cosmology - Evolution of the Hubble parameter

The solution to the Friedmann equation in a flat universe is
$$H^2 = frac{8pi G}{3}rho + frac{Lambda}{3},$$
where $rho$ is the matter density (including dark matter) and $Lambda$ is the cosmological constant.

As the universe expands, $rho$ of course decreases, but $Lambda$ remains constant.

Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = sqrt{Lambda/3}$ as time tends towards infinity.

As $Lambda = 3H_0^{2} Omega_Lambda$, and measurements suggest that $Omega_{Lambda} simeq 2/3$, then $Lambda simeq 2H_0^2$, and the Hubble parameter will therefore decrease to approximately $sqrt{2/3}$ of its present value if the cosmological constant stays constant.

Of course if $Lambda = Lambda(t)$, (ie not the basic $Lambda$-CDM model) then the behaviour will be different.

EDIT: Another useful form of the solution (for the case of a constant vacuum energy density) is

$$H^2 = H_0^2 left( frac{Omega_r}{a^4} + frac{Omega_M}{a^3} + frac{Omega_k}{a^2} + Omega_{Lambda}right),$$
where $H_0$ is the Hubble parameter now, $a(t)$ is the scale factor of the universe, $Omega_r$ is the current (i.e. $a=1$) ratio of the radiation density to the critical density and $Omega_M$, $Omega_k$ and $Omega_{Lambda}$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.

As $a$ increases you can see that all three of the leading terms get smaller and the Hubble parameter decreases at all times. When $a$ is very large, $H$ approaches $sqrt{Omega_{Lambda}} H_0$ as before.