
Citation: Foday Conteh, Shota Tobaru, Mohamed E. Lotfy, Atsushi Yona, Tomonobu Senjyu. An effective Load shedding technique for micro-grids using artificial neural network and adaptive neuro-fuzzy inference system[J]. AIMS Energy, 2017, 5(5): 814-837. doi: 10.3934/energy.2017.5.814
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Korteweg and de-Vries developed the classical KdV equation in 1895 as a nonlinear PDE to investigate the waves that occurs on the surfaces of shallow water. Many studies have been conducted on this exactly solvable model. Many scholars have proposed novel applications of the classical KdV equation, such as acoustic waves that produces in a plasma in ion form, and acoustic waves which are produces on a crystal lattice. In [1], the classical KdV equation is as follows
Ut+β1Uxxx+β2UUx=0. | (1.1) |
Different variations of Eq (1.1) have been published in the literature, including [2,3]. In literature [2], various efficient methods have been used to get solitons solutions of different kinds of KdV equations. Wang [3] used a quadratic function Ansatz to obtain lump solutions for the (2+1)-dimensional KdV equation. In [4,5,6], several systematic techniques were utiltized to investigate the different types of KdV equations. Wang and Kara introduced a new 2D-mKdV in 2019 [7]. The new (2+1)-dimensional mKdV equation is given by
ft=6f2fx−6f2fy+fxxx−fyyy−3fxxy+3fxyy. | (1.2) |
After 1695, the non-integer order or fractional-order derivative (FOD) was described as a simple academic generalization of the classical derivative. A FOD is an operator that extends the order of differentiation from Natural numbers (N) to a set of real numbers (R) or even to a set of complex numbers (C). Fractional calculus has emerged as one of the most effective methods for describing long-memory processes over the last decade. Engineers and physicists, as well as pure mathematicians, are interested in such models. The models that are represented by differential equations with fractional-order derivatives are the most interesting [8,9,10]. Their evolutions are much more complicated than in the classical integer-order case, and deciphering the underlying principle is a difficult task. There are many fractional operators regarding to the kernels involved in the integration. The most popular fractional operator is Caputo-Liouville which is based on power-law kernel, but this kernel has issue about the singularity of the kernel. To tackle the limitation of derivative operators with power-law kernels, new types of nonlocal FOD have been introduced in recent literature. For example, in literature [11], Caputo and Fabrizio (CF) introduced a FOD that is focused on the exponential kernel. However, the CF derivative, on the other hand, has some issues with the kernel's locality. In 2016, Atangana and Baleanu constructed an updated version of FOD that is based on the Mittag-Leffler function [12]. This derivative solves the issues of locality and singularity. The FOD of Atangana-Baleanu in Caputo sense (ABC) accurately describes the memory. The ABC operator's most significant applications are available in [13,14,15,16,17].
Certain important techniques have been used to solve fractional order differential equations (FODEs). Some of these includes the homotopy perturbation method (HPM), Laplace Adomian decomposition method (LADM) fractional operational matrix method (FOMM), homotopy analysis method (HAM) and many more [18,19,20,21,22]. In contrast to these techniques, the Laplace-Adomian decomposition method (LADM) is an important tool for solving non linear FODEs. The ADM and the Laplace transformation are two essential methods that are combined in LADM. Furthermore, unlike a Runge-Kutta process, LADM does not require a predefine size declaration. Every numerical or analytical approach has its own set of benefits and drawbacks. For instance, discretization of data is used in collocation techniques that require extra memory and a longer operation. Since the Laplace-Adomian approach has less parameters than all other methods, it is a useful tool that does not necessitate discretion or linearization [23]. With the aid of LADM, the smoke model was successfully solved in [24]. LADM was utilized by the authors to solve third order dispersive PDE defined by FOD in [22].
Inspired by above literature, in this paper, we study Eq (1.2) under ABC-operator. We use LADM to solve the proposed equation. Consider Eq (1.2) under ABC-operator as
ABCDΨtft=6f2fx−6f2fy+fxxx−fyyy−3fxxy+3fxyy. | (1.3) |
Definition 2.1. [12] Let 0<Ψ≤1 and f(t)∈H1. Then ABC FOD of order Ψ is expressed as
ABCDΨtf(t)=c(Ψ)(1−Ψ)∫t0f′(η)EΨ(−(t−η)ΨΨ(1−Ψ))dη, |
where c(Ψ) is the normalization function such that c(0)=c(1)=1. The symbol EΨ denotes the Mittag-Leffler kernel which is defined as:
EΨ(t)=∞∑k=0tkΓ(Ψk+1). |
Definition 2.2. [12] Let 0<Ψ≤1 and f(t)∈H1(0,T). Then AB fractional integral of order Ψ is defined as
ABIΨtf(t)=(1−Ψ)c(Ψ)f(t)+(Ψ)c(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1f′(η)dη. |
Definition 2.3. [12] The formula for the Laplace transform of ABC FOD of f(t) is defined by
L[ABCDΨ0f(t)]=c(Ψ)sΨ(1−Ψ)+Ψ[sΨL[f(t)]−sΨ−1u(0)]. |
Theorem 2.4. [25] Let H be a Banach space and X:H→Hbe a mapping. Then X is said to be Picard's X-stable, if ∀ξ,m∈H,
‖Xξ−Xm‖≤a‖ξ−Xξ‖+b‖ξ−m‖, |
where a≥0, and b∈[0,1]. Further, X has a fixedpoint.
Theorem 3.1. [12] The ABC derivative holds the following Lipschitztype condition for 0<ϝ<∞.
‖ABCDΨtf(t)−ABCDΨtg(t)‖≤ϝ‖f(t)−g(t)‖. |
Proof. By using definition of ABC, wehave
‖ABCDΨtf(t)−ABCDΨtg(t)‖=‖c(Ψ)(1−Ψ)t∫0f′(η)EΨ(−(t−η)ΨΨ(1−Ψ))dη−c(Ψ)(1−Ψ)t∫0g′(η)EΨ(−(t−η)ΨΨ(1−Ψ))dη‖=‖c(Ψ)(1−Ψ)[t∫0f′(η)EΨ(−(t−η)ΨΨ(1−Ψ))dη−t∫0g′(η)EΨ(−(t−η)ΨΨ(1−Ψ))dη]‖. |
Using the Lipschitz condition for the first order derivative, we can find a small positive constant ρ1 such that
‖ABCDΨtf(t)−ABCDΨtg(t)‖=‖c(Ψ)ρ1(1−Ψ)EΨ(−ΨtΨ(1−Ψ))[t∫0f′(η)dη−t∫0g′(η)dη]‖≤c(Ψ)ρ1(1−Ψ)EΨ(−ΨtΨ(1−Ψ))‖t∫0f′(η)dη−t∫0g′(η)dη‖≤ϝ‖t∫0f′(η)dη−t∫0g′(η)dη‖≤ϝ‖f(t)−g(t)‖, |
where ϝ=c(Ψ)ρ1(1−Ψ)EΨ(−ΨtΨ(1−Ψ)). Thus Lipschitz condition holds for ABC derivative.
Let
Φ(x,y,t;f)=6f2fx−6f2fy+fxxx−fyyy−3fxxy+3fxyy, | (4.1) |
Eq (1.3) can be written as
ABCDΨtf(x,y,t)=Φ(x,y,t;f). | (4.2) |
Applying the ABC FOI to Eq (4.2), we have
f(x,y,t)−f(x,y,0)=(1−Ψ)c(Ψ)Φ(x,y,t;f)+(Ψ)c(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1Φ(x,y,t;f)dη. |
First we have to verify that the Lipschitz condition holds for the kernel Φ(x,y,t;f). For this, let us take two bounded functions, f and g, i.e., ‖f‖≤Δ1, and ‖g‖≤Δ2 where Δ1,Δ2>0, and consider
‖Φ(x,y,t;f)−Φ(x,y,t;g)‖={‖(6f2fx−6g2gx)−(6f2fy−6g2gy)+(fxxx−gxxx)−(fyyy−gyyy)−(3fxxy−3gxxy)+(3fxyy−3gxyy)‖={‖2∂∂x(f3−g3)−2∂∂y(f3−g3)+∂3∂x3(f−g)−∂3∂y3(f−g)−3∂3∂y∂x2(f−g)+3∂3∂y2∂x(f−g)‖≤{2‖∂∂x(f3−g3)‖+2‖∂∂y(f3−g3)‖+‖∂3∂x3(f−g)‖+‖∂3∂y3(f−g)‖+3‖∂3∂y∂x2(f−g)‖+3‖∂3∂y2∂x(f−g)‖. |
Since f and g are bounded functions. Therefore, their partial derivatives satisfy the Lipschitz conditions and there exists non-negative constants K,L,M,N,O,P such that
‖Φ(x,y,t;f)−Φ(x,y,t;g)‖≤{2K‖(f3−g3)‖+2L‖(f3−g3)‖+M‖(f−g)‖+N‖(f−g)‖+3O‖(f−g)‖+3P‖(f−g)‖≤{(2K+2L)(f2+fg+g2)‖(f−g)‖+M‖(f−g)‖+N‖(f−g)‖+3O‖(f−g)‖+3P‖(f−g)‖=((2K+2L)(Δ21+Δ1Δ2+Δ22)+M+N+3O+3P)‖(f−g)‖=κ‖(f−g)‖. |
Let
κ=((2K+2L)(Δ21+Δ1Δ2+Δ22)+M+N+3O+3P), |
thus
‖Φ(x,y,t;f)−Φ(x,y,t;g)‖≤κ‖f−g‖. |
For further analysis, we make an iterative scheme as
fξ+1(x,y,t)=(1−Ψ)c(Ψ)Φ(x,y,t;fξ)+(Ψ)c(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1Φ(x,y,η;fξ)dη, |
where f0(x,y,t)=f(x,y,0). Now the difference between two consecutive terms can be taken as
eξ(x,y,t)=fξ(x,y,t)−fξ−1(x,y,t)=(1−Ψ)c(Ψ)[Φ(x,y,t;fξ−1)−Φ(x,y,t;fξ−2)]+(Ψ)c(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1[Φ(x,y,η;fξ−1)−Φ(x,y,η;fξ−2)]dη. |
Also, we have
fξ(x,y,t)=ξ∑k=0ek(x,y,t), | (4.3) |
with f−1=0.
Theorem 4.1. Assume that f(x,y,t) is bounded a function. Then
‖eξ(x,y,t)‖≤((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))ξ‖f(x,y,0)‖. | (4.4) |
Proof. Consider
eξ(x,y,t)=fξ(x,y,t)−fξ−1(x,y,t). | (4.5) |
The Eq (4.5) gets the form under norm as
‖eξ(x,y,t)‖=‖fξ(x,y,t)−fξ−1(x,y,t)‖. |
To get the required result, we use the concept of mathematical induction. For ξ=1, one can get
‖e1(x,y,t)‖=‖f1(x,y,t)−f0(x,y,t)‖≤(1−Ψ)c(Ψ)‖Φ(x,y,t;f0)−Φ(x,y,t;f−1)‖+Ψc(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1‖Φ(x,y,η;f0)−Φ(x,y,η;f−1)‖dη≤(1−Ψ)c(Ψ)κ‖f0−f−1‖+Ψc(Ψ)Γ(Ψ)κ∫t0(t−η)Ψ−1‖f0−f−1‖dη=(1−Ψ)c(Ψ)κ‖f(x,y,0)‖+Ψc(Ψ)Γ(Ψ)κ‖f(x,y,0)‖∫t0(t−η)Ψ−1dη=(1−Ψ)c(Ψ)κ‖f(x,y,0)‖+Ψc(Ψ)Γ(Ψ)κ‖f(x,y,0)‖t=((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))‖f(x,y,0)‖. |
Now assume that the result is true for ξ−1, i.e.,
‖eξ−1(x,y,t)‖≤((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))ξ−1‖f(x,y,0)‖. | (4.6) |
Next, we have to show that
‖eξ(x,y,t)‖≤((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))ξ‖f(x,y,0)‖. | (4.7) |
To get the result (4.4), consider
‖eξ(x,y,t)‖=‖fξ(x,y,t)−fξ−1(x,y,t)‖≤(1−Ψ)c(Ψ)‖Φ(x,y,t;fξ−1)−Φ(x,y,t;fξ−2)‖+Ψc(Ψ)Γ(Ψ)∫t0(t−η)Ψ−1‖Φ(x,y,η;fξ−1)−Φ(x,y,η;fξ−2)‖dη≤(1−Ψ)c(Ψ)κ‖fξ−1−fξ−2‖+Ψc(Ψ)Γ(Ψ)κ∫t0(t−η)Ψ−1‖fξ−1−fξ−2‖dη=(1−Ψ)c(Ψ)κ‖eξ−1‖+κtΨ−1c(Ψ)Γ(Ψ)‖eξ−1‖=((1−Ψ)c(Ψ)κ+κtΨ−1c(Ψ)Γ(Ψ))‖eξ−1‖=((1−Ψ)c(Ψ)κ+κtΨ−1c(Ψ)Γ(Ψ))((1−Ψ)c(Ψ)κ+κtΨ−1c(Ψ)Γ(Ψ))ξ−1‖f(x,y,0)‖=((1−Ψ)c(Ψ)κ+κtΨ−1c(Ψ)Γ(Ψ))ξ‖f(x,y,0)‖. |
This ends the proof.
Theorem 4.2. If the following relation holds at t=t0≥0, where
0≤((1−Ψ)c(Ψ)κ+κtΨ−10c(Ψ)Γ(Ψ))<1. | (4.8) |
Then at least one solution of the new 2D KdV equation under the ABCfractional derivative exists.
Proof. With the help of Eq (4.3), we have
‖fξ(x,y,0)‖≤ξ∑k=0‖ek(x,y,t)‖≤ξ∑k=0(((1−Ψ)c(Ψ)κ+κtΨ−1c(Ψ)Γ(Ψ))k‖f(x,y,0)‖), |
for t=t0, we get
‖fξ(x,y,0)‖≤‖f(x,y,0)‖ξ∑k=0((1−Ψ)c(Ψ)κ+κtΨ−10c(Ψ)Γ(Ψ))k. |
From the above relation, we can say that
limξ→∞‖fξ(x,y,0)‖≤‖f(x,y,0)‖limξ→∞ξ∑k=0((1−Ψ)c(Ψ)κ+κtΨ−10c(Ψ)Γ(Ψ))k. |
Since
0≤((1−Ψ)c(Ψ)κ+κtΨ−10c(Ψ)Γ(Ψ))<1, | (4.9) |
this implies that sequence fξ(x,y,t) is convergent and therefore the sequence is bounded for each ξ. Further, assume that
Rξ(x,y,t)=f(x,y,t)−fξ(x,y,t). |
Since fξ(x,y,t) is bounded. It follows that for λ>0, we have ‖fξ(x,y,t)‖≤λ. After simple manipulation like we did in Theorems 4.1 and 4.2, we obtain
‖Rξ(x,y,t)‖≤((1−Ψ)c(Ψ)κ+κtΨ−10c(Ψ)Γ(Ψ))ξ+1λ. |
Using Eq (4.9), one can get
limξ→∞‖Rξ(x,y,t)‖=0, |
it follows that limξ→∞fξ(x,y,t)=f(x,y,t). This finish the proof.
Theorem 4.3. If the inequality (4.8) holds at t=t0≥0.Then the unique solution of proposed equation exists.
Proof. On contrary, suppose that there are two solutions f and g of the proposed equation such that f≠g. Now
f(x,y,t)−g(x,y,t)=(1−Ψ)c(Ψ)[Φ(x,y,t;f)−Φ(x,y,t;g)]+Ψc(Ψ)Γ(Ψ)×[∫t0(t−η)Ψ−1[Φ(x,y,η;f)−Φ(x,y,η;g)]dη]. |
Taking norm both side
‖f(x,y,t)−g(x,y,t)‖≤(1−Ψ)c(Ψ)‖Φ(x,y,t;f)−Φ(x,y,t;g)‖+Ψc(Ψ)Γ(Ψ)×[∫t0(t−η)Ψ−1‖Φ(x,y,η;f)−Φ(x,y,η;g)‖dη]≤(1−Ψ)c(Ψ)κ‖f−g‖+Ψc(Ψ)Γ(Ψ)∫t0κ(t−η)Ψ−1‖f−g‖dη≤((1−Ψ)c(Ψ)κ+Ψc(Ψ)Γ(Ψ)κ)‖f−g‖∫t0(t−η)Ψ−1dη=((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))‖f−g‖, |
but
0≤((1−Ψ)c(Ψ)κ+κtΨc(Ψ)Γ(Ψ))<1. |
Using the above inequality, we achieve
‖f(x,y,t)−g(x,y,t)‖=0, |
thus, our supposition is wrong. Hence, the solution is unique.
In this section, we briefly discuss the solution of the model by applying Ansatz method. For this purpose, we will consider a test function as
f(x,y,t)=β0+β1sech(b1x+b2y+b3t). | (5.1) |
By putting above equation into classical form of the model, we obtain
6β20a1−6β20b2+b31−3b21b2+3b22b1−b32−b3=0,12β0β1b1−12β0b1b2=0,6β21b1−6β21b2−6b31+18b21b2−18b22b1+6b32=0, |
solution becomes as
β0=0,β1=∓b1±b2,b3=b31−3b21b2+3b22b1−b32, |
solution of classical model becomes as
f1,2(x,y,t)=(∓b1±b2)sech(b1x+b2y+(b31−3b21b2+3b22b1−b32)t). |
For b1=1 and b2=−1, the above solution becomes
f1,2(x,y,t)=∓4exp(x−y+8t)1+exp(2x−2y+16t). | (5.2) |
In this section, to obtain analytic solution we applying Laplace transform (LT) on both sides of equations f(x,y,t) is the source term. Subject to the initial condition f(x,y,0)=f0(x,y,0). On utilizing LT, one can get
L[ABCDΨtf]=L[6f2fx−6f2fy+fxxx−fyyy−3fxxy+3fxyy], |
L[f(x,y,t)]=f(x,y,0)s+[s(1−Ψ)+Ψc(Ψ)L[6f2fx−6f2fy]+fxxx−fyyy−3fxxy+3fxyy]. | (6.1) |
The approximate solution is represented by
f(x,y,t)=∞∑ξ=0fξ(x,y,t), | (6.2) |
and the nonlinear term is represented by Adomain polynomials, i.e., G(f)=f2=∑∞ξ=0Aξ, where Aξ is defined as follows for any ξ=0,1,2,⋯
Aξ=1Γ(ξ+1)dξdλξ[G(∞∑ξ=0(λξfξ))]λ=0. |
Using Eq (6.2), we obtain
L[∞∑ξ=0fξ(x,y,t)]=f(x,y,0)s+s(1−Ψ)+Ψc(Ψ)L[6∞∑ξ=0Aξ∞∑ξ=0∂∂xfξ−6∞∑ξ=0Aξ∞∑ξ=0∂∂yfξ+∞∑ξ=0(∂3∂x3fξ−∂3∂y3fξ−3∂3∂x2∂yfξ+3∂3∂y∂x2fξ)]. |
The following can be obtain by comparing terms
L[f0(x,y,t)]=f(x,y,0)s,L[f1(x,y,t)]=sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6A0f0x−6A0f0y+f0xxx+f0yyy−3f0xxy+3f0xyy],L[f2(x,y,t)]=sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6A1f1x−6A1f1y+f1xxx+f1yyy−3f1xxy+3f1xyy],⋮L[fξ+1(x,y,t)]=sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6Aξfξx−6Aξfξy+fξxxx+fξyyy−3fξxxy+3fξxyy]. |
Applying L−1, we get
f0(x,y,t)=L−1[f(x,y,0)s],f1(x,y,t)=L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6A0f0x−6A0f0y+f0xxx+f0yyy−3f0xxy+3f0xyy]]f2(x,y,t)=L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6A1f1x−6A1f1y+f1xxx+f1yyy−3f1xxy+3f1xyy]]⋮fξ+1(x,y,t)=L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6Aξfξx−6Aξfξy+fξxxx+fξyyy−3fξxxy+3fξxyy]]. |
The required series solution is given as
f(x,y,t)=∞∑ξ=0fξ(x,y,t). | (6.3) |
Here, we present two special cases of the proposed equation.
For the first case, we take the initial condition as
f(x,y,0)=−4exp(x+y)1+exp(2(x−y)). |
Using the detail procedure as discussed above, we achieve
f0(x,y,t)=f(x,y,0)=−4exp(x+y)1+exp(2(x−y)),f1(x,y,t)=L−1[sΨ(1−Ψ)+ΨsΨc(Ψ)L[6f20∂∂xf0−6f20∂∂yf0+∂3∂x3f0−∂3∂y3f0−3∂2∂x2(∂∂yf0)+3∂∂x(∂2∂y2f0)]], |
using Mathematica, we obtain
f1(x,y,t)=(1−Ψ+ΨtΨΓ(Ψ+1))(32texp(x+y)(−exp(2x)+exp(2y))(exp(2x)+exp(2y))2c(Ψ)), |
similarly, other terms can be calculated with the help of Mathematica. The required series solution is given as
{f(x,y,t)=−4exp(x+y)1+exp(2(x−y))+32t(1−Ψ+ΨtΨΓ(Ψ+1))×(exp(x+y)(−exp(2x)+exp(2y))(exp(2x)+exp(2y))2c(Ψ))+… | (6.4) |
Remark 6.1. When we put Ψ=1 in Eq (6.4), the solution rapidly converges to the exact classical solution, i.e.,
f(x,y,t)=−4exp(x−y+8t)1+exp(2(x−y+8t)). | (6.5) |
For the first case, we take the initial condition as
f(x,y,0)=4exp(x+y)1+exp(2(x−y)). |
Using the detail procedure as discussed above, we get
f0(x,y,t)=f(x,y,0)=4exp(x+y)1+exp(2(x−y)),f1(x,y,t)=L−1[sΨ(1−Ψ)+ΨsΨc(Ψ)L[6f20∂∂xf0−6f20∂∂yf0+∂3∂x3f0−∂3∂y3f0−3∂2∂x2(∂∂yf0)+3∂∂x(∂2∂y2f0)]], |
using Mathematica, we obtain
f1(x,y,t)=(1−Ψ+ΨtΨΓ(Ψ+1))(−32texp(x+y)(−exp(2x)+exp(2y))(exp(2x)+exp(2y))2c(Ψ)), |
similarly, other terms can be calculated with the help of Mathematica. The required series solution is given as
{f(x,y,t)=4exp(x+y)1+exp(2(x−y))−32t(1−Ψ+ΨtΨΓ(Ψ+1))×(exp(x+y)(−exp(2x)+exp(2y))(exp(2x)+exp(2y))2c(Ψ))+… | (6.6) |
Remark 6.2. When we put Ψ=1 in Eq (6.6), the solution rapidly converges to the exact classical solution, i.e.,
f(x,y,t)=4exp(x−y+8t)1+exp(2(x−y+8t)). | (6.7) |
Here, we derive some results regarding to the convergence and stability of the proposed scheme with the help of functional analysis. The convergence of the proposed scheme of the is presented in the following theorem.
Theorem 7.1. Let H be a Banach space and T:H→Hbe an operator. Suppose that f be the exact solution of the proposedequation. If ∃ϖ such that 0≤ϖ<1 and ‖fξ+1‖≤ϖ‖fξ‖,∀ξ∈N∪{0}, thenthe approximate solution ∑∞ξ=0fξ convergesto the exact solution f.
Proof. We construct a series as
S0=f0,S1=f0+f1,S2=f0+f1+f2,⋮Sξ=f0+f1+⋯fξ. |
We want to prove that the sequence {Sξ}∞ξ=0 is a Cauchy sequence in H. Let us consider
‖Sξ+1−Sξ‖=‖fξ‖≤ϖ‖fξ‖≤ϖ2‖fξ−1‖≤ϖ3‖fξ−2‖⋮≤ϖξ+1‖f0‖. |
Now for every ξ,m∈N, we have
‖Sξ−Sm‖=‖(Sξ−Sξ−1)+(Sξ−1−Sξ−2)+⋯+(Sm+1−Sm)‖≤‖Sξ−Sξ−1‖+‖Sξ−1−Sξ−2‖+⋯+‖Sm+1−Sm‖≤ϖξ‖f0‖+ϖξ−1‖f0‖+⋯+ϖm+1‖f0‖≤(ϖξ+1+ϖξ+2+⋯)‖f0‖=ϖξ+11−ϖ‖f0‖. |
Now, limξ,m→∞‖Sξ−Sm‖=0. This implies that {Sξ}∞ξ=0 is Cauchy sequence in H. So ∃f∈H such that limξ→∞Sξ=f. This ends the proof.
Next, we present the Picard's X-stability of the proposed scheme in the following theorem.
Theorem 7.2. Let X be a self-mapping which is definedas
X(fξ(x,t))=fξ+1(x,t)=fξ(x,t)+L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[6f2ξfξx−6f2ξfξy+fξxxx+fξyyy−3fξxxy+3fξxyy]]. | (7.1) |
The iteration is X-stable in L1(a,b), ifthe condition
‖(2K+2L)(Φ21+Φ1Φ2+Φ2)ℵ1+Mℵ2−Nℵ3+3Oℵ4+3Pℵ5‖<1, | (7.2) |
is satisfied.
Proof. With the help of Banach contraction theorem, first we show that the mapping X possesses a unique fixed point. For this, assume that the bounded iterations for (ξ,m)∈N×N. Let Φ1,Φ2>0 such that ‖fξ‖≤Φ1, and ‖fm‖≤Φ2. Consider
X(fξ(x,t))−X(fm(x,t))=fξ(x,t)−fm(x,t)+L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[2f2ξfξx−2f2ξfξy+fξxxx+fξyyy−3fξxxy+3fξxyy]]−L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[2f2mfmx−2f2mfmy+fmxxx+fmyyy−3fmxxy+3fmxyy]]=fξ(x,t)−fm(x,t)+L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[2K(f3ξ−f3m)−6L(f3ξ−f3m)+M(fξ−fm)−N(fξ−fm)−3O(fξ−fm)+3P(fξ−fm)]]. |
Now, using triangle inequality, we have
‖X(fξ(x,t))−X(fm(x,t))‖=‖fξ(x,t)−fm(x,t)‖+‖L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[2K(f3ξ−f3m)−2L(f3ξ−f3m)+M(fξ−fm)−N(fξ−fm)−3O(fξ−fm)+3P(fξ−fm)]]‖. |
Using boundedness of fξ and fm, we have
‖X(fξ(x,t))−X(fm(x,t))‖≤[(2K+2L)(Φ21+Φ1Φ2+Φ2)ℵ1+Mℵ2−Nℵ3+3Oℵ4+3Pℵ5]‖fξ−fm‖, |
where ℵi,i=1,2,3,4,5, are functions obtained from L−1[sΨ(1−Ψ)+Ψ)sΨc(Ψ)L[∗]]. Using assumption (7.2), the mapping X fulfills the contraction condition. Hence by Banach fixed point result, X has a unique fixed point. Also, the mapping X satisfies the condition of Theorem 2.4 with
a=0,b=(2K+2L)(Φ21+Φ1Φ2+Φ2)ℵ1+Mℵ2−Nℵ3+3Oℵ4+3Pℵ5. |
Thus, the mapping X fulfills all conditions of Picard's X-stable. Hence our proposed scheme is Picard's X-stable.
Thanks to the Mittag-Leffler kernel, which solves the singularity and locality problems with the Caputo and Caputo-Fabrizio FOD kernels. Since ABC-derivative is based on the Mittag-Leffler kernel, it has recently become popular for investigating the dynamics of a mathematical model that governs a physical process. We use ABC-derivative to investigate the soliton solution of the new modified KdV equation in (2+1) dimension in the current paper. Since the presence of a solution is essential for the study of a model, we have deduced some results using fixed point theory that guarantees at least one solution and the unique solution of the proposed equation. There are several techniques for solving FDEs, but among the analytical methods, LADM is the most effective and accurate. Under ABC-derivative, we used the LADM to obtain the solution of the proposed equation. Graphs of the solution are used to observe the method's convergence. The exact solution and the approximate solution obtained with the aid of LADM are in good agreement (See Figures 1 and 2). The dynamics of the solution under the ABC-derivative have been investigated. We can see from Figures 3 and 4 that the fractional-order solution curves are approaching the integer-order curve when fractional-order equals 1. Figure 5 shows that the ABC-derivative solution curve is much closer to the exact solution than the Caputo-Fabrizio derivative curve. As a result, the proposed model is better than Caputo-Fabrizio's. Figures 6–9 denote the graphical representation of solution obtained in the Case B. The considered equation will be studied under more generalized fractional operators in the next paper.
The authors wish to express their sincere thanks to the honorable referees for their valuable comments and suggestions to improve the quality of the paper. In addition, authors would like to express their gratitude to the United Arab Emirates University, Al Ain, UAE for providing the financial support with Grant No. 12S005-UPAR 2020.
In this research, there are no conflicts of interest.
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