If $x(t)$ is a solution to $ \dot{x}(t)=f(x)$ then $x(t)$approaches infinity or a fixed point, and if $x(t_1)=x(t_2)$ then $x(t)$ is constant This is an easy problem, but I would like to know a formal solution using calculus and not the Picard–Lindelöf theorem, and also without using the (obvious) fact that solutions are monotonic for the second question.
Let $x$ be a solution of $\dot x = f(x)$ where $f$ is continuously differentiable.
I would like to show that:


*

*Every solution approaches $\pm \infty$ or a fixed point.

*If $x(t_1)=x(t_2)$ for some $t_1 \neq t_2$ then $x(t_1)$ is a constant solution.


My attempt, please let me how can I improve it:


*

*The solution is monotonic so it approaches a limit. If the limit is $\pm \infty$, we are done. Otherwise, it approaches some point. I think we can show that the derivative approaches 0, and then we get that the solution approaches a fixed point.

*Using the Rolle theorem we get that there exists $t_1<t_0<t_2$ such that $\dot x(t_0)=0$. We get that $y\equiv x(t_0)$ is a constant solution since $f(y(t))=f(x(t_0))=\dot x(t_0)=0$.
At first I wanted to say that since the equation is autonomous then if the derivative is 0 at one point it is always 0, but that's not exactly an appropriate argument, I think.


Thanks in advance!

1.) $x(n+1)-x(n)=x'(n+\theta_n)=f(x(n+\theta_n))$, $θ_n\in(0,1)$, gives you the derivative in the limit, since the first difference tends to $0$.
2.) You got two solutions passing through the same point. By local uniqueness, this is impossible if they are not identical. You can get the local uniqueness also from the Gronwall lemma, however this is rather close to Picard-Lindelöf.
