Let $f:\mathbb{R}\to \mathbb{R}$ be a Lipschitz function such that $f(x)=0$ if and only if $x=\pm n^2$ where $n\in \mathbb{N}$. Consider the initial value problem $$y'(t)=f(y(t)),\ y(0)=y_0$$ Then which of the following are true?
$y$ is a monotone function for all $y_0\in \mathbb{R}$.
For any $y_0\in \mathbb{R}$, there exists $M_{y_0}>0$ such that $|y(t)|\leq M_{y_0}$ for all $t\in \mathbb{R}$.
there exists a $y_0\in \mathbb{R}$, such that the corresponding solution $y$ is unbounded
$\sup_{t, s\in \mathbb{R}}|y(t)-y(s)|=2n+1$ if $y_0\in (n^2, (n+1)^2), \ n\geq 1$
I was trying it by considering a linear function on subintervals $(n^2, (n+1)^2)$ and $0$ on end points. But not able to conclude anything. Plese help.
Suppose for a solution $y(t)$ where $y(0) = y_0 \neq \pm n^2$, there existed a $T$ such that $y(T) = \pm n^2$. Then $y'(T) = 0$. Since $f$ is Lipschitz, existence and uniqueness tells us that this implies $y$ must be constant, because it coincides with the constant $y(t) = \pm n^2$ solution. In other words, $$\pm n^2 < y(0) < \pm (n\pm 1)^2 \implies \pm n^2 < y(t) < \pm (n\pm 1)^2$$ for all $t$. Can you make other deductions from here?