Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 1 / ii / Solution

Away from $z=0,\pm 1$, the base profile is linear or constant, so $U^{\prime \prime }=0$ and ${\varphi }^{\prime \prime }-{\alpha }^2\varphi =0$. Across a corner $z=z_0$, integration of the Rayleigh equation gives the jump conditionFor the even mode, at $z=0$ this givesFor $0<z<1$, therefore,For $z>1$, decay requires $\varphi \propto e^{-\alpha (z-1)}$. The jump at $z=1$, where $[U^{\prime }]=1$, givesSubstitution and elementary simplification yield

The coefficients are real, so instability occurs when the quadratic discriminant is negative. At $\alpha =1$ it iswhereas at $\alpha =2$, using $e^{-4}\simeq 1/55$, it is positive. By continuity there is a first threshold $1<{\alpha }_s<2$ at which the discriminant vanishes. Hence one conjugate root has $c_i>0$ for