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For the linear evolution from part (iii), maximize the energy ratio at each time. The optimal energy amplification of a linear system isThus the optimizing initial state is a right singular vector, not generally an eigenvector of $L$.

There is a convenient exact way to find the maximizing time. Let $x_0$ be a unit optimizing initial vector and $y=M(t)x_0$. The singular value decomposition gives $M^{T}y=G_{\mathrm{opt}}{x}_0$. For $t>0$, the top singular value is simple. At an interior time maximum,Since $LM=ML$, this also equals $G_{\mathrm{opt}}{x}_0^{T}L{x}_0$. Thus both initial and final orientations are neutral energy-growth directions. The positive entries of $M^{T}M$ let the top right singular vector be chosen with positive components. Its orientation decreases through the growth sector, so it must start at ${\rho }^{+}$ and end at ${\rho }^{-}$.

Put $q=a/d=e^{-3t/(4R)}$ and $K=4R/3$. The solution givesSetting this equal to ${\rho }^{-}$ yieldsTherefore the exact optimal time for $R>1$ isHere and below $\log$ is the natural logarithm. This is the unique positive stationary time: $q$ decreases monotonically from one to zero, and the neutral-direction condition determines one $q_{\ast }\in (0,1)$. Also $G_{\mathrm{opt}}(0)=1$, $G_{\mathrm{opt}}(\infty )=0$, and $R>1$ admits initial energy growth, so this stationary point is the global maximum. For $R\le 1$, the instantaneous energy-growth criterion for a linear system gives no amplification, and the maximum is one at time zero.

Differentiate the quadratic energy:For $x_1\ne 0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when ${\lambda }_2{r}^2+r+{\lambda }_1>0$. With both eigenvalues negative, such directions exist iff $1-4{\lambda }_1{\lambda }_2>0$. Writing $\delta =\sqrt{1-4{\lambda }_1{\lambda }_2}$, the growing initial conditions are the two opposite open wedgesIf $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly whenIndeed, this makes the symmetric part of a matrix $H$ a negative semidefinite matrix, so $\dot{E}\le 0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the instantaneous energy-growth criterion for a linear system.