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An equatorial inertia--gravity wave is a high-frequency equatorially trapped mode restored jointly by buoyancy and the Coriolis force. Its dispersion relation has eastward and westward branches lying outside the low-frequency equatorial Rossby wave branch.

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 $\omega$.

At $\omega =0$, the Kelvin relation gives $k=\pm i\sqrt{\alpha \gamma }/c$. Meridional trapping accepts onlywhich is the response east of the forcing and has zonal decay length $c/\sqrt{\alpha \gamma }$. The Rossby relations giveTrapping now accepts onlyso the response west of the forcing is a sum of equatorial Rossby waves; its longest zonal scale is $c/(3\sqrt{\alpha \gamma })$ for $n=1$.

Both families haveThus the damped equatorial Kelvin and Rossby response extends farther east than west. Increasing either damping rate shortens the zonal tails, while their ratio controls the common meridional trapping width.

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 a vertical plane wave $\chi =e^{imz}$, the vertical-mode equation givesConsequently the Boussinesq equatorial Kelvin wave and $n=1$ equatorial Rossby wave haveTheir meridional structures are those in part b with $L_e=[N/(\beta |m|){]}^{1/2}$.

Let the positive forcing frequency be $\omega =N\sin \theta$ with $\theta \ll 1$. Upward radiation into $z>0$ requires $m<0$ for both responses. The Kelvin wave has $k>0$ andIts group velocity points up and right along a ray making angle approximately $\theta$ above the horizontal, so it occupies the region to the right of the localized source. Its phase propagates down and right.

The $n=1$ Rossby wave has $k<0$ andIts group velocity points up and left along a ray making angle approximately $3\theta$ above the negative horizontal direction, so it occupies the region to the left. Its phase propagates down and left. Thus energy radiates upward and away from the source on both sides, while the vertical phase propagation is downward in both wave beams.

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.

Every forced propagating mode must have the imposed zonal frequencyFor $C>0$, only the equatorial Kelvin wave has the correct sign. Upward group velocity in the fluid $z>0$ selectsWith $K(y)=\exp [-\beta y^2/(2C)]$, its pressure field can be writtenHere $v_K=0$, $u_K={\varphi }_K/C$, and the complex vertical-velocity amplitude isThe lower boundary is matched only ifafter choosing the phase of $a_K$ to make the prescribed cosine amplitude real.

For $C<0$, the upward-radiating equatorial Rossby waves haveChoose each pair $({\hat{v}}_n,{\hat{\varphi }}_n)$ as in part c for this $m_n$. The propagating sum isThe other fields follow mode by mode from part c and ${\hat{w}}_n=-kC{m}_n{\hat{\varphi }}_n/N^2$. Thus the boundary matching condition is

For $C>0$, the space of upward-radiating wave profiles is only the one-dimensional Gaussian Kelvin profile, so a generic $W(y)$ cannot be matched by propagating waves. Its Kelvin projection radiates upward; the remaining forcing produces a balanced, vertically evanescent response trapped near the lower boundary. This is the equatorial analogue of the fact that quasi-geostrophic Rossby waves have westward rather than eastward phase propagation.