= Solution
In the northern hemisphere the coast is on the wave's right when looking in its direction of propagation. In the ocean-$x<0$ convention above it travels northward, with land east of the flow; for the opposite ocean side it travels southward. This follows from combining $g\eta_x=f_0v$ with a sea-level bulge that decreases offshore, rather than choosing propagation independently of trapping.
For a positive northward bulge, $v=(c/H)\eta$ is northward and strongest at its crest, and there is no cross-shore <velocity> in the ideal mode. Ahead of the crest the current decreases northwards, so $v_y<0$: <horizontal flow convergence> increases sea level, $\eta_t=-Hv_y>0$. Behind the crest $v_y>0$: <divergence> lowers sea level. At the crest itself the alongshore <derivative> and instantaneous <horizontal flow convergence> vanish. These signs reverse appropriately for a trough or the opposite coast, while the trapping rule remains the same.
The first panel below sketches a positive bulge, alongshore <velocity> and the leading convergence/trailing <divergence> regions. Its cross-shore decay is the <pressure>-gradient/Coriolis balance, not a current directed into the coast.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2018-coastal-kelvin-wave.png]
{title=Coastal Kelvin-wave trapping, convergence and amplitude under conserved energy transport}
{description=Original normalized model. The ocean lies left of the coast, the bulge travels north, and the amplitude curves assume slow changes with negligible dissipation and reflection.}
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