= Solution
In a <steady state>, <mass conservation> makes $\mathcal F$ independent of radius. Taking the <accretion rate> $\dot M>0$ for inward flow gives $\mathcal F=-\dot M$. The <conservation of angular momentum> equation then makes $-\dot M h+\mathcal G$ constant. The <zero-torque inner boundary condition> fixes this constant to $-\dot M h_{\rm in}$, so
$$
\mathcal G=\dot M(h-h_{\rm in}).
$$
The <viscous torque in an accretion disk> is $\mathcal G=-2\pi r^3\bar\nu\Sigma\Omega'=2\pi qh\bar\nu\Sigma$. Combining these expressions gives the <steady accretion disk with arbitrary rotation law>
$$
\bar\nu\Sigma=\frac{\dot M}{2\pi q}
\left(1-\frac{h_{\rm in}}h\right).
$$
For $x=r/R_g$, the <Paczyński-Wiita circular orbit> formulas give
$$
\frac3{2q}=\frac{x-2}{x-2/3},\qquad
\frac{h_{\rm in}}h=\frac{3\sqrt3(x-2)}{\sqrt2\,x^{3/2}}.
$$
Consequently
$$
\boxed{\bar\nu\Sigma=\frac{f\dot M}{3\pi}},\qquad
\boxed{f=\frac{x-2}{x-2/3}
\left[1-\frac{3\sqrt3(x-2)}{\sqrt2\,x\sqrt x}\right]}.
$$
In particular, $f(6)=0$ and $f\to1$ far from the black hole, recovering the outer <Keplerian accretion disk>.
Inside the <innermost stable circular orbit>, the gas enters the <plunging region of a black-hole accretion disk>. Its rapid inward motion leaves little time for stresses to exchange <angular momentum>, motivating the <zero-torque inner boundary condition>. This is a thin-disc approximation; a strong magnetic stress could change it.
Matter supplied from very large radius has negligible <specific orbital energy>, while matter crossing the inner edge carries $\varepsilon_{\rm in}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by <radiative transfer>, rather than being lost through inward <advection>, the integrated <conservation of energy> balance gives
$$
\boxed{L_{\rm disc}=-\dot M\varepsilon_{\rm in}=\eta\dot M c^2=\frac{\dot M c^2}{16}}.
$$
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