= Solution
For $x_1\ne0$, the energy balance is
$$
\dot E=x_1^2\left[-\frac1R+\rho-\frac{\rho^2}{4R}\right].
$$
The concave quadratic changes sign at the roots of $\rho^2-4R\rho+4=0$. For $R>1$, \b[the neutral orientations and the growth interval are]
$$
\boxed{\rho^\pm=2R\left(1\pm\sqrt{1-R^{-2}}\right),\qquad
\dot E>0\iff\rho^-<\rho<\rho^+.}
$$
Outside 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: \b[growth begins at $\rho^+$ and ends at $\rho^-$], as in the question.
For the full nonlinear model, however,
$$
\dot\rho=|x|(1+\rho^2)+\frac{3\rho}{4R}-\rho^2.
$$
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.
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