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

122 CHAPTER 6. LINEAR MULTISTEP METHODS
that interpolates the points
P (t 0 ) = f(t 0 ) (6.40)
P (t 1 ) = f(t 1 ) (6.41)
.
P (t n−1 ) = f(t n−1 ) (6.42)
P (t n ) = f(t n ) (6.43)
We define the backward difference operator ∇ for an element f n of a sequence
as
∇f n = f n − f n−1 (6.44)
∇ 2 f n = ∇f n − ∇f n−1 = f n − 2f n−1 + f n−2 (6.45)
∇ 3 f n = ∇ 2 f n − ∇ 2 f n−1 = f n − 3f n−1 + 3f n−2 − f n−3 (6.46)
.
∇ k f n = ∇ k−1 f n − ∇ k−1 f n−1 (6.47)
Letting f n = f(t n ) we have by substituting 6.43 into 6.39 that
From 6.42 we get
Substituting at t = t n−2 = t n − 2h gives
f n = a 0 (6.48)
f n−1 = f n + a 1 (t n−1 − t n ) (6.49)
= f n − a 1 h (6.50)
a 1 = 1 h (f n − f n−1 ) = 1 h ∇f n (6.51)
f n−2 = a 0 + a 1 (t n−2 − t n ) + a 2 (t n−2 − t n )(t n−2 − t n−2 ) (6.52)
= f n + 1 h (f n − f n−1 )(−2h) + a 2 (−2h)(−h) (6.53)
= 2f n−1 − f n + 2h 2 a 2 (6.54)
a 2 = 1
2h 2 (f n − 2f n−1 + f n−2 ) = 1
2h 2 ∇2 f n (6.55)
Continuing the process we find in general that
Next we define the polynomials Q k by
a k = 1
k!h k ∇k f n (6.56)
k−1
∏
Q k (t) = (t − t n−j ) (6.57)
j=0
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007