Let $f:\mathbb{R} \to \mathbb{R} $ be a continuous function. Let $x'(t)=f(x(t))$ be a maximal solution to the first order ODE. Show that the solution is a monotonic function.
This is the classical problem when you have the intuition, but you don't know how to write it down. My first idea was to show that if it's not monotonic, then you will have two points, say $a$ and $b$ with $x(a)=x(b)$ but $x'(a)> 0$, $x'(b)<0$.
I repeat then the proof given by Robert Israel as an answer to Proof by contradiction that there are no periodic solutions to x˙=f(x). Suppose to the contrary that for some solution $x(\cdot)$ one has $x(a) = x(b)$ for some $a < b$. Consider the integral $$ \int\limits_{a}^{b} f(x(t)) x'(t) \, dt. $$ On the one hand, as $x'(t) = f(x(t))$ for all $t \in [a,b]$, we have $$ \int\limits_{a}^{b} f(x(t)) x'(t) \, dt = \int\limits_{a}^{b} (x'(t))^2 \, dt \ge 0. $$ On the other hand, integrating by substitution we obtain $$ \int\limits_{a}^{b} f(x(t)) x'(t) \, dt = \int\limits_{x(a)}^{x(b)} f(\xi) \, d\xi, $$ which equals zero. Therefore, $x' \equiv 0$ on $[a,b]$, so the solution is constant between $a$ and $b$.
