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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.