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Lecture #7 Stability and convergence of ODEs

Hybrid Control and Switched Systems. Lecture #7 Stability and convergence of ODEs. NO CLASSES on Oct 18 & Oct 20. Jo ã o P. Hespanha University of California at Santa Barbara. Summary. Lyapunov stability of ODEs epsilon-delta and beta-function definitions Lyapunov’s stability theorem

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Lecture #7 Stability and convergence of ODEs

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

  2. Summary • Lyapunov stability of ODEs • epsilon-delta and beta-function definitions • Lyapunov’s stability theorem • LaSalle’s invariance principle • Stability of linear systems

  3. 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) 2Qsig£Xsigsatisfiesp if p(q, x) = true A hybrid automatonHsatisfiesp ( write H²p ) 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…

  4. Lyapunov stability (ODEs) equilibrium point´xeq2 Rn for which f(xeq) = 0 thus x(t) = xeq8t¸ 0 is a solution to the ODE E.g., pendulum equation l q two equilibrium points: x1 = 0, x2 = 0 (down) and x1 = p, x2 = 0 (up) m

  5. Lyapunov stability (ODEs) equilibrium point´xeq2 Rn for which f(xeq) = 0 thus x(t) = xeq8t¸ 0 is a solution to the ODE Definition (e–d definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 8e > 0 9d >0 : ||x(t0) – xeq|| ·d) ||x(t) – xeq|| ·e 8t¸t0¸ 0 e xeq d • if the solution starts close to xeq it will remain close to it forever • e can be made arbitrarily small by choosing d sufficiently small x(t)

  6. 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

  7. Lyapunov stability – continuity definition e xeq d x(t) 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 x02 Rn into the solution that starts at x(0) = x0 Definition (continuity definition): The equilibrium point xeq2 Rn is (Lyapunov) stable if T is continuous at xeq: 8e > 0 9d >0 : ||x0 – xeq|| ·d) ||T(x0) – T(xeq)||sig·e supt¸0 ||x(t) – xeq|| ·e can be extended to nonequilibrium solutions

  8. 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 x02 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., 8e > 0 9d >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

  9. Example #2: Van der Pol oscillator x*Lyapunov stable vdp.m

  10. Stability of arbitrary solutions E.g., Van der Pol oscillator x*unstable vdp.m

  11. Lyapunov stability equilibrium point ´xeq2 Rn for which f(xeq) = 0 class K´ set of functions a:[0,1)![0,1) that are1. continuous2. strictly increasing3. a(0)=0 a(s) s Definition (class K function definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 9a 2 K: ||x(t) – xeq|| ·a(||x(t0) – xeq||) 8t¸t0¸ 0, ||x(t0) – xeq||· c the function a can be constructed directly from the d(e) in the e–d (or continuity) definitions x(t) a(||x(t0) – xeq||) ||x(t0) – xeq|| xeq t

  12. Asymptotic stability equilibrium point ´xeq2 Rn for which f(xeq) = 0 class K´ set of functions a:[0,1)![0,1) that are1. continuous2. strictly increasing3. a(0)=0 a(s) s Definition: The equilibrium point xeq2 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) a(||x(t0) – xeq||) ||x(t0) – xeq|| xeq t

  13. Asymptotic stability b(s,t) (for each fixed t) equilibrium point ´xeq2 Rn 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) 2K2. 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) t Definition (class KL function definition): The equilibrium point xeq2 Rn is (globally) asymptotically stable if 9b2KL: ||x(t) – xeq|| ·b(||x(t0) – xeq||,t – t0) 8t¸t0¸ 0 We have exponential stability when b(s,t) = c e –l t s with c,l > 0 b(||x(t0) – xeq||,t) b(||x(t0) – xeq||,0) x(t) ||x(t0) – xeq|| xeq t linear in s and negative exponential in t

  14. Example #1: Pendulum k = 0 (no friction) k > 0 (with friction) x2 x1 xeq=(0,0) asymptoticallystable xeq=(0,0) stable but not asymptotically xeq=(p,0) unstable xeq=(p,0) unstable pend.m

  15. Example #3: Butterfly Why was Mr. Lyapunov so picky? Why shouldn’t boundedness and convergence to zero suffice? Convergence by itself does not imply stability, e.g., 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

  16. Lyapunov’s stability theorem Definition (class K function definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 9a 2 K: ||x(t) – xeq|| ·a(||x(t0) – xeq||) 8t¸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|| 8t¸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 8x 2 Rn with = 0 only for x = 0 V: 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)

  17. Lyapunov’s stability theorem V: Rn!R positive definite ´V(x) ¸ 0 8x 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? gradient of V A: V(x(t) – xeq) will decrease if can be computed without actually computing x(t)(i.e., solving the ODE)

  18. Lyapunov’s stability theorem Definition (class K function definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 9a 2 K: ||x(t) – xeq|| ·a(||x(t0) – xeq||) 8t¸t0¸ 0, ||x(t0) – xeq||· c Lyapunov function Theorem (Lyapunov): Suppose there exists a continuously differentiable, positive definite function V: Rn!R such that V(z – xeq) (cup-likefunction) Then xeq is a Lyapunov stable equilibrium. z Why? V non increasing )V(x(t) – xeq) ·V(x(t0) – xeq) 8t ¸t0 Thus, by making x(t0) – xeq small we can make V(x(t) – xeq) arbitrarily small 8t ¸t0 So, by making x(t0) – xeq small we can make x(t) – xeq arbitrarily small 8t ¸t0 (we can actually compute a from V explicitly and take c = +1).

  19. Example #1: Pendulum l q 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) For xeq = (0,0) Therefore xeq=(0,0) is Lyapunov stable pend.m

  20. Example #1: Pendulum l q 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) For xeq = (p,0) Cannot conclude that xeq=(p,0) is Lyapunov stable (in fact it is not!) pend.m

  21. Lyapunov’s stability theorem Definition (class K function definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 9a 2 K: ||x(t) – xeq|| ·a(||x(t0) – xeq||) 8t¸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

  22. Lyapunov’s stability theorem Definition (class K function definition):The equilibrium point xeq2 Rn is (Lyapunov) stable if 9a 2 K: ||x(t) – xeq|| ·a(||x(t0) – xeq||) 8t¸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?

  23. Example #1: Pendulum l q m For xeq = (0,0) not strict for (x1¹ 0, x2=0 !) pend.m

  24. LaSalle’s Invariance Principle M 2 Rn is an invariant set ´x(t0) 2M ) x(t)2 M8 t¸ t0(in the context of hybrid systems: Reach(M) ½M…) 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 ›{ z2 Rn : W(z) = 0 } • Note that: • When W(z) = 0 only for z = xeq then E = {xeq }.Since M½E, M = {xeq } and therefore x(t) !xeq) asympt. stability • Even when E is larger then {xeq } we often have M = {xeq } and can conclude asymptotic stability. Lyapunovtheorem

  25. Example #1: Pendulum l q m For xeq = (0,0) E› { (x1,x2): x12 R , x2=0} Inside E,the ODE becomes define set M for which system remains inside E Therefore x converges to M› { (x1,x2): x1 = kp2 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

  26. Linear systems Solution to a linear ODE: • Theorem: The origin xeq = 0 is an equilibrium point. It is • 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. • Asymptotically stable if and only if all eigenvalues of A have negative real parts. In this case the origin is actually exponentially stable

  27. Linear systems linear.m

  28. 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 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 positiveP positive definite )x’ P x > 0 8x¹ 0 P is positive semi-definite ´ all eigenvalues are positive or zeroP positive semi-definite )x’ P x¸ 0 8x

  29. 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 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: thus system is asymptotically stable by Lyapunov Theorem

  30. 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 has a unique solutions P that is symmetric and positive definite Lyapunov equation 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 integrationvariable t = T – s

  31. Next lecture… Lyapunov stability of hybrid systems

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