 Instantaneous Lorenz error growth at a zero-y crossing

Consider the Lorenz system with $\sigma >0$ and $\beta >0$, at a state where $y=0$. An infinitesimal error $\eta$ evolves by $\dot{\eta }=J\eta$, where $J$ is the Jacobian matrix. Its squared Euclidean norm satisfiesAt that crossing, the skew couplings involving $x$ cancel, leavingThe planar block has negative trace and determinant $4\sigma -(\sigma +\rho -z{)}^2$. It has a positive eigenvalue exactly whenThus some error direction initially grows for either $z<\sigma +\rho -2\sqrt{\sigma }$ or $z>\sigma +\rho +2\sqrt{\sigma }$. The vertical direction initially contracts.

For $\sigma =10$, $\rho =28$, $z=26$ and $\beta =3$, the eigenvalues of $J+J^T$ are $4,-26,-6$. The most rapidly growing unit direction is $(1,2,0)/\sqrt{5}$, with instantaneous logarithmic norm growth $4/2=2$. The factor of two distinguishes squared-norm growth from norm growth.

This is a local criterion at the specified state. Later rotation of the perturbation direction can change its growth, so it does not prove a positive asymptotic Lyapunov exponent.