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

Table of contents
- Solution b

Solution

Put

ϕ = (U - c) F

. The Rayleigh equation for inviscid shear flow becomes

[(U - c)^{2} F^{'}]^{'} - α^{2} (U - c)^{2} F = 0.

Multiplication by

F^{*}

and integration gives

\int_{0}^{1} (U - c)^{2} Q d z = 0, Q = ∣ F^{'} ∣^{2} + α^{2} ∣ F ∣^{2} \geq 0.

For

c_{i} > 0

, its real and imaginary parts imply

c_{r} = \frac{\int U Q d z}{\int Q d z}, c_{i}^{2} = \frac{\int ( U - c _{r} ) ^{2} Q d z}{\int Q d z} .

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:

(c_{r} - \frac{U _{+} + U _{-}}{2})^{2} + c_{i}^{2} \leq (\frac{U _{+} - U _{-}}{2})^{2} .

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