1 Chapter 4A

4.1 Definitions

A dynamical system is an evolution rule that defines trajectories in phase space:

.

is a differentiable map that is parameterized by time. is the solution of the

initial value problem discussed in chapter 3. However, this chapter emphasizes a geometrical

point of view and first discusses without reference to the initial value problem. It then

comes back to the initial value problem through the properties of .

The orbit or trajectory of a point is . Sketch ,

The forward orbit is .

The preorbit or backward orbit is .

Example. If is an equilibrium point then . ■

Example. If , a periodic orbit with period , then and . ■

We think of the map as operating on sets as well as points. Sketch

Set is invariant under rule if for all . This means that, for each

for all .

is forward invariant if for .

is backward invariant if for .

4.2 Flows

Recall that if is on an open set then the initial value problem

has a solution that is differentiable with respect to both and .

A complete flow is a differentiable mapping such that

(a)

(b) for all

The symbol denotes composition: . Property (b) is the group

property, and implies

2 Chapter 4A

The mapping is invertible: .

Any time may be viewed as the initial time: .

Trajectories cannot cross. sketch , trajectories starting on and and then crossing

If , then apply to both sides to obtain . That is, and

are on the same trajectory.

A flow may refer to a complete flow, but may instead refer to a mapping for which property (b)

does not hold for all because trajectories are not defined for all times.

A vector field is a function that defines a vector at each point . The

vector field associated with a given flow is

.

Lemma 4.1 If is a flow, then it is a solution of the initial value problem

,

Proof. In text. Don’t give.

If the flow is complete, the maximum interval of existence is .

Example. Show that the function defined by

is a complete flow on . Find the vector field associated with the flow.

First note that is defined for all and . satisfies property (a) since

.

To verify that satisfies property (b), simplify the composition:

.

The vector field associated with the flow is

3 Chapter 4A

. ■

Lemma 4.2. Let be an open subset of , and a vector field such that the

initial value problem , has a unique solution that exists for all

and all . Then is a complete flow.

Proof. In text. Don’t give. □

4.3 Global Existence of Solutions

The solutions of initial value problems for ODEs sometimes have finite maximum intervals of

existence. How can we make use of the elegant framework of complete flow? Consider

, [1]

. [2]

Theorem 4.3’ (Bounded Global Existence). If and bounded, then the solution of

[1] defines a complete flow.

Proof. Denote the maximum interval of existence . By assumption, there is an such

that . For , [2] gives

.

If were finite we would have . This contradicts the theorem on

escape from compact sets. So cannot be finite. Similarly, cannot be finite. Therefore the

solution defines a flow by lemma 4.2. □

Remark. If a physical system were exactly described by an initial value problem of form [1], its

vector field would presumably be bounded. ■

Example. The overdamped pendulum, , defines a flow by theorem 4.3’. ■

Theorem 4.4’ (Reparameterization of Time). If then is equivalent to

, [3]

4 Chapter 4A

upon reparameterization of time. The vector field of defines a flow since it is and

bounded.

Proof. The solution of [1] has a maximum interval of existence . Define

using

Since

is increasing and the transformation is one-to-one. By the

chain rule

. Therefore

.

Clearly is bounded. It remains to show that . It is sufficient to show that all of

the partial derivatives

are . View as a function of , fixing the other components of

. Let , and

. Then

We need to show . We are given . Note and

. *3’+

is except at values where . Then except possibly at those values.

Suppose Observe that

.

Then

5 Chapter 4A

Further, if ,

From the last expression in *3’+, remains bounded as . Therefore

Thus is continuous at and . □

Example. The undamped pendulum

,

defines a complete flow on upon reparameterization of time. ■

Example. The reparameterization of time [3] also converts some vector fields on domains with

boundaries into flows defined for all times. The Lotka-Volterra system for the competitive

interaction of two species is

,

.

The domain is . The right hand of this system is in the interior of

. Although the domain has boundaries, trajectories do not escape the domain by approaching

the boundaries. Sketch the domain and the flow on the boundaries. However, for certain

initial conditions they may escape to in finite time. (Suppose and are positive, , and

. The trajectory will escape to in finite negative time.) Upon reparameterization of

time by [3] this system defines a flow for all time. ■

There are also cases where where is a proper open subset of . The following

theorem and its proof are found in Perko (3rd edition, theorem 2 in section 3.2).

Theorem (Global Existence on Open Domains with Boundaries). Suppose is an open subset

of and . Then there is a function such that is equivalent to

[4]

6 Chapter 4A

upon reparameterization of time. The vector field defines a complete flow.

Idea of Proof. where is given by [3]. Define the closed set

and let

.

is a distance from to the boundary of :

.

The effect of is to slow trajectories that approach any boundary of . The maximum interval

of existence for the reparameterized system is . □

Sketch bo ndary and tra e tory approaching boundary

Example. The initial value problem

, which we analyzed in chapter 3. is

on and . Upon reparameterization of time by [4], this system defines a complete

flow. ■

4.4 Linearization

Let , where and is an dimensional manifold. Consider

. [1]

Suppose has an equilibrium point . Let and expand sing Taylor’s theorem

about :

.

Use to find

.

These expressions use notation (“little-oh” notation). In general:

as if there is a neighborhood of such that

.

There is also notation (“big-oh” notation):

7 Chapter 4A

as if there is a neighborhood of and a such that

.

Example. If then Taylor’s theorem wo ld give

. ■

The linearization of [1] at is

[2]

where . Here and .

Recall the decomposition of into unstable, center and stable eigenspaces of :

, where is the unstable eigenspace, and so forth.

An equilibrium point is hyperbolic if none of the eigenvalues of have zero real part.

Then is empty.

Hyperbolic equilibrium points are important because, by the Hartman-Grobman theorem

(section 4.8), the flow of [1] in some neighborhood of a hyperbolic equilibrium point is

topologically equivalent to the flow of [2].

Distinguish three classes of hyperbolic equilibrium points :

is a sink if all of the eigenvalues of have negative real parts. Then .

is a source if all of the eigenvalues of have positive real parts. Then .

is a saddle if some of the eigenvalues of have negative real parts and some

have positive real parts. Then .

Often very useful information about the behavior of can be obtained by finding the

equilibrium points and analyzing the associated linearized systems using the methods of

chapter 2.

Example. The Lotka-Volterra Competition Model.

,

.

The equilibrium points are , (3,0), (0,2) and (1,1).

sketch -, -, eq. pts.

Linearized systems have the form

, where

.

8 Chapter 4A

The Jacobean matrix is

.

Linearize about Obtain

. Notice that linearizing about the origin

is equivalent to dropping the nonlinear terms in the Lotka-Volterra equations. See that ,

and

. This is a source. Add to sketch.

Linearize about Obtain

. See , . See

. This is a sink. Add to sketch.

Linearize about Obtain

. See , and find

.

See . This is also a sink. Add to sketch.

Linearize about Obtain

. Find and the eigenvalues

. Find the corresponding eigenvectors

and

. This

is a saddle. add to sketch. ■

Example. The Lorenz model with .

,

.

Linearize about the equilibrium point at . This is equivalent to dropping the nonlinear

terms. Obtain

.

This has block diagonal form. The block has eigenvalues . For all of

the eigenvalues of the Lorenz model are negative and the origin is a sink. For one

eigenvalue is positive and the origin is a saddle. The origin loses stability as increases through

1. A bifurcation occurs at . ■

9 Chapter 4A

4.5 Stability

An equilibrium of a flow is Lyapunov stable or stable if such that

.

sketch , , , , trajectory based at stays in

If is not stable then it is unstable.

An equilibrium is asymptotically stable if it is Lyapunov stable and, in addition, there is a

neighborhood of such that .

sketch , , , trajectory based at converges to

Remarks

Meiss’ example, eq. 4.14 and Fig. 4.6, shows why it is necessary to use the two closed

balls and in the definition of Lyapunov stability. It shows a system for which

initial conditions must be confined to a with in order to guarantee that

trajectories stay within .

Meiss’ example, eq. 4.16 and Fig. 4.7, shows why it is necessary to require Lyapunov

stability explicitly in the definition of asymptotic stability. It shows a system for which

there is a neighborhood of such that and yet the system is

not stable.

Instead of reiterating these examples, let’s see how the definitions apply to systems we are

already familiar with.

Stability of 1D systems: , .

asymptotically stable.

-, -, decreasing through with negative slope, flow on axis.

-, -, decreasing through with zero slope, flow on axis. unstable

-, -, increasing through with positive slope, flow on axis. semi-stable (a type of unstable)

-, -, concave up with a minimum at

If asymptotically stable unstable

10 Chapter 4A

need more information to make a conclusion on stability

Example. The Logistic Equation. .

-, -, Equilibrium points: unstable, asymptotically stable. ■

Stability of 2D systems: , .

Example. The Harmonic Oscillator. or where .

Let obtaining

or

and

or

Observe , , . This is a center.

-, -, circular trajectories surrounding the origin. is stable but not asymptotically stable. ■

Example. Stable Degenerate Node

, ,

.

, , , find , .

Find and

.

-, -, is a line of degenerate equilibria, exponential decay along . The equilibrium points in are stable but not asymptotically stable. ■

Example. Lotka-Volterra Competition Model. -, -, equilibria , putative trajectories connecting

these The source and saddle (1,1) are unstable. The sinks and are asymptotically stable. ■

Example. The Lorenz Model with . , , . The equilibrium point has three eigenvalues:

, , . It is asymptotically stable for and unstable for . ■

11 Chapter 4A

Recall theorem 2.10 (Asymptotic Linear Stability) Let . all

eigenvalues of have negative real parts.

Theorem 4.6 (Asymptotic Linear Stability Implies Asymptotic Stability) Let be

and have an equilibrium such that all of the eigenvalues of have negative real

parts. Then is asymptotically stable.

Do not give proof. (The statement is plausible and the proof depends on detailed estimates.

Re ommend for reading; it is an exer ise in sing Grönwall’s ineq ality. This is also implied by

the Hartman-Grobman theorem, although the text does not prove that in detail. The proof of

theorem 4.21 is a discrete analog of this.)

Remark. For nonlinear systems, the condition that have eigenvalues with negative

real parts is sufficient but not necessary for asymptotic stability.

4.6 Lyapunov Functions

A continuous function is a strong Lyapunov function for an equilibrium of a flow

on if there exists an open neighborhood of such that for

and for all and .

is a weak Lyapunov function if the same holds except for all

and .

is positive definite on a neighborhood of if and . A

strong Lyapunov function is positive definite in some neighborhood of an equilibrium point .

Theorem 4.7 (Lyapunov Functions) Let be an equilibrium point of a flow . If is a

weak Lyapunov function in some neighborhood of , then is stable. If is a strong

Lyapunov function, then is asymptotically stable.

Proof. First prove that is stable if is a weak Lyapunov function. Choose so small that

Let Then . Choose so small that

. Then if , for all This means

and is stable.

Next prove that is asymptotically stable if is a strong Lyapunov function. Note that is

still stable as argued above. Consider a trajectory whose initial point is any . We will

show that is the only limit point of the flow. Together with the fact that is stable, this

shows that .

12 Chapter 4A

Finally prove that is the only limit point of the flow. Consider any sequence of times .

Then has a convergent subsequence. For simplicity, denote the subsequence of times

too. Suppose . Because is a decreasing function

.

However, because is not an equilibrium point, for any . Fix By

continuity of , for large enough, we must also have . But then for

, , which contradicts the chain of inequalities displayed above.

□

Example. Show that the origin is an asymptotically stable state for the system

,

.

For a candidate Lyapunov function, try a very simple positive definite function with minimum at

the origin. Let . Then by the chain rule

for .

is a strong Lyapunov function. The origin is an asymptotically stable equilibrium. ■

Recall Hamiltonian systems. Let and consider

and .

often represents the energy of a system. It is constant along trajectories:

.

If is an equilibrium, then

Is zero at the equilibrium and constant along trajectories. If is positive definite in some

neighborhood of the equilibrium, then it is a weak Lyapunov function.

Example. The simple pendulum. The Hamiltonian is

, leading to

,

13 Chapter 4A

.

sketch support, rod, bob, angle

is an equilibrium point. . Consider

.

is positive definite in a neighborhood of the origin, so is a weak Lyapunov function.

Conclude that is a stable equilibrium point.

sketch phase portrait near origin. Nearly circular orbits surround the origin.

The linearized system is a center. ■

Example. The damped pendulum. The pendulum experiences friction proportional to velocity.

,

.

We now expect the system to lose energy. The system is no longer Hamiltonian. is still

an equilibrium point. Is still a Lyapunov function?

.

Since when , is still a weak Lyap nov f n tion. Lyap nov’s theorem g arantees

that the origin is stable. But can we say more? (To be continued.) ■

Theorem 4.9 (LaSalle’s Invariance Principle). Suppose is an equilibrium for and

suppose is a weak Lyapunov function on some compact forward invariant neighborhood of

. Let

be the set where is not decreasing. Then if is the largest

forward invariant subset of , it is asymptotically stable and attracts every point of .

Proof. For any , suppose is a limit point of the trajectory . Then

for all , since if we could find a sequence and

argue , for sufficiently large , as in the proof of Lyap nov’s theorem and

leading to a contradiction.

Consequently must be a forward invariant subset of . Therefore, by

assumption, . □

Example. The damped pendulum (continued). Let and

14 Chapter 4A

is connected, }. Note that on .

Then is closed and bounded and therefore compact. We can see that is forward invariant

since is always directed out on the boundary of ( has the same sign as

in ) and . is a weak Lyapunov function on . is the

subset of where is non-decreasing. If , the equations of motion give

,

Thus the only subset of that is forward invariant is . By LaSalle’s invarian e prin iple,

is asymptotically stable. ■

Example. The Lorenz system

,

,

.

Consider

. Then if we have and for

. Furthermore

,

where the square is completed from the first two terms of the second line to get the third line.

When and , is a strong Lyapunov function and the Lyapunov function theorem

implies that the origin is an asymptotically stable state. When ,

on the set

and . However, when the Lorenz equations are evaluated on we

find unless . The origin is the largest forward invariant subset of

, and by LaSalle’s invarian e prin iple it is asymptotically stable. ■

Theorem 4.8. If is an asymptotically stable equilibrium that attracts a neighborhood ,

then the function

15 Chapter 4A

is a strong Lyapunov function on .

Do not give proof.

Remark. Lyapunov functions provide a method for showing that equilibria are stable or

asymptotically stable. If the system has an equilibrium such that all of the

eigenvalues of have negative real parts, then theorem 4.6 implies that is

asymptotically stable. The challenge is to construct Lyapunov functions when this is not true. If

all the eigenvalues of do not have negative real parts, there is no general method to

obtain practical expressions for Lyapunov functions.