Startseite Discovering optical solutions to a nonlinear Schrödinger equation and its bifurcation and chaos analysis
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Discovering optical solutions to a nonlinear Schrödinger equation and its bifurcation and chaos analysis

  • Shami A. M. Alsallami EMAIL logo
Veröffentlicht/Copyright: 1. August 2024
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Abstract

The pursuit of solitary wave solutions to complex nonlinear partial differential equations is gaining significance across various disciplines of nonlinear science. This study seeks to uncover the solutions to the perturbed nonlinear Schrödinger equation using a robust and efficient analytical method, namely, the generalized exponential rational function technique. This equation is a fundamental tool used in various fields, including fluid mechanics, nonlinear optics, plasma physics, and optical communication systems, and has numerous practical applications across multiple disciplines. The employed method in this study stands out from existing approaches by being more comprehensive and straightforward. It offers a broader range of symbolic structures, surpassing the capabilities of some previously known methods. By applying this method to the perturbed nonlinear Schrödinger equation, we obtain a variety of exact solutions that significantly expand the existing literature and provide a fresh understanding of the model’s properties. Through numerical simulations, we demonstrate the dynamic characteristics of the system, including bifurcation and chaos analysis, and validate our findings by adjusting parameter settings to match expected behaviors.

1 Introduction

The development of innovative analytical techniques is gaining importance as it enables us to understand complex phenomena that arise in various fields of science and physics, including plasma physics, ocean physics, and fluid dynamics [17]. The significance of this topic has led to the introduction of numerous efficient techniques, which have been widely discussed in the literature. In the work of Tariq et al. [1], three methods including 1 G -expansion, Bernoulli, and modified Kudryashov methods were used to obtain novel analytical solutions for the (3+1)-dimensional Boiti–Leon–Manna–Pempinelli model,

(1) 3 x ( x y + x z ) 3 x x ( y + z ) + x x x y x x x z + y t + z t = 0 ,

where = ( x , y , z ) represents wave propagation in incompressible fluids. Linearization of shallow water waves demands stricter conditions compared to deep water ones, and nonlinear properties are evident. Novel traveling wave solutions, including solitons, kinks, periodic, and rational ones, have been observed using tools such as Mathematica and Maple. Fahim et al. [2] explored the Kadomtsev–Petviashvili and modified Korteweg–De Vries–Zakharov–Kuznetsov (KdV–ZK) equations’ applications in various fields. It analyzes the effects of wave speed and parameters on wave contours. Different standard wave configurations, such as kink, breather soliton, bell-shaped soliton, and periodic waves, are examined. The sine-Gordon expansion method is used to derive soliton solutions for high-dimensional nonlinear evolution equations (HNEEs). Some of these models are

  1. The Kadomtsev–Petviashvili equation

    (2) ( t + 6 x + x x x ) x 3 y y 3 z z = 0 .

  2. The modified KdV–ZK equation

    (3) t + β 2 x + x x x + x y y + x z z = 0 .

Their study demonstrates the efficiency of the method in solving nonlinear HNEEs, providing closed-form solutions. Sulaiman et al. [3] used the Hirota bilinear scheme to analyze a nonlinear model:

(4) x t t + x x x x t + 6 x x x t + 3 x x x t + 3 x x x x t + σ x y t + δ x x t = 0 ,

obtaining two-wave and breather-wave solutions. Breathers, localized pulsating structures, can survive in complex environments like random seas. The solutions’ authenticity matches the original equation. 3D and contour graphs illustrate the solutions with suitable values. The model’s stability contributes to the ocean engineering challenges’ understanding. Zekavatmand et al. [4] aimed to discover soliton solutions for the long- and short-wave interaction system:

(5) i t + x x v = 0 , v t + v x + ( 2 ) x = 0 ,

using extended rational sine-cosine and rational sinh-cosh methods. Assuming hypothetical soliton solutions, the methods lead to a system of equations solved with Maple software. New soliton solutions, including bright, kink type, bright periodic, and dark solutions, are obtained. Some 3-D figures illustrate the solutions, and computational results suggest the method’s superiority over other literature methods for solving the system equations. Zahran et al. [5] explored new ideal optical soliton solutions for the nonlinear Schrödinger equation:

(6) i t + 1 2 x x + 2 λ 2 γ x t = 0 ,

in the context of two-layer baroclinic instability and lossless symmetric two-stream plasma instability models. Two distinct methods, the solitary wave ansatz and the modified extended mapping, are employed. Both methods are applied in parallel, with some results aligning with previous findings and others being novel. Moreover, the unstable nonlinear Schrödinger equation with β -time derivative

(7) i D t β 0 A + D x 2 β 0 A + 2 λ 2 2 γ = 0

is studied by Razzaq et al. [6] using the new sub-equation method to derive soliton solutions and other types such as dark, bright-singular, and singular solitons for unstable and modified unstable nonlinear Schrödinger equations with time derivative. Additionally, periodic, rational, and exponential solutions are reported. By assigning suitable parameter values, novel solution structures are visually depicted. Graphical representations of physical surfaces for various solutions in different forms help in understanding complex physical phenomena related to these dynamical models. The method’s superiority is demonstrated, making it applicable to numerous other nonlinear physical equations. The work of Yusuf et al. [7] solves the Ostrovsky equation without rotation effects using the Hirota bilinear method and symbolic calculations. It explores unique interaction phenomena between lump, breather, periodic, kink, and two-wave solutions. Physical characteristics are visually represented. The novel conservation theory generates conserved vectors for the governing equation. These findings contribute to understanding specific physical phenomena in fluid dynamics, nonlinear wave fields, and computational physics, benefiting ocean engineering and related fields.

In this study, we explore the in-depth analysis of a nonlinear partial differential equation (PDE) in fluid dynamics as follows [813]:

(8) i t + ρ 1 2 t x + ρ 2 2 x 2 + ρ 3 2 = i ρ 4 x + ρ 5 ( 2 p ) x + ρ 6 ( 2 p ) x ,

where ( x , t ) symbolizes a complex field envelope, with x and t representing spatial and temporal variables, respectively. In this model, the first term illustrates the linear progression of pulses within nonlinear optical fibers. The coefficient ρ 1 pertains to spatiotemporal dispersion, and ρ 2 corresponds to group velocity dispersion. On the right-hand side of the model, the first term introduces the perturbation terms, while ρ 4 stands for the inter-model dispersion, and ρ 5 signifies the perturbation term of self-steepening. In the last part, ρ 6 indicates the nonlinear dispersion. Moreover, p represents the coefficient of the full nonlinearity. And finally, the parameter ρ 3 adheres to the complete law nonlinearity of P .

Uncovering and categorizing the diverse optical soliton forms in relation to Eq. (8) is crucial for understanding and harnessing the potential of this equation. For example, Zafar et al. [8] explored solitary wave solutions to Model (8) across various nonlinearities, such as Kerr law, power law. By employing a novel derivative operator known as the beta derivative, the study successfully retrieves dark, bright, singular, and combined solutions using a simple integration scheme. Using the advanced exp ( ϕ ( ξ ) ) -expansion method, the study by Arshed [9] successfully identified various forms of explicit solutions, including dark, singular, rational, and periodic solitary wave solutions to (8). The existence criteria for these newly discovered solutions are thoroughly discussed. Ozisik et al. [10] explored some analytical solutions to the nonlinear Schrödinger equation offered in (8). Through the implementation of the modified extended tanh expansion method in conjunction with new Riccati solutions, a wide range of solutions have been derived. These solutions encompass singular, dark, singular periodic waves, combined dark–singular soliton, and rational function solutions. The solutions are visualized through contour, 2D, and 3D plots created using Matlab, while Maple is utilized for computations. Saha Ray and Das [11] focused on studying the space-time fractional perturbed nonlinear Schrödinger equation in nanofibers. The improved tan ( ϕ 2 ) expansion method is used to explore new exact solutions. The method proves to be efficient in obtaining solutions for nonlinear differential equations. Using the modified Riemann–Liouville derivative, an equivalent ordinary differential equation is derived from the original fractional differential equation. The proposed method not only provides new exact solutions to the fractional perturbed NLSE but also has the potential to be applied in other engineering and physics fields. The obtained soliton solutions are visually represented through 3D graphs to observe their behavior. Moreover, Gepreel [12] focused on studying the model various considered nonlinear media laws, including Kerr law, power law, quadratic–cubic law, and dual-power law. The perturbed nonlinear Schrödinger equation (8) has been also studied by Akram et al. [13], who have obtained new soliton solutions with Kerr law nonlinearity not found in the existing literature, and by implementing an improved F-expansion method for modulation instability. The research utilizes 2D, 3D, contour, and density plots to validate the results and demonstrate the effectiveness and efficiency of the proposed technique for solving complex models in mathematical physics.

This article explores novel optical solutions to Eq. (8) along with its bifurcation and chaos analysis. For this objective, we utilize a systematic analytical method, which is derived from the adaptations of the generalized exponential rational function methods (mGERFM). This article is divided into various parts. In Section 2, we obtain a sneak peek into the various stages required by the method. In the subsequent sections, we first delve into the mathematical background of handling the model. Then, we obtain analytical solutions for the model in Section 4. This part also demonstrates several 2D plots to enrich our understanding of the proposed results. Furthermore, the bifurcation and chaos analysis of the model are presented in Section 5. The last section of the article will also discuss the conclusions.

2 Analytical method

Numerous research articles [1426] have investigated the application of the GERFM method to resolve various PDEs. Within the work presented in the study of Ghanbari [27], the author has introduced a novel adaptation of the GERFM technique called mGERFM. Here, we review a step-by-step guide on how to implement this approach.

  1. Consider the following nonlinear equation:

    (9) N ( x , t ) , ( x , t ) x , ( x , t ) t , 2 ( x , t ) x 2 , 2 ( x , t ) t 2 , = 0 .

    Taking ( x , t ) = ( ζ ) and ζ = μ 1 x μ 2 t in (9), we can transfer model into the following new problem:

    (10) N ( ζ ) , μ 1 d ( ζ ) d ζ , μ 2 d ( ζ ) d ζ , μ 1 2 d 2 ( ζ ) d ζ 2 , μ 2 2 d 2 ( ζ ) d ζ 2 , = 0 .

  2. Taking into account this approach, a carefully structured layout is provided, showcasing the solution for the model in a formal manner:

    (11) ( ζ ) = 0 + j = 1 n j ( ζ ) ( ζ ) j + j = 1 n ϱ j ( ζ ) ( ζ ) j ,

    where

    (12) ( ζ ) = r 1 exp ( s 1 ζ ) + r 2 exp ( s 2 ζ ) r 3 exp ( s 3 ζ ) + r 4 exp ( s 4 ζ ) .

    In this framework, 0 , j , ϱ j ( 1 j n ) , j , and ϱ j (where 1 j 4 ) are the undetermined parameters. Moreover, the balance number n is determined using some known existing balance rules.

  3. Employing Eq. (11) within Eq. (10) generates a polynomial equation containing m = exp ( s m ζ ) , with m ranging from 1 to 4.

  4. Having obtained the outcomes of the resolved system, the analytical solutions to Eq. (9) are achieved.

3 Mathematical analysis

Within this article, the primary focus lies in tackling the equation’s solutions by employing the subsequent wave transformations

(13) ( x , t ) = ( ζ ) × exp ( i [ x + θ t ] ) , ζ = x ρ 3 t ,

where ρ 3 , , and θ are the disposal parameters. These transformations hold substantial influence on the derivation of the analytical solution for the equation.

Substituting (13) into Eq. (8), from real part, we obtain

(14) ( ρ 2 ρ 1 ρ 3 ) ( ζ ) ( θ + ρ 4 ρ 1 θ + ρ 2 2 ) ( ζ ) ρ 5 2 p + 1 ( ζ ) + ρ 3 P ( 2 ( ζ ) ) ( ζ ) = 0 .

Also, the imaginary part is given by

(15) ρ 3 ( ρ 1 1 ) + ρ 1 θ 2 ρ 2 ρ 4 + ( ( 2 p + 1 ) ρ 5 + 2 p ρ 6 ) 2 p ( ζ ) = 0 .

From Eq. (15), the speed of the wave solution is obtained as follows:

ρ 3 = ρ 1 θ 2 ρ 2 ρ 4 1 ρ 1 .

Furthermore, we obtain

(16) ( 2 p + 1 ) ρ 5 + 2 p ρ 6 = 0 ,

which implies ρ 6 = 3 2 ρ 5 .

After performing certain manipulations while considering the scenario of Kerr law nonlinear medium P ( ) = along with p = 1 in Eq. (14), it is simplified to the following Duffing oscillator [28]:

(17) Λ 1 ( ζ ) + Λ 2 ( ζ ) + Λ 3 3 ( ζ ) = 0 ,

where

(18) Λ 1 = ρ 1 2 θ ρ 2 ρ 1 ρ 4 ρ 1 ρ 2 , Λ 2 = 3 ρ 2 ρ 1 2 ρ 1 2 θ + ρ 4 2 ρ 1 ρ 2 2 + 2 ρ 1 θ ρ 4 θ , Λ 3 = ρ 5 2 ρ 1 ρ 1 ρ 3 ρ 5 + ρ 3 .

4 Discovering optical solutions to the model

Using the well-known principles with Eq. (17), specifically examining the balance number of the model is obtained by comparing between 3 ( ζ ) and ( ζ ) . Thus, it demonstrates that 3 n = n + 2 holds. As a result, the value for n is found to be n = 1 . In this way, the following standard framework is considered for addressing Eq. (8)’s solution approach:

(19) ( ζ ) = 0 + 1 ( ζ ) ( ζ ) + ϱ 1 ( ζ ) ( ζ ) 1 .

In Section 2, we provide an in-depth exploration of our utilized methodology, allowing us to offer a variety of novel analytical solutions for Eq. (8) as follows.

  1. Taking [ r 1 , r 2 , r 3 , r 4 ] = [ 1 , 1 , 2 , 0 ] and [ s 1 , s 2 , s 3 , s 4 ] = [ i , i , 0 , 0 ] in Eq. (12) offers

    (20) ( ζ ) = cos ( ζ ) .

    • – If one takes

      (21) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 1 ρ 2 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 ρ 1 2 2 2 ρ 1 2 ρ 1 2 + 1 , 0 = 0 , 1 = 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 4 ρ 1 + ρ 2 ( ρ 1 2 2 2 ρ 1 2 ρ 1 2 + 1 ) ρ 3 ρ 5 , ϱ 1 = 0

      where is a free parameter, provided that ρ 3 ρ 5 > 0 .

    • Incorporating these results into the structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 tan ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (22) 1 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 tan ( ζ ) × exp ( i φ ) ,

      where

      (23) ζ = x ρ 2 ( 2 2 ) ρ 1 2 ρ 2 ρ 4 1 + ( 2 2 ) ρ 1 2 2 ρ 1 t , φ = x 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 1 ρ 2 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 ρ 1 2 2 2 ρ 1 2 ρ 1 2 + 1 t .

      Figure 1 demonstrates the contour plots of solution (22) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.1 .

    • – If one takes

      (24) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 8 ρ 2 ρ 1 ρ 4 ) 8 ρ 4 ρ 1 8 ρ 2 2 ρ 1 2 2 ρ 1 8 ρ 1 2 + 1 , 0 = 0 , 1 = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 8 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 , ϱ 1 = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 8 ) ρ 1 2 2 ρ 1 ( 2 ρ 1 2 2 ρ 1 8 ρ 1 2 + 1 ) ρ 3 ρ 5 ,

      where is a free parameter, provided that ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 ( 2 cot ( ζ ) + sec ( ζ ) csc ( ζ ) ) 1 + ( 2 8 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (25) 2 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 ( 2 cot ( ζ ) + sec ( ζ ) csc ( ζ ) ) 1 + ( 2 8 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 × exp ( i φ ) ,

      where

      (26) ζ = x ρ 2 ( 2 8 ) ρ 1 2 ρ 2 ρ 4 1 + ( 2 8 ) ρ 1 2 2 ρ 1 t , φ = x 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 8 ρ 2 ρ 1 ρ 4 ) 8 ρ 4 ρ 1 8 ρ 2 2 ρ 1 2 2 ρ 1 8 ρ 1 2 + 1 t .

      Figure 2 depicts the contour plots of solution (25) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.7 .

    • – If one takes

      (27) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 2 ρ 1 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 , 0 = 0 , 1 = 0 , ϱ 1 = 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 4 ρ 1 + ρ 2 ( 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 ) ρ 3 ρ 5 ,

      where is a free parameter, provided that ρ 3 ρ 5 > 0 . Incorporating these results into structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 tan ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (28) 3 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 tan ( ζ ) × exp ( i φ ) ,

      where

      (29) ζ = x ρ 2 ( 2 2 ) ρ 1 2 ρ 2 ρ 4 1 + ( 2 2 ) ρ 1 2 2 ρ 1 t , φ = x 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 2 ρ 1 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 t .

      Figure 3 depicts the contour plots of solution (28) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.7 .

    • – If one takes

      (30) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 2 ρ 1 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 , 0 = 0 , 1 = 2 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 4 ρ 1 + ρ 2 ( 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 ) ρ 3 ρ 5 , ϱ 1 = 0 ,

      where is a free parameter, provided that ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 tan ( ζ ) 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (31) 4 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 tan ( ζ ) 1 + ( 2 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 × exp ( i φ ) ,

      where

      (32) ζ = x ρ 2 ( 2 2 ) ρ 1 2 ρ 2 ρ 4 1 + ( 2 2 ) ρ 1 2 2 ρ 1 t , φ = x 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 2 ρ 1 ρ 4 ) 2 ρ 4 ρ 1 2 ρ 2 2 ρ 1 2 2 ρ 1 2 ρ 1 2 + 1 t .

      Figure 4 depicts the contour plots of solution (31) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.3 .

    • – If one takes

      (33) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 4 ρ 2 ρ 1 ρ 4 ) + 4 ρ 4 ρ 1 + 4 ρ 2 2 ρ 1 2 2 ρ 1 + 4 ρ 1 2 + 1 , 0 = 0 , 1 = 2 ρ 4 ρ 1 + ρ 2 ρ 3 ρ 5 1 + ( 2 + 4 ) ρ 1 2 2 ρ 1 , ϱ 1 = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 + 4 ) ρ 1 2 2 ρ 1 ( 2 ρ 1 2 2 ρ 1 + 4 ρ 1 2 + 1 ) ρ 3 ρ 5 ,

      where is a free parameter, provided that ρ 3 ρ 5 > 0 . Incorporating these results into structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 sec ( ζ ) csc ( ζ ) 1 + ( 2 + 4 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (34) 5 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 sec ( ζ ) csc ( ζ ) 1 + ( 2 + 4 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 × exp ( i φ ) ,

      where

      (35) ζ = x ( ρ 2 ( 2 + 4 ) ρ 1 2 ρ 2 ρ 4 ) 1 + ( 2 + 4 ) ρ 1 2 2 ρ 1 t , φ = x ( 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 4 ρ 2 ρ 1 ρ 4 ) + 4 ρ 4 ρ 1 + 4 ρ 2 ) 2 ρ 1 2 2 ρ 1 + 4 ρ 1 2 + 1 t .

      Figure 5 demonstrates the contour plots of solution (34) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.3 .

  1. Taking [ r 1 , r 2 , r 3 , r 4 ] = [ 1 , 1 , 1 , 0 ] and [ s 1 , s 2 , s 3 , s 4 ] = [ 0 , 1 , 0 , 0 ] in Eq. (12) offers

    (36) ( ζ ) = 1 + exp ( ζ ) .

    • – If one takes

      (37) θ = 2 3 ρ 2 ρ 1 + ( 2 ρ 4 ρ 1 2 ρ 2 ) 2 + ( ρ 1 ρ 2 2 ρ 4 ) + ρ 4 ρ 1 + ρ 2 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 , 0 = ρ 4 ρ 1 + ρ 2 ρ 3 ρ 5 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 , 1 = 2 ρ 4 ρ 1 + ρ 2 ρ 3 ρ 5 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 , ϱ 1 = 0

      where is a free parameter, provided ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = ρ 4 ρ 1 + ρ 2 ( exp ( ζ ) 1 ) 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 ρ 3 ρ 5 ( exp ( ζ ) + 1 ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (38) 6 ( x , t ) = ρ 4 ρ 1 + ρ 2 ( exp ( ζ ) 1 ) 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 ρ 3 ρ 5 ( exp ( ζ ) + 1 ) × exp ( i φ ) ,

      where

      (39) ζ = x ρ 2 ( 2 2 + 1 ) ρ 1 4 ρ 2 2 ρ 4 2 + ( 2 2 + 1 ) ρ 1 2 4 ρ 1 t , φ = x 2 3 ρ 2 ρ 1 + ( 2 ρ 4 ρ 1 2 ρ 2 ) 2 + ( ρ 1 ρ 2 2 ρ 4 ) + ρ 4 ρ 1 + ρ 2 2 ρ 1 2 2 4 ρ 1 + ρ 1 2 + 2 t .

      Figure 6 depicts the contour plots of solution (38) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.7 .

    • – If one takes

      (40) = ρ 3 ρ 5 , θ = ( ρ 4 ρ 1 ρ 2 ) ρ 5 3 ρ 3 ( ρ 2 ρ 1 + ρ 4 ) ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 , 0 = ϱ 1 , 1 = 0 ,

      where and ϱ 1 are the free parameters, provided ρ 3 ρ 5 > 0 . Incorporating these results into structure (19) allows us to derive

      ( ζ ) = ϱ 1 exp ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (41) 7 ( x , t ) = ϱ 1 exp x ( ρ 5 2 ρ 2 ρ 1 ρ 2 ρ 1 ρ 3 2 + ρ 5 2 ρ 4 + 2 ρ 5 ρ 2 ρ 3 ) ρ 5 2 ρ 1 2 ρ 3 2 ρ 1 2 + 2 ρ 3 ρ 5 ρ 1 ρ 5 2 t × exp ( i φ ) ,

      where

      (42) φ = ρ 3 ρ 5 x ( ( ρ 4 ρ 1 ρ 2 ) ρ 5 3 ρ 3 ( ρ 2 ρ 1 + ρ 4 ) ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ) ( ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 t .

      Figure 7 demonstrates the contour plots of solution (41) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with ϱ 1 = 1 , = 0.7 .

  1. Taking [ r 1 , r 2 , r 3 , r 4 ] = [ 1 , 1 , 1 , 0 ] and [ s 1 , s 2 , s 3 , s 4 ] = [ 0 , 1 , 0 , 0 ] in Eq. (12) offers

    (43) ( ζ ) = 1 + exp ( ζ ) .

    1. If one takes

      (44) = ρ 3 ρ 5 , θ = ( ρ 4 ρ 1 ρ 2 ) ρ 5 3 ρ 3 ( ρ 2 ρ 1 + ρ 4 ) ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 , 0 = 1 , 1 = 0 ,

      where and ϱ 1 are the free parameters, provided ρ 3 ρ 5 > 0 . Incorporating these results into structure (19) allows us to derive

      ( ζ ) = ϱ 1 exp ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (45) 8 ( x , t ) = ( ϱ 1 exp ( ζ ) ) × exp ( i φ ) ,

      where

      (46) ζ = x ρ 5 2 ρ 2 ρ 1 ρ 2 ρ 1 ρ 3 2 + ρ 5 2 ρ 4 + 2 ρ 5 ρ 2 ρ 3 ρ 5 2 ρ 1 2 ρ 3 2 ρ 1 2 + 2 ρ 3 ρ 5 ρ 1 ρ 5 2 t , φ = ρ 3 x ρ 5 ( ρ 4 ρ 1 ρ 2 ) ρ 5 3 ρ 3 ( ρ 2 ρ 1 + ρ 4 ) ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 t .

  1. Taking [ r 1 , r 2 , r 3 , r 4 ] = [ 1 , 1 , 2 , 0 ] and [ s 1 , s 2 , s 3 , s 4 ] = [ 2 , 0 , 0 , 0 ] in Eq. (12) offers

    (47) ( ζ ) = sinh ( ζ ) exp ( ζ ) .

    • – If one takes

      (48) θ = 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 1 ρ 2 ρ 4 ) + 2 ρ 4 ρ 1 + 2 ρ 2 ρ 1 2 2 2 ρ 1 + 2 ρ 1 2 + 1 , 0 = 2 1 + ( 2 + 2 ) ρ 1 2 2 ρ 1 ρ 4 ρ 1 + ρ 2 ( ρ 1 2 2 2 ρ 1 + 2 ρ 1 2 + 1 ) ρ 3 ρ 5 , 1 = 2 1 + ( 2 + 2 ) ρ 1 2 2 ρ 1 ρ 4 ρ 1 + ρ 2 ( ρ 1 2 2 2 ρ 1 + 2 ρ 1 2 + 1 ) ρ 3 ρ 5 , ϱ 1 = 0 ,

      where is a free parameter, provided ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 + 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 coth ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (49) 9 ( x , t ) = 2 ρ 4 ρ 1 + ρ 2 1 + ( 2 + 2 ) ρ 1 2 2 ρ 1 ρ 3 ρ 5 coth ( ζ ) × exp ( i φ ) ,

      where

      (50) ζ = x ρ 2 ( 2 + 2 ) ρ 1 2 ρ 2 ρ 4 1 + ( 2 + 2 ) ρ 1 2 2 ρ 1 t , φ = x 3 ρ 2 ρ 1 + ( ρ 4 ρ 1 ρ 2 ) 2 + ( 2 ρ 1 ρ 2 ρ 4 ) + 2 ρ 4 ρ 1 + 2 ρ 2 ρ 1 2 2 2 ρ 1 + 2 ρ 1 2 + 1 t .

      Figure 8 shows the contour plots of solution (49) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with = 0.7 .

    • – If one takes

      (51) = ρ 3 ρ 5 , θ = ( 4 ρ 4 ρ 1 4 ρ 2 ) ρ 5 3 4 ρ 2 ρ 1 + ρ 4 4 ρ 3 ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( 4 ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 , 0 = ϱ 1 2 , 1 = 0 ,

      where and ϱ 1 are the free parameters, provided ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = ϱ 1 ( coth ( ζ ) 1 ) 2 coth ( ζ ) + 2 .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (52) 10 ( x , t ) = ϱ 1 ( coth ( ζ ) 1 ) 2 coth ( ζ ) + 2 × exp ( i φ ) ,

      where

      (53) ζ = x ( 4 ρ 5 2 ρ 2 ρ 1 ρ 2 ρ 1 ρ 3 2 + ρ 5 2 ρ 4 + 2 ρ 5 ρ 2 ρ 3 ) 4 ρ 5 2 ρ 1 2 ρ 3 2 ρ 1 2 + 2 ρ 3 ρ 5 ρ 1 ρ 5 2 t , φ = ρ 3 x ρ 5 ( 4 ρ 4 ρ 1 4 ρ 2 ) ρ 5 3 4 ρ 2 ρ 1 + ρ 4 4 ρ 3 ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( 4 ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 t .

      Figure 9 demonstrates the contour plots of solution (52) by taking ρ 1 = 0.1 , ρ 2 = 0.2 , ρ 3 = 0.3 , ρ 4 = 0.4 , ρ 5 = 0.4 along with ϱ 1 = 1 , = 0.8 .

Figure 1 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 1
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{1}(x,t)
                     
                  .
Figure 1

Contour plots of 1 ( x , t ) .

Figure 2 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 2
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{2}(x,t)
                     
                  .
Figure 2

Contour plots of 2 ( x , t ) .

Figure 3 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 3
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{3}(x,t)
                     
                  .
Figure 3

Contour plots of 3 ( x , t ) .

Figure 4 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 4
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{4}(x,t)
                     
                  .
Figure 4

Contour plots of 4 ( x , t ) .

Figure 5 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 5
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{5}(x,t)
                     
                  .
Figure 5

Contour plots of 5 ( x , t ) .

  1. Taking [ r 1 , r 2 , r 3 , r 4 ] = [ 2 , 0 , 1 , 1 ] and [ s 1 , s 2 , s 3 , s 4 ] = [ 1 , 0 , 1 , 1 ] in Eq. (12) offers

    (54) ( ζ ) = sech ( ζ ) exp ( ζ ) .

    • – If one takes

      (55) = ρ 3 ρ 5 , θ = ( 4 ρ 4 ρ 1 4 ρ 2 ) ρ 5 3 4 ρ 2 ρ 1 + ρ 4 4 ρ 3 ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( 4 ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 , 0 = ϱ 1 2 , 1 = 0 ,

      where and ϱ 1 are the free parameters, provided ρ 3 ρ 5 > 0 .

    • Incorporating these results into structure (19) allows us to derive

      ( ζ ) = ϱ 1 ( 1 + tanh ( ζ ) ) 2 2 tanh ( ζ ) .

      As a result, an analytical solution to the model Eq. (8) is derived as

      (56) 11 ( x , t ) = ϱ 1 ( 1 + tanh ( ζ ) ) 2 2 tanh ( ζ ) × exp ( i φ ) ,

      where

      (57) ζ = x ( 4 ρ 5 2 ρ 2 ρ 1 ρ 2 ρ 1 ρ 3 2 + ρ 5 2 ρ 4 + 2 ρ 5 ρ 2 ρ 3 ) 4 ρ 5 2 ρ 1 2 ρ 3 2 ρ 1 2 + 2 ρ 3 ρ 5 ρ 1 ρ 5 2 t , φ = ρ 3 ρ 5 x ( 4 ρ 4 ρ 1 4 ρ 2 ) ρ 5 3 4 ρ 2 ρ 1 + ρ 4 4 ρ 3 ρ 5 2 ρ 3 2 ( ρ 4 ρ 1 + ρ 2 ) ρ 5 + ρ 2 ρ 1 ρ 3 3 ( 4 ρ 1 2 + 1 ) ρ 5 3 2 ρ 3 ρ 5 2 ρ 1 + ρ 3 2 ρ 1 2 ρ 5 t .

Figure 6 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 6
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{6}(x,t)
                     
                  .
Figure 6

Contour plots of 6 ( x , t ) .

Figure 7 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 7
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{7}(x,t)
                     
                  .
Figure 7

Contour plots of 7 ( x , t ) .

Figure 8 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 9
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{9}(x,t)
                     
                  .
Figure 8

Contour plots of 9 ( x , t ) .

Figure 9 
               Contour plots of 
                     
                        
                        
                           
                              
                                 ℰ
                              
                              
                                 10
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal E} }}}_{10}(x,t)
                     
                  .
Figure 9

Contour plots of 10 ( x , t ) .

5 Bifurcation, and chaos analysis of the model

To elaborate further, we utilize the Galilean transformation method to convert the given differential equation, which can be found in Eq. (17), into a system of equations presented as

(58) d ( ζ ) d ζ = P ( ζ ) , d P ( ζ ) d ζ = 1 Λ 1 ( Λ 2 ( ζ ) + Λ 3 3 ( ζ ) ) ,

where Λ 1 = η 2 ( ε 1 4 ζ 2 ) , Λ 2 = ( 2 + ε 1 2 + ζ 3 ) , and Λ 3 = ε 2 ( > 0 ) .

The primary focus of our research is to undertake a detailed examination of the system’s bifurcation behavior, which involves the investigation of phase portraits associated with the equation presented in Eq. (17). To this end, the Hamiltonian function of Eq. (58) is determined as

(59) H ( , P ) = P 2 2 + 1 Λ 1 Λ 2 2 2 + Λ 3 4 4 = h ,

where h is called the Hamiltonian constant.

From (59), we arrive at the following system:

(60) P = 0 , Λ 2 ( ζ ) + Λ 3 3 ( ζ ) = 0 .

The resulted system in (60) possesses these three equilibrium points

1 = ( 0 , 0 ) , 2 = Λ 2 Λ 3 , 0 , and 3 = Λ 2 Λ 3 , 0 . 6

The Jacobian matrix of the system (58) has a determinant equal to

D ( , P ) = 0 1 1 Λ 1 ( Λ 2 + 3 Λ 3 2 ) 0 = 1 Λ 1 ( Λ 2 + 3 Λ 3 2 ) .

Now, any of the following scenarios may occur

  • ( , P ) is saddle point, if D ( , P ) < 0 ;

  • ( , P ) is center point, if D ( , P ) > 0 ;

  • ( , P ) is cuspid point, if D ( , P ) = 0 ;

Now, we have

i. If Λ 1 , Λ 2 > 0 holds, then ( 0 , 0 ) the only equilibrium point of (58), and

D ( 0 , 0 ) = Λ 2 Λ 1 > 0 .

As shown in Figure 10 (left), the equilibrium point of ( 0 , 0 ) is a center point for (58).

Figure 10 
               Dynamics of (58) bifurcations when: (Left) 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.6
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.1
                        
                        {\Lambda }_{1}=0.6,{\Lambda }_{2}=0.1
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.2
                        
                        {\Lambda }_{3}=0.2
                     
                  . The only equilibrium point is 
                     
                        
                        
                           
                              
                                 ℳ
                              
                              
                                 1
                              
                           
                           =
                           
                              (
                              
                                 0
                                 ,
                                 0
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal M} }}}_{1}=\left(0,0)
                     
                  . (Right) 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.3
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           −
                           0.3
                        
                        {\Lambda }_{1}=0.3,{\Lambda }_{2}=-0.3
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.9
                        
                        {\Lambda }_{3}=0.9
                     
                  . The equilibrium points are 
                     
                        
                        
                           
                              
                                 ℳ
                              
                              
                                 1
                              
                           
                           =
                           
                              (
                              
                                 0
                                 ,
                                 0
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal M} }}}_{1}=\left(0,0)
                     
                  , and 
                     
                        
                        
                           
                              
                                 ℳ
                              
                              
                                 2
                                 ,
                                 3
                              
                           
                           ≈
                           
                              (
                              
                                 ±
                                 0.577
                                 ,
                                 0
                              
                              )
                           
                        
                        {{\mathcal{ {\mathcal M} }}}_{2,3}\approx (\pm 0.577,0)
                     
                  .
Figure 10

Dynamics of (58) bifurcations when: (Left) Λ 1 = 0.6 , Λ 2 = 0.1 , and Λ 3 = 0.2 . The only equilibrium point is 1 = ( 0 , 0 ) . (Right) Λ 1 = 0.3 , Λ 2 = 0.3 , and Λ 3 = 0.9 . The equilibrium points are 1 = ( 0 , 0 ) , and 2 , 3 ( ± 0.577 , 0 ) .

ii. The case Λ 1 > 0 , Λ 2 < 0 : in this case, System (58) possesses three equilibrium points. At 1 = ( 0 , 0 ) , we have

D ( 0 , 0 ) = Λ 2 Λ 1 < 0 ,

which shows ( 0 , 0 ) is a saddle point of (58), as plotted in Figure 10 (right). Also, from

D ± Λ 2 Λ 3 , 0 = 2 Λ 2 Λ 1 > 0 ,

we conclude that 2 and 3 are the center points of (58).

iii. If Λ 2 = 0 holds, (58) possesses only one equilibrium point at 1 = ( 0 , 0 ) . Furthermore, we obtain

D ( 0 , 0 ) = 0 ,

which means ( 0 , 0 ) is a cuspid point of (58).

To analyze the chaotic behavior of (58), let us consider the dynamics of the following Duffing oscillatory system [29]:

(61) d ( ζ ) d ζ = P ( ζ ) , d P ( ζ ) d ζ = 1 Λ 1 ( Λ 2 ( ζ ) + Λ 3 3 ( ζ ) ) + σ cos ( Ω t ) .

In this system, the perturbation is characterized by two parameters: σ , which measures its amplitude, and Ω , which represents its frequency.

To study the potential for chaotic behavior in the system described by (61), we introduce these parameters into the system, as shown in Figures 11, 12, 13, 14, 15.

Figure 11 
               Dynamics of (61) when 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.7
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.2
                        
                        {\Lambda }_{1}=0.7,{\Lambda }_{2}=0.2
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.9
                        
                        {\Lambda }_{3}=0.9
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           1
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=1
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           −
                           1
                        
                        {\mathcal{P}}\left(0)=-1
                     
                   and 
                     
                        
                        
                           σ
                           =
                           1.3
                           ,
                           Ω
                           =
                           1.5
                        
                        \sigma =1.3,\Omega =1.5
                     
                  .
Figure 11

Dynamics of (61) when Λ 1 = 0.7 , Λ 2 = 0.2 , and Λ 3 = 0.9 , using ( 0 ) = 1 and P ( 0 ) = 1 and σ = 1.3 , Ω = 1.5 .

Figure 12 
               Dynamics of (61) when 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.7
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.2
                        
                        {\Lambda }_{1}=0.7,{\Lambda }_{2}=0.2
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.9
                        
                        {\Lambda }_{3}=0.9
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           −
                           1
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=-1
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           −
                           1
                        
                        {\mathcal{P}}\left(0)=-1
                     
                   and 
                     
                        
                        
                           σ
                           =
                           1.3
                           ,
                           Ω
                           =
                           1.5
                        
                        \sigma =1.3,\Omega =1.5
                     
                  .
Figure 12

Dynamics of (61) when Λ 1 = 0.7 , Λ 2 = 0.2 , and Λ 3 = 0.9 , using ( 0 ) = 1 and P ( 0 ) = 1 and σ = 1.3 , Ω = 1.5 .

Figure 13 
               Dynamics of (61) when 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.7
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.2
                        
                        {\Lambda }_{1}=0.7,{\Lambda }_{2}=0.2
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.9
                        
                        {\Lambda }_{3}=0.9
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.75
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=0.75
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.75
                        
                        {\mathcal{P}}\left(0)=0.75
                     
                   and 
                     
                        
                        
                           σ
                           =
                           1.3
                           ,
                           Ω
                           =
                           1.5
                        
                        \sigma =1.3,\Omega =1.5
                     
                  .
Figure 13

Dynamics of (61) when Λ 1 = 0.7 , Λ 2 = 0.2 , and Λ 3 = 0.9 , using ( 0 ) = 0.75 and P ( 0 ) = 0.75 and σ = 1.3 , Ω = 1.5 .

Figure 14 
               Dynamics of (61) when 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.6
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           −
                           0.8
                        
                        {\Lambda }_{1}=0.6,{\Lambda }_{2}=-0.8
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.3
                        
                        {\Lambda }_{3}=0.3
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.75
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=0.75
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.75
                        
                        {\mathcal{P}}\left(0)=0.75
                     
                   and 
                     
                        
                        
                           σ
                           =
                           0.9
                           ,
                           Ω
                           =
                           0.6
                        
                        \sigma =0.9,\Omega =0.6
                     
                  .
Figure 14

Dynamics of (61) when Λ 1 = 0.6 , Λ 2 = 0.8 , and Λ 3 = 0.3 , using ( 0 ) = 0.75 and P ( 0 ) = 0.75 and σ = 0.9 , Ω = 0.6 .

Figure 15 
               Dynamics of (61) when 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           1.6
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           −
                           0.2
                        
                        {\Lambda }_{1}=1.6,{\Lambda }_{2}=-0.2
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.6
                        
                        {\Lambda }_{3}=0.6
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.5
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=0.5
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           1.3
                        
                        {\mathcal{P}}\left(0)=1.3
                     
                   and 
                     
                        
                        
                           σ
                           =
                           2.6
                           ,
                           Ω
                           =
                           3.4
                        
                        \sigma =2.6,\Omega =3.4
                     
                  .
Figure 15

Dynamics of (61) when Λ 1 = 1.6 , Λ 2 = 0.2 , and Λ 3 = 0.6 , using ( 0 ) = 0.5 and P ( 0 ) = 1.3 and σ = 2.6 , Ω = 3.4 .

Figures 16 and 17 display the bifurcation diagram of System (61), which is presented in terms of two variables, σ and Ω . The model’s chaotic attractors emerged naturally from these figures.

Figure 16 
               Bifurcation of System (61) in respect to 
                     
                        
                        
                           σ
                        
                        \sigma 
                     
                   the case 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.7
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.3
                        
                        {\Lambda }_{1}=0.7,{\Lambda }_{2}=0.3
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.4
                        
                        {\Lambda }_{3}=0.4
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.2
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=0.2
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.7
                        
                        {\mathcal{P}}\left(0)=0.7
                     
                   and 
                     
                        
                        
                           Ω
                           =
                           0.3
                        
                        \Omega =0.3
                     
                  .
Figure 16

Bifurcation of System (61) in respect to σ the case Λ 1 = 0.7 , Λ 2 = 0.3 , and Λ 3 = 0.4 , using ( 0 ) = 0.2 and P ( 0 ) = 0.7 and Ω = 0.3 .

Figure 17 
               Bifurcation of System (61) in respect to 
                     
                        
                        
                           Ω
                        
                        \Omega 
                     
                   the case 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 1
                              
                           
                           =
                           0.7
                           ,
                           
                              
                                 Λ
                              
                              
                                 2
                              
                           
                           =
                           0.3
                        
                        {\Lambda }_{1}=0.7,{\Lambda }_{2}=0.3
                     
                  , and 
                     
                        
                        
                           
                              
                                 Λ
                              
                              
                                 3
                              
                           
                           =
                           0.4
                        
                        {\Lambda }_{3}=0.4
                     
                  , using 
                     
                        
                        
                           ℰ
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.2
                        
                        {\mathcal{ {\mathcal E} }}\left(0)=0.2
                     
                   and 
                     
                        
                        
                           P
                           
                              (
                              
                                 0
                              
                              )
                           
                           =
                           0.7
                        
                        {\mathcal{P}}\left(0)=0.7
                     
                   and 
                     
                        
                        
                           σ
                           =
                           0.8
                        
                        \sigma =0.8
                     
                  .
Figure 17

Bifurcation of System (61) in respect to Ω the case Λ 1 = 0.7 , Λ 2 = 0.3 , and Λ 3 = 0.4 , using ( 0 ) = 0.2 and P ( 0 ) = 0.7 and σ = 0.8 .

6 Conclusion

This study primarily delves into discovering novel findings concerning the perturbed nonlinear Schrödinger equation, a significant instrument for analyzing pulse propagation in optical fibers. Our study presents an efficient approach that unveils various types of optical wave solutions within monomode optical fibers. These include trigonometric, singular, bright, and dark solitons, forming distinct families. These results exhibit exceptional qualities, such as periodicity or singular amplitude profiles. Additionally, these phenomena are brought about by particular physical processes. To convey our findings effectively, we have employed suitable parameter values and created multiple plots. These visual depictions clearly showcase the dynamic characteristics of the obtained solutions, including their modulus, real, and imaginary components. Examining these graphs provides a deep understandings of the complex soliton behaviors within the model. These visualizations also demonstrate the effectiveness of our proposed method. We acknowledge that the entire computations and illustrations presented in this study have been efficiently executed using Mathematica. The findings presented in this study have the potential to make a significant impact on the advancement of pulse propagation in optical fibers, opening doors for future developments and practical applications.

  1. Funding information: The author states no funding involved.

  2. Author contributions: The author has accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The author states no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-03-16
Revised: 2024-06-07
Accepted: 2024-06-10
Published Online: 2024-08-01

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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