Dynamics of new truncated M-fractional derivative wave structures to the nonlinear Zhiber-Shabat equation arising in variety of fields

1. Introduction

Nonlinearity in nature is fascinating, and many researchers see nonlinear science as the most promising field for enhancing our understanding of the universe [1,2]. The mathematical characterization of complex systems with time-varying parameters necessitates the examination of a variety of nonlinear partial differential equations (NLPDEs). Since the mid-18th century, scholars have endeavored to simplify intricate physical phenomena by employing NLPDEs. Researchers frequently use NLPDEs to reconstruct various dynamic phenomena [3]. As a result, the study of NLPDEs has continued to attract a substantial quantity of attention in recent years. To explore the exact solutions to nonlinear equations is a critical component of the investigation of nonlinear physical phenomena [4]. The significance of obtaining exact solutions for nonlinear equations is paramount, as these solutions are essential for numerical solution verification and stability analysis. A mathematical structure must be used to express the physical characteristics of any nonlinear physical system in order to obtain an exact solution. This mathematical structure is frequently represented by NLPDEs. Therefore, the investigation of NLPDEs is essential; these equations provide a fundamental framework that facilitates the highly precise understanding of a diverse array of nonlinear physical phenomena. NLPDEs are basic mathematical models that help us understand many different types of phenomena in physics, photonics, optics, fluid dynamics and nonlinear fiber optics. They are used in fields like plasma physics and nonlinear optics [5].

The soliton theory has garnered significant attention in trial designs due to its function as a research domain in the disciplines of media transmission, numerical physical science, designing, and various nonlinear sciences [6]. Specifically, the current period has seen a concentration of optical solitons. Optical solitons are waves that possess the ability to generate waves without dispersing over a significant distance, i.e., they maintain their shape over a significant distance [7, 8]. The basic role of solitons in the media communications society underscores their significance in nonlinear optics. In the system of single wave-based interchanges, optical pulse generators, fiber-optic amplifiers, and many others, solitons models are extensively useful. Examining the motion of solitons through nonlinear optical fibers, extreme laser radiation into plasmas, and the hypothesis of optical solitons are among the most interesting topics [9]. The scholars have adopted a variety of nonlinear models to predict and comprehend the nature of soliton waves.

Researchers have developed a number of approaches and methods to deal with these problems and get solutions in a wide range of systems. Among the most well-known and widely used techniques developed recently are: the Darboux transformation [10], the enhanced modified extended tanh expansion method [11], the advanced F-expansion function approach [12], the truncated Painlevé approach [13], the modified F-expansion method [14], the modified simple equation method [15], the iterative transform method [16], the Lie classical approach [17], the new sub equation approach [18], the Adomian decomposition technique [19], the simplest equation technique [20], the $\tan (\frac{\varphi }{2})$ technique [21], the multiple exp-function approach [22], the Bernoulli$(\frac{G^{\prime }}{G})$expansion method [23], $(\frac{G^{\prime }}{G})$-expansion method [24], the tanh-coth method [25], the inverse scattering approach [26], the Bäcklund transformation [27], the Riccati equation mapping technique [28], the bifurcation analysis [29, 30] and so on [31–36].

Moreover, the model under consideration has not been investigated using these novel sophisticated methodologies in the existing literature. So, this paper has the aim to study various kinds of soliton solutions of the M-fractional nonlinear ZSE by applying the new integration methods known as modified generalized Riccati equation mapping method (MGREMM) [37], KMM [38] and multivariate generalized exponential rational integral function method (MGERIFM) [39]. These methods can be employed to obtain novel and exact solutions, including exponential, trigonometric, and rational functions. The proposed solutions will markedly improve the comprehension of diverse physical processes in associated domains.

The rest of the article is organized as follows: The definition of the new truncated M-fractional derivative of $\alpha$ is presented in Section 2. The applications of methods together with graphs are presented in Section 3. The conclusion obtained in this paper is reported in Section 4.

2. Preliminaries

In order to explain the complex nonlinear systems and provide a more accurate representation of system dynamics, fractional derivatives are a more sophisticated approach. Compared to their integer-order counterparts, the fractional models demonstrate superior precision and accuracy, closely aligning with the experimental data. Additionally, the system's overall efficacy is improved by the implementation of sophisticated optimization and control strategies, which are made possible by the fractional form. The truncated M-fractional derivatives, a novel idea of fractional order derivatives, is presented contemporaneously with the development of this process. Scholars have given these fractional-order derivatives a lot of attention.

3. Applications

In this section, we apply the powerful methods such as MGREMM, KMM, and MGERIFM to obtain desired solutions for the nonlinear ZSE [41–43] in truncated M-fractional derivative given as:

In this work, we study the exact solutions of the proposed model using the suggested techniques with truncated M-fractional derivatives. First of all, let's consider the following transformation as:

Furthermore, for solving the Eq.(4), the following assumption is used as:

Manipulating Eq.(5) into Eq.(4), we get

Now, applying the homogeneous balance principle between the terms $v^3$ and $v{v}^{\prime \prime }$ in Eq.(6) gives $n=2$.

3.1. Application of the MGREMM

The solution for MGREMM [37] is expressed as:

7

$$\nu (\xi )={\varphi }_0+\sum _{r=1}^n{\varphi }_r{\Omega }^r(\xi )+\sum _{r=1}^n{\psi }_r(\xi ){(\frac{{\Omega }^{\prime }(\xi )}{\Omega (\xi )})}^r.$$

For $n=2$, we get

8

$$\nu (\xi )={\varphi }_0+{\varphi }_1\Omega (\xi )+{\varphi }_2(\Omega (\xi ){)}^2+{\psi }_1(\xi )\frac{{\Omega }^{\prime }(\xi )}{\Omega (\xi )}+{\psi }_2(\xi ){\left(\frac{{\Omega }^{\prime }(\xi )}{\Omega (\xi )}\right)}^2,$$

and ${\Omega }^{\prime }(\xi )={\delta }_0+{\delta }_1\Omega (\xi )+{\delta }_2\Omega (\xi {)}^2$. On putting Eq.(8) into Eq.(6). the general solutions are given as:

(I): When $\Delta ={\delta }_1^2-4{\delta }_0{\delta }_2>0,{\delta }_1{\delta }_2\ne 0,{\delta }_0{\delta }_2\ne 0,\text{and}{\varphi }_1=-{\delta }_1{\delta }_2{\psi }_2,p=\frac{2c}{{\psi }_2},q=\frac{c({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)(({\delta }_1^2-10{\delta }_0{\delta }_2){\psi }_2-6{\varphi }_0}{{\psi }_2},)r=\frac{c({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0{)}^2(4{\varphi }_0-({\delta }_1^2-8{\delta }_0{\delta }_2){\psi }_2)}{{\psi }_2},{\varphi }_2=-{\delta }_2^2{\psi }_2,{\psi }_1=-{\delta }_1{\psi }_2$ the soliton solutions are

9

$$u_1(x,t)=ln\left(\frac{{\delta }_0{\delta }_2{\psi }_2{\text{sech}}^2(\frac{\sqrt{\Delta }\xi }{2})({\delta }_1\sqrt{\Delta }sinh(\sqrt{\Delta }\xi )+({\delta }_1^2-2{\delta }_0{\delta }_2)cosh(\sqrt{\Delta }\xi )-{\delta }_1^2+6{\delta }_0{\delta }_2)}{{({\delta }_1+\sqrt{\Delta }tanh(\frac{\sqrt{\Delta }\xi }{2}))}^2}+{\varphi }_0\right),$$

10

$$u_2(x,t)=ln\left(\frac{{\delta }_0{\delta }_2{\psi }_{2}csc{h}^2\left(\frac{\sqrt{A}\xi }{2}\right)\left({\delta }_1\sqrt{\Delta }sinh\right(\sqrt{\Delta }\xi )+({\delta }_1^2-2{\delta }_0{\delta }_2\left)cosh\right(\sqrt{\Delta }\xi )+{\delta }_1^2-6{\delta }_0{\delta }_2)}{{({\delta }_1+\Delta coth(\frac{\sqrt{\Delta }\xi }{2}\left)\right)}^2}+{\varphi }_0\right),$$

11

$$u_3(x,t)=\ln ({\varphi }_0+{\delta }_0{\delta }_2{\psi }_2(1+\frac{i({\delta }_1^2-4{\delta }_0{\delta }_2)}{{\delta }_1^2\sinh \left(\sqrt{\Delta \xi }\right)-2{\delta }_0{\delta }_2\left(\sinh \right(\sqrt{\Delta \xi })+i)+{\delta }_1\sqrt{\Delta }\cosh \left(\sqrt{\Delta \xi }\right)}),)$$

12

$$u_4(x,t)=\ln \left(\frac{{\delta }_1\sqrt{\Delta }({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)\sinh \left(\sqrt{\Delta }\xi \right)+({\delta }_1^2-2{\delta }_0{\delta }_2)({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)\cosh \left(\sqrt{\Delta }\xi \right)+{\delta }_0{\delta }_2({\delta }_1^2-6{\delta }_0{\delta }_2){\psi }_2-2{\varphi }_0}{\left({\delta }_1\sinh \right(\frac{\sqrt{\Delta }\xi }{2})+\sqrt{\Delta }\cosh (\frac{\sqrt{\Delta }\xi }{2}){)}^2}\right),$$

When ${\varphi }_1=\frac{2c{\delta }_1{\delta }_2}{p},q=-\frac{(c{\delta }_1^2+2c{\delta }_0{\delta }_2-3p{\varphi }_0)(2c{\delta }_0{\delta }_2-p{\varphi }_0)}{p},r=\frac{{(p{\varphi }_0-2c{\delta }_0{\delta }_2)}^2(2p{\varphi }_0-c{\delta }_1^2)}{p^2},{\mathit{\psi }}_2=0,{{\varphi }}_2=\frac{2c{\delta }_2^2}{p},{{\psi }}_1=0$, we have the solitary wave solutions:

13

$$u_5(x,t)=\ln \left(\frac{-4c{\delta }_0{\delta }_2+c\Delta {\mathrm{csch}}^2\left(\frac{\sqrt{\Delta }\xi }{2}\right)+2p{\varphi }_0}{2p}\right),$$

14

$$u_6(x,t)=\ln (\frac{1}{p(d+esinh(\sqrt{\Delta }\xi ){)}^2}\left(c{\delta }_1^{2}e\right(e-d\sinh \left(\sqrt{\Delta }\xi \right))-c\sqrt{\Delta }e\sqrt{\Delta (d^2+e^2)}\cosh (\sqrt{\Delta }\xi )p{\varphi }_0(d+e\sinh (\sqrt{\Delta }\xi ){)}^2-c{\delta }_0{\delta }_2(2d^2+e^2\cosh (2\sqrt{\Delta }\xi )+3e^2))),$$

15

$$u_7(x,t)=\ln (-\frac{4c{\delta }_0{\delta }_1{\delta }_2}{{\delta }_{1}p-\sqrt{\Delta }p\tanh \left(\frac{\sqrt{\Delta }\xi }{2}\right)}+\frac{4c{\delta }_0^2{\delta }_2^2(\cosh (\sqrt{\Delta }\xi )+1)}{p{({\delta }_1\cosh \left(\frac{\sqrt{\Delta }\xi }{2}\right)-\sqrt{\Delta }\sinh \left(\frac{\sqrt{\Delta }\xi }{2}\right))}^2}+{\varphi }_0).$$

When ${\varphi }_1=-{{\delta }}_1{{\delta }}_2{{\psi }}_2,q=-\frac{1}{2}p({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)(6{\varphi }_0-({\delta }_1^2-10{\delta }_0{\delta }_2){\psi }_2),r=\frac{1}{2}p{({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)}^2(4{\varphi }_0-({\delta }_1^2-8{\delta }_0{\delta }_2){\psi }_2),c=\frac{p{\psi }_2}{2},{\varphi }_2=-{\delta }_2^2{\psi }_2,{\psi }_1=-{\delta }_1{\psi }_2,$we have the following solutions:

16

$$u_8(x,t)=\ln (\frac{1}{4}{\psi }_2(4{\delta }_0{\delta }_2+\Delta {\mathrm{csch}}^2\left(\frac{\sqrt{\Delta }\xi }{2}\right))+{\varphi }_0),$$

17

$$u_9(x,t)=\ln ({\delta }_0{\delta }_2{\psi }_2-\frac{\Delta {\psi }_2}{2+2\sinh \left(\sqrt{\Delta }\xi \right)}+{\varphi }_0),$$

18

$$u_{10}(x,t)=\ln (\frac{1}{4}{\psi }_2(4{\delta }_0{\delta }_2+\Delta {\mathrm{csch}}^2\left(\frac{\sqrt{\Delta }\xi }{2}\right))+{\varphi }_0),$$

(II): When $\Delta ={\delta }_1^2-4{\delta }_0{\delta }_2<0,{\delta }_1{\delta }_2\ne 0,{\delta }_0{\delta }_2\ne 0,\text{and}{\varphi }_1=-{\delta }_1{\delta }_2{\psi }_2,q=-\frac{1}{2}p({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)(6{\varphi }_0-({\delta }_1^2-10{\delta }_0{\delta }_2){\psi }_2),r=\frac{1}{2}p{({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)}^2(4{\varphi }_0-({\delta }_1^2-8{\delta }_0{\delta }_2){\psi }_2),c=\frac{p{\psi }_2}{2},{\varphi }_2=-{\delta }_2^2{\psi }_2,{\psi }_1=-{\delta }_1{\psi }_2$, we have the following periodic solutions:

19

$$u_{11}(x,t)=ln\left(\frac{{\varphi }_0{\left(\sqrt{-\Delta }tan\right(\frac{\sqrt{-\Delta }\epsilon }{2})-{\delta }_1)}^2}{4 {\delta }_0{\delta }_2}+\frac{{\delta }_1{\varphi }_0\left(\sqrt{-\Delta }tan\right(\frac{\sqrt{-\Delta }\epsilon }{2})-{\delta }_1)}{2{\widehat{\delta }}_0{\delta }_2}+{\varphi }_0\right),,$$

20

$$u_{12}(x,t)=\ln (\frac{{\delta }_0{\delta }_2{\psi }_2{\csc }^2\left(\frac{\sqrt{-\Delta }\xi }{2}\right)({\delta }_1\sqrt{-\Delta }\sin (\sqrt{-\Delta }\xi )-{\delta }_1^2+6{\delta }_0{\delta }_2-\Delta \cos (\sqrt{-\Delta }\xi \left)\right)}{{({\delta }_1+\sqrt{-\Delta }\cot \left(\frac{\sqrt{-\Delta }\xi }{2}\right))}^2}+{\varphi }_0),$$

21

$$u_{13}(x,t)=\ln (\frac{{\delta }_0{\delta }_1{\delta }_2\sqrt{-\Delta }{\psi }_2\cosh \left(\sqrt{\Delta }\xi \right)+{\delta }_1\sqrt{-\Delta }{\varphi }_0\cos \left(\sqrt{-\Delta }\xi \right)+\sin \left(\sqrt{-\Delta }\xi \right)}{({\delta }_1^2-2{\delta }_0{\delta }_2)\sin \left(\sqrt{-\Delta }\xi \right)+{\delta }_1\sqrt{-\Delta }\cos \left(\sqrt{-\Delta }\xi \right)-2{\delta }_0{\delta }_2}({\delta }_1^2-2{\delta }_0{\delta }_2)({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)+{\delta }_0{\delta }_2(({\delta }_1^2-6{\delta }_0{\delta }_2){\psi }_2-2{\varphi }_0)),$$

22

$$u_{14}(x,t)=\ln \left(\frac{{\delta }_1\sqrt{\Delta }({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)\sin \left(\sqrt{-\Delta \xi }\right)+({\delta }_1^2-2{\delta }_0{\delta }_2)({\delta }_0{\delta }_2{\psi }_2+{\varphi }_0)\cosh \left(\sqrt{\Delta }\xi \right)+{\delta }_0{\delta }_2(({\delta }_1^2-6{\delta }_0{\delta }_2){\psi }_2-2{\varphi }_0)}{{({\delta }_1\sin \left(\frac{\sqrt{-\Delta }\xi }{2}\right)+\sqrt{\Delta }\cosh \left(\frac{\sqrt{\Delta }\xi }{2}\right))}^2}\right),$$

When ${\varphi }_1=\frac{2c{\delta }_1{\delta }_2}{p},q=-\frac{(c{\delta }_1^2+2c{\delta }_0{\delta }_2-3p{\varphi }_0)(2c{\delta }_0{\delta }_2-p{\varphi }_0)}{p},r=\frac{{(p{\varphi }_0-2c{\delta }_0{\delta }_2)}^2(2p{\varphi }_0-c{\delta }_1^2)}{p^2},{\psi }_2=0,{\varphi }_2=\frac{2c{\delta }_2^2}{p},{\psi }_1=0,$ we have:

23

$$u_{15}(x,t)=\ln (-\frac{2c{\delta }_0{\delta }_2}{p}+\frac{c\Delta }{p(\cos (\sqrt{-\Delta }\xi )-1)}+{\varphi }_0),$$

24

$$u_{16}(x,t)=\ln (\frac{1}{p(d+e\text{sin}{\left(\sqrt{-∆}\xi \right))}^2}(-c{\delta }_1^{2}e(d\text{sin}\left(\sqrt{-\Delta }\xi \right)+e)-c\sqrt{-\Delta }e\sqrt{\Delta (d-e)(d+e)}\cos (\sqrt{-\Delta }\xi )+c{\delta }_0{\delta }_2(-2d^2+e^2\cosh (2\sqrt{\Delta }\xi )+3e^2)+\frac{1}{2}p{\varphi }_0(2d^2+4de\sin (\sqrt{-\Delta }\xi )-e^2\cos (2\sqrt{-\Delta }\xi )+e^2))),$$

25

$$u_{17}(x,t)=\ln (-\frac{4c{\delta }_0{\delta }_1{\delta }_2}{{\delta }_{1}p+\sqrt{-\Delta }p\tan \left(\frac{\sqrt{-\Delta }\xi }{2}\right)}+\frac{4c{\delta }_0^2{\delta }_2^2(\cos (\sqrt{-\Delta }\xi )+1)}{p{({\delta }_1\cos \left(\frac{\sqrt{-\Delta }\xi }{2}\right)+\sqrt{-\Delta }\sin \left(\frac{\sqrt{-\Delta }\xi }{2}\right))}^2}+{\varphi }_0),$$

26

$$u_{18}(x,t)=\ln (-\frac{4c{\delta }_0{\delta }_1{\delta }_2}{{\delta }_{1}p-\sqrt{-\Delta }p\cot \left(\frac{\sqrt{-\Delta }\xi }{2}\right)}-\frac{4c{\delta }_0^2{\delta }_2^2(\cos (\sqrt{-\Delta }\xi )-1)}{p{({\delta }_1\sin \left(\frac{\sqrt{-\Delta }\xi }{2}\right)-\sqrt{-\Delta }\cos \left(\frac{\sqrt{-\Delta }\xi }{2}\right))}^2}+{\varphi }_0),$$

where ξ is expressed in Eq.(3). All the obtained solutions are written in the form of Eq.(5), such that u = lnv. We simulate the wave behaviours of the solutions of the governing model.

Fig. 1

Fig. 2

3.2. Application of the KMM

For n = 2, the general solution forms in KMM [38] can be described as

27

$$v(\xi )={\rho }_0+{\rho }_1\Omega (\xi )+{\rho }_2(\Omega (\xi ){)}^2,$$

with

28

$${\Omega }^{\prime }(\xi )=\sqrt{{\gamma }_1\Omega (\xi {)}^4+{\gamma }_2\Omega (\xi {)}^3+{\gamma }_3\Omega (\xi {)}^2+{\gamma }_4\Omega (\xi )+{\gamma }_5}.$$

The following solutions are obtained by employing (27) along with (28) in (6):

Case-1: For ${\gamma }_5=\frac{{(4{\gamma }_1{\gamma }_3-{\gamma }_2^2)}^2}{64{\gamma }_1^3},{\gamma }_4=\frac{{\gamma }_2(4{\gamma }_1{\gamma }_3-{\gamma }_2^2)}{8{\gamma }_1^2}$ offers ${\rho }_2=\frac{2{\gamma }_1{\rho }_1}{{\gamma }_2},p=\frac{c{\gamma }_2}{{\rho }_1},r=\frac{c(8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1){({\gamma }_2^2{\rho }_1+4{\gamma }_1{\gamma }_2{\rho }_0-4{\gamma }_1{\gamma }_3{\rho }_1)}^2}{64{\gamma }_1^3{\gamma }_2{\rho }_1},q=-\frac{c(3{\gamma }_2{\rho }_0-{\gamma }_3{\rho }_1)({\gamma }_2^2{\rho }_1+4{\gamma }_1{\gamma }_2{\rho }_0-4{\gamma }_1{\gamma }_3{\rho }_1)}{4{\gamma }_1{\gamma }_2{\rho }_1}$ As a result we have the following solutions:

When ${\gamma }_1>0$ and $8{\gamma }_1{\gamma }_3-3{\gamma }_2^2<0$, we get the solutions as follows:

The dark type soliton solution

29

$$u_1(x,t)=\ln \left(\frac{\frac{(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3){\rho }_1{\tanh }^2\left(\frac{\sqrt{3{\gamma }_2^2-8{\gamma }_1{\gamma }_3}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4\sqrt{{\gamma }_1\epsilon }}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}{8{\gamma }_1}\right).$$

The singular soliton solution

30

$$u_2=\ln (-\frac{{\gamma }_3{\rho }_1{\coth }^2\left(\frac{\sqrt{{\gamma }_1(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3)}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4{\gamma }_1\epsilon }\right)}{{\gamma }_2}+\frac{{\gamma }_2{\rho }_1(3{\coth }^2\left(\frac{\sqrt{{\gamma }_1(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3)}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4{\gamma }_1\epsilon }\right)-1)}{8{\gamma }_1}+{\rho }_0).$$

When ${\gamma }_1>0$ and $8{\gamma }_1{\gamma }_3-3{\gamma }_2^2>0$, we get the periodic solutions as follows:

31

$$u_3(x,t)=\ln \left(\frac{\frac{(8{\gamma }_1{\gamma }_3-3{\gamma }_2^2){\rho }_1{\tan }^2\left(\frac{\sqrt{8{\gamma }_1{\gamma }_3-3{\gamma }_2^2}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4\sqrt{{\gamma }_1\epsilon }}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}{8{\gamma }_1}\right),$$

32

$$u_4(x,t)=\ln (\frac{{\gamma }_3{\rho }_1{\cot }^2\left(\frac{\sqrt{{\gamma }_1(8{\gamma }_1{\gamma }_3-3{\gamma }_2^2)}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4{\gamma }_1\epsilon }\right)}{{\gamma }_2}-\frac{{\gamma }_2{\rho }_1(3{\cot }^2\left(\frac{\sqrt{{\gamma }_1(8{\gamma }_1{\gamma }_3-3{\gamma }_2^2)}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{4{\gamma }_1\epsilon }\right)+1)}{8{\gamma }_1}+{\rho }_0),$$

Case-2 For ${\gamma }_5=\frac{{\gamma }_2^2(16{\gamma }_1{\gamma }_3-5{\gamma }_2^2)}{256{\gamma }_1^3},{\gamma }_4=\frac{{\gamma }_2(4{\gamma }_1{\gamma }_3-{\gamma }_2^2)}{8{\gamma }_1^2}$ offers ${\rho }_2=\frac{2{\gamma }_1{\rho }_1}{{\gamma }_2},p=\frac{c{\gamma }_2}{{\rho }_1},r=-\frac{c{(8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1)}^2(-5{\gamma }_2^2{\rho }_1-8{\gamma }_1{\gamma }_2{\rho }_0+16{\gamma }_1{\gamma }_3{\rho }_1)}{256{\gamma }_1^3{\rho }_1},q=\frac{c(8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1)(-9{\gamma }_2^2{\rho }_1-24{\gamma }_1{\gamma }_2{\rho }_0+32{\gamma }_1{\gamma }_3{\rho }_1)}{64{\gamma }_1^2{\rho }_1}$.As a result we have the following solutions:

When ${\gamma }_1<0$ and $8{\gamma }_1{\gamma }_3-3{\gamma }_2^2<0$, we get: The bright soliton solution

33

$$u_5(x,t)=\ln \left(\frac{\frac{2(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3){\rho }_1{\mathrm{sech}}^2\left(\frac{\sqrt{8{\gamma }_1{\gamma }_3-3{\gamma }_2^2}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{2\sqrt{2}\sqrt{{\gamma }_1\epsilon }}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}{8{\gamma }_1}\right).$$

When ${\gamma }_1>0$ and $8{\gamma }_1{\gamma }_3-3{\gamma }_2^2<0$, we get the hyperbolic solution as follows:

34

$$u_6(x,t)=\ln \left(\frac{{\displaystyle \frac{2(8{\gamma }_1{\gamma }_3-3{\gamma }_2^2){\rho }_1{\mathrm{csch}}^2\left(\frac{\sqrt{8{\gamma }_1{\gamma }_3-3{\gamma }_2^2}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{2\sqrt{2}\sqrt{{\gamma }_1\epsilon }}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}}{8\gamma 1}\right).$$

When ${\gamma }_1>0$ and $8{\gamma }_1{\gamma }_3-3{\gamma }_2^2<0$, we get the periodic solutions as follows:

35

$$u_7(x,t)=\ln \left(\frac{\frac{2(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3){\rho }_1{\sec }^2\left(\frac{\sqrt{3{\gamma }_2^2-8{\gamma }_1{\gamma }_3}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{2\sqrt{2}\sqrt{{\gamma }_1\epsilon }}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}{8{\gamma }_1}\right),$$

36

$$u_8(x,t)=\ln \left(\frac{\frac{2(3{\gamma }_2^2-8{\gamma }_1{\gamma }_3){\rho }_1{\csc }^2\left(\frac{\sqrt{3{\gamma }_2^2-8{\gamma }_1{\gamma }_3}\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{2\sqrt{2}\sqrt{{\gamma }_1}}\right)}{{\gamma }_2}+8{\gamma }_1{\rho }_0-{\gamma }_2{\rho }_1}{8{\gamma }_1}\right),$$

Case-3 For ${\gamma }_2=0,{\gamma }_4=0,{\gamma }_5=0$ and ${\gamma }_3>0$ offers ${\rho }_1=0,c=\frac{p{\rho }_2}{2{\gamma }_1},r=\frac{2({\gamma }_{1}p{\rho }_0^3-{\gamma }_{3}p{\rho }_0^2{\rho }_2)}{{\gamma }_1},q=\frac{2{\gamma }_{3}p{\rho }_0{\rho }_2-3{\gamma }_{1}p{\rho }_0^2}{{\gamma }_1}$. As a result we have the following exponential solution:

37

$$u_9(x,t)=\ln (\frac{16{\gamma }_3^2{\rho }^2{\rho }_2{e}^{\frac{2\sqrt{{\gamma }_3}\Gamma (\beta +1)(x^{\epsilon }-\frac{p{\rho }_2{t}^{\epsilon }}{2{\gamma }_1})}{\epsilon }}}{({\gamma }_1{\gamma }_3-4{\rho }^2{e}^{\frac{2\sqrt{{\gamma }_3}\Gamma (\beta +1)(x^{\epsilon }-\frac{p{\rho }_2{t}^{\epsilon }}{2{\gamma }_1})}{\epsilon }}){}^2}+{\rho }_0).$$

Taking ${\gamma }_1=-\frac{4{\rho }^2}{{\gamma }_3}$ in Eq. (37), we have bright soliton solution

38

$$u_{10}(x,t)=\ln (\frac{{\gamma }_3^2{\rho }_2{\mathrm{sech}}^2\left(\frac{\sqrt{{\gamma }_3\Gamma }\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{\epsilon }\right)}{4{\rho }^2}+{\rho }_0).$$

Similarly, taking ${\gamma }_1=\frac{4{\rho }^2}{{\gamma }_3}$ Eq. (37), we have

39

$$u_{11}(x,t)=\ln (\frac{{\gamma }_3^2{\rho }_2{\mathrm{csch}}^2\left(\frac{\sqrt{{\gamma }_3\Gamma }\Gamma (\beta +1)(x^{\epsilon }-c{t}^{\epsilon })}{\epsilon }\right)}{4{\rho }^2}+{\rho }_0).$$

Remark: All the obtained solutions are written in the form of Eq. (5), such that $u=\ln v$.

Fig. 3

Fig. 4

Fig. 5