URL:http://www.emath.fr/cocv/ DOI: 10.1051/cocv:2002001
UNIQUE LOCALIZATION OF UNKNOWN BOUNDARIES
IN A CONDUCTING MEDIUM FROM BOUNDARY MEASUREMENTS
Bruno Canuto
1Abstract. We consider the problem of localizing an inaccessible pieceI of the boundary of a con- ducting medium Ω, and a cavity D contained in Ω, from boundary measurements on the accessible part A of ∂Ω. Assuming that g(t, σ) is the given thermal flux for (t, σ) ∈ (0, T)×A, and that the corresponding output datum is the temperatureu(T0, σ) measured at a given timeT0forσ∈Aout⊂A, we prove thatIandDare uniquely localized from knowledge of all possible pairs of input-output data (g, u(T0)|Aout). The same result holds when a mean value of the temperature is measured over a small interval of time.
Mathematics Subject Classification. 35R30.
Received March 3, 2000. Revised February 12, 2001.
1. Introduction and main results
In the present paper we are concerned with the study of some problems in thermal imaging. This is a technique used to determine some physical proprieties of a thermic conducting mediumvia boundary measurements of temperature. More precisely we denote by Ω the medium,i.e. a bounded and sufficiently smooth domain inRN, N ≥2. Suppose that a piece I of the boundary of Ω is unknown and inaccessible to direct inspections. On the other hand we have access to the remaining partA:= (∂Ω)\I of ∂Ω. Letg be the thermal flux assigned on (0, T)×A, andu(T0)|Aout the corresponding temperature measured at a given timeT0 >0 on a pieceAout
ofA. The goal is to identify the unknown partI, by knowing all possible pairs of data (g, u(T0)|Aout). In a similar problem we might suppose that a cavity D, of which neither the form nor the position is known, is contained in Ω (i.e. D is a domain contained in Ω), and the whole boundary of Ω is known and accessible to measurements. In this case the goal is to identify the cavityD via the same previous data. In fact we are concerned with the problem in which one tries to identify both a pieceI of the boundary of Ω and a cavity D in its interior from all pairs of data (g, u(T0)|Aout). This problem can occur in nondestructive tests of materials, for example in detecting the corrosion parts of an aircraft which are not accessible to direct inspections. In this caseI andDrepresent the damaged and inaccessible parts of the aircraft, and u(T0)|Aout the measurements of temperature that one disposes to attempt to recoverIandD(see Bryan and Caudill [1], and their references).
We denote by u(t, x) the temperature at the time t and at the point x∈ Ω\D, u0 the initial temperature in Ω\D, ϕ,ψ, andg the flux on (0, T)×∂D, (0, T)×I, and (0, T)×Arespectively, and κ(x) the anisotropic
Keywords and phrases:Inverse boundary value problems, cavities, corrosion, uniqueness.
1 Laboratoire de Math´ematiques Appliqu´ees, UMR 7641, Universit´e de Versailles, 45 avenue des ´Etats-Unis, 78035 Versailles Cedex, France; e-mail:[email protected]
c EDP Sciences, SMAI 2002
the inaccessible part of the boundary
D Ω Ω
D Ω
(i) (ii) (iii)
the cavity
the accessible part of the boundary
Figure 1. The pictures (i), (ii), (iii) show the three different situations.
thermal diffusion coefficient, that isκis a symmetricN×N matrix-valued function in Ω satisfying the following conditions:
(i) there exists a constantα >0, such that for all x∈Ω, and for allξ∈RN,
κ(x)ξ·ξ≥α|ξ|2 (ellipticity), (1.1) (ii) there exists a constantC≥0, such that for allx, y∈Ω,
|κ(x)−κ(y)| ≤C|x−y| (Lipschitz continuity). (1.2) For Ω, D, κ, u0, ϕ, ψ, g assigned, suppose that u solves the following parabolic problem, which we call the direct problem:
∂tu−div(κ(x)∇u) = 0 in (0, T)×Ω\D, u(0) = u0 in Ω\D,
κ∇u·n = ϕ(t, σ) on (0, T)×∂D, κ∇u·n = ψ(t, σ) on (0, T)×I, κ∇u·n = g(t, σ) on (0, T)×A,
(1.3)
wherendenotes the outer unit normal at∂(Ω\D). Here and in the sequelIis a relativelyopenpiece of∂Ω. It is well-known that, under reasonable assumptions on the data, problem (1.3) has a unique solution, and that the temperatureu(t, σ) is well-defined for (t, σ) ∈(0, T)×∂Ω. In the present paper we are interested in the following problem:
Let Λ be the so-called input-output map, that is
Λ :g7−→u(T0)|Aout, (1.4)
where T0∈(0, T] is a given fixed time andAout is a relatively open piece ofA, and let Φ be the operator
Φ : (I, D)7−→Λ. (1.5)
Is Φinjective?
We point out that to prove the injectivity of the operator Φ is equivalent to show the uniqueness ofI andD from knowledge of all possible pairs of input-output data
(g, u(T0)|Aout)
of the solutionuof (1.3), that is from all possible measurements of temperatureu(T0)|Aout at a given timeT0
onAout. We observe moreover that the temperature is measured at a given timeT0only, instead of measuring it over a whole interval of time such as [0, T0]. We study also the problem in which ameanvalue of the temperature is measured over a small interval of time. We note finally that the initial temperature u0, and the boundary dataϕ,ψin (1.3) are givenarbitrarily. This assumption corresponds to a real situation in which the data u0, ϕ, ψarea prioriunknown.
A similar problem has been investigated by Vessella [8]. He proves the unique localization of a thermic insulating regionDin Ω (in (1.3) he supposesκ∇u·n= 0 on (0, T)×∂D,Iknown, andA=∂Ω) from asingle measurement of temperatureu|(t0,t1)×Aout on [t0, t1]×Aout, where (t0, t1) is a subinterval of [0, T], and Aout
is a relatively open piece of∂Ω, provided that Ω is a contractible domain, the initial temperature u0 in (1.3) is constant, and the inputg is monotone with respect to the time variablet. Vessella’s proof is based on the unique continuation principle and the maximum principle for parabolic equations. Moreover Vessella shows, whenN = 3, and κ=I3 (I3 is the 3×3 identity matrix), a continuous dependence oflogarithmictype of the domainDfrom the temperature u|(t0,t1)×Aout.
In order to prove the injectivity of the operator Φ in (1.5), we will inspire with the so-called boundary spectral data method, introduced in [2] by the author and Kavian to show the identifiability of coefficients in a class of heat equationsvia boundary measurements. This method consists in studying the identifiability of the boundary spectral data for the underlying elliptic operator in (1.3) from the input-output map Λ. More precisely let (λk)∞k=1, (ϕk)∞k=1 be respectively the nondecreasing sequence of eigenvalues and the corresponding eigenfunctions of the problem (with Neumann boundary conditions):
−div(κ∇ϕk) = λkϕk in Ω\D, κ∇ϕk·n = 0 on∂D, κ∇ϕk·n = 0 on∂Ω, Z
Ω\D
|ϕk|2dx = 1.
(1.6)
Let us denote by DBSD(I, D) the so-calledDirichlet Boundary Spectral Data, i.e.
DBSD(I, D) := (λk, ϕk|A)∞k=1. (1.7)
The question we ask is the following:
Does the input-output mapΛ determine the Dirichlet Boundary Spectral DataDBSD(I, D)uniquely?
The first result of the present paper is the following:
Theorem 1.1. For an integerN ≥2, letΩj, forj ∈ {0,1}, be two bounded domains inRNof classC0,1, having common boundary A := (∂Ωj)\Ij, Ij being a relatively open piece of ∂Ωj, and letDj be aC0,1 domains such
A:
D 0
D Ω 1
Ω 0
1
to observations the part accessible
Figure 2. The domains Ω0, Ω1, and the cavities D0,D1.
that the closureDj ⊂Ωj. Let us denote byκa symmetric N×N matrix-valued function in Ω0∪Ω1 satisfying conditions (1.1, 1.2) in Ω0∪Ω1. Let uj0 ∈ H1(Ωj\Dj), ϕj ∈ C([0, T];L2(∂Dj)), ψj ∈ C([0, T];L2(Ij)), g∈C([0, T];L2(A))be such thatuj∈C((0, T];H1(Ωj\Dj))∩C1([0, T];L2(Ωj\Dj))solve (1.3), whenΩ := Ωj, and D:=Dj. Suppose that
Λ0(g) = Λ1(g) inL2(Aout), (1.8)
whereΛj(g) :=uj(T0)|Aout, for allg ∈C([0, T];L2(A)) such that the supp(g(t,·))⊂A for t∈[0, T]. Then the Dirichlet Boundary Spectral DataDBSD(Ij, Dj)(1.7) (when Ω := Ωj, andD:=Dj in (1.6)) coincide, that is, up to an appropriate choice of the eigenfunctionsϕ0k, for all k≥1, one has
λ0k=λ1k, and ϕ0k=ϕ1k a.e on A.
We point out that the conclusion of Theorem 1.1 remains valid if we replace hypothesis (1.8) by equality of the mean values of the temperatures in the interval [τ0−T0, T0], that is the following result holds:
Theorem 1.2. Let 0< τ0< T0 be given. Under the assumptions of Theorem 1.1 assume that
T0
Z
T0−τ0
u0(t)|Aoutdt=
T0
Z
T0−τ0
u1(t)|Aoutdt inL2(Aout), (1.9)
for allg∈C([0, T] ;L2(A))such that the supp(g(t,·))⊂Afor t∈[0, T]. Then the Dirichlet Boundary Spectral DataDBSD(Ij, Dj)(1.7) coincide, that is, up to an appropriate choice of the eigenfunctionsϕ0k, for allk≥1, one has that
λ0k=λ1k, and ϕ0k=ϕ1k a.e. onA.
We now consider the Dirichlet case. More precisely let (µk)∞k=1, (ψk)∞k=1 be respectively the nondecreasing sequence of eigenvalues and the corresponding eigenfunctions of the following problem (with Dirichlet boundary conditions):
−div(κ∇ψk) = µkψk in Ω\D, ψk = 0 on∂D, ψk = 0 on∂Ω, Z
Ω\D
|ψk|2dx = 1.
(1.10)
Let us denote by NBSD(I, D) the so-calledNeumann Boundary Spectral Data,i.e.
NBSD(I, D) := (µk, κ∇ψk·n|A)∞k=1. (1.11) Then the conclusion of Theorem 1.1 remains valid if in (1.3) we substitute the Neumann boundary conditions with the corresponding Dirichlet boundary conditions. More precisely the following result holds:
Theorem 1.3. Under the hypothesis of Theorem 1.1, assume that A := (∂Ωj)\Ij is of class C1,1. Let uj0 ∈ H1(Ωj\Dj), ϕj ∈ C([0, T];H12(∂Dj)), ψj ∈ C([0, T];H12(Ij)), f ∈ C([0, T];H32(A)) be such that uj ∈C((0, T];H1(Ωj\Dj))∩C1([0, T];L2(Ωj\Dj))solve
∂tuj−div(κ(x)∇uj) = 0 in (0, T)×Ωj\Dj, uj(0) = uj0 in Ωj\Dj,
uj = ϕj(t, σ) on (0, T)×∂Dj, uj = ψj(t, σ) on (0, T)×Ij, uj = f(t, σ) on (0, T)×A.
(1.12)
We denote by
Λej(f) :=κ∇uj(T0)·n|Aout (1.13) the thermal fluxes measured at a given timeT0∈(0, T] onAout. Suppose that
Λe0(f) =Λe1(f) in L2(Aout) (1.14) for allf ∈C([0, T];H32(A))such that thesupp(f(t,·)⊂Afor t∈[0, T]. Then the Neumann Boundary Spectral Data NBSD(Ij, Dj) (1.11) (when Ω := Ωj, and D := Dj in (1.10)) coincide, that is, up to an appropriate choice of the eigenfunctionsψ0k, for allk≥1, one has:
µ0k=µ1k, and κ∇ψ0k·n=κ∇ψ1k·n a.e on A.
Remark 1.4. Theorem 1.3 holds true if we replace hypothesis (1.14) by equality of the mean values of the fluxes in the interval [τ0−T0, T0],i.e. we suppose that, for a fixed τ0, 0< τ0< T0,
T0
Z
T0−τ0
κ∇u0(t)·n|Aoutdt=
T0
Z
T0−τ0
κ∇u1(t)·n|Aoutdt in L2(Aout),
for allf ∈C([0, T];H32(A)) such that the support supp(f(t,·))⊂Afort∈[0, T].
Once the result of Theorem 1.1 is at hand, we can prove the injectivity of the operator Φ defined in (1.5).
This is proved in the following:
Theorem 1.5. Under the assumptions of Theorem 1.1, for N = 2, 3, suppose that the Dirichlet Boundary Spectral DataDBSD(Ij, Dj)(1.7) (when Ω := Ωj, andD:=Dj in (1.6)) coincide, that is, for all k≥1,
λ0k =λ1k, and ϕ0k=ϕ1k a.e. onA.
Then I0=I1 andD0=D1.
The conclusion of Theorem 1.5 remains valid if we assume that the Neumann Boundary Spectral Data NBSD(Ij, Dj) coincide:
Theorem 1.6. Under the assumptions of Theorem 1.3, for N = 2, 3, suppose that the Neumann Boundary Spectral DataNBSD (Ij, Dj)(1.11) (whenΩ := Ωj, andD:=Dj in (1.10)) coincide, that is, for all k≥1,
µ0k=µ1k, κ∇ψ0k·n=κ∇ψ1k·n a.e. onA.
Then I0=I1 andD0=D1.
The remainder of the paper is organized as follows: in Section 2 we gather some preliminary results and the notations used throughout; in Section 3 we prove Theorems 1.1–1.3; in Section 4 we prove Theorems 1.5 and 1.6.
2. Preliminary results
We denote byΩ a bounded domain ine RN,N ≥2, with boundary of classC0,1.
Byκ(x) we mean a symmetricN×N matrix-valued function inΩ satisfying conditions (1.1, 1.2) ine Ω.e We denote by (L,D(L)) the elliptic operator
Lv:=−div(κ(x)∇v), (2.1)
with domain
D(L) :=n
v∈L2(eΩ); Lv∈L2(eΩ), κ∇v·n|∂Ωe= 0o
· (2.2)
Actually whenκ∈C0,1(Ω) ande ∂Ωe∈C0,1, then D(L) =n
v∈Hloc2 (eΩ)∩H1(eΩ); κ∇v·n|∂Ωe= 0o
·
The operatorL possesses a sequence of eigenvalues (λk)∞k=1 (which we suppose in a nondecreasing order) and corresponding eigenfunctions (ϕk)∞k=1 satisfying:
−div(κ∇ϕk) = λkϕk inΩ,e κ∇ϕk·n = 0 on∂Ω,e Z
Ωe
|ϕk|2dx = 1, (2.3)
which form a Hilbert basis of L2(eΩ). We recall that the first eigenvalue λ1 = 0, and the corresponding eigenfunction ϕ1(x) = eΩ−
1
2, where|·| denotes the Lebesgue measure ofΩ. It is also known that the domaine D(L) can be characterized by
D(L) = (
v∈L2(eΩ);
X∞ k=1
λ2k|(v|ϕk)|2<+∞ )
, (2.4)
where (· | ·) is the inner product in L2(eΩ).
We denote bymk the geometric multiplicity ofλk. We recall that the eigenvaluesλk behave like λk ∼C0kN2 ask→+∞,
where the constant C0 depends on κ, eΩ, N (see Courant and Hilbert [5], pp. 442-443). Moreover there exist two positive constantsC1,C2 such that, for allk≥1, one has:
C1λk ≤ kϕkk2H1(eΩ)≤C2λk. (2.5) The following three lemmas, which are identical to Lemmas 2.1–2.3 respectively in Canuto and Kavian [2], are a tools to prove Theorem 1.1 later on. First we shall need the following result concerning the linear independence, or linear dependence, of the family (ϕk|∂Ωe)k≥1. In general these functions are not linearly independent. However one can show that the traces on the boundary of Ω of eigenfunctions corresponding to a given eigenvaluee λk0
are actually independent. More precisely ifλk is an eigenvalue of Lhaving multiplicity mk ≥1, let us denote byϕk,i for 1≤i≤mk the eigenfunctions corresponding to the eigenvalueλk which form a Hilbert basis of the kernelN(L−λkI). We may state the following:
Lemma 2.1. For an integerN ≥2, letΩe be a bounded domain inRN of classC0,1. If, for a fixed k≥1,λk is an eigenvalue of multiplicitymk ≥1of L, and ifΓis a relatively open piece of ∂Ω, then the dimension of thee subspace spanned inL2(Γ) by(ϕk,i|∂Ωe)1≤i≤mk is exactlymk.
Proof of Lemma 2.1. Indeed if there exists (ci)mi=1k ∈Rmk such that
mk
X
i=1
ciϕk,i = 0 on Γ,
then, setting ϕ:=
mk
P
i=1
ciϕk,i, one checks that
Lϕ=λkϕ in Ω,e κ∇ϕ·n= 0 on∂Ω,e ϕ= 0 on Γ.
Now, following a standard argument, we extend ϕ ≡ 0 in an exterior neighborhood of Γ0 ⊂ Γ. The unique continuation principle (see Garofalo and Lin [4]) implies then that ϕ ≡ 0 in Ω. Due to the fact that thee functions ϕk,i are linearly independent, we conclude that ci = 0 for 1 ≤i≤mk. The proof of Lemma 2.1 is
complete.
From this we conclude the following:
Lemma 2.2. Under the assumptions of Lemma 2.1, letΓ1 andΓ2 be two relatively open pieces of∂Ω. For ae fixedk≥1consider the function Ξk defined by
Ξk(σ0, σ) :=
mk
X
i=1
ϕk,i(σ0)ϕk,i(σ) onΓ1×Γ2.
Then Ξk(σ0, σ)is not identically zero on any relatively open subset ofΓ1×Γ2.
Proof of Lemma 2.2. By contradiction, let Γ01 be a relatively open piece of Γ1, and let Γ02 be a relatively open piece of Γ2 such that
Ξk(σ0, σ)≡0 on Γ01×Γ02. (2.6)
By Lemma 2.1, we have that (2.6) implies that ϕk,i ≡ 0 on Γ01 for i = 1,· · · , mk, and so by the unique continuation principle it follows thatϕk,i ≡0 onΩ, which leads to a contradiction.e
We shall also need the following algebraic lemma:
Lemma 2.3. For two arbitrary integersm,n≥1, letZ be a non empty set, and letX,Y be two subsets ofZ.
Assume thatfi:X∪Y →R(for1≤i≤m) andg`:X∪Y →R(for1≤`≤n) are functions such that (i)
Xm i=1
fi(x)fi(y) = Xn
`=1
g`(x)g`(y) for(x, y)∈X×Y; (2.7) (ii) X∩Y contains infinitely many points;
(iii) fi, for 1≤i≤m, (resp. g`, for 1≤`≤n,) are not identically zero in X∩Y; (iv) {f1, . . . , fm}(resp.{g1, . . . , gn}) are linearly independent in X∩Y.
Then m=n, and denoting
F(x) :=
f1(x)
... fm(x)
and G(x) :=
g1(x)
... gn(x)
,
there exists anm×morthogonal matrixM such that for allz∈X∪Y one hasF(z) =M G(z).
(Recall that by an orthogonal matrixM we mean M M∗ =M∗M = Im, whereIm is them×m identity matrix.) For the reader’s convenience we give the proof of Lemma 2.3, although is identical to which of Lemma 2.3 in [2].
Proof of Lemma 2.3. Let us denote byV0 (resp. V1) the space spanned by{f1,· · ·, fm}(resp. {g1,· · ·, gn}).
Sincef1 is not identically zero inX∩Y, there existsx1∈X∩Y such thatf1(x1)6= 0; then,f2being linearly independent off1, and X∩Y containing infinitely many points, there existsx2∈X∩Y such that:
det
f1(x1) f2(x1) f1(x2) f2(x2)
6
= 0.
By induction one sees that we may find pointsx1, x2, . . . , xmin X∩Y such that them×mmatrix
P:=
f1(x1) f2(x1) · · · fm(x1) f1(x2) f2(x2) · · · fm(x2)
... ... ... ... f1(xm) f2(xm) · · · fm(xm)
is invertible. So, setting x=xj in (2.7), it follows thatP F(y) =P G(y) ine Y, wherePe is the following m×n matrix
Pe:=
g1(x1) g2(x1) · · · gn(x1) g1(x2) g2(x2) · · · gn(x2)
... ... ... ... g1(xm) g2(xm) · · · gn(xm)
.
From this it follows thatF(y) =P−1P G(y) for alle y∈Y, whereP−1 is the inverse matrix ofP. Similarly, changing the role of the variablesxandy, we obtain thatF(x) =P−1P G(x) ine X, that is
F(z) =M G(z) inX∪Y
whereM:=P−1P. Therefore, recalling that the functionse {f1, . . . , fm}and{g1, . . . , gn}are linearly indepen- dent inX∩Y, it follows thatV0⊆V1, that ism≤n. In the same way one may prove thatn≤m, and so we conclude thatm=n.
Finally we prove that the matrixM = P−1Pe is orthogonal. Indeed, recalling that F(z) = M G(z) for all z∈X∪Y and using (2.7), we obtain
(M∗M−Im)G(x)·G(y) = 0 in X×Y,
where a · b denotes the euclidean scalar product in Rm, M∗ is the transpose matrix of M, and Im is the m×m identity matrix. Since the functions {g1,· · · , gm} are linearly independent in X ∩Y it follows that M∗M=M M∗=Im, that isM is orthogonal.
The proof of Lemma 2.3 is complete.
3. Proof of Theorems 1.1, 1.2 and 1.3
The first task in this section is to prove Theorem 1.1. Before doing so we need to establish some preliminary lemmas.
Lemma 3.1. For an integerN≥2, letΩe be a bounded domain inRN of classC0,1. Forg∈C([0, T];L2(∂Ω)),e let u∈C((0, T];H1(eΩ))∩C1([0, T];L2(eΩ))solve
∂tu−div(κ(x)∇u) = 0 in(0, T)×Ω,e u(0) = 0 inΩ,e
κ∇u·n = g on(0, T)×∂Ω,e
(3.1)
whereκsatisfies assumptions (1.1, 1.2) inΩ. Thene ucan be written in the following Fourier expansion:
u(t) = X∞ k=1
αk(t)ϕk inL2(Ω),e (3.2)
whereϕk is defined in (2.3), and
αk(t) :=− Zt 0
Z
∂Ωe
ϕk(σ0)e−λk(t−τ)g(τ, σ0)dσ0dτ. (3.3)
Proof of Lemma 3.1. Since (ϕk)∞k=1 is a Hilbert basis in L2(eΩ), we can write u in the following Fourier expansion:
u(t) = X∞ k=1
αk(t)ϕk in L2(Ω),e
whereαk(t) := (u(t)|ϕk). Multiplying the equation in (3.1) byϕk, and integrating by parts overΩ, we obtain,e for allk≥1,
α0k(t) +λkαk(t) = −Z
∂eΩ
ϕk(σ0)g(t, σ0)dσ0 in (0, T), αk(0) = 0,
(3.4)
where α0k(t) denotes the derivative ofαk(t). The thesis of the lemma follows then trivially.
Lemma 3.2. Under the assumptions of Lemma 3.1, letg satisfy the following condition:
g≡0 on [T0−ε0, T0]×∂Ω,e (3.5)
whereT0∈(0, T], and0< ε0< T0. Thenu(T0)|∂Ωe can be written in the following way
u(T0)|∂eΩ=−
TZ0−ε0 0
Z
∂Ωe
X∞ k=1
ϕk(σ0) e−λk(T0−τ)g(τ, σ0)dσ0dτ
ϕk|∂Ωe inH12(∂Ω).e
Proof of Lemma 3.2. We divide the proof into two steps:
Step 1. Ifg satisfies (3.5), thenu(T0)∈ D(L), where D(L) is defined in (2.2). So it follows that (see (2.4)) X∞
k=1
λ2k|αk(T0)|2<+∞,
whereαk(T0) := (u(T0)|ϕk). Settingum(T0) :=
Pm k=1
αk(T0)ϕk, then
um(T0)→u(T0) inL2(Ω)e asm→+∞, and
Lum(T0)−Lu(T0)→0 in L2(eΩ) asm→+∞. By the fact that kum(T0)kH1(Ω)e ≤CkLum(T0)kL2(eΩ), we have
u(T0) = X∞ k=1
αk(T0)ϕk in H1(Ω).e (3.6)
Now, since the trace operatorγ:u(T0)→u(T0)|∂Ωe is continuous fromH1(eΩ) toH12(∂Ω), one has, in the sensee ofH12(∂Ω):e
u(T0)|∂Ωe =− X∞ k=1
TZ0−ε0 0
Z
∂eΩ
ϕk(σ0)e−λk(T0−τ)g(τ, σ0)dσ0dτ
ϕk|∂eΩ. (3.7)
Step 2. In this step we prove that we can commute the series sign with the integral signs in the right hand side of (3.7). By Fubini’s theorem it is sufficient, for example, to show that
I:=
TZ0−ε0 0
X∞ k=1
kϕkkH12(∂Ω)e
Z
∂Ωe
ϕk(σ0)e−λk(T0−τ)g(τ, σ0)dσ0dτ <+∞.
In fact, denoting byh·,·ithe dualityH−12(∂Ω),e H12(∂Ω), we havee I ≤
TZ0−ε0
0
X∞ k=1
|hϕk, g(τ)i| kϕkkH12(∂Ω)e e−λkε0dτ
≤
TZ0−ε0 0
kg(τ)kH−1 2(∂Ω)e dτ
X∞ k=1
kϕkk2H12(∂Ω)e e−λkε0
≤ C X∞ k=1
λke−λkε0,
where the last inequality is obtained upon using the fact that, by the trace inequality, and (2.5), we have:
kϕkkH12(∂Ω)e ≤CkϕkkH1(eΩ)≤Cλk12.
Note thatP∞
k=1λke−λkε0 <+∞sinceλk ∼kN2, ask→+∞. Therefore we may write equation (3.7) as:
u(T0)|∂eΩ=−
T0Z−ε0
0
Z
∂Ωe
X∞ k=1
ϕk(σ0)e−λk(T0−τ)g(τ, σ0)dσ0dτ
ϕk|∂Ωe inH12(∂Ω).e (3.8)
The proof of Lemma 3.2 is complete.
Lemma 3.3. Under the assumptions of Theorem 1.1, forj∈ {0,1}, letuj be solutions of (1.3) whenΩ := Ωj, and D:=Dj, with initial data uj0≡0inΩj\Dj, and boundary data ϕj ≡0on (0, T)×∂Dj, and ψj ≡0on (0, T)×Ij. Then
Λ0(g) = Λ1(g) inL2(Aout),
whereΛj(g) :=uj(T0)|Aout, for all g∈([0, T];L2(A))such that the supp(g(t,·))⊂A for t∈[0, T].
Proof of Lemma 3.3. Following Rakesh and Symes [6], we putuj(t, x) :=uj(t, x)−vj(t, x), wherevj solve (1.3) when Ω := Ωj, D :=Dj, with data vj(0) = uj0 in Ωj\Dj, vj = ϕj on (0, T)×∂Dj, vj =ψj on (0, T)×Ij,
andvj≡0 on (0, T)×A. Thenuj0≡0 in Ωj\Dj,ϕj≡0 on (0, T)×∂Dj, andψj ≡0 on (0, T)×Ij. So, if we denote by Λj(g) :=uj(T0)|Aout, it follows that
Λj(g) = Λj(g)−Λj(0).
Hence Λ0(g) = Λ1(g) for allg∈C([0, T];L2(A)) such that the support supp(g(t,·))⊂Afort∈[0, T].
The proof of Lemma 3.3 is complete.
Lemma 3.4. Under the assumptions of Theorem 1.1, for allk≥1, we have λ0k =λ1k,
and
mX0k
i=1
ϕ0k,i(σ0)ϕ0k,i(σ) =
m1k
X
`=1
ϕ1k,`(σ0)ϕ1k,`(σ) a.e. onA×Aout.
(As we have mentioned in Sect. 2,mjk is the multiplicity of the eigenvalueλjk.)
Proof of Lemma 3.4. First of all, using Lemma 3.3, we can always reduce to the case where the initial data uj0≡0 in Ωj\Dj, and the boundary dataϕj ≡0 on (0, T)×∂Dj, andψj≡0 on (0, T)×Ij. We recall that by hypothesis we have
u0(T0)|Aout =u1(T0)|Aout inL2(Aout), (3.9) for all g∈C([0, T];L2(A)) such that the support supp(g(t,·))⊂A fort∈[0, T]. By Lemma 3.2 we know that if g ≡ 0 on [T0−ε0, T0]×A, then uj(T0)|Aout on Aout can be written in the following way (in the sense of H12(Aout)):
uj(T0)|Aout=−
TZ0−ε0 0
Z
A
X∞ k=1
Φjk(σ0, T0−τ)g(τ, σ0)dσ0dτ,
where
Φjk(σ0, τ) :=
X∞ k=1
ϕjk(σ0) e−λjkτϕjk|Aout. (3.10) Then, from (3.9), it follows that
TZ0−ε0 0
Z
A
X∞ k=1
(Φ0k(σ0, T0−τ)−Φ1k(σ0, T0−τ))g(τ, σ0)dσ0dτ= 0 inL2(Aout). (3.11)
In particular we may assume thatg(τ, σ0)≡0 forτ 6= [T0−ε0, T0+ε0], and σ0 ∈A, where T0 is a fixed time, T0 ∈(0, T0−ε0), and 0< ε0< T0. Then (3.11) becomes
TZ0+ε0 T0−ε0
Z
A
X∞ k=1
(Φ0k(σ0, T0−τ)−Φ1k(σ0, T0−τ))g(τ, σ0)dσ0dτ= 0
for all such functionsg. Hence we have X∞ k=1
Φ0k(σ0, τ) = X∞ k=1
Φ1k(σ0, τ) inL2(Aout)
for allσ0 ∈A,τ ∈[T0−ε0, T0+ε0]. By the unique continuation principle for analytic functions of the variableτ, we obtain
X∞ k=1
Ξ0k(σ0)e−λ0kτ = X∞ k=1
Ξ1k(σ0)e−λ1kτ in L2(Aout) (3.12)
for allσ0 ∈A,τ∈(0,∞), where Ξjk(σ0) :=
mPjk
i=1
ϕjk,i(σ0)ϕjk,i|Aout,j ∈ {0,1}. By Lemma 2.2 we know that, for all fixedk≥1, Ξjk(σ0) is not identically zero on any relatively open subset ofA. Therefore, using the classical results on Dirichlet’s series, equation (3.12) yields that, for allk≥1,
λ0k =λ1k, and
Ξ0k= Ξ1k inL2(A)×L2(Aout), that is
m0k
X
i=1
ϕ0k,i(σ0)ϕ0k,i(σ) =
m1k
X
i=1
ϕ1k,i(σ0)ϕ1k,i(σ) (3.13)
a.e. onA×Aout. The proof of Lemma 3.4 is complete.
We are now in a position to prove Theorem 1.1.
Proof of Theorem 1.1. We prove that (3.13) implies that, for allk≥1,m0k =m1k, and, up to an appropriate choice of the eigenfunctionsϕ0k, ϕ0k =ϕ1k a.e. onA.
For a fixedk ≥1, let us note that, by Lemma 2.1, ϕjk,i, for i = 1,· · ·, mjk, and j ∈ {0,1}, are linearly independent on L2(A). Now, applying the algebraic Lemma 2.3 with m := m0k, n := m1k, Z = X := A, Y :=Aout,
fi:=ϕ0k,i|A for 1≤i≤m0k, g`:=ϕ0k,`|A for 1≤i≤m1k, andF andGrespectively the vectors
F :=
f1
... fm0k
and G:=
g1
... gm1k
,
we derive thatm0k=m1k, and that there exists anm×morthogonal matrixM, wherem:=m0k =m1k, such that
F(z) =M G(z) forz∈A. (3.14)
We prove now thatϕ0k,i =ϕ1k,i a.e. on A, fori= 1,· · ·, m, up to an appropriate choice of the eigenfunctions ϕ0k,i. To prove this, let us define the vector
f
ϕ0:=M∗ϕ∗0,
where M∗ is the transpose matrix of M, that is Mir∗ = Mri, and ϕ∗0 is the transpose vector of ϕ0 = (ϕ0k,1,· · ·, ϕ0k,m). First let us note that
(ϕ]0k,i|ϕ]0k,`) =δi` for 1≤i, `≤m,
where (· | ·) denotes the scalar product in L2(Ω0\D0), and δi` is the Kronecker’s symbol. In fact ϕ]0k,i = Pm
r=1
Mir∗ϕ0k,r, andϕ]0k,`= Pm s=1
M`s∗ϕ0k,s= Pm s=1
ϕ0k,sMs`, so
(ϕ]0k,i|ϕ]0k,`) = Xm r=1
Xm s=1
Mir∗Ms`δrs= Xm r=1
Mir∗Mr`=δi`, where the last equality follows since the matrixM is orthogonal.
Nowϕ]0k,i = Pm
`=1
Mi`∗ϕ0k,`, and, by (3.14), we know thatϕ0k,`= Pm j=1
M`jϕ1k,j onA, so, substituting inϕ]0k,i, we obtain
] ϕ0k,i=
Xm
`=1
Mi`∗
Xm j=1
M`jϕ1k,j= Xm j=1
ϕ1k,j
Xm
`=1
Mi`∗M`j=ϕ1k,i
onA, for 1≤i≤m, where the last equality follows since the matrixM is orthogonal.
The proof of Theorem 1.1 is complete.
Now we prove Theorem 1.2.
Proof of Theorem 1.2. First of all, by Lemma 3.3, we can suppose that the initial data uj0 ≡ 0 in Ωj\Dj, and the boundary data ϕj ≡ 0 on (0, T)×∂Dj, and ψj ≡ 0 on (0, T)×Ij. Choosing g such that g ≡ 0 on [T0−ε0, T0]×A, whereε0 is such that τ0 < ε0 < T0, we can write uj in the following Fourier expansion (see (3.2)):
uj(t) =− X∞ k=1
T0Z−ε0 0
Z
A
ϕjk(σ0)e−λjk(t−τ)g(τ, σ0)dσ0dτ ϕjk inL2(Ωj\Dj),
fort∈[T0−τ0, T0]. By Lemma 3.2 we obtain
uj(t)|Aout=−
TZ0−ε0
0
Z
A
X∞ k=1
Φjk(σ0, t−τ)g(τ, σ0)dσ0dτ inH12(Aout) (3.15)
fort ∈[T0−τ0, T0], where Φjk are defined in (3.10). Now (1.9), and the change of variablet−τ =sin the right hand side of (3.15), imply
Z T0 T0−τ0
Zt t−T0+ε0
Z
A
X∞ k=1
(Φ0k(σ0, s)−Φ1k(σ0, s))g(t−s, σ0)dσ0dsdt= 0. (3.16)
