Can an Ellipsoidal Region Contain All Trajectories of the Lorenz Equations?

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Homework Statement



Show that there is a certain ellipsoidal region E of the form rx2 + σy2 + σ(z-2r)2 ≤ C such that all trajectories of the Lorenz equations eventually enter E and stay in there forever.

Homework Equations



Lorenz Equations:

\dot{x} = \sigma (y - x)

\dot{y} = rx - xz - y

\dot{z} = xy - bz


V is the volume.

The Attempt at a Solution



V = rx^2 + \sigma y^2 + \sigma(z-2r)^2

\dot{V} = 2rx\dot{x} + 2\sigma y \dot{y} + 2 \sigma(z - 2r) \dot{z}

\dfrac{1}{2} \dot{V} = rx(\sigma y - \sigma x) + \sigma y (rx - xz - y) + (\sigma z - 2 \sigma r)(xy - bz)

\Rightarrow \dfrac{1}{2 \sigma} \dot{V} = -rx^2 - y^2 -bz^2 + 2bz


\Rightarrow \dot{V} < 0

Therefore, all trajectories of the Lorenz equations eventually enter E and stay in there forever.



I am not sure I approached this problem correctly, and the additional challenge problem states to try and obtain the smallest possible value of C with this attracting property. Did I miss something in my workthrough above involving C? If I subtract it from both sides of the original inequality and involve it in the workthrough involving the change in volume, isn't the value of C irrelevant given that its derivative is zero (being that it's a constant) ?
 
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That seems to help quite a bit. Now I'm only going half as crazy as I was. Thanks a lot!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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