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>:
$$
\boxed{
\left(c_r-\frac{U_++U_-}{2}\right)^2+c_i^2
\leq\left(\frac{U_+-U_-}{2}\right)^2}.
$$

Solution

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

{c}

If an inviscid shear-flow mode has complex wave speed $c$ and $U_-\leq U\leq U_+$, then
$$
\left(\operatorname{Re}c-\frac{U_++U_-}{2}\right)^2
+(\operatorname{Im}c)^2
\leq\left(\frac{U_+-U_-}{2}\right)^2.
$$


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 Incoming links: Howard's semicircle theorem