# Novel Optical Soliton Waves in Metamaterials with Parabolic Law of Nonlinearity via the IEFM and ISEM

Authors: Xiaoyan Li; Jalil Manafian; Mostafa Abotaleb; Onur Alp Ilhan; Atheer Y. Oudah; A. S. Prakaash
Journal: Journal of Function Spaces
 (2022)

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## Abstract

Here, the miscellaneous soliton solutions of the generalized nonlinear Schrödinger equation are considered that describe the model of few-cycle pulse propagation in metamaterials with parabolic law of nonlinearity. The novel analytical wave solutions to the mentioned nonlinear equation in the sense of the nonlinear ordinary differential transform equation are obtained. The techniques are the improvedexp−Γϖ function method and the improved simple equation method. The nonlinear ordinary transform to concern the generalized Schrodinger equation to convert it for a solvable integer-order differential equation is used. After the successful implementation of the presented methods, the exact solitary wave solutions in the form of trigonometric, rational, and hyperbolic functions are obtained. Hence, the presented methods are relatable and efficient to solve nonlinear problems in mathematical physics.


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## Body

## 1. Introduction

In nonlinear engineering, physics, and mathematics, solving the nonlinear models would still be a useful approach for comprehending the complicated systems. The usefulness of explicit traveling wave solutions for nonlinear partial differential equations (NLPDEs) is noteworthy in the current situation [1]. NLPDEs have recently become quite popular in various topics such as optical fiber, fractional dynamics, fluid mechanics, control theory, chemical kinematics, signal transmission control theory, plasma physics, earthquakes, relativistic, solid-state physics, chemical physics, geochemistry, ecosystem, biomechanics, biophysics, gas dynamics, and so forth [2–10].Furthermore, the scientific community has taken an interest in extraction for dynamic wave propagation recognized by many other types of nonlinear evolution equations (NLEEs). It is operating several sorts of population models in atmospheric engineering and is also used to represent wave propagation and heat movement in physics [11, 12].Many academics have focused on finding the optimum nonlinear differential equation solution. In the last decade, the development of the computer algebraic systems such as Mathematica and Maple has been utilized for obtaining the exact solutions to the NLPDEs that help a lot to identify solutions to the traveling wave of NLPDEs, and various strategies have been created and established [13].There are numerous effective approaches available, notably the extended tanh-coth method and extended rational sinh-cosh method [14], the improved tan(ϕ/2) method [15], the Riccati equation expansion approach [16], the csch-function method [17], the multiple exp-function scheme [18], the exp-function method [19], the machine learning method [20], the bilinear transformation [21], the molecular dynamic simulation method [22], the generalized Darboux transformation scheme [23], the binary Darboux transformation method [24], a partial least square structural equation modeling [25], the modified simple equation (MSE) method [26], the modified Kudryashov method [27], the Jacobi elliptic function method [28], a CFD-based simulation approach by using nanofluids [29], the modified auxiliary expansion method [30], the Lie symmetry analysis approach [31], the sine-Gordon expansion method [32], the direct algebraic method [33], understanding by a design model as a useful tool for a meaningful and permanent learning [34], the generalized exp(−ϕξ)-expansion method [35, 36], and residual power series method [37] which have been created and effectively developed in order to attain the exact solution to NLPDEs. Mixed Morrey spaces and their applications were studied for partial differential equations of parabolic type [38] and boundedness of fractional integral operator Morrey-type and anisotropic spaces to the KdV equation [39] by Benia and Scapellato.The nonlinear dynamics that describes the propagation of pulses in optical metamaterials is modeled by the nonlinear Schrödinger equation [40] which is shown as follows:
(1)iut+auxx+b1u−4+b2u2+b3u4u=iαux+βu2ux+νu2xu+θ1u2uxx+θ2u2uxx+θ3u2uxx∗,and has been noted by the numerous researchers of authors. Zhou et al. investigated the optical metamaterials with parabolic law nonlinearity and spatiotemporal dispersion [41]. Ekici and coauthors determined the exact solutions for the nonlinear negative-indexed materials with anticubic nonlinearity [42]. Also, Foroutan and coworkers probed soliton perturbation in optical metamaterials, with anticubic nonlinearity, by implementing the improved tanϕ/2-expansion method [43]. Take the optical metamaterials modeled by the nonlinear Schrödinger equation with power law nonlinearity [44] which reads as
(2)iut+auxx+bu2+σ1u4u=iα1ux+λ1u2ux+νu2xu+θ1u2uxx+θ2u2uxx+θ3u2uxx∗.In [45], Li and coauthors found new types of solitary wave solutions for the higher-order nonlinear Schrödinger equation describing propagation of femtosecond light pulses in an optical fiber. Biswas et al. [46] probed the bright and dark soliton solutions for equation (2). Moreover, Yakadaa and coworkers [47] implemented the new extended direct algebraic method to determine more general and new soliton solutions for equation (2). In equation (2), u=ux,t and a shows the group velocity dispersion, b stands for the self-modulation term, and σ1 is the coefficient of quintic nonlinearity. And constant λ1 shows the self-steepening effect term, α1 shows the intermodal dispersion, ν is the nonlinear dispersion, and the constants θk for k=1,2,3 are the perturbation terms that appear in the context of metamaterials [43, 46]. The authors of [48–50] have worked on the Kundu-Eckhaus, Vakhnenko-Parkes, and unsteady Korteweg-de Vries equations, respectively, and obtained the exact solutions.The main goal of this research is to find a well-formed exact solution for the NLPDEs that are broad-ranging and well-known, such as soliton, kinks, bells, and other sorts of solutions. The solutions of the generalized nonlinear Schrödinger equation were solved using the improvedexp−Γϖ function method and improved simple equation method. So, solving this proposed equation utilizing the mentioned methods with the nonlinear ordinary differential transform equation is completely new. These methods are utilized in this article to obtain some more recent and broad results that are simple to implement, as well as versatile, flexible, configurable, extensible, and faster to simulate. On the other hand, we introduced three new graphical styles, lislpointplot3D, density, and vector plot, which also have an important impact on this type of research. Additionally, we visualize a three-dimensional plotline and density and two-dimensional plots. Those eight types of graphical description help us to describe the physical sketch a lot more explicitly.The remainder of the article will be organized in the following manner: the governing equation and the base configuration are introduced in Section2; the execution of the methodology about the improved exp−Γϖ function method has been depicted in Section 3. In Section 4, we used the second mentioned method to construct the specific solution for the proposed equation. Section 5 contains a short discussion, as well as a graphical description and classification, which are part of the visual explanation. The final thoughts are presented in the last section.


## 2. The Governing Equation

Pulse propagation in a one-core fiber has distinction from continuous wave propagation. In a conventional one-core fiber, pulse propagation has been investigated extensively by solving the model equation. To determine soliton waves of equation (2), we commence the following wave transformations [46, 51] as
(3)ux,t=Vξx,texpiϕx,t,where
(4)ξx,t=x+νt.Here,Vξx,t is an amplitude component of the wave profile, while ϕx,t is a phase component of the profile where it is given as
(5)ϕx,t=−kx+ωt+ϕ0.The parametersk, ω, and ϕ0 are the wave number, frequency, and phase constant, respectively, while ν represents the velocity of soliton. Substitute equation (3) and its derivatives into equation (2), and decompose it into real and imaginary parts. The real part and imaginary part equation for the component is given below, for
(6)ω+α1k+ak2V−aV″+kλ1−kθ1+θ2+θ3−bV3−bσ1V5−6θ1VV′2−3θ1+θ2+θ3V2V″=0,(7)ν−α1+2akV=3λ1+2ν−6kθ1−2kθ2+2kθ3V2V′.Then, we obtain(8)ν=α1+2ak.


## 3. Description of the First Proposed Method

In this section, we seek to utilize a suitable traveling wave transformation for the NLPDEs to translate it into nonlinear-order ordinary differential equations. Furthermore, we study the special ODE and then determine the obtained solutions. Therefore, consider the improvedexp−Γϖ function method. Let the NLPDEs
(9)SU,DxU,DtU,DxxU,DttU,⋯=0.Let(10)U=Uϖ,ϖ=x+νt.Substitute equation (10) into equation (9):
(11)WU,U′,U″,U′″,⋯=0.Let the solution of equation (11) be
(12)Uϖ=∑i=0P1Hie−Γϖi∑i=0P2Nie−Γϖi.LetΓ satisfy
(13)Γ′=ρ1eΓϖ+e−Γϖ+ρ2.Put equation (12) with equation (11) into equation (14). Then, the solutions are presented asProduct 1: withρ1≠0 and ρ22−4ρ1>0, then we have
(14)Γϖ=ln−ρ22−4ρ12ρ1tanhρ22−4ρ12ϖ+ϖ0−ρ22ρ1,where ϖ0 is the integral constant.Product 2: withρ1≠0 and ρ22−4ρ1<0, then we have
(15)Γϖ=ln−ρ22+4ρ12ρ1tan−ρ22+4ρ12ϖ+ϖ0−ρ22ρ1.Product 3: withρ1=0, ρ2≠0, and λ2−4μ>0, then we get
(16)Γϖ=−lnρ2expρ2ϖ+ϖ0−1.Product 4: withρ1≠0, ρ2≠0, and ρ22−4ρ1=0, then we get
(17)Γϖ=ln−2λϖ+ϖ0+4ρ22ϖ+ϖ0.Product 5: withρ1=0, ρ2=0, and ρ22−4ρ1=0, then we get
(18)Ωη=lnϖ+ϖ0,where Hi0≤i≤P1, Ni0≤i≤P2, and ρ1 and ρ2 are also the values to be explored later.ConsiderG=Gϖ=e−Γϖ, and we have
(19)Vϖ≃δFN−M,V′ϖ≃δGP1−P2−1G′=−δρ1GP1−P2−1+ρ2GP1−P2+GP1−P2+1≃−δGP1−P2+1,V″ϖ≃δGP1−P2−1,Vϖ4≃δ4G5P1−5P2,where δ=HP1/NP2. Balancing V2V″ with V5 in equation (6) concludes
(20)G3P1−3P2+2≃Vϖ2V″ϖ=Vϖ5≃G5P1−5P2.We can determine values ofN and M as follows:
(21)3P1−3P2+2=5P1−5P2⟹P1=P2+1.


### 3.1. Case I:P1=1,P2=0

The IEFM allows us to recruit the substitutions:(22)vϖ=H0+H1GϖN0.Plugging (35) along with (12) into equation (6), we get
(23)−bH15σ1−2H13N02Θ2−6H13N02θ1=0−5bH0H14σ1−3H13N02Θ2ρ2−12H13N02ρ2θ1−4H0H12N02Θ2−6H0H12N02θ1=0−k2H13N02Θ1−H13N02Θ2ρ22−6H13N02ρ22θ1−10bH02H13σ1+kH13N02λ1−6H0H12N02Θ2ρ2−12H0H12N02ρ2θ1−2H13N02Θ2ρ1−12H13N02ρ1θ1−2aH1N04−bH13N02−2H02H1N02Θ2=0−3k2H0H12N02Θ1−2H0H12N02Θ2ρ22−6H0H12N02ρ22θ1−H13N02Θ2ρ1ρ2−12H13N02ρ1ρ2θ1−3aH1N04ρ2−10bH03H12σ1+3kH0H12N02λ1−3H02H1N02Θ2ρ2−4H0H12N02Θ2ρ1−12H0H12N02ρ1θ1−3bH0H12N02=0ak2H1N04−3k2H02H1N02Θ1−H02H1N02Θ2ρ22−2H0H12N02Θ2ρ1ρ2−12H0H12N02ρ1ρ2θ1−6H13N02ρ12θ1+−aH1N04ρ22−2aH1N04ρ1−5bH04H1σ1+3kH02H1N02λ1+kH1N04α1−2H02H1N02Θ2ρ1−3bH02H1N02+ωH1N04=0ak2H0N04−aH1N04ρ1ρ2−k2H03N02Θ1−H02H1N02Θ2ρ1ρ2−6H0H12N02ρ12θ1−bH05σ1+kH03N02λ1+kH0N04α1−bH03N02+ωH0N04=0..Solving the above nonlinear system, we get the following.Product I:(24)ρ1=12N02−k2Θ1Θ2−3k2Θ1θ1+abσ1+3kλ1θ1−3bθ1−Θ2bH02σ1−kN02λ1+bN02−6bH02σ1θ1N02Θ2+3θ1Θ2+6θ1,k=k,H0=H0,H1=−2bσ1Θ2+3θ1N0bσ1,N0=N0,ρ2=2H0H1,ρ22−4ρ1=−2−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1Θ2+3θ1Θ2+6θ1,ω=Λ1+Λ2+Λ3Θ2+6θ12bσ1,Λ1=−3k4θ1Θ12Θ2−9k4θ12Θ12−k2Θ1Θ2abσ1−6kλ1θ1+6bθ1−bkΘ22σ1ak+α1−18k2Θ1θ12−kλ1+b,Λ2=−Θ212abk2σ1θ1−abkλ1σ1+12bkα1σ1θ1+3k2λ12θ1+ab2σ1−6bkλ1θ1+3b2θ1,Λ3=−36abk2σ1θ12+a2b2σ12−36bkα1σ1θ12−9k2λ12θ12+18bkλ1θ12−9b2θ12.Through (35) and (36), then we find the below cases:Product IA:By using (14), we have
(25)qx,t=H0N0+−2bσ1Θ2+3θ1/bσ1ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1ei−kx+ωt,where H0,N0,ϖ0 are free amounts. This solution is valid for the following case:
(26)−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ19θ1+θ2+θ36θ1+θ2+θ3>0.Product IB:By utilizing (15), we have
(27)q2x,t=H0N0−−2bσ1Θ2+3θ1/bσ1−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1ei−kx+ωt,where H0,N0,ϖ0 are free amounts. This solution is valid for the following case:
(28)−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ19θ1+θ2+θ36θ1+θ2+θ3<0.Product IC:By using (16), we get
(29)q3x,t=H0N0+−2bσ1Θ2+3θ1/bσ1ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx+ωt,in which
(30)H0=N0−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1bσ1Θ2+6θ1−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1,where H0,ϖ0 are free values. This solution is valid for the following case:
(31)9θ1+θ2+θ3>0,6θ1+θ2+θ3>0,H0−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1>0.Product ID:By utilizing (17), we conclude
(32)q4x,t=H0N0−−2bσ1Θ2+3θ1/bσ1ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx+ωt,in which
(33)k=Θ2λ1+3λ1θ1+4abΘ1Θ2σ1+12abΘ1σ1θ1−4bΘ1Θ22−24bΘ1Θ2θ1−36bΘ1θ12+Θ22λ12+6Θ2λ12θ1+9λ12θ122Θ1Θ2+3θ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(34)ρ2,ρ1≠0,4abΘ1Θ2σ1+12abΘ1σ1θ1−4bΘ1Θ22−24bΘ1Θ2θ1−36bΘ1θ12+Θ22λ12+6Θ2λ12θ1+9λ12θ12>0.


### 3.2. Case II:P1=2,P2=1

By using (21) and P1=P2+1, we get P1=2,P2=1. Therefore, we get
(35)vϖ=H0+H1Fϖ+H2Fϖ2N0+N1Fϖ,where the values H0,H1,H2,N0,N1 will be found. To get to the exact solutions, put (35) along with (12) into equation (6), and equating all the coefficients of powers of Fkϖ,k=0:10 to be zero, one obtains a system of algebraic equations. Solving the related system concludes the following.Set I:(36)b=−2N12Θ2+3θ1H22σ1,H0=N0H1N1−H2N0N12,H1=H1,H2=H2,N0=N0,N1=N1,ρ2=2H1N1−H2N0H2N1,ρ1=−12σ1kH22kΘ1−λ1−2σ1Θ2+6θ1H1N1−H2N02+2N14aσ1−Θ2−3θ1H22N12σ1Θ2+6θ1,ρ22−4ρ1=22aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1H22N12σ1Θ2+6θ1,k=k,ω=Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22,Λ1=−3k2H24σ12θ1kΘ1−λ12+2ak2H22N12σ12Θ1Θ2+Θ22+12Θ2θ1+36θ12,Λ2=−2kH22N12σ1aΘ2λ1σ1−6kΘ1Θ2θ1−18kΘ1θ12−Θ22α1σ1−12Θ2α1σ1θ1−36α1σ1θ12,Λ3=4N12Θ2+3θ1a2N12σ12−3kH22λ1σ1θ1−aN12Θ2σ1−3N12Θ2θ1−9N12θ12,where H0,N0, and N1 are arbitrary amounts. From the first set, (36) into (35), we have the following:Product IA:By utilizing (14), we gain
(37)q1x,t=N0H1N1−H2N0/N12−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+H2/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12N0−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where H0,H1,ϖ0 are free amounts. This solution is valid for the following case:
(38)2aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1σ19θ1+θ2+θ3>0.Product IB:By utilizing (15), we get
(39)q2x,t=N0H1N1−H2N0/N12+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+H2/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12N0+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where H0,H1,ϖ0 are free amounts. This solution is valid for the following case:
(40)2aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1σ19θ1+θ2+θ3<0.Product IC:By using (16), we gain
(41)q3x,t=N0H1N1−H2N0/N12+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+H2/ρ2/expρ2x+α1+2akt+ϖ0−12N0+N1/ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(42)k=λ1σ1H2+Π2Θ1σ1H2,Π=−8aN14Θ1σ12+8H12N12Θ1Θ2σ12+48H12N12Θ1σ12θ1−16H1H2N0N1Θ1Θ2σ12−96H1H2N0N1Θ1σ12θ1+8H22N02Θ1Θ2σ12+48H22N02Θ1σ12θ1+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22,where H0,ϖ0 are free arbitraries. This solution is valid for the following case:
(43)−8N14Θ1σ1aσ1−Θ2−3θ1+8Θ1σ12H1N1−H2N02Θ2+6θ1+λ12σ12H22>0.Product ID:By utilizing (17), we obtain
(44)q4x,t=N0H1N1−H2N0/N12+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+H2/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42N0+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(45)k=12λ1σ1H2+−8aN14Θ1σ12+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22Θ1σ1H2,where H0,ϖ0 are free arbitraries. This solution is valid for the following case:
(46)−8aN14Θ1σ12+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22>0.Insertingθ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3, the solution ux,t given by (37) shows the soliton wave solution. Moreover, the solution finds to the soliton solution as suggested in Figures 1(a)–1(c). Likewise, by taking θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10, the solution ux,t given by (38) denotes the periodic wave solution. Moreover, the solution transforms to the periodic solution as mentioned in Figures 2(a)–2(c). Also, by plugging θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10, the solution ux,t given by (38) shows the kink-singular wave solution. Moreover, the solution finds to the soliton solution as found in Figures 3(a)–3(c). Finally, by putting θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3, the solution ux,t given by (38) finds the cupson-singular wave solution. Moreover, the solution finds to the rational-singular solution as offered in Figures 4(a)–4(c).Figure 1
Graph of (37) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 2
Graph of (38) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 3
Graph of (39) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 4
Graph of (60) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Set II:(47)H0=−aN12H22Θ2−3θ1,H1=0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=0,N1=N1,ρ2=0,ρ1=−aN122Θ2−3θ1H22,ρ22−4ρ1=4aN122Θ2−3θ1H22,k=2H2Θ2λ1σ1−3H2λ1σ1θ1+4H22Θ22λ12σ12−12H22Θ2λ12σ12θ1+9H22λ12σ12θ12−8Λ1Θ1Θ2σ1+12Λ1Θ1σ1θ12Θ1σ12Θ2−3θ1,Λ1=8aN12Θ2σ1+18aN12σ1θ1−4N12Θ22−6N12Θ2θ1+18N12θ12,ω=−Λ2+Λ3H22Θ1σ12Θ2−3θ1,Λ2=2Θ2kH22Θ1α1σ1+2aN12Θ2−Θ1σ13kH22α1θ1+4a2N12,Λ3=−2aΘ2−kH22λ1σ1+4aN12σ1−3B12θ1−3aθ1kH22λ1σ1+6aN12σ1+6N12θ1,where H0, N0, and N1 are arbitrary amounts. Imposing the solution set (36) into (35), we get the following.Product IIA:By using (14), one gains
(48)q5x,t=−2bσ1Θ2+3θ1bσ1aN122Θ2−3θ1H22cothaN122Θ2−3θ1H22x+α1+2akt+ϖ0−aN1aN12/2Θ2−3θ1H22H22Θ2−3θ1tanhaN122Θ2−3θ1H22x+α1+2akt+ϖ0ei−kx−Λ2+Λ3H22Θ1σ12Θ2−3θ1t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(49)a3θ1+2θ2+2θ3>0.Product IIB:By utilizing (15), we get
(50)q6x,t=−2bσ1Θ2+3θ1bσ1−aN122Θ2−3θ1H22cot−aN122Θ2−3θ1H22x+α1+2akt+ϖ0−aN1−aN12/2Θ2−3θ1H22H22Θ2−3θ1tan−aN122Θ2−3θ1H22x+α1+2akt+ϖ0ei−kx−Λ2+Λ3H22Θ1σ12Θ2−3θ1t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(51)a3θ1+2θ2+2θ3<0.Set III:(52)H1=2−2σ1Θ2+6θ1k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−8A0H2Θ2σ1−48H0H2σ1θ1−2N12Θ2−6N12θ14σ1Θ2+6θ1,H0=H0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12A2,N1=N1,k=k,ρ2=H1H2,ρ1=−k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ14H22σ1Θ2+6θ1,ρ22−4ρ1=k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ12H22σ1Θ2+6θ1,ω=−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22,Λ1=−3k4H24σ12θ1Θ12+72ak2H22N12σ12θ12−3k2H24λ12σ12θ1+72kH22N12α1σ12θ12+12a2N14σ12θ1−36kH22N12λ1σ1θ12−108N14θ13,Λ2=2N12Θ212ak2H22σ12θ1−akH22λ1σ12+12kH22α1σ12θ1+2a2N12σ12−6kH22λ1σ1θ1−6aN12σ1θ1−36N12θ12,Λ3=6k2H22Θ1σ1θ1kH22λ1σ1+6N12θ1+2N12Θ22ak2H22σ12+kH22α1σ12−2aN12σ1−6N12θ1+2k2H22N12Θ1Θ2σ1aσ1+6θ1,where H0, N0, and N1 are free values. Imposing the solution set (36) into (35), one gains the following.Product IIIA:By using (14), we gain
(53)q7x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where N0,N1,ϖ0 are free amounts. This solution is valid for the following case:
(54)k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ1σ19θ1+θ2+θ3>0.Product IIIB:Via utilizing (15), we get
(55)q8x,t=H0+H1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(56)k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ1σ19θ1+θ2+θ3<0.Product IIIC:By the help of (16), we get
(57)q9x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx−Λ1+Λ2+Λ32B12σ12Θ2+6θ12H22t,in which
(58)k=H22λ1σ1+−H22σ18aN12Θ1σ1−16H0H2Θ1Θ2σ1−96H0H2Θ1σ1θ1−H22λ12σ1−8N12Θ1Θ2−24N12Θ1θ1H22Θ1σ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(59)H0N1−2bσ1Θ2+3θ1bσ1>0.Product IIID:Via using (17), one obtains
(60)q10x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(61)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24N12Θ1σ1θ12Θ1σ1H2,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(62)24N12Θ1σ1θ1−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1>0,bσ19θ1+θ2+θ3<0.Product IV:(63)H1=−σ1Θ2+6θ1k2H22Θ1σ1−kH22λ1σ1+2N12σ1a−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ1σ1Θ2+6θ1,H0=H0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=−k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ14H22σ1Θ2+6θ1,ρ22−4ρ1=H12−4H0H2H22,Λ1=ak2H22σ1+kH22α1σ1−2aN12,ω=−ak2H22σ1Θ1+Θ2Λ1+6θ1ak2H22σ1+kH22α1σ1−aN12+aσ1−kH22λ1+2aN12σ1H22Θ2+6θ1,where H0, N0, and N1 are free amounts. Imposing the solution set (36) into (35), we conclude the following.Product IVA:Via utilizing (14), one gets
(64)q11x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(65)H12−4H0H2>0.Product IVB:Via utilizing (15), one concludes
(66)q12x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free amounts. This solution is valid for the following case:
(67)H12−4H0H2<0.Product IVC:Via utilizing (16), we obtain
(68)q12x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2A2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(69)k=H22λ1σ1+−H22σ18aN12Θ1σ1−16H0H2Θ1Θ2σ1−96H0H2Θ1σ1θ1−H22λ12σ1−8N12Θ1Θ2−24N12Θ1θ1H22Θ1σ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(70)H0H1−2bσ1Θ2+3θ1bσ1>0.Product IVD:Via employing (17), one obtains
(71)q13x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/λ2x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(72)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24N12Θ1σ1θ12Θ1σ1H2,where H0,E are free amounts. This solution is valid for the following case:
(73)24N12Θ1σ1θ1−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1>0,bσ19θ1+θ2+θ3<0.Set V:(74)H1=−6kH22σ12Θ2−3θ1kΘ1−λ1+12N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12σ1Θ22−36θ126σ1Θ22−36θ12,H0=kH22σ14Θ2−33θ1kΘ1−λ1+2N122aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ1224H2σ1Θ22−36θ12,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=kH22σ14Θ2−33θ1kΘ1−λ1+2N122aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ1224Θ2+6θ1σ1H22Θ2−6θ1,ρ22−4ρ1=−k2H22Θ1σ1−kH22λ1σ1+2N12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1H22,ω=−kH22σ1akΘ1+akΘ2+6akθ1−aλ1+Θ2α1+6α1θ1+2aN12aσ1−Θ2−3θ1σ1H22Θ2+6θ1,where H0, N0, and N1 are free values. Imposing the solution set (36) into (35), we conclude the following.Product VA:Via using (14), one gains
(75)q14x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(76)k2H22Θ1σ1−kH22λ1σ1+2B12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1<0.Product VB:Via utilizing (15), we get
(77)q15x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(78)k2H22Θ1σ1−kH22λ1σ1+2B12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1>0.Product VC:Via employing (16), one obtains
(79)q16x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(80)k=4H22Θ2λ1σ1−33H22λ1σ1θ1−−H22σ14Θ2−33θ18N12Θ1Λ1−H22λ12σ14Θ2−33θ12H22Θ1σ14Θ2−33θ1,Λ1=2aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ12,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(81)a4Θ2−33θ1<0.Product VD:Via using (17), we conclude
(82)q17x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(83)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24B12Θ1σ1θ12Θ1σ1H2,where H0,ϖ0 are free values. This solution is valid for the following case:
(84)−8aN12Θ1σ12+H22λ12σ12+8N12Θ1Θ2σ1+24N12Θ1σ1θ1>0,σ1H23θ1+θ2+θ3≠0.Set VI:(85)H1=12−2σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12σ1Θ22−36θ12,H0=−189σ1kH22θ1kΘ1−λ1+2N122aΘ2σ1+21aσ1θ1−9Θ2θ1−27θ12σ1H2Θ22−36θ12,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=H0H2,ρ22−4ρ1=−4H0H2−H12H22,ω=−σ1kH22akΘ1+akΘ2+6akθ1−aλ1+Θ2α1+6α1θ1+2aN12aσ1−Θ2−3θ1σ1H22Θ2+6θ1,where N1 is a free value. Imposing the solution set (36) into (35), we conclude the following.Product VIA:Via using (14), one gains
(86)q18x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(87)σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12<0,4H0H2−H12<0.Product VIB:Via utilizing (15), we get
(88)q19x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(89)σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12<0,4H0H2−H12>0.Product VIC:Via employing (16), one obtains
(90)q20x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(91)k=163Θ2λ1θ1+9λ1θ12+4bΘ1Θ22θ1Δ1+1108bΘ1Θ2θ12Δ2+36bΘ1θ13Δ3+9λ12θ12Θ2+3θ12Θ1θ1Θ2+3θ1,Δ1=2aσ1−9θ1,Δ2=aσ1−2θ1,Δ3=7aσ1−9θ1,where N0,ϖ0 are free amounts. This solution is valid for the following case:
(92)4bΘ1Θ22θ1Δ1+1108bΘ1Θ2θ12Δ2+36bΘ1θ13Δ3+9λ12θ12Θ2+3θ12>0.Product VID:Via using (17), we conclude
(93)q21x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(94)k=12λ1σ1H2+λ12σ12H22−8N12Θ1σ1aσ1−Θ2−3θ1Θ1σ1H2,where N0,ϖ0 are free values. This solution is valid for the following case:
(95)λ12σ12H22−8N12Θ1σ1aσ1−Θ2−3θ1>0.Insertingθ1=−1,λ1=k=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,σ1=−3, the solution ux,t given by (86) shows the soliton wave solution. Moreover, the solution finds to the soliton solution as suggested in Figures 5(a)–5(c). Likewise, by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (87) denotes the periodic wave solution. Moreover, the solution transforms to the periodic solution as mentioned in Figures 6(a)–6(c). Also, by plugging θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (88) shows the kink-singular wave solution. Moreover, the solution finds to the soliton solution as found in Figures 7(a)–7(c). Finally, by putting θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (89) finds the cupson-singular wave solution. Moreover, the solution finds to the rational-singular solution as offered in Figures 8(a)–8(c).Figure 5
Graph of (86) by taking θ1=−1,λ1=k=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 6
Graph of (87) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 7
Graph of (88) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 8
Graph of (89) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.


## 3.1. Case I:P1=1,P2=0

The IEFM allows us to recruit the substitutions:(22)vϖ=H0+H1GϖN0.Plugging (35) along with (12) into equation (6), we get
(23)−bH15σ1−2H13N02Θ2−6H13N02θ1=0−5bH0H14σ1−3H13N02Θ2ρ2−12H13N02ρ2θ1−4H0H12N02Θ2−6H0H12N02θ1=0−k2H13N02Θ1−H13N02Θ2ρ22−6H13N02ρ22θ1−10bH02H13σ1+kH13N02λ1−6H0H12N02Θ2ρ2−12H0H12N02ρ2θ1−2H13N02Θ2ρ1−12H13N02ρ1θ1−2aH1N04−bH13N02−2H02H1N02Θ2=0−3k2H0H12N02Θ1−2H0H12N02Θ2ρ22−6H0H12N02ρ22θ1−H13N02Θ2ρ1ρ2−12H13N02ρ1ρ2θ1−3aH1N04ρ2−10bH03H12σ1+3kH0H12N02λ1−3H02H1N02Θ2ρ2−4H0H12N02Θ2ρ1−12H0H12N02ρ1θ1−3bH0H12N02=0ak2H1N04−3k2H02H1N02Θ1−H02H1N02Θ2ρ22−2H0H12N02Θ2ρ1ρ2−12H0H12N02ρ1ρ2θ1−6H13N02ρ12θ1+−aH1N04ρ22−2aH1N04ρ1−5bH04H1σ1+3kH02H1N02λ1+kH1N04α1−2H02H1N02Θ2ρ1−3bH02H1N02+ωH1N04=0ak2H0N04−aH1N04ρ1ρ2−k2H03N02Θ1−H02H1N02Θ2ρ1ρ2−6H0H12N02ρ12θ1−bH05σ1+kH03N02λ1+kH0N04α1−bH03N02+ωH0N04=0..Solving the above nonlinear system, we get the following.Product I:(24)ρ1=12N02−k2Θ1Θ2−3k2Θ1θ1+abσ1+3kλ1θ1−3bθ1−Θ2bH02σ1−kN02λ1+bN02−6bH02σ1θ1N02Θ2+3θ1Θ2+6θ1,k=k,H0=H0,H1=−2bσ1Θ2+3θ1N0bσ1,N0=N0,ρ2=2H0H1,ρ22−4ρ1=−2−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1Θ2+3θ1Θ2+6θ1,ω=Λ1+Λ2+Λ3Θ2+6θ12bσ1,Λ1=−3k4θ1Θ12Θ2−9k4θ12Θ12−k2Θ1Θ2abσ1−6kλ1θ1+6bθ1−bkΘ22σ1ak+α1−18k2Θ1θ12−kλ1+b,Λ2=−Θ212abk2σ1θ1−abkλ1σ1+12bkα1σ1θ1+3k2λ12θ1+ab2σ1−6bkλ1θ1+3b2θ1,Λ3=−36abk2σ1θ12+a2b2σ12−36bkα1σ1θ12−9k2λ12θ12+18bkλ1θ12−9b2θ12.Through (35) and (36), then we find the below cases:Product IA:By using (14), we have
(25)qx,t=H0N0+−2bσ1Θ2+3θ1/bσ1ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1ei−kx+ωt,where H0,N0,ϖ0 are free amounts. This solution is valid for the following case:
(26)−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ19θ1+θ2+θ36θ1+θ2+θ3>0.Product IB:By utilizing (15), we have
(27)q2x,t=H0N0−−2bσ1Θ2+3θ1/bσ1−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1ei−kx+ωt,where H0,N0,ϖ0 are free amounts. This solution is valid for the following case:
(28)−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ19θ1+θ2+θ36θ1+θ2+θ3<0.Product IC:By using (16), we get
(29)q3x,t=H0N0+−2bσ1Θ2+3θ1/bσ1ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx+ωt,in which
(30)H0=N0−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1bσ1Θ2+6θ1−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1,where H0,ϖ0 are free values. This solution is valid for the following case:
(31)9θ1+θ2+θ3>0,6θ1+θ2+θ3>0,H0−k2Θ1Θ2−3k2Θ1θ1+abσ1+kΘ2λ1+3kλ1θ1−bΘ2−3bθ1>0.Product ID:By utilizing (17), we conclude
(32)q4x,t=H0N0−−2bσ1Θ2+3θ1/bσ1ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx+ωt,in which
(33)k=Θ2λ1+3λ1θ1+4abΘ1Θ2σ1+12abΘ1σ1θ1−4bΘ1Θ22−24bΘ1Θ2θ1−36bΘ1θ12+Θ22λ12+6Θ2λ12θ1+9λ12θ122Θ1Θ2+3θ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(34)ρ2,ρ1≠0,4abΘ1Θ2σ1+12abΘ1σ1θ1−4bΘ1Θ22−24bΘ1Θ2θ1−36bΘ1θ12+Θ22λ12+6Θ2λ12θ1+9λ12θ12>0.


## 3.2. Case II:P1=2,P2=1

By using (21) and P1=P2+1, we get P1=2,P2=1. Therefore, we get
(35)vϖ=H0+H1Fϖ+H2Fϖ2N0+N1Fϖ,where the values H0,H1,H2,N0,N1 will be found. To get to the exact solutions, put (35) along with (12) into equation (6), and equating all the coefficients of powers of Fkϖ,k=0:10 to be zero, one obtains a system of algebraic equations. Solving the related system concludes the following.Set I:(36)b=−2N12Θ2+3θ1H22σ1,H0=N0H1N1−H2N0N12,H1=H1,H2=H2,N0=N0,N1=N1,ρ2=2H1N1−H2N0H2N1,ρ1=−12σ1kH22kΘ1−λ1−2σ1Θ2+6θ1H1N1−H2N02+2N14aσ1−Θ2−3θ1H22N12σ1Θ2+6θ1,ρ22−4ρ1=22aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1H22N12σ1Θ2+6θ1,k=k,ω=Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22,Λ1=−3k2H24σ12θ1kΘ1−λ12+2ak2H22N12σ12Θ1Θ2+Θ22+12Θ2θ1+36θ12,Λ2=−2kH22N12σ1aΘ2λ1σ1−6kΘ1Θ2θ1−18kΘ1θ12−Θ22α1σ1−12Θ2α1σ1θ1−36α1σ1θ12,Λ3=4N12Θ2+3θ1a2N12σ12−3kH22λ1σ1θ1−aN12Θ2σ1−3N12Θ2θ1−9N12θ12,where H0,N0, and N1 are arbitrary amounts. From the first set, (36) into (35), we have the following:Product IA:By utilizing (14), we gain
(37)q1x,t=N0H1N1−H2N0/N12−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+H2/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12N0−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where H0,H1,ϖ0 are free amounts. This solution is valid for the following case:
(38)2aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1σ19θ1+θ2+θ3>0.Product IB:By utilizing (15), we get
(39)q2x,t=N0H1N1−H2N0/N12+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+H2/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12N0+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where H0,H1,ϖ0 are free amounts. This solution is valid for the following case:
(40)2aN14σ1+k2H22Θ1σ1−kH22λ1σ1−2N14Θ2−6N14θ1σ19θ1+θ2+θ3<0.Product IC:By using (16), we gain
(41)q3x,t=N0H1N1−H2N0/N12+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+H2/ρ2/expρ2x+α1+2akt+ϖ0−12N0+N1/ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(42)k=λ1σ1H2+Π2Θ1σ1H2,Π=−8aN14Θ1σ12+8H12N12Θ1Θ2σ12+48H12N12Θ1σ12θ1−16H1H2N0N1Θ1Θ2σ12−96H1H2N0N1Θ1σ12θ1+8H22N02Θ1Θ2σ12+48H22N02Θ1σ12θ1+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22,where H0,ϖ0 are free arbitraries. This solution is valid for the following case:
(43)−8N14Θ1σ1aσ1−Θ2−3θ1+8Θ1σ12H1N1−H2N02Θ2+6θ1+λ12σ12H22>0.Product ID:By utilizing (17), we obtain
(44)q4x,t=N0H1N1−H2N0/N12+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+H2/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42N0+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx+Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(45)k=12λ1σ1H2+−8aN14Θ1σ12+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22Θ1σ1H2,where H0,ϖ0 are free arbitraries. This solution is valid for the following case:
(46)−8aN14Θ1σ12+8N14Θ1Θ2σ1+24N14Θ1σ1θ1+λ12σ12H22>0.Insertingθ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3, the solution ux,t given by (37) shows the soliton wave solution. Moreover, the solution finds to the soliton solution as suggested in Figures 1(a)–1(c). Likewise, by taking θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10, the solution ux,t given by (38) denotes the periodic wave solution. Moreover, the solution transforms to the periodic solution as mentioned in Figures 2(a)–2(c). Also, by plugging θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10, the solution ux,t given by (38) shows the kink-singular wave solution. Moreover, the solution finds to the soliton solution as found in Figures 3(a)–3(c). Finally, by putting θ1=θ2=θ3=N0=N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3, the solution ux,t given by (38) finds the cupson-singular wave solution. Moreover, the solution finds to the rational-singular solution as offered in Figures 4(a)–4(c).Figure 1
Graph of (37) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 2
Graph of (38) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 3
Graph of (39) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 4
Graph of (60) by taking θ1=1,θ2=1,θ3=1,N0=1,N1=1,H1=3,b=5,α1=1,a=2,k=3,b=1,σ1=−1,λ1=10: (a) 3D graph, (b) density plot, and (c) 2D plot.Set II:(47)H0=−aN12H22Θ2−3θ1,H1=0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=0,N1=N1,ρ2=0,ρ1=−aN122Θ2−3θ1H22,ρ22−4ρ1=4aN122Θ2−3θ1H22,k=2H2Θ2λ1σ1−3H2λ1σ1θ1+4H22Θ22λ12σ12−12H22Θ2λ12σ12θ1+9H22λ12σ12θ12−8Λ1Θ1Θ2σ1+12Λ1Θ1σ1θ12Θ1σ12Θ2−3θ1,Λ1=8aN12Θ2σ1+18aN12σ1θ1−4N12Θ22−6N12Θ2θ1+18N12θ12,ω=−Λ2+Λ3H22Θ1σ12Θ2−3θ1,Λ2=2Θ2kH22Θ1α1σ1+2aN12Θ2−Θ1σ13kH22α1θ1+4a2N12,Λ3=−2aΘ2−kH22λ1σ1+4aN12σ1−3B12θ1−3aθ1kH22λ1σ1+6aN12σ1+6N12θ1,where H0, N0, and N1 are arbitrary amounts. Imposing the solution set (36) into (35), we get the following.Product IIA:By using (14), one gains
(48)q5x,t=−2bσ1Θ2+3θ1bσ1aN122Θ2−3θ1H22cothaN122Θ2−3θ1H22x+α1+2akt+ϖ0−aN1aN12/2Θ2−3θ1H22H22Θ2−3θ1tanhaN122Θ2−3θ1H22x+α1+2akt+ϖ0ei−kx−Λ2+Λ3H22Θ1σ12Θ2−3θ1t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(49)a3θ1+2θ2+2θ3>0.Product IIB:By utilizing (15), we get
(50)q6x,t=−2bσ1Θ2+3θ1bσ1−aN122Θ2−3θ1H22cot−aN122Θ2−3θ1H22x+α1+2akt+ϖ0−aN1−aN12/2Θ2−3θ1H22H22Θ2−3θ1tan−aN122Θ2−3θ1H22x+α1+2akt+ϖ0ei−kx−Λ2+Λ3H22Θ1σ12Θ2−3θ1t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(51)a3θ1+2θ2+2θ3<0.Set III:(52)H1=2−2σ1Θ2+6θ1k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−8A0H2Θ2σ1−48H0H2σ1θ1−2N12Θ2−6N12θ14σ1Θ2+6θ1,H0=H0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12A2,N1=N1,k=k,ρ2=H1H2,ρ1=−k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ14H22σ1Θ2+6θ1,ρ22−4ρ1=k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ12H22σ1Θ2+6θ1,ω=−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22,Λ1=−3k4H24σ12θ1Θ12+72ak2H22N12σ12θ12−3k2H24λ12σ12θ1+72kH22N12α1σ12θ12+12a2N14σ12θ1−36kH22N12λ1σ1θ12−108N14θ13,Λ2=2N12Θ212ak2H22σ12θ1−akH22λ1σ12+12kH22α1σ12θ1+2a2N12σ12−6kH22λ1σ1θ1−6aN12σ1θ1−36N12θ12,Λ3=6k2H22Θ1σ1θ1kH22λ1σ1+6N12θ1+2N12Θ22ak2H22σ12+kH22α1σ12−2aN12σ1−6N12θ1+2k2H22N12Θ1Θ2σ1aσ1+6θ1,where H0, N0, and N1 are free values. Imposing the solution set (36) into (35), one gains the following.Product IIIA:By using (14), we gain
(53)q7x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where N0,N1,ϖ0 are free amounts. This solution is valid for the following case:
(54)k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ1σ19θ1+θ2+θ3>0.Product IIIB:Via utilizing (15), we get
(55)q8x,t=H0+H1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/ρ22−4ρ1/2ρ1tanρ22−4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(56)k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−2N12Θ2−6N12θ1σ19θ1+θ2+θ3<0.Product IIIC:By the help of (16), we get
(57)q9x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1ei−kx−Λ1+Λ2+Λ32B12σ12Θ2+6θ12H22t,in which
(58)k=H22λ1σ1+−H22σ18aN12Θ1σ1−16H0H2Θ1Θ2σ1−96H0H2Θ1σ1θ1−H22λ12σ1−8N12Θ1Θ2−24N12Θ1θ1H22Θ1σ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(59)H0N1−2bσ1Θ2+3θ1bσ1>0.Product IIID:Via using (17), one obtains
(60)q10x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4ei−kx−Λ1+Λ2+Λ32N12σ12Θ2+6θ12H22t,in which
(61)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24N12Θ1σ1θ12Θ1σ1H2,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(62)24N12Θ1σ1θ1−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1>0,bσ19θ1+θ2+θ3<0.Product IV:(63)H1=−σ1Θ2+6θ1k2H22Θ1σ1−kH22λ1σ1+2N12σ1a−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ1σ1Θ2+6θ1,H0=H0,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=−k2H22Θ1σ1−kH22λ1σ1+2aN12σ1−4H0H2Θ2σ1−24H0H2σ1θ1−2N12Θ2−6N12θ14H22σ1Θ2+6θ1,ρ22−4ρ1=H12−4H0H2H22,Λ1=ak2H22σ1+kH22α1σ1−2aN12,ω=−ak2H22σ1Θ1+Θ2Λ1+6θ1ak2H22σ1+kH22α1σ1−aN12+aσ1−kH22λ1+2aN12σ1H22Θ2+6θ1,where H0, N0, and N1 are free amounts. Imposing the solution set (36) into (35), we conclude the following.Product IVA:Via utilizing (14), one gets
(64)q11x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(65)H12−4H0H2>0.Product IVB:Via utilizing (15), one concludes
(66)q12x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free amounts. This solution is valid for the following case:
(67)H12−4H0H2<0.Product IVC:Via utilizing (16), we obtain
(68)q12x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2A2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(69)k=H22λ1σ1+−H22σ18aN12Θ1σ1−16H0H2Θ1Θ2σ1−96H0H2Θ1σ1θ1−H22λ12σ1−8N12Θ1Θ2−24N12Θ1θ1H22Θ1σ1,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(70)H0H1−2bσ1Θ2+3θ1bσ1>0.Product IVD:Via employing (17), one obtains
(71)q13x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/λ2x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(72)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24N12Θ1σ1θ12Θ1σ1H2,where H0,E are free amounts. This solution is valid for the following case:
(73)24N12Θ1σ1θ1−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1>0,bσ19θ1+θ2+θ3<0.Set V:(74)H1=−6kH22σ12Θ2−3θ1kΘ1−λ1+12N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12σ1Θ22−36θ126σ1Θ22−36θ12,H0=kH22σ14Θ2−33θ1kΘ1−λ1+2N122aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ1224H2σ1Θ22−36θ12,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=kH22σ14Θ2−33θ1kΘ1−λ1+2N122aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ1224Θ2+6θ1σ1H22Θ2−6θ1,ρ22−4ρ1=−k2H22Θ1σ1−kH22λ1σ1+2N12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1H22,ω=−kH22σ1akΘ1+akΘ2+6akθ1−aλ1+Θ2α1+6α1θ1+2aN12aσ1−Θ2−3θ1σ1H22Θ2+6θ1,where H0, N0, and N1 are free values. Imposing the solution set (36) into (35), we conclude the following.Product VA:Via using (14), one gains
(75)q14x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(76)k2H22Θ1σ1−kH22λ1σ1+2B12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1<0.Product VB:Via utilizing (15), we get
(77)q15x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(78)k2H22Θ1σ1−kH22λ1σ1+2B12σ1a−2N12Θ2−6N12θ1Θ2+6θ1σ1>0.Product VC:Via employing (16), one obtains
(79)q16x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(80)k=4H22Θ2λ1σ1−33H22λ1σ1θ1−−H22σ14Θ2−33θ18N12Θ1Λ1−H22λ12σ14Θ2−33θ12H22Θ1σ14Θ2−33θ1,Λ1=2aΘ2σ1−45aσ1θ1−4Θ22+21Θ2θ1+99θ12,where H0,ϖ0 are free amounts. This solution is valid for the following case:
(81)a4Θ2−33θ1<0.Product VD:Via using (17), we conclude
(82)q17x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(83)k=λ1σ1H2+−8aN12Θ1σ12+λ12σ12H22+8N12Θ1Θ2σ1+24B12Θ1σ1θ12Θ1σ1H2,where H0,ϖ0 are free values. This solution is valid for the following case:
(84)−8aN12Θ1σ12+H22λ12σ12+8N12Θ1Θ2σ1+24N12Θ1σ1θ1>0,σ1H23θ1+θ2+θ3≠0.Set VI:(85)H1=12−2σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12σ1Θ22−36θ12,H0=−189σ1kH22θ1kΘ1−λ1+2N122aΘ2σ1+21aσ1θ1−9Θ2θ1−27θ12σ1H2Θ22−36θ12,H2=−2bσ1Θ2+3θ1N1bσ1,N0=H1N12H2,N1=N1,k=k,ρ2=H1H2,ρ1=H0H2,ρ22−4ρ1=−4H0H2−H12H22,ω=−σ1kH22akΘ1+akΘ2+6akθ1−aλ1+Θ2α1+6α1θ1+2aN12aσ1−Θ2−3θ1σ1H22Θ2+6θ1,where N1 is a free value. Imposing the solution set (36) into (35), we conclude the following.Product VIA:Via using (14), one gains
(86)q18x,t=H0−H1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ12H1N1/2H2−N1/ρ22−4ρ1/2ρ1tanhρ22−4ρ1/2ϖ+ϖ0+ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free values. This solution is valid for the following case:
(87)σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12<0,4H0H2−H12<0.Product VIB:Via utilizing (15), we get
(88)q19x,t=H0+H1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1+−2bσ1Θ2+3θ1N1/bσ1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ12H1N1/2H2+N1/−ρ22+4ρ1/2ρ1tan−ρ22+4ρ1/2ϖ+ϖ0−ρ2/2ρ1×ei−kx+ωt,where N0,N1,ϖ0 are free arbitraries. This solution is valid for the following case:
(89)σ1Θ22−36θ12σ1kH222Θ2−3θ1kΘ1−λ1+2N124aΘ2σ1+9aσ1θ1−2Θ22−3Θ2θ1+9θ12<0,4H0H2−H12>0.Product VIC:Via employing (16), one obtains
(90)q20x,t=H0+H1/ρ2/expρ2x+α1+2akt+ϖ0−1+−2bσ1Θ2+3θ1N1/bσ1/ρ2/expρ2x+α1+2akt+ϖ0−12H1N1/2H2+N1/ρ2/expρ2x+α1+2akt+ϖ0−1×ei−kx+ωt,in which
(91)k=163Θ2λ1θ1+9λ1θ12+4bΘ1Θ22θ1Δ1+1108bΘ1Θ2θ12Δ2+36bΘ1θ13Δ3+9λ12θ12Θ2+3θ12Θ1θ1Θ2+3θ1,Δ1=2aσ1−9θ1,Δ2=aσ1−2θ1,Δ3=7aσ1−9θ1,where N0,ϖ0 are free amounts. This solution is valid for the following case:
(92)4bΘ1Θ22θ1Δ1+1108bΘ1Θ2θ12Δ2+36bΘ1θ13Δ3+9λ12θ12Θ2+3θ12>0.Product VID:Via using (17), we conclude
(93)q21x,t=H0+H1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4+−2bσ1Θ2+3θ1N1/bσ1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+42H1N1/2H2+N1/ρ22x+α1+2akt+ϖ0/2ρ2x+α1+2akt+ϖ0+4×ei−kx+ωt,in which
(94)k=12λ1σ1H2+λ12σ12H22−8N12Θ1σ1aσ1−Θ2−3θ1Θ1σ1H2,where N0,ϖ0 are free values. This solution is valid for the following case:
(95)λ12σ12H22−8N12Θ1σ1aσ1−Θ2−3θ1>0.Insertingθ1=−1,λ1=k=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,σ1=−3, the solution ux,t given by (86) shows the soliton wave solution. Moreover, the solution finds to the soliton solution as suggested in Figures 5(a)–5(c). Likewise, by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (87) denotes the periodic wave solution. Moreover, the solution transforms to the periodic solution as mentioned in Figures 6(a)–6(c). Also, by plugging θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (88) shows the kink-singular wave solution. Moreover, the solution finds to the soliton solution as found in Figures 7(a)–7(c). Finally, by putting θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3, the solution ux,t given by (89) finds the cupson-singular wave solution. Moreover, the solution finds to the rational-singular solution as offered in Figures 8(a)–8(c).Figure 5
Graph of (86) by taking θ1=−1,λ1=k=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 6
Graph of (87) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 7
Graph of (88) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.Figure 8
Graph of (89) by taking θ1=1,λ1=b=θ2=θ3=1,N1=2,b=5,α1=1,a=−2,k=3,σ1=−3: (a) 3D graph, (b) density plot, and (c) 2D plot.


## 4. Description of the Second Proposed Method

Take the improved simple equation method (ISEM). Let the solution of equation (11) be
(96)Uϖ=∑i=0P1HiΓϖi∑i=0P1NiΓϖi.LetΓ=Γϖ satisfy
(97)Γ′=ρ0+ρ1Γϖ+ρ2Γϖ2+ρ3Γϖ3.The solutions of equation (97) will be as follows:Product 1: withρ3=0, then we have
(98)Γϖ=12tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1ρ2,where ϖ0 is the integral constant.Product 2: withρ0=ρ3=0, then we have
(99)Γϖ=ρ1eρ1ϖ+ϖ0C0ρ1−ρ2eρ1ϖ+ϖ0,where C0 is the integral constant.Product 3: withρ0=ρ2=0, then we have
(100)Γϖ=C0e−2ρ1ϖ+ϖ0ρ1−ρ3ρ1C0e−2ρ1ϖ+ϖ0ρ1−ρ3.Product 4: withρ0=ρ1=ρ2=0, then we have
(101)Γϖ=1−2ϖ+ϖ0ρ3+C0.where Hi,Ni0≤i≤P1 are also the values to be explored later.


### 4.1. Application of ISEM on Equation (1)

The IEFM allows us to recruit the substitutions:(102)vϖ=H0+H1Γϖ+H2Γϖ2N0+N1Γϖ+N2Γϖ2.Plug (35) along with (97) into equation (6).With case (98), we have the following:Product IA:(103)b=−2aN12ρ2−N22ρ02+H22N12k2Θ1+2Θ2ρ0ρ2+12ρ0ρ2θ1−kλ1−2H22N22ρ02Θ2+6θ1H22N12,k=k,ω=−N12ak2−2aρ0ρ2+kα1+2ρ02aN22−3H22θ1N12,H0=H1=N0=0,N1=N1,ρ1=2N2ρ0N1,N2=N2,σ1=2N12ρ2−N22ρ02Θ2+3θ12aN12ρ2−N22ρ02+H22N12k2Θ1+2Θ2ρ0ρ2+12ρ0ρ2θ1−kλ1−2H22N22ρ02Θ2+6θ1.Through (102) and (103), then we can find the below solution:By using (98), we have
(104)q1x,t=H21/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ22N11/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ2+N21/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ22×ei−kx+ωt,where N1,N2,ϖ0 are free amounts.Product IB:(105)b=−2aN0N1ρ02H0N1−H1N0+kH03N0kΘ1−λ1−H0H1ρ02Θ2+6θ1H0N1−H1N0H03N0,k=k,ω=−ak2H02N0+aH0H1N1ρ02−aH12N0ρ02+kH02N0α1N0H02,H0=H0,H1=H1,N0=N0,N1=N1,N2=14H1N1H0,ρ1=H1ρ0H0,ρ2=14H1N1ρ0N0H0,σ1=2N1ρ02Θ2+3θ1H0N1−H1N0N02aH0N0N12ρ02−2aH1N02N1ρ02+kH03N0kΘ1−λ1−H0H1ρ02Θ2+6θ1H0N1−H1N0.Through (102) and (103), then we can find the below solution:By using (98), we have
(106)q2x,t=ei−kx+ωt×H0H1N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ22+4H1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ2N0+4N0H0H1N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ22+4N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ2H0+4N0H0N0,where N0,H0,N1,H1,ϖ0 are free amounts.With case (99), we have the following:Product IIA:(107)b=8N0ρ12Θ2+3θ1H12N2−H22N0H14σ1,ω=−ak2+aρ12−kα1,H0=0,N1=2H2N0H1,λ1=−8aN0ρ12σ1H12N2−H22N0−H14σ1k2Θ1+Θ2ρ12+6ρ12θ1−8N0ρ12Θ2+3θ1H12N2−H22N0kH14σ1,k=k,H1=H1,H2=H2,N2=N2,ρ2=H2ρ1H1.Through (102) and (107), then we can find the below solution:By using (99), we have
(108)q1x,t=H1ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0H2ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0+H1H1N2ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ02+2H2N0ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0+H1N0ei−kx+−ak2+aρ12−kα1t,where N0,H1,N2,ϖ0 are free amounts.With case (100), we have the following:Product IIIA:(109)a=−18H22k2Θ1−kλ1+b−8H02ρ32Θ2+6θ1N02ρ32,H1=N1=N2=0,ρ1=2ρ3H0H2,σ1=−8N02ρ32Θ2+3θ1bH22,ω=−64H04ρ34Θ2+3θ1+8k2H02H22ρ32Θ1−Θ2−6θ1+k2H24S1+8H02H22ρ32S2−8kH22N02α1ρ328N02ρ32H22,k=k,H0=H0,N0=N0,S1=k2Θ1−kλ1+b,S2=−kλ1+b.Through (102) and (109), then we can find the below solution:By using (100), we have
(110)q1x,t=H2C0e−2ρ1ϖ+ϖ0ρ1−ρ3ρ1/C0e−2ρ1ϖ+ϖ0ρ1−ρ32+H0N0ei−kx+ωt,where C0,N0,H0,ϖ0 are free amounts.In reference to case (101) and the solution function Γϖ=1/−2ϖ+ϖ0ρ3+C0, we can get to the below results:Product IVA:(111)a=−18kH22σ1kΘ1−λ1−8N02ρ32Θ2+3θ1N02ρ32σ1,b=−8N02ρ32Θ2+3θ1H22σ1,H0=H1=N1=N2=0,k=k,ω=18kk2H22σ1kΘ1−λ1−8N02ρ32kΘ2+3kθ1+α1σ1N02ρ32σ1.Through (102) and (111), then we can find the below solution:By using (101), we have
(112)q1x,t=H21/−2ϖ+ϖ0ρ3+C02N0ei−kx+1/8kk2H22σ1kΘ1−λ1−8N02ρ32kΘ2+3kθ1+α1σ1/N02ρ32σ1t,where C0,N0,ϖ0 are free amounts.Product IVB:(113)a=−18H22k2Θ1−kλ1+bN02ρ32,H0=H1=N1=0,σ1=−8N02ρ32Θ2+3θ1bH22,k=k,ω=18kk3H22Θ1−k2H22λ1−8N02α1ρ32+bkH22N02ρ32.Through (102) and (113), then we can find the below solution:By using (101), we have
(114)q2x,t=H21/−2ϖ+ϖ0ρ3+C02N0+N21/−2ϖ+ϖ0ρ3+C02ei−kx+1/8kk3H22Θ1−k2H22λ1−8N02α1ρ32+bkH22/N02ρ32t,where C0,N0,N2,ϖ0 are free amounts.


## 4.1. Application of ISEM on Equation (1)

The IEFM allows us to recruit the substitutions:(102)vϖ=H0+H1Γϖ+H2Γϖ2N0+N1Γϖ+N2Γϖ2.Plug (35) along with (97) into equation (6).With case (98), we have the following:Product IA:(103)b=−2aN12ρ2−N22ρ02+H22N12k2Θ1+2Θ2ρ0ρ2+12ρ0ρ2θ1−kλ1−2H22N22ρ02Θ2+6θ1H22N12,k=k,ω=−N12ak2−2aρ0ρ2+kα1+2ρ02aN22−3H22θ1N12,H0=H1=N0=0,N1=N1,ρ1=2N2ρ0N1,N2=N2,σ1=2N12ρ2−N22ρ02Θ2+3θ12aN12ρ2−N22ρ02+H22N12k2Θ1+2Θ2ρ0ρ2+12ρ0ρ2θ1−kλ1−2H22N22ρ02Θ2+6θ1.Through (102) and (103), then we can find the below solution:By using (98), we have
(104)q1x,t=H21/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ22N11/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ2+N21/2tan1/2ϖ+ϖ04ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/ρ22×ei−kx+ωt,where N1,N2,ϖ0 are free amounts.Product IB:(105)b=−2aN0N1ρ02H0N1−H1N0+kH03N0kΘ1−λ1−H0H1ρ02Θ2+6θ1H0N1−H1N0H03N0,k=k,ω=−ak2H02N0+aH0H1N1ρ02−aH12N0ρ02+kH02N0α1N0H02,H0=H0,H1=H1,N0=N0,N1=N1,N2=14H1N1H0,ρ1=H1ρ0H0,ρ2=14H1N1ρ0N0H0,σ1=2N1ρ02Θ2+3θ1H0N1−H1N0N02aH0N0N12ρ02−2aH1N02N1ρ02+kH03N0kΘ1−λ1−H0H1ρ02Θ2+6θ1H0N1−H1N0.Through (102) and (103), then we can find the below solution:By using (98), we have
(106)q2x,t=ei−kx+ωt×H0H1N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ22+4H1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ2N0+4N0H0H1N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ22+4N1tan1/2ξ4ρ0ρ2−ρ124ρ0ρ2−ρ12−ρ1/2ρ2H0+4N0H0N0,where N0,H0,N1,H1,ϖ0 are free amounts.With case (99), we have the following:Product IIA:(107)b=8N0ρ12Θ2+3θ1H12N2−H22N0H14σ1,ω=−ak2+aρ12−kα1,H0=0,N1=2H2N0H1,λ1=−8aN0ρ12σ1H12N2−H22N0−H14σ1k2Θ1+Θ2ρ12+6ρ12θ1−8N0ρ12Θ2+3θ1H12N2−H22N0kH14σ1,k=k,H1=H1,H2=H2,N2=N2,ρ2=H2ρ1H1.Through (102) and (107), then we can find the below solution:By using (99), we have
(108)q1x,t=H1ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0H2ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0+H1H1N2ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ02+2H2N0ρ1eρ1ϖ+ϖ0/C0ρ1−ρ2eρ1ϖ+ϖ0+H1N0ei−kx+−ak2+aρ12−kα1t,where N0,H1,N2,ϖ0 are free amounts.With case (100), we have the following:Product IIIA:(109)a=−18H22k2Θ1−kλ1+b−8H02ρ32Θ2+6θ1N02ρ32,H1=N1=N2=0,ρ1=2ρ3H0H2,σ1=−8N02ρ32Θ2+3θ1bH22,ω=−64H04ρ34Θ2+3θ1+8k2H02H22ρ32Θ1−Θ2−6θ1+k2H24S1+8H02H22ρ32S2−8kH22N02α1ρ328N02ρ32H22,k=k,H0=H0,N0=N0,S1=k2Θ1−kλ1+b,S2=−kλ1+b.Through (102) and (109), then we can find the below solution:By using (100), we have
(110)q1x,t=H2C0e−2ρ1ϖ+ϖ0ρ1−ρ3ρ1/C0e−2ρ1ϖ+ϖ0ρ1−ρ32+H0N0ei−kx+ωt,where C0,N0,H0,ϖ0 are free amounts.In reference to case (101) and the solution function Γϖ=1/−2ϖ+ϖ0ρ3+C0, we can get to the below results:Product IVA:(111)a=−18kH22σ1kΘ1−λ1−8N02ρ32Θ2+3θ1N02ρ32σ1,b=−8N02ρ32Θ2+3θ1H22σ1,H0=H1=N1=N2=0,k=k,ω=18kk2H22σ1kΘ1−λ1−8N02ρ32kΘ2+3kθ1+α1σ1N02ρ32σ1.Through (102) and (111), then we can find the below solution:By using (101), we have
(112)q1x,t=H21/−2ϖ+ϖ0ρ3+C02N0ei−kx+1/8kk2H22σ1kΘ1−λ1−8N02ρ32kΘ2+3kθ1+α1σ1/N02ρ32σ1t,where C0,N0,ϖ0 are free amounts.Product IVB:(113)a=−18H22k2Θ1−kλ1+bN02ρ32,H0=H1=N1=0,σ1=−8N02ρ32Θ2+3θ1bH22,k=k,ω=18kk3H22Θ1−k2H22λ1−8N02α1ρ32+bkH22N02ρ32.Through (102) and (113), then we can find the below solution:By using (101), we have
(114)q2x,t=H21/−2ϖ+ϖ0ρ3+C02N0+N21/−2ϖ+ϖ0ρ3+C02ei−kx+1/8kk3H22Θ1−k2H22λ1−8N02α1ρ32+bkH22/N02ρ32t,where C0,N0,N2,ϖ0 are free amounts.


## 5. Discussion and Remark

The physical explanation of the established traveling wave solutions to the generalized nonlinear Schrödinger equation with parabolic law of nonlinearity will be discussed in this section. The collected traveling wave solutions of this equation are discussed in the three-dimensional plotline, the plot of density, three-dimensional list point plotline, and plot of vector which are designed via Maple. Using those eight sorts of pictorial descriptions, we may more explicitly characterize the physical sketch. As can be seen from the diagrams above, various structures, including those of the well-known form of solitons, such as the solutions ofq1x,t (37) in Figure 1 within the duration −30<x<30,−2<t<2 which is the solution of the generalized nonlinear Schrödinger equation, represented shape traveling wave solution. Moreover, the solution transforms to the periodic solution as mentioned in q2x,t (38) in Figure 2 within the duration −5<x<5,0<t<5. Also, by plugging the suitable values, the solution q3x,t given by (39) in Figure 3 shows the kink-singular wave solution within the duration −3<x<3,−5<t<5. Finally, by putting the specified amounts, the solution q4x,t given by (60) in Figure 4 finds the cupson-singular wave solution within the duration −15<x<15,−1<t<1. In addition, the solutions of q18x,t (86) in Figure 5 within the duration −20<x<20,−10<t<10 which is the solution of the generalized nonlinear Schrödinger equation represented shape traveling wave solution. Moreover, the solution transforms to the periodic solution as mentioned in q19x,t (87) in Figure 6 within the duration −20<x<20,−10<t<10. Also, by plugging the suitable values, the solution q20x,t given by (88) in Figure 7 shows the kink-singular wave solution within the duration −10<x<10,−5<t<5. Finally, by putting the specified amounts, the solution q21x,t given by (89) in Figure 8 finds the cupson-singular wave solution within the duration −20<x<20,−1<t<1.The graphics for the generalized nonlinear Schrödinger equation solutions are developed here. We represent the different types of the picture which are gotten by those equations and also avoid the same shape which overlaps with the presented picture. We will discover all of the most recent exact solutions in our paper. However, for the sake of convenience, we will ignore it in this post. We have seen that the solutions to the aforementioned equation are relatively novel in literature as compared to previous solutions. Figures1–8 are generated utilizing a computation package program, namely, Maple.


## 6. Conclusion

We proposed the novel schemes, namely, the improvedexp−Γϖ function method and the improved simple equation method, which are utilized to build the exact solutions of the generalized nonlinear Schrödinger equation. In addition, multifarious transformation was operated to renovate the nonlinear ordinary transformation equation. We found a slide of the closed form of solutions to the suggested equations, including kink-shaped, anti-kink-shaped, bell-shaped, singular bell-shaped, singular periodic-shaped, multiple soliton-shaped, single soliton-shaped, and other types of solutions with a variety of free parameters. These free parameters have significant consequences, for example, the ability to find specific solutions by changing the free parameter values of an individual solution. The activities of solitary waves were graphically illustrated concerning space and time, revealing the higher efficiency and validity of the claimed schemes. The retrieved solutions could be used to study the transmission of metamaterial waves, nonlinear optic, and so on. The constructed results have extensive potential to comprehend the interior configurations of the usual manifestation that arise in physics, mathematics, and other different fields. Hence, it is worth declaring that the execution of our schemes is extremely steady and well organized and for nonlinear models. The adopted method is direct, trustworthy, conformable, and effective, and it provides many novel physical model solutions to NLPEEs that arise in mathematical physics, applied mathematics, and engineering.



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*Source: 1351377-2022-04-26.xml*