 Solution

Differentiate the Friedmann equation and use the cosmological continuity equation to obtain $\dot{H}+H^2=-(4\pi G/3)(\rho +3P/c^2)$. The relation extends to turning points by continuity. Radiation obeys ${\rho }_R={\rho }_{R0}{a}^{-4}$, while dark energy with $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ has constant density. Substitution at $a(t_0)=1$ gives $k{c}^2=\beta H_0^2$, and at arbitrary $a$ givesPut $x=a^2$. For $\beta >4$, the upward-opening polynomial $P(x)=x^2-\beta x+\beta$ has roots $x_{\pm }=(\beta \pm \sqrt{\beta (\beta -4)})/2$, with $1<x_{-}<2<x_{+}$. The expanding Big Bang branch reaches $x_{-}$ in finite time. There $H=0$ but $\dot{H}<0$, so expansion turns to contraction; it returns to $a=0$ in finite time, a Big Crunch. A sketch of $a^4{H}^2/H_0^2=P(x)$ has the forbidden negative interval between the roots, while $a^4\dot{H}/H_0^2=\beta (x-2)$ is a line crossing zero at two.

For $\beta =4$, $P(x)=(x-2{)}^2$. The branch starting at zero has $\dot{x}=2H_0(2-x)$, givingAt small $t$, $a\sim 2\sqrt{H_{0}t}$, the radiation-dominated scale factor. At large $t$, $a=\sqrt{2}[1-\frac{1}{2}{e}^{-2H_{0}t}+O(e^{-4H_{0}t})]$. It approaches an Einstein static universe with equal radiation and dark-energy densities, rather than recollapsing or entering unbounded expansion.