A Dimension-Breaking Phenomenon for Steady Water Waves with Weak Surface Tension

A Dimension-Breaking Phenomenon for Steady Water Waves with Weak Surface Tension Erik Wahlén, Lund University, Sweden Joint with Mark Groves, Saarland University and Shu-Ming Sun, Virginia Tech Nonlinear Waves and Interface Problems, Lund, 2012

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Dimension breaking A family of (n + 1)-dimensional solutions bifurcates from an n- dimensional solution. Kirchgässner & Schaaf 94, Haragus & Kirchgässner 95, 96, Dias & Haragus 00.

Dimension breaking A family of (n + 1)-dimensional solutions bifurcates from an n- dimensional solution. Kirchgässner & Schaaf 94, Haragus & Kirchgässner 95, 96, Dias & Haragus 00. Solitary water waves with strong surface tension Groves, Haragus & Sun 02. Model equation: KP-I (integrable, explicit solutions). Related to transverse instability. (Groves, Haragus & Sun 01, Pego & Sun 04, Rousset & Tzvetkov 11).

Water waves Air y = h + (t, x, z) Water r y (x, y, z) z x y =0

Equations of motion φ = 0, 0 < y < h + η, φ y = 0, y = 0, η t = φ y η x φ x η z φ z, y = h + η, ( ) φ t + 1 η 2 φ 2 + gη σ = 0, y = h + η. 1 + η 2

Equations of motion φ = 0, 0 < y < h + η, φ y = 0, y = 0, η t = φ y η x φ x η z φ z, y = h + η, ( ) φ t + 1 η 2 φ 2 + gη σ = 0, y = h + η. 1 + η 2 Steady waves η(x, z, t) = η(x ct, z), φ(x, y, z, t) = φ(x ct, y, z).

Equations of motion φ = 0, 0 < y < h + η, φ y = 0, y = 0, η t = φ y η x φ x η z φ z, y = h + η, ( ) φ t + 1 η 2 φ 2 + gη σ = 0, y = h + η. 1 + η 2 Steady waves η(x, z, t) = η(x ct, z), φ(x, y, z, t) = φ(x ct, y, z). Line solitary waves: η(x, z) = η(x) 0 as x ±.

Line solitary waves with weak surface tension There exist two distinct solitary waves of the following form (Iooss & Kirchgässner 90, Iooss & Pérouème 93): η(x) = ±εa sech(εbx) cos(ωx) + O(ε 2 ) }{{} NLS soliton for α = α 0 + ε 2 and 0 < β < 1/3. Steady (cubic, focusing) NLS: ζ ζ xx 2 ζ 2 ζ = 0.

Stability and finite-amplitude solutions studied numerically and for model equations: Akers, Akylas, Calvo, Dias, Iooss, Kim, Longuet-Higgins, Berger, Menasce, Milewski, Parau, Vanden-Broeck, Wang,... Existence and stability by variational methods (for Euler): Buffoni, Groves and Wahlén.

Benney-Roskes/Davey-Stewartson Water wave problem (3D): g ε (D)η = N(η), (WW) where N represents nonlinear (and nonlocal) terms and g ε (k) = α 0 + ε 2 + β k 2 k coth k + k2 2 k coth k, k 2 D = 1 i x,z, k = (k 1, k 2 ). g 0 (k) vanishes precisely at k = (±ω, 0). Ansatz: η(x, z) = ε(ζ(εx, εz)e iωx + ζ(εx, εz)e iωx ) + ε 2 (ζ 2 (εx, εz)e 2iωx + ζ 2 (εx, εz)e 2iωx ) + ε 2 ζ 0 (εx, εz) +....

Plug into (WW) and retain terms of order O(ε 3 ): ζ ζ xx ζ zz ζ 2 ζ ζψ x = 0, ψ xx ψ zz + ( ζ 2 ) x = 0. (normalised).

Plug into (WW) and retain terms of order O(ε 3 ): ζ ζ xx ζ zz ζ 2 ζ ζψ x = 0, ψ xx ψ zz + ( ζ 2 ) x = 0. (normalised). Linear part comes from expanding g ε around (ω, 0).

Plug into (WW) and retain terms of order O(ε 3 ): (normalised). ζ ζ xx ζ zz ζ 2 ζ ζψ x = 0, ψ xx ψ zz + ( ζ 2 ) x = 0. Linear part comes from expanding g ε around (ω, 0). Nonlocal term related to discontinuity of g ε (and N) at origin. It disappears in 2D: ζ ζ xx ζ 2 ζ ζψ x = 0, ψ xx + ( ζ 2 ) x = 0.

Heuristic argument for Davey-Stewartson Linearisation around (ζ 0, ψ 0 ): ζ ζ xx ζ zz 3ζ0ζ 2 ζ0 2 ζ ζ 0 ψ x = 0, ψ xx ψ zz + [2 Re(ζ 0 ζ)] x = 0.

Heuristic argument for Davey-Stewartson Linearisation around (ζ 0, ψ 0 ): ζ ζ xx ζ zz 3ζ0ζ 2 ζ0 2 ζ ζ 0 ψ x = 0, ψ xx ψ zz + [2 Re(ζ 0 ζ)] x = 0. Look for harmonic solution e iκz ζ(x), e iκz ψ(x): ζ ζ xx 3ζ 2 0ζ ζ 2 0 ζ ζ 0 ψ x = κ 2 ζ, ψ xx + [2 Re(ζ 0 ζ)] x = κ 2 ψ.

Divide into real and imaginary parts (ζ = ζ 1 + iζ 2, ψ real, λ real): L 1 ζ 2 := ζ 2 ζ 2xx 2ζ0ζ 2 2 = λζ 2 ( ) ( ζ1 ζ1 ζ L 2 := 1xx 4ζ0 2ζ ) ( ) 1 ζ 0 ψ x ζ1 = λ ψ ψ xx + [2ζ 0 ζ 1 ] x ψ σ(l 1 ): σ(l 2 ) 0 1?? 0 1 Theorem L 2 has precisely one negative eigenvalue κ 2.

Proof. Idea: Compare with L 3 ζ 1 := ζ 1 ζ 1xx 6ζ 2 0 ζ 1.

Proof. Idea: Compare with L 3 ζ 1 := ζ 1 ζ 1xx 6ζ 2 0 ζ 1. σ(l 3 ): 3 0 1

Proof. Idea: Compare with L 3 ζ 1 := ζ 1 ζ 1xx 6ζ 2 0 ζ 1. σ(l 3 ): ζ 2 0 3 eigenvector corresponding to λ = 3. 0 1

Proof. Idea: Compare with L 3 ζ 1 := ζ 1 ζ 1xx 6ζ 2 0 ζ 1. σ(l 3 ): ζ 2 0 3 eigenvector corresponding to λ = 3. Quadratic form (complete the square!) L 2 U, U = 1 R(ζ 2 + ζ1x 2 6ζ0ζ 2 1) 2 dx + 1 (ψ x 2ζ 0 ζ 1 ) 2 dx. 2 R 0 if ζ 1 ζ0 2 at most one negative eigenvalue. Choose ζ 1 = ζ 0 and ψ = 0: L 2 U, U = (ζ0 2 + (ζ 0x ) 2 4(ζ 0 ) 4 ) dx = 8 3. R Hence, at least one negative eigenvalue. 0 1

How do we make this rigorous?

Lyapunov centre theorem Classical form: q = p H(q, p), ṗ = q H(q, p), or du = Lu + N(u), u = (q, p) X. dt ±iω eigenvalue of L, inω, n ±1 not an eigenvalue. There exists a family of 2π/ω s periodic, small-amplitude solutions, ω s ω, u s 0 as s 0. Devaney: Reversibility instead of Hamiltonian structure.

Spatial dynamics for water waves Luke s variational principle { 1+η(x,z) δ ( φ x + 12 ) φ 2 dy R 2 0 + 1 2 αη2 + β[ } 1 + η 2 1] dx dz = 0.

Spatial dynamics for water waves Luke s variational principle { 1+η(x,z) δ ( φ x + 12 ) φ 2 dy R 2 0 + 1 2 αη2 + β[ } 1 + η 2 1] dx dz = 0. Set ỹ = y, Φ(x, ỹ, z) = φ(x, y, z). 1 + η δl = 0, L = L(η, Φ, η z, Φ z ) dz R

Spatial dynamics for water waves Luke s variational principle { 1+η(x,z) δ ( φ x + 12 ) φ 2 dy R 2 0 + 1 2 αη2 + β[ } 1 + η 2 1] dx dz = 0. Set Legendre-transform ỹ = y, Φ(x, ỹ, z) = φ(x, y, z). 1 + η δl = 0, L = L(η, Φ, η z, Φ z ) dz R ρ = δl δη z, Ψ = δl δφ z.

Hamiltonian H(η, ρ, Φ, Ψ) = R 1 0 Φ z Ψ dy dx+ η z ρ dx L(η, Φ, η z, Φ z ). R

Hamiltonian H(η, ρ, Φ, Ψ) = R Hamilton s equations 1 0 Φ z Ψ dy dx+ η z ρ dx L(η, Φ, η z, Φ z ). R du dz = f(u), u = (η, ρ, Φ, Ψ).

Hamiltonian H(η, ρ, Φ, Ψ) = R Hamilton s equations 1 0 du dz = f(u), Φ z Ψ dy dx+ η z ρ dx L(η, Φ, η z, Φ z ). R u = (η, ρ, Φ, Ψ). Evolutionary equation in a Hilbert space X of functions which decay as x ±.

Hamiltonian H(η, ρ, Φ, Ψ) = R Hamilton s equations 1 0 Φ z Ψ dy dx+ η z ρ dx L(η, Φ, η z, Φ z ). R du dz = f(u), u = (η, ρ, Φ, Ψ). Evolutionary equation in a Hilbert space X of functions which decay as x ±. Equilibrium solutions u correspond to line solitary waves. Seek solutions u(x, z) = u (x) + w(x, z)

Reversible Hamiltonian system: R(η, ρ, Φ, Ψ) = (η, ρ, Φ, Ψ).

Reversible Hamiltonian system: R(η, ρ, Φ, Ψ) = (η, ρ, Φ, Ψ). σ(l): i"apple?," i"apple?,"

Reversible Hamiltonian system: R(η, ρ, Φ, Ψ) = (η, ρ, Φ, Ψ). σ(l): i"apple?," i"apple?," Spectral problem: Lw = µw + w.

Reversible Hamiltonian system: R(η, ρ, Φ, Ψ) = (η, ρ, Φ, Ψ). σ(l): i"apple?," i"apple?," Spectral problem: Lw = µw + w. Main step: reduce to a perturbation of Davey-Stewartson for µ = iεκ with κ = O(1). 0 belongs to the essential spectrum. Lw = N(v) has a unique solution w for each v, but only the derivatives of Φ are determined. L and N depend upon Φ only through its derivatives. Need special function spaces.

Reduction procedure Solve for (ρ, Ψ) as functions of (η, Φ, w ) (explicit) and Φ as function of (η, w ) (elliptic BVP).

Reduction procedure Solve for (ρ, Ψ) as functions of (η, Φ, w ) (explicit) and Φ as function of (η, w ) (elliptic BVP). Decompose η = η 1 + η 2 according to wave numbers near and far away from ω. Solve for η 2 as a function of η 1 : g ε (D x, εκ)η 1 = χ(d x )F (η 1 + η 2, w ) g ε (D x, εκ)η 2 = (1 χ(d x ))F (η 1 + η 2, w ) χ cut-off near ±ω, g ε (k) c > 0 when k 1 ω δ > 0.

In the end ζ ζ xx 3ζ 2 0ζ ζ 2 0 ζ ζ 0 ψ x = κ 2 ζ + εr(ζ, κ, ε) + f (w, ε), ψ xx + [2 Re(ζ 0 ζ)] x = κ 2 ψ. Conclusion For each sufficiently small ε > 0 we obtain two families of periodically modulated solitary waves one of elevation and one of depression. Future work Transverse instability. Fully localised waves.

Thank you for your attention!