ch 9.6: liapunov’s second method
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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.
22)2/1()cos1(),( ymLxmgLyxV
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)/(/,/
22)2/1()cos1(),( ymLxmgLyxV
,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
),,(/),,(/ yxGdtdyyxFdtdx
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.
),,(/),,(/ yxGdtdyyxFdtdx
),(),(),(),( yxGyxVyxFyxVV yx
V
Theorem 9.6.2
Suppose that the origin is an isolated critical point of the autonomous system
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
),(),(),(),( yxGyxVyxFyxVV yx
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/:),(
22)2/1()cos1(),( ymLxmgLyxV
,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
Suppose that the origin is an isolated critical point of the autonomous system
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
),(),(),(),( yxGyxVyxFyxVV yx
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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