Semi-smooth Newton method for Solving Unilateral Problems in Fictitious Domain Formulations

Transkript

1 Semi-smooth Newton method for Solving Unilateral Problems in Fictitious Domain Formulations Jaroslav Haslinger, Charles University, Prague Tomáš Kozubek, VŠB TU Ostrava Radek Kučera, VŠB TU Ostrava SNA 08, Liberec

2 OUTLINE Motivation Fictitious domain formulation Algorithm Semi-Smooth Newton method Inner solvers Numerical experiments

3 OUTLINE Motivation Fictitious domain formulation Algorithm Semi-Smooth Newton method Inner solvers Numerical experiments

4 FDM for Dirichlet problem: Main idea of FDM: u + u = f u = g on ω in γ ω PDE is solved on the fictitious domain Ω, ω Ω, with a simple geometry. The corresponding stiffness matrix A is structured. The original boundary conditions on γ are enforced by Lagrange multipliers or control variables. } Ω δ ω γ Γ Ξ Classical FDM with Γ γ ( ) ( ) ( A B γ u f = B γ 0 λ γ g ) ( A B Γ B γ 0 Smooth FDM with Γ γ ) ( ) u = λ Γ ( f g )

5 FDM for unilateral problem: u g 0, Equivalent form of BC: Discretized FDM u u Au + B Γ λ Γ f = 0 u + u = f on ω 0, (u g) u = 0 in γ } = max {0, u ρ(u g)}, ρ > 0 C γ u max {0, C γ u ρ(b γ u g)} = 0 } F (u, λ Γ ) = 0 where B γ, B Γ and C γ are Dirichlet and Neumann trace matrices, respectively. This system may be solved by a semi-smooth Newton method with: Jacobian ( A B Γ G(u) 0 )

6 OUTLINE Motivation Fictitious domain formulation Algorithm Semi-Smooth Newton method Inner solvers Numerical experiments

7 Formulation of the problem: u g 0, u where f L 2 loc (R2 ), g H 1/2 (γ). u + u = f on ω 0, (u g) u = 0 in γ } Weak formulation: Find u H 1 (ω) such that (u, v) 1,ω = (f, v) 0,ω + u, v γ v H 1 (ω) u H 1/2 + (γ) µ u, u g γ 0 µ H 1/2 + (γ) It is equivalent to a convex minimization on a convex set. Here, λ := u the role of the Lagrange multiplier. plays

8 Reformulation of the inequality: for u L 2 +(γ), ρ > 0 (µ u, u µ u, u g γ 0 µ H 1/2 + (γ) (µ u, u g) 0,γ 0 µ L 2 +(γ) ρ(u g) u ) 0,γ 0 µ L 2 +(γ) u = P ( u ρ(u g)) Projection: P : L 2 (γ) L 2 +(γ) : P ϕ = max{0, ϕ(x)}, x γ

9 Reformulation of the weak formulation: (u, v) 1,ω = (f, v) 0,ω + ( u, v) 0,γ u = P ( u ρ(u g)) if u L 2 +(γ) v H 1 (ω) Fictitious domain formulation: Find (û, λ) Hper(Ω) 1 H 1/2 (Γ) such that (û, v) 1,Ω = (f, v) 0,Ω + λ, v Γ v H0(Ω) 1 û ω L 2 (γ) û ω = P ( û ω ρ(û ω g)) Here, λ represents the jump of normal derivatve on Γ.

10 Discretization: Let V h H 1 per(ω), Λ H (γ) L 2 (γ), Λ H (Γ) L 2 (Γ) be such that dim V h = n, dim Λ H (γ) = dim Λ H (Γ) = m. Find (û h, λ H ) V h Λ H (Γ) such that (û h, v h ) 1,Ω = (f, v h ) 0,Ω + (λ H, v h ) 0,Γ v h V h δ H û h = P (δ H û h ρ(τ H û h g H )) where δ H û h, τ H û h and g H are appropriate approximations of û h ω, û h γ and g, respectively, in Λ H (γ). Algebraic form: Au B Γ λ Γ f = 0 C γ u max {0, C γ u ρ(b γ u g)} = 0 } F (u, λ Γ ) = 0

11 OUTLINE Motivation Fictitious domain formulation Algorithm Semi-Smooth Newton method Inner solvers Numerical experiments

12 Definition of slanting functions (X. Chen, Z. Nashed, L. Qi 2000) A function F : Y Z is said to be slantly differentiable at y if F o : Y L(Y, Z): F (y + h) F (y) F o (y + h)h lim h 0 h F o is uniformly bounded in an open neighbourhood of y. F o is said to be a slanting function for F at y. = 0. Example: ψ : R R, y max{0, y} 0 if y < 0 ψ o (y) = 1 if y > 0 σ R if y = 0

13 Theorem (X. Chen, Z. Nashed, L. Qi 2000) Let F be slantly differentiable in an open subset U Y with a slanting function F o : U L(Y, Z), and y U be a solution to the nonlinear equation F (y) = 0. If F o (y) is non-singular for all y U and { F o (y) 1 : y U} is bounded by a constant M 0, then the Newton iteration y k+1 = y k F o (y k ) 1 F (y k ) converges super-linearly to y, provided that y 0 y is sufficiently small. Super-linear convergence: y k+1 y lim k y k y = 0 Linear convergence: y k+1 y η y k y, 0 η < 1 Quadratic convergence: y k+1 y C y k y 2, 0 C

14 Idea of the proof: y k+1 = y k F o (y k ) 1 F (y k ) y k+1 y = y k y F o (y k ) 1 (F (y k ) F (y )), h = y k y y k+1 y = F (y + h) 1 (F o (y + h)h F (y + h) + F (y )) y k+1 y M F o (y + h)h F (y + h) + F (y ) y k+1 y y k y M F (y + h) F (y ) F o (y + h)h h if it converges, then it converges super-linearly it converges, provided that y 0 y is sufficiently small (by induction)

15 Slanting function for our problem: F : R n+m R n+m, y := (u, λ ) F (y) := ( Au B Γ λ f G(u) ), G(u) := (G 1 (u),..., G m (u)) G i (u) := C γ,i u max {0, C γ,i u ρ(b γ,i u g i )} F o (y) = ( A B Γ G o (u) 0 ), G o (u) = (G o 1(u),..., G o m(u)) G o i(u) = C γ,i ψ o (C γ,i u ρ(b γ,i u g i ))(C γ,i ρb γ,i ) Active/Inactive set terminology: M := {1, 2,..., m} A := {i M : C γ,i u ρ(b γ,i u g i )) > 0} I := {i M : C γ,i u ρ(b γ,i u g i )) 0} G o i(u) = { ρbγ,i for i A C γ,i for i I

16 Active set algorithm: (0) Set k := 0, ρ > 0, ε u > 0, u 0 R n, λ 0 R m. (1) Define the inactive and active sets by: I k := {i : C γ,i u k ρ(b γ,i u k g i ) 0} A k := {i : C γ,i u k ρ(b γ,i u k g i ) > 0} (2) Solve: A B Γ ρb γ,a k 0 C γ,i k 0 ( u k+1 λ k+1 Γ ) = f ρg A k 0 (3) If u k+1 u k / u k+1 ε u, return u := u k. (4) Set k := k + 1, and go to step (1). Remark: The mixed Dirichlet-Neumann problem is solved in each Newton step.

17 OUTLINE Motivation Fictitious domain formulation Algorithm Semi-Smooth Newton method Inner solvers: Schur complement + Null-space method Projected BiCGSTAB Circulant matrices Numerical experiments

18 Non-symmetric sadle-point system ( A B 1 B 2 0 ) ( u λ ) = ( f g ) General assumptions A... non-symmetric (n n) matrix... singular with p = dim Ker A B 1, B 2... full rank (m n) matrices... B 1 B 2 Special FDM assumptions n is large (n = ) m n (m = 360) p m (p = 1, periodic BC on the fictitious boundary Ω for ) A is structured (circulant) so that actions of A (or A 1 ) are cheap B 1, B 2 are highly sparse so that their actions are cheap

19 Schur complement reduction a generalized inverse A columns of (n p) matrices N, M span Ker A, Ker A, respectively Au + B 1 λ = f f B 1 λ Im A Ker A u = A (f B 1 λ) + Nα M (f B 1 λ) = 0 & B 2 u = g B 2 A B 1 λ B 2Nα = B 2 A f g The reduced system: ( B2 A B 1 B 2 N M B 1 0 )( λ α ) ( B2 A = f g M f )

20 Null-space method for the reduced system ( ) ( ) F G 1 λ G 2 0 α = ( d e Two orthogonal projectors P 1 and P 2 onto Ker G 1 and Ker G 2 : P k : R m Ker G k, P k := I G k (G k G k ) 1 G k, k = 1, 2 Property: Ker P k = Im G k P k G k = 0 ) P 1 splits the saddle-point structure: P 1 Fλ + P 1 G 1 α = P 1d P 1 Fλ = P 1 d, G 2 λ = e, α := (G 1 G 1 ) 1 (G 1 d G 1 Fλ) P 2 decomposes λ = λ Im + λ Ker, λ Im Im G 2, λ Ker Ker G 2 At first: G 2 λ = G 2 λ Im = e = λ Im := G 2 (G 2 G 2 ) 1 e At second: P 1 Fλ Ker = P 1 (d Fλ Im ) on Ker G 2

21 Algorithm PSCM Step 1.a: Assemble G 1 := N B 2, G 2 := M B 1. Step 1.b: Assemble d := B 2 A f g, e := M f. Step 1.c: Assemble H 1 := (G 1 G 1 ) 1, H 2 := (G 2 G 2 ) 1. Step 1.d: Assemble λ Im := G 2 H 2e, d := P1 (d Fλ Im ). Step 1.e: Solve P 1 Fλ Ker = d on Ker G2. Step 1.f: Assemble λ := λ Im + λ Ker. Step 2: Step 3: Assemble α := H 1 G 1 (d Fλ). Assemble u := A (f B 1 λ) + Nα. an iterative projected Krylov subspace method for non-symmetric operators can be used in Step 1.e matrix-vector products Fµ := ( B 2 ( A ( B 1 µ))), P k µ := µ ( G k (H k (G k µ)) )

22 Idea: keep the iterative process for solving P 1 Fλ Ker = d on Ker G2. Algorithm ProjBiCGSTAB [ɛ, λ 0, F, P 1, P 2, d ] λ Initialize: λ 0 Ker G 2, r 0 := P 2 d P2 P 1 Fλ 0, p 0 := r 0, r 0, k := 0 While r k > ɛ 1 p k := P 2 P 1 Fp k 2 α k := (r k ) r 0 /( p k ) r 0 3 s k := r k α k p k 4 s k := P 2 P 1 Fs k 5 ω k := ( s k ) s k /( s k ) s k 6 λ k+1 := λ k + α k p k + ω k s k 7 r k+1 := s k ω k s k 8 β k+1 := (α k /ω k )(r k+1 ) r 0 /(r k ) r 0 9 p k+1 := r k+1 + β k+1 (p k ω k p k ) 10 k := k + 1 end

23 ( A B ) Theorem 2 The saddle-point matrix A := 1 is invertible iff B 2 0 B 1 has full row-rank Ker A Ker B 2 = {0} (NSC) A Ker B 2 Im B 1 = {0} The generalized Schur complement: the matrix of the reduced system ( ) B2 A B 1 B 2 N S := M B 1 0 Theorem 3 The following three statements are equivalent: The necessary and sufficient condition (NSC) holds. A is invertible. S is invertible. Remark: The generalized Schur complement S is not defined uniquely.

24 Theorem 4 Let A be invertible. The linear operator P 1 F: Ker G 2 Ker G 1 is invertible. Proof: Notice dim Ker G 1 = dim Ker G 2. Let µ Ker G 2 be such that P 1 Fµ = 0. Then Fµ Ker P 1 = Im G 1 and, therefore, there is β R p so that Fµ = G 1 β and G 2µ = 0. We obtain ( F G 1 G 2 0 ) ( µ β ) = ( 0 0 ), where the matrix is S (i.e. the generalized Schur complement) that is invertible iff A is invertible. Therefore µ = 0.

25 Circulant matrices and Fourier transform A = a 1 a n... a 2 a 2 a 1... a 3 a 3 a 2... a a n a n 1... a 1 = ( a, Ta, T 2 a,, T n 1 a ) T k f(ω) = R f(x k)e ixω dx = e ikω f(ω) XA = (Dx 0, Dx 1, Dx 2,, Dx n 1 ) = DX Lemma: Let A be circulant. Then A = X 1 DX, where X is the DFT matrix and D = diag(â), â = Xa, a = A(:, 1).

26 Multiplying procedure: A v := X 1 ( D (Xv) )... Moore-Penrose 0 d := fft(a) 1 v := fft(v) 2 v := v. d 1 3 A v := ifft(v) O(2n log 2 n) Multiplying procedures: Nα, N v (and Mα, M v) As AN = 0, the matrix N may be formed by eigenvectors corresponding to zero eigenvalues. ( ) I DD = diag(1, 1, 1, 0,..., 0) = X 1 = (N, Y), X 1 N = Y ( ) α Therefore we can define the operation: ind(α) = R 0 n 1 v α := ind(α) 1 v := ifft(v) 2 Nα := ifft(v α ) 2 N v := ind 1 (v) } O(n log 2 n)

27 Kronecker product of matrices: A x A y = A x R n x n x, A y R n y n y a y 11A x... a y 1n y A x..... a y n y 1A x... a y n y n y A x Lemma 1: (A x A y )(B x B y ) = A x B x A y B y (A x A y ) = A x A y N = N x N y Lemma 2: (A x A y )v = vec(a x VA y ), where V = vec 1 (v). V = (v 1,..., v ny ) R n x n y vec(v) = v 1. R n xn y v ny

28 Kronecker product and circulant matrices: Let A x, A y be circulant then: A = A x I y + I x A y = X 1 x D xx x X 1 X y + X 1 X x X 1 D yx y y = (X 1 x X 1 y )(D x I y + I x D y )(X x X y ) = X 1 DX x y with X = X x X y (DFT matrix in 2D) D = D x I y + I x D y (diagonal matrix) where X x, X y are the DFT matrices, D x = diag(x x a x ), D y = diag(x y a y ) and a x = A x (:, 1), a y = A y (:, 1), respectively.

29 Multiplying procedure: A v := X 1 ( D (Xv) ) 0 d x := fft(a x ), d y := fft(a y ) V := vec 1 (v) 1 V := fft(v) 2 V := fft(v ) 3 V := vec 1 (D vec(v)) 4 V := ifft(v) 5 V := ifft(v ) A v := vec(v) Number of arithmetic operations : O(2n(log 2 n x + log 2 n y ) + n) O(n log 2 n), n = n x n y Multiplying procedures: Nα, N v, Mα, M v... analogous

30 REFERENCES J. Haslinger, T. Kozubek, R.K. (2008): Fictitious domain formulation of unilateral problems: analysis and algorithms, in preparing. J. Haslinger, T. Kozubek, R.K., G. Peichel (2007): Projected Schur complement method for solving non-symmetric systems arising from a smooth fictitious domain approach, Num. Lin. Algebra Appl., 14, Farhat, C., Mandel, J., Roux, F., X. (1994): Optimal convergence properties of the FETI domain decomposition method, Comput. Meth. Appl. Mech. Engrg., 115, R. Glowinski, T. Pan, J. Periaux (1994): A fictitious domain method for Dirichlet problem and applications. Comput. Meth. Appl. Mech. Engrg. 111, R. K. (2005): Complexity of an algorithm for solving saddle-point systems with singular blocks arising in wavelet-galerkin discretizations, Appl. Math. 50, H. A. Van der Vorst (1992): BiCGSTAB: a fast and smoothly converging variant of BiCG for solution of nonsymmetric linear systems, SIAM J. Sci. Statist. Comput., 13, M. Benzi, G. H. Golub, J. Liesen(2005): Numerical solution of saddle point systems, Acta Numerica, pp