I'm trying to prove a result about differential equations and maximal solutions (for me, a maximal solution of a diferential equation is a solution that can not be extended beyond its domain, i.e., if $f:I\to\mathbb{R}$ is a maximal solution of certain ode then there's no other solution $g:I'\to\mathbb{R}$ such that $I\subseteq I'$ and $f(x)=g(x)$ for every $x\in I$). This is the exercise:
If $f:I\to \mathbb{R}^n$ is maximal solution of the differential equation $x'=s(x)$, then one of the next statements hold:
- $f$ is inyective.
- $I=\mathbb{R}$ and $f$ is constant.
- $I=\mathbb{R}$ and $f$ is periodic (non-constant).
I think the next is a natural procedure: first I assume $f$ is not injective and then i prove that $I=\mathbb{R}$. Finally i prove that if $f$ is not constant, it must be periodic. But then i get stuck without even beginning. Any hints about a good direction will be appreciated.
Suppose that 1. does not hold. Then there exist $x_1,x_2$ with $x_1<x_2$ in the domain of the solution such that $y(x_1)=y(x_2)$. Let $T=x_2-x_1$. Then $z(t)=y(x+T)$ is a solution and $z(x_1)=y(x_1+T)=y(x_2)=y(x_1)$. By uniqueness, $y$ and $z$ coincide. This means that either $y$ is constant or periodic of period $T$.
Note: the above uses uniqueness of solutions. The equation $y'=2\sqrt{|y|}$ has solutions of the form $$ y_a(x)=\begin{cases}0&x\le a\\(x-a)^2&x>a\end{cases} $$ for any $a>0$. They do not satisfy any of 1., 2., 3.