4

Let $f : \mathbb R \longrightarrow \mathbb R$ be a $C^{\infty}$-function such that $f(x) = 0$ iff $x \in \mathbb Z.$ Suppose that the function $x : \mathbb R \longrightarrow \mathbb R$ satisfies $x'(t) = f(x(t)),$ for all $t \in \mathbb R.$ If $\mathbb Z \cap \{x(t)\ |\ t \in \mathbb R\}$ is non-empty then $x$ is a constant.

This question appeared in one of the entrance examinations. It is clear that $x'(t_0) = 0,$ for some $t_0 \in \mathbb R.$ But how do I show that $x'(t) = 0,$ for all $t \in \mathbb R\ $? Any help would be much appreciated.

Thanks!

daw
  • 54,637
  • 2
  • 44
  • 85
Fanatics
  • 211
  • I am trying to proceed by the density of $\mathbb R\setminus \mathbb Z$. Like, for $t\neq t_0$ let $x(t)\in \mathbb R\setminus \mathbb Z$. So for each $a\in \mathbb Z$ $\exists$ $x_n(t)$ such that $x_n(t) \to a$. – Aritra Jul 01 '21 at 10:51
  • Does something like this serves you as counterexample? – Joako Apr 10 '25 at 19:22

1 Answers1

3

This is a direct consequence of the local existence and uniqueness theorem for ordinary differential equations.

Take $t_0$ such that $x(t_0)=:z_0\in \mathbb Z$. Since $f$ is locally Lipschitz, there is $\delta>0$ (depending on $f$ and $x(t_0)$ but not on $t_0$) such that the ODE $x'(t)=f(x(t))$, $x(t_0)=z_0$ is uniquely solvable for $t\in [t_0-\delta, t_0+\delta]$. This solution is trivially $x(t)=z_0$.

Now we can repeat the argument for $x(t_0 \pm \delta)$ and extend the interval, where $x(t)=z_0$ to $[t-2\delta,t+2\delta]$. Now proceed by induction to show $x(t)=z_0$ for all $t$. Here it is important that $\delta$ is independent of $t_0$.

daw
  • 54,637
  • 2
  • 44
  • 85
  • Could you please explain why $\delta$ has to be independent of $t_0\ $? Local Lipschitzness of $f$ only says that corresponding to every point in the domain of $f$ we can get hold of some neighborhood around it on which $f$ is Lipschitz. But when you are demanding that we can continue the argument regardless of whatever point we randomly pick from the domain without altering the diameter of the neighborhood aren't you implicitly assuming some sort of uniformity in the definition of locally Lipschitz maps? – Fanatics Jul 01 '21 at 15:47
  • Because we always pick the same point of the domain of $f$. First at $x(t_0)=z_0$, then at $x(t_0\pm \delta)$, which is also equal to $z_0$, etc. – daw Jul 02 '21 at 06:05
  • Oh! Yes. You are right. I get it now. Thanks. – Fanatics Jul 02 '21 at 07:38