How to find an appropriate Lyapunov function to show the stability of a system:

$\dfrac{dx}{dt} = -(x-1)(x-2)^2$

around its equilibrium point $x = 1$?

By linearizing this system, it can be shown that the $x=1$ point is locally asymptotically stable.

What is the appropriate Lyapunov function?


$V(x)=(x-1)^2$ is a (strict) Lyapunov function.

Added: Indeed, $V\ge0$ and $V(x)=0$ if and only if $x=1$. Moreover, $$ \dot V(x)=-2(x-1)^2(x-2)^2 $$ and so for $x\in(0,2)\setminus\{1\}$ we have $\dot V(x)<0$. Hence, $V$ is a strict Lyapunov function, which shows that the equilibrium point $x=1$ is asymptotically stable.

  • $\begingroup$ I think that V(x) can show Xe = 1 is stable, but not asymptotically stable. $\endgroup$ – yyzr Feb 29 '16 at 6:14
  • $\begingroup$ I don't think you could find a class K function f(x) satisfying dV/dt <= -f(x) $\endgroup$ – yyzr Feb 29 '16 at 6:18
  • $\begingroup$ Since you don't seem to be convinced, I added some details. $\endgroup$ – John B Feb 29 '16 at 15:16

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