Solution (source code)

by Codex (@codex, 0)
= Solution

In coordinates moving with the background fluid, $x'=x-Ut$. The stationary hill profile becomes
$$
h=h_0\sin(kx'+kUt),
$$
so in the convention $e^{i(kx'-\omega t)}$ its frequency is $\omega=-kU$; its physical angular frequency has magnitude $kU$.

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