I have this system:

$\dot x=-(x-1)(x-2)^2$

I'm asked to find the equilibria and to study the stability using:

i) linearization

ii) appropriate Lyapunov function

How should I linearize the system? And how should I proceed with the stability analysis? Is there a variable change involved (e.g. $x_2=x-1$, $x_3=x-2$)? I'm lost...

  • $\begingroup$ "i) linearization ii) appropriate Lyapunov function" Neither is useful, a phase portrait gives everything (the fixed points are 1 (stable) and 2 (unstable)). $\endgroup$
    – Did
    Oct 25, 2014 at 15:09

1 Answer 1


Equilibrium points of the given system are $x_{eq} = 1,2$. The Jacobian is $J(x) = -[2(x-1)(x-2) + (x-2)^2]$. Evaluating the Jacobian at $x = 1, J(1) = -1$ and at $x = 2, J(2) = 0$. $\therefore$ the linearized system is $\dot{x}_2 = J(1)x_2 = -x_2$ about the point $x = 1$ and $\dot{x}_3 = J(2)x_3 = J(2)x_3 = 0$(Note: $x_2$ and $x_3$ are as defined in the question). The point $x_{eq} = 1$ is stable because $J(1)$ is negative while nothing can be commented on $x_{eq} = 0$ because $J(2) = 0$.

The system equation in terms of $x_2$ is $\dot{x_2} = -x_2(x_2 - 1)^2$ and in terms of $x_3$ is $\dot{x_3} = -(x_3 + 1)x_3^2$. Consider $ V_2(x_2) = (1/2)x_2^2, \dot{V}_2 = -x_2^2(x_2 - 1)^2 \leq 0$. Can't figure out from here onwards.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.