I'm classifying the equilibrium points of the system $\dot{x}=f(x)=x(x-1)^2$. $x=1$ is one of them, and looking at the graphic of $\dot{x}$ versus $x$, I concluded is not stable. Nonetheless, I would like to conclude the same answer from it's derivative like in this theorem . In my case, $f'(1)=0$, so I can't conclude anything, how can I do it without plotting the system?
Is it possible to conclude unstability if derivative of equilibrium point is zero
control theorydynamical systemsordinary differential equationsstability-in-odesstability-theory
Related Solutions
Hints: This will guide you through the process and you can figure out how to do this in Matlab.
To find the critical points, you want to simultaneously solve $x' = 0, y' = 0$. You will get two critical points at $$(x,y) = \left(\dfrac{1}{2}, \dfrac{1}{6}\right), (1, 0)$$
You can then determine the types of critical points these are by finding the Jacobian, $J(x, y)$, and evaluating the eigenvalues of the $2x2$ Jacobian. In the phase portrait below, you can see we have a stable and an unstable critical point.
The Phase portrait will show these two critical points and should look something like:
Best Answer
But $f''(1)\ne 0$, so locally $x=1+u$ behaves similar to $$\dot u=cu^2,$$ this gives semi-stable behavior.
You could also just argue by the signs in the intervals between and outside the roots.