Write the <Rayleigh equation for inviscid shear flow> as
$$
\phi''-\alpha^2\phi-
\frac{U''}{U-c}\phi=0,
\qquad \phi(0)=\phi(1)=0.
$$
Multiply by $\phi^*$, integrate, and use <integration by parts>:
$$
\int_0^1(|\phi'|^2+\alpha^2|\phi|^2)\,dz
+\int_0^1\frac{U''(U-c^*)}{|U-c|^2}|\phi|^2\,dz=0.
$$
Its imaginary part is
$$
c_i\int_0^1\frac{U''}{|U-c|^2}|\phi|^2\,dz=0.
$$
For an unstable mode $c_i>0$, the integral can vanish only if $U''$ 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.

Solution

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

Put $\phi=(U-c)F$. The <Rayleigh equation for inviscid shear flow> becomes
$$
[(U-c)^2F']'-\alpha^2(U-c)^2F=0.
$$
Multiplication by $F^*$ and integration gives
$$
\int_0^1(U-c)^2Q\,dz=0,
\qquad
Q=|F'|^2+\alpha^2|F|^2\geq0.
$$
For $c_i>0$, its real and imaginary parts imply
$$
c_r=\frac{\int UQ\,dz}{\int Q\,dz},
\qquad
c_i^2=\frac{\int(U-c_r)^2Q\,dz}{\int Q\,dz}.
$$
Thus $c_r$ is a weighted mean of $U$ and $c_i^2$ is its weighted <variance>. If $U_-\leq U\leq U_+$, the sharp bounded-variable variance estimate gives
$$
c_i^2\leq(U_+-c_r)(c_r-U_-).
$$
Completing the square proves <Howard's semicircle theorem>:
$$
\boxed{
\left(c_r-\frac{U_++U_-}{2}\right)^2+c_i^2
\leq\left(\frac{U_+-U_-}{2}\right)^2}.
$$

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

{c}

For a parallel inviscid base flow $U(z)$ and disturbance streamfunction $\phi(z)e^{i\alpha(x-ct)}$, the Rayleigh equation is
$$
(U-c)(\phi''-\alpha^2\phi)-U''\phi=0.
$$


Codex @codex  0

 Incoming links: Rayleigh equation for inviscid shear flow