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isothermalwind

-              v4
+              v5
 Substitute (3) and (2) into (1) and assume the wind is in steady state, we can get:
 $\frac{1}{u}\frac{du}{dt}=(\frac{2a^2}{r}-\frac{GM}{r^2})/(u^2-a^2)$
+$\frac{1}{u}\frac{du}{dt}=(\frac{2a^2}{r}-\frac{GM}{r^2})/(u^2-a^2)$  (4)
 $u=a$ is a singular point and $\frac{2a^2}{r}-\frac{GM}{r^2}=0$ is critical point. If the wind is to experience constant acceleration, u will exceed a at some radius such that the equation become singular. Therefore the reasonable picture for the constant accelerating wind is that numerator goes to 0 as the denominator goes to 0.
 …
 $\left.\frac{du}{dt}\right|_{r=r_c}=\frac{\pm 2a^2}{GM}$
+We can also solve (4)
+$u e^{-\frac{u^2}{2a^2}}=a(\frac{r_c^2}{r})e^{(-\frac{2r_c}{r}+1.5)}$
 Remark: We can use mass conservation law to get $\rho(r)$ profile.