Stability and Stabilization of polynomial dynamical systems. Hadi Ravanbakhsh Sriram Sankaranarayanan University of Colorado, Boulder

Transcription

1 Stability and Stabilization of polynomial dynamical systems Hadi Ravanbakhsh Sriram Sankaranarayanan University of Colorado, Boulder

2 Proving Asymptotic Stability: Lyapunov Functions Lyapunov Function: V (x) Positive Definite Function of the State: (8 x) (x 6= x ) ) V (x) > 0 V (x )=0 Negative Definite Derivative: (8 x) (x 6= x ) ) V 0 (x) < 0

3 Verifying Stability Find (Guess) a Lyapunov function: V(x) Check the conditions for the function: (8 x) (x 6= x ) ) V (x) > 0 V (x )=0 (8 x) (x 6= x ) ) V 0 (x) < 0

4 Finding a Lyapunov Function Guess a parametric form of the function: V (x 1,x 2 ): c 0 + c 1 x 1 + c 2 x 2 + c 3 x 1 x 2 + c 4 x c 5 x 2 2 Encode Lyapunov conditions: Derive constraints on the unknown coefficients. Solve constraints.

5 Encoding constraints for Lyapunov QE for reals: [Tarski; Collins; ] Nonlinear constraints over template parameters. Sum-of-Squares Relaxations: [Shor; Lasserre; Tibken; Parrillo; ] Polynomial positivity strengthened to sum of squares (SOS). Reduced to solving a semi-definite optimization problem. LP relaxations: Positive Semi-definiteness to diagonal dominance [Ali Ahmadi+Parrillo 2014] Use properties of Bernstein polynomials. [Ben Sassi + Girard 12, Ben Sassi+S+Chen+Abraham 2015]

6 Lyapunov Function Synthesis: State-of-the-art Linear Systems: Solve LMI constraints. Polynomial Systems: up to 10 variables, low degree. Beyond polynomials: limited progress. Systems involving switching (Hybrid Systems). Lyapunov conditions extended [Branicky et al. 96] Stability proofs that do not involve Lyapunov functions [Prabhakar et al. 13]

7 Synthesis for Stabilization

8 Problem Setup Plant Model x 0 = f(x, u),u2 U x u Controller u = g(x) Find a suitable g(x)

9 Problem Setup Plant Model x 0 = f(x, u),u2 U x u Controller u = g(x) Find g(x) s.t. closed loop x 0 = f(x, g(x)) is asymp. stable

10 Option #1 : Static Feedback Plant Model x 0 = f(x, u),u2 U x u Controller u = g(x) g(x) : P i ix i [Tan & Packard, El Ghaoui & Balakrishnan, Ben Sassi & S]

11 Option #2 : Switched Stabilization Plant Model x 0 = f(x, u),u2 U Choose Appropriate Control Control Mode #1 u = g 1 (x) Control Mode #2 u = g 2 (x) No single mode stabilizes Goal: Synthesize this logic. Control Mode #k u = g k (x)

12 CONTROL LYAPUNOV FUNCTIONS

13 Control Lyapunov Function Lyapunov function: V(x) [Artstein; Sontag; ] V(x) is positive definite. A control input chosen s.t. derivative is negative definite. (8 x 6= x )(9u 2 U) V 0 (x) =r x V (x) f(x, u) < 0

14 Control Lyapunov Function: Interpretation V(x) Lyapunov Function. x Control u 2 Control u 1

15 Switched Stabilization Problem Plant Model x 0 = f(x, u),u2 U Choose Appropriate Control Control Mode #1 u = g 1 (x) Control Mode #2 u = g 2 (x) No single mode stabilizes Goal: Synthesize this logic. Control Mode #k u = g k (x)

16 Choosing Control Modes Idea #1: Sample control values. u 1 Space of Control Inputs: U u 2 u m u m 1 Idea #2: Combine feedback laws that may stabilize for a subset of the states.

17 CLF V(x) Control Lyapunov Function 1. Positive Definite. 2. For all x x * x 0 = f(x, u),u2 U u = g 1 (x) u = g 2 (x) u = g k (x) Exists mode g i Choosing mode g i causes V to decrease. V 0 i :(r xv )f(x, g i (x)) is negative?

18 From CLF to controller x 0 = f(x, u),u2 U? u = g 1 (x) u = g 2 (x) From g 1,, g n choose a mode that causes V(x) to decrease u = g k (x)

19 Issue #1 : Zenoness x 0 = f(x, u),u2 U? u = g 1 (x) u = g 2 (x) From g 1,, g n choose a mode that causes V(x) to decrease u = g k (x) Potentially Switch Infinitely Often in a finite time interval

20 Minimum Dwell Time Switching x 0 = f(x, u),u2 U? u = g 1 (x) u = g 2 (x) u = g k (x) T min T min T min g 5 g 4 g 7 g 10

21 Contribution #1 Define restrictions on CLF V s.t. V(x) allows a switching strategy with a minimal dwell time. Provide a lower bound for the minimal dwell time. V (x) x 2 2 V 0 (x) apple ˆ x 2 2 Extra conditions on V 00 [Ravanbaksh+S 15]

22 CLF-based Synthesis If we can find a CLF V(x) with additional restrictions, From g 1,, g n choose a mode that causes? V(x) to decrease x 0 = f(x, u),u2 U u = g 1 (x) u = g 2 (x) u = g k (x) Minimum dwell time guaranteed.

23 Discovering CLFs Counter-Example Guided Inductive Synthesis.

24 CLF Conditions for the Switched Case x 0 = f(x, u),u2 U? u = g 1 (x) u = g 2 (x) u = g k (x) (8 x 6= x )V (x) x 2 2 (8 x 6= x )(9i 2 {1,...,k})(rV )f(x, g i (x)) apple x 2 2 V can be made to decrease by some choice of g i

25 Synthesizing CLFs Fix a template (ansatz) for the CLF with unknown coefficients. V (x 1,x 2 ): c 0 + c 1 x 1 + c 2 x 2 + c 3 x 1 x 2 + c 4 x c 5 x 2 2 Enforce CLF constraints on the unknown form. (9 ~c) (8 ~x) (9i 2 [1,k]) SOS relaxations cannot be used. More Complex Constraints.

26 Counterexample Guided Inductive Synthesis. Program Sketching [Solar-Lezama + Others] int search(int [] a,int n, int s){ //@pre: if (n <=??){ return 0; } for i = 0 to?? if (a[i] ==??){ return 1; } } //@post: end return??; Find appropriate program expressions.

27 CEGIS Approach Constraints to be solved: (9 c) (8 y)'(c, y) Template Parameters Program Behaviors Iterative Procedure: Finite set Y i : {y 1,, y k } Instatiate the Quantifier 8

28 Basic CEGIS Loop 1. Check SATisfiability of the formula: (9 c) (8 y)'(c, y) (c, y 1 ) ^ (c, y 2 ) ^ ^ (c, y k ) SAT, c k SAT, y k+1 2. Check UNSATisfiability of (c k,y) UNSAT, Failure UNSAT, Success

29 Applying CEGIS to synthesis CLFs (9c) (8 x 6= x ) apple V (x) x 2 2 W k i=1 (rv )f(x, g i(x)) apple x When x is instantiated: Linear Arithmetic over c. 2. When c is instantiated, non-linear arithmetic over x.

30 Integrating SMT solvers CEGIS-based CLF synthesis procedure. instantiate x SMT-solver Z3 Value for c New value for x Instantiate c Nonlinear solver dreal.

31 CEGIS: novelties CEGIS procedure is out-of-the-shelf. But Prove eventual termination of the CEGIS procedure for our problem. Provide heuristics to speedup termination by choosing the right instantiations.

32 Controller Synthesis Support synthesis of MATLAB function (time triggered). Plant Model (Predictive) Control Mode Control Logic Reside Time

33 RESULTS

34 Inverted Pendulum: Bang-Bang Control Synthesis =!,! = g l sin( ) h ml 2! + 1 ml cos( )u, g =9.8, h = 2, l = 2 and m =0.5 Control Mode #1: u = +30 Control Mode #2: u = - 30 V ([!] T ) = ! ! 2

35 Inverted Pendulum! time time! V time (a) Inverted Pendulum time

36 2-5 System Variables 2-6 control modes Overall Results Problem Results ID n q itr z3 T dreal T Tot. Time Status >1hr >1hr >1hr 6 17 out of 20 benchmarks up to 3 variables

37 IMPROVING NONLINEAR SOLVING

38 Use Relaxations Use LMI/SDP relaxations. Replace dreal with a mixed cone optimization problem. Polynomial time but numerical procedures. Tricky to formulate. Solve all but one benchmark. Results need post-verification due to possible numerical errors.

39 FUTURE WORK

40 More Work Safety + Stability Synthesis. Reach While Avoid Properties. Handling state spaces with obstacles. Handle more general temporal objectives Future: Stochastic system stabilization. Control super-martingales.

41 Thank You Supported in part by US NSF CAREER Award #