# natural symmetry of the Lorenz equation

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

 $R=\begin{bmatrix}-1&0&0\\ 0&-1&0\\ 0&0&1\end{bmatrix}.$ (2)

Let

 $\dot{\textbf{x}}=f(\textbf{x})=\begin{bmatrix}\sigma(y-x)\\ x(\tau-z)-y\\ xy-\beta z\end{bmatrix}$ (3)

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

 $\displaystyle Rf(\textbf{x})$ $\displaystyle=$ $\displaystyle\begin{bmatrix}-1&0&0\\ 0&-1&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}\sigma(y-x)\\ x(\tau-z)-y\\ xy-\beta z\end{bmatrix}$ $\displaystyle=$ $\displaystyle\begin{bmatrix}\sigma(x-y)\\ x(z-\tau)+y\\ xy-\beta z\end{bmatrix}$

and now looking at the right hand side

 $\displaystyle f(R\textbf{x})$ $\displaystyle=$ $\displaystyle f(\begin{bmatrix}-1&0&0\\ 0&-1&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix})$ $\displaystyle=$ $\displaystyle f(\begin{bmatrix}-x\\ -y\\ z\end{bmatrix})$ $\displaystyle=$ $\displaystyle\begin{bmatrix}\sigma(x-y)\\ x(z-\tau)+y\\ xy-\beta z\end{bmatrix}.$

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 NaturalSymmetryOfTheLorenzEquation 2013-03-22 13:44:12 2013-03-22 13:44:12 Daume (40) Daume (40) 5 Daume (40) Result msc 34-00 msc 65P20 msc 65P30 msc 65P40 msc 65P99