Solving linear and nonlinear Abel fuzzy integral equations by homotopy analysis method

Peer review under responsibility of Taibah University

Abstract

The main purpose of this article is to present an approximation method for solving Abel fuzzy integral equation in the most general form. The proposed approach is based on homotopy analysis method. This method transforms linear and nonlinear Abel fuzzy integral equations into two crisp linear and nonlinear integral equations. The convergence analysis for the proposed method is also introduced. We give some numerical applications to show efficiency and accuracy of the method. All of the numerical computations have been performed on a computer with the aid of a program written in Matlab.

Keywords:

• Fuzzy number
• Fuzzy integral equation
• Abel fuzzy integral equations
• Homotopy analysis method

1 Introduction

Fuzzy integral equations are important in studying and solving a large proportion of the problems in many topics in applied mathematics, in particular in relation to physics, geographic, medical and biology. Usually in many applications, some of the parameters in our problems are represented by fuzzy number rather than crisp, and hence it is important to develop mathematical models and numerical procedures that would appropriately treat general fuzzy integral equations and solve them.

The concept of integration of fuzzy functions was first introduced by Dubois and Prade [1]. Alternative approaches were later suggested by Goetschel and Voxman [2], Kaleva [3], Nanda [4] and others. While Goetschel and Voxman [2] preferred a Riemann integral type approach, Kalva [3] defined the integral of fuzzy function, using the Lebesgue type concept for integration. One of the first applications of fuzzy integration was given by Wu and Ma [5], who investigated the fuzzy Fredholm integral equation of the second kind (FFIE-2). This work which established the existence of a unique solution for (FFIE-2) was followed by other works such as Mirzaee et al. [6] and Nguyen [7] where an original fuzzy differential equation is replaced by a fuzzy integral equation. Recently Liao, in his Ph.D. thesis [8], has proposed the homotopy analysis method (HAM) to solve some classes of nonlinear equations. Step by step, the method was developed and its effectiveness was proved in handling nonlinear equations [8–11].

Abel integral equations occur in many branches of scientific fields, such as microscopy, seismology, radio astronomy, electron emission, atomic scattering, radar ranging, plasma diagnostics, X-ray radiography, and optical fiber evaluation [12].

Recently, Mirzaee et al. [13–15] have studied the numerical solutions of the Fredholm fuzzy integral equations. Since the homotopy analysis method is a powerful device for solving a wide variety of problems arising in many scientific applications, we will develop the numerical methods for the approximate solutions of linear and nonlinear Abel fuzzy integral equations.

The structure of this paper is organized as follows: in Section 2, some basic definitions and results which will be used later are given. In Section 3, Abel fuzzy integral equations are introduced. In Section 4, we apply homotopy analysis method to solve Abel fuzzy integral equations, then the proposed method is implemented for solving three illustrative examples in Section 5 and finally, conclusion is drawn in Section 6.

2 Preliminaries

We now recall some definitions needed through the paper.

Definition 1

(Kaleva [3]). A fuzzy number is a fuzzy set $v:R^1\to I=[0,1]$ which satisfies

•	$v$ is upper semi continuous,

•	$v(x)=0$ outside some interval [c, d],

•	There are real numbers a,b : c ≤ a ≤ b ≤ d for which

•	$\underset{\_}{v}(x)$ is monotonic increasing on [c, a],

•	$\overline{v}(x)$ is monotonic decreasing on [b, d],

•	$v(x)=1$, a ≤ x ≤ b.

The set of all such fuzzy number is denoted by R_F.

Definition 2

(Kaleva [3]). Let V be a fuzzy set on R. V is called a fuzzy interval if:

•	V is normal: there exists x0 ∈ R such that V(x0) = 1.

•	V is convex: for all x, t ∈ R and 0 ≤ λ ≤ 1, it holds that V(λx + (1 − λ)t) ≥ min{V(x), V(t)},

•	V is upper semi-continuous: for any x0 ∈ R, it holds that $V(x_0)\ge \underset{x\to 0^{\pm }}{lim}V(x)$,

•	[V]α = Cl{x ∈ R|V(x) > 0} is a compact subset of R.

The α-cut of a fuzzy interval V with 0 < α ≤ 1 is the crisp set [V]^α = {x ∈ R|V(x) > 0}. For a fuzzy interval V, its α-cut are closed intervals in R. They will be denoted by them by ${[V]}^{\alpha }=[\underset{\_}{V}(\alpha ),\overline{V}(\alpha )]$. An alternative definition or parametric form of a fuzzy number which yields the same R_F is given by Kaleva [8] as follows:

Definition 3

(Ma et al. [16]). An arbitrary fuzzy number $\tilde{u}$ in the parametric form is represented by an ordered pair of functions $(\underset{\_}{u}(r),\overline{u}(r))$ which satisfy the following requirements:

•	$\underset{\_}{u}(r)$ is a bounded left-continuous non-decreasing function over [0, 1],

•	$\overline{u}(r)$ is a bounded right-continuous non-increasing function over [0, 1],

•	$\underset{\_}{u}(r)\le \overline{u}(r)$, for all 0 ≤ r ≤ 1 .

For arbitrary fuzzy numbers $\tilde{v}=(\underset{\_}{v}(r),\overline{v}(r))$, $\tilde{w}=(\underset{\_}{w}(r),\overline{w}(r))$ and real number λ, one may define the addition and the scalar multiplication of the fuzzy numbers by using the extension principle as follows:

•	$\tilde{v}=\tilde{w}$ if and only if $\underset{\_}{v}(r)=\underset{\_}{w}(r)$ and $\overline{v}(r)=\overline{w}(r)$,

•	$\tilde{v}\oplus \tilde{w}=(\underset{\_}{v}(r)+\underset{\_}{w}(r),\overline{v}(r)+\overline{w}(r))$,

•	$(\lambda \odot \tilde{v})=\left\{\begin{array}{cc}(\lambda \underset{\_}{v}(r),\lambda \overline{v}(r))\hfill & \lambda \ge 0\hfill \\ (\lambda \overline{v}(r),\lambda \underset{\_}{v}(r))\hfill & \lambda <0\hfill \end{array}.\right.$

Definition 4

(Ga [17]). For arbitrary numbers $\tilde{v}=(\underset{\_}{v}(r),\overline{v}(r))$ and $\tilde{w}=(\underset{\_}{w}(r),\overline{w}(r))$$D(\tilde{v},\tilde{w})=max\{\underset{0\le r\le 1}{sup}|\overline{v}(r)-\overline{w}(r)|,\underset{0\le r\le 1}{sup}|\underset{\_}{v}(r)-\underset{\_}{w}(r)|\},$in the distance between $\tilde{v}$ and $\tilde{w}$. It is proved that (R_F, D) is a complete metric space with following properties [5]

•	$D(\tilde{u}+\tilde{w},\tilde{v}+\tilde{w})=D(\tilde{u},\tilde{v});\forall \tilde{u},\tilde{v},\tilde{w}\in R_F$,

•	$D(k\odot \tilde{u},k\odot \tilde{v})=|k|D(\tilde{u},\tilde{v});\forall \tilde{u},\tilde{v}\in R_F\forall k\in R,$

•	$D(\tilde{u}\oplus \tilde{v},\tilde{w}\oplus \tilde{e})\le D(\tilde{u},\tilde{w})+D(\tilde{v},\tilde{e});\forall \tilde{u},\tilde{v},\tilde{w},\tilde{e}\in R_F$.

Definition 5

(Anastassiou [18]). Let $\tilde{f},\tilde{g}:[a,b]\to R_F$, be fuzzy real number valued functions. The uniform distance between $\tilde{f},\tilde{g}$ is defined by$D(\tilde{f},\tilde{g})=sup\{D(\tilde{f}(x),\tilde{g}(x))|x\in [a,b]\},$In Goetschel and Voxman [2] the authors proved that if the fuzzy function $\tilde{f}(x)$ is continuous in the metric D, its definite integral exists and also,$\underset{\_}{{\int }_a^{b}f(x,r)\mathrm{dx}}={\int }_a^b\underset{\_}{f}(x,r)\mathrm{dx},\overline{{\int }_a^{b}f(x,r)d}x={\int }_a^b\overline{f}(x,r)\mathrm{dx}.$Where $(\underset{\_}{f}(x,r),\overline{f}(x,r))$ is the parametric form of $\tilde{f}(x)$. It should be noted that the fuzzy integral can be also defined using the Lebesgue-type approach [3]. However, if $\tilde{f}(x)$ be continuous, both approaches yield the same value. Moreover, the representation of the fuzzy integral is more convenient for numerical calculations. More details about the properties of the fuzzy integral are given in [6,8].

Definition 6

(Wu [19]). A fuzzy real number valued function $\tilde{f}:[a,b]\to R_F$, is said to be continuous in x₀ ∈ [a, b], if for each ϵ > 0 there is δ > 0 such that $D(\tilde{f}(x),\tilde{f}(x_0))<ϵ$, whenever x ∈ [a, b] and |x − x₀| < δ. We say that f is fuzzy continuous on [a, b] if f is continuous at each x₀ ∈ [a, b] and denote the space of all such functions by C_F([a, b]).

Lemma 1

(Anastassiou [18]). If $\tilde{f},\tilde{g}:[a,b]\subseteq R\to R_F$ are fuzzy continuous function, then the function F : [a, b] → R⁺ by $\tilde{F}(x)=D(\tilde{f}(x),\tilde{g}(x))$ is continuous on [a, b], and$D\left({\int }_a^b\tilde{f}(x)\mathrm{dx},{\int }_a^b\tilde{g}(x)\mathrm{dx}\right)\le {\int }_a^{b}D(\tilde{f}(x),\tilde{g}(x))\mathrm{dx}.$

Theorem 1

(Hc [20]). Let $\tilde{f}(x)$ be a fuzzy value function on [a, ∞) and it is represented by $(\underset{\_}{f}(x,r),\overline{f}(x,r))$. For any fixed r ∈ [0, 1], assume that $\underset{\_}{f}(x,r)$ and $\overline{f}(x,r)$ are Riemann-integrable on [a, b] for every b ⩾ a and assume there are two positive functions $\underset{\_}{M}(r)$ and $\overline{M}(r)$ such that ${\int }_a^b|\underset{\_}{f}(x,r)|\mathrm{dx}⩽\underset{\_}{M}(r)$ and ${\int }_a^b|\overline{f}(x,r)|\mathrm{dx}⩽\overline{M}(r)$ for every b ⩾ a. Then $\tilde{f}(x)$ is improper fuzzy Riemann-integrable on [a, ∞) and the improper fuzzy Riemann-integral is a fuzzy number. Further, we have:${\int }_a^{\infty }\tilde{f}(x)\mathrm{dx}=\left({\int }_a^{\infty }\underset{\_}{f}(x,r)\mathrm{dx},{\int }_a^{\infty }\overline{f}(x,r)\mathrm{dx}\right).$

3 Abel fuzzy integral equations

The Abel integral equation is [21,22]

(1) $f(x)={\int }_a^x\frac{u(t)}{{(x-t)}^{\alpha }}\mathrm{dt};a\le x\le b,$(1) where α is a known constant such that 0 < α < 1, f(x) is a predetermined data function and u(x) is unknown function that will be determined. The expression (x − t)^−α is called the kernel of Abel integral equation, or simply Abel kernel, that is singular as t → x.

If f(x) is a crisp function, then the solutions of Eq. (1)(1) $f(x)={\int }_a^x\frac{u(t)}{{(x-t)}^{\alpha }}\mathrm{dt};a\le x\le b,$(1) are crisp too. However, if f(x) is a fuzzy function, these equations may only possess fuzzy solutions. In this paper, the Abel fuzzy integral equations are discussed. Introducing the parametric forms of f(x) and u(x), we have the parametric form of fuzzy Able integral equation as follows:

(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) where 0 ≤ r ≤ 1 and α is a known constant such that 0 < α < 1, $\tilde{f}(x)=(\underset{\_}{f}(x,r),\overline{f}(x,r))$ is a predetermined data function and $\tilde{u}(x)=(\underset{\_}{u}(x,r),\overline{u}(x,r))$ is the solution that will be determined.

By putting α = 1/2 in Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) , we obtain the standard form of the nonlinear Abel fuzzy integral equation as

(3) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{F(\underset{\_}{u}(t,r))}{\sqrt{x-t}}\mathrm{dt},{\int }_a^x\frac{F(\overline{u}(t,r))}{\sqrt{x-t}}\mathrm{dt}\right),$(3) where the function $(\underset{\_}{f}(x,r),\overline{f}(x,r))$ is a given real-valued function, and $(F(\underset{\_}{u}(x,r)),F(\overline{u}(x,r)))$ is a nonlinear function of $(\underset{\_}{u}(x,r),\overline{u}(x,r))$. Recall that the unknown function $(\underset{\_}{u}(x,r),\overline{u}(x,r))$ occurs only inside the integral sign for the Abel fuzzy integral Eq. (3)(3) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{F(\underset{\_}{u}(t,r))}{\sqrt{x-t}}\mathrm{dt},{\int }_a^x\frac{F(\overline{u}(t,r))}{\sqrt{x-t}}\mathrm{dt}\right),$(3) .

4 The homotopy analysis method for solving Abel fuzzy integral equations

Let us consider the Abel fuzzy integral Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) . We first remark that Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) is not written in the canonical form of HAM, necessary for calculating the decomposition solution series. Furthermore, the linear operator defined by Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) generally does not have an inverse so it is difficult to obtain a precise numerical solution by HAM. For these considerations, we begin our analysis by putting α = 1/2 and writing Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) as:(4) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(x,r)+\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},{\int }_a^x\frac{\overline{u}(x,r)+\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(4) thus(5) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}+{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},{\int }_a^x\frac{\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}+{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(5) so(6) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left(2\sqrt{x-a}\underset{\_}{u}(x,r)+{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},2\sqrt{x-a}\overline{u}(x,r)+{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(6) therefore, it is clear that Eq. (2)(2) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left({\int }_a^x\frac{\underset{\_}{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt},{\int }_a^x\frac{\overline{u}(t,r)}{{(x-t)}^{\alpha }}\mathrm{dt}\right),$(2) can be replaced by a suitable equivalent expression (6)(6) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left(2\sqrt{x-a}\underset{\_}{u}(x,r)+{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},2\sqrt{x-a}\overline{u}(x,r)+{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(6) , which is written in the canonical form and then it can be solved by means of the HAM decomposition method. Prior to applying HAM for Eq. (6)(6) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left(2\sqrt{x-a}\underset{\_}{u}(x,r)+{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},2\sqrt{x-a}\overline{u}(x,r)+{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(6) we rewrite Eq. (6)(6) $(\underset{\_}{f}(x,r),\overline{f}(x,r))=\left(2\sqrt{x-a}\underset{\_}{u}(x,r)+{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},2\sqrt{x-a}\overline{u}(x,r)+{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(6) in the following form(7) $(\underset{\_}{u}(x,r),\overline{u}(x,r))=\left(\frac{\underset{\_}{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},\frac{\overline{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(7) Eq. (7)(7) $(\underset{\_}{u}(x,r),\overline{u}(x,r))=\left(\frac{\underset{\_}{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},\frac{\overline{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(7) is a system of linear Abel integral equations in crisp case for each 0 ≤ r ≤ 1. To solve system (7)(7) $(\underset{\_}{u}(x,r),\overline{u}(x,r))=\left(\frac{\underset{\_}{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{u}(t,r)-\underset{\_}{u}(x,r)}{\sqrt{x-t}}\mathrm{dt},\frac{\overline{f}(x,r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{u}(t,r)-\overline{u}(x,r)}{\sqrt{x-t}}\mathrm{dt}\right),$(7) by HAM, we construct the zero-order deformation equation(8) $\begin{array}{c}(1-p)L[\underset{\_}{U}(x,p;r)-\underset{\_}{Z_0}(x;r)]=\mathrm{pc}\left(\underset{\_}{U}(x,p;r)-\frac{\underset{\_}{f}(x,r)}{2\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{U}(t,p;r)-\underset{\_}{U}(x,p;r)}{\sqrt{x-t}}\mathrm{dt}\right),\hfill \\ [(1-p)L[\overline{U}(x,p;r)-\overline{Z_0}(x;r)]=\mathrm{pc}\left(\overline{U}(x,p;r)-\frac{\overline{f}(x,r)}{2\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{U}(t,p;r)-\overline{U}(x,p;r)}{\sqrt{x-t}}\mathrm{dt}\right),\hfill \end{array}$(8) where p ∈ [0, 1] is the embedding parameter, c is non-zero auxiliary parameter, L is an auxiliary linear operator, ${\underset{\_}{Z}}_0(x,r)$ and ${\overline{Z}}_0(x,r)$ are initial guesses of $\underset{\_}{u}(x,r)$ and $\overline{u}(x,r)$ respectively and $\underset{\_}{U}(x,p;r)$ and $\overline{U}(x,p;r)$ are unknown function depend on the variable p. Using the above zero-order deformation equation, with assumption L[u] = u, we have(9) $\begin{array}{c}(1-p)[\underset{\_}{U}(x,p;r)-\underset{\_}{Z_0}(x;r)]=\mathrm{pc}\left(\underset{\_}{U}(x,p;r)-\frac{\underset{\_}{f}(x,r)}{\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{U}(t,p;r)-\underset{\_}{U}(x,p;r)}{\sqrt{x-a}}\mathrm{dt}\right),\hfill \\ (1-p)[\overline{U}(x,p;r)-\overline{Z_0}(x;r)]=\mathrm{pc}\left(\overline{U}(x,p;r)-\frac{\overline{f}(x,r)}{\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{U}(t,p;r)-\overline{U}(x,p;r)}{\sqrt{x-a}}\mathrm{dt}\right).\hfill \end{array}$(9) Obviously, when p = 0 and p = 1, it holds(10) $\left\{\begin{array}{c}\underset{\_}{U}(x,0;r)={\underset{\_}{Z}}_0(x;r)\hfill \\ \overline{U}(x,0;r)={\overline{Z}}_0(x;r)\hfill \\ \hfill \end{array}\right.,$(10) and(11) $\begin{array}{c}\underset{\_}{U}(x,1;r)=\left(\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{U}(t,1;r)-\underset{\_}{U}(x,1;r)}{\sqrt{x-t}}\mathrm{dt}\right),\hfill \\ \overline{U}(x,1;r)=\left(\frac{\overline{f}(x;r)}{2\sqrt{x-a}}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{U}(t,1;r)-\overline{U}(x,1;r)}{\sqrt{x-t}}\mathrm{dt}\right).\hfill \end{array}$(11) Thus, as p increases from 0 to 1, the solution $(\underset{\_}{U}(x,p;r),\overline{U}(x,p;r))$ varies from initial guess $({\underset{\_}{Z}}_0(x;r),{\overline{Z}}_0(x;r))$ to the solution $(\underset{\_}{u}(x;r),\overline{u}(x;r))$. Expanding $\underset{\_}{U}(x,p;r)$ and $\overline{U}(x,p;r)$ in Taylor series with respect p, we have(12) $\begin{array}{c}\underset{\_}{U}(x,p;r)={\underset{\_}{Z}}_0(x;r)+\sum _{m=1}^{\infty }{\underset{\_}{u}}_m(x;r)p^m,\hfill \\ \overline{U}(x,p;r)={\overline{Z}}_0(x;r)+\sum _{m=1}^{\infty }{\overline{u}}_m(x;r)p^m,\hfill \end{array}$(12) where(13) $\begin{array}{c}{\underset{\_}{u}}_m(x;r)=\frac{1}{m!}\frac{d^m\underset{\_}{U}(x,p;r)}{{\mathrm{dp}}^m}{|}_{p=0},\hfill \\ {\overline{u}}_m(x;r)=\frac{1}{m!}\frac{d^m\overline{U}(x,p;r)}{{\mathrm{dp}}^m}{|}_{p=0}.\hfill \end{array}$(13) It should be noted that $\underset{\_}{U}(x,0;r)={\underset{\_}{Z}}_0(x;r)$ and $\overline{U}(x,0;r)={\overline{Z}}_0(x;r)$. Differentiating the zero-order deformation Eq. (9)(9) $\begin{array}{c}(1-p)[\underset{\_}{U}(x,p;r)-\underset{\_}{Z_0}(x;r)]=\mathrm{pc}\left(\underset{\_}{U}(x,p;r)-\frac{\underset{\_}{f}(x,r)}{\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\underset{\_}{U}(t,p;r)-\underset{\_}{U}(x,p;r)}{\sqrt{x-a}}\mathrm{dt}\right),\hfill \\ (1-p)[\overline{U}(x,p;r)-\overline{Z_0}(x;r)]=\mathrm{pc}\left(\overline{U}(x,p;r)-\frac{\overline{f}(x,r)}{\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{\overline{U}(t,p;r)-\overline{U}(x,p;r)}{\sqrt{x-a}}\mathrm{dt}\right).\hfill \end{array}$(9) m times with respect to the embedding parameter p and then setting p = 0 and finally dividing them by m !, we have(14) $\begin{array}{c}{\underset{\_}{u}}_m(x;r)-{\chi }_m{\underset{\_}{u}}_{m-1}(x;r)=\hslash \left({\underset{\_}{u}}_{m-1}(x;r)+({\chi }_m-1)\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)-{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}\right),\hfill \\ {\overline{u}}_m(x;r)-{\chi }_m{\overline{u}}_{m-1}(x;r)=\hslash \left({\overline{u}}_{m-1}(x;r)+({\chi }_m-1)\frac{\overline{f}(x;r)}{2\sqrt{x-a}}+\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)-{\overline{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}\right).\hfill \end{array}$(14) Where m ≥ 1 and(15) ${\chi }_m=\left\{\begin{array}{cc}0\hfill & m\le 1\hfill \\ 1\hfill & m\ge 2\hfill \end{array},\right.$(15) and ${\underset{\_}{u}}_0(x;r)={\underset{\_}{Z}}_0(x;r)$ and ${\overline{u}}_0(x;r)={\overline{Z}}_0(x;r)$. If we take ${\underset{\_}{Z}}_0(x;r)={\overline{Z}}_0(x;r)=0$, then we have(16) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)=(1+c){\underset{\_}{u}}_{m-1}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)-{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)=(1+c){\overline{u}}_{m-1}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)-{\overline{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \end{array}$(16) where m ≥ 2. Using the fact(17) $\begin{array}{c}\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)-{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}=\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ \frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)-{\overline{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}=\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}-\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \end{array}$(17) and(18) $\begin{array}{c}\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}=\frac{{\underset{\_}{u}}_{m-1}(x;r)}{2\sqrt{x-a}}{\int }_a^x\frac{1}{\sqrt{x-t}}\mathrm{dt},\hfill \\ \frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(x;r)}{\sqrt{x-t}}\mathrm{dt}=\frac{{\overline{u}}_{m-1}(x;r)}{2\sqrt{x-a}}{\int }_a^x\frac{1}{\sqrt{x-t}}\mathrm{dt},\hfill \end{array}$(18) and using the fact(19) ${\int }_a^x\frac{1}{\sqrt{x-t}}\mathrm{dt}=2\sqrt{x-a},$(19) for Eqs. (16)–(19), we obtain(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20) where m ≥ 2.

Proposition 1

Consider the following Abel fuzzy integral equations(21) $\left(\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\underset{\_}{g}(r)x^{n+(1/2)},\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\overline{g}(r)x^{n+(1/2)}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt}.$(21) and(22) $\left(\sqrt{\pi }\frac{\Gamma ((n+2)/2)}{\Gamma ((n+3)/2)}\underset{\_}{g}(r)x^{(n+1)/2},\sqrt{\pi }\frac{\Gamma ((n+2)/2)}{\Gamma ((n+3)/2)}\overline{g}(r)x^{(n+1)/2}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt},$(22) The exact solutions in those cases are given by(23) $\tilde{u}(x)=(\underset{\_}{g}(r)x^n,\overline{g}(r)x^n),$(23) and(24) $\tilde{u}(x)=(\underset{\_}{g}(r)x^{n/2},\overline{g}(r)x^{n/2}),$(24) respectively. For c = −1, in those cases the series will converge to the exact solutions.

Proof

We consider Eq. (21)(21) $\left(\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\underset{\_}{g}(r)x^{n+(1/2)},\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\overline{g}(r)x^{n+(1/2)}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt}.$(21) , for n = 1, we have(25) $\left(\frac{4}{3}\underset{\_}{g}(r)x^{3/2},\frac{4}{3}\overline{g}(r)x^{3/2}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt},$(25)

the exact solution in this case is given by(26) $\tilde{u}(x)=(\underset{\_}{g}(r)x,\overline{g}(r)x).$(26)

By substituting c = −1 in (20)(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20) , we have(27) $\begin{array}{c}{\underset{\_}{u}}_1(x,r)=\frac{\underset{\_}{f}(x,r)}{2\sqrt{x}}=\frac{4/3\underset{\_}{g}(r)x^{3/2}}{2\sqrt{x}}=\frac{2}{3}\underset{\_}{g}(r)x,\hfill \\ {\underset{\_}{u}}_2(x,r)=\frac{2}{3}\underset{\_}{g}(r)x-\frac{1}{2\sqrt{x}}\left(\frac{2}{3}\underset{\_}{g}(r)\right)\left(\frac{4}{3}{x}^{\frac{3}{2}}\right)=\frac{2}{3}\underset{\_}{g}(r)\left(1-\frac{2}{3}\right)x,\hfill \\ {\underset{\_}{u}}_3(x,r)=\frac{2}{3}\underset{\_}{g}(r){\left(1-\frac{2}{3}\right)}^{2}x,\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_n(x,r)=\frac{2}{3}\underset{\_}{g}(r){\left(1-\frac{2}{3}\right)}^{n}x.\hfill \end{array}$(27) Then$\underset{\_}{u}(x,r)=\sum _{n=0}^{\infty }\frac{2}{3}\underset{\_}{g}(r){\left(1-\frac{2}{3}\right)}^{n}x=\underset{\_}{g}(r)x.$Similarly, we have(28) $\begin{array}{c}{\overline{u}}_1(x,r)=\frac{\overline{f}(x,r)}{2\sqrt{x}}=\frac{4/3\overline{g}(r)x^{\frac{3}{2}}}{2\sqrt{x}}=\frac{2}{3}\overline{g}(r)x,\hfill \\ {\overline{u}}_2(x,r)=\frac{2}{3}\overline{g}(r)x-\frac{1}{2\sqrt{x}}\left(\frac{2}{3}\overline{g}(r)\right)\left(\frac{4}{3}{x}^{3/2}\right)=\frac{2}{3}\overline{g}(r)\left(1-\frac{2}{3}\right)x,\hfill \\ {\overline{u}}_3(x,r)=\frac{2}{3}\overline{g}(r){\left(1-\frac{2}{3}\right)}^{2}x,\hfill \\ ⋮\hfill \\ {\overline{u}}_n(x,r)=\frac{2}{3}\overline{g}(r){\left(1-\frac{2}{3}\right)}^{n}x.\hfill \end{array}$(28) Then$\overline{u}(x,r)=\sum _{n=0}^{\infty }\frac{2}{3}\overline{g}(r){\left(1-\frac{2}{3}\right)}^{n}x=\overline{g}(r)x.$Which is the exact solution is $\tilde{u}(x)=(\underset{\_}{g}(r)x,\overline{g}(r)x)$. Now, we assume that Eq. (21)(21) $\left(\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\underset{\_}{g}(r)x^{n+(1/2)},\frac{2^{n+1}\Gamma (n+1)}{1\times 3\times 5\cdots \times (2n+1)}\overline{g}(r)x^{n+(1/2)}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt}.$(21) is true for n = m − 1. We prove the relations for n = m. We consider the following equation(29) $\left(\frac{2^{m+1}\Gamma (m+1)}{1\times 3\times 5\cdots \times (2m+1)}\underset{\_}{g}(r)x^{m+(1/2)},\frac{2^{m+1}\Gamma (m+1)}{1\times 3\times 5\cdots \times (2m+1)}\overline{g}(r)x^{m+(1/2)}\right)={\int }_0^x\frac{u(t,r)}{\sqrt{x-t}}\mathrm{dt},$(29) which is the exact solution(30) $\tilde{u}(x)=(\underset{\_}{g}(r)x^m,\overline{g}(r)x^m),$(30) where m is an integer number. Using c = −1 in Eq. (20)(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20) we have(31) $\begin{array}{c}{\underset{\_}{u}}_1(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\underset{\_}{g}(r)x^m,\hfill \\ {\underset{\_}{u}}_2(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\underset{\_}{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^2{x}^m,\hfill \\ {\underset{\_}{u}}_3(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\underset{\_}{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^3{x}^m,\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_n(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\underset{\_}{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^n{x}^m.\hfill \end{array}$(31) Then$\underset{\_}{u}(x,r)=\sum _{n=0}^{\infty }\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\underset{\_}{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^n{x}^m=\underset{\_}{g}(r)x^m.$Similarly, we have(32) $\begin{array}{c}{\overline{u}}_1(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\overline{g}(r)x^m,\hfill \\ {\overline{u}}_2(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\overline{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^2{x}^m,\hfill \\ {\overline{u}}_3(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\overline{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^3{x}^m,\hfill \\ ⋮\hfill \\ {\overline{u}}_n(x,r)=\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\overline{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^n{x}^m.\hfill \end{array}$(32) Then$\overline{u}(x,r)=\sum _{n=0}^{\infty }\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\overline{g}(r){\left(1-\frac{2^m\Gamma (m+1)}{1\times 3\times 5\times \cdots \times (2m+1)}\right)}^n{x}^m=\overline{g}(r)x^m.$which is the exact solution $\tilde{u}(x)=(\underset{\_}{g}(r)x^m,\overline{g}(r)x^m)$. The same trend holds for Eq. (22)(22) $\left(\sqrt{\pi }\frac{\Gamma ((n+2)/2)}{\Gamma ((n+3)/2)}\underset{\_}{g}(r)x^{(n+1)/2},\sqrt{\pi }\frac{\Gamma ((n+2)/2)}{\Gamma ((n+3)/2)}\overline{g}(r)x^{(n+1)/2}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt},$(22) .

5 Numerical examples

Here, we consider three examples to illustrate the homotopy analysis method for solving Abel fuzzy integral equations.

Example 1

Consider the following Abel fuzzy integral equation$\left(\frac{4}{3}{\mathrm{rx}}^{(3/2)},\frac{4}{3}(2-r)x^{(3/2)}\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt}.$The exact solution in this case is given by$\tilde{u}(x)=(\mathrm{rx},(2-r)x)\mathrm{and}0\le r\le 1.$

By substituting c = −1 in Eq. (20)(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20)

$\begin{array}{c}{\underset{\_}{u}}_1(x,r)=\frac{2}{3}\mathrm{rx},\hfill \\ {\underset{\_}{u}}_2(x,r)=\frac{2}{3}r\left(1-\frac{2}{3}\right)x,\hfill \\ {\underset{\_}{u}}_3(x,r)=\frac{2}{3}r{\left(1-\frac{2}{3}\right)}^{2}x,\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_n(x,r)=\frac{2}{3}r{\left(1-\frac{2}{3}\right)}^{n}x,\hfill \end{array}$and also

$\begin{array}{c}{\overline{u}}_1(x,r)=\frac{2}{3}(2-r)x,\hfill \\ {\overline{u}}_2(x,r)=\frac{2}{3}(2-r)\left(1-\frac{2}{3}\right)x,\hfill \\ {\overline{u}}_3(x,r)=\frac{2}{3}(2-r){\left(1-\frac{2}{3}\right)}^{2}x,\hfill \\ ⋮\hfill \\ {\overline{u}}_n(x,r)=\frac{2}{3}(2-r){\left(1-\frac{2}{3}\right)}^{n}x.\hfill \end{array}$Thus,$(\underset{\_}{u}(x,r),\overline{u}(x,r))=\left(\sum _{n=0}^{\infty }\frac{2}{3}r{\left(1-\frac{2}{3}\right)}^{n}x,\sum _{n=0}^{\infty }\frac{2}{3}(2-r){\left(1-\frac{2}{3}\right)}^{n}x\right),$where the above summation yields to the exact solution$\tilde{u}(x)=(\mathrm{rx},(2-r)x).$

Example 2

Consider the following Abel fuzzy integral equation$\left(\frac{5}{16}(r^2+r)\pi x^3,\frac{5}{16}(4-r^3-r)\pi x^3\right)={\int }_0^x\frac{\tilde{u}(t)}{\sqrt{x-t}}\mathrm{dt}.$The exact solution in this case is given by$\tilde{u}(x)=((r^2+r)x^{(5/2)},(4-r^3-r)x^{(5/2)})\mathrm{and}0\le r\le 1.$

By substituting c = −1 in Eq. (20)(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20) $\begin{array}{c}{\underset{\_}{u}}_1(x,r)=\frac{5}{32}(r^2+r)\pi x^{(5/2)},\hfill \\ {\underset{\_}{u}}_2(x,r)=\frac{5}{32}(r^2+r)\left(1-\frac{5}{32}\pi \right)\pi x^{(5/2)},\hfill \\ {\underset{\_}{u}}_3(x,r)=\frac{5}{32}(r^2+r){\left(1-\frac{5}{32}\pi \right)}^2\pi x^{(5/2)},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_n(x,r)=\frac{5}{32}(r^2+r){\left(1-\frac{5}{32}\pi \right)}^n\pi x^{(5/2)},\hfill \end{array}$and also$\begin{array}{c}{\overline{u}}_1(x,r)=\frac{5}{32}(4-r^3-r)\pi x^{(5/2)},\hfill \\ {\overline{u}}_2(x,r)=\frac{5}{32}(4-r^3-r)\left(1-\frac{5}{32}\pi \right)\pi x^{(5/2)},\hfill \\ {\overline{u}}_3(x,r)=\frac{5}{32}(4-r^3-r){\left(1-\frac{5}{32}\pi \right)}^2\pi x^{(5/2)},\hfill \\ ⋮\hfill \\ {\overline{u}}_n(x,r)=\frac{5}{32}(4-r^3-r){\left(1-\frac{5}{32}\pi \right)}^n\pi x^{(5/2)}.\hfill \end{array}$Thus,$\tilde{u}(x)=\left(\sum _{n=0}^{\infty }\frac{5}{32}(r^2+r){\left(1-\frac{5}{32}\pi \right)}^n\pi x^{(5/2)},\sum _{n=0}^{\infty }\frac{5}{32}(4-r^3-r){\left(1-\frac{5}{32}\pi \right)}^n\pi x^{(5/2)}\right),$where the above summation yields to the exact solution$\tilde{u}(x)=((r^2+r)x^{(5/2)},(4-r^3-r)x^{(5/2)}).$

Example 3

Consider the following nonlinear Abel fuzzy integral equation(33) $\left(\frac{2043}{3003}{(r^2+2r)}^3{x}^{(13/2)},\frac{2043}{3003}{(6-3r^3)}^3{x}^{(13/2)}\right)={\int }_0^x\frac{{\tilde{u}}^3(t)}{\sqrt{x-t}}\mathrm{dt}.$(33) The exact solution in this case is given by$\tilde{u}(x)=((r^2+2r)x^2,(6-3r^3)x^2)\mathrm{and}0\le r\le 1.$The transformation(34) $\tilde{v}(x)={\tilde{u}}^3(x),\tilde{u}(x)=\sqrt[3]{\tilde{v}(x)},$(34) carries Eq. (34)(34) $\tilde{v}(x)={\tilde{u}}^3(x),\tilde{u}(x)=\sqrt[3]{\tilde{v}(x)},$(34) into(35) $\left(\frac{2043}{3003}(r^2+2r)x^{(13/2)},\frac{2043}{3003}(6-3r^3)x^{(13/2)}\right)={\int }_0^x\frac{\tilde{v}(t)}{\sqrt{x-t}}\mathrm{dt}.$(35) Substituting Eq. (35)(35) $\left(\frac{2043}{3003}(r^2+2r)x^{(13/2)},\frac{2043}{3003}(6-3r^3)x^{(13/2)}\right)={\int }_0^x\frac{\tilde{v}(t)}{\sqrt{x-t}}\mathrm{dt}.$(35) in Eq. (20)(20) $\begin{array}{c}{\underset{\_}{u}}_1(x;r)=-c\frac{\underset{\_}{f}(x;r)}{2\sqrt{x-a}},\hfill \\ {\overline{u}}_1(x;r)=-c\frac{\overline{f}(x;r)}{2\sqrt{x-a}},\hfill \\ ⋮\hfill \\ {\underset{\_}{u}}_m(x;r)={\underset{\_}{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\underset{\_}{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt},\hfill \\ {\overline{u}}_m(x;r)={\overline{u}}_{m-1(x;r)}+c\frac{1}{2\sqrt{x-a}}{\int }_a^x\frac{{\overline{u}}_{m-1}(t;r)}{\sqrt{x-t}}\mathrm{dt}\hfill \end{array}$(20) gives:$\begin{array}{c}{\underset{\_}{v}}_1(x,r)=\frac{1024}{3003}{(r^2+2r)}^3{x}^6,\hfill \\ {\underset{\_}{v}}_2(x,r)=\frac{1024}{3003}{(r^2+2r)}^3\left(1-\frac{1024}{3003}\right)x^6,\hfill \\ {\underset{\_}{v}}_3(x,r)=\frac{1024}{3003}{(r^2+2r)}^3{\left(1-\frac{1024}{3003}\right)}^2{x}^6,\hfill \\ ⋮\hfill \\ {\underset{\_}{v}}_n(x,r)=\frac{1024}{3003}{(r^2+2r)}^3{\left(1-\frac{1024}{3003}\right)}^n{x}^6,\hfill \\ \hfill \end{array}$and also$\begin{array}{c}{\overline{v}}_1(x,r)=\frac{1024}{3003}{(6-3r^3)}^3{x}^6,\hfill \\ {\overline{v}}_2(x,r)=\frac{1024}{3003}{(6-3r^3)}^3\left(1-\frac{1024}{3003}\right)x^6,\hfill \\ {\overline{v}}_3(x,r)=\frac{1024}{3003}{(6-3r^3)}^3{\left(1-\frac{1024}{3003}\right)}^2{x}^6,\hfill \\ ⋮\hfill \\ {\overline{v}}_n(x,r)=\frac{1024}{3003}{(6-3r^3)}^3{\left(1-\frac{1024}{3003}\right)}^n{x}^6.\hfill \end{array}$Thus,$\tilde{v}(x)=(\underset{\_}{v}(x,r),\overline{v}(x,r))=\left(\sum _{n=0}^{\infty }\frac{1024}{3003}{(r^2+2r)}^3{\left(1-\frac{1024}{3003}\right)}^n{x}^6,\sum _{n=0}^{\infty }\frac{1024}{3003}{(6-3r^3)}^3{\left(1-\frac{1024}{3003}\right)}^n{x}^6\right),$where the above summation yields to the exact solution$\tilde{v}(x)=({(r^2+2r)}^3{x}^6,{(6-3r^3)}^3{x}^6).$Then, by using Eq. (34)(34) $\tilde{v}(x)={\tilde{u}}^3(x),\tilde{u}(x)=\sqrt[3]{\tilde{v}(x)},$(34) , we have:$\tilde{u}(x)=((r^2+2r)x^2,(6-3r^3)x^2).$

6 Conclusion

In this paper, linear and nonlinear Abel fuzzy integral equations were converted into two crisp linear and nonlinear Abel integral equations based on the embedding method. Then, we applied homotopy analysis method to obtain the unique solution of Abel fuzzy integral equations. It was shown that this new technique is easy to implement and produces accurate results. A considerable advantage of the method is that the approximate solutions are found very easily by using computer programs such as Matlab. The method can also be extended to the system of linear integro-differential equations with variable coefficients, but some modifications are needed.

Acknowledgement

We appreciate the editors and the reviewers for their careful reading, valuable suggestions and timely review.

Notes

Peer review under responsibility of Taibah University