# natural symmetry of the Lorenz equation

The Lorenz equation^{} has a natural symmetry defined by

$$(x,y,z)\mapsto (-x,-y,z).$$ | (1) |

To verify that (1) is a symmetry of an ordinary differential equation^{} (Lorenz equation) there must exist a $3\times 3$ matrix which commutes with the differential equation. This can be easily verified by observing that the symmetry is associated with the matrix $R$ defined as

$$R=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right].$$ | (2) |

Let

$$\dot{\text{\mathbf{x}}}=f(\text{\mathbf{x}})=\left[\begin{array}{c}\hfill \sigma (y-x)\hfill \\ \hfill x(\tau -z)-y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right]$$ | (3) |

where $f(\text{\mathbf{x}})$ is the Lorenz equation and ${\text{\mathbf{x}}}^{T}=(x,y,z)$. We proceed by showing that $Rf(\text{\mathbf{x}})=f(R\text{\mathbf{x}})$. Looking at the left hand side

$Rf(\text{\mathbf{x}})$ | $=$ | $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill \sigma (y-x)\hfill \\ \hfill x(\tau -z)-y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right]$ | ||

$=$ | $\left[\begin{array}{c}\hfill \sigma (x-y)\hfill \\ \hfill x(z-\tau )+y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right]$ |

and now looking at the right hand side

$f(R\text{\mathbf{x}})$ | $=$ | $f(\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right])$ | ||

$=$ | $f(\left[\begin{array}{c}\hfill -x\hfill \\ \hfill -y\hfill \\ \hfill z\hfill \end{array}\right])$ | |||

$=$ | $\left[\begin{array}{c}\hfill \sigma (x-y)\hfill \\ \hfill x(z-\tau )+y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right].$ |

Since the left hand side is equal to the right hand side then (1) is a symmetry of the Lorenz equation.

Title | natural symmetry of the Lorenz equation |
---|---|

Canonical name | NaturalSymmetryOfTheLorenzEquation |

Date of creation | 2013-03-22 13:44:12 |

Last modified on | 2013-03-22 13:44:12 |

Owner | Daume (40) |

Last modified by | Daume (40) |

Numerical id | 5 |

Author | Daume (40) |

Entry type | Result |

Classification | msc 34-00 |

Classification | msc 65P20 |

Classification | msc 65P30 |

Classification | msc 65P40 |

Classification | msc 65P99 |