 Solution

Place a straight coast at $x=0$, ocean at $x<0$, and impose no normal flow at the wall. A coastal Kelvin mode has $u=0$ throughout. The zonal momentum equation becomes $g{\eta }_x=f_{0}v$, while the remaining two equations are $v_t+g{\eta }_y=0$ and ${\eta }_t+H{v}_y=0$. Eliminating $v$ givesFor a northward disturbance $\eta =A(x)F(y-ct)$, the nontrivial solution has $c^2=gH$ and $v=(c/H)\eta$. In the northern hemisphere choose $c=+\sqrt{gH}$; then cross-shore geostrophic balance requiresIt decays into this ocean half-plane because $x<0$ and $f_0>0$. The decay length is the external Rossby deformation radius $L_R=c/f_0$. A Fourier component has $\omega =ck$, so both phase and alongshore group speed are $c$, independent of wavenumber: the ideal shallow-water Kelvin wave is nondispersive.

The signed choice is part of the boundary condition. For ocean at $x>0$ in the same hemisphere, decay instead selects southward speed $-\sqrt{gH}$ and $A\propto e^{-f_{0}x/\sqrt{gH}}$. Without stating the ocean side, an exponential growth into the sea could be mistaken for a trapped coastal mode.