Internal (Lyapunov) Stability

Home , Lyapunov stability

Lectures on Dynamic Systems and

Control

Mohammed Dahleh Munther A. Dahleh George Verghese

Department of Electrical Engineering and Computer Science

Massachuasetts Institute of Technology

c �

Chapter 13

Internal (Lyapunov) Stability

13.1 Intro duction

We have already seen some examples of b oth stable and unstable systems. The ob jective of

chapter is to formalize the notion of internal stability for general nonlinear state-space this

mo dels. Apart from de�ning the various notions of stability, we de�ne an entity known as a

various stability notions. Lyapunov function and relate it to these

13.2 Notions of Stability

For a general undriven system

x_ (t) � f (x(t)� 0� t) (C T ) (13.1)

x(k + 1) � f (x(k )� 0� k ) (D T )� (13.2)

we say that a p oint x is an equilibrium point from time t for the CT system ab ove if f (x� 0� t) �

, and is an equilibrium 0� 8t � t p oint from time k for the DT system ab ove if f ( x� 0� k ) �

0 0

, it will remain there for all x at time t or k � . If the system is started in the state � x� 8k � k

0 0 0

have multiple equilibrium p oints (or equilibria). (Another class time. Nonlinear systems can

sp ecial solutions for nonlinear systems are periodic solutions, but we shall just fo cus on of

We would like to characterize the stability of the equilibria in some fashion. equilibria here.)

For example, do es the state tend to return to the equilibrium p oint after a small p erturbation

away from it� Do es it remain close to the equilibrium p oint in some sense� Do es it diverge�

The most fruitful notion of stability for an equilibrium p oint of a nonlinear system is

given by the de�nition b elow. We shall assume that the equilibrium p oint of interest is at

the origin, since if 6� 0, a simple translation can always b e applied to obtain an equivalent x

system with the equilibrium at 0.

De�nition 13.1 A system is called asymptotical ly stable around its equilibrium p oint at the

following two conditions: origin if it satis�es the

, then : 1. Given any � � 0� 9� � 0 such that if kx(t )k � � kx(t)k � �� 8 t � t

0 1 0 1

, then 2. 9� � 0 such that if kx(t )k � � x(t) ! 0 as t ! 1.

0 2 2

The �rst condition requires that the state tra jectory can b e con�ned to an arbitrarily

centered at the equilibrium p oint and of radius �, when released from an arbitrary small \ball"

. This is called stability in su�ciently small (but p ositive) radius � initial condition in a ball of

p ossible to have stability in the sense of Lyapunov without the sense of Lyapunov (i.s.L.). It is

having asymptotic stability, in which case we refer to the equilibrium p oint as marginal ly

stable. Nonlinear systems also exist that satisfy the second requirement without b eing stable

following example shows. An equilibrium p oint that is not stable i.s.L. is termed i.s.L., as the

unstable.

Example 13.1 (Unstable Equilibrium Point That Attracts All Tra jectories)

Consider the second-order system with state variables x and x whose dynamics

1 2

describ ed in p olar co ordinates via the equations are most easily

r_ � r (1 ; r )

� � sin (� �2) (13.3)

2 2

where the radius r is given by r � x + x and the angle � by 0 � � �

1 2

) � (x �x 2� . (You might try obtaining a state-space description directly arctan

2 1

.) It is easy to see that there are precisely involving x and x two equilibrium

2 1

p oints: one at the origin, and the other at r � 1, � � 0. We leave you to verify

simulation from various initial conditions) with rough calculations (or computer

tra jectories of the system have the form shown in the �gure b elow. that the

Evidently all tra jectories (except the trivial one that starts and stays at the origin)

However, this equilibrium p oint is not stable i.s.L., end up at r � 1, � � 0.

b ecause these tra jectories cannot b e con�ned to an arbitrarily small ball around

p oint when they are released from arbitrary p oints with any ball the equilibrium

how small) around this equilibrium. (no matter y

unit circle

Figure 13.1: System Tra jectories

13.3 Stability of Linear Systems

We may apply the preceding de�nitions to the LTI case by considering a system with a

p oint diagonalizable A matrix (in our standard notation) and u � 0. The unique equilibrium

0, provided A has no eigenvalue at 0 (resp ectively 1) in the CT (resp ectively DT) is at x �

every p oint in the entire eigenspace corresp onding to this eigenvalue is an case. (Otherwise

Now equilibrium.)

x_ (t) � e x(0)

2 3

t �

6 7

� V W x(0) (C T ) (13.4)

4 5

� t

x(k ) � A x(0)

2 3

�

6 7

� V W x(0) (D T ) (13.5)

4 5

�

Hence, it is clear that in continuous time a system with a diagonalizable A is asymptotically

stable i�

) � Re(� 0� i 2 f1� : : : � ng� (13.6)

while in discrete time the requirement is that

j � 1 j� i 2 f1� : : : � ng� (13.7)

) j � 1 (DT), the system is not asymptotically stable, but Note that if Re(� � 0 (CT) or j�

i i

is marginally stable.

Exercise: For the nondiagonalizable case, use your understanding of the Jordan form to show

stability are the same as in the diagonalizable case. For that the conditions for asymptotic

) � 0, with marginal stability, we require in the CT case that Re(� equality holding for at

eigenvalue� furthermore, every eigenvalue whose real part equals 0 should have its least one

multiplicity equal to its algebraic multiplicity, i.e., all its asso ciated Jordan blo cks geometric

b e of size 1. (Verify that the presence of Jordan blo cks of size greater than one for should

eigenvalues would lead to the state variables growing polynomial ly with these imaginary-axis

stability in the DT case. time.) A similar condition holds for marginal

Stability of Linear Time-Varying Systems

Recall that the general unforced solution to a linear time-varying system is

x(t) � �(t� t )x(t )�

0 0

where �(t� � ) is the state transition matrix. It follows that the system is

1. stable i.s.L. at ) � k�(t� t )k � m(t 1. x � 0 if sup

0 0

2. asymptotically stable at . x � 0 if lim k�(t� t )k ! 0� 8t

0 0

t!1

These conditions follow directly from De�nition 13.1.

13.4 Lyapunov's Direct Metho d

General Idea

Consider the continuous-time system

x_ (t) � f (x(t)) (13.8)

with an equilibrium p oint at x � 0. This is a time-invariant (or \autonomous") system, since f

do es not dep end explicitly on t. The stability analysis of the equilibrium p oint in such a system

we cannot write a simple formula is a di�cult task in general. This is due to the fact that

tra jectory to the initial state. The idea b ehind Lyapunov's \direct" metho d is to relating the

prop erties of the equilibrium p oint (or, more generally, of the nonlinear system) by establish

how certain carefully selected scalar functions of the state evolve as the system state studying

evolves. (The term \direct" is to contrast this approach with Lyapunov's \indirect" metho d,

which attempts to establish prop erties of the equilibrium p oint by studying the b ehavior of

linearized system at that p oint. We shall study this next Chapter.) the

Consider, for instance, a continuous scalar function V (x) that is 0 at the origin and

p ositive elsewhere in some ball enclosing the origin, i.e. V (0) � 0 and V (x) � 0 for x 6� 0 in

this ball. Such a V (x) may b e thought of as an \energy" function. Let V (x) denote the time

derivative of V (x) along any tra jectory of the system, i.e. its rate of change as x(t) varies

according to (13.8). If this derivative is negative throughout the region (except at the origin),

over time. In this case, b ecause the then this implies that the energy is strictly decreasing

lower b ounded by 0, the energy must go to 0, which implies that all tra jectories energy is

converge to the zero state. We will formalize this idea in the following sections.

Lyapunov Functions

De�nition 13.2 Let V b e a continuous map from R to R . We call V (x) a local ly positive

(lp d) function around x � 0 if de�nite

1. V (0) � 0.

2. V (x) � 0� 0 � kxk � r for some r .

Similarly, the function is called lo cally positive semide�nite (lpsd) if the strict inequality on

by V (x) � 0. The function V (x) is lo cally the function in the second condition is replaced

negative de�nite (lnd) if ;V (x) is lp d, and lo cally negative semide�nite (lnsd) if ;V (x) is

may b e useful in forming a mental picture of an lp d function V (x) is to think of lpsd. What

having \contours" of constant V that form (at least in a small region around the origin) it as

a nested set of closed surfaces surrounding the origin. The situation for n � 2 is illustrated

in Figure 13.2.

V(x)=c 2

V(x)=c 3

V(x)=c 1

Figure 13.2: Level lines for a Lyapunov function, where c � c � c .

1 2 3

Throughout our treatment of the CT case, we shall restrict ourselves to V (x) that have

continuous �rst partial derivatives. (Di�erentiability will not b e needed in the DT case |

continuity will su�ce there.) We shall denote the derivative of such a V with resp ect to time

along a trajectory of the system (13.8) by V (x(t)). This derivative is given by

dV (x) (x) dV

V (x(t)) � x_ � f (x)

dx dx

dV (x)

where is a row vector | the gradient vector or Jacobian of V with resp ect to x |

@ V

containing the comp onent-wise partial derivatives .

@ x

De�nition 13.3 Let V b e an lp d function (a \candidate Lyapunov function"), and let V b e

its derivative along tra jectories of system (13.8). If V is lnsd, then V is called a Lyapunov

function of the system (13.8).

Lyapunov Theorem for Lo cal Stability

Theorem 13.1 If there exists a Lyapunov function of system (13.8), then x � 0 is a stable

equilibrium p oint in the sense of Lyapunov. If in addition V (x) � 0, 0 � kxk � r for some

, i.e. if V is lnd, then x � 0 is an asymptotically stable equilibrium r p oint.

Pro of: First, we prove stability in the sense of Lyapunov. Supp ose � � 0 is given. We need

�nd a � � 0 such that for all kx(0)k � � , it follows that kx(t)k � �� 8t � 0. The Figure to

19.6 illustrates the constructions of the pro of for the case n � 2. Let � � min(�� r ). De�ne

ε1

r δ

Figure 13.3: Illustration of the neighb orho o ds used in the pro of

m � min V (x):

kxk��

Since V (x) is continuous, the ab ove m is well de�ned and p ositive. Cho ose � satisfying

� � � � such that for all kxk � � , V (x) � m. Such a choice is always p ossible, again 0

b ecause of the continuity of V (x). Now, consider any x(0) such that kx(0)k � � , V (x(0)) � m,

and let x(t) b e the resulting tra jectory. V (x(t)) is non-increasing (i.e. V (x(t)) � 0) which

. (x(t)) � m. We will show that this implies that kx(t)k � � Supp ose there results in V

t such that kx(t )k � � , then by continuity we must have that at an earlier time t , exists

1 1 1 2

kV (x)k � m � V (x(t )), which is a contradiction. Thus stability kx(t )k � � , and min

2 2 1

kxk��

Lyapunov holds. in the sense of

To prove asymptotic stability when V is lnd, we need to show that as t ! 1, V (x(t)) !

by continuity of V , kx(t)k ! 0. Since V (x(t)) is strictly decreasing, and V (x(t)) � 0 0� then,

we know that V (x(t)) ! c, with c � 0. We want to show that c is in fact zero. We can argue

by contradiction and supp ose that c � 0. Let the set S b e de�ned as

S � fx 2 R jV (x) � cg �

and let B b e a ball inside S of radius �,

�

B � f x 2 S jkxk � �g :

�

Supp ose x(t) is a tra jectory of the system that starts at x(0), we know that V (x(t)) is

� recall that 2 B (x(t)) � c for all t. Therefore, x(t) � decreasing monotonically to c and V

�

for which V (x) � c. In the �rst part of B � S which is de�ned as all the elements in R

�

pro of, we have established that if kx(0)k � � then kx(t)k � �. We can de�ne the largest the

derivative of V (x) as

;� � max V (x):

��kxk��

Clearly ;� � 0 since V (x) is lnd. Observe that,

V (x(t) � V (x(0)) + V (x(� ))d�

� V (x(0)) ; � t�

which implies that V (x(t)) will b e negative which will result in a contradiction establishing

must b e zero. the fact that c

Example 13.2 Consider the dynamical system which is governed by the di�er-

ential equation

x_ � ;g (x)

where g (x) has the form given in Figure 13.4. Clearly the origin is an equilibrium

p oint. If we de�ne a function

V (x) � g (y )dy

then it is clear that V (x) is lo cally p ositive de�nite (lp d) and

V (x) � ;g (x)

which is lo cally negative de�nite (lnd). This implies that x � 0 is an asymptotically

p oint. stable equilibrium g(x)

-1 1 x

Figure 13.4: Graphical Description of g (x)

Lyapunov Theorem for Global Asymptotic Stability

The region in the state space for which our earlier results hold is determined by the region

over which V (x) serves as a Lyapunov function. It is of sp ecial interest to determine the

p oint, i.e. the set of initial \basin of attraction" of an asymptotically stable equilibrium

subsequent tra jectories end up at this equilibrium p oint. An equilibrium conditions whose

p oint is global ly asymptotical ly stable (or asymptotically stable \in the large") if its basin of

entire state space. attraction is the

If a function V (x) is p ositive de�nite on the entire state space, and has the additional

prop erty that jV (x)j % 1 as kxk % 1, and if its derivative V is negative de�nite on the

entire state space, then the equilibrium p oint at the origin is globally asymptotically stable.

We omit the pro of of this result. Other versions of such results can b e stated, but are also

omitted.

Example 13.3

Consider the nth-order system

x_ � ;C (x)

with the prop erty that C (0) � 0 and x C (x) � 0 if x 6� 0. Convince yourself that

p oint of the system is at 0. Now consider the candidate the unique equilibrium

Lyapunov function

V (x) � x x

which satis�es all the desired prop erties, including jV (x)j % 1 as kxk % 1.

Evaluating its derivative along tra jectories, we get

0 0

V (x) � 2x x_ � ;2x C (x) � 0 for x �6 0

Hence, the system is globally asymptotically stable.

Example 13.4 Consider the following dynamical system

x_ � ;x + 4x

1 1 2

x_ � ;x ; x :

2 1

The only equlibrium p oint for this system is the origin x � 0. To investigate the

2 2

stability of the origin let us prop ose a quadratic Lyapunov function V � x + ax ,

1 2

p ositive constant to b e determined. It is clear that V is p ositive where a is a

de�nite on the entire state space R . In addition, V is radially unb ounded, that

jV (x)j % 1 as kxk % 1. The derivative of V along the tra jectories is it satis�es

given by of the system is

" #

h i

;x + 4x

1 2

V � 2x 2ax

1 2

;x ; x

2 4

� ;2x + (8 ; 2a)x x ; 2ax :

1 2

, and the If we cho ose a � 4 then we can eliminate the cross term x x derivative

1 2

b ecomes of V

2 4

V � ;2x ; 8x �

1 2

which is clearly a negative de�nite function on the entire state space. Therefore

we conclude that x � 0 is a globally asymptotically stable equilibrium p oint.

Example 13.5 A highly studied example in the area of dynamical systems and

chaos is the famous Lorenz system, which is a nonlinear system that evolves in R

whose equations are given by

x_ � � (y ; x)

y_ � r x ; y ; xz

z_ � xy ; bz �

where � , r and b are p ositive constants. This system of equations provides an

approximate mo del of a horizontal �uid layer that is heated from b elow. The

warmer �uid from the b ottom rises and thus causes convection currents. This

approximates what happ ens in the atmosphere. Under intense heating this mo del

b ehaviour. However, in this example we would like to exhibits complex dynamical

stability of the origin under the condition r � 1, which is known not to analyze the

2 2 2

, , and + � y + � z , where � � b ehaviour. Le us de�ne V � � x lead to complex

2 3 1 2 1

� are p ositive constants to b e determined. It is clear that V is p ositive de�nite

on R and is radially unb ounded. The derivative of V along the tra jectories of the

given by system is

2 3

� (y ; x)

h i

6 7

y z V � 2� x 2� 2� r x ; y ; xz

4 5

1 2 3

xy ; bz

2 2 2

y � ;2� � x ; 2� ; 2� bz

1 2 3

� ) + +xy (2� + 2r � (2� ; 2� )xy z :

1 2 2 3

If we cho ose � � � � 1 and � � then the V b ecomes

2 3 1

�

� �

2 2 2

V � ;2 x + y + 2bz ; (1 + r )xy

" #

� � � �

1 1 + r

2 2 2

� ;2 x ; (1 + r )y + 1 ; ( ) y + bz :

2 2

1+r

Since 0 � r � 1 it follows that 0 � � 1 and therefore V is negative de�nite on

the entire state space R . This implies that the origin is globally asymptotically

stable.

Example 13.6 (Pendulum)

The dynamic equation of a p endulum comprising a mass M at the end of a rigid

ro d of length R is but massless

�

M R � + M g sin � � 0

where � is the angle made with the downward direction, and g is the acceleration

due to gravity. To put the system in state-space form, let x � � , and x � � �

1 2

then

x_ � x

1 2

x_ � ; sin x

2 1

Take as a candidate Lyapunov function the total energy in the system. Then

2 2

V (x) � ) � kinetic + M R x + M g R (1 ; cos x p otential

" #

V � ] f (x) � [M g R sin x M R x

2 1

; sin x

� 0

Hence, V is a Lyapunov function and the system is stable i.s.L. We cannot conclude

asymptotic stability with this analysis.

Consider now adding a damping torque prop ortional to the velo city, so that the

state-space description b ecomes

x_ � x

1 2

x_ � ;D x ; sin x

2 2 1

With this change, but the same V as b efore, we �nd

2 2

V � ;D M R x � 0:

From this we can conclude stability i.s.L. We still cannot directly conclude asymp-

totic stability. Notice however that V � 0 ) � � 0. Under this condition,

� �

� ;(g �R ) sin � : Hence, � 6� 0 if � �6 k � for integer k , i.e. if the p endulum is not �

vertically down or vertically up. This implies that, unless we are at the b ottom or

� _

�6 0 when V � 0, so � will not remain at velo city, we shall have � top with zero

Lyapunov function will b egin to decrease again. The only place 0, and hence the

velo city, hanging vertically down or the system can end up, therefore, is with zero

vertically up, i.e. at one of the two equilibria. The formal pro of of this standing

invariant set theorem") is b eyond the scop e result in the general case (\LaSalle's

of this course.

The conclusion of lo cal asymptotic stability can also b e obtained directly through

- alternative choice of Lyapunov function. Consider the Lyapunov function can an

didate

1 1

2 2

V (x) � x + (x + x ) + 2(1 ; cos x ):

2 1 1

2 2

It follows that

2 2

V � ;(x + x sin x ) � ; ; (� + � sin � ) � 0:

1 1

2 2

_ _ _

Also, � + � sin � � 0 ) � � 0� � sin � � 0 ) � � 0� � � 0: Hence, V is strictly

negative in a small neighb orho o d around 0. This proves asymptotic stability.

Discrete-Time Systems

Essentially identical results hold for the system

x(k + 1) � f (x(k )) (13.9)

provided we interpret V as

V (x) � V (f (x)) ; V (x) �

i.e. as

state ) ; V (present state ) V (next

Example 13.7 (DT System)

Consider the system

x (k )

x (k + 1) �

1 + x (k )

x (k )

x (k + 1) �

1 + x (k ) 2

which has its only equilibrium at the origin. If we cho ose the quadratic Lyapunov

function

2 2

V (x) � x + x

1 2

we �nd

� �

V (x(k )) � V (x(k )) ; 1 � 0

[1 + x (k )]

from which we can conclude that the equilibrium p oint is stable i.s.L. In fact,

ab ove relations more carefully (in the same style as we did for the examining the

p endulum with damping), it is p ossible to conclude that the equilibrium p oint is

global ly asymptotical ly stable. actually

Notes

The system in Example 2 is taken from the eminently readable text by F. Verhulst, Nonlinear

Di�erential Equations and Dynamical Systems, Springer-Verlag, 1990.

Exercises

Exercise 13.1 Consider the horizontal motion of a particle of unit mass sliding under the in�uence

gravity on a frictionless wire. It can b e shown that, if the wire is b ent so that its height h is given of

by h(x) � V (x), then a state-space mo del for the motion is given by

�

x_ � z

z_ � ; V (x)�

�

4 2

Supp ose V (x) � x ; �x .

�

(a) Verify that the ab ove mo del has (z � x) � (0� 0) as equilibrium p oint for any � in the interval

� �

�

;1 � � � 1, and it also has (z � x) � 0� � p oints when � is in the interval as equilibrium

0 � � � 1.

(b) Verify that the linearized mo del ab out any of the equilibrium p oints is neither asymptotically

any � in the interval ;1 � � � 1. stable nor unstable for

Exercise 13.2 Consider the dynamic system describ ed b elow:

y + cy y� + a y_ + a � u + u_ �

1 2

where y is the output and u is the input.

(a) Obtain a state-space realization of dimension 2 that describ es the ab ove system.

(b) If a � 3� a � 2� c � 2, show that the system is asymptotically stable at the origin.

1 2

converges to zero as t approaches in�nity. This is known as initial state starting in this region,

a region of attraction.

Exercise 13.3 Consider the system

(x) dP

x_ (t) � ;

(x) has continuous �rst partial derivatives. The function P (x) is referred to as the potential where P

x b e a gradient system. Let � b e an isolated lo cal function of the system, and the system is said to

) minimum of P (x), i.e. P (x � P (x) for 0 � kx ; � � x � r , some r . k

(a) Show that x� is an equilibrium p oint of the gradient system.

(b) Use the candidate Lyapunov function

V (x) � P (x) ; P (x� )

to try and establish that x� is an asymptotically stable equilibrium p oint.

Exercise 13.4 The ob jective of this problem is to analyze the convergence of the gradient algorithm

�

! R and assume that x is a lo cal minimum� i.e., lo cal minimum of a function. Let f : R for �nding a

� �

) � f (x) for all x close enough but not equal to x . Assume that f is continuously di�erentiable. f (x

: R ! R b e the gradient of f : g Let

@ g @ g

g � ( : : : ) :

@ x @ x

1 n

�

It follows from elementary Calculus that g (x ) � 0.

�

, then it is argued that the solution to the dynamic system: go o d estimate of x If one has a

x_ � ;g (x) (13.10)

�

with x(0) close to x will give x(t) such that

�

lim x(t) � x :

t!1

(a) Use Lyapunov stability analysis metho ds to give a precise statement and a pro of of the ab ove

argument.

(b) System 13.10 is usually solved numerically by the discrete-time system

x(k + 1) � x(k ) ; �(x )g (x )� (13.11)

k k

) is some function from where �(x R ! R . In certain situations, � can b e chosen as a constant

choice is not always go o d. Use Lyapunov stability analysis metho ds for function, but this

) so that give a p ossible choice for �(x discrete-time systems to

�

lim x(k + 1) � x :

k !1

f (x) � x Qx� Q Symmetric, Positive De�nite:

�

Show directly that system 13.10 converges to zero (� x ). Also, show that � in system 13.11

b e chosen as a real constant, and give tight b ounds on this choice. can

Exercise 13.5 (a) Show that any (p ossibly complex) square matrix M can b e written uniquely as

0 0

� H and S � ;S . skew-Hermitian matrix S , i.e. H the sum of a Hermitian matrix H and a

.) Note that if M is real, then this decomp osition (Hint: Work with combinations of M and M

skew-symmetric matrix. expresses the matrix as the sum of a symmetric and

0 0

(b) With M , H , and S as ab ove, show that the real part of the quadratic form x M x equals x H x,

0 0 0

M x equals x S x. (It follows that if M and x are real, then x M x � x and the imaginary part of

H x.) x

row vector in this case | whose j th entry is @ V (x)�@ x , show that matrix | actually just a

dV (x)

� 2x H

where H is the symmetric part of M , as de�ned in part (a).

(d) Show that a Hermitian matrix always has real eigenvalues, and that the eigenvectors asso ciated

eigenvalues are orthogonal to each other. with distinct

Exercise 13.6 Consider the (real) continuous-time LTI system x_ (t) � Ax(t).

(a) Supp ose the (continuous-time) Lyapunov equation

P A + A P � ;I (3:1)

has a symmetric, p ositive de�nite solution P . Note that (3.1) can b e written as a linear system

entries of P , so solving it is in principle straightforward� go o d numerical of equations in the

algorithms exist.

Show that the function V (x) � x P x serves as a Lyapunov function, and use it to deduce the

stability of the equilibrium p oint of the LTI system ab ove, i.e. to deduce that global asymptotic

eigenvalues of A are in the op en left-half plane. (The result of Exercise 13.5 will b e helpful the

in computing V (x).)

What part (a) shows is that the existence of a symmetric, p ositive de�nite solution of (3.1) is

given LTI system is asymptotically stable. The existence of such su�cient to conclude that the

b e necessary, as we show in what follows. [Instead of ;I on the a solution turns out to also

right side of (3.1), we could have had ;Q for any p ositive de�nite matrix Q. It would still b e

p ositive true that the system is asymptotically stable if and only if the solution P is symmetric,

We leave you to mo dify the arguments here to handle this case.] de�nite.

(b) Supp ose the LTI system ab ove is asymptotically stable. Now de�ne

A t At

P � R (t)dt � R (t) � e e (3:2)

The reason the integral exists is that the system is asymptotically stable | explain this in more

Show that P is symmetric and p ositive de�nite, and that it is the unique solution of the detail!

Lyapunov equation (3.1). You will �nd it helpful to note that

dR (t)

R (1) ; R (0) � dt

The results of this problem show that one can decide whether a matrix A has all its eigenvalues

op en left-half plane without solving for all its eigenvalues. We only need to test for the in the

p ositive de�niteness of the solution of the linear system of equations (3.1). This can b e simpler.

Exercise 13.7 This problem uses Lyapunov's direct metho d to justify a key claim of his indirect

method: if the linearized mo del at an equilibrium p oint is asymptotically stable, then this equilibrium

p oint of the nonlinear system is asymptotically stable. (We shall actually only consider an equilibrium

p oint at the origin, but the approach can b e applied to any equilibrium p oint, after an appropriate

change of variables.)

time-invariant continuous-time nonlinear system given by Consider the

x_ (t) � Ax(t) + h(x(t)) (4:1)

where A has all its eigenvalues in the op en left-half plane, and h(:) represents \higher-order terms", in

kh(x)k�kxk ! 0 as kxk ! 0. the sense that

(a) Show that the origin is an equilibrium p oint of the system (4.1), and that the linearized mo del at

x_ (t) � Ax(t). the origin is just

(b) Let P b e the p ositive de�nite solution of the Lyapunov equation in (3.1). Show that V (x) � x P x

quali�es as a candidate Lyapunov function for testing the stability of the equilibrium p oint at

the origin in the system (4.1). Determine an expression for V (x), the rate of change of V (x)

tra jectories of (4.1) along

0 2

kP k) of the ratio kh(x)k�kxk will allow you to conclude that V (x(t)) � 0 for x(t) �6 0� Now

you can indeed limit kh(x)k�kxk to this small a value by cho osing a small enough argue that

neighb orho o d of the equilibrium. In this neighb orho o d, therefore, V (x(t)) � 0 for x(t) �6 0. By

Lyapunov's direct metho d, this implies asymptotic stability of the equilibrium p oint.

Exercise 13.8 For the discrete-time LTI system x(k + 1) � Ax(k ), let V (x) � x P x, where P is a

p ositive de�nite matrix. What condition will guarantee that V (x) is a Lyapunov function symmetric,

involving A and P will guarantee asymptotic stability of the system� for this system� What condition

your answers in terms of the p ositive semide�niteness and de�niteness of a matrix.) (Express MIT OpenCourseWare http://ocw.mit.edu

6.241J / 16.338J Dynamic Systems and Control Spring 2011

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