Towards a Dimensionally Adaptive Inversion of the ... · Towards a Dimensionally Adaptive Inversion...

Towards a Dimensionally Adaptive Inversionof the Magnetotelluric Problem

J. Alvarez-Aramberri 1 D. Pardo 2 H. Barucq3

1University of the Basque Country (UPV/EHU), Bilbao, Spain,Inria team-project Magique-3D, France

2University of the Basque Country (UPV/EHU), BCAM, andIkerbasque, Bilbao, Spain.

3EPC Magique-3D, Inria, LMA, University of Pau, France

Jul 18, 2014The Third BCAM Workshop on Computational Mathematics, BCAM.

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The Magnetotelluric Method

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The solar wind produces a radiation pressure that causes acompression on the day-side and a tail on the night-side ontothe magnetosphere and due to this interaction, hydromagneticwaves are created. When those waves reach the ionosphere,they induce an EM field that works as power source inmagnetotellurics.

Range (distance): From few meters to hundreds of kilometers.

Range (frequency): Between 10−5 − 10 Hz.

Applications: Hydrocarbon (oil and gas) exploration,geothermal exploration, earthquake precursor prediction,mining exploration, as well as hydrocarbon and groundwatermonitoring.

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Figure: Typical MT problem.

4


Figure: Sketch of the MT method.

5

Motivation

Figure: 2D MT problem. Blue rectangle: Natural source. Red crosses:Receivers.

6

Motivation

Figure: Primary Problem: 1D layered media with known analyticalsolution E1D .

7

Motivation

Figure: Secondary Problem: 2D media. Blue rectangle: New source.Red crosses: Receivers.

8

Outline

1 Mathematical Modelization: Direct Problem

2 Inverse Problem: Dimensionally Adaptive Algorithm

3 Numerical Results

9

Outline



3 Numerical Results

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Direct Problem

Maxwell’s equations in frequency domain

∇× E = −iωµH

∇×H = (ρ−1 + iωε)E + J imp

∇ · (εE) = ρf

∇ · (µH) = 0,

When E = E(x , z) and H = H(x , z) → 2 uncoupled modes:

Transverse Electric (TE): (Ey ,Hx ,Hz)

Transverse Magnetic (TM): (Hy ,Ex ,Ez)

∇×(µ−1∇× E

)− k2E = −jωJ imp.

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Direct Problem

We focus on the TE mode and we solve the equation for Ey with ahp-Finite Element Method (FEM):

−µ−1∆Ey − k2Ey = −jωJ impy , (1)

where k2 = ω2ε− jωρ−1.

Multiply equation (1) by the complex conjugate of a test functionF ∈ V (Ω) = H1

ΓD (Ω) = F ∈ L2(Ω) : F |ΓD = 0,∇F ∈ L2(Ω) isthe space of admissible test functions and integrating by parts(omitting the subscript y from now on):

Find E ∈ E D + V (Ω), such that:∫Ω

(∇F )Tµ−1∇E −∫

ΩF k2E = −jω

∫Ω

F J imp ∀F ∈ V (Ω),

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Direct ProblemWith the state scalar-valued solution function E , we compute theQuantity of Interest Li (E ), a linear and continuous functional in Eassociated to the i-th receiver and defined as:

Li (E ) =1|ΩR i |

∫ΩRi

E dΩ,

where ΩR i is the domain occupied by the i-th receiver. From

Maxwell’s equations:Hx =

1jωµ

∂E∂z .

The impedance Z i = Z iyx and the apparent resistivity ρi

ap are:

Z i =Li (E )

Li (Hx )ρi

ap =|Z i |ωµ

.

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Outline



3 Numerical Results

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Definition of the Inverse Problem

We define the following Misfit Function:

C(σ) =1

2M

M∑i|g i (ρ)− d i

obs |2

g(ρ) = g i (ρ)Mi=1 is the vector with the i-th componentbeing equal to the quantity of interest corresponding to thei-th receiver. In particular,

g i (ρ) = Z i (ρ) or g i (ρ) = ρiap(σ)

d = d iobsMi=1 is the vector of EM measurements obtained at

different receivers.

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Definition of the Inverse Problem

Piecewise constant distribution of conductivity:

ρ(x) =P∑iρiχi (x), where χi (x) are Heaviside shape functions.

Model Parameters: ρ = ρ1, ..., ρP

Problem to Solve:

ρ∗ = arg minl≤ρ≤u

C(ρ),

where the vectors l and u represent lower and upper bounds on theresistivity.

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Solving the Inverse Problem

ρNumericalsolution atreceivers

Experimentaldata at

receivers

ρ∗: Solutionto the IP

Solve DirectMT Problem

Comparison

Update the value of ρ

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Solving the Inverse Problem

ρNumericalsolution atreceivers

Experimentaldata at

receivers

ρ∗: Solutionto the IP

Solve DirectMT Problem

Comparison

Update the value of ρ

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Dimensionally Adaptive Algorithm

Consider a reference 1D resistivity model ρ1D, with known(analytical) solution.

Let Ω2D be the domain where the 2D inhomogeneities arelocated. Then, the conductivity distribution can berepresented as the sum ρ = ρ1D + ρ2D, being ρ2D zero outsideΩ2D.

Divide the solution into the primary E 1D and secondary E 2D

fields : E = E 1D + E 2D.

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Figure: 2D MT problem. Blue rectangle: Natural source. Red crosses:Receivers.

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Figure: Primary Problem: 1D layered media with known analyticalsolution E1D .

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Figure: Secondary Problem: Solution E2D computed with hp-FEM.Blue rectangle: New source. Red crosses: Receivers.

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Compute E 1D with an analytic solution.

Compute E 2D solving the equation for the secondary fieldwith the hp-FEM.

∇×(µ−1∇× E 2D

)− k2E 2D = (ρ2D)−1E 1D,

Its variational formulation is given by (E 1D already known):Find E 2D ∈ E D + V , such that:∫

Ω(∇F )Tµ−1∇E 2D −

∫Ω

F k2E 2D =

∫Ω

F (ρ2D)−1E 1D,∀F ∈ V .

We obtain E = E 1D + E 2D.

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Define a new Misfit Function:

C(ρ) =1

2M

M∑i|g i (ρ)− d i

obs |2 + β(ρ− ρ∗1D)

where ρ∗1D is the solution of the 1D Inverse Problem.

Step1: Solve the Inverse Problem for the 1D layered media.

Step2: Solve the Inverse Problem for the 2D problem.Use the secondary field formulation for the direct problem.Use the 1D solution of the Inverse Problem as a regularizationparameter.

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Advantages of the New Approach

MT Direct Problem is computationally cheaper.

We only need to accurately solve the secondary field variations,which in general are more localized than the total fields.Therefore, it is generally possible to use a coarser grid.

Smaller computational domain: With this formulation, weavoid modeling the original source, which in MT implies todeal with big computational domains.

Only the 46% of elements needed.

Nonlinear Inverse Problems are usually ill-posed. It allows thedesign of more robust algorithms.

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Outline



3 Numerical Results

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Model Problem: Magnetotelluric Problem

Figure: TE mode driven by the impressed electric current Jimp = ωy

ρ1 ρ2 ρ3Model 1 1 1 1Model 2 1 10 3Model 3 1 10 10Model 4 1 100 3

Table: Different models for the formation of the subsurface (Ohm-m).27

Validation of the Solution

10−5

10−4

10−3

10−2

10−1

100

0

2

4

6

8

10

Frequency (Hz.)

Ap

pa

ren

t re

sis

tivity (

Oh

m−

m)

Model 1Model 2Model 3Model 4

10−5

10−4

10−3

10−2

10−1

100

0

0.5

1

1.5

Frequency (Hz.)

Re

lative

err

or

in p

erc

en

t

Model 1

Model 2

Model 3

Model 4

Figure: Apparent resistivity against frequency for the numerical solution(left) and the relative error for different subsurface formations for a rangeof frequencies (right).

ρ1 ρ2 ρ3Model 1 1 1 1Model 2 1 10 3Model 3 1 10 10Model 4 1 100 3

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Formation with an Inhomogeneity (Target)

Figure: TE mode driven by Jimp = ωy. Target’s width: 10km.

ρ1 ρ2 ρ3 ρ4Layered 1 1 2 3 -Layered 2 1 10 3 -

Target 1 2 3 10

Table: Different models of the formation of the subsurface. (Ohm-m.)29

Formation with an Inhomogeneity (Target)

Figure: TE mode driven by Jimp = ωy. Target’s width: 10km.

ρ1 ρ2 ρ3 ρ4Layered 1 1 2 3 -Layered 2 1 10 3 -

Target 1 2 3 10

Table: Different models of the formation of the subsurface. (Ohm-m.)30

Solution with the Secondary Field Formulation

−20 −10 0 10 201

1.5

2

2.5

3

3.5

4

Position (km.)

Appare

nt re

sis

tivity (

Ohm

−m

)

Layered 1

Layered 2

Target

Figure: Apparent resistivity at different distances for three differentsubsurface formations. In all cases ω = 10−3Hz.

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Conclusions / Future (Current) Work

The secondary field formulation provides accurate solutions tothe MT problem.

These solutions are obtained with lower computational cost(only the half of the elements needed).

Implement the adaptive dimensional algorithm for theinversion using the secondary field formulation for the DirectProblem.

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