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A Keplerian disk with surface density $\Sigma \sim r^{-p}$ at infinity has finite outer mass for $p>2$ and finite outer angular momentum for $p>5/2$, since the relevant integrals are $\int \Sigma rdr$ and $\int \Sigma r^{3/2}dr$. The borderline $p=5/2$ has logarithmically divergent angular momentum even though its mass converges.

The aspect ratio of an astrophysical disk is its vertical thickness scale divided by cylindrical radius. A thin disk has $H/r\ll 1$. In gas-supported vertical hydrostatic equilibrium with harmonic external gravity $-{\Omega }_z^{2}z$, $H/r\sim c_s/(r{\Omega }_z)$, where ${\Omega }_z$ is the vertical epicyclic frequency and $c_s$ is the sound speed. In a centrally dominated Keplerian disk, ${\Omega }_z=\Omega$, so this compares sound speed with orbital speed. If disk self-gravity supplies the vertical restoring force, the relation $H\sim c_s/\Omega$ requires replacement. A polytropic free-surface height and an isothermal scale height differ by model-dependent constants even though their aspect ratios have the same basic role.

In a star-dominated Keplerian disk with $H\simeq c_s/\Omega$ and characteristic disk mass $M_D\sim \pi r^2\Sigma$, the Toomre stability criterion becomes $Q\sim (H/r)(M_{\ast }/M_D)$. Thus instability can occur at a disk-to-star mass ratio comparable to the small disk aspect ratio. The numerical coefficient depends on whether $M_D$ means a local annular estimate or integrated enclosed mass.

For $0<A<2\Omega$ and equal kinematic viscosity and magnetic diffusivity, a vertical wavevector admits $\mathbf{v}=(V,V,0)$ and $\mathbf{b}=(i{\omega }_{A}V/A,-i{\omega }_{A}V/A,0)$ in Alfvén-speed units. Substitution into the magnetorotational instability equations gives the displayed frequency and growth rate. In a Keplerian disk, growth of this polarization requires $B_z^2/({\mu }_0\rho )>5\nu \Omega /4$. This is a bound for this selected channel and its wavelength, not a universal threshold for every MRI mode in an unbounded domain.

Let $\mathcal{R}=k/({\mu }_m{m}_H)$, take constant opacity $\kappa$, and prescribe the dynamic viscosity $\mu =\alpha p_g/\Omega =\alpha \mathcal{R}\rho T/\Omega$ in a Keplerian disk. The pressure supporting the disk may contain both gas pressure and radiation pressure. Combining viscous dissipation and radiative diffusion gives ${\partial }_m^2{T}^4+A\Omega T=0$ in the vertical mass coordinate of a disk, where $A=27\alpha \kappa \mathcal{R}/(16\sigma )$. Set $\zeta =2m/\Sigma$, $T=T_{\ast }t(\zeta )$ and $T_{\ast }^3=A\Omega {\Sigma }^2/4$. With a radiative-zero disk surface, $t$ satisfies $(t^4{)}^{\prime \prime }+t=0$, $t^{\prime }(0)=t(1)=0$. Writing $I_t={\int }_0^{1}td\zeta$, the density-weighted mean kinematic viscosity is $\bar{\nu }=\alpha \mathcal{R}{T}_{\ast }{I}_t/\Omega$, giving the displayed formula. This calculation does not assume gas-pressure domination: the closure depends on gas pressure while hydrostatic support can be mixed.

In an inviscid locally isothermal Keplerian disk, a radial normal mode of wavenumber $k$ has pressure variable $\xi$ proportional to a Probabilists' Hermite polynomial in $x=z/H$. Gaussian-weighted finite energy excludes the exponentially growing vertical branch. Polynomial termination gives $n=0,1,\dots$ and the local dispersion relation $({\hat{\omega }}^2-n{\Omega }_K^2)({\hat{\omega }}^2-{\Omega }_K^2)=k^2{c}_s^2{\hat{\omega }}^2$, where $\hat{\omega }=\omega -m{\Omega }_K$. At singular values introduced by eliminating velocity components, the original equations must also be checked. The $n=0$ propagating branch is an inertial-acoustic wave; $n=1$ has vertically uniform vertical velocity and odd pressure perturbation, permitting bending disturbances.

A steady Keplerian disk can transport angular momentum by viscous stress with zero mass flux. The radial drift vanishes, but the torque is constant and nonzero. An inner source and outer sink must supply its boundary angular-momentum exchange. This is distinct from a decretion flow, which transports mass outward.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

The additional conditions are $\boxed{\nabla \cdot \mathbf{u}=0,\nabla \cdot \mathbf{B}=0}$: incompressible flow and the solenoidal magnetic field constraint. The scalar $\Psi$ combines pressure per unit density with magnetic pressure per unit density and any conservative gravitational and centrifugal potentials. In particular, the Lorentz force identity separates magnetic tension from $-\nabla (B^2/2{\mu }_0\rho )$. Adding this latter term to the rotating-flow pressure potential gives the pressure appearing in the stated equation; the Coriolis force remains explicit.

Work in magnetic velocity units $\mathbf{B}/\sqrt{{\mu }_0\rho }$, and write $\mathbf{U}=-2Ax{\mathbf{e}}_y$ for the background shearing sheet flow. Its advective acceleration is zero, and $2\Omega {\mathbf{e}}_z\times \mathbf{U}=4\Omega Ax{\mathbf{e}}_x=-\nabla (-2\Omega A{x}^2)$, so the stated quadratic background pressure potential gives an exact balance. A spatially uniform mean magnetic field must evolve asThis is stretching of radial field into azimuthal field by differential rotation.

For the shearing wave phase $\Theta =\mathbf{k}(t)\cdot \mathbf{x}$, advection by the background givesIt vanishes identically ifThe changing wavevector describes the shear of initially planar wavefronts. The mean-field stretching and wavevector evolution cancel in the Alfvén frequency:Hence $\boxed{{\omega }_A=\mathbf{k}\cdot \bar{\mathbf{b}}\text{is constant}}$.

These solutions are exact finite-amplitude disturbances, not merely a linearization. To see this, let ${\mathbf{u}}^{\prime }=\mathrm{Re}(\mathbf{v}{e}^{i\Theta })$ and ${\mathbf{B}}^{\prime }=\mathrm{Re}(\mathbf{b}{e}^{i\Theta })$ in magnetic velocity units. The divergence constraints give $\mathbf{k}\cdot \mathbf{v}=\mathbf{k}\cdot \mathbf{b}=0$. Thus both real disturbances are pointwise perpendicular to k, whereas all their spatial derivatives are along k. Every quadratic term ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$, ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$, ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$ and ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$ vanishes, including terms involving conjugate amplitudes. This extends the exact incompressible shearing wave to an exact magnetic shearing wave. The gas pressure can compensate quadratic magnetic-pressure terms even though the combined $\Psi$ contains only its stated background and wave harmonic.

The remaining wave terms are therefore preciselyThe shear terms arise from perturbations acting on $\nabla \mathbf{U}$, the magnetic terms from the uniform field acting on the wave, and the last terms from diffusion. Pressure enforces the moving-wavevector velocity constraint: differentiating it givesUsing this pressure leaves $\mathbf{k}\cdot \mathbf{v}=0$ invariant. Differentiating $\mathbf{k}\cdot \mathbf{b}$ in the induction equation similarly leaves it zero. Thus the evolution is consistent with both constraints for initially transverse amplitudes.

Now take positive $\Omega$, so the stipulated condition implies $0<A<2\Omega$, and take magnetic Prandtl number $\nu /\eta =1$, with $\nu =\eta >0$. For a vertical wavevector $\mathbf{k}=k{\mathbf{e}}_z$, use the equal-diffusivity magnetorotational channel formThe two horizontal momentum equations, after adding them, give $\dot{V}=(A-\nu k^2)V$. Either component then gives $i{\omega }_{A}B=(A-2\Omega )V$. The first induction equation gives $({\partial }_t+\nu k^2)B=i{\omega }_{A}V$. These equations are simultaneously solved byThe second induction component gives the same relation, since $B_y=-B_x$. Exponential growth occurs when $\boxed{\nu k^2<A}$. The quarter-cycle phase between magnetic and velocity perturbations is required by the factor i in the Fourier equations.

For Keplerian rotation, expanding the local shear gives $2A=-(r{\Omega }^{\prime })=3\Omega /2$, hence $A=3\Omega /4$. Therefore ${\omega }_A^2=15{\Omega }^2/16$. With $v_{Az}={\bar{B}}_z/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{v}_{Az}$, the growth condition for this prescribed polarization becomesA weaker field would require a shorter wavelength to retain this mode's selected Alfvén frequency, making diffusive damping too strong. The required wavelength must also fit any actual vertical domain; no vertical boundary conditions are specified here.

The field bound is for the requested equal-component mode, rather than a universal necessary condition for all magnetorotational instability. This distinction matters if all vertical wavelengths are admitted. Indeed, let $s=\gamma +\nu k^2$ for a general horizontal polarization. Eliminating its two magnetic amplitudes gives $(s+{\omega }_A^2/s)v_x-2\Omega v_y=0$ and $(s+{\omega }_A^2/s)v_y+[2(\Omega -A)-2A{\omega }_A^2/s^2]v_x=0$. Their determinant givesFor a Keplerian disk, ${\kappa }^2={\Omega }^2$. As a concrete counterexample to a universal reading of the bound, in consistent dimensionless units take $\Omega =1$, $A=3/4$, $\nu =\eta =1$, $v_{Az}=0.1$, and $k=0.01$. The positive root gives $s\simeq 0.00173205$ and $\gamma =s-0.0001>0$, even though $v_{Az}^2=0.01<1.25$. This mode has a different polarization and a much longer wavelength. Real finite-height discs restrict such wavelengths; the PDF supplies no such restriction.

Finally, the horizontally averaged radial flux of azimuthal momentum is the sum of Reynolds stress and magnetic tension stress:Multiplication by the local radius gives an outward angular-momentum flux. Equal radial and azimuthal velocities give the largest positive product at fixed horizontal kinetic energy, while opposite magnetic components give the largest positive $-B_x{B}_y$ at fixed horizontal magnetic energy. Both follow from $2|ab|\le |a{|}^2+|b{|}^2$. This stress-maximizing magnetorotational channel polarization therefore makes both contributions reinforce outward angular momentum transport. It extracts shear energy at rate $-T_{xy}{\partial }_x{U}_y=2A{T}_{xy}>0$, explaining why these correlations are especially effective in driving accretion.

Reflection symmetry gives ${\Phi }_z(R,0)=0$. In a vertically stable thin disk, expand the vertical force as ${\Phi }_z(R,z)={\Omega }_z^{2}z+O(z^3)$, where ${\Omega }_z^2={\Phi }_{zz}(R,0)$ is the vertical epicyclic frequency squared. For local hydrostatic equilibrium and height-independent isothermal sound speed, $p=c_s^2\rho$ andThus $\rho ={\rho }_0\exp [-z^2/(2H^2)]$ with $H=c_s/{\Omega }_z$. Integrating the pressure and mass density through the disk gives the vertically integrated pressure of an isothermal disk:The force expansion is approximate in $H/R$, while $P=c_s^2\Sigma$ is exact for a height-independent sound speed.

In a nearly Keplerian disk, the vertical oscillation and radial epicyclic motion frequencies are close to the orbital frequency. The exact leading orbital differences are $\Omega -{\Omega }_z$ and $\Omega -\kappa$, soThus ${\omega }_1$ is the nodal precession rate of a slightly tilted orbit, and ${\omega }_2$ is the apsidal precession rate of a slightly eccentric orbit. Their signs distinguish prograde from retrograde precession. This also agrees with the sign convention in the complex tilt: a freely precessing ring obeys $W_t=i{\omega }_{1}W$.

For exactly Keplerian motion, ${\Omega }_z=\kappa =\Omega$, so ${\omega }_1={\omega }_2=0$ and $P/\Sigma =c_s^2$. In the local approximation, freeze the slowly varying coefficients and take $W,G\propto e^{i(kR-\omega t)}$. The inviscid warp equations of a nearly Keplerian disk becomeEliminating $G$ gives ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$. HenceThese are bending waves of a Keplerian disk, carrying changes in orbital tilt. The local approximation needs wavelength small compared with $R$; the secular thin-disk warp equations also describe the long-vertical-wavelength regime $H\ll \lambda \ll R$.

For a nonspinning hole, $a=0$ gives ${\omega }_1=0$. A steady disk warp then satisfies $G_R=0$. The zero inner torque forces $G=0$ throughout, and the second warp equation, with positive $P$, forces $W_R=0$. The outer orientation fixes $W(R)=W_{\infty }$ everywhere. An apsidal precession term by itself therefore does not require a tilt warp.

For nonzero spin, set the time derivatives to zero. The two equations giveEliminating the internal torque yieldsUse ${\omega }_2=3\Omega /r$, ${\omega }_1=2a\Omega r^{-3/2}$, ${\Omega }^2{R}^3=GM$, and $P\simeq {\Omega }^2\Sigma H^2$. The first coefficient becomes $GM\Sigma H^{2}r/12$ and the second becomes $2aGM\Sigma r^{-3/2}$. ThereforeThis is the stationary disk warp equation under the stated weak-precession approximations. The inner condition is $W_R(R_{\mathrm{in}})=0$ when its coefficient is finite and nonzero, because $G=iGM\Sigma H^{2}r{W}_R/12$.

Let $R_g=GM/c^2$, so $R=R_{g}r$, and impose $H=ϵR$, $\Sigma \propto R^{-3/4}$. Converting both radial derivatives carefully to $r$ and dividing out the common positive factor givesor $W_{rr}+9W_r/(4r)+(24a/{ϵ}^2)r^{-9/2}W=0$. For $a>0$, introduceSince $x_r$ is proportional to $r^{-9/4}$, the first-derivative term cancels exactly, leaving $W_{xx}+W=0$. Thus $W=A\cos x+B\sin x$.

The outer boundary $r\to \infty$ is $x\to 0$, giving $A=W_{\infty }$. The specified inner-radius relation is exactly $x_{\mathrm{in}}=\pi$. The inner zero-torque condition $W_x(\pi )=0$ then gives $B=0$. Hence the stationary bending warp of a Kerr disk isThe tilt magnitude is $\beta (r)=|W_{\infty }||\cos x|$. Far out it tends to the imposed tilt. Moving inward, it decreases to zero atthen grows again with opposite signed tilt, reaching $W(r_{\mathrm{in}})=-W_{\infty }$. All tilt vectors lie on the same line in the complex $W$ plane. Their azimuth is $\arg W_{\infty }$ outside the node and differs by $\pi$ inside; azimuth is undefined at the aligned node itself. The tilt vector remains smooth there. Thus this is an oscillatory standing bending warp, with equal outer and inner tilt magnitudes and reversed inner orientation, rather than monotone inner alignment.

The prescribed real inner radius assumes $a>0$: for $a<0$, its square root is not real, so that particular boundary prescription does not apply. With a separately chosen positive inner radius and negative spin, put $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$. The equation becomes $W_{xx}-W=0$, and the same outer tilt and zero inner torque giveThis negative-spin solution decreases smoothly toward the inner disk without a sign reversal. The specified positive-spin solution is the cosine profile above; the $a=0$ case is the flat solution already derived.

In a thin disk dominated by a central stellar mass, the vertical gravitational potential is harmonic to leading order, so its vertical acceleration is $-{\Omega }_0^{2}z$. For a locally isothermal background, $p=c_s^2\rho$ with $c_s$ independent of height. Vertical hydrostatic equilibrium therefore givesThis is the isothermal vertical structure in stellar gravity. On the upper half of the disk the downward gravity magnitude is $g={\Omega }_0^{2}z>0$; the density profile extends evenly through the midplane.

For an adiabatically displaced parcel kept in pressure equilibrium, the parcel's logarithmic density gradient is ${\gamma }^{-1}d\log p/dz$, whereas the environmental density gradient is $d\log \rho /dz$. A small upward displacement $\eta$ consequently produces a density contrast $\delta \rho /\rho =[{\gamma }^{-1}d\log p/dz-d\log \rho /dz]\eta$. Its buoyancy acceleration is $-g\delta \rho /\rho =-N^2\eta$. Thus the Brunt–Väisälä frequency isBoth background gradients equal $-z/H^2$ in the isothermal atmosphere, soAn isothermal background is therefore stably stratified for adiabatic perturbations with $\gamma >1$, but not buoyant for isothermal perturbations. In applying the adiabatic equation of state to a stratified atmosphere, its proportionality relates Lagrangian parcel changes: $\Delta p=(\gamma p/\rho )\Delta \rho$. Eulerian perturbations generally obey $p^{\prime }+\eta p_z=(\gamma p/\rho )({\rho }^{\prime }+\eta {\rho }_z)$ instead. Thus the printed simple proportionality must be interpreted as a parcel relation for the buoyancy calculation. For $\gamma =1$ in this isothermal background, the gradient terms cancel and the Eulerian proportionality holds too.

Now take isothermal perturbations, so $\gamma =1$ and $N^2=0$. The two supplied wave equations reduce toFor the nonsingular oscillatory case $\omega \ne 0$ and ${\omega }^2\ne {\Omega }_0^2$, substitute $u_z={\xi }_z/{\omega }^2$. With $x=z/H$, the resulting Hermite differential equation isThe exceptional frequencies must be treated through the original velocity-pressure equations rather than division by these factors.

The physical vertical condition is finite perturbation energy when integrated with the Gaussian density. In particular the pressure contribution is proportional to $\int e^{-x^2/2}|\xi {|}^{2}dx$ for nonzero frequency. Generic nonterminating solutions have an exponentially growing component at at least one vertical end and fail this condition; acceptable modes are the terminating Hermite polynomials. This is a finite-energy condition, not pointwise boundedness of the velocity in the vanishing-density atmosphere.

To prove the allowed values of $\alpha$, let the nonzero polynomial have degree $n$ and leading coefficient $a_n$. The highest power in ${\xi }_{xx}-x{\xi }_x+\alpha \xi$ has coefficient $(\alpha -n)a_n$. Hence $\alpha =n\in \{0,1,2,\dots \}$. Conversely, writing $\xi ={\sum }_j{a}_j{x}^j$ gives the recurrenceFor $\alpha =n$, choose the parity of $n$ and iterate until the series terminates at degree $n$. Thus all allowed Hermite vertical modes of an isothermal disk havewhere ${\mathrm{He}}_n$ is the Probabilists' Hermite polynomial. In particular, ${\mathrm{He}}_0=1$ and ${\mathrm{He}}_1=x$.

Let $K=kH$ and $s={\omega }^2/{\Omega }_0^2$. Eliminating the denominator for regular modes givesFor $n=0$, the nonzero-frequency solution isIts pressure variable is independent of height and $u_z=0$. These are inertial-acoustic waves, restored by epicyclic motion and pressure; at short radial wavelength they become ordinary acoustic waves. The multiplied dispersion polynomial also has $s=0$. This is not an extra oscillatory solution represented by nonzero $\xi$, since $\xi =i\omega p^{\prime }/\rho$ vanishes at zero frequency for finite pressure perturbation. The full horizontal equations have a zero-frequency balanced mode of an axisymmetric disk, with no radial velocity and azimuthal Coriolis acceleration balancing a radial pressure gradient. It must be recovered without dividing by $\omega$, rather than counted as a propagating wave. At $k=0$, the propagating branch limits to an epicyclic oscillation at $|\omega |={\Omega }_0$, another value at which the eliminated form must be interpreted by a limit.

For $n=1$, solve $(s-1{)}^2=K^{2}s$. The two branches areBoth signs of frequency occur on each squared-frequency branch. The upper branch is acoustic in the large-$|K|$ limit, $|{\omega }_{+}|\sim |k|c_s$; the lower branch is an inertial wave with $|{\omega }_{-}|\sim {\Omega }_0/|K|$. At $|K|\ll 1$, both frequencies approach ${\Omega }_0$, reflecting the equality of the radial and vertical epicyclic frequencies in a Keplerian disk. Here $\xi \propto z$, so $u_z={\xi }_z/{\omega }^2$ is independent of height: these modes have a coherent vertical displacement of the disk column, rather than a vertical breathing motion.

For a local non-axisymmetric disturbance proportional to $e^{i(kR+m\varphi -\omega t)}$, the material frequency is $\hat{\omega }=\omega -m{\Omega }_0$. In the same local radial-wave approximation, replace $\omega$ by $\hat{\omega }$. For $n=1,m=1$, this givesThe four frequencies can equivalently be written $\omega /{\Omega }_0=1\pm (\sqrt{K^2+4}\pm |K|)/2$, with independent choices of the two signs. This is a local dispersion relation, not a claim that Doppler substitution supplies an exact global non-axisymmetric solution in a differentially rotating disk. The radial-wave ordering neglects azimuthal gradients compared with radial gradients; $|k|R\gg 1$ can coexist with $|k|H\ll 1$ in a thin disk.

Write $w=\omega /{\Omega }_0$. For the two low inertial-frame frequency branches, $|w|\ll 1$, while the material frequency remains near $-{\Omega }_0$. Expanding the dispersion relation givesThereforeFor signed $k$, the two leading branches are conventionally labelled by $\pm kH/2$. These are bending waves of a Keplerian disk: an $m=1$ vertical displacement varies once around an annulus and corresponds to a small tilt of its orbital plane. A radial variation of the tilt is a warp. The waves transport the warp radially at group speeds $c_s/2$ in opposite directions. At zero radial wavenumber their low-frequency limit is a stationary rigid tilt; the apparent slow pattern does not mean that the orbiting gas has slow vertical motion in its own frame.