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Advances in numerical projection methods for

MOR of large-scale linear dynamical systems

V. Simoncini

Dipartimento di Matematica, Universita di Bologna

valeria@dm.unibo.it

1

Model Order Reduction

Given the continuous-time system

Σ =

A B

C

, A ∈ Cn×n

Analyse the construction of a reduced system

Σ =

A B

C

with A of size m ≪ n, and issues associated with its accuracy.

Applications: signal processing, system and control theory

2

Projection methods and Linear Dynamical Systems

• Solvers for the Lyapunov matrix equation

• Approximation of the matrix Transfer function

• Approximation of Hankel singular values by balanced truncation

3

Solving the Lyapunov equation. The problem

Approximate soln X to:

AX + XA⊤ + BB⊤ = 0

A ∈ Rn×n positive real B ∈ Rn×s, s ≥ 1

————————————

Time-invariant linear system:

x′(t) = Ax(t) + Bu(t), x(0) = x0

Analytic solution:

X =

∫ ∞

0

e−tABB⊤e−tA⊤

dt =

∫ ∞

0

xx⊤dt with x = exp(−tA)B.

see, e.g., Antoulas ’05, Benner ’06

4

Solving the Lyapunov equation. The problem

Approximate soln X to:

AX + XA⊤ + BB⊤ = 0

A ∈ Rn×n positive real B ∈ Rn×s, s ≥ 1

————————————

Time-invariant linear system:

x′(t) = Ax(t) + Bu(t), x(0) = x0

Analytic solution:

X =

∫ ∞

0

e−tABB⊤e−tA⊤

dt =

∫ ∞

0

xx⊤dt with x = exp(−tA)B

see, e.g., Antoulas ’05, Benner ’06

5

Standard Krylov subspace projection

X ≈ Xm Xm ∈ K

Galerkin condition: R := AXm + XmA⊤ + bb⊤ ⊥ K

V ⊤

m RVm = 0 K = range(Vm)

————————————

Assume V ⊤m Vm = Im and let Xm := VmYmV ⊤

m .

Projected Lyapunov equation:

(V ⊤

m AVm)Ym + Ym(V ⊤

m A⊤Vm) + V ⊤

m bb⊤Vm = 0

mTmYm + YmT⊤

m + e1e⊤

1 = 0

with b = Vme1 (Saad, ’90, for K = Km(A, b) = span{b, Ab, . . . , Am−1b})

6

Standard Krylov subspace projection

X ≈ Xm Xm ∈ K

Galerkin condition: R := AXm + XmA⊤ + bb⊤ ⊥ K

V ⊤

m RVm = 0 K = range(Vm)

————————————

Assume V ⊤m Vm = Im and let Xm := VmYmV ⊤

m .

Projected Lyapunov equation:

(V ⊤

m AVm)Ym + Ym(V ⊤

m A⊤Vm) + V ⊤

m bb⊤Vm = 0

mTmYm + YmT⊤

m + e1e⊤

1 = 0

with b = Vme1 (Saad, ’90, for K = Km(A, b) = span{b, Ab, . . . , Am−1b})

7

Standard Krylov projection. In quest of error bounds

AX + XA⊤ + BB⊤ = 0, X ≈ Xm ∈ Km(A, B)

‖X − Xm‖ ≤ ??

————————————

Simoncini & Druskin ’09. Analytic solution:

X =

∫ ∞

0

e−tABB⊤e−tA⊤

dt =

∫ ∞

0

xx⊤dt

with x = exp(−tA)B, B = b, ‖b‖ = 1

Let αmin = λmin((A + A⊤)/2) > 0. Then

‖x‖ ≤ exp(−tαmin)‖B‖

8

Standard Krylov projection. In quest of error bounds

AX + XA⊤ + BB⊤ = 0, X ≈ Xm ∈ Km(A, B)

‖X − Xm‖ ≤ ??

————————————

(Simoncini & Druskin ’09). Analytic solution:

X =

∫ ∞

0

e−tABB⊤e−tA⊤

dt =

∫ ∞

0

xx⊤dt

with x = exp(−tA)B, B = b, ‖b‖ = 1

Let αmin = λmin((A + A⊤)/2) > 0. Then

‖x‖ ≤ exp(−tαmin)‖B‖

9

First (key) step

Krylov subspace proj.: Xm = VmYmV ⊤m , range(Vm) = Km(A, b)

TmYm + YmT⊤

m + e1e⊤

1 = 0

Clearly,

Xm = Vm

(∫ ∞

0

e−tTme1e⊤

1 e−tT⊤

m dt

)

V ⊤

m =

∫ ∞

0

xmx⊤

mdt

————————————

II step: ‖X − Xm‖ = ‖∫∞

0(xx⊤ − xmx⊤

m)dt‖, so that

‖X − Xm‖ ≤∫ ∞

0

‖xx⊤ − xmx⊤

m‖dt ≤ 2

∫ ∞

0

e−tαmin ‖x − xm‖dt

10

First (key) step

Krylov subspace proj.: Xm = VmYmV ⊤m , range(Vm) = Km(A, b)

TmYm + YmT⊤

m + e1e⊤

1 = 0

Clearly,

Xm = Vm

(∫ ∞

0

e−tTme1e⊤

1 e−tT⊤

m dt

)

V ⊤

m =

∫ ∞

0

xmx⊤

mdt

————————————

II step: ‖X − Xm‖ = ‖∫∞

0(xx⊤ − xmx⊤

m)dt‖, so that

‖X − Xm‖ ≤∫ ∞

0

‖xx⊤ − xmx⊤

m‖dt ≤ 2

∫ ∞

0

e−tαmin ‖x − xm‖dt

11

The case of A symmetric

A symmetric ⇒ αmin = λmin(A)

Let 0 < λmin ≤ . . . ≤ λmax eigs of A + λminI, κ := λmax

λmin

Then

‖X − Xm‖ ≤√

κ + 1

λmin

√κ

(√κ − 1√κ + 1

)m

Note: same rate as CG for (A + λminI)z = b

12

The case of A symmetric. An example

0 5 10 15 20 25 3010

−12

10−10

10−8

10−6

10−4

10−2

100

dimension of Krylov subspace

ab

so

lute

err

or

no

rm

error norm ||X−X

m||

estimate of Proposition 3.1

A: 400 × 400 diagonal with uniformly distributed eigenvalues in [1, 10]

(αmin = λmin = 1)

13

The case of W (A) in an ellipse

Assume W (A) ⊆ E ⊂ C+

(E ellipse of center (c, 0), foci (c ± d, 0) and major semi-axis a)

Then

‖X − Xm‖ ≤ 4

αmin

r2

r2 − r

(r

r2

)m

where

r =a

d+

√(a

d

)2

− 1, r2 =c + αmin

d+

√(

c + αmin

d

)2

− 1

Note: same rate as FOM for (A + αminI)z = b

14

The case of W (A) in an ellipse.

0 2 4 6 8 10 12 14 1610

−14

10−12

10−10

10−8

10−6

10−4

10−2

100

dimension of Krylov subspace

ab

so

lute

err

or

no

rm

error norm ||X−X

m||

estimate of Corollary 4.2

A normal with eigenvalues on an elliptic curve

15

The case of W (A) in a wedge-shaped set. An example

Generalization to a wedge-shaped convex set of C+.

2 3 4 5 6 7 8 9−1.5

−1

−0.5

0

0.5

1

1.5

ℜ (λ)

ℑ(λ

)

0 5 10 15 2010

−12

10−10

10−8

10−6

10−4

10−2

100

dimension of Krylov subspace

ab

so

lute

err

or

no

rm

error norm ||X−X

m||

asymp. estimate of Corollary 4.6

A: diagonal (normal) matrix on the wedge-shaped curve.

16

Cyclic low rank Smith method

(ADI made efficient)

(see, e.g., Li 2000, Penzl 2000)

X0 = 0, Xj = −2pj(A + pjI)−1BB⊤(A + pjI)−⊤ j = 1, . . . , ℓ

+(A + pjI)−1(A − pjI)Xj−1(A − pjI)⊤(A + pjI)−⊤

with

rℓ(t) = Πℓj=1(t − pj), {p1, . . . , pℓ} = argmin max

t∈Λ(A)

˛

˛

˛

˛

rℓ(t)

rℓ(−t)

˛

˛

˛

˛

Convergence considerations:

Convergence depends on choice of {pj}. For A spd:

‖X − Xℓ‖ ≈(√

κadi − 2√κadi + 2

)ℓ

, κadi =λmax

λmin

17

Cyclic low rank Smith method

(ADI made efficient)

(see, e.g., Li 2000, Penzl 2000)

X0 = 0, Xj = −2pj(A + pjI)−1BB⊤(A + pjI)−⊤ j = 1, . . . , ℓ

+(A + pjI)−1(A − pjI)Xj−1(A − pjI)⊤(A + pjI)−⊤

with

rℓ(t) = Πℓj=1(t − pj), {p1, . . . , pℓ} = argmin max

t∈Λ(A)

˛

˛

˛

˛

rℓ(t)

rℓ(−t)

˛

˛

˛

˛

Convergence considerations:

Convergence depends on choice of {pj}. For A spd:

‖X − Xℓ‖ ≈(√

κadi − 2√κadi + 2

)ℓ

, κadi =λmax

λmin

18

Extended Krylov subspace method

Galerkin condition: Xm ∈ K s.t.

R := AXm + XmA⊤ + bb⊤ ⊥ K

K = Km(A, B) + Km(A−1, A−1B), range(Vm) = K(Druskin & Knizhnerman ’98, Simoncini ’07) Xm = VmYmV⊤

m

Projected Lyapunov equation:

(V⊤

mAVm)Ym + Ym(V⊤

mA⊤Vm) + V⊤

mbb⊤Vm = 0

mTmYm + YmT ⊤

m + e1e⊤

1 = 0

19

Performance evaluation. I

x′ = xxx + xyy + xzz − 10xxx − 1000yxy − 10zxz + b(x, y)u(t)

A matrix 183 × 183

0 5 10 15 20 2510

−11

10−10

10−9

10−8

10−7

10−6

10−5

10−4

10−3

10−2

CPU Time

no

rm o

f re

latv

e r

esid

ua

l

Extended Krylov

Standard Krylov

approximation space dim.: 146 (Standard Krylov) 112 (Extended Krylov)

20

Performance evaluation. II

Stopping criterion: norm of difference in solution

s EKSM CF-ADI

time(#its) dim.space time (#its) dim.space

Example 1 5.95 (12) 24 31.66 (6) 120

rail 5177 2 8.08 (10) 40 30.83 (5) 200

tol=10−5 4 11.11 ( 7) 56 40.20 (5) 400

7 18.12 ( 6) 84 54.22 (5) 700

Example (*) 1 38.95 (34) 68 588.68 (5) 150

tol=10−8 2 50.50 (33) 132 633.41 (5) 300

4 90.69 (33) 264 722.92 (5) 600

7 204.91 (32) 448 857.57 (5) 1050

x′ = xxx + xyy + xzz − 10xxx − 1000yxy − 10zxz + b(x, y)u(t) (∗)

21

Convergence analysis of Extended Krylov

General considerations

AX + XA⊤ + BB⊤ = 0

A−1X + XA−⊤ + A−1BB⊤A−⊤ = 0

Summing up for any γ ∈ R, we obtain yet a Lyapunov equation:

(A+γA−1)X+X(A⊤+γA−⊤)+[B,√

γA−1B][B⊤;√

γB⊤A−⊤] = 0

with Km(A + γA−1, [B,√

γA−1B]) ( Km(A, B) + Km(A−1, A−1B)

AX + XA⊤ + BB⊤ = 0

22

Convergence analysis of Extended Krylov

General considerations

AX + XA⊤ + BB⊤ = 0

A−1X + XA−⊤ + A−1BB⊤A−⊤ = 0

Summing up for any γ ∈ R, we obtain yet a Lyapunov equation:

(A+γA−1)X+X(A⊤+γA−⊤)+[B,√

γA−1B][B⊤;√

γB⊤A−⊤] = 0

with Km(A + γA−1, [B,√

γA−1B]) ( Km(A, B) + Km(A−1, A−1B)

AX + XA⊤ + BB⊤ = 0

23

Convergence analysis of Extended Krylov: A symmetric pos.def.

Kressner & Tobler tr’09:

‖X − Xm‖ .

(4√

κ − 14√

κ − 1

) 1

2

︸ ︷︷ ︸

ρ

m

κ =λmax

λmin

Knizhnerman & Simoncini: ‖X − Xm‖ . ρm

m1

2

24

A symmetric. An example

True error norm and asymptotic estimates for A symmetric.

0 2 4 6 8 10 12 14 16 18 2010

−12

10−10

10−8

10−6

10−4

10−2

100

number of iterations

no

rm o

f e

rro

r

true errorρm

ρm/m1/2

0 10 20 30 40 50 6010

−12

10−10

10−8

10−6

10−4

10−2

100

number of iterations

no

rm o

f e

rro

r

true error

ρm

ρm/m1/2

Left: Spectrum in [0.1, 10]. Right: Spectrum in [0.01, 100]

Knizhnerman & Simoncini (in prep.)

25

Convergence analysis of Extended Krylov: A nonsymmetricW (A) ⊂ C+ a disk of center c and radius r.

‖X − Xm‖ .1√m

(r2

4c2 − 3r2

)m

0 5 10 15 20 25 30 35 4010

−12

10−11

10−10

10−9

10−8

10−7

10−6

10−5

10−4

10−3

number of iterations

no

rm o

f e

rro

r

||X−Xm

||

ρm/m1/2

Knizhnerman & Simoncini (in prep.)

26

Comparison of convergence rates: A symmetric

ADI iteration: εadi,j ≈“√

κadi−2√κadi+2

”jStandard Krylov: εkr,j ≈

“√κ−1√κ−1

”j

Extended Krylov: εek,ℓ ≈

4√

κ−14√

κ−1

”1/2«ℓ

0 1 2 3 4 5 6 7 8 9 10

x 104

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

value of λmax

co

nve

rge

nce

ra

te

ADIKrylovExtended Krylov

λmin = 1, λmax ∈ [102, 105]

27

Transfer function approximation (cf. MOR)

h(σ) = c∗(A − iσI)−1b, σ ∈ [α, β]

Given space K and V s.t. K=range(V ),

h(σ) ≈ (V ∗c)∗(V ∗AV − σI)−1(V ∗b)

For K = Km(A, b) (standard Krylov):

hm(σ) = (V ∗mc)∗(Hm − σI)−1e1‖b‖

For K = Km(A, b) + Km(A−1, A−1b) (EKSM):

hm(σ) = (U∗mc)∗(Tm − σI)−1e1‖b‖

Alternative: Rational Krylov (Grimme-Gallivan-VanDooren etc. )

choosing the poles unresolved issue (A nonsymmetric)

28

An example: CD Player, |h(σ)| = |C∗:,i(A − iσI)−1B:,j |

10−1

100

101

102

103

104

105

106

10−8

10−6

10−4

10−2

100

102

104

106

108

CDplayer, m=20, (1,1)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−8

10−6

10−4

10−2

100

102

104

CDplayer, m=20, (1,2)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

101

102

CDplayer, m=20, (2,1)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−5

10−4

10−3

10−2

10−1

100

101

102

103

104

CDplayer, m=20, (2,2)

truearnoldiextended

29

An example: CD Player, |h(σ)| = |C∗:,i(A − iσI)−1B:,j |

10−1

100

101

102

103

104

105

106

10−8

10−6

10−4

10−2

100

102

104

106

108

CDplayer, m=20, (1,1)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−8

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

101

102

CDplayer, m=50, (1,2)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

101

102

CDplayer, m=20, (2,1)

truearnoldiextended

10−1

100

101

102

103

104

105

106

10−5

10−4

10−3

10−2

10−1

100

101

102

103

104

CDplayer, m=20, (2,2)

truearnoldiextended

30

Other related problems

• Projected generalized Lyapunov equation

EXA⊤ + AXE⊤ = −PlBB⊤P⊤l , X = PrXP⊤

r

(Stykel & Simoncini (in prep.))

• Sylvester equation: AX + XB + C = 0 (Heyouni ’08)

• Riccati equation: AX + XA⊤ − XGX + C = 0

(Heyouni & Jbilou ’08)

• Shifted systems: (A − σI)x = b with many σ’s

(..., Simoncini tr’09)

• Special Sylvester equation: AX + XΣ = [b(σ1), ..., b(σs)]

(Simoncini tr’09)

Approximation space: Extended Krylov subspace

31

Balanced reduction.

Balancing matrix transformation. Given

AP + PA⊤ + BB⊤ = 0, QA + A⊤Q + C⊤C = 0.

Find Tr, Tℓ such that T⊤ℓ PTℓ = Σ = T⊤

r QTr

The matrix

Σ = diag(σ1, σ2, σ3, . . .)

contains the Hankel singular values of the system

——————————–

Large body of literature, and various possibilities, even in the

small-scale case (cf., e.g., Antoulas ’05)

Error estimate for the reduced system:

‖Σ − Σ‖H∞≤ 2(σk+1 + · · · + σn),

32

An iterative procedure. Joint work in progress with T. Stykel

Given K0, L0.

For k = 1, 2, . . .

1. Update approx. spaces Kk−1 → Kk=range(Vk), Lk−1 → Lk=range(Wk)

2. Compute approximate Gramians Xk, Yk s.t.

P ≈ Pk = VkXkV⊤

k , Q ≈ Qk = WkYkW⊤k

with W⊤k Vk = I

3. Approximate Hankel singular values:p

λj(PQ) ≈ σj(L⊤XLY ), Xk = LXL

⊤X , Yk = LY L

⊤Y

UΣZ⊤ = svd(L⊤XLY )

4. If satisfied, compute truncated balancing transformation matrices:

Tr = VkLXUΣ−1/2, Tℓ = WkLY ZΣ−1/2 and stop

What spaces Kk, Lk to obtain accurate and small size Tr, Tℓ ?

33

Truncated balancing

What spaces Kk, Lk to obtain accurate and small size Tr, Tℓ ?

Two possible choices we are exploring:

⋆ Kk = Lk = EKk(A, [B, C⊤])

(Related to cross-Gramians for A symmetric)

⋆ Kk = EKk(A, B) Lk = EKk(A⊤, C⊤)

bi-orthogonal bases (a la Lanczos)

EKk: Extended Krylov subspace

34

Example

Penzl’s example (408 × 408): A = blkdiag(A1, A2, A3, A4, D)

A1 =

2

4

−0.01 −200

200 0.001

3

5 A2 =

2

4

−0.2 −300

300 −0.1

3

5

A3 =

2

4

−0.02 −500

500 0

3

5 A4 =

2

4

−0.01 −520

520 −0.01

3

5

and D =diag(1:400)

B = C⊤. Vector (s = 1) with large projection onto nonsym part.

35

Example. cont’ed. Error |H(σ) − Hk(σ)|

10−4

10−2

100

102

104

106

108

10−25

10−20

10−15

10−10

10−5

100

105

Frequency w

Ma

gn

itu

de

Absolute error

Error ELanczError EKSMError INV−Arnoldi

Balancing Ext’d Krylov: Tr of size 19 (space of max size 40)

Balancing Ext’d Lanczos: Tr of size 17 (left-right spaces of max size 20 each)

Inverted-Arnoldi: space of size 40

36

Convergence of Hankel singular values

0 2 4 6 8 10 12 14 16 18 2010

−8

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

101

102

number of iterations

Qu

an

tity

τk f

or

sto

pp

ing

te

st

EK: At each iteration k

τk =X

j∈Jk

|σ(k−1)j − σ

(k)j |

σ(k)1

where Jk = {index j : σj/σ1 > 10−10}

Residuals of Lyapunov equations: ‖Rk‖ = ‖Sk‖ = O(10−3)

37

One more example. Error |H(σ) − Hk(σ)|iss case. (tiny: 270 × 270), B 6= C⊤ vectors

10−4

10−2

100

102

104

106

108

10−25

10−20

10−15

10−10

10−5

100

Frequency w

Ma

gn

itu

de

Absolute error

Error ELanczError EKSMError INV−Arnoldi

Balancing Ext’d Krylov: Tr of size 35 (space of max size 40)

Balancing Ext’d Lanczos: Tr of size 20 (left-right spaces of max size 20 each)

Inverted-Arnoldi: space of size 40

38

Convergence of Hankel singular values

2 4 6 8 10 12 14 16 18 2010

−3

10−2

10−1

100

101

number of iterations

Qu

an

tity

τk f

or

sto

pp

ing

te

st

EK Lanczos: At each iteration k

τk =X

j∈Jk

|σ(k−1)j − σ

(k)j |

σ(k)1

where Jk = {index j : σj/σ1 > 10−10}

Ext’d Lanczos. Residuals of Lyapunov equations: ‖Rk‖ = O(1), ‖Sk‖ = O(103)

39

Conclusions

• Great potential of enriched projection spaces

• Exploit low cost of using A and A−1

• Projection combined with matrix function theory for proofs

valeria@dm.unibo.it

http://www.dm.unibo.it/~ simoncin

40

References

1. Leonid Knizhnerman and V. Simoncini, Convergence analysis of the

Extended Krylov Subspace Method for the Lyapunov equation, In

preparation.

2. V. Simoncini and Tatjana Stykel, Iterative methods for balanced truncation

of large-scale linear dynamical systems, In preparation.

3. V. Simoncini, The Extended Krylov subspace for parameter dependent

systems, August 2009, pp.1-13.

4. V. Simoncini and Vladimir Druskin, Convergence analysis of projection

methods for the numerical solution of large Lyapunov equations, SIAM J.

Numerical Analysis. Volume 47, Issue 2,pp. 828-843 (2009).

5. V. Simoncini, A new iterative method for solving large-scale Lyapunov

matrix equations, SIAM J. Scient. Computing, v.29, n.3 (2007), pp.

1268-1288.

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