Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 331 / 4 / b / ii / Solution

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.