= Solution
Let $e$ be <specific internal energy>, $Q$ the <heat> supplied per unit volume per unit time, and $E=\rho(e+|\mathbf u|^2/2)$. For an inviscid <perfect gas> in a prescribed <Newtonian gravitational potential> $\Phi$, the <fluid total-energy equation> is
$$
\partial_tE+\nabla\cdot[(E+p)\mathbf u]=-\rho\mathbf u\cdot\nabla\Phi+Q.
$$
There is no gravitational term if no external body is present. For a time-independent $\Phi$, adding <potential energy> gives the equivalent conservative form $\partial_t(E+\rho\Phi)+\nabla\cdot[(E+p+\rho\Phi)\mathbf u]=Q$. Subtracting the kinetic-energy equation and using <mass conservation> gives
$$
\rho\frac{De}{Dt}=-p\nabla\cdot\mathbf u+Q.
$$
For a <perfect gas> with constant <specific-heat ratio> $\gamma$, $e=p/[(\gamma-1)\rho]$. The <isothermal equation of state> therefore makes $e=c_s^2/(\gamma-1)$ constant. Hence \b[the required <heat> supply and <heat> loss are]
$$
\boxed{Q=p\nabla\cdot\mathbf u,\qquad \mathcal L=-Q=-p\nabla\cdot\mathbf u.}
$$
Compression requires cooling; expansion requires heating. The <isothermal sound speed> $c_s$ differs from the <adiabatic sound speed> $\sqrt\gamma\,c_s$.
Define the isothermal <Mach numbers> by $\mathcal M_i=u_i/c_s$. Orient the normal to a stationary <isothermal shock> along the flow, so the mass flux $j=\rho_1u_1=\rho_2u_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal <momentum> conservation gives
$$
\rho_1u_1^2+c_s^2\rho_1=\rho_2u_2^2+c_s^2\rho_2,
\qquad u_1+\frac{c_s^2}{u_1}=u_2+\frac{c_s^2}{u_2}.
$$
For a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1u_2=c_s^2$. Thus \b[the <isothermal shock> jump relations] are
$$
\boxed{\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{u_1^2}{c_s^2}=\mathcal M_1^2=\frac1{\mathcal M_2^2},\qquad \mathcal M_1\mathcal M_2=1.}
$$
Across a thin shock, gravitational potential and tangential <kinetic energy> are unchanged. The <specific enthalpy> $h=\gamma c_s^2/(\gamma-1)$ is the same on both sides, so the energy removed per unit area per unit time is
$$
F_{E,1}-F_{E,2}=\frac j2(u_1^2-u_2^2).
$$
Positive net cooling requires $u_1>u_2$, equivalently $\rho_2>\rho_1$ and $\mathcal M_1>1>\mathcal M_2$. \b[<Cooling across an isothermal shock> permits only compression shocks]; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.
For steady radial flow, <mass conservation> gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the <mass density> gradient from radial <momentum> balance gives
$$
\left(u_r-\frac{c_s^2}{u_r}\right)\frac{du_r}{dr}=\frac{2c_s^2}{r}-\frac{GM}{r^2}.
$$
A regular <sonic point> has $|u_r|=c_s$, so the right-hand side must vanish there as well. Therefore
$$
\boxed{r_s=\frac{GM}{2c_s^2}.}
$$
This is the <Bondi sonic point> for accretion; the same critical radius appears in an isothermal <Parker wind>. A general steady solution need not pass through a <sonic point>, but a smooth transonic one must obey both critical conditions.
Put $x=r/r_s$ and $\mathcal M=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomes
$$
\left(\mathcal M-\frac1{\mathcal M}\right)\frac{d\mathcal M}{dx}=\frac2x-\frac2{x^2}.
$$
Integrating yields \b[the <integrated isothermal Bondi flow relation>]:
$$
\boxed{\frac12\mathcal M^2-\ln\mathcal M=\frac2x+2\ln x+C.}
$$
For a <transonic branch>, $x=\mathcal M=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal M-1)^2=(x-1)^2+O((x-1)^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.
For <Isothermal Bondi accretion>, integrate radial <momentum> balance once more, using $u_r\to0$ and $\rho\to\rho_\infty$ at infinity:
$$
\frac12u_r^2+c_s^2\ln\frac\rho{\rho_\infty}-\frac{GM}{r}=0.
$$
At the <Bondi sonic point>, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so $\rho_s=e^{3/2}\rho_\infty$. The positive inward <isothermal Bondi accretion rate> is consequently
$$
\boxed{\dot M=-4\pi r^2\rho u_r=4\pi r_s^2\rho_s c_s=\pi e^{3/2}\frac{G^2M^2\rho_\infty}{c_s^3}.}
$$
The critical <mass density> and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.
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