Download Lecture #7 Stability and convergence of ODEs Hybrid Control and

Hybrid Control and Switched Systems
Lecture #7
Stability and convergence of ODEs
João P. Hespanha
University of California
at Santa Barbara
Summary
Lyapunov stability of ODEs
• epsilon-delta and beta-function definitions
• Lyapunov’s stability theorem
• LaSalle’s invariance principle
• Stability of linear systems
1
Properties of hybrid systems
?sig ≡ set of all piecewise continuous signals x:[0,T) → Rn, T∈(0,∞]
8sig ≡ set of all piecewise constant signals q:[0,T)→ 8, T∈(0,∞]
Sequence property ≡ p : 8sig × ?sig → {false,true}
E.g.,
A pair of signals (q, x) ∈ 8sig × ?sig satisfies p if p(q, x) = true
A hybrid automaton H satisfies p ( write H ² p ) if
p(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
l
two equilibrium points:
x1 = 0, x2 = 0 (down)
and
x1 = π, x2 = 0 (up)
θ
m
2
Lyapunov stability (ODEs)
equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0
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
e
xeq
δ
x(t)
1. if the solution starts close to xeq
it will remain close to it forever
2. e can be made arbitrarily small
by choosing δ sufficiently small
Example #1: Pendulum
l
θ
xeq=(0,0)
stable
xeq=(π,0)
unstable
m
pend.m
3
Lyapunov stability – continuity definition
?sig ≡ set of all piecewise continuous signals taking values in Rn
Given a signal x∈?sig, ||x||sig ú supt≥0 ||x(t)||
signal norm
ODE can be seen as an operator
T : Rn → ?sig
n
that maps x0 ∈ R into the solution that starts at x(0) = x0
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
supt≥0 ||x(t) – xeq|| · e
e
xeq
can be extended to
nonequilibrium solutions
δ
x(t)
Stability of arbitrary solutions
?sig ≡ set of all piecewise continuous signals taking values in Rn
Given a signal x∈?sig, ||x||sig ú supt≥0 ||x(t)||
signal norm
ODE can be seen as an operator
T : Rn → ?sig
that maps x0 ∈ Rn into the solution that starts at x(0) = x0
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
supt≥0 ||x(t) – x*(t)|| · e
δ
e
x(t)
x*(t)
pend.m
4
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
5
Lyapunov stability
equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0
α(s)
class 2 ≡ set of functions α:[0,∞)→[0,∞) that are
1. continuous
2. strictly increasing
3. α(0)=0
s
||x(t0) – xeq||
α(||x(t0) – xeq||)
Definition (class 2 function definition):
The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ 2:
||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c
x(t)
xeq
t
the function α can be constructed
directly from the δ(e) in the e–δ
(or continuity) definitions
Asymptotic stability
equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0
α(s)
class 2 ≡ set of functions α:[0,∞)→[0,∞) that are
1. continuous
2. strictly increasing
3. α(0)=0
s
||x(t0) – xeq||
α(||x(t0) – xeq||)
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→∞.
x(t)
xeq
t
6
Asymptotic stability
β(s,t)
(for each fixed t)
equilibrium point ≡ xeq ∈ Rn for which f(xeq) = 0
class 23 ≡ set of functions β:[0,∞)×[0,∞)→[0,∞) s.t.
1. for each fixed t, β(·,t) ∈ 2
2. for each fixed s, β(s,·) is monotone
decreasing and β(s,t) → 0 as t→∞
s
β(s,t)
(for each fixed s)
||x(t0) – xeq||
β(||x(t0) – xeq||,0)
t
Definition (class 23 function definition):
The equilibrium point xeq ∈ Rn is (globally) asymptotically stable if ∃ β∈23:
||x(t) – xeq|| · β(||x(t0) – xeq||,t – t0) ∀ t≥ t0≥ 0
We have exponential stability
when
β(s,t) = c e-λ t s
with c,λ > 0
β(||x(t0) – xeq||,t)
xeq
x(t)
t
linear in s and negative
exponential in t
Example #1: Pendulum
k = 0 (no friction)
k > 0 (with friction)
x2
x1
xeq=(0,0)
asymptotically
stable
xeq=(π,0)
unstable
xeq=(0,0)
stable but not
asymptotically
xeq=(π,0)
unstable
pend.m
7
Example #3: Butterfly
Convergence by itself does not imply stability, e.g.,
Why was Mr. Lyapunov so
picky? Why shouldn’t
boundedness and convergence
to zero suffice?
equilibrium point ≡ (0,0)
all solutions converge to zero but xeq= (0,0) system is not stable
converge.m
Lyapunov’s stability theorem
Definition (class 2 function definition):
The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ 2:
||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c
Suppose we could show that ||x(t) – xeq|| always decreases along solutions to
the ODE. Then
||x(t) – xeq|| · ||x(t0) – xeq|| ∀ t≥ t0≥ 0
we 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 = 0
V: Rn → R radially unbounded ≡ x→ ∞ ⇒ V(x)→ ∞
provides a measure of
how far x is from xeq
(not necessarily a metric–may
not satisfy triangular inequality)
8
Lyapunov’s stability theorem
V: Rn → R positive definite ≡ V(x) ≥ 0 ∀ x ∈ Rn with = 0 only for x = 0
provides a measure of
how 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
gradient of V
can be computed without
actually computing x(t)
(i.e., solving the ODE)
Lyapunov’s stability theorem
Definition (class 2 function definition):
The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ 2:
||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c
Lyapunov function
Theorem (Lyapunov):
Suppose there exists a continuously differentiable, positive definite function V:
Rn → R such that
Then xeq is a Lyapunov stable equilibrium.
(cup-like
function)
V(z – xeq)
z
Why?
V non increasing ⇒ V(x(t) – xeq) · V(x(t0) – xeq) ∀ t ≥ t0
Thus, by making x(t0) – xeq small we can make V(x(t) – xeq) arbitrarily small ∀ t ≥ t0
So, 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 = +∞).
9
Example #1: Pendulum
l
θ
m
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)
For xeq = (0,0)
Therefore xeq=(0,0) is Lyapunov stable
pend.m
Example #1: Pendulum
l
θ
For xeq = (π,0)
m
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
10
Lyapunov’s stability theorem
Definition (class 2 function definition):
The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ 2:
||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c
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 xeq
but V must stop decreasing because it cannot become negative
Thus, x(t) must converge to xeq
Lyapunov’s stability theorem
Definition (class 2 function definition):
The equilibrium point xeq ∈ Rn is (Lyapunov) stable if ∃ α ∈ 2:
||x(t) – xeq|| · α(||x(t0) – xeq||) ∀ t≥ t0≥ 0, ||x(t0) – xeq||· c
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.
What if
for other z then xeq ? Can we still claim some form of convergence?
11
Example #1: Pendulum
l
θ
m
For xeq = (0,0)
not strict for (x1≠ 0, x2=0 !)
pend.m
LaSalle’s Invariance Principle
M ∈ Rn is an invariant set ≡ x(t0) ∈ M ⇒ x(t)∈ M∀ t≥ t0
(in the context of hybrid systems: Reach(M) ⊂ M…)
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 }
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.
Lyapunov
theorem
12
Example #1: Pendulum
l
θ
m
For xeq = (0,0)
E ú { (x1,x2): x1∈ R , x2=0}
Inside E, the ODE becomes
define set M for which
system remains inside E
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.
pend.m
Linear systems
Solution to a linear ODE:
Theorem: The origin xeq = 0 is an equilibrium point. It is
1. 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
13
Linear systems
linear.m
Lyapunov equation
Solution to a linear ODE:
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 = – Q
Lyapunov equation
has a unique solutions P that is symmetric and positive definite
Recall: given a symmetric matrix P
P is positive definite ≡ all eigenvalues are positive
P positive definite ⇒ x’ P x > 0 ∀ x ≠ 0
P is positive semi-definite ≡ all eigenvalues are positive or zero
P positive semi-definite ⇒ x’ P x ≥ 0 ∀ x
14
Lyapunov equation
Solution to a linear ODE:
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 = – Q
Lyapunov equation
has a unique solutions P that is symmetric and positive definite
Why?
1. asympt. stable ⇒ P exists and is unique (constructive proof)
A is asympt. stable ⇒ eAt decreases to
zero exponentiall fast ⇒ P is well
defined (limit exists and is finite)
change of integration
variable τ = T – s
Lyapunov equation
Solution to a linear ODE:
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 = – Q
Lyapunov equation
has a unique solutions P that is symmetric and positive definite
Why?
2. P exists ⇒ asymp. stable
Consider the quadratic Lyapunov equation: V(x) = x’ P x
V is positive definite & radially unbounded because P is positive definite
V is continuously differentiable:
thus system is asymptotically stable by Lyapunov Theorem
15
Next lecture…
Lyapunov stability of hybrid systems
16