Stability of equilibrium points advice

Question

How do I check the stability of the equilibrium points for the equation: $\frac{dy}{ds}=Y^{*}\Big(1-Y^{*}\Big)$

Obviously I understand that the equilibrium points are $Y_{1}=1$ and $Y_{2}=0$ but how can I work out the stability?

Answer 1

You can visualize equilibrium point stability in terms of "flow fields." If solutions of the ODE tend toward an equilibrium point, it is stable. If solutions move away from the equilibrium point, it is unstable. Thus, we need to look at the behavior of solutions close to the equilibrium point. Let's consider $y = 0$ first. When $y < 0$, $dy/ds < 0$, so $y$ decreases. When $0 < y < 1$, $dy/ds > 0$, so $y$ increases. Thus, on both sides of $y = 0$, the solutions move away from the point. This means it is unstable. Let's look at $y = 1$ now. When $y > 1$, $dy/ds < 0$, so $y$ decreases. Thus, on both sides of $y = 1$, the solutions tend toward the point. This means it is stable. You can see this more clearly if you graph $dy/ds$ versus $y$. This article is a great resource: https://web.ma.utexas.edu/users/davis/375/popecol/lec9/equilib.html

Answer 2

The simplest way is to utilize qualitative methods to determine the stability. Draw phase plane for the system:

The equilibrium point $y_e=(1,0)$ is stable or attractive because the two blue arrows are heading toward the equilibrium point. The direction of the arrows are determined based on the positivity or negativity of $\frac{dy}{ds}$. If the differential equation is positive as the case with this example when $0<\frac{dy}{ds}<1$, the trajectory moves to the right and vise versa. While the equilibrium point $y_e=(0,0)$ is unstable or repeller because the two blue arrows are move away from the equilibrium point.