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For the $a=1$ compact self-similar disk with density-linear viscosity, radial drift is $\dot{r}=r/(2t)-R/(10t)$. An integrating factor gives the displayed trajectories. From $0<r_i<R(t_i)$, the arrival time at the origin is $t_i[1-r_i/R(t_i){]}^{-10}$, finite for every interior point. The outer-edge trajectory has zero surface density of a disk and carries no mass.

An aligned eccentric ring consists of nested Kepler orbits with nearly common longitude of periapsis. Its width and surface density of a disk depend on both the distribution of semi-major axis values and the radial gradient of forced eccentricity.

A two-dimensional disk equation with pressure acceleration $-{\Sigma }^{-1}\nabla P$ requires an equation of state for vertically integrated pressure $P$ in terms of surface density of a disk $\Sigma$. A barotropic fluid closure is $P=P(\Sigma )$, with linear enthalpy perturbation $h^{\prime }=c_s^2{\Sigma }^{\prime }/{\Sigma }_0$. Merely giving $P=P(\rho )$ for a volume density does not close the two-dimensional equations unless a vertical conversion $\rho (\Sigma )$ is specified. Positive $c_s^2$ gives ordinary pressure restoration of an inertial-acoustic wave.

The midplane galaxy rotation curve fixes the radial derivative of the Newtonian gravitational potential, not every exterior boundary condition. Adding $a|z|$ leaves that radial derivative unchanged but adds a uniform sheet of surface density of a disk $a/(2\pi G)$. The pure scale-free Mestel disk excludes such extra sources.

Let ${\Sigma }_{d0}$ be the uniform background dust surface density of a disk and ${\delta }_d=({\Sigma }_d-{\Sigma }_{d0})/{\Sigma }_{d0}$ its fractional perturbation. Here $k$ is the radial wavenumber, $\Omega$ is the epicyclic frequency and $u$ is a real velocity amplitude. For that uniform background and late velocity $v_x=u\cos (kx-\Omega t)$, linear dust continuity equation gives ${\partial }_t{\delta }_d=-{\partial }_x{v}_x=ku\sin (kx-\Omega t)$. Integrating yields the displayed formula. The coherent travelling-wave fractional excess is $|ku|/\Omega$, with validity requiring this number much smaller than one. The static term $C(x)$ records initial density and transient displacement: drag damps velocity, but supplies no density diffusion. The long-wavelength gas wave has the same fractional density amplitude, so entrainment alone does not produce an additional dust-to-gas ratio enhancement at this order.

A disk flares when its disk aspect ratio increases outward. This is stronger than the statement that its height increases: a disk with $H\propto r$ has constant aspect ratio and is not flaring by this definition. The radial temperature and surface density of a disk profiles determine flaring in a specified vertical equilibrium model.

For cylindrical rotation independent of height, write $W_{r\varphi }=\int T_{r\varphi }dz$. Define the effective kinematic viscosity by $W_{r\varphi }=\bar{\nu }\Sigma r{\Omega }^{\prime }$, where $\Sigma$ is the surface density of a disk. For Keplerian rotation, $\bar{\nu }=-2W_{r\varphi }/(3\Sigma \Omega )$, and the outward viscous torque in an accretion disk is $-2\pi r^2{W}_{r\varphi }=3\pi \bar{\nu }\Sigma \sqrt{GMr}$. A Newtonian fluid stress tensor gives the density-weighted viscosity of a disk; the stress definition can also parameterize an effective turbulent or magnetic transport stress.

// Target: astrophysics.bigb

The reflection-symmetric Newtonian gravitational potential $\Phi =v_0^2\log [(\sqrt{R^2+z^2}+|z|)/R_{\ast }]$ is harmonic off the plane. Its normal-derivative jump is $2v_0^2/R$, so the Poisson equation for Newtonian gravity gives surface density of a disk $S=v_0^2/(2\pi GR)$ and volume mass density $\rho =S\delta (z)$. The midplane circular speed is constant. The scale-free model has infinite total mass and does not use a finite-mass boundary condition at infinity.

A narrow surface density of a disk source approaching the outer edge from inside, with $S=S_0\delta (r-r_2)$ normalized to have full weight, gives $\dot{M}=2\pi r_2{S}_0$ and the displayed steady interior profile. This follows by replacing the inner source integral by $r_2{S}_0$ and integrating $u^{-1/2}$. The flux jumps from $\dot{M}$ just inside to zero outside the injecting edge. Thus a zero exterior flux is consistent, whereas a continuous zero interior boundary trace is not. A half-weight convention for a symmetric endpoint Dirac delta function represents a different total injected mass; it must not be silently mixed with the full-weight boundary source.

Using the radial-drift expression already derived, the factor $X-1$ cancels and givesFor $a=1$, $X=R/r$ and $\sigma =1/(15At)$, soThe instantaneous flow is inward for $r<R/5$ and outward for $R/5<r<R$. Nevertheless outward motion need not prevent eventual accretion. The trajectories in accretion trajectories in a compact viscous similarity disk satisfyMultiplication by $t^{-1/2}$ and integration giveA parcel at $r_i\in (0,R_i)$ at time $t_i>0$ has $K=(r_i-R_i)/\sqrt{t_i}<0$. It reaches the origin atBefore arrival, $r/R=1+K{t}^{1/10}/C$ decreases monotonically. Even a parcel initially moving outward eventually crosses $r/R=1/5$, turns inward, and is accreted. Only $K=0$, the outer-edge trajectory, never arrives; its surface density of a disk is zero. Thus almost every mass-carrying parcel is accreted in finite time.

For $a=5/4$, the coefficient $5/2-2a$ vanishes, leavingThe trajectories are $r(t)=r_i\sqrt{t/t_i}$, so each parcel keeps a fixed ratio $r/R$. This is an expanding nonaccreting similarity disk supplied by an inner torque, not an inward accretion solution.

Assume an axisymmetric thin disk in the fixed Newtonian potential of a point mass, negligible disk self-gravity, and slow evolution compared with an orbital period. Radial pressure and radial inertia are small compared with gravity and centrifugal acceleration, so $u_{\varphi }=r\Omega (r)$ with $\Omega =(GM/r^3{)}^{1/2}$, independent of height and time. Assume no mass loss or torque through the upper and lower disk faces. Define the surface density of a disk, mass-weighted radial velocity and integrated shear stress byVertical integration of the continuity equation givesThe azimuthal momentum equation, including the geometric term $u_r{u}_{\varphi }/r$, gives conservation of angular momentum in terms of $j=r^2\Omega$:Subtracting $j$ times mass conservation eliminates the time derivative and leaves $\Sigma {\bar{u}}_r{j}^{\prime }=r^{-1}(r^{2}W{)}^{\prime }$.

Define the effective density-weighted viscosity of a disk through the actual stress:If the local viscous stress is $T_{r\varphi }=\rho \nu r{\Omega }^{\prime }$, this reduces to $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$. The definition can also describe an effective turbulent transport stress, provided the same shear closure is appropriate. Since ${\Omega }^{\prime }<0$, outward transport corresponds to $W<0$ and positive $\bar{\nu }$.

For Keplerian rotation, $j=\sqrt{GMr}$ and $r^{2}W=-(3/2)\sqrt{GM}{r}^{1/2}\bar{\nu }\Sigma$. ThereforeSubstitution into mass conservation yields the Keplerian viscous diffusion equationThe absence of surface fluxes, fixed rotation law and leading thin-disk approximation are essential: a wind, appreciable radial inertia or evolving gravitational potential would alter this reduction.

Both profiles have integrable central cusps. Directly integrating the surface density of a disk and its specific angular momentum $j=\sqrt{GMr}$ givesFor $a=1$, substitution of the similarity functions givesThe inward accretion rate is $\dot{M}(r)=-2\pi r\Sigma {\bar{u}}_r$. At the origin it tends to $M_d/(5t)=-d{M}_d/dt$. The outward viscous torque in an accretion disk isFor this profile $\mathcal{G}\to 0$ as $r\to 0$, and the advected angular-momentum flux $\dot{M}j$ also vanishes there. Both outer fluxes vanish at $R$. Thus disk mass decreases through central accretion while disk angular momentum is conserved. Including the accreted central mass restores total mass conservation for the disk-plus-sink system.

For $a=5/4$, instead,Here the mass flux tends to zero at the origin, despite positive radial velocity everywhere else. The inner torque has the nonzero limitThus the nonaccreting similarity disk supplied by an inner torque conserves disk mass but gains angular momentum from an external central torque. Its vanishing edge flux cannot supply that gain. The disk is not globally isolated in angular momentum; the combined disk and torque source may of course conserve it.

These profiles illustrate different boundary-controlled similarity classes of the Keplerian viscous diffusion equation. The $a=1$ solution is compatible with a zero-torque inner boundary condition and conserved finite angular momentum, and can describe a late-time spreading accretion disk after the detailed initial profile has been forgotten. Its mass diverges as $t\downarrow 0$, so the formal zero-radius origin is not a finite-mass initial disk. The $a=5/4$ solution has finite initial mass concentrated at the origin and initially zero disk angular momentum, but requires the specified inner torque for $t>0$; that torque's $t^{-3/4}$ singularity is time-integrable. It may describe a spreading nonaccreting disk for compatible torque-driven boundary conditions, but cannot be an isolated angular-momentum-conserving disk's generic asymptotic state. Similarity scaling suggests their possible asymptotic roles; it does not establish convergence for arbitrary initial and boundary data.

Far from the inner edge of a steady Keplerian accretion disk, the inner angular-momentum flux is negligible compared with $\dot{M}j$. Since $\Omega \propto R^{-3/2}$, the viscous torque in an accretion disk becomes $\mathcal{G}=3\pi \nu \Sigma j$. The steady balance $\mathcal{G}\simeq \dot{M}j$ therefore gives $\nu \Sigma =\dot{M}/(3\pi )$. With the stated surface density of a disk,Steady conservation of mass gives $\dot{M}=-2\pi R\Sigma v_R$, and consequentlyBoth relations use the far-inner-edge approximation, rather than extrapolating the stress-free boundary profile to the inner edge.

For the trace passive scalar, put $\sigma =\Sigma C$. Substitute this into the contaminant conservation equation and expand its advective flux:The braces multiplying $C$ vanish by background conservation of mass. Hence the concentration satisfiesThe small contaminant fraction ensures it does not alter the ambient disk dynamics. The surface-density model already averages vertically; an axisymmetric background and ring-like injection give $C=C(R,t)$. More generally, rapid differential orbital motion winds azimuthal structure and facilitates mixing, but radial-only dependence remains a symmetry or mixing assumption, not a property of every initial injection.

For the axisymmetric advection-diffusion equation, cylindrical geometry givesHere $k=\zeta \nu$ and $k\Sigma =\zeta {\nu }_0{\Sigma }_0$ are independent of radius. ThusLet $Y=\ln (R/R_{\ast })$, with any fixed reference length $R_{\ast }$. Since $C_Y=R{C}_R$ and $C_{YY}=R^2{C}_{RR}+R{C}_R$, this is the tracer diffusion in a power-law accretion disk equationAssume $\zeta >0$, ${\nu }_0>0$, and write $v=3{\nu }_0/2$, $D=\zeta {\nu }_0$. The change $X=Y+vt$ turns this into the heat equation $C_t=D{C}_{XX}$. The radial Dirac delta function must be transformed with its Jacobian:Convolution with the Gaussian heat kernel givesThis full-line logarithmic solution extends the far-edge power-law model to $0<R<\infty$. An actual finite inner boundary would require its own boundary condition and modified kernel.

The fraction must be computed with the contaminant mass measure, rather than with $CdR$. IndeedThe total contaminant mass is the conserved quantity $M_c=2\pi {\Sigma }_0{C}_0/R_0$. Its normalized logarithmic radius is therefore a normal distribution with mean $Y_0-vt$ and variance $2Dt$. Integrating this tail gives, for $r>0$,The complementary error function accounts for the one-time mass fraction outside the specified radius.

For an outward target $r_1>1$, put $\ell =\ln r_1>0$. Since $\mathrm{erfc}$ is decreasing, maximize $F$ by minimizing $z(t)=(\ell +vt)/\sqrt{4Dt}$. Its derivative has the sign of $vt-\ell$, so the unique maximum occurs at $t_{\ast }=\ell /v=2\ln r_1/(3{\nu }_0)$. At this time $z(t_{\ast })=\sqrt{v\ell /D}$. Thus the printed expression is correctly derived as the maximum instantaneous exterior fraction:

It is not, however, the probability of ever reaching the radius. A tracer can pass through the target and later return; its history is not counted by a one-time exterior fraction. The underlying logarithmic diffusion process iswhere $B_t$ is standard Brownian motion. To compute its actual hitting probability, first put absorbing levels at $-A$ and $\ell$ relative to $Y_0$. Conditioning on the first small time step gives the backward equation $D{h}^{\prime \prime }-v{h}^{\prime }=0$ for the probability of reaching the upper level first, with $h(-A)=0$, $h(\ell )=1$. Solving givesLetting the lower level tend to $-\infty$ exhausts all finite-time upward-hitting paths. Therefore the correct eventual probability in this idealized model isFor example, if $\ln r_1=2\zeta /3$, the maximum exterior fraction is $\frac{1}{2}\mathrm{erfc}(1)\simeq 0.07865$, whereas the actual hitting probability is $e^{-1}\simeq 0.36788$. This is an explicit counterexample to the last inference in the question and an instance of maximum occupation is not a hitting probability.

For $r_1=1$, the target is the initial radius and the hitting probability is one, while ${\sup }_{t>0}F(1,t)=1/2$ with no attained maximum. For $0<r_1<1$, the inward drift ensures eventual hitting with probability one, and ${\sup }_{t>0}F(r_1,t)=1$ as $t\downarrow 0$. The square-root expression involving $\ln r_1$ is therefore only meaningful for the outward-target case, and even there it describes occupancy rather than first passage.

Take $\dot{M}>0$ to mean inward accretion rate, and let $S$ count loss through both faces per unit projected area of the accretion disk. An annulus of width $dR$ loses $2\pi RSdR$ to the wind. In steady state, its incoming mass flux at its outer edge equals the inward flux at its inner edge plus this loss. Hence mass conservation givesIf a loss rate is instead defined per individual face, it must be doubled before substituting for $S$ here.

Use Keplerian rotation ${\Omega }^2=GM/R^3$. The vertically integrated viscous dissipation, summed over both disk faces, isHere $\nu$ is the kinematic viscosity, and $\Sigma$ is the surface density of a disk; for height-dependent viscosity the product means $\int \rho \nu dz$. Material already in a circular orbit has specific orbital energy $v_K^2/2-GM/R=-GM/(2R)$. Giving it zero total binding energy therefore costs an additional $GM/(2R)$ per unit mass. The total launch speed is the escape velocity $\sqrt{2GM/R}$, but one must not charge the wind for the orbital kinetic energy it already possesses. Equating wind power to $f{Q}^{+}$ givesCombining this with mass conservation yields the wind-powered mass loss from a steady accretion disk relation

Define the outward viscous torque in an accretion disk asThe net inward angular momentum flux is $\dot{M}h-\mathcal{G}$. Because each unit mass lost to the wind carries the local specific angular momentum $h$, annular conservation of angular momentum readsIn the wind region, $\mathcal{G}=Rh{\dot{M}}^{\prime }/(3f)$. For constant $f$, differentiation and cancellation of $h$ givesThus the requested Cauchy-Euler differential equation is

For $f=1/3$, a power-law trial $\dot{M}\propto R^p$ gives $p^2+p/2-1/2=0$, with roots $p=1/2,-1$. Therefore $\dot{M}=A\sqrt{R}+B/R$ in $R_{\ast }\le R\le R_0$. A zero inner viscous torque in an accretion disk implies ${\dot{M}}^{\prime }(R_{\ast })=0$, so $B=A{R}_{\ast }^{3/2}/2$. There is no point mass sink at $R_0$, so the accretion rate is continuous there and equals ${\dot{M}}_0$. The complete result isIts derivative is nonnegative throughout the wind region, as required for a nonnegative loss rate $S$. The derivative can jump to zero outside $R_0$: the energy relation applies only inside the wind zone and does not impose ${\dot{M}}^{\prime }(R_0)=0$ on its inner limit. The viscous torque in an accretion disk remains continuous and extends outside via ${\mathcal{G}}^{\prime }={\dot{M}}_0{h}^{\prime }$.

At the inner boundary the exact accreted fraction isThe last expression is a leading-order asymptotic equality, not the exact finite-radius fraction. Most of the supplied mass is expelled before reaching the star when $R_{\ast }/R_0$ is small.

Take the retained disc mass to be the cosmic baryon fraction of the total dark matter halo mass: $M_d=f_b{M}_h$, where $f_b=0.05/0.25=0.2$. Adopt $v_{\mathrm{vir}}^2=G{M}_h/r_{\mathrm{vir}}$ and describe the disc gravity by $v_d^2=C_{d}G{M}_d/R_d$. The structure coefficient $C_d$ must be stated: the scale-free Mestel disc gives $C_d=1$ when its flat rotation speed and enclosed mass are used, while a finite disc requires boundary and thickness information. The equality of edge specific angular momentum gives $R_d{v}_d=q{r}_{\mathrm{vir}}{v}_{\mathrm{vir}}$, with $q=0.1$. Combining these relations gives the angular-momentum-conserving self-gravitating disc radius:With the commonly intended $C_d\simeq 1$ approximation,The corresponding masses are $M_h\simeq 2.49\times 10^{11}{M}_{\odot }$ and $M_d\simeq 4.97\times 10^{10}{M}_{\odot }$, where $M_{\odot }$ is the solar mass.

For a definite formation estimate, assume the dark matter halo acquired this mass and radius at virial equilibrium, retained them afterward, and had mean density ${\Delta }_v{\bar{\rho }}_m(z_f)$ with ${\Delta }_v=18{\pi }^2$. This uses the matter-dominated spherical-collapse model as an approximation to the specified matter-plus-vacuum cosmology. Since ${\bar{\rho }}_m(z_f)=3H_0^2{\Omega }_{m0}(1+z_f{)}^3/(8\pi G)$,The approximate distance therefore gives formation modestly before observation. Using the numerical rather than approximate distance integral gives $z_f\simeq 1.22$, within the accuracy of these structural assumptions. The calculation also assumes inclination-corrected spectroscopy, negligible disc pressure support, conserved edge specific angular momentum, no substantial later accretion or mergers, and no significant baryon loss.

This formation redshift is model-dependent, not uniquely fixed by the stated disc data. Keeping $C_d$ explicit gives $1+z_f\simeq 2.209C_d^{-4/3}({\Delta }_v/18{\pi }^2{)}^{-1/3}$. In particular, replacing the mean-density convention by $200$ times the critical density gives $H(z_f)=v_{\mathrm{vir}}/(10r_{\mathrm{vir}})$ and $z_f\simeq 0.87$, later than the observation. That inconsistent chronological result cannot be silently used as the formation epoch.

There is a further idealization in applying $C_d=1$ exactly at the edge: an abruptly truncated razor-thin disc with nonzero edge surface density of a disk has a logarithmically divergent in-plane force at that edge. Locally the available mass occupies a half-plane, and its radial contribution is proportional to $G{\Sigma }_d{\int }_{ϵ}^{\ell }ds/s$, with no opposite exterior half-plane to cancel it. Finite thickness or a smooth taper removes that sharp-edge force singularity of a thin disk. Thus $C_d=1$ is an explicitly qualified Mestel disc or monopole-scale approximation, not an exact finite-disc calculation; enclosed mass does not determine a disc rotation curve.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Interpret the outer Dirac delta function as a boundary source carrying its full mass, or as the limit of a narrow source lying just inside $r_2$. Then ${\int }_u^{r_2}vS(v)dv=r_2{S}_0$ for $r_1<u<r_2$, andThus the interior is the usual steady Keplerian accretion disk with a constant inward accretion rate. The product $\bar{\nu }\Sigma$ vanishes at the stress-free inner edge, and the surface density of a disk does too if the viscosity has a finite nonzero boundary value; far from that edge $\bar{\nu }\Sigma$ approaches $\dot{M}/(3\pi )$. Its density need not itself be constant because the viscosity may vary.

The outer-edge source of a viscous disk requires care about the boundary trace. The flux just inside the injecting edge is $\mathcal{F}(r_2^{-})=\dot{M}$, while the no-flux value outside the source is $\mathcal{F}(r_2^{+})=0$; the source produces the intervening jump. A simultaneously continuous interior condition $\mathcal{F}(r_2^{-})=0$ would contradict nonzero steady injection. A symmetrically regularized delta integrated only up to its center can instead give half its full weight; that would mean a different supplied mass, not a different interior solution for the same $\dot{M}$. The physically normalized one-sided limiting source removes this convention ambiguity.

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.

Assume the vertical dynamical timescale of a disk is shorter than the thermal timescale of an accretion disk, which is itself shorter than the viscous timescale. This ordering is controlled in a thin disk with $\alpha \ll 1$: $t_{\mathrm{dyn}}\sim {\Omega }^{-1}$, $t_{\mathrm{th}}\sim (\alpha \Omega {)}^{-1}$ and $t_{\nu }\sim (r/H{)}^2{t}_{\mathrm{th}}$. Therefore compare heating and cooling at fixed surface density of a disk, radius, $\alpha$ and composition. Vertical hydrostatic equilibrium adjusts on the vertical dynamical timescale of a disk, giving $H\propto T^{1/2}$ and $\rho \propto T^{-1/2}$. Consequently the stipulated negative hydrogen ion opacity varies as $\kappa \propto T^{10-1/6}=T^{59/6}$. The rates per disk face then scale asThus heating increases while cooling decreases after a positive temperature perturbation. At equilibrium $Q^{+}=Q^{-}=Q_0$, a small change gives $\delta (Q^{+}-Q^{-})=(41/6)(Q_0/T)\delta T$, which drives the perturbation further from equilibrium for positive effective heat capacity. A negative temperature perturbation likewise increases the net cooling. This proves thermal instability from negative hydrogen ion opacity within the local closure; it is the temperature-slope test of thermal stability of an accretion disk. Holding volume density rather than surface density of a disk fixed would give a different cooling exponent, but still the same qualitative instability; the fixed-column comparison is the relevant thin-disk thermal test.