξ2+4c2+3c4A0+6c6A02ψξ2A12=0,2c2+3c4A0+4c6A02A0A1=0.Solving the system of equations in (43),(44)A1=∓ik23αγ.Family  1
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α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.Family  11
(55)
A
0
=
-
3
β
+
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
11
ξ
=
ε
111
+
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
11
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
111
+
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.Family  12
(56)
A
0
=
-
3
β
-
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
12
ξ
=
ε
121
+
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
12
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
121
+
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.


## 4.1. Csch Function Method

The solution of many nonlinear equations can be expressed in the form [11](16)uξ=Acschτμξand their derivative(17)u′ξ=-Aτμcschτμξ·coth⁡μξ,u′′ξ=Aτμ2τ+1cschτ+2μξ+τcschτμξ,where A, μ, and τ are parameters to be determined and μ and λ are the wave number and the wave speed, respectively. We substitute (16)-(17) into the reduced equation (15); we get(18)A2τ2μ2csch2τμξ+A2τ2μ2csch2τ+2μξ+c2A2csch2τμξ+c4A4csch4τμξ+c6A6csch6τμξ=0.Balance the terms of the Csch functions to find τ(19)2τ+2=6τ,Thenτ=12.We next collect all terms in (18) with the same power in cschkμξ and set their coefficients to zero to get a system of algebraic equations among the unknowns A, μ, and τ and solve the subsequent system(20)A214μ2+c2A2=0,A214μ2+c6A6=0.Solving the system of equations in (20), we get(21)μ=2ic2=2ikβααk13-k12-k2,A=∓c2c64=∓3βγαk13-k12-k24;then(22)uξ=∓3βγαk13-k12-k24cschμξtherefore(23)qx,t=∓3βγαk13-k12-k24eik1x+k2tcsch⁡2ikβααk13-k12-k2kx-λt;Figure 1 represents the solitary wave in (23) for k=k1=k2=1,λ=-1,γ=-1, δ=-10/3, β=3/2, α=3/4, and then qx,t=1.54csch⁡3.1622(x+t).Figure 1
The solitary wave in (23) for 0≤t≤1,0≤x≤10.


## 4.2. Tanh-Coth Method

The key step is to introduce the ansatz, the new independent variable [12, 13] (24)Y=tanh⁡ξthat leads to the change of variables:(25)dUdξ=1-Y2dUdY,(26)d2Udξ2=-2Y1-Y2dUdY+1-Y22d2UdY2,(27)d3Udξ3=21-Y23Y2-1dUdY-6Y1-Y22d2UdY2+1-Y23d3UdY3.Assume(28)Vξ=u2ξ.Equation (15) can be written as(29)V′2+4c2V2+4c4V3+4c6V4=0.The next step is that the solution of (29) is expressed in the form(30)Vξ=∑i=0maiYi+∑i=1mbiY-i,where the parameter m can be found by balancing the highest-order linear term with the nonlinear terms in (29).We balanceV4 with dV/dY2, to obtain 4m=m+12; then m=1. The Tanh-Coth method admits the use of the finite expansion for(31)V=a0+a1Y+b1Y-1,V′=a1-b1Y-2.λ,k,a0,a1,b1 are to be determined. Substituting (31) into (25) and then into (29) will yield a set of algebraic equations because all coefficients of Y have to vanish(32)1-Y22a1-b1Y-22+4c2a0+a1Y+b1Y-12+4c4a0+a1Y+b1Y-13+4c6a0+a1Y+b1Y-14=0.Equation (32) can be written as(33)a121-2Y2+Y4-2a1b1Y-2-2+Y2+b12Y-4-2Y-2+1+4c2a02+2a1b1+2a0a1Y+a12Y2+2a0b1Y-1+b12Y-2+4c4a03+3a02a1Y+3a0a12Y2+a13Y3+3a02Y-1+2a0a1+a12Yb1+3a0Y-2+a1Y-1b12+b13Y-3+4c6a04+4a03a1Y+6a02a12Y2+4a0a13Y3+a14Y4+4a03Y-1+3a02a1+3a0a12Y+a13Y2b1+6a02Y-2+2a0a1Y-1+a12b12+4a0Y-3+a1Y-2b13+b14Y-4=0.Equating expressions at Yi,i=-4,-3,-2,-1,0,1,2,3,4 to zero, we have the following system of equations:(34)CoefficientsofY-4:1+4c6b12b12=0CoefficientsofY-3:c4+4c6a0b13=0CoefficientsofY-2:-2a1b1-2b12+4c2b12+12c4a0b12+4c66a02+4a1b1b12=0CoefficientsofY-18c2a0+12c4a02+a1b1+4c64a03+12a0a1b1b1=0CoefficientsofY0:a12+4a1b1+b12+4c2a02+4c4a03+6a0a1b1+4c6a04+12a02a1b1+6a12b12=0CoefficientsofY:8c2a0a1+12c4a02+a1b1a1+16c6a0a1a02+3a1b1=0CoefficientsofY2:-2a12-2a1b1+4c2a12+12c4a0a12+4c6a126a02+4a1b1=0CoefficientsofY3:c4+4c6a0a13=0CoefficientsofY4:1+4c6a12a12=0.Solving the system of equations (34), we get(35)b1=∓i2c6,a0=-12c6,a1=0,c2=1,c4=2c6.Substitute for c6 from (14), and then(36)b1=∓ik2,a0=-k2,a1=0,c4=1k1k,k=2αk13-k12-k2,γ=3α4k2,δ=31+2αk14k2;therefore(37)Vξ=k2-1∓icoth⁡ξ,uξ=k-1∓icoth⁡ξ.Then(38)qx,t=eik1x+k2tk-1∓icoth⁡kx-λtfor k=1/2, k1=k2=1,λ=3k, α=1/4, and then (39)qx,t=12-1+icoth⁡12x-3t.Figure 2 represents the solitary wave in (39).Figure 2
The solitary wave in (39) for 0≤t≤1,0≤x≤10.


## 4.3. The Modified Simple Equation Method

We look for solutions of (29) in the form [14](40)V=A0+A1ψξψ,Vξ=A1ψξξψ-ψξ2ψ2.Then (29) can be written as(41)A12ψξξψ-ψξ2ψ22+4c2A0+A1ψξψ2+4c4A0+A1ψξψ3+4c6A0+A1ψξψ4=0.Then (41) can be written as(42)A12ψξξ2ψ2-2ψξξψξ2ψ3+ψξ4ψ4+4c2A02+2A0A1ψξψ+A12ψξ2ψ2+4c4A03+3A02A1ψξψ+3A0A12ψξ2ψ2+A13ψξ3ψ3+4c6A04+4A03A1ψξψ+6A02A12ψξ2ψ2+4A0A13ψξ3ψ3+A14ψξ4ψ4=0.Equating expressions in (42) at ψ-1,ψ-2,ψ-3, and ψ-4 to zero, we have the following system of equations:(43)1+4c6A12A12=0,ψξξ-2A1c4+4c6A0ψξA12=0,ψξξ2+4c2+3c4A0+6c6A02ψξ2A12=0,2c2+3c4A0+4c6A02A0A1=0.Solving the system of equations in (43),(44)A1=∓ik23αγ.Family  1
(45)
A
0
=
-
β
+
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
1
ξ
=
ε
11
+
ε
12
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
1
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
k
ε
12
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
11
+
ε
12
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  2
(46)
A
0
=
-
β
-
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
2
ξ
=
ε
21
+
ε
22
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
2
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
k
ε
22
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
21
+
ε
22
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  3
(47)
A
0
=
-
3
β
+
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
3
ξ
=
ε
31
+
ε
32
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
3
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
k
ε
32
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
31
+
ε
32
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  4
(48)
A
0
=
-
3
β
-
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
4
ξ
=
ε
41
+
ε
42
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
4
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
k
ε
42
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
41
+
ε
42
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  5
(49)
A
0
=
-
β
+
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
5
ξ
=
ε
51
+
ε
52
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
5
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
+
k
ε
52
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
51
+
ε
52
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  6
(50)
A
0
=
-
β
-
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
6
ξ
=
ε
61
+
ε
62
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
6
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
+
k
ε
62
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
61
+
ε
62
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  7
(51)
A
0
=
-
3
β
+
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
7
ξ
=
ε
71
+
ε
72
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
7
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
+
k
ε
72
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
71
+
ε
72
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  8
(52)
A
0
=
-
3
β
-
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
8
ξ
=
ε
81
+
ε
82
e
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
ξ
,
q
8
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
+
k
ε
82
3
α
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
γ
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
ε
81
+
ε
82
e
-
2
i
c
2
+
3
c
4
A
0
+
6
c
6
A
0
2
k
x
-
λ
t
1
/
2
.Family  9
(53)
A
0
=
-
β
+
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
9
ξ
=
ε
91
+
ε
92
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
9
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
92
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
91
+
ε
92
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.Family  10
(54)
A
0
=
-
β
-
β
2
-
16
γ
β
α
k
1
3
-
k
1
2
-
k
2
/
3
4
γ
/
3
,
ψ
10
ξ
=
ε
101
+
ε
102
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
10
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
102
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
101
+
ε
102
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.Family  11
(55)
A
0
=
-
3
β
+
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
11
ξ
=
ε
111
+
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
11
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
111
+
ε
112
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.Family  12
(56)
A
0
=
-
3
β
-
9
β
2
-
128
β
γ
α
k
1
3
-
k
1
2
-
k
2
/
3
16
γ
/
3
,
ψ
12
ξ
=
ε
121
+
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
ξ
,
q
12
x
,
t
=
e
i
k
1
x
+
k
2
t
A
0
-
3
k
2
α
γ
c
4
+
2
c
6
A
0
2
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
ε
121
+
ε
122
e
i
k
3
α
/
γ
c
4
+
2
c
6
A
0
k
x
-
λ
t
1
/
2
.


## 5. Conclusion

In this paper, series of new traveling wave solutions have been obtained. The Csch method and the Extended Tanh-Coth method and modified simple equation method are used to carry out the integration of the generalized NLS equation, which is RKL. These methods can be also applied to solve other types of the generalized nonlinear evolution equations with complex coefficients. The solitary waves in Figures1 and 2 obtained by the Csch and Tanh-Coth methods, respectively, are identical in form and behavior. The obtained solutions are very useful and may be important to explain some physical phenomena and find applications in the nonlinear pulse propagation through optical fibers.



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*Source: 5137946-2017-10-18.xml*