Z. Omar. Department of Mathematics School of Quantitative Sciences College of Art and Sciences Univeristi Utara Malaysia, Malaysia. Ra ft.

Transcription

1 International Journal of Mathematical Analysis Vol. 9, 015, no. 46, 57-7 HIKARI Ltd, Developing a Single Step Hybrid Block Method with Generalized Three Off-step Points for the Direct Solution of Second Order Ordinary Differential Equations Z. Omar Department of Mathematics School of Quantitative Sciences College of Art and Sciences Univeristi Utara Malaysia, Malaysia Ra ft. Abdelrahim Department of Mathematics School of Quantitative Sciences College of Art and Sciences Univeristi Utara Malaysia, Malaysia Copyright c 015 Z. Omar and Ra ft. Abdelrahim. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract A new one step hybrid block method with generalized three off-step points using interpolation and collocation approach for solving initial value problems of second order ordinary differential equation directly is proposed. In deriving this method, the power series used as basis function to approximate the solution is interpolated at the last two points of the three off-step points while its second derivative is collocated at all points in the selected interval. The method is proven to be zero stable, consistent, convergent and of order five. For the purpose of testing, specific points of the developed method are chosen to solve second order initial value problem directly. The numerical results obtained indicate

2 58 Z. Omar and Ra ft. Abdelrahim that the new method outperforms the previous methods in terms of accuracy. Mathematics Subject Classification: 65L06, 65L05 Keywords: Hybrid method, Block method, Second order differential equation, Power series, Three off-step points 1 Introduction In this paper, we consider the direct solution to general second order initial value problem(ivps) of the form y = f(x, y, y ), y(a) = η 0, y (a) = η 1, x [a, b]. (1) Many scholars have been developed numerical methods for solving second order IVPs directly. These techniques have been introduced in many literature such as [7] [8], [9], [5] and others. The idea of hybrid method which involves the use off step points was introduced to overcome the zero stability barrier in linear multistep method. This barrier implies that the highest order of zero stability of linear multistep method when steplength k is odd is k +1 and k + when k is even[6]. [] proposed a hybrid block method in which two off-step points were considered in the development the method but the accuracy of the method are low. [9] developed three step with one off-step points. However, this method is also of low accuracy when it was applied to solve (1). Subsequently, one step with one off-step was developed by [1] but the accuracy of the method is still not encouraging. Hybrid block method depends on the off-step points chosen. To the best of our knowledge, one step hybrid block methods have been developed only to specific point(s). This paper, therefore, attempts to develop one step hybrid block method with generalized three off-step points for solving second initial value problems of second order ordinary differential equations(odes) directly. Derivation of the Method In this section, a hybrid one step block method with generalized three off-step points x n+s1, x n+s and x n+s3 for solving (1) is derived. Let the approximate solution to equation(1) be the power series of the form y(x) = v+m 1 i=0 ( ) x xn i a i, x [xn, x n+1] () h

3 Developing a single step hybrid block method 59 where n = 0, 1,,..., N 1, v denotes of the number of interpolation points which is equal to the order of ODE, m represents the number of collocation points, h = x n x n 1 is constant step size of partition of interval [a, b] which is given by a = x 0 < x 1 <... < x N 1 < x N = b. The second derivative of () is given by y (x) = f(x, y, y ) = v+m 1 i= i(i 1) a h i ( ) x xn i. (3) where v = and m = 5. Interpolating () at x n+s and x n+s3, and collocating (3) at all points i.e x n, x n+s1, x n+s, x n+s3, x n+1 in the selected interval yields the following equations which can be written in matrix form. 1 s s s 3 s 4 s 5 s 6 1 s 3 s 3 s 3 3 s 4 3 s 5 3 s h s h h 1 s h s h h s h 6 1 s h h 3 s h 3 0s 3 h 1 0s 3 h 0s 3 h 3 30s 4 h 1 30s 4 h 30s 4 h h h h h h h a 0 a 1 a a 3 a 4 a 5 a 6 = y n+s y n+s3 +s1 +s +s3 +1 Gaussian elimination method is then applied to find the coefficients of a i for i = 0(1)6. Then, the value of a i are substituted into equation () to give a continuous implicit scheme of the form y(x) = α si y n+si + β si +i + β si +si (5) i= i=0 i=1 The first derivative of equation (5) is where y (x) = 3 i= α s3 = (x n x + hs ) (h(s s 3 α s = (x x n hs 3 ) (h(s s 3 x α s i (x)y n+si + 1 i=0 x β i(x)+i + 3 i=1 (4) x β s i (x)+si (6) β 0 = (x n x + hs 3 )(x n x + hs ) (1x x (60s 1 s s 3 h 4 n 3hx 3 + 3hx 3 n 8xx 3 n + x 4 ) +x 4 n + h 4 s 3 h 4 s 4 + h 4 s 3 3 h 4 s 4 3 3hs 1 x 3 hs x 3 hs 3 x 3 + 3hs 1 x 3 n +hs 3 x 3 n 9hxx n + 9hx x n 5h 4 s 1 s + h 4 s 1 s 3 5h 4 s 1 s 3 4h s 1 s 3 xx n

4 60 Z. Omar and Ra ft. Abdelrahim 3h 4 s s 3 + h 4 s s h 4 s 3 s 3 + 5h s 1 x + h s x + h 3 s x 5h 3 s 1 s x +h 3 s 3x h 3 s 3 3x + 5h s 1 x n + h s x n h 3 s x n + h 3 s 3 x n h s s 3 xx n +h 3 s 3 3x n + h 4 s s 3 h s x h s 3x h s x n h s 3x n + 15h 4 s 1 s s 3 5h 3 s 1 s 3 x 3h 3 s s 3 x + 5h 3 s 1 s x n + 5h 3 s 1 s 3 x n + 3h 3 s s 3 x n 9hs 1 xx n 10h s 1 xx n 3hs xx n + 3hs x x n 4h s xx n 3hs 3 xx n + 3hs 3 x x n 3h 4 s 1 s s 3 3h 4 s 1 s s 3 + h s 1 s x + h 3 s 1 s x + h s 1 s 3 x + h s 1 s x n +h 3 s 1 s 3x + h 3 s s 3x + h 3 s s 3 x h 3 s 1 s x n + h s 1 s 3 x n h 3 s 1 s 3x n h 3 s s 3 x n + h s xx n + h s 3xx n 3h 3 s 1 s s 3 x + 3h 3 s 1 s s 3 x n + hs x 3 n +h s 3 x n + h s 3 x h 3 s s 3x n + h 4 s 1 s 3 3 3h 4 s s 3 h 3 s 3x n h 3 s 3 x +9hs 1 x x n 4h s 3 xx n + h s s 3 x n 4h s 1 s xx n + h s s 3 x 8x 3 x n ) β s1 = (x n x + hs 3 )(x n x + hs ) 60h 4 s 1 (s 1 1)(s 1 s 3 )(s 1 s ) (h4 s 4 h 4 s 3 s 3 h 4 s 3 h 4 s s 3 h 4 s s h 4 s s 3 + h 4 s 4 3 h 4 s h 3 s 3 x h 3 s 3 x n h 3 s s 3 x + h 3 s s 3 x n +h 3 s x n h 3 s s 3x + h 3 s s 3x n + 3h 3 s s 3 x 3h 3 s s 3 x n + h 3 s 3 3x h 3 s 3 3x n +h 3 s 3x n + h s x h s xx n + h s x n h s s 3 x + h s s 3 xx n 3hx 3 n +4h s xx n h s x n + h s 3x h s 3xx n + h s 3x n h s 3 x hs 3 x 3 n h s 3 x n + hs x 3 3hs x x n + 3hs xx n hs x 3 n + hs 3 x 3 x 4 + 8x 3 x n 3hs 3 x x n + 4h s 3 xx n + 3hs 3 xx n h s s 3 x n h 3 s 3x h 3 s x + 8xx 3 n +3h 4 s s 3 h s x + 3hx 3 9hx x n + 9hxx n 1x x n x 4 n) β s = (x n x + hs 3 )(x n x + hs ) 60h 4 s (s 1)(s s 3 )(s 1 s ) (3hx3 1x x n + 8xx 3 n + 8x 3 x n h 4 s s 3 + 3h 4 s 3 h 4 s 4 h 4 s hs 1 x 3 hs x 3 3hs 1 x 3 n h 4 s 1 s 3 3 h s s 3 xx n x 4 n + 9hxx n 9hx x n 5h 4 s 1 s + 3h 4 s 1 s 3 + 5h 4 s 1 s 3 +h 4 s s h 4 s 3 s 3 5h s 1 x + 3h s x + 3h 3 s x h 3 s 3 x + 5h 3 s 1 s x n 5h s 1 x n + 3h s x n 3h 3 s x n + h 3 s 3 x n h s 3 x n + h 3 s 3x n h 3 s 3 3x n h s x h s x n + h s 3x n + 5h 4 s 1 s s 3 5h 3 s 1 s x + 5h 3 s 1 s 3 x 3hx 3 n 5h 3 s 1 s 3 x n + h 3 s s 3 x n + 9hs 1 xx n 9hs 1 x x n + 10h s 1 xx n 6hs xx n 6h s xx n + 3hs 3 xx n 3hs 3 x x n + 4h s 3 xx n h 4 s 1 s s 3 h 4 s 1 s s 3 +3h 3 s 1 s x h s 1 s 3 x h 3 s 1 s 3x + h s s 3 x + h 3 s s 3x + h 3 s s 3 x x 4 3h 3 s 1 s x n h s 1 s 3 x n + h 3 s 1 s 3x n + h s s 3 x n h 3 s s 3x n h 3 s s 3 x n +4h s xx n hs 3 x 3 n h s 3xx n h 3 s 1 s s 3 x + h 3 s 1 s s 3 x n + 3h s 1 s x n 6h s 1 s xx n h 3 s s 3 x h 4 s s 3 + 3h s 1 s x + 6hs x x n + 4h s 1 s 3 xx n h s 3 x + h 4 s s 3 + hs x 3 n + h s 3x + h 3 s 3 3x h 3 s 3x + hs 3 x 3 + h 4 s 4 3) β s3 = (x n x + hs 3 )(x n x + hs ) (60h 4 s 3 (s 3 1)(s s 3 )(s 1 s 3 (1x x n 3hx 3 + 3hx 3 n 8xx 3 n

5 Developing a single step hybrid block method 61 +6h s 1 s 3 xx n + h 4 s 3 h 4 s 4 3h 4 s h 4 s 4 3 3hs 1 x 3 + hs 3 x 3 + 3hs 1 x 3 n hs 3 x 3 n 9hxx n + 9hx x n 5h 4 s 1 s + h 4 s 1 s 3 + 5h 4 s 1 s 3 4h s 1 s xx n h 4 s s 3 3 h 4 s 3 s 3 + 5h s 1 x + h s x + h 3 s x + h 4 s 1 s s 3 + 5h 3 s 1 s 3 x +5h s 1 x n + h s x n h 3 s x n + h 3 s 3 x n 3h s 3 x n + 3h 3 s 3x n h 3 s 3 3x n h s x + h s 3x h s x n + h s 3x n 5h 4 s 1 s s 3 5h 3 s 1 s x h 4 s s 3 +5h 3 s 1 s x n 5h 3 s 1 s 3 x n h 3 s s 3 x n 9hs 1 xx n + 9hs 1 x x n 10h s 1 xx n +3hs x x n 4h s xx n + 6hs 3 xx n 6hs 3 x x n + 6h s 3 xx n + h 4 s 1 s s 3 +h s 1 s x + h 3 s 1 s x 3h s 1 s 3 x 3h 3 s 1 s 3x h s s 3 x + h 3 s s 3 x +h s 1 s x n h 3 s 1 s x n 3h s 1 s 3 x n + 3h 3 s 1 s 3x n h s s 3 x n 3h 4 s 1 s 3 3 +h 3 s s 3 x n h 3 s 1 s s 3 x n + h s xx n 4h s 3xx n + h 3 s 3 3x 3hs xx n +h 4 s s 3 h 3 s 3 x 3h s 3 x 3h 3 s 3x + h 4 s s 3 + hs x 3 n + x 4 h 3 s s 3x 8x 3 x n + h 3 s s 3 x + h 3 s s 3x n + h 3 s 1 s s 3 x + h s s 3 xx n hs x 3 + x 4 n) β 1 = (x n x + hs 3 )(x n x + hs ) (60h 4 (s 3 1)(s 1)(s 1 1 (s 1h 4 s 3 + h 4 s s 3 3s 1 h 4 s s 3 h 3 s s 3 x n + h 4 s 3 s 3 + h 4 s s 3 3 3s 1 h 4 s s 3 h 4 s s 1 h 4 s 3 3 h 3 s 3 x +h 3 s 3 x n + h 3 s s 3 x h 4 s 4 + s 1 h 3 s x s 1 h 3 s x n + h 3 s s 3x h 3 s s 3x n 3s 1 h 3 s s 3 x + 3s 1 h 3 s s 3 x n h 3 s 3 3x + h 3 s 3 3x n + s 1 h 3 s 3x s 1 h 3 s 3x n +h s s 3 x n + s 1 h s x 4s 1 h s xx n + s 1 h s x n h s 3x + h s 3xx n 8xx 3 n + s 1 h s 3 x 4s 1 h s 3 xx n + s 1 h s 3 x n hs x 3 + 3hs x x n +3hs 3 x x n 3hs 3 xx n + hs 3 x 3 n 3s 1 hx 3 + 9s 1 hx x n 9s 1 hxx n +3s 1 hx 3 n 8x 3 x n + hs x 3 n + x 4 h s 3x n 3hs xx n h s s 3 xx n +x 4 n + 1x x n h s x + h s xx n h s x n + h s s 3 x hs 3 x 3 ) Evaluating (5) at the non-interpolating point i.e x n, x n+s1, x n+1 and (6) at all points produces the discrete schemes and its derivative. Combining the discrete schemes and its derivative at x n gives the equation in a matrix form as follows where A [3] Y [3] m = B [3] 1 R [3] 1 + B [3] R [3] + h [ D [3] R [3] 3 + E [3] R [3] 4 s 0 3 s s s 3 s s 3 0 (s A [3] 1 1 s 3 ) (s 1 s ) (s = s 3 ) (s s 3 0 ) (s 0 3 1) (s 1) (s s 3 1 ) (s s3) (h(s s3 (h(s s B [3] = , R[3] 1 = , Y [3] m = y n 3 y n y n 1 y n y n+s1 y n+s y n+s3 y n+1, B[3] = ] (7)

6 6 Z. Omar and Ra ft. Abdelrahim R [3] = E [3] = y n 3 y n y n 1 y n, D[3] = E [3] 11 E [3] 1 E [3] 13 E [3] 14 E [3] 1 E [3] E [3] 3 E [3] 4 E [3] 31 E [3] 3 E [3] 33 E [3] 34 E [3] 41 E [3] 4 E [3] 43 E [3] D [3] D [3] D [3] D [3] 44, R [3] 3 3 = and R [4] 3 = The non-zero elements of D [3] and E [3] are given by D [3] 14 = D [3] 4 = +s1 +s +s (60s 1 ) (s s 3 5s 1 s + s 1 s 3 5s 1 s 3 + s 1 s 3 3 3s s 3 3s s 3 + s s 3 3 +s 3 s 3 + s 3 s 4 + s 3 3 s 4 3 3s 1 s s 3 3s 1 s s s 1 s s 3 ) 1 (60s 1 s s 3 ) (s 1 s )(s 1 s 3 )(s 4 1 s 3 1s s 3 1s 3 s 3 1 s 1s + s s 1s s 1s 3 + 3s 1s 3 s 1 s 3 + s 1 s s 3 + 3s 1 s + s 1 s s 3 + s 1s s 3 s 3 s 1 s s 1 s 3 + s 4 s 3 s 3 s s 3 + 3s s 3 s s s s 3 s 3 3 1s 1 s s 3 ) D [3] 1 34 = (60s 1 s s 3 ) (s 1)(s 3 1)(s 1 + s + s 3 + s s 3 3s 1 s 3s 1 s 3 3s 1 s + s 1 s 3 3s 1 s 3 + s 1 s 3 3 s s 3 s s 3 + s s s 3 s 3 + s +s 3 s s 3 s 4 + s 3 + s 3 3 s 4 3 3s 1 s s 3 3s 1 s s 3 + 1s 1 s s 3 1) D [3] 1 44 = h(60s 1 s s 3 ) (s s 3 3 3s s 3 + s 3 s 3 5s 1 s 3 + s 1 s 4 5s 1 s s 1 s 4 3 3s s 3 3 3s 3 s 3 + s s s 4 s 3 + s 4 + s 4 3 s s 1 s s 3 3s 1 s s s 1 s s 3 3s 1 s 3 s 3 3s 1 s s 3 s 5 ) E [3] s s 3 11 = 60s 1 (s 1 s )(s 1 s 3 )(s 1 1) (s3 s 3 s 4 + s 3 + s s 3 3s s 3 + s s 3 3 s 4 3 3s s 3 + s 3 3) E [3] s 3 1 = 60(s 1 s )(s s 3 )(s 1) (s s 3 5s 1 s + 3s 1 s 3 + 5s 1 s 3 s 1 s 3 3 s 3 3 s s 3 s s 3 + s s s 3 s 3 + 3s 3 + s 4 3 s 1 s s 3 s 1 s s 3 +5s 1 s s 3 s 4 ) E [3] s 13 = 60(s 1 s 3 )(s s 3 )(s 3 1) (s s 3 + 5s 1 s s 1 s 3 5s 1 s 3 + 3s s 3 3 +s s s 3 s 3 s 3 + s 4 + 3s 3 3 s 4 3 s 1 s s 3 s 1 s s 3 + 5s 1 s s 3 s s 3 s s 3 ) E [3] s s 3 14 = (60s 1 60)(s 1)(s 3 1) ( s4 + s 3 s 3 + s 1 s 3 + s s 3 3s 1 s s 3

7 Developing a single step hybrid block method 63 E [3] 1 = E [3] = E [3] 3 = E [3] 4 = E [3] 31 = E [3] 3 = E [3] 33 = E [3] 34 = E [3] 41 = E [3] 4 = +s s 3 3 3s 1 s s 3 s s 1 s 3 3) 1 (60s 1 (s 1 1 (s4 1 s 3 1s s 3 1s 3 3s 3 1 s 1s + s 1s s 3 + s 1s s 4 +s 1s 3 s 1 s 3 + s 1 s s 3 + s 1 s + s 1 s s 3 3s 1 s s 3 s 1 s s 1 s 3 +s s 3 3 s 1s 3 + s 3 s 3 + s 3 + s s 3 3s s 3 3s s 3 s s 3 3) (s 1 s 3 ) (60s (s s 3 )(s 1 (s 1s s 4 1 s 3 1s + s 3 1s 3 + s 3 1 s 1s + s 1s 3 +s 1s s 3 3s 1s 3 s 1 s 3 + s 1 s s 3 + s 1 s + s 1 s s 3 3s 1 s s 3 + s s 3 +s 1 s 3 3 3s 1 s 3 + s 4 s 3 s 3 3s 3 s s 3 + s s 3 s s s 3 3 s 4 3) (s 1 s ) (60s 3 (s s 3 )(s 3 1 (s3 1 s s 3 1s s 3 1s 3 + s 1s + s 1s s 3 + s 3 s 1s 3 + s 1s 3 + s 1 s 3 + s 1 s s 3 3s 1 s + s 1 s s 3 3s 1 s s 3 s 4 + s 4 3 s 1 s 3 3 3s 1s + s 1 s 3 s 3 s 3 s s 3 + s s 3 s s s s 3 3s 3 3) (s 1 s )(s 1 s 3 ) (60(s 1 1)(s 1)(s 3 1 (s 1s 3 s s 3 1s + s 3 1s 3 + s 1s + s 3 s 3 s 1s s 3 + s 1 s 3 s 1 s s 3 s 1 s s 3 + s 1 s 3 3 s 4 + s s 3 + s s 3 3 s 4 3) (s 1)(s 3 1) (60s 1 (s 1 s )(s 1 s 3 )(s 1 1) ( s4 + s 3 s 3 + s 3 + s s 3 s s 3 + s +s s 3 3 s s 3 s s 3 + s s s s 3 + s 3 1) (s 3 1) (60s (s 1 s )(s s 3 ) (s 1 s + s 3 s s 3 + s 1 s 3s 1 s 3 + s s 3 3s 1 s 3 3s 1 s 3 + s 1 s s s 3 + s s 3 s s 3 3 s 3 s 3 s s 3 + s 4 +s 3 + s s 1 s + s 1 s s 3 + s 1 s s 3 s 4 3 3s 1 s s 3 1) (s 1) (60s 3 (s 1 s 3 )(s s 3 ) (s 1 + s s 3 s s 3 3s 1 s + s 1 s 3 + s s 3 +s 1 s 3 + s 1 s 3 3s 1 s s s 3 + s s 3 s s 3 3 s 3 s 3 + s + s 3 s 4 s 3 s 3 3 3s 1 s + s s 1 s s 3 + s 1 s s 3 3s 1 s s 3 1) 1 (60s 1 60) (s s 3 s s 3 3s 1 + s 1 s + s 1 s 3 + s s 3 + s 1 s +s 1 s 3 + s 1 s s s 3 + s s 3 + s s s 3 s 3 s s 3 s 4 s 3 +s 1 s 3 3s 1 s s 3 3s 1 s s 3 3s 1 s s 3 s 3 3 s ) 1 (60hs 1 (s 1 s )(s 1 s 3 )(s 1 1 ( s5 + s 4 s 3 + s 4 + s 3 s 3 3s 3 s 3 +s s 3 3 3s s 3 + s s 4 3 3s s 3 3 s s 4 3) 1 (60hs (s 1 s )(s s 3 )(s 1 (s s 3 s s 3 3 s 3 s 3 + 5s 1 s 3 3s 1 s 4 +s 1 s s s s 3 s 3 s s 4 3 s 4 s 3 3s 4 + s 4 3 5s 1 s s 3 + s 5 5s 1 s s 3 5s 1 s s 1 s s s 1 s 3 s 3 + s 1 s s 3 s 5 3)

8 64 Z. Omar and Ra ft. Abdelrahim E [3] 43 = E [3] 44 = 1 (60hs 3 (s 1 s 3 )(s s 3 )(s 3 1 (s s 3 s s 3 3 s 3 s 3 5s 1 s 3 + s 4 3s 1 s s s s 3 s 3 s s 4 3 s 4 s 3 3s s 5 3 5s 1 s s 3 + s 1 s 4 +5s 1 s 3 3 5s 1 s s 3 + s 1 s s s 1 s 3 s 3 + s 1 s s 3 s 5 ) 1 (h(60s 1 60)(s 1)(s 3 1 ( s5 + s 4 s 3 + s 1 s 4 + s 3 s 3 3s 1 s 3 s 3 +s s 3 3 3s 1 s s 3 + s s 4 3 3s 1 s s 3 3 s s 1 s 4 3) Multiplying equation (7) by the inverse of A [3] gives the following one step with generalized three off-step points block method. I [] 3 Y [3] m = B [3] 1 R [3] 1 + B [3] R [3] + h D[3] R [3] 3 + h Ē [3] R [3] 4 (8) where I [] 3 = with D [3] = D[3] D[3] D[3] D[3] 44, B[3] 1 =, Ē [3] = Ē [3] 11 Ē [3] 1 Ē [3] 31 Ē [3] 41, B[3] = Ē [3] 1 Ē [3] Ē [3] 3 Ē [3] 4 Ē [3] 13 Ē [3] 3 Ē [3] 33 Ē [3] s 1 h s h s 3 h h Ē [3] 14 Ē [3] 4 Ē [3] 34 Ē [3] 44 D [3] 14 = s 1(5s 1 s + 5s 1 s 3 0s s 3 s 1s s 1s 3 s 1 + s s 1 s s 3 ) (60s s 3 ) D [3] 4 = s (0s 1 s 3 5s 1 s 5s s 3 + s 1 s + s s 3 + s s 3 5s 1 s s 3 ) (60s 1 s 3 ) D [3] 34 = s 3(5s 1 s 3 0s 1 s + 5s s 3 s 1 s 3 s s 3 s 3 + s s 1 s s 3 ) (60s 1 s ) D [3] 44 = (s 1 + s + s 3 5s 1 s 5s 1 s 3 5s s 3 + 0s 1 s s 3 1) (60s 1 s s 3 ) Ē [3] 11 = s 1(5s 1 s + 5s 1 s 3 10s s 3 3s 1s 3s 1s 3 3s 1 + s s 1 s s 3 ) (60(s 1 1)(s 1 s 3 )(s 1 s Ē [3] s 4 1 = 1(5s 3 s 1 s 1 s 3 + s 1) (60s (s 1)(s s 3 )(s 1 s Ē [3] s 4 13 = 1(5s s 1 s 1 s + s 1) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 Ē [3] 14 = s4 1(5s s 3 s 1 s 3 s 1 s + s 1) (60(s 3 1)(s 1)(s 1 1

9 Developing a single step hybrid block method 65 Ē [3] 1 = s4 (5s 3 s s s 3 + s ) 60s 1 (s 1 1)(s 1 s 3 )(s 1 s Ē [3] = s (10s 1 s 3 5s 1 s 5s s 3 + 3s 1 s + 3s s 3 + 3s s 3 5s 1 s s 3 ) (60(s 1)(s s 3 )(s 1 s Ē [3] 3 = s 4 (s 5s 1 + s 1 s s ) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 Ē [3] 4 = s4 (s 1 s 5s 1 s 3 + s s 3 s ) (60(s 3 1)(s 1 1)(s 1 Ē [3] 31 = Ē [3] 3 = h s 4 3(s 3 5s + s s 3 s 3) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s s 4 3(s 3 5s 1 + s 1 s 3 s 3) (60s (s 1)(s s 3 )(s 1 s Ē [3] 33 = s 3(5s 1 s 3 10s 1 s + 5s s 3 3s 1 s 3 3s s 3 3s 3 + s s 1 s s 3 ) (60(s 3 1)(s s 3 )(s 1 s 3 Ē [3] 34 = s4 3(5s 1 s s 1 s 3 s s 3 + s 3) (60(s 1)(s 1 1)(s 3 1 Ē [3] 41 = Ē [3] 4 = Ē [3] 43 = (5s s 3 s 3 s + 1) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s (5s 1 s 3 s 3 s 1 + 1) (60s (s 1)(s s 3 )(s 1 s (5s 1 s s s 1 + 1) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 Ē [3] 44 = (3s 1 + 3s + 3s 3 5s 1 s 5s 1 s 3 5s s s 1 s s 3 ) (60(s 3 1)(s 1)(s 1 1 Equation (8) can also be written as y n+s1 = y n + hs 1 y n + h s 1(5s 1 s + 5s 1 s 3 0s s 3 s 1s s 1s 3 s 1 + s s 1 s s 3 ) (60s s 3 ) + h s 1(5s 1 s + 5s 1 s 3 10s s 3 3s 1s 3s 1s 3 3s 1 + s s 1 s s 3 ) +s1 (60(s 1 1)(s 1 s 3 )(s 1 s + h s 4 1(5s 3 s 1 s 1 s 3 + s 1) 60s (s 1)(s s 3 )(s 1 s ) +s h s 4 1(5s s 1 s 1 s + s 1) 60s 3 (s 3 1)(s s 3 )(s 1 s 3 ) +s 3 + h s 4 1(5s s 3 s 1 s 3 s 1 s + s 1) +1 (60(s 3 1)(s 1)(s 1 1 y n+s = y n + hs y n + h s (0s 1 s 3 5s 1 s 5s s 3 + s 1 s + s s 3 + s s 3 5s 1 s s 3 ) (60s 1 s 3 )

10 66 Z. Omar and Ra ft. Abdelrahim h s 4 (5s 3 s s s 3 + s ) 60s 1 (s 1 1)(s 1 s 3 )(s 1 s ) +s 1 + h s 4 (s 5s 1 + s 1 s s ) 60s 3 (s 3 1)(s s 3 )(s 1 s 3 ) +s 3 + h s (10s 1 s 3 5s 1 s 5s s 3 + 3s 1 s + 3s s 3 + 3s s 3 5s 1 s s 3 ) +s (60(s 1)(s s 3 )(s 1 s h s 4 (s 1 s 5s 1 s 3 + s s 3 s ) +1 (9) (60(s 3 1)(s 1 1)(s 1 y n+s3 = y n + hs 3 y n + h s 3(5s 1 s 3 0s 1 s + 5s s 3 s 1 s 3 s s 3 s 3 + s s 1 s s 3 ) (60s 1 s ) + h s 4 3(s 3 5s + s s 3 s 3) 60s 1 (s 1 1)(s 1 s 3 )(s 1 s ) +s 1 + h s 4 3(5s 1 s s 1 s 3 s s 3 + s 3) 60(s 1)(s 1 1)(s 3 1) + h s 4 3(s 3 5s 1 + s 1 s 3 s 3) (60s (s 1)(s s 3 )(s 1 s +s + h s 3(5s 1 s 3 10s 1 s + 5s s 3 3s 1 s 3 3s s 3 3s 3 + s s 1 s s 3 ) +s3 (60(s 3 1)(s s 3 )(s 1 s 3 y n+1 = y n + hy n + h (s 1 + s + s 3 5s 1 s 5s 1 s 3 5s s 3 + 0s 1 s s 3 1) (60s 1 s s 3 ) + h (5s s 3 s 3 s + 1) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s +s 1 + h (5s 1 s 3 s 3 s 1 + 1) 60s (s 1)(s s 3 )(s 1 s ) +s + h (5s 1 s s s 1 + 1) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 +s 3 + h (3s 1 + 3s + 3s 3 5s 1 s 5s 1 s 3 5s s s 1 s s 3 ) +1 (60(s 3 1)(s 1)(s 1 1 The first derivative of (8)is given by y n+s 1 = y n +1 hs 1(10s 1 s + 10s 1 s 3 30s s 3 5s 1s 5s 1s 3 5s 1 + 3s s 1 s s 3 ) 60s s 3 + hs 1(0s 1 s + 0s 1 s 3 30s s 3 15s 1s 15s 1s 3 15s 1 + 1s s 1 s s 3 ) +s1 60(s 1 1)(s 1 s 3 )(s 1 s ) + hs3 1(10s 3 5s 1 5s 1 s 3 + 3s 1) (60s (s 1)(s s 3 )(s 1 s +s + hs3 1(10s 5s 1 5s 1 s + 3s 1) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 +s 3 + hs3 1(10s s 3 5s 1 s 3 5s 1 s + 3s 1) +1 (60(s 3 1)(s 1)(s 1 1

11 Developing a single step hybrid block method 67 y n+s = y n + s (30s 1 s 3 10s 1 s 10s s 3 + 5s 1 s + 5s s 3 + 5s 3s 3 10s 1 s s 3 ) (60s 1 s 3 ) + s3 (10s 3 5s 5s s 3 + 3s ) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s +s 1 + s (30s 1 s 3 0s 1 s 0s s s 1 s + 15s s s 1s 3 0s 1 s s 3 ) +s (60(s 1)(s s 3 )(s 1 s + s3 (5s 10s 1 + 5s 1 s 3s ) (60s 3 (s 3 1)(s s 3 )(s 1 s 3 + s3 (5s 1 s 10s 1 s 3 + 5s s 3 3s ) (60(s 3 1)(s 1 1)(s 1 y n+s 3 = y n s 3(10s 1 s 3 30s 1 s + 10s s 3 5s 1 s 3 5s s 3 5s 3 + 3s s 1 s s 3 ) (60s 1 s ) + s3 3(5s 3 10s + 5s s 3 3s 3) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s +s 1 s3 3(5s 3 10s 1 + 5s 1 s 3 3s 3) (60s (s 1)(s s 3 )(s 1 s +s + s 3(0s 1 s 3 30s 1 s + 0s s 3 15s 1 s 3 15s s 3 15s 3 + 1s s 1 s s 3 ) +s3 (60(s 3 1)(s s 3 )(s 1 s 3 + s3 3(10s 1 s 5s 1 s 3 5s s 3 + 3s 3) +1 (60(s 1)(s 1 1)(s 3 1 y n+1 = y n + (5s 1 + 5s + 5s 3 10s 1 s 10s 1 s 3 10s s s 1 s s 3 3) (60s 1 s s 3 ) (10s s 3 5s 3 5s + 3) (60s 1 (s 1 1)(s 1 s 3 )(s 1 s +s 1 (10s 1 s 3 5s 3 5s 1 + 3) + (60s (s 1)(s s 3 )(s 1 s f (10s 1 s 5s 5s 1 + 3) n+s (60s 3 (s 3 1)(s s 3 )(s 1 s 3 +s 3 + (15s s + 15s 3 0s 1 s 0s 1 s 3 0s s s 1 s s 3 1) +1 (60(s 3 1)(s 1)(s 1 1 (10)

12 68 Z. Omar and Ra ft. Abdelrahim 3 Properties of the Method 3.1 Order of Method The linear difference operator L associated with (8) is defined as L[y(x); h] = Y [3] m B [3] 1 R [3] 1 B [3] R [3] 3 h [ D[3] R [3] 3 + Ē[3] R [3] ] 4 (11) where y(x) is an arbitrary test function continuously differentiable on [a,b]. The elements of Y m and R [] 3 are expanded in Taylors series respectively and their terms are collected in powers of h to give j=0 (s 1 ) j h j y j y n s 1 hy n + h s 1 (5s 1s +5s 1 s 3 0s s 3 s 1 s s 1 s 3 s 1 +s3 1 +5s 1s s 3 ) (60s s 3 y ) n h s 4 1 (5s s 3 s 1 s 3 s 1 s +s 1 ) j=0 h j+ (60(s 3 1)(s 1)(s 1 y j+ 1 s4 1 (5s 3 s 1 s 1 s 3 +s 1 ) j=0 s j hj+ (60s (s 1)(s s 3 )(s 1 s y j+ s 1 (5s 1s +5s 1 s 3 10s s 3 3s 1 s 3s 1 s 3 3s 1 +s3 1 +5s 1s s 3 ) j=0 (s 1 ) j h j+ (60(s 1 1)(s 1 s 3 )(s 1 s y j+ + s4 1 (5s s 1 s 1 s +s 1 ) j=0 (s 3 ) j h j+ (60s 3 (s 3 1)(s s 3 )(s 1 s 3 y j+ 0 j=0 (s ) j h j y j y n s hy n h s (0s 1s 3 5s 1 s 5s s 3 +s 1 s +s s 3+s s3 5s 1s s 3 ) (60s 1 s 3 y ) n + s4 (s 1s 5s 1 s 3 +s s 3 s ) j=0 h j+ (60(s 3 1)(s 1 1)(s y j+ 1 s (10s 1s 3 5s 1 s 5s s 3 +3s 1 s +3s s 3+3s s3 5s 1s s 3 ) j=0 (s ) j h j+ (60(s 1)(s s 3 )(s 1 s y j+ + s4 (5s 3 s s s 3 +s ) j=0 (s 1 ) j h j+ 60s 1 (s 1 1)(s 1 s 3 )(s 1 s y j+ s4 (s 5s 1 +s 1 s s ) j=0 (s 3 ) j h j+ (60s 3 (s 3 1)(s s 3 )(s 1 s 3 y j+ = 0 j=0 (s 3 ) j h j y j y n s 3 hy n + h s 3 (5s 1s 3 0s 1 s +5s s 3 s 1 s 3 s s 3 s 3 +s3 3 +5s 1s s 3 ) (60s 1 s y ) n s4 3 (5s 1s s 1 s 3 s s 3 +s 3 ) j=0 h j+ (60(s 1)(s 1 1)(s 3 y j+ 1 + s4 3 (s 3 5s 1 +s 1 s 3 s 3 ) j=0 (s ) j h j+ (60s (s 1)(s s 3 )(s 1 s y j+ s4 3 (s 3 5s +s s 3 s 3 ) j=0 (s 1 ) j h j+ (60s 1 (s 1 1)(s 1 s 3 )(s 1 s y j+ s 3 (5s 1s 3 10s 1 s +5s s 3 3s 1 s 3 3s s 3 3s 3 +s3 3 +5s 1s s 3 ) j=0 (s 3 ) j h j+ (60(s 3 1)(s s 3 )(s 1 s 3 y j+ j=0 h j j! yj n y n hy 0 n h (s 1 +s +s 3 5s 1 s 5s 1 s 3 5s s 3 +0s 1 s s 3 1) (60s 1 s s 3 y ) n (3s 1+3s +3s 3 5s 1 s 5s 1 s 3 5s s 3 +10s 1 s s 3 ) j=0 h j+ (60(s 3 1)(s 1)(s 1 y j+ 1 (5s 1s 3 s 3 s 1 +1) j=0 (s ) j h j+ (60s (s 1)(s s 3 )(s 1 s y j+ + (5s s 3 s 3 s +1) j=0 (s 1 ) j h j+ (60s 1 (s 1 1)(s 1 s 3 )(s 1 s y j+ + (5s 1s s s 1 +1) j=0 (s 3 ) j h j+ (60s 3 (s 3 1)(s s 3 )(s 1 s 3 y j+ 0 Comparing the coefficients of h j and y j gives the order of the method to be

13 Developing a single step hybrid block method 69 [5, 5, 5, 5] T with vector of error constant C 7 = s (5s 1s + 5s 1 s 3 10s s 3 3s 1s 3s 1s 3 3s 1 + s s 1 s s 3 ) s (10s 1s 3 5s 1 s 5s s 3 + 3s 1 s + 3s s 3 + 3s s 3 5s 1 s s 3 ) s (5s 1s 3 10s 1 s + 5s s 3 3s 1 s 3 3s s 3 3s 3 + s s 1 s s 3 ) (3s 1 + 3s + 3s 3 5s 1 s 5s 1 s 3 5s s s 1 s s 3 ) for all s 1, s, s 3 (0, 1) \ {s = 5s 1s 3 +3s 1 s 3+3s 1 s3 1 5s 1 10s 3 3s 1 +5s 1s 3 } {s 1 = 5s 3 3s s 3 3s +s3 10s 3 5s +3s 5s s 3 } {s 1 = 5s 1s 3 +3s s 3 +3s 3 s3 3 5s 3 10s 3s 3 +5s s 3 } {s 1 = 3s 3s 3 +5s s s 5s 3 +10s s 3 }. 3. Zero stability Definition 3.1 The hybrid block method (8) is said to be zero stable if the first characteristic polynomial π(r) having roots such that r z 1, and if r z = 1 then the multiplicity of r z must not greater than two. In order to fined the zero-stability of (8), we only consider the first characteristic polynomial of the method according to Definition(3.1) as follows Π(r) = r I [] 3 B [3] 3 1 = r = r 3 (r 1) which implies r = 0, 0, 0, 1. Hence, our method is zero stable since r z 1, and the multiplicity when r z = 1 is one for all s 1, s, s 3 (0, 1). 3.3 Consistency Definition 3. The one step hybrid block method (8) is said to be consistent if the order of the method is greater than or equal to one i.e. P 1 By above definition the block method (8) is consistent because its order is five. 3.4 Convergence Theorem 3.3 (Henrici, 196) Consistency and zero stability are sufficient conditions for a linear multistep method to be convergent Since the method (8) is consistent and zero stable, it implies the method is convergent for all off step points.

14 70 Z. Omar and Ra ft. Abdelrahim 3.5 Region of absolute stability In this article, the locus method was adopted to determine the function of stability region. The method (8) is said to be absolutely stable if for a given h all roots of the characteristic polynomial π(z, h) = ρ(z) h σ(z), satisfies z t < 1. The test equation y = λ y is substituted in (8) where h = λ h and λ = df. Substituting r = cos θ i sin θ and considering real part yields dy h(θ, h) = 4300(cos θ 1)/(s 1 s s 3 (4s 1 + 4s + 4s 3 3s 1 s 3s 1 s 3 3s s 3 + s 1 s s 3 + s 1 s s 3 cos θ 5 (1) Numerical Examples In order to find the accuracy of our methods, the following second order ODEs are tasted. The new block methods solved the same problems the existing methods solved in order to compare results in terms of error. Problem 1: y x(y ) = 0, y(0) = 1, y (0) = 1. Exact solution: y(x) = 1 + 1ln ( ) +x x with h = 0.05 Table 1: Comparison of the new method with Jator(007) for solving Problem 1 x Exact solution Computed solution in our method with three off-step points s 1 = 3 10, Error in our method, P = 5 Errors in Jator(007), P = 6 s = = e e e e e e e e e e e e e e e e e e e e 9 Problem : y + ( 6 x )y + ( 4 )y = 0, y(1) = 1, y (1) = 1. x Exact solution: y(x) = 5 with h = 1. 3x 3x 4 30

15 Developing a single step hybrid block method 71 Table : Comparison of the new method with Badmus (014) for solving Problem. x Exact solution Computed solution in our method with one off-step points s1 = Error in our method, P = 5 Badmus (014), P = 8 s3 = e e e e e e e e e e 1.896e e e e e e e e e 5 4 Conclusion A new single step hybrid block method with generalized three off-step points for the direct solution of second order ordinary differential equation has been developed successfully. The developed method is zero-stable, consistent and also convergent. When solving the same problems, the numerical results confirm that the new method produces better accuracy if compared to the existing methods. References [1] T. A. Anake, D. O. Awoyemi and A. A. Adesanya, A one step method for the solution of general second order ordinary differential equations, International Journal of Science and Technology, (01), no. 4, [] T. A. Anake, D. O. Awoyemi and A. A. Adesanya, One-step implicit hybrid block method for the direct solution of general second order ordinary differential equations, IAENG International Journal of Applied Mathematics, 4 (01), no. 4, 4-8. [3] K. M. Fasasi, A. O. Adesanya and S. O. Adee, One Step Continuous Hybrid Block Method for the Solution of y = f(x, y, y, y ), Journal of Natural Sciences Research, 4 (014), no. 10, 55-6.

16 7 Z. Omar and Ra ft. Abdelrahim [4] P. Henrici, Discrete Variable Methods in ODE, John Wiley and Sons, New York, 196. [5] S. N. Jator, Solving second order initial value problems by a hybrid multistep method without predictors, Applied Mathematics and Computation, 17 (010), no. 8, [6] J. D. Lambert, Computational Methods in Ordinary Differential Equations, [7] Z. Omar and M. Sulaiman, Parallel r-point implicit block method for solving higher order ordinary differential equations directly, Journal of ICT, 3 (004), no. 1, [8] Z. Omar and M. Sulaiman, Parallel r-point explicit block method for solving second order ordinary differential equations directly, International Journal of Computer Mathematics, 79 (00), no. 3, [9] A. Sagir, An accurate computation of block hybrid method for solving stiff ordinary differential equations, Journal of Mathematics, 4 (01), Received: August, 015; Pubslihed: September 5, 015