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An Edgeworth expansion approximates a probability density function by a normal distribution multiplied by Hermite polynomials encoding higher cumulants. For a centered variable of variance ${\sigma }^2$ and third cumulant ${\kappa }_3$,The correction follows from inverse-transforming the cubic term of the cumulant-generating function. It is an asymptotic approximation near the bulk of the distribution; truncation can give a negative result in far tails.

With $c=\sqrt{gH}$, beta plane parameter $\beta >0$, and Hermite polynomial index $n\ge 0$, meridional trapping of the nonzero meridional-velocity branch requiresIts meridional velocity is proportional to $H_n(Y)e^{-Y^2/2}$, with $Y=y\sqrt{\beta /c}$. Exceptional roots with ${\omega }^2=c^2{k}^2$ must be checked in the original equations, since eliminating velocity and height divides by this factor.

An equatorial wave is trapped near the equator by the sign change of the Coriolis parameter. Its meridional structures are Gaussian-weighted Hermite polynomials.

Gauss-Hermite quadrature uses the roots of a degree-$N$ Hermite polynomial as nodes for an integral with Gaussian weight. Its positive interpolation weights make it exact for every polynomial of degree at most $2N-1$.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.

For the quantum harmonic oscillator, the Time-independent Schrodinger equation iswith $m,\omega ,\hslash >0$. The given dimensionless coordinate yields ${\psi }_{\xi \xi }+(2E/(\hslash \omega )-{\xi }^2)\psi =0$. Substituting the Gaussian factor givesBy the supplied normalizability fact, $f$ is a nonzero polynomial. If its degree is $n$, comparison of the leading coefficient forces $2E/(\hslash \omega )-1=2n$. Conversely, at this value the coefficient recurrenceproduces a degree-$n$ polynomial with parity $n$: start with a nonzero constant for even $n$ or a nonzero linear coefficient for odd $n$, and the recurrence terminates at degree $n$. This is a multiple of the Hermite polynomial. Its product with the Gaussian is square integrable and can be normalized. Therefore all and only the allowed energies are

Assume $\beta >0$, $c=\sqrt{gH}$, and use complex normal mode amplitudes. The zonal momentum and continuity equations giveWhen ${\omega }^2\ne k^2{c}^2$, solving these two linear equations gives

For $\hat{v}=0$, a nonzero solution requires $\omega =\pm kc$ and $\hat{\eta }=\pm c\hat{u}/g$. The remaining momentum equation gives ${\hat{u}}^{\prime }=\mp \beta y\hat{u}/c$. Only the positive sign decays at both infinities, so the equatorial Kelvin wave isIts phase and group velocities are both eastward with speed $c$.

For nonzero meridional velocity, substitute the eliminated amplitudes into $-i\omega \hat{v}+\beta y\hat{u}=-g{\hat{\eta }}^{\prime }$. The result isSet $Y=y\sqrt{\beta /c}$. The decaying Hermite polynomial modes have $\hat{v}=V_n{H}_n(Y)e^{-Y^2/2}$, up to normalization, and their eigenvalue gives the equatorial shallow-water dispersion relation

For $n=0$, the quadratic in $k$ factors asThe admissible Yanai wave root is $k=\omega /c-\beta /\omega$. With $\hat{v}(0)=1$, substitution and cancellation giveThese expressions satisfy the original equations even at the isolated point where the two algebraic roots coincide and the earlier elimination formula has a zero denominator. The other root is not an independent admissible branch: its generic singular reconstruction fails the trapping conditions.

In the zonal momentum equation, the acceleration term is $-i\omega \hat{u}=({\omega }^{2}y/c)\hat{v}$, the Coriolis term is $-\beta y\hat{v}$, and the pressure term is $-ikg\hat{\eta }=k\omega y\hat{v}$. When ${\omega }^2\ll \beta c$, acceleration is small and the leading balance is between Coriolis acceleration and the pressure gradient. When ${\omega }^2\gg \beta c$, acceleration balances pressure and the Coriolis term is small; the branch tends to an eastward gravity wave.

For $n\ge 1$, the fixed-frequency zonal roots areThe radicand is nonnegative exactly when $s={\omega }^2/(\beta c)$ satisfies $s^2-(2n+1)s+1/4\ge 0$. This gives the frequency gap for higher equatorial wave modes:

For the forcing frequency $\omega =\frac{1}{2}\sqrt{\beta c}$, we have $s=1/4$. Already the $n=1$ low-frequency upper bound is $3/2-\sqrt{2}<1/4$, and that bound decreases with $n$, while the high-frequency bound increases. Hence no $n\ge 1$ mode propagates. The available real-wavenumber branches areFor the equatorial Kelvin wave, $c_{g,K}=c$. For the Yanai wave, differentiating $k=\omega /c-\beta /\omega$ givesThus both propagating responses are detected east of the localized forcing, provided it has nonzero projection onto their meridional structures. The Yanai phase travels westward at this frequency, but its wave packet and energy travel eastward. Higher modes contribute only an evanescent response near the forcing.

LetSolving zonal momentum and incompressibility for $\hat{u}$ and $\hat{\varphi }$ in terms of $\hat{v}$ givesSubstitution into meridional geostrophic balance cancels the terms involving $D$ and yields

SetThe equation becomesUsing the Hermite polynomial eigenvalues givesandThe associated pressure amplitude isFor $n=0$, the denominator used to solve for $\hat{u}$ and $\hat{\varphi }$ vanishes because ${\omega }_0^2{m}^2/N^2=k^2$. The inversion therefore assumed precisely the condition that excludes that degenerate case; it must be analyzed separately and is not a member of this equatorial Rossby wave family.