Convergence analysis for the fractional decomposition method applied to class of nonlinear fractional Fredholm integro-differential equation

Abstract

For scientists conducting research, fractional integral differential equation analysis is crucial. Therefore, in this study, we investigate analysis utilizing a novel method called the fractional decomposition method, which is applicable to fractional nonlinear fractional Fredholm integro-differential equations. Then, we apply the approach to five test problems for a general fractional derivative $\beta$ involving fractional Fredholm integro-differential equations. To the best of our knowledge, we are the first to ever do so because of the very complicated calculations involved when dealing with the general case $\beta$ . For fractional Fredholm integral-differential equations, we provide both exact and approximate solutions. Throughout this work, the fractional Caputo derivative is discussed. This technique leads us to say that the method is precise, accurate, and efficient, according to the theoretical analysis.

Introduction

Fractional integro-differential equation (FIDE) is used in a variety of scientific fields, including demography, insurance mathematics, and physics.^1–5 Any IVP or BVP that is converted to an IE typically produces an integral-differential equation (IDE), which may be seen in a variety of scientific models.^6,7 Many integro-differential equations contain both integral and differential operators. As a result, we must look for an effective method for locating analytical solutions to fractional differential equations.

Due to their usefulness in several disciplines of science and engineering, FIDEs have attracted the attention of numerous researchers in recent years.^8–17 in particular, fractional Volterra and Fredholm FIDEs.

The research community is aware of many methods to handle FIDE. To name a few, the fractional decomposition method (FDM),^18–22 the fractional adomian decomposition method,^23,24 the interpolation correction for collocation, ⁶ the second-kind Chebyshev wavelet,^7,8 the least squares method and shifted Laguerre polynomials pseudo-spectral method, ¹⁵ the spectral-collocation method, ¹¹ the Taylor expansion method, ¹² the cubic b-spline operational matrix, ¹³ and the ¹³ are recent examples of effective and dependable techniques^9,16 are recent examples of effective and dependable techniques. ¹⁵ Additionally, M. Rawashdeh has produced additional theorems that support utilizing the FNDM to obtain analytical approximations of solutions to fractional nonlinear PDEs. They were the first researchers to integrate both the natural decomposition technique and the adaptive decomposition technique to solve linear and nonlinear ODEs and PDEs in a thesis written by S. Maitama in 2014 and the first author of the current work.

We use the newly developed approaches (FDM) to five linear and nonlinear Fredholm integro-differential equations in order to demonstrate the viability and efficiency of our new approach, which will also demonstrate the ease and simplicity of the current procedure.

First, we explore the linear fractional Fredholm integro-differential equations (FFIDE) with $0<\beta \le 1$ given by:

$$D_t^{\beta }(v(t))=2\sqrt{\frac{t}{\pi }}+\frac{3\sqrt{\pi }t}{4}-\frac{9}{10}+{\int }_0^{1}v\left(\tau \right)d\tau ,$$

(1)

a company with its IC:

$$v(0)=0.$$

(2)

It is known that $v(t)=t+t^{\frac{3}{2}}$ is the exact solution when $\beta =\frac{1}{2}$ of the above equation (1).

Second, we take a look at the linear FFIDE given by:

$$\begin{array}{rl}{D}_t^{\beta }\left(v\left(t\right)\right)=& \frac{3\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)t^{\frac{1}{3}}}{\pi }-\frac{t^2}{5}-\frac{t}{4}\\ & +{\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau ,\\ & 0\le t,\tau \le 1,1<\beta \le 2,\end{array}$$

(3)

a company with its I.C:

$$v^{\prime }(0)=v(0)=0.$$

(4)

Third, we examine the nonlinear FFIDE with $1<\beta \le 2$ given as:

$$\begin{array}{rl}{D}_t^{\beta }\left(v\left(t\right)\right)=& \frac{6t^{\frac{1}{3}}}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}-\frac{t^2}{7}-\frac{t}{4}-\frac{1}{9}\\ & +{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau ,0\le t\le 1,\end{array}$$

(5)

a company with its I.C:

$$v^{\prime }(0)=v(0)=0.$$

(6)

Fourth,we explore the nonlinear FFIDE given as:

$$\begin{array}{r}{D}_t^{\beta }\left(v\left(t\right)\right)=1-\frac{t}{4}+{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau ,\\ 0\le t\le 1,0<\beta \le 1,\end{array}$$

(7)

a company with its I.C:

$$v(0)=0.$$

(8)

Finally, we inspect the nonlinear FFIDE given as:

$$\begin{array}{rl}{D}_t^{\frac{1}{2}}\left(v\left(t\right)\right)=& \frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}(\frac{8t^{\frac{3}{2}}}{3}-2t^{\frac{1}{2}})-\frac{t}{1260}\\ & +{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau ,0\le t\le 1,\end{array}$$

(9)

a company with its I.C:

$$v(0)=0.$$

(10)

The current research work is presented as follows: Definitions and some background information about fractional calculus are presented in the ‘Fractional Calculus Background’ section. ‘The natural-adomian method’ section is devoted to some theories of N-transformation as well as some significant properties of the natural transform. The convergence study of the FDM used with the nonlinear FFIDE is thoroughly examined in the ‘Convergence Analysis of the FDM for nonlinear FFIDE’ section. In the ‘Numerical results and applications’ section, we apply the FDM to five linear and nonlinear FFIDEs. The ‘Concluding Remarks’ section will serve as the conclusion to the current work.

Fractional Calculus Background

We begin by going over some crucial concepts and properties that are helpful whenever someone discusses the topic of fractional calculus.^1–5

Definition 1

If $h(t)\in \mathbb{R}$ , where $t>0$ . Then $h(t)$ is in the space $C_{\mu },\mu \in \mathbb{R}$ if $\mathrm{\exists }p\in \mathbb{R}$ such that $h(t)=t^{p}g(t)$ , where $g(t)$ in $C[0,\mathrm{\infty })$ , and $h(t)\in C_{\mu }^m$ if $h^{(m)}\in C_{\mu },m=1,2,\dots$ .

Definition 2

The Riemann-Liouville fractional integral for a function $g$ of order $\beta \ge 0$ , is defined as:

$$\left\{\begin{array}{c}{J}^{\beta }(g(s))=\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s(s-t{)}^{\beta -1}g(t)dt,\beta >0,s>0\\ J^0(g(s))=g(s)\end{array}\right\}.$$

(11)

Definition 3

The Caputo fractional derivative of $g$ is given by:

$${}^c{D}^{\beta }(g(s))=J^{k-\beta }{D}^k(g(s))=\frac{1}{\mathrm{\Gamma }(k-\beta )}{\int }_0^s(s-t{)}^{k-\beta -1}{g}^{(k)}(t)dt,$$

(12)

for $k-1<\beta \le k,k=1,2,\dots ,s>0,g\in C_{-1}^k$ .

Definition 4

The Gamma function can be defined as:

$$\mathrm{\Gamma }(w)={\int }_0^{\mathrm{\infty }}{e}^{-s}{s}^{w-1}ds,w>0.$$

(13)

Definition 5

Given a complete $(Y,\rho )$ metric space. Then $F:Y\to Y$ is called a contraction mapping on $Y$ if we can find a $C>0$ such that $\rho (F(x),F(y))\le C\rho (x,y)$ , for all $x,y\in Y$ .

Theorem 1

Given complete $Y$ nonempty metric space $\rho$ , and $F:Y\to Y$ is a contraction mapping, then $F$ has a unique fixed-point, such that $F(y^{*})=y^{*}$ .

Theorem 2

Given a non-empty complete metric space $(Y,\rho )$ with a map $F:Y\to Y$ which is of a contraction type. Then the mapping has a unique fixed-point $y^{*}\in Y$ (i.e. $F(y^{*})=y^{*}$ ). Furthermore, $y^{*}$ can be found as follows: start with an arbitrary element $y_0\in Y$ and define a sequence $\{y_n\}$ by $F(y_{n-1})=y_n$ for $n\ge 1$ . Then $y_n\to y^{*}$ .

The natural-adomian method

We recommend that readers learn more about the history of the general integral transform, the Laplace, Sumudu, and natural transform methods, as well as their related properties, for any given function $\zeta (x)$ , $x\in \mathbb{R}$ , see for example.^25–27

Definition 6

Let $\zeta (x)$ be a piece-wise continuous function on $\mathbb{R}$ . If $C_1,C_2,a,b>0$ with $a<b$ , define $A=\{\zeta (x):|\zeta (x)|<C_1{e}^{ax}{\chi }_{(x_2,\mathrm{\infty })}(x)+C_2{e}^{bx}{\chi }_{(-\mathrm{\infty },x_1)}(x)\}.$ So, $|\zeta (x)|\le C_1{e}^{ax}$ for $x⟶\mathrm{\infty }$ i.e. $x>x_2$ and $|\zeta (x)|\le C_2{e}^{bx}$ for $x⟶-\mathrm{\infty }$ i.e. $x<x_1$ .

Note that for any $\zeta (x)$ in the class $A$ with $r,w>0$ we have:

$$\begin{array}{cc}\left|{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-rx}\zeta \left(xw\right)dx\right|& \le C_1{\int }_0^{\mathrm{\infty }}{e}^{-rx}{e}^{a\left|xw\right|}dx+C_2{\int }_{-\mathrm{\infty }}^0{e}^{-rx}{e}^{b\left|xw\right|}dx\\ & =C_1{\int }_{x_2}^{\mathrm{\infty }}{e}^{(aw-r)x}dx+C_2{\int }_{-\mathrm{\infty }}^{x_1}{e}^{(bw-r)x}dx.\end{array}$$

Which is convergent provided that $aw-r<0$ and $bw-r>0$ , which implies that $aw<r<bw$ i.e. $a<\frac{r}{w}<b$ . Loosely speaking, $\zeta (x)$ is a function of exponential order.

Then, one can define the natural transformation (N-transformation) as:

$$\mathrm{\aleph }\left(\zeta \left(x\right)\right)=L(r,w)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-rx}\zeta (wx)dx,r,w>0,$$

(14)

where $\mathrm{\aleph }$ is the N-transform of $\zeta (x)$ and $r$ and $u$ are the N-transform variables. Note that one can write Equation (refeq3.1) as,

$$\mathrm{\aleph }\left(\zeta \left(x\right)\right)={\mathrm{\aleph }}^{+}\left(\zeta \left(x\right)\right)+{\mathrm{\aleph }}^{-}\left(\zeta \left(x\right)\right)=L^{+}(r,w)+L^{-}(r,w),$$

where,

$${\mathrm{\aleph }}^{+}\left(\zeta \left(x\right)\right)=L^{+}(r,w)={\int }_0^{\mathrm{\infty }}{e}^{-rx}\zeta (wx)dx,r,w\in (0,\mathrm{\infty }).$$

(15)

Moreover,

$${\mathrm{\aleph }}^{-1}\left[L(r,w)\right]=\zeta \left(x\right)=\frac{1}{2\pi i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{\frac{rx}{w}}L(r,w)dr.$$

(16)

Thus, equation (15) is the natural transformation and equation (16) is the inverse natural transformation.

We will use the following helpful N-transforms properties throughout this paper, see Abassy ²⁴ , Belgacem ²⁵ and Bulut et al. ²⁶ :

${\mathrm{\aleph }}^{+}[M]=\frac{M}{r}.$

${\mathrm{\aleph }}^{+}[x^{\beta }]=\frac{\mathrm{\Gamma }(\beta +1)w^{\beta }}{r^{\beta +1}},\beta >-1$ .

${\mathrm{\aleph }}^{+}[e^{bx}]=\frac{1}{(r-bw)}.$

Suppose that $k>0$ , where $k-1<\beta \le k$ and $L(r,w)$ is the natural transformation of the function $\zeta (x)$ , then the natural transformation of the fractional derivative in the Caputo sense of the function $\zeta (x)$ of order $\beta$ denoted by ${}^c{D}^{\beta }\zeta (x)$ is given by:

$${\mathrm{\aleph }}^{+}[{}^c{D}^{\beta }\zeta (x)]=\frac{r^{\beta }}{w^{\beta }}L(r,w)-\sum _{n=0}^{k-1}\frac{r^{\beta -(n+1)}}{w^{\beta -n}}{(D^n\zeta (x))}_{x=0}.$$

Convergence analysis of the FDM for nonlinear FFIDE

In this section, we first give proofs for the uniqueness and convergence theorems, and then we give an error estimate using the FDM. Consider the general nonlinear nonhomogeneous FFIDEs:

$$\begin{array}{rl}{D}_t^{\beta }\left(v\left(t\right)\right)& =g\left(t\right)v\left(t\right)+s\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ,\\ & \mathrm{\forall }t\in [0,T],\end{array}$$

(17)

where $0<\beta \le 1$ .

Along with I.C:

$$v\left(0\right)=c$$

(18)

Note that the fractional derivative of the function $v(t)$ is in the sense of Caputo, $F(v(\tau ))$ is a nonlinear continuous function and $s(t)$ is the non-homogeneous term and $|\sigma (\tau ,t)|\le M$ .

Employ the N-transformation and property 4 to equation (17) to find:

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[D_t^{\beta }\left(v\left(t\right)\right)\right]& ={\mathrm{\aleph }}^{+}\left[s\left(t\right)\right]\\ +& {\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].\end{array}$$

(19)

So,

$$\begin{array}{rl}\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& -\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}\\ & =L(r,w)+{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].\end{array}$$

(20)

Thus,

$${\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]=\frac{a{w}^{\beta }}{r^{\beta +1}}+\frac{w^{\beta }}{r^{\beta }}L(r,w)+\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].$$

(21)

Take the inverse of N-transform to find:

$$\begin{array}{r}v(t)=S\left(t\right)+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ]\right].\end{array}$$

(22)

Suppose our solution of $v(t)$ given by:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(23)

Moreover, the nonlinear part is written as:

$$F\left(v\left(\tau \right)\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right).$$

(24)

Note that the Adomian polynomial of $v_0,v_1,\dots ,v_n$ are represented by $A_n$ ’s which can be computed by:

$$A_n=\frac{1}{n!}\frac{d^n}{d{\delta }^n}\left[F\left(\sum _{i=0}^n{\delta }^i{v}_i\right)\right],n=0,1,2,\dots .$$

(25)

Note that we can simplify the formula in equation (25) to be:

$$\begin{array}{c}{A}_0=F\left(v_0\right)\\ A_1=v_1{F}^{\prime }\left(v_1\right)\\ A_2=v_2{F}^{\prime }\left(v_0\right)+\frac{1}{2!}{v}_1^2{F}^{″}\left(v_0\right)\end{array}$$

(26)

We can continue in this manner to get the other polynomials.

Substitute equation (23) into equation (22) to arrive at:

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)& =S\left(t\right)+{\mathrm{\aleph }}^{-1}[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)\sum _{n=0}^{\mathrm{\infty }}{v}_n(t)\\ & +\delta {\int }_a^b\sigma (\tau ,t)\sum _{n=0}^{\mathrm{\infty }}{A}_n(\tau )d\tau ]].\end{array}$$

(27)

From equation (27) to obtain:

$$v_0(t)=S(t).$$

(28)

In general, one can find:

$$\begin{array}{r}{v}_{n+1}\left(t\right)={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v_n\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)A_n\left(\tau \right)d\tau ]\right],n\ge 0.\end{array}$$

(29)

In this case, our solution will be:

$$v\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right).$$

(30)

Again $F(v(t))=\sum _{i=0}^{\mathrm{\infty }}{A}_i$ , where $A_i$ ’s are the Adomian polynomials.

Note that $S(t)$ represents the initial conditions and the non-homogeneous part. We shall use the new form of Adomian polynomials, see Adomian ²³ and Abassy ²⁴ to find:

$A_n=F(s_n)-\sum _{j=0}^{n-1}{A}_j$ , where $s_n=\sum _{i=0}^n{v}_i(t)$ .

Theorem 3

(Uniqueness Theorem). Let $0<\gamma <1$ with $\gamma =\frac{(C_1+\lambda MT{C}_2)}{\mathrm{\Gamma }(\beta +1)}{t}^{\beta },\mathrm{\forall }t\in [0,T]$ . Then equation (17) has a unique solution.

Proof.

Consider the Banach space of all functions on $J=[0,T]$ which are also continuous, say $A=(C[J],\Vert .\Vert )$ having a norm $\Vert .\Vert$ . Let $G:A\to A$ be define by:

$$v_{n+1}\left(t\right)={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v_n\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)A_n\left(\tau \right)d\tau ]\right].$$

Assume $L[g(t)v(t)]=v(t)$ and $F[v(t)]=\sum _{n=0}^{\mathrm{\infty }}{A}_n$ . Further, suppose $|L(v)-L(v^{*})|<C_1|v-v^{*}|$ and $|F(v)-F(v^{*})|<C_2|v-v^{*}|$ , where $C_1,C_2$ are the Lipschitz constants and $v,v^{*}$ are two different solutions of equation (17).

$$\begin{array}{r}\Vert G(v)-G(v^{*})\Vert \\ & =\underset{t\in J}{max}\left|\begin{array}{c}{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}[L(v)+\delta {\int }_a^b\sigma (\tau ,t)F(v)d\tau ]\right]\\ -{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}[L(v^{*})+\delta {\int }_a^b\sigma (\tau ,t)F(v^{*})d\tau )]\right]\end{array}\right|\\ & =\underset{t\in J}{max}\left|\begin{array}{c}{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}[L(v)-L(v^{*})]\right]\\ +{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[\delta {\int }_a^b\sigma (\tau ,t)(F(v)-F(v^{*}))d\tau \right]\right]\end{array}\right|\\ & \le \underset{t\in J}{max}[C_1{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[|v-v^{*}|\right]\right]\\ & +\left(\delta M{C}_{2}T\right){\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[|v-v^{*}|\right]\right]]\\ & \le \underset{t\in J}{max}(C_1+\delta MT{C}_2)\left[{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[|v-v^{*}|\right]\right]\right]\\ & \le (C_1+\delta MT{C}_2)\left[{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[\Vert v(t)-v^{*}(t)\Vert \right]\right]\right]\\ & =\Vert v(t)-v^{*}(t)\Vert \frac{(C_1+\delta MT{C}_2)}{\mathrm{\Gamma }(\beta +1)}{t}^{\beta }.\end{array}$$

Since $0<\gamma <1$ , then $G$ is contraction mapping and it follows by Theorem (2), then there exists unique solution to equation (17).  □

Theorem 4

(Theorem of Convergence). The series solution in equation (30) of equation (17) using the FDM converges provided that $0<\gamma <1$ and $|v_1|<\mathrm{\infty }$ .

Proof.

Consider the $s_k$ to be the m-th partial sum, i.e. $s_k=\sum _{i=0}^k{v}_i(t)$ . We need to show that $\{s_k\}$ is a Cauchy sequence in the Banach space $A$ . Consider the new formulation of the Adomian polynomials mentioned: $F(s_k)=A_k^{*}+\sum _{i=0}^{k-1}{A}_i^{*}$ . Let $s_n$ and $s_k$ be any two partial sums with $k\ge n$ . Then,

$$\begin{array}{rl}\Vert s_k-s_n\Vert & =\underset{t\in J}{max}|s_k-s_n|\\ & =\underset{t\in J}{max}\left|{\displaystyle \sum _{i=n+1}^k{v}_i^{*}(t)}\right|,k=1,2,\dots \\ & \le \underset{t\in J}{max}|{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[{\displaystyle \sum _{i=n+1}^{k}L(v_{i-1}(t))}\right]\right]+{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[{\displaystyle \sum _{i=n+1}^k\delta {\int }_a^b\sigma (\tau ,t)F(v)d\tau }\right]\right]|\\ & =\underset{t\in J}{max}|{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[{\displaystyle \sum _{i=n}^{k-1}L(v_i(t))}\right]\right]+{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[\delta {\int }_a^b\sigma (\tau ,t){\displaystyle \sum _{i=n}^{k-1}{A}_i(t)d\tau }\right]\right]|\\ & \le \underset{t\in J}{max}|{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}[L(s_{k-1})-L(s_{n-1})]\right]+{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[\delta {\int }_a^b\sigma (\tau ,t)d\tau (F(p_{k-1})-F(p_{n-1}))\right]\right]|\\ & \le C_1\underset{t\in J}{max}{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[|s_{k-1}-s_{n-1}|\right]\right]+\left(\delta MT{C}_2\right)\underset{t\in J}{max}{\mathrm{\aleph }}^{-1}\left[{\left(\frac{w}{r}\right)}^{\beta }{\mathrm{\aleph }}^{+}\left[|s_{k-1}-s_{n-1}|\right]\right]\\ & =\frac{(C_1+\delta MT{C}_2)t^{\beta }}{\mathrm{\Gamma }(\beta +1)}\Vert s_{m-1}-s_{n-1}\Vert .\end{array}$$

Thus, $\Vert r_m-r_n\Vert \le \gamma \Vert s_{k-1}-s_{n-1}\Vert$ . Choose $k=n+1$ , then

$$\Vert s_{n+1}-s_n\Vert \le \gamma \Vert s_n-s_{n-1}\Vert \le {\gamma }^2\Vert s_{n-1}-s_{n-2}\Vert \le \cdots \le {\gamma }^n\Vert s_1-s_0\Vert ,$$

where $\gamma =\frac{(C_1+\delta MT{C}_2)t^{\beta }}{\mathrm{\Gamma }(\beta +1)}$ .

Similarly, using the triangle inequality

$$\begin{array}{rl}\Vert s_k-s_n\Vert & \le \Vert s_{n+1}-s_n\Vert +\Vert s_{n+2}-s_{n+1}\Vert +\cdots +\Vert s_k-s_{k-1}\Vert \\ \\ & \le [{\gamma }^n+{\gamma }^{n+1}+\cdots +{\gamma }^{k-1}]\Vert s_1-s_0\Vert \\ \\ & \le {\gamma }^n\left[\frac{1-{\gamma }^{k-n}}{1-\gamma }\right]\Vert v_1\Vert .\end{array}$$

But, $0<\gamma <1$ , then $1-{\gamma }^{k-n}<1$ . Thus,

$$\Vert s_k-s_n\Vert \le \frac{{\gamma }^n}{1-\gamma }\underset{t\in J}{max}\left|v_1\right|.$$

(31)

Since $v(t)$ is bounded, then $\Vert v_1\Vert <\mathrm{\infty }$ . So, as $n\to \mathrm{\infty }$ , then $\Vert s_k-s_n\Vert \to 0$ . Thus, the sequence $\{s_k\}$ is a Cauchy in $A$ . Hence, $v(t)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n(t)}$ converges.  □

Theorem 5

(Estimating the Error). The maximum absolute truncation error of the series solution in Equations (17) to (30) is estimated to be

$$\underset{t\in J}{max}|v(t)-\sum _{k=0}^n{v}_k(t)|\le \frac{{\gamma }^n}{1-\gamma }\underset{t\in J}{max}\left|v_1\right|.$$

Proof.

From Equation (31) in Theorem 4 we conclude that $\Vert s_k-s_n\Vert \le \frac{{\gamma }^n}{1-\gamma }{max}_{t\in J}|v_1|$ . So as $k\to \mathrm{\infty }$ , we have $s_k\to v(t)$ . Then $\Vert v(t)-s_n\Vert \le \frac{{\gamma }^n}{1-\gamma }{max}_{t\in J}|v_1(t)|$ . Therefore, the maximum absolute truncation error in $J$ is:

$$\underset{t\in J}{max}|v(t)-\sum _{k=0}^n{v}_k(t)|\le \underset{t\in J}{max}\frac{{\gamma }^n}{1-\gamma }|v_1(t)|=\frac{{\gamma }^n}{1-\gamma }\Vert v_1(t)\Vert .$$

 □

Numerical results and applications

In this section, we apply the FDM to five examples and compare the approximate solutions to the exact solutions that are already in place. First, we present the methodology of the FDM: Consider the general nonlinear FFIDEs with initial conditions given by:

$$\begin{array}{r}{D}_t^{\beta }\left(v\left(t\right)\right)=g\left(t\right)v\left(t\right)+s\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ,\\ \mathrm{where}k-1<\beta \le k,n=1,2,3,\dots .\end{array}$$

(32)

A company with I.Cs:

$$v^{\left(i\right)}\left(0\right)=c_i,i=0,1,2,3,\dots$$

(33)

Note that the fractional derivative is in the Caputo sense of $v(t)$ , $F(v(\tau ))$ is a nonlinear continuous function and $s(t)$ is the non-homogeneous term.

Employing the natural transformation and property (4) to equation (32), we arrive at:

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[D_t^{\beta }\left(v\left(t\right)\right)\right]& ={\mathrm{\aleph }}^{+}\left[s\left(t\right)\right]\\ & +{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].\end{array}$$

(34)

$$\begin{array}{rl}\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& -\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}=L(r,w)\\ & +{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].\end{array}$$

(35)

Thus,

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& =\frac{w^{\beta }}{r^{\beta }}\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}+\frac{w^{\beta }}{r^{\beta }}L(r,w)\\ & +\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ].\end{array}$$

(36)

Substituting equation (33) into equation (36) and applying the inverse transform to equation (36), we arrive at:

$$\begin{array}{rl}v\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[{\displaystyle \sum _{k=0}^{n-1}\frac{c_k{r}^{-(k+1)}}{w^{-k}}}\right]+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}L(r,w)\right]\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ]\right]=S\left(t\right)\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)F\left(v\left(\tau \right)\right)d\tau ]\right].\end{array}$$

(37)

Here the non-homogeneous part and the I.Cs are represented by $S(t)$ . Suppose our intended solution is given by:

$$v\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right).$$

(38)

Also, the nonlinear term is:

$$F\left(v\left(\tau \right)\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right).$$

(39)

Here, $A_n$ ’s are the polynomials of $v_0,v_1,\dots ,v_n$ which can be calculated using:

$$A_n=\frac{1}{n!}\frac{d^n}{d{\delta }^n}\left[F\left(\sum _{i=0}^n{\delta }^i{v}_i\right)\right],n=0,1,2,\dots .$$

(40)

Note that, equation (25) can be written as:

$$\begin{array}{cc}{A}_0& =F\left(v_0\right)\\ A_1& =v_1{F}^{\prime }\left(v_1\right)\\ A_2& =v_2{F}^{\prime }\left(v_0\right)+\frac{1}{2!}{v}_1^2{F}^{″}\left(v_0\right)\end{array}$$

(41)

We can continue in this manner to get the other polynomials. Now, we substitute equation (38) and equation (39) into equation (37) to get:

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)& =S\left(t\right)\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)\sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right)d\tau ]\right].\end{array}$$

(42)

Going through equation (42) we find:

$$\begin{array}{rl}& v_0\left(t\right)=S\left(t\right)\\ & v_1\left(t\right)={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v_0\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)A_0\left(\tau \right)d\tau ]\right]\\ & v_2\left(t\right)={\mathrm{\aleph }}^{-1}\left[\frac{w^{\alpha }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v_1\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)A_1\left(\tau \right)d\tau ]\right].\end{array}$$

One can form the general recursive relation as follows:

$$\begin{array}{rl}{v}_{n+1}\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[g\left(t\right)v_n\left(t\right)+\delta {\int }_a^b\sigma (\tau ,t)A_n\left(\tau \right)d\tau ]\right],\\ & n\ge 0\end{array}.$$

(43)

Finally, the exact solution is given by:

$$v\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right).$$

(44)

Example 1

Given the linear FFIDE:

$$D_t^{\beta }v\left(t\right)=2\sqrt{\frac{t}{\pi }}+\frac{3\sqrt{\pi }t}{4}-\frac{9}{10}+{\int }_0^{1}v\left(\tau \right)d\tau ,$$

(45)

together with I.C:

$$v(0)=0.$$

(46)

Solution. Using the natural transformation of equation (45) to find:

$${\mathrm{\aleph }}^{+}\left[D_t^{\beta }v\left(t\right)\right]={\mathrm{\aleph }}^{+}[2\sqrt{\frac{t}{\pi }}+\frac{3\sqrt{\pi }t}{4}-\frac{9}{10}+{\int }_0^{1}v\left(\tau \right)d\tau ].$$

(47)

Now using property 4 we get:

$$\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]-\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}=\frac{w^{1/2}}{r^{3/2}}+\frac{3\sqrt{\pi }w}{4r^2}-\frac{9}{10r}+{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}v\left(\tau \right)d\tau \right].$$

(48)

Substitute equation (46) into equation (48) to find:

$${\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]=\frac{w^{\beta +\frac{1}{2}}}{r^{\beta +\frac{3}{2}}}+\frac{3\sqrt{\pi }{w}^{\beta +1}}{4r^{\beta +2}}-\frac{9w^{\beta }}{10r^{\beta +1}}+\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}v\left(\tau \right)d\tau \right].$$

(49)

Employ ${\mathrm{\aleph }}^{-1}$ to equation (49) to find:

$$v\left(t\right)=\frac{t^{\beta +\frac{1}{2}}}{\mathrm{\Gamma }(\beta +\frac{3}{2})}+\frac{3\sqrt{\pi }{t}^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{9t^{\beta }}{10\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}v\left(\tau \right)d\tau \right]\right].$$

(50)

Suppose our intended solution that is given as:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(51)

From equation (51), then equation (50) become:

$$\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)=\frac{t^{\beta +\frac{1}{2}}}{\mathrm{\Gamma }(\beta +\frac{3}{2})}+\frac{3\sqrt{\pi }{t}^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{9t^{\beta }}{10\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(\tau \right)d\tau \right]\right].$$

(52)

Looking at equation (52) above, we calculate these iterations:

$$\begin{array}{rl}{v}_0(t)& =\frac{t^{\beta +\frac{1}{2}}}{\mathrm{\Gamma }(\beta +\frac{3}{2})}+\frac{3\sqrt{\pi }{t}^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}\\ & .v_1\left(t\right)=\frac{-9t^{\beta }}{10\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{v}_0\left(\tau \right)d\tau \right]\right]\\ & =\frac{-9t^{\beta }}{10\mathrm{\Gamma }(\beta +1)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(\frac{{\tau }^{\beta +\frac{1}{2}}}{\mathrm{\Gamma }(\beta +\frac{3}{2})}+\frac{3\sqrt{\pi }{\tau }^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)})d\tau \right]\right]\\ & =\frac{-9t^{\beta }}{10\mathrm{\Gamma }(\beta +1)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[\frac{1}{\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +3)}]\right]\\ & =(\frac{-9}{10\mathrm{\Gamma }(\beta +1)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +3)})t^{\beta }.\end{array}$$

And,

$$\begin{array}{rl}{v}_2\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{v}_1\left(\tau \right)d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(\frac{-9}{10\mathrm{\Gamma }(\beta +1)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +3)}){\tau }^{\beta }d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[\frac{-9}{10\mathrm{\Gamma }(\beta +2)}+\frac{1}{\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +3)}]\right]\\ & =(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +3)})t^{\beta }.\end{array}$$

So,

$$\begin{array}{rl}{v}_3\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{v}_2\left(\tau \right)d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +3)}){\tau }^{\beta }d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[\frac{-9}{10\mathrm{\Gamma }{(\beta +2)}^2}+\frac{1}{\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +3)}]\right]\\ & =(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +3)})t^{\beta }.\end{array}$$

Following this direction to find out:

$$\begin{array}{rl}{v}_4\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{v}_3\left(\tau \right)d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^2\mathrm{\Gamma }(\beta +3)}){\tau }^{\beta }d\tau \right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[\frac{-9}{10\mathrm{\Gamma }{(\beta +2)}^3}+\frac{1}{\mathrm{\Gamma }{(\beta +2)}^3\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }{(\beta +2)}^3\mathrm{\Gamma }(\beta +3)}]\right]\\ & =(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^3}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^3\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }{(\beta +2)}^3\mathrm{\Gamma }(\beta +3)})t^{\beta }.\end{array}$$

Hence,

$$\begin{array}{rl}v\left(t\right)& ={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right).}\\ & =v_0\left(t\right)+v_1\left(t\right)+v_2\left(t\right)+\dots \\ & =\frac{t^{\beta +\frac{1}{2}}}{\mathrm{\Gamma }(\beta +\frac{3}{2})}+\frac{3\sqrt{\pi }{t}^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+(\frac{-9}{10\mathrm{\Gamma }(\beta +1)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +3)})t^{\beta }\\ & +(\frac{-9}{10\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)}+\frac{1}{\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +\frac{5}{2})}+\frac{3\sqrt{\pi }}{4\mathrm{\Gamma }(\beta +1)\mathrm{\Gamma }(\beta +2)\mathrm{\Gamma }(\beta +3)})t^{\beta }+\dots \end{array}$$

Now when $\beta =\frac{1}{2}$ , we get

$$\begin{array}{rl}v\left(t\right)& =\frac{t}{\mathrm{\Gamma }\left(2\right)}+\frac{3\sqrt{\pi }{t}^{\frac{3}{2}}}{4\mathrm{\Gamma }\left(\frac{5}{2}\right)}+0+0+\dots \\ & =t+t^{\frac{3}{2}}.\end{array}$$

Which is in fact our intended solution to equation (45) when $\beta =\frac{1}{2}$ .

Choosing $\beta =\{0.25,0.5,0.75,1\}$ in the above equation, we find Figure 1 and Table 1 below:

Figure 1.

Numerical values for $v(t)$ of Ex. (1) for multiple values of $\beta$ when $0\le t\le 1$ .

Table 1.

The exact solutions and approximate of $v(t)$ for Ex. (1) for multiple values of $\beta$ .

$t$	$\beta =$ 0.25	$\beta =$ 0.75	$\beta =$ 1	$\beta =0.5$			Absolute error
				Appr FND	Appr Chebyshev- $K=M=4$	Exact $\beta =0.5$
0.2	1.0833	0.0159	$-$ 0.0477	0.2894	0.2911	0.2894	0
0.4	1.6352	0.1921	0.0135	0.653	0.6549	0.653	0
0.6	2.1523	0.4586	0.1642	1.0648	1.0670	1.0648	0
0.8	2.658	0.7983	0.3973	1.5155	1.5180	1.5155	0
1	3.16	1.2022	0.709	2	2.0027	2	0

Example 2

Given the linear FFIDE :

$$D_t^{\beta }v\left(t\right)=\frac{3\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)t^{\frac{1}{3}}}{\pi }-\frac{t^2}{5}-\frac{t}{4}+{\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau ,0\le t,\tau \le 1,1<\beta \le 2,$$

(53)

together with I.C:

$$v^{\prime }(0)=v(0)=0.$$

(54)

Solution. Taking N-transformation of equation (53) we find:

$${\mathrm{\aleph }}^{+}[D_t^{\beta }v(t)]={\mathrm{\aleph }}^{+}[\frac{3\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)t^{\frac{1}{3}}}{\pi }-\frac{t^2}{5}-\frac{t}{4}+{\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau ].$$

(55)

Now using property 4 we get:

$$\begin{array}{rl}\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& -{\displaystyle \sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}}\\ & =\frac{3\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)w^{\frac{1}{3}}}{\pi r^{\frac{4}{3}}}-\frac{\mathrm{\Gamma }\left(3\right)w^2}{5r^3}-\frac{w}{4r^2}\\ & +{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau }\right].\end{array}$$

(56)

Substitute equation (54) into equation (56) to find:

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& =\frac{3\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)w^{\beta +\frac{1}{3}}}{\pi r^{\beta +\frac{4}{3}}}-\frac{\mathrm{\Gamma }\left(3\right)w^{\beta +2}}{5r^{\beta +3}}\\ & -\frac{w^{\beta +1}}{4r^{\beta +2}}+\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau \right].\end{array}$$

(57)

Employ ${\mathrm{\aleph }}^{-1}$ of equation (57) we find:

$$\begin{array}{rl}v\left(t\right)=& \frac{\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)t^{\beta +\frac{1}{3}}}{\pi \mathrm{\Gamma }(\beta +\frac{4}{3})}-\frac{\mathrm{\Gamma }\left(3\right)t^{\beta +2}}{5\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)v\left(\tau \right)d\tau \right]\right].\end{array}$$

(58)

Suppose our intended solution is given by:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(59)

From equation (59), one can rewrite equation (58) as:

$$\begin{array}{rl}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)}& =\frac{\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)t^{\beta +\frac{1}{3}}}{\pi \mathrm{\Gamma }(\beta +\frac{4}{3})}-\frac{\mathrm{\Gamma }\left(3\right)t^{\beta +2}}{5\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1(t\tau +t^2{\tau }^2){\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n\left(\tau \right)d\tau }}\right]\right].\end{array}$$

(60)

Then looking at equation (60) above, we calculate the iterations:

$$v_0\left(t\right)=\frac{\sqrt{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)t^{\beta +\frac{1}{3}}}{\pi \mathrm{\Gamma }(\beta +\frac{4}{3})}=\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}.$$

So,

$$\begin{array}{rl}{v}_1\left(t\right)& =-\frac{\mathrm{\Gamma }\left(3\right)t^{\beta +2}}{5\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)v_0\left(\tau \right)d\tau \right]\right]\\ & =-\frac{\mathrm{\Gamma }\left(3\right)t^{\beta +2}}{5\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)\left(\frac{2{\tau }^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}\right)d\tau \right]\right]\\ & =-\frac{\mathrm{\Gamma }\left(3\right)t^{\beta +2}}{5\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}[\frac{2t}{(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}+\frac{2t^2}{(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}]\right]\\ & =\left(\frac{8-(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}{4(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +2)}\right)t^{\beta +1}+\left(\frac{20-2(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}{5(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +3)}\right)t^{\beta +2}.\end{array}$$

And,

$$\begin{array}{rl}{v}_2\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)v_1\left(\tau \right)d\tau \right]\right]\\ & =\left(\frac{4(\beta +\frac{7}{3})(\beta +3)\mathrm{\Gamma }(\beta +2)(20-2(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))+5(\beta +\frac{10}{3})(\beta +4)\mathrm{\Gamma }(\beta +3)(8-(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))}{20(\beta +\frac{7}{3})(\beta +\frac{10}{3})(\beta +3)(\beta +4)\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +3){\left(\mathrm{\Gamma }(\beta +2)\right)}^2}\right)t^{\beta +1}\\ & +\left(\frac{4(\beta +\frac{7}{3})(\beta +4)\mathrm{\Gamma }(\beta +2)(40-4(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))+5(\beta +\frac{10}{3})(\beta +5)\mathrm{\Gamma }(\beta +3)(16-2(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))}{20(\beta +\frac{7}{3})(\beta +\frac{10}{3})(\beta +4)(\beta +5)\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +2){\left(\mathrm{\Gamma }(\beta +3)\right)}^2}\right)t^{\beta +2}.\end{array}$$

Following this direction to find out:

$$\begin{array}{rl}{v}_3\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1(t\tau +t^2{\tau }^2)v_2\left(\tau \right)d\tau \right]\right]\\ & =\left(\begin{array}{c}\frac{(20-2(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))((\beta +5)\mathrm{\Gamma }(\beta +3)+2(\beta +3)\mathrm{\Gamma }(\beta +2))}{5(\beta +\frac{10}{3})(\beta +3)(\beta +4)(\beta +5)\mathrm{\Gamma }(\beta +\frac{4}{3}){\left(\mathrm{\Gamma }(\beta +3)\right)}^2{\left(\mathrm{\Gamma }(\beta +2)\right)}^2}\\ +\frac{(8-(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))({(\beta +4)}^2\mathrm{\Gamma }(\beta +3)+2{(\beta +3)}^2\mathrm{\Gamma }(\beta +2))}{4(\beta +\frac{7}{3}){(\beta +3)}^2{(\beta +4)}^2\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +3){\left(\mathrm{\Gamma }(\beta +2)\right)}^3}\end{array}\right)t^{\beta +1}\\ & +\left(\begin{array}{c}\frac{(40-4(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))({(\beta +5)}^2\mathrm{\Gamma }(\beta +3)+2(\beta +4)\mathrm{\Gamma }(\beta +2))}{5(\beta +\frac{10}{3}){(\beta +4)}^2{(\beta +5)}^2\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +2){\left(\mathrm{\Gamma }(\beta +3)\right)}^3}\\ +\frac{(16-2(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3}))((\beta +5)\mathrm{\Gamma }(\beta +2)+2(\beta +3)\mathrm{\Gamma }(\beta +2))}{4(\beta +\frac{7}{3})(\beta +3)(\beta +4)(\beta +5)\mathrm{\Gamma }(\beta +\frac{4}{3}){\left(\mathrm{\Gamma }(\beta +2)\right)}^2{\left(\mathrm{\Gamma }(\beta +3)\right)}^2}\end{array}\right)t^{\beta +2}.\end{array}$$

Hence,

$$\begin{array}{rl}v\left(t\right)& ={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right).=}\\ & v_0\left(t\right)+v_1\left(t\right)+v_2\left(t\right)+\cdots \\ & =\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}+\left(\frac{8-(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}{4(\beta +\frac{7}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +2)}\right)t^{\beta +1}+\left(\frac{20-2(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})}{5(\beta +\frac{10}{3})\mathrm{\Gamma }(\beta +\frac{4}{3})\mathrm{\Gamma }(\beta +3)}\right)t^{\beta +2}+\cdots \end{array}$$

When $\beta =\frac{5}{3}$ , we get

$$\begin{array}{rl}v\left(t\right)& =\frac{2t^2}{\mathrm{\Gamma }\left(3\right)}+0+0+\dots =t^2.\end{array}$$

Which is in fact our intended solution to equation (53) for $\beta =\frac{5}{3}$ .

Choosing $\beta =\{1.25,1.5,1.75,2\}$ in the above equation, we find Figure 2 and Table 2 below:

Figure 2.

Numerical values for $v(t)$ of Ex. (2) for multiple values of $\beta$ when $0\le t\le 1$ .

Table 2.

The exact solutions and approximate of $v(t)$ for Ex. (2) for multiple values of $\beta$ .

t	$\beta =$ 1.25	$\beta =$ 1.5	$\beta =$ 1.75	$\beta =$ 2	$\beta =\frac{5}{3}$		Absolute error
					Approximate	Exact
0	0	0	0	0	0	0	0
0.2	0.11277948	0.0610099	0.03227304	0.01671345	0.04001513	0.04	1.51282062 $\times$ 10 ${}^{-6}$
0.4	0.34230479	0.21831973	0.13655442	0.08378542	0.16001921	0.16	1.92116243 $\times$ 10 ${}^{-5}$
0.6	0.65866176	0.46100645	0.31733181	0.21469377	0.36008496	0.36	8.49634298 $\times$ 10 ${}^{-5}$
0.8	1.05199191	0.78448572	0.57697021	0.41791392	0.64024397	0.64	2.43972421 $\times$ 10 ${}^{-4}$
1	1.51753466	1.1861593	0.91703951	0.69966246	1.0005529	1	5.5293929 $\times$ 10 ${}^{-4}$

Example 3

Given the nonlinear FFIDE with $1<\beta \le 2$ of the form:

$$\begin{array}{rl}{D}_t^{\beta }v\left(t\right)& =\frac{6t^{\frac{1}{3}}}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}-\frac{t^2}{7}-\frac{t}{4}-\frac{1}{9}\\ & +{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau ,0\le t\le 1,\end{array}$$

(61)

together with I.C:

$$v^{\prime }(0)=v(0)=0.$$

(62)

Solution. Using the natural transform of equation (61) to find:

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[D_t^{\beta }v\left(t\right)\right]& ={\mathrm{\aleph }}^{+}[\frac{6t^{\frac{1}{3}}}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}-\frac{t^2}{7}-\frac{t}{4}-\frac{1}{9}\\ & +{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau ].\end{array}$$

(63)

Now using property 4 we get:

$$\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]-\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}[D^{(k)}v(0)]=\frac{2w^{\frac{1}{3}}}{r^{\frac{4}{3}}}-\frac{2w^2}{7r^3}-\frac{w}{4r^2}-\frac{1}{9r}+{\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau \right].$$

(64)

Substitute equation (62) into equation (64) to find:

$$\begin{array}{rl}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]& =\frac{2w^{\beta +\frac{1}{3}}}{r^{\beta +\frac{4}{3}}}-\frac{2w^{\beta +2}}{7r^{\beta +3}}-\frac{w^{\beta +1}}{4r^{\beta +2}}-\frac{w^{\beta }}{9r^{\beta +1}}+\frac{w^{\beta }}{r^{\beta }}\\ & {\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau \right].\end{array}$$

(65)

Applying ${\mathrm{\aleph }}^{-1}$ of equation (65) to find:

$$\begin{array}{rl}v\left(t\right)& =\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}-\frac{2t^{\beta +2}}{7\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{t^{\beta }}{9\mathrm{\Gamma }(\beta +1)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{v}^3\left(\tau \right)d\tau \right]\right].\end{array}$$

(66)

Suppose our intended solution is given by:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(67)

From equation (67), one can write equation (66) as:

$$\begin{array}{cc}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)}& =\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}-\frac{2t^{\beta +2}}{7\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{t^{\beta }}{9\mathrm{\Gamma }(\beta +1)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right)d\tau }}\right]\right].\end{array}$$

(68)

Note that $\sum _{n=0}^{\mathrm{\infty }}{A}_n(\tau )=v^3(\tau )$ is the Adomian polynomial which represents the nonlinear part.

Note that:

$$\begin{array}{rl}{A}_0\left(\tau \right)& =v_0^3\left(\tau \right)\\ A_1\left(\tau \right)& =3v_0^2\left(\tau \right)v_1\left(\tau \right)\\ A_2\left(\tau \right)& =3v_0^2\left(\tau \right)v_2\left(\tau \right)+3v_0\left(\tau \right)v_1^2\left(\tau \right)\\ A_3\left(\tau \right)& =3v_0^2\left(\tau \right)v_3\left(\tau \right)+6v_0\left(\tau \right)v_1\left(\tau \right)v_2\left(\tau \right)+v_1^3\left(\tau \right)\\ & ⋮\end{array}$$

Looking at equation (68) above, one can calculate the iterations as:

$$\begin{array}{rl}{v}_0(t)& =\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}.\\ v_1\left(t\right)& =-\frac{2t^{\beta +2}}{7\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{t^{\beta }}{9\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{A}_0\left(\tau \right)d\tau \right]\right]\\ & =-\frac{2t^{\beta +2}}{7\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{t^{\beta }}{9\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{v}_0^3\left(\tau \right)d\tau \right]\right]\\ & =-\frac{2t^{\beta +2}}{7\mathrm{\Gamma }(\beta +3)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}-\frac{t^{\beta }}{9\mathrm{\Gamma }(\beta +1)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^1{(t+\tau )}^2{\left(\frac{2{\tau }^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}\right)}^{3}d\tau \right]\right]\\ & =\frac{(112-2(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta +2}}{7(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +3)}+\frac{(64-(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta +1}}{4(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +2)}+\frac{(72-(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta }}{9(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +1)}.\end{array}$$

And,

$$\begin{array}{rl}{v}_2\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2{A}_1\left(\tau \right)d\tau }\right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2\left(3v_0^2\left(\tau \right)v_1\left(\tau \right)\right)d\tau }\right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2\left(3{\left(\frac{2{\tau }^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}\right)}^2\left(\frac{(112-2(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3){\tau }^{\beta +2}}{7(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +3)}\right)\right)d\tau }\right]\right]\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2\left(3{\left(\frac{2{\tau }^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}\right)}^2(\frac{(64-(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3){\tau }^{\beta +1}}{4(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +2)}+\frac{(72-(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3){\tau }^{\beta }}{9(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +1)})\right)d\tau }\right]\right]\\ & =\frac{2(\frac{72(\frac{56}{3\beta +2}-\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)}{9\beta +11}+\frac{21(\beta +2)(\frac{64}{\beta +1}-3\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)}{9\beta +8}-\frac{28((3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3-72)\mathrm{\Gamma }(\beta +3)}{(3\beta +4)(9\beta +5)\mathrm{\Gamma }(\beta +1)})t^{\beta +2}}{7\mathrm{\Gamma }{(\beta +\frac{4}{3})}^5\mathrm{\Gamma }(\beta +3{)}^2}\\ & +\frac{2(\frac{28(\beta +1)(\frac{72}{3\beta +4}-\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)}{9\beta +8}+\frac{21(\frac{64}{\beta +1}-3\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)}{9\beta +11}-\frac{72((3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3-56)\mathrm{\Gamma }(\beta +2)}{(3\beta +2)(9\beta +14)\mathrm{\Gamma }(\beta +3)})t^{\beta +1}}{7\mathrm{\Gamma }{(\beta +\frac{4}{3})}^5\mathrm{\Gamma }(\beta +2{)}^2}\\ & -\frac{(\frac{84(\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3-\frac{72}{3\beta +4})}{9\beta +11}+\frac{63(3(\beta +1)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3-64)}{(\beta +1{)}^2(9\beta +14)}+\frac{216\mathrm{\Gamma }(\beta +1)((3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3-56)}{(3\beta +2)(9\beta +17)\mathrm{\Gamma }(\beta +3)})t^{\beta }}{21\mathrm{\Gamma }(\beta +1{)}^2\mathrm{\Gamma }{(\beta +\frac{4}{3})}^5}.\end{array}$$

(69)

Thus,

$$\begin{array}{rl}{v}_3\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2{A}_2\left(\tau \right)d\tau }\right]\right]\\ & ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\displaystyle {\int }_0^1{(t+\tau )}^2(3v_0^2\left(\tau \right)v_2\left(\tau \right)+3v_0\left(\tau \right)v_1^2\left(\tau \right))d\tau }\right]\right].\end{array}$$

(70)

It is worth mentioning here because this iteration, $v_3(t)$ , is almost three pages; we didn’t include it in the text of the paper, but when we sketched the graph and did the table, we included it in the Mathematica file, and it is available.

Hence,

$$\begin{array}{rl}v(t)& =\sum _{n=0}^{\mathrm{\infty }}{v}_n(t)=v_0(t)+v_1(t)+v_2(t)+\cdots \\ & =\frac{2t^{\beta +\frac{1}{3}}}{\mathrm{\Gamma }(\beta +\frac{4}{3})}+\frac{(112-2(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta +2}}{7(3\beta +2)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +3)}\\ & +\frac{(64-(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta +1}}{4(3\beta +3)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +2)}\\ & +\frac{(72-(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3)t^{\beta }}{9(3\beta +4)\mathrm{\Gamma }{(\beta +\frac{4}{3})}^3\mathrm{\Gamma }(\beta +1)}+\dots .\end{array}$$

When $\beta =\frac{5}{3}$ , we get

$$\begin{array}{rl}v\left(t\right)& =\frac{2t^2}{\mathrm{\Gamma }\left(3\right)}+0+0+\dots =t^2.\end{array}$$

Which is in fact the exact solution for equation (61) in the special case $\beta =\frac{5}{3}$ .

Choosing β = {1.25, 1.5, 1.75, 2} in the above equation, we find Figure 3 and Table 3 below:

Figure 3.

Numerical values for $v(t)$ of Ex. (3) for multiple values of $\beta$ when $0\le t\le 1$ .

Table 3.

The exact solutions and approximate of $v(t)$ for Ex. (3) for multiple values of $\beta$ .

t	$\beta =$ 1.25	$\beta =$ 1.5	$\beta =$ 1.75	$\beta =$ 2	$\beta =\frac{5}{3}$		Absolute error
					Approximate	Exact
0	0	0	0	0	0	0	0
0.2	0.31753144	0.07477132	0.0307145	0.01485997	0.04	0.04	0
0.4	0.92358041	0.26336384	0.13067947	0.07574266	0.16	0.16	0
0.6	1.79832602	0.55615655	0.30397263	0.19507664	0.36	0.36	0
0.8	2.96129706	0.95184456	0.55231534	0.38014599	0.64	0.64	0
1	4.44094665	1.45169382	0.87651912	0.63583408	1	1	0

Example 4

Given the nonlinear FFIDE:

$$D_t^{\beta }v\left(t\right)=1-\frac{t}{4}+{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau ,0\le t\le 1,$$

(71)

together with I.C:

$$v(0)=0.$$

(72)

Solution: Using N-transformation of equation (71) to find:

$${\mathrm{\aleph }}^{+}\left[D_t^{\beta }v\left(t\right)\right]={\mathrm{\aleph }}^{+}[1-\frac{t}{4}+{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau ].$$

(73)

Now using property 4 we get:

$$\frac{r^{\beta }}{w^{\beta }}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]-\sum _{k=0}^{n-1}\frac{r^{\beta -(k+1)}}{w^{\beta -k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}=\frac{1}{r}-\frac{w}{4r^2}+{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau \right].$$

(74)

Substitute equation (72) into equation (74) to find:

$${\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]=\frac{w^{\beta }}{r^{\beta +1}}-\frac{w^{\beta +1}}{4r^{\beta +2}}+\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau \right].$$

(75)

Apply ${\mathrm{\aleph }}^{-1}$ of equation (75) to obtain:

$$\begin{array}{r}v\left(t\right)=\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^2\left(\tau \right)d\tau \right]\right].\end{array}$$

(76)

Suppose our intended solution is given by:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(77)

From equation (77), one can write equation (76) as:

$$\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)=\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau \sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right)d\tau \right]\right].$$

(78)

Note that $\sum _{n=0}^{\mathrm{\infty }}{A}_n(\tau )=v^2(\tau )$ is the Adomian polynomial which represents the nonlinear part.

Here,

$$\begin{array}{rl}{A}_0\left(\tau \right)& =v_0^2\left(\tau \right)\\ A_1\left(\tau \right)& =2v_0\left(\tau \right)v_1\left(\tau \right)\\ A_2\left(\tau \right)& =2v_0\left(\tau \right)v_2\left(\tau \right)+v_1^2\left(\tau \right)\\ A_3\left(\tau \right)& =2v_0\left(\tau \right)v_3\left(\tau \right)+2v_1\left(\tau \right)v_2\left(\tau \right)\\ ⋮\end{array}$$

Looking at equation (78) above, one can calculate the remaining iterations as:

$$v_0(t)=\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}.$$

So,

$$\begin{array}{rl}{v}_1\left(t\right)& =-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_0\left(\tau \right)d\tau \right]\right]=-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v_0^2\left(\tau \right)d\tau \right]\right]=-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}\\ & +{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau \left(\frac{{\tau }^{\beta }}{\mathrm{\Gamma }(\beta +1)}\right)d\tau \right]\right]=-\frac{t^{\beta +1}}{4\mathrm{\Gamma }(\beta +2)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[\frac{t}{(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2}\right]\right]\\ & =\frac{(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)t^{\beta +1}}{4(2\beta +2)\mathrm{\Gamma }(\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2}.\end{array}$$

And,

$$\begin{array}{rl}{v}_2\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_1\left(\tau \right)d\tau \right]\right]=\frac{(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)t^{\beta +1}}{2(2\beta +2)(2\beta +3){\left(\mathrm{\Gamma }(\beta +2)\right)}^2{\left(\mathrm{\Gamma }(\beta +1)\right)}^3}.\end{array}$$

Thus,

$$\begin{array}{rl}{v}_3\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_2\left(\tau \right)d\tau \right]\right]\\ & =\frac{(16(2\beta +2)(2\beta +4)(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)+{(2\beta +3)}^2{(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)}^2)t^{\beta +1}}{16(2\beta +4){(2\beta +3)}^2{(2\beta +2)}^2{\left(\mathrm{\Gamma }(\beta +2)\right)}^3{\left(\mathrm{\Gamma }(\beta +1)\right)}^4}.\end{array}$$

We continue to find:

$$\begin{array}{rl}{v}_4\left(t\right)& ={\mathrm{\aleph }}^{-1}\left[\frac{w^{\beta }}{r^{\beta }}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_3\left(\tau \right)d\tau \right]\right]\\ & =\frac{(16(2\beta +2)(2\beta +4)(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)+3{(2\beta +3)}^2{(4-(2\beta +2){\left(\mathrm{\Gamma }(\beta +1)\right)}^2)}^2)t^{\beta +1}}{8(2\beta +4){(2\beta +3)}^3{(2\beta +2)}^2{\left(\mathrm{\Gamma }(\beta +2)\right)}^4{\left(\mathrm{\Gamma }(\beta +1)\right)}^5}.\end{array}$$

Then using equation (78), one can calculate the other iterations.

Choosing $\beta =1$ the above solution becomes:

$$\begin{array}{rl}v\left(t\right)& =\frac{t}{\mathrm{\Gamma }\left(2\right)}+0+0+\dots \\ & =t.\end{array}$$

Which is in fact the solution for equation (71).

Choosing $\beta =\{0.25,0.5,0.75,1\}$ in equation (77), the graphs of $v(t)$ is (Figure 4 and Table 4):

Figure 4.

Numerical Solutions of $v(t)$ for Ex. (4) for multiple values of $\beta$ when $0\le t\le 1$ .

Table 4.

Numerical results for approximate and exact solutions of $v(t)$ for Ex. (4)

t	$\beta =$ 0.25	$\beta =$ 0.5	$\beta =$ 0.75	$\beta =$ 1		Absolute error
				Approximate	Exact
0	0	0	0	0	0	0
0.2	0.81468187	0.52845592	0.33047747	0.2	0.2	0
0.4	1.06025845	0.78104941	0.56432142	0.4	0.4	0
0.6	1.27455608	0.99786002	0.77644018	0.6	0.6	0
0.8	1.47833239	1.19988833	0.97775229	0.8	0.8	0
1	1.67811708	1.39480013	1.17282903	1	1	0

Example 5

Given the nonlinear FFIDE:

$$D_t^{\frac{1}{2}}v\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}\left(\frac{8t^{3/2}}{3-2t^{\frac{1}{2}}}\right)-\frac{t}{1260}+{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau ,0\le t\le 1,$$

(79)

together with I.C:

$$v(0)=0.$$

(80)

Solution: Using N-transformation of equation (79) we find:

$${\mathrm{\aleph }}^{+}\left[D_t^{\frac{1}{2}}v\left(t\right)\right]={\mathrm{\aleph }}^{+}[\frac{8t^{1/2}}{3\mathrm{\Gamma }\left(\frac{1}{2}\right)}-\frac{2t^{\frac{1}{2}}}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}-\frac{t}{1260}+{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau ].$$

(81)

Now using property 4 we get:

$$\frac{r^{\frac{1}{2}}}{w^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]-\sum _{k=0}^{n-1}\frac{r^{\frac{1}{2}-(k+1)}}{w^{\frac{1}{2}-k}}{\left[D^{(k)}v\left(t\right)\right]}_{t=0}=\frac{2w^{\frac{3}{2}}}{r^{\frac{5}{2}}}-\frac{w^{\frac{1}{2}}}{r^{\frac{3}{2}}}-\frac{w}{1260r^2}+{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau \right].$$

(82)

Plug in equation (80) into equation (82) to find:

$${\mathrm{\aleph }}^{+}\left[v\left(t\right)\right]=\frac{2w^2}{r^3}-\frac{w}{r^2}-\frac{w^{\frac{3}{2}}}{1260r^{\frac{5}{2}}}+\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau \right].$$

(83)

Apply ${\mathrm{\aleph }}^{-1}$ of equation (83) to obtain:

$$v\left(t\right)=\frac{2t^2}{\mathrm{\Gamma }\left(3\right)}-\frac{t}{\mathrm{\Gamma }\left(2\right)}-\frac{t^{3/2}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v^4\left(\tau \right)d\tau \right]\right].$$

(84)

Suppose our intended solution is given by:

$$v(t)=\sum _{n=0}^{\mathrm{\infty }}{v}_n(t).$$

(85)

From equation (85), one can write equation (84) as:

$$\sum _{n=0}^{\mathrm{\infty }}{v}_n\left(t\right)=t^2-t-\frac{t^{\frac{3}{2}}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau \sum _{n=0}^{\mathrm{\infty }}{A}_n\left(\tau \right)d\tau \right]\right].$$

(86)

Note that, $\sum _{n=0}^{\mathrm{\infty }}{A}_n(\tau )=v^4(\tau )$ is the Adomian polynomial which represents the nonlinear part.

Note that

$$\begin{array}{rl}{A}_0\left(\tau \right)& =v_0^4\left(\tau \right)\\ A_1\left(\tau \right)& =4v_0^3\left(\tau \right)v_1\left(\tau \right)\\ A_2\left(\tau \right)& =4v_0^3\left(\tau \right)v_2\left(\tau \right)+6v_0^2\left(\tau \right)v_1^2\left(\tau \right)\\ A_3\left(\tau \right)& =4v_0^3\left(\tau \right)v_3\left(\tau \right)+4v_0\left(\tau \right)v_1^3\left(\tau \right)\\ & +12v_0^2\left(\tau \right)v_1\left(\tau \right)v_2\left(\tau \right)\\ & ⋮\end{array}$$

Looking at equation (75) above, one can calculate the remaining iterations as:

$$v_0(t)=t^2-t.$$

And,

$$\begin{array}{rl}{v}_1\left(t\right)& =-\frac{t^{\frac{3}{2}}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_0\left(\tau \right)d\tau \right]\right]\\ & =-\frac{t^{\frac{3}{2}}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau v_0^4\left(\tau \right)d\tau \right]\right]\\ & =-\frac{t^{\frac{3}{2}}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau {({\tau }^2-\tau )}^{4}d\tau \right]\right]\\ & =-\frac{t^{\frac{3}{2}}}{1260\mathrm{\Gamma }\left(\frac{5}{2}\right)}+{\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[\frac{t}{1260}\right]\right]\\ & =0.\end{array}$$

So,

$$v_2\left(t\right)={\mathrm{\aleph }}^{-1}\left[\frac{w^{\frac{1}{2}}}{r^{\frac{1}{2}}}{\mathrm{\aleph }}^{+}\left[{\int }_0^{1}t\tau A_1\left(\tau \right)d\tau \right]\right]=0.$$

Thus, one can concludes that $v_1(t)=v_2(t)=\cdots =v_n(t)=0.$

Hence,

$$\begin{array}{rl}v(t)& =\sum _{n=0}^{\mathrm{\infty }}{v}_n(t)\\ & =v_0(t)+v_1(t)+v_2(t)+\cdots \\ & =t^2-t.\end{array}$$

Which is in fact the exact solution for equation (79).

Concluding remarks

The FDM for nonlinear FFIDE convergence analysis was successfully adopted in the current study work. Additionally, we discovered approximate and analytical solutions to the fractional linear and nonlinear Fredholm integro-differential equations. The new method reduces the calculation challenges of some of the well-known and famous methods and allows for simple computations. Utilizing the FDM, some well-known FFIDE applications were investigated, and the results revealed distinct disparities. Without any linearization, discretization, or perturbation, the employed technique can be applied to a variety of linear and nonlinear FFIDEs. Our long-term goals are to use FDM to solve various linear and nonlinear FFIDEs that appear in many branches of applied research, such as physics and engineering.