Unformatted text preview: 4, we have
R(z) = 1 + z + z2 + : : : + z !
s = p 4:
The exact solution of the test equation (3.2.8) is y(tn) = eh y(tn 1)
; 15 (3.2.10b) (3.2.10c) thus, as expected, a pth-order Runga-Kutta formula approximates a Taylor's series expansion of the exact solution through terms of order p.
Using De nition 2.1.7 and (3.2.10), the region of absolute stability of an explicit
Runge-Kutta method is
z2 + : : : + zp + X zj bT Aj 1l 1:
jR(z)j = 1 + z + 2
; In particular, jR(z)j = 1 + z + z2 + : : : + z !
p p 2 1 s = p 4: (3.2.11b) Since no Runge-Kutta coe cients appear in (3.2.11b), we have the interesting result. Lemma 3.2.1. All p-stage explicit Runge-Kutta methods of order p 4 have the same
region of absolute stability. Since jei j = 1, 0
2 , we can determine the boundary of the absolute stability
regions (3.2.11a,b) by solving the nonlinear equation R(z) = ei : (3.2.12) Clearly, (3.2.12) implies that jyn=yn 1j = 1. For p = 1 (i.e., for Euler's method), the
boundary of the a...
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- Spring '14
- Numerical Analysis, yn, Tn, Numerical ordinary differential equations