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

Write the Rayleigh equation for inviscid shear flow asMultiply by ${\varphi }^{\ast }$, integrate, and use integration by parts:Its imaginary part isFor an unstable mode $c_i>0$, the integral can vanish only if $U^{\prime \prime }$ changes sign somewhere in the flow. Thus the velocity profile must have an inflection point. This is Rayleigh's inflection-point theorem; it is necessary, not sufficient, for inviscid instability.

Put $\varphi =(U-c)F$. The Rayleigh equation for inviscid shear flow becomesMultiplication by $F^{\ast }$ and integration givesFor $c_i>0$, its real and imaginary parts implyThus $c_r$ is a weighted mean of $U$ and $c_i^2$ is its weighted variance. If $U_{-}\le U\le U_{+}$, the sharp bounded-variable variance estimate givesCompleting the square proves Howard's semicircle theorem:

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