= Solution
Use $v=|u_r|>0$ for the radial <speed> and distinguish the central mass $M$ from the <Mach number> $\mathcal M=v/c$. In each smooth nonzero-flow region, <mass conservation>, the <adiabatic equation of state>, and the <Bernoulli equation> give
$$
j=r^2\rho v=\frac{|\dot M|}{4\pi}>0,\qquad p=K\rho^{3/2},\qquad c^2=\frac32K\rho^{1/2},\qquad
\mathcal E=\frac{v^2}{2}+2c^2-\frac{GM}{r}.
$$
Here $K>0$ is constant on a smooth <isentropic flow> branch, $c$ is the <adiabatic sound speed>, and $\mathcal E$ is the <Bernoulli function>. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal E>0$; flows with $\mathcal E\leq0$ do not have these positive $r_s,x,\lambda$ and require another scaling.
Indeed, the supplied <adiabatic process> equations give $D(p\rho^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial <fluid flow>. The <adiabatic sound speed> is $c^2=(\partial p/\partial\rho)_s$, and the <specific enthalpy> is $h=c^2/(\gamma-1)=2c^2$. The radial <Euler momentum equation>, $v\,dv/dr=-\rho^{-1}dp/dr-GM/r^2$, integrates to $\mathcal E$ because $dh=dp/\rho$ on this <isentropic flow> branch.
Put $D=[j(3K/2)^2]^{2/5}$. Eliminating the <mass density> from the <continuity equation> gives
$$
c^2=D r^{-4/5}\mathcal M^{-2/5}.
$$
With $y=\mathcal M^{2/5}$, the <Bernoulli equation> becomes
$$
\mathcal E+\frac{GM}{r}=\frac D2 r^{-4/5}(y^4+4/y).
$$
Therefore the physical constants of the <spherical polytropic flow with adiabatic exponent three halves> are
$$
\boxed{r_s=\frac{GM}{4\mathcal E},\qquad \lambda=\frac{[j(3K/2)^2]^{2/5}}{2\mathcal E r_s^{4/5}},\qquad
\lambda F(y)=F(x),\quad F(t)=t^4+\frac4t,\quad x=(r/r_s)^{1/5}.}
$$
For $t>0$, $F'(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the <algebraic curves>. For $0<\lambda<1$, there are two disconnected branches at every $x>0$, one <subsonic flow> with $y<1$ and one <supersonic flow> with $y>1$. For $\lambda=1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta)=5+10\eta^2+O(\eta^3)$. For $\lambda>1$, solutions exist only where $F(x)\geq5\lambda$: if $x_-<1<x_+$ solve $F(x_\pm)=5\lambda$, there is a forbidden interval $x_-<x<x_+$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.
A smooth <sonic point> needs $F'(x)=F'(y)=0$; hence \b[a regular sonic transition requires $\lambda=1$ and $r=r_s$]. The <critical speed of a polytropic flow> is then $c_s^2=GM/(2r_s)=2\mathcal E$, and the allowed <mass flux> is $j=(2\mathcal E)^{5/2}r_s^2/(3K/2)^2$. Here the subscript on $c_s$ denotes evaluation at the <sonic point>, rather than the <isothermal sound speed> convention used in the next solution.
The simple <transonic branch> is $y=x$. Direct substitution gives
$$
\boxed{\mathcal M=(r/r_s)^{1/2},\quad v=\sqrt{2\mathcal E},\quad c^2=\frac{GM}{2r},\quad
\rho=\left(\frac{GM}{3Kr}\right)^2,\quad p=K\left(\frac{GM}{3Kr}\right)^3.}
$$
Thus the radial <velocity> has constant magnitude, while the <mass density> and <pressure> scale as $r^{-2}$ and $r^{-3}$. For an outward <fluid flow>, this is a <subsonic flow> becoming a <supersonic flow> as its <adiabatic sound speed> decreases.
On the other <transonic branch>, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim4^{1/4}x^{-1/4}$. The resulting <asymptotic expansions> are
$$
\begin{aligned}
r\gg r_s:\quad&\mathcal M\sim32(r_s/r)^2,\qquad c^2\longrightarrow\mathcal E/2,\qquad
\rho\longrightarrow\left(\frac{\mathcal E}{3K}\right)^2,\qquad
v\sim32\sqrt{\mathcal E/2}(r_s/r)^2,\\
r\ll r_s:\quad&\mathcal M\sim2^{5/4}(r_s/r)^{1/8},\qquad
c^2\sim\sqrt2\,\mathcal E(r_s/r)^{3/4},\qquad
v\sim\sqrt{\frac{2GM}{r}},\\
&\rho\sim\frac{j}{\sqrt{2GM}}r^{-3/2},\qquad
p\sim K\left(\frac{j}{\sqrt{2GM}}\right)^{3/2}r^{-9/4}.
\end{aligned}
$$
For inward <fluid flow>, these are the nearly static reservoir at infinity and the inner <free fall> region of <Bondi accretion>. The <specific enthalpy> is smaller than $GM/r$ in the inner limit, which explains <free fall>. Reversing the radial <velocity> gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.
A stationary, nonradiative <normal shock wave> conserves the <mass flux> and the sum $v^2/2+h$, while the <Newtonian gravitational potential> is continuous across its thin layer. Hence $\mathcal E$ and $r_s$ are unchanged. The <entropy production in a perfect-gas shock> increases $K=p/\rho^{3/2}$, so
$$
\boxed{\frac{\lambda_2}{\lambda_1}=\left(\frac{K_2}{K_1}\right)^{4/5}>1.}
$$
More explicitly, the <Rankine-Hugoniot condition> for $\gamma=3/2$ gives, with upstream $m=\mathcal M_1^2>1$,
$$
\frac{\rho_2}{\rho_1}=\frac{5m}{m+4},\qquad
\frac{p_2}{p_1}=\frac{6m-1}{5},\qquad
\mathcal M_2^2=\frac{m+4}{6m-1}<1.
$$
Thus the <shock wave> jumps vertically, at the same $x$, from the <supersonic flow> to a <subsonic flow> on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a <shock wave> at $x=1.4$; the downstream <subsonic flow> has $\lambda\simeq1.224$ and continues to larger $x$. The <shock wave> radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-314-flow-curves.png]
{title=Transonic curves and a stationary shock}
{description=Solid blue curves have $\lambda=1$; grey curves have $\lambda=0.7$; orange curves have the downstream shock value $\lambda>1$. The vertical arrow is a supersonic-to-subsonic jump at fixed radius.}
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