= Solution
For an infinitesimal perturbation $\boldsymbol\eta$, $\dot{\boldsymbol\eta}=J\boldsymbol\eta$, and
$$
\frac{d}{dt}\log|\boldsymbol\eta|=\frac{\boldsymbol\eta^TS\boldsymbol\eta}{|\boldsymbol\eta|^2},\qquad
S=\frac{J+J^T}{2}=\begin{pmatrix}0&0&(z-1)/2\\0&a&0\\(z-1)/2&0&x-c\end{pmatrix}.
$$
Thus the principal <instantaneous stretching rates>, called local <Lyapunov exponents> in the question, are
$$
\boxed{\lambda_y=a,\qquad
\lambda_\pm=\frac{x-c\pm\sqrt{(x-c)^2+(z-1)^2}}{2}.}
$$
For $z\ne1$, $\lambda_+>0$ 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 $s=\operatorname{sgn}(z-1)$. The <eigenvectors> and rates are
$$
\boxed{\mathbf e_+=\frac{(1,0,s)}{\sqrt2},\quad\lambda_+=\frac{|z-1|}{2};\qquad
\mathbf e_-=\frac{(1,0,-s)}{\sqrt2},\quad\lambda_-=-\frac{|z-1|}{2}.}
$$
The 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.
Back to article page
