A hybrid method for solving fuzzy Volterra integral equations of separable type kernels

1 Introduction

The study of fuzzy integral equations is of more interest and fast growing, especially the relation to fuzzy control, which has been developed recently. Mostly, mathematical models are applied in various problems of Chemistry, Engineering, Biology, Physics, and other many fields are based on integral equations, should be Mostly, mathematical models are applied in various problems of Chemistry, Engineering, Biology, Physics, and other many fields are based on differential, fractional order differential and integral equations see Atangana (2018), Atangana (2020, 2017). Particularly, the Volterra linear integral equations of second kind (Bede, 2004; Congxin and Ma, 1990; Chang and Zadeh, 1972; Friedman et al., 1996, 1997) is one of the most powerful types of integral equations to deal with such type of problems. Clearly, each model has some parameters which may be inherited to some vagueness. To solve these models we should mention such studied problems under uncertainty which leads to the presentation of fuzzy concept. The numerical idea for solving the Fredholm fuzzy linear integral equations of second kind by using the Adomian method were presented by Babolian et al. (2005). Also Friedman et al. (1999) presented the embedding method for the solution of fuzzy Volterra and Fredholm integral equations. In another numerical scheme, which was proposed by Abbasbandy et al. (2007) for the solutions of second kind fuzzy linear Fredholm integral equations, the authors applied the parametric form of fuzzy number to transform the fuzzy linear Fredholm integral equations into two systems of linear integral equation in deterministic cases. Recently, to solve the second kind of non-linear fuzzy Volterra–Fredholm integral equations by using the algorithm of homotopy perturbation method was presented by Attari et al. (2011). Molabahrami et al. (2011) used fuzzy parametric pair to transform the fuzzy Fredholm linear integral equation into two systems of linear integral equations as in crisp form, then they applied homotopy analysis method for obtaining the approximate solutions of the systems. To get the solutions of fuzzy Volterra and Fredholm integral equations, there are many research papers which presented the integration methods numerically for the solution of fuzzy-valued function (Friedman et al., 1999; Goetschel et al., 1986; Maltok, 1987; Nanda, 1989; Park et al., 1995; Wu, 1999, 2000). Moreover, there are some analytical and computational methods for the solution of fuzzy Volterra and fuzzy Volterra integro-differential equation (Salahshour and Allahviranloo, 2013; Alikhani et al., 2012). The classical Volterra integral equation is given by

Here we remark that we investigate the given problem in Eq. (1) under fuzzy concept for analytical solution. The fuzzy form of the given equation is provided by

This paper is arranged as follow: Section 1 is the introduction part, Section 2 consists of some basic necessary preliminaries about the fuzzy calculus. Section 3, contains the main work and in Section 4, some examples are handled by using the suggested method. Finally, a brief conclusion is given in Section 5.

2 Preliminaries

Some basic definitions are given which used throughout the paper.

Let $E_1$ be the set of all fuzzy numbers on R. The ${\alpha }_1$ -level set of $\nu$ , is denoted by $${\left[\nu \right]}^{{\alpha }_1}=\left\{\begin{array}{cc}\{x\in R:\nu (x)⩾{\alpha }_1\}\hfill & \mathit{if}0<{\alpha }_1\le 1,\hfill \\ \left(\mathit{cl}\right(\mathit{supp}\nu )\hfill & \mathit{if}{\alpha }_1=0,\hfill \end{array}\right.$$ and is a closed bounded interval $\left[\underline{\nu }\right({\alpha }_1),\bar{\nu }({\alpha }_1\left)\right]$ , where, $\underline{\nu }\left({\alpha }_1\right)$ is the left-hand endpoint and $\bar{\nu }\left({\alpha }_1\right)$ the right-hand endpoint of ${\left[\nu \right]}^{{\alpha }_1}$ respectively.

For any different fuzzy numbers $$\nu =\left(\underline{\nu }\right({\alpha }_1),\bar{\nu }({\alpha }_1\left)\right),\omega =\left(\underline{\omega }\right({\alpha }_1),\bar{\omega }({\alpha }_1\left)\right),$$ and for arbitrary scaler ${\kappa }_1$ , the various operations are defined as follow,

1. Addition: $(\underline{\nu \left({\alpha }_1\right)+\omega \left({\alpha }_1\right)},\overline{\nu \left({\alpha }_1\right)+\omega \left({\alpha }_1\right)})=(\underline{\nu \left({\alpha }_1\right)}+\underline{\omega \left({\alpha }_1\right)},\overline{\nu \left({\alpha }_1\right)}+\overline{\omega \left({\alpha }_1\right)})$ .

2. Subtraction: $(\underline{\nu \left({\alpha }_1\right)-\omega \left({\alpha }_1\right)},\overline{\nu \left({\alpha }_1\right)-\omega \left({\alpha }_1\right)})=(\underline{\nu \left({\alpha }_1\right)}-\underline{\omega \left({\alpha }_1\right)},\overline{\nu \left({\alpha }_1\right)}-\overline{\omega \left({\alpha }_1\right)})$ .

3. Scaler multiplication: ${\kappa }_1·\nu \left({\alpha }_1\right)=\left\{\begin{array}{cc}\left({\kappa }_1\underline{\nu }\right({\alpha }_1),{\kappa }_1\bar{\nu }({\alpha }_1\left)\right)\hfill & {\kappa }_1⩾0,\hfill \\ \left({\kappa }_1\bar{\nu }\right({\alpha }_1),{\kappa }_1\underline{\nu }({\alpha }_1\left)\right)\hfill & {\kappa }_1<0\text{.}\hfill \end{array}\right.$

In $E_1$ , a metric $D_1$ as defined above have following properties (see Puri and Ralescu, 1986):

1. $D_1(\nu +\upsilon ,\omega +\upsilon )=D_1(\nu ,\omega )$   for all $\nu ,\upsilon ,\omega \in E_1$ ;

2. $D_1({\kappa }_1·\nu ,{\kappa }_1·\omega )=\left|{\kappa }_1\right|D_1(\nu ,\omega )$   for all ${\kappa }_1\in R,\nu ,\omega \in E_1$ ;

3. $D_1(\nu +\mu ,\omega +\upsilon )⩽D_1(\nu ,\omega )+D_1(\mu ,\upsilon )$ for all $\nu ,\omega ,\mu ,\upsilon \in E_1$ ;

4. $(E_1,D_1)$ is a complete metric space.

3 Basic idea

In this section, we propose a basic idea to solve the fuzzy Volterra linear integral equations with convolution class kernel applying LADM. Suppose that a kernel function $k(y,s)$ is in the form of $(y-s)$ which is a separable kernel i.e. $k(y,s)={\sum }_{i=0}^n{h}_i\left(y\right)g_i\left(s\right)$ and satisfies the conditions of Theorem 1 such that the solution of Eq. (2) exists and may be unique. We assume that for any ${\beta }_1⩽y,s⩽{\beta }_2$ in $k(y,s)$ . Let the parametric form of Eq. (2) are written as

Now applying the Laplace transform and Theorem 2 to Eq. (5), one has $$\left\{\begin{array}{c}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\underline{h}\right(y,{\alpha }_1\left)\right]+L\left[{\int }_{{\beta }_1}^y(y-s)\underline{\nu }(s,{\alpha }_1)\mathit{ds}\right],\hfill \\ L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\bar{h}\right(y,{\alpha }_1\left)\right]+L\left[{\int }_{{\beta }_1}^y(y-s)\bar{\nu }(s,{\alpha }_1)\mathit{ds}\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\underline{h}\right(y,{\alpha }_1\left)\right]+L\left[y\right]·L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right],\hfill \\ L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\bar{h}\right(y,{\alpha }_1\left)\right]+L\left[y\right]·L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right],\hfill \end{array}\right.$$

Now applying the inverse Laplace transform on Eq. (6), and simplify one has

Let the solution of Eq. (2) can be written in series as

Putting Eq. (8) in Eq. (7), we get

Comparing both side of Eq. (9) termwise respectively, we get $$\left\{\begin{array}{c}{\underline{\nu }}_0(y,{\alpha }_1)=\underline{h}(y,{\alpha }_1),\hfill \\ {\bar{\nu }}_0(y,{\alpha }_1)=\bar{h}(y,{\alpha }_1),\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_1\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_1\right(y,{\alpha }_1\left)\right]\right]\text{.}\hfill \end{array}\right.$$ and so on and we write general terms as

4 Application

In this section, three examples are given to described the application of proposed algorithm.

Parametric form of Eq. (11) can be written as

Applying the Laplace transform and convolution Theorem 2 to Eq. (12), one has $$\left\{\begin{array}{c}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(3+{\alpha }_1\left)\right]+L\left[{\int }_0^y(y-s)\underline{\nu }(s,{\alpha }_1)\mathit{ds}\right],\hfill \\ L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(8-2{\alpha }_1\left)\right]+L\left[{\int }_0^y(y-s)\bar{\nu }(s,{\alpha }_1)\mathit{ds}\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(3+{\alpha }_1\left)\right]+L\left[y\right]·L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right],\hfill \\ L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(8-2{\alpha }_1\left)\right]+L\left[y\right]·L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(3+{\alpha }_1\left)\right]+\frac{1}{s^2}L\left[\underline{\nu }\right(y,{\alpha }_1\left)\right],\hfill \\ L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]=L\left[\right(8-2{\alpha }_1\left)\right]+\frac{1}{s^2}L\left[\bar{\nu }\right(y,{\alpha }_1\left)\right]\text{.}\hfill \end{array}\right.$$

Now applying the inverse Laplace transform, and simplify, we get

Let the series solution of Eq. (11) be as

Putting Eq. (14) in Eq. (13) $$\left\{\begin{array}{c}{\displaystyle \sum _{i=0}^{\infty }}{\underline{\nu }}_i(y,{\alpha }_1)=(3+{\alpha }_1)+L^{-1}\left[\frac{1}{s^2}L\left[{\displaystyle \sum _{i=0}^{\infty }}{\underline{\nu }}_i(y,{\alpha }_1)\right]\right],\hfill \\ {\displaystyle \sum _{i=0}^{\infty }}{\bar{\nu }}_i(y,{\alpha }_1)=(8-2{\alpha }_1)+L^{-1}\left[\frac{1}{s^2}L\left[{\displaystyle \sum _{i=0}^{\infty }}{\bar{\nu }}_i(y,{\alpha }_1)\right]\right],\hfill \end{array}\right.$$ which may written as

Comparing termwise of Eq. (15) on both side respectively, we obtain $$\left\{\begin{array}{c}{\underline{\nu }}_0(y,{\alpha }_1)=3+{\alpha }_1,\hfill \\ {\bar{\nu }}_0(y,{\alpha }_1)=8-2{\alpha }_1,\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_1\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_1\right(y,{\alpha }_1\left)\right]\right]\text{.}\hfill \end{array}\right.$$

The general terms are given by

Next in Fig. 1, we provide graphical presentation of fuzzy solutions at the given values of uncertainty ${\alpha }_1$ as.

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Fig. 1

Graphical presentation of fuzzy solutions at the given values of ${\alpha }_1$ for Example 1.

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To solve the Eq. (19) by LADM.

Considered the parametric form of Eq. (19) which is given by

Applying the Laplace transform together with convolution Theorem 2 and after the inverse Laplace transform, we get

Let the solution of Eq. (19) in the form of infinite series as

Putting Eq. (22) in Eq. (21), one has

Comparing termwise of Eq. (23) both side respectively $$\left\{\begin{array}{c}{\underline{\nu }}_0(y,{\alpha }_1)={\alpha }_1\left(1-y-\frac{y^2}{2}\right),\hfill \\ {\bar{\nu }}_0(y,{\alpha }_1)=(2-{\alpha }_1)\left(1-y-\frac{y^2}{2}\right),\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_1(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_0\right(y,{\alpha }_1\left)\right]\right],\hfill \end{array}\right.$$ $$\left\{\begin{array}{c}{\underline{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\underline{\nu }}_1\right(y,{\alpha }_1\left)\right]\right],\hfill \\ {\bar{\nu }}_2(y,{\alpha }_1)=L^{-1}\left[\frac{1}{s^2}L\left[{\bar{\nu }}_1\right(y,{\alpha }_1\left)\right]\right],\hfill \end{array}\right.$$ the remaining terms may be in same way computed. The general terms are written as