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JST Vol. 21 (1) Jan. 2013 - Pertanika Journal - Universiti Putra ...

JST Vol. 21 (1) Jan. 2013 - Pertanika Journal - Universiti Putra ...

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Solving Delay Differential Equations by Using Implicit 2-Point Block Backward Differentiation Formula<br />

and then solving,<br />

U× E= X<br />

(<strong>21</strong>)<br />

to get the matrix value of E . Finally, the approximation values of y n+<br />

1 and y n+<br />

2 can be<br />

obtained. Furthermore, the existence of the delay term is the main difference between ODEs<br />

and DDEs. Thus, in order to get the numerical solutions for DDEs, the delay term needs to be<br />

solved in advance. For a situation where α ≤ t , the initial function φ 0<br />

( t)<br />

is used to calculate<br />

the delay term, y ( α ) . Otherwise, the delay term is calculated by using Newton Divided<br />

Difference interpolation.<br />

Meanwhile, the formula to calculate the delay term (Richard et al., 1993) can be written as:<br />

where zx [ ] = zx ( ) .<br />

n<br />

P( x) = zx [ ] + å zx [ , x, ¼, x]( x-x) ¼( x-x - )<br />

(22)<br />

n 0 0 1 k o k 1<br />

k=<br />

1<br />

0 0<br />

Equation [22] is known as the Newton’s interpolatory divided-difference formula.<br />

Normally, the delay terms are obtained by interpolation on the values of the chosen support<br />

points. In order to obtain these support points, the delay argument should be close to the centre<br />

of the points. However, if the delay argument under consideration is outside the points, it will<br />

be treated as a special case and it can only be obtained through extrapolation.<br />

PROBLEM TESTED<br />

The followings are some of the tested problems to get the numerical results. All the tested<br />

problems are stiff DDEs. The numerical results are obtained by using C language with constant<br />

stepsize, H.<br />

Problem 1:<br />

( )<br />

( ) ( ) ( )<br />

'<br />

y t yt yt t<br />

( )<br />

=- 1000 + -(ln 1000- 1 ) , 0 £ £ 3,<br />

-t<br />

yt= e , t£<br />

0.<br />

t<br />

Exact solution is yt ( ) e .<br />

-<br />

=<br />

Problem 2:<br />

- -<br />

( ) 1000 ( ) 997 ( 1) ( 1000 99 )<br />

' 3 3<br />

y t yt e yt e t<br />

=- + - + - 7 , 0 £ £ 3,<br />

( ) ( )<br />

yt= 1+ exp - 3 t, t£<br />

0.<br />

<strong>Pertanika</strong> J. Sci. & Technol. <strong>21</strong> (1): 283 - 298 (<strong>2013</strong>)<br />

41

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