= Solution
Let $\Delta M=M_1-M_0$ and retain only terms first order in $\mu/M_0$ and $\rho/M_0$. The outgoing <event horizon> equation expands to
$$
4M_0\dot\rho=\rho-2\mu,
\qquad \dot\rho-\kappa_0\rho=-\frac{\mu}{2M_0},
\qquad \kappa_0=\frac1{4M_0}.
$$
Here $\kappa_0$ is the initial <Schwarzschild surface gravity>. The <integrating factor> $e^{-\kappa_0v}$ gives $d(e^{-\kappa_0v}\rho)/dv=-e^{-\kappa_0v}\mu/(2M_0)$. Future stationarity imposes $\rho=2\Delta M$ for $v>v_0$ and removes the growing homogeneous solution. Integrating from $v$ to infinity yields the <teleological response of a Vaidya event horizon>:
$$
\boxed{\rho(v)=\frac1{2M_0}\int_v^\infty e^{-\kappa_0(s-v)}\mu(s)\,ds.}
$$
The integral is finite because $\mu$ is bounded. Integrating by parts also gives
$$
\rho(v)=2\mu(v)+2\int_v^\infty e^{-\kappa_0(s-v)}\dot\mu(s)\,ds.
$$
For $v>v_0$, this reduces to $2\Delta M$. For $v<0$, the first form becomes
$$
\rho(v)=C e^{\kappa_0v},\qquad C=\frac1{2M_0}\int_0^\infty e^{-\kappa_0s}\mu(s)\,ds.
$$
Consequently \b[$r_H\to2M_0$ as $v\to-\infty$]. The positive constant $C$ shows that this <event horizon> is already expanding in the initially vacuum region. For monotone accretion, $\rho-2\mu\geq0$, so it lies outside the instantaneous <apparent horizon> to first order. All displayed radius corrections are understood up to $O((\Delta M)^2/M_0)$.
Back to article page
