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Short notes on Lyapunov stability

by Sanand D“Sometimes I need only you can provide, your absence.”- Anonymous

1 Introduction

All the discussion here is borrowed predominantly from Haddad, [2], Sastry and Khalil. We considerthe following form for a general non linear dynamical system (continuous/discrete)

x(t) = f(t,x,u),x(t0) = x0, t ∈ [t0, t1]. (1)

x(t+ 1) = f(x(k),u(k)), k = 0, 1, . . . (2)

where f for continuous time systems may assumed to be continuous in t and x, x ∈ Rn, u ∈ Rm, xbeing the state of the system and u being the input. In the absence of inputs or if inputs are fixedto some specified value, then one obtains

x(t) = f(t,x),x(t0) = x0, t ∈ [t0, t1]. (3)

1.1 Basic definitions for stability and invariance

In the following definitions, properties are said to be true

• locally if they are true for all x ∈ Bε(0) for some ε.

• globally if they are true for all x ∈ Rn.

• uniformly if they are true for all t0 ≥ 0.

• semi globally if they are true for all x ∈ Bε(0) for an arbitrary ε.

We may replace s(t, t0,x0) with x(t) in all the following definitions.

Local stability definitions:

Definition 1.1 (stability). The equilibrium point 0 is said to be stable if, for every ε > 0, thereexists a δ = δ(ε, t0) such that ‖x0‖ < δ(ε, t0)⇒ ‖s(t, t0,x0)‖ < ε for all t ≥ t0.

Definition 1.2 (uniform stability). The equilibrium point 0 is said to be uniformly stable if, forevery ε > 0, there exists a δ = δ(ε) such that ‖x0‖ < δ(ε)⇒ ‖s(t, t0,x0)‖ < ε for all t ≥ t0.

An equilibrium point is unstable if it is not stable. For autonomous systems stability anduniformly stability is the same.

Definition 1.3 (attractivity). The equilibrium point 0 is attractive, if for each t0 ∈ R+, thereexists η(t0) > 0 such that ‖x0‖ < η(t0)⇒ s(t+ t0, t0,x0)→ 0 as t→∞. It is said to be uniformlyattractive if there exists η > 0 such that ‖x0‖ < η ⇒ s(t + t0, t0,x0) → 0 as t → ∞ uniformly int0,x0.

Definition 1.4 (asymptotic stability). The equilibrium point 0 is asymptotically stable if it isstable and attractive. It is uniformly asymptotically stable if it is uniformly stable and uniformlyattractive.

Definition 1.5 (exponential stability). The equilibrium point 0 is exponentially stable there existsconstants c, γ, ε such that ‖s(t0 + t, t0,x0)‖ ≤ c‖x0‖e−γt, ∀t, t0 ≥ 0, ∀x0 ∈ Bε(0). The constant γis called an estimate of the rate of convergence.

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All the definitions above are local since they are concerned with neighborhoods of the equilibriumpoint.Global stability definitions:

Definition 1.6 (Global asymptotic stability). The equilibrium point 0 is globally asymptoticallystable if it is stable and limt→∞s(t, t0,x0) = 0 for all x0 ∈ Rn.

Definition 1.7. The equilibrium point 0 is said to be globally uniformly asymptotically stable if it isuniformly stable and for each pair of positive numbers M, ε with M arbitrarily large and ε arbitrarilysmall, there exists a finite number T = T (M, ε) such that ‖x0‖ < M, t0 ≥ 0⇒ ‖s(t+ t0, t0,x0)‖ <ε, ∀t ≥ T (M, ε).

Definition 1.8. The equilibrium point 0 is said to be globally exponentially stable if there existsconstants c, γ such that ‖s(t0 + t, t0,x0)‖ ≤ c‖x0‖e−γt, ∀t, t0 ≥ 0, ∀x0 ∈ Rn.

For an equilibrium point to be either globally uniformly asymptotically stable or globally ex-ponentially stable, a necessary condition is that it is the only equilibrium point.

Energy-like functions:

Definition 1.9 (class K, KR, L, KL, KRL functions). A function φ : R+ → R+ is of classK if it is continuous, strictly increasing, and φ(0) = 0. It is said to belong to class KR if it is inclass K and in addition, φ(x)→∞ as x→∞.

It is of class L if it is continuous on [0,∞], strictly decreasing, φ(0) < ∞ and φ(x) → 0 asx→∞. A continuous function β : [0,∞)× [0,∞)→ [0,∞) is said to belong to class KL if, for afixed s, β(r, s) ∈K with respect to r, and for each fixed r, β(r, s) ∈L with respect to s. Similarly,we define class KRL functions.

In the definition of class K functions, if we change the domain from [0,∞) to [0, a), we getclass Ka functions. Then we can define class KaR, KaL, KaRL analogously.

Definition 1.10 (locally positive definite function (lpdf)). A continuous function V : Rn ×R+ →R+ is called locally positive definite if for some ε > 0 and some α(.) of class K functions,

V (0, t) = 0 and V (x, t) ≥ α(‖x‖) ∀x ∈ Bε, t ≥ 0. (4)

Definition 1.11 (positive definite function (pdf)). A continuous function V : Rn × R+ → R+ iscalled positive definite if (4) holds ∀x ∈ Rn. It is said to be radially unbounded if for some α(.) ofclass KR functions,

V (0, t) = 0 and V (x, t) ≥ α(‖x‖) ∀x ∈ Rn, t ≥ 0. (5)

Definition 1.12 (decrescent functions). A continuous function V : Rn × R+ → R+ is calleddecrescent if there exists some β(.) of class KR functions and an ε > 0, such that

V (x, t) ≤ β(‖x‖) ∀x ∈ Bε(0), t ≥ 0. (6)

Example 1.13. V (t,x) = (t+ 1)‖x‖2 is a pdf but not decrescent.

Derivative along a trajectory: Let V (x(t)) be a real valued function on the solution trajectoriesof (3). Then by chain rule, V = ∇V (x).x = ∇V (x).f . This is also called the Lie derivative of Valong f .

Theorem 1.14 (Basic equivalence). The equilibrium point 0 of (3) is

1. stable ⇔ for each t0 ∈ R+, there exists d(t0) > 0 and φt0 ∈K such that

‖s(t, t0,x0)‖ ≤ φt0(‖x0‖), ∀t ≥ t0, ∀x0 ∈ Bd(t0)(0).

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2. uniformly stable ⇔ there exists d > 0 and φ ∈K such that

‖s(t, t0,x0)‖ ≤ φ(‖x0‖), ∀t ≥ t0, ∀x0 ∈ Bd(0).

3. attactive ⇔ for each t0 ∈ R+, there exists r(t0) > 0, and for each x0 ∈ Br(t0)(0), there existsa function σt0,x0 ∈L such that

‖s(t, t0,x0)‖ ≤ σt0,x0(t), ∀t ≥ 0, ∀x0 ∈ Br(t0)(0).

4. uniformly attactive ⇔ there exists r > 0 and a function σt0,x0 ∈L such that

‖s(t, t0,x0)‖ ≤ σ(t), ∀t, t0 ≥ 0, ∀x0 ∈ Br(t0)(0).

5. asymptotically stable ⇔ there exists a number r(t0) > 0, a function φt0 ∈ K and for eachx0 ∈ Br(t0)(0), there exists a function σt0,x0 ∈L such that

‖s(t, t0,x0)‖ ≤ φt0(‖x0‖)σt0,x0(t), ∀t, t0 ≥ 0, ∀x0 ∈ Br(t0)(0).

6. i. uniformly asymptotically stable ⇔ there exists a number r > 0, a function φ ∈ K and afunction σ ∈L such that

‖s(t, t0,x0)‖ ≤ φ(‖x0‖)σ(t), ∀t, t0 ≥ 0, ∀x0 ∈ Br(0).

ii. uniformly asymptotically stable ⇔ there exists a class KL function β and a positiveconstant c independent of t0 such that

‖x(t)‖ ≤ β(‖x(t0)‖, t− t0), ∀t ≥ t0 ≥ 0, ∀‖x(t0)‖ ≤ c.

Proof. ([2] and Khalil) 1. (⇐) Given ε > 0, t0 ∈ R+, choose δ(ε, t0) = mind(t0), φ−1t0

(ε).(⇒) Fix t0 and ε > 0. There exists δ(t0, ε) > 0 such that

‖x(t0)‖ < δ ⇒ ‖x(t)‖ < ε, ∀t ≥ t0.

Let δ(t0, ε) be the supremum of all applicable δ. The function δ is positive and non decreasing, butneed not be continuous. Choose θt0(ε) ∈K such that θt0(ε) ≤ δ(ε) for all ε > 0 (need justificationfor existence of such function to be rigorous). Now choose φt0 = θ−1t0 .

2. For uniform stability, we apply similar arguments without any t0 dependence.The proof of remaining cases is based on similar arguments. Intuitively, the first case says that

the trajectories remain bounded where the bound is given by a level surface of a class K function.Same for the second case where the level surface is independent of the initial time. For the third case,the criteria says that the magnitude of the trajectories (“energy” inside the trajectory) decreases,in other words, it is bounded by a decreasing function which goes to zero at infinity. This meansthe magnitude of the trajectory goes to zero as t→∞ i.e., it approaches the origin. This explainsattractivity of the origin. For asymptotic stability, the trajectory must be bounded and the originmust be attractive too. This explains the fifth and the sixth (i) case. The case 6(ii) (Khalil) givesan alternate formulation of uniform asymptotic stability using class KL functions. The trajectoryis bounded depending up on the first argument of β and it decreases to zero since β → 0 as thesecond argument in β goes to infinity. Refer Khalil/[2] for rigorous arguments of all cases.

Definition 1.15 (Domain of attraction). Suppose that 0 is an asymptotically stable equilibriumpoint of x = f(x). Then the domain of attraction D0 is given by

D0 := x0 ∈ S ⊂ Rn | if x(0) = x0, then limt→∞x(t) = 0.

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Table 1: Basic Lyapunov stability theorems for autonomous systems (Theorem 2.1)

Conditions on V (x) Conditions on −V (x) Conclusion

V (x) > 0, x 6= 0 locally ≥ 0 locally stableV (x) > 0, x 6= 0 locally > 0, ∀x 6= 0 locally asymptotically stable

V (x) > 0, x 6= 0, V (x)→∞ as ‖x‖ → ∞ > 0, ∀x 6= 0 globally asymptotically stableα‖x‖p ≤ V (x) ≤ β‖x‖p locally ≥ εV (x) locally locally exponentially stableα‖x‖p ≤ V (x) ≤ β‖x‖p globally ≥ εV (x) globally globally exponentially stable

2 Stability theory for Autonomous/time invariant systems

stability, asymptotic stability, exponential stability, global asymptotic stability, global exponentialstability Lyapunov stability, invariant set stability, constructing Lyapunov functions, ConverseLyapunov theorems, instability theorems, linear systems and Lyapunov’s linearization

We consider non linear dynamical systems of the form

x = f(x(t)). (7)

2.1 Basic stability theorems using Lyapunov’s direct method

Theorem 2.1. Consider an autonomous system x = f(x(t)), x(0) = x0 where x(t) ∈ S ⊂ Rn. LetV : S → R. Let α, β, ε > 0 and p ≥ 1. Then, Table 1 holds.

Proof. Let ε > 0 such that Bε(0) ⊂ S. Observe that since ∂Bε(0) is compact and V (x) is continuous,V (∂Bε(0)) is compact hence, α = minx∈∂Bε(0)V (x) exists and α > 0 since 0 /∈ ∂Bε(0) and V (x) > 0for x ∈ S,x 6= 0. (Draw a picture of S and Bε inside S.)

Consider β ∈ (0, α) where α is the minimum value of V on ∂Bε(0). Let Ωβ = x ∈ S | V (x) ≤β. Then, Ωβ ⊂ Bε(0) \ ∂Bε(0). This is true because if there exists y ∈ Ωβ such that y ∈ ∂Bε(0),then V (y) ≥ α > β which is a contradiction. (Draw this “ellipsoid” inside Bε.)

Since V ≤ 0, V is non increasing and V (x(t)) ≤ V (x(0)) ≤ β. Therefore, x(t) ∈ Ωβ for allx(0) ∈ Ωβ. Consider Bδ(0) ⊂ Ωβ for some δ(ε) > 0. Therefore, for all x(0) ∈ Bδ(0), V (x(0)) < βand V (x(t)) ≤ V (x(0)) ⇒ x(t) ∈ Ωβ ⊂ Bε(0). Thus, x(t) ∈ Bε(0) ⇒ ‖x(t)‖ < ε. This provesLyapunov stability.

For the second case, V < 0. Therefore, V is decreasing and it is bounded from below by 0. Weneed to show that x(t)→ 0 as t→∞ i.e., for every ε > 0, there exists T > 0 such that ‖x(t)‖ < εfor all t > T . By previous arguments, for every ε > 0, we can choose β > 0 such that Ωβ ⊂ Bε.Therefore, it is enough to show that V (x(t))→ 0 as t→∞.

Since V is decreasing and it is bounded from below by 0, let V (x(t)) → c ≥ 0 as t → ∞.To show c = 0, we use contradiction. Suppose c > 0. By continuity of V , there exists d > 0such that Bd ⊂ Ωc. Since V (x(t)) → c ≥ 0 as t → ∞, x(t) lies outside Bd for all t ≥ 0. Let−γ =maxd≤‖x‖≤εV (x). Since V < 0,−γ < 0. Moreover,

V (x(t)) = V (x(0)) +

ˆ t

0V (x(τ))dτ ≤ V (x(0))− γt.

Observe that the rhs eventually becomes negative contracting the assumption that c > 0. Therefore,c = 0 which proves asymptotic stability.

The third case is exactly similar to the second case. For any x ∈ Rn, let a = V (x). Since‖x‖ → ∞ ⇒ V (x) → ∞, for any a > 0, there exists ε > 0 such that V (x) > a when ‖x‖ > ε.Therefore, Ωa ⊂ Bε(0) which implies that Ωa is bounded. Remaining arguments are exactly similaras to the second case.

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Since V ≤ −εV (x), V (x(t)) ≤ V (x(0))e−εt, t ≥ 0. By assumption, V (x(0)) ≤ β‖x(0)‖p andα‖x(t)‖p ≤ V (x(t)). Therefore,

α‖x(t)‖p ≤ β‖x(0)‖pe−εt, t ≥ 0⇒ ‖x(t)‖ ≤ (β

α)1p ‖x(0)‖e(

−εp)t, t ≥ 0

which proves local exponential stability.The last case follows similarly.

Example 2.2 (Simple pendulum). Consider a simple pendulum where the constants are chosensuch that θ + sin(θ) = 0. Choosing x1 = θ and x2 = θ, x1 = x2, x2 = − sin(x1). The totalenergy is 1

2 θ2 − cos(θ) = 1

2x22 − cos(x1). Consider V (x1, x2) = 1 + 1

2x22 − cos(x1)0. Since this is

continuously differentiable and lpdf, it is a Lyapunov function candidate. Moreover, V (x1, x2) =sin(x1)x1 + x2x2 = sin(x1)x2 + x2(− sinx1) = 0. Therefore, the equilibrium point is stable.

Consider a simple pendulum with damping i.e., θ + sin(θ) + bθ = 0 (Khalil). Hence, the stateequations are x1 = x2, x2 = − sin(x1) − bx2. Let V (x1, x2) be the same function as above. Thus,V = sin(x1)x1 + x2x2 = sin(x1)x2 + x2(− sinx1) − bx22 = −bx22 ≤ 0. Thus, the equilibriumpoint is stable for the damped pendulum. Consider a different Lyapunov function V (x1, x2) =xTPx − cos(x1) where P > 0. Need to choose P such that V < 0. Choose p11 = bp12, p12 =b/2, p22 = 1. Now, V = −1

2abx1 sin(x1) − 12bx

22 and x1 sin(x1) > 0 for −π < x1 < π. Thus,

choosing x ∈ R2 | − π < x1 < π as an open set around 0, V < 0 on this set ⇒ asymptoticstability.

Example 2.3 (Non linear series RLC circuit/Non linear mass spring damper). Non linear seriesRLC circuit (Sastry)/Mass-spring-damper. Suppose the inductor is linear but the resistor andcapacitor are non linear (or mass-spring- damper system with non linear spring and damping). Letx1 be the charge on the capacitor and x2 be the current through the inductor. Therefore,

x1 = x2, x2 = −f(x2)− g(x1) (8)

where f is a continuous function modelling resistor current-voltage characteristics and g modelscapacitor charge-voltage characteristics. Suppose both f, g model locally passive elements i.e., thereexists σ0 such that

σf(σ) ≥ 0, σg(σ) ≥ 0, ∀σ ∈ [−σ0, σ0].

A Lyapunov function candidate is the total energy of the system i.e.,

V (x) =x222

+

ˆ x1

0g(σ)dσ

where the first term represents the energy stored in the inductor (kinetic energy) and the secondterm represents the energy stored in the capacitor (potential energy). By passivity of g, V is lpdf.Moreover,

V (x) = x2(−f(x2)− g(x1)) + g(x1)x2 = −x2f(x2) ≤ 0 (9)

when |x2| < σ0. Thus, this shows local stability of the origin.

Example 2.4 (Rigid body and rotational motion). Consider a rotational motion of a rigid bodyin 3−D space. Let ω be the angular velocity and I be the Inertia matrix. Then, in the absence ofexternal torques, the motion is described by

Iω + ω × Iω = 0 (10)

where × dentoes cross product in R3. Do a change of basis such that I is a diagonal matrix. Letωx, ωy, ωz be the components of the angular velocity. Equation (10) reduces to

Ixωx = −(Iz − Iy)ωyωz, Iyωy = −(Ix − Iz)ωxωz, Izωz = −(Iy − Ix)ωxωx. (11)

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Without loss of generality assume that Ix ≥ Iy ≥ Iz > 0 and replace ωx, ωy, ωz by x, y, z. Define

a =Iy−IzIx

, b = Ix−IzIy

, c =Ix−IyIz

. Therefore, (11) becomes

x = ayz, y = −bxz, z = cxy.

For simplicity, suppose Ix > Iy > Iz ⇒ a, b, c > 0. For equilibrium, at least two quantities amongx, y, z must be zero. Thus, the equilibria are union of the three axes. Therefore, none of theequilibria is isolated. Consider the origin as an equilibrium point. Define a Lyapunov functioncandidate V (x, y, z) = px2 + qy2 + rz2 where p, q, r > 0. Then V is an lpdf.

V = 2(pxx+ qyy + rzz) = 2xyz(ap− bq + cr).

Choose p, q, r such that ap− bq + cr = 0. Hence, the origin is stable.

Example 2.5 (Non linear RC circuit). non linear resistive and linear capacitive circuit. Considera bank of capacitors (linear characteristics) with capacitances C1, . . . , Cn connected to a bank ofresistors (non linear). Let x = [x1 · · · , xn] denote the voltages accross each capacitor. Then thecurrent through the capacitors is Cx where C =diag(C1, . . . , Cn). Let i(x) denote the current vectorin the resistive network when the voltage vector x is applied accross its terminals such that i(0) = 0.Let i(x) = G(x)x where G(.) is the non linear version of the conductance matrix. Therefore,

Cx = G(x)x⇒ x = C−1G(x)x. (12)

Clearly, 0 is an equilibrium point. Consider the total energy stored in the capacitors i.e., V (x) =12xTCx. Therefore,

V (x) =1

2(xTCx + xTCx) = −1

2xT[GT(x) +G(x)]x.

Let M(x) := GT(x)+G(x). If M(x) is lpdf, then 0 is asymptotically stable. Since M is continuous,M is lpdf if M(0) > 0.

Suppose M(0) > 0. Therefore, by continuity, there exists ε > 0 such that M(x) > 0 for allx ∈ Bε(0). Let d := infx∈Bε(0)λminM(x) and choose ε small enough such that d > 0. Then,

V = −xTM(x)x ≤ −d‖x‖2, ∀x ∈ Bε(0). Since, V (x) = 12xTCx,

λmax(C)‖x‖2 ≥ V (x)⇒ −d‖x‖2 ≤ −dλmax(C)

V (x).

Therefore, V ≤ −dλmax(C)V (x) and from Theorem 2.1 case 4, 0 is locally exponentially stable. If

d = infx∈RnλminM(x) > 0, then 0 is globally exponentially stable. It is also globally asymptoticallystable since, V < 0.

Example 2.6. not radially unbounded and finite escape time: Ex. on p.174 of [2]. It showsthat if V is not radially unbounded and satisfies all the remaining properties, then trajectoriesstarting sufficiently far from the origin may have a finite escape time and the origin is not globallyexponentially stable.

Example 2.7 (Phase locked loop). [2] p.181.

2.2 Invariant sets and stability theorems

Lemma 2.8 (limit sets). If a solution x(t) of x = f(x) is bounded and belongs to S ⊂ Rn for t ≥ 0,then its limit set L is nonempty, compact and invariant. Moreover, x(t)→ L as t→∞.

Proof. The proof is slightly technical and we refer the reader to Haddad Theorem 2.41.

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Relaxing strict negative definite condition on V :

Theorem 2.9 (LaSalle’s invariance principle (Barbashin-Krasovskii-LaSalle Theorem)). Let Ω ⊂ Sbe a compact invariant set of x = f(x). Let V : S → R be a continuously differentiable functionsuch that V ≤ 0 in Ω. Let E be the set of points where V (x) = 0. Let M be the largest invariantset in E. If x(0) ∈ Ω, then x(t)→M as t→∞.

Proof. Let x(t) be a solution of x = f(x) starting in Ω. Since V ≤ 0 in Ω, V is a decreasingfunction of t. Furthermore, since V is continuous on the compact set Ω, it is bounded from belowon Ω. Therefore, limt→∞V (x(t)) = a exists. The limit set L of the trajectory x(t) lies in Ω sinceΩ is invariant and closed. For any p ∈ L, there exists a sequence tn such that tn → ∞ andx(tn) → p as n → ∞. By continuity of V (x), V (p) = limn→∞V (x(tn)) = a. Hence, V (x) = a onL. By the previous lemma, L is an invariant set and since V is a constant function on L, V = 0on L. Therefore, L ⊂ M ⊂ E ⊂ Ω. Since x(t) is bounded, x(t) approaches L as t → ∞. Hence,x(t)→M as t→∞.

Observe that there is no positiveness assumption on V in the above theorem. However, invari-ance of Ω is assumed an the initial condition is assumed to belong to this invariant set. One canconstruct invariant sets using level sets of V (x) > 0 with V ≤ 0. So one can observe that one canhave any of these two hypotheses to obtain invariance.

Corollary 2.10 (asymptotic stability using LaSalle). Let x = 0 be an equilibrium point of x = f(x).Let V : S → R be a continuously differentiable positive definite function on S (i.e., V is lpdf)containing x = 0 such that V ≤ 0 on S. Let E be the set of points where V (x) = 0 and supposethat no solution can stay identically in E other than the trivial solution x = 0. Then, the origin isasymptotically stable.

Proof. Follows as a consequence of the previous theorem.

Corollary 2.11 (global asymptotic stability using LaSalle). Let x = 0 be an equilibrium point ofx = f(x). Let V : Rn → R be a continuously differentiable radially unbounded positive definitefunction on Rn \ 0 such that V ≤ 0 on Rn. Let E be the set of points where V (x) = 0 andsuppose that no solution can stay identically in E other than the trivial solution x = 0. Then, theorigin is globally asymptotically stable.

Proof. Again follows from the theorem.

Example 2.12. Consider the case of damped simple pendulum in Example 2.2. Since V = −bx22,V = 0 on the line x2 = 0. To maintain V = 0, the trajectory must be confined to the line x2 = 0.Now x2 = 0⇒ x2 = 0⇒ sin(x1) = 0⇒ x1 = nπ for n = 0,±1,±2, . . . . Therefore, on the segment−π < x1 < π, the largest invariant set is the origin and this implies asymptotic stability by LaSalle.

Example 2.13. Non linear RLC circuit (Example 2.3), asymptotic stability of the origin usingLaSalle: Recall from Example 2.3 that V (x1, x2) = −x2f(x2) ≤ 0 for x2 ∈ [−σ0, σ0]. Let c =min(V (−σ0, 0), V (σ0, 0)). Then, V ≤ 0 for x ∈ Ωc := (x1, x2) | V (x1, x2) ≤ c. By LaSalle’s in-variance principle, the trajectory approaches the largest invariant set in Ωc∩(x1, x2), | V (x1, x2) =0 = Ωc ∩ x1, 0 (since f(x2) = 0 only when x2 = 0). Note that x2 = 0 in the largest invariantset which implies that x1 = 0 ⇒ x1 = x10. Furthermore x2 = 0 ⇒ x2 = 0 = −f(0) − g(x10) ⇒g(x10) = 0 ⇒ x10 = 0 (since, g(x1) = 0 only when x1 = 0). Therefore, the origin is the largestinvariant set in Ωc ∩ (x1, x2), | V (x1, x2) = 0. Hence, it is locally asymptotically stable.

Example 2.14 (Stabilization of a rigid robot). [2] p.183. Consider a rigid robot where componentsof q represent generalized co-ordinates of the robot. Let u represent generalized forces. Suppose thesystem of equations are

M(q)q + C(q, q)q = u

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Let x = q and y = q. Therefore, x = y and y = [M(x)]−1[u − C(x,y)y]. Let qd represent thedesired value of the generalized co-ordinates. Suppose

u = −Kp(q− qd)−Kdq = −Kp(x− qd)−Kdy,

where Kp,Kd > 0. Now the system equations are

x = y, y = −[M(x)]−1[Kp(x− qd) +Kdy + C(x,y)y]

The equilibrium point is (x,y) = (qd,0). Consider a Lyapunov function

V (x,y) =1

2[yTM(x)y + (x− qd)

TKp(x− qd)]

⇒ V = yTM(x)y +1

2yTM(x)y + xTKp(x− qd)

= −yT[Kp(x− qd) +Kdy + C(x,y)y] +1

2yTM(x)y + yTKp(x− qd)

= −yTKdy +1

2yT[M(x,y)− 2C(x,y)]y

It turns out that M − 2C is skew symmetric. Therefore, V ≤ 0 which implies stability. ApplyingKrasovskii-LaSalle principle, for the set of points where V = 0 i.e., to points (x,y) ∈ Rn × 0, thelargest invariant set is given by the set of points in Rn×0 where (x, y) = (0,0). Now x = 0 impliesthat y = 0 and y = 0 implies [M(x)]−1[Kp(x − qd) + Kdy + C(x,y)y] = 0. Since y = 0 and Mis invertible, this implies that Kp(x− qd) = 0 and since Kp > 0, x = qd. Therefore, (qd,0) formsthe largest invariant set and by LaSalle (Corollary 2.11), it is globally asymptotically stable.

Example 2.15 (Stability of a limit cycle using the invariance principle). Consider a system

x1 = x2 + x1(r2 − x21 − x22)

x2 = −x1 + x2(r2 − x21 − x22), r > 0

which has an equilibrium point at the origin. Observe that if x21 + x22 = r2, then, x1 = x2 andx2 = −x1 and the evolution remains on the circle x21 + x22 = r2 and it forms an invariant set. LetV (x) = 1

4(x21 + x22 − r2)2. Thus, V (x) ≥ 0 on R2. Note that V (x) = ∇V.f = −(x21 + x22)(x21 +

x22 − r2)2 ≤ 0. Note that V = 0 either at the origin or on the circle x21 + x22 = r2. Let c > r andMc := x ∈ R2 | V (x) ≤ c. Since V ≤ 0 on Mc and V ≥ 0, Mc is an invariant set containing theorigin and the circle x21 +x22 = r2. On both these sets, V = 0 and the largest invariant set in Mc isthe union of the origin with the circle (they are not comparable as neither is a subset of the other).Therefore, all trajectories converge to either one of them by the invariance principle.

Observe that at the origin, V (0) = r4

4 . Suppose r < r4

4 and let r < c < r4

4 . Let Mc be asdefined before. Note that Mc excludes the origin in this case but contains the circle x21 + x22 = r2.Now applying the invariance principle, all trajectories converge to the circle implying a stable limitcycle.

2.3 Constructing Lyapunov functions

Observe that by chain rule, V =∑n

i=1∂V∂xixi =

∑ni=1

∂V∂xifi(x). Let g(x) = ∂V

∂x . Therefore, V =g(x).f(x). We need to construct g such that it is a gradient of a positive definite function andV (x) = g(x).f(x) < 0. Note that

V (x) =

ˆ x

0g(s)ds =

ˆ x

0

n∑i=1

gi(s)dsi. (13)

8

Since line integral of a gradient is path independent, take a path from 0 to x formed by linesparallel to the co-ordinate axes. Therefore, (13) becomes

V (x) =

ˆ x1

0g1(s1, 0, . . . , 0)ds1 +

ˆ x2

0g2(x1, s2, 0, . . . , 0)ds2 + . . .+

ˆ xn

0gn(x1, x2, . . . , xn−1, sn)dsn.(14)

Alternatively, choosing a straight line path joining x to 0 with parametrization s = σx, whereσ ∈ [0, 1], (13) can also be written as

V (x) =

ˆ 1

0g(σx).xdσ =

ˆ 1

0

n∑i=1

gi(σx)xidσ. (15)

Proposition 2.16. A function g : Rn → Rn is the gradient vector of a scalar valued functionV : Rn → R if and only if ∂gi

∂xj=

∂gj∂xi, i, j = 1, . . . , n.

Proof. (⇒) trivial. (⇐) Suppose ∂gi∂xj

=∂gj∂xi, i, j = 1, . . . , n. Let

V (x) =

ˆ 1

0g(σx)xdσ =

ˆ 1

0

n∑j=1

gj(σx)xjdσ (16)

Let y = σx, therefore,

∂V

∂xi=

ˆ 1

0

n∑j=1

∂gj∂xi

(σx)xjdσ +

ˆ 1

0gi(σx)dσ

=

ˆ 1

0

n∑j=1

∂gi∂xj

(σx)xjdσ +

ˆ 1

0gi(σx)dσ

=

ˆ 1

0

n∑j=1

∂gi∂yj

(y)σxjdσ +

ˆ 1

0gi(σx)dσ

=

ˆ 1

0

d(σgi(σx))

dσdσ (17)

= gi(x), i = 1, . . . , n

Proposition 2.17. Let f ,g : Rn → Rn be continuously differentiable functions such that f(0) = 0.Then for every x ∈ Rn, there exists α ∈ [0, 1] such that

g(x).f(x) = gT(x)∂f

∂x(αx)x. (18)

Proof. (MVT for vector valued functions (Theorem 2.16 of Haddad): Let D ⊂ Rm and f : D → Rnwhich is continuously differentiable. Let x,y ∈ D such that L := z | z = µx + (1 − µ)y, µ ∈(0, 1) ⊂ D. Then, for every v ∈ Rn, there exists z ∈ L such that vT[f(y)− f(x)] = vT[f ′(z)(y −x)].) Let p ∈ Rn. Then by mean value theorem, for every x ∈ Rn, there exists α ∈ [0, 1] such that

gT(p)f(x) = gT(p)[f(x)− f(0)]

= gT(p)[∂f

∂x(αx)x].

Hence, there exists α ∈ [0, 1] such that g(p).f(p) = gT(p) ∂f∂p(αp)p. Since p is arbitrary, the resultfollows.

9

Theorem 2.18 (Krasovskii’s theorem). Let 0 be an equilibrium point of x = f(x(t)), x(0) = x0,t ≥ 0 where f : S → Rn is continuously differentiable and S is an open set containing 0. LetP,R ∈ Rn×n, P,R > 0 such that

[∂f(x)

∂x]TP + P [

∂f(x)

∂x] ≤ −R, x ∈ S, x 6= 0. (19)

Then the zero solution is a unique asymptotically stable equilibrium point with Lyapunov functionV (x) = fT(x)P f(x). If S = Rn, then the zero solution is a unique globally asymptotically stableequilibrium.

Proof. Suppose there exists xe ∈ S, xe 6= 0 and f(xe) = 0. By the previous proposition, for every

xe ∈ S, there exists α ∈ [0, 1] such that 0 = xTe P f(xe) = xT

e P∂f(αxe)∂x xe. Hence,

xTe [

∂f(αxe)

∂x]TP + P [

∂f(αxe)

∂x]xe = 0

which contradicts (19). Therefore, S does not contain any other equilibrium point of f . Note thatV (x) = fT(x)P f(x) ≥ λmin(P )‖f(x)‖22 ≥ 0, x ∈ S. This implies that V (x) = 0 if and only iff(x) = 0 i.e., x = 0. Therefore, V > 0 on S \ 0. Moreover,

V = V ′(x)f(x) = 2fT(x)P∂f(x)

∂xf(x) = fT(x)[∂f(x)

∂x]TP + P [

∂f(x)

∂x]f(x)

≤ −fT(x)Rf(x) ≤ −λmin(R)‖f(x)‖22 ≤ 0, x ∈ S.

Since f(x) = 0 iff x = 0 in S, V < 0 is S \ 0 which proves that 0 is asymptotically stableequilibrium point with Lyapunov function V (x) = fT(x)P f(x).

Now suppose S = Rn. We need to show that V (x) → ∞ as ‖x‖ → ∞. By the previousproposition, for every x ∈ Rn and some α ∈ (0, 1),

xTP f(x) = xTP∂f(x)

∂x(αx)x =

1

2xT[∂f(αx)

∂x]TP + P [

∂f(αx)

∂x]x

≤ −1

2xTRx ≤ −1

2λmin(R)‖x‖2. (20)

This implies that xTP f(x) < 0 since rhs above is negative. Therefore, |xTP f(x)| ≥ 12λmin(R)‖x‖2.

Moreover, |xTP f(x)| ≤ λmax(P )‖x‖‖f(x)‖ ⇒ ‖f(x)‖ ≥ |xTP f(x)|λmax(P )‖x‖ ≥

λmin(R)2λmax(P )‖x‖. This implies by

definition of V that V (x)→∞ as ‖x‖ → ∞.

Example 2.19. Example 3.9 Haddad.

The following theorem allows us to find the domain of attraction as well.

Theorem 2.20 (Zubov’s theorem). Let 0 be an equilibrium point of x = f(x(t)). Let S ⊂ Rn bebounded and suppose there exists a continuously differentiable function V : S → R and a continuousfunction h : Rn → R such that V (0) = 0, h(0) = 0 and

0 < V (x) < 1, x ∈ S, x 6= 0 (21)

V (x)→ 1 asx→ ∂S (22)

h(x) > 0, x ∈ Rn, x 6= 0 (23)

V ′(x)f(x) = −h(x)[1− V (x)]. (24)

Then, 0 is asymptotically stable with domain of attraction S.

10

Proof. It follows from (21), (23) and (24) that in the neighborhood Bε(0) of the origin, V (x) > 0and V < 0. Hence, 0 is locally asymptotically stable. To show that S is the domain of attraction,we need to show that x(0) ∈ S implies that x(t)→ 0 as t→∞ and x(0) /∈ S implies that x(t) 6→ 0as t→∞.

Let x(0) ∈ S. Then from (21), V (x(0)) < 1. Let β > 0 be such that V (x(0)) ≤ β < 1 anddefine Ωβ := x ∈ S | V (x) ≤ β. Since Ωβ ⊂ S, Ωβ is bounded. Since V < 0 on Ωβ, it is clearthat Ωβ forms an invariant set. If V = 0, then h(x) = 0 which further implies that x = 0. Now byTheorem 2.9, x(t)→ 0 as t→∞.

Let x(0) /∈ S and suppose x(t)→ 0 as t→∞. Thus, x(t)→ S for some t ≥ 0. Therefore, thereexists t1, t2 such that x(t1) ∈ ∂S and x(t) ∈ S for all t ∈ (t1, t2]. Let W (x) = 1 − V (x). Hence,W = h(x)W (x). Therefore,

ˆ W (x(t))

W (x(t0))

dW

W=

ˆ t

t0

h(x(s))⇒ 1− V (x(t0)) = [1− V (x(t))]e−´ tt0h(x(s))ds

. (25)

Let t = t2 and t0 → t1. Using (22), it follows that limt0→t1 [1 − V (x(t0))] = 0 and limt0→t1 [1 −V (x(t))]e

−´ tt0h(x(s))ds

> 0 which is a contradiction. Therefore, for x(0) /∈ S, x(t) 6→ 0 as t→∞.

Example 2.21. Haddad: Example 3.10

Constants of motion or first integrals or integrals of motion or dynamic invariants or Casimirfunctions. A function C : S → R is an integral of motion if it is conserved along the flow of thedynamical system i.e., C ′(x).f(x) = 0. Let Ci : S → R i = 1, . . . , r be twice differentiable Casimirfunctions. Define

E(x) :=r∑i=1

µiCi(x) (26)

for µi ∈ R, i = 1, . . . , r.

Theorem 2.22 (Energy-Casimir theorem). Consider a non linear dynamical system x = f(x)where f : S → Rn is Lipschitz. Let xe be an equilibrium point and let Ci : S → R, i = 1, . . . , r beCasimir functions of the dynamical system. Suppose C ′i(xe) i = 2, . . . , r are linearly independentand suppose there exists µ = [µ1, . . . , µr]

T ∈ Rr such that µ1 6= 0, E′(xe) = 0 and xTE′′(xe)x > 0for x ∈M where M = x ∈ S | C ′i(xe)x = 0, i = 2, . . . , r. Then, there exists α ≥ 0 such that

E′′(xe) + αr∑i=2

(∂Ci(xe)

∂x)T(

∂Ci(xe)

∂x) > 0. (27)

Furthermore, the equilibrium point xe is Lyapunov stable with Lyapunov function

V (x) = E(x)− E(xe) +α

2

r∑i=2

[Ci(x)− Ci(xe)]2. (28)

Proof. Note that since Ci are Casimir functions, C ′i(x).f(x) = 0. Moreover,

V (x) = V ′(x).f(x)

= E′(x).f(x) + α

r∑i=2

[Ci(x)− Ci(xe)]C ′i(x).f(x)

=

r∑i=1

µiC′i(x).f(x) + α

r∑i=2

[Ci(x)− Ci(xe)]C ′i(x).f(x) = 0 (29)

11

for x ∈ S. We need to show that V (xe) = 0 and V (x) > 0, x ∈ S \ xe. Clearly, V (xe) = 0 and

V ′(x) = E′(x) + αr∑i=2

[Ci(x)− Ci(xe)]C ′i(x)⇒ V ′(xe) = 0.

V ′′(x) = E′′(x) + αr∑i=2

[C ′i(x)]TC ′i(x) + [Ci(x)− Ci(xe)]C ′′i (x)

⇒ V ′′(xe) = E′′(xe) + αr∑i=2

(∂Ci(xe)

∂x)T(

∂Ci(xe)

∂x). (30)

Recall that by hypothesis, if x ∈ M, then C ′i(xe).x = 0, i = 2, . . . , r. Consider a subspace V1formed by linearly independent elements of M which is a r − 1 dimensional subpace and extendthe basis of this subspace to a basis of Rn by taking linearly independent vectors in the orthogonalcomplement V ⊥1 . With respect to the new basis, elements of V1 can be written as [∗ 0]T andelements of V ⊥1 can be written as [0 ∗]T. Therefore, there exists an invertible matrix T ∈ Rn×nsuch that [0 Ir−1]Tx = 0 when x ∈M and [In−r−1 0]Tx = 0 when x ∈M⊥. Hence, for xe ∈ S,

E′′(xe) = TT

[E1 E12

ET12 E2

]T (31)

where E1 > 0 and

r∑i=2

(∂Ci(xe)

∂x)T(

∂Ci(xe)

∂x) = TT

[0 00 N

]T (32)

where N > 0. Substituting (31) and (32) in (30),

V ′′(x) = TT

[E1 E12

ET12 E2 + αN

]T.

By choosing appropriate α, V ′′(xe) > 0. Since V is twice differentiable and V (xe) = V ′(xe) = 0, itfollows from V ′′(xe) > 0 that V > 0 in the neighborhood of xe.

Thus, using integrals of motions, one can construct a Lyapunov function.

Example 2.23. Example 3.11 Haddad.

2.4 Instability theorems

Theorem 2.24 (instability theorem). Let 0 be an equilibrium point of x = f(x) and let V : S → Rbe a continuously differentiable function such that V (0) = 0 and V (x0) > 0 for an x0 with arbitrarilysmall ‖x0‖. Let r > 0 and U = x ∈ Br(0) | V (x) > 0 such that V > 0 on U . Then, x = 0 isunstable.

Proof. By continuity of V if V (x0) > 0, then V > 0 in a neighborhood of x0 hence, x0 lies in theinterior of U . Let V (x0) = a > 0. The trajectory starting at x0 must eventually leave U . Onejustifies this claim as follows. As long as x(t) is inside U , V (x(t)) ≥ a since, V > 0 in U . Let

γ = minV (x) | x ∈ U, V (x) ≥ a

which exists since, the continuous function V has a minimum over the compact set x ∈ U | V (x) ≥a. Clearly, γ > 0. Note that

V (x(t)) = V (x0) +

ˆ t

0V (x(s))ds ≥ a+ γt.

12

This inequality shows that x(t) can not stay in U forever because V (x) is bounded on U . Moreover,x(t) can not leave through the surface x | V (x) = 0 since V (x(t)) ≥ a. Hence, it must leaveU through the sphere ‖x‖ = r. Beause this can happen for arbitrarily small ‖x0‖, the origin isunstable.

Theorem 2.25 (second instability theorem). Let 0 be an equilibrium point of x = f(x) and letV : S → R be a continuously differentiable function, W : S → R and λ, ε > 0 such that V (0) = 0and W (x) ≥ 0 for all x ∈ Bε(0) and V = V ′(x).f(x) = λV (x) +W (x). Furthermore, assume thatfor every sufficiently small δ > 0, there exists x0 ∈ S such that ‖x0‖ < δ and V (x0) > 0. Then,x = 0 is unstable.

Proof. Suppose there exists δ > 0 such that if x0 ∈ Bδ(0), then x(t) ∈ Bε(0), t ≥ 0. Note thatV (x0) > 0. From the hypothesis, V = λV (x)+W (x) and W (x) ≥ 0. This implies that V ≥ λV (x)for x ∈ Bε(0) i.e., V − λV (x) ≥ 0 for x ∈ Bε(0). Therefore,

e−λtV ′(x(t)).f(x(t))− λe−λtV (x(t)) ≥ 0, t ≥ 0

⇒ d

dt[e−λtV (x(t))] ≥ 0, t ≥ 0.

Integrating both sides, e−λtV (x(t))− V (x(0)) ≥ 0⇒ V (x(t)) ≥ eλtV (x(0)), t ≥ 0. Since V (x0) >0, x(t) /∈ Bε(0) as t→∞ which is a contradiction. Hence, the zero solution is unstable.

Theorem 2.26 (Chetaev’s instability theorem). Let 0 be an equilibrium point of x = f(x) and letV : S → R be a continuously differentiable function. Let ε > 0 and an open set O ⊂ Bε(0) suchthat

V (x) > 0 ∀x ∈ O, supx∈OV (x) <∞, 0 ∈ ∂O (33)

V (x) = 0 ∀x ∈ ∂O ∩ Bε(0), (34)

V ′(x).f(x) > 0, x ∈ O. (35)

Then, the zero solution is unstable.

Proof. Let P ⊂ O be a closed set such that for x0 ∈ O, x(t) ∈ P ⊂ O ⊂ Bε(0) for t ≥ 0. Note that

V (x(t)) = V (x(0)) +

ˆ t

0V (x(s))ds

= V (x(0)) +

ˆ t

0V ′(x(s)).f(x(s))ds ≥ V (x(0)) + αt (36)

where α = minx∈PV′(x).f(x) > 0. This implies that V (x(t))→∞ as t→∞ contradicting one of

the hypothesis supx∈OV (x) < ∞. Therefore, there exists T > 0 such that x(T ) ∈ ∂O (i.e., eitherx(t) escapes O at some finite time) or x(t) → ∂O as t → ∞ (i.e., x(t) escapes O at infinity). Oneof these two cases must be true since we have shown by contradiction that there is no closed set inO that contains limt→∞x(t).

Consider the first case i.e., suppose there exists T > 0 such that x(T ) ∈ ∂O. Since V isstrictly increasing on O, V (x(T )) > 0. By hypothesis, V (x) = 0 on ∂O ∩ Bε(0). This implies thatx(T ) /∈ ∂O ∩ Bε(0) hence, x(T ) ∈ ∂O \ Bε(0). Observe that ∂O = O ∩ ∂O and O ⊂ Bε(0). Thisimplies that

∂O \ Bε(0) = (O ∩ ∂O) \ Bε(0) ⊂ (Bε(0) ∩ ∂O) \ Bε(0) = ∂O ∩ ∂Bε(0).

Therefore, x(T ) ∈ ∂Bε(0). Similarly for the second case, if x(t) → ∂O as t → ∞, then x(t) →∂Bε(0) as t → ∞. Thus, there does not exists a δ > 0 such that if x0 ∈ Bδ(0), then x(t) ∈ Bε(0)for t ≥ 0. This implies that the zero solution is unstable.

13

Thus, it can be observed that Chetaev’s instability theorem requires milder conditions (milderin the sense that the properties need to be satisfied on a subset O ⊂ S and not on the whole of S)than Lyapunov instability theorems to conclude about instability of an equilibrium point.

Example 2.27.

2.5 Linear autonomous/LTI systems, linearization

Consider an autonomous linear systemx = Ax. (37)

Theorem 2.28 (Stability for LTI). The zero solution of (37) is Lyapunov stable if and only ifevery eigenvalue of A has a real part strictly less than zero or equal to zero and eigenvalues withzero real part have trivial Jordan structure (i.e., they are semi-simple).

Proof. Note that eAt is bounded iff the condition on the eigenvalues of A mentioned above issatisfied.

Theorem 2.29 (Lyapunov asymptotic stability for LTI). Following are equivalent.

1. The system (37) is asymptotically stable.

2. The system (37) is exponentially stable.

3. All eigenvalues of A have strictly negative real parts.

4. For every Q > 0, there exists a unique solution P > 0 to the following Lyapunov equation

ATP + PA = −Q. (38)

5. There exists P > 0 which satisfies the following Lyapunov matrix inequality

ATP + PA < 0. (39)

6. There exists C ∈ Rm×n such that the pair (C,A) is observable, and there exists P > 0 whichsatisfies

ATP + PA+ CTC = 0. (40)

7. For all C ∈ Rm×n such that the pair (C,A) is observable, and there exists P > 0 whichsatisfies (40).

Proof. The equivalence of first five statements is shown in Hespanha. Equivalence of the remainingtwo statements with the remaining onces is given in Sastry. (7) ⇒ (6). To show (6) ⇒ (1),let V (x) = xTPx and −V = xTCTCx. Then, by LaSalle’s invariance principle, the trajectoriesstarting from arbitrary initial conditions converge to the largest invariant set in the null space ofC. Since (C,A) is observable, largest invariant set in the null space of C is the origin. Therefore,the origin is asymptotically stable.

(1) ⇒ (7) Just as in (2) ⇒ (4), let P =´∞0 eA

TtCTCeAtdt. Clearly, P ≥ 0. To show thatP > 0, let xTPx = 0. This implies that CeAtx = 0. Differentiating n times at the origin, we obtainCx = CAx = . . . = CAn−1x = 0. Since (C,A) is observable, x = 0 ⇒ P > 0 and P also satisfies(40).

Theorem 2.30 (Lyapunov stability for LTI). The zero solution of (37) is Lyapunov stable ⇔ thereexists P > 0 and Q ≥ 0 such that (38) holds.

14

Proof. (⇐) Follows by choosing V (x) = xTPx. (⇒) Suppose the zero solution is stable. FromTheorem 2.28, eigenvalues of A lie in the left half complex plane with those having zero real partsbeing semi-simple. Without loss of generality, suppose A is in Jordan canonical form with first rJordan blocks having eigenvalues strictly in the left half plane and the remaining ones with zero

real parts and semi-simple. Let A =

[Jl 00 J0

]. It is clear that J0 is skew symmetric. Since

Jl has eigenvalues in the LHP, there exists P1 > 0, Q1 > 0 such that JTl P1 + P1Jl = Q1. Let

P =

[P1 00 I

]and Q =

[Q1 00 0

]such that (38) holds.

Remark 2.31. In the above theorem, if C =√Q and (C,A) is observable, then the origin is

asymptotically stable.

2.6 Indirect method of Lyapunov and local linearization

These methods work only locally i.e., they tell about local stability of an equilibrium point. Noconclusion can be made about global stability.

Theorem 2.32 (Lyapunov’s indirect theorem for autonomous systems using linearization). Letzero be an equilibrium point of (7) where f : S → Rn is continuously differentiable and S is openwith 0 ∈ S. Let A = ∂f

∂x |x=0. Then following holds:

1. (Local exponential stability of linearization) If Re(λ) < 0 where λ ∈ spec(A), then the zerosolution is exponentially stable.

2. (Unstability of linearization) If there exists λ ∈ spec(A) such that Re(λ) > 0, then the zerosolution is unstable.

Proof. Suppose Re(λ) < 0, λ ∈ spec(A). Therefore, there exists a unique P > 0 which satisfies(38). Let V (x) = xTPx. Note that using the Taylor expansion of f around 0,

V = V ′.f = 2xTP [Ax + r(x)]

= xT(ATP + PA)x + 2xTPr(x)

= −xTQx + 2xTPr(x). (41)

Using Cauchy-Schwarz inequality and using −xTQx ≤ −λmin(Q)‖x‖22,

V ≤ −λmin(Q)‖x‖22 + 2λmax(P )‖x‖2‖r(x)‖2. (42)

Since ‖r(x)‖2 goes to zero faster than ‖x‖, for every γ > 0, there exists ε > 0 such that ‖r(x)‖2 <γ‖x‖2. Therefore, for all x ∈ Bε(0),

V < −[λmin(Q)− 2γλmax(P )]‖x‖22. (43)

Choosing γ ≤ λmin(Q)/2λmax(P ), it follows that V (x) < 0 for all x ∈ Bε(0) \ 0. This showslocal asymptotic stability. Local exponential stability follows from the inequality λmin(P )‖x‖22 ≤V (x) ≤ λmax(P )‖x‖22.

For the proof of the second statement we refer the reader to Haddad p. 178.

For the converse of the first statement, refer Hespanha

Example 2.33. Consider damped simple pendulum from Example 2.2. Linearizing around theorigin, it can be checked that the linearization matrix has eigenvalues in the left half plane whichimplies asymptotic stability. For a pendulum without damping, eigenvalues of the linearizationmatrix are purely imaginary. Therefore, we can not conclude about stability by linearization. Butwe have already checked stability for this case using Lyapunov function.

Consider another equilibrium point x1 = π, x2 = 0. It can be checked that the linearizationmatrix has an eigenvalue in the right half plane which implies unstable equilibrium point.

15

Example 2.34 (Van der Pol oscillator). Consider the Van der Pol oscillator ([2]) described by

x1 = x2, x2 = −µ(1− x21)x2 − x1.

Linearize around 0. It turns out that if µ > 0, then both eigenvalues of the linearized matrix havepositive real parts which implies instability.

Theorem 2.35 (Lyapunov’s indirect theorem for control systems using linearization). Consider anon linear control system x = F(x,u) such that F(0,0) = 0 and F is continuously differentiable. LetA := ∂F

∂x |(x,u)=(0,0), B := ∂F∂u |(x,u)=(0,0) such that (A,B) is stabilizable i.e., there exists K ∈ Rm×n

such that spec(A + BK) lies in the left half complex plane. With the linear control law u = Kx,the zero solution of the closed loop non linear system x = F(x,u) is locally exponentially stable.

Proof. The closed loop system x = F(x,Kx) = f(x) and

∂f

∂x|x=0 = [

∂F(x,Kx)

∂x+∂F(x,Kx)

∂uK]|x=0 = A+BK.

Since A + BK is asymptotically stable, by the first statement of the previous theorem, the resultfollows.

One can locally linearize the system and use separation principle to construct a local controllerwhich locally asymptotically stabilizes the non linear control system.

Applications in singularly perturbed systems ([2])

2.7 Converse theorems

Does there exists a Lyapunov function for stable, asymptotically stable, exponentially stable dy-namical systems? These are converse theorems. For time varying systems, such functions existunder some conditions. For time invariant or autonomous systems, Lyapunov stability does notgurantee existence of a continuously differentiable or continuous time-independent Lyapunov func-tion. (Existence of lower semi-continuous Lyapunov function Haddad.) However, for asymptoticallystable equilibrium points of autonomous systems, a continuously differentiable time-independentLyapunov function exists.

Lemma 2.36 (Massera’s lemma). We refer the reader to Haddad p. 162.

Theorem 2.37 (asymptotic stability converse). Suppose the zero solution of (7) is asymptoticallystable, f : S → Rn is continuously differentiable and let δ > 0 be such that Bδ(0) ⊂ S is containedin the domain of attraction of (7). Then, there exists a continuously differentiable function V :Bδ(0) → R such that V (0) = 0, V (x) > 0 for all x ∈ Bδ(0) \ 0 and V ′(x).f(x) < 0 for allx ∈ Bδ(0) \ 0.

Proof. Theorem 3.9 Haddad.

Theorem 2.38 (exponential stability converse). Suppose the zero solution of (7) is exponentiallystable, f : S → Rn is continuously differentiable and let δ > 0 be such that Bδ(0) ⊂ S is containedin the domain of attraction of (7). Then, for every p > 1, there exists a continuously differentiablefunction V : S → R and scalars α, β, ε > 0 such that α‖x‖p ≤ V (x) ≤ β‖x‖p for all x ∈ Bδ(0) andV ′(x).f(x) < −εV (x) for all x ∈ Bδ(0).

Proof. Theorem 3.10 Haddad.

Corollary 2.39. Suppose the zero solution of (7) is exponentially stable, f : S → Rn is continuouslydifferentiable and let δ > 0 be such that Bδ(0) ⊂ S is contained in the domain of attraction of (7).Then, there exists a continuously differentiable function V : S → R and scalars α, β, ε > 0 suchthat α‖x‖2 ≤ V (x) ≤ β‖x‖2 for all x ∈ Bδ(0) and V ′(x).f(x) < −εV (x) for all x ∈ Bδ(0).

16

Table 2: Basic Lyapunov stability theorems for non autonomous systems (Theorem 3.1)

Conditions on V (x, t) Conditions on −V (x, t) Conclusion

lpdf ≥ 0 locally stablelpdf, decrescent ≥ 0 locally uniformly stable

lpdf lpdf asymptotically stablelpdf, decrescent lpdf uniformly asymptotically stablepdf, decrescent pdf globally uniformly asymptotically stable

a‖x‖p ≤ V (t,x) ≤ b‖x‖p locally ≥ c‖x‖p locally locally exponentially stablea‖x‖p ≤ V (t,x) ≤ b‖x‖p globally ≥ c‖x‖p globally globally exponentially stable

Proof. Take p = 2 in the previous theorem.

Theorem 2.40 (global exponential stability converse). Suppose the zero solution of (7) is globallyexponentially stable, f : Rn → Rn is continuously differentiable and globally Lipshitz. Then, forevery p > 1, there exists a continuously differentiable function V : S → R and scalars α, β, ε > 0such that α‖x‖2 ≤ V (x) ≤ β‖x‖2 for all x ∈ Bδ(0) and V ′(x).f(x) < −εV (x) for all x ∈ Bδ(0).

Proof. Theorem 3.11 of Haddad.

3 Stability theory for non autonomous/time varying systems

3.1 Basic stability theorems for time varying systems

Theorem 3.1. Consider a non autonomous system x = f(x(t), t), x(0) = x0 where x(t) ∈ S ⊂ Rn.Let V : S × R→ R. Let a, b, c > 0 and p ≥ 1. Then, Table 2 holds.

Proof. Consider the first case. Since V is an lpdf, there exists α ∈K such that V (x, t) ≥ α(‖x‖),∀x ∈ Bs(0). Moreover, since V ≤ 0 locally, V (x, t) ≤ 0 ∀t ≥ t0, ∀x ∈ Br(0). To show localstability of 0, we need to show that for given ε > 0, t0 ≥ 0, there exists δ = δ(ε, t0) such that forall ‖x(t0)‖ < δ, ‖x(t)‖ < ε, ∀t ≥ t0.

Define ε1 := min(ε, s, r). Choose δ > 0 such that

β(t0, δ) := sup‖x‖≤δV (x, t0) < α(ε1).

Such δ always exists because, α(ε1) > 0 and limδ→0β(t0, δ) = limδ→0sup‖x‖≤δV (x, t0) = 0. Notethat

α(‖x(t0)‖) ≤ V (x(t0), t0) < α(ε1).

Since α is increasing, this implies that ‖x(t0)‖ < ε1. We need to show that ‖x(t)‖ < ε1 for allt ≥ t0. Suppose this is not true. Let t1 > t0 be the first instant such that ‖x(t)‖ ≥ ε1. Then

V (x(t1), t1) ≥ α(ε1) > V (x(t0), t0).

But this is a contradiction since V (x, t) ≤ 0 for all ‖x‖ < ε1. Thus, ‖x(t)‖ < ε1 for all t ≥ t0.For the second case, since V is decrescent, define

β(δ) := sup‖x‖≤δsupt≥t0V (x, t).

Note that β is non decreasing and there exists d > 0 such that β(δ) <∞ for 0 ≤ δ ≤ d. Choosingδ such that β(δ) < α(ε1), using similar arguments used in the previous case, the statement follows.

For the third case, it is clear from the first case that 0 is stable. We need to show thatlimt→∞x(t) = 0 i.e., limt→∞‖x(t)‖ = 0. To show this, we need to show that for every ε > 0,

17

there exists δ = δ(ε, t0) > 0 and T (ε, t0) < ∞ such that ‖x(t0)‖ < δ ⇒ ‖s(t, t0,x(t0))‖ < ε for allt > T (ε, t0) ≥ t0. For the sake of simplicity, we avoid giving further details here. Case 4 is provedin [2] and the proof of case 3 follows on similar lines. For cases 5, 6, 7, we refer the reader to [2].

Converses two first two cases are true ([2] p.160, Remark pt. 1).Alternate geometric proof (Khalil):

Proof. 1. Since V is an lpdf, there exists α ∈ K such that α(‖x‖) ≤ V (t,x) for all x ∈ Bs(0).Moreover, V = ∂V

∂t + ∂V∂x f(t,x) ≤ 0 for all x ∈ Br(0). Given ε < min (r, s), Bε(0) ⊂ S. Choose

c > 0 such that c < min (min‖x‖=εα(‖x‖), r, s). Then, x ∈ Bε(0) | α(‖x‖) ≤ c is in the interiorof Bε(0). Define

Ωt,c := x ∈ Bε(0) | V (t,x) ≤ c.

Since V (t,x) ≤ c ⇒ α(‖x‖) ≤ c, Ωt,c ⊂ x ∈ Bε(0) | α(‖x‖) ≤ c. Observe that V ≤ 0 on Ωt,c forall t ≥ t0. Therefore, solution starting in Ωt0,c, stays in Ωt0,c for all t ≥ t0. Now choose a δ ballaround 0 which lies inside Ωt0,c. For all x0 lying in this δ ball, x(t) remains inside Ωt0,c ⊂ Bε(0).This shows the stability of 0.2. Now suppose V is decrescent. Hence, α1(‖x(t)‖) ≤ V (t,x(t)) ≤ α2(‖x(t)‖). Therefore,

‖x(t)‖ ≤ α−11 (V (t,x(t))) ≤ α−11 (V (t0,x(t0))) ≤ α−11 (α2(‖x(t0)‖))

Therefore, given ε > 0, let δ = α−12 (α1(ε)). This shows uniform stability.3. Since, V < 0, Ωt,c shrinks as t→∞. This implies asymptotic stability.4. Follows from 2 and 3.5. Since V is a pdf, follows from 4.6, 7. Follows similarly using extension of arguments used for the time invariant case. Using thegiven inequalities, it follows that V ≤ − c

bV = εV . Now using comparison lemma and similararguments used for time invariant case, the statements follow.

Example 3.2.

For Lyapunov stability of periodic systems, refer [2]

Theorem 3.3 (Exponential stability and its converse). Sastry, Theorem 5.17.

Lemma 3.4 (Barbalat’s lemma). Suppose that x(t) is bounded, x(t) is bounded, w is uniformlycontinuous, and

´∞t0w(x(t))dt <∞. Then, w(x(t))→ 0 as t→∞.

Proof. Since w is uniformly continuous, for every ε > 0, there exists δx > 0 such that ‖x − y‖ <δx ⇒ ‖w(x)−w(y))‖ < ε

2 . Since x is bounded, x(t) is uniformly continuous i.e., given any δx > 0,there exists δt such that |t1 − t2| < δt ⇒ ‖x − y‖ < δx ⇒ ‖w(x) − w(y)‖ < ε

2 . Therefore, w isuniformly continuous w.r.t. t.

Suppose w(x(t)) 6→ 0 as t → ∞, i.e., there is an ε > 0 and a sequence tk → ∞ such that|w(x(tk))| ≥ ε for all k. Let t ∈ [tk, tk + δ]. Then,

|w(x(t))| = |w(x(tk))− (w(x(tk))− w(x(t)))| ≥ |w(x(tk))| − |w(x(tk))− w(x(t))| ≥ ε− ε

2=ε

2.

Therefore,

|ˆ tk+δ

tk

w(x(t))dt| =ˆ tk+δ

tk

|w(x(t))|dt ≥ ε

2δ > 0 (44)

where the equality holds since w(x(t)) retains its sign for tk ≤ t ≤ tk + δ. Therefore,´ st0w(x(t))dt

cannot converge to a finite limit as s→∞, a contradiction.

18

Theorem 3.5 (Generalization of LaSalle (LaSalle-Yoshizawa)). Suppose that the function f of (3)is Lipshitz continuous in x, uniformly continuous in t in Br(0). Let α1, α2 ∈ K and V (t,x) suchthat

α1(‖x‖) ≤ V (t,x) ≤ α2(‖x‖).

Suppose there exists a continuous non negative function W (x) such that

V (t,x) =∂V

∂t+∂V

∂x.f(t,x) ≤ −W (x) ≤ 0. (45)

Then for all ‖x(t0)‖ ≤ α−12 (α1(r)), the trajectories x(.) are bounded and limt→∞W (x(t)) = 0.

Proof. (Sastry, Theorem 5.27.) Consider V (t,x) ≤ −W (x), and integrate from t0 to t. Therefore,

V (t,x)− V (t0,x0) ≤ −ˆ t

t0

W (x(s))ds⇒ˆ t

t0

W (x(s))ds ≤ V (t0,x0)− V (t,x) ≤ V (t0,x0)

where the last inequality holds because V > 0. Therefore,´ tt0W (x(s))ds exists and is bounded.

Claim: ‖x(t0))‖ ≤ α−12 (α1(ρ))⇒ ‖x(t)‖ ≤ ρ for all t ≥ t0.Let ρ < min( min‖x‖=rα1(r), r). Then, x ∈ Br(0) | α1(‖x‖ ≤ ρ) ⊂ Br(0). Let µ := α−12 (α1(ρ)).Suppose x(t0) ∈ Bµ(0). Therefore, α1(‖x(t0)‖) ≤ V (t0,x(t0)) ≤ α2(‖x(t0)‖).

Let Ωt,ρ = x ∈ Br(0) | V (t,x) ≤ ρ. Therefore, since V is decreasing, if x(t0) ∈ Ωt0,ρ,x(t) ∈ Ωt0,ρ. Since V is decreasing,

V (t,x(t)) ≤ V (t0,x(t0)).

Moreover, α1(‖x(t)‖) ≤ V (t,x(t)) ≤ α2(‖x(t)‖). Therefore,

‖x(t)‖ ≤ α−11 (V (t,x(t))) ≤ α−11 (V (t0,x(t0))) ≤ α−11 (α2(‖x(t0)‖))

Therefore, if ‖x(t0)‖ ≤ µ, then ‖x(t)‖ ≤ α−11 (α2(µ)) = ρ. Thus, x(t) is bounded.

We need to show that W (x(s)) → 0 as s → ∞. Since x(t) is bounded, it is contained in somecompact set D. Since W is continuous, it is uniformly continuous over D (continuous functionsover compact sets are uniformly continuous) i.e., for every ε > 0, there exists δx > 0 such that‖x− y‖ < δx ⇒ |W (x)−W (y)| < ε.

Since f is Lipshitz, ‖x‖ = ‖f(t,x) − f(t,y)‖ ≤ L(‖x − y‖) (y in some neighborhood of x), forall x ∈ D and for all t. Thus, x is bounded and now by Barbalat’s lemma, theorem follows.

3.2 Instability theorems

Theorem 3.6 (non autonomous). Consider a system (1) such that there exists a continuouslydifferentiable function V : R+ × Rn → R and a time t0 ≥ 0 such that

• V is lpdf,

• V (0, t) = 0, ∀t ≥ t0,

• there exists points x0 in a neighborhood around 0 such that V (t0,x0) ≥ 0.

Then 0 is an unstable equilibrium point.

Proof. [2]

Theorem 3.7. Consider a system (1) such that there exists a continuously differentiable functionV : R+ × Rn → R and a constant r > 0 such that

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Table 3: Basic Lyapunov stability theorems for non autonomous systems (Theorem 3.9)

Conditions on Φ(t, t0) Conclusion

supt≥t0Φ(t, t0) = m(t0) <∞ stablesupt0≥0 supt≥t0Φ(t, t0) = m <∞ uniformly stable

limt→∞‖Φ(t, t0)‖ = 0 asymptotically stablesupt0≥0 supt≥t0Φ(t, t0) = m <∞, limt→∞‖Φ(t, t0)‖ = 0 uniformly asymptotically stable

‖Φ(t, t0)‖ ≤ me−λ(t−t0) exponentially stable

• V is decrescent,

• V (0, 0) = 0 and V (., 0) is positive in a neighborhood around 0,

• there exists λ > 0 and a function W : R+ × Rn → R such that V (x, t) = λV (x, t) + W (x, t)and W (x, t) ≥ 0, ∀t ≥ 0, ∀x ∈ Br(0).

Then 0 is an unstable equilibrium point.

Proof. [2]

Theorem 3.8 (Chetaev’s instability theorem). Consider a system (1) such that there exists acontinuously differentiable function V : R+ × Rn → R an open ball Bε(0), an open set O ⊂ Bε(0)and a function γ ∈K such that

V (x, t) > 0, ∀t ≥ 0, ∀x ∈ O, supt≥0supx∈OV (t,x) <∞, 0 ∈ ∂O (46)

V (t,x) = 0, ∀t ≥ 0, ∀x ∈ ∂O ∩ Bε(0), (47)

V ′(x).f(x) > γ(‖x‖), ∀t ≥ 0, ∀x ∈ O. (48)

Then, the zero solution is unstable.

Proof. [2]

3.3 Linear autonomous/LTV systems, linearization

Consider an LTV systemx(t) = A(t)x(t). (49)

Let x(t0) = x0 and let Φ(t, t0) be the state transition matrix (refer Hespanha or short notes onlinear systems). Then x(t) = Φ(t, t0)x(t0). Note that for linear systems, local and global stabilityis the same.

Theorem 3.9. Consider an LTV system (49). Then for the equilibrium point 0, conditions inTable 3 hold. Moreover, the last two cases are equivalent.

Proof. Suppose supt≥t0Φ(t, t0) = m(t0) < ∞. For an ε > 0 and t0 ≥ 0, define δ(ε, t0) := εm(t0)

.Then,

‖x(t0)‖ < δ ⇒ ‖x(t)‖ = ‖Φ(t, t0)x(t0)‖ ≤ ‖Φ(t, t0)‖‖x(t0)‖ < m(t0)δ = ε.

This shows that 0 is stable.Suppose ‖Φ(t, t0)‖ is unbounded function of t for some t0 ≥ 0. We need to show that one can

choose x(t0) ∈ Bδ(0) such that ‖x(t)‖ ≥ ε for some t ≥ t0. Select δ1 ∈ (0, δ). Since ‖Φ(t, t0)‖ is

20

unbounded function of t, there exists t ≥ t0 such that ‖Φ(t, t0)‖ > εδ1

. Choose v such that ‖v‖ = 1and ‖Φ(t, t0)v‖ = ‖Φ(t, t0)‖ and let x(t0) = δ1v. Thus, x(t0) ∈ Bδ(0). Furthermore,

‖x(t)‖ = ‖Φ(t, t0)x(t0)‖ = ‖δ1Φ(t, t0)v‖ = δ1‖Φ(t, t0)‖ > ε.

Hence, 0 is unstable.We refer the reader to Sastry and [2] for proofs of other statements.

Quadratic Lyapunov functions for LTV systems:

Lemma 3.10. Suppose Q : R+ → Rn×n is continuous and bounded, and that the equilibrium 0 of(49) is uniformly asymptotically stable. Then,for each t ≥ 0, the matrix

P (t) =

ˆ ∞t

ΦT(τ, t)Q(t)Φ(τ, t)dτ (50)

is well-defined; moreover, P (t) is bounded as a function of t.

Proof. Uniform asymptotic stability is equivalent to exponential stability for LTV systems (Theo-rem 3.9). Therefore, ‖Φ(τ, t)‖ ≤ me−λ(τ−t), ∀τ ≥ t ≥ 0. The boundedness of Q and exponentialboundedness of Φ implies that P is bounded.

Lemma 3.11. Suppose that, in addition to the hypotheses of the previous lemma, the followingconditions also hold:

1. Q(t) is symmetric and positive definite for each t ≥ 0; moreover, there exists a constant α > 0such that α‖x‖2 ≤ xTQ(t)x,∀t ≥ 0, ∀x ∈ Rn.

2. The matrix A(.) is bounded i.e., m0 = supt≥0‖A(t)‖ <∞.

Under these conditions, the matrix P (t) defined in (50) is positive definite for each t ≥ 0; moreover,there exists a constant β > 0 such that β‖x‖2 ≤ xTP (t)x, ∀t ≥ 0, ∀x ∈ Rn.Proof. [2], Lemma 56, p.203.

Theorem 3.12. Suppose Q(.) and A(.) satisfy the hypotheses of Lemmas 3.10 and 3.11. Then,for each function Q(.) satisfying the hypotheses, the function V (x, t) = xTP (t)x is a Lyapunovfunction establishing the exponential stability of the equilibrium point 0.

Proof. [2] Theorem 64, p.216, Sastry, Theorem 5.40, p.213.

periodic systems: [2].

3.4 Indirect method of Lyapunov and local linearization

For a non autonomous system (3), define

A(t) =∂f(x, t)

∂x|x=0 and f1(x, t) = f(x, t)−A(t)x. (51)

Then it follows that limx→0‖f1(x,t)‖‖x‖ = 0. However, it may not be true that

lim‖x‖→0supt≥0‖f1(x, t)‖‖x‖

= 0. (52)

Theorem 3.13. Consider (3) where f is continuously differentiable. Define A(t), f1(x, t) as in(51). Suppose (52) holds and A(.) is bounded. If 0 is an exponentially stable equilibrium point ofthe linearized system x(t) = A(t)x(t), then it is also an exponentially stable equilibrium point of(3).

Proof. [2], Theorem 15, Section 5.5, p.211 or Sastry, Theorem 5.41, p.215.

Example 3.14. For LTV systems, exponential stability/uniform asymptotic stability is not char-acterized by the location of eigenvalues of A(t) for all t, Example 4.22, Khalil.

21

3.5 Converse theorems

Converse theorems for uniform asymptotic stability, exponential stability and and global exponen-tial stability are stated in [2] Section 5.7. The essence of these theorems is that they state that thesufficient conditions obtained before for stability are necessary as well.

3.5.1 Applications of converse theorems

Theorem 3.15. local exponential stability of autonomous systems (Section 5.8.1, [2]). This showsthat the converse of the first statement of Theorem 2.32.

slowly varying systems (Section 5.8.2, [2]), Observer-Controller stabilization for time varyingsystems (separation principle): local stabilizing controllers and local separation principle for nonlinear systems via linearization (Section 5.8.3, [2]), stabilizing triangular/hierarchical systems (iso-lated subsystems) (Section 5.8.4, [2]).

4 Stability of periodic solutions and tracking trajectories

Let γ(t) be a periodic solution of x = f(t,x). We want to investigate stability of this periodicsolution for trajectories starting from nearby points. Let y(t) = x(t) − γ(t). Thus, y = 0 isan equilibrium point of the time varying system y = f(t,y + γ(t)) − f(t, γ(t)). Thus, one cancharacterize stability properties of γ(t) (e.g., Lyapunov stability, asymptotic stability etc.) basedon stability properties of y = 0.

Example 4.1.

Suppose we are interested in tracking a trajectory r(t). Again, let y(t) = x(t) − r(t). Now,one needs to choose an input such that y = 0 is an equilibrium point of the time varying systemy = f(t,y + r(t)) − r(t). To achieve asymptotic tracking, one needs an input which makes y = 0asymptotically stable.

Example 4.2. Recall the example of stabilization of a rigid robot. Suppose one wants to track a

trajectory r(t). Thus, one wants (x,y) → (r, r). Let z =

[z1z2

]=

[xy

]−[

rr

]. One needs to

find u such that 0 is an asymptotically stable equilibrium point of

z =

[z1z2

] [y − r

M(x)−1[u− C(x,y)y]− r

].

Let u = −Kp(x− r)−Kd(y − r) + C(x,y)r +M(x)r where Kp,Kd > 0. Therefore,

z =

[y − r

M(x)−1[−Kp(x− r)−Kd(y − r)]

]=

[z2

M(x)−1[−Kpz1 −Kdz2 − C(x,y)z2]

]and z has the required equilibrium point. Consider a Lyapunov function

V (z1, z2) =1

2[zT2M(x)z2 + zT1Kpz1]

⇒ V = zT2M(x)z2 +1

2zT2 M(x)z2 + zT1Kpz1

= zT2 (−Kpz1 −Kdz2 − C(x,y)z2) +1

2zT2 M(x)z2 + zT1Kpz2

= −zT2Kdz2

since M − 2C is skew symmetric. Now we find the largest invariant set. Observe that V vanisheson Rn × 0. Note that z1 = 0 ⇒ z2 = 0 and z1 = 0 ⇒ M(x)−1[−Kpz1 −Kdz2 − C(x,y)z2] = 0.Since z2 = 0 and M is invertible, Kpz1 = 0⇒ z1 = 0. Thus, (0,0) forms the largest invariant setand global asymptotic stability follows from LaSalle.

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5 Input to state stability

(Haddad Khalil) Boundedness of solutionsConsider an LTI system x = Ax +Bu where A is Hurwitz. Recall that

x(t) = eA(t−t0)x(t0) +

ˆ t

t0

eA(t−τ)Bu(τ)dτ.

Using ‖eA(t−t0)‖ ≤ ke−λ(t−t0),

‖x(t)‖ ≤ ke−λ(t−t0)‖x(t0)‖+

ˆ t

t0

ke−λ(t−τ)‖B‖‖u(τ)‖dτ

≤ ke−λ(t−t0)‖x(t0)‖+ k‖B‖supt0≤τ≤t‖u(τ)‖e−λtˆ t

t0

eλτdτ

= ke−λ(t−t0)‖x(t0)‖+k‖B‖λ

supt0≤τ≤t‖u(τ)‖(1− e−λ(t−t0))

≤ ke−λ(t−t0)‖x(t0)‖+k‖B‖λ

supt0≤τ≤t‖u(τ)‖.

This shows that the zero input response decays to zero exponentially whereas; the zero stateresponse is bounded for bounded inputs. Thus, A Hurwitz implies that state remains bounded foran LTI system if the input is bounded.

We want to understand conditions under which x(t) remains bounded starting from an initialcondition.

Definition 5.1 (Haddad/Khalil). The solutions of (3) are

1. uniformly bounded if there exists γ > 0, independent of t0 ≥ 0 such that for every δ ∈ (0, γ),there exists ε = ε(δ) > 0, independent of t0, such that

‖x(t0)‖ < δ ⇒ ‖x(t)‖ < ε, ∀t ≥ t0. (53)

2. globally uniformly bounded if (53) holds for arbitrarily large δ.

3. uniformly ultimately bounded with an ultimate bound ε if there exists γ > 0, independent oft0 ≥ 0, such that for every δ ∈ (0, γ), there exists T = T (δ, ε), independent of t0 such that

‖x(t0)‖ < δ ⇒ ‖x(t)‖ < ε, ∀t ≥ t0 + T. (54)

4. globally uniformly ultimately bounded if (54) holds for arbitrarily large δ.

For solutions of (7), we drop the word uniformly since the solution depends only on t− t0.

We refer the reader to generalized LaSalle theorem (Theorem 3.5).

Theorem 5.2 (boundedness). Let S ⊂ Rn be an open set containing 0 and V : [0,∞)× S → R bea continuously differentiable function such that

α1(‖x‖) ≤ V (t,x) ≤ α2(‖x‖) (55)

∂V

∂t+∂V

∂xf(t,x) ≤ −W3(x), ∀‖x‖ ≥ µ > 0 (56)

∀t ≥ 0 and ∀x ∈ S where α1, α2 ∈K and W3 is continuous positive definite function. Take r > 0such that Br ⊂ S and suppose that µ < α−12 (α1(r)). Then, there exists a class KL function β and

23

for every initial state x(t0), satisfying ‖x(t0)‖ ≤ α−12 (α1(r)), there exists T = T (x(t0), µ) such thatthe solution to (3) satisfies

‖x(t)‖ ≤ β(‖x(t0)‖, t− t0), ∀t0 ≤ t ≤ t0 + T (57)

‖x(t)‖ ≤ α−11 (α2(µ)), ∀t ≥ t0 + T. (58)

Moreover, if S = Rn, and α1 ∈ KR, then (57) and (58) hold for any initial state x(t0) with norestriction on how large µ is.

Proof. (Sketch): Observe that if µ = 0, then recalling conditions for uniform asymptotic stability ofthe origin (Theorem 1.14, Theorem 3.1) and comparing with the conditions of the present theorem,it is clear that one gets uniform asymptotic stability of the origin. The proof is based on extendingsimilar ideas to the case where we can have uniform asymptotic stability w.r.t. a set rather thanjust a point. We refer the reader to Khalil for the complete proof.

Let ρ = α1(r), hence, α2(µ) < α1(r) = ρ. Since ‖x(t0)‖ ≤ α−12 (α1(r)), α2(‖x(t0)‖) ≤ α1(r) = ρ.Let η = α2(µ) and define Ωt,η := x ∈ Br(0) | V (t,x) ≤ η and Ωt,ρ := x ∈ Br(0) | V (t,x) ≤ ρ.Note that if x(t0) lies outside Bµ(0) and inside Br(0), then for t ≥ t0,

α1(‖x(t)‖) ≤ V (t,x(t)) ≤ V (t,x(t0))α2(‖x(t0)‖)⇒ ‖x(t)‖ ≤ α−11 α2(‖x(t0)‖).

Note that Bµ ⊂ Ωt,η ⊂ Ωt,ρ ⊂ Br. The fact that these sets are contained in Br follows fromdefinition. To show that Bµ ⊂ Ωt,η, note that ‖x‖ < µ ⇒ α2(‖x‖) < α2(µ). Moreover, V (t,x) ≤α2(‖x‖) < α2(µ) = η. Thus, Bµ ⊂ Ωt,η. Since η < ρ, Ωt,η ⊂ Ωt,ρ. Sets Ωt,η,Ωt,ρ have a propertythat a solution starting inside it can not leave it since V is negative on the boundary. A solutionstarting in Ωt,ρ must enter Ωt,η in finite time because in the set Ωt,ρ \Ωt,η, V satisfies V ≤ −k < 0where k = minW (x) over the set Bµ ≤ ‖x‖ ≤ Br. Therefore,

V (t,x) ≤ V (t0,x(t0))− k(t− t0) ≤ ρ− k(t− t0) (59)

which shows that V (t,x) reduces to η within the time interval [t0, t0 + (ρ − η)/k]. For a solutionstarting in Ωt,η, (58) is satisfied for all t ≥ t0. For a solution starting in Ωt,ρ, it enters Ωt,η aftertime T . Now (57) can be obtained by applying arguments used for checking uniform asymptoticstability (Theorem 1.14).

Definition 5.3 (Input to state stability). The system (3) is said to be input-to-state stable if thereexists a class KL function β and a class K function γ such that for any initial state x(t0) andany bounded input u(t), the solution x(t) exists for all t ≥ t0 and satisfies

‖x(t)‖ ≤ β(‖x(t0)‖, t− t0) + γ(supt0≤τ≤t‖u(τ)‖). (60)

Theorem 5.4 (ISS for time independent systems). A system x = F(x,u) is input-to-state stable⇔ there exists a continuously differentiable radially unbounded pdf function V : Rn → R andcontinuous functions γ1, γ2 ∈K such that for every u ∈ Rm,

V ′(x)F(x,u) ≤ −γ1(‖x‖), ‖x‖ ≥ γ2(‖u‖) (61)

Proof. Sketch: (Haddad) (⇐) Let f(t,x) = F(x,u), V (t,x) = V (x) and W (x) = γ1(‖x‖) in theprevious theorem. One obtains that there exists t1 such that ‖x‖ ≤ γ(‖u‖), t ≥ t1 (this follows fromthe proof of (58)). And ‖x‖ ≤ η(‖x0‖, t), t ≤ t1 (this follows from the proof of (57)). Therefore,‖x‖ ≤ η(‖x0‖, t) + γ(‖u‖), t ≥ 0.

(⇒) involves technicalities.

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Theorem 5.5 (ISS for time varying systems). Let V : [0,∞) × Rn → R be a continuously differ-entiable function such that

α1(‖x‖) ≤ V (t,x) ≤ α2(‖x‖) (62)

∂V

∂t+∂V

∂x.f(t,x,u) ≤ −W (x), ∀ ‖x‖ ≥ ρ(‖u‖) > 0 (63)

∀(t,x,u) ∈ [0,∞) × Rn × Rm, where α1, α2 ∈ KR, ρ ∈ K and W > 0 on Rn. Then the systemx = F(t,x,u) is input-to-state stable with γ = α−11 α2 ρ.

Proof. Sketch: Again using f(t,x) = F(x,u), and using the same method used in the proof ofTheorem 5.2, the proof can be constructed.

6 Center manifold theory

A manifold M is said to be invariant w.r.t. a dynamical system if x(0) ∈ M ⇒ x(t) ∈ Mfor all t. As seen in the Introduction of nonlinear control, at an equilibrium x0 of a nonlineardynamical system, if one computes the spectrum of the linearization A and finds a number ofeigenvalues in the left half-plane, then there is an invariant manifold (i.e., a manifold that isinvariant under the flow and that is simply the graph of a mapping in this case) that is tangent tothe corresponding (generalized) eigenspace; it is called the local stable manifold. All trajectorieson this stable manifold are asymptotic to the point x0 as t → ∞. Similarly, associated with theeigenvalues in the right half-plane is an unstable manifold.

If none of the eigenvalues associated with an equilibrium are on the imaginary axis, then theequilibrium is called hyperbolic. In this case, the tangent spaces to the stable and unstable manifoldsspan the whole of Rn. When there are zero eigenvalues or eigenvalues on the imaginary axis, oneneeds the notion of the center manifold as one cannot conclude about the local stability fromlinearization.

Consider a time invariant system (7). Let A = ∂f∂x |x=0 and suppose f is twice continuously

differentiable. Then, (7) can be rewritten as

x = Ax + [f(x)−Ax] = Ax + f(x)

where f(x) = f(x) − Ax such that f(0) = 0 and ∂ f∂x(0) = 0. Since we are interested in the case

when the linearization fails, suppose A has k eigenvalues with zero real parts and n−k eigenvalueswith negative real parts. We can assume without loss of generality using a similarity transform

T that A is in block diagonal form

[A1 00 A2

]where eigenvalues of A1 has zero real parts and

eigenvalues of A2 have negative real parts. Let Tx =

[yz

]which transforms (7) into the form

y = A1y + g1(y, z) (64)

z = A2z + g2(y, z) (65)

where g1 and g2 inherit properties of f . They are twice continuously differentiable and

gi(0,0) = 0,∂gi∂z

(0,0) = 0 (66)

for i = 1, 2.

Definition 6.1. If z = h(y) is an invariant manifold for (64) − (65) and h is smooth, then it iscalled a center manifold if

h(0) = 0,∂h

∂y(0) = 0.

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Theorem 6.2 (Existence (Khalil, Theorem 8.1)). If gi, i = 1, 2 are twice continuously differentiableand satisfy (66), all eigenvalues of A1 have zero real parts, and all eigenvalues of A2 have negativereal parts, then there exists a constant δ > 0 and a continuously differentiable function h(y) definedfor all ‖y‖ < δ, such that z = h(y) is a center manifold for (64)− (65).

Due to its invariance property, if the initial condition lies on the center manifold, then thesolution remains on the center manifold. Since z(t) = h(y(t)), the evolution of the system in thecenter manifold is described by the k−th order differential equation

y = A1y + g1(y,h(y)) (67)

which is refered as reduced system.Suppose z(0) 6= h(y(0)). Then z(t) − h(y(t)) represents the deviation of the trajectory from

the center manifold at any time t. Let w = z− h(y). This transforms (64)− (65) into

y = A1y + g1(y,w + h(y)) (68)

w = A2(w + h(y)) + g2(y,w + h(y))− ∂h

∂y[A1y + g1(y,w + h(y))]. (69)

In the new coordinates, w = 0 is the center manifold. The dynamics on the center manifold arecharacterized by w(t) = 0⇒ w = 0. Substituting this in (69),

0 = A2h(y) + g2(y,h(y))− ∂h

∂y[A1y + g1(y,h(y))]. (70)

Since the above equation must be satisfied for any solution lying on the center manifold, we concludethat h(y) must satisfy the pde (70). This provides a condition that the center manifold z = h(y)must satisfy.

Adding and subtracting g1(y,h(y)) to rhs of (68) and subtracting (70) from (69),

y = A1y + g1(y,h(y)) +N1(y,w) (71)

w = A2(w + h(y)) +N2(y,w) (72)

whereN1(y,w) = g1(y,w+h(y))−g1(y,h(y)), N2(y,w) = g2(y,w+h(y))−g2(y,h(y))∂h∂yN1(y,w).

One can check that N1, N2 are twice differentiable and Ni(y,0) = 0, ∂Ni∂w (0,0) = 0 for i = 1, 2.

Theorem 6.3 (Stability condition). Under assumptions of Theorem 6.2, if the origin y = 0 ofthe reduced order system (67) is asymptotically stable (unstable), then the origin of the full system(64)− (65) is also asymptotically stable (unstable).

Proof. Theorem 8.2 of Khalil, the proof makes use of the converse Lyapunov theorem for asymptoticstability to construct a Lyapunov candidate for the reduced order system.

Corollary 6.4. Under assumptions of Theorem 6.2, if the origin y = 0 of the reduced order system(67) is stable and there is a continuously differentiable Lyapunov function V (y) such that

∂V

∂y[A1y + g1(y,h(y))] ≤ 0

in some neighborhood of y = 0, then the origin of the full system (64)− (65) is stable.

Proof. Note that unlike asymptotic stability, converse theorem does not hold for Lyapunov stability.Therefore, if the origin of the reduced order system is known to be Lyapunov stable, we need anexistence of Lyapunov candidate as well in our hypothesis to make the previous proof work.

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Corollary 6.5. Under assumptions of Theorem 6.2, the origin y = 0 of the reduced order system(67) is asymptotically stable ⇔ the origin of the full system (64)−(65) is also asymptotically stable.

Proof. (⇒) Follows from the theorem above. (⇐) Stability of the full system implies the stabilityof the reduced order system as well, so this implication is trivial.

To use Theorem 6.3, one needs to find the center manifold z = h(y) for which one needs tosolve the pde (70) with boundary conditions h(0) = 0, ∂h

∂y (0) = 0. This is a difficult pde in generaland one uses Taylor series approximation of the solution. We refer the reader to Khalil, Section8.1 for more details.

7 Control Lyapunov functions and feedback stabilization

Consider non linear controlled dynamical system

x(t) = F(x(t),u(t)), x(t0) = x0, t ≥ 0 (73)

where x(t) ∈ S ⊂ Rn, u(t) ∈ U ⊂ Rm and F : S × U → Rn satisfies F(0,0) = 0. Suppose F isLipshitz continuous in the neighborhood of the origin in S×U . We need to find a feedback controllaw u(t) = φ(x(t)) such that the closed loop system is stable.

Definition 7.1 (Control Lyapunov function (Haddad)). Consider the controlled nonlinear dy-namical system given by (73). A continuously differentiable positive-definite function V : S → Rsatisfying

infu∈UV′(x).F(x,u) < 0, x ∈ S, x 6= 0 (74)

is called a control Lyapunov function.

If (74) holds, then there exists a feedback control law φ : S → U such that V ′(x).F(x, φ(x)) < 0,then by Theorem 2.1 case 2, the origin is asymptotically stable. Conversely, if there exists a feedbacklaw such that the origin is asymptotically stable, then by Theorem 2.37, there exists a Lyapunovfunction V such that V ′(x).F(x, φ(x)) < 0 which implies the existence of a control Lyapunovfunction. Thus, (73) is feedback stabilizable iff there exists a control Lyapunov function satisfying(74). The analogues of other cases of Theorem 2.1 also hold for control Lyapunov functions.

Consider an affine non linear control system

x(t) = f(x(t)) +G(x(t))u(t), x(t0) = x0, t ≥ 0 (75)

where f : Rn → Rn, G : Rn → Rn×m.

Theorem 7.2 (Haddad). Consider the controlled nonlinear system given by (75). Then a continu-ously differentiable positive-definite, radially unbounded function V : Rn → R is a control Lyapunovfunction of (75) if and only if

V ′(x)f(x) < 0,x ∈R (76)

where R = x ∈ Rn,x 6= 0 : V ′(x)G(x) = 0.

Proof. (⇐) Obvious. (⇒) If x /∈R, then infu∈UV′(x).[f(x(t))+G(x(t))u(t)] = −∞. When x ∈R,

(76) must hold if V is a control Lyapunov function.

We now construct an explicit feedback control law which is a function of the control Lyapunovfunction V (Haddad). Let α(x) := V ′(x).f(x) and β(x) := GT(x)(V ′(x))T and c0 ≥ 0. Let

φ(x) =

−(c0 +α(x)+

√α2(x)+(βT(x)β(x))2

βT(x)β(x))β(x) if β(x) 6= 0,

0 if β(x) = 0.(77)

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In this case, the control Lyapunov function turns out to be the Lyapunov function for the closedloop system with u = φ(x). Observe that

V (x) = V ′(x)[f(x) +G(x)φ(x)]

= α(x) + βT(x)φ(x)

=

−c0βT(x)β(x)−

√α2(x) + (βT(x)β(x))2 if β(x) 6= 0,

α(x) if β(x) = 0

< 0, ∀x ∈ Rn \ 0 (78)

provided that Theorem 7.2 is satisfied. This implies that V is indeed a Lyapunov function for theclosed loop system. Since f and G are smooth, the feedback law (77) is smooth everywhere exceptthe origin.

Theorem 7.3. Consider the nonlinear dynamical system G given by (75) with a radially unboundedcontrol Lyapunov function V : Rn → R. Then the following statements hold:

1. The control law φ(x) given by (77) is continuous at x = 0 if and only if for every ε > 0,there exists δ > 0 such that for all 0 < ‖x‖ < δ, there exists u ∈ Rm such that ‖u‖ < ε andα(x) + βT(x)u < 0.

2. There exists a stabilizing control law φ(x) such that α(x) + βT(x)φ(x) < 0,x ∈ Rn,x 6= 0and φ(x) is Lipschitz continuous at x = 0 if and only if the control law φ(x) given by (77) isLipschitz continuous at x = 0.

Proof. We refer the reader to Theorem 6.8 of Haddad.

8 Periodic systems

8.1 Linear periodic systems

The following discussion is from [7]. Consider a linear periodic system x(t) = A(t)x(t) such thatA(t + T ) = A(t), T > 0. Let Φ(t, τ) be the state transition matrix. Recall that for general linearsystems,

Φ(t, t0) := I +

ˆ t

t0

A(s1)ds1 +

ˆ t

t0

A(s1)

ˆ s1

t0

A(s2)ds2ds1 + . . . .

and ddtΦ(t, t0) = A(t)Φ(t, t0). Hence, d

dtΦ(t+ T, t0) = A(t+ T )Φ(t+ T, t0). Therefore, for periodic

systems, ddtΦ(t+T, t0) = A(t)Φ(t+T, t0) since A(t+T ) = A(t). Furthermore, by the concatenation

property of the state transition matrix, Φ(t+T, t0) = Φ(t+T, t)Φ(t, t0). Let C(t) = Φ−1(t, t0)Φ(t+T, t0). Therefore

C = −(Φ−1(t, t0)A(t)Φ(t, t0)Φ−1(t, t0))Φ(t+ T, t0) + Φ−1(t, t0)A(t)Φ(t+ T, t0) = 0.

Hence, C is a constant matrix and C = C(t0) = Φ(t0 + T, t0). Therefore, Φ(t+ T, t0) = Φ(t, t0)C.Define R := 1

T ln(C), hence, eRT = C . Let P (t) = Φ(t, 0)e−Rt. Clearly P (t) is non-singular.

P (t+ T ) = Φ(t+ T, 0)e−R(t+T ) = Φ(t, 0)Ce−R(t+T ) = Φ(t, 0)Ce−RT e−Rt = Φ(t, 0)e−Rt = P (t).

Therefore, Φ(t, 0) = P (t)e−Rt where P (t) is non singular and periodic. This is called Floquetdecomposition.

The state evolution is given by x(t) = Φ(t, 0)x0 = P (t)e−Rtx0. Since P (t) is periodic andcontinuous, it is bounded. Therefore the periodic linear system is globally exponentially stable ⇔

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R has eigenvalues in the strict LHP⇔ C has eigenvalues strictly inside the unit circle. The matrixC = Φ(t+ T, t) is called monodromy matrix and eigenvalues of C are called Floquet multipliers.

Let y(t) := P−1(t)x(t). Therefore,

x = Py + P y⇒ Ax = Py + P y⇒ APy = Py + P y⇒ y = P−1(APy − Py).

Using P (t) = Φ(t, 0)e−Rt in the above equation,

y = P−1(AΦ(t, 0)e−Rt −AΦ(t, 0)e−Rt + Φ(t, 0)Re−Rt)y = eRtΦ(t, 0)−1(Φ(t, 0)Re−Rt)y = eRtRe−Rty = Ry.(79)

Therefore the linear periodic system is globally exponentially stable iff the LTI system obtainedabove is exponentially stable.

8.2 Nonlinear periodic systems

Consider a nonlinear periodic system x = f(t,x) such that f(t+T,x) = f(t,x) for all x ∈ Rn, T > 0.Let s(t, 0,x0) be the flow of this ode. Hence, d

dts(t, 0,x0) = f(t,x(t)). Consider s(t + T, T,x0).Because the time varying vector field f(t,x) is periodic with period T , by existence and uniqueness,s(t+ T, T,x0) = s(t, 0,x0).

Consider a fixed point or a periodic solution x∗(t) of this periodic system. Let z = x − x∗ bethe perturbation from this trajectory. Therefore,

z = f(t,x)− t,x∗.

Using linearization, one obtains a periodic LTV system. If the linear system is exponentially stable,then the nonlinear periodic system is locally exponentially stable at the solution x∗.

It turns out that for periodic systems, 0 is stable/asymptotically stable iff is is stable/asymptoticallystable ([2]). Lyapunov analysis for stability is applicable to periodic systems as well. One looksfor a periodic lpdf Lyapunov function V with the same period as that of the system such thatV ≤ 0 in some open neighborhood N of 0. Then, one can apply LaSalle-Krasovskii theorem toconclude convergence of trajectories towards the invariant set M of V where M ⊂ S ⊂ N suchthat S := x ∈ N | V (x) = 0 and M is the largest invariant set of S. If S contains no invarianttrajectory other than the zero trajectory, then 0 is uniformly asymptotically stable. If in addition,V is a pdf and radially unbounded, then 0 is globally uniformly asymptotically stable equilibriumpoint of the periodic system ([2]).

9 Discrete non linear systems

Consider an autonomous discrete time non linear system

x(k + 1) = f(x(k)), x(0) = x0, k ∈ Z+ (80)

where x ∈ S ⊂ Rn (S open), 0 ∈ S and f(0) = 0. Assume that f is continuous on S. The flow mapis given by s : Z+×S → S where s(0,x0) = x0, s(1,x0) = f(s(0,x0)) = f(x0), s(2,x0) = f(s(1,x0))and so on. There is an analogue of existence and uniqueness for discrete non linear system ([1],Theorem 13.1)

There are similar definitions on stability and asymptotic stability as we had in the continuoustime case. Instead of exponential stability for continuous time systems, we define geometric stabilityfor discrete systems where ‖x(k)‖ ≤ α‖x(0)‖β−k, α, β > 1. The discrete time varying systems areof the form

x(k + 1) = f(k,x(k)), x(0) = x0, k ∈ Z+. (81)

There are natural discrete time counterparts of all results stated in the continuous time case. Werefer the reader to Chapter 13 of Haddad ([1]) for complete details.

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10 Advanced topics

10.1 Partial stability

Partial stability means stability with respect to part of the system’s state. The state vector x isdivided into two components x1 and x2 and one asks for stability w.r.t. one of the two components.The sufficient stability conditions given in previous sections can be extended to obtain sufficientconditions for partial stability (Haddad [1], Section 4.2).

10.2 Finite time stability

In some applications, one needs to reach the equilibrium point in finite time rather than reachingit asymptotically.

10.3 Semistability

For systems with continuum of equilibria, one need an appropriate notion of stability instead ofasymptotic stability.

10.4 Generalized Lyapunov theorems

Here the differentiability criteria is weakened and one only assumes lower semi-continuity of theLyapunov function to obtain sufficient conditions for stability, asymptotic stability, invariance ofsets and so on. Lyapunov function V must be positive definite in the neighborhood of an equilibriumpoint but it should be decreasing i.e., V (x(t)) decreases as t increases. These results allow usto patch different Lyapunov functions over different sets together without being worried aboutdifferentiability conditions to obtain sufficient conditions for stability. We refer the reader toHaddad ([1]), Section 4.8

10.5 Lyapunov and asymptotic stability of sets and periodic orbits

We can define Lyapunov and asymptotic stability of sets by replacing the equilibrium point in thecorresponding definitions with appropriate sets say S0. One needs a Lyapunov function V which iszero on the say S0, positive on points in an open set containing S0 which lie outside S0 such thatV (x(t)) is decreasing as x(t) evolves in time. For extension of the results stated for equilibriumpoints, we refer the reader to Haddad ([1], 4.9).

Poincare’s theorem provides necessary and sufficient conditions for stability of periodic orbits(Section 4.10, [1]).

10.6 Vector Lyapunov functions

Here one uses vector Lyapunov function which requires less rigid requirements compared to a scalarLyapunov function. One can obtain sufficient conditions for stability using the vector Lyapunovfunctions (Section 4.11, [1]).

10.7 Applications

Stability of switched systems, hybrid systems..

11 Appendix

11.1 lpdf, pdf and decrescent functions

The discussion here is from [2].

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Lemma 11.1. Suppose φ : R+ → R+ is continuous, non decreasing and φ(0) = 0, φ(r) > 0 ∀r > 0.Then, there is a class K function α such that α(r) ≤ φ(r), ∀r. Moreover, if φ(r)→∞, as r →∞,then α can be chosen to be a class KR function.

Proof. [2], Lemma 1, Section 5.2.

Lemma 11.2. A continuous function V : Rn → R is an lpdf ⇔

• V (0) = 0.

• there exists r > 0 such that V (x) > 0, ∀x ∈ Br \ 0.

V is a pdf ⇔

• V (0) = 0.

• V (x) > 0, ∀x ∈ Rn \ 0.

• there exists r > 0 such that inf‖x‖≥rV (x) > 0.

V is radially unbounded ⇔ V (x)→∞ as ‖x‖ → ∞ uniformly in x.

Proof. (⇒) Obvious. (⇐) Define φ(p) := infp≤‖x‖≤rV (x). Then, φ(0) = 0, φ is continuous andnon decreasing because as p increases, the infimum is taken over a smaller region. Moreover,φ(p) > 0 when p > 0. Now by the previous lemma, one can chosen a class K function α such thatα(‖x‖) ≤ φ(‖x‖) ≤ V (x), ∀x ∈ Br.

Necessary conditions for pdf follow from the definition. The sufficient conditions follows fromthe above arguments (using the third condition). The last statement also follows.

A function V (x) = x2

1+x4> 0 on R \ 0 but it is not a pdf. Note that it violates the third

condition in the above lemma. If V is pointwise positive everywhere except the origin and radiallyunbounded, then V is a pdf.

Lemma 11.3. A continuous function V : Rn × R+ → R is an lpdf ⇔

1. V (t, 0) = 0, ∀t.

2. there exists an lpdf W : Rn → R and a constant r > 0 such that V (t,x) ≥ W (x), ∀t ≥ 0,∀x ∈ Br(0).

V is a pdf ⇔

1. V (t, 0) = 0, ∀t.

2. there exists a pdf W : Rn → R such that V (t,x) ≥W (x), ∀t ≥ 0, ∀x ∈ Rn.

V is radially unbounded ⇔ there exists a radially unbounded function W : Rn → R such thatV (t,x) ≥W (x), ∀t ≥ 0, ∀x ∈ Rn.

Proof. Consider lpdfs. If V is an lpdf, then there exists a class K function α such that V (t,x) ≥α(‖x‖). Define W (x) := α(‖x‖). Conversely, suppose there exists W satisfying given conditions.Then using the previous lemma, one can show that V is an lpdf.

The necessary conditions for V being pdf follow from the definition. For sufficiency, we can usethe previous lemma. The last statement also follows similarly.

Example 11.4. Consider a quadratic function V (x) = xTPx where P is a square real symmetricmatrix. V is pdf ⇔ P is positive definite. Thus, the term positive definite matrix and positivedefinite function are consistent. Note that V is radially unbounded.

31

Example 11.5. A function V1(x1, x2) = x21+x22 is an example of a radially unbounded pdf. Anotherfunction V2(t, x1, x2) = (t + 1)(x21 + x22) is a pdf since it dominates time invariant V1. A functionV3(t, x1, x2) = e−t(x21 + x22) is not a pdf but it is decrescent.

A function W1(x1, x2) = x21 + sin2(x2) is an lpdf but not a pdf. A function W2(x1, x2) =x21 + tanh2(x2) is a pdf but not radially unbounded since, tanh2 x2 → 1 as |x2| → ∞.

References

[1] W. Haddad and V. Chellaboina, Non linear dynamical systems and control, Princeton universitypress, 2008.

[2] Mathukumali Vidyasagar, Non linear systems analysis, Dover, 2009.

[3] S. Sastry, Non linear systems, Springer, 1999.

[4] H. Khalil, Non linear systems, Pearson, 2009.

[5] D. Liberzon, Non linear systems analysis, Lecture notes, 2012.

[6] J. Hespanha, Linear systems theory, Princeton university press, 2009.

[7] H. Taha, Geometric control theory, Lecture notes, 2019.

“I am returning your nose dear, found it in my business.”- Anonymous

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