hybrid systems course lyapunov stability...stability of linear continuous systems theorem (necessary...

Hybrid Systems CourseLyapunov stability

OUTLINE

Focus: stability of an equilibrium point

• continuous systems decribed by ordinary differentialequations (brief review)

• hybrid automata

ORDINARY DIFFERENTIAL EQUATIONS

An ordinary differential equation is a mathematical model of acontinuous state continuous time system:

X = <n ´ state spacef: <n! <n ´ vector field (assigns a “velocity” vector to each x)

Given an initial value x0 2 X,an execution (solution in the sense of Caratheodory) overthe time interval [0,T) is a function x: [0,T) ! <n such that:

• x(0) = x0

• x is continuous and piecewise differentiable

•

ODE SOLUTION: WELL-POSEDNESS

Theorem [global existence and uniqueness non-blocking,deterministic, non-Zeno]If f: <n! <n is globally Lipschitz continuous, then 8 x0 thereexists a single solution with x(0)=x0 defined on [0,1).

STABILITY OF CONTINUOUS SYSTEMS

with f: <n! <n globally Lipschitz continuous

Definition (equilibrium):xe 2 <n for which f(xe)=0

Remark: {xe} is an invariant set

Definition (stable equilibrium):

Graphically:

δxe

equilibrium motion

perturbed motion

small perturbations lead to small changes in behavior

execution startingfrom x(0)=x0

Definition (asymptotically stable equilibrium):

and can be chosen so that

Graphically:

δxe

equilibrium motion

perturbed motion

small perturbations lead to small changes in behaviorand are re-absorbed, in the long run

Definition (asymptotically stable equilibrium):

and can be chosen so that

Graphically:

small perturbations lead to small changes in behaviorand are re-absorbed, in the long run

δxe

EXAMPLE: PENDULUM

m

l

frictioncoefficient (α)

EXAMPLE: PENDULUM

unstable equilibrium

m

EXAMPLE: PENDULUM

as. stable equilibriumm

EXAMPLE: PENDULUM

m

l

Let xe be asymptotically stable.

Definition (domain of attraction):The domain of attraction of xe is the set of x0 such that

Definition (globally asymptotically stable equilibrium):xe is globally asymptotically stable (GAS) if its domain ofattraction is the whole state space <n

execution startingfrom x(0)=x0

EXAMPLE: PENDULUM

as. stable equilibrium

small perturbations areabsorbed, not allperturbations not GAS

m

m

Let xe be asymptotically stable.

Definition (exponential stability):xe is exponentially stable if 9 , , >0 such that



Definition (equilibrium):xe 2 <n for which f(xe)=0

Without loss of generality we suppose that

xe = 0if not, then z := x -xe! dz/dt = g(z), g(z) := f(z+xe) (g(0) = 0)



How to prove stability?find a function V: <n! < such that

V(0) = 0 and V(x) >0, for all x 0V(x) is decreasing along the executions of the system

V(x) = 3

V(x) = 2

x(t)


execution x(t)

candidate function V(x)

behavior of V along theexecution x(t): V(t): = V(x(t))

Advantage with respect to exhaustive check of all executions?


V: <n! < continuously differentiable (C1) function

Rate of change of V along the execution of the ODE system:

(Lie derivative of V with respect to f)


gradient vector


V: <n! < continuously differentiable (C1) function

Rate of change of V along the execution of the ODE system:

(Lie derivative of V with respect to f)


gradient vector

No need to solve the ODE for evaluating if V(x) decreasesalong the executions of the system

LYAPUNOV STABILITY

Theorem (Lyapunov stability Theorem):Let xe = 0 be an equilibrium for the system and D½ <n an open

set containing xe = 0.If V: D! < is a C1 function such that

Then, xe is stable.

V positive definite on D

V non increasing alongsystem executions in D(negative semidefinite)

EXAMPLE: PENDULUM

m

l

frictioncoefficient (α)

energy function

xe stable

LYAPUNOV STABILITY

Theorem (Lyapunov stability Theorem):Let xe = 0 be an equilibrium for the system and D½ <n an open

set containing xe = 0.If V: D! < is a C1 function such that

Then, xe is stable.

If it holds also that

Then, xe is asymptotically stable (AS)

LYAPUNOV GAS THEOREM

Theorem (Barbashin-Krasovski Theorem):Let xe = 0 be an equilibrium for the system.

If V: <n! < is a C1 function such that

Then, xe is globally asymptotically stable (GAS).

V positive definite on <n

V decreasing alongsystem executions in <n

(negative definite)

V radially unbounded

STABILITY OF LINEAR CONTINUOUS SYSTEMS

• xe = 0 is an equilibrium for the system

• the elements of matrix eAt are linear combinations of tKei(A)t,k=0,…,hi, i=1,2,…,n



• xe =0 is asymptotically stable if and only if A is Hurwitz (alleigenvalues with real part <0)

• asymptotic stability GAS




• asymptotic stability GAS

Alternative characterization…


Theorem (necessary and sufficient condition):

The equilibrium point xe =0 is asymptotically stable if and onlyif for all matrices Q = QT positive definite (Q>0) the

ATP+PA = -Q

has a unique solution P=PT >0.

Lyapunov equation




ATP+PA = -Q


Remarks:Q positive definite (Q>0) iff xTQx >0 for all x 0Q positive semidefinite (Q¸ 0) iff xTQx ¸ 0 for all x andxT Q x = 0 for some x 0

Lyapunov equation




ATP+PA = -Q


Proof.

(if) V(x) =xT P x is a Lyapunov function

Lyapunov equation




ATP+PA = -Q


Proof.

(only if) Consider

Lyapunov equation




ATP+PA = -Q


Proof.

(only if) Consider

P = PT and P>0 easy to show

P unique can be proven by contradiction

Lyapunov equation


Remarks: for a linear system

• existence of a (quadratic) Lyapunov function V(x) =xT P x is anecessary and sufficient condition for asymptotic stability

• it is easy to compute a Lyapunov function since the Lyapunovequation

ATP+PA = -Q

is a linear algebraic equation


Theorem (exponential stability):

Let the equilibrium point xe =0 be asymptotically stable. Then,the rate of convergence to xe =0 is exponential:

for all x(0) = x0 2 <n, where 0 2 (0, mini |Re{i(A)}|) and >0is an appropriate constant.





Re

Im

o

o

o o

eigenvalues of A


Proof (exponential stability):

A + 0 I is Hurwitz (eigenvalues are equal to (A) + 0)

Then, there exists P = PT >0 such that

(A + 0I)T P + P (A + 0I) <0

which leads to

x(t)T[AT P + P A]x(t) < - 2 0 x(t)T P x(t)

Define V(x) = xT P x, then

from which


(cont’d) Proof (exponential stability):

thus finally leading to





Remark:




• asymptotic stability GAS exponential stability GES

OUTLINE

Focus: stability of an equilibrium point

• continuous systems decribed by ordinary differentialequations (brief review)

• hybrid automata

HYBRID AUTOMATA: FORMAL DEFINITION

A hybrid automaton H is a collection

H = (Q,X,f,Init,Dom,E,G,R)• Q = {q1,q2, …} is a set of discrete states (modes)

• X = <n is the continuous state space

• f: Q£ X! <n is a set of vector fields on X

• Init µ Q£ X is a set of initial states

• Dom: Q! 2X assigns to each q2 Q a domain Dom(q) of X

• E µ Q£ Q is a set of transitions (edges)

• G: E! 2X is a set of guards (guard condition)

• R: E£ X! 2X is a set of reset maps

q = q1

q = q2

HYBRID TIME SET

A hybrid time set is a finite or infinite sequence of intervals

= {Ii, i=0,1,…, M } such that

• Ii = [i, i’] for i < M• IM = [M, M’] or IM = [M, M’) if M<1• i’ = i+1

• i · i’

[ ]

[ ]

[ ]

τ0

I0

τ0’

τ1 τ1’I1

I2 τ2 = τ2’

τ3 τ3’I3

t1

t2

t3

t4

t1 Á t2 Á t3 Á t4

time instants in arelinearly ordered

HYBRID TIME SET: LENGTH

Two notions of length for a hybrid time set = {Ii, i=0,1,…, M }:

• Discrete extent:<> = M+1

• Continuous extent:|||| = i=0,1,..,M |i’-i|

number of discrete transitions

total durationof intervals in

<> = 4|||| = 3’ - 0

[ ]

[ ]

[ ]

τ0

I0

τ0’

τ1 τ1’I1

I2 τ2 = τ2’

τ3 τ3’I3

HYBRID TIME SET: CLASSIFICATION

A hybrid time set = {Ii, i=0,1,…, M } is

• Finite: if <> is finite and IM = [M, M’]

• Infinite: if <> is infinite or |||| is infinite

• Zeno: if <> is infinite but |||| is finite

finite infinite

infinite ZenoZeno

HYBRID TRAJECTORY

A hybrid trajectory (, q, x) consists of:

• A hybrid time set = {Ii, i=0,1,…, M }• Two sequences of functions q = {qi(¢), i=0,1,…, M } and x =

{xi(¢), i=0,1,…, M } such that

qi: Ii! Q

xi: Ii! X

HYBRID AUTOMATA: EXECUTION

A hybrid trajectory (, q, x) is an execution (solution) of thehybrid automaton H = (Q,X,f,Init,Dom,E,G,R) if it satisfiesthe following conditions:

• Initial condition: (q0(0), x0(0)) 2 Init

• Continuous evolution:for all i such that i < i’

qi: Ii! Q is constantxi:Ii! X is the solution to the ODE associated with qi(i)xi(t) 2 Dom(qi(i)), t2 [i,i’)

• Discrete evolution:(qi(i’),qi+1(i+1)) 2 E transition is feasiblexi(i’) 2 G((qi(i’),qi+1(i+1))) guard condition satisfiedxi+1(i+1) 2 R((qi(i’),qi+1(i+1)),xi(i’)) reset condition satisfied

HYBRID AUTOMATA: EXECUTION

Well-posedness?

Problems due the hybrid nature:

for some initial state (q,x)• infinite execution of finite duration Zeno• no infinite execution blocking• multiple executions non-deterministic

We denote by

H(q,x) the set of (maximal) executions of H starting from (q,x)

H(q,x)1 the set of infinite executions of H starting from (q,x)

STABILITY OF HYBRID AUTOMATA

H = (Q,X,f,Init,Dom,E,G,R)

Definition (equilibrium):xe =0 2 X is an equilibrium point of H if:• f(q,0) = 0 for all q 2 Q• if ((q,q’)2 E) Æ (02 G((q,q’)) ) R((q,q’),0) = {0}

Remarks:

• discrete transitions are allowed out of (q,0) but only to (q’,0)• if (q,0) 2 Init and (, q, x) is an execution of H starting from

(q,0), then x(t) = 0 for all t2

EXAMPLE: SWITCHED LINEAR SYSTEM


• Q = {q1, q2} X = <2

• f(q1,x) = A1x and f(q2,x) = A2x with:

• Init = Q £ {x2 X: ||x|| >0}

• Dom(q1) = {x2 X: x1x2 · 0} Dom(q2) = {x2 X: x1x2 ¸ 0}

• E = {(q1,q2),(q2,q1)}• G((q1,q2)) = {x2 X: x1x2 ¸ 0} G((q2,q1)) = {x2 X: x1x2 · 0}

• R((q1,q2),x) = R((q2,q1),x) = {x}

x1

x2




• Q = {q1, q2} X = <2


• Init = Q £ {x2 X: ||x|| >0}



• R((q1,q2),x) = R((q2,q1),x) = {x}

xe = 0 is an equilibrium: f(q,0) = 0 & R((q,q’),0) = {0}


Definition (stable equilibrium):Let xe = 0 2 X be an equilibrium point of H. xe = 0 is stable if


set of (maximal) executionsstarting from (q0, x0) 2 Init



Remark:

• Stability does not imply convergence

• To analyse convergence, only infinite executions should beconsidered





Definition (asymptotically stable equilibrium):Let xe = 0 2 X be an equilibrium point of H. xe = 0 is asymptotically

stable if it is stable and >0 that can be chosen so that



set of infinite executionsstarting from (q0, x0) 2 Init

1 := i(i’-i)continuous extent1 <1 if Zeno



Question:

xe = 0 stable equilibrium for each continuous systemdx/dt = f(q,x) implies that xe = 0 stable equilibrium for H?




• Q = {q1, q2} X = <2


• Init = Q £ {x2 X: ||x|| >0}



• R((q1,q2),x) = R((q2,q1),x) = {x}

xe = 0 is an equilibrium: f(q,0) = 0 & R((q,q’),0) = {0}

x1

x2



Swiching between asymptotically stable linear systems.

q1: quadrants 2 and 4q2: quadrants 1 and 3

Switching between asymptotically stable linear systems, butxe = 0 unstable equilibrium of H

q1

q1

q2

q2


||x(i+1)|| > ||x(i)||

overshootssum up


q1: quadrants 1 and 3q2: quadrants 2 and 4

q1

q1

q2

q2


||x(i+1)|| < ||x(i)|||


LYAPUNOV STABILITY


Theorem (Lyapunov stability):Let xe = 0 be an equilibrium for H with R((q,q’),x) = {x}, 8 (q,q’)2 E, and

D½ X=<n an open set containing xe = 0.Consider V: Q£ D! < is a C1 function in x such that for all q 2 Q:

If for all (, q, x) 2 H(q0,x0) with (q0,x0) 2 Init Å (Q£ D), and all q’2 Q, thesequence {V(q(i),x(i)): q(i) =q’} is non-increasing (or empty),

then, xe = 0 is a stable equilibrium of H.

LYAPUNOV STABILITY




If for all (, q, x) 2 H(q0,x0) with (q0,x0) 2 Init Å (Q£ D), and all q’2 Q, thesequence {V(q(i),x(i)): q(i) =q’} is non-increasing (or empty),


V(q,x) Lyapunov functionfor continuous system q) xe =0 is stableequilibrium for system q

LYAPUNOV STABILITY




If for all (, q, x) 2 H(q0,x0) with (q0,x0) 2 Init Å (Q£ D), and all q’2 Q,the sequence {V(q(i),x(i)): q(i) =q’} is non-increasing (or empty),


V(q,x) Lyapunov functionfor continuous system q) xe =0 is stableequilibrium for system q

LYAPUNOV STABILITY


Sketch of the proof.

V(q(t),x(t))V(q1,x(t))

[ ][ ][ ][q(t)= q1 q(t)= q1V(q2,x(t))

0 0’=1 1’=2 2’=3

Lyapunov function forsystem q1! decreaseswhen q(t) = q1, but canincrease when q(t) q1

LYAPUNOV STABILITY




[ ]0 0’=1

[ ][ ]1’=2 2’=3

[q(t)= q1 q(t)= q1

{V(q1,x(i))}non-increasing

LYAPUNOV STABILITY




[ ]0 0’=1

[ ][ ]1’=2 2’=3

[q(t)= q1 q(t)= q1

LYAPUNOV STABILITY


V(q(t),x(t)) Lyapunov-like function


• Q = {q1, q2} X = <2


• Init = Q £ {x2 X: ||x|| >0}

• Dom(q1) = {x2 X: Cx ¸ 0} Dom(q2) = {x2 X: Cx · 0}

• E = {(q1,q2),(q2,q1)}• G((q1,q2)) = {x2 X: Cx · 0} G((q2,q1)) = {x2 X: Cx ¸ 0}, CT2 <2

• R((q1,q2),x) = R((q2,q1),x) = {x}



q1q2


x1

x2


Cx = 0

Proof that xe = 0 is a stable equilibrium of H for any CT2 <2 :

• xe = 0 is an equilibrium: f(q1,0) = f(q2,0) = 0

R((q1,q2),0) = R((q2,q1),0) = {0}


Proof that xe = 0 is a stable equilibrium of H for any CT2 <2 :


R((q1,q2),0) = R((q2,q1),0) = {0}

• xe = 0 is stable:

consider the candidate Lyapunov-like function:

V(qi,x) = xT Pi x,

where Pi =PiT >0 solution to Ai

T Pi + Pi Ai = - I

(V(qi,x) is a Lyapunov function for the asymptotically stablelinear system qi)

In each discrete state, the continuous system is as. stable.


Proof that xe = 0 is a stable equilibrium of H for any CT2 <2:


R((q1,q2),0) = R((q2,q1),0) = {0}

• xe = 0 is stable:

consider the candidate Lyapunov-like function:

V(qi,x) = xT Pi x,

where Pi =PiT >0 solution to Ai

T Pi + Pi Ai = - I


Test for non-increasing sequence condition

Let q(i)=q1 and x(i)=z.


Cx = 0

zi


Since V(q1,x(t)) is not increasing during [i,i’], then, when x(t)intersects the switching line at i’, it does at z with 2 (0,1],hence ||x(i+1)|| = ||x(i’)|| · ||x(i)||. Let q(i+1)=q2

Cx = 0

-z

i


i’=i+1

z


Since V(q2,x(t)) is decreasing during [i+1,i+1’], then, when x(t)intersects the switching line at i+1’,||x(i+2)|| = ||x(i+1’)|| · ||x(i+1)|| · ||x(i)||

Cx = 0

-z

i

i’=i+1

i+2


z

Test for non-increasing sequence conditionFrom this, it follows that V(q1,x(i+2)) · V(q1,x(i))

Cx = 0

-z

i

i’=i+1

i+2


z

LYAPUNOV STABILITY


Drawbacks of the approach based on Lyapunov-like functions:

• In general, it is hard to find a Lyapunov-like function

• The sequence {V(q(i),x(i)): q(i) =q’} must be evaluated, whichmay require solving the ODEs

LYAPUNOV STABILITY


Corollary (common Lyapunov function):Let xe = 0 be an equilibrium for H with R((q,q’),x) = {x}, 8

(q,q’)2 E, and D½ X=<n an open set containing xe = 0.

If V: D! < is a C1 function such that for all q 2 Q:


LYAPUNOV STABILITY






independent of q

LYAPUNOV STABILITY






V(x) common Lyapunovfunction for all systems q

LYAPUNOV STABILITY






Proof: Define W(q,x) = V(x), 8 q 2 Q and apply the previous theorem

V(x) common Lyapunovfunction for all systems q

t

))(( txVsame V function+ identity reset map

COMPUTATIONAL LYAPUNOV METHODS

HPL = (Q,X,f,Init,Dom,E,G,R)

non-Zeno and such that for all qk2 Q:• f(qk,x) = Ak x

(linear vector fields)• Init ½ [q2 Q {q } £ Dom(q)

(initialization within the domain)• for all x2 X, the set

Jump(qk,x):= {(q’,x’): (qk,q’)2 E, x2G((qk,q’)), x’2R((qk,q’),x)}has cardinality 1 if x 2 Dom(qk), 0 otherwise(discrete transitions occur only from the boundary of the domains)

• (q’,x’) 2 Jump(qk,x)! x’2 Dom(q’) and x’ = x(trivial reset for x)






Jump(qk,x):= {(q’,x’): (qk,q’)2 E, x2G((qk,q’)), x’2R((qk,q’),x)}has cardinality 1 if x 2 Dom(qk), 0 otherwise(discrete transitions occur only from the boundary of the domains)


For this class of (non-blocking, deterministic) Piecewise Linear hybridautomata computationally attractive methods exist to construct theLyapunov-like function

GLOBALLY QUADRATIC LYAPUNOV FUNCTION


Theorem (globally quadratic Lyapunov function):

Let xe = 0 be an equilibrium for HPL.

If there exists P=PT >0 such thatAk

T P+ PAk < 0, 8 k

Then, xe = 0 is asymptotically stable.

Remark:

V(x)=xTPx is a common Lyapunov function xe = 0 is stable


Proof (showing exponential stability):There exists >0 such that Ak

T P+ PAk + I · 0, 8 k

There exists a unique, infinite, non-Zeno execution (,q,x) forevery initial state with x: ! <n satisfying

where k: ! [0,1] is such that k k(t)=1, t2 [i,i’].

Let V(x) = xT Px. Then, for t2 [i,i’).


Proof. (cont’d)Since min ||x||2 · V(x) · max ||x||2, then

and, hence,

which leads to

Then,

Since 1 =1 (non-Zeno), then ||x(t)|| goes to zeroexponentially as t! 1

min and max eigenvalues of P


q1q2

conditions of the theorem satisfied with P = I


q1q2



Theorem (globally quadratic Lyapunov function):


If there exists P=PT >0 such thatAk

T P+ PAk < 0, 8 k


Remark:

A set of LMIs to solve. This problem can be reformulated as aconvex optimization problem. Efficient solvers exist.


Suppose that Ak , k=1,2,…,N, are Hurwitz matrices.

Then, the set of linear matrix inequalities

AkT P+ PAk < 0, k=1,2,…,N,

where P is positive definite symmetric is not feasible if andonly if there exist positive definite symmetric matrices Rk,k=1,2,…,N, such that

Proof. Based on results in convex analysis


q1q2

stable node stable focus


q1q2

no globally quadratic Lyapunov function exists although xe = 0 stableequilibrium

stable node stable focus

PIECEWISE QUADRATIC LYAPUNOV FUNCTION

• Idea:consider different quadratic Lyapunov functions on eachdomain and glue them together so as to provide a (non-quadratic) Lyapunov function for H that is continuous at theswitching times


• Idea:consider different quadratic Lyapunov functions on eachdomain and glue them together so as to provide a (non-quadratic) Lyapunov function for H that is continuous at theswitching times

• Developed for piecewise linear systems with structureddomain and reset

• LMIs characterization






Jump(qk,x):= {(q’,x’): (qk,q’)2 E, x2G((qk,q’)), x’2R((qk,q’),x)}has cardinality 1 if x 2 Dom(q), 0 otherwise(discrete transitions occur only from the boundary of the domains)




satisfies the following additional assumptions:• Dom(q) = {x 2 X: Eq1x ¸ 0, Eq2 x ¸ 0, … , Eqn x¸ 0}

(each domain is a polygon)Eq = [Eq1

T Eq2T … Eqn

T] T 2 <n£ n defines the domain.

• (q’,x’) 2 Jump(q,x) Fq’x = Fqx, q’q, x’=x(matching condition at the boundaries of the domain)



Theorem (piecewise quadratic Lyapunov function):


If there exists Uk=UkT, Wk=Wk

T, with all non-negative elements and

M=MT, such that Pk=FkTMFk satisfies

AkT Pk+ PAk + Ek

TUkEk < 0

Pk – EkTWkEk > 0









AkT Pk+ PAk + Ek

TUkEk < 0

Pk – EkTWkEk > 0


Proof based on the fact that V(x)=xTPkx, xDom(qk) is aLyapunov-like function for HPL, strictly decreasing along itsexecutions








AkT Pk+ PAk + Ek

TUkEk < 0

Pk – EkTWkEk > 0 Pk positive definite within Dom(k)










AkT Pk+ PAk + Ek

TUkEk < 0 AkT Pk+ PAk < 0 within Dom(k)

Pk – EkTWkEk > 0 Pk positive definite within Dom(k)










AkT Pk+ PAk + Ek

TUkEk < 0

Pk – EkTWkEk > 0



continuitity of V(x)


q1q2

level curves of thepiecewise quadraticLyapunov function(red lines)

phase plot of somecontinuous statetrajectories(blue lines)

REFERENCES

• H.K. Khalil.Nonlinear Systems.Prentice Hall, 1996.

• S. Boyd, L. El Ghaoui, E. Feron, and V. Balakrishnan.Linear Matrix Inequalities in System and Control Theory.SIAM, 1994.

• M. Branicky.Multiple Lyapunov functions and other analysis tools for switched andhybrid systems.IEEE Trans. on Automatic Control, 43(4):475-482, 1998.

• H. Ye, A. Michel, and L. Hou.Stability theory for hybrid dynamical systems.IEEE Transactions on Automatic Control, 43(4):461-474, 1998.

• M. Johansson and A. Rantzer.Computation of piecewise quadratic Lyapunov function for hybridsystems.IEEE Transactions on Automatic Control, 43(4):555-559, 1998.

• R.A. Decarlo, M.S. Branicky, S. Petterson, and B. Lennartson.Perspectives and results on the stability and stabilization of hybridsystems.Proceedings of the IEEE, 88(7):1069-1082, 2000.

hybrid systems course lyapunov stability...stability of linear continuous systems theorem (necessary...

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