= Solution
For a steady flow independent of $y$, <geostrophic balance> gives
$$
u=0,
\qquad
v=\frac gf\frac{d\eta}{dx}.
$$
Hence $\zeta=v_x=(g/f)\eta''$, and the definition of $q$ becomes
$$
q=\frac gf\eta''-\frac fh\eta.
$$
Equivalently,
$$
\boxed{\eta''-\frac{f^2}{gh}\eta=\frac fgq}.
$$
With the <Rossby deformation radius> $R=\sqrt{gh}/f$, the bounded solution for the stated potential-vorticity step is
$$
\eta(x)=
\begin{cases}
-\dfrac{hq_1}{f}+A e^{x/R},&x<0,\\
-\dfrac{hq_2}{f}+B e^{-x/R},&x>0.
\end{cases}
$$
Continuity of $\eta$ and $v=(g/f)\eta'$ at $x=0$ gives
$$
A=\frac{h(q_1-q_2)}{2f},
\qquad
B=-\frac{h(q_1-q_2)}{2f}.
$$
Therefore
$$
\boxed{
\eta(x)=
\begin{cases}
-\dfrac{hq_1}{f}+\dfrac{h(q_1-q_2)}{2f}e^{x/R},&x<0,\\
-\dfrac{hq_2}{f}-\dfrac{h(q_1-q_2)}{2f}e^{-x/R},&x>0,
\end{cases}}
$$
and
$$
\boxed{v(x)=\frac{R(q_1-q_2)}2e^{-|x|/R}}.
$$
When $q_1>q_2$, the surface height increases smoothly from $-hq_1/f$ to $-hq_2/f$, while $v$ is a positive eastward jet with its maximum at the discontinuity and <exponential decay> on the Rossby-radius scale:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4-potential-vorticity-step.png]
{title=Free surface and jet for a potential-vorticity step}
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