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In my Differential Equations course, we defined the equilibrium point $x_0$ of a dynamical system $\frac{dx}{dt} = f(x(t))$ (for $f$ defined on an open subset of $\mathbb R^n$, say $\mathbb R^n$ itself) to be stable if it is:

  1. Lyapunov Stable
  2. There is an $\epsilon$ ball around $x_0$ such that the solutions $\varphi$ of this differential equation with initial conditions in this ball satisfy $\lim_{t \to \infty} \varphi(t) = x_0$.

I am trying to find an example of the case where the property (2) holds while the point $x_0$ is not Lyapunov stable.

After some searching, I ran across Homoclinic Bifurcation, which is intuitively how I would expect Lyapunov Stability to fail, but have been unable to find examples of Homoclinic Bifurcation where property (2) holds as well.

Any help would be appreciated.

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Hint: Consider the system $$\begin{cases} \dot x = x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}}\\ \dot y = x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}} \end{cases}$$ and its fixed point $(1,0)$. (Converting into polar coordinates might help.)

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    $\begingroup$ Thank you, this is exactly what I was looking for. Incidentally, what motivated this? From the equation in polar coordinates, I can see why this works. Did you just cook up this example by phase velocity considerations, or was there some deeper motivation? $\endgroup$ – Tim The Enchanter Apr 17 at 16:19
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    $\begingroup$ I can guess the idea. The simplest example for the behaviour that you wanted in OP is a saddle-node equilibrium with a homoclinic trajectory. Polar form is a quite easy way to get such a saddle-node: throw away terms with square roots, and you'll just get a limit cycle at $r =1$. If you add these terms, the invariant circle at $r =1$ still exists, but now there is an equilibrium at $(1, 0)$ which happens to be a saddle-node. Similar system is often used as an illustration for SNIC-bifurcation. $\endgroup$ – Evgeny Apr 17 at 16:56
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    $\begingroup$ @TimTheEnchanter I guess that Evgeny's comment answers your question much better than I could, but since you asked: I did not cook up this example myself; this system was given to me as an exercise in my ODE 2 course and the point of the exercise was precisely to show that it is necessary to include the additional stability assumption in the definition of asymptotic stability. $\endgroup$ – MisterRiemann Apr 17 at 18:44
  • $\begingroup$ I see, thank you. $\endgroup$ – Tim The Enchanter Apr 18 at 12:57
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    $\begingroup$ It appears that for pedagogical reasons the decoupled system in polar coordinates $\dot{r}=r(1-r)$, $\dot{\theta}=1-\cos{\theta}$, as considered in Lyaponov function in dynamic system (in polar cordonates), would be slightly better (on the other hand, it doesn't have so simple form in the Cartesian coordinates, at least according to Mathematica). $\endgroup$ – user539887 Apr 19 at 8:11

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