Numerical solution of system of linear integral equations via improvement of block-pulse functions

Journal of Mathematical Modeling Vol. 4, No., 16, pp. 133-159 JMM Numerical solution of system of linear integral equations via improvement of block-pulse functions Farshid Mirzaee Faculty of Mathematical Sciences and Statistics, Malayer University, P.O. Box 65719-95863, Malayer, Iran Email: f.mirzaee@malayeru.ac.ir Abstract. In this article, a numerical method based on improvement of block-pulse functions (IBPFs) is discussed for solving the system of linear Volterra and Fredholm integral equations. By using IBPFs and their operational matrix of integration, such systems can be reduced to a linear system of algebraic equations. An efficient error estimation and associated theorems for the proposed method are also presented. Some examples are given to clarify the efficiency and accuracy of the method. Keywords: system of linear integral equations, improvement of block-pulse functions, operational matrix, vector forms, error analysis. AMS Subject Classification: 45A5, 45D5, 45F5, 65R, 65D3. 1 Introduction Systems of linear integral equations and their solutions have great importance in science and engineering. Most physical problems, such as biological applications in population dynamics and genetics where impulses arise naturally or are caused by control, can be modeled by an integral equation or a system of these equations. The systems of integral equations are usually difficult to solve analytically, thus some numerical methods are applied to approximately solve them. Systems of linear Volterra and Fredholm inte- Corresponding author. Received: 18 April 16 / Revised: 1 August 16 / Accepted: September 16. c 16 University of Guilan http://jmm.guilan.ac.ir

134 F. Mirzaee gral equations can be appeared in the following form, respectively and f(x) = g(x) + x f(x) = g(x) + k(x, y)f(y)dy, x D = [, 1), (1) k(x, y)f(y)dy, x D, () where f(x) is an unknown function and g(x) and k(x, y) are analytical function on D and D D, respectively, as follows f(x) = [f 1 (x), f (x),..., f m (x)] T, (3) g(x) = [g 1 (x), g (x),..., g m (x)] T, (4) k(x, y) = [k ij (x, y)], i, j = 1,,..., m. (5) Recently, several numerical methods such as the homotopy perturbation method [4, 5], the Lagrange method [3], the modified homotopy perturbation method [8], the rationalized Haar functions method [1, 11], the differential transformation method [1], the Tau method [16], the variational iteration method [19], the Legendre matrix method [], the Adomian method [5], the Galerkin method [1], the Bessel matrix method [1], and other methods [14 19] have been used for solving integral and integrodifferential equations systems. In this paper, IBPFs are introduced. Also, some theorems are proved for IBPFs method that show the results of this numerical expansions are more precise than the results of block pulse expansions. This functions are disjoint, orthogonal and complete. According to the disjointness of IBPFs, the joint terms will disappear in each subinterval when multiplication, division and some other operations are applied. Also, the orthogonality property of IBPFs will cause that the operational matrix to be a sparse matrix. The completeness of IBPFs guarantees that an arbitrary small mean square error can be obtained for a real bounded function, which has only a finite number of discontinuous point in the interval x [, 1), by increasing the number of terms in the improved block pulse series. The rest of paper is organized as follows: In Section, we describe IBPFs and their properties. In Section 3, we apply IBPFs for approximating the solution of system of linear Volterra and Fredholm integral equations. Convergence analysis is discussed in Section 4. Numerical results are given in Section 5 to illustrate the efficiency and accuracy of proposed method. Finally, Section 6 concludes the paper.

Numerical solution of system of linear integral equations 135 IBPFs and their properties.1 Definition of IBPFs An (n + 1)-set of IBPFs consists of (n + 1) functions which are defined over district D as follows { 1, x [, h φ (x) = ),, otherwise, φ i (x) = { 1, x [(i 1)h + h, ih + h ),, otherwise, i = 1,,..., n 1, φ n (x) = { 1, x [1 h, 1),, otherwise, where n is an arbitrary positive integer and h = 1 n. The IBPFs are disjoint φ i (x)φ j (x) = { φi (x), i = j,, otherwise, where i, j =, 1,..., n and are orthogonal to each other h 1, i = j {, n}, φ i (x)φ j (x)dx = h, i = j {1,,..., n 1},, otherwise, (6) (7) where x D.. Vector forms Consider the first (n + 1) terms of IBPFs and write them concisely as (n + 1)-vector Φ n (x) = [φ (x), φ 1 (x),..., φ n (x)] T, x D. (8) Eq. (6) implies that Φ n (x)φ T n (x) = φ (x)... φ 1 (x)......... = diag(φ n(x)). (9)... φ n (x)

136 F. Mirzaee Now suppose that X be an (n + 1)-vector. Hence by using Eq. (9) we obtain Φ n (x)φ T n (x)x = XΦ n (x), where X = diag(x) is an (n + 1) (n + 1) diagonal matrix..3 Operational matrix We have x { x, x [, h φ (y)dy = ), h, otherwise. Note that x = h 4, at mid-point of [, h ). x [, h ), by h 4. x φ i (x) = So we can approximate x, for, x [, (i 1)h + h ), x (i 1)h h, x [(i 1)h + h, ih + h ), h, otherwise, for i = 1,,..., n 1. Also, we can approximate x (i 1)h h, for x [(i 1)h + h, ih + h ), by h. x { x 1 + h φ n (x) =, x [1 h, 1), otherwise. So, we can approximate x 1 + h, for x [1 h, 1), by h 4. Therefore, the integration of the vector Φ n (x) defined in Eq. (8) can be approximated by where Using Eq. (9), we get x Φ n (y)dy P 1Φ n (x), (1) P 1 = h 4. 1... 4... 4 4... 4 4.......... 4... 1. Φ n (y)φ T n (y)dy P, (11)

Numerical solution of system of linear integral equations 137 where P = h. 1...................... 1..4 IBPFs expansions A continues function f(x) L (D) may be expanded by the IBPFs as n f(x) f n (x) = f i φ i (x) = Fn T Φ n (x) = Φ T n (x)f n, (1) i= where F n is an (n + 1) 1 vector given by F n = [f, f 1,..., f n ] T, and Φ n (x) is defined in Eq. (8), and f i is obtained as n h f(x)dx, i =, f i = n ih+ h f(x)dx, i = 1,,..., n 1, (i 1)h+ h n 1 h f(x)dx, i = n. (13) Similarly a function of two variables, k(x, y) L (D D) can be approximated by IBPFs as follows k(x, y) k n (x, y) = Φ T n (x)k n Φ n (y), (14) where Φ n (x) and Φ n (y) are IBPFs vector of dimension (n + 1), and K n = [k ij ] is the (n + 1) (n + 1) IBPFs coefficients matrix of k(x, y). 3 Method of solution 3.1 System of linear Volterra integral equations In this section, we solve system of linear Volterra integral equations of the form Eq. (1) by using IBPFs. We can rewrite Eq. (1) as follows f i (x) = g i (x) + m j=1 x k ij (x, y)f j (y)dy, i = 1,,..., m. (15)

138 F. Mirzaee We now approximate functions f i (x), g i (x) and k ij (x, y), i, j = 1,,..., m, by IBPFs as follows f i (x) Φ T n (x)f i, g i (x) Φ T n (x)g i, (16) k ij (x, y) Φ T n (x)k ij Φ n (y), where i, j = 1,,..., m, Φ n (x) is defined by (8), F i, G i and K ij are IBPFs coefficients of f i (x), g i (x) and k ij (x, y), respectively. By substituting Eq. (16) in Eq. (15), we get Φ T n (x)f i = Φ T n (x)g i + m j=1 x therefore by Eq. (9), we have Φ T n (x)f i = Φ T n (x)g i + From Eq.(1), we have where Φ T n (x)k ij Φ n (y)φ T n (y)f j dy, i = 1,,..., m, m x Φ T n (x)k ij diag(φ n (y))dyf j, i = 1,,..., m. j=1 Φ T n (x)f i = Φ T n (x)g i + m Φ T n (x)k ij AF j, i = 1,,..., m, (17) j=1 P 1 Φ n (x)... P 1 1 Φ n (x)... A =......,... P 1 n Φ n (x) and P 1 i is ith row of P 1. On the other hand, using (6), we have Kij P 1 Φ n (x) Kij 1P 1 1Φ n (x)... Kij np 1 nφ n (x) Φ T n (x)k ij A = Φ T Kij 1 n (x) P 1 Φ n (x) Kij 11P 1 1Φ n (x)... Kij 1nP 1 nφ n (x)...... Kij np 1 Φ n (x) Kij n1p 1 1Φ n (x)... Kij nnp 1 nφ n (x) [ n ] n n = Kij l P 1 l φ l (x) Kij l1 P 1 1l φ l (x)... Kij ln P 1 nl φ l (x) l= = Φ T n (x)b ij, l= l=

Numerical solution of system of linear integral equations 139 where B ij = K ij P 1 K 1 ij P 1 1... K n ij P 1 n K 1 ij P 1 1 K 11 ij P 1 11... K 1n ij P 1 n1. K n ij P 1 n..... K n1 ij P 1 1n... K nn ij P 1 nn. So, we can rewrite Eq. (17) as follows or Let m Φ T n (x)f i = Φ T n (x)g i + Φ T n (x)b ij F j, i = 1,,..., m, j=1 m F i = G i + B ij F j, i = 1,,..., m. (18) j=1 F = [F 1, F,..., F m ] T, G = [G 1, G,..., G m ] T, (19) B = [B ij ], i, j = 1,,..., m. Therefore, Eq. (18) can be written as F = G + BF. After solving above linear system, we can find F and accordingly find F i, i = 1,,..., m, so f i (x) Φ T n (x)f i. 3. System of linear Fredholm integral equations In this section, we solve system of linear Fredholm integral equations of the form Eq. () by using IBPFs. We can rewrite Eq. () as follows f i (x) = g i (x) + m j=1 Substituting Eq. (16) in Eq. (), we get Φ T n (x)f i = Φ T n (x)g i + m j=1 k ij (x, y)f j (y)dy, i = 1,,..., m. () Φ T n (x)k ij Φ n (y)φ T n (y)f j dy, i = 1,,..., m.

14 F. Mirzaee Using Eq. (11), we get Φ T n (x)f i = Φ T n (x)g i + m Φ T n (x)k ij P F j, i = 1,,..., m. j=1 Letting C = [K ij P ], from Eq.(19) we have F = G + CF. After solving above linear system, we can find F and accordingly find F i ; i = 1,,..., m, so f i (x) Φ T n (x)f i. 4 Convergence analysis In this section, we show that the method discussed in the previous section is convergent and its order of convergence is O( 1 n ). We define and ( f(x) = f(x) dx) 1, f(x) = ( m i=1 where f(x) L (D) and f(x) is defined in Eq. (3) and f i (x) ) 1, (1) and ( k(x, y) = i=1 j=1 k(x, y) dxdy) 1, m m k(x, y) = k ij (x, y) 1, () where k(x, y) L (D D) and k(x, y) is defined in Eq. (5). Also, for our purposes we will need the following theorems. Theorem 1. Let f(x) L (D) and f n (x) be the IBPFs expansion of f(x) that is defined as n f n (x) = f i φ i (x), i=

Numerical solution of system of linear integral equations 141 where f i ; i =, 1,..., n, are defined in Eq. (13). Then the criterion of this approximation is that the mean square error between f(x) and f n (x) in the interval x D (f(x) f n (x)) dx, achieves its minimum value and also we have f (x)dx = fi φ i (x). i= Proof. It is an immediate consequence of theorem which is proved in [9]. Theorem. Suppose f(x) is continuous on D, differentiable on (, 1), and there exists a positive scalar M such that f (x) M, for every x D. Then f(b) f(a) M b a, a, b D. Proof. See [18]. Theorem 3. Suppose f n (x) is the IBPFs expansions of f(x) that is defined as Eq. (1) and f(x) is differentiable on D such that f (x) M. Also, assume that e n (x) = f(x) f n (x), then e n (x) = O(h). Proof. Suppose x =, x i = ih h, i = 1,..., n and x n+1 = 1. We define the error between f(x) and its IBPFs expansion over every subinterval I i = [x i, x i+1 ) as follows e n,i (x) = f(x) f i (x), x I i, where i =, 1,..., n. By using mean value theorem for integral, we have e n, (x) = h e n,(x)dx = h where ξ I. Also, for i = 1,,..., n 1, we have e n,i (x) = ih+ h ih h e n,i(x)dx = where ξ i I i. Furthermore, we have e n,n (x) = 1 h e n,n(x)dx = ih+ h ih h 1 h (f(x) f ) dx = h (f(ξ ) f ), (f(x) f i ) dx = h (f(ξ i ) f i ), (f(x) f n ) dx = h (f(ξ n) f n ),

14 F. Mirzaee where ξ n I n. Using Eq. (13) and the mean value theorem, we have f i = n h f(x)dx = n h f(η ) = f(η ), i =, n ih+ h f(x)dx = nhf(η (i 1)h+ h i ) = f(η i ), i = 1,,..., n 1, n 1 h f(x)dx = n h f(η n) = f(η n ), i = n, where η i I i, i =, 1,..., n. From the above equations and Theorem, we get e n,i (x) = h (f(ξ ) f(η )) M h ξ η M h 3 8, i =, h (f(ξ i ) f(η i )) M h ξ i η i M h 3, i = 1,,..., n 1, h (f(ξ n) f(η n )) M h ξ n η n M h 3 8, i = n. (3) We have e n (x) = = e n(x)dx = i= ( n e n,i (x)) dx i= ( n ) e n,i(x) dx + i j e n,i (x)e n,j (x)dx. Since for i j, I i I j =, then e n (x) = ( n ) e n,i(x) dx = i= n e n,i (x). (4) i= Substituting Eq. (3) into Eq. (4), we get which completes the proof. e n (x) M h 3M h 3, 4 Suppose that e n(x) is the error between f(x) and its BPFs expansion. From [9], it is clear that e n (x) e n(x).

Numerical solution of system of linear integral equations 143 Lemma 1. Let g(x) be as defined in Eq. (4), g n (x) be the IBPFs of g(x) and e g (x) = g(x) g n (x). Then Proof. From Eq. (1), we have e g (x) = e g (x) = O(h). ( m i=1 g i (x) g n,i (x) ) 1, and from Theorem, g i (x) g n,i (x) C i h. Then e g (x) ( m which completes the proof. i=1 C i h ) 1 = ( m i=1 C i ) 1 h = Ch, (5) Theorem 4. Let k n (x, y) be the IBPFs expansions of k(x, y) defined as Eq. (14) and k(x, y) be differentiable on D D such that k (x, y) M. Also, assume that e n (x, y) = k(x, y) k n (x, y), then e n (x, y) = O(h). Proof. Suppose x = y =, x i = y i = ih h, i = 1,..., n and x n+1 = y n+1 = 1. We define the error between k(x, y) and its IBPFs expansion over every subinterval I i,j = [x i, x i+1 ) [y j, y j+1 ) as follows e n,ij (x, y) = k(x, y) k ij (x, y), x I i,j, i, j =, 1,..., n. By using the mean value theorem for integral and similar to the proof of Theorem 3, we get e n,ij (x, y) { M h 4 8, j =, n, 5M h 4 8, j = 1,,..., n 1, (6) for i =, n and e n,ij (x, y) { 5M h 4 8, j =, n, M h 4, j = 1,,..., n 1, (7)

144 F. Mirzaee for i = 1,,..., n 1. We have e n (x, y) = = e n(x, y)dxdy = n i= j= + i k j l n e n,ij(x, y) dxdy n i= j= e n,ij (x, y)e n,kl (x, y)dxdy. n e n,ij (x, y) Since for i k and j l, we have I i I k = and I j I l =, then e n (x, y) = = n i= j= i= j= n e n,ij(x, y) dxdy dxdy n n e n,ij (x, y). (8) Substituting Eqs. (6) and (7) into Eq. (8), we get e n (x, y) M h 3M h 3. Suppose e n(x, y) be the error between k(x, y) and its BPFs expansion. From [14], it is clear that e n (x, y) e n(x, y). Lemma. Let k(x, y) be as defined in Eq. (5), k n (x, y) be the IBPFs of k(x, y) and e k (x, y) = k(x, y) k n (x, y). Then Proof. From Eq. (), we have e k (x, y) = O(h). 1 m m e k (x, y) = k ij (x, y) k n,ij (x, y). i=1 j=1

Numerical solution of system of linear integral equations 145 From Theorem 3, we conclude that k ij (x, y) k n,ij (x, y) C ij h. Therefore m m e k (x, y) i=1 j=1 C ijh 1 m m = i=1 j=1 C ij 1 h = Ch. (9) Let the error of IBPFs be denoted by E n = f(x) f n (x), x D, where f(x) was defined in Eq. (3). Furthermore, assume the following hypotheses: (M1) Let f(x) N for x D; (M) Let k(x, y) N for (x, y) D D; (M3) According to lemma 1 and, let and E g = e g (x) Ch, E k = e k (x, y) C h, where C and C are coefficients defined in Eqs. (5) and (9) and g(x) and k(x, y) were defined in Eqs. (4) and (5), respectively; (M4) Let N + C h < 1. Theorem 5. Let f(x) and f n (x) be the exact and approximate solutions of Eq. (1) or Eq. (), respectively. Also assumptions (M1)-(M4) are satisfied. Then we have Proof. For the first case, from Eq. (1), we have f(x) f n (x) = g(x) g n (x) + and therefore E n (C + C N)h 1 N C h. (3) x (k(x, y)f(y) k n (x, y)f n (y)) dy, E n E g + x k(x, y)f(y) k n (x, y)f n (y).

146 F. Mirzaee It is clear that x 1, So E n E g + k(x, y)f(y) k n (x, y)f n (y). (31) Also for the second case, from Eq. (), we have f(x) f n (x) = g(x) g n (x) + and therefore (k(x, y)f(y) k n (x, y)f n (y)) dy, E n E g + k(x, y)f(y) k n (x, y)f n (y). So Eq. (31) is true in the both cases. Now, according to assumptions (M1)-(M3), we have k(x, y)f(y) k n (x, y)f n (y) k(x, y) E n + E k (E n + f(x) ) N E n + C h (E n + N). (3) Also from assumptions (M3), Eqs. (31) and (3), we have E n (C + C N)h + (N + C h)e n. Therefore according to (M4), Eq. (3) is satisfied and this completes the proof. Also we have E n = O(h). Lemma 3. Suppose f(x) and f n (x) are the exact and approximate solution of Eq. (1) or Eq. (), respectively, where f(x) was defined in Eq. (3) and f n (x) = [f 1,n (x), f,n (x),..., f m,n (x)] T. Then e i,n = f i (x) f i,n (x) = O(h). (33) Proof. From Theorem 4, we have E n Ch and according to Eq. (1) we have e i,n E n Ch.

Numerical solution of system of linear integral equations 147 Table 1: Numerical results for Example 1 (n = 8, 16). Nodes x Exact solution Present method n = 8 n = 16. (1.,.,.) (1.68,.948,.3384) (1.17,.4696,.167).1 (1.,.3,.1) (1.33,.37638,.15868) (1.8,.37535,.15686). (1.,.6,.8) (1.391,.75169,.37768) (1.87,.5687,.5843).3 (1.,.9,.48) (1.391,.75169,.37768) (1.15,.93799,.5857).4 (1.,3.,.7) (1.477,3.1733,.65979) (1.117,3.1558,.65713).5 (1.,3.5,1.) (1.61,3.5347,1.56) (1.15,3.586,1.131).6 (1.,3.8,1.3) (1.831,3.8835,1.41441) (1.5,3.8763,1.488).7 (1.,4.1,1.68) (1.1195,4.5841,1.88776) (1.44,4.6416,1.63535).8 (1.,4.4,.8) (1.1195,4.5841,1.88776) (1.359,4.4411,.13678).9 (1.,4.7,.5) (1.1799,4.6384,.463) (1.443,4.683,.411) 5 Numerical experiments In this section, several examples are presented to demonstrate the effectiveness of our approach. In this regard, we have reported the values of the exact solution f(x), the approximate solution f n (x); n = 8; 16; 3; 64 and the L -norms of errors that have been calculated by using ( E,n = f(x) f n (x) dx) 1, x D. at the selected points of the given interval in tables and figures. All computations is preformed in Matlab. In order to show the convergence rate of this numerical method, we introduce the following notation E n,n = log f(x) f n (x) f(x) f n (x). The following problems have been tested. Example 1. Consider the following system of linear Volterra integral equations [] x x f 1 (x) = 1 x x 3 1 5 x5 1 3 x6 + yf (y)dy + y 3 f 3 (y)dy, x x f (x) = + 3x 1 x 1 4 x4 5 x5 + yf 1 (y)dy + y f 3 (y)dy, f 3 (x) = x + x x 3 1 x4 3 5 x5 + x x f 1 (y)dy + x y 3 f (y)dy, where x [, 1). The exact solution is f 1 (x) = 1, f (x) = + 3x and f 3 (x) = x+x. Table 1 and illustrates the comparison between the exact

148 F. Mirzaee solution and numerical solution given by the proposed method (IBPFs) for different values of n. Also, we compare the infinity-norm of absolute error computed by the present method, block pulse functions method (BPFs) [13] and rationalized Haar functions method (RHFs) [15] in Table 3. Furthermore, for this example, numerical convergence rate for proposed method is tabulated in Table 4. Graph of the absolute errors for f 1 (x), f (x) and f (x) are shown in Figures 1, and 3, respectively. Table : Numerical results for Example 1 (n = 3, 64). Nodes x Present method n = 3 n = 64. (1.4,.346,.797) (1.1,.117,.395).1 (1.19,.8133,.11148) (1.5,.817,.11137). (1.,.5659,.5797) (1.5,.694,.8569).3 (1.6,.9376,.58) (1.6,.8965,.47319).4 (1.31,3.1891,.73657) (1.8,3.1879,.73639).5 (1.37,3.51,1.33) (1.9,3.55,1.8).6 (1.47,3.78155,1.998) (1.1,3.7813,1.9894).7 (1.61,4.691,1.63345) (1.16,4.1948,1.696).8 (1.89,4.43815,.1338) (1.1,4.3978,.6713).9 (1.13,4.71967,.551) (1.31,4.71898,.54917) Table 3: Approximate L -norm of absolute error for Example 1. Methods E,n Method of [15] n = 8 (1.5e-,1.7e-1,1.31e-1) n = 16 (3.e-3,5.68e-,6.11e-) n = 3 (6.93e-4,3.1e-,3.16e-) n = 64 (1.87e-4,.1e-,1.96e-) Method of [13] n = 8 (1.6e-,1.9e-1,1.16e-1) n = 16 (.67e-3,5.4e-,5.8e-) n = 3 (6.67e-4,.71e-,.9e-) n = 64 (1.67e-4,1.4e-,1.48e-) Present method n = 8 (8.89e-3,1.3e-1,1.9e-1) n = 16 (.44e-3,5.9e-,5.63e-) n = 3 (6.37e-4,.67e-,.86e-) n = 64 (1.63e-4,1.35e-,1.44e-)

Numerical solution of system of linear integral equations 149 Table 4: Numerical convergence rate for Example 1. m E n,n for f 1 (x) for f (x) for f 3 (x) 4 1.736.967.9116 8 1.8685.9656.9578 16 1.934.9831.979 3 1.967.9916.9894..18.16.14 Error for m=8 Error for m=16 Error for m=3 Error for m=64.1 Error.1.8.6.4..1..3.4.5.6.7.8.9 1 x Figure 1: Absolute value of error, Example 1 for f 1 (x). Example. Consider the following system of linear Volterra integral equations [4]: x f 1 (x) = 1 x + (f 1 (y) + ye y f (y)) dy, f (x) = 1 + x + x ( ye y f 1 (y) f (y) ) dy, where x [, 1). The exact solution is f 1 (x) = e x and f (x) = e x. Table 5 illustrates the comparison between the exact solution and numerical solution given by the proposed method (IBPFs) for different values of n. Also, we compare the infinity-norm of absolute error computed by the present method, BPFs [4] and RHFs [15] method in Table 6. Furthermore, for this example, numerical convergence rate for proposed method is tabulated in Table 7. Graph of the absolute errors for f 1 (x) and f (x) are shown in Figures 4 and 5, respectively.

15 F. Mirzaee.7.4.1.18 Error for m=8 Error for m=16 Error for m=3 Error for m=64 Error.15.1.9.6.3.1..3.4.5.6.7.8.9 1 x Figure : Absolute value of error, Example 1 for f (x)..3.5. Error for m=8 Error for m=16 Error for m=3 Error for m=64 Error.15.1.5.1..3.4.5.6.7.8.9 1 x Figure 3: Absolute value of error, Example 1 for f 3 (x). Table 5: Numerical results for Example. Nodes x, Exact Present method n = 8 n = 16 n = 3 n = 64., (1.,1.) (1.36,.96939) (1.1596,.98454) (1.789,.993) (1.393,.9961).1, (1.1517,.9484) (1.14747,.877) (1.141,.8778) (1.118,.9839) (1.9984,.9931)., (1.14,.81873) (1.31,.763) (1.1968,.8151) (1.138,.8471) (1.9,.81371).3, (1.34986,.748) (1.31,.763) (1.396,.71933) (1.37969,.7491) (1.35167,.73977).4, (1.4918,.673) (1.516,.6693) (1.48767,.6797) (1.51917,.65795) (1.5114,.6619).5, (1.6487,.6653) (1.747,.57499) (1.69669,.58868) (1.674,.5975) (1.6649,.6165).6, (1.81,.54881) (.8,.49749) (1.93339,.5144) (1.8465,.54164) (1.856,.54681).7, (.1375,.49659) (.9115,.494) (.631,.486) (.531,.4911) (.3877,.4891).8, (.554,.44933) (.9115,.494) (.3475,.4191) (.9969,.4391) (.486,.4446).9, (.4596,.4657) (.61538,.36847) (.5313,.3997) (.5876,.3937) (.5165,.3979)

Numerical solution of system of linear integral equations 151 Table 6: Approximate L -norm of absolute error for Example. Methods E,n Method of [15] n = 8 (6.47e-,7.89e-) n = 16 (8.13e-,4.7e-) n = 3 (4.1e-,.7e-) n = 64 (.34e-,6.8e-) Method of [4] n = 8 (5.61e-,7.17e-) n = 16 (7.94e-,.89e-) n = 3 (3.71e-,1.35e-) n = 64 (.3e-,5.97e-3) Present method n = 8 (1.44e-1,4.e-) n = 16 (7.13e-,.17e-) n = 3 (3.54e-,1.13e-) n = 64 (1.77e-,5.74e-3) Table 7: Numerical convergence rate for Example. m for f 1 (x) for f (x) 4 1.16.7896 8 1.159.8916 16 1.96.9449 3 1.5.97 E n,n Example 3. Consider the following system of linear Fredholm integral equations [13]: f 1 (x) = 11 6 x + 11 15 (x + y)f 1 (y)dy f (x) = 5 4 x + 1 4 x xy f 1 (y)dy (x y )f (y)dy, x yf (y)dy, where x [, 1). The exact solution is f 1 (x) = x and f (x) = x. Table 8 illustrates the comparison between the exact solution and numerical solution given by the proposed method (IBPFs) for different values of n. Also,

15 F. Mirzaee.35.3.5 Error for m=8 Error for m=16 Error for m=3 Error for m=64 Error..15.1.5.1..3.4.5.6.7.8.9 1 x Figure 4: Absolute value of error, Example for f 1 (x)..1.8 Error for m=8 Error for m=16 Error for m=3 Error for m=64.6 Error.4..1..3.4.5.6.7.8.9 1 x Figure 5: Absolute value of error, Example for f (x). we compare the infinity-norm of absolute error computed by the present method, BPFs [13] and RHFs [15] method in Table 9. Furthermore, for this example, numerical convergence rate for proposed method is tabulated in Table 1. Graph of the absolute errors for f 1 (x) and f (x) are shown in Figures 6 and 7 respectively. Example 4. Consider the following system of linear Fredholm integral equations [13]: f 1 (x) = e x + ex+1 1 x+1 e x y f 1 (y)dy e (x+)y f (y)dy, f (x) = e x + e x + ex+1 1 x+1 e xy f 1 (y)dy e x+y f (y)dy,

Numerical solution of system of linear integral equations 153.1.8 Error for m=8 Error for m=16 Error for m=3 Error for m=64.6 Error.4..1..3.4.5.6.7.8.9 1 x Figure 6: Absolute value of error, Example 3 for f 1 (x)..1.8 Error for m=8 Error for m=16 Error for m=3 Error for m=64.6 Error.4..1..3.4.5.6.7.8.9 1 x Figure 7: Absolute value of error, Example 3 for f (x). Table 8: Numerical results for Example 3. Nodes x, Exact Present method n = 8 n = 16 n = 3 n = 64., (.,.) (.34,.133) (.1638,.33) (.81,.8) (.396,.).1, (.1,.1) (.1755,.174) (.1568,.1598) (.9393,.888) (.938,.881)., (.,.4) (.56,.645) (.18815,.3553) (.18767,.355) (.317,.418).3, (.3,.9) (.56,.645) (.3137,.987) (.3165,.9776) (.9691,.8816).4, (.4,.16) (.37697,.1433) (.37553,.1416) (.4638,.16515) (.468,.1657).5, (.5,.5) (.5168,.5189) (.545,.548) (.51,.51) (.53,.53).6, (.6,.36) (.6639,.397) (.6537,.39116) (.59385,.3567) (.59378,.3557).7, (.7,.49) (.7511,.5648) (.68784,.473) (.68759,.478) (.7315,.4944).8, (.8,.64) (.7511,.5648) (.8176,.6678) (.8157,.6631) (.79689,.6355).9, (.9,.81) (.87581,.768) (.875,.7668) (.963,.8146) (.966,.8133)

154 F. Mirzaee Table 9: Approximate L -norm of absolute error for Example 3. Methods E,n Method of [15] n = 8 (3.98e-,4.44e-) n = 16 (.1e-,.58e-) n = 3 (1.4e-,1.53e-) n = 64 (1.77e-,5.74e-3) Method of [13] n = 8 (3.61e-,4.16e-) n = 16 (1.8e-,.8e-) n = 3 (9.e-3,1.4e-) n = 64 (4.51e-3,5.1e-3) Present method n = 8 (3.44e-,3.88e-) n = 16 (1.76e-,.1e-) n = 3 (8.9e-3,1.e-) n = 64 (4.48e-3,5.16e-3) Table 1: Numerical convergence rate for Example 3. m for f 1 (x) for f (x) 4.958.8896 8.9651.9471 16.989.9741 3.9915.987 E n,n where x [, 1). The exact solution is f 1 (x) = e x and f (x) = e x. Table 11 illustrates the comparison between the exact solution and numerical solution given by the proposed method (IBPFs) for different values of n. Also, we compare the infinity-norm of absolute error computed by the present method, BPFs [13] and RHFs [15] method in Table 1. Furthermore, for this example, numerical convergence rate for proposed method is tabulated in Table 13. Graph of the absolute errors for f 1 (x) and f (x) are shown in Figures 8 and 9, respectively.

Numerical solution of system of linear integral equations 155 Table 11: Numerical results for Example 4. Nodes x, Exact Present method n = 8 n = 16 n = 3 n = 64., (1.,1.) (1.197,.975) (1.157,.9863) (1.73,.9961) (1.371,.996).1, (1.1517,.9484) (1.116,.88873) (1.199,.88415) (1.9747,.9193) (1.988,.916)., (1.14,.81873) (1.7114,.78496) (1.9,.8367) (1.537,.8945) (1.51,.8168).3, (1.34986,.748) (1.7114,.78496) (1.3633,.7333) (1.3659,.733) (1.3454,.7434).4, (1.4918,.673) (1.44131,.69334) (1.45134,.68889) (1.5,.66655) (1.593,.6665).5, (1.6487,.6653) (1.63418,.6143) (1.64484,.689) (1.6477,.6693) (1.64847,.6663).6, (1.81,.54881) (1.8579,.5495) (1.8641,.53677) (1.897,.5564) (1.815,.5535).7, (.1375,.49659) (.16,.47776) (1.98448,.5431) (1.98764,.531) (.1977,.49513).8, (.554,.44933) (.16,.47776) (.491,.44514) (.537,.44411) (.183,.4583).9, (.4596,.4657) (.38148,.4189) (.394,.418) (.4738,.4437) (.47471,.441) Table 1: Approximate L -norm of absolute error for Example 4. Methods E,n Method of [15] n = 8 (6.94e-,.85e-) n = 16 (3.46e-,1.33e-) n = 3 (1.88e-,6.17e-3) n = 64 (8.36e-3,3.1e-3) Method of [13] n = 8 (6.69e-,.45e-) n = 16 (3.6e-,1.e-) n = 3 (1.6e-,5.94e-3) n = 64 (8.7e-3,.98e-3) Present method n = 8 (6.6e-,.9e-) n = 16 (3.15e-,1.16e-) n = 3 (1.59e-,5.85e-3) n = 64 (8.e-3,.94e-3) Table 13: Numerical convergence rate for Example 4. m for f 1 (x) for f (x) 4 1.16.9954 8.989.9831 16.9871.9855 3.9913.999 E n,n

156 F. Mirzaee.16.14.1 Error for m=8 Error for m=16 Error for m=3 Error for m=64.1 Error.8.6.4..1..3.4.5.6.7.8.9 1 x Figure 8: Absolute value of error, Example 4 for f 1 (x)..6.5.4 Error for m=8 Error for m=16 Error for m=3 Error for m=64 Error.3..1.1..3.4.5.6.7.8.9 1 x Figure 9: Absolute value of error, Example 4 for f (x). 6 Conclusion The IBPFs, together with operational matrices of integration P 1 and P, are used to obtain the solution of system of linear Volterra and Fredholm integral equations. The present method reduces a system of linear Volterra and Fredholm integral equations into a system of algebraic equations. The matrices P 1 and P have many zeros, hence the method is computationally very attractive. Also, we have shown that our approach is convergence and its order of convergent is O( 1 n ). For the approximate solution of the given examples, we plot absolute value of error to discover applicability and accuracy of the proposed method. Comparisons of the results of applying IBPFs, BPFs and RHFs methods reveals that the new

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