Ch 9.6: Liapunov’s Second Method

Ch 9.6: Liapunov’s Second Method

In Section 9.3 we showed how the stability of a critical point of an almost linear system can usually be determined from a study of the corresponding linear system.

However, no conclusion can be drawn when the critical point is a center of the corresponding linear system.

Also, for an asymptotically stable critical point, we may want to investigate the basin of attraction, for which the localized almost linear theory provides no information.

In this section we discuss Liapunov’s second method, or direct method, in which no knowledge of the solution is required.

Instead, conclusions about the stability of a critical point are obtained by constructing a suitable auxiliary function.

Physical Principles

Liapunov’s second method is a generalization of two physical principles for conservative systems.

The first principle is that a rest position is stable if the potential energy is a local minimum, otherwise it is unstable.

The second principle states that the total energy is a constant during any motion.

To illustrate these concepts, we again consider the undamped pendulum, which is a conservative system.

Undamped Pendulum Equation (1 of 5)

The governing equation for the undamped pendulum is

To convert this equation into a system of two first order equations, we let x = and y = d /dt, obtaining

The potential energy U is the work done in lifting pendulum above its lowest position:

0sin2

2

L

g

dt

d

xL

g

dt

dyy

dt

dxsin,

)cos1(),( xmgLyxU

Undamped Pendulum System: Potential Energy (2 of 5)

The critical points of our system

are x= n , y = 0, for n = 0, 1, 2,….

Physically, we expect the points (2n , 0) to be stable, since the pendulum bob is vertical with the weight down, and the points ((2n+1) , 0) to be unstable, since the pendulum bob is vertical with the weight up.

Comparing this with the potential energy U,

we see that U is a minimum equal to zero at (2n , 0) and U is a maximum equal to 2mgL at ((2n+1) , 0).

xL

g

dt

dyy

dt

dxsin,

),cos1(),( xmgLyxU

Undamped Pendulum System: Total Energy (3 of 5)

The total energy V is the sum of potential and kinetic energy:

On a solution trajectory x = (t), y = (t), V is a function of t.

The derivative of V((t), (t)) with respect to t is called the rate of change of V following the trajectory.

For x = (t), y = (t), and using the chain rule, we obtain

Since x and y satisfy the differential equations

it follows that dV(, )/dt = 0, and hence V is constant.

22)2/1()cos1(),( ymLxmgLyxV

dt

dyymL

dt

dxxmgL

dt

dV

dt

dV

dt

dVyx

2sin),(),(,

,/sin/,/ Lxgdtdyydtdx

Undamped Pendulum System: Small Energy Trajectories (4 of 5)

Observe that we computed the rate of change dV(, )/dt of the total energy V without solving the system of equations.

It is this fact that enables us to use Liapunov’s second method for systems whose solution we do not know.

Note that V = 0 at the stable critical points (2n , 0), where we recall

If the initial state (x1, y1) of the pendulum is sufficiently near a stable critical point, then the energy V(x1, y1) is small, and the corresponding trajectory will stay close to the critical point.

It can be shown that if V(x1, y1) is sufficiently small, then the trajectory is closed and contains the critical point.


Undamped Pendulum System: Small Energy Elliptical Trajectories (5 of 5)

Suppose (x1, y1) is near (0,0), and that V(x1, y1) is very small. The energy equation of the corresponding trajectory is

From the Taylor series expansion of cos x about x = 0, we have

Thus the equation of the trajectory is approximately

This is an ellipse enclosing the origin. The smaller V(x1, y1) is, the smaller the axes of the ellipse are. Physically, this trajectory corresponds to a periodic solution, whose motion is a small oscillation about equilibrium point.

2211 )2/1()cos1(),( ymLxmgLyxV

2/)!4/!2/1(1cos1 242 xxxx

1/),(2/),(2 2

11

2

11

2

mLyxV

y

mgLyxV

x

Damped Pendulum System: Total Energy (1 of 2)

If damping is present, we may expect that the amplitude of motion decays in time and that the stable critical point (center) becomes an asymptotically stable critical point (spiral point).

Recall from Section 9.3 that the system of equations is

The total energy is still given by

Recalling

it follows that dV/dt = -cLy2 0.

yLmcxLgdtdyydtdx )/(sin)/(/,/


,sin

, 2

dt

dyymL

dt

dxxmgL

dt

dV

Damped Pendulum System: Nonincreasing Total Energy (2 of 2)

Thus dV/dt = -cLy2 0, and hence the energy is nonincreasing along any trajectory, and except for the line y = 0, the motion is such that the energy decreases.

Therefore each trajectory must approach a point of minimum energy, or a stable equilibrium point.

If dV/dt < 0 instead of dV/dt 0, we can expect this to hold for all trajectories that start sufficiently close to the origin.

General Autonomous System: Total Energy

To pursue these ideas further, consider the autonomous system

and suppose (0,0) is an asymptotically stable critical point.

Then there exists a domain D containing (0,0) such that every trajectory that starts in D must approach (0,0) as t .

Suppose there is an “energy” function V such that V(x, y) 0 for (x, y) in D with V = 0 only at (0,0).

Since each trajectory in D approaches (0,0) as t , then following any particular trajectory, V approaches 0 as t .

The result we want is the converse: If V decreases to zero on every trajectory as t , then the trajectories approach (0,0) as t , and hence (0,0) is asymptotically stable.

),,(/),,(/ yxGdtdyyxFdtdx

Definitions: Definiteness

Let V be defined on a domain D containing the origin. Then we make the following definitions.

V is positive definite on D if V(0,0) = 0 and V(x, y) > 0 for all other points (x, y) in D.

V is negative definite on D if V(0,0) = 0 and V(x, y) < 0 for all other points (x, y) in D.

V is positive semi-definite on D if V(0,0) = 0 and V(x, y) 0 for all other points (x, y) in D.

V is negative semi-definite on D if V(0,0) = 0 and V(x, y) 0 for all other points (x, y) in D.

Example 1

Consider the function

Then V is positive definite on the domain

since V(0,0) = 0 and V(x, y) > 0 for all other points (x, y) in D.

22sin),( yxyxV

2/:),( 22 yxyxD

Example 2

Consider the function

Then V is only positive semi-definite on the domain

since V(x, y) = 0 on the line y = -x.

2),( yxyxV

2/:),( 22 yxyxD

Derivative of V With Respect to System

We also want to consider the function

where F and G are the functions given in the system

The function can be identified as the rate of change of V along the trajectory that passes through (x, y), and is often referred to as the derivative of V with respect to the system.

That is, if x = (t), y = (t) is a solution of our system, then

),,(),(),(),( yxGyxVyxFyxVV yx


V

yxGyxVyxFyxVdt

dV

dt

dV

dt

dV

yx

yx

),(),(),(),(

),(),(,

V

Theorem 9.6.1

Suppose that the origin is an isolated critical point of the autonomous system

If there is a function V that is continuous and has continuous first partial derivatives, is positive definite, and for which

is negative definite on a domain D in the xy-plane containing (0,0), then the origin is an asymptotically stable critical point.

If negative semidefinite, then (0,0) is a stable critical point.

See the text for an outline of the proof for this theorem.


),(),(),(),( yxGyxVyxFyxVV yx

V

Theorem 9.6.2


Let V be a function that is continuous and has continuous first partial derivatives.

Suppose V(0,0) = 0 and that in every neighborhood of (0,0) there is at least one point for which V is positive (negative).

If there is a domain D containing the origin such that

is positive definite (negative definite) on D, then the origin is an unstable critical point.

See the text for an outline of the proof for this theorem.

),(/),,(/ yxGdtdyyxFdtdx


Liapunov Function

The function V in Theorems 9.6.1 and 9.6.2 is called a Liapunov function.

The difficulty in using these theorems is that they tell us nothing about how to construct a Liapunov function, assuming that one exists.

In the case where the autonomous system represents a physical problem, it is natural to consider first the actual total energy of the system as a possible Liapunov function.

However, Theorems 9.6.1 and 9.6.2 are applicable in cases where the concept of physical energy is not pertinent.

In these cases, a trial-and-error approach may be necessary.

Example 3: Undamped Pendulum (1 of 3)

For the undamped pendulum system

use Theorem 9.6.1 show that (0,0) is a stable critical point, and use Theorem 9.6.2 to show (, 0) is an unstable critical point.

Let V be the total energy function

and let

Thus V is positive definite on D, since V > 0 on D, except at the origin, where V(0,0) = 0.

yxyxD ,2/2/:),(


,sin//,/ uLgdtdvvdtdu

Example 3: Critical Point at (0,0) (2 of 3)

Thus V is positive definite on D,

Further, as we have seen,

for all x and y. Thus is negative semidefinite on D.

Thus by Theorem 9.6.1, it follows that the origin is a stable critical point for the undamped pendulum.

To examine the critical point (, 0) using Theorem 9.6.2, we cannot use the same Liapunov function

since is not positive or negative definite.

yxyxD ,2/2/:),(

0sin 2 xymLyxmgLV

V

,)2/1()cos1(),( 22 ymLxmgLyxV

V

Example 3: Critical Point at (, 0) (3 of 3)

Consider the change of variable x = + u, and y = v. Then our system of differential equations becomes

with critical point (0, 0) in the uv-plane. Let V be defined by

and let D be the domain

Then V(u, v) > 0 in the first and third quadrants, and

is positive definite on D.

Thus (0, 0) in the uv-plane, or (, 0) in xy-plane, is unstable.

vuyxD ,4/4/:),(

uvvuV sin),(

,sin//,/ uLgdtdvvdtdu

uLguvuLguvuvV 22 sin/cossin/sincos

Theorem 9.6.3


Let V be a function that is continuous and has continuous first partial derivatives.

If there exists a bounded domain DK containing the origin on which V(x, y) < K, with V is positive definite and

negative definite, then every solution of the system above that starts at a point in DK approaches the origin as t .

Thus DK gives a region of asymptotic stability, but may not be the entire basin of attraction.

),(/),,(/ yxGdtdyyxFdtdx


Liapunov Function Discussion

Theorems 9.6.1 and 9.6.2 give sufficient conditions for stability and instability, respectively.

However, these conditions are not necessary, nor does our failure to determine a suitable Liapunov function mean that there is not one.

Unfortunately, there are no general methods for the construction of Liapunov functions.

However, there has been extensive work on the construction of Liapunov functions for special classes of equations.

An algebraic result that is often useful in constructing positive or negative definite functions is stated in the next theorem.

Theorem 9.6.4

Let V be the function defined by

Then V is positive definite if and only if

and is negative definite if and only if

22 cybxyaxV

,04 and 0 2 baca

.04 and 0 2 baca

Example 4

Consider the system

We try to construct a Liapunov function of the form

Then

If we choose b = 0, and a, c to be any positive numbers, then

is negative definite, and V positive definite by Theorem 9.6.4.

Hence (0,0) is asymptotically stable, by Theorem 9.6.1.

yxydtdyxyxdtdx 22 /,/

22 cybxyaxV

)(2)2()(2

))(2())(2(22233222

22

yxycyxxyxybyxxa

yxycybxxyxbyaxV

)(2)(2 222222 yxycyxxaV

Example 5: Competing Species System (1 of 7)

Consider the system

In Example 1 of Section 9.4 we found that this system models a certain pair of competing species, and that the point (0.5,0.5) is asymptotically stable. We confirm this conclusion by finding a suitable Liapunov function.

We transform (0.5,0.5) to the origin by letting x = 0.5 + u, and y = 0.5 + v. Our system then becomes

)5.075.0(/),1(/ xyydtdyyxxdtdx

2

2

5.05.025.0/

,5.05.0/

vuvvudtdv

uvuvudtdu

Example 5: Liapunov Function (2 of 7)

We have

Consider a possible Liapunov function of the form

Then V is positive definite, so we only need to determine whether there is region of the origin in the uv-plane where

is negative definite.

22),( vuvuV

322322

22

2225.1

5.05.025.025.05.02

vuvvuuvuvu

vuvvuvuvuvuuV

2

2

5.05.025.0/

,5.05.0/

vuvvudtdv

uvuvudtdu

Example 5: Derivative With Respect to System (3 of 7)

To show that

is negative definite, it suffices to show that

is positive definite, at least for u and v sufficiently small.

Observe that the quadratic terms of H can be written as

and hence are positive definite.

The cubic terms may be of either sign. We show that in some neighborhood of the origin, the following inequality holds:

322322 2225.1 vuvvuuvuvuV

322322 2225.1),( vuvvuuvuvuvuH

,75.025.05.1 22222 vuvuvuvu

2223223 75.025.0222 vuvuvuvvuu

Example 5: Negative Definite (4 of 7)

To estimate the left side of the desired inequality

we introduce polar coordinates u = rcos and v = rsin.

Then

since |cos |, |sin | 1. It is then sufficient to require

2223223 75.025.0222 vuvuvuvvuu

,7

sin2sincossincos2cos2

sin2sincossincos2cos2

222

3

32233

32233

3223

r

r

r

vuvvuu

0357.028/125.025.07 2223 rrvur

Example 5: Asymptotically Stable Critical Point (5 of 7)

Thus for the domain D defined by

the hypotheses of Theorem 9.6.1 are satisfied, so the origin is an asymptotically stable critical point of the system

It follows that the point (0.5,0.5) is an asymptotically stable critical point of the original system of equations

,28/1:),( 22 vuvu

2

2

5.05.025.0/

,5.05.0/

vuvvudtdv

uvuvudtdu

)5.075.0(/

),1(/

xyydtdy

yxxdtdx

Example 5: Region of Asymptotic Stability (6 of 7)

Recall that the Liapunov function for this example is

If we refer to Theorem 9.6.3, then the preceding argument also shows that the disk

is a region of asymptotic stability for the system of equations)5.075.0(/),1(/ xyydtdyyxxdtdx

22 )5.0()5.0(),( yxyxV

28/1)5.0()5.0(:),( 2228/1 yxyxD

Example 5: Estimating Basin of Attraction (7 of 7)

The disk

is a severe underestimate of the full basin of attraction.

To obtain a better estimate of the actual basin of attraction from Theorem 9.6.3, we would need to estimate the terms in

more accurately, use a better (and possibly more complicated) Liapunov function, or both.

28/1)5.0()5.0(:),( 2228/1 yxyxD

2223223 75.025.0222 vuvuvuvvuu

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Ch 9.6: Liapunov’s Second Method