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Article

Solution for Time-Fractional Coupled Burgers Equations by Generalized-Laplace Transform Methods

Mathematics Department, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
*
Author to whom correspondence should be addressed.
Fractal Fract. 2024, 8(12), 692; https://doi.org/10.3390/fractalfract8120692
Submission received: 27 October 2024 / Revised: 20 November 2024 / Accepted: 22 November 2024 / Published: 25 November 2024

Abstract

:
In this work, nonlinear time-fractional coupled Burgers equations are solved utilizing a computational method, which is called the double and triple generalized-Laplace transform and decomposition method. We discuss the proof of triple generalized-Laplace transform for a Caputo fractional derivative. We have given four examples to show the precision and adequacy of the suggested approach. The results show that this method is easy and accurate when compared to the A domain decomposition method (ADM), homotopy perturbation method (HPM), and generalized differential transform method (GDTM). Finally, we have sketched the graphics for all these examples.

1. Introduction

Generalized classical calculus is a fractional calculus. Several research areas such as science and engineering have used fractional differential equations to solve these problems. For instance, Burgers equation is widely utilized to explain numerous physical problems, namely, unidirectional propagation of weakly nonlinear acoustic waves, shock waves in the viscous medium, flow systems, and electromagnetic waves, etc. The time-fractional Burgers equation is a type of diffusion convection equation. The Burgers equation was presented by the author of [1] in 1915 and subsequently discussed by the researcher in [2]. Many studies have been conducted by mathematicians to find out the numerical or analytical solution of time-space fractional Burgers equations such as, e.g., [3,4]; the authors in [4], implemented the A domian decomposition method and Pade approximation technique, and the homotopy perturbation method was utilized to obtain exact solutions for nonlinear Burgers equation [5]. The space–time-fractional Burgers equation for various initial conditions were solved by the variational iteration method (VIM) in [6,7]. The time-fractional coupled Burgers equations were discussed using the homotopy perturbation Sumudu transform method (HPSTM) in [8]. The authors in [9] studied a series solution for two-dimensional Burgers equations, employing the homotopy perturbation method. The author in [10] investigated the solutions of the two-dimensional nonlinear Burgers differential equations employing the Laplace decomposition method (LDM). The authors in [11] used the G-Laplace transform method to find the exact solution of the Burgers equations. In [12], the approximate solutions of the coupled Burgers equations were obtained by applying the homotopy perturbation method and Laplace transform. The authors in [13,14] obtained the approximate solution of the viscous coupled Burgers equation using cubic and cubic B-spline collocation method.
This study aims to introduce a new technique named the double and triple generalized-Laplace transform decomposition method. This technique is used to obtain the solution of fractional coupled Burgers equation and to the assist mathematicians in solving numerous equations related to physics and engineering in the future.
Remarks: Through this study, we use the following abbreviations:
(1)
(GLT) in place of “Generalized-Laplace transform”.
(2)
(DGLT) in place of “double Generalized-Laplace transform”.
(3)
(IDGLT) in place of “inverse double Generalized-Laplace transform”.
(4)
(TGLT) in place of “triple Generalized-Laplace transform”.
(5)
(ITGLT) in place of “inverse triple Generalized-Laplace transform”.
(6)
(DM) “decomposition method”.
(7)
(DGLTDM) in place of “double Generalized-Laplace transform decomposition method”.
(8)
(TGLTDM) in place of “triple Generalized-Laplace transform decomposition method”.
Now, we recall the following definitions which are useful in this work.
Definition 1. 
The fractional derivative of f ( χ , t ) in the Caputo sense is denoted by
D t σ f ( χ , t ) = 1 Γ k σ 0 t t τ k σ 1 k f ( χ , τ ) τ k d τ , k 1 < σ < k , k f ( χ , t ) t k , for k = σ N
For more details, see [15].
Definition 2 
([16]). The partial fractional integrals and caputo derivatives of a function
f χ , t , where χ , t R + × R + are granted by
D t β f χ , t = 1 Γ n β 0 t t λ n β 1 n λ n f χ , λ d λ ,
where n 1 < β n , 0 < β .
Definition 3 
([17]). Let f ( t ) be integrable, for all t 0 . The (GLT) G α of the function f ( t ) is given by
F s = G α f = s α 0 f t e t s d t ,
for s C and α Z .
The following symbols are using in this work:
G 2 = G χ G t , G 3 = G χ G γ G t , G 2 1 = G p 1 G s 1 , G 3 1 = G p 1 G q 1 G s 1 .
Definition 4 
([18]). The (DGLT) of the function f ( χ , t ) is defined as
G 2 f ( χ , t ) = F ( p , s ) = p α s α 0 0 e χ p t s f ( χ , t ) d t d χ ,
where α Z , p , s C and the symbol G 2 indicate transform of χ and t, respectively, and the function F p , s is denoted by the (DGLT) of the f χ , t .
Definition 5 
([18]). The (IDGLT) is given by
G 2 1 F p , s = f χ , t = 1 2 π i 2 τ i τ + i ς i ς + i e 1 p χ + 1 s t F p , s d s d p ,
where G 2 1 indicates (IDGLT).
Definition 6 
([19]). The (TGLT) of the function f ( χ , γ , t ) is defined as
G 3 f ( χ , γ , t ) = F ( p , q , s ) = p α q α s α 0 0 0 e χ p γ q t s f ( χ , γ , t ) d t d γ d χ ,
where α Z , p , q , s C and the symbol G 3 indicate the transform of χ , γ and t, respectively, and the function F p , q , s is denoted as the (TGLT) of the f χ , γ , t .
Definition 7 
([19]). The inverse (TGLT)
G 3 1 F p , q , s = 1 2 π i 3 τ i τ + i ς i ς + i η i δ + i e 1 p χ + 1 q γ + 1 s t F p , q , s d s d q d p ,
where G 3 1 denotes (ITGLT).
The advantage of the (TGLT) is that it is useful to produce some transformations from Definition 6 as follows:
  • If we set α = 0 , s = 1 s , q = 1 q and p = 1 p , we obtain triple Laplace transform:
    L χ L γ L ν f χ , γ , t = F p , q , s = 0 0 0 f χ , γ , t e p χ + q γ + s t d t d γ d χ ,
  • If we set α = 0 , σ = 1 p , ρ = 1 q and substituting s by ϖ , we obtain double Laplace– Yang Transform:
    L χ L γ γ t f χ , γ , t = F σ , ρ , ϖ = 0 0 0 f χ , γ , t e ( σ χ + ρ γ + t ϖ ) d t d γ d χ ,
  • At α = 1 and substituting p , q , s by u , v , μ , respectively, we obtain a triple Sumudu Transform:
    S χ S γ S t f χ , γ , t = F u , v , μ = 1 u v μ 0 0 0 f χ , γ , t e ( χ u + γ v + t μ ) d t d γ d χ .
Theorem 1. 
If (DGLT) of the function f ( χ , t ) is given by G 2 [ f ( χ , t ) ] = F p , s , then ( D G L T ) of f ( χ , t ) t and 2 f χ , t t 2 are given by
G 2 f ( χ , t ) t = F ( p , s ) s s α F ( p , 0 ) ,
and
G 2 2 f ( χ , t ) 2 t = F ( p , s ) s 2 s α 1 F ( p , 0 ) s α F ( p , 0 ) t .
Proof. 
By using definition of (DGLT) for f ( χ , t ) t , we have
G χ G t f ( χ , t ) t = p α 0 e χ p s α 0 e t s f χ , t t d t d χ ,
and the integral inside the bracket is calculated as follows:
s α 0 e t s f χ , t t d t = s α e t s f χ , t 0 + s α s 0 e t s f χ , t d t = 1 s F p , t s α f χ , 0
By substituting Equation (11) into Equation (10), we have
G 2 f ( χ , t ) t = p α 0 e χ p 1 s F p , t s α f χ , 0 d t ,
and thus,
G 2 f ( χ , t ) t = 1 s F p , s s α F p , 0 .
In a similar way, one can easily see that
G 2 2 f ( χ , t ) 2 t = F ( p , s ) s 2 s α 1 F ( p , 0 ) s α F ( p , 0 ) t .
Theorem 2. 
The (DGLT) of the Caputo fractional derivative D t β f χ , t is denoted by
G χ G t D t β f χ , t = F p , s s β s α k = 1 n 1 s β k G χ k 1 f χ , 0 t k 1
where n 1 < β n , 0 < β .
Proof. 
Utilizing Equation (3), we obtain
G χ G t D t β f χ , t = p α s α 0 0 e χ p t s D t β f χ , t d t d χ = p α s α 0 0 e χ p t s 1 Γ n β 0 t t λ n β 1 n f χ , λ λ n d λ d t d χ = p α s α Γ n β 0 0 e χ p t s λ t λ n β 1 n f χ , λ λ n d t d χ d λ .
  Assuming r = t λ in Equation (22), we gain
G χ G t D t β f χ , t = p α s α Γ n β 0 e χ p 0 n f χ , λ λ n e λ s 0 r n β 1 e r s d r d λ d χ .
  The gamma function is defined by the integral inside the bracket as
0 r n β 1 e r s d r = Γ n β 1 s n β
Hence, Equation (23) becomes
G χ G t D t β f χ , t = p α s α Γ n β 0 e χ p 0 n f χ , λ λ n e λ s Γ n β 1 s n β d λ d χ .
By rewriting Equation (24), we obtain
G χ G t D t β f χ , t = s n β p α + 1 s α + 1 0 e χ p 0 n f χ , λ λ n e λ s d λ d χ .
We define the integral inside the bracket in the following form:
G χ G t n f χ , t t n = p α + 1 s α + 1 0 e χ p 0 n f χ , λ λ n e λ s d λ d χ = f p , s s n s α k = 1 n 1 s n k G χ k 1 f χ , 0 t k 1 .
By substituting Equation (26) into Equation (25), we obtain
G χ G t D t β f χ , t = F p , s s β s α k = 1 n 1 s β k G χ k 1 f χ , 0 t k 1 .
The proof is complete. □
One can easily obtain the (GDLT) of the functions β f χ , t t β and 2 β f χ , t t 2 β from Theorem 2 as follows:
G 2 β f χ , t t β = F p , s s β s α β + 1 F p , 0 , 0 < β 1 G 2 2 β f χ , t t 2 β = F p , s s 2 β s α 2 β + 1 F p , 0 s α 2 β + 2 F t p , 0 ,
0 < β 1 .
Definition 8. 
The (TGLT) of the partial derivative n ω χ , γ , t t n is denoted by
G 3 n ω χ , γ , t t n = F p , q , s s n s α k = 1 n 1 s n k G 2 k 1 ω χ , γ , 0 t k 1 .
Theorem 3. 
The (TGLT) of th Caputo fractional derivative D t β ω χ , γ , t is given by
G 3 D t β ω χ , γ , t = F p , q , s s β s α k = 1 n 1 s β k G 2 k 1 ω χ , γ , 0 t k 1 ,
where n 1 < β n , 0 < β .
Proof. 
Using the definition of (TGLT) for D t β ω χ , γ , t , we have
G 3 D t β ω χ , γ , t = p α q α s α 0 0 0 e χ p γ q t s D t β ω χ , γ , t d t d γ d χ = p α q α s α 0 0 0 e χ p γ q t s 1 Γ n β 0 t t λ n β 1 n ω χ , γ , λ λ n d λ d t d γ d χ = p α q α s α Γ n β 0 0 0 e χ p γ q t s λ t λ n β 1 n ω χ , γ , λ λ n d t d γ d χ d λ .
By putting r = t λ in Equation (22), we obtain
G 3 D t β ω χ , γ , t = p α q α s α Γ n β 0 0 e χ p γ q 0 n ω χ , γ , λ λ n e λ s 0 r n β 1 e r s d r d λ d γ d χ .
The gamma function is defined by the integral inside the bracket as follows:
0 r n β 1 e r s d r = Γ n β 1 s n β .
Hence, Equation (23) becomes
G 3 D t β ω χ , γ , t = p α q α s α Γ n β 0 0 e χ p γ q 0 n ω χ , γ , λ λ n e λ s Γ n β 1 s n β d λ d γ d χ .
By rewriting Equation (24), one can obtain
G 3 D t β ω χ , γ , t = s n β p α q α s α 0 0 e χ p γ q 0 n ω χ , γ , λ λ n e λ s d λ d γ d χ .
The integral inside the bracket in the above equation is defined by
G 3 n ω χ , γ , t t n = p α q α 0 0 e χ p γ q s α 0 n ω χ , γ , λ λ n e λ s d λ d γ d χ = F p , q , s s n s α k = 1 n 1 s n k G χ G γ k 1 ω χ , γ , 0 t k 1 .
By replacing Equation (26) into Equation (25), we have
G 3 D t β ω = F p , q , s s β s α k = 1 n 1 s β k G 2 k 1 ω χ , γ , 0 t k 1 .
The proof is complete. □

2. Double Generalized-Laplace Transform Decomposition Method and Time-Fractional Burgers Equation (DGLTDM)

Here, we explain the solutions to two problems by utilizing the (DGLTDM).
In this work, we deal with the time-fractional Burgers equation containing the initial condition granted by
D t β ω ω χ χ + η ω ω χ + ζ ω 2 χ = h χ , t , ω χ , 0 = h 1 χ .
The first problem: To show the basic idea of (DGLTDM), we consider Equation (27) can be written in the following form:
D t β ω L ω χ , t + N ω χ , t = f χ , t , ω χ , 0 = f 1 χ .
Here, L and N are linear and nonlinear differential operators, respectively. The following steps are needed to solve Equation (28).
  • Step 1: Using the (DGLT) on both sides of Equation (28), it becomes:
    G 2 D t β ω G 2 L ω χ , t N ω χ , t = G 2 f χ , t .
    Employing Theorem 1 and Equation (19), in the above equation, we obtain
    ω p , s s β = s α β + 1 ω p , 0 + G 2 L ω χ , t N ω χ , t + G 2 f χ , t .
  • Step 2: Multiplying Equation (30) by s β , and using the inverse of (IDGLT), we acquire
    ω χ , t = f 1 χ + G 2 1 s β G 2 L ω χ , t N ω χ , t + G 2 1 s β F p , s .
  • Step 3: The solution of Equation (28) is given by infinite series as follows:
    ω χ , t = n = 0 ω n χ , t .
    By replacing Equation (32) into Equation (31), we obtain
    n = 0 ω n χ , t = f 1 χ + G 2 1 s β F p , s + G 2 1 s β G 2 L n = 0 ω n χ , t n = 0 N ω n χ , t .
  • Step 4: Now, by using Equation (33), we derive the following:
    ω 0 = f 1 χ + G 2 1 s β F p , s ,
    and
    ω 1 = G 2 1 s β G 2 L ω 0 N ω 0 χ , t ω 2 = G 2 1 s β G 2 L ω 1 N ω 1 χ , t ω 3 = G 2 1 s β G 2 L ω 2 N ω 2 χ , t . . . ω n = G 2 1 s β G 2 L ω n 1 N ω n 1 χ , t .
    The solution of Equation (28) is obtained by substituting Equation (35) into Equation (32) as follows:
    ω χ , t = lim n 1 ω n χ , t = ω 0 χ , t + ω 1 χ , t + ω 2 χ , t + . . . + ω n χ , t .
We assume the inverse of Equations (34) and (35) exists.
  • Convergence:
Theorem 4. 
Let B be a Banach space. The series solution of Equation (35) is convergent if there exists k , 0 k < 1 , such that ω n χ , t k ω n 1 χ , t for all n N .
Proof. 
By defining sequence S n of partial sums of the series of Equation (35) as follows:
S 0 = ω 0 χ , t S 1 = ω 0 χ , t + ω 1 χ , t S 2 = ω 0 χ , t + ω 1 χ , t + ω 2 χ , t . . . S n = ω 0 χ , t + ω 1 χ , t + ω 2 χ , t + . . . + ω n χ , t ,
we show that S n n = 1 is a Cauchy sequence in Banach space B. Therefore, we consider
S n + 1 S n = ω n + 1 χ , t k ω n χ , t k 2 ω n 1 χ , t . . . k n + 1 ω 0 χ , t .
By using the above triangle inequality for n m , we have
S n S m = S n S n 1 + S n 1 S n 2 + . . . + S m + 1 S m , S n S n 1 + S n 1 S n 2 + . . . + S m + 1 S m , k n ω 0 χ , t + k n 1 ω 0 χ , t + . . . + k m + 1 ω 0 χ , t , = k m + 1 1 k n m 1 k ω 0 χ , t .
From 0 k < 1 , we see that 1 k n m 1 ; therefore,
S n S m k m + 1 1 k ω 0 χ , t .
Since ω 0 χ , t is bounded, S n S m 0 at n , m . Therefore, the sequence S n n = 1 is a Cauchy sequence in the Banach space B, and the series solution of Equation (35) converges. □
To illustrate this technique for time-fractional Burgers equations, we offer the following example:
Example 1. 
Consider the following time-fractional Burgers equation contains the initial condition
D t β ω = ω χ χ + 2 ω ω χ ω 2 χ , ω χ , 0 = sin χ , 0 < β 1 .
By using the previous steps, we obtain
ω 0 = sin χ ,
and
ω n + 1 = G 2 1 s β G 2 ω n χ χ + 2 A n B n .
The Adomain polynomials of the nonlinear term F ( u ) can be estimated by the following expression:
Q n = 1 n ! d n d λ n F i = 0 n λ i u i λ = 0 , n = 0 , 1 , 2 , · · · ,
where Adomian polynomials Q n of the nonliear term F ( u ) are defined by
Q 0 = F ( u 0 ) , Q 1 = u 1 F ( u 0 ) , Q 2 = u 2 F ( u 0 ) + 1 2 ! u 1 F ( u 0 ) , Q 3 = u 3 F ( u 0 ) + u 1 u 2 F ( u 0 ) + 1 3 ! u 1 3 F ( u 0 ) , Q 4 = u 4 F ( u 0 ) + 1 2 ! u 2 2 + u 1 u 3 F ( u 0 ) + 1 2 ! u 1 2 u 2 F ( u 0 ) + 1 4 ! u 1 4 F 4 ( u 0 ) .
Now, it is easy to generate A n and B n from the above equation, as follows:
A 0 = ω 0 ω 0 χ , A 1 = ω 0 ω 1 χ + ω 1 ω 0 χ , A 2 = ω 0 ω 2 χ + ω 1 ω 1 χ + ω 2 ω 0 χ ,
and
B 0 = ω 0 χ 2 , B 1 = 2 ω 0 ω 1 χ + 2 ω 1 ω 0 χ , B 2 = 2 ω 0 ω 2 χ + 2 ω 1 ω 1 χ + 2 ω 2 ω 0 χ .
According to the GDLTDM, we obtain the following:
ω 1 = G 2 1 s β G 2 ω 0 χ χ + 2 A 0 B 0 = G 2 1 s β G 2 sin χ = G 2 1 p α + 2 1 + p 2 s α + β + 1 , ω 1 = t β Γ β + 1 sin χ .
At n = 1 , we have
ω 2 = G 2 1 s β G 2 t β Γ β + 1 sin χ = G 2 1 p α + 2 1 + p 2 s α + 2 β + 1 . = t 2 β Γ 2 β + 1 sin χ ,
In a similar way, we obtain
ω 3 = G 2 1 s β G 2 t 2 β Γ 2 β + 1 sin χ , = G 2 1 p α + 2 1 + p 2 s α + 3 β + 1 , = t 3 β Γ 3 β + 1 sin χ .
The series solution of Equation (36) can be found in the following form:
ω χ , t = sin χ t β Γ β + 1 sin χ + t 2 β Γ 2 β + 1 sin χ t 3 β Γ 3 β + 1 sin χ + · · · .
By putting β = 1 , the exact solution of Equation (36) is given by the following:
ω χ , t = sin χ e t .
Figure 1: We demonstrate the comparison between the exact solution and the obtained numerical solution for Equation (36). At t = 1 and β = 1 , we obtained the accurate solution. By taking various values of β , for instance, ( β = 0.75 , β = 0.85 and β = 0.95 ), we obtained the estimate solutions.
Figure 2: We demonstrate the result of the functions ω χ , t in three-dimensional space.
Table 1 Shows the numerical solution for different values of β for the function ω x , t .
  • The second problem:
To illustrate the essential idea of this method, we consider the following system of time-fractional coupled Burgers equation with the initial conditions of the form:
D t β u u χ χ + η u u χ + ζ u ω χ = f χ , t , D t β ω ω χ χ + η ω ω χ + μ u ω χ = h χ , t ,
and
u χ , 0 = f 1 χ , ω χ , 0 = h 1 χ
for t > 0 . Here, f ( χ , t ) , h ( χ , t ) , f 1 χ and h 1 χ are given functions, η is a real constant, ζ and μ are arbitrary constants depending on the system parameters such as Peclet number, Stokes velocity of particles due to gravity and Brownian diffusivity [20]. The next steps are needed to solve Equation (39).
  • Step 1: Taking the (DGLT) on both sides of Equation (39) and single generalized-Laplace transform for Equation (40) and multiplying the outcome by s β , we obtain
    U ( p , s ) = s α + 1 F 1 ( p ) + s β F ( p , s ) + s β G 2 u χ χ η u u χ ζ u ω χ ,
    and
    W ( p , s ) = s α + 1 H 1 ( p ) + s β H ( p , s ) + s β G 2 ω χ χ η ω ω χ μ u ω χ .
    where U ( p , s ) and W ( p , s ) are double generalized-Laplace transform of u ( χ , t ) and ω ( χ , t ) , respectively.
  • Step 2: The (DGLTDM) defined the solution of Equation (39) according to the following forms:
u χ , t = n = 0 u n χ , t , ω χ , t = n = 0 ω n χ , t .
We can obtain Adomian’s polynomials A n , C n and D n , respectively, as follows:
C n = n = 0 u n u n χ ,
and
D n = n = 0 u n ω n ,
where A n is mentioned in Example 1. The Adomian polynomials for the nonlinear term ω ω χ , u u χ , and u ω are given by
A 0 = ω 0 ω 0 χ , A 1 = ω 0 ω 1 χ + ω 1 ω 0 χ , A 2 = ω 0 ω 2 χ + ω 1 ω 1 χ + ω 2 ω 0 χ , A 3 = ω 0 ω 3 χ + ω 1 ω 2 χ + ω 2 ω 1 χ + ω 3 ω 0 χ , A 4 = ω 0 ω 4 χ + ω 1 ω 3 χ + ω 2 ω 2 χ + ω 3 ω 1 χ + ω 4 ω 0 χ .
C 0 = u 0 u 0 χ , C 1 = u 0 u 1 χ + u 1 u 0 χ , C 2 = u 0 u 2 χ + u 1 u 1 χ + u 2 u 0 χ , C 3 = u 0 u 3 χ + u 1 u 2 χ + u 2 u 1 χ + u 3 u 0 χ , C 4 = u 0 u 4 χ + u 1 u 3 χ + u 2 u 2 χ + u 3 u 1 χ + u 4 u 0 χ .
and
D 0 = u 0 ω 0 , D 1 = u 0 ω 1 + u 1 ω 0 , D 2 = u 0 ω 2 + u 1 ω 1 + u 2 ω 0 . D 3 = u 0 ω 3 + u 1 ω 2 + u 2 ω 1 + u 3 ω 0 , D 3 = u 0 ω 4 + u 1 ω 3 + u 2 ω 2 + u 3 ω 1 + u 4 ω 0 .
Operating with the (IDGLT) on both sides of Equations (41) and (42) and using Equations (43)–(45), we obtain
n = 0 u n χ , t = f 1 χ + G 2 1 s β F ( p , s ) + G 2 1 s β G 2 n = 0 u n χ χ G 2 1 s β G 2 η n = 0 C n G 2 1 s β G 2 ζ n = 0 D n χ ,
and
n = 0 ω n χ , t = h 1 χ + G 2 1 s β H ( p , s ) + G 2 1 s β G 2 n = 0 ω n χ χ G 2 1 s β G 2 η n = 0 A n G 2 1 s β G 2 μ n = 0 D n χ .
On matching both sides of Equations (49) and (50), we can obtain
u 0 = f 1 χ + G 2 1 s β F ( p , s ) , ω 0 = h 1 χ + G 2 1 s β H ( p , s ) .
In general, the recursive relations are given by
u n + 1 = G 2 1 s β G 2 u n χ χ G 2 1 s β G 2 η C n G 2 1 s β G 2 ζ D n χ ,
and
ω n + 1 = G 2 1 s β G 2 ω n χ χ G 2 1 s β G 2 η A n G 2 1 s β G 2 μ D n χ .
Here, we assume that the (IDGLT) concerning p and s exists for each term on the right-hand side of the above equations. To demonstrate this approach for time-fractional coupled Burgers equations, we offer the following examples:
Example 2. 
Consider the following homogeneous form of a coupled Burgers equation [12]
D t β u u χ χ 2 u u χ + u ω χ = 0 , D t β ω ω χ χ 2 ω ω χ + u ω χ = 0 ,
with initial condition
u χ , 0 = sin χ , ω χ , 0 = sin χ .
By using Equations (51)–(53), we have
u 0 = sin χ , ω 0 = sin χ , u 1 = G 2 1 s β G 2 2 u 0 χ 2 + 2 u 0 u 0 χ u 0 ω 0 χ , u 1 = G 2 1 s β G 2 sin χ = G 2 1 s α + β + 1 p α + 2 1 + p 2 = t β Γ β + 1 sin χ , ω 1 = G 2 1 s β G 2 sin χ = G 2 1 s α + β + 1 p α + 2 1 + p 2 = t β Γ β + 1 sin χ , u 2 = G 2 1 s β G 2 2 u 1 χ 2 + 2 u 0 u 1 χ + u 1 u 0 χ u 0 ω 1 + u 1 ω 0 χ , = G 2 1 s β G 2 t β Γ β + 1 sin χ 2 t β Γ β + 1 sin 2 χ + 2 t β Γ β + 1 sin 2 χ , = G 2 1 s α + 2 β + 1 p α + 2 1 + p 2 , u 2 = t 2 β Γ 2 β + 1 sin χ .
In a similar manner,
ω 2 = t 2 β Γ 2 β + 1 sin χ .
We continue in the same way to obtain the following solutions:
u χ , t = sin χ t β Γ β + 1 sin χ + t 2 β Γ 2 β + 1 sin χ 2 t 3 β Γ 3 β + 1 sin χ + · · · , ω χ , t = sin χ t β Γ β + 1 sin χ + t 2 β Γ 2 β + 1 sin χ 2 t 3 β Γ 3 β + 1 sin χ + · · · .
By putting β = 1 , we obtain
u χ , t = u 0 + u 2 + u 3 + · · · = 1 t + t 2 2 ! t 3 3 ! + · · · sin χ , ω χ , t = ω 0 + ω 2 + ω 3 + · · · = 1 t + t 2 2 ! t 3 3 ! + · · · sin χ ,
and thus, the exact solutions become
u χ , t = e t sin χ , ω χ , t = e t sin χ .
Figure 3: This shows the comparison between the exact solution and the gained numerical solution for Equation (54). At t = 1 and β = 1 , we obtained the accurate solution; by taking diverse values of β such as ( β = 0.75 , β = 0.85 and β = 0.95 ), we obtained the estimated solutions.
Figure 4: We demonstrate the result of the functions ω χ , t = u χ , t in three-dimensional space.
Table 2 Shows the numerical solution for different values of β for the function ω χ , t = u χ , t .
Example 3 
([21]). Consider the following system of singular fractional coupled Burgers equation with the initial conditions of the form:
D t β u 1 χ χ u χ χ 2 u u χ + u ω χ = χ 2 e t 4 e t , D t β ω 1 χ χ ω χ χ 2 ω ω χ + u ω χ = χ 2 e t 4 e t ,
and
u χ , 0 = χ 2 , ω χ , 0 = χ 2 .
By applying the above-mentioned steps, we have
U ( p , s ) = 2 s α + 1 p α + 3 + s β G 2 1 χ χ u χ χ + 2 u u χ u ω χ s β G 2 χ 2 + 4 1 t + t 2 2 ! t 3 3 ! + · · · ,
and
ω ( p , s ) = 2 s α + 1 p α + 3 + s β G 2 ω χ χ + 2 ω ω χ u ω χ s β G 2 χ 2 + 4 1 t + t 2 2 ! t 3 3 ! + · · · .
Hence,
U ( p , s ) = 2 s α + 1 p α + 3 + s β G 2 1 χ χ u χ χ + 2 u u χ u ω χ 2 p α + 3 + 4 p α + 1 s β + α + 1 s β + α + 3 + s β + α + 5 · · · ,
and
ω ( p , s ) = 2 s α + 1 p α + 3 + s β G 2 ω χ χ + 2 ω ω χ u ω χ 2 p α + 3 + 4 p α + 1 s β + α + 1 s β + α + 3 + s β + α + 5 · · · .
Operating the (IDGLT) for Equations (60) and (61), and applying Equations (43) and (33), we obtain
n = 0 u n χ , t = χ 2 + G 2 1 s β G 2 1 χ χ n = 0 u n χ χ + 2 n = 0 C n n = 0 D n χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · ,
and
n = 0 ω n χ , t = χ 2 + G 2 1 s β G 2 1 χ χ n = 0 ω n χ χ + 2 n = 0 A n n = 0 D n χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · .
Then, we determine the iteration components as
u 0 = χ 2 χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · ω 0 = χ 2 χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · ,
and
u n + 1 = G 2 1 s β G 2 1 χ χ u n χ χ + 2 C n D n , ω n + 1 = G 2 1 s β G 2 1 χ χ ω n χ χ + 2 A n D n ,
where n 0 . The rest terms are given by, at n = 0 ,
u 1 = G 2 1 s β G 2 1 χ χ u 0 χ χ + 2 u 0 u 0 χ u 0 ω 0 χ , = G 2 1 s β G 2 4 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · , = G 2 1 4 p α + 1 s α + β + 1 4 p α + 1 s 2 β + α + 1 s 2 β + α + 1 + s 2 β + α + 1 · · · , u 1 = 4 t β Γ β + 1 4 t 2 β Γ 2 β + 1 t 2 β + 1 Γ 2 β + 2 + t 2 β + 2 Γ 2 β + 3 · · · .
In a similar manner,
ω 1 = G 2 1 s β G 2 1 χ χ ω 0 χ χ + 2 ω 0 ω 0 χ u 0 ω 0 χ , = G 2 1 s β G 2 4 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · , = G 2 1 4 p α + 1 s α + β + 1 4 p α + 1 s 2 β + α + 1 s 2 β + α + 1 + s 2 β + α + 1 · · · , ω 1 = 4 t β Γ β + 1 4 t 2 β Γ 2 β + 1 t 2 β + 1 Γ 2 β + 2 + t 2 β + 2 Γ 2 β + 3 · · · .
Similarly, at n = 1
u 2 = G 2 1 s β G 2 0 = 0 , ω 2 = 0 .
In a similar manner,
u 3 = 0 , ω 3 = 0 ·
As a collection of all terms, we obtain
u χ , t = u 0 + u 1 + u 2 + u 3 + · · · , ω χ , t = ω 0 + ω 1 + ω 2 + ω 3 + · · · .
Thus, the approximation solution of Equation (56) is defined by
u χ , t = χ 2 χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · + 4 t β Γ β + 1 4 t 2 β Γ 2 β + 1 t 2 β + 1 Γ 2 β + 2 + t 2 β + 2 Γ 2 β + 3 · · · , ω χ , t = χ 2 χ 2 + 4 t β Γ β + 1 t β + 1 Γ β + 2 + t β + 2 Γ β + 3 · · · + 4 t β Γ β + 1 4 t 2 β Γ 2 β + 1 t 2 β + 1 Γ 2 β + 2 + t 2 β + 2 Γ 2 β + 3 · · · .
Our exact solution can be obtained by putting β = 1 , as follows:
u χ , t = χ 2 1 t + t 2 2 ! t 3 3 ! + t 4 4 ! · · · ,
and
ω χ , t = χ 2 1 t + t 2 2 ! t 3 3 ! + t 4 4 ! · · · .
Hence,
u χ , t = χ 2 e t , ω χ , t = χ 2 e t .
Figure 5: This illustrates the contrast between the exact solution and the obtained numerical solution for Equation (54). At t = 1 and β = 1 , we achieved the accurate solution. By taking different values of β , for instance, ( β = 0.75 , β = 0.85 and β = 0.95 ), we obtained the estimated solutions.
Figure 6: We demonstrate the result of the functions ω χ , t = u χ , t in three-dimensional space.
Table 3 Explain the numerical solution for different values of β for the function ψ ( x , t ) .

3. Triple Generalized-Laplace Transform Decomposition Method and Fractional Coupled Burgers Equation (TGLTDM)

Here, we explain the solutions of the singular time-fractional coupled system of Burgers equation by applying the triple Generalized- Laplace transform decomposition method (TGLTDM):
To demonstrate the essential idea of this method, we consider the following singular time-fractional coupled system of Burgers equation with the initial conditions of the form:
D t β u 1 χ χ u χ χ 1 γ γ u γ γ + η χ u u χ + ζ γ u γ ω = f χ , γ , t , D t β ω 1 χ χ ω χ χ 1 γ γ ω γ γ + η χ u ω χ + ζ γ ω γ ω = h χ , γ , t ,
and
u χ , 0 = f 1 χ , γ , ω χ , 0 = h 1 χ , γ
for t > 0 . Here, f ( χ , γ , t ) , h ( χ , γ , t ) , f 1 χ , γ and h 1 χ , γ are known functions, η is a real constant, ζ and μ are arbitrary constants. To obtain the solution of Equation (64), the following steps are applied.
  • Step 1: Applying the (TGLT) on both sides of Equation (64) and (DGLT) for Equation (65) and multiplying the outcome by s β , one can obtain
    U ( p , q , s ) = s α + 1 F 1 ( p , q ) + s β F ( p , q , s ) + s β G 3 1 χ χ u χ χ + 1 γ γ u γ γ η χ u u χ ζ γ u γ ω ,
    and
    ω ( p , q , s ) = s α + 1 H 1 ( p , q ) + s β H ( p . q , s ) + s β G 3 1 χ χ ω χ χ + 1 γ γ ω γ γ η χ u ω χ ζ γ ω γ ω .
  • Step 2: The (TGLTDM) defined the solution of the time-space fractional coupled Burgers equation in the following forms:
    u χ , γ , t = n = 0 u n χ , γ , t , ω χ , γ , t = n = 0 ω n χ , γ , t .
    We can obtain Adomian’s polynomials F n , E n and R n , respectively, as follows:
    E n = n = 0 u n γ ω n , F n = n = 0 u n ω n χ , R n = n = 0 ω n ω n γ
    where C n is mentioned in Equation (47). The Adomian polynomials for the nonlinear term u γ ω , u ω χ , and ω ω γ are defined by
    E 0 = u 0 ω 0 γ E 1 = u 0 ω 1 γ + u 1 ω 0 γ , E 2 = u 0 ω 2 γ + u 1 ω 1 γ + u 2 ω 0 γ , E 3 = u 0 ω 3 γ + u 1 ω 2 γ + u 2 ω 1 γ + u 3 ω 0 γ ,
    F 0 = u 0 ω 0 χ F 1 = u 0 ω 1 χ + u 1 ω 0 χ , F 2 = u 0 ω 2 χ + u 1 ω 1 χ + u 2 ω 0 χ , F 3 = u 0 ω 3 χ + u 1 ω 2 χ + u 2 ω 1 χ + u 3 ω 0 χ ,
    and
    R 0 = ω 0 ω γ R 1 = ω 0 ω 1 γ + ω 1 ω 0 γ R 2 = ω 0 ω 2 γ + ω 1 ω 1 γ + ω 2 ω 0 γ . R 3 = ω 0 ω 3 γ + ω 1 ω 2 γ + ω 2 ω 1 γ + ω 3 ω 0 γ .
  • Step 3: Applying the (ITGLT) on both sides of Equations (66) and (67) and using Equations (44) and (69), we obtain
    n = 0 u n χ , γ , t = f 1 χ , γ + G 3 1 s β F ( p , q , s ) + G 3 1 s β G 3 1 χ χ n = 0 u n χ χ + G 3 1 s β G 3 1 γ γ n = 0 u n γ γ G 3 1 s β G 3 η χ n = 0 C n G 3 1 s β G 3 ζ n = 0 E n χ ,
    and
    n = 0 ω n χ , γ , t = h 1 χ , γ + G 3 1 s β H ( p , q , s ) + G 3 1 s β G 3 1 χ χ n = 0 ω n χ χ + G 3 1 s β G 3 1 γ γ n = 0 ω n γ γ G 3 1 s β G 3 η χ n = 0 F n G 3 1 s β G 3 ζ n = 0 R n χ .
  • Step 4: On comparing both sides of Equations (73) and (74), we can obtain
    u 0 χ , γ , t = f 1 χ , γ + G p 1 G s 1 s β F ( p , q , s ) , ω 0 χ , γ , t = h 1 χ , γ + G p 1 G s 1 s β H ( p , q , s ) .
In general, the rest of the recursive relations are given by
u n + 1 = G 3 1 s β G 3 1 χ χ u χ χ + 1 γ γ u γ γ η χ C n ζ γ E n , ω n + 1 = G 3 1 s β G 3 1 χ χ ω χ χ + 1 γ γ ω γ γ η χ F n ζ γ R n ,
We assume that the (ITGLT) concerning p , q , and s exists for each term on the right-hand side of Equations (75) and (76). In the next example, we apply (TGLTDM) to solve the singular time-fractional coupled Burgers equation.
Example 4 
([22]). Consider the following system of singular fractional coupled Burgers equation with the initial conditions of the form:
D t β u 1 χ χ u χ χ 1 γ γ u γ γ + 1 χ u u χ + 1 γ u γ ω = χ 2 γ 2 e t D t β ω 1 χ χ ω χ χ 1 γ γ ω γ γ + 1 χ u ω χ + 1 γ ω γ ω = χ 2 γ 2 e t ,
and
u χ , γ , 0 = χ 2 γ 2 , ω χ , γ , 0 = χ 2 γ 2 .
By applying the above-mentioned method, we have
U ( p , q , s ) = s α + 1 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 + s β G 3 1 χ χ u χ χ + 1 γ γ u γ γ 1 χ u u χ 1 γ u γ ω + s χ β G 3 χ 2 γ 2 1 + t + t 2 2 ! + t 3 3 ! + · · · ,
and
ω ( p , q , s ) = s α + 1 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 + s β G 3 1 χ χ ω χ χ + 1 γ γ ω γ γ 1 χ u ω χ 1 γ ω γ ω + s χ β G 3 χ 2 γ 2 1 + t + t 2 2 ! + t 3 3 ! + · · · .
Therefore,
U ( p , q , s ) = s α + 1 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 + s β G 3 1 χ χ u χ χ + 1 γ γ u γ γ 1 χ u u χ 1 γ u γ ω + 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 s β + α + 1 + s β + α + 2 + s β + α + 3 + · · · ,
and
ω ( p , q , s ) = s α + 1 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 + s β G 3 1 χ χ ω χ χ + 1 γ γ ω γ γ 1 χ u ω χ 1 γ ω γ ω + 2 ! p α + 3 q α + 1 2 ! q α + 3 p α + 1 s β + α + 1 + s β + α + 2 + · · · .
Operating an (ITGLT) for Equations (81) and (82), and using the polynomial series solution, we obtain the following:
n = 0 u n χ , γ , t = χ 2 γ 2 + G 3 1 s β G 3 1 χ χ n = 0 u n χ χ + 1 γ γ n = 0 u n γ χ + G 3 1 s β G 3 1 χ n = 0 C n 1 γ n = 0 E n + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + . . . ,
and
n = 0 ω n χ , γ , t = χ 2 γ 2 + G 3 1 s β G 3 1 χ χ n = 0 ω n χ χ + 1 γ γ n = 0 ω n γ χ + G 3 1 s β G 3 1 χ n = 0 F n 1 γ n = 0 R n + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + · · · .
Then, we determine the iteration components as
u 0 = χ 2 γ 2 + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + · · · , ω 0 = χ 2 γ 2 + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + · · · ,
and
u n + 1 = G 3 1 s β G 3 1 χ χ u n χ χ + 1 γ γ u n γ χ C n E n , ω n + 1 = G 3 1 s β G 3 1 χ χ ω n χ χ + 1 γ γ ω n γ χ F n R n ,
where n 0 . The remaining terms are given by the following, at n = 0 :
u 1 = G 3 1 s β G 3 1 χ χ u 0 χ χ + 1 γ γ u 0 γ γ 1 χ u 0 u 0 χ 1 γ u 0 γ ω 0 , = G 3 1 s β G 3 0 . u 1 = 0 ,
In a similar manner,
ω 1 = G 3 1 s β G 3 1 χ χ ω 0 χ χ + 1 γ χ γ ω 0 γ γ 1 χ u ω χ 1 γ ω 0 ω 0 γ , = G 3 1 s β G 3 0 , ω 1 = 0 .
Similarly, at n = 1 ,
u 2 = G 3 1 s β G 3 0 = 0 , ω 2 = 0 .
In a similar manner,
u 3 = 0 , ω 3 = 0 .
As a collection of all terms, we obtain
u χ , γ , t = u 0 + u 1 + u 2 + u 3 + · · · , ω χ , γ , t = ω 0 + ω 1 + ω 2 + ω 3 + · · · .
Thus, the approximate solution of Equation (77) is defined by
u χ , γ , t = χ 2 γ 2 + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + · · · , ω χ , γ , t = χ 2 γ 2 + χ 2 γ 2 t β Γ β + 1 + t β + 1 Γ β + 2 + t β + 2 Γ β + 3 + · · · .
The exact solution can be obtained by substituting with β = 1 , as follows:
u χ , γ , t = χ 2 γ 2 1 + t + t 2 2 ! + t 3 3 ! + t 4 4 ! + · · · , ω χ , γ , t = χ 2 γ 2 1 + t + t 2 2 ! + t 3 3 ! + t 4 4 ! + · · · .
Hence,
ω χ , γ , t = χ 2 γ 2 e t , ω χ , γ , t = χ 2 γ 2 e t .
Figure 7: This illustrates the contrast between the exact solution and the obtained numerical solution for Equation (54). At t = 1 and β = 1 , we obtained the accurate solution. By taking different values of β , for instance, ( β = 0.75 , β = 0.85 and β = 0.95 ), we obtained the estimated solutions.
Figure 8: We demonstrate the result of the functions ω χ , t = u χ , t in three-dimensional space.
Table 4 Indicates the numerical solution for different values of β for the function ψ ( x , t ) .

4. Conclusions

This study proposes a new double and triple generalized-Laplace transform decomposition method (DGLTDM and TGLTDM), a novel method combining DGLT, TGLT, and DM to obtain the solutions of regular and singular fractional Burgers equations. This combination produces a strong method. The ability and precision of the proposed plan are confirmed through examples. This technique can be used for many difficult linear and nonlinear FPDEs and systems of FPDE that do not require linearity. Furthermore, in the future, we plan to employ this technique to solve various scientific problems related to our research field.

Author Contributions

Methodology, H.E. and S.M.; Software, H.E.; Formal analysis, H.E.; Investigation, H.E.; Writing—original draft, H.E.; Writing—review and editing, S.M. All authors have read and agreed to the published version of the manuscript.

Funding

The authors would like to extend their sincere appreciation to the Researchers Supporting Project (RSPD 2024R948), King Saud University, Riyadh, Saudi Arabia.

Data Availability Statement

Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

Conflicts of Interest

The authors declare that they have no competing interests.

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Figure 1. Comparison between exact and numerical solutions.
Figure 1. Comparison between exact and numerical solutions.
Fractalfract 08 00692 g001
Figure 2. The surface of the function ω x , t .
Figure 2. The surface of the function ω x , t .
Fractalfract 08 00692 g002
Figure 3. Comparison between exact and numerical solutions.
Figure 3. Comparison between exact and numerical solutions.
Fractalfract 08 00692 g003
Figure 4. The surface of the function ω x , t = u x , t .
Figure 4. The surface of the function ω x , t = u x , t .
Fractalfract 08 00692 g004
Figure 5. Comparison between exact and numerical solutions.
Figure 5. Comparison between exact and numerical solutions.
Fractalfract 08 00692 g005
Figure 6. The surface of the function ψ ( x , t ) .
Figure 6. The surface of the function ψ ( x , t ) .
Fractalfract 08 00692 g006
Figure 7. Comparison between exact and numerical solutions.
Figure 7. Comparison between exact and numerical solutions.
Fractalfract 08 00692 g007
Figure 8. The surface of the function ψ ( x , t ) .
Figure 8. The surface of the function ψ ( x , t ) .
Fractalfract 08 00692 g008
Table 1. Comparison between exact and approximation solutions.
Table 1. Comparison between exact and approximation solutions.
tx β = 0.75 β = 0.85 β = 0.95 β = 1 Exact
0.001.28321.255281.231571.22141.2214
0.251.647671.611811.581371.568311.56831
0.10.502.115642.069612.030522.013752.01375
0.752.716542.657432.607242.585712.58571
1.003.488113.412213.347763.320113.320112
Table 2. Comparison between exact and approximation solutions.
Table 2. Comparison between exact and approximation solutions.
tx β = 0.75 β = 0.85 β = 0.95 β = 1 Exact
0.000.000.000.000.0000.00
0.250.2043280.2132460.2206390.2238180.22386
0.10.500.3959520.4132330.427560.433720.433802
0.750.5629570.5875280.6078970.6166560.616772
1.000.6949610.7252930.7504380.7612510.761394
Table 3. Comparison between exact and approximation solutions.
Table 3. Comparison between exact and approximation solutions.
tx β = 0.75 β = 0.85 β = 0.95 β = 1 Exact
0.00−0.0509492−0.0200888−0.00492833−0.00083330.00
0.250.00009855640.03354430.05075720.05570830.0565523
0.10.500.1532420.19444430.2178140.2253330.226209
0.750.4084810.4626090.4962410.5080420.508971
1.000.7658150.8380410.886040.9038330.904837
Table 4. Comparison between exact and approximation solutions.
Table 4. Comparison between exact and approximation solutions.
ytx β = 0.75 β = 0.85 β = 0.95 β = 1 Exact
0.00−0.0118283−0.0114159−0.0110883−0.0109517−0.0110517
0.250.06209880.05993330.05821370.05749630.0580215
0.100.100.500.283880.2739810.266120.262840.265241
0.750.6535160.6307270.612630.605080.610607
1.001.171011.130171.097741.084221.09412
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Eltayeb, H.; Mesloub, S. Solution for Time-Fractional Coupled Burgers Equations by Generalized-Laplace Transform Methods. Fractal Fract. 2024, 8, 692. https://doi.org/10.3390/fractalfract8120692

AMA Style

Eltayeb H, Mesloub S. Solution for Time-Fractional Coupled Burgers Equations by Generalized-Laplace Transform Methods. Fractal and Fractional. 2024; 8(12):692. https://doi.org/10.3390/fractalfract8120692

Chicago/Turabian Style

Eltayeb, Hassan, and Said Mesloub. 2024. "Solution for Time-Fractional Coupled Burgers Equations by Generalized-Laplace Transform Methods" Fractal and Fractional 8, no. 12: 692. https://doi.org/10.3390/fractalfract8120692

APA Style

Eltayeb, H., & Mesloub, S. (2024). Solution for Time-Fractional Coupled Burgers Equations by Generalized-Laplace Transform Methods. Fractal and Fractional, 8(12), 692. https://doi.org/10.3390/fractalfract8120692

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