Graphically, where are the equilibrium points of the Lorenz system?

Question

Given the Lorenz equations $$ \frac{dx}{dt} = \sigma(y-x);\\ \frac{dy}{dt} = x(\rho-z)-y;\\ \frac{dz}{dt} = xy - \beta z $$

It is well known that the equilibrium points are: $(0,0,0)$, $(\pm\sqrt{\beta(\rho-1)},\pm\sqrt{\beta(\rho-1)},\rho-1)$

It is clear that the trajectory first takes flight from the equilibrium point at the origin. Are the other equilibrium points visible on computer generated plots of the solution for the Lorenz system?

http://homepages.math.uic.edu/~kjerland/Lorenz/lorenz_attractor.html

Answer 1

The other fixed points are in the centre of the two “butterfly wings”. E.g., here they are for the canonical parameter set ($σ=10; β=\tfrac{8}{3}; ρ=28$):

If you want an interactive 3D plot or change parameter values, here is a Python 3 script that generates this plot (and allows you to rotate it):

#!/usr/bin/python3
# -*- coding: utf-8 -*-

from jitcode import jitcode, provide_basic_symbols
import numpy as np
from sympy import Rational, solve
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

σ = 10
β = Rational(8,3)
ρ = 28

t, y = provide_basic_symbols()

lorenz = [
    σ*(y(1)-y(0)),
    y(0)*(ρ-y(2)) - y(1),
    y(0)*y(1)-β*y(2)
    ]

fixed_points = [[sol[y(i)].n() for sol in solve(lorenz)] for i in range(3)]

ODE = jitcode(lorenz)
ODE.set_integrator("dopri5", nsteps=1e10)
ODE.set_initial_value(np.array([0.5,0.6,0.7]),0.0)

data = np.vstack(ODE.integrate(t) for t in np.arange(1000,1100,0.01))

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(*(data.T))
ax.scatter(*fixed_points)
plt.show()

(To run this, you need this module by yours truly, which, on a Unixoid system, you can probably install with pip3 install jitcode.)