9
$\begingroup$

I need to find a Lyapunov function for $(0,0)$ in the system:

\begin{cases} x' = -2x^4 + y \\ y' = -2x - 2y^6 \end{cases}

Graph built using this tool showed that there should be stability but not asymptotic stability. I'd like to prove that fact by means of Lyapunov function but cannot find the appropriate one among the most common ones such as $Ax^{2n}+By^{2m}$, $Ax^2+By^2+Cxy$, etc.

Please, give me some hint about how to proceed with the search of the suitable function or even the desired function itself it you know the easy way of finding it in this case.

Thanks in advance

$\endgroup$
  • $\begingroup$ Simulated phase diagrams do not seem conclusive enough to decide whether (0,0) is stable. Why do you think it is? $\endgroup$ – Did May 25 '15 at 12:33
  • $\begingroup$ @Did As I was told by my groupmates, the situation when trajectories don't go away from 0 to the infinity leads to stability. And this is exactly the case for my system: s9.postimg.org/ywsij9kbj/image.png $\endgroup$ – mik May 25 '15 at 13:07
  • $\begingroup$ Sorry but, from the phase diagrams, I cannot determine whether trajectories cycle or spiral outwardly or spiral inwardly (and I would be cautious about approximation errors in simulations, if I were you). $\endgroup$ – Did May 25 '15 at 14:27
  • 2
    $\begingroup$ If your guess (stable but not asymptotically stable) is correct, then the equation has periodic trajectories. Consequently, a Lyapunov function $V$ would have to be constant on those trajectories, which means that $V=c$ is an implicit equation of a trajectory. The upshot is that in such a case, finding a Lyapunov function is the same as finding the trajectories explicitly -- a task that is unlikely to have a closed form answer. $\endgroup$ – user147263 May 25 '15 at 21:17
  • 1
    $\begingroup$ Related question. $\endgroup$ – Did Jun 21 '15 at 19:00
2
$\begingroup$

As explained in the comments, even though simulated phase diagrams seem to exhibit cycling trajectories near the origin, they are not conclusive enough to decide whether the origin is stable or not, that is, whether trajectories cycle or spiral outwardly or spiral inwardly. Caution is advised about approximation errors in simulations.

streamplot{{-2x^4+y,-2x-2y^6},{x,-1,1},{y,-1,1}}

enter image description here

$\endgroup$
0
$\begingroup$

Dynamical system

$$ % \begin{align} % \dot{x} &= -2x^{4} + y \\ % \dot{y} &= -2x - 2y^{6} \\ % \end{align} % $$

There are two critical points: the origin $(0,0)$ and $$\left\{-\frac{1}{2^{6/23}},2 \left(\frac{1}{2^{6/23}}\right)^4\right\} =\left\{ -\frac{1}{2^{6/23}}, \frac{1}{2^{1/23}} \right\} \approx \{-0.834585,0.970313\}$$

Phase portrait

The phase portrait below displays the nullcline $\dot{x}=0$ with a red, dashed line and $\dot{x}=0$ in purple.

flow

Stability

The polar transformations $$ x = r \cos \theta, \qquad y = r \sin \theta $$ lead to $$ r^{2} = x^{2} + y^{2} $$ which implies $$ \dot{r} = \frac{x\dot{x}+y\dot{y}}{\sqrt{x^2+y^2}} = \frac{-2 x^5-x y-2 y^7}{\sqrt{x^2+y^2}}, $$ plotted below.

rdot

The problem is that the sign of $\dot{r}$ changes twice in every neighborhood of the origin.

$\endgroup$

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.