The Bramble-Pasciak + preconditioner and combination preconditioning - PowerPoint PPT Presentation

The Bramble-Pasciak + preconditioner and combination preconditioning (Relaxing Conditions for the Bramble Pasciak CG) Martin Stoll & Andy Wathen Computational Methods with Applications,Harrachov, Czech Republic, August 19 - 25, 2007

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments The linear system The Problem We want to solve A x = b where � � B T A B − C (1) � �� � A with A ∈ R n , n symmetric and positive definite and C ∈ R m , m symmetric positive semi-definite. B ∈ R m , n is assumed to have full rank.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Saddle point problems Saddle point problems arise in a variety of applications such as ❼ Mixed finite element methods for Fluid and Solid mechanics ❼ Interior point methods in optimisation See Benzi, Golub, Liesen (2005), Elman, Silvester, Wathen (2005), Brezzi, Fortin (1991), Nocedal, Wright (1999) Saddle point problems provide due to their indefiniteness and often poor spectral properties a challenge for people developing solvers. Benzi, Golub, Liesen (2005)

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Saddle point problems Saddle point problems arise in a variety of applications such as ❼ Mixed finite element methods for Fluid and Solid mechanics ❼ Interior point methods in optimisation See Benzi, Golub, Liesen (2005), Elman, Silvester, Wathen (2005), Brezzi, Fortin (1991), Nocedal, Wright (1999) Saddle point problems provide due to their indefiniteness and often poor spectral properties a challenge for people developing solvers. Benzi, Golub, Liesen (2005)

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Some Background – Basic relations We introduce the bilinear form induced by H � x , y � H := x T H y which is an inner product iff H is positive definite. A matrix A ∈ R n × n is self-adjoint in �· , ·� H if and only if �A x , y � H = � x , A y � H for all x , y which can be reformulated to A T H = HA .

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Some Background – Solvers ❼ cg needs symmetry in �· , ·� H plus positive definiteness in �· , ·� H ❼ minres needs the symmetry �· , ·� H but no definiteness in �· , ·� H Spectral properties of A can be enhanced by preconditioning, ie. considering A = P − 1 A � instead of A . Original matrix A is symmetric in �· , ·� I ⇒ minres can be used. What about the symmetry of � A ?

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments The Bramble-Pasciak CG We consider saddle point problem � � B T A A = B − C with a block-triangular preconditioner � � A 0 0 P = B − I which results in � � A − 1 A − 1 0 B T 0 A A = P − 1 A = � . 0 B T + C BA − 1 BA − 1 0 A − B

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments The Bramble-Pasciak CG The preconditioned matrix � � A − 1 A − 1 0 B T 0 A A = P − 1 A = � 0 B T + C BA − 1 BA − 1 0 A − B is self-adjoint in the bilinear form defined by � � A − A 0 0 H = . 0 I Under certain conditions for A 0 H defines an inner product and � A is also positive definite in this inner product, e.g. A 0 = . 5 A . The condition for A 0 usually involves the solution of an eigenvalue problem which can be expensive.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments The Bramble-Pasciak + CG We always want an inner product for symmetric and positive definite A 0 � � A + A 0 H + = . I Therefore, new preconditioner P + � � A 0 0 P + = − B I is required. The preconditioned matrix � � � P + � − 1 A = A − 1 A − 1 0 B T 0 A � A = BA − 1 BA − 1 0 B T − C 0 A + B is self-adjoint in this inner product.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Definiteness in H + If we split � AA − 1 � 0 B T + B T AA − 1 0 A + A A T H + = � 0 B T − C BA − 1 BA − 1 0 A + B as � � � � � � AA − 1 A − 1 B T 0 A + A I I 0 B T − C BA − 1 − BA − 1 I I we see that since this is a congruence transformation the matrix is always indefinite. This means: ❼ No reliable CG can be applied ❼ In practice CG quite often works fine ❼ Augmented methods might be used.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments An H + -inner product implementation of minres Use that � A symmetric in H -inner product and therefore implement a version of Lanczos process with H -inner product which gives � A V k = V k T k + β k v k +1 e T k with V T k H + V k = I . The following condition holds for the residual � r k � H + = �� r 0 � e 1 − T k +1 y k � H + . Minimizing �� r 0 � e 1 − T k +1 y k � H + can be done by the standard updated-QR factorization technique. Implementation details can be found in Greenbaum (1997).

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments The ideal transpose-free QMR method ( itfqmr ) The matrix formulation of the non-symmetric Lanczos process � A V k = V k +1 H k gives r k = V k +1 ( � r 0 � e 1 − H k y k ) . Ignoring the term V k +1 gives qmr method. A T H + = H + � Using � A a simplified Lanczos process can be implemented and we obtain w j = γ j H + v j . This is the basis for the ideal transpose-free qmr ( itfqmr ) see Freund (1994).

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Eigenvalue analysis for A 0 = A To get some insight into the convergence behaviour we the eigenvalues of � � � P + � − 1 A = A − 1 B T I � A = . BA − 1 B T 2 B � x � ) of � For the eigenpair ( λ, A we know that y � � � � � � � � A − 1 B T x + A − 1 B T y I x x = = λ BA − 1 B T 2 Bx + BA − 1 B T y 2 B y y For λ = 1 we get Ax + B T y = Ax which gives B T y = 0 and y = 0 iff Bx = 0. Since dim(ker( B )) = n − m multiplicity of λ = 1 is n − m .

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Eigenvalue analysis for A 0 = A λ − 1 A − 1 B T y which gives 1 For λ � = 1, we get that x = BA − 1 B T y = λ ( λ − 1) y . λ + 1 For an eigenvalue σ of BA − 1 B T we get σ = λ ( λ − 1) . λ + 1 Eigenvalues of � A become � λ 1 , 2 = 1 + σ (1 + σ ) 2 ± + σ. 2 4 Since σ > 0 we have m negative eigenvalues.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Eigenvalue analysis for A 0 = A λ − 1 A − 1 B T y which gives 1 For λ � = 1, we get that x = BA − 1 B T y = λ ( λ − 1) y . λ + 1 For an eigenvalue σ of BA − 1 B T we get σ = λ ( λ − 1) . λ + 1 Eigenvalues of � A become � λ 1 , 2 = 1 + σ (1 + σ ) 2 ± + σ. 2 4 Since σ > 0 we have m negative eigenvalues.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Eigenvalue analysis for A 0 = A λ − 1 A − 1 B T y which gives 1 For λ � = 1, we get that x = BA − 1 B T y = λ ( λ − 1) y . λ + 1 For an eigenvalue σ of BA − 1 B T we get σ = λ ( λ − 1) . λ + 1 Eigenvalues of � A become � λ 1 , 2 = 1 + σ (1 + σ ) 2 ± + σ. 2 4 Since σ > 0 we have m negative eigenvalues.

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Basic properties I Lemma 1 If A 1 and A 2 are self-adjoint in �· , ·� H then for any α, β ∈ R , α A 1 + β A 2 is self-adjoint in �· , ·� H . Lemma 2 If A is self-adjoint in �· , ·� H 1 and in �· , ·� H 2 then A is self-adjoint in �· , ·� α H 1 + β H 2 for every α, β ∈ R . Lemma 3 For symmetric A , � A = P − 1 A is self-adjoint in �· , ·� H if and only if P − T H is self-adjoint in �· , ·� A .

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Basic properties I Lemma 1 If A 1 and A 2 are self-adjoint in �· , ·� H then for any α, β ∈ R , α A 1 + β A 2 is self-adjoint in �· , ·� H . Lemma 2 If A is self-adjoint in �· , ·� H 1 and in �· , ·� H 2 then A is self-adjoint in �· , ·� α H 1 + β H 2 for every α, β ∈ R . Lemma 3 For symmetric A , � A = P − 1 A is self-adjoint in �· , ·� H if and only if P − T H is self-adjoint in �· , ·� A .

The Bramble-Pasciak+ CG Introduction Combination Preconditioning Numerical Experiments Basic properties I Lemma 1 If A 1 and A 2 are self-adjoint in �· , ·� H then for any α, β ∈ R , α A 1 + β A 2 is self-adjoint in �· , ·� H . Lemma 2 If A is self-adjoint in �· , ·� H 1 and in �· , ·� H 2 then A is self-adjoint in �· , ·� α H 1 + β H 2 for every α, β ∈ R . Lemma 3 For symmetric A , � A = P − 1 A is self-adjoint in �· , ·� H if and only if P − T H is self-adjoint in �· , ·� A .