A 1D universe with Hubble expansion I'm thinking about a 1D universe model to understand the expansion of the universe in terms of special relativity. At the origin there's the planet earth and the sun at $\,x_0=0\,$ at rest. There are an infinite line of other stars moving at constant speeds according to
$$x_n=a_n+\beta_nct,\quad n\in\mathbb{Z},$$
where $\,a_0=\beta_0=0$, $\,a_{\pm1}=\pm a$, $\,\beta_{\pm1}=\pm\beta$. So $a$ is the current distance the two nearest stars are from us at $\,t=0\,$ and $\,\beta c\,$ is their recession speed. Following the cosmological principle, let us now assume that other stars are seeing the same picture in their reference frames. We can use Lorentz transformation to go to the reference frame of the $n$-th star and obtain
$$\begin{pmatrix}
ct^{(n)}\\x_m^{(n)}
\end{pmatrix}=\begin{bmatrix}
\gamma_n & -\beta_n\gamma_n\\
-\beta_n\gamma_n & \gamma_n
\end{bmatrix}\begin{pmatrix}
ct\\a_m+\beta_mct
\end{pmatrix}.$$
And we would like to require for $\,m=n\pm 1\,$ that
$$\frac{dx_{n\pm1}^{(n)}}{dt^{(n)}}=\pm\beta c,\quad
\left.x_{n+1}^{(n)}+x_{n-1}^{(n)}-2x_n^{(n)}\right|_{t^{(n)}}=0.$$
After solving this model, I find that
$$\beta_n=\frac{(1+\beta)^n-(1-\beta)^n}{(1+\beta)^n+(1-\beta)^n},\quad a_n=\frac{a}{\beta}\beta_n.$$
So Hubble's law applies to the whole 1D universe with Hubble's constant $\,H=\beta c/a$. The age of the universe is $\,1/H=a/\beta c$. At $\,t=-a/\beta c$, all stars were together, which was the moment of the "big bang". At $t=0$, the universe has a boundary with $a_{\pm\infty}=\pm a/\beta$. The boundary does not violate the cosmological principle because it is expanding at the speed of $c$. As one approaches the speed of $c$, relativistic effects would unfold the contracted length to keep the boundary from being reached.
My questions: How well does this model resemble our current understanding of the universe? What does the accelerated expansion mean? Does it mean the recession speeds of the stars are increasing, or does it mean Hubble's constant $H$ is increasing with time?
 A: You are correct, except in the Hubble law, the velocity is proportional to the distance (in your case, $\beta\propto a$) for non-relativistic velocities. The accelerated expansion means that you should add a small second derivative (acceleration) term to your equations.
The Hubble law is only a part of "our current understanding of the universe" based on the $\Lambda\text{-CDM}$ Model. This model is based on Friedmann's solution of the General Relativity equations (with some add-ons like inflation). This is not the only possible model based on the Hubble law. For example, the model of Milne (whole student was Walker as W in FLRW) also is based on the Hubble law. The Milne Model is based on Special Relativity and is ruled out, but it is a great educational tool to study the principles of cosmology and Special Relativity.
A: Your model only approximates to our universe for small, non-relativistic velocities.
Your model is equivalent to saying that the recession velocity of a galaxy is given by
$$ v = c \left(\frac{(1+z)^2 -1}{(1+z)^2 +1}\right)$$
where $z$ is a measured redshift.
In this model, the space between the galaxies is not expanding, they are simply moving away from each other with a speed proportional to their distance away from us. SR then ensures that the recession velocities cannot exceed $c$.
In the expanding universe, governed by GR, this isn't the case. Hubble's law relating the recession velocity now with the proper distance to a galaxy applies for all distances and hence recession velocities greater than the speed of light are permitted. Indeed, the relationship between recession velcity and redshift is not that provided by SR. It is
$$v = c\frac{\dot{R}(t)}{R_(t_0)} \int_0^{z} \frac{dz'}{H(z')},  $$
where $H(z)$ is the Hubble parameter at redshift $z$, $R$ is the scale factor of the universe and $t$ is the epoch for which the recession velocity is calculated.
Essential Reading - Davis & Lineweaver (2003).
