Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 31C / b

Solution

Near

x = λ

1 - \frac{λ ^{2}}{x ^{2}} = \frac{2 ( x - λ )}{λ} + O ((x - λ)^{2}) .

With

t = (\frac{2}{λ})^{1/3} \frac{x - λ}{ϵ ^{2/3}},

the leading equation is

y_{tt} - t y = 0

. Decay for

t > 0

selects

y = C Ai (t)

y \sim C^{'} p (x)^{- 1/2} sin (\frac{1}{ϵ} \int_{x}^{λ} p (s) d s + \frac{π}{4}) .

For this to be proportional to the inner expression from part (a), the two phases must differ by an integer multiple of

π

\int_{1}^{λ} (\frac{λ ^{2}}{x ^{2}} - 1)^{1/2} d x = ϵ (n + \frac{3}{4}) π, n = 0, 1, 2, \dots .

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