Numerical solution of fuzzy fractional integro-differential equation via two-dimensional Legendre Wavelet method

Abstract

In this paper, the two-dimensional Legendre wavelet is studied to approximated the solution of fuzzy fractional integro-differential equation under Caputo generalized Hukuhara differentiability. The existence and uniqueness theorems for a fuzzy fractional integro-differential equation by considering the type of differentiability of solution are proved. Furthermore, the accuracy and efficiency of the proposed method are demonstrated is a series of numerical experiments.

Keywords

Fuzzy fractional integro-differential equation Caputo generalized Hukuhara differentiability Legendre wavelet method

1 Introduction

The application of fractional calculus has attracted more attention in the past few years because of significant advantage of the fractional order models in comparison with integer order models. Also, the fuzzy set theory is a powerful tool for modeling uncertain problems. In fractional equations, these vagueness may appear in each part of the equation like initial condition, boundary condition, etc. So solving fractional equations in the sense of real conditions leads to the use of interval or fuzzy calculations.

The concept of the fuzzy derivative was first introduced by Chang and Zadeh [13], it was followed by many authors [8 , 20]. The starting point of the topic in the set valued differential equation and also fuzzy differential equation is Hukuharas paper [17]. The Hukuhara derivative was the starting point for the topic of set differential equations and later also for fuzzy fractional differential equations. By the concept of Hukuhara differentiability, the fuzzy Riemann- Liouville fractional differential equations is introduced by Agarwal et al. in [1], that was the starting point of the topic in fuzzy fractional derivative. They have considered the Riemann-Liouville differentiability concept based on the Hukuhara differentiability to solve uncertain fractional differential equations. The existence and uniqueness of solutions of Riemann- Liouville fuzzy fractional differential equations is proved in [8, 9]. Allahviranloo et al. in [5] presented the explicit solutions of uncertain fractional differential equations under Riemann-Liouville H-differentiability using Mittag-Leffler functions and in [25] introduced the fuzzy fractional differential equations under Riemann-Liouville H-differentiability and obtained the solution of this equation by fuzzy Laplace transforms. They shown two new uniqueness results for fuzzy fractional differential equations involving Riemann-Liouville generalized H-differentiability with fuzzy version of Nagumo and Krasnoselskii-Krein conditions [6]. Consequently, the Caputo generalized Hukuhara derivative is introduced in [4], the authors introduced an ordinary fractional differential equation under the generalized Hukuhara differentiability, and studied the existence and uniqueness of the solution. The nonlinear fuzzy fractional integro-differential equation under generalized fuzzy Caputo derivative is introduced in [2, 3] and proved the existence and uniqueness of the solutions of this set of equation by considering the type of differentiability. Recently, P.K. Sahu et al. [22] applied the two dimensional Legendre wavelet method to solve the fuzzy integro-differential equations and they developed the Bernoulli wavelet method to solve the nonlinear fuzzy Hammerste in Volterra integral equations with constant delay [23].

The purpose of the present paper is two topics. One of our intentions is to prove the uniqueness of the solution of a fuzzy fractional integro-differential equation. The other aim is solving a fuzzy fractional integro-differential equation under Caputo generalized Hukuhara differentiability by two dimensional Legendre wavelet method. For this purpose, we convert a fuzzy fractional integro-differential equation into a system of algebraic equations that can be solved by any numerical method.

The importance of this study, from a theoretical point of view, is that the present two dimensional Legendre wavelet method is developed for a general form of fuzzy fractional integro-differential equation under Caputo generalized Hukuhara differentiability. This can be great help in the numerical study of fuzzy fractional integro-differential equation and other equations in this form.

The paper is organized as follows. Section 2 collects some definitions of basic notions and notations concerning fuzzy calculus, and then in Section 3, we introduce a fuzzy fractional integro-differential equation under Caputo generalized Hukuhara differentiability and we study the existence and uniqueness of the solutions of this set of equations. In Section 4, we discuss the properties of Legendre wavelets. To determine the approximate solution for the fuzzy fractional integro-differential equation, two-dimensional Legendre wavelet method has been applied in Section 5. Moreover according to the type of differentiability, solutions of a fuzzy fractional integro-differential equations are investigated in different scenarios. The convergence of two-dimensional Legendre wavelet is discussed in Section 6. In Section 7, some examples are given to show the efficiency of the proposed method and conclusions are drawn in Section 8.

2 Preliminaries

In this section, we introduce notation, definitions, and preliminary results, which will be used throughout this paper. We denote $C (J, ℝ)$ the Banach space of all real-valued continuous functions from J into $ℝ$ . For measurable real-valued function $m : J \to ℝ^{+}$ , define the norm $| | m | |_{L^{p} (J, ℝ^{+})} = (\int_{J} | m (t) |^{p})^{\frac{1}{p}}, 1 \leq p < \infty$ . We denote $L^{p} (J, ℝ^{+})$ the Banach space of all Lebesgue measurable real-valued functions m with $| | m | |_{L^{p} (J, ℝ^{+})} < \infty$ .

We denote by $ℝ_{F}$ , the set of fuzzy numbers, that is, normal, fuzzy convex, upper semi-continuous and compactly supported fuzzy sets which defined over the real line. For 0 < r ≤ 1, set $[u]_{r} = {t \in ℝ^{n} | u (t) \geq r}$ , and $[u]_{0} = cl {t \in ℝ^{n} | u (t) > 0}$ . We represent $[u]_{r} = [\underline{u} (r), \bar{u} (r)]$ , so if $u \in ℝ_{F}$ , the r-level set [u] _r is a closed interval for all r ∈ [0, 1]. For arbitrary $u, v \in ℝ_{F}$ and $k \in ℝ$ , the addition and scalar multiplication are defined by [u + v] _r = [u] _r + [v] _r , [ku] ^r = k [u] _r respectively.

A triangular fuzzy number defined as a fuzzy set in $ℝ_{F}$ , that is specified by an ordered triple $u = (a, b, c) \in ℝ^{3}$ with a ≤ b ≤ c such that $\underline{u} (r) = a + (b - a) r$ and $\bar{u} = c - (c - b) r$ are the endpoints of r-level sets for all r ∈ [0, 1]. The support of fuzzy number u is defined as follows: $\begin{matrix} supp (u) = cl {t \in ℝ^{n} | u (t) > 0}, \end{matrix}$ where cl is closure of set ${t \in ℝ^{n} | u (t) > 0}$ . The Hausdorff distance between fuzzy numbers is given by $D : ℝ_{F} \times ℝ_{F} \to ℝ^{+} \cup {0}$ as in [15] $\begin{matrix} D (u, v) \\ = sup_{t \in [0, 1]} d_{H} ([u]_{r}, [v]_{r}) \\ = sup_{r \in [0, 1]} max {| \underline{u} (r) - \underline{v} (r) |, | \bar{u} (r) - \bar{u} (r) |}, \end{matrix}$ where d_H is the Hausdorff metric. The metric space $(ℝ_{F}, D)$ is complete, separable and locally compact and the following properties from [15] for metric D are valid:

D (u ⊕ w, v ⊕ w) = D (u, v) ,

$D (λ u, λ v) = | λ | D (u, v), v \in ℝ_{F};$

D (u ⊕ v, w ⊕ z) ≤ D (u, w) + D (v, z) ,

D (u⊝v, w⊝z) ≤ D (u, w) + D (v, z), as long as u⊝v and w⊝z exist, where $u, v, w, z \in ℝ_{F}$ .

where, ⊝ is the Hukuhara difference(H-difference), it means that w⊝v = u if and only if u ⊕ v = w.

Definition 2.1. (See [10]) Let $f, g : [a, b] \to ℝ_{F}$ , be fuzzy real number valued functions. The uniform distance between f, g is defined by $D^{*} (f, g) = sup {D (f (x), g (x)) ∣ x \in [a, b]} .$

Definition 2.2. (See [11]) The generalized Hukuhara difference of two fuzzy numbers $u, v \in ℝ_{F}$ is defined as follows: $\begin{matrix} u ⊝_{gH} v = w \Leftrightarrow {\begin{matrix} (i) . u = v + w; \\ or (ii) . v = u + (- 1) w . \end{matrix} \end{matrix}$

In terms of r-levels we have $[u ⊝_{gH} v]_{r} = [min {\underline{u} (r) - \underline{v} (r), \bar{u} (r) - \bar{u} (r)}, max {\underline{u} (r) - \underline{v} (r), \bar{u} (r) - \bar{v} (r)}]$ and if the H-difference exists, then u⊝v = u_⊝gHv; the conditions for the existence of $w = u ⊝_{gH} v \in ℝ_{F}$ are $\begin{matrix} i . {\begin{matrix} \underline{w} (r) = \underline{u} (r) - \underline{v} (r) and \bar{w} (r) = \bar{u} (r) - \bar{v} (r), \\ with \underline{w} (r) increasing, \bar{w} (r) decreasing, \\ \underline{w} (r) \leq \bar{w} (r), \end{matrix} \\ ii . {\begin{matrix} \underline{w} (r) = \bar{u} (r) - \bar{v} (r) and \bar{w} (r) = \underline{u} (r) - \underline{v} (r) . \\ with \underline{w} (r) increasing, \bar{w} (r) decreasing, \\ \underline{w} (r) \leq \bar{w} (r) . \end{matrix} \end{matrix}$

It is easy to show that (i) and (ii) are both valid if and only if w is a crisp number.

Remark 2.3. Throughout the rest of this paper, we assume that $u ⊝_{gH} v \in ℝ_{F}$ .

Definition 2.4. (See [7]) A fuzzy valued function $f : [a, b] \to ℝ_{F}$ is said to be continuous at x₀ ∈ [a, b] iff whenever x_n → x₀, then D (f (x_n) , f (x₀)) →0 as n→ ∞.

Definition 2.5. (See [11]) The generalized Hukuhara derivative of a fuzzy-valued function $f : (a, b) \to ℝ_{F}$ at x₀ ∈ (a, b) is defined as $f_{gH}^{'} (x_{0}) = lim_{h \to 0} \frac{f (x_{0} + h) ⊝_{gH} f (x_{0})}{h} .$ (1)

If $f_{gH}^{'} (x_{0}) \in ℝ_{F}$ satisfying (1) exists, we say that f is generalized Hukuhara differentiable (gH-differentiable for short) at x₀.

Definition 2.6. (See [14]) Let $f : [a, b] \to ℝ_{F}$ . We say that f (t) is fuzzy Riemann integrable in $𝕀 \in ℝ_{F}$ if for any ɛ > 0, there exists δ > 0 such that for any division P = {[u, v] ; ξ} with the norms Δ (P) < δ, we have $\begin{matrix} D (\sum_{p}^{*} (v - u) ⊙ f (ξ), 𝕀) < ɛ, \end{matrix}$ where $\sum_{p}^{*}$ denotes the fuzzy summation. We choose to write $𝕀 : = \int_{a}^{b} f (t) dt .$

Note that if the fuzzy-valued function $f (t, r) = [\underline{f} (t, r), \bar{f} (t, r)]$ is continuous in the metric D, the Lebesgue integral and the Riemann integral yield the same value, and also

$\int_{a}^{b} f (t, r) dt = [\int_{a}^{b} \underline{f} (t, r) dt, \int_{a}^{b} \bar{f} (t, r) dt]$ (2) for all 0 ≤ r ≤ 1.

Throughout this paper, we consider the notation $A^{𝔽} [a, b]$ for the space of the fuzzy-valued functions from [a, b] into $ℝ_{F}$ that are absolutely continuous on [a, b]. Also, $C^{𝔽} [a, b]$ denote the set of the fuzzy-valued function which are fuzzy continuous on all of [a, b] such that the continuity is one-sided at endpoints a, b. Also, we denote the space of all Lebesgue integrable the fuzzy-valued functions on the bounded interval $[a, b] \subset ℝ$ by $L^{𝔽} [a, b]$ .

Definition 2.7. (See [21]) Let $f \in L^{𝔽} [a, b]$ . The fuzzy Riemann-Liouville integral of a fuzzy-valued function f is defined as follows: $(I_{a}^{q} f) (t) = \frac{1}{Γ (q)} \int_{a}^{t} (t - s)^{q - 1} f (s) d s$ (3) for a ≤ t, and 0 < q ≤ 1.

Theorem 2.8. (See [21]) Let $f \in L^{𝔽} [a, b]$ is a fuzzy-valued function, The fuzzy Riemann-Liouville integral of a fuzzy-valued function f can be expressed as follows $\begin{matrix} I_{a}^{q} f (t, r) = [I_{a}^{q} \underline{f} (t, r), I_{a}^{q} \bar{f} (t, r)] \end{matrix}$ where $I_{a}^{q} \underline{f} (t, r) = \frac{1}{Γ (q)} \int_{a}^{t} (t - s)^{q - 1} \underline{f} (s, r) ds,$ (4) $I_{a}^{q} \bar{f} (t, r) = \frac{1}{Γ (q)} \int_{a}^{t} (t - s)^{q - 1} \bar{f} (s, r) d s$ (5)

Definition 2.9. (See [4]) Let $f \in A^{𝔽} [a, b]$ . The fuzzy gH-fractional Caputo differentiability of the fuzzy-valued function f is defined as following:

$(_{gH} D_{*}^{q} f) (t) = \frac{1}{Γ (1 - q)} \int_{a}^{t} \frac{(f_{gH}^{'}) (s) d s}{(t - s)^{q}} .$ (6) where a < s < t, q ∈ (0, 1].

Definition 2.10. (See [4]) Let $f : [a, b] \to ℝ_{F}$ be differentiable at t₀ ∈ (a, b). We say that f is differentiable at t₀ if $(i) (_{gH} D_{*}^{q} f) (t_{0}, r) = [(D_{*}^{q} \underline{f}) (t_{0}, r), (D_{*}^{q} \bar{f}) (t_{0}, r)],$ (7) and that f is differentiable at t₀ if $(ii) (_{gH} D_{*}^{q} f) (t_{0}, r) = [(D_{*}^{q} \bar{f}) (t_{0}, r), (D_{*}^{q} \underline{f}) (t_{0}, r)] .$ (8)

Definition 2.11. (See [4]) Let $f : [a, b] \to ℝ_{F}$ be a fuzzy function. A point t₀ ∈ (a, b) is said to be a switching point for the differentiably of f, if in any neighborhood V of t₀ there exist points t₁ < t₀ < t₂ such that

Type(I). at t₁ (7) holds while (8) does not hold and at t₂ (8) holds and (7) does not hold, or

Type(II). at t₁ (8) holds while (7) does not hold and at t₂ (7) holds and (8) does not hold.

Theorem 2.12. (See [16]) (Hölder’s Inequality) If p and q are positive numbers satisfying the relation $\frac{1}{p} + \frac{1}{q} = 1$ and if f ∈ L^p (a, b), g ∈ L^q (a, b), then fg ∈ L (a, b) and $\begin{matrix} \int_{a}^{b} | f (x) g (x) | d x \\ \leq (\int_{a}^{b} | f (x) |^{p} d x)^{\frac{1}{p}} (\int_{a}^{b} | g (x) |^{q} d x)^{\frac{1}{q}} . \end{matrix}$

3 Fuzzy fractional integro-differential equation

In this section, first we introduce a fuzzy fractional integro-differential equation and then we prove that this equation has a uniqueness solution. Consider the following fuzzy initial value problem: ${\begin{matrix} (_{gH} D_{*}^{q} y) (t) = f (t, y (t), (S y) (t)), \\ y (0) = y_{0} \in ℝ_{F}, \end{matrix}$ (9) where t ∈ J = [0, T] , q ∈ (0, 1] and $_{g H} D_{*}^{q}$ is the fuzzy Caputo fractional derivative of order q, $f : J \times ℝ_{F} \times ℝ_{F} \to ℝ_{F}$ is a given function satisfying some assumptions that will be specified later and y₀ is an element of $ℝ_{F}$ and $S$ is a nonlinear integral operator given by:

$(S y) (t) = \int_{0}^{t} k (x, t) y (x) d x,$ (10) where $k : J \times J \to ℝ$ , with $γ_{0} = max {\int_{0}^{t} k (x, t) d x : (x, t) \in J \times J}$ and the sign of k (s, t) does not change in J.

In rest of the paper, the existence and uniqueness of solutions for the fuzzy initial value problem (9) under -differentiability is proved. For this purpose, we need the following Lemma and assumptions.

Lemma 3.1. (See [4]) Let $f : [0, T] \to ℝ_{F}$ be a fuzzy-valued function such that $f \in A^{𝔽} [0, T]$ , then

$I_{0}^{q} (_{gH} D_{*}^{q} f) (t) = f (t) \underset{gH}{⊖} f (0) .$ (11)

Lemma 3.2. The fuzzy initial value problem (ref3.1) is equivalent to one of the following integral equations:

Case 1. If y (t) be ^cf [(i) - gH]-differentiable, then $y (t) = y_{0} + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s .$

Case 2. If y (t) be ^cf [(ii) - gH]-differentiable, hence $y (t) = y_{0} ⊝ \frac{- 1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s .$

Case 3. If there exists a point c ∈ (0, T) such that y (t) is ^cf [(i) - gH]-differentiable on [0, c] and ^cf [(ii) - gH]-differentiable on (c, T] (i.e. c is a switching point type (I)); therefore, $y (t) = {\begin{matrix} y_{0} + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s, \\ t \in [0, c]; \\ y_{0} ⊝ \frac{- 1}{Γ (q)} \int_{c}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s, \\ t \in [c, T] . \end{matrix}$

Proof. Using integral operator $I_{0}^{q}$ to both side of (9), we have $I_{0}^{q} (_{gH} D_{*}^{q} y (t)) = I_{0}^{q} (f (t, y (t), (S y) (t)) .$

Thus by Lemma 3.1 we obtain

$\begin{matrix} y (t) \underset{gH}{⊖} y (0) \\ = \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s . \end{matrix}$ (12)

According Definition 2.2, if y (t) be ^cf [(i) - gH]-differentiable, $y (t) = y_{0} + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s,$ (13) and if y (t) be ^cf [(ii) - gH]-differentiable $y (t) = y_{0} ⊖ \frac{- 1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s .$ (14)

Also, if we have a switching point c ∈ (a, b) of type (I) the ^cf [gH]-differentiability changes from type (I) to type (II) at t = c. Then by Equations (13) and (14), the proof is straightforward. □

Our result is based on Banach contraction principle. For brevity, γ₀ = max {|k (x, t) | : (x, t) ∈ J × J} and $M = | | m_{1} (t) + γ_{0} T m_{2} (t) | |_{L^{\frac{1}{q_{1}}} (J, ℝ^{+})}$ .

Theorem 3.3. Assume that the following conditions hold:

(H₁). $f : J \times ℝ_{F} \times ℝ_{F} \to ℝ_{F}$ is fuzzy continuous.

(H₂). There exists a constant q₁ ∈ (0, q) and real-valued positive functions $m_{1} (t), m_{2} (t) \in L^{\frac{1}{q_{1}}} (J, ℝ^{+})$ such that $\begin{matrix} D (f (t, x (t), (S x) (t)), f (t, y (t), (S y) (t))) \\ \leq m_{1} (t) D (x (t), y (t)) + m_{2} (t) D ((S x) (t), (S) y (t)), \end{matrix}$ for each t ∈ J, and all $x (t), y (t), (S x) (t), (S y) (t) \in C^{𝔽} [0, T]$ .

$Ω_{q, q_{1}, T} = \frac{{MT}^{q - q_{1}}}{Γ (q) (\frac{q - q_{1}}{1 - q_{1}})^{1 - q_{1}}} < 1,$ (15) then the Equation (9) has a unique solution on J.

Proof. Consider the operator $F : C^{𝔽} [0, T] \to C^{𝔽} [0, T]$ defined by

$\begin{matrix} (Fy) (t) & = & y_{0} + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \\ f (s, y (s), (S y) (s)) d s \end{matrix}$ (16)

First, we proof that F is a fuzzy continuous operator. Let us assume that {y_n} be a sequence such that y_n → y as n→ ∞ in C [0, T]. Then for each t ∈ [0, T] $\begin{matrix} D (({Fy}_{n}) (t), (Fy) (t)) \\ \leq \frac{1}{Γ (q)} (\int_{0}^{t} (t - s)^{q - 1} \\ D (f (s, y_{n} (s), (S y_{n}) (s)), f (s, y (s), S y (s))) d s) \\ \leq \frac{D^{*} (f (s, y_{n} (s), (S y_{n}) (s)), f (s, y (s), S y (s)))}{Γ (q)} \\ \int_{0}^{t} (t - s)^{q - 1} d s \\ \leq \frac{t^{q} D^{*} (f (s, y_{n} (s), (S y_{n}) (s)), f (s, y (s), (S y) (s)))}{Γ (q)} \end{matrix}$

Since f is a fuzzy continuous function, we have $\begin{matrix} D (({Fy}_{n}) (t), (Fy) (t)) \\ \leq \frac{t^{q} D^{*} (f (s, y_{n} (s), (S y_{n}) (s)), f (s, y (s), (S y) (s)))}{Γ (q)} \\ \to 0 as n \to \infty \end{matrix}$

Hence, F is a fuzzy continuous operator.

Now, Transform the problem (9) into a fixed point problem. Suppose that y (t) is a -differentiable and $y_{0} \in ℝ_{F}$ be fixed. We shall use the Banach contraction principle to prove that F defined by (16) has a fixed point. For this, let $x (t), y (t) \in ℝ_{F}$ . Using the properties of Hausdorff distance, Definition 2.1, H₂ and the well known Hölder inequality we have that

$\begin{matrix} D (Fx (t), Fy (t)) \leq \frac{1}{Γ (q)} \int_{0}^{t} | (t - s)^{q - 1} | \\ D (f (s, x (s), (S x) (s)), f (s, y (s), (S y) (s))) d s \\ \leq \frac{1}{Γ (q)} \int_{0}^{t} | (t - s)^{q - 1} | \\ [m_{1} (s) D (x (t), y (t)) + m_{2} (s) D ((S x) (s), (S y) (s))] d s \\ \leq \frac{D^{*} (x (t), y (t))}{Γ (q)} \\ (\int_{0}^{t} | t - s |^{q - 1} [m_{1} (s) + γ_{0} T m_{2} (s)] d s) \\ \leq \frac{D^{*} (x (t), y (t))}{Γ (q)} (\int_{0}^{t} (t - s)^{\frac{q - 1}{1 - q_{1}}} d s)^{1 - q_{1}} \\ (\int_{0}^{t} [m_{1} (s) + γ_{0} T m_{2} (s)]^{\frac{1}{q_{1}}} d s)^{q_{1}} \\ \leq (\frac{M T^{q - q_{1}}}{Γ (q) (\frac{q - q_{1}}{1 - q_{1}})^{1 - q_{1}}}) D^{*} (x (t), y (t)) . \end{matrix}$ (17)

So we obtain D (Fx (t) , Fy (t)) ≤ Ω_q,q₁,TD^* (x (t) , y (t)). Thus, F is a contraction due to the condition (15). By Banach contraction principle, we can deduce that F has a unique fixed point which is just the unique -differentiable solution of the fractional initial value problem (9).

Let y (t) be differentiable. In this case, we define a operator $F : C^{F} [0, T] \to C^{F} [0, T]$ , by $\begin{matrix} (Fy) (t) \\ = y_{0} ⊖ \frac{- 1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} f (s, y (s), (S y) (s)) d s . \end{matrix}$

The proof for this type of differentiability will be obtained in similar manner and hence is omitted. □

Now that the existence and uniqueness of the solution for the Equation (9) were proved. In the following, we will use the proposed method for a particular case of Equation (9) in the form of Equation (23).

4 Legendre wavelets [22]

Wavelets constitute a family of functions constructed from dilation and translation of single function called the mother wavelet ψ (t). They are defined by $\begin{matrix} ψ_{a, b} (t) = \frac{1}{\sqrt{| a |}} ψ (\frac{t - b}{a}), a, b \in ℝ, \end{matrix}$ where a is dilation parameter and b is a translation parameter.

The Legendre polynomial of order m, denoted by L_m (t) are defined on the interval [-1, 1] and can be determined with aid of the following recurrence formula [12]:

$\begin{matrix} L_{0} (t) & = & 1, \\ L_{1} (t) & = & t, \\ L_{m + 1} (t) & = & \frac{2 m + 1}{m + 1} {tL}_{m} (t) - \frac{m}{m + 1} L_{m - 1} (t), \\ m = 1, 2, . . . \end{matrix}$ (18)

Relevant properties are: $\begin{matrix} | L_{m} (t) | & \leq & 1, - 1 \leq t \leq 1, \\ L_{m} (\pm 1) & = & (\pm 1)^{k}, \\ | L_{m}^{'} (t) | & \leq & \frac{1}{2} m (m + 1), - 1 \leq t \leq 1 \\ \int_{- 1}^{1} L_{m}^{2} (t) d t & = & (m + \frac{1}{2})^{- 1} . \end{matrix}$

Legendre wavelets ψ_n,m (t) = ψ (k, n, m, t) have four arguments, defined on interval [0, 1) by:

$\begin{matrix} ψ_{n, m} (t) \\ = {\begin{matrix} (m + \frac{1}{2})^{\frac{1}{2}} 2^{\frac{k}{2}} & \frac{n - 1}{2^{k - 1}} \leq t < \frac{n}{2^{k - 1}}, \\ L_{m} (2^{k} t - 2 n + 1), \\ 0, & elsewhere . \end{matrix} \end{matrix}$ (19) where $k \in ℤ^{+}$ , n = 1, 2, 3, . . . , 2^k-1 and m = 0, 1, . . . , M - 1 is the order of the Legendre polynomials and M is a fixed positive integer. The coefficient $\sqrt{m + \frac{1}{2}}$ is for orthonormality and the set of Legendre wavelets forms an orthogonal basis of $L^{2} (ℝ)$ . It may be noted that {ψ_n,i (x) , ψ_l,j (t) : n, l = 1, 2, 3, . . . , 2^k-1, i, j = 1, 2, 3, . . . , M - 1} is an orthonormal set over [0, 1) × [0, 1).

The two dimensional Legendre wavelets are defined as

$\begin{matrix} ψ_{n, i, l, j} (x, t) \\ = {\begin{matrix} A L_{i} (2^{k_{1}} x - 2 n + 1) L_{j} & \frac{n - 1}{2^{k_{1} - 1}} \leq x < \frac{n}{2^{k_{1} - 1}}, \\ (2^{k_{2}} t - 2 l + 1), \\ \frac{l - 1}{2^{k_{2} - 1}} \leq t < \frac{l}{2^{k_{2} - 1}}, \\ 0, & elsewhere . \end{matrix} \end{matrix}$ (20) where $A = \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} 2^{\frac{k_{1} + k_{2}}{2}}$ , n = 1, 2, . . . , 2^k₁-1, l = 1, 2, . . . , 2^k₂-1, k₁ and k₂ are any positive integers, i and j are the order for Legendre polynomials and ψ_n,i,l,j (x, t) forms a basis for L² ([0, 1) × [0, 1)).

4.1 Function approximation by Legendre wavelets

A function y (x, t) defined over [0, 1) × [0, 1) can be expanded in terms of Legendre wavelets as

$y (x, t) = \sum_{n = 1}^{\infty} \sum_{i = 0}^{\infty} \sum_{l = 1}^{\infty} \sum_{j = 0}^{\infty} c_{n, i, l, j} ψ_{n, i, l, j} (x, t) .$ (21)

If the infinite series in Equation (21) is truncated, then it can be written as:

$\begin{matrix} y (x, t) & = & \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (x, t) \\ = & C^{T} Ψ (x, t), \end{matrix}$ (22) where Ψ (x, t) is (2^k₁-12^k₂-1M₁M₂ × 1) matrix, given by $\begin{matrix} Ψ (x, t) \\ = [Ψ_{1, 0, 1, 0} (x, t), Ψ_{1, 0, 1, 1} (x, t), . . ., \\ Ψ_{1, 0, 1, M_{2} - 1} (x, t), . . ., Ψ_{1, 0, 2^{k_{2} - 1}, M_{2} - 1} (x, t), . . ., \\ Ψ_{2^{k_{1} - 1}, M_{1} - 1, 2^{k_{2} - 1}, M_{2} - 1} (x, t)] . \end{matrix}$

Also, C is (2^k₁-12^k₂-1M₁M₂ × 1) matrix whose elements can be calculated from the formula $\begin{matrix} c_{n, i, l, j} = \int_{0}^{1} \int_{0}^{1} ψ_{n, i} (x) ψ_{l, j} (t) y (x, t) d x d t, \end{matrix}$ and $\begin{matrix} C & = & [c_{1, 0, 1, 0}, c_{1, 0, 1, 1}, . . ., c_{1, 0, 1, M_{2} - 1}, \\ . . ., c_{1, 0, 2^{k_{2} - 1}, M_{2} - 1}, . . ., c_{2^{k_{1} - 1}, M_{1} - 1, 2^{k_{2} - 1}, M_{2} - 1}]^{T} \end{matrix}$

4.2 Operational matrix of the fractional order integration [24]

The Legendre wavelet operational matrix Q^q for integration of the fractional order q is given by $\begin{matrix} I_{a}^{q} Ψ (t) = [I_{a}^{q} ψ_{1, 0} (t), . . ., I_{a}^{q} ψ_{1, M - 1} (t), I_{a}^{q} ψ_{2, 0} (t), . . ., \\ I_{a}^{q} ψ_{2, M - 1} (t), . . ., I_{a}^{q} ψ_{2^{k - 1}, 0} (t), . . ., I_{a}^{q} ψ_{2^{k - 1}, M - 1} (t)]^{T} \end{matrix}$ where $\begin{matrix} I_{a}^{q} ψ_{n, m} (t) \\ = {\begin{matrix} (m + \frac{1}{2})^{\frac{1}{2}} 2^{\frac{k}{2}} I_{a}^{q} \\ (L_{m} (2^{k} t - 2 n + 1)), \\ \frac{n - 1}{2^{k - 1}} \leq t < \frac{n}{2^{k - 1}} \\ 0, & elsewhere . \end{matrix} \end{matrix}$ for $k \in ℤ^{+}$ , n = 1, 2, . . . , 2^k-1 and m = 0, 1, 2, . . . , M - 1 is the order of the Legendre polynomial and M is a fixed positive integer.

5 Legendre wavelet method for fuzzy fractional integro-differential equation

Consider the following fuzzy fractional integro-differential equation ${\begin{matrix} (_{gH} D_{*}^{q} y) (t) = y (t) + \int_{0}^{t} k (x, t) y (x) d x + g (t), \\ t \in 𝕁 = [0, 1) \\ y (0) = y_{0} \in ℝ_{F}, \end{matrix}$ (23) where y (t) and g (t) are fuzzy functions and $k (x, t) : 𝕁 \times 𝕁 \to ℝ$ and the sign of k (s, t) does not change in $𝕁$ . Applying $I_{0}^{q}$ on both sides of Equation (23), Using Lemma 3.1, we obtian: $\begin{matrix} y (t) \underset{gH}{⊖} y (0) \\ = I_{0}^{q} (y (t) + \int_{0}^{t} k (x, t) y (x) d x + g (t)), \end{matrix}$

By Lemma 3.2 we get:

$\begin{matrix} y (t) & = & y (0) + I_{0}^{q} \\ (y (t) + \int_{0}^{t} k (x, t) y (x) d x + g (t)), \end{matrix}$ (24) if y (t) is -differentiable, and:

$\begin{matrix} y (t) & = & y (0) ⊖ (- 1) I_{0}^{q} \\ (y (t) + \int_{0}^{t} k (x, t) y (x) d x + g (t)), \end{matrix}$ (25) if y (t) is -differentiable.

To find the numerical solution of the initial value problem (23), we consider the following expressions:

Case 1. Let assume that y (t) is -differentiable so by Theorem 2.8 and Equation (24), we have the following fractional integro-differential equations system: $\begin{matrix} Y (t, r) \\ = Y (0, r) + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} Y (s, r) d s \\ + \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) Y (x, r) d x d s \\ + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} G (s, r) d s \end{matrix}$ (26) where $Y (t, r) = (\begin{matrix} \underline{y} (t, r) \\ \bar{y} (t, r) \end{matrix})$ .

Case 1.1: Consider k (x, t) be a positive real-valued function.

Using Equation (26) clears we have: $\begin{matrix} \underline{y} (t, r) & = & \underline{y} (0, r) + \frac{1}{Γ (q)} (\int_{0}^{t} (t - s)^{q - 1} \underline{y} (s, r) d s \\ + \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \underline{y} (x, r) d x d s \\ + \int_{0}^{t} (t - s)^{q - 1} \underline{g} (s, r) d s) \end{matrix}$ (27) $\begin{matrix} \bar{y} (t, r) & = & \bar{y} (0, r) + \frac{1}{Γ (q)} (\int_{0}^{t} (t - s)^{q - 1} \bar{y} (s, r) d s \\ + \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \bar{y} (x, r) d x d s \\ + \int_{0}^{t} (t - s)^{q - 1} \bar{g} (s, r) d s) \end{matrix}$ (28) Now, we explain the propose method to solve Equation (27). In order to apply the Legendre wavelets defined in Equation (27), we first approximate the unknown function $\underline{y} (t, r)$ as: $\begin{matrix} \underline{y} (t, r) \\ = \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (t, r) \end{matrix}$ (29) $= C^{T} Ψ (t, r),$ (30) Putting Equation (29) in Equation (27) we obtain: $\begin{matrix} \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (t, r) \\ = \underline{y} (0, r) \frac{1}{Γ (q)} (\sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} \\ \int_{0}^{t} (t - s)^{q - 1} ψ_{n, i, l, j} (s, r) d s \\ \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} \\ \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) ψ_{n, i, l, j} (x, r) d x d s \\ + \int_{0}^{t} (t - s)^{q - 1} \underline{g} (s, r) d s) \end{matrix}$ (31) By calculating Equation (31) at (2^k₁-12^k₂-1M₁M₂) collocation points $t_{i} = \frac{2 i - 1}{2^{k_{1}} M_{1}}$ , $r_{j} = \frac{2 j - 1}{2^{k_{2}} - M_{2}}$ for i = 1, 2, . . . , 2^k₁-1M₁, j = 1, 2, . . . , 2^k₂-1M₂, we conclude that: $\begin{matrix} \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (t_{i}, r_{j}) \\ = \underline{y} (0, r_{j}) + \frac{1}{Γ (q)} (\sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} \\ \int_{0}^{t_{i}} (t_{i} - s)^{q - 1} ψ_{n, i, l, j} (s, r_{j}) d s \\ + \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} \\ \int_{0}^{t_{i}} \int_{0}^{s} (t_{i} - s)^{q - 1} k (x, s) ψ_{n, i, l, j} (x, r_{j}) d x d s \\ + \int_{0}^{t_{i}} (t_{i} - s)^{q - 1} \underline{g} (s, r_{j}) d s) \end{matrix}$ (32) In a similar way, for $\bar{y} (t, r)$ we have $\begin{matrix} \bar{y} (t, r) = \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j}^{'} ψ_{n, i, l, j} (t, r) . \end{matrix}$ (33) According to the process described above, we obtain the following equation: $\begin{matrix} \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j}^{'} ψ_{n, i, l, j} (t_{i}, r_{j}) \\ = \bar{y} (0, r_{j}) + \frac{1}{Γ (q)} (\sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j}^{'} \\ \int_{0}^{t_{i}} (t_{i} - s)^{q - 1} ψ_{n, i, l, j} (s, r_{j}) d s \\ + \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j}^{'} \\ \int_{0}^{t_{i}} \int_{0}^{s} (t_{i} - s)^{q - 1} k (x, s) ψ_{n, i, l, j} (x, r_{j}) d x d s \\ + \int_{0}^{t_{i}} (t_{i} - s)^{q - 1} \bar{g} (s, r_{j}) d s) . \end{matrix}$ (34) Equations (32) and (34) yield 2 (2^k₁-1M₁) (2^k₂-1M₂) equations in 2 (2^k₁-1M₁) (2^k₂-1M₂) unknowns in c_n,i,l,j and $c_{n, i, l, j}^{'}$ . By solving this system of equations using mathematical software, the Legendre wavelet coefficients c_n,i,l,j and $c_{n, i, l, j}^{'}$ can be obtained and hence substituting them in Equations (29) and (261), the approximate solutions can be obtained.

Case 1.2: Consider k (x, t) be a negative real-valued, then $\begin{matrix} \int_{0}^{t} k (x, t) Y (x, r) d x \\ = [\int_{0}^{t} k (x, t) \bar{y} (x, r) d x, \int_{0}^{t} k (x, t) \underline{y} (x, r) d x] \end{matrix}$ (35) Using Equations (26) and (35), we observe that: $\begin{matrix} (\begin{matrix} \underline{y} (t, r) \\ \bar{y} (t, r) \end{matrix}) = (\begin{matrix} \underline{y} (0, r) \\ \bar{y} (0, r) \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{y} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{y} (s, r) d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \bar{y} (x, r) d x d s \\ \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \underline{y} (x, r) d x d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{g} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{g} (s, r) d s \end{matrix}) . \end{matrix}$ According to the process described above, we get the values of unknown vectors c_n,i,l,j and $c_{n, i, l, j}^{'}$ and then obtain the solutions $\underline{y} (t, r)$ and $\bar{y} (t, r)$ .

Case 2. Now, consider y (t) is -differentiable, we have the following fractional integro-differential equations system: $\begin{matrix} Y (x, r) = Y (0, r) ⊖ (- 1) \\ (\frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} Y (s, r) d s \\ + \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) Y (x, r) d x d s \\ + \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} G (s, r) d s) . \end{matrix}$ (36)

Case 2.1: Let us consider k (x, t) be a positive real-valued function.

Using Equation (25) and definition of Hukuhara difference, system (36) can be written in the form $\begin{matrix} (\begin{matrix} \underline{y} (t, r) \\ \bar{y} (t, r) \end{matrix}) = (\begin{matrix} \underline{y} (0, r) \\ \bar{y} (0, r) \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{y} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{y} (s, r) d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \bar{y} (x, r) d x d s \\ \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \underline{y} (x, r) d x d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{g} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{g} (s, r) d s \end{matrix}) . \end{matrix}$ Then in similar way to previous case, we get the values of unknown vectors c_n,i,l,j and $c_{n, i, l, j}^{'}$ and then obtain the solutions $\underline{y} (t, r)$ and $\bar{y} (t, r)$ from Equations (29) and (261), respectively.

Case 2.2: Now consider k (x, t) is a negative real-valued function.

In accordance with the process described in Case 1.2 and by Equation (36), we observe that $\begin{matrix} (\begin{matrix} \underline{y} (t, r) \\ \bar{y} (t, r) \end{matrix}) = (\begin{matrix} \underline{y} (0, r) \\ \bar{y} (0, r) \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{y} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{y} (s, r) d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \underline{y} (x, r) d x d s \\ \frac{1}{Γ (q)} \int_{0}^{t} \int_{0}^{s} (t - s)^{q - 1} k (x, s) \bar{y} (x, r) d x d s \end{matrix}) \\ + (\begin{matrix} \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \bar{g} (s, r) d s \\ \frac{1}{Γ (q)} \int_{0}^{t} (t - s)^{q - 1} \underline{g} (s, r) d s \end{matrix}) . \end{matrix}$ So, by applying the method which is discussed in detail in previous case, we obtain a square system which, when solved, gives unknowns $\underline{y} (t, r)$ and $\bar{y} (t, r)$ .

6 Convergence analysis

The aim of this section is to analyze the numerical scheme Legendre wavelet.

Theorem 6.1. If y (x, t) defined on [0, 1) × [0, 1) and |y (x, t) | ≤ K, then the Legendre wavelets expansion of y (x, t) defined in Equation (22) converges uniformly and also $\begin{matrix} | c_{n, i, l, j} | \leq \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{4 K}{2^{\frac{k_{1} + k_{2}}{2}}} . \end{matrix}$

Proof. y (x, t) is a function defined on [0, 1) × [0, 1) and |y (x, t) | ≤ K, where K is a positive constant. The Legendre wavelet coefficients of continuous functions y (x, t) are defined as: $\begin{matrix} c_{n, i, l, j} \\ = \int_{0}^{1} \int_{0}^{1} y (x, t) ψ_{n, i} (x) ψ_{l, j} (t) d x d t \\ = \int_{I_{2}} \int_{I_{1}} y (x, t) \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} 2^{\frac{k_{1} + k_{2}}{2}} \\ L_{i} (2^{k_{1}} x - 2 n + 1) L_{j} (2^{k_{2}} t - 2 n + 1) d x d t, \end{matrix}$ where $I_{1} = [\frac{n - 1}{2^{k_{1} - 1}}, \frac{n}{2^{k_{1} - 1}})$ and $I_{2} = [\frac{l - 1}{2^{k_{2} - 1}}, \frac{l}{2^{k_{2} - 1}})$ . Now by change of variable 2^{k
₁}x - 2n + 1 = u we obtain: $\begin{matrix} c_{n, i, l, j} \\ = \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{2^{\frac{k_{1} + k_{2}}{2}}}{2^{k_{1}}} \\ \int_{I_{2}} \int_{- 1}^{1} y (\frac{u + 2 n - 1}{2^{k_{1}}}, t) L_{i} (u) {duL}_{j} \\ (2^{k_{2}} t - 2 n + 1) dt . \end{matrix}$

Similarly, changing the variable for t as 2^{k
₂}t - 2n + 1 = v, we get: $\begin{matrix} c_{n, i, l, j} & = & \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{2^{\frac{k_{1} + k_{2}}{2}}}{2^{k_{1} + k_{2}}} \\ \int_{- 1}^{1} \int_{- 1}^{1} y (\frac{u + 2 n - 1}{2^{k_{1}}}, \frac{v + 2 n - 1}{2^{k_{2}}}) \\ L_{i} (u) du L_{j} (v) dv . \end{matrix}$

Therefore, using the property of Legendre polynomials: $\begin{matrix} | c_{n, i, l, j} | & \leq & \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{1}{2^{\frac{k_{1} + k_{2}}{2}}} \\ \int_{- 1}^{1} \int_{- 1}^{1} | y (\frac{u + 2 n - 1}{2^{k_{1}}}, \frac{v + 2 n - 1}{2^{k_{2}}}) | \\ | L_{i} (u) | du | L_{j} (v) | dv \\ \leq & \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{K}{2^{\frac{k_{1} + k_{2}}{2}}} \\ (\int_{- 1}^{1} | L_{i} (u) | d u) (\int_{- 1}^{1} | L_{j} (v) | d v) \\ \leq & \sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{4 K}{2^{\frac{k_{1} + k_{2}}{2}}} . \end{matrix}$

Therefore $\sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j}$ is absolutely convergent and hence the series: $\begin{matrix} \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (x, t), \end{matrix}$ is uniformly convergent. □

Theorem 6.2. (Error estimate). If a continuous function $y (t, r) \in L^{2} (ℝ \times ℝ)$ defined on [0, 1) × [0, 1) be bounded via |y (t, r) | ≤ K, then $\begin{matrix} | | E_{n, i, l, j} (t, r) | |_{L^{2} ([0, 1) \times [0, 1))}^{2} \\ \leq \sum_{n = 2^{k_{1} - 1} + 1}^{\infty} \sum_{i = M_{1}}^{\infty} \sum_{l = 2^{k_{2} - 1} + 1}^{\infty} \sum_{j = M_{2}}^{\infty} \\ (\sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{4 K}{2^{\frac{k_{1} + k_{2}}{2}}})^{2} \end{matrix}$

Proof. Any function y (t, r) can be expressed by the Legendre polynomial. Now suppose that, we can clearly express the function y (t, r) as:

$y (t, r) = \sum_{n = 1}^{\infty} \sum_{i = 0}^{\infty} \sum_{l = 1}^{\infty} \sum_{j = 0}^{\infty} c_{n, i, l, j} ψ_{n, i, l, j} (t, r),$ (37)

$\begin{matrix} y^{*} (t, r) = \sum_{n = 1}^{2^{k_{1} - 1}} \sum_{i = 0}^{M_{1} - 1} \sum_{l = 1}^{2^{k_{2} - 1}} \sum_{j = 0}^{M_{2} - 1} c_{n, i, l, j} ψ_{n, i, l, j} (t, r), \end{matrix}$ (38) be the trucated series of (37), let: $\begin{matrix} E_{n, i, l, j} (t, r) = y (t, r) - y^{*} (t, r) \\ = \sum_{n = 2^{k_{1} - 1} + 1}^{\infty} \sum_{i = M_{1}}^{\infty} \sum_{l = 2^{k_{2} - 1} + 1}^{\infty} \sum_{j = M_{2}}^{\infty} c_{n, i, l, j} ψ_{n, i, l, j} (t, r) \end{matrix}$

Hence: $\begin{matrix} | | E_{n, i, l, j} (t, r) | |_{L^{2} ([0, 1) \times [0, 1))}^{2} \\ = \int_{0}^{1} \int_{0}^{1} | \sum_{n = 2^{k_{1} - 1} + 1}^{\infty} \sum_{i = M_{1}}^{\infty} \sum_{l = 2^{k_{2} - 1} + 1}^{\infty} \\ \sum_{j = M_{2}}^{\infty} c_{n, i, l, j} ψ_{n, i} (t) ψ_{l, j} (r) |^{2} dtdr \\ \leq \sum_{n = 2^{k_{1} - 1} + 1}^{\infty} \sum_{i = M_{1}}^{\infty} \sum_{l = 2^{k_{2} - 1} + 1}^{\infty} \sum_{j = M_{2}}^{\infty} | c_{n, i, l, j} |^{2} \\ \int_{0}^{1} \int_{0}^{1} | ψ_{n, i} (x) ψ_{l, j} (r) |^{2} dtdr . \end{matrix}$

Using Theorem 6.1 we have: $\begin{matrix} | | E_{n, i, l, j} (t, r) | |_{L^{2} ([0, 1) \times [0, 1))}^{2} \\ \leq \sum_{n = 2^{k_{1} - 1} + 1}^{\infty} \sum_{i = M_{1}}^{\infty} \sum_{l = 2^{k_{2} - 1} + 1}^{\infty} \sum_{j = M_{2}}^{\infty} \\ (\sqrt{(i + \frac{1}{2}) (j + \frac{1}{2})} \frac{4 K}{2^{\frac{k_{1} + k_{2}}{2}}})^{2} . \end{matrix}$

That is establishing the claim. □

7 Numerical experiments

In this section, we will use the above proposed method to solve two different examples. The computations associated with the examples are performed using Mathematica software.

Example 7.1. Consider the following fuzzy fractional integro differential equation:

${\begin{matrix} (_{gH} D_{*}^{\frac{1}{2}} y) (t) = \frac{[r + 1, 5 - 3 r]}{15 \sqrt{π}} t^{\frac{5}{2}} (48 - \sqrt{π} t^{\frac{7}{2}}) \\ + \int_{0}^{t} \frac{xt}{3} y (x) d x, \\ y (0) = 0, \end{matrix}$ (39) where 0 denotes the crisp set {0} and $\tilde{2} = (1, 2, 5)$ is a triangular fuzzy number such that $[\tilde{2}]_{r} = [r + 1, 5 - 3 r]$ . The exact solution of Equation (39) is given by $y (t) = \tilde{2} t^{3}$ . The exact solution is plotted in Fig. 1 and its $^{g H} D_{*}^{\frac{1}{2}} y (t)$ is plotted in Fig. 2. As it is seen, y (t) is -differentiable, so by applying the method discussed in detail in Section 5, this problem has been solved by Legendre polynomial for M₁ = M₂ = 4, k₁ = k₂ = 1 and calculate the absolute errors as |E_r| = |y (t, r) - y^* (t, r) |. Table 1 shows the approximate solutions obtained by Legendre wavelets method.

Fig.1

The level sets of y (t) of Example 7.1.

Fig.2

The level sets of $^{g H} D_{*}^{\frac{1}{2}} y (t)$ (Right) of Example 7.1.

Table 1

Numerical results for Example 7.1

r	t	$\underline{y} (t, r)$	$\bar{y} (t, r)$
		M₁ = M₂ = 4	M₁ = M₂ = 4
0.3	0.3	3.81639 × 10^–16	6.38378 × 10^–16
	0.6	9.71445 × 10^–17	7.63278 × 10^–16
	0.9	8.18789 × 10^–16	1.31839 × 10^–16
0.6	0.3	1.66533 × 10^–16	9.99201 × 10^–16
	0.6	3.33067 × 10^–16	4.44089 × 10^–16
	0.9	5.55112 × 10^–17	4.99693 × 10^–16
0.9	0.3	2.22045 × 10^–16	1.77636 × 10^–15
	0.6	0	0
	0.9	2.22045 × 10^–16	2.22045 × 10^–16

It is evident from the Table 1, that the numerical solution converge to the exact solution. It is also concluded that the proposed method is very efficient for numerical solutions of these problems.

Example 7.2. Consider the following initial value problem $\begin{matrix} {\begin{matrix} (_{gH} D_{*}^{\frac{1}{3}} y) (t) = \frac{1}{6} e^{t} t (6 + 3 t + t^{2}) [8 - 3 r, 2 + 3 r] \\ + \frac{3 (5 + 3 t) t^{\frac{2}{3}}}{10 Γ [\frac{2}{3}]} + \int_{0}^{t} - e^{x} y (x) d x, \\ y (0) = [2 + 3 r, 8 - 3 r] . \end{matrix} \end{matrix}$

The exact solution is given by $y (t) = [2 + 3 r, 8 - 3 r] (\frac{x^{2}}{2} + x + 1)$ . Since y (t) is differentiable and k (x, t) = - e^x, therefore by applying the method which is discussed in detail Case 1.2, we have Table 2.

Table 2

Numerical results for Example 7.2

r	t	$\underline{y} (t, r)$	$\bar{y} (t, r)$
		M₁ = M₂ = 4	M₁ = M₂ = 4
0.3	0.3	8.88178 × 10^–16	0
	0.6	1.77636 × 10^–15	8.88178 × 10^–15
	0.9	5.32907 × 10^–15	8.88175 × 10^–15
0.6	0.3	4.44089 × 10^–15	3.55271 × 10^–15
	0.6	2.66454 × 10^–15	3.55271 × 10^–15
	0.9	8.88178 × 10^–15	5.32907 × 10^–15
0.9	0.3	1.77636 × 10^–15	0
	0.6	1.95399 × 10^–14	1.06581 × 10^–14
	0.9	1.42109 × 10^–14	5.50671 × 10^–14

Example 7.3. Consider the following fractional integro-differential equation $\begin{matrix} {\begin{matrix} (_{gH} D_{*}^{\frac{1}{2}} y) (t) = - [0.25 + 0.75 r, 2.5 - 1.5 r] \\ \frac{e^{- t}}{\sqrt{π}} \int_{0}^{t} \frac{e^{s}}{\sqrt{s}} d s \\ ⊖ [0.25 + 0.75 r, 2.5 - 1.5 r] t + \int_{0}^{t} e^{x} y (x) d x, \\ y (0) = [0.25 + 0.75 r, 2.5 - 1.5 r] . \end{matrix} \end{matrix}$

The exact solution of this equation is y (t, r) = [0.25 + 0.75r, 2.5 - 1.5r] e^–t By considering to Figs. 3 and 4, y (t) is -differentiable.

Fig.3

The level sets of y (t) of Example 7.2.

Fig.4

The level sets of $^{g H} D_{*}^{\frac{1}{2}} y (t)$ of Example 7.3.

By applying the Legendre wavelet method discussed in detail in Section 5, we approximate the -differentiable solution with k₁ = k₂ = 1 and M₁ = M₂ = 8. The numerical result are shown in Table 3.

Table 3

Numerical results for Example 7.3

r	t	$\underline{y} (t, r)$	$\bar{y} (t, r)$
		M₁ = M₂ = 8	M₁ = M₂ = 8
0.3	0.3	1.85211 × 10^–10	1.03556 × 10^–10
	0.6	1.54845 × 10^–10	1.08187 × 10^–10
	0.9	1.24485 × 10^–10	1.12842 × 10^–10
0.6	0.3	2.43887 × 10^–10	5.17273 × 10^–11
	0.6	1.89542 × 10^–10	7.97329 × 10^–11
	0.9	1.35185 × 10^–10	1.07727 × 10^–10
0.9	0.3	1.11655 × 10^–10	8.68369 × 10^–10
	0.6	6.46392 × 10^–11	6.24644 × 10^–10
	0.9	2.40913 × 10^–10	3.80909 × 10^–10

8 Conclusion

In the present paper, the two dimensional Legendre wavelet method was applied to approximate the solution of fuzzy fractional integro-differential equation. We transformed our problem to a system of algebraic equations that by solving this system, we obtained the solution this kind equation by considering the type of differentiability. Finally, we established the convergence of the Legendre wavelet approximation. The numerical results confirmed that the two dimensional Legendre wavelet method is a powerful mathematical tool for the exact and numerical solutions of the fuzzy fractional integro-differential equation in terms of accuracy and efficiency.

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