Lorenz equations and find a minimal trapping region. 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.

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 $$ C = \begin{cases}br^2 & r\le \frac{\sigma^{-2}}{3^{\frac13}}\\\ \frac{27}{16}br^4\sigma^2 &r > \frac{\sigma^{-2}}{3^{\frac13}}\end{cases}$$