= Solution
Put $y=\widetilde c^2$ and $s=J/(2\alpha)>0$. The <dispersion relation for two density interfaces in uniform shear> is
$$
y^2-2(1+s)y+(1-s)^2-s^2e^{-4\alpha}=0.
$$
Its <discriminant> is $16s+4s^2e^{-4\alpha}>0$, so both roots $y_\pm$ are real. Their sum is positive; therefore one is negative precisely when their product is negative. That condition is
$$
(1-s)^2<s^2e^{-4\alpha},\qquad |1-s|<se^{-2\alpha}.
$$
Solving for $J$ gives
$$
\boxed{\frac{2\alpha}{1+e^{-2\alpha}}<J<\frac{2\alpha}{1-e^{-2\alpha}}.}
$$
In this interval $y_-<0$, so $\widetilde c=i\sqrt{-y_-}$ is the growing <normal mode>. At either endpoint the corresponding root is zero and the mode is neutral.
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