Fast Inversion of Logging-While-Drilling (LWD) … Inversion of Logging-While-Drilling (LWD)...

Fast Inversion of Logging-While-Drilling(LWD) Resistivity Measurements

David Pardo 1 Carlos Torres-Verdın 2

1University of the Basque Country (UPV/EHU) andIkerbasque, Bilbao, Spain.

2The University of Texas at Austin, USA

17 Oct. 2013BCAM Workshop on

Computational Mathematics, Bilbao, Spain

1

formation evaluationSurface measurements on the sea.

Marine seismic measurements.

2

formation evaluationSurface measurements on the sea.

Marine controlled source electromagnetic (CSEM) measurements.

3

formation evaluationSurface measurements on land.

Seismic measurements.

4

formation evaluationSurface measurements on land.

Magnetotelluric (MT) measurements.

5

formation evaluation

Logging measurements

Logging while drilling in a deviated well.

Multiphysics

Dip Angle

Borehole eccentricity

Invasion

Anisotropy

Fractures

Different Logging Devices

6




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


7




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


8




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


9




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


10




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


11




Multiphysics

Dip Angle


Invasion

Anisotropy

Fractures


12

main areas of expertise

Resistivity Measurements:Marine CSEM measurements.Magnetotelluric (MT) measurements.Galvanic and induction devices.Cased wells.Cross-well and borehole-to-surface measurements.Deviated wells.Borehole eccentered tools.Hydrofracture characterization.

Sonic Measurements:Wireline and logging-while-drilling.Borehole-eccentered tools.

Inversion of Resistivity Measurements:One-dimensional model reduction.Rapid inversion of logging-while-drilling measurements.

13

main areas of expertise

Resistivity Measurements:Marine CSEM measurements.Magnetotelluric (MT) measurements.Galvanic and induction devices.Cased wells.Cross-well and borehole-to-surface measurements.Deviated wells.Borehole eccentered tools.Hydrofracture characterization.

Sonic Measurements:Wireline and logging-while-drilling.Borehole-eccentered tools.

Inversion of Resistivity Measurements:One-dimensional model reduction.Rapid inversion of logging-while-drilling measurements.

14

inversion of LWD measurements

Motivation and objectives.

Assumptions.

Forward Problem.

Inverse Problem.

Numerical Results.

Conclusions.

15

motivation and objectives

Goal: Inversion of LWD resistivity measurements.

We want the inversion algorithm to be:

Efficient. Inversion in real time using 1D model reduction.

Flexible. It should enable the dynamic selection of a subset ofmeasurement and/or unknowns during inversion.

Robust. It should always converge to physically meaningfulsolutions.

Reliable. It should provide error bars.

Useful. It should work for any commercial LWD instrumentwith actual field measurements.

16


Goal: Inversion of LWD resistivity measurements.We want the inversion algorithm to be:






17








18








19








20








21

assumptions

We assume a planarly TI layered media with piecewiseconstant resistivities.We assume no borehole effects and no mandrel effects.

1000 1050 1100 1150

0

1

2

3

4

Model Problem and Well Trajectory

Horizontal Depth (m)

True

Ver

tical

Dep

th (m

)

1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

1000 1050 1100 1150

0

1

2

3

4



True

Ver

tical

Dep

th (m

)

We know the bedboundaries a priori.

We know the dip andazimuthal angles ofintersection a priori.

22

forward problem

Magnetic field H produced by a magnetic dipole is obtained usinga semi-analytical solution for a 1D planarly layered TI media(Kong, 1972).

A) Hankel transform in the horizontal plane.

B) Analytical solution of the resulting ordinary differentialequation in the vertical direction.

C) Numerical inverse Hankel transform (integration).

Result: Magnetic field H.

23

forward problem






24

forward problem






25

forward problem






26

forward problem






27

forward problem

CASE I: Triaxial Induction.

H =

Hxx Hxy HxzHyx Hyy HyzHzx Hzy Hzz

.

28

forward problem

CASE II: Conventional LWD resistivity tool.

Hq := log|HRX1

zz ||HRX2zz |︸︷︷︸

ATTENUATION

+ i [ph(HRX1zz ) − ph(HRX2

zz )]︸︷︷︸PHASE DIFFERENCE

10−2

100

102

1.2

2.5

5

Resistivity (Ohm−m) −− Log Scale −−

Atte

nuat

ion

−−

Log

Sca

le −

−

10−2

100

102

10−2

10−1

100


Pha

se D

iff. −

− L

og S

cale

−−

29

forward problem

CASE II: Conventional LWD resistivity tool.

Hq := log log|HRX1

zz ||HRX2zz |︸︷︷︸

ATTENUATION

+ i log [ph(HRX1zz ) − ph(HRX2

zz )]︸︷︷︸PHASE DIFFERENCE

10−2

100

102

1.2

2.5

5


Atte

nuat

ion

−−

Log

Sca

le −

−

10−2

100

102

10−2

10−1

100


Pha

se D

iff. −

− L

og S

cale

−−

30

forward problemTo accelerate computations, we employ a WINDOWING system:

1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

31


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

hhhhhh

32


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

33


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

) hhhhhhhhhh

34


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

35


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

) hhh

36


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

) h

37


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

38


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

39


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

40


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

hh

41


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

h

42


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

hh

43


1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

hhh

44

inverse problem (formulation)Cost Functional:

CW (s) = ‖ H(s) − M ‖2l2WM︸︷︷︸

MISFIT

,

wheres is either the conductivity σ, the resistivity ρ, or log ρ,H(s) is the set of simulated measurement for s,M is the set of actual (or synthetic) field measurements,

HD

(m)

110

1000

1020

1040

1060

Resis

tivity

(Ohm

-m)

Goal: To find s∗ := arg mıns

CW (s).

45


CW (s) = ‖ H(s) − M ‖2l2WM︸︷︷︸

MISFIT

+ λ ‖ s − s0 ‖2L2

Ws0︸︷︷︸REGULARIZATION

,

wheres is either the conductivity σ, the resistivity ρ, or log ρ,H(s) is the set of simulated measurement for s,M is the set of actual (or synthetic) field measurements,

λ is a regularization parameter, ands0 is an a priori distribution of s.


CW (s).

46

inverse problem (sol. method)We select the following deterministic iterative scheme:

s(n+1) = s(n) + δs(n).

Using a Taylor’s series expansion of first order of H:

H(s(n+1)) ≈ H(s(n)) +(

∂H(s(n))∂s

)︸︷︷︸

J

δs(n).

Solving∂CW (s(n+1))

∂δs(n)= 0, we obtain Gauss-Newton’s method:

δs(n) := −Re(J, H(s(n)) − M)l2

WM+ λ(I, s(n) − s0)L2

Ws0

(J, J)l2WM

+ λ(I, I)L2Ws0

.

47

inverse problem (jacobian)

To compute the Jacobian, we employ:

The chain rule:

J =∂H(s)

∂sj=

∂H(s)∂ρj︸︷︷︸Jρ

∂ρj

∂sj.

The definition of derivative:

Jρ =∂H(s)

∂ρj≈

H(ρ + hδρj) − H(ρ)

h(h small).

Only one Jacobian matrix is computed for any variable s.

48

inverse problem (jacobian)Misfit( %) Dip Angle = 82◦. Thinnest Bed: 0.37 m.

ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��mnn

mnnmnn

mnnmnn

49


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��

mnnmnn

mnnmnn

mnn

50


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��

mnnmnn

mnnmnn

mnn

51


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��mnn

mnnmnn

mnnmnn

52


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��mnn

mnnmnn

mnnmnn

53


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��mnn

mnnmnn

mnnmnn

54


ρ

σ

log ρ

Best

11,35%

11,32%

7,87%

6,58%

HD

(m) 1

10

1000

1020

1040

1060

Resist

ivity (O

hm-m)

110

100

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

110

1000

1020

1040

1060

Resistiv

ity (Oh

m-m)

��mnn

mnnmnn

mnnmnn

55

inverse problem (error bars)

Once we achieve convergence, we have:

δs(n) ≈ 0.

Considering new noisy measurements of the type:

M := M + N

and using these new measurements in our Gauss-Newton method,we obtain the following new correction δs(n):

δs(n) :=Re(J, N)l2

WM

(J, J)l2WM

+ λ(I, I)L2Ws0

Error bars: [s(n) − |δs(n)|, s(n) + |δs(n)|].

56

numerical results (synthetic 1)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)T

rue

Ver

tical

Dep

th (

m)

1000 1020 1040 1060

0

2

4

6

8

10



Tru

e V

ertic

al D

epth

(m

)

Dip Angle: 82◦.

Thinnest bed: 0.37m

110

1000

1020

1040

1060

Resis

tivity

(Ohm

-m)

57

numerical results (synthetic 1)TOOL 1 TOOL 2 TOOL 3 TOOL 4 TOOL 5

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l De

pth

(m

)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

Dip Angle = 82◦. Thinnest bed: 0.37m.

58

numerical results (synthetic 1)TOOL 1 TOOL 2 TOOL 3 TOOL 4 TOOL 5

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l De

pth

(m

)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

mnn mnn mnnmnn mnn

mnn mnnmnnDip Angle = 82◦. Thinnest bed: 0.37m.

59

numerical results (synthetic 1)Sensitivity with respect to the Dip Angle. Thinnest bed: 0.37m.

30◦ 45◦ 60◦ 82◦ 89◦

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l De

pth

(m

)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0

2

4

6

8

10

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

60


1 10 100

0

2

4

6

8

10


rue

Ver

tical

Dep

th (

m)

1000 1020 1040 1060

0

2

4

6

8

10



Tru

e V

ertic

al D

epth

(m

)

Dip Angle: 82◦.

Thinnest bed: 0.37m

Anisotropy. 110

100

1000

1020

1040

1060

Resis

tivity

(Ohm

-m)

61


30◦

HD

(m)

110

100

1000

1002

1004

1006

Resis

tivity

(Ohm

-m)

Vertical well –> Rh.

� ��

62


30◦

82◦

HD

(m)

110

100

1000

1002

1004

1006

Resis

tivity

(Ohm

-m)

Vertical well –> Rh.

� ��

110

100

1000

1020

1040

1060

Resis

tivity

(Ohm

-m)

Horizontal well –> Rv .

� ��

� ��

63


1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4


rue

Ver

tical

Dep

th (

m)

1000 1002 1004 1006 1008 1010

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4



Tru

e V

ertic

al D

epth

(m

)

Dip Angle: 82◦.

Thinnest bed: 0.05m.

110

1000

1002

1004

1006

1008

1010

Resis

tivity

(Ohm

-m)

64


TOOL 1 TOOL 2 TOOL 3 TOOL 4 TOOL 5

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l De

pth

(m

)

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

65

numerical results (synthetic 2)TOOL 1 TRIAXIAL TRIAXIAL TRIAXIAL TRIAXIALNO NOISE NO NOISE 5 % NOISE 10 % NOISE NO NOISE

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l De

pth

(m

)

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

1 10 100

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Resistivity (Ohm-m)

Tru

e V

ert

ica

l D

ep

th (

m)

66

numerical results (field 1)

1 10

0

1

2

3

4


rue

Ver

tical

Dep

th (

m)

1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

Almost horizontal.

Field data. 110

1000

1050

1100

1150

Resis

tivity

(Ohm

-m)

67


1 10

0

1

2

3

4


rue

Ver

tical

Dep

th (

m)

1000 1050 1100 1150

0

1

2

3

4



Tru

e V

ertic

al D

epth

(m

)

Almost horizontal.

Field data. 110

1000

1050

1100

1150

Resis

tivity

(Ohm

-m)

&%'$

68


1 10

1.5

2.0

2.5

3.0

3.5

4.0


rue

Ver

tical

Dep

th (

m)

1110 1120 1130 1140 1150

1.5

2.0

2.5

3.0

3.5

4.0



Tru

e V

ertic

al D

epth

(m

)

Almost horizontal.

Field data.

Zoom.

110

1110

1120

1130

1140

1150

Resis

tivity

(Ohm

-m)

69


1 10

0

2

4

6

8


rue

Ver

tical

Dep

th (

m)

1000 1010 1020 1030 1040

0

2

4

6

8



Tru

e V

ertic

al D

epth

(m

)

Dip Angle: 79.3◦.

Field data. 110

1000

1010

1020

1030

1040

Resis

tivity

(Ohm

-m)

70

numerical results (field 2)H

D(m

)

110

1000

1010

1020

1030

1040

Res

istiv

ity (

Ohm

-m)

Inversion results (blue) are similar to those obtained by Dr.Olabode Ijasan (red).

71

conclusions

We have developed a library for the fast inversion of LWDresistivity measurements.

The library enables any well trajectory and any logginginstrument. We assume a 1D planarly layered TI media.

The library automatically selects the regularization parameter,stopping criteria, and inversion variable.

It enables to first invert a subset of measurements and/or asubset of resistivities.

Numerical results illustrate the stability of the proposedinversion algorithm.

72

conclusions






73

conclusions






74

conclusions






75

conclusions






76

future work

Computational cost of one forward simulation:

COST = C ∗ NPOSITIONS ∗ NLAYERS ∗ NFREQ. ∗ NTX ∗ NRX

Computational cost of building the Jacobian:

COST = C ∗ NPOSITIONS ∗ NLAYERS2 ∗ NFREQ. ∗ NTX ∗ NRX

Can we eliminate the factor NRX? I think so!Can we eliminate the factor NTX? To some extend!Can we eliminate the square on the factor NLAYERS? Perhaps!

77

future work






78

future work






79

change of coordinates

To employ a Model Reduction algorithm based on Cartesian (C)coordinates and obtain results for Borehole (B) coordinates, weemploy:

HBB = JBC HCC JCB,

where:HCC and HBB are the model reduction algorithms for theCartesian and Borehole systems of coordinates, respectively,

JCB =

cos θ 0 − sin θ

0 1 0sin θ 0 cos θ

·

cos φ − sin φ 0sin φ cos φ 0

0 0 1

JBC = J−1

CB , θ is the dip angle, and φ is the azimuthal angle.

80

inverse problem (reg. param.)

We have:

C (n)W (s) = ‖ H(s) − M ‖2

l2WM︸︷︷︸

MISFIT

+ λ(n) ‖ s − s0 ‖2L2


,

90% 10%

We want the regularization term to contribute with 10 % to thetotal cost functional. Then:

λ(n) := 0,1 ∗‖ H(s(n)) + Jδs(n)

λ(n) − M ‖2l2WM

‖ s(n) + δs(n)λ(n) − s0 ‖2

L2Ws0

We perform a fixed-point iteration to obtain the value of λ(n).

81


We have:

C (n)W (s) = ‖ H(s) − M ‖2

l2WM︸︷︷︸

MISFIT

+ λ(n) ‖ s − s0 ‖2L2


,

90% 10%

We want the regularization term to contribute with 10 % to thetotal cost functional.

Then:

λ(n) := 0,1 ∗‖ H(s(n)) + Jδs(n)


‖ s(n) + δs(n)λ(n) − s0 ‖2

L2Ws0


82


We have:

C (n)W (s) = ‖ H(s) − M ‖2

l2WM︸︷︷︸

MISFIT

+ λ(n) ‖ s − s0 ‖2L2


,

90% 10%


λ(n) := 0,1 ∗‖ H(s(n)) + Jδs(n)


‖ s(n) + δs(n)λ(n) − s0 ‖2

L2Ws0


83


We have:

CW (s(n+1)λ(n) ) =‖ H(s(n+1)

λ(n) )−M ‖2l2WM

+λ(n+1) ‖ s(n+1)λ(n) −s0 ‖2

L2Ws0

≈‖ H(s(n))+Jδs(n)λ(n) −M ‖2

l2WM

+λ(n) ‖ s(n)+δs(n)λ(n) −s0 ‖2

L2Ws0

.


λ(n) := 0,1 ∗‖ H(s(n)) + Jδs(n)


‖ s(n) + δs(n)λ(n) − s0 ‖2

L2Ws0


84

inverse p. (stopping criteria)

We stop the inversion process when both the relative data misfitand regularization term do not vary significantly. Mathematically,we require the following two conditions to be satisfied:

100∗| ‖ H(s(n+1)) − M ‖2

l2WM

− ‖ H(s(n)) − M ‖2l2WM

|

‖ M ‖2l2WM

≤ 0,5%

And:

100λ(n) ∗| ‖ s(n+1) − s0 ‖2

L2Ws0

− ‖ s(n) − s0 ‖2L2

Ws0

|

‖ s0 ‖2L2

Ws0

≤ 5%.

85


C(s) =‖ H(s) − M ‖2l2 +λ ‖ s − s0 ‖2

L2 .

We want to weight all measurements and resistivities so equalrelative errors will contribute equally to the cost functional.

Weighted cost functional:

CW (s) =‖ H(s) − M ‖2l2WM

+λ ‖ s − s0 ‖2L2

Ws0,


CW (s).

86


C(s) =‖ H(s) − M ‖2l2 +λ ‖ s − s0 ‖2

L2 .

We want to weight all measurements and resistivities so equalrelative errors will contribute equally to the cost functional.

Weighted cost functional:

CW (s) =‖ H(s) − M ‖2l2WM

+λ ‖ s − s0 ‖2L2

Ws0,


CW (s).

87

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