T =0 Loewner framework Lecture 8 - Hébergement...

Introduction Reminders Lagrange rational interpolation The Loewner framework Summary

Model Reduction (Approximation) of Large-Scale Systems

Loewner frameworkLecture 8

C. Poussot-Vassal, P. Vuillemin & I. Pontes Duff

EDSYS, April 4-7th, 2016 (Toulouse, France)

moremoreΣ

(A,B,C,D)i

Σ

Σ

(A, B, C, D)i

model reduction toolbox

Kr(A,B)

AP + PAT + BBT = 0

WTV

DAE/ODE

State x(t) ∈ Rn, n large orinfinite

Data

ReducedDAE/ODE

Reduced state x(t) ∈ Rrwith r � n(+) Simulation(+) Analysis(+) Control(+) Optimization

Case 1u(f) = [u(f1) . . . u(fi)]y(f) = [y(f1) . . . y(fi)]

Case 2Ex(t) = Ax(t) +Bu(t)y(t) = Cx(t) +Du(t)

Case 3H(s) = e−τs

C. Poussot-Vassal, P. Vuillemin & I. Pontes Duff [Onera - DCSD] Model Reduction (Approximation) of Large-Scale Systems


IntroductionThe big picture and classification of the linear system problemsInterpolatory conditions

RemindersDescriptor realizationLagrange interpolation

Lagrange rational interpolationThe Loewner matrixProperties of the Loewner matrix

The Loewner frameworkThe Loewner pencilRedundant dataThe Loewner algorithmIllustrative examples

Summary



Outlines

IntroductionThe big picture and classification of the linear system problemsInterpolatory conditions

Reminders

Lagrange rational interpolation

The Loewner framework

Summary



IntroductionThe big picture and classification of the linear system problems


DAE/ODE


Data

ReducedDAE/ODE







Dynamical system Hx1(.)x2(.)...

xn(.)

y1(.)y2(.)...

yny (.)

u1(.)u2(.)...

unu (.)

Case 1 Data-driven case: ui(.), and yi(.) are given (marginally treated in this lecture)Case 2 Finite order case: (E,A,B,C,D) is given (first part of the course)Case 3 Infinite order case: H(s) is given (second part of the course)


Input-outputfrequency data

Finite orderlarge-scale linear model

Infinite orderlinear model

moremoreΣ

(A,B,C,D)i

Σ

Σ

(A, B, C, D)i


Kr(A,B)

AP + PAT + BBT = 0

WTV A reduced-orderlinear dynamical system



Projector based model approximationGiven H, full order LTI dynamical system (order n):

H :{

Ex(t) = Ax(t) +Bu(t)y(t) = Cx(t) +Du(t) (1)

choose Vk = range(V ), Wk = range(W ). Then the reduced order model H is given by

H :{

WTEV ˙x(t) = WTAV x(t) +WTBu(t)y(t) = CV x(t) +Du(t) (2)


E,A B

C




H :{



H :{



WTEV ,WTAV WTB

CV

ΠV,W =⇒C

E,A B




H :{



H :{


What about models that cannot be represented by (E, A, B, C, D) ?

Example: H(s) =1

s+ e−s.

This lecture was largely inspired byI A. C. Antoulas lectures on Loewner framework.I "A tutorial introduction to the Loewner framework for model reduction", A. C.

Antoulas, S. Lefteriu and A. C. Ionita (2015)



IntroductionInterpolatory conditions

SISO model: Given H, seek a reduced-order system H , such that the associatedtransfer function H(s) satisfies the interpolation conditions:

H(µi) = H(µi), i = 1, . . . , q, H(λj) = H(λj), j = 1, . . . , k.

If instead of the transfer function data, we are given input/output data, the resultingproblem is to find a (low order) model H such that:

H(µi) = vi, i = 1, . . . , q, H(λj) = wj, j = 1, . . . , k







H(µi) = vi, i = 1, . . . , q, H(λj) = wj, j = 1, . . . , k

MIMO tangential interpolation: In a similar way, given H, seek H s.t.

l∗i H(µi) = l∗i H(µi), i = 1, . . . , q, H(λj)rj = H(λj)rj, j = 1, . . . , k

and the corresponding data-based problem is

l∗H(µi) = v∗i , i = 1, . . . , q, H(λj)rj = wj , j = 1, . . . , k







H(µi) = vi, i = 1, . . . , q, H(λj) = wj, j = 1, . . . , k

MIMO tangential interpolation: In a similar way, given H, seek H s.t.

l∗i H(µi) = l∗i H(µi), i = 1, . . . , q, H(λj)rj = H(λj)rj, j = 1, . . . , k

and the corresponding data-based problem is

l∗H(µi) = v∗i , i = 1, . . . , q, H(λj)rj = wj , j = 1, . . . , k

Use the data information to construct H = (E,A,B,C).



Outlines

Introduction

RemindersDescriptor realizationLagrange interpolation



Summary



RemindersDescriptor realization

Descriptor realizationA LTI model H with nu inputs ny outputs, is in the descriptor realization if

H :{

Ex(t) = Ax(t) +Bu(t)y(t) = Cx(t) . (3)

where x(t) ∈ Rn and the matrices have the coherent dimension. The transfer functionof H is given by

H(s) = C(sE −A)−1B ∈ Cny×nu .




Descriptor realizationA LTI model H with nu inputs ny outputs, is in the descriptor realization if

H :{

Ex(t) = Ax(t) +Bu(t)y(t) = Cx(t) . (3)

where x(t) ∈ Rn and the matrices have the coherent dimension. The transfer functionof H is given by

H(s) = C(sE −A)−1B ∈ Cny×nu .

if det(E) 6= 0, one could inverse to get rid algebraic part BUTI Expensive to inverse E in the large-scale setting.I Dedicated algorithms were developed for descriptor systems.I H(s) is a strictly proper rational function, i.e., H(s) = n(s)

d(s) , deg(n) < deg(d).I On matlab : sys = dss(A,B,C, 0,E).




Q: What happens if det(E) = 0??I Non removable algebraic constraint.I H(s) can have a polynomial behavior, i.e., H(s) = N(s)

d(s) + P (s).

Example of polynomial behaviorLet us consider the descriptor system H(s) = C(sE −A)−1B such that

E =

[0 1 00 0 10 0 0

], A =

[1 0 00 1 00 0 1

], B = [0, 0,−1] CT = [1, 0, 0]. (4)




Q: What happens if det(E) = 0??I Non removable algebraic constraint.I H(s) can have a polynomial behavior, i.e., H(s) = N(s)

d(s) + P (s).

Example of polynomial behaviorLet us consider the descriptor system H(s) = C(sE −A)−1B such that

E =

[0 1 00 0 10 0 0

], A =

[1 0 00 1 00 0 1

], B = [0, 0,−1] CT = [1, 0, 0]. (4)

ThenH(s) = C(sE −A)−1B = s2.



RemindersRational function to descriptor realization

Given a rational function of order n

φ(x) =βnxn + . . . β2x2 + β1x+ β0

αnxn + . . . α2x2 + α1x+ α0,

where αn 6= 0 or βn 6= 0, the numerator and denominator have no common roots.Then

I if φ is strictly proper of order n, there is a state-space of dimension n,(E,A,B,C), with rank(E) = n.

I if φ is not strictly proper of order n, there is a state-space of dimension n+ 1,(E,A,B,C), with rank(E) = n.

Example:

φ(s) =3x2 + 6x+ 4

x+ 2=

4x+ 2

+ 3x= C(sE −A)−1B.

(5)

C =[1 −1 0

], E =

[1 0 00 0 10 0 0

], A =

[−2 0 00 −1 00 0 −1

]B =

[403

]



RemindersGeneralized eigenvalues

Generalized eigenvalue problemλ is generalized eigenvalues (v respectiv. generalized eigenvector) of the pencil (E,A)if

Av = λEv.

Moreover, the characteristic equation of the pencil (E,A) is

p(s) = det(sE −A) = 0.

On matlab: eig(E,A);

Remark: The poles of H(s) = C(sE −A)−1B are the zeros of p(s).



RemindersControllabillity and observabillity

Controllabillity and observabillity(E,A,B,C) is controllable if

rank(Rn) = rank[(λ1E −A)−1B . . . (λnE −A)−1B

]= n,

where λi are distinct and are not eigenvalues of (E,A). It is observable if

rank(On) = rank

C(µ1E −A)−1

...C(µnE −A)−1

= n.

where µi are distinct and are not eigenvalues of (E,A).

H = (E, A, B, C) is a minimal realization if it is controllable and observable ⇒Nopoles and zeros cancellation on H(s).



RemindersLagrange interpolation

Board development.Given some data P{(xi,yi), i = 1, . . . , N}, seek a polynomial p(s) satisfying theinterpolation conditions:

p(xi) = yi1) Define a Lagrange basis, i.e., given xi ∈ P , we define

Li(s) = Πi′ 6=i(s− λi′ ), i = 1, . . . , q.

2) Then

p(s) =N∑i=1

yiLi(s)Li(xi)

3) Barycentric representation

p(s) =

∑N

k=11

Li(xi)yis−xi∑N

k=11

Li(xi)1

s−xi



Outlines

Introduction

Reminders

Lagrange rational interpolationThe Loewner matrixProperties of the Loewner matrix


Summary



Lagrange rational interpolationThe Loewner matrix

Problem setting : Given some data (xi,yi), seek a minimal order r(s) = n(s)d(s)

satisfying the interpolation conditions:

r(xi) = yi, i = 1, . . . , N.




The Loewner matrixGiven a row array of pairs of complex numbers (µj ,vj), j = 1, . . . , k and a columnarray of pairs of complex numbers (λi,wi), i = 1, . . . , q, the associated Loewner matrixis:

L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q (6)

I Named after Charles Loewner (1893 - 1968), German mathematician.I First appearance: Uber monotone Matrixfunctionen, Mathematische Zeitschrift,

38: 177-216 (1934).I Play a very importante role in rational interpolation of data.I It encodes information about minimal admissible complexity of solution as we will

see later).



Lagrange rational interpolationThe Loewner matrix - The rational interpolation problem

Consider the array of pairs points

P = {(xi,yi) : i = 1, . . . , N, xi 6= xj , i 6= j}

We are looking for rational functions

φ(s) =n(s)d(s)

, greatest common divisor(n,d) = 1, (7)

which interpolate the points of the array P, i.e.

φ(xi) = yi, i = 1, . . . , N

Remark: If we seek a polynomial instead of a rational function, this problem can beeasily solved with the Lagrange polynomials :

Li(s) = Πj 6=i(s− xj) and φLag(x) =N∑i=1

yiLi(s)Li(xi)

=

Barycentric representation︷︸︸︷∑N

i=11

Li(λi)yis−xi∑N

i=11

Li(λi)1

s−xi

.




Lagrange-type formulaIdea: 1) We split P in two disjoint ensembles :

Pc = {(λi,wi), i = 1, . . . , q} and Pr = {(µi,vi), i = 1, . . . , p}

wherePc ∪ Pr = P.




Lagrange-type formulaIdea: 1) We split P in two disjoint ensembles :

Pc = {(λi,wi), i = 1, . . . , q} and Pr = {(µi,vi), i = 1, . . . , p}

wherePc ∪ Pr = P.

2) Define a Lagrange basis for the interpolation points from Pc, i.e., given λi ∈ C, wedefine

Li(s) = Πi′ 6=i(s− λi′ ), i = 1, . . . , q.

Li(s) from a basis for polynomials of degree at most q − 1.




Lagrange-type formula3) For constants αi, wi, i = 1, . . . , q, consider φ(s) defined implicitly by

q∑i=1

αiφ(s)−wi

s− λi= 0, αi 6= 0.

Thus, solving for φ we obtain

φ(s) =

∑q

i=1αiwis−λi∑q

i=1αis−λi

=

∑q

i=1 αiwiLi(s)∑q

i=1 αiLi(s), αi 6= 0.





q∑i=1

αiφ(s)−wi

s− λi= 0, αi 6= 0.


φ(s) =

∑q

i=1αiwis−λi∑q

i=1αis−λi

=

∑q

i=1 αiwiLi(s)∑q

i=1 αiLi(s), αi 6= 0.

Hence, since Li(λk) = 0, if k 6= i, it follows that φ(λi) = wi. !!and φ(s) interpolates already PcRemark that if

αi =1

Li(λi)⇒ φ(s) =

∑q

i=1αiwis−λi∑q

i=1αis−λi

we obtain the Lagrange polynomial interpolation of Pc





q∑i=1

αiφ(s)−wi

s− λi= 0, αi 6= 0.


φ(s) =

∑q

i=1αiwis−λi∑q

i=1αis−λi

=

∑q

i=1 αiwiLi(s)∑q

i=1 αiLi(s), αi 6= 0.

Hence, since Li(λk) = 0, if k 6= i, it follows that φ(λi) = wi. !!and φ(s) interpolates already Pc

How to interpolate Pu ?? Use the free parameters αi.




Lagrange-type formula4) The parameters αi can be used to interpolate the data from Pr, ie.

φ(µj) = vj, j = 1, . . . , p.

For this to holdLc = 0,

where

L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q , c =

α1...αq .

∈ C1×q (8)

L is the Loewner matrix constructed by means of Pc = (µj ,vj , j = 1, . . . , p) andPu = (λi,wi, i = 1, . . . , q).




If there is c =

α1...αq .

solution of

Lc = 0

then we can construct φL(s) = nL(s)dL(s) using the barycentric formula.

nL(s) =q∑j=1

αiwj

s− λjdL(s) =

q∑j=1

αi

s− λj.



Lagrange rational interpolationProperties of the Loewner matrix - From rational function to Loewner matrix

Main propertyGiven φ rational function that can be represented by the minimal descriptor realizationφ(s) = C(sE −A)−1B and the array of points P , where yi = φ(xi), let L be a p× qLoewner matrix for some partitioning Pc, Pr of P . Then

p, q ≥ degφ⇒ rank(L) = rank(E) = deg(φ).

Consequently, every square sub-Loewner matrix of size deg(φ) is non-singular.

The proof is based on:PropositionLet (E,A,B) be a controllable descriptor system, and λi, i = 1, . . . , r scalars whichare not eigenvalues of (E,A). Then

rank(Rr) = rank[(λ1E −A)−1B . . . (λrE −A)−1B] = n

provided that r ≥ n.



Lagrange rational interpolationProperties of the Loewner matrix

Proof.Let (E,A,B,C) be a minimal descriptor realization of size n, i.e., φ(s) = C(sE −A)−1B and (E,A,B,C) is observable and controllable. Then, vi = C(µiE−A)−1B,wj = C(λjE −A)−1B and it follows :

[L]ij =vi −wj

µi − λj=

C(µiE −A)−1B − C(λjE −A)−1B

µi − λj= −C(µiE −A)−1E(λjE −A)−1B.

(9)

HenceL = −OpERq ,

where Rq = [(λ1E −A)−1B . . . (λqE −A)−1B] and Op =

C(µ1E −A)...

C(µpE −A)

Since rank(Rq) = n and rank(Op) = n ⇒ rank(L) = rank(E) . Finally, rank(E) isequal to the degree of φ(s).




ExampleLet φ(s) = 1

s+1 , deg(φ) = 1. Suppose we have

µ1 = 0 →︸︷︷︸φ

v1 = 1;µ2 = −12→︸︷︷︸φ

v2 = 2}

andλ1 = −

23→︸︷︷︸φ

w1 = 3;λ2 = −34→︸︷︷︸φ

w2 = 2}

Then

L =[v1−w1µ1−λ1

v1−w2µ1−λ2v2−w1

µ2−λ1v2−w2µ2−λ2

]=[−3 −4−6 −8

].

Hence,rank(L) = 1 = deg(φ)




From Loewner matrix to rational function

Definition The rank of the array P:

rank(P ) := maxL

(rankL) = n,

where the maximum is taken over all possible Loewner matrices.Assume now that 2n < N . For any Loewner matrix with rank(L) = n and n rowa,there exists a column vector c, of appropriate dimension, N − n satisfying

Lc = 0 c ∈ Ck, k = N − n (10)

Then, we attach to L a rational function φL using the barycentric formula. ThenProposition(a) deg(φL) ≤ n < N.

(b) There is a unique φL, attached to all L and c satisfying (10), as long asrank(L) = n.

(c) φL interpolates exactly N − n+ deg(φL), points of the array P .




CorollaryThe rational function φL interpolates all given points if, and only if, deg(φ) = n, if andonly if, all n×n Loewner matrices which can be formed from the data are non-singular.

The main theoremGiven the array of N pairs of points P and let rank(P ) = n.(a) If 2n < N and all square Loewner matrices of size n which can be formed from P

are non-singular, there is a unique interpolating function of minimal degreedenoted by φmin(s) and deg(φmin) = n

(b) Otherwise, φmin(s) is not unique and deg(φmin) = N − n.




Lagragian barycentric algorithmInput: array P = (xi,yi), i = 1, . . . , N .

1: Partition of P into Pc = {(λi,wi), i = 1, . . . , q} and Pr = {(µi,vi), i =1, . . . , p}, where p = q = N/2.

2: Construct L.

3: Compute n = rankL.

4: New partition: Choose P = Pc ∪ Pu, where Pc = {(λi, wi), i =1, . . . , n + 1} and Pr = {(µi, vi), i = 1, . . . , N − n − 1}. Constructnew Loewner matrix L.

5: Compute c = [α1, . . . , αn+1]T as the null space of Lc = 0.

6: Then, the unique rational interpolant of minimal order n is the following :

φ(s) =

∑n+1i=1

αiwi

s−λi∑n+1i=1

αi

s−λi

.



Outlines

Introduction

Reminders


The Loewner frameworkThe Loewner pencilRedundant dataThe Loewner algorithmIllustrative examples

Summary



The Loewner frameworkTangential interpolation conditions

MIMO tangential interpolation problem:We are given the right data and the left data:

(λj ; rj ,wj), j = 1, . . . , k (µi; lTi ,vTi ), i = 1, . . . , q

and we seek H = (E,A,B,C), whose transfer function is H(s) = C(sE −A)−1B s.t.

lTi H(µi) = vTi , i = 1, . . . , q, H(λj)rj = wj , j = 1, . . . , k

We assume that all λi and µj are distinct.



The Loewner frameworkTangential interpolation conditions

MIMO tangential interpolation problem:We are given the right data and the left data:

(λj ; rj ,wj), j = 1, . . . , k (µi; lTi ,vTi ), i = 1, . . . , q

and we seek H = (E,A,B,C), whose transfer function is H(s) = C(sE −A)−1B s.t.

lTi H(µi) = vTi , i = 1, . . . , q, H(λj)rj = wj , j = 1, . . . , k

We assume that all λi and µj are distinct.

Q: How to find E,A,B,C ?



The Loewner frameworkThe Loewner pencil

The right data can be expressed as:

Λ = diag [λ1, . . . , λk] ∈ Ck×k,R =

[r1 r2 . . . rk

]∈ Cm×k

W =[w1 w2 . . . wk

]∈ Cp×k

and the left data can be expressed as:

M = diag [µ1, . . . , µq ] ∈ Cq×q

LT =[l1 l2 . . . lq

]∈ Cp×q

VT =[v1 v2 . . . vq

]∈ Cm×q



The Loewner frameworkThe Loewner pencil

The right data can be expressed as:

Λ = diag [λ1, . . . , λk] ∈ Ck×k,R =

[r1 r2 . . . rk

]∈ Cm×k

W =[w1 w2 . . . wk

]∈ Cp×k

and the left data can be expressed as:

M = diag [µ1, . . . , µq ] ∈ Cq×q

LT =[l1 l2 . . . lq

]∈ Cp×q

VT =[v1 v2 . . . vq

]∈ Cm×q

The Loewner matrix in this case is

L =

vT

1 r1−lT1 w1

µ1−λ1. . .

vT1 rk−lT

1 wk

µ1−λk

.... . .

...vT

q r1−lTq w1

µ1−λ1. . .

vTq rk−lT

q wk

µ1−λk

∈ Cq×k

With this notation L satisfy the Sylvester equation : ML− LΛ = VR − LW.



The Loewner frameworkThe Loewner matrix

The Loewner matrix is (and satisfies the Sylvester equation):

L =

vT

1 r1−lT1 w1

µ1−λ1. . .

vT1 rk−lT

1 wk

µ1−λk

.... . .

...vT

q r1−lTq w1

µ1−λ1. . .

vTq rk−lT

q wk

µq−λk

∈ Cq×k

ML− LΛ = VR − LW.



The Loewner frameworkThe Loewner matrix

The Loewner matrix is (and satisfies the Sylvester equation):

L =

vT

1 r1−lT1 w1

µ1−λ1. . .

vT1 rk−lT

1 wk

µ1−λk

.... . .

...vT

q r1−lTq w1

µ1−λ1. . .

vTq rk−lT

q wk

µq−λk

∈ Cq×k

ML− LΛ = VR − LW.

The shifted Loewner matrix is defined as follows (and satisfies the Sylvester equation):

Lσ =

µ1vT

1 r1−lT1 w1λ1

µ1−λ1. . .

µ1vT1 rk−lT

1 wkλk

µ1−λk

.... . .

...µqvT

q r1−lTq w1λ1

µ1−λ1. . .

µqvTq rk−lT

q wkλk

µ1−λk

∈ Cq×k

MLσ − LσΛ = MVR − LWΛ.



The Loewner frameworkthe Loewner pencil

If data are sample from a system whose transfer function is H(s) = C(sE −A)−1B,let us define the matrices with :

Oq =

lT1 C(µ1E −A)−1

...lTq C(µqE −A)−1

, Rk =[(λ1E −A)−1Br1, . . . , (λkE −A)−1Brk,

](11)

of size q × n and n× k respectively (generalized tangential observability andcontrollability matrices). Then,

[L]ji =vTj ri − lTj wi

µj − λi= −lTj C(µjE −A)−1E(λiE −A)−1Bri,

(12)




and similarly

[Lσ ]ji =µjvTj ri − lTj wiλi

µj − λi= −ljTC(µjE −A)−1A(λiE −A)−1Bri.

(13)




and similarly

[Lσ ]ji =µjvTj ri − lTj wiλi

µj − λi= −ljTC(µjE −A)−1A(λiE −A)−1Bri.

(13)

Hence,L = −OqERk and Lσ = −OqARk

Main theorem: Minimal amount of dataAssume that k = q, and let, be a regular pencil with no being an eigenvalue. Then

E = −L, A = −Lσ , , B = V, C = W,

is a descriptor realization of minimal interpolant of the data, i.e., the rational function

H(s) = W(Lσ − sL)−1V

interpolates the data.



The Loewner frameworkRedundant data

Suppose that we have more data then necessary. The problem has a solution if

rank[ωL− Lσ ] rank[L, Lσ ] = rank[LLσ

]= r, ω ∈ {λi} ∪ {µj}.

[L, Lσ ] = Y ΣlXT ,

[LLσ

]= Y ΣrX∗, Y ,X ∈ CN×n.

Redundant data - SVDA realization (E,A,B,C) of an (approximate) interpolant is given by:

E = −Y ∗LX, A = −Y ∗LσX,B = −Y ∗V, C = WX.

Arbitrary projectionLet Φ, Ψ be s.t. Φ∗Ψ, X∗Φ and Ψ∗Y are non singular. Then

(Y ∗LX,−Y ∗LσX,−Y ∗V,WX) and (Y ∗LX,−Y ∗LσX,−Y ∗V,WX)

are equivalent minimal descriptor realizations.



The Loewner frameworkThe Loewner algorithm

The Loewner algorithm

1: Build the Loewner matrix pencil (Lσ ,L).

2: Compute the SVDs

[L, Lσ ] = Y ΣlXT ,

[LLσ

]= Y ΣrX∗, Y ,X ∈ CN×n.

3: Then, the descriptor model H(s) = C(sE − A)B is (approximate) interpolant ofminimal, where

E = −Y ∗LX, A = −Y ∗LσX, B = −Y ∗V, C = WX.



The Loewner frameworkIllustrative examples

Let H(s) = 1s2+1 . Let us choose

λ1 = 1, λ1 = 2

andµ1 = −1, µ2 = −2.

ThenW =

[12

15

]V = WT ; R =

[1 1

]; L = RT .

Then

L =[

0 − 1101

10 0

],Lσ =

[12

3103

1015

],

and, since the pencil ((L,Lσ) is regular, we recover

H(s) = W(Lσ − sL)−1V.




Let H(s) = 1s2+1 . Let us choose Now let us suppose

λ1 = 1, λ1 = 2, λ3 = 3

andµ1 = −1, µ2 = −2, µ3 = −3.

ThenW =

[12

15

110

]V = WT ; R =

[1 1 1

]; L = RT .

Then

L =

[0 − 1

10−1101

10 0 − 1501

101

50 0

],Lσ =

[ 12

310

153

1015

7501

57

501

10

],

but rank[L Ls] = 2⇒ not regular.




L =

[0 − 1

10−1101

10 0 − 1501

101

50 0

],Lσ =

[ 12

310

153

1015

7501

57

501

10

],

but rank([L Ls]) = 2⇒ not regular.We can choose

X =

[1 11 00 0

]and Y = XT

and L = Y LX =[

0 110

− 110 0

], Lσ = Y LσX =

[1310

454

512

], W = WX and V = YV.

Then we recover once again

H(s) = W(Lσ − sL)−1V.



Outlines

Introduction

Reminders



Summary



SummaryTo sum up

I Loewner matrix L encodes information of rational interpolation.

L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q (14)



SummaryTo sum up


L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q (14)

I Barycentric rational interpolation via Lc = 0 ⇒ φL(s) = nL(s)dL(s)where:

nL(s) =q∑j=1

αiwj

s− λjdL(s) =

q∑j=1

αi

s− λj.



SummaryTo sum up


L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q (14)


nL(s) =q∑j=1

αiwj

s− λjdL(s) =

q∑j=1

αi

s− λj.

I Loewner pencil (L,Lσ) ⇒ construct interpolant ⇒ H(s) = W(Lσ − sL)−1V.



SummaryTo sum up


L =

v1−w1µ1−λ1

. . .v1−wq

µ1−λq

.... . .

...vk−w1µk−λ1

. . .vk−wq

µk−λq

∈ Ck×q (14)


nL(s) =q∑j=1

αiwj

s− λjdL(s) =

q∑j=1

αi

s− λj.

I Loewner pencil (L,Lσ) ⇒ construct interpolant ⇒ H(s) = W(Lσ − sL)−1V.

I Next presentation: Application to model reduction.



Model Reduction (Approximation) of Large-Scale Systems

Loewner frameworkLecture 8

C. Poussot-Vassal, P. Vuillemin & I. Pontes Duff

EDSYS, April 4-7th, 2016 (Toulouse, France)

moremoreΣ

(A,B,C,D)i

Σ

Σ

(A, B, C, D)i


Kr(A,B)

AP + PAT + BBT = 0

WTV

DAE/ODE


Data

ReducedDAE/ODE






T =0 Loewner framework Lecture 8 - Hébergement...

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