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In dimensionless equatorial wave units with gravity wave speed and beta plane coefficient one, a Fourier mode $e^{i(kx-\omega t)}$ obeys $\omega u-k\eta =iyv$ and $\omega \eta -ku=-i{v}^{\prime }$. Inverting this two-by-two linear system gives the displayed wave polarization, provided ${\omega }^2\ne k^2$. Inserting it in the meridional momentum equation gives the Hermite differential equation after a Gaussian factor is removed. At singular values the original equations, rather than these quotient formulae, determine whether a trapped mode exists.

For the whole-line linear complex Ginzburg-Landau equation with real positive diffusion $\gamma$ and $\mu (x)={\mu }_0-\nu {ϵ}^2{x}^2$, removing drift and scaling $\xi =\sqrt{ϵ}(\nu /\gamma {)}^{1/4}x$ gives the Hermite differential equation. Decay at both infinities quantizes the spectral parameter to $2j+1$. The strongest mode is $j=0$, giving global instability iff ${\mu }_0>U^2/(4\gamma )+ϵ\sqrt{\gamma \nu }$. The confinement correction vanishes in the slow-profile limit, recovering the homogeneous absolute wave-packet instability threshold.

Use the unitary Fourier transform with phase $e^{-i\lambda x}$. It commutes with $-d^2/d{x}^2+x^2$, because differentiation transforms to multiplication by $i\lambda$ and multiplication by $x$ transforms to $id/d\lambda$. The Hermite differential equation has a unique polynomial solution of each degree up to scale, and multiplying by the Gaussian gives a Hermite function. Transforming a polynomial times this Gaussian again gives a polynomial times the Gaussian of the same degree; comparison of leading coefficients gives the eigenvalue $(-i{)}^n$.

For the Hermite differential equation $f^{\prime \prime }-2\xi f^{\prime }+(ϵ-1)f=0$ with real $ϵ$, the wavefunction $e^{-{\xi }^2/2}f$ is square integrable on the real line only if $f$ is a polynomial. To prove this without assuming that even and odd tails cannot cancel, project a square-integrable solution onto its even and odd parts first. Each projection remains square integrable and satisfies the same equation.

In either nonterminating parity sequence, $b_j=a_{2j+p}$ obeysThe tail has constant complex phase. For any fixed $0<\delta <1/2$, comparison with $(1-\delta )/(j+1)$ bounds its coefficient magnitudes below by $K(1-\delta {)}^j/j!$ for sufficiently large $j$. Thus on the positive real axis the tail dominates a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of finitely many terms. The Gaussian function factor leaves exponential growth $e^{(1/2-\delta ){\xi }^2}$, contradicting the square-integrable condition. Hence each nonzero parity sequence terminates. Its top coefficient forces $ϵ=2n+1$, giving Hermite oscillator quantization and a Hermite polynomial of degree $n$.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

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.

Substitution of the power series, with termwise differentiation, gives the coefficient identityThus $a_0$ and $a_1$ are arbitrary, and all power series solutions of the Hermite differential equation areAn empty product is one. For a nonterminating branch the ratio of consecutive terms in its own parity subsequence is asymptotic to $|x{|}^2/j$, hence tends to zero for every fixed $x$. The ratio test proves convergence everywhere; the resulting power series can be differentiated termwise and indeed solve the differential equation. A terminating branch is already a polynomial.

The boundary condition $W(0)=0$ forces $a_0=0$; the solution is odd. Apart from the zero solution, which works for every $E$, a polynomial occurs exactly when an odd index $d$ satisfies $2d-E=0$. Before that index the odd coefficients are nonzero, and after it they vanish. Thus the odd polynomial solutions of the Hermite differential equation haveEquivalently, comparison of leading polynomial coefficients gives $E=2\mathrm{deg}W$, proving that no other nonzero polynomial is possible. Normalizing by $a_1=1$, the polynomials of degree below six areEach can be multiplied by an arbitrary nonzero constant; these are normalized odd Hermite polynomials.

Let $L=-d^2/d{x}^2+2xd/dx$. The forcing in the final inhomogeneous linear differential equation is precisely the displayed degree-five eigenfunction $P_5$, with $L{P}_5=10P_5$. HenceDirect differentiation gives $LW=x-4x^3/3+4x^5/15$, and $W(0)=0$. Among polynomial solutions this choice is unique: a polynomial in the kernel of $L$ must be constant, and its value at zero eliminates that constant.

First assume the usual diffusion regime $\gamma >0$, with $U$ real; the complex-diffusion extension is described below. Substitution of a global normal mode givesSet $a=e^{Ux/(2\gamma )}b$ to remove the first derivative, then put $\xi =\sqrt{\epsilon }(\nu /\gamma {)}^{1/4}x$. The resulting oscillator equation isWriting $b=e^{-{\xi }^2/2}H(\xi )$ gives the Hermite differential equation, $H^{\prime \prime }-2\xi H^{\prime }+(\Lambda -1)H=0$. For a power series $H={\sum }_{j\ge 0}{c}_j{\xi }^j$, its recurrence isTermination at degree $n$ requires $\Lambda =2n+1$. If it does not terminate in the appropriate parity, the series has a growing Gaussian component at one infinity; the solution decaying at one end then fails at the other. The Hermite oscillator quantization can also be seen from the self-adjoint operator $-d^2/d{\xi }^2+{\xi }^2$, whose square-integrable eigenfunctions are the Hermite functions. Their spectrum is $2n+1$ with $n=0,1,2,\dots$. The ground state $H_0=1$ shows that the paper's phrase “positive integer” must include zero.

The global modes of a quadratically confined Ginzburg-Landau equation are thereforeTheir spatial functions are proportional to $e^{Ux/(2\gamma )}{e}^{-{\xi }^2/2}{H}_n(\xi )$; the quadratic Gaussian decay dominates the linear drift factor at both infinities. For real coefficients, the most dangerous mode is $n=0$. The system has global hydrodynamic instability precisely whenEquality is marginal and smaller ${\mu }_0$ gives decay of all these modes. The finite-width confinement correction explains why a locally absolutely unstable region need not immediately support a growing global mode.

If complex diffusion is intended, impose the parabolic condition $\mathrm{Re}\gamma >0$ and choose $\sqrt{\nu /\gamma }$ with positive real part. The same formulas continue along the rotated $\xi$ ray, with the corresponding branch of $\sqrt{\gamma \nu }$; instability is determined by the real part of the square bracket. Arbitrary constant $\gamma$, especially negative diffusion, does not justify the asserted whole-line Hermite decay or a well-posed evolution.

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.

Retain the factor $ϵ$ in the last temporal phase, as printed in the original PDF. Let $a=U/(2\gamma )$ and put $\eta =e^{ax}b(\xi )e^{st}$ with $s=-i{\omega }_0-iϵ{\omega }_1$. The drift-removing substitution for a Dirichlet Sturm-Liouville problem cancels the first derivative:With $\xi =\sqrt{ϵ}fx$, this becomesChoose the positive scale $f=(\nu /\gamma {)}^{1/4}$. To leave an order-one spectral parameter, cancel the numerator's order-one term. ThenThe assumed decaying solutions of the Hermite differential equation require $\lambda =2j+1$, $j=0,1,\dots$. They are proportional to $e^{-{\xi }^2/2}{H}_j(\xi )$, with physicists' Hermite polynomials. Therefore the global modes of a quadratically confined Ginzburg-Landau equation haveFor every fixed positive $ϵ$, multiplying the Hermite-Gaussian by $e^{Ux/(2\gamma )}$ still gives decay at both infinities. The strongest mode is $j=0$, so the whole-line system is globally unstable precisely whenEquality is a neutral ground mode. This spectrum is exact for the stated quadratic profile, even though the notation organizes it as an expansion in $ϵ$.

The homogeneous absolute-instability threshold in part (a) is $\mu =U^2/(4\gamma )$. The global threshold approaches that value as the profile varies increasingly slowly, but at finite $ϵ$ the quadratic confinement adds the loss $ϵ\sqrt{\gamma \nu }$. A locally absolutely unstable patch at the profile maximum is necessary but must be strong and broad enough to overcome that loss. Merely having positive local temporal growth ${\mu }_0>0$, which can be only convective wave-packet instability, is insufficient for a growing decaying global normal mode. This connects the local impulse classification to the discrete global spectrum.

The Time-dependent Schrödinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.

For the equatorial Kelvin wave, set $\hat{V}=0$. The zonal momentum and mass equations give ${\omega }^2=k^2{c}^2$. Meridional trapping selects the eastward branchfor whichThe westward algebraic branch would grow away from the equator and is rejected.

For $\hat{V}\ne 0$, let $D={\omega }^2-k^2{c}^2$. The zonal momentum and mass equations giveSubstitution in meridional geostrophic balance and use of the stated Hermite differential equation gives the trapped Equatorial Rossby-wave dispersion relationFor $n=1$, define $L_e=(c/\beta {)}^{1/2}$, $Y=y/L_e$, and takeThen

Without imposing meridional geostrophic balance, the equatorial shallow-water modes satisfy the Matsuno cubicA dispersion diagram therefore contains high-frequency eastward and westward equatorial inertia--gravity waves as well as westward equatorial Rossby waves, plus the separate straight equatorial Kelvin wave branch $\omega =kc$. The Rossby curves approach $-kc/(2n+1)$ only in the long-wave limit and bend toward zero like $-\beta /k$ at short wavelength. The geostrophic model retains the Kelvin and long-wave Rossby lines but filters the inertia--gravity modes and misses Rossby-wave dispersion at larger $|k|$.

For the normalized second-order linear differential equation $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, an ordinary point $z_0$ has analytic $P,Q$ near $z_0$. At a regular singular point, which is not ordinary, $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ are analytic. Ordinary points admit two independent convergent power series solutions. Regular singular points instead call for the Frobenius method, which permits leading powers and, in some cases, logarithms.

Here zero is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{z}^n$ into the Hermite differential equation and equate coefficients:The even and odd initial coefficients are independent, givingThese are the power series solutions of the Hermite differential equation, with the present parameter equal to twice the parameter in that article. The ratio of successive terms within each nonterminating parity is asymptotically $2|z{|}^2/n$, so both series have infinite radius of convergence.

For a nonzero polynomial of degree $m$, its highest coefficient in $-2z{y}^{\prime }+\lambda y$ forces $\lambda =2m$. Conversely at that value the matching parity terminates. HenceThe polynomial solutions are scalar multiples of the Hermite polynomials. Two examples are $\boxed{\lambda =0:y=1;\lambda =2:y=z}$; direct substitution verifies both. The other independent solution generally remains nonpolynomial.

The product rule gives ${\psi }^{\prime }=e^{-x^2/2}(y^{\prime }-xy)$ and ${\psi }^{\prime \prime }=e^{-x^2/2}[y^{\prime \prime }-2x{y}^{\prime }+(x^2-1)y]$. Substituting in the Schrödinger equation cancels the $x^2$ and constant terms, leaving the Hermite differential equationFor $y={\sum }_{n\ge 0}{a}_n{x}^n$, matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data $(y(0),y^{\prime }(0))=(1,0)$ and $(0,1)$, respectively, their first three terms areEach nonterminating parity series has infinite radius of convergence: the ratio of successive terms of that parity has magnitude asymptotic to $2|x{|}^2/n$. Their initial data prove linear independence. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.