FOUR STEPS IMPLICIT METHOD FOR THE
SOLUTION OF GENERAL SECOND ORDER ORDINARY
DIFFERENTIAL EQUATIONS
*1OSILAGUN, J. A., 2ADESANYA, A.O. 2ANAKE, T. A. AND 2OGHOYON, G. J. 1Department of Mathematical Sciences, Olabisi Onabanjo University,
Ago Iwoye, Ogun State, Nigeria.
2Department of Mathematics, Covenant University, Sango Ota, Ogun State, Nigeria.
*Corresponding author: [email protected] Tel: +2348086049694
plicit block method based on Newton’s
backward divided difference formula.
Adesanya et al. (2009) proposed a two step method for the general solution of second order which is self stating and adopted New-ton’s polynomial to generate the starting value. Awoyemi et al. (2009) recently pro-posed a self starting Numerov’s method. This method solves both initial and bound-ary value problems of ordinbound-ary differential equation. Yahaya (2007) constructed a Nu-merov method from a quadratic continuous polynomial solution. This process led to method which is applied to both initial and boundary value problems.
In this research work, we propose a block method for step length of four. This method adopts Newton’s polynomial approximation to generate the starting value and solves gen-eral second order ordinary differential
equa-ABSTRACT
Four steps implicit scheme for the solution of second order ordinary differential equation was derived through interpolation and collocation method. Newton polynomial approximation method was used to generate the unknown parameters in the corrector. The method was tested with numerical examples and it was found to be efficient in solving second order ordinary differential equations.
Keyword: collocation, interpolation, continuous, implicit, Newton’s polynomial, corrector.
INTRODUCTION
The second order initial value ordinary
dif-ferential equation of the form
,
(1) is considered in this paper. Differential equation is often used to model scientific and technological problems which most times, these equations do not have analytic solution, hence an approximate numerical method is required to solve these problems. Block methods for numerical solution of ordinary differential equations have been proposed by several researchers such as Milne (1953), Shampine and Watt (1969), Worland (1976) and Omar (1999). Rosser (1967) introduced the 3 point implicit block method based on integration formulae which is basically Newton’s cote type. Zanariah et al. (2006) proposed 3 points
im-) , ,
( '
''
y y x f
y 1
' 0, ( )
)
(a y a y
Journal of Natural Sciences, Engineer-ing and Technology ISSN - 2277 - 0593
tion directly.
METHODOLOGY
We consider an approximate solution to (1) in power series
(2.2)
are constants to be determined. We Consider a linear multistep method of the form
k
j j
j x
a x
y
0
) ( )
(
1 2 ) 1 ( 0 ,
,
j aj j k
j
(2)
where , m=the distinct
collo-cation point, t is the interpolation point, for our method, the step length k=4
1
0
1
0 2
) ( )
( )
( t
r
m
r
r n r r
n
r x y h x f x
y
] , [xn xn r
x
, (3)
, r=0, 1, 2…, m-1 (4)
, r=0, 1, m-1 (5)
To get and , According to Yahaya (2007), he derived a matrix of the form
where: is an identity matrix of dimension .
1
0 , 1 ( )
) (
m i
i
i r i
r x p x
1 ,...., 2 , 1 ,
0
m
r
1
0 ,
1 ( ) ( )
) (
m t
i
i r
i x p x
x
r n r
n y
x
y( ) r[0,1,2...,t1] ( ) ,
''
r n f x
y
) (x j
j(x)
I
DC
I (tm)(tm)
.
. .
. .
.
. .
. .
.
. .
2 0
0
. 1
. .
. .
.
. 1
. 1
1 1 2
1 1
1 1 2
1 1
1 2
m t
t n t
n t n
i t n n
n
i t n n
n
x x
x
x x
x
x x
x
Development value for the unknown
Theorem (3.0):
Assuming that and for k=0, 1, n are distinct values, then
, where is a polynomial that can be used to approximate
For Newton’s polynomial
, is the remainder and has the form
(Awoyemi et al.,2009)
Development of four steps method
In developing the method with step length k=4, we consider
D= 1 , 2 0 , 2 1 , 1 , 0 , 1 , 2 2 0 , 2 2 1 , 2 1 , 2 0 , 2 1 , 1 2 0 , 1 2 1 , 1 1 , 1 0 , 1 . . . . . . . . . . . . . . . . . . . . m m t m t t m t m t m t m t m t h h h h h h c ] , [ 1 b a c f n
xk [a,b]
) ( ) ( )
(x y x R x
f n y(x)
) (x f ) )...( )( ( ... ) )( ( ) ( )
(x a0 a1 xx0 a2 xx0 xx1 an xx0 xx1 xxn1 y
) ( ) (x y x
f Rn(x)
) )( )....( )( ( )! ( )
( 0 1 1
1 1 n n n n
n x x x x x x x x
x f x
R
4 4 3 4 2 4 4 4 3 3 3 2 3 3 4 2 3 2 2 2 2 4 1 3 1 2 1 1 4 3 2 6 3 5 3 4 3 3 3 2 3 3 6 2 5 2 4 2 3 2 2 2 2 30 20 12 6 2 0 0 30 20 12 6 2 0 0 30 20 12 6 2 0 0 30 20 12 6 2 0 0 30 20 12 6 2 0 0 1 1 n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x (7) (8) (9) (10)
This gives a continuous scheme
(11)
where:
Evaluating (11) at i.e. when , gives
(12)
(13)
(14) Evaluating the first derivative of (4,1) at t = -3, gives
t
2
) 1 (
3 t
t t t t t
h
t 2 6 5 20 11
1440 )
( 6 5 4 3
2
0
t t t t t
h
t 2 9 5 30 18
360 )
( 6 5 4 3
2
1
t t t t t
ht 2 12 5 60 55
240 )
( 6 5 4 3
2
2
t t t t t t
h
t 2 15 25 50 180 118
360 )
( 6 5 4 3 2
2
3
t t t t t
ht 2 18 55 60 21
1440 )
( 6 5 4 3
2
4
h x x t n3
4
n
xx t1,2,3
) 4
14 204
19 ( 240
480
240 4 3 2 1
2 2 3
4 n n n n n n n
n y y h f f f f f
y
) 24
194 24
( 240
480
240yn3 yn2 yn1 h2 fn4 fn3 fn2 fn1 fn
) 17 252
402 52
3 ( 240
720
480yn3 yn2 yn1 h2 fn4 fn3 fn2 fn1 fn
) 475 1908
966 284
33 ( 1440
1440
Solving (12)-(16) using block method gives
(16) Hence, from (16)
' ' 3 ' 2 ' 1 3 2 1 2 4 3 2 1 2 3 2 1 4 3 2 1 4 0 0 0 3 0 0 0 2 0 0 0 1 0 0 0 720 869 0 0 0 160 147 0 0 0 90 53 0 0 0 1440 367 0 0 0 80 57 45 577 120 191 12 49 160 9 8 3 80 27 40 117 30 1 45 8 3 1 5 8 480 7 360 29 240 47 8 3 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 n n n n n n n n n n n n n n n n n n n n y y y y h f f f f h f f f f h y y y y y y y y
' 4 3 2 1 21 540 282 116 21 367
1440 n n n n n
n
n f f f f f hy
h y
y
' 4 3 2 1 22 144 30 16 3 53 2
90 n n n n n
n
n f f f f f hy
h y
y
' 4 3 2 1 23 468 54 60 9 147 3
160 n n n n n n
n
n f f f f f hy
h y
y
'3 2
1 2
4 192 48 64 56 4
45 n n n n n
n
n f f f f hy
h y
y
(17)
(18)
(19)
Developing the unknown for k=4
We evaluate the first derivative of (8), and neglecting and higher values.
Solving (24)-(28) for , i = 1(1)4 gives,
Substituting for in (15) and substituting in (16)-(19) and solving for ,
gives,
6
a
n n
n n
n n
n y y y y y y
hy 12 75 120 220 300 137
60 ' 5 4 3 2 1
n n
n n
n n
n y y y y y y
hy 3 20 60 120 65 12
60 '1 5 4 3 2 1
n n
n n
n n
n y y y y y y
hy 2 15 60 20 30 3
60 ' 5 4 3 2 1
2
n n
n n
n n
n y y y y y y
hy 3 30 20 60 15 2
60 '3 5 4 3 2 1
n n
n n
n n
n y y y y y y
hy 15 65 120 60 20 3
60 '4 5 4 3 2 1
) 2259 247
2382 420
5386 ( 6348
' '
4 '
3 '
2 '
1
1 n n n n n n
n y y y y y
h y
y
) 174 12
135 288
705 ( 529
' '
4 '
3 '
2 '
1
2 n n n n n n
n y y y y y
h y
y
) 729 105
1578 2520
2598 ( 2116
' '
4 '
3 '
2 '
1
3 n n n n n n
n y y y y y
h y
y
) 504 476
2580 1272
2188 ( 1587
' '
4 '
3 '
2 '
1
4 n n n n n n
n y y y y y
h y
y
2
n
y fni i1(1)4
) 48 37 48
1 6
1 4
3 6
1 ( 1 4
1 ' '
4 '
3 '
2 '
1
1 n n n n n n
n y y y y y
h f
f
(21) (22)
(23)
(24)
(25)
(26)
(27)
(28)
Substituting (25)-(28) into the second derivative of (8) and comparing with (29)-(32) give
Analysis of the scheme Convergence
The convergence analysis of this scheme is determined using the approach of Fatunla (1992), where each block integrator is represented as a single step block r- point multistep method of the form
where: are r by r matrix respectively with element
specifically, is an identity matrix,
are vectors of numerical estimate describe below. With r-vector
) 48 29 16
3 6
7 4
9 2
3 ( 1 41
1 ' '
4 '
3 '
2 '
1
3 n n n n n n
n y y y y y
h f
f
) 3 7 3
7 3
16 6
3 16 (
1 ' '
4 '
3 '
2 '
1
4 n n n n n n
n y y y y y
h f
f
n n
n y hf
y'1 ' 0.222193
n n
n y hf
y'2 ' 0.111768
n n
n y hf
y'3 ' 1.188485
n n
n y hf
y'4 ' 1.311393
(31)
(32)
(33) (34)
(35)
Awoyemi (2009) for details
Analysis of the block
det
det
Hence the block is zero stable.
Numerical example
We test the efficiency of our scheme on linear and non linear second order differential equation.
Problem 1:
0 1 0 0 0
1 0 0 0
1 0 0 0
1 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
0 1 0
0 0
1 0
0
1 0 0
1 0 0
0 ) 1 (
3
1 , 0 , 0 , 0
0 ) ( ' 2
''
x y y
' 1
(0) 1, (0) , 0.1/ 40
2
S/N Expected solution New method error
1 1.00125000065104 1.001250000186 4.650D-10
2 1.00250000520835 1.0025000039971 1.211D-09
3 1.00375001757828 1.003750013474 4.030D-09
4 1.00500004166729 1.005000047982 6.314D-09
5 1.00625008138212 1.0062500803586 8.462D-09
6 1.00750014062974 1.00750002579038 1.148D-09
7 1.00875022331755 1.008750239662 1993D-09
8 1.01000033335333 1.0100000783829 2.550D-09
9 1.01125047464542 1.0112504890376 4.256D-09
10 1.01250065110271 1.0125006101014 4.100D-08
Problem 11
x
xe y y'' 3
40 / 1 . 0 , 32
5 ) 0 ( , 32
3 ) 0
( y' h
y
) 3 exp( 32
3 4 ) (
x x x
y
Exact solution
S/N Expected result y(x) New method yn(x) Error y(x)-yn(x)
1 -0.094140915761848 -0.0941409131568 2.61D-09
2 -0.094532404142338 -0.0945324074753 3.33D-09
3 -0.094924451608388 -0.0949244224215 2.92D-08
4 -0.095317044390700 -0.0953170247449 2.02D-08
5 -0.095710168480980 -0.0957101793552 1.08D-08
6 -0.096103809629113 -0.0961039982252 1.88D-07
7 -0.09649533403163 0.0964952920353 4.23D-08
8 -0.096892584872264 -0.0968923659413 2.18D-07
9 -0.097289689232184 -0.0972874625827 2.22D-06
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