= Rigid-lid resonance of stationary topographic internal waves
{title2=$mH=n\pi$}
For a sinusoidal small-amplitude bed, constant current $U>0$ and <buoyancy frequency> $N$, put $m=\sqrt{N^2/U^2-k^2}>0$. The linear vertical <velocity> amplitude is $W(z)=ikU\eta_0\sin[m(H-z)]/\sin(mH)$. The upward-wave coefficient is $A_+=-kU\eta_0e^{-imH}/[2\sin(mH)]$. At $mH=n\pi$, the prescribed nonzero bed motion and homogeneous rigid-lid condition are incompatible with a bounded stationary inviscid solution. With angle $\theta$ between intrinsic <group velocity> and vertical, $\tan\theta=m/k$, giving $kH\tan\theta=n\pi$.
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