Extraction of Exact Solutions of Higher Order Sasa-Satsuma Equation in the Sense of Beta Derivative

Abstract

Nearly every area of mathematics, natural, social, and engineering now includes research into finding exact answers to nonlinear fractional differential equations (NFDES). In order to discover the exact solutions to the higher order Sasa-Satsuma equation in the sense of the beta derivative, the paper will discuss the modified simple equation (MSE) and exponential rational function (ERF) approaches. In general, symmetry and travelling wave solutions of the Sasa-Satsuma equation have a common correlation with each other, thus we reduce equations from wave transformations to ordinary differential equations with the help of Lie symmetries. Actually, we can say that wave moves are symmetrical. The considered procedures are effective, accurate, simple, and straightforward to compute. In order to highlight the physical characteristics of the solutions, we also provide 2D and 3D plots of the results.

1. Introduction

Nowadays, academic researchers deal with many physical phenomena in plasma physics, physical chemistry, geophysics, fluid mechanics, nonlinear optics, electromagnetic theory, and fluid motion, and their mathematical models are expressed by NFDEs [1,2]. These equations are commonly used in various scientific disciplines and have been investigated from different viewpoints [3]. The exact solutions of these equations have gained more and more interest. For this reason, a lot of different techniques have been dealt with by researchers. Among this research, the following ones can be listed, for instance, Hosseini et al. derived new exact solutions of a new (4+1)-dimensional Burgers equation [4]. Dehingia et al. applied the Hopf-bifurcation [5]. Mirzazadeh et al. worked on the second-order nonlinear Schrödinger equation with weakly nonlocal and parabolic laws [6]. Aktar et al. obtained the soliton solutions of a biological model [7]. Mathanaranjan et al. applied the sinh-Gordon procedure [8]. Jannat et al. constructed different types of travelling wave solutions for the considered equation by employing the Auto-Backlund transformations [9]. Ala et al. applied the improved Bernoulli sub-equation function procedure [10]. Kaplan and Akbulut utilized the modified Kudryashov approach [11], Hosseini et al. used the homotopy analysis method (HAM) [12], Mahmood et al. utilized discrete fractional operators [13], Mohammed et al. employed discrete CF-fractional operators [14], Arshed et al. utilized the modified auxiliary equation (MAE) method [15].

The limit relationships between solitary wave solutions and periodic wave solutions, which correspond to the outer and inner trajectories of asymmetric homoclinic trajectories, were explored by the authors in [16]. We deal with the wave transformations, which are connected to the Lie symmetries of the equations, as a symmetry of an equation is a transformation that does not change the set of solutions to the equation [17].

In the current work, we will take into consideration the generalized (3+1)-dimensional nonlinear Sasa-Satsuma (3DNSS) equation in the sense of the beta derivative,

$$\begin{array}{ll}i\frac{{\partial }^{\alpha }q(x,y,z,t)}{\partial t^{\alpha }}& +{\alpha }_1(q_{xx}+q_{yy}+q_{zz})+{\alpha }_2{\left|q\right|}^{2}q+i({\alpha }_3(q_{xxx}+q_{yyy}+q_{zzz})\\ & +{\alpha }_4{\left|q\right|}^2(q_x+q_y+q_z)+{\alpha }_5({\left({\left|q\right|}^2\right)}_x+{\left({\left|q\right|}^2\right)}_y+{\left({\left|q\right|}^2\right)}_z)q)& \hfill =0,\end{array}$$

(1)

where $i=\sqrt{-1},q=q(x,y,z,t)$ is the soliton profile, and ${\alpha }_{\vartheta },\vartheta =1,\dots ,5$ are real constants. This equation includes self-steepening, third order dispersion, and self-frequency shift, and is an integrable generalization of the NFDE. Yao et al. applied the mapping method, extended hyperbolic function technique, Bäcklund transformation, and improved generalized Riccati equation to find optical solitons of this equation [18]. Wazwaz and Mehanna obtained exponential, bright, singular, and dark soliton solutions of this equation [19].

This work is arranged as follows. We provide some fundamental details and certain features of the beta derivative in Section 2. Then, in Section 3, the preliminary information is given. In Section 4, we describe the MSE technique. In Section 5, we present the ERF technique. Then, in Section 6, we represent the applications of the adopted procedure. Finally, we provide a discussion on the results.

2. Some Useful Properties of the Beta Derivative

In the current part of the paper, we are going to give some definitions and basic results:

Definition 1.

Let $ϵ\left(t\right)$ be a function defined for all non-negative t. The β derivative of $f\left(t\right)$ of order β is given by [20,21]:

$$T^{\beta }\left(ϵ\left(t\right)\right)=\underset{\epsilon \to 0}{lim}\frac{ϵ\left(t+\epsilon {\left(t+\frac{1}{\Gamma \left(\beta \right)}\right)}^{1-\beta }\right)-ϵ\left(t\right)}{\epsilon },0<\beta \le 1,$$

and $T^{\beta }\left(ϵ\left(t\right)\right)=\frac{d^{\beta }ϵ\left(t\right)}{d{t}^{\beta }}$.

Some properties are given for $\beta$ derivative in the following theorem [22,23,24]:

Theorem 1.

Let $ϵ\left(t\right)$ and $\zeta \left(t\right)$ be β-differentiable functions for all $t>0$ and $\beta ϵ(0,1]$. Some basic properties are discussed as foloows:

• $T^{\beta }(\gamma ϵ\left(t\right)+\delta \zeta \left(t\right))=\gamma T^{\beta }\left(ϵ\left(t\right)\right)+\delta T^{\beta }\left(\zeta \left(t\right)\right),\forall \gamma ,\delta \in R$
• $T^{\beta }\left(ϵ\left(t\right)\zeta \left(t\right)\right)=\zeta \left(t\right)T^{\beta }\left(ϵ\left(t\right)\right)+ϵ\left(t\right)T^{\beta }\left(\zeta \left(t\right)\right),$
• $T^{\beta }\left(\frac{ϵ\left(t\right)}{\zeta \left(t\right)}\right)=\frac{\zeta \left(t\right)T^{\beta }\left(ϵ\left(t\right)\right)-ϵ\left(t\right)T^{\beta }\left(\zeta \left(t\right)\right)}{\zeta {\left(t\right)}^2},$
• $T^{\beta }\left(ϵ\left(t\right)\right)={\left(t+\frac{1}{\Gamma \left(\beta \right)}\right)}^{1-\beta }\frac{dϵ\left(t\right)}{dt}.$

3. Preliminary Information

In this piece of the work, we first introduce the background data for employing to convert a nonlinear partial differential problem to an ordinary differential equation (ODE). Therefore, the following time fractional NFDE will be taken into account:

$$P(q,D_t^{\beta }q,D_{x}q,D_{y}q,D_{z}q,D_t^{\beta }{D}_t^{\beta }q,D_t^{\beta }{D}_{x}q,D_x{D}_{x}q,\dots )=0,0<\beta \le 1,$$

(2)

where P inherits q and its various partial derivatives include $\beta$ fractional derivatives with respect to t and classical derivatives with respect to $x,y,z$.

The wave transformation is defined as:

$$q(x,y,z,t)=u\left(\epsilon \right)e^{i\left({\tau }_{1}x+{\tau }_{2}y+{\tau }_{3}z+{\tau }_{4}t\right)},\epsilon ={\varrho }_{1}x+{\varrho }_{2}y+{\varrho }_{3}z+\frac{{\varrho }_4}{\beta }{\left(t+\frac{1}{\Gamma \left(\beta \right)}\right)}^{\beta },$$

(3)

where ${\varrho }_i$ and ${\tau }_i\left(i=0,1,\dots ,4\right)$ are constants to be calculated. Then, (3) is used in Equation (2) and equates the imaginary and real parts to zero; we discover an equation system. After that, we solve the resulting system to determine the parameters’ conditions and use those results. As a result, we obtain the following ODE, as follows, which is going to be integrated with respect to the $\epsilon$ possible times.

$$Q\left(u,u^{\prime },u^{″},u^{‴},\dots \right)=0,$$

(4)

where the partial derivative is given with respect to $\epsilon$.

4. Modified Simple Equation Technique

We can summarize the MSE technique, as follows [25,26]:

Firstly, the exact solution of Equation (4) can used to demonstrate a polynomial in $\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)$ as follows

$$u\left(\epsilon \right)=\underset{j=0}{\sum ^m}{\sigma }_j{\left[\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right]}^j,{\sigma }_j=\mathrm{const}.,{\sigma }_m\ne 0.$$

(5)

Here, m is a positive integer, and it is the balancing number. It can be evaluated by applying the homogeneous balance principle between the highest order derivative term with the highest order nonlinear term, which is shown in Equation (4). Moreover, ${\sigma }_j$ are arbitrary real constants to be calculated.

We obtain a polynomial of order ${\varsigma }^{-j}\left(\epsilon \right)$ with the derivatives of $\varsigma \left(\epsilon \right)$ by inserting Equation (5) into Equation (4). After that, we obtain an algebraic equation system to be solved for finding ${\sigma }_j\left(j=0,1,2,\dots ,m\right),$ ${\varrho }_i$ and ${\tau }_i\left(i=0,1,\dots ,4\right)$ and $\varsigma \left(\epsilon \right),$ by equating all the coefficients of ${\varsigma }^{-j}\left(\epsilon \right)$ to zero $(j\ge 0).$ Finally, we substitute the values of ${\sigma }_j,$c and $\varsigma \left(\epsilon \right)$ into Equation (5) and we obtain the exact solutions of Equation (1).

5. The Exponential Rational Function Technique

Assume that Equation (4) has a solution of the following type [27,28]:

$$u\left(\epsilon \right)=\sum _{j=0}^m\frac{{\sigma }_j}{{(1+e^{\epsilon })}^j},{\rho }_m\ne 0$$

(6)

where ${\sigma }_j(j=0,1,2,\dots ,m)$ are constants to be evaluated, and m is the balancing number. If we substitute Equation (6) into Equation (4), we discover a collection of algebraic equations in parameters ${\sigma }_j(j=0,1,2,\dots ,m),{\varrho }_i$ and ${\tau }_i\left(i=0,1,\dots ,4\right).$ Hence, we can derive fresh answers to Equation (4).

6. Application of the Given Methods from the Solutions of This System

We shall look for the novel traveling wave solutions to Equation (1) in this piece of the work. Due to this, we first provide a mathematical examination of the problem under consideration.

6.1. Mathematical Analysis

We split Equation (1) into its real and imaginary components using the transformation specified in Equation (3):

$$\begin{array}{c}\left(\left({\varrho }_1^2+{\varrho }_2^2+{\varrho }_3^2\right){\alpha }_1-3{\alpha }_3\left({\varrho }_1^2{\tau }_1+{\varrho }_2^2{\tau }_2+{\varrho }_3^2{\tau }_3\right)\right)u^{″}\\ -\left({\tau }_4+\left({\tau }_1^2+{\tau }_2^2+{\tau }_3^2\right){\alpha }_1-\left({\tau }_1^3+{\tau }_2^3+{\tau }_3^3\right){\alpha }_3\right)u+\left({\alpha }_2-\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4\right)u^3=0,\end{array}$$

(7)

and

$$\begin{array}{c}\left({\varrho }_4+2{\alpha }_1\left({\varrho }_1{\tau }_1+{\varrho }_2{\tau }_2+{\varrho }_3{\tau }_3\right)-3{\alpha }_3\left({\varrho }_1{\tau }_1^2+{\varrho }_2{\tau }_2^2+{\varrho }_3{\tau }_3^2\right)\right)u^{\prime }\\ +\left({\varrho }_1+{\varrho }_2+{\varrho }_3\right)\left({\alpha }_4+2{\alpha }_5\right)u^2{u}^{\prime }+{\alpha }_3\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right)u^{‴}=0.\end{array}$$

(8)

If we integrate Equation (8) and take the integration constant as zero, we find the resultant equation as follows:

$$\begin{array}{c}{\alpha }_3\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right)u^{″}+\left({\varrho }_4+{\varrho }_3{\tau }_3^2\right)+2{\alpha }_1\left({\varrho }_1{\tau }_1+{\varrho }_2{\tau }_2+{\varrho }_3{\tau }_3\right)\\ -3{\alpha }_3\left({\varrho }_1{\tau }_1^2+{\varrho }_2{\tau }_2^2\right)u+\frac{1}{3}\left({\varrho }_1+{\varrho }_2+{\varrho }_3\right)\left({\alpha }_4+2{\alpha }_5\right)u^3=0.\end{array}$$

(9)

If we compare the coefficients of Equations (7) and (9), we obtain the following relations:

$${\alpha }_1=\frac{{\alpha }_3\left(3\left({\varrho }_1^2{\tau }_1+{\varrho }_2^2{\tau }_2+{\varrho }_3^2{\tau }_3\right)+\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right)\right)}{{\varrho }_1^2+{\varrho }_2^2+{\varrho }_3^2},$$

(10)

$${\alpha }_5=\frac{3{\alpha }_2-{\alpha }_4\left(\left({\varrho }_1+{\varrho }_2+{\varrho }_3\right)+3\left({\tau }_1+{\tau }_2+{\tau }_3\right)\right)}{2\left({\varrho }_1+{\varrho }_2+{\varrho }_3\right)},$$

(11)

and

$$\begin{array}{cc}{\varrho }_4=\hfill & -{\tau }_4-{\alpha }_1\left(2\left({\varrho }_1{\tau }_1+{\varrho }_2{\tau }_2+{\varrho }_3{\tau }_3\right)+{\tau }_1^2+{\tau }_2^2+{\tau }_3^2\right)\hfill \\ \hfill & +{\alpha }_3\left(3\left({\varrho }_1{\tau }_1^2+{\varrho }_2{\tau }_2^2+{\varrho }_3{\tau }_3^2\right)+{\tau }_1^3+{\tau }_2^3+{\tau }_3^3\right).\hfill \end{array}$$

(12)

If we balance $u^{″}$ and $u^3$ Equation (7), we obtain $m=1.$

6.2. MSE Technique

According to the adopted technique, the exact solution of the Equation (7) is given by:

$$u\left(\epsilon \right)={\sigma }_0+{\sigma }_1\left[\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right],{\sigma }_1\ne 0.$$

(13)

Here, we substiture the Equation (13) into Equation (7), and we obtain:

$$\begin{array}{cc}{\varsigma }^0\left(\epsilon \right):\hfill & {\alpha }_3{\tau }_2^3{\sigma }_0-{\alpha }_1{\tau }_3^2{\sigma }_0-{\alpha }_1{\tau }_1^2{\sigma }_0-{\tau }_4{\sigma }_0+{\alpha }_2{\sigma }_0^3+{\alpha }_3{\tau }_1^3{\sigma }_0\hfill \\ \hfill & -{\alpha }_1{\tau }_2^2{\sigma }_0-{\alpha }_4{\tau }_2{\sigma }_0^3-{\alpha }_4{\tau }_3{\sigma }_0^3+{\alpha }_3{\tau }_3^3{\sigma }_0-{\alpha }_4{\tau }_1{\sigma }_0^3=0,\hfill \end{array}$$

(14)

$$\begin{array}{cc}{\varsigma }^1\left(\epsilon \right):\hfill & {\alpha }_1{\varrho }_1^2{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-{\alpha }_1{\tau }_3^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_3{\varrho }_3^2{\tau }_3{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}+3{\alpha }_2{\sigma }_0^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & -{\alpha }_1{\tau }_2^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}+{\alpha }_1{\varrho }_3^2{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-{\alpha }_1{\tau }_1^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-{\tau }_4{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & -3{\alpha }_3{\varrho }_2^2{\tau }_2{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_4{\tau }_1{\sigma }_0^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}+{\alpha }_3{\tau }_1^3{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & +{\alpha }_3{\tau }_2^3{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_4{\tau }_2{\sigma }_0^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_3{\varrho }_1^2{\tau }_1{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & +{\alpha }_1{\varrho }_2^2{\sigma }_1\frac{{\varsigma }^{‴}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}+{\alpha }_3{\tau }_3^3{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_4{\tau }_3{\sigma }_0^2{\sigma }_1\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}=0\hfill \end{array}$$

(15)

$$\begin{array}{cc}{\varsigma }^2\left(\epsilon \right):\hfill & 9{\alpha }_3{\varrho }_1^2{\tau }_1{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_4{\tau }_3{\sigma }_0{\sigma }_1^2{\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)}^2-3{\alpha }_4{\tau }_1{\sigma }_0{\sigma }_1^2{\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)}^2\hfill \\ \hfill & +3{\alpha }_2{\sigma }_0{\sigma }_1^2{\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)}^2-3{\alpha }_1{\varrho }_3^2{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}+9{\alpha }_3{\varrho }_2^2{\tau }_2{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & -3{\alpha }_1{\varrho }_1^2{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}-3{\alpha }_4{\tau }_2{\sigma }_0{\sigma }_1^2{\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)}^2-3{\alpha }_1{\varrho }_2^2{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\hfill \\ \hfill & +9{\alpha }_3{\varrho }_3^2{\tau }_3{\sigma }_1\frac{{\varsigma }^{″}\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}=0,\hfill \end{array}$$

(16)

$$\begin{array}{cc}{\varsigma }^3\left(\epsilon \right):\hfill & (-6{\alpha }_3{\varrho }_2^2{\tau }_2{\sigma }_1-6{\alpha }_3{\varrho }_3^2{\tau }_3{\sigma }_1-{\alpha }_4{\tau }_1{\sigma }_1^3-{\alpha }_4{\tau }_2{\sigma }_1^3-{\alpha }_4{\tau }_3{\sigma }_1^3+2{\alpha }_1{\varrho }_1^2{\sigma }_1\hfill \\ \hfill & +{\alpha }_2{\sigma }_1^3-6{\alpha }_3{\varrho }_1^2{\tau }_1{\sigma }_1+2{\alpha }_1{\varrho }_2^2{\sigma }_1+2{\alpha }_1{\varrho }_3^2{\sigma }_1){\left(\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right)}^3=0\hfill \end{array}$$

(17)

From the solutions of Equations (14) and (17), we find the following results for ${\sigma }_0$ and ${\sigma }_1:$

$${\sigma }_0=\frac{\sqrt{\varpi (-{\tau }_4-{\alpha }_1{\tau }_1^2-{\alpha }_1{\tau }_2^2-{\alpha }_1{\tau }_3^2+{\alpha }_3{\tau }_1^3+{\alpha }_3{\tau }_2^3+{\alpha }_3{\tau }_3^3)}}{\varpi },$$

(18)

and

$${\sigma }_1=\frac{\sqrt{-2\varpi (-{\alpha }_1{\varrho }_2^2+3{\alpha }_3{\varrho }_2^2{\tau }_2+3{\alpha }_3{\varrho }_1^2{\tau }_1-{\alpha }_1{\varrho }_3^2+3{\alpha }_3{\varrho }_3^2{\tau }_3-{\alpha }_1{\varrho }_1^2)}}{\varpi }$$

(19)

where $\varpi ={\alpha }_4{\tau }_3+{\alpha }_4{\tau }_1+{\alpha }_4{\tau }_2-{\alpha }_2$. Then, if we substitute Equations (18) and (19) into Equations (15) and (16), we find a differential equation system for $\varsigma \left(\epsilon \right)$. After that, we solve this differential equation system and find the resultant equation

$$\begin{array}{cc}\varsigma \left(\epsilon \right)=\hfill & c_1\hfill \\ \hfill & +c_{2}exp(\sqrt{(+{\alpha }_4({\alpha }_3-{\alpha }_1)({\tau }_2{\pi }_4+{\tau }_1{\pi }_5)-{\alpha }_4{\tau }_3{\alpha }_1({\tau }_1^2+{\tau }_2^2)+{\alpha }_4{\tau }_3{\alpha }_3{\pi }_4+{\alpha }_2{\tau }_4+{\pi }_7}\hfill \\ \hfill & \sqrt{-\sqrt{2}\left(({\alpha }_4{\pi }_2-{\alpha }_2)\right)(-{\alpha }_1{\pi }_3+3{\alpha }_3({\varrho }_3^2{\tau }_3+{\varrho }_1^2{\tau }_1+{\varrho }_2^2{\tau }_2))}\hfill \\ \hfill & \epsilon /({\alpha }_4{\pi }_2-{\alpha }_2)(-{\alpha }_1{\pi }_3+3{\alpha }_3({\varrho }_3^2{\tau }_3+{\varrho }_1^2{\tau }_1+{\varrho }_2^2{\tau }_2))\hfill \end{array}$$

where $$\begin{gathered} {\pi }_1={\tau }_1^2+{\tau }_2^2+{\tau }_3^2,{\pi }_2={\tau }_1+{\tau }_2+{\tau }_3,{\pi }_3={\varrho }_1^2+{\varrho }_2^2+{\varrho }_3^2,{\pi }_4={\tau }_1^3+{\tau }_3^3,{\pi }_5 \\ ={\tau }_2^3+{\tau }_3^3,{\pi }_6={\tau }_1^3+{\tau }_2^3+{\tau }_3^3,{\pi }_7=-{\alpha }_4{\tau }_1({\tau }_4+{\tau }_1^2{\alpha }_1+{\tau }_1^3{\alpha }_3)-{\alpha }_4{\tau }_4({\tau }_2+{\tau }_3)-{\alpha }_4{\alpha }_1{\pi }_5+ \\ {\alpha }_4{\alpha }_3({\tau }_2^4+{\tau }_3^4)+{\alpha }_2{\alpha }_1{\pi }_1-{\alpha }_2{\alpha }_3{\pi }_6 \end{gathered}$$

Therefore, the solution of $u\left(\epsilon \right)={\sigma }_0+{\sigma }_1\left[\frac{{\varsigma }^{\prime }\left(\epsilon \right)}{\varsigma \left(\epsilon \right)}\right].$

We plotted the above result in Figure 1.

Figure 1. Figures plotted when ${\tau }_1$ = 0.3, ${\tau }_2$ = 0.1, ${\tau }_3$ = 0.2, ${\tau }_4$ = 0.1, ${\varrho }_1$ = 0.2, ${\varrho }_2$ = 0.2, ${\varrho }_3$ = 0.1, ${\alpha }_1$ = 1, ${\alpha }_2$ = 1, ${\alpha }_3$ = 3, ${\alpha }_4$ = 1, y = z = $c_1$ = $c_2$ = 1, $\beta$ = 0.99.

6.3. ERF Technique

We will use the ERF method to solve Equation (1) in this subsection. The preferred response for $u\left(\epsilon \right)$ has the form:

$$u\left(\epsilon \right)={\sigma }_0+\frac{{\sigma }_1}{1+e^{\epsilon }}.$$

(20)

Substituting the Equation (20) into Equation (7), then gathering all of the coefficients of ${\left(e^{\epsilon }\right)}^k$ to zero, we obtain:

$$\begin{array}{c}-{\sigma }_0\left(\left(\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2\right){\sigma }_0^2+\left({\tau }_1^2+{\tau }_2^2+{\tau }_3^2\right){\alpha }_1\right)\\ -{\sigma }_0\left(\left(-{\tau }_1^3-{\tau }_2^3-{\tau }_3^3\right){\alpha }_3+{\tau }_4\right)=0,\end{array}$$

(21)

$$\begin{array}{c}{\sigma }_0((-3{\alpha }_4{\tau }_1-3{\alpha }_4{\tau }_2-3{\alpha }_4{\tau }_3+3{\alpha }_2){\sigma }_0^2+({\varrho }_1^2+{\varrho }_2^2+{\varrho }_3^2-{\tau }_1^2-{\tau }_2^2-{\tau }_3^2){\alpha }_1\\ +(-3{\varrho }_1^2{\tau }_1-3{\varrho }_2^2{\tau }_2-3{\varrho }_3^2{\tau }_3+{\tau }_1^3+{\tau }_2^3+{\tau }_3^3){\alpha }_3-{\tau }_4){\sigma }_1+3\left(\right(-{\alpha }_4{\tau }_1-{\alpha }_4{\tau }_2\\ -{\alpha }_4{\tau }_3+{\alpha }_2){\sigma }_0^2+(-{\tau }_1^2-{\tau }_2^2-{\tau }_3^2){\alpha }_1+({\tau }_1^3+{\tau }_2^3+{\tau }_3^3){\alpha }_3-{\tau }_4)=0,\end{array}$$

(22)

$$\begin{array}{c}3{\sigma }_0((-{\tau }_1-{\tau }_2-{\tau }_3){\alpha }_4+{\alpha }_2){\sigma }_0{\sigma }_1^2+(((-6{\tau }_1-6{\tau }_2-6{\tau }_3){\alpha }_4+6{\alpha }_2){\sigma }_0^2\\ +(-{\varrho }_1^2-{\varrho }_2^2-{\varrho }_3^2-2{\tau }_1^2-2{\tau }_2^2-2{\tau }_3^2){\alpha }_1+(3{\varrho }_1^2{\tau }_1+3{\varrho }_2^2{\tau }_2+3{\varrho }_3^2{\tau }_3+2{\tau }_1^3\\ +2{\tau }_2^3+2{\tau }_3^3){\alpha }_3-2{\tau }_4){\sigma }_1+3(((-{\tau }_1-{\tau }_2-{\tau }_3){\alpha }_4+{\alpha }_2){\sigma }_0^2+(-{\tau }_1^2-{\tau }_2^2-{\tau }_3^2){\alpha }_1\\ +({\tau }_1^3+{\tau }_2^3+{\tau }_3^3){\alpha }_3-{\tau }_4)=0,\end{array}$$

(23)

$$\begin{array}{c}-({\sigma }_0+{\sigma }_1)({({\sigma }_0+{\sigma }_1)}^2({\tau }_1+{\tau }_2+{\tau }_3){\alpha }_4-{\tau }_1^3{\alpha }_3-{\tau }_2^3{\alpha }_3-{\tau }_3^3{\alpha }_3\\ -{\sigma }_0^2{\alpha }_2-2{\sigma }_0{\sigma }_1{\alpha }_2-{\sigma }_1^2{\alpha }_2+{\tau }_1^2{\alpha }_1+{\tau }_2^2{\alpha }_1+{\tau }_3^2{\alpha }_1+{\tau }_4)=0.\end{array}$$

(24)

If we solve Equations (21)–(24), we obtain values of constants, as follows:

$${\sigma }_0=\pm \frac{\sqrt{2}\sqrt{\frac{\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right){\alpha }_3}{\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2}}}{2},{\sigma }_1=\mp \frac{\sqrt{2}\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right){\alpha }_3}{\sqrt{\frac{\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right){\alpha }_3}{\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2}}\left(\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2\right)},$$

$${\tau }_4=\frac{{\alpha }_1\left(-{\varrho }_1^2-{\varrho }_2^2-{\varrho }_3^2-2{\tau }_1^2-2{\tau }_2^2-2{\tau }_3^2\right)}{2}+\frac{3{\alpha }_3\left({\varrho }_1^2{\tau }_1+{\varrho }_2^2{\tau }_2+{\varrho }_3^2{\tau }_3+\frac{2}{3}\left({\tau }_1^3+{\tau }_2^3+{\tau }_3^3\right)\right)}{2}$$

As a result, we may verify the following solutions:

$$\begin{array}{cc}q(x,y,z,t)=\hfill & \pm \left(\frac{\sqrt{2}{\alpha }_3\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right)\left(e^{\left({\varrho }_{1}x+{\varrho }_{2}y+{\varrho }_{3}z+\frac{{\varrho }_4{t}^{\alpha }}{\Gamma (1+\alpha )}\right)}-1\right)}{2\sqrt{\frac{\left({\varrho }_1^3+{\varrho }_2^3+{\varrho }_3^3\right){\alpha }_3}{\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2}}\left(\left({\tau }_1+{\tau }_2+{\tau }_3\right){\alpha }_4-{\alpha }_2\right)\left(1+e^{{}^{\left({\varrho }_{1}x+{\varrho }_{2}y+{\varrho }_{3}z+\frac{{\varrho }_4}{\beta }{\left(t+\frac{1}{\Gamma \left(\beta \right)}\right)}^{\beta }\right)}}\right)}\right)\hfill \\ \hfill & \times e^{i\left({\tau }_{1}x+{\tau }_{2}y+{\tau }_{3}z+{\tau }_{4}t\right)}\hfill \end{array}$$

In Figure 2, we gave figure of the above solution.

Figure 2. Figures plotted when ${\tau }_1$ = 0.3, ${\tau }_2$ = 0.1, ${\tau }_3$ = 0.2, ${\varrho }_1$ = 0.2, ${\varrho }_2$ = 0.2, ${\varrho }_3$ = 0.1, ${\alpha }_0$ = 1, ${\alpha }_2$ = 1, ${\alpha }_3$ = 3, ${\alpha }_4$ = 1, y = z = 0.1, $\beta$ = 0.6.

In Figure 3, we gave figures of the above solution.

Figure 3. Figures plotted when ${\tau }_1$ = 3, ${\tau }_2$ = 2, ${\tau }_3$ = −3, ${\varrho }_1$ = −0.6, ${\varrho }_2$ = −0.8, ${\varrho }_3$ = 0.9, ${\alpha }_0$ = 1, ${\alpha }_2$ = 1, ${\alpha }_3$ = 1, ${\alpha }_4$ = 1, y = z = 1, $\beta$ = 0.9.

7. Discussion on the Results

In this work, the higher order Sasa-Satsuma problem with the $\beta$ derivative is solved by the MSE and ERF method, leading to novel and important results. The periodic wave solutions that correspond to periodic trajectories surrounded by symmetric homogeneous orbits satisfy the concept of symmetry, making it likely that these solutions will be essential for understanding the dynamics of this equation and pinpointing some of its physical characteristics. In a variety of situations where the equation in question is pertinent, the validated solutions could be useful for gaining a greater knowledge of the interacting wave phenomena. In order to verify the accuracy of the recovered solution, it is also helpful to know that the exact answers retrieved are fed back into the original equation. In order to highlight the physical characteristics of the solutions, we also provide 3D and 2D plots of the obtained results. To the best of our understanding, the results offer a lot of promise for mathematical physics. To ensure that the outcomes were accurate, we employed Maple software.