On finding an obstacle with the Leontovich boundary condition via the time domain enclosure method

An inverse obstacle scattering problem for the wave governed by the Maxwell system in the time domain, in particular, over a finite time interval is considered. It is assumed that the electric field $\mbox{\boldmath $E$}$ and magnetic field $\mbox{\boldmath $H$}$ which are solutions of the Maxwell system are generated only by a current density at the initial time located not far a way from an unknown obstacle. The obstacle is embedded in a medium like air which has constant electric permittivity $\epsilon$ and magnetic permeability $\mu$. It is assumed that the fields on the surface of the obstacle satisfy the impedance-or the Leontovich boundary condition $\mbox{\boldmath $\nu$}\times\mbox{\boldmath $H$} -\lambda\,\mbox{\boldmath $\nu$}\times(\mbox{\boldmath $E$}\times\mbox{\boldmath $\nu$})=\mbox{\boldmath $0$}$ with $\lambda$ an unknown positive function and $\mbox{\boldmath $\nu$}$ the unit outward normal. The observation data are given by the electric field observed at the same place as the support of the current density over a finite time interval. It is shown that an indicator function computed from the electric fields corresponding two current densities enables us to know: the distance of the center of the common spherical support of the current densities to the obstacle; whether the value of the impedance $\lambda$ is greater or less than the special value $\sqrt{\epsilon/\mu}$.


Introduction
In this paper, we consider an inverse obstacle scattering problem for the wave governed by the Maxwell system in the time domain, in particular, over a finite time interval.The formulation of the problem basically follows that of [9].
We assume that the electric field E and magnetic field H are generated only by the current density J at the initial time located not far a way from an unknown obstacle.The obstacle is embedded in a medium like air which has constant electric permittivity ǫ( > 0) and magnetic permeability µ(> 0).On the surface of the obstacle unlike [9] it is assumed that the fields E and H satisfy the impedance-or the Leontovich boundary condition ( [4,12]).
Let us formulate the problem more precisely.We denote by D the unknown obstacle.We assume that D is a non empty bounded open set of R 3 with C 2 -boundary such that R 3 \ D is connected.
Let 0 < T < ∞.The governing equations of E and H over the time interval ]0, T [ take the form Note that ν denotes the unit outward normal to ∂D.To ensure the solvability of the initial boundary value problem (1.1) and (1.2) by the theory of C 0 contraction semigroup [13] we assume that λ ∈ C 1 (∂D) and satisfies λ(x) ≥ 0 for all x ∈ ∂D.Under the condition J ∈ C 1 ([0, T ], L 2 (R 3 \ D) 3 ) we have the unique solution (E, H) which belongs to 2) is satisfied in the sense of the trace [5].The boundary condition in (1.2) is called the Leontovich boundary condition and equivalent to the condition In what follows we use these equivalent forms without mentioning explicitly.
The mathematical role of the existence of the impedance λ can be seen by formally differentiating the energy By the help of the divergence theorem and (1. 2), we have Thus if J (t) = 0 for t ∈ [δ, T ] with a small δ > 0, we have E ′ (t) ≤ 0 therein.The solution loses its energy on the surface of the obstacle.
There should be several choices of current density J as a model of antenna [1,2].In this paper, as considered in [9] we assume that J takes the form J (x, t) = f (t)χ B (x)a, (1.3) where a is an arbitrary unit vector; B is a (very small) open ball satisfying B ∩ D = ∅ and χ B denotes the characteristic function of B; f ∈ C 1 [0, T ] with f (0) = 0. We consider the following problem.
Problem.Fix a large (to be determined later) T < ∞.Generate the solutions E and H of system (1.1) and (1.2) by the source J having the form (1.3) with f = 0 and observe E on B over the time interval ]0, T [.The unit vector a on (1.3) should be taken from a set of three linearly independent vectors.Extract information about the geometry of D and the qualitative state of the distribution of λ over ∂D from the observed data.
As the author knows there is no result for this problem.Note that in [9] the perfect conductive boundary condition ν × E = 0 on ∂D which is the extreme case λ = +∞ of the Leontovich boundary condition has been considered.Thus, therein extracting the geometry of D is the main interest.Here we wish to know the qualitative state of the surface of the obstacle which is described by the unknown function λ on ∂D.

The statement of the results
Let E 0 and H 0 be the solutions of Using E and E 0 on B over time interval ]0, T [, we introduce the indicator function of the enclosure method in this paper where ) and where f (τ ) = T 0 e −τ t f (t)dt. (1.10) Before describing the main result we introduce two conditions (A.I) and (A.II) listed below: Theorem 1.1.Let a j , j = 1, 2 be two linearly independent unit vectors.Let Let f j , j = 1, 2 denote the f given by (1.9) and (1.10) with f above and a = a j .Then, we have: Moreover, in case of both (ii) and (iii) we have, for all T > 2 √ µǫdist (D, B) (1.12) Theorem 1.1 says that the T in the problem should be all T satisfying In particular, if we have a known upper bound M such that M > dist (D, B), then we can choose T = 2 √ µǫM and know the exact value of dist (D, B) via formula (1.12).
Moreover, we have another characterization of dist (D, B).Let M be the same positive constant as above.Assume that λ satisfies (A.I) or (A.II).Given We assume that (1.11) is satisfied with f = f2 √ µǫM .Generate E and H over time interval ]0, 2 √ µǫM[ using J(x, t) = f (t)χ B (x)a j .Measure E on B over time interval ]0, 2 √ µǫM[.Compute also E 0 on B for J above over the same time interval.Using those fields, for each T ∈]0, 2 √ µǫM[ compute W e and V e given by (1.7) and (1.8), respectively.Denote by f T j the f given by (1.6) and compute I f T j (τ, T ) given by (1.5) for f = f T j .
Then, it is easy to see that (1.11) is satisfied also with f = fT for each Formula (1.13) has a similarity in the style as that of the originall enclosure method applied to, for example, the Laplace equation, see (1.3) in [6].See also [8] for similar statements to (ii) and (iii) in Theorem 1.1 for scalar wave equations in the whole space.We think that (i), (ii) and (iii) implicitly represent the finite propagation property of the wave governed by the Maxwell system.It means that if T ≤ 2 √ µǫdist (D, B), then one can not get a qualitative information about the state of the surface of the obstacle (by using the enclosure method).This is consistent with that the propagation speed of the Maxwell system is given by 1/ √ µǫ.However, note that in the proof presented in this paper we do not make use of the finite propagation property.
The results as stated in (ii) and (iii) together with formula (1.12) can be considered as an extension of the corresponding results in [7].More precisely, therein we considered an inverse obstacle scattering problem for the wave governed by the classical wave equation ∂ 2 t u − △u = 0 outside an obstacle which we denote by D again.The wave u as a solution of the equation satisfies ∂u/∂ν − γ∂ t u − βu = 0 on ∂D× ]0, T [.It is assumed that γ and β are essentially bounded functions on ∂D and γ ≥ 0. Using an indicator function which can be computed from a wave generated by a single set of initial data and a special solution of the modified Helmholtz equation, we found that γ ≡ 1 is the special value as same as λ ≡ ǫ/µ of (A.I) and (A.II).It means that the indicator function therein changes the behaviour according to whether γ is greater or less than 1.
For the computation of indicator function (1.5) we need E 0 on B over time interval ]0, T [.This can be done by solving explicitly and analytically the initial value problem (1.4).
Here we present another way which introduces an indicator function using an approximation of (1.8).
Let V 0 e be the weak solution of (1.15) The theoretical advantage of this indicator function compared with (1.5) is that one has no need of computing V e which requires the space time computation of E 0 and H 0 on B over time interval ]0, T [.Instead, in (1.15)  (τ, T ).
A brief outline of this paper is as follows.Theorems 1.1 and 1.2 are proven in Section 2 by using Lemmas 2.1-2.3.Lemma 2.1 gives lower and upper estimates for I f (τ ) as τ −→ ∞ with a single f and the proof is described in Section 3. The proof is based on a rough asymptotic formula of I f (τ ) as τ −→ ∞ which is proved in Subsection 3.2.Lemma 2.2 is concerned with an estimation of the sum of the two indicator functions corresponding to two input current sources in the case when λ satisfies (A.I) or (A.II).It is proved in Section 4. The statements (ii) and (iii) in Theorem 1.1 are direct consequences of Lemmas 2.1 and 2.2.Lemma 2.3 describes a simple estimate of the absolute value of I f (τ ) as τ −→ ∞ which needs for the proof of (i) and (1.12) in Theorem 1.1.The proof is given in Subsection 4.5.The final section is devoted to conclusions and some of problems to be solved in the future.
2 Proof of Theorems 1.1 and 1.2 One can obtain immediately the validity of the statements (ii) and (iii) in Theorem 1.1 from the following two lemmas. and where Let V e, j and V m, j with j = 1, 2 denote V e and V m given by (1.8) and (2.3) using E 0 and H 0 with J = J j .Lemma 2.2.Let Je, j with j = 1, 2 denote the Je given by (2.4) for V e = V e, j and V m = V m, j .
(i) If λ satisfies (A.I), then there exist positive numbers ρ, C ′ and τ 0 such that, for all (ii) If λ satisfies (A.II), then there exist positive numbers ρ, C ′ and τ 0 such that, for all τ ≥ τ 0 For the proof of (i) and formula (1.12) in Theorem 1.1 we need the following lemma.
Lemma 2.3.We have, as τ −→ ∞ This yields (i) of Theorem 1.1.Furthermore a combination of this and Lemma 2.1 gives (1.12).Note that ρ in (i) and (ii) of Lemma 2.2 can be chosen as ρ = 2γ + 5. See the end of the proof of (i) in Subsection 4.3.Note that γ in (1.11) has to be γ ≥ 3/2 by the asymptotics of f (τ ) mentioned above.Theorem 1.2 is a transplantation of Theorem 1.1 via the following simple estimate: ).This together with (2.6) gives Now from this and Lemmas 2.1-2.3 we obtain Theorem 1.2.Thus everything is reduced to giving the proof of Lemmas 2.1-2.3.
3 Proof of Lemma 2.1

Preliminaries
From (1.1) and (1.2) we see that W e and W m satisfy and Taking the rotation of equations on (3.1), we obtain where Vector valued functions V e and V m satisfy Taking the rotation of equations on (3.6), we obtain where From (3.1) and (3.6) we see that R e and R m satisfy where Taking the differences of (3.4) from (3.7), we see that R ⋆ with ⋆ = e, m satisfy (3.12)

Rough asymptotic formula of the indicator function
We start with having the following asymptotic formula of the indicator function.
where Je (τ ) is given by (2.4) and Proof.The proof is divided into three steps.
Step 1.First we show that where This is proved as follows.Integration by parts gives We have Thus Substituting the first equations on (3.4) and (3.7) into this left-hand side, we obtain (3.18) Using W e = V e + R e , one has (3.19) Note that from (3.3), we have and thus From these we obtain Therefore we obtain (3.11) with ⋆ = e and (3.12) and integration by parts give (3.23) Define We have also from the first equation on (3.7) and (3.24) (3.26) Here we make an order of the integrals over ∂D in (3.26).It follows from the second equation on (3.6) that (3.28) It follows from the second equation on (3.9) that and (3.30) A combination of (3.27) and (3.29) gives (3.31) A combination of (3.28) and (3.30) gives It follows from (2.4) and (3.25) that Step 2. We have, as This is proved as follows.
Recall V 0 e is the weak solution of (1.14).Since ), it is easy to see that we have Then from the first equation on (3.7) and (1.14) we have Then applying the first equation on (3.8) to this right-hand side, we conclude that and and Then, the trace theorem [5] yields (3.35) with ⋆ = m.
Step 3. We have, as This is proved as follows.
From (3.9) we obtain (3.44) Here we note that and from (3.20) we have and hence This yields and hence where Thus we have another expression for (3.45): Therefore (3.44) becomes and hence In fact, using the explicit expression of V 0 e outside B, we see that V 0 e together with its derivatives is exponentially decaying as τ −→ ∞ uniformly for all x ∈ D. This yields better estimates than (3.35), (3.42) and (3.43).However, at this stage we do not need such a detailed information about V 0 e .This is a reason why we call (3.13) the rough asymptotic formula.Note that (3.39) is nothing but (2.6).

Finishing the proof of Lemma 2.1
In this subsection we derive the following estimate for Ẽe (τ ) given by (3.14): where V em is given by (3.47).Note that from this together with (3.13) we obtain (2.1) and (2.2).It follows from (3.12) and (3.23) that It follows from (3.50) that Then, from the second equation on (3.9) together with this yields Thus we obtain Now a combination of (3.51) and (3.52) gives Then this together with (3.8) and (3.10) yields (3.49).

Proof of Lemmas 2.2 and 2.3
First we describe the proof of Lemma 2.2.

A reduction
Lemma 4.1 We have and where c = c(x, τ ) ≡ τ µλ(x) and V em is given by (3.47).
4.2 Preliminary computation for the proof of (i) and (ii) 18) in [9] we know that V 0 e is smooth outside B and has the form where The expression (4.10) is a simple application of the mean value theorem [3] for the modified Helmholtz equation and a special form of the fundamental solution of equation (1.14).
From (4.10) we have 4.2.2Computation of (∇ × V 0 e ) × ν By (23) in [9], from (4.10) we have already derived the expression of ∇ × V 0 e outside B: Thus we obtain Combining these with expression where and Thus we obtain Therefore we obtain (4.12)

Proof of (i)
In this and next subsections, to emphasize the dependence on space variable x we write First we prepare an elementary lemma which can be proved by using a contradiction argument.
Lemma 4.2 Given η > 0 there exists a positive number δ such that, for all x ∈ ∂D ∩ B d ∂D (p)+δ (p) Since it follows from Lemma 4.2 that, for all x ∈ ∂D ∩ B d ∂D (p)+δ (p) Since a 1 and a 2 are linearly independent, it is easy to see that, for all x ∈ ∂D we have Since ∂D is compact, this yields that min x∈∂D 2 j=1 |ν x × (a j × ν x )| 2 > 0. Choose a sufficiently small η > 0 in such a way that We have (4.14) Write (4.15) Since we have, for all x ∈ ∂D λ it follows from these and Lemma 4.2 for η < 1/2 we obtain, for all τ > 0 and x ∈ ∂D ∩ B d ∂D (p)+δ (p) Using (4.13) and simply estimating from above, we have Applying this together with (4.16) to (4.14), we have, for all κ > 0 Thus, letting κ = C/(2C ′ ), we obtain, for all x ∈ ∂D ∩ B d ∂D (p)+δ (p) and τ > 0 where Now re-choosing η in such a way that we have Therefore this together with (4.17) yields where Using the same argument done in the proof of Lemma 2.2 in [7], we know that there exists a positive constant C ′′ such that, for all τ >> 1 And it is easy to see that, as τ −→ ∞ where C ′′′ is a positive constant and τ >> 1.Since V 0 e together with its derivatives on ∂D \ B d ∂D (p)+δ (p) is decaying as e −τ √ µǫdist (D,B) e −τ √ µǫδ (see (4.10) and note that d ∂D (p) − η = dist (D, B)), (4.1), (4.18), (4.20) and (1.11) yields (i) with ρ = 2γ + 5.

Write
Thus, choosing η in such a way that Therefore from (4.15) we obtain Thus, choosing τ 0 > 0 in such a way that we have, for all τ ≥ τ 0 and all x ∈ ∂D ∩ B d ∂D (p)+δ (p) Since A ≥ 1, this yields It is easy to see that we have Applying this together with (4.22) to (4.21), we obtain where κ is an arbitrary positive number.Thus letting κ = C/(4C 4 ), we obtain Hereafter the procedure is completely same as that of the proof of (i).Then, it follows from these, (2.1) and (2.2) that (2.5) is valid.

Conclusions and further problems
We have succeeded in extending the previous result in [7] for the scalar wave equation with the dissipative boundary condition to the Maxwell system with the Leontovich boundary condition.It can be also considered as an extension of Theorem 1.1 in [9] for the Maxwell system with the perfect conductivity condition to the Leontovich boundary condition.The main difference is to introduce an indicator function which employs two sets of electric fields corresponding to input sources oriented to two independent directions: I f j (τ ). (5.1) Technically the previous indicator function of Theorem 1.1 in [9] is based on the behaviour of a volume integral of V 0 e over the obstacle which is reduced to that of the fundamental solution of the modified Helmholz equation.In contrast to the previous case, as a consequence of the presence of the impedance λ on the obstacle, the asymptotic behaviour of each of two components of the new indicator function is governed by the behaviour of a surface integral of functions involving V 0 e together with derivatives.Since V 0 e has a directivity, it will be difficult to obtain the same result as Theorem 1.1 in [9] by using only a single component of the indicator function without assuming some restrictive condition for the source direction like (9) in [9].
Our finding is: to obtain the distance of a given point to an unknown obstacle together with a qualitative state of the surface of the obstacle without any restriction on the orientation of the source it is enough to make use of two sets of electric fields generated by two linearly independent input sources supported on a common open ball.
In future research [11] we will consider: extract information about the curvatures and impedance at the points on the surface of the obstacle nearest to the center of the support of input sources from the asymptotic behaviour of indicator function (5.1).This can be considered as an extension of Theorem 1.2 in [9].And also it would be interested to consider the extension of the results in [8] to the Maxwell system.See [10] for a survey on other results using the enclosure method in the time domain together with other open problems.
Note that dist (D, B) = d ∂D (p) − η where p and η are the center and radius of B, respectively and d ∂D (p) = inf x∈∂D |x − p|.Thus knowing dist (D, B) is equivalent to knowing d ∂D (p).
one can compute V 0 e on B in advance by solving only(1.14).Our result on this indicator function is the following.