51 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 Abstract The main aim of this paper is to apply a new technique suggested by Temimi and Ansari namely (TAM) for solving higher order Integro-Differential Equations. These equations are commonly hard to handle analytically so it is request numerical methods to get an efficient approximate solution. Series solutions of the problem under consideration are presented by means of the Iterative Method (IM). The numerical results show that the method is effective, accurate and easy to implement rapidly convergent series to the exact solution with minimum amount of computation. The MATLAB is used as a software for the calculations. Keywords: Integro-Differential Equations, Iterative Method, Temimi and Ansari method. 1. Introduction Integro-Differential Equations (IDEs) include in many mathematical formulations of physical phenomena, these problems have a major role of interest and arise in many applications in various fields of science, such as chemical kinetics, fluid dynamics, engineering problems and biological models. In recent years, there exist numerical techniques gained a great interest by many authors such as Lagrange Interpolation Method [1]. Wavelet- Galerkin Method (WGM) [2]. Adomian's Decomposition Method (ADM) [3-5]. Modified Adomian Decomposition Method (MADM) [6]. Variational Iteration Method (VIM) [7, 8]. Differential Transform Method [9]. Tau Method [10]. Generalized Spline Method [11], and Semi Analytical-Numerical Techniques such that Taylor polynomials [12]. and Rationalized Haar Functions Method [13]. However, none of a fore mentioned methods are successfully solved higher order IDEs. Furthermore, prior studies need more effort to realize the outcomes, they are not precise and commonly they are improved for specific sorts IDEs. Temimi and Ansari (TAM) have been suggested a new iterative method, i.e., Semi Analytic Iterative Method (SAIM) for solving linear and nonlinear functional equations [14]. This method has been extensively studied by many researchers recently; it has been successfully applied for Application of Iterative Method for Solving Higher Order Integro- Differential Equations Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Samaher M. Yassein Department of Mathematics, College of Education for Pure Science Ibn Al-Haitham, University of Baghdad, Baghdad, Iraq. 10.30526/32.2.2139 Doi: samamarez@yahoo.com Article history: Received 13 November 2018, Accepted 20 January 2019, Publish May 2019 file:///F:/مجلة/2019-5-20/رياضيات/تحتاج%20تصحيح%20مصادر/samamarez@yahoo.com file:///F:/مجلة/2019-5-20/رياضيات/تحتاج%20تصحيح%20مصادر/samamarez@yahoo.com 52 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 solving some linear and nonlinear partial and ordinary differential equations [15-18]. It is worth mentioning, SAIM is not yet used to solve higher order IDEs. This method is accurate and powerful technique, needn’t to impose any additional restrictions to get the numerical solution of these problems. It is qualified method for extremely the number of calculations will be reduced while still maintaining the solution is more accurate and efficient. Higher Order Linear Fredholm IDEs was solved by Power Series and Chebyshev Series Approximation Methods in [19]. IDEs was solved by Modified Taylor Expansion Method in [20]. And Voltera IDEs was solved by using the Laplace Transform Method in [21]. In this research, we applied this reliable technique to solve higher order IDEs. Generally, Voltera – integro – differential equation [21]. Given in the form: y(n)(x) = h(x) + 𝜆 ∫ k(x, t)𝑦(t) x 0 d(t), 𝜆 ≠ 0, (1) And Fredholm – integro – differential equation [19]. given in the form: y(n)(x) = h(x) + ∫ k(x, t)𝑦(t) b a d(t) (2) Both of Equations. (1) & (2) with Initial Conditions (ICs). y(k)(0) = ωk, 0 ≤ k ≤ n − 1 (3) where y(n)(x) is the n th derivatives, k(x, t)and h(x), are given continuous smooth functions, 𝜆 is a parameter, 𝑦(x) unknown function to be determined and 𝑎, 𝑏, ωk are constants. Since the results of equation (1) and equation (2) combine the differential and integral operators, then it is necessary to define ICs as in (3).The proposed method was applied to establish series solutions for Equation (1) or Equation (2). We illustrated that this method is effective and perfect in handling to solve higher order IDEs in scientific and engineering problems. Several numerical examples are introduced and comparison with existing methods, the results reveal that the method is accurate and easy to implement. 2. Fundamental Idea for the Iterative Method The main steps of Iterative Method (IM). It is rewrite that any differential equation can be written as [14]. L(y(x)) + N(y(x)) + h(x) = 0 (4) With Boundary Conditions B (y, 𝑑𝑦 𝑑𝑥 ) = 0, Where x is the independent variable, L is a Linear operator, N is a non-linear operator and the boundary operator is B. The method which proposed as the following way. The initial approximation is the primary step in the IM, by assuming that the initial guess 𝑦0(x) is solution of problem y(x) and solution of equation can be solving: L(𝑦0(x)) + h(x) = 0, B (𝑦0, 𝑑𝑦0 𝑑𝑥 ) = 0 (5) To generate the next iteration of the solution as follows: L(𝑦1(x)) +h(x)+N (𝑦0 (x)) = 0, B (𝑦1 , 𝑑𝑦1 𝑑𝑥 ) = 0 (6) 53 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 After several simple iterative steps of the solution, the general form of this Equation which is: L(yn+1 (x)) + h(x) + N(yn (x)) = 0 , B(yn+1, dyn+1 dx ) = 0 (7) Evidently each iteration of the function 𝑦𝑛 (x) represent effectively alone solution for equation (4). We will implement the steps of method at the equation (1), so Equation (1) can be express as: L(y (x)) = h(x) + 𝜆 ∫ k(x, t)𝑦(t) x 0 d(t) , λ ≠ 0, (8) The differential operator L(y (x)) is the highest order derivative in the equation (8), we assume that L is invertible by using the given ICs in equation (3) and applying the inverse operator L−1 in both sides of equation (8), we get the following Equation: y (x) = ψ0 + L −1(h(x)) + L−1(λ ∫ k(x, t)y(t) x 0 d(t)) , λ ≠ 0, (9) Where the function ψ0 is arising from integrating the source term, from applying the given ICs in Equation (3) which are prescribed. Now, we illustrate the method as the following steps: Step 1: to get 𝑦0(𝑡) solving L(𝑦0(x)) - h(x) = 0 (10) with ICs in Equation (3) and applying the inverse operator L−1 in both sides of Equation (10), we obtain: y0(𝑥) = ψ0 + L −1(h(x)) Step 2: The next iterate is: L(y1(x)) − h(x) − ∫ k(x, t)y0(t) x 0 d(t) = 0 with ICs in Eq. (3) , (11) solving this equation and applying the inverse operator L−1 in both sides of Equation (11), leads to get 𝑦1(𝑥) as: y1(𝑥) = ψ0 + L −1(h(x)) + L−1(∫ k(x, t)y0(t) x 0 d(t)) Step (3): After several simple iterative steps of the solution, the general form of this equation given as L(yn+1(t)) − h(x) − ∫ k(x, t)yn(t) x 0 d(t) = 0 with ICs in Eq. (3) , (12) Solving this equation and applying the inverse operator L−1 in both sides of Equation (12), leads to get 𝑦𝑛+1(𝑥): yn+1(𝑥) = ψ0 + L −1(h(x)) + L−1(∫ k(x, t)yn(t) x 0 d(t)) evidently each iteration of the function 𝑦𝑛 (x) represents effectively solution for Equation (8). Similarly, by the same steps we solve Fredholm IDEs. 54 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 3. Numerical Results We will be applying the SAIM for solving some examples of the Fredholm IDEs and Voltera IDEs. Example 1 Consider third- order Fredholm IDE [19]: y(3)(x) = 1 − e + ex + ∫ y(t)dt 1 0 , (13) with ICs y(0) = y′(0) = y′′(0) = 1, the Exact solution is Y(x) = ex Solution Via implementing same steps as described in the previous section, we first begin by solving the following initial problem to find the initial approximation 𝑦0 (x), the SAIM will be applied as L(y0) = 1 − e + e x with 𝑦0 (0)= 𝑦0′ (0) = 𝑦0 ′′(0) =1 where h(x) = e − ex − 1, L(y) = d3y dx3 , N(y) = 0. So, the primary step is: L(y0) = 1 − e + e x with 𝑦0 (0)= 𝑦0′ (0) = 𝑦0 ′′(0) =1 (14) Then, the general relation as follows: L(yn+1) − h(x) − ∫ k(x, t)yn(t) x 0 d(t) = 0, yn+1(0) = y′n+1(0) = y′′n+1(0) = 1 (15) By solving the problem defined in Equation (14), we have y0 = e x − 1934613350591413 6755399441055744 x3 . The first iteration can be gotten as: y1 (3)(x) = 1 − e + ex + ∫ y0(t)dt 1 0 with 𝑦1 (0)= y′1(0) = y′′1(0) =1 (16) Thus, the solution of Equation (16) as: y1 = e x − 644871116863805 54043195528445952 x3 . The second iteration is: y2 (3)(x) = 1 − e + ex + ∫ y1(t)dt 1 0 with 𝑦2 (0)= 𝑦2′ (0)= 𝑦′′2 (0)=1 (17) Then, the solution of Eqaution (17) as y2 = e x − 26869629869327 54043195528445952 x3. Also, by same steps, the other solutions can be generated from calculating these problems via using MATLAB, we obtain y3 , y4, … , we get the solution till y11 . y11 = e x − 11 54043195528445952 x3 . 55 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 Table 1. Numerical results of the illustrative example above of Y11. 𝒙 Exact solution SAIM Error 0 1.000000000000000 1.000000000000000 0 0.1 1.105170918075648 1.105170918075648 0 0.2 1.221402758160170 1.221402758160170 0 0.3 1.349858807576003 1.349858807576003 0 0.4 1.491824697641270 1.491824697641270 0 0.5 1.648721270700128 1.648721270700128 0 0.6 1.822118800390509 1.822118800390509 0 0.7 2.013752707470477 2.013752707470477 0 0.8 2.225540928492468 2.225540928492468 0 0.9 2.459603111156950 2.459603111156950 0 1 2.718281828459046 2.718281828459046 0 Figure 1. Exact and approximate solution of the illustrative example. Example 2 Consider third- order Voltera IDE [20]. y(3)(x) − xy′′(x) = 4 7 x9 − 8 5 x7 − x6 + 6x2 − 6 + 4 ∫ x2 x 0 t3y(t)dt (18) With ICs y(0) = 1, y′(0) = 2, y′′(0) = 0, x≥ 0, 𝑡 ≤ 1, The Exact solution is Y(x) = −x2 + 2x + 1, Solution Applying same steps as in the previous example, we first begin by solving the following initial problem in order to find the initial approximation 𝑦0 (x), the SAIM will be applied: as L(y0) = 4 7 x9 − 8 5 x7 − x6 + 6x2 − 6 with 𝑦0(0)= 1, 𝑦0 ′(0) = 2, 𝑦0 ′′(0) = 0 where h(x) = − 4 7 x9 + 8 5 x7 + x6 − 6x2 + 6 , L(y) = d3y dx3 , N(y) = 0. So, the primary step is: L(y0) = 4 7 x9 − 8 5 x7 − x6 + 6x2 − 6 with 𝑦0 (0)=1, 𝑦0 ′ (0) =, 𝑦0 ′′(0) = 0 (19) Then, the general relation as follows: L(yn+1) − h(x) − ∫ k(x, t)yn(t) x 0 d(t) = 0, yn+1(0) = 1, y′n+1(0) = 2, y′′n+1(0) = 0 (20) By solving the problem defined in Equation (19) we have: 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 The solution at n=12 x-axis y -a x is Approximate Y11 of Ex1 Exact Solution 56 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 y0 = x12 2310 − x 10 450 − x 9 504 + x 5 10 − x3 + 2x + 1 by same way in the previous example. The next step is to find y1 as follows: The first iteration can be gotten as: y1 (3)(x) = 4 7 x9 − 8 5 x7 − x6 + 6x2 − 6 + xy0 ′′(x) + 4 ∫ x2 x 0 t3y0(t)dt (21) and has solution y1 = x21 73735200 − x 19 9157050 − x 18 8019648 + 4x 14 85995 − x 12 6600 − x 11 6930 + x7 105 − x3 + 2x + 1 and y2 = 𝑥30 11226184200000 − 𝑥28 1034948105100 − 𝑥27 774096523200 + 20747𝑥23 13750604627760 − 709x21 75209904000 − 1657𝑥20 147891744000 + 31𝑥16 8731800 − 𝑥14 109200 − 𝑥13 108108 + 𝑥9 1260 − x3 + 2x + 1 . Hence, in iteration steps, we have: 𝑦12 = 3.062817340869851 × 10−82 × 𝑥120 − 1.101677695323877 × 10−80 × 𝑥118 − 2.725562818772886 × 10−80 × 𝑥117 + 1.778958719423882 × 10−75 × 𝑥113 − 5.682824661052544 × 10−74 × 𝑥111 − 1.370653337095081 × 10−73 × 𝑥110 + 3.984172272372894 × 10−69 × 𝑥106 − 1.119818251782719 × 10−67 × 𝑥104 − 2.628843074816043 × 10−67 × 𝑥103 + 4.503028638092516 × 10−63 × 𝑥 99 − 1.101417450504745 × 10−61 × 𝑥 97 − 2.511985562347862 × 10−61 × 𝑥 96 + 2.827909775701136 × 10−57 × 𝑥 92 − 5.940775524869664 × 10−56 × 𝑥 90 − 1.313503431540083 × 10−55 × 𝑥 89 + 1.026299389853779 × 10−51 × 𝑥 85−1.822336772542073 × 10−50 × 𝑥 83 − 3.896422990960605 × 10−50 × 𝑥 82 + 2.175609109040211 × 10−46 × 𝑥 78 − 3.200678509627877 × 10−45 × 𝑥 76 − 6.598849111192931 × 10−45 × 𝑥 75 + 2.671403340873855 × 10−41 × 𝑥 71 − 3.173731777895746 × 10−40 × 𝑥 69 − 6.287576987469821 × 10−40 × 𝑥 68 + 1.85332923203967 × 10−36 × 𝑥 64 − 1.718197579672121 × 10−35 × 𝑥 62 − 3.257269778930915 × 10−35 × 𝑥 61 + 1.027480158702644 × 10−49 × 𝑥 59 + 6.961493020520868 × 10−32 × 𝑥 57 − 4.798561225077398 × 10−31 × 𝑥 55 − 8.659823953316347 × 10−31 × 𝑥 54 + 3.114021333121303 × 10−44 × 𝑥 52 + 1.32543948187801 × 10−27 × 𝑥 50 − 6.307772348188415 × 10−27 × 𝑥 48 − 1.076556205716984 × 10−26 × 𝑥 47 + 2.671419265527734 × 10−39 × 𝑥 45 + 1.154719813291063 × 10−23 × 𝑥 43 − 3.330521444892846 × 10−23 × 𝑥 41 − 5.329596586395636 × 10−23 × 𝑥 40 + 7.831725504493517 × 10−35 × 𝑥 38 + 3.873151195225192 × 10−20 × 𝑥 36 − 4.991934045359734 × 10−20 × 𝑥 34 − 7.402102900997027 × 10−20 × 𝑥 33 + 7.642681309794397 × 10−31 × 𝑥 31 + 3.461646293785214 × 10−17 × 𝑥 29 + 2.098836527787579 × 10−27 × 𝑥 24 + 1.058401144288997 × 10−24 × 𝑥17 − 𝑥 3 + 2𝑥 + 1 Table 2. Numerical results of the illustrative example above of Y12. 𝒙 Exact solution SAIM Error 0 1.000000000000000 1.000000000000000 0 0.1 1.199000000000000 1.199000000000000 0 0.2 1.392000000000000 1.392000000000000 0 0.3 1.573000000000000 1.573000000000000 0 0.4 1.736000000000000 1.736000000000000 0 0.5 1.875000000000000 1.875000000000000 0 0.6 1.984000000000000 1.984000000000000 0 0.7 2.057000000000000 2.057000000000000 0 0.8 2.088000000000000 2.088000000000000 0 0.9 2.071000000000000 2.071000000000000 0 1 2.000000000000000 2.000000000000000 0 57 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 Figure 2. Exact and approximate solution of the illustrative example. Example 3 Consider fourth – order Fredholm IDE [19, 22]. y(4)(x) = 1 4 + (1 − 2ln2)x − 6 (1+x)4 + ∫ (x − t)y(t)dt 1 0 (21) with ICs y(0) = 0 , y′(0) = 1, y′′(0) = −1, y′′′(0) = 2, The exact solution Y(x) = ln(1 + x). Solution To find y0 we take L(y0) = 1 4 + (1 − 2ln2)x − 6 (1+x)4 with y0 (0) = 0, y0 ′(0) = 1, y0 ′′(0) = −1,y0 ′′′(0) = 2 where h(x) = 1 4 + (1 − 2ln2)x − 6 (1+x)4 , L(y) = d4y dx4 , N(y) = 0. So, the primary step is: L(y0) = 1 4 + (1 − 2ln2)x − 6 (1+x)4 with 𝑦0(0)=0, 𝑦0 ′(0) = 1, 𝑦0 ′′(0) = −1,𝑦0 ′′′(0) = 2 (22) Then, the general relation as: L(yn+1) − h(x) − ∫ k(x, t)yn(t) x 0 d(t) = 0, yn+1(0) = 0, y ′ n+1 (0) = 1, yn+1 ′′ (0) = −1, y′′′n+1(0) = 2 (23) By solving the problem defined in Equation (22) we have: y0 = ln(x + 1) + x4 96 − 579905046931621 180143985094819840 x5 By same way the next step is to find 𝑦1 as follows: The first iteration can be gotten as: 𝑦1 (4)(x) = 1 4 + (1 − 2ln2)x − 6 (1+x)4 + ∫ (x − t)𝑦0(t)dt 1 0 (24) and has a solution: 𝑦1 = ln(x + 1) − 3621025736840333 68094426365841899520 𝑥4 + 104493422922097 8106479329266892800 𝑥5 𝑦2 = ln(x + 1) + 398426818305727 1361888527316837990400 𝑥4 − 2407976480303 34047213182920949760 𝑥5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 1.2 1.4 1.6 1.8 2 2.2 2.4 The solution at n=13 x-axis y -a x is Approximate Y12 of Ex2 Exact Solution 58 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 Hence, in iteration steps, we have: 𝑦7 = ln(x + 1) − 1446180158675437 980744084899180960278380544000000 𝑥4 + 977445743089 1539253760479322796195840000000 𝑥5. Table 3. Numerical results of the illustrative example above of Y7. 𝒙 Exact solution SAIM Error 0 0 0 0 0.1 0.095310179804325 0.095310179804325 0 0.2 0.182321556793955 0.182321556793955 0 0.3 0.262364264467491 0.262364264467491 0 0.4 0.336472236621213 0.336472236621213 0 0.5 0.405465108108164 0.405465108108164 0 0.6 0.470003629245736 0.470003629245736 0 0.7 0.530628251062170 0.530628251062170 0 0.8 0.587786664902119 0.587786664902119 0 0.9 0.641853886172395 0.641853886172395 0 1 0.693147180559945 0.693147180559945 0 Figure 3. Exact and approximate solution of the illustrative example. Example 4 Finally, consider fourth – order Voltera–IDE [21, 23]. y(4)(𝑥) = sinx + cosx + 2 ∫ sin(x − t) y(t)dt x 0 (25) with ICs y(0) = y′(0) = y′′(0) = y′′ ′(0) = 1, the Exact solution Y(x) = ex Solution To find 𝑦0 we take L(y0) = sinx + cosx with y(0) = y ′(0) = y′′(0) = y′′′(0) = 1, where g(x) = − sinx − cosx , L(y) = d4y dx4 , N(y) = 0. So, the primary step is taken: L(y0) = sinx + cosx , y0(0) = y0 ′(0) = y0 ′′(0) = y0 ′′′(0) = 1 (26) Then, the general relation as: L(yn+1) − h(x) − ∫ k(x, t)yn(t) x 0 d(t) = 0, yn+1(0) = y′n+1(0) = y′′n+1(0) = y′′′n+1(0) = 1 (27) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 The solution at n=8 x-axis y -a x is Approximate Y7 of Ex3 Exact Solution 59 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 By solving the problem defined in Eq. (26) we have y0 = cosx + sinx + x 2 + x3 3 as same way, the next step is to find y1 as follows: The first iteration can be gotten as: y1 (4)(𝑥) = sinx + cosx + 2 ∫ sin(x − t) y(t)dt x 0 (28) and has a solution: y1 = 9cosx − 8x + 10sinx − xcosx + xsinx + 4x 2 + 4x3 3 − x4 6 − x5 30 + x6 180 + x7 1260 − 8 y2 = 97cosx − 96x + 223sinx 2 − x 2cosx 2 − x2sinx 2 − 29𝑥𝑐𝑜𝑠𝑥 2 + 27𝑥𝑠𝑖𝑛𝑥 2 + 36𝑥2 + 12𝑥3 − 2𝑥4 − 2𝑥5 5 + 2𝑥6 45 + 2𝑥7 315 − 𝑥8 2520 − 𝑥9 22680 + 𝑥10 453600 + 𝑥11 4989600 − 96 . Table 4. Numerical results of the illustrative example above of Y2. 𝑥 Exact solution SAIM Error 0 1.000000000000000 1.000000000000000 0 0.1 1.105170918075648 1.105170918075658 1.0E -014 0.2 1.221402758160170 1.221402758160139 3.0E -014 0.3 1.349858807576003 1.349858807575984 1.9E -014 0.4 1.491824697641270 1.491824697641278 7.E -015 0.5 1.648721270700128 1.648721270700122 6.E -015 0.6 1.822118800390509 1.822118800390498 1.0E -014 0.7 2.013752707470477 2.013752707470502 2.5 E -014 0.8 2.225540928492468 2.225540928492492 2.4 E -014 0.9 2.459603111156950 2.459603111156937 1.2 E -014 1 2.718281828459046 2.718281828459041 4. E -015 Figure 4. Exact and approximate solution of the illustrative example. The essence of this method, SAIM in comparison with the other analytical methods does not need large computations such as Lagrange multiplier in the VIM or any complex assumptions like nonlinear Adomian’s polynomials in the ADM. It also does not need to constrict Homotopy in the HPM. Furthermore, this method proved that it is efficient in overcoming the difficulties in calculating and solving high -order integro-differential equations with easier steps. 4. Conclusion Most integro-differential equations are difficult to be solved analytically. For this purpose, the efficient Iterative analytic method (Temimi and Ansari) is applied to solve such 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 The solution at n=3 x-axis y -a x is Approximate Y2 of Ex5 Exact Solution 60 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (2) 2019 mentioned type of these equations. The present iteration method has strength features among other analytic methods. The obtained solutions can be shown as a series form that converges to the exact solution with simple computations. No complicated calculations have been shown with the presented method. 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