Lecture #7 Stability and convergence of ODEs Hybrid Control and Switched Systems Summary

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Lecture #7Stability and convergence of ODEs

João P. Hespanha

University of Californiaat Santa Barbara

Hybrid Control and Switched Systems

Summary

Lyapunov stability of ODEs• epsilon-delta and beta-function definitions• Lyapunov’s stability theorem• LaSalle’s invariance principle• Stability of linear systems

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Properties of hybrid systems

sig ≡ set of all piecewise continuous signals x:[0,T) → Rn, T∈(0,∞]sig ≡ set of all piecewise constant signals q:[0,T)→ , T∈(0,∞]

Sequence property ≡ p : sig × sig → {false,true}E.g.,

A pair of signals (q, x) ∈ sig × sig satisfies p if p(q, x) = true

A hybrid automaton H satisfies p ( write H ² p ) ifp(q, x) = true, for every solution (q, x) of H

“ensemble properties” ≡ property of the whole family of solutions(cannot be checked just by looking at isolated solutions)e.g., continuity with respect to initial conditions…

Lyapunov stability (ODEs)

equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0

thus x(t) = xeq ∀ t ≥ 0 is a solution to the ODE

E.g., pendulum equation

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l

two equilibrium points:x1 = 0, x2 = 0 (down)

and x1 = π, x2 = 0 (up)

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Lyapunov stability (ODEs)


thus x(t) = xeq ∀ t ≥ 0 is a solution to the ODE

Definition (e–δ definition):The equilibrium point xeq ∈ Rn is (Lyapunov) stable if

∀ e > 0 ∃ δ >0 : ||x(t0) – xeq|| · δ ⇒ ||x(t) – xeq|| · e ∀ t≥ t0≥ 0

xeq

δ

e

x(t)1. if the solution starts close to xeq

it will remain close to it forever2. e can be made arbitrarily small

by choosing δ sufficiently small

Example #1: Pendulum

pend.m

xeq=(0,0)stable

xeq=(π,0)unstable

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Lyapunov stability – continuity definition

Definition (continuity definition):The equilibrium point xeq ∈ Rn is (Lyapunov) stable if T is continuous at xeq:

∀ e > 0 ∃ δ >0 : ||x0 – xeq|| · δ ⇒ ||T(x0) – T(xeq)||sig · e

sig ≡ set of all piecewise continuous signals taking values in Rn

Given a signal x∈ sig, ||x||sig ú supt≥0 ||x(t)||

ODE can be seen as an operatorT : Rn → sig

that maps x0 ∈ Rn into the solution that starts at x(0) = x0

signal norm

supt≥0 ||x(t) – xeq|| · e

xeq

δ

e

x(t)

can be extended to nonequilibrium solutions

Stability of arbitrary solutions

Definition (continuity definition):A solution x*:[0,T)→Rn is (Lyapunov) stable if T is continuous at x*

0ú x*(0), i.e.,∀ e > 0 ∃ δ >0 : ||x0 – x*

0|| · δ ⇒ ||T(x0) – T(x*0)||sig · e

sig ≡ set of all piecewise continuous signals taking values in Rn

Given a signal x∈ sig, ||x||sig ú supt≥0 ||x(t)||

ODE can be seen as an operatorT : Rn → sig

that maps x0 ∈ Rn into the solution that starts at x(0) = x0

signal norm

supt≥0 ||x(t) – x*(t)|| · eδ

ex(t) x*(t)

pend.m

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Example #2: Van der Pol oscillator

x* Lyapunov stable

vdp.m

Stability of arbitrary solutions

E.g., Van der Pol oscillator

x* unstable

vdp.m

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Lyapunov stability


class ≡ set of functions α:[0,∞)→[0,∞) that are1. continuous2. strictly increasing3. α(0)=0

Definition (class function definition):The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ :

||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c

the function α can be constructed directly from the δ(e) in the e–δ

(or continuity) definitions

s

α(s)

xeq

α(||

x(t 0)

–x e

q||)

||x(t 0

) –x e

q||

x(t)

t

Asymptotic stability

Definition:The equilibrium point xeq ∈ Rn is (globally) asymptotically stable if it is Lyapunov stable and for every initial state the solution exists on [0,∞) and

x(t) → xeq as t→∞.

xeq

α(||

x(t 0)

–x e

q||)

||x(t 0

) –x e

q||

x(t)

s

α(s)equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0

class ≡ set of functions α:[0,∞)→[0,∞) that are1. continuous2. strictly increasing3. α(0)=0

t

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Asymptotic stability

Definition (class function definition):The equilibrium point xeq ∈ Rn is (globally) asymptotically stable if ∃ β∈ :

||x(t) – xeq|| · β(||x(t0) – xeq||,t – t0) ∀ t≥ t0≥ 0

xeq

β(||x

(t 0) –

x eq||

,0)

||x(t 0

) –x e

q||

x(t)


class ≡ set of functions β:[0,∞)×[0,∞)→[0,∞) s.t.1. for each fixed t, β(·,t) ∈2. for each fixed s, β(s,·) is monotone decreasing and β(s,t) → 0 as t→∞

s

β(s,t)

(for each fixed t)

t

β(s,t)(for each fixed s)

β(||x(t0) – xeq||,t)

t

We have exponential stabilitywhen

β(s,t) = c e-λ t swith c,λ > 0

linear in s and negative exponential in t


x2

x1

xeq=(0,0)asymptotically

stable

xeq=(π,0)unstable

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k > 0 (with friction) k = 0 (no friction)

xeq=(0,0)stable but not

asymptotically

xeq=(π,0)unstable

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Example #3: ButterflyConvergence by itself does not imply stability, e.g.,

all solutions converge to zero but xeq= (0,0) system is not stable

equilibrium point ≡ (0,0)

converge.m

Why was Mr. Lyapunov so picky? Why shouldn’t

boundedness and convergence to zero suffice?

Lyapunov’s stability theorem



Suppose we could show that ||x(t) – xeq|| always decreases along solutions to the ODE. Then

||x(t) – xeq|| · ||x(t0) – xeq|| ∀ t≥ t0≥ 0we could pick α(s) = s ⇒ Lyapunov stability

We can draw the same conclusion by using other measures of how far the solution is from xeq:

V: Rn → R positive definite ≡ V(x) ≥ 0 ∀ x ∈ Rn with = 0 only for x = 0V: Rn → R radially unbounded ≡ x→ ∞ ⇒ V(x)→ ∞

provides a measure ofhow far x is from xeq

(not necessarily a metric–may not satisfy triangular inequality)

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V: Rn → R positive definite ≡ V(x) ≥ 0 ∀ x ∈ Rn with = 0 only for x = 0

provides a measure ofhow far x is from xeq

(not necessarily a metric–may not satisfy triangular inequality)

Q: How to check if V(x(t) – xeq) decreases along solutions?

A: V(x(t) – xeq) will decrease if

can be computed without actually computing x(t)(i.e., solving the ODE)

gradient of V




Theorem (Lyapunov):Suppose there exists a continuously differentiable, positive definite function V: Rn → R such that

Then xeq is a Lyapunov stable equilibrium.

Why?V non increasing ⇒ V(x(t) – xeq) · V(x(t0) – xeq) ∀ t ≥ t0Thus, by making x(t0) – xeq small we can make V(x(t) – xeq) arbitrarily small ∀ t ≥ t0So, by making x(t0) – xeq small we can make x(t) – xeq arbitrarily small ∀ t ≥ t0(we can actually compute α from V explicitly and take c = +∞).

V(z – xeq)

z

(cup-likefunction)

Lyapunov function

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For xeq = (0,0)

positive definite because V(x) = 0 only for x1 = 2kπ k∈Z & x2 = 0

(all these points are really the same because x1 is an angle)

Therefore xeq=(0,0) is Lyapunov stable pend.m


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For xeq = (π,0)

positive definite because V(x) = 0 only for x1 = 2kπ k∈Z & x2 = 0

(all these points are really the same because x1 is an angle)

Cannot conclude that xeq=(π,0) is Lyapunov stable (in fact it is not!) pend.m

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Theorem (Lyapunov):Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: Rn → R such that

Then xeq is a Lyapunov stable equilibrium and the solution always exists globally. Moreover, if = 0 only for z = xeq then xeq is a (globally) asymptotically stable equilibrium.

Why?V can only stop decreasing when x(t) reaches xeqbut V must stop decreasing because it cannot become negativeThus, x(t) must converge to xeq


What if

for other z then xeq ? Can we still claim some form of convergence?



Theorem (Lyapunov):Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: Rn → R such that

Then xeq is a Lyapunov stable equilibrium and the solution always exists globally. Moreover, if = 0 only for z = xeq then xeq is a (globally) asymptotically stable equilibrium.

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For xeq = (0,0)

not strict for (x1≠ 0, x2=0 !)

pend.m

LaSalle’s Invariance Principle

Theorem (LaSalle Invariance Principle):Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: Rn → R such that

Then xeq is a Lyapunov stable equilibrium and the solution always exists globally.Moreover, x(t) converges to the largest invariant set M contained in

E ú { z ∈ Rn : W(z) = 0 }

M ∈ Rn is an invariant set ≡ x(t0) ∈ M ⇒ x(t)∈ M∀ t≥ t0(in the context of hybrid systems: Reach(M) ⊂ M…)

Note that:1. When W(z) = 0 only for z = xeq then E = {xeq }.

Since M ⊂ E, M = {xeq } and therefore x(t) → xeq ⇒ asympt. stability

2. Even when E is larger then {xeq } we often have M = {xeq } and can conclude asymptotic stability.

Lyapunovtheorem

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For xeq = (0,0)

E ú { (x1,x2): x1∈ R , x2=0}Inside E, the ODE becomes

Therefore x converges to M ú { (x1,x2): x1 = k π ∈ Z , x2=0}

However, the equilibrium point xeq=(0,0) is not (globally) asymptotically stable because if the system starts, e.g., at (π,0) it remains there forever.

define set M for which system remains inside E

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Linear systems

Solution to a linear ODE:

Theorem: The origin xeq = 0 is an equilibrium point. It is1. Lyapunov stable if and only if all eigenvalues of A have negative or zero real

parts and for each eigenvalue with zero real part there is an independent eigenvector.

2. Asymptotically stable if and only if all eigenvalues of A have negative real parts. In this case the origin is actually exponentially stable

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Linear systems

linear.m

Lyapunov equation


Theorem: The origin xeq = 0 is an equilibrium point. It is asymptotically stable if and only if for every positive symmetric definite matrix Q the equation

A’ P + P A = – Qhas a unique solutions P that is symmetric and positive definite

Lyapunov equation

Recall: given a symmetric matrix PP is positive definite ≡ all eigenvalues are positive

P positive definite ⇒ x’ P x > 0 ∀ x ≠ 0P is positive semi-definite ≡ all eigenvalues are positive or zero

P positive semi-definite ⇒ x’ P x ≥ 0 ∀ x

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Lyapunov equation



Lyapunov equation

Why?1. asympt. stable ⇒ P exists and is unique (constructive proof)

change of integrationvariable τ = T – s

A is asympt. stable ⇒ eAt decreases to zero exponentiall fast ⇒ P is well defined (limit exists and is finite)


Lyapunov equation



Lyapunov equation

Why?2. P exists ⇒ asymp. stable

Consider the quadratic Lyapunov equation: V(x) = x’ P xV is positive definite & radially unbounded because P is positive definiteV is continuously differentiable:


thus system is asymptotically stable by Lyapunov Theorem

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Next lecture…

Lyapunov stability of hybrid systems

Lecture #7 Stability and convergence of ODEs Hybrid Control and Switched Systems Summary

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