How can this system not be asymptotically stable? I am currently studying stability of nonautonomous systems using the book Applied Nonlinear Control by Slotine & Li. On page 125, there is example 4.13:
$$
\begin{align}
\dot{e} &= -e + \theta \, w(t) \\
\dot{\theta} &= -e \, w(t)
\end{align} \tag{1}
$$
with $w(t)$ a bounded, continuous but otherwise arbitrary time-varying function. They consider the Lyapunov function 
$$V(e, \theta) = e^2 + \theta^2$$
with derivative
$$\dot{V}(e, \theta) = -2 e^2 \leq 0 \tag{2}$$
so $e$ and $\theta$ are bounded. Then, they use Barbalat's lemma to show that $\dot{V}(e, \theta) \rightarrow 0$ as $t \rightarrow \infty$, so also $e \rightarrow 0$. Then they say: 

Note that, although $e$ converges to zero, the system is not asymptotically stable, because $\theta$ is only guaranteed to be bounded.

However, isn't it like this: If $e \rightarrow 0$ then this implies that $\dot{e} \rightarrow 0$ as well. So, for $t \rightarrow \infty$, system $(1)$ reduces to
$$
\begin{align}
0 &= \theta \, w(t) \\
\dot{\theta} &= 0
\end{align} \tag{3}
$$
Because $w(t)$ can be arbitrary for all time, the first equation of $(3)$ is only true if $\theta \rightarrow 0$ as well. The second equation of $(3)$ also confirms that $\theta$ doesn't change anymore for $t \rightarrow \infty$.
So, my conclusion would be: Since $e \rightarrow 0$ and $\theta \rightarrow 0$, the system is actually asymptotically stable. However, this is in contradiction to the citation above.
Question: Basically two questions:


*

*Where is the mistake in my argument? Or is it actually correct and the book is wrong?

*Is system $(1)$ now asymptotically stable or not?
Note: I also tried some functions like $w(t) = \sin(t)$ with different initial conditions in simulation, and at least for those examples, the system always seemed to converge to $(0,0)$.
 A: First of all, $e \to 0$ does not in general imply that $\dot e \to 0$ (see the warning about this on p. 122).
But I don't think that's the main issue with your argument. The problem seems to be that you're going to the limit $t \to \infty$ in some terms, while keeping $t$ in other terms. But it's the same $t$ everywhere, so if $t \to \infty$ in one place, it has to do so everywhere.
One can imagine a scenario where $w(t) \to 0$ as $t \to \infty$, and where $(e,\theta)\to (0,\theta_0)$ for some constant $\theta_0 \neq 0$ (in such a way that $\dot e \to 0$ and $\dot \theta \to 0$). I haven't thought deeply about showing that this is really possible, but at least it is consistent with the ODEs: in $\dot e = - e + \theta \, w(t)$ all three terms tend to zero, and similarly in $\dot \theta = -e \, w(t)$ both terms tend to zero.
A: It is a standard problem in adaptive control and estimation. To ensure that $\theta$ converges you have to assume something about $w$. The required condition is known as persistency of excitation: there exist $T>0$ and $a>0$ such that for all $t$ $$\int_{t}^{t+T}w(s)w^\top(s)ds \ge aI.$$
For example, $w(t)\equiv 0$ or $w(t) = e^{-t}$ are not persistently exciting. 
A: We have
$$
e\dot e+e^2-e\theta w(t) =0\\
\theta\dot\theta +\theta e w(t) = 0
$$
after adding
$$
\frac 12\frac{d}{dt}(e^2+\theta^2)=-e^2
$$
so as long as $e \ne 0$ supposing that $\theta = \theta_0\ne 0$ we have that the movement goes to the equilibrium point $(0,\theta_0)\ne (0,0)$ hence no asymptotic stability is possible.
