1550078481-Ordinary_Differential_Equations__Roberts_

(jair2018) #1
The Initial Value Problem y' = l(x, y); y(e) = d 89

(18) kz = l(xn + ehn, Yn + dk1hn) = l(xn, Yn) + ehnlx + dk1hnly+
ezh2 d2h2 kz
T lxx + edh;kifxy + 2n^1 lyy + O(h^3 ).

Substituting ki = l into (18), substituting the resulting equation into (16),
and rearranging in ascending powers of hn, we find

Comparing (17) with (19), we see that for the corresponding coefficients of
hn and h~ to agree, we must have

(20) a+ b = 1,

1
be=
2
,

1
and bd =
2
.

Thus, we have three equations in four unknowns. Hence, we might hope to
be able to choose the constants in such a manner that the coefficients of h~
in (17) and (19) agree. However, for these coefficients to agree we must have

be^2
2

1
6'

1
bed= -
3'

bd^2
2

1

6'

and lxly + l;l = 0.

Obviously, the last equality is not satisfied by a ll functions f.

There are an infinite number of solutions to the simultaneous equations (20).

The choice a = ~, b = ~, e = 1, and d = 1 yields the improved Euler's

method (14). Choosing a= 0, b = 1, e = ~' and d = ~results in the following

recursion which is known as the modified Euler's method:


(21)

Fourth Order Runge-Kutta Method If one tries to develop a general
recursion of the form


(22) Yn+i = Yn + hn[aif(xn, Yn) + a2l(xn + bihn, Yn + bihnk1)+
a3l(xn + b2hn, Yn + b2hnk2) + a4l(xn + b3hn, Yn + b3hnk3)]

where ki = l(xn, Yn) and ki = l(xn + bi-1hn, Yn + bi-1hnki-1) for i = 2, 3, 4

by determining the constants ai, az, a3, a4, bi, b2, and b3 in such a manner that
(22) will agree with a Taylor series expansion of as high an order as possible,
one obtains a system of algebraic equations in the constants. In this case, as
before, there are an infinite number of solutions to the system of equations.

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