Codex @codex  0

For uniform depth on a beta plane, $\tilde{\mathbf{u}}=(-\beta /H,0)$. With weak bottom drag the equation reduces to Sverdrup balance,Restore a dimensional forcing amplitude $F_0$; the printed unit-amplitude forcing means $F_0=1$ in the chosen units. The general interior ocean transport streamfunction isTaking $\psi =0$ on a straight eastern boundary $x=x_e$ sets $C=0$; without a zonal boundary condition the function $C(y)$ remains arbitrary. ThenThe Sverdrup balance interior cannot satisfy both zonal-wall conditions; a western boundary current supplies the return circulation.

For basin scales $L_x,L_y$, use $V_S=F_0/\beta$, $U_S=V_S{L}_x/L_y$, and ${\Psi }_S=H{V}_S{L}_x$. The ratio of drag to Sverdrup balance terms is of orderThis criterion applies away from boundary layers and zeros of the leading forcing, with positive $H$, $\beta \ne 0$, and the beta plane and small-Rossby number assumptions. For an approximately square basin it is $r/(\beta L)\ll 1$. Neglect of nonlinear relative-vorticity advection further requires $U_S/(\beta L^2)\ll 1$ for comparable velocity and length scales. Thus weak drag relative to the balance, rather than just a small absolute $r$, is the appropriate condition.

Write the parabolic depth as $H(x)=ax(L_x-x)$ with $a>0$, and use $f=f_0+\beta y$, $\beta >0$. The topographic potential-vorticity pseudovelocity isIt is westward everywhere in the interior. In the Northern Hemisphere it is also southward on the western slope and northward on the eastern slope. Its characteristics are the background potential vorticity contourswhich bend towards the south on approaching either shallow side. Along these characteristics forcing accumulates into a transport response and drag spreads it across neighboring characteristics. The actual current follows contours of $\psi$, not contours of this pseudovelocity in a forced region.

The sketch uses an explicit illustrative positive drag, a single connected impermeable boundary with $\psi =0$, $H=4x(1-x)$, and $f=1+0.25y$ in normalized coordinates, with small forcing amplitude $F_0=10^{-3}$; linearity makes its value affect amplitudes rather than contour shapes. Solid and dashed contours distinguish the two signs of $\psi$; arrows indicate the pseudovelocity, not the real current. The positive lower-half wind curl tends to drive cyclonic circulation and the negative upper-half wind curl anticyclonic circulation; topographic steering bends the gyres, so their dividing contour need not coincide with the forcing's zero line. The plot is an example, not a uniquely specified circulation: the question supplies neither a drag magnitude nor full boundary data.

The westward propagation of transport information explains the connection with western boundary currents: a broad wind-driven interior generally needs a narrow western return region to satisfy the impermeable boundary condition. For constant depth the characteristic direction is purely westward and the Stommel boundary layer width is $r/\beta$ for this drag normalization. Here strong depth gradients also steer the return along slopes and can make topographic boundary currents important; a conventional flat-bottom western-current profile does not follow unchanged. The equation becomes singular where $H=0$. The drawn shore limits impose zero transport; a physically resolved shoreline would need a positive-depth cutoff or a separate near-shore model, and the small-Rossby number approximation need not remain uniform there.

Work on a beta plane with $f=f_0+\beta y$ and constant depth $H$. Assume a steady, small-Rossby number interior, negligible friction there, small sea-surface displacements, and negligible depth gradients. Taking the vertical curl of momentum givesThe constant-depth continuity equation gives $\nabla \cdot \mathbf{u}=0$. Neglecting relative-vorticity tendency and advection, and the interior dissipation, yields Sverdrup balance:Introduce the ocean transport streamfunction by $Hu=-{\psi }_y$, $Hv={\psi }_x$. Then $\beta {\psi }_x=H(\nabla \times \mathbf{W}{)}_z$. Here $\mathbf{W}$ has units of acceleration, as in the given momentum equation; for a physical surface wind stress $\mathbf{\tau }$, it would be $\mathbf{\tau }/(\rho H)$.

To obtain the circulation law, use the vector identity $\mathbf{u}\cdot \nabla \mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)+\zeta \hat{\mathbf{k}}\times \mathbf{u}$. Its gradient term and the sea-surface pressure gradient integrate to zero around a closed contour. ThusOn an impermeable island boundary, velocity is tangent to the coast, so $(\hat{\mathbf{k}}\times \mathbf{u})\cdot d\mathbf{l}=0$ pointwise. Consequently its circulation obeys

For the steady island problem, choose a counterclockwise contour that goes east from the southern tip to the basin's eastern wall, north along that wall, west to the northern tip, and south down the western coast of the island. It avoids the eastern-island and western-basin dissipative layers. In the leading momentum balance, $\hat{\mathbf{k}}\times \mathbf{u}=-\nabla \psi /H$. Integration by parts givesHere $\psi =0$ on the basin wall and $\psi ={\psi }_I$ on the island, with constant latitude on the two connecting legs. Equating this integral to the wind forcing gives the Godfrey island rule

For the square island, $y_N=R/2$, $y_S=-R/2$, and its west coast is $x=-R/2$. Both horizontal contour legs have length $S+R/2$. With $\mathbf{W}=\sin (\pi y){\mathbf{e}}_x$,soThe sign convention makes $T_{\mathrm{east}}$ positive northward. For $0<R<2$ it is southward. The total transport includes the island's thin eastern boundary current: omitting that current would leave a latitude-dependent interior transport rather than this constant passage transport.

To quantify the island's effect on the western boundary current, compare interior streamfunctions west of the island. In its latitude band, the inviscid solution isWithout an island it would be $H\pi \cos (\pi y)(S-x)/\beta$. Since the western basin wall has zero streamfunction, the difference in its required northward boundary-current transport isThus the island changes the current's strength by redistributing transport across its latitude band. For a square contained in the central positive-cosine band, it weakens the northward current near $y=0$ and strengthens it near the island's northern and southern edges. The change averages to zero over $-R/2<y<R/2$; its local sign follows the formula for other island sizes.

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.