physics and applied mathematics contain boundary value problems ...... Murugesan K. , Numerical. Strategies for the ... journal of modern physics B, vol. 20, no.
Eng. & Tech. Journal, Vol. 28 , No. 2 , 2010
Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods Anwar Ja'afar Mohamad – Jawad * Recived on: 5/2/2009 Accepted on: 13/8/2009
Abstract The boundary value problems for the 2nd order non-linear ordinary differential equations are solved with four numerical methods. These numerical methods are Rung-Kutta of 4th order, Rung–Kutta Butcher of 6th order, differential transformation method, and the Homotopy perturbation method. Three physical problems from the literature are solved by the four methods for comparing results. Results were presented in tables and figures. The differential transformation method appeared to be effective and reliable to finding the semi numerical-analytical solutions for such type of boundary value problems. Keywords: Boundary value problem, Rung-Kutta RK4, Rung–Kutta Butcher, Differential transformation method, Homotopy perturbation method.
Introduction Several models of mathematical physics and applied mathematics contain boundary value problems BVPs in the 2nd order non-linear ordinary differential equations (ODEs) [1].
Consider the boundary value problem of 2nd order nonlinear (ODEs):
d2y f ( x, y ) 0, dx 2 y (0) 0 , y (1) c
*Electrical Engineering Department, University of Technology / Baghdad
0 x 1 (1) (2)
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
where
0 , c is constant and
f ( x, y ) : ,continuous
k2 h. f n1 ( xn h / 2, yn k1 / 2, pn L1 / 2)
(0,1] [0, ) (0,1] and
f 0 y
exists
L2 h. f n 2 ( xn h / 2, yn k1 / 2, pn L1 / 2) k3 h. f n1 ( xn h / 2, yn k2 / 2, pn L2 / 2)
and
L3 h. f n 2 ( x h / 2, y k2 / 2, p L2 / 2)
continuous. Different analytical and numerical methods were used to solve the 2nd order (ODEs) this can be concerned Agrawal et al. [2-4], Ha and Lee [5] and O' Regan [6,7] using different techniques. In this paper four numerical methods are applied containing: Rung-Kutta (RK4) [8], (RK-Butcher) [9], differential transformation method [10] and homotopy perturbation method [11] to solve BVPs.
k4 h. f n1 ( xn h, yn k3 , pn L3 ) L4 h. f n 2 ( xn h, yn k3 , pn L3 ) Rung –Kutta Butcher of 6th order method: The BVPs of second order (ODEs) is solved by Rung – Kutta Butcher of 6th order (RK-Butcher). Applying this method to solve system of 1st order differential equations eqs.(3) and (4) : y n1 y n
Rung-Kutta 4th Order Method: Consider the boundary value problem in Eq.(1) under the condition in Eq.(2) , it will be transformed to two 1st (ODEs) by assuming:
dy p f n1 ( x, y, p) dx
(6) h p n 1 p n (7 L1 32 L3 12 L4 32 L5 7 L6 ) 90
(7) The Constants k1 , k2 , k3 , k4 , L1 , L2 , L3 , L4 in Eqs.(6) and (7) are calculated from the following :
(3)
then Eq.(1) becomes:
k1 f n1 ( xn , yn , pn )
dp ( f ( x, y) f n 2 ( x, y, p) (4) dx The solution y (x ) for this problem from x0 0 to xn 1 after assuming h and n where: x x0 (5) h n n
L1 f n 2 ( xn , yn , pn )
h L1 4 h hk hL L2 f n 2 ( xn , yn 1 , pn 1 ) 4 4 4 h h k3 f n1 ( xn , yn , pn ) L1 L2 8 8 h hk1 hk2 hL hL L3 f n 2 ( xn , yn , pn 1 2 ) 4 8 8 8 8 h k4 f n1 ( xn , yn , pn ) L2 hL3 2
k2 f n1 ( xn , yn , pn )
The set of 1st order (ODEs) (3) and (4) are solved together from the following:
1 y n1 y n (k1 2k2 2k3 k4 ) 6 1 p n1 p n ( L1 2 L2 2 L3 L4 ) 6 k1 , k2 , k3 , k4 , L1 , L2 , L3 , L4
where calculated as follows :
h (7k1 32k3 12k4 32k5 7k6 ) 90
h hk hL L4 f n 2 ( xn , yn 2 hk3 , pn 2 hL3 ) 2 2 2
are
k5 f n1 ( xn , yn , pn )
k1 h. f n1 ( xn , yn , pn ) L1 h. f n 2 ( xn , yn , pn )
2
3h 9h L1 L4 16 16
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
3h 3hk1 9hk4 3hL1 9hL4 , yn , pn ) Theorem 4.2. If 4 16 16 16 16 Y (k ) .G(k ) . h k6 f n1 ( xn , yn , pn ) (3L1 2 L2 12 L3 12 L4 8L5 ) Theorem 4.3. If 7 L5 f n 2 ( xn
y( x) .g ( x) ,then y ( x)
dg ( x) dx
,then
Y (k ) (k 1).G(k 1) . h m f [ x h , y ( 3 k 2 k 12 k 12 k 8 k ), n 1 2 3 4 5 Theorem 4.4. If y ( x) d g m( x) ,then n 2 n 7 L6 dx h Y (k ) ((k m)!/ k!).G(k m) . pn (3L1 2 L2 12 L3 12 L4 8L5 )] 7 Theorem 4.5. If y ( x) g ( x).h( x) ,then k Y (k ) l 0 G(l ) H (k l ) . p( 0) is predicted from: Theorem 4.6. If y ( x) x m ,then r p( 0) [ y(1) y( 0) ] / h (8) if k m . 1 Y ( k ) ( k m) 0 if k m Then p( 0) is modified by : Theorem 4.7. If y ( x) exp( .x) ,then p(r0)1 p(r0) ( y(r1) y(1) ) / h Y ( k ) k / k! . (9) Theorem 4.8. If y ( x) sin( .x ) Error is accepted as: k ,then Y (k ) ( / k!) sin( k / 2 ) . r 5 y(1) y(1) 10 Theorem 4.9. If y( x) cos( .x ) ( or any tolerance needed ) ,then Y (k ) ( k / k!) cos(k / 2 ) . Theorem 4.10. If y ( x) exp( u ( x)) , Differential Transformation Method then: DTM : .... r The differential transformation of the 1 k k Y (k ) k 3 k 2 U (k1)U (k2 k1)... kth derivatives of function y (x ) is k 0 n 0 n! k 0 k 0 k 2 0k 1 0 defined as follows [10]: k n 1
n 1
Y (k )
1 d k y k! dx k x x
n2
U (kn 1 kn 2 )U (k kn 1 )( x k0 )
(10)
Homotopy Perturbation Method HPM: Consider the following system of the integral equation [11]:
0
and y (x ) is the differential inverse transformation of Y (k ) defined as follows:
t
y ( x) Y (k ).( x x0 ) k
F (t ) G (t ) K (t , s ) F ( s ) ds
k 0
(11) for finite series of k = N , eq.(11) ca n be written as:
0
(13)
F(t) (f1(t),f2(t), . . . . ,fn(t))
T
N
y ( x) Y (k ).( x x0 ) k
G(t) (g1(t),g2(t), . . . . ,gn(t))T
k 0
(12) The following theorems that can be deduced from eqs.(10) and (12) [10]:
K (t , s) [K ij(t,s)] i 1,2 ,3, . . . ,n:j 1,2,3, . . . ,n
Theorem 4.1. If y ( x) g ( x) h( x) ,then Y (k ) G(k ) H (k ) .
Let
L( y ) 0
3
(14)
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
Where L an integral or differential operator, and then a convex homotopy is
i 0
H ( y, p) defines by:
,
f 2 (t ) p i vi i 0
,
f 3 (t ) p i wi
H ( y, p) (1 p) F ( y ) pL( y )
, …… (19) The comparison of like powers of p gives solution of various orders. i 0
(15) where F ( y ) is a functional operator with known solution v0 , which can be obtained easily. It is clear that H ( y, p) 0 (16) From which we have H ( y,0) F ( y ) , and H ( y,1) L( y ) This shows that H ( y, p) continuously traces an implicitly defined curve from a starting point H (v0 ,0) to a solution
Numerical Examples Some of the physical problems are solved to assign the effectiveness and accuracy of the four numerical methods. Results are presented in tables and figures. Computer programs are written to implement the procedures of the four numerical methods.
H ( f ,1) .The
embedding parameter increases monotonically from zero to unit at the problem F ( y ) 0 is continuously deforms the original problem L( y ) 0 . The embedding parameter can be considered as an expanding parameter [11]. The homotopy perturbation method uses the homotopy parameter p as an expanding parameter to obtain:
Example 1 Consider the following boundary value problem of 2nd order non-homogenous non-linear (ODEs)[1]:
d2y 2( y / ) 2 8 y 0 dx 2
0 x 1
(20) subject to the boundary conditions y (0) 0 , y (1) 0 (21) This problem was studied by Vedat [1] by applying the differential transformation method to eq.(20) then the recurrence relation can be evaluated using theorems 4.1, 4.2, 4.4, and 4.5 as follows:
y pi yi y0 y1 p y2 p 2 i 0
y3 p y4 p 4 y5 p5 ..... 3
(17) If p 1 then (17) corresponds to (12) becomes the approximate solution of the form
k
Y (k 2)
[8Y (k ) 2 (l 1)(k l 1)Y (l 1)Y (k l 1)] l 0
(k 1)(k 2)
(22) The boundary conditions in eq.(21) can be transformed at x0 0 as:
f lim y yi p 1
f1 (t ) p i yi
i 0
(18) It is well known that the series (4.7) is convergent for most of the cases and also the rate of convergent is dependent on L( y ) , see [12]. We assume that problem (10) has a unique solution. Consider the ith equation of (10), take
Y (0) 0, Y (1) A
(23) where according to eq.(10) , y / (0) A . For N =8 and by using the recurrence relations in Eq.(22) and the transformed boundary conditions in Eq.(23), the 9 following series solution up to O( x ) is obtained :
4
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
4 A 4 A3 3 ) x ( 2 A2 2 A4 ) x 4 3 3 8 A 56 A3 16 A 5 88 A2 328 A4 16 A6 6 ( ) x ( )x 15 15 5 45 45 3 32 A 1696 A3 4544 A5 64 A7 7 ( )x 315 315 315 7 344 A2 1448 A4 1808 A6 ( 16 A8 ) x8 O( x 9 ) 315 105 63
y ( x) Ax A2 x 2 (
(24) Applying the boundary conditions at x=1, then A= 1 and equation (24) can be written as:
y ( x) x x 2
(25) This was the exact solution of the problem. The same result found by Wazwaz [13]. If we want to solve the BVPs in eq(20) by applying the Homotopy perturbation method, we can rewrite the second order boundary value problem as a system of two differential equations:
dy / dx q( x) dq / dx 2q 2 ( x) 8 y ( x)
4x 2 2 y2 A x 3 p ( 2) : 3 q 4 A3 x 2 16 Ax 2 3 3 3 4 A x 4 Ax 4 y 3 3 3 ( 3) p : 2 4 5 q 8 A4 x 3 40 A x 32 x 3 3 5 2 5 8 A x 16 x 6 4 4 y 2 A x 4 3 15 ( 4) p : 6 q 16 A5 x 4 32 A3 x 5 832 Ax 4 45
(29) Adding all the terms, (29) gives:
x
y ( x) 0 q(t ).dt
y ( x) Ax A2 x 2 (
0
q( x) A [2q 2 (t ) 8t ]dt
(27)
(30) Using the boundary conditions at x=1, we have A 0.99998 then:
0
Using (16) and (18) for (27) we have: x
y ( x) 0.99998 x 0.99996 x 2 0.000079998 x 3
y0 py1 p 2 y2 p3 y3 ........ 0 p (q0
0.666533 x 4 0.533121x 5
0
pq1 p q2 p q3 ........).dt 3
(31) Results are presented in table (1), while errors of the four numerical methods are summarized in table (2). Figure (1) presents the solution of problem in example (1) by the four numerical methods, RK4, RK-Butcher, DTM, and HPM.
x
q0 pq1 p 2 q2 p 3q3 ........ A p 0
[2(q0 pq1 p q2 p q3 .....) 8t ]dt (28) Comparing the coefficient of like powers of p, we have 3
4 A3 4 3 4 A )x ( 2 A4 ) x 4 3 3 3
16 A5 8 A2 5 ( ) x O( x 6 ) 5 3
x
2
,
3
16 A5 x 5 16 A3 x 6 832 Ax 7 y ) p ( 5) : 5 5 3 315 q ..... 5
(26)
with y (0) 0 and q(0) A This can be written as a system of integral equations:
2
y0 0 p (0) : q0 A y1 Ax p (1) : 2 2 q1 2 A x 4 x ,
2
5
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
Example 2
dy / dx q( x)
Consider the following boundary value problem of 2nd order non-homogenous non-linear (ODEs)[1]:
dq / dx y 2 ( x) x 4 2
2
d y ( x) 2 y ( x) x 4 2 0 2 dx
(38)
with y (0) 0 and q(0) A This can be written as a system of integral equations:
0 x 1
x
(32) subject to the boundary conditions
y ( x) 0 q (t ).dt
(33) This problem was studied by Vedat [1]. By applying the differential transformation method of eq.(32) using theorems 4.1, 4.2, 4.4, and 4.5 as follows:
q ( x) A [ y 2 (t ) t 4 2]dt
0
y (0) 0 , y (1) 1
x
0
(39) Using (16) and (18) for (39) we have: y0 py1 p 2 y2 p 3 y3 . . . . x
k
Y (k 2)
[ Y (l )Y (k l ) (k 4) 2 (k )]
0 p (q0 pq1 p 2 q2 p 3q3 .. . ) dt
l 0
0
(k 1)(k 2)
q0 pq1 p 2 q2 p 3q3 . . . .
(34) The boundary conditions in eq.(33) can be transformed at x0 0 as: Y (0) 0, Y (1) A
x
A p [(y0 py1 p 2 y2 p 3 y3 . . . .)2 0
t 2]dt
(35)
4
/ where according to Eq.(10) , y (0) A .
For k = 9 and by using the recurrence relations in Eq.(34) and the transformed boundary conditions in Eq.(35), the 10 following series solution up to O ( x ) is obtained : A2 4 A 5 A3 7 y ( x) Ax x 2 x x x 12 10 252 (36) 11A2 8 A 9 x x O( x10 ) 1680 360 Applying the boundary conditions at x=1, then A= 0 and eq.(36) can be written as:
Comparing the coefficient powers of p, we have
y0 0 p (0) : q0 A y1 Ax (1) p : x5 q 2x 1 5 , ,
x6 y x2 p ( 2) : 2 30 q 0 2 , y3 0 ( 3) p : A2 x 3 q 3 3 ,
y ( x) x 2
(37) Which is the exact of Eq.(32). This result was agreement with Jafari [14]. If we want to solve the BVPs in eq.(32) by applying the Homotopy perturbation method, we can rewrite the second order boundary value problem as a system of two differential equations:
A2 x 4 y4 12 p ( 4) : 8 4 q Ax Ax 4 24 2 ,
6
(40) of like
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
Ax9 Ax 5 y 5 216 10 ( 5) p : 13 x9 x5 q x 5 11700 135 5 x14 x10 x6 y p ( 6) : 6 163800 1350 30 q ... 6
where:
sinh( y( x))
then eq(44) became:
d 2 y ( x) 1 1 y y3 y5 2 dx 3! 5!
A2 4 A 5 A 9 x x x 12 10 1080
Y (k 2)
1 10 1 x x14 O( x15 ) 1350 163800
1 k k4 k3 k2 Y ( k1 ) 5! k 4 0 k 30 k 2 0 k10 Y (k 2 k1 )Y (k 3 k 2 )Y (k 4 k 3 )Y (k k 4 )]
0.000166162 x5 0.00000153 x9 0.00047 x10 0.0000061x14 O( x15 )
(47) The boundary conditions in eq (45) can be transformed at x0 0 as: Y (0) 0, Y (1) A
(48) where according to eq (10) , y (0) A . For N = 9 and by using the recurrence relations in eq(47) and the transformed boundary conditions in eq.(48), the following series solution up to O ( x10 ) is obtained : /
(43) By using the special computer programs solutions for the four numerical are evaluated and presented in table (3), while errors of the four numerical methods are summarized in table (4). Figure (2) presents the solution of problem in example (2) by the four numerical methods RK4, RK-Butcher, DTM, and HPM . Results by the four methods are compatible, so that solution by Homotopy perturbation method in eq(43) is agreed with the exact solution in eq.(37) determined by the differential transformation method. Example 3 Consider Troesch's problem of 2nd order non-homogenous non-linear (ODEs) [1]:
A 3 ( A A3 ) 5 ( A 11A3 A5 ) 7 x x x 3! 5! 7! 9 A 102 A3 57 A5 9 ( ) x O( x11) 9! y ( x) Ax
(49) Applying the boundary conditions at x=1, then A= 0.84524 and equation (49) can be written as: y ( x) 0.84524 x 0.140873 x3 0.0120759 x 5 0.00157126 x 7 0.000239832 x 9 O( x11) (50) If we want to solve the BVPs in eq.(44) by applying the Homotopy perturbation method, we can rewrite the second order boundary value problem as a system of two differential equations:
0 x 1
(44)
subject to the boundary conditions
y (0) 0 , y (1) 1
1 1 k k2 [Y (k ) Y (k1 ) (k 1)(k 2) 3! k 2 0 k10
Y (k 2 k1 )Y (k k 2 )
(42) Using the boundary conditions at x=1, we have A 0.00166162 then: y ( x) .00166162 x x 2 0.00000023 x 4
d 2 y ( x) sinh( y ( x)) dx 2
(46)
This problem was studied by Vedat [1] by applying the differential transformation method to eq.(46) using theorems 4.1, 4.2, 4.4, and 4.5 as follows:
(41) Adding all the terms, (41) gives: y ( x) Ax x 2
e y e y 1 1 y y3 y5 2 3! 5!
(45)
7
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
dy / dx q( x) dq / dx sinh( y( x))
A 3 A A3 5 A 61A3 7 x ( )x ( )x 6 120 20 5040 1 A2 13 9 ( ) x O( x11) 72 60 180 y ( x) Ax
(51)
q(0) A With y (0) 0 and
(55) Using the boundary conditions at x=1, we have A 0.821162 then:
This can be written as a system of integral equations: x
y ( x) 0 q (t ).dt
y ( x) 0.821162 x 0.1368603333 x 3 0.0345287 x 5
0
0.00686464 x 7 0.0011591x 9 O( x11)
x
q ( x) A [sinh( y (t ))]dt
(56) By using the special computer programs solutions for the four numerical are evaluated and presented in table (5), while errors of the four numerical methods are summarized in table (6). Figure (3) presents the solution of problem in example (3) by the four numerical methods, RK4, RK-Butcher, DTM, and HPM.
(52)
0
If sinh[y(x)] is expanded around y0 0 ,
1 1 5 sinh( y) y y 3 y 6 120 then Using (16) and (18) for (52) we have:
y0 py1 p 2 y2 p3 y3 . . . . x
0 p (q0 pq1 p 2 q2 . . . . ).dt
Conclusion In this paper, four numerical methods RK4, RK-Butcher, DTM, and HPM, were applied for finding the solution of a class of two-point nonlinear boundary value problems. It may be concluded that all methods gave the same trend of solutions. It is obvious that all solution in the three examples are positive solutions in 0 < x < 1. That was appeared from figures(1) - (3), the differential transformation method is an effective tool in finding the semi numerical-analytical solutions to this type of boundary value problems.
0
q0 pq1 p 2 q2 p 3q3 . . . . x
1 3 1 5 y y ]dt 6 120 0 (53) Comparing the coefficient of like powers of p, we have A p [y
y0 0 p (0) : q0 A
y Ax p (1) : 1 q1 0 , , 3 y2 0 Ax ( 3) y3 ( 2) 2 p : p : 6 Ax q 0 q2 2 , 3 y4 0 ( 4) p : ( A A3 ) x 4 q 4 24 ,
References [1]
( A A3 ) x 5 y p ( 5) : 5 120 q 0 5 (54) Adding all the terms, (54) gives:
8
Vedat Suat S. Momani, Differential transformation method for obtaining positive solutions for two-point nonlinear boundary value problems, 2007, international journal mathematical manuscripts. Vol.1 number 1 6572.
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
[2]
R. P. Agarwal, Boundary value problems for High Ordinary Differential Equations, 1986, World Scientific Singapore.
[3]
R. P. Agarwal and D.O' Regan, Boundary value problems for super linear second order Ordinary Delay Differential Equations, 1996, J. Differential Equations. vol. 130 pp. 333-335.
[4]
[5]
[10]
[11] Sayed Tauseef Mohyud-Din , Muhammad A. N., Homotopy perturbation method for solving fourth order Boundary value problem, 2007, Mathematical Problems in Engineering, Article ID 98602
R. P. Agarwal, F.H.Wong, and W.C.Lian, positive solutions of nonlinear singular Boundary value problems, 1999, Applied math. Lett. 12 (2)115-120.
[12] J. H. He , Some asymptotic methods for strongly nonlinear equations, 2006, international journal of modern physics B, vol. 20, no. 10, 1141-1199.
K.S.Ha and Y.H.Lee, Existence of Multiple positive solutions of singular Boundary value problems, 1997, nonlinear Anal, 28: 1429-1438.
[6]
D.O' Regan, Theory of Boundary value problems, 1994, World Scientific Singapore.
[7]
D.O' Regan, Existence Theory for ordinary differential equations, 1997, Kluwer Boston, MA.
[8]
Stroud K. A., Advanced Engineering Mathematics, 2003, Palgrave MacMillan.
[9]
Sekar S., Murugesh V. ,and Murugesan K. , Numerical Strategies for the system of Second order IVPs Using the RKButcher Algorithms, 2004, International journal of Computer Science & Application Vol 1. No. 11 pp. 96-117.
Vedat Suat , Application of differential transformation method to linear sixth-order boundary value problems, 2207, applied mathematical Sciences, vol. 1, No. 2, pp. 51-58.
[13] A.M.
Wazwaz , A reliable algorithm for obtaining positive solutions for nonlinear boundary value problems, 2001, Comp. Math. Appl., 41: 1237-1244.
[14] H.Jafari and V. Daftardar-Gejji, Positive solutions of nonlinear Fractional boundary value problems Using Adomian Decomposition method, 2006, Appl. Math. Comp. 180: 700-706
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Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
Table (1) Results of the problem in example (1). X
( DTM )Exact Solution
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0.09 0.16 0.21 0.24 0.25 0.24 0.21 0.16 .08999992 0
RK4 solution
RK-Butcher solution
HPM solution
9.000599E-2 0.1600111 0.2100154 0.2400192 0.2500225 0.2400254 0.2100277 0.1600295 9.003057E-2 3.074731E-5
9.000006E-2 0.1600001 0.2100001 0.2400001 0.2500001 0.2400001 0.2100001 0.1600001 9.000009E-2 3.607737E-8
0.089972 0.1596373 0.208596 0.236928 0.2458333 0.238272 0.219604 0.1638 0.09212 3.60773E-5
Table (2) Errors of the four numerical methods of example (1). X
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
( DTM )Exact Solution
RK4 error
RK-Butcher Error
0.09 0.16 0.21 0.24 0.25 0.24 0.21 0.16 .08999992 0
5.990267E-6 1.105666E-5 1.54078E-5 1.92225E-5 2.253056E-5 2.537668E-5 2.773106E-5 2.95341E-5 3.065169E-5 3.074731E-5
5.960464E-8 1.043081E-7 1.192093E-7 1.490116E-7 1.490116E-7 1.490116E-7 1.639128E-7 1.788139E-7 1.713634E-7 3.607737E-8
HPM Error 0 2.799928E-5 3.626645E-4 1.404002E-3 3.071994E-3 4.166663E-3 1.727998E-3 9.604007E-3 3.822401E-3 9.622807E-2 3.607731E-5
Table (3) Results of the problem in example (2). X
( DTM )Exact Solution
RK4 solution
RK-Butcher solution
HPM solution
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0.01 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.81 1
9.999979E-3 3.999992E-2 8.999982E-2 0.1599997 0.2499995 0.3599993 0.489999 0.6399987 0.8099983 0.9999979
0.01 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.810001 1
0.0101661 0.0403322 0.0904980 0.1606626 0.2508220 0.3609615 0.4910293 0.6408696 0.8100682 0.9976317
Table (4) Errors of the four numerical methods of example (2). X
( DTM )Exact Solution
RK4 Error
RK-Butcher Error
HPM Error
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0.01 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.81 1
2.142042E-8 8.568168E-8 1.862645E-7 3.427267E-7 5.066395E-7 7.450581E-7 1.072884E-6 1.430511E-6 1.907349E-6 2.384186E-6
0 0 0 0 0 0 5.960464E-8 5.960464E-8 1.192093E-7 0
1.661601E-4 3.322698E-4 4.980639E-4 6.625503E-4 8.219779E-4 9.614825E-4 1.029253E-3 8.69453E-4 6.80089E-5 2.368569E-3
10
Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
Table (5) Results of the problem in example (3). X 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
( DTM )Exact Solution 0 8.466499E-2 0.1701789 0.2574054 0.347239 0.4406234 0.5385728 0.6421996 0.7527484 0.8716399 1.000527
RK4 solution 0 8.466491E-2 0.1701787 0.257405 0.3472379 0.4406189 0.5385576 0.6421563 0.7526405 0.8714001 1.000043
RK-Butcher solution 0 8.466499E-2 0.1701789 0.2574053 0.3472382 0.4406193 0.5385582 0.642157 0.7526415 0.8714013 1.000045
HPM solution 0 8.466525E-2 0.170187 0.2574677 0.3475057 0.4414536 0.5406895 0.6469099 0.7622502 0.8894458 1.032048
Table (6) Errors of the four numerical methods of example (3). 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
( DTM )Exact Solution 0 8.466499E-2 0.1701789 0.2574054 0.347239 0.4406234 0.5385728 0.6421996 0.7527484 0.8716399 1.000527
RK4 error 0 8.195639E-8 1.639128E-7 3.576279E-7 1.192093E-6 4.529953E-6 1.519918E-5 4.333258E-5 1.078844E-4 2.397895E-4 4.839897E-4
RK-Butcher Error 0 0 0 1.192093E-7 8.642673E-7 4.11272E-6 1.466274E-5 4.261732E-5 1.069307E-4 2.385974E-4 4.825592E-4
HPM Error 0 2.533197E-7 8.121133E-6 6.234646E-5 2.66701E-4 8.301735E-4 2.116621E-3 4.710257E-3 9.501755E-3 1.780593E-2 3.152049E-2
y(x) for example(1) Exact RK4 RK-Butcher
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Fig(1) Exact solution of problem in example (1)
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Eng. & Tech. Journal, Vol. , No. , 2009 Solving Second Order Non-Linear Boundary Value Problems by Four Numerical Methods
y(x) for example(2) Exact RK4 RK-Butcher
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Fig(2) Exact solution of problem in example(2) y(x) for example(3) Exact RK4 RK-Butcher
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Fig(3) Exact solution of problem in example (3)
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