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Example System: Predator Prey Model
https://www.math.duke.edu/education/webfeatsII/Word2HTML/Predator-prey.doc
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Example System: Predator Prey Modelx ax bxy
y cy pxy
x y
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
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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.
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Preview
x1
x2
x(0)
System stops moving, “stable”
How can we know which way our system behaves?
Answer: Design a function ( ( )) 0
with the property ( ( )) 0
V x t
V x t
x1
x2
x(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
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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
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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
x1
x2
may depend on
the value of
Starting time (typically 0)
Evolution of the state
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x1 ' = x2 x2 ' = - a sin(x1)
a = 10
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x1
x2 x1 ' = x2 x2 ' = - a sin(x1)
a = 10
-6 -4 -2 0 2 4 6
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
t
x1
Pendulum without friction.Is (0,0) a stable equilibrium point in the sense of Definition 2?
YesInitial condition is ,0
6
Give me an
We can find a
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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
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x1 ' = x2 x2 ' = - a sin(x1)
a = 10
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x1
x2 x1 ' = x2 x2 ' = - a sin(x1)
a = 10
-6 -4 -2 0 2 4 6
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
t
x1
Pendulum without friction.Is (0,0) convergent?Is (0,0) asymptotically stable?
NoNo
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x1 ' = x2 x2 ' = - a sin(x1) - 5 x2
a = 10
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x1
x2
x1 ' = x2 x2 ' = - a sin(x1) - 5 x2
a = 10
0 0.5 1 1.5 2 2.5 3
0
0.1
0.2
0.3
0.4
0.5
0.6
t
x1
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?
Yes
YesYes
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Includes a sense of “how fast” the system converges.
Exponential is “stricter” than asymptotic
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Can perform this shift to any equilibrium point of interest (a system may have multiple equilibrium points)
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V is a scalar function
x1
V(x1)
Could be x1(t)
Could be V(x1(t))
x1
V(x1)
x1
V(x1)
x1
V(x1)
Example: Are each of these PSD, PD, ND, or NSD?
PSD PD
None
ND
Notation:PSD write as V0PD write as V>0NSD write as V0 ND write as V<0
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x1
V(x1)
Example: T
211 1
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 T
2=x x x
"if and only if"
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Don’t loose generality by restricting Q to be symmetric in quadratic form
i.e., Q symmetric
2 2
2 2
2 21 1 2 2
0
0
( ) x x
b c b cT T
b c b c
symmetric asymmetric
aa bV x x x x
dc d
ax b c dx
See chapter 2
i.e. only the symmetric part contributes to the quadratic
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V is a general function of x1 and x2
V(x) is specific to this
system of interest
Substitute original system
( )x f x
1
2
xx
x
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Where is this going?
x1
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
t
V(x1(t))
V(x1(t)) will not increase
1V(x )x
System equations
1(x )f
Gradient
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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
0
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Every convergent sequence has a convergent subsequence in Br
Br is a ball about the origin. Our Theorem only applies at the origin
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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):
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mg
lq
Not required to be able to derive these equations for this class.
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Example 3 (cont)
mg
lq
h
l-h
cos( )
cos( )
1 cos( )
l h
ll h l
h l
v
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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)
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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
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Example: LC Oscillator
Khalil, Nonlinear Systems 3rd Ed, p59
1div L i vdt
dt L
dvi C
dt
into node
1 2
1 2
2 1
Kirchoff's current law at node 1
10
Differentiate, define ,
1
dvi C vdt
dt L
x v x v
x x
x xLC
1
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Example: Oscillator
Khalil, Nonlinear Systems 3rd Ed, p59
iR1iR2
i=h(v)
v
vsys
vD
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Example: Oscillator
1
2 1
x v
x x v
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ND?
NSD?
We would expect that the friction will take energy out of the system, thus the system will stop moving.
friction
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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.
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This is a Local stability result because we have limited the range of the state variables for which the result applies.
+
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Have we learned anything useful?
x , constants , 0
input designed to stabilize system as
( )
p
p p
x ax bu a b
u k x
x ax bk x a bk x f x
Given control design problem:
2
2
What values (if any) of will work?
2 2 ( ) 2( )
0 2( ) 0
p
p p
p p
k
V x
V xx x a bk x a bk x
aV if a bk k
b
Given the actual system a=1, b=2:
1
2pk
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)
x
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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 V
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
x
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1 1
1
Not radially unbounded because the term in 1 as
. . small V does not imply small
x V x
i e x
Evolution of the state (green line)
Constant V(x1,x2) contours
State x1 gets large but it is trapped by the constant Lyapunov function
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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
:
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To be radially unbounded the condition must be verified along any path that results in:
1 2
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.
V x x
x x V x x x x
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scalar function
like radially unbounded
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V is PD iif there are class K functions that upper and lower bound the Lyapunov function.
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Raleigh-Ritz Theorem.
class functions
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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
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Reminder of Theorem 2 (need on next slide)
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Solve the differential inequality
Solve rhs and substitute a new upper bound
Solve lhs
Find pth root
2
pVx
K
00
2
( ) pV xx
K
(0) 0
( ) 0
( ) 0
V
V x
V x
( ) is less than an exponential boundx t p
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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.
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An oscillator has a limit cycle
Slotine and Li, Applied Nonlinear Control
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This is a condition
that we can try to testThe system ( ) defines the trajectory
Use knowledge of the trajectory to test this condition
x f x
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The only place when ( ) 0 is when the system is at the equilibriumV x
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"vanish identically" -> exactly =0 (vs. approaching zero)
The system ( ) defines the trajectoryx f x
iv)
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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
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Different from Lyapunov Theorem because:1) V does not need to be Positive Definite2) Applies to multiple equilibrium points and limit cycles
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Example of LaSalle’s Theorem
System:
Equilibrium Points:
, are positive constantsa
2Any point (0, ) is an equilibrium point.x
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2 21 2
2 2 21 1 2 2 2 1 1 1 2
2 22 1 1
1 1
2 2
1
0 (PD), 0 ??? (no)
V x x
V x x x x x x ax x xx x ax
V V
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
1 1
2 2
V
V x x
Example of LaSalle’s Theorem (cont)
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2 21 2
2 2 21 1 2 2 2 1 1 1 2
21
1 1
2 21
0 (PD), 0 ??? (no)
V x x
V x x x x x x ax x x
axV V
Lyapunov Function Candidate:
Can we further change the Lyapunov function to yield a better derivative?
21
2 21 1
Would be nice if we could turn intowhere the is more negative than is positive
bxbx ax
Example of LaSalle’s Theorem (cont)
2
2
2 2 2 2 2
2 22 1 1
Suggests the possibility of changing further
1...
21 1 1
... ..
..
V
V x b
V x b x x x bx
V x x bx
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Example of LaSalle’s Theorem (cont)
2
"Checklist"
Set
) invariant wrt system
) 0 in
) where 0
) largest invariant set in
D
i M D
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
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Lyapunov Function Candidate:
Example of LaSalle’s Theorem (cont)
M
N
2
"Checklist"
Set
) invariant wrt system
) 0 in
) where 0
) largest invariant set in
D
i M D
ii V M
iii E V
iv N E
Conclusion from Theorem
2( ) constant as x t t
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Example of LaSalle’s Theorem (cont)
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x
x
4x2x
Not negative beyond this point
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-4 -3 -2 -1 0 1 2 3 4-5
0
5
10
15
20
25
30
35
40
( )V x
?
?
Is this correct yes
Why includes sy mV ste
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x1 ' = 3 x2 x2 ' = - 5 x1 + x13 - 2 x2
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x1
x2
D
However, it does look like there should be a region such that the system converges.Can we define that region?
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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?
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Original assumption , V is decreasing
We are assuming we know there is a bound on x2
Solve differential inequality
0
k
k
k
We have now shown there is a bound on x2
1x
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x ' = - k x + x3
y ' = 0 k = 1
-4 -3 -2 -1 0 1 2 3 4
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
x
y
01
k=2
2
2
(0) system is ES
for the specific case of 2 :
(0) 2
(0) 2
2 (0) 2
A
k x
R k
x
x
x
Find RA for k=2:
22x
Phase portrait:
This is our RA:
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Estimate Region of Attraction (RA) Using LaSalle’s Theorem:
Doesn’t mean it is the entire RA
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Finish Example 14
Could be larger,
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-4-2
02
4
-4
-2
0
2
40
50
100
150
x1 ' = 3 x2 x2 ' = - 5 x1 + x13 - 2 x2
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x1
x2Contour lines of V(x)
V(x)
x1
x2
Finish Example 14
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T T Tx Ax x x A
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PA
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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.
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Example: Is the following system AS?
1 1
2 2
0 1
1 1
x x
x x
1 2 11 2
2 3 2
p p xV x x
p p x
2 2
1 2 1 2
2 3 2 3
( )
Choose an arbitrary that is PD and symetric
1 0easy pick
0 1
now use Lyapunov equation ( - ( )) to find
1 0 0 1 0 1
0 1 1 1 1 1
T
x
T
Q A P PA
Q
Q I
Q A P PA P
p p p 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
1 0
0 1
21 0
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
1
2
3
1
2
3
1 2
solve 3 equations with 3 unknowns
0 2 0 1
1 1 1 0
0 2 2 1
1.5
0.5
1
1.5 0.5
0.5 1
with 0.6910, 1.8090
p
p
p
p
p
p
P
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.
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Linearization of a Nonlinear System
Assume small= 0 by assumption of equilibrium point
Change of variables to make the origin the equilibrium
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Example: Linearize System
2
12 1 2
22 1 1 1 2
2 2 1 2
1 2 1 2
[0,0]
cos( ),
1 sin( )
cos( ) 2 sin( )
2 1 sin( ) 1 cos( )
1 0
1 1Tx
xx x xf x
xx x x x x
x x x xf
x x x xx
f
x
Eigenvalues = 1,1
Origin is unstable
Reminder of Jacobian:
2x2
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x1 ' = x1 x2 ' = x1 + x2
-2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
x1
x2
x1 ' = x22 + x1 cos(x2) x2 ' = x2 + (x1 + 1) x1 + x1 sin(x1)
-2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
x1
x2
x1 ' = x22 + x1 cos(x2) x2 ' = x2 + (x1 + 1) x1 + x1 sin(x1)
-0.1 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 0.1
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
x1
x2
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
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
x1
x2
Compare favorably close to the equilibrium point
LinearizedOriginal System
May compare less favorably further away from equilibrium point
Example : Linearize System (cont)
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8080
Example System: Predator Prey Modelx ax bxy
y cy pxy
x y
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
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Example System: Predator Prey Model
( ) 0e
Lost idea of extinction
(100,100)
Lost idea of oscillating populations
(0,0)
Linearization at the equilibrium points
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ex
0x
Unstable equilibrium
In general, proving a system is unstable is not very “constructive” in designing control systems.
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22( )( )
0 if
V x
V x
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1 2
2 2 1
2 21 2
21 2 2
1
2
2
x x
x x x
V x x
V x x x
Example
Can we define a set U V>0 ?
1 2
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
0x
Not conclusive, doesn’t mean it is stable
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Summary
x=0 stable
x=0 exponentially stable
x=0 asymptotically stable
convergent
x=0 stable
time
X(t)
time
X(t)
time
X(t)
x=0 asymptotically stable x=0 exponentially stableconvergent
time
X(t)
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x=0 stablex=0 asymptotically stable
Locally, NSD dV/dtLocally, state could grow while V shrinks
x=0 asymptotically stable
globally
1 2 3
x=0 exponentially stable
global, local depend on conditions
7
V is class K8
x=0 asymptotically stable local
Global, asymptotically stable
Conditions true in entire state space9
10
N is asymptotically stable
Local, global
Summary
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Summary
Linearization
11
x=0 exponentially stable
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
-4
-3
-2
-1
0
1
2
x1
x2
x1 ' = x22 + x1 cos(x2) x2 ' = x2 + (x1 + 1) x1 + x1 sin(x1)
-2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
x1
x2
system Linear approximation
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f2(x)
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)
f(x)u
f(x)u
2( )
is still autonomous
x f x
g(x)
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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
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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
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Homework 3-A2
1
0
0 sin( )
Equilibrium points at (0,0), ( ,0) for 1, 2,3...
x
a x
n n
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
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Homework 3-A
2 22 1
1(1 cos( ))
2V ml y mgl y
1
21 2
22
1 21
2 1 2 1
2 1
( )( ) ,
( )
sin( ) ,sin( )
sin( ) sin( )
2 sin( )
f yV VV y
f yy y
ymgl y ml y g
yl
mgly y mgly y
mgly y
1 1 1 1
2 2 2 2
then
then
y x y x
y x y x
1 2
2 1sin( )
y y
gy y
l
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)
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Homework 3-A
2 22 1
1(1 cos( ))
2V ml y mgl y
1
21 2
22
1 21
2 1 2 1
( )( ) ,
( )
sin( ) ,sin( 2 )
sin( ) sin( 2 )
0
f yV VV y
f yy y
ymgl y ml y g
yl
mgly y mgly y
1 1 1 1
2 2 2 2
2 then
then
y x y x
y x y x
1 2
2 1sin( 2 )
y y
gy y
l
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)
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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
2 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
x x x x
6)
specify ( )
x x u
u f x
1 1 2
2 1 2
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
1 2
22 1 1 2
4)
1
x x
x x x x
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Homework 3-B
2
2
1)
Eq pt at 0
1
2
0 (PD), 0 (ND) AS (at x=0)
x x
x
V x
V xx x x x
V V
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Homework 3-B
2
2
2) 5
Eq pt at 5
Define change of variables
New system
1
2
0 (PD), 0 (ND) AS at y=0 AS at
5 then
x=-5
x x
x
x
y
V y
V yy y y y
V V
y x y
y
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9797
Homework 3-B
1 2
2 1 2
1 2
1 2 21 2 1 2
2
1 1 2 2 1 2 2 1 2
22
3)
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
Con
x x
x x x
x x
xV x x x x
x
V x x x x x x x x x
x
V V
1 2 1 2
2 2 2
1 2 1
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
Th
Tx x x x
V x x x
x x x
V x x
eorem 9: ince V is also radially unbounded, Origin is Globally Asymptotically Stable (GAS)s
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Homework 3-B
1 22
2 1 1 2
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
1
4) 4)
1
Eq pt at 0
1 1
2 2
1
1
0 (PD) , 0 when -1 1 (NSD) Locally Stable ISL
Can we get a "better" r
x x
x x x x
x x
V x x
V x x x x x x x x x x
x x x x x x x x x
V V x
1 2 1 2
2 2 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
V
1 2ther than = =0 Asymptotically Stable
Theorem 9: since V is also radially unbounded, Origin is Locally Asymptotically Stable (LAS)
x x
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Homework 3-B
1 2 3 1
2 1 3 2
3 1 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
5)
Eq pt at 0
1 1 1
2 2 2
( )
0 (PD), 0 (ND) AS
x x x x
x x x x
x x x 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
x x x
V V
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100100
Homework 3-B
2
2
2
2
2 2 2
6) Eq pt at 0
1
2
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 x
V xx x xux
xu x u Kx
V x x xV V
x
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.
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101101
Homework 3-B
1 1 2
2 1 2
1 2
2 21 2
1 1 2 22 2
1 1 2 1 2 22 2
1 2 1 2
7)
Eq pt at 01 1
2 2
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
V x x x xax bx x cx x dxax dx b c x x
a d a db c
2
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
0
( ) 0
b c
V V
a bA I a d bc
c d
a d ad bc
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102102
Homework #3.CChapter 3 - Problems 3.1, 3.3, 3.6, 3.7
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Homework 3.C
1 2
3 21 1 1 2 1 1 1
0
0 1 0 0,1, 1
equilibrium points at (0,0), (1,0), and (-1,0)
x x
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
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104104
Homework 3.C
See that (0,0) which corresponds to (-1,0) in the original system is an equilibrium point
(-1,0)
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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.
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Homework 3.CEmphasizing that we are looking for a constant x3
x1=yo
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Homework 3.3 (sol)
This is why linearization can be so powerful tool - you get to use all of the linear analysis tools.
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Homework 3.C
a) Verify that the origin is an equilibriumSubstitute (0,0)
1
2
00
thus (0,0) is an equilibrium point
xx
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Homework 3.C
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Homework 3.C
1 1
0 From pplane8:
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Homework 3.CTest this first
V can be zero other than at the origin
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Homework 3.C
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Homework 3.DChapter 3 - Problems 3.4,3.5,3.13, 3.14
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Homework 3.D (Sol)
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Homework 3.D (Sol)
For small x2
Because we assumed small x2
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Homework 3.D (Sol)
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Homework 3.D (Sol)
1 2
1 2
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
x x
x x
t any equilibrium point is an invariant set.
11 2
2
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
1 1
Sub from system
dset
dtxd
x x x x xxdt x
2 22 1 2 1 2
1
2
2: = 1 3 2
3
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
x dt
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Homework 3.D (Sol)
2 2 2 2 21 2 1 2 1 2 1 2
22 2
1b) The set defines a relationship between x and x as: 1 (3 2 ) 0 3 1 2 1 2 ;
31
i.e. points ( 1 2 , )3
x x x x x x
x x
12 2 2 21 2 1 2 1 2 1 1 2 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
3
dset
dtxd
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
x x
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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)
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Homework 3.D (Sol)Why choose this? Because someone tried a lot of other V’s that did not work.
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Homework 4.E
• Pendulum with friction AS
1 2
2 1 2
1
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( )
x x
g kx x x
l m
gV x
l
11 3 1
3 22 2
211 22 11 22 3
11 22 3
1, with ,
2
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 x
P
p p p p p
p p p V
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1 2
2 1 2
1
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( )
x x
g kx x x
l m
gV x
l
11 3 1
3 22 2
211 22 11 22 3
11 22 3
2 1
1, with ,
2
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 x
P
p p p p p
p p p V
gV x x
l
11 2 3 1 3 2
2 1
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
T 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
m
23 1 1 22 3 2
3
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
l m
p
k kp p p p p p
m mk k
V p p pm m
13 3 2
2
11 3 22 1 12
1 1
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
kp p
m
k kp p p V x x
m m
V V V x x
rigin is locally asymptotically stable.
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Additional Examples
30 1
0 1
1 2
32 2 2 0 1 1 1
1 2
3.AE1: Exam stability at the origin of
0
where , >0.
State space model:
1
eq points: ( , )=(0,0) (only consider the real roots).
Lyapunov function can
mx bx x k x k x
k k
x x
x bx x k x k xm
x x
2 2
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
22 2
3
2
didate
1 1 (P.D., radially unbounded)
2 2 4
(N.S.D.)
x x
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
bx x
b x
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Additional Examples
1 2 1
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
p p
p
V x x
x
x x x
x k x k x x xm
1
2
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
x
V V
E x M x
( ,0))
(0,0) is the largest invariant set in .
( ) 0 as
px
N E E
x t t