Detection and range estimation of a hidden object

Talk Slide, Oct. 6. 2017, Geometry and Inverse Problems in cooperation with A3 FORESIGHT PROGRAM Tohoku University, Sendai, JAPAN

Detection and range estimation of a hidden object using the time domain enclosure method Masaru IKEHATA*1 Laboratory of Mathematics, Graduate School of Engineering, Hiroshima University, Higashihiroshima 739-8527, JAPAN

*1

[email protected]

Abstract Two inverse obstacle problems using waves governed by the wave equations over a finite time interval are considered. The problems are concerned with detection and range estimation of an unknown obstacle embedded in a general rough background medium or placed behind a known impenetrable obstacle. It is shown that the time domain enclosure method enables us to know whether the obstacle exists or not by using a singel wave over a finite time interval on an open ball where the wave is generated. Moreover, if the obstacle exists, the method yields also information about the Euclidean distance between the obstacle and the center of the open ball.

目次 1

Introduction Problem A. Detecting an obstacle embedded in a rough background medium . . . . . . . . . . . . . . . . Problem B. Detecting an obstacle behind a impenetrable obstacle . . . . . . . . . . . . . . . . . . . . . . .

3

2.1 2.2 2.3

On Problem A Main Result . . . . . . . . . . . . . . . . . . . . . . . . . Outline of Proof . . . . . . . . . . . . . . . . . . . . . . Further remarks on Problem A . . . . . . . . . . . . . .

14 20 27 35

3.1 3.2 3.3

On Problem B Main result . . . . . . . . . . . . . . . . . . . . . . . . . Outline of Proof . . . . . . . . . . . . . . . . . . . . . . Further remarks on Problem B . . . . . . . . . . . . . .

37 42 49 62

1.1 1.2

2

3

4 9

1 Introduction

1.1 Problem A. Detecting an obstacle embedded in a rough background medium

The first part of this talk is concerened with through-wall imaging via wave phenomena in the time domain.

Assume that there is a large wall between an observer and an unknown object. The wall is opaque and thus the observer can not see the object directly by using visible ray.

How can the observer find the object? the case: the object is scilent (acoustically or electromagnetically and not moving).

Consider the case when the wall is electromagnetically penetrable.

Generate the electromagnetic wave at the place where the observer is. And observe the wave at the same place over a finite time interval. The observed wave should include information about the unknown object behind the several walls (or embedded in a complicated medium). Free-space assumptions no longer apply after the electromagnetic waves propagate through the walls. How can one extract the information from the observed wave? This is a typical problem in through-wall imaging.

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Engineering approach Introducing various processing of the reflected signal from the wall (having a special geometry) and targets behind the wall, e.g., Amin, M. G. and Ahmad, F., Compressive sensing for through-the-wall radar imaging, J. Electron. Imaging, 22(2013), 030901. The linear sampling method in the frequency domain under a simple geometry assumption of the wall Catapano, I. and Crocco, L., A qualitative inverse scattering method for through-the-wall imaging, IEEE Geosci. Remote Sens.Lett. 7(2010), 685-689

For a survey (several engineering approach) Baranoski, E. J., Through-wall imaging: historical perspective and future directions, J. Franklin Inst. 345(2008) 556-69. The purpose Develop a mathematical method for through-wall imaging based on the governing equation of the wave in time domain from the beginning to end (PDE approach). Original. The Maxwell system in time domain This talk. A scalar wave equation in time domain

1.2 Problem B. Detecting an obstacle behind a impenetrable obstacle

The purpose

Develop a mathematical method of imaging an unknow obstacle behind a known impenetrable obstacle form a single wave generated and observed at the same place where one can not see the unknown obstacle by using visible ray.

じ

1

ヽ V

Some of experimental studies appear in engineering, for example, robotics for object tracking.

[NOKO] Niwa, H., Ogata, T., Komatani, K. and Okuno, H. G., Detection and range finder of intercepted object by AH-based method using audible sound, IEICE Tech. Rep., 106(2006), no. 267, EA2006-48, 1-6, Sept. 2006, in Japanese. [NKOO] Niwa, H., Komatani, K., Ogata, T. and Okuno, H. G., Distance estimation of hidden objects based on acoustical holography by applying acoustic diffraction of audible sound, Proceedings of IEEE-RAS International Conference on Robotics and Automation (ICRA-2007), pp.423-428, (Apr. 2007). doi:10.1109/ROBOT.2007.363823

They considered the detection problem of an unknown object behind another object (occlusion as called by them) motivated in robotics for object tracking. They made use of audible sound (3.2kHz) generated by a speaker as a wave and catches the reflected sound by microphones placed where the object is line-of-site because of the occlusion. They found that it is possible to give some estimate of the distance to the object using the observed data when the object and occlusion are consists of two plates or balls and placed in a simple configuration.

Their study is purely experimental and does not based on the governing equation of the wave.

In this talk, we formulate the problem as an inverse problem for the wave equation.

2 On Problem A

The governing equation  2 α(x)∂  t u − ∆u = 0     u(x, 0) = 0      ∂t u(x, 0) = f (x)

in R3 × ]0, T [, in R3 , in R3 ,

where • 0 < T < ∞ (to be determined later). • α ∈ L∞ (R3 ) and ess. inf x∈R3 α(x) > 0. • for suimplicity, we choose f (x) = χB (x), where χB denotes the characteristic function of an open ball B with a very small radius.

Assume that   α0 (x), α(x) =



if x ∈ R3 \ D,

α0 (x) + h(x), if x ∈ D,

where • D is a bounded open subset of R3 with Lipschitz boundary. • α0 ∈ L∞ (R3 ) and satisfies m20 ≤ α0 (x) ≤ M02 with m0 , M0 > 0. (the distribution of α0 can be complicated) • h ∈ L∞ (D) satisfies one of two conditions α >> α0 and α > α0 means ∃C > 0 h(x) ≥ C 

α 0 − h(x) ≥ C

• B in χB (x) satisfies B ∩ D = ∅.

a.e. x ∈ D.

Problem A.

Fix T (to be determined later) and B. Assume that α0 is known. Can one know the existence of D by observing u(x, t) given at all (x, t) ∈ B× ]0, T [ ? If possible, extract information about the geometry of D together with property of h from the back-scattering data u(x, t) given at all (x, t) ∈ B× ]0, T [.

Method: The Time Domain Enclosure Method (2007, Ik) (Kawashita, Itou)

⇒ its origin goes back to The Enclosure Method (1999, 2000, Ik) (Siltanen, Ohe, Itou, Ide, Nakata, Isozaki, Uhlmann, Nakamura, Wang, Nagayasu, Lin, Sini, Kar, Brander, Salo, Cao, Xu,...) One of direct (non-iterative) methods for inverse obstacle problems: others, The Linear Sampling Method (1996, Colton-Kirsh), The Factorization Method (1998, Kirsh), The Probe Method (1998, Ik), The Singular Sources Method (2000, Potthast), The Monotonicity Method (2002, Tamburrino-Rubinacci, 2013, Harrach-Ullrich)

Survey on the time domain enclosure method Ikehata, M., New development of the enclosure method for inverse obstacle scattering, Capter 6 in Inverse Problems and Computational Mechanics (eds. Marin, L., Munteanu, L., Chiroiu, V.), Vol. 2, 123-147, Editura Academiei, Bucharest, Romania, 2016. We present a result from Ikehata, M., On finding an obstacle embedded in the rough background medium via the enclosure method in the time domain, Inverse Problems, 31(2015) 085011(21pp).

2.1 Main Result

Let u solve

 2 3 α(x)∂ u − ∆u = 0 in R × ]0, T [,  t     u(x, 0) = 0 in R3 ,      ∂t u(x, 0) = χB (x) in R3 .

Define the indicator function ∫ IB (τ ) = α0 (w − v)dx, τ > 0 B

where

∫ w(x, τ ) =

T

e−τ t u(x, t)dt

0

and v = v(x, τ ) ∈ H 1 (R3 ) solves ∆v − α0 (x)τ 2 v + α0 (x)χB (x) = 0

in R3 .

Define dist(D, B) =

inf

x∈B, y∈D

|x − y|.

if D 6= ∅. Theorem 1 (Ik, Theorem We have  0      −∞ lim eτ T IB (τ ) = τ →∞      ∞

1.1, Inverse Problems, 2015). if D = ∅ and T > 0, if D 6= ∅, α >> α0 , T > 2M0 dist (D, B), if D 6= ∅, α 2M0 dist (D, B).

Moreover, if α >> α0 or α 2M0 dist (D, B)  1   log |IB (τ )|, inf   −M0 dist (D, B) ≤ lim τ −→∞ 2τ  1    lim sup log |IB (τ )| ≤ −m0 dist (D, B). τ −→∞ 2τ

Remarks. 1. Let p and η denote the center and radius of B, respectively. We have dist (D, B) = d∂D (p) − η, where d∂D (p) = inf |p − q|. q∈∂D

Thus knowing dist (D, B) is equivalent to knowing d∂D (p). Thus, from the observed data we obtain the upper and lower estimation of the sphere |x − p| = d∂D (p) whose exterior encloses the unknown object.

2. Theorem 1 suggests a new direction of the enclosure method in the case when the background medium is inhomogeneous and quite complicated. • give up to find a precise quantity in the observation data which is related to the exact location of unknown obstacles: • instead give lower and/or upper estimates rigorously for dist (D, B) (ranging). 3. If we have the a priori information dist (D, B) < M for a known constant M , then we can know the waiting time for collecting the observation data in advance as T ≥ 2M0 M .

Let V solve

 2 3 α (x)∂ V − ∆V = 0 in R × ]0, T [,  0 t     V (x, 0) = 0 in R3 ,      ∂t V (x, 0) = χB (x) in R3 .

Define 0 IB (τ ) =

∫

α0 (w − v 0 )dx, B

where

 ∫ T  −τ t  w(x, τ ) = e u(x, t)dt,    0 ∫      v 0 (x, τ ) =

T

e−τ t V (x, t)dt.

0 0 Theorem 1 remains valid if IB is replaced with IB .

Assume that α0 = α = 1 a.e. on B. Let T be a large positive numer. On 2017.10.1 Send a wave in a medium α = α0 and measure u = u0 on B over ]0, T [. On 2017.10.6 Send a wave in a medium α = α1 and measure u = u1 on B over ]0, T [. The asymptotic behaviour of the indicator function ∫

∫

τ 7−→

(w1 − w0 )dx, B

wj =

T

e−τ t uj (x, t) dx, j = 0, 1.

0

tells us a deviation of α1 from α0 without knowing αj outside B.

2.2 Outline of Proof

Proposition. We have, as τ −→ ∞ ∫  α0 2 2 −1 −τ T  I (τ ) ≤ τ (α − α)v dx + O(τ e ),  0   B R3 α ∫     IB (τ ) ≥ τ 2 (α0 − α)v 2 dx + O(τ −1 e−τ T ). R3

A sketch of the proof. Set   R = w − v, 

F = ∂t u(x, T ) + τ u(x, T ).

Note that kF kL2 (R3 ) = O(τ ).

We have

∫ f {(α0 − α)v + αR} dx R3

∫

∫ (α0 − α)v 2 dx +

= τ2 R3

+e−τ T

(|∇R|2 + ατ 2 R2 ) dx R3

(∫

)

∫ αF Rdx − R3

αF vdx , R3

The governing equation of v ∆v − α0 τ 2 v + α0 f = 0 yields

in R3

kvkL2 (R3 ) = O(τ −2 ).

(A)

Rewrite (A) as ∫

{

∫

} ( ) 2 −τ t f − e F 2 |∇R| + α τ R − dx 2τ

(α0 − α)v 2 dx +

τ2

R3

R3

∫

1 = (α0 − α)f vdx + 2 4τ R3 This yields

∫

∫

α(f − e−τ t F )2 dx. R3

(|∇R|2 + τ 2 R2 )dx = O(τ −2 ) R3

and hence

kRkL2 (R3 ) = O(τ −2 ).

Thus (A) becomes ∫ f {(α0 − α)v + αR} dx R3

∫

∫ (α0 − α)v 2 dx +

= τ2

(|∇R|2 + ατ 2 R2 ) dx

(A)

R3

R3

+O(τ −1 e−τ T ). Since α = α0 on B and supp f ⊂ B, we obtain the lower estimate ∫ (α0 − α)v 2 dx + O(τ −1 e−τ T ). IB (τ ) ≥ τ 2 R3

The proof of the upper estimate ∫ α0 2 IB (τ ) ≤ τ (α0 − α)v 2 dx + O(τ −1 e−τ T ) R3 α is more technical. Using another integral identity, we have ∫ ∫ (α − α0 )2 2 2 2 2 2 (|∇R| + ατ R )dx ≤ τ v dx + O(τ −1 e−τ T ). α R3 R3 Then (A) yields ∫ f {(α0 − α)v + αR} dx R3

∫ ≤τ

2 R3

{

(α − α0 ) (α0 − α) + α

2

}

v 2 dx + O(τ −1 e−τ T ).

Remark. Another integral identity is

∫

f {(α − α0 )w − α0 R} dx R3

∫ = τ2 R3

+e−τ T

α0 (α − α0 )v 2 dx + α

(∫

R3

¯ ( ) ¯2 } α0 ¯ ¯ |∇R|2 + ατ 2 ¯R + 1 − v¯ dx α

)

∫ αF R dx + R3

Compare

{

∫

αF v dx . R3

∫ f {(α0 − α)v + αR} dx R3

∫

∫ (α0 − α)v 2 dx +

= τ2 R3

+e−τ T

(|∇R|2 + ατ 2 R2 ) dx R3

(∫

)

∫ αF Rdx −

R3

αF vdx , R3

(A)

Thus everything is reduced to studying the leading profile of v. We have Lemma.

 ∫ −M0 τ |x−y| 1 e   v(x) ≥ α0 (y) dy,    4π B |x − y| ∫  −m0 τ |x−y|   1 e   v(x) ≤ α0 (y) dy. 4π B |x − y|

Reduced to the asymptotic behaviour of the integral as τ −→ ∞ ∫ B

where γ > 0 constant.

e−γτ |x−y| dy, x ∈ D. |x − y|

2.3 Further remarks on Problem A

• If α0 is a piecewise constant function, then we have a positive result in one-space dimension case (Ik, Theorem 5.2, Inverse Problems, 2015). • In three-space dimension case, for the signal governed by the equation ∂t2 u − ∇ · γ∇u = 0 with a piecewise constant background we have a positive result (Ik-Kawashita, submitted). • Under the assumption α0 ∈ L∞ (R3 ), show that 1 lim log |IB (τ )| τ −→∞ 2τ = the first arrival time of the signal governed by α0 (x)∂t2 u − ∆u = 0 started from ∂B at t = 0 and arrived at ∂D.

3 On Problem B

The preprint is available for everyone.

Ikehata, M., Detecting a hidden obstacle via the time doman enclosure method. A scalar wave case, arXiv:1709.02118v3 [math.AP]

Unknown obstacle D behind a known obstacle D0 • D and D0 bounded open subsets of R3 with C 2 -boundary • D0 ∩ D = ∅ • R3 \ D0 and R3 \ (D0 ∪ D) are connected • 0 < T < ∞ (to be determined later). • ν the outward normal to both D0 on ∂D0 and D on ∂D.

The governing equation  2 ∂t u − ∆u = 0           u(x, 0) = 0  ∂t u(x, 0) = f (x)          ∂u = 0 ∂ν

in (R3 \ (D0 ∪ D))× ]0, T [, in R3 \ (D0 ∪ D), in R3 \ (D0 ∪ D), on (∂D0 ∪ ∂D)× ]0, T [,

  (η − |x − p|)2 g(x),

where f (x) =



0,

x ∈ B, x ∈ R3 \ B

and g ∈ C 2 (B) with inf x∈B g(x) > 0; p and η the center and radius of B. f belongs to H 2 (R3 ) with supp f = B.

Problem B.

Fix T (to be determined later) and B. Assume that D0 is known. Can one know the existence of D by observing u(x, t) given at all (x, t) ∈ B× ]0, T [ ? If possible, extract information about the geometry of D from the back-scattering data u(x, t) given at all (x, t) ∈ B× ]0, T [.

3.1 Main result

∫

Define

f (w − v)dx, τ > 0,

IB (τ ; T ) = B

where ∫ w(x, τ ) =

T

e−τ t u(x, t)dt,

x ∈ R3 \ (D0 ∪ D),

0

and v ∈ H 1 (R3 \ D0 ) is the unique solution of  2 (∆ − τ )v + f = 0  

in R3 \ D0 ,

  ∂v = 0 ∂ν

on ∂D0 .

Theorem 2. (i) If D = ∅, then for all T > 0 we have lim eτ T IB (τ ; T ) = 0;

τ −→∞

(ii) Assume that there exists an open ball W such that D0 ⊂ W and B ∪ D ⊂ R3 \ W . Assume that D 6= ∅ and satifies M > dist (D, B) for a positive number M . Fix T ≥ 2CM with √( ) √ π 2 C= 2 + 1. 4 Then, we have limτ −→∞ eτ T IB (τ ; T ) = ∞. Moreover, we have 1 log IB (τ ; T ) ≥ −C dist (D, B). 0 > lim inf τ −→∞ 2τ

O Q

● 口鳳

Remarks. 1. Under the assumptions in Theorem 2, we have 1 1 dist (D, B) > − lim inf log IB (τ ; T ) ( > 0 ), C τ −→∞ 2τ where

√( ) √ π 2 + 1. C= 2 4

2. If D0 is an open ball, one can always choose the ball W in Theorem 2 and thus all the conclusions in Theorem 2 are valid. Thus, we can detect any unknown obstacle D behind an open ball D0 by using a single wave over the finite time interval ]0, T [ with an arbitrary fixed T satisfying T ≥ 2 CM under a priori information dist (D, B) < M.

3. However, there are some situtation that D ⊂ W \ D0 for any choice of ball W with constraint D0 ⊂ W and B ⊂ R3 \ W . For example, consider the case when D0 is an ellipsoid and D is closed to ∂D0 . The next result covers this case under some a priori assumtion on the possible location of D relative to D0 . We assume that D0 is convex and thus D0 is also convex. Then, given x ∈ R3 \ D there exists a unique point q(x) ∈ ∂D0 such that |x − q(x)| = d∂D0 (x). Given α ∈ ] − 1, 0] and y ∈ B define Vα (B; D0 ) = ∩y∈B Vα (y; D0 ), where

{

}

Vα (y; D0 ) = x ∈ R \ D0 | νq(x) · νq(y) ≥ α . 3

Theorem 3. Let D0 be convex. Let B satisfy B ∩ D0 = ∅. Let −1 < α ≤ 0 and assume that D(6= ∅) satisfies D ⊂ Vα (B; D0 ) and dist (D, B) < M with a positive constant M . Fix T satisfying T ≥ 2 C(α) M with √ 2 C(α) = . 1+α Then, we have limτ −→∞ eτ T IB (τ ; T ) = ∞. Moreover, we have 1 0 > lim inf log IB (τ ; T ) ≥ −C(α) dist (D, B). τ −→∞ 2τ

Remark. There should be the cases when • D is placed in R3 \ Vα (B; D) or a nearest point on D to B is therein which is the case when D is too close to the backside of D0 from B; • both B and D are inside the smallest ball that contains D0 . For these cases at the present time the speaker can not say the behaviour of the indicator function which may be useful for the detection and range estimatation.

3.2 Outline of Proof

Proposition. We have, as τ −→ ∞ IB (τ ; T ) = O(τ 2 JB (τ ; D) + τ −1 e−τ T ) and

JB (τ ; D) + O(τ −1 e−τ T ) ≤ IB (τ ; T ), ∫

where

(|∇v|2 + τ 2 |v|2 )dx.

JB (τ ; D) = D

Remark. This proposition says: the asymptotic behaviour of the indicator function is governed by that of v on D.

Proof. This is a result of the combination of the following two facts. 1. We have, as τ −→ ∞ IB (τ ; T ) = JB (τ ; D) + E(τ ) + O(τ −1 e−τ T ), ∫

where E(τ ) =

R3 \(D0 ∪D)

(|∇²|2 + τ 2 |²|2 )dx

and ² = w − v. 2. We have, as τ −→ ∞ E(τ ) = O(τ 2 JB (τ ; D) + e−2τ T ). Remark. If ∂D is C 3 , then we can say more E(τ ) = O(JB (τ ; D) + e−2τ T ).

From Proposition we have: (i) if D = ∅, then, for all T > 0 eτ T IB (τ ; T ) = O(τ −1 ). (ii) if D 6= ∅ and T satisfies lim eτ T JB (τ ; D) = ∞,

τ −→∞

then lim eτ T IB (τ ; T ) = ∞

τ −→∞

and

 1 1 −2    2τ log JB (τ ; D) + O(τ ) ≤ 2τ log IB (τ ; T ),    1 log I (τ ; T ) ≤ 1 log J (τ ; D) + O( log τ ). B B 2τ 2τ τ

Thus, it suffices to consider

• find T as explicit as possible such that lim eτ T JB (τ ; D) = ∞.

τ −→∞

• give lower and upper estimates for 1 log JB (τ ; D). 2τ Remark. If ∂D is C 3 , then (?) is equivalent to lim eτ T IB (τ ; T ) = ∞.

τ −→∞

(?)

Reduction Using the semigroup theory, given δ > 0 one can find Z such that  3 (∂ − ∆)Z = 0 in (R \ D0 )× ]0, δ[, t       ∂Z =0 on ∂D0 × ]0, δ[,  ∂ν      Z(x, 0) = f (x) in R3 \ D0 . Note that f ∈ H 2 (R3 ) with sup f = B. Set ∫

δ

−τ 2 t

e

vδ (x, τ ) =

Z(x, t)dt,

x ∈ R3 \ D0 , τ > 0.

0

We have kvk2L2 (D)

1 2 −4 −2τ 2 δ ≥ kvδ kL2 (D) + O(τ e ). 2

Let K1 = K1 (x, y; t) be the Neumann heat kernel for the domain R3 \ D0 . One has the expression ∫ K1 (x, y; t)f (y)dy. Z(x, t) = R3 \D0

The following lemma can be derived from a general result due to Ouhabaz (Potential Anal., 5(1996), 611-625). Lemma. Let K0 = K0 (x, y; t) be the Dirichlet heat kernel for the domain R3 \ D0 . We have, for all t > 0 and x ∈ R3 \ D0 ∫ ∫ K1 (x, y; t)f (y)dy. K0 (x, y; t)f (y)dy ≤ B

B

Thus, we have

∫ Z(x, t) ≥

K0 (x, y; t)f (y) dy. B

Two results due to van den Berg. Given ² > 0 define (D0 )² = {x ∈ R3 \ D0 | d∂D0 (x) ≥ ²}. Given (x, y) ∈ (D0 )2² let d² (x, y) denote the infimum of the length of arcs in (D0 )² with endpoints x and y. If there is no arc in (D0 )² with endpoints x and y, define d² (x, y) = ∞. Lemma(van den Berg, Bull. Lond. Math. Soc., 24(1992), 475-477). For (x, y) ∈ ((D0 )² )2 , t > 0 and ² > 0 we have ( ) 2 d² (x, y) K0 (x, y; t) ≥ exp − K² (0, 0; t), 4t where K² (x, y; t) denotes the Dirichlet heat kernel for the open ball with radius ² centered at the origin.

Lemma(van den Berg, J. Func. Anal., 88(1990), 267-278). We have, for all t > 0 K² (0, 0; t) ≥ (4πt)

−3/2 −9π 2 t/(4²2 )

e

.

Using these lemmas and choosing ² sufficiently small in such a way that B ∪ D ⊂ (D0 )² , we obtain for all x ∈ D vδ (x, τ ) ∫ (∫ ≥ Cδ,² B

) ( ) δ 2 d (x, y) ² −3/2 −τ 2 t (4πt) exp − e dt (η − |y − p|)2 dy. 4t 0

Now ready to give a proof of (ii) in Theorem 2. Choose ² again smaller in such a way that R3 \ W ⊂ (D0 )² . Let x ∈ D and y ∈ B. Since B ∪ D ⊂ R3 \ W , (x, y) ∈ R3 \ W . Apply the following lemma to W = U : Lemma. Let U be an open ball. Then, given (x, y) ∈ (R3 \ U )2 there exists an arc γ in R3 \ U with endpoints x and y such that √( ) √ π 2 L(γ) ≤ 2 + 1 |x − y|. 4 This gives d² (x, y) ≤ C|x − y|, where

√( ) √ π 2 C= 2 + 1. 4

Thus, we obtain vδ (x, τ ) ∫ (∫ ≥ Cδ,² B

) ( ) δ 2 |x − y| −3/2 2 −τ 2 t (4πt) exp −C e dt (η − |y − p|)2 dy. 4t 0

It is not difficult to deduce the estimate ( ) ∫ −Cτ |x−y| 1 e 2 −τ 2 δ vδ (x, τ ) ≥ Cδ,² (η − |y − p|) dy + O(e ) . 4πC B |x − y| From this it is a due course to find a lower estimate for kvδ kL2 (D) and finally have the lower estimate for kvkL2 (D) as τ −→ ∞: τ 8 kvk2L2 (D) ≥ C1 e−2C dist (D,B) τ . Since JB (τ ; D) ≥ τ 2 kvk2L2 (D) , we conclude 1 lim inf log JB (τ ; D) ≥ −C dist (D, B). τ −→∞ 2τ

The statement

1 0 > lim inf log JB (τ ; D) τ −→∞ 2τ

is not also trivial. This can be done by using an upper bound of the heat kernel together with its space variables derivative. See the book by Ouhabaz, Analysis of heat equations on domains, London Math. Soc. Monographs, vol. 31, Princeton University Press, 2004, Princeton and Oxford.

Remark. For the proof of Theorem 3 instead of the lemma above we make use of the following lemma Lemma. Assume that D0 is convex. Let −1 < α ≤ 0. Then, for all ² > 0 and (x, y) ∈ ((D0 )² )2 satisfying νq(x) · νq(y) ≥ α we have d² (x, y) ≤ C(α) |x − y|.

For the proof we employ a part of an argument done in the proof of Proposition 6.16 in Grya, P. and Saloff-Coste, L., Ast´ erisque, 336(2011). In the proposition they showed that the complement of an arbitrary closed convex set is inner uniform.

3.3 Further remarks on Problem B

Extension to the Maxwell system D0 and D perfect conductive bodies Instead of v one should consider the solution of  1  ∇ × ∇ × V + τ 2 V + f = 0 in R3 \ D0 ,  ²µ   ν×V =0 on ∂D0 . Instead of JB (τ ; D) one should consider ∫ ∫ 1 |∇ × V |2 dx + τ 2 |V |2 dx. τ 7−→ ²µ D D See the case when D0 = ∅: Ikehata, M., The enclosure method for inverse obstacle scattering using a single elecrtromagnetic wave in time domain, Inverse Problems and Imaging, 10, No.1, 2016, 131-163.

A possible approach: ∫

δ

V (x, τ ) ∼

−τ 2 t

e

Z(x, t) dt, δ > 0 and small,

0

where  1 3  ∇ × ∇ × Z + ∂ Z = 0 in (R \ D0 )× ]0, δ[,  t  ²µ    ν×Z =0       Z(x, 0) = F (x)

on ∂D0 × ]0, δ[, in R3 \ D0 .

and F ∼ −f . How about the Gaussian lower/upper bound for the heat kernel?

Thank you!