System with a Lyapunov function over $\mathbb{R}^n$ but not globally asymptotically stable

I'd like to find an example of a system $\dot{\mathbf{x}} = F(\mathbf{x})$, where $\mathbf{x} = \mathbf{0}$ is an equilibrium point, with a corresponding Lyapunov function $V(\mathbf{x})$ that satisfies:

(1) $V(\mathbf{x}) > 0 \; \forall \mathbf{x} \in \mathbb{R}^n - \{\mathbf{0}\}$, $V(\mathbf{0}) = 0$

(2) $\dot V (\mathbf{x}) < 0 \; \forall \mathbf{x} \in \mathbb{R}^n - \{\mathbf{0}\}$, $\dot V(\mathbf{0}) = 0$

But where the origin is only locally asymptotically stable.

Context:

According to every text book that I've read on the subject, in order to guarantee global asymptotic stability, (1) and (2) are not enough: I also need $\lim_{||\mathbf{x}|| \to \infty} V(\mathbf{x}) = \infty$ (or, equivalently, that every subset of V is bounded). However, when I read the proof of local asymptotic stability, it seems to me that if (1) and (2) apply to all of $\mathbb{R}^n$, then asymptotic stability should also apply to all of $\mathbb{R}^n$.

• I have seen an example but I have to look it up in my documents. I will post an answer as soon as I have found it. Commented Oct 15, 2017 at 15:06
• @Evgeny after careful consideration, I think you are in fact correct. Read any proof on G.A.S. (e.g., stanford.edu/class/ee363/lectures/lyap.pdf , pages 9-10, taking into account that the author considers the fact that $V \to \infty$ as part of the definition of a function being positive definite -see page 5-). The key step where the boundness of sublevel sets comes into play is where the set $C =\left \{ z | \epsilon \leq V(z) \leq V(x(0)) \right \}$ can be known to be compact. Commented Oct 17, 2017 at 13:55
• However, in order to satisfy that, it's not really necessary that every sublevel set of V is bounded, i.e., that $S = \left \{ z | V(z) \leq c \right \}$ is bounded $\forall c \in \mathbb{R}$ (a condition that your function $V_2$ does not satisfy, as ilustrated by taking $c=1$), but only that $S$ is bounded $\forall c \in \operatorname{Im}(V)$ (a condition which $V_2$ does indeed satisfy). So, to summarize, you cannot use $V_2$ to prove G.A.S. by invoking the standard theorem, but you can still use it by introducing a little modification to the theorem's hypotheses. Commented Oct 17, 2017 at 13:55
• @MrYouMath If I'm not mistaken the function $e^{-\frac{1}{x^2+y^2}}$ behaves just like $e^{-\frac{1}{x^2}}$ — and the latter (if I'm not mistaken again) decays to zero faster than any polynomial function. This is why you can extend this function from $\mathbb{R} \setminus \lbrace 0 \rbrace$ to the whole $\mathbb{R}$. Moreover, this function is infinitely differentiable at $0$ but not analytic. Because of these properties $V_2$ can be very naturally extended to the origin and it'll be a nice function. Commented Oct 31, 2017 at 12:42
• @MrYouMath, you are correct. Throughout this discussion I assumed that the function we were really talking about was $V(x,y) = e^{-\frac{1}{x^2+y^2}}$ for $(x,y) \neq (0,0)$, $V(0,0) = 0$. This function is of type $C^1$ (at least), so it constitutes a valid candidate for a Lyapunov function. Commented Oct 31, 2017 at 13:42

Interesting question!

The issue is that some trajectories may tend to infinity.

I don't think I've ever seen an explicit example, but (with some effort!) I managed to come up with one myself. Take $$V(x,y) = \frac{x^2}{1+x^2} + y^2 ,$$ which is clearly positive definite, but doesn't tend to infinity as $\sqrt{x^2+y^2}\to \infty$ (indeed, $V(x,0)\to 1$ as $x \to \pm\infty$).

The level set $\{ V(x,y)=1 \}$ is given by $y=\pm 1/\sqrt{1+x^2}$, a pair of curves which extend out to infinity, approaching the $x$ axis. Between these two curves, you have level sets $\{ V(x,y)=c \}$ for $c \in (0,1)$ which are closed curves encircling the origin, and $\{ V(x,y)=0 \}$ which is just the origin.

Now consider the system $$\dot x = x \, \frac{3x^2y^2-1}{x^2 y^2+1} ,\qquad \dot y = -y .$$ The idea here is that on the curve $xy=1$ (which for large positive $x$ lies just above the level curve $y=1/\sqrt{1+x^2}$) the system becomes $\dot x=x$, $\dot y=-y$, which has a solution that stays on that curve. That is, $$x(t)=e^t ,\qquad y(t)=e^{-t}$$ is a particular solution of our system, and it doesn't tend to the equilibrium $(x,y)=(0,0)$, so the system is not globally asymptotically stable.

And on the other hand, the system becomes $\dot x \approx -x$, $\dot y=-y$ when we are close to the origin, so it's locally asymptotically stable.

It remains to show that $\dot V$ is really negative definite: $$\begin{split} \dot V & = \frac{\partial V}{\partial x} \, \dot x + \frac{\partial V}{\partial y} \, \dot y \\ & = \frac{2x}{(1+x^2)^2} \, x \, \frac{3x^2 y^2-1}{x^2 y^2+1} + 2y \, (-y) \\ & = \frac{-2}{(1+x^2)^2 (1+x^2 y^2)} \biggl( x^2 (1-3x^2 y^2) + y^2 (1+x^2)^2 (1+x^2 y^2) \biggr) \\ & = \frac{-2}{(1+x^2)^2 (1+x^2 y^2)} \biggl( x^2 - 2 x^4 y^2 + x^6 y^4 + y^2 + 2 x^2 y^2 + x^2 y^4 + 2 x^4 y^4 \biggr)\\ & = \frac{-2}{(1+x^2)^2 (1+x^2 y^2)} \biggl( x^2 (1 - x^2 y^2)^2 + y^2 (1+2x^2)(1+x^2 y^2) \biggr) , \end{split}$$ which is clearly negative away from the origin. Done!

(Remark: The rewriting of $\dot V$ in the last step is due to a helpful comment by @SampleTime, which simplified the argument greatly. See the edit history for my own original version, which was much uglier!)

• It would be interesting to see if @MrYouMath 's example is simpler (mainly for educational purposes), but yours is indeed a great example Hans. Kudos! Commented Oct 17, 2017 at 6:02
• Very nice example. Just an alternative for showing $\dot{V} < 0$ for $(x, y) \neq (0, 0)$ is to use the fact that the above term (with the $-3x^2y^2$) can be represented as a SOS as $2\, x^2\, y^2 + x^2\, y^4 + 2\, x^4\, y^4 + 2\, {\left(\frac{\sqrt{2}\, x}{2} - \frac{\sqrt{2}\, x^3\, y^2}{2}\right)}^2 + y^2$ which, in this form, is obviously positive definite, thus $\dot{V}$ is negative definite. Commented Oct 20, 2017 at 21:59
• @SampleTime: Awesome, thank you! I can't believe I missed that, despite wrestling with the expression for $\dot V$ for so long... Commented Oct 21, 2017 at 5:14
• A tour de force! +1. Would you mind to disclose in some detail how you massage out the ODE system?
– Hans
Commented May 11, 2019 at 22:36
• @Hans: I'm not sure I understand what “massage out” means. But if you're asking how I found those ODEs, I don't quite remember anymore. Lots of trial and error, I suppose... Commented May 12, 2019 at 7:39

Just for the record, here's another example that I just found on p. 109 of the book Stability of Motion by Wolfgang Hahn (1967), where it is credited to a 1952 paper in Russian by Barbashin & Krasovskii: $$\dot x = -\frac{6x}{(1+x^2)^2} + 2y ,\qquad \dot y = -\frac{2(x+y)}{(1+x^2)^2}.$$ With the same(!) $V$ as in the example I came up with in my other answer, $$V(x,y) = \frac{x^2}{1+x^2} + y^2 ,$$ one computes $$\dot V = -\frac{4}{(1+x^2)^2} \left( \frac{3x^2}{(1+x^2)^2} + y^2 \right) ,$$ so $V$ is positive definite and $\dot V$ is negative definite, and hence $V$ is a strong Lyapunov function on all of $\mathbf{R}^2$.

But Hahn shows that trajectories of this system cannot cross the curve $$y=\frac{2}{x-\sqrt2} \quad (x>\sqrt2)$$ in the direction towards the origin. (He compares $\dot y/\dot x$ to the slope of the curve.)

Hence not all solutions tend to the origin.

• I prefer your original example, as you could provide an explicit solution which didn't tend to the origin. Commented Dec 20, 2017 at 19:37
• +1. Informative quote and reference thereof.
– Hans
Commented May 12, 2019 at 3:13
• This example is also given as Exercice 4.8 (with reference to Hahn's book) in Khalil's Nonlinear Systems. Moreover, Khalil calls his Theorem 4.2 (about global asymptotic stability) the Barbashin–Krasovskii theorem. Commented Aug 12, 2019 at 4:58