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!
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 $t_0$.