[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
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 abouttransient/'tr?nz??nt/短暂的
) - ability to return to the desired asymptotic behavior (nout just convergence)
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˙=[x2?lg?cosx1??]=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??x1?x2?x1?2?]?{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
? ? 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
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 -
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˙=[x2?lg?cosx1??]=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??x1?x2?x1?2?]?{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?=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
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))=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 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)=[x1?x2??]TP[x1?x2??]=x1?2+x1?x2?+3x2?2 -
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)=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)?{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∥=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,?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∥?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:D→R is called anExponential 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∥αLf?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?∥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)
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)
z∈R1,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?t∥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?+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)=∥x∥2=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
- 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
- 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
- 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) 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 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 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
- ? 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 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\} 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∥αandΔf?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
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