an interesting initial value problem involving function whose zeroes are squares of nonzero integers 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 : $y'(t)=f(y(t))$ with $y(0)=y_0$. Then which of the following are true?
$(a)$ $y$ is a monotone function for all $y_0\in \mathbb{R}$.
$(b)$ for any $y_0\in \mathbb{R}$, there exists $M_{y_0}>0$ such that $|y(t)|\leq M_{y_0}$ for all $t\in \mathbb{R}$.
$(c)$ there exists a $y_0\in \mathbb{R}$, such that the corresponding solution $y$ is unbounded.
$(d)$ $\sup_{t,s\in \mathbb{R}}|y(t)-y(s)|=2n+1$ if $y_0\in (n^2,(n+1)^2)$ where $n\geq 1$.
========================================================
I have no specific idea about how to tackle this problem. I just observed that the last option has the supremum as the length of the interval around $y_0$, i.e. $2n+1=(n+1)^2-n^2$. But that's just a silly observation and I don't think it has to do something for the actual logic. Also $(b)$ and $(c)$ are contradictory to each other. What is the general idea to think for this type of problem?
 A: Since $f$ is Lipschitz, we have that given an initial value, the solution exists and it is unique. We also know that the function $y(t) = y_0$ is an equilibrium solution if, and only, $y_0 = \pm n^2$ for $n \in \mathbb{N}$, therefore $f$ is either positive or negative in $(n^2, (n+1)^2)$ or $(-(n+1)^2,-n^2)$, which means that the derivative of solutions is either positive or negative in those intervals and this shows that solutions are either equilibriums or monotone, which answers (a) (I don't know if you consider a constant function monotone). Also, if a solution starts in one of those intervals I mentioned above, it is trapped there for all time since it cannot cross an equilibrium solution (by uniqueness), which implies that all solutions are bounded and answers (b). It also answers (c) because it shows that solutions can't be unbounded. Finally, any solution that stars in one of the intervals I mentioned, since it's monotone and bounded, is such that it's limit as time goes to infinity is one of the points $n^2$ or $(n+1)^2$ and it's limit as time goes to minus infinity is the other point, therefore (d) is true, because we can make $s$ go to infinity and $t$ go to minus infinity, so that this difference approaches the size of the interval, which is, as you observed, $(n+1)^2 - n^2 = 2n +1$.
