An Example of a Nonlinear Differential Equation
R. C. Daileda
In class we mentioned the following theorem, whose proof the interested reader can find in Section 2.8 of
[1].
Theorem. 1.
Consider the initial value problem (IVP)
y
′
=
f
(
t, y
)
,
y
(
t
0
) =
y
0
.
(1)
If
f
and
∂f/∂y
are both continuous on a disk centered at
(
t
0
, y
0
)
then (1) has a unique solution defined on
some interval
t
0
−
h < t < t
0
+
h
,
h >
0
.
Several points should be made here. While this theorem does give us an effective way of determining if a
solution to an IVP exists, it gives us no way of determining this solution nor does it give any information on
the interval of definition of that solution. This should be compared to the analogous fact for IVPs involving
linear first order ODEs (Theorem 2.4.1 of [1]). If
f
is linear then we can write down an explicit solution,
and the interval of definition of that solution can be determined
from
f
and
t
0
alone
.
The moral is that
the behavior of solutions to nonlinear differential equations can be drastically different than that of linear
equations, as the following example is meant to illustrate.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '11
 Arubi
 Quadratic equation, Elementary algebra, Boundary value problem

Click to edit the document details