Transcript Document

Hybrid Control and Switched Systems
Lecture #7
Stability and convergence of ODEs
NO CLASSES
on Oct 18 & Oct 20
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
Properties of hybrid systems
Xsig ´ set of all piecewise continuous signals x:[0,T) ! Rn, T2(0,1]
Qsig ´ set of all piecewise constant signals q:[0,T)! Q, T2(0,1]
Sequence property ´ p : Qsig £ Xsig ! {false,true}
E.g.,
A pair of signals (q, x) 2 Qsig £ Xsig 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 2 Rn for which f(xeq) = 0
thus x(t) = xeq 8 t ¸ 0 is a solution to the ODE
E.g., pendulum equation
l
two equilibrium points:
x1 = 0, x2 = 0 (down)
and
x1 = p, x2 = 0 (up)
q
m
Lyapunov stability (ODEs)
equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
thus x(t) = xeq 8 t ¸ 0 is a solution to the ODE
Definition (e–d definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if
8 e > 0 9 d >0 : ||x(t0) – xeq|| · d ) ||x(t) – xeq|| · e 8 t¸ t0¸ 0
e
xeq
d
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 d sufficiently small
Example #1: Pendulum
l
q
m
x1 is an angle
so you must
“glue” left to
right extremes
of this plot
xeq=(0,0)
stable
xeq=(p,0)
unstable
pend.m
Lyapunov stability – continuity definition
Xsig ´ set of all piecewise continuous signals taking values in Rn
Given a signal x2Xsig, ||x||sig supt¸0 ||x(t)||
signal norm
ODE can be seen as an operator
T : Rn ! Xsig
that maps x0 2 Rn into the solution that starts at x(0) = x0
Definition (continuity definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if T is continuous at xeq:
8 e > 0 9 d >0 : ||x0 – xeq|| · d ) ||T(x0) – T(xeq)||sig · e
e
supt¸0 ||x(t) – xeq|| · e
d
can be extended to
nonequilibrium solutions
xeq
x(t)
Stability of arbitrary solutions
Xsig ´ set of all piecewise continuous signals taking values in Rn
Given a signal x2Xsig, ||x||sig supt¸0 ||x(t)||
signal norm
ODE can be seen as an operator
T : Rn ! Xsig
that maps x0 2 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*0x*(0), i.e.,
8 e > 0 9 d >0 : ||x0 – x*0|| · d ) ||T(x0) – T(x*0)||sig · e
supt¸0 ||x(t) – x*(t)|| · e
d
e
x(t)
x*(t)
pend.m
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
Lyapunov stability
equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
class K ´ set of functions a:[0,1)![0,1) that are
1. continuous
2. strictly increasing
3. a(0)=0
a(s)
s
||x(t0) – xeq||
a(||x(t0) – xeq||)
Definition (class K function definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 a 2 K:
||x(t) – xeq|| · a(||x(t0) – xeq||) 8 t¸ t0¸ 0, ||x(t0) – xeq||· c
x(t)
xeq
t
the function a can be constructed
directly from the d(e) in the e–d
(or continuity) definitions
Asymptotic stability
equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
a(s)
class K ´ set of functions a:[0,1)![0,1) that are
1. continuous
2. strictly increasing
3. a(0)=0
s
||x(t0) – xeq||
a(||x(t0) – xeq||)
Definition:
The equilibrium point xeq 2 Rn is (globally) asymptotically stable if
it is Lyapunov stable and for every initial state the solution exists on [0,1) and
x(t) ! xeq as t!1.
x(t)
xeq
t
Asymptotic stability
b(s,t)
equilibrium point ´ xeq 2
Rn
(for each fixed t)
for which f(xeq) = 0
class KL ´ set of functions b:[0,1)£[0,1)![0,1) s.t.
1. for each fixed t, b(¢,t) 2 K
2. for each fixed s, b(s,¢) is monotone
decreasing and b(s,t) ! 0 as t!1
s
b(s,t)
(for each fixed s)
||x(t0) – xeq||
b(||x(t0) – xeq||,0)
t
Definition (class KL function definition):
The equilibrium point xeq 2 Rn is (globally) asymptotically stable if 9 b2KL:
||x(t) – xeq|| · b(||x(t0) – xeq||,t – t0) 8 t¸ t0¸ 0
We have exponential stability
when
b(s,t) = c e –l t s
with c,l > 0
b(||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=(p,0)
unstable
xeq=(0,0)
stable but not
asymptotically
xeq=(p,0)
unstable
pend.m
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
x1 is an angle
so you must
“glue” left to
right extremes
of this plot
converge.m
Lyapunov’s stability theorem
Definition (class K function definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 a 2 K:
||x(t) – xeq|| · a(||x(t0) – xeq||) 8 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|| 8 t¸ t0¸ 0
we could pick a(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 8 x 2 Rn with = 0 only for x = 0
V: Rn ! R radially unbounded ´ x! 1 ) V(x)! 1
provides a measure of
how far x is from xeq
(not necessarily a metric–may
not satisfy triangular inequality)
Lyapunov’s stability theorem
V: Rn ! R positive definite ´ V(x) ¸ 0 8 x 2 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 K function definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 a 2 K:
||x(t) – xeq|| · a(||x(t0) – xeq||) 8 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) 8 t ¸ t0
Thus, by making x(t0) – xeq small we can make V(x(t) – xeq) arbitrarily small 8 t ¸ t0
So, by making x(t0) – xeq small we can make x(t) – xeq arbitrarily small 8 t ¸ t0
(we can actually compute a from V explicitly and take c = +1).
Example #1: Pendulum
l
q
For xeq = (0,0)
Therefore xeq=(0,0) is Lyapunov stable
m
positive definite because V(x) = 0
only for x1 = 2kp k2Z & x2 = 0
(all these points are really the same
because x1 is an angle)
pend.m
Example #1: Pendulum
l
q
For xeq = (p,0)
m
positive definite because V(x) = 0
only for x1 = 2kp k2Z & x2 = 0
(all these points are really the same
because x1 is an angle)
Cannot conclude that xeq=(p,0) is Lyapunov stable (in fact it is not!)
pend.m
Lyapunov’s stability theorem
Definition (class K function definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 a 2 K:
||x(t) – xeq|| · a(||x(t0) – xeq||) 8 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 K function definition):
The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 a 2 K:
||x(t) – xeq|| · a(||x(t0) – xeq||) 8 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?
Example #1: Pendulum
l
q
m
For xeq = (0,0)
not strict for (x1 0, x2=0 !)
pend.m
LaSalle’s Invariance Principle
M 2 Rn is an invariant set ´ x(t0) 2 M ) x(t)2 M8 t¸ t0
(in the context of hybrid systems: Reach(M) ½ 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 { z 2 Rn : W(z) = 0 }
Note that:
1. When W(z) = 0 only for z = xeq then E = {xeq }.
Since M ½ 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
Example #1: Pendulum
l
q
m
For xeq = (0,0)
Inside E, the ODE becomes
E { (x1,x2): x12 R , x2=0}
define set M for which
system remains inside E
Therefore x converges to M { (x1,x2): x1 = k p 2 Z , x2=0}
However, the equilibrium point xeq=(0,0) is not (globally) asymptotically stable because if the system
starts, e.g., at (p,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
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 8 x  0
P is positive semi-definite ´ all eigenvalues are positive or zero
P positive semi-definite ) x’ P x ¸ 0 8 x
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. 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
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. 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 t = T – s
Next lecture…
Lyapunov stability of hybrid systems