Parameter identification and state estimation for linear parabolic - - PowerPoint PPT Presentation

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Parameter identification and state estimation for linear parabolic - - PowerPoint PPT Presentation

Parameter identification and state estimation for linear parabolic equations Sergiy Zhuk IBM Research - Ireland Joint work with J.Frank (Utrecht University), I.Herlin (INRIA), R.Shorten (IBM) and S.McKenna (IBM) Stochastic Modelling of


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Parameter identification and state estimation for linear parabolic equations

Sergiy Zhuk

IBM Research - Ireland Joint work with J.Frank (Utrecht University), I.Herlin (INRIA), R.Shorten (IBM) and S.McKenna (IBM)

Stochastic Modelling of Multiscale Systems

NDNS+ workshop, Eindhoven Multiscale Institute December 5, 2013

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Outline

Minimax projection method Projection coefficients as a solution of DAE Bounding set for the projection error Ellipsoid containing the projection coefficients State estimation for a linear transport equation Parameter identification for linear Darcy equation

1 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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Problem statement

Assume a > 0 and I(·, t) ∈ H1

0(Ω) satisfies for almost all

t ∈ (0, T) the following equation: ∂tI + M · ∇I − a∆I = f , I(x, 0) = f0(x) , where

  • x ∈ Ω ⊂ Rn, n ≥ 2, Ω is an open bounded convex set;
  • M(x, t) = (M1(x, t) . . . Mn(x, t))′ with Mi ∈ L∞(0, T, H1

0(Ω))

for all i = 1, . . . , n;

  • f ∈ L2(0, T, L2(Ω)) and f0 ∈ H2(Ω) ∩ H1

0(Ω).

2 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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Galerkin projection

We expand the solution I into the following series: I(x, t) =

  • i∈N

ai(t)ϕi(x) , ai(t) := I(·, t), ϕiL2(Ω) , (1) where {ϕk}k∈N is the orthonormal set of eigenfunctions of −∆: −∆ϕk = λkϕk , ϕk ∈ C ∞(Ω) ∩ H1

0(Ω) ,

ϕk = 0 on ∂Ω .

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Galerkin projection

We expand the solution I into the following series: I(x, t) =

  • i∈N

ai(t)ϕi(x) , ai(t) := I(·, t), ϕiL2(Ω) , (1) where {ϕk}k∈N is the orthonormal set of eigenfunctions of −∆: −∆ϕk = λkϕk , ϕk ∈ C ∞(Ω) ∩ H1

0(Ω) ,

ϕk = 0 on ∂Ω . Define projection operator: PNI(·, t) = a(t) := (a1(t) . . . aN(t))′ , and reconstruction operator: P+

Na(t) = N

  • i=1

ai(t)ϕi

3 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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Galerkin projection

We expand the solution I into the following series: I(x, t) =

  • i∈N

ai(t)ϕi(x) , ai(t) := I(·, t), ϕiL2(Ω) , (1) where {ϕk}k∈N is the orthonormal set of eigenfunctions of −∆: −∆ϕk = λkϕk , ϕk ∈ C ∞(Ω) ∩ H1

0(Ω) ,

ϕk = 0 on ∂Ω . Define projection operator: PNI(·, t) = a(t) := (a1(t) . . . aN(t))′ , and reconstruction operator: P+

Na(t) = N

  • i=1

ai(t)ϕi and the vector of the exact projection coefficients: atrue

N

:= PNI.

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DAE for the projection coefficents

Define a differential operator Aϕ = M · ∇ϕ − a∆ϕ and a commutation error: e(x, t) := AP+

NPNI(x, t) − P+ NPNAI(x, t) .

(2)

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DAE for the projection coefficents

Define a differential operator Aϕ = M · ∇ϕ − a∆ϕ and a commutation error: e(x, t) := AP+

NPNI(x, t) − P+ NPNAI(x, t) .

(2) Since atrue

N

(t) = PNI(·, t) it follows that atrue

N

solves: ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf .

(3)

4 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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DAE for the projection coefficents

Define a differential operator Aϕ = M · ∇ϕ − a∆ϕ and a commutation error: e(x, t) := AP+

NPNI(x, t) − P+ NPNAI(x, t) .

(2) Since atrue

N

(t) = PNI(·, t) it follows that atrue

N

solves: ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf .

(3) As PNP+

N = I, we get, multiplying (3) by PN, that atrue N

solves da dt = −PNAP+

Na + PNe + PNf

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DAE for the projection coefficents

On the other hand, ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf

(4) has a solution if and only if −AP+

Na + e is in the range of P+ N.

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DAE for the projection coefficents

On the other hand, ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf

(4) has a solution if and only if −AP+

Na + e is in the range of P+ N.

This holds true, in turn, if (I − P+

NPN)AP+ Na = (I − P+ NPN)e .

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DAE for the projection coefficents

On the other hand, ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf

(4) has a solution if and only if −AP+

Na + e is in the range of P+ N.

This holds true, in turn, if (I − P+

NPN)AP+ Na = (I − P+ NPN)e .

Noting that (I − P+

NPN)e(t) = (I − P+ NPN)AP+ Natrue N

and, recalling that (P+

N)′ = PN, we compute:

(I−P+

NPN)AP+ Natrue N

2

L2(Ω) = (SN−A′ NAN)atrue N

·atrue

N

= HNatrue

N

2

RN ,

where SN = {Aϕi, Aϕj}N

i,j=1, AN = PNAP+ N and

HN := (SN − A′

NAN)

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DAE for the projection coefficents

On the other hand, ∂tP+

Na = P+ NPN∂tI = −AP+ Na + e + P+ NPNf

(4) has a solution if and only if −AP+

Na + e is in the range of P+ N.

This holds true, in turn, if (I − P+

NPN)AP+ Na = (I − P+ NPN)e .

Noting that (I − P+

NPN)e(t) = (I − P+ NPN)AP+ Natrue N

and, recalling that (P+

N)′ = PN, we compute:

(I−P+

NPN)AP+ Natrue N

2

L2(Ω) = (SN−A′ NAN)atrue N

·atrue

N

= HNatrue

N

2

RN ,

where SN = {Aϕi, Aϕj}N

i,j=1, AN = PNAP+ N and

HN := (SN − A′

NAN)

1 2 .

Thus atrue

N

solves the algebraic equation: 0 = HNa + eo for eo = −HNatrue

N

.

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DAE for the projection coefficients

Finally we find that if ∂tI + AI = f , I(0) = f0 then atrue

N

= P+

NP+ NI

solves the following DAE: da dt = −PNAP+

Na + em + PNf ,

0 = HNa + eo , a(0) = PNf0 , em = PNe = PNA(P+

NPNI − I)

eo = −HNPNI (5) where SN = {Aϕi, Aϕj}N

i,j=1 and HN := (SN − A′ NAN)

1 2 . 6 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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Outline

Minimax projection method Projection coefficients as a solution of DAE Bounding set for the projection error Ellipsoid containing the projection coefficients State estimation for a linear transport equation Parameter identification for linear Darcy equation

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A priori estimates

For Aϕ = M · ∇ϕ − a∆ϕ we get an estimate: em · em = PNA(P+

NPNI − I)2 RN = N

  • k=1

ϕk, M · ∇(P+

NPNI − I)2 L2(Ω)

≤ ρ1(·, t)L∞(Ω)∇(P+

NPNI − I)2 L2(Ω)

≤ ρ1(·, t)L∞(Ω)λ−1

N+1∆I(·, t)2 L2(Ω)

where ρ1(x, t) := M(x, t)2

Rn.

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A priori estimates

For Aϕ = M · ∇ϕ − a∆ϕ and I N := P+

NPNI we get an estimate:

eo · eo = HNatrue

N

2

RN = (I − P+ NPN)AP+ NPNI(·, t)L2(Ω)

=

  • k>N

ϕk, AP+

NPNI2 L2(Ω) =

  • k>N

ϕk, M · ∇P+

NPNI2 L2(Ω)

=

  • k>N

λ−2

k −∆ϕk, M · ∇P+ NPNI2 L2(Ω)

≤ 2λ−1

N+1λ−1 1 ρ2(·, t) + ρ1(·, t)L∞(Ω)∆I(·, t)2 L2(Ω)

where ρ1(x, t) := M(x, t)2

Rn, ρ2(x, t) := JM(x, t)2, JM is the

Jacobian of M.

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Outline

Minimax projection method Projection coefficients as a solution of DAE Bounding set for the projection error Ellipsoid containing the projection coefficients State estimation for a linear transport equation Parameter identification for linear Darcy equation

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Uncertain DAE for the projection coefficents

Finally we find that if ∂tI + AI = f , I(0) = f0 then atrue

N

= P+

NP+ NI

solves the following DAE: da dt = −PNAP+

Na + em + PNf ,

0 = HNa + eo , a(0) = PNf0 , em = PNe = PNA(P+

NPNI − I)

eo = −HNPNI (6) and λN+1 T em2

RN + eo2 RNdt ≤ C

T ∆I(·, t)2

L2(Ω)dt

≤ C1(∇f02

L2(Ω) +

T f (x, t)2

L2(Ω)dt)

≤ C ∗ where C ∗ = C ∗(M, f0, f ) is a constant.

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Ellipsoid for the projection coefficients

Let ˆ a solve the following ODE: dˆ a dt = − PNAP+

a − λN+1 C ∗ K(HN)′HNˆ a + PNf , ˆ a(0) = PNf0 , (7) where C ∗ = C ∗(M, f0, f ) is a constant, K = VU−1 and the matrix-valued functions V, U solve the following linear Hamiltonian ODE: ˙ U = (PNAP+

N)′U + λN+1

C ∗ (HN)′HNV , U(0) = I , ˙ V = −PNAP+

NV +

C ∗ λN+1 U , V(t0) = 0 . (8)

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Ellipsoid for the projection coefficients

Let ˆ a solve the following ODE: dˆ a dt = − PNAP+

a − λN+1 C ∗ K(HN)′HNˆ a + PNf , ˆ a(0) = PNf0 , (7) where C ∗ = C ∗(M, f0, f ) is a constant, K = VU−1 and the matrix-valued functions V, U solve the following linear Hamiltonian ODE: ˙ U = (PNAP+

N)′U + λN+1

C ∗ (HN)′HNV , U(0) = I , ˙ V = −PNAP+

NV +

C ∗ λN+1 U , V(t0) = 0 . (8) Assume that I solves ∂tI + AI = f , I(0) = f0.

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Ellipsoid for the projection coefficients

Let ˆ a solve the following ODE: dˆ a dt = − PNAP+

a − λN+1 C ∗ K(HN)′HNˆ a + PNf , ˆ a(0) = PNf0 , (7) where C ∗ = C ∗(M, f0, f ) is a constant, K = VU−1 and the matrix-valued functions V, U solve the following linear Hamiltonian ODE: ˙ U = (PNAP+

N)′U + λN+1

C ∗ (HN)′HNV , U(0) = I , ˙ V = −PNAP+

NV +

C ∗ λN+1 U , V(t0) = 0 . (8) Assume that I solves ∂tI + AI = f , I(0) = f0. Then K− 1

2 (t)(PNI(·, t) − ˆ

a(t))2 ≤ 1

12 / 56 Estimation and identification for parabolic PDEs (Sergiy Zhuk) NDNS+ workshop, EMI

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Outline

Minimax projection method Projection coefficients as a solution of DAE Bounding set for the projection error Ellipsoid containing the projection coefficients State estimation for a linear transport equation Parameter identification for linear Darcy equation

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Problem statement

We assume that I solves the following linear hyperbolic equation: ∂tI + u∂xI + v∂yI = 0 , I(x, y, 0) = I0(x, y) , I = 0 on ∂Ω , (9) where Ω = (0, 2π)2 and the fluid flow M = (u(x, y, t), v(x, y, t))′ is modelled by: ∂tω + u∂xω + v∂yω = 0 , u = −∂yψ , v = ∂xψ , − ∆ψ = ω , ψ(x, y) = 0 , (x, y) ∈ ∂Ω , ω(x, y, 0) = ω0(x, y) , ω(x, y, t) = 0 on ∂Ω . (10) We aim at solving the following problem: given incomplete sparse observations of I estimate I.

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Numerical experiment: setup

  • 75x75 basis functions ϕks := sin( kx

2 ) sin( sy 2 ) to represent

vorticity and advected quantity I;

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Numerical experiment: setup

  • 75x75 basis functions ϕks := sin( kx

2 ) sin( sy 2 ) to represent

vorticity and advected quantity I;

  • strongly occluded observations every 20 timesteps;

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Ground truth

∂tI + u∂xI + v∂yI = e ∂tω + u∂xω + v∂yω = f u = −∂yψ, v = ∂xψ, −∆ψ = ω

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Observations

Truth Observed data

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Observations

Truth Observed data

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Observations

Truth Observed data

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Observations

Truth Observed data

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Observations

Truth Observed data

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Reconstruction results

Truth Estimate

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Worst-case estimation error

Observation noise pattern Worst-case error pattern

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Relative observation error

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Estimation of the projection coefficients

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Outline

Minimax projection method Projection coefficients as a solution of DAE Bounding set for the projection error Ellipsoid containing the projection coefficients State estimation for a linear transport equation Parameter identification for linear Darcy equation

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Problem statement

Let h(·, t) ∈ H1

0(Ω) solve the following linear parabolic PDE:

∂th = ∂x(u∂xh) + ∂y(u∂yh) + W , h(0, x, y) = h0(x, y), h(t, 0, y) = h(t, a, y) = 0 , ∂yh(t, x, 0) = ∂yh(t, x, b) = 0 . where Ω := [0, a] × [0, b]

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Problem statement

Let h(·, t) ∈ H1

0(Ω) solve the following linear parabolic PDE:

∂th = ∂x(u∂xh) + ∂y(u∂yh) + W , h(0, x, y) = h0(x, y), h(t, 0, y) = h(t, a, y) = 0 , ∂yh(t, x, 0) = ∂yh(t, x, b) = 0 . where Ω := [0, a] × [0, b] and Ykl is observed in the following form: Ykl(ts) =

gkl(x, y)h(ts, x, y)dxdy + ηs

kl, k = 1, px, l = 1, py ,

where gkl ∈ L2(Ω) is an averaging kernel supported in a point (xk, yl) ∈ Ω and ηkl ∈ L2(t0, T) is an observation error.

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Problem statement

Let h(·, t) ∈ H1

0(Ω) solve the following linear parabolic PDE:

∂th = ∂x(u∂xh) + ∂y(u∂yh) + W , h(0, x, y) = h0(x, y), h(t, 0, y) = h(t, a, y) = 0 , ∂yh(t, x, 0) = ∂yh(t, x, b) = 0 . where Ω := [0, a] × [0, b] and Ykl is observed in the following form: Ykl(ts) =

gkl(x, y)h(ts, x, y)dxdy + ηs

kl, k = 1, px, l = 1, py ,

where gkl ∈ L2(Ω) is an averaging kernel supported in a point (xk, yl) ∈ Ω and ηkl ∈ L2(t0, T) is an observation error. We aim at solving the following problem:

px,py,M

  • k,l,s=1

Rkl(Ykl(ts)−

gkl(x, y)h(ts, x, y)dxdy)2+u2

L2(Ω) → min u>0 .

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Weak formulation

dh, ϕkl dt = − u∂xh, ∂xϕkl − u∂yh, ∂yϕkl + W , ϕkl , (11) with initial condition h(0, ·, ·) − h0, ϕkl = 0 where ·, · denotes the inner product in L2(0, a) × L2(0, b) and ϕkl(x, y) = ϕk(x)ϕl(y) = sin(kπx a ) cos(lπy b ) , k = 1 . . . Nx, l = 0 . . . Ny.

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Weak formulation

dh, ϕkl dt = − u∂xh, ∂xϕkl − u∂yh, ∂yϕkl + W , ϕkl , (11) with initial condition h(0, ·, ·) − h0, ϕkl = 0 where ·, · denotes the inner product in L2(0, a) × L2(0, b) and ϕkl(x, y) = ϕk(x)ϕl(y) = sin(kπx a ) cos(lπy b ) , k = 1 . . . Nx, l = 0 . . . Ny. We further assume that the permeability field u is represented as follows: u(x, y) =

Mx,My

  • m,n=1

umnψx

m(x)ψy n(y) > 0 ,

(12) where {ψx

m}m∈N and {ψy n}n∈N are total non-negative systems in

L2(0, a) and L2(0, b) respectively.

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SLIDE 59

DAE for the projection coefficients

Substituting the approximation hN = Nx,Ny

i=1,j=0 hijϕij into the weak

formulation and taking into account the projection error we get: dh dt = −

Mx,My

  • m,n=1

umnAmnh + W(t) + em , 0 = HNh + eo , hkl(0) = h0, ϕkl , k = 1, Nx, l = 0, Ny , (13) where h = (h11 . . . hNx,Ny )T and W = (W11 . . . WNxNy )T and Amn := 4 ab{ψx

mψy n∂xϕij, ∂xϕkl + ψx mψy n∂yϕij, ∂yϕkl}Nx,Ny k,i=1,l,j=0 .

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SLIDE 60

Reduced observation equation

We rewrite the observation equation as follows: Y(t) = Ch + η , (14) where Y = (Y11 . . . Ypxpy )T, η absorbs the projection and

  • bservation errors and

C = {

gkl(x, y)ϕij(x, y)dxdy}px,py,Nx,Ny

k,l,i=1,j=0 .

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SLIDE 61

Reduced control problem

M

  • s=1

R

1 2 (Y(ts) − Ch(ts))2 +

Mx,My

  • m,n=1

(umn)2ψx

mψy n2 L2(Ω) → min umn ,

dh dt = −

Mx,My

  • m,n=1

umnAmnh + W + em , 0 = HNh + eo , hkl(0) = h0, ϕkl , k = 1, Nx, l = 0, Ny . (15) This problem is solved in two steps:

  • optimization step: em, eo are dropped and umn are

approximated using Newton method;

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SLIDE 62

Reduced control problem

M

  • s=1

R

1 2 (Y(ts) − Ch(ts))2 +

Mx,My

  • m,n=1

(umn)2ψx

mψy n2 L2(Ω) → min umn ,

dh dt = −

Mx,My

  • m,n=1

umnAmnh + W + em , 0 = HNh + eo , hkl(0) = h0, ϕkl , k = 1, Nx, l = 0, Ny . (15) This problem is solved in two steps:

  • optimization step: em, eo are dropped and umn are

approximated using Newton method;

  • filtering step: for the fixed umn the estimate of h is

constructed.

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SLIDE 63

Optimization problem

For the optimization problem: J(u) : =

M

  • s=1

R

1 2 (Y(ts) − Ch(ts))2 + Ψ 1 2 u2 → min

u ,

dh dt = −

Mx,My

  • m,n=1

umnAmnh + W , h(0) = h0 , the gradient and Jacobian are computed analytically: ∇J(u) = 2JT(u)F(u) , F(u) = (R

1 2 (Y(t1) − Ch(t1))...R 1 2 (Y(tM) − Ch(tM)), Ψ 1 2 u)T ,

J(u) =

  • −CA11(t1h(t1)−z(t1))

... −CAMx My (t1h(t1)−z(t1)) ... ... ... −CA11(tMh(tM)−z(tM)) ... −CANx Ny (tMh(tM)−z(tM))

  • ,

dz dt = −

Mx,My

  • m,n=1

umnAmnz + tW(t) , z(0) = 0 .

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SLIDE 64

Optimization problem

The Newton method reads as follows: ui+1 := ui −

  • JT(ui)J(ui) + αI

−1 ∇J(ui) , u0 = 0 . where α > 0 is obtained through the line search: J(ui+1(α)) → min

α>0

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SLIDE 65

Numerical experiment: setup

  • 50x50 shifted Chebyshev polynomials to represent

permeability u

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SLIDE 66

Numerical experiment: setup

  • 50x50 shifted Chebyshev polynomials to represent

permeability u

  • 40x20 basis functions to represent h

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SLIDE 67

Numerical experiment: setup

  • 50x50 shifted Chebyshev polynomials to represent

permeability u

  • 40x20 basis functions to represent h
  • 300 observations over space

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SLIDE 68

Numerical experiment: setup

  • 50x50 shifted Chebyshev polynomials to represent

permeability u

  • 40x20 basis functions to represent h
  • 300 observations over space
  • 100 time-steps

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SLIDE 69

Numerical experiment: setup

  • 50x50 shifted Chebyshev polynomials to represent

permeability u

  • 40x20 basis functions to represent h
  • 300 observations over space
  • 100 time-steps
  • the problem is strongly ill-posed: given 30000 data points find

2500x800 parameters!

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SLIDE 70

Numerical experiment: true u

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SLIDE 71

Numerical experiment: estimate

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SLIDE 72

Numerical experiment

Truth Estimate: relative error ≈ 25%

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SLIDE 73

Numerical experiment: L2-estimation error for h

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SLIDE 74

Numerical experiment: summary

  • 25% relative error in estimating u

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SLIDE 75

Numerical experiment: summary

  • 25% relative error in estimating u
  • less than 2% relative error in estimating h

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SLIDE 76

Numerical experiment: summary

  • 25% relative error in estimating u
  • less than 2% relative error in estimating h
  • J(ˆ

u) ≈ 0.06

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SLIDE 77

Numerical experiment: summary

  • 25% relative error in estimating u
  • less than 2% relative error in estimating h
  • J(ˆ

u) ≈ 0.06

  • relative error in estimating the state transition matrix

umnAmn is ≈ 13%

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SLIDE 78

Numerical experiment: summary

  • 25% relative error in estimating u
  • less than 2% relative error in estimating h
  • J(ˆ

u) ≈ 0.06

  • relative error in estimating the state transition matrix

umnAmn is ≈ 13%

  • multiple global minima?

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SLIDE 79

Conclusions

  • robust extension of Galerkin projection method allows to

model the projection coefficents in the closed form and produce worst-case estimation error estimate;

  • method was applied to:
  • filtering problem for linear transport equation with strongly
  • ccluded observations (Zhuk, Frank, Herlin, Shorten, 2013,

submitted);

  • inversion problem for 2D linear Darcy equation with

heterogenious diffusion coefficient (Zhuk, McKenna, 2013, in preparation).

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