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Article

Analytical Techniques for Studying Fractional-Order Jaulent–Miodek System Within Algebraic Context

1
Mathematics Department, College of Basic Education, Public Authority for Applied Education and Training (PAAET), Ardiya 70654, Kuwait
2
Laboratory Technology Department, College of Technological Studies, Public Authority for Applied Education and Training (PAAET), Shuwaikh 70654, Kuwait
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Fractal Fract. 2025, 9(1), 50; https://doi.org/10.3390/fractalfract9010050
Submission received: 29 November 2024 / Revised: 8 January 2025 / Accepted: 13 January 2025 / Published: 17 January 2025
(This article belongs to the Special Issue Fractional Systems, Integrals and Derivatives: Theory and Application)

Abstract

The proposed study seeks to investigate various analytical and numerical techniques for solving fractional differential equations, with a particular focus on their applications in mathematical modeling and scientific research within the field of algebra. This study intends to investigate methods such as the Aboodh transform iteration method and the Aboodh residual power series method, specifically for addressing the Jaulent–Miodek system of partial differential equations. By analyzing the behavior of fractional-order differential equations and their solutions, this research seeks to contribute to a deeper understanding of complex mathematical phenomena. Furthermore, this study examines the role of the Caputo operator in fractional calculus, offering insights into its significance in modeling real-world systems within the algebraic context. Through this research, novel approaches for solving fractional differential equations are developed, offering essential tools for researchers in diverse fields of science and engineering, including algebraic applications.

1. Introduction

Fractional calculus (FC), which examines integrals and derivatives of fractional orders [1,2], can be utilized to determine the fundamental characteristics and memory of a wide variety of complex structures. A considerable number of recent FC applications have incorporated the conversion of classical to non-integer-order derivatives and integrals in order to examine the dynamics of large-scale physical events. Several areas of engineering and physical research make use of it, including signal processing, mathematical biology, flow models, relaxation, and viscoelasticity [3,4,5,6]. By delving into fractional mathematics, many previously opaque ideas have become clear. Several examples have shown that fractional operators surpass operators of the integer order, such as logistic regression, Malthusian growth, and blood alcohol concentration [7,8].
A few of the fractional derivatives that have been developed recently include Caputo, Hilfer, Riemann–Liouville, Caputo–Fabrizio, Atangana–Baleanu, and Grunwald–Letnikov [9,10,11,12]. Since all fractional derivatives may be reduced to Caputo’s meaning with small parametric modifications, the Caputo fractional derivative is the fundamental concept of fractional calculus and is used in the analysis of fractional differential equations (FDEs). The power-law kernel is an essential aspect of Caputo’s operator, a popular tool for modeling many physical systems. The authors suggested using an alternate fractional differential operator [,13] to address this problem. A Mittag–Leffler kernel plus an exponentially decaying kernel make up this operator. Two operators, Atangana–Baleanu and Caputo–Fabrizio (C-F), can be told apart by the fact that their kernels are not singular. The operators stated above have been widely used to analyze several types of issues, including those in bioengineering, economics, geophysics, and biology [14,15,16]. Fractional calculus and advanced nonlinear analysis methods have become increasingly significant in addressing complex mathematical models across science and engineering [17,18,19,20,21]. Recent studies have provided innovative approaches to solving diverse problems, such as traveling wave solutions in nonlinear Schrodinger equations [22], fractional damping in asymmetric Duffing systems [23], and nonlinear flutter dynamics in wind-excited bridges [24]. Moreover, advanced techniques like few-shot identification for stochastic systems [25] and neural ordinary differential equations for robust parameter estimation [26] have shown great promise. The integration of fractional-order control [27] and algebraic structures [28] further underscores the critical role of modern mathematical and computational tools in solving real-world challenges. These studies collectively highlight the potential of interdisciplinary approaches to advancing knowledge in nonlinear dynamics and fractional systems [29,30,31,32,33].
In nonlinear science, integrable nonlinear evolution equations have garnered significant attention because of their ability to describe various nonlinear phenomena in natural sciences [34]. It is common practice to use the Jaulent–Miodek equation to describe the energy-dependent Schrodinger ability [35]. To learn more about how these equation systems are utilized as a model to solve real-world issues in many technical and natural science domains, see [36,37,38] and its references. Understanding the numerical and analytic-approximate solutions to these nonlinear partial differential equations (NPDEs) is crucial because of the paramount importance of these equations in the aforementioned domains. Plasma physics [39] and condensed matter physics [40,41] are only two examples of the several scientific and technical fields that rely on the nonlinear fractional-order coupled Jaulent–Miodek equations, which are the subject of ongoing comprehensive mathematical research.
The following Jaulent–Miodek time fractional partial differential systems are under consideration:
D ψ p Φ 1 ( σ , ψ ) + 3 Φ 1 ( σ , ψ ) σ 3 + 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 + 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ = 0 , D ψ p Φ 2 ( σ , ψ ) + 3 Φ 2 ( σ , ψ ) σ 3 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ = 0 , where 0 < p 1 ,
with the following initial conditions:
Φ 1 ( σ , 0 ) = N ( σ ) , Φ 2 ( σ , 0 ) = M ( σ ) .
The Jaulent–Miode Equation (1) describes the dispersive long wave in shallow water, where Φ 1 ( σ , ψ ) is the field of horizontal velocity, Φ 2 ( σ , ψ ) is the height which deviates from the equilibrium position of liquid, ψ respresents time, and p is the fractional derivative. The following methods can be used to approximate solutions of linear and nonlinear first-order partial differential equations: the Adomian decomposition method (ADM) [42,43], the homotopy perturbation method (HPM) [44,45,46,47,48], the variational iteration method (VIM) [49], the q-homotopy analysis transform method (q-HATM) [50,51], the fractional natural decomposition method (FNDM) [52], the fractional multi-step differential transformed method (FMsDTM) [53], the new iterative method (NIM) [54,55,56], and the homotopy analysis method (HAM) [57,58,59,60].
Omar Abu Arqub founded the residual power series method (RPSM) in 2013, as pointed out in [61]. It is produced by the multiplication of the residual error function with the Taylor series. Some of the aspects that are solved by an infinite convergence series include DEs [57]. This has led to the development of several DEs, including Boussinesq DEs, Fuzzy DEs, KdV Burger’s equation, and many others, which, in turn, created new RPSM algorithms. Essentially, these algorithms seek to provide the so-called “best guess” approximations.
A novel approach to solving fractional-order differential equations (FODEs) was developed by combining two effective approaches. When combined with the Sumudu transform, this category includes methods such as the Shehu transformation and the Adomian decomposition method [62], the Laplace transform with RPSM [63,64,65], the natural transform [66], and the homotopy perturbation approach [67]. Other examples fall into this category as well. See [68,69,70] for further information on how the two approaches are combined. For time-fractional nonlinear PDEs, we found both approximative and exact solutions using a new combination approach called the ARPSM in this research. This approach is noteworthy because of its novel combination of the Aboodh transform methodology with the RPSM [71,72].
The computing effort and complexity needed by the aforementioned approaches are major limitations. The distinguishing aspect of our study is the iterative Aboodh transform technique (ATIM) that we devised [73] for solving the Jaulent–Miodek system of partial differential equations. This approach successfully reduces the computational complexity and effort required by integrating the Aboodh transform with the new iterative method. The convergent series solution is generated by the suggested approach, as mentioned in [74,75].
The Aboodh residual power series method (ARPSM) and the Aboodh transform iterative technique (ATIM) are the two simplest ways to solve fractional DEs, according to the literature [71,72,73,74,75]. Not only do these approaches provide discretization-free and nonlinear numerical solutions to PDEs, but they also define all the symbolic concepts used in analytical solutions instantly and thoroughly. This work aims to compare and contrast the effectiveness of ARPSM and ATIM in solving the Jaulent–Miodek system of PDEs. Notably, many fractional differential problems, both linear and nonlinear, have been solved using these two approaches.
This work uses the ARPSM and ATIM to solve the partial differential equations of the Jaulent–Miodek system. The numerical solutions produced by these approaches are more accurate than those of other numerical techniques. There is also a comparison study of the numerical data. The results of the suggested methods are consistent with one another, which is a strong sign of how well and reliably they work. As their values grow, fractional-order derivatives become more visually striking. Hence, the approaches are quick, accurate, easy to apply, and resistant to computational error phases. Mathematicians are able to solve a wide variety of partial differential equations with ease because of this breakthrough.
The rest of the study is structured as follows: We describe the fractional calculus definitions in Section 2. We construct the general methodologies of the proposed methods in Section 3; the application of the proposed methods is presented in Section 4. Section 5 provides results and discussion; finally, Section 6 concludes our study.

2. Fractional Calculus Fundamental Concepts

Definition 1
([76]). Consider the piecewise continuous function Φ ( σ , ψ ) to be exponentially ordered. According to the definition of the Aboodh transform (AT), τ 0 for Φ ( σ , ψ ) .
A [ Φ ( σ , ψ ) ] = Λ ( σ , ϵ ) = 1 ϵ 0 Φ ( σ , ψ ) e ψ ϵ d ψ , r 1 ϵ r 2 .
The Aboodh inverse transform (AIT) is precisely defined as follows:
A 1 [ Λ ( σ , ϵ ) ] = Φ ( σ , ψ ) = 1 2 π i u i u + i Λ ( σ , ψ ) ϵ e ψ ϵ d ψ .
where σ = ( σ 1 , σ 2 , , σ p ) R and p N .
Lemma 1
([77,78]). The two piecewise continuous functions that are ordered exponentially are supposed to be Φ 1 ( σ , ψ ) and Φ 2 ( σ , ψ ) in [ 0 , ] as an interval. Let A [ Φ 1 ( σ , ψ ) ] = Λ 1 ( σ , ψ ) , A [ Φ 2 ( σ , ψ ) ] = Λ 2 ( σ , ψ ) , and χ 1 , χ 2 are constants. Thus, the following characteristics are true:
  • A [ χ 1 Φ 1 ( σ , ψ ) + χ 2 Φ 2 ( σ , ψ ) ] = χ 1 Λ 1 ( σ , ϵ ) + χ 2 Λ 2 ( σ , ψ ) ,
  • A 1 [ χ 1 Λ 1 ( σ , ψ ) + χ 2 Λ 2 ( σ , ψ ) ] = χ 1 Φ 1 ( σ , ϵ ) + χ 2 Φ 2 ( σ , ψ ) ,
  • A [ J ψ p Φ ( σ , ψ ) ] = Λ ( σ , ϵ ) ϵ p ,
  • A [ D ψ p Φ ( σ , ψ ) ] = ϵ p Λ ( σ , ϵ ) K = 0 r 1 Φ K ( σ , 0 ) ϵ K p + 2 , r 1 < p r , r N .
Definition 2
([79]). Caputo’s fractional derivative for any order p may be found for the function Φ ( σ , ψ ) is given as
D ψ p Φ ( σ , ψ ) = J ψ m p Φ ( m ) ( σ , ψ ) , r 0 , m 1 < p m ,
where σ = ( σ 1 , σ 2 , , σ p ) R p and m , p R , J ψ m p is the R-L integral of Φ ( σ , ψ ) .
Definition 3
([80]). The power series representation is comprised of the following form:
r = 0 r ( σ ) ( ψ ψ 0 ) r p = 0 ( ψ ψ 0 ) 0 + 1 ( ψ ψ 0 ) p + 2 ( ψ ψ 0 ) 2 p + ,
where σ = ( σ 1 , σ 2 , , σ p ) R p and p N . The series about ψ 0 with ψ as a variable and the series coefficients r ( σ ) is referred to as a multiple fractional power series (MFPS).
Lemma 2.
Let Φ ( σ , ψ ) be the exponential order function.For A [ Λ ( σ , ϵ ) ] = Φ ( σ , ψ ) , the AT in this case is described by the following equation:
A [ D ψ r p Φ ( σ , ψ ) ] = ϵ r p Λ ( σ , ϵ ) j = 0 r 1 ϵ p ( r j ) 2 D ψ j p Φ ( σ , 0 ) , 0 < p 1 ,
where σ = ( σ 1 , σ 2 , , σ p ) R p and p N and D ψ r p = D ψ p . D ψ p . . D ψ p ( r t i m e s ) .
Proof. 
Equation (2) may be proven via induction. When r = 1 is inserted into Equation (2), the following outcomes are produced:
A [ D ψ 2 p Φ ( σ , ψ ) ] = ϵ 2 p Λ ( σ , ϵ ) ϵ 2 p 2 Φ ( σ , 0 ) ϵ p 2 D ψ p Φ ( σ , 0 ) .
According to the fourth part of Lemma 1, Equation (2) holds for r = 1 . With r = 2 replaced in Equation (2), we obtain the following:
A [ D r 2 p Φ ( σ , ψ ) ] = ϵ 2 p Λ ( σ , ϵ ) ϵ 2 p 2 Φ ( σ , 0 ) ϵ p 2 D ψ p Φ ( σ , 0 ) .
By using the L.H.S, of Equation (4), we can obtain the following:
L . H . S = A [ D ψ 2 p Φ ( σ , ψ ) ] .
Equation (5) may be expressed as follows:
L . H . S = A [ D ψ p Φ ( σ , ψ ) ] .
Consider
z ( σ , ψ ) = D ψ p Φ ( σ , ψ ) .
As a result, Equation (6) becomes
L . H . S = A [ D ψ p z ( σ , ψ ) ] .
The use of the Caputo derivative leads to the modification of Equation (8).
L . H . S = A [ J 1 p z ( σ , ψ ) ] .
Equation (9) provides the R-L integral for AT, which is used to derive the following:
L . H . S = A [ z ( σ , ψ ) ] ϵ 1 p .
Equation (10) is transformed into the following form by using the differentiability characteristic of the AT:
L . H . S = ϵ p Z ( σ , ϵ ) z ( σ , 0 ) ϵ 2 p .
We obtain the following result from Equation (7):
Z ( σ , ϵ ) = ϵ p Λ ( σ , ϵ ) Φ ( σ , 0 ) ϵ 2 p ,
where A [ z ( σ , ψ ) ] = Z ( σ , ϵ ) . Consequently, Equation (11) is transform to
L . H . S = ϵ 2 p Λ ( σ , ϵ ) Φ ( σ , 0 ) ϵ 2 2 p D ψ p Φ ( σ , 0 ) ϵ 2 p ,
when r = K . Equation (2) and Equation (12) are compatible. The validity of Equation (2) for r = K is assumed. This replaces r = K in Equation (2).
A [ D ψ K p Φ ( σ , ψ ) ] = ϵ K p Λ ( σ , ϵ ) j = 0 K 1 ϵ p ( K j ) 2 D ψ j p D ψ j p Φ ( σ , 0 ) , 0 < p 1 .
The last step is to demonstrate the validity of Equation (2) for the value r = K + 1 . To write the following, we use Equation (2):
A [ D ψ ( K + 1 ) p Φ ( σ , ψ ) ] = ϵ ( K + 1 ) p Λ ( σ , ϵ ) j = 0 K ϵ p ( ( K + 1 ) j ) 2 D ψ j p Φ ( σ , 0 ) .
The following is obtained by examining the left side of Equation (14):
L . H . S = A [ D ψ K p ( D ψ K p ) ] .
Let
D ψ K p = g ( σ , ψ ) .
From Equation (15), we obtain the following:
L . H . S = A [ D ψ p g ( σ , ψ ) ] .
The following result may be obtained from Equation (16) by using the R-L integral and the Caputo derivative:
L . H . S = ϵ p A [ D ψ K p Φ ( σ , ψ ) ] g ( σ , 0 ) ϵ 2 p .
Using Equation (13) to obtain Equation (17),
L . H . S = ϵ r p Λ ( σ , ϵ ) j = 0 r 1 ϵ p ( r j ) 2 D ψ j p Φ ( σ , 0 ) .
In addition, the outcome that follows is obtained from Equation (18):
L . H . S = A [ D ψ r p Φ ( σ , 0 ) ] .
Equation (2) is valid for r = K + 1 , therefore. Equation (2) is valid for all positive integers when the mathematical induction method is used. □
As a definition of MFTS, consider the following lemma. We go into more depth about the ARPSM here, but this formula is useful throughout.
Lemma 3.
Suppose we have an exponential order function Φ ( σ , ψ ) . A [ Φ ( σ , ψ ) ] = Λ ( σ , ϵ ) represents the AT of Φ ( σ , ψ ) . Thefollowingis the notation that is used for AT in MFTS:
Λ ( σ , ϵ ) = r = 0 r ( σ ) ϵ r p + 2 , ϵ > 0 ,
where σ = ( s 1 , σ 2 , , σ p ) R p , p N .
Proof. 
Consider the fractional order Taylor series:
Φ ( σ , ψ ) = 0 ( σ ) + 1 ( σ ) ψ p Γ [ p + 1 ] + + 2 ( σ ) ψ 2 p Γ [ 2 p + 1 ] + .
When the AT is applied to Equation (20), the following equality is obtained as a result:
A Φ ( σ , ψ ) = A 0 ( σ ) + A 1 ( σ ) ψ p Γ [ p + 1 ] + A 1 ( σ ) ψ 2 p Γ [ 2 p + 1 ] +
We obtain the following by taking advantage of the characteristics of the AT:
A Φ ( σ , ψ ) = 0 ( σ ) 1 ϵ 2 + 1 ( σ ) Γ [ p + 1 ] Γ [ p + 1 ] 1 ϵ p + 2 + 2 ( σ ) Γ [ 2 p + 1 ] Γ [ 2 p + 1 ] 1 ϵ 2 p + 2
Hence, (19) is product, a version of the Taylor series that is particular to AT. □
Lemma 4.
The MFPS may be represented as A [ Φ ( σ , ψ ) ] = Λ ( σ , ϵ ) , using the revised version of the Taylor series (19).
0 ( σ ) = lim ϵ ϵ 2 Λ ( σ , ϵ ) = Φ ( σ , 0 ) .
Proof. 
Consider the new form of the Taylor series:
0 ( σ ) = ϵ 2 Λ ( σ , ϵ ) 1 ( σ ) ϵ p 2 ( σ ) ϵ 2 p
The necessary solution may be obtained by applying l i m x to Equation (21) and performing a brief calculation, as shown in (22). □
Theorem 1.
The following is the representation of the function A [ Φ ( σ , ψ ) ] = Λ ( σ , ϵ ) in MFPS notation:
Λ ( σ , ϵ ) = 0 r ( σ ) ϵ r p + 2 , ϵ > 0 ,
where σ = ( σ 1 , σ 2 , , σ p ) R p and p N . Then, we have
r ( σ ) = D r r p Φ ( σ , 0 ) ,
where D ψ r p = D ψ p . D ψ p . . D ψ p ( r t i m e s ) .
Proof. 
The new variant of the Taylor series is as follows:
1 ( σ ) = ϵ p + 2 Λ ( σ , ϵ ) ϵ p 0 ( σ ) 2 ( σ ) ϵ p 3 ( σ ) ϵ 2 p
The lim ϵ is taken on both sides of (23) to obtain
1 ( σ ) = lim ϵ ( ϵ p + 2 Λ ( σ , ϵ ) ϵ p 0 ( σ ) ) lim ϵ 2 ( σ ) ϵ p lim ϵ 3 ( σ ) ϵ 2 p
Taking the limit leads to the equality shown below:
1 ( σ ) = lim ϵ ( ϵ p + 2 Λ ( σ , ϵ ) ϵ p 0 ( σ ) ) .
Lemma 2 is applied to Equation (24) to provide the following result:
1 ( σ ) = lim ϵ ( ϵ 2 A [ D ψ p Φ ( σ , ψ ) ] ( ϵ ) ) .
Furthermore, Lemma 3 is used to transform Equation (25).
1 ( σ ) = D ψ p Φ ( σ , 0 ) .
Assuming limit ϵ and using the new form of the Taylor series again, we obtain
2 ( σ ) = ϵ 2 p + 2 Λ ( σ , ϵ ) ϵ 2 p 0 ( σ ) ϵ p 1 ( σ ) 3 ( σ ) ϵ p
The outcome is given by Lemma 3.
2 ( σ ) = lim ϵ ϵ 2 ( ϵ 2 p Λ ( σ , ϵ ) ϵ 2 p 2 0 ( σ ) ϵ p 2 1 ( σ ) ) .
For the purpose of transforming Equation (26), the Lemmas 2 and 4 are utilized.
2 ( σ ) = D ψ 2 p Φ ( σ , 0 ) .
Applying the same method to the new Taylor series results in the following outcomes being achieved:
3 ( σ ) = lim ϵ ϵ 2 ( A [ D ψ 2 p Φ ( σ , p ) ] ( ϵ ) ) .
Through the utilization of Lemma 4, the ultimate equation is acquired.
3 ( σ ) = D ψ 3 p Φ ( σ , 0 ) .
In general,
r ( σ ) = D ψ r p Φ ( σ , 0 ) .
The proof is concluded. □
A description and demonstration of the concepts that govern the convergence of the Taylor series in its new form is provided in the following theorem.
Theorem 2.
A formula for MFTS is presented in Lemma 3, and it can be written in the following manner: A [ Φ ( ψ , σ ) ] = Λ ( σ , ϵ ) . When | ϵ a A [ D ψ ( K + 1 ) p Φ ( σ , ψ ) ] | T for all 0 < ϵ s and 0 < p 1 , the residual R K ( σ , ϵ ) of the new MFTS is satisfied with the following inequality:
| R K ( σ , ϵ ) | T ϵ ( K = 1 ) p + 2 , 0 < ϵ s .
Proof. 
Assume A [ D ψ r p Φ ( σ , ψ ) ] ( ϵ ) is defined on 0 < ϵ s for r = 0 , 1 , 2 , , K + 1 . Let | ϵ 2 A [ D ψ K + 1 Φ ( σ , t a u ) ] | T o n 0 < ϵ s . Determine the following relationship using the new Taylor series:
R K ( σ , ϵ ) = Λ ( σ , ϵ ) r = 0 K r ( σ ) ϵ r p + 2 .
In order to convert Equation (27), Theorem 1 is utilized.
R K ( σ , ϵ ) = Λ ( σ , ϵ ) r = 0 K D ψ r p Φ ( σ , 0 ) ϵ r p + 2 .
This equation may be solved by multiplying ϵ ( K + 1 ) a + 2 on both sides in order to obtain
ϵ ( K + 1 ) p + 2 R K ( σ , ϵ ) = ϵ 2 ( ϵ ( K + 1 ) p Λ ( σ , ϵ ) r = 0 K ϵ ( K + 1 r ) p 2 D ψ r p Φ ( σ , 0 ) ) .
When Lemma 2 is applied to Equation (29), the result is
ϵ ( K + 1 ) p + 2 R K ( σ , ϵ ) = ϵ 2 A [ D ψ ( K + 1 ) p Φ ( σ , ψ ) ] .
Taking the absolute value of Equation (30), we obtain
| ϵ ( K + 1 ) p + 2 R K ( σ , ϵ ) | = | ϵ 2 A [ D ψ ( K + 1 ) p Φ ( σ , ψ ) ] | .
The result that is obtained by applying the condition that is presented in Equation (31) is as follows:
T ϵ ( K + 1 ) p + 2 R K ( σ , ϵ ) T ϵ ( K + 1 ) p + 2 .
Using Equation (32), the required outcome is achieved.
| R K ( σ , ϵ ) | T ϵ ( K + 1 ) p + 2 .
Consequently, the convergence of a series is established in accordance with a new condition. □

3. Methodologies

3.1. ARPSM Technique

Here, we outline how the ARPSM rules formed the foundation for our general model solution.
Step1. Consider the following general equation:
D ψ q p Φ ( σ , ψ ) + ϑ ( σ ) N ( Φ ) δ ( σ , Φ ) = 0 .
Step 2. AT is applied to both sides of Equation (33) in order to obtain
A [ D ψ q p Φ ( σ , ψ ) + ϑ ( σ ) N ( Φ ) δ ( σ , Φ ) ] = 0 .
The following version of Equation (34) may be obtained by using Lemma 2.
Λ ( σ , s ) = j = 0 q 1 D ψ j Φ ( σ , 0 ) s q p + 2 ϑ ( σ ) Y ( s ) s q p + F ( σ , s ) s q p ,
where A [ δ ( σ , Φ ) ] = F ( σ , s ) , A [ N ( Φ ) ] = Y ( s ) .
Step 3. Examine the form that is obtained after solving Equation (35).
Λ ( σ , s ) = r = 0 r ( σ ) s r p + 2 . s > 0 ,
Step 4. Observe these procedures:
0 ( σ ) = lim s s 2 Λ ( σ , s ) = Φ ( σ , 0 ) .
The following outcome is obtained when Theorem 2 is applied:
1 ( σ ) = D ψ p Φ ( σ , 0 ) , 2 ( σ ) = D ψ 2 p Φ ( σ , 0 ) , w ( σ ) = D ψ w p Φ ( σ , 0 ) .
Step 5. The following formula may be used to obtain the K t h truncated series Λ ( σ , s ) :
Λ K ( σ , s ) = r = 0 K r ( σ ) s r p + 2 , s > 0 ,
Λ K ( σ , s ) = 0 ( σ ) s 2 + 1 ( σ ) s p + 2 + + w ( σ ) s w p + 2 + r = w + 1 K r ( σ ) s r p + 2 .
Step 6. Evaluate the Aboodh residual function (ARF) from (35) and the K t h -truncated Aboodh residual function independently in order to deduce the following:
A R e s ( σ , s ) = Λ ( σ , s ) j = 0 q 1 D ψ j Φ ( σ , 0 ) s j p + 2 + ϑ ( σ ) Y ( s ) s j p F ( σ , s ) s j p ,
and
A R e s K ( σ , s ) = Λ K ( σ , s ) j = 0 q 1 D ψ j Φ ( σ , 0 ) s j p + 2 + ϑ ( σ ) Y ( s ) s j p F ( σ , s ) s j p .
Step 7. The expansion form of Equation (36) may be substituted with Λ K ( σ , s ) .
A R e s K ( σ , s ) = 0 ( σ ) s 2 + 1 ( σ ) s p + 2 + + w ( σ ) s w p + 2 + r = w + 1 K r ( σ ) s r p + 2 j = 0 q 1 D ψ j Φ ( σ , 0 ) s j p + 2 + ϑ ( σ ) Y ( s ) s j p F ( σ , s ) s j p .
Step 8. The solution to Equation (37) may be obtained by multiplying s K p + 2 on both sides of the equation.
s K p + 2 A R e s K ( σ , s ) = s K p + 2 ( 0 ( σ ) s 2 + 1 ( σ ) s p + 2 + + w ( σ ) s w p + 2 + r = w + 1 K r ( σ ) s r p + 2 j = 0 q 1 D ψ j Φ ( σ , 0 ) s j p + 2 + ϑ ( σ ) Y ( s ) s j p F ( σ , s ) s j p ) .
Step 9: Evaluating Equation (38) with lim s yields the following result:
lim s s K p + 2 A R e s K ( σ , s ) = lim s s K p + 2 ( 0 ( σ ) s 2 + 1 ( σ ) s p + 2 + + w ( σ ) s w p + 2 + r = w + 1 K r ( σ ) s r p + 2 j = 0 q 1 D ψ j Φ ( σ , 0 ) s j p + 2 + ϑ ( σ ) Y ( s ) s j p F ( σ , s ) s j p ) .
Step 10. In order to obtain the value of K ( σ ) , one may solve the equation shown above.
lim s ( s K p + 2 A R e s K ( σ , s ) ) = 0 ,
where K = w + 1 , w + 2 , .
Step 11. To find the K-approximate solution of Equation (35), replace the values of K ( σ ) with a K-truncated series of Λ ( σ , s ) .
Step 12. For the purpose of obtaining the K-approximate solution using the AIT, find Λ K ( σ , s ) as Φ K ( σ , ψ ) .

3.2. ATIM Technique

Consider the general PDE of fractional order:
D ψ p Φ ( σ , ψ ) = Υ Φ ( σ , ψ ) , D σ ψ Φ ( σ , ψ ) , D σ 2 ψ Φ ( σ , ψ ) , D σ 3 ψ Φ ( σ , ψ ) , 0 < p , ψ 1 .
Having the ICs,
Φ ( k ) ( σ , 0 ) = h k , k = 0 , 1 , 2 , , m 1 .
The function Φ ( σ , ψ ) is an unknown function, while Υ ( Φ ( σ , ψ ) , D σ ψ Φ ( σ , ψ ) , D σ 2 ψ Φ ( σ , ψ ) , D σ 3 ψ Φ ( σ , ψ ) ) may be the linear operator or nonlinear operator of Φ ( σ , ψ ) , D σ ψ Φ ( σ , ψ ) , D σ 2 ψ Φ ( σ , ψ ) , and D σ 3 ψ Φ ( σ , ψ ) . By using the AT on both sides of Equation (39), the following expression is produced.
A [ Φ ( σ , ψ ) ] = 1 s p k = 0 m 1 Φ ( k ) ( σ , 0 ) s 2 p + k + A Υ Φ ( σ , ψ ) , D σ ψ Φ ( σ , ψ ) , D σ 2 ψ Φ ( σ , ψ ) , D σ 3 ψ Φ ( σ , ψ ) .
In this case, the AIT provides the following solution:
Φ ( σ , ψ ) = A 1 1 s p k = 0 m 1 Φ ( k ) ( σ , 0 ) s 2 p + k + A Υ Φ ( σ , ψ ) , D σ ψ Φ ( σ , ψ ) , D σ 2 ψ Φ ( σ , ψ ) , D σ 3 ψ Φ ( σ , ψ ) .
The solution found by ATIM is shown by the infinite series.
Φ ( σ , ψ ) = i = 0 Φ i .
Since Υ Φ , D σ ψ Φ , D σ 2 ψ Φ , D σ 3 ψ Φ is a nonlinear or linear operator, it can be decomposed as follows:
Υ Φ , D σ ψ Φ , D σ 2 ψ Φ , D σ 3 ψ Φ = Υ Φ 0 , D σ ψ Φ 0 , D σ 2 ψ Φ 0 , D σ 3 ψ Φ 0 + i = 0 Υ k = 0 i Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k Υ k = 1 i 1 Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k .
The following equation is obtained by replacing the values of Equations (43) and (44) in the original Equation (42).
i = 0 Φ i ( σ , ψ ) = A 1 1 s p k = 0 m 1 Φ ( k ) ( σ , 0 ) s 2 p + k + A [ Υ ( Φ 0 , D σ ψ Φ 0 , D σ 2 ψ Φ 0 , D σ 3 ψ Φ 0 ) ] + A 1 1 s p A i = 0 Υ k = 0 i ( Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k ) A 1 1 s p A Υ k = 1 i 1 ( Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k )
Φ 0 ( σ , ψ ) = A 1 1 s p k = 0 m 1 Φ ( k ) ( σ , 0 ) s 2 p + k , Φ 1 ( σ , ψ ) = A 1 1 s p A [ Υ ( Φ 0 , D σ ψ Φ 0 , D σ 2 ψ Φ 0 , D σ 3 ψ Φ 0 ) ] , Φ m + 1 ( σ , ψ ) = A 1 1 s p A i = 0 Υ k = 0 i ( Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k ) A 1 1 s p A Υ k = 1 i 1 ( Φ k , D σ ψ Φ k , D σ 2 ψ Φ k , D σ 3 ψ Φ k ) , m = 1 , 2 , .
The analytical approximation for the m-term of Equation (39) may be obtained using the following expression:
Φ ( σ , ψ ) = i = 0 m 1 Φ i .

4. Application of the Proposed Methods

4.1. Solution of the Problem Using ARPSM

Examine the system of Jaulent–Miodek equations of fractional order:
D ψ p Φ 1 ( σ , ψ ) + 3 Φ 1 ( σ , ψ ) σ 3 + 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 + 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ = 0 , D ψ p Φ 2 ( σ , ψ ) + 3 Φ 2 ( σ , ψ ) σ 3 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ = 0 , where 0 < p 1 .
Having ICs,
Φ 1 ( σ , 0 ) = 1 8 q 2 1 sec h 2 q σ 2 , Φ 2 ( σ , 0 ) = q sec h q σ 2 ,
and exact solution
Φ 1 ( σ , ψ ) = 1 8 q 2 1 sec h 2 1 2 q σ q 2 ψ 2 , Φ 2 ( σ , ψ ) = q sec h 1 2 q σ q 2 ψ 2 ,
Using Equation (49), AT is applied to Equation (48) in order to obtain
Φ 1 ( σ , s ) 1 8 q 2 1 sec h 2 q σ 2 s 2 + 1 s p 3 Φ 1 ( σ , s ) σ 3 + 3 2 s p A ψ A ψ 1 Φ 2 ( σ , s ) × 3 A ψ 1 Φ 2 ( σ , s ) σ 3 + 9 2 s p A ψ [ A ψ 1 Φ 2 ( σ , s ) σ × 2 A ψ 1 Φ 2 ( σ , s ) σ 2 6 s p A ψ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ 6 s p A ψ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ 3 2 s p A ψ A ψ 1 Φ 2 2 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ = 0 , Φ 2 ( σ , s ) q sec h q σ 2 s 2 + 1 s p 3 Φ 2 ( σ , s ) σ 3 6 s p A ψ A ψ 1 Φ 2 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ 6 s p A ψ [ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ ] 15 2 s p A ψ A ψ 1 Φ 2 2 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ = 0 .
Thus, the following are the k t h -truncated term series:
Φ 1 ( σ , s ) = 1 8 q 2 1 sec h 2 q σ 2 s 2 + r = 1 k f r ( σ , s ) s r p + 1 , r = 1 , 2 , 3 , 4 , Φ 2 ( σ , s ) = q sec h q σ 2 s 2 + r = 1 k g r ( σ , s ) s r p + 1 , r = 1 , 2 , 3 , 4 .
The residual function of Aboodh is as follows:
A ψ R e s ( σ , s ) = Φ 1 ( σ , s ) 1 8 q 2 1 sec h 2 q σ 2 s 2 + 1 s p 3 Φ 1 ( σ , s ) σ 3 + 3 2 s p A ψ A ψ 1 Φ 2 ( σ , s ) × 3 A ψ 1 Φ 2 ( σ , s ) σ 3 + 9 2 s p A ψ [ A ψ 1 Φ 2 ( σ , s ) σ × 2 A ψ 1 Φ 2 ( σ , s ) σ 2 6 s p A ψ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ 6 s p A ψ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ 3 2 s p A ψ A ψ 1 Φ 2 2 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ = 0 , A ψ R e s ( σ , s ) = Φ 2 ( σ , s ) q sec h q σ 2 s 2 + 1 s p 3 Φ 2 ( σ , s ) σ 3 6 s p A ψ A ψ 1 Φ 2 ( σ , s ) × A ψ 1 Φ 1 ( σ , s ) σ 6 s p A ψ A ψ 1 Φ 1 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ 15 2 s p A ψ A ψ 1 Φ 2 2 ( σ , s ) × A ψ 1 Φ 2 ( σ , s ) σ = 0 ,
and the k t h -LRFs are
A ψ R e s k ( σ , s ) = Φ 1 k ( σ , s ) 1 8 q 2 1 sec h 2 q σ 2 s 2 + 1 s p 3 Φ 1 k ( σ , s ) σ 3 + 3 2 s p A ψ A ψ 1 Φ 2 k ( σ , s ) × 3 A ψ 1 Φ 2 k ( σ , s ) σ 3 + 9 2 s p A ψ [ A ψ 1 Φ 2 k ( σ , s ) σ × 2 A ψ 1 Φ 2 k ( σ , s ) σ 2 6 s p A ψ A ψ 1 Φ 1 k ( σ , s ) × A ψ 1 Φ 1 k ( σ , s ) σ 6 s p A ψ A ψ 1 Φ 1 k ( σ , s ) × A ψ 1 Φ 2 k ( σ , s ) × A ψ 1 Φ 2 k ( σ , s ) σ 3 2 s p A ψ A ψ 1 Φ 2 k 2 ( σ , s ) × A ψ 1 Φ 1 k ( σ , s ) σ = 0 , A ψ R e s k ( σ , s ) = Φ 2 k ( σ , s ) q sec h q σ 2 s 2 + 1 s p 3 Φ 2 k ( σ , s ) σ 3 6 s p A ψ A ψ 1 Φ 2 k ( σ , s ) × A ψ 1 Φ 1 k ( σ , s ) σ 6 s p A ψ A ψ 1 Φ 1 k ( σ , s ) × A ψ 1 Φ 2 k ( σ , s ) σ 15 2 s p A ψ A ψ 1 Φ 2 k 2 ( σ , s ) × A ψ 1 Φ 2 k ( σ , s ) σ = 0 .
Finding the values of f r ( σ , s ) and g r ( σ , s ) for r = 1 , 2 , 3 , requires some computation. Following this procedure, take the r t h -Aboodh residual function Equation (54) and substitute it for the r t h -truncated series in Equation (52), and we solve the following relation after multiplying the resultant equation by s r p + 1 : lim s ( s r p + 1 A t R e s Φ 1 , r ( σ , s ) ) = 0 and lim s ( s r p + 1 A t R e s Φ 2 , r ( σ , s ) ) = 0 , and r = 1 , 2 , 3 , . A few of the terms that we obtain are as follows:
f 1 ( σ , s ) = 1 64 q 5 ( 11 cosh ( q σ ) 79 ) tanh q σ 2 sec h 4 q σ 2 , g 1 ( σ , s ) = 1 8 q 4 ( cosh ( q σ ) + 28 ) tanh q σ 2 sec h 3 q σ 2 ,
f 2 ( σ , s ) = q 8 ( 319221 cosh ( q σ ) + 23400 cosh ( 2 q σ ) + 13 cosh ( 3 q σ ) + 345568 ) sec h 8 q σ 2 4096 , g 2 ( σ , s ) = q 7 ( 31411 cosh ( q σ ) 488 cosh ( 2 q σ ) + cosh ( 3 q σ ) 41972 ) sec h 7 q σ 2 512 ,
and so on.
For each r = 1 , 2 , 3 , , we put the values of f r ( σ , s ) and g r ( σ , s ) in Equation (52) and obtain
Φ 1 ( σ , s ) = q 2 1 sec h 2 q σ 2 8 s 2 + q 5 ( 11 cosh ( q σ ) 79 ) tanh q σ 2 sec h 4 q σ 2 64 s p + 1 q 8 ( 319221 cosh ( q σ ) + 23400 cosh ( 2 q σ ) + 13 cosh ( 3 q σ ) + 345568 ) sec h 8 q σ 2 4096 s 2 p + 1 + , Φ 2 ( σ , s ) = q sec h q σ 2 s 2 q 4 ( cosh ( q σ ) + 28 ) tanh q σ 2 sec h 3 q σ 2 8 s p + 1 + q 7 ( 31411 cosh ( q σ ) 488 cosh ( 2 q σ ) + cosh ( 3 q σ ) 41972 ) sec h 7 q σ 2 512 s 2 p + 1 + .
Apply AIT to obtain
Φ 1 ( σ , ψ ) = q 2 1 sec h 2 q σ 2 8 + q 5 ψ p ( 11 cosh ( q σ ) 79 ) tanh q σ 2 sec h 4 q σ 2 64 Γ ( p + 1 ) q 8 ψ 2 p ( 319221 cosh ( q σ ) + 23400 cosh ( 2 q σ ) + 13 cosh ( 3 q σ ) + 345568 ) sec h 8 q σ 2 4096 Γ ( 2 p + 1 ) + , Φ 2 ( σ , ψ ) = q sec h q σ 2 q 4 ψ p ( cosh ( q σ ) + 28 ) tanh q σ 2 sec h 3 q σ 2 8 Γ ( p + 1 ) + q 7 ψ 2 p ( 31411 cosh ( q σ ) 488 cosh ( 2 q σ ) + cosh ( 3 q σ ) 41972 ) sec h 7 q σ 2 512 Γ ( 2 p + 1 ) + .

4.2. Solution of the Problem Using ATIM

Examine the following system of Jaulent–Miodek equations of fractional order:
D ψ p Φ 1 ( σ , ψ ) = 3 Φ 1 ( σ , ψ ) σ 3 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 + 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ , D ψ p Φ 2 ( σ , ψ ) = 3 Φ 2 ( σ , ψ ) σ 3 + 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ , where 0 < p 1 .
with the initial conditions listed below:
Φ 1 ( σ , 0 ) = 1 8 q 2 1 sec h 2 q σ 2 , Φ 2 ( σ , 0 ) = q sec h q σ 2 .
Both sides of Equation (59) are evaluated using the Aboodh transform, and the following equations are produced as a result:
A [ D ψ p Φ 1 ( σ , ψ ) ] = 1 s p ( k = 0 m 1 Φ 1 ( k ) ( σ , 0 ) s 2 p + k + A [ 3 Φ 1 ( σ , ψ ) σ 3 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 + 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ ) ] ) , A [ D ψ p Φ 2 ( σ , ψ ) ] = 1 s p ( k = 0 m 1 Φ 2 ( k ) ( σ , 0 ) s 2 p + k + A [ 3 Φ 2 ( σ , ψ ) σ 3 + 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ ) ] ) .
For Equation (61), the application of the inverse Aboodh transform results in the following equations:
Φ 1 ( σ , ψ ) = A 1 [ 1 s p ( k = 0 m 1 Φ 1 ( k ) ( σ , η , 0 ) s 2 p + k + A [ 3 Φ 1 ( σ , ψ ) σ 3 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 + 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ ] ) ] , Φ 2 ( σ , ψ ) = A 1 [ 1 s p ( k = 0 m 1 Φ 2 ( k ) ( σ , η , 0 ) s 2 p + k + A [ 3 Φ 2 ( σ , ψ ) σ 3 + 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ ] ) ] .
Utilizing the Aboodh transform in an iterative manner results in the extraction of the following equation:
Φ 1 0 ( σ , ψ ) = A 1 1 s p k = 0 m 1 Φ 1 ( k ) ( σ , 0 ) s 2 p + k = A 1 Φ 1 ( σ , 0 ) s 2 = 1 8 q 2 1 sec h 2 q σ 2 , Φ 2 0 ( σ , ψ ) = A 1 1 s p k = 0 m 1 Φ 2 ( k ) ( σ , 0 ) s 2 p + k = A 1 Φ 2 ( σ , 0 ) s 2 = q sec h q σ 2 .
By applying the RL integral to Equation (59), we perform the objective of obtaining the equivalent form.
Φ 1 ( σ , ψ ) = 1 8 q 2 1 sec h 2 q σ 2 + A [ 3 Φ 1 ( σ , ψ ) σ 3 3 2 Φ 2 ( σ , ψ ) 3 Φ 2 ( σ , ψ ) σ 3 9 2 Φ 2 ( σ , ψ ) σ 2 Φ 2 ( σ , ψ ) σ 2 + 6 Φ 1 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 3 2 Φ 2 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ ] , Φ 2 ( σ , ψ ) = q sec h q σ 2 + A [ 3 Φ 2 ( σ , ψ ) σ 3 + 6 Φ 2 ( σ , ψ ) Φ 1 ( σ , ψ ) σ + 6 Φ 1 ( σ , ψ ) Φ 2 ( σ , ψ ) σ + 15 2 Φ 2 2 ( σ , ψ ) Φ 2 ( σ , ψ ) σ ] .
The following few terms are produced by the NITM method:
Φ 1 0 ( σ , ψ ) = 1 8 q 2 1 sec h 2 q σ 2 , Φ 2 0 ( σ , ψ ) = q sec h q σ 2 ,
Φ 1 1 ( σ , ψ ) = q 5 ψ p ( 11 cosh ( q σ ) 79 ) tanh q σ 2 sec h 4 q σ 2 64 Γ ( p + 1 ) , Φ 2 1 ( σ , ψ ) = q 4 ψ p ( cosh ( q σ ) + 28 ) tanh q σ 2 sec h 3 q σ 2 8 Γ ( p + 1 ) ,
Φ 1 2 ( σ , ψ ) = q 8 ψ 2 p sec h 8 σ q 2 131072 ( ( 3 q 3 ψ p tanh σ q 2 sec h 2 σ q 2 ( 8 Γ ( p + 1 ) Γ ( 2 p + 1 ) ( 1998149 cosh ( σ q ) + 137904 cosh ( 2 σ q ) + 1413 cosh ( 3 σ q ) + cosh ( 4 σ q ) + 2510223 ) ( q 3 ψ p Γ ( 3 p + 1 ) 2 ( 4372722 cosh ( σ q ) + 107724 cosh ( 2 σ q ) + 5242 cosh ( 3 σ q ) + 33 cosh ( 4 σ q ) + 6547243 ) tanh σ q 2 sec h 2 σ q 2 ) / ( Γ ( 4 p + 1 ) ) ) ) / Γ ( p + 1 ) 3 Γ ( 3 p + 1 ) ( 32 ( 321813 cosh ( σ q ) + 23352 cosh ( 2 σ q ) + 13 cosh ( 3 σ q ) + 348208 ) ) / Γ ( 2 p + 1 ) ) , Φ 2 2 ( σ , ψ ) = q 7 ψ 2 p sec h 7 σ q 2 65536 ( ( 30 q 6 ψ 2 p Γ ( 3 p + 1 ) 55 sinh σ q 2 + sinh 3 σ q 2 2 ( 158 cosh ( σ q ) + cosh ( 2 σ q ) 275 ) sec h 6 σ q 2 ) / Γ ( p + 1 ) 3 Γ ( 4 p + 1 ) + ( 24 q 3 ψ p Γ ( 2 p + 1 ) ( 363823 sinh σ q 2 101835 sinh 3 σ q 2 + 925 sinh 5 σ q 2 + 23 sinh 7 σ q 2 ) sec h 3 σ q 2 ) / Γ ( p + 1 ) 2 Γ ( 3 p + 1 ) + 128 ( 24931 cosh ( σ q ) 608 cosh ( 2 σ q ) + cosh ( 3 σ q ) 35372 ) / Γ ( 2 p + 1 ) ) .
The final solution through the NITM algorithm is presented in the following manner:
Φ 1 ( σ , ψ ) = Φ 1 0 ( σ , ψ ) + Φ 1 1 ( σ , ψ ) + Φ 1 2 ( σ , ψ ) + , Φ 2 ( σ , ψ ) = Φ 2 0 ( σ , ψ ) + Φ 2 1 ( σ , ψ ) + Φ 2 2 ( σ , ψ ) + .
Φ 1 ( σ , ψ ) = 1 8 q 2 1 sec h 2 q σ 2 + q 5 ψ p ( 11 cosh ( q σ ) 79 ) tanh q σ 2 sec h 4 q σ 2 / 64 Γ ( p + 1 ) + q 8 ψ 2 p sec h 8 σ q 2 ( ( 3 q 3 ψ p tanh σ q 2 sec h 2 σ q 2 ( 8 Γ ( p + 1 ) Γ ( 2 p + 1 ) ( 1998149 cosh ( σ q ) + 137904 cosh ( 2 σ q ) + 1413 cosh ( 3 σ q ) + cosh ( 4 σ q ) + 2510223 ) ( q 3 ψ p Γ ( 3 p + 1 ) 2 ( 4372722 cosh ( σ q ) + 107724 cosh ( 2 σ q ) + 5242 cosh ( 3 σ q ) + 33 cosh ( 4 σ q ) + 6547243 ) tanh σ q 2 sec h 2 σ q 2 ) / ( Γ ( 4 p + 1 ) ) ) ) / Γ ( p + 1 ) 3 Γ ( 3 p + 1 ) ( 32 ( 321813 cosh ( σ q ) + 23352 cosh ( 2 σ q ) + 13 cosh ( 3 σ q ) + 348208 ) ) / 131072 Γ ( 2 p + 1 ) ) + ,
Φ 2 ( σ , ψ ) = q sec h q σ 2 q 4 ψ p ( cosh ( q σ ) + 28 ) tanh q σ 2 sec h 3 q σ 2 / 8 Γ ( p + 1 ) + q 7 ψ 2 p sec h 7 σ q 2 ( ( 30 q 6 ψ 2 p Γ ( 3 p + 1 ) 55 sinh σ q 2 + sinh 3 σ q 2 2 ( 158 cosh ( σ q ) + cosh ( 2 σ q ) 275 ) sec h 6 σ q 2 ) / Γ ( p + 1 ) 3 Γ ( 4 p + 1 ) + ( 24 q 3 ψ p Γ ( 2 p + 1 ) ( 363823 sinh σ q 2 101835 sinh 3 σ q 2 + 925 sinh 5 σ q 2 + 23 sinh 7 σ q 2 ) sec h 3 σ q 2 ) / Γ ( p + 1 ) 2 Γ ( 3 p + 1 ) + 128 ( 24931 cosh ( σ q ) 608 cosh ( 2 σ q ) + cosh ( 3 σ q ) 35372 ) / 65536 Γ ( 2 p + 1 ) ) + .

5. Results and Discussion

Figure 1 provides a graphical comparison of the fractional-order p = 0.7 and q = 0.7 in the function Φ 1 ( σ , ψ ) using the ARPSM (analytical residual power series method). The graph demonstrates the effect of varying p on the solution profile of Φ 1 ( σ , ψ ) . The trends observed in the figure indicate that as p decreases, the solution exhibits notable changes in amplitude and shape, highlighting the sensitivity of Φ 1 ( σ , ψ ) to the fractional order. This sensitivity underlines the utility of fractional-order parameters in accurately modeling dynamic systems. Figure 2 showcases a 2D comparison of the fractional-order p = 0.7 and q = 0.7 of Φ 1 ( σ , ψ ) at different values of ψ . In Figure 2a, at ψ = 0.1 , the graph reveals a smooth variation in Φ 1 ( σ , ψ ) , showing a steady transition in behavior as p changes. In Figure 2b, at ψ = 0.05 , the solution exhibits more pronounced fluctuations, reflecting the increasing impact of fractional-order dynamics at smaller ψ values. In Figure 2c, at ψ = 0.01 , the solution further intensifies its dependency on p, with sharper transitions observed. These subplots collectively demonstrate the interplay between the parameters ψ and p, emphasizing the method’s precision in capturing complex solution behaviors under varying conditions. Figure 3 illustrates the comparison of fractional-order p for 0.7 and q = 0.7 in Φ 2 ( σ , ψ ) using the ARPSM. The results follow a similar pattern to Figure 1, with notable deviations as p varies. However, the changes in Φ 2 ( σ , ψ ) appear more pronounced compared to Φ 1 ( σ , ψ ) , reflecting the distinct nature of Φ 2 ( σ , ψ ) as influenced by the chosen fractional order. Figure 4 presents a 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using the ARPSM. Here, p reaches its peak, with sharp transitions evident in the solution profile. This set of subplots reaffirms the adaptability of the ARPSM in addressing fractional-order complexities and capturing nuanced solution behaviors. Figure 5 shows a comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) using ATIM. Figure 6 shows a 2D comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM. Figure 7 shows a comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) using ATIM. Figure 8 shows a 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM.
Table 1 lists the ARPSM solutions for the fractional-order p of Φ 1 ( σ , ψ ) . The tabulated results corroborate the graphical findings, showcasing precise numerical values for different p values. The data confirm that as p varies, the corresponding changes in Φ 1 ( σ , ψ ) align well with the trends depicted in Figure 1 and Figure 2. Table 2 provides the ARPSM solutions for the fractional-order p of Φ 2 ( σ , ψ ) . Similar to Table 1, the results validate the graphical observations from Figure 3 and Figure 4. The numerical data highlight the dependency of Φ 2 on p, with greater sensitivity noted compared to Φ 1 . Table 3 shows the ATIM solution for the fractional order p of Φ 1 ( σ , ψ ) . Table 4 shows the ATIM solution for the fractional order p of Φ 2 ( σ , ψ ) . Table 5 shows the ARPSM and ATIM solution error comparison of Φ 1 ( σ , ψ ) . Table 6 shows the ARPSM and ATIM solution error comparison of Φ 2 ( σ , ψ ) .
The graphical and tabular results collectively demonstrate the effectiveness of the ARPSM in solving fractional-order systems. The visual trends and numerical solutions provide a comprehensive understanding of the dynamic behavior of Φ 1 ( σ , ψ ) and Φ 2 ( σ , ψ ) under varying fractional-order parameters. The findings underscore the versatility of fractional calculus in modeling and analyzing complex systems.

6. Conclusions

In conclusion, this study explored the numerical resolution of the fractional-order Jaulent–Miodek system within the Caputo operator framework using the Aboodh Transform Iterative Method and the Aboodh Residual Power Series Method. These advanced techniques have proven effective and reliable in analyzing and solving the system’s dynamic behavior. The results highlight the applicability of these methods to complex mathematical models, demonstrating their relevance to science, engineering, and algebraic applications. This work not only enhances the understanding of fractional partial differential equations but also emphasizes the importance of innovative numerical methods in tackling real-world challenges. The insights gained from this study lay a strong foundation for future research, offering potential for further applications of advanced numerical techniques to increasingly intricate mathematical and physical models, thereby driving progress in scientific and engineering disciplines.

Author Contributions

The authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by the Public Authority for Applied Education and Training (PAAET) under project No. BE-24-03.

Data Availability Statement

The original contributions presented in this study are included in the article. Further inquiries can be directed to the corresponding author.

Acknowledgments

The authors would like to express their gratitude to the referees for their valuable comments and suggestions, which greatly improved this work. We also acknowledge the generous financial support provided by the Public Authority for Applied Education and Training (PAAET) under project No. BE-24-03, which made this research possible.

Conflicts of Interest

The authors declare no conflicts of interest.

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Figure 1. Comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) using the ARPSM.
Figure 1. Comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) using the ARPSM.
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Figure 2. 2D comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using the ARPSM.
Figure 2. 2D comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using the ARPSM.
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Figure 3. Comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) using the ARPSM.
Figure 3. Comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) using the ARPSM.
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Figure 4. 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using the ARPSM.
Figure 4. 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using the ARPSM.
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Figure 5. Comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) using ATIM.
Figure 5. Comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) using ATIM.
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Figure 6. 2D comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM.
Figure 6. 2D comparison of fractional order p for q = 0.7 of Φ 1 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM.
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Figure 7. Comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) using ATIM.
Figure 7. Comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) using ATIM.
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Figure 8. 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM.
Figure 8. 2D comparison of fractional order p for q = 0.7 of Φ 2 ( σ , ψ ) at (a) ψ = 0.1 , (b) ψ = 0.05 , and (c) ψ = 0.01 using ATIM.
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Table 1. ARPSM solution for the fractional order p of Φ 1 ( σ , ψ ) .
Table 1. ARPSM solution for the fractional order p of Φ 1 ( σ , ψ ) .
ψ σ ARPSM p = 0.48 ARPSM p = 0.68 ARPSM p = 1.00 Exact ARPSM Error p = 1.00
0.01020.02210980.02119570.02203520.02233192.967220 × 10 4
040.04782960.04802870.04802210.04799053.152320 × 10 5
060.05790540.05779270.05770840.05767793.045560 × 10 5
080.06044110.06038790.06035860.06034929.415964 × 10 6
100.06105190.06103690.06102910.06102662.447684 × 10 6
0.05020.03219430.02200650.02102410.02217071.146590 × 10 3
040.04676220.04780490.04803990.04790981.301040 × 10 4
060.05794580.05794300.05779940.05765401.453130 × 10 4
080.06052750.06045610.06038950.06034304.649000 × 10 5
100.06107930.06105600.06103730.06102511.218940 × 10 5
0.10020.04670140.02635250.02051830.02196901.450630 × 10 3
040.04539320.04730460.04799970.04780831.914300 × 10 4
060.05782650.05800340.05789720.05762402.732150 × 10 4
080.06057790.06051120.06042680.06033529.150590 × 10 5
100.06109810.06107270.06104740.06102312.425630 × 10 5
Table 2. ARPSM solution for the fractional order p of Φ 2 ( σ , ψ ) .
Table 2. ARPSM solution for the fractional order p of Φ 2 ( σ , ψ ) .
ψ σ ARPSM p = 0.48 ARPSM p = 0.68 ARPSM p = 1.00 Exact ARPSM Error p = 1.00
0.01020.5321070.5454680.5549640.5579833.019230 × 10 3
040.3148780.3209590.3244830.3256931.209400 × 10 3
060.1658050.1676960.1686550.1690473.916230 × 10 4
080.0837460.0844020.0847330.0848921.587480 × 10 4
100.0417700.0420550.0421990.0422747.408930 × 10 5
0.05020.5241150.5277840.5447500.5591381.438730 × 10 2
040.3061400.3131250.3207500.3266825.932430 × 10 3
060.1622080.1652960.1676520.1696101.958690 × 10 3
080.0824880.0835700.0843870.0851827.945760 × 10 4
100.0412310.0416930.0420480.0424193.705760 × 10 4
0.10020.5331860.5193640.5335760.5605782.700240 × 10 2
040.3021760.3072770.3163440.3279231.157840 × 10 2
060.1595800.1631300.1663980.1703173.918810 × 10 3
080.0815600.0828150.0839540.0855451.591250 × 10 3
100.0408390.0413680.0418590.0426007.414770 × 10 4
Table 3. ATIM solution for the fractional order p of Φ 1 ( σ , ψ ) .
Table 3. ATIM solution for the fractional order p of Φ 1 ( σ , ψ ) .
ψ σ ARPSM p = 0.48 ARPSM p = 0.68 ARPSM p = 1.00 Exact ARPSM Error p = 1.00
0.01020.02202780.02119970.02203550.02233192.963860 × 10 4
040.04785450.04803060.04802210.04799053.156600 × 10 5
060.05790720.05779280.05770840.05767793.045750 × 10 5
080.06044120.06038790.06035860.06034929.416050 × 10 6
100.06105190.06103690.06102910.06102662.447689 × 10 6
0.05020.03086510.02190020.02102790.02217071.142720 × 10 3
040.04696280.04783450.04804140.04790981.315680 × 10 4
060.05796190.05794510.05779940.05765401.453970 × 10 4
080.06052830.06045620.06038950.06034304.649390 × 10 5
100.06107930.06105600.06103730.06102511.218970 × 10 5
0.10020.04278230.02575350.02051130.02196901.457670 × 10 3
040.04590340.04740970.04800760.04780831.992630 × 10 4
060.05786860.05801150.05789770.05762402.737310 × 10 4
080.06058000.06051160.06042680.06033529.153030 × 10 5
100.06109830.06107280.06104740.06102312.425800 × 10 5
Table 4. ATIM solution for the fractional order p of Φ 2 ( σ , ψ ) .
Table 4. ATIM solution for the fractional order p of Φ 2 ( σ , ψ ) .
ψ σ ARPSM p = 0.48 ARPSM p = 0.68 ARPSM p = 1.00 Exact ARPSM Error p = 1.00
0.01020.5286150.5450310.5549500.5579833.032920 × 10 3
040.3142840.3208830.3244810.3256931.211790 × 10 3
060.1657610.1676900.1686550.1690473.918050 × 10 4
080.0837420.0844020.0847330.0848921.587640 × 10 4
100.0417700.0420550.0421990.0422747.409110 × 10 5
0.05020.5069510.5237600.5444040.5591381.473360 × 10 2
040.3032960.3124410.3206900.3266825.992400 × 10 3
060.1620000.1652450.1676470.1696101.963240 × 10 3
080.0824700.0835660.0843870.0851827.949810 × 10 4
100.0412290.0416930.0420480.0424193.706200 × 10 4
0.10020.4987440.5087600.5321710.5605782.840760 × 10 2
040.2965730.3055010.3161030.3279231.181980 × 10 2
060.1591790.1629990.1663800.1703173.937010 × 10 3
080.0815250.0828030.0839520.0855451.592860 × 10 3
100.0408360.0413670.0418590.0426007.416530 × 10 4
Table 5. ARPSM and ATIM solution error comparison of Φ 1 ( σ , ψ ) .
Table 5. ARPSM and ATIM solution error comparison of Φ 1 ( σ , ψ ) .
ψ σ Exact ARPSM p = 1 ATIM p = 1 ARPSM Error p = 1.00 ATIM Error p = 1.00
0.01020.02233190.02203520.02203552.967220 × 10 4 2.963860 × 10 4
040.04799050.04802210.04802213.152320 × 10 5 3.156600 × 10 5
060.05767790.05770840.05770843.045560 × 10 5 3.045750 × 10 5
080.06034920.06035860.06035869.415964 × 10 6 9.416050 × 10 6
100.06102660.06102910.06102912.447684 × 10 6 2.447689 × 10 6
0.05020.02217070.02102410.02102791.146590 × 10 3 1.142720 × 10 3
040.04790980.04803990.04804141.301040 × 10 4 1.315680 × 10 4
060.05765400.05779940.05779941.453130 × 10 4 1.453970 × 10 4
080.06034300.06038950.06038954.649000 × 10 5 4.649390 × 10 5
100.06102510.06103730.06103731.218940 × 10 5 1.218970 × 10 5
0.10020.02196900.02051830.02051131.450630 × 10 3 1.457670 × 10 3
040.04780830.04799970.04800761.914300 × 10 4 1.992630 × 10 4
060.05762400.05789720.05789772.732150 × 10 4 2.737310 × 10 4
080.06033520.06042680.06042689.150590 × 10 5 9.153030 × 10 5
100.06102310.06104740.06104742.425630 × 10 5 2.425800 × 10 5
Table 6. ARPSM and ATIM solution error comparison of Φ 2 ( σ , ψ ) .
Table 6. ARPSM and ATIM solution error comparison of Φ 2 ( σ , ψ ) .
ψ σ Exact ARPSM p = 1 ATIM p = 1 ARPSM Error p = 1.00 ATIM Error p = 1.00
0.01020.5579830.5549640.5549503.019230 × 10 3 3.032920 × 10 3
040.3256930.3244830.3244811.209400 × 10 3 1.211790 × 10 3
060.1690470.1686550.1686553.916230 × 10 4 3.918050 × 10 4
080.0848920.0847330.0847331.587480 × 10 4 1.587640 × 10 4
100.0422740.0421990.0421997.408930 × 10 5 7.409110 × 10 5
0.05020.5591380.5447500.5444041.438730 × 10 2 1.473360 × 10 2
040.3266820.3207500.3206905.932430 × 10 3 5.992400 × 10 3
060.1696100.1676520.1676471.958690 × 10 3 1.963240 × 10 3
080.0851820.0843870.0843877.945760 × 10 4 7.949810 × 10 4
100.0424190.0420480.0420483.705760 × 10 4 3.706200 × 10 4
0.10020.5605780.5335760.5321712.700240 × 10 2 2.840760 × 10 2
040.3279230.3163440.3161031.157840 × 10 2 1.181980 × 10 2
060.1703170.1663980.1663803.918810 × 10 3 3.937010 × 10 3
080.0855450.0839540.0839521.591250 × 10 3 1.592860 × 10 3
100.0426000.0418590.0418597.414770 × 10 4 7.416530 × 10 4
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Alkhezi, Y.; Shafee, A. Analytical Techniques for Studying Fractional-Order Jaulent–Miodek System Within Algebraic Context. Fractal Fract. 2025, 9, 50. https://doi.org/10.3390/fractalfract9010050

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Alkhezi Y, Shafee A. Analytical Techniques for Studying Fractional-Order Jaulent–Miodek System Within Algebraic Context. Fractal and Fractional. 2025; 9(1):50. https://doi.org/10.3390/fractalfract9010050

Chicago/Turabian Style

Alkhezi, Yousuf, and Ahmad Shafee. 2025. "Analytical Techniques for Studying Fractional-Order Jaulent–Miodek System Within Algebraic Context" Fractal and Fractional 9, no. 1: 50. https://doi.org/10.3390/fractalfract9010050

APA Style

Alkhezi, Y., & Shafee, A. (2025). Analytical Techniques for Studying Fractional-Order Jaulent–Miodek System Within Algebraic Context. Fractal and Fractional, 9(1), 50. https://doi.org/10.3390/fractalfract9010050

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