Hindawi Journal of Applied Mathematics Volume 2017, Article ID 2417195, 12 pages https://doi.org/10.1155/2017/2417195
Research Article Computational Methods for Solving Linear Fuzzy Volterra Integral Equation Jihan Hamaydi and Naji Qatanani Department of Mathematics, An-Najah National University, Nablus, State of Palestine Correspondence should be addressed to Naji Qatanani;
[email protected] Received 21 February 2017; Accepted 18 April 2017; Published 28 May 2017 Academic Editor: Mehmet Sezer Copyright © 2017 Jihan Hamaydi and Naji Qatanani. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Two numerical schemes, namely, the Taylor expansion and the variational iteration methods, have been implemented to give an approximate solution of the fuzzy linear Volterra integral equation of the second kind. To display the validity and applicability of the numerical methods, one illustrative example with known exact solution is presented. Numerical results show that the convergence and accuracy of these methods were in a good agreement with the exact solution. However, according to comparison of these methods, we conclude that the variational iteration method provides more accurate results.
1. Introduction Fuzzy integral equations of the second kind have attracted the attention of many scientists and researchers in recent years. These equations appear frequently in fuzzy control, fuzzy finance, approximate reasoning, and economic systems [1]. The concept of fuzzy sets was originally introduced by Zadeh [2] and led to the definition of fuzzy numbers and its implementation in fuzzy control [3] and approximate reasoning problems [4]. Dubois and Prade [5] were the first to introduce the concept of fuzzy functions. Alternative approaches were later suggested by Goetschel and Voxman [6], Kaleva [7], Nanda [8], and others. In recent years, numerous methods have been proposed for solving Volterra integral equations [9]. Tricomi [10] was the first to introduce the successive approximations method for nonlinear integral equations. Liao [11] employed the homotopy analysis method to solve nonlinear problems and then it has been applied by Abbasbandy [12] to solve fuzzy Volterra integral equations of the second kind. Babolian et al. [13] have used the orthogonal triangular functions as a direct method for numerically solving integral equations system. Jafarian et al. [14] have solved systems of linear integral equations by using Legendre wavelets. Kanwal and Liu [15] implemented the Taylor expansion approach for
solving integral equations while Xu [16] solved integral equations by the variational iteration method. In addition, Amawi and Qatanani [17, 18] have investigated the analytical and numerical treatment of fuzzy linear Fredholm integral equation. Hamaydi [19] has used various analytical and numerical methods to solve fuzzy Volterra integral equations. The paper is organized as follows: in Section 2, fuzzy Volterra integral equation of the second kind is introduced. The Taylor expansion method used to approximate solution of fuzzy Volterra integral equation is addressed in Section 3. In Section 4, we present the variational iteration method (VIM) which provides a sequence of functions that converges to the exact solution to the problem. The proposed methods are implemented using a numerical example with known exact solution by applying the MAPLE software in Section 5. Conclusions are given in Section 6.
2. Fuzzy Volterra Integral Equation A standard form of the Volterra integral equation of the second kind is given by [14] 𝑥
𝑢 (𝑥) = 𝑓 (𝑥) + 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡) 𝑑𝑡, 𝑎
(1)
2
Journal of Applied Mathematics
where 𝜆 is a positive parameter, 𝑘(𝑥, 𝑡) is an arbitrary function called the kernel of the integral equation defined over square 𝐺 : [𝑎, 𝑏] ∗ [𝑎, 𝑏], 𝑘(𝑥, 𝑡) = 0, 𝑎 ≤ 𝑡, 𝑥 ≤ 𝑏, 𝑡 > 𝑥, and 𝑓(𝑥) is a given function of 𝑥 ∈ [𝑎, 𝑏]. If𝑓(𝑥) is a fuzzy function, then (1) is called a fuzzy Volterra integral equation of the second kind. Definition 1 (see [20]). The second kind fuzzy Volterra integral equations system is of the following form:
From (2), (3), and (4) we have the following system: 𝑖
𝑐𝑖,𝑗
𝑢𝑖 (𝑥, 𝛼) = 𝑓𝑖 (𝑥, 𝛼) + ∑ 𝜆 𝑖,𝑗 (∫ 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) 𝑑𝑡 𝑗=1
𝑎
𝑥
+ ∫ 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) 𝑑𝑡) , 𝑐𝑖,𝑗
𝑖
𝑚
𝑢𝑖 (𝑥, 𝛼) = 𝑓𝑖 (𝑥, 𝛼) + ∑𝜆 𝑖,𝑗 (∫ 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) 𝑑𝑡
𝑥
𝑢𝑖 (𝑥) = 𝑓𝑖 (𝑥) + ∑ 𝜆 𝑖𝑗 ∫ 𝑘𝑖𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡) 𝑑𝑡, 𝑗=1
(5)
𝑐𝑖,𝑗
𝑎
(2)
𝑎
𝑗=1
𝑥
+ ∫ 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) 𝑑𝑡) . 𝑐𝑖,𝑗
where 𝑎 ≤ 𝑡 ≤ 𝑥 ≤ 𝑏 and 𝜆 𝑖𝑗 ≠ 0 (for 𝑖, 𝑗 = 1, 2, 3, . . . , 𝑚) are real constants. Moreover, in system (2), the fuzzy function 𝑓𝑖 (𝑥) and kernel 𝑘𝑖𝑗 (𝑥, 𝑡) are given and assumed to be sufficiently differentiable with respect to all their arguments on the interval 𝑎 ≤ 𝑡, 𝑥 ≤ 𝑏, and we assume that the kernel function 𝑘𝑖𝑗 (𝑥, 𝑡) ∈ 𝐺 : [𝑎, 𝑏] ∗ [𝑎, 𝑏], and 𝑢𝑗 (𝑥) = [𝑢1 , 𝑢2 , . . . , 𝑢𝑚 ]𝑡 is the solution to be determined. Now let (𝑓𝑖 (𝑥, 𝛼), 𝑓𝑖 (𝑥, 𝛼)) and (𝑢𝑖 (𝑥, 𝛼), 𝑢𝑖 (𝑥, 𝛼)), 0 ≤ 𝛼 ≤ 1, 𝑥 ∈ [𝑎, 𝑏], be the parametric forms of 𝑓𝑖 (𝑥) and 𝑢𝑖 (𝑥), respectively; then the parametric forms of the fuzzy Volterra integral equations system are as follows: 𝑚
𝑥
𝑢𝑖 (𝑥, 𝛼) = 𝑓𝑖 (𝑥, 𝛼) + ∑ (𝜆 𝑖𝑗 ∫ 𝑔 (𝑡, 𝛼) 𝑑𝑡) , 𝑗=1 𝑚
𝑎
𝑖𝑗
𝑥
𝑢𝑖 (𝑥, 𝛼) = 𝑓𝑖 (𝑥, 𝛼) + ∑ (𝜆 𝑖𝑗 ∫ 𝑔𝑖𝑗 (𝑡, 𝛼) 𝑑𝑡) , 𝑗=1
(3)
𝑎
𝑖 = 1, 2, . . . , 𝑚,
We seek the solution of system (5) in the form of (𝑟) 1 𝜕 𝑢𝑗𝑠 (𝑥, 𝛼) (𝑡 − 𝑧)𝑟 ) , 𝑢𝑗,𝑠 (𝑡, 𝛼) = ∑ ( 𝑟! 𝑟 𝑥=𝑧 𝑟=0 (𝑟) 𝑠 1 𝜕 𝑢𝑗𝑠 (𝑥, 𝛼) 𝑟 𝑢𝑗,𝑠 (𝑡, 𝛼) = ∑ ( (𝑡 − 𝑧) ) 𝑟! 𝜕𝑥𝑟 𝑟=0 𝑥=𝑧 𝑠
𝑎 ≤ 𝑥, 𝑧 ≤ 𝑏, for 𝑗 = 1, . . . , 𝑚, which are the Taylor expansion of degree 𝑠 at 𝑥 = 𝑧 for the unknown functions 𝑢𝑗𝑠 (𝑥, 𝛼) and 𝑢𝑗𝑠 (𝑥, 𝛼), respectively. To obtain the solution in the form of expression (6) we find the 𝑛th derivative of each equation in system (5) with respect to 𝑥 by using the Leibniz rule, (𝑛 = 0, 1, . . . , 𝑠) and obtain [21] (𝑛) 𝜕(𝑛) 𝑢𝑖𝑠 (𝑥, 𝛼) 𝜕 𝑓𝑖 (𝑥, 𝛼) = 𝜕𝑥𝑛 𝜕𝑥𝑛
where {𝑘𝑖𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) , 𝑘𝑖𝑗 (𝑥, 𝑡) ≥ 0 𝑔𝑖𝑗 (𝑡, 𝛼) = { 𝑘 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) , 𝑘𝑖𝑗 (𝑥, 𝑡) < 0, { 𝑖𝑗
𝑚 { 𝑐𝑖,𝑗 𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡) + ∑ 𝜆 𝑖,𝑗 {∫ 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡 𝜕𝑥𝑛 𝑎 𝑗=1 {
(4)
{𝑘𝑖𝑗 (𝑥, 𝑡) 𝑢𝑗 (𝑡, 𝛼) , 𝑘𝑖𝑗 (𝑥, 𝑡) ≥ 0 𝑔 (𝑡, 𝛼) = { 𝑖𝑗 𝑘 𝑡) 𝑢𝑗 (𝑡, 𝛼) , 𝑘𝑖𝑗 (𝑥, 𝑡) < 0. { 𝑖𝑗 (𝑥,
3. Taylor Expansion Method This method depends on differentiating the fuzzy integral equation of the second kind 𝑛 – times; then substitute the Taylor series expansion for the unknown function into the integral equation. As a result, we get a linear system for which the solution of this system yields the unknown Taylor coefficients of the solution functions.
(6)
(𝑛−𝑙−1) 𝜕(𝑙) 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗𝑠 (𝑥, 𝛼)) +∑( 𝑡=𝑥 𝜕𝑥𝑙 𝑙=0 𝑛−1
+∫
𝑥
𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡)
𝑐𝑖,𝑗
𝜕(𝑛) 𝑢𝑗𝑠 (𝑥, 𝛼) 𝜕𝑥𝑛
𝜕𝑥𝑛 =
} 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡} , }
𝜕(𝑛) 𝑓𝑖 (𝑥, 𝛼) 𝜕𝑥𝑛
𝑚 { 𝑐𝑖,𝑗 𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡) + ∑ 𝜆 𝑖,𝑗 {∫ 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡 𝜕𝑥𝑛 𝑎 𝑗=1 {
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3
(𝑛−𝑙−1) 𝜕(𝑙) 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑢𝑗𝑠 (𝑥, 𝛼)) +∑( 𝑡=𝑥 𝜕𝑥𝑙 𝑙=0 𝑛−1
𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡)
𝑥
+∫
𝜕𝑥𝑛
𝑐𝑖,𝑗
Substituting (9) into (8) gives
} 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡} , } (7)
for 𝑛 = 0, 1, . . . , 𝑠 and 𝑖 = 1, 2, . . . , 𝑚. Using the Leibniz rule which is dealing with differentiation of product of functions, system (7) becomes 𝜕(𝑛) 𝑓𝑖 (𝑥, 𝛼)
(𝑛)
𝜕 𝑢𝑖𝑠 (𝑥, 𝛼) = 𝜕𝑥𝑛
𝜕𝑥𝑛
𝑚
𝑛−1
𝑗=1
𝑟=0
(𝑛) (𝑖,𝑗) (𝑟) 𝑢𝑗𝑆 (𝑧, 𝛼) 𝑢(𝑛) (𝑧, 𝛼) + ∑ { ∑ 𝐷𝑛,𝑟 𝑖,𝑠 (𝑧, 𝛼) = 𝑓𝑖 𝑠
𝑠
𝑟=0
𝑟=0
(𝑖,𝑗)
(𝑖,𝑗) (𝑟) + ∑𝐸𝑛,𝑟 𝑢𝑗𝑆 (𝑧, 𝛼) + ∑𝐸 𝑛,𝑟 𝑢(𝑟) 𝑗𝑆 (𝑧, 𝛼)} ,
𝑢(𝑛) 𝑖,𝑠 (𝑧, 𝛼) = 𝑓𝑖
𝑚
+ ∑ 𝜆 𝑖,𝑗
(𝑛)
𝑚
𝑛−1
𝑗=1
𝑟=0
(10)
(𝑖,𝑗) (𝑟) 𝑢𝑗𝑆 (𝑧, 𝛼) (𝑧, 𝛼) + ∑ { ∑ 𝐷𝑛,𝑟
𝑆
𝑠
𝑟=0
𝑟=0
(𝑖,𝑗)
(𝑖,𝑗) (𝑟) + ∑𝐸𝑛,𝑟 𝑢𝑗𝑆 (𝑧, 𝛼) + ∑𝐸 𝑛,𝑟 𝑢(𝑟) 𝑗𝑆 (𝑧, 𝛼)} ,
𝑗=1
(𝑛)
{ 𝑐𝑖,𝑗 𝜕 𝑘𝑖,𝑗 (𝑥, 𝑡) ⋅ {∫ 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡 𝜕𝑥𝑛 𝑎 {
where
(𝑛−𝑙−𝑟−1) 𝜕(𝑙) 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑛−𝑙−1 ) +∑ ∑ ( )( 𝑟 𝑡=𝑥 𝜕𝑥𝑙 𝑟=0 𝑙=0 𝑛−1 𝑛−𝑟−1
(𝑟)
⋅ (𝑢𝑗𝑠 (𝑥, 𝛼))
+∫
𝑥
(𝑛)
𝜕 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝜕𝑥𝑛
𝑐𝑖,𝑗
(𝑛)
(𝑛)
} 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡} , }
𝑚
(𝑖,𝑗) 𝐸𝑛,𝑟
(8)
𝜕 𝑢𝑖𝑠 (𝑥, 𝛼) 𝜕 𝑓𝑖 (𝑥, 𝛼) = + ∑𝜆 𝑖,𝑗 𝜕𝑥𝑛 𝜕𝑥𝑛 𝑗=1
(𝑖,𝑗) 𝐸 𝑛,𝑟
=
=
𝜆 𝑖,𝑗 𝑟! 𝜆 𝑖,𝑗 𝑟!
∫
𝑐𝑖,𝑗
𝜕(𝑛) 𝑘𝑖,𝑗 𝜕𝑥𝑛
𝑎 𝑍
𝜕(𝑛) 𝑘𝑖,𝑗
𝑐𝑖,𝑗
𝜕𝑥𝑛
∫
(𝑥, 𝑡) (𝑡 − 𝑧)𝑟 𝑑𝑡, 𝑥=𝑧 (𝑥, 𝑡) (𝑡 − 𝑧)𝑟 𝑑𝑡, 𝑥=𝑧 (11)
(𝑖,𝑗) 𝐷𝑛,𝑟
{ 𝑐𝑖,𝑗 𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡) ⋅ {∫ 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡 𝜕𝑥𝑛 𝑎 {
𝑛−𝑙
𝑛−𝑙−1
𝑟=0
𝑟
= 𝜆 𝑖,𝑗 ∑ (
(𝑛−𝑙−𝑟−1) 𝜕(𝑙) 𝑘𝑖,𝑗 (𝑥, 𝑡) 𝑛−𝑙−1 ) +∑ ∑ ( )( 𝑟 𝑡=𝑥 𝜕𝑥𝑙 𝑟=0 𝑙=0
)(
𝜕(𝑟) 𝑘𝑖,𝑗 𝜕𝑥𝑟
(𝑛−𝑙−𝑟−1) , (𝑥, 𝑡) ) 𝑥=𝑧
𝑛−1 𝑛−𝑟−1
(𝑟)
⋅ (𝑢𝑗𝑠 (𝑥, 𝛼))
+∫
𝑥
𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝑡)
𝑐𝑖,𝑗
𝜕𝑥
𝑛
𝑖, 𝑗 = 1, 2, . . . , 𝑚,
} 𝑢𝑗𝑠 (𝑡, 𝛼) 𝑑𝑡} . }
𝑛−1 (𝑖,𝑗) (𝑟) for 𝑛 = 0, ∑𝑚 𝑗=1 ∑𝑟=0 𝐷𝑛,𝑟 𝑢𝑗𝑆 (𝑧, 𝛼) = 0, for 𝑛 ≤ 𝑟, we have
Our objective is to determine the coefficients 𝑢𝑗𝑠 (𝑛) (𝑧, 𝛼) and
(𝑖,𝑗) = 0, and 𝑛, 𝑟 = 0, 1, . . . , 𝑠. 𝐷𝑛,𝑟 Consequently, (10) can be written in the matrix form:
𝑢𝑗𝑠 (𝑛) (𝑧, 𝛼), for 𝑛 = 0, . . . , 𝑠 and 𝑗 = 1, . . . , 𝑚 in system (7); thus we expand 𝑢𝑗𝑠 (𝑡, 𝛼) and 𝑢𝑗𝑠 (𝑡, 𝛼) in Taylor's series at arbitrary point 𝑧: 𝑎 ≤ 𝑧 ≤ 𝑏 (𝑟) 1 𝜕 𝑢𝑗,𝑠 (𝑥, 𝛼) ⋅ (𝑡 − 𝑧)𝑟 ) , 𝑟 𝑟! 𝜕𝑥 𝑥=𝑧 𝑟=0 (𝑟) 𝑠 1 𝜕 𝑢𝑗,𝑠 (𝑥, 𝛼) 𝑟 𝑢𝑗,𝑠 (𝑡, 𝛼) = ∑ ( 𝑟 ⋅ (𝑡 − 𝑧) ) , 𝑟! 𝜕𝑥 𝑥=𝑧 𝑟=0 𝑠
where
𝑢𝑗,𝑠 (𝑡, 𝛼) = ∑ (
𝑎 ≤ 𝑥, 𝑧 ≤ 𝑏, 0 ≤ 𝛼 ≤ 1, for 𝑗 = 1, . . . , 𝑚.
(𝐷 + 𝐸) 𝑈 = F,
(9)
𝐹 = [−𝑓 (𝑧, 𝛼) , . . . , −𝑓(𝑆) (𝑥, 𝛼) , −𝑓1 (𝑧, 𝛼) , . . . , 1 1 𝑥=𝑧 (𝑆) −𝑓1 (𝑥, 𝛼) , . . . , −𝑓 (𝑧, 𝛼) , . . . , −𝑓(𝑆) (𝑥, 𝛼) , 𝑚 𝑚 𝑥=𝑧 𝑥=𝑧 𝑡 (𝑆) −𝑓𝑚 (𝑧, 𝛼) , . . . , −𝑓𝑚 (𝑥, 𝛼) ] , 𝑥=𝑧
(12)
4
Journal of Applied Mathematics 𝑈 = [𝑢1𝑆 (𝑧, 𝛼) , . . . , 𝑢(𝑆) 1𝑆 (𝑥, 𝛼)𝑥=𝑧 , 𝑢1𝑆 (𝑧, 𝛼) , . . . , 𝑢(𝑆) 1𝑆
(𝑥, 𝛼)𝑥=𝑧 , . . . , 𝑢𝑚𝑆 (𝑧, 𝛼) , . . . , 𝑢(𝑆) 𝑚𝑆 (𝑥, 𝛼)𝑥=𝑧 ,
(𝑖,𝑗)
𝑡 𝑢𝑚𝑆 (𝑧, 𝛼) , . . . , 𝑢(𝑆) 𝑚𝑆 (𝑥, 𝛼)𝑥=𝑧 ] ,
𝐷(1,1) ⋅ ⋅ ⋅ 𝐷(1,𝑚) [ ] [ ... ] 𝐷 = [ ... ], d [ ] (𝑚,1) (𝑚,𝑚) ⋅⋅⋅ 𝐷 [𝐷 ] 𝐸
(1,1)
(𝑖,𝑗)
𝐷1,2 = 𝐷2,1
(𝑖,𝑗)
(𝑖,𝑗)
𝐷1,1 = 𝐷2,2
(1,𝑚)
⋅⋅⋅ 𝐸
[ ] [ .. ] . 𝐸 = [ ... ] d . [ ] (𝑚,1) (𝑚,𝑚) ⋅⋅⋅ 𝐸 [𝐸 ]
0 0 [ (𝑖,𝑗) [𝑑 0 [ 1,0 [ [ . .. . =[ . [ . [ (𝑖,𝑗) [ (𝑖,𝑗) [𝑑 𝑠−1,0 𝑑 𝑠−1,1 [ (𝑖,𝑗) (𝑖,𝑗) [ 𝑑 𝑠,0 𝑑 𝑠,1 0 [ [0 [ [. =[ [ .. [ [0 [
⋅⋅⋅
0
⋅⋅⋅
0
d
.. .
0
⋅⋅⋅
0
⋅⋅⋅
(𝑖,𝑗) 𝑑 𝑠,𝑠−1
0 ⋅⋅⋅ 0 0 ] 0 ⋅ ⋅ ⋅ 0 0] ] .. .. .. ] ] . d . .] ] 0 ⋅ ⋅ ⋅ 0 0] ]
] 0] ] ] .. ] .] ], ] ] 0] ] 0]
.
[0 0 ⋅ ⋅ ⋅ 0 0](𝑠+1)×(𝑠+1)
(14) (13)
The Parochial matrices 𝐷(𝑖,𝑗) are defined by the following elements (see [14]):
(𝑖,𝑗)
(𝑖,𝑗)
𝐷1,1 𝐷1,2 𝐷(𝑖,𝑗) = [ (𝑖,𝑗) (𝑖,𝑗) ] , [𝐷2,1 𝐷2,2 ] 𝐸
(𝑖,𝑗)
(𝑖,𝑗)
(𝑖,𝑗)
(𝑖,𝑗)
(𝑖,𝑗)
𝐸1,1 𝐸1,2
=[
𝑖, 𝑗 = 1, . . . , 𝑚, (𝑖,𝑗)
𝐸1,1 = 𝐸2,2 (𝑖,𝑗) 𝑒 − [ 0,0 [ (𝑖,𝑗) [ 𝑒1,0
This method provides a sequence of functions, which converges to the exact solution of the problem and is based on the use of Lagrange multipliers for identification of optimal value of a parameter in a functional. Consider the following general nonlinear system [22]: 𝐿 [𝑢 (𝑥)] + 𝑁 [𝑢 (𝑡)] = ℎ (𝑥) ,
1
[ [ . =[ [ .. [ [ (𝑖,𝑗) [ 𝑒𝑠−1,1 [ (𝑖,𝑗) [ 𝑒𝑠,0 (𝑖,𝑗)
Theorem 2 (see [14]). Let the kernel be bounded and belong to 𝐺:[𝑎, 𝑏] ∗ [𝑎, 𝑏] and 𝑢𝑗,𝑠 (𝑥, 𝛼), 𝑢𝑗,𝑠 (𝑥, 𝛼) (for 𝑗 = 1, . . . , 𝑚) are Taylor polynomials of degree 𝑠, and their coefficients are computed by solving linear system (12); then these polynomials converge to the exact solution of fuzzy system (2), when 𝑠 → +∞.
4. Variational Iteration Method
],
[𝐸2,1 𝐸2,2 ]
(𝑖,𝑗)
3.1. Convergence Analysis. One can show that the above numerical method converges to the exact solution of the fuzzy system (2) (see [14] for more details).
(𝑖,𝑗) 𝑒0,1 (𝑖,𝑗) 𝑒1,1 −
⋅⋅⋅ 1 ⋅⋅⋅
.. .
(𝑖,𝑗) 𝑒0,𝑠−1 (𝑖,𝑗) 𝑒1,𝑠−1
(𝑖,𝑗) 𝑒0,𝑠 (𝑖,𝑗) 𝑒1,𝑠
.. .
.. .
d
(𝑖,𝑗)
⋅ ⋅ ⋅ 𝑒𝑠−1,𝑠−1 − 1
(𝑖,𝑗)
⋅⋅⋅
𝑒𝑠−1,1 𝑒𝑠,1
(𝑖,𝑗)
(𝑖,𝑗)
𝑒𝑠,𝑠−1
(𝑖,𝑗)
𝑒𝑠−1,𝑠
] ] ] ] ] ], ] ] ] ] ]
(𝑖,𝑗) 𝑒𝑠,𝑠 − 1]
[ [ (𝑖,𝑗) [ 𝑒 [ 1,0 [ [ .. =[ . [ [ (𝑖,𝑗) [ 𝑒 [ 𝑠−1,0 [ (𝑖,𝑗) [ 𝑒 𝑠,0
𝑢𝑝+1 (𝑥) = 𝑢𝑝 (𝑥)
(16)
𝑢𝑝 (𝑧)] − ℎ (𝑧)] 𝑑𝑧, + ∫ 𝜆 (𝑧) [𝐿 [𝑢𝑝 (𝑧)] + 𝑁 [̃
(𝑖,𝑗)
(𝑖,𝑗)
where 𝐿 is a linear operator, 𝑁 is a nonlinear operator, and ℎ(𝑥) is a given continuous function. According to the variational iteration method, we can construct the following correction functional:
𝑥
𝐸1,2 = 𝐸2,1
𝑒 0,0 − 1
(15)
0
(𝑖,𝑗)
𝑒 0,1
(𝑖,𝑗)
(𝑖,𝑗)
⋅⋅⋅
𝑒 0,𝑠−1
𝑒 0,𝑠
𝑒 1,1 − 1 ⋅ ⋅ ⋅
𝑒 1,𝑠−1
(𝑖,𝑗)
𝑒 1,𝑠
(𝑖,𝑗)
.. . (𝑖,𝑗)
𝑒 𝑠−1,1 (𝑖,𝑗) 𝑒 𝑠,1
] ] ] ] ] ] .. .. ], d . . ] ] (𝑖,𝑗) (𝑖,𝑗) ] ⋅ ⋅ ⋅ 𝑒 𝑠−1,𝑠−1 − 1 𝑒 𝑠−1,𝑠 ] ] (𝑖,𝑗) (𝑖,𝑗) ⋅⋅⋅ 𝑒 𝑠,𝑠−1 𝑒 𝑠,𝑠 − 1] (𝑖,𝑗)
where 𝑝 = 0, 1, 3, . . ., 𝜆 is Lagrange multiplier which can be identified optimally via variational theory, 𝑢𝑝 is the 𝑝th approximate solution, and 𝑢̃𝑝 is a restricted variation (i.e., 𝛿̃ 𝑢𝑝 = 0). Next, we take the partial derivative to both sides of the Volterra integral equation (1) with respect to 𝑥, and we get 𝑢 (𝑥) = 𝑓 (𝑥) +
𝑑 𝑥 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡) 𝑑𝑡. 𝑑𝑥 𝑎
(17)
Journal of Applied Mathematics
5
Take
Now, using the variational iteration method and (22), we get the following iteration formulas:
𝑥
𝑑 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡) 𝑑𝑡, 𝑑𝑥 𝑎
(18)
as a restricted variation and use the variational iteration method in direction 𝑥, then we obtain the following iteration sequence:
𝑥
𝑢𝑝+1 (𝑥, 𝛼) = 𝑢𝑝 (𝑥, 𝛼) ∫ [𝑢𝑝 (𝑧, 𝛼) − 𝑓 (𝑧, 𝛼) 0
−∫
𝑎
𝑥
𝑧
𝑢𝑝+1 (𝑥) = 𝑢𝑝 (𝑥) + ∫ 𝜆 (𝑧) [𝑢𝑝 (𝑧) − 𝑓 (𝑍) 0
−
𝑧
𝑑 ∫ 𝑘 (𝑧, 𝑡) 𝑢𝑝 (𝑡) 𝑑𝑡] 𝑑𝑧, 𝑑𝑧 𝑎
−∫ (19)
(20)
therefore, we get the following stationary conditions:
1 + 𝜆 (𝑧)|𝑧=𝑥 = 0;
(21)
hence, the Lagrange multiplier can be identified: 𝜆(𝑧) = −1. If we substitute the value of the Lagrange multiplier into (19), we obtain the following iteration formula: 𝑥
𝑢𝑝+1 (𝑥) = 𝑢𝑝 (𝑥) − ∫ [𝑢𝑝 (𝑧) − 𝑓 (𝑧) 0
𝑧
𝑑 − ∫ 𝑘 (𝑧, 𝑡) 𝑢𝑝 (𝑡) 𝑑𝑡] 𝑑𝑧. 𝑑𝑧 𝑎
(22)
−∫
𝐶
𝑎
−∫
𝑧
𝑐
𝑢 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼)𝑑𝑡,
𝜕𝑘 (𝑧, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡 − 𝑘 (𝑧, 𝑧) 𝑢 (𝑧, 𝛼) 𝜕𝑧 𝜕𝑘 (𝑧, 𝑡) (𝑡, 𝛼) 𝑢𝑘 (𝑡, 𝛼) 𝑑𝑡] 𝑑𝑧, 𝜕𝑧
5. Numerical Examples and Results In this section, in order to examine the accuracy of the proposed methods, we have chosen one example of linear fuzzy integral equation of the second kind. Moreover, the numerical results will be compared with the exact solution. Example 3 (Taylor expansion method). Consider the following fuzzy linear Volterra integral equations: 𝑢 (𝑥, 𝛼) = (𝛼) cot (𝑥) + ∫ 𝑒𝑥−𝑡 𝑢 (𝑡, 𝛼) 𝑑𝑡, 0
𝑥
(23)
(26)
0 ≤ 𝛼 ≤ 1, 0 ≤ 𝑥 ≤
𝜋 . 4
The analytical solution of the above equation is given as
Suppose that the kernel 𝑘(𝑥, 𝑡) > 0 for 𝑎 ≤ 𝑡 ≤ 𝑐 and 𝑘(𝑥, 𝑡) < 0 for 𝑐 ≤ 𝑡 ≤ 𝑥, and then (23) becomes [12] 𝑐
3 1 2 𝑢 (𝑥, 𝛼) = (𝛼) ( cos (𝑥) + sin (𝑥) + 𝑒2𝑥 ) , 5 5 5 3 1 2 𝑢 (𝑥, 𝛼) = (2 − 𝛼) ( cos (𝑥) + sin (𝑥) + 𝑒2𝑥 ) , 5 5 5
𝑢 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼) 𝑑𝑡 𝑎
𝑥
(24)
Expand the unknown functions in Taylor series at 𝑧 = 𝜋/12 and implement the following algorithm.
𝑎
𝑥
+ 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼) 𝑑𝑡, 𝑐
for each 0 ≤ 𝛼 ≤ 1 and 𝑎 ≤ 𝑥 ≤ 𝑏.
(27)
0 ≤ 𝛼 ≤ 1.
+ 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼) 𝑑𝑡, 𝑢 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼) 𝑑𝑡
(𝑧, 𝛼) − 𝑓 (𝑧, 𝛼)
0
𝑎
𝑐
(25)
where 𝑝 = 0, 1, 2, . . . In virtue of (25), we can find a solution of (24) and hence obtain a fuzzy solution of the linear fuzzy Volterra integral equation of the second kind.
𝑢 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + 𝜆 ∫ 𝑘 (𝑥, 𝑡) 𝑢 (𝑡, 𝛼)𝑑𝑡.
𝑐
[𝑢𝑝
𝑢 (𝑥, 𝛼) = (2 − 𝛼) cot (𝑥) + ∫ 𝑒𝑥−𝑡 𝑢 (𝑡, 𝛼) 𝑑𝑡,
𝑥
𝑥
𝑥
𝑥
We apply the variational iteration method on fuzzy Volterra integral equation of the second kind:
𝑎
𝜕𝑘 (𝑧, 𝑡) (𝑡, 𝛼) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡] 𝑑𝑧, 𝜕𝑧 0
𝑥
𝜆 (𝑧)𝑧=𝑥 = 0,
𝑐
𝜕𝑘 (𝑧, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡 − 𝑘 (𝑧, 𝑧) 𝑢𝑝 (𝑧, 𝛼) 𝜕𝑧
𝑢𝑝+1 (𝑥, 𝛼) = 𝑢𝑝 (𝑥, 𝛼) − ∫
and finally we calculate variation with respect to 𝑢𝑝 ; notice that 𝛿𝑢𝑝 (0) = 0 yields 𝛿𝑢𝑝+1 = 𝛿𝑢𝑝 + 𝜆𝛿𝑢𝑝 𝑧=𝑥 − ∫ 𝜆 (𝑧) 𝛿𝑢𝑝 𝑑𝑧 = 0; 0
𝐶
Algorithm 4. (1) Input 𝑎, 𝑏, 𝑧, 𝑚, 𝜆, 𝑘𝑖,𝑗 (𝑥, 𝑡), 𝑓𝑖 (𝑥, 𝛼), 𝑓𝑖 (𝑥, 𝛼). (2) Input the Taylor expansion degree (𝑠).
6
Journal of Applied Mathematics (3) Calculate
(9) Put 𝜕(𝑛) 𝑘𝑖,𝑗 (𝑥, 𝛼) 𝜕𝑥𝑛 𝜕(𝑛) 𝑓 (𝑥, 𝛼) 𝑖
𝜕𝑥𝑛
, (𝑖,𝑗)
,
(28)
𝜕(𝑛) 𝑓𝑖 (𝑥, 𝛼) , 𝜕𝑥𝑛
=
𝜕𝑟 𝑘𝑖,𝑗
𝑐𝑖,𝑗
∫
𝑟!
𝜕𝑥𝑟
𝑎
(𝑖,𝑗)
𝑟
(𝑥, 𝑡) ⋅ (𝑡 − 𝑧) 𝑑𝑡
(29)
(5) Calculate 𝜆 𝑖,𝑗
=
𝑟!
∫
𝜕𝑟 𝑘𝑖,𝑗
𝑐𝑖,𝑗
𝜕𝑥𝑟
𝑎
𝑟
(𝑥, 𝑡) ⋅ (𝑡 − 𝑧) 𝑑𝑡
(30)
𝑖, 𝑗 = 1, . . . , 𝑚.
(𝑖,𝑗)
0 0 [ (𝑖,𝑗) [𝑑 0 [ 2,1 [ [ . .. . =[ . [ . [ (𝑖,𝑗) [ (𝑖,𝑗) [ 𝑑 𝑘,1 𝑑 𝑘,2 [ (𝑖,𝑗) (𝑖,𝑗) [𝑑 𝑘+1,1 𝑑 𝑘+1,2
(𝑖,𝑗)
𝐷
(𝑖,𝑗)
𝐸1,1 = 𝐸2,2 (𝑖,𝑗)
(𝑖,𝑗)
𝑒 − 1 𝑒1,2 [ 1,1 (𝑖,𝑗) [ (𝑖,𝑗) 𝑒2,2 − 1 [ 𝑒2,1 [ [ . .. =[ . [ .. [ [ (𝑖,𝑗) (𝑖,𝑗) [ 𝑒𝑠,1 𝑒𝑠,2 [ (𝑖,𝑗)
(𝑖,𝑗)
[ 𝑒𝑠+1,1
𝑒𝑠+1,2
(𝑖,𝑗)
𝑒1,𝑠+1
𝑒2,𝑠
(𝑖,𝑗)
𝑒2,𝑠+1
.. .
.. .
⋅⋅⋅
𝑒1,𝑠
⋅⋅⋅ d
(𝑖,𝑗)
(𝑖,𝑗) ⋅ ⋅ ⋅ 𝑒𝑠,𝑠 −1
⋅⋅⋅
(𝑖,𝑗)
𝑒𝑠+1,𝑠
𝑒𝑠,𝑠+1 (𝑖,𝑗)
0
⋅⋅⋅
0
d
.. .
⋅⋅⋅
0
0
(𝑖,𝑗)
⋅ ⋅ ⋅ 𝑑 𝑘+1,𝑘
] 0] ] ] .. ] . .] ] ] ] 0] ] 0](𝑠+1)×(𝑠+1)
(35)
(𝑖,𝑗)
(𝑖,𝑗)
𝐷1,1 𝐷1,2
=[
]. (𝑖,𝑗) (𝑖,𝑗) 𝐷 𝐷 2,2 ] [ 2,1
(36)
(12) Put
(𝑖,𝑗) (𝑖,𝑗)
⋅⋅⋅
(11) Denote
(6) Put (𝑖,𝑗)
(34)
𝐷 1,2 = 𝐷 2,1
𝑖, 𝑗 = 1, . . . , 𝑚.
(𝑖,𝑗) 𝑒 𝑛+1,𝑟+1
.
(10) Put
(4) Calculate 𝜆 𝑖,𝑗
0 ⋅⋅⋅ 0 0 ] 0 ⋅ ⋅ ⋅ 0 0] ] .. .. .. ] ] . d . .] ] 0 ⋅ ⋅ ⋅ 0 0] ]
[0 0 ⋅ ⋅ ⋅ 0 0](𝑠+1)×(𝑠+1) 𝑛 = 0, 1, . . . , 𝑠.
(𝑖,𝑗) 𝑒𝑛+1,𝑟+1
(𝑖,𝑗)
𝐷1,1 = 𝐷2,2
0 [ [0 [ [. =[ [ .. [ [0 [
] ] ] ] ] ] ] ] ] ] ]
F = [−𝑓 (𝑧, 𝛼) , . . . , −𝑓(𝑘) (𝑧, 𝛼) , 1
1
.
(31)
(𝑠)
−𝑓1 (𝑧, 𝛼) , . . . , −𝑓1 (𝑧, 𝛼) , . . . , −𝑓 (𝑧, 𝛼) , . . . , −𝑓(𝑠) (𝑧, 𝛼) , 𝑚
𝑒𝑠+1,𝑠+1 − 1](𝑠+1)×(𝑠+1)
(37)
𝑚
(𝑠)
𝑡
−𝑓𝑚 (𝑧, 𝛼) , . . . , −𝑓𝑚 (𝑧, 𝛼)] .
(7) (𝑖,𝑗)
(𝑖,𝑗)
𝐸 1,2 = 𝐸 2,1
(13) Put
(𝑖,𝑗)
(𝑖,𝑗)
𝑒 1,1 − 1
[ [ (𝑖,𝑗) [ 𝑒 [ 2,1 [ [ .. =[ . [ [ (𝑖,𝑗) [ 𝑒 [ 𝑠,1 [ (𝑖,𝑗) [ 𝑒 𝑠+1,1
𝑒 1,2
(𝑖,𝑗)
⋅⋅⋅
𝑒 1,𝑠
𝑒 2,2 − 1 ⋅ ⋅ ⋅
𝑒 2,𝑠
(𝑖,𝑗)
.. . (𝑖,𝑗)
𝑒 𝑠,2
(𝑖,𝑗)
𝑒 𝑠+1,2
(𝑖,𝑗)
𝑒 1,𝑠+1
] ] ] ] (32) ] ] .. .. ], d . . ] ] (𝑖,𝑗) (𝑖,𝑗) ⋅ ⋅ ⋅ 𝑒 𝑠,𝑠 − 1 𝑒 𝑠,𝑠+1 ] ] ] (𝑖,𝑗) (𝑖,𝑗) ⋅ ⋅ ⋅ 𝑒 𝑠+1,𝑠 𝑒 𝑠+1,𝑠+1 − 1] (𝑖,𝑗)
(𝑖,𝑗)
𝑒 2,𝑠+1
where 𝑖, 𝑗 = 1, 2, . . . , 𝑚. (8) Calculate 𝑑𝑛+1,𝑙+1 (𝑛−𝑙−𝑟−1) (33) 𝜕(𝑟) 𝑘𝑖,𝑗 (𝑥, 𝑡) ) = ∑ ( )( . 𝑟 𝜕𝑥 𝑛−𝑙−𝑟−1 𝑡=𝑥 𝑟=0 𝑥=𝑧 𝑛−𝑙−1
𝑛−𝑟−1
(𝑠) 𝑢 = [𝑢1𝑠 (𝑧, 𝛼) , . . . , 𝑢(𝑠) 1𝑠 (𝑧, 𝛼) , 𝑢1𝑠 (𝑧, 𝛼) , . . . , 𝑢1𝑠 (𝑧, 𝛼) ,
. . . , 𝑢𝑚𝑠 (𝑧, 𝛼) , . . . , 𝑢(𝑠) 𝑚𝑠 (𝑧, 𝛼) , 𝑢𝑚𝑠 (𝑧, 𝛼) , . . . ,
(38)
𝑡
𝑢(𝑠) 𝑚𝑠 (𝑧, 𝛼)] .
(14) Solve the following linear system (𝐸 + 𝐷)𝑢 = F. (15) Estimate 𝑢(𝑧, 𝛼), 𝑢(𝑧, 𝛼) by computing Taylor expansion for 𝑢 (𝑟) 𝑠 1 𝜕 𝑢𝑗 (𝑥, 𝛼) (𝑥 − 𝑧)𝑟 ) 𝑢𝑗,𝑠 = ∑ ( 𝑟 𝑟! 𝜕𝑥 𝑟=0 𝑥=𝑧 (𝑟) 𝑠 (39) 1 𝜕 𝑢𝑗 (𝑥, 𝛼) 𝑟 𝑢𝑗,𝑠 = ∑ ( (𝑥 − 𝑧) ) , 𝑟 𝑟! 𝜕𝑥 𝑟=0 𝑥=𝑧 𝑎 ≤ 𝑥, 𝑧 ≤ 𝑏, 0 ≤ 𝛼 ≤ 1, 𝑗 = 1, 2, . . . , 𝑚.
Journal of Applied Mathematics
7
We get the following results:
(1,1) (1,1) 𝐸1,1 = 𝐸2,2
(1,1) (1,1) = 𝐸2,1 𝐸1,2
−.04088 .00364 −.00024 1.275 × 10−5 −5.598 × 10−7
−.70073 [ [ .29927 [ [ [ .29927 [ =[ [ .29927 [ [ [ .29927 [ [ .29927
] −1.0409 .00364 −.00024 1.275 × 10−5 −5.598 × 10−7 ] ] ] −.04088 −.99636 −.00024 1.275 × 10−5 −5.598 × 10−7 ] ] ], −5 −7 ] −.04088 .00364 −1.0002 1.275 × 10 −5.598 × 10 ] ] −.04088 .00364 −.00024 −0.99999 −5.598 × 10−7 ] ] −.04088 .00364 −.00024 1.275 × 10−5
𝐷=[
0 [ [1 [ [ [1 =[ [1 [ [ [1 [
0 0 0 0 0
𝜋 −𝛼 cos ( ) [ ] 12 [ ] [ ] 𝜋 [ ] ) 𝛼 sin ( [ ] 12 [ ] [ ] [ 𝛼 cos ( 𝜋 ) ] [ ] [ ] 12 [ ] [ ] 𝜋 [ −𝛼 sin ( ) ] [ ] 12 [ ] [ ] [ −𝛼 cos ( 𝜋 ) ] [ ] 12 [ ] [ ] [ ] 𝜋 [ ] ) 𝛼 sin ( [ ] 12 [ ]. F= [ ] [− (2 − 𝛼) cos ( 𝜋 )] [ 12 ] [ ] [ ] 𝜋 ] [ [ (2 − 𝛼) sin ( ) ] [ 12 ] [ ] [ ] [ (2 − 𝛼) cos ( 𝜋 ) ] [ 12 ] [ ] [ ] 𝜋 ] [ [ − (2 − 𝛼) sin ( ) ] [ 12 ] [ ] [ ] [− (2 − 𝛼) cos ( 𝜋 )] [ 12 ] [ ] [ 𝜋 ] (2 − 𝛼) sin ( ) [ 12 ]
0 [ [0 [ [ [0 =[ [0 [ [ [0 [ [0
0 0 0 0 0
] 0 0 0 0 0] ] ] 1 0 0 0 0] ], 1 1 0 0 0] ] ] 1 1 1 0 0] ] [1 1 1 1 1 0 ]
(1,1) (1,1) 𝐷1,2 𝐷1,1 (1,1) (1,1) 𝐷2,1 𝐷2,2
] 0 0 0 0 0] ] ] 0 0 0 0 0] ], 0 0 0 0 0] ] ] 0 0 0 0 0] ] 0 0 0 0 0]
],
(41)
F (𝑧, 𝛼) = [−𝑓 (𝑧, 𝛼) , −𝑓 (𝑧, 𝛼) , −𝑓 (𝑧, 𝛼) , −𝑓 (𝑧, 𝛼) , −𝑓(4) (𝑧, 𝛼) , −𝑓(5) (𝑧, 𝛼) − 𝑓 (𝑧, 𝛼) ,
(4)
−𝑓 (𝑧, 𝛼) , −𝑓 (𝑧, 𝛼) , −𝑓 (𝑧, 𝛼) , −𝑓 (5)
−𝑓
𝑡
(𝑧, 𝛼)] ,
(40)
0 0 0 0 0 ] 0 0 0 0 0] ] ] 0 0 0 0 0] ]; 0 0 0 0 0] ] ] 0 0 0 0 0] ] [0 0 0 0 0 0]
(1,1) (1,1) 𝐸1,1 𝐸1,2 𝐸 = [ (1,1) (1,1) ] , 𝐸2,1 𝐸2,2
(1,1) (1,1) 𝐷1,2 = 𝐷2,1
]
0 [ [0 [ [ [0 =[ [0 [ [ [0 [
hence
(1,1) (1,1) = 𝐷2,2 𝐷1,1
−1.000
Solving the linear system
(𝑧, 𝛼) , (𝐷 + 𝐸) 𝑢 = F,
(42)
8
Journal of Applied Mathematics 1
we obtain 1.30655494406977𝛼 [ ] [ 1.38836501660070𝛼 ] [ ] [ ] [ 2.06962325214024𝛼 ] [ ] [ 5.36399137514409𝛼 ] [ ] [ ] [ 11.4350895320246𝛼 ] [ ] [ ] [ 21.4350895320246𝛼 ] [ ] 𝑢 (𝑧, 𝛼) = [ ], [1.30655494406977 (2 − 𝛼)] [ ] [ ] [1.38836501660070 (2 − 𝛼)] [ ] [ ] [2.06962325214024 (2 − 𝛼)] [ ] [5.36399137514409 (2 − 𝛼)] [ ] [ ] [11.4350895320246 (2 − 𝛼)] [ ] [21.4350895320246 (2 − 𝛼)]
0.8 0.6 훼 0.4 0.2 0
1
2
3
U(x, 훼) Exact Approximate
(43)
Figure 1: Exact and numerical solutions at 𝑥 = 𝜋/8. 3.4 × 10−6 3.3 × 10−6 3.2 × 10−6 3.1 × 10−6 3. × 10−6 2.9 × 10−6 2.8 × 10−6 2.7 × 10−6 2.6 × 10−6 2.5 × 10−6 2.4 × 10−6 2.3 × 10−6 2.2 × 10−6 2.1 × 10−6 2. × 10−6 1.9 × 10−6 1.8 × 10−6
(𝑟) 𝑠 1 𝜕 𝑢𝑗 (𝑥, 𝛼) 𝜋 𝑟 (𝑥 − ) ), 𝑢 (𝑥, 𝛼) = ∑ ( 𝑟! 𝜕𝑥𝑟 12 𝑥=𝑧 𝑟=0 (𝑟) 1 𝜕 𝑢𝑗 (𝑥, 𝛼) 𝜋 𝑟 𝑢 (𝑥, 𝛼) = ∑ ( (𝑥 − ) ) , 𝑟 𝑟! 𝜕𝑥 12 𝑟=0 𝑥=𝑧 𝑠
0≤𝑥≤
0
𝜋 , 0 ≤ 𝛼 ≤ 1. 4
The approximate solution is 𝑢 (𝑥, 𝛼) = 0.9999816737𝛼 + 1.000399233𝛼𝑥
0
+ 0.4962378250 + 0.5186782119𝛼𝑥3
0.1
0.2
0.3
0.4
0.5 훼
0.6
0.7
0.8
0.9
1
Figure 2: Absolute error between the exact and numerical solutions.
+ 0.2403470047𝛼𝑥4 + 0.1803786182𝛼𝑥5 , 𝑢 (𝑥, 𝛼) = 2.000407015𝑥 + 1.738708618𝛼 + 0.3524026447𝑥3 + 0.1975243281𝑥4 + 0.1905848255𝑥5 − 0.7387269443
(44)
4
+ 0.04282267662𝛼𝑥
5
− 0.01020620726𝛼𝑥
− 1.000007782𝛼𝑥 + 0.9961625156𝑥2 − 0.4999246903𝛼𝑥2 + 0.1662755673𝛼𝑥3 . Figure 1 compares both the exact and numerical solutions for the fuzzy integral equation (26) using the Taylor expansion method at 𝑥 = 𝜋/8. Moreover, Figure 2 shows the absolute error between the exact and numerical solutions of this example.
Example 5 (variation alliteration method). Consider the linear fuzzy Volterra integral equation (26) with the exact solution (27). In the view of the variational iteration method, we construct a correction functional in the following form: 𝑥
𝑢𝑝+1 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + ∫ 𝑘 (𝑥, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡, 0
(45)
𝑥
𝑢𝑝+1 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + ∫ 𝑘 (𝑥, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡, 0
where 𝑝 = 0, 1, 2, . . . , 𝑠. Start with the initial approximation in (45); that is, 𝑢0 (𝑥, 𝛼) = (𝛼) cos 𝑥, 𝑢0 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥,
(46) 0 ≤ 𝛼 ≤ 1.
Then successive approximations 𝑢𝑝 (𝑥, 𝛼) 𝑠 will be generated. In this example we calculate the 6th order of approximate
Journal of Applied Mathematics
9
solution using the variational iteration method by MAPLE software. We have carried out the following algorithm.
1 1 1 𝑢3 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒𝑥 − 𝛼𝑒𝑥 − cos 𝑥 2 4 2 +
Algorithm 6.
1 1 1 − 𝛼𝑒𝑥 𝑥 + 𝑒𝑥 𝑥2 − 𝛼𝑒𝑥 𝑥2 , 2 2 4
(1) Input 𝑎, 𝑏, 𝜆, 𝑘(𝑥, 𝑡), 𝑠, 𝑓(𝑥, 𝛼), 𝑓(𝑥, 𝛼). (2) Input 𝑢0 (𝑥, 𝛼), 𝑢0 (𝑥, 𝛼).
1 1 𝑢4 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒𝑥 − 𝛼𝑒𝑥 − 𝛼𝑒𝑥 𝑥3 2 12
(3) For 𝑝 = 1 to 𝑠 + 1, compute
1 1 1 + 𝑒𝑥 𝑥 + 𝑒𝑥 𝑥2 + 𝑒𝑥 𝑥3 − cos 𝑥 2 2 6
𝑥
𝑢𝑝+1 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + ∫ 𝑘 (𝑥, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡, 0
𝑥
𝑢𝑝+1 (𝑥, 𝛼) = 𝑓 (𝑥, 𝛼) + ∫ 𝑘 (𝑥, 𝑡) 𝑢𝑝 (𝑡, 𝛼) 𝑑𝑡. 0
We obtain the following results: 1 1 1 𝑢1 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼 sin 𝑥, 2 2 2 1 1 1 𝑢2 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼𝑒𝑥 𝑥, 2 2 2 3 1 1 1 𝑢3 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼 sin 𝑥 + 𝛼𝑒𝑥 𝑥 4 4 4 2 1 + 𝛼𝑒𝑥 𝑥2 , 2 1 1 1 1 𝑢4 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼 sin 𝑥 + 𝛼𝑒𝑥 𝑥 2 2 4 4 1 1 + 𝛼𝑒𝑥 𝑥2 + 𝛼𝑒𝑥 𝑥3 , 4 12 5 1 3 1 𝑢5 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼 sin 𝑥 + 𝛼𝑒𝑥 𝑥 8 8 8 2
1 1 1 sin 𝑥 + 𝛼 cos 𝑥𝑒 − 𝛼 sin 𝑥 + 𝑒𝑥 𝑥 2 4 4
(47)
+
1 1 1 sin 𝑥 − 𝛼𝑒𝑥 𝑥 − 𝛼𝑒𝑥 𝑥2 2 4 4
1 1 + 𝛼 cos 𝑥 − 𝛼 sin 𝑥, 2 4 3 3 1 𝑢5 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒𝑥 − 𝛼𝑒𝑥 + 𝑒𝑥 𝑥2 4 8 4 3 1 + 𝛼 cos 𝑥 + 𝑒𝑥 𝑥3 8 6 +
1 3 1 sin 𝑥 + 𝑒𝑥 𝑥 − cos 𝑥 − 𝛼𝑒𝑥 𝑥4 4 4 48
1 1 1 − 𝛼 sin 𝑥 − 𝛼𝑒𝑥 𝑥3 − 𝛼𝑒𝑥 𝑥2 8 12 8 1 1 − 𝛼𝑒𝑥 𝑥 + 𝑒𝑥 𝑥4 , 2 24 3 3 1 𝑢6 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒𝑥 − 𝛼𝑒𝑥 + 𝑒𝑥 𝑥2 4 8 2 3 1 + 𝛼 cos 𝑥 + 𝑒𝑥 𝑥3 8 12 +
1 3 3 1 𝑥 5 sin 𝑥 + 𝑒𝑥 𝑥 − cos 𝑥 + 𝑒 𝑥 2 4 4 120
−
1 1 1 𝛼𝑒𝑥 𝑥5 − 𝛼𝑒𝑥 𝑥4 − 𝛼 sin 𝑥 240 48 4
1 1 1 + 𝛼𝑒𝑥 𝑥2 + 𝛼𝑒𝑥 𝑥3 + 𝛼𝑒𝑥 𝑥4 4 24 48
−
1 𝑥 3 1 𝑥 2 3 𝑥 𝛼𝑒 𝑥 − 𝛼𝑒 𝑥 − 𝛼𝑒 𝑥 24 4 8
1 + 𝛼𝑒𝑥 𝑥5 , 240
+
1 𝑥 4 𝑒 𝑥. 24
1 1 1 + 𝛼𝑒𝑥 𝑥2 + 𝛼𝑒𝑥 𝑥3 + 𝛼𝑒𝑥 𝑥4 , 8 12 48 5 1 3 3 𝑢6 (𝑥, 𝛼) = 𝛼 cos 𝑥 + 𝛼𝑒𝑥 + 𝛼 sin 𝑥 + 𝛼𝑒𝑥 𝑥 8 8 4 8
1 𝑢1 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒 − 𝛼𝑒𝑥 − cos 𝑥 + sin 𝑥 2 𝑥
1 1 + 𝛼 cos 𝑥 − 𝛼 sin 𝑥, 2 2 1 𝑢2 (𝑥, 𝛼) = (2 − 𝛼) cos 𝑥 + 𝑒𝑥 − 𝛼𝑒𝑥 − cos 𝑥 2 1 1 + 𝛼 cos 𝑥 + 𝑒𝑥 𝑥 − 𝛼𝑒𝑥 𝑥, 2 2
(48) Figure 3 shows a comparison between the exact and the numerical solutions for the fuzzy integral equation (26) by the variational iteration method (VIM) at 𝑥 = 𝜋/8. Moreover, Figure 4 shows the absolute error between the exact and numerical solutions of this example. Table 1 shows a comparison between the exact and the numerical solutions for the fuzzy integral equation (26) using the Taylor expansion method and the variational iteration method (VIM) at 𝑥 = 𝜋/8.
Taylor numerical solution 𝑢(𝑥, 𝛼)
0 0.15081747 0.30163494 0.45245241 0.60326988 0.75408735 0.90490482 1.05572230 1.20653977 1.35735724 1.50817471
Exact solution 𝑢(𝑥, 𝛼)
0 0.15081764 0.30163528 0.45245292 0.60327057 0.75408821 0.90490585 1.05572349 1.20654114 1.35735878 1.50817642
𝛼
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0 0.15081760 0.30163520 0.45245280 0.60327040 0.75408800 0.90490560 1.05572320 1.20654080 1.35735840 1.50817600
Variation numerical solution 𝑢(𝑥, 𝛼) 3.01635285 2.86553521 2.71471756 2.56389992 2.41308228 2.26226464 2.11144699 1.96062935 1.80981171 1.65899406 1.50817642
Exact solution 𝑢(𝑥, 𝛼) 3.016349432 2.865531960 2.714714489 2.563897017 2.413079546 2.262262074 2.111444602 1.960627131 1.809809659 1.658992188 1.508174716
Taylor numerical solution 𝑢(𝑥, 𝛼) 3.01635200 2.86553440 2.71471680 2.56389920 2.41308160 2.26226400 2.11144640 1.96062880 1.80981120 1.65899360 1.50817600
Variation numerical solution 𝑢(𝑥, 𝛼)
Error = 𝐷(𝑢exact , 𝑢approx ) (using Taylor expansion) 3.4210 × 10−6 3.2500 × 10−6 3.0790 × 10−6 2.9080 × 10−6 2.7360 × 10−6 2.5660 × 10−6 2.3950 × 10−6 2.2230 × 10−6 2.0520 × 10−6 1.8806 × 10−6 1.7104 × 10−6
Table 1: Absolute error between the exact and numerical solutions using the Taylor expansion method and the variational iteration method. Error = 𝐷(𝑢exact , 𝑢approx ) (using variational iteration) 8.4600 × 10 −7 8.0400 × 10−7 7.6100 × 10−7 7.1900 × 10−7 6.7700 × 10−7 6.3600 × 10−7 5.9300 × 10−7 5.5000 × 10−7 5.0700 × 10−7 4.6500 × 10−7 4.2400 × 10−7
10 Journal of Applied Mathematics
Journal of Applied Mathematics
11
1
References
0.8 0.6 훼 0.4 0.2 0
0
1
2 u6 (x, 훼)
3
4
Exact Approximate
Figure 3: Exact and numerical solutions at 𝑥 = 𝜋/8.
× − 7
8. × 10
7
.5 10 .5
10
7
.
10
7
.5
10
7
0.1
0.5
0.9
Figure 4: Absolute error between the exact and numerical solutions.
6. Conclusion In this article, Taylor expansion and variational iteration methods are proposed to solve a fuzzy linear Volterra integral equation of the second kind. The results of the example show that the convergence and accuracy of both methods were in a good agreement with the analytical solution. According to comparison of numerical results, mentioned in tables and figures, we conclude that the variational iteration method provides more accurate results and therefore is more advantageous.
Conflicts of Interest The authors declare that they have no conflicts of interest.
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Hindawi Publishing Corporation http://www.hindawi.com
Discrete Dynamics in Nature and Society
Journal of
Function Spaces Hindawi Publishing Corporation http://www.hindawi.com
Abstract and Applied Analysis
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of
Journal of
Stochastic Analysis
Optimization
Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
