Re: Lorenz equations and equilibria

Archie <kohmover123@xxxxxxxxx> writes:

The Lorenz equations are

x' = sigma (y - x)
y' = rx - y - xz
z' = xy - bz

The equilibria are the origin plus

x = y = sqrt(b (r-1)) and z = r - 1

My question is this:

I can see that these are definitely equilibria. But I'm wondering how
they would be derived from the Lorenz equations. Maybe there could be
other equilibria that are equally obvious (i.e., not at all). How
would anyone know, since these equilibria are so unlikely, given the
base equations?

This is not a homework problem. I'm just trying to understand this.

An equilibrium (x,y,z) must satisfy the equations

1) sigma (y - x) = 0
2) rx - y - xz = 0
3) xy - bz = 0

From (1), y=x. Plug in to (2) and you see that x = 0 or z = r-1.
If x = 0, (3) says z = 0. If z = r-1, (3) says x = y = (+/-) sqrt(b (r-1)).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.