= Solution
For $k>0$, put $q=\sqrt{k^2+l^2}$ and define the <Squire transformation>
$$
\boxed{k'=q,\qquad l'=0,\qquad R'=\frac{k}{q}R,\qquad c'=c,\qquad \omega'=qc.}
$$
The two-dimensional <Orr-Sommerfeld equation> then has the same transverse operator because $k'^2=q^2$, and the same viscous coefficient because $k'R'=kR$. Its <boundary conditions> on $v$ and $v'$ are unchanged. Thus it has the same <eigenfunction> and the same phase-speed <eigenvalue>.
If the original disturbance grows, $\operatorname{Im}c>0$, so the transformed disturbance also grows. Since $R'\leq R$, every unstable three-dimensional mode maps to an unstable two-dimensional mode at a no larger <Reynolds number>. It follows that \b[the minimum critical Reynolds number is attained among two-dimensional disturbances]. This is the <Squire theorem> for temporal normal-mode onset, not a statement that three-dimensional transient amplification is impossible. The case $k=0$ is treated separately without dividing by $k$; its streamwise-uniform viscous <normal modes> decay, although coupling can produce <transient growth>.
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