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For an island in a constant-depth ocean with Sverdrup balance, choose a counterclockwise contour running from the island's southern tip east to the basin wall, north along that wall, west to the northern tip, and south along the island's western coast. If that contour avoids dissipative boundary layers, the ocean transport streamfunction on the island isHere $\mathbf{W}$ is the wind acceleration appearing in the depth-averaged momentum equation. The basin-boundary streamfunction is zero. This integral fixes the otherwise undetermined constant on the island without requiring a friction law, as in Godfrey's original island-circulation model.

On a beta plane, $f=f_0+\beta y$, the preceding vorticity equation isIn the broad ocean interior, assume a statistically steady flow, weak advection of relative vorticity, and negligible friction compared with planetary-vorticity advection. The constant-depth, constant-density assumptions of 2(a) exclude topographic and baroclinic vorticity generation contributions. More precisely, require $|{\xi }_t|,|\mathbf{u}\cdot \nabla \xi |,|r\xi |\ll |\beta v|$. Then Sverdrup balance isSmall Rossby number and slow forcing help justify these conditions in the interior; they do not justify them inside an intense western boundary current. For fluctuations, the mean eddy contribution to relative-vorticity advection must also be negligible.

The depth-integrated northward volume flux per unit zonal distance is $V=Hv$. For a basin $0<x<L$ and wind stress independent of $x$, the northward interior volume flux isA negative value denotes southward transport. The cancellation of $H$ is why this interior volume flux is insensitive to the chosen constant depth. In a closed, steady basin the net volume flux across a full latitude section is zero: a boundary return must compensate $T$. Thus the displayed formula cannot represent the total closed-basin transport including that return.

For a concrete circulation sketch, use a northern-hemisphere subtropical basin $0<y<B$ andThis represents westward trade winds at the southern edge and eastward westerlies at the northern edge. The negative wind stress curl drives southward interior flow. Define an ocean transport streamfunction by $Hu=-{\Psi }_y$, $Hv={\Psi }_x$ and impose no normal flow at the eastern wall through $\Psi (L,y)=0$. Integrating Sverdrup balance westward givesSince $V=0$ at $y=0,B$, this also gives no normal flow at the north and south walls. Interior water moves westward in the southern part, southward across the basin, and eastward in the northern part. A northward western boundary current completes the clockwise ocean gyre.

Pure Sverdrup balance cannot satisfy the western impermeability condition as well: ${\Psi }_I(0,y)=-LV(y)$ varies along that wall. Indeed imposing no normal flow on both east and west walls would require their constant streamfunction values to agree in this simply connected closed basin, whereas their difference here is the nonzero $T(y)$. The shaded return in the sketch represents a boundary region where the interior approximation breaks down. It is supplied by friction in 2(c), rather than inferred from Sverdrup balance alone. A southern-hemisphere subtropical ocean gyre has the reflected, anticlockwise circulation.

For an order-of-magnitude estimate, use ${\tau }_0=0.1\mathrm{N}{\mathrm{m}}^{-2}$, ${\rho }_0=10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $\beta =2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$, $L=5\times 10^6\mathrm{m}$ and a wind stress variation scale ${\ell }_{\tau }=10^6\mathrm{m}$. ThenHere $1\mathrm{Sv}=10^6{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is a Sverdrup. The sinusoidal example has maximum $|{\tau }_{s,y}|=\pi {\tau }_0/B$; taking $B=\pi \times 10^6\mathrm{m}$ makes it agree with the chosen variation scale. Other reasonable basin widths and wind profiles change the coefficient, but retain a transport of order tens of Sverdrups.

Let $V_b$ be the northward speed in the western boundary current, whose zonal width $\delta$ is much less than its meridional scale. Its dominant relative vorticity is $\xi \simeq v_x\sim V_b/\delta$. Neglecting local wind stress curl and relative-vorticity advection in the narrow return region, the linear stationary vorticity equation becomesConsequentlyThis decay into the interior from the western wall explains why the Stommel boundary layer lies on the west. The sign also matters: a northward return has $v_x<0$, so its negative relative vorticity supplies a positive drag contribution $-r\xi$ to balance $\beta v$. Here $r$ has units ${\mathrm{s}}^{-1}$; no extra factor of $f_0$ belongs in $r/\beta$ with the paper's definition of linear drag.

For example, at fixed $y$ the leading correction to the interior ocean transport streamfunction isIt cancels the western value of ${\Psi }_I$ and carries ${\int }_0^{\infty }H{v}_{B}dx=-LV=-T$. Exponentially small eastern-wall corrections and the turning regions near the northern and southern walls do not affect the leading width estimate.

Using one year $=3.156\times 10^7\mathrm{s}$, the representative $\beta$ of 2(b), and $H=10^3\mathrm{m}$ givesTo return the $25\mathrm{Sv}$ interior volume flux, conservation of mass requiresFor the exponential profile this is the boundary speed associated with its effective transport width $\delta$. It is much narrower and faster than the familiar large-scale western boundary currents, whose energetic cores span tens of kilometres and have speeds of order $1\mathrm{m}{\mathrm{s}}^{-1}$. The simplified constant-depth linear drag model with a one-year decay time therefore does not give a realistic quantitative boundary structure. Lateral viscosity, eddies, stratification and inertia can all affect a real current.

The corresponding relative vorticity scale isThus relative vorticity cannot be discarded relative to planetary vorticity in this predicted return. Even a more realistic $1\mathrm{m}{\mathrm{s}}^{-1}$ across $50\mathrm{km}$ gives $|\xi |\sim 2\times 10^{-5}{\mathrm{s}}^{-1}$, an appreciable fraction of $|f_0|$.

Magnitude alone does not prove that its material advection is large: a perfectly parallel current $v=v(x)$, $u=0$ has $\mathbf{u}\cdot \nabla \xi =0$ exactly. For a gyre return varying over meridional scale $B$, however, incompressibility gives $u\sim \delta V_b/B$, and both generic contributions to $\mathbf{u}\cdot \nabla \xi$ scale as $V_b^2/(B\delta )$. Relative to $\beta V_b$, this isAbsent special cancellations, the assumed neglect of relative-vorticity advection is therefore not self-consistent for these parameters. Its natural inertial width is instead ${\delta }_I\sim \sqrt{V_I/\beta }$, where $V_I=|T|/(HL)\simeq 5\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$; this gives about $16\mathrm{km}$. That comparison diagnoses the missing inertia rather than supplying a complete nonlinear boundary-current solution.

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.

Let $Q_b=f/H$. At sufficiently small Rossby number, retain advection of this full background shallow-water potential vorticity while dropping the higher-order advection of $\zeta /H$. For uniform-depth Sverdrup balance, this also requires relative-vorticity advection to be small against planetary-vorticity advection; small Rossby number alone does not guarantee this when $\beta$ is very small. Multiplying the resulting steady potential vorticity equation by $H$ givesDefine the topographic potential-vorticity pseudovelocityIts sign is important: it is opposite to the coefficient vector on the left of the preceding equation. Substituting the ocean transport streamfunction expression for relative vorticity yieldsThe pseudovelocity follows contours of $f/H$ and is not the fluid velocity; its units are inverse length squared per time because $\psi$ is a transport streamfunction. If $r=F=0$, the ocean transport streamfunction is constant along each connected background potential vorticity contour. Locally, at regular points,with arbitrary differentiable $\Phi$. Different disconnected components of a level set may carry different functions until boundary conditions identify them. If $f/H$ is constant over an open region, the leading equation gives no restriction there; critical points similarly require regularity and global matching rather than division by a vanishing gradient.

For steady depth and no sources of fluid mass, the continuity equation gives ${\nabla }_h\cdot (H\mathbf{u})=0$. In a simply connected region this defines an ocean transport streamfunction:A multiply connected basin additionally needs its circulation or boundary-constant data. Taking the vertical curl of $\mathbf{u}=(-{\psi }_y,{\psi }_x)/H$ gives the exact relative vorticity from a variable-depth transport streamfunction relationNo small-depth-variation expansion has been made. No normal flow at a basin wall means that $\psi$ is constant along that connected wall.

Put $X=x-x_0$, $Y=y-y_0$ and take $\beta ,r,H>0$. The constant-depth ocean transport streamfunction equation isIt is a steady advection-diffusion equation for the transport response, with westward pseudovelocity. For an unbounded domain the point-forced Sverdrup–drag Green function can make its contours precise. Set $a=\beta /(2r)$ and $\psi =e^{-aX}\varphi$. Thenwhere $K_0$ is the Modified Bessel function of the second kind. This optional explicit expression is used only to produce the requested sketch; no weak-drag assumption is imposed.

Near the source the logarithmic singularity produces almost circular contours. At distances large compared with $\ell =r/\beta$, $K_0(aR)\sim \sqrt{\pi /(2aR)}{e}^{-aR}$, so the response is proportional to $e^{-a(R+X)}/\sqrt{R}$. It decays exponentially eastward, but only algebraically on the western axis; its broad western wake has transverse width of order $\sqrt{\ell |X|}$. Contours are stretched westward, with closed finite-level contours around the source and sharper eastern gradients. Larger drag broadens the scale $\ell$ and makes any fixed nearby view more nearly circular. For $r=0$ the elliptic smoothing disappears; for $\beta =0$ the infinite-plane response is logarithmic up to a gauge and cannot be fixed by a zero-at-infinity condition.

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.

The ocean transport streamfunction convention $Hu=-{\psi }_y$, $Hv={\psi }_x$ gives relative vorticity $\zeta =H^{-1}{\nabla }^2\psi -H^{-2}\nabla H\cdot \nabla \psi$. This identity is exact for variable positive depth and does not assume that depth variations are small.

In the steady small-Rossby number transport equation, this coefficient field advects the ocean transport streamfunction along contours of background shallow-water potential vorticity $f/H$. It is not a physical velocity: when $\psi$ has units of volume transport, its units are inverse length squared per time. For uniform depth on a beta plane it points westward, $\tilde{\mathbf{u}}=(-\beta /H,0)$. With no forcing or drag, $\psi$ is constant on each connected characteristic; locally $\psi =\Phi (f/H)$ at regular points.

With $f=f_0+\beta y$ and parabolic depth, the topographic potential-vorticity pseudovelocity is $(-\beta /H,-f{H}_x/H^2)$. For positive $f,\beta$, it points west and turns south on the western slope, north on the eastern slope. Its characteristics are the parabolas $y=Qax(L_x-x)/\beta -f_0/\beta$, where $Q=f/H$. Wind-driven ocean gyres are steered by these depth gradients; drag and boundary data determine the actual ocean transport streamfunction. The zero-depth shoreline is a singular limit that requires a separate physical regularization.