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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 $\omega$.

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.

Put $A=\gamma -i\omega$ and $B=\alpha -i\omega$. For $\hat{v}=0$,Continuity gives the damped equatorial Kelvin wave dispersion relationOnly roots satisfying $\mathrm{Re}(i\beta k/A)<0$ are trapped; for weak damping this is the eastward branch $k\mathrm{Re}\omega >0$. The other algebraic root makes the Gaussian grow with $|y|$ and must be rejected.

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.

For an equatorial Kelvin wave, set $\hat{v}=0$. The zonal and meridional momentum equations givesoHydrostatic balance and buoyancy evolution give $\hat{\sigma }=im\hat{\varphi }$ and $\hat{w}=-\omega m\hat{\varphi }/N^2$. Continuity then givesBecause the mode must decay as $|y|\to \infty$, one needs $k/\omega >0$. With the stipulated $\omega >0$, the acceptable branch is thereforeThe branch $\omega =-Nk/|m|$ would require $k<0$ when $\omega >0$, making the Gaussian exponent positive and the mode unbounded; it is not an equatorially trapped solution.

For real $k,m,\omega$,Upward group propagation has $m<0$, while a Kelvin wave has $k>0$, so $\overline{u^{\prime }{w}^{\prime }}>0$. Dissipation makes this eastward momentum flux decrease with height; therefore $-{\partial }_z\overline{u^{\prime }{w}^{\prime }}>0$ and the wave exerts an eastward force on the mean flow.

Put $v=0$ and use amplitudes proportional to $e^{i(kx+mz-\omega t)}$. Zonal momentum giveswhile incompressibility, buoyancy evolution, and hydrostatic balance giveHenceMeridional geostrophic balance requiressoDecay as $|y|\to \infty$, together with $\beta >0$ and $k>0$, requires $\omega >0$. ThereforeThe negative-frequency root makes the Gaussian exponent positive and the solution diverge away from the equator. The acceptable branch is the eastward equatorial Kelvin wave.

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.