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.

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.

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.

The variables $u,v,w$ are eastward, northward, and upward velocity, $\varphi =p^{\prime }/{\rho }_0$ is the pressure perturbation divided by reference density, $\sigma$ is buoyancy, $N$ is the constant buoyancy frequency, and $\beta y$ is the equatorial approximation to the Coriolis parameter. The equations express, respectively:

Together they are the hydrostatic Boussinesq approximation for long equatorial waves.

Because $v={\psi }_x=(g/f_0){\eta }_x$, the mean boundary-current velocity is exactly related to the height change byFor the e-folding convention used in part d,Its robust narrow-layer scaling isThus weaker drag makes the current proportionally narrower and faster. Their product is independent of $\gamma$ to leading order:Mass conservation requires the narrow return transport to cancel the broad Sverdrup balance transport. The meridional gradient of planetary potential vorticity makes a frictional closure possible on the western side and produces western intensification; bottom drag supplies the vorticity sink that permits fluid parcels to cross potential-vorticity contours there.

No normal flow makes $\psi$ constant on the connected basin boundary; set that constant to zero. PutThe interior solution is $A(L-x)$. The decaying boundary correction that enforces $\psi (0,y)=0$ isThus, to leading order in $\delta /L$,

If $\delta$ denotes the e-folding width literally, thenSince a boundary-layer width is only defined up to an order-one factor, matching at any point satisfying $\delta \ll x\ll L$ gives the convention-independent leading jumpIt is the height difference needed to return the broad Sverdrup transport in a narrow western boundary current.

In a narrow western layer, $x$ derivatives dominate $y$ derivatives. Subtracting the forced interior balance leavesIf the layer has width $\delta$, then ${\psi }_x=O(\psi /\delta )$ and ${\psi }_{xx}=O(\psi /{\delta }^2)$. Balancing the beta effect with linear bottom drag givesand therefore the Stommel boundary layer width isThis scaling assumes $\delta \ll L$ and that along-boundary variations occur on the basin scale.

The eastern wall has no normal flow, so the quasi-geostrophic streamfunction is constant there. Choose $\psi (L,y)=0$. Integrating the Sverdrup relation givesSince geostrophic balance gives $\psi =g\eta /f_0$,If this interior solution were extended to both walls, the west-minus-east height difference would be

The northward volume transport per unit meridional distance isThusThis is the basin-integrated form of Sverdrup balance.

Assume a steady, small-Rossby-number interior in which drag is negligible and the nonlinear advection of relative vorticity,is small compared with advection of planetary vorticity. The stretching term proportional to $\psi$ has zero Jacobian with $\psi$. On the beta plane,Hence the forced quasi-geostrophic equation reduces to Sverdrup balanceFor the zonal wind stress $\mathbf{\tau }=\tau (y)\hat{\mathbf{x}}$,so