Solving large Hamiltonian eigenvalue problems David S. Watkins - PowerPoint PPT Presentation

Example (our application) � � � � � � � � 0 − K v G M v = 0 − λ 0 − M w − M 0 w � � G M N = 0 − M N = R T JR = Adaptivity and Beyond, Vancouver, August 2005 – p.10

Example (our application) � � � � � � � � 0 − K v G M v = 0 − λ 0 − M w − M 0 w � � G M N = 0 − M � � � � � � − 1 0 0 I 2 G I I N = R T JR = 1 0 0 M − I 2 G M Adaptivity and Beyond, Vancouver, August 2005 – p.10

Example (our application) � � � � � � � � 0 − K v G M v = 0 − λ 0 − M w − M 0 w � � G M N = 0 − M � � � � � � − 1 0 0 I 2 G I I N = R T JR = 1 0 0 M − I 2 G M cost: zero flops Adaptivity and Beyond, Vancouver, August 2005 – p.10

H = J − 1 R − T AR − 1 Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Do not form the product explicitly Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Do not form the product explicitly Krylov subspace methods Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Do not form the product explicitly Krylov subspace methods We just need to apply the operator. Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Do not form the product explicitly Krylov subspace methods We just need to apply the operator. M = LL T (done once) Adaptivity and Beyond, Vancouver, August 2005 – p.11

H = J − 1 R − T AR − 1 After some algebra . . . � � � � � � M − 1 I 0 0 I 0 H = − 1 − 1 0 2 G I − K 2 G I Do not form the product explicitly Krylov subspace methods We just need to apply the operator. M = LL T (done once) backsolve Adaptivity and Beyond, Vancouver, August 2005 – p.11

However, . . . Adaptivity and Beyond, Vancouver, August 2005 – p.12

However, . . . Adaptivity and Beyond, Vancouver, August 2005 – p.12

However, . . . . . . we really want H − 1 . Adaptivity and Beyond, Vancouver, August 2005 – p.12

However, . . . . . . we really want H − 1 . � � � � � � M − 1 0 0 0 I I H = − 1 − 1 0 2 G I − K 2 G I Adaptivity and Beyond, Vancouver, August 2005 – p.12

However, . . . . . . we really want H − 1 . � � � � � � M − 1 0 0 0 I I H = − 1 − 1 0 2 G I − K 2 G I � � � � � � ( − K ) − 1 I 0 0 I 0 H − 1 = 1 1 0 2 G I M 2 G I Adaptivity and Beyond, Vancouver, August 2005 – p.12

However, . . . . . . we really want H − 1 . � � � � � � M − 1 0 0 0 I I H = − 1 − 1 0 2 G I − K 2 G I � � � � � � ( − K ) − 1 I 0 0 I 0 H − 1 = 1 1 0 2 G I M 2 G I − K = LL T Adaptivity and Beyond, Vancouver, August 2005 – p.12

Shift and Invert? Adaptivity and Beyond, Vancouver, August 2005 – p.13

Shift and Invert? ( H − τI ) − 1 Adaptivity and Beyond, Vancouver, August 2005 – p.13

Shift and Invert? ( H − τI ) − 1 is not Hamiltonian Adaptivity and Beyond, Vancouver, August 2005 – p.13

Shift and Invert? ( H − τI ) − 1 is not Hamiltonian Structure is lost. Adaptivity and Beyond, Vancouver, August 2005 – p.13

Shift and Invert? ( H − τI ) − 1 is not Hamiltonian Structure is lost. How can we recover it? Adaptivity and Beyond, Vancouver, August 2005 – p.13

Exploitable Structures Adaptivity and Beyond, Vancouver, August 2005 – p.14

Exploitable Structures symplectic (first idea) ( H − τI ) − 1 ( H + τI ) Adaptivity and Beyond, Vancouver, August 2005 – p.14

Exploitable Structures symplectic (first idea) ( H − τI ) − 1 ( H + τI ) skew-Hamiltonian (easiest?) H − 2 Adaptivity and Beyond, Vancouver, August 2005 – p.14

Exploitable Structures symplectic (first idea) ( H − τI ) − 1 ( H + τI ) skew-Hamiltonian (easiest?) H − 2 ( H − τI ) − 1 ( H + τI ) − 1 Adaptivity and Beyond, Vancouver, August 2005 – p.14

Exploitable Structures symplectic (first idea) ( H − τI ) − 1 ( H + τI ) skew-Hamiltonian (easiest?) H − 2 ( H − τI ) − 1 ( H + τI ) − 1 Hamiltonian H − 1 Adaptivity and Beyond, Vancouver, August 2005 – p.14

Exploitable Structures symplectic (first idea) ( H − τI ) − 1 ( H + τI ) skew-Hamiltonian (easiest?) H − 2 ( H − τI ) − 1 ( H + τI ) − 1 Hamiltonian H − 1 H − 1 ( H − τI ) − 1 ( H + τI ) − 1 Adaptivity and Beyond, Vancouver, August 2005 – p.14

Structured Lanczos Processes Adaptivity and Beyond, Vancouver, August 2005 – p.15

Structured Lanczos Processes Unsymmetric Lanczos Process = u k +1 b k d k Bu k − u k a k d k − u k − 1 b k − 1 d k B T w k − w k d k a k − w k − 1 d k − 1 b k − 1 = w k +1 d k +1 b k Adaptivity and Beyond, Vancouver, August 2005 – p.15

Hamiltonian Lanczos Process u k +1 b k +1 = Hv k − u k a k − u k − 1 b k − 1 = v k +1 d k +1 Hu k +1 Adaptivity and Beyond, Vancouver, August 2005 – p.16

Symplectic Lanczos Process v k +1 d k +1 b k = Sv k − v k d k a k − v k − 1 d k − 1 b k − 1 + u k d k S − 1 v k +1 = u k +1 d k +1 Adaptivity and Beyond, Vancouver, August 2005 – p.17

Symplectic Lanczos Process v k +1 d k +1 b k = Sv k − v k d k a k − v k − 1 d k − 1 b k − 1 + u k d k S − 1 v k +1 = u k +1 d k +1 many interesting relationships Adaptivity and Beyond, Vancouver, August 2005 – p.17

Implicit Restarts Adaptivity and Beyond, Vancouver, August 2005 – p.18

Implicit Restarts short Lanczos runs (breakdowns!!, no look-ahead) Adaptivity and Beyond, Vancouver, August 2005 – p.18

Implicit Restarts short Lanczos runs (breakdowns!!, no look-ahead) Restart implicitly as in IRA (Sorensen 1991), ARPACK Adaptivity and Beyond, Vancouver, August 2005 – p.18

Implicit Restarts short Lanczos runs (breakdowns!!, no look-ahead) Restart implicitly as in IRA (Sorensen 1991), ARPACK Restart with HR , not QR Adaptivity and Beyond, Vancouver, August 2005 – p.18

Remarks on Stability Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Are the answers worth anything? Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Are the answers worth anything? right and left eigenvectors Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Are the answers worth anything? right and left eigenvectors residuals Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Are the answers worth anything? right and left eigenvectors residuals condition numbers for eigenvalues Adaptivity and Beyond, Vancouver, August 2005 – p.19

Remarks on Stability Both Hamiltonian and symplectic Lanczos processes are potentially unstable. Breakdowns can occur. Are the answers worth anything? right and left eigenvectors residuals condition numbers for eigenvalues Don’t skip these tests. Adaptivity and Beyond, Vancouver, August 2005 – p.19

Example Adaptivity and Beyond, Vancouver, August 2005 – p.20

Example λ 2 Mv + λGv + Kv = 0 Adaptivity and Beyond, Vancouver, August 2005 – p.20

Example λ 2 Mv + λGv + Kv = 0 n = 3423 Adaptivity and Beyond, Vancouver, August 2005 – p.20

Example λ 2 Mv + λGv + Kv = 0 n = 3423 � � � � � � M − 1 0 0 0 I I = H − 1 − 1 2 G I − K 0 2 G I Adaptivity and Beyond, Vancouver, August 2005 – p.20

Example λ 2 Mv + λGv + Kv = 0 n = 3423 � � � � � � M − 1 0 0 0 I I = H − 1 − 1 2 G I − K 0 2 G I � � � � � � ( − K ) − 1 I 0 0 I 0 H − 1 = 1 1 0 2 G I M 2 G I Adaptivity and Beyond, Vancouver, August 2005 – p.20