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Article

Fractional Analysis of Nonlinear Boussinesq Equation under Atangana–Baleanu–Caputo Operator

1
King Abdulaziz University, College of Science & Arts, Department of Mathematics, Rabigh, Saudi Arabia
2
Department of Mathematics, Abdul Wali Khan University Mardan, Mardan 23200, Pakistan
3
Department of Mechanical Engineering, Sejong University, Seoul 05006, Republic of Korea
4
Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work and are co-first authors.
Submission received: 13 October 2022 / Revised: 7 November 2022 / Accepted: 8 November 2022 / Published: 15 November 2022
(This article belongs to the Section Mathematics)

Abstract

:
This article proposed two novel techniques for solving the fractional-order Boussinesq equation. Several new approximate analytical solutions of the second- and fourth-order time-fractional Boussinesq equation are derived using the Laplace transform and the Atangana–Baleanu fractional derivative operator. We give some graphical and tabular representations of the exact and proposed method results, which strongly agree with each other, to demonstrate the trustworthiness of the suggested methods. In addition, the solutions we obtain by applying the proposed approaches at different fractional orders are compared, confirming that as the value trends from the fractional order to the integer order, the result gets closer to the exact solution. The current technique is interesting, and the basic methodology suggests that it might be used to solve various fractional-order nonlinear partial differential equations.

1. Introduction

Fractional calculus (FC) is considered to be the best tool for modeling a variety of models accurately; it may be utilized to solve major problems in chemistry, physics, and other applied sciences. The fractional-order control society model has gotten a lot of attention recently. Another example is the movement of a loaded and fixed elastic column on both sides. Other FC theories with strong mathematical backgrounds [1] have been discussed. The primary motivation for exhausting this calculus is that it accurately describes many real-world systems. Heat conduction, water flow, and infinite lossy transmission lines are examples of fractional phenomena. Fractional calculus is used in a variety of industrial applications, including electrical circuits, signal processing, chaos theory, chemical processes, and control systems [2,3]. Many writers have looked into the relationship between the ideas of stability and symmetry in differential equations for many different types of differential equations, including irreversible thermodynamics in [4] and nonequilibrium statistical mechanics in [5]. In [6], it is examined how stability and symmetries interact with the study and control of nonlinear dynamical systems and networks. As a result, this problem is a current one in differential equations with fractional derivatives. One of the most crucial characteristics of functional differential equations is recognized to be the stability of the solutions [7,8,9].
Fractional differential and integral equations have been the best tool for accurately modeling diverse physical events throughout the last few decades. The rheological properties of numerous polymers can be represented using fractional differential equations (FDEs), according to the models that have been studied. In the high-population polymers of nuclear mechanics, mechanical vibrations, thermo-elasticity, and biological tissues, fractional phenomena have been developed as damping phenomena. Diffusion and electrolytic coating methods have been experimentally validated. Fractional diffusion is a fractal structure that emerges from a rough metallic covering. Quantum fractional differential models have been examined by Xiao et al., Yang et al., and Herrmann in his book [10,11,12]. Nature is nonlinear; hence, many systems in it are nonlinear as well. Interestingly, FDEs may be used to explain some of these nonlinear systems. FDEs have recently attracted much attention due to their multiple benefits in applied sciences. Several important topics in electrochemistry, electro genetics, acoustics, and material science are described in the FDEs. The importance and popularity of FDEs can also be shown in other branches of science, such as nonlinear earthquake oscillation and so on [13,14,15,16,17,18,19,20,21,22].
In this article, we implemented two analytical methods in addition to the Laplace transform and Atangana–Baleanu fractional derivative operator [23,24,25,26] to solve fractional-order Bs equations. Atangana–Baleanu’s fractional derivative modeling has been demonstrated to have a short random walk. The Atangana–Baleanu fractional derivative in the sense of Caputo is a powerful mathematical tool for simulating more challenging real-world problems [27,28], and it has been found that the Mittag-Leffler function is a more significant and useful filter tool than the power and exponential law functions. The first technique demonstrates how Adomian’s decomposition method can be implemented to approximate the solution of a nonlinear differential using the Laplace transform. This approach for solving the Diffing equation was used by Agadjanov [29] in 2006. The Laplace decomposition method is consistent with the varied nature of physical problems and has been implemented in a large number of functional equations [30,31,32,33]. The second is the Laplace variational iteration method (LVIM), in which the Laplace transform (LT) and the variational iteration method are combined. In this method, Lagrange multipliers are taken for solving nonlinear equations based on the nature of the issue [34,35]. He [36] was the first to present the VIM, which was an effective approach for a wide range of problems in applied sciences [37,38,39]. This technique has recently become popular among researchers for solving linear and nonlinear equations. Unlike other approaches, the VIM does not require linearization, discretization, or perturbation. It allows for a rapid convergence and successive approximations to the exact answer [40,41,42].
PDEs have recently attracted the interest of several scientists due to their fascination with showing particle movement in clustered media and complex processes. The BsEqs identify the wave propagation in nonlinear and dissipation media in hydrodynamics [43]. In this study, we plan to use the LTDM and VITM to find approximate-analytical solutions for the fourth-order time-fractional BsEq. Bear [44] first introduced the classical BsEq in 1978, as follows:
ψ ψ ψ + ϖ ϖ ϖ + W = Q
where ψ denotes the saturated hydraulic conductivity in the ψ direction ( U T ) ; ϖ denotes the saturated hydraulic conductivity in the ϖ direction ( U T ) ; indicates the hydraulic head (U); Q is the precise yield (dimensionless); and W is the recharge/discharge rate ( U T ) .
The fractional BsEq is built by taking into account the following three assumptions:
1.
Assume that the Dupuit–Forchhimer assumptions are correct, as well as Darcy’s law.
2.
In the control volume, the fluid (water) is non-compressible.
3.
The flux variations in the control volume are governed by a power-law function.
The BsEq is a set of equations developed in hydrodynamics to describe the wave propagation in nonlinear and dissipative media [45]. They are commonly employed in coastal and ocean engineering to solve difficulties with water percolation in porous subsurface strata. The BsEq also serves as the foundation for many models that describe unconfined groundwater flow and subsurface drainage issues [46,47,48].
The sections of this work are structured as follows. Some fundamental FC concepts connected to our current investigation are presented in Section 2. In Section 3, the LTDM is used to solve the fractional partial differential equations, and Section 4 presents the VITM methodology. Using the suggested methods, we arrive at the time-fractional Boussinesq equation’s solution in Section 5. A concise conclusion is provided in Section 6.

2. Preliminaries

In this part, we present the fractional calculus basic definitions related to our present work.

Definition

Definition 1.
The fractional-order Caputo derivative is defined by [49,50]
L C D ρ { g ( ) } = 1 Γ ( n ρ ) 0 ( k ) n ρ 1 g n ( k ) d k , w h e r e n < ρ n + 1 .
Definition 2.
The Laplace transformation in combination with Caputo derivative L C D ρ { g ( ) } is given as [51,52]
L { L C D ρ { g ( ) } } ( ω ) = 1 ω n ρ ω n L { g ( ) } ( ω ) ω n 1 g ( 0 ) g n 1 ( 0 ) .
Definition 3.
In the Caputo manner, the Atangana–Baleanu derivative is defined as [51,52]
A B C D ρ { g ( ) } = A ( ρ ) 1 ρ a g ( k ) E ρ ρ 1 ρ ( 1 k ) ρ d k
where A ( γ ) is a normalization function with A ( 0 ) = A ( 1 ) = 1 , g H 1 ( a , b ) , b > a , ρ [ 0 , 1 ] and E γ is the Mittag-Leffler function.
Definition 4.
In the Riemann–Liouville manner, the Atangana–Baleanu derivative is defined as [51,52]
A B C D ρ { g ( ) } = A ( ρ ) 1 ρ d d a g ( k ) E ρ γ 1 ρ ( 1 k ) ρ d k
Definition 5.
The Atangana–Baleanu operator in connection with the Laplace transform is defined as [51,52]
A B D ρ { g ( ) } ( ω ) = A ( γ ) ω ρ L { g ( ) } ( ω ) ω ρ 1 g ( 0 ) ( 1 ρ ) ω ρ + ρ 1 γ
Definition 6.
Let g be a function of ρ, with 0 < ρ < 1 , and then the integral operator having fractional order of ρ is defined as [51,52]
A B C I ρ { g ( ) } = 1 ρ A ( ρ ) g ( ) + ρ A ( ρ ) Γ ( ρ ) a g ( k ) ( k ) ρ 1 d k

3. Methodology of LTDM

We will show the general implementation of LTDM to solve fractional-order PDE.
D ρ μ ( ψ , ) + G ¯ 1 ( ψ , ) + N 1 ( ψ , ) = F ( ψ , ) , 0 < ρ 1 ,
having initial sources
μ ( ψ , 0 ) = ξ ( ψ ) , μ ( ψ , 0 ) = ζ ( ψ ) .
where D ρ = ρ ρ is the fractional AB operator having order ρ ; G ¯ 1 , N 1 are linear and non-linear operators; and F ( ψ , ) is the source term.
By employing Laplace transform, we have
L [ D ρ μ ( ψ , ) + G ¯ 1 ( ψ , ) + N 1 ( ψ , ) ] = L [ F ( ψ , ) ] .
On utilizing differentiation property of Laplace, we obtain
L [ μ ( ψ , ) ] = Θ ( ψ , ω ) ω ρ + ρ ( 1 ρ ) ω ρ L [ G ¯ 1 ( ψ , ) + N 1 ( ψ , ) ] ,
where Θ ( ψ , ω ) = 1 ω ρ + 1 [ ω ρ g 0 ( ψ ) + ω ρ 1 g 1 ( ψ ) + + g 1 ( ψ ) ] + ω ρ + ρ ( 1 ρ ) ω ρ F ( ψ , ) .
On taking Laplace inverse transform of (4), we have
μ ( ψ , ) = Θ ( ψ , ω ) L 1 ω ρ + ρ ( 1 ρ ) ω ρ L [ G ¯ 1 ( ψ , ) + N 1 ( ψ , ) ] ,
where Θ ( ψ , ω ) is the term that comes as a result of source term. The solution by means of LTDM for μ ( ψ , ) in terms of infinite sequence is given as
μ ( ψ , ) = m = 0 μ m ( ψ , ) .
and decomposing the nonlinear operator N 1 as
N 1 ( ψ , ) = m = 0 A m .
where Adomian polynomials are represented by A m and determined as
A m = 1 m ! m m N 1 k = 0 k ψ k , k = 0 k k = 0 ,
putting Equations (6) and (8) into Equation (5) gives
m = 0 μ m ( ψ , ) = Θ ( ψ , ω ) L 1 ω ρ + ρ ( 1 ρ ) ω ρ L [ G ¯ 1 ( m = 0 ψ m , m = 0 m ) + m = 0 A m ] ,
The given terms are derived.
μ 0 ( ψ , ) = Θ ( ψ , ω ) ,
μ 1 ( ψ , ) = L 1 ω ρ + ρ ( 1 ρ ) ω ρ L [ G ¯ 1 ( ψ 0 , 0 ) + A 0 ] ,
The remaining terms for m 1 are determined as
μ m + 1 ( ψ , ) = L 1 ω ρ + ρ ( 1 ρ ) ω ρ L [ G ¯ 1 ( ψ m , m ) + A m ] .

4. Methodology of VITM

We will show the general implementation of VITM to solve fractional-order PDE.
D δ μ ( ψ , ) + M μ ( ψ , ) + N μ ( ψ , ) P ( ψ , ) = 0 , m 1 < δ m ,
having initial source
μ ( ψ , 0 ) = g 1 ( ψ ) .
where D δ = δ δ is the fractional AB operator of order. M , N are linear and non-linear operators and P is the source term.
On employing Laplace transform, we have
L [ D δ μ ( ψ , ) ] + L [ M μ ( ψ , ) + N μ ( ψ , ) P ( ψ , ) ] = 0 .
By utilizing differentiation property of Laplace, we obtain
L [ μ ( ψ , ) ] = ω ρ ω ρ + ρ ( 1 ρ ) L M μ ( ψ , ) + N μ ( ψ , ) P ( ψ , ) .
The iteration technique for Equation (14)
μ m + 1 ( ψ , ) = μ m ( ψ , ) + ρ ( ω ) ω ρ ω ρ + ρ ( 1 ρ ) L M μ ( ψ , ) + N μ ( ψ , ) P ( ψ , ) .
ρ ( ω ) is Lagrange multiplier and
ρ ( ω ) = ω ρ + ρ ( 1 ρ ) ω ρ ,
with the application of inverse Laplace transform, Equation (15) series form solution is given by
μ 0 ( ψ , ) = μ ( 0 ) + L 1 ρ ( ω ) L [ P ( ψ , ) ] ,
μ 1 ( ψ , ) = L 1 ρ ( ω ) L [ M μ ( ψ , ) + N μ ( ψ , ) ] ,
μ n + 1 ( ψ , ) = L 1 ρ ( ω ) L M μ 0 ( ψ , ) + μ 1 ( ψ , ) + , μ n ( ψ , ) + N μ 0 ( ψ , ) + μ 1 ( ψ , ) , , μ n ( ψ , ) .

5. Applications

In this part, we implemented LTDM and HPTM to solve different time-fractional Boussinesq equation.

5.1. Example 1

Consider the general fourth-order time-fractional Boussinesq equation with one-dimensional space variable
D ρ μ ( ψ , ) = δ D ψ 4 μ ( ψ , ) + γ D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) 1 < ρ 2 , > 0 ,
having initial condition
μ ( ψ , 0 ) = exp ( ψ ) , μ ( ψ , 0 ) = 0 .
On employing Laplace transform, we have
ω ρ L [ μ ( ψ , ) ] ω 1 μ ( ψ , 0 ) ω ρ + ρ ( 1 ω ρ ) = L δ D ψ 4 μ ( ψ , ) + γ D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) .
By inverse Laplace transform, we have
μ ( ψ , ) = exp ( ψ ) + L 1 ω ρ + ρ ( 1 ω ρ ) ω ρ L δ D ψ 4 μ ( ψ , ) + γ D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) .
The solution by means of LTDM for μ ( ψ , ) in terms of infinite sequence is given as
μ ( ψ , ) = m = 0 μ m ( ψ , ) .
where μ ( μ ) ψ ψ = m = 0 A m and μ 2 = m = 0 B m are Adomian polynomials which denote the nonlinear terms, so Equation (19) can be rewritten as
m = 0 μ m ( ψ , ) = exp ( ψ ) + L 1 ω ρ + ρ ( 1 ω ρ ) ω ρ L δ D ψ 4 μ ( ψ , ) + γ D ψ 2 μ ( ψ , ) + θ m = 0 A m 4 θ m = 0 B m .
The decomposition of nonlinear terms with the help of Adomian polynomials are calculated as
A 0 = ( μ 0 2 ) ψ ψ , A 1 = 2 ( μ 0 ) ψ ψ ( μ 1 ) ψ ψ , A 2 = 2 ( μ 0 ) ψ ψ ( μ 2 ) ψ ψ + ( μ 1 2 ) ψ ψ , B 0 = μ 0 2 , B 1 = 2 μ 0 μ 1 , B 2 = 2 μ 0 μ 2 + ( μ 1 ) 2 .
Thus, in comparison with Equation (21) both sides,
μ 0 ( ψ , ) = exp ( ψ ) ,
For m = 0 ,
μ 1 ( ψ , ) = ( δ + γ ) exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
For m = 1 ,
μ 2 ( ψ , ) = ( δ + γ ) 2 exp ( ψ ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
For m = 2 ,
μ 3 ( ψ , ) = ( δ + γ ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
Thus, series form solution is determined as
μ ( ψ , ) = m = 0 μ m ( ψ , ) = μ 0 ( ψ , ) + μ 1 ( ψ , ) + μ 2 ( ψ , ) + μ 3 ( ψ , ) +
μ ( ψ , ) = exp ( ψ ) + ( δ + γ ) exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( δ + γ ) 2 exp ( ψ ) [ ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ] + ( δ + γ ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( ψ + ( δ + γ ) ) .
Now, implementing VITM.
Now, by iteration formula, Equation (17) is written as
μ m + 1 ( ψ , ) = μ m ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) δ D ψ 4 μ m ( ψ , ) + γ D ψ 2 μ m ( ψ , ) + θ D ψ 2 μ m 2 ( ψ , ) 4 θ μ m 2 ( ψ , ) } ] ,
where
μ 0 ( ψ , ) = exp ( ψ ) .
For m = 0 , 1 , 2 , ,
μ 1 ( ψ , ) = μ 0 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) δ D ψ 4 μ 0 ( ψ , ) + γ D ψ 2 μ 0 ( ψ , ) + θ D ψ 2 μ 0 2 ( ψ , ) 4 θ μ 0 2 ( ψ , ) } ] , μ 1 ( ψ , ) = ( δ + γ ) exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
μ 2 ( ψ , ) = μ 1 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) δ D ψ 4 μ 1 ( ψ , ) + γ D ψ 2 μ 1 ( ψ , ) + θ D ψ 2 μ 1 2 ( ψ , ) 4 θ μ 1 2 ( ψ , ) } ] , μ 2 ( ψ , ) = ( δ + γ ) 2 exp ( ψ ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
μ 3 ( ψ , ) = μ 2 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) δ D ψ 4 μ 2 ( ψ , ) + γ D ψ 2 μ 2 ( ψ , ) + θ D ψ 2 μ 2 2 ( ψ , ) 4 θ μ 2 2 ( ψ , ) } ] , μ 3 ( ψ , ) = ( δ + γ ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
μ ( ψ , ) = m = 0 μ m ( ψ , ) = exp ψ + ( δ + γ ) exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( δ + γ ) 2 exp ( ψ ) [ ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ] + ( δ + γ ) 3 exp ( ψ ) [ ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ] +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( ψ + ( δ + γ ) ) .
The graphs in Figure 1a,b show the behavior of the exact and proposed methods’ solution in the (AB fractional derivative) sense at ρ = 2 .  Figure 1c shows our method’s solution at different fractional-orders of ρ = 2 , 1.9 , 1.8 , 1.7 , δ = 0.1 , γ = 0.3 , and 0 ψ 1 for problem 1 and Figure 1d, respectively, at [ 0 , 1 ] and 0 ψ 1 . The graphical representation shows that exact solution and proposed methods’ solution are in good agreement. Moreover, in Table 1, the comparison of the exact solution and our method’s solution in addition with absolute error at various fractional orders.

5.2. Example 2

Consider the n-dimensional time-fractional fourth-order Boussinesq equation
D ρ μ ( ψ , ) = i = 0 n δ i D ψ i 4 μ ( ψ , ) + i = 0 n γ i D ψ i 2 μ ( ψ , ) + i = 0 n θ i D ψ i 2 μ 2 ( ψ , ) i = 0 n 4 θ i μ 2 ( ψ , ) 1 < ρ 2 , > 0 ,
having initial condition
μ ( ψ , 0 ) = exp i = 0 n ( ψ i ) , μ ( ψ , 0 ) = 0 .
where 1 < ρ 2 , δ i , γ i , θ i , ( i = 1 , 2 , , n ) , and ψ = ( ψ 1 , ψ 2 , , ψ n ) n , > 0 .
On employing Laplace transform, we have
ω ρ L [ μ ( ψ , ) ] ω 1 μ ( ψ , 0 ) ω ρ + ρ ( 1 ω ρ ) = L [ i = 0 n δ i D ψ i 4 μ ( ψ , ) + i = 0 n γ i D ψ i 2 μ ( ψ , ) + i = 0 n θ i D ψ i 2 μ 2 ( ψ , ) i = 0 n 4 θ i μ 2 ( ψ , ) ] .
By inverse Laplace transform, we have
μ ( ψ , ) = exp i = 0 n ( ψ i ) + L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L [ i = 0 n δ i D ψ i 4 μ ( ψ , ) + i = 0 n γ i D ψ i 2 μ ( ψ , ) + i = 0 n θ i D ψ i 2 μ 2 ( ψ , ) i = 0 n 4 θ i μ 2 ( ψ , ) ] ] .
The solution by means of LTDM for μ ( ψ , ) in terms of infinite sequence is given as
μ ( ψ , ) = m = 0 μ m ( ψ , ) .
where μ ( μ ) ψ ψ = m = 0 A m and μ 2 = m = 0 B m are Adomian polynomials which denote the nonlinear terms, so Equation (30) can be rewritten as
m = 0 μ m ( ψ , ) = exp i = 0 n ( ψ i ) + L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L [ i = 0 n δ i μ ( ψ , ) ψ ψ ψ ψ + i = 0 n γ i μ ( ψ , ) ψ ψ + i = 0 n θ i m = 0 A m 4 i = 0 n θ i m = 0 B m ] ] .
As the nonlinear terms can be determined with the help of Adomian’s polynomials given in (22), we acquire
μ 0 ( ψ , ) = exp i = 0 n ( ψ i ) ,
For m = 0 ,
μ 1 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) exp i = 0 n ( ψ i ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
For m = 1 ,
μ 2 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) 2 exp i = 0 n ( ψ i ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
For m = 2 ,
μ 3 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) 3 exp i = 0 n ( ψ i ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
Thus, series form solution is determined as
μ ( ψ , ) = m = 0 μ m ( ψ , ) = μ 0 ( ψ , ) + μ 1 ( ψ , ) + μ 2 ( ψ , ) + μ 3 ( ψ , ) +
μ ( ψ , ) = exp i = 0 n ( ψ i ) + ( i = 0 n δ i + i = 0 n γ i ) exp i = 0 n ( ψ i ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( i = 0 n δ i + i = 0 n γ i ) 2 exp i = 0 n ( ψ i ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 + ( i = 0 n δ i + i = 0 n γ i ) 3 exp i = 0 n ( ψ i ) [ ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ] +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( i = 0 n ( ψ i ) + ( i = 0 n δ i + i = 0 n γ i ) ) .
Now, implementing VITM.
Now, by iteration formula, Equation (28) is written as
μ m + 1 ( ψ , ) = μ m ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) i = 0 n δ i D ψ i 4 μ m ( ψ , ) + i = 0 n γ i D ψ i 2 μ m ( ψ , ) + i = 0 n θ i D ψ i 2 μ m 2 ( ψ , ) i = 0 n 4 θ i μ m 2 ( ψ , ) } ] ,
where
μ 0 ( ψ , ) = exp i = 0 n ( ψ i ) .
For m = 0 , 1 , 2 , ,
μ 1 ( ψ , ) = μ 0 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) i = 0 n δ i D ψ i 4 μ 0 ( ψ , ) + i = 0 n γ i D ψ i 2 μ 0 ( ψ , ) + i = 0 n θ i D ψ i 2 μ 0 2 ( ψ , ) i = 0 n 4 θ i μ 0 2 ( ψ , ) } ] , μ 1 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) exp i = 0 n ( ψ i ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
μ 2 ( ψ , ) = μ 1 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) i = 0 n δ i D ψ i 4 μ 1 ( ψ , ) + i = 0 n γ i D ψ i 2 μ 1 ( ψ , ) + i = 0 n θ i D ψ i 2 μ 1 2 ( ψ , ) i = 0 n 4 θ i μ 1 2 ( ψ , ) } ] , μ 2 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) 2 exp i = 0 n ( ψ i ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
μ 3 ( ψ , ) = μ 2 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) i = 0 n δ i D ψ i 4 μ 2 ( ψ , ) + i = 0 n γ i D ψ i 2 μ 2 ( ψ , ) + i = 0 n θ i D ψ i 2 μ 2 2 ( ψ , ) i = 0 n 4 θ i μ 2 2 ( ψ , ) } ] , μ 3 ( ψ , ) = ( i = 0 n δ i + i = 0 n γ i ) 3 exp i = 0 n ( ψ i ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
μ ( ψ , ) = m = 0 μ m ( ψ , ) = exp i = 0 n ( ψ i ) + ( i = 0 n δ i + i = 0 n γ i ) exp i = 0 n ( ψ i ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( i = 0 n δ i + i = 0 n γ i ) 2 exp i = 0 n ( ψ i ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 + ( i = 0 n δ i + i = 0 n γ i ) 3 exp i = 0 n ( ψ i ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( i = 0 n ( ψ i ) + ( i = 0 n δ i + i = 0 n γ i ) ) .
The graphs in Figure 2a,b show the behavior of the exact and approximate solution in the (AB fractional derivative) sense at δ 1 = 0.1 , δ 2 = 0.3 , γ 1 = 0.5 , γ 2 = 0.2 , ρ = 2 , and 0 ψ 3 . The graphical representation shows that exact solution and proposed methods’ solution are in good agreement. In Table 2, the comparison of exact solution and our method’s solution is shown which confirm the validity of the proposed methods. Moreover, in Table 2, the comparison of absolute errors at different fractional orders is taken which tells us that results get better as fractional order goes to integer order.

5.3. Example 3

Consider the 2nth-order time-fractional Boussinesq equation
D ρ μ ( ψ , ) = D ψ ( 2 n ) μ ( ψ , ) + D ψ ( 2 n 1 ) μ ( ψ , ) + + D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) 1 < ρ 2 , > 0 ,
having initial condition
μ ( ψ , 0 ) = exp ( ψ ) , μ ( ψ , 0 ) = 0 .
On employing Laplace transform, we have
ω ρ L [ μ ( ψ , ) ] ω 1 μ ( ψ , 0 ) ω ρ + ρ ( 1 ω ρ ) = L [ D ψ ( 2 n ) μ ( ψ , ) + D ψ ( 2 n 1 ) μ ( ψ , ) + + D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) ] ,
equivalently, we have
L [ μ ( ψ , ) ] = exp ( ψ ) + L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L [ D ψ ( 2 n ) μ ( ψ , ) + D ψ ( 2 n 1 ) μ ( ψ , ) + + D ψ 2 μ ( ψ , ) + θ D ψ 2 μ 2 ( ψ , ) 4 θ μ 2 ( ψ , ) ] ] ,
The solution by means of LTDM for μ ( ψ , ) in terms of infinite sequence is given as
μ ( ψ , ) = m = 0 μ m ( ψ , ) .
where μ ( μ ) ψ ψ = m = 0 A m and μ 2 = m = 0 B m are Adomian polynomials which denote the nonlinear terms, so Equation (40) can be rewritten as
m = 0 μ m ( ψ , ) = exp ( ψ ) + L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L [ D ψ ( 2 n ) μ ( ψ , ) + D ψ ( 2 n 1 ) μ ( ψ , ) + + D ψ 2 μ ( ψ , ) + θ m = 0 A m 4 θ m = 0 B m ] ] .
As the nonlinear terms can be determined with the help of Adomian’s polynomials given in (22), we acquire
μ 0 ( ψ , ) = exp ( ψ ) ,
For m = 0 ,
μ 1 ( ψ , ) = n exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
For m = 1 ,
μ 2 ( ψ , ) = ( n ) 2 exp ( ψ ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
For m = 2 ,
μ 3 ( ψ , ) = ( n ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
Thus, series form solution is determined as
μ ( ψ , ) = m = 0 μ m ( ψ , ) = μ 0 ( ψ , ) + μ 1 ( ψ , ) + μ 2 ( ψ , ) + μ 3 ( ψ , ) +
μ ( ψ , ) = exp ( ψ ) + n exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( n ) 2 exp ( ψ ) [ ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ] + ( n ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( ψ + n ) .
Now, implementing VITM.
Now, by iteration formula, Equation (38) is written as
μ m + 1 ( ψ , ) = μ m ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) D ψ ( 2 n ) μ m ( ψ , ) + D ψ ( 2 n 1 ) μ m ( ψ , ) + + D ψ 2 μ ( ψ , ) + θ D ψ 2 μ m 2 ( ψ , ) 4 θ μ m 2 ( ψ , ) } ] ,
where
μ 0 ( ψ , ) = exp ( ψ ) .
For m = 0 , 1 , 2 , ,
μ 1 ( ψ , ) = μ 0 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) D ψ ( 2 n ) μ 0 ( ψ , ) + D ψ ( 2 n 1 ) μ 0 ( ψ , ) + + D ψ 2 μ 0 ( ψ , ) + θ D ψ 2 μ 0 2 ( ψ , ) 4 θ μ 0 2 ( ψ , ) } ] , μ 1 ( ψ , ) = n exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) ,
μ 2 ( ψ , ) = μ 1 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) D ψ ( 2 n ) μ 1 ( ψ , ) + D ψ ( 2 n 1 ) μ 1 ( ψ , ) + + D ψ 2 μ 1 ( ψ , ) + θ D ψ 2 μ 1 2 ( ψ , ) 4 θ μ 1 2 ( ψ , ) } ] , μ 2 ( ψ , ) = ( n ) 2 exp ( ψ ) ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ,
μ 3 ( ψ , ) = μ 2 ( ψ , ) L 1 [ ω ρ + ρ ( 1 ω ρ ) ω ρ L { ω ρ ω ρ + ρ ( 1 ω ρ ) D ψ ( 2 n ) μ 2 ( ψ , ) + D ψ ( 2 n 1 ) μ 2 ( ψ , ) + + D ψ 2 μ 2 ( ψ , ) + θ D ψ 2 μ 2 2 ( ψ , ) 4 θ μ 2 2 ( ψ , ) } ] , μ 3 ( ψ , ) = ( n ) 3 exp ( ψ ) ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ,
μ ( ψ , ) = m = 0 μ m ( ψ , ) = exp ( ψ ) + n exp ( ψ ) ρ ρ Γ ( ρ + 1 ) + ( 1 ρ ) + ( n ) 2 exp ( ψ ) [ ρ 2 2 ρ Γ ( 2 ρ + 1 ) + 2 ρ ( 1 ρ ) ρ Γ ( ρ + 1 ) + ( 1 ρ ) 2 ] + ( n ) 3 exp ( ψ ) [ ρ 3 3 ρ Γ ( 3 ρ + 1 ) + 3 ρ 2 ( 1 ρ ) 2 ρ Γ ( 2 ρ + 1 ) + 3 ρ ( 1 ρ ) 2 ρ Γ ( ρ + 1 ) + ( 1 ρ ) 3 ] +
The problem’s integer-order ρ = 2 solution μ ( ψ , ) = exp ( ψ + n ) .
The graphs in Figure 3a,b show the behavior of the exact and approximate solution in the (AB fractional derivative) sense at ρ = 2 .  Figure 1c gives 3D representation of our method’s solution at different fractional-orders for ρ = 2 , 1.9 , 1.8 , 1.7 , δ = 0.1 , γ = 0.3 , and 0 ψ 5 , for example, Figure 3 and Figure 1d, respectively, at [ 0 , 1 ] and 0 ψ 1 . The graphical representation shows that our solution converges quickly to exact solution as fractional order converges to integer order. Moreover, in Table 3, the exact solution and our method’s solution comparison in terms of absolute error at various fractional orders are shown.

6. Conclusions

The HPTM and VITM were used to study the significance of the thoroughly studied fractional-order AB derivative in the mathematical analysis that enforces soil moisture in a gradient unconstrained aquifer with an impenetrable base. To solve the fractional-order BsEq, the presented approaches are put to the test. The method creates solutions in a series that quickly converge to the exact solutions. These techniques have a user-friendly and easy-to-understand computation efficiency. The graphs and tables confirm the strong relationship between the exact and analytical results. The convergence phenomenon has also confirmed the reliability of the suggested techniques. Moreover, the LTDM and VITM are very effective and efficient tools for solving a wide range of real-world problems arising in science and engineering.

Author Contributions

Conceptualization, S.A. and R.S.; methodology, R.S.; software, N.A.S.; validation, A.K.; formal analysis, K.N.; investigation, R.S.; resources, N.A.S.; data curation, K.N.; writing—original draft preparation, A.K.; writing—review and editing, N.A.S.; visualization, A.K.; supervision, K.N.; project administration, N.A.S.; funding acquisition, K.N. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. Solution graph of Example 1: (a) exact solution, (b) analytical solution at ρ = 2 , (c) analytical solution at various fractional orders of ρ , (d) ϕ = 0.5 .
Figure 1. Solution graph of Example 1: (a) exact solution, (b) analytical solution at ρ = 2 , (c) analytical solution at various fractional orders of ρ , (d) ϕ = 0.5 .
Symmetry 14 02417 g001
Figure 2. Solution graph of Example 2: (a) exact solution, (b) analytical solution at ρ = 2 .
Figure 2. Solution graph of Example 2: (a) exact solution, (b) analytical solution at ρ = 2 .
Symmetry 14 02417 g002
Figure 3. Solution graph of Example 3: (a) exact solution, (b) analytical solution at ρ = 2 , (c) analytical solution at various fractional orders of ρ , (d) ϕ = 0.5 .
Figure 3. Solution graph of Example 3: (a) exact solution, (b) analytical solution at ρ = 2 , (c) analytical solution at various fractional orders of ρ , (d) ϕ = 0.5 .
Symmetry 14 02417 g003
Table 1. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional orders of Example 1.
Table 1. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional orders of Example 1.
= 0.01 Exact SolutionProposed TechniquesProposed TechniquesProposed TechniquesProposed Techniques
ψ ρ = 2 ρ = 2 ρ = 2 ρ = 1.9 ρ = 1.8
01.0040080110000001.0040080110000000.000000000 × 10 + 0 4.1959900000 × 10 4 8.4706360000 × 10 3
0.11.1096004550000001.1096004550000000.000000000 × 10 + 0 4.6372900000 × 10 4 9.3615010000 × 10 3
0.21.2262981530000001.2262981540000001.000000000 × 10 9 5.1250000000 × 10 4 1.0346059000 × 10 2
0.31.3552690560000001.3552690570000001.000000000 × 10 9 5.6640000000 × 10 4 1.1434164000 × 10 2
0.41.4978039470000001.4978039480000001.000000000 × 10 9 6.2596900000 × 10 4 1.2636705000 × 10 2
0.51.6553293630000001.6553293640000001.000000000 × 10 9 6.9180300000 × 10 4 1.3965719000 × 10 2
0.61.8294218720000001.8294218720000000.000000000 × 10 + 0 7.6455900000 × 10 4 1.5434505000 × 10 2
0.72.0218238500000002.0218238500000000.000000000 × 10 + 0 8.4496900000 × 10 4 1.7057766000 × 10 2
0.82.2344609200000002.2344609210000001.000000000 × 10 9 9.3383500000 × 10 4 1.8851748000 × 10 2
0.92.4694612270000002.4694612270000000.000000000 × 10 + 0 1.0320470000 × 10 3 2.0834403000 × 10 2
1.02.7291767310000002.7291767310000000.000000000 × 10 + 0 1.1405890000 × 10 3 2.3025576000 × 10 2
Table 2. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional-orders of Example 2.
Table 2. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional-orders of Example 2.
= 0.01 Exact SolutionProposed TechniquesProposed TechniquesProposed TechniquesProposed Techniques
ψ ρ = 2 ρ = 2 ρ = 2 ρ = 1.9 ρ = 1.8
01.0011006050000001.0011006050000000.0000000000 × 10 + 0 1.1470000000 × 10 4 2.3021230000 × 10 3
0.11.1063872750000001.1063872750000000.0000000000 × 10 + 0 1.2676300000 × 10 4 2.5442390000 × 10 3
0.21.2227470400000001.2227470400000000.0000000000 × 10 + 0 1.4009500000 × 10 4 2.8118190000 × 10 3
0.31.3513444690000001.3513444690000000.0000000000 × 10 + 0 1.5482900000 × 10 4 3.1075410000 × 10 3
0.41.4934666080000001.4934666080000000.0000000000 × 10 + 0 1.7111200000 × 10 4 3.4343640000 × 10 3
0.51.6505358620000001.6505358620000000.0000000000 × 10 + 0 1.8910800000 × 10 4 3.7955590000 × 10 3
0.61.8241242340000001.8241242330000001.0000000000 × 10 9 2.0899600000E × 10 4 4.1947410000 × 10 3
0.72.0159690540000002.0159690530000001.0000000000 × 10 9 2.3097700000 × 10 4 4.6359060000 × 10 3
0.82.2279903700000002.2279903690000001.0000000000 × 10 9 2.5526900000 × 10 4 5.1234680000 × 10 3
0.92.4623101630000002.4623101620000001.0000000000 × 10 9 2.8211600000 × 10 4 5.6623080000 × 10 3
1.02.7212735840000002.7212735830000001.0000000000 × 10 9 3.1178500000 × 10 4 6.2578180000 × 10 3
Table 3. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional orders of Example 3.
Table 3. Exact solution, proposed techniques solution, and absolute error (AE) at different fractional orders of Example 3.
= 0.01 Exact SolutionProposed TechniquesProposed TechniquesProposed TechniquesProposed Techniques
ψ ρ = 2 ρ = 2 ρ = 2 ρ = 1.9 ρ = 1.8
01.0100501670000001.0100501670000000.0000000000 × 10 + 0 1.0620730000 × 10 3 2.1703277000 × 10 2
0.11.1162780700000001.1162780700000000.0000000000 × 10 + 0 1.1737720000 × 10 3 2.3985831000 × 10 2
0.21.2336780600000001.2336780600000000.0000000000 × 10 + 0 1.2972190000 × 10 3 2.6508442000 × 10 2
0.31.3634251140000001.3634251140000000.0000000000 × 10 + 0 1.4336490000 × 10 3 2.9296360000 × 10 2
0.41.5068177850000001.5068177850000000.0000000000 × 10 + 0 1.5844270000 × 10 3 3.2377485000 × 10 2
0.51.6652911950000001.6652911950000000.0000000000 × 10 + 0 1.7510620000 × 10 3 3.5782655000 × 10 2
0.61.8404313990000001.8404313980000001.0000000000 × 10 9 1.9352220000 × 10 3 3.9545948000 × 10 2
0.72.0339912590000002.0339912580000001.0000000000 × 10 9 2.1387510000 × 10 3 4.3705032000 × 10 3
0.82.2479079870000002.2479079860000001.0000000000 × 10 9 2.3636860000 × 10 3 4.8301530000 × 10 2
0.92.4843225330000002.4843225330000000.0000000000 × 10 + 0 2.6122780000 × 10 3 5.3381448000 × 10 2
1.02.7456010150000002.7456010140000001.0000000000 × 10 9 2.8870130000 × 10 3 5.8995623000 × 10 2
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Alyobi, S.; Shah, R.; Khan, A.; Shah, N.A.; Nonlaopon, K. Fractional Analysis of Nonlinear Boussinesq Equation under Atangana–Baleanu–Caputo Operator. Symmetry 2022, 14, 2417. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14112417

AMA Style

Alyobi S, Shah R, Khan A, Shah NA, Nonlaopon K. Fractional Analysis of Nonlinear Boussinesq Equation under Atangana–Baleanu–Caputo Operator. Symmetry. 2022; 14(11):2417. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14112417

Chicago/Turabian Style

Alyobi, Sultan, Rasool Shah, Adnan Khan, Nehad Ali Shah, and Kamsing Nonlaopon. 2022. "Fractional Analysis of Nonlinear Boussinesq Equation under Atangana–Baleanu–Caputo Operator" Symmetry 14, no. 11: 2417. https://0-doi-org.brum.beds.ac.uk/10.3390/sym14112417

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