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Article

Mathematical and Physical Analysis of the Fractional Dynamical Model

by
Mohammed Ahmed Alomair
1 and
Haitham Qawaqneh
2,*
1
Department of Quantitative Methods, School of Business, King Faisal University, Al-Ahsa 31982, Saudi Arabia
2
Department of Mathematics, Faculty of Science and Information Technology, Al-Zaytoonah University of Jordan, Amman 11733, Jordan
*
Author to whom correspondence should be addressed.
Fractal Fract. 2025, 9(7), 453; https://doi.org/10.3390/fractalfract9070453
Submission received: 5 June 2025 / Revised: 23 June 2025 / Accepted: 27 June 2025 / Published: 11 July 2025

Abstract

This paper consists of various kinds of wave solitons to the mathematical model known as the truncated M-fractional FitzHugh–Nagumo model. This model explains the transmission of the electromechanical pulses in nerves. Through the application of the modified extended tanh function technique and the modified ( G / G 2 ) -expansion technique, we are able to achieve the series of exact solitons. The results differ from the current solutions because of the fractional derivative. These solutions could be helpful in the telecommunication and bioscience domains. Contour plots, in two and three dimensions, are used to describe the results. Stability analysis is used to check the stability of the obtained solutions. Moreover, the stationary solutions of the focusing equation are studied through modulation instability. Future research on the focused model in question will benefit from the findings. The techniques used are simple and effective.

1. Introduction

Equations involving fractional derivatives or integrals are known as fractional differential equations (FDEs). They are utilized to model complex phenomena in various fields, such as physics, engineering, etc. Various kinds of FDEs have been developed, such as the fractional paraxial nonlinear Schrödinger equation [1], the fractional Westervelt model [2], the fractional generalized Pochhammer–Chree equation [3], the fractional Riemann wave solution [4], the fractional density-dependent diffusion reaction model [5], etc. One crucial aspect of the nonlinear physical phenomenon is determining the exact wave solutions of nonlinear fractional differential equations (FDEs). It is true that exact wave solitons offer a wealth of physical data and aid in comprehending the mechanisms underlying a number of physical models, including solid-state physics, chemical physics, biology, plasma physics, optical fibers, and so forth. Many effective techniques for determining the exact wave solutions have emerged in recent years.
In our research, we used the effective and useful techniques, the modified extended tanh function technique and modified ( G / G 2 ) -expansion technique. The concerned techniques are utilized for various equations. For instance, the modified extended tanh function technique is utilized for the nonlinear coupled plasma equations [6], the Bogoyavlenskii equation [7], the modified Volterra’s equations [8], the concatenation equation [9], the Biswas–Milovic equation [10], etc. The modified ( G / G 2 ) -expansion scheme is utilized for the coupled nonlinear Higgs equations [11], Wazwaz Kaur Boussinesq equation [12], (1+1)-dimensional classical Boussinesq model [13], etc.
Our concerned model is the FitzHugh–Nagumo equation, is given as [14].
h t h x x h ( 1 h ) ( h θ ) = 0 .
Here, h = h ( x , t ) represents the wave function, and θ is a parameter. Equation (1) is used to explain the transmission of the electromechanical pulses in nerves. The concerned model is also used in the fields of electromechanics and biology. In the literature, distinct methods are used to solve Equation (1), including the extended tanh–coth function method [14], the Exp-function method [15], the auto-Bäcklund transformation method [16], the improved Riccati expansion method [17], etc.
There are different sections in this paper. Descriptions of the modified extended tanh function scheme and modified ( G / G 2 ) -expansion scheme are given in Section 2; the mathematical analysis and exact wave solutions are mentioned in Section 3; graphical representation is mentioned in Section 4; modulation instability (MI) analysis is provided in Section 5; and the conclusion is provided in Section 6.

Truncated M-Fractional Derivative (TMFD)

Definition 1.
Consider w ( y ) : [ 0 , ) , so TMFD of w of order ϵ is given as [18]:
D M , y ϵ , ϱ w ( y ) = lim ϵ 0 w ( y E ϱ ( ϵ y 1 ϵ ) ) w ( y ) ϵ , ϵ ( 0 , 1 ] , ϱ > 0 ,
where E ϱ ( . ) shows a TML profile that is given as [19]:
E ϱ ( z ) = j = 0 i z j Γ ( 1 + ϱ j ) , ϱ   is positive and z C .
Theorem 1.
Suppose a,b are real numbers, and g , f are both differentiable ϵ times for y > 0 , from [18]:
( a ) D M , y ϵ , ϱ ( a g ( y ) + b f ( y ) ) = a D M , y ϵ , ϱ g ( y ) + b D M , y ϵ , ϱ f ( y ) .
( b ) D M , y ϵ , ϱ ( g ( y ) . f ( y ) ) = g ( y ) D M , y ϵ , ϱ f ( y ) + f ( y ) D M , y ϵ , ϱ g ( y ) .
( c ) D M , y ϵ , ϱ ( g ( y ) f ( y ) ) = f ( y ) D M , y ϵ , ϱ g ( y ) g ( y ) D M , y ϵ , ϱ f ( y ) ( f ( y ) ) 2 .
( d ) D M , y ϵ , ϱ ( C ) = 0 , where C is a constant .
( e ) D M , y ϵ , ϱ g ( y ) = y 1 ϵ Γ ( ϱ + 1 ) d g ( y ) d y .
This definition of the fractional derivative fulfills the characteristics of both fractional derivatives as well as the integer order derivative. Fractional differential equations represent the phenomenon in a more prominent form. This fractional derivative provides the solutions closer to the numerical solutions of the concerned equation.

2. Methodologies

2.1. Explanation of Modified Extended tanh Function Technique

In this subsection, we mention the basic steps of the technique. Assuming a fractional nonlinear partial differential equation (NLPDE):
χ ( v , v 2 , , ϵ v t ϵ , 2 ϵ v t 2 ϵ , , v x , 2 v x 2 , ) = 0 .
Here v = v ( x , t ) is a wave function. Assuming the following wave transformation:
v ( x , t ) = V ( δ ) , δ = x μ Γ ( 1 + ϱ ) ϵ t ϵ .
where Γ is a function, ϵ is a parameter, δ is a variable, x and t are the dimensions, and μ denotes the soliton velocity. Putting Equation (3) into Equation (2), we obtain a nonlinear ordinary differential equation:
χ ( V , V 2 , μ V , μ 2 V , ) = 0 .
Assuming the root of Equation (4) given below:
V ( δ ) = α 0 + i = 1 m α i ψ i ( δ ) + i = 1 m β i ψ i ( δ ) .
Here α 0 , α i , β i , ( i = 1 , 2 , 3 , , m ) are undetermined.
By applying the homogeneous balance method to Equation (4), we obtain a value of m, where ψ ( ζ ) fulfills the following equation:
ψ ( δ ) = Ω + ψ 2 ( δ ) .
Here, Ω is a constant, and solutions of Equation (6) are mentioned as [20]:
Case 1: when Ω < 0 :
ψ ( δ ) = Ω tanh ( Ω δ ) ,
or
ψ ( δ ) = Ω coth ( Ω δ ) .
Case 2: when Ω > 0 :
ψ ( δ ) = Ω tan ( Ω δ ) ,
or
ψ ( δ ) = Ω cot ( Ω δ ) .
Case 3: when Ω = 0 :
ψ ( δ ) = 1 δ .
Put Equation (5) into Equation (4) with Equation (6). By collecting the coefficient of ψ ( δ ) for each order and putting each order equal to 0, we achieve sets of algebraic equations having α 0 , α i , β i ( i = 1 , 2 , 3 , m ) , and Ω with the use of Mathematica software. By solving the sets, we obtain the solutions of Equation (2).

2.2. Explanation of Modified ( G / G 2 ) —Expansion Technique

Now, the main steps of this technique are explained as [21].
Step 1:
Consider the Equations (2)–(4).
Step 2:
Consider the root of Equation (4) represented as;
Q ( δ ) = i = 0 m α i ( G G 2 ) i ,
where α i ( i = 0 , 1 , 2 , 3 , , m ) are to be found where α j 0 . A new function G = G ( δ ) fulfills the equation,
( G G 2 ) = λ 0 + λ 1 ( G G 2 ) 2 ,
where λ 0 and λ 1 denote the constants. We achieve the following solutions to Equation (13) depending on λ 0 :
Case 1: λ 0 λ 1 < 0 , which yields
( G G 2 ) = | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ,
Case 2: if λ 0 λ 1 > 0 , then
( G G 2 ) = λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ,
Case 3: if λ 0 = 0 and λ 1 0 , then
( G G 2 ) = C 1 λ 1 ( C 1 δ + C 2 ) .
Here C 1 and C 2 are constants.
Step 3:
Put Equation (12) in Equation (4) along with Equation (13), and collect the coefficients of every power of ( G G 2 ) i equal to 0, now manipulate the achieved set having α i , λ 0 , λ 1 , μ .
Step 4: Equation (12) of that α i , μ are obtained in step 3 in Equation (12), and one can gain the results of Equation (2).

3. Mathematical Analysis

Rewrite Equation (1) in the sense of a truncated M-fractional derivative (TMFD) as;
D M , t ϵ , ϱ h D M , x 2 ϵ , ϱ h h ( 1 h ) ( h θ ) = 0 .
applying the given transformation
h ( x , t ) = H ( δ ) , δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ,
D M , t ϵ , ϱ h = λ H .
D M , x 2 ϵ , ϱ h = μ 2 H .
By using Equations (18) in Equation (17), we obtain the following ODE;
μ 2 H + λ H + H ( 1 H ) ( H θ ) = 0 .
By applying the homogeneous balance method, we obtain m = 1 .

3.1. Exact Solitons Through mEThF Technique

Equation (5) reduces into following form for m = 1.
V ( δ ) = α 1 ϕ ( δ ) + α 0 + β 1 ϕ ( δ ) .
Inserting Equation (22) with Equation (6) in Equation (21), and solving the system, we obtain the following sets:
  • Set 1:
α 0 = 1 μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω 2 , θ = 1 2 μ 2 Ω .
Case 1:
h ( x , t ) = 1 μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = 1 μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 2:
α 0 = 1 μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 2 μ Ω , θ = 1 2 μ 2 Ω .
Case 1:
h ( x , t ) = 1 μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = 1 μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 3:
α 0 = 1 μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 2 μ Ω , θ = 1 2 μ 2 Ω .
Case 1:
h ( x , t ) = 1 μ 2 Ω 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 μ 2 Ω 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 μ 2 Ω 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 μ 2 Ω 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 4:
α 0 = 1 μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω 2 , θ = 1 2 μ 2 Ω .
Case 1:
h ( x , t ) = 1 μ 2 Ω + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 μ 2 Ω + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 μ 2 Ω + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 μ 2 Ω + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 5:
α 0 = 1 + μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = 1 + μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 6:
α 0 = 1 + μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω + 2 , θ = 2 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = 1 + μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 7:
α 0 = 1 + μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω + 2 , θ = 2 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + μ 2 Ω 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 + μ 2 Ω 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 8:
α 0 = 1 + μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + μ 2 Ω + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 + μ 2 Ω + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + μ 2 Ω + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 9:
α 0 = 1 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω 2 , θ = 1 4 μ Ω .
Case 1:
h ( x , t ) = 1 2 μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 2 μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 2 μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 2 μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 10:
α 0 = 1 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 2 4 μ Ω , θ = 1 4 μ 2 Ω .
Case 1:
h ( x , t ) = 1 2 μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 2 μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 2 μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 2 μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 11:
α 0 = 1 + 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω 2 , θ = 4 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + 2 μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + 2 μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 + 2 μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + 2 μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 12:
α 0 = 1 + 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω + 2 , θ = 4 μ 2 Ω + 1 .
Case 1:
h ( x , t ) = 1 + 2 μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + 2 μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 1 + 2 μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 1 + 2 μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 13:
α 0 = μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω + 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 14:
α 0 = μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 15:
α 0 = μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = μ 2 Ω 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 16:
α 0 = μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω + 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 17:
α 0 = μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 2 μ Ω , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 18:
α 0 = μ 2 Ω , α 1 = 2 μ , β 1 = 0 , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Case 2:
h ( x , t ) = μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) .
Set 19:
α 0 = μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 μ Ω 2 , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = μ 2 Ω 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 20:
α 0 = μ 2 Ω , α 1 = 0 , β 1 = 2 μ Ω , λ = μ 2 2 μ Ω , θ = 2 μ 2 Ω .
Case 1:
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = μ 2 Ω + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 21:
α 0 = 2 μ Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω + 2 , θ = 4 μ 2 Ω .
Case 1:
h ( x , t ) = 2 μ Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 2 μ Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 22:
α 0 = 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω 2 , θ = 4 μ 2 Ω .
Case 1:
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 23:
α 0 = 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 2 4 μ Ω , θ = 4 μ 2 Ω .
Case 1:
h ( x , t ) = 2 μ 2 Ω 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 2 μ 2 Ω 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) + 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Set 24:
α 0 = 2 μ 2 Ω , α 1 = 2 μ , β 1 = 2 μ Ω , λ = μ 4 μ Ω 2 , θ = 4 μ 2 Ω .
Case 1:
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tanh ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω coth ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .
Case 2:
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω tan ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ,
or
h ( x , t ) = 2 μ 2 Ω + 2 μ ( Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) ) 2 μ Ω Ω cot ( Ω Γ ( ϱ + 1 ) ( μ x ϵ ϵ λ t ϵ ϵ ) ) .

3.2. Exact Wave Solitons via Modified ( G / G 2 ) Expansion Scheme

For m = 1 , Equation (12) reduces into:
V ( δ ) = α 0 + α 1 G ( δ ) G 2 ( δ ) .
Here, α 0 and α 1 are unknowns. Putting Equation (143) along Equation (13) in Equation (21) yields the given solutions:
  • Set 1:
{ α 0 = 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ + 1 , θ = 2 μ 4 λ 0 λ 1 μ + 2 λ 0 λ 1 2 2 λ 0 λ 1 μ + 1 } .
Case 1:
h ( x , t ) = 2 λ 0 λ 1 μ 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 2 λ 0 λ 1 μ 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ + 1 t ϵ ϵ ) .
  • Set 2:
{ α 0 = 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ + 1 , θ = 2 μ 4 λ 0 λ 1 μ + 2 λ 0 λ 1 2 2 λ 0 λ 1 μ + 1 } .
Case 1:
h ( x , t ) = 2 λ 0 λ 1 μ + 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 2 λ 0 λ 1 μ + 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ + 1 t ϵ ϵ ) .
  • Set 3:
{ α 0 = 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 , θ = 2 μ 4 λ 0 λ 1 μ + 2 λ 0 λ 1 2 2 λ 0 λ 1 μ 1 } .
Case 1:
h ( x , t ) = 2 λ 0 λ 1 μ 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 2 λ 0 λ 1 μ 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 t ϵ ϵ ) .
  • Set 4:
{ α 0 = 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ 1 , θ = 2 μ 4 λ 0 λ 1 μ + 2 λ 0 λ 1 2 2 λ 0 λ 1 μ 1 } .
Case 1:
h ( x , t ) = 2 λ 0 λ 1 μ + 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 2 λ 0 λ 1 μ + 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ 1 t ϵ ϵ ) .
  • Set 5:
{ α 0 = 1 + 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ + 1 , θ = 8 λ 0 λ 1 μ 2 + 1 + 4 2 λ 0 λ 1 μ 2 2 λ 0 λ 1 μ + 1 } .
Case 1:
h ( x , t ) = 1 + 2 λ 0 λ 1 μ 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 1 + 2 λ 0 λ 1 μ 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ + 1 t ϵ ϵ ) .
  • Set 6:
{ α 0 = 1 + 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ + 1 , θ = 8 λ 0 λ 1 μ 2 + 1 + 4 2 λ 0 λ 1 μ 2 2 λ 0 λ 1 μ + 1 } .
Case 1:
h ( x , t ) = 1 + 2 λ 0 λ 1 μ + 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 1 + 2 λ 0 λ 1 μ + 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ + 1 2 μ 2 2 λ 0 λ 1 μ + 1 t ϵ ϵ ) .
  • Set 7:
{ α 0 = 1 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 , θ = 8 λ 0 λ 1 μ 2 + 4 2 λ 0 λ 1 μ 1 2 2 λ 0 λ 1 μ 1 } .
Case 1:
h ( x , t ) = 1 2 λ 0 λ 1 μ 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 1 2 λ 0 λ 1 μ 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ + 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 t ϵ ϵ ) .
  • Set 8:
{ α 0 = 1 2 λ 0 λ 1 μ , α 1 = 2 λ 1 μ , λ = 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 , θ = 8 λ 0 λ 1 μ 2 + 4 2 λ 0 λ 1 μ 1 2 2 λ 0 λ 1 μ 1 , } .
Case 1:
h ( x , t ) = 1 2 λ 0 λ 1 μ + 2 λ 1 μ ( | λ 0 λ 1 | λ 1 + | λ 0 λ 1 | 2 ( C 1 sinh ( λ 0 λ 1 δ ) + C 2 cosh ( λ 0 λ 1 δ ) C 1 cosh ( λ 0 λ 1 δ ) + C 2 sinh ( λ 0 λ 1 δ ) ) ) .
Case 2:
h ( x , t ) = 1 2 λ 0 λ 1 μ + 2 λ 1 μ ( λ 0 λ 1 ( C 1 cos ( λ 0 λ 1 δ ) + C 2 sin ( λ 0 λ 1 δ ) C 1 sin ( λ 0 λ 1 δ ) C 2 sin ( λ 0 λ 1 δ ) ) ) .
where, δ = Γ ( ϱ + 1 ) ( μ x ϵ ϵ 4 λ 0 λ 1 μ 2 + 3 2 λ 0 λ 1 μ 1 2 μ 2 2 λ 0 λ 1 μ 1 t ϵ ϵ ) .

4. Physical Interpretation

This section presents the physical interpretation of the obtained solutions. Figure 1 shows the physical behavior of the appearance of solution Abs(h(x,t)) in Equation (24) with ϵ = 0.6 , ϱ = 2 , μ = 0.5 , Ω = 0.1 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 . Figure 2 shows the physical behavior of the appearance of solution Abs(h(x,t)) in Equation (25) with ϵ = 0.6 , ϱ = 2 , μ = 1.5 , Ω = 1 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 . Figure 3 shows the physical behavior of the appearance of solution Abs(h(x,t)) in Equation (26) with ϵ = 0.8 , ϱ = 2 , μ = 0.5 , Ω = 0.1 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 . Figure 4 shows the physical behavior of that appearance of solution Abs(h(x,t)) in Equation (27) with ϵ = 0.6 , ϱ = 2 , μ = 1.5 , Ω = 1 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 . Figure 5 shows the physical behavior of the appearance of solution Abs(h(x,t)) in Equation (145) with ϵ = 0.8 , λ 0 = 0.2 , λ 1 = 3 , μ = 0.5 , C 1 = 1 , C 2 = 0.3 , ϱ = 2 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 . Figure 6 shows the physical behavior of the appearance of solution Abs(h(x,t)) in Equation (146) with ϵ = 0.8 , λ 0 = 3 , λ 1 = 3 , μ = 0.5 , C 1 = 1 , C 2 = 0.3 , ϱ = 2 in (a) 3D, in (b) 2D, in (c) contour, in (d) polar form, and in (e) with fractional order ϵ = 0.2 , 0.4 , 0.6 , 0.8 , 1 .

5. MI Analysis

Examination of a steady-state method for solving the FitzHugh–Nagumo model is shown in [22]:
h ( x , t ) = H ( x , t ) + τ e ι τ t ,
where τ shows the optical normalizing power.
By using Equation (168) in Equation (1), by linearity we get
H ( 0 , 1 ) ( x , t ) H ( 2 , 0 ) ( x , t ) + θ H ( x , t ) + ι τ H ( x , t ) + θ τ + ι τ 3 / 2 = 0 .
Suppose the solution of Equation (169) is shown in [22], mentioned as:
H ( x , t ) = A 1 e ι ( p x r t ) + A 2 e ι ( p x r t ) .
Here, “p” and “r” are the constants. Equation (170) is substituted into Equation (169). After solving the determinant of the coefficient matrix, we obtain the dispersion relation by adding the coefficients of e ι ( p x r t ) and e ι ( p x r t ) .
θ 2 + p 4 + 2 θ p 2 + r 2 τ 2 = 0 .
The dispersion relation can be found from Equation (171) for “r”, results
r = ± θ 2 p 4 2 θ p 2 + τ 2 .
The steady-state stability is demonstrated by the gained dispersion relation. The steady-state outcome will not be stable when wave number q is not real because the perturbation increases exponentially. However, a steady state becomes stable against minor perturbations if q is not imaginary. The result is in the unstable steady state if
θ 2 p 4 2 θ p 2 + τ 2 < 0 .
It is possible to acquire the MI gain spectrum G ( p ) and shown in the Figure 7:
G ( p ) = 2 I m ( r ) = ± θ 2 p 4 2 θ p 2 + τ 2 .

6. Conclusions

This paper contained the several kinds of exact wave solitons of the nonlinear FitzHugh–Nagumo model with a truncated M-fractional derivative. By using the modified extended tanh function technique and the modified ( G / G 2 ) -expansion technique, we are able to obtain many more exact wave solitons, having kink, dark-bright, singular-periodic, periodic, bright, dark, singular, and complexiton solitons. The solutions differ from the current solutions due to the use of fractional derivatives. By using the fractional derivative, we obtained the solutions closer to the numerical solutions. The fractional derivative represents the concerned model in a more prominent form. In the very different domains of biosciences and telecommunications, especially in neural signal simulation and pulse control for optical communication, these solutions might be helpful. Contour graphs, two- and three-dimensional plots, are used to illustrate the obtained results. In addition, the stationary results of the governing equation are studied through modulation instability. Findings aid in the development of the relevant system. It is suggested that the techniques employed have applicability for other nonlinear equations in different scientific and engineering domains.

Author Contributions

Both authors equally contribute in the work. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the Deanship of Scientific Research, Vice Presidency for Graduate Studies and Scientific Research, King Faisal University, Saudi Arabia [KFU252427].

Data Availability Statement

Data Availability Statements are available in the manuscript.

Conflicts of Interest

The authors declare no conflicts of interest.

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Figure 1. (Kink-like soliton) Physical behavior of solution’s appearance in Equation (24).
Figure 1. (Kink-like soliton) Physical behavior of solution’s appearance in Equation (24).
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Figure 2. (Dark soliton) Physical behavior of solution’s appearance in Equation (25).
Figure 2. (Dark soliton) Physical behavior of solution’s appearance in Equation (25).
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Figure 3. (Periodic soliton) Physical behavior of solution’s appearance in Equation (26).
Figure 3. (Periodic soliton) Physical behavior of solution’s appearance in Equation (26).
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Figure 4. (Singular soliton) Physical behavior of solution’s appearance in Equation (27).
Figure 4. (Singular soliton) Physical behavior of solution’s appearance in Equation (27).
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Figure 5. (Kink soliton) Physical behavior of solution’s appearance in Equation (145).
Figure 5. (Kink soliton) Physical behavior of solution’s appearance in Equation (145).
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Figure 6. (Periodic wave solution) Physical behavior of solution’s appearance in Equation (146).
Figure 6. (Periodic wave solution) Physical behavior of solution’s appearance in Equation (146).
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Figure 7. MI gain spectrum at the values p = 1 , 2 , 3 , 4 and θ = 0.5 for τ [ 5 , 5 ] .
Figure 7. MI gain spectrum at the values p = 1 , 2 , 3 , 4 and θ = 0.5 for τ [ 5 , 5 ] .
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Alomair, M.A.; Qawaqneh, H. Mathematical and Physical Analysis of the Fractional Dynamical Model. Fractal Fract. 2025, 9, 453. https://doi.org/10.3390/fractalfract9070453

AMA Style

Alomair MA, Qawaqneh H. Mathematical and Physical Analysis of the Fractional Dynamical Model. Fractal and Fractional. 2025; 9(7):453. https://doi.org/10.3390/fractalfract9070453

Chicago/Turabian Style

Alomair, Mohammed Ahmed, and Haitham Qawaqneh. 2025. "Mathematical and Physical Analysis of the Fractional Dynamical Model" Fractal and Fractional 9, no. 7: 453. https://doi.org/10.3390/fractalfract9070453

APA Style

Alomair, M. A., & Qawaqneh, H. (2025). Mathematical and Physical Analysis of the Fractional Dynamical Model. Fractal and Fractional, 9(7), 453. https://doi.org/10.3390/fractalfract9070453

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