week11 - 6 Nonlinear waves Note: this material will not be...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 Nonlinear waves Note: this material will not be examined 6.1 Quadratic corrections So far we have restricted ourselves to the case of waves of small amplitude, which means we only took terms linear in the wave amplitude into account. Now we will consider quadratic terms in the boundary conditions. Comparing their size to that of the leading linear calculation will permit us to give a more precise meaning to what we shall call “small amplitude”. As an example, consider the kinematic condition (53), where the velocities u and w are to be evaluated at the free surface z = ζ . As discussed before, this can be reduced to quantities evaluated at a fixed position z = 0 by performing a Taylor expansion: ∂φ ∂x vextendsingle vextendsingle vextendsingle vextendsingle z = ζ = ∂φ ∂x vextendsingle vextendsingle vextendsingle vextendsingle z =0 + ζ ∂ 2 φ ∂x∂z vextendsingle vextendsingle vextendsingle vextendsingle z =0 + . . ., ∂φ ∂z vextendsingle vextendsingle vextendsingle vextendsingle z = ζ = ∂φ ∂z vextendsingle vextendsingle vextendsingle vextendsingle z =0 + ζ ∂ 2 φ ∂z 2 vextendsingle vextendsingle vextendsingle vextendsingle z =0 + . . .. Thus (53), including terms up to second order, becomes ∂ζ ∂t + ∂φ ∂x ∂ζ ∂x = ∂φ ∂z + ζ ∂ 2 φ ∂z 2 , on z = 0 , (60) which means there are two new quadratic terms. Now we insert the solution found in section 5.3, and compare the relative size of linear and quadratic terms. For example, we find ∂φ ∂z = Z ′ (0)cosΦ = − gHk ω tanh kh cos Φ = − Hω cosΦ , where we wrote Φ = kx − ωt for the phase factor for brevity. On the other hand, the nonlinear term on the left of (60) is ∂φ ∂x ∂ζ ∂x = − k 2 HZ (0)sinΦcos Φ = gH 2 k 2 2 ω sin2Φ . Thus the condition for the amplitude of the nonlinear term to be small relative to the linear one, assuming formally cosΦ and sin2Φ assume their maximum values, is gH 2 k 2 2 ω 2 H = Hk 2tanh kh ≪ 1 . (61) In the case of infinite depth h → ∞ this simplifies to Hk/ 2 ≪ 1. In other words, the small parameter that measures the size of the nonlinearity is Hk , staying with the case of infinite depth for the rest of this chapter....
View Full Document

This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.

Page1 / 6

week11 - 6 Nonlinear waves Note: this material will not be...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online