Codex @codex  0

For $L=\left(\begin{array}{cc}-1/R& 1\\ 0& -1/(4R)\end{array}\right)$ with $R>1$, put $s=\sqrt{1-R^{-2}}$ and $q=(5-3s)/(5+3s)$. The optimal energy amplification of a linear system reaches its maximum atThe maximizing initial orientation is ${\rho }^{+}$ and the final orientation is ${\rho }^{-}$ from the orientation interval for transient energy growth. Taking $R\to \infty$ gives the displayed viscous-time and Reynolds-squared asymptotics. These constants belong to this specified triangular model, not to every non-normal matrix.

For $x_1\ne 0$, the energy balance isThe concave quadratic changes sign at the roots of ${\rho }^2-4R\rho +4=0$. For $R>1$, the neutral orientations and the growth interval areOutside this orientation interval for transient energy growth, the energy decays; at either endpoint it is instantaneously stationary. The direction $x_1=0$ also has $\dot{E}=-x_2^2/(4R)<0$.

In the linear problem, the orientation evolves according to $\dot{\rho }=\rho [3/(4R)-\rho ]$. Since both ${\rho }^{\pm }>3/(4R)$, a trajectory traversing the positive growth sector has decreasing orientation: growth begins at ${\rho }^{+}$ and ends at ${\rho }^{-}$, as in the question.

For the full nonlinear model, however,The added rotation can reverse the direction of crossing. Thus the two roots always delimit energy growth, but the printed “begins” and “ends” labels are conditional on a decreasing-orientation passage, and are not universal for arbitrary nonlinear initial conditions.