On finding a cavity in a thermoelastic body using a single displacement measurement over a finite time interval on the surface of the body

Masaru Ikehata

doi:10.1515/jiip-2017-0066

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On finding a cavity in a thermoelastic body using a single displacement measurement over a finite time interval on the surface of the body

Masaru Ikehata

Published/Copyright: January 30, 2018

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From the journal Journal of Inverse and Ill-posed Problems Volume 26 Issue 3

Abstract

A mathematical formulation of an estimation problem of a cavity inside a three-dimensional thermoelastic body by using time domain data is considered. The governing equation of the problem is given by a system of equations in the linear theory of thermoelasticity which is a coupled system of the elastic wave and heat equations. A new version of the enclosure method in the time domain which is originally developed for the classical wave equation is established. For a comparison, the results in the decoupled case are also given.

Keywords: Enclosure method; inverse obstacle problem; dynamic theory of thermoelasticity, displacement-temperature equation of motion; coupled heat equation; coupled system; cavity, non-destructive testing

MSC 2010: 74J25; 35Q79; 74F05; 35R30; 35L05; 35K05; 35B40

Funding statement: The author was partially supported by Grant-in-Aid for Scientific Research (C) (no. 17K05331) of Japan Society for the Promotion of Science. Some part of this work was initiated during the author’s visit at ICUB (Research Institute of the University of Bucharest), Bucharest, Romania in November 2016, with support by ICUB’s visiting professors fellowship program.

A Appendix

A.1 Explicit computation of integrals of (2.38) and (2.41)

The following formulae have been derived in [22, (A.5), (A.6)]. Note that the first formula of (A.1) below is also a result of the mean value theorem for the modified Helmholtz equation [2].

Proposition A.1 ([22]).

Let x∈R3∖B¯. Then we have

(A.1){I0⁢(x;τ)=4⁢π⁢φ0⁢(τ⁢η)τ3⁢e-τ⁢|x-p||x-p|,I1⁢(x;τ)=4⁢π⁢φ1⁢(τ⁢η)τ4⁢e-τ⁢|x-p||x-p|,

where

(A.2){φ0⁢(s)=s⁢cosh⁡s-sinh⁡s,φ1⁢(s)=(s2+2)⁢cosh⁡s-2⁢s⁢sinh⁡s-2.

In this subsection, we add the following formulae.

Proposition A.2.

We have

(A.3){𝑰0⁢(x;τ)=π⁢e-τ⁢|x-p||x-p|2⁢Kτ0⁢(|x-p|,τ⁢η)⁢x-p|x-p|,𝑰1⁢(x;τ)=4⁢πτ2⁢e-τ⁢|x-p||x-p|2⁢Kτ1⁢(|x-p|,τ⁢η)⁢x-p|x-p|,I2⁢(x;τ)=4⁢π⁢φ2⁢(τ⁢η)τ5⁢e-τ⁢|x-p||x-p|,

where

Kτ0⁢(ξ,s)=2τ⁢{(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3}⁢(cosh⁡s-1)+2τ3⁢(1-1τ)⁢{(s2+2)⁢cosh⁡s-2⁢s⁢sinh⁡s-2}

(A.4)+4τ3⁢(s⁢sinh⁡s-cosh⁡s),

Kτ1⁢(ξ,s)={(1-1τ)⁢ξ2-2τ2⁢(ξ+1τ)}⁢(s⁢cosh⁡s-sinh⁡s)+1τ2⁢(1-1τ)⁢{s2⁢(s+6)⁢cosh⁡s-3⁢(s2+2)⁢sinh⁡s}

(A.5)+2τ2⁢(ξ+1τ)⁢{(s2+1)⁢sinh⁡s-2⁢s⁢cosh⁡s}

and

(A.6)φ2⁢(s)=s2⁢(s+6)⁢cosh⁡s-3⁢(s2+2)⁢sinh⁡s.

Proof.

It suffices to consider the case when p=0. Let 𝑨x be the orthogonal matrix such that (𝑨x)T⁢x=|x|⁢𝒆3. The change of variables y=rω(0<r<η,ω∈S2) and a rotation give us

𝑰0⁢(x;τ)=∫0ηr2⁢𝑑r⁢∫S2e-τ⁢|x-r⁢ω||x-r⁢ω|⁢ω⁢𝑑ω

=∫0ηr2⁢𝑑r⁢∫S2e-τ⁢||x|⁢𝒆3-r⁢ω|||x|⁢𝒆3-r⁢ω|⁢𝑨x⁢ω⁢𝑑ω

=∫0ηr2⁢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⁢𝑨x⁢(sin⁡φ⁢cos⁡θsin⁡φ⁢sin⁡θcos⁡φ)

=2⁢π⁢∫0ηr2⁢𝑑r⁢∫0πsin⁡φ⁢d⁢φ⁢e-τ⁢|x|2-2⁢r⁢|x|⁢cos⁡φ+r2|x|2-2⁢r⁢|x|⁢cos⁡φ+r2⁢cos⁡φ⁢𝑨x⁢𝒆3

(A.7)=2⁢π⁢∫0ηU⁢(|x|,r)⁢r2⁢𝑑r⁢x|x|,

where

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

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

s=ξ2-2⁢r⁢ξ⁢cos⁡φ+r2,φ∈]0,π[,

gives

sin⁡φ⁢cos⁡φ⁢d⁢φ=ξ2+r2-s22⁢r2⁢ξ2⁢s⁢d⁢s.

Thus we have

U⁢(ξ,r)=12⁢r2⁢ξ2⁢∫ξ-rξ+re-τ⁢s⁢(ξ2+r2-s2)⁢𝑑s=12⁢r2⁢ξ2⁢{(ξ2+r2)⁢(-e-τ⁢(ξ+r)+e-τ⁢(ξ-r))-∫ξ-rξ+re-τ⁢s⁢s2⁢𝑑s}.

Using the formula

∫ξ-rξ+re-τ⁢s⁢s2⁢𝑑s=-1τ⁢{(ξ+r)2+2τ⁢(ξ+r)+2τ2}⁢e-τ⁢(ξ+r)+1τ⁢{(ξ-r)2+2τ⁢(ξ-r)+2τ2}⁢e-τ⁢(ξ-r),

we obtain

U⁢(ξ,r)=e-τ,ξ2⁢ξ2⁢r2⁢(A⁢(ξ,r)⁢eτ⁢r-B⁢(ξ,r)⁢e-τ⁢r)

and

A⁢(ξ,r)=(ξ2+r2)-1τ⁢(ξ-r)2-2τ2⁢(ξ-r)-2τ3,

B⁢(ξ,r)=(ξ2+r2)-1τ⁢(ξ+r)2-2τ2⁢(ξ+r)-2τ3.

From this and (A.7) we obtain

(A.8)𝑰0(x;τ)=πe-τ⁢ξ∫0η(A(ξ,r)eτ⁢r-B(ξ,r)e-τ⁢r)dr|ξ=|x|x|x|3.

We have

∫0ηe±τ⁢r⁢r⁢𝑑r=±1τ⁢(η∓1τ)⁢e±τ⁢η+1τ2,

∫0ηe±τ⁢r⁢r2⁢𝑑r=±1τ⁢(η2∓2⁢ητ+2τ2)⁢e±τ⁢η∓2τ3,

∫0ηe±τ⁢r⁢r3⁢𝑑r=±1τ⁢e±τ⁢η⁢η3-3τ2⁢(η2∓2⁢ητ+2τ2)⁢e±τ⁢η+6τ4,

and thus

(A.9){∫0η(eτ⁢r-e-τ⁢r)⁢r⁢𝑑r=2τ2⁢(τ⁢η⁢cosh⁡τ⁢η-sinh⁡τ⁢η),∫0η(eτ⁢r+e-τ⁢r)⁢r⁢𝑑r=2τ2⁢(τ⁢η⁢sinh⁡τ⁢η-cosh⁡τ⁢η),∫0η(eτ⁢r+e-τ⁢r)⁢r2⁢𝑑r=2τ3⁢{(η2⁢τ2+1)⁢sinh⁡τ⁢η-2⁢η⁢τ⁢cosh⁡τ⁢η},∫0η(eτ⁢r-e-τ⁢r)⁢r2⁢𝑑r=2τ3⁢{(η2⁢τ2+2)⁢cosh⁡τ⁢η-2⁢η⁢τ⁢sinh⁡τ⁢η-2},∫0η(eτ⁢r-e-τ⁢r)⁢r3⁢𝑑r=2τ4⁢{η2⁢τ2⁢(η⁢τ+6)⁢cosh⁡τ⁢η-3⁢(η2⁢τ2+2)⁢sinh⁡τ⁢η}.

Now writing

(A.10){A⁢(ξ,r)=(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3+(1-1τ)⁢r2+2τ⁢(ξ+1τ)⁢r,B⁢(ξ,r)=(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3+(1-1τ)⁢r2-2τ⁢(ξ+1τ)⁢r,

and using (A.9), we obtain

∫0η(A⁢(ξ,r)⁢eτ⁢r-B⁢(ξ,r)⁢e-τ⁢r)⁢𝑑r

={(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3}⁢∫0η(eτ⁢r-e-τ⁢r)⁢𝑑r+(1-1τ)⁢∫0η(eτ⁢r-e-τ⁢r)⁢r2⁢𝑑r+2τ⁢∫0η(eτ⁢r+e-τ⁢r)⁢r⁢𝑑r

=2τ⁢{(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3}⁢(cosh⁡τ⁢η-1)+2τ3⁢(1-1τ)⁢{(η2⁢τ2+2)⁢cosh⁡τ⁢η-2⁢η⁢τ⁢sinh⁡τ⁢η-2}

+4τ3⁢(τ⁢η⁢sinh⁡τ⁢η-cosh⁡τ⁢η).

A combination of this and (A.8) gives the desired formula for 𝑰0⁢(x;τ).

Using the same changes of variable as used in the computation of 𝑰0⁢(x;τ), we have

(A.11)𝑰1(x;τ)=πe-τ⁢ξ∫0η(A(ξ,r)eτ⁢r-B(ξ,r)e-τ⁢r)rdr|ξ=|x|x|x|3.

Then from (A.9) and (A.10) we obtain

∫0η(A⁢(ξ,r)⁢eτ⁢r-B⁢(ξ,r)⁢e-τ⁢r)⁢r⁢𝑑r=2τ2⁢{(1-1τ)⁢ξ2-2τ2⁢ξ-2τ3}⁢(τ⁢η⁢cosh⁡τ⁢η-sinh⁡τ⁢η)

+2τ4⁢(1-1τ)⁢{η2⁢τ2⁢(η⁢τ+6)⁢cosh⁡τ⁢η-3⁢(η2⁢τ2+2)⁢sinh⁡τ⁢η}

+4τ4⁢(ξ+1τ)⁢{(η2⁢τ2+1)⁢sinh⁡τ⁢η-2⁢η⁢τ⁢cosh⁡τ⁢η}.

A combination of this and (A.11) gives the dersired formula for 𝑰1⁢(x;τ).

Similarily to the above, one obtains the expression

I2⁢(x;τ)=2⁢πτ⁢e-τ⁢|x||x|⁢∫0η(eτ⁢r-e-τ⁢r)⁢r3⁢𝑑r.

Now (A.9) gives the desired formula for I2⁢(x;τ). ∎

A.2 Lower estimates for volume integrals

Recall that B is an open ball centered at p with radius η and satisfies B¯∩D¯=∅; we have dist⁡(D,B)=d∂⁡D⁢(p)-η. In the following lemma, it is assumed that ∂⁡D is C2.

Lemma A.3.

There exist positive constants C, τ0 and κ such that

(A.12)τκ⁢e2⁢τ⁢ρ/μ⁢dist⁡(D,B)⁢∫De-2⁢τ⁢ρ/μ⁢(|x-p|-η)⁢|x-p|x-p|×𝒂|2⁢𝑑x≥C

for all τ≥τ0.

Proof.

Choose a point q∈∂⁡D such that |q-p|=d∂⁡D⁢(p). Since ∂⁡D is C2, one can find an open ball B′ with radius δ and centered at q-δ⁢𝝂q such that B′⊂D and ∂⁡B′∩∂⁡D={q}. Then dist⁡(B′,B)=dist⁡(D,B). Thus, it suffices to prove (A.12) in the case when D=B′.

Set d=d∂⁡D⁢(p) and -𝝂q=𝝎0. Let B′′ be the open ball with radius d+δ centered at p. First we give a parametrization of the domain B′′∩B′. Given s∈]0,δ[, we find the set of all unit vectors 𝝎 satisfying p+(d+s)⁢𝝎∈B′. Since the center of B′ has the expression p+(d+δ)⁢𝝎0, this condition is equivalent to the equation

|(d+s)⁢𝝎-(d+δ)⁢𝝎0|<δ.

This is equivalent to the condition

𝝎⋅𝝎0>(d+δ)2+(d+s)2-δ22⁢(d+s)⁢(d+δ).

Thus we have

B′′∩B′=⋃0<s<δ{p+(d+s)⁢ω∣ω∈S⁢(s)},

where

S⁢(s)={𝝎∈S2|𝝎⋅𝝎0>(d+δ)2+(d+s)2-δ22⁢(d+s)⁢(d+δ)}.

Choose two linearly independent vectors 𝒃 and 𝒄 in such a way that 𝒃⋅𝒄=0 and 𝒃×𝒄=𝝎0. We denote by θ(s)∈]0,π2[ the unique solution of

(A.13)cos⁡θ=(d+δ)2+(d+s)2-δ22⁢(d+s)⁢(d+δ).

Given s∈]0,δ[, r∈]0,(d+s)sinθ(s)[ and γ∈[0,2π[, set

𝚼⁢(s,r,γ)=p+(d+s)⁢cos⁡θ⁢(s)⁢𝝎0+r⁢(cos⁡γ⁢𝒃+sin⁡γ⁢𝒄)+h⁢𝝎0,

where h is an unknown parameter to be determined by the equation

|𝚼⁢(r,s,γ)-p|=d+s.

By solving this, we obtain

h=-(d+s)⁢cos⁡θ⁢(s)+(d+s)2-r2.

Thus we have

𝚼⁢(s,r,γ)=p+(d+s)2-r2⁢𝝎0+r⁢(cos⁡γ⁢𝒃+sin⁡γ⁢𝒄).

Note that |𝚼⁢(s,r,γ)-p|=d+s and the unit vector (𝚼⁢(s,r,γ)-p)/(d+s) belongs to S⁢(s) for each fixed s. It is easy to check also that the map

G∋(s,r,γ)↦𝚼⁢(s,r,γ)∈B′′∩B′

is bijective, where

G={(s,r,γ)∣(s,r)∈G′,γ∈[0,2π[}

and

G′={(s,r)∣0<s<δ, 0<r<(d+s)⁢sin⁡θ⁢(s)}.

A simple computation gives

det⁡𝚼′⁢(s,r,γ)=r⁢(d+s)(d+s)2-r2.

Here we have

(A.14)(𝚼⁢(s,r,γ)-p)×𝒂=(d+s)2-r2⁢𝝎0×𝒂+r⁢(cos⁡γ⁢𝒃×𝒂+sin⁡γ⁢𝒄×𝒂)≡𝑨⁢(s,r,γ)⁢𝒂.

Using the change of variables formula and B′′∩B′⊂B′, we obtain

e2⁢τ⁢ρ/μ⁢(d-η)⁢∫B′e-2⁢τ⁢ρ/μ⁢(|x-p|-η)⁢|x-p|x-p|×𝒂|2⁢𝑑x

(A.15)≥∫0δ𝑑s⁢∫0(d+s)⁢sin⁡θ⁢(s)𝑑r⁢∫02⁢π𝑑γ⁢e-2⁢τ⁢ρ/μ⁢sd+s⁢|𝑨⁢(s,r,γ)⁢𝒂|2⁢r(d+s)2-r2.

It is easy to see that from (A.14) one gets

∫02⁢π|𝑨⁢(s,r,γ)⁢𝒂|2⁢𝑑γ=2⁢π⁢{(d+s)2-r2}⁢|𝝎0×𝒂|2+π⁢r2⁢(|𝒃×𝒂|2+|𝒄×𝒂|2).

Also,

∫0(d+s)⁢sin⁡θ⁢(s)r⁢(d+s)2-r2⁢𝑑r=(d+s)33⁢(1-cos3⁡θ⁢(s)),

∫0(d+s)⁢sin⁡θ⁢(s)r3⁢d⁢r(d+s)2-r2=(d+s)3⁢{(1-cos⁡θ⁢(s))+1-cos3⁡θ⁢(s)3}.

Thus,

∫0(d+s)⁢sin⁡θ⁢(s)r⁢d⁢r(d+s)2-r2⁢∫02⁢π|𝑨⁢(s,r,γ)⁢𝒂|2⁢𝑑γ

=2⁢π3⁢(d+s)3⁢(1-cos3⁡θ⁢(s))⁢|𝝎0×𝒂|2+π⁢(d+s)3⁢{(1-cos⁡θ⁢(s))-1-cos3⁡θ⁢(s)3}⁢(|𝒃×𝒂|2+|𝒄×𝒂|2).

From (A.13) we have

(d+s)⁢(1-cos⁡θ⁢(s))=s⁢(2⁢δ-s)2⁢(d+δ)≥δ⁢s2⁢(d+δ).

This gives

(d+s)3⁢(1-cos3⁡θ⁢(s))≥δ⁢(d+s)2⁢s2⁢(d+δ)⁢(1+cos⁡θ⁢(s)+cos2⁡θ⁢(s))≥δ⁢d⁢(d+s)⁢s2⁢(d+δ).

Also,

(d+s)3⁢{(1-cos⁡θ⁢(s))-1-cos3⁡θ⁢(s)3}≥δ⁢(d+s)2⁢s2⁢(d+δ)⁢(1-1+cos⁡θ⁢(s)+cos2⁡θ⁢(s)3)

=δ⁢(d+s)2⁢s6⁢(d+δ)⁢(1-cos⁡θ⁢(s))⁢(2+cos⁡θ⁢(s))

≥δ2⁢(d+s)⁢s212⁢(d+δ)2⁢(2+cos⁡θ⁢(s))

≥δ2⁢(d+s)⁢s26⁢(d+δ)2.

Thus we have

∫0s𝑑s⁢e-2⁢τ⁢ρ/μ⁢sd+s⁢∫0(d+s)⁢sin⁡θ⁢(s)r⁢d⁢r(d+s)2-r2⁢∫02⁢π|𝑨⁢(s,r,γ)⁢𝒂|2⁢𝑑γ

(A.16)≥π3⁢δ⁢dd+δ⁢∫0δs⁢e-2⁢τ⁢ρ/μ⁢s⁢𝑑s⁢|𝝎0×𝒂|2+π6⁢(δd+δ)2⁢∫0δs2⁢e-2⁢τ⁢ρ/μ⁢s⁢𝑑s⁢(|𝒃×𝒂|2+|𝒄×𝒂|2).

Here we have, for j=1,2,

∫0δsj⁢e-2⁢τ⁢ρ/μ⁢s⁢𝑑s=1(2⁢τ⁢ρ/μ)j+1+O⁢(τ-1⁢e-2⁢τ⁢ρ/μ⁢δ).

Therefore, from these and (A.15) and (A.16) one can conclude that if 𝝎0×𝒂≠𝟎, then one can choose κ=2 in (A.12), and if 𝝎0×𝒂=𝟎, then 𝒂=±𝝎0 and thus |𝒃×𝒂|2+|𝒄×𝒂|2>0, and one can choose κ=3 in (A.12). ∎

Lemma A.4.

There exist positive constants C and τ0 such that

τ2⁢e2⁢τ⁢dist⁡(D,B)⁢∫De-2⁢τ⁢(|x-p|-η)⁢𝑑x≥C

for all τ≥τ0.

Proof.

From the proof of Lemma A.3 we have

e2⁢τ⁢(d-η)⁢∫De-2⁢τ⁢(|x-p|-η)⁢𝑑x≥e2⁢τ⁢(d-η)⁢∫B′e-2⁢τ⁢(|x-p|-η)⁢𝑑x

≥2⁢π⁢∫0δ𝑑s⁢∫0(d+s)⁢sin⁡θ⁢(s)𝑑r⁢e-2⁢τ⁢s⁢r⁢(d+s)(d+s)2-r2

=2⁢π⁢∫0δ(d+s)⁢e-2⁢τ⁢s⁢𝑑s⁢∫0(d+s)⁢sin⁡θ⁢(s)r(d+s)2-r2⁢𝑑r

=π⁢∫0δ(d+s)2⁢(1-cos⁡θ⁢(s))⁢e-2⁢τ⁢s⁢𝑑s

≥π⁢∫0δ(d+s)⋅δ⁢s2⁢(d+δ)⋅e-2⁢τ⁢s⁢𝑑s

≥π⁢δ⁢d2⁢(d+δ)⁢∫0δs⁢e-2⁢τ⁢s⁢𝑑s.

Since

∫0δs⁢e-2⁢τ⁢s⁢𝑑s=14⁢τ2⁢∫02⁢τ⁢δξ⁢e-ξ⁢𝑑ξ,

we obtain the desired conclusion.∎

Acknowledgements

The author thanks Hiromichi Itou for pointing out the reference [4] and for useful discussions.

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Received: 2017-6-25

Accepted: 2018-1-6

Published Online: 2018-1-30

Published in Print: 2018-6-1

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Articles in the same Issue

https://doi.org/10.1515/jiip-2017-0066

Keywords for this article

Enclosure method; inverse obstacle problem; dynamic theory of thermoelasticity, displacement-temperature equation of motion; coupled heat equation; coupled system; cavity, non-destructive testing