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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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2. Does the sequence converge?
3. What are the properties of the limit function? In particular, does it satisfy the
integral equation (3), and hence the initial value problem (2)?
4. Is this the only solution, or may there be others?
We first show how these questions can be answered in a specific and relatively simple example, and then comment on some of the difficulties that may be encountered in the general case.
Solve the initial value problem
= 2t (1 + y), y(0) = 0. (8)
by the method of successive approximations.
Note first that if y = (), then the corresponding integral equation is
( = f 25[1 + (5)] d5. (9)
Jo

If the initial approximation is 0(t) = 0, it follows that
1() = f 25[1 + 0(5)] d5 = j 25 d5 = t2.
Jo Jo
Similarly,
2(t) = f 25[1 + 1 (5)] d5 = f 25[1 + 52] d5 = t2 + l
o o 2
(10)
(11)
and
3(() = f 25[1 + 2(5)] d5 = f 25
Jo Jo
1 + 52 +
A t6
d5 = t+ 2 + ^3 . (12)
4
5
2
108
Chapter 2. First Order Differential Equations
Equations (1o), (11), and (12) suggest that
2 t4 t6
(t) = t + 2! + 3! +
+
2
!
(13)
for each > 1, and this result can be established by mathematical induction. Equation (13) is certainly true for = 1; see Eq. (). We must show that if it is true for = , then it also holds for = + 1. We have
+1() = f 25[1 + (5)] d5
o
t
5
4
= 2511 + ^ + 2! +
2
+
!
d5
4
6
2
^2+2
t + 2! + 3! + + ( + 1)!
(14)
and the inductive proof is complete.
A plot of the first four iterates, (),..., 4() is shown in Figure 2.8.1. As increases, the iterates seem to remain close over a gradually increasing interval, suggesting eventual convergence to a limit function.
-1.5 -1 - 0.5 0.5 1 1.5 1
FIGURE 2.8.1 Plots of (), 4(t) for Example 1.
It follows from Eq. (13) that ( t) is the rth partial sum of the infinite series
,2
? ; (15)
= 1
hence lim (t) exists if and only if the series (15) converges. Applying the ratio test, we see that, for each t,
/2+2
!
( + 1)! t
2
2
+ 1
o as oo;
(16)
2.8 The Existence and Uniqueness Theorem
109
thus the series (15) converges for all t, and its sum () is the limit of the sequence {(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
() = t2k / k! is a solution ofthe integral equation (9). Alternatively, by substituting
k=1
() 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 in terms of
2
elementary functions, namely, () = el 1. However, this is not necessary for the discussion of existence and uniqueness.
Explicit knowledge of () does make it possible to visualize the convergence of the sequence of iterates more clearly by plotting () (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 (t) (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 and ty. Since and ty both satisfy the integral equation (9), we have by subtraction that
() ty(t) = ( 2s[(s) ty(s)] ds.
0
Taking absolute values of both sides, we have, if t > 0,
() ty(t) | =
f 2s^(s) ty(s)] ds < f 2s(s) ty(s) ds. 00
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
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