# 64 have unbounded absolute stability regions as reh 1

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ximal order of a k-step LMM is 2k since there are 2k +1 free parameters in the general formula (5.1.2) however, these maximal-order methods are typically unstable. Theorem 5.6.2. No k-step LMM satisfying the root condition can have order exceeding k + 1 when k is odd or order k + 2 when k is even. Proof. cf. Hairer et al. 12]. The k-step methods of order k +2 are called optimal-order methods. All roots of ( ) are on the unit circle, so they are only weakly stable. Example 5.6.5. The highest order two-step LMM yn = ;4yn;1 + 5yn;2 + 2h(2fn;1 + fn;2) has order four. Using (5.6.7b), the rst characteristic polynomial for this method is ( )= 2 + 4 ; 5: 39 The roots of this polynomial, 1 = 1 and 2 = ;5, do not satisfy the root condition hence, the method is unstable. Let us illustrate this by applying the method to the very simple equation y0 = 0 y(0) = 0: The exact solution of this IVP is, of course, the trivial solution y(t) = 0. For starting values, let us use y0 = 0 y1 = where represents a small rounding error. Successive solutions satisfy y2 = ;4y1 + 5y0 = ;4 y3 = ;4y2 + 5y1 = 21 y4 = ;4y3 + 5y2 = ;104 ::: : The step size h does not explicitly appear in the method for this IVP hence, the solution cannot be bounded as h ! 0 and, therefore, the method is unstable. The method also does not converge as h ! 0 thus, consistency alone is not su cient for convergence. Necessary and su cient conditions for convergence are given in the following theorem which is due to G. Dahlquist. Theorem 5.6.3. The necessary and su cient conditions for a LMM to be convergent are that it be consistent and satisfy the root condition. Proof. Complete proofs of this important theorem are given in Gear 10], Chapter 10, and Henrici 14], Chapter 5. We'll prove that convergence implies stability and consistency. Applying the LMM (5.1.2) to the IVP y0 = 0, y(0) = 0 yields k X i=0 i yn;i =0 Seeking to use a contradiction argument, suppose that ( ) has a root 2 with j 2j > 1 and/or a root 3 on the unit circle whose multiplicity exceeds unity. Fu...
View Full Document

## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online