
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)
ẋ(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
ẋ(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 ∈ B�(0) for some �.
• 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 ∈ B�(0) for an
arbitrary �.
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 � > 0, thereexists a δ = δ(�, t0) such
that ‖x0‖ < δ(�, t0)⇒ ‖s(t, t0,x0)‖ < � for all t ≥ t0.
Definition 1.2 (uniform stability). The equilibrium point 0 is
said to be uniformly stable if, forevery � > 0, there exists a δ
= δ(�) such that ‖x0‖ < δ(�)⇒ ‖s(t, t0,x0)‖ < � 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, � such that
‖s(t0 + t, t0,x0)‖ ≤ c‖x0‖e−γt, ∀t, t0 ≥ 0, ∀x0 ∈ B�(0). The
constant γis called an estimate of the rate of convergence.
1

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, � with M arbitrarily large and �
arbitrarilysmall, there exists a finite number T = T (M, �) such
that ‖x0‖ < M, t0 ≥ 0⇒ ‖s(t+ t0, t0,x0)‖ <�, ∀t ≥ T (M,
�).
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 exponentially stable, a
necessary condition is that it is the only equilibrium point.
Energylike 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 � > 0 and some α(.) of class K
functions,
V (0, t) = 0 and V (x, t) ≥ α(‖x‖) ∀x ∈ B�, 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 � > 0, such that
V (x, t) ≤ β(‖x‖) ∀x ∈ B�(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, V̇
= ∇V (x).ẋ = ∇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).
2

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 � > 0, t0 ∈ R+, choose
δ(�, t0) = min{d(t0), φ−1t0 (�)}.(⇒) Fix t0 and � > 0. There
exists δ(t0, �) > 0 such that
‖x(t0)‖ < δ ⇒ ‖x(t)‖ < �, ∀t ≥ t0.
Let δ̄(t0, �) be the supremum of all applicable δ. The function
δ̄ is positive and non decreasing, butneed not be continuous.
Choose θt0(�) ∈K such that θt0(�) ≤ δ̄(�) for all � > 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 ẋ = 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 −V̇ (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 ≥ �V
(x) locally locally exponentially stableα‖x‖p ≤ V (x) ≤ β‖x‖p
globally ≥ �V (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
ẋ = f(x(t)). (7)
2.1 Basic stability theorems using Lyapunov’s direct method
Theorem 2.1. Consider an autonomous system ẋ = f(x(t)), x(0) =
x0 where x(t) ∈ S ⊂ Rn. LetV : S → R. Let α, β, � > 0 and p ≥ 1.
Then, Table 1 holds.
Proof. Let � > 0 such that B�(0) ⊂ S. Observe that since
∂B�(0) is compact and V (x) is continuous,V (∂B�(0)) is compact
hence, α = minx∈∂B�(0)V (x) exists and α > 0 since 0 /∈ ∂B�(0)
and V (x) > 0for x ∈ S,x 6= 0. (Draw a picture of S and B�
inside S.)
Consider β ∈ (0, α) where α is the minimum value of V on ∂B�(0).
Let Ωβ = {x ∈ S  V (x) ≤β}. Then, Ωβ ⊂ B�(0) \ ∂B�(0). This is
true because if there exists y ∈ Ωβ such that y ∈ ∂B�(0),then V (y)
≥ α > β which is a contradiction. (Draw this “ellipsoid” inside
B�.)
Since V̇ ≤ 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(�) > 0. Therefore, for all x(0) ∈ Bδ(0), V (x(0)) < βand V
(x(t)) ≤ V (x(0)) ⇒ x(t) ∈ Ωβ ⊂ B�(0). Thus, x(t) ∈ B�(0) ⇒ ‖x(t)‖
< �. This provesLyapunov stability.
For the second case, V̇ < 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 � > 0, there exists T > 0 such that ‖x(t)‖
< �for all t > T . By previous arguments, for every � > 0,
we can choose β > 0 such that Ωβ ⊂ B�.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‖≤�V̇ (x). Since V̇ < 0,−γ < 0.
Moreover,
V (x(t)) = V (x(0)) +
ˆ t0V̇ (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 � > 0 such that V (x) > a when ‖x‖ >
�.Therefore, Ωa ⊂ B�(0) which implies that Ωa is bounded. Remaining
arguments are exactly similaras to the second case.
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Since V̇ ≤ −�V (x), V (x(t)) ≤ V (x(0))e−�t, 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−�t, t ≥ 0⇒ ‖x(t)‖ ≤ (βα
)1p ‖x(0)‖e(
−�p)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 θ̈ + sin(θ) = 0. Choosing x1 = θ
and x2 = θ̇, ẋ1 = x2, ẋ2 = − sin(x1). The totalenergy is 12
θ̇
2 − cos(θ) = 12x22 − cos(x1). Consider V (x1, x2) = 1 + 12x
22 − cos(x1)0. Since this is
continuously differentiable and lpdf, it is a Lyapunov function
candidate. Moreover, V̇ (x1, x2) =sin(x1)ẋ1 + x2ẋ2 = sin(x1)x2 +
x2(− sinx1) = 0. Therefore, the equilibrium point is stable.
Consider a simple pendulum with damping i.e., θ̈ + sin(θ) + bθ̇
= 0 (Khalil). Hence, the stateequations are ẋ1 = x2, ẋ2 = −
sin(x1) − bx2. Let V (x1, x2) be the same function as above.
Thus,V̇ = sin(x1)ẋ1 + x2ẋ2 = 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 V̇ < 0.
Choose p11 = bp12, p12 =b/2, p22 = 1. Now, V̇ = −12abx1 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,
V̇ < 0 on this set ⇒ asymptoticstability.
Example 2.3 (Non linear series RLC circuit/Non linear mass
spring damper). Non linear seriesRLC circuit
(Sastry)/Massspringdamper. Suppose the inductor is linear but the
resistor andcapacitor are non linear (or massspring damper system
with non linear spring and damping). Letx1 be the charge on the
capacitor and x2 be the current through the inductor.
Therefore,
ẋ1 = x2, ẋ2 = −f(x2)− g(x1) (8)
where f is a continuous function modelling resistor
currentvoltage characteristics and g modelscapacitor
chargevoltage 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 x10
g(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,
V̇ (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
Iω̇ + 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
Ixω̇x = −(Iz − Iy)ωyωz, Iyω̇y = −(Ix − Iz)ωxωz, Izω̇z = −(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. Definea =
Iy−IzIx
, b = Ix−IzIy , c =Ix−IyIz
. Therefore, (11) becomes
ẋ = ayz, ẏ = −bxz, ż = 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.
V̇ = 2(pxẋ+ qyẏ + rzż) = 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 Cẋ 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,
Cẋ = G(x)x⇒ ẋ = C−1G(x)x. (12)
Clearly, 0 is an equilibrium point. Consider the total energy
stored in the capacitors i.e., V (x) =12x
TCx. Therefore,
V̇ (x) =1
2(ẋTCx + xTCẋ) = −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 �
> 0 such that M(x) > 0 for allx ∈ B�(0). Let d :=
infx∈B�(0)λminM(x) and choose � small enough such that d > 0.
Then,V̇ = −xTM(x)x ≤ −d‖x‖2, ∀x ∈ B�(0). Since, V (x) = 12x
TCx,
λmax(C)‖x‖2 ≥ V (x)⇒ −d‖x‖2 ≤−d
λmax(C)V (x).
Therefore, V̇ ≤ −dλmax(C)V (x) and from Theorem 2.1 case 4, 0 is
locally exponentially stable. Ifd = infx∈RnλminM(x) > 0, then 0
is globally exponentially stable. It is also globally
asymptoticallystable since, V̇ < 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 ẋ = 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 V̇ :
Theorem 2.9 (LaSalle’s invariance principle
(BarbashinKrasovskiiLaSalle Theorem)). Let Ω ⊂ Sbe a compact
invariant set of ẋ = f(x). Let V : S → R be a continuously
differentiable functionsuch that V̇ ≤ 0 in Ω. Let E be the set of
points where V̇ (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 ẋ = f(x) starting in Ω. Since
V̇ ≤ 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, V̇ = 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, invariance 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
V̇ ≤ 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 ẋ = 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 V̇ ≤ 0 on S. Let E be the set
of points where V̇ (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 ofẋ = f(x). Let V : Rn → R be a
continuously differentiable radially unbounded positive
definitefunction on Rn \ {0} such that V̇ ≤ 0 on Rn. Let E be the
set of points where V̇ (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 V̇ = −bx22,V̇ = 0 on the line x2 = 0. To
maintain V̇ = 0, the trajectory must be confined to the line x2 =
0.Now x2 = 0⇒ ẋ2 = 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
V̇ (x1, x2) = −x2f(x2) ≤ 0 for x2 ∈ [−σ0, σ0]. Let c =min(V (−σ0,
0), V (σ0, 0)). Then, V̇ ≤ 0 for x ∈ Ωc := {(x1, x2)  V (x1, x2) ≤
c}. By LaSalle’s invariance principle, the trajectory approaches
the largest invariant set in Ωc∩{(x1, x2),  V̇ (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 ẋ1 = 0 ⇒ x1 = x10.
Furthermore x2 = 0 ⇒ ẋ2 = 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),  V̇ (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
coordinates of the robot. Let u represent generalized forces.
Suppose thesystem of equations are
M(q)q̈ + C(q, q̇)q̇ = u
7

Let x = q and y = q̇. Therefore, ẋ = y and ẏ = [M(x)]−1[u −
C(x,y)y]. Let qd represent thedesired value of the generalized
coordinates. Suppose
u = −Kp(q− qd)−Kdq̇ = −Kp(x− qd)−Kdy,
where Kp,Kd > 0. Now the system equations are
ẋ = y, ẏ = −[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)]
⇒ V̇ = yTM(x)ẏ + 12yTṀ(x)y + ẋTKp(x− qd)
= −yT[Kp(x− qd) +Kdy + C(x,y)y] +1
2yTṀ(x)y + yTKp(x− qd)
= −yTKdy +1
2yT[Ṁ(x,y)− 2C(x,y)]y
It turns out that Ṁ − 2C is skew symmetric. Therefore, V̇ ≤ 0
which implies stability. ApplyingKrasovskiiLaSalle principle, for
the set of points where V̇ = 0 i.e., to points (x,y) ∈ Rn × 0,
thelargest invariant set is given by the set of points in Rn×0
where (ẋ, ẏ) = (0,0). Now ẋ = 0 impliesthat y = 0 and ẏ = 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
ẋ1 = x2 + x1(r2 − x21 − x22)
ẋ2 = −x1 + x2(r2 − x21 − x22), r > 0
which has an equilibrium point at the origin. Observe that if
x21 + x22 = r
2, then, ẋ1 = x2 andẋ2 = −x1 and the evolution remains on the
circle x21 + x22 = r2 and it forms an invariant set. LetV (x) =
14(x
21 + x
22 − r2)2. Thus, V (x) ≥ 0 on R2. Note that V̇ (x) = ∇V.f =
−(x21 + x22)(x21 +
x22 − r2)2 ≤ 0. Note that V̇ = 0 either at the origin or on the
circle x21 + x22 = r2. Let c > r andMc := {x ∈ R2  V (x) ≤ c}.
Since V̇ ≤ 0 on Mc and V ≥ 0, Mc is an invariant set containing
theorigin and the circle x21 +x
22 = r
2. On both these sets, V̇ = 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 x
21 + x
22 = r
2.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, V̇ =∑n
i=1∂V∂xiẋi =
∑ni=1
∂V∂xifi(x). Let g(x) =
∂V∂x . Therefore, V̇ =
g(x).f(x). We need to construct g such that it is a gradient of
a positive definite function andV̇ (x) = g(x).f(x) < 0. Note
that
V (x) =
ˆ x0
g(s)ds =
ˆ x0
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 coordinate axes.
Therefore, (13) becomes
V (x) =
ˆ x10
g1(s1, 0, . . . , 0)ds1 +
ˆ x20
g2(x1, s2, 0, . . . , 0)ds2 + . . .+
ˆ xn0
gn(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 10
g(σx).xdσ =
ˆ 10
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 10
g(σx)xdσ =
ˆ 10
n∑j=1
gj(σx)xjdσ (16)
Let y = σx, therefore,
∂V
∂xi=
ˆ 10
n∑j=1
∂gj∂xi
(σx)xjdσ +
ˆ 10gi(σx)dσ
=
ˆ 10
n∑j=1
∂gi∂xj
(σx)xjdσ +
ˆ 10gi(σx)dσ
=
ˆ 10
n∑j=1
∂gi∂yj
(y)σxjdσ +
ˆ 10gi(σx)dσ
=
ˆ 10
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 ẋ = 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 everyxe ∈ S, there exists α ∈ [0, 1]
such that 0 = xTe P f(xe) = xTe 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,
V̇ = 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, V̇ < 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
≤ −12xTRx ≤ −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)‖ ≥ x
TP f(x)λmax(P )‖x‖ ≥
λmin(R)2λmax(P )
‖x‖. This implies bydefinition 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
ẋ = 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 B�(0) of the origin, V (x) > 0and V̇ < 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 V̇ < 0 on Ωβ, it is clearthat
Ωβ forms an invariant set. If V̇ = 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,Ẇ = h(x)W
(x). Therefore,
ˆ W (x(t))W (x(t0))
dW
W=
ˆ tt0
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 (EnergyCasimir theorem). Consider a non linear
dynamical system ẋ = 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,
V̇ (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 E12ET12 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 E12ET12 E2 + αN
]T.
By choosing appropriate α, V ′′(xe) > 0. Since V is twice
differentiable and V (xe) = V′(xe) = 0, it
follows 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 ẋ = 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 V̇ > 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, V̇ > 0 in U . Let
γ = min{V̇ (x)  x ∈ U, V (x) ≥ a}
which exists since, the continuous function V̇ has a minimum
over the compact set {x ∈ U  V (x) ≥a}. Clearly, γ > 0. Note
that
V (x(t)) = V (x0) +
ˆ t0V̇ (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 ẋ = f(x) and letV : S → R be a continuously
differentiable function, W : S → R and λ, � > 0 such that V (0)
= 0and W (x) ≥ 0 for all x ∈ B�(0) and V̇ = 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) ∈ B�(0), t ≥ 0. Note thatV (x0) > 0. From the
hypothesis, V̇ = λV (x)+W (x) and W (x) ≥ 0. This implies that V̇ ≥
λV (x)for x ∈ B�(0) i.e., V̇ − λV (x) ≥ 0 for x ∈ B�(0).
Therefore,
e−λtV ′(x(t)).f(x(t))− λe−λtV (x(t)) ≥ 0, t ≥ 0
⇒ ddt
[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) /∈ B�(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 ẋ = f(x) and letV : S → R be a continuously
differentiable function. Let � > 0 and an open set O ⊂ B�(0)
suchthat
V (x) > 0 ∀x ∈ O, supx∈OV (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 ⊂ B�(0) for t ≥ 0. Note that
V (x(t)) = V (x(0)) +
ˆ t0V̇ (x(s))ds
= V (x(0)) +
ˆ t0V ′(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 ∩ B�(0). This implies thatx(T )
/∈ ∂O ∩ B�(0) hence, x(T ) ∈ ∂O \ B�(0). Observe that ∂O = Ō ∩ ∂O
and Ō ⊂ B̄�(0). Thisimplies that
∂O \ B�(0) = (Ō ∩ ∂O) \ B�(0) ⊂ (B̄�(0) ∩ ∂O) \ B�(0) = ∂O ∩
∂B�(0).
Therefore, x(T ) ∈ ∂B�(0). Similarly for the second case, if
x(t) → ∂O as t → ∞, then x(t) →∂B�(0) as t → ∞. Thus, there does
not exists a δ > 0 such that if x0 ∈ Bδ(0), then x(t) ∈ B�(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 systemẋ = 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
semisimple).
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 −V̇ = 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 e
ATtCTCeAtdt. 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
semisimple. 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 semisimple. 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,
V̇ = V ′.f = 2xTP [Ax + r(x)]
= xT(ATP + PA)x + 2xTPr(x)
= −xTQx + 2xTPr(x). (41)
Using CauchySchwarz inequality and using −xTQx ≤
−λmin(Q)‖x‖22,
V̇ ≤ −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 � > 0 such that ‖r(x)‖2 <γ‖x‖2. Therefore, for
all x ∈ B�(0),
V̇ < −[λmin(Q)− 2γλmax(P )]‖x‖22. (43)
Choosing γ ≤ λmin(Q)/2λmax(P ), it follows that V̇ (x) < 0
for all x ∈ B�(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
ẋ1 = x2, ẋ2 = −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 ẋ =
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 ∈ R
m×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 ẋ = F(x,u) is locally exponentially stable.
Proof. The closed loop system ẋ = F(x,Kx) = f(x) and
∂f
∂xx=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 dynamical 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 timeindependent Lyapunov function.
(Existence of lower semicontinuous Lyapunov function Haddad.)
However, for asymptoticallystable equilibrium points of autonomous
systems, a continuously differentiable timeindependentLyapunov
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, � > 0 such that α‖x‖p ≤ V (x) ≤ β‖x‖p for all x ∈
Bδ(0) andV ′(x).f(x) < −�V (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, � > 0 suchthat α‖x‖2 ≤ V
(x) ≤ β‖x‖2 for all x ∈ Bδ(0) and V ′(x).f(x) < −�V (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 −V̇ (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, � > 0such that α‖x‖2 ≤ V
(x) ≤ β‖x‖2 for all x ∈ Bδ(0) and V ′(x).f(x) < −�V (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 ẋ = 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 V̇ ≤ 0
locally, V̇ (x, t) ≤ 0 ∀t ≥ t0, ∀x ∈ Br(0). To show localstability
of 0, we need to show that for given � > 0, t0 ≥ 0, there exists
δ = δ(�, t0) such that forall ‖x(t0)‖ < δ, ‖x(t)‖ < �, ∀t ≥
t0.
Define �1 := min(�, s, r). Choose δ > 0 such that
β(t0, δ) := sup‖x‖≤δV (x, t0) < α(�1).
Such δ always exists because, α(�1) > 0 and limδ→0β(t0, δ) =
limδ→0sup‖x‖≤δV (x, t0) = 0. Notethat
α(‖x(t0)‖) ≤ V (x(t0), t0) < α(�1).
Since α is increasing, this implies that ‖x(t0)‖ < �1. We
need to show that ‖x(t)‖ < �1 for allt ≥ t0. Suppose this is not
true. Let t1 > t0 be the first instant such that ‖x(t)‖ ≥ �1.
Then
V (x(t1), t1) ≥ α(�1) > V (x(t0), t0).
But this is a contradiction since V̇ (x, t) ≤ 0 for all ‖x‖ <
�1. Thus, ‖x(t)‖ < �1 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) 0,
17

there exists δ = δ(�, t0) > 0 and T (�, t0) < ∞ such that
‖x(t0)‖ < δ ⇒ ‖s(t, t0,x(t0))‖ < � for allt > T (�, 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, V̇ = ∂V∂t +
∂V∂x f(t,x) ≤ 0 for all x ∈ Br(0). Given � < min (r, s),
B�(0) ⊂ S. Choose
c > 0 such that c < min (min‖x‖=�α(‖x‖), r, s). Then, {x ∈
B�(0)  α(‖x‖) ≤ c} is in the interiorof B�(0). Define
Ωt,c := {x ∈ B�(0)  V (t,x) ≤ c}.
Since V (t,x) ≤ c ⇒ α(‖x‖) ≤ c, Ωt,c ⊂ {x ∈ B�(0)  α(‖x‖) ≤ c}.
Observe that V̇ ≤ 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 ⊂ B�(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 � > 0, let δ = α−12 (α1(�)). This shows
uniform stability.3. Since, V̇ < 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 V̇ ≤ − cbV = �V . 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,
ẋ(t) is bounded, w is uniformlycontinuous, and
´∞t0w(x(t))dt 0, there exists δx > 0 such that ‖x − y‖ <δx
⇒ ‖w(x)−w(y))‖ < �2 . Since ẋ 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)‖ < �2 .
Therefore, w isuniformly continuous w.r.t. t.
Suppose w(x(t)) 6→ 0 as t → ∞, i.e., there is an � > 0 and a
sequence {tk} → ∞ such thatw(x(tk)) ≥ � 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)) ≥ �−�
2=�
2.
Therefore,
ˆ tk+δtk
w(x(t))dt =ˆ tk+δtk
w(x(t))dt ≥ �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 (LaSalleYoshizawa)).
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
V̇ (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 V̇ (t,x) ≤ −W (x), and
integrate from t0 to t. Therefore,
V (t,x)− V (t0,x0) ≤ −ˆ tt0
W (x(s))ds⇒ˆ tt0
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 V̇ is
decreasing, if x(t0) ∈ Ωt0,ρ,x(t) ∈ Ωt0,ρ. Since V̇ 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 � > 0, there exists δx > 0 such that‖x− y‖ < δx ⇒ W
(x)−W (y) < �.
Since f is Lipshitz, ‖ẋ‖ = ‖f(t,x) − f(t,y)‖ ≤ L(‖x − y‖) (y in
some neighborhood of x), forall x ∈ D and for all t. Thus, ẋ 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
• V̇ 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
19

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) 0 and t0 ≥ 0, define δ(�, t0) := �m(t0)
.Then,
‖x(t0)‖ < δ ⇒ ‖x(t)‖ = ‖Φ(t, t0)x(t0)‖ ≤ ‖Φ(t, t0)‖‖x(t0)‖
< m(t0)δ = �.
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)‖ ≥ � 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)‖ > �δ1 . 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)‖ >
�.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 welldefined; moreover, P (t) is bounded as a function of
t.
Proof. Uniform asymptotic stability is equivalent to exponential
stability for LTV systems (Theorem 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)‖ 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)
∂xx=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 ẋ(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 characterized 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 exponential 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]), ObserverController
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 (isolated subsystems)
(Section 5.8.4, [2]).
4 Stability of periodic solutions and tracking trajectories
Let γ(t) be a periodic solution of ẋ = 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 ẏ = 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 systemẏ = f(t,y
+ r(t)) − ṙ(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, ṙ). Let z =[
z1z2
]=
[xy
]−[
rṙ
]. One needs to
find u such that 0 is an asymptotically stable equilibrium point
of
ż =
[ż1ż2
] [y − ṙ
M(x)−1[u− C(x,y)y]− r̈
].
Let u = −Kp(x− r)−Kd(y − ṙ) + C(x,y)ṙ +M(x)r̈ where Kp,Kd >
0. Therefore,
ż =
[y − ṙ
M(x)−1[−Kp(x− r)−Kd(y − ṙ)]
]=
[z2
M(x)−1[−Kpz1 −Kdz2 − C(x,y)z2]
]and ż has the required equilibrium point. Consider a Lyapunov
function
V (z1, z2) =1
2[zT2M(x)z2 + z
T1Kpz1]
⇒ V̇ = zT2M(x)ż2 +1
2zT2 Ṁ(x)z2 + z
T1Kpż1
= zT2 (−Kpz1 −Kdz2 − C(x,y)z2) +1
2zT2 Ṁ(x)z2 + z
T1Kpz2
= −zT2Kdz2since M − 2C is skew symmetric. Now we find the
largest invariant set. Observe that V̇ vanisheson Rn × 0. Note that
ż1 = 0 ⇒ z2 = 0 and ż1 = 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.
22

5 Input to state stability
(Haddad Khalil) Boundedness of solutionsConsider an LTI system
ẋ = Ax +Bu where A is Hurwitz. Recall that
x(t) = eA(t−t0)x(t0) +
ˆ tt0
eA(t−τ)Bu(τ)dτ.
Using ‖eA(t−t0)‖ ≤ ke−λ(t−t0),
‖x(t)‖ ≤ ke−λ(t−t0)‖x(t0)‖+ˆ tt0
ke−λ(t−τ)‖B‖‖u(τ)‖dτ
≤ ke−λ(t−t0)‖x(t0)‖+ k‖B‖supt0≤τ≤t‖u(τ)‖e−λtˆ tt0
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 � = �(δ) > 0,
independent of t0, such that
‖x(t0)‖ < δ ⇒ ‖x(t)‖ < �, ∀t ≥ t0. (53)
2. globally uniformly bounded if (53) holds for arbitrarily
large δ.
3. uniformly ultimately bounded with an ultimate bound � if
there exists γ > 0, independent oft0 ≥ 0, such that for every δ
∈ (0, γ), there exists T = T (δ, �), independent of t0 such
that
‖x(t0)‖ < δ ⇒ ‖x(t)‖ < �, ∀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 V̇ 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, V̇ satisfies V̇ ≤ −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 inputtostate 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 ẋ =
F(x,u) is inputtostate 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.
24

Theorem 5.5 (ISS for time varying systems). Let V : [0,∞) × Rn →
R be a continuously differentiable 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 systemẋ = F(t,x,u) is inputtostate 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 halfplane, 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 halfplane 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 continuouslydifferentiable. Then, (7) can be
rewritten as
ẋ = Ax + [f(x)−Ax] = Ax + f̂(x)
where f̂(x) = f(x) − Ax such that f̂(0) = 0 and ∂ f̂∂x(0) = 0.
Since we are interested in the casewhen 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
ẏ = A1y + g1(y, z) (64)
ż = A2z + g2(y, z) (65)
where g1 and g2 inherit properties of f̂ . 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.
25

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
ẏ = 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
ẏ = A1y + g1(y,w + h(y)) (68)
ẇ = 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⇒ ẇ =
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),
ẏ = A1y + g1(y,h(y)) +N1(y,w) (71)
ẇ = 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.
26

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
ẋ(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 dynamical system given by (73). A
continuously differentiable positivedefinite 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
ẋ(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 continuously differentiable
positivedefinite, 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)
27

In this case, the control Lyapunov function turns out to be the
Lyapunov function for the closedloop system with u = φ(x). Observe
that
V̇ (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 � > 0,there exists δ > 0 such that for
all 0 < ‖x‖ < δ, there exists u ∈ Rm such that ‖u‖ < �
andα(x) + βT(x)u < 0.
2. There exists a stabilizing control law φ̂(x) such that α(x) +
βT(x)φ̂(x) < 0,x ∈ Rn,x 6= 0and φ̂(x) is Lipschitz continuous at
x = 0 if and only if the control law φ̂(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 ẋ(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 tt0
A(s1)ds1 +
ˆ tt0
A(s1)
ˆ s1t0
A(s2)ds2ds1 + . . . .
and ddtΦ(t, t0) = A(t)Φ(t, t0). Hence,ddtΦ(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 concatenationproperty 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
Ċ = −(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 := 1T ln(C), hence,
e
RT = C . Let P (t) = Φ(t, 0)e−Rt. Clearly P (t) is
nonsingular.
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 and
continuous, it is bounded. Therefore the periodic linear system
is globally exponentially stable ⇔
28

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,
ẋ = Ṗy + P ẏ⇒ Ax = Ṗy + P ẏ⇒ APy = Ṗy + P ẏ⇒ ẏ = P−1(APy
− Ṗy).
Using P (t) = Φ(t, 0)e−Rt in the above equation,
ẏ = 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 ẋ = 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,
ddts(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,
ż = 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 thatV̇ ≤ 0 in some
open neighborhood N of 0. Then, one can apply LaSalleKrasovskii
theorem toconclude convergence of trajectories towards the
invariant set M of V where M ⊂ S ⊂ N suchthat S := {x ∈ N  V̇ (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.
29

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 semicontinuity 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].
30

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 radiall