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

d

2 

dt

2 

g L

sin   0 To convert this equation into a system of two first order equations, we let

x

=  and

y

=

d

 /

dt

, obtaining

dx



dt y

,

dy dt

 

g L

sin

x

The potential energy

U

is the work done in lifting pendulum above its lowest position:

U

(

x

,

y

) 

mgL

( 1  cos

x

)

Undamped Pendulum System: Potential Energy

(2 of 5) The critical points of our system

dx



dt

are

x

= 

n y

 , ,

dy dt y

 

g

sin

L

= 0, for

n x

= 0, 1, 2,…. Physically, we expect the points (  2

n

 , 0) to be stable, since the pendulum bob is vertical with the weight down, and the points (  (2

n+

1

)

 , 0) to be unstable, since the pendulum bob is vertical with the weight up. Comparing this with the potential energy

U

,

U

(

x

,

y

) 

mgL

( 1  cos

x

), we see that

U

is a minimum equal to zero at (  2

n

is a maximum equal to 2

mgL

at (  (2

n+

1

)

 , 0).

 , 0) and

U

Undamped Pendulum System: Total Energy

(3 of 5) The total energy

V

is the sum of potential and kinetic energy:

V

(

x

,

y

) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 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 dV

=  (

t

),

y

 

dt

,   =  (

t

), and using the chain rule, we obtain 

V x

(  ,  )

d



dt



V y

(  ,  )

d



dt



mgL

sin

x dx dt



mL

2

y dy dt

Since

x

and

y

satisfy the differential equations

dx

/

dt



y

,

dy

/

dt

 

g

sin it follows that

dV

(  ,  )/

dt x

/

L

, = 0, and hence

V

is constant.

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 (  2

n

 , 0), where we recall

V

(

x

,

y

) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 If the initial state (

x

1 ,

y

1 ) of the pendulum is sufficiently near a stable critical point, then the energy

V

(

x

1 ,

y

1 ) is small, and the corresponding trajectory will stay close to the critical point.

It can be shown that if

V

(

x

1 ,

y

1 ) is sufficiently small, then the trajectory is closed and contains the critical point.

Undamped Pendulum System: Small Energy Elliptical Trajectories

(5 of 5) Suppose (

x

1 ,

y

1 ) is near (0,0), and that

V

(

x

1 ,

y

1 ) is very small. The energy equation of the corresponding trajectory is

V

(

x

1 ,

y

1 ) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 From the Taylor series expansion of cos

x

about

x

= 0, we have 1  cos

x

 1  ( 1 

x

2 / 2 !



x

4 / 4 !

  ) 

x

2 / 2 Thus the equation of the trajectory is approximately 2

V

(

x

1 ,

x

2

y

1 ) /

mgL



y

2 2

V

(

x

1 ,

y

1 ) /

mL

2  1 This is an ellipse enclosing the origin. The smaller

V

(

x

1 ,

y

1 ) 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.

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

dx

/

dt



y

,

dy

/

dt

  (

g

/

L

) sin

x

 (

c

/

Lm

)

y

The total energy is still given by

V

(

x

,

y

) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 Recalling

dV

  , 

dt

 

mgL

sin

x dx dt



mL

2

y dy

,

dt

it follows that

dV

/

dt

= -

cLy

2  0.

Damped Pendulum System: Nonincreasing Total Energy

(2 of 2) Thus

dV

/

dt

= -

cLy

2  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

dx

/

dt



F

(

x

,

y

),

dy

/

dt



G

(

x

,

y

), and suppose (0,0) is an asymptotically stable critical point. Then there exists a domain

D

trajectory that starts in

D

containing (0,0) such that every 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

following any particular trajectory,

V

  , then approaches 0 as

t

  . The result we want is the converse: If

V

every trajectory as

t

as

t

   decreases to zero on , then the trajectories approach (0,0)  , and hence (0,0) is asymptotically stable.

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

V

(

x

,

y

)  sin 

x

2 

y

2  Then

V

is positive definite on the domain

D

  (

x

,

y

) :

x

2 

y

2   / 2  since

V

(0,0) = 0 and

V

(

x

,

y

) > 0 for all other points (

x

,

y

) in

D

.

Example 2

Consider the function

V

(

x

,

y

)  

x



y

 2 Then

V

is only positive semi-definite on the domain

D

  (

x

,

y

) :

x

2 

y

2   / 2  since

V

(

x

,

y

) = 0 on the line

y

= -

x

.

Derivative of

V

With Respect to System

We also want to consider the function

V

 

V x

(

x

,

y

)

F

(

x

,

y

) 

V y

(

x

,

y

)

G

(

x

,

y

), where

F

and

G dx

/

dt

 are the functions given in the system

F

(

x

,

y

),

dy

/

dt



G

(

x

,

y

), 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 dV

  ,   

dt

=  (

t

),

y V x

(  ,  ) 

V x

(

x

,

y

)

F

=  (

t

) is a solution of our system, then

d



dt

(

x

, 

V y y

)  (  , 

V y

( )

x

,

d



dt y

)

G

(

x

,

y

) 

V



Theorem 9.6.1

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

dx

/

dt



F

(

x

,

y

),

dy

/

dt



G

(

x

,

y

), If there is a function

V

that is continuous and has continuous first partial derivatives, is positive definite, and for which

V

 

V x

(

x

,

y

)

F

(

x

,

y

) 

V y

(

x

,

y

)

G

(

x

,

y

) is negative definite on a domain

D

in the

xy

-plane containing (0,0), then the origin is an asymptotically stable critical point.

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

Theorem 9.6.2

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

dx

/

dt



F

(

x

,

y

),

dy

/

dt



G

(

x

,

y

) 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 V

 

V x

(

x

,

y

)

F

(

x

,

y

) containing the origin such that 

V y

(

x

,

y

)

G

(

x

,

y

) 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.

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

du

/

dt



v

,

dv

/

dt

  

g

/

L

 sin

u

, 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

V

(

x

,

y

) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 and let

D

  (

x

,

y

) :   / 2 

x

  / 2 ,   

y

   Thus

V

is positive definite on

D

, since

V

> 0 on

D

, except at the origin, where

V

(0,0) = 0.

Example 3: Critical Point at (0,0)

(2 of 3) Thus

V D

  ( is positive definite on

D

,

x

,

y

) :   / 2 

x

  / 2 ,   

y

   Further, as we have seen,

V

  

mgL

sin

x



y

 

mL

2

y



x

 0 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

V

(

x

,

y

) 

mgL

( 1  cos

x

)  ( 1 / 2 )

mL

2

y

2 ,

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

du

/

dt



v

,

dv

/

dt

 

g

/

L

 sin

u

, with critical point (0, 0) in the

uv

-plane. Let

V

be defined by

V

(

u

,

v

) 

v

sin

u

and let

D D

  ( be the domain

x

,

y

) :   / 4 

u

  / 4 ,   

v

   Then

V

(

u

,

v

) > 0 in the first and third quadrants, and

V

  

v

cos

u

  sin

u



g

/

L

 sin

u

 

v

2 cos

u

 

g

/

L

 sin 2

u

is positive definite on

D

. Thus (0, 0) in the

uv

-plane, or (  , 0) in

xy

-plane, is unstable.

Theorem 9.6.3

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

dx

/

dt



F

(

x

,

y

),

dy

/

dt



G

(

x

,

y

) Let

V

be a function that is continuous and has continuous first partial derivatives. If there exists a bounded domain

D K

which

V

(

x

,

y

) <

K

, with

V

containing the origin on is positive definite and

V

 

V x

(

x

,

y

)

F

(

x

,

y

) 

V y

(

x

,

y

)

G

(

x

,

y

) negative definite, then every solution of the system above that starts at a point in

D K

approaches the origin as

t

  . Thus

D K

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

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

V



ax

2 

bxy



cy

2 Then

V a

is positive definite if and only if  0 and 4

ac



b

2  0 , and is negative definite if and only if

a

 0 and 4

ac



b

2  0 .

Example 4

Consider the system

dx

/

dt

 

x



xy

2 ,

dy

/

dt

 

y



x

2

y

We try to construct a Liapunov function of the form

V



ax

2 

bxy



cy

2 Then

V

   ( 2

ax

  2

a

( 

x

2

by

 )(

x

 2

x y

2  )

xy

2  ) 

b

( 2

xy

(

bx

  2

cy

)( 

y



xy

3 

x

2

y

)

x

3

y

)  2

c

(

y

2 

x

2

y

2 )  If we choose

b V

    2

a

(

x

2 = 0, and

a

,

c



x

2

y

2 )  2

c

( to be any positive numbers, then

y

2 

x

2

y

2 )  is negative definite, and

V

positive definite by Theorem 9.6.4.

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

Example 5: Competing Species System

(1 of 7) Consider the system

dx

/

dt



x

( 1 

x



y

),

dy

/

dt



y

( 0 .

75 

y

 0 .

5

x

) 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

du

/

dt

  0 .

5

u

 0 .

5

v



u

2 

uv

,

dv

/

dt

  0 .

25

u

 0 .

5

v

 0 .

5

uv



v

2

Example 5: Liapunov Function

(2 of 7) We have

du

/

dt

  0 .

5

u

 0 .

5

v



u

2 

uv

,

dv

/

dt

  0 .

25

u

 0 .

5

v

 0 .

5

uv



v

2 Consider a possible Liapunov function of the form

V

(

u

,

v

) 

u

2 

v

2 Then

V

is positive definite, so we only need to determine whether there is region of the origin in the

uv

-plane where

V

    2

u

  

u

 2   0 .

5

u

 1 .

5

uv

0 .

5

v



v

2   

u

2 2

u

 3

uv

   2

u

2

v

  0 .

25

u uv

2   2

v

3  0 .

5

v

  0 .

5

uv



v

2  is negative definite.

Example 5: Derivative With Respect to System

(3 of 7) To show that

V

    

u

2  1 .

5

uv



v

2 2

u

3  2

u

2

v



uv

2  2

v

3   is negative definite, it suffices to show that

H

(

u

,

v

)  

u

2  1 .

5

uv



v

2   2

u

3  2

u

2

v



uv

2  2

v

3  is positive definite, at least for

u

and

v

sufficiently small.

Observe that the quadratic terms of

H u

2  1 .

5

uv



v

2  0 .

25 

u

2 

v

2   0 .

75 

u

can be written as 

v

 2 , 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: 2

u

3  2

u

2

v



uv

2  2

v

3  0 .

25 

u

2 

v

2   0 .

75 

u



v

 2

Example 5: Negative Definite

(4 of 7) To estimate the left side of the desired inequality 2

u

3  2

u

2

v



uv

2  2

v

3  0 .

25 

u

2 

v

2   0 .

75 

u



v

 2 we introduce polar coordinates

u

=

r

cos  and

v

=

r

sin  . Then 2

u

3  2

u

2

v



uv

2  2

v

3  

r r

3 3 2  2 cos 3 cos 3    2 cos 2  sin   2 cos 2  sin   cos  sin 2   cos  sin 2   2 sin 3   2 sin 3    7

r

3 , since |cos  7

r

3  |, |sin  0 .

25 

u

2  | 

v

2   1. It is then sufficient to require 0 .

25

r

2 

r

 1 / 28  0 .

0357

Example 5: Asymptotically Stable Critical Point

(5 of 7) Thus for the domain

D

 (

u

,

v

) :

u

2 

v

2  1 / defined by 28  , the hypotheses of Theorem 9.6.1 are satisfied, so the origin is an asymptotically stable critical point of the system

du

/

dt

  0 .

5

u

 0 .

5

v



u

2 

uv

,

dv

/

dt

  0 .

25

u

 0 .

5

v

 0 .

5

uv



v

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

dx

/

dt dy

/

dt

 

x

( 1 

x



y

( 0 .

75 

y

),

y

 0 .

5

x

)

Example 5: Region of Asymptotic Stability

(6 of 7) Recall that the Liapunov function for this example is

V

(

x

,

y

)  (

x

 0 .

5 ) 2  (

y

 0 .

5 ) 2 If we refer to Theorem 9.6.3, then the preceding argument also shows that the disk

D

1 / 28   (

x

,

y

) : (

x

 0 .

5 ) 2  (

y

 0 .

5 ) 2  1 / 28  is a region of asymptotic stability for the system of equations

dx

/

dt



x

( 1 

x



y

),

dy

/

dt



y

( 0 .

75 

y

 0 .

5

x

)

Example 5: Estimating Basin of Attraction

(7 of 7) The disk

D

1 / 28   (

x

,

y

) : (

x

 0 .

5 ) 2  (

y

 0 .

5 ) 2  1 / 28  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 2

u

3  2

u

2

v



uv

2  2

v

3  0 .

25 

u

2 

v

2   0 .

75 

u



v

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