# 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:

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 $$(0,0)$$.

• LaSalle is not applicable for non-autonomous systems. And this one is not autonomous due to $w(t)$. Commented May 26, 2019 at 7:59
• 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. Commented May 26, 2019 at 9:16

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.

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.

• Thats really counter intuitive but makes sense, thank you. Commented May 25, 2019 at 21:01
• 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). Commented May 26, 2019 at 0:08
• @SampleTime: OK, sounds good! Commented May 26, 2019 at 11:51

We have

$$e\dot e+e^2-e\theta w(t) =0\\ \theta\dot\theta +\theta e w(t) = 0$$

$$\frac 12\frac{d}{dt}(e^2+\theta^2)=-e^2$$
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.