THE EFFECT OF INVOLVING MORE DERIVATIVE PROPERTIES, ON THE ACCURACY OF AN IMPLICIT ONE-STEP METHOD

(1)

Leena and Luna International, Oyama, Japan.

((((株株株株) ) ) ) リナリナリナリナアンドアンドアンドアンドルナインターナショナルルナインターナショナルルナインターナショナル , , , , 小山市、日本ルナインターナショナル小山市、日本小山市、日本....小山市、日本

www.ajsc.leena-luna.co.jp P a g e | 42

THE EFFECT OF INVOLVING MORE DERIVATIVE PROPERTIES, ON THE ACCURACY OF AN IMPLICIT ONE-STEP METHOD

Famurewa O. K. E.¹, Olorunsola S. A.²

1Department of Mathematical Sciences, Wesley University of Science and Technology, Ondo,

2 Department of Mathematical Sciences, Ekiti State University, Ado-Ekiti, NIGERIA.

1[email protected]

ABSTRACT

This work describes the development of a class of implicit one-step multi-derivative methods of the form;

j in j

ij j

l

i i j

n

j y h β y

⁺

=

= =

+

∑

⁼

∑

¹

0 1

α

’

1=+1 α

with local truncation error

y β h y

α T

j

j in ij l

i i j

n j

j

n

∑ ∑ ∑

=

+

= +

=

+ = −

1

0 1 1

0

1

Also, it investigates the pattern of the order of accuracy of the methods with varying order of derivative. The development of the methods adopts the Taylor Series expansion of the functionsy_n₊_j,y'n+j

,y''n+j

, ---- . Accuracy of order P is imposed on Tn+1 and the resulting equations are solved for parameters αj’^s and βij’^sto generate the required methods (schemes). The order of accuracy and error constants of the methods are determined by substituting the results of parameters αj’^s and βij’^s into the original equations. The methods were implemented by using them to solve some initial value problems of first order ordinary differential equations in order to establish the pattern of the accuracy. The result showed that the order of accuracy of the methods increased as the order of the derivative increased with optimum accuracy obtained at l =3.

Keywords: Implicit, Multi-derivative, One-step, Order of accuracy, Derivative property

INTRODUCTION

According to Burden (2004), many problems encountered in the various branches of Science, Engineering and Management give rise to differential equations which can either be ordinary or partial (O .D. E. or P. D .E.). O.D.Es. occur when the dependent variable y is a function of a single independent variable x e.g

= − . Partial differential equations occur when the dependent variable v is a function of two or more independent variables, x and y e.g + = 0.

Various approximation methods for solving ordinary differential equations have been developed, they include; implicit family of linear multistep methods by the Adams (Hairer

(2)

www.ajsc.leena-luna.co.jp P a g e | 43 Ernst and Gerhard Wanner, 1996), defined as: n j

k

j j k

j

j y +

= +

=

∑

⁼ ^'

0 j n 0

h

y β

α

,

backward difference formula by Fatunla, (1988), explicit Runge-kutta method of the form; − = ℎ∅ ( , , ℎ), Famurewa et al, (2011), implicit family of linear multiderivative multistep

methods by Famurewa (2011), defined as in j

j ij j

l

i i j

n

j y h β y

+

=

= =

+

∑

⁼

∑

¹

0 1

α

’

1 =+1

α _and

so on.

Explicit one-step method is the easiest to implement, it requires no additional starting values and it readily permits a change of step-length during the computation (Lambert, 1991). Its low order of accuracy of course makes it of limited practical value. This is why linear multistep methods achieve higher order by sacrificing the one-step nature of the algorithm while retaining linearity with respect to , , ….

This study will show that higher order of accuracy can be achieved in a one-step method, by involving more analytical properties of the differential equation by way of more of the derivative properties of y (Famurewa et al., 2011).

This study will consider the development of a class of implicit one-step methods, for which the order of the derivative () is varied from = 1,2, … 5. Also, attempt will be made to determine the effect of increasing the order of derivative, on the order of accuracy of the one- step methods.

DERIVATION OF THE METHODS

A class of linear one-step methods of the form:

j in j

ij j

l

i i j

n

j y h β y

+

=

= =

+

∑

⁼

∑

¹

0 1

α

’

1 =+1 α

- - - (1.1) is developed by using the local truncation error formula:

y β h y

α T

j

j in ij l

i i j

n j

j

n

∑ ∑ ∑

=

+

= +

=

+ = −

1

0 1 1

0

1

- - - (1.2) to determine parameters αj’^s and βij’^sfor = 1,2,… ,5.

Consequently, it was assumed that the local truncation error for one-step application of the formula to equation (1.1) can be defined as;

β y h α y

T

j

j in ij l

i i j

n j

j

n

∑ ∑ ∑

=

+

= +

=

+ = −

1

0 1 1

0

1

Adopting the Taylor series expansion of in j

y + , j=0

( )

1, defined by

( ) ( )

^l

r y y

r

i r r j

n

i , i 01

! jh

0

=

∑

^∞

=

+ +

(Barrio, 2006)

in equation (1.2) and combine terms in equal powers of h to have:

i n i

i i n

n n

o

n C y Chy C h y Ch y

T

∑

^∞

=

+ = + + =

0 2 ''

2 ' 1

1 .... ……… (1.3)

(3)

where

∑

=

−

=

k

j j k

j

j j

C

0 1 1

1 α β

∑

=

−

=

k

j j k

j

j j j

j C

0 2 1

1 1

2

2 2!

1 α β β

∑

=

−

=

k

j j k

j

j j j

j C

0 2 1

1 2 1

3

3 2!

1

! 3

1 α β β

∑

=

−

=

k

j

j k

j j k

j

j j j

j C

0 2 2 1

1 3 1

4

4 2!

1

! 3 1

! 4

1 α β β

∑

=

−

=

k

j

j k

j

j k

j

j j j

j C

0 2 3 1

1 1

5

5 3!

4 1

! 4 1

! 5

1 α β β

( )

∑

=

−

=

−

= − −

− −

=

k

j

j P k

j

j P k

j j P

P j

j P j P

C P

1

2 2 1

1 1

1 2!

1

! 1 1

!

1 α β β

and

( ) ( )

( )

∑

=

−

=

= +

+ − − −

= +

k

j

j P k

j

j P k

j

j P

P j

j P j P

C P

1

2 1 1

1 1

1

1 1!

1

! 1

1 α β β

- - -(1.4) Setting l =1 in equation (1.1) gives

1 ) 1 ( 11 )

1 ( 10 1

1

0y_n + α y_n+ = hβ y _n + hβ y _n+

α - - - (1.5)

with local truncation error

1 ) 1 ( 11 )

1 ( 10 1

1 0

1 + +

+ = _n + _n − _n − _n

n y y h y h y

T α α β β - - (1.6)

Adopting the Taylor’s expansion of yn+1 and y¹n+1 given as ) ( 0 . .

! . 3

! 2

4 )

111 3 ( )

11 2 ( ) 1 (

1 h y h y h

hy y

y_n₊ = _n + _n _n + ⁿ + ⁿ + +

) ( 0 . .

! . 3

! 2

4 )

1 ( ) 3

111 2 ( ) 11 ) (

1 ( 1

1 h y h y h

hy y

y

V n

n n

n₊ = + + + + +

in equation (1.6) and combining terms in equal powers of h gives

( ) ( )

0( )

2 6 2

4 ) 111 ( 3 11 1 ) 11 ( 2 11 1 (1) 11 10 1 1

0

1 y hy h y h y h

T_n _n _n _n  _n+



 



 −

 +



 



 −

+

−

− + +

+ =

β β α

β α β α α

α - (1.7)

that is;

) ( 0 . .

. ⁴

) 111 ( 3 3 ) 11 ( 2 2 ) 1 ( 1 0

1 C y C hy C h y C h y h

T_n₊ = _n + _n + _n + _n + + where

(4)

11 1

3

11 1 2

11 10 1 1

1 0 0

12 6 2 α β α β

β β α

α α

−

=

−

=

−

= +

=

C C C C

- - - (1.8)

Imposing accuracy of order 2 on Tn+1 to have C0 = C1 = C2 =0 and Tn+1 = 0(h³) that is;

2 0 1

0 0

11 1

11 10 1

1 0

=

−

=

−

= +

β α

β β α

α α

where α₁ =+1

gives

















−

=

































−

2 1 11

10 0

1 1

1 0 0

1 1 0

0 0 1

β β α

Solving gives

α0 = -1, β10 = ½, and β11 = ½.

putting these values into equation (1.5) gives the one – step, first derivative method of the form;

= +(⁽⁾ + ⁽⁾) - - - - (1.9) which coincides with the trapezoidal rule.

Setting l=2 in equation (1.1) gives

( ) (

21 ⁽¹¹⁾1

)

) 11 ( 20 2 1 ) 1 ( 11 )

1 ( 10 1

1

0y_n + α y_n+ = hβ y n + β y _n+ + h β y _n + β y_n+

α - - (1.10)

( ) (

21 ⁽¹¹1⁾

)

) 11 ( 20 2 ) 1 (

1 11 )

1 ( 10 1

1 0

1 + + +

+ = _n + _n − _n + _n − _n + _n

n y y h y y h y y

T α α β β β β - -

(1.11)

Adopting Taylor’s expansion of , ⁽¹⁾ !" ⁽⁾ in equation (1.11) and combining terms in equal powers of h gives

(5)

( ) ( )

) ( 6 0

24 120 2

6 24

2 6 2

6 ) ( 21 5 11 ) 1

( 21 4 11 1

) 111 ( 3 21 11 ) 1

11 ( 2 20 11 ) 1

1 ( 11 10 1 1

0 1

h y

h

y h y

h hy

y T

V n IV

n

n n

n

 +



 



 − −

 +



 



 − −

+



 



 − −

 +



 



 − −

+

−

− + +

+ =

β β α β

β α

β β β α

α β β

β α α

α

that is

) ( 0 . .

. ⁴

) 111 ( 3 3 ) 11 ( 2 2 ) 1 ( 1 0

1 C y C hy C h y C h y h

T_n₊ = _n + _n + _n + _n + +

where

6 24 5 120

2 1 6

1 24 4 1

12 6 1 12

21 11

1

21 11

1

21 11 1

3

21 20 11 1 2

11 10 1 1

1 0 0

β β

α

β β

α

β β α

β β β α

β β α

α α

−

=

−

=

−

=

−

=

−

= +

=

C C C C C C

- - - - - (1.13)

Imposing accuracy of order 5 on Tn+1 to have C0 = C1 = C2 = C3 = C4 = 0 and Tn+1= 0(h⁵) This gives













−

=





































−

24 1 6 1 2 1

21 20 11 10 0

2 1 6

1 2 1

1 1

0 0

0

1 0 0

0

1 1 1 0 0

0 0 1 1 0

0 0 0 0 1

β β β β α

Solving gives

112 112

12 12 ,

, 1

1 ₁₀ ₁₁ ₂₀ ₂₁

0=− β = β = β = β = −

α and

Putting these values into equation (1.10) gives a one – step, second derivative method of the form;

(

n n

) (

n n

)

n

n h y y

y h y

y

y (11) 1 ⁽¹¹⁾

2 )

1 ( 1 ) 1 (

1 = + 2 + + − 12 + −

+ … (1.14)

( ) ( ) (

31⁽¹¹¹1⁾

)

) 111 ( 30 3 ) 11 (

1 21 ) 11 ( 20 2 ) 1 (

1 11 ) 1 ( 10 1

1

0

y

_n

+ α y

_n+

= h β y

_n

+ β y

_n+

+ h β y

_n

+ β y

_n+

+ h β y

_n

+ β y

_n+

α

..(1.15)

- - - - (1.12)

(6)

( ) ( ) (

31 ⁽¹¹¹1⁾

)

) 111 ( 30 3 ) 11 (

1 21 ) 11 ( 20 2 ) 1 (

1 11 ) 1 ( 10 1

1 0

1 + + + +

+

=

_n

+

_n

−

_n

+

_n

−

_n

+

_n

−

_n

+

_n

n

y y h y y h y y h y y

T α α β β β β β β

..(1.16) Adopting Taylor’s expansion of _, _, _!" in equation (1.1) and combining terms in equal powers of h gives

( ) ( )

) 17 . 1 ( )

(

! 0 4

1

! 5

1

! 6

1

! 7

1

! 3

1

! 4 1

! 5

1

! 6

1 2

1

! 3

1

! 4

1

! 5

1

2 1

! 3

1

! 4 1 2

1

! 3

1

2 1

8 )

( 7 31 21

11 1

) ( 6 31 21

11 1

) ( 5 31 21

11 1

) ( 4 31 21 11

1 )

111 ( 3 31 30 21 11 1

) 11 ( 2 21 20 11 1 )

1 ( 11 10 1 1

0 1

−

 +







 − − −

+







 − − −

 +







 − − −

+







 − − −

 +







 − − − −

+



 



 − − −

+

−

− + +

+ =

h y

h

y h y

h

y h y

h

y h hy

y T

VII n

VI n V

n

IV n n

n n

β β

β α

β β

β α

β β

β α

β β β

α β

β β β α

β β β α β

β α α

α

That is

) ( 0 . . .

| ₂ ² ⁽¹¹⁾ ₃ ³ ⁽¹¹¹⁾ ⁴ ⁴ ⁽ ⁾ ⁵

) 1 ( 1 0

1 C y C hy C h y C h y C h y h

T_n₊ = _n + _n + _n + _n + _n^IV +

Where

31 21

11 1

7

31 21

11 1

6

31 21

11 1

5

31 21 11

1 4

31 30 21 11 1

3

21 20 11 1 2

11 10 1 1

1 0 0

! 4 1

! 5

1

! 6

1

! 7 1

! 3 1

! 4

1

! 5 1

! 6

1

2 1

! 3

1

! 4 1

! 5 1

2 1

! 3 1

! 4 1

2 1

! 3 1 2 1

β β

β α

β β

β α

β β

β α

β β β

α

β β β β α

β β β α

β β α

α α

−

=

−

=

−

=

−

=

−

=

−

=

−

= +

=

C C C C C C C C

- - - (1.18)

Imposing accuracy of order 7 on Tn+1 to have C0 = C1 = C2 = C3 = C4 =C5 =C6= 0 and Tn+1 = 0(h⁷) That is

(7)





























−

=





















































−

720 1 120

1 24 1 6 1 2 1

31 30 21 20 11 10 0

6 1 24

1 120

1

2 1 6

1 24

1

2 1 6

1 2 1

1 1

0 0

1 0

0 0

0

1 10 1

0 0

0

0 0

1 1 1 0 0

0 0

0 0 1 1 0

0 0

0 0 0 0 1

β β β β β β α

Solving gives

α0 = -1, β10 = 0.5, β11 = 0.5, β20 = 0.1, β21 = - 0.1, β30 = 0.0083 and β31 = 0.0083.

Putting these values into equation (1.15) gives the one – step third derivative method of the form;

( ) ( ) (

⁽¹¹¹1⁾ ⁽¹¹¹⁾

)

3 )

11 ( ) 11 (

1 2 ) 1 ( ) 1 (

1

1 ⁿ 2 ⁿ ⁿ 10 ⁿ ⁿ 120 ⁿ ⁿ

n h y y

y h y

y

y ₊ = + ₊ + − ₊ − + ₊ + - - - (1.19)

( ) ( ) ( )

(

41 ⁽ 1⁾

)

⁽¹^.²⁰⁾

) ( 40 4

) 111 (

1 31 ) 111 ( 30 3 ) 11 (

1 21 ) 11 ( 20 2 1

) 1 ( 11 )

1 ( 10 1

1 0

−

− +

−

+

− +

= +

+

+ +

IV n IV

n

n n

n

y y

h

y y

h y

y h

y n

y h y y

β β

α α

With local truncation error

( ) ( ) ( )

(

41 ⁽ 1⁾

)

⁽¹^.²¹⁾

) ( 40 4

) 111 (

1 31 ) 111 ( 30 3 ) 11 (

1 21 ) 11 ( 20 2 ) 1 (

1 11 )

1 ( 10 1

1 0

1

−

− +

−

+

− +

=

+

+ +

+

IV n IV

n

n n

n

y y

h

y y

h y

y h y y

h y

y T

β β

α α

Adopting Taylor’s expansion of , , , !" #$ in equation (1.21) and combining terms in equal powers of h gives

( ) ( )

) 22 . 1 ( )

(

! 0 5 1

! 6 1

! 7 1

! 8

1

! 9

1

! 4 1

! 5

1

! 6 1

! 7

1

! 8

1

! 3 1

! 4

1

! 5

1

! 6

1

! 7

1

! 2 1

! 3

1

! 4 1

! 5

1

! 6

1

! 2 1

! 3

1

! 4

1

! 5

1

! 2 1

! 3

1

! 4 1 2

1

! 3

1

2 1

10 )

( 9 41 31

21 11

1

) ( 8 41 31

21 11

1 )

( 7 41 31

21 11

1

) ( 6 41 31

21 11

1 )

( 5 41 31 21

11 1

) ( 4 41 40 31 21 11

1 )

111 ( 3 31 30 21 11 1

) 11 ( 2 21 20 11 1 )

1 ( 11 10 1 1

0 1

−

 +







 − − − −

+







 − − − −

 +







 − − − −

+







 − − − −

 +







 − − − −

+







 − − − − −

 +







 − − − −

+



 



 − − −

+

−

− + +

+ =

h y

h

y h y

h

y h y

h

y h y

h

y h hy

y T

IX n

VIII n VII

n

VI n V

n

v IV n n

n n

β β

α

β β

α β

β β

β α

β β

α β

β β

β α

β β β β β

α β

β β β α

β β β α β

β α α

α

that is

) ( 0 . . .

| ₂ ² ⁽¹¹⁾ ₃ ³ ⁽¹¹¹⁾ ⁴ ⁴ ⁽ ⁾ ⁵

) 1 ( 1 0

1 C y C hy C h y C h y C h y h

T_n₊ = _n + _n + _n + _n + _n^IV +

(8)

www.ajsc.leena-luna.co.jp P a g e | 49 where

41 31

21 11

1 9

41 31

21 11

1 8

41 31

21 11

1 7

41 31

21 11

1 6

41 31 21

11 1

5

41 40 31 21 11

1 4

31 30 21

11 1

3

21 20 11 2 1

1 2

11 10 1 1

1 0 0

! 5 1

! 6 1

! 7

1

! 8

1

! 9 1

! 4 1

! 5

1

! 6

1

! 7 1

! 8 1

! 3

1

! 4 1

! 5

1

! 6 1

! 7 1

! 2 1

! 3 1

! 4

1

! 5

1

! 6 1

) 23 . 1

! ( 2 1

! 3

1

! 4 1

! 5 1

2 1

! 3 1

! 4 1

2 1

! 3 1

β β

α

β β

α

β β

α

β β

α

β β β

β α

β β β β β

α

β β β

β α

β β β α

β β α

α α

−

=

−

=

−

=

−

=

−

=

−

=

−

=

−

=

−

= +

=

C C C C C C C C C C

Imposing accuracy of order 8 on Tn+1 to have

C0 = C1 = C2 = C3 = C4 =C5 =C6 =C7 =C8 = 0 and Tn+1 = 0(h⁹) which gives

















−

=

















































−

! 8 1

! 7 1

! 6 1

! 5 1

! 4 1

! 3 1 2 1

41 40 31 30 21 20 11 10 0

! 4 1

! 5 1

! 6 1

! 7 1

! 3 1

! 4 1

! 5 1

! 6 1

2 1

! 3 1

! 4 1

! 5 1

2 1

! 3 1

! 4 1

2 1

! 3 1 2 1

1 1

0 0

0

0 0

0

0 0

0

1 0 0

0 0

0

1 1 1 0 0

0 0

0 0 1 1 1 0 0

0

0 0 0 0 1 1 1 0 0

0 0 0 0 0 0 1 1 0

0 0 0 0 0 0 0 0 1

β β β β β β β β α

solving gives

α0 = -1, β10 = 0.5, β11 = 0.5, β20 = 0.1071, β21 = - 0.1071, β30 = 0.0119, β31 = 0.0119 β40 = 0.0006, β41 = -0.0006.

Putting these values into equation (1.20) gives the one – step, fourth derivative method of the form;

( ) ( ) (

⁽¹¹¹1⁾ ⁽¹¹¹⁾

)

3 )

11 ( ) 11 (

1 2 )

1 ( ) 1 (

1

1 _n 0.5 _n _n 0.11 _n _n 0.01 _n _n

n y h y y h y y h y y

y ₊ = + ₊ + − ₊ − + ₊ +

- - (1.24)