At the critical point, both the coefficient of $du/dr$ and the remaining right-hand side of the <Isothermal Bondi equation> vanish, allowing a finite derivative through the sonic transition.

Critical point of the isothermal Bondi equation

Let $u(r)>0$ denote the inward <radial velocity>. Steady <spherical symmetry> and <mass conservation> give
$$
\dot M=4\pi r^2\rho u=\text{constant}.
$$
The radial <Euler momentum equation> for the <globally isothermal equation of state> $p=c_s^2\rho$ is
$$
u\frac{du}{dr}=-c_s^2\frac1\rho\frac{d\rho}{dr}-\frac{GM}{r^2}.
$$
The logarithmic derivative of mass conservation is $\rho'/\rho=-u'/u-2/r$. Substitution gives the <Isothermal Bondi equation>
$$
\boxed{\left(u-\frac{c_s^2}{u}\right)\frac{du}{dr}
=\frac{2c_s^2}{r}-\frac{GM}{r^2}}.
$$
At a smooth <sonic point>, $u=c_s$ makes the coefficient of $u'$ vanish, so the right-hand side must vanish too. Therefore
$$
\boxed{r_s=\frac{GM}{2c_s^2}}.
$$
The solution that crosses this <critical point of the isothermal Bondi equation> continuously is the transonic accretion solution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 2 / a / Solution

{c}

For inward speed $u>0$, steady spherical mass conservation and radial momentum balance give
$$
\left(u-\frac{c_s^2}{u}\right)\frac{du}{dr}
=\frac{2c_s^2}{r}-\frac{GM}{r^2}.
$$


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