inverse problems in lithospheric exure and viscoelasticity€¦ · viscoelasticity axel osses in...

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Inverse problems in lithospheric flexure and viscoelasticity Axel Osses in coll with M. de Buhan (CNRS, France), E. Contreras (Geophysics Department, Chile), B. Palacios (DIM, Chile) DIM - Departamento de Ingenier´ ıa Matem´ atica CMM - Centro de Modelamiento Matem´ atico Facultad de Ciencias F´ ısicas y Matem´ aticas, Universidad de Chile, Santiago. [matematika mugaz bestalde] - [bcam] Bilbao, July 4 th 2011

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Page 1: Inverse problems in lithospheric exure and viscoelasticity€¦ · viscoelasticity Axel Osses in coll with M. de Buhan (CNRS, France), E. Contreras (Geophysics Department, Chile),

Inverse problems in lithospheric flexure andviscoelasticity

Axel Ossesin coll with M. de Buhan (CNRS, France),

E. Contreras (Geophysics Department, Chile),

B. Palacios (DIM, Chile)

DIM - Departamento de Ingenierıa Matematica

CMM - Centro de Modelamiento Matematico

Facultad de Ciencias Fısicas y Matematicas, Universidad de Chile, Santiago.

[matematika mugaz bestalde] - [bcam]Bilbao, July 4th 2011

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Outline

1 IntroductionThe Maxwell viscoelastic modelRelationship to other viscoelastic models

2 Inverse problems

3 Motivation

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)

6 Numerical resolution

7 Examples

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IntroductionThe Maxwell viscoelastic model

σ → stress, ε→ strain

springs: σ0 = E0ε, σe = Eεe , dashpot: σv = ηε′v

(E ,E0 = Young’s modulus, η = viscosity)

in parallel: σe = σv , ε = εe + εv ⇒ ηε′v = Eεe = E (ε− εv )

⇒ εv =

∫ t

0

1

τe−

t−sτ ε(s) ds, τ =

η

E(relaxation time)

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IntroductionThe Maxwell viscoelastic model

σ = σ0 + σe = E0ε+ Eεe

= E0ε+ E (ε− εv )

= (E0 + E )ε− Eεv

= E 0ε︸︷︷︸elasticity

−E

∫ t

0

1

τe−

t−sτ ε(s) ds︸ ︷︷ ︸

viscoelasticity

PDE: ∂2t u −∇ · σ = f

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IntroductionThe Maxwell viscoelastic model

Let x ∈ Ω bounded domain in R3, t ≥ 0, we consider the 3D-viscoelasticmodel (density ρ = 1), stress tensor: ε(u) = ∇u +∇uT

(1)

Pu := ∂2t u −∇ ·

(µ0ε(u) + λ0(∇ · u)I

)︸ ︷︷ ︸elasticity

+ ∇ ·∫ t

0

(µ(t − s)ε(u(s)) + λ(t − s)(∇ · u)(s)I )ds︸ ︷︷ ︸viscoelasticity (one branch)

= 0,

u(0) = u0, ∂tu(0) = u1 in Ω,

u = 0, on ∂Ω× (0,+∞).

µ(x , t) = µ(x)h(t), λ(x , t) = λ(x)h(t) (e.g. h(t) = τ−1e−t/τ ).

u(x , t) : displacement vector,(λ0(x), µ0(x), λ(x), µ(x)) : coefficients.

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IntroductionRelationship to other viscoelastic models

Changing u by us in the viscoelastic part...

(1′)

Pu := ∂2t u −∇ · (µ0ε(u) + λ0(∇ · u)I )︸ ︷︷ ︸

elasticity

− ∇ ·∫ t

0

(µ(t − s)ε(us(s)) + λ(t − s)(∇ · us)(s)I )ds︸ ︷︷ ︸viscoelasticity (one branch)

= 0,

u(0) = u0, ∂tu(0) = u1 in Ω,

u = 0, on ∂Ω× (0,+∞).

µ(x , t) = µ(x)h(t), λ(x , t) = λ(x)h(t) (e.g. h(t) = e−t/τ ).

µ0(x) = µ0(x) + µ(x), λ0(x) = λ0(x) + λ(x),

(λ0(x), µ0(x), λ(x), µ(x)) : coefficients.

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IntroductionRelationship to other viscoelastic models

generalized Maxwell model...

(1′′)

Pu := ∂2t u −∇ ·

(µ0ε(u) + λ0(∇ · u)I

)︸ ︷︷ ︸elasticity

+ ∇ ·∫ t

0

N∑j=1

µj(t − s)ε(u(s))ds︸ ︷︷ ︸viscoelasticity (N branchs)

= 0,

u(0) = u0, ∂tu(0) = u1 in Ω,

u = 0, on ∂Ω× (0,+∞).

µ(x , t) =N∑j=1

µj(x)hj(t), (e.g. hj(t) = τ−1j e−t/τj ).

(λ0(x), µ0(x), µj(x)Nj=1) : coefficients.

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IntroductionRelationship to other viscoelastic models

fractional Maxwell models (α ∈ (0, 1))...

(1′′)

Pu := ∂2t u −∇ ·

(µ0ε(u) + λ0(∇ · u)I

)︸ ︷︷ ︸elasticity

+ ∇ ·∫ t

0

∫ ∞0

µ(ξ, t − s)dξ ε(u(s))ds︸ ︷︷ ︸viscoelasticity (∞ branchs)

= 0,

u(0) = u0, ∂tu(0) = u1 in Ω,

u = 0, on ∂Ω× (0,+∞).

µα(x , t) =

∫ ∞0

µα(ξ, x)h(ξ, t) dξ, (e.g. h(ξ, t) = ξ e−ξ t , ξ = 1/τ).

(λ0(x), µ0(x), µα(ξ, x)) : coefficients.

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IntroductionRelationship to other viscoelastic models

viscoelastic Kirchhoff plates... (D, u) = Dyyuxx − 2Dxyuxy + Dxxuyy

(2)

Qu :=∂2ttu − ε2∂2

tt∆u + D0∆2u(s) + (D0, u)︸ ︷︷ ︸elasticity

+

∫ t

0

D(t − s)∆2u + (D(t − s), u(s))ds︸ ︷︷ ︸viscoelasticity

= 0,

+i.c and b.c., on ∂Ω× (0,+∞).

D(x , t) = D(x)h(t), (e.g. hj(t) = τ−1e−t/τ ).

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Outline

1 Introduction

2 Inverse problemsRecover viscoelastic parameter from local displacementRecover viscoelastic parameter for plates (open problem)

3 Motivation

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)

6 Numerical resolution

7 Examples

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Inverse problemsRecover viscoelastic parameter from local displacement

Recover p := µ(x) or p := λ(x) from measurements of the solution :

u(x , t) in ω × (0,T )︸ ︷︷ ︸(single time dependent interior measurements)

where ω is a (small) subset of Ω and T > 0 or

u(x , t) on Γ× (0,T )︸ ︷︷ ︸(single time dependent boundary measurements)

where Γ is a (small) part of ∂Ω and T > 0.

boundary measure

ω

Γ

T

internal measure

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Inverse problemsRecover viscoelastic parameter for plates (open problem)

• Recover (uniqueness, stability) p := D0(x) (or p := D(x) for t →∞) inthe stationary flexure plate model from the Cauchy data:(

u,∂u

∂n,∆u,

∂∆u

∂n

)|∂Ω

• Recover (uniqueness, stability) p := D(x) in the stationary flexure platemodel from the Cauchy data: from internal (ω × (0,T )) or boundary(Γ× (0,T )) local measurements.

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Outline

1 Introduction

2 Inverse problems

3 MotivationMotivation: elastographyLithospheric flexure

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)

6 Numerical resolution

7 Examples

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MotivationMotivation: elastography

McLaughlin and Yoon, Inverse Problems 2004 [1]elastography: diagram for identification of stiffness profile

M. Fink, presentation in Valparaiso, 2010quantitative shear wave elastography, Airexplorer 2008

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MotivationLithospheric flexure

Watts and Zhong, Geophys. J. Int. 2000flexure of a two-layer viscoelastic plate model, relaxation time=1Myr

Contreras and Osses, Geophys. J. Int. 2010 [2]Nazca plate variable thickness flexure (1D)

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MotivationLithospheric flexure

Contreras, Manriquez, Osses, 2011, work in progressBathymetry

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MotivationLithospheric flexure

Contreras, Manriquez, Osses, 2011, work in progressNazca plate variable thickness flexure (2D)

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Outline

1 Introduction

2 Inverse problems

3 Motivation

4 Stability Result (interior measurement)SettingTheoremAssumptions

5 Stability Result (boundary measurement)

6 Numerical resolution

7 Examples

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Stability Result (interior measurement)Setting

p in Ω ?

u in ω

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Stability Result (interior measurement)Theorem

Theorem (2) (Stability) [3] Let u (resp. u) be the solution of (1)associated to the coefficient p (resp. p). Under Hypothesis 1, 2 and 3,there exists κ ∈ (0, 1) such that:

‖p − p‖H2(Ω) ≤ C‖u − u‖κH2(ω×(0,T )),

where C depends on the W 2,∞(Ω)-norm of p and p and on the W 8,∞(Ω×(0,T ))-norm of u and u.

Corollary (Uniqueness from interior measurement)

u = u in ω × (0,T ) =⇒ p = p in Ω

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Stability Result (interior measurement)Assumptions

Hypothesis 1 on the coefficients and trajectories

(λ0, µ0) ∈ (W 2,∞(Ω))2 and (λ, µ) ∈ (W 2,∞(Ω× (0,T )))2 are suchthat u ∈W 8,∞(Ω× (0,T )),

µ0, λ0 + 2µ0, µ0 − h(0)µ, λ0 + 2µ0 − h(0)(λ+ 2µ) satisfyCondition 1,

p is known in a neighborhood V of ∂Ω,

h(0) 6= 0, u1 = − h′(0)h(0) u0, (u1 = τ−1u0).

Condition 1 The scalar function q satisfies Condition 1 if :

it exists K > 0 such that ∀x ∈ Ω, q(x) ≥ K ,

it exists x0 ∈ R3 \ Ω such that ∀x ∈ Ω :

1

2q(x)− |∇q(x) · (x − x0)| ≥ 0.

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Stability Result (interior measurement)Assumptions

Hypothesis 2 on the observation part

T > T0 =d√β︸ ︷︷ ︸

large enough

and ω ⊃ V︸ ︷︷ ︸arbitrarily narrow, non trapping

,

d = supx∈Ω|x − x0|, β > 0 small.

Ω

V

∂Ω

Hypothesis 3 on the initial dataWe suppose that u0 is such that there exists M > 0 such that :

|ε(u0(x))(x − x0)| ≥ M, ∀x ∈ Ω.

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Stability Result (interior measurement)Proof: Carleman estimate for the viscoleastic system (1) u(0)∂t u(0) ≤ 0

1) Carleman weight: ϕ(x , t) = |x − x0|2 − βt2 → level sets → cones.

weighted energy ≤ source terms + internal measurements + ...

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Stability Result (interior measurement)Proof: Carleman estimate for the viscoleastic system (1) u(0)∂t u(0) ≤ 0

Write (1) as a system of 7 scalar equations for u, ∇ · u and ∇∧ u(see McLaughlin-Yoon 2004 [1], Imanuvilov-Yamamoto 2005 [4]),

∂2t ui (x , t)−pi (x)∆ui (x , t)+

∫ t

0

h(t−s) pi (x)∆ui (x , s)ds = Li (u,∇·u,∇∧u)

χ = 0

εε

χ = 1Ω

Change of variables to drop the integral term (similar to Cavaterraet al. 2005 [5]):

ui (x , t) = (1+αt)

(pi (x)ui (x , t)−

∫ t

0

h(t − s) pi (x)ui (x , s)ds

), α large.

=⇒ ∂2t ui (x , t)− qi (x)∆ui (x , t) = Li (u,∇ · u,∇∧ u)

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Stability Result (interior measurement)Proof: Carleman estimate for the viscoleastic system (1) u(0)∂t u(0) ≤ 0

Apply a modified pointwise Carleman inequality for a scalarhyperbolic equation (Klibanov and Timonov, 2004 [6]) in a cone.

weigth = eσϕ(x,t), ϕ(x , t) = |x−x0|2−βt2, x0 ∈ R3\Ω, β, σ > 0

0x

t

ϕ(x , t) = 0

Q

δ2δ

χ = 1

χ=

0

εε

Return to the initial variable and bound/absorb the integral terms:∫Q

(∫ t

0

|u(x , s)|ds

)2

e2σϕ(x,t)dx dt ≤ C

σ

∫Q

|u(x , t)|2e2σϕ(x,t)dx dt

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Stability Result (interior measurement)Proof: Bukhgeim and Klibanov method

2) We apply the method of Bukhgeim and Klibanov 1981 [7] :

w = ∂2t (u − u), w = ∂2

t u,

Pw = −∫ t

0

h(t − s)∇ · (2(p − p) ε(w(s))) ds

−h′(t)∇ · (2(p − p) ε(u0)) , in Ω× (0,+∞),

w(0) = 0, in Ω,

∂tw(0) = −h(0)∇ · (2(p − p) ε(u0)), in Ω,

w = 0, on ∂Ω× (0,+∞).

+ Apply Carleman inequality :

weighted initial energy ≤ interior measurement + sources

=⇒ the stability result : ‖p − p‖H2(Ω) ≤ C‖u − u‖κH2(ω×(0,T ))

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Outline

1 Introduction

2 Inverse problems

3 Motivation

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)SettingTheoremAssumptionsProof: unique continuationMain scheme of the proof - elastic caseMain scheme of the proof - viscoelastic case

6 Numerical resolution

7 Examples

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Stability Result (boundary measurement)Setting

p in Ω ?

u on Γ

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Stability Result (boundary measurement)Theorem

Theorem (3) (Logarithmic stability) [8] Let u (resp. u) be the solutionof (1) associated to the coefficient p (resp. p). Under Hypothesis 1’, 2’and 3’, there exists κ ∈ (0, 1) such that:

‖p − p‖H2(Ω) ≤ C

[log

(2 +

C∑1≤|α|≤2 ‖δαx (u − u)‖2

L2(Γ×(0,T ))

)]−κ

where C depends on the W 2,∞(Ω)-norm of p and p and on the W 8,∞(Ω×(0,T ))-norm of u and u.

Corollary (Uniqueness from boundary measurement)

∂αx u = ∂αx u on Γ× (0,T ), |α| = 1, 2 =⇒ p = p in Ω.

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Stability Result (boundary measurement)Assumptions

Hypothesis 1’ on the coefficientsthe same.

Hypothesis 2’ on the observation part

T > T0︸ ︷︷ ︸large enough

and Γ ⊂ ∂Ω︸ ︷︷ ︸arbitrarily small

.

Hypothesis 3’ on the initial datathe same.

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Stability Result (boundary measurement)Proof: unique continuation

Theorem (1) (Unique continuation) [8] Let u be a regular solution of(1) starting from rest and with right hand side vanishing in a neighborhoodV of ∂Ω. Then

‖u‖H2(V×(0,T/3)) ≤ C

[log

(2 +

C∑1≤|α|≤2 ‖δαx u‖2

L2(Γ×(0,3T ))

)]−1

where C depends on the W 2,∞(Ω)-norm of coefficients and on the W 4,∞(Ω×(0,T ))-norm of u.

Corollary (Unique continuation)

∂αx u = 0 on Γ× (0, 3T ), |α| = 1, 2 =⇒ u = 0 in V × (0,T/3).

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Stability Result (boundary measurement)Proof: unique continuation

This is done by

using a variant of Fourier-Bros-Iagolnitzer transform [9] for changingequations from hyperbolic to elliptic character as in Bellassoued2008 [4].

applying a Carleman inequality for the resulting elliptic sytem.

using interpolation results as in Robbiano 1995 [10].

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Stability Result (boundary measurement)Main scheme of the proof - elastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Stability Result (boundary measurement)Main scheme of the proof - viscoelastic case

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Outline

1 Introduction

2 Inverse problems

3 Motivation

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)

6 Numerical resolution

7 Examples

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Numerical resolutionDirect problem: finite elements

We discretize the system (1):

in space by P1 Lagrange Finite Elements,

in time by a θ-scheme with θ = 0.5 (implicit centered scheme),

by using the Trapezium Formula for the integral term.

We consider :

µ0(x) = λ0(x) = 3,

λ(x) = µ(x) = 1,

h(t) = e−t/τ with τ = 0.3.

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Numerical resolutionInverse problem: variational approach

We are looking for the minimizer of the non-quadratic functional :

J(p) =1

2

∫ T

0

∫ω

(|u(p)− uobs |2 + |∇(u(p)− uobs)|2

)dxdt

with u(p) =M(p), M being the nonlinear operator associated to system(1). We solve the minimization problem by a BFGS algorithm, so we need:

∇J(p, δp) = limε→0

1

ε(J(p + εδp)− J(p))

=

∫ T

0

∫ω

Mpδp · ((u − uobs)−∆(u − uobs))

with Mp the linearized operator ofM around p, i.e. Mpδp = δu satisfies :

(2)

Pδu = −∇ ·

∫ t

0

2δp h(t − s)ε(u(s))ds

δu(0) = 0, in Ω,

∂tδu(0) = 0, in Ω,

δu = 0, on ∂Ω× (0,+∞).

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Numerical resolutionInverse problem: adjoint problem and sensitivity

We introduce the adjoint problem of (2) :

(2)∗

P∗δu∗ := ∂2t δu∗ −∇ ·

(2µ0ε(δu∗) + λ0(∇ · δu∗)I

)+

∫ T

t

∇ ·(

2p h(s − t)ε(δu∗(s)) + λ(s)(∇ · δu∗)(s)I)

ds

P∗δu∗ =

(u(p)− uobs)−∆(u(p)− uobs) in ω × (0,T )0 in (Ω \ ω)× (0,T )

δu∗(T ) = 0, in Ω,

∂tδu∗(T ) = 0, in Ω,

δu∗ = 0, on ∂Ω× (0,+∞).

So we can compute :

∇J(p, δp) =

∫ T

0

∫Ω

δu · P∗δu∗ =

∫ T

0

∫Ω

Pδu · δu∗

=

∫Ω

δp

(∫ T

0

∫ t

0

2h(t − s)ε(u(s)) : ε(δu∗(t))ds dt

)dx

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Outline

1 Introduction

2 Inverse problems

3 Motivation

4 Stability Result (interior measurement)

5 Stability Result (boundary measurement)

6 Numerical resolution

7 ExamplesExample 1: Reference dataA first result in recovering the coefficientExample II: More realistic domain

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ExamplesExample 1: Reference data

Resolution with FreeFem 2D, Visualization with Medit,

µ(x) = λ(x) = 3, λ(x , t) = h(t) = e−t/τ with τ = 0.3,

p(x) =

1, in the healthy tissu,3, in the tumor,

T = 4, N =T

δt= 20, K = 25, δ = 0.

tumor measurement region mesh

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ExamplesA first result in recovering the coefficient

p ΠK p p∗

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ExamplesChanging the tumor size

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

50

Error in the parameter in L2!norm

number of eigenvectors in the basis

% o

f erro

r

big tumormedium tumorsmall tumor

=⇒ K optimal depends on the tumor size

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ExamplesChanging the noise level

δ

2%

5%

10%

‖uobs − u

u‖L∞(ω×(0,T )) ≤ δ

0 1 2 3 4 5 6 7 8 9 100

5

10

15

20

25

30

35

40

45

50

Error in the parameter in L2!norm

% error in the data

% e

rror i

n th

e pa

ram

eter

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ExamplesChanging the time of observation

T

2

4

6

Error in the parameter with respect toT = Nδt with N fixed

1 2 3 4 5 6 70

5

10

15

20

25

30

35

40

45

50

Error in the parameter in L2!norm

observation time

% e

rror i

n th

e pa

ram

eter

=⇒ T0

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ExamplesChanging the measurement region

Error in the parameter withrespect to ω

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90

5

10

15

20

25

30

35

40

45

50

Error in the parameter in L2!norm

observation zone

% e

rror i

n th

e pa

ram

eter

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ExamplesExample II: More realistic domain

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ExamplesInitial condition and trajectory

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ExamplesRegularization: finite number K of eigenfunctions

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ExamplesRecovering with fixed eigenfunctions

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ExamplesEigen and mesh adaptation

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ExamplesRecovering with adaptative eigenfunctions

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ExamplesComparison

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C. R. Acad. Sci. Paris, Ser. 347:1373–1378, 2009.

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Inverse problem of determining the density and the Lame coefficients by boundary data.

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Carleman estimates for coefficient inverse problems and numerical applications.

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A.L. Bukhgeim and M.V. Klibanov.

Global uniqueness of class of multidimensional inverse problems.

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M. de Buhan and A. Osses.

Logarithmic stability in determination of a 3d viscoelastic coefficient and numerical examples.

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