The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 5. RUNGE-KUTTA METHODS 107
Consider for example the following 2-stage method:
0 1/4 −1/4
2/3 1/4 5/12
1/4 3/4
This gives the following iteration formulas:
(
K 1 = f t n−1 , y n−1 + h )
4 (K 1 − K 2 )
(
K 2 = f t n−1 + 2 3 h, y n−1 + h )
12 (3K 1 + 5K 2 )
(5.149)
(5.150)
(5.151)
y n = y n−1 + h 4 (K 1 + 3K 2 ) (5.152)
It is sufficient to assume that the differential equation is autonomous, that is, that
f(t, y) = f(y) only and does not depend explicitly on t. This is because it is
always possible to increase the dimension of a differential system by adding an extra
variable, whose derivative is 1. This variable is equal to t and including it makes the
system autonomous. Hence it is sufficient to study just autonomous systems. We will
treat a scalar autonomous differential equation; the corresponding derivations for
the vector system are analogous. The corresponding autonomous iteration formulas
are
(
K 1 = f y n−1 + h )
4 (K 1 − K 2 )
(5.153)
(
K 2 = f y n−1 + h )
12 (3K 1 + 5K 2 )
(5.154)
Expanding in a Taylor Series about y n−1 ,
y n = y n−1 + h 4 (K 1 + 3K 2 ) (5.155)
K 1 = f(y n−1 ) + h 4 (K 1 − K 2 )f y + h2
32 (K 1 − K 2 )f yy + O(h 3 ) (5.156)
K 2 = f(y n−1 ) + h 12 (3K 1 + 5K 2 )f y + h2
288 (3K 1 + 5K2)f yy + O(h 3 ) (5.157)
Hence K 1 , K 2 = f(y n−1 ) + O(h); substituting this back into 5.156 gives
K 1 = f(y n−1 ) + O(h 2 ) (5.158)
K 2 = f(y n−1 ) + 2 3 hf yf + O(h 2 ) (5.159)
Substituting equations 5.158 and 5.159 back into the right-hand sides of equations
5.156 and 5.157 gives
K 1 = f(y n−1 ) − 1 6 h2 f 2 y f + O(h 3 ) (5.160)
K 2 = f(y n−1 ) + 2 3 hf yf + h 2 ( 5
18 f 2 y f + 2 9 f yyf 2 )
+ O(h 3 ) (5.161)
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge