Upload
oberon
View
52
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Instabilities in the Forced Truncated NLS. Eli Shlizerman and Vered Rom-Kedar Weizmann Institute of Science. 1. ES & RK, Characterization of Orbits in the Truncated NLS Model, ENOC-05. 2. ES & RK, Hierarchy of bifurcations in the truncated and forced NLS model, CHAOS-05. - PowerPoint PPT Presentation
Citation preview
Instabilities in the Forced Instabilities in the Forced Truncated NLSTruncated NLS
Eli Shlizerman and Vered Rom-KedarWeizmann Institute of Science
SNOWBIRD, 2005
1. ES & RK, Characterization of Orbits in the Truncated NLS Model, ENOC-05 2. ES & RK, Hierarchy of bifurcations in the truncated and forced NLS model, CHAOS-05
3. ES & RK, Energy surfaces and Hierarchies of bifurcations - Instabilities in the forced truncated NLS, Cargese-03
Near-integrable NLSNear-integrable NLS
Conditions Periodic Boundary u (x , t) = u (x + L , t) Even Solutions u (x , t) = u (-x , t)
Parameters Forcing Frequency Ω2
Wavenumber k = 2π / L
ixxt eiBBBiB )( 22
(+) focusing
dispersion
Homoclinic OrbitsHomoclinic Orbits
For the unperturbed eq. B(x , t) = c (t) + b (x,t)
Plain Wave Solution Bpw(0 , t) = |c| e i(ωt+φ₀)
Homoclinic Orbit to a PW Bh(x , t) t±∞ Bpw(0 , t)
Bpw
[McLaughlin, Cai, Shatah]
Bh
Resonance – Circle of Fixed PointsResonance – Circle of Fixed Points
When ω=0 – circle of fixed points occur Bpw(0 , t) = |c| e i(φ₀)
Heteroclinic Orbits!
[Haller, Kovacic]
Bpw
φ₀
Bh
Two Mode ModelTwo Mode Model Consider two mode Fourier truncation
B(x , t) = c (t) + b (t) cos (kx)
Substitute into the unperturbed eq.:
2222222224224 cb+c b8
1 |c|
2
1-|b|k+Ω
2
1-|b|
16
3|c||b|
2
1|c|
8
1
=H
[Bishop, McLaughlin, Forest, Overman]
=I
)|b||c(|2
1 22 )cΓ(c2
εiH *
p
General Action-Angle Coordinates General Action-Angle Coordinates for cfor c≠≠00
Consider the transformation: c = |c| eiγ b = (x + iy) eiγ I = ½(|c|2+x2+y2)
[Kovacic]
3y41y2x
43y2kx
2xy433x
47x2I2ky
2xI2Ωγ
0Iγ
H
Plain Wave StabilityPlain Wave Stability
Then the 2 mode model is plausible for I < 2k2
Plain wave: B(0,t)= c(t)
Introduce x-dependence of small magnitude B (x , t) = c(t) + b(x,t)
Plug into the integrable equation and solve the linearized equation. From dispersion relation get instability for:
0 < k2 < |c|2
Hierarchy of BifurcationsHierarchy of Bifurcations
Level 1 Single energy surface - EMBD, Fomenko
Level 2 Energy bifurcation values - Changes in EMBD
Level 3 Parameter dependence of the energy
bifurcation values - k, Ω
Preliminary step - Local StabilityPreliminary step - Local Stability
Fixed Point Stable Unstable
x=0 y=0 I > 0 I > ½ k2
x=±x2 y=0 I > ½k2 -
x =0 y=±y3 I > 2k2 -
x =±x4 y=±y4 - I > 2k2
[Kovacic & Wiggins]
B(x , t) = [|c| + (x+iy) coskx ] eiγ
Level 1: Singularity SurfacesLevel 1: Singularity Surfaces
Construction of the EMBD -(Energy Momentum Bifurcation Diagram)
Fixed Point H(xf , yf , I; k=const, Ω=const)
x=0 y=0 H1
x=±x2 y=0 H2
x =0 y=±y3 H3
x =±x4 y=±y4 H4
[Litvak-Hinenzon & RK]
EMBDEMBD
Parameters: k=1.025 , Ω=1
H2
H1
H3
H4
Dashed – Unstable
Full – Stable
Fomenko Graphs and Energy Fomenko Graphs and Energy SurfacesSurfaces
Example: H=const (line 5)
Level 2: Energy Bifurcation ValuesLevel 2: Energy Bifurcation Values
4 65**
Possible Energy BifurcationsPossible Energy Bifurcations Branching surfaces – Parabolic Circles Crossings – Global Bifurcation Folds - Resonances
H
I
0I
H0θ
pI
31 HH
Finding Energy BifurcationsFinding Energy Bifurcations
Resonance Parabolic GB
What happens when energy What happens when energy bifurcation values coincide?bifurcation values coincide?
Example: Parabolic Resonance for (x=0,y=0)
Resonance IR= Ω2
hrpw = -½ Ω4
Parabolic Circle Ip= ½ k2
hppw = ½ k2(¼ k2 - Ω2)
Parabolic Resonance: IR=IP k2=2Ω2
Level 3: Bifurcation ParametersLevel 3: Bifurcation Parameters
Example of a diagram:
Fix k
Find Hrpw(Ω)
Find Hppw(Ω)
Find Hrpwm(Ω)
Plot H(Ω) diagram
Perturbed motion classification Perturbed motion classification
Close to the integrable motion
“Standard” dyn. phenomena Homoclinic Chaos, Elliptic Circles
Special dyn. phenomena PR, ER, HR, GB-R
Homoclinic ChaosHomoclinic Chaos
k=1.025, Ω=1, ε ~ 10-4 i.c. (x, y, I, γ) = (0,0,1.5,π/2)
Model
PDE
Hyperbolic ResonanceHyperbolic Resonance
k=1.025, Ω2=1, ε ~ 10-4 i.c. (x, y, I, γ) = (0,0,1,π/2)
Model
PDE
k=1.025, Ω2=k2/2, ε ~ 10-4 i.c. (x, y, I, γ) = (0,0,k2/2,π/2)
Model
PDE
Parabolic ResonanceParabolic Resonance
ClassificationClassification
x
y
Measure: σmax = std( |B0j| max)
Measure Dependence on Measure Dependence on εε
p is the power of the order: O(εp)
DiscussionDiscussion
Solutions close to HR
Stability of solutions
Applying measure to PDE results