Kelvin Wave
\[ \begin{aligned} &\underbrace{\frac{\partial \eta}{\partial t} + H \frac{\partial u}{\partial x} \quad \xrightarrow[\text{Assume } \eta = \hat{\eta}(y) e^{i(kx - \omega t)}]{\partial_t \to -i\omega,\; \partial_x \to ik} \quad -i\omega \hat{\eta} + ikH \hat{u} = 0}_{\text{Contributes to dispersion relation between } \omega \text{ and } k} \\[1.5ex] &\underbrace{\frac{\partial u}{\partial t} = -g \frac{\partial \eta}{\partial x} \quad \xrightarrow[\text{Assume } u = \hat{u}(y) e^{i(kx - \omega t)}]{\partial_t \to -i\omega,\; \partial_x \to ik} \quad -i\omega \hat{u} = -ikg \hat{\eta}}_{\text{Contributes to dispersion relation between } \omega \text{ and } k} \\[1.5ex] &\underbrace{fu = -g \frac{\partial \eta}{\partial y} \quad \xrightarrow[\text{Use } u, \eta = \hat{u}(y), \hat{\eta}(y) e^{i(kx - \omega t)}]{\partial_y \text{ acts only on } \hat{\eta}(y)} \quad f \hat{u} = -g \frac{d\hat{\eta}}{dy}}_{\text{Determines transverse dependence}} \end{aligned} \]

\( \begin{aligned} &\hat{u} = \frac{-ikg \hat{\eta}}{-i\omega} = \frac{\cancel{-} \cancel{i} k g \hat{\eta}}{\cancel{-} \cancel{i} \omega} = \frac{k g \hat{\eta}}{\omega} \quad \text{(from } -i\omega \hat{u} = -ikg \hat{\eta} \text{)} \\ &-i\omega \hat{\eta} + ikH \hat{u} = 0 \quad \Rightarrow \quad -i\omega \hat{\eta} + ikH \cdot \frac{k g \hat{\eta}}{\omega} = 0 \quad \Rightarrow \quad \hat{\eta} (\omega^2 - gHk^2) = 0 \end{aligned} \)
A non-trivial solution is therefore possible only if \(\boxed{ \omega = \pm k \sqrt{gH}} \)
so that the wave propagates with a non-dispersive speed $$ c = \sqrt{gH} $$ The propagation speed of a Kelvin wave is therefore identical to that of non-rotating gravity waves. Its dispersion equation is a straight lineDispersionRelationsPoincareKelvin All frequencies are possible.