= Solution
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> becomes
$$
\beta v\simeq-r\xi\simeq-rv_x.
$$
Consequently
$$
\boxed{\delta\sim\frac r\beta,\qquad v(x,y)\propto e^{-x/\delta}.}
$$
This 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> is
$$
\Psi_B=LV(y)e^{-x/\delta},\qquad
Hv_B=\partial_x\Psi_B=-\frac{LV(y)}\delta e^{-x/\delta}.
$$
It cancels the western value of $\Psi_I$ and carries $\int_0^\infty Hv_B\,dx=-LV=-T$. 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 $=3.156\times10^7\ \mathrm s$, the representative $\beta$ of 2(b), and $H=10^3\ \mathrm m$ gives
$$
r=3.17\times10^{-8}\ \mathrm{s^{-1}},\qquad
\boxed{\delta\simeq1.58\times10^3\ \mathrm m=1.6\ \mathrm{km}.}
$$
To return the $25\ \mathrm{Sv}$ interior <volume flux>, <conservation of mass> requires
$$
\boxed{V_b\sim\frac{|T|}{H\delta}
\simeq15.8\ \mathrm{m\,s^{-1}}.}
$$
For 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 $1\ \mathrm{m\,s^{-1}}$. 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 is
$$
\boxed{|\xi|\sim\frac{V_b}\delta\simeq9.96\times10^{-3}\ \mathrm{s^{-1}},\qquad
\frac{|\xi|}{|f_0|}\sim100\quad\text{for }|f_0|\sim10^{-4}\ \mathrm{s^{-1}}.}
$$
Thus <relative vorticity> cannot be discarded relative to <planetary vorticity> in this predicted return. Even a more realistic $1\ \mathrm{m\,s^{-1}}$ across $50\ \mathrm{km}$ gives $|\xi|\sim2\times10^{-5}\ \mathrm{s^{-1}}$, 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 is
$$
\frac{|\mathbf u\cdot\nabla\xi|}{\beta V_b}
\sim\frac{V_b}{\beta B\delta}
\simeq1.6\times10^2\qquad(B=\pi\times10^6\ \mathrm m).
$$
Absent 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 $V_I=|T|/(HL)\simeq5\times10^{-3}\ \mathrm{m\,s^{-1}}$; this gives about $16\ \mathrm{km}$. That comparison diagnoses the missing inertia rather than supplying a complete nonlinear boundary-current solution.
Back to article page
