By J. D. Lambert

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Show that the method is zero-unstable if b = -1. Illustrate the resulting divergence of the method with b = -I by applying it to the initial value problem y' = Y, y(O) = I and solving exactly the resulting difference equation when the starting values are Yo = 1, Yl = 1. ) 10. A given linear multistep method is defined by its first and second characteristic polynomials p(C) and ufO. Sequences of polynomials {piC)} and {Uj(C)},j = 1,2,3, .. , are constructed as follows: j= 1,2, .... Prove that the given linear IDultistep method will have order p if and only if p,(I) = 0,.

From inspection of the equations D, ~ D, ~ 0, it is clear that a solution for the set of equations DJ ~ 0, j ~ 0, 1,2,3, exists for all a, so that an order of at least 3 is always attainable. Moreover,by inspection it is clear that Dl and D4 cannot both vanish unless a = -1. SO' that when a #= -I, the maximum attainable order ·is 3. When a ~ -I, D, ~ 0 implies D. ~ 0, so that order of at least 4 is attainable. That order 5 is not attainable when a. ~ -1 follows from observing that D, and D, cannot both vanish unless" ~ + I.

1 - s). ~ 0, (0 - s). ~ 0, -,s)' - (3 - a)(1 - s) = _as' + (I + a)s, and for s E [1, 2], G(s) ~ (2 - s)'. Now _as' + (1 + a)s vanishes at s ~ 0 and at s ~ (I + a)/a, and the latter root lies in the interval (0, I) if and only if a < -I. Thus, for a;' -I, G(s) is of constant sign in [0, I] and is, in fact, non-negative there; since it is clearly non-negative in [1,2], it follows that G(s) does not change sign in [0,2] provided a ;. -I. For this range of a So' G(s)ds = s: [_as' + (I + a)s]ds + f (2 - s)'ds = i(5 + a).

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