= Solution
Assume positive fluid density, $\rho_0>0$, and write $m=1+3w>0$. A <perfect fluid> with constant <equation of state> obeys the <cosmological perfect-fluid continuity equation>, giving
$$
\dot\rho+3H(\rho+P)=0,\qquad\rho=\rho_0a^{-3(1+w)}.
$$
In the stated units the <Friedmann equation> is $H^2=\rho/3-k/a^2+\Lambda/3$. Multiplication by $a^2/2$ yields the <Friedmann effective potential for a constant-equation-of-state fluid>,
$$
\boxed{\tfrac12\dot a^2+V(a)=0,\qquad V(a)=-\frac{\rho_0}{6}a^{-m}+\frac{k}{2}-\frac{\Lambda}{6}a^2}.
$$
The allowed region has $V\leq0$. The <Friedmann acceleration equation> is equivalently $\ddot a=-V'(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.
For $k=0$, $\Lambda<0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero gives
$$
\boxed{a_{\max}=\left(\frac{\rho_0}{|\Lambda|}\right)^{1/[3(1+w)]}}.
$$
Expansion from the <Big Bang> stops there, with negative acceleration, then reverses into a <Big Crunch>. Both the turning point and the final singularity occur in finite <proper time>: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.
For $\Lambda=0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero at
$$
\boxed{a_{\max}=\left(\frac{\rho_0}{3}\right)^{1/(1+3w)}}.
$$
This is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot a\to1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.
For $k=0$, $\Lambda>0$, $V$ tends to negative infinity at both ends. It has a maximum at
$$
a_*^{m+2}=\frac{m\rho_0}{2\Lambda},\qquad
V(a_*)=-\frac{(m+2)\rho_0}{12}a_*^{-m}<0.
$$
There is no turning point. Expansion initially decelerates, then accelerates once $a>a_*$, and approaches <de Sitter spacetime> expansion $a\propto e^{\sqrt{\Lambda/3}\,t}$. Thus \b[positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-48-friedmann-potentials.png]
{title=Zero-energy Friedmann potentials showing recollapse, curvature-dominated expansion and positive-cosmological-constant expansion}
{height=350}
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