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If $r^3\gg {\beta }^2\nu$, the broad western Stommel boundary layer has thickness $r/\beta$, but small viscosity supplies no-slip skins of thickness $(\nu /r{)}^{1/2}$ at both east and west walls. The western solution therefore contains two decaying exponentials, while the east has the rapidly decaying skin. Setting viscosity identically to zero removes the modes needed to satisfy all four lateral no-slip conditions.

With depth transport $(-{\psi }_y,{\psi }_x)$ and a depth-mean bottom-drag closure, the wind-driven equation is $\beta {\psi }_x=S-r{\nabla }^2\psi +\nu {\nabla }^4\psi$, where $S$ is wind-stress curl divided by density and $r=\gamma /({\rho }_{0}H)$ is the drag rate. Its zonal boundary modes solve $\nu {\lambda }^3-r\lambda -\beta =0$. The Munk boundary layer and Stommel boundary layer are different limits of this combined equation.

The nonconstant exponential modes of the leading basin equation satisfyThus a leading uniformly valid form is $\bar{\psi }\simeq S(y)x/\beta +C_0(y)+{\sum }_{j=1}^3{C}_j(y)e^{{\lambda }_{j}x}$, with exponentials referenced to the wall where they decay. The coefficients enforce the four east/west no-slip conditions; their values are not needed to identify the scales.

If $r^3\ll {\beta }^2\nu$, horizontal viscosity dominates drag in the boundary layers. The Munk boundary layer scale is ${\delta }_M=(\nu /\beta {)}^{1/3}$, assumed much smaller than basin width. The roots are ${\delta }_M^{-1}$ and $(-1\pm i\sqrt{3})/(2{\delta }_M)$ to leading order, soThe east layer has exponential thickness ${\delta }_M$ and the west oscillatory layer has envelope thickness $2{\delta }_M$, both of order ${\delta }_M$. The west correction supplies the order-one return transport, while the east no-slip correction is typically smaller in amplitude. Setting $r=0$ still leaves enough viscous modes to impose no-slip.

If $r^3\gg {\beta }^2\nu$, linear drag dominates the broad vorticity layer. Definewith ${\delta }_S$ small compared with basin width. The roots are approximately $-{\delta }_S^{-1}$ and $\pm {\delta }_v^{-1}$; the fast roots have smaller correction $\beta /(2r)$. The drag-dominated basin solution with no-slip layers isThere is a broad western Stommel boundary layer of thickness ${\delta }_S$, a thinner western no-slip layer of thickness ${\delta }_v$, and an eastern no-slip layer of thickness ${\delta }_v$. The narrow layers can have small streamfunction amplitude while supplying an order-one change in wall velocity.

If viscosity is set identically to zero, only the particular solution, a constant and the western Stommel exponential remain. That second-order model cannot generally satisfy both streamfunction and derivative conditions at both walls. Small nonzero viscosity must be retained in the wall skins when no-slip is required. In the question's stress notation, the limits compare $[\gamma /({\rho }_{0}H){]}^3$ with ${\beta }^2\nu$. Neither limit applies at the crossover, where all cubic terms contribute. These composite forms are leading asymptotic solutions of the basin problem; meridional derivatives and interior diffusion supply higher-order corrections.

To close a depth-integrated bottom-drag model, approximate the bottom horizontal velocity by the depth mean, ${\mathbf{u}}_b\simeq \mathbf{T}/H$. This is the barotropic closure needed to express the stated stress solely in terms of the transport. DefineThe coefficient $\gamma$ in a physical stress law is not itself a drag rate; $r$ has units of inverse time. The ocean basin equation with bottom drag and lateral viscosity follows by taking curl of the integrated momentum equation and yields the combined Stommel boundary layer and Munk boundary layer equation,The bottom contribution is negative because the vertically integrated stress difference is surface stress minus bottom stress. Set the constant wall streamfunction to zero. No normal flow and no tangential slip impose $\bar{\psi }=0$ and ${\partial }_n\bar{\psi }=0$ on the walls.

The inviscid interior solution has ${\bar{\psi }}_I=S(y)x/\beta +C_0(y)$. In a thin east/west layer, $x$ derivatives dominate the smaller $y$ derivatives. Holding $y$ as a parameter givesEquivalently the boundary correction $\varphi =\bar{\psi }-{\bar{\psi }}_I$ obeys $\nu {\varphi }^{\prime \prime \prime \prime }-r{\varphi }^{\prime \prime }-\beta {\varphi }^{\prime }=0$. In a stretched inward coordinate $X=x/\delta$ at the west wall and $X=(1-x)/\delta$ at the east wall, this becomes respectivelyThe first-derivative sign reversal distinguishes east from west. Vanishing $W$ and its derivatives at the north/south boundaries makes leading forcing-dependent solutions compatible with those boundaries, avoiding extra leading meridional layers. These are leading boundary-layer equations, not an exact deletion of every meridional derivative in the full basin PDE.

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.

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.