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 x1(0) = 0 and x2(0) = 2. In this case x1(t) = 0 and x2(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) x1(0) = 1 and x2(0) = 2. In this case x1(t) = e2t and x2(t) = 2e-t. In other words, x1(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 x1(0) = 1 and x2(0) = 0. In this case x1(t) = 1 and x2(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) x1(0) = 1 and x2(0) = 2. In this case x1(t) = 1+2t and x2(t) = 2. In other words, x1(t) approaches infinity with time, which proves that the system is Lyapunov unstable.

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Last Revised: August 5, 2002