= Solution
Use pressure per unit constant density, $\pi$, with the <gravitational potential> and <centrifugal potential> absorbed into it. <Strict geostrophic balance> is the exact balance
$$
2\boldsymbol\Omega\times\mathbf u=-\nabla\pi,\qquad \nabla\cdot\mathbf u=0,
$$
with <material acceleration> omitted, for a homogeneous <incompressible flow>. Writing $f=2\Omega\ne0$, its components are $fv=\pi_x$, $fu=-\pi_y$ and $\pi_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the <vector calculus> identity
$$
\nabla\times(2\boldsymbol\Omega\times\mathbf u)=2\boldsymbol\Omega(\nabla\cdot\mathbf u)-(2\boldsymbol\Omega\cdot\nabla)\mathbf u
$$
shows, on taking the <curl> of the balance, that $(\boldsymbol\Omega\cdot\nabla)\mathbf u=0$. Thus $w_z=0$ too, proving the <Taylor–Proudman theorem>:
$$
\boxed{\pi_z=0,\qquad \mathbf u_z=0.}
$$
The physical pressure still includes its background <hydrostatic pressure>; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.
For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, <impermeability condition> gives $w_b=\mathbf u_H\cdot\nabla_Hz_b$ and $w_t=\mathbf u_H\cdot\nabla_Hz_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq\mathbf u_H\cdot\nabla_HH\ne0$. \b[Exact columnarity must give way to small corrections from <ageostrophic flow>.] The leading horizontal <geostrophic flow> can remain nearly independent of $z$, and the leading reduced pressure has $\pi_z=0$, but $w_z$ is now nonzero and exact three-dimensional <strict geostrophic balance> no longer holds. For small slopes, the dominant vertical <vorticity equation> is
$$
\frac{D(q+f)}{Dt}\simeq(q+f)w_z,\qquad w_z\simeq\frac1H\mathbf u_H\cdot\nabla_HH,
$$
where $q=v_x-u_y$ is <relative vorticity>. Equivalently, $D[(q+f)/H]/Dt\simeq0$. Crossing <geostrophic contours> stretches or squashes columns: increasing $H$ produces cyclonic relative <vorticity>, decreasing $H$ produces anticyclonic relative <vorticity>. The full <vorticity equation> also contains tilting terms, neglected at this columnar order.
For the circular shallowing, fluid arriving from upstream has <potential vorticity> $f/h$. Its height is $h(1-\epsilon)$ inside, so conservation gives
$$
\frac{f+q}{h(1-\epsilon)}=\frac fh,\qquad \boxed{q=-f\epsilon\quad(r<a),\qquad q=0\quad(r>a).}
$$
This uses the stated column-stretching description and initially open <streamlines>. To obtain the leading disturbance <velocity field>, use the conventional <streamfunction> definition $u=-\Psi_y$, $v=\Psi_x$, and write $\Psi=-Uy+\psi(r)$. Then $q=\nabla_H^2\psi=r^{-1}(r\psi_r)_r$. Regularity at $r=0$ gives $\psi_r=-f\epsilon r/2$ inside. Outside, $\psi_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-f\epsilon a^2/2$. Choose the irrelevant additive constant so that
$$
\psi=C\begin{cases}
-r^2/2,&r<a,\\
a^2[\log(a/r)-1/2],&r>a,
\end{cases}
\qquad \boxed{C=\frac{f\epsilon}{2}=\Omega\epsilon.}
$$
Both $\psi$ and $\psi_r$ match at $a$, so there is no extra <vorticity sheet> on the rim. This is the <circular topographic anticyclone>. The radial expression specifies the disturbance; the \b[complete streamfunction is $\Psi=-Uy+\psi$], since a radial function alone cannot represent the required uniform far-field current. Reversing the <streamfunction> sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.
Differentiation gives the full relative <velocity field> in the rotating frame:
$$
(u,v)=\begin{cases}
(U+Cy,-Cx),&r<a,\\
(U+Ca^2y/r^2,-Ca^2x/r^2),&r>a,
\end{cases}
\qquad w\simeq0
$$
away from the narrow transition across the change of depth. The disturbance turns clockwise for $f\epsilon>0$ and decays as $1/r$. The horizontal field is the leading small-$\epsilon$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.
Take $U>0$ and $\epsilon>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed <streamlines>. If $Ca>U$, the interior <stagnation point> $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior <stagnation point> $(0,-Ca^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a <separatrix> through the saddle. The two points meet at the rim at onset, giving the <closed-streamline threshold for a circular topographic anticyclone>:
$$
\boxed{Ca=U,\qquad \left(\frac\epsilon U\right)_{\rm crit}=\frac2{fa}=\frac1{\Omega a}.}
$$
At this threshold $U/(fa)=\epsilon/2\ll1$, consistent with small <Rossby number>. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the <potential vorticity> of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.
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