Abstract

In this paper, we present a novel numerical approach for solving nonlinear problems with a singular kernel. We prove the existence and uniqueness of the solution for these models as well as the uniform convergence of the function sequence produced by our novel approach to the unique solution. Additionally, we offer a closed form and prove these results for a specific class of these problems where the free term is a fractional polynomial, an exponential, or a trigonometric function. These findings are new to the best of our knowledge. To demonstrate the effectiveness of our numerical method and how to apply our theoretical findings, we solved a number of physical problems. Comparisons with various researchers are reported. Findings demonstrate that our approach is more effective and accurate. In addition, compared to methods that address this type of problems, our approach is simple to implement and has lower computing costs.

1 Introduction

Fractional calculus (FC) is a branch of mathematics that deals with derivatives and integrals of noninteger order. Although the foundations of FC can be found in the work of mathematicians like Liouville et al., it has recently attracted new attention and found use in a number of other areas, including physics, engineering, finance, biology, and signal processing, among others. Many phenomena, including anomalous diffusion, viscoelastic materials, fractals, and fractional-order control systems, have been modeled using it [13].

There are several different types of fractional derivatives (FDs) that have been proposed and studied in the field of fractional calculus. Some of the commonly used types of fractional derivatives include Riemann–Liouville derivative, Caputo derivative (CD), and Grünwald–Letnikov derivative [46]. In this article, we use CD with a singular kernel since it is a powerful tool in fractional calculus that allows for modeling and analyzing systems with memory-dependent behavior. Its unique properties and applications make it a valuable tool in the problem under consideration.

Several scientists strive for numerical techniques to tackle fractional initial value problems (FIVP) because finding the exact solution is difficult and occasionally impossible. It can be difficult to create a suitable numerical method that is efficient. For specific types of problems, direct generalization to the numerical methods for the integer case will work, but not in the general case. It is obvious why that is. The traditional methods are made for local derivative and do not take function memory into account. But nevertheless, for the fractional case, the memory effects make it crucial to think in fractions rather than assuming the common methods will work always. The numerical techniques currently in use fall into two main categories: The first class is an analytical class, including HPM [7] and residual method [8]. The second class is a numerical class, including the Adams Bashforth method [9] and spline methods [10].

In our paper, we consider the following FIVP:
Dξϕ(x)=Ξ(x,ϕ(x)),ϕ(0)=γ,0<x1,0<ξ1
(1.1)

The above problem is covered by numerous researchers. In Ref. [9], the Adams Bashforth method (ABM) is used to solve this problem based on the predictor corrector idea. Their accuracy does not exceed 105 in most of the cases and usually this technique needs a huge calculations. For this reason, we observe that they sometimes take N =1000. Also, In Ref. [11], a special case of this problem is solved when Ξ(x,ϕ(x)) is a quadratic polynomial of ϕ. The variational iteration method (VIM) is used. The problem of this type of methods is when x is far from the initial condition, the approximation will be inaccurate. This is clear from the graphs provided in his paper. This type of method depends on the initial condition so either they give the exact solution or the solution will not be accurate when x is far from the starting point. In Ref. [7], the HPM is used for the same special case. This approach is also depending on the initial condition and a similar conclusion can be drawn. In Ref. [10], spline method (SM) is used. Generalization of the classical SM is used to fit with the fractional case. Their absolute error in this technique is 103 or less. In Ref. [12], they consider only the linear case of Problem (1.1) while in Ref. [13], they use the operational matrix method to solve the case when Ξ(x,ϕ(x)) is a quadratic polynomial of ϕ.

In order to address the FIVP (1.1), we develop a novel numerical technique in this paper. We take into account the memory effects on the problem at hand and iteratively produce the coefficients of an approximate solution. When compared to other methods like ABM or SM, the computational cost is extremely low. We first prove the existence and uniqueness of Problem (1.1), and then we prove that the series of functions produced by our new strategy uniformly converges to this solution. We also think about the scenario in which Ξ(x,ϕ(x))=αϕ(x)+r(x). When r(x) is a fractional polynomial, a fractional exponential function, or a fractional trigonometric function, we find the closed form of the solution. Additionally, by using the integral form, we are able to determine the solution for any function of r(x). To demonstrate the effectiveness of the new numerical method, we also offer a number of examples and comparisons with work by other researchers. Additionally, in order to make our conclusions more apparent and applicable, we apply our theoretical results to a number of examples.

Following is how the remainder of the paper is organized. We provide the fundamental definitions and theorems that are required for this paper in Sec. 2. Describe a thorough study for the scenario where Ξ(x,ϕ(x))=αϕ(x)+r(x) in Sec. 3. We state and prove the results in all instances of r(x). We construct our new numerical technique in Sec. 4. We have shown the theorem's existence, uniqueness, and uniformly convergent properties. In Sec. 5, a number of illustrations and comparisons with other researchers are provided. Finally, we conclude this article with a few closing remarks limitations for the proposed method.

2 Preliminaries

In this section, we define some concepts and we state some results that we use in this paper.

Definition 2.1. Letξ(0,1)andν>0. Then, the CD ofβ(ν)is given by [13]
Dξβ(ν)=1Γ(1ξ)0ν(να)ξβ(α)dα
and the fractional integral (FI) operator is given by
Iξβ(ν)=1Γ(ξ)0ν(να)ξ1β(α)dα

The following are some properties of the CD and the FI which are givenLemmas 2.1–2.3.

Lemma 2.1. Ifξ1,ξ2,b>0, andr(x)Lq(0,b),1q, then
Iξ1Iξ2r(x)=Iξ1+ξ2r(x)

almost everywhere in[0,b]. Ifξ1+ξ2>1, then relation is true for allx[0,b].

Lemma 2.2. Ifξ>0, then
Iξxμ=Γ(μ+1)Γ(μ+ξ+1)xμ+ξ,μ0
and
Dξxμ={0,μ<ξ,μ{0,1,2,}Γ(μ+1)Γ(μξ+1)xμξotherwise}
Lemma 2.3. Ifξ>0, andr(x)C[0,b], then
IξDξr(x)=r(x)r(0)
and
DξIξr(x)=r(x)

In the next three definitions, we define some important functions which we use in this paper.

Definition 2.2. Letξ1,ξ2>0, then we define Mittag–Leffler of one and two parameters functions by
Eξ1(x)=l=0xlΓ(lξ1+1)
and
Eξ1,ξ2(x)=l=0xlΓ(lξ1+ξ2)

respectively.

Definition 2.3. Letξ>0andn{0,1,2,}, then we define the ξ-polynomial of degree n as
Pξ,n(x)=a0+a1xξ+a2x2ξ++anxnξ
In addition, we define the ξ-sine and ξ-cosine functions as
sinξ(x)=Eξ(ix)Eξ(ix)2i
and
cosξ(x)=Eξ(ix)+Eξ(ix)2
Definition 2.4. Letxl=lΔ,l{0,1,2,,L1},Δ=1/L, andL, then the l block pulse function (BPF) is given by [13]
πl(x)={1,xlx<xl+10,otherwise,0l<L

We end this section by the completeness property.

Lemma 2.4. IfβL2[0,1), then [14]
β(x)=j=0βjπj(x)
(2.1)
where
βj=1ΔjΔ(j+1)Δβ(x)dx
(2.2)

3 Exact Solutions for a Class of Fractional Initial Value Problems

In this section, we study the solution of the following problem:
Dξϕ(x)+αϕ(x)=r(x),x>0,0<ξ<1
(3.1)
with
ϕ(0)=γ
(3.2)

First, we study the case when r(x)=0. The result is given in the following theorem.

Theorem 3.1. For0<ξ<1andα, the solution of the following IVP:
Dξϕ(x)+αϕ(x)=0,ϕ(0)=γ,x>0
(3.3)
is
ϕ(x)=γEξ(αxξ)
Proof. By taking the fractional integral for both sides and using Lemma 2.3, we get
ϕ(x)ϕ(0)+αIξϕ(x)=0
which can be written as
ϕ(x)=γαΓ(ξ)0x(xs)ξ1ϕ(s)ds
Using the method of successive approximation with
ϕk+1(x)=γαΓ(ξ)0x(xs)ξ1ϕk(s)ds,ϕ0(x)=γ,k=0,1,2,
we get
ϕ1(x)=γαΓ(ξ)0x(xs)ξ1ϕ0(s)ds=γαγxξξΓ(ξ)=γαγxξΓ(ξ+1)
since Γ(ξ+1)=ξΓ(ξ). Similarly,
ϕ2(x)=γαΓ(ξ)0x(xs)ξ1ϕ1(s)ds=γαΓ(ξ)0x(xs)ξ1(γαγsξΓ(ξ+1))ds=γαγxξΓ(ξ+1)+α2γx2ξΓ(2ξ+1)=γk=02(α)kxkξΓ(kξ+1)
If we continue in this process, we find the nth term
ϕn(x)=γk=0n(α)kxkξΓ(kξ+1)
If we take the limit as n approaches to infinity, we find the solution of Problem (3.3), which is given as
ϕ(x)=limnϕn(x)=γk=0(α)kxkξΓ(kξ+1)=γEξ(αxξ)

Now, we will use the same technique to prove the general case when rL(0,b) where b >0.

Theorem 3.2. For0<ξ<1,α, andrL(0,b),b>0, then the solution of the following IVP:
Dξϕ(x)+αϕ(x)=r(x),ϕ(0)=γ,x>0
(3.4)
is
ϕ(x)=γEξ(αxξ)+0x(xs)ξ1Eξ,ξ(α(xs)ξ)r(s)ds
Proof. By taking the fractional integral for both sides and using Lemma 2.3, we get
ϕ(x)ϕ(0)+αIξϕ(x)=Iξr(x)
which can be written as
ϕ(x)=γ+1Γ(ξ)0x(xs)ξ1r(s)dsαΓ(ξ)0x(xs)ξ1ϕ(s)ds
Using the method of successive approximation, we have
ϕk+1(x)=γ+1Γ(ξ)0x(xs)ξ1r(s)dsαΓ(ξ)0x(xs)ξ1ϕk(s)ds=γ+Iξr(x)αIξϕk(x),k=0,1,2,
where
ϕ0(x)=γ+Iξr(x)
Using Lemmas 2.1 and 2.2
ϕ1(x)=γ+Iξr(x)αIξϕ0(x)=γ+Iξr(x)αIξ(γ+Iξr(x))=γαIξ(γ)+Iξr(x)αIξ(Iξr(x))=γγαxξΓ(ξ+1)+Iξr(x)αI2ξr(x)
Similarly,
ϕ2(x)=γ+Iξr(x)αIξ(γγαxξΓ(ξ+1)+Iξr(x)αI2ξr(x))=γαIξ(γ)+αIξ(γαxξΓ(ξ+1))+Iξr(x)αI2ξr(x)+α2I3ξr(x)=γγαxξΓ(ξ+1)+γα2x2ξΓ(2ξ+1)+Iξr(x)αI2ξr(x)+α2I3ξr(x)=γk=02(α)kxkξΓ(kξ+1)+k=02(α)kI(k+1)ξr(x)=γk=02(α)kxkξΓ(kξ+1)+0xk=02(α)kΓ((k+1)ξ)(xs)(k+1)ξ1g(s)ds=γk=02(α)kxkξΓ(kξ+1)+0x(xs)ξ1(k=02(α)k(xs)kξΓ(kξ+ξ))g(s)ds
If we continue in this process, we find the nth term is
ϕn(x)=γk=0n(α)kxkξΓ(kξ+1)+0x(xs)ξ1(k=0n(α)k(xs)kξΓ(kξ+ξ))g(s)ds
If we take the limit as n approaches to infinity, we find the solution of Problem (3.4) which is given by
ϕ(x)=limnϕn(x)=γEξ(αxξ)+0x(xs)ξ1Eξ,ξ(α(xs)ξ)r(s)ds

Theorem 3.3. Let0<ξ<1,α, andPξ,n=k=0nakxkξbe a ξ-polynomial of degreen0, then the solution of the following IVP:
Dξϕ(x)+αϕ(x)=Pξ,n(x),ϕ(0)=γ,x>0
(3.5)
is
ϕ(x)=γEξ(αxξ)+k=0nakΓ(kξ+1)x(k+1)ξEξ,ξ(k+1)+1(αxξ)
Proof. For any k{0,1,2,,n}, let s=tx to have
0x(xs)ξ1(xs)mξskξds=01(xtx)ξ1(xtx)mξtkξxkξxdt=x(m+k+1)ξ01tkξ(1t)(m+1)ξ1dt=x(m+k+1)ξB(kξ+1,(m+1)ξ)
(3.6)
where B is beta function. Theorem 3.2 implies that
ϕ(x)=γEξ(αxξ)+0x(xs)ξ1Eξ,ξ(α(xs)ξ)Pξ,n(s)ds=γEξ(αxξ)+k=0nakm=0(α)mΓ(mξ+ξ)0x(xs)ξ1(xs)mξskξds=γEξ(αxξ)+k=0nakm=0(α)mΓ(mξ+ξ)x(m+k+1)ξB(kξ+1,(m+1)ξ)=γEξ(αxξ)+k=0nakΓ(kξ+1)x(k+1)ξm=0(α)mxmξΓ(mξ+(k+1)ξ+1)=γEξ(αxξ)+k=0nakΓ(kξ+1)x(k+1)ξEξ,(k+1)ξ+1(αxξ)

Theorem 3.4. Let0<ξ<1, α,β be constants such thatβ0andα+β0. Then the solution of the following IVP:
Dξϕ(x)+αϕ(x)=Eξ(βxξ),ϕ(0)=γ,x>0
(3.7)
is
ϕ(x)=γEξ(αxξ)+βα+βxξEξ,ξ+1(βxξ)+αα+βxξEξ,ξ+1(αxξ)
Proof. The Cauchy product of two infinite series yields
Eξ,ξ(α(xs)ξ)Eξ(βsξ)=(i=0(α)i(xs)iξΓ(iξ+ξ))(j=0βjsjξΓ(jξ+1))=k=0(l=0k(α)lβ(kl)(xs)lξs(kl)ξΓ(lξ+ξ)Γ((kl)ξ+1))
Using formula (3.6), the particular solution is
ϕp(x)=k=0(l=0k(α)lβkl0x(xs)ξ1(xs)lξs(kl)ξdsΓ(lξ+ξ)Γ((kl)ξ+1))=k=0(l=0k(α)lβklxkξ+ξB((kl)ξ+1,lξ+ξ)Γ(lξ+ξ)Γ((kl)ξ+1))=k=0(l=0k(α)lβklxkξ+ξΓ((k+1)ξ+1))=k=0βkxkξ+ξΓ((k+1)ξ+1)l=0k(αβ)l=k=0βkxkξ+ξΓ((k+1)ξ+1)1(αβ)k+11+αβ=βα+βxξ(k=0βkxkξΓ(1+kξ+ξ)+αβk=0(α)kxkξΓ(1+kξ+ξ))=βα+βxξEξ,ξ+1(βxξ)+αα+βxξEξ,ξ+1(αxξ)
Therefore, the solution of problem (3.7) is
ϕ(x)=γEξ(αxξ)+βα+βxξEξ,ξ+1(βxξ)+αα+βxξEξ,ξ+1(αxξ)

Theorem 3.5. Let0<ξ<1, α,β be constants such thatβ0andα+β0. Then the solution of the following IVP:
Dξϕ(x)+αϕ(x)=sinξ(βzξ),ϕ(0)=γ,x>0
(3.8)
is
ϕ(x)=γEξ(αxξ)+1α2+β2(αsinξ(βxξ)+β(1cosξ(βxξ))αβxξEξ,ξ+1(αxξ))
Proof. Since
sinξ(βxξ)=Eξ(iβxξ)Eξ(iβξ)2i
using Theorem 3.4 and superposition principle, the particular solution is
ϕp(x)=xξ2i(iβα+iβEξ,ξ+1(iβxξ)+αα+iβEξ,ξ+1(αxξ))xξ2i(iβαiβEξ,ξ+1(iβxξ)+ααiβEξ,ξ+1(αxξ))=iβxξ2i(α2+β2)((αiβ)Eξ,ξ+1(iβxξ)+(α+iβ)Eξ,ξ+1(iβxξ))+αxξEξ,ξ+1(αxξ)2i(α2+β2)((αiβ)(α+iβ))=αβxξ2(α2+β2)(Eξ,ξ+1(iβxξ)+Eξ,ξ+1(iβxξ))iβ2xξ2(α2+β2)(Eξ,ξ+1(iβxξ)Eξ,ξ+1(iβxξ))αβα2+β2xξEξ,ξ+1(αxξ)
When x >0, then
Eξ,ξ+1(λxξ)=k=0λkxkξΓ(ξk+ξ+1)=l=1λ(l1)x(l1)ξΓ(lξ+1)=1λxξ(Eξ(λxξ)1)
Also
limx01λxξ(Eξ(λxξ)1)=1Γ(ξ+1)
which is the value of Eξ,ξ+1(βxξ) at x =0. Then
ϕp(x)=α2i(α2+β2)(Eξ(iβxξ)Eξ(iβxξ))β2(α2+β2)(Eξ(iβxξ)+Eξ(iβxξ)2)αβα2+β2xξEξ,ξ+1(αxξ)
Using Definition 2.3, we have
ϕp(x)=1α2+β2(αsinξ(βxξ)+β(1cosξ(βxξ))αβxξEξ,ξ+1(αxξ))
and the solution to Problem (3.9) is
ϕ(x)=γEξ(αxξ)+1α2+β2(αsinξ(βxξ)+β(1cosξ(βxξ))αβxξEξ,ξ+1(αxξ))

Theorem 3.6. Let0<ξ<1, α,β be constants such thatβ0andα+β0. Then the solution of the following IVP:
Dξϕ(x)+αϕ(x)=cosξ(βzξ),ϕ(0)=γ,x>0
(3.9)
is
ϕ(x)=γEξ(αxξ)+1α2+β2(βsinξ(βxξ)+α(cosξ(βxξ)1)+α2xξEξ,ξ+1(αxξ))

Proof. The proof is similar to the proof of Theorem 3.5.

We end this section by the following remark.

Remark 3.1. We should note that if r(t) is a linear combination of tkξ,Eξ(βxξ),sinξ(βxξ), or cosξ(βxξ), then the solution of Problem (3.1) is
ϕ(x)=γEξ(αxξ)+ϕp(x)

where ϕp(x) can be computed using the superposition principle and Theorems 3.1–3.6.

4 A Numerical Scheme for the Nonlinear Fractional Initial Value Problems

In the following, we study the nonlinear problem:
Dξϕ(x)=Ξ(x,ϕ(x)),ϕ(0)=γ,0<x1,0<ξ1
(4.1)
Let us define the space H=C((0,1),) with the norm
||q(x)||(0,1)=supx(0,1)|g(x)|

Then, H is a Banach space. For simplicity, we denote this norm by ||.||. First, we prove the existence and uniqueness of the solution of Problem (4.1).

Theorem 4.1. LetΞ(x,ϕ(x))Hbe a function that satisfy the Lipschitz condition with respect toϕwith Lipschitz constant θ, then the FIVP(4.1)has a unique solution if
θΓ(ξ+1)<1
(4.2)
Proof. Using Lemma 2.2, we can rewrite Eq. (4.1) as
ϕ(x)=γ+1Γ(ξ)0x(xs)ξ1Ξ(s,ϕ(s))ds
(4.3)
Define the operator ϒ as
ϒ(ϕ)=γ+1Γ(ξ)0x(xs)ξ1Ξ(s,ϕ(s))ds
Then
|ϒ(ϕ1)ϒ(ϕ2)|=1Γ(ξ)|0x(xs)ξ1(Ξ(s,ϕ1(s))Ξ(s,ϕ2(s)))ds|1Γ(ξ)0x(xs)ξ1|Ξ(s,ϕ1(s))Ξ(s,ϕ2(s)))|ds1Γ(ξ)0x(xs)ξ1ds||Ξ(x,ϕ1(x))Ξ(x,ϕ2(x)))||θxξξΓ(ξ)||ϕ1(x)ϕ2(x)||θΓ(ξ+1)||ϕ1(x)ϕ2(x)||

Since θΓ(ξ+1)<1, then ϒ is contraction on H. Hence, Banach fixed point theorem will guarantee the existence and uniqueness of Problem (4.1).

To find the unique solution to Problem (4.1), Eq. (4.3) proposed the following sequence of functions:
ϕ0(x)=γ
(4.4)
ϕk(x)=1Γ(ξ)0x(xs)ξ1Ξ(s,ϕk1(s))ds,k=1,2,
(4.5)
which is based on the method of successive approximation. However, modification for it is necessary to reduce the computational cost. A numerical method will be derived to solve
Dξϕ(x)=Ξ(x,ϕ(x)),ϕ(0)=γ,x>0
(4.6)
where ΞH. By taking the fractional integral, Lemma 2.3 gives
ϕ(x)=ϕ(0)+Iξ(Ξ(x,ϕ(x)))=γ+1Γ(ξ)0x(xs)ξ1Ξ(s,ϕ(s))ds
(4.7)
Approximate ϕ in terms of the BPFs as
ϕL(x)=l=0L1ϕlπl(x)
(4.8)
Substitute Eq. (4.8) in Eq. (4.7) to get
l=0L1ϕlπl(x)=γ+1Γ(ξ)0x(xs)ξ1Ξ(s,l=0L1ϕlπl(s))ds
(4.9)
Collocate Eq. (4.9) at xj,1j<L, to get
ϕj=γ+1Γ(ξ)0xj(xjs)ξ1Ξ(s,l=0L1ϕlπl(s))ds=γ+1Γ(ξ)k=1jxk1xk(xjs)ξ1Ξ(s,l=0L1ϕlπl(s))ds
(4.10)
since
πl(xk)={1,l=k0,lk
Using Definition 2.4, Eq. (4.10) yields
ϕj=γ+1Γ(ξ)k=1jxk1xk(xjs)ξ1Ξ(s,ϕk1)ds
(4.11)
Note that γ=ϕ(0)=ϕ0. Then, Eq. (4.11) becomes
ϕj=γ+1Γ(ξ)k=1jTj,k
(4.12)
where
Tj,k=xk1xk(xjs)ξ1Ξ(s,ϕk1)ds
(4.13)

Since ϕj depends on the previous values of ϕk,k=0,1,,j1, then we can find the solution in direct way without solving nonlinear algebraic system. The solution will have cheap computational cost. We can summarize our numerical approach in the following Algorithm.

The next theorem, we prove the convergence of the proposed solution in Eqs. (4.8) and (4.11).

Theorem 4.2. LetΞ(x,ϕ(x))Hbe a function that satisfy the Lipschitz condition with respect toϕwith Lipschitz constant θ, then the sequence{ϕL(x)}L=0converges uniformly to the unique solution of the FIVP(4.1).

Proof. Assume that ϕ(x) is the exact solution of Problem (4.1). Then, it satisfies the integral Eq. (4.3)
Φ(x)=γ+1Γ(ξ)0x(xs)ξ1Ξ(s,ϕ(s))ds
(4.14)
Also, using Lemma 2.4, we have
Φ(x)=k=0λkπk(x),λ0=γ
(4.15)
For any 0j<L1, then
Φ(xj)=k=0λkπk(xj)=λj
Then
λj=Φ(xj)=γ+1Γ(ξ)0xj(xjs)ξ1Ξ(s,ϕ(s))ds
(4.16)
Subtract Eq. (4.10) from Eq. (4.16) to have
|λjϕj|=1Γ(ξ)0xj(xjs)ξ1|Ξ(s,ϕ(s))Ξ(s,ϕL(s))|dsθ||ϕϕL||Γ(ξ)0xj(xjs)ξ1dsxjθ||ϕϕL||ξΓ(ξ)θ||ϕϕL||Γ(ξ+1)
(4.17)
Then
|Φ(x)ϕL(x)|=|k=0L1(λjϕj)π(x)+k=Lλkπk(x)|k=0L1|λjϕj|+|k=Lλkπk(x)|
From Eq. (4.17), we get
|Φ(x)ϕL(x)|k=0L1θ||ϕϕL||Γ(ξ+1)+|k=Lλkπk(x)|Lθ||ϕϕL||Γ(ξ+1)+|k=Lλkπk(x)|
(4.18)
Since the series in Eq. (4.15) converges uniformly, then for ϵ = 1, there exists N such that
|k=Lλkπk(x)|1,LN
(4.19)
Hence, Inequalities (4.18) and (4.19) imply that
|Φ(x)ϕL(x)|Lθ||ϕϕL||Γ(ξ+1)+1,LN
(4.20)
Take the supreme on [0,1] for both sides of inequality (4.20) to get
||ΦϕL||Lθ||ϕϕL||Γ(ξ+1)+1,LN
which gives
limL||ΦϕL||limLΓ(ξ+1)Γ(ξ+1)Lθ=0

Therefore, the sequence {ϕL(x)}L=0 converges uniformly to the unique solution of the FIVP (4.1).

5 Illustrative Examples

In the following, we solve several examples to show the efficiency of the proposed method. Also, compassion with other methods will be presented.

Example 5.1. Consider the following nonlinear Riccati equation [11]:
Dξϕ(x)=1ϕ2(x),ϕ(0)=0
Let us approximate ϕ(x) as in Eq. (4.8)
ϕL(x)=j=0L1ϕjπj(x)
where ϕl is given by formula (4.11) as
ϕj=γ+1Γ(ξ)k=1jxk1xk(xjs)ξ1Ξ(s,ϕk1)ds
Since Ξ(x,ϕ(x))=1ϕ2(x) and γ = 0, then
ϕj=1Γ(ξ)k=1jxk1xk(xjs)ξ1(1ϕk12)ds=ΔξΓ(ξ+1)k=1j(1ϕk12)((jk+1)ξ(jk)ξ)
As a special case, if we take ξ = 1, then
ϕj=Δk=1j(1ϕk12)
Direct calculations imply that
ϕ0=0,ϕ1=Δ,ϕ2=2ΔΔ3,ϕ3=3Δ5Δ3+4Δ5Δ7,.

The exact solution when ξ = 1 is ϕe(x)=e2x1e2x+1. In Fig. 1, we plot the graph of the exact and approximate solutions when ξ = 1. Now, we compare between the absolute error in our results with Adams Bashforth method of order 2 (AM2) [9], variational iteration method (VIM) [11], and modified homotopy perturbation method (MHPM) [7] in Table 1. In Fig. 2, we plot the graph of approximate solutions when ξ=0.75,0.85,0.90,0.95,1, and the exact solution when ξ = 1.

Example 5.2. Consider the following FIVP [10]:
Dξϕ(x)=(1ϕ(x))4,ϕ(0)=0
Let us approximate ϕ(x) as in Eq. (4.8)
ϕL(x)=j=0L1ϕjπj(x)
where ϕl is given by formula (4.11) as
ϕj=1Γ(ξ)k=1jxk1xk(xjs)ξ1(1ϕk1)4ds=ΔξΓ(ξ+1)k=1j(1ϕk1)4((jk+1)ξ(jk)ξ)
As a special case, if we take ξ = 1, then
ϕj=Δk=1j(1ϕk1)4=ϕj1+Δ(1ϕj1)4
Direct calculations imply that
ϕ0=0,ϕ1=Δ,ϕ2=2Δ4Δ2+6Δ34Δ4+Δ5,.

The exact solution when ξ = 1 is ϕe(x)=111+3x3. In Fig. 3, we plot the graph of the exact and approximate solutions when ξ = 1. Now, we compare between the absolute error in our results with Adams Bashforth method of order 2 (AM2) [9], spline method (SM) [10] in Table 2. In Fig. 4, we plot the graph of approximate solutions when ξ=0.75,0.85,0.90,0.95,1, and the exact solution when ξ = 1.

In the next example, we will implement our numerical technique to system of two equations to show the possibility of generalize it to system of fractional equations.

Example 5.3. Consider the following system of equation [9,12]:
Dξϕ(x)=ϕ(x)+ψ(x),ϕ(0)=0Dξψ(x)=ϕ(x)+ψ(x),ψ(0)=1
Let us approximate ϕ(x) and ψ(x) as in Eq. (4.8)
ϕL(x)=j=0L1ϕjπj(x),ψL(x)=j=0L1ψjπj(x)
where ϕj and ψj are given by formula (4.11) as
ϕj=1Γ(ξ)k=1j(ϕk1+ψk1)xk1xk(xjs)ξ1dsψj=1+1Γ(ξ)k=1j(ψk1ϕk1)xk1xk(xjs)ξ1ds
Perform the integration in the last two equations to get
ϕj=ΔξΓ(ξ+1)k=1j(ϕk1+ψk1)((jk+1)ξ(jk)ξ)ψj=1+ΔξΓ(ξ+1)k=1j(ψk1ϕk1)((jk+1)ξ(jk)ξ)
As a special case, if we take ξ = 1, then
ϕj=Δk=1j(ϕk1+ψk1)=ϕj1+Δ(ϕj1+ψj1)ψj=1+Δk=1j(ψk1ϕk1)=ψj1+Δ(ψj1ϕj1)
Direct calculations implies that
ϕ0=0,ϕ1=Δ,ϕ2=2Δ+2Δ2,ϕ3=3Δ+6Δ2+4Δ3,ψ0=1,ψ1=1+Δ,ψ2=1+2Δ+2Δ2,ϕ3=1+3Δ+6Δ2+4Δ3,.
The exact solutions when ξ = 1 are ϕe(x)=exsin(x) and ψe(x)=excos(x). In Fig. 5, we plot the graph of the exact and approximate solutions when ξ = 1. In Figs. 6 and 7, we plot the graph of approximate solutions of ϕ and ψ, respectively, when ξ=0.75,0.85,0.90,0.95,1, and the exact solution when ξ = 1. Let us define the maximum error as follows:
error=max{||(Dξϕ(xi)ϕ(xi)ψ(xj)Dξψ(xi)+ϕ(xi)ψ(xj))||:xj=j1m,j=1,2,,11}

Then, we compare the error in our approach with Refs. [9] and [12] and results are reported in Table 3.

Example 5.4. Consider the following FIVP [10]:
D12ϕ(x)+ϕ(x)=r(x),ϕ(0)=0,x>0
(5.1)
where
r(x)=x4x323x52Γ(72)+24x72Γ(92)
is 12-polynomial of degree eight, α = 1, and γ = 0. Theorem 3.3 yields to
a5=3Γ(72),a6=12,a7=24Γ(92),a8=1,ak=0,k=0,1,,4
and
ϕ(x)=3Γ(72)Γ(72)x3E12,4(x12)+12Γ(4)x72E12,92(x12)+24Γ(92)Γ(92)x4E12,5(x12)+Γ(5)x92E12,112(x12)=3x3E12,4(x12)+3x72E12,92(x12)+24x4E12,5(x12)+24x92E12,112(x12)=3x3(k=0(1)kxk2Γ(12k+4)k=0(1)kxk+12Γ(12(k+1)+4))+24x4(k=0(1)kxk2Γ(12k+5)k=0(1)kxk+12Γ(12(k+1)+5))=3x3(k=0(1)kxk2Γ(12k+4)l=1(1)l1xl2Γ(12l+4))+24x4(k=0(1)kxk2Γ(12k+5)l=1(1)l1xl2Γ(12(l)+5))
Cancel the similar terms to get
ϕ(x)=3x3Γ(4)+24x4Γ(5)=x32+x4
It is easy to verify that ϕ(x)=x32+x4 is a solution to the IVP (5.1). In Ref. [10], the authors solved the same Problem and their absolute error is given in Table 4.
Example 5.5. Consider the following FIVP:
Dξϕ(x)+3ϕ(x)=r(x),ϕ(0)=1,x>0
(5.2)

In the following, we study FIVP (5.2) for different choices of r(x). Case 1

If r(x)=Eξ(2xξ), then using Theorem 3.4, we have
ϕ(x)=Eξ(3xξ)+25xξEξ,ξ+1(2xξ)+35xξEξ,ξ+1(3xξ)
Then
limξ1ϕ(x)=limξ1(Eξ(3xξ)+25xξEξ,ξ+1(2xξ)+35xξEξ,ξ+1(3xξ))=e3x+25(e2x12)35(e3x13)=e2x+4e3x5
since
limξ1xξEξ,ξ+1(μz)=xE1,2(μz)=k=0μkzk+1Γ(k+2)=n=1μn1znn!=eμz1μ
One can easily verify that limξ1ϕ(x) is a solution to the IVP (5.2) when ξ approaches to one which is
ϕ(x)+3ϕ(x)=e2x,ϕ(0)=1,x>0

Case 2

If r(x)=sinξ(2xξ), then using Theorem 3.5, we have
ϕ(x)=Eξ(3xξ)+113(3sinξ(2xξ)+2(1cosξ(2xξ))6xξEξ,ξ+1(3xξ))
Then
limξ1ϕ(x)=limξ1(Eξ(3xξ)+113(3sinξ(2xξ)+2(1cosξ(2xξ))6xξEξ,ξ+1(3xξ)))=e3x+113(3sin(2x)+2(1cos(2x))+6e3x13)=1513e3x+313sin(2x)213cos(2x)
since
limξ1sinξ(μxξ)=sin(μx),limξ1cosξ(μxξ)=cos(μx)
One can easily verify that limξ1ϕ(x) is a solution to the IVP (5.2) when ξ approaches to one which is
ϕ(x)+3ϕ(x)=sin(2x),ϕ(0)=1,x>0

6 Conclusions and Limitations

In this paper, we present a novel numerical approach for solving nonlinear problems with a singular kernel. We prove the existence and the uniqueness of the solution for these models as well as the uniform convergence of the function sequence produced by our novel approach to the unique solution. Additionally, we offer a closed form and prove these results for a specific class of these problems where the free term is a fractional polynomial, a fractional exponential, or a fractional trigonometric function. These findings are new to the best of our knowledge. To demonstrate the effectiveness of our numerical method and how to apply our theoretical findings, we solved a number of examples. Comparisons with various researchers are reported. We end this article by the following remarks:

  1. We test our new numerical technique by three examples. Since the exact solution is unknown for any 0<ξ<1, we plot the exact and approximate solutions for ξ = 1 in Figs. 1 and 3. We notice that they are coincide for ξ = 1. This gives us a numerical evidence that the new approach is working efficiently.

  2. We compare our results with the results in Refs. [7] and [911] using AM2, VIM, MHPM, and SM methods. These results are reported in Tables 1 and 2 and they show that our method has more accuracy and less computational cost.

  3. We sketch the graph of the approximate solutions for ξ=0.75,0.85.0.95,1, and the exact solution at ξ = 1. We notice that the approximate solutions converges to the exact solution at ξ = 1 when ξ approaches to one. This fact is clear from Figs. 2 and 4.

  4. We generalize the idea of the numerical method into a system of two equations. This is just to show that the method can be generalized and will be suitable for future work and investigation. We compare our error with the errors in Refs. [10] and [12] as in Table 3. Results show that our method has more accuracy and less computational cost. In addition, we sketch the graphs of ϕ and ψ for several values of ξ. We notice that the approximate solution converges to the exact solution at ξ = 1 when ξ approaches to one. Also, the approximate and exact solutions are coincide when ξ = 1. This fact is clear from Figs. 57.

  5. In Example 5.4, we compare between our results with Refs. [10] and [12]. We use Theorem 3.3 to find the exact solution while the max error in Ref. [10] is of order 106 and in Ref. [12] is of order 103 when ξ=0.3,0.7 and of order 1014 when ξ=0.5. These results are reported in Table 4.

  6. We implement Theorems 3.4 and 3.5 in the last example. We find the exact solution for two problems mentioned in this example.

  7. As we see from the previous discussion that we can find the exact solution for a class of FIVP using Theorems 3.1–3.6. Also, our numerical technique is accurate and can be generalize to other problems such as fractional partial equations and system of FIVP.

6.1 Limitations of the Proposed Method.

The proposed method, particularly in the context of solving differential and integral equations, is a powerful mathematical tool. However, like any method, it has its limitations, and there are situations where it may not be the most effective or suitable approach. Here are some limitations and situations where the operational matrix method might face challenges or fail:

  1. The method may become less effective when dealing with highly complex or nonlinear problems, as the operational matrix method is often more suitable for problems that are not highly nonlinear systems.

  2. Certain boundary conditions may not be easily incorporated into the operational matrix framework, leading to difficulties in solving problems with nonstandard or complex boundary conditions. We attempted to use the linear shooting method, but sometimes it is not possible.

  3. The method might struggle when the solution contains discontinuities or singularities, as it relies on approximating functions using basis functions that may not capture such behavior accurately.

  4. Additionally, the proposed method may also struggle when the nonlinear term is not Lipschitz continuous or when the minimum value of the Lipschitz constant is very large. This will affect either the speed of convergence or the existence of the numerical solution.

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