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

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²ô(³) — ty(t)² < A ¥ ²ô(s) — ty(s)² ds. (17)

0

It is now convenient to introduce the function U defined by

U(t) =f ²ô(s) — ty(s)² ds. (18)

0

Then it follows at once that

U (0) = 0, (19)

U(t) > 0, for t > 0. (20)

Further, Uis differentiable, and U'(t) = |ô(¿) — ty(t)². Hence, by Eq. (17),

U'(t) — AU(t) < 0. (21)

Multiplying Eq. (21) by the positive quantity e~At gives

[e— AtU(t)]' < 0. (22)

Then, upon integrating Eq. (22) from zero to t and using Eq. (19), we obtain

e~AtU(t) < 0 for t > 0.

Hence U(t) < 0 for t > 0, and in conjunction with Eq. (20), this requires that U(t) = 0

for each t > 0. Thus U(t) = 0, and therefore ty(t) = ô(), which contradicts the

original hypothesis. Consequently, there cannot be two different solutions of the initial value problem for t > 0. A slight modification of this argument leads to the same conclusion for t < 0.

Returning now to the general problem of solving the integral equation (3), let us consider briefly each of the questions raised earlier:

1. Do all members of the sequence {ôï} exist? In the example f and d f/äy were continuous in the whole ty-plane, and each member of the sequence could be explicitly calculated. In contrast, in the general case, f and ä f/äy are assumed to be continuous only in the rectangle R: 111 < a, ²ó² < b (see Figure 2.8.3). Furthermore, the members of the sequence cannot as a rule be explicitly determined. The danger is that at some stage, say for n = k, the graph of ó = ôê(t) may contain points that lie outside of the rectangle R. Hence at the next stage—in the computation of ôê+ 1(t)—it would be necessary to evaluate f (t, y) at points where it is not known to be continuous or even to exist. Thus the calculation of ôê+1 (t) might be impossible.

To avoid this danger it may be necessary to restrict t to a smaller interval than

² 11 < a. To find such an interval we make use of the fact that a continuous function on a closed bounded region is bounded. Hence f is bounded on R; thus there exists a positive number M such that

f (t, y) ² < M,

(t, y) in R.

(23)

2.8 The Existence and Uniqueness Theorem

111

y (a,b)

(-a, b)

R

t

(-a, -b) (a, -b)

FIGURE 2.8.3 Region of definition for Theorem 2.8.1.

We have mentioned before that

ôï(0) = 0

for each n. Since f[t,ôê(t)] is equal to ô'ê+1(³), the maximum absolute slope of the graph of the equation y = ôê 1(t) is M. Since this graph contains the point (0, 0) it must lie in the wedge-shaped shaded region in Figure 2.8.4. Hence the point [t,ôê+ 1(t)] remains in R at least as long as R contains the wedgeshaped region, which is for ² t ² < b/ M. We hereafter consider only the rectangle D: ² t² < h, ²ó³ < b, where h is equal either to a or to b/M, whichever is smaller. With this restriction, all members of the sequence {ôï(t)} exist. Note that if b/M < a, then a larger value of h can be obtained by finding a better bound for If (t, y) ² , provided that M is not already equal to the maximum value of

If (t, y) I.

2. Does the sequence {ôï(t)} converge? As in the example, we can identify ôï(t) = ô() + [ô2(0 — ô()] + ••• + [ôï(t) — ôï—1() as the nth partial sum of the series

TO

ô() + ? [ôê+() — ôê(0]. (24)

ê=1

The convergence of the sequence {ôï(t)} is established by showing that the series (24) converges. To do this, it is necessary to estimate the magnitude

² ôê+ 1(t) — ôê(t) I of the general term. The argument by which this is done is

(a) (b)

FIGURE 2.8.4 Regions in which successive iterates lie. (a) b/M < a; (b) b/M > a.

112

Chapter 2. First Order Differential Equations

indicated in Problems 15 through 18 and will be omitted here. Assuming that the sequence converges, we denote the limit function by ô, so that

ô() = lim ô (t). (25)

3. What are the properties of the limit function ô? In the first place, we would like to know that ô is continuous. This is not, however, a necessary consequence of the convergence of the sequence {ôï(t)}, even though each member of the sequence is itself continuous. Sometimes a sequence of continuous functions converges to a limit function that is discontinuous. A simple example of this phenomenon is given in Problem 13. One way to show that ô is continuous is to show not only that the sequence {ôï} converges, but also that it converges in a certain manner, known as uniform convergence. We do not take up this question here but note only that the argument referred to in paragraph 2 is sufficient to establish the uniform convergence of the sequence {ôï} and, hence, the continuity of the limit function ô in the interval | t | < h.

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