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Differentiate the Friedmann equationand use the cosmological perfect-fluid continuity equationAfter dividing by $2H$,Since $\ddot{a}/a=\dot{H}+H^2$, eliminating the curvature term gives the Friedmann acceleration equation

Under the strong energy condition in a Friedmann universe, $\rho +3P\ge 0$, so $\ddot{a}/a\le 0$. Thereforeand wherever $H\ne 0$,

If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0{)}^{-1}$. Thus $H\to +\infty$ and $a\to 0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to -\infty$ and $a\to 0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a Big Crunch. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.

The generalized momenta conjugate to the two ignorable coordinates areandSince the Lagrangian contains neither $\varphi$ nor $\psi$, both momenta are constant in time.

Eliminating the two angular velocities givesConservation of energy then yields the heavy symmetric top reductionwhere

As $\theta \to 0$, the first term diverges unlessWriting their common value as $p$, the small-angle expansion isBy the effective potential stability criterion, the vertical state is stable exactly whenAt the vertical state $p=I_3(\dot{\psi }+\dot{\varphi })$, so this is the gyroscopic stabilization of an inverted symmetric top by sufficiently rapid spin.

Taking the Laplace transform in $t$ and using the zero initial condition givesThe solution bounded as $x\to \infty$ isThe transformed boundary condition isso this is the sinusoidally forced heat equation on a half-line. Integrating in $x$ gives

Use the principal square root, with a branch cut along the negative real axis. The Bromwich contour is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contributeHence

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt{p}$ are $i\sqrt{s}$ and $-i\sqrt{s}$. Their jump and the opposite contour orientations combine to giveThis is precisely Bromwich inversion with a square-root branch cut: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.

The log density isIts kink occurs at the unknown location $\mu$, so its parameter dependence cannot be separated into a fixed finite-dimensional statistic of $y$ and the smooth canonical form required of an exponential dispersion family. Thus the two-parameter location-scale Laplace distribution is not such a family.

For independent observations, the log-likelihood isFor every fixed $\sigma >0$, maximizing this expression in $\beta$ is equivalent to minimizing $S(\beta )$. Hencethe least absolute deviations estimator. At this value,so, in the nondegenerate case $S(\hat{\beta })>0$,