39B (Waves)

Use the Fourier transform convention

η (k, t) = \int_{- \infty}^{\infty} η (x, t) e^{- ik x} d x, η (x, t) = \frac{1}{2 π} \int_{- \infty}^{\infty} η (k, t) e^{ik x} d k .

(1)

The deep-water capillary-wave dispersion relation makes each Fourier mode satisfy

\partial_{t}^{2} η + S^{2} ∣ k ∣^{3} η = 0.

(2)

The transformed initial data are

η (k, 0) = 0, \partial_{t} η (k, 0) = - W \int_{- ϵ}^{ϵ} e^{- ik x} d x = - \frac{2 W s i n ( k ϵ )}{k} .

(3)

Therefore

η (k, t) = - \frac{2 W s i n ( k ϵ )}{k S ∣ k ∣ ^{3/2}} sin (S ∣ k ∣^{3/2} t) .

(4)

The integrand is even, so Fourier inversion gives the surface response to a localized impulsive velocity

η (x, t) = - \frac{2 W}{π S} \int_{0}^{\infty} \frac{sin ( k ϵ ) sin ( S k ^{3/2} t ) cos ( k x )}{k ^{5/2}} d k .

(5)

ii

 0  0

Solution

 0  0

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 .

(1)

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)

iii

 0  0

Solution

 0  0

The stationary contribution travels with group velocity

V = ω^{'} (k_{0}) = \frac{3 S}{2} k_{0},

(1)

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

. The condition

\frac{ϵ V ^{2}}{S ^{2}} ≪ 1

(2)

is therefore equivalent, up to a constant factor, to

k_{0} ϵ ≪ 1

: the dominant wavelength is much longer than the impacted interval. Its Fourier transform can then be replaced by

- \frac{2 W s i n ( k ϵ )}{k} \sim - 2 W ϵ,

(3)

which is the total initial vertical velocity integrated over space. The far field cannot resolve

W

and

ϵ

separately and depends only on their product

W ϵ

 Ancestors (9)

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 39B (Waves)

 Discussion (0)  Subscribe (1)

 Discussion (0)