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A smooth cosmological bounce is a positive minimum of the scale factor: contraction changes to expansion, with $a(t_b)>0$, $\dot{a}(t_b)=0$ and $\ddot{a}(t_b)>0$ in the nondegenerate case. The closed slicing of de Sitter spacetime gives $a=H_{\Lambda }^{-1}\cosh [H_{\Lambda }(t-t_b)]$, whose minimum is $H_{\Lambda }^{-1}$. The Friedmann acceleration equation with zero separate cosmological constant requires $\varrho +3p<0$ there.

Set $H=\dot{a}/a$, the Hubble parameter. The first supplied component of the Einstein field equations gives the Friedmann acceleration equationSubstitution into the second supplied equation givesTherefore the first Friedmann equation isDifferentiate this equation with respect to cosmic time:Insert the Friedmann acceleration equation on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This givesThis is the cosmological continuity equation, expressing stress-energy conservation for the perfect fluid in general relativity. No division by $\dot{a}$ was made, so the derivation remains valid at a cosmological bounce or an instant of maximum expansion.

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.