25

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 $\mathbb{R}^n$.

Rodrigo de Azevedo
  • 23,223
LGenzelis
  • 677
  • 1
    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. – MrYouMath Oct 15 '17 at 15:06
  • @MrYouMath Would be interesting to see it. Must be a quite cheeky counter-example. I mean, these conditions guarantee that $V(x)$ has no critical points except at the origin. I'm not very sure but for me these conditions also imply that level-sets of $V(x)$ are homeomorphic to hyperspheres that surround the origin. And vector field also strictly points to the inside of these hyperspheres. – Evgeny Oct 16 '17 at 08:21
  • @Evgeny: The level sets don't have to be bounded. (See the answer that I just posted.) – Hans Lundmark Oct 16 '17 at 08:42
  • @HansLundmark So the key factor is boundedness of level sets, not the unboundedness of the function, right? and if I had compactified the $\mathbb{R}^n$ I could have seen that level sets that pass through $\infty$ can be more complicated? – Evgeny Oct 16 '17 at 08:59
  • @Evgeny: I haven't thought very deeply about it, but I guess so, yes. Or maybe it's rather the boundedness of the sublevel sets ${ V(x,y) \le a }$ that's important. (But if you assume $V \to \infty$ you don't need to worry about the possibility that the sublevel set may be the whole space if $a$ is too large.) – Hans Lundmark Oct 16 '17 at 09:55
  • @Evgeny note that the condition is not only that $V(x)$ is unbounded, but that $V(x) \to \infty$ (e.g., $h(u) = |u \sin u|$ is unbounded but it is not correct to state $h(u) \to \infty$). The condition $V(x) \to \infty$ is a necessary and sufficient one for all of the sublevel sets of V to be bounded. – LGenzelis Oct 16 '17 at 14:34
  • I don't really get why $V(x)$ has to tend to $+\infty$. If you take system $\dot{x} = -x, ; \dot{y} = -y$ then (besides usual candidate like $V_1(x, y) = x^2 + y^2$ which behaves exactly as you expect) it also has Lyapunov function $V_2(x, y) = e^{-\frac{1}{x^2+y^2}}$ if I'm not mistaken. It has exactly the same geometry of level sets as $V_1(x, y)$ but it's bounded. – Evgeny Oct 16 '17 at 18:27
  • I don't understand your point. $V_2$ is bounded, yes, but as a consequence it has unbounded sublevel sets: the sublevel set given by $V(x) \leq 1$ is the whole state space. Therefore, you can use $V_2$ to prove that the origin is locally asymptotically stable, but not globally. As as side note, note that $\nexists V_2(0,0)$, so you'd had to redefine it to make it a valid Lyapunov function. – LGenzelis Oct 16 '17 at 20:23
  • Why can't I use $V_2$ to prove that equilibrium is globally asymptotically stable? I don't see the reason and I want to understand. All level sets of this function are compact and vector field points exactly inside. The origin is globally asymptotically stable. If you have the reference for your statement about compact sublevel sets, that would be nice. This example is different from @HansLundmark because of different structure of level sets. – Evgeny Oct 17 '17 at 08:53
  • 1
    @Evgeny after careful consideration, I think you are in fact correct. Read any proof on G.A.S. (e.g., https://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. – LGenzelis Oct 17 '17 at 13:55
  • 1
    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. – LGenzelis Oct 17 '17 at 13:55
  • @LGenzelis The funny thing is that after googling I found the same lecture slides as yours, read the proof and came to the same conclusions and modification :) still thanks for an explanation! – Evgeny Oct 17 '17 at 14:06
  • @Evgeny: Very interesting discussion. But how can you use $V_2$ (without further modifications) if it is not even defined at the origin? I thought that a Lyapunov candidate function needs to be continuously differentiable in a region around the origin? – MrYouMath Oct 31 '17 at 10:22
  • 1
    @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. – Evgeny Oct 31 '17 at 12:42
  • 1
    @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. – LGenzelis Oct 31 '17 at 13:42
  • @Evgeny & LGenzelis: Thank you learned something new again. I always rejected such functions. I thought that extending the function was forbidden (I don't know why I thought this strange way :D). – MrYouMath Oct 31 '17 at 13:54
  • 1
    @MrYouMath This kind of functions occurs in analysis from times to times. For example, bump function is a similar concept ($C^{\infty}$-function, but not analytic; good for some kind of smoothing and for local changes of function). Also, as I've already said, $e^{-\frac{1}{x^2}}$ is a nice example of $C^{\infty}$ function which is not analytic. – Evgeny Oct 31 '17 at 14:06
  • What about the discontinuous Lyapunov functions? – Yaosheng Deng Oct 22 '23 at 15:32

2 Answers2

22

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.

Level sets of V(x,y)

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!)

Hans Lundmark
  • 55,038
  • 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! – LGenzelis Oct 17 '17 at 06:02
  • 2
    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. – SampleTime Oct 20 '17 at 21:59
  • 1
    @SampleTime: Awesome, thank you! I can't believe I missed that, despite wrestling with the expression for $\dot V$ for so long... – Hans Lundmark Oct 21 '17 at 05:14
  • A tour de force! +1. Would you mind to disclose in some detail how you massage out the ODE system? – Hans May 11 '19 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... – Hans Lundmark May 12 '19 at 07:39
  • @HansLundmark: "massage out" means manipulate with trial and error to produce. Fair enough... – Hans May 12 '19 at 18:27
  • This example is also given in the book 'nonlinear systems - Hassan K. Khaili' page 123, section 4.1. – happyle Sep 27 '23 at 17:03
  • @happyle: Well, that's only half true. He gives the same function $V(x,y)$, but not the same system of ODEs, and actually he doesn't mention any system at all at that point in the text. However, he gives an example in the exercises with another system; see the comments to my other answer. – Hans Lundmark Sep 27 '23 at 20:23
4

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.

Hans Lundmark
  • 55,038
  • I prefer your original example, as you could provide an explicit solution which didn't tend to the origin. – LGenzelis Dec 20 '17 at 19:37
  • +1. Informative quote and reference thereof. – Hans May 12 '19 at 03:13
  • 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. – Hans Lundmark Aug 12 '19 at 04:58