The enclosure method for inverse obstacle scattering over a finite time interval: V. Using time-reversal invariance

Masaru Ikehata

doi:10.1515/jiip-2018-0046

Article

The enclosure method for inverse obstacle scattering over a finite time interval: V. Using time-reversal invariance

Published/Copyright: December 18, 2018

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From the journal Journal of Inverse and Ill-posed Problems Volume 27 Issue 1

Abstract

The wave equation is time-reversal invariant. The enclosure method, using a Neumann data generated by this invariance, is introduced. The method yields the minimum ball that is centered at a given arbitrary point and encloses an unknown obstacle embedded in a known bounded domain from a single point on the graph of the so-called response operator on the boundary of the domain over a finite time interval. The occurrence of the lacuna in the solution of the free space wave equation is positively used.

Keywords: Enclosure method; time-reversal invariance; inverse obstacle problem; wave equation; non-destructive testing

MSC 2010: 35R30; 35L05

Funding source: Japan Society for the Promotion of Science

Award Identifier / Grant number: 17K05331

Award Identifier / Grant number: 18H01126

Funding statement: The author was partially supported by Grant-in-Aid for Scientific Research (C) (No. 17K05331) and (B) (No. 18H01126) of Japan Society for the Promotion of Science.

A Appendix

In this appendix we give an explicit computation result for the potential

vj⁢(x)=∫Be-τ⁢|x-y||x-y|⁢|y|j⁢𝑑y,x∈B,

where B={y∈ℝ3∣|y|<η}, with η>0 and j=-1,0,1,2.

Proposition A.1.

For all x∈B∖{0}, we have

v-1⁢(x)=4⁢πτ2⁢(1-e-τ⁢|x||x|-e-τ⁢η⁢sinh⁡τ⁢|x||x|),

v0⁢(x)=4⁢πτ2⁢{1-(η+1τ)⁢e-τ⁢η⁢sinh⁡τ⁢|x||x|},

v1⁢(x)=4⁢πτ2⁢{|x|+2τ2⁢1-e-τ⁢|x||x|-e-τ⁢η⁢(η2+2τ⁢η+2τ2)⁢sinh⁡τ⁢|x||x|},

v2⁢(x)=4⁢πτ2⁢{|x|2+6τ2-e-τ⁢η⁢(η3+3⁢η2τ+6⁢ητ2+6τ3)⁢sinh⁡τ⁢|x||x|}.

Proof.

The change of variables y=r⁢ω (0<r<η,ω∈S2) and a rotation give us

vj⁢(x)=∫0ηr2+j⁢𝑑r⁢∫S2e-τ⁢|x-r⁢ω||x-r⁢ω|⁢𝑑ω=∫0ηr2+j⁢𝑑r⁢∫S2e-τ⁢||x|⁢𝐞3-r⁢ω|||x|⁢𝐞3-r⁢ω|⁢𝑑ω

=∫0ηr2+j⁢𝑑r⁢∫02⁢π𝑑θ⁢∫0πsin⁡φ⁢d⁢φ⁢e-τ⁢|x|2-2⁢r⁢|x|⁢cos⁡φ+r2|x|2-2⁢r⁢|x|⁢cos⁡φ+r2

=2⁢π⁢∫0ηQ⁢(|x|,r)⁢r2+j⁢𝑑r,

where 𝐞3=(0,0,1) and

Q⁢(ξ,r)=∫0πe-τ⁢ξ2-2⁢r⁢ξ⁢cos⁡φ+r2ξ2-2⁢r⁢ξ⁢cos⁡φ+r2⁢sin⁡φ⁢d⁢φ,0≤ξ<η, 0<r<η.

Fix ξ∈]0,η[ and r∈]0,η[. The change of variable

s=ξ2-2⁢r⁢ξ⁢cos⁡φ+r2,φ∈]0,π[,

gives

s2=ξ2-2⁢r⁢ξ⁢cos⁡φ+r2 and s⁢d⁢s=r⁢ξ⁢sin⁡φ⁢d⁢φ.

Hence, we have

Q⁢(ξ,r)=1r⁢ξ⁢∫|ξ-r|ξ+re-τ⁢s⁢𝑑s=-1r⁢ξ⁢τ⁢(e-τ⁢(ξ+r)-e-τ⁢|ξ-r|).

Therefore, we obtain

(A.1)vj⁢(x)=2⁢π⁢∫0ηQ⁢(|x|,r)⁢r2+j⁢𝑑r=2⁢πξ⁢τ⁢∫0η(e-τ⁢|ξ-r|-e-τ⁢(ξ+r))⁢r1+j⁢𝑑r|ξ=|x|.

Thus, everything is reduced to computing the integral

Kj=∫0η(e-τ⁢|ξ-r|-e-τ⁢(ξ+r))⁢r1+j⁢𝑑r,j=-1,0,1,2.

A direct computation yields

∫0ηe-τ⁢|ξ-r|⁢𝑑r=2τ-e-τ⁢ξτ-e-τ⁢(η-ξ)τ,

∫0ηe-τ⁢|ξ-r|⁢r⁢𝑑r=2⁢ξτ+e-τ⁢ξτ2-eτ⁢(ξ-η)τ⁢(η+1τ),

∫0ηe-τ⁢|ξ-r|⁢r2⁢𝑑r=1τ3⁢{(2⁢τ2⁢ξ2+4)-2⁢e-τ⁢ξ-(τ2⁢η2+2⁢τ⁢η+2)⁢e-τ⁢(η-ξ)},

∫0ηe-τ⁢|ξ-r|⁢r3⁢𝑑r=2⁢ξ3τ-1τ⁢e-τ⁢(η-ξ)⁢η3+6τ4⁢e-τ⁢ξ-3τ4⁢{(τ2⁢η2+2⁢τ⁢η+2)⁢e-τ⁢(η-ξ)-4⁢τ⁢ξ}.

Also we have

∫0ηe-τ⁢(ξ+r)⁢𝑑r=e-τ⁢ξτ-e-τ⁢(ξ+η)τ,

∫0ηe-τ⁢(ξ+r)⁢r⁢𝑑r=e-τ⁢ξτ2-e-τ⁢(ξ+η)τ⁢(η+1τ),

∫0ηe-τ⁢(ξ+r)⁢r2⁢𝑑r=1τ3⁢e-τ⁢ξ⁢{-e-τ⁢η⁢(τ2⁢η2+2⁢τ⁢η+2)+2},

∫0ηe-τ⁢(ξ+r)⁢r3⁢𝑑r=-1τ⁢η3⁢e-τ⁢(ξ+η)+3τ4⁢e-τ⁢ξ⁢{-e-τ⁢η⁢(τ2⁢η2+2⁢τ⁢η+2)+2}.

From these, we obtain

K-1=2τ⁢(1-e-τ⁢ξ-e-τ⁢η⁢sinh⁡τ⁢ξ),

K0=2τ⁢{ξ-(η+1τ)⁢e-τ⁢η⁢sinh⁡τ⁢ξ},

K1=2τ3⁢{(τ2⁢ξ2+2)-2⁢e-τ⁢ξ-(τ2⁢η2+2⁢τ⁢η+2)⁢e-τ⁢η⁢sinh⁡τ⁢ξ},

K2=2⁢ξ3τ+12τ3⁢ξ-2τ⁢(η3+3⁢η2τ+6⁢ητ2+6τ3)⁢e-τ⁢η⁢sinh⁡τ⁢ξ.

Substituting these into (A.1), we obtain the desired formulas. ∎

Acknowledgements

The author would like to thank anonymous referees for giving valuable comments on the improvement of the presentation of the results.

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Received: 2018-05-17

Revised: 2018-11-23

Accepted: 2018-11-23

Published Online: 2018-12-18

Published in Print: 2019-02-01

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https://doi.org/10.1515/jiip-2018-0046

Keywords for this article

Enclosure method; time-reversal invariance; inverse obstacle problem; wave equation; non-destructive testing