Coercivity of high-frequency scattering problems

Transcription

1 Coercivity of high-frequency scattering problems Valery Smyshlyaev Department of Mathematics, University College London Joint work with: Euan Spence (Bath), Ilia Kamotski (UCL); Comm Pure Appl Math Also with: Ivan Graham (Bath), Simon Chandler-Wilde (Reading) Spence, Chandler-Wilde, Graham, S, Comm Pure Appl Math Durham, 12 July 2016

2 Summary 1960 s s: tremendous interest in rigorous aspects of scattering/ diffraction: decay at t of wave equation 2 w t 2 c2 w = 0 asymptotics as k of Helmholtz equation u + k 2 u = 0 (related in subtle way)

3 Summary 1960 s s: tremendous interest in rigorous aspects of scattering/ diffraction: decay at t of wave equation 2 w t 2 c2 w = 0 asymptotics as k of Helmholtz equation u + k 2 u = 0 (related in subtle way) 2000 s - present: interest in Numerical Analysis of Helmholtz for k 1 e.g. hybrid asymptotic-numerical methods This talk: boundary integral equations One analysis question: prove relevant operator is coercive Surprise (?) appears that for coercivity need stronger results than those obtained classically (at least in the context of one classic tool Morawetz multipliers)

4 Obstacle scattering problem: acoustically soft/ TE Maxwell (2d) perfectly conducting boundary u i = e ikx â u s u s + k 2 u s = 0 in Ω ext := R d \ Ω int Ω int Γ u s + u i = 0 on Γ k > 0 Radiation conditions: u s r iku s = o = Uniqueness and existence. ) (r d 1 2 as r.

5 Boundary integral equations Green s Integral Representation: ( ) u s Φk (x) = n(y) (x, y)us (y) Φ k (x, y) us n (y) ds(y), where Γ Φ k (x, y) := i 4 H(1) 0 (k x y ) (d = 2) e ik x y 4π x y (d = 3) Hence boundary integral equations for v := u n : Single layer: S k v(x) := Φ k (x, y)v(y) ds(y) = u i (x) Γ x Ω ext (uniqueness fails for k 2 = interior Dirichlet eigenvalues) (Adjoint) double layer: ( ) 1 2 I + D k v(x) := 12 v(x) + Φ k ui (x, y)v(y) ds(y) = n(x) n (x) (uniqueness fails for k 2 = interior Neumann eigenvalues) Γ

6 Combined boundary integral equations Try a combination of a double layer and of a single layer: (Double Layer) iη (Single Layer) with a coupling constant η k (k 1). I.e. let A k = 1 2 I + D k i ηs k. Combined boundary integral equation: ( ) u A k = f n ( ) f = ui n (x) i η ui (x) At high frequencies (k 1) kernel of A k highly oscillatory (and non-linear) in k.

7 The operator A k ( ) u A k = f n For a fixed k, η > 0: A k : L 2 (Γ) L 2 (Γ) bounded and invertible and u/ n L 2 (Γ) if Γ is Lipschitz (Nečas) Q. What do we want to know about A k? 1. bound on A k (explicit in k) 2. bound on A 1 k (explicit in k) 3. coercivity: γ> 0 such that (A k φ, φ) L 2 (Γ) γ φ 2 L 2 (Γ), φ L 2 (Γ) (γ explicit in k)

8 The operator A k ( ) u A k = f n For a fixed k, η > 0: A k : L 2 (Γ) L 2 (Γ) bounded and invertible and u/ n L 2 (Γ) if Γ is Lipschitz (Nečas) Q. What do we want to know about A k? 1. bound on A k (explicit in k) relatively easy 2. bound on A 1 k (explicit in k) 3. coercivity: γ> 0 such that (A k φ, φ) L 2 (Γ) γ φ 2 L 2 (Γ), φ L 2 (Γ) (γ explicit in k)

9 The operator A k ( ) u A k = f n For a fixed k, η > 0: A k : L 2 (Γ) L 2 (Γ) bounded and invertible and u/ n L 2 (Γ) if Γ is Lipschitz (Nečas) Q. What do we want to know about A k? 1. bound on A k (explicit in k) relatively easy 2. bound on A 1 k (explicit in k) much harder (but still not enough!) 3. coercivity: γ> 0 such that (A k φ, φ) L 2 (Γ) γ φ 2 L 2 (Γ), φ L 2 (Γ) (γ explicit in k)

10 The operator A k ( ) u A k = f n For a fixed k, η > 0: A k : L 2 (Γ) L 2 (Γ) bounded and invertible and u/ n L 2 (Γ) if Γ is Lipschitz (Nečas) Q. What do we want to know about A k? 1. bound on A k (explicit in k) relatively easy 2. bound on A 1 k (explicit in k) much harder (but still not enough!) 3. coercivity: γ> 0 such that (A k φ, φ) L 2 (Γ) γ φ 2 L 2 (Γ), φ L 2 (Γ) (γ explicit in k) even harder!

11 Why is bounding A 1 not enough? k A k v = f ( v = u ) n Solving numerically using Galerkin method: choose S N L 2 (Γ) (N-dimensional subspace), find v N S N such that (A k v N, φ N ) L 2 (Γ) = (f, φ N ) L 2 (Γ), φ N S N Want quasi-optimality : (Lax-Milgramm + Cea s Lemma) : v v N L2 (Γ) C(k) inf v φ N L2 (Γ) φ N S N ( ) in some sense numerical well-posedness C(k) = A k /γ k S N Bound on A 1 can t give ( ) for important k

12 Plan Multiplier Methods

13 Multiplier methods Helmholtz equation u + k 2 u = 0, where u(x), x D R 3, k > 0

14 Multiplier methods Helmholtz equation D M where u(x), x D R 3, k > 0 integrate by parts ( ) u + k 2 u = 0, M u = (M u) M u get D M u n M u + k 2 M u = 0 D D

15 Some famous (and not so famous) multipliers radiation condition for u s : us r D ) M ( u s + k 2 u s = 0, iku s = o ) (r d 1 2, d = 2, 3 u s (x) eikr f (ˆx) as r = x r d 1 2

16 Some famous (and not so famous) multipliers radiation condition for u s : us r D ) M ( u s + k 2 u s = 0, iku s = o ) (r d 1 2, d = 2, 3 u s (x) eikr f (ˆx) as r = x r d 1 2 Green (1828), M = u s Rellich (1940), Morawetz (1968), e.g. M = r us r e.g. M = r us r = x u s ikru s + d 1 u s 2

17 Classic (high frequency) scattering/ diffraction theory Enormous interest from 1960 s onwards, e.g., USA Keller, Lax, Philips, Morawetz (@ Courant), Melrose... USSR/ Russia - Fock, Buslaev, Babich, Vainberg... 3 main problems 1. Wave equation: behaviour as t 2. Wave equation: propagation of singularities 3. Helmholtz: behaviour as k related in subtle way: 1+2=3 [Vainberg, 1975]

18 Key concept: (non-)trapping as k Helmholtz in trapping domains has almost eigenvalues/eigenfunctions (resonances)

19 Key concept: (non-)trapping as k Helmholtz in trapping domains has almost eigenvalues/eigenfunctions (resonances) Classic theory can be translated into results about A 1 k. Expect that 1. For Ω ext certain trapping domains A 1 k n e αkn, 0 < k 1 < k 2 <... some α > 0 2. If Ω ext is non-trapping then A 1 k 1, k k 0 Find 1. Proved 2. Proved ( for star-shaped domains [Chandler-Wilde, Monk 2008] using Rellich M = u ) (N.B. needed extra work to deal with ) r

20 The operator A k ( ) u A k = f n Spaces: A k : L 2 (Γ) L 2 (Γ) bounded and invertible and u/ n L 2 (Γ) if Γ is Lipschitz Q. What do we want to know about A k? 1. bound on A k (explicit in k) relatively easy 2. bound on A 1 k (explicit in k) use classic high-frequency scattering theory 3. coercivity: γ> 0 such that (A k φ, φ) L 2 (Γ) γ φ 2 L 2 (Γ), φ L 2 (Γ) (γ explicit in k) why?

21 Quasi-optimality v v N L 2 (Γ) C(k) inf v φ N L φ N S 2 (Γ) N ( ) Want to establish (with explicit k dependence of C) for 1. S N piecewise polynomials 2. S N,k hybrid subspace incorporating asymptotics of v = us n For 1. need N = O(k d 1 ) as k, possibility of 2. giving N = O(1). k-explicit ( ) for 1. classic problem, solved by [Melenk, 2011] (needs bound on A 1 k ) Coercivity (+bound on A k easy) gives ( ) for 1. and 2. k-explicit.

22 Coercivity γ > 0 such that (A k φ, φ) L2 (Γ) γ φ 2 L 2 (Γ), φ L2 (Γ) Not obvious will hold standard approach to formulations of Helmholtz: prove operator k = coercive + compact k Coercivity for circle (2d) and sphere (3d) k k 0, γ = 1 [Domínguez, Graham, Smyshlyaev, 2007] (Fourier analysis)

23 Two Coercivity Results using Morawetz Multipliers Result 1. Ω int Lipschitz star-shaped, a specially constructed star-combined A k is coercive k, γ = O(1). [Spence, Chandler-Wilde, Graham, S., Comm Pure Appl Math 2011] Result 2. Ω int smooth convex, the classical combined A k is coercive k k 0, η > η 0 k, γ = 1 2 ε [Spence, I.Kamotski, S. Comm Pure Appl Math 2015 ] (N.B. 1. is first quasi-optimality result for 2nd kind integral equations on L 2 (Γ) for Γ Lipschitz, even for Laplace (k = 0 )

24 Two Coercivity Results using Morawetz Multipliers Result 1. Ω int Lipschitz star-shaped, a specially constructed star-combined A k is coercive k, γ = O(1). [Spence, Chandler-Wilde, Graham, S., Comm Pure Appl Math 2011] Result 2. Ω int smooth convex, the classical combined A k is coercive k k 0, η > η 0 k, γ = 1 2 ε [Spence, I.Kamotski, S. Comm Pure Appl Math 2015 ] (N.B. 1. is first quasi-optimality result for 2nd kind integral equations on L 2 (Γ) for Γ Lipschitz, even for Laplace (k = 0 ) How? A k arose from us n iηus on Γ M = r us r ikru s + d 1 u s star-shaped coercivity 1. 2 M = Z(x) u s iη(x)u s + α(x)u s smooth convex coercivity 2.

25 Morawetz - 1 Morawetz & Ludwig (1968): take as a multiplier Mu := x u ikru + d 1 u, 2 Then (the Morawetz-Ludwig identity): 2Re ( Mu( u + k 2 u) ) = [ 2Re ( Mu u ) + ( k 2 u 2 u 2) ] ( x u 2 u r 2) u r iku 2.A corollary: [ ( 2Re Mu u ) + ( k 2 u 2 u 2) ] x ν ds = ν D D [ 2Re ( Mu( u + k 2 u) ) ( + u 2 u r 2) + u r iku 2] dx Let Ω in star-shaped x n β > 0 (n = ν). Take D = Ω ext B(0, R), let R, use radition conditions. = bounds for D-t-N; error bounds for GO/ GTD, etc. k-uniform

26 Morawetz - 1 Star-combined BIE (Spence, Chandler-Wilde, Graham, S., Comm Pure Appl Math 2011): In the Morawetz identity, choose u = S k φ. Take D = (Ω ext B(0, R)) Ω int, let R. Then, for ( 1 A k := (x n) 2 I + D k ( ) u A k = f n Coercivity: k 0, ) + x Γ S k iηs k, η := kr + i d 1. 2 ( f = x u i (x) i η(x) u i (x) ) Re (A k φ, φ) γ φ 2 L 2, γ = 1 2 ess inf (x n(x)) > 0. x Γ

27 Classical combined (Spence, I. Kamotski, S., Comm Pure Appl Math 2015) Try as a multiplier, with appropriate vector field Z(x), and scalar functions α(x) and β(x): Mu := Z(x) u ikβ(x)u + α(x)u. Then the following Morawetz-type identity holds: 2Re ( Mu ( + k 2 )u ) = [ 2Re ( Mu u ) + ( k 2 u 2 u 2) ] Z + ( )( 2α Z k 2 u 2 u 2) - 2 Re ( i Z j i u j u ) 2Re (u (ik β + α) u)

28 For wave equation Morawetz needed: For coercivity of A k we need: Z(x), x Ω ext Z(x), x Ω ext Ω int R ( j Z i ξ i ξ j ) 0, ξ C d, Z.n > 0 on Γ, Z(x) c x as x R ( j Z i ξ i ξ j ) θ ξ 2, ξ C d, Z= n on Γ, Z(x) c x as x (almost enough for A 1 k bound) have for Ω ext non-trapping in 2-d have for Ω int smooth convex in 2 & 3-d...non-trapping?

29 Summary 1960 s s: tremendous interest in rigorous aspects of scattering: decay at t of wave equation 2 w t 2 c2 w = 0 asymptotics as k of Helmholtz equation u + k 2 u = 0 (related in subtle way) 2000 s - present: interest in Numerical Analysis of Helmholtz for k 1 e.g. hybrid asymptotic-numerical methods This talk: boundary integral equations Analysis question: prove relevant operator is coercive Surprise (?) appears that for coercivity need stronger results than those obtained classically (at least in the context of one classic tool Morawetz multipliers)