Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 2 / b / ii

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