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Suppose I have an autonomous differential equation $x'(t)=f(x(t)) $, with $x(t) \in [0,1]^d$, such that $x'(t)$ vanishes at $+ \infty$ in the following sense : $$‖x'(t)‖_2 \to 0 \text{ when } t \to + \infty.$$ For $‖ \cdot ‖ $ the euclidean norm in $\mathbb{R}^d$. Let us also assume that $f$ is analytic. Could it be that $x(t)$ does not converge at $t = + \infty$ ?

I have seen in other posts (When does gradient flow not converge?) that for gradient system it can't be due to the Łojasiewicz inequality, but I have trouble using it for more general systems. Also the common counter-examples of bounded functions that do not converge while their derivative vanishes like $x \mapsto sin(\sqrt{x})$ do not solve an autonomous system of equation. Are you aware of any theorem or examples that could help me tackle this issue ?

My attempt : I introduce some terminology For $U \subset [0,1]^d $ an open set, $f$ analytic on $U$, we consider the differential equation $$x'(t) = f(x(t))$$

We denote by $\phi^t(\xi)$ the maximal solution of the equation satisfying $\phi^0(\xi) = \xi $ on the interval $I_\xi$.

  • The orbit $\gamma(x)$ of $x$ is the set in $\mathbb{R}^d$ defined by $\gamma(x)=\{x(t) : t \in I\}$.We also designate the orbit of $x$ as $\gamma(\xi)$ where $\xi$ is the initial value of our problem.

  • If $\gamma(\xi) = \{ \xi \}$, then $\xi$ is called a critical point or equilibrium point. We can notice that $\xi$ is a critical point if and only if $f( \xi )=0$.

  • We say that a set $A \subset \mathbb{R}^d$ is a positively invariant set of the equation if $\phi^t(A) \subset A$ for all $t \geq 0$.

  • We define the limit set $\omega(\xi) = \cap_{\tau \geq 0} \overline{\cup_{t \geq \tau} \phi^t(\xi)}$ of an orbit containing $\xi$.

    Theorem : If $\phi^t(\xi)$ takes its value in a compact set, then $\omega(\xi)$ is a connected, non empty compact set.

My attempt : If $f$ is analytic in $\mathbb{R}^d$, then its zeros cannot be such a set, unless this set is a singleton. For example, on this discussion (Zeros set of analytic functions over complex plane with several variables) , we have the following fact :

In particular, the zero set of a nonconstant analytic function on an open connected set has empty interior.

So the limit set is a connected compact set with empty interior. It can be any $d-1$-dimensional surface... I don't see how to make progress after that point

cocojar
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2 Answers2

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I think I have a counter-example. Take the system

$$ \begin{align*}\dot x &= zy \\ \dot y &= -zx \\ \dot z &= - z^3\end{align*} $$

This system can be solved for exactly. One such solution is $$(x(t), y(t), z(t)) = \left(\sin(\sqrt{2t+2}), \cos(\sqrt{2t+2}), \frac 1 {\sqrt{2t+2}}\right).$$ The speed of the solution goes to $0$ as $t\to\infty$. However, the solution does not converge to a point since $\sqrt{2t+2} \to \infty$ as $t\to\infty$.

Trevor Norton
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  • Thanks a lot for your answer. It has helped me refining my thoughts. The problem is, $\sin( \sqrt(t))$ is not analytic on $\mathbb{R}$, and the function $z(t)$ is not even defined on $\mathbb{R}$ (let alone bounded). Actually the function $\sin( \sqrt(t))$ does not provide an example of an analytic function whose derivative goes to 0 while not converging. I am wondering if there are any. – cocojar Jun 26 '24 at 15:00
  • @cocojar Sorry, but the question only specifies that $f$ is analytic. I believe this is a counter-example to your claim as it is written. It might be worth noting that the solution is analytic on its domain of existence ($t\in (-1,\infty)$) but is not defined for all backward time. If you also want the solution to be defined for all time, then adjusting $-z^3$ might get you that as well. I'd expect $\sin(-z^3)$ would do the trick, but the solution might be harder to write down analytically. – Trevor Norton Jun 26 '24 at 17:54
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A non-autonomous system can be transformed into an autonomous one by including time as an additional variable. Starting from your example (although strictly speaking it is not analytic), let $(x,y)=\left(\sin(\sqrt t),1-\exp(-t)\right).$ Then $(x'(t),y'(t))=\left(\frac{\cos(\sqrt t)}{2\sqrt t},\exp(-t)\right)$ can be expressed as an autonomous equation because $t=-\log(1-y).$

Lieven
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  • Thanks a lot for the answer. Actually, you are write, the function is not analytic and hence the counter example provided by $\sin(\sqrt(t))$ does not work. – cocojar Jun 26 '24 at 15:01