1 Example System: Predator Prey Model

Example System: Predator Prey Model

https://www.math.duke.edu/education/webfeatsII/Word2HTML/Predator-prey.doc

Example System: Predator Prey Modelx ax bxy

y cy pxy

Populations OscillateWithout predators the prey grows unbounded

Without prey the predators become extinct

Two equilibrium:

(0,0) (x= ,y= )c a

p b(Chose a=100,b=1,c=100,p=1 to fix equilibrium at (100,100))

0x axy y

0x xy cy

Chapter 3Lyapunov Stability – Autonomous Systems

System of differential equations

Question: Is this system “well behaved”?Follow Up Question: What does well behaved mean?• Do the states go to fixed values?• Do the states stay bounded ?• Do we know limits on the size of the states?

Bottom line in this chapter is that we want to know if a differential equation, which we can’t solve, is “well behaved”?

If we could “solve” the system then theses questions may be easy to answer.• We are assuming we can’t solve our

systems of interest.

Preview

System stops moving, “stable”

How can we know which way our system behaves?

Answer: Design a function ( ( )) 0

with the property ( ( )) 0

System state grows, “unstable”

This can be an open- or closed-loop system.

For a closed-loop system it provides a tool:

design to mak

( ( ) 0e )V xg t tu x

No explicit time dependence.The solution will evolve with time, i.e. x(t)

Open or closed-loop system

We will work to quantify “well behaved”

Can have multiple equilibrium points

Three main issues:• System is nonlinear• f is a vector• Can’t find a solution

Khalil calls this the “Challenge and answer form to demonstrate stability”• Challenger proposes an ε bound for the

final state• The answerer has to produce a bound on

the initial condition so that the state always stays in the ε bound.

• Answerer has to provide an answer for every ε proposed

Bottom line: If we start close enough to xe we stay close to xe

may depend on

the value of

Starting time (typically 0)

Evolution of the state

x1 ' = x2 x2 ' = - a sin(x1)

a = 10

-4 -3 -2 -1 0 1 2 3 4

x2 x1 ' = x2 x2 ' = - a sin(x1)

a = 10

-6 -4 -2 0 2 4 6

Pendulum without friction.Is (0,0) a stable equilibrium point in the sense of Definition 2?

YesInitial condition is ,0

Give me an

We can find a

A equilibrium point could be Stable and not ConvergentAn equilibrium point could be Convergent and Not Stable

x(t) goes to xe as t goes to

x1 ' = x2 x2 ' = - a sin(x1)

a = 10

-4 -3 -2 -1 0 1 2 3 4

x2 x1 ' = x2 x2 ' = - a sin(x1)

a = 10

-6 -4 -2 0 2 4 6

Pendulum without friction.Is (0,0) convergent?Is (0,0) asymptotically stable?

x1 ' = x2 x2 ' = - a sin(x1) - 5 x2

a = 10

-4 -3 -2 -1 0 1 2 3 4

x1 ' = x2 x2 ' = - a sin(x1) - 5 x2

a = 10

0 0.5 1 1.5 2 2.5 3

Pendulum with friction.Is (0,0) a stable equilibrium point in the sense of Definition 2?Is (0,0) convergent?Is (0,0) asymptotically stable?

YesYes

Includes a sense of “how fast” the system converges.

Exponential is “stricter” than asymptotic

Can perform this shift to any equilibrium point of interest (a system may have multiple equilibrium points)

V is a scalar function

Could be x1(t)

Could be V(x1(t))

Example: Are each of these PSD, PD, ND, or NSD?

PSD PD

Notation:PSD write as V0PD write as V>0NSD write as V0 ND write as V<0

Example: T

Example: V(x)=x Qx where x

then Q (i.e. a scalar)

we can write a general example as V(x)=q x

11PSD or NSD if q 0

11PD if q 0

11ND q 0

2=x x x

"if and only if"

Don’t loose generality by restricting Q to be symmetric in quadratic form

i.e., Q symmetric

2 21 1 2 2

( ) x x

b c b cT T

b c b c

symmetric asymmetric

aa bV x x x x

ax b c dx

See chapter 2

i.e. only the symmetric part contributes to the quadratic

V is a general function of x1 and x2

V(x) is specific to this

system of interest

Substitute original system

( )x f x

Where is this going?

V(x)21PD: V=x 0

Let’s say that x1 is the state of our system

x12 could be an abstraction of the

energy stored in the system• Analogy would work for potential

energy of a spring or charge on a capacitor

Now if we find: V(x(t)) 0

How could that happen?

Could be negative

V(x1(t))

V(x1(t)) will not increase

1V(x )x

System equations

1(x )f

Gradient

Eventually we will perform the control design to make this true

Can take as many derivatives as you need

Recall that we are considering the equilibrium point at the origin; thus, we already know

Every convergent sequence has a convergent subsequence in Br

Br is a ball about the origin. Our Theorem only applies at the origin

Closeness of x to x=0 implies closeness of V(x) to V(0)

(0) 0V 0ex

Definition of stability

By the

Proof of Theorem 1 (cont):

Not required to be able to derive these equations for this class.

Example 3 (cont)

cos( )

1 cos( )

ll h l

Example 3 (cont)

Note: open interval that does not include 2 , -p2p

Note: •Stable implies that the system (pendulum position and velocity) remains bounded•Does not mean that the system has stopped moving•The energy is constant (V=E) but the system continuously moves exchanging kinetic and potential energy Limit Cycle)

Limit CyclesOscillation leads to a closed path in the phase plane, called periodic orbits. For example, the pendulum has a continuum of closed paths

If there is a single, isolated periodic orbit such that all trajectories tend to that periodic orbit then it is called a stable limit cycle.

Can also have unstable limit cycles.

Khalil, Nonlinear Systems 3rd Ed, p59

Example: LC Oscillator

1div L i vdt

into node

Kirchoff's current law at node 1

Differentiate, define ,

dvi C vdt

x v x v

Example: Oscillator

iR1iR2

i=h(v)

Example: Oscillator

We would expect that the friction will take energy out of the system, thus the system will stop moving.

friction

Example 4 (cont)

Physically, we know that the system will stop because of the friction but our analysis does not show this (we only show it is Stable but can’t show Asymptotic Stability). We know the system is asymptotically stable even if that is not illustrated by this specific Lyapunov analysis.

Note: We will return to this later and fix using the Invariant Set Theorem.

This is a Local stability result because we have limited the range of the state variables for which the result applies.

Have we learned anything useful?

x , constants , 0

input designed to stabilize system as

x ax bu a b

x ax bk x a bk x f x

Given control design problem:

What values (if any) of will work?

2 2 ( ) 2( )

0 2( ) 0

V xx x a bk x a bk x

aV if a bk k

Given the actual system a=1, b=2:

Eq point

0 1 1 x x anything

Eq point

0 1 1.1 0x x

Eq point

0 1 0.9 0x x

t(sec)

There is a missing piece in our original proof that prevents us from applying the result globally:

e0 (PD) and 0 (ND) AS at x

that is, ( ) ( (0))

V x V x

Problem: ( ) might be decreasing (as to the Theorem)but ( ) may not be sufficientto capture the behavior of the states (globablly)

V xV x

( ) decreasingV x

Not radially unbounded because the term in 1 as

. . small V does not imply small

Evolution of the state (green line)

Constant V(x1,x2) contours

State x1 gets large but it is trapped by the constant Lyapunov function

Fixes the problem that the Theorem 2 was local.

Add this condition to AS Theorem:

Global AS: The states will go to zero as time increases from any finite starting state

To be radially unbounded the condition must be verified along any path that results in:

2 21 2 1 22

is Not radially unbounded because along the

line , 0 but as .

Thus we have a path that does not meet the condition

for radially unbounded.

x x V x x x x

scalar function

like radially unbounded

V is PD iif there are class K functions that upper and lower bound the Lyapunov function.

Raleigh-Ritz Theorem.

class functions

Summary: If the state starts within some ball, then the state remains within some other ball for all times.

We will eventually use t here

Reminder of Theorem 2 (need on next slide)

Solve the differential inequality

Solve rhs and substitute a new upper bound

Solve lhs

Find pth root

( ) pV xx

( ) is less than an exponential boundx t p

Now want to address the problem demonstrated in the pendulum with friction example, couldn’t show function was negative definite using energy arguments.

“What happens in M stays in M”

Now want to address the problem demonstrated in the pendulum with friction example, couldn’t show function was negative definite using energy arguments.

An oscillator has a limit cycle

Slotine and Li, Applied Nonlinear Control

This is a condition

that we can try to testThe system ( ) defines the trajectory

Use knowledge of the trajectory to test this condition

The only place when ( ) 0 is when the system is at the equilibriumV x

"vanish identically" -> exactly =0 (vs. approaching zero)

The system ( ) defines the trajectoryx f x

i) PD iv) Radially Unbounded

ii) NSD

Stays at x2=0 forever, i.e. doesn’t move

iii) Doesn’t vanish along a trajectory except x1=0

trajectory

Different from Lyapunov Theorem because:1) V does not need to be Positive Definite2) Applies to multiple equilibrium points and limit cycles

Example of LaSalle’s Theorem

System:

Equilibrium Points:

, are positive constantsa

2Any point (0, ) is an equilibrium point.x

2 21 2

2 2 21 1 2 2 2 1 1 1 2

2 22 1 1

0 (PD), 0 ??? (no)

V x x x x x x ax x xx x ax

Lyapunov Function Candidate:

Lyapunov function design is an iterative process!

Can we change the Lyapunov function to yield a better derivative?: It would be nice if this was a “1”

2 21 2

Suggests the possibility of adding in

Example of LaSalle’s Theorem (cont)

2 21 2

2 2 21 1 2 2 2 1 1 1 2

0 (PD), 0 ??? (no)

V x x x x x x ax x x

Can we further change the Lyapunov function to yield a better derivative?

2 21 1

Would be nice if we could turn intowhere the is more negative than is positive

bxbx ax

2 2 2 2 2

2 22 1 1

Suggests the possibility of changing further

21 1 1

... ..

V x b x x x bx

V x x bx

"Checklist"

) invariant wrt system

) 0 in

) where 0

) largest invariant set in

ii V M

iii E V

iv N E

We have enough information to conclude the system is “stable” in some region-> set is invariant wrt the system

Is V PD? no

"Checklist"

) invariant wrt system

) 0 in

) where 0

) largest invariant set in

ii V M

iii E V

iv N E

Conclusion from Theorem

2( ) constant as x t t

Not negative beyond this point

-4 -3 -2 -1 0 1 2 3 4-5

( )V x

Is this correct yes

Why includes sy mV ste

x1 ' = 3 x2 x2 ' = - 5 x1 + x13 - 2 x2

-4 -3 -2 -1 0 1 2 3 4

However, it does look like there should be a region such that the system converges.Can we define that region?

Define Region of Attraction (RA):

Question we want to answer: Where can the system start (Initial Conditions at t=0) so that we know it will move to the equilibrium point?

Original assumption , V is decreasing

We are assuming we know there is a bound on x2

Solve differential inequality

We have now shown there is a bound on x2

x ' = - k x + x3

y ' = 0 k = 1

-4 -3 -2 -1 0 1 2 3 4

(0) system is ES

for the specific case of 2 :

2 (0) 2

Find RA for k=2:

Phase portrait:

This is our RA:

Estimate Region of Attraction (RA) Using LaSalle’s Theorem:

Doesn’t mean it is the entire RA

Finish Example 14

Could be larger,

x1 ' = 3 x2 x2 ' = - 5 x1 + x13 - 2 x2

-4 -3 -2 -1 0 1 2 3 4

x2Contour lines of V(x)

Finish Example 14

T T Tx Ax x x A

Proven the “If” part of the Theorem. If Q>0 and found P then AS real part of eigenvalues is >0

Now prove the “only If” part of the Theorem.

Example: Is the following system AS?

1 2 11 2

p p xV x x

1 2 1 2

2 3 2 3

Choose an arbitrary that is PD and symetric

1 0easy pick

now use Lyapunov equation ( - ( )) to find

1 0 0 1 0 1

0 1 1 1 1 1

Q A P PA

Q A P PA P

p p p p

2 3 2 1 2

1 2 2 3 3 2 3

2 3 1 2

3 1 2 2 3

2 20 1

solve 3 equations with 3 unknowns

p p p p p

p p p p p p p

p p p p

p p p p p

solve 3 equations with 3 unknowns

0 2 0 1

1 1 1 0

0 2 2 1

1.5 0.5

with 0.6910, 1.8090

In MATLAB: >> P=lyap([0 -1;1 -1],[1 0;0 1])

Found a PD, symmetric P from an arbitrary Q

eigenvalues of A satify e( ) 0

the origin of the system is ASi

Wouldn’t it be easier just to find the eigenvalues of A? Yes, but having P (and hence a Lyapunov function) will be useful later.

Linearization of a Nonlinear System

Assume small= 0 by assumption of equilibrium point

Change of variables to make the origin the equilibrium

Example: Linearize System

12 1 2

22 1 1 1 2

2 2 1 2

1 2 1 2

cos( ),

1 sin( )

cos( ) 2 sin( )

2 1 sin( ) 1 cos( )

xx x xf x

xx x x x x

x x x xf

x x x xx

Eigenvalues = 1,1

Origin is unstable

Reminder of Jacobian:

x1 ' = x1 x2 ' = x1 + x2

-2 -1 0 1 2 3 4

x1 ' = x22 + x1 cos(x2) x2 ' = x2 + (x1 + 1) x1 + x1 sin(x1)

-2 -1 0 1 2 3 4

-0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 0.1

x1 ' = x1 x2 ' = x1 + x2

-0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 0.1

Compare favorably close to the equilibrium point

LinearizedOriginal System

May compare less favorably further away from equilibrium point

Example : Linearize System (cont)

Example System: Predator Prey Modelx ax bxy

y cy pxy

Populations OscillateWithout predators the prey grows unbounded

Without prey the predators become extinct

Two equilibrium:

(0,0) (x= ,y= )c a

p b(Chose a=100,b=1,c=100,p=1 to fix equilibrium at (100,100))

0x axy y

0x xy cy

Example System: Predator Prey Model

( ) 0e

Lost idea of extinction

(100,100)

Lost idea of oscillating populations

Linearization at the equilibrium points

Unstable equilibrium

In general, proving a system is unstable is not very “constructive” in designing control systems.

22( )( )

2 21 2

21 2 2

V x x x

Example

Can we define a set U V>0 ?

Doesn't appear possible because in region about the origin

would have to include a point x 0 and x >0

which would make V 0

Not conclusive, doesn’t mean it is stable

Summary

x=0 stable

x=0 exponentially stable

x=0 asymptotically stable

convergent

x=0 stable

x=0 asymptotically stable x=0 exponentially stableconvergent

x=0 stablex=0 asymptotically stable

Locally, NSD dV/dtLocally, state could grow while V shrinks

x=0 asymptotically stable

globally

global, local depend on conditions

V is class K8

x=0 asymptotically stable local

Global, asymptotically stable

Conditions true in entire state space9

N is asymptotically stable

Local, global

Summary

Linearization

Inherently local result since it is an approximation of a nonlinear system at a point

x1 ' = x1 x2 ' = x1 + x2

-2 -1 0 1 2 3 4

system Linear approximation

Summary

if we use feedback of the state i.e. ( ) ( ) then

x f x u

u x g x

the analysis tools in this

chapter will be the basis

for designing the

control u(x)

is still autonomous

Homework

• Set 3.A: Pendulum without friction• Set 3.B: Equilibrium points, quadratic

Lyapunov function candidate to determine stability

• Set 3.C: Book problems 3.1, 3.3, 3.6, 3.7• Set 3.D: Book problems 3.4,3.5,3.13, 3.14• Set 3.E Pendulum with friction AS

Homework #3-A

• “Pend w/out friction” - Find all of the equilibrium points for the pendulum without friction. – Plot the phase portrait from -3<x1<3 and -

10<x2<10. – From inspection of the phase portrait , does it

appear that the equilibrium points are stable?– Use the Lyapunov function candidate from the

notes to examine stability at x1= and 2

Homework 3-A2

0 sin( )

Equilibrium points at (0,0), ( ,0) for 1, 2,3...

Pendulum spins

Pendulum swings

Pendulum doesn’t move(stable EQ point)

Pendulum doesn’t move at exactly that point(Unstable EQ point)

Pend w/out fric

Homework 3-A

2 22 1

1(1 cos( ))

2V ml y mgl y

2 1 2 1

( )( ) ,

sin( ) ,sin( )

sin( ) sin( )

2 sin( )

f yV VV y

f yy y

ymgl y ml y g

mgly y mgly y

mgly y

1 1 1 1

2 2 2 2

y x y x

2 1sin( )

Change of variable Shifted system

1Is EQ at stable?x

( ) 0 ?V y No

Is the EQ point stable or unstable?

No conclusion can be made

yy y y y

Same Lyapunov function candidate that we used in notes

Pend w/out fric (cont)

Homework 3-A

2 22 1

1(1 cos( ))

2V ml y mgl y

2 1 2 1

( )( ) ,

sin( ) ,sin( 2 )

sin( ) sin( 2 )

f yV VV y

f yy y

ymgl y ml y g

mgly y mgly y

1 1 1 1

2 2 2 2

2 then

y x y x

2 1sin( 2 )

Change of variable Shifted system

1Is EQ at 2 stable?x

( ) 0 ?V y Yes

Is the EQ point stable or unstable?

Stable

Same Lyapunov function candidate that we used in notes

yy y y y

Pend w/out fric (cont)

Homework #3-B

1) x x

Find the equilibrium points for each of the following and use a quadratic Lyapunov function candidate to determine stability of each equilibrium point.

2) 5x x 1 2

3) x xx x x

1 2 3 1

2 1 3 2

3 1 2 3

5) x x x x

x x x x

specify ( )

7) x ax bx

x cx dx

Using a quadratic Lyapunov function, find the conditions on a,b,c,d such that the system is stable at x=0

Using a quadratic Lyapunov function, find u=f(x) such that the system is stable at x=0

22 1 1 2

x x x x

Homework 3-B

Eq pt at 0

0 (PD), 0 (ND) AS (at x=0)

V xx x x x

Homework 3-B

Eq pt at 5

Define change of variables

New system

0 (PD), 0 (ND) AS at y=0 AS at

5 then

V yy y y y

Homework 3-B

1 2 21 2 1 2

1 1 2 2 1 2 2 1 2

Eq pt at 0

1 01 1 1[ ]

0 12 2 2

0 (PD), 0 (NSD) Stable

Can we get a "better" result? Probably, let's try Theorem 8

xV x x x x

V x x x x x x x x x

1 2 1 2

sider R=[ , ] ,

0 0 thus 0 t 0

then by the second system equation 0= 0

0 does not vanish identically along any trajectory other than = =0 Asymptotically Stable

Tx x x x

V x x x

eorem 9: ince V is also radially unbounded, Origin is Globally Asymptotically Stable (GAS)s

Homework 3-B

2 1 1 2

2 21 2

21 1 2 2 1 2 2 1 1 2

2 2 2 2 21 2 1 2 2 1 2 2 1

Eq pt at 0

0 (PD) , 0 when -1 1 (NSD) Locally Stable ISL

Can we get a "better" r

x x x x

V x x x x x x x x x x

x x x x x x x x x

1 2 1 2

21 1 2 1

esult? Probably, let's try Theorem 8

Consider R=[ , ] ,

0 0 thus 0 t 0

then by the second system equation 0= 1 0

0 does not vanish identically along any trajectory o

Tx x x x

V x x x

x x x x

1 2ther than = =0 Asymptotically Stable

Theorem 9: since V is also radially unbounded, Origin is Locally Asymptotically Stable (LAS)

Homework 3-B

1 2 3 1

2 1 3 2

3 1 2 3

2 2 21 2 3

1 1 2 2 3 3 1 2 3 1 2 1 3 2 3 1 2 3

2 2 21 2 3

Eq pt at 0

0 (PD), 0 (ND) AS

x x x x

V x x x

V x x x x x x x x x x x x x x x x x x

100100

Homework 3-B

6) Eq pt at 0

would like to get rid of the unhelpful and add a stabilizing termlet -2 (or even - with K>1)

20 (PD), 0 (ND) AS

Note that

x x uAssume x

V xx x xux

xu x u Kx

V x x xV V

has an EQ point at zero as we assumed.x

We just designed a feedback control to stabilize the systemWe used the Lyapunov function to design the control.

101101

Homework 3-B

2 21 2

1 1 2 22 2

1 1 2 1 2 22 2

1 2 1 2

Eq pt at 01 1

choose and to stabilize the system ie 0 and 0choose and to make the las

x ax bx

x cx dxAssume x x

V x x x xax bx x cx x dxax dx b c x x

a d a db c

t term "go away" ie 0 then

0 (PD), 0 (NSD) Stable

Since this is a linear system, compare to finding conditions for positive eigenvalues:0

a bA I a d bc

a d ad bc

102102

Homework #3.CChapter 3 - Problems 3.1, 3.3, 3.6, 3.7

103103

Homework 3.C

3 21 1 1 2 1 1 1

0 1 0 0,1, 1

equilibrium points at (0,0), (1,0), and (-1,0)

x x x x x x x

System doesn’t move, i.e. derivatives are zero

Notation means the second eq point, not the square of the eq point

See that (0,0) which corresponds to (1,0) in the original system is an equilibrium point

104104

Homework 3.C

See that (0,0) which corresponds to (-1,0) in the original system is an equilibrium point

(-1,0)

105105

Homework 3.C

Solve for conditions on v (note that this is the voltage on the coil not a Lyapunov function candidate) so that x1=yo and the system is at rest, i.e. solve v so that there is an equilibrium point at x1=yo. Approach: set derivatives to zero, x1 to the constant yo solve for v.

106106

Homework 3.CEmphasizing that we are looking for a constant x3

107107

Homework 3.3 (sol)

This is why linearization can be so powerful tool - you get to use all of the linear analysis tools.

108108

Homework 3.C

a) Verify that the origin is an equilibriumSubstitute (0,0)

thus (0,0) is an equilibrium point

109109

Homework 3.C

110110

Homework 3.C

0 From pplane8:

111111

Homework 3.CTest this first

V can be zero other than at the origin

112112

Homework 3.C

113113

Homework 3.DChapter 3 - Problems 3.4,3.5,3.13, 3.14

114114

Homework 3.D (Sol)

115115

Homework 3.D (Sol)

For small x2

Because we assumed small x2

116116

Homework 3.D (Sol)

117117

Homework 3.D (Sol)

a) Substitute (0,0) in the system, then 0 the system can't move from the point

0 for all time which means in stays in the set (0,0). The equilibrium point is an invariant set.

Note tha

t any equilibrium point is an invariant set.

a) Alternate solution - A more general way to consider this question

Invariance implies that there is no deviation from the set over time, ie ( ) 0

Sub from system

x x x x xxdt x

2 22 1 2 1 2

2: = 1 3 2

Sub from our set definition (using 0): =0

Thus the set is invariant w.r.t the system.

x x x x x

x dx x

118118

Homework 3.D (Sol)

2 2 2 2 21 2 1 2 1 2 1 2

1b) The set defines a relationship between x and x as: 1 (3 2 ) 0 3 1 2 1 2 ;

i.e. points ( 1 2 , )3

x x x x x x

12 2 2 21 2 1 2 1 2 1 1 2 2

1 2 2 1 2

Invariance implies that there is no deviation from the set over time, ie ( ) 0

1 (3 2 ) 1 (3 2 ) 6 4 6 4

2Sub from system: = 6 4 1 3

x x x x x x x x x x xxdt x

x x x x x

2 2 2 2 21 2 2 1 2

2 21 2

2 4 1 3 2

Sub from our set definition (using 3 1 2 ): =0

Thus the set is invariant w.r.t the system.

x x x x x

119119

Homework 3.D (Sol)

This is a clever way to add a degree of freedom to a quadratic Lyapunov function. a,b don’t change the basic nature of the quadratic function, ie still PD and radially unbounded.

a,b provide the opportunity to cancel these cross terms which might have otherwise stopped our analysis (ie had we chosen a=b=1)

120120

Homework 3.D (Sol)Why choose this? Because someone tried a lot of other V’s that did not work.

Homework 4.E

• Pendulum with friction AS

Remember the pendulum with friction?

sin( )

It is possible to prove asymptotic stability without needing the invariant set theorem

using the Layounov function candidate

1 cos( )

g kx x x

11 3 1

3 22 2

211 22 11 22 3

11 22 3

1, with ,

and where must be P.D. implies

0, 0, and 0.

Show what additional constraints are needed on , ,and to make N.D.?

T p p xx Px P x

p p p p p

p p p V

Remember the pendulum with friction?

sin( )

It is possible to prove asymptotic stability without needing the invariant set theorem

using the Layounov function candidate

1 cos( )

g kx x x

11 3 1

3 22 2

211 22 11 22 3

11 22 3

1, with ,

and where must be P.D. implies

0, 0, and 0.

Show what additional constraints are needed on , ,and to make N.D.?

sin( )

T p p xx Px P x

p p p p p

p p p V

gV x x

11 2 3 1 3 2

3 2 22 1 22 2

23 1 1 2 1 22 2 1 22 3 2 11 3 1 2

11 3 22

sin( )sin( )

sin( )

sin( ) sin( ) sin( )

and 1 (to zero t

g kp x p x p x

g l mx Px x x xg kl

p x p x p xl m

g g g k kp x x x x p x x p p x p p x x

l l l m m

klet p p p

23 1 1 22 3 2

2 211 22 3 3 3 3

22 3 3

he unfriendly terms)

sin( )

requires 0

from PD requirement 0 0

Now look at the last term in the last : 0 .

At this point a lot of

g kV p x x p p x

k kp p p p p p

m mk k

V p p pm m

13 3 2

11 3 22 1 12

will work, we can pick for convenience.

Then , , 1 ensures is ND for : .2 2

(0) 0, is PD and radially unbounded and is ND for :

guarantees that the o

k kp p p V x x

V V V x x

rigin is locally asymptotically stable.

Additional Examples

32 2 2 0 1 1 1

3.AE1: Exam stability at the origin of

where , >0.

State space model:

eq points: ( , )=(0,0) (only consider the real roots).

Lyapunov function can

mx bx x k x k x

x bx x k x k xm

2 2 42 0 1 1 1

32 2 0 1 1 1 1 1

2 3 32 2 0 1 2 1 1 2 0 1 2 1 1 2

didate

1 1 (P.D., radially unbounded)

(N.S.D.)

mV x k x k x

V x mx k x x k x x

bx x k x x k x x k x x k x x

Additional Examples

32 0 1 1 1 1 2

3.AE1: (cont) Use Invariant Set Theorem

0 for all ( ,0) where .

Look at original dynamics to see for ( ,0) :

0 is constant

1 where (0,0) is only 0

for the othe

x k x k x x xm

r the system continues to move.

Thus the is only identically zero, meaning it will always stay at 0,

at the origin (0,0).

GAS at origin.

Could use LaSalles Theorem

{ : 0} (not invariant in

E x M x

( ,0))

(0,0) is the largest invariant set in .

( ) 0 as

1 Example System: Predator Prey Model

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