Considering small perturbation determine whetherit is a stable or unstable node

Question

This is the part of the question I do not understand. I'm given an nonlinear equation for a population, let's say $$ \dot{x}= x^{4}-3x^{3}+2 $$ So we are asked to find the equilibrium points, one is x=1 and considering small perturbation determine it's stability which is where I am lost. If someone can explain this I will really appreciate it.

Answer 1

I'll write $\dot x = f(x) = x^4-3x^3+2$.

Consider the value of $f$ in a neighbourhood of 1.

If $f(a) > 0$ when $a > 1$ (for $a$ still sufficiently close to 1), then the values of $x$ will increase further since $\dot x > 0$. Hence the point can't be stable.

Similar reasoning applies when $f(b) < 0$ for $b < 1$ and sufficiently close to 1.

If, on the other hand, $f(a) < 0$ when $a > 1$, then after a small perturbation upwards, $x$ would tend to return to the value 1. Likewise for small perturbations downwards if $f(b) > 0$ for $b < 0$.

I would say that the previous paragraph is a reasonable definition of stability. Can you provide the definition you are using?

There's also the case where $f(a) = 0$ when $a > 1$. How to handle this depends on the definition of stability you have, but since $x$ would not tend to return to the initial value after perturbation, I would not call this stable.