# Friedmann equation and model universes

I was studying Barbara Ryden book on modern cosmology, on it's 5th chapter, she introduces the following image:

This image summarises the general behaviour of the Hubble constant. I was wondering, what is the best way to reproduce this plot. For instance, taking the Big bounce case, disregarding radiation, one could solve the Friedmann equation in a grid $$(\Omega_m,\Omega_\Lambda)$$ and test for each point if there is a minimum point for the scale factor such that $$a_{min}>0$$.

Is there any other easier method to reproduce this plot?

Assuming (as this chart does) that the universe contains only dust and dark energy, the scale factor satisfies $$\dot a^2 = H_0^2 \left( Ω_{Λ,0}\, a^2 + Ω_{k,0} + Ω_{m,0}\, a^{-1} \right)$$
where $$Ω_k = 1 - Ω_m - Ω_Λ$$. (The exponents are $$-1{-}3w$$ where $$w$$ is the equation of state parameter.)
The boundary of the $$κ=\pm1$$ regions is just the line $$Ω_{k,0}=0$$.
The expansion is presently speeding up if the derivative of the right hand side at $$a=1$$ is positive, i.e. if $$2\,Ω_{Λ,0} - Ω_{m,0} > 0$$.
If $$\dot a = 0$$ for some $$a$$ then the expansion stops and reverses at that scale, either from below (big crunch) or above (big bounce), so the big crunch and big bounce regions are those where the cubic $$Ω_{Λ,0}\, a^3 + Ω_{k,0}\, a + Ω_{m,0}$$ has a positive real root. The boundary of these regions for $$κ=+1$$ is given by $$4\,Ω_{k,0}^3 + 27\,Ω_{Λ,0}\,Ω_{m,0}^2=0$$.
Edit: here's a derivation of that, skipping boundary cases where one of the coefficients is zero. Let $$f(a) = Ω_{Λ,0}\, a^3 + Ω_{k,0}\, a + Ω_{m,0}$$. Note that $$f(1)=1$$, $$f(0) = Ω_{m,0} > 0$$, and the sum of the roots is $$0$$. If $$Ω_{Λ,0}<0$$ then $$f$$ has a real root at some $$a>1$$. If $$Ω_{Λ,0}>0$$ then $$f$$ has a real root at some $$a<0$$, and the roots can't all be negative, so it has a positive real root iff it has three real roots, that is, if the discriminant is nonnegative. The discriminant is $$-\left( 4\left(\frac{Ω_{k,0}}{Ω_{Λ,0}}\right)^3 + 27\left(\frac{Ω_{m,0}}{Ω_{Λ,0}}\right)^2 \right)$$.