Dynamical systems and equilibrium points Recently I have been trying to understand dynamical systems and I came up with   the following question:
Consider the system $y' =f(y)$ , $t\ge0$ with only two equilibrium points $0, 1$ and $f(y)\le0$ in $[0,1]$, $f'(0)\lt0$,  $f'(1)\gt0$. Is there any solution  $y$ with $y(0)\in(0,1)$ such that $lim_{t\to \infty}y$ is not $0$ ? 
Remark:    As $y$ is decreasing, it will have an infimum $s$ ,  $s\ge0$. If $s\ne0$ then it cannot be an equilibrium point....
Thanks in advance. 
 A: No.  Any zero of $f(y)$ is an equilibrium, so the hypothesis that the only equilibria are $0$ and $1$ forces $f(y) < 0$ in $(0, 1)$.  This in turn implies that for any $0 < \epsilon < 1/2$ we have $f(y) < 0$ on the closed interval 
$I_\epsilon = [\epsilon, 1 - \epsilon], \tag 1$
and since $I_\epsilon$ is compact, $f(y)$ attains a global maximum $m < 0$ on $I_\epsilon$; then for
$y(t_0) = y_0 \in I_\epsilon, \tag 2$
$y(t)$ obeys
$\dot y = f(y) \le m < 0 \tag 3$
on $I_\epsilon$; therefore, $y(t)$ satisfying (3) will reach the value $\epsilon$ within time
$\Delta t = \displaystyle \int_{y_0}^\epsilon \dfrac{dy}{f(y)} = -\int_\epsilon^{y_0} \dfrac{dy}{f(y)} \le -\int_\epsilon^{y_0} \dfrac{dy}{m} = -\dfrac{y_0 - \epsilon}{m} = \dfrac{\epsilon - y_0}{m}; \tag 4$
since this holds for every $0 < \epsilon < 1/2$, we see that for every $y_0 \in (0, 1)$, $y(t)$ becomes arbitrarily small for large enough $t$; but this implies 
$\displaystyle \lim_{t \to \infty} y(t) = 0. \tag 5$
The result is false if we remove the condition that $f(y)$ have only two equilibria in $[0, 1]$; consider
$f(y) = y \left (y - 1 \right ) \left ( y - \dfrac{1}{2} \right )^2, \tag 6$
and set
$y(0) = \dfrac{3}{4}; \tag 7$
$f(y)$ satisfies the requisite criteria, but 
$\displaystyle \lim_{t \to \infty} y(t) = \dfrac{1}{2}, \tag 8$
which may be proved in a manner similar to the above.
