For $\epsilon=0$, the quantity under the square root is real:
$$
F(\mu)=\frac1{\mu^2}-\frac{\coth\mu}{\mu}+\frac14.
$$
It is negative for small $\mu$ and positive for large $\mu$. Therefore the <Eady instability> occurs for $\mu<\mu_c$, where
$$
\boxed{
\frac1{\mu_c^2}-\frac{\coth\mu_c}{\mu_c}+\frac14=0},
$$
and modes are neutral for $\mu>\mu_c$.

For $\mu\gg1$, $\coth\mu=1+O(e^{-2\mu})$ and $\operatorname{csch}\mu$ is exponentially small. Hence
$$
\boxed{
\widetilde c_+=1-\frac1\mu+O(e^{-2\mu}),
\qquad
\widetilde c_-=\frac1\mu+O(e^{-2\mu})}.
$$
These are two decoupled <Boundary Rossby waves>, one localized near each boundary. Their laboratory phase speeds lie just inside the basic velocities $\Lambda H$ and $0$ because each propagates intrinsically against the local flow.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 333 / 3 / i / Solution

{c}

A boundary Rossby wave is supported by a buoyancy or potential-temperature anomaly on a rigid horizontal boundary. Its intrinsic propagation opposes the local basic flow in the Eady model.


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 Incoming links: Boundary Rossby wave