 Solution

For $l=0$ put $L_D=NH/f_0$, $\eta =k{L}_D$ with $k>0$, and write the discriminant asThe second factor is positive for every $\eta >0$, since $x>\tanh x$. The first changes sign at the positive solution of $x=\coth x$, $x=\eta /2$. Numerically $x_c=1.19968$, so ${\eta }_c=2.39936$. The growing branch exists for $0<k{L}_D<{\eta }_c$, givingEquality is neutral, so this is an infimum wavelength, rather than an attained shortest growing mode.

In the unstable interval,Both phase-speed branches have the same real part there; the imaginary parts have opposite signs. At and beyond the short-wave cutoff the growth rate is zero and the two real speeds areThey split continuously from $1/2$ and approach the boundary-flow speeds zero and $\Lambda H$ at large $\eta$. As $\eta \to 0$, $D=-{\eta }^2/12+O({\eta }^4)$, so $\sigma \to 0$ linearly with $k$, despite a finite limiting $c_i=\Lambda H/\sqrt{12}$. A zonally uniform disturbance has no exponential growth from this mechanism.

The maximum growth minimizes $D$. Its derivative isThe positive interior minimum is at ${\eta }_m\simeq 1.606$, yieldingThe supplied phase-speed formula therefore predicts fastest growth at a wavelength of order four deformation lengths, not at the short-wave cutoff.