 Solution

Put $\eta =h-h_0$ and $c=\sqrt{g^{\prime }{h}_0}$. With $u=0$ and complex dependence $e^{i({\omega }_{K}t-{\ell }_{K}y)}$, the northward momentum equation and continuity equation giveThus ${\omega }_K^2=c^2{\ell }_K^2$. Write the signed phase velocity as $c_y={\omega }_K/{\ell }_K=\pm c$, so $v=(c_y/h_0)\eta$. The eastward momentum equation is the transverse geostrophic balance $g^{\prime }{\eta }_x=f_{0}v$. HenceThe decaying sign is selected by which side of $x_b$ contains the ocean. For an eastern boundary the ocean occupies $x<x_b$, so decay offshore requires $f_0/c_y>0$. Consequently the Kelvin wave moves northwards in the northern hemisphere and southwards in the southern hemisphere, with the coast respectively on its right and left. The positive offshore decay length isThe sign of $f_0$ reverses the propagation direction and the sign of $v$ relative to $\eta$, but not the speed or offshore decay length. The real physical fields are the real parts of the complex solution. At the equator itself this local f-plane coastal solution loses its finite trapping scale.