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FOUR STEPS IMPLICIT METHOD FOR THE SOLUTION OF GENERAL SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS

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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

(2)

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...,t1] ( ) ,

''

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

(3)

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 ) )...( )( ( ... ) )( ( ) ( )

(xa0a1 xx0a2 xx0 xx1  an xx0 xx1 xxn1 y

) ( ) (x y x

fRn(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)

(4)

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

h

t 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

h

t 2 18 55 60 21

1440 )

( 6 5 4 3

2

4     

h x x t  n3

4

n

xx t1,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

240yn3yn2yn1h2 fn4fn3fn2fn1fn

) 17 252

402 52

3 ( 240

720

480yn3yn2yn1h2  fn4fn3fn2fn1fn

) 475 1908

966 284

33 ( 1440

1440

(5)

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 2

1 540 282 116 21 367

1440 n n n n n

n

n f f f f f hy

h y

y    

' 4 3 2 1 2

2 144 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 2

3 468 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)

(6)

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  54321

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  54321

n n

n n

n n

n y y y y y y

hy 15 65 120 60 20 3

60 '454321

) 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 i1(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)

(7)

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)

(8)

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

(9)

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

(10)

REFERENCES

Adesanya, A.O., Anake, T.A., Bishop S. A., Osilagun, J.A. 2009. Two steps block

method for the solution of general second order initial value problems of ordinary dif-ferential equation. Asset International Jour-nal, University of Agriculture, Abeokuta

Awoyemi, D.O., Adesanya, A.O., Ogun-yebi, S.N. 2009. Construction of self

Start-ing Numerov method for the solution of initial value problems of general second or-der ordinary differential equation. Int. j. Num. Math, University of Ado Ekiti, 4(2): 269-278

Milne, W.E. 1953. Numerical solution of

differential equation, Wiley, New York.

Omar, Z. 1999. Developing parallel block

method for solving higher orders ODES directly. Thesis, University putra Malaysia

Rosser, J.B. 1967. A Runge Kutta for all

season, SIAM, P. 417-452.

Sham pine, L.F., Watts, H. 1969. Block

implicit one step method, international journal

of Math-comp 23: 731-740.

Worland, P.B. 1976. Parallel method for the

numerical solution of ordinary differential equation, IEEE trans. Computer C-25: 1045-1048.

Yahaya, Y. A., 2007. A note on the

con-struction of Numerov method through a quadratic continuous polynomial for the general second order ordinary differential equation. Journal of Nigerian Association of

mathematical physics, II: 253-260

Zanariah Abdul, Mohamed Bin, Zurni Omar 2006. 3 point implicit block method

for solving ordinary differential equation. Bulletin of the Malaysian Mathematical Sci-ence 2, 29(1):23-31

References

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