Lecture #7 Stability and convergence of ODEs João P. Hespanha University of California at Santa Barbara Hybrid Control and Switched Systems NO CLASSES on Oct 18 & Oct 20

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  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…

Lyapunov stability (ODEs) equilibrium point ´ x eq 2 R n for which f(x eq ) = 0 thus x(t) = x eq 8 t ¸ 0 is a solution to the ODE E.g., pendulum equation  m l two equilibrium points: x 1 = 0, x 2 = 0 (down) andx 1 = , x 2 = 0 (up)

Lyapunov stability (ODEs) equilibrium point ´ x eq 2 R n for which f(x eq ) = 0 thus x(t) = x eq 8 t ¸ 0 is a solution to the ODE Definition ( e –  definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 8 e > 0 9  >0 : ||x(t 0 ) – x eq || ·  ) ||x(t) – x eq || · e 8 t ¸ t 0 ¸ 0 x eq  e x(t)x(t) 1. if the solution starts close to x eq it will remain close to it forever 2. e can be made arbitrarily small by choosing  sufficiently small

Example #1: Pendulum pend.m x eq =(0,0) stable x eq =( ,0) unstable  m l x 1 is an angle so you must “glue” left to right extremes of this plot

Lyapunov stability – continuity definition Definition (continuity definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if T is continuous at x eq : 8 e > 0 9  >0 : ||x 0 – x eq || ·  ) ||T(x 0 ) – T(x eq )|| sig · e  sig ´ set of all piecewise continuous signals taking values in R n Given a signal x 2  sig, ||x|| sig › sup t ¸ 0 ||x(t)|| ODE can be seen as an operator T : R n !  sig that maps x 0 2 R n into the solution that starts at x(0) = x 0 signal norm sup t ¸ 0 ||x(t) – x eq || · e x eq  e x(t)x(t) can be extended to nonequilibrium solutions

Stability of arbitrary solutions Definition (continuity definition): A solution x * :[0,T) ! R n is (Lyapunov) stable if T is continuous at x * 0 › x * (0), i.e., 8 e > 0 9  >0 : ||x 0 – x * 0 || ·  ) ||T(x 0 ) – T(x * 0 )|| sig · e  sig ´ set of all piecewise continuous signals taking values in R n Given a signal x 2  sig, ||x|| sig › sup t ¸ 0 ||x(t)|| ODE can be seen as an operator T : R n !  sig that maps x 0 2 R n into the solution that starts at x(0) = x 0 signal norm sup t ¸ 0 ||x(t) – x * (t)|| · e  e x(t)x(t) 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 * un stable vdp.m

Lyapunov stability equilibrium point ´ x eq 2 R n for which f(x eq ) = 0 class  ´ set of functions  :[0, 1 ) ! [0, 1 ) that are 1. continuous 2. strictly increasing 3.  (0)=0 Definition (class  function definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||) 8 t ¸ t 0 ¸ 0, ||x(t 0 ) – x eq || · c the function  can be constructed directly from the  ( e ) in the e –  (or continuity) definitions s (s)(s) x eq  (||x(t 0 ) – x eq ||) ||x(t 0 ) – x eq || x(t)x(t) t

Asymptotic stability Definition: The equilibrium point x eq 2 R n is (globally) asymptotically stable if it is Lyapunov stable and for every initial state the solution exists on [0, 1 ) and x(t) ! x eq as t !1. x eq  (||x(t 0 ) – x eq ||) ||x(t 0 ) – x eq || x(t)x(t) s (s)(s) equilibrium point ´ x eq 2 R n for which f(x eq ) = 0 class  ´ set of functions  :[0, 1 ) ! [0, 1 ) that are 1. continuous 2. strictly increasing 3.  (0)=0 t

Asymptotic stability Definition (class  function definition): The equilibrium point x eq 2 R n is (globally) asymptotically stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||,t – t 0 ) 8 t ¸ t 0 ¸ 0 x eq  (||x(t 0 ) – x eq ||,0) ||x(t 0 ) – x eq || x(t)x(t) equilibrium point ´ x eq 2 R n for which f(x eq ) = 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)(s,t) (for each fixed t) t (s,t)(s,t) (for each fixed s)  (||x(t 0 ) – x eq ||,t) t We have exponential stability when  (s,t) = c e – t s with c, > 0 linear in s and negative exponential in t

Example #1: Pendulum x2x2 x1x1 x eq =(0,0) asymptotically stable x eq =( ,0) unstable pend.m k > 0 (with friction) k = 0 (no friction) x eq =(0,0) stable but not asymptotically x eq =( ,0) unstable

Example #3: Butterfly Convergence by itself does not imply stability, e.g., all solutions converge to zero but x eq = (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? x 1 is an angle so you must “glue” left to right extremes of this plot

Lyapunov’s stability theorem Definition (class  function definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||) 8 t ¸ t 0 ¸ 0, ||x(t 0 ) – x eq || · c Suppose we could show that ||x(t) – x eq || always decreases along solutions to the ODE. Then ||x(t) – x eq || · ||x(t 0 ) – x eq || 8 t ¸ t 0 ¸ 0 we could pick  (s) = s ) Lyapunov stability We can draw the same conclusion by using other measures of how far the solution is from x eq : V: R n ! R positive definite ´ V(x) ¸ 0 8 x 2 R n with = 0 only for x = 0 V: R n ! R radially unbounded ´ x ! 1 ) V(x) ! 1 provides a measure of how far x is from x eq (not necessarily a metric–may not satisfy triangular inequality)

Lyapunov’s stability theorem V: R n ! R positive definite ´ V(x) ¸ 0 8 x 2 R n with = 0 only for x = 0 provides a measure of how far x is from x eq (not necessarily a metric–may not satisfy triangular inequality) Q: How to check if V(x(t) – x eq ) decreases along solutions? A: V(x(t) – x eq ) will decrease if can be computed without actually computing x(t) (i.e., solving the ODE) gradient of V

Lyapunov’s stability theorem Definition (class  function definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||) 8 t ¸ t 0 ¸ 0, ||x(t 0 ) – x eq || · c Theorem (Lyapunov): Suppose there exists a continuously differentiable, positive definite function V: R n ! R such that Then x eq is a Lyapunov stable equilibrium. Why? V non increasing ) V(x(t) – x eq ) · V(x(t 0 ) – x eq ) 8 t ¸ t 0 Thus, by making x(t 0 ) – x eq small we can make V(x(t) – x eq ) arbitrarily small 8 t ¸ t 0 So, by making x(t 0 ) – x eq small we can make x(t) – x eq arbitrarily small 8 t ¸ t 0 (we can actually compute  from V explicitly and take c = + 1 ). V(z – x eq ) z (cup-like function) Lyapunov function

Example #1: Pendulum  m l For x eq = (0,0) positive definite because V(x) = 0 only for x 1 = 2k  k 2 Z  x 2 = 0 (all these points are really the same because x 1 is an angle) Therefore x eq =(0,0) is Lyapunov stable pend.m

Example #1: Pendulum  m l For x eq = ( ,0) positive definite because V(x) = 0 only for x 1 = 2k  k 2 Z  x 2 = 0 (all these points are really the same because x 1 is an angle) Cannot conclude that x eq =( ,0) is Lyapunov stable (in fact it is not!) pend.m

Lyapunov’s stability theorem Definition (class  function definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||) 8 t ¸ t 0 ¸ 0, ||x(t 0 ) – x eq || · c Theorem (Lyapunov): Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: R n ! R such that Then x eq is a Lyapunov stable equilibrium and the solution always exists globally. Moreover, if = 0 only for z = x eq then x eq is a (globally) asymptotically stable equilibrium. Why? V can only stop decreasing when x(t) reaches x eq but V must stop decreasing because it cannot become negative Thus, x(t) must converge to x eq

Lyapunov’s stability theorem What if for other z then x eq ? Can we still claim some form of convergence? Definition (class  function definition): The equilibrium point x eq 2 R n is (Lyapunov) stable if 9  2  : ||x(t) – x eq || ·  (||x(t 0 ) – x eq ||) 8 t ¸ t 0 ¸ 0, ||x(t 0 ) – x eq || · c Theorem (Lyapunov): Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: R n ! R such that Then x eq is a Lyapunov stable equilibrium and the solution always exists globally. Moreover, if = 0 only for z = x eq then x eq is a (globally) asymptotically stable equilibrium.

Example #1: Pendulum  m l For x eq = (0,0) not strict for (x 1  0, x 2 =0 !) pend.m

LaSalle’s Invariance Principle Theorem (LaSalle Invariance Principle): Suppose there exists a continuously differentiable, positive definite, radially unbounded function V: R n ! R such that Then x eq 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 R n : W(z) = 0 } M 2 R n is an invariant set ´ x(t 0 ) 2 M ) x(t) 2 M 8 t ¸ t 0 (in the context of hybrid systems: Reach(M) ½ M…) Note that: 1.When W(z) = 0 only for z = x eq then E = {x eq }. Since M ½ E, M = {x eq } and therefore x(t) ! x eq ) asympt. stability 2.Even when E is larger then {x eq } we often have M = {x eq } and can conclude asymptotic stability. Lyapunov theorem

Example #1: Pendulum  m l For x eq = (0,0) E › { (x 1,x 2 ): x 1 2 R, x 2 =0} Inside E, the ODE becomes Therefore x converges to M › { (x 1,x 2 ): x 1 = k  2 Z, x 2 =0} However, the equilibrium point x eq =(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

Linear systems Solution to a linear ODE: Theorem: The origin x eq = 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 x eq = 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 has 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 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 Theorem: The origin x eq = 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 has 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 x V is positive definite & radially unbounded because P is positive definite V is continuously differentiable: Solution to a linear ODE: thus system is asymptotically stable by Lyapunov Theorem

Lyapunov equation Theorem: The origin x eq = 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 has 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 integration variable  = T – s A is asympt. stable ) e At decreases to zero exponentiall fast ) P is well defined (limit exists and is finite) Solution to a linear ODE:

Next lecture… Lyapunov stability of hybrid systems