Applied Mathematical Sciences, Vol. 7, 2013, no. 53, 2629 - 2645 HIKARI Ltd, www.m-hikari.com
Solving Third Order Boundary Value Problem Using Fourth Order Block Method Ahmad Shah Abdullah1, *Zanariah Abdul Majid1,2 and Norazak Senu1,2 1
Institute for Mathematical Research Universiti Putra Malaysia, 43400 Serdang, Malaysia 2
Department of Mathematics, Faculty of Science Universiti Putra Malaysia, 43400 Serdang, Malaysia *
[email protected] Copyright © 2013 Ahmad Shah Abdullah et al. This is an open access article 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 fourth order two-point block method is developed for solving nonlinear third order boundary value problems (BVPs) directly. The two-point block method will solve the nonlinear third order BVPs at two points simultaneously within the block. The shooting technique will use the Newton’s method for checking of the convergent and the guessing values for the next block. The proposed method will be implemented using constant step size and the PECE mode. Numerical examples is presented to illustrate the applicability of the propose method. The results clearly show that the proposed block method is able to produce good results compared to the existing method.
Mathematics Subject Classification: 65L06, 65L10 Keywords: Boundary value problem, Linear shooting method, Block method
1
Introduction
Boundary value problems (BVPs) manifest themselves in many branches of science. Some of them are in the field of engineering, technology and optimization
Ahmad Shah Abdullah et al
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theory. Since the boundary value problem has wide application in science research, therefore faster and accurate numerical solution of boundary value problem are very importance. In literature contains several methods has been proposed to solve BVPs. Loghmani and Ahmadinia [4] use a third degree B-spline function to construct an approximate solution for third order linear and nonlinear boundary value problems coupled with the least square method. Quartic nonpolynomial spline method was proposed by El-Danaf [7] for the numerical solution of third order two point boundary value problems. El-Salam et al.[3] are presented second and fourth order convergent methods based on quartic nonpolynomial spline function for the numerical solution of a third order two-point boundary value problem. While Phang et al. [5] had solved second order boundary value problem using two step direct method by shooting technique. The fourth order two point block method also use shooting technique to solve the boundary value problem directly. In this paper, we propose a fourth order block method for solving boundary value problems of the form as follows y ′′′ = f ( x, y, y ′, y ′′)
a≤ x≤b
(1)
y ' (b) = β
(2)
with boundary conditions y (a ) = γ ,
y ' (a) = α ,
where a, b, α, β, γ are the given constant. The Newton method will be implemented as the iterative method to estimate the guessing values. The advantage of these methods is to solve BVPs without reduce it to the system of first order ordinary differential equations (ODEs). The given equations in (1) will be treated in their original second order form and therefore the requirement of the storage is lower.
2
Formulation of the method
h
h
h
h
Figure 1: Two-Point Block Method
Solving third order boundary value problem
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The interval [a, b] is divided into a series of blocks with each block containing two points as shown in Figure 1. Two value points will be found simultaneously using the same back value i.e. y n+1 and y n + 2 .The point y n+1 at x n+1 can be obtained by integrating Eq. (1) over the interval [xn , xn +1 ] once, twice and thrice that shown in Eq. (3) – (5): Integrate once: xn +1
xn +1
∫x y ′′′( x)dx = x∫ f ( x, y, y ′, y ′′)dx n
(3)
n
Integrate twice: xn +1 x
∫∫ x x n
xn +1 x
y ′′′( x)dxdx =
n
∫x x∫ f ( x, y, y' , y ′′)dxdx n
(4)
n
Integrate thrice: xn +1 x x
xn +1 x x
∫x x∫ x∫ y ′′′( x)dxdxdx = x∫ x∫ x∫ f ( x, y, y' , y ′′)dxdxdx . n
n n
n
(5)
n n
The same process will be applied to find the second point y n + 2 . Eq. (1) will be integrated over the interval [x n , x n + 2 ] once, twice and thrice gives, Integrate once: xn + 2
xn + 2
∫x y ′′′( x)dx = x∫ f ( x, y, y' , y ′′)dx n
(6)
n
Integrate twice: xn + 2 x
xn + 2 x
∫x x∫ y ′′′( x)dxdx = x∫ x∫ f ( x, y, y' , y ′′)dxdx n
n
n
(7)
n
Integrate thrice: xn + 2 x x
xn + 2 x x
∫x x∫ x∫ y ′′′( x)dxdxdx = x∫ x∫ x∫ f ( x, y, y' , y ′′)dxdxdx n
n n
n
n n
(8)
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The function f ( x, y, y ' , y ′′) in Eq. (3) – (8) will be approximated using Lagrange interpolating polynomial, P as in Eq. (9). The interpolation points involved are ( xn−3 , f n−3 ), ( xn−2 , f n−2 ), ( xn−1 , f n−1 ), ( xn , f n ) and ( xn+1 , f n+1 ) we will obtain the Lagrange interpolating polynomial: (x - x n-1 )(x - x n )(x - x n +1 ) f n +2 (x n +2 - x n -1 )(x n +2 - x n )(x n +2 - x n +1 )
P3 =
Taking s =
+
(x - x n-1 )(x - x n )(x - x n +2 ) f n +1 (x n +1 - x n -1 )(x n +1 - x n )(x n +1 - x n +2 )
+
(x - x n -1 )(x - x n +1 )(x - x n +2 ) fn (x n - x n-1 )(x n - x n +1 )(x n - x n +2 )
+
(x - x n )(x - x n +1 )(x - x n +2 ) f n-1 (x n-1 - x n )(x n -1 - x n +1 )(x n -1 - x n +2 )
(9)
x − x n +1 and replacing dx = hds , changing the limit of integration from -2 h
to -1 for Eq. (3) – (5) can be written as: −1
yn′′ +1 − yn′′ =
∫ P hds
(10)
3
−2 −1
y′n +1 − yn′ − hy ′n′ =
∫ − (s + 1) P h ds 2
(11)
3
−2
h2 yn +1 − y n − hy n′ − y ′′ = 2!
−1
∫ (− s − 1)
2
P3h 3ds
(12)
−2
and from -2 to 0 for Eq. (6) – (8) can be written as: 0
∫
y ′n′ + 2 − yn′′ = P3hds 2
(13)
Solving third order boundary value problem
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0
∫
y ′n +1 − y n′ − 2 hy n′′ = − sP3 h 2 ds
(14)
2 0
yn + 2 − yn − 2hy ′n −
h2 y ′n′ = ( − s ) 2 P3h 3ds 2!
∫
(15)
2
Evaluate Eq. (10) – (15) using MAPLE and the corrector formulae can be obtained. The method is the combination of predictor and corrector. The predictor is one order less than the corrector. The same process as above can be applied to find the predictor formulae. Predictor First point: h y n'' +1 − y n'' = (5 f n −2 − 16 f n−1 + 23 f n ) 12 h2 y n' +1 − y n' − hy n'' = (3 f n −2 − 10 f n −1 + 19 f n ) 24 y n +1 − y n − hy n' −
(16)
h 2 '' h3 yn = (7 f n − 2 − 24 f n −1 + 57 f n ) 2 240
Second point: y n'' + 2 − y n'' =
h (7 f n−2 − 20 f n−1 + 19 f n ) 3
y n' + 2 − y n' − 2hy n'' =
y n + 2 − y n − hy n' −
Corrector First point:
h2 ( 4 f n −2 − 12 f n −1 + 14 f n ) 3
( 2h) 2 '' h 3 yn = (9 f n − 2 − 28 f n −1 + 39 f n ) 2 15
(17)
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h (− f n−1 + 13 f n + 13 f n+1 − f n + 2 ) y n'' +1 − y n'' = 24 h2 y n' +1 − y n' − hy n'' = ( −7 f n −1 + 66 f n + 129 f n +1 − 8 f n + 2 ) 360 y n +1 − y n − hy n' −
(18)
h 2 '' h3 yn = ( −4 f n −1 + 33 f n + 96 f n +1 − 5 f n + 2 ) 2 720
Second point: y n'' + 2 − y n'' =
h ( f n + 4 f n +1 + f n+ 2 ) 3
yn' + 2 − yn' − 2hyn'' =
h2 (−2 f n −1 + 36 f n + 54 f n +1 + 2 f n + 2 ) 45
y n + 2 − y n − hy n' −
( 2h) 2 '' h 3 yn = ( −2 f n −1 + 33 f n + 30 f n +1 − f n + 2 ) . 2 45
(19)
The formulae (18) and (19) may be rewritten in the form of matrix difference equation as follows:
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 y n −2 0 0 0 0 y n −1 0 − 1 1 0 y n = h 0 0 0 0 y n+1 0 0 0 0 y n+ 2 0 − 1 0 1 0
0 0 0 1 0 1 0 0 0 1 0 2
0 0 y n − 2 0 − 1 0 y n −1 0 0 2 0 yn + h 0 0 0 y n +1 0 − 1 0 y n + 2 0 0 0 0
−1 0
0 0
1 1 1 2 1 2
0
2
0
0 0 0
0 0 0
0 y n−2 0 y n−1 0 yn − 1 y n+1 0 y n+ 2 0
Solving third order boundary value problem
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1 0 − 24 0 − 8 360 5 0 − 720 + h3 0 0 2 0 − 45 0 − 2 45
(20)
13 24 129 360 96 720 1 3 36 45 33 45
13 24 66 360 33 720 4 3 54 45 30 45
1 24 7 − f n − 2 360 4 f − n−1 720 f . 1 n 3 f n+1 2 f n+2 45 1 − 45 −
The order of this developed method is identified by referring to Majid et al. [8]. The two-point block method for ODEs can be written in a matrix difference equation as follows: αYm = hβYm′ + h 2 γYm′′ + h 3 δFm ,
(21)
where α, β, γ, and δ are the coefficients with the m-vector Ym , Ym′ , Ym′′ and Fm be defined as Ym = [ y n −2 , y n −1 , y n , y n +1 , y n + 2 ] T , Ym′ = [ y ′n −2 , y n′ −1 , y ′n , y ′n +1 , y ′n + 2 ] T , Ym′′ = [ y n′′−2 , y n′′−1 , y n′′ , y n′′+1 , y n′′+ 2 ] T , Fm = [Fn −2 , Fn −1 , Fn , Fn +1 , Fn + 2 ] T
(22)
By applying the formulae for the constants C q , in Fatunla [6], the formulae is defined as k
C0 =
∑α j , j =0 k
C1 =
∑ ( jα j − β j ) , j =0
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k
j2 α j − jβ j − γ j ) , 2! j =0
C2 =
∑
C3 =
∑ ( 3! α j −
(
k
j3
j =0
j2 β j − jγ j − δ j ) , 2!
(23)
M jq j q −1 j q−2 j q −3 αj − βj − γj − δ j), q! ( q − 1)! ( q − 2)! ( q − 3)! j =0 k
Cq =
∑
(
where q = 4,5,6, K
Therefore, the order and error constant of the two-point block method can be obtained by using (23). For q = 0 , 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 − 1 1 0 0 1 0 C0 = j α j = α 0 + α1 + α 2 + α 3 + α 4 = + + + + = . 0! 0 0 0 0 0 0 j =0 0 0 0 0 0 0 0 0 − 1 0 1 0
∑
(24)
For q = 1 , 4
C1 =
1
4
1
∑ 1! j1α j − ∑ 0! j 0β j = 11 ⋅ α1 + 21 ⋅ α 2 + 31 ⋅ α 3 + 41 ⋅ α 4 − (β 0 + β1 + β 2 + β 3 + β 4 ) j =0
j =0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 − 1 0 0 0 − 1 1 0 0 0 1 0 0 0 =1 + 2 + 3 + 4 − + + + + = . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 − 1 0 − 0 1 0 1 0 0 2 0 0 0
(25)
Solving third order boundary value problem
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For q = 2 , 4
C2 =
4
4
1 2 1 1 1 0 j αj − j βj − j γj 2! 1! 0! j =0 j =0 j =0
∑
=
∑
∑
(
(26)
) (
1 2 1 ⋅ α1 + 2 2 ⋅ α 2 + 32 ⋅ α 3 + 42 ⋅ α 4 − 11 ⋅ β1 + 21 ⋅ β 2 + 31 ⋅ β 3 + 41 ⋅ β 4 2 − (γ 0 + γ1 + γ 2 + γ 3 + γ 4 )
1 = 1 2 2
0 0 0 2 +2 0 0 0
0 0 − 1 2 +3 0 0 − 1
0 0 1 2 +4 0 0 0
)
0 0 0 0 0 0 0 0 1 − 1 0 0 1 0 0 − + 2 3 + 4 1 + 0 0 0 0 0 0 1 0 − 1 0 1 2 0 0 0
0 0 1 − 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 − + + 2 + + = . 0 −1 0 1 0 0 0 0 2 0 0 0 0 0 2 0 0 0
For q = 3 , 4
C3 =
1
j =0
=
4
1
4
4
1
1
∑ 3! j 3α j − ∑ 2! j 2β j − ∑ 1! j1γ j − ∑ 0! j 0 δ j j =0
j =0
j =0
(
) (
1 3 1 1 ⋅ α1 + 23 ⋅ α 2 + 33 ⋅ α 3 + 43 ⋅ α 4 − 12 ⋅ β1 + 2 2 ⋅ β 2 + 32 ⋅ β3 + 4 2 ⋅ β 4 2 6
(
)
− 11 ⋅ γ1 + 21 ⋅ γ 2 + 31 ⋅ γ 3 + 41 ⋅ γ 4 − (δ 0 + δ1 + δ 2 + δ3 + δ 4 )
(27)
)
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1 3 = 1 6
0 0 0 3 +2 0 0 0
0 0 − 1 3 +3 0 0 − 1
0 0 1 3 +4 0 0 0
0 1 − 1 0 0 0 0 11 0 0 0 − 1 + 2 2 + 3 + 4 0 1 0 − 1 0 0 0 2 0 0 2 0
0 0 0 = . 0 0 0
For q = 4 ,
0 0 0 1 2 − 1 0 2 0 1
0 0 0 2 +2 0 0 0
0 1 1 2 +3 0 1 2
0 − 1 0 2 +4 0 0 0
0 0 0 0 − 1 0
13 13 1 1 − − 24 24 24 24 0 129 66 7 − 8 360 360 − 360 0 360 96 33 4 0 5 − − + − 720 + 720 + 720 + 720 0 0 1 4 1 2 3 3 3 0 − 36 54 2 0 45 − − 2 45 45 45 45 33 30 − 1 45 45 45
Solving third order boundary value problem
0 0 4 4 4 4 0 1 4 1 3 1 2 1 1 C4 = j αj − j βj − j γj − j δj = . 4! 3! 2! 1! 0 j =0 j =0 j =0 j =0 0 0
∑
∑
∑
∑
2639
(28)
For q = 5 , 0 0 4 4 4 4 0 1 5 1 4 1 3 1 2 C5 = j αj − j βj − j γj − j δj = . 5! 4! 3! 2! 0 j =0 j =0 j =0 j =0 0 0
∑
∑
∑
∑
(29)
For q = 6 , 0 0 4 4 4 4 0 1 6 1 5 1 4 1 3 C6 = j αj − j βj − j γj − j δj = . 6! 5! 4! 3! 0 j =0 j =0 j =0 j =0 0 0
∑
For q = 7 ,
∑
∑
∑
(30)
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11 720 11 0 1440 0 23 4 4 4 4 0 1 7 1 6 1 5 1 4 C7 = j αj − j βj − j γj − j δ j = 10080 ≠ . 1 7! 6! 5! 4! − 0 j =0 j =0 j =0 j =0 90 0 1 90 0 1 70
∑
∑
∑
∑
(31)
The method is order p if C0 = C1 = K = C p + 2 = 0 and C p + 3 ≠ 0 is the error constant. Thus, we conclude that the method in (16) to (19) is of order 4 and the error constant is 11 C p +3 = C 7 = 720
11 1440
23 10080
−
1 90
1 90
T
1 . 70
(32)
The starting initial points in the code will consider the direct AdamsBashforth explicit method of one-step to compute the starting values, but the method will solved the problem directly without reducing to first order equations. The initial points will be compute only once at the beginning of the step. Then, the initial points will be used for starting the predictor and corrector block method. The predictor and corrector block method can be applied until the end of interval. This block method will be adapted for solving the boundary value problems via shooting techniques. Shooting technique will allow for new guessing and for each new guessing of the direct Adams-Bashforth explicit method will be used again to find the starting initial points. In order to get better approximation for the initial points when using the direct Adams-Bashforth explicit method, the value of h will be reduced to
h 128
. While the predictor and corrector block method will remain using the
choosing step size h.
Solving third order boundary value problem
3
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Implementation of the method
Shooting technique will be applied in the propose block method. The code will start with the initial guess, that determines solution of the derivative ′′ gives, y ′′′ = f ( x, y, y ' , y ′′)
a≤ x≤b
y (a ) = γ ,
y ' (b) = β ,
y ' (a) = α ,
(33) y ′′(a, t ) = t 0 .
(34)
Differentiate Eq. (33) with respect to t, and it is simplified follows: ∂y ' ' ' ∂ ( x, t ) = f ( x, y( x, t ), y ' ( x, t ), y' ' ( x, t )) ∂t ∂t ∂ ∂x ∂t ∂y = ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) + ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) ∂x ∂t ∂y ∂t ∂f ∂y ' ∂f ∂y ' ' ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) + ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) ∂y ' ∂t ∂y ' ' ∂t ∂t ∂y ∂f ∂y ' = ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) + ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) ∂y ∂t ∂y ' ∂t ∂f ∂y ' ' + ( x, y ( x, t ), y ' ( x, t ), y ' ' ( x, t )) ( x, t ) . ∂y ' ' ∂t Using z ( x, t ) to denote (∂y / ∂t )( x, t ) , we have the initial-value problem +
z ′′′ =
∂f ∂f ∂f ( x, y, y' , y ′′) z + ( x, y, y' , y ′′) z '+ ( x, y, y ' , y ′′) z ′′ ∂y ∂y' ∂y' '
z (a) = 0,
z ′′(a) = 1.
z ' ( a) = 0,
a≤ x≤b,
(35) (36)
For the first initial guessing, we considered t0 =
β−α . b−a
(37)
See Faires and Burden (1998). The solution of y' from Eq. (19) is determined when, ϕ(t ) = y ' (b, t ) − β = 0.
(38)
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2642
Newton method will be used to get a very rapidly converging iteration. We compute the {t k } defined as: t k +1 = t k −
ϕ(t ) . ϕ' (t )
(39)
From Eq. (33) and (35), we may obtain the solution for y ' (b, t k ) and z ' (b, t k ) respectively. The solutions were applied in Newton’s method to find the next guess, t k +1 . y ' (b, t k ) − β . t k +1 = t k − z ' ( b , t ) k
(40)
Both Eq. (33) and Eq. (35) will be solved simultaneously using the block method. The process will stop until the error β − y ' (b, t k ) ≤ tolerance , where tolerance = 10 −5. The algorithm of the proposed method were developed in C language.
4
Results and discussion
We now consider three numerical example illustrating the comparative performance of the propose method over other existing methods. All calculations are implemented by Microsoft Visual C++ 6.0. In problem 1 and 2, we obtained the maximum errors at different values of step size i.e. h =
1
,
1
,
1
16 32 64
and
1 . The 128
maximum errors were compared with Al-Said and Noor [2]. For problem 3, the results were compared with El-Danaf [7].
Problem 1: y ′′′ = xy + ( x 3 − 2 x 2 − 5 x − 3)e x , y (0) = 0, y ' (1) = 0, y ' (0) = −e
Exact solution: y ( x ) = x (1 − x )e x Source: El-Salam et al. (2010).
0 ≤ x ≤ 1,
Solving third order boundary value problem
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Problem 2: y ′′′ = − y + ( x − 4) sin x + (1 − x) cos x, y (0) = 0, y ' (1) = 0, y ' (0) = sin 1
0 ≤ x ≤ 1,
Exact solution: y ( x) = x( x − 1) sin x Source: El-Salam et al. (2010).
Problem 3: y ′′′ = − y + (7 − x 2 ) cos x + ( x 2 − 6 x − 1) cos x, y (0) = 0, y ' (1) = −1, y ' (0) = 2 sin 1
0 ≤ x ≤ 1,
Exact solution: y ( x ) = ( x 2 − 1) sin x Source: El-Danaf (2008). The following notations are used in the tables: FOBM Fourth order block method proposed in this research h Step size TS Total step Table 1: Comparison of maximum errors for Problem 1 at different values of h h 1 16 1 32 1 64 1 128
FOBM
TS
Al-Said and Noor[2]
TS
6.5304 x10-6
8
8.1224 x10-4
16
8.5985 x10-7
16
2.1812 x10-4
32
1.1113 x10-7
32
5.5859 x10-5
64
1.4152 x10-8
64
1.4091 x10-5
128
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Table 2: Comparison of maximum errors for Problem 2 at different values of h h 1 16 1 32 1 64 1 128
FOBM
TS
Al-Said and Noor[2]
TS
4.1815 x10-6
8
4.5978 x10-5
16
5.4620 x10-7
16
1.2530 x10-5
32
6.9679 x10-8
32
3.2356 x10-6
64
8.7950 x10-9
64
8.1999 x10-7
128
Table 3: Comparison of maximum errors for Problem 3 at different values of h h
FOBM
TS
El-Danaf [7]
TS
2-3
3.2723 x10-5
250
1.6501 x10-4
500
2-4
2.3921 x10-6
2500
9.8380 x10-6
5000
2-5
1.5905 x10-7
25000
5.8773 x10-7
50000
2-6
1.0190 x10-8
250000
3.5687 x10-8
500000
2-7
6.4399 x10-10
2500000
2.1968 x10-9
5000000
In Table 1 and 2 shown the maximum errors of FOBM are better compared to the results in Al-Said and Noor [2]. The results even better as the step size decrease. In Table 3, the maximum errors for both methods are comparable, but the result of FOBM is better than the result in El-Danaf [7] when h = 2-3 and 2-7. The FOBM only need half total steps compared to the other two methods because FOBM will solved the problem at two steps simultaneously in a block.
5
Conclusion
In this research, we conclude that fourth order block method with shooting technique using constant step size is suitable to solve third order nonlinear boundary value problems. This proposed method is simple, efficient and economically.
Solving third order boundary value problem
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Acknowledgement The author gratefully acknowledged the financial support of Graduate Research Fund (GRF) from Universiti Putra Malaysia and MyMaster from the Ministry of Higher Education.
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D. Faires and R.L. Burden, Numerical Methods.2nd Ed. Pacific Grove: International Thomson Publishing Inc, 1998.
[2]
E.A. Al-Said and M.A Noor, Numerical solutions of third-order system of boundary value problems. Applied Mathematics and Computation, 190, 2007, pp. 332- 338.
[3]
F.A. Abd El-Salam, A.A. El-Sabbagh, and Z.A. Zaki, The Numerical Solution of Linear Third Order Boundary Value Problems using Nonpolynomial Spline Technique. Journal of American Science, 6(12), 2010, pp. 303-309.
[4]
G. B. Loghmani and M. Ahmadinia, Numerical Solution of Third-order Boundary Value Problems, Iranian Journal of Science & Technology, Tran. A, Volume 30, Number A3, 2006, pp. 291-295.
[5]
P. S. Phang, Z. A. Majid and M. Suleiman, Solving Nonlinear Two Point Boundary Value Problem using Two Step Direct Method. Journal of Quality Measurement and Analysis, 7(1), 2011, pp. 129-140.
[6]
S. O. Fatunla, Block methods for second order ODEs. International Journal of Computer Mathematics, vol. 41, 1991, pp. 55-63.
[7]
S. Talaat El-Danaf, Quartic Nonpolynomial Spline Solutions for Third Order Two-Point Boundary Value Problem. World Academy of Science, Engineering and Technology, 45, 2008, pp. 453-456.
[8]
Z. A. Majid N. Z. Mokhtar and M. Suleiman, Direct Two-Point Block OneStep Method for Solving General Second-Order Ordinary Differential Equations. Mathematical Problems in Engineering, 2012, pp. 1-16.
Received: February 10, 2013
