Fourier mode dynamics for NLS Synchronization in fiber lasers arrays

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Fourier mode dynamics for NLS Synchronization in fiber lasers arrays Nonlinear Schrodinger workshop Heraklio, May 21 2013 Jean-guy Caputo Laboratoire de Mathématiques INSA de Rouen, France A. Aceves, N. K. Efremidis, Chao Hang

Motivation : NLS in finite systems Fermi, Pasta, Ulam (1947) Masses and nonlinear springs recurrence in Fourier space Dissipative systems Cascade for Burgers (Muraki) Wave turbulence (Newell, Zakharov, Dias..) Nonlinear Schrodinger, dynamics of transfer in optical communications (Zharnitsky) arrays of fiber lasers real engineering boundary conditions : Dirichlet, Neuman

Nonlinear Schrodinger in 1D domain iψ t +ψ xx + ψ 2d ψ = 0 Simple model d = 1 standard cubic d = 2 critical NLS, collapse Fourier machinery : analysis

Nonlinear Schrodinger in 1D domain iψ t +ψ xx + ψ 2d ψ = 0 Dirichlet boundary conditions ψ(x = 0) = ψ(x = π) = 0 Neuman boundary conditions ψ x (x = 0) = ψ x (x = π) = 0 Simple model : collapse, energy transfer between Fourier modes L 2 norm and Hamiltonian conserved P = π 0 ψ 2 dx, H = ( ) π 0 ψ x 2 1 d+1 ψ 2d+2 Momentum flux Π = i π 0 (ψψ x ψ ψ x )dx, Π t = [ 4[ ψ x 2 ] π 0 Dirichlet ] π Π t = ( ψ 2 ) xx + 2d d+1 ψ 2d+2, Neuman 0

Collapse : virial Dirichlet d 2 ( π dt 2 0 ψ 2 x 2 dx) = 4π[ ψ x 2 ] x=π +8H 4 π 0 H < 0 sufficient for collapse d 2 d+1 ψ 2d+2 dx Neuman [ ] d 2 I 1 = 2π ( ψ 2 ) dt 2 xx + 2d d+1 ψ 2d+2 +8H 4 π d 2 x=π 0 d+1 ψ 2d+2 dx H < 0 not sufficient for collapse

Numerical procedure Linear part iu t +u xx = 0, û(dt) = e ik2 dtû(0), non linear part u(2dt) = e i u(dt) 2d dt u(dt). discrete sine Fourier transform (Matlab) û(k) = N πkn n=1u(n)sin( N+1 ), k = 1,...N, u(n) = N πkn k=1û(k)sin( N+1 ), n = 1,...N. N = 2 11 1 or N = 2 12 1, step dt = 10 4. L 2 norm conserved up to 10 10 in absolute value.

Numerical simulations cubic / quintic qualitative difference collapse occurs on fast time scales

Numerical results cubic (d = 1) c m (t), m = 1,3,5 two different initial conditions ψ(0,x) = sin(x)+2sin(3x) (a) ψ(0,x) = sin(x)+sin(3x) (b) No long time transfer

Numerical results cubic (d = 2) left panels, top to bottom ψ(0,x) = 1.0sin(x) (P = 1.57), ψ(0,x) = 1.3sin(x) (P = 2.65), ψ(0,x) = 1.31sin(x) right panels, Fourier spectra at t = 4

Numerical results cubic (d = 2) ψ(0,x) = 1.3sin(x) (P = 2.65) left panels ψ, right panels Fourier spectra Top to bottom t = 0, 0.6, 1.2,1.8

Evolution of the Fourier modes (d = 2) c 1 (0) = A = 1,1.1,1.2 Bottom left max c m, m = 1,3,5,7,9 vs. A

Resonant coupling of Fourier modes : cubic nonlinearity ψ(x,t) = m=1 c m(t)sin(mx) iċ q q 2 c q + 2 π k,l,m c kc l cm klm q = 0, klm q π 0 sin(kx)sin(lx)sin(mx)sin(qx)dx. Transform c q = a q e iq2 t iȧ q + 2 π k,l,m a ka l am klm q e i(k2 +l 2 m 2 q 2 )t = 0. Resonant condition k 2 +l 2 m 2 q 2 = 0 Maple ( ) iȧ q +a q P aq 2 = 0 P = j=1 a j 2 conserved 4 d dt ( a q 2 ) = 0 no transfer of energy same for Neuman

Fourier modes : Collapse Parseval π 0 ψ 2 dx = π 2 m=1 c m(t) 2 when collapse, series ceases to converge pointwise

Resonant coupling of Fourier modes : Dirichlet quintic 1 Assume odd terms ψ(x,t) = a 1 e it sin(x)+a 3 e i9t sin(3x)+a 5 e i25t sin(5x) iȧ q = 2 π k,l,m,n,p a ka l a m ana p klmnp q e i(k2 +l 2 +m 2 n 2 p 2 q 2 )t Resonant condition k 2 +l 2 +m 2 n 2 p 2 q 2 = 0 Maple [ iȧ 1 +a 9 1 4 a 1 2 P 13 8 a ( [ 1 4 +3 a 3 2 a 5 2 + 9 8 a3 4 + a 5 4)] = 3 8 a 1 a3 3 a 5, iȧ 3 +a 9 3 4 a 3 2 P 13 8 a ( [ 3 4 +3 a 1 2 a 5 2 + 9 8 a1 4 + a 5 4)] = 9 16 a2 1 a 2 3 a 5, iȧ 5 +a 9 5 4 a 5 2 P 13 8 a ( 5 4 +3 a 1 2 a 3 2 + 9 8 a1 4 + a 3 4)] = 3 16 a 2 1 a3 3 where P = a 1 2 + a 3 2 + a 5 2 and dp dt = 0

Resonant coupling of Fourier modes : Dirichlet quintic 2 Remarks If a 7 present additional terms in ȧ 1 a 3 2 a 2 5 a 7, a 2 5 a 7 2 a 7, a2 5 a 5 2 a 7. If even modes are included extra resonant term a 2 2 a 2 4 a 5 in ȧ 1

Resonant coupling of Fourier modes : Dirichlet quintic 3 Reduction I j = a j 2 İ 1 = 3 4 I 1I 3/2 3 I 1/2 5 sin θ, İ 3 = + 9 8 I 1I 3/2 3 I 1/2 5 sin θ, İ 5 = 3 8 I 1I 3/2 3 I 1/2 5 sin θ, where θ = 2θ 1 +3θ 3 θ 5 with θ j = arga j. or İ 1 = 3 4 I 1I 3/2 3 I 1/2 5 sin θ, θ = 1 4 (13I2 1 +3I2 3 +7I2 5 +21I 1I 3 +27I 1 I 5 +12I 3 I 5 ) + cosθ 8 [6I1/2 1 I 3/2 3 I 1/2 5 27I 1 I 3 I 1/2 5 3I 1 I 3/2 3 I 1/2 5 ]

Resonant coupling of Fourier modes : Dirichlet quintic 4 Comparison numerical solution of NLS vs reduced model Initial conditions I 1 = 0.64,I 3 = 0.36,I 5 = 0.16 and θ = π

Resonant coupling of Fourier modes : Neuman quintic zero mode couples to a i, i > 0 iȧ 0 +a 0 [...] = 3 4 a 0 a 1a2 2a 3, iȧ 1 +a 1 [...] = 3 4 a2 0 a 2 2 a 3, iȧ 2 +a 2 [...] = 3 2 a2 0 a 1 a 2 a 3 iȧ 3 +a 3 [...] = 3 4 a 2 0 a 1a2 2 The 4 modes a i, i = 1 4 need to be present for coupling Route to collapse changed

Partial conclusion Analyzed and solved numerically 1D NLS on a finite interval evolution of Fourier modes Cubic nonlinearity : no coupling for long times, trace of integrability Quintic : no collapse : resonant terms, good reduced model Quintic collapse : quick time, series breaks technique does not see modulational instability Extend approach to other systems : NLS on a graph

The model : NLS on a graph 3 3 1 2 1 2 4 4 1 1 2 2 α 3 4 β 3 4 iu t = Gu + u 2 u G graph Laplacian U = u 1 u 2 u 3 u4, G = 1 1 0 0 1 2 α α 1 0 α α β β 0 1 β 1 β, N(U) = u 1 2 u 1 u 2 2 u 2 u 3 2 u 3 u 4 2 u 4.

Eigenmodes of Laplacian, amplitude equations G symmetric, eigenvectors z i Gz i = ω 2 i z i. GZ = ZD Z orthogonal D diagonal D(i,i) = ω 2 i New coordinates U = ZΓ iγ k = ωk 2γ k + 4 j=1 z jk u j 2 u j Expanding u i γ k = ω 2 k γ k + jlmn z jlz jm z jn γ l γ m γ n Eliminate natural frequency ω 2 k γ k = e iω2 k t β k, iβ k = jlmn z jlz jm z jn β l β m βn e i(ω2 l +ω2 m ωn ω 2 k 2)t

Rotating wave approximation Resonant condition : ω 2 l +ω 2 m ω 2 n ω 2 k = 0 If all eigenvalues ω j are distinct, resonant condition satisfied if l = n and m = k Resonant evolution of β k iβ k = β k jl z2 jl z jk β l 2 Intensity in each mode is constant, I k = β k 2, I k = 0. β k (t) = β k (0)exp( i t Ω k ), Ω k = jl z2 jl z jk β l 2 γ k (t) = γ k (0)exp [ i t (ω 2 k Ω k) ]

Case study : a graph with a swivel 1 1 2 2 3 α 3 4

Eigenmodes 3 4 ω 2 1 0 3 2 1 0 1 2 3 4 5 α index i 1 2 3 4 λ i = ωi 2 0-1 1 2 (δ 2α 3) 2 1 (δ + 2α + 3) 1/2 1/ 2 1/n 3 1/n 4 z i 1/2 0 1 (δ 2α 1) 2n 3 2n 1 (δ + 2α + 1) 4 1/2 0 2n 1 (δ 2α + 3) 1 (δ + 2α 3) 3 2n 4 1/2 1/ 2 1/n 3 1/n 4 δ = 4α 2 4α + 9, n 3 = 2 + (δ 2α 1) 2 /4 + (δ 2α + 3) 2 /4, n 4 = 2 + (δ + 2α + 1) 2 /4 + (δ + 2α 3) 2 /4.

Eigenmodes 2 1 0.5 1,4 1 0.5 3 z 3 i 0 z 4 i 0 1,4-0.5 2 3-0.5 2-1 0 1 2 3 4 5 α -1 0 1 2 3 4 5 α

NLS on interval Numerical results β1 = 0, β2 = 1.3, β3 = 0 and β4 = 2. α = 0.5, 3.5 4 4 4 Ik Ik 4 2 2 0 0 3 1 2 2 3 1 0 5 10 t 15 20 0 50 100 t 150 200

Numerical results 2 : Coherence factor C = i u i i u i γ 1 = γ 2 = γ 3 = 0, γ 4 0. α >> 1 u 2 = z 24 γ 4 0.8γ 4, u 3 = z 34 γ 4 0.6γ 4, u 1 = u 4 = 0 C = u 2+u 4 u 2 + u 4 = 0.14. if instead of u we consider γ then C 1

Numerical results 3 : Other direction : Goldstone mode γ 1 = 2, γ 4 = 0, γ 2 = 0.1, γ 3 = 0 γ 2 = 0, γ 3 = 0.1 4 4 1 1 2 I k 2 I k 2 3 0 0 10 20 30 t 3 0 0 10 20 30 40 50 t

Conclusion Importance of cubic nonlinearity New type of design of lasers Enables synchronization Refs : J. Phys. A 2013 and arxiv