Consider Lorenz's equations

$x^{'}= \sigma (y-x)$

$y^{'}= (rx-y-xz)$

$z^{'}= (xy-bz)$

$\sigma, r, b>0$ are parameters of the system.

The question is as follows Show that there is a certain ellipsoidal trapping region E of the form $rx^{2} + \sigma y^{2} + \sigma (z-2r)^{2} < or = C$ Find the minimal value of constant C

basically i am given a trapping function for the equations an im asked to find the minimal value for this region.

using some notes for class i have manged to Deduce an approximate Value of C that will satisfy this equation but i haven't an idea how to find the minimal one.

To be honest is someone could show how to do this from the start that would probably be better as what i did i don't entirely understand and im not sure if its right.


1 Answer 1


We first see that the time derivative of the function $$V = rx^2+\sigma y^2+\sigma(z−2r)^2$$ is

$$\dot V = -2\sigma(rx^2+y^2+bz^2-2brz)$$

We know that a trapping region exists. Because $\dot V$ goes to $-\infty$ in all directions, the region where $\dot V \ge 0$ is bounded. In fact, if $C$ is the maximum value of $V$ in the region where $\dot V \ge 0$, then it's easy to see that any point outside of $rx^2+\sigma y^2+\sigma(z−2r)^2 \le C + \epsilon$ enters this region in finite time. So we try to find the maximum value obtained by $V$ in this region. $\dot V = 0$ gives $$rx^2+y^2+bz^2 \le 2brz$$

Here the computation got difficult, so I gave up and consulted Mathematica. This give the solution for $C$ as $$ C = \begin{cases}br^2 & r\le \frac{\sigma^{-2}}{3^{\frac13}}\\ \frac{27}{16}br^4\sigma^2 &r > \frac{\sigma^{-2}}{3^{\frac13}}\end{cases}$$

  • $\begingroup$ Your name is beyond priceless! and thank you it seems i was on the right track but once i got to that derivative well... $\endgroup$
    – Faust
    Mar 24, 2013 at 4:28

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