An iterative algorithm for a Cauchy problem in eddy-current modelling
Transcript of An iterative algorithm for a Cauchy problem in eddy-current modelling
Applied Mathematics and Computation 217 (2010) 237–246
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Applied Mathematics and Computation
journal homepage: www.elsevier .com/ locate/amc
An iterative algorithm for a Cauchy problem in eddy-current modelling
Marián Slodicka *, Valdemar MelicherDepartment of Mathematical Analysis, Ghent University, Galglaan 2, 9000 Gent, Belgium
a r t i c l e i n f o a b s t r a c t
Keywords:Cauchy problemEddy-currentIterative algorithm
0096-3003/$ - see front matter � 2010 Elsevier Incdoi:10.1016/j.amc.2010.05.054
* Corresponding author.E-mail addresses: [email protected] (M.URLs: http://www.cage.ugent.be/~ms/ (M. Slodic
We consider a linear steady-state eddy-current problem for a magnetic field in a boundeddomain. The boundary consists of two parts: reachable with prescribed Cauchy data andunreachable with no data on it. We design an iterative (Landweber type) algorithm forsolution of this problem. At each iteration step two auxiliary mixed well-posed boundaryvalue problems are solved. The analysis of temporary problems is performed in suitablefunction spaces. This creates the basis for the convergence argument. The theoreticalresults are supported with numerical experiments.
� 2010 Elsevier Inc. All rights reserved.
1. Introduction
Let X � R3 be a bounded domain with Lipschitz continuous boundary C, which is split into two complementary, non-empty and non-overlapping parts Cd � C (d – determined) and Cid ¼ C n Cd (id – identify). We would like to solve a lineareddy-current problem for the magnetic field H in X. We suppose that Cid is not reachable and there is no available informa-tion about the solution on this part of the boundary. On the other hand, Cd is well reachable and there is over-determinedboundary data available. The problem is to determine the magnetic field H in X for the following steady-state eddy-currentproblem.
Problem 1. Find H 2 H(curl,X) such that
Hþr�r�H ¼ f in X;
along with
H� m ¼ ~H� m
r�H� m ¼ ~C� m
)on Cd;
where ~H and ~C are given.Data completion problems like Problem 1 are known as Cauchy problems. They have applications in corrosion detection,
cf. [14], soil venting [12,22], the temperature reconstruction [15], crack detection [5], medical imaging [11] etc.One possible application of Problem 1 is passive shielding, see [21]. Induction heater equipment is exploited for the ther-
mal treatment of metallic specimen. This thermal treatment is obtained by huge eddy currents, induced in the conductivespecimen. The eddy currents result from the time dependent magnetic field, generated by an excitation coil. This gives riseto magnetic field levels that may be significantly higher than the allowed reference levels. Passive shielding is one of the
. All rights reserved.
Slodicka), [email protected] (V. Melicher).ka), http://www.cage.ugent.be/~valdo/ (V. Melicher).
238 M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246
techniques for minimizing the magnetic field inside a given area which otherwise could have negative effects on operator orequipment.
It can happen that one side of the shield is not accessible for measurements for some reasons. But one has this missinginformation at the accessible part of the boundary where both the magnetic field H � m and the electric fieldr� H � m couldbe measured and then Problem 1 could be solved. However, we have to pay the price that this data completion problem is ill-posed.
The ill-posedness of Cauchy problems, particularly of the Cauchy problem for the Laplace equation, was for the first timestudied by Hadamard [13]. The amount of literature concerning Cauchy problems is vast. Some recent theoretical results in-clude [2,3], where the reader may find further references.
Cauchy problems enjoy also a considerable interest of numerical analysts. We refer the reader to the interestingarticle [1], where the authors propose to solve symmetric linear elliptic Cauchy problems by minimizing an energy-like functional. This delivers exceptionally robust method when compared with the well-known alternating KMF(Kozlov–Maz’ya–Fomin) algorithm [16]. See [1] for the overview of other numerical methods used to solve Cauchyproblems.
When analyzing Problem 1, one encounter a crucial difference between D and r�r� operators – the regularity of thecorresponding solutions. In the case of standard linear elliptic problems, the H1(X) regularity of a solution is usually ob-tained. However, the energy space of Maxwell’s system is only H(curl,X). It affects also the trace of a solution. H1/2(C) reg-ularity on the boundary cannot be justified, see Theorem 1.
The definitions of the function spaces under consideration can be found in Buffa and Ciarlet Jr. [6,7]. The characterizationof trace spaces for Lipschitz polyhedra is far more complicated than that for smooth domains. First, some of the definitionsmake no sense anymore. Since the normal unit vector m is discontinuous, only m 2 L1(X), the scalar product v � m is not de-fined for v 2 H�1/2(oX), thus the condition v � m = 0 does not make sense. Consequently, all the results gained in the case of asmooth domain need a careful reinterpretation. This has been done in [6,7].
Melicher and Slodicka [17] have solved Problem 1 by minimizing a cost-functional
FðxÞ :¼ kHðxÞ � m � ~H� mk2H�1=2k;00ðdivCd
;CdÞþ gkHðxÞk2
Hðcurl;XÞ; ð1Þ
where the second term is a Tikhonov-type regularization with a constant g > 0 and H(x) solves the following problem
HðxÞ þ r� ðr�HðxÞÞ ¼ f in X;
r�HðxÞ � m ¼ ~C� m on Cd;
r�HðxÞ � m ¼ x� m on Cid:
ð2Þ
The authors showed that the steepest-descent method can be applied to generate a minimizing sequence xk
xkþ1 ¼ xk � akF 0ðxkÞ ð3Þ
starting from an initial guess x0 with a controlled step ak. The Frechét derivative F0(x) is understood in the weak sense. Here,we prefer a different approach.
1.1. Iterative procedure for Problem 1
Let ~h be the unique solution to the following temporary problem
eh þr�r� ~h ¼ f in X;
r� ~h� m ¼ ~C� m on Cd;
~h� m ¼ 0 on Cid:
ð4Þ
The Problem 1 is equivalent (setting H ¼ hþ ~h) to the following one
Problem 2. Find h 2 H(curl,X) such that
hþr�r� h ¼ 0 in X;
along with
h� m ¼ ~H� m � ~h� m ¼ ~h� m
r� h� m ¼ 0
)on Cd;
where ~h is given.We suggest the following iterative algorithm to find a solution to Problem 2.
Step 1: Choose g = gk�1 and solve
1 We1 and c
M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246 239
uþr�r� u ¼ 0 in X;
r� u� m ¼ 0 on Cd;
u� m ¼ g� m on Cid:
ð5Þ
In this way we define an operator T1: Cid ? Cd such that T1(csg) = csu.1
Step 2: Solve
v þr�r� v ¼ 0 in X;
r� v � m ¼ 0 on Cid;
v � m ¼ ðu�~hÞ � m on Cd:
ð6Þ
In this way we define an operator T2: Cd ? Cid such that T2ðcsu� cs~hÞ ¼ csv.
Step 3: Set
gk ¼ gk�1 � bv; ð7Þ
with a sufficiently small positive b.Step 4: Increase k and go back to Step 1.
We seek an g such that T1ðcsgÞ ¼ cs~h. Thus
csgk ¼ csgk�1 � bcsv ¼ csgk�1 � bT2 csu� cs~h
� �¼ csgk�1 � bT2 T1ðcsgk�1Þ � T1ðcsgÞð Þ
¼ csgk�1 � bT2T1ðcsgk�1 � csgÞ: ð8Þ
This can be also written as
csðgk � gÞ ¼ ðI � bT2T1Þcsðgk�1 � gÞ:
Let us note that the procedure (8) is a Landweber type algorithm. The convergence of this method has been analyzed in ab-stract Hilbert spaces, e.g., Engl et al. [10] or Rieder [20] under the condition that T1 2 LðX;YÞ; T2 ¼ T�1 and 0 < b < 1
kT1k2 with
appropriate norms and starting from a reasonable start condition. We are in a different situation because T2–T�1. First, wehave to choose suitable function spaces with their norms adapted for our purposes. Then we have to show the boundednessof both operators T1 and T2. Further we need to prove convergence of a generalized Landweber iteration (8) – see Theorem 4and/or the more general result formulated in Theorem 3.
Remark 1. We have designed our algorithm as Dirichlet-to-Dirichlet map, i.e., we solve problems with mixed BCs. The caseDirichlet-to-Neumann map is also possible, but then one has to solve Dirichlet problem on the whole boundary, which maybring some troubles with the regularity of a solution at the boundary points, where Cd and Cid meet each other. Let us notethat the steepest-descent method of [17], when g = 0, can also be seen as a relaxed Landweber method.
2. Notation
We are going to work in a variational framework. We will use the following notation – cf. Cessenat [9], Monk [19]
L2ðXÞ :¼ u 2 L2ðXÞ3n o
; ð�; �Þ its scalar product:
Hðcurl;XÞ :¼ u 2 L2ðXÞ : r� u 2 L2ðXÞn o
;
Hðdiv;XÞ :¼ u 2 L2ðXÞ : r � u 2 L2ðXÞn o
;
L2t ðCÞ :¼ u 2 L2ðCÞ : u � mjC ¼ 0
n o; h�; �it its scalar product:
As we suppose only Lipschitz regularity of the domain X, we have to use theory developed in Buffa and Ciarlet Jr. [6,7]. Thelast two articles treat the problem of traces for Maxwell’s equations in Lipschitz polyhedra in more detail than it has beendone in Buffa et al. [8] for general Lipschitz domains. Thus, we follow the terminology and use the results in Buffa and CiarletJr. [6,7]. The case of a Lipschitz polyhedron is general enough for our purposes. In the next we review some of the most nec-essary notations and results from two above-given articles. We paraphrase these in notation used here. Let us set
H1=2� ðCÞ :¼ u 2 L2
t ðCÞ : ujCj2 H1=2ðCjÞ; 1 6 j 6 N
n o;
where Cj are faces of the polyhedron.
apply here the notation, that g can be either a function from a suitable trace space or its prolongation to X, which exists according to the trace Theoreman be found using Lemma 1. The meaning is clear from the context.
240 M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246
Definition 1. The ‘‘tangential components trace” mapping ps : C1ðXÞ ! H1=2� ðCÞ and the ‘‘tangential trace” mapping
cs : C1ðXÞ ! H1=2� ðCÞ are defined as u ´ m � (u � m)jC and u ´ u � mjC, respectively.
These mappings can be extended by continuity to linear continuous mappings from H1(X) to H1=2� ðCÞ.
Let us set
2 InH�1=2k ðd
H1=200 ðCdÞ :¼ u 2 H1=2ðCdÞ : ~u 2 H1=2ðCÞ
n o;
H�1=200 ðCdÞ :¼ ðH1=2
00 ðCdÞÞ0;
H0;Cidðcurl;XÞ :¼ u 2 Hðcurl;XÞ : u� mjCid
¼ 0 in H�1=200 ðCÞ
3n o
;
where ~u is the prolongation of u by zero to C. Now, we state Buffa and Ciarlet Jr. [7, Theorem 6.6].
Theorem 1. The trace mapping cds : Hðcurl;XÞ ! H�1=2
k;00 ðdivCd;CdÞ ( respectively its restriction cd;0
s : H0;Cidðcurl;XÞ !
H�1=2k ðdiv0
Cd;CdÞ) which associates to a vector field u 2 H(curl,X) (resp. to u 2 H0;Cid
ðcurl;XÞ) its tangential components onCd, that is u� mjCd
, is linear continuous and admits a continuous inverse.For definitions of the Hilbert spaces H�1=2
k ðdiv0Cd;CdÞ and H�1=2
k;00 ðdivCd;CdÞ, as suitable subspaces of H1=2
� ðCÞ, we refer thereader to Buffa and Ciarlet Jr. [6,7], because this exceeds the scope of our paper. Analogical theorem holds true also for psand indeed for the case of the whole boundary as well.
2.1. The quotient norm
First, we state the characterization of elements from H�1=2k;00 ðdivCd
;CdÞ in terms of their boundary values - see Melicher andSlodicka [17].2
Lemma 1. The minimum of the quotient norm of a m 2 H�1=2k;00 ðdivCd
;CdÞ
kmk2H�1=2k;00ðdivCd
;CdÞ¼ inf
u2Hðcurl;XÞu�mjCd
¼m
kuk2Hðcurl;XÞ
is reached in the solution of the following problem
uþr�r� u ¼ 0 in X;
r� u� m ¼ 0 on Cid;
u� m ¼m on Cd:
ð9Þ
From the fact that u 2 H(curl,X) we can also deduce that r�r� u 2 L2(X). This follows from the following reasoning.The formula u +r�r� u = 0 is valid in a weak sense for any u 2 H0(curl,X). So, we can write
r�r� u;uð Þ ¼ ð�u;uÞ:
We can see that the functional on the right-hand side can be prolonged according to Hahn–Banach theorem to a functionalon L2(X) because of
ð�u;uÞ 6 kukL2ðXÞkukL2ðXÞ:
Thus, there also exists the extension gr�r� u of r�r� u to L2(X). Moreover we have
k gr�r� ukL2ðXÞ ¼ kukL2ðXÞ:
From now on we identify gr�r� u and r�r� u.On the basis of the last lines we have v =r� u 2 H(curl,X). This means that we can apply the integration by parts for-
mula from [6] to system (9) and similar ones.
3. Stability
We adopt the following assumption on the data from Problem 2
cs~h 2 H�1=2
k;00 ðdivCd;CdÞ: ð10Þ
Let us note that (10) can also be reformulated in terms of conditions for data from Problem 1, which can be of the form
f 2 L2ðXÞ; cs~C 2 H�1=2
k;00 ðdivCd;CdÞ; cs
~H 2 H�1=2k;00 ðdivCd
;CdÞ:
a similar way we can give a characterization of elements from H�1=2k;00 ðdivCid
;CidÞ. The lemma also holds if H�1=2k;00 ðdivCd
;CdÞ is substituted byiv0
Cd;CdÞ.
M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246 241
In some applications the solution of Problem 1 should be divergence free. This can be assured by an additional postulate thatf is divergence free.
Further we suppose that the first guess in the Step 1 obeys
csg0 2 H�1=2k;00 ðdivCd
;CdÞ: ð11Þ
Note, that the structure of systems (5) and (6) is the same, only the location (Cd or Cid) of Dirichlet and Neumann boundarydata is swapped. Both Dirichlet data belong to the qualitatively same space. Since g is generated by (7), being thus the traceof a H(curl,X) function, it certainly belongs to the space H�1=2k;00 ðdivCid;CidÞ. The Dirichlet data of (6), belongs to the space
H�1=2k;00 ðdivCd
;CdÞ. So it is sufficient to prove, that only one of the systems is well-posed, e.g. (5).
Lemma 2. For any csg 2 H�1=2k;00 ðdivCid
;CidÞ there exists a unique solution to (5) u 2 H(curl,X). Moreover we have
kukHðcurl;XÞ ¼ kcsgkH�1=2k;00ðdivCid
;CidÞ: ð12Þ
Proof. We know from Theorem 1 that cs is surjective, i.e., there exists g 2 H(curl,X) such that g ¼ cids ðgÞ. We seek the solu-
tion u in the form u = w + g, where
ðw;uÞ þ ðr �w;r�uÞ ¼ �ðg;uÞ � ðr � g;r�uÞ ð13Þ
for any u 2 H0;Cidðcurl;XÞ. We see that for a fixed g, the map l : H0;Cid
ðcurl;XÞ ! R, defined by
lðuÞ ¼ �ðg;uÞ � ðr � g;r�uÞ;
is a linear functional on H0;Cidðcurl;XÞ. Moreover, l is bounded since
jlðuÞj 6 CkgkHðcurl;XÞkukHðcurl;XÞ:
By the Riesz representation, we know that there exists a unique solution w 2 H0;Cidðcurl;XÞ to (13). Thus, we have the exis-
tence of the solution u to (5). Note, that the form of (5) is identical to the form of (9), which means that the solution to (5) isthe H(curl,X) function, where the infimum of the quotient norm is achieved, i.e., (12) holds. This also gives theuniqueness. h
Thus, the operator T1: Cid ? Cd defined by the Step 1 of our algorithm can be also seen as an operator
T1 : H�1=2k;00 ðdivCid
;CidÞ ! H�1=2k;00 ðdivCd
;CdÞcsg # csu:
Moreover, the operator T1 is continuous because of Theorem 1 and Lemma 2:
kcsuk2H�1=2k;00ðdivCd
;CdÞ6 Ckuk2
Hðcurl;XÞ ¼ Ckcsgk2H�1=2k;00ðdivCid
;CidÞ:
The operator T1 is defined on the whole space H�1=2k;00 ðdivCid
;CidÞ. We have already shown that T1 2 L H�1=2k;00 ðdivCid
;�
CidÞ;H�1=2k;00 ðdivCd
;CdÞÞ. This implies that R(T1) is closed, i.e., T1 is an closed operator.Let us denote by Ty1 the Moore–Penrose (see Engl et al. [10, p. 33]) generalized inverse of T1. This is defined as the unique
extension of eT�11 to DðTy1Þ ¼ RðT1Þ � RðT1Þ? with NðTy1Þ ¼ RðT1Þ?, where
eT 1 :¼ T1jNðT1Þ? : NðT1Þ? ! RðT1Þ:Using Engl et al. [10, p. 34]) we can see that the Moore–Penrose inverse Ty1 has a closed graph and moreover Ty1 is bounded.Now, let us look on the operator T2: Cd ? Cid. Applying an analogous way for deduction as we used for the operator T1 we
can see the operator T2 as
T2 : H�1=2k;00 ðdivCd
;CdÞ ! H�1=2k;00 ðdivCid
;CidÞ;
csu� cs~h # csv:
According to the fact that cs~h is a fixed element from H�1=2
k;00 ðdivCd;CdÞ – see (10) – we also deduce that
T2 2 LðH�1=2k;00 ðdivCd
;CdÞ;H�1=2k;00 ðdivCid
;CidÞÞ.
4. Convergence
The standard proof of convergence for Landweber’s iterations is given for T2 ¼ T�1 and T1 2 LðX;YÞ – see, e.g., Engl et al.[10, Theorem 6.1]. The proof relies on spectral theory for self-adjoint operators and on the interpretation
ðT�TÞ�1 ¼Z
1k
dEk;
242 M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246
where Ek is a spectral family for T*T. Our proof of convergence for generalized Landweber’s iterations will be based on anoperational calculus for linear bounded operators, cf. Taylor [23, Section 5.6]. If we denote BR = {k;jkj 6 R}, then one can de-fine a function of an operator A 2 LðX;YÞ as
f ðAÞ ¼ 12pi
ZC
f ðkÞðkI � AÞ�1dk;
where f is a holomorphic function in a domain containing BkAk (and thus also the spectrum r(A)) and C is a simple closedcontour oriented counterclockwise, enclosing BkAk (and thus C lies in the resolvent set q(A)). Let us note that the integralis independent of the choice of C � q(A), which follows from the Cauchy formula. Therefore we may consider the followingsituation – see Fig. 1, with some small e > 0. We want to solve
T1x ¼ y: ð14Þ
If the operator T2T1 is continuously invertible, then the solution can be characterized as
xy ¼ ðT2T1Þ�1T2y ¼ 12pi
ZC
1kðkI � T2T1Þ�1T2ydk:
The problem is that the function 1k is not holomorphic in 0. But this integral can be interpreted as a limit of a sequence of well-
posed integrals if y 2 DðTy1Þ, namely
xa ¼ gaðT2T1ÞT2y ¼ gaðT2T1ÞT2T1xy ¼ 12pi
ZC
gaðkÞkðkI � T2T1Þ�1xydk;
assuming that ga(k)k are nice functions. More exactly:
Theorem 2. Let X, Y be Banach spaces, T1 2 LðX;YÞ; T2 2 LðY;XÞ, A = T2T1. Assume that for all a > 0, ga(k) are holomorphicfunctions for jkj < kAk + e, for some e > 0 along with
jkgaðkÞj 6 C 8a > 0;
and
lima!0
kgaðkÞ ¼ 1 for jkj < kAk þ e:
Then for all y 2 DðTy1Þ we have
lima!0
xa ¼ lima!0
gaðT2T1ÞT2y ¼ xy:
If y R DðTy1Þ then
lima!0kgaðT2T1ÞT2yk ¼ 1:
Γ
εσ(A)
||A|| ||A|| +
Fig. 1. A curve C in the resolvent set q(A).
M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246 243
Proof. It holds
xy � xa ¼ xy � gaðT2T1ÞT2y ¼ xy � gaðT2T1ÞT2T1xy ¼ ðI � gaðAÞAÞxy ¼1
2pi
ZCð1� gaðkÞkÞðkI � AÞ�1xydk:
A simple calculation yields that
kxy � xak 6 CkxykZ
Cj1� gaðkÞkj jdkj;
which together with the Dominated Convergence Theorem gives
lima!0
xa ¼ xy:
The fact that kxak?1 if y R DðTy1Þ follows from [10, Proposition 3.6]. We apply the method of successive approximationsto
x ¼ xþ bT2ðy� T1xÞ;
as follows
xk :¼ xk�1 þ bT2ðy� T1xk�1Þ ð15Þ
in order to solve (14). Without loss of generality we put x0 = 0. Getting rid of the recursion we obtain
xk ¼Xk�1
j¼0
ðI � bT2T1ÞjbT2y:
Assuming y 2 DðTy1Þ we see that T2y = T2T1x� for xy ¼ Ty1y. Therefore
xk ¼Xk�1
j¼0
ðI � AÞjAxy;
where A = bT2T1. Let us suppose that 0 < b < 1kT1kkT2k
. Then clearly kAk < 1. Using the identity
I � ðI � AÞk ¼ AXk�1
j¼0
ðI � AÞj;
we deduce that
xy � xk ¼ ðI � AÞkxy:
We introduce the following function gkðkÞ :¼Pk�1
j¼0 ð1� kÞj. We see that gk(k) is holomorphic for jkj 6 1 and
limk!1
kgkðkÞ ¼ 1;
and jkgk(k)j 6 C for jkj < 1. Then the following theorem is a consequence of Theorem 2. h
Theorem 3. Let X, Y be Banach spaces, T1 2 LðX;YÞ; T2 2 LðY;XÞ;0 < b < 1kT1kkT2k
and y 2 DðTy1Þ. Then the Landweber iterations(15) fulfills
limk!1
xk ¼ Ty1y:
If y R DðTy1Þ, then kxkk?1 as k ?1.Now, we can state the convergence result for our iteration scheme (8).
Theorem 4. Let (10) be satisfied. Moreover we assume that ~h 2 DðTy1Þ and the relaxation parameter fulfils 0 < b < 1kT1kkT2k. Then
the iteration scheme (8) converges, more exactly, gk ! Ty1~h. If ~h R DðTy1Þ, then the scheme diverges.
5. Numerical experiments
We employ the finite element method for the space discretization in all Eqs. (5), (6) etc., particularly we use Whitney’sedge elements of the first order [19].
Let us define a relative residuum
rðgÞ :¼kT g�~hdkH�1=2
k;00ðdivCd
;CdÞ
k~hdkH�1=2k;00ðdivCd
;CdÞ
;
244 M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246
where ~hd are noisy data with a noise level d. We applied the discrepancy principle as the stopping rule for the algorithm, i.e.,
Fig. 2.
rðgkþ1Þ < sd; s > 0; ð16Þ
together with the control of the reduction speed of the residuum
rðgkÞ � rðgkþ1ÞrðgkÞ
> e; e > 0: ð17Þ
If any of the conditions is fulfilled, the algorithm stops. The e and s are given constants here. When no noise is present, thesecond condition ensures the finite number of iterations. We put e = 10�3 and s = 1.3.
Remark 2. One often asks why not to transform the Eq. (5) using the well-known vector formula r�r� u =�D � u +r(r � u)? Then, taking into account that r � u = 0 in many situations, one can substitute the problematicr�r� �operator with the Laplacian. First, one has in fact r� (r�1r� u) or similar, where r is the conductivity of material. If wesuppose, that r is smooth enough, we could use a bit more complicated formula to obtain the term r � (r�1ru). However,we would obtain also other terms and we would find out, that the symmetry is broken. Second, when one would perform thetransformation, one would very probably use the Lagrange elements to discretize the resulting system. These enforce thecontinuity of the resulting solution. However, the Maxwell equations require the continuity of the tangential component,only. The normal part has natural jumps across interfaces. Together, r� � operator is natural for Maxwell’s system and itsburden should be carried. A interested reader should reach for [4,19].
Let the unite cube (0,1)3 be the computational domain. Let us take
Hexact ¼z� yx� z
y� x
0B@1CA
as an exact solution. We suppose that the boundary conditions are known everywhere except the face x = 1. We take a meshwith 4905 degrees of freedom (DOFs). The iterative process for different noise levels is depicted in Fig. 3. Fig. 2 shows thecorresponding recovered solution at face x = 1.
Recovered solutions at face x = 1; (a)–(d) depict the exact solution and results with 0%, 1%, 10% noise level, respectively; colour denotes a magnitude.
0 20 40 600
0.2
0.4
0.6
0.8
0%1%10%
Fig. 3. Iterative process for different noise levels; number of iterations versus relative H(curl,X) error.
0 20 40 60 800
0.2
0.4
0.6
0.8
1
errorres
Fig. 4. The plot of relative H(curl,X) error against the residuum.
M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246 245
Note, that the convergence of the Landweber scheme is very slow. But, in practice usually we are not interested in findingthe best-approximate solution (the least square solution of minimal norm), to which the method converges. If we are search-ing for a reasonably smooth function, a good approximation is usually obtained after a few iterations, see Fig. 3. Fig. 4, wherethe residuum and the relative error are plotted against the number of iterations, gives one a good heuristic stopping crite-rion. When the character of the residuum becomes linear, the minimum is achieved. It can be easily justified in one-dimen-sional case.3
6. Conclusion
The Cauchy problem can be generally attacked by various numerical methods. Most of them are based on an appropriateregularization. All of them have a different speed of convergence and accuracy – cf. [18]. We have studied an identificationproblem for the steady-state eddy-current model on a polyhedral domain X. The over-determined boundary data are knownon the reachable part Cd of the boundary C. The task was to determine the tangential component of the magnetic field on theunreachable part Cid of C together with the magnetic field inside the domain.
We have worked in the variational framework in natural spaces for magnetic fields, thus no additional regularity had tobe assumed. We designed an iterative algorithm of the Landweber type, which acted between Cd and Cid. At each step of thisscheme two auxiliary problems were solved. These were described in terms of abstract operators T1 and T2. The suitablechoice of function spaces enabled to prove appropriate properties of T1 and T2. The convergence of the proposed schemewas discussed.
3 Take the operator Tx = ax + b, take x = x + h and look at it as a function of h.
246 M. Slodicka, V. Melicher / Applied Mathematics and Computation 217 (2010) 237–246
There arises a natural question about the compactness of the operator T2T1. This is closely coupled with the regularity ofsolutions of direct problems. We recall that – in comparison to steady-state heat equation – we do not have enough infor-mation about the trace of a solution. The normal component is missing! Therefore one can expect lower regularity of a solu-tion. That is why we have not been able to overcome some difficulties by the proof of compactness. This remains for us as achallenge in the future work.
Acknowledgements
This work has been sponsored by the BOF/GOA – Project No. 01G006B7 of Ghent University, Belgium and the Grant No.3G008206 of the Research Foundation – Flanders.
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