5
$\begingroup$

I'm trying to prove that a solution of globally Lipschitz continuous system of ODEs cannot intersect any hyperplane infinitely many times in a finite amount of time. So for example, something like a spiral which converges to it's centre in finite time is not possible, because the centre is an equilibrium and therefore the solution for that point is not unique (either a constant trajectory which stays in the equilibrium, or the spiral itself). Also, the solution can't blow up to infinity because the system is globally Lipschitz.

My intuition so far is that even in more complex cases, this follows from the fact that Lipschitz continuous equations have unique solutions, which would break if the trajectory converges "too fast" to a specific point - like in the case of the spiral above. But I can't quite grasp how to show this for trajectories which do not converge to an equilibrium. For example, take the 2-dimensional above mentioned spiral, but set it into a 3-dimensional system such that $z' = 1$. The spiral still converges to it's centre in finite time, but the centre is not an equilibrium any more. How do I show that the solution is still not unique? Or is it also just "obvious" consequence of Picard–Lindelöf? :)

$\endgroup$
1
  • $\begingroup$ In the following papers "Finite Time Controllers" and "Finite Time Differential Equations" by V. T. Haimo are studied 1st and 2nd order scalar autonomous ODEs that achieve solutions of finite duration, where is explained uniqueness of solutions is not hold since is required for the ODE to have at least one point in time where is not-Lipschitz. With this, no Linear ODE could have a solution that reach zero in finite time and stays there forever after. As counter example x'=-sgn(x)sqrt(|x|) , x(0)=1 could stand the solution x(t)=1/4*(1-t/2+|1-t/2|)^2. I believe is related, so I hope it helps. $\endgroup$
    – Joako
    Commented May 20, 2022 at 11:17

1 Answer 1

7
$\begingroup$

The statement is not correct. You’ll find below counterexamples.

Consider $$\begin{array}{l|rcl} f : & \mathbb R \times \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (t, (x,y)) & \longmapsto & (1, 3 t^2 \sin(1/t) - t \cos(1/t))\end{array}$$

$f$ is continuous in $t$. $f$ is also independent of $(x,y)$ and therefore uniformly Lipschitz continuous in $(x,y)$.

Consequently, Picard–Lindelöf theorem applies to the IVP $$(x^\prime(t),y^\prime(t)) = f(t,(x,y)) \text{ and } (x(0),y(0))=(0,0)$$

The map $t \mapsto (t, t^3 \sin(1/t))$ is a (unique) solution. However this solution intersects infinitely many times the hyperplane $y=0$ in the neighborhood of $t=0$.

And if you prefer an autonomous ODE, you can use $$\begin{array}{l|rcl} g : & \mathbb R \times \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (t, (x,y)) & \longmapsto & (1, 5 x^4 \sin(1/x) - x^3 \cos(1/x))\end{array}$$

$t \mapsto (t,t^5 \sin(1/t))$ is the unique solution of the IVP $$(x^\prime(t),y^\prime(t)) = g(t,(x,y)) \text{ and } (x(0),y(0))=(0,0).$$ It intersects infinitely many times the hyperplane $y=0$ in the neighborhood of $t=0$.

$\endgroup$
5
  • 1
    $\begingroup$ Awesome! Just the thing I was looking for. Yes, you are right, that disproves my hypothesis :D On the other hand, for my current application, it would actually suffice to have the ODEs autonomous (I didn't mention it in the original question because I hoped it would hold universally). Do you think this changes the result? (I could make the t in your conterexample a variable with derivative 1, but then it wouldn't be Lipschitz continuous in t, right?) $\endgroup$
    – daemontus
    Commented Jan 7, 2018 at 8:42
  • $\begingroup$ But I guess I can always just replace the t^2 with a piecewise function which is constant on (-inf, -1) and (1, inf) and t^2 on (-1, 1) to get a globally Lipschitz continuous function even though t would be a variable. Which would also make this a counterexample for the autonomous case, right? $\endgroup$
    – daemontus
    Commented Jan 7, 2018 at 9:16
  • $\begingroup$ See the update of my answer. $\endgroup$ Commented Jan 7, 2018 at 10:08
  • $\begingroup$ Awesome, thank you very much! $\endgroup$
    – daemontus
    Commented Jan 7, 2018 at 10:41
  • $\begingroup$ @mathcounterexamples.net Does your solution could be an example to this question? $\endgroup$
    – Joako
    Commented May 20, 2022 at 11:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .