Quadratic stability of linear time varying system

Question

(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.

Answer 1

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 :)