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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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(1 + hL)B - 1
|En |< ----------y----- h. (ii)
Equation (ii) gives a bound for | En | in terms of h, L, n, and . Notice that for a fixed h, this error bound increases with increasing n; that is, the error bound increases with distance from the starting point t0.
(c) Showthat (1 + hL)B < enhL; hence
eBhL - 1 e(tn-t0)L - 1
I En | < L h = ----------l------h.
For a fixed point t = t0 + nh [that is, nh is constant and h = (t - t0)/n] this error bound is of the form of a constant times h and approaches zero as h ^ 0. Also note that for nhL = (t - t0)L small the right side of the preceding equation is approximately nh2 = (t - ^)h, which was obtained in Eq. (24) by an intuitive argument.
24. Derive an expression analogous to Eq. (21) for the local truncation error for the backward Euler formula.
Hint: Construct a suitable Taylor approximation to () about t = tn+1.
> 25. Using a step size h = 0.05 and the Euler method, but retaining only three digits throughout the computations, determine approximate values of the solution at t = 0.1, 0.2, 0.3, and
0.4 for each of the following initial value problems.
(a) = 1 - t + 4 y, y(0) = 1
(b) = 3 + t - y, y(0) = 1
(c) = 2 - 3t, y(0) = 1
Compare the results with those obtained in Example 1 and in Problems 1 and 3. The small differences between some of those results rounded to three digits and the present results are due to round-off error. The round-off error would become important if the computation required many steps.
26. The following problem illustrates a danger that occurs because of round-off error when nearly equal numbers are subtracted, and the difference then multiplied by a large number. Evaluate the quantity
1000-
6.010 18.04
2.004 6.000
as follows.
(a) First round each entry in the determinant to two digits.
(b) First round each entry in the determinant to three digits.
(c) Retain all four digits. Compare this value with the results in parts (a) and (b).
27. The distributive law a(b - c) = ab - ac does not hold, in general, if the products are
rounded off to a smaller number of digits. To show this in a specific case take a = 0.22,
b = 3.19, and c = 2.17. After each multiplication round off the last digit.
430
Chapter 8. Numerical Methods
8.2 Improvements on the Euler Method
Since for many problems the Euler method requires a very small step size to produce sufficiently accurate results, much effort has been devoted to the development of more efficient methods. In the next three sections we will discuss some of these methods. Consider the initial value problem
= f (t, y), y(t0) = (1)
and let y = () denote its solution. Recall from Eq. (10) of Section 8.1 that by
integrating the given differential equation from tn to tn+1 we obtain
f n+i
(+1) = () + / f[t,(t)] dt. (2)
Jtn
The Euler formula
+1 = + hf(tn, ) (3)
is obtained by replacing f [t, ()] in Eq. (2) by its approximate value f (tn, yn) at the
left endpoint of the interval of integration.
Improved Euler Formula. A better approximate formula can be obtained if the integrand in Eq. (2) is approximated more accurately. One way to do this is to replace the integrand by the average of its values at the two endpoints, namely, { f [tn, ()] + f [tn+1, (+1)]}/2. This is equivalent to approximating the area under the curve in Figure 8.2.1 between t = tn and t = tn+1 by the area of the shaded trape-zoid. Further, we replace () and (^+1) by their respective approximate values yn and yn 1 . In this way we obtain from Eq. (2)
, f (tn, ) + f(tn+1 +1) u +1 = + ----------------2-------------- (4)
Since the unknown yn+1 appears as one of the arguments of f on the right side of Eq. (4), this equation defines yn+1 implicitly rather than explicitly. Depending on the nature of the function f, it may be fairly difficult to solve Eq. (4) for yn+1. This
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