27 yn rzyn 3210a 1 where 2 p x jt j1 rz

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4, we have 2 p R(z) = 1 + z + z2 + : : : + z ! s = p 4: p 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 s z2 + : : : + zp + X zj bT Aj 1l 1: jR(z)j = 1 + z + 2 (3.2.11a) p! j=p+1 ; 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...
View Full Document

Ask a homework question - tutors are online