Topic 3 Lyapunov.pdf

Lyapunov Stability Theory

Aleksandr Mikhailovich Lyapunov Born: 6 June 1857 in Yaroslavl, Russia Died: 3 Nov 1918 in Odessa, Russia

His doctoral thesis The general problem of the stability of motion at the University of Moscow (he was awarded his doctorate on 12 October, 1892) established the Lyapunov Stability Theory

On 31 October 1918 Lyapunov's wife died and later that day Lyapunov shot himself. He died three days later in hospital.

Autonomous Systems Consider a nonlinear autonomous system with state space representation

x  f ( x)

where x is an n-dimensional state vector. Suppose x0 is an equilibrium point of the system. Want: Analyze the behavior of the system around x0 Defining

e  x  x0

The perturbation system

e  x  f ( x )  f (e  x0 )  f1 (e) e = 0 is an equilibrium point of the perturbation system. We will always assume that 0 is the equilibrium point of study.

Norms The Euclidean norm of an n-dimensional vector x, is defined as 1/ 2

 2 x 2    xi   i 1  n

L2 norm:

A norm is a measure of the size of x. A norm must satisfy the following properties: (1) For any constant c,

cx | c| x

(2) For any n-dimensional vectors x and y,

xy  x  y

(triangular inequality) (3)

x 0

and if

x 0

then x = 0.

Other Examples of Norms n

x 1   | xi |

L1 norm:

i 1

L∞ norm:

x



 max | xi | i 1, , n

x 2 1 1

x 1 1

x



Balls of radius 1 in different norms

1

Equivalence of Norms x x



1

 x2 n x

x





 x1n x

x2 2

2

x 2 1

1

1 x

1



1

x

x

Implications: If

x





0





x

1

then

then

x 1 0

x 1  n

and

and



1

x 2 0

x2 n

Hence, conclusions of convergence, closeness, and continuity established in one norm will be valid for other norms also.

Stability in the Lyapunov Sense of an Isolated Equilibrium Point Definition: The equilibrium point x = 0 of the system (1) is said to be stable in the Lyapunov sense if

for any R > 0, there exists r > 0 such that whenever

x(0)  r we have

x(t )  R

for all

Otherwise, the equilibrium point is unstable.

Notation:

Ballr   x : x  r

t0

x

x

x

x

x   x3

x  x3

x

x

x

x   sin x

x

x   x4

Lyapunov stability means that the maximum deviation of the whole trajectory from 0 depends continuously on the deviation of the initial state.

R  0, r  0 such that x(0)  Ballr  x(t )  BallR , t x(t) Answer: Start in smaller r

R

Challenge: reduced R

t r

Examples For linear systems, all stable and marginally stable systems are stable in the Lyapunov sense.

x x 0 x1  x, x2  x

A Mechanical System: No friction State Variables:

State Space Model:

x1  x2 , x2   x1 Note: For any R (challenge), just start at any point inside the disk of radius r < R (answer)

x2

x1

Observe that the Lyapunov stability requires for all (not some) initial states in the ball of radius r, the trajectories stay in a ball of radius R. Example: Saddle Point in a Second-Order System

Example:

x  x

2

x(t)

x

x

t

Lyapunov stability is different from the condition “bounded initial condition implies bounded state trajectories”

Bounded Initial Condition Implies Bounded State Trajectories:

r  0, R  0 such that x(0)  Ballr  x(t )  BallR , t That is, it goes “given any r > 0, there exists R > 0” Compare to Lyapunov: “given any R > 0, there exists r > 0” This is called Lagrange Stability. This requires basically that state trajectories stay bounded.

Nonlinear Mechanical System: Nonlinear Friction

mx  2c( x  1) x  kx  0 2

Van del Pol Equation

x 2  1:

positive friction



stable system

x 2  1:

negative friction



unstable system

Limit Cycle:

Unstable Equilibrium Point In Lyapunov

Stable Equilibrium Point In Lagrange

Definition: The equilibrium point x = 0 of the system (1) is said to be asymptotically stable in the Lyapunov sense if 1.

it is Lyapunov stable;

2. there exists  > 0, such that

x(0)  

implies

x(t )  0, t   x2

Remarks: A system may be Lyapunov stable but not asymptotically stable, e.g., marginally stable linear systems.

x1

State convergence = attraction of the equilibrium point

Asymptotic stability: Stability + Attraction

x 2 ( y  x)  y 5 x  2 ( x  y 2 )(1  ( x 2  y 2 ) 2 )

y 2 ( y  2 x) y  2 ( x  y 2 )(1  ( x 2  y 2 ) 2 ) Denominator always positive



x  0 if x  0; y  0 

y  0 if x  0; y  0

y Switches on y  2 x

Pendulum Equation

ml  mg sin   kl x1   , x2    x1    x2



x2

g k   sin x1  x2 l m

Equilibrium points:

(n ,0), n  0, 1, 2,

Definition:

The equilibrium point x = 0 of the system (1) is said to be exponentially stable in the Lyapunov sense if (1) it is Lyapunov stable; (2) there exists  > 0, such that x(0)  

implies

x(t )  c x(0) e  t t  0

where c and  are some positive constants.

A system may be asymptotically stable but not exponentially stable. Exponential stability requires that the rate of convergence to 0 be exponential.

2t xx 2 t 1



Equilibrium point: x = 0

It is Lyapunov stable. It is asymptotically stable. It is not exponentially stable.

x(0) x(t )  2 t 1

For linear systems, however, stable systems are always exponentially stable, with the rate of convergence determined by eigenvalues: Stable modes:

e t , te t , t 2e t , e

 t

sin(t   ), te

 t

sin(t   ),

Definition:

If asymptotic stability holds for all initial states, then the equilibrium point x = 0 of the system (1) is said to be globally asymptotically stable in the Lyapunov sense. Local asymptotic stability does not imply global asymptotic stability. x

x

x   sin x

A system with multiple equilibrium points cannot be globally asymptotically stable.

Local instability does not imply that trajectories will go to infinity. They may go to other stable equilibrium points or limit cycles.

Lyapunov Linearization Methods

x  f ( x)

(1)

with f(0) = 0. Assume that the nonlinear function f(x) is continuously differentiable (so that partial derivatives which we are going to use are well defined). Jacobian Linearization

x  Ax where A is the Jacobian matrix of f(x) at 0

f ( x ) A x x  0

(2)

Lyapunov Stability: Local Linearization Theorem (Lyapunov Linearization Method for Isolated Equilibrium Points) If the linear system (2) is exponentially stable, then the equilibrium point x = 0 of the nonlinear system (1) is asymptotically stable. If the linear system (2) is unstable, then the equilibrium point x = 0 of the nonlinear system (1) is unstable. If the linear system (2) has poles on the imaginary axis and has no poles in the right half plane, then no conclusion can be reached about the equilibrium point.

Advantages of this theorem: 1. It is convenient to apply. 2. It is general, regardless the structure of the nonlinear function f(x). 3. It implies a local design method: Locally linearize the plant; design a linear controller that stabilizes the linearized plant; this will lead to a locally stable closed-loop system. Limitations of this theorem: 1. It is inconclusive when the linear system (2) is marginally stable. 2. It cannot deal with some hard nonlinearities since f(x) is not differentiable. 3. It is valid for local stability only.

The nonlinear system:

x  ax  bx5 x  ax

The linearized system at 0:

a < 0: the equilibrium point x = 0 is asymptotically stable a > 0: the equilibrium point x = 0 is unstable a = 0: no conclusion from local linearization. We need to look at further details beyond the linear part:

x  bx

5

b < 0, it is actually globally asymptotically stable. b > 0, it is unstable.

Linearization: Keep the linear terms only:

Unstable Node

Lyapunov Direct Methods

Lyapunov Direct Methods Main Ideas: 1. It is usually difficult to solve a nonlinear differential equation. 2. Lyapunov's linearization method may be invalid (e.g., for hard nonlinearities), or inconclusive (e.g., marginally stable linearized systems), and can only reach local conclusions. 3. For the analysis of stability we don't need to obtain exact system trajectories. 4. Stability properties of system trajectories can sometimes be related to the properties of some other functions which can be studied without solving the differential equations.

Example:Energy of a Mechanical System

mx  bx | x | k0 x  k1 x3  0

x1  x, x2  x  x1   x2  



x2

k0 k1 3 b   x2 | x2 |  x1  x1 m m m

Linearized System at 0:  x1   x2  



x2  0 k0  A   k0   x1  m  m

1  0 

Poles on the imaginary axis: Inconclusive!

Total Mechanical Energy of the System: x 1 2 V ( x, x)  mx    k0 x  k1 x 3 dx 0 2 1 2 1 1 4 2  mx  k0 x  k1 x 2 2 4

Some properties of the energy:

V ( x, x) is continuously differentiable V (0,0)  0

no energy at the equilibrium point

V ( x, x)  0 if ( x, x)  0 non-zero energy nearby dV ( x1 , x2 )  V V   x1  3  ,   x   b x2  0 dt  x1 x2   2  total energy is not increasing along the system trajectories

Positive and Negative Definite Functions Definition: Positive Definite Functions A scalar continuous function V(x) of n-dimensional vector x is said to be locally positive definite if V(0)= 0 There exists R  0 such that 0

x  R0 , x  0  V ( x )  0 If this holds only for V ( x )  0 , instead of V ( x )  0 , then V(x) is called positive semi-definite.

V(x) is said to be negative definite (or negative semi-definite) if -V(x) is positive definite (or positive semi-definite). If the above properties hold for all x, then V(x) is said to be globally positive definite (positive semi-definite, negative definite, negative semi-definite).

Examples x is a scalar V ( x)  x 2 V ( x)  x sin x

globally positive definite locally positive definite

V ( x)  1  cos x

locally positive definite

V ( x)  x cos x

not a definite function

x is a vector:

x  [ x1 , x2 , x3 ] V ( x)  x12

globally positive semi-definite

V ( x)  x12  x24  | x3 | globally positive definite

Lyapunov Functions Consider a system

x  f ( x )

where x is an n-dimensional state vector, and f(0) = 0. Definition: Lyapunov Functions A scalar continuously differentiable function V(x) is said to be a Lyapunov function for the equilibrium point x = 0 of the system if there is a constant R0  0 such that (a) V(x) is locally positive definite for x  R0 ; (b)

dV ( x ) 0 dt

along any trajectories x of the system in x  R0 (it is locally negative semi-definite)

A locally positive definite function V(x) is only a Lyapunov Candidate

The total Mechanical Energy of the Previous System: 1 1 1 V ( x1 , x2 )  mx22  k0 x12  k1 x14 2 2 4

is a Lyapunov function for the system at (0,0):  x1   x2  



x2

k0 k1 3 b   x2 | x2 |  x1  x1 m m m

V ( x, x) is continuously differentiable V (0,0)  0 V ( x, x)  0 if ( x, x)  0 dV ( x1 , x2 )  V V   x1  3  ,   b x 2 0  x  dt  x1 x2   2 

Theorem: Lyapunov Stability

If for a constant R0  0 , there exists a Lyapunov function V(x) for the system, then the equilibrium point x = 0 is Lyapunov stable.

Hence, the equilibrium point (0,0) of the mechanical system is Lyapunov stable.

A Simple Pendulum System:

x1   , x2  

 x1   x2

    sin   0

 x2   sin x1  x2

Consider the equilibrium point (0,0) Define

1 2 V ( x)  (1  cos x1 )  x2 2

It is locally positive definite

dV ( x)  x2 sin x1  x2 x2   x22  0 dt So, the equilibrium point is Lyapunov stable.

Main Ideas of the Proof V(x)

For trajectories to go beyond [-R,R], V(x) must increase to pass this line

Smallest v(x) on the boundary dV ( x) 0 will dt

make sure that this will not happen

x [-r,r]

answer [-R,R]

challenge

Illustration on the Phase Plane x2

level sets of v(x) V(x)=v1

ball of radius R

V(x)=v2 < v1 x1

start inside a ball of radius r

Definition: Strict Lyapunov Functions A scalar continuous function V(x) is said to be a strict Lyapunov function for the equilibrium point x = 0 of the system if there is a constant R0  0 such that (a)

V(x) is positive definite for

(b)

dV ( x ) 0 dt

in

x  R0

;

along any trajectories x of the system x  R0 and x  0

Theorem: Asymptotic Lyapunov Stability If for a constant R0  0, there exists a strict Lyapunov function V(x) for the system at 0, then the equilibrium point x = 0 is asymptotically Lyapunov stable.

Basic Ideas of the Proof on Attraction (State Convergence)

level sets of V(x)

x2

V(x)=v1 V(x)=v2 < v1 x1

V(x)=v3 < v2

local region of attraction: ball of radius R0 dV ( x)  0 and V ( x)  0 dt implies V ( x(t ))  C  0

x(t )  0, t  

C  0, since V ( x) is negative definite

Examples x   g ( x), g (0)  0, xg ( x)  0 for x  0 and | x | r Equilibrium point: x = 0 x Define: V ( x)  g ( s) ds



0

0

V (0)   g ( s) ds  0 0

V ( x)  0, for x  0 and | x | r

Lyapunov Candidate

dV ( x)  g ( x) x  ( g ( x)) 2  0, for x  0 and | x | r dt This is a strict Lyapunov function at x = 0. So, x = 0 is asymptotically stable.

Examples of such g(x)

x  x x   x n , n is an odd integer x   sin x x   x cos x

x

x

x   x3e 2 x x   g ( x)

xg ( x)  0 simply implies: x  0  g ( x)  0  x   g ( x)  0  x decreases x  0  g ( x)  0  x   g ( x)  0  x increases

Example

Lyapunov candidate

For x12  x22  1 and x  0, V  0 This is a (locally) strict Lyapunov function for (0,0). (0,0) is asymptotically stable

1. The theorem gives a region of local stability, or a region of contraction. 2. Lyapunov functions are not unique. Some may give a larger region of contraction than others. 3. There is no standard method for finding or constructing a Lyapunov function. Hence, Lyapunov theorems are not constructive. 4. For practical systems, a good understanding of physical systems is of great importance in selecting Lyapunov candidates. For mechanical systems, total energy and norms are usually tried first as potential Lyapunov candidates. 5. Lyapunov functions are sufficient conditions only. Failure of finding a Lyapunov function does not imply that the equilibrium point x = 0 is unstable.

Definition: Strict Radially Unbounded Lyapunov Functions

A scalar continuously differentiable function V(x) is said to be a strict radially unbounded Lyapunov function for the equilibrium point x = 0 of the system if (a) V(x) is globally positive definite; (b)

dV ( x ) 0 dt

along any trajectories x of the system for all x  0

(c) V ( x)   as

x 

Theorem: Global Asymptotic Lyapunov Stability If there exists a strict radially unbounded Lyapunov function V(x) for the system, then the equilibrium point x = 0 is globally asymptotically Lyapunov stable.

Motivations for the Radial Unboundedness Condition V(x) V(x) decreases

x

but x goes to infinity This V(x) is not radially unbounded

Illustration on the Phase Plane

Examples

x   x,

V ( x)  x 2

x   x n , n is an odd integer,

V ( x)  x 2

x   sin x, multiple EP, not global x   x cos x, multiple EP, not global 3 x

x  x e ,

V ( x)  x

2

x   x  sin x 5

Define V ( x)  x2 ,

6

this is a Lyapunov candidate

dV ( x)  2 x( x5  sin 6 x)  0, x  0 dt V ( x)  x 2  , x   This is a strict radially unbounded Lyapunov function.

x = 0 is globally asymptotically stable.

Linear System Analysis Using Lyapunov Methods Consider a linear system

x  Ax

We know that the system is exponentially stable if and only all eigenvalues of A lie in the open left half plane. In this case, the (unique) equilibrium point x = 0 is globally asymptotically stable in the Lyapunov sense. Can we reach the same conclusion by constructing a Lyapunov function for the system? Yes!!

Fact 1: If for a given positive definite matrix Q > 0 the following matrix linear equation, called Lyapunov Equation, has a positive definite solution P > 0, then x = 0 of the linear system is globally asymptotically stable in the Lyapunov sense.

A P  PA  Q T

Proof: Define V ( x)  xT Px  0, x  0, this is a Lyapunov candidate

V  xT Px  xT Px  xT AT Px  xT PAx   xT Qx  0, x  0

V ( x)  , x   This is a strict radially unbounded Lyapunov function. x = 0 is globally asymptotically stable.

Fact 2: The Lyapunov Equation has a unique positive definite solution P > 0 for any positive definite matrix Q >0 if and only if all eigenvalues of A have negative real parts.

Hence, we can obtain the same stability results by solving the Lyapunov Equation and test for positive definiteness of P, instead of calculating eigenvalues.

4  0 A   8 12 

1 0  Choose Q  I    0 1 

AT P  PA  Q

Robustness Analysis x  Ax   ( x),  ( x) is a small nonlinear perturbation and  (0)  0 Suppose that A is stable. Then, there is P > 0 such that AT P  PA   I Define V ( x)  xT Px  0

V  xT Px  xT Px  ( xT AT   T ( x)) Px  xT P( Ax   ( x))   xT x  2 xT P ( x) If the second term is sufficiently small:  ( x)   x  2 xT P ( x)  2 k x , where k   max ( P) 2

2 k  1 

dV ( x)  0, x  0 dt

x = 0 is asymptotically stable, robustly for any small perturbations.

Recall: Local Linearization Method x  f ( x), f (0)  0 Local linearization: x  Ax  ( x), A  the Jacobian matrix ( x)  higher order terms ( x) f ( x) is continuously differentable  x

 0,x  0

 x  Ax   ( x),

 ( x) is a small nonlinear perturbation and  (0)  0

Construction of Lyapunov Functions Try something as potential Lyapunov candidates and hope that they are working. Try Energy Functions for Mechanical Systems  x1   x2  



x2

k0 k b   x2 | x2 |  x1  1 x13 m m m

1 2 1 1 4 2 V ( x1 , x2 )  mx2  k0 x1  k1 x1 2 2 4

Try Some Functions of Norms of the State:

x   x5  sin 6 x,  V ( x)  x 2

Lyapunov candidate

Krasovski's Methods

1   3  1 1  3x 2   2

2   6  2 2  6x 2   2

Invariant Set Theorem

Motivations Issue: Lyapunov functions we construct often can only conclude Lyapunov stability, rather than asymptotic stability. Typical examples are energy functions. Searching for strict Lyapunov functions can be very difficult. Question: Can we use the Lyapunov functions to study asymptotic stability also?

Example:Energy of a Mechanical System

mx  bx | x | k0 x  k1 x3  0

x1  x, x2  x  x1   x2  



x2

k0 k1 3 b   x2 | x2 |  x1  x1 m m m

Linearized System at 0:  x1   x2  



x2  0 k0  A   k0   x1  m  m

1  0 

Poles on the imaginary axis: Inconclusive!

Total Mechanical Energy of the System: x 1 2 V ( x, x)  mx    k0 x  k1 x 3 dx 0 2 1 2 1 1 4 2  mx  k0 x  k1 x 2 2 4

Some properties of the energy:

V ( x, x) is continuous V (0,0)  0 no energy at the equilibrium point V ( x, x)  0 if ( x, x)  0 non-zero energy nearby dV ( x1 , x2 )  V V   x1  3  ,   x   b x2  0 dt  x1 x2   2  total energy is not increasing along the system trajectories

Issue: Lyapunov functions work for isolated equilibrium points. Question: How can we study stability of limit cycles and continuous equilibrium points?

Issue: For practical purposes, it is important to obtain regions of contraction (how close is close enough for the system to maintain local stability?). Lyapunov functions may not be the best functions to obtain such regions.

Invariant Sets Definition: Invariant Sets A set G in the state space is a positively invariant set of the system if every system trajectory which starts from a point in G remains in G for all future time.

x(t0 )  G  x(t )  G, t  t0

Typical Invariant Sets Equilibrium points. x2

0 0 x x  0 1 

Limit cycles.

x1

If V(x) is a Lyapunov function of the system x  f ( x ) , then the level set l  {x : V ( x)  l} note: V ( x)  0 in l is an invariant set of the system.  x1    x2

  

x2 k k b x2 | x2 |  0 x1  1 x13 m m m

x 1 2 V ( x, x)  mx    k0 x  k1 x 3 dx 0 2 1 1 1  mx 2  k0 x 2  k1 x 4 2 2 4

100  ( x, x) : V ( x, x)  100 (the set of the states that carry energy no more than 100)

Invariant Set Theorems Theorem: Local Invariant Set Theorem For the system x  f ( x ) , suppose V(x) is a continuously differentiable function of x satisfying (a) for some l > 0, the region  l is bounded; dV ( x ) (b) dt  0

system.

l  {x: V ( x )  l}

for all x in

l

along the trajectories of the dV ( x ) 0 dt

Let R be the set of all points within  l at which , and M be the largest invariant set in R. Then, every trajectory x(t) of the system starting in  l tends to M as t   .

 x1    x2

  

x2 k k b x2 | x2 |  0 x1  1 x13 m m m

1 1 1 V ( x, x)  mx 2  k0 x 2  k1 x 4 2 2 4

dV ( x1 , x2 )  V V   x1  3  ,   x   b x2  0  x2  0 dt  x1 x2   2 

100  ( x, x) : V ( x, x)  100 R  ( x,0) : V ( x,0)  100 M  (0, 0) By the Invariant Set Theorem:

Starting at any x(0)  100 , x(t )  (0, 0), t  

(0,0) is asymptotically stable!



x2

x1

R At a point in R other than (0,0):

M

 x1    x2



0

k0 k1 3  it will leave R   x1  x1 m m

M is the only invariant set in R

A Generalization x  b( x)  c( x)  0 xb( x )  0, for x  0 xc( x )  0, for x  0

x1  x, x2  x

 x1   x2

 x2  b( x2 )  c( x1 )

x 1 2 V ( x, x)  x   c( y)dy 0 2

V ( x, x) is continuous V (0,0)  0 V ( x, x)  0 if ( x, x)  0

V  xx  c( x) x   xb( x)  xc( x)  c( x) x   xb( x)  0 V is a Lyapunov function of the system at (0,0). So the equilibrium point is Lyapunov stable.

L  ( x, x) : V ( x, x)  L R  ( x,0) : V ( x,0)  L M  (0, 0) By the Invariant Set Theorem:

Starting at any x(0)   L , x(t )  (0, 0), t   (0,0) is asymptotically stable!

Stability of a Limit Cycle

 x1   x2

 x2  x1 ( x  2 x  10)   x13  3x25 ( x14  2 x22  10) 4 1

x14  2 x22  10 is a limit cycle Verify:

d ( x14  2 x22  10) 0 dt

and it satisfies the equations.

2 2

x2

x1

To test stability of the limit cycle, define

V  ( x  2 x  10) 4 1

2 2

2

dV  8( x14  3x26 )( x14  2 x22  10) 2  0 dt

dV  8( x14  3x26 )( x14  2 x22  10) 2  0 dt  (0, 0) (the equilibrium point)  x14  2 x22  10 (the limit cycle)

20  ( x1 , x2 ) : V ( x1 , x2 )  ( x14  2 x22  10)2  20 R  ( x1 , x2 ) : x14  2 x22  10 M  ( x1 , x2 ) : x14  2 x22  10

x2

By the Invariant Set Theorem:

Starting at any x(0)  20 , x(t )  M , t  

x1

The limit cycle is asymptotically stable!

 20

A Simple Adaptive Control System

Plant: y  ay  u, a is unknown Controller: u  ky

Adaptation Law: k   y

2

k

Overall System  y  ay  ky  2 k   y  set of equilibrium points

y

Define:

1 2 1 V y  ( k  b) 2 2 2 dV 1  yy  (k  b)k  (b  a) y 2  0, if b  a dt 

For any a (unknown), there is a value b (I don’t need to know its actual value) such that

V 0

  1 2 1 2  L  ( y , k ) : y  ( k  b)  L  2 2     1 2 R  ( y, k ) : V  0   L  (0, k ) : (k  b)  L  2   M R k

L

By the Invariant Set Theorem:

Starting at any x(0)   L , x(t )  M , t    y0

M R y

The output will go to zero; The gain k will approach some constant