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For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

The Keplerian viscous diffusion equation with the displayed kinematic viscosity has two such nontrivial finite-support self-similar solutions: $a=1$, $R=C{t}^{2/5}$, and $a=5/4$, $R=C{t}^{1/2}$. Both have $\sigma =(15At{)}^{-1}$ for $A,C,t>0$. Matching powers of radius eliminates the third power only when $(5/2-2a)(2-2a)=0$. The first solution conserves disk angular momentum; the second conserves disk mass but requires an inner torque.

For $\nu ={\nu }_0(\Sigma /{\Sigma }_0{)}^{1/2}(R/R_0{)}^{9/4}$, the variables $x=\sqrt{R/R_0}$, $S=(\Sigma /{\Sigma }_0)x^3$ and $\tau =3{\nu }_{0}t/(4R_0^2)$ reduce the source-free Keplerian viscous diffusion equation to $S_{\tau }=(x{S}^{3/2}{)}_{xx}$. The displayed self-similar solution has an outer edge $R_e=36R_0\tau$ and a surface-density cusp proportional to $R^{-3/2}$ at the origin. The positive-part cutoff is essential: beyond the edge, $S^{3/2}$ is the cube of the absolute value, so continuing the squared polynomial without truncation does not solve the same equation.

The disk mass is $8\pi {\Sigma }_0{R}_0^2{\tau }^{-1/2}$ and its angular momentum is $12\pi {\Sigma }_0{R}_0^2\sqrt{GM{R}_0}$, independent of time. It therefore describes accretion with outward spreading and conserved disk angular momentum, under the formal zero-torque condition at the origin. The inward mass flux changes sign at $R=9R_0\tau /4$, so the outer portion moves outward while the inner portion accretes.

Two integrations of the steady Keplerian viscous diffusion equation give this general transported quantity. $A$ fixes the mass accretion rate, while $B$ fixes the additive angular-momentum flux or inner torque. A zero-torque inner boundary condition gives $B=-A\sqrt{r_{\mathrm{in}}}$; a nonaccreting constant-torque disk has $A=0$.

For source-free Keplerian viscous diffusion equation, the variables $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$ give $g_t=(3{\nu }_0/4)g_{yy}$ when $\bar{\nu }={\nu }_{0}r$. Rescaling y gives the displayed heat equation. Initial and radial boundary conditions must also be transformed; the reduction does not specify them.

For an axisymmetric thin disc supplied at $R_0$ with the local Keplerian rotation specific angular momentum, the displayed source augments the Keplerian viscous diffusion equation. The radial velocity remains $v_R=-3(\Sigma R^{1/2}{)}^{-1}{\partial }_R(\nu \Sigma R^{1/2})$: subtracting local specific angular momentum times conservation of mass from conservation of angular momentum cancels the source term. The normalization of the Dirac delta function makes the integrated supplied mass per unit time equal to $\dot{M}(t)$.

Freezing coefficients locally in the Keplerian viscous diffusion equation gives ${\Sigma }_{1,t}\simeq 3{\partial }_{\Sigma }(\bar{\nu }\Sigma ){\Sigma }_{1,rr}$. A radial Fourier mode grows at rate $-3{\partial }_{\Sigma }(\bar{\nu }\Sigma )k_r^2$. Negative response reverses ordinary smoothing and gives a backward heat equation, within the wavelength range where the thin disc transport closure applies.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

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.

Set $S=T=0$ and choose $\boxed{\bar{\nu }(r)={\nu }_{0}r}$, with ${\nu }_0>0$ constant. Define $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$. Then $\sqrt{r}\bar{\nu }\Sigma ={\nu }_{0}g$, and ${\partial }_r=(2y{)}^{-1}{\partial }_y$. Directly transforming the Keplerian viscous diffusion equation givesThe heat-equation transform for linear-radius disk viscosity is completed by $x=2y/\sqrt{3{\nu }_0}$:Equivalently one can use $g=\sqrt{r}\bar{\nu }\Sigma$, which differs only by a constant. This is the viscosity law making the transformed diffusivity constant under this square-root-radius reduction. Initial and boundary conditions transform too; the change of variables does not supply unspecified disc-edge data.

The physical diffusion arises from angular momentum transport. Differential rotation creates a viscous shear stress, which transfers angular momentum between neighboring annuli. Gas that loses angular momentum moves inward, while gas that gains it can move outward. Combining conservation of angular momentum with mass conservation turns the stress gradient into a radial mass flux and its divergence into a second radial derivative. Positive kinematic viscosity smooths the appropriately weighted profile; matter need not all move in one direction, and an accreting disc can simultaneously spread outward. Viscosity dissipates orbital energy but redistributes angular momentum internally except for boundary losses.

Use a characteristic midplane temperature $T$, density $\rho$ and disk scale height $H$. For an ideal gas of fixed composition, dominant gas pressure gives $c_s^2\propto T$, where $c_s$ is the sound speed. Vertical hydrostatic equilibrium and the alpha disk prescription giveNumerical factors in the vertical averaging do not affect the exponents. The stipulated opacity becomesFor an optically thick, radiatively efficient disk, radiative diffusion gives a cooling rate per unit area proportional to $T^4/(\kappa \Sigma )$. The viscous heating rate is proportional to $\nu \Sigma {\Omega }^2$. ThusEquating heating and cooling gives $T\propto \alpha {\Sigma }^{1/2}{\Omega }^{-1/2}$, soAt fixed $\alpha$ and central mass, $\Omega \propto R^{-3/2}$. Therefore the gas-pressure alpha disk with inverse-density opacity satisfies

For thermal stability of an accretion disk, perturb $T$ at fixed $R$ and $\Sigma$, since the thermal adjustment is faster than radial mass redistribution. The logarithmic heating and cooling slopes are respectively one and two. At equilibrium, a fractional temperature increase therefore makes the excess heating $Q^{+}-Q^{-}$ decrease, and the disk cools back. Thus the local thermal equilibrium is stable within the assumed gas-pressure, radiative-diffusion closure.

For viscous stability of an accretion disk, use the thermally equilibrated relation between viscosity and surface density. It givesLocally freezing the background radius in the Keplerian viscous diffusion equation, a short radial Fourier mode has growth rate $-3k_R^2{\partial }_{\Sigma }(\nu \Sigma )<0$. Density enhancements diffuse rather than grow, so the disk is locally viscously stable as well.

This part uses source-free viscous evolution of an accretion disk. Put $R=R_0{x}^2$ and $\Sigma ={\Sigma }_0{x}^{-3}S$. The prescribed kinematic viscosity then becomesso $\nu \Sigma R^{1/2}={\nu }_0{\Sigma }_0{R}_0^{1/2}x{S}^{3/2}$. Since ${\partial }_R=(2R_{0}x{)}^{-1}{\partial }_x$, the Keplerian viscous diffusion equation givesCanceling the prefactor and setting $\tau =3{\nu }_{0}t/(4R_0^2)$ proves

Use the similarity ansatz $S={\tau }^{-1}f(\xi )$ with $\xi =x/\sqrt{\tau }$. Derivatives at fixed $x$ giveOn the branch $1-k\xi \ge 0$, substitute $f=(1-k\xi {)}^2$. The two expressions reduce toEquality of the constant coefficients requires $\boxed{k=1/6}$, and then the other coefficients agree as well.

The physical compact self-similar disk with square-root density viscosity uses this branch up to its outer edge and vacuum beyond it:Here $(u{)}_{+}=\max (u,0)$ is the positive part. The cutoff matters: beyond $\xi =6$, the untruncated squared expression would have $f^{3/2}=|1-\xi /6{|}^3$, and its right-hand side would no longer be the polynomial used above. At the moving edge, $S$ vanishes quadratically and $x{S}^{3/2}$ vanishes cubically. The latter and its first derivative match the vacuum values continuously, so joining to zero creates no distributional mass-flux source. The joined profile satisfies the equation, including in the weak sense at the edge.

Returning to the surface density gives the explicit resultwith $\Sigma =0$ for $R\ge 36R_0\tau$. Thus the outer edge is $\boxed{R_e=36R_0\tau }$ and moves linearly in physical time, with speed $27{\nu }_0/R_0$. The density decreases monotonically with radius, has an integrable $R^{-3/2}$ central cusp, and drops quadratically to zero at the outer edge. The formal origin boundary has zero viscous torque in an accretion disk, because $\nu \Sigma \sqrt{R}\to 0$ there, even though the surface density diverges.

To characterize the spreading, let $a=6\sqrt{\tau }$ and integrate in $x$. The disk mass and angular momentum areThe mass decreases while the angular momentum remains constant: matter accretes at the origin and the remaining disk spreads outward. More explicitly, with $u=x/(6\sqrt{\tau })$, the inward mass flux isIt is inward for $u<1/4$ and outward for $u>1/4$. The dividing radius is $9R_0\tau /4$, and the origin flux $3\pi {\nu }_0{\Sigma }_0{\tau }^{-3/2}$ equals $-d{M}_d/dt$. The outer edge carries no flux beyond itself. This family is a self-similar accreting disk with finite mass at each positive time, rather than a regular finite-mass initial profile at $\tau =0$.

The required pressure integral uses ${\int }_0^1(1-x^2{)}^{4}dx=128/315$. With $\Sigma =(32/35){\rho }_{0}H$ and $p_0/{\rho }_0={\Omega }^2{H}^2/8$,The vertically averaged alpha closure for a mixed-pressure layer therefore givesThe height-integrated dynamic viscosity is $2H\mu =8cH\lambda /[9\kappa (1+\lambda )]$. Equate it to $\bar{\nu }\Sigma$ to obtainThese are integrated equalities, not an imposed pointwise relation $\mu (z)\propto p(z)$.

Now keep $\alpha$, the opacity and molecular constants independent of radius, and use Keplerian rotation $\Omega \propto r^{-3/2}$. In the gas-dominated limit, $\lambda \ll 1$ but positive, the last equality gives $\lambda \propto H\Omega \Sigma$. The column-density relation gives $\lambda \Sigma \propto H^7{\Omega }^6$. Eliminating $\lambda$ yields $H^6\propto {\Sigma }^2{\Omega }^{-5}$, whenceThis is the gas-pressure branch of a mixed-pressure alpha disk. In the radiation-dominated limit, $\lambda /(1+\lambda )\to 1$, so $H\propto (\Omega \Sigma {)}^{-1}$ andFor viscous stability of an accretion disk, linearize the Keplerian viscous diffusion equation at fixed radius: a local mass density perturbation has diffusion coefficient $3{\partial }_{\Sigma }(\bar{\nu }\Sigma )$. The gas branch has $\bar{\nu }\Sigma \propto {\Sigma }^{5/3}$ and a positive derivative, so it smooths perturbations. The radiation branch has $\bar{\nu }\Sigma \propto {\Sigma }^{-1}$ and a negative derivative, giving radiation-pressure viscous instability. The gas-pressure branch is viscously stable; the radiation-pressure branch is viscously unstable in this closure. This concerns radial mass-transport stability on wavelengths where the vertically averaged thin-disc model applies.

Assume an axisymmetric thin disc rotating in the fixed potential of a dominant central mass, with $\Omega =(G{M}_{\ast }{)}^{1/2}{r}^{-3/2}$ independent of time and height. Neglect vertical mass loss and vertical angular-momentum flux at the two faces, as well as self-gravity and radial pressure corrections to the rotation law. Define the surface density and density-weighted kinematic viscosity byand let $v_r={\Sigma }^{-1}\int \rho u_{r}dz$. These assumptions give the vertically averaged viscous disk equationswhere $l=r^2\Omega$ is specific angular momentum. Subtract $l$ times conservation of mass from conservation of angular momentum. Since $l$ is fixed in time,Substitution into conservation of mass proves the Keplerian viscous diffusion equationNo assumption of height-independent kinematic viscosity is needed; its density-weighted average is the one appearing in the integrated stress. A wind or surface magnetic stress would add terms and must not be silently discarded.

The negative-diffusion criterion for viscous disk instability is ${\partial }_{\Sigma }(\bar{\nu }\Sigma )<0$. A surface density enhancement then transports angular momentum less effectively and grows. Freezing the Keplerian viscous diffusion equation locally gives growth proportional to the square of radial wavenumber, within the thin-disc transport range; wavelengths must still be sufficiently long compared with the disk scale height for that closure. The formal short-wave limit is a backward heat equation.

Within an instantaneous vertically averaged transport closure, the local Keplerian viscous diffusion equation has effective diffusion coefficient $3{\partial }_{\Sigma }(\bar{\nu }\Sigma )$. A positive coefficient smooths short radial Fourier modes in the allowed thin-disc wavelength range; a negative coefficient gives the negative-diffusion criterion for viscous disk instability. A zero coefficient is linearly degenerate and does not establish strict decay. This local response criterion alone does not settle thermal, self-gravitating or boundary-driven stability.