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

CHAPTER 5. RUNGE-KUTTA METHODS 113
=⇒
Proof. When applied to the test equation y ′ = λy the general Runge-Kutta method
has
ν∑
K i = λy n−1 + z a ji K i , = 1, 2, . . . , ν (5.212)
i=1
y n = y n−1 + h ∑ b i K i (5.213)
where z = hλ. Let us define the following vector in R ν :
⎛ ⎞
K 1
⎜ ⎟
ξ = ⎝ . ⎠ (5.214)
K ν
Then we can rewrite equation 5.212 as
ξ = λ1y n−1 + zAξ ⇐= (5.215)
Rearranging and solving for ξ, ⇐=
ξ − zAξ = λ1y n−1 (5.216)
(I − zA)ξ = λ1y n−1 (5.217)
ξ = λ(I − zA) −1 1y n−1 (5.218)
hence ⇐=
ν∑
y n = y n−1 + h b j K j (5.219)
which is equivalent to the desired result.
j=1
Theorem 5.3. The stability function satisfies
= y n−1 + hb T ξ (5.220)
= ( 1 + zb T (I − zA) −1 1 ) y n−1 (5.221)
R(z) = det(I − zA + z1bT )
det(I − zA)
(5.222)
Proof. For the test problem y ′ = λy, ⇐=
(I − zA)ξ = λ1y n−1 (5.223)
−hb T ξ + y n = y n−1 (5.224)
hence
( ) ( ) ( )
I − zA 0 ξ
λ1
−hb T = y
1 y n−1
n 1
(5.225)
The result follows from Cramer’s Rule (solve for y n ).
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge