1550078481-Ordinary_Differential_Equations__Roberts_

48 Ordinary Differential Equations

If f(x, y) and fy(x, y) are both defined and continuous on a finite rectangle,

the rectangle can be enlarged until either f(x, y) or fy(x, y) is not defined or
not continuous on each bounding side of the rectangle or until the bounding
side of the rectangle approaches infinity. Thus, by the continuation theorem
the solution to the IVP y' = f(x, y); y(c) = d has a unique solution which
extends from one boundary of the enlarged rectangle to another (although
perhaps, the same) boundary of the enlarged rectangle. Suppose, for instance,

the enlarged rectangle R for a particular differential equation y' = f(x, y) has

left boundary x = L, right boundary x = +oo, bottom boundary y = B, and

top boundary y = T, where L , B, and Tare real numbers and B < T. That

is, suppose f(x, y) and fy(x, y) are both defined and continuous inside the
"infinite" rectangle

R = {(x, y) I L < x and B < y < T}

and that either f(x, y) or fy(x, y) is not defined or not continuous at some
point or set of points on all of the lines x = L, y = B , and y = T. Displayed
in Figure 2.7 is the "infinite" rectangle Rand the solutions Yi(x) of the initial

value problems y' = f(x, y); y(c) =di for i = 1, 2, 3, 4.

e a

X=L

c b

y (x)

2

Figure 2.7 Solutions to an Initial Value Problem on an Infinite Rectangle

Notice that the solution y 1 (x) exists and is unique on the interval (a, b) and
that both endpoints (a, B) and (b, B) of y 1 (x) lie on the bottom boundary of
R. If f(x, y) and fy(x, y) are both defined and continuous at (a, B), then the
solution y 1 (x) can be extended uniquely to the left. If f(x, y) is continuous
at (a, B) but fy(x, y) is not defined or not continuous at (a, B), then the
solution can be extended to the left, but the extension may not be unique.
Furthermore, if f(x, y) is not continuous at (a, B), the solution may or may
not be extendable to the left. And if extendable, the solution may or may not
be unique. Similar comments apply regarding extending y 1 ( x) to the right