Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

# 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
Previous << 1 .. 48 49 50 51 52 53 < 54 > 55 56 57 58 59 60 .. 486 >> Next

0(t) = f f[s,0(s)] ds, (3)
Jo
where we have made use of the initial condition 0 (0) = 0. We also denote the dummy variable of integration by s.
Since Eq. (3) contains an integral of the unknown function 0, it is called an integral equation. This integral equation is not a formula for the solution of the initial value problem, but it does provide another relation satisfied by any solution of Eqs. (2). Conversely, suppose that there is a continuous function y = 0(t) that satisfies the integral equation (3); then this function also satisfies the initial value problem (2). To show this, we first substitute zero for t in Eq. (3), which shows that the initial condition is satisfied. Further, since the integrand in Eq. (3) is continuous, it follows from the fundamental theorem of calculus that 0'(t) = f [t, 0(t)]. Therefore the initial value problem and the integral equation are equivalent in the sense that any solution of one is also a solution of the other. It is more convenient to show that there is a unique solution
of the integral equation in a certain interval It| < h. The same conclusion will then
hold also for the initial value problem.
One method of showing that the integral equation (3) has a unique solution is known as the method of successive approximations, or Picard’s12 iteration method. In using this method, we start by choosing an initial function 00, either arbitrarily or to approximate in some way the solution of the initial value problem. The simplest choice is
00 (t) = 0; (4)
then 00 at least satisfies the initial condition in Eqs. (2), although presumably not the differential equation. The next approximation 01 is obtained by substituting 00(s) for 0 (s) in the right side of Eq. (3), and calling the result of this operation 01 (t). Thus
01(t) = f f[s, 00(s)] ds. (5)
0
12Charles-Emile Picard (1856-1914), except for Henri Poincare, perhaps the most distinguished French mathematician of his generation, was appointed professor at the Sorbonne before the age of 30. He is known for important theorems in complex variables and algebraic geometry as well as differential equations. A special case of the method of successive approximations was first published by Liouville in 1838. However, the method is usually credited to Picard, who established it in a general and widely applicable form in a series of papers beginning in 1890.
2.8 The Existence and Uniqueness Theorem
107
EXAMPLE
1
Similarly, 02 is obtained from 0X:
= f f[s>0i(s)] ds, (6)
Jo
and, in general,
tn+i(t) = f f[s,0n(s)] ds. (7)
o
In this manner we generate the sequence of functions {0n} = 00, 01,..., 0n,. Each
member of the sequence satisfies the initial condition, but in general none satisfies the
differential equation. However, if at some stage, say for n = k, we find that 0k+ 1(t) = 0k (t), then it follows that 0k is a solution of the integral equation (3). Hence 0k is also a solution of the initial value problem (2), and the sequence is terminated at this point. In general, this does not occur, and it is necessary to consider the entire infinite sequence. To establish Theorem 2.8.1 four principal questions must be answered:
1. Do all members of the sequence {0n} exist, or may the process break down at some stage?
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
y = 2t (1 + y), y(0) = 0. (8)
by the method of successive approximations.
Note first that if y = 0(t), then the corresponding integral equation is
0(t) = f 2s[1 + 0(s)] ds. (9)
Jo
10
If the initial approximation is 0o(t) = 0, it follows that
01(t) = f 2s[1 + 0o(s)] ds = f 2s ds = t2.
Jo Jo
Similarly,
02(t) = f 2s[1 + 01(s)] ds = f 2s[1 + s2] ds = t2 + t—
o o 2
(10)
(11)
and
03(t) = f 2s[1 + 02(s)] ds = f 2s
J0 J0
1 + s2 +
A t6
ds = t+ 2 + W3 . (12)
4
s
2
108
Chapter 2. First Order Differential Equations
Equations (10), (11), and (12) suggest that
2 t4 t6
0n(t) = t + 2! + 3! +
+
/In
n!
(13)
for each n > 1, and this result can be established by mathematical induction. Equation (13) is certainly true for n = 1; see Eq. (10). We must show that if it is true for n = k, then it also holds for n = k + 1. We have
0k+1(t) = f 2s[1 + 0k(s)] ds Jo
~t
s
,4
= L 2s i1 + / + 2! +
2k
+
k!
ds
4
2
t2k+2
t + 2! + 3! + ••• + (k + 1)! ’
(14)
and the inductive proof is complete.
A plot of the first four iterates, 0l(t),..., 04(t) is shown in Figure 2.8.1. As k 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 01(t), 04(t) for Example 1.
Previous << 1 .. 48 49 50 51 52 53 < 54 > 55 56 57 58 59 60 .. 486 >> Next