International Journal of Analysis and Applications Volume 17, Number 2 (2019), 303-310 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-303 APPROXIMATE SOLUTION OF FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS BY LEAST SQUARES METHOD D. JABARI SABEG1,∗, R. EZZATI2 AND K. MALEKNEJAD2 1Department of Mathematics, Bonab Branch, Islamic Azad University, Bonab, Iran 2Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran ∗Corresponding author: davood.jabari@bonabiau.ac.ir Abstract. In this paper, least squares approximation method is developed for solving a class of linear fractional integro-differential equations comprising Volterra and Fredhlom cases. This method is based on a polynomial of degree n to compute an approximate solution of these equations. The convergence analysis of the proposed method is proved. In addition, to show the accuracy and the efficiency of the proposed method, some examples are presented. 1. Introduction Fractional calculus is a significant branch of mathematics that is used in many fields of science and engineering [2–4]. Many researchers have investigated the analytic results on the existence and uniqueness of solutions of the fractional differential equations [5–8]. As we know, for most fractional differential equations, there are not method to obtain analytic solutions, so numerical techniques must be used. During the past years, methods for solving fractional differential equations are developed. Additionally, some methods have recently been emerged, such as the Adomian decomposition method [10, 11], the operational matrix [12, 13], the collocation method [14, 16], etc. Received 2018-10-06; accepted 2018-11-27; published 2019-03-01. 2010 Mathematics Subject Classification. 26A33, 45A05. Key words and phrases. fractional differential equation; least squares; approximation method; convergence analysis. c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 303 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-303 Int. J. Anal. Appl. 17 (2) (2019) 304 In this paper, by using least squares approximation, a numerical method has been shown to linear fractional integro-differential equations in the following form Dαy(t) = p(t)y(t) + f(t) + λ1 ∫ 1 0 k1(t,x)y(x) dx + λ2 ∫ t 0 k2(t,x)y(x) dx t ∈ I = [0, 1] (1.1) with the initial conditions y(k)(0) = dk i = 0, ..,m− 1, m− 1 < α ≤ m, (1.2) where yk(t) stands for the kth-order derivative of y(t) and Dα denotes the Riemann-Liouville fractional derivative of order α. Clearly, when λ1 = 0, λ2 = 0, the above equation reduces to a linear fractional differential equation. The rest of the paper is organized as follows: In section 2, we will briefly review some notations and definitions of the fractional calculus theory are used in the paper. In Section 3, we introduce the least squares approximation method for solving Eq. (1.1), and discuss its convergence. In Section 4, we show the efficiency of the proposed method with some numerical examples. Section 5, as the final section, presents a conclusion. 2. Brief review of fractional calculus In this section, notations and definitions of the fractional calculus theory, which are going to be used in this paper, are presented [1]. 2.1. Definition. The Riemann-Liouville fractional integral operator Iα of order α ≥ 0 of a function f(x), is defined as Iαf(x) = 1 Γ(α) ∫ x 0 (x− t)(α−1)f(t) dt ,α > 0, (2.1) where x > 0 and Γ(.) is the Euler gamma function. The Riemann-Liouville fractional derivative of order α will be denoted by Dα and defined by Dαf(x) = dm dxm (Im−αf(x)), (2.2) where m − 1 < α ≤ m,m ∈ N and m is the smallest integer order greater than α. We just mention the following property Dαxβ = Γ(β + 1) Γ(β + 1 −α) xβ−α,β > −1. (2.3) Int. J. Anal. Appl. 17 (2) (2019) 305 3. Method of solution In this section, we apply the least squares approximation method for solving Eq. (1.1). We define the following operator T(t,y(t)) = Dαy(t) −p(t)y(t) −f(t) −λ1 ∫ 1 0 k1(t,x)y(x) dx−λ2 ∫ t 0 k2(t,x)y(x) dx. (3.1) We construct Taylor-series expansion for the solution y(t) in Eq.(1.1) as y(t) ' yn(t) = n∑ r=0 y(r)(0) r! tr = n∑ r=0 dr r! tr. (3.2) Substituting (3.2) into ( 3.1), we have T(t,yn(t)) = D αyn(t) −p(t)yn(t) −f(t) −λ1 ∫ 1 0 k1(t,x)yn(x) dx−λ2 ∫ t 0 k2(t,x)yn(x) dx = Dα( n∑ r=0 dr r! tr) −p(t) n∑ r=0 dr r! tr −f(t) − n∑ r=0 dr r! λ1 ∫ 1 0 k1(t,x)x r dx− n∑ r=0 dr r! λ2 ∫ t 0 k2(t,x)x r dx = n∑ r=0 drΓ(r + 1) r!Γ(r −α + 1) tr−α −p(t) n∑ r=0 dr r! tr −f(t) − n∑ r=0 dr r! λ1 ∫ 1 0 k1(t,x)x r dx− n∑ r=0 dr r! λ2 ∫ t 0 k2(t,x)x r dx = n∑ r=0 drγr(t) −f(t), where γr(t) = Γ(r + 1) r!Γ(r −α + 1) tr−α −p(t) tr r! − λ1 r! ∫ 1 0 k1(t,x)x r dx− λ2 r! ∫ t 0 k2(t,x)x r dx. Let Rn(t) := T(t,yn(t)) −T(t,y(t)), t ∈ [0, 1]. Remark 3.1. If Rn(t) = 0, then y(t) = yn(t); if limn→∞Rn(t) = 0, then limn→∞yn(t) = y(t). Remark 3.2. For any t ∈ [0, 1], if Rn(t) ≡ 0, then yn(t) is an exact solution of Eqs. (1.1) and (1.2); if limn→∞Rn(t) = 0, then yn(t) converges to the exact solution of Eqs. (1.1) and (1.2). Let J = J(dm,dm+1, ...,dn) = ∫ 1 0 T2(t,yn(t))dt. (3.3) The problem is to find real constants dm,dm+1, ...,dn such that these constants will minimize J. A necessary condition for the constants dm,dm+1, ...,dn to minimize J is that Int. J. Anal. Appl. 17 (2) (2019) 306 ∂J ∂dj = 0, for each j = m,m + 1, ...,n. By referring (3.3), we get ∂J ∂dj = 2[ n∑ r=0 dr ∫ 1 0 γr(t)γj(t)dt− ∫ 1 0 f(t)γj(t)dt] = 0. (3.4) Thus, we have n∑ r=m dr ∫ 1 0 γr(t)γj(t)dt = ∫ 1 0 f(t)γj(t)dt−βj, (3.5) where βj = m−1∑ r=0 dr ∫ 1 0 γr(t)γj(t)dt (3.6) for each j = m,m + 1, ...,n. In order to find yn(t), we have to solve (n−m) a system of linear equations while assuming (n−m) unknowns dr. The system (3.5) can be written in the form: Gd = F (3.7) where G =   (γm,γm) (γm,γm+1) . . . (γm,γn) (γm+1,γm) (γm+1,γm+1) . . . (γm+1,γn) ... ... . . . ... (γn,γm) (γn,γm+1) . . . (γn,γn)   , (3.8) d = [dm,dm+1, ...,dn] T , and F = [(γm,f) −βm, (γm+1,f) −βm+1, ..., (γn,f) −βn]T Definition 3.3. If Eq. (3.7) has a unique solution d, then yn(t) = ∑n r=1 dr r! tr is called an optimal squared approximation solution of Eqs. (1.1)-(1.2) defined on a set as span{1, t, t2, ...tn}, t ∈ [0, 1]. Remark 3.4. If limn→∞ ∫ 1 0 T2(t,yn(t))dt = 0, then the optimal squared approximation solution yn(t) con- verges to the exact solution y(t) of Eqs. (1.1) and (1.2). We are interested to know that as n → ∞ the optimal squared approximation solution yn(t) will converge to the exact solution y(t) of Eqs. (1.1) and (1.2). This conception is proven in Theorem 3.5. Int. J. Anal. Appl. 17 (2) (2019) 307 Theorem 3.5. Suppose y(t), t ∈ [0, 1] is an exact solution and yn(t) is an optimal squared approximation solution of Eqs. (1.1) and (1.2). If ∃pn(t) = n∑ r=1 drt r such that ∀t ∈ [0, 1], limn→∞pn(t) = y(t) then lim n→∞ ∫ 1 0 T2(t,yn(t))dt = 0. Proof. The proof is similar to proof of Theorem 3 in [19]. 4. Illustrative examples In this section, we use the presented method in Section 3 for solving two examples. Example 4.1. For first example, consider the fractional integro-differential equation D0.75y(t) + 1 5 t2ety(t) − ∫ t 0 xety(x) dx = 6t2.25 Γ(3.25) , y(0) = 0, where the exact solution is given by y(t) = t3. We applied the presented method with n = 3 for solving this example and achieved the corresponding absolute errors in Table 1. Table 1: Absolute errors for Example 1 for n = 3. t = 0 t = 0.2 t = 0.4 t = 0.6 t = 0.8 t = 1.0 0 1.314×10−15 1.045×10−15 1.914×10−16 2.475×10−16 6.717×10−16 Example 4.2. Consider the equation Dαy(t) + 7t2 12 y(t) − ∫ 1 0 txy(x) dx− ∫ t 0 (x + t)y(x) dx = 2t2−α Γ(3 −α) − t 4 , y(0) = 0, with the exact solution y(t) = t2. By the presented method in section 3 for n = 2 and different values of α absolute errors are reported in Table 2. Example 4.3. Consider the equation [20] Dαy(t) + y(t) = t4 − 1 2 t3 − 3 Γ(4 −α) t3−α + 24 Γ(5 −α) t4−α, y(0) = 0, 0 ≤ α ≤ 1 whose exact solution is given by y(t) = t4 − 1 2 t3. Int. J. Anal. Appl. 17 (2) (2019) 308 Table 2: Absolute errors for Example 2. t α = 0.2 α = 0.5 α = 0.7 α = 0.9 α = 1.0 0.0 0 0 0 0 0 0.1 0 1.734 × 10−17 1.387 × 10−17 0 0 0.2 0 3.469 × 10−17 2.775 × 10−17 0 0 0.3 0 4.163 × 10−17 4.163 × 10−17 0 0 0.4 0 5.551 × 10−17 5.551 × 10−17 0 0 0.5 0 5.551 × 10−17 5.551 × 10−17 0 0 0.6 0 5.551 × 10−17 5.551 × 10−17 0 0 0.7 0 5.551 × 10−17 1.110 × 10−16 0 0 0.8 0 0 1.110 × 10−16 0 0 0.9 0 1.110 × 10−16 1.110 × 10−16 0 0 1.0 0 0 2.220 × 10−16 0 0 By taking different values of α, we solved the above problem by means of the presented method. The maximum absolute error with the presented method and SCT method [20] for n = 4 are compared in Table 3. Table 3: Comparison of maximum absolute error for example 3. α n=4 Present method Method of [20] 0.01 1.26×10−14 1.2×10−5 0.1 1.41×10−14 1.3×10−4 0.5 5.30 ×10−15 7.8 ×10−4 0.99 1.30 ×10−15 8.6 ×10−4 1 1.14 ×10−15 8.6 ×10−4 Example 4.4. Consider the equation D 3 2 y(t) + y(t) = 6 Γ(2.5) t1.5 + 6 Γ(1.5) t0.5 + t3 + t2, y(0) = 0,y′(t) = 0, whose exact solution is given by y(t) = t3 + t2. By applying the technique described in section 3 with m=4, we approximate solution as y(t) = 4∑ r=0 y(r)(0) r! tr = 4∑ r=0 dr r! tr Int. J. Anal. Appl. 17 (2) (2019) 309 Here, by using Eq. (3.7), we obtain   1.00901 0.422206 0.123762 0.422206 0.19103 0.0586763 0.123762 0.0586763 0.0186265     d2 d3 d4   =   4.55127 1.9906 0.599582   (4.1) Finally by solving Eq.(4.14), we get d2 = 2,d3 = 6,d4 = 0. Thus we can write y(t) = d0 + d1t + d2 t2 2! + d3 t3 3! + d4 t4 4! = t3 + t2, which is the exact solution. 5. Conclusion In this paper, we proposed least squares approximation method to solve a class of linear fractional integro-differential equations comprising of Fredholm and Volterra cases based on a polynomial of degree n. 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