An <Orr-Sommerfeld mode> has $\widehat v\ne0$ satisfying the <Orr-Sommerfeld equation>, with accompanying vorticity satisfying the forced <Squire equation>. A <Squire mode> has $\widehat v=0$ and $\widehat\eta\ne0$, so its Squire equation is homogeneous.

For a Squire mode multiply that homogeneous equation by $\overline\eta$ and integrate over the channel. Under the no-slip condition $\eta=0$ at the walls, <integration by parts> gives
$$
-i\omega\int|\eta|^2dy+i\alpha\int U|\eta|^2dy
=-\frac1{Re}\int(|\eta'|^2+\kappa^2|\eta|^2)dy.
$$
Since $U,\alpha$ are real, taking the real part proves
$$
\boxed{\operatorname{Im}\omega=-\frac{\int(|\eta'|^2+\kappa^2|\eta|^2)dy}{Re\int|\eta|^2dy}<0.}
$$
The numerator is strictly positive for a nontrivial finite-channel mode, and $Re>0$. Thus Squire modes are exponentially damped in the convention $e^{-i\omega t}$. This does not rule out <transient growth from non-normal modes> in the coupled system.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 3 / b / Solution

{c}

An Orr-Sommerfeld mode has nonzero wall-normal velocity satisfying the <Orr-Sommerfeld equation>. Its wall-normal vorticity satisfies the accompanying forced <Squire equation>. The triangular coupling separates the velocity eigenproblem from the vorticity forcing; the vorticity need not vanish in three dimensions.

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