= 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 $g\eta_x=f_0v$, while the remaining two equations are $v_t+g\eta_y=0$ and $\eta_t+Hv_y=0$. Eliminating $v$ gives
$$
\eta_{tt}=gH\eta_{yy}.
$$
For 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> requires
$$
\frac{dA}{dx}=\frac{f_0}{c}A,\qquad
\boxed{A(x)=A_0e^{f_0x/c},\quad c=\sqrt{gH}.}
$$
It 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 $A\propto e^{-f_0x/\sqrt{gH}}$. Without stating the ocean side, an exponential growth into the sea could be mistaken for a trapped coastal mode.
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