0
$\begingroup$

(This question was originally asked at Math.SE, where it didn't receive any answers.)

Consider the linear time-varying system $$ \dot{x} = A(t) x, $$

where $x \in \mathbb{R}^n$ and $A: [0,+\infty) \rightarrow \mathbb{R}^{n\times n}$ is continuous.

It is known (see for instance, [1, Example 4.21]) that if there exists a positive definite matrix valued function $P:[0,+\infty) \rightarrow \mathbb{R}^{n\times n}$ that is continuously differentiable, bounded and satisfying the matrix differential equation

$$ - \dot{P}(t) = P(t)A(t) + A^{T}(t)P(t) + Q(t),$$

where $Q$ is a continuous, symmetric and positive definite matrix valued function, then the origin of the system is exponentially stable.

Supposing that $Q$ is also bounded, then the converse is valid (see [1, Theorem 4.12]). It is known any counter-example for the case that $Q$ is not bounded?

(1): Hassan K. Khalil (2002). Nonlinear Systems. Prentice Hall.

$\endgroup$
2
$\begingroup$

Sure: take the scalar ODE $\dot x=-x$, with $A=-1$, which is exponentially stable and the state transition matrix is $\Phi(t)=e^{-t}$ (i.e. $x(t)=\Phi(t) x_0$). Assume $Q=e^{2t}$, which is not bounded and the definition of $P$ in the proof of Theorem 4.12 leads to unbounded $P$.

Now, the practical implication of this is questionable. The intuition behind the theorem is that, if you carefully choose a suitable $Q$, you can construct a Lyapunov function. If you fail to select a good enough $Q$, even if the system is stable, you have got nothing, so... well, try again :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.