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:

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

  2. 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 $(0,0)$.

  • 1
    $\begingroup$ LaSalle is not applicable for non-autonomous systems. And this one is not autonomous due to $w(t)$. $\endgroup$
    – Arastas
    May 26, 2019 at 7:59
  • $\begingroup$ See also my second comment to the answer by Hans Lundmark. With $w(t)=exp(−t)$, the system is not asymptotically stable even though V˙≤0. $\endgroup$
    – SampleTime
    May 26, 2019 at 9:16

3 Answers 3


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.

  • $\begingroup$ Thats really counter intuitive but makes sense, thank you. $\endgroup$
    – SampleTime
    May 25, 2019 at 21:01
  • 1
    $\begingroup$ Just tested in simulation, $w(t) = \exp(-t)$ shows exactly the behavior you described. So system $(1)$ with this $w(t)$ is not asymptotically stable (at least not globally). $\endgroup$
    – SampleTime
    May 26, 2019 at 0:08
  • $\begingroup$ @SampleTime: OK, sounds good! $\endgroup$ May 26, 2019 at 11:51

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.


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.


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