-\infty$ by expressing equation (\[phip\]) with the normalised variables and considering its asymptotical limit near equilibrium. It comes: $$\label{phiAs}
\dot{\phi}=\frac{2\phi^2U_\phi}{(3+2\omega)U}$$ This important equation allows us to get the asymptotical behaviour of $\phi$ near equilibrium and consequently that of $\ell$.

From the asymptotical behaviour of $H$ and the expression (\[lapse\]) for the lapse function, it is possible to get the isotropic part of the metric, $e^{-\Omega}$, as a function of the proper time. If $\ell^2$ tends toward a non vanishing constant, then $e^{-\Omega}\rightarrow t^{\ell^{-2}}$. If $\ell^2$ tends to $0$ faster than $(-\Omega)^{-1}$, $e^{-\Omega}$ behaves like an exponential. Let us show that it is always the case. Equation (\[phiAs\]) can be rewritten as $d\phi/d\Omega=2\ell^2U(U_\phi)^{-1}$ and then $U\propto e^{2\int\ell^2 d\Omega}$ when $\Omega\rightarrow -\infty$. Introducing this expression and the expression for $H$ into (\[v2\]), we get $y\propto e^{-\ell^2\Omega+\int\ell^2 d\Omega}$. Thus, if $\ell^2$ vanishes slower than $(-\Omega)^{-1}$, $y$ diverges or tends to $0$ instead of a non vanishing constant and there is no equilibrium.

Hence, when an isotropic equilibrium state is reached with $\ell\rightarrow 0$, $\ell^2$ always vanishes faster than $(-\Omega)^{-1}$ and Universe always tends toward a De Sitter one. This proof for $\ell=0$ relies on the asymptotic form of $\phi$, $H$ and the definition 