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Lyapunov Stability Theory
M. S. Fadali
Professor of EE
1
Outline
Stability of an equilibrium of a nonlinear
system .
Lyapunov’s (first, indirect) linearization
method.
Lyapunov’s (second) direct method.
Linear time-invariant case.
2
Lyapunov’s Linearization Method Linearize nonlinear system in vicinity
of equilibrium :
. Find the eigenvalues of the linearized system.
The equilibrium of the nonlinear system is:◦ asymptotically stable if all the eigenvalues are in the
open LHP.
◦ unstable if one or more of its eigenvalues is in theopen RHP.
◦ Inconclusive for LHP eigenvalues and one ormore eigenvalues on the imaginary axis.
3
Example Determine the stability of the equilibrium
of the mechanical system at the origin
Equilibrium with
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Nonlinear State Equations Physical state variables
State Equations
5
Linearization and Stability
Equilibrium state
Linearized model with
Characteristic polynomial and stability
,
Stable ,
6
Lyapunov’s Direct Method
Examine stability of nonlinear system
directly.
Generalize concept of energy function.
Gives sufficient stability or instability
conditions (in general).
Possible difficulty, choice of suitable
Lyapunov (generalized energy) function.
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Lyapunov Stability
there exists a such that
Defined for an equilibrium point (origin)
Start near the equilibrium and stay near
the equilibrium.8
x 1
x 2
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Asymptotic Stability
For an equilibrium point
1. Lyapunov stabilty
2. Convergence to the origin, i.e. there
exists a such that
_|Å →
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Directional Derivative
Consider a continuous function f with
continuous partial derivatives. Directional derivative of at in the
direction
_|Å →
Some authors assume
;
10
Lyapunov Function
Positive definite:
Decreasing (or non-increasing) along the
trajectories of the system.
Derivative negative (or semidefinite)
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Quadratic form
Replacing terms in the first summation by
gives
Assume a symmetric matrix with no lossof generality
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Positive Definite Matrix P>0
for any nonzero
All its eigenvalues are positive
/ /
for any nonzero
Common choice of Lyapunov Function
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Negative Definite Matrix All its eigenvalues are negative
If is positive definite, then is negativedefinite / /
for any nonzero
Negative semi-definite Matrix: eigenvalues
are negative or zero .
Look for negative definite
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Lyapunov Stability Theorem
Given a positive definite function
If the derivative of along the
trajectories of the system isa) negative semi-definite then the equilibrium is
stable in the sense of Lyapunov.
b) negative definite then the equilibrium is
asymptotically stable.
c) positive definite then the equilibrium is
unstable.
15
Justification
Lypunov function gets smaller with the
length of .
If the function has a negative derivativealong the trajectories, it is getting smaller.
The function continues to get smaller
along the trajectories until it reaches a
minimum: convergence.
The minimum value of the function is zero
(at the equilibrium point).
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La Salle’s Principle
If the derivative of along the
trajectories of the system isnegative semi-definite but is only zero for
trivial trajectories (ones that imply )
then the equilibrium is asymptotically
stable.
17
Remarks
The theorem provides sufficient
conditions for stability and sufficientconditions for instability.
If the test fails, there is no conclusion.
It is often difficult to find a suitable
Lyapunov function for nonlinear systems.
For linear systems, the theorem canprovide a necessary and sufficient
condition.
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Scalar System (Slotine & Li)
Show that the system is asymptoticallystable.
Lyapunov function
,
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Nonlinear Spring-Mass-Damper
Show that the system is asymptotically
stable.
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Solution Use energy as the Lyapunov function
(stable i.s. Lyap.)
(asymptot. stable)
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Linear Time-invariant Case
The LTI system
is asymptotically stable if and only if for
any positive definite matrix there exists
a positive definite symmetric solution to
the Lyapunov equation
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Proof
Use a quadratic Lyapunov function
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Remarks
Recall that the original Lyapunov theorem
only gives a sufficient condition.
If we start with (i.e. with Lyapunovfunction) and solve for , the condition
the test may or may not work.
If we start with (i.e. with the derivative
and we find a the condition is necessary
and sufficient.
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Example
Determine the stability of the system with
state matrix
using the Lyapunov equation with .
Note: The system is clearly stable byinspection since is in companion form.
Choosing P = I will not work! No conclusion.
25
Solution
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0 6
1 5
0 1
6 5
1 0
0 1
• Multiply
12 6 56 5 2 10
1 0
0 1
• Equate to obtain three equations in three unknowns.
Equivalent Linear System
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12 6 56 5 2 10
1 0
0 1
0 12 0
1 5 60 2 10
1
01
1/12
1 2 /10 7/60
6 5 7/10 5/12 67/60
67/60 1/12 1/12 7/60 1.1167 0.08333 0.08333 0.1167
MAPLE
Compute:
with(LinearAlgebra):
Transpose(A).P+P.A
Solve the equivalent linear system:
LinearSolve(M,B)
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MATLAB
Solve a different equation.
Identical to our equation with
replaced by .
Eigenvalues are the same!
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MATLAB Example
>> A=[0,1;-6,-5];
>> Q=eye(2)>> P=lyap(A,eye(2))
P =
0.5333 -0.5000
-0.5000 0.7000
>> eig(P)
ans =
0.1098
1.1236
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To Get Earlier Answer
>> P=lyap(A',eye(2))
P =
1.1167 0.0833
0.0833 0.1167
31
1167.008333.0
08333.01167.1 P
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