 Solution

Assume a steady spherical outflow of fixed composition. Let $\bar{m}$ be its mean particle mass, and define the constant isothermal sound speed by $a^2=k_B{T}_0/\bar{m}$. Subsonic here refers to this isothermal closure, so $0<u<a$. Mass conservation and radial momentum balance giveLogarithmically differentiating the first equation yields $d\ln \rho /dr=-2/r-u^{\prime }/u$. Substitution eliminates the density gradient and gives the isothermal Parker wind equationAt a smooth speed maximum, $u^{\prime }=0$, while the left prefactor is finite and negative. The right side must vanish, henceFor $r<r_M$ both the right side and the prefactor are negative, so $u^{\prime }>0$; for $r>r_M$, $u^{\prime }<0$. This is a true maximum on a subsonic isothermal breeze, not the accelerating transonic branch.

There is an underdetermination in the requested density estimate. Introduce the peak Mach number ${\mathcal{M}}_M=u_M/a$, where $0<{\mathcal{M}}_M<1$. The density at the maximum speed of an isothermal breeze is exactlyThe hypotheses do not fix ${\mathcal{M}}_M$. A boundary density or pressure, or an equivalent wind integration constant, is needed. The definite statement following from subsonicity is the lower boundIf a maximum speed of order $a$ is intended as an extra estimate, then ${\rho }_M\sim {\rho }_{\mathrm{char}}\propto \dot{M}{T}_0^{3/2}{M}^{-2}$. A very slow breeze can have a much larger density, so setting $u_M=a$ as an exact equality would contradict the stated always-subsonic assumption. The mean particle mass is another implicit fixed-composition input needed to express a temperature as a squared sound speed.

One can see the free parameter directly by integrating the wind equation:Different $C$ give different subsonic maxima at the same radius. The original plot illustrates this nonuniqueness. These breeze branches require nonzero ambient pressure at large radius; they are not the usual transonic escape into a low-pressure exterior.

Maintaining an isothermal expanding wind generally requires heat exchange; isothermal motion is not adiabatic motion.^[ref]