Semi-analytical methods for solving fuzzy impulsive fractional differential equations

Abstract

In this paper, we study semi-analytical methods entitled fuzzy fractional differential transform method to solve fuzzy impulsive fractional differential equations using Caupto fractional derivative. At the end first of all fuzzy differential transform method in fractional case is defined and its properties are considered completely. Then existence and uniquness theorem for the solution are proved and convergence of the proposed method is considered in details. Some examples indicate that this method can be easily applied to many linear and nonlinear problems.

Keywords

Fractional differential transform method fuzzy impulsive fractional differential generalized Hukuhara differentiability

1 Introduction

Fuzzy set theory is the significant tool for modeling unknown problems and can be found in many branches of regional, physical, mathematical and engineering sciences. As a result many things can happen in the real world has a fuzzy meaning. The concept of the fuzzy set theory was first proposed by Zadeh, S. Chang, Zimmerman and Kaleva (see [13, 17 , 36]). One of the very important branches of the fuzzy theory is fuzzy differential equations. Then Kandel and Byatt [18] have contributed greatly to the development of fuzzy differential equations. Later many scientists, including M. Friedman et al. [15] applied numerical methods to solve this equations. Also the idea of fuzzy differential equations has been studied by scientists and engineers such as T. Allaviranloo et al. [2, 3] and Dong Qiu et al. [23, 24]. They have considered new method to solve fuzzy differential equation based on fuzzy Taylor expansion as one of the branches of fuzzy differential equations. The idea of the theory of fuzzy impulsive differential equation has been emerging as an effective tool area of investigation in recent years (see [25]). Mouffak Benchohra et al. [11] proposed fuzzy solutions for impulsive differential equations. Subsequently, the basic result for fuzzy impulsive differential equation was defined by S. Vengataasalam et al. [30].

Although the fuzzy fractional differential equations have many branches and many applications and in this research we will restrict our attention to fuzzy impulsive fractional differential equations while fuzzy impulsive fractional differential equations are usually hard to solve analytically and the exact solution is rather difficult to be obtained. There are not too many papers on fuzzy impulsive fractional differential equation up to now. Our aim in this paper is to study the semi-analytical methods for solving fuzzy impulsive fractional differential equations. We will use the fuzzy fractional differential transform method (FFDTM) based on generalized Hukuhara differentiability to solve a nonlinear and linear fuzzy impulsive fractional differential equations given by $^{c} D^{\frac{1}{α}} y (t) = f (t, y), t \in J = [0, T], t \neq t_{k}$ (1.1) $Δ y |_{t = t_{k}} = I_{k} (y (t_{k}^{-})) t = t_{k} k = 0, 2, \dots, m$ (1.2) $y (0) = y_{0}$ (1.3)

In this paper the set of all fuzzy real numbers is denoted R_F. It is clear that R ⊂ R_F. Where $k = 1, 2, \dots, m, 0 < \frac{1}{α} \leq 1$ , $_{c} D^{\frac{1}{α}}$ denote the Caputo fractional generalized derivative of order $\frac{1}{α}$ , and f : J × R_F → R_F, is continuous fuzzy function, I_k : R → R, is continuous function, $y_{0} \in R_{F}, 0 = t_{0} < t_{1} < \dots < t_{m} < t_{m + 1} = T, Δ |_{t = t_{k}} = y (t_{k}^{+}) \underset{gH}{⊖} y (t_{k}^{-}), y (t_{k}^{+}) = \lim_{h \to 0^{+}} y (t_{k} + h), and y (t_{k}^{-}) = \lim_{h \to 0^{-}} y (t_{k} + h)$ represent the right and left limits of y (t) at t = t_k. This paper shows that some difficult fuzzy impulsive fractional differential equations can be easily calculated by fuzzy fractional differential transform method.

The paper is organized as follows: We describe the basic notation and prelimition in Section 2. We define the fuzzy fractional differential transform method expansion such that f is generalized Hukuhara differentiable in Section 3. In Section 4, we shall provide extremely useful basic properties of generalized fuzzy fractional differential transform. We describe the fuzzy fractional impulsive differential equation in Section 5. Some numerical examples are given to clarify the details and efficiency of the method in Section 6. This paper ends with conclusion in Section 7.

2 Preliminaries

In this section, we introduce Definitions, Propositions, Lemmas, Theorems and provided the new Theorem will be needed throughout the paper.

Lemma 1. ∀ α > 0 andγ > -1 $\int_{0}^{t} (t - s)^{\frac{k}{α} - 1} s^{γ} ds = \frac{Γ (\frac{k}{α}) Γ (γ + 1)}{Γ (\frac{k}{α} + γ + 1)} t^{\frac{k}{α} + γ} .$

Proof 1. Lemma 2.6 [31].

We denote C^F [a, b] as the space of all continuous fuzzy-valued function on [a, b]. Also, we denote the space of all Lebesgue integrable fuzzy-valued function on the bounded interval [a, b] ⊂ R by L^F [a, b].

Definition 2.1. [7] Let f (x) ∈ C^F [a, b] ∩ L^F [a, b], and $x_{0} \in (a, b), \frac{1}{α} > 0$ , and $(n = [\frac{1}{α}] + 1)$ such that for all 0 ≤ r ≤ 1, then the Caputo definition of fractional differential operator is given by $D_{a^{+}}^{\frac{1}{α}} f (t, r) = {\begin{matrix} \frac{d^{n} f (t, r)}{{dt}^{n}}, \\ \frac{1}{α} = n \in Z^{+}, \\ \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{f^{(n)} (t, r) d t}{(x - t)^{\frac{1}{α} - n + 1}}, \\ m - 1 < \frac{1}{α} < m, m \in N . \end{matrix}$ (2.1) where

$\begin{matrix} D_{a^{+}}^{\frac{1}{α}} \underline{f} (x_{0}, r) & = & \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}^{(n)} (t, r) d t}{(x - t)^{\frac{1}{α} - n + 1}}, \\ m - 1 < \frac{1}{α} < m, m \in N . \end{matrix}$ (2.2) and

$\begin{matrix} D_{a^{+}}^{\frac{1}{α}} \bar{f} (x_{0}, r) & = & \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\bar{f}}^{(n)} (t, r) d t}{(x - t)^{\frac{1}{α} - n + 1}}, \\ m - 1 < \frac{1}{α} < m, m \in N . \end{matrix}$ (2.3) $D_{a}^{\frac{1}{α}}$ is the Caputo fractional derivative of order $\frac{1}{α}$ .

Definition 2.2. Let f : [a, b] → R_F and x₀ ∈ (a, b), with $\underline{f} (x, r)$ and $\bar{f} (x, r)$ both differentiable at x₀ for all r ∈ [0, 1]. We say that f is [(i) - gH]-differentiable at x₀ if $[D_{a | x}^{\frac{1}{α}} f_{i . gH}] (x_{0}, r) = [D_{a | x}^{\frac{1}{α}} \underline{f} (x_{0}, r), D_{a | x}^{\frac{1}{α}} \bar{f} (x_{0}, r)]$ (2.4) -f is [(ii) - gH]-differentiable at x₀ if $[D_{a | x}^{\frac{1}{α}} f_{ii . gH}] (x_{0}, r) = [D_{a | x}^{\frac{1}{α}} \bar{f} (x_{0}, r), D_{a | x}^{\frac{1}{α}} \underline{f} (x_{0}, r)]$ (2.5)

Definition 2.3. [10] We say that a point t₀ ∈ (a, b), is a switching point for the differentiability of f, if in any neighborhood v of t₀ there exist points t₁ < t₀ < t₂ such that

type(I): at t₁ (2.4) holds while (2.5) does not hold and at t₂ (2.5) holds while (2.4) does not hold, or

type(II): at t₁ (2.5) holds while (2.4) does not hold and at t₂ (2.4) holds while (2.5) does not hold.

Here, we introduced the fuzzy fractional Taylor formula under the Caputo-type derivatives as follow:

Definition 2.4. Let f (x) ∈ C^F [a, b] ∩ L^F [a, b]. And suppose that $(D_{a | x}^{\frac{1}{α}})^{k} f (x) \in C^{F} [a, b] for k = 0, 1, \dots, (n α - 1),$ where $0 < \frac{1}{α} < 1, a < ξ \leq x and a < x \leq b,$ Then we have $\begin{matrix} [f (x, r)] & = & [\underline{f} (x, r), \bar{f} (x, r)], r \in [0, 1], \\ \underline{f} (x, r) & = & \sum_{k = 0}^{n} [(_{c} D^{\frac{1}{α}})^{k} \underline{f} (a, r)] \frac{(x - a)^{\frac{k}{α}}}{Γ (\frac{(n + 1)}{α} + 1)} \\ + \frac{(_{c} D^{\frac{1}{α}})^{(n + 1)} \underline{f} (ξ, r)}{\frac{(n + 1)}{α}} (x - a)^{\frac{(n + 1)}{α}} \end{matrix}$ (2.6) $\begin{matrix} \bar{f} (x, r) & = & \sum_{k = 0}^{n} [(_{c} D^{\frac{1}{α}})^{k} \bar{f} (a, r)] \frac{(x - a)^{\frac{k}{α}}}{Γ (\frac{(n + 1)}{α} + 1)} \\ + \frac{(_{c} D^{\frac{1}{α}})^{(n + 1)} \bar{f} (ξ, r)}{\frac{(n + 1)}{α}} (x - a)^{\frac{(n + 1)}{α}} \end{matrix}$ (2.7)

Lemma 2. [5] Let f_gH ∈ C_f (R, R_F), n ∈ N. Then the following integrals

$\begin{matrix} (I_{x_{0} | x}^{\frac{1}{α}} D_{x_{0} | x}^{\frac{1}{α}} f_{gH}) (x, r), \\ (I_{x_{0} | x}^{\frac{1}{α}} (I_{x_{0} | x_{1}}^{\frac{1}{α}} (D_{x_{0} | x_{1}}^{\frac{1}{α}})^{2} f_{gH})) (x, r), \\ (I_{x_{0} | x}^{\frac{1}{α}} (I_{x_{0} | x_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (D_{x_{0} | x_{2}}^{\frac{1}{α}})^{3} f_{gH}))) (x, r), \\ ⋮ \\ I_{x_{0} | x}^{\frac{1}{α}} (I_{x_{0} | x_{1}}^{\frac{1}{α}} \dots (I_{x_{0} | x_{n - 2}}^{\frac{1}{α}} (I_{x_{0} | x_{n - 1}}^{\frac{1}{α}} ((D_{x_{0} | x_{n - 1}}^{\frac{1}{α}})^{n} f_{gH})) \dots) (x, r) . \end{matrix}$ (2.8)

are continuous functions in x_n-1, x_n-2, …, x respectively. Here x_n-1, x_n-2, …, x > x₀ and all are real numbers. ${D_{a}}^{\frac{1}{α}}$ is the Caputo fractional derivative of order $\frac{1}{α}$ , and $({D_{a}}^{\frac{1}{α}})^{k} = {D_{a}}^{\frac{1}{α}} . {D_{a}}^{\frac{1}{α}} \dots {D_{a}}^{\frac{1}{α}} .$

Lemma 3. Consider f_gH : [a, b] → R_F is gH-differentiable and $({D_{a}}^{\frac{1}{α}} f_{gH}) (x)$ continuous on [a, b]. Then

$\begin{matrix} (I_{a | x}^{\frac{1}{α}} D_{a | x}^{\frac{1}{α}} f_{i . gH}) (x, r) \\ = (- 1) ⊙ (I_{a | x}^{\frac{1}{α}} D_{a | x}^{\frac{1}{α}} f_{ii . gH}) (x, r), r \in [0, 1] . \end{matrix}$ (2.9)

Proof. The proof is obvious using the definition.

Theorem 2.1. Let $\frac{1}{α} \in (0, 1]$ and f (x) ∈ C^F [a, b] ∩ L^F [a, b]. Consider f_gH (t) : [a, b] → R_F is gH-differentiable such that type of differentiability f_gH (t) in [a, b] does not change. Then for a < t < b and

If f_gH (t) is [(i) - gH]-differentiable then $(D_{a}^{\frac{1}{α}} f_{i . gH}) (t)$ is fuzzy Riemann-integralable over [a, b] and

$\begin{matrix} f (x, r) & = & f (a, r) \oplus (I_{a | x}^{\frac{1}{α}}^{c} D_{a | x}^{\frac{1}{α}} f_{i . gH}) (t, r), \\ r \in [0, 1] \end{matrix}$ (2.10)

For case [(ii)-gH]-differentialablity we have

$\begin{matrix} f (a, r) & = & f (x, r) \oplus (- 1) ⊙ (I_{a | x}^{\frac{1}{α}}^{c} D_{a | x}^{\frac{1}{α}} f_{ii . gH}) (t, r), \\ r \in [0, 1] . \end{matrix}$ (2.11)

Proof 3. The proof is clear.

3 Fuzzy fractional differential transform method

In this section, we are going to introduce the fuzzy fractional differential transform method. The proposed method is based on generalized Taylor’s formula. We define the generalized fractional differential transform of the k-th derivative of function f (x) in one varible as follows: $F_{\frac{1}{α}} (k) = {\begin{matrix} \frac{1}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} f (x)]_{x = x_{0}}, & if \frac{k}{α} \in Z^{+} \\ 0, & if \frac{k}{α} \notin Z^{+} \end{matrix}$ (3.1) where $(D_{a}^{\frac{1}{α}})^{k} = D_{a}^{\frac{1}{α}} . D_{a}^{\frac{1}{α}} \dots D_{a}^{\frac{1}{α}}$ k-times and the differential inverse transform of $F_{\frac{1}{α}} (x)$ is defined as follow: $f (x) = \sum_{k = 0}^{\infty} F_{\frac{1}{α}} (k) (x - x_{0})^{\frac{k}{α}}$ (3.2) if we substitute Equation (3.1) into Equation (3.2), using the generalized Taylor’s formula, then we get $f (x) = \sum_{k = 0}^{\infty} \frac{(x - x_{0})^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} f] (x_{0})$ (3.3)

Where $\frac{1}{α}$ is he order of fractional and F (k) is the fractional differential transform of f (x). We define the fuzzy fractional differential transform for any positive integer k and any $\frac{1}{α}, 0 < \frac{1}{α} < 1$ , then the follwing equality holds, which is

if f (x) is [(i)-gH]-differentiability then: $\begin{matrix} f (x, r) & = & [\sum_{k = 0}^{\infty} \frac{(x - x_{0})^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} \underline{f}] (x_{0}, r), \\ \sum_{k = 0}^{\infty} \frac{(x - x_{0})^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} \bar{f}] (x_{0}, r)] \end{matrix}$

And

if f (x) is [(ii)-gH]-differentiability then: $\begin{matrix} f (x, r) & = & [\sum_{k = 0}^{\infty} \frac{(x - x_{0})^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} \bar{f}] (x_{0}, r), \\ \sum_{k = 0}^{\infty} \frac{(x - x_{0})^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} [(D_{x_{0}}^{\frac{1}{α}})^{k} \underline{f}] (x_{0}, r)] \end{matrix}$

3.1 Fuzzy fractional differential transform theorem

In section, we are going to prove fuzzy fractional differential transform expansion for fuzzy valued function in four different cases by using concept of generalized Hukuhara differentiability.

Theorem 3.1. Let $T = [a, a + \frac{1}{α}] \subset R,$ with $\frac{1}{α} > 0$ and $f \in C_{gH}^{n} ([a, b], R_{f}) .$ For x ∈ T

If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiability has no change. Then

$\begin{matrix} f (x) & = & f_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{1}{α}}) f_{i . gH}] (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ [D_{x_{0}}^{\frac{2}{α}} f_{i . gH}] (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ [D_{x_{0}}^{\frac{3}{α}} f_{i . gH}] (x_{0}) \\ \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ [D_{x_{0}}^{\frac{n}{α}} f_{i . gH}] (x_{0}) \\ \oplus R_{n + 1} (x_{0}, x) \end{matrix}$ (3.4) where $R_{n + 1} (x_{0}, x) : = \frac{[D_{a}^{\frac{n + 1}{α}} f_{i . gH}] (ξ)}{Γ (\frac{n + 1}{α} + 1)} (x - x_{0})^{\frac{n + 1}{α}} .$ (3.5) and

$\begin{array}{l} ^{c} (D_{a^{+}}^{\frac{1}{α}} f_{i . g H}) (x; r) \\ = [\frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}, \\ \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\bar{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}] \end{array}$ (3.6) where $\frac{1}{α} \leq ξ \leq x$ ,for each x ∈ (a, b] and $D_{a}^{\frac{1}{α}}$ is the Caputo fractional derivative of order $\frac{1}{α}$ , and $(D_{a}^{\frac{1}{α}})^{k} = D_{a}^{\frac{1}{α}} . D_{a}^{\frac{1}{α}} \dots D_{a}^{\frac{1}{α}}$ (3.7)

If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(ii)-gH]-differentiability has no change. Then

$\begin{matrix} f (x) & = & f_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{1}{α}}) f_{ii . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2}{α}} f_{ii . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{3}{α}} f_{ii . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ [D_{x_{0}}^{\frac{n}{α}} f_{ii . gH}] (x_{0}) \\ \underset{gH}{⊖} (- 1) R_{n + 1} (x_{0}, x) \end{matrix}$ (3.8)

If $((D_{a}^{\frac{1}{α}})^{i} f_{gH})$ are [(i)-gH]-differentiable for i = 2k - 1, k ∈ N, and $((D_{a}^{\frac{1}{α}})^{i} f_{gH})$ are [(ii) - gH]-differentiable for i = 2k, k ∈ N ∪ {0}

$\begin{matrix} f (x) & = & f_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{1}{α}} f_{ii . gH}] (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2}{α}} f_{i . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2 k - 1}{2 α}} f_{ii . gH}] (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2 k}{2 α}} f_{i . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) R_{n + 1} (x_{0}, x) \end{matrix}$ (3.9) where

$\begin{matrix} R_{n + 1} (x_{0}, x) \\ : = \frac{(D_{a}^{\frac{n + 1}{α}} f_{i . gH}) (ξ)}{Γ (\frac{n + 1}{α} + 1)} (x - x_{0})^{\frac{n + 1}{α}} \underset{gH}{⊖} (- 1) \\ \frac{[D_{a}^{\frac{n + 1}{α}} f_{ii . gH}] (ξ)}{Γ (\frac{n + 1}{α} + 1)} (x - x_{0})^{\frac{n + 1}{α}} . \end{matrix}$ (3.10) and

$\begin{array}{l} ^{c} (D_{a^{+}}^{\frac{1}{α}} f) (x; r) \\ = [\frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}} \\ + \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\bar{f}}_{i i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}, \\ \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}_{i i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}} \\ + \frac{1}{Γ (n - \frac{1}{α}} \int_{a}^{x} \frac{{\bar{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}] \end{array}$ (3.11)

If $((D_{a}^{\frac{1}{α}})^{i} f_{gH})$ are [(i)-gH]-differentiable for i = 2k, k ∈ N, and $((D_{a}^{\frac{1}{α}})^{i} f_{gH})$ are [(ii) - gH]-differentiable for i = 2k - 1, k ∈ N ∪ {0}

$\begin{matrix} f (x) & = & f_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{1}{α}} f_{ii . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2}{α}} f_{i . gH}] (x_{0}) \oplus \dots \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2 k - 1}{2 α}} f_{ii . gH}] (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} \\ ⊙ [D_{x_{0}}^{\frac{2 k}{2 α}} f_{i . gH}] (x_{0}) \oplus \dots \oplus R_{n + 1} (x_{0}, x) \end{matrix}$ (3.12) where

$\begin{matrix} R_{n + 1} (x_{0}, x) \\ : = \frac{(D_{a}^{\frac{n + 1}{α}} f_{ii . gH}) (ξ)}{Γ (\frac{n + 1}{α} + 1)} (x - x_{0})^{\frac{n + 1}{α}} \\ \underset{gH}{⊖} (- 1) \frac{[D_{a}^{\frac{n + 1}{α}} f_{i . gH}] (ξ)}{Γ (\frac{n + 1}{α} + 1)} (x - x_{0})^{\frac{n + 1}{α}} . \end{matrix}$ (3.13) and

$\begin{array}{l} ^{c} (D_{a^{+}}^{\frac{1}{α}} f) (x; r) \\ = [\frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}_{i i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}} \\ + \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\bar{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}, \\ \frac{1}{Γ (n - \frac{1}{α})} \int_{a}^{x} \frac{{\underline{f}}_{i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}} \\ + \frac{1}{Γ (n - \frac{1}{α}} \int_{a}^{x} \frac{{\bar{f}}_{i i . g H}^{(n)} d t}{{(x - t)}^{\frac{1}{α} - n + 1}}] \end{array}$ (3.14)

Suppose that $f \in C_{gH}^{n} ([a, b], R_{F}), 3 \leq n .$ Furthermore let f in [a, ξ] is [(i) - gH]-differentiable and in [ξ, b] is [(ii) - gH]- differentiable, in fact ξ is switching point type I for first order derivative of f and t₀ ∈ [a, ξ] in neighborhood of ξ. Moreover suppose that second order derivative of f in ζ₁ ∈ [t₀, ξ] has switching point type II. Moreover type of differentiability for $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i \leq n$ on [ξ, b], does not change. So $\begin{matrix} f (x) \\ = f (t_{0}) \oplus [D_{x_{0}}^{\frac{1}{α}} f_{i . gH}] (t_{0}) ⊙ \frac{(ξ - t_{0})}{Γ (\frac{1}{α} + 1)} \\ \underset{gH}{⊖} (D_{x_{0}}^{\frac{2}{α}} f_{ii . gH}) (t_{0}) ⊙ (t_{0} - ζ_{1}) ⊙ (ξ - t_{0}) \\ \oplus (D_{x_{0}}^{\frac{2}{α}} f_{i . gH}) (ζ_{1}) ⊙ (\frac{(ξ - ζ_{1})^{2}}{2} - \frac{(t_{0} - ζ_{1})^{2}}{2}) \\ \underset{gH}{⊖} (- 1) (D_{x_{0}}^{\frac{1}{α}} f_{ii . gH}) (ξ) ⊙ \frac{(x - ξ)}{Γ (\frac{1}{α} + 1)} \underset{gH}{⊖} (- 1) \\ [D_{x_{0}}^{\frac{2}{α}} f_{ii . gH}] (ξ) ⊙ \frac{(x - ξ)^{2}}{Γ (\frac{2}{α} + 1)} \underset{gH}{⊖} (- 1) \\ I_{t_{0} | ξ}^{α} (I_{t_{0} | ζ}^{α} (I_{t_{0} | x_{2}}^{α} D_{x_{0}}^{\frac{3}{α}} f_{ii . gH} (x_{4}))) \\ \oplus I_{t_{0} | ξ}^{α} (I_{ζ_{1} | ζ x_{1}}^{α} (I_{ζ_{1} | x_{3}}^{α} D_{x_{0}}^{\frac{3}{α}} f_{i . gH} (x_{5}))) ⊖ (- 1)_{gH} \\ I_{ξ | x}^{α} (I_{ξ | t_{1}}^{α} (I_{t_{0} | t_{2}}^{α} D_{x_{0}}^{\frac{3}{α}} f_{ii . gH} (t_{3}))) \end{matrix}$ (3.15)

Proof 4. We prove only the case (iv) and the proof of other cases are similar to this case.

(iv) According to the hypothesis f is [(i) - gH]-differentiable in [x₀, ξ], by Theorem 2.1 $f (ξ) = f (x_{0}) \oplus I_{x_{0} | ξ}^{\frac{1}{α}} ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{1})),$ (3.16) and f is [(ii) - gH]-differentiable in [ξ, b], so for x ∈ [ξ, b] $f (x) = f (ξ) \underset{gH}{⊖} (- 1) I_{ξ | x}^{\frac{1}{α}} (((D_{ξ}^{\frac{1}{α}})^{1} f_{ii . gH}) (t_{1})),$ (3.17)

The point ξ is a switching point for differentiability so from Equations (3.16) and (3.17) we obtain

$\begin{matrix} f (x) & = & f (x_{0}) \oplus I_{x_{0} | ξ}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{1})) \\ \underset{gH}{⊖} (- 1) I_{ξ | x}^{\frac{1}{α}} (((D_{ξ}^{\frac{1}{α}})^{1} f_{ii . gH}) (t_{1})) \end{matrix}$ (3.18)

Consider the first integral on the right of the above equation. By hypothesis ζ is switching point type II for gH-derivative of the function f. So $D_{t_{0}}^{\frac{1}{α}} f_{i . gH}$ is [(ii) - gH]-differentiable on [x₀, ζ₁], then type of differentiability change. Using Theorem 2.1 we deduce that

$\begin{matrix} ((D_{a}^{\frac{1}{α}})^{1} f_{i . gH}) (ζ_{1}) = ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}) (x_{2})) \end{matrix}$ (3.19)

Since $D_{t_{0}}^{\frac{1}{α}} f_{i . gH}$ is [(i) - gH]-differentiable on [ζ₁, ξ], such that the type of differentiability does not change. Hence, for x₁ ∈ [ζ₁, ξ], we have

$\begin{matrix} ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{1}) = ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (ζ_{1}) \\ \oplus I_{ζ | x_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}) (x_{3})) \end{matrix}$ (3.20)

Substituting Equation (3.19) into Equation (3.20) gets

$\begin{matrix} ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{1}) = ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}) (x_{2})) \\ \oplus I_{ζ_{1} | x_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}) (x_{3})), \end{matrix}$ (3.21)

In this case,

$\begin{matrix} ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{2}) = ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{0}) \\ \oplus I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH})) (x_{4}) \\ \Rightarrow \\ I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{2}) = (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}) (x_{0})) \\ ⊙ (ζ_{1} - x_{0}) \oplus I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH})) (x_{4})) \end{matrix}$ (3.22) and

$\begin{matrix} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH})) (x_{3}) = (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}))) (ζ_{1}) \\ \oplus I_{ζ_{1} | x_{3}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{i . gH})) (x_{5}) \\ \Rightarrow \\ I_{ζ_{1} | x_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH})) (x_{3}) = (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH})) (ζ_{1}) \\ ⊙ (x_{1} - ζ_{1}) \oplus I_{ζ_{1} | x_{1}}^{\frac{1}{α}} (I_{ζ_{1} | x_{3}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{i . gH}))) (x_{5}) . \end{matrix}$ (3.23)

The insertion of the Equations (3.23) and (3.22), in Equation (3.21) leads to obtain

$\begin{matrix} \frac{((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH})}{Γ (\frac{1}{α} + 1)} (x_{1}) = \frac{1}{Γ (\frac{1}{α} + 1)} ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{0}) \\ \underset{gH}{⊖} \frac{((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (x_{0}) ⊙ (x_{0} - ζ_{1}) \\ \oplus \frac{((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH})}{Γ (\frac{2}{α} + 1)} (ζ_{1}) ⊙ (x_{1} - ζ_{1}) \underset{gH}{⊖} (- 1) \\ I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}))) (x_{4}) \\ \oplus I_{ζ_{1} | x_{1}}^{\frac{1}{α}} (I_{ζ_{1} | x_{3}}^{\frac{1}{α}} ((D_{x_{0}}^{\frac{1}{α}})^{3} f_{i . gH}))) (x_{5}) \end{matrix}$ (3.24)

So, the first integral on the rigth hand side of Equation (3.18) is as follow

$\begin{matrix} I_{x_{0} | ξ}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH})) (x_{1}) \\ = \frac{1}{Γ (\frac{1}{α} + 1)} (((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH})) (x_{0}) ⊙ (ξ - x_{0}) \underset{gH}{⊖} \\ \frac{(((D_{x_{0}}^{α})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (x_{0}) ⊙ (x_{0} - ζ_{1}) ⊙ (ξ - x_{0}) \\ \oplus \frac{(((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}))}{Γ (\frac{2}{α} + 1)} (ζ_{1}) \\ ⊙ (\frac{(ξ - ζ_{1})^{2}}{2} - \frac{(t_{0} - ζ_{1})^{2}}{2}) \underset{gH}{⊖} (- 1) \\ I_{x_{0} | ξ}^{\frac{1}{α}} (I_{x_{0} | ζ_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{α})^{3} f_{ii . gH})))) (x_{4}) \\ \oplus I_{x_{0} | ξ}^{\frac{1}{α}} (I_{ξ | x_{1}}^{\frac{1}{α}} (I_{ξ | x_{3}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{i . gH})))) (x_{5}) \end{matrix}$ (3.25)

Consider the second integral on the rigth side of the Equation (3.18). According to the hypothesis f⁽ⁱ⁾, i = 2, 3 are [(ii) - gH]-differentiable on [ξ, b]. We have

$\begin{matrix} \frac{(((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}))}{Γ (\frac{1}{α} + 1)} (x_{1}) \\ = (\frac{(((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}))}{Γ (\frac{1}{α} + 1)}) (ξ) \\ \oplus I_{ξ | x_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{2}) \end{matrix}$ (3.26) and $\begin{matrix} \frac{(((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{2})}{Γ (\frac{2}{α} + 1)} \\ = \frac{(((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (ξ) \\ \oplus I_{ξ | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH})) (x_{3}) \end{matrix}$ (3.27) $\begin{matrix} \Rightarrow I_{ξ | x_{1}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH})) (x_{2}) \\ = \frac{((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (ξ) ⊙ (x_{1} - ξ) \\ \oplus I_{ξ | x_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}))) (x_{3}) \end{matrix}$ (3.28) thus substituting Equation (3.28) into Equation (3.26), gives

$\begin{matrix} \frac{(((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}))}{Γ (\frac{1}{α} + 1)} (x_{1}) \\ = (\frac{(((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}))}{Γ (\frac{1}{α} + 1)}) (ξ) \\ \oplus \frac{(((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (ξ) ⊙ (x_{1} - ξ) \\ \oplus I_{ξ | x_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}))) (x_{3}) \end{matrix}$ (3.29)

Therefor the second integral on the right hand side of Equation (3.18) obtains as follow

$\begin{matrix} I_{ξ | x}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH})) (x_{1}) \\ = (\frac{(((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}))}{Γ (\frac{1}{α} + 1)}) (ξ) \\ ⊙ (x - ξ) \oplus \frac{(((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}))}{Γ (\frac{2}{α} + 1)} (ξ) \\ ⊙ (t_{1} - ξ)^{2} \\ \oplus I_{ξ | x}^{\frac{1}{α}} (I_{ξ | x_{1}}^{\frac{1}{α}} (I_{x_{0} | x_{2}}^{\frac{1}{α}} (((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}))) (x_{3}) . \end{matrix}$ (3.30)

By substituting Equations (3.30) and (3.25) into Equation (3.18)

$\begin{matrix} f (x) \\ = f (t_{0}) \oplus ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{0}) ⊙ \frac{(ξ - x_{0})}{Γ (\frac{1}{α} + 1)} \\ \underset{gH}{⊖} [(D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}] (x_{0}) ⊙ (x_{0} - ζ_{1}) ⊙ (ξ - x_{0}) \\ \oplus ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}) (ζ_{1}) \\ ⊙ (\frac{(ξ - ζ_{1})^{2}}{2} - \frac{(x_{0} - ζ_{1})^{2}}{2}) \underset{gH}{⊖} (- 1) \\ ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}) (ξ) ⊙ \frac{(x - ξ)}{Γ (\frac{1}{α} + 1)} \underset{gH}{⊖} (- 1) \\ (D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}] (ξ) ⊙ \frac{(x - ξ)^{2}}{Γ (\frac{2}{α} + 1)} \\ ⊖ (- 1) I_{x_{0} | ξ}^{\frac{1}{α}} (I_{ξ | x_{1}}^{\frac{1}{α}} (I_{x_{0} | ζ_{1}}^{\frac{1}{α}} ((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}))) (x_{4}) \\ \oplus I_{x_{0} | ξ}^{\frac{1}{α}} (I_{x_{1} | ζ_{1}}^{\frac{1}{α}} (I_{ζ_{1} | x_{3}}^{\frac{1}{α}} ((D_{x_{0}}^{α})^{3} f_{i . gH}))) (x_{5}) \\ \underset{gH}{⊖} (- 1) \\ I_{xi | x} (I_{xi | x_{1}} (I_{x_{0} | x_{2}} \frac{((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}) (x_{3})}{Γ (\frac{3}{α} + 1)} d x_{3}) {dx}_{2}) {dx}_{1} . \end{matrix}$

This establishes the formula.

4 Main results

In this section we prove all theorems about generalized fuzzy fractional differential transform by using Theorem 3.1. If $F_{\frac{1}{α}}, G_{\frac{1}{α}}, H_{\frac{1}{α}} and W_{\frac{1}{α}}$ are the differential transforms of functions f (x), g (x), h (x) and w (x) respectively, the following theorems are established.

Theorem 4.1. If f (x) = g (x) ± h (x), then $F_{\frac{1}{α}} (k) = G_{\frac{1}{α}} (K) \pm H_{\frac{1}{α}} (K)$ .

Theorem 4.2. If f (x) = ag (x), then $F_{\frac{1}{α}} (k) = {aG}_{\frac{1}{α}} (k) .$

Proof 5. The proof is similar to proof Theorem 4.1.

Theorem 4.3. If f (x) = g (x) h (x), then $F_{\frac{1}{α}} (k) = \sum_{l = 0}^{k} G_{\frac{1}{α}} (l) H_{\frac{1}{α}} (K - l) .$

Proof 6. (i). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiability has no change. Then $\begin{matrix} f_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} f_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} f_{i . gH}) (x_{0}) = (g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} g_{i . gH}) (x_{0}) \oplus \dots \\ \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{i . gH}) (x_{0})) \\ (h_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} h_{i . gH}) (x_{0}) \oplus \dots \\ \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{i . gH}) (x_{0})) \\ = (g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} g_{i . gH}) (x_{0}) \oplus \dots \\ \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{i . gH}) (x_{0})) \\ (\frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{i . gH}) (x_{0}) \\ \oplus \dots \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} h_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{i . gH}) (x_{0}) \\ \oplus h_{i . gH} (x_{0})) = (g_{i . gH} (x_{0}) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{i . gH}) (x_{0}) \oplus \dots \oplus \\ \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{i . gH}) (x_{0}) h_{i . gH} (x_{0})) \\ \Rightarrow \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α}} f_{i . gH}}{Γ (\frac{k}{α} + 1)} (x - x_{0})^{\frac{k}{α}} \\ = \sum_{l = o}^{k} (\sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α}} g_{i . gH}}{Γ (\frac{k}{α} + 1)} \frac{D_{x_{0}}^{\frac{n - k}{α}} h_{i . gH}}{Γ (\frac{n - k}{α} + 1)}) (x - x_{0})^{\frac{k}{α}} \\ \Rightarrow F_{\frac{1}{α}} (k) = \sum_{l = 0}^{k} G_{\frac{1}{α}} (l) H_{\frac{1}{α}} (K - l) \end{matrix}$

(ii). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(ii)-gH]-differentiability has no change. Then $\begin{matrix} f_{ii . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} f_{ii . gH}) (x_{0}) \\ = (g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{ii . gH}) (x_{0})) \\ (h_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} h_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{ii . gH}) (x_{0})) \\ = (g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ ⊖ (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{ii . gH}) (x_{0})) \\ (\frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} h_{ii . gH}) (x_{0}) \\ ⊖ (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) h_{ii . gH} (x_{0})) \\ = (g_{ii . gH} (x_{0}) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{ii . gH}) (x_{0}) h_{i . gH} (x_{0})) \\ \Rightarrow F_{\frac{1}{α}} (k) = \sum_{l = 0}^{k} G_{\frac{1}{α}} (l) H_{\frac{1}{α}} (K - l) \end{matrix}$

(iii). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiable for i = 2k - 1, k ∈ N, and fⁱ are [(ii) - gH]-differentiable for i = 2k, k ∈ N ∪ {0} $\begin{matrix} f_{ii . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} f_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} f_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} f_{i . gH}) (x_{0}) \\ = (g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} g_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} g_{i . gH}) (x_{0})) (h_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} h_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} h_{i . gH}) (x_{0})) \\ = (g_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} g_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} h_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} h_{i . gH}) (x_{0})) \\ (\frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{3} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{1} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) h_{ii . gH} (x_{0})) \\ = (g_{ii . gH} (x_{0}) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{n} g_{ii . gH}) (x_{0}) h_{i . gH} (x_{0})) \\ \Rightarrow F_{\frac{1}{α}} (k) = \sum_{l = 0}^{k} G_{\frac{1}{α}} (l) H_{\frac{1}{α}} (K - l) \end{matrix}$

Theorem 4.4. If f (x) = g₁ (x) g₂ (x) g₃ (x) … g_n-1 (x) g_n (x) then $\begin{matrix} F_{\frac{1}{α}} (k) = \sum_{k_{n - 1 = 0}}^{k} \sum_{k_{n - 2 = 0}}^{k_{n - 1}} \dots \sum_{k_{2} = 0}^{k_{3}} \sum_{k_{1}}^{k_{2}} G_{1} (k_{1}) G_{2} (k_{2} - k_{1}) \\ \dots G_{n - 1} (k_{n - 1} - k_{n - 2}) G_{n} (k - k_{n - 1}) \end{matrix}$

Proof 7. The proof is based on the same way as Theorem 4.3.

Theorem 4.5. If $f (x) = D_{x_{0}}^{β} [g (x)]$ , then $F_{\frac{1}{α}} (k) = \frac{Γ (\frac{k}{α} + β + 1)}{Γ (\frac{k}{α} + 1)} G_{\frac{1}{α}} (k + β α)$

Proof 8. (i). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiability has no change. Then $\begin{matrix} f_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) f_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α}} f_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} f_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} f_{i . gH}) (x_{0}) \\ = D_{x_{0}}^{β} (g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{i . gH}) (x_{0})) \\ = (D_{x_{0}}^{β} g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α} + β} g_{i . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α} + β} g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α} + β} g_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α} + β} g_{i . gH}) (x_{0})) \\ \Rightarrow \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α}} f_{i . gH} (x_{0})}{Γ (\frac{k}{α} + 1)} (x - x_{0})^{\frac{k}{α}} \\ = \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α} + β} g_{i . gH} (x_{0})}{Γ (\frac{k}{α} + β + 1)} (x - x_{0})^{\frac{k}{α}} \Rightarrow F_{\frac{1}{α}} (k) \\ = \frac{Γ (\frac{k}{α} + β + 1)}{Γ (\frac{k}{α} + 1)} \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α} + β} g_{i . gH} (x_{0})}{Γ (\frac{k}{α} + β + 1)} (x - x_{0})^{\frac{k}{α} + β} \\ \Rightarrow F_{\frac{1}{α}} (k) = \frac{Γ (\frac{k}{α} + β + 1)}{Γ (\frac{k}{α} + 1)} G_{\frac{1}{α}} (k + β α) \end{matrix}$

(ii). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(ii)-gH]-differentiability has no change. Then $\begin{matrix} f_{ii . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α}} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{3}{α}} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{n}{α}} f_{ii . gH}) (x_{0}) \\ = D_{x_{0}}^{β} (g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{ii . gH}) (x_{0})) \\ = (D_{x_{0}}^{β} g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1) L} \\ ⊙ (D_{x_{0}}^{\frac{1}{α} + β} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α} + β} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{3}{α} + β} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{n}{α} + β} g_{ii . gH}) (x_{0})) \Rightarrow \\ \underset{gH}{⊖} (- 1) \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α}} f_{ii . gH} (x_{0})}{Γ (\frac{k}{α} + 1)} (x - x_{0})^{\frac{k}{α}} \\ = \underset{gH}{⊖} (- 1) \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α} + β} g_{ii . gH} (x_{0})}{Γ (\frac{k}{α} + β + 1)} (x - x_{0})^{\frac{k}{α}} \\ \Rightarrow F_{\frac{1}{α}} (k) = \frac{Γ (\frac{k}{α} + β + 1)}{Γ (\frac{k}{α} + 1)} \\ \sum_{k = 0}^{n} \frac{D_{x_{0}}^{\frac{k}{α} + β} g_{ii . gH} (x_{0})}{Γ (\frac{k}{α} + β + 1)} (x - x_{0})^{\frac{k}{α} + β} \\ \Rightarrow F_{\frac{1}{α}} (k) = \frac{Γ (\frac{k}{α} + β + 1)}{Γ (\frac{k}{α} + 1)} G_{\frac{1}{α}} (k + β α) \end{matrix}$

Theorem 4.6. If $f (x) = \int_{x_{0}}^{x} g (t) dt$ , then $\begin{matrix} F_{\frac{1}{α}} (k) = \frac{1}{α} \frac{G_{\frac{1}{α}} (k - \frac{1}{α})}{k} where \frac{1}{α} \leq k . \end{matrix}$

Proof 9. The proof is based on the same way as Theorem 4.5.

Theorem 4.7. If $f (x) = g (x) \int_{x_{0}}^{x} h (t) dt$ then $\begin{matrix} F_{\frac{1}{α}} (k) & = & \sum_{k_{1} = α}^{k} \frac{H_{\frac{k}{α}} (k_{1} - \frac{1}{α})}{k / α} G_{\frac{k}{α}} (k - k_{1}) \\ where k ⩾ \frac{1}{α} \end{matrix}$

Proof 10. (i). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiability has no change. Then $\begin{matrix} f_{i . gH} (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) f_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} f_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} f_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} f_{i . gH}) (x_{0}) = g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{i . gH}) (x_{0}) (\int_{x_{0}}^{x} (h_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) h_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} h_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} h_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} h_{i . gH}) (x_{0})) dt) \\ = (g_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{i . gH}) (x_{0})) (h_{i . gH} (x_{0}) \oplus \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) h_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} h_{i . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} h_{i . gH}) (x_{0}) \oplus \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} h_{i . gH}) (x_{0})) (\int_{x_{0}}^{x} (t - x_{0})^{\frac{k}{α}} dt) \end{matrix}$

By using Theorems 4.3 and 4.6, we get $\begin{matrix} F_{\frac{1}{α}} (k) & = & \sum_{k_{1} = α}^{k} \frac{H_{\frac{1}{α}} (k_{1} - \frac{1}{α})}{k / α} G_{\frac{1}{α}} (k - k_{1}) \\ where k ⩾ \frac{1}{α} \end{matrix}$

(ii). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(ii)-gH]-differentiability has no change. Then $\begin{matrix} f_{ii . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α}} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{3}{α}} f_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} f_{ii . gH}) (x_{0}) \\ = g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{ii . gH}) (x_{0}) \\ (\int_{x_{0}}^{x} (h_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) h_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} h_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} h_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{n}{α}} h_{ii . gH}) (x_{0})) dt) \\ = (g_{ii . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{1}{α}}) g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{2}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{3}{α}} g_{ii . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} g_{ii . gH}) (x_{0})) (h_{ii . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{1}{α}}) h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{2}{α}} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{3}{α}}}{Γ (\frac{3}{α} + 1)} ⊙ (D_{x_{0}}^{\frac{3}{α}} h_{ii . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \oplus \frac{(x - x_{0})^{\frac{n}{α}}}{Γ (\frac{n}{α} + 1)} \\ ⊙ (D_{x_{0}}^{\frac{n}{α}} h_{ii . gH}) (x_{0})) (\int_{x_{0}}^{x} (t - x_{0})^{\frac{k}{α}} dt) \end{matrix}$

By using Theorems 4.3 and 4.6, we get $\begin{matrix} F_{\frac{1}{α}} (k) & = & \sum_{k_{1} = α}^{k} \frac{H_{\frac{1}{α}} (k_{1} - α)}{k / α} G_{\frac{1}{α}} (k - k_{1}) where k ⩾ \frac{1}{α} \end{matrix}$

(iii). If $((D_{a}^{\frac{1}{α}})^{i} f_{gH}), i = 0, 1, \dots, n$ are [(i)-gH]-differentiable for i = 2k - 1, k ∈ N, and fⁱ are [(ii) - gH]-differentiable for i = 2k, k ∈ N ∪ {0}

$\begin{matrix} f_{i . gH} (x_{0}) \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}}) f_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} f_{i . gH}) (x_{0}) \\ \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} f_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} f_{i . gH}) (x_{0}) = (g_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}}) g_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{i . gH}) (x_{0}) ⊖ (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} g_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} g_{i . gH}) (x_{0})) \\ (\int_{x_{0}}^{x} (h_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}}) h_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{i . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} h_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} f_{i . gH}) (x_{0}))) dt \\ = (g_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}}) g_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} g_{i . gH}) (x_{0}) \underset{gH}{⊖} (- 1) \dots \\ \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} g_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} g_{i . gH}) (x_{0})) \\ (h_{i . gH} (x_{0}) \underset{gH}{⊖} (- 1) \frac{(x - x_{0})^{\frac{1}{α}}}{Γ (\frac{1}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}}) h_{ii . gH}) (x_{0}) \oplus \frac{(x - x_{0})^{\frac{2}{α}}}{Γ (\frac{2}{α} + 1)} \\ ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{2} h_{i . gH}) (x_{0}) ⊖ (- 1) \dots \underset{gH}{⊖} (- 1) \\ \frac{(x - x_{0})^{\frac{2 k - 1}{2 α}}}{Γ (\frac{2 k - 1}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k - 1}{2}} h_{ii . gH}) (x_{0}) \\ \oplus \frac{(x - x_{0})^{\frac{2 k}{2 α}}}{Γ (\frac{2 k}{2 α} + 1)} ⊙ ((D_{x_{0}}^{\frac{1}{α}})^{\frac{2 k}{2}} f_{i . gH}) (x_{0})) \\ (\int_{x_{0}}^{x} (t - x_{0})^{\frac{k}{α}} dt) \end{matrix}$ (4.1)

Theorem 4.8. If f (x) = (x - x₀) ^p, then $F_{\frac{1}{α}} (k) = δ (k - p α)$ where $δ (k) = {\begin{matrix} 1, & if k = 0, \\ 0, & if k . \neq 0 \end{matrix}$

Proof. The proof is left to the reader.

Theorem 4.9. $w (x) = \int_{0}^{x} (x - t)^{\frac{k}{α} - 1} g (t, y (t)) dt$ , then by choosing a suitable β ∈ z⁺ such that $\frac{k}{α} β \in z^{+}$ we have $W_{\frac{1}{α}} (k) = \sum_{l = 0}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) G_{\frac{1}{α}} (k - l)$

Proof 12. The proof is based on the same way as Theorem 4.6.

5 Fuzzy impulsive fractional differential equation

Consider the following fuzzy impulsive fractional differential equation $\begin{array}{l} ^{c} D \frac{k}{α} y (t) = f (t, y), t \in J = [0, T], \\ t \neq t_{k}, m - 1 < \frac{1}{α} < m, m \in N \end{array}$ (5.1) $Δ y |_{t = t_{k}} = I_{k} (y (t_{k}^{-}))$ (5.2) $y (0) = y_{0}$ (5.3) where $0 < \frac{1}{α} < 1$ is a real number and the operator $_{c} D^{\frac{1}{α}}$ denote the Caputo fractional generalized derivative of order $\frac{1}{α}$ , and f : J × R_F → R_F, is continuous fuzzy function. Also I : R → R is continuous function. In this section, using fractional differential transform method for fuzzy impulsive fractional differential Equation eqrefk : 1 under the conditions eqrefe : 2 and eqrefe : 3 with fuzzy initial conditions is solved under _cf [gH]-differentiability.

Lemma 4. [8, 16] The initial value problem eqrefk : 1 under the conditions eqrefe : 2 and eqrefe : 3 is equivalent to one of the following integral equations:

$\begin{matrix} y (t) & = & y_{0} \oplus \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds, \\ if t \in [0, t_{1}] \end{matrix}$ (5.4) if y (t) be ^cf [(i) - gH]-differentiable,

$\begin{matrix} y (t) & = & y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds, \\ if t \in [0, t_{1}] \end{matrix}$ (5.5) if y (t) be ^cf [(ii) - gH]-differentiable, $y (t) = {\begin{matrix} y_{0} \oplus \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds, \\ if t \in [0, t_{1}] \\ y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (y (t_{i}^{-})), \\ if t \in (t_{k}, t_{k + 1}] \end{matrix}$ (5.6)if there exists a point t₁ ∈ (0, t_k+1) such that y (t) is [(i) - gH]-differentiable on [0, t₁] and [(ii) - gH]-differentiable on (t₁, t_k+1).

Theorem 5.1. Assume that

There exists a constant 0 ≤ l such that

$\begin{matrix} d (f (t, u), f (t, \bar{u})) & \leq & ld (u, \bar{u}), for each lt \in [0, T], \\ and for each u, \bar{u} \in R_{F} \end{matrix}$ (5.7)

There exists a constant 0 ≤ l^* such that

$\begin{matrix} d (I_{k} (u), I_{k} (\bar{u})) & \leq & l^{*} d (u, \bar{u}), for each u, \bar{u} \in R_{F}, \\ and k = 1, 2, \dots, m . \end{matrix}$ (5.8) if $[\frac{T^{\frac{k}{α}} l (m + 1)}{Γ (\frac{k}{α} + 1)} + {ml}^{*}] < 1$ (5.9)

Such that T is very small numbers therefore, Equations (5.1)–(5.3) has a unique solution on [0, T].

Proof 13. We transform the problems (5.1)–(5.3) into a fixed point problem. Now we introduce $\begin{matrix} pc (J, R_{F}) & = & {y : J \to R_{F} : y \in c ((t_{k}, t_{k + 1}], R_{F}}, \\ k = 0, 1, \dots, m \end{matrix}$ and there exist $y (t_{k}^{-})$ and $y (t_{k}^{+}), k = 1, \dots, m$ with $y (t_{k}^{-}) = y (t_{k})}$ , that is a closed and convex subset of the Banach space of all continuous function on (0, k + 1]. Therefore, pc is a Banach space, too. We suppose that the solution of problems (5.1)–(5.3) is in case [i - gH]-differentiability and [ii - gH]-differentiability is equivalent to integral Equation (5.6). So, we define a mapping $F : pc (J, R_{F}) \to pc (J, R_{F}),$ that given by

$\begin{matrix} F (y) (t) = y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} f (s, y (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \\ \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds) \underset{gH}{⊖} (- 1) \\ \sum_{i = 1}^{k} I_{i} (y (t_{i}^{-})) \end{matrix}$ (5.10)

Therefore, the fixed point of the operator F is the solution of the problems (5.1)–(5.3). We shall use the Banach contraction principle to prove that F has a fixed point. We will show that F is a contraction map. Let x, y ∈ pc (J, R_F). Then, for each t ∈ [0, T], and using the Lemma 1 we have

$\begin{matrix} d (F (x) (t), F (y) (t)) \leq \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ d (f (s, x (s)), f (s, y (s))) ds \underset{gH}{⊖} (- 1) \\ \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} d (f (s, x (s)), f (s, y (s))) ds \\ \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} d (I (s, x (t_{k}^{-})), I (s, y (t_{k}^{-}))) \\ \leq \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} ld (x (s), y (s)) ds \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} ld (x (s), y (s)) ds \\ \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} l^{*} d (x (t_{k}^{-}), y (t_{k}^{-})) \\ = \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} l (\int_{0}^{t} (d (x (s), y (s)))^{α} ds)^{\frac{1}{α}} (\frac{Γ (\frac{k}{α}) T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ + \frac{1}{Γ (\frac{k}{α})} l (\int_{0}^{t} (d (x (s), y (s)))^{α} ds)^{\frac{1}{α}} (\frac{Γ (\frac{k}{α}) T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ + {ml}^{*} (\sum_{i = 0}^{k} (d (x (s), y (s)))^{α} ds)^{\frac{1}{α}} \\ \leq \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} l | | d (x, y) | |_{α} (\frac{Γ (\frac{k}{α}) T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ + \frac{1}{Γ (\frac{k}{α})} l | | d (x, y) | |_{α} (\frac{Γ (\frac{k}{α}) T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ + {ml}^{*} | | d (x, y) | |_{0} \\ = \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} l | | d (x, y) | |_{α} (T^{\frac{k}{α}} \frac{Γ (\frac{k}{α})}{Γ (\frac{k}{α} + 1)}) \\ + \frac{1}{Γ (\frac{k}{α})} l | | d (x, y) | |_{α} (T^{\frac{k}{α}} \frac{Γ (\frac{k}{α})}{Γ (\frac{k}{α} + 1)}) \\ + {ml}^{*} | | d (x, y) | |_{α} = Ml | | d (x, y) | |_{α} (\frac{T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ + l | | d (x, y) | |_{α} (\frac{T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) + {ml}^{*} | | d (x, y) | |_{α} \\ = [\frac{T^{\frac{k}{α}} l (m + 1)}{Γ (\frac{k}{α} + 1)} + {ml}^{*}) | | d (x, y) | |_{α} . \end{matrix}$

Therefore,

$\begin{matrix} | | d (F (x), F (y)) | |_{α} \\ \leq [\frac{T^{\frac{k}{α}} l (m + 1)}{Γ (\frac{k}{α} + 1)} + {ml}^{*}] | | d (x, y) | |_{α} . \end{matrix}$ (5.11)

Consequently by (5.9), F is a contraetion. As a consequence of Banach fixed point theorem, we deduce that F has a fixed point which is a solution of problems (5.1)–(5.3).

5.1 Solving fuzzy impulsive fractional differential equation by fuzzy fractional differential transform method

Let us define $w (t) = \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds$ . Suppose that Y (K), W (K) and F (K) are the fractional differential transform of the function y (t), w (t) and f (t, y (t)), respectively. Then, by choosing β ∈ z⁺ such that $\frac{k}{α} β \in Z^{+}$ and using the Theorem 4.9, we obtain

$\begin{matrix} Y_{\frac{1}{α}} (k) \\ = {\begin{matrix} I_{k} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \oplus \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)), \\ t \in [0, t_{1}], k \geq l \\ I_{0} (Y_{\frac{1}{α}} 0)) \oplus Y_{\frac{1}{α}} (0) \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \\ (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \\ \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (Y_{\frac{1}{α}} (K_{i})), \\ t \in (t_{k}, t_{k + 1}], k \geq l \end{matrix} \end{matrix}$ (5.12) with $F_{\frac{1}{α}} (k - l) = {\begin{matrix} \frac{1}{Γ (\frac{k}{α} + 1)} [(D_{t_{0}}^{\frac{k}{α}}) f (k - l)]_{t = t_{0}}, \\ if \frac{k}{α} \in Z^{+} \\ 0, if \frac{k}{α} \notin Z^{+} \end{matrix}$ (5.13)

Therefore, the solution of (5.1)–(5.3) falls short of providing an explicit farmula $y (t) = \sum_{k = 0}^{\infty} Y_{\frac{1}{α}} (K) t^{\frac{k}{α}}$ (5.14) for k = 0, 1, 2, …, (nα - 1), where $\frac{1}{α}$ is the order of the fuzzy impulsive fractional differential equation.

Theorem 5.2. Let $0 < \frac{1}{α} \leq 1,$ then the exact solution of Equations (5.1)–(5.3) could be represented by $y (t) = \sum_{k = 0}^{\infty} Y_{\frac{1}{α}} (K) t^{\frac{k}{α}}$ (5.15)

Also, the approximate solution y (t) can be obtained by taking finitely many terms in the series representation of y (t) so, $y_{N} (t) = \sum_{k = 0}^{N} Y_{\frac{1}{α}} (K) t^{\frac{k}{α}}$ (5.16)

And $y_{N} (t), y_{N}^{\frac{1}{α}} (t)$ convergence uniformly to the exact solution $y (t), y^{\frac{1}{α}} (t)$ .

Proof 14. Let y (t) be solution of Equations (5.1)–(5.3). From Equation (3.1). y (t) could be expressed by Taylor series as follow $y (t) = \sum_{k = 0}^{\infty} \frac{1}{Γ (\frac{k}{α} + 1)} [(D_{t_{0}}^{\frac{k}{α}}) y (t)]_{t = t_{0}} t^{\frac{k}{α}}$ (5.17)

Substiting Equation (3.1) into Equation (5.17), $y (t) = \sum_{k = 0}^{\infty} Y_{\frac{1}{α}} (k) t^{\frac{k}{α}}$ (5.18)

Where

$\begin{matrix} Y_{\frac{1}{α}} (k) = I_{k} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \oplus \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)), \\ t \in [0, t_{1}], k \geq l \\ Y_{\frac{1}{α}} (k) = I_{0} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) \\ F_{\frac{1}{α}} (k - l)) \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (Y_{\frac{1}{α}} (K_{i})), \\ t \in (t_{k}, t_{k + 1}], k \geq l \end{matrix}$ (5.19) is the exact solution of Equations (5.1)–(5.3). Also let us y (t) and y_N (t) be exact and approximate solution of the problems (5.1)–(5.3). We will show that, $d (y (t), y_{N} (t)) \leq M_{1} | | d (y (t), y_{N} (t)) | |_{α}$ (5.20)

Also $d (y^{\frac{1}{α}} (t), y_{N}^{\frac{1}{α}} (t)) \leq M_{2} | | d (y^{\frac{1}{α}} (t), y_{N}^{\frac{1}{α}} (t)) | |_{α}$ (5.21)

Therefore if y (t) and y_N (t) be exat and approximate solution of the problems (5.1)–(5.3), without loss of generality suppose that there exists a point t₁ ∈ (0, t_k+1) such that y (t) is [(i) - gH]-differentiable on [0, t₁] and [(ii) - gH]-differentiable on (t₁, t_k+1). Using Lemma (4), the solution of problems (5.1)–(5.3) in this case is equivalent to integral Equation (5.6). By using Lemma 1 we have, $y (t) = {\begin{matrix} y_{0} \oplus \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds, \\ t \in [0, t_{1}] \\ y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (y (t_{i})), \\ t \in (t_{k}, t_{k + 1}] \end{matrix}$ (5.22)

And $y_{N} (t) = {\begin{matrix} y_{0} \oplus \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y_{N} (s)) ds, \\ t \in [0, t_{1}], \\ y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ f (s, y_{N} (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} \\ f (s, y_{N} (s)) ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (y (t_{i}^{-})), \\ t \in (t_{k}, t_{k + 1}] \end{matrix}$ (5.23)

Thus

$\begin{matrix} d (y (t), y_{N} (t)) \\ = {\begin{matrix} d (\frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y (s)) ds, \\ \frac{1}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} f (s, y_{N} (s)) ds) \\ \leq \frac{l}{Γ (\frac{k}{α})} \int_{0}^{t} (t - s)^{\frac{k}{α} - 1} d (y (s), y_{N} (s)) ds \\ = \frac{l}{Γ (\frac{k}{α})} (\int_{0}^{t} (t - s)^{\frac{k - α}{α}} ds) \\ (\int_{0}^{t} (d (y (s), y_{N} (s))) ds)^{α})^{\frac{1}{α}} = \frac{l}{Γ (\frac{k}{α})} (\frac{Γ (\frac{k}{α}) T^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) \\ (\int_{0}^{t} (d (fy (s), y_{N} (s))) ds)^{α})^{\frac{1}{α}} \\ = \frac{{lT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} | | d (y (t), y_{N} (t)) | |_{α}, t \in [0, t_{1}] \\ d (y_{0} ⊖ (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} \\ f (s, y (s)) ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (y (t_{i}^{-})), \\ y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ f (s, y_{N} (s)) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} \\ f (s, y_{N} (s)) ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (y (t_{i}^{-}))) \\ \leq d (y_{0} \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} \\ d (f (s, y (s)), f (s, y_{N} (s))) ds \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \\ \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} d (f (s, y (s)), f (s, y_{N} (s))) \\ ds \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} d (I_{i} (y (t_{i}^{-})), I_{i} (y_{N} (t_{i}^{-})))) \\ \leq \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \int_{t_{i - 1}}^{t_{i}} (t_{i} - s)^{\frac{k}{α} - 1} ld (x (s), y (s)) ds \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \int_{t_{k}}^{t} (t - s)^{\frac{k}{α} - 1} ld (x (s), y (s)) ds \\ \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} l^{*} d (x (t_{k}^{-}), y (t_{k}^{-})) = \underset{gH}{⊖} (- 1) \\ \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} (\frac{Γ (\frac{k}{α}) {lT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) (\int_{0}^{t} (d (y (s), y_{N} (s)) ds)^{α})^{\frac{1}{α}} \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} (\frac{Γ (\frac{k}{α}) {lT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)}) (\int_{0}^{t} (d (y (s), \\ y_{N} (s)) ds)^{α})^{\frac{1}{α}} \underset{gH}{⊖} (- 1) l^{*} ((d (y (t_{i}^{-}), y_{N} (t_{i}^{-}))^{α})^{\frac{1}{α}} \\ = \underset{gH}{⊖} (- 1) \frac{{MlT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} | | y (t), y_{N} (t) | |_{α} \underset{gH}{⊖} (- 1) \\ \frac{{MlT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} | | y (t), y_{N} (t) | |_{α} \underset{gH}{⊖} (- 1) l^{*} | | y (t), y_{N} (t) | |_{α} \\ = \underset{gH}{⊖} (- 1) (\frac{{MlT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} + \frac{{lT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} + l^{*}) \\ | | y (t), y_{N} (t) | |_{α} = (\frac{(M + 1) {lT}^{\frac{k}{α}}}{Γ (\frac{k}{α} + 1)} + l^{*}) | | y (t), y_{N} (t) | |_{α}, \\ t \in (t_{k}, t_{k + 1}] \end{matrix} \end{matrix}$ (5.24)

It follows that $d (y (t), y_{N} (t)) \leq M_{1} | | d (y (t), y_{N} (t)) | |_{α}$

Also $d (y^{\frac{1}{α}} (t), y_{N}^{\frac{1}{α}} (t)) \leq M_{2} | | d (y (t), y_{N} (t)) | |_{α}$ where M₁, M₂ are constants.

Hence ||d (y (t), y_N (t)) ||₀→ 0 as n → ∞, the approximate solution $y_{N} (t), y_{N}^{\frac{1}{α}} (t)$ convergence uniformly to the exact solution y (t) and its fractional derivative, $y^{\frac{1}{α}} (t)$ respectively. The proof for othere case is similar.

Now we claim that solving fuzzy impulsive fractional differential Equations (5.1)–(5.3) by fuzzy fractional differential transform method is convregece.

Theorem 5.3. Cansider the fuzzy impulsive fractional differential Equations (5.1)–(5.3) if there exists a fixed n such that n ≤ k, |f (t, y (t)) | ≤ r^k, 0 < r ≤ 1 and $\sum_{i = 1}^{n} I_{i} (Y_{\frac{1}{α}} (n_{i})) \leq r^{n},$ then the numerical solution of the present method absolutely convergence.

Proof 15. Without loss of generality suppose the exists a point t₁ ∈ (0, t_k+1) such that y (t) is [(i) - gH]-differentiable on [0, t₁] and [(ii) - gH]-differentiable on (t₁, t_k+1), proof of the other cases are left to the reader. For simplicity, let $| I_{k} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) | = g_{0} r^{k}$ . We suppose that the solution of problems (5.1)–(5.3) is [(i) - gH]-differentiable on [0, t₁] and [(ii) - gH]-differentiable on (t₁, t_k+1). Using Lemma (4), the solution of problem (5.1)–(5.3) is equivalent to integral Equation (5.6). So the fuzzy fractional differential transformation of (5.6) is

$\begin{matrix} Y (k) \\ = {\begin{matrix} I_{k} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \oplus \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)), \\ t \in [0, t_{1}], k \geq l \\ I_{0} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) \\ δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (Y_{\frac{1}{α}} (K_{i})), \\ t \in (t_{k}, t_{k + 1}], k \geq l \end{matrix} \end{matrix}$ (5.25)

Suppose $| F_{\frac{1}{α}} (k - l) | \leq r^{k}$ is true for all n ≤ k. From (5.25), we have

$\begin{matrix} | Y (k) | \\ = {\begin{matrix} | I_{k} (Y_{\frac{1}{α}} (0)) \oplus Y_{\frac{1}{α}} (0) \oplus \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) | \\ \leq g_{0} r^{k} \oplus | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) | r^{k} \\ = r^{k} (g_{0} \oplus | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} (\sum_{l = t_{i - 1}}^{k} \\ B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) |), \\ | I_{0} (y (0)) \oplus Y (0) \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) F_{\frac{1}{α}} (k - l)) \\ ⊖ (- 1) \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α}) \\ F_{\frac{1}{α}} (k - l)) \underset{gH}{⊖} (- 1) \sum_{i = 1}^{k} I_{i} (Y_{\frac{1}{α}} (K_{i})) | \\ \leq g_{0} r^{k} \underset{gH}{⊖} (- 1) | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) ⊖ \\ (- 1) \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) \\ \underset{gH}{⊖} (- 1) 1) | r^{k} = r^{k} (g_{0} \underset{gH}{⊖} (- 1) | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) \underset{gH}{⊖} (- 1) \\ \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) \\ \underset{gH}{⊖} (- 1) 1 |), \end{matrix} \end{matrix}$ (5.26)

Let $\begin{matrix} h_{1} = g_{0} \oplus \\ | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α} | \end{matrix}$

And $\begin{matrix} h_{2} = g_{0} \underset{gH}{⊖} (- 1) | \frac{1}{Γ (\frac{k}{α})} \sum_{i = 1}^{k} \\ (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) \underset{gH}{⊖} (- 1) \\ \frac{1}{Γ (\frac{k}{α})} (\sum_{l = 1}^{k} B (\frac{k - l}{k} β + 1, \frac{k}{α}) δ (l - β \frac{k}{α})) \\ \underset{gH}{⊖} (- 1) 1 | \end{matrix}$ k is large enough, we now proceed by induction, the hypothesis is true. So we have

$\begin{matrix} | y (t) | \\ = {\begin{matrix} | \sum_{k = 0}^{\infty} Y_{\frac{1}{α}} (K) t^{\frac{k}{α}} | \leq \sum_{k = 0}^{\infty} | Y_{\frac{1}{α}} (K) | t^{\frac{k}{α}} \\ \leq h_{1} \sum_{k = 0}^{\infty} r^{k} = \frac{h_{1}}{1 - r}, t \in [0, t_{1}], \\ | \sum_{k = 0}^{\infty} Y_{\frac{1}{α}} (K) t^{\frac{k}{α}} | \leq \sum_{k = 0}^{\infty} | Y_{\frac{1}{α}} (K) | t^{\frac{k}{α}} \\ \leq h_{2} \sum_{k = 0}^{\infty} r^{k} = \frac{h_{2}}{1 - r}, t \in (t_{k}, t_{k + 1}] . \end{matrix} \end{matrix}$ (5.27)

6 Numerical examples

We demonstrate the effectiveness of the fuzzy fractional differential transform method,for solving fuzzy impulsive fractional differential equations by the following some examples.

Example 6.1. Let us consider the fuzzy impulsive fractional equation, $\begin{matrix} ^{c} D^{\frac{k}{α}} y (t) = \frac{{ty}^{2} (t)}{(3 + t) (1 + y^{2} (t))}, \\ t \in J : = [0, 1], t \neq \frac{1}{2}, \\ m - 1 < \frac{1}{α} < m, m \in N, \end{matrix}$ (6.1) $Δ y |_{t = \frac{1}{2}} = \frac{| y (\frac{1^{-}}{2}) |}{2 + | y (\frac{1^{-}}{2}) |},$ (6.2) $y (0) = [\tilde{0}, \tilde{0}]$ (6.3) $Y_{\frac{1}{α}} (0) = [\tilde{0}, \tilde{0}]$ (6.4)

Set $I_{k} (t) = \frac{t}{t + 2}, t \in [0, \infty)$ (6.5) $\begin{matrix} f (t, y (t)) \\ = [\frac{t^{3} (r - 1)}{(3 + t) (1 + t^{2})}, \frac{t^{3} (1 - r)}{(3 + t) (1 + t^{2})}] \end{matrix}$ (6.6)

Using the Equations (5.12) and (5.13) and taking β = 2 and $α = \frac{1}{2}$ , Equation (6.1) can be transformed to:

$\begin{matrix} Y_{\frac{1}{α}} (k) \\ = {\begin{matrix} 0 \oplus \frac{1}{Γ (\frac{k}{\frac{1}{2}})} \sum_{i = 1}^{k} (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} 2 + 1, \frac{k}{\frac{1}{2}}) \\ δ (l - k) F_{\frac{1}{α}} (k - l)), if t \in [0, \frac{1}{2}], k \geq l \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{\frac{1}{2}})} \sum_{i = 1}^{k} (\sum_{l = t_{i - 1}}^{k} B (\frac{k - l}{k} 2 + 1, \frac{k}{\frac{1}{2}}) \\ δ (l - k) F_{\frac{1}{α}} (k - l)) \underset{gH}{⊖} (- 1) \frac{1}{Γ (\frac{k}{\frac{1}{2}})} \\ (\sum_{l = 1}^{k} B (\frac{k - l}{k} 2 + 1, \frac{k}{\frac{1}{2}}) δ (l - k) F_{\frac{1}{α}} (k - l)) \underset{gH}{⊖} \\ (- 1) \sum_{i = 1}^{k} I_{i} (Y_{\frac{1}{α}} (K_{i})), if t \in (\frac{1}{2}, 1], k \geq l \end{matrix} \end{matrix}$ (6.7)

For case [(i)-gH]-differentialablity by solving (6.7) for k = l and N = 3 we obtain

$\begin{matrix} F_{\frac{1}{α}} (0) & = & [\frac{r - 1}{Γ (0 + 1)} [(D_{t_{0}}^{0}) \frac{t^{3}}{(3 + t) (1 + t^{2})}]_{t = 0}, \\ \frac{1 - r}{Γ (0 + 1)} [(D_{t_{0}}^{0}) \frac{t^{3}}{(3 + t) (1 + t^{2})}]_{t = 0}] \\ = & [\tilde{0}, \tilde{0}], if t \in [0, \frac{1}{2}] \end{matrix}$ (6.8) $\begin{matrix} F_{\frac{1}{α}} (0) & = & [\frac{r - 1}{Γ (0 + 1)} [(D_{t_{0}}^{0}) \frac{t^{3}}{(3 + t) (1 + t)}]_{t = \frac{1}{2}}, \\ \frac{1 - r}{Γ (0 + 1)} [(D_{t_{0}}^{0}) \frac{t^{3}}{(3 + t) (1 + t)}]_{t = \frac{1}{2}}] \\ = & [\frac{r - 1}{35}, \frac{1 - r}{35}], if t \in (\frac{1}{2}, 1] \end{matrix}$ (6.9)

And

$Y_{\frac{1}{α}} (0) = [\tilde{0}, \tilde{0}]$ (6.10) $\begin{matrix} Y_{\frac{1}{α}} (1) \\ = {\begin{matrix} \frac{1}{Γ (2)} \sum_{l = 0}^{1} B (\frac{k - l}{k} \times 2 + 1, 2) δ (l - k) F_{\frac{1}{α}} (k - l) \\ = \frac{1}{Γ (2)} (B (3, 2) δ (- 1) F_{\frac{1}{α}} (1) + B (1, 2) δ (0) F_{\frac{1}{α}} (0)) \\ = \frac{1}{2} δ (0) F_{\frac{1}{α}} (0), t \in [0, \frac{1}{2}], k \geq l \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (2)} \sum_{l = 0}^{2} B (\frac{k - l}{k} \times 2 + 1, 2) δ (l - k) \\ F_{\frac{1}{α}} (k - l) \underset{gH}{⊖} (- 1) \frac{1}{Γ (2)} \sum_{l = 1}^{1} B (\frac{k - l}{k} \times 2 + 1, 2) \\ δ (l - k) F_{\frac{1}{α}} (k - l) = \underset{gH}{⊖} (- 1) \frac{1}{Γ (2)} \\ (B (3, 2) δ (- 1) F_{\frac{1}{α}} (1) + B (1, 2) δ (0) F_{α} (0)) \underset{gH}{⊖} (- 1) \\ \frac{1}{Γ (2)} (B (1, 2) δ (0) F_{\frac{1}{α}} (0)) = \underset{gH}{⊖} (- 1) δ (0) F_{\frac{1}{α}} (0),] \\ t \in (\frac{1}{2}, 1], k \geq l \end{matrix} \end{matrix}$ (6.11)

$\begin{matrix} Y_{\frac{1}{α}} (2) \\ = {\begin{matrix} \frac{1}{Γ (4)} \sum_{l = 0}^{2} B (\frac{k - l}{k} \times 2 + 1, 4) δ (l - k) F_{\frac{1}{α}} (k - l) \\ \oplus \frac{1}{Γ (4)} \sum_{l = \frac{1}{2}}^{2} B (\frac{k - l}{k} \times 2 + 1, 4) δ (l - k) F_{\frac{1}{α}} (k - l) \\ = \frac{1}{Γ (4)} (B (3, 4) δ (- 2) F_{\frac{1}{α}} (2) + B (2, 4) δ (- 1) F_{\frac{1}{α}} (1) \\ + B (1, 4) δ (0) F_{\frac{1}{α}} (0)) \oplus \frac{1}{Γ (4)} (B (\frac{5}{2}, 4) δ (\frac{- 3}{2}) F_{\frac{1}{α}} (\frac{3}{2}) \\ + B (2, 4) δ (- 1) F_{\frac{1}{α}} (1) + B (\frac{3}{2}, 4) δ (\frac{- 1}{2}) F_{\frac{1}{α}} (\frac{3}{2}) \\ + B (1, 4) δ (0) F_{\frac{1}{α}} (0)) = \frac{1}{Γ (4)} 2 B (1, 4) δ (0) F_{α} (0) \\ = \frac{1}{12} δ (0) F_{\frac{1}{α}} (0), t \in [0, \frac{1}{2}], k \geq l \\ \underset{gH}{⊖} (- 1) \frac{1}{Γ (4)} \sum_{l = 0}^{2} B (\frac{k - l}{k} \times 2 + 1, 4) δ (l - k) \\ F_{\frac{1}{α}} (k - l) \underset{gH}{⊖} (- 1) \frac{1}{Γ (4)} \sum_{l = \frac{1}{2}}^{2} B (\frac{k - l}{k} \times 2 + 1, 4) \\ δ (l - k) F_{\frac{1}{α}} (k - l) \underset{gH}{⊖} (- 1) \frac{1}{Γ (4)} \sum_{l = 1}^{2} B (\frac{k - l}{k} \times 2 \\ + 1, 4) δ (l - k) F_{\frac{1}{α}} (k - l) \oplus \frac{δ (0) F_{\frac{1}{α}} (0)}{δ (0) F_{\frac{1}{α}} (0) + 2} \\ = \underset{gH}{⊖} (- 1) \frac{1}{Γ (4)} (B (3, 4) δ (- 2) F_{\frac{1}{α}} (2) + B (2, 4) δ (- 1) \\ F_{\frac{1}{α}} (1) + B (1, 4) δ (0) F_{\frac{1}{α}} (0)) \underset{gH}{⊖} (- 1) \frac{1}{Γ (4)} \\ (B (\frac{5}{2}, 4) δ (\frac{- 3}{2}) F_{\frac{1}{α}} (\frac{3}{2}) + B (2, 4) δ (- 1) F_{\frac{1}{α}} (1) \\ + B (\frac{3}{2}, 4) δ (\frac{- 1}{2}) F_{\frac{1}{α}} (\frac{3}{2}) + B (1, 4) δ (0) F_{\frac{1}{α}} (0)) \underset{gH}{⊖} \\ (- 1) \frac{1}{Γ (4)} (B (2, 4) δ (- 1) F_{\frac{1}{α}} (1) + B (1, 4) δ (0) F_{\frac{1}{α}} (0)) \\ \oplus \frac{δ (0) F_{\frac{1}{α}} (0)}{δ (0) F_{\frac{1}{α}} (0) + 2} = \frac{δ (0) F_{\frac{1}{α}} (0)}{δ (0) F_{\frac{1}{α}} (0) + 2} \underset{gH}{⊖} (- 1) \frac{3}{24} δ (0) F_{\frac{1}{α}} (0) \\ = \frac{F_{\frac{1}{α}}^{2} (0) + 3 F_{\frac{1}{α}} (0)}{8 (F_{\frac{1}{α}} (0) + 2)}, t \in (\frac{1}{2}, 1], k \geq l \end{matrix} \end{matrix}$ (6.12)

Substituting Equations (6.8), and (6.9) into Equation (6.10)–(6.12) gets $Y_{\frac{1}{α}} (0) = [\tilde{0}, \tilde{0}]$ (6.13) $Y_{\frac{1}{α}} (1) = {\begin{matrix} [\tilde{0}, \tilde{0}], & if t \in [0, \frac{1}{2}], k \geq l \\ [\frac{r - 1}{35}, \frac{1 - r}{35}], & if t \in (\frac{1}{2}, 1], k \geq l \end{matrix}$ (6.14) $\begin{matrix} Y_{\frac{1}{α}} (2) \\ = {\begin{matrix} [\tilde{0}, \tilde{0}], & if t \in [0, \frac{1}{2}], k \geq l \\ [\frac{r^{2} + 103 r - 104}{1225}, \frac{r^{2} - 107 r + 106}{1225}] / & if t \in (\frac{1}{2}, 1], k \geq l \\ ([\frac{8 r - 8}{35}, \frac{8 - 8 r}{35}] + 16), \end{matrix} \end{matrix}$ (6.15)

Thus the solution of (6.1) is

$y_{i . gH} (t) = {\begin{matrix} \sum_{k = 0}^{2} Y_{\frac{1}{α}} (K) t^{\frac{k}{\frac{1}{2}}} = \sum_{k = 0}^{2} Y_{\frac{1}{α}} (K) t^{2 k}, & if t \in [0, \frac{1}{2}], k \geq l \\ \sum_{k = 0}^{2} Y_{\frac{1}{α}} (K) t^{\frac{k}{\frac{1}{2}}} = \sum_{k = 0}^{2} Y_{\frac{1}{α}} (K) t^{2 k}, & if t \in (\frac{1}{2}, 1], k \geq l \end{matrix}$ (6.16)

Therefore,

$\begin{matrix} y_{i . gH} (t) \\ = {\begin{matrix} [\tilde{0}, \tilde{0}], & if t \in [0, \frac{1}{2}], k \geq l \\ [\frac{r - 1}{35}, \frac{1 - r}{35}] t^{2} + [\frac{r^{2} + 103 r - 104}{1225}, \frac{r^{2} - 107 r + 106}{1225}] / & if t \in (\frac{1}{2}, 1], k \geq l \\ ([\frac{8 r - 8}{35}, \frac{8 - 8 r}{35}] + 16) t^{4}, \end{matrix} \end{matrix}$ (6.17)

7 Conclusions

In this paper a new approach by presenting fuzzy fractional differential transform expansion based on gH-differentiability was introduced. The numerical results show that the present method is an accurate and reliable analytical technique for fuzzy impulsive fractional differential equation. Moreover, the method can be effective for solving other nonlinear initial value problems.

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