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

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B站：CLEAR_LAB
笔者带更新-运动学
课程主讲教师：
Prof. Wei Zhang

南科大高等机器人控制课 Ch10 Bascis of Stability Analysis

[1. Background](#1. Background)
- [1.1 What is Stability Analysis](#1.1 What is Stability Analysis)
- [1.2 General ODE Models for Dynamical Systems](#1.2 General ODE Models for Dynamical Systems)
- [1.3 Example](#1.3 Example)
- - [1.3.1 Pendulum](#1.3.1 Pendulum)
  - [1.3.2 Adaptive Control](#1.3.2 Adaptive Control)
- [1.4 Equilibrium Point of Dynamical Systems](#1.4 Equilibrium Point of Dynamical Systems)
- [1.5 Invariant Set of Dynamical Systems](#1.5 Invariant Set of Dynamical Systems)
[2. Lyapunov Stability Definitions](#2. Lyapunov Stability Definitions)
- [2.1 Lyapunov Stability Definitions](#2.1 Lyapunov Stability Definitions)
- [2.2 Stability Examples using 2D Phase Portrait](#2.2 Stability Examples using 2D Phase Portrait)
[3. Lyapunov Stability Theorem](#3. Lyapunov Stability Theorem)
- [3.1 How to verify stability of a system?](#3.1 How to verify stability of a system?)
- [3.2 Sign Definite Functions](#3.2 Sign Definite Functions)
- [3.3 Lyapunov Stability Theorem](#3.3 Lyapunov Stability Theorem)
- [3.4 Proof of Lyapunov Stability Theorem](#3.4 Proof of Lyapunov Stability Theorem)
- [3.5 Exponential Lyapunov Function](#3.5 Exponential Lyapunov Function)
- [3.6 Stability Analysis Examples](#3.6 Stability Analysis Examples)
[4. Lyapunov Stability of Linear Systems](#4. Lyapunov Stability of Linear Systems)
- [4.1 Stability of Linear Systems](#4.1 Stability of Linear Systems)
- [4.1 Lyapunov Function of Linear Systems](#4.1 Lyapunov Function of Linear Systems)
- [4.2 Stability Conditions for Linear Systems](#4.2 Stability Conditions for Linear Systems)
[5. Converse Lyapunov Function](#5. Converse Lyapunov Function)
- [When there is a Lyapunov Function?](#When there is a Lyapunov Function?)
[6. Extension of Discrete-Time System](#6. Extension of Discrete-Time System)
- [6.1 What about Discrete Time Systems?](#6.1 What about Discrete Time Systems?)
- [6.2 Concluding Remarks](#6.2 Concluding Remarks)

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

1. Background

1.1 What is Stability Analysis

system asymptotic/ˌæsimp'tɔtik,-kəl/ 渐进的 behavior (not too much about transient/'trænzɪənt/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 x∈X⊆Rn : state
u ∈ U ⊆ R m u\in \mathcal{U} \subseteq \mathbb{R} ^m u∈U⊆Rm : control input
f : R n × R m → R n f:\mathbb{R} ^n\times \mathbb{R} ^m\rightarrow \mathbb{R} ^n f:Rn×Rm→Rn : (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:X→U , u = μ ( x ) u=\mu \left( x \right) u=μ(x)
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

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

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˙=[x2lgcosx1]=0⇒{x2=0cosx1=0,x1=22kπ+π,k∈Z

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 :

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−x1x2x12]⇒{x1=0x2=arbitrary⇒E={x∈R2,x1=0}

2. Lyapunov Stability Definitions

Stability :

about equilibrium
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),x∈Rn , 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 :

The equilibrium x = 0 x=0 x=0 is called stable(stay close to equilibrium) in the sense of Lyapunov , if
ϵ − δ \epsilon -\delta ϵ−δ argument ------ ∀ ϵ > 0 , ∃ δ > 0 , s . t . ∥ x ( 0 ) ∥ ⩽ δ ⇒ ∥ x ( t ) ∥ ⩽ ϵ , ∀ 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 ∀ϵ>0,∃δ>0,s.t.∥x(0)∥⩽δ⇒∥x(t)∥⩽ϵ,∀t⩾0

Objective: For any ϵ > 0 \epsilon >0 ϵ>0 , ensure ∥ x ( t ) ∥ ⩽ ϵ \left\| x\left( t \right) \right\| \leqslant \epsilon ∥x(t)∥⩽ϵ 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
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)
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)∥⩽c∥x(0)∥e−λt,∀∥x(0)∥⩽δ
globallt asymptomtotically / exponentially stable (G.A.S / G.E.S) if the above conditions holds for all δ > 0 \delta >0 δ>0
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≜{x∈Rn: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˙=[x2lgcosx1]=0⇒{x2=0cosx1=0,x1=22kπ+π,k∈Z
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−x1x2x12]⇒{x1=0x2=arbitrary⇒E={x∈R2,x1=0}
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=x12−x22x˙2=2x1x2 ------ 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

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:Rn→R 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))=dtdV(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 0∈D⊆Rn

g : D → R g:D\rightarrow \mathbb{R} g:D→R 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,∀x∈D
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)=[x1x2]TP[x1x2]=x12+x1x2+3x22
g : D → R g:D\rightarrow \mathbb{R} g:D→R 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,∀x∈D\{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:Rn→R is radically unbounded if g ( x ) → ∞ g\left( x \right) \rightarrow \infty g(x)→∞ as ∥ x ∥ → ∞ \left\| x \right\| \rightarrow \infty ∥x∥→∞

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:D→R (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)=x12,x=[x1x2]=[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)≜{x∈Rn: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

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)
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^
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={x∈Rn: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=min∥x∥=εV(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,∀x∈B(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={x∈Rn: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∥⩽εV˙(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))+∫0tV˙(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:D→R 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∥x∥α⩽V(x)⩽k2∥x∥αLfV(x)⩽−k3V(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∥x∥α⩽V(x)⩽k2∥x∥α ⇒ 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) LfV(x)⩽−k3V(x) ⇒ V ˙ \Rightarrow \dot{V} ⇒V˙ is ND, V ˙ ⩽ − k 3 V \dot{V}\leqslant -k_3V V˙⩽−k3V

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) z∈R1,z˙=−k3z⇒z(t)=e−k3tz(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˙⩽−k3V⇒V(t)⩽e−k3tV(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⩽k11V(x(t))⩽k11e−k3tV(x(0))⩽k1k2e−k3t∥x(0)∥α⇒∥x(t)∥α⩽ce−βt∥x(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+x1x2x˙2=x1−x2−x12−x23 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)=∥x∥2=x12+x22
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⇒(x1x2)=(00)
check Lyapunov conditions

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)=x12+x22=xT[1001]x is PD and C 1 \mathcal{C} ^1 C1
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 LfV(x)=(∂x∂V)Tf(x)=[2x12x2]T[f1(x)f2(x)]=2x1(−x1+x2+x1x2)+2x2(x1−x2−x12−x23)=−2(x1−x2)2−2x24⇒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+x1x2x˙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)=x12+x22

V V V is PD
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) LfV(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+x12)+x22

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= λ1λ2⋱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 eλ1teλ2t⋱eλnt 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 eAt→0 ⇒ 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} P∈Rn×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 LfV 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 LfV≜(∂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=[1−111],g(x)=xTQx=[x1x2]T[1−111][x1x2]=x12+x22⇒[x1x2]T[1001][x1x2] ， Q ^ = 1 2 Q + 1 2 Q T \hat{Q}=\frac{1}{2}Q+\frac{1}{2}Q^{\mathrm{T}} Q^=21Q+21QT

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

∃ 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

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 LfV 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) ΔfV(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) LfV(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 ΔfV is ND ------ Δ f V ( x ) ≺ 0 \varDelta _{\mathrm{f}}V\left( x \right) \prec 0 ΔfV(x)≺0 for all x ∈ R n / { 0 } x\in \mathbb{R} ^n/\left\{ 0 \right\} x∈Rn/{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∥x∥α⩽V(x)⩽k2∥x∥αandΔfV(x)⩽−k3V(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
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