lecture #7 stability and convergence of odes joo p. hespanha university of california at santa...
DESCRIPTION
Properties of hybrid systems sig ´ set of all piecewise continuous signals x:[0,T) ! R n, T 2 (0, 1 ] sig ´ set of all piecewise constant signals q:[0,T) ! , T 2 (0, 1 ] Sequence property ´ p : sig £ sig ! {false,true} E.g., A pair of signals (q, x) 2 sig £ 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…TRANSCRIPT
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Lecture #7Stability and convergence of ODEs
João P. Hespanha
University of Californiaat Santa Barbara
Hybrid Control and Switched Systems
NO CLASSES on Oct 18 & Oct 20
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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, T2(0,1]sig ´ set of all piecewise constant signals q:[0,T)! , T2(0,1]
Sequence property ´ p : sig £ sig ! {false,true}E.g.,
A pair of signals (q, x) 2 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…
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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
m
l
two equilibrium points:x1 = 0, x2 = 0 (down)
and x1 = , x2 = 0 (up)
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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 (– definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if
8 > 0 9 >0 : ||x(t0) – xeq|| · ) ||x(t) – xeq|| · 8 t¸ t0¸ 0
xeq
x(t)1. if the solution starts close to xeq
it will remain close to it forever2. can be made arbitrarily small
by choosing sufficiently small
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Example #1: Pendulum
pend.m
xeq=(0,0)stable
xeq=(,0)unstable
m
l
x1 is an angle so you must “glue” left to
right extremes of this plot
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Lyapunov stability – continuity definition
Definition (continuity definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if T is continuous at xeq:
8 > 0 9 >0 : ||x0 – xeq|| · ) ||T(x0) – T(xeq)||sig ·
sig ´ set of all piecewise continuous signals taking values in Rn
Given a signal x2sig, ||x||sig › supt¸0 ||x(t)||
ODE can be seen as an operatorT : Rn ! sig
that maps x0 2 Rn into the solution that starts at x(0) = x0
signal norm
supt¸0 ||x(t) – xeq|| ·
xeq
x(t)
can be extended to nonequilibrium solutions
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Stability of arbitrary solutions
Definition (continuity definition):A solution x*:[0,T)!Rn is (Lyapunov) stable if T is continuous at x*
0 x*(0), i.e.,8 > 0 9 >0 : ||x0 – x*
0|| · ) ||T(x0) – T(x*0)||sig ·
sig ´ set of all piecewise continuous signals taking values in Rn
Given a signal x2sig, ||x||sig › supt¸0 ||x(t)||
ODE can be seen as an operatorT : Rn ! sig
that maps x0 2 Rn into the solution that starts at x(0) = x0
signal norm
supt¸0 ||x(t) – x*(t)|| ·
x(t) x*(t)
pend.m
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Example #2: Van der Pol oscillator
x* Lyapunov stable
vdp.m
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Stability of arbitrary solutions
E.g., Van der Pol oscillator
x* unstable
vdp.m
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Lyapunov stability
equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
class ´ set of functions :[0,1)![0,1) that are1. continuous2. strictly increasing3. (0)=0
Definition (class function definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 2 :
||x(t) – xeq|| · (||x(t0) – xeq||) 8 t¸ t0¸ 0, ||x(t0) – xeq||· c
the function can be constructed directly from the () in the –
(or continuity) definitions
s
(s)
xeq
(||
x(t 0)
– x
eq||)
||x(t 0)
– x
eq||
x(t)
t
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Asymptotic stability
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.
xeq
(||
x(t 0)
– x
eq||)
||x(t 0)
– x
eq||
x(t)
s
(s)equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
class ´ set of functions :[0,1)![0,1) that are1. continuous2. strictly increasing3. (0)=0
t
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Asymptotic stability
Definition (class function definition):The equilibrium point xeq 2 Rn is (globally) asymptotically stable if 9 2:
||x(t) – xeq|| · (||x(t0) – xeq||,t – t0) 8 t¸ t0¸ 0
xeq
(||x
(t 0) –
xeq
||,0)
||x(t 0)
– x
eq||
x(t)
equilibrium point ´ xeq 2 Rn for which f(xeq) = 0
class ´ set of functions :[0,1)£[0,1)![0,1) s.t.1. for each fixed t, (¢,t) 2 2. for each fixed s, (s,¢) is monotone decreasing and (s,t) ! 0 as t!1
s
(s,t)
(for each fixed t)
t
(s,t)(for each fixed s)
(||x(t0) – xeq||,t)
t
We have exponential stability when
(s,t) = c e – t swith c, > 0
linear in s and negative exponential in t
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Example #1: Pendulum
x2
x1
xeq=(0,0)asymptotically
stable
xeq=(,0)unstable
pend.m
k > 0 (with friction) k = 0 (no friction)
xeq=(0,0)stable but not
asymptotically
xeq=(,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?
x1 is an angle so you must “glue” left to
right extremes of this plot
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Lyapunov’s stability theorem
Definition (class function definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 2 :
||x(t) – xeq|| · (||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¸ 0we could pick (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 = 0V: Rn ! R radially unbounded ´ x! 1 ) V(x)! 1
provides a measure ofhow far x is from xeq
(not necessarily a metric–may not satisfy triangular
inequality)
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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 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
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Lyapunov’s stability theorem
Definition (class function definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 2 :
||x(t) – xeq|| · (||x(t0) – xeq||) 8 t¸ t0¸ 0, ||x(t0) – xeq||· c
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) 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 from V explicitly and take c = +1).
V(z – xeq)
z
(cup-likefunction)
Lyapunov function
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Example #1: Pendulum
m
l
For xeq = (0,0)
positive definite because V(x) = 0 only for x1 = 2k k2Zx2 = 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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Example #1: Pendulum
m
l
For xeq = (,0)
positive definite because V(x) = 0 only for x1 = 2k k2Zx2 = 0
(all these points are really the same because x1 is an angle)
Cannot conclude that xeq=(,0) is Lyapunov stable (in fact it is not!) pend.m
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Lyapunov’s stability theorem
Definition (class function definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 2 :
||x(t) – xeq|| · (||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 negativeThus, x(t) must converge to xeq
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Lyapunov’s stability theorem
What if
for other z then xeq ? Can we still claim some form of convergence?
Definition (class function definition):The equilibrium point xeq 2 Rn is (Lyapunov) stable if 9 2 :
||x(t) – xeq|| · (||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.
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Example #1: Pendulum
m
l
For xeq = (0,0)
not strict for (x1 0, x2=0 !)
pend.m
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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 › { z 2 Rn : W(z) = 0 }
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…)
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.
Lyapunovtheorem
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Example #1: Pendulum
m
l
For xeq = (0,0)
E › { (x1,x2): x12 R , x2=0}Inside E, the ODE becomes
Therefore x converges to M › { (x1,x2): x1 = k 2 Z , x2=0}However, the equilibrium point xeq=(0,0) is not (globally) asymptotically stable because if the system starts, e.g., at (,0) it remains there forever.
define set M for which system remains inside E
pend.m
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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
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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 = – Qhas a unique solutions P that is symmetric and positive definite
Lyapunov equation
Recall: given a symmetric matrix P
P is positive definite ´ all eigenvalues are positiveP positive definite ) x’ P x > 0 8 x 0
P is positive semi-definite ´ all eigenvalues are positive or zeroP positive semi-definite ) x’ P x ¸ 0 8 x
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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
Why?1. 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:
Solution to a linear ODE:
thus system is asymptotically stable by Lyapunov Theorem
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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
Why?2. asympt. stable ) P exists and is unique (constructive proof)
change of integrationvariable = T – s
A is asympt. stable ) eAt decreases to zero exponentiall fast ) P is well defined (limit exists and is finite)
Solution to a linear ODE:
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Next lecture…
Lyapunov stability of hybrid systems