Error estimation and numerical solution of nonlinear fuzzy Fredholm integral equations of the second kind using triangular functions

Abstract

In this paper we present an efficient iterative procedure based on the triangular functions (TFs) to obtain the numerical solution of the specific nonlinear fuzzy Fredholm integral equations of the second kind. The error estimation of the proposed method is given in terms of uniform modulus of continuity. Also, we prove the stability of the method with respect to the choice of the first iteration. Finally, in order to demonstrate the accuracy and the convergence of the method, some numerical examples are included.

Keywords

Nonlinear fuzzy Fredholm integral equation iterative procedure triangular functions numerical stability

1 Introduction

Fuzzy integral equations are important for studying and solving a large proportion of the problems in many topics in applied mathematics, particularly in relation to fuzzy control. The study of fuzzy integral equations began with the investigations of Kaleva [30] and Siekkala [47] for the fuzzy Volterra integral equations which were equivalent to the initial value problem for the first order fuzzy differential equations. These concepts were developed by many researchers [12 , 50]. The Banach fixed point theorem is the main tool in studying existence and uniqueness of the solution for fuzzy Fredholm integral equations (FFIE), which is carried out in [9 , 40–42].

The numerical methods for solving fuzzy integral equations based on the method of successive approximations and other iterative techniques are applied in [12 , 43]. Bica proved the convergence method of successive approximations used to approximate the solution of nonlinear Hammerstein fuzzy integral equations in [16]. In [19], an iterative procedure based on the trapezoidal quadrature for solving nonlinear FFIE is proposed. Recently, in [8], the authors presented a new approach based on the hybrid of block-pulse functions and Taylor series for solving nonlinear FFIE.

Some noticeable techniques which used in the construction of the numerical methods for solving fuzzy integral equations are: quadrature rules and Nyström methods [2, 31], Lagrange interpolation [4], Chebyshev interpolation [11], divided and finite differences [39] and Galerkin type techniques [33]. Also, in this manner, there are some methods that used Bernstein polynomials [20, 37], Legendre wavelets [46], fuzzy Haar wavelets [21, 52] and block pulse functions [44]. Analytic-numeric methods like Adomian decomposition, homotopy analysis and homotopy perturbation are used in [1 , 35].

TFs have been presented by Deb et al. [17] in 2006, which studied and used by Babolian et al. [7] for solving integral equations. In [34], a numerical method to solve linear FFIE of the second kind by using TFs is proposed. As we know, most of the methods to solve integral equations lead us to solve the linear systems, but the singularity of these systems may be caused problems, so using successive approximations based on TFs can be very useful for solving such equations. In the present paper, we introduce an iterative procedure based on TFs for solving nonlinear FFIE $F (t) = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y (F (s)) ds, t \in [0, 1] .$ (1) In addition, we prove the convergence and the numerical stability of the proposed method with respect to the choice of the first iteration.

The rest of this paper is organized as follows: In Section 2, we will review some elementary concepts of the fuzzy set theory and it presents some properties for fuzzy-number-valued functions and modulus of continuity. Section 3 is devoted to the study of TFs and approximation of fuzzy function by TFs. In Section 4, we present not only an iterative method to obtain numerical solutions of Equation.(1) based on TFs but also an error estimation of the introduced method in terms of modulus of continuity to prove the convergence of the method. Finally, we will prove the numerical stability of the presented method with respect to the choice of the first iteration. Some numerical experiments confirm the theoretical results and illustrate the accuracy of the method, which are presented in Section 5.

2 Preliminaries

From the various definitions of the concept of fuzzy number, we choose for this paper the following one:

Definition 1. [18] A fuzzy number is a function $u : ℝ \to [0, 1]$ satisfying the following properties:

u is normal, i.e. $\exists x_{0} \in ℝ$ with u (x₀) = 1.

u is a convex fuzzy set, i.e. $u (λ x + (1 - λ) y) \geq min {u (x), u (y)}, \forall x, y \in ℝ, λ \in [0, 1]$ .

u is upper semi-continuous on $ℝ$ .

the $[u]_{0} = \bar{{x \in ℝ : u (x) > 0}}$ is compact, here $\bar{A}$ denotes the closure of A.

The set of all fuzzy real numbers is denoted by $ℝ_{ℱ}$ . Any real number $a \in ℝ$ can be interpreted as a fuzzy number $\tilde{a} = χ_{{a}}$ and therefore $ℝ \subset ℝ_{ℱ}$ . For 0 < r ≤ 1, we denote the r-level (or simply the r-cut) set ${[u]}_{r} = {x \in ℝ : u (x) \geq r}$ , that is a closed interval (see [49]) and ${[u]}_{r} = [{\underline{u}}_{r}, {\bar{u}}_{r}], \forall r \in [0, 1]$ . These lead to the usual parametric representation of a fuzzy number. According to [13], for any 0 < r ≤ 1 the fuzzy number u is determined by an ordered pair of function $(\underline{u} (r), \bar{u} (r))$ which $\underline{u}, \bar{u} : [0, 1] \to ℝ$ satisfies the three conditions:

$\underline{u} (r)$ is a bounded monotonic non decreasing left-continuous function ∀r ∈]0, 1] and right-continuous for r = 0;

$\bar{u} (r)$ is a bounded monotonic non increasing left-continuous function ∀r ∈]0, 1] and right-continuous for r = 0;

$\underline{u} (r) \leq \bar{u} (r), \forall r \in [0, 1]$ .

For $u, v \in ℝ_{ℱ}, k \in ℝ$ , the addition and the scalar multiplication operations of fuzzy number are defined as follow:

${[u \oplus v]}_{r} = {[u]}_{r} + {[v]}_{r} = [{\underline{u}}_{r} + {\underline{v}}_{r}, {\bar{u}}_{r} + {\bar{v}}_{r}] \forall r \in [0, 1]$

${[k ⊙ u]}_{r} = k \cdot {[u]}_{r} = {\begin{matrix} [k {\underline{u}}_{r}, k {\bar{u}}_{r}], k \geq 0, \\ [k {\bar{u}}_{r}, k {\underline{u}}_{r}], k < 0 . \end{matrix}$

The subtraction of fuzzy numbers u ⊖ v is defined as the addition u ⊕ (- v) where (- v) = (- 1) ⊙ v. The standard Hukuhara difference (H-difference ⊖_H) is defined by u ⊖ _Hv = w ⇔ u = v ⊕ w; if u ⊖ _Hv exists, its r-cuts are ${[u ⊖_{H} v]}_{r} = [{\underline{u}}_{r} - {\underline{v}}_{r}, {\bar{u}}_{r} - {\bar{v}}_{r}]$ . It is well-known that $u ⊖_{H} u = \tilde{0}$ for all fuzzy numbers u, but $u ⊖ u \neq \tilde{0}$ .

According to [12, 49], the following algebraic properties hold:

u ⊕ (v ⊕ w) = (u ⊕ v) ⊕ w and u ⊕ v = v ⊕ u for any $u, v, w \in ℝ_{ℱ}$ ,

$u \oplus \tilde{0} = \tilde{0} \oplus u = u$ for any $u \in ℝ_{ℱ}$ ,

with respect to $\tilde{0}$ , none $u \in ℝ_{ℱ} ∖ ℝ, u \neq \tilde{0}$ has opposite in $(ℝ_{ℱ}, \oplus)$ ,

for any $a, b \in ℝ$ with ab ≥ 0 and any $u \in ℝ_{ℱ}$ , (a + b) ⊙ u = a ⊙ u ⊕ b ⊙ u,

for any $a \in ℝ$ and any $u, v \in ℝ_{ℱ}$ , a ⊙ (u ⊕ v) = a ⊙ u ⊕ a ⊙ v,

for any $a, b \in ℝ$ and any $u \in ℝ_{ℱ}$ , a ⊙ (b ⊙ u) = (ab) ⊙ u and 1 ⊙ u = u.

As a distance between fuzzy numbers, we use the Hausdorff metric (see [2]) defined by $D (u, v) = sup_{r \in [0, 1]} {max (| {\underline{u}}_{r} - {\underline{v}}_{r} |, | {\bar{u}}_{r} - {\bar{v}}_{r} |)}$ for any $u, v \in ℝ_{ℱ}$ . The Hausdorff metric has the following properties [27, 49]:

$(ℝ_{ℱ}, D)$ is a complete metric space,

$D (u \oplus w, v \oplus w) = D (u, v) \forall u, v, w \in ℝ_{ℱ}$ ,

D (u ⊕ v, w ⊕ e) ≤ D (u, w) + D (v, e) ∀ u, v, $w, e \in ℝ_{ℱ}$ ,

$D (u \oplus v, \tilde{0}) \leq D (u, \tilde{0}) + D (v, \tilde{0})$ $\forall u, v \in ℝ_{ℱ}$ ,

$D (k ⊙ u, k ⊙ v) = | k | D (u, v) \forall u, v \in ℝ_{ℱ}$ $\forall k \in ℝ$ .

$D (k_{1} ⊙ u, k_{2} ⊙ u) = | k_{1} - k_{2} | D (u, \tilde{0}) \forall k_{1},$ $k_{2} \in ℝ$ with k₁ · k₂ ≥ 0 and $\forall u \in ℝ_{ℱ}$ .

The property (iv) suggest the definition of a function $∥ \cdot ∥ : ℝ_{ℱ} \to ℝ$ by $∥ u ∥ = D (u, \tilde{0})$ that has the properties of usual norms. In [12], the properties of this function are presented as follows:

$∥ u ∥ \geq 0 \forall u \in ℝ_{ℱ}$ and ∥u ∥ = 0 iff $u = \tilde{0}$ ,

∥λ⊙ u ∥ = |λ| · ∥ u ∥ and ∥u⊕ v ∥ ≤ ∥ u ∥ + $∥ v ∥ \forall u, v \in ℝ_{ℱ} \forall λ \in ℝ$ ,

| ∥ u ∥ - ∥ v ∥ | ≤ D (u, v) and $D (u, v) \leq ∥ u ∥ + ∥ v ∥ \forall u, v \in ℝ_{ℱ}$ .

We see that $(ℝ_{ℱ}, \oplus, ⊙, ∥ \cdot ∥)$ is not a normed space because $(ℝ_{ℱ}, \oplus)$ it is not a group.

Definition 2. Regarding to [14] and with our notation, if $u = (\underline{u} (r), \bar{u} (r))$ , $v = (\underline{v} (r), \bar{v} (r))$ are fuzzy numbers, then

$(u \oplus v) = (\underline{u} (r), \bar{u} (r)) \oplus (\underline{v} (r), \bar{v} (r)) = (\underline{u} (r) + \underline{v} (r), \bar{u} (r) + \bar{v} (r)) = ((\underline{u} + \underline{v}) (r), (\bar{u} + \bar{v}) (r))$ ,

$(u ⊙ v) = (\underline{u} (r), \bar{u} (r)) ⊙ (\underline{v} (r), \bar{v} (r)) = (\underline{u} (r) \cdot \underline{v} (r), \bar{u} (r) \cdot \bar{v} (r)) = ((\underline{u} \cdot \underline{v}) (r), (\bar{u} \cdot \bar{v}) (r))$ , for all u, v with positive supports.

$ψ (u) = ψ (\underline{u} (r), \bar{u} (r)) = (ψ (\underline{u} (r)), ψ (\bar{u} (r)))$ , for all strictly increasing and positive real function ψ.

Definition 3. For any fuzzy-number-valued function $f : I \subset ℝ \to ℝ_{ℱ}$ we can define the functions ${\underline{f}}_{r}, {\bar{f}}_{r} : I \subset ℝ \to ℝ, r \in [0, 1]$ by ${\underline{f}}_{r} (t) = {\underline{(f (t))}}_{r}, {\bar{f}}_{r} (t) = {\bar{(f (t))}}_{r}, \forall t \in [0, 1]$ . These functions are the left and right r-level functions of f.

Definition 4. [51] A fuzzy-number-valued function $f : [a, b] \to ℝ_{ℱ}$ is said to be continuous at t₀ ∈ [a, b] if for each ɛ > 0 there is δ > 0 such that D (f (t) , f (t₀)) < ɛ whenever |t - t₀| < δ. If f is continuous for each t ∈ [a, b] then we say that f is fuzzy continuous on [a, b], and denote the space of all such functions by C_ℱ [a, b]. $f : [a, b] \to ℝ_{ℱ}$ is bounded iff there is M ≥ 0 such that $D (f (t), \tilde{0}) = ∥ f (t) ∥ \leq M$ for all t ∈ [a, b].

Definition 5. [22] A fuzzy-number-valued function $f : [a, b] \to ℝ_{ℱ}$ is said to be uniformly continuous on [a, b], if for each ɛ > 0 there is δ > 0 such that D (f (t) , f (t′)) < ɛ whenever t, t′ ∈ [a, b] with|t - t′| < δ.

Lemma 1. [3] If $f, g : [a, b] \subseteq ℝ \to ℝ_{ℱ}$ are fuzzy continuous functions, then the function $F : [a, b] \to ℝ_{+}$ defined by F (x) = D (f (x) , g (x)) is continuous on [a, b]. Also $D (f (x), \tilde{0}) \leq M, \forall x \in [a, b], M > 0$ , that is f is fuzzy bounded.

Corollary 1. [45] If $g : ℝ \to ℝ_{ℱ}$ and $f : ℝ_{ℱ} \to ℝ_{ℱ}$ are fuzzy continuous functions, then the function $F : ℝ \to ℝ_{ℱ}$ defined by F (t) = (fog) (t) = f (g (t)) is continuous on $ℝ$ .

Let C_ℱ [a, b], be the space of fuzzy continuous functions with the metric $D^{*} (f, g) = sup_{a \leq t \leq b} D (f (t), g (t)), \forall f, g \in C_{ℱ} [a, b]$ that it is called the uniform distance between fuzzy-number-valued functions. We see that (C_ℱ [a, b] , D^*) is a complete metric space.

Definition 6. [12] Let $f : [a, b] \to ℝ_{ℱ}$ be a bounded mapping. Then the function $ω_{[a, b]} (f, .) : ℝ_{+} \cup 0 \to ℝ_{+}$ $ω_{[a, b]} (f, δ) = sup {D (f (x), f (y)) : x, y \in [a, b], | x - y | \leq δ}$ is said to be the modulus of oscillation of f on [a, b].

If f ∈ C_ℱ [a, b], then ω_[a,b] (f, δ) is called uniform modulus of continuity of f. Some properties of the modulus of oscillation are given below: According to [12] the following statements are true:

D (f (x) , f (y)) ≤ ω_[a,b] (f, |x - y|) for any x, y ∈ [a, b],

ω_[a,b] (f, δ) is a non-decreasing mapping in δ,

ω_[a,b] (f, 0) = 0,

ω_[a,b] (f, δ₁ + δ₂) ≤ ω_[a,b] (f, δ₁) + ω_[a,b] (f, δ₂) for any δ₁, δ₂ ≥ 0,

ω_[a,b] (f, nδ) ≤ nω_[a,b] (f, δ) for any δ ≥ 0 and $n \in ℕ$ ,

ω_[a,b] (f, λδ) ≤ (λ + 1) ω_[a,b] (f, δ) for any δ, λ ≥ 0,

If [c, d] ⊆ [a, b] then ω_[c,d] (f, δ) ≤ ω_[a,b](f, δ).

In [49] the notion of Henstock integral for fuzzy-number-valued functions is defined as follows:

Definition 7. [12] Let $f : [a, b] \to ℝ_{ℱ}$ . For ▵_n : a = x₀ < x₁ < ⋯ < x_n-1 < x_n = b a partition of the interval [a, b], we consider the points ξ_i ∈ [x_i-1, x_i] , i = 1, ⋯ , n, and function $δ : [a, b] \to ℝ_{+}$ . The partition P = { ([x_i-1, x_i] ; ξ_i) ; i = 1, ⋯ , n } denoted by P = (▵ _n, ξ) is called δ-fine iff [x_i-1, x_i] ⊆ (ξ_i - δ (ξ_i) , ξ_i + δ (ξ_i)). For $I \in ℝ_{ℱ}$ , the function f is fuzzy Henstock integrable on [a, b] if for any ɛ > 0 there is a function $δ : [a, b] \to ℝ_{+}$ such that for any partition δ-fine P, $D (\sum_{i = 1}^{n} (x_{i} - x_{i - 1}) ⊙ f (ξ_{i}), I) < ɛ$ . The fuzzy number I is named the fuzzy Henstock integral of f and will be denoted by $(FH) \int_{a}^{b} f (t) dt$ .

When the function $δ : [a, b] \to ℝ_{+}$ is constant, then we obtain the Riemann integrability for fuzzy-number-valued functions (see [28]). In this case, $I \in ℝ_{ℱ}$ is called the fuzzy Riemann integral of f on the interval [a, b], being denoted by $(FR) \int_{a}^{b} f (t) dt$ . Consequently, the fuzzy Riemann integrability is a particular case of the fuzzy Henstock integrability, and therefore the properties of the integral (FH) will be valid for the integral (FR), too.

Lemma 2. [29] Let $f : [a, b] \to ℝ_{ℱ}$ . Then f is (FH) integrable if and only if ${\underline{f}}_{r}$ and ${\bar{f}}_{r}$ are Henstock integrable for any r ∈ [0, 1]. Furthermore, for any r ∈ [0, 1], ${[(FH) \int_{a}^{b} f (t) dt]}_{r} = [(H) \int_{a}^{b} {\underline{f}}_{r} (t) dt, (H) \int_{a}^{b} {\bar{f}}_{r} (t) dt] .$

Remark 1. If $f : [a, b] \to ℝ_{ℱ}$ is fuzzy continuous, then ${\underline{f}}_{r}$ and ${\bar{f}}_{r}$ are continuous for any r ∈ [0, 1] and consequently, they are Henstock integrable. According to Lemma 2 we infer that f is (FH) integrable.

Lemma 3. [12] If f and g are fuzzy Henstock integrable functions and if the function given by D (f (t) , g (t)) is Lebesgue integrable, then $\begin{matrix} D ((FH) \int_{a}^{b} f (t) dt, (FH) \int_{a}^{b} g (t) dt) \\ \leq (L) \int_{a}^{b} D (f (t), g (t)) dt . \end{matrix}$ Lemma 4. [28] If $f, g : [a, b] \to ℝ_{ℱ}$ are (FR) integrable fuzzy functions, and α, β are real numbers, then $\begin{matrix} (FR) \int_{a}^{b} (α ⊙ f (t) \oplus β ⊙ g (t)) dt \\ = α ⊙ (FR) \int_{a}^{b} f (t) dt \oplus β ⊙ (FR) \int_{a}^{b} g (t) dt . \end{matrix}$

Lemma 5. [16] If f ∈ C_ℱ ([a, b] × [a, b]) , g ∈ C_ℱ [a, b], and $α \in C ([a, b], ℝ_{+})$ then the functions $α \cdot g : [a, b] \to ℝ_{ℱ}$ and $F : [a, b] \to ℝ_{ℱ}$ given by (α · g) (t) = α (t) · g (t) , ∀ t ∈ [a, b] and $F (t) = (FH) \int_{a}^{b} f (t, s) ds$ are continuous.

Remark 2. If $f : [a, b] \to ℝ_{ℱ}$ is fuzzy continuous, for a partition ▵ : a = x₀< x₁ < ⋯ < x_n-1 <x_n = b, according to [49], fuzzy-Riemann integral has the property $(FR) \int_{a}^{b} f (t) dt = \sum_{i = 0}^{n - 1} * (FR) \int_{x_{i}}^{x_{i + 1}} f (t) dt .$ where Σ^* means addition with respect to ⊕ in $ℝ_{ℱ}$ .

Definition 8. [12] For L ≥ 0, a function $f : [a, b] \to ℝ_{ℱ}$ is L-Lipschitz if $D (f (x), f (y)) \leq L ∣ x - y ∣,$ for any x, y ∈ [a, b] .

3 Review of Triangular functions

In this section, we recall some definitions, notations and facts of the TFs. Also, we generalize them into fuzzy setting.

Definition 9. [17] Two m-sets of TFs are defined over the interval [0, 1) as: $\begin{matrix} φ_{i} (t) = {\begin{matrix} 1 + i - mt, \frac{i}{m} \leq t < \frac{i + 1}{m} \\ 0, otherwise \end{matrix} \\ ψ_{i} (t) = {\begin{matrix} mt - i, \frac{i}{m} \leq t < \frac{i + 1}{m} \\ 0, otherwise \end{matrix} \end{matrix}$ where m is an arbitrary positive integer and i = 0, 1, ⋯ , m - 1. Also, φ_i as the ith left-hand Triangular function and ψ_i as the ith right-hand Triangular function. From the definition of TFs, we can write $\int_{0}^{1} φ_{i} (t) φ_{j} (t) dt = \int_{0}^{1} ψ_{i} (t) ψ_{j} (t) dt = {\begin{matrix} \frac{1}{3 m} i = j \\ 0 i \neq j . \end{matrix}$

Therefore TFs are disjoint, orthogonal and complete, [17].

The TFs vectors functions Φ (t) , Ψ (t) left-handed TFs vector and right-hand TFs vector, respectively, are defined as the following $Φ (t) = {[φ_{0} (t), φ_{1} (t), \dots, φ_{m - 1} (t)]}^{T}$ $Ψ (t) = {[ψ_{0} (t), ψ_{1} (t), \dots, ψ_{m - 1} (t)]}^{T} .$

For f ∈ C_ℱ [0, 1], we consider fuzzy polynomial as follow $\begin{matrix} {TF}_{m}^{ℱ} (f) (t) & = \sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t) + \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t) \\ = c^{T} ⊙ Φ (t) + d^{T} ⊙ Ψ (t), \end{matrix}$ where $c_{i} = f (\frac{i}{m})$ and $d_{i} = f (\frac{i + 1}{m})$ for i = 0, 1, ⋯ , m - 1.

Theorem 1. If $f : [0, 1] \to ℝ_{ℱ}$ is a continuous function, then ${TF}_{m}^{ℱ} (f) : [0, 1] \to ℝ_{ℱ}$ is continuous. In addition, the following inequalities holds: $D ({TF}_{m}^{ℱ} (f) (t), f (t)) \leq ω_{[0, 1]} (f, \frac{1}{m}) \forall t \in [0, 1],$ (2) $D ({TF}_{m}^{ℱ} (f) (t), \tilde{0}) \leq max_{0 \leq t \leq 1} D (f (t), \tilde{0}) .$ (3)

Proof. For t₀ ∈ [0, 1] we have the following two possibilities: $(i) t_{0} \in (\frac{j}{m}, \frac{j + 1}{m}), (ii) t_{0} = \frac{j}{m} (j = 0, 1, \dots, m - 1)$ Since f ∈ C_ℱ [0, 1], by Lemma 1, f is bounded and there is M ≥ 0 such that $D (f (t), \tilde{0}) \leq M$ for all t ∈ [0, 1]. Furthermore for all arbitrary ɛ > 0, there exists δ (ɛ) >0 which can be $δ (ɛ) < \frac{ɛ}{2 Mm}$ such for any t ∈ [0, 1] with |t - t₀| < δ (ɛ) it follows that D (f (t) , f (t₀)) < ɛ.

For the case (i), we get that $t, t_{0} \in (\frac{j}{m}, \frac{j + 1}{m})$ and |t - t₀| < δ. Then $\begin{matrix} D ({TF}_{m}^{ℱ} (f) (t), {TF}_{m}^{ℱ} (f) (t_{0})) \\ = D (\sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t) \oplus \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t), \\ \sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t_{0}) \oplus \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t_{0})) \\ = D (c_{j} ⊙ φ_{j} (t) \oplus d_{j} ⊙ ψ_{j} (t), c_{j} ⊙ φ_{j} (t_{0}) \oplus d_{j} ⊙ ψ_{j} (t_{0})) \\ = D (c_{j} ⊙ φ_{j} (t), c_{j} ⊙ φ_{j} (t_{0})) + D (d_{j} ⊙ ψ_{j} (t), d_{j} ⊙ ψ_{j} (t_{0})) \\ \leq | φ_{j} (t) - φ_{j} (t_{0}) | D (c_{j}, \tilde{0}) + | ψ_{j} (t) - ψ_{j} (t_{0}) | D (d_{j}, \tilde{0})_ \\ = m | t - t_{0} | D (c_{j}, \tilde{0}) + m | t - t_{0} | D (d_{j}, \tilde{0}) \\ = 2 Mm | t - t_{0} | \leq 2 Mm δ < ɛ \end{matrix}$

For the case (ii), first we get $t \in (\frac{j}{m}, \frac{j + 1}{m})$ and |t - t₀| < δ, then as the proof of above case (i), we have: $D ({TF}_{m}^{ℱ} (f) (t), {TF}_{m}^{ℱ} (f) (t_{0})) < ɛ .$

As the second possible, we consider $t \in (\frac{j - 1}{m}, \frac{j}{m})$ , then we have $\begin{matrix} D ({TF}_{m}^{ℱ} (f) (t), {TF}_{m}^{ℱ} (f) (t_{0})) \\ = D (\sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t) \oplus \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t), \\ \sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t_{0}) \oplus \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t_{0})) \\ = D (c_{j - 1} ⊙ φ_{j - 1} (t) \oplus d_{j - 1} ⊙ ψ_{j - 1} (t), \\ c_{j} ⊙ φ_{j} (t_{0}) \oplus d_{j} ⊙ ψ_{j} (t_{0})) \\ \leq D (c_{j} ⊙ φ_{j - 1} (t), d_{j} ⊙ ψ_{j} (t_{0})) \\ + D (d_{j - 1} ⊙ ψ_{j - 1} (t), c_{j} ⊙ φ_{j} (t_{0})) \\ = D (f (\frac{j - 1}{m}) ⊙ (j - mt), f (\frac{j + 1}{m}) ⊙ (m t_{0} - j)) \\ + D (f (\frac{j}{m}) ⊙ (mt - j + 1), f (\frac{j}{m}) ⊙ (1 + j - m t_{0})) \\ = D (f (\frac{j - 1}{m}) ⊙ (j - mt), \tilde{0}) \\ + D (f (\frac{j}{m}) ⊙ (mt - j + 1), f (\frac{j}{m})) \\ = (j - mt) D (f (\frac{j - 1}{m}), \tilde{0}) + (j - mt) D (f (\frac{j}{m}), \tilde{0}) \\ \leq 2 M (j - mt) = 2 Mm (t_{0} - t) \leq 2 Mm δ \leq ɛ . \end{matrix}$ that is the continuity of ${TF}_{m}^{ℱ} (f)$ .

In order to prove the inequalities (2) and (3), let $\frac{j}{m} \leq t < \frac{j + 1}{m}$ , so $\begin{matrix} D ({TF}_{m}^{ℱ} (f) (t), f (t)) = D (\sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t) \\ + \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t), f (t)) \\ = D (\sum_{i = 0}^{m - 1}^{*} f (\frac{i}{m}) ⊙ φ_{i} (t) + \sum_{i = 0}^{m - 1}^{*} f (\frac{i + 1}{m}) ⊙ ψ_{i} (t), f (t)) \\ = D (f (\frac{j}{m}) ⊙ (1 + j - mt) \oplus f (\frac{j + 1}{m}) \\ ⊙ (mt - j), f (t) ⊙ (φ_{j} (t) + ψ_{j} (t))) \\ \leq D (f (\frac{j}{m}) ⊙ (1 + j - mt), f (t) ⊙ (1 + j - mt)) \\ + D (f (\frac{j + 1}{m}) ⊙ (mt - j), f (t) ⊙ (mt - j)) \\ = (1 + j - mt) \cdot D (f (\frac{j}{m}), f (t)) \\ + (mt - j) \cdot D (f (\frac{j + 1}{m}), f (t)) \\ \leq ω_{[\frac{j}{m}, \frac{j + 1}{m})} (f, \frac{1}{m}) \leq ω_{[0, 1]} (f, \frac{1}{m}) . \end{matrix}$

Consequently $D ({TF}_{m}^{ℱ} (f) (t), f (t)) \leq ω_{[0, 1]} (f, \frac{1}{m}),$ for all t ∈ [0, 1], where $ω_{[0, 1]} (f, \frac{1}{m}) = sup {D (f (x),$ $sf (y)) : x, y \in [0, 1], | x - y | \leq \frac{1}{m}} .$

Finally, we obtain $\begin{matrix} D ({TF}_{m}^{ℱ} (f) (t), \tilde{0}) \leq D (\sum_{i = 0}^{m - 1}^{*} c_{i} ⊙ φ_{i} (t) \\ + \sum_{i = 0}^{m - 1}^{*} d_{i} ⊙ ψ_{i} (t), \tilde{0}) \\ = D (c_{j} ⊙ φ_{j} (t) \oplus d_{j} ⊙ ψ_{j} (t), \tilde{0}) \leq φ_{j} (t) D (c_{j}, \tilde{0}) \\ + ψ_{j} (t) D (d_{j}, \tilde{0}) \\ \leq (φ_{j} (t) + ψ_{j} (t)) M = M . \end{matrix}$

4 Numerical solution to fuzzy integral equations

Consider the following conditions:

$k \in C ([0, 1] \times [0, 1], ℝ), X \in C_{ℱ} [0, 1]$ and k (s, t) ⩾0, ∀ (s, t) ∈ [0, 1] × [0, 1],

$Y : ℝ_{ℱ} \to ℝ_{ℱ}$ is continuous, and there exist ℒ ⩾ 0 with

$\begin{matrix} D (Y (F_{1} (s)), Y (F_{2} (t))) & ⩽ ℱ \cdot D (F_{1} (s), F_{2} (t)), \\ \forall s, t \in [0, 1] \end{matrix}$ (4)We define the nonlinear integral operator $A : C_{ℱ} [0, 1] \to {F | F : [0, 1] \to ℝ_{ℱ}}$ by: $\begin{matrix} A (F) (t) & = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y (F (s)) ds, \\ \forall t \in [0, 1], \forall F \in C_{ℱ} [0, 1] \end{matrix}$

and the sequence of successive approximations (F_n) _n⩾1 given by:

F_{0} (t) = X (t), F_{n} (t) = A (F_{n - 1}) (t), \forall t \in [0, 1], \forall n \in ℕ .

Theorem 2. [19] Under the conditions (i) , (ii), if𝒞 . = 𝒦ℒ< 1 where $𝒦 = \max_{0 \leq t, s \leq 1} | k (s, t) |$ , we have A (C_ℱ [0, 1]) ⊂ C_ℱ [0, 1], i.e. the operator A is well-defined and the fuzzy integral equation $F (t) = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y (F (s)) ds, t \in [0, 1]$ has a unique solution F^* ∈ C_ℱ [0, 1], and $lim_{n \to \infty}$ D^* (F_n, F^*) =0. Moreover, the following inequality holds: $D (F^{*} (t), F_{n} (t)) ⩽ \frac{𝒞^{n + 1}}{ℱ (1 - 𝒞)} M_{0} \forall t \in [0, 1], n ⩾ 1,$ (5)where M₀ = sup _0⩽t⩽1∥ Y (X (t)) ∥.

Now, we are introducing the numerical method to find the approximate solution of Eq. (1). In this way, we will define the uniform partition of the interval [0, 1]: $▵ : 0 = t_{0} < t_{1} < \dots < t_{m - 1} < t_{m} = 1$ with t_i = ih, where $h = \frac{1}{m}$ . Then the following iterative procedure gives the approximate solution of Equation.(1) in point t ∈ [0, 1], $\begin{matrix} f_{0} (t) & = X (t), \\ f_{n} (t) & = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (f_{n - 1})) (s) ds, \\ \forall t \in [0, 1], n ⩾ 1 . \end{matrix}$ (6)

4.1 Convergence analysis

We obtain an error estimation between the exact solution and the approximate solution for Equation.(1).

Theorem 3. Consider the nonlinear fuzzy Fredholm integral equation Equation.(1). Under conditions (i) , (ii), if 𝒦ℒ . < 1 where $𝒦 = \max_{0 \leq t, s \leq 1} | k (s, t) |$ , then the iterative procedure (6) converges to the unique solution of Eq.(1), F, and the following error estimation holds true: $D^{*} (F, f_{n}) \leq \frac{𝒦}{1 - 𝒦 ℒ} (M_{0} {(𝒦 ℒ)}^{n} + ω)$ (7)where $M_{0} = sup_{0 \leq t \leq 1} ∥ Y (X (t)) ∥$ and $ω = max {ω_{[0, 1]} (Y (f_{i}), h), i = 0, 1, \dots, n - 1}$

Proof. At the first, we prove that the sequence of functions $(f_{n})_{n \in ℕ}$ is bounded in [0, 1]. According to Lemma 1, if $f_{n} (n \in ℕ)$ is continuous in [0, 1] then f_n is bounded. Assume that f_n-1 is a fuzzy continuous on [0, 1]. As a result of using Corollary 1 the function Y (f_n-1) is continuous, and by Theorem 1, ${TF}_{m}^{ℱ} (Y (f_{n - 1})) (.) : [0, 1] \to ℝ_{ℱ}$ is continuous. But according to Lemma 5, the function $k (., t) ⊙ {TF}_{m}^{ℱ} (Y (f_{n - 1})) (.) : [0, 1] \to ℝ_{ℱ}$ is a continuous function too. So using Lemma 5 again, we conclude that the function $H : [0, 1] \to ℝ_{ℱ}$ , defined by $H (t) = (FR) \int_{0}^{1} k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (f_{n - 1})) (s) ds$ is continuous, and also since X (t) is continuous on [0, 1], we conclude that f_n is continuous on [0, 1]. Now, we can prove error estimation (7). Since $F_{1} (t) = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y (F_{0} (s)) ds$ , we have $\begin{matrix} D (F_{1} (t), f_{1} (t)) = D (X (t), X (t)) \\ + D ((FR) \int_{0}^{1} k (s, t) ⊙ Y (F_{0} (s)) ds, \\ (FR) \int_{0}^{1} k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (f_{0})) (s) ds) \\ \leq \int_{0}^{1} D (k (s, t) ⊙ Y (X (s)), k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (X)) (s)) ds \\ \leq 𝒦 \int_{0}^{1} D (Y (X (s)), {TF}_{m}^{ℱ} (Y (X)) (s)) ds \\ = 𝒦 \cdot \sum_{i = 0}^{m - 1} \int_{ih}^{(i + 1) h} D (Y (X (s)), {TF}_{m}^{ℱ} (Y (X)) (s)) ds . \end{matrix}$ Because Y (X) is fuzzy continuous, according to (2), we have $D ({TF}_{m}^{ℱ} (Y (X)) (t), Y (X (t))) \leq ω_{[0, 1]} (Y (X), h)$ for all t ∈ [ih, (i + 1) h] that i = 0, 1, ⋯ , m - 1; therefore, $D (F_{1} (t), f_{1} (t)) \leq 𝒦 \cdot ω_{[0, 1]} (Y (X), h)$ (8)

Now, since $F_{2} (t) = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y$ (F₁ (s)) ds, we conclude that $\begin{matrix} D (F_{2} (t), f_{2} (t)) \\ = D ((FR) \int_{0}^{1} k (s, t) ⊙ Y (F_{1} (s)) ds, \\ (FR) \int_{0}^{1} k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (f_{1})) (s) ds) \\ \leq 𝒦 \int_{0}^{1} D (Y (F_{1} (s)), {TF}_{m}^{ℱ} (Y (f_{1})) (s)) ds \\ \leq 𝒦 \int_{0}^{1} D (Y (F_{1} (s)), Y (f_{1} (s))) ds \\ + 𝒦 \int_{0}^{1} D (Y (f_{1} (s)), {TF}_{m}^{ℱ} (Y (f_{1})) (s)) ds . \end{matrix}$

According to (4), and from Equations.(2) and (8) we have $\begin{matrix} D (F_{2} (t), f_{2} (t)) & \leq 𝒦^{2} \cdot ℒ ω_{[0, 1]} (Y (X), h) \\ + 𝒦 \cdot ω_{[0, 1]} (Y (f_{1}), h) \end{matrix}$

By induction, it is prove that $\begin{matrix} D (F_{n} (t), f_{n} (t)) \\ \leq (𝒦^{n - 1} ℒ^{n - 1} + 𝒦^{n - 2} ℒ^{n - 2} + \dots + 𝒦 ℒ + 1) \cdot 𝒦 ω \end{matrix}$ for each t ∈ [0, 1], such that $ω = max {ω_{[0, 1]} (Y (f_{i}), h), i = 0, 1, \dots, n - 1} .$

Consequently $D^{*} (F_{n}, f_{n}) \leq \frac{1 - (𝒦 ℒ)^{n}}{1 - 𝒦 ℒ} ω 𝒦 .$

Since 𝒦ℒ < 1, we get: $D^{*} (F_{n}, f_{n}) \leq \frac{𝒦 ω}{1 - 𝒦 ℒ} .$

Using (5), we obtain $D (F (t), F_{n} (t)) \leq \frac{{(𝒦 ℒ)}^{n + 1}}{ℒ (1 - 𝒦 ℒ)} M_{0}, \forall t \in [0, 1], n ⩾ 1 .$ So we have $\begin{matrix} D^{*} (F, f_{n}) \leq D^{*} (F, F_{n}) + D^{*} (F_{n}, f_{n}) \\ \leq \frac{(𝒦 ℒ)^{n + 1}}{ℒ (1 - 𝒦 ℒ)} M_{0} + \frac{𝒦 ω}{1 - 𝒦 ℒ} . \\ \leq \frac{𝒦}{1 - 𝒦 ℒ} (M_{0} {(𝒦 ℒ)}^{n} + ω) □ \end{matrix}$

Remark 3. Since 𝒦 ℒ < 1, it is easy to prove that $lim_{\begin{matrix} n \to \infty \\ m \to \infty \end{matrix}} D^{*} (F, f_{n}) = 0$

that shows the convergence of the method.

4.2 The numerical stability analysis

In order to investigate the numerical stability of the computed values with respect to small perturbations in the first iteration, we consider another first iteration term G₀ ∈ C_ℱ [0, 1] such that ∃ɛ > 0 for which D (F₀ (t) , G₀ (t)) < ɛ, ∀ t ∈ [0, 1] . Suppose that there exist M_G with $D (G_{0} (t), \tilde{0}) ⩽ M_{G}, \forall t \in [0, 1] .$ The obtained sequence of successive approximations is: $\begin{matrix} G_{n} (t) & = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ Y (G_{n - 1} (s)) ds, \\ \forall t \in [0, 1], \forall n \in ℕ \end{matrix}$ and applying the same iterative procedure (6), the computed values are: g₀ (t) = G₀ (t) and $\begin{matrix} g_{n} (t) & = X (t) \oplus (FR) \int_{0}^{1} k (s, t) ⊙ {TF}_{m}^{ℱ} (Y (g_{n - 1})) (s) ds, \\ \forall t \in [0, 1], \forall n \in ℕ \end{matrix}$ As in [16], we give the following definition and derive the following numercal stability result.

Definition 10. We say that the iterative procedure (6) applied to integral equation (1) is numerically stable with respect to the choice of the first iteration, iff there exist two costants K₁, K₂ which are independent by $h = \frac{1}{m}$ , and a function α (h), such that $lim_{h \to 0} α (h) = 0$ and $D (f_{n} (t), g_{n} (t)) < K_{1} ɛ + K_{2} α (h), \forall n \in ℕ, t \in [0, 1] .$

Theorem 4. Under the conditions (i),(ii) and 𝒦 ℒ < 1, the iterative procedure (6) is numerically stable with respect to the choice of the first iteration.

Proof. Similarly as Theorem 3, it follows that $D^{*} (G_{n}, g_{n}) \leq \frac{𝒦 ω^{'}}{1 - 𝒦 ℒ} .$ where $ω^{'} = max {ω_{[0, 1]} (Y (g_{i}), h), i = 0, 1, \dots, n - 1}$ and we obtain $\begin{matrix} D (f_{n} (t), g_{n} (t)) & ⩽ D (f_{n} (t), F_{n} (t)) + D (F_{n} (t), G_{n} (t)) \\ + D (G_{n} (t), g_{n} (t)) \end{matrix}$ $⩽ \frac{𝒦 ω}{1 - 𝒦 ℒ} + \frac{𝒦 ω^{'}}{1 - 𝒦 ℒ} + D (F_{n} (t), G_{n} (t)) .$

Now, we get: $D (F_{1} (t), G_{1} (t)) ⩽ D (X (t), X (t))$ $+ \int_{0}^{1} D (k (s, t) ⊙ Y (F_{0} (s)), k (s, t) ⊙ Y (G_{0} (t))) ds$ $⩽ 𝒦 ℒ \int_{0}^{1} D (F_{0} (s), G_{0} (s)) ds ⩽ 𝒦 ℒ ɛ, \forall t \in [0, 1] .$

By induction, for $n \in ℕ, n ⩾ 2$ , according to the condition 𝒦 ℒ < 1, we obtain $(F_{2} (t), G_{2} (t)) ⩽ 𝒦 ℒ \int_{0}^{1} D (F_{1} (t), G_{1} (t)) ds < {(𝒦 ℒ)}^{2} ɛ,$ $(F_{n} (t), G_{n} (t)) ⩽ 𝒦 ℒ \int_{0}^{1} D (F_{n - 1} (t), G_{n - 1} (t)) ds < {(𝒦 ℒ)}^{n} ɛ < ɛ,$ for all t ∈ [0, 1] and $n \in ℕ$ . Then $\begin{matrix} D (f_{n} (t), g_{n} (t)) & ⩽ \frac{𝒦 (ω + ω^{'})}{1 - 𝒦 ℒ} + ɛ = K_{1} ɛ + K_{2} α (h), \\ \forall n \in ℕ, t \in [0, 1], \end{matrix}$ where $K_{1} = 1, K_{2} = \frac{𝒦}{1 - 𝒦 ℒ}, α (h) = ω + ω^{'}$ obtaining the numerical stability.□

5 Numerical examples

In this section, we applied the presented iterative method in Section 4 for solving fuzzy integral equation (1) in two examples. The approximate solution is calculated for different values of m,n. Also, we compare the numerical solution obtained by using the proposed method with the exact solution. The computations associated with the examples were performed using Mathematica 7.

Example 1. Consider the following nonlinear fuzzy Fredholm integral equation: $F (t) = X (t) \oplus (FR) \int_{0}^{1} \frac{ts}{3} ⊙ {(F (s))}^{2} ds$ where $k (s, t) = \frac{ts}{3}, \forall s, t \in [0, 1]$ , $\begin{matrix} \underline{X} (t, r) & = \frac{11}{12} t - \frac{1}{6} (1 - r) + \frac{1}{27} t (1 - r) \\ - \frac{1}{216} t {(1 - r)}^{2}, t, r \in [0, 1], \end{matrix}$ $\begin{matrix} \bar{X} (t, r) & = \frac{11}{12} t + \frac{1}{6} (1 - r) - \frac{1}{27} t (1 - r) \\ - \frac{1}{216} t {(1 - r)}^{2}, t, r \in [0, 1] . \end{matrix}$

The exact solution is $(\underline{F} (t, r), \bar{F} (t, r)) = (t - \frac{1}{6} (1 - r), t + \frac{1}{6} (1 - r)) .$

Table 1 presents a comparison of the numerical solution in [19] with the TFs solution.

Example 2. [8] Consider the following nonlinear fuzzy Fredholm integral equation: $F (t) = X (t) \oplus (FR) \int_{0}^{1} s^{3} exp (- t) ⊙ {(F (s))}^{2} ds$ where k (s, t) = s³ exp(- t) , ∀ s, t ∈ [0, 1], $\begin{matrix} \underline{X} (t, r) & = cosh t + exp (- t) (\frac{475 exp (- 2)}{512} - \frac{exp (2)}{32} \\ - \frac{139}{512} + \frac{19 exp (- 2) r}{128} + \frac{9 r}{640} \\ + \frac{19 exp (- 2) r^{2}}{3200} - \frac{3 r^{2}}{3200}), t, r \in [0, 1], \\ \bar{X} (t, r) & = cosh t + exp (- t) (\frac{15979 exp (- 2)}{12800} - \frac{exp (2)}{32} \\ - \frac{3163}{12800} - \frac{551 exp (- 2) r}{3200} - \frac{33 r}{3200} \\ + \frac{19 exp (- 2) r^{2}}{3200} - \frac{3 r^{2}}{3200}), t, r \in [0, 1] . \end{matrix}$

The exact solution in this case is given by $\begin{matrix} (\underline{F} (t, r), \bar{F} (t, r)) & = (cosh t + \frac{1}{40} (5 + 2 r) exp (- t), \\ cosh t + \frac{1}{40} (9 - 2 r) exp (- t)) . \end{matrix}$

The comparison among the TFs solution and the HBT solution in [8] is shown in Table 2.

6 Conclusion

In this paper, we have suggested an iterative procedure by utilizing fuzzy TFs to solve the nonlinear Fredholm fuzzy integral equation (1). In Theorem 3, we obtain the boundedness and the error estimate of the sequence of successive approximations. The numerical stability with respect to the choice of the initial iteration have been proved in Theorem 4. The error estimation of the present method is proved; for getting the best approximating solution of the equation, the number m should be chosen sufficiently large. The implementation of this method needs no complex calculations; therefore, the operation speed is high. The analyzed examples illustrate the ability and reliability of fuzzy TFs method for Equation.(1).

Acknowledgments

The authors are grateful to anonymous referees for their constructive comments and suggestions.

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