Lyapunov
Stability

A Lyapunov stable system is a system for which the states will remain
bounded for all time, for any finite initial condition. A continuous-time
linear time-invariant system is Lyapunov stable (internally stable) if and
only if all the eigenvalues of A have real parts less than or equal
to 0, and those with real parts equal to 0 are nonrepeated.

For example, consider the following system:

This *A* matrix has eigenvalues at 2 and -1. One of the eigenvalues
has its real part in the right-half plane, so this system is Lyapunov unstable.
In fact, it is easy to derive the solution of this system as follows:

Consider the initial conditions *x*_{1}(0) = 0 and *x*_{2}(0)
= 2. In this case *x*_{1}(t) = 0 and *x*_{2}(t) = 2e^{-t}.
In other words, *x(t)* is bounded for all time. However, this does not
mean that the system is Lyapunov stable! Lyapunov stability requires that the
state remain bounded for all time, for *all* initial conditions - not
just for some specific initial condition. In other words, if we can find even
one initial condition that causes one of the states to approach infinity with
time, then the system is Lyapunov unstable. For the above system, we can choose
(for example) *x*_{1}(0) = 1 and *x*_{2}(0) = 2. In
this case *x*_{1}(t) = e^{2t} and *x*_{2}(t) = 2e^{-t}.
In other words, *x*_{1}(t) approaches infinity with time, which
proves that the system is Lyapunov unstable.

Now consider the following system:

Both of the eigenvalues of *A* are at 0. The system has repeated
eigenvalues on the imaginary axis (with real parts equal to 0) so the system is
Lyapunov unstable. In fact, it is easy to derive the solution of this system as
follows:

Consider the initial conditions *x*_{1}(0) = 1 and *x*_{2}(0)
= 0. In this case *x*_{1}(t) = 1 and *x*_{2}(t) = 0.
In other words, *x(t)* is bounded for all time. (In fact, *x(t)* is
constant for all time.) However, this does not mean that the system is Lyapunov
stable! We reiterate here that Lyapunov stability requires that the state
remain bounded for all time, for *all* initial conditions - not
just for some specific initial condition. If we can find even one initial
condition that causes one of the states to approach infinity with time, then
the system is Lyapunov unstable. For this system, we can choose (for example) *x*_{1}(0)
= 1 and *x*_{2}(0) = 2. In this case *x*_{1}(t) =
1+2t and *x*_{2}(t) = 2. In other words, *x*_{1}(t)
approaches infinity with time, which proves that the system is Lyapunov
unstable.

Stability
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Last Revised: August 5, 2002