New Abundant Analytical Solitons to the Fractional Mathematical Physics Model via Three Distinct Schemes

Abstract

New types of truncated M-fractional wave solitons to the simplified Modified Camassa–Holm model, a mathematical physics model, are obtained. This model is used to explain the unidirectional propagation of shallow water waves. The required solutions are obtained by utilizing the simplest equation, the Sardar subequation, and the generalized Kudryashov schemes. The obtained results consist of the dark, singular, periodic, dark-bright, and many other analytical solitons. Dynamical behaviors of some obtained solutions are represented by two-dimensional (2D), three-dimensional (3D), and Contour graphs. An effect of fractional derivative is shown graphically. The results are newer than the existing results of the governing equation. Obtained solutions have much importance in the various areas of applied science as well as engineering. We concluded that the utilized methods are helpful and applicable for other partial fractional equations in applied science and engineering.

1. Introduction

Soliton theory, which originated in the sixties and seventies of the past century, has played a crucial rule in many areas, including plasmas and optical fibers. It has much importance in applied mathematics with various applications in optical fibers, ferromagnetic dynamics, optics, etc. The applications of this theory extend its influence across a wide spectrum of natural sciences, such as telecommunication, engineering, etc. In modern days, phenomena that occur in nature are represented as fractional nonlinear partial differential Equations (FNLPDEs) [1,2].

Obtaining exact solutions to NLPDEs is achieved via various techniques, which encompass methods such as the modified direct algebraic technique [3], the modified Khater scheme [4], the Kudryashov technique [5], novel $(G^{\prime }/G)$-expansion technique [6], extended FAN subequation technique [7,8] and extended mapping scheme [9], the ${\partial }^{-}$-dressing technique [10], Riemann–Hilbert approach [11], the ${\partial }^{-}$-generalized Deift–Zhou nonlinear steepest descent method [12], extended $(G^{\prime }/G)$-expansion technique [13], extended Sinh-Gordon expansion technique [14], modified extended tanh function method [15], Hirota method [16], and generalized Eigenfunction technique [17].

In this research, we consider three simple yet useful methods: the modified simplest equation technique, the Sardar subequation technique, and the generalized Kudryashov technique. Different kinds of models are solved by using these techniques. Modified simplest equation techniques are employed to derive exact solitons of coupled Higgs equation, with Maccari’s model in [18], different solitary wave solitons of Gardner model in [19], and optical soliton solutions of the Radha–Lakshmanan equation in [20]. The Sardar subequation technique has been utilized to obtain the many kinds of wave solitons for the Sawada–Kotera (SK) equation [21], optical solitons for the Fokas–Lenells equation were achieved in [22], and some new kinds of exact soliton solutions for the nonlinear Akbota equation were obtained in [23]. Similarly, the generalized Kudryashov scheme has been instrumental in finding various types of wave solutions, including the bell, dark, kink, and others for the Fokas–Lenelles equation [24], exact solitons for the KdV-Burger model [25], and some general solutions for the Kolmogorov–Petrovskii–Piskunov Equation [26], among other applications.

Our primary mathematical physics model is the nonlinear simplified Modified Camassa–Holm model (SMCHM). Various techniques have been employed to derive different types of exact solitons for this model. The generalized $(G^{\prime }/G)$-expansion technique was used to derive specific types of traveling wave solitons in [27], while the Exp-function technique yielded solitary wave solitons in [28]. Moreover, other forms of exact solutions were obtained by utilizing the Riccati–Bernoulli sub-ODE scheme in [29], while some new kinds of soliton solutions, including the smooth kink, flat kink, and singular periodic kink, were gained by using the extended Kudryashov technique in [30], and different soliton solutions, including the bell-shaped, anti-bell shaped, compacton, and singular bell-shaped solitons, were obtained by applying the $(G^{\prime }/G,1/G)$-expansion method in [31], among many others.

The purpose of this research is to explore novel types of exact wave solitons to the (1+1)-D simplified Modified Camassa–Holm model by using three distinct techniques: the modified simplest equation, the Sardar subequation, and the generalized Kudryashov technique.

This paper is organized into several sections. Section 2 gives an explanation of the equation and its mathematical treatment. In Section 3, we give the details of the modified simplest equation scheme and its use in obtaining wave solitons. Section 4 focuses on the Sardar subequation scheme and its implementation to derive new wave solitons. Section 5 delves into the generalized Kudryashov scheme and its implementation in obtaining wave solutions. In Section 6, we showcase some of the derived results by two-dimensional, three-dimensional, and contour plots. Finally, we present our conclusions in Section 7.

2. The Concerning Equation and Its Mathematical Treatment

The simplified modified Camassa–Holm (SMCH) equation is a nonlinear model that is important for understanding and classifying wave phenomena in ocean engineering and science. The equation is used to model the propagation of shallow water waves that are characterized by nonlinearities and weak dispersion. The simplified MCH, a developed Korteweg–de Vries equation, is an integrable nonlinear dispersive water wave equation with potential applications in diverse fields, such as shallow water waves, liquid drop patterning, water surface waves in channels, tsunamis in coastal regions, oceans, etc. Assume a nonlinear truncated M-fractional simplified modified Camassa–Holm model, which belong to the class of significant equations known as modified $\beta$-equations, as described by Wazwaz [32]:

$$D_{M,t}^{\alpha ,\gamma }g-D_{M,t}^{\alpha ,\gamma }(D_{M,2x}^{2\alpha ,\gamma }g)+(\beta +1)g^2{D}_{M,2x}^{2\alpha ,\gamma }g-\beta D_{M,x}^{\alpha ,\gamma }g{D}_{M,2x}^{2\alpha ,\gamma }g-D_{M,3x}^{3\alpha ,\gamma }g=0.\beta >0.$$

(1)

where $g=g(x,t)$ shows a wave-profile where $\Omega$ and $\theta$ are nonzero parameters. Substituting $\beta =2$ into Equation (1) implies:

$$D_{M,t}^{\alpha ,\gamma }g-D_{M,t}^{\alpha ,\gamma }(D_{M,2x}^{2\alpha ,\gamma }g)+3g^2{D}_{M,2x}^{2\alpha ,\gamma }g-2D_{M,x}^{\alpha ,\gamma }g{D}_{M,2x}^{2\alpha ,\gamma }g-D_{M,3x}^{3\alpha ,\gamma }g=0.$$

(2)

Equation (2) is called the modified Camassa–Holm model. A more simplified form of Equation (2) is shown by [33]:

$$D_{M,t}^{\alpha ,\gamma }g+2\Omega D_{M,x}^{\alpha ,\gamma }g-D_{M,t}^{\alpha ,\gamma }(D_{M,2x}^{2\alpha ,\gamma }g)+\theta g^2{D}_{M,x}^{\alpha ,\gamma }g=0.\Omega \in \Re \beta >0$$

(3)

where

$$D_{M,t}^{\alpha ,{\rm Y}}g(t)=\underset{\tau \to 0}{lim}{\displaystyle \frac{g(t{E}_{{\rm Y}}(\tau t^{1-\alpha }))-g(t)}{\tau }},\alpha \in (0,1],{\rm Y}>0,$$

(4)

in which $E_{{\rm Y}}(.)$ indicates a truncated Mittag–Leffler (ML) profile [34,35].

Consider the following wave transformation:

$$g(x,t)=G(\xi ),\xi ={\displaystyle \frac{1}{\alpha }}\Gamma ({\rm Y}+1)(\mu x^{\alpha }+\delta t^{\alpha }),$$

(5)

Here, $\mu$, and $\delta$ are the constants. Putting Equation (5) in Equation (3), we attain

$$(\delta +2\Omega \mu )G-\delta {\mu }^2{G}^{\prime \prime }+{\displaystyle \frac{\theta \mu }{3}}{G}^3=0.$$

(6)

3. Modified Simplest Equation Technique

Here, there is a basic concept of this scheme given as follows:

• Step 1: Suppose a nonlinear partial differential equation:

  $$Y(g,g^2,g^2{g}_x,g_{xx},g_{xt},\dots )=0,$$

  (7)

  where $g=g(x,t)$ denotes a real wave-profile. Suppose the given relation:

  $$g(x,t)=G(\Theta ),\Theta =x+\lambda t.$$

  (8)
  Inserting Equation (8) in Equation (7), a nonlinear ordinary differential equation is obtained

  $$Z(G,G^2{G}^{\prime },G^{″},...)=0.$$

  (9)
• Step 2: Consider Equation (9) has solution given as:

  $$G(\Theta )=\sum _{s=1}^m{b}_s{\varphi }^s(\Theta ),$$

  (10)

  where $b_s(s=1,2,3,...,m)$ represent the undetermined. A novel profile $\varphi (\Theta )$ satisfies the ODE:

  $${\varphi }^{\prime }(\Theta )={\varphi }^2(\Theta )+\mho ,$$

  (11)

  where ℧ indicates a constant. Notice that Equation (11) has the distinct results based on ℧:
  If $\mho <0$, then:

  $$\varphi (\Theta )=-\sqrt{-\mho }tanh(\sqrt{-\mho }\Theta ),$$

  (12)

  $$\varphi (\Theta )=-\sqrt{-\mho }coth(\sqrt{-\mho }\Theta ),$$

  (13)

  $$\varphi (\Theta )=\sqrt{-\mho }(-tanh(2\sqrt{-\mho }\Theta )\pm isech(2\sqrt{-\mho }\Theta )),$$

  (14)

  $$\varphi (\Theta )=\sqrt{-\mho }(-coth(2\sqrt{-\mho }\Theta )\pm csch(2\sqrt{-\mho }\Theta )),$$

  (15)

  $$\varphi (\Theta )=-{\displaystyle \frac{\sqrt{-\mho }}{2}}(tanh({\displaystyle \frac{\sqrt{-\mho }}{2}}\Theta )+coth({\displaystyle \frac{\sqrt{-\mho }}{2}}\Theta )).$$

  (16)
  If $\mho >0$:

  $$\varphi (\Theta )=\sqrt{\mho }tan(\sqrt{\mho }\Theta ),$$

  (17)

  $$\varphi (\Theta )=-\sqrt{\mho }cot(\sqrt{\mho }\Theta ),$$

  (18)

  $$\varphi (\Theta )=\sqrt{\mho }(tan(2\sqrt{\mho }\Theta )\pm sec(2\sqrt{\mho }\Theta )),$$

  (19)

  $$\varphi (\Theta )=\sqrt{\mho }(-cot(2\sqrt{\mho }\Theta )\pm csc(2\sqrt{\mho }\Theta )),$$

  (20)

  $$\varphi (\Theta )={\displaystyle \frac{\sqrt{\mho }}{2}}(tan({\displaystyle \frac{\sqrt{\mho }}{2}}\Theta )-cot({\displaystyle \frac{\sqrt{\mho }}{2}}\Theta )).$$

  (21)
  If $\mho =0$,

  $$\varphi (\Theta )=-{\displaystyle \frac{1}{\Theta }}.$$

  (22)
• Step 3: Using Equation (10) and Equation (11) in Equation (9). Summing up the coefficients of every ${\varphi }^j$, taking the coefficients of same order equal to 0 to obtain a system involving $b_s$ and $\lambda$. By solving the system, solutions for the undetermined are obtained.

Step 4: Putting Equation (9), where $b_s$ and $\lambda$ are obtained in Equation (10) will result in Equation (7).

Application of MSE Scheme

Equation (10) changes into the given form in our case:

$$G(\xi )=b_0+b_1\psi (\xi ).$$

(23)

Using Equation (23) and Equation (11) in Equation (6), one obtains the given sets. Set 1:

$$\left\{b_0=0,b_1=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}},\delta ={\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}}\right\}.$$

(24)

Case 1: if $\mho <0$:

$$g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{-\mho }tanh(\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(25)

$$g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{-\mho }coth(\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(26)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{-\mho }(-tanh(2\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hfill \\ \hfill \pm isech(2\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(27)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{-\mho }(-coth(2\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hfill \\ \hfill \pm csch(2\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(28)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-{\displaystyle \frac{\sqrt{-\mho }}{2}}(tanh({\displaystyle \frac{\sqrt{-\mho }}{2}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hfill \\ \hfill +coth({\displaystyle \frac{\sqrt{-\mho }}{2}}{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\omega -1}})t^{\alpha })))).\end{array}$$

(29)

Case 2: if $\mho >0$:

$$g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\mho }tan(\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(30)

$$g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{\mho }cot(\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(31)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\mho }(tan(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \hfill \\ \hfill \pm sec(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(32)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\mho }(-cot(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \pm csc(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(33)

$$\begin{array}{c}g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}({\displaystyle \frac{\sqrt{\mho }}{2}}(tan({\displaystyle \frac{\sqrt{\mho }}{2}}{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em} \hfill \\ \hfill -cot({\displaystyle \frac{\sqrt{\mho }}{2}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(34)

Case 3: if $\mho =0$:

$$g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{-\theta }\frac{\Gamma (1+{\rm Y})}{\alpha }(\mu x^{\alpha }-2\mu \Omega t^{\alpha })}}.$$

(35)

Set 2:

$$\left\{b_0=0,b_1={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}},\delta ={\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}}\right\}.$$

(36)

Case 1:if $\mho <0$:

$$g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{-\mho }tanh(\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(37)

$$g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{-\mho }coth(\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(38)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{-\mho }(-tanh(2\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hfill \\ \hfill \pm isech(2\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(39)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{-\mho }(-coth(2\sqrt{-\mho }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hfill \\ \hfill \pm csch(2\sqrt{-\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(40)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-{\displaystyle \frac{\sqrt{-\mho }}{2}}(tanh({\displaystyle \frac{\sqrt{-\mho }}{2}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hfill \\ \hfill +coth({\displaystyle \frac{\sqrt{-\mho }}{2}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(41)

Case 2: if $\mho >0$:

$$g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\mho }tan(\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(42)

$$g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(-\sqrt{\mho }cot(\sqrt{\omega }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))).$$

(43)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\mho }(tan(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \hspace{1em}\hfill \\ \hfill \pm sec(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(44)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}(\sqrt{\omega }(-cot(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em} \hspace{1em}\hfill \\ \hfill \pm csc(2\sqrt{\mho }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(45)

$$\begin{array}{c}g(x,t)={\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{2\theta {\mu }^2\mho -\theta }}}({\displaystyle \frac{\sqrt{\mho }}{2}}(tan({\displaystyle \frac{\sqrt{\mho }}{2}}{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em} \hspace{1em}\hfill \\ \hfill -cot({\displaystyle \frac{\sqrt{\mho }}{2}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{2{\mu }^2\mho -1}})t^{\alpha })))).\end{array}$$

(46)

Case 3: if $\mho =0$:

$$g(x,t)=-{\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{-\theta }\frac{1}{\alpha }\Gamma (1+{\rm Y})(\mu x^{\alpha }-2\mu \Omega t^{\alpha })}}$$

(47)

4. Explanation of Sardar Subequation Technique

Here, there are main steps of this technique [36] by supposing the nonlinear PDE:

$$J(g,g_z,g_{zz},g_{zt},g{g}_{tt},g_{zzt},\dots )=0,$$

(48)

where $g=g(x,t)$ indicates a wave-profile. Substituting a wave relation given as:

$$g(z,t)=G(\zeta ),\zeta =\lambda z+\mu t$$

(49)

results into the nonlinear ODE:

$$Y(G,G^{″},G{G}^{″},G^{\prime }{G}^2,\dots )=0.$$

(50)

Consider the solution of Equation (50) is as follows:

$$G(\zeta )=\sum _{j=0}^m{b}_j{\psi }^j(\zeta ),$$

(51)

where, $\psi (\zeta )$ satisfies the ODE given as

$${\psi }^{\prime }(\zeta )=\sqrt{\sigma +\kappa {\psi }^2(\zeta )+{\psi }^4(\zeta )},$$

(52)

where, $\sigma$ and $\kappa$ denote the constants.

Substituting Equations (51) and (52) in Equation (50). Summing up the coefficients of every power of ${\psi }^j$. Taking them equal to 0 to obtain a system in $b_j$, $\lambda$, and $\mu$. By solving the system, one finds the undetermined values.

• Case 1: when $\kappa$ is positive and $\sigma$ is zero, then:

  $${\psi }_1^{\pm }=\pm \sqrt{\kappa ab}{sech}_{ab}(\sqrt{\kappa }\zeta ),$$

  (53)

  $${\psi }_2^{\pm }=\pm \sqrt{\kappa ab}{csch}_{ab}(\sqrt{\kappa }\zeta ),$$

  (54)

  where ${sech}_{ab}(\zeta )={\displaystyle \frac{2}{a{e}^{\zeta }+b{e}^{-\zeta }}}$, ${csch}_{ab}(\zeta )={\displaystyle \frac{2}{a{e}^{\zeta }-b{e}^{-\zeta }}}$
• Case 2: when $\kappa$ is negative and $\sigma$ is zero, we have

  $${\psi }_3^{\pm }=\pm \sqrt{-\kappa ab}{sec}_{ab}(\sqrt{-\kappa }\zeta ),$$

  (55)

  $${\psi }_4^{\pm }=\pm \sqrt{-\kappa ab}{csc}_{ab}(\sqrt{-\kappa }\zeta ),$$

  (56)

  where ${sec}_{ab}(\zeta )={\displaystyle \frac{2}{a{e}^{\iota \zeta }+b{e}^{-\iota \zeta }}}$, ${csc}_{ab}(\zeta )={\displaystyle \frac{2\iota }{a{e}^{\iota \zeta }-b{e}^{-\iota \zeta }}}$
• Case 3: when $\kappa$ is negative and $\sigma ={\displaystyle \frac{{\kappa }^2}{4}}$, then:

  $${\psi }_5^{\pm }=\pm \sqrt{-{\displaystyle \frac{\kappa }{2}}}{tanh}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}\zeta ),$$

  (57)

  $${\psi }_6^{\pm }=\pm \sqrt{-{\displaystyle \frac{\kappa }{2}}}{coth}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}\zeta ),$$

  (58)

  $${\psi }_7^{\pm }=\pm \sqrt{-{\displaystyle \frac{\kappa }{2}}}({tanh}_{ab}(\sqrt{-2\kappa }\zeta )\pm \iota \sqrt{ab}sec{h}_{ab}(\sqrt{-2\kappa }\zeta )),$$

  (59)

  $${\psi }_8^{\pm }=\pm \sqrt{-{\displaystyle \frac{\kappa }{2}}}({coth}_{ab}(\sqrt{-2\kappa }\zeta )\pm \sqrt{ab}csc{h}_{ab}(\sqrt{-2\kappa }\zeta )),$$

  (60)

  $${\psi }_9^{\pm }=\pm \sqrt{-{\displaystyle \frac{\kappa }{8}}}({tanh}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{8}}}\zeta )+{coth}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{8}}}\zeta )),$$

  (61)

  where ${tanh}_{ab}(\zeta )={\displaystyle \frac{a{e}^{\zeta }-b{e}^{-\zeta }}{a{e}^{\zeta }+b{e}^{-\zeta }}}$, ${coth}_{ab}(\zeta )={\displaystyle \frac{a{e}^{\zeta }+b{e}^{-\zeta }}{a{e}^{\zeta }-b{e}^{-\zeta }}}$
• Case 4: when $\kappa$ is positive and $\sigma ={\displaystyle \frac{{\kappa }^2}{4}}$, we have

  $${\psi }_{10}^{\pm }=\pm \sqrt{{\displaystyle \frac{\kappa }{2}}}{tan}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{2}}}\zeta ),$$

  (62)

  $${\psi }_{11}^{\pm }=\pm \sqrt{{\displaystyle \frac{\kappa }{2}}}{cot}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{2}}}\zeta ),$$

  (63)

  $${\psi }_{12}^{\pm }=\pm \sqrt{{\displaystyle \frac{\kappa }{2}}}({tan}_{ab}(\sqrt{2\kappa }\zeta )\pm \sqrt{ab}{sec}_{ab}(\sqrt{2\kappa }\zeta )),$$

  (64)

  $${\psi }_{13}^{\pm }=\pm \sqrt{{\displaystyle \frac{\kappa }{2}}}({cot}_{ab}(\sqrt{2\kappa }\zeta )\pm \sqrt{ab}{csc}_{ab}(\sqrt{2\kappa }\zeta )),$$

  (65)

  $${\psi }_{14}^{\pm }=\pm \sqrt{{\displaystyle \frac{\kappa }{8}}}({tan}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{8}}}\zeta )+{cot}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{8}}}\zeta )),$$

  (66)

  where ${tan}_{ab}(\zeta )=-\iota {\displaystyle \frac{a{e}^{\iota \zeta }-b{e}^{-\iota \zeta }}{a{e}^{\iota \zeta }+b{e}^{-\iota \zeta }}}$, ${cot}_{ab}(\zeta )=\iota {\displaystyle \frac{a{e}^{\iota \zeta }+b{e}^{-\iota \zeta }}{a{e}^{\iota \zeta }-b{e}^{-\iota \zeta }}}$

New Wave Solitons via the Sardar Subequation Method

For $m=1$, Equation (51) changes into:

$$G(\zeta )=b_0+b_1\psi (\zeta ).$$

(67)

Using Equation (67) in Equation (6) along Equation (52), we attain the solution

Set:

$$\left\{b_0=0,b_1=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}},\delta ={\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}}\right\}.$$

(68)

• Case 1: when $\kappa$ is positive and $\sigma$ is zero, then:

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{\kappa ab}{sech}_{ab}(\sqrt{\kappa }{\displaystyle \frac{1}{\alpha }}\Gamma (1+{\rm Y})(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (69)

  $$g(x,t))=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{\kappa ab}{csch}_{ab}(\sqrt{\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (70)
• Case 2: when $\kappa$ is negative and $\sigma$ is zero, we have:

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-\kappa ab}{sec}_{ab}(\sqrt{-\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (71)

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-\kappa ab}{csc}_{ab}(\sqrt{-\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (72)
• Case 3: when $\kappa$ is negative and $\sigma ={\displaystyle \frac{{\kappa }^2}{4}}$, then:

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}{tanh}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (73)

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}{coth}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (74)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}({tanh}_{ab}(\sqrt{-2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em} \hfill \\ \hfill \pm \iota \sqrt{ab}{sech}_{ab}(\sqrt{-2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (75)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-{\displaystyle \frac{\kappa }{2}}}({coth}_{ab}(\sqrt{-2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em} \hfill \\ \hfill \pm \sqrt{ab}{csch}_{ab}(\sqrt{-2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (76)

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{{\displaystyle \frac{\kappa }{2}}}{tan}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{2}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (77)
• Case 4: when $\kappa$ is positive and $\sigma ={\displaystyle \frac{{\kappa }^2}{4}}$, we have:

  $$g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{{\displaystyle \frac{\kappa }{2}}}{cot}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{2}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))).$$

  (78)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{{\displaystyle \frac{\kappa }{2}}}({tan}_{ab}(\sqrt{2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \pm \sqrt{ab}{sec}_{ab}(\sqrt{2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (79)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{-{\displaystyle \frac{\kappa }{8}}}({tanh}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{8}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill +{coth}_{ab}(\sqrt{-{\displaystyle \frac{\kappa }{8}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (80)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{{\displaystyle \frac{\kappa }{2}}}({cot}_{ab}(\sqrt{2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \pm \sqrt{ab}{csc}_{ab}(\sqrt{2\kappa }{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (81)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{2\sqrt{3}\mu \sqrt{\Omega }}{\sqrt{\theta \kappa {\mu }^2-\theta }}}(\sqrt{{\displaystyle \frac{\kappa }{8}}}({tan}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{8}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha }))\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill +{cot}_{ab}(\sqrt{{\displaystyle \frac{\kappa }{8}}}{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }+({\displaystyle \frac{2\mu \Omega }{\kappa {\mu }^2-1}})t^{\alpha })))).\end{array}$$

  (82)

5. Generalized Kudryashov Technique

This technique [37,38] is performed through the following steps.

• Step 1: Assume a nonlinear PDE:

  $$Y(q,q^2{q}_x,q_t,q_{tt},q_{xx},q_{xt},\dots )=0.$$

  (83)

  where q is a wave-profile. Substituting the given relation:

  $$q(x,t)=Q(\xi ),\xi =x-\nu t.$$

  (84)

  into Equation (83) yields the NLODE:

  $$F(Q,Q^{\prime },Q^2{Q}^{\prime },Q^{″},Q^2{Q}^{\prime },\dots )=0.$$

  (85)
• Step 2: Considering roots of Equation (85) are:

  $$Q(\xi )={\alpha }_0+\sum _{p=1}^m{\displaystyle \frac{{\alpha }_p}{{(1+\psi (\xi ))}^p}},$$

  (86)

  where ${\alpha }_p$, $(p=0,2,3,\cdots ,m)$ are unknowns and $\psi$ is also a wave profile of $\xi$, which is a root of the general Riccati equation defined by:

  $${\psi }^{\prime }(\xi )=a+b\psi (\xi )+c{\psi }^2(\xi ).$$

  (87)
  The coefficients $a,b$ and c are the parameters. Determining the roots for Equation (87) is discussed through the given cases [39]:
  Case 1: when all $a,b$ and c are all nonzero, we have $\psi (\xi )$ shown as

  $$\psi (\xi )={\displaystyle \frac{1}{2c}}\left(\sqrt{4ca-b^2}tan\left({\displaystyle \frac{1}{2}}\sqrt{4ca-b^2}\left(d_0+\xi \right)\right)-b\right),4ac>b^2.$$

  (88)

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(\sqrt{4ca-b^2}cot\left({\displaystyle \frac{1}{2}}\sqrt{4ca-b^2}\left(d_0+\xi \right)\right)+b\right),4ca>b^2.$$

  (89)

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(\sqrt{4ca-b^2}tanh\left({\displaystyle \frac{1}{2}}\sqrt{4ca-b^2}\left(d_0+\xi \right)\right)+b\right),4ca<b^2.$$

  (90)

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(\sqrt{4ca-b^2}coth\left({\displaystyle \frac{1}{2}}\sqrt{4ca-b^2}\left(d_0+\xi \right)\right)+b\right),4ca<b^2.$$

  (91)

  $$\psi (\xi )={\displaystyle \frac{-1}{c}}\left({\displaystyle \frac{1}{d_0+\xi }}+{\displaystyle \frac{b}{2}}\right),4ca=b^2.$$

  (92)
  Case 2: when $a=0$, and $c\ne 0$, we have:

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(btanh\left({\displaystyle \frac{b}{2}}\left(d_0+\xi \right)\right)+b\right),b^2>0.$$

  (93)

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(bcoth\left({\displaystyle \frac{b}{2}}\left(d_0+\xi \right)\right)+b\right),b^2>0.$$

  (94)

  $$\psi (\xi )={\displaystyle \frac{1}{2c}}\left(\sqrt{-b^2}tan\left({\displaystyle \frac{\sqrt{-b^2}}{2}}\left(d_0+\xi \right)\right)-b\right),b^2<0.$$

  (95)

  $$\psi (\xi )={\displaystyle \frac{-1}{2c}}\left(\sqrt{-b^2}cot\left({\displaystyle \frac{\sqrt{-b^2}}{2}}\left(d_0+\xi \right)\right)+b\right),b^2<0.$$

  (96)

  $$\psi (\xi )={\displaystyle \frac{b}{bexp(-b(d_0+\xi ))-c}},b\ne 0.$$

  (97)

  $$\psi (\xi )={\displaystyle \frac{-1}{c\xi }},b=0.$$

  (98)
  Case 3: when b is zero and c is nonzero, we have:

  $$\psi (\xi )={\displaystyle \frac{\sqrt{ca}}{c}}tan\left(\sqrt{ca}\left(d_0+\xi \right)\right),ca>0.$$

  (99)

  $$\psi (\xi )=-{\displaystyle \frac{\sqrt{ca}}{c}}cot\left(\sqrt{ca}\left(d_0+\xi \right)\right),ca>0.$$

  (100)

  $$\psi (\xi )=-{\displaystyle \frac{\sqrt{ca}}{c}}tanh\left(\sqrt{-ca}\left(d_0+\xi \right)\right),ca<0.$$

  (101)

  $$\psi (\xi )=-{\displaystyle \frac{\sqrt{ca}}{c}}coth\left(\sqrt{-ca}\left(d_0+\xi \right)\right),ca<0.$$

  (102)

  $$\psi (\xi )={\displaystyle \frac{-1}{c(d_0+\xi )}},a=0.$$

  (103)
  Case 4: when c is zero and b is nonzero, we have:

  $$\psi (\xi )={\displaystyle \frac{1}{b}}\left(exp\left(b(d_0+\xi )\right)-a\right).$$

  (104)
• Step 3: Substitute Equation (86) in Equation (85), then find the sum of $\psi (\xi )$ terms, and take the coefficients of same orders equal to 0 to obtain a system in ${\alpha }_j$ for $(j=0,1,2,\cdots ,m)$. By manipulating this obtained system, the values of the undetermined parameters are found.
  Set 4: By putting the found results for ${\alpha }_p$, $(p=0,1,2,\cdots ,m)$ into Equation (86) and using Equations (88)–(104), we obtain the solutions to Equation (6).

Novel Wave Solitons via the Generalized Kudryashov Method

Equation (86) changes into following form in our case:

$$G(\xi )={\alpha }_0+{\displaystyle \frac{{\alpha }_1}{1+\psi (\xi )}},$$

(105)

where ${\alpha }_0$ and ${\alpha }_1$ are unknowns. Inserting Equation (105) and (87) into Equation (6) gives the following set of parameters:

$$\begin{array}{c}\{{\alpha }_0=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }(b-2c)}{\sqrt{\theta (-4ac{\mu }^2+b^2{\mu }^2+2)}}},{\alpha }_1=\pm {\displaystyle \frac{2i\mu \sqrt{6\Omega }(a-b+c)}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}},\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \delta =-{\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}},\Delta =\sqrt{4ac-b^2}\}.\end{array}$$

(106)

which in return leads to the following cases:

• Case 1: when all $a,b$ and c are all nonzero, we have $\psi (\xi )$ shown as.

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}}((b-2c)+(2(a-b+c))/(1+({\displaystyle \frac{1}{2c}}(\Delta tan({\displaystyle \frac{1}{2}}\Delta (d_0+\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))-b)))).\hspace{1em}\hspace{1em}\end{array}$$

  (107)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}}((b-2c)+(2(a-b+c))/(1+({\displaystyle \frac{-1}{2c}}(\Delta cot({\displaystyle \frac{1}{2}}\Delta (d_0+\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))+b)))).\hspace{1em}\hspace{1em}\end{array}$$

  (108)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}}((b-2c)+(2(a-b+c))/(1+({\displaystyle \frac{-1}{2c}}(\Delta tanh({\displaystyle \frac{1}{2}}\Delta (d_0+\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))+b)))).\hspace{1em}\hspace{1em} \end{array}$$

  (109)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}}((b-2c)+(2(a-b+c))/(1+({\displaystyle \frac{-1}{2c}}(\Delta coth({\displaystyle \frac{1}{2}}\Delta (d_0+\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))+b)))).\end{array}$$

  (110)

  $$g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (2-4ac{\mu }^2+b^2{\mu }^2)}}}(b-2c+{\displaystyle \frac{2(a-b+c)}{1-\frac{1}{c}(\frac{1}{d_0+\frac{\Gamma (1+{\rm Y})}{\alpha }(\mu x^{\alpha }-(\frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2})t^{\alpha })}+\frac{b}{2})}}).$$

  (111)
• Case 2: when $a=0$, and $c\ne 0$, we have:

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}((b-2c)+(2(c-b))/(1+({\displaystyle \frac{-1}{2c}}(btanh({\displaystyle \frac{b}{2}}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))+b)))).\hspace{1em}\hspace{1em} \end{array}$$

  (112)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\mu \sqrt{6\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}((b-2c)+(2(c-b))/(1+({\displaystyle \frac{-1}{2c}}(bcoth({\displaystyle \frac{b}{2}}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{2-4ac{\mu }^2+b^2{\mu }^2}})t^{\alpha })))+b)))).\hspace{1em}\hspace{1em} \end{array}$$

  (113)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}((b-2c)+(2(-b+c))/(1+({\displaystyle \frac{1}{2c}}(\sqrt{-b^2}tan({\displaystyle \frac{\sqrt{-b^2}}{2}}(d_0+\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))-b)))). \end{array}$$

  (114)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}((b-2c)+(2(-b+c))/(1+({\displaystyle \frac{-1}{2c}}(\sqrt{-b^2}cot({\displaystyle \frac{\sqrt{-b^2}}{2}}(d_0+\hfill \\ \hfill {\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }-({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))+b)))). \end{array}$$

  (115)

  $$g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}(b-2c+{\displaystyle \frac{2(c-b)}{1+\frac{b}{bexp(-b(d_0+\frac{1}{\alpha }\Gamma (1+{\rm Y})(\mu x^{\alpha }-(\frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2})t^{\alpha })))-c}}}).$$

  (116)

  $$g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{2\theta }}}(-2c+{\displaystyle \frac{2c}{1+(\frac{-1}{c\frac{\Gamma (1+{\rm Y})}{\alpha }(\mu x^{\alpha }-(\frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2})t^{\alpha })})}}).$$

  (117)
• Case 3: when b is zero and c is nonzero, we have:

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{\sqrt{-6}\mu \sqrt{\Omega }}{\sqrt{\theta (-4ac{\mu }^2+2)}}}(-2c+(2(a+c))/(1+({\displaystyle \frac{\sqrt{ac}}{c}}tan(\sqrt{ac}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))))).\hspace{1em}\hspace{1em}\hspace{1em}\end{array}$$

  (118)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{\theta (-4ac{\mu }^2+2)}}}(-2c+(2(a+c))/(1+(-{\displaystyle \frac{\sqrt{ac}}{c}}cot(\sqrt{ac}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))))).\hspace{1em}\hspace{1em}\hspace{1em}\end{array}$$

  (119)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{\sqrt{-6}\mu \sqrt{\Omega }}{\sqrt{\theta (-4ac{\mu }^2+2)}}}(-2c+(2(a+c))/(1+(-{\displaystyle \frac{\sqrt{ac}}{c}}tanh(\sqrt{-ac}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))))).\end{array}$$

  (120)

  $$\begin{array}{c}g(x,t)=\pm {\displaystyle \frac{\sqrt{-6}\mu \sqrt{\Omega }}{\sqrt{\theta (-4ac{\mu }^2+2)}}}(-2c+(2(a+c))/(1+(-{\displaystyle \frac{\sqrt{ac}}{c}}coth(\sqrt{-ac}(d_0+{\displaystyle \frac{\Gamma (1+{\rm Y})}{\alpha }}(\mu x^{\alpha }\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill -({\displaystyle \frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2}})t^{\alpha })))))).\end{array}$$

  (121)

  $$g(x,t)=\pm {\displaystyle \frac{i\sqrt{6}\mu \sqrt{\Omega }}{\sqrt{2\theta }}}(-2c+{\displaystyle \frac{2c}{1+(\frac{-1}{c(d_0+\frac{\Gamma (1+{\rm Y})}{\alpha }(\mu x^{\alpha }-(\frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2})t^{\alpha }))})}}).$$

  (122)
• Case 4: when c is zero and b is nonzero, we have:

  $$g(x,t)=\pm {\displaystyle \frac{\sqrt{-6}\mu \sqrt{\Omega }}{\sqrt{\theta (b^2{\mu }^2+2)}}}(b+{\displaystyle \frac{2(a-b)}{1+\frac{1}{b}(exp(b(d_0+\frac{\Gamma (1+{\rm Y})}{\alpha }(\mu x^{\alpha }-(\frac{4\mu \Omega }{-4ac{\mu }^2+b^2{\mu }^2+2})t^{\alpha })))-a)}}).$$

  (123)

6. Physical Explanation

Here, we will explain few of the gained solutions by 2D, 3D, and contour graphs (Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6, Figure 7, Figure 8, Figure 9 and Figure 10).

7. Conclusions

We have successfully derived novel kinds of M-fractional wave solutions for the nonlinear simplified Modified Camassa–Holm model through the application of the modified simplest equation, the Sardar subequation, and the generalized Kudryashov techniques. These results open new forums for further research in this model, with potential applications in various areas in partial differential equations.

Our application of the modified simplest equation, the Sardar Subequation, and the generalized Kudryahov techniques has established their efficiency as straightforward, precise, and helpful tools for tackling nonlinear FPDEs. The simplicity of the provided solutions guarantees our understanding of the fundamental dynamics of this important mathematical physics model.

The obtained solutions have been thoroughly verified and further explained through the visualization of 2D, 3D, and contour plots, which were generated using the powerful Mathematica. These graphical representations not only confirm the cogency of our derived results but also offer valuable insight into the behavior of M-fractional wave solitons.