Put $Y=y^2$. On the inner supersonic branch, the equation from part (d) is
$$
Y-\ln Y=\frac4x+4\ln x-3.
$$
Its <asymptotic expansion> as $x\to0^+$ begins
$$
Y=\frac4x+3\ln x+\ln4-3+O(x\ln x).
$$
Taking the positive <square root> gives
$$
y=\frac2{\sqrt x}
\left[1+\frac x8\left(3\ln x+\ln4-3\right)
+O\!\left(x^2\ln^2x\right)\right].
$$
Since $x=r/r_s$ and $y=u/c_s$,
$$
\boxed{u(r)=2c_s\sqrt{\frac{r_s}{r}}
\left[1+\frac{r}{8r_s}
\left(3\ln\frac r{r_s}+\ln4-3\right)
+O\!\left(\frac{r^2}{r_s^2}\ln^2\frac r{r_s}\right)\right]}.
$$
The leading term is the <free-fall speed> $\sqrt{2GM/r}$; the logarithmic term is the first pressure correction to the inner <Isothermal Bondi accretion> flow.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-314/2/e/solution