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Nonlinear Systems Theory Nonlinear Systems Theory Matthew M. Peet Arizona State University Lecture 02: Nonlinear Systems Theory 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 02: 1 / 56 Ordinary Nonlinear Differential Equations Computing Stability and Domain of Attraction Consider: A System of Nonlinear Ordinary Differential Equations x_(t) = f(x(t)) Problem: Stability Given a specific polynomial f : Rn ! Rn, find the largest X ⊂ Rn such that for any x(0) 2 X, limt!1 x(t) = 0. M. Peet Lecture 02: 2 / 56 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. x_ = σ(y − x)_y = rx − y − xz z_ = xy − bz M. Peet Lecture 02: 3 / 56 Stability and Periodic Orbits The Poincar´e-BendixsonTheorem and van der Pol Oscillator 3 2 An oscillating circuit model: 1 2 y_ = −x − (x − 1)y y 0 x_ = y −1 −2 Domain−of−attraction −3 −3 −2 −1 0 1 2 3 x Figure : The van der Pol oscillator in reverse Theorem 1 (Poincar´e-Bendixson). Invariant sets in R2 always contain a limit cycle or fixed point. M. Peet Lecture 02: 4 / 56 Stability of Ordinary Differential Equations Consider x_(t) = f(x(t)) n with x(0) 2 R . Theorem 2 (Lyapunov Stability). Suppose there exists a continuous V and α; β; γ > 0 where βkxk2 ≤ V (x) ≤ αkxk2 −∇V (x)T f(x) ≥ γkxk2 for all x 2 X. Then any sub-level set of V in X is a Domain of Attraction. M. Peet Lecture 02: 5 / 56 Mathematical Preliminaries Cauchy Problem The first question people ask is the Cauchy problem: Autonomous System: Definition 3. The system x_(t) = f(x(t)) is said to satisfy the Cauchy problem if there exists a n continuous function x : [0; tf ] ! R such that x_ is defined and x_(t) = f(x(t)) for all t 2 [0; tf ] If f is continuous, the solution must be continuously differentiable. Controlled Systems: • For a controlled system, we have x_(t) = f(x(t); u(t)). • At this point u is undefined, so for the Cauchy problem, we take x_(t) = f(t; x(t)) • In this lecture, we consider the autonomous system. I Including t complicates the analysis. I However, results are almost all the same. M. Peet Lecture 02: 6 / 56 Lecture 1++, 2006 • Nonlinear Phenomena and Stability theory ◮ Nonlinear phenomena [Khalil Ch 3.1] ◮ existence and uniqueness Nonlinear Control Theory 2006 ◮ finite escape time ◮ peaking ◮ 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 Ordinary Differential Equations◮ quadratic stability ◮ time-varying systems Existence of Solutions ◮ invariant sets ◮ center manifold theorem There exist many systems for which no solution exists or for which a solution Existence problemsonly of exists solutions over a finite time interval. Finite Escape Time Even for something as simple as 2 Finite escape time of dx/dt = x Example: The differential equation 5 2 x_(t) = x(t) x(0) = x 4.5 dx 2 0 x , x 0 x0 dt = ( )= 4 has the solution 3.5 has the solution 3 x0 x0 1 x(t) = 2.5 x t , 0 t x(t) ( )= 1 x t ≤ < x 1 − x0t − 0 0 2 1.5 Finite escape time which clearly has escape time 1 1 t f = 1 0.5 x0 t = e 0 x0 0 1 2 3 4 5 Time t Figure : Simulation of x_ = x2 for several x(0) Uniqueness Problems Previous problem is like the water-tank problem in backward M. Peet time Lecture 02: 7 / 56 Example: The equation x˙ √x, x 0 0 has many solutions: = ( )= (Substitute τ t in differential equation). =− t C 2 4 t C x t ( − ) / > ( ) = 0 t C ≤ 2 1.5 1 0.5 x(t) 0 −0.5 −1 0 1 2 3 4 5 Time t dh dt a√h, h : height water level / =− ( ) Compare with water tank: Change to backward-time: “If I see it empty, when was it full?”) Existence and Uniqueness The peaking phenomenon Theorem Let Ω R denote the ball Example: Controlled linear system with right-half plane zero Feedback can change location of poles but not location of zero Ω z; z a R R = { q − q≤ } (unstable pole-zero cancellation not allowed). If f is Lipschitz-continuous: s 1 ω 2 o (1) Gcl s 2(− + ) 2 f z f y K z y , for all z, y Ω ( )= s 2ω os ω o q ( )− ( )q ≤ q − q ∈ + + then x˙ t f x t , x 0 a has a unique solution in ( )= ( ( )) ( )= 0 t R C , A step response will reveal a transient which grows in amplitude ≤ < / R for faster closed loop poles s ω o, see Figure on next slide. where C max f x =− R = ΩR q ( )q see [Khalil Ch. 3] 1 Ordinary Differential Equations Non-Uniqueness A classical example of a system without a unique solution is x_(t) = x(t)1=3 x(0) = 0 For the given initial condition, it is easy to verify that 2t 3=2 x(t) = 0 and x(t) = 3 both satisfy the differential equation. 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 Figure : Matlab simulation of 1=3 1=3 x_(t) = x(t) with x(0) = 0 x_(t) = x(t) with x(0) = :000001 M. Peet Lecture 02: 8 / 56 Lecture 1++, 2006 • Nonlinear Phenomena and Stability theory ◮ Nonlinear phenomena [Khalil Ch 3.1] ◮ existence and uniqueness Nonlinear Control Theory 2006 ◮ finite escape time ◮ peaking ◮ 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 ◮ quadratic stability ◮ time-varying systems ◮ invariant sets ◮ center manifold theorem Existence problems of solutions Finite Escape Time Finite escape time of dx/dt = x2 Example: The differential equation 5 4.5 dx 2 x , x 0 x0 dt = ( )= 4 3.5 has the solution 3 x0 1 2.5 x t , 0 t x(t) ( )= 1 x t ≤ < x − 0 0 2 1.5 Finite escape time 1 1 t f = x0 0.5 0 0 1 2 3 4 5 Time t Ordinary Differential Equations Uniqueness Problems Previous problem is like the water-tank problem in backward Non-Uniqueness time Example: The equation x˙ √x, x 0 0 has many solutions: = ( )= (Substitute τ t in differential equation). Another Example of a system with several solutions is given by =− t C 2 4 t C x_(t) = x(t) x t x(0) = 0 ( − ) / > ( ) = 0 t C p ≤ For the given initial condition, it is easy 2 to verify that for any C, 1.5 (t−C)2 t > C 1 x(t) = 4 0.5 (0 t ≤ C x(t) 0 satisfies the differential equation. −0.5 −1 0 1 2 3 4 5 Time t √ p dh dt a h, h : height water level Figure : Several solutions of x_ = x / =− ( ) Compare with water tank: Change to backward-time: “If I see it empty, when was it full?”) M. Peet Lecture 02: 9 / 56 Existence and Uniqueness The peaking phenomenon Theorem Let Ω R denote the ball Example: Controlled linear system with right-half plane zero Feedback can change location of poles but not location of zero Ω z; z a R R = { q − q≤ } (unstable pole-zero cancellation not allowed). If f is Lipschitz-continuous: s 1 ω 2 o (1) Gcl s 2(− + ) 2 f z f y K z y , for all z, y Ω ( )= s 2ω os ω o q ( )− ( )q ≤ q − q ∈ + + then x˙ t f x t , x 0 a has a unique solution in ( )= ( ( )) ( )= 0 t R C , A step response will reveal a transient which grows in amplitude ≤ < / R for faster closed loop poles s ω o, see Figure on next slide. where C max f x =− R = ΩR q ( )q see [Khalil Ch. 3] 1 Ordinary Differential Equations Customary Notions of Continuity Definition 4. For normed linear spaces X; Y , a function f : X ! Y is said to be continuous at the point x0 ⊂ X if for any > 0, there exists a δ > 0 such that kx − x0k < δ implies kf(x) − f(x0)k < . M. Peet Lecture 02: 10 / 56 Ordinary Differential Equations Customary Notions of Continuity Definition 5. For normed linear spaces X; Y , a function f : A ⊂ X ! Y is said to be continuous on B ⊂ A if it is continuous for any point x0 2 B. A function is said to be simply continuous if B = A. Definition 6. For normed linear spaces X; Y , a function f : A ⊂ X ! Y is said to be uniformly continuous on B ⊂ A if for any > 0, there exists a δ > 0 such that for any points x; y 2 B, kx − yk < δ implies kf(x) − f(y)k < .A function is said to be simply uniformly continuous if B = A.
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