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I would like to show that the quantity:

$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right)$

is negative on the surface:

$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$

for some sufficiently large value of $C$.

I was not able to massage the first quantity any more in order to make it look like the second. I also considered a change of coordinates, but had no luck. $\sigma, b, r$ are positive parameters.

This is a step in exercise 9.2.2 from Strogatz Nonlinear Dynamics and Chaos.

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  • $\begingroup$ I was a little confused at first because usually $r = \sqrt{x^2 + y^2 + z^2}$, but you really mean that $r$ is just some given parameter, right? $\endgroup$ Nov 23, 2010 at 12:17
  • $\begingroup$ In this problem you could try to use Lagrange multipliers to find the maximum of your quantity for any given $C$, and then take the limit as $C \rightarrow \infty$. I was just trying to work it out though, and because of all those parameters it becomes a pain to deal with all the special cases. What assumptions are you allowed to make in those parameters? Like, can one assume that $\sigma \neq 1$ or $b \neq \sigma$. $\endgroup$ Nov 23, 2010 at 12:32
  • $\begingroup$ I may assume that $r>1$ but I really should just use that they are positive parameters. I tried using Lagrange multipliers but also had a problem with the tractability of the math. $\endgroup$
    – Gus
    Nov 23, 2010 at 17:48

2 Answers 2

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Since the parameter $\sigma$ is positive, the quantity $$ -2\sigma \left ( rx^2 + y^2 + b(z-r)^2 - br^2 \right ) $$ is negative if $$ rx^2 + y^2 + b(z-r)^2 > br^{2}. $$ This inequality defines the exterior of an ellipsoid (call it $E_1$); note that the size of this ellipsoid is fixed by the parameters (that's a hint).

Now the equation $$ rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2 = C $$ defines a different ellipsoid, $E_2$, the size of which is determined by your choice of $C$ (another hint).

At this point, remind yourself what it is that you want to show. Typically, the goal is to show that there exists a $C$ such that $E_2$ defines a trapping region for the Lorenz equations, in which case it suffices to show that $E_2$ can be made large enough so that it contains $E_1$. There's really no additional calculation necessary to do this -- you just need to understand what you've done so far.

A related (but different) question is to find an explicit lower bound on $C$ in terms of the parameters. In this case, you can find bounds on each of $x$, $y$, and $z$ separately for points inside of $E_1$. This will then give you a bound on the quantity $$ rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2 $$ which then defines $C$.

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Just simplifying:

$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right) < 0$

$=> rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2} > 0 ... \sigma > 0$

$=> rx^{2}+y^{2}+bz^{2}-2brz+br^{2}-br^{2} > 0 $

$=> rx^{2}+y^{2}+bz^{2}-2brz > 0 $

$=> rx^{2}+y^{2}+bz^{2} > 2brz $

$=> z<=0$ or $z>=2r$ for any $C$ or

$ rx^{2}+y^{2} > 2brz - bz^{2}, 0<z<2r $

Now, $rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$

$=> rx^{2}+y^{2}+(\sigma-1) y^{2}+\sigma\left(z-2r\right)^{2}=C$

$=> rx^{2}+y^{2}= C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}$

So the original expression is negative when $z$ is outside $(0,2r)$ or

$ 2brz - bz^{2} < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0<z<2r $

$=> bz(2r - z) < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0<z<2r $

$=> 0 < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}+bz(z - 2r), 0<z<2r $

$=> 0 < C -(\sigma-1) y^{2}-\sigma\left(z(1-b/2)-2r\right)^{2}+(bz/2)^{2}, 0<z<2r $ ... completing the square

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