1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 81

on the interval [O, l].

1 1

IEI :S:

4

, max lfC^3 l(x, y)l(.1)^4 < 1 9(.1)^4 < .0000375.

. o::;x:s:i 4.

The following example illustrates how to estimate an appropriate stepsize
for a Taylor series approximation given a specific accuracy requirement per
step.

EXAMPLE 2 Stepsize Selection for the Third Order

Taylor Series Method

When using a Taylor series expansion of order 3 with constant stepsize h

to approximate the solution of the IVP (7) y' = y + x; y(O) = 1 on the

interval [O, 1], how small must the stepsize be in order to ensure six decimal
place accuracy per step?

SOLUTION

Setting m = 3 in equation (6), taking the absolute value of the resulting

equation, and using the inequality (8), we find on the interval [O, 1] that the
loca l discretization error satisfies

1 1

IEI :S: - max IJC^3 l(x y)lh^4 < -(9)h^4.

4! o::;x:s:i ' 4!

If we require h to satisfy

(9)

1

JEI <

4

! (9)h^4 < .5 x 10 -^5 ,

then the local discretization error will have six decimal place accuracy. Solving
the right-hand inequality in (9) for h, we find

4 6 1

h < (3 x 10-)• < 0.033.

Hence, any constant stepsize less than 0.033 will achieve the desired accuracy
per step.

The computational disadvantage of using a Taylor series expansion to ap-
proximate the solution to an initial value problem is fairly obvious. For any
given function f(x, y) one must calculate the derivatives jCll, j<^2 ), ... , f(m-l)
and then evaluate all of theses derivatives at (xn, Yn)· However, the Taylor
series expansion is of theoretical value, since most other numerical approxima-
tion schemes were derived by attempting to achieve a given order of accuracy