On finding a buried obstacle in a layered medium via the time domain enclosure method

An inverse obstacle problem for the wave equation in a two layered medium is considered. It is assumed that the unknown obstacle is penetrable and embedded in the lower half-space. The wave as a solution of the wave equation is generated by an initial data whose support is in the upper half-space and observed at the same place as the support over a finite time interval. From the observed wave an indicator function in the time domain enclosure method is constructed. It is shown that, one can find some information about the geometry of the obstacle together with the qualitative property in the asymptotic behavior of the indicator function.


Introduction
The problem of finding an obstacle embedded or hidden in a complicated environment by using the electromagnetic wave appears in, for example, ground penetrating or subsurface radar [4] and the through-wall imaging [2].
In this paper, we consider such type of the problems in a simplest, however, important mathematical model which employs a wave governed by a scalar wave equation in a two homogeneous layered medium over a finite time interval.
Let D be a bounded open subset of R 3 − with a C 2 -boundary and satisfy D ⊂ R 3 − . Consider γ ∈ L ∞ (R 3 ) given by where h = h(x), x ∈ D is a real symmetric 3 × 3-matrix valued function and satisfies that: all the components of h are essentially bounded on D ; there exists a positive constant C such that (γ 0 (x)I 3 + h(x))ξ · ξ ≥ C|ξ| 2 for all ξ ∈ R 3 and a.e. x ∈ D.
Let 0 < T < ∞. Given f ∈ L 2 (R 3 ) let u = u f (x, t) be the weak solution of the initial value problem: (1.1) Note that the solution class is taken from [5]. See also [10] for its detailed description.
We consider the following problem: Problem. Assume that γ + = γ − . Fix a large T (to be determined later). Assume that γ 0 is known and that both D and h are unknown. Let B be an open ball whose closure is contained in R 3 + . Fix a f ∈ L 2 (R 3 ) with supp f ⊂ B and satisfying that there exists a positive constant C 0 such that f (x) ≥ C 0 a.e. x ∈ B or −f (x) ≥ C 0 a.e.
x ∈ B. Generate u = u f of the solution of (1.1) by the f . Extract information about the location and shape of D from the measured data u on B over the time interval ]0, T [. It should be emphasized that the problem asks us to extract information about unknown obstacle D from a single wave observed over a finite time interval at the same place where the wave is generated. There are some studies which consider the time harmonic reduced case in a two layered medium. See [17] for uniqueness issue of impenetrable obstacles using infinitely many incident fields; [18] a reconstruction scheme of an impenetrable obstacle using a far-field pattern corresponding to a single incident plane wave; [6] study of a direct problem with an application to mine detection and propose a numerical reconstruction scheme using near field measurements corresponding to finitely many incident sources. Clearly our problem formulation is different from their one and to our best knowledge there is no result for the problem.
In [10] Ikehata has considered the case when γ + = γ − (= 1) and the wave is observed on a closed surface S over a finite time interval which encloses the obstacle. He assumed that γ satisfies one of the following two conditions: (A1) there exists a positive constant C ′ such that −h(x)ξ · ξ ≥ C ′ |ξ| 2 for all ξ ∈ R 3 and a.e. x ∈ D; (A2) there exists a positive constant C ′ such that h(x)ξ · ξ ≥ C ′ |ξ| 2 for all ξ ∈ R 3 and a.e. x ∈ D.
In Theorem 1.2 of [10] he showed that if B is outside surface S, then one can extract the distance dist (D, B) = inf x∈D,y∈B |x − y| from the observed wave and also can distin-guish whether (A1) or (A2) is satisfied by using the signature of an indicator function computed from the observed wave. This is the beginning of the multi-dimensional version of the time domain enclosure method [9] for inverse obstacle scattering in the time domain. In [11] this idea has been extended to the case when the wave is observed on the same place as the support of an initial data. This is a version of the near field inverse back-scattering problem. One can easily transplant the results in [10] to this case as pointed out in Subsection 1.3 of [11]. However, the case when γ + = γ − is not trivial. Clearly this is a quite interesting case from practical and mathematical point of view. The unknown obstacle is embedded in the lower half-space which has a different refractive index from the upper half-space. Thus the wave generated by an initial data produces reflected and refracted waves at the interface. The produced refracted wave hits the surface of the obstacle and generates reflected and refracted waves. How can one extract information about the geometry of the obstacle from the observed wave? The aim of this paper is to extend the previous result to the case when γ + = γ − using the enclosure method in the time domain. Now let us describe our main result. Let τ > 0. Let u be the solution of (1.1). Define We call the function τ −→ I f (τ, T ) the indicator function. Note that this symbol follows from that of [12]. and The quantity l(D, B) corresponds to the optical distance or optical path length between B and D in optics and it is easy to see that we have where p and η denote the center and radius of B, respectively and Thus the unknown obstacle D is contained in the set The following theorem is the main result of this paper.
Theorem 1.1 Assume that γ + < γ − . Then, we have: Moreover, if γ satisfies (A1) or (A2), then for all T > 2l(D, B) From Theorem 1.1 we see that the T in the problem should be an arbitrary number satisfying T > 2l(D, B). We think that this is optimal. From the indicator function one gets the value l(D, B) and thus the set E(D; B, γ + , γ − ) which encloses D. Moreover, one can distinguish whether unknown obstacle D satisfies (A1) or (A2) which is a qualitative property of D relative to the surrounding background medium, by checking the asymptotic behavior of the indicator function. Remark 1.2 Intuitively, any signal emanating from B reaches D. For these signals to go back to the upper side, we need to catch the refracted waves of the reflected waves by D. Hence we need to take measurement in B at least till time 2l(D, B) for which the fastest signals may come back. To check whether signals are exactly coming back, we need to take account of total reflection waves. Assumption γ + < γ − means that the propagation speed of waves in the upper side is slower than that of the lower side. Hence, there is no total reflected wave for the incident waves from the lower side. This is the case that we do not need to take care of it. Mathematically, this is appeared as a difficulty for obtaining asymptotics of the refracted wave. As is in (1.14) and (1.15) below, it is relatively simple since it does not contain waves for total reflection.
The proof of Theorem 1.1 employs two important facts. The first one is the following lemma.
For the proof see Appendix. Combining (1.6) and (1.7) under the assumption (A1) or (A2), we can easily see that Theorem 1.1 can be proved if one has the following fact concerning with the asymptotic behavior of ∇v on D as τ −→ ∞. Theorem 1.4 Assume that ∂D is C 1 and that γ + < γ − . Then, there exist positive numbers C and τ 0 such that, for all τ ≥ τ 0 we have This is the second important fact. We found that the proof of Theorem 1.4 is not a simple matter and thus the remaining part of this paper is devoted to the proof. In this sense, the main contribution of this paper to the enclosure method in the time domain is the establishment of the estimate (1.8). Note that in [16,3] one can find some formal asymptotic computation of the solution of (1.3), however, we do not know whether or not their formal theory enables us to derive estimate (1.8).
In [12], Ikehata considered a mathematical model of the through-wall imaging by using the enclosure method in the time domain. Originally the governing equation should be the Maxwell system, however, as a first step, it is assumed that the governing equation is given by the single wave equation The assumption on the function α ∈ L ∞ (R 3 ) is that: α has a positive lower bound in R 3 and takes the form where the function α 0 is essentially bounded in R 3 with a positive lower bound; D is an arbitrary bounded open set of the whole space with a Lipschitz boundary; the function h(x) or −h(x) has a positive essential infimum on D.
Remarkably enough, in [12] a higher regularity more than the essential boundedness of α 0 is not assumed. Thus the model covers various background media such as multilayered media with complicated interfaces or unions of various domains with different refractive indexes. He showed that an indicator function computed from the wave observed on the same place as the support of an initial data yields lower and upper estimates of the distance dist (D, B) together with a criterion whether ess. inf x∈D h(x) > 0 or ess. inf x∈D (−h(x)) > 0 provided α 0 is known. The result is based on a system of inequalities similar to (1.6) and (1.7) in Lemma 1.3 and explicit upper and lower estimates of the solution of the equation In contrast to this result, Theorem 1.1 tells us that one can extract the exact value of the optical distance l(D, B) from the asymptotic behavior of the indicator function under the assumption that the background medium consists of two isotropic homogeneous layered media. It would be possible to apply the idea of the derivation of the estimate in Theorem 1.4 to the case when α 0 takes two different constants, α + in x 3 > 0 and α − in x 3 < 0 provided α + > α − which corresponds to γ + < γ − . However, a typical case to be considered for the Maxwell system is: the upper layer consists of air and the lower of material, like soil, wall, etc., see [4] and [2]. In our problem setting this corresponds to the case when γ + > γ − . Developing an analysis that covers this case together with application to the Maxwell system belongs to our next project. See also [13] for a survey on recent results for inverse obstacle scattering via the time domain enclosure method and [15] for applications to the inverse boundary value problems for the heat equation in three-dimensional space.
The outline of this paper is as follows. Let Φ τ (x, y) be the fundamental solution of (1.3), which satisfies Since the solution v of (1.3) is written by the convolution of f and Φ τ (x, y) as Thus, we need to know an asymptotic behavior of ∇ x Φ τ (x, y) as τ → ∞ for x = (x ′ , x 3 ) with x 3 < 0, x ′ ∈ R 2 and y ∈ B.
The first step for obtaining the asymptotic behavior of (1.9) is to show that the fundamental solution Φ τ (x, y) for x 3 < 0 is given by which is given in Section 2. In (1.10), E γ − τ (x, z ′ ) is a function given by is the point on the transmission boundary ∂R 3 ± and R − (|ξ ′ |) is a function of |ξ ′ | standing for the transmission coefficient given by Note that E γ − τ (x, z ′ ) can be interpreted as the refracted part of the fundamental solution. We put which is a fundamental solution for the equation corresponding to the case that there is no transmission boundary, i.e. γ − = γ + , and given by Thus, (1.10) stands for the refraction phenomena by the transmission boundary ∂R 3 ± . As in Proposition 3.1, for any N ∈ N, the refracted wave E γ − τ (x, z ′ ) is of the form: where each E j is a C ∞ function in R 3 − , andẼ N (x, z ′ ; τ ) is a continuous for x ∈ R 3 − and z ′ ∈ R 2 , and satisfies , as is in Proposition 3.1 and Remark 3.2, we have where each G j is a C ∞ function in R 3 − , andG N (x, z ′ ; τ ) is continuous for x ∈ R 3 − and z ′ ∈ R 2 , and satisfies The main task of Section 3 is to show (1.14) and (1.15).
From (1.14), (1.15) and (1.10), the original problem can be reduced to finding asymptotics of the Laplace integral of the form: By usual Laplace's method, the main part of the asymptotics for (1.16) is given by points z ′ (x, y) ∈ R 2 attaining the minimum l(x, y) of l x,y (z ′ ). We can check the point z ′ (x, y) uniquely exists, which corresponds to Snell's law in geometrical optics. Further, we can show that z ′ (x, y) is a C ∞ function for (x, y) ∈ R 3 − × R 3 + and Hess(l x,y )(z ′ (x, y)) is positive definite, where Hess(l x,y )(z ′ ) = (∂ z i ∂ z j l x,y (z ′ )) (cf. Lemma 4.1). We put H(x, y) = Hess(l x,y )(z ′ (x, y)) and Take φ ∈ C ∞ 0 (R 2 ) with 0 ≤ φ ≤ 1 and φ = 1 near the set {z ′ (x, y) ∈ R 2 |x ∈ D, y ∈ B}, and divide (1.16) into two parts, (1.17) Note that usual Laplace's method (cf. Theorem 7.7.5 of Hörmander [7], which is for oscillatory integrals, however, the proof also works for this case) can be applied for the first integral of (1.17). For the second integral of (1.17), integration by parts implies that this term is negligible. Hence we obtain where L j is a differential operator of order less than or equal to 6j given by and for any N ∈ N ∪ {0}, there exists a constant C N > 0 depending also on a such that From (1.18), we can obtain the asymptotic expansion of ∇ x Φ τ (x, y) of the form: have the following asymptotics: (1.20) Note that in (1.19), E 0 is the function appeared in (1.14). The form of E 0 is given by (3.22) in Section 3.
Proposition 1.5 is crucial to obtain Theorem 1.4. A proof of Proposition 1.5 is given in Section 4. In Section 5, we show Theorem 1.4 by using Proposition 1.5. This is the outline of this paper.

The refracted part of the fundamental solution
In what follows, we only treat the case y = (y ′ , y 3 ), y ′ = (y 1 , y 2 ), y 3 > 0 for large τ > 0. Note that the fundamental solution Φ τ (x, y) is given by In what follows, we also write We put which is the representation by the partial Fourier transform To obtain v ± , we take the partial Fourier transform which satisfy the partial Fourier transform of (2.1), that is, Since v ± are bounded, from (2.3), the solutions of (2.4) are given bŷ .
We can also obtain the formula of Φ τ (x, y) for x 3 > 0, which is for the reflected phenomena. In this case, for x 3 > 0 similarly. In this paper, we do not use this formula.

Asymptotics of the refracted waves
In this section, we show the asymptotics (1.14) and (1.15) for the refracted wave defined by (1.11). We put R(ρ) Note that from (2.2), it follows that by rotating the coordinate. Thus, we obtain We change the variable ζ 1 = 1 + ζ 2 2ζ 1 , and have To obtain the asymptotics of E γ − τ (x, z ′ ) and ∇ x E γ − τ (x, z ′ ), we need to study the asymptotics of (3.3). We use the steepest decent method, which is similar to getting the distribution kernel for the usual wave equations in the two dimensional half-space by Hankel functions (cf. [1], p. 286 for example).

From (3.4) and (3.15), it follows that which yields
where the relations between θ and x −z ′ is given by (3.5).
(2) If x 0 ∈ D, y 0 ∈ B satisfy l 0 = l(x 0 , y 0 ), then x 0 ∈ ∂D and y 0 ∈ ∂B. Further, where ν x 0 and ν y 0 are the unit outer normal of ∂D and ∂B at x 0 and y 0 , respectively.
Next, we show e is a unit outer normal of ∂D at x 0 . Take any C 1 class curve c : (−ε, ε) → ∂D with c(0) = x 0 . Since which yields e is a unit normal of ∂D.

Acknowledgements
It is easy to derive the following decomposition formula of the indicator function which formally corresponds to the case when Ω = R 3 on (3.2) of Proposition 3.1 in [10].
Thus, changing the role of v and w in (A.3), we obtain wdx .
This is noting but the following formula.

Rewrite (A.3) as
For the proof of (1.7) we recall the following inequality (see [8]) where A and B are real vectors. Applying this to the first and second term in the right-hand side on (A.4), we obtain