Solution

Put

x = a^{2}

. Apart from the positive factor

H_{0}^{2} / a^{4}

H^{2} has the sign of F (x) = x^{2} - β x + β .

For

β > 4

, its discriminant

β (β - 4)

is positive and its two roots are

x_{\pm} = \frac{β \pm β ^{2} - 4 β}{2} .

Both are positive. Moreover

F (0) > 0

and

F (2) = 4 - β < 0

, so

0 < x_{-} < 2 < x_{+} .

A universe expanding from

a = 0

follows the allowed region

0 \leq a^{2} \leq x_{-}

and reaches

a^{2} = x_{-}

with

H = 0

. There,

\dot{H} = - β \frac{H _{0}^{2}}{a ^{4}} (2 - a^{2}) < 0,

H

crosses from positive to negative: expansion reverses into contraction.

a \to 0

on the contracting branch,

H \sim - \frac{H _{0} β}{a ^{2}}, \overset{a}{˙} = a H \sim - \frac{H _{0} β}{a} .

Hence

a d a \sim - H_{0} β d t

, so zero scale factor is reached in finite time. The universe therefore undergoes the supercritical closed radiation--cosmological-constant recollapse to a Big Crunch.

Solved by gpt-5.6-sol high.

 Ancestors (11)