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

CHAPTER 5. RUNGE-KUTTA METHODS 95
5.3 Traditional Runge-Kutta Methods
Simpson’s quadrature rule gives
∫ t
f(s, y(s))ds ≈ t − t 0
t 0
6
[
( t + t0
f(t 0 , y 0 ) + 4f , y
2
( t + t0
2
))
]
+ f(t, y(t))
(5.41)
and therefore
y n = y n−1 + h 6
[
f(tn−1 , y n−1 ) + 4f(t n−1/2 , y(t n−1/2 )) + f(t n , y n ) ] (5.42)
We can derive a three-stage explicit method by setting
k 1 = f(t n−1 , y n−1 ) (5.43)
k 2 = y n−1 + h 2 f(t n−1/2, k 1 ) (5.44)
k 3 = y n−1 + hf(t n , k 2 ) (5.45)
y n = y n−1 + h (k1 + 4k2 + k3) (5.46)
6
A better approximation performs a second round of function evaluations:
y n = y n−1 + h 6 (f(t n−1, k1) + 4f(t n−1/2 , k 2 ) + f(t n , k 3 )) ⇐= (5.47)
We obtain the traditional explicit 4-stage Runge-Kutta Method by splitting
up the term in the center:
y n = y n−1 + h 6
[
f(tn−1 , y n−1 ) + 2f(t n−1/2 , y(t n−1/2 )) + (5.48)
2f(t n−1/2 , y(t n−1/2 )) + f(t n , y n ) ] (5.49)
The iteration formulas are
k 1 = y n−1 (5.50)
k 2 = y n−1 + h 2 f(t n−1, k 1 ) (5.51)
k 3 = y n−1 + h 2 f(t n−1/2, k 2 ) (5.52)
k 4 = y n−1 + hf(t n−1/2 , k 3 ) (5.53)
y n = y n−1 + h (
f(tn−1 , k 1 ) + 2f(t
6
n−1/2 , k 2 ) + 2f(t n−1/2 , k 3 ) + f(t n , k 4 )) ) (5.54)
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge