Let $x(t)$ be the solution of the initial value problem: $$ \dot{x}(t) = f(x(t)); \; \; x(0) = x_0 $$
I have made the following asumption during my work:
If $x_0 \geq 0$, and $f(0) \geq 0$, then $x(t) \geq 0$ for $t \geq 0$.
It intuitivelly makes sense: starting with x positive, it might decrease until $x=0$. But as soon as it reaches that point, $\dot{x} = f(0) \geq 0$, so it either "grows back up" or just stay at $0$.
I feel like there should be a simple proof for this, but I cannot find one.
Also, if it not true, what additional conditions should I add? Maybe Lipschitz continuity of $f$?
Thanks in advance.
This is untrue if $f$ is not locally Lipschitz-continuous. Take e.g. $f(x) = 0$ for $x \geq 0$ and $f(x) = -\sqrt{-x}$ for $x \leq 0$. Then $$ x(t) = \begin{cases} 0 & t \leq 0 \\ -\tfrac{1}{4}t^2 & t > 0 \end{cases} $$ is a solution to $x' = f(x)$, $x(0)= 0$. Check that this contradicts your hypothesis.
If you assume however that $f$ is locally Lipschtz, then your claim is true. First note, that solutions to $x' = f(x)$, $x(0) = x_0$ are then unique! Then let $y$ be the solution to the IVP $y' = f(y) - f(0)$, $y(0) = x_0$. Note that $0$ is a constant solution to $v' = f(v)-f(0)$, so due to uniqueness, either $y =0$ (if $x_0=0$) or $y>0$ if $x_0>0$ (the constant solution $0$ can't be intersected!).
Now, since $x' = f(x) \geq f(x) - f(0)$ (remember $f(0) \geq 0$) we can use the ODE comparison principle (Teschl, ordinary differential equations p.24, Lemma 1.2) to conclude that $x \geq y \geq 0$ for $t \geq 0$. Note very carefully that Teschl indeed requires uniqueness to the ODE (see text after (1.61)) which is not the case in my above example.
I guess that this result is hard to find, since many consider this result intuitive, but "annoying to prove rigorously".