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Let $g=G(\xi )$ with $\xi ={\tau }^{-1}{x}^{-1/2}$. The derivatives of this similarity solution areThe first-derivative terms cancel in the dimensionless diffusion equation, leaving $G^{\prime \prime }+G^{\prime }=0$ for $\xi >0$. ThusAt the inner boundary $x\to 0$, $\xi \to \infty$. Interpreting the named variable as vanishing there sets $A=0$. Even if only the true viscous torque in an accretion disk $x^{1/2}g$ is required to vanish, a nonzero $A$ would give infinite total mass near $r=0$; finite mass again selects $A=0$. With nonnegative surface density, choose $B=C>0$; its value fixes the initial mass normalization. ThereforeThe surface density vanishes exponentially at the origin and has an $x^{-5/2}$ tail at infinity. Its peak occurs at $x_{\max }=1/(25{\tau }^2)$, with ${\Sigma }_{\max }/{\Sigma }_0=C(5\tau {)}^5{e}^{-5}$. The peak moves inward and grows as time increases.

The plot uses $C=1$ and shows the actual surface density, with logarithmic radius to resolve both peaks. The later profile is larger at every fixed radius. Its total mass is finite; its total angular momentum is not, as the next calculation establishes.

Write $\Sigma ={\Sigma }_0{x}^{-5/2}g$ because the prescribed kinematic viscosity is $\bar{\nu }={\bar{\nu }}_0{x}^{5/2}$. Substituting $r=r_{0}x$ and ${\partial }_t=(3{\bar{\nu }}_0/4r_0^2){\partial }_{\tau }$ into the viscous evolution of an accretion disk givesHence the dimensionless diffusion equation isFor the useful inverse-radius coordinate $y=x^{-1/2}$, this becomes simply $g_{\tau }=y{g}_{yy}$. The positive factor in $\tau =1+3{\bar{\nu }}_{0}t/(4r_0^2)$ makes increasing $\tau$ correspond to increasing physical time.

There is a distinction between the named torque variable and the physical viscous torque in an accretion disk. For Keplerian rotation, the outward torque magnitude isHere $M_{\star }$ denotes the central mass. The zero-torque inner boundary condition therefore requires $x^{1/2}g\to 0$ as $x\to 0$. Requiring $g\to 0$ is stronger. The nontrivial finite-mass profile used below satisfies both; this distinction matters when counting integration constants.

Write $W(\Sigma )=\bar{\nu }\Sigma$. In a small radial patch $r=r_c+\chi$ with disturbance scale much smaller than $r_c$, the leading disk diffusion equation is ${\partial }_t\Sigma \simeq 3{\partial }_{\chi }^{2}W$. Taking the linearization around a uniform surface density ${\Sigma }_c$ givesFor a local Fourier mode $\delta \Sigma \propto e^{st+ik\chi }$,because $W\propto {\Sigma }^{-1}$ has $W^{\prime }<0$. Thus the effective density diffusion coefficient is negative and the disk has a viscous instability of an accretion disk.

A localized overdensity is a maximum of $\Sigma$ but a minimum of $W$. At its center ${\partial }_{\chi }^{2}W>0$, so ${\partial }_t\Sigma >0$: it grows. The local radial mass flux is proportional to $-{\partial }_{\chi }W$, directing matter toward that minimum from both sides. This is the physical feedback: an overdense patch has a smaller viscous torque in an accretion disk and attracts further accumulation. The arbitrarily rapid $k\to \infty$ growth in this ideal negative-diffusion model is not a resolved prediction below the validity scale of the thin-disk and local viscous closure.

Use characteristic, vertically averaged quantities at fixed radius. Let $H$ be a disk thickness, $\rho \sim \Sigma /H$, and let the alpha disk prescription give $\mu \sim \alpha p/\Omega$, up to a convention-dependent constant. From the preceding radiation pressure and constant-opacity balance, $\mu \sim c/\kappa$ is independent of $\Sigma$; thusVertical hydrostatic equilibrium gives $p\sim \rho {\Omega }^2{H}^2\sim \Sigma {\Omega }^{2}H$, henceThe vertically averaged kinematic viscosity satisfies $\bar{\nu }\Sigma =\int \mu dz\sim \mu H$. ConsequentlyAt fixed radius the omitted factors are independent of $\Sigma$, so this is the requested inverse-density scaling of the viscous torque in an accretion disk.

This is an order-of-magnitude vertical model, as requested. Imposing both exact pointwise $\mu =\text{constant}$ and exact pointwise $\mu \propto p(z)$ would force $p$ to be constant, contrary to nontrivial vertical hydrostatic equilibrium. The characteristic-pressure prescription supplies a scaling, not an exact simultaneous vertical solution to all those idealized equations.

Take $0<r_1<r_2$ and positive mean kinematic viscosity $\bar{\nu }$. The steady state of the supplied viscous evolution of an accretion disk equation satisfies ${\mathcal{F}}^{\prime }=-2\pi rS(r)$. The outer no-flux condition therefore givesPositive $S$ produces positive inward accretion rate in this sign convention. Set $Y=r^{1/2}\bar{\nu }\Sigma$. In a Keplerian disk, the viscous torque in an accretion disk is proportional to $Y$, so the zero-torque inner boundary condition gives $Y(r_1)=0$. Since $Y^{\prime }=\mathcal{F}/(6\pi r^{1/2})$, a second integration yieldsThis is the steady viscous disk supplied by a radial source. It does not require $\bar{\nu }$ to be independent of radius: the equation directly determines $\bar{\nu }\Sigma$. If viscosity depends on the local thermodynamic state, the displayed formula is an implicit relation completed by a separate closure. For nonnegative integrable $S$ it gives nonnegative surface density of a disk, and the inner accretion rate equals the total supplied mass per unit time, $2\pi {\int }_{r_1}^{r_2}rS(r)dr$. Differentiating both nested integrals recovers the flux and the differential equation.

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 $\dot{M}$, soThe boundary condition is $\mathcal{G}(r_{\mathrm{in}})=T_m$. 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 hasIn particular, $\Sigma (r_{\mathrm{in}})=\lambda \dot{M}/(3\pi \bar{\nu })$. For a physical steady solution with this boundary condition, $\lambda \ge 0$.

For $0\le \lambda \ll 1$, 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 $\lambda \gg 1$, 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 $r/r_{\mathrm{in}}\ll (\lambda -1{)}^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

The constant net inward angular-momentum flux isFor $\lambda >1$, 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 $r_{\mathrm{in}}=r_m$, the Maxwell stress tensor estimate givesUsing the magnetospheric truncation radius scaling in part (b), all dimensional parameters cancel:If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi )$; retaining $\Sigma \simeq 2H\rho$ and exact free-fall speed throughout gives $\sqrt{2}/(4\pi )$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda \gg 1$.

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 $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ 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 ${\epsilon }_{\mathrm{in}}=-\eta c^2$. 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

Let $l(R)=R^2\Omega =\sqrt{GMR}$ be the Keplerian specific angular momentum. In a source-free steady interval, the sum of advected and viscously transported angular momentum is constant:Inside the injection radius, $\dot{m}={\dot{m}}_0$. The zero-torque inner boundary condition at $R_{\mathrm{ISCO}}$ sets $C={\dot{m}}_0{l}_{\mathrm{ISCO}}$, and therefore

Outside $R_0$ there is no net mass flow in the stated steady distribution, but it must carry outward the angular momentum deposited by matter moving from $R_0$ to the ISCO. Its constant viscous torque in an accretion disk is thereforeThusThe two expressions agree at $R_0$; the jump in mass flux there is exactly the injected rate.

For Keplerian angular velocity, $R^2(d\Omega /dR{)}^2=9GM/(4R^3)$. The stated $F_{\mathrm{diss}}$ is the dissipation summed over both disk faces, so $F_{\mathrm{diss}}=2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. Inside $R_0$ this givesDifferentiating the factor $R^{-3}[1-(R_{\mathrm{ISCO}}/R{)}^{1/2}]$ shows that the maximum occurs atwhen this radius lies below $R_0$. In the ordinary inflowing region far from its inner edge, $T_{\mathrm{eff}}\propto R^{-3/4}$.

For the static angular-momentum sink outside $R_0$, part c instead givesso the genuinely large-radius behavior of the complete injected disk is $T_{\mathrm{eff}}\propto R^{-7/8}$.

Integrating $2\pi R{F}_{\mathrm{diss}}dR$ over both regions givesandHencewhich is exactly the loss of Keplerian orbital energy as matter moves from its injection orbit to the inner edge.

For a standard disk extending through a radius $R\gg R_{\mathrm{ISCO}}$,This is three times the binding-energy release $GM\dot{m}/(2R)$ available outside $R$. The excess is energy carried outward by the viscous torque in an accretion disk and dissipated at larger radii. In the injected model the nonaccreting outer disk is an especially direct example: it radiates despite having zero mean radial mass flux because it absorbs the angular momentum and mechanical work exported by the inner disk.

As $\lambda \to 0$, the exponential function satisfiesConsequentlywhich is the purely viscous result from part (c)(i).

For $\lambda (\sqrt{x}-1)\gg 1$, the exponential is negligible andSince the magnetic prescription is $v_m=\beta r^{-1/2}$, this is precisely $\nu$ times the magnetically driven profile in part (b)(i).

The dimensionless cumulative competition in the exact profile is $\lambda (\sqrt{x}-1)$. The surface magnetic torque dominates wherewhereas the viscous torque in an accretion disk controls the inner transition where this quantity is at most order one. Thus the crossover is approximatelyFor $\lambda \ll 1$ this is $r/r_{\mathrm{in}}\sim {\lambda }^{-2}$; for $\lambda \gg 1$, magnetic transport dominates outside a narrow inner layer of fractional width $O({\lambda }^{-1})$.

Steady mass conservation gives $\dot{m}=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed viscous torque in an accretion diskThenIntegration from $R_1$ to $R_2$ gives

The viscous power generated in an annulus is $2\pi R{F}_{\mathrm{diss}}dR=\mathcal{G}d\Omega$. Taking the inner torque to vanish and writing $l_{\mathrm{in}}=l(R_{\mathrm{in}})$ gives $\mathcal{G}=-\dot{m}(l-l_{\mathrm{in}})$, henceIntegration by parts yieldsFor steady circular force balance, $de=\Omega dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\mathrm{out}}-e_{\mathrm{in}}$. Therefore

When $R_{\mathrm{out}}\gg R_{\mathrm{in}}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\mathrm{in}}=-R_{\mathrm{in}}^2{\Omega }_{\mathrm{in}}^2/2$, soComparison with part (c) givesA Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because photon trapping in an accretion flow carries part of that generated energy through the inner edge.

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

For inward accretion rate $\dot{M}>0$, the steady conservation of angular momentum equation is $\mathcal{G}=\dot{M}(h-h_{\mathrm{in}})$ when the inner torque vanishes. The viscous torque in an accretion disk is $\mathcal{G}=2\pi qh\bar{\nu }\Sigma$, where $h=r^2\Omega$ and $q$ is the orbital shear parameter. ThusThe familiar Keplerian accretion disk follows when $q=3/2$ and $h\propto r^{1/2}$. The general form also applies to the non-Keplerian Paczyński-Wiita circular orbit rotation law.

With constant kinematic viscosity and zero inner viscous torque in an accretion disk, $\nu \Sigma \sqrt{r}=(6\pi {)}^{-1}{\int }_0^r{F}_{\mathrm{acc}}(r^{\prime })/\sqrt{r^{\prime }}d{r}^{\prime }$. For the steady photoevaporating-disk mass flux, the density is $({\dot{M}}_d-{\dot{M}}_w)/(3\pi \nu )$ inside $r_g$ and the displayed logarithmic profile outside. At equal feeding and wind rates the inner density vanishes, and the outer normalized profile rises from zero with zero slope and tends to one. Its near-edge expansion is $(r/r_g-1{)}^2/8$.

For steady injection into a Keplerian disk between two zero-torque boundaries, conservation of mass gives $\dot{M}={\dot{M}}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}$. Conservation of angular momentum gives $\dot{M}{h}_0={\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$, where $h(R)=\sqrt{GMR}$. Eliminating the outward mass flux gives the displayed inward fraction. This fraction does not depend on the viscosity law: the two boundary fluxes must carry away the injected mass and angular momentum. Equivalently, continuity of the viscous torque in an accretion disk joins $\nu \Sigma \sqrt{R}={\dot{M}}_{\mathrm{in}}(\sqrt{R}-\sqrt{R_{\mathrm{in}}})/(3\pi )$ inside the source to ${\dot{M}}_{\mathrm{out}}(\sqrt{R_{\mathrm{out}}}-\sqrt{R})/(3\pi )$ outside.

For a steady accretion disk with arbitrary rotation law and a zero-torque inner boundary condition at $R=3R_S$, the outward viscous torque in an accretion disk is $\mathcal{G}=\dot{M}(j-j_{\mathrm{in}})$. The dissipation per unit planar area summed over both disk faces is $D=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)=\nu \Sigma (R{\Omega }^{\prime }{)}^2$. Integrating $2\pi RD$ gives $L=\dot{M}GM/(8R_S)=\dot{M}{c}^2/16$ in the Paczyński-Wiita potential. The radiative flux from one face is $D/2$, preventing a factor-of-two error in the luminosity.

A thin disk with alpha disk viscosity $\nu \sim \alpha H^2\Omega$ has a viscous timescale much longer than its dynamical infall time in the plunging region of a black-hole accretion disk. At fixed nonzero accretion rate, rapid inflow reduces its surface density of a disk. The resulting viscous torque in an accretion disk is smaller by the displayed factor than in the slowly drifting circular disk. This estimate concerns radii outside the narrow matching layers.

Subtract local specific angular momentum times mass conservation from conservation of angular momentum including wind losses S,T. With outward radial mass flux $\mathcal{F}$ and outward viscous torque in an accretion disk $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$, this gives $\mathcal{F}{h}^{\prime }+{\mathcal{G}}^{\prime }=2\pi r(Sh-T)$. For Keplerian rotation and $T=qSh$, substitution in mass conservation yields the displayed equation. A wind with q=1 removes local angular momentum but supplies no additional drift term; q>1 can drive accretion even with zero viscosity.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.