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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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It follows from Eq. (13) that 0n ( t) is the nth partial sum of the infinite series
to ,2k
? kr; (15)
k1
hence lim 0n (t) exists if and only if the series (15) converges. Applying the ratio test, we see that, for each t,
t2k+2
k!
(k + 1)! t
2k
2
k + 1
—>- 0 as k oo;
(16)
2.8 The Existence and Uniqueness Theorem
109
thus the series (15) converges for all t, and its sum 0(t) is the limit of the sequence {0n(t)}. Further, since the series (15) is a Taylor series, it can be differentiated or integrated term by term as long as t remains within the interval of convergence, which in this case is the entire t-axis. Therefore, we can verify by direct computation that
TO
0(t) = t2k / k! is a solution of the integral equation (9). Alternatively, by substituting
k=1
0(t) for y in Eqs. (8), we can verify that this function satisfies the initial value problem.
In this example it is also possible, from the series (15), to identify 0 in terms of
2
elementary functions, namely, 0( t) = el — 1. However, this is not necessary for the discussion of existence and uniqueness.
Explicit knowledge of 0(t) does make it possible to visualize the convergence of the sequence of iterates more clearly by plotting 0(t) — 0k(t) for various values of k. Figure 2.8.2 shows this difference for k = 1,..., 4. This figure clearly shows the gradually increasing interval over which successive iterates provide a good approximation to the solution of the initial value problem.
-1.5 -1 -0.5 0.5 1 1.5 t
FIGURE 2.8.2 Plots of 0 (t) — 0k(t) for Example 1 for k = 1,..., 4.
Finally, to deal with the question of uniqueness, let us suppose that the initial value problem has two solutions 0 and ?. Since 0 and ? both satisfy the integral equation (9), we have by subtraction that
0(t) — ?(t) = ( 2s[0(s) — ?(s)] ds.
Jo
Taking absolute values of both sides, we have, if t > 0,
\0(t) — ?(t)\ =
f 2s[0(s) — ?(s)] ds < f 2s\0(s) — ?(s)\ ds. Jo Jo
110
Chapter 2. First Order Differential Equations
If we restrict t to lie in the interval 0 < t < A/2, where A is arbitrary, then 2t < A, and
I0(t) - f(t)\ < A \0(s) - &(s)\ ds. (17)
J0
It is now convenient to introduce the function U defined by
U(t) =f |0(s) - ^(s)\ ds. (18)
J0
Then it follows at once that
U(0) = 0, (19)
U(t) > 0, for t > 0. (20)
Further, Uis differentiable, and U'(t) = \0(t) - &(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) = 0(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 {0n} exist? In the example f and d f/dy 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 d f/dy are assumed to be continuous only in the rectangle R: \t\< a, \y\ < 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 y = 0k(t) may contain points that lie outside of the rectangle R. Hence at the next stage—in the computation of 0k+ 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 0k+1 (t) might be impossible.
To avoid this danger it may be necessary to restrict t to a smaller interval than \t \ < 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
If (t, y) I < 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
0n (0) = 0
for each n. Since f[t,0k(t)] is equal to 0'k+1(t), the maximum absolute slope of the graph of the equation y = 0k+ 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,0k+ 1(t)] remains in R at least as long as R contains the wedgeshaped region, which is for ItI < b/M. We hereafter consider only the rectangle D: It I < h, I yl < b, where h is equal either to a or to b/ M, whichever is smaller. With this restriction, all members of the sequence {0n(t)} exist. Note that if b/M < a, then a larger value of h can be obtained by finding a better bound for | f (t, y)|, provided that M is not already equal to the maximum value of
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