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Rotate the contour for the Gaussian integral through angle

- π /4

; a small positive damping factor justifies the deformation and can then be removed. After scaling by

t^{- 1/2}

, the oscillatory Gaussian integral with negative phase gives

R \to \infty lim \int_{- R}^{R} e^{- i t u^{2}} d u = e^{- iπ /4} \frac{π}{t} .

Similarly, first scale

u = t^{- 1/3} v

and then rotate the positive ray through angle

- π /6

. Since

\int_{0}^{\infty} e^{- v^{3}} d v = Γ (4/3)

, the cubic oscillatory Gamma integral gives

R \to \infty lim \int_{0}^{R} e^{- i t u^{3}} d u = e^{- iπ /6} t^{- 1/3} Γ (\frac{4}{3}) .

(2)

For fixed

α

, consider the complex phase

Φ (θ) = x sin θ - α θ .

(3)

Its unique interior stationary point is

θ_{0} = arccos (α / x) = \frac{π}{2} - \frac{α}{x} + O (x^{- 3}),

(4)

where

Φ (θ_{0}) = x - \frac{π α}{2} + O (x^{- 1}), Φ^{''} (θ_{0}) = - x + O (x^{- 1}) .

(5)

The endpoint contributions are

O (x^{- 1})

, while the one-dimensional stationary-phase formula has order

x^{- 1/2}

. Taking its real part proves the stationary-phase asymptotic for the generalized Bessel cosine integral:

Q_{α} (x) \sim \frac{2}{π x} cos (x - \frac{π α}{2} - \frac{π}{4}) .

(6)

For

Q_{x} (x)

the phase is

x (sin θ - θ)

. Its stationary point has merged with the endpoint

θ = 0

, where

sin θ - θ = - \frac{θ ^{3}}{6} + O (θ^{5}) .

(7)

Use the cubic endpoint stationary-phase scaling

θ = (6/ x)^{1/3} u

. The other endpoint and the region away from zero contribute only

O (x^{- 1})

, so

Q_{x} (x) \sim \frac{1}{π} (\frac{6}{x})^{1/3} \int_{0}^{\infty} cos (u^{3}) d u = \frac{6 ^{1/3}}{π x ^{1/3}} Γ (\frac{4}{3}) cos \frac{π}{6} .

(8)

Therefore

Q_{x} (x) \sim \frac{3 6 ^{1/3}}{2 π} Γ (\frac{4}{3}) x^{- 1/3} .

(9)

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 39B / ii / Solution Created 2026-09-24 Updated 2026-09-29

Put

x = V t

with fixed

V > 0

. The relevant stationary point will have

k = O (V^{2} / S^{2})

, so the assumption

ϵ V^{2} / S^{2} ≪ 1

allows the long-wave approximation

sin (k ϵ) \sim k ϵ .

Using

sin A cos B = [sin (A + B) + sin (A - B)] /2

, the phase

S k^{3/2} + Vk

has no stationary point for

k > 0

. The other phase is

ϕ (k) = S k^{3/2} - Vk,

(2)

with

ϕ^{'} (k_{0}) = 0, k_{0} = \frac{4 V ^{2}}{9 S ^{2}},

(3)

and

ϕ (k_{0}) = - \frac{4 V ^{3}}{27 S ^{2}}, ϕ^{''} (k_{0}) = \frac{9 S ^{2}}{8 V} > 0.

(4)

The stationary part of the integral is therefore

- \frac{W ϵ}{π S} \int_{0}^{\infty} k^{- 3/2} sin [tϕ (k)] d k .

(5)

By the one-dimensional stationary-phase formula, this is asymptotic to

- \frac{W ϵ}{π S} k_{0}^{- 3/2} \frac{2 π}{t ϕ ^{''} ( k _{0} )} sin (tϕ (k_{0}) + \frac{π}{4}) .

(6)

Since

k_{0}^{- 3/2} \frac{2 π}{t ϕ ^{''} ( k _{0} )} = \frac{9 S ^{2} π}{2 V ^{5/2} t},

(7)

we obtain

η (x, t) \sim \frac{9 W ϵ S}{2 π V ^{5/2} t} sin (\frac{4 V ^{3} t}{27 S ^{2}} - \frac{π}{4}) .

(8)

Using

V = x / t

, this has the requested form

η (x, t) \sim \frac{W ϵ S t ^{2}}{x ^{5/2}} F (\frac{x ^{3}}{S ^{2} t ^{2}}),

(9)

where

F (q) = \frac{9}{2 π} sin (\frac{4 q}{27} - \frac{π}{4}) .

(10)

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 39C / b / i / Solution Created 2026-09-24 Updated 2026-09-25

The Fourier modes evolve according to the dispersion relation found in part (a), so

ϕ (V t, t) = \int_{- \infty}^{\infty} A (k) e^{i t ψ (k)} d k, ψ (k) = (V - U) k - \frac{k ^{9}}{9} .

For

V > U

, the stationary-point equation is

ψ^{'} (k) = V - U - k^{8} = 0.

(2)

It has the two real solutions

k = \pm k_{0}, k_{0} = (V - U)^{1/8} .

(3)

Define

Ω = ψ (k_{0}) = \frac{8}{9} k_{0}^{9}, ∣ ψ^{''} (\pm k_{0}) ∣ = 8 k_{0}^{7} .

(4)

The one-dimensional stationary-phase formula gives one contribution from each stationary point:

ϕ (V t, t) \sim \frac{π}{4 t k _{0}^{7}} [A (k_{0}) e^{i (Ω t - π /4)} + A (- k_{0}) e^{- i (Ω t - π /4)}] .

(5)

Because the initial field is real, its Fourier transform has the conjugate symmetry

A (- k) = A (k)^{*}

. Hence the same result can be written

ϕ (V t, t) \sim \frac{π}{t k _{0}^{7}} Re [A (k_{0}) e^{i (Ω t - π /4)}] .

(6)

Stationary-phase wake of an impulsive capillary wave 2026-09-29

Along

x = V t > 0

, if

ϵ V^{2} / S^{2} ≪ 1

, the stationary wavenumber

k_{0} = 4 V^{2} / (9 S^{2})

obeys

k_{0} ϵ ≪ 1

. The one-dimensional stationary-phase formula then gives

η (x, t) \sim \frac{W ϵ S t ^{2}}{x ^{5/2}} \frac{9}{2 π} sin (\frac{4 x ^{3}}{27 S ^{2} t ^{2}} - \frac{π}{4}) .

Only the integrated initial velocity

- 2 W ϵ

survives in this long-wave far field.