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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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() ty(t) < A (s) ty(s) ds. (17)
It is now convenient to introduce the function U defined by
U(t) =f (s) ty(s) ds. (18)
Then it follows at once that
U (0) = 0, (19)
U(t) > 0, for t > 0. (20)
Further, Uis differentiable, and U'(t) = |() ty(t). 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) = (), 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 {} exist? In the example f and d f/y 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 f/y are assumed to be continuous only in the rectangle R: 111 < a, < 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 = (t) may contain points that lie outside of the rectangle R. Hence at the next stagein the computation of + 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 +1 (t) might be impossible.
To avoid this danger it may be necessary to restrict t to a smaller interval than
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) < M,
(t, y) in R.
2.8 The Existence and Uniqueness Theorem
y (a,b)
(-a, b)
(-a, -b) (a, -b)
FIGURE 2.8.3 Region of definition for Theorem 2.8.1.
We have mentioned before that
(0) = 0
for each n. Since f[t,(t)] is equal to '+1(), the maximum absolute slope of the graph of the equation y = 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,+ 1(t)] remains in R at least as long as R contains the wedgeshaped region, which is for t < b/ M. We hereafter consider only the rectangle D: t < h, < b, where h is equal either to a or to b/M, whichever is smaller. With this restriction, all members of the sequence {(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) , provided that M is not already equal to the maximum value of
If (t, y) I.
2. Does the sequence {(t)} converge? As in the example, we can identify (t) = () + [2(0 ()] + + [(t) 1() as the nth partial sum of the series
() + ? [+() (0]. (24)
The convergence of the sequence {(t)} is established by showing that the series (24) converges. To do this, it is necessary to estimate the magnitude
+ 1(t) (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.
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 , so that
() = lim (t). (25)
3. What are the properties of the limit function ? In the first place, we would like to know that is continuous. This is not, however, a necessary consequence of the convergence of the sequence {(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 is continuous is to show not only that the sequence {} 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 {} and, hence, the continuity of the limit function in the interval | t | < h.
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