 Solution

At the optimum, $d=q_{\ast }^{1/3}$ and the initial direction is $(1,{\rho }^{+})$. Since the final direction has ratio ${\rho }^{-}$,Using ${\rho }^{\pm }=2R(1\pm s)$, $R^2(1-s^2)=1$, and $q_{\ast }=(5-3s)/(5+3s)$ simplifies this to the exact maximum gain:As $R\to \infty$, $s=1-1/(2R^2)+O(R^{-4})$, so $q_{\ast }\to 1/4$ and $(1+s)/(1-s)\sim 4R^2$. Consequently the requested large-Reynolds-number transient growth scaling isThese are the optimal time and gain of a Reynolds-scaled triangular model. The leading gain can also be read from the off-diagonal matrix exponential entry: with $\tau =t/R$, $b=(4R/3)(e^{-\tau /4}-e^{-\tau })$ is maximized at $\tau =4\log 4/3$ and has leading maximum $R{4}^{-1/3}$.