The Vaidya geometry written in ingoing null coordinate $v$ is given by the metric,
\begin{equation} ds^2 = -\left(1 - \frac{2 m(v)}{r} \right) dv^2 + 2dvdr + r^2 (d\theta^2+ \sin^2{\theta} d\phi^2) \tag{1} \end{equation}
where $m(v)$ is a mass function given in general by,
\begin{equation} m(v) = m_0 \left(1 - \frac{v_0}{v_f} \right)^{-\alpha} \left(1 - \frac{v}{v_f} \right)^{\alpha} \tag{2} \end{equation}
for constants $v_0$, $v_f$, $m_0 > 0$ and $0 < \alpha < 1$. This geometry can describe black hole formation starting at time $v_0$ and reaching an initial mass $m_0$, while the evaporation ends at time $v_f$, which means that $v_0 < v_f$. Notice that the null coordinate $v$ plays the role of time instead of the typical coordinate time $t$.
The radial null geodesic on the event horizon can be written by setting $ds^2 = 0$,
\begin{equation} -\left(1 - \frac{2 m(v)}{r} \right) dv^2 + 2dvdr = 0 \tag{3} \end{equation}
where $r$ is the radial direction and notice we have dropped the transverse direction of $\theta$ and $\phi$ (since we're interested only in the radial direction). One of the solutions is $dv = 0$, while the other is given by,
\begin{equation} \frac{dr}{dv} = \frac{1}{2} \left(1 - \frac{2 m(v)}{r} \right) \tag{4} \label{4} \end{equation}
Question 1
Is there a solution to this differential equation? If yes, how to solve it? I believe there should be a solution for arbitrary mass $m(v)$.
Question 2
The event horizon of a black hole quantified by the radius $r_{EH}$ is quite popular so I shouldn't explain it here. Another quantity that is lesser known but is still known in the literature is the apparent horizon (for brevity purposes and the only important thing in this post is mathematical, there's no need to explain it here). An inequality between the two is given by $r_{EH} < r_{AH} = 2 m(v)$. Near the final point of the evaporation $v \approx v_f$, we have $r_{EH} << r_{AH}$, and the approximate solution to eq.$\eqref{4}$ is given by,
\begin{equation} r_{EH}(v) \approx \sqrt{\frac{2 m_0 v_f}{\alpha+1}} \left(1 - \frac{v_0}{v_f} \right)^{-\alpha/2} \left(1 - \frac{v}{v_f} \right)^{(\alpha+1)/2} \tag{5} \label{5} \end{equation}
How do you obtain eq.\eqref{5}?
All of these are obtained from section III-A of the paper 2111.05151. The discussion in that section is quite independent of the rest of the paper and it's quite brief.