Modulation and water waves

Transcription

1 Summer School on Water Waves Newton Institute August 2014 Part 2 Thomas J. Bridges, University of Surrey

2 Outline Examples defocussing NLS to KdV KP equation via modulation Whitham modulation theory from the viewpoint of modulation! Modulation in 3(2+1) dimensions + dimension breaking modulation of elliptic PDEs: KdV planforms multi-dimensional conservation laws: e.g. modulation equations for interfacial waves

3 Classical NLS dark solitary waves Defocussing NLS, has an exact DSW solution iψ t + Ψ xx Ψ 2 Ψ = 0, Ψ(x, t) = 2 e i(kx ωt)( k + iβ tanh(βx) ), when β 2 = 1 2 (ω 3k 2 ) > 0 Bi-asymptotic to a periodic state with a phase shift Ψ(x, t) 2 e i(kx ωt)( k ± iβ ) as x ±. Gen solitary waves: Wahlén (on Tuesday) DSWs in Euler: Milewski, Vanden-broeck, Wang (2013) PRSLA

4 Emergence of DSWs Ψ(x, t) = 2 e i(kx ωt)( k + iβ tanh(βx) ) Periodic state at infinity exists for but the DSW exists only for k 2 < ω, k 2 < 1 3 ω. Kivshar s theory: emergence of the DSW at k 2 = ω/3 governed by a KdV equation. Y. Kivshar [1990] Phys. Rev. A

5 KIVSHAR s theory Kivshar proposes a solution of NLS the form where Ψ(x, t) = (Ψ 0 + ε 2 q(x, T, ε))e i(kx+εφ(x,t,ε)), X = εx and T = ε 3 t, and shows that q(x, T, 0) satisfies (to leading order) a KdV equation c 0 q T + c 1 qq X + c 2 q XXX = 0. NB. Works, but better to modulate wavenumber, not amplitude.

6 (A, B) calculation for NLS DSWs Defocussing NLS (setting ω = 1), Basic spatially periodic state iψ t + Ψ xx + Ψ Ψ 2 Ψ = 0. Ψ(x) = A 0 e iθ, θ = kx + θ 0, A k 2 = 1. The components of the conservation law are A = 1 2 Ψ 2 and B = Im(ΨΨ x ), and so A (k) = 1 2 (1 k 2 ) and B(k) = k(1 k 2 ). Differentiating A (k) = 2k, B (k) = 1 3k 2, B (k) = 6k. and K = 1 2. Hence the emergent KdV is 2kq T 6kqq X 1 2 q XXX (same as Kivshar).

7 Periodic states DSWs via KdV Given a 1+1 PDE generated by a Lagrangian and a branch of periodic solutions, or periodic travelling waves, e.g. branch of Stokes waves modulate the periodic solution; look for B (k) = 0 in this case q(x, T, ε) is the perturbation wave number, with appropriate conditions q satisfies KdV A (k) is the wave action evaluated on periodic state, B(k) is the wave action flux K is a Krein index of the periodic orbit

8 Finding bifurcation points for KdV Parameterize using (A, B) and look for points where B k = 0, or look for eigenvalues meeting at zero (or Floquet multipliers at +1). This picture occurs along branches of Stokes waves Vanden-Broeck (1983), Zufuria (1987 finite depth) Baesens & MacKay (1992) infinite depth TJB & Donaldson (2006) finite depth with mean flow NB. in this view, c is fixed and wavelength varies along a branch.

9 NLS shallow water equations: Madelung transformation Madelung transformation: in NLS let and define then h and u satisfy Ψ = ρe iφ, gh = 2ρ 2 and u = 2φ x, h t + uh x + hu x = 0 u t + uu x + gh x = h 1 h xxx +. Emergence of DSWs in NLS KdV emerges in shallow water pointed out by Roger Grimshaw

10 h 0 v 0 u 0 Modulation of uniform flows and KP Q 1 = h 0 u 0 and Q 2 = h 0 v 0 R = gh ( u v 2 0 ). Consider the generalized criticality conditions Q 1 u 0 R fixed = 0 and Q 2 u 0 R fixed = 0, and the coefficients 2 Q 1 u0 2 R fixed and Q 2 v 0 R fixed.

11 Modulation of uniform flows and KP Modulate the uniform flow: h 0 v 0 u u 0 u 0 + ε 2 q(x, Y, T, ε), X = εx, Y = ε 2 x, T = ε 3 t. Claim: q satisfies the KP equation with ( ) q 2M u T + 2 Q 1 R fixed q q X + K 3 q X 3 u 2 0 X = Q 2 v 0 R fixed q YY. It is a modulation of the mass CLAW for the full WW problem M t + (Q 1 ) x + (Q 2 ) y = 0, where M is M evaluated on the uniform flow.

12 Symmetry, modulation, KdV Lagrangian L dxdydzdt (e.g. Luke s Lagr) One-parameter symmetry group CLAW A t + B x + C y = 0 basic state: Ẑ (θ, k, l) with θ = kx + ly + θ 0 modulate: Z (x, y, t) = Ẑ (θ + εψ, k + ε2 q, l) + ε 3 W (θ, X, Y, T ) Evaluate CLAW on basic state: A (k, l), B(k, l) and C (k, l) If B k = 0 and C k = 0 (gen criticality) then KP emerges (2A k q T + B kk qq X + K q XXX ) X = C l q YY. K : Krein signature, sign of momentum flux, dispersion relation,... (same as KdV)

13 Substitute and expand in powers of ε ε 2 : q = φ X ε 3 : B k = 0 ε 4 : C k = 0 ε 5 : C l 0, A k 0 and B kk 0. Then 2A k q T + B kk qq X + K q XXX + C l φ YY = 0. or, by differentiating with respect to X, (2A k q T + B kk qq X + K q XXX ) X + C l q YY = 0.

14 NLS KP-I NLS 2+1 model for moderate values of surface tension in deep water iψ t + Ψ xx + Ψ yy + Ψ Ψ 2 Ψ = 0, RE: Ψ = Ψ 0 e iθ, θ = kx + ly + θ 0 CL: A t + B x + C y = 0. 2kq T 6kqq X 1 2 q XXX + (1 k 2 )φ YY = 0, Necessary conditions: k 2 = 1 3 and l = 0 KP is a model for water waves on infinite depth!

15 Ansatz Given a basic state Ẑ (θ, k) the modulation is an ansatz Z (x, t) = Ẑ (θ + εa φ, k + ε b q, ω + ε c Ω) + ε d W (θ, X, T, ε), with X = ε α x and T = ε β t. Whitham modulation theory Z (x, t) = Ẑ (θ + φ, k + εq, ω + εω) + ε2 W (θ, X, T, ε), with X = εx and T = εt. A ω B k A k B ω 0. Emergence of the Boussinesq equation Z (x, t) = Ẑ (θ + εφ, k + ε2 q, ω) + ε 3 W (θ, X, T, ε), with X = εx and T = ε 2 t. Also need B (k) = 0.

16 Whitham modulation theory Consider a PDE generated by a Lagrangian L (Z ), and suppose there is a steady (RE) solution of the form Z (x, t) = Ẑ (θ, ω, k), θ = kx ωt + θ 0, associated with a conservation law A t + B x = 0. In Whitham theory the basic state is a periodic TW and the conservation law is conservation of wave action. Modulate Z (x, t) = Ẑ (θ + φ, ω + εω, k + εq) + ε2 W (θ, X, T, ε), with φ(x, T, ε), q(x, T, ε), Ω(X, T, ε) and scaling T = εt and X = εx. Need A ω B k A k B ω 0. Modulation generates the Whitham modulation equations which are quasilinear hyperbolic or elliptic no dispersion.

17 Generalizing Whitham modulation theory Consider a PDE generated by a Lagrangian L (Z ). The ansatz Z (x, t) = Ẑ (θ + φ, k + εq, ω + εω) + ε2 W (θ, X, T, ε), with T = εt and X = εx gives A ω Ω T + (A k + B ω )q T + B k q X = 0, A ω B k A k B ω 0. The ansatz Z (x, t) = Ẑ (θ + εφ, k + ε2 q, ω + ε 4 Ω) + ε 3 W (θ, X, T, ε), with T = ε 3 t and X = εx and B k = 0 gives (A k + B ω )q T + B kk qq X + K q XXX = 0.

18 Dimension breaking via modulation The dimension breaking of patterns bif from localised 1D pattern to periodic in y Kirchgässner, Hărăguş, Groves, B & Derks, etc NB. Figure credit: Mark Groves

19 Converse dimension breaking problem look for mechanisms for multi-dimensional patterns begin with a 1D spatially periodic pattern advantage: spatially periodic patterns can be modulated derive modulation equation(s)

20 Modulation: from rolls planforms Swift Hohenberg equation u t + 2 u + p u + f (u) = 0, for (real scalar-valued) u(x, y, t) and some given f (u), and real Ginzburg-Landau equation Ψ t = Ψ + Ψ Ψ 2 Ψ, for (complex scalar-valued) Ψ(x, y, t) where := 2 x y 2.

21 Rolls Swift-Hohenberg equation u t + 2 u + p u + f (u) = 0. 1D steady SH: u xxxx Pu xx + u u 2 = 0, has a family of roll solutions, û(θ, k), with a sequence of points where B (k) = 0. Real Ginzburg Landau equation Ψ t = Ψ + Ψ Ψ 2 Ψ, 1D steady real GL: rolls: Ψ(x) = Ψ 0 e iθ, θ = kx + θ 0. Can compute everything explicitly.

22 Modulate rolls u(x, y, t) = û(θ + εφ, k + ε 2 q, l) + ε 3 w(θ, X, Y, T, ε), with X = εx, Y = ε 2 y and T = ε 4 t. Substitute into equation and expand. Fifth order terms give q T = B l q YY + ( 1 2 A kkq 2 + K q XX ) XX, a gradient-like KP equation (replace q T by q XT to get KP). The steady part is a Boussinesq equation (completely integrable). The condition B (k) = 0 is a generalization of the Ekhaus instability threshold in Swift-Hohenberg. TJB PhysicaD (2014)

23 Gradient in time elliptic in space Boussinesq q T = B l q YY + ( 1 2 A kkq 2 + K q XX ) XX. The PDE is gradient in time (can show that the right-hand side is the derivative of a functional), and is well posed if B l > 0 and K < 0 in which case the steady part is an elliptic Boussinesq equation (the bad Boussinesq equation). Related to KP and (2+1) Boussinesq. and q XT = (aqq X + bq XXX ) X + cq YY, q TT = (aqq X + bq XXX ) X + cq YY, (both of which can be derived via modulation).

24 Explicit Multi-bump solutions Scale the steady part v YY v XX 3(v 2 ) XX v XXXX = 0. (1) This is the canonical form for the bad Boussinesq equation and it has an exact solution DAI ET AL. 1, with v(x, Y ) = v 0 + 2φ XX, φ(x, Y ) = lnf(x, Y ), (2) F(X, Y ) = 1 + 2a cos(px)e ΩY +θ 0 + be 2ΩY +2θ 0, The solution is periodic in the X direction and localized in the Y direction. Are these solutions stable? 1. Dai et al, Chaos Solitons & Fractals 26 (2005)

25 Multi-bump solutions of Dai et al

26 N-pulse web-like multi-dimensional patterns For w tt w xx 6(w 2 ) xx w xxxx = 0, Hirota 2 shows that there are explicit multi-pulse N web solutions for every natural number N w = 2 logf (x, t), x 2 with explicit expressions for f for every N in terms of sums of exponentials. Replacing time in Hirota by space, they become steady web-like multi-pulse patterns. Biondini et al 3 call these Boussinesq solutions web solitons. 2. Hirota, J Math Phys 14 (1973) 3. Biondini et al, Stud Appl Math. 122 (2009)

27 Canonical form for the steady part C(Z xx + Z yy ) + F(Z ) = 0, Z R n (n 2), where C is a symmetric (not necessarily positive definite) matrix, and F(Z ) is a specified function. Both real Ginzburg Landau and Swift-Hohenberg fit into this framework. Time can be brought in by adding a term MZ t where M is a positive semi-definite matrix MZ t + C(Z xx + Z yy ) + F(Z ) = 0, Z R n (n 2),

28 Conservation law on loops C(Z xx + Z yy ) + F(Z ) = 0, Z R n (n 2), Suppose there exists a family of solutions, parameterised by µ, Z (x, y, µ) with Z (x, y, µ + 2π) = Z (x, y, µ). A loop of solutions (which may or may not exist). Then this family satisfies the conservation law with A x + B y = 0, A = CZ µ, Z x and B = CZ µ, Z y. (A steady version of wave action conservation.)

29 q Boussinesq an example For real GL, Ψ t = Ψ + Ψ Ψ 2 Ψ, take the simple family of steady rolls Ψ = Ψ 0 e iθ with θ = kx + ly + θ 0 giving Ψ k 2 + l 2 = 1. Evaluate the loop conservation law on this two-parameter family A (k, l) = kψ 2 0 = k(1 k 2 l 2 ), B(k, l) = lψ 2 0 = l(1 k 2 l 2 ). The conditions A k = 0 and B k = 0 are both satisfied with l = 0. For the second derivative A kk (k, 0) = 6k. The coefficient K = 1 2, giving the parabolic PDE ( ) q T = Ψ 2 0 q YY 3kq q XX XX

30 Multi-dimensional conservation laws N-component (abelian) symmetry group N CLAWS (A j ) t + (B j ) x = 0, j = 1,..., N RE: Ẑ (θ 1,..., θ N, k 1,..., k N ), θ j = k j x + θ o j. criticality: define B 1 k 1 DB(k) =..... B N k 1 B 1 k N B N k N. non-degenerate multi-cwa; critical: det[db(k)] = 0 with simple zero eigenvalue [DB(k)] n = 0.

31 Multi-dimensional conservation laws B 1 k 1 DB(k) =..... B N k 1 B 1 k N B N k N. Singularities of this matrix generate modulation equations. simple zero eigenvalue generates KdV double zero eigenvalue generates coupled KdV simple zero + secondary singularity (Thom-Boardman classification) KdV with cubic nonlinearity Strategy: modulate and see what happens...

32 Multi-dimensional CLAWS at criticality Modulate ansatz at criticality Z (x, t) = Ẑ (θ 1+εφ 1,..., θ N +εφ N, k 1 +ε 2 q 1,..., k N +ε 2 q N )+ε 3 W. Linear theory: 0 has geom mult N and algebraic mult 2N + 2 Modulation equation for ˆq(X, T, ε) defined by q = ˆqn + p aˆq T + κˆqˆq X + K ˆq XXX = 0. where a = n T [ DA(k) + DA(k) T ] n, and κ is the intrinsic second derivative of the map B(k) in the direction n κ = d 2 n, B(k + sn) ds2. s=0 K is as before (Krein signature, dispersion relation, excess Momentum flux, top of the Jordan chain)

33 Example: two fluids with a rigid lid ρ 2 u 2 h 2 ρ 1 u 1 h 1 Uniform flow: h 1, h 2 and u 1, u 2, with h 1 + h 2 = d. Three conserved quantities (R, Q 1, Q 2 ) where R is the Bernoulli energy and Q j are the mass flux in each layer. Emergence of KdV at criticality from the uniform flow

34 Criticality of two layer flow Let c = (h 1, u 1, u 2 ) and B(c) = (R(c), Q 1 (c), Q 2 (c)), with R(c) = 1 2 ρ 1u ρ 2u (ρ 1 ρ 2 )gh 1 Q 1 (c) = ρ 1 h 1 u 1 Q 2 (c) = ρ 2 (d h 1 )u 2. The derivative of the mapping is (ρ 1 ρ 2 )g ρ 1 u 1 ρ 2 u 2 DB(c) = ρ 1 u 1 ρ 1 h 1 0, ρ 2 u 2 0 ρ 2 (d h 1 ) and there exists n satisfying DB(c)n = 0 when f (c) = 0 where f (c) = det(db(c)) = ρ 1 ρ 2 (ρ 1 ρ 2 )gh 1 (d h 1 ) [ 1 F 2 1 rf 2 2 ], where F 2 j = u 2 j /((1 r)gh j ) and r = ρ 2 /ρ 1.

35 Mapping (h 1, u 1, u 2 ) (R, Q 1, Q 2 )

36 Criticality and df (c) n Now f (c) := det[db(c)] = C [ ] (1 r) u2 1 r u2 2, C = ρ 2 gh 1 gh 1ρ 2 gh 1 h 2. 2 The criticality surface in (h 1, u 1, u 2 ) space is defined by f 1 (0) and a vector v is tangent to this surface if df v = 0. Now, and so f, n = 3C ρ 1 g f = C g ( u 2 1 h 2 1 ru2 2 h2 2, 2u 1, 2ru ) 2, h 1 h 2 ( u1 2 u 2 ) 2 ρ 1 h1 2 ρ 2 h2 2 = C ρ 1 g d 2 ds n, B(c + sn 2. s=0 Criticality surfaces: TJB & Donaldson [2007] Phys. Fluids

37 Remarks Start with a Lagrangian with symmetry Construct families of relative equilibria Given a relative equilibrium it depends on a phase (or phases) and each phase has a parameter (speed, wavenumber, uniform flow): Ẑ (θ, k) Modulation ansatz Z (x, t) = Ẑ (θ + εa φ, k + ε b q) + ε c W (θ, X, T, ε), with X = ε α x and T = ε β t. Substitute and determine modulation equation via a sequence of solvability conditions generalizations: higher space dimension, higher dimensional group, non-abelian groups