2
$\begingroup$

Consider system:

$\dot x = ax - {e^{\tanh \left( k \right)}}x$

$\dot k = - \frac{1}{{{{{\mathop{\rm sech}\nolimits} }^2}\left( k \right)}}\frac{1}{{{e^{\tanh \left( k \right)}}}}a{x^2}$

where, $a>e=2.71828\dots$ Choose a Lyapunov function as

$V(x,k) = \frac{1}{2}{x^2} + {e^{{\mathop{\rm tanh}\nolimits} \left( k \right)}}$

Its time derivative is

$\begin{aligned} \dot{V} &=x\left(a x-e^{\tanh ({k})} x\right)+e^{\operatorname{tanh}({k})} \operatorname{sech}^{2}({k}) \dot{{k}} \\ &=a x^{2}-e^{\tanh ({k})} x^{2}-a x^{2} \\ &=-e^{\tanh ({k})} x^{2} \leq 0 \end{aligned}$

This implies the asymptotic stability of $x$. However, since $a$ is larger than ${e^{\tanh \left( k \right)}}$, this system is clearly unstable. To be more specific, consider another "Lyapunov" function (clearly this should not be called Lyapunov function, see answer for clearification):

$U(x,k)=\frac{1}{2}x^2$

$\begin{aligned} \dot U &=x\left(a x-e^{\tanh ({k})} x\right) \\ &=\left(a-e^{\tanh ({k})}\right) x^{2} \geq 0 \end{aligned}$

According to Chetaev instability theorem, $x=0$ is an unstable equilibrium.

Why comes this contradiction?

$\endgroup$

1 Answer 1

5
$\begingroup$

Your Lyapunov function is invalid. You want to analyse the equilibrium $x = 0$, but your Lyapunov function does not have an isolated minimum there.

For example, $V(0,0) = \frac{1}{2}0^2 + e^{\tanh(0)} = 0 + 1 = 1$.

Now, for $x = 0$, $k < 0$ you will have $V(0, k) < V(0, 0)$.

Note 1: Your function $U$ is usually called Chetaev function rather than Lyapunov function. By definition, your $U$ is not a Lyapunov function as it is only positive semi-definite.

Edit: Expanded the answer for better clarity.

Note 2: One thing to note first is that your system has an equilibrium set with $x = 0$ and $k \in \mathbb{R}$ arbitrary. So, stability cannot be checked with the direct method, as it works only for isolated equilibria, not for a continuum of equilibria.

Note 3: To analyze stability of an isolated equilibrium, your Lyapunov function should have that point as an isolated minimum. So even if this system would have an isolated equilibrium, the analysis would be invalid because this $V$ does not have an isolated minimum at all.

$\endgroup$
3
  • 1
    $\begingroup$ This is so simple answer. I did not even realize to check this... Thank you! You just save my day. $\endgroup$
    – hory c.
    Commented Apr 25, 2019 at 17:59
  • 1
    $\begingroup$ $V(0,0)=0$ does not have to hold because one can always alter the Lyapunov function to $V'(x,k)=V(x,k)-V(0,0)$. However this does not solve the problem that $V'(0,k)<V'(0,0)$ when $k<0$. $\endgroup$ Commented Apr 26, 2019 at 11:28
  • $\begingroup$ @KwinvanderVeen True. I wrote "minimum" to reflect that, but the $≠0$ was a bit misleading. I expanded the answer for better clarity. $\endgroup$
    – SampleTime
    Commented Apr 26, 2019 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .