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

112 CHAPTER 5. RUNGE-KUTTA METHODS
The remaining coefficients are left as an exercise. The result is
1/2 − √ 3/6 1/4 1/4 − √ 36
1/2 + √ 3/6 1/4 + √ 3/6 1/4
1/2 1/2
(5.205)
The method just computed is A-stable, because the region of stability is precisely
the left-hand plane. It is also a fourth-order method, because it has two stages and
was derived using the Gauss-Legendre method. The absolute stability condition
gives
∣ y n+1 ∣∣∣ 1 + z/2 + (1 + √ 3)z 2 /12
∣ =
y n ∣ 1 − z/2 + z 2 /12 ∣ < 1 (5.206)
Because of the extra computation time needed for implicit methods they are more
expensive than comparable order/stage explicit methods. On the other hand, they
are much easier to derive and many of them are suitable for stiff problems. According
to [13] the following three-stage sixth order method is “probably the largest that is
consistent with reasonable implementation costs:”
1/2 − √ 15/10 5/36 2/9 − √ 15/15 5/36 − √ 15/30
1/2 5/36 + √ 15/24 2/9 5/36 − √ 15/24
1/2 + √ 15/10 5/36 + √ 15/30 2/9 + √ 15/15 5/36
5/18 4/9 5/18
(5.207)
5.9 Páde Approximants and A-Stability
Definition 5.1. Let
c
A
b T (5.208)
be a Runge Kutta method. Then
R(z) =
y n
y n−1
(5.209)
is called the Stability Function for the method. In terms of this definition, the
requirement for absolute stability is that |R(z)| ≤ 1.
Theorem 5.2. The stability function for any Runge-Kutta method c A b T is
R(z) = 1 + zb T (I − zA) −1 1 (5.210)
z = hλ, and
⎛ ⎞
1
⎜ ⎟
1 = ⎝.
⎠ (5.211)
1
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007