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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.

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.

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.

Choose the positive-frequency branch and real radial amplitude $u$. The azimuthal amplitude of the inertial-acoustic wave is $U_y=-i\Omega u/(2\omega )$, so taking real parts givesFor $|k|c_s/\Omega \ll 1$, $\omega =\Omega [1+O((k{c}_s/\Omega {)}^2)]$. Thus, to leading long-wavelength order,The quarter-cycle phase shift and the factor $1/2$ are the epicyclic motion relation in a Keplerian shearing sheet. This approximation drops the small frequency detuning; over times $|\omega -\Omega |t\gtrsim 1$ that phase approximation is not uniform. The associated fractional surface density of a disk perturbation is $\sigma /{\Sigma }_0=(ku/\omega )\cos (kx-\omega t)$, so linearization also requires $|ku/\omega |\ll 1$. The source's arbitrary real $u$ sets the wave phase/amplitude within that small-perturbation regime.

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

When $k_y=0$, $k_x$ is constant and the system becomesThusFor time dependence $e^{-i\omega t}$, the dispersion relation isExplicitly,and $\tilde{\delta }=2i{k}_x{\tilde{u}}_y/\Omega$. This is an axisymmetric inertial-acoustic wave: pressure supplies the $c_s^2{k}_x^2$ restoring term and epicyclic motion supplies the ${\Omega }^2$ term.

If the prescribed gas velocity is itself a free epicyclic motion solution, subtract its two momentum equations from the forced dust epicycle with aerodynamic drag equations. The velocity difference obeys a damped epicycle and decays as $e^{-ϵ\Omega t}$. Therefore dust approaches the gas velocity with identical phase and amplitude for every positive drag rate. At zero drag the difference does not decay, so the late-time limit is nonuniform as drag vanishes. A finite-wavelength inertial-acoustic wave is detuned from the free epicycle; matching its phase and amplitude in the weak-drag limit additionally requires $|\omega -\Omega |\ll ϵ\Omega$. For a gas density wave of frequency $\omega$, the exact radial transfer amplitude isobtained by solving the two dust equations and using $U_y=-i\Omega U_x/(2\omega )$. At $\omega =\Omega$ it is exactly one.

Besides the two propagating inertial-acoustic wave branches, a uniform barotropic Keplerian shearing sheet has a stationary branch balancing pressure against Coriolis acceleration. For radial wavenumber $k\ne 0$, setting $\omega =0$ in continuity gives $U_x=0$, and radial momentum gives the displayed azimuthal balance. The full determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. At $k=0$ the stationary branch is a uniform column-density change; it is not a travelling density wave.