RBF-FD Approximation to Solve Poisson Equation in 3D

Transcription

1 RBF-FD Approximation to Solve Poisson Equation in 3D Jagadeeswaran.R March 14, / 28

2 Overview Problem Setup Generalized finite difference method. Uses numerical differentiations generated by Gaussian RBF interpolation on irregular centers. Dirichlet problem for the Poisson Equation in 2D and 3D with smooth solution. RBF-QR method used when the Interpolation matrix is ill conditioned due to small Shape parameters 2 / 28

3 RBF-FD Problem Definition Dirichlet Problem for the Poisson equation: u = f on Ω u δω = g We want to find a Linear numerical differentiation formula for the Differential operator: Du(x) n w i (x)u(x i ) i=1 where w i (x) called weights, The vector w = [w 1,..., w n ] T called stencil. 3 / 28

4 RBF-FD Problem Definition... Let Discretization centers: Ξ Ω, δξ := Ξ δω, Ξ int := Ξ \ δξ Assume for each ζ Ξ int a set Ξ ζ Ξ is chosen such that ζ Ξ ζ and Ξ = ζ Ξ int Ξ ζ For each ζ Ξ int, choose a linear numerical differentiation formula for Laplace operator: u(ζ) Ξ Ξ ζ w ζ,ξ u(ξ) So we get the approximation: Ξ Ξ ζ w ζ,ξ û(ξ) = f (ζ), ζ Ξ int ; û(ξ) = g(ξ), Ξ δξ (1) 4 / 28

5 RBF-FD My Goals Add Stable Computation Algorithm from G.E. Fasshauer, M.J. McCourt, Add Support for 3 Dimensional Problems Generalize Stencil Selection for n Dimensional Domains Experiment Stable computation for n Dimensional Problems CVT Points for n Dimensions 5 / 28

6 Interpolation Problem Let a Positive(or Conditionally) definite function κ : R + R and a continuous function u : R d R Given Ξ ζ = {x 0,..., x n} R d, ζ = x 0, K j (x) = K(x x j ), K(x) = κ( x ) the RBF interpolant with a constant term is sought in the form n s(x) = a j K j (x) + c, s(x i ) = u(x i ), i = 0,..., n, j=0 n a j = 0 (2) j =0 Then the coefficients a j and c are uniquely determined by the following conditions n a j K(x i x j ) + c = u(x i ), i = 0,..., n, j=0 n a j = 0 j =0 written in matrix form as [ ] KX 1 [ ] [ ac u x 1 T = 0 0 ], K X := [e ɛ2 x i x j 2 ] n i,j=0, 1 := [1,..., 1]T (3) 6 / 28

7 RBF Differentiation Matrix Consider the approximation of a differential operator D by using RBF n w j K(x x j ) + v = DK(x x i ), i = 0,..., n, Note: This is indept ofu j=0 These weights can be found by solving the symmetric positive definite linear system [ ] KX 1 [ ] [ ] wv [Dki (x 1 T = 0 )] n i=0 0 0 (4) where, κ j (x) = K(x x j ), K(x) := κ( x ), K X := [e ɛ2 x i x j 2 ] n i,j=0, For Gaussian RBF κ(r) = e ɛr 2. Its Laplacian is: k(x) = 2ɛ 2 e ɛ2 x i x j 2 (2ɛ 2 x i x j 2 d), d := num of dimensions 7 / 28

8 Differentiation Problem The derivatives of s are good approximations of the derivatives of u if k is sufficiently smooth. An approximation of Du(x) at x 0, where D is a linear differential operator annihilating constants, may be considered in the form. Apply D to (2) Du(x 0 ) Ds(x 0 ) = n a j DK(x 0 x j ) = j=0 n w i u(x i ) where the weights w i (depending on x) exist because the coefficients a j of the interpolation function s defined as before depend linearly on the data u(x i ). This can be verified as below using (4) n j=0 [ ] [ ac T [K(x xi )] a j DK(x x j ) = n i=0 0 i=0 ] [ ] [ ac T KX 1 = 1 T 0 ] [ wv ] [ u x = 0 ] T [ wv ] = n w i u(x i ) i=0 8 / 28

9 Example Stencil Figure: typical Stencil Points in 2D Stencil used to compute Laplacian, in case of uniform points with h fill distance u(ζ) = 1 (u(ζ + (h, 0)) + u(ζ (h, 0)) + u(ζ + (0, h)) + u(ζ (0, h)) 4u(ζ)) h2 9 / 28

10 Stencil Selection algorithm in R 2 from Davydov Given ζ Ξ \ δξ, we will select a set Ξ ζ containing ζ. Let Ξ ζ = {x 0, x 1,..., x k }, ζ := x 0, where the points x 0, x 1,..., x k are ordered counterclockwise with respect to ζ. Consider the following cost function: µ(x 1,..., x k ) := k i=1 α2 i where α i denotes the angle between the rays ζx i, ζx i+1 in the counterclockwise direction. We will also need the minimum and the maximum angle: α min = min{α 1,..., α k } α max = max{α 1,..., α k } Since k i=1 α i = 2π, the expression k i=1 α2 i achieves its unique minimum for α 1 =... = α k = 2π/k, that is for the uniformly spaced directions ζx i if x 1,..., x k were chosen freely in R 2. So the goal of the algorithm below is to choose {x 1,..., x k } Ξ such that µ(x 1,..., x k is minimised while keeping the distances x i ζ as small as possible. To achieve a balance between the goals of a small µ and small distances, we introduce the restriction that x i must be among m closest points to ζ, and terminate the algorithm if the set {ζ, x 1,..., x k } satisfies α max <= uα min, where m > k and u > 1.0 are parameters to be determined empirically. 10 / 28

11 Stencil Selection algorithm in R 2 from Davydov... Input: Ξ, ζ. Output: Ξ ζ. Parameters: k (the target number of points x i ), m > k (the number of points in the local cloud) and u > 1 (the angle uniformity tolerance). Parameter values used in our numerical experiments: k = 6, m = 30, u = 3.0. I Find m nearest points {x 1,..., x m} closer to ζ in Ξ \ ζ, sorted by increasing distance to ζ, and initialise Ξ ζ := {ζ, x 1,..., x k }. If α max <= uα min, then STOP: return Ξ ζ. II For i = n + 1,..., m: 1 Compute the angles α 1,..., α k+1 formed by the extended set {x 1,..., x k+1 } = {x 1,..., x k, x i }. 2 If both angles between ζx i and its two neighbouring rays are greater than the minimum angle α min := α min(x 1,..., x k, x i ): i Find j such that α j = α. Choose p = j or p = j +1 depending on whether α j 1 < α j+1 orα j 1 >= α j+1. ii If µ(x 1,..., x k+1 x p ) < µ(x 1,..., x k ): a. Update {x 1,..., x k } = {x 1,..., x k+1 } \ {x p }. b. If α max <= uα min STOP: return the current set Ξ ζ = {ζ, x 1,..., x k }. III Observe that α max > uα min must hold for the current set Ξ ζ = {ζ, x 1,..., x k } if the algorithm has not been terminated earlier. Find j such that α j = α min (x 1,..., x k ). Choose p = j or p = j + 1 depending on whether α j 1 < α j+1 or α j 1 >= α j+1. STOP: return Ξ ζ = {ζ, x 1,..., x k } \ {x p}. 11 / 28

12 Stencil Selection algorithm in R 3 Input: Ξ, ζ. Output: Ξ ζ. Parameter values used in our numerical experiments: k = 12, m = 30, u = Cost function: Standard Deviation in Sum of Squared Distance between each other points in the Stencil (Energy). µ(x 1,..., x k ) := std({e i : i = 1,..., k}) where E i := E(x i ) = j=1,i j x i x j 2 Another option: Projected Distance: shift the current set of stencil point to make zeta as the center and normalize, i.e. {x i = (x i ζ) : i = 1,..., k}, then x i ζ compute the Squared Distance between each other points I Find m nearest points {x 1,..., x m} closer to ζ in Ξ \ ζ, sorted by increasing distance to ζ, and initialise Ξ ζ := {ζ, x 1,..., x k }. If E max <= ue min, then STOP: return Ξ ζ. II For i = n + 1,..., m: 1 Compute the angles E 1,..., E k+1 formed by the extended set {x 1,..., x k+1 } = {x 1,..., x k, x i }. 2 Find p = j: such that E j == min({e i }) 3 If µ(x 1,..., x k+1 x p ) < µ(x 1,..., x k ): a. Update {x 1,..., x k } = {x 1,..., x k+1 } \ {x p}. b. If E max <= ue min STOP: return the current set Ξ ζ = {ζ, x 1,..., x k }. III STOP: return Ξ ζ = {ζ, x 1,..., x k }. 12 / 28

13 CVT: Centroidal Voronoi Points Algorithm based iterative voronoi tessellation There is an initial assignment for the generators [Halton, random, uniform...] In each iteration, estimates is made of the volume and location of the Voronoi cells Sampling done by Monte Carlo sampling The accuracy of the resulting CVT depends in part on the number of sampling points and the number of iterations taken. Lloyd s algorithm Can be customized to generate points for any dimensions, regions Boundary points placements Algorithm called: Constrained CVT CCVTBOX software 13 / 28

14 CVT points generation 14 / 28

15 RBF-QR Since the matrix K x is extremely ill-conditioned for small ɛ, we need to use Stable computation in this case. Hilbert-Schmidt series: K(x, z) = n=1 λ nϕ n (x)ϕ n (z) When we have such an Expansion, we truncate the series to M such that K(x i, x j ) = M n=1 λ nϕ n (x i )ϕ n (x j ) and write as a finite matrix: So we get K = ΦΛΦ T 15 / 28

16 New Basis Note: λ n = since λ decays in power n, it leads to ill conditioned interpolation matrix when ɛ is small ( ) n 1 α 2 ɛ 2 α 2 +δ 2 +ɛ 2 α 2 +δ 2 +ɛ 2 We want to find a new Basis which improves the conditioning of the problems. Compute Φ = QR = Q (R 1 R 2 ) K = (QR)ΛΦ T = Q (R 1 R 2 ) ΛΦ T With more Simplifications we get a new basis: Ψ as shown next. 16 / 28

17 Finding New Basis Ψ K = ΦΛΦ T = Q (R 1 R 2 ) ( Λ1 Λ 2 ) Φ T = Q ( R 1 Λ 1 R 2 Λ 2 ) Φ T ( = QR 1 Λ 1 IN Λ 1 ) }{{} 1 R 1 1 R 2 Λ 2 Φ T }{{} ( = QR 1 Λ 1 IN Λ 1 ) ( Φ T ) }{{} 1 R R 2 Λ 2 Φ T 2 }{{} ( = QR 1 Λ 1 Φ T }{{} 1 + Λ 1 ) 1 R 1 1 R 2 Λ 2 Φ T 2 }{{} = X = Ψ So Now we have a new basis, more stable due to Λ 1 1 Λ 2 : Ψ = X 1 K = Φ T 1 + Λ 1 1 R 1 1 R 2 Λ 2 Φ T 2 18 / 28

18 Computation of RBF-FD Stencils weights by QR Method Better Conditioned basis With the new basis, [ modify the weight ] computation matrix. By left-multiplying both sides of (4) by: (QR1 Λ 1 ) [ (QR1 Λ 1 ) ] [ KX 1 1 T 0 ] [ ] [ wv = (QR1 Λ 1 ) ] [ [ ki (x 0 )] n i=0 0 ] [ ΨX h 1 T 0 ] [ ] [ wv [ ψi (x = 0 )] n i=0 0 ] (5) where, h = (QR 1 Λ 1 ) 1 1 Are We done yet? Computing h is still problematic, since it involves Λ / 28

19 Numerical Computation of RBF-FD Stencils weights by QR Method For numerical implementation Ψ is truncated: [ ΨX h 1 T 0 ] [ ] [ wv = [ ψi (x 0 )] n i=0 0 ] (6) where, Ψ X := ψ 0 (x 0 )... ψ0 (x n).. ψ n(x 0 )... ψn(x n) To improve the stability, following are done: replace h by h = (QR 1 Λ 1 ) 1 1 where Λ 1 is obtained by zero replacing the elements in Λ 1 that exceed the reciprocal of machine epsilon. The above equation is replaced by n ( ψ i (x j ) ψ i (x 0 ))w j + h i v = ψ n i (x 0 ), i = 0,..., n. w 0 = w j j=1 j=1 20 / 28

20 Numerical Computation of RBF-FD Stencils weights by QR Method ψ 0 (x 1 ) ψ 0 (x 2 )... ψ 0 (x n) h(0) ψ 1 (x 1 ) ψ1 (x 2 )... ψ1 (x n) h(1).. ψ n(x 1 ) ψn(x 2 )... ψn(x n) h(n) w 1 w 2... w n = ψ 0 (x 0 ) ψ 1 (x 0 ). ψ n(x 0 ) Also the Stencil points {x 0, x 1, x 2,..., x n} are centered to x 0 : {x 0 x 0, x 1 x 0, x 2 x 0,..., x n x 0 }. So that we have {0, x 1, x 2,..., x n}. This simplifies the computation of [ ψ i (x 0 )] n i=0 as x 0 = 0 (7) 21 / 28

21 Open Questions Points Selection? it does not work for Quasi random points (Sobol, Halton)!! Adaptive points placements, more points closer boundary, corners Boundary points placements? How many points required for the Stencil? Stencil selection criterion? 22 / 28

22 What it does? Computes Mesh Triangulations upto 5 refinements, either in Square or Circle Domain Computes Stencil for each Interior point in the domain For each ɛ compute the Stencil weights using Direct Method if the condition number of the interpolation is < Else use the Stable Method to compute the weights For the given Problem computes the Finite Difference Matrix solves the Problem using the above matrix Computes RMS Error between the actual function and approximation Plots the Error values the defined range of ɛ 23 / 28

23 RMS Error for the 2D test function Figure: RMS Error in Function approx, RMS Error in Differentiation 24 / 28

24 RMS Error for the 3D test function Figure: RMS Error in Function approx, RMS Error in Differentiation 25 / 28

25 RMS Error for the 4D test function Figure: RMS Error in Function approx, RMS Error in Differentiation 26 / 28

26 Conclusion Implemented Davydov s 2D Poisson solver method Integrated with Stable computation of Fasshauer Extended to 3D support Next planning to work on Survey of similar work done so far Work on answering some theoretical questions?? Adaptive meshless refinement Hermite interpolation Solve some real and difficult PDEs 27 / 28

27 Reference O. Davydov, D.T. Oanh, On the optimal shape parameter for Gaussian radial basis function finite difference approximation of the Poisson equation, Comput. Math. Appl. 62 (2011) G.E. Fasshauer, M.J. McCourt, Stable evaluation of Gaussian RBF interpolants, SIAM J. Sci. Comput. 34 (2) (2012) A737A CVT Points: http: //people.sc.fsu.edu/~jburkardt/m_src/cvt/cvt.html 28 / 28