Computational electrodynamics in geophysical applications

Epov M. I., Shurina E.P., Arhipov D.A., Mikhaylova E.I., Kutisheva A. Yu., Shtabel N.V.

The main features of the geological media

• Heterogeneous media, fluid-saturated rocks.

• The complex geometry of objects.

• The complex configuration of interface boundary.

• The electrophysical properties: the contrast between separate fragments of media, anisotropy, polarization, the dispersion of the conductivity, permittivity and permeability.

The Maxwell’s equations

0,

,

, 0.

t

t

B E

H J D

D B

0

, ,

.

B H D E

J J E

The Faraday's law

The Maxwell – Ampere law

The Gauss’s laws for electric and magnetic flux densities

The Second order equations

• Hyperbolic equation

• Parabolic equation

21 0

2rot( rot )

t t t

JE E

E

1 0rot( rot )t t

JE

E

n 0

E0 gtE Ε

0t

gt

E

n 0

E0 gtE Ε

Frequency domain. Helmholtz equation

1 2 2 20( ) ( )

( ) 0

k i k i

i

E E J

E

The boundary conditions

The charge conservation law

0

0

, PEC

e

n E 0

n E E

1 rot 0 PMCm

E n

i j

The interface conditions

0,

0,

,

0,

0.

ij

ij

ij

ij

ij

E n

H n

D n

B n

E ni j i j

12 2

2 2

2 2

{ : grad }

(rot, ) { ( ) : rot ( )}

(div, ) { ( ) : div ( )}

H u L u

H x x

H w w L

L

L L

L

2 2 2( ; )

2 2 2( ; )

2 2 2( ; )

,

,

.

H grad

H

H div

u u u

rotu u u

u u u

The functional spaces

The functional subspaces and de Rham’s complex

0

0

0

(grad; ) (grad; ) | | 0 ,

(rot; ) (rot; ) | | 0 ,

(div; ) (div; )| | 0 .

H u H u

H H

H H

u u n

u u n

2( ; ) (rot; ) (div; ) ( )H grad H H L

20 0 0( ; ) (rot; ) (div; ) ( )H grad H H L

Variational Formulations

For 30 2( ( )) J L find 0 (rot; ) E H such

that 0 (rot; ) W H the following is held2

1 02

( rot , rot ) ( , ) ( , ) ( , )t t t

JΕ E

Ε W W W W

1 0( rot , rot ) ( , ) ( , )t t

JE

E W W W

For 30 2( ( )) J L find 0 (rot; ) E H such

that 0 (rot; ) W H the following is held

Parabolic equation

Hyperbolic equation

Time ApproximationWe introduce the following partition of the time ),0( T

Tttt M ...0 10

and function on -th time step

where ),,( zyxEu hjj is a solution on j-th time step

j step on j-th step of time scheme.

,))((

)(,)(

))(()(

1

21

1

120

jj

jjj

jjj

jjj ttttt

ttttt

j

11

12 )(

))(()(

jjj

jjj ttttt

Then the function of interest is

)()()(),,,( 22110 tutututzyxE jj

jj

jj

h

Newmark-beta Scheme

where jJ0 the value of right hand side on j-th

time step, parameter of the scheme. 4/1

)W,)21(

()W,)(

)(

2()W,

)(

22

)(

2()W),)21(((

h2

01

0

0h2

111

1

1

1

1h2

111

1

1

h21

1

t

J

t

J

t

Juu

uuu

uuuu

jj

j

jjjj

jj

jj

jj

jjjj

jjj

jjjj

jj

jjjj

jjj

The variational formulation

The following propertyallows to fulfill the variational analog of the charge conservation law

0 0( ; ) , ( ; )H grad H rot

1 20( , ) ( , ) ( , )k i

E W E W J W

20(( ) , ) 0 (grad, )i H E

For 30 2( ( )) J L to find 0 (rot; ) E H such

that 0 (rot; ) W H the following is held

Geometric domain decomposition

Difficulties:• Local source of the field (the source should be in one

subdomain and can’t touch its boundaries)• Balancing the dimensions of subdomains matrices

(CPU time should be comparable in different subdomains)

• The geometry of the computational domain should be taken into account

Decomposition approaches:• Custom decomposition (effective, but time-consuming)• Automatic decomposition

Automatic Decomposition

• Decomposition by enclosed “spheres”

• Decomposition by layers

EM Logging

Borehole - Inclined bed

1 – 3-coil probe, 2 – borehole with mud, 3 – host formation, 4 – low-conductive bed, Г – generator coil, И1, И2 – receiver coils

ElectroPhysical Properties

Operating frequency 14 МHz,

amperage J=1 А.

Domain

1 1 1 0

2 1 1 5

3 1 1 0.1

4 1 1 0.01

Re Ex (X0Y)

00

Zenith angle

450 750

Re Ez (X0Y)

00

Zenith angle

450 750

Ez=0

Surface Soundings

Transmitter loop 40 x 40 m²Receiver loop 20 x 20 m²Impulse length 5 µsSimulation time 10 msMesh: 335666 edges, 49244 nodes, 281342 tetrahedrons Computation one time step 30 sec, after current is turn offSolver: Multilevel iterative solver with V-cycle

Anisotropic layer

Isotropic layer

Zenith Angle 0, 30, 60, 902 2

11 z

12 21 23 32

22 y

13 31

2 233

' = cos θ sin θ

' = ' =0 ' = ' =0

' =

' = ' =( - )sinθcosθ

' = sin θ+ cos θ

z x

x z

x

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600

Transversal isotropic medium θ=0°

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600

Ex, z=-50

Ey, z=0

Ez, z=-50Ey, z=-50

Ex, z=0

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600Ex z=0 Ey z=0

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600

x

y

-500 0 500-600

-400

-200

0

200

400

600Ex z=-50 Ey z=-50 Ez z=-50

Transversal isotropic medium rotated for zenith angle θ=60°

The anisotropic object in the isotropic halfspace

5.12 -0.75 0.27

= -0.75 1.03 -0.05

0.27 -0.05 1.02

10 0 0

0 1 0

0 0 1

1 0 0

0 1 0

0 0 10

24

Re Ex, Ez for vertical object

The cross-section x=3.4 m

The isotropic object The anisotropic object

The conductivity of the medium is=0.01 Sm

Re Ex, Ez for horizontal object

The cross-section z= -1 m

The isotropic object The anisotropic object

The conductivity of the medium is=0.01 Sm

The multiscale modeling in media with microinclusions

10div grad u

0 1 1 2

1| gu u

2

10

u

n

1 , 0u

1 , 1u

21

0

1 , 0u

2

10

1 , 1u The problem is stated in the domain

and governed by

the following equation:

1

electric potential

electrical conductivity

electrical resistivity

u

Problem definition

1 2( ) , ( ) : ( ( ), ( )) ( ) ( ) ,H u v L u x v x u x v x d x

1

1 10( ) ( ) : | 0H v H v

1 10 0 0

1

( )

( ) ( ) , ( ) held

( ) ( ) 0x

Find u H u such that v H the following is

u x v x d in

Variational problem

We introduce the Hilbert space

Then the variational problem of the homogeneous elliptic problem states:

Discrete variational problem

Let's consider a partition in the area Ω. Element is a tetrahedron.h hK

Let's introduce the spaces

Then the variational problem of the homogeneous elliptic problem states:

10 ( )( ) : 1,...,n ;K

i Kh hV Hspan i K

0, 10 ( )( ) : 1,...,n ;K

i Kh hW Hspan i K

,

,

0

1

( )

( ) ( ) , ( )

( ) ( ) 0

h h h

h h

x

Find u V u such that v V the following is held

u x v x d in

Taking into account the partition we introduce the following statements:h

where and – quadrature points and weights respectively. lx l

Discrete variational problem

1

( )

1( ) ( ) 0

( )

( ) ( ) 0, ,

, ,

K Ki j

K

K Kl i l j l

K l l

x

K x xx

x x d x K

x K

The basic principles

The local functions 0 , 1,...,3j j

The local multiscale “form functions” , 1,...,3i i

The global multiscale “form functions” , 1,...,i i N

FEM

Assemble according degrees offreedom associated with nodes of the coarse mesh

The integration points

Heterogeneous Finite Element Method

015 mm

40 mm

X

ZY

15 mm

Scalability

InclusionsVolume of inclusions, %

Number of Cores

1 2 45х10х10 3.81 823 455 246

CPU time (sec)

1.4e 7inclusion Оhm m 3.13matrix Оhm m

Method The error

Physical experiment

2.77 -

Maxwell's approach

2.84 2.35%

Bruggeman's approach

2.82 1.64%

Approach of coherent potential

2.80 1.24%

Numerical Modeling

2.79 0.72%

,eff Оhm m

2мм

Comparison with the physical experiment

1.8e 8inclusion Оhm m

6.15matrix Оhm m

Method The error

Physical experiment

4.25 -

Numerical Modeling

4.00 5.80%

,eff Оhm m

Comparison with the physical experiment

В)horizontal

The size of the inclusions: _*100%

_

V inclusionsC

V domain

a) vertical b) arbitrary directed

{1e 3, 1e 3}

1

inclusion

matrix

Оhm m

Оhm m

г) spheres

5.8мм

The cylinder with inclusions

The influence of the geometry and orientation of the inclusions

Horizontal plates

Arbitrary oriented plates

Vertical plates

Spheres

Horizontal plates

Arbitrary oriented plates

Vertical plates

Spheres

0

81

1.391

1.8 10

Ohm m the matrix

Ohm m theinslusions

The percolation

The size of the inclusions

The calculation of the effective tensor coefficients

E. Shurina, M. Epov, N. Shtabel and E. Mikhaylova. The Calculation of the Effective Tensor Coefficient of the Medium for the Objects with Microinclusions // Engineering, Vol. 6 No. 3, 2014, pp. 101-112.

The main steps of the algorithm

Mathematical modelThe Helmholtz equation

in Ω

Boundary conditions

0

0

, PEC

n E 0

n E Ee

1 PMCE n gm

i

2 2k i is the wave number

1 2( 0E) + Ek

The direct problem

Calculation of the effective coefficient

Z is a complex-valued second rank tensor, which can be interpreted as the analog of i

Scalar

Tensor

(i ) rot E H

rotZ E H

11 12 13

21 22 23

31 32 33

z z z

Z z z z

z z z

The 1-st method of calculating tensor Z

2 2( ) ( )/ij i jZ RH E

L L

Re( ) Re( ) / Re( )ij i jZ RH E

Im( ) Im( ) / Im( )ij i jZ RH E

where , , ,i j x y z

1

2

3

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

x i y i z i l l i

x j y j z j l l j

x k y k z k l l k

E x E x E x z RH x

E x E x E x z RH x

E x E x E x z RH x

where , , 1,i j k N , ,l x y z

The 2-nd method of calculating tensor Z

Fields E and rot H are calculated in N points of the domain (for example, in barycentres of tetrahedral finite elements). We obtain the set of tensors Z {Zm, m=1,..,N-2}, by running over the points xi, xj, xk. The effective tensor coefficient of the medium is calculated as an average of {Zm, m=1,..,N-2}.

Variational formulation

1( 0i Z E) + E

Helmholtz equation in anisotropic media

Variational formulation:

1 0d i Z d

E W E W

The problem in anisotropic media

0 0H ( , ) E rot E

Find such that 0 (rot; ) W H

the following is held

Boundary conditions

The domain with one side boundary

conditions

0

0

,

e

n E 0

n E E

1 0m

E n

0e

n E E

1 0m

E n

The domain with boundary conditions

given by the closed path

• The size of the computational domain:

15 mm 40 mm 15 mm• The diameter of the

inclusions d = 2 mm• The number of the inclusions

is different

Domains

The electrophysical properties of the computational domain

The matrix The inclusions

ε [F/m] 4.5 ε0 1 ε0

σ [Sm/m] 0.001 0.1

µ [H/m] 1 µ0 1 µ0ε0 = 8,85 ×10-12 F/m µ0 = 4π ×10-7 H/m

The mesh (40 inclusions)

The results of numerical experiment

Number of the inclusions

Volume of the

inclusions

The size of SLAE

40 regular 2% 171 872

40 chaotically 2% 169 412

176 regular 27% 296 070

The homogeneous medium. The one size boundary conditions. The frequency 10 kHz

1

0.001 5.352 8 0.001

1.868 1 0.001 1.869 1

9.999 4 5.352 8 0.001

9.293 2 9.328 4 9.294 2

9.963 2 0.001 9.963 2 .

9.385 2 9.42 4 9.386 2

E

Z E E

E E

E E E

i E E

E E E

2

1 3 9.023 21 2.297 26

1.91 21 1 3 3.404 27

3.007 16 1.62 15 1 3

2.502 6 1.457 21 1.196 26

6.404 21 2.502 6 2.8 26 .

1.25 16 5.38 16 2.502 6

e e e

Z e e e

e e e

e e e

i e e e

e e e

EzR – Re Ez computed for homogeneous medium (=0.001Sm/m) with inclusionsEzR tensor – Re Ez computed for the medium with tensor coefficient Z2

1

2.103 3 1.880 6 2.112 3

3.987 3.563 3 4.004

2.096 3 1.873 6 2.105 3

2.701 3 1.425 2.710 3

2.130 6 1.124 3 2.137 6

2.690 3 1.419 2. 3

.

699

E E E

Z E

E E E

E E E

i E E E

E E E

2

1 3 1.405 16 2.322 20

5.211 18 1 3 1.001 20

1.824 14 1.567 13 1 3

1.7516 1.7957 16 2.843 20

9.857 18 1.7516 2.353 21 .

4.418 14 4.998 13 1.7516

e e e

Z e e e

e e e

e e

i e e

e e

The homogeneous medium. The one size boundary conditions. The frequency 7 GHz

EzR – Re Ez computed in homogeneous medium (=0.001Sm/m) with inclusionsEzR tensor - Re Ez computed in the medium with tensor coefficient Z2

176 inclusions

10 kHz 7 GHz

In the medium with inclusionsIn anisotropic mediumIn uniform medium, 0.001 Sm/m

In the medium with inclusionsIn anisotropic mediumIn uniform medium, 0.001 Sm/mIn uniform medium, 0.1 Sm/m