Slotine and Li, Applied Nonlinear Control

Slotine and Li, Applied Nonlinear Control

LYAPUNOV STABILITY THEORY:

Given a control system, the first and most important question about its various properties is whether it is stable, because an unstable control system is typically useless and potentially dangerous. Qualitatively, a system is described as stable if starting the system somewhere near its desired operating point implies that it will stay around the point ever after. Every control system, whether linear or nonlinear, involves a stability problem which should be carefully studied.

The most useful and general approach for studying the stability of nonlinear control systems is the theory introduced in the late 19th century by the Russian mathematician Alexandr Mikhailovich Lyapunov. Lyapunov’s work, The General Problem of Motion Stability, includes two methods for stability analysis (the so-called linearization method and direct method) and was published in 1892. The linearization method draws conclusions about a nonlinear system’s local stability around an equilibrium point from the stability properties of its linear approximation. The direct method is not restricted to local motion, and determines the stability properties of a nonlinear system by constructing a scalar “energy-like” function for the system and examining the function’s time variation.

Today, Lyapunov’s linearization method has come to respresent the theoretical justification of linear control, while Lyapunov’s direct method has become the most important tool for nonlinear system analysis and design. Together, the linearization method and direct method constitute the so-called Lyapunov stability theorem.

A few simplifying notations are defined at this point. Let BR denote the spherical region (or ball) defined by ║x║<R in state-space, and SR the sphere itself, defined by ║x║=R.

STABILITY AND INSTABILITY:

Definition 1. The equilibrium state x=0 is said to be stable if, for any R>0, there exists r>0, such that if ║x(0)║<r, then ║x(t)║<R for all t≥0. Otherwise, the equilibrium point is unstable.

"thatimplies"for

"setthein"for

"existsthere"for

"anyfor"meanto

R)t(x,0tr)0(x,0r,0R

or equivalently

Rr B)t(x,0tB)0(x,0r,0R

0x0

1

2

3

Curve 1: asymptotically stable

Curve 2: marginally stable

Curve 3: unstable

Figure 1. Concepts of stability.

POSITIVE DEFINITE FUNCTIONS: The core of the Lyapunov stability theory is the analysis and construction of a class of functions to be defined and its derivative along the trajectories of the system under study. We start with the positive definite functions. In the following definition, D represents an open and connected subset of Rn.

Definition: A function V:D R is said to be positive semi definite in D if it satisfies the following conditions:

0Dinx,0)x(Vii

.0)0(VandD0i

V:D R is said to be positive definite in D if condition (ii) is replaced by (ii’)

(ii’) V(x)>0 in D-{0}.

Finally, V:D R is said to be negative definite (semi definite) in D if –V is positive definite (semi definite).

We will often abuse the notation slightly and write V>0, V≥0, and V<0 in D to indicate that V is positive definite, semi definite, and negative definite in D, respectively.

Example: The simplest and perhaps more important class of positive function is defined as follows,

TnxnTn QQ,Q,xQx:)x(V RRR

In this case, V(.) defines a quadratic form. Since by assumption, Q is symmetric (i.e., Q=QT), we have that its eigenvalues i, i=1,...n, are all real.Thus we have that

n,,1i,00x,0Qxxdefiniteseminegative)(V

n,,1i,00x,0Qxxdefinitenegative)(V

n,,1i,00x,0Qxxdefinitesemipositive)(V

n,,1i,00x,0Qxxdefinitepositive)(V

iT

iT

iT

iT

Thus for example:

.0a,0x

x

00

0ax,xax:)x(V

0b,a,0x

x

b0

0ax,xbxax:)x(V

2

121

21

22

2

121

22

21

21

RR

RR

V2(.) is not positive definite since for any x2≠0, any x of the form x*=[0,x2]T≠0;however, V2(x*)=0.

Positive definite function (PDFs) constitute the basic building block of the Lyapunov theory. PDFs can be seen as an abstraction of the total energy stored in a system, as we will see. All of the Lyapunov stability theorems focus on the study of the time derivative of a positive definite function along the trajectory of the system. In other words, given an autonomous system of the form dx/dt=f(x), we will first construct a positive definite function V(x) and study dV(x)/dt given by

)x(f

)x(f

x

V,,

x

V,

x

V

)x(fVdt

dx

x

V

dt

dV)x(V

n

1

n21

The following definition introduces a useful and very common way of representing this derivative.

Definition: Let V:D R and f:DRn. The Lie derivative of V along f,denoted by LfV, is defined by

)x(fx

V)x(VL

1f

Thus, according to this definition, we have that

).x(VL)x(fV)x(fx

V)x(V f

Example: Consider the system

12

1

xcosbx

axx

and define 22

21 xxV

Thus, we have

1222

21

12

121f

xcosx2bx2ax2

xcosbx

axx2,x2)x(VL)x(V

It is clear from this example that dV(x)/dt depends on the system’s equation f(x) and thus it will be different for different systems.

Stability Theorems:

Theorem 1. (Lyapunov Stability Theorem) Let x=0 be an equilibrium point of dx/dt=f(x), f:DRn, and let V:DR be a continuously differentiable function such that

,0Din0)x(V)iii(

,0Din0)x(V)ii(

,0)0(V)i(

Thus x=0 is stable.

In other words, the theorem implies that a sufficient condition for the stability of the equilibrium point x=0 is that there exists a continuously differentiable-positive definite function V(x) such that dV(x)/dt is negative semi definite in a neighborhood of x=0.

Theorem 2. (Asymptotic Stability Theorem) Under the conditions of Theorem 1, if V(.) is such that

,0Din0)x(V)iii(

,0Din0)x(V)ii(

,0)0(V)i(

Thus x=0 is asymptotically stable.

In other words, the theorem says that asymptotic stability is achieved if the conditions of Theorem 1 are strengthened by requiring dV(x)/dt to be negative definite, rather than semi definite.

Marquez, HJ, Nonlinear Control Systems

Examples:

Pendulum without friction

θl

mg

The equation of motion of the system is

0sing

0sinmgm

l

l

Choosing state variables

2

1

x

xwe have

12

21

xsing

x

xx

l

To study the stability of the equilibrium at the origin, we need to propose a Lyapunov function candidate (LFC) V(x) and show that satisfies the properties of one of the stability theorems seen so far. In general, choosing this function is rather difficult; however, in this case we proceed inspired by our understanding of the physical system. Namely, we compute the total energy of the pendulum (which is a positive function), and use this quantity as our Lyapunov function candidate. We have

mghm2

1E

PKE

2

l

where

122

2

1

2

cosx-1mgxm2

1E

Thus

xcos1cos1h

x

ll

ll


We now define V(x)=E and investigate whether V(.) and its derivative dV(.)/dt satisfies the conditions of theorem 1 and/or 2. Clearly V(0)=0; thus defining property (i) is satisfied in both theorems. With respect to (ii) we see that because of the periodicity of cos(x1), we have that V(x)=0 whenever x=(x1,x2)T=(2k,0)T, k=1,2,.... Thus V(.) is not positive definite. This situation, however, can be easily remedied by restricting the domain of x1 to the interval (-2,2);i.e., we take V:DR, with D=((-2,2), R)T.

There remains to evaluate the derivative of v(.) along the trajectories of f(t). We have

0xsinmglxxsinxmg

xsingx

xm,xsinmg

)x(f),x(fx

V,

x

V

)x(fV)x(V

1212

1

2

22

1

T21

21

ll

ll

-8 -6 -4 -2 0 2 4 6 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1-cosx1

Thus dV(x)/dt=0 and the origin is stable by theorem 1.

This result is consistent with our physical observations. Indeed, a sipmle pendulum without friction is a conservative system. This means that the sum of the kinetic and potential energy remains constant. The pendulum will continue to balance without changing the amplitude of the oscillations and thus constitutes a stable system.

Example:

Pendulum with friction.

θl

mg

Viscous friction, b0sin

g

m

c

0sinmgcm

l

ll

212

21

xm

cxsin

gx

xx

l

2

1

x

x


Again x=0 is an equilibrium point. The energy is the same as previous example

0Din0xcos1mgxm2

1)x(VE 1

22

2 ll

22

2

21

2

22

1

T21

21

xc

xm

cxsin

gx

xm,xsinmg

)x(f),x(fx

V,

x

V

)x(fV)x(V

l

lll

Thus dV(x)/dt is negative semi-definite. It is not negative definite since dV(x)/dt=0 for x2=0, regardless of the value x1 (thus dV(x)/dt=0 along the x1 axis).

According to this analysis, we conclude that the origin is stable by theorem 1, but cannot conclude asymptotic stability as suggested by our intuitive analysis, since we were not able to establish the conditions of theorem 2. namely, dV(x)/dt is not negative definite in a neighborhood of x=0. The result is indeed disappointing since we know that a pendulum with friction has an asymptotically stable equilibrium point at origin.

Example:

Consider the following system

22

221212

222

22111

xxxxx

xxxxx

To study the equilibrium point at the origin we define

2221 xx

2

1)x(V


22

221

22

21

222

21

222121

222

21

21

222

2121

222

2211

21

T21

21

xxxx

xxxxxxxxxx

xxxx

xxxxx,x

)x(f),x(fx

V,

x

V

)x(fV)x(V

Thus, V(x)>0 and and dV(x)/dt<0, provided that 222

21 xx

and it follows that the origin is an asymptotically equilibrium point.

Asymptotic Stability in the Large:

When the equilibrium is asymptotically stable, it is often important to know under what conditions an initial state will converge to the equilibrium point. In the best possible case, any initial state will converge to the equilibrium point. An equilibrium point that has this property is said to be globally asymptotically stable, or asymptotically stable in the large.

Definition: A function V(x) is said to radially unbounded if

xas)x(V

The origin x=0 is globally asymptotically stable (stable in the large) if the following conditions are satisfied

0x0)x(V)iiii(

unboundedradiallyis)x(V)iii(

0x0)x(V)ii(

,0)0(V)i(


Example:

Consider the following system

2221121 xxxxx

2221212 xxxxx

222

21

22

2121

22

2112

21

T21

21

xx2

xxxx

xxxxx,x2

)x(f),x(fx

V,

x

V

)x(fV)x(V

Thus, V(x)>0 and dV(x)/dt<0 for all x. Morover, since V(.) is radially unbounded, it follows that the origin is globally asymptotically stable.

01

23

45

0

2

4

60

5

10

15

20

25

30

x1x2

V(x

)

22

21 xx)x(V

Construction of Lyapunov Functions:

The main shortcoming of the Lyapunov theory is the difficulty associated with the construction of suitable Lyapunov function. The “variable gradient” method is used for this purpose. This method is applicable to autonomous systems and often but not always leads to a desired Lyapunov function for a given system.

The Variable Gradient: The essence of this method is to assume that the gradient of the (unkown) Lyapunov function V(.) is known up to some adjustable parameters, and then finding V(.) itself by integrating the assumed gradient. In other words, we start out by assuming that

)x(g)x(V )x(f)x(g)x(V)x(V

The power of this method relies on the following fact. Given that

)x(g)x(V )x(dVdx)x(Vdx)x(g Thus, we have that

b

a

b

a

x

x

x

xab dx)x(gdx)x(V)x(V)x(V

2221

122

211

1121 xhxh,xhxhg,g)x(g Example function:

that is, the difference V(xb)-V(xa) depends on the initial and final states xa and xb and not on the particular path followed when going from xa to xb. This property is often used to obtain V by integrating along the coordinate axis: )x(V

n

2

1

x

0 nn21n

x

0 2212

x

0 111

x

0

ds)s,,x,x(g

ds)0,,0,s,x(g

ds)0,,0,s(gdx)x(g)x(V

The free parameters in the function g(x) are constrained to satisfy certain symmetry conditions, satisfied by all gradients of a scalar function.

Theorem: A function g(x) is the gradient of a scalar function V(x) if and only if the matrix

n

n

n

2

n

1

2

n

2

2

2

1

1

n

1

2

1

1

x

g

x

g

x

g

x

g

x

g

x

gx

g

x

g

x

g

)x(g

is symmetric.

Example:

Consider the following system:

22122

11

xxxbx

xax

Clearly the origin is an equilibrium point. To study the stability of the equilibrium point, we proceed to find a Lyapunov function as follows,

Step 1: Assume that has the form)x(g)x(V

2221

122

211

11 xhxh,xhxh)x(g

Step 2: Impose the symmetry conditions

i

j

j

i

ij

2

ji

2

x

g

x

glyequivalent,or

xx

V

xx

V


In our case we have

1

22

21

12

112

1

2

2

21

221

2

11

12

1

x

hx

x

hxh

x

g

x

hxh

x

hx

x

g

To simplify the solution, we attempt to solve the problem assuming that the hij’s

are constant. If this is the case, then

0x

h

x

h

x

h

x

h

1

22

1

12

2

21

2

11

and we have that

222121

11

12

21

1

2

2

1

xhkx,kxxhxg

khhx

g

x

g

2221

1121 xh,xhg,g)x(g

In particular, choosing k=0, we have

Step 3: VFind

2

22122

21

11

2212

12

221

11

xxxbhxah

xxbx

axxh,xh

)x(f)x(g

)x(fV)x(V


Step 4: Find V from by integrationV

Integration along the axes, we have that

22

22

21

11

22

x

0

2211

x

0

11

x

0

x

0 2212111

xh2

1xh

2

1

dsshdssh

dss,xgds0,sg)x(V

21

1 2

Step 5: Verify that V>0 and dV/dt<0. We have that

2221

22

21

11

22

22

21

11

xxxbhxah)x(V

xh2

1xh

2

1)x(V

V(x)>0 if and only if

.1hhthatthenAssume.0handh 22

11

22

11

2221

21 xxxbax)x(V

Assume now that a>0, and b<0. In this case

2221

21 xxxbax)x(V

And we conclude that, under these conditions, the origin is (locally) asymptotically stable.


Slotine and Li, Applied Nonlinear Control

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