Identification of small flaws in conductors using magnetostatic measurements

Mathematics and Computers in Simulation 50 (1999) 457±471

Identification of small flaws in conductors using magnetostatic measurements Sailing Hea,*, Vladimir G. Romanovb,1 a

State Key Laboratory of Modern Optical Instrumentation, Centre for Optical and Electromagnetic Research, Zhejiang University, Yuquan, 310027 Hangzhou, PR China b Department of Electromagnetic Theory, Royal Institute of Technology, S-100 44 Stockholm, Sweden

Abstract The identification of small flaws in a conducting half-space or rectangular specimen using boundary measurements of the static magnetic field is considered. The perturbed magnetic field due to the presence of a small flaw is assumed to be generated by an equivalent current dipole located at the flaw position. Explicit formulas are given for the identification of a single flaw or a set of flaws in a conducting half-space or rectangular specimen. # 1999 IMACS/Elsevier Science B.V. All rights reserved. Keywords: Flaw; Conductor; Magnetostatic; Explicit identification

1. Introduction In maintenance of structures, conducting components are often tested in order to detect any flaws or cracks, in particular, in industries where very high quality is mandatory, such as aerospace and nuclear power generation. Identification of flaws or cracks is a special type of inverse problems (see e.g. [1±3]). If the flaw does not present a risk of failure, the component should not be discarded even though the flaw may be of a considerable size. On the other hand, some smaller flaws may present a greater risk of failure due to their position. Various electromagnetic non-destructive testing methods have been used to identify the cracks in conducting materials (see e.g. [4±8]). However, research works on identification of small flaws in conducting materials are quite few. When a field is created in a conducting object, the presence of a flaw will cause a distortion of the field. Such a field distortion can be measured at the boundary or in an exterior region of the object, and used to predict the location and size of the flaw. If the flaw is small, the distortion of the field in the exterior region may be very small. A modern multichannel SQUID (Superconducting Quantum ÐÐÐÐ * Corresponding author. E-mail address: [email protected] (S. He) 1 Institute of Mathematics, Universitetskij prospekt 4, 630090 Novosibirsk, Russian Federation. 0378-4754/99/$20.00 # 1999 IMACS/Elsevier Science B.V. All rights reserved. PII: S 0 3 7 8 - 4 7 5 4 ( 9 9 ) 0 0 0 9 8 - 1

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Interference Device) can detect a tiny magnetic flux of the order of 10ÿ15 T/m2 (the so-called magnetic flux quantum in the quantum mechanical theory) [9]. With such a highly-sensitive device, one can detect bolt-hole flaws as small as 0.1 mm. This is impossible for any non-destructive testing method using other types of wave fields. If the flaw or crack is situated deep inside the conductor, one should use a low frequency measurement in order to allow the electromagnetic wave to penetrate down to this deep level. In the static or quasi-static case, the distortion (due to the presence of a flaw) of the magnetic field in the exterior vacuum region is independent of the stratification of the conducting structure (cf. Section 2.2). One can thus identify directly the small flaw without knowing the stratification of the structure. This is another advantage to use the magnetic field measurement for the identification of flaws. The problem of recovering of small delta-like inclusions, rather than the slow changing background has been considered in other applications (e.g. [10,11]). In the present paper, we consider the identification of small flaws in conductors using magnetostatic measurements. The perturbed magnetic field (due to the presence of a small flaw) is approximated by a magnetic field generated by an internal current dipole (located at the flaw position). Under this assumption we give uniqueness theorems for the identification of a single flaw or a set of flaws in a conducting half-space or rectangular specimen. Explicit formulas for the flaw identification are given in the present paper. 2. Identification of flaws in a conducting half-space 2.1. Problem formulation and the equivalent dipole model for a flaw Consider the electromagnetic fields {E0, H0} for a perfect specimen and fields {Ef, Hf} for a flawed specimen excited by the same exterior current source J. Thus, one has the following Maxwell's equations:  r  E0 ˆ i!0 H0 ; (1) r  H0 ˆ ÿi!E0 ‡ E0 ‡ J;  r  Ef ˆ i!0 Hf ; (2) r  Hf ˆ ÿi!f Ef ‡ f Ef ‡ J; where (, ) and (f, f) are the permittivity and conductivity profiles for cases with or without flaws, respectively. The permeability is assumed to be constant and equal to its vacuum value 0 in the whole space. The time-dependence is assumed to be eÿi!t. Denote the differences of the fields by: ~ ˆ E0 ÿ Ef ; E

(3)

~ ˆ H0 ÿ H f : H

(4)

Subtracting the systems (1) and (2), one obtains:  ~ ~ ˆ i!0 H; rE ~ ~ ‡ E ~ ÿ i!… ÿ f †Ef ‡ … ÿ f †Ef ; r  H ˆ ÿi!E

(5)

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459

In the present paper we only consider the static case, which means that we will take ! ˆ 0 in the above system of equations, i.e. ~ ˆ 0; rE

(6)

~ ˆ E ~ ‡ j; rH

(7)

where the equivalent current source j is given by: j ˆ … ÿ f †Ef :

(8)

In the case of a single flaw, the profile f differs from the profile  only at the flaw area (where f ˆ 0). We now assume that the flaw is very small in the sense that one can approximate the equivalent current source by a current dipole: j ˆ Q…x ÿ x…0† †;

(9)

where x(0) is the position of the flaw center (note that here x ˆ (x1, x2, x3)) and: Z Q ˆ  Ef …x† dx;

(10)

V

and where V is the volume of the flaw. Eq. (9) indicates that the perturbed fields caused by a small flaw can be considered as the fields generated by an equivalent current dipole (located at the flaw position) inside a flaw-free conducting object. Therefore, the identification of a small flaw is equivalent to the identification of a current dipole in a conducting object in the static case. Taking the divergence of the first equation of the system (5), one also obtains: ~ ˆ 0: r
H

(11)

For the simplicity of notation, we will omit the ~ sign over the fields in the rest of the paper (one should then keep in mind that hereafter E, H stand for the perturbed electric and magnetic field, respectively). In this subsection we consider the case when there is only one flaw inside a conducting half-space 3 R‡ ˆ fx 2 R3 jx3 > 0g (here the x3 axis is pointing downward). The half-space R3ÿ ˆ fx 2 R3 jx3 < 0g is a free space (with zero conductivity). The interface between the half-spaces R3ÿ and R3‡ will be denoted by S, i.e., S ˆ fx 2 R3 jx3 ˆ 0g. In this section we will use only the third component of the magnetic field H ˆ (H1, H2, H3) on S as the input information for the flaw identification. The third component H3 is the most effective for use for this type of flaw identification as will be shown below. The direct problem can be reformulated as the problem of finding H from the following system of equation (cf. Eqs. (6),(7),(9) and (11)): r2 H ˆ ÿr  j; ‰HŠS ˆ 0;   @H3 ˆ 0; @x3 S

x 2 R3 nS;

(12) (13) (14)

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  @Hk @H3 ÿ ˆ 0; @x3 @xk Sÿ

k ˆ 1; 2;

(15)

with the far field condition that H approaches zero as x ! 1. Here the jump [u]S  u|S‡ ÿ u|Sÿ, where u|S‡ and u|Sÿ denote the limiting values of the function u on S from the half-spaces R3‡ and R3ÿ , respectively. Note that the continuity of H3 in Eq. (13) is due to the fact r
H ˆ 0 (applying this equation to a small pillbox which includes the interface [12]), the condition (14) is due to r
H ˆ 0 on S, the condition (15) is due to the zero curl of the magnetic field on Sÿ (cf. Eq. (7)). For the case of a single flaw, the equivalent current j are given by Eq. (9). The above direct problem is well-posed for a given j. One can easily prove that the solution H to the system (12)±(14) satisfies the relation r  H ˆ 0 in R3ÿ . Eqs. (13) and (14) indicate that the third component H3 and its normal derivative are continuous across the surface S. Therefore, H3 is the solution of the following equation: r2 H3 ˆ ÿ…r  j†3 ;

x 2 R3 ;

(16)

with the zero field condition at the infinity. The solution can be given explicitly as: 0 1 Z Z 1 B @ j2 …y† @ j1 …y† C dy ÿ dyA: H3 …x† ˆ @ 4 @x1 jx ÿ yj @x2 jx ÿ yj R3

(17)

R3

Note that the above solution is based upon the free space Green's function (instead of the Green's function for the two-media problem) for the Laplace's equation, and this is because the third component H3 and its normal derivative are continuous across the surface S (cf. Eqs. (13) and (14)). In other words, the discontinuities of the conductivity along the x3 direction are `invisible' to the x3 component of the magnetic field in the static case. The transverse components H1(x) and H2(x) in R3 can be determined by the following systems of equations (cf. Eqs. (12) and (15)): 8 2 x 2 R3ÿ ; < r H k ˆ 0; @Hk @H3 (18) ˆ ; k ˆ 1; 2: : @x3 Sÿ @xk S From the above systems one sees that H1 and H2 are completely determined in R3ÿ by H3|S. Therefore, one can conclude that the complete information in the magnetic field H in R3ÿ is contained in the third component only (that is why we will not use any transverse component of the magnetic field to solve the inverse problem in this section). Furthermore, since H3 in R3ÿ can be determined by the boundary value (note that r2H3 ˆ 0 in R3ÿ ): H3 jS ˆ F…x1 ; x2 †;

(19)

the boundary information given by the above equation contains all the information in the magnetic field H in R3ÿ . This information can be used to reconstruct the current components j1(x) and j2(x) in R3‡ .

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Remark 1. Since H3 in R3ÿ does not depend on the third current component j3, this component of j cannot be found from the given H in R3ÿ . Note that even the transverse component of the magnetic field is independent of j3 (this is quite different from the case for a homogeneous full space). Remark 2. If j1  0 and j2  0, then H  0 in R3ÿ . This means that the current flowing in the normal direction x3 is a non-radiating source for the magnetic field in R3ÿ . 2.2. Identification of a single flaw For the case of a single flaw, one has j ˆ Q…x ÿ x…0† †, where x…0† 2 R3‡ . The inverse problem is to find x(0), Q1 and Q2 from the boundary information given by Eq. (19). Theorem 1. If the function F(x1, x2)  0, then Q1 ˆ Q2 ˆ 0. If F(x1, x2) 6 0, then Q1, Q2 and x(0) are uniquely determined from F(x1, x2). Proof. Let: ~ F…k† ˆ

Z F… x†exp…ik
 x†d x;

(20)

R2

x ˆ (x1, x2). Then Eqs. (17) and (19) lead to the following formula: where k ˆ (1, 2) and  i …0† ~ F…k† ˆ …1 Q2 ÿ 2 Q1 †exp…ik
 x…0† ÿ jkjx3 †; 2jkj …0†

(21)

…0†

where  x…0† ˆ …x1 ; x2 †). This formula can also be obtained by applying Fourier transform with respect to x1 and x2 directly to Eq. (16) and using the special expression for the current j. Alternatively, Eqs. (17) and (21) can be derived by using either an image-type approach (see e.g. [13]) or a spectral integral ~ method (see e.g. [14]). From Eq. (21) it follows that F…k†  0 for all k 2 R2 if and only if Q1 ˆ Q2 ˆ 0. ~  0 and then Q1 ˆ Q2 ˆ 0. Therefore, if F(x1, x2)  0, one has F…k† ~  6 0 and Q21 ‡ Q22 6ˆ 0. One can consider the following Now if we assume F(x1, x2) 6 0, then F…k† quantities (by choosing 2 ˆ 0 or 1 ˆ 0 in Eq. (21): …0† ~ 1 ; 0† ˆ i Q2 exp…i1
x…0† F… 1 ÿ 1 x3 †; 2

1 > 0;

(22)

…0† ~ 2 † ˆ i Q2 exp…i2
x…0† F…0; 2 ÿ 2 x3 †; 2

2 > 0;

(23)

~ 2 † is not identical zero. Without loss of ~ 1 ; 0† and F…0; It is obvious that at least one of the functions F… ~ generality, we assume that F…1 ; 0† 6 0. Then Q2 6ˆ 0 and: …0†

x3 ˆ

~ 1 ; 0†j 1 jF… ; ln ~ 1 jF…2 1 ; 0†j

8 1 > 0;

~ 1 ; 0†jexp…1 x…0† Q2 ˆ 2jF… 3 †;

8 1 > 0:

(24) (25)

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S. He, V.G. Romanov / Mathematics and Computers in Simulation 50 (1999) 457±471 …0†

One then uses Eq. (22) to obtain x1 through: …0†

…0†

~ exp…i1 x1 † ˆ F1 …1 †  ÿ2iQÿ1 2 F…1 ; 0†exp…1 x3 †;

1 > 0:

(26)

From the above equation, one has: @F1 …1 † …0† : jx1 j ˆ @1

(27)

…0† ~ 2 …0; 2 † 6 0, The sign of x1 (real value) can be determined by Eq. (26). Now consider Eq. (23). If F …0† ~ 2 †  0, then Q1 ˆ 0. To find x2…0† in the latter then one can find Q1 and x2 in a similar way. If F…0; case one can use the relation (21) with 2 6ˆ 0 and 1 6ˆ 0. Since Q1 ˆ 0, Eq. (21) can be rewritten in the following form: …0†

…0†

…0†

~ 1 ; 2 †exp…jkjx3 ÿ i1 x1 †; exp‰i…2 x2 †Š ˆ F2 …2 †  ÿ2jkj…1 Q2 †ÿ1 F… where F2(2) is a given function for an arbitrary 2. Thus, one has: @F2 …2 † …0† ; jx2 j ˆ @2

(28)

(29)

…0†

and the sign of x2 (real value) can be found from Eq. (28). & In the above proof, we not only prove the uniqueness theorem, but more importantly we give an explicit algorithm for reconstructing the flaw position x(0) and the tangential components Q1, Q2 of the moment of the equivalent current dipole. This is also the case in the proofs for the theorems in the rest of the paper (i.e., all the reconstruction algorithms for various situations are given explicitly in the proofs for the theorems). The explicit reconstruction algorithms are the main and most important results of the present paper. Remark 3. The same result holds for the identification of a flaw inside a conductive slab D ˆ fx 2 R3 j0 < x3 < dg using the boundary data H3 on the plane x3 ˆ 0. The slab can be multilayered as long as the equivalent current (or the flaw) is inside a homogeneous layer. This is because Eq. (16) is true in each layer, and H3 and its normal derivative are continuous across any interface (i.e., the field H3 is independent of the stratification of the multi-layered conducting structure). Therefore, one can localize the small flaw directly without knowing the stratification of the conducting structure. This is one of the advantages of using magnetic field measurements for the identification of flaws. 2.3. Identification of multiple flaws In this subsection we consider the case when there are a set of flaws in the conducting half-space R3‡ . For such a case, the equivalent current has the following form: jˆ

N X

Q…n† …x ÿ x…n† †;

(30)

nˆ1

where N is the total number of the flaws. Below we prove a similar uniqueness theorem and give an …n† …n† algorithm for reconstructing x…n† ; Q1 and Q2 .

S. He, V.G. Romanov / Mathematics and Computers in Simulation 50 (1999) 457±471 …n†

463

…n†

Theorem 2. If H3  0 on S, then Q1 ˆ Q2 ˆ 0 for all n. If H3 6 0 on S, the boundary values H3|S …n† …n† …n† …n† determine the set x…n† ; Q1 ; Q2 uniquely if jQ1 j2 ‡ jQ2 j2 6ˆ 0 for n ˆ 1, 2, ..., N. Proof. For an equivalent current j given by Eq. (30), one obtains the following formula from Eq. (17):  N  1 X 1 …n† @ …n† @ H3 …x† ˆ Q2 ÿ Q1 : 4 nˆ1 @x1 @x2 jx ÿ x…n† j

(31)

Taking Fourier transform of the above equation on the surface x3 ˆ 0 and using the boundary values H|S ˆ F(x1, x2), one obtains (cf. Eq. (21)): N i X …n† …n† …n† ~ F…k† ˆ …1 Q2 ÿ 2 Q1 †exp…ik
 x…n† ÿ jkjx3 †: 2jkj nˆ1

(32)

…n† …n† ~ Hence the first part of Theorem 2 is quite obvious, i.e., F…†  0 if and only if Q1 ˆ Q2 ˆ 0. …n† …n† 2 2 ~ Now we assume that F…† 6 0 and jQ1 j ‡ jQ2 j 6ˆ 0 for n ˆ 1, 2, ..., N. Introduce the notations:

 ˆ jkj;

m  …1 ; 2 † ˆ

k ; jkj

and consider the following function: X N …n† …n† …n† ~ F1 …; m†  2jF…m†j ˆ …1 Q2 ÿ 2 Q1 † exp…ÿx3 †: nˆ1

(33)

(34)

Assume that: …n†

(35)

n ˆ 1; 2; :::; N1 ;

(36)

h1 ˆ min x3 ; 1nN

and …n†

x3 ˆ h1 ;

where N1 is the number of the flaws on the upmost flaw plane x3 ˆ h1. Note that N1  N. One then has the following lemma. Lemma 1. The function F1 …p; m† has the following property: for almost all m, except for two values of m (opposite to each other), the relation:  0; if z < h1 ; z lim F1 …; m†e ˆ (37) !1 1; if z > h1 is valid for all z 2 R.

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Indeed, the factor: N X 1 …n† …n† A  …1 Q2 ÿ 2 Q1 † nˆ1 is zero if and only if the vector ( 1,  2) is parallel to (i.e., along the direction or the opposite direction of) the vector: ! N1 N1 X X …n† …n† Q1 ; Q2 : nˆ1

nˆ1

Hence A 6ˆ 0 for all other values of m and in these cases Eq. (37) is a simple corollary of Eq. (34). An idea (that certain function is infinite numerically on the boundary of the abnormality) similar to the formula (37) is used in [10]. In [10] the authors are concerned only with the location (but the actual physical properties of the abnormalities). Unlike the present method, the numerical method in [10] does not allow one to find those physical properties since the abnormalities are not modeled as delta functions. One can use Lemma 1 to find h1:  h1 ˆ supfzg : (38) fzg; ˆ fz 2 Rj lim!1 F1 …; m†ez ˆ 0; 8 mg Now consider the function: Z 1 ~ ^ F…k†exp…ÿik
 x ‡ jkjz† dk; F… x; z† ˆ …2†2 R2

for z < h1. It follows from Eq. (32) that:  N  1 X 1 …n† @ …n† @ ^ q : F… x; z† ˆ Q2 ÿ Q1 4 nˆ1 @x1 @x2 …n† jx ÿ x…n† j2 ‡ jz ÿ x3 j2

(39)

The above function is analytic for all  x 2 R2 and z
Q1 ˆ lim

^ …n† lim F x; z†; 1 …

(40)

z!h1 ÿ0 x!x…n†

…n†

Q2 ˆ ÿ lim

^ …n† lim F x; z†; 2 …

z!h1 ÿ0 x!x…n†

n ˆ 1; 2; :::; N1 ;

(41)

where: @ ^ …n† ^ …n† F x; z† ˆ 4…j xÿ x…n† j2 ‡ jz ÿ x3 j2 †3=2 F…x; z†; 1 … @x2

(42)

@ ^ …n† ^ …n† x; z† ÿ 4…j xÿ x…n† j2 ‡ jz ÿ x3 j2 †3=2 F F…x; z†: 2 … @x1

(43)

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465

Hence, the flaws on the upmost layer have been identified. In order to exclude the already identified ~ and consider the flaws at the depth h1, we subtract the contribution of the upmost N1 flaws from F…k† following function: n i X …n† ~ …1 Q2 ÿ 2 Q1 †exp…ik
 x…n† ÿ jkjx3 †: F…k† ˆ 2jkj nˆ1

In a similar way one can identify the flaws which have: …n†

…n†

x3 ˆ h2  min x3 ; N1 nN

n ˆ N1 ‡ 1; :::; N2

N2  N:

…n† …n† Following this way, one can, layer by layer, determine x…n† ; Q1 and Q2 for all n (n ˆ 1, 2, 3,..., N). &

3. Identification of a flaw in a conducting rectangular specimen In this section we consider the identification of a small flaw inside a conducting box D ˆ fx 2 R3 j0 < xi < di ; i ˆ 1; 2; 3g. Here di ; i ˆ 1; 2; 3, give the sizes of the rectangular specimen in the xi directions. Outside the box there is vacuum. Let S denote the boundary of D. We wish to find the flaw position x(0) 2 D and the moment Q of the equivalent current dipole (given by Eq. (9)) using the following measurement of the magnetic field on S: HjS ˆ g…x†:

(44)

We prove that this inverse problem has a unique solution. A numerical algorithm for the flaw identification is also given in the section. The magnetic field H satisfies: r2 H ˆ ÿr  j;

r
H ˆ 0;

x 2 D:

(45)

Introduce an auxiliary vector function u ˆ (u1, u2, u3) which is a solution of the following system:  2 x 2 D; r u ˆ 0; (46) u  n ˆ 0; x 2 S; where n is the outward normal to S. Obviously the above system of equations has many solutions. One has the following lemma for a more general case when the domain D has a piecewise smooth boundary S and the current j has an arbitrary form.  Lemma 2. Assume that S is a piecewise smooth closed surface containing D, H 2 C2(D) \ C1(D), 2 1  1   u 2 C (D) \ C (D) and j 2 C (D) \ C(D), where D ˆ D [ S. Furthermore, H and u satisfy the systems Eqs. (45) and (46), respectively. Then the following relation holds Z Z ÿ j
…r  u†dx ˆ ‰…H  n†
…r  u† ÿ …H
n†…u
n†ŠdS: (47) D

S

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Proof. Consider the following identity: u
r2 H ÿ H
r2 u ˆ r
…u…r
H† ÿ H…r
u† ‡ u  …r  H† ÿ H  …r  u††:

(48)

Using the systems (45) and (46), one can write the left-hand side of the above identity in the following form: u
r2 H ÿ H
r2 u ˆ ÿu
…r  j† ˆ ÿr
…j  u† ÿ j
…r  u†: Applying the Gauss formula to Eq. (48), one then obtains: Z Z Z ÿ …j  u†
n dS ÿ j
…r  u† dx ˆ fÿu…H
n† ‡ u  …r  H† ÿ H  …r  u†g
n dS: D

S

S

(49) One can check easily that …j  u†
n ˆ j
…u  n†; …u  …r  H††
n ˆ ÿ…r  H†
…u  n† and …H  …r  u††
n ˆ ÿ…H  n†
…r  u†. Thus the relation (47) follows from (49) (note that u n ˆ 0 on S). & Remark 4. The relation (47) can be extended easily for more general forms of functions j and H, which can be distributions in D (i.e., functionals over a space of infinitely differentiable functions supported in any D0  D). The above remark indicates that one can apply the relation (47) for the present problem. Thus, as a corollary of relation (47), one obtains the following formulas for finding Q and x(0) (cf. Eqs. (9) and (44)): ÿQ
…r  u†…x…0† † ˆ F…u†; where

(50)

Z

F…u† ˆ

‰…g  n†
…r  u† ÿ …g
n†…u
n†Š dS

(51)

S

is a known quantity for any function u satisfying Eq. (46). The set of all possible solutions to the system (46) can be found easily by the method of separation of variables. The complete set of linearly independent functions for u can be written in the following forms: …1† …1† u…11† n12 n13 …x† ˆ V n12 n13 …x2 ; x3 †exp…x1 n12 n13 †; …1† …1† u…12† n12 n13 …x† ˆ V n12 n13 …x2 ; x3 †exp…ÿx1 n12 n13 †; …2† …2† u…21† n23 n21 …x† ˆ V n23 n21 …x3 ; x1 †exp…x2 n23 n21 †; …2† …2† u…22† n23 n21 …x† ˆ V n23 n21 …x3 ; x2 †exp…ÿx2 n23 n21 †;

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467

…3† …3† u…31† n31 n32 …x† ˆ V n31 n32 …x1 ; x2 †exp…x3 n31 n32 †; …3† …3† u…32† n31 n32 …x† ˆ V n31 n32 …x1 ; x2 †exp…ÿx3 n31 n32 †;

where nij (i 6ˆ j) are arbitrary positive integers and: ÿ1 ÿ1 V …1† n12 n13 ˆ …1; 0; 0†sin…d2 n12 x2 †sin…d3 n13 x3 †; ÿ1 ÿ1 V …2† n23 n21 ˆ …0; 1; 0†sin…d3 n23 x3 †sin…d1 n21 x1 †; ÿ1 ÿ1 V …3† n31 n32 ˆ …0; 0; 1†sin…d1 n31 x1 †sin…d2 n32 x2 †; s  2  2 n12 n13 ; ‡ …1† n12 n13 ˆ d2 d3 s  2  2 n23 n21 …2† ‡ ; n23 n21 ˆ d3 d1 s  2  2 n31 n32 ‡ …3† : n31 n32 ˆ d1 d2

Introduce the following notations: …0†

dkÿ1 xk ˆ k ; 1 …ks† …nki ; nkj †; F…u…ks† nki nkj ˆ F 

(52) i 6ˆ j

(53)

for all possible values of indices k, i, j, s, and define: 8 < P1 …n12 ; n13 † ˆ Q3 d2ÿ1 n12 sin…n13 3 †cos…n12 2 † ÿ Q2 d3ÿ1 n13 sin…n12 2 †cos…n13 3 †; P …n ; n † ˆ Q1 d3ÿ1 n23 sin…n21 1 †cos…n23 3 † ÿ Q3 d1ÿ1 n13 sin…n23 3 †cos…n21 1 †; : 2 23 21 P3 …n31 ; n32 † ˆ Q2 d1ÿ1 n31 sin…n32 2 †cos…n31 1 † ÿ Q1 d2ÿ1 n32 sin…n31 1 †cos…n32 2 †: Then it follows from Eq. (50) that: 8 …11† > P …n ; n †exp…1 d1 …1† …n12 ; n13 †; > n12 n13 † ˆ F > 1 12 13 > …1† …12† > P …n ; n †exp…ÿ1 d1 n12 n13 † ˆ F …n12 ; n13 †; > > > 1 12 13 < …21† …n23 ; n21 †; P2 …n23 ; n21 †exp…2 d2 …2† n23 n21 † ˆ F …2† …22† > P …n ; n †exp…ÿ d  † ˆ F …n23 ; n21 †; 2 23 21 2 2 n23 n21 > > > …3† …31† > P3 …n31 ; n32 †exp…3 d3 n31 n32 † ˆ F …n31 ; n32 †; > > > : P …n ; n †exp…ÿ d …3† † ˆ F …31† …n ; n †; 3 31 32 3 3 n31 n32 32 32

(54)

for all positive integer nij, i 6ˆ j. Note that the values for the right-hand sides of the above system are known from the measurements.

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The following lemma is useful for identification of Q and x(0) from the relation (54). Lemma 3. If Pk …nki ; nkj † ˆ 0;

8 nki ; nkj ˆ 1; 2; :::;

for a fixed integer k and i 6ˆ j, then Qj ˆ 0 for all j 6ˆ k. Proof. Without loss of generality, we assume that: P1 …n12 ; n13 † ˆ 0;

8 n12 ; n13 ˆ 1; 2; :::

Then, in particular, one has: 8 < Q3 d2ÿ1 sin3 cos2 ÿ Q2 d3ÿ1 sin2 cos3 ˆ P1 …1; 1† ˆ 0; Q dÿ1 sin23 cos2 ÿ Q2 d3ÿ1 2sin2 cos23 ˆ P1 …1; 2† ˆ 0; : 3 2ÿ1 Q3 d2 2sin3 cos22 ÿ Q2 d3ÿ1 sin22 cos3 ˆ P1 …2; 1† ˆ 0; One can show that: 0 sin3 cos2 rank@ sin23 cos2 2sin3 cos22

1 sin2 cos3 2sin2 cos23 A ˆ 2; sin22 cos3

(55)

(56)

for all 2 2 (0, ), 3 2 (0, ). Indeed, one has:   sin3 cos2 sin2 cos3 I11 …2 ; 3 †  det ˆ ÿsin3 3 sin22 ; sin23 cos2 2sin2 cos23   sin3 cos2 sin2 cos3 I13 …2 ; 3 †  det ˆ sin3 2 sin23 : 2sin23 cos22 sin22 cos3 Obviously, I11(2, 3) ˆ 0 and I13(2, 3) ˆ 0 are true only if 2 ˆ 3 ˆ /2. But in this case:     sin23 cos2 2sin2 cos23  det I23 ; ˆ ÿ4 6ˆ 0: 2sin3 cos22 sin22 cos3 2 ˆ3 ˆ 2 2 2

Thus, Eq. (56) is true for all 2 2 (0, ), 3 2 (0, ), and Eq. (55) can be satisfied only when Q2 ˆ Q3 ˆ 0. & Theorem 3. If H|S  0, then Q ˆ 0. Proof. If H|S  0, then F(u) ˆ 0, which means that Pk(nki, nkj) ˆ 0, i 6ˆ j for all k ˆ 1, 2, 3 and nki, nkj ˆ 1, 2, 3... Thus it follows from Lemma 2 that Q1 ˆ Q2 ˆ Q3 ˆ 0. & Corollary 1. H|S 6 0 if and only if Q 6ˆ 0. Theorem 4. If H|S 6 0, then Q and x…0† are uniquely determined by H|S.

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Proof. Since H|S 6 0, one has Q 6ˆ 0. If Pk(nki, nkj) ˆ 0 for two different k (and for all (nki, nkj), i 6ˆ j) then it follows from Lemma 3 that Q1 ˆ Q2 ˆ Q3 ˆ 0. Therefore, there exist only the following two possibilities: 1. for every fixed k ˆ 1, 2, 3 there exists a set Nk of (nki, nkj), i 6ˆ j, such that Pk …nki ; nkj † 6ˆ 0; if

…nki ; nkj † 2 Nk ;

(57)

2. for one fixed k, Pk …nki ; nkj † ˆ 0;

i 6ˆ j;

…nki ; nkj † ˆ 1; 2; :::;

but for the other two values of k there exist a set Nk of (nki, nkj) (i 6ˆ j) for which the relation (57) holds. …0†

Consider the first possibility. Then one can find xk by the following formula (cf. Eq. (54)) "   #ÿ1=2 X nkj 2 d F …k1† …nki ; nkj † k …0† ; k ˆ 1; 2; 3:::; ln …k2† xk ˆ 2 j6ˆk dj F …nki ; nkj † where (nki, nkj) 2 Nk. Then one can compute 2 3  2 !…1=2† X  n kj …0† 5; Pk …nki ; nkj † ˆ F …k1† …nki ; nkj †exp4 xk dk d j j6ˆk

k ˆ 1; 2; 3; :::;

(58)

(59)

for all integers nki and nkj (i 6ˆ j). Using values Pk(1, 1), Pk(1, 2) and Pk(2, 1) for a fixed k, one can obtain uniquely Qj, for j 6ˆ k. For example, if k ˆ 1 one then has the following algebraic system for Q3 and Q2: 8 < Q3 d2ÿ1 sin3 cos2 ÿ Q2 d3ÿ1 sin2 cos3 ˆ P1 …1; 1†; (60) Q dÿ1 sin23 cos2 ÿ Q2 d3ÿ1 2sin2 cos23 ˆ P1 …1; 2†; : 3 2ÿ1 ÿ1 Q3 d2 2sin3 cos22 ÿ Q2 d3 sin22 cos3 ˆ P1 …2; 1†; where the values for P1(1, 1), P1(1, 2) and P1(2, 1) are determined from Eq. (54) and …ÿ1† …0† k ˆ dk xk ; k ˆ 1; 2; 3. Since Eq. (56) is valid for all 2 2 (0,) and 3 2 (0, ), one can uniquely find Q2 and Q3 from the above system of linear equations. One can obtain a similar result for the other values of the integer k. Therefore, one can obtain Q uniquely by using two different values of k. Now consider the second possibility. Without loss of generality, we assume that P3 …n31 ; n32 † ˆ 0;

n31 ; n32 ˆ 1; 2: …0†

…0†

Then from Lemma 2 one has Q1 ˆ Q2 ˆ 0. The coordinates x1 and x2 can be found from Eq. (58) for …0† some suitable values of n12, n13, n23 and n21. Hence, one only needs to find Q3 and x3 . If …ÿ1† …0† 2  d2 x2 6ˆ 2 one considers the following relations Q3 d1ÿ1 sin3 cos2 ˆ P1 …1; 1†; 2Q3 d1ÿ1 sin23 cos2 ˆ P1 …2; 1†:

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S. He, V.G. Romanov / Mathematics and Computers in Simulation 50 (1999) 457±471 …0†

Since Q3 6ˆ 0 (otherwise H|S ˆ 0), one has P1(1, 1) 6ˆ 0. Thus, Q3, 3 and x3 can be found by the following formulas 8 P1 …2; 1† > > > 3 ˆ arccos 4P …1; 1† ; > > 1 < 3 d3 …0† (61) ; x3 ˆ >  > > > d1 > : Q3 ˆ P1 …1; 1†: sin3 cos2 If 2 ˆ /2, one can consider the following relations ÿ2Q3 d1ÿ1 sin3 ˆ P1 …1; 2†; ÿ2Q3 d1ÿ1 sin23 ˆ P1 …2; 2†: Then Q, 3 and x0 can be determined by the following formulas 8 P1 …2; 2† > > ; 3 ˆ arccos > > 2P > 1 …1; 2† < 3 d3 …0† ; x3 ˆ >  > > > > : Q3 ˆ d1 P1 …1; 2†: 2sin3

(62)

Therefore, Q and x(0) are also uniquely determined by H|S for this case. & Remark 5. From the proof of Theorem 3 one sees that it is sufficient to use F(ks)(nki, nkj), i 6ˆ j, only for the values s ˆ 1, 2, k ˆ 1, 2, 3 and nki, nkj ˆ 1, 2 in order to determine x(0) and Q. 4. Discussion and conclusion In the present paper, the identification of small flaws in a conducting halfspace or rectangular specimen using boundary measurements of the static magnetic field has been considered. The perturbed magnetic field due to the presence of a small flaw is treated as a magnetic field generated by an equivalent current dipole located at the flaw position. Uniqueness theorems have been given for the identification of a single flaw or a set of flaws in a conducting half-space or rectangular specimen. Explicit formulas for the flaw identification in various cases have been given in the proofs of these theorems. The goal of this paper is to image small inclusions, rather than the background. Locations can be imaged accurately using the present explicit formulas. But the physical properties can be imaged only roughly since the inclusions can be modeled only approximately as delta functions. The explicit reconstruction algorithms are the main and most important results of the present paper. These explicit formulas are quite different from the ones given by the authors in [15] where the time-harmonic (! 6ˆ 0) magnetic field at the boundary is measured. The explicit formulas derived there are based on the fact that one can determine the normal derivative …@n H†jS‡ uniquely if …@n H†jSÿ is known by using (cf.

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471

[15]):   n  …r  H† ˆ 0:  ÿ i! S This, however, is not true in the static case since the denominator  ÿ i! in the above expression becomes zero on the vacuum side Sÿ when ! ˆ 0 and one can not extend the normal derivative of the magnetic field from outside to inside (crossing the boundary surface). This makes the explicit identification formulas more complicated to derive in the magnetostatic case. The present analysis and methods can also be used for the quasi-static case (when the frequency is very low), for which the static solution gives a good approximation [16]. Acknowledgements The partial support of the Swedish Research Council for Engineering Sciences is gratefully acknowledged. References [1] V.G. Romanov, Inverse Problems of Mathematical Physics, VNU Science Press, Utrecht, 1987. [2] V. Isakov, Inverse Source Problems, American Mathematical Society, Providence, Rhode Island, 1990. [3] S. He, S. StroÈm, V.H. Weston, Time Domain Wave-splitting and Inverse Problems, Oxford University Press, Oxford, 1998. [4] W.D. Dover, F.D.W. Charlseworth, K.A. Taylor, R. Collins, D.H. Michael, in: G. Birnbaum, G. Free (Eds.), The use of AC field measurements to determine the shape and size of a crack in a metal, Eddy Current Characterization of Materials and Structures, American Society for Testing and Materials, 1981, pp. 401±427. [5] M. Mclver, An inverse problem in electromagnetic crack detection, IMA J. Appl. Math. 47 (1991) 127±145. [6] H. Saotome, T. Doe, S. Hayano, Y. Saito, Crack identification in Metallic Materials, IEEE Trans. Magnetics 29(2) (1993) 1861±1865. [7] R.R. Gans, R.M. Rose, Crack detection in conducting materials using SQUID magnetometry, J. Nondestr. Eval. 12(4) (1993) 199±207. [8] J.R. Bowler, Review of eddy current inversion with application to nondestructive evaluation, Int. J. Appl. Electromag. Mech. 8 (1997) 3±16. [9] M. Hoke, in: B. Kramer (Ed.), SQUID-based measuring techniques ± a challenge for the functional diagnostics in medicine, The Art of Measurement: Metrology in Fundamental and Applied Physics, VCH Verlagsgesellschaft mbH, Weinheim, 1998, pp. 287±335. [10] D. Colton, A. Kirsch, A simple method for solving inverse scattering problems in the resonance region, Inverse Problems 12 (1996) 383±393. [11] M.V. Klibanov, T.R. Lucas, R.M. Frank, A fast and accurate imaging algorithm in optical/diffusion tomography, Inverse Problems 13 (1997) 1341±1361. [12] D.K. Cheng, Field and Wave Electromagnetics, Addison-Wesley, Reading, 1983. [13] A. Sezginer, T.M. Habashy, J.R. Wait, An image method to compute the static magnetic field due to currents injected into a homogeneous, conducting, and magnetically polarizable half space, Radio Sci. 23(1) (1988) 41±45. [14] J.A. Kong, Electromagnetic Wave Theory, Wiley-Interscience, New York, 1986. [15] S. He, V.G. Romanov, Identification of dipole sources in a bounded domain for Maxwell's equations, Wave Motion 28 (1998) 25±40. [16] S. He, Frequency series expansion of an explicit solution for a dipole inside a conducting sphere at low frequency, IEEE Trans. Bio-Med. Engng. 45(10) (1998) 1249±1258.