The Initial Value Problem y' = f(x, y); y(c) = d 49
from t he point (b, B). The solution Y2(x) is unique on the interval (e, +oo).
The endpoint (e, B) of Y2(x) li es on the bottom boundary of R. If f(x, y)and f y ( x, y) are both defined and continuous at ( e, B), then y 2 ( x) can b e
extended uniquely to the left. If f(x, y) is continuous at (e, B) but fy(x, y)
is not defined or not continuous at ( e, B), then the solution can be extended
to the left, but the extension may not be unique. If f(x, y) is not continuous
at (e, B), then the solution Y2(x) may or may not be extendable to the left.
Moreover, ifthe solution y 2 (x) can be extended to the left, the extension may
or may not be unique. Notice that t he solution y 3 (x).extends from the bottom
boundary of R to the top boundary and may or may not be extendable to
the left or r ight in a unique or nonunique fashion. Likewise, the solution
y 4 ( x) extends from the left boundary of R to t he top boundary and may or
may not be extendable to t he left or right in a unique or nonunique manner.
The following two examples should help further clarify the results that can be
obtained by using the continuation theorem.
EXAMPLE 3 At Every Point in the xy-plane a Unique Solution
Exists Yet the Interval of Existence of the Solution
Is FiniteAnalyze the initial value problem(9) y' = 1 + y^2 ; y(n'/4) = 1.
SOLUTION
Here f(x, y) = 1 +y^2 and fy(x, y) = 2y are both defined and continuous on
any finite rectangle in the x y-plane which contains the point (n/4, 1). By the
fundamental existence and uniqueness theorem, there exists a unique solution
to the IVP (9) on some interval with center n/4. By the continuation theorem
the solution can be extended uniquely until the boundary of the rectangle--
the xy-plane, in this case- is reached. Thus, the solution can be extended
until two of the following four things occur: x--> - oo, x--> oo, y(x)--> - oo,
y(x) __, oo. It is tempting to erroneously jump to the conclusion that since
the functions f(x, y) and fy(x, y) are continuous on the entire plane, the
solution to t he IVP (9) should be vali d for all real x. However , this is not
what the continuation theorem states. Verify that y(x) =tan x is the unique
solution of the IVP (9). Notice tha t y(x) = tanx is defined, continuous, and
differentiable on the interval (-n/2,n/2), but it is not defined at x = -n/2
or at x = n/2. So (-n/2,n/2) is the largest interval on which the IVP (9)
has a so lut ion. Observe that as x --> -n+ /2, y(x) = tanx --> -oo and as
x --> n -/2, y(x) --> +oo. Notice that these results satisfy the conclusion of
the continuation theorem. The direction field for the differential equation
y' = 1 + y^2 and the solution of the IVP (9) are displayed in Figure 2.8.