Assume there is driving force f(r,v) and g(r,v)v\frac{dv}{dr} and the wind is in steady state.

\rho v\frac{dv}{dr}=-\frac{dp}{dr}-\frac{GM}{r^2}+f(r,v)+g(r,v)v\frac{dv}{dr}

Use p=\frac{\rho RT}{\mu} and \frac{RT}{\mu}=a^2 become:

v\frac{dv}{dr}=-\frac{1}{\rho}\frac{d\rho}{dr}-\frac{1}{\rho}\frac{GM}{r^2}+f'(r,v)+g'(r,v)v\frac{dv}{dr} (1)

Where f'(r,v)=\frac{f(r,v)}{\rho} g'(r,v)=\frac{g(r,v)}{\rho}

Differentiate \dot{M}=4\pi r^2\rho v with respect to r, get

\frac{1}{\rho}\frac{d\rho}{dr}=-\frac{1}{v}\frac{dv}{dr}-\frac{2}{r} (2)

Substitute (2) into (1)

\frac{1}{v}\frac{dv}{dr}=\frac{\frac{2a^2}{r}-\frac{GM}{r^2}+f'(r,v)}{(1-g'(r,v))v^2-a^2} (3)

Critical point will happen at:

\frac{2a^2}{r_c}-\frac{GM}{r_c^2}+f'(r_c,v_c)=0

With velocity determined by:

(1-g'(r_c,v_c))v_c^2-a^2=0

Different wind initial speed, f(r,v) and g(r,v) will give very different v profile. One solution of interest is: subsonic initial wind accelerate all the way to the critical point and become supersonic when at very large r with \frac{2a^2}{r_0}-\frac{GM}{r_0^2}+f'(r_0,v_0)>0 and (1-g'(r_0,v_0))v_0^2-a^2>0

Where r_0,v_0 are initial radius and velocity.

These driving force will affect mass loss rate significantly.