On the Inverse MEG Problem with a 1 D Current Distribution

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http://dx.doi.org/10.4236/am.2015.61010

On the Inverse MEG Problem with a 1-D

Current Distribution

George Dassios, Konstantia Satrazemi

Department of Chemical Engineering, University of Patras and ICE/HT-FORTH, Patras, Greece Email: [email protected]

Received 29 October 2014; revised 20 November 2014; accepted 8 December 2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract

The inverse problem of magnetoencephalography (MEG) seeks the neuronal current within the conductive brain that generates a measured magnetic flux in the exterior of the brain-head system. This problem does not have a unique solution, and in particular, it is not even possible to identify the support of the current if it extends over a three-dimensional set. However, a localized current supported on a zero-, one- or two-dimensional set can in principle be identified. In the present work, we demonstrate an analytic algorithm that is able to recover a one-dimensional distribution of current from the knowledge of the exterior magnetic flux field. In particular, we consider a neuronal current that is supported on a small line segment of arbitrary location and orientation in space, and we reduce the identification of its characteristics to a nonlinear algebraic system. A se-ries of numerical tests show that this system has a unique real solution. A special case is easily solved via the use of trivial algebraic operations.

Keywords

Magnetoencephalography, Current Identification

1. Introduction

The brain is a conducting material and therefore, every generated neuronal current is accompanied by an induc-tion current. Consequently, when we measure the magnetic flux density outside the head we actually measure the effects of both the neuronal as well as the induction current. This is the main problem with the inverse prob-lem of magnetoencephalography, the fact that the induction current “hides” somehow the primary neuronal ex-citation. An excellent review of the electromagnetic activity of the human brain can be found in [1], as well as in the book by Malmivuo and Plonsey [2].

(2)

com-plete quantitative characterization of what part of the current is possible to be identified was a topic of intense investigation during the last two decades and the main results can be found in [4]. Fokas proved that, indepen-dently of the geometry of the conductor, we cannot recover more than one out of the three functions that define the current, in the case of electroencephalography, and no more than two such functions in the case of magne-toencephalography. Even in the case that we have complete data from both modalities, still one out of the three functions is not recoverable. Another related question concerns localized neuronal currents. If the current is re-stricted to a small subset of the conducting brain tissue, is it possible to identify the characteristics of this current and especially its extent and its location? Albanese and Monk [5] proved that such localization is not possible. More precisely they showed that it is impossible to find the support of the current if the current occupies a three- dimensional subset of the brain. However, if the current is distributed over a surface, which is a two-dimensional subset, a curve, which is a one-dimension subset, or on isolated points, which form zero-dimensional subsets, then it is possible to identify it. It is the purpose of the present work to demonstrate that this is true for a one- dimensional current distribution. In particular, we consider a dipolar current distribution over a small line seg-ment, and we develop an algorithm that reduces the identification of the position, the length and the orientation of the line segment, as well as the average dipolar moment of the current, to the solution of a nonlinear algebraic system. The solution of this system can be handled numerically.

2. The MEG Problem for a Single Dipole

Within the Quasi-Static Theory of Electromagnetism Magnetoencephalography [6]-[8] the magnetic field, gen-erated by a dipolar current at the point r0 having the moment Q, is given by the Geselowitz formula [9]

(

)

0 0 0

(

) ( )

( )

0 3 0 3

0

ˆ

; ; d ,

4π 4π

c

Su s

µ − µ σ _′ _′ − ′ _′

= × − × ∈ Ω

′ −

−



∫

r r r r

B r r Q r r n r r r

r r

r r (1)

where u is the electric potential on the boundary S of the conducting medium Ω representing the brain-head system. In Formula (1), ΩC

denotes the exterior domain, σ is the constant conductivity of the brain tissue, 0

µ is the magnetic permeability both inside and outside Ω and n stands for the outward unit normal on the boundary S.

When Ω is a sphere of radius a we know from the solution of the corresponding electroencephalography problem that the electric potential on the boundary of the sphere is given by [10] [11]

(

)

(

₀

)

0

(

)

(

₀

)

0

( ) ( )

0 1 0 1 0

=1 1

1 2 1 1 1

ˆ ˆ ˆ ˆ

; .

4π

n n n

m m

n n n

n n

n n m n

r r

n

u P Y Y

n n

σ α σ α

∞ ∞ _∗

+ +

= =−

+

′ = ⋅∇_r

∑

′⋅ = ⋅∇_r

∑ ∑

′

r r Q r r Q r r (2)

where m n

Y stands for the normalized complex spherical harmonics

( )

2 1

(

₍

)

₎

!

(

)

cos e 4π !

m

m im

n n

n m

n

Y P

n m

φ

ϑ

− + =

+

r (3)

and m n

P denotes the Legendre functions of the first kind.

Inserting expression (2) in the Formula (1) and performing the indicated integration we can obtain the mag-netic field outside the sphere. However, since the magmag-netic field B in the exterior to the sphere is both sole-noidal and irrotational it follows that there exists a scalar magnetic potential U , which is also harmonic, such that [8]

(

; 0

)

= ∇U

(

; 0

)

, r>a

B r r r r (4) Then, a series of calculations lead to the following expression for the magnetic potential [10] [11],

(

)

(

)

₍

₎₍

₎

( ) ( )

(

)

(

)

0 0

0

0 0 0 1 0

1

0 0

0 1 0

1

ˆ ˆ

;

1 2 1 1

ˆ ˆ

.

4π 1

n n

m m

n n

n n m n

n n n n

r

U Y Y

n n r

r P

n r

µ

∞ _∗

+

= =−

∞

+ =

= × ⋅∇

+ +

= × ⋅∇ ⋅

+

∑ ∑

∑

r

r r Q r r r

Q r r r

(3)

The above expression provides the magnetic potential in the exterior of the sphere due to a single current di-pole

{

r Q0,

}

. Therefore, it can be considered as the fundamental solution of the MEG problem for the spherical

geometry [12]. Consequently, any discrete, or continuous, current distribution can be obtained through summa-tion, or integrasumma-tion, respectively, of the above fundamental solution [13].

3. The Field of a Linearly Distributed Current

We consider here the special case where the neuronal current is supported on a small segment of a smooth curve which is parametrically centered at the point r0. Let this curve be represented by the equation

( )

t , t

[

L L,

]

,

( )

0 0

= ∈ − =

r r r r (6) The neuronal current is then described by the function p

( )

t

J r , t∈ −

[

L L,

]

. Since the support curve is taken to be small we can approximate the current p

( )

t

J r by the linear part of its Taylor expansion, that is

( )

2

0 0

d 0 d

p p p

t t O t

t  

= + r ⋅ ∇ ⊗_ _+

J r J r J r (7)

where the symbol ⊗ denotes tensor product.

In particular, if the curve is a small line segment of length 2L, centered at r0 and oriented along the direc-tion αˆ=

(

α α α1, 2, 3

)

, that is

( )

t = +0 tˆ, t∈ −

[

L L,

]

r r α (8) then representation (7) is written as

( )

p

t ≈ +t

J r Q l (9) where

(

1, 2, 3

)

( )

0

p

Q Q Q

= =

Q J r provides an average moment, and

(

1, ,2 3

)

ˆ

( )

0

p

l l l

= = ⋅∇ ⊗

l α J r provides an average directional derivative of the current along the direction ˆα .

Next we calculate the total potential which is generated by the approximate current (9). We recall that our ul-timate goal is to invert the MEG data in order to identify the quantities Q, r0, ˆα and L, which are nine par-ticular numbers, considering that the direction ˆα has two independent components. Therefore, we should be able to obtain these nine numbers from a few initial terms of the expansion (5).

Formula (5), for the excitation dipole

{

r J r′,

( )

′

}

, is written as

(

)

0

(

( )

)

(

)

(

2

(

)

(

3

(

)

( )

5

1 2 3

2 3 4

1 1 1

ˆ ˆ ˆ ˆ ˆ ˆ

;

4π 2 3 4

p

U r P r P r P O r

r r r

µ −

′ ′ ′

 

′ = ′ × ′ ⋅_ ∇ ′ ⋅ ′ + ∇ ′ ⋅ ′ + ∇ ′ ⋅ ′ _+

 r r r 

r r J r r r r r r r r (10)

Using the standard expressions of the Legendre polynomials [14] and performing the indicated calculation we obtain the following relations, which are written in dyadic form [15] in order to isolate the factors that are going to be integrated

(

)

(

r P1 ˆ ˆ

)

ˆ

′ ′ ′

∇r r r⋅ =r (11)

(

)

(

2

) (

)

(

)

(

)

2 ˆ ˆ 3ˆ ˆ 3ˆ ˆ . 0 ˆ ,

r P t

′ ′ ′ ′

∇_r r r⋅ = r⊗ − ⋅ =r I r r⊗ −r I r + α (12)

(

)

(

)

(

)

(

)

(

) (

)

3 3

0 0

3

ˆ ˆ 5ˆ ˆ ˆ 2 ˆ ˆ : 2

3

ˆ ˆ ˆ ˆ ˆ ˆ ˆ

5 2 : .

2

r P

t t

′ ′ ′ ′ ′

∇ ⋅ = ⊗ ⊗ − ⊗ − ⊗ ⊗

= ⊗ ⊗ − ⊗ − ⊗ + ⊗ +

 

r r r r r r I r r I r r

r r r I r r I r α r α

(13)

The symbol I denotes the identity dyadic, “:” defines the double contraction [15]

(4)

(

a⊗ ⊗b c

) (

 d⊗ ⊗e f

) (

= ⋅c d

)(

b e⋅

)(

a f⋅

)

(15) The exterior potential, given in (10), can be written in its Cartesian form [11] [13] as follows

(

)

0 1

( )

2

( )

3

( )

5

3 5 7

; 4π

H H H

U O r

r r r

µ   −

′ = _ + + _+

 

r r r

r r (16)

where the coefficients

( )

(

( )

)

1

1 2

p

H r = r⋅ J r′ ×r′ (17)

( )

(

( )

)

2 :

p

H r = ⊗r r r′⊗ J r′ ×r′ (18)

( )

(

2

)

(

( )

)

3 3

5 8

p

H r = r⊗ ⊗ −r r r r⊗I r′⊗ ⊗r′ J r′ ×r′ (19) are homogeneous harmonic functions [13].

In what follows we insert the expressions (8) and (9) in (17), (18) and (19) and integrate the resulting equa-tions with respect to t from −L to L. Performing these calculations we arrive at the expressions

( )

(

) (

)

(

)

3

(

)

1 0 0

1 1

ˆ d ˆ ,

2 3

L L

H r = r⋅

∫

₋ Q+tl × r +tα t=Lr⋅ Q r× + Lr⋅ ×l α (20)

( )

(

) (

)

(

)

(

)

(

)

(

)

2 0 0

2 2 2

0 0 0 0

ˆ ˆ

: d

2

ˆ ˆ ˆ ˆ

: 3 ,

3

L L

H t t t t

L L L L

−

= ⊗ + ⊗ + × +

 

= ⊗ _ ⊗ × + ⊗ × + ⊗ × + ⊗ × _

∫

r r r r Q l r

r r r Q r r l Q l r

α α

α α α α (21)

( )

(

)

(

) (

)

(

)

(

)

(

) (

)

(

)

2

3 0 0 0

2 2

0 0 0

2 2 4

0 0 0 0

3

ˆ ˆ ˆ

5 d

8

ˆ

5 5 3

20

ˆ ˆ ˆ ˆ ˆ ˆ

5 5 3 .

L L

H r t t t t t

L

r L

L L L

−

= ⊗ ⊗ − ⊗ + ⊗ + ⊗ + × +



= ⊗ ⊗ − ⊗ _ ⊗ ⊗ × + ×



+ ⊗ + ⊗ ⊗ × + × + ⊗ ⊗ × + × _

∫

 

r r r r r I r r Q l r

r r r r I r r Q r l

r r Q l r Q r l

α α α

α

α α α α α α

(22)

Finally, we replace the above expressions of the harmonic functions H1, , H2 H3 in the expansion (16) and obtain the Cartesian representation of the exterior potential U up to the terms of order 5

r− . That solves the relative forward MEG problem for a neuronal excitation that is supported on a small line segment.

4. Determination of the Current

The harmonic functions H1, H2 and H3 are homogeneous polynomials of degrees 1, 2 and 3, respectively, that is

( )

1 1 1 2 2 3 3

H r =A x +A x +A x (23)

( )

2 2 2

2 1 1 2 2 3 3 12 1 2 23 2 3 31 3 1

H r =B x +B x +B x +B x x +B x x +B x x (24) where, because of harmonicity, we should have the constrain

1 2 3 0

B +B +B = (25) and

( )

3 3 3 2 2 2 2 2 2

3 1 1 2 2 3 3 12 1 2 21 2 1 23 2 3 32 3 2 31 3 1 13 1 3 123 1 2 3 H r =C x +C x +C x +C x x +C x x +C x x +C x x +C x x +C x x +C x x x (26) together with the constrains

1 21 31

(5)

12 3 2 32 0

C + C +C = (28)

13 23 3 3 0

C +C + C = (29) In the idealized case where the exterior magnetic potential U is known, the expansion (16) is known and therefore the coefficients A, B and C are also known. Hence, if we rewrite the polynomials H1, H2 and

3

H in terms of the Cartesian monomials that appear in (23), (24) and (26), then we can utilize their linear in-dependence to equate each monomial with the corresponding known coefficients A, B or C.

Equations (20) and (23) imply immediately that

(

)

(

)

3

(

)

1 2 3 0

1 ˆ , ,

3

A A A L L

= = × + ×

A Q r l α (30) Then, from Equations (30) and (33) we obtain the six relations

(

)

(

)

(

)

(

)

3

1 01 2 3 3 2 1 2 03 3 02 1 2 3 3 2 01 2 03 3 02

2

2 3

B = L _x lα −lα +α l x −l x +α Qα −Qα _+ Lx Q x −Q x (31)

(

)

(

)

(

)

(

)

3

2 02 3 1 1 3 2 3 01 1 03 2 3 1 1 3 02 3 01 1 03

2

2 3

(

)

(

)

(

)

(

)

3

3 03 1 2 2 1 3 1 02 2 01 3 1 2 2 1 03 1 02 2 01

2

2 3

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

3

12 01 3 1 1 3 1 3 01 1 03 1 3 1 1 3

3

02 2 3 3 2 2 2 03 3 02 2 2 3 3 2

01 3 01 1 03 02 2 03 3 02 2

3 2

3

2 2 ,

B L x l l l x l x Q Q

L x l l l x l x Q Q

Lx Q x Q x Lx Q x Q x

α α α α α α

 

= _ − + − + − _

 

+ _ − + − + − _

+ − + −

(34)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

3

23 02 1 2 2 1 2 1 02 2 01 2 1 2 2 1

3

03 3 1 1 3 3 3 01 1 03 3 3 1 1 3

02 1 02 2 01 03 3 01 1 03 2

3 2

3

2 2 ,

B L x l l l x l x Q Q

L x l l l x l x Q Q

α α α α α α

 

= _ − + − + − _

 

+ _ − + − + − _

+ − + −

(35)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

3

31 03 2 3 3 2 3 2 03 3 02 3 2 3 3 2

3

01 1 2 2 1 1 1 02 2 01 1 1 2 2 1

03 2 03 3 02 01 1 02 2 01 2

3 2

3

2 2 .

B L x l l l x l x Q Q

L x l l l x l x Q Q

α α α α α α

 

= _ − + − + − _

 

+ _ − + − + − _

+ − + −

(36)

where it is easily shown that condition (25) holds. Similarly, from Equations (22) and (26) we obtain

(

) (

)

(

)

(

)

(

)

(

)

(

) (

)

2 2 3 2 3 2 2 5 2

1 01 0 1 2 03 3 02 01 0 1 2 3 3 2

3

01 1 0 2 3 3 2 2 03 3 02

1 1

3 5 5 1 5 5 3 5 1

4 20

1

ˆ

5 ,

2

C L x r L Q x Q x L x r L l l

L x Q Q l x l x

α α α α

α α α

   

= _ − + − _ − + _ − + − _ −

 

+ _ − r a⋅ _ − + −

(37)

(

) (

)

(

)

(

)

(

)

(

)

(

) (

)

2 2 3 2 3 2 2 5 2

2 02 0 2 3 01 1 03 02 0 2 3 1 1 3

3

02 2 0 3 1 1 3 3 01 1 03

1 1

3 5 5 1 5 5 3 5 1

4 20

1

ˆ

5 ,

2

L x Q Q l x l x

α α α α

α α α

   

= _ − + − _ − + _ − + − _ −

 

+ _ − r a⋅ _ − + −

(6)

(

) (

)

(

)

(

)

(

)

(

)

(

) (

)

2 2 3 2 3 2 2 5 2

3 03 0 3 1 02 2 01 03 0 3 1 2 2 1

3

03 3 0 1 2 2 1 1 02 2 01

1 1

3 5 5 1 5 5 3 5 1

4 20

1

ˆ

5 .

2

L x Q Q l x l x

α α α α

α α α

   

= _ − + − _ − + _ − + − _ −

 

+ _ − r a⋅ _ − + −

(39)

for the cubic terms 3 1

x , 3 2

x and 3 3

x , respectively. For the cross-terms 2 1 2

x x and 2 1 2

x x we obtain the expres-sions

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

12 01 0 1 3 01 1 03 01 0 1 3 1 1 3

3 3 5

01 02 1 2 2 03 3 02 01 02 1 2 2 3 3 2

3 3

1 02 2 01 2 3 3 2 2 03 3 02 1

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

Lx x L Q x Q x L x x L l l

L x x Q Q l x l x L x

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

₀₁− ⋅r a₀ ˆ

)(

Q₃α₁−Q₁α₃+l x_{3 01}−l x_{1 03}

)

,

(40)

and

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

21 02 0 2 2 03 3 02 02 0 2 2 3 3 2

3 3 5

01 02 1 2 3 01 1 03 01 02 1 2 3 1 1 3

3 3

1 02 2 01 3 1 1 3 3 01 1 03 2

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

₀₂− ⋅r a₀ ˆ

)(

Q₂α₃−Q₃α₂+l x_{2 03}−l x_{3 02}

)

,

(41)

Similarly, for the cross-terms 2 2 3

x x and 2 2 3

x x we obtain

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

23 02 0 2 1 02 2 01 02 0 2 1 2 2 1

3 3 5

02 03 2 3 3 01 1 03 02 03 2 3 3 1 1 3

3 3

3 02 2 03 3 1 1 3 3 01 1 03 2

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

₀₂− ⋅r a₀ ˆ

)(

Q₁α₂−Q₂α₁+l x_{1 02}−l x_{2 01}

)

,

(42)

and

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

32 03 0 3 3 01 1 03 03 0 3 3 1 1 3

3 3 5

02 03 2 3 1 02 2 01 02 03 2 3 1 2 2 1

3 3

3 02 2 03 1 2 2 1 1 02 2 01 3

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

03− ⋅r a0 ˆ

)(

Q3α1−Q1α3+l x3 01−l x1 03

)

,

(43)

while, for the cross-terms 2 3 1

x x and 2 3 1

x x we obtain

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

31 03 0 3 2 03 3 02 03 0 3 2 3 3 2

3 3 5

01 03 1 3 1 02 2 01 01 03 1 3 1 2 2 1

3 3

1 03 3 01 1 2 2 1 1 02 2 01 3

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

₀₃− ⋅r a₀ ˆ

)(

Q₂α₃−Q₃α₂+l x_{2 03}−l x_{3 02}

)

, (44)

(7)

(

) (

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

2 2 3 2 3 2 2 5 2

13 01 0 1 1 02 2 01 01 0 1 1 2 2 1

3 3 5

01 03 1 3 2 03 3 02 01 03 1 3 2 3 3 2

3 3

1 03 3 01 2 3 3 2 2 03 3 02 1

1 1

3 5 5 1 5 5 3 5 1

4 20

5 1

3 5 3

2 2

5 1

5

2 2

α α α α

α α α α α α

α α α α α

   

= _ − + − _ − + _ − + − _ −

+ + − + + −

+ + − + − +

(

01− ⋅r a0 ˆ

)(

Q1α2−Q2α1+l x1 02−l x2 01

)

, (45)

Finally for the product term x x x1 2 3 we obtain

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

3 3

123 02 03 2 3 2 03 3 02 01 03 1 3 3 01 1 03

3 3 5

01 02 1 2 1 02 2 01 02 03 2 3 2 3 3 2

3 5 3 5

01 03 1 3 3 1 1 3 01 02 1 2 1 2 2 1

5 5

3 3

2 2

5 1

3 5 3

2 2

1 1

5 3 5 3

2 2

C Lx x L Q x Q x Lx x L Q x Q x

L x x L l l L x x L l l

α α α α

α α α α α α

α α α α α α α α

= + − + + −

+ + − + + −

(

)(

)

(

)(

)

(

)(

)

3

2 03 3 02 2 3 3 2 2 03 3 02 3

1 03 3 01 3 1 1 3 3 01 1 03 3

1 02 2 01 1 2 2 1 1 02 2 01

5

2 5

.

2

L x x Q Q l x l x

α α α α

+ + − + −

(46)

It is straightforward to verify that the three constrains (27)-(29) are satisfied.

The set of the 16 equations, which are the 20 scalar equations appearing in (30)-(46) minus the four constrains (25) and (27)-(29), defines a nonlinear system for the determination of the 12 independent variables r0, ˆa, l, Q and L, three components for each one of the vectors r0, l, Q, two components for the direction vector

ˆ

a and one for the length L. In fact, we can simplify this system as follows. In view of Equation (30) the three components of the exterior product Q r× ₀ provide the relations

(

)

2 1

2 03 3 02 2 3 3 2

3

A L

Q x Q x l l

L α α

− = − − (47)

(

)

2 2

3 01 1 03 3 1 1 3

3

A L

Q x Q x l l

L α α

− = − − (48)

(

)

2 3

1 02 2 01 1 2 2 1

3

A L

Q x Q x l l

L α α

− = − − (49)

and these relations reduce the Equations (31)-(36) to

(

)

3

1 1 01 1 2 03 3 02 2 3 3 2

2 2

3

B = A x + Lα l x −l x +Qα −Qα (50)

(

)

3

2 2 02 2 3 01 1 03 3 1 1 3

2 2

3

B = A x + Lα l x −l x +Qα −Qα (51)

(

)

3

3 3 03 3 1 02 2 01 1 2 2 1

2 2

3

B = A x + Lα l x −l x +Qα −Qα (52)

(

)

3

(

)

3

(

)

12 1 02 2 01 1 3 01 1 03 3 1 1 3 2 2 03 3 02 2 3 3 2

2 2

2

3 3

B = A x +A x + Lα l x −l x +Qα −Qα + Lα l x −l x +Qα −Qα (53)

(

)

3

(

)

3

(

)

23 2 03 3 02 3 3 01 1 03 3 1 1 3 2 1 02 2 01 1 2 2 1

2 2

2

3 3

(8)

(

)

3

(

)

3

(

)

31 3 01 1 03 1 1 02 2 01 1 2 2 1 3 2 03 3 02 2 3 3 2

2 2

2

3 3

B = A x +A x + Lα l x −l x +Qα −Qα + Lα l x −l x +Qα −Qα (55) Furthermore, utilizing the Equations (50)-(52) we arrive at the relations

(

)

01 02 2 1

1 2 2 1 12 1 2

1 2 1 2

2 x x α A α A B α B α B

α α α α

 

− − = − −

 

  (56)

(

)

02 03 2 3

2 3 3 2 23 3 2

2 3 3 2

α α α α

 

− − = − −

 

  (57)

(

)

03 01 1 3

3 1 1 3 31 3 1

3 1 3 1

α α α α

 

− − = − −

 

  (58)

which allow rewriting Equations (37)-(46) as follows

(

)

5

(

2

)

(

)

2

(

2

) (

2 2

)

01

(

)

1 1

1 01 1 0 1 2 3 3 2 1 01 0 0

1 1

3 3

ˆ ˆ

5 5 1 5 1 5 2

4 15 4 3

x

B L A L

C x α α lα lα α x r

α α

 

= − ⋅ + − − + _ − − + + ⋅ _

 

r a r a (59)

(

)

5

(

2

)

(

)

2

(

2

) (

2 2

)

02

(

)

2 2

2 02 2 0 2 3 1 1 3 2 02 0 0

2 2

3 3

ˆ ˆ

5 5 1 5 1 5 2

4 15 4 3

x

B L A L

α α

 

= − ⋅ + − − + _ − − + + ⋅ _

 

r a r a (60)

(

)

5

(

2

)

(

)

2

(

2

) (

2 2

)

(

)

3 3 03

3 03 3 0 3 1 2 2 1 3 03 0 0

3 3

ˆ ˆ

5 5 1 5 1 5 2

4 15 4 3

B L A L x

α α

 

= − ⋅ + − − + _ − − + + ⋅ _

 

r a r a (61)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

12 1 01 02 1 2 2 01 0 1 1 1 01 1 02 2 01

1

5 5 2

2 2 02 1 01 0 1 2 2 3 3 2 1 3 1 1 3

2

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 ,

4 3 15

C A x x L A x r L B A x x x

B A x x L l l L l l

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(62)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

21 2 01 02 1 2 1 02 0 2 2 2 02 1 02 2 01

2

5 5 2

1 1 01 2 02 0 1 2 3 1 1 3 2 2 3 3 2

1

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 ,

4 3 15

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(63)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

23 2 02 03 2 3 3 02 0 2 2 2 02 2 03 3 02

2

5 5 2

3 3 03 2 02 0 2 3 3 1 1 3 2 1 2 2 1

3

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 ,

4 3 15

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(64)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

32 3 02 03 2 3 2 03 0 3 3 3 03 2 03 3 02

3

5 5 2

2 2 02 3 03 0 2 3 1 2 2 1 3 3 1 1 3

2

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 ,

4 3 15

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(65)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

31 3 01 03 1 3 1 03 0 3 3 3 03 1 03 3 01

3

5 5 2

1 1 01 3 03 0 1 3 1 2 2 1 3 2 3 3 2

1

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 ,

4 3 15

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(9)

(

)

(

) (

)

(

)(

)

(

)(

)

(

)

(

)

(

)

2 2 2 2 2

13 1 01 03 1 3 3 01 0 1 1 1 01 1 03 3 01

1

5 5 2

3 3 03 1 01 0 1 3 2 3 3 2 1 1 2 2 1

3

5 1 15

3 3 5 5 1 2

2 4 4

3 2 1

ˆ

2 5 5 1 .

4 3 15

α α α α α

α

 

= + + _ − + − _+ − +

+ − − ⋅ +r a − + − −

(67)

and

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

(

)(

)

(

)(

)

2 2 2 5

123 1 02 03 2 3 2 01 03 1 3 3 01 02 1 2 2 3 2 3 3 2

5 5

1 3 3 1 1 3 1 2 1 2 2 1 1 1 01 2 03 3 02

1

2 2 02 1 03 3 01 3 3 03 1 02 2 01

2 3

5 5 5 2

3 3 3

2 2 2 3

2 2 15

2

3 3 4

15 15

2 2 .

4 4

C A x x L A x x L A x x L L l l

L l l L l l B A x x x

B A x x x B A x x x

α α α α α α α α α α

α

α α α α

α α

= + + + + + + −

+ − + − + − +

+ − + + − +

(68)

Because of the constrains (27)-(29), only 7 out of the 10 equations (59)-(68) are independent. Then, the re-duced set of these 7 independent equations, plus the 6 equations (47)-(49) and (53)-(55) provides a nonlinear system for the determination of the unknown quantities r0, ˆa, Q, l and L. However, since some of these quantities enter the system through the components of exterior products, it follows that the above quantities cannot be completely specified. For example, from Equations (47)-(49) it follows that it is not possible to iden-tify the three components of Q from the exterior product of Q with the position vector r0, since the com-ponent of Q that is parallel to r0 gives a vanishing term. Hence, this component of Q forms a “silent” source [8]. The solution of this system can easily be obtained with the use of classical computational methods.

To illustrate the inversion algorithm we consider the following special case.

Special Case. Let us assume that we have the a-priori information that the line segment is oriented along the 1

x -axis and that its middle point is r0 =

(

0, 0,r0

)

. This choice leads to

1 2 0

A =LQ r (69)

3

3 2

1 3

A = − L l (70)

3 2

1 2 0 0

3 4 C =Q r _L − Lr _

  (71)

(

3 2

)

3 2 5

2 1 0 0 3 0

1 1 3

3

4 4 5

C = Q r L + Lr − l _L r + L _

  (72)

3 3 5 3 2

12 1 0 0 3 0

3 3 1

4 5 4

C =Q _ Lr −L r _+l _ L − L r _

    (73)

(

2 3

)

31 2 0 0

1

12 11 4

C = Q r Lr − L (74) Inserting the expression of A1 in the equations for C1 and C31 we obtain the following 2 2× system for the determination of 2

L and 2 0

r

2 2

1 1 0 1

4A L −3A r =4C (75)

2 2

1 1 0 31

11A L 12A r 4C

− + = (76) which immediately gives the values of L and r0

1 31

1 4 2

5

C C

L

A

+

(10)

1 31 0

1 11 4 2

15

C C

r

A

+

= (78)

Then from (69) we obtain

1 2

0 A Q

Lr

= (79)

and from (70) we obtain

3 2 1

2 3

1 31 5 3

8 4

A

l A

C C

 

= − _ _

+

  (80) Finally, from (72) and (73) we obtain a 2 2× system for the unknowns Q1 and l3, from which we obtain

(

2 2

)

(

2 2

)

0 2 0 12

1 3 3

0

12 5 3 5

5

L r C L r C

Q

L r

− + +

= (81)

(

2 2

)

(

2 2

)

0 2 0 12

3 5 2

0

4L 3r C L 3r C l

L r

− + +

= (82)

with L and r0 given by (77) and (78), respectively. Note that the only constants that remain unspecified are 1

l and Q3, but the constant l1 is not needed, and the constant Q3 can not be determined since the component 3ˆ3

Q x is parallel to the position vector r0, and therefore their exterior product vanishes.

Acknowledgements

The present work is part of the project “Functional Brain”, which is implemented within the “ARISTEIA” Ac-tion of the “OPERATIONAL PROGRAMME EDUCATION AND LIFELONG LEARNING” and is co-funded by the European Social Fund (ESF) and National Resources.

References

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[3] Helmholtz, H. (1853) Ueber einige Gesetze der Vertheilung elektrischer Str ome in k orperlichen Leitern mit Anwen-dung auf die thierisch-elektrischen Versuche. Annalen der Physik und Chemie, 89, 211-233, 353-377.

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[8] Sarvas, J. (1987) Basic Mathematical and Electromagnetic Concepts of the Biomagnetic Inverse Problems. Physics in

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http://dx.doi.org/10.1088/0266-5611/25/3/035001

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