In this problem we discuss the global truncation error associated with the Euler

Chapter 8, Problem 23

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In this problem we discuss the global truncation error associated with the Euler methodfor the initial value problem y = f(t, y), y(t0) = y0. Assuming that the functions f and fyare continuous in a closed, bounded region R of the ty-plane that includes the point(t0, y0),it can be shown that there exists a constant L such that |f(t, y) f(t, y| < L|y y|, where(t, y) and (t, y) are any two points in R with the same t coordinate (see ofSection 2.8). Further, we assume that ft is continuous, so the solution has a continuoussecond derivative.(a) Using Eq. (20), show that|En+1||En| + h|f[tn, (tn)] f(tn, yn)| + 12h2|(tn)| |En| + h2, (i)where = 1 + hL and = max |(t)|/2 on t0 t tn.(b) Assume that if E0 = 0, and if |En| satisfies Eq. (i), then |En| h2(n 1)/( 1)for = 1. Use this result to show that|En| (1 + hL)n 1L h. (ii)Equation (ii) gives a bound for |En| in terms of h, L, n, and . Notice that for a fixed h, thiserror bound increases with increasing n; that is, the error bound increases with distancefrom the starting point t0.(c) Show that (1 + hL)n enhL; hence|En| enhL 1L h.If we select an ending point T greater than t0 and then choose the step size h so that nsteps are required to traverse the interval [t0, T], then nh = T t0, and|En| e(Tt0)L 1L h = Kh,which is Eq. (25). Note that K depends on the length T t0 of the interval and on theconstants L and that are determined from the function f

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