For fixed $h\in(0,1)$, $X=\alpha h+O(\alpha^3)$ and $Y=\alpha(1-h)+O(\alpha^3)$. The factored <dispersion relation> therefore gives
$$
\lim_{\alpha\to0}c^2=2h-1,\qquad
\lim_{\alpha\to\infty}c^2=1,
$$
where the latter follows from $X,Y\to1$ and $c^2\sim(1-1/(2\alpha h))^2$. A real negative $c^2$ yields a growing branch $c=i\sqrt{-c^2}$ for positive $k$. Thus $h<1/2$ gives instability at sufficiently small nonzero wavenumber.

For the converse, monotonicity toward the limiting value one would suffice, as permitted in the question. In fact a direct sign argument avoids this extra assumption. With $q=\alpha h>0$, $X=\tanh q<q$, so $J-XY=qX+(q-X)Y>0$. If $h\ge1/2$, then $q\ge d=\alpha(1-h)$ and $Y=\tanh d<d\le q$. Hence $H-Y=q(1+XY)-Y>0$, so $c^2>0$ for every $\alpha>0$. We obtain the exact <instability threshold for a bounded piecewise-linear shear layer>:
$$
\boxed{0<\zeta_L<\frac12\quad\text{is the condition for exponential linear instability}.}
$$
At $\zeta_L=1/2$ the long-wave limiting value is zero, but every finite positive wavenumber has $c^2>0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 2 / c / Solution

{title2=$L_s/L<1/2$}

The <bounded piecewise-linear shear layer> has exponentially growing modes exactly when $h<1/2$. Its squared phase speed tends to $2h-1$ as $kL\to0$ and to $1$ as $kL\to\infty$. For $h<1/2$, sufficiently long nonzero waves are therefore unstable. Conversely $J-XY>0$, and for $h\ge1/2$, $q\ge\alpha(1-h)>Y$, so $H-Y>0$ for every positive wavenumber and $c^2>0$. This proves the threshold without assuming monotonicity of the dispersion curve.

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