Vtyjkyuk

The solution to the Landau equation

$$fracd vert Arvert^2d t=2sigma lvert A rvert^2-l lvert A rvert^4$$

is

$$lvert A rvert ^2= fracA_0^2fracl2sigma A_0^2+left(1-fracl2 sigma A_0^2 right) e^-2sigma t ,$$
where $l$ and $sigma$ are constant, $A_0$ is the initial value of $lvert A rvert$.

If $l>0$ and $sigma>0$, the textbook showed that the above solution gives

$$lvert A rvert sim A_0 e^sigma t$$ as $tto-infty$ and $A_0to 0$. But I think simply
$$lim _t to -infty , A_0 to 0 lvert A rvert ^2=frac0infty=0$$

Can anybody suggest me how to show this asymptotic relation? Thank you in advance!

Note
begineqnarray
lvert A rvert ^2 &=& fracA_0^2fracl2sigma A_0^2+left(1-fracl2 sigma A_0^2 right) e^-2sigma t \
&=& fracA_0^2e^2sigma tfracl2sigma A_0^2e^2sigma t+left(1-fracl2 sigma A_0^2 right)
endeqnarray
and hence
begineqnarray
lim_tto-inftyfraclvert A rvert ^2e^2sigma t
&=& lim_tto-inftyfracA_0^2fracl2sigma A_0^2e^2sigma t+left(1-fracl2 sigma A_0^2 right) =A_0^2.
endeqnarray
So $|A|sim A_0e^sigma t$ as $tto-infty$.