Lecture 4 Lyapunov Stability

Transcription

1 Lecture 4 Lyapunov Stability Material Glad & Ljung Ch Khalil Ch Lecture notes

2 Today s Goal To be able to prove local and global stability of an equilibrium point using Lyapunov s method show stability of a set (e.g., an equilibrium, or a limit cycle) using La Salle s invariant set theorem.

3 Alexandr Mihailovich Lyapunov ( ) Master thesis On the stability of ellipsoidal forms of equilibrium of rotating fluids, St. Petersburg University, Doctoral thesis The general problem of the stability of motion, 1892.

4 Main idea Lyapunov formalized the idea: If the total energy is dissipated, then the system must be stable. Main benefit: By looking at how an energy-like function V (a so called Lyapunov function) changes over time, we might conclude that a system is stable or asymptotically stable without solving the nonlinear differential equation. Main question: How to find a Lyapunov function?

5 Examples Start with a Lyapunov candidate V to measure e.g., "size" 1 of state and/or output error, "size" of deviation from true parameters, energy difference from desired equilibrium, weighted combination of above... Example of common choice in adaptive control V= 1 2 (e 2 + γ a ã 2 + γ b b2 ) (here weighted sum of output error and parameter errors) 1 Often a magnitude measure or (squared) norm like e 2 2,...

6 Analysis: Check if V is decreasing with time Continuous time: Discrete time: dv dt <0 V(k+1) V(k)<0 Synthesis: Choose, e.g., control law and/or parameter update law to satisfy V 0 dv dt =eė+γ aã ã+ γ b b b= = x( a x ã x+ bu)+ γ a ã ã+ γ b b b=... Ifais constant andã=a â then ã= â. Choose update law dâ in a "good way" to influencedv dt dt. (more on this later...)

7 A Motivating Example x m mẍ= bẋ ẋ k }{{} 0 x k 1 x }{{ 3 } damping spring b,k 0,k 1 >0 Total energy = kinetic + pot. energy:v= mv2 2 + x 0 F sprin ds V(x,ẋ)=mẋ 2 /2+k 0 x 2 /2+k 1 x 4 /4>0, V(0,0)=0 d dt V(x,ẋ)=mẍẋ+k 0xẋ+k 1 x 3 ẋ={pluginsystemdynamics 2 } = b ẋ 3 <0,forẋ =0 What does this mean? 2 Also referred to evaluate along system trajectories.

8 Stability Definitions An equilibrium pointx ofẋ=f(x) (i.e.,f(x )=0) is locally stable, if for everyr>0there existsr>0, such that x(0) x <r x(t) x <R, t 0 locally asymptotically stable, if locally stable and x(0) x <r lim t x(t)=x globally asymptotically stable, if asymptotically stable for allx(0) R n.

9 Lyapunov Theorem for Local Stability Theorem Letẋ=f(x),f(x )=0wherex is in the interior of Ω R n. Assume thatv: Ω Ris ac 1 function. If (1) V(x )=0 (2) V(x)>0, for allx Ω,x =x (3) V(x) 0along all trajectories of the system in Ω = x is locally stable. Furthermore, if also (4) V(x)<0for allx Ω,x =x = x is locally asymptotically stable.

10 Lyapunov Functions ( Energy Functions) A function V that fulfills (1) (3) is called a Lyapunov function. Condition (3) means that V is non-increasing along all trajectories in Ω: V(x)= V xẋ= i where V [ V x =, V,... V ] x 1 x 2 x n V x i f i (x) 0 V level sets wherev=const. x 1 x 2

11 Conservation and Dissipation V Conservation of energy: V(x)= xf(x)=0, i.e., the vector fieldf(x) is everywhere orthogonal to the normal V x to the level surfacev(x)=c. Example: Total energy of a lossless mechanical system or total fluid in a closed system. Dissipation of energy: V(x)= V xf(x) 0, i.e., the vector fieldf(x) and the normal V x to the level surface{z: V(z)=c} make an obtuse angle (Sw. trubbig vinkel ). Example: Total energy of a mechanical system with damping or total fluid in a system that leaks.

12 Geometric interpretation V(x)=constant f(x) gradient V x x(t) Vector field points into sublevel sets Trajectories can only go to lower values ofv(x)

13 Boundedness: For any trajectory x(t) t V(x(t))=V(x(0))+ 0 V(x(τ))dτ V(x(0)) which means that the whole trajectory lies in the set {z V(z) V(x(0))} For stability it is thus important that the sublevel sets {z V(z) c} bounded c 0 V(x) as x.

14 2 min exercise Pendulum Show that the origin is locally stable for a mathematical pendulum. ẋ 1 =x 2, ẋ 2 = l sinx 1 Use as a Lyapunov function candidate V(x)=(1 cosx 1 ) l+l 2 x 2 2 / x x 2 2

15 Example Pendulum (1) V(0)=0 (2) V(x)>0for 2π<x 1 <2π and(x 1,x 2 ) =0 (3) V(x)=ẋ 1 sinx 1 l+l 2 x 2 ẋ 2 =0, for allx Hence,x=0is locally stable. Note that x = 0 is not asymptotically stable, because V(x) = 0 and not<0for allx =0.

16 Positive Definite Matrices Definition: Symmetric matrixm=m T is positive definite (M>0) if x T Mx>0, x =0 positive semidefinite (M 0) if x T Mx 0, x Lemma: M=M T >0 λ i (M)>0, i M=M T 0 λ i (M) 0, i M=M T >0 V(x):=x T Mx V(0)=0, V(x)>0, x =0 V(x) candidate Lyapunov function

17 More matrix results for symmetric matrixm=m T λ min (M) x 2 x T Mx λ max (M) x 2, x Proof idea: factorizem=u ΛU T, unitaryu (i.e., Ux = x x), Λ= diag(λ 1,...,λ n ) for any matrixm Mx λ max (M T M) x, x

18 Example- Lyapunov function for linear system ẋ=ax = [ ][ x1 x 2 ] (1) Eigenvalues of A :{ 1, 3} (global) asymptotic stability. Find a quadratic Lyapunov function for the system (1): V(x)=x T Px= [ x 1 x 2 ] [ p 11 p 12 p 12 p 22 ][ x1 x 2 ], P=P T >0 Take anyq=q T >0,sayQ=I 2 2. SolveA T P+PA= Q.

19 Example cont d A T P+PA= I [ ][ ] [ ][ ] 1 0 p11 p 12 p11 p = 4 3 p 12 p 22 p 12 p [ ] [ ] (2) 2p 11 4p 12 +4p = 4p 12 +4p 11 8p 12 6p Solving forp 11,p 12 andp 22 gives 2p 11 = 1 4p 12 +4p 11 =0 8p 12 6p 22 = 1 [ ] p11 p = 12 = p 12 p 22 [ ] 1/2 1/2 >0 1/2 5/6

20 x1 = x1 + 4 x2 x2 = 3 x2 x 1 2 +x2 2 8 = x x 1 Phase plot showing that V= 1 2 (x2 1 +x2 2 )=[ x 1 x 2 ] [ ][ x1 x 2 ] does NOT work.

21 x1 = x1 + 4 x2 x2 = 3 x2 (1/2 x 1 +1/2 x 2 ) x 1 +(1/2 x 1 +5/6 x 2 ) x 2 7 = x x 1 Phase plot with level curvesx T Px=c forp found in example.

22 Lyapunov Stability for Linear Systems Linear system: ẋ = Ax Lyapunov equation: LetQ=Q T >0. Solve PA+A T P= Q with respect to the symmetric matrix P. Lyapunov function:v(x)=x T Px, V(x)=x T Pẋ+ẋ T Px=x T (PA+A T P)x= x T Qx<0 Asymptotic Stability: IfP=P T >0, then the Lyapunov Stability Theorem implies (local=global) asymptotic stability, hence the eigenvalues ofa must satisfyre λ k (A)<0, k

23 Converse Theorem for Linear Systems IfRe λ k (A)<0 k, then for everyq=q T >0there exists P=P T >0such thatpa+a T P= Q Proof: ChooseP= 0 e ATt Qe At dt. Then t ( A T P+PA = lim A T e ATτ Qe Aτ +e ATτ QAe Aτ) dτ t 0 [ = lim e ATτ Qe Aτ] t t 0 = Q

24 Interpretation Assumeẋ=Ax,x(0)=z. Then ( ) x T (t)qx(t)dt=z T e ATt Qe At dt z=z T Pz 0 ThusV(z)=z T Pz is the cost-to-go fromz (with no input) and integral quadratic cost function with weighting matrix Q. 0

25 Lyapunov s Linearization Method Recall from Lecture 2: Theorem Consider ẋ=f(x) Assume thatf(0)=0. Linearization ẋ=ax+ (x), (x) =o( x ) asx 0. (1) Re λ k (A)<0, k x=0 locally asympt. stable (2) k: Reλ k (A)>0 x=0 unstable

26 Proof of (1) in Lyapunov s Linearization Method PutV(x):=x T Px. Then,V(0)=0,V(x)>0 x =0, and V(x)=x T Pf(x)+f T (x)px =x T P[Ax+ (x)]+[x T A T + T (x)]px =x T (PA+A T P)x+2x T P (x)= x T Qx+2x T P (x) x T Qx λ min (Q) x 2 and for all γ >0there existsr>0such that (x) < γ x, x <r Thus, choosing γ sufficiently small gives V(x) ( λ min (Q) 2γ λ max (P) ) x 2 <0

27 Lyapunov Theorem for Global Asymptotic Stability Theorem Letẋ=f(x) andf(x )=0. If there exists ac 1 functionv:r n Rsuch that (1) V(x )=0 (2) V(x)>0, for allx =x (3) V(x)<0for allx =x (4) V(x) as x thenx is a globally asymptotically stable equilibrium.

28 Radial Unboundedness is Necessary If the conditionv(x) as x is not fulfilled, then global stability cannot be guaranteed. Example AssumeV(x)=x1 2/(1+x2 1 )+x2 2 is a Lyapunov function for a system. Can have x even if V(x)<0. 2 Contour plotv(x)=c: 1.5 Example [Khalil]: x ẋ 1 = 6x 1 (1+x 2 1 )2+2x 2 ẋ 2 = 2(x 1+x 2 ) (1+x 2 1 )2 1.5 x

29 Somewhat Stronger Assumptions Theorem: Letẋ=f(x) andf(x )=0. If there exists ac 1 functionv: R n R such that (1) V(x )=0 (2) V(x)>0for allx =x (3) V(x) αv(x) for allx (4) V(x) as x thenx is globally exponentially stable.

30 Proof Idea Assumex(t) =0(otherwise we havex(τ)=0for all τ>t). Then V(x) V(x) α Integrating from0totgives logv(x(t)) logv(x(0)) αt V(x(t)) e αt V(x(0)) Hence,V(x(t)) 0,t. Using the properties ofv it follows thatx(t) 0,t.

31 Invariant Sets Definition: A setm is called invariant if for the system ẋ=f(x), x(0) M implies thatx(t) M for allt 0. x(0) x(t) M

32 LaSalle s Invariant Set Theorem Theorem Let Ω R n compact invariant set forẋ=f(x). LetV: Ω Rbe ac 1 function such that V(x) 0, x Ω, E:={x Ω: V(x)=0},M:=largest invariant subset ofe = x(0) Ω,x(t) approachesm ast + V(x) E M x Ω E M Note thatv must not be a positive definite function in this case.

33 Special Case: Global Stability of Equilibrium Theorem: Letẋ=f(x) andf(0)=0. If there exists ac 1 functionv: R n R such that (1) V(0)=0,V(x)>0for allx =0 (2) V(x) 0for allx (3) V(x) as x (4) The only solution ofẋ=f(x), V(x)=0isx(t)=0 t = x=0is globally asymptotically stable.

34 A Motivating Example (cont d) mẍ= bẋ ẋ k 0 x k 1 x 3 V(x)=(2mẋ 2 +2k 0 x 2 +k 1 x 4 )/4>0, V(0,0)=0 V(x)= b ẋ 3 Assume that there is a trajectory withẋ(t)=0,x(t) =0. Then d dtẋ(t)= k 0 m x(t) k 1 m x3 (t) =0, which means that ẋ(t) can not stay constant. Hence, V(x)=0 x(t) 0, and LaSalle s theorem gives global asymptotic stability.

35 Example Stable Limit Cycle Show thatm={x: x =1} is a asymptotically stable limit cycle for (almost globally, except for starting at x=0): ẋ 1 =x 1 x 2 x 1 (x 2 1 +x2 2 ) ẋ 2 =x 1 +x 2 x 2 (x 2 1 +x2 2 ) LetV(x)=(x 2 1 +x2 2 1)2. dv dt =2(x2 1 +x2 2 1)d dt (x2 1 +x2 2 1) = 2(x1 2 +x2 2 1)2 (x1 2 +x2 2 ) 0 forx Ω Ω={0< x R} is invariant forr=1.

36 Example Stable Limit Cycle E={x Ω: V(x)=0}={x: x =1} M=E is an invariant set, because d dt V= 2(x2 1 +x2 2 1)(x2 1 +x2 2 )=0 forx M We have shown thatm is a asymtotically stable limit cycle (globally stable inr {0})

37 A Motivating Example (revisited) mẍ= bẋ ẋ k 0 x k 1 x 3 V(x,ẋ)=(2mẋ 2 +2k 0 x 2 +k 1 x 4 )/4>0, V(0,0)=0 V(x,ẋ)= b ẋ 3 givese={(x,ẋ): ẋ=0}. Assume there exists( x, x) M such that x(t 0 ) =0. Then m x(t 0 )= k 0 x(t 0 ) k 1 x 3 (t 0 ) =0 so x(t 0 +) =0so the trajectory will immediately leavem. A contradiction to that M is invariant. Hence,M={(0,0)} so the origin is asymptotically stable.

38 Adaptive Noise Cancellation by Lyapunov Design u b s+a x + x b s+â x ẋ+ax=bu x+â x= bu Introduce x=x x, ã=a â, b=b b. Want to design adaptation law so that x 0

39 Let us try the Lyapunov function V= 1 2 ( x2 + γ a ã 2 + γ b b2 ) V= x x+ γ a ã ã+ γ b b b= 2 = x( a x ã x+ bu)+ γ a ã ã+ γ b b b= a x where the last equality follows if we choose ã= â= 1 γ a x x b= b= 1 γ b xu Invariant set: x=0. This proves that x 0. (The parametersãand b do not necessarily converge:u 0.) Demonstration if time permits

40 Results â time [s] ˆb time [s] Estimation of parameters starts at t=10 s.

41 Results x ˆx time [s] Estimation of parameters starts at t=10 s.

42 Next Lecture Stability analysis using input-output (frequency) methods