# 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.

• Added the "differential-equations" tag! Oct 14, 2018 at 6:09
• "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. Oct 14, 2018 at 6:23
• @GerryMyerson: No, equilibrium means that $y(t)$ is constant, i.e., $y'(t)=0$. Oct 14, 2018 at 6:45

## 1 Answer

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.

• What is $y_0$ in your nice proof? Oct 14, 2018 at 9:03
• @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! Oct 14, 2018 at 15:42