Finding sequence of closed orbits in dynamical system

Question

Consider the dynamical system in polar coordinates \begin{cases} \dot{r}=r^2\sin(\frac{1}{r}) \\ \dot{\theta}=1 \end{cases} I need to show there is a sequence of closed orbits $$\gamma_n=\{(r,\theta) \mid r=r_n,\theta=t \}$$ such that $r_n \rightarrow 0$ as $n \rightarrow \infty$ and that every trajectory between $\gamma_n$ and $\gamma_{n+1}$ is attracted by either $\gamma_n$ or $\gamma_{n+1}$, depending on whether $n$ is odd or even.
If I consider the equation $\dot{r}=r^2\sin(\frac{1}{r})=0$, I obtain $$r=\frac{1}{n\pi},$$ where $n \in \mathbb{Z}$. So I can take $r_n=\frac{1}{n\pi}$, but how do I show the second part of the question?

Answer 1

More generally, consider a differential equation $\dot{y} = f(y)$ where $f$ is a continuously differentiable function on interval $[a,b]$ with $f(b) = 0$ and $f'(y) > 0$ for $a < y < b$. Then any solution with initial point $a < y(0) < b$ has $\lim_{t \to \infty} y(t) = b$.

Proof: $y(t)$ is an increasing function, and $y(t) < b$ for all $t > 0$ (by the Existence and Uniqueness Theorem for differential equations). Therefore $\lim_{t \to \infty} y(t) = y_\infty$ exists and is $\le b$. But if $z \in (a,b)$, take $\epsilon > 0$ so $a < z - \epsilon < z + \epsilon < b$. Then there is $\delta > 0$ so $f(y) > \delta$ for $z - \epsilon \le y \le z + \epsilon$. That implies that any solution that enters the interval $[z-\epsilon, z+\epsilon]$ leaves it at the top end after time at most $2\epsilon/\delta$ (and can't come back). That implies $y_\infty$ can't be $z$. So it can only be $b$.

Similarly if $f(a) = 0$ and $f'(y) < 0$ for $a < y < b$, you'll have $\lim_{t \to \infty} y(t) = a$.