# Analytical Soliton Solutions of the Coupled Radhakrishnan-Kundu-Lakshmanan Equation via Three Techniques

Authors: Khalid K. Ali; M. S. Mehanna; M. Ali Akbar; Prasun Chakrabarti
Journal: Journal of Mathematics
 (2022)

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## Abstract

The analytical soliton solutions to the coupled Radhakrishnan-Kundu-Lakshmanan (RKL) model are greatly important for birefringent fibers without the effect of four-wave mixing (4WM). A significant number of general and standard analytical soliton solutions to this model have been extracted using three powerful techniques, namely the generalized Kudryashov’s method, the extended tanh method, and theG′/G-expansion method in this article. The schematic profiles of the solitons are sketched using the symbolic mathematical program Mathematica and are presented in two and three dimensions. The reported solutions might be helpful in explaining the RKL equation’s physical significance as well as some other related nonlinear phenomena that appear in engineering and nonlinear sciences.


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## Body

## 1. Introduction

The theory and investigation of soliton solutions is one of the important research fields relating to nonlinear partial differential equations ascending in telecommunication engineering, optics, mathematical physics, and other domains of nonlinear sciences. Therefore, diverse academics and researchers developed a number of numerical and analytical techniques, namely, them+1/G′-expansion technique [1], the truncated M-fractional derivative scheme [2], the q-homotopy analysis technique [3], Atangana-Baleanu operator scheme [4], the improved Bernoulli subequation function process [5], the sine-Gordon expansion approach [6], the Haar wavelet technique [7], the biframelet systems process [8], the Lie symmetry technique [9], the generalized exponential rational function mode [10], the Painlevé analysis [11], the extended subequation method [12], the improved G′/G-expansion scheme [13], the Hirota simplified method [14], the one-dimensional subalgebra system [15], Painlevé analysis and multi-soliton solutions technique [16], the one-parameter Lie group of transformations approach [17], etc.Optical solitons are one of the kinds of solitary waves where the waves propagate without scattering across a vast distance. The optical solitons were discovered since 1971 when Zhakarov and Sabat investigated the nonlinear Schrödinger (NLS) equation using the inverse scattering method. Hasegawa and Tappert realized in 1973 that the same NLS equation governs pulse propagation inside optical fibers. Fiber-optic solitons are light pulses achieve stabilization in their shape when the balance between the fibers dispersion and its nonlinearity occur; this enables them to be convenient as signalling light pulses in optical data transmission. The Radhakrishnan-Kundu-Lakshmanan (RKL) model represents dispersive nonlinear waves in polarization-preserving fibers with the Kerr law(1)ipt+apxx+bp2p=iλp2px−iγpxxx,where px,t represent complex-valued wave profile, a is the coefficient of chromatic dispersion (CD), λ is the coefficient of the self-steepening considered for short pulses to eschew the formation of shock waves, γ is the coefficient of third order dispersion (3OD), and b is Kerr nonlinearity. The Radhakrishnan-Kundu-Lakshmanan equation is one of the well-known models to deal with dispersive optical solitons with Kerr law nonlinearity. Most of the optical fibers that have become reliable at present obey this law of nonlinearity. Also, this medium indicates itself as self-phase modulation, a self-encouraged phase, and frequency-shift of a pulse of light as it travels toward a fiber nonlinearity. Therefore, in this work, we study the couple Radhakrishnan-Kundu-Lakshmanan model for birefringent fibers without the effect of four wave mixing (4WM) named by the basic case of fiber nonlinearity through three efficient methods, notably the G′/G-expansion method, the generalized Kudryashov method, and the extended tanh method to find different types of soliton solutions. In the following, it is considered the couple Radhakrishnan-Kundu-Lakshmanan model [18, 19]:
(2)ipt+a1pxx+b1p2+c1q2p=iλ1p2px+γ1q2px−iβ1pxxx,iqt+a2qxx+b2q2+c2p2q=iλ2q2qx+γ2p2qx−iβ2qxxx.The complex valued wave potentialspx,t and qx,t present the wave profiles, bj accounts for self-phase modulation (SPM), cj is the cross-phase modulation terms, and λj and γj associated with self steepening terms where the impact of four-wave mixing is dismissed, for j=1,2. Some strategies for solving the RKL model are found in the literature, such as the extended rational sine-cosine and sinh-cosh techniques applied by Rehman and Ahmad [20], Bilal et al. [21] used the generalized exponential rational function method, the advanced generalized auxiliary equation method (NGAEM) was used by Abbagari et al. [22], Seadawy et al. applied the Fan-extended subequation (FESE) approach [23], the Riccati equation method, the Sine-Gordon equation method, the functional variable technique, the F-expansion principle, and the exp-expansion function are applied by Yildirim et al. [18], the extended auxiliary equation scheme and the unified Riccati equation approaches were implemented by Zayed et al. [24], and Yildirim et al. [19] used the trial equation method and the modified simple equation method.This work is arranged as, in Section1, we analyze briefly about the fundamental techniques. The G′/G-expansion method [25–27], the general form of Kudryashov method [28, 29], and the extended tanh method [30–34], the analysis of solutions is given in Section 3, in Section 4, we present some illustrative graphs, and a definitive conclusion is presented in Section 5.


## 2. Fundamental Techniques

### 2.1. TheG′/G-Expansion Method

We take into account the governing equation in the form(3)Fu,ux,ut,uxx,utt,uxt=0,where F is a polynomial u=ux,t and its partial derivatives. Applying the traveling wave transformation, Equation (3) can be converted to an ordinary differential equation:
(4)Hv,v′,v′′,v′,′′,v′′′′=0.The essential steps of theG′/G-expansion method are:Step 1.
Assume the solution of (4) as follows:
(5)vη=∑i=0NfiG′Gi,where G=Gη fulfill the linear ODE:
(6)G′′η+λG′η+μGη=0,
wherefii=0,1,2,⋯,N,fN≠0,λ and μ are constants to be determined.Step 2.
In (5), N is a positive integer to be calculated by balancing the highest order derivative term with the highest power nonlinear term in (4).Step 3.
We acquire three different cases of solutions of (5):Case 1.
Hyperbolic function solutions, whenλ2−4μ>0,(7)G′G=−λ2+12λ2−4μh1sinh1/2λ2−4μη+h2cosh1/2λ2−4μηh1cosh1/2λ2−4μη+h2sinh1/2λ2−4μη.Case 2.
Trigonometric function solutions, whenλ2−4μ<0,(8)G′G=−λ2+124μ−λ2−h1sin1/24μ−λ2η+h2cos1/24μ−λ2ηh1cos1/24μ−λ2η+h2sin1/24μ−λ2η.Case 3.
Rational function solutions, whenλ2−4μ=0,(9)G′G=−λ2+h2h1+h2η.Step 4.
Putting (5) into (4), and using (6), gathering all terms with the like power of G′/G with each other and set each coefficient equal to zero, we get a system of algebraic equations, which can be solved by the aid of Mathematica program.


### 2.2. The Generalized Kudryashov Method

In this section, we give a brush up of the steps of the generalized Kudryashov method.Step 1.
Suppose the exact solutions of (4) can be expressed as follows:
(10)vη=∑i=0NAiQηi,where Aii=0,1,2,⋯,N,AN≠0 are constants to be determined with Ai≠0. N will be determined by homogeneous balance principle.Step 2.
Qη has the following definition:
(11)Qη=δϕηδ+ρϕη−1,ρ≠0,where A0,A1,A2,⋯,AN,δ and ρ are arbitrary constants to be calculated. The function ϕη fulfill the next differential equation:
(12)dϕdη=blogmϕη−1ϕη.
The solution of (12) is given by
(13)ϕη=11+dmbη.Step 3.
Putting (10), (11), and (13) into (4), we acquire a polynomial of Qη. Gathering all terms with the like powers of Qη with each other’s and setting each coefficient to zero, we get a set of algebraic equations.Step 4.
Solving this system with the aid of Mathematica program, we attain the exact soliton solution of (4).


### 2.3. The Extended Tanh-Function Method

Step 1.
The modified extended tanh method present the solution of (4) by the following finite series:
(14)vη=A0+∑i=1NAiϕηi+Biϕη−i,where ϕ=ϕη satisfy the Riccati equation
(15)dϕdη=w1+ϕ2.
The solution of the Riccati Equation (15) has the following cases of solutions:
Ifω1<0, then
(16)ϕη=−−ω1tanh−ω1η,ϕη=−−ω1coth−ω1η.
Ifω1=0, then
(17)ϕη=−1η.
Ifω1>0, then
(18)ϕη=ω1tanω1η,ϕη=−ω1cotω1η.Step 2.
N can be determined by balancing the highest order derivative term with the highest power nonlinear term in (4).Step 3.
Substituting (14) and (15) into (4), then gather all coefficients of the same powers of ϕηi and put them equal to zero, we get a system of algebraic equations for ω1,A0,⋯,AN,B1,⋯,BN, solving this equations we get all constants.


## 2.1. TheG′/G-Expansion Method

We take into account the governing equation in the form(3)Fu,ux,ut,uxx,utt,uxt=0,where F is a polynomial u=ux,t and its partial derivatives. Applying the traveling wave transformation, Equation (3) can be converted to an ordinary differential equation:
(4)Hv,v′,v′′,v′,′′,v′′′′=0.The essential steps of theG′/G-expansion method are:Step 1.
Assume the solution of (4) as follows:
(5)vη=∑i=0NfiG′Gi,where G=Gη fulfill the linear ODE:
(6)G′′η+λG′η+μGη=0,
wherefii=0,1,2,⋯,N,fN≠0,λ and μ are constants to be determined.Step 2.
In (5), N is a positive integer to be calculated by balancing the highest order derivative term with the highest power nonlinear term in (4).Step 3.
We acquire three different cases of solutions of (5):Case 1.
Hyperbolic function solutions, whenλ2−4μ>0,(7)G′G=−λ2+12λ2−4μh1sinh1/2λ2−4μη+h2cosh1/2λ2−4μηh1cosh1/2λ2−4μη+h2sinh1/2λ2−4μη.Case 2.
Trigonometric function solutions, whenλ2−4μ<0,(8)G′G=−λ2+124μ−λ2−h1sin1/24μ−λ2η+h2cos1/24μ−λ2ηh1cos1/24μ−λ2η+h2sin1/24μ−λ2η.Case 3.
Rational function solutions, whenλ2−4μ=0,(9)G′G=−λ2+h2h1+h2η.Step 4.
Putting (5) into (4), and using (6), gathering all terms with the like power of G′/G with each other and set each coefficient equal to zero, we get a system of algebraic equations, which can be solved by the aid of Mathematica program.


## 2.2. The Generalized Kudryashov Method

In this section, we give a brush up of the steps of the generalized Kudryashov method.Step 1.
Suppose the exact solutions of (4) can be expressed as follows:
(10)vη=∑i=0NAiQηi,where Aii=0,1,2,⋯,N,AN≠0 are constants to be determined with Ai≠0. N will be determined by homogeneous balance principle.Step 2.
Qη has the following definition:
(11)Qη=δϕηδ+ρϕη−1,ρ≠0,where A0,A1,A2,⋯,AN,δ and ρ are arbitrary constants to be calculated. The function ϕη fulfill the next differential equation:
(12)dϕdη=blogmϕη−1ϕη.
The solution of (12) is given by
(13)ϕη=11+dmbη.Step 3.
Putting (10), (11), and (13) into (4), we acquire a polynomial of Qη. Gathering all terms with the like powers of Qη with each other’s and setting each coefficient to zero, we get a set of algebraic equations.Step 4.
Solving this system with the aid of Mathematica program, we attain the exact soliton solution of (4).


## 2.3. The Extended Tanh-Function Method

Step 1.
The modified extended tanh method present the solution of (4) by the following finite series:
(14)vη=A0+∑i=1NAiϕηi+Biϕη−i,where ϕ=ϕη satisfy the Riccati equation
(15)dϕdη=w1+ϕ2.
The solution of the Riccati Equation (15) has the following cases of solutions:
Ifω1<0, then
(16)ϕη=−−ω1tanh−ω1η,ϕη=−−ω1coth−ω1η.
Ifω1=0, then
(17)ϕη=−1η.
Ifω1>0, then
(18)ϕη=ω1tanω1η,ϕη=−ω1cotω1η.Step 2.
N can be determined by balancing the highest order derivative term with the highest power nonlinear term in (4).Step 3.
Substituting (14) and (15) into (4), then gather all coefficients of the same powers of ϕηi and put them equal to zero, we get a system of algebraic equations for ω1,A0,⋯,AN,B1,⋯,BN, solving this equations we get all constants.


## 3. Analysis of Solutions

We first propose the following assumptions in order to generate the traveling wave solutions to the RKLE (2):
(19)px,t=eiξv1η,qx,t=eiξv2η,where
(20)ξ=−kx+ωt,η=x−αt.where vjη is the amplitude (j=1,2), α gives the soliton velocity, k is the frequency of the soliton, and ω is the wave number of the soliton.Applying the assumptions (19) and (20) into (2), we acquire real and imaginary parts as:
(21)−ajk2vj−k3βjvj−ωvj+bjvj3+cjvjvj^2−kλjvj3−kγjvjvj^2+ajvj′′+3kβjvj′′=0,(22)−2ajkvj′−αvj′−3k2βjvj′−3λjvj2vj′−γjvj^2vj′−2γjvjvj^vj^′+βjvj′′′=0.For bothj=1,2 and j^=3−j with the balance principle vj=vj^, Equations (21) and (22) become:
(23)−ajk2vj−k3βjvj−ωvj+bjvj3+cjvj3−kλjvj3−kγjvj3+ajvj′′+3kβjvj′′=0,(24)−2ajkvj′−αvj′−3k2βjvj′−3λjvj2vj′−3γjvj2vj′+βjvj′′′=0.Integrating (24) with respect to η and putting the constant of the integration equal to zero, we obtain
(25)−2ajk+α+3k2βjvj−λj+γjvj3+βjvj′′=0.The functionvj satisfies both Equations (23) and (25) in the following restriction relation:
(26)aj+3kβjβj=−ajk2+k3βj+ω−2ajk+α+3k2βj=bj+cj−kλj−kγj−λj+γj,where
(27)bj=−cjβj+ajλj+2kβjλj+ajγj+2kβjγjβj,ω=2aj2k+ajα+8ajk2βj+3kαβj+8k3βj2βj.Applying the balance principle in (23) between vj′′ and vj3, we get N+2=3N⇒N=1.


### 3.1. Solutions through theG′/G-Expansion Method

From (5), the solution of (23) can be presented as:
(28)vjη=f0+f1G′G.Substituting (28) into (23), setting the coefficient of like power of G′/G equal to zero, we acquire the following system:
(29)−2aj2kf0−ajαf0−9ajk2βjf0−3kαβjf0−9k3βj2f0−ajλjf03−3kβjλjf03−ajγjf03−3kβjγjf03+ajβjλμf1+3kβj2λμf1=0,−2aj2kf1−ajαf1−9ajk2βjf1−3kαβjf1−9k3βj2f1+ajλ2f1+3kβj2λ2f1+2ajβjμf1+6kβj2μf1−3ajλjf02f1−9kβjλjf02f1−3ajγjf02f1−9kβjγjf02f1=0,3ajβjλf1+9kβj2λf1−3ajλjf0f12−9kβjλjf0f12−3ajγjf0f12−9kβjγjf0f12=0,2ajβjf1+6kβj2f1−ajλjf13−3kβjλjf13−ajγjf13−3kβjγjf13=0.Solving the upwards set of equations with the aid of Mathematica program, we obtain the following solutions. Substituting (7), (8), and (9) with the different classes of solutions into (28), the traveling wave solutions of (2) are listed in the following:Class 1:(30)f0=−λβj2λj+γj,f1=−2βjλj+γj,aj=−2α−6k2βj−βjλ2+4μβj4k.Hyperbolic function solutions, whenλ2−4μ>0,(31)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+12λ2−4μh1sinh1/2λ2−4μx−αt+h2cosh1/2λ2−4μx−αth1cosh1/2λ2−4μx−αt+h2sinh1/2λ2−4μx−αt.Trigonometric function solutions, whenλ2−4μ<0,(32)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+124μ−λ2−h1sin1/24μ−λ2x−αt+h2cos1/24μ−λ2x−αth1cos1/24μ−λ2x−αt+h2sin1/24μ−λ2x−αt.Rational function solutions, whenλ2−4μ=0,(33)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+h2h1+h2x−αt.Class 2:(34)f0=λβj2λj+γj,f1=2βjλj+γj,aj=−2α−6k2βj−βjλ2+4μβj4k,Hyperbolic function solutions, whenλ2−4μ>0,(35)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+12λ2−4μh1sinh1/2λ2−4μx−αt+h2cosh1/2λ2−4μx−αth1cosh1/2λ2−4μx−αt+h2sinh1/2λ2−4μx−αt.Trigonometric function solutions, whenλ2−4μ<0,(36)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+124μ−λ2−h1sin1/24μ−λ2x−αt+h2cos1/24μ−λ2x−αth1cos1/24μ−λ2x−αt+h2sin1/24μ−λ2x−αt.Rational function solutions, whenλ2−4μ=0,(37)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+h2h1+h2x−αt.


### 3.2. Solutions through the Generalized Kudryashov Method

From (10), the solution of (23) can be written in the form:
(38)vη=A0+A1Qη.Substituting (38) into (23), setting the coefficient of like power of Qη equal to zero, we acquire the following set of equations:
(39)2ajk−aj−3kβjA0+α−aj−3kβjA0+3k2βj−aj−3kβjA0+−aj−3kβjλj+γjA03=0,2ajk−aj−3kβjA1+α−aj−3kβjA1+3k2βj−aj−3kβjA1−b2βj−aj−3kβjlogm2A1+3−aj−3kβjλj+γjA02A1=0,3b2βj−aj−3kβjlogm2A1+3−aj−3kβjλj+γjA0A12=0,−2b2βj−aj−3kβjlogm2A1+−aj−3kβjλj+γjA13=0.Solving this system, we get the following solution classes:Class 1:(40)A0=−blogmβj2λj+γj,A1=blogm2βjλj+γj,aj=−2α−6k2βj−b2βjlogm24k.Substituting (40) in (38) with (11),(13), and (21), we obtain the following solution of (2):
(41)px,t=qx,t=ei−kx+ωt−blogmβj2λj+γj+blogm2βjλj+γj×δ1/1+dmbx−αtδ+ρ1/1+dmbx−αt−1.Class 2:(42)A0=blogmβj2λj+γj,A1=−blogm2βjλj+γj,aj=−2α−6k2βj−b2βjlogm24k.We obtain the following solution:(43)px,t=qx,t=ei−kx+ωtblogmβj2λj+γj−blogm2βjλj+γj×δ1/1+dmbx−αtδ+ρ1/1+dmbx−αt−1.


### 3.3. Solutions through the Extended Tanh-Function Method

From (15), the solutions of (24) are given in the form
(44)vjη=A0+A1ϕη+B1ϕη.Substituting from (44) into (24) and use (16), grouping the coefficients of the same ϕiη, we obtain the following system:
(45)−2aj2kA0−ajαA0−9ajk2βjA0−3kαβjA0−9k3βj2A0−ajλjA03−3kβjλjA03−ajγjA03−3kβjγjA03−6ajλjA0A1B1−18kβjλjA0A1B1−6ajγjA0A1B1−18kβjγjA0A1B1=0,2ajβjω12B1+6kβj2ω12B1−ajλjB13−3kβjλjB13−ajγjB13−3kβjγjB13=0,3ajλjA0B12+9kβjλjA0B12+3ajγjA0B12+9kβjγjA0B12=0,−2aj2kB1−ajαB1−9ajk2βjB1−3kαβjB1−9k3βj2B1+2ajβjω1B1+6kβj2ω1B1−3ajλjA02B1−9kβjλjA02B1−3ajγjA02B1−9kβjγjA02B1−3ajλjA1B12−9kβjλjA1B12−3ajγjA1B12−9kβjγjA1B12=0,−2aj2kA1−ajαA1−9ajk2βjA1−3kαβjA1−9k3βj2A1+2ajβjω1A1+6kβj2ω1A1−3ajλjA02A1−9kβjλjA02A1−3ajγjA02A1−9kβjγjA02A1−3ajλjA12B1−9kβjλjA12B1−3ajγjA12B1−9kβjγjA12B1=0,−3ajλjA0A12−9kβjλjA0A12−3ajγjA0A12−9kβjγjA0A12=0,2ajβjA1+6kβj2A1−ajλjA13−3kβjλjA13−ajγjA13−3kβjγjA13=0.Setting the coefficients equal to zero, we obtain the following results after solving the system:Class 1:(46)A1=±2βjλj+γj,B1=0,A0=0,aj=−α−3k2βj+2βjω12k.From (46) into (44) with (20) and (21), we get the following solutions:Ifω1<0, then
(47)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt,(48)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1tanh−ω1x−αt.Ifω1>0, then
(49)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt,(50)px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt.Class 2:(51)A1=±2βjλj+γj,B1=±2βjω1λj+γj,A0=0,aj=−α−3k2βj−4βjω12k.Ifω1<0, then
(52)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt∓2βjω1λj+γj−ω1tanh−ω1x−αt,px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1tanh−ω1x−αt∓2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(53)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt∓2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt±2βjω1λj+γjω1cotω1x−αt.Class 3:(54)A1=±2βjλj+γj,B1=∓2βjω1λj+γj,A0=0,aj=−α−3k2βj+8βjω12k.Ifω1<0, then
(55)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt±2βjω1λj+γj−ω1tanh−ω1x−αt,px,t=qx,t=ei−kx+ωt+θ1∓2βjλj+γj−ω1tanh−ω1x−αt±2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(56)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt±2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt∓2βjω1λj+γjω1cotω1x−αt.Class 4:(57)A1=0,B1=±2βjω1λj+γj,A0=0,aj=−α−3k2βj+2βjω12k.Ifω1<0, then
(58)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γj−ω1tanh−ω1x−αt,(59)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(60)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjω1λj+γjω1cotω1x−αt.


## 3.1. Solutions through theG′/G-Expansion Method

From (5), the solution of (23) can be presented as:
(28)vjη=f0+f1G′G.Substituting (28) into (23), setting the coefficient of like power of G′/G equal to zero, we acquire the following system:
(29)−2aj2kf0−ajαf0−9ajk2βjf0−3kαβjf0−9k3βj2f0−ajλjf03−3kβjλjf03−ajγjf03−3kβjγjf03+ajβjλμf1+3kβj2λμf1=0,−2aj2kf1−ajαf1−9ajk2βjf1−3kαβjf1−9k3βj2f1+ajλ2f1+3kβj2λ2f1+2ajβjμf1+6kβj2μf1−3ajλjf02f1−9kβjλjf02f1−3ajγjf02f1−9kβjγjf02f1=0,3ajβjλf1+9kβj2λf1−3ajλjf0f12−9kβjλjf0f12−3ajγjf0f12−9kβjγjf0f12=0,2ajβjf1+6kβj2f1−ajλjf13−3kβjλjf13−ajγjf13−3kβjγjf13=0.Solving the upwards set of equations with the aid of Mathematica program, we obtain the following solutions. Substituting (7), (8), and (9) with the different classes of solutions into (28), the traveling wave solutions of (2) are listed in the following:Class 1:(30)f0=−λβj2λj+γj,f1=−2βjλj+γj,aj=−2α−6k2βj−βjλ2+4μβj4k.Hyperbolic function solutions, whenλ2−4μ>0,(31)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+12λ2−4μh1sinh1/2λ2−4μx−αt+h2cosh1/2λ2−4μx−αth1cosh1/2λ2−4μx−αt+h2sinh1/2λ2−4μx−αt.Trigonometric function solutions, whenλ2−4μ<0,(32)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+124μ−λ2−h1sin1/24μ−λ2x−αt+h2cos1/24μ−λ2x−αth1cos1/24μ−λ2x−αt+h2sin1/24μ−λ2x−αt.Rational function solutions, whenλ2−4μ=0,(33)px,t=qx,t=ei−kx+ωt−λβj2λj+γj−2βjλj+γj−λ2+h2h1+h2x−αt.Class 2:(34)f0=λβj2λj+γj,f1=2βjλj+γj,aj=−2α−6k2βj−βjλ2+4μβj4k,Hyperbolic function solutions, whenλ2−4μ>0,(35)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+12λ2−4μh1sinh1/2λ2−4μx−αt+h2cosh1/2λ2−4μx−αth1cosh1/2λ2−4μx−αt+h2sinh1/2λ2−4μx−αt.Trigonometric function solutions, whenλ2−4μ<0,(36)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+124μ−λ2−h1sin1/24μ−λ2x−αt+h2cos1/24μ−λ2x−αth1cos1/24μ−λ2x−αt+h2sin1/24μ−λ2x−αt.Rational function solutions, whenλ2−4μ=0,(37)px,t=qx,t=ei−kx+ωtλβj2λj+γj+2βjλj+γj−λ2+h2h1+h2x−αt.


## 3.2. Solutions through the Generalized Kudryashov Method

From (10), the solution of (23) can be written in the form:
(38)vη=A0+A1Qη.Substituting (38) into (23), setting the coefficient of like power of Qη equal to zero, we acquire the following set of equations:
(39)2ajk−aj−3kβjA0+α−aj−3kβjA0+3k2βj−aj−3kβjA0+−aj−3kβjλj+γjA03=0,2ajk−aj−3kβjA1+α−aj−3kβjA1+3k2βj−aj−3kβjA1−b2βj−aj−3kβjlogm2A1+3−aj−3kβjλj+γjA02A1=0,3b2βj−aj−3kβjlogm2A1+3−aj−3kβjλj+γjA0A12=0,−2b2βj−aj−3kβjlogm2A1+−aj−3kβjλj+γjA13=0.Solving this system, we get the following solution classes:Class 1:(40)A0=−blogmβj2λj+γj,A1=blogm2βjλj+γj,aj=−2α−6k2βj−b2βjlogm24k.Substituting (40) in (38) with (11),(13), and (21), we obtain the following solution of (2):
(41)px,t=qx,t=ei−kx+ωt−blogmβj2λj+γj+blogm2βjλj+γj×δ1/1+dmbx−αtδ+ρ1/1+dmbx−αt−1.Class 2:(42)A0=blogmβj2λj+γj,A1=−blogm2βjλj+γj,aj=−2α−6k2βj−b2βjlogm24k.We obtain the following solution:(43)px,t=qx,t=ei−kx+ωtblogmβj2λj+γj−blogm2βjλj+γj×δ1/1+dmbx−αtδ+ρ1/1+dmbx−αt−1.


## 3.3. Solutions through the Extended Tanh-Function Method

From (15), the solutions of (24) are given in the form
(44)vjη=A0+A1ϕη+B1ϕη.Substituting from (44) into (24) and use (16), grouping the coefficients of the same ϕiη, we obtain the following system:
(45)−2aj2kA0−ajαA0−9ajk2βjA0−3kαβjA0−9k3βj2A0−ajλjA03−3kβjλjA03−ajγjA03−3kβjγjA03−6ajλjA0A1B1−18kβjλjA0A1B1−6ajγjA0A1B1−18kβjγjA0A1B1=0,2ajβjω12B1+6kβj2ω12B1−ajλjB13−3kβjλjB13−ajγjB13−3kβjγjB13=0,3ajλjA0B12+9kβjλjA0B12+3ajγjA0B12+9kβjγjA0B12=0,−2aj2kB1−ajαB1−9ajk2βjB1−3kαβjB1−9k3βj2B1+2ajβjω1B1+6kβj2ω1B1−3ajλjA02B1−9kβjλjA02B1−3ajγjA02B1−9kβjγjA02B1−3ajλjA1B12−9kβjλjA1B12−3ajγjA1B12−9kβjγjA1B12=0,−2aj2kA1−ajαA1−9ajk2βjA1−3kαβjA1−9k3βj2A1+2ajβjω1A1+6kβj2ω1A1−3ajλjA02A1−9kβjλjA02A1−3ajγjA02A1−9kβjγjA02A1−3ajλjA12B1−9kβjλjA12B1−3ajγjA12B1−9kβjγjA12B1=0,−3ajλjA0A12−9kβjλjA0A12−3ajγjA0A12−9kβjγjA0A12=0,2ajβjA1+6kβj2A1−ajλjA13−3kβjλjA13−ajγjA13−3kβjγjA13=0.Setting the coefficients equal to zero, we obtain the following results after solving the system:Class 1:(46)A1=±2βjλj+γj,B1=0,A0=0,aj=−α−3k2βj+2βjω12k.From (46) into (44) with (20) and (21), we get the following solutions:Ifω1<0, then
(47)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt,(48)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1tanh−ω1x−αt.Ifω1>0, then
(49)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt,(50)px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt.Class 2:(51)A1=±2βjλj+γj,B1=±2βjω1λj+γj,A0=0,aj=−α−3k2βj−4βjω12k.Ifω1<0, then
(52)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt∓2βjω1λj+γj−ω1tanh−ω1x−αt,px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1tanh−ω1x−αt∓2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(53)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt∓2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt±2βjω1λj+γjω1cotω1x−αt.Class 3:(54)A1=±2βjλj+γj,B1=∓2βjω1λj+γj,A0=0,aj=−α−3k2βj+8βjω12k.Ifω1<0, then
(55)px,t=qx,t=ei−kx+ωt∓2βjλj+γj−ω1coth−ω1x−αt±2βjω1λj+γj−ω1tanh−ω1x−αt,px,t=qx,t=ei−kx+ωt+θ1∓2βjλj+γj−ω1tanh−ω1x−αt±2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(56)px,t=qx,t=ei−kx+ωt∓2βjλj+γjω1cotω1x−αt±2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjλj+γjω1tanω1x−αt∓2βjω1λj+γjω1cotω1x−αt.Class 4:(57)A1=0,B1=±2βjω1λj+γj,A0=0,aj=−α−3k2βj+2βjω12k.Ifω1<0, then
(58)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γj−ω1tanh−ω1x−αt,(59)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γj−ω1coth−ω1x−αt.Ifω1>0, then
(60)px,t=qx,t=ei−kx+ωt∓2βjω1λj+γjω1tanω1x−αt,px,t=qx,t=ei−kx+ωt±2βjω1λj+γjω1cotω1x−αt.


## 4. Graphical Illustrations

Herein, we present some figures in the two-dimensional, and three-dimensional to clarify the solutions that we presented. Some of the analytical solutions are presented in Figures1–8. In Figure 1, we introduce the graph of (31) using the G′/G-expansion method at λj=0.2,βj=0.7,γj=0.2,k=0.4,α=0.7,λ=1,μ=0.1,h1=0.5,h2=0.3.. Also, the graph of (32) using the G′/G-expansion method at θ1=0,θ1=0,λj=0.2,βj=0.7,γj=0.2,k=0.4,α=0.7,λ=0.1,μ=0.1,h1=0.6,h2=0.7, is presented in Figure 2. Moreover, we present the graph of (41) using the new Kudryashov’s method at λj=0.1,βj=0.1,γj=0.1,k=0.6,A0=0.1,m=0.01,d=0.5,ρ=0.4,δ=2,α=0.4,b=0.1 in Figure 3. In Figure 4, we introduce the graph of (50) using the extended tanh-function method at λj=0.4,βj=0.3,γj=0.4,k=0.2,α=0.3,ω1=0.001. The graph of (47) using the extended tanh-function method at λj=0.1,βj=0.1,γj=0.1,k=0.5,α=0.4,ω1=−0.1 is presented in Figure 5. Also, the graph of (58) using the extended tanh-function method at λj=0.1,βj=0.1,γj=0.1,k=0.5,α=0.4,ω1=−0.1 is presented in Figure 6. In Figure 7, we introduce the graph of (58) using the extended tanh-function method at λj=0.1,βj=0.1,γj=0.1,k=0.5,α=0.4,ω1=−0.1. Finally, in Figure 8, we introduce the graph of (47) using the extended tanh-function method at λj=0.4,βj=0.3,γj=0.4,k=0.2,α=0.3,ω1=0.001.Figure 1
Graph of (31) using the G′/G-expansion method at λj=0.2,βj=0.7,γj=0.2,k=0.4,α=0.7,λ=1,μ=0.1,h1=0.5,h2=0.3.Figure 2
Graph of (32) using the G′/G-expansion method at λj=0.2,βj=0.7,γj=0.2,k=0.4,α=0.7,λ=0.1,μ=0.1,h1=0.6,h2=0.7.Figure 3
Graph of (41) using the new Kudryashov’s method at λj=0.1,βj=0.1,γj=0.1,k=0.6,A0=0.1,m=0.01,d=0.5,ρ=0.4,δ=2,α=0.4,b=0.1.Figure 4
Graph of (42) using the new Kudryashov’s method at λj=0.1,βj=0.1,γj=0.1,k=0.6,A0=0.1,m=0.01,d=0.5,ρ=0.4,δ=1.2,α=0.4,b=0.1.Figure 5
Graph of (47) using the extended tanh-function method at λj=0.1,βj=0.1,γj=0.1,k=0.5,α=0.4,ω1=−0.1.Figure 6
Graph of (50) using the extended tanh-function method at λj=0.4,βj=0.3,γj=0.4,k=0.2,α=0.3,ω1=0.001.Figure 7
Graph of (58) using the extended tanh-function method at λj=0.1,βj=0.1,γj=0.1,k=0.5,α=0.4,ω1=−0.1.Figure 8
Graph of (47) using the extended tanh-function method at λj=0.4,βj=0.3,γj=0.4,k=0.2,α=0.3,ω1=0.001.


## 5. Discussion

Graphs are effective visual tools because they present information quickly. In the previous section, we depict some of the obtained solutions with several values of the parameters. The 2D and 3D profile of the solution (31) are presented in Figure 1 exhibit the transfer of the wave to the right with time progress. In Figure 2, the soliton solution (32) present bell shape soliton, and we notice the movement of the wave towards right. As displayed in Figures 3 and 4, the waves travels to the right as time increasing. As we see in Figures 5 and 7, as time proceed, the wave also proceed towards right. Finally, the wave moves to the right in Figures 6 and 8 with progress of time.


## 6. Conclusion

Soliton radiation presented by the Radhakrishnan-Kundu-Lakshmanan equation, which is known also by dispersive optical solitons. The results are fascinating, and highly influential in the field of optical fiber and useful for improving the performance capacity of transmission networks in the telecommunications industry. In this article, we have efficaciously ascertained diverse soliton solutions to the Radhakrishnan-Kundu-Lakshmanan equation by implementing three dynamic techniques, for instance, theG′/G-expansion method, the generalized Kudryashov method, and the extended tanh method. We attain scores of solutions in different shapes, including rational, hyperbolic, and trigonometric functions. The attained solutions are inclusive and for definite values of the constrains well-known optical solitons are produced. To illustrate the context, the results have also been explained with the help of suitable 2D and 3D graphs. The gained solutions may be applied to explain the model simply and straightforwardly.



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