Use $\dot M>0$ for inward <mass accretion rate>, and $T_m>0$ for <magnetic torque> adding <angular momentum> to the disk. Combining the <continuity equation> with <conservation of angular momentum> identifies the inward mass flux as $(dh/dr)^{-1}d\mathcal G/dr$. In a steady <Keplerian accretion disk>, this is the supplied constant $\dot M$, so
$$
\frac{d\mathcal G}{dr}=\dot M\frac{dh}{dr},\qquad
\mathcal G(r)=T_m+\dot M(h-h_{\rm in}),\qquad h=\sqrt{GMr}.
$$
The <boundary condition> is $\mathcal G(r_{\rm in})=T_m$. Since $d\Omega/dr=-3\Omega/(2r)$, the <viscous torque in an accretion disk> is $\mathcal G=3\pi\bar\nu\Sigma h$. Hence the <steady Keplerian accretion disk with an inner torque> has
$$
\boxed{\Sigma(r)=\frac{\dot M}{3\pi\bar\nu}\left[1+(\lambda-1)\left(\frac{r_{\rm in}}r\right)^{1/2}\right],\qquad
\lambda=\frac{T_m}{\dot M h_{\rm in}}.}
$$
In particular, $\Sigma(r_{\rm in})=\lambda\dot M/(3\pi\bar\nu)$. For a physical steady solution with this <boundary condition>, $\lambda\geq0$.

For $0\leq\lambda\ll1$, the <surface density of a disk> rises from a small inner value toward $\dot M/(3\pi\bar\nu)$: this is the usual <Keplerian accretion disk> with an approximately <zero-torque inner boundary condition>. For $\lambda\gg1$, the inner <surface density of a disk> is large and decreases approximately as $r^{-1/2}$, giving a <torque-dominated accretion disk>. This approximation applies where $r/r_{\rm in}\ll(\lambda-1)^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-321-surface-density.png]
{title=Steady surface-density profiles with weak and strong inner torques}
{height=380}

The constant net inward <angular-momentum flux> is
$$
\dot M h-\mathcal G=\dot M h_{\rm in}(1-\lambda).
$$
For $\lambda>1$, net <angular momentum> moves outward. The strong-torque profile is therefore <decretion disk>-like in its stress-dominated structure, but the specified mass flux remains inward: a genuine net <decretion disk> requires an outward mass flux, not merely large $\lambda$.

At $r_{\rm in}=r_m$, the <Maxwell stress tensor> estimate gives
$$
T_m\sim\frac{\epsilon\mu^2}{4\pi r_m^3},\qquad
\lambda\sim\frac{\epsilon\mu^2}{4\pi\dot M\sqrt{GM}\,r_m^{7/2}}.
$$
Using the <magnetospheric truncation radius> scaling in part (b), all dimensional parameters cancel:
$$
\boxed{\lambda\sim1\quad\text{up to geometric numerical factors}.}
$$
If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi)$; retaining $\Sigma\simeq2H\rho$ and exact <free-fall speed> throughout gives $\sqrt2/(4\pi)$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda\gg1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 1 / e / Solution

{title2=$\bar\nu\Sigma=\frac{\dot M}{3\pi}[1+(\lambda-1)\sqrt{r_{\rm in}/r}]$}

For a positive inner <magnetic torque> $T_m$, define $\lambda=T_m/(\dot M h_{\rm in})$. Integrating steady <conservation of angular momentum> gives $\mathcal G=T_m+\dot M(h-h_{\rm in})$. With <Keplerian rotation>, this determines the <surface density of a disk> in the displayed formula.

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 Incoming links: Steady Keplerian accretion disk with an inner torque