 Solution

The outward mass loss rate is $\dot{M}=4\pi r^2\rho u>0$. Put $\beta =4/(\gamma +1)$, $\alpha =4(\gamma -1)/(\gamma +1)$ and $y=\rho r^{\beta }$. Both terms in the Bernoulli equation then have the same radial scaling: $\beta (\gamma -1)=4-2\beta =\alpha$. Eliminate $u$ and multiply by $r^{\alpha }$ to obtainFor positive terminal speed, $C>0$. Differentiation gives $F^{\prime }=r^{\alpha -2}[\alpha Cr-(1-\alpha )GM]$. It has an interior minimum precisely when $0<\alpha <1$, or $1<\gamma <5/3$, atAt a sonic point, the Bernoulli equation and $GM/r_c=2c_{s,c}^2$ give $C=c_{s,c}^2(5-3\gamma )/[2(\gamma -1)]$, recovering this radius. The limiting $C=0$, $\gamma =5/3$ case has no isolated interior minimum selected by these formulas.

The function $G$ diverges at both ends of $y>0$ and has a unique minimum whereThis condition is $u_c^2=c_{s,c}^2$. Thus a smooth crossing requires minimum matching at a polytropic sonic point: $F(r_c)=G(y_c)$. If the minimum of $F$ is lower, an interval has no real positive solution; if it is higher, the two $y$ branches stay separate and cannot cross the sonic value. When the minima agree, their positive second derivatives give $G^{\prime \prime }(y_c)(y-y_c{)}^2\sim F^{\prime \prime }(r_c)(r-r_c{)}^2$, allowing a finite-slope branch to pass between them. The wind with finite positive terminal speed takes the lower-$y$ branch outside $r_c$, because $\rho \sim \dot{M}/(4\pi \sqrt{2C}{r}^2)$ and $\beta <2$ make $y\to 0$.

At the matching point,Insert these into $\dot{M}=4\pi r_c^2{\rho }_c{c}_{s,c}$:The positive-$C$, $1<\gamma <5/3$ range is part of this finite-radius transonic wind result, rather than an unrestricted formula for every $\gamma >1$.