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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
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I f (t, y)|.
2. Does the sequence {0n(t)} converge? As in the example, we can identify 0n(t) =
01(t) + [02(t)  01(t)] + ? ?? + [0n(t)  0n1(t)] as the nth partial sum of the series
TO
01(t) + ? [0k+1(t)  0k(I)]. (24)
k=1
The convergence of the sequence {0n(t)} is established by showing that the series (24) converges. To do this, it is necessary to estimate the magnitude I0k+ 1(t)  0k(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 0, so that
0(t) = lim 0 (t). (25)
3. What are the properties of the limit function 0? In the first place, we would like to know that 0 is continuous. This is not, however, a necessary consequence of the convergence of the sequence {0n (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 0 is continuous is to show not only that the sequence {0n} 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 {0n} and, hence, the continuity of the limit function 0 in the interval It| < h.
Now let us return to Eq. (7),
0n+1(t) = f f[s,0n(s)] ds.
Jo
Allowing n to approach to on both sides, we obtain
0(t) = lim f f[s,0n(s)] ds. (26)
n^<xJ o n
We would like to interchange the operations of integrating and taking the limit on the right side of Eq. (26) so as to obtain
0(t) = f lim f[s,0n(s)] ds. (27)
Jo n^TO
In general, such an interchange is not permissible (see Problem 14, for example), but once again the fact that the sequence {0n(t)} converges uniformly is sufficient to allow us to take the limiting operation inside the integral sign. Next, we wish to take the limit inside the function f, which would give
0(t) = f f[s, lim 0 (s)] ds (28)
Jo n^TO
and hence
0(t) = f f[s,0(s)] ds. (29)
o
The statement that lim f [s,0n(s)] = f [s, lim 0n(s)] is equivalent to the statement that f is continuous in its second variable, which is known by hypothesis. Hence Eq. (29) is valid and the function 0 satisfies the integral equation (3). Therefore 0 is also a solution of the initial value problem (2).
4. Are there other solutions of the integral equation (3) beside y = 0(t)? To show
the uniqueness of the solution y = 0(t), we can proceed much as in the example.
2.8 The Existence and Uniqueness Theorem
113
First, assume the existence of another solution y = y(t). It is then possible to show (see Problem 19) that the difference 0(t)  y(t) satisfies the inequality
I0(t)  f(t)I < A I0(s)  f(s)I ds (30)
Jo
for 0 < t < h and a suitable positive number A. From this point the argument is identical to that given in the example, and we conclude that there is no solution of the initial value problem (2) other than the one generated by the method of successive approximations.
PROBLEMS In each of Problems 1 and 2 transform the given initial value problem into an equivalent problem s with the initial point at the origin.
1. dy/dt = t2 + y2, y(1) = 2 2. dy/dt = 1  y3, y(1) = 3
In each of Problems 3 through 6 let 00(t) = 0 and use the method of successive approximations to solve the given initial value problem.
(a) Determine 0n(t) for an arbitrary value of n.
(b) Plot 0n (t) for n = 14. Observe whether the iterates appear to be converging.
(c) Express lim 0 (t) = 0(t) in terms of elementary functions; that is, solve the given
nto> n
initial value problem.
(d) Plot I0(t)  0n (t) | for n = 1,..., 4. For each of 01(t), .04 (t) estimate the interval in which it is a reasonably good approximation to the actual solution.
? 3. y = 2(y + 1), y(0) = 0 ? 4. y = y  1, y(0) = 0
? 5. y = y/2 + t, y(0) = 0 ? 6. y = y + 1  t, y(0) = 0
In each of Problems 7 and 8 let 00(t) = 0 and use the method of successive approximations to solve the given initial value problem.
(a) Determine 0n(t) for an arbitrary value of n.
(b) Plot 0n (t) for n = 1,..., 4. Observe whether the iterates appear to be converging.
? 7. y = ty + 1, y(0) = 0 ? 8. y= t2y  t, y(0) = 0
In each of Problems 9 and 10 let 00(t) = 0 and use the method of successive approximations to approximate the solution of the given initial value problem.
(a) Calculate 0x(t),..., 03(t).
(b) Plot 0(),... ,03(t) and observe whether the iterates appear to be converging.
? 9. y = t2 + y2, y(0) = 0 ? 10. y= 1  y3, y(0) = 0
In each of Problems 11 and 12 let 00(t) = 0 and use the method of successive approximations to approximate the solution of the given initial value problem.
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