With $\mathcal G=0$ and $\mathcal T=2\pi v_mh'\Sigma$, the evolution equation reduces to the radial <continuity equation>
$$
\partial_t\Sigma=-\frac1r\partial_r(rv_m\Sigma),
$$
or, for the mass per unit radius $M_r=2\pi r\Sigma$,
$$
\partial_tM_r+\partial_r(v_mM_r)=0.
$$
The outward mass flux is $F_M=2\pi rv_m\Sigma$. If $\dot M>0$ denotes the inward <accretion rate>, steady mass conservation requires $F_M=-\dot M$. Hence
$$
\boxed{2\pi r v_m\Sigma=-\dot M},
\qquad
\boxed{\Sigma(r)=-\frac{\dot M}{2\pi r v_m(r)}}.
$$
This is positive because the magnetic drift velocity satisfies $v_m<0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / b / i / Solution

Write $y=\nu\Sigma$. A steady inward <accretion rate> has $F_M=-\dot M$, so the integrated angular-momentum equation is
$$
\partial_r\mathcal G-r\mathcal T=\dot M\frac{dh}{dr}.
$$
For $\beta=0$, $\mathcal T=0$ and therefore
$$
\mathcal G=\dot M(h-h_{\rm in})
$$
after imposing the <zero-torque inner boundary condition> at $r_{\rm in}$. In a <Keplerian accretion disk>, $h\propto r^{1/2}$ and $\mathcal G=3\pi hy$, so
$$
\boxed{\nu\Sigma
=\frac{\dot M}{3\pi}
\left[1-\left(\frac{r_{\rm in}}r\right)^{1/2}\right]}.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / c / i / Solution

{title2=$\dot M$}

The accretion rate is the mass crossing an inward-oriented cylindrical surface per unit time. A common convention takes $\dot M>0$ for inward flow, so the outward radial mass flux is $-\dot M$.


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 Incoming links: Accretion rate