High Frequency Scattering by Convex Polygons Stephen Langdon

Transcription

1 High Frequency Scattering by Convex Polygons Stephen Langdon University of Reading, UK Joint work with: Simon Chandler-Wilde, Steve Arden Funded by: Leverhulme Trust University of Maryland, September

2 Contents The scattering problem The Burton & Miller integral equation Understanding solution behaviour Approximating the solution efficiently Galerkin method Collocation method Numerical Results 2

3 The Scattering Problem u i, incident wave u + k 2 u = 0 obstacle u = 0 D Γ x 2 x 1 3

4 u i, incident wave u + k 2 u = 0 obstacle u = 0 D Γ x 2 Green s representation theorem: u(x) = u i (x) Φ(x, y) u (y)ds(y), x D, n Γ x 1 where Φ(x, y) := i 4 H(1) 0 (k x y ). 4

5 u i, incident wave u + k 2 u = 0 obstacle u = 0 D Γ x 2 x 1 From Green s representation theorem (Burton & Miller 1971): ( ) 1 u Φ(x, y) u 2 n (x) + + iηφ(x, y) (y)ds(y) = f(x), x Γ, n(x) n where Γ f(x) := ui n (x) + iηui (x). 5

6 u i, incident wave u + k 2 u = 0 2D polygon u = 0 D Γ x 2 x 1 From Green s representation theorem: ( ) 1 u Φ(x, y) u 2 n (x) + + iηφ(x, y) (y)ds(y) = f(x), x Γ. n(x) n Γ Theorem (follows from Burton & Miller 1971, Selepov 1969) If η R, η 0, then this integral equation is uniquely solvable in L 2 (Γ). 6

7 1 2 u n (x) + Γ ( Φ(x, y) n(x) ) u + iηφ(x, y) (y)ds(y) = f(x), x Γ. n Conventional BEM: Apply a Galerkin method, approximating u/ n by a piecewise polynomial of degree P, leading to a linear system to solve with N degrees of freedom. Problem: N of order of kl, where L is linear dimension, so cost is O(N 2 ) to compute full matrix and apply iterative solver... or close to O(N) if a fast multipole method (e.g. Amini & Profit 2003, Darve 2004) is used. This is fantastic but still infeasible as kl. 7

8 1 2 u n (x) + Alternative: Γ ( Φ(x, y) n(x) ) u + iηφ(x, y) (y)ds(y) = f(x), x Γ. n Reduce N by using new basis functions, e.g. (i) approximate u/ n by taking a large number of plane waves and multiplying these by conventional piecewise polynomial basis functions (Perrey-Debain et al. 2003, 2004). This is very successful (in 2D, 3D, for acoustic/elastic waves and Neumann/impedance b.c.s), reducing number of degrees of freedom per wavelength from e.g to close to 2. However N still increases proportional to kl. 8

9 1 2 u n (x) + Alternative: Γ ( Φ(x, y) n(x) ) u + iηφ(x, y) (y)ds(y) = f(x), x Γ. n Reduce N by using new basis functions, e.g. (ii) for convex scatterers, remove some of the oscillation by factoring out the oscillation of the incident wave, i.e. writing u ui (y) = (y) F (y) n n and approximating F by a conventional BEM (e.g. Abboud et al. 1994, Darrigrand 2002, Bruno et al 2004). 9

10 Alternative: Reduce N by using new basis functions, e.g. (ii) for convex scatterers, remove some of the oscillation by factoring out the oscillation of the incident wave, i.e. writing u ui (y) = (y) F (y) n n ( ) and approximating F by a conventional BEM. For smooth obstacles this works well: equation ( ) holds with F (y) 2 on the illuminated side (physical optics) and F (y) 0 in the shadow zone. 10

11 (ii) for convex scatterers, remove some of the oscillation by factoring out the oscillation of the incident wave, i.e. writing u ui (y) = (y) F (y) n n ( ) and approximating F by a conventional BEM. Not very effective for non-smooth scatterers. 11

12 Understanding solution behaviour Let G(x, y) := Φ(x, y) Φ(x, y ) be the Dirichlet Green function for the left half-plane Ω. By Green s representation theorem, u(x) = u i (x) + u r G(x, y) (x) + u(y)ds(y), x Ω. n(y) Ω\Γ 12

13 Understanding solution behaviour In the left half-plane Ω, u(x) = u i (x) + u r (x) + u (x) = 2 ui n n (x)+2 Ω\Γ Ω\Γ G(x, y) n(y) u(y)ds(y) 2 Φ(x, y) n(x) n(y) u(y)ds(y), x γ = Ω Γ. 13

14 Explicitly, where s is distance along γ, and φ(s) and ψ(s) are k 1 u/ n and u, at distance s along γ, φ(s) = P.O. + i 2 [ e iks v + (s) + e iks v (s) ] where v + (s) := k 0 and F (z) := e iz H (1) 1 (z)/z F ( k(s s 0 ) ) e iks 0 ψ(s 0 )ds 0. 14

15 where φ(s) = P.O. + i 2 v + (s) := k 0 [ e iks v + (s) + e iks v (s) ] F ( k(s s 0 ) ) e iks 0 ψ(s 0 )ds 0. Now F (z) := e iz H (1) 1 (z)/z which is non-oscillatory, in that F (n) (z) = O(z 3/2 n ) as z. 15

16 where φ(s) = P.O. + i 2 v + (s) := k 0 [ e iks v + (s) + e iks v (s) ] F ( k(s s 0 ) ) e iks 0 ψ(s 0 )ds 0. Now F (z) := e iz H (1) 1 (z)/z which is non-oscillatory, in that F (n) (z) = O(z 3/2 n ) as z. v (n) + (s) = O(k n (ks) 1/2 n ) as ks. 16

17 where φ(s) = P.O. + i [ e iks v + (s) + e iks v (s) ] 2 ( k n v (n) + (s) = O (ks) 1/2 n) as ks and (by separation of variables local to the corner), k n v (n) + (s) = O ( (ks) α n) as ks 0, where α < 1/2 depends on the corner angle. 17

18 φ(s) = P.O. + i 2 [ e iks v + (s) + e iks v (s) ] where k n v (n) + (s) = O ( (ks) 1/2 n) as ks O ((ks) α n ) as ks 0, where α < 1/2 depends on the corner angle. Thus approximate φ(s) P.O. + i 2 [ e iks V + (s) + e iks V (s) ], where V + and V are piecewise polynomials on graded meshes. 18

19 Thus approximate φ(s) P.O. + i 2 [ e iks V + (s) + e iks V (s) ], where V + and V are piecewise polynomials on graded meshes. Figure 1: Scattering by a square 19

20 Thus approximate φ(s) P.O. + i 2 [ e iks V + (s) + e iks V (s) ], where V + and V are piecewise polynomials on graded meshes. Figure 2: Scattering by a square 20

21 Thus approximate φ(s) P.O. + i 2 [ e iks V + (s) + e iks V (s) ], where V + and V are piecewise polynomials on graded meshes. Figure 3: Scattering by a square 21

22 Approximation error Theorem: If V + is the best L 2 approximation from the approximation space, then where k 1/2 v + V + 2 C p n 1/2 (1 + log 1/2 (kl)) N p+1, N degrees of freedom p = polynomial degree L = max side length n = number of sides of polygon 22

23 Boundary integral equation method Integral equation in parametric form ϕ(s) + Kϕ(s) = F (s), where ϕ(s) := 1 u (x(s)) P.O.. k n Theorem. The operator (I + K) : L 2 (Γ) L 2 (Γ) is bijective with bounded inverse (I + K) 1 2 C, so that the integral equation has exactly one solution. 23

24 Boundary integral equation method Integral equation in parametric form ϕ(s) + Kϕ(s) = F (s), where ϕ(s) := 1 u (x(s)) P.O.. k n Difficulty 1 The operator (I + K) : L 2 (Γ) L 2 (Γ) is bijective with bounded inverse (I + K) 1 2 C(k), where the dependence of C(k) on k is not clear. 24

25 Approximation space: seek ϕ N (s) = M v j ρ j (s) V N, j=1 where ρ j (s) := e ±iks piecewise polynomial supported on graded mesh. Question: how do we compute v j? 25

26 Galerkin method To solve seek ϕ NG V N such that ϕ(s) + Kϕ(s) = F (s), (I + P NG K)ϕ NG = P NG F, where P NG is the orthogonal projection onto the approximation space. Equivalently (ϕ NG, ρ) + (Kϕ NG, ρ) = (F, ρ), ρ V N, M v j [(ρ j, ρ m ) + (Kρ j, ρ m )] = (F, ρ m ). j=1 If ρ j, ρ m supported on same side of polygon, integrals not oscillatory. 26

27 Galerkin method Theorem. For N N, the operator (I + P NG K) : L 2 (Γ) V N is bijective with bounded inverse (I + P NG K) 1 2 C s. 27

28 Galerkin method Difficulty 2. For N N (k), the operator (I + P NG K) : L 2 (Γ) V N is bijective with bounded inverse (I + P NG K) 1 2 C s (k), where the dependence of N (k) and C s (k) on k is not clear. 28

29 Collocation method To solve seek ϕ NC V N such that ϕ(s) + Kϕ(s) = F (s), (I + P NC K)ϕ NC = P NC F, where P NC is the interpolatory projection onto the approximation space. Equivalently ϕ NC (s m ) + Kϕ NC (s m ) = F (s m ), m = 1,..., M, M v j [ρ j (s m ) + Kρ j (s m )] = F (s m ). j=1 If ρ j supported on same side of polygon as s m, integrals not oscillatory. 29

30 Collocation method We have not shown that (I + P NC K) : L 2 (Γ) V N is bijective with bounded inverse. 30

31 Galerkin vs. Collocation: error analysis Theorem There exists a constant C p > 0, independent of k, such that for N N k 1/2 ϕ ϕ NG 2 C p C s sup u(x) n1/2 (1 + log 1/2 (kl/n)) x D N p+1, k 1/2 u(x) u NG (x) C p C s sup u(x) n1/2 (1 + log 1/2 (kl/n)) x D N p+1. Stability and convergence not proven for collocation scheme. 31

32 Galerkin vs. Collocation: conditioning Galerkin: mass matrix M G := [(ρ j, ρ m )] has condm (1 + σ)/(1 σ), where min(y + j σ max, y m) max(y + j 1, y m 1 ) (y + j y+ j 1 )(y m ym 1 ) < 1, and if side lengths and angles are equal we can prove σ < ( ) 1/2N log k 1. kl Collocation: difficulty with choice of collocation points, M C := [ρ j (s m )] may be ill conditioned. 32

33 Galerkin vs. Collocation: implementation Galerkin: need to evaluate numerically many integrals of form a [ ] d H (1) 0 (k s 2 + t 2 ) + ith(1) 1 (k s 2 + t 2 ) e ik(σ jt σ m s) dt ds. s2 + t 2 b c Collocation: need to evaluate numerically many integrals of form [ ] b H (1) 0 (k s 2 m + t 2 ) + ith(1) 1 (k s 2 m + t 2 ) e ikσjt dt. s 2 m + t 2 a Collocation method easier to implement 33

34 scattering by a square, k = 5 scattering by a square, k = 10 Numerical results 34

35 Numerical results (scattering by a square) Solution minus P.O. approximation; 35

36 Numerical results (scattering by a square) Solution minus P.O. approximation; 36

37 Numerical results (scattering by a square) Solution minus P.O. approximation; 37

38 Numerical results (scattering by a square) Solution minus P.O. approximation; 38

39 Numerical results (scattering by a square) Solution minus P.O. approximation; 39

40 Numerical results (scattering by a square) Solution minus P.O. approximation; 40

41 Numerical results (scattering by a square) Solution minus P.O. approximation; 41

42 Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 5; 42

43 Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 10; 43

44 Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 20; 44

45 Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 40; 45

46 Table 1: Relative errors, k = 10 k N dof ϕ ϕ NG 2 ϕ 2 ϕ ϕ NC 2 ϕ

47 Table 2: Relative errors, k = 160 k N dof ϕ ϕ NG 2 ϕ 2 ϕ ϕ NC 2 ϕ

48 k M N ϕ ϕ N 2 ϕ ϕ N 2 / ϕ 2 COND Table 3: Relative L 2 errors, various k, N = 32 48

49 k N u N u 256 u 256 ( π, 3π) u N u 256 u 256 (3π, 3π) u N u 256 u 256 (3π, π) Table 4: Relative errors, for u N (x) 49

50 k N u N u 256 u 256 ( π, 3π) u N u 256 u 256 (3π, 3π) u N u 256 u 256 (3π, π) Table 5: Relative errors, for u N (x) 50

51 What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 4: square, k = 5 51

52 What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 5: square, k = 10 52

53 What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 6: square, k = 20 53

54 What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 7: square, k = 40 54

55 Summary and Conclusions Using Green s representation theorem in a half-plane we can understand behaviour of the field on the boundary and its derivatives for scattering by a convex polygon (extends to convex polyhedron in 3D) For a convex polygon, design of an optimal graded mesh for piecewise polynomial approximation is then straightforward The number of degrees of freedom need only grow logarithmically with the wavenumber to maintain a fixed accuracy Ongoing considerations Galerkin vs. Collocation - stability and convergence analysis Better schemes for evaluating oscillatory integrals hp ideas 55