LMI Methods in Optimal and Robust Control

LMI Methods in Optimal and Robust Control Matthew M. Peet Arizona State University Lecture 15: Nonlinear Systems and Lyapunov Functions

Overview Our next goal is to extend LMI s and optimization to nonlinear systems analysis. Today we will discuss 1. Nonlinear Systems Theory 1.1 Existence and Uniqueness 1.2 Contractions and Iterations 1.3 Gronwall-Bellman Inequality 2. Stability Theory 2.1 Lyapunov Stability 2.2 Lyapunov s Direct Method 2.3 A Collection of Converse Lyapunov Results The purpose of this lecture is to show that Lyapunov stability can be solved Exactly via optimization of polynomials. M. Peet Lecture 15: 1 / 44

Ordinary Nonlinear Differential Equations Computing Stability and Domain of Attraction Consider: A System of Nonlinear Ordinary Differential Equations ẋ(t) = f(x(t)) Problem: Stability Given a specific polynomial f : R n R n, find the largest X R n such that for any x(0) X, lim t x(t) = 0. M. Peet Lecture 15: 2 / 44

Nonlinear Dynamical Systems Long-Range Weather Forecasting and the Lorentz Attractor A model of atmospheric convection analyzed by E.N. Lorenz, Journal of Atmospheric Sciences, 1963. ẋ = σ(y x) ẏ = rx y xz ż = xy bz M. Peet Lecture 15: 3 / 44

Stability and Periodic Orbits The Poincaré-Bendixson Theorem and van der Pol Oscillator 3 An oscillating circuit model: ẏ = x (x 2 1)y ẋ = y y 2 1 0 1 2 Domain of attraction 3 3 2 1 0 1 2 3 x Figure: The van der Pol oscillator in reverse Theorem 1 (Poincaré-Bendixson). Invariant sets in R 2 always contain a limit cycle or fixed point. M. Peet Lecture 15: 4 / 44

Stability of Ordinary Differential Equations Consider ẋ(t) = f(x(t)) with x(0) R n. Theorem 2 (Lyapunov Stability). Suppose there exists a continuous V and α, β, γ > 0 where β x 2 V (x) α x 2 V (x) T f(x) γ x 2 for all x X. Then any sub-level set of V in X is a Domain of Attraction. A Sublevel Set: Has the form V δ = {x : V (x) δ}. M. Peet Lecture 15: 5 / 44

Do The Equations Have a Solution? The Cauchy Problem The first question people ask is the Cauchy problem: For Autonomous (Uncontrolled) Systems: Definition 3 (Cauchy Problem). The Cauchy problem is to find a unique, continuous x : [0, t f ] R n for some t f such that ẋ is defined and ẋ(t) = f(t, x(t)) for all t [0, t f ]. If f is continuous, the solution must be continuously differentiable. Controlled Systems: For a controlled system, we have ẋ(t) = f(x(t), u(t)) and assume u(t) is given. This precludes feedback In this lecture, we focus on the autonomous system. Including t complicates the analysis. However, results are almost all the same. M. Peet Lecture 15: 6 / 44

Ordinary Differential Equations Existence of Solutions There exist many systems for which no solution exists or for which a solution lems onlyof exists solutions over a finite time interval. Finite Escape Time exponential stability quadratic stability time-varying systems invariant sets center manifold theorem Even for something as simple as ation ẋ(t) = x(t) 2 x(0) = x 0 x(0)= x 0 has the solution t, 0 t< 1 x 0 x(t) = x 0 1 x 0 t which clearly has escape time = 1 x 0 x(t) 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 Finite escape time of dx/dt = x 2 t e = 1 x 0 0 0 1 2 3 4 5 Time t ssproblems Figure: Simulation of ẋ = x 2 for several x(0) Previous problem is like the water-tank problem in backwar time M. Peet Lecture 15: 7 / 44

Ordinary Differential Equations Non-Uniqueness A classical example of a system without a unique solution is ẋ(t) = x(t) 1/3 x(0) = 0 For the given initial condition, it is easy to verify that ( 2t x(t) = 0 and x(t) = 3 both satisfy the differential equation. ) 3/2 1.6 1.6 1.4 1.4 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure: Matlab simulation of ẋ(t) = x(t) 1/3 with x(0) = 0 Figure: Matlab simulation of ẋ(t) = x(t) 1/3 with x(0) =.000001 M. Peet Lecture 15: 8 / 44

Ordinary Differential Equations Non-Uniqueness Uniqueness Problems Example: The equation ẋ= x, x(0)=0has man An Example of a system with several solutions is given by ẋ(t) = { (t C) x(t) x(t) x(0) = 2 /4 t> C = 0 0 t C For the given initial condition, it is easy to verify that for any C, { (t C) 2 x(t) = 4 t > C 0 t C satisfies the differential equation. x(t) 2 1.5 1 0.5 0 0.5 1 0 1 2 3 4 5 Time t Compare with water tank: Figure: Several solutions of ẋ = x M. Peet Lecture 15: 9 / 44

Continuity of a Function Customary Notions of Continuity Nonlinear Stability requires some additional Math Definitions. Definition 4 (Continuity at a Point). For normed spaces X, Y, a function f : X Y is continuous at the point x 0 X if for any ɛ > 0, there exists a δ > 0 such that x x 0 < δ (U) implies f(x) f(x 0 ) < ɛ (V ). M. Peet Lecture 15: 10 / 44

Ordinary Differential Equations Customary Notions of Continuity Definition 5 (Continuity on a Set of Points (B)). For normed spaces X, Y, a function f : A X Y is continuous on B if it is continuous at any point x 0 B. A function is simply continuous if B = A. Dropping some of the notation, Definition 6 (Uniform Continuity on a Set of Points (B)). f : A X Y is uniformly continuous on B if for any ɛ > 0, there exists a δ > 0 such that for x, y B, x y < δ implies f(x) f(y) < ɛ. A function is simply uniformly continuous if B = A. Example: f(x) = x 3 is uniformly continuous on B = [0, 1], but not B = R f (x) = 3x 2 < 3 for x [0, 1] hence f(x) f(y) 3 x y. So given δ, choose ɛ < 1 3 δ. M. Peet Lecture 15: 11 / 44

Lipschitz Continuity A Quantitative Notion of Continuity Definition 7 (Lipschitz Continuity). The function f is Lipschitz continuous on X if there exists some L > 0 such that f(x) f(y) L x y for any x, y X. The constant L is referred to as the Lipschitz constant for f on X. Definition 8 (Local Lipschitz Continuity). The function f is Locally Lipschitz continuous on X if for every x X, there exists a neighborhood, B of x such that f is Lipschitz continuous on B. Definition 9. The function f is Globally Lipschitz if it is Lipschitz on its entire domain. Example: f(x) = x 3 is Locally Lipschitz on [ 1, 1] with L = 3. But f(x) = x 3 is NOT Globally Lipschitz on R L is typically just a bound on the derivative. M. Peet Lecture 15: 12 / 44

A Theorem on Existence of Solutions Existence and Uniqueness Let B(x 0, r) be the unit ball, centered at x 0 of radius r. Theorem 10 (A Typical Existence Theorem). Suppose x 0 R n, f : R n R n and there exist L, r such that for any x, y B(x 0, r), f(x) f(y) L x y and f(x) c for x B(x 0, r). Let t f < min{ 1 L, r c }. Then there exists a unique differentiable x : [0, t f ] R n, such that x(0) = x 0, x(t) B(x 0, r) and ẋ(t) = f(x(t)). M. Peet Lecture 15: 13 / 44

Examples on Existence of Solutions Theorem 11 (A Typical Existence Theorem). Suppose x 0 R n, f : R n R n and there exist L, r such that peaking for any x, y B(x 0, r), f(x) f(y) L x y and f(x) c for x B(x 0, r). Let t f < min{ 1 L, r c }. Then quadratic there stability time-varying systems exists a invariant sets unique differentiable solution on interval [0, t f ]. center manifold theorem Recall: has the solution Nonlinear Control Theory 2006 Existence problems of solutions ẋ(t) Example: = x(t) The differential 2 equation x(0) = x 0 has the solution dx dt = x2, x(0)= x0 x(t) = x 0 x(t)= x0 1 x 0 t 1 x0t, 0 t< 1 x0 Lets take r = 1, xfinite 0 = escape 1. time Then L = sup x [0,2] f (x) = 4. t f= c = sup x [0,2] f(x) = 4. Then we 1 x0have a solution for t f < min{ 1 L, r c } = min{ 1 4, 1 4 } = 1 4 and where x(t) < 2 for t [0, t f ]. We can verify that the solution Uniqueness x(t) Problems = 1 1 t < 4 3 for t < t f. Lecture 1++, 2006 Nonlinear Phenomena and Stability theory Nonlinear phenomena [Khalil Ch 3.1] existence and uniqueness finite escape time Linear system theory revisited Second order systems [Khalil Ch 2.4, 2.6] periodic solutions / limit cycles Stability theory [Khalil Ch. 4] Lyapunov Theory revisited exponential stability x(t) 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 Finite Escape Time Finite escape time of dx/dt = x 2 0 0 1 2 3 4 5 Time t Previous problem is like the water-tank problem in backw time M. Peet Example: The equation ẋ= x, x(0)=0has many solutions: Lecture 15: (Substitute τ = t in differential equation). 14 / 44

Examples on Existence of Solutions Non-Uniqueness Theorem 12 (A Typical Existence Theorem). Suppose x 0 R n, f : R n R n and there exist L, r such that for any x, y B(x 0, r), f(x) f(y) L x y and f(x) c for x B(x 0, r). Let t f < min{ 1 L, r c }. Then there exists a unique differentiable solution on interval [0, t f ]. Recall the system without a unique solution is ẋ(t) = x(t) 1/3 x(0) = 0 The problem here is that f (x) = 1 3 x 2 3 = 1 L = 3x 2 3 sup f (x) = sup x [0,2] x [0,2] Since 1 0 =. So there is no Lipschitz Bound.. 1 3x 2 3 = M. Peet Lecture 15: 15 / 44

Typical Existence of Solutions Proof Approach Step 1: Contraction Mapping Theorem Theorem 13 (Contraction Mapping Principle). Let (X, ) be a complete normed space and let P : X X. Suppose there exists a ρ < 1 such that P x P y ρ x y for all x, y X. Then there is a unique x X such that P x = x. Furthermore for y X, define the sequence {x i } as x 1 = y and x i = P x i 1 for i > 2. Then lim i x i = x. Some Observations: Proof of CM: Show that P k y is a Cauchy sequence for any y X. For existence of solutions, X is the space of solutions. The contraction is and x := sup t [0,tf ] x(t). (P x)(t) = x 0 + t 0 f(x(s))ds, M. Peet Lecture 15: 16 / 44

Ordinary Differential Equations Contraction Mapping Theorem Key Point: P x = x if and only if x is a solution! (P x)(t) = x 0 + t 0 f(x(s))ds, Theorem 14 (Fundamental Theorem of Calculus). Suppose x and f : R n R n are continuous. Then the following are equivalent. 1. x is differentiable at any t [0, t f ] and ẋ(t) = f(x(t)) at all t [0, t f ] x(0) = x 0 2. x(t) = x 0 + t 0 f(x(s))ds for all t [0, t f ] M. Peet Lecture 15: 17 / 44

Ordinary Differential Equations Existence and Uniqueness First recall what we are trying to prove: Theorem 15 (A Typical Existence Theorem). Suppose x 0 R n, f : R n R n and there exist L, r such that for any x, y B(x 0, r), f(x) f(y) L x y and f(x) c for x B(x 0, r). Let t f < min{ 1 L, r c }. Then there exists a unique differentiable solution on interval [0, t f ]. We will show that (P x)(t) = x 0 + t 0 f(x(s))ds is a contraction if t f is sufficiently small. Then P x = x has a solution! M. Peet Lecture 15: 18 / 44

Proof of Existence Theorem Proof. For given x 0, define the complete normed space B = {x C[0, t f ] : x(0) = x 0, x(t) B(x 0, r)} with norm sup t [0,tf ] x(t) (Proof of Completeness omitted). Recall we define P as P x(t) = x 0 + t 0 f(x(s))ds Step 1: We first show that if x B, then P x B. Suppose x B. To see that P x is continuous in t, we note that P x(t 2 ) P x(t 1 ) = t2 t 1 f(x(s))ds t2 t 1 f(x(s)) ds c(t 2 t 1 ) Which is the definition of uniform continuity. Clearly P x(0) = x 0. Finally, we need to show (P x)(t) B(x 0, r). Since f(x) c when x B(x 0, r), we have sup P x(t) x 0 = sup t [0,t f ] t [0,t f ] t 0 f(x(s))ds tf where the final inequality is from the constraint t f < min{ 1 L, r c }. 0 f(x(s)) ds t f c < r M. Peet Lecture 15: 19 / 44

Proof of Existence Theorem Proof. We have shown that P : B B. Step 2: Now show that P is a contraction. For any two x, y B, we have P x P y = sup t [0,t f ] tf L 0 sup t [0,t f ] t ( 0 t 0 f(x(s)) f(y(s))ds f(x(s)) f(y(s)) ds) = tf x(s) y(s) ds Lt f x y < x y 0 f(x(s)) f(y(s)) ds Where the last inequality comes from the constraint t f < min{ 1 L, r c }. Thus P is a contraction which means there exists a solution x B such that ẋ = f(x(t)) M. Peet Lecture 15: 20 / 44

Picard Iteration Constructing the Solution This proof provides a way of actually constructing the solution. Picard-Lindelöf Iteration: From the proof, unique solution of P x = x is a solution of ẋ = f(x ), where t (P x)(t) = x 0 + f(x(s))ds From the contraction mapping theorem, the solution P x = x can be found as x = lim P k z for ANY z B k Note that this existence theorem only guarantees existence on the interval [ t 0, 1 ] [ or t 0, r ] L c Where L is a Lipschitz constant for f in the ball of radius r centered at x 0 c is a bound for f in the ball of radius r centered at x 0. Q: What does a Lyapunov function prove if there is no guaranteed solution? A: Nothing. (Particularly problematic for PDEs) 0 M. Peet Lecture 15: 21 / 44

Illustration of Picard Iteration This is a plot of Picard iterations for the solution map of ẋ = x 3. z(t, x) = 0; P z(t, x) = x; P 2 z(t, x) = x tx 3 ; P 3 z(t, x) = x tx 3 + 3t 2 x 5 3t 3 x 7 + t 4 x 9 1.5 1.0 0.5 0.2 0.4 0.6 0.8 1.0 Figure: The solution for x 0 = 1 Convergence is only guaranteed on interval t [0,.333]. M. Peet Lecture 15: 22 / 44

Extension of Existence Theorem For Lyapunov Functions to work, we need to extend the solution Indefinitely! Step 1: Prove that solutions exist for a fixed r (yielding L, c, and t f ). Step 2: Use a Lyapunov function to show that there exists a δ < r, such that any solution which begins in B(0, δ) will stay in B(0, r). Step 3: Since x(0) < δ < r, a solution exists on [0, t f ]. Further, x(t f ) r. Step 4: Since x(t f ) < r, a solution exists on [0, 2t f ]. Further, Step 2 proves x(2t f ) r. Repeat Step 4 Indefinitely For a given set W, define W r := {x : x y r, y W }. Theorem 16 (Formal Solution Extension Theorem). Suppose that there exists a domain D and L > 0 such that f(x) f(y) L x y for all x, y D R n and t > 0. Suppose there exists a compact set W and r > 0 such that W r D. Furthermore suppose that it has been proven that for x 0 W, any solution to ẋ(t) = f(x), x(0) = x 0 must lie entirely in W. Then, for x 0 W, there exists a unique solution x with x(0) = x 0 such that x lies entirely in W. M. Peet Lecture 15: 23 / 44

Illustration of Extended Picard Iteration We take the previous approximation to the solution map and extend it. x t 1.0 0.9 0.8 0.7 0.6 0.5 0.2 0.4 0.6 0.8 1.0 t Figure: The Solution map φ and the functions G k i for k = 1, 2, 3, 4, 5 and i = 1, 2, 3 for the system ẋ(t) = x(t) 3. The interval of convergence of the Picard Iteration is T = 1 3. M. Peet Lecture 15: 24 / 44

Stability Definitions Whenever you are trying to prove stability, Please define your notion of stability! Denote the set of bounded continuous functions by C := {x C : x(t) r, r 0} with norm x = sup t x(t). We define g : D C to be the solution map: g(x 0, t) if t g(x 0, t) = f(g(x 0, t)) and g(x 0, 0) = x 0 x 0 D Definition 17. The system is locally Lyapunov stable on D where D contains an open neighborhood of the origin if it defines a unique map g : D C which is continuous at the origin (x 0 = 0). The system is locally Lyapunov stable on D if for any ɛ > 0, there exists a δ(ɛ) such that for x(0) δ(ɛ), x(0) D we have x(t) ɛ for all t 0 M. Peet Lecture 15: 25 / 44

Stability Definitions Definition 18. The system is globally Lyapunov stable if it defines a unique map g : R n C which is continuous at the origin. We define the subspace of bounded continuous functions which tend to the origin by G := {x C : lim t x(t) = 0} with norm x = sup t x(t). Definition 19. The system is locally asymptotically stable on D where D contains an open neighborhood of the origin if it defines a map g : D G which is continuous at the origin. M. Peet Lecture 15: 26 / 44

Stability Definitions Definition 20. The system is globally asymptotically stable if it defines a map g : R n G which is continuous at the origin. Definition 21. The system is locally exponentially stable on D if it defines a map g : D G where g(x, t) Ke γt x for some positive constants K, γ > 0 and any x D. Definition 22. The system is globally exponentially stable if it defines a map g : R n G where g(x, t) Ke γt x for some positive constants K, γ > 0 and any x R n. M. Peet Lecture 15: 27 / 44

Lyapunov Theorem Consider the system: ẋ = f(x), f(0) = 0 Theorem 23. Let V : D R be a continuously differentiable function and D compact such that V (0) = 0 V (x) > 0 for x D, x 0 V (x) T f(x) 0 for x D. Then ẋ = f(x) is well-posed and locally Lyapunov stable on the largest sublevel set V γ = {x : V (x) γ} of V contained in D. Furthermore, if V (x) T f(x) < 0 for x D, x 0, then ẋ = f(x) is locally asymptotically stable on the largest sublevel set V γ = {x : V (x) γ} contained in D. M. Peet Lecture 15: 28 / 44

Lyapunov Theorem Sublevel Set: For a given Lyapunov function V and positive constant γ, we denote the set V γ = {x : V (x) γ}. Proof. Existence: Denote the largest bounded sublevel set of V contained in the interior of D by V γ. Because V (x(t)) = V (x(t)) T f(x(t)) 0, if x(0) V γ, then x(t) V γ for all t 0. Therefore since f is locally Lipschitz continuous on the compact V γ, by the extension theorem, there is a unique solution for any initial condition x(0) V γ. Lyapunov Stability: Given any ɛ > 0, choose ɛ < ɛ with B(ɛ) V γ, choose γ i such that V γi B(ɛ). Now, choose δ > 0 such that B(δ) V γi. Then B(δ) V γi B(ɛ) and hence if x(0) B(δ), we have x(0) V γi B(ɛ) B(ɛ ). Asymptotic Stability: V monotone decreasing implies lim t V (x(t)) = 0. V (x) = 0 implies x = 0. Proof omitted. M. Peet Lecture 15: 29 / 44

Lyapunov Theorem Exponential Stability Theorem 24. Suppose there exists a continuously differentiable function V and constants c 1, c 2, c 3 > and radius r > 0 such that the following holds for all x B(r). c 1 x p V (x) c 2 x p V (x) T f(x) c 3 x p Then ẋ = f(x) is exponentially stable on any ball contained in the largest sublevel set contained in B(r). Exponential Stability allows a quantitative prediction of system behavior. M. Peet Lecture 15: 30 / 44

The Gronwall-Bellman Inequality Exponential Stability Lemma 25 (Gronwall-Bellman). Let λ be continuous and µ be continuous and nonnegative. Let y be continuous and satisfy for t b, Then y(t) λ(t) + If λ and µ are constants, then y(t) λ(t) + t a t a µ(s)y(s)ds. [ t ] λ(s)µ(s) exp µ(τ)dτ ds s y(t) λe µt. For λ(t) = y 0, the condition is equivalent to ẏ(t) µ(t)y(t), y(0) = y(t). M. Peet Lecture 15: 31 / 44

Lyapunov Theorem Exponential Stability Proof. We begin by noting that we already satisfy the conditions for existence, uniqueness and asymptotic stability and that x(t) B(r). Now, observe that V (x(t)) c 3 x(t) 2 c 3 c 2 V (x(t)) Which implies by the Gronwall-Bellman inequality (µ = c3 c 2, λ = V (x(0))) that V (x(t)) V (x(0))e c 3 t c2. Hence x(t) 2 1 V (x(t)) 1 e c 3 t c2 V (x(0)) c 2 e c t 2 x(0) 2. c 1 c 1 c 1 c 3 M. Peet Lecture 15: 32 / 44

Lyapunov Theorem Invariance Sometimes, we want to prove convergence to a set. Recall Definition 26. V γ = {x, V (x) γ} A set, X, is Positively Invariant if x(0) X implies x(t) X for all t 0. Theorem 27. Suppose that there exists some continuously differentiable function V such that V (x) > 0 for x D, x 0 V (x) T f(x) 0 for x D. for all x D. Then for any γ such that the level set X = {x : V (x) = γ} D, we have that V γ is positively invariant. Furthermore, if V (x) T f(x) 0 for x D, then for any δ such that X V δ D, we have that any trajectory starting in V δ will approach the sublevel set V γ. M. Peet Lecture 15: 33 / 44

Problem Statement 1: Global Lypunov Stability Given: Vector field, f(x) Find: function V, scalars α, β such that i α i =.01, i β i =.01 and V (x) i α i (x T x) i for all x V (x) i β i (x T x) i for all x V (x) T f(x) 0 for all x Conclusion: Lyapunov stability for any x(0) R n. Can replace V (x) i β i(x T x) i with V (0) = 0 if it is well-behaved. M. Peet Lecture 15: 34 / 44

Examples of Lyapunov Functions Mass-Spring: Pendulum: ẍ = c mẋ k m x ẋ 2 = g l sin x 1 ẋ 1 = x 2 V (x) = 1 2 mẋ2 + 1 2 kx2 V (x) = (1 cos x 1 )gl + 1 2 l2 x 2 2 V (x) = ẋ( cẋ kx) + kxẋ = cẋ 2 kẋx + kxẋ V (x) = glx 2 sin x 1 glx 2 sin x 1 = 0 = cẋ 2 0 M. Peet Lecture 15: 35 / 44

Problem Statement 2: Global Exponential Stability Given: Vector field, f(x), exponent, p Find: function V, scalars α, β, δ such that i α i =.01, i β i =.01 and V (x) i V (x) i α i x 2p i β i x 2p i for all x for all x V (x) T f(x) δv (x) for all x Conclusion: Exponential stability for any x(0) R n. Convergence Rate: x(t) 2p βmax α min x(0) e δ 2p t M. Peet Lecture 15: 36 / 44

Problem Statement 2: Global Exponential Stability Example Consider: Attitude Dynamics of a rotating Spacecraft: What about: V (x) T f(x) = ω 1 ω 2 ω 3 J 1 ω 1 = (J 2 J 3 )ω 2 ω 3 J 2 ω 2 = (J 3 J 1 )ω 3 ω 1 J 3 ω 3 = (J 1 J 2 )ω 1 ω 2 V (x) = ω 2 1 + ω 2 2 + ω 2 3? T J 2 J 3 J 1 ω 2 ω 3 J 3 J 1 J 2 ω 3 ω 1 J 1 J 2 J 3 ω 1 ω 2 OK, maybe not. Try u i = k i ω i. ( J2 J 3 = + J 3 J 1 + J ) 1 J 2 ω 1 ω 2 ω 3 J 1 J 2 J 3 ( J 2 = 2 J 3 J3 2 J 2 + J3 2 J 1 J1 2 J 3 + J 2 J1 2 J2 2 J 1 J 1 J 2 J 3 ) ω 1 ω 2 ω 3 M. Peet Lecture 15: 37 / 44

Problem Statement 3: Local Exponential Stability Given: Vector field, f(x), exponent, p Ball of radius r, Find: function V, scalars α, β, δ such that i α i =.01, i β i =.01 and V (x) i V (x) i α i x 2p i for all x : x T x r 2 β i x 2p i for all x : x T x r 2 V (x) T f(x) δv (x) for all x : x T x r 2 Conclusion: A Domain of Attraction! of the origin Exponential stability for x(0) {x : V (x) γ} if V γ B r. Sub-Problem: Given, V, r, max γ γ such that V (x) γ for all x {x T x r} M. Peet Lecture 15: 38 / 44

Domain of Attraction The van der Pol Oscillator An oscillating circuit: (in reverse time) ẋ = y Choose: Derivative V (x) T f(x) = ẏ = x + (x 2 1)y V (x) = x 2 + y 2, r = 1 = xy + xy + (x 2 1)y 2 0 for x 2 1 [ ] T [ ] 2x y 2y x (x 2 1)y Level Set: V γ=1 = {(x, y) : x 2 + y 2 1} = B 1 y 3 2 1 0 1 2 Domain of attraction 3 3 2 1 0 1 2 3 x Figure: The van der Pol oscillator in reverse So B 1 = V γ=1 is a Domain of Attraction! M. Peet Lecture 15: 39 / 44

Problem Statement 4: Invariant Regions/Manifolds Given: Vector field, f(x), exponent, p Ball of radius r, Find: function V, scalars α, β, δ such that i α i =.01, i β i =.01 and V (x) i V (x) i α i x 2p i for all x : x T x r 2 β i x 2p i for all x : x T x r 2 V (x) T f(x) δv (x) for all x : x T x r 2 Conclusion: There exist a T such that x(t) {x : V (x) γ} for all t T if B r V γ. Sub-Problem: Given, V, r, min γ γ such that x T x r 2 for all x {V (x) γ} M. Peet Lecture 15: 40 / 44

Nonlinear Dynamical Systems Long-Range Weather Forecasting and the Lorentz Attractor A model of atmospheric convection analyzed by E.N. Lorenz, Journal of Atmospheric Sciences, 1963. ẋ = σ(y x) ẏ = rx y xz ż = xy bz M. Peet Lecture 15: 41 / 44

Problem Statement 5: Controller Synthesis (Local) Suppose ẋ(t) = f(x) + g(x)u(t) Given: Vector fields, f(x), g(x), exponent, p u(t) = k(x) Find: function functions, k, V, scalars α 0, β 0 such that i α i =.01, i β i =.01 and V (x) i V (x) i α i x 2p i for all x : x T x r 2 β i x 2p i for all x : x T x r 2 V (x) T f(x) + V (x) T g(x)k(x) 0 for all x : x T x r 2 Conclusion: Controller u(u) = k(x(t)) stabilizes the system for x(0) {x : V (x) γ} if V γ B r. M. Peet Lecture 15: 42 / 44

Problem Statement 6: Output Feedback Controller Synthesis (Global Exponential ) Suppose ẋ(t) = f(x) + g(x)u(t) y(t) = h(x(t)) Given: Vector fields, f(x), g(x), h(x) exponent, p u(t) = k(y(t)) Find: function functions, k, V, scalars α 0, β 0, δ > 0 such that i α i =.01, i β i =.01 and V (x) α i x 2p i for all x i V (x) i β i x 2p i for all x V (x) T f(x) + V (x) T g(x)k(h(x)) δv (x) for all x Conclusion: Controller u(t) = k(y(t)) exponentially stabilizes the system for any x(0) R n. M. Peet Lecture 15: 43 / 44

How to Solve these Problems? General Framework for solving these problems Convex Optimization of Functions: Variables V C[R n ] and γ R max V,γ γ subject to V (x) x T x 0 x X V (x) T f(x) + γx T x 0 x X Going Forward Assume all functions are polynomials or rationals. Assume X := {x : g i (x) 0} (Semialgebraic) M. Peet Lecture 15: 44 / 44