Let [itex]f: \mathbb{R}^n \to \mathbb{R}^n[/itex] and [itex]A : \mathbb{R}^n \to \mathbb{R}^n[/itex] be an invertible linear map. Then [itex]A[/itex] is a symmetry of the ODE [tex]
\dot x = f(x) [/tex] if and only if [itex]y(t) = Ax(t)[/itex] is also a solution of the ODE, ie. [itex]\dot y = f(y)[/itex]. This requires that [itex]f = A^{-1} \circ f \circ A[/itex].
If you plotted all of the solution curves in phase space, then the resulting diagram would have [itex]A[/itex] as a symmetry.
Example 1: [tex]\begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = \begin{pmatrix} y \\ -x \end{pmatrix}[/tex] is symmetric with respect to [tex]
\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}[/tex] for any [itex]\theta[/itex].
Example 2: [tex]
\dot x_1 = x_1 + x_2x_3, \qquad \dot x_2 = x_2 + x_1x_3, \qquad \dot x_3 = x_3 + x_1x_2[/tex] is symmetric with respect to any permutation of [itex](x_1, x_2, x_3)[/itex].
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