Orbit of a vector field on $\Bbb R^2$

Question

Let $f\in C^\infty(\Bbb R)$ satisfy $f(0) = 0$ and $f'(x) > 0$ for all $x$. Consider the vector field $X = (x^2-x)\dfrac{\partial}{\partial x}+f(y)\dfrac{\partial}{\partial y}$ on $\Bbb R^2$. Let $\Phi_t$ denote the time-$t$ flow of $X$. For $p\in \Bbb R^2$, we define the orbit of $p$ to be the set $O_p=\{\Phi_t(p): t \in \Bbb R\}$. I am asked to show that there is one and only one orbit whose closure contains $(1, 0)$ and $(0, 0)$, but I have no idea for this. Can I get some hints?

Answer 1

From the conditions on $f$ we easily get that $f$ is strictly increasing and that $f(y) < 0$ for $y < 0$ and $f(y) > 0$ for $y > 0$. Thus if we start at any point $p=(u,v) \in \mathbb R^2$ with $v \neq 0$, the $y$-coordinate of $\Phi_t(p)$ will tend to $\mp \infty$ as $t \to +\infty$ according to $v < 0$ or $v > 0$. Though, it goes to $0$ for $t \to -\infty$ but then the $x$-coordinate of $\Phi_t(p)$ will tend, in principle, to exactly one of $-\infty$, $0$, $1$ or $+\infty$ depending on $u$ (in fact, it can go only to $-\infty$ or $1$) $-$ use here the fact that $(0,0)$ and $(1,0)$ are the only fixed points of the flow.

Thus we get that if we start at a point with non-zero $y$-coordinate, we can get (at best) one of the required points in the orbit closure. So the only candidates for the points with orbit closures containing both $(0,0)$ and $(1,0)$ are the points on the $x$-axis. It is easy to check that the only orbit actually satisfying this closure condition is the orbit of any point $(x,0)$ with $x \in (0,1)$ (look also at the end of the paragraph above). Done.