general relativity - Can the Cosmological Constant explain an accelerated expansion?

From what I've learned so far, it appears that all models that attempt to explain the expansion of the universe are either based on Lambda-CDM or quintessence. The former support a big bang with rapid expansion, then deceleration of the expansion and then expansion again (non accelerated expansion) with $w=-1$. The latter (quintessence) do not support big bang, but support accelerated expansion with $w<-1$. The two schools of thought appear to box you in one way or the other, depending whether $w=-1$ or $w<-1$.

Why doesn't Lambda-CDM have a model that explain an accelerated expansion (i.e. $w < -1$) ? Or do they have one? Does Lambda-CDM maintain that $\Lambda$ has to be constant and so you're stuck with quintessence whenever $w<-1$? If that is the case, why couldn't $\Lambda$ increase with time?

In summary, is there any model that support a universe with:

1. 
   

   Big Bang

   
   


2. 
   

   Inflationary period with rapid expansion

   
   



3. 
   

   Deceleration of expansion

   
   


4. 
   

   Linear Expansion

   
   


5. 
   

   Future acceleration of expansion?

   
   

That is, we should be able to see in that model that $H_t > H_0$ for any $t_i >> t_0$ when $w<-1$.

Answer

(Disclaimer: this is a follow-up question to Equation for Hubble Value as a function of time)

You still have a few misconceptions:

First, a model with a cosmological constant does lead to accelerated expansion. Look at the second derivative of $a(t)$ in my post: $$ \ddot{a}(t) = \frac{1}{2}H_0^2\left(-2\,\Omega_{R,0}\,a(t)^{-3} - \Omega_{M,0}\,a(t)^{-2} +2\,\Omega_{\Lambda,0}\,a(t)\right). $$ You see that $\ddot{a}(t)>0$ if $a(t)$ is sufficiently large. In particular, with the values $$ H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} \approx 0,\qquad\Omega_{M,0} = 0.315,\qquad\Omega_{\Lambda,0} = 0.685,\qquad\Omega_{K,0} = 0, $$ you can work out that $\ddot{a}(t) > 0$ for $a(t)> 0.6$, which corresponds with $t > 7.7$ billion years. That is, the expansion began to accelerate when the universe was 7.7 billion years old, and will continue to do so. See also this post for more details: Can space expand with unlimited speed?

Second, a model with quintessence does have a big bang. The most general equation for $t(a)$, which I hadn't posted in my answer to your other question, is $$ \begin{align} t(a) &= \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^{1-3w}}}. \end{align} $$ This is a well-behaved function when $a\rightarrow 0$, because $\Omega_{R,0}>0$. This means that in the early universe, radiation was the dominant factor (the other terms in the denominator of the integrand go to zero for $a\rightarrow 0$), and we get $t(0)=0$ or conversely $a(0)=0$.

In your previous question, you used a simplified model with $\Omega_{R,0} = \Omega_{M,0} = \Omega_{K,0} =0$. Those models don't have a big bang, because then the integrand becomes infinite for $a\rightarrow 0$. But of course, those toy models do not correspond with our actual universe. You need the general model.

So, both a model with a cosmological constant and one with quintessence produce a universe with a big bang and an accelerated expansion. Does the data suggest a non-constant dark energy? It's too soon to tell, the error-bars are still too large; a cosmological constant is still consistent with the data. It's obvious that quintessence provides a better fit, because it has the extra parameter $w$, but that itself doesn't mean that this extra parameter is necesary to explain the observations. But time will tell...