Gen. Math. Notes, Vol. 4, No. 2, June 211, pp. 37-48 ISSN 2219-7184; Copyright c ICSRS Publication, 211 www.i-csrs.org Available free online at http://www.gean.in Nuerical Solution of Volterra-Fredhol Integro-Differential Equation by Block Pulse Functions and Operational Matrices Leyla Rahani 1, Bijan Rahii 1 and Mohaad Mordad 2 1 Departent of Matheatics, Faculty of science, Islaic Azad University, Takestan branch, Takestan, Iran E-ail: l.rahani@tiau.ac.ir E-ail: bigrahii@gail.co 2 Departent of Matheatics, Faculty of science, Islaic Azad University, Islashahr branch, Islashahr, Tehran, Iran E-ail: ordad25@yahoo.co (Received:11-5-11/Accepted:2-6-11) Abstract In this paper, Block Pulse Functions and their operational atrices are used to solve Volterra-Fredhol integro-differential equation (VFIDE). First the equation is integrated over interval [, x] and then Block Pulse functions are used to obtain nuerical solution. Soe theores will prove convergence of the ethod. Soe nuerical exaples are included to illustrate accuracy of the ethod. Keywords: Volterra-Fredhol integro-differential equation, Block Pulse Functions, Operational atrix. 1 Introduction Integro-differential equations have been discussed in any applied fields, such as biological, physical and engineering probles. Therefore, their nuerical treatent is desired. Many nuerical ethods used to solve such equations [1, 3, 4, 6, 7, 8, 1, 12, 14]. In this paper, we propose a new nuerical ethod
38 Bijan Rahii et al. to solve the linear VFIDE: x f (x) = g(x) + k(x, t)f(t) dt + f() = a, l(x, t)f(t) dt x, t < 1, (1) where f(x) and g(x) are in L 2 ([, 1)) and k(x, t) and l(x, t) belong to L 2 ([, 1) [, 1)). Moreover k(x, t), l(x, t) and g(x) are known and f(x) is unknown. We assue Eq. (1.1) has a unique solution. The paper is organized as follows: In section 2, we describe Block Pulse functions, their properties and their operational atrices [5, 15]. In section 3, we propose a new nuerical ethod for solving linear VFIDE. In section 4 we analysis the error. Finally in section 5 we apply the proposed ethod on soe exaples to show the accuracy and efficiency of the ethod. 2 Review of Block Pulse Functions A set of Block Pulse functions (BPFs) are usually defined in [,1) as: i 1, t < (i+1), φ i (t) =, otherwise, (2) where i =, 1, 2,..., 1 and is an arbitrary positive integer. According to (2.1), the unit interval [,1) is divided into equidistant subintervals and the ith Block Pulse function φ i has only one rectangular pulse of unit hight in the subinterval [ i, i+1 ). One of the iportant properties of BPFs is the disjointness of the, which can directly be obtained fro the definition of BPFs. Indeed: φ i (t), i = j, φ i (t)φ j (t) = (3), i j, where i, j =, 1,..., 1. The orthogonality of BPFs is derived iediately fro: where δ ij is the Kronecker delta and h = 1. φ i (t)φ j (t) dt = h δ ij, (4)
Nuerical Solution of Volterra-Fredhol... 39 The set of BPFs is coplete, i.e. for every f L 2 ([, 1)), Parseval s identity holds: f 2 (t) dt = fi 2 φ i (t) 2, (5) where f i = 1 h i= f(t)φ i (t) dt, i =, 1,..., 1. (6) An arbitrary function can be expanded in vector for as: f(t) F T Φ(t) = Φ T (t)f. (7) where F = [f, f 1,..., f 1 ] T and Φ(t) = [φ (t), φ 1 (t),..., φ 1 (t)] T. Now let k(x, t) be arbitrary L 2 function of two variables on [, 1) [, 1). It can be expanded by BPFs as: k(x, t) Φ T (x)kψ(t), (8) where Φ(x) and Ψ(t) are 1 and 2 diensional BPF vectors, respectively, and K is the ( 1 2 ) Block Pulse coefficient atrix with entries k ij, i =, 1,..., 1 1, j =, 1,..., 2 1, as follows: k ij = 1 2 In this paper for convenience, we put 1 = 2 =. k(x, t)φ i (x)φ j (t) dx dt. (9) Moreover, fro the disjointness property of BPFs, follows: φ (t) Φ(t)Φ T φ 1 (t) (t) =......, (1) φ 1 (t) Φ(t)Φ T (t) dt = h I, (11) Φ T (t)φ(t) = 1, (12) Φ(t)Φ T (t)v = Ṽ Φ(t), (13)
4 Bijan Rahii et al. where V is an diensional vector and Ṽ = diag(v ). Moreover: Φ T (t)bφ(t) = Φ T (t) B, (14) where B is an ( ) atrix, ˆB is an diensional vector with eleents equal to the diagonal entries of atrix B. As represented in [2, 9]: t Φ(τ) dτ P Φ(t), (15) where P is the following ( ) atrix: 1 2 2 2 P = h 1 2 2 1 2, (16) 2....... 1 called operational atrix of integration. So, the integral of every function f can be approxiated as follows: t 3 A Method to Solve VFIDE Consider the following VFIDE: f(τ) dτ F T P Φ(t). (17) f (x) = g(x) + f() = a, k(x, t)f(t) dt + l(x, t)f(t) dt x, t < 1, (18) First, we integrate fro (3.1) in the interval [, x], we will have: f(x) = a + g(u)du + u k(u, t)f(t)dtdu+ l(u, t)f(t) dtdu. (19) Now, we approxiate functions f, g, k and l with respect to BPFs as follows: f(x) F T Φ(x) = Φ T (x)f, g(x) Φ T (x)g, k(x, t) Φ T (2) (x)kφ(t), l(x, t) Φ T (x)lφ(t),
Nuerical Solution of Volterra-Fredhol... 41 where vectors F and G and atrices K and L are BPFs coefficients of f, g, k and l, respectively. Substituting (3.3) into (3.2) we will have: Φ T (x)f a + Φ T (x)p T G + ( u Φ T (u)k ( Φ T (u)l ) Φ(t)Φ T (t)f dt du+ ) Φ(t)Φ T (t)dt F du. (21) Substituting (2.1), (2.11), (2.12) and (2.14) into (3.4), results: Φ T (x) F a Φ T (x) Φ(x) + Φ T (x) P T G+ Put B = K F P and apply (2.13), we obtain: Φ T (u)bφ(u)du = Φ T (u)k F P Φ(u)du + hφ T (x)p T LF. (22) Φ T (u) Bdu Φ T (x)p T B, putting  = P T B and using (2.13), above relation converts to: Φ T (x)p T B = Φ T (x)â = ΦT (x)aφ(x), (23) where A = diag(â). Substituting (3.6) into (3.5), results: Φ T (x)f Φ T (x)(a I + A)Φ(x) + Φ T (x)(p T G + hp T LF ). Now, with C = a I + A and using(2.13), we have: Φ T (x)f Φ T (x)ĉ + ΦT (x)(p T G + hp T LF ), or F Ĉ + P T G + hp T LF, (24) where Ĉ can be written as follows: Ĉ = a 1 + h 2 K F (25) where 1 = (1, 1,, 1) T and
42 Bijan Rahii et al. K = 1 k 4 11 2 i=1 3 i=1 i=1. k i1 1 4 k 22 k i1 k i1 3 i=2. k i2 i=2 k i2 1 4 k 33. i=3 k i3.... i= 1 1 k i, 1 k 4. and eans that the first and last ters have factor 1 2. Substituting (3.8) into (3.7) and replacing with =, we will have: (I hp T L h 2 K )F = (P T G + a 1). (26) Solving the syste of equations (3.9) we obtain unknowns f, f 1,, f 1. Also, structure of K shows that we do not need to evaluate k ij for j > i. 4 Error Analysis In this section, we analyse the error when a differentiable function f(x) is represented in a series of BPFs over the interval I = [, 1). We need the following theore. Theore: Suppose f is continuous in I, is differentiable in (, 1), and there is a nuber M such that f (x) M, for every x I. Then for all a, b I. f(b) f(a) M b a, Proof. See [11].
Nuerical Solution of Volterra-Fredhol... 43 Now, we assue that f(x) is a differentiable function on I such that f (x) M. We define the error between f(x) and its BPFs expansion over every subinterval I i as follows: e i (x) = f i f(x), x I i where I i = [ i, i+1 ). It can be shown that: e i 2 = i+1 i e i 2 (x) dx = i+1 i ( ) f i f(x) 2 dx = 1 ( ) f i f(η) 2, η I i, (27) where we used ean value theore for integral. Using Eq.(2.5) and the ean value theore, we have: f i = i+1 i f(x) dx = 1 f(ζ) = f(ζ), ζ I i. (28) Substituting (4.2) into (4.1) and using Theore, we will have: This leads to: e i 2 = 1 = e(x) 2 = ( 1 i= ( ) f(ζ) f(η) 2 M 2 ζ η 2 M 2. (29) 3 Since for i j, I i I j =, then e(x) 2 = e 2 (x) dx = ) e 2 i (x) dx + 2 i j 1 i= ( Substituting (4.3) into(4.4), we have ( 1 i= ) e i (x) 2 dx e i (x) e j (x) dx. ) 1 e 2 i (x) dx = e i 2. (3) e(x) 2 M 2 2, hence, e(x) = O( 1 ), where e(x) = f (x) f(x) and f (x) = i= 1 i= f i φ i (x).
44 Bijan Rahii et al. 5 Nuerical Exaples In this section, we present soe exaples and their nuerical results. Exaples 2 and 3 were used in [13] and [1], respectively. All coputations are perfored by the Maple 9.5 software package. Exaple 1: Consider the following VIDE: f (x) = f(t) dt, f() = 1 with the exact solution f(x) = cosh(x). g(x) =, l(x, t) = and k(x, t) = 1. Indeed, in this exaple, we have The nuerical results are shown in table 1. Table 1: Nuerical results for Exaple 1 x Exact solution Approxiate solution Approxiate solution = 32 = 64 1 1.244 1.61.1 1.54 1.6111 1.5193.2 1.267 1.2829 1.19166.3 1.45339 1.44527 1.46811.4 1.8172 1.77413 1.8467.5 1.127626 1.13667 1.131772.6 1.185465 1.191664 1.18656.7 1.255169 1.257743 1.251676.8 1.337435 1.334886 1.341668.9 1.43386 1.423773 1.431547 Exaple 2: Consider the following VIDE [13]: f (x) = 1 f(t) dt, f() = with the exact solution f(x) = sin(x). In this exaple, we have g(x) = 1, l(x, t) = and k(x, t) = 1.
Nuerical Solution of Volterra-Fredhol... 45 Table 2: Nuerical results for Exaple 2 x Exact solution Approxiate solution Approxiate solution = 64 = 128.7812.396.1.99833.11383.975.2.198669.19463.19791.3.29552.29998.296263.4.389418.387959.391571.5.479426.486243.482844.6.564642.56594.5627.7.644218.64595.646312.8.717356.7258.717892.9.783327.782319.784773 The nuerical results are shown in table 2. Exaple 3: Consider the following linear FIDE [1]: f (x) = e x + e 1 1 + f(t) dt, f() = 1 with the exact solution f(x) = e x. We have g(x) = e x +e 1 1, l(x, t) = 1 and k(x, t) =. This exaple was used in [1] by using spline functions with = 1 and = 2, where equals the order of spline functions. The absolute values of error in [1] with stepsize h =.1 at the point.4 are given as 1.5 1 2 and 1.8 1 3 for = 1 and = 2, respectively. The nuerical results are shown in table 3. Exaple 4: Consider the following linear VFIDE: f (x) = 2sin(x) x 2 sin(2x) + 2sin(2x) 2xcos(2x) 2e x f() = +5e x 1 + 2x + cos(x + t)f(t) dt + e x t f(t) dt, with the exact solution f(x) = x 2. In this exaple, we have g(x) = 2sin(x) x 2 sin(2x) + 2sin(2x) 2xcos(2x) 2e x + 5e x 1 + 2x, l(x, t) = e x t and
46 Bijan Rahii et al. Table 3: Nuerical results for Exaple 3 x Exact solution Approxiate solution Approxiate solution = 16 = 64 1.969719.992248.1.94837.91993.93455.2.818731.845.82268.3.74818.755324.737384.4.6732.666636.671399.5.6653.588375.61842.6.548812.552766.547986.7.496585.487894.498951.8.449329.458378.447261.9.4657.4466.4724 k(x, t) = cos(x + t). The nuerical results are shown in table 4. Table 4: Nuerical results for Exaple 4 x Exact solution Approxiate solution Approxiate solution = 32 = 64.493.123.1.1.1224.1384.2.4.41571.38226.3.9.88486.92926.4.16.152985.158856.5.25.266336.257994.6.36.371862.36212.7.49.494971.48361.8.64.635662.647692.9.81.793932.87378 6 Conclusion The proposed ethod for solving linear VFIDE was based on BPFs and their operational atrix. This ethod converts equation to a linear syste of algebraic equations. It, s accuracy is shown on soe exaples. As exaples
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