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For the equation of state $p=-\varrho$, the cosmological continuity equation gives $a\dot{\varrho }=0$. Hence $\varrho ={\varrho }_0$ is constant. First take the physically intended positive energy density and defineThe Friedmann equation and the original Friedmann acceleration equation becomeSolve the second equation as $a=A{e}^{H_{\Lambda }t}+B{e}^{-H_{\Lambda }t}$. The first equation then requires $4H_{\Lambda }^{2}AB=k$. Choosing the time origin separately in each solution gives all positive branches:For $k=-1$, there is also the time reverse $a=H_{\Lambda }^{-1}\sinh [H_{\Lambda }(t_B-t)]$ on $t<t_B$, which contracts to a future zero. For $k=+1$, the cosmological bounce joins contraction to expansion at the positive minimum $a(t_b)=H_{\Lambda }^{-1}$. Retaining $\ddot{a}=H_{\Lambda }^{2}a$ at the minimum excludes a spurious permanently static solution that would satisfy the first-order constraint alone.

These are vacuum-energy FLRW solutions: the expanding flat branch, closed slicing of de Sitter spacetime, and open slicing of de Sitter spacetime. To compare them at the same energy density, the figure sets $T=H_{\Lambda }(t-t_c)$, takes $A_0=H_{\Lambda }^{-1}$, and aligns the open boundary and closed minimum at $T=0$. Thus the three plotted functions are $H_{\Lambda }a=e^T$, $\cosh T$, and $\sinh T$, the last only for $T>0$.

Only the expanding $k=-1$ branch has a finite past zero. The expanding $k=0$ branch tends to zero only as $t\to -\infty$, and the $k=+1$ branch is bounded below. All three describe portions of de Sitter spacetime, with Ricci scalar $R=12H_{\Lambda }^2$ and Kretschmann scalar $24H_{\Lambda }^4$. In particular the $k=-1$ zero is a boundary of that slicing, not a curvature singularity. The inequality in part (d) does not apply, because $\varrho +3p=-2{\varrho }_0<0$.

For zero energy density, the remaining solutions are $a=A_0>0$ for $k=0$, and the expanding or contracting Milne model $a=\pm (t-t_B)>0$ for $k=-1$; no positive real solution exists for $k=+1$. If negative energy density is admitted mathematically, write ${\alpha }^2=-8\pi G{\varrho }_0/3>0$. Only $k=-1$ is possible, with $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ for $0<\alpha (t-t_B)<\pi$, an open slicing of anti-de Sitter spacetime. These limiting and negative-density cases do not alter the positive-density comparison above.

Write $H_{\Lambda }=\sqrt{\Lambda /3}$ and $\kappa =\sqrt{-k}>0$. With matter and radiation absent, the Friedmann equation isChoose the expanding branch and put its coordinate origin at $a(0)=0$. Separating variables,so the open slicing of de Sitter spacetime isFor $a(t_0)=1$ the equivalent normalized form isIt has $H(t)=H_{\Lambda }\coth (H_{\Lambda }t)$ and $H_0^2=H_{\Lambda }^2-k$, as required. In terms of a prescribed present Hubble parameter,At early times $a\simeq \sqrt{-k}t$ resembles the Milne universe, while at late times it grows exponentially. The empty model still has vacuum energy density; the vanishing of $a$ at the open-patch origin is not a curvature singularity of de Sitter spacetime.