High Frequency Wave Scattering

Transcription

1 High Frequency Wave Scattering University of Reading March 21st, 2006

2 - Scattering theory What is the effect of obstacles or inhomogeneities on an incident wave?

3 - Scattering theory What is the effect of obstacles or inhomogeneities 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.

4 Applications High frequency scattering High frequency acoustics aircraft noise, road vehicle noise Geophysics prospecting for hydrocarbons, understanding seismic processes Non destructive testing scanning for tumours, archaeological surveys Electromagnetic scattering radar cross sections, mobile telephones

5 Mathematical formulation for time harmonic waves With time harmonic waves, propagation of sound in a homogeneous media is mathematically described by the Helmholtz equation: u + k 2 u = 0, where the wavenumber k is proportional to frequency.

6 Difficulties at high frequencies Standard schemes for scattering problems become prohibitively expensive as k.

7 Difficulties at high frequencies Standard schemes for scattering problems become prohibitively expensive as k. For standard finite element or boundary element methods, the rule of thumb in the literature is a requirement for ten elements per wavelength.

8 Difficulties at high frequencies Standard schemes for scattering problems become prohibitively expensive as k. For standard finite element or boundary element methods, the rule of thumb in the literature is a requirement for ten elements per wavelength. The total number of degrees of freedom thus increases with at least O(k) in 2D, O(k 2 ) in 3D.

9 Difficulties at high frequencies Standard schemes for scattering problems become prohibitively expensive as k. For standard finite element or boundary element methods, the rule of thumb in the literature is a requirement for ten elements per wavelength. The total number of degrees of freedom thus increases with at least O(k) in 2D, O(k 2 ) in 3D. This leads to large systems of equations to be solved when the wavelength is small.

10 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.

11 Second difficulty - pollution errors Additional error caused by innacurate modelling of wavelength - can build up across numerical model.

12 To date High frequency scattering Many ingenious schemes in literature Improved performance at high frequencies. Very little error analysis.

13 To date High frequency scattering Many ingenious schemes in literature Improved performance at high frequencies. Very little error analysis. Recently: For a specific problem of acoustic scattering by a planar surface of piecewise constant impedance, we developed a numerical method, complete with rigorous error analysis, demonstrating a convergence rate independent of the wavenumber k. We have extended the ideas to scattering by convex polygons, with logarithmic dependence on k.

14 u + k 2 u = 0 u i, incident wave u = 0 x 2 obstacle D Γ x 1

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

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

17 Scattering by a square, side length 2π Plot of 1 k u n for k = 10 and k = 640

18 Step 1 - subtract off leading order behaviour as k On each side of polygon u = u i + u r + u d.

19 Step 1 - subtract off leading order behaviour as k On each side of polygon u = u i + u r + u d. For a straight line segment we know u r explicitly.

20 Step 1 - subtract off leading order behaviour as k On each side of polygon u = u i + u r + u d. For a straight line segment we know u r explicitly. As k, { u u i n Ψ := n + ur n = 2 ui n on lit sides 0 on shadow sides.

21 Scattering by a square, side length 2π Plot of φ := 1 k ( u n Ψ) for k = 10 and k = 640

22 Step 2 - express φ as a product of oscillatory and non-oscillatory functions Where Γ = n =1 Γ, with n the number of sides of the polygon, we write φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n.

23 Step 2 - express φ as a product of oscillatory and non-oscillatory functions Where Γ = n =1 Γ, with n the number of sides of the polygon, we write φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n. We can show that k m v + 1 (m) (s) Cm (ks) 1/2 m, for ks 1,

24 Step 2 - express φ as a product of oscillatory and non-oscillatory functions Where Γ = n =1 Γ, with n the number of sides of the polygon, we write φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n. We can show that k m v + 1 (m) (s) Cm (ks) 1/2 m, for ks 1, and (by separation of variables local to corner) k m v + 1 (m) (s) Cm (ks) α m, for ks 1, where α < 1/2 depends on the corner angle.

25 Approximation scheme Thus approximate φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n,

26 Approximation scheme Thus approximate φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n, where V + e iks V + (s) + e iks V (s), x(s) Γ, = 1,..., n, and V are piecewise polynomials on graded meshes.

27 Approximation scheme Thus approximate φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n, where V + e iks V + (s) + e iks V (s), x(s) Γ, = 1,..., n, and V are piecewise polynomials on graded meshes.

28 Approximation scheme Thus approximate φ(s) = e iks v + (s) + e iks v (s), x(s) Γ, = 1,..., n, where V + e iks V + (s) + e iks V (s), x(s) Γ, = 1,..., n, and V are piecewise polynomials on graded meshes.

29 Error estimate for Galerkin boundary element method 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 sup u(x) n1/2 (1 + log 1/2 (kl/n)) x D N p+1. N degrees of freedom p =polynomial degree L = max side length n = number of sides of polygon

30 Numerical results High frequency scattering Example: scattering by a square of side length 2π, so each side is k wavelengths long. video for the case k = 5

31 Relative L 2 errors, various k, N = 64 k M N φ φ NG 2 / φ 2 COND

32 Ongoing research High frequency scattering Multiple scattering (many obstacles) Non-convex polygons Obstacles with curved surfaces Three-dimensional scattering problems Electromagnetic scattering.