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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.

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    $\begingroup$ Added the "differential-equations" tag! $\endgroup$ Oct 14, 2018 at 6:09
  • $\begingroup$ "Equilibrium point" means $y(x)=x$, right? But if $y'\le0$, then there can't be two of these, there can be at most one. $\endgroup$ Oct 14, 2018 at 6:23
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    $\begingroup$ @GerryMyerson: No, equilibrium means that $y(t)$ is constant, i.e., $y'(t)=0$. $\endgroup$ Oct 14, 2018 at 6:45

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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.

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    $\begingroup$ What is $y_0$ in your nice proof? $\endgroup$
    – dmtri
    Oct 14, 2018 at 9:03
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    $\begingroup$ @dmtri: Sorry it took me so long to get back to you; sleep and all. Anyway, $y_0$ is simply an arbitrary initial condition for $\dot y = f(y)$, lying within the interval $I_\epsilon$. Hope this helps! $\endgroup$ Oct 14, 2018 at 15:42

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