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Unformatted text preview: hen if
must satisfy. □ we also require leading to the modified inequality that Now prove the important Grönwall inequality. As a preliminary remark, note that if
then is increasing less rapidly that . Integrate both sides to obtain
, provided
.
Lemma 3.13 (Grönwall). Suppose
obeys the inequality are continuous, , and [2]
for all . Then for all
. Proof. Notice that is differentiable.
an integrating factor to give
, or . Multiply by 13
Chapter 3
.
Integrate both sides between and : ,
from which the result follows. ■
Example. Suppose two solutions of
with initial conditions
compared on a common time interval of existence. They satisfy If is Lipschitz with constant
the inequalities and and are being , then the distance between the trajectories satisfies .
This has the form of [2] with and . Then obtain .■
Theorem 3.14 (Lipschitz Dependence on Initial Conditions). Let
a such that
is Lipschitz with constant and that
interval of existence for solutions
. Then
in with Lipschitz constant . Proof. Based on the example above and a similar example with Recall that , and suppose there is
is the common
is uniformly Lipschitz is differentiable at
. This can be rewritten
. if the following limit exists .□ 14
Chapter 3
is differentiable at if there is such that .
This means where
. Th Ty [3] ’
u wh
such that Let d Note that [3] is equivalent to:
. be the solution of the initial value problem
[4] The linearization of about satisfies .
Example. If
has period , obtain
subject of Floquet theory in section 2.8. ■
The Fundamental Matrix Solution , where has period . This was the satisfies
. [5] Theorem 3.15 (Smooth Dependence on Initial Conditions). Suppose
is
on an
open set . Then there exists an
such that the solution
of [1] is a
function of
for
.
Proof. Do not give. See in text.
Remark. The proof is an excellent illustration of the use of G ö w ’ inequality. It shows that
is
by demonstrating that
is its derivative:
.
If we assume that , this can be demonstrated directly. . 15
Chapter 3
Thus satisfies [5]. ■ Let be the solution of the initial value problem
, where [6] is a vector of parameters. [6] is equivalent to
. In the integrand, depends on [7] in two ways. Theorem 3.16 (Continuous Dependence on Parameters). Suppose
is
Lipschitz in
and uniformly continuous in
. Then [6] has a unique solution
for
that is a uniformly continuous function of on some interval
.
Proof. By theorem 3.10,
and
for some
. Consider will have a common interval of existence
. From [7]
. Write Since is a uniformly continuous in , for any there is a
. Then assume
to obtain such that implies ,
wh ch g v , by G w’ qu y, .
This expression can be made as small as we wish by making sufficiently small. A similar
argument holds for
.□ 16
Chapter 3 3.5 Maximal Interval of Existence
Consider the initial value problem
. [1] If is locally Lipschitz, then theorem 3.10 gua...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .
 Fall '14
 MarcEvans

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