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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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First of all, we note that it is sufficient to consider the problem in which the initial point (t0, y0) is the origin; that is, the problem
/= f(t, ), (0) = 0. (2)
If some other initial point is given, then we can always make a preliminary change of variables, corresponding to a translation of the coordinate axes, that will take the given point (t0, y0) into the origin. The existence and uniqueness theorem can now be stated in the following way.
106
Chapter 2. First Order Differential Equations
Theorem 2.8.1
If f and f/ are continuous in a rectangle R: \t \ < a, \ y\ < b, then there is some interval \t \< h < a in which there exists a unique solution = () of the initial value problem (2).
To prove this theorem it is necessary to transform the initial value problem (2) into a more convenient form. If we suppose temporarily that there is a function = () that satisfies the initial value problem, then f [t, (t)] is a continuous function of t only. Hence we can integrate = f (t, y) from the initial point t = 0 to an arbitrary value of t, obtaining
( = f f[s^(s)] ds, (3)
0
where we have made use of the initial condition (0) = 0. We also denote the dummy variable of integration by s.
Since Eq. (3) contains an integral of the unknown function , 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 = (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 '() = f [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 \ t\ < 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 Picards12 iteration method. In using this method, we start by choosing an initial function 0, either arbitrarily or to approximate in some way the solution of the initial value problem. The simplest choice is
0^) = 0; (4)
then 0 at least satisfies the initial condition in Eqs. (2), although presumably not the differential equation. The next approximation 1 is obtained by substituting 0 (s) for (s) in the right side of Eq. (3), and calling the result of this operation 1 (t). Thus
1() = f f[s^0(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, 2 is obtained from 1:
Գ({) = f f[5,Գ(5)] d5, (6)
Jo
and, in general,
+() = f [5,(5)] d5. (7)
o
In this manner we generate the sequence of functions {} = 0, 1,... ,,.... 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 +1(t) = (t), then it follows that is a solution of the integral equation (3). Hence 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 {} exist, or may the process break down at some stage?
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