nonlinear control systems lecture 1:...

Nonlinear Control SystemsLecture 1: Basics

Claudio De Persis

31-03-2014ver. 1.1

1

Course outline

Aim Introduce basic concepts of nonlinear control systems

Schedule

Lecture 1 (31-3)Examples of nonlinear control systemsfundamental propertiesbasics of stability theory

Lecture 2 (7-4)Converse Lyapunov TheoryInput-to-state stability

Lecture 3 (14-4)Relative degreeZero dynamicsFeedback linearization

Lecture 4 (28-4)Backstepping approachto control design

Textbook H. Khalil, Nonlinear Systems, 3rd edition, Prentice Hall, 2002,Chapter 1, 2, and 3, Section 4.7–4.9, Chapter 13, Section 14.3

2

Lecture outline

Nonlinear systems

Existence of solutions

Stability

Lyapunov theorem

Invariance principle

3

Lecture outline

Nonlinear systems


Stability

Lyapunov theorem


4

Nonlinear systems

Systems of nonlinear equation with forcing terms (inputs)

x1 = f1(t, x1, . . . , xn, u1, . . . , up)x2 = f2(t, x1, . . . , xn, u1, . . . , up)

...xn = fn(t, x1, . . . , xn, u1, . . . , up)

and outputsy1 = h1(t, x1, . . . , xn, u1, . . . , up)y2 = h2(t, x1, . . . , xn, u1, . . . , up)

...yn = hn(t, x1, . . . , xn, u1, . . . , up)

5

Nonlinear systems

In compact formx = f (t, x , u)y = h(t, x , u)

If the maps f , h are independent of t, that is

x = f (x , u)y = h(x , u)

then the system is autonomous or time-invariant.

If in addition the input u appears linearly

x = f (x) + g(x)uy = h(x) + k(x)u

then the system is input affine.

If no forcing term is present (u = 0), then

x = f (t, x)

is the unforced state equation.6

Nonlinear vs. linear systems

Nonlinear autonomous systems

x = f (x , u)y = h(x , u)

Linear autonomous systems

x = Ax + Buy = Cx + Du

An analytical solution to a linear system is available

x(t) = eAtx(0) +

∫ t

0eA(t−s)Bu(s)ds

y(t) = CeAtx(0) +

∫ t

0CeA(t−s)Bu(s)ds + Du(t)

Finding an analytical expression for nonlinear ODEs is usually difficult

7


Nonlinear autonomous systems

x = f (x , u)y = h(x , u)

Linear autonomous systems

x = Ax + Buy = Cx + Du

A solution to a linear systems exists for all t.Solutions to nonlinear systems can have finite escape time

x = −x2, x(0) = −1

has a solution

x(t) =1

t − 1

Hence x(t)→ −∞ as t → 1−.8


Nonlinear/linear autonomous systems

x = f (x), x = Ax

Equilibrium point

x∗ is an equilibrium point for x = f (x) if it solves the system of algebraicequations

f (x) = 0

Isolated equilibrium point

x∗ is an isolated equilibrium if any neighborhood Ix∗ of x∗ contains noother equilibrium than x∗.

Linear systems have only one isolated system – the origin (because thenull space of A is either the origin or a subspace of dimension ≥ 1)Nonlinear systems have multiple isolated equilibria

9

Pendulum equation

Pendulum equation with friction1.2. EXAMPLES 5

mg

Figure 1.1: Pendulum.

1.2 Examples

1.2.1 Pendulum Equation

Consider the simple pendulum shown in Figure 1.1, where l denotes the length of the rod and m denotes the mass of the bob. Assume the rod is rigid and has zero mass. Let () denote the angle subtended by the rod and the vertical axis through the pivot point. The pendulum is free to swing in the vertical plane. The bob of the pendulum moves in a circle of radius Z. To write the equation of motion of the pendulum, let us identify the forces acting on the bob. There is a downward gravitational force equal to mg, where 9 is the acceleration due to gravity. There is also a frictional force resisting the motion, which we assume to be proportional to the speed of the bob with a coefficient of friction k. Using Newton's second law of motion, we can write the equation of motion in the tangential direction as

mZB = -mg sin () - kZiJ

Writing the equation of motion in the tangential direction has the advantage that the rod tension, which is in the normal direction, does not appear in the equation. We could have arrived at the same equation by writing the moment equation about the pivot point. To obtain a state model for the pendulum, let us take the state variables as Xl = () and X2 = iJ. Then, the state equations are

9 . k - -Smxl- -X2 l m

To find the equilibrium points, we set Xl = X2 = 0 and solve for Xl and X2:

o o 9 . k - - smxl - -X2

Z m

(1.5)

(1.6)

m`θ = −mg sin θ − k`θ

State variables x1 = θ, x2 = θ. State space form

x1 = x2

x2 = −g

`sin x1 −

k

mx2

10

Pendulum equation

Equilibria are the solutions to

0 = x2

0 = −g

`sin x1 −

k

mx2

that isx1 = ±kπ, k = 0,±1,±2, . . . , x2 = 0,

Infinite number of isolated equilibrium points50 CHAPTER 2. SECOND-ORDER SYSTEMS

8

6

4

2

-2

-6

-8 -8 -6 -4 -2 o 2 4 6 8 10

Figure 2.16: Phase portrait of the pendulum equation of Example 2.2.

X2 -10 sinxI - X2

A computer-generated phase portrait is shown in Figure 2.16. The phase portrait is periodic in Xl with period 27r. Consequently, all distinct features of the system's qualitative behavior can be captured by drawing the portrait in the vertical strip -7r :S Xl :S 7r. As we noted earlier, the equilibrium points (0,0), (27r,0), (-27r, 0), etc., correspond to the downward equilibrium position (0,0). Trajectories near these equilibrium points have the pattern of a stable focus. On the other hand, the equilibrium points at (7r, 0), (-7r, 0), etc., correspond to the upward equilibrium position (7r,0). Trajectories near these equilibrium points have the pattern of a saddle. The stable trajectories of the saddles at (7r, 0) and (-7r, 0) form separatrices which contain a region with the property that all trajectories in its interior approach the equilibrium point (0,0). This picture is repeated periodically. The fact that trajectories could appro.ach different equilibrium points correspond to the number of full swings a trajectory would take before it settles at the downward equilibrium position. For example, the trajectories starting at points A and B have the same initial position, but different speeds. The trajectory starting at A oscillates with decaying amplitude until it settles down at equilibrium. The trajectory starting at B, on the other hand, has more initial kinetic energy. It makes a full swing before it

11

Lecture outline

Nonlinear systems


Stability

Lyapunov theorem


12

Fundamental properties – Existence and uniqueness ofsolutions

Cauchy problemx = f (t, x), x(t0) = x0 (1)

f (t, x) is continuous in x and piece-wise continuous

Solution

A solution to the Cauchy problem over the interval [t0, t1] ⊂ R is anypiecewise continuously differentiable function x(·) : [t0, t1]→ Rn whichsatisfies (5) for all t ∈ [t0, t1].

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Theorem

∃ t1 > t0, a neighborhood Ix0 and a constant L > 0 such that

‖f (t, x)− f (t, y)‖ ≤ L‖x − y‖, ∀x , y ∈ Ix0 , ∀t ∈ [t0, t1] (3)

Then there exists and is unique a solution to (2) over [t0, t0 + δ] ⊂ [t0, t1].

Terminology – Locally Lipschitz

f (t, x) is Lipschitz at x0 if ∃Ix0 where (3) holds

f (t, x) is locally Lipschitz on a domain D ⊂ Rn if it is Lipschitz ateach x ∈ D

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Theorem

∃ t1 > t0, a neighborhood Ix0 and a constant L > 0 such that

‖f (t, x)− f (t, y)‖ ≤ L‖x − y‖, ∀x , y ∈ Ix0 , ∀t ∈ [t0, t1]

Then there exists and is unique a solution to (4) over [t0, t0 + δ] ⊂ [t0, t1].

Example

x = x1/3, x(0) = 0

Two solutions exist: (i) x(t) = 0, (ii) x(t) = (2t/3)3/2

f (x) not Lipschitz at x = 0.

15


Sufficient conditions for Lipschitz property

[a, b] ⊂ R time interval and W ⊂ Rn a convex set such that

∂f

∂x(t, x) ∈ C0([a, b]×W )

Then f (t, x) is locally Lipschitz on W , i.e.

‖f (t, x)− f (t, y)‖ ≤ L‖x − y‖, ∀x , y ∈W , ∀t ∈ [a, b]

Proof Let x , y ∈W . Then λx + (1− λ)y ∈W for all λ ∈ [0, 1] (byconvexity).Let φ(λ) := f (t, λx + (1− λ)y) and notice that

φ(1)− φ(0) = f (t, x)− f (t, y)

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Proof (continued) By the Fundamental Theorem of Calculus

φ(1)− φ(0) =

∫ 1

0φ′(λ)dλ

Recall that

φ′(λ) =d

dλf (t, λx + (1− λ)y)

=∂f

∂z(t, z)

∣∣∣∣z=λ(x−y)+y

(x − y)

Hence

f (t, x)− f (t, y) =

∫ 1

0

∂f

∂z(t, z)

∣∣∣∣z=λ(x−y)+y

dλ (x − y)

(Hadamard’s lemma)

17


Proof (continued)

f (t, x)− f (t, y) =

∫ 1

0

∂f

∂z(t, z)

∣∣∣∣z=λ(x−y)+y

dλ (x − y)

Taking the norm of both sides

‖f (t, x)− f (t, y)‖ =

∥∥∥∥∥∫ 1

0

∂f

∂z(t, z)

∣∣∣∣z=λ(x−y)+y

dλ (x − y)

∥∥∥∥∥≤

∫ 1

0

∥∥∥∥∥ ∂f

∂z(t, z)

∣∣∣∣z=λ(x−y)+y

∥∥∥∥∥ dλ ‖x − y‖

≤ L‖x − y‖

as claimed.

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Global existence and uniqueness of solutions


Theorem

f (t, x) piecewise continuous in t and

‖f (t, x)− f (t, y)‖ ≤ L‖x − y‖, ∀x , y ∈ Rn, ∀t ∈ [t0, t1]

Then there exists and is unique a solution to (5) for all t ∈ [t0, t1].

Example

x = A(t)x(t) + b(t)

A(t), b(t) piecewise continuous, ‖A(t)‖ ≤ a, ‖b(t)‖ ≤ b, for t ∈ [t0, t1].Then a solution exists and is unique for all t ∈ [t0, t1].

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A useful lemma

Comparison principle

Scalar differential equation

u = f (t, u), u(t0) = u0

with f (t, u) continuous in t and locally Lipschitz in u.

[t0,T ) be the maximal interval of existence of the solution u(t).

v(t) : [t0,T )→ R a C1 function such that

v(t) = f (t, v(t)), v(t0) ≤ u0

for all t ∈ [t0,T ).

Thenv(t) ≤ u(t), ∀t ≥ t0.

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Lecture outline

Nonlinear systems


Stability

Lyapunov theorem


21

A review of basic stability theory

Nonlinear systemx = f (t, x), x ∈ Rn, (6)

with map f : [0,∞)× Rn → Rn piecewise continuous in t and locallyLipschitz in x .

Equilibrium

A point x ∈ Rn is an equilibrium point for (6) if

f (t, x) = 0, ∀t ≥ t0

22

Basic stability theory

(Asymptotic) stability

The equilibrium x = 0 is

stable if, for any ε > 0, there exists δ(ε, t0) > 0 such that

||x(t0)|| < δ(ε, t0)⇒ ||x(t, x(t0))|| < ε ∀t ≥ t0

asymptotically stable if it is stable and there exists c(t0) > 0 such that

||x(t0)|| < c(t0)⇒ limt→+∞

||x(t, x(t0))|| = 0

102 CHAPTER 4. DYNAMIC BEHAVIOR

0 1 2 3 4 5 60

2

4

Statex

Time t

Figure 4.6: Illustration of Lyapunov’s concept of a stable solution. The solution representedby the solid line is stable if we can guarantee that all solutions remain within a tube ofdiameter ε by choosing initial conditions sufficiently close the solution.

space that satisfy the dynamics of the system. In many situations, stable limit cy-cles can be found by simulating the system with different initial conditions.

4.3 Stability

The stability of a solution determines whether or not solutions nearby the solutionremain close, get closer or move further away. We now give a formal definition ofstability and describe tests for determining whether a solution is stable.

Definitions

Let x(t;a) be a solution to the differential equation with initial condition a. Asolution is stable if other solutions that start near a stay close to x(t;a). Formally,we say that the solution x(t;a) is stable if for all ε > 0, there exists a δ > 0 suchthat

!b"a! < δ =# !x(t;b)" x(t;a)! < ε for all t > 0.

Note that this definition does not imply that x(t;b) approaches x(t;a) as time in-creases but just that it stays nearby. Furthermore, the value of δ may depend onε , so that if we wish to stay very close to the solution, we may have to start very,very close (δ $ ε). This type of stability, which is illustrated in Figure 4.6, is alsocalled stability in the sense of Lyapunov. If a solution is stable in this sense and thetrajectories do not converge, we say that the solution is neutrally stable.An important special case is when the solution x(t;a) = xe is an equilibrium

solution. Instead of saying that the solution is stable, we simply say that the equi-librium point is stable. An example of a neutrally stable equilibrium point is shownin Figure 4.7. From the phase portrait, we see that if we start near the equilibriumpoint, then we stay near the equilibrium point. Indeed, for this example, given anyε that defines the range of possible initial conditions, we can simply choose δ = εto satisfy the definition of stability since the trajectories are perfect circles.A solution x(t;a) is asymptotically stable if it is stable in the sense of Lyapunov

and also x(t;b) % x(t;a) as t % ∞ for b sufficiently close to a. This correspondsto the case where all nearby trajectories converge to the stable solution for largetime. Figure 4.8 shows an example of an asymptotically stable equilibrium point.

23


Uniform (asymptotic) stability

The equilibrium x = 0 is

uniformly stable if, for any ε > 0, there exists δ(ε) > 0 such that

||x(t0)|| < δ(ε)⇒ ||x(t, x(t0))|| < ε ∀t ≥ t0

uniformly asymptotically stable if it is uniformly stable and thereexists c > 0 such that

||x(t0)|| < c ⇒ limt→+∞

||x(t, x(t0))|| = 0

uniformly in t0, i.e.

∀ε > 0, ∃T (ε) > 0 s.t. ‖x(t)‖ < ε, ∀t ≥ t0 + T (ε), ∀‖x(t0)‖ < c

24


Example (non uniform asymptotic stability) The nonautonomousdifferential equation

x = − x

1 + t

has solution

x(t) =1 + t01 + t

x(t0)

It shows (global) asymptotic stability, but not uniform asymptotic stabilityconvergence, since

‖x(t)‖ < ε⇔ |1 + t| > 1

ε|x(t0)||1 + t0|

There exists no T independent of t0 to bound t(as t0 grows, ‖t‖ must grow as well)

25

Comparison functions

Class-K function

A function α : [0, a)→ R≥0 is a class-K function if it is continuous,strictly increasing and zero at zero (α(0) = 0).

Example α(r) = r1−r (restricted to [0, 1)) is a class-K function defined on

[0, 1).

Class-K∞ function

If a =∞ and limr→+∞ α(r) = +∞, then α is said to be a class-K∞function.

Example α(r) = r , r2, r12 are class-K∞ functions.

26


Properties

(Composition) If α1 : [0, a)→ R≥0, α2 : [0, b)→ R≥0 are class-Kfunctions and b = limr→a α1(r), then

α2 α1(r) := α2(α1(r))

is a class-K function defined on [0, a).The definition also applies to class-K∞ functions (verify this).

(Inverse) if α : [0, a)→ R≥0 is a class-K function, then there exists aunique class-K function denoted α−1 and defined on [0, b), withb = limr→a α(r), such that

α−1(α(r)) = r , ∀r ∈ [0, a).

Example The inverse of α(r) = r1−r is the class-K function

α−1(s) = s1+s defined on [0,∞) (verify this).

The definition also applies to class-K∞ functions. In this case α−1 isa class-K∞ function (verify this).

27


Class-KL function

A continuous function β : [0, a)× R≥0 → R≥0 is a class-KL function if(i) for each fixed s ∈ R≥0, the function β(r , s) is a class-K functiondefined on [0, a)(ii) for each fixed r ∈ [0, a), the function β(r , s) is a class-L function, i.e.it is a continuous decreasing function such that lims→∞ β(r , s) = 0.

Exampleβ(r , s) = α1(α2(r) exp(−s))

with α1, α2 class-K functions, is a class-KL function.

28

Stability and Comparison functions

Lemma

x = 0 is

1 uniformly stable iff ∃α ∈ K and c > 0 such that

‖x(t)‖ ≤ α(‖x(t0)‖), ∀t ≥ t0, ∀‖x(t0)‖ < c

2 uniformly asymptotically stable iff ∃β ∈ KL and c > 0 such that

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

If c =∞, then x = 0 is globally uniformly asymptotically stable.

Proof

1 If Let ε < α(c) wlog. Fix accordingly δ(ε) = α−1(ε) < c. Then‖x(t0)‖ < δ(ε) = α−1(ε) and

‖x(t)‖ ≤ α(‖x(t0)‖) < α(α−1(ε)) = ε

29


Lemma

x = 0 is


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


Proof

2 If Let ‖x(t0)‖ < c . Then

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0) ≤ β(‖x(t0)‖, 0) =: α(‖x(t0)‖)

and this shows uniform stability.

30


Lemma

x = 0 is


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


Proof

2 Since β ∈ KL and ‖x(t0)‖ < c, then for any ε > 0, ∃Tε such that

β(c,Tε) < ε.

Hence, for any ε > 0, ∃Tε such that t > t0 + Tε implies

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0) < β(c ,Tε) < ε

thus showing uniform attractivity of the origin.

31

Lecture outline

Nonlinear systems


Stability

Lyapunov theorem


32

Lyapunov theorem

Theorem

Let V : [0,∞)× Bd → R be a C 1 function α1, α2 be class-K functionsdefined on [0, d) which satisfy

α1(||x ||) ≤ V (t, x) ≤ α2(||x ||), ∀(t, x) ∈ [0,∞)× Bd

If∂V

∂t+∂V

∂xf (t, x) ≤ 0, ∀(t, x) ∈ [0,∞)× Bd

then x = 0 is a uniformly stable equilibrium.

If for some class-K function α defined on [0, d),

∂V

∂t+∂V

∂xf (t, x) ≤ −α(||x ||), ∀(t, x) ∈ [0,∞)× Bd

then x = 0 is a uniformly asymptotically stable equilibrium point.

33

Lyapunov theorem

Examplex = −(1 + g(t))x3, x ∈ R

g continuous and such that g(t) ≥ 0 for all t ≥ 0.Let

V (x) =1

2x2, α1(r) = α2(r) =

1

2r2 ∈ K∞

Then∂V

∂xf (t, x) = −(1 + g(t))x4 ≤ −x4, α(r) = r4

The origin is uniformly asymptotically stable.

34

Lyapunov theorem (cont’d)

Uniform stability Let ‖x(t0)‖ ≤ c , with c < α−12 α1(d), such that

Bc ⊂ x ∈ Rn : V (t0, x) < α1(d) ⊂ Bd

Compute the time derivative of V along the solution x(t) starting fromx(t0), namely

V (t, x(t)) =∂V

∂t+∂V

∂xf (t, x(t)) ≤ 0

ThenV (t, x(t)) ≤ V (t0, x(t0)), ∀t ≥ t0

Hence,

α1(||x(t)||) ≤ V (t, x(t)) ≤ V (t0, x(t0)) ≤ α2(||x(t0)||)

and therefore

||x(t)|| ≤ α−11 (α2(||x(t0)||)) = α(||x(t0)||)

Remark α2(||x(t0)||), α1(d); hence, α2(||x(t0)||) is in the domain of α−11 .

35


A useful result

Lemma

Consider the scalar differential equation

y(t) = −α(y(t))

with α a locally Lipschitz class-K function defined on [0, a). For anyy0 ∈ [0, a), the unique solution y(t) to the Cauchy problem

y(t) = −α(y(t)), y(0) = y0

satisfiesy(t) = β(y0, t)

where β : [0, a)× R≥0 → R≥0 is a class-KL function.

By the Comparison Lemma, the result also holds if “=” is replaced by “≤”.

36

Lyapunov theorem (cont’d)

Uniform asymptotic stability Let ‖x(t0)‖ ≤ c , with c < α−12 α1(d),such that

Bc ⊂ x ∈ Rn : V (t0, x) < α1(d) ⊂ Bd

Compute the time derivative of V along the solution x(t), namely

V (t, x(t)) =∂V

∂t+∂V

∂xf (t, x(t)) ≤ −α(||x(t)||) ≤ −α α−12 (V (t, x(t)))

If α α−12 is locally Lipschitz, then ∃β ∈ KL such that

V (t, x(t)) ≤ β(V (t0, x(t0)), t − t0)

Hence,α1(||x(t)||) ≤ V (t, x(t)) ≤ β(α2(||x(t0)||), t)

and therefore

||x(t)|| ≤ α−11 (β(α2(||x(t0)||), t)) = β(||x(t0)||, t)

37

Lyapunov theorem - Global uniform asymptotic stability

Theorem

Let V : [0,∞)× Rn → R be a C 1 function α1, α2 be class-K∞ functionswhich satisfy

α1(||x ||) ≤ V (t, x) ≤ α2(||x ||), ∀(t, x) ∈ [0,∞)× Rn

If for some class-K function α defined on Rn,

∂V

∂t+∂V

∂xf (t, x) ≤ −α(||x ||), ∀(t, x) ∈ [0,∞)× Rn

then x = 0 is a uniformly globally asymptotically stable equilibriumpoint.

The inequalityV (t, x(t)) ≤ −α α−12 (V (t, x(t)))

now holds for all x(t0) ∈ Rn and all t ≥ t0.38

Lyapunov theorem

Example Linear time-varying system

x = A(t)x , x ∈ Rn

A(t) continuous for all t ≥ 0.

∃P(t) = PT (t) > 0, P ∈ C1, and Q(t) = QT (t) > 0, Q ∈ C0, such thatfor all t ≥ t0

0 < c1I ≤ P(t) ≤ c2I , Q(t) ≥ c3I > 0

and−P(t) = AT (t)P(t) + P(t)A(t) + Q(t)

39

Lyapunov theorem

Example Lyapunov function

V (t, x) = xTP(t)x(t)

It satisfiesc1‖x‖2 ≤ V (t, x) ≤ c2‖x‖2

and

∂V

∂t+∂V

∂xf (t, x(t)) = xT P(t) + xT (AT (t)P(t) + P(t)A(t))x

= −xTQ(t)x≤ −c3‖x‖2

which shows uniform global asymptotic stability.

40

Lyapunov theorem

Example The result actually shows more

∂V

∂t+∂V

∂xf (t, x(t)) ≤ −c3‖x‖2 ≤ −

c3c2

V (t, x)

By the comparison lemma, along the trajectories x(t) of the system

V (t, x(t)) ≤ e− c3

c2(t−t0)V (t0, x(t0))

Hencec1‖x(t)‖2 ≤ e

− c3c2(t−t0)c2‖x(t0)‖2

showing uniform global exponential stability.

41

Exponential stability

Theorem

Let V : [0,∞)× Bd → R be a C 1 function, k1, k2, k3, c be positivenumbers.If

k1||x ||c ≤ V (t, x) ≤ k2||x ||c , ∀(t, x) ∈ [0,∞)× Bd

and∂V

∂t+∂V

∂xf (t, x) ≤ −k3||x ||c , ∀(t, x) ∈ [0,∞)× Bd

then x = 0 is an exponentially stable equilibrium.If d =∞, then x = 0 is a globally exponentially stable equilibrium.

42

Lyapunov theorem for autonomous systems

x = f (x)

Theorem

Let V : Bd → R be a C 1 function α1, α2 be class-K functions defined on[0, d) which satisfy

α1(||x ||) ≤ V (x) ≤ α2(||x ||), ∀x ∈ Bd

If∂V

∂xf (x) ≤ 0, ∀x ∈ Bd

then x = 0 is a stable equilibrium.

If for some class-K function α defined on [0, d),

∂V

∂xf (x) ≤ −α(||x ||), ∀x ∈ Bd

then x = 0 is an asymptotically stable equilibrium point.43

Lyapunov theorem

A few comments are in order

The requirement

α1(||x ||) ≤ V (x) ≤ α2(||x ||), ∀x ∈ Bd

is equivalent to requiring that V (x) is positive definite on anyBr ⊂ Bd (i.e. V (0) = 0 and V (x) > 0 for all x ∈ Br \ 0).

In fact, suppose the latter is true (the converse is trivial). Letr ∈ [0, d) and define

α(s) = mins≤||y ||≤r

V (y), 0 ≤ s ≤ r < d

Take x ∈ Br and set s = ||x ||. Then V (x) ≥ α(||x ||). Moreover,α(0) = 0, α is continuous, α in non-decreasing. To make it strictlyincreasing, take any class-K function α which satisfies

α(s) < α(s), ∀s ∈ [0, r ].

44

Lyapunov theorem

Similarly

The requirement

∂V

∂xf (x) ≤ −α(||x ||), ∀x ∈ Bd (7)

is equivalent to requiring ∂V∂x f (x) < 0 for all x ∈ Br \ 0 and

∂V∂x f (x) = 0 for x = 0.

45


This leads to a more “familiar” form of Lyapunov’s stability theorem

x = f (x)

Theorem

Let V : Bd → R be a C 1 function positive definite on [0, d), i.e.

V (0) = 0 and V (x) > 0 ∀x ∈ Bd \ 0

IfV (x) < 0 ∀x ∈ Bd \ 0

then the origin is an asymptotically stable equilibrium.

46


Examplex = −g(x)

g locally Lipschitz and such that

g(0) = 0, xg(x) > 0, ∀x ∈ (−a, a) \ 0118 CHAPTER 4.

g(x)

Figure 4.3: A possible nonlinearity in Example 4.2.

functions. In some cases, there are natural Lyapunov function candidates like energy functions in electrical or mechanical systems. In other cases, it is basically a matter of trial and error. The situation, however, is not as bad as it might seem. As we go over various examples and applications throughout the book, some ideas and approaches for searching for Lyapunov functions will be delineated.

Example 4.2 Consider the first-order differential equation

x = -g(x)

where g(x) is locally Lipschitz on (-a,a) and satisfies

g(O) = 0; xg(x) > 0, '\I x =1= 0 and x E (-a,a)

A sketch of a possible g(x) is shown in Figure 4.3. The system has an isolated equilibrium point at the origin. It is not difficult in this simple example to see that the origin is asymptotically stable, because solutions starting on either side of the origin will have to move toward the origin due to the sign of the derivative x. To arrive at the same conclusion using Lyapunov's theorem, consider the function

V(x) = fox g(y) dy

Over the domain D = (-a, a), V(x) is continuously differentiable, V(O) = 0, and V(x) > 0 for all x =1= O. Thus, V(x) is a valid Lyapunov function candidate. To see whether or not V(x) is indeed a Lyapunov function, we calculate its derivative along the trajectories of the system.

. oV 2 V(x) = ox [-g(x)] = -g (x) < 0, '\I xED - O

Hence, by Theorem 4.1 we conclude that the origin is asymptotically stable. D

Example 4.3 Consider the pendulum equation without friction, namely,

Xl x2

X2 - asinxI

47


Examplex = −g(x)

g locally Lipschitz and such that

g(0) = 0, xg(x) > 0, ∀x ∈ (−a, a) \ 0

Consider

V (x) =

∫ x

0g(y)dy

which is C1, positive definite on (−a, a) and such that

V (x) = g(x) · (−g(x)) = −g2(x) < 0 ∀x ∈ (−a, a) \ 0.

Hence the equilibrium x = 0 is asymptotically stable.

48

Global asymptotic stability for autonomous systems

This leads to a more “familiar” form of Lyapunov’s stability theorem

x = f (x)

Theorem

Let V : Rn → R be a C 1 function positive definite on Rn, i.e.

V (0) = 0 and V (x) > 0 ∀x ∈ Rn \ 0

Assume additionally that V (x) is radially unbounded, that is

lim‖x‖→+∞

V (x) = +∞.

IfV (x) < 0 ∀x ∈ Rn \ 0

then the origin is an asymptotically stable equilibrium.

49


The radial unboundedness of V(x) plays a key role.Example Consider the system

x1 =−6x1

(1 + x21 )2

+ 2x2

x2 =−2(x1 + x2)

(1 + x21 )2

and the candidate Lyapunov function

V (x) =x21

1 + x21

+ x22

The function is positive definite on R2. Moreover

V (x) = − 12x21

(1 + x21 )4− 4x2

2

(1 + x21 )2

< 0, ∀x ∈ R2 \ 0

50


Example The system is not globally asymptotically stable: solutions to theright of the branch in the 1st quadrant of the hyperbola

x2 =2

x1 −√

2

cannot cross that branch ([Khalil, Exercise 4.8]).4.1. AUTONOMOUS SYSTEMS 123

Figure 4.4: Lyapunov surfaces for V(x) = xr/(1 + xi) +

as t -7 00, no matter how large Ilxll is. If an asymptotically stable equilibrium point at the origin has this property, it is said to be globally asymptotically stable. Recalling again the proof of Theorem 4.1, we can see that global asymptotic stability can be established if any point x E Rn can be included in the interior of a bounded set nco It is obvious that for this condition to hold, the conditions of the theorem must hold globally, that is, D = Rn; but, is that enough? It turns out that we need more conditions to ensure that any point in Rn can be included in a bounded set nco The problem is that for large c, the set nc need not be bounded. Consider, for

. example, the function xi 2 V(x) = --2 + x2

1 + Xl

Figure 4.4 shows the surfaces V(x) = c for various positive values of C. For small c, the surface V(x) = c is closed; hence, nc is bounded since it is contained in a closed ball Br for some r > O. This is a consequence of the continuity and positive definiteness of V (x). As c increases, a value is reached after which the surface V(x) = c is open and nc is unbounded. For nc to be in the interior of a ball Bn c must satisfy c < infllxll2:r V(x). If

l = lim inf V(x) < (Xl r--+CX) IlxlI2:r

then nc will be bounded if c < l. In the preceding example,

. . [xi 2] . xi l = hm mm ---2 + x2 = hm --2 = 1 r->CX) Ilxll=r 1 + Xl IXll->CX) 1 + Xl

Thus, nc is bounded only for c < 1. An extra condition that ensures that nc is bounded for all values of c > 0 is

V(x) -7 (Xl as Ilxll -7 (Xl

A function satisfying this condition is said to be radially unbounded.

An intuitive explanation of the need for radial unboundedness can beobtained from the contour lines of V (x)

51

Lecture outline

Nonlinear systems


Stability

Lyapunov theorem


52


A motivating example Consider again the equations of a pendulum withfriction1.2. EXAMPLES 5

mg

Figure 1.1: Pendulum.

1.2 Examples

1.2.1 Pendulum Equation

Consider the simple pendulum shown in Figure 1.1, where l denotes the length of the rod and m denotes the mass of the bob. Assume the rod is rigid and has zero mass. Let () denote the angle subtended by the rod and the vertical axis through the pivot point. The pendulum is free to swing in the vertical plane. The bob of the pendulum moves in a circle of radius Z. To write the equation of motion of the pendulum, let us identify the forces acting on the bob. There is a downward gravitational force equal to mg, where 9 is the acceleration due to gravity. There is also a frictional force resisting the motion, which we assume to be proportional to the speed of the bob with a coefficient of friction k. Using Newton's second law of motion, we can write the equation of motion in the tangential direction as

mZB = -mg sin () - kZiJ

Writing the equation of motion in the tangential direction has the advantage that the rod tension, which is in the normal direction, does not appear in the equation. We could have arrived at the same equation by writing the moment equation about the pivot point. To obtain a state model for the pendulum, let us take the state variables as Xl = () and X2 = iJ. Then, the state equations are

9 . k - -Smxl- -X2 l m

To find the equilibrium points, we set Xl = X2 = 0 and solve for Xl and X2:

o o 9 . k - - smxl - -X2

Z m

(1.5)

(1.6)

x1 = x2

x2 = −g

`sin x1 −

k

mx2

where x1 = θ, x2 = θ.

53


Energy often serves as a valid Lyapunov function

V (x) =g

`(1− cos x1)︸︷︷︸

potential energy

+1

2x22︸︷︷︸

kinetic energy

V (x) is positive definite. On the other hand,

V (x) =g

`sin(x1) x2 + x2(−g

`sin x1 −

k

mx2) = − k

mx22

is only negative semi-definite. Stability is guaranteed but not asymptoticstability.

Energy is dissipated until x2 6= 0 but Lyapunov theorem does not capturethe fact that x1 = 0 when x2 = 0. The invariance principle does.

54


A few useful notions x = f (x) (autonomous system)Let x(t) be a solution to the system.

Limit points and positive limit set

A point p is a positive limit point of x(t) if there exists a sequence tnsuch that limn→+∞ tn = +∞ and

limn→+∞

‖x(tn)− p‖ = 0

The set of all positive limit points of x(t) is the positive limit set

Example The stable limit cycle is the positive limit set of any solutionconverging to it

55


A few useful notions

x = f (x), (autonomous system)

Invariant set

A set S is an invariant set with respect to x = f (x) if for any initialcondition x0 ∈ S , the corresponding solution x(t) satisfies

x(t) ∈ S , ∀t ∈ R

A set S is a positively invariant set with respect to x = f (x) if

x(t) ∈ S , ∀t ≥ 0

Example Consider the compact level set Ωc = x ∈ Rn : V (x) ≤ c andassume that V (x) = ∂V

∂x f (x) ≤ 0 for all x ∈ Ωc . Ωc is a positivelyinvariant set for x = f (x).

56

Birchoff’s lemma

Lemma

Let x(t) be a solution tox = f (x)

that exists and bounded for all t ≥ 0.Then is positive limit set L+ is non-empty compact and invariant.Moreovera

x(t)→ L+, as t → +∞ax(t) is said to approach a set S as t approaches infinity, x(t)→ S as

t → +∞, if

∀ε > 0, ∃Tε > 0 s.t. dist(x(t), S) < ε, ∀t > Tε

wheredist(x(t),S) = infp∈S‖x(t)− p‖

57

LaSalle’s theorem

Theorem

Ω a positively invariant set wrt

x = f (x)

V , C1 such thatV (x) ≤ 0, ∀x ∈ Ω

Let E ⊂ Ω be the set of points where V (x) = 0

E = x ∈ Ω : V (x) = 0

Then every solution to x = f (x) starting from Ω asymptoticallyapproaches the largest invariant set contained in E .

58

LaSalle’s theorem

Proof

1 x(t) ∈ Ω for all t ≥ 0. Hence, V (x(t)) is non-increasing for all t ≥ 0

2 Ω is a compact set and V (x) is a continuous function. Hence thereexists a minimum for V (x) on Ω (Weierstrass extremum theorem).

3 Since V (x(t)) is non-decreasing and bounded from below, then thereexists c ∈ R such that

limt→+∞

V (x(t)) = c

4 Since x(t) is bounded, its positive limit set L+ is nonempty. Take anyp ∈ L+. By definition, there exists a sequence tk such thatx(tk)→ p. Hence by continuity

limtk→+∞

V (x(tk)) = V ( limtk→+∞

x(tk)) = V (p) = c

Since p is any, then V (p) = c for all p ∈ L+.

59

LaSalle’s theorem

4 Since L+ is an invariant, then any trajectory x(t) starting in L+

remains in L+, that is V (x(t)) = c for all t. This implies that

d

dtV (x(t)) = 0, ∀t

Since this applies to any trajectory in L+, then it is true thatV (p) = 0 for all p ∈ L+, that is

L+ ⊂ E .

5 Since L+ is an invariant and x(t) ∈ L+, then it is proven that thesolutions converge to the largest invariant set contained in E .

60

LaSalle’s theorem

Examplex1 = x2

x2 = −g

`sin x1 −

k

mx2

Energy

V (x) =g

`(1− cos x1)︸︷︷︸

potential energy

+1

2x22︸︷︷︸

kinetic energy

yields

V (x) = − k

mx22

V (x) has a local minimum around the origin. Hence, around the origin,level sets of V (x) are compact and invariant. We can apply LaSalletheorem and conclude that every solution converges to the largestinvariant set for the system where x2 = 0.

61

LaSalle’s theorem

Example

x1 = x2

x2 = −g

`sin x1 −

k

mx2

x2=0⇒x1 = 0

0 = −g

`sin x1

which implies that any invariant set around the origin such that x2 = 0must necessarily satisfy x1 = 0. This proves convergence to the origin.

62

Next lecture

Lecture 2: Converse Lyapunov theorems and ISS

(Prof. B. Jayawardhana)

nonlinear control systems lecture 1:...

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