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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
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
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
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...
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