= Solution
Take the <accretion rate> $\dot M>0$ to mean inward flow. In a <steady state>, <conservation of mass> makes the inward <mass flux> independent of <radius>, so $\mathcal F=\dot M$. Integrating its radial expression gives
$$
r^{1/2}\bar\nu\Sigma=\frac{\dot M}{3\pi}r^{1/2}+C.
$$
For <Keplerian rotation>, $\Omega=(GM_*/r^3)^{1/2}$ and the <viscous torque in an accretion disk> is $\mathcal G=3\pi\bar\nu\Sigma\sqrt{GM_*r}$. The <zero-torque inner boundary condition> therefore sets $\bar\nu\Sigma=0$ at $r_*$. It fixes $C=-\dot M\sqrt{r_*}/(3\pi)$, giving \b[the steady density profile]
$$
\boxed{\Sigma(r)=\frac{\dot M}{3\pi\bar\nu(r)}\left(1-\sqrt{\frac{r_*}{r}}\right).}
$$
This is the <Keplerian accretion disk> solution on $r_*<r<r_{\rm out}$. It determines the <kinematic viscosity>–<surface density> product; a separate viscosity closure is needed to turn it into an explicit power law. If the <kinematic viscosity> is finite and nonzero at the inner edge, the <surface density> tends to zero there.
Back to article page
