Towards a Dimensionally Adaptive Inversion of the Magnetotelluric Problem

Transcription

1 1 Towards a Dimensionally Adaptive Inversion of the Magnetotelluric Problem J. Alvarez-Aramberri 1 D. Pardo 2 H. Barucq 3 1 University of the Basque Country (UPV/EHU), Bilbao, Spain, Inria team-project Magique-3D, France 2 University of the Basque Country (UPV/EHU), BCAM, and Ikerbasque, Bilbao, Spain. 3 EPC Magique-3D, Inria, LMA, University of Pau, France Jul 18, 2014 The Third BCAM Workshop on Computational Mathematics, BCAM.

2 The Magnetotelluric Method 2

3 The Magnetotelluric Method 3 The solar wind produces a radiation pressure that causes a compression on the day-side and a tail on the night-side onto the magnetosphere and due to this interaction, hydromagnetic waves are created. When those waves reach the ionosphere, they induce an EM field that works as power source in magnetotellurics. Range (distance): From few meters to hundreds of kilometers. Range (frequency): Between Hz. Applications: Hydrocarbon (oil and gas) exploration, geothermal exploration, earthquake precursor prediction, mining exploration, as well as hydrocarbon and groundwater monitoring.

4 The Magnetotelluric Method 4 Figure: Typical MT problem.

5 The Magnetotelluric Method 5 Figure: Sketch of the MT method.

6 Motivation 6 Figure: 2D MT problem. Blue rectangle: Natural source. Red crosses: Receivers.

7 Motivation 7 Figure: Primary Problem: 1D layered media with known analytical solution E 1D.

8 Motivation 8 Figure: Secondary Problem: 2D media. Blue rectangle: New source. Red crosses: Receivers.

9 Outline 9 1 Mathematical Modelization: Direct Problem 2 Inverse Problem: Dimensionally Adaptive Algorithm 3 Numerical Results

10 Outline 10 1 Mathematical Modelization: Direct Problem 2 Inverse Problem: Dimensionally Adaptive Algorithm 3 Numerical Results

11 Direct Problem 11 Maxwell s equations in frequency domain E = iωµh H = (ρ 1 + iωɛ)e + J imp (ɛe) = ρ f (µh) = 0, When E = E(x, z) and H = H(x, z) 2 uncoupled modes: Transverse Electric (TE): (E y, H x, H z ) Transverse Magnetic (TM): (H y, E x, E z ) ( ) µ 1 E k 2 E = jωj imp.

12 Direct Problem 12 We focus on the TE mode and we solve the equation for E y with a hp-finite Element Method (FEM): where k 2 = ω 2 ε jωρ 1. µ 1 E y k 2 E y = jωj imp y, (1) Multiply equation (1) by the complex conjugate of a test function F V (Ω) = H 1 Γ D (Ω) = {F L 2 (Ω) : F ΓD = 0, F L 2 (Ω)} is the space of admissible test functions and integrating by parts (omitting the subscript y from now on): Find E E D + V (Ω), such that: ( F ) T µ 1 E F k 2 E = jω Ω Ω Ω F J imp F V (Ω),

13 Direct Problem 13 With the state scalar-valued solution function E, we compute the Quantity of Interest L i (E), a linear and continuous functional in E associated to the i-th receiver and defined as: L i (E) = 1 Ω R i Ω R i E dω, where Ω R i is the domain occupied by the i-th receiver. From Maxwell s equations: H x = 1 E jωµ z. The impedance Z i = Z i yx and the apparent resistivity ρ i ap are: Z i = Li (E) L i (H x ) ρ i ap = Zi ωµ.

14 Outline 14 1 Mathematical Modelization: Direct Problem 2 Inverse Problem: Dimensionally Adaptive Algorithm 3 Numerical Results

15 Definition of the Inverse Problem 15 We define the following Misfit Function: C(σ) = 1 2M M g i (ρ) dobs i 2 i g(ρ) = {g i (ρ)} M i=1 is the vector with the i-th component being equal to the quantity of interest corresponding to the i-th receiver. In particular, g i (ρ) = Z i (ρ) or g i (ρ) = ρ i ap(σ) d = {dobs i }M i=1 is the vector of EM measurements obtained at different receivers.

16 Definition of the Inverse Problem 16 Piecewise constant distribution of conductivity: P ρ(x) = ρ i χ i (x), where χ i (x) are Heaviside shape functions. i Model Parameters: ρ = {ρ 1,..., ρ P } Problem to Solve: ρ = arg min l ρ u C(ρ), where the vectors l and u represent lower and upper bounds on the resistivity.

17 Solving the Inverse Problem 17 Solve Direct MT Problem Experimental data at receivers ρ Numerical solution at receivers Comparison Update the value of ρ ρ : Solution to the IP

18 Solving the Inverse Problem 18 Solve Direct MT Problem Experimental data at receivers ρ Numerical solution at receivers Comparison Update the value of ρ ρ : Solution to the IP

19 Dimensionally Adaptive Algorithm 19 Consider a reference 1D resistivity model ρ 1D, with known (analytical) solution. Let Ω 2D be the domain where the 2D inhomogeneities are located. Then, the conductivity distribution can be represented as the sum ρ = ρ 1D + ρ 2D, being ρ 2D zero outside Ω 2D. Divide the solution into the primary E 1D and secondary E 2D fields : E = E 1D + E 2D.

20 Dimensionally Adaptive Algorithm 20 Figure: 2D MT problem. Blue rectangle: Natural source. Red crosses: Receivers.

21 Dimensionally Adaptive Algorithm 21 Figure: Primary Problem: 1D layered media with known analytical solution E 1D.

22 Dimensionally Adaptive Algorithm 22 Figure: Secondary Problem: Solution E 2D computed with hp-fem. Blue rectangle: New source. Red crosses: Receivers.

23 Dimensionally Adaptive Algorithm 23 Compute E 1D with an analytic solution. Compute E 2D solving the equation for the secondary field with the hp-fem. ) (µ 1 E 2D k 2 E 2D = (ρ 2D ) 1 E 1D, Its variational formulation is given by (E 1D already known): Find E 2D E D + V, such that: ( F ) T µ 1 E 2D F k 2 E 2D = Ω Ω Ω F (ρ 2D ) 1 E 1D, F V. We obtain E = E 1D + E 2D.

24 Dimensionally Adaptive Algorithm 24 Define a new Misfit Function: C(ρ) = 1 2M M g i (ρ) dobs i 2 + β(ρ ρ 1D) i 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 regularization parameter.

25 Advantages of the New Approach 25 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, we avoid modeling the original source, which in MT implies to deal with big computational domains. Only the 46% of elements needed. Nonlinear Inverse Problems are usually ill-posed. It allows the design of more robust algorithms.

26 Outline 26 1 Mathematical Modelization: Direct Problem 2 Inverse Problem: Dimensionally Adaptive Algorithm 3 Numerical Results

27 Model Problem: Magnetotelluric Problem 27 Figure: TE mode driven by the impressed electric current J imp = ωŷ ρ 1 ρ 2 ρ 3 Model Model Model Model Table: Different models for the formation of the subsurface (Ohm-m).

28 Validation of the Solution 28 Apparent resistivity (Ohm m) Model 1 Model 2 Model 3 Model Frequency (Hz.) 10 0 Relative error in percent Model 1 Model 2 Model 3 Model Frequency (Hz.) 10 0 Figure: Apparent resistivity against frequency for the numerical solution (left) and the relative error for different subsurface formations for a range of frequencies (right). ρ 1 ρ 2 ρ 3 Model Model Model Model

29 Formation with an Inhomogeneity (Target) Figure: TE mode driven by J imp = ωŷ. Target s width: 10km. ρ 1 ρ 2 ρ 3 ρ 4 Layered Layered Target Table: Different models of the formation of the subsurface. (Ohm-m.) 29

30 Formation with an Inhomogeneity (Target) Figure: TE mode driven by J imp = ωŷ. Target s width: 10km. ρ 1 ρ 2 ρ 3 ρ 4 Layered Layered Target Table: Different models of the formation of the subsurface. (Ohm-m.) 30

31 Solution with the Secondary Field Formulation 31 Apparent resistivity (Ohm m) Layered 1 Layered 2 Target Position (km.) Figure: Apparent resistivity at different distances for three different subsurface formations. In all cases ω = 10 3 Hz.

32 Conclusions / Future (Current) Work 32 The secondary field formulation provides accurate solutions to the MT problem. These solutions are obtained with lower computational cost (only the half of the elements needed). Implement the adaptive dimensional algorithm for the inversion using the secondary field formulation for the Direct Problem.