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

Potential-vorticity conservation still determines the final surface throughThe unbounded problem only required decay or matching as $|y|$ and $x$ become large. The coast adds a boundary condition. In the adjusted geostrophic flow,so $\eta$ must be constant along the connected wall. Odd symmetry and the arbitrary common height datum set that constant to zero:This is a Dirichlet condition on $\eta$, even though it originated as no normal flow.

The initial disturbance does not instantaneously know this along the whole coast. A southward coastal Kelvin wave for $f>0$ carries the pressure signal and establishes the constant wall height behind its wavefront. In the final streamline sketch, the quasi-geostrophic streamfunction is proportional to ${\eta }_f$. Far from the wall the contours and current resemble the unbounded east--west front; near $x=0$ those contours bend through a right angle and run along the coast, so the incident geostrophic current turns into a southward boundary current instead of crossing the wall.

Set $c=\sqrt{g{H}_0}$ and seekThe two remaining prognostic equations implyThe $x$ momentum equation is now geostrophic:ThereforeBoundedness for $x>0$ selects $\gamma =\mathrm{sgn}(f)c$ andBecause $F$ depends on $y+\gamma t$, the wave travels with meridional phase speed $-\gamma$. Along this western boundary it travels southward in the Northern Hemisphere and northward in the Southern Hemisphere, keeping the coast on the dynamically required side. This is a coastal Kelvin wave.

Moreover,Since $u=0$, its linear potential-vorticity disturbance is

Initially $\zeta =0$, so conservation of the linear potential vorticity givesThe final flow is in geostrophic balance, henceConsequently the adjusted height solves the modified Helmholtz equationwhere $L_R$ is the Rossby deformation radius.

The initial condition is independent of $x$ and odd in $y$. The bounded solution that is continuously differentiable at $y=0$ isIt givesDuring geostrophic adjustment, the part of the initial energy incompatible with the conserved potential-vorticity distribution radiates away as inertia-gravity waves. The remaining current has width $O(L_R)$ and is geostrophically balanced.

The horizontal velocities $u$ and $v$ point east and north, $\eta$ is the displacement of the free surface from its mean level, $H_0$ is the undisturbed depth, $g$ is gravitational acceleration, and $f$ is the constant Coriolis parameter on an f-plane. The three linearized shallow water equations are horizontal momentum balance and mass conservation:They follow from the rotating Navier-Stokes equation by assuming an inviscid homogeneous layer, hydrostatic pressure, horizontal scales much larger than $H_0$, depth-independent horizontal velocity, a flat impermeable bottom, constant $f$, and small surface displacement and velocity so that nonlinear products are neglected.

In a steady state, geostrophic balance givesThe relative vorticity is $\zeta =v_x-u_y$. Expanding the shallow-water potential vorticity $(f+\zeta )/(H_0+\eta )$ to first order givesThus one convenient normalization of its disturbance isTaking the curl of momentum and using continuity shows $q_t=0$.

For finite mantle depth, retain all four vertical solutions of the biharmonic equation,rather than discarding the terms that grow as $z\to -\infty$. Apply the plate conditions at $z=0$ and matching conditions at $z=-H$. At the mantle--core interface these include continuity of normal velocity and normal stress, an appropriate tangential-stress condition for the liquid core, and a kinematic boundary condition for interface displacement. The normal-stress balance gains the stable buoyancy termbecause ${\rho }_c>\rho$.

The resulting homogeneous linear system for the mode amplitudes has a nontrivial solution only when its determinant vanishes; that condition replaces the half-space decay rate. Modes with $kH\gg 1$ decay before sensing the core and recover the half-space result. Modes with $kH\lesssim 1$ involve the full mantle depth, feel the basal density contrast and core mobility, and acquire a modified relaxation time. This is the planar finite-depth extension of flexural isostasy.

The kinematic boundary condition is ${\zeta }_t=w$ at $z=0$. Part c therefore givesand henceIn terms of wavelength $\lambda =2\pi /k$,Without elasticity, $\tau \sim 2\mu k/(\rho g)$: shorter wavelengths penetrate less deeply but relax more slowly because their viscous gradients are larger. At short wavelength, elastic bending dominates andso the strong $B{k}^4$ restoring stress makes very short waves relax rapidly. The graph therefore tends to zero at both wavelength extremes and has one maximum. Differentiating with respect to $k$ places that maximum atThis exponential relaxation is the simplest half-space model of postglacial rebound and glacial isostatic adjustment.

To first order, vanishing tangential velocity on the elastic plate is simply $u(0)=0$. In the mode from part a this givesConsequentlyso $u(0)=0$, $w(0)=ikA{e}^{ikx}$, and $w_z(0)=0$. The dynamic pressure at the surface is

Evaluating the hydrostatic pressure at $z=\zeta$ contributes the restoring normal stress $\rho g\zeta$. Since ${\nabla }^4\zeta =k^4\zeta$ for this mode, the linearized normal-stress balance isIt follows thatThe two restoring effects are buoyancy and the plate's bending stiffness.

For the surface $F(x,z,t)=z-\zeta (x,t)=0$, an outward normal vector and positively oriented tangent vector areWith $\zeta =ϵe^{ikx}$ and $|kϵ|\ll 1$, linearization gives

For $\mathbf{u}=\nabla \times (\psi \hat{\mathbf{y}})$,Hence $u_x+w_z=-{\psi }_{zx}+{\psi }_{xz}=0$, so mass conservation holds identically. Removing the hydrostatic part by writingand taking the curl of the Stokes flow equation gives the biharmonic equation

For a Fourier mode proportional to $e^{ikx}$ with $k>0$, decay as $z\to -\infty$ selectsThe velocity and pressure fields are thereforeDirect substitution verifies the momentum equation.