= Solution
Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced <shallow water equations> on the equatorial <beta plane> are
$$
u_t+uu_x+vu_y-\beta yv=-gh_x,\qquad
v_t+uv_x+vv_y+\beta yu=-gh_y,\qquad
h_t+(hu)_x+(hv)_y=0.
$$
Linearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{gh_0}$ 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> become
$$
u_t-yv=-\eta_x,\qquad
v_t+yu=-\eta_y,\qquad
\eta_t+u_x+v_y=0.
$$
The 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 are
$$
\omega u-k\eta=iyv,\qquad
\omega\eta-ku=-iv',\qquad
-i\omega v+yu=-\eta'.
$$
For $v=0$ and nonzero $k$, the first two require $\omega^2=k^2$. The root $\omega=k$ gives $u=\eta$, and meridional balance becomes $u'=-yu$. Hence the <equatorial Kelvin wave> is
$$
\boxed{\omega=k,\qquad v=0,\qquad u=\eta=C e^{-y^2/2}.}
$$
It travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta=-u$ and $u'=yu$, so $u\propto e^{y^2/2}$ is not trapped.
For $v\ne0$, $\omega\ne0$ and $\omega^2\ne k^2$, inversion of the first two equations gives the <equatorial wave velocity polarization>
$$
u=\frac{i(\omega yv-kv')}{\omega^2-k^2},\qquad
\eta=\frac{i(kyv-\omega v')}{\omega^2-k^2}.
$$
Substitute in the remaining momentum equation. Multiplication by $(\omega^2-k^2)/i$ leaves
$$
-\omega(\omega^2-k^2)v+\omega y^2v+kv-\omega v''=0,
$$
so
$$
v''+(\lambda-y^2)v=0,\qquad
\lambda=\omega^2-k^2-\frac k\omega.
$$
Writing $v=e^{-y^2/2}H(y)$ yields $H''-2yH'+(\lambda-1)H=0$, 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 $a=d/dy+y$, $a^\dagger=-d/dy+y$ and $\mathcal H=-d^2/dy^2+y^2=a^\dagger a+1$. <Integration by parts> gives $\lambda\ge1$ and $\|av\|^2=(\lambda-1)\|v\|^2$. If $av\ne0$, 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 by
$$
\boxed{
v=C H_n(y)e^{-y^2/2},\qquad
\omega^2-k^2-\frac k\omega=2n+1,\qquad n=0,1,2,\ldots.}
$$
Insert this $v$ and its <derivative> in the polarization formulae to obtain $u$ and $\eta$. For $n\ge1$, 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> as
$$
(\omega+k)(\omega^2-k\omega-1)=0.
$$
The physically admissible quadratic gives the <mixed Rossby-gravity wave>, equivalently the <Yanai wave>:
$$
\boxed{\omega-k-\frac1\omega=0,\qquad
v=C e^{-y^2/2},\qquad u=\eta=i\omega yv.}
$$
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 $\omega=-k\ne0$. The original equations give $u+\eta=-iyv/k$ and $v'=-yv$. Put $\delta=u-\eta$; meridional momentum then gives
$$
\delta'-y\delta=i(2k-k^{-1})v,\qquad
(e^{-y^2/2}\delta)'=i(2k-k^{-1})C e^{-y^2}.
$$
Decay 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 $\omega$.
Back to article page
