Asymptotic stability of an equilibrium solution of an autonomous differential equation

Question

I'm having trouble understanding how I would prove the asymptotic stability of an equilibrium solution for the following problem. I understand what stability is and how to show it with other functions. However, I'm having trouble doing it with this specific function:

$$dy/dt=r(1-y/K)y-Ey$$ where the equilibrium solutions are

$$ Y_1=0,\quad Y_2=K(1-E/R) > 0 $$ and it is also given that ${E < r}$

I have proven both equilbrium solutions thus far. I also know that I can treat the original function as F(y) and then take $f'(y)=F'(y)f(y)$ since $f(y)$ = $dy/dt$ and I can use f(y) and f'(y) to sketch a solution family. At least, I know I should be able to do those things. For some reason I just don't see the solution here (bad pun completely intended).

Can anyone help me out here? If anyone needs additional information please leave a comment. I'm just not sure what all a person might need to help guide someone through such a problem.

Answer 1

A standard method to determine the asymptotic stability of an equilibrium $y_*$ of an autonomous ODE $y'=f(y)$ is to compute $f'(y_*)$.

• If $f'(y_*)<0$, the equilibrium is asymptotically stable
• If $f'(y_*)>0$, the equilibrium is unstable
• If $f'(y_*)=0$, further considerations are needed. Such further considerations involve the sign of $f$ on both sides of equilibrium. If it changes sign from negative to positive, the equilibrium is unstable. If from positive to negative, it is asymptotically stable. If it has the same sign on both sides, we have some sort of semi-stable equilibrium.

Some online resources that expand on this:

• Paul's Online Math Notes
• Equilibrium: Stable or Unstable?

In your case, $f(y) = r(1-y/K)y-Ey$. Hence, $$f'(y) = r(1-y/K) - ry/K - E$$ Plug equilibrium value into $f'$, and determine the sign.