Model Order Reduction for Wave Equations Rob F. Remis and J orn T. - - PowerPoint PPT Presentation

General solution and symmetry Krylov MOR

Model Order Reduction for Wave Equations

Rob F. Remis and J¨

• rn T. Zimmerling

DCSE Fall School, Delft, November 4 – 8, 2019 1

General solution and symmetry Krylov MOR

General solution and symmetry

Semidiscrete Maxwell/wave field system (D + S + M∂t) f(t) = −q′(t), t > 0 Field and source vector vanish for t < 0 Time dependence of source can be factored out: q′(t) = w(t)˜ q w(t) is called the source wavelet or source signature and vanishes for t < 0

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General solution and symmetry Krylov MOR

General solution and symmetry

Semidiscrete Maxwell/wave field system (D + S + M∂t) f(t) = −w(t)˜ q, t > 0 Symmetry properties of matrix D:

Matrix D is skew-symmetric w.r.t. W: DTW = −WD Matrix D is symmetric w.r.t. Wδ−: DTWδ− = Wδ−D

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General solution and symmetry Krylov MOR

General solution and symmetry

Multiply system by M−1 to obtain (A + I∂t) f(t) = −w(t)q, t > 0 with q = M−1˜ q and A = M−1(D + S) solution can be written in terms of the matrix exponential function (evolution operator) f(t) = −w(t) ∗ U(t) exp(−At)q, t > 0

U(t): Heaviside unit step function ∗: convolution in time

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General solution and symmetry Krylov MOR

General solution and symmetry

Initial-value problem: (A + I∂t) f(t) = 0, t > 0 with f(0) = f0. Solution: f(t) = exp(−At)f0 t ≥ 0

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General solution and symmetry Krylov MOR

General solution and symmetry

Lossless media: A = M−1D

• A. Matrix A is skew-symmetric w.r.t. WM

ATWM = −WMA

General case: A = M−1(D + S)

• B. Matrix A is symmetric w.r.t. WMδ−

ATWMδ− = WMδ−A

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General solution and symmetry Krylov MOR

General solution and symmetry

Symmetry property A is related to energy conservation Solution initial value problem: f(t) = exp(−At)f0, t ≥ 0 Stored energy in initial field is given by E0 = 1 2fT

0 WMf0

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General solution and symmetry Krylov MOR

General solution and symmetry

Stored energy at time instant t: E(t) = 1 2fT(t)WMf(t) = 1 2fT

0 exp(−ATt)WM exp(−At)f0

= 1 2fT

0 WM exp(+At) exp(−At)f0

= E0

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General solution and symmetry Krylov MOR

General solution and symmetry

WM is diagonal positive definite Energy inner product x, yen = yTWMx Inner product induces the energy norm fen = f, f1/2

en

E(t) = 1 2f2

en

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General solution and symmetry Krylov MOR

General solution and symmetry

Symmetry property B is related to reciprocity Introduce the matrices δe = 1 2(I + δ−) and δh = 1 2(I − δ−) We have δeδ− = δ−δe = δe and δhδ− = δ−δh = −δh

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General solution and symmetry Krylov MOR

General solution and symmetry

Electric-type vector: u = δeu Magnetic-type vector: u = δhu Source vector is of the electric-type: q = δeq

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General solution and symmetry Krylov MOR

General solution and symmetry

Solution: fq(t) = −w(t) ∗ U(t) exp(−At)q Measurement: rTWMfq(t) for some receiver vector r Using symmetry property B, we have

Electric-field measurement, r = δer: rTWMfq(t) = qTWMfr(t) Magnetic-field measurement, r = δhr: rTWMfq(t) = −qTWMfr(t)

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General solution and symmetry Krylov MOR

General solution and symmetry

Recall 1 2fTWMf = stored field energy in the domain 1 2fTWMf = sum of electric and magnetic field energy WMδ− is not positive definite 1 2fTWMδ−f = difference of electric and magnetic field energy Free-field Lagrangian: L(t) = 1 2fT(t)WMδ−f(t)

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General solution and symmetry Krylov MOR

General solution and symmetry

Bilinear form x, yla = yTWMδ−x (not an inner product) Energy and Lagrangian: E(t) = 1 2f, fen and L(t) = 1 2f, fla

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General solution and symmetry Krylov MOR

General solution and symmetry

Symmetry:

Lossless case (S = 0): A is skew-symmetric w.r.t. ·, ·en General case: A is symmetric w.r.t. ·, ·la

General solution (again): f(t) = −w(t) ∗ U(t) exp(−At)q, t > 0

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General solution and symmetry Krylov MOR

Krylov MOR

Power series expansion matrix exponential function exp(−At) =

∞

• k=0

(−At)k k! Solution consists of a superposition of powers of A acting on q It makes sense to look for approximations that belong to the Krylov subspace Km = span{q, Aq, ..., Am−1q}

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General solution and symmetry Krylov MOR

Krylov MOR

Lossless case: A = M−1D Let v1, v2, ...,vm be a basis of Km orthonormal w.r.t. WM vi, vjen = δi,j Expand approximation fm(t) in terms of these basis vectors fm(t) = α1(t)v1 + α2(t)v2 + ... + αm(t)vm = Vmam(t) with Vm = [v1, v2, ..., vm] and am(t) = [α1(t), α2(t), ..., αm(t)]T

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General solution and symmetry Krylov MOR

Krylov MOR

Vm is a tall matrix having the basis vectors as its columns Residual of the field approximation rm(t) = −w(t)q − Afm(t) − ∂tfm(t) = −w(t)q − AVmam(t) − Vm∂tam(t) We determine the expansion coefficients from the Galerkin condition VT

mWMrm(t) = 0

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General solution and symmetry Krylov MOR

Krylov MOR

Using the orthonormality of the basis vectors w.r.t. WM, we

• btain

(Tm + Im∂t) am(t) = −w(t)VT

mWMq,

t > 0 with Im identity matrix order m, and Tm = VT

mWMAVm

Note that Tm is skew-symmetric

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General solution and symmetry Krylov MOR

Krylov MOR

In addition, we take v1 = q−1

en q

Note that v1 is first column of matrix Vm v1 = Vme1 Consequently, q = qenVme1

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General solution and symmetry Krylov MOR

Krylov MOR

and we have (Tm + Im∂t) am(t) = −w(t)qene1, t > 0 Solution: am(t) = −qen w(t) ∗ U(t) exp(−Tmt)e1, t > 0 Field approximation or ROM: fm(t) = −qen w(t) ∗ U(t)Vm exp(−Tmt)e1, t > 0

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General solution and symmetry Krylov MOR

Krylov MOR

The basis vectors can be generated using the algorithm βi+1vi+1 = Avi + βivi−1, i = 1, 2, ..., m with v0 = 0, β1 = qen, and the βi, i ≥ 2, are determined from the condition vi, vien = 1, i = 1, 2, .. Proof is by induction This is the Lanczos algorithm for skew-symmetric matrices

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General solution and symmetry Krylov MOR

Krylov MOR

After m steps of this algorithm we have the summarizing equation AVm = VmTm + βm+1vm+1eT

m

with Tm =        −β2 β2 ... −βm βm        = VT

mWMAVm

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General solution and symmetry Krylov MOR

Krylov MOR

Compared with an explicit leap-frog time-stepping scheme 1/βi act as a time step For the general case proceed in a similar manner Basis vectors are “orthonormal” w.r.t. ·, ·la Matrix Tm is tridiagonal and (complex/sign) symmetric in this case Remark: the resulting Lanczos algorithm may break down, since ·, ·la is not an inner product

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General solution and symmetry Krylov MOR

Krylov MOR

Frequency-domain modeling A Laplace transform gives ˆ f(s) = − ˆ w(s)(A + sI)−1q Matrix resolvent instead of matrix exponential needs to be evaluated ROM: ˆ fm(s) = −qen ˆ w(s)Vm(Tm + sI)−1e1

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General solution and symmetry Krylov MOR

Krylov MOR

Convergence, what to expect? High-frequency expansion of resolvent: (A + sI)−1 = s−1(I + s−1A)−1 = s−1

∞

• k=0

(−s−1A)k Powers of A: Early times/high frequencies are approximated first

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General solution and symmetry Krylov MOR

Krylov MOR

When it exists, the group inverse A# of A is uniquely defined by the conditions AA#A = A, A#AA# = A, and AA# = A#A AA# projector onto the range of A Source vector belongs to the range of A: q = AA#q ˆ f(s) = − ˆ w(s)(A + sI)−1q = − ˆ w(s)(A + sI)−1AA#q = − ˆ w(s)A#(I + sA#)−1q = − ˆ w(s)A#

∞

• k=0

(−sA#)kq Low-frequency expansion Inverse powers of A: late-times/low-frequencies are approximated first

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General solution and symmetry Krylov MOR

Krylov MOR

Construct ROMs belonging to the Krylov space Km = span{q, A#q, ..., (A#)m−1q} A# inherits symmetry properties of A Lanczos algorithms with A# A# can be determined explicitly Action of A# on a vector requires solution of Poisson equation(s)

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General solution and symmetry Krylov MOR

Krylov MOR

Early- and late-time field approximations Construct a ROM that belongs to the extended Krylov space Kk,m = span{(A#)k−1q, (A#)k−2q, ..., A#q, q, Aq, ..., Am−1q} By exploiting symmetry, a basis of this space can again be generated via short recurrence relations

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General solution and symmetry Krylov MOR

Krylov MOR

Standard Krylov: field approximated by a polynomial in A Extended Krylov: field approximated by a Laurent polynomial in A Rational Krylov: field approximated by a rational function in A

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