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A plane-parallel atmosphere with isothermal sound speed $c_s$ in gravity $-{\Omega }^{2}z$ has mass density ${\rho }_0{e}^{-z^2/(2H^2)}$, $H=c_s/\Omega$. Above a rigid base, vertically directed adiabatic fluid perturbations satisfy $w^{\prime \prime }-z{w}^{\prime }/H^2+({\sigma }^2/{\Omega }^2-1)w/(\gamma H^2)=0$ and $w(0)=0$. Set $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The substitution $y=e^{-x^2/4}w$ converts finite Gaussian-weighted kinetic energy into square integrability for $-y^{\prime \prime }+x^{2}y/4=(\lambda +1/2)y$. Odd extension across the base and the oscillator lowering operator force $\lambda$ to be an odd positive integer. Thus $w_n$ is proportional to the Probabilists' Hermite polynomial ${\mathrm{He}}_{2n+1}(z/H)$. These are vertical acoustic waves; a general stably stratified atmosphere also admits internal gravity waves with horizontal structure.

Fix the sign convention first: depth $z$ increases downward, while the vertical displacement amplitude $\xi$ is positive upward. Write $\Pi =\delta p$ for the Lagrangian pressure perturbation, and $p_1,{\rho }_1$ for Eulerian fluid perturbations. Hydrostatic equilibrium gives $dp/dz=\rho g$, and the adiabatic relation gives $\delta \rho =\Pi /c^2$. Since the vertical displacement vector has component $-\xi$ along increasing $z$,In the Cowling approximation, horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$. The mass conservation equation $\delta \rho =-\rho (i\mathbf{k}\cdot {\mathbf{\xi }}_h-{\xi }^{\prime })$ and vertical momentum ${\omega }^2\rho \xi =-p_1^{\prime }+g{\rho }_1$ then eliminate the horizontal displacement. The plane-parallel Cowling displacement-pressure equations areFor ${\omega }^2=gk$, the second equation admits $\Pi =0$ and the first reduces to ${\xi }^{\prime }=-k\xi$. Thus the plane-parallel stellar f-mode hasIts Eulerian eigenfunctions are $p_1=\rho g\xi$ and ${\rho }_1={\rho }^{\prime }\xi$, while $\delta \rho =0$. Hence it has no Lagrangian compression even in a stratified star. It decays inward but grows upward, explaining the failure of a pure small-amplitude continuation to $z\to -\infty$.

For the lower part of the matched chromospheric interfacial mode, put ${\omega }^2=gk(1-\epsilon )$, $\xi =e^{-k(z-z_0+\epsilon \varphi )}$ and $\Pi =\epsilon f\xi$. The pressure amplitude is of order $\epsilon$ because the limiting f-mode has identically zero Lagrangian pressure amplitude; the frequency correction perturbs that zero solution. Expand $g{k}^2/{\omega }^2=k(1+\epsilon )+O({\epsilon }^2)$ and ${\omega }^2-g^2{k}^2/{\omega }^2=-2gk\epsilon +O({\epsilon }^2)$. The leading equations areExclude the lower homogeneous component $f\propto e^{2kz}$, which would make $\Pi$ grow into the star. Integrating the remaining equation with its integrating factor givesThe accompanying correction is $\varphi (z)={\int }_{z_0}^z[1+(f/\rho )(1/g-1/(k{c}^2))]ds$, with the integrand evaluated at $s$. This fixes $\varphi (z_0)=0$ and determines the lower eigenfunctions to the stated order.

For the upper part, distinguish the rescaled sound speed by writing $\rho =\epsilon \hat{\rho }$ and $c^2={\epsilon }^{-1}{\hat{c}}^2$. Set $\Pi =\epsilon \psi$, $\psi =A{e}^{k(z-z_0+\epsilon \varphi )}$ and $\xi =\mu \psi$. To leading order the first displacement equation becomes ${\xi }^{\prime }=-k\xi -k\psi /(g\hat{\rho })$, while the pressure equation gives ${\psi }^{\prime }=k\psi +\epsilon [k\psi +2gk\hat{\rho }\xi ]+O({\epsilon }^2)$. ConsequentlyThe component $\mu \propto e^{-2kz}$ would give $\xi \propto e^{-kz}$, which diverges upward. Remove it by the upper boundary condition. The leading upper solution isIntegrating ${\varphi }^{\prime }$ with $\varphi (z_0)=0$ gives the first correction to the upper exponential. The displayed $\xi =\mu \psi$ satisfies the required decay only for compatible upper density profiles; integral convergence alone is not the full boundary condition.

To match, define the positive weighted integralsThe lower normalization is $\xi (z_0)=1$ and $f(z_0)=-2gk{I}_b$. Continuity of Lagrangian pressure perturbation makes the upper $A=f(z_0)$; continuity of fluid displacement then requires $\mu (z_0)A=1$. Since $1/\hat{\rho }=\epsilon /\rho$, the upper integral gives $\mu (z_0)=-(k/g)\epsilon I_a$. The two matching conditions therefore giveAs a sign and normalization check, two incompressible constant-density layers give $I_a=(2k{\rho }_a{)}^{-1}$ and $I_b={\rho }_b/(2k)$, so $\epsilon \simeq 2{\rho }_a/{\rho }_b$. This agrees with expansion of the exact interfacial gravity-wave dispersion relation ${\omega }^2=gk({\rho }_b-{\rho }_a)/({\rho }_b+{\rho }_a)$.

For validity, require $\epsilon \ll 1$, convergent weighted integrals, a matching level separating the low-density high-sound-speed atmosphere from the denser envelope, and small accumulated corrections $\epsilon k|\varphi |$ over the regions carrying the matching integrals. The upper expansion also needs $g/(k{c}^2)\ll 1$. The resulting upper displacement must tend to zero: for example if $\hat{\rho }\propto e^{a(z-z_0)}$ at great height, its leading amplitude is proportional to $e^{(k-a)(z-z_0)}$, requiring $a<k$; the weaker condition $a<2k$ merely makes the upper integral converge. Plane geometry, constant $g$ and the Cowling approximation must be appropriate on the wavelength scale. The physical mode amplitude must remain small compared with both its wavelength and local background scale heights, so the adiabatic fluid perturbation remains linear. If these requirements fail, neither the matching formula nor continuation of the leading expansions is controlled.

The equilibrium isothermal equation of state is $p=c_s^2\rho$. Hydrostatic equilibrium gives $p_z=-\rho {\Omega }^{2}z$, henceHere $\Omega \ne 0$ is assumed. This is the same local harmonic-gravity balance as isothermal vertical structure in stellar gravity.

For $\gamma >1$, this equilibrium is stably stratified under adiabatic fluid perturbations. Its squared buoyancy frequency isA general perturbation can therefore support acoustic waves restored by compression and internal gravity waves restored by buoyancy, with nonoscillatory entropy or vortical disturbances also possible. The prescribed velocity is purely vertical and horizontally uniform. It describes vertical acoustic waves; it does not contain the horizontally varying circulation required for an oscillatory internal-gravity branch. For zero horizontal wavenumber, that branch has zero frequency. An isothermal equilibrium must not be confused with an isothermal perturbation: the latter would remove the buoyancy restoring force.

Introduce the fluid Lagrangian displacement $\xi (z)e^{i\sigma t}$, so $w=i\sigma \xi$. Use $\delta \rho ,\delta p$ for fluid Eulerian perturbations and $\Delta \rho ,\Delta p$ for fluid Lagrangian perturbations. The continuity equation and the adiabatic relation areThe fixed external gravitational potential is not perturbed. The vertical linearized Euler equations consequently giveSubstituting the preceding relations, and using $p_z=-\rho g$ and $p_{zz}=-{\rho }_{z}g-\rho g_z$, reduces its right-hand side toSince $g={\Omega }^{2}z$ and $g_z={\Omega }^2$, the same equation holds for $w$:The rigid fixed base supplies the essential condition $w(0)=0$. At infinity the condition is finite kinetic energy, not bounded velocity amplitude; a polynomial can grow while the Gaussian mass density makes its energy finite.

Put $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The equation becomes the Hermite differential equation in its probabilists' convention,For $w={\sum }_{j\ge 0}{a}_j{x}^j$, its recurrence isThe base condition makes $a_0=0$ and therefore every even coefficient zero. An odd series terminates at degree $2n+1$ when $\lambda =2n+1$. It remains to prove that a nonterminating odd solution cannot have finite energy; termination alone would not establish the requested exhaustion of modes.

For that purpose, set $y=e^{-x^2/4}w$. The finite-energy condition is $y\in L^2(0,\infty )$, and its equation isBecause the coefficients are even and the solution is odd at the base, extend $y$ oddly to the whole real line. Define lowering and raising operatorsIntegration by parts gives $\Vert ay{\Vert }^2=\lambda \Vert y{\Vert }^2$, so $\lambda \ge 0$. These identities apply to the square-integrable solution: multiplying its equation by a compactly cut off conjugate and taking larger cutoffs bounds $y^{\prime }$ and $xy$ in $L^2$, and removes the endpoint terms. The same argument applies to each lowered eigenfunction.

If $\lambda$ is not a nonnegative integer, repeatedly apply $a$. Each application lowers $\lambda$ by one and remains nonzero whenever its current value is positive, by the norm identity. After finitely many applications it would yield a nonzero square-integrable eigenfunction with negative $\lambda$, contradicting that identity. Thus $\lambda =N$ is a nonnegative integer. At $N=0$, $ay=0$ gives $y\propto e^{-x^2/4}$, which is even. Lowering reverses parity, so an initially odd eigenfunction can reach this even ground state only if $N$ is odd. Consequently $N=2n+1$. The recurrence then terminates and uniquely determines the odd solution from $a_1$.

The complete adiabatic vertical modes of a Gaussian atmosphere are thereforeHere ${\mathrm{He}}_j$ is the Probabilists' Hermite polynomial. For example the first shapes are $x$, $x^3-3x$, and $x^5-10x^3+15x$, with frequencies squared ${\Omega }^2(1+\gamma )$, ${\Omega }^2(1+3\gamma )$, and ${\Omega }^2(1+5\gamma )$. Either sign of each real frequency gives a time-harmonic solution. Gaussian moments make every displayed polynomial have finite kinetic energy. Without the fixed-base condition, finite energy alone would also admit even polynomials, which is why that condition is necessary.