1 Chapter 3

3.1 Set and Topological Preliminaries

Notation

natural numbers

integers

reals

element of

for all

there exists

such that

subset of (whether proper or not)

union

intersection

set subtraction

implies

if and only if

Euclidean norm of .

Notions from analysis

Open ball:

Closed ball:

Let . is an interior point of if an open ball centered on such that .

is open if it consists only of interior points. An arbitrary union of open sets is open. A finite

intersection of open sets is open.

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is a neighborhood of if contains an open set containing .

A sequence is an ordered list: .

Sequence converges to if , an integer such that

.

This is often abbreviated as .

A subsequence of has form: .

is a limit point of if a subsequence of that converges to .

Example. Consider the sequence . has subsequence

that converges to , and subsequence that converges to . and are

limit points of . ■

Set is closed if it contains all of its limit points. An arbitrary intersection of closed sets is

closed. A finite union of closed sets is closed.

Closure of : .

Boundary of : .

Example. The closed ball is closed. The corresponding open ball

is open. . . ■

is bounded if for some . If not, is unbounded.

In , compact closed and bounded. Every sequence contained in a compact set has a

convergent subsequence.

is continuous if such that whenever

. means f is continuous on .

is uniformly continuous if such that whenever

. Note that for a given there must be a that suffices for all .

Lemma. A continuous function on a compact set is uniformly continuous.

3 Chapter 3

Lemma. A continuous function on a compact set attains a maximum and a minimum value on

that set.

Example. Let is compact but is not. is continuous but not

uniformly continuous on . is uniformly continuous on . does not attain a

maximum value on but does attain a maximum value on . ■

A sequence of functions converges uniformly to a limit function if

such that for all .

- , -tube about , with the tube

Lemma 3.2 The limit of a uniformly convergent sequence of continuous functions is

continuous.

Proof. Similar to text (use and ).

3.2 Function Space Preliminaries

Let and . The derivative of at a point is given by the Jacobian matrix:

.

Note that .

A function is continuously differentiable on if the elements of are continuous on the

open set . We then write . If the k^th partial derivatives of are continuous

functions on write . If is continuous with domain and range we can write

.

Example. Suppose and . Then the chain rule gives . ■

The sup norm of is given by

. We will sometimes use the second

notation to emphasize that the sup norm is being used.

4 Chapter 3

Metric Spaces

We will often regard functions as points in abstract spaces. These spaces will all be metric

spaces.

A metric space is a set of elements and a metric . Let . The metric satisfies:

and iff (positivity).

(symmetry)

(triangle inequality).

Example. A normed linear space is a metric space with . (A linear space is the

same thing as a vector space.) ■

A sequence converges to if as . If

, then

convergence corresponds to uniform convergence of a sequence of functions.

A sequence is Cauchy if, an integer such that

. From Analysis, every convergent sequence is Cauchy. However, not every Cauchy sequence

converges.

Example. The set of numbers with . The sequence

is

Cauchy but does not converge (in the sense that its limit does not belong to ). ■

A metric space is complete if every Cauchy sequence converges to an element of . A complete

normed linear space is a Banach space.

Contraction Maps

Meiss uses map as a synonym for operator.

Let be a map on a complete metric space . is a contraction if there is a constant

such that for all .

Remark. The following little exercise in geometric series is used in the following theorem. Let

integers and .

,

5 Chapter 3

,

subtract

,

.

Theorem 3.4 (Contraction Mapping). Let be a contraction on a complete metric

space . Then has a unique fixed point .

Proof. As in the text. Plan to give it.

Example (source: http://www.math.uconn.edu/~kconrad/blurbs/analysis/contraction.pdf)

Consider the complete metric space with . The mean value

theorem applied to leads to

for and some between and . We then have

.

Since for , is a contraction. Starting from any , repeated applications

of gives a sequence that converges to a unique fixed point. Test this by entering on your

calculator and repeatedly pressing the cosine key. The unique fixed point ■

Lipschitz Functions

Let . A function is Lipschitz if for all there is a constant such that

. The smallest such constant is the Lipschitz constant for in .

Lemma 3.5. A Lipschitz function is uniformly continuous.

Proof. is uniformly continuous means such that

. But Lipschitz implies . Therefore is uniformly

continuous with . □

The following example shows that need not be differentiable in order to be Lipschitz.

6 Chapter 3

Example. . sketch Define piecewise by

. For

. The same statement holds if and are both less than or equal to

zero. For , . is Lipschitz with

constant =1. ■

A function is locally Lipschitz if for all there is a neighborhood such that is

Lipschitz on .

Example.

is not locally Lipschitz at . For any there is with

sufficiently small that

, implying

. ■

Lemma 3.6 Let be a function on a compact, convex set . Then satisfies a Lipschitz

condition on with Lipschitz constant .

Proof. For let . parameterizes the straight line between

and . Since is convex, this line is within . sketch A, x, y, line

By the Fundamental Theorem of Calculus and the chain rule:

Since is compact and is continuous, has a maximum value on . Thus

. □

Corollary 3.7 If is on an open set , then it is locally Lipschitz on .

Proof. For each there is a closed ball . Since each is compact and

convex, by lemma 3.6 is Lipschitz on it. □

3.3 Existence and Uniqueness Theorem

Let be open and . Consider the initial value problem

7 Chapter 3

. [1]

Integrate to obtain

. [2]

The initial value problem [1] is equivalent to the integral equation [2]. We will solve Eq. [2] on

for some , which will give .

There is a relationship between the smoothness of and . Assume only , Then if

we are given , we have from [2] that .

Now assume and . From the reasoning above obtain . But then

[2] gives , which means . This can be generalized to obtain

Lemma 3.9. Suppose for and is a solution of [2]. Then

.

Proof. By induction in Meiss.

Let be an element of a suitable function space . Define an operator

. [3]

A solution of [2] is a fixed point of :

.

The contraction mapping theorem implies that if is a complete metric space and is a

contraction then exists and is unique.

Theorem (3.10) (Existence and Uniqueness) Suppose that for there is a such that

is Lipschitz with constant . Then the initial value problem [1] has a unique

solution, for , provided that

where

Partial Proof. Let . Divide the proof into three parts.

8 Chapter 3

(A). Show complete. Use the sup norm. Consider a Cauchy sequence of functions in

. Since convergence is uniform, the limit is a continuous function .

Since is closed, . Therefore and is complete.

(B) Show that . From [3], is continuous. Since is continuous and is

closed, attains a maximum on . Let

. From [3]

,

where it is understood in expressions like these that if the limits of integration should be

reversed. If then .

(C) Show is a contraction. Consider . Then since is Lipschitz

.

Taking the max of both sides over obtain . Then is a

contraction provided that .

Since the hypotheses of the contraction mapping theorem are satisfied, has a unique fixed

point that solves [2] and hence [1]. □

Remark. The only deficiency in this proof is the additional condition .

Remark. If and is locally Lipschitz on , then for every there is a such

that is Lipschitz on and theorem 3.10 applies.

Remark. Geometrically, this theorem implies that trajectories never cross. sketch trajectories

crossing at . This can never happen.

Remark. A non-autonomous system may be converted into an autonomous one.

Is equivalent to

, ,

, .

where , and . Then theorem 3.10 gives existence and uniqueness as long as

is Lipschitz in as well as . However, the requirement that is Lipschitz in may be relaxed.

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Theorem 3.11 (Nonautonomous Existence and Uniqueness). Suppose is a

uniformly Lipschitz function of with constant , and a continuous function on

. Then the initial value problem

,

has a unique solution for and , where

Where the maximum is taken over and .

The contraction mapping theorem helps prove the existence and uniqueness of a solution to [1]

but does not show us how to find a solution. Picard iteration constructs successive

approximations to the solution of [1] by repeatedly applying the operator . Start with an

initial guess to generate the sequence

,

Example. , where and from [3]

Let . Apply to to get

.

Apply to to get

).

This procedure generates successive approximations to the solution

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. ■

Our text presents a constructive proof of the Existence and Uniqueness theorem by analyzing

the convergence of Picard iterates. The proof is longer than that using the Contraction

Mapping theorem, but it has the advantage of dropping the restriction .

If is not Lipschitz, a solution to [1] may exist but fail to be unique.

Example (The Leaky Bucket) (Strogatz, Nonlinear Dynamics and Chaos)

Sketch partially full bucket with water running out of hole

Symbols:

height of water remaining at time .

cross-section of bucket.

area of hole

velocity of water passing through hole

density of water

Suppose that in a short time interval the height of the water in the bucket decreases by .

The mass of the water that escaped is . Assume that the potential energy of is

entirely converted into kinetic energy of escaping water.

Solve for

,

where the minus sign reflects the fact that the escaping water is falling. By conservation of

volume

.

Combine the last two equations to obtain

where the constant .

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I w ub u h ‘b ckw d ’ v b for we obtain

. Notice that

the right hand side is not Lipschitz at .

Now suppose we are taking a walk and that at we notice an empty bucket with a hole in

h b d w u d h. L ’ u u d g k d wh

the water ran out by integrating backwards in time! Consider the initial value problem

, . [4]

Our initial condition corresponds to the fact that we found the bucket empty. We notice

immediately that one solution is

,

that is, there never was any water in the bucket. Undaunted, we try separating variables

,

and integrate, applying the initial condition, to find

.

This means it is possible that the water has just finished running out of the bucket. Finally we

notice that we can combine these two solutions to obtain for any :

.

The initial value problem [4] admits an infinite family of solutions! sketch solutions for

and . ■

3.4 Dependence on Initial Conditions and Parameters

Here we modify section 3.3 eq. [1] by considering

, [1]

for some close to . Write the solution

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,

showing the dependence on the initial condition. Operator has initial condition :

.

Lemma 3.12 (Neighborhood Existence). Suppose that for a given there is a such that

satisfies a Lipschitz condition with constant , and .

Then the family of solutions of [1] exists for each

and is unique for

providing

Proof. Very similar to that of theorem 3.10. Steps (A) and (C) are essentially identical. Step

(B), which shows is modified slightly. We now have

.

Use the triangle inequality and the maximum value of :

.

Then if we also require leading to the modified inequality that

must satisfy. □

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 are continuous, , and

obeys the inequality

[2]

for all . Then for all

.

Proof. Notice that is differentiable. or . Multiply by

an integrating factor to give

,

13 Chapter 3

.

Integrate both sides between and :

,

from which the result follows. ■

Example. Suppose two solutions of with initial conditions and are being

compared on a common time interval of existence. They satisfy

If is Lipschitz with constant and , then the distance between the trajectories satisfies

the inequalities

.

This has the form of [2] with and . Then obtain

.■

Theorem 3.14 (Lipschitz Dependence on Initial Conditions). Let , and suppose there is

a such that is Lipschitz with constant and that is the common

interval of existence for solutions

. Then is uniformly Lipschitz

in with Lipschitz constant .

Proof. Based on the example above and a similar example with . □

Recall that is differentiable at if the following limit exists

.

This can be rewritten

.

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is differentiable at if there is such that

.

This means where

. [3]

Th T y ’ u w h d Note that [3] is equivalent to:

such that .

Let be the solution of the initial value problem

[4]

The linearization of about satisfies

.

Example. If has period , obtain , where has period . This was the

subject of Floquet theory in section 2.8. ■

The Fundamental Matrix Solution 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.

.

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Thus satisfies [5]. ■

Let be the solution of the initial value problem

, [6]

where is a vector of parameters. [6] is equivalent to

. [7]

In the integrand, depends on 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 will have a common interval of existence

for some . Consider . From [7]

.

Write

Since is a uniformly continuous in , for any there is a such that implies

. Then assume to obtain

,

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 guarantees there is an such that a solution

exists and is unique for . The maximal interval of existence is the largest

interval of time that includes for which a solution to [1] exists.

Example. The initial value problem

.

is locally Lipschitz on and Separate variables and apply the

initial condition to find

.

The maximal interval of existence is . Note that is not in the domain of . ■

Theorem 3.17 (Maximal Interval of Existence). Let be open and be locally

Lipschitz. Then there is a maximal open interval containing such that the initial

value problem [1] has a unique solution .

Proof. Let . Theorem 3.10 guarantees there is a ball , a time interval

and a unique solution of [1] such that

.

Let Then and we can apply theorem 3.10 again. Let .

Then there is a ball , a time interval and a unique solution

such that

.

is not empty and uniqueness implies on Continue this

construction both backwards and forwards in time. Note that every endpoint lies in .

The maximal interval of existence is the union of all such intervals. Let be the unique

solution constructed.

17 Chapter 3

Suppose that has a closed endpoint, for example . Then and the solution

could be extended further, contradicting the claim that is maximal. Therefore is open. □

Example (continued). Notice that is open. ■

Theorem 3.18 (Escape from any compact set). Suppose is an open set and is

locally Lipschitz. Let be the maximal interval of existence for [1]. If is finite, then

for any compact set there is a such that . Similarly, if is finite, then

for any compact there is a such that .

Idea of Proof. If is finite and , then we can define , with

, and extend using theorem 3.10.

sketch , , , , extension □

Example (continued) sketch -, -, both branches of The

solution escapes from any compact set in or as or as . ■

Corollary 3.19. If is finite then either does not exist or

Proof. If the limit exists it cannot be in , since then the solution could be extended. However,

every point for . Therefore . □

Example (continued). sketch -, - indicate or and the flow

. ■

Example. The initial value problem

.

is locally Lipschitz on . is the solution corresponding to Assume ,

separate variables and apply the initial condition to find

.

For sketch -, -, vertical asymptote, . The maximal

interval of existence is and

does not exist. ■