y = h + η(x,t) Δϕ = 0

Transcription

1 HYDRODYNAMIC PROBLEM y = h + η(x,t) Δϕ = 0 y ϕ y = 0 y = 0 Kinematic boundary condition: η t = ϕ y η x ϕ x Dynamical boundary condition: z x ϕ t ϕ 2 + gη + D 1 1 η xx ρ (1 + η 2 x ) 1/2 (1 + η 2 x ) 1/2 (1 + η 2 x ) 3/2 x x+ 1 3 η xx = 0 2 (1 + η 2 x ) 3/2 Difficulties: A free-boundary value problem Nonlinear boundary conditions

2 HYDRODYNAMIC PROBLEM y = 1+ η(x,t) Δϕ = 0 y ϕ y = 0 y = 0 Kinematic boundary condition: η t = ϕ y η x ϕ x Dynamical boundary condition: z x ϕ t ϕ 2 + η 1 1 η xx +γ (1 + η 2 x ) 1/2 (1 + η 2 x ) 1/2 (1 + η 2 x ) 3/2 x x+ 1 3 η xx = 0 2 (1 + η 2 x ) 3/2 Difficulties: A free-boundary value problem Nonlinear boundary conditions Parameter: γ = D/ρgh 4

3 SOLITARY WAVES η(x, t) = η(x ct) η(x ct) 0 as x ct ± x x

4 MODELLING Dispersion relation for periodic wave trains (η cos k x ): c 2 c k 0 k The Ansatz c 2 = c 2 0(1 μ 2 ), η(x) = μ ζ(μx)e ik 0x + ζ(μx)e ik 0x + O(μ 2 ) leads to the nonlinear Schrödinger equation ζ xx ζ ± ζ 2 ζ = 0 Focussing (+) with solitary waves e iωx ζ NLS (x + x 0 ) for γ > γ Typical values: γ 10 5 (McMurdo sound) γ 10 2 (Lake Saroma)

5 VARIATIONAL PRINCIPLE Minimise the energy H(η, ϕ) = 1+η subject to fixed momentum I(η, φ) = (ϕ2 x + ϕ2 y ) dy + 1 η 2 2 η2 xx + γ dx (1 + η 2 x ) 5/2 η x ϕ y=η dx = 2c 0 μ, 0 < μ 1; the Lagrange multiplier is the wave speed. H and I are conserved quantities Yields conditional, energetic stability of the set of minimisers

6 DIRICHLET-NEUMANN OPERATOR Use a Dirichlet-Neumann operator: ϕ y=1+η = ξ G(η)ξ = 1 + η 2 x ϕ n y=1+η Δϕ = 0 Minimise H(η, ξ) = subject to fixed I(η, ξ) = where ξ = φ y=1+η ϕ y y=0 = ξg(η)ξ + 1 η 2 2 η2 xx + γ dx (1 + η 2 x ) 5/2 η x ξ dx = 2c 0 μ,

7 Minimise subject to fixed REFORMULATION H(η, ξ) = 1 2 ξg(η)ξ + 1 η 2 2 η2 xx + γ (1 + η 2 x )5/2 I(η, ξ) = η x ξ dx = 2c 0 μ Fix η and minimise H(η, ξ) over I(η, ξ) = 2c 0 μ. There is a unique minimiser ξ η with G(η)ξ η = c η η x Minimise where K(η) = J(η) = H(η, ξ η ) = K(η) + μ2 L(η), 1 η 2 2 η2 xx + γ, L(η) = 1 (1 + η 2 x ) 5/2 2 η x G(η) 1 η x We show that J(η) has a minimiser Minimising sequences converge (up to subsequences/translations)

8 ANALYTICITY K(η)ξ = x (G(η) 1 ξ x ) (η, ξ) K(η)ξ is analytic at the origin Flatten the domain: y = 1 + η y = 0 y = y 1 + η u(x, y ) = ϕ(x, y) y = 1 y = 0 K(η)ξ = (ϕ y=η ) x K(η)ξ = u x y =1 Δϕ= 0, Δ u x F 1 (η, u) y F 2 (η, u) = 0, ϕ y η x ϕ x ξ x = 0, u y F 2 (η, u) ξ x = 0, ϕ y = 0, u y = 0

9 ANALYTICITY Δ u x F 1 (η, u) y F 2 (η, u) u y F 2 (η, u) ξ x y =1 u y y =0 = 0 The left-hand side is an analytic function of u, η and ξ (u, η, ξ) = (0, 0, 0) is a solution The linearisation of the left-hand side with respect to u at (η, ξ) = (0, 0) is invertible: Δ u = G u y y =1 = g 1, u y y =0 = g 0 has a unique solution for each G, g 1, g 0 By the analytic implicit function theorem u is an analytic function of (η, ξ) at the origin. Hence (η, ξ) K(η)ξ is also analytic at the origin (since K(η)ξ = u x y =1)

10 MINIMISATION PROCEDURE Pretend R is bounded! Work in a ball of radius R in H 2 (R) R η 2 = 1/2 (η 2 + η 2 xx) A standard argument shows that 1 η 2 J(η) = 2 η2 xx + γ + μ2 (1 + η 2 x ) 5/2 L(η) has a minimiser η min over B R (0) η min is obviously not zero How to show that η min is not on the boundary of B R (0)? η 2 2 J(η) in B R (0) The test function η (x) = μζ NLS (μx) cos k 0 x Aμ 2 ζ NLS (μx) 2 cos(2k 0 x) Bμ 2 ζ NLS (μx) 2 satisfies J(η ) < 2c 0 μ

11 SOLITARY WAVES Use concentration-compactness minimising sequences undergo concentration, vanishing or dichotomy Rule out dichotomy by showing that i μ = inf J satisfies i μ1 +μ 2 < i μ1 + i μ2 ( strict sub-additivity ) Difficulties: Nonlocal equations Inhomogeneous nonlinearities Our equations are almost local We show that the functions in a minimising sequence {η n } scale like the test function η (x) = μζ NLS (μx) cos k 0 x Aμ 2 ζ NLS (μx) 2 cos(2k 0 x) Bμ 2 ζ NLS (μx) 2 for which J(η ) = 2c 0 μ Cμ 3 (η ) 4 + o(μ 3 )

12 SOLITARY WAVES J (η n ) 0 and η n 2 2 μ Write η n,1 = χ(d)η n, η n,2 = (1 χ(d))η n, where χ is the characteristic function of this set: k k k k k 0 < δ Variational reduction: J (η n ) = 0 χ(d)j (η n,1 + η n,2 ) = 0, (1 χ(d))j (η n,1 + η n,2 ) = 0 Solve for η 2,n = η 2,n (η n,1 ), set J(η n,1 ) = J(η n,1 + η 2,n (η n,1 )) and consider J (η n,1 ) = 0 Write η 1.n (x) = and show that ζ n 2 μ, η 2,n 2 μ μζ n(μx)e ik0x μζ n(μx)e ik 0x

13 CONVERGENCE TO NLS The set D μ of minimisers of J μ satisfies sup inf ζ η e iω ζ NLS ( + x 0 ) 1 0 ω [0,2π],x 0 R η D μ as μ 0, where we write η 1 (x) = 1 2 μζ η(μx)e ik 0x μζ η(μx) e ik 0x Furthermore, the wave speed for a minimiser η satisfies c η = c 0 C NLS μ 2 + o(μ 2 ) uniformly over D μ.