[足式机器人]Part4 南科大高等机器人控制课 CH10 Bascis of Stability Analysis

发布时间:2023年12月23日

本文仅供学习使用
本文参考:
B站:CLEAR_LAB
笔者带更新-运动学
课程主讲教师:
Prof. Wei Zhang


This lecture introduces basic concepts and results on Lyapunov stability of nonlinear systems

1. Background

1.1 What is Stability Analysis

  • system asymptotic/??simp't?tik,-k?l/ 渐进的 behavior (not too much about transient/'tr?nz??nt/短暂的)
  • ability to return to the desired asymptotic behavior (nout just convergence)

示例-小球平衡系统与反馈控制

1.2 General ODE Models for Dynamical Systems

  • ODE: x ˙ = f ( x , u ) \dot{x}=f\left( x,u \right) x˙=f(x,u) , with x ( 0 ) = x 0 x\left( 0 \right) =x_0 x(0)=x0?
    x ∈ X ? R n x\in \mathcal{X} \subseteq \mathbb{R} ^n xX?Rn : state
    u ∈ U ? R m u\in \mathcal{U} \subseteq \mathbb{R} ^m uU?Rm : control input
    f : R n × R m → R n f:\mathbb{R} ^n\times \mathbb{R} ^m\rightarrow \mathbb{R} ^n f:Rn×RmRn : (time-invariant) vector field
  • System output y = g ( x , u ) y=g\left( x,u \right) y=g(x,u)
  • (static)Control law : μ : X → U \mu :\mathcal{X} \rightarrow \mathcal{U} μ:XU , u = μ ( x ) u=\mu \left( x \right) u=μ(x)
    在这里插入图片描述
  • Closed-loop dynamics under μ \mu μ : x ˙ = f ( x , μ ( x ) ) \dot{x}=f\left( x,\mu \left( x \right) \right) x˙=f(x,μ(x)) ? x ˙ = f c l ( x ) \Rightarrow \dot{x}=f_{\mathrm{cl}}\left( x \right) ?x˙=fcl?(x)
  • Autonomous system :
    x ˙ = f ( x , u ) \dot{x}=f\left( x,u \right) x˙=f(x,u) , with x ( 0 ) = x 0 x\left( 0 \right) =x_0 x(0)=x0?

1.3 Example

1.3.1 Pendulum

在这里插入图片描述

1.3.2 Adaptive Control

Closed-loop dynamics under adaptive control:
{ y ˙ = y + u u = ? k y , k ˙ = y 2 \begin{cases} \dot{y}=y+u\\ u=-ky,\dot{k}=y^2\\ \end{cases} {y˙?=y+uu=?ky,k˙=y2?
在这里插入图片描述

1.4 Equilibrium Point of Dynamical Systems

Definition 1 (Equilibrium Point) - 平衡点
A state x ? x^* x? is an equilibrium point of system (1) if once x ( t ) = x ? x\left( t \right) =x^* x(t)=x? , it remains equal to x ? x^* x? at all future time.

  • Mathematically : f ( x ? ) = 0 f\left( x^* \right) =0 f(x?)=0
  • e.g. undamped pendulum with no driving force: f ( x ) = x ˙ f\left( x \right) =\dot{x} f(x)=x˙ velocity
    x ˙ = [ x 2 g l cos ? x 1 ] = 0 ? { x 2 = 0 cos ? x 1 = 0 , x 1 = 2 k π + π 2 , k ∈ Z \dot{x}=\left[ \begin{array}{c} x_2\\ \frac{g}{l}\cos x_1\\ \end{array} \right] =0\Rightarrow \begin{cases} x_2=0\\ \cos x_1=0,x_1=\frac{2k\pi +\pi}{2},k\in \mathbb{Z}\\ \end{cases} x˙=[x2?lg?cosx1??]=0?{x2?=0cosx1?=0,x1?=22+π?,kZ?

1.5 Invariant Set of Dynamical Systems

Definition 2 (Invariant Set) - 不变集
A set E E E is an invariant set of system (1) if every trajectory which starts from a point E E E remains in E E E at all future time.

  • Mathematically : If x ( t 0 ) ∈ E x\left( t_0 \right) \in E x(t0?)E , then x ( t ) ∈ E , ? t ? t 0 x\left( t \right) \in E,\forall t\geqslant t_0 x(t)E,?t?t0?
  • e.g. Closed-loop dynamics under adaptive control :
    在这里插入图片描述
    f ( x ) = x ˙ = [ x 1 ? x 1 x 2 x 1 2 ] ? { x 1 = 0 x 2 = a r b i t r a r y ? E = { x ∈ R 2 , x 1 = 0 } f\left( x \right) =\dot{x}=\left[ \begin{array}{c} x_1-x_1x_2\\ {x_1}^2\\ \end{array} \right] \Rightarrow \begin{cases} x_1=0\\ x_2=arbitrary\\ \end{cases}\Rightarrow E=\left\{ x\in \mathbb{R} ^2,x_1=0 \right\} f(x)=x˙=[x1??x1?x2?x1?2?]?{x1?=0x2?=arbitrary??E={xR2,x1?=0}

2. Lyapunov Stability Definitions

Stability :

  1. about equilibrium
  2. ability to stay close or return to equilibrium

2.1 Lyapunov Stability Definitions

Consider a time-invariant autonomous (with no control) nonlinear system : (on closed-loop system : x ˙ = f ( x , u ( x ) ) = f c l ( x ) \dot{x}=f\left( x,u\left( x \right) \right) =f_{\mathrm{cl}}\left( x \right) x˙=f(x,u(x))=fcl?(x))
x ˙ = f ( x ) , x ∈ R n \dot{x}=f\left( x \right) ,x\in \mathbb{R} ^n x˙=f(x),xRn , with I.C. x ( 0 ) = x 0 x\left( 0 \right) =x_0 x(0)=x0?
f ( x ) f\left( x \right) f(x) - vector field

  • Assumption :
    (i) f f f Lipshitz continuous —— existence & uniqueness of ODE
    (ii) origin is an isolated equilibrium f ( 0 ) = 0 f\left( 0 \right) =0 f(0)=0 —— f ( x ? ) = 0 f\left( x^* \right) =0 f(x?)=0
    If equilibrium x ? x^* x? is not at the origin define x ~ = x ? x ? \tilde{x}=x-x^* x~=x?x? , x ~ ˙ = x ˙ ? 0 = f ( x ~ + x ? ) \dot{\tilde{x}}=\dot{x}-0=f\left( \tilde{x}+x^* \right) x~˙=x˙?0=f(x~+x?)
  • Stability Definitions :
  1. The equilibrium x = 0 x=0 x=0 is called stable(stay close to equilibrium) in the sense of Lyapunov , if
    ? ? δ \epsilon -\delta ??δ argument —— ? ? > 0 , ? δ > 0 , s . t . ∥ x ( 0 ) ∥ ? δ ? ∥ x ( t ) ∥ ? ? , ? t ? 0 \forall \epsilon >0,\exists \delta >0,s.t.\left\| x\left( 0 \right) \right\| \leqslant \delta \Rightarrow \left\| x\left( t \right) \right\| \leqslant \epsilon ,\forall t\geqslant 0 ??>0,?δ>0,s.t.x(0)?δ?x(t)??,?t?0
    在这里插入图片描述
    Objective: For any ? > 0 \epsilon >0 ?>0 , ensure ∥ x ( t ) ∥ ? ? \left\| x\left( t \right) \right\| \leqslant \epsilon x(t)?? for all t t t
    our choice : selecting initial state x ( 0 ) x\left( 0 \right) x(0)
    stability : objective can be ensure by choosing I.C. sufficiently small

  2. asymptotically stable (stay close + convergence) if it is stable and δ \delta δ can be chosen so that
    ∥ x ( 0 ) ∥ ? δ ? ∥ x ( t ) ∥ → 0 \left\| x\left( 0 \right) \right\| \leqslant \delta \Rightarrow \left\| x\left( t \right) \right\| \rightarrow 0 x(0)?δ?x(t)0 as t → ∞ t\rightarrow \infty t (convergence)

  3. exponentially stable if there exist positive constants δ , λ , c \delta ,\lambda ,c δ,λ,c such that
    ∥ x ( t ) ∥ ? c ∥ x ( 0 ) ∥ e ? λ t , ? ∥ x ( 0 ) ∥ ? δ \left\| x\left( t \right) \right\| \leqslant c\left\| x\left( 0 \right) \right\| e^{-\lambda t},\forall \left\| x\left( 0 \right) \right\| \leqslant \delta x(t)?cx(0)e?λt,?x(0)?δ
    在这里插入图片描述

  4. globallt asymptomtotically / exponentially stable (G.A.S / G.E.S) if the above conditions holds for all δ > 0 \delta >0 δ>0

  5. Region of Attraction - 吸引域 : R A ? { x ∈ R n : w h e v e r ?? x ( 0 ) = x , t h e n ?? x ( t ) → 0 } R_A\triangleq \left\{ x\in \mathbb{R} ^n:whever\,\,x\left( 0 \right) =x,then\,\,x\left( t \right) \rightarrow 0 \right\} RA??{xRn:wheverx(0)=x,thenx(t)0}
    Globaly asymptotically stable R A ? R n R_A\triangleq \mathbb{R} ^n RA??Rn

2.2 Stability Examples using 2D Phase Portrait

  • Undamped pendulum with no driving force :
    x ˙ = [ x 2 g l cos ? x 1 ] = 0 ? { x 2 = 0 cos ? x 1 = 0 , x 1 = 2 k π + π 2 , k ∈ Z \dot{x}=\left[ \begin{array}{c} x_2\\ \frac{g}{l}\cos x_1\\ \end{array} \right] =0\Rightarrow \begin{cases} x_2=0\\ \cos x_1=0,x_1=\frac{2k\pi +\pi}{2},k\in \mathbb{Z}\\ \end{cases} x˙=[x2?lg?cosx1??]=0?{x2?=0cosx1?=0,x1?=22+π?,kZ?
    在这里插入图片描述
  • Closed-loop dynamics under adaptive control :
    f ( x ) = x ˙ = [ x 1 ? x 1 x 2 x 1 2 ] ? { x 1 = 0 x 2 = a r b i t r a r y ? E = { x ∈ R 2 , x 1 = 0 } f\left( x \right) =\dot{x}=\left[ \begin{array}{c} x_1-x_1x_2\\ {x_1}^2\\ \end{array} \right] \Rightarrow \begin{cases} x_1=0\\ x_2=arbitrary\\ \end{cases}\Rightarrow E=\left\{ x\in \mathbb{R} ^2,x_1=0 \right\} f(x)=x˙=[x1??x1?x2?x1?2?]?{x1?=0x2?=arbitrary??E={xR2,x1?=0}
    在这里插入图片描述
  • Does attractiveness implies stable in Lyapunov sense?
    Answer is No —— e.g. { x ˙ 1 = x 1 2 ? x 2 2 x ˙ 2 = 2 x 1 x 2 \begin{cases} \dot{x}_1={x_1}^2-{x_2}^2\\ \dot{x}_2=2x_1x_2\\ \end{cases} {x˙1?=x1?2?x2?2x˙2?=2x1?x2?? —— asymptotically stable : 1. stable 2. convergence
    By inspection of its vector field, we see that x ( t ) → 0 x\left( t \right) \rightarrow 0 x(t)0 for all x ( 0 ) ∈ R 2 x\left( 0 \right) \in \mathbb{R} ^2 x(0)R2
    However, there is no δ \delta δ-ball satisfying the Lyapunov stability condition
    在这里插入图片描述
    convergence but not stable

3. Lyapunov Stability Theorem

3.1 How to verify stability of a system?

  • Find explicit solution of the ODE x ( t ) x\left( t \right) x(t) and check stability definitions (typically not possible for nonlinear systems) —— e.g. x ( t ) = e ? t x 0 x\left( t \right) =e^{-t}x_0 x(t)=e?tx0?
  • Numerical simulations of ODE do not provide stability guarantees and offer limited insights
  • Need to determine stability without explicitly solving the ODE
  • Perferably, analysis only depends on the vector field
  • The most powerful tool is : Lyapunov function
  • State trajectory x ( t ) x\left( t \right) x(t) governed by complex dynamics in R n \mathbb{R} ^n Rn —— x ˙ ( t ) = f ( x ( t ) ) \dot{x}\left( t \right) =f\left( x\left( t \right) \right) x˙(t)=f(x(t))
  • Lyapunov function V : R n → R V:\mathbb{R} ^n\rightarrow \mathbb{R} V:RnR maps x ( t ) x\left( t \right) x(t) to a scalar function of time V ( x ( t ) ) V\left( x\left( t \right) \right) V(x(t))
    scalar - V ( x ( t ) ) ? V ˙ ( x ( t ) ) = d d t V ( x ( t ) ) = g ( V ( ) ) V\left( x\left( t \right) \right) \leftrightarrow \dot{V}\left( x\left( t \right) \right) =\frac{\mathrm{d}}{\mathrm{d}t}V\left( x\left( t \right) \right) =g\left( V\left( \right) \right) V(x(t))?V˙(x(t))=dtd?V(x(t))=g(V()) - scalar ODE
  • If the function is designed such that : [ x ( t ) → e q u i l i b r i u m ] ? [ V ( x ( t ) ) → 0 ] \left[ x\left( t \right) \rightarrow equilibrium \right] \Leftrightarrow \left[ V\left( x\left( t \right) \right) \rightarrow 0 \right] [x(t)equilibrium]?[V(x(t))0]. Then we can study V ( x ( t ) ) V\left( x\left( t \right) \right) V(x(t)) as function of time t t t to infer stability of the state trajectory in R n \mathbb{R} ^n Rn

3.2 Sign Definite Functions

Assume that 0 ∈ D ? R n 0\in D\subseteq \mathbb{R} ^n 0D?Rn

  • g : D → R g:D\rightarrow \mathbb{R} g:DR is called positive semidefinite (PSD) on D D D if g ( 0 ) = 0 g\left( 0 \right) =0 g(0)=0 and g ( 0 ) ? 0 , ? x ∈ D g\left( 0 \right) \geqslant 0,\forall x\in D g(0)?0,?xD
    For quadratic function : g ( x ) = x T P x : [ g ?? i s ?? P S D ] ? [ P ?? i s ?? a ?? P S D ?? m a t r i x ] g\left( x \right) =x^{\mathrm{T}}Px:\left[ g\,\,is\,\,PSD \right] \Leftrightarrow \left[ P\,\,is\,\,a\,\,PSD\,\,matrix \right] g(x)=xTPx:[gisPSD]?[PisaPSDmatrix]
    e.g. : g ( x ) = [ x 1 x 2 ] T P [ x 1 x 2 ] = x 1 2 + x 1 x 2 + 3 x 2 2 g\left( x \right) =\left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] ^{\mathrm{T}}P\left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] ={x_1}^2+x_1x_2+3{x_2}^2 g(x)=[x1?x2??]TP[x1?x2??]=x1?2+x1?x2?+3x2?2
  • g : D → R g:D\rightarrow \mathbb{R} g:DR is called positive definite (PD) on D D D if g ( 0 ) = 0 g\left( 0 \right) =0 g(0)=0 and g ( x ) > 0 , ? x ∈ D \ { 0 } g\left( x \right) >0,\forall x\in D\backslash\{0\} g(x)>0,?xD\{0}
    Similarly, if g ( x ) = x T P x g\left( x \right) =x^{\mathrm{T}}Px g(x)=xTPx is quadratic, then [ g ?? i s ?? P D ] ? [ P ?? i s ?? a ?? P D ?? m a t r i x ] \left[ g\,\,is\,\,PD \right] \Leftrightarrow \left[ P\,\,is\,\,a\,\,PD\,\,matrix \right] [gisPD]?[PisaPDmatrix]
  • g g g is negative semidefinite (NSD) if ? g -g ?g is PSD
  • g : R n → R g:\mathbb{R} ^n\rightarrow \mathbb{R} g:RnR is radically unbounded if g ( x ) → ∞ g\left( x \right) \rightarrow \infty g(x) as ∥ x ∥ → ∞ \left\| x \right\| \rightarrow \infty x
    在这里插入图片描述

3.3 Lyapunov Stability Theorem

[Lyapunov Theorem] : Let D ? R n D\subseteq \mathbb{R} ^n D?Rn be a set containing an open neighborhood of the origin. If there exists a C 1 \mathcal{C} ^1 C1 (continuous differentiable) function V : D → R V:D\rightarrow \mathbb{R} V:DR (observable condition - e.g. V ( x ) = x 1 2 , x = [ x 1 x 2 ] = [ 0 100 ] ?? , V ( x ) = 0 i s ?? P S D V\left( x \right) ={x_1}^2,x=\left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] =\left[ \begin{array}{c} 0\\ 100\\ \end{array} \right] \,\,,V\left( x \right) =0 is\,\,PSD V(x)=x1?2,x=[x1?x2??]=[0100?],V(x)=0isPSD) such that
{ V ?? i s ?? P D V ˙ ( x ) ? ? V ( x ) T f ( x ) ?? i s ?? N S D \begin{cases} V\,\,is\,\,PD\\ \dot{V}\left( x \right) \triangleq \nabla V\left( x \right) ^{\mathrm{T}}f\left( x \right) \,\,is\,\,NSD\\ \end{cases} {VisPDV˙(x)??V(x)Tf(x)isNSD?
the value of V V V along sys state trajectory nonincreasing V ˙ ( x ( t ) ) = ( ? V ? x ) T ? x ? t = ? V ( x ) T f ( x ) , ? V ( x ) [ ? V ? x 1 ? V ? x 2 ? ? V ? x n ] \dot{V}\left( x\left( t \right) \right) =\left( \frac{\partial V}{\partial x} \right) ^{\mathrm{T}}\frac{\partial x}{\partial t}=\nabla V\left( x \right) ^{\mathrm{T}}f\left( x \right) ,\nabla V\left( x \right) \left[ \begin{array}{c} \frac{\partial V}{\partial x_1}\\ \frac{\partial V}{\partial x_2}\\ \vdots\\ \frac{\partial V}{\partial x_{\mathrm{n}}}\\ \end{array} \right] V˙(x(t))=(?x?V?)T?t?x?=?V(x)Tf(x),?V(x) ??x1??V??x2??V???xn??V?? ? , ? V ( x ) T f ( x ) ? L f [ V ] \nabla V\left( x \right) ^{\mathrm{T}}f\left( x \right) \triangleq Lf\left[ V \right] ?V(x)Tf(x)?Lf[V] Lie derivative of V ( ? ) V\left( \cdot \right) V(?) with vetor field f f f

then the origin is stable. If in addition ,
V ˙ ( x ) ? ? V ( x ) T f ( x ) ?? i s ?? N D \dot{V}\left( x \right) \triangleq \nabla V\left( x \right) ^{\mathrm{T}}f\left( x \right) \,\,is\,\,ND V˙(x)??V(x)Tf(x)isND
then the origin is asymptotically stable —— Value of V V V along sys state trajectory is decreasing

Remarks:
A PD C 1 \mathcal{C} ^1 C1 function satisfying above equation will be called a Lyapunov function (1+2 or 1+3)
Under condition 3 , if V V V is also radially unbounded —— globally asympotically stable (G.A.S)

3.4 Proof of Lyapunov Stability Theorem

Main idea : 1+2 —— stability

  • Fact : suppose V V V function satisfies 1+2 , then the sub level set Ω b ( V ) ? { x ∈ R n : V ( x ) ? b } \varOmega _{\mathrm{b}}\left( V \right) \triangleq \left\{ x\in \mathbb{R} ^n:V\left( x \right) \leqslant b \right\} Ωb?(V)?{xRn:V(x)?b} is forward invariant
    Proof Fact : if x ( 0 ) ∈ Ω b x\left( 0 \right) \in \varOmega _{\mathrm{b}} x(0)Ωb? fro some b ? 0 b\geqslant 0 b?0 , we have V ( x ( t ) ) ? V ( x ( 0 ) ) ? b V\left( x\left( t \right) \right) \leqslant V\left( x\left( 0 \right) \right) \leqslant b V(x(t))?V(x(0))?b ? x ( t ) ∈ Ω b \Rightarrow x\left( t \right) \in \varOmega _{\mathrm{b}} ?x(t)Ωb?

  • Proof of stability : Given ε > 0 \varepsilon >0 ε>0 , goal is to find δ > 0 \delta >0 δ>0, such that ∥ x ( 0 ) ∥ ? δ ? ∥ x ( t ) ∥ ? ε \left\| x\left( 0 \right) \right\| \leqslant \delta \Rightarrow \left\| x\left( t \right) \right\| \leqslant \varepsilon x(0)?δ?x(t)?ε
    在这里插入图片描述

  1. Ω b { 0 } \varOmega _{\mathrm{b}}\left\{ 0 \right\} Ωb?{0} if b = 0 b=0 b=0 (due to P.D. of V V V)
  2. As b b b increases , level set Ω b \varOmega _{\mathrm{b}} Ωb? grows in size until bitting B a l l ? ε Ball-\varepsilon Ball?ε then fix b = b ^ b=\hat{b} b=b^
  3. Find B a l l ? δ Ball-\delta Ball?δ inside Ω b \varOmega _{\mathrm{b}} Ωb? (because V V V is continuous at 0 0 0) then x 0 ∈ B a l l ? δ ? x 0 ∈ Ω b ? x ( t ) ∈ Ω b x_0\in Ball-\delta \Rightarrow x_0\in \varOmega _{\mathrm{b}}\Rightarrow x\left( t \right) \in \varOmega _{\mathrm{b}} x0?Ball?δ?x0?Ωb??x(t)Ωb? ( Ω b \varOmega _{\mathrm{b}} Ωb? is invariant)

Sketch of proof of Lyapunov Stability theorem:

  • First show stability under condition 2

Define sublevel set Ω b = { x ∈ R n : V ( x ) ? b } \varOmega _{\mathrm{b}}=\left\{ x\in \mathbb{R} ^n:V\left( x \right) \leqslant b \right\} Ωb?={xRn:V(x)?b}. Condition 2 implies V ( x ( t ) ) V\left( x\left( t \right) \right) V(x(t)) nonincreasing along system trajectory ? \Rightarrow ? if x 0 ∈ Ω b x_0\in \varOmega _{\mathrm{b}} x0?Ωb? , then x ( t ) ∈ Ω b x\left( t \right) \in \varOmega _{\mathrm{b}} x(t)Ωb?, ? t \forall t ?t
Given arbitrary ε > 0 \varepsilon >0 ε>0 , if we can find δ , b \delta ,b δ,b such that B ( 0 , δ ) ? Ω b ? B ( 0 , ε ) B\left( 0,\delta \right) \subseteq \varOmega _{\mathrm{b}}\subseteq B\left( 0,\varepsilon \right) B(0,δ)?Ωb??B(0,ε). Then the Lyapunov stability conditions are satisfied. Below is to show how we can find such b b b and δ \delta δ
V V V is continuous ? \Rightarrow ? m = min ? ∥ x ∥ = ε V ( x ) m=\min _{\left\| x \right\| =\varepsilon}V\left( x \right) m=minx=ε?V(x) exists (due to Weierstrass theorem). In addition, V V V is PD ? \Rightarrow ? m > 0 m>0 m>0. Therefore, if we choose b ∈ ( 0 , m ) b\in \left( 0,m \right) b(0,m) , then Ω b ? B ( 0 , ε ) \varOmega _{\mathrm{b}}\subseteq B\left( 0,\varepsilon \right) Ωb??B(0,ε)
V ( x ) V\left( x \right) V(x) is continuous at origin ? \Rightarrow ? for any b > 0 b>0 b>0 , there exists δ > 0 \delta >0 δ>0 such that ∣ V ( x ) ? V ( 0 ) ∣ = V ( x ) < b , ? x ∈ B ( 0 , δ ) \left| V\left( x \right) -V\left( 0 \right) \right|=V\left( x \right) <b,\forall x\in B\left( 0,\delta \right) V(x)?V(0)=V(x)<b,?xB(0,δ) . This implies that B ( 0 , δ ) ? Ω b B\left( 0,\delta \right) \subseteq \varOmega _{\mathrm{b}} B(0,δ)?Ωb?

  • Second, show asymptotic stability under condition 3:

We know V ( x ( t ) ) V\left( x\left( t \right) \right) V(x(t)) decreases monotonically as t → ∞ t\rightarrow \infty t and V ( x ( t ) ) ? 0 V\left( x\left( t \right) \right) \geqslant 0 V(x(t))?0, ? t \forall t ?t. Therefore, c = lim ? t → ∞ V ( x ( t ) ) c=\lim _{t\rightarrow \infty}V\left( x\left( t \right) \right) c=limt?V(x(t)) exists . So it suffices to show c = 0 c=0 c=0. Let us use a contradiction argument.
Suppose c ≠ 0 c\ne 0 c=0. Then c > 0 c>0 c>0. Therefore, x ( t ) ? Ω c = { x ∈ R n : V ( x ) ? c } x\left( t \right) \notin \varOmega _{\mathrm{c}}=\left\{ x\in \mathbb{R} ^n:V\left( x \right) \leqslant c \right\} x(t)/Ωc?={xRn:V(x)?c} , ? t \forall t ?t . We can choose β > 0 \beta >0 β>0 such that B ( 0 , β ) ? Ω c B\left( 0,\beta \right) \subseteq \varOmega _{\mathrm{c}} B(0,β)?Ωc? (due to continuity of V V V at 0 0 0)
Now let a = ? max ? β ? ∥ x ∥ ? ε V ˙ ( x ) a=-\max _{\beta \leqslant \left\| x \right\| \leqslant \varepsilon}\dot{V}\left( x \right) a=?maxβ?x?ε?V˙(x). Since V V V is ND, then a > 0 a>0 a>0
V ( x ( t ) ) = V ( x ( 0 ) ) + ∫ 0 t V ˙ ( x ( s ) ) d s ? V ( x ( 0 ) ) ? a ? t < 0 V\left( x\left( t \right) \right) =V\left( x\left( 0 \right) \right) +\int_0^t{\dot{V}\left( x\left( s \right) \right)}\mathrm{d}s\leqslant V\left( x\left( 0 \right) \right) -a\cdot t<0 V(x(t))=V(x(0))+0t?V˙(x(s))ds?V(x(0))?a?t<0 for sufficiently large t t t. ? \Rightarrow ? contradiction !

3.5 Exponential Lyapunov Function

Definition 3 (Exponential Lyapunov Function) —— Important for application
V : D → R V:D\rightarrow \mathbb{R} V:DR is called an Exponential Lyapunov Function (ELF) on D ? R n D\subset \mathbb{R} ^n D?Rn if ? k 1 , k 2 , k 3 , α > 0 \exists k_1,k_2,k_3,\alpha >0 ?k1?,k2?,k3?,α>0 such that
{ k 1 ∥ x ∥ α ? V ( x ) ? k 2 ∥ x ∥ α L f V ( x ) ? ? k 3 V ( x ) \begin{cases} k_1\left\| x \right\| ^{\alpha}\leqslant V\left( x \right) \leqslant k_2\left\| x \right\| ^{\alpha}\\ \mathcal{L} _{\mathrm{f}}V\left( x \right) \leqslant -k_3V\left( x \right)\\ \end{cases} {k1?xα?V(x)?k2?xαLf?V(x)??k3?V(x)?

Lyapunov stability ? C 1 \exists \mathcal{C} ^1 ?C1 func V V V
V V V is PD - deserable ; V ˙ \dot{V} V˙ is ND/NSD
k 1 ∥ x ∥ α ? V ( x ) ? k 2 ∥ x ∥ α k_1\left\| x \right\| ^{\alpha}\leqslant V\left( x \right) \leqslant k_2\left\| x \right\| ^{\alpha} k1?xα?V(x)?k2?xα ? V \Rightarrow V ?V is PD (radially unbounded)
L f V ( x ) ? ? k 3 V ( x ) \mathcal{L} _{\mathrm{f}}V\left( x \right) \leqslant -k_3V\left( x \right) Lf?V(x)??k3?V(x) ? V ˙ \Rightarrow \dot{V} ?V˙ is ND, V ˙ ? ? k 3 V \dot{V}\leqslant -k_3V V˙??k3?V

Droof sketch :
recall : z ∈ R 1 , z ˙ = ? k 3 z ? z ( t ) = e ? k 3 t z ( 0 ) z\in \mathbb{R} ^1,\dot{z}=-k_3z\Rightarrow z\left( t \right) =e^{-k_3t}z\left( 0 \right) zR1,z˙=?k3?z?z(t)=e?k3?tz(0)
By comparison theorem : V ˙ ? ? k 3 V ? V ( t ) ? e ? k 3 t V ( 0 ) \dot{V}\leqslant -k_3V\Rightarrow V\left( t \right) \leqslant e^{-k_3t}V\left( 0 \right) V˙??k3?V?V(t)?e?k3?tV(0)
? ∥ x ( t ) ∥ α ? 1 k 1 V ( x ( t ) ) ? 1 k 1 e ? k 3 t V ( x ( 0 ) ) ? k 2 k 1 e ? k 3 t ∥ x ( 0 ) ∥ α ? ∥ x ( t ) ∥ α ? c e ? β t ∥ x ( 0 ) ∥ α \Rightarrow \left\| x\left( t \right) \right\| ^{\alpha}\leqslant \frac{1}{k_1}V\left( x\left( t \right) \right) \leqslant \frac{1}{k_1}e^{-k_3t}V\left( x\left( 0 \right) \right) \leqslant \frac{k_2}{k_1}e^{-k_3t}\left\| x\left( 0 \right) \right\| ^{\alpha}\Rightarrow \left\| x\left( t \right) \right\| ^{\alpha}\leqslant ce^{-\beta t}\left\| x\left( 0 \right) \right\| ^{\alpha} ?x(t)α?k1?1?V(x(t))?k1?1?e?k3?tV(x(0))?k1?k2??e?k3?tx(0)α?x(t)α?ce?βtx(0)α

Theorem 1 (ELF Theorem)
If system 2 has an ELF, then it is exponentially stable

3.6 Stability Analysis Examples

  • Example 1 :
    { x ˙ 1 = ? x 1 + x 2 + x 1 x 2 x ˙ 2 = x 1 ? x 2 ? x 1 2 ? x 2 3 \begin{cases} \dot{x}_1=-x_1+x_2+x_1x_2\\ \dot{x}_2=x_1-x_2-{x_1}^2-{x_2}^3\\ \end{cases} {x˙1?=?x1?+x2?+x1?x2?x˙2?=x1??x2??x1?2?x2?3? Try V ( x ) = ∥ x ∥ 2 = x 1 2 + x 2 2 V\left( x \right) =\left\| x \right\| ^2={x_1}^2+{x_2}^2 V(x)=x2=x1?2+x2?2
    equilibrium : x ˙ = 0 ? ( x 1 x 2 ) = ( 0 0 ) \dot{x}=0\Rightarrow \left( \begin{array}{c} x_1\\ x_2\\ \end{array} \right) =\left( \begin{array}{c} 0\\ 0\\ \end{array} \right) x˙=0?(x1?x2??)=(00?)
    check Lyapunov conditions
  1. V ( x ) = x 1 2 + x 2 2 = x T [ 1 0 0 1 ] x V\left( x \right) ={x_1}^2+{x_2}^2=x^{\mathrm{T}}\left[ \begin{matrix} 1& 0\\ 0& 1\\ \end{matrix} \right] x V(x)=x1?2+x2?2=xT[10?01?]x is PD and C 1 \mathcal{C} ^1 C1
  2. L f V ( x ) = ( ? V ? x ) T f ( x ) = [ 2 x 1 2 x 2 ] T [ f 1 ( x ) f 2 ( x ) ] = 2 x 1 ( ? x 1 + x 2 + x 1 x 2 ) + 2 x 2 ( x 1 ? x 2 ? x 1 2 ? x 2 3 ) = ? 2 ( x 1 ? x 2 ) 2 ? 2 x 2 4 ? N D \mathcal{L} _{\mathrm{f}}V\left( x \right) =\left( \frac{\partial V}{\partial x} \right) ^{\mathrm{T}}f\left( x \right) =\left[ \begin{array}{c} 2x_1\\ 2x_2\\ \end{array} \right] ^{\mathrm{T}}\left[ \begin{array}{c} f_1\left( x \right)\\ f_2\left( x \right)\\ \end{array} \right] =2x_1\left( -x_1+x_2+x_1x_2 \right) +2x_2\left( x_1-x_2-{x_1}^2-{x_2}^3 \right) =-2\left( x_1-x_2 \right) ^2-2{x_2}^4\Rightarrow ND Lf?V(x)=(?x?V?)Tf(x)=[2x1?2x2??]T[f1?(x)f2?(x)?]=2x1?(?x1?+x2?+x1?x2?)+2x2?(x1??x2??x1?2?x2?3)=?2(x1??x2?)2?2x2?4?ND

? \Rightarrow ? system is asymptotically stable

  • Example 2 :
    { x ˙ 1 = ? x 1 + x 1 x 2 x ˙ 2 = ? x 2 \begin{cases} \dot{x}_1=-x_1+x_1x_2\\ \dot{x}_2=-x_2\\ \end{cases} {x˙1?=?x1?+x1?x2?x˙2?=?x2??
    Can we find a simple quadratic Lyapunov function ? First try V ( x ) = x 1 2 + x 2 2 V\left( x \right) ={x_1}^2+{x_2}^2 V(x)=x1?2+x2?2
  1. V V V is PD
  2. L f V ( x ) = ? 2 ( ( x 2 ? 4 ) 2 ? 8 ) \mathcal{L} _{\mathrm{f}}V\left( x \right) =-2\left( \left( x_2-4 \right) ^2-8 \right) Lf?V(x)=?2((x2??4)2?8) Not ND

In fact the system does not have any (global polynomial Lyapunov function.) But it is GAS with a Lyapunov function V ( x ) = ln ? ( 1 + x 1 2 ) + x 2 2 V\left( x \right) =\ln \left( 1+{x_1}^2 \right) +{x_2}^2 V(x)=ln(1+x1?2)+x2?2

4. Lyapunov Stability of Linear Systems

4.1 Stability of Linear Systems

Consider autonomous linear system : x ˙ = f ( x ) = A x \dot{x}=f\left( x \right) =Ax x˙=f(x)=Ax

  • Recall solution to the linear system : x ( t ) = e A t x ( 0 ) x\left( t \right) =e^{At}x\left( 0 \right) x(t)=eAtx(0)
  • If isolated equilibrium only possible equilibrium is origin x = 0 x=0 x=0
    f ( x ) = 0 ? A x = 0 f\left( x \right) =0\Rightarrow Ax=0 f(x)=0?Ax=0 : 1. if A A A is nonsingular ? x = 0 \Rightarrow x=0 ?x=0 . 2. If A A A is singular, ?? A A A is the set of equilibrium
  • Fact : Origin asympt. stable ? \Leftrightarrow ? R e ( λ i ) < 0 \mathrm{Re}\left( \lambda _{\mathrm{i}} \right) <0 Re(λi?)<0 for all eigenvalues λ i \lambda _{\mathrm{i}} λi? of A A A
    suppose we have isolated equilibrium x ? = 0 x^*=0 x?=0
    For simplicity, consider a simple case when A A A is diagonalizable : A = T D T ? 1 A=TDT^{-1} A=TDT?1 where D = [ λ 1 λ 2 ? λ n ] D=\left[ \begin{matrix} \lambda _1& & & \\ & \lambda _2& & \\ & & \ddots& \\ & & & \lambda _{\mathrm{n}}\\ \end{matrix} \right] D= ?λ1??λ2????λn?? ? ? e A t = T e D t T ? 1 = T [ e λ 1 t e λ 2 t ? e λ n t ] T ? 1 \Rightarrow e^{At}=Te^{Dt}T^{-1}=T\left[ \begin{matrix} e^{\lambda _1t}& & & \\ & e^{\lambda _2t}& & \\ & & \ddots& \\ & & & e^{\lambda _{\mathrm{n}}t}\\ \end{matrix} \right] T^{-1} ?eAt=TeDtT?1=T ?eλ1?t?eλ2?t???eλn?t? ?T?1 . If R e ( λ i ) < 0 \mathrm{Re}\left( \lambda _{\mathrm{i}} \right) <0 Re(λi?)<0 , for all i i i , every entry of e A t → 0 e^{At}\rightarrow 0 eAt0 ? e A t x ( 0 ) \Rightarrow e^{At}x\left( 0 \right) ?eAtx(0) expenentially
  • Discrete time system : x ( k + 1 ) = A x ( k ) x\left( k+1 \right) =Ax\left( k \right) x(k+1)=Ax(k) os asymptotically stable if e i g ( A ) eig\left( A \right) eig(A) inside unit circle

4.1 Lyapunov Function of Linear Systems

  • Consider a quadratic Lyapunov function candidate : V ( x ) = x T P x V\left( x \right) =x^{\mathrm{T}}Px V(x)=xTPx , with P ∈ R n × n P\in \mathbb{R} ^{n\times n} PRn×n
    V is PD ? P ? 0 \Rightarrow P\succ 0 ?P?0 ( P P P is a PD matrix)
    L f V \mathcal{L} _{\mathrm{f}}V Lf?V is ND ? \Rightarrow ? L f V ? ( ? V ? x ) T A x = ( 2 P x ) T A x = 2 x T P T A x \mathcal{L} _{\mathrm{f}}V\triangleq \left( \frac{\partial V}{\partial x} \right) ^{\mathrm{T}}Ax=\left( 2Px \right) ^{\mathrm{T}}Ax=2x^{\mathrm{T}}P^{\mathrm{T}}Ax Lf?V?(?x?V?)TAx=(2Px)TAx=2xTPTAx or equirelatly, V ˙ ( x ( t ) ) = x ˙ T P x + x T P x ˙ = x T A T P x + x T P A x \dot{V}\left( x\left( t \right) \right) =\dot{x}^{\mathrm{T}}Px+x^{\mathrm{T}}P\dot{x}=x^{\mathrm{T}}A^{\mathrm{T}}Px+x^{\mathrm{T}}PAx V˙(x(t))=x˙TPx+xTPx˙=xTATPx+xTPAx —— Is 2 P T A = A T P + P A 2P^{\mathrm{T}}A=A^{\mathrm{T}}P+PA 2PTA=ATP+PA ? —— x T P T A x = x T A T P x x^{\mathrm{T}}P^{\mathrm{T}}Ax=x^{\mathrm{T}}A^{\mathrm{T}}Px xTPTAx=xTATPx

? V \Rightarrow V ?V is LF if P P P is PD and A T P + P A A^{\mathrm{T}}P+PA ATP+PA is ND

Fact : for Linear System , quadratic form of LF , ai all we need to consider. —— A A A is asym stable if and only if ??

In proof of the above function , we assumed P P P is symmetric so P T A = P A P^{\mathrm{T}}A=PA PTA=PA
e.g. P T A = P A P^{\mathrm{T}}A=PA PTA=PA , Q = [ 1 1 ? 1 1 ] , g ( x ) = x T Q x = [ x 1 x 2 ] T [ 1 1 ? 1 1 ] [ x 1 x 2 ] = x 1 2 + x 2 2 ? [ x 1 x 2 ] T [ 1 0 0 1 ] [ x 1 x 2 ] Q=\left[ \begin{matrix} 1& 1\\ -1& 1\\ \end{matrix} \right] , g\left( x \right) =x^{\mathrm{T}}Qx=\left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] ^{\mathrm{T}}\left[ \begin{matrix} 1& 1\\ -1& 1\\ \end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] ={x_1}^2+{x_2}^2\Rightarrow \left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] ^{\mathrm{T}}\left[ \begin{matrix} 1& 0\\ 0& 1\\ \end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ \end{array} \right] Q=[1?1?11?],g(x)=xTQx=[x1?x2??]T[1?1?11?][x1?x2??]=x1?2+x2?2?[x1?x2??]T[10?01?][x1?x2??] Q ^ = 1 2 Q + 1 2 Q T \hat{Q}=\frac{1}{2}Q+\frac{1}{2}Q^{\mathrm{T}} Q^?=21?Q+21?QT

Fact : A A A is asym stable if and only if

  1. ? P ? 0 \exists P\succ 0 ?P?0 , such that A T P + P A ? 0 A^{\mathrm{T}}P+PA\prec 0 ATP+PA?0
  2. equivalently , for any Q ? 0 , ? P Q\succ 0,\exists P Q?0,?P such that A T P + P A = ? Q A^{\mathrm{T}}P+PA=-Q ATP+PA=?Q (Lyapunov equation)

4.2 Stability Conditions for Linear Systems

Theorem 1 (Stability Conditions for Linear System)
For an autonomous Linear system x ˙ = A x \dot{x}=Ax x˙=Ax. The following statements are equivalent.

  • (Linear) System is (globally) asmptotically stable
  • (Linear) System is (globally) exponentially stable
  • R e ( λ i ) < 0 \mathrm{Re}\left( \lambda _{\mathrm{i}} \right) <0 Re(λi?)<0 for all eigenvalues λ i \lambda _{\mathrm{i}} λi? of A A A —— lie on open left half complex plane (OLHP)
  • System has a quadratic Lyapunov function V ( x ) = x T P x V\left( x \right) =x^{\mathrm{T}}Px V(x)=xTPx
  • For ant symmetric Q ? 0 Q\succ 0 Q?0 , there exists a symmetric P ? 0 P\succ 0 P?0 that solves the following Lyapunov equation :
    A T P + P A = ? Q A^{\mathrm{T}}P+PA=-Q ATP+PA=?Q
    Q ? 0 Q\succ 0 Q?0 is given , P P P is the variale to be solved , and V ( x ) = x T P x V\left( x \right) =x^{\mathrm{T}}Px V(x)=xTPx is Lyapunov function of the system

5. Converse Lyapunov Function

When there is a Lyapunov Function?

  • Converse Lyapunov Theorem for Asymptotic Stability
    origin asymptotically stable ; f f f is locallt Lipschitz on D with region of attration R A R_A RA? ? V ?? s . t . \Rightarrow V\,\,s.t. ?Vs.t. V V V is continuuos and PD on R A R_A RA? ; L f V L_{\mathrm{f}}V Lf?V is ND on R A R_A RA? ; V ( x ) → ∞ V\left( x \right) \rightarrow \infty V(x) as x → ? R A x\rightarrow \partial R_{\mathrm{A}} x?RA?
    convex result that is not constructive

  • Converse Lyapunov Theorem for Exponential Stability
    origin exponentially stable on D D D ; f f f is C 1 \mathcal{C} ^1 C1 ? ? \Rightarrow \exists ?? an ELF V V V on D D D

  • For nonlinear sys , ? V ? \exists V\Rightarrow ?V? stability (sufficient condition)

  • Proofs are involved especially for the converse theorem for asymptotic stability

  • Important : proofs of converse theorems often assume the knowledge of system solution and hence are not constructive

6. Extension of Discrete-Time System

6.1 What about Discrete Time Systems?

  • So far, all our definitions, results, examples are given using continuous dynamical system models.
  • All of them have discrete-time counterparts. The ideas and conclusions are the “same” (in sprit)
  • For example, given autonomous discrete-time system : x ( k + 1 ) = f ( x ( k ) ) x\left( k+1 \right) =f\left( x\left( k \right) \right) x(k+1)=f(x(k)) with f ( 0 ) = 0 f\left( 0 \right) =0 f(0)=0 (origin is an isolated equilibrium)
    Rate of change of a function V ( x ) V\left( x \right) V(x) along system trajectory can be defined as : Δ f V ( x ) = V ( f ( x ) ) ? V ( x ) ? V ( x ( k + 1 ) ) ? V ( x ( k ) ) \varDelta _{\mathrm{f}}V\left( x \right) =V\left( f\left( x \right) \right) -V\left( x \right) \Leftarrow V\left( x\left( k+1 \right) \right) -V\left( x\left( k \right) \right) Δf?V(x)=V(f(x))?V(x)?V(x(k+1))?V(x(k)) , where L f V ( x ) = ( ? V ? x ) T f ( x ) L_{\mathrm{f}}V\left( x \right) =\left( \frac{\partial V}{\partial x} \right) ^{\mathrm{T}}f\left( x \right) Lf?V(x)=(?x?V?)Tf(x)
    Asymptotically stable requires : V V V is PD(observable all the bad behavior of ‘x’ shows up in V V V) and Δ f V \varDelta _{\mathrm{f}}V Δf?V is ND —— Δ f V ( x ) ? 0 \varDelta _{\mathrm{f}}V\left( x \right) \prec 0 Δf?V(x)?0 for all x ∈ R n / { 0 } x\in \mathbb{R} ^n/\left\{ 0 \right\} xRn/{0}
    Exponentially stable requires : k 1 ∥ x ∥ α ? V ( x ) ? k 2 ∥ x ∥ α ?? a n d ?? Δ f V ( x ) ? ? k 3 V ( x ) k_1\left\| x \right\| ^{\alpha}\leqslant V\left( x \right) \leqslant k_2\left\| x \right\| ^{\alpha}\,\,and\,\,\varDelta _{\mathrm{f}}V\left( x \right) \leqslant -k_3V\left( x \right) k1?xα?V(x)?k2?xαandΔf?V(x)??k3?V(x)

6.2 Concluding Remarks

  • We have learned different notions of internal stability, e.g. stability in Lyapunov sense, asymptotic stability, globally asymptotic stability (G.A.S), exponential stability, globally exponential stability(G.E.S)
  • Sufficient condition to ensure stability is often the existence of a properly defined Lyapunov function
  • Key requirements for a Lyapunov function :
    Positive definite and is zero at the system equilibrium
    Descease along system trajectory
  • For linear system : G.A.S ? \Leftrightarrow ? G.E.S ? \Leftrightarrow ? Existence of a quadratic Lyapunov function
  • The definitions and results in this lecture have sometimes been stated in simplified form to facilitate presentation. More general version can be found in standard textbooks on nonlinear systems
  • Next Lecture : Semidefinite Programming and computational stability analysis
文章来源:https://blog.csdn.net/LiongLoure/article/details/135085434
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