High Frequency Scattering by Convex Polygons Stephen Langdon

Transcription

1 Bath, October 28th 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 Bath, October 28th 2005

2 Bath, October 28th Contents Background - high frequency scattering problems Scattering by a convex polygon Understanding solution behaviour Approximating the solution efficiently Galerkin method Collocation method Numerical Results

3 Bath, October 28th Background - scattering problems

4 Bath, October 28th Scattering theory Concerned with the effect obstacles or inhomogeneities have on an incident wave. Direct problem: Compute scattered field when incident field and scattering obstacle are known. Inverse problem: Compute shape / properties of scatterer when scattered field is known.

5 Bath, October 28th Applications Prospecting for hydrocarbons Searching for archaeological remains Scanning for tumours Modelling road / rail / aircraft noise Defence applications Identifying objects via radar / sonar

6 Bath, October 28th Mathematical formulation for time harmonic waves Propagation of sound in a homogeneous media is mathematically described by the wave equation ũ(x, t) 1 c 2 2 ũ(x, t) = 0, (1) t2 where c denotes the speed of sound, and ũ is the induced pressure. Considering only time harmonic waves of the form ũ(x, t) = u(x)e iωt, where ω denotes frequency, the wave equation (1) then reduces to the Helmholtz equation: u + k 2 u = 0, where k = ω/c is the wavenumber.

7 Bath, October 28th Analytical solution possible only for very simple obstacles. In general we need to use a numerical method. Finite Element Method Need to cover whole domain in a mesh Leads to large sparse systems Boundary Element Method Only need to cover boundary in a mesh Leads to smaller full systems For each approach - difficulties when k is large.

8 Bath, October 28th Problem: Find u such that FEM for a model 1D problem u + k 2 u = 0, 0 < x < 1 u(0) = 1 u (1) iku(1) = 0 Approximate by piecewise linear FEM on a uniform grid of size h. THEOREM (Ihlenburg and Babuška) Let u h denote the FE solution. Then for h sufficiently small This is VERY BAD when k is large. u u h H 1 (0,1) C(hk + h 2 k 3 ).

9 Bath, October 28th First difficulty - local errors To accurately model a wave potential the rule of thumb is 10 elements per wave length. This is expensive for high wavenumbers. Figure 1: Influence of the number of elements per wavelength on the accuracy of numerical wave modelling

10 Bath, October 28th Second difficulty - pollution errors Additional error caused by innacurate modelling of wavelength - can build up across numerical model. Figure 2: Pollution error in numerical wave modelling

11 Bath, October 28th Scattering by convex polygons u i, incident wave u + k 2 u = 0 obstacle u = 0 D Γ x 2 x 1

12 Bath, October 28th 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 ).

13 Bath, October 28th 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).

14 Bath, October 28th 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).

15 Bath, October 28th 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.

16 Bath, October 28th 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.

17 Bath, October 28th 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.

18 Bath, October 28th 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.

19 Bath, October 28th 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, Domínguez, Graham and Smyshlyaev 2005).

20 Bath, October 28th 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.

21 Bath, October 28th (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.

22 Bath, October 28th 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

23 Bath, October 28th 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.

24 Bath, October 28th 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.

25 Bath, October 28th φ(s) = P.O. + i 2 [ e iks v + (s) + e iks v (s) ] where v + (s) := k 0 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.

26 Bath, October 28th φ(s) = P.O. + i 2 [ e iks v + (s) + e iks v (s) ] where v + (s) := k 0 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.

27 Bath, October 28th 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.

28 Bath, October 28th 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.

29 Bath, October 28th 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

30 Bath, October 28th 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 4: Scattering by a square

31 Bath, October 28th 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 5: Scattering by a square

32 Bath, October 28th 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

33 Bath, October 28th 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.

34 Bath, October 28th 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.

35 Bath, October 28th 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?

36 Bath, October 28th 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.

37 Bath, October 28th 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.

38 Bath, October 28th 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.

39 Bath, October 28th 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.

40 Bath, October 28th Collocation method We have not shown that (I + P NC K) : L 2 (Γ) V N is bijective with bounded inverse.

41 Bath, October 28th 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.

42 Bath, October 28th 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.

43 Bath, October 28th 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

44 Bath, October 28th Efficient evaluation of highly oscillatory integrals Consider Conventional wisdom... b a g(x)e ikf(x) dx. oscillatory integrals are difficult. oscillatory nature of solution increases errors. If we use interpolatory quadrature, this is certainly true - we need at least N quadrature points per wavelength, N 1, to achieve a reasonable level of accuracy.

45 Bath, October 28th Figure 6: Interpolation of e ikx2, k = 10.

46 Bath, October 28th Figure 7: Interpolation of e ikx2, k = 10.

47 Bath, October 28th Figure 8: Interpolation of e ikx2, k = 10.

48 Bath, October 28th Figure 9: Interpolation of e ikx2, k = 10.

49 Bath, October 28th Figure 10: Interpolation of e ikx2, k = 100.

50 Bath, October 28th Figure 11: Interpolation of e ikx2, k = 100.

51 Bath, October 28th Figure 12: Interpolation of e ikx2, k = 100.

52 Bath, October 28th Figure 13: Interpolation of e ikx2, k = 100.

53 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, N orsett)

54 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, Nörsett)... rather than interpolating the whole function g(x)e ikf(x), we just interpolate the non-oscillatory part, i.e. b a g(x)e ikf(x) dx b a [ gj χ j (x)] e ikf(x) dx, = b g j χ j (x)e ikf(x) dx. a

55 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, Nörsett)... rather than interpolating the whole function g(x)e ikf(x), we just interpolate the non-oscillatory part, i.e. b a g(x)e ikf(x) dx g j b a χ j (x)e ikf(x) dx. If hk 1, error O(h p+1 ). If hk 1, error O(h p+1 /(hk) 2 ). So performance actually improves as k increases.

56 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, Nörsett)... rather than interpolating the whole function g(x)e ikf(x), we just interpolate the non-oscillatory part, i.e. b a g(x)e ikf(x) dx g j b a χ j (x)e ikf(x) dx. If hk 1, error O(h p+1 ). If hk 1, error O(h p+1 /(hk) 2 ). So performance actually improves as k increases. However, we need to know the moments, hence for scattering problems application of Filon quadrature is difficult.

57 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, Nörsett) 2. Steepest descents (Huybrechs, Vanderwalle)

58 Bath, October 28th Some alternative approaches 1. Filon quadrature (Iserles, Nörsett) 2. Steepest descents (Huybrechs, Vanderwalle) 3. Asymptotic methods (Bruno, Reitich et. al., Iserles, Nörsett)

59 Bath, October 28th Asymptotic methods - the good news Oscillation leads to cancellation, and we can take advantage of this. Provided f (x) 0, b a g(x)e ikf(x) dx = 1 k [boundary terms] + i k b a g 1 (x)e ikf(x) dx, = 1 k [boundary terms] + 1 [boundary terms] k2 1 b k 2 g 2 (x)e ikf(x) dx, = n j=1 a boundary terms k n + O ( ) 1 k n+1.

60 Bath, October 28th Asymptotic methods - the bad news b a g(x)e ikf(x) dx = n j=1 boundary terms k n + ( ) n i b g n e ikf(x) dx, k a where g 0 = g, g n = ( gn 1 f ) =... = O ( ) 2n 1 f. As an asymptotic estimate - no problems. As an error estimate, care is needed, especially if f is small.

61 Bath, October 28th scattering by a square, k = 5 scattering by a square, k = 10 Numerical results

62 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

63 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

64 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

65 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

66 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

67 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

68 Bath, October 28th Numerical results (scattering by a square) Solution minus P.O. approximation;

69 Bath, October 28th Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 5;

70 Bath, October 28th Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 10;

71 Bath, October 28th Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 20;

72 Bath, October 28th Numerical results (scattering by a square) Exact solution minus P.O. approximation, k = 40;

73 Bath, October 28th Table 1: Relative errors, k = 10 k N dof ϕ ϕ NG 2 ϕ 2 ϕ ϕ NC 2 ϕ

74 Bath, October 28th Table 2: Relative errors, k = 160 k N dof ϕ ϕ NG 2 ϕ 2 ϕ ϕ NC 2 ϕ

75 Bath, October 28th k M N ϕ ϕ N 2 ϕ ϕ N 2 / ϕ 2 COND Table 3: Errors and relative L 2 errors, various k, N = 64

76 Bath, October 28th 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)

77 Bath, October 28th 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)

78 Bath, October 28th What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 14: square, k = 5

79 Bath, October 28th What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 15: square, k = 10

80 Bath, October 28th What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 16: square, k = 20

81 Bath, October 28th What we actually are computing... The difference between the exact solution and a leading order approximation; Figure 17: square, k = 40

82 Bath, October 28th 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