Transparent boundary conditions for the elastic Transparent boundary conditions for the elastic waves in anisotropic media waves in anisotropic media Ivan L. Sofronov, Nickolai A. Zaitsev Keldysh Institute of Applied Mathematics RAS, Moscow Outline Outline • Problem formulation • Generation of LRBC operator, main steps • Numerical tests • Conclusions HYP 2006, Lyon, July 17-21 1
Motivation Motivation Generation of low-reflecting boundary conditions on open boundaries for anisotropic elastodynamics is a challenging problem. The PML method is not stable in this case, see � Beaches E., Fauqueux S., Joly P., Stability of perfectly matched layers group velocities and anisotropic waves, JCP, 188, 2003, 399 – 433; � D. Appel ö and G. Kreiss, A New Absorbing Layer for Elastic Waves, JCP, 215, 2006, 642 – 660. HYP 2006, Lyon, July 17-21 2
Low- -reflecting boundary conditions (LRBC) reflecting boundary conditions (LRBC) Low LRBC on “open boundaries” are needed for modeling of wave propagation using a bounded computational domain. Typical setup: Γ Domain of interest Computational domain C Generation of LRBC on G is an additional problem which is set up by considering auxiliary external Initial Boundary Value Problems (IBVPs) outside C . Remark: In the domain of interest the governing equations and geometry can be much more complex than outside C . HYP 2006, Lyon, July 17-21 3
Anisotropic elasticity, 2D orthotropic media Anisotropic elasticity, 2D orthotropic media We consider 2D elastodynamic equations: ∂ ∂ ∂ ∂ 2 2 2 2 v v v v ( ) ρ = + + + 11 33 33 12 1 c 1 c 1 c c 2 , ∂ ∂ ∂ ∂ ∂ 2 2 2 t x x x x 1 2 1 2 ∂ ∂ ∂ ∂ 2 2 2 2 v v v v ( ) ρ = + + + 33 22 33 12 2 c 2 c 2 c c 1 . ∂ ∂ ∂ ∂ ∂ 2 2 2 t x x x x 1 2 1 2 { } ρ v v , ( ) nm c Here is density; is the Cartesian velocities; are (constant) elastic 1 2 σ = ε n nm c coefficients of the Hook’s law written in the matrix form, . m = f v v θ , ( ) In polar coordinates for the velocity vector the system reads: r ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 2 f f f f f f = + + + + + 11 22 12 1 2 0 A A A A A A f ∂ ∂ ∂ θ ∂ ∂ θ ∂ ∂ θ 2 2 2 t r r r θ θ i j i A ( , ), r A r ( , ) where are 2x2 matrices HYP 2006, Lyon, July 17-21 4
Generation of LRBC operator, main steps Generation of LRBC operator, main steps − = f Lf 0 Governing equation in 2D space: tt Main steps: Stage 1: consider a set of auxiliary external IBVPs outside C (set wrt “ m ”): − = m m 2 E L E 0 in R / C t t Γ = m E | 0 (1) = t 0 = δ ϕ ϕ θ θ m m | ( t ) ( ) E Γ δ ( ) t where is Dirac’s delta function; = ∑ ϕ θ ∞ m m m θ ϕ θ { ( )} is a basis on G , i.e. f t ( , ) c ( ) t ( ) = m 0 m θ θ {sin k ,cos k } (in 2D it is constructed, e.g., by using ) HYP 2006, Lyon, July 17-21 5
Generation of LRBC operator (cont.) Generation of LRBC operator (cont.) s Stage 2: make Laplace transform and pass to elliptic BVPs (parameterized by ): ˆ ˆ − = 2 m m 2 s E L E 0 in R / C ˆ = ϕ θ m m | ( ) (2) E Γ Γ ˆ C = m | 0 E →∞ r ∂ m r ˆ ( , ) on θ Γ E Stage 3: solve (numerically) the problems and evaluate ∂ n Thus we obtain the Dirichlet-to-Neumann maps ∂ ≡ ˆ m m m ϕ θ ψ θ θ = ( ) ( )[ ] s E ( , ), r r R a Γ ∂ n HYP 2006, Lyon, July 17-21 6
Generation of LRBC operator (cont.) Generation of LRBC operator (cont.) Stage 4: form matrix of the Poincare-Steklov operator: we take arbitrary data on G = ∑ ˆ m m θ ϕ θ ˆ f s ( , ) c ( ) s ( ) m and write the representation of its normal derivative on G ∂ ∑ ∑ ∑ ˆ m m m m n θ = ψ θ = ϕ θ f s ( , ) c ˆ ( ) s ( )[ ] s c ˆ ( ) s P ( ) s ( ) n ∂ n m m n Thus we obtain the Poincare-Steklov operator in space of Fourier coefficients: = ∑ ∂ ˆ ∑ ˆ ˆ ( , ) ˆ n m m n n θ = ϕ θ ˆ f s d ( ) s ( ) d ( ) s P ( ) s c ( ) s ∂ n n n n or, in matrix form: { } T 0 1 ˆ ˆ = ˆ ˆ ˆ = c c , c , ... ˆ d ( ) s P ( ) ( ), s c s HYP 2006, Lyon, July 17-21 7
Generation of LRBC operator (cont.) Generation of LRBC operator (cont.) Stage 5: make inverse Laplace transform for the P-S operator. ˆ ( ) P s First we represent matrix by sum of three matrices to take into account → ∞ s : asymptotic at ˆ ˆ ˆ = + + = P ( ) s P s P K ( ); s P P , are consts , K ( ) s o (1) 1 0 1 0 ˆ ( ) K s Then we calculate rational approximations to each entry in such that all poles have negative real parts, i.e. m L m α n ˆ ∑ ˆ ( ) m % m m ≈ ≡ n l β ≤ δ < K s K ( ) s , Re( ) 0 n n n l m − β s = l 1 n l ˆ ˆ = ˆ d ( ) s P ( ) ( ) s c s Finally the inverse Laplace transform of gives: ∂ c ( ) t % = + + ∗ d ( ) t P P c ( ) t K ( ) t c ( ) t 1 0 ∂ t HYP 2006, Lyon, July 17-21 8
Generation of LRBC operator (cont.) Generation of LRBC operator (cont.) % m Remark: the explicit form of kernels is K ( ) t n m L n ∑ % m m m m = α β β ≤ δ < K ( ) t exp( t ), Re( ) 0 n n l n l n l = l 1 that permits to treat convolutions by stable recurrent formulas. Stage 6: compose azimuth modes: = ∑ m m θ ϕ θ f t ( , ) c ( ) t ( ) Q Denote by the operator of Fourier decomposition for m { } ∂ { } { } f 1 : m − m m θ → → Q : f t ( , ) c ( ) . t Q c ( ) , t d ( ) t f , i.e. Consequently, ∂ n The LRBC in the physical space reads: ∂ ∂ f f { } - - - 1 1 1 % − + + ∗ = Q PQ Q P Q f Q K ( ) t Q f 0 1 0 ∂ ∂ t n HYP 2006, Lyon, July 17-21 9
Setup of the test problem Setup of the test problem Governing equations are implemented in the polar system of coordinates. We consider the task in a circle: R G ( LRBC ) R 0 Γ R = • At 2 we prescribe Dirichlet data (pulse) to initiate elastic waves; 0 • G with radius R G = 10 is the external boundary where we put LRBC (with 36 azimuth harmonics). HYP 2006, Lyon, July 17-21 10
Test problem: reference solution Test problem: reference solution R G ( LRBC ) R Extended The verification of LRBC is made by comparing with the reference solution of a second problem having 8 x bigger external radius (extended domain). Comparison of two solutions in C- norm is made at r < R G. HYP 2006, Lyon, July 17-21 11
Test calculations Test calculations Parameters of anisotropic media are taken from Beaches E., Fauqueux S., Joly P., Stability of perfectly matched layers group velocities and anisotropic waves, JCP, 188, 2003, 399 – 433: anisotropic medium, case IV: ∂ ∂ ∂ ∂ 2 2 2 2 v v v v ( ) ρ = + + + 11 33 33 12 1 c 1 c 1 c c 2 , ∂ ∂ ∂ ∂ ∂ 2 2 2 t x x x x 1 2 1 2 ∂ ∂ ∂ ∂ 2 2 2 2 v v v v ( ) ρ = + + + 33 22 33 12 2 c 2 c 2 c c 1 . ∂ ∂ ∂ ∂ ∂ 2 2 2 t x x x x 1 2 1 2 HYP 2006, Lyon, July 17-21 12
Anisotropic case- -IV, t=0.52 IV, t=0.52 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 13
Anisotropic case- -IV, t=1.92 IV, t=1.92 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 14
Anisotropic case- -IV, t=2.62 IV, t=2.62 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 15
Anisotropic case- -IV, t=3.32 IV, t=3.32 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 16
Anisotropic case- -IV, t=4.01 IV, t=4.01 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 17
Anisotropic case- -IV, t=4.71 IV, t=4.71 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 18
Anisotropic case- -IV, t=5.41 IV, t=5.41 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 19
Anisotropic case- -IV, t=6.11 IV, t=6.11 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 20
Anisotropic case- -IV, t=6.81 IV, t=6.81 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 21
Anisotropic case- -IV, t=7.50 IV, t=7.50 Anisotropic case LRBC Reference Ur Utheta HYP 2006, Lyon, July 17-21 22
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