= Solution
Taking the vertical component of the curl of the <momentum> equation gives
$$
\omega_t+f\nabla\cdot\mathbf u=0.
$$
The continuity equation gives $\eta_t+h_0\nabla\cdot\mathbf u=0$, and hence
$$
\boxed{\frac\partial{\partial t}
\left(\omega-\frac f{h_0}\eta\right)=0}.
$$
Since initially $\mathbf u=0$ and $\eta=\eta_0$,
$$
\omega=\frac f{h_0}(\eta-\eta_0).
$$
Taking the divergence of the <momentum> equation and using
$$
\nabla\cdot(\mathbf f\times\mathbf u)=-f\omega
$$
gives
$$
(\nabla\cdot\mathbf u)_t-f\omega=-g\nabla^2\eta.
$$
Differentiate continuity in time and substitute the last two identities to obtain
$$
\boxed{\eta_{tt}-gh_0\nabla^2\eta+f^2\eta=f^2\eta_0}.
$$
Let the <Rossby deformation radius> be
$$
L_R=\frac{\sqrt{gh_0}}{|f|}.
$$
The even, decaying steady solution of
$$
-L_R^2\eta_\infty''+\eta_\infty=\eta_0
$$
with continuous value and <derivative> at $x=\pm a$ is
$$
\boxed{
\eta_\infty(x)=
\begin{cases}
\epsilon\left[1-e^{-a/L_R}\cosh(x/L_R)\right],&|x|<a,\\
\epsilon\sinh(a/L_R)e^{-|x|/L_R},&|x|>a.
\end{cases}}
$$
The steady <momentum> balance is a <geostrophic balance>:
$$
\boxed{u=0,
\qquad v=\frac g f\frac{d\eta_\infty}{dx}}.
$$
Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.
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