12 applied to the ivp 511 we begin by applying the

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Unformatted text preview: 1 : : : n ; 1, using (5.6.2), and using the normalization 0 = 1 (cf. (5.1.2c)), we nd y(tn) ; yn = y(tn) + Using (5.6.1) k X i=1 i y (tn;i) ; h 0 f (tn yn) ; h k X i=1 i f (tn;i y(tn) ; yn = h 0 f (tn y(tn)) ; f (tn yn)] + h n: 27 y(tn;i)): Assuming that f is continuously di erentiable and using the mean value theorem f (tn y(tn)) ; f (tn yn) = fy (tn n) y(tn) ; yn] is between yn and y(tn). Using (5.6.2), we have dn = 1 ; h hf n(t ) : 0y n n Thus, the local error is proportional to the local discretization error for all LMMs and the local error is the product of the step size and the local discretization errors for all explicit LMMs ( 0 = 0). where n De nition 5.6.3. A LMM is of order p if p is the largest integer for which Lh(y(tn)) = O(hp). De nition 5.6.4. A LMM is consistent if it is at least of order one. From De nition 5.6.4, consistency implies that the LMM must be exact when y(t) is a linear polynomial (Problem 5.1.1). Using (5.1.2) with y(t) = 1 and y(t) = t yields the explicit consistency conditions k X i=0 i =0 k X i=1 i i+ k X i=0 i = 0: (5.6.3) De nition 5.6.5. Consider solving (5.1.1) using (5.1.2) on 0 < t T with a sequence of meshes f0 = t0 < t1 < : : : < tN = T g such that N ! 1 and h = max1 n N (tn ; tn;1 ) ! 0. A LMM is convergent if yn ! y(tn), n 2 0 N ] uniformly as N ! 1 (h ! 0). Stability and its relation to consistency and convergence is not as simple for LMMs as for one-step methods. While a convergent multistep method is necessarily consistent, consistency alone is not su cient for convergence. Before tackling the general question of stability (absolute stability, etc.), let us consider a simple example. Example 5.6.1. Consider the solution of the test problem y0 = y y(0) = 1 by the leap frog method yn = yn;2 + 2hfn;1 = yn;2 + 2h yn;1 28 y0 = 1: A solution, calculated on 0 < t 4:5 with = ;2 and h = 1=32, is shown in Figure 5.6.1. The solution appears to be correct initially, but oscillations of increasing amplitude develop and, at the very least, the solution is not absolutely stable. The local discretization error of the leap...
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