Direct solution technique for frequency-domain scattering problems 1 - PowerPoint PPT Presentation

Direct solution technique for frequency-domain scattering problems 1 Konik Kothari CS598 Course Project, Fall 2017 1 Gillman, A., Alex H. B., and Martinsson P.G. ”A spectrally accurate direct solution technique for frequency-domain scattering problems with variable media.” BIT Numerical Mathematics 55.1 (2015): 141-170 CS598 Course Project Fall 2017 1 / 11

Introduction incident field u i scattered field u s artificial domain Ω Figure: Schematic of the problem, u = u s + u i Compute the scattered wave u s , given incident wave u i Mathematically, the scattered field u s satisfies the variable coefficient Helmholtz equation ∆ u s ( x ) + κ 2 (1 − b ( x )) u s ( x ) = κ 2 b ( x ) u i ( x ) , x ∈ R 2 , (1) Sommerfeld radiation condition ∂ u s ∂ r − i κ u s = o ( r − 1 / 2 ) , r := | x | → ∞ , (2) CS598 Course Project Fall 2017 2 / 11

Motivation Time-harmonic wave equations are relevant for practical applications: photonics, acoustics, placing your WiFi router! 2 Figure: Adventures in the acoustics of movie theaters 3 Solution method proposed is spectrally accurate, robust and computationally efficient. 2 http://www.caam.rice.edu/ gillmana/Wi-Fly.html 3https://i.pinimg.com/236x/9f/18/50/9f1850ce9a989ed19b1f9f86afebaacd–acoustic-asparagus.jpg CS598 Course Project Fall 2017 3 / 11

Overview Split problems into the interior Ω, and the exterior R 2 \ Ω problems Prepare solution operators for both, ’glue’ them at ∂ Ω to get solution Use a tree of boxes to solve the interior variable coefficient problem (hierarchical merges) Interior Dirichlet problem with variable coefficient b ( x ) DtN map: T int : T int u = u n ∀ x ∈ ∂ Ω Issue: Discrete domain difference operator → norm scales as N! Exterior Dirichlet problem with Sommerfeld condition: DtN map: T ext : T ext u s = u s n ∀ x ∈ ∂ Ω Issue: Same as T int Combine the two: ( T int − T ext ) u s | ∂ Ω = u i n − T int u i | ∂ Ω Order of ( T int − T ext ) still +1 → Ill-conditioned system → use Impedance-to-impedance maps! CS598 Course Project Fall 2017 4 / 11

Impedance-to-Impedance maps We solve the interior variable coefficient problem: [∆ + κ 2 (1 − b ( x ))] u ( x ) = 0 x ∈ Ω , (3) u n + i η u | ∂ Ω = f on ∂ Ω , (4) Use incoming and outgoing impedance boundary conditions (different from mixed boundary conditions!): f := u n + i η u | ∂ Ω (5) := u n − i η u | ∂ Ω (6) g Define R : L 2 ( ∂ Ω) → L 2 ( ∂ Ω) s.t. Rf = g R = ( T int − i η )( T int + i η ) − 1 (7) For real η , real b ( x ) and self-adjoint T int , R is unitary! CS598 Course Project Fall 2017 5 / 11

ItI maps 21 23 29 31 5 7 10 11 14 15 20 22 28 30 1 2 3 17 19 25 27 4 6 8 9 12 13 16 18 24 26 Figure: Ω split into boxes. J n (a) (b) τ Ω J i J J e w q+2 4 q q q+1 12 3 J s Figure: Operators on a leaf box CS598 Course Project Fall 2017 6 / 11

Leaf box operations Constructs 4 q Gauss-Legendre edge grid and an internal p × p Chebyshev grid (careful indexing) Construct discretized PDE (4) operators: (D (1) ) 2 + (D (2) ) 2 + diag { κ 2 (1 − b ( x j )) } p 2 A = (8) j =1 N + i η I 2 F = p ( J b , :) → Impedance operator (9) � F � B = (10) A( J i , :) Construct a ”solution matix”, X (basis) for the B operator: � I 4 p − 4 � BX = 0 ( p − 2) 2 × (4 p − 4) Interpolate X from Chebyshev to Gauss points using P, Y = XP. Define G similar to F (but on Gauss points), which gives: R = QGY , Q → Gauss to Chebyshev (11) CS598 Course Project Fall 2017 7 / 11

Merging leaf boxes Ω α Ω β J 1 J 3 J 2 Figure: Merging operators for children α and β If f α,β and g α,β are the impedance traces: � � f α � g α � 13 WR β 13 WR β R α 11 + R α 33 R α − R α � � 31 1 1 32 = f β g β − R β R β 22 + R β 33 QR β 23 (R α 31 + R α 33 WR α 33 R α 23 R α 31 ) 2 2 32 (12) � − 1 � I − R β 33 R α Here, W := 33 R 1 − I � − 1 � R 1 + I � � T int = − i η (13) CS598 Course Project Fall 2017 8 / 11

Exterior constant coefficient problem Any solution may be written as (Green’s formula) 4 : for x ∈ Ω c , u s ( x ) = ( D u s | ∂ Ω ) ( x ) − ( S u s n ) ( x ) , (14) � i 4 H (1) ∂ � � where ( D φ ) ( x ) := 0 ( κ | x − y | ) φ ( y ) ds y and ∂ n y ∂ Ω 4 H (1) � i ( S φ ) ( x ) := 0 ( κ | x − y | ) φ ( y ) ds y ∂ Ω Final formulation: � 1 � u s | ∂ Ω = S ( u i n − T int u i | ∂ Ω ) 2 I − D + ST int (15) Use Nystr¨ om method with composite (panel-based) quadrature with √ n ≈ N nodes in total. 4 Colton, Kress, Inverse acoustic and Electromagnetic Scattering Theory CS598 Course Project Fall 2017 9 / 11

Complexity Leaf solution matrix ∼ O ( p 6 ) × k boxes Compute R ∼ O ( N 3 / 2 ) Applying T int ∼ O ( N ) Approximating T int ∼ O ( N 3 / 2 ) Quadrature ∼ O ( N 3 / 2 ) (GMRES convergence in O (1) iterations) CS598 Course Project Fall 2017 10 / 11

Results (to come soon!) On a Gaussian bump scattering potential: (a) Bump scattering potential, (b) Re( u ) b ( x ) = 1 . 5 e − 160 r 2 Figure: Result for toy problem, with N = 231361 , n = 1760, error ≈ 5 e − 10 CS598 Course Project Fall 2017 11 / 11