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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...

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  • $\begingroup$ "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

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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.

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