 Solution

For an infinitesimal perturbation $\mathbf{\eta }$, $\dot{\mathbf{\eta }}=J\mathbf{\eta }$, andThus the principal instantaneous stretching rates, called local Lyapunov exponents in the question, areFor $z\ne 1$, ${\lambda }_{+}>0$ and ${\lambda }_{-}<0$, while $a>0$. There are two instantaneously expanding orthogonal directions and one contracting direction. These are instantaneous norm-growth rates; the directions rotate along a trajectory, so this does not imply two positive long-time Lyapunov exponents or indefinitely exponential growth along fixed axes.

At $x=c$, put $s=\mathrm{sgn}(z-1)$. The eigenvectors and rates areThe condition $|z-1|>2a$ makes ${\mathbf{e}}_{+}$ the fastest-growing direction, faster than ${\mathbf{e}}_y$; ${\mathbf{e}}_{-}$ is the fastest-contracting direction. Both lie at $45^{\circ }$ to the $x,z$ axes.