daniel ratli & tom bridges august 9th, siam nwcs 2016 · daniel ratli & tom bridges august 9th,...

Phase Dynamics of Multiphase Wavetrains


Daniel Ratliff & Tom Bridges

August 9th, SIAM NWCS 2016

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Introduction

Motivation

How does modulation work for systems with multiple conservation laws?

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Introduction

Lagrangian and Conservation Laws I

We consider systems generated by a Lagrangian density,

L (Z) =

∫∫L(U,Ux ,Ut) dx dt (1.1)

which generates the Euler-Lagrange equation for the system of interest.Assume the existence of a two-phased periodic travelling wave solution

Z(x , t) = Ẑ(θ1, θ2) ≡ Ẑ(θ), Ẑ(θ1 + 2π, θ2) = Ẑ(θ) = Ẑ(θ1, θ2 + 2π),

θi =kix + ωi t + θ0i

(1.2)

for wavenumbers ki and frequencies ωi . This solution is a relative equilibriumfor the system.

In the symmetry case, small divisors are avoided.

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Introduction

Lagrangian and Conservation Laws II

The Lagrangian can be averaged over these phases:

L̂ =1

4π2

∫ 2π0

∫ 2π0

L(Ẑ , ω1Ẑθ1 , ω2Ẑθ2 , k1Ẑθ1 , k2Ẑθ2) dθ1 dθ2.

We can then extract the conservation law components evaluated along thewave as

A(k,ω) =

(L̂ω1L̂ω2

), B(k,ω) =

(L̂k1L̂k2

)with k =

(k1k2

), ω =

(ω1ω2

)(1.3)

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Introduction

Modulation Approach

Idea: Find wavetrain Ẑ , then consider an ansatz by perturbing the independentvariables (modulation) as

Z = Ẑ(θ + φ(X ,T ), k + εq(X ,T ),ω + εΩ(X ,T )

)+ ε2W (θ,X ,T )

with X = εx , T = εt and ε� 1. Method is to substitute the ansatz into theEuler-Lagrange equation, expand around ε = 0 and solve at each order.

Strengths of the approach:

Do asymptotics on general Lagrangian once, then result applies to allsystems that can be put in that form.

Coefficients are related to properties of the basic state - can be determineda-priori and are simple to compute.

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Degeneracies of the Vector Whitham System

The Linear Whitham System and Degeneracy

Analysis leads to the linear Whitham system

qT = ΩX , DωAΩT + DkAqT + DωBΩX + DkBqX = 0. (2.4)

(Ablowitz and Benney in the symmetry case)Write the system as(qΩ

)T

+W(k,ω)

(qΩ

)X

= 0, W =

(0 −I

DωA−1DkB DωA

−1(DkA + DkB)

).

When W has zero eigenvalues, system is degenerate.

The number of zero eigenvalues of W determines the nonlinear correction:

simple: |DkB| = 0 - KdV,double nonsemisimple - Boussinesq.

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Simple Zero Eigenvalue

In this case|DkB| = 0 ⇒ ∃ ζ with DkBζ = 0

and as a consequence in the theory it must be that

q = ζU.

Modulate as

Z = Ẑ(θ + εφ, k + ε2q,ω + ε4Ω) + ε3W (θ,X ,T )

and X = εx , T = ε3t. Result is[(DkA + DωB)ζ

]UT + D

2kB(ζ, ζ)UUX + KUXXX + DkBαXX = 0, (2.5)

Project using ζ[ζT (DkA + DωB)ζ

]UT + ζ

TD2kB(ζ, ζ)UUX + ζTKUXXX = 0. (2.6)

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Double Nonsemisimple Zero Eigenvalue

The additional requirement needed for this to occur is

∃ γ with (DkA + DωB)ζ = DkBγ.

When this and the previous condition are met, modulate as

Z = Ẑ(θ + εφ, k + ε2q,ω + ε3Ω) + ε3W (θ,X ,T )

and X = εx , T = ε2t. The scalar PDE emerging is the two-way Boussinesq:[ζT(DωAζ − (DkA + DωB)γ

)]UTT +

[1

2ζTD2kB(ζ, ζ)U

2 + ζTKUXX

]XX

= 0.

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Examples

Example 1 - Stratified Shallow water

Consider the potential stratified shallow water system

(ρ1η)t +(ρ1η(ψ2)x

)x= 0,

(ρ2χ)t +(ρ2χ(ψ2)x

)x= 0,

(ρ1ψ1)t +ρ1

2(ψ1)

2x + gρ1η + gρ2χ = R1 + a11ηxx + a12χxx ,

(ρ2ψ2)t +ρ2

2(ψ2)

2x + gρ2η + gρ2χ = R2 + a21ηxx + a22χxx

(3.7)

for velocity potentials ψi , free surface heights η, χ and Bernoulli constants Ri .

One solution to the above is the constant velocity profile:

ψi = θi , η = η0(k,ω), χ = χ0(k,ω)

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Examples

Criticality 1: KdV

The conservation vectors are

A =

(ρ1η0ρ2χ0

), B =

(ρ1k1η0ρ2k2χ0

)Criticality in B occurs when

(1− F 21 )(1− F 22 ) =ρ2ρ1≡ r ,

with Fi is the Froude number in the respective layer. Matches the literature(Lawrence, Benton).The modulation theory gives the resulting KdV as

ρ2χ0

(k1gη0

(1− F 22 ) +k2gχ0

(1− F 21 ))UT

− 32ρ1ρ2k2

(χ0r(1− F 22 )F 21 − η0(1− F 21 )2F 22

)UUX

− χ02g

(a11r(1− F 22 )− 2ra12 + (1− F 21 )a22)UXXX = 0 .

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Examples

Criticality 2: Boussinesq

The time term vanishes exactly at the double nonsemisimple point, and so thetheory predicts the two-way Boussinesq as

χ0

(1− F 22gη0

+1− F 21gχ0

− 4k1k2g 2η0χ0

)UTT

+

[3

4ρ1ρ2k2

(χ0r(1− F 22 )F 21 − η0(1− F 21 )2F 22

)U2

+χ02g

(a11r(1− F 22 )− 2ra12 + (1− F 21 )a22)UXX

]XX

= 0 .

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Examples

Example 2 - Coupled NLS equations

Consider

iAt + Axx + α|A|2A + β|B|2A = 0,

iBt + Bxx + α|B|2B + β|A|2B = 0.(3.8)

Models Bose-Einstein condensates, freak wave formation. (e.g. see Roskes,Onorato et al., Salman and Berloff)

We can find relative equilibria of the form

A = A0eiθ1 , B = B0e

iθ2 with

(α ββ α

)(|A0|2|B0|2

)=

(ω1 + k

21

ω2 + k22

). (3.9)

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Examples

Coupled NLS to KdV

The conservation law vectors are

A =1

2

(|A0|2|B0|2

), B =

(k1|A0|2k2|B0|2

),

and the required criticality condition is((α2 − β2)|A0|2 + 2αk21

)((α2 − β2)|B0|2 + 2αk22

)= 4β2k21k

22 ,

. When this is met, the KdV equation arises:[(α|B0|2 + 2k22 )k1 + (α|A0|2 + 2k21 )k2

]UT

+3k2

α2 − β2

((|A0|2 + 2αk21 )(α|A0|2 + 2k21 )− β|A0|2(1 + 2αk22 )

)UUX

+

(|A0|2(2α3k21k22 + β2k21 + α2k22 ) + |B0|2(2α3k21k22 + α2k21 + β2k22 )

)2|A0|2|B0|2(α2 − β2)2

UXXX = 0 ,

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Examples

Summary and Next Steps

qT = ΩX , DωAΩT + DkAqT + DωBΩX + DkBqX = 0.

Simple Zero (|DkB| = 0)Double zero (SS)(3+ phases)

Coupled set of KdVs[ζT (DkA + DωB)ζ

]UT + ζ

TD2kB(ζ, ζ)UUX + ζTKUXXX = 0.

Double Zero(NSS)

Two-Way Boussinesq

ζTD2kB(ζ, ζ) = 0

mKdV?

ζTK = 0

Fifth order KdV?

Extensions to N spatial dimensions, arbitrary no. phaseand Gradient reaction-diffusion systems

(in progress)

Thanks for listening!

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IntroductionDegeneracies of the Vector Whitham SystemExamples

daniel ratli & tom bridges august 9th, siam nwcs 2016 · daniel ratli & tom bridges august 9th,...

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