# Inverse Problem for a Curved Quantum Guide

Authors: Laure Cardoulis; Michel Cristofol
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We consider the Dirichlet Laplacian operator−Δ on a curved quantum guide in ℝn(n=2,3) with an asymptotically straight reference curve. We give uniqueness results for the inverse problem associated to the reconstruction of the curvature by using either observations of spectral data or a boot-strapping method.


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## Body

## 1. Introduction and Main Results in Dimensionn=2

The spectral properties of curved quantum guides have been studied intensively for several years, because of their applications in quantum mechanics electron motion. We can cite among several papers [1–7].However, inverse problems associated with curved quantum guides have not been studied to our knowledge, except in [8]. Our aim is to establish uniqueness results for the inverse problem of the reconstruction of the curvature of the quantum guide: the data of one eigenpair determines uniquely the curvature up to its sign and similar results are obtained by considering the knowledge of a solution of Poisson's equation in the guide.We consider the Laplacian operator on a nontrivially curved quantum guideΩ⊂ℝ2 which is not self-intersecting, with Dirichlet’s boundary conditions, denoted by -ΔDΩ. We proceed as in [1]. We denote by Γ=(Γ1,Γ2) the function C3-smooth (see [7, Remark 5]) which characterizes the reference curve and by N=(N1,N2) the outgoing normal to the boundary of Ω. We denote by d the fixed width of Ω and by Ω0:=ℝ×]-d/2,d/2[. Each point (x,y) of Ω is described by the curvilinear coordinates (s,u) as follows:
(1.1)f^:Ω0⟶Ωwith(x,y)=f^(s,u)=Γ(s)+uN(s).
We assume Γ1′(s)2+Γ2′(s)2=1 and we recall that the signed curvature γ of Γ is defined by
(1.2)γ(s)=-Γ1′′(s)Γ2′(s)+Γ2′′(s)Γ1′(s),
named so because |γ(s)| represents the curvature of the reference curve at s. We recall that a guide is called simply bent if γ does not change sign in ℝ. We assume throughout this paper the following.Assumption 1.1.
One has the following.(i)
f^ is injective.(ii)
γ∈C2(ℝ)∩L∞(ℝ),γ≢0, (i.e., Ω is nontrivially curved).(iii)
d/2<1/∥γ∥∞, where ∥γ∥∞:=sups∈ℝ|γ(s)|=∥γ∥L∞(ℝ).(iv)
γ(s)→0 as |s|→+∞ (i.e., Ω is asymptotically straight).Note that by the inverse function theorem, the mapf^ (defined by (1.1)) is a local diffeomorphism provided 1-uγ(s)≠0, for all u,s, which is guaranteed by Assumption 1.1 and since f^ is assumed to be injective, the map f^ is a global diffeomorphism. Note also that 1-uγ(s)>0 for all u and s. More precisely, 0<1-(d/2)∥γ∥∞≤1-uγ(s)≤1+(d/2)∥γ∥∞ for all u,s. The curvilinear coordinates (s,u) are locally orthogonal, so by virtue of the Frenet-Serret formulae, the metric in Ω is expressed with respect to them through a diagonal metric tensor, (e.g., [4])
(1.3)(gij)=((1-uγ(s))2001).
The transition to the curvilinear coordinates represents an isometric map of L2(Ω) to L2(Ω0,g1/2dsdu) where
(1.4)(g(s,u))1/2:=1-uγ(s)
is the Jacobian ∂(x,y)/∂(s,u). So we can replace the Laplacian operator -ΔDΩ acting on L2(Ω) by the Laplace-Beltrami operator Hg acting on L2(Ω0,g1/2dsdu) relative to the given metric tensor (gij) (see (1.3) and (1.4)) where
(1.5)Hg:=-g-1/2∂s(g-1/2∂s)-g-1/2∂u(g1/2∂u).
We rewrite Hg (defined by (1.5)) into a Schrödinger-type operator acting on L2(Ω0,dsdu). Indeed, using the unitary transformation
(1.6)Ug:L2(Ω0,g1/2dsdu)⟶L2(Ω0,dsdu)ψ⟼g1/4ψ,
setting
(1.7)Hγ:=UgHgUg-1,
we get
(1.8)Hγ=-∂s(cγ(s,u)∂s)-∂u2+Vγ(s,u),
with
(1.9)cγ(s,u)=1(1-uγ(s))2,(1.10)Vγ(s,u)=-γ2(s)4(1-uγ(s))2-uγ′′(s)2(1-uγ(s))3-5u2γ′2(s)4(1-uγ(s))4.We will assume throughout all this paper that the following assumption is satisfied.Assumption 1.2.
γ
∈
C
2
(
ℝ
) and γ(k)∈L∞(ℝ) for  each k=0,1,2 where γ(k) denotes the kth derivative of γ.Remarks 1.
SinceΩ is nontrivially curved and asymptotically straight, the operator -ΔDΩ has at least one eigenvalue of finite multiplicity below its essential spectrum (see [4, 7]; see also [1] under the additional assumptions that the width d is sufficiently small and the curvature γ is rapidly decaying at infinity; see [3] under the assumption that the curvature γ has a compact support).
Furthermore, note that such operatorHγ admits bound states and that the minimum eigenvalue λ1 is simple and associated with a positive eigenfunction ϕ1 (see [9, Section 8.17]). Then, note that by [10, Theorem 7.1] any eigenfunction of Hγ is continuous and by [11, Remark 25 page 182] any eigenfunction of Hγ belongs to H2(Ω0).
Finally, note also that(λ,ϕ) is an eigenpair (i.e., an eigenfunction associated with its eigenvalue) of the operator Hγ acting on L2(Ω0,dsdu) means that (λ,Ug-1ϕ) is an eigenpair of -ΔDΩ acting on L2(Ω). So the data of one eigenfunction of the operator Hγ is equivalent to the data of one eigenfunction of -ΔDΩ.
We first prove that the data of one eigenpair determines uniquely the curvature.Theorem 1.3.
LetΩ be the curved guide in ℝ2 defined as above. Let γ be the signed curvature defined by (1.2) and satisfying Assumptions 1.1 and 1.2. Let Hγ be the operator defined by (1.8) and (λ,ϕ) be an eigenpair of Hγ.
Then(1.11)γ2(s)=-4Δϕ(s,0)ϕ(s,0)-4λ,
for all s when ϕ(s,0)≠0.Note that the conditionϕ(s,0)≠0 in Theorem 1.3 is satisfied for the positive eigenfunction ϕ1 and for all s∈ℝ. Then, we prove later in the paper under the following assumption.Assumption 1.4.
γ
∈
C
5
(
ℝ
) and γ(k)∈L∞(ℝ) for each k=0,…,5, that one weak solution ϕ of the problem
(1.12)Hγϕ=finΩ0ϕ=0on⁡∂Ω0,
(where f is a known given function) is in fact a classical solution and the data of ϕ determines uniquely the curvature γ.Theorem 1.5.
LetΩ be the curved guide in ℝ2 defined as above. Let γ be the signed curvature defined by (1.2) and satisfying Assumptions 1.1 and 1.4. Let Hγ be the operator defined by (1.8). Let f∈H3(Ω0)∩C(Ω0) and let ϕ∈H01(Ω0) be a weak solution of (1.12).
Then we haveγ2(s)=-4(Δϕ(s,0)/ϕ(s,0))-4(f(s,0)/ϕ(s,0)) for all s when ϕ(s,0)≠0.In the case of a simply bent guide (i.e., whenγ does not change sign in ℝ), we can restrain the hypotheses upon the regularity of γ. We obtain the following result.Theorem 1.6.
LetΩ be the curved guide in ℝ2 defined as above. Let γ be the signed curvature defined by (1.2) and satisfying Assumptions 1.1 and 1.2. We assume also that γ is a nonnegative function. Let Hγ be the operator defined by (1.8). Let f∈L2(Ω0) be a non null function and let ϕ be a weak solution in H01(Ω0) of (1.12). Assume that there exists a positive constant M such that |f(s,u)|≤M|ϕ(s,u)| a.e. in Ω0. Then (f,ϕ) determines uniquely the curvature γ.Note that the above result is still valid for a nonpositive functionγ.This paper is organized as follows. In Section2, we prove Theorems 1.3, 1.5, and 1.6. In Sections 3 and 4, we extend our results to the case of a curved quantum guide defined in ℝ3.


## 2. Proofs of Theorems1.3, 1.5, and 1.6

### 2.1. Proof of Theorem1.3

Recall thatϕ is an eigenfunction of Hγ, belonging to H2(Ω0). Since ϕ is continuous and Hγϕ=λϕ, then Hγϕ is continuous too. Thus, noticing that cγ(s,0)=1, we deduce the continuity of the function (s,0)↦Δϕ(s,0) and from (1.8) to (1.10), we get
(2.1)-Δϕ(s,0)-γ2(s)4ϕ(s,0)=λϕ(s,0),
and equivalently,
(2.2)γ2(s)=-4Δϕ(s,0)ϕ(s,0)-4λifϕ(s,0)≠0.


### 2.2. Proof of Theorem1.5

First, we recall from [11, Remark 25 page 182] the following lemma.Lemma 2.1.
For a second-order elliptic operator defined in a domainω⊂ℝn, if ϕ∈H01(ω) satisfies
(2.3)∫ω‍∑i,j‍aij∂ϕ∂xi∂ψ∂xj=∫ω‍fψ∀ψ∈H01(ω)
then if ω is of class C2(2.4)(f∈L2(ω),aij∈C1(ω¯),Dαaij∈L∞(ω)∀i,j,∀α,|α|≤1)imply(ϕ∈H2(ω))
and for m≥1, if ω is of class Cm+2(2.5)(f∈Hm(ω),aij∈Cm+1(ω¯),Dαaij∈L∞(ω)∀i,j,∀α,|α|≤m+1)imply(ϕ∈Hm+2(ω)).Now we can prove Theorem1.5.We haveHγϕ=f, so
(2.6)∫Ω0‍[cγ(∂sϕ)(∂sψ)+(∂uϕ)(∂uψ)]=∫Ω0‍[f-Vγϕ]ψ,∀ψ∈H01(Ω0)
with cγ defined by (1.9) and Vγ defined by (1.10).Using Assumption1.4, since γ(k)∈L∞(Ω0) for k=0,1,2 then Vγ∈L∞(Ω0) and f-Vγϕ∈L2(Ω0). From the hypotheses γ∈C1(ℝ) and γ′∈L∞(ℝ), we get that cγ∈C1(Ω0¯),Dαcγ∈L∞(Ω0) for any α,|α|≤1, and so, using Lemma 2.1 for (2.6), we obtain that ϕ∈H2(Ω0).By the same way, we get thatf-Vγϕ∈H1(Ω0),cγ∈C2(Ω0¯) and Dαcγ∈L∞(Ω0) for any α,|α|≤2 (from γ∈C3(ℝ),γ(k)∈L∞(ℝ) for any k=0,…,3). Using Lemma 2.1, we obtain that ϕ∈H3(Ω0).We apply again Lemma2.1 to get that ϕ∈H4(Ω0) (since f-Vγϕ∈H2(Ω0),cγ∈C3(Ω0¯),Dαcγ∈L∞(Ω0) for all α,|α|≤3, from the hypotheses γ∈C4(ℝ) and γ(k)∈L∞(ℝ) for k=0,…,4).Finally, using Assumption1.4 and Lemma 2.1, we obtain that ϕ∈H5(Ω0).Due to the regularity ofΩ0, we have ϕ∈H5(ℝ2) and Δϕ∈H3(ℝ2). Since ∇(Δϕ)∈(H2(ℝ2))2 and H2(ℝ2)⊂L∞(ℝ2), we can deduce that Δϕ is continuous (see [11, Remark 8 page 154]).Therefore, we can conclude by using the continuity of the function(2.7)(s,0)↦-∂s(cγ(s,0)∂sϕ(s,0))-∂u2ϕ(s,0)=f(s,0)-Vγ(s,0)ϕ(s,0).
Therefore, we get -Δϕ(s,0)-(γ2(s)/4)ϕ(s,0)=f(s,0) and equivalently,
(2.8)γ2(s)=-4Δϕ(s,0)ϕ(s,0)-4f(s,0)ϕ(s,0)ifϕ(s,0)≠0.


### 2.3. Proof of Theorem1.6

We prove here that(f,ϕ) determines uniquely γ when γ is a nonnegative function.For that, assume thatΩ1 and Ω2 are two quantum guides in ℝ2 with same width d. We denote by γ1 and γ2 the curvatures, respectively, associated with Ω1 and Ω2, and we suppose that each γi satisfies Assumption 1.2 and is a nonnegative function. Assume that Hγ1ϕ=f=Hγ2ϕ.Thenϕ satisfies
(2.9)-∂s((cγ1(s,u)-cγ2(s,u))∂sϕ(s,u))+(Vγ1(s,u)-Vγ2(s,u))ϕ(s,u)=0.
Assume that γ1≢γ2.Step 1.
First, we consider the case where (for example)γ1(s)<γ2(s) for all s∈ℝ.
Letϵ>0,ωϵ:=ℝ×Iϵ with Iϵ=]-ϵ,0[. Multiplying (2.9) by ϕ and integrating over ωϵ, we get
(2.10)∫ωϵ‍(cγ1-cγ2)(∂sϕ)2-∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs+∫ωϵ‍(Vγ1-Vγ2)ϕ2=0.
Since ϵ≪1, Vγi(s,u)≃-γi2(s)/4 for i=1,2, and so Vγ1(s,u)-Vγ2(s,u)>0 in ωϵ.
Moreover, since(2.11)cγ1(s,u)-cγ2(s,u)=u(γ1(s)-γ2(s))(2-u(γ1(s)+γ2(s)))(1-uγ1(s))2(1-uγ2(s))2,
we have cγ1(s,u)>cγ2(s,u) in ωϵ.
Since(2.12)∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs=0
Thus, from (2.10)−(2.12), we get
(2.13)∫ωϵ‍(cγ1-cγ2)(∂sϕ)2+∫ωϵ‍(Vγ1-Vγ2)ϕ2=0,
with cγ1-cγ2>0 in ωϵ and Vγ1-Vγ2>0 in ωϵ. We can deduce that ϕ=0 in ωϵ.
Using a unique continuation theorem (see [12, Theorem XIII.63 page 240]), from Hγϕ=f, noting that -Δ(Ug-1ϕ)=Ug-1f=g-1/4f, (recall that Ug is defined by (1.6)) and so by |f|≤M|ϕ| we have |Δ(Ug-1ϕ)|≤M|g-1/4ϕ| with g>0 a.e., and we can deduce that ϕ=0 in Ω0. So we get a contradiction (since Hγϕ=f and f is assumed to be a non null function).Step 2.
From Step1, we obtain that there exists at least one point s0∈ℝ such that γ1(s0)=γ2(s0). Since γ1≢γ2, we can choose a∈ℝ and b∈ℝ∪{+∞} such that (for example) γ1(a)=γ2(a),γ1(s)<γ2(s) for  all s∈]a,b[ and γ1(b)=γ2(b)ifb∈ℝ.
We proceed as in Step1, considering, in this case, ωϵ:=]a,b[×Iϵ. We study again (2.10) and as in Step 1, we have
(2.14)∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs=0.
Indeed from (2.11) and γ1(a)=γ2(a) we have cγ1(a,u)=cγ2(a,u) and so
(2.15)∫-ϵ0‍(cγ1(a,u)-cγ2(a,u))∂sϕ(a,u)ϕ(a,u)du=0.
By the same way, if b∈ℝ, we also have cγ1(b,u)=cγ2(b,u). Thus, (2.10) becomes (2.13) with cγ1-cγ2>0 in ωϵ and Vγ1-Vγ2>0inωϵ. So ϕ=0 in ωϵ and as in Step 1, by a unique continuation theorem, we obtain that ϕ=0 in Ω0. Therefore, we get a contradiction.
Note that the previous theorem is true if we replace the hypothesis “γ is nonnegative” by the hypothesis “γ is nonpositive.” Indeed, in this last case, we just have to take Iϵ=]0,ϵ[ and the proof rests valid.


## 2.1. Proof of Theorem1.3

Recall thatϕ is an eigenfunction of Hγ, belonging to H2(Ω0). Since ϕ is continuous and Hγϕ=λϕ, then Hγϕ is continuous too. Thus, noticing that cγ(s,0)=1, we deduce the continuity of the function (s,0)↦Δϕ(s,0) and from (1.8) to (1.10), we get
(2.1)-Δϕ(s,0)-γ2(s)4ϕ(s,0)=λϕ(s,0),
and equivalently,
(2.2)γ2(s)=-4Δϕ(s,0)ϕ(s,0)-4λifϕ(s,0)≠0.


## 2.2. Proof of Theorem1.5

First, we recall from [11, Remark 25 page 182] the following lemma.Lemma 2.1.
For a second-order elliptic operator defined in a domainω⊂ℝn, if ϕ∈H01(ω) satisfies
(2.3)∫ω‍∑i,j‍aij∂ϕ∂xi∂ψ∂xj=∫ω‍fψ∀ψ∈H01(ω)
then if ω is of class C2(2.4)(f∈L2(ω),aij∈C1(ω¯),Dαaij∈L∞(ω)∀i,j,∀α,|α|≤1)imply(ϕ∈H2(ω))
and for m≥1, if ω is of class Cm+2(2.5)(f∈Hm(ω),aij∈Cm+1(ω¯),Dαaij∈L∞(ω)∀i,j,∀α,|α|≤m+1)imply(ϕ∈Hm+2(ω)).Now we can prove Theorem1.5.We haveHγϕ=f, so
(2.6)∫Ω0‍[cγ(∂sϕ)(∂sψ)+(∂uϕ)(∂uψ)]=∫Ω0‍[f-Vγϕ]ψ,∀ψ∈H01(Ω0)
with cγ defined by (1.9) and Vγ defined by (1.10).Using Assumption1.4, since γ(k)∈L∞(Ω0) for k=0,1,2 then Vγ∈L∞(Ω0) and f-Vγϕ∈L2(Ω0). From the hypotheses γ∈C1(ℝ) and γ′∈L∞(ℝ), we get that cγ∈C1(Ω0¯),Dαcγ∈L∞(Ω0) for any α,|α|≤1, and so, using Lemma 2.1 for (2.6), we obtain that ϕ∈H2(Ω0).By the same way, we get thatf-Vγϕ∈H1(Ω0),cγ∈C2(Ω0¯) and Dαcγ∈L∞(Ω0) for any α,|α|≤2 (from γ∈C3(ℝ),γ(k)∈L∞(ℝ) for any k=0,…,3). Using Lemma 2.1, we obtain that ϕ∈H3(Ω0).We apply again Lemma2.1 to get that ϕ∈H4(Ω0) (since f-Vγϕ∈H2(Ω0),cγ∈C3(Ω0¯),Dαcγ∈L∞(Ω0) for all α,|α|≤3, from the hypotheses γ∈C4(ℝ) and γ(k)∈L∞(ℝ) for k=0,…,4).Finally, using Assumption1.4 and Lemma 2.1, we obtain that ϕ∈H5(Ω0).Due to the regularity ofΩ0, we have ϕ∈H5(ℝ2) and Δϕ∈H3(ℝ2). Since ∇(Δϕ)∈(H2(ℝ2))2 and H2(ℝ2)⊂L∞(ℝ2), we can deduce that Δϕ is continuous (see [11, Remark 8 page 154]).Therefore, we can conclude by using the continuity of the function(2.7)(s,0)↦-∂s(cγ(s,0)∂sϕ(s,0))-∂u2ϕ(s,0)=f(s,0)-Vγ(s,0)ϕ(s,0).
Therefore, we get -Δϕ(s,0)-(γ2(s)/4)ϕ(s,0)=f(s,0) and equivalently,
(2.8)γ2(s)=-4Δϕ(s,0)ϕ(s,0)-4f(s,0)ϕ(s,0)ifϕ(s,0)≠0.


## 2.3. Proof of Theorem1.6

We prove here that(f,ϕ) determines uniquely γ when γ is a nonnegative function.For that, assume thatΩ1 and Ω2 are two quantum guides in ℝ2 with same width d. We denote by γ1 and γ2 the curvatures, respectively, associated with Ω1 and Ω2, and we suppose that each γi satisfies Assumption 1.2 and is a nonnegative function. Assume that Hγ1ϕ=f=Hγ2ϕ.Thenϕ satisfies
(2.9)-∂s((cγ1(s,u)-cγ2(s,u))∂sϕ(s,u))+(Vγ1(s,u)-Vγ2(s,u))ϕ(s,u)=0.
Assume that γ1≢γ2.Step 1.
First, we consider the case where (for example)γ1(s)<γ2(s) for all s∈ℝ.
Letϵ>0,ωϵ:=ℝ×Iϵ with Iϵ=]-ϵ,0[. Multiplying (2.9) by ϕ and integrating over ωϵ, we get
(2.10)∫ωϵ‍(cγ1-cγ2)(∂sϕ)2-∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs+∫ωϵ‍(Vγ1-Vγ2)ϕ2=0.
Since ϵ≪1, Vγi(s,u)≃-γi2(s)/4 for i=1,2, and so Vγ1(s,u)-Vγ2(s,u)>0 in ωϵ.
Moreover, since(2.11)cγ1(s,u)-cγ2(s,u)=u(γ1(s)-γ2(s))(2-u(γ1(s)+γ2(s)))(1-uγ1(s))2(1-uγ2(s))2,
we have cγ1(s,u)>cγ2(s,u) in ωϵ.
Since(2.12)∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs=0
Thus, from (2.10)−(2.12), we get
(2.13)∫ωϵ‍(cγ1-cγ2)(∂sϕ)2+∫ωϵ‍(Vγ1-Vγ2)ϕ2=0,
with cγ1-cγ2>0 in ωϵ and Vγ1-Vγ2>0 in ωϵ. We can deduce that ϕ=0 in ωϵ.
Using a unique continuation theorem (see [12, Theorem XIII.63 page 240]), from Hγϕ=f, noting that -Δ(Ug-1ϕ)=Ug-1f=g-1/4f, (recall that Ug is defined by (1.6)) and so by |f|≤M|ϕ| we have |Δ(Ug-1ϕ)|≤M|g-1/4ϕ| with g>0 a.e., and we can deduce that ϕ=0 in Ω0. So we get a contradiction (since Hγϕ=f and f is assumed to be a non null function).Step 2.
From Step1, we obtain that there exists at least one point s0∈ℝ such that γ1(s0)=γ2(s0). Since γ1≢γ2, we can choose a∈ℝ and b∈ℝ∪{+∞} such that (for example) γ1(a)=γ2(a),γ1(s)<γ2(s) for  all s∈]a,b[ and γ1(b)=γ2(b)ifb∈ℝ.
We proceed as in Step1, considering, in this case, ωϵ:=]a,b[×Iϵ. We study again (2.10) and as in Step 1, we have
(2.14)∫∂ωϵ‍(cγ1-cγ2)(∂sϕ)ϕνs=0.
Indeed from (2.11) and γ1(a)=γ2(a) we have cγ1(a,u)=cγ2(a,u) and so
(2.15)∫-ϵ0‍(cγ1(a,u)-cγ2(a,u))∂sϕ(a,u)ϕ(a,u)du=0.
By the same way, if b∈ℝ, we also have cγ1(b,u)=cγ2(b,u). Thus, (2.10) becomes (2.13) with cγ1-cγ2>0 in ωϵ and Vγ1-Vγ2>0inωϵ. So ϕ=0 in ωϵ and as in Step 1, by a unique continuation theorem, we obtain that ϕ=0 in Ω0. Therefore, we get a contradiction.
Note that the previous theorem is true if we replace the hypothesis “γ is nonnegative” by the hypothesis “γ is nonpositive.” Indeed, in this last case, we just have to take Iϵ=]0,ϵ[ and the proof rests valid.


## 3. Uniqueness Result for aℝ3-Quantum Guide

Now, we apply the same ideas for a tubeΩ in ℝ3. We proceed here as in [7]. Let s↦Γ(s),Γ=(Γ1,Γ2,Γ3), be a curve in ℝ3. We assume that Γ:ℝ→ℝ3 is a C4-smooth curve satisfying the following hypothesesAssumption 3.1.
Γ possesses a positively oriented Frenet frame {e1,e2,e3} with the following properties (i)
e1=Γ′,(ii)
for alli∈{1,2,3},ei∈C1(ℝ,ℝ3),(iii)
for alli∈{1,2}, for all s∈ℝ,ei′(s) lies in the span of e1(s),…,ei+1(s).Recall that a sufficient condition to ensure the existence of the Frenet frame of Assumption3.1 is to require that for all s∈ℝ the vectors Γ′(s),Γ′′(s) are linearly independent.Then we define the moving frame{e~1,e~2,e~3} along Γ by following [7]. This moving frame better reflects the geometry of the curve and it is still called the Tang frame because it is a generalization of the Tang frame known from the theory of three-dimensional waveguides.Given aC5 bounded open connected neighborhood ω of (0,0)∈ℝ2, let Ω0 denote the straight tube ℝ×ω. We define the curved tube Ω of cross-section ω about Γ by
(3.1)Ω:=f~(R×ω)=f~(Ω0),f~(s,u2,u3)≔Γ(s)+∑i=23‍ui∑j=23‍Rij(s)ej(s)=Γ(s)+∑i=23‍uie~i(s),
with u=(u2,u3)∈ω and
(3.2)R(s):=(Rij(s))i,j∈{2,3}=(cos⁡(θ(s))-sin⁡(θ(s))sin⁡(θ(s))cos⁡(θ(s))),θ being a real-valued differentiable function such that θ′(s)=τ(s) the torsion of Γ. This differential equation is a consequence of the definition of the moving Tang frame (see [7, Remark 3]).Note thatR is a rotation matrix in ℝ2 chosen in such a way that (s,u2,u3) are orthogonal “coordinates” in Ω. Let k be the first curvature function of Ω. Recall that since Ω⊂ℝ3,k is a nonnegative function. We assume throughout all this section that the following hypothesis holds:Assumption 3.2.
One has the following.
k
∈
C
2
(
ℝ
)
∩
L
∞
(
ℝ
)
,
a
:
=
sup
u
∈
ω
∥
u
∥
ℝ
2
<
1
/
∥
k
∥
∞
,
k
(
s
)
→
0
as
|
s
|
→
+
∞
Ω does not overlap.Assumption3.2 assures that the map f~ (defined by (3.1)) is a diffeomorphism (see [7]) in order to identify Ω with the Riemannian manifold (Ω0,(gij)) where (gij) is the metric tensor induced by f~, that is, (gij)≔tJ(f~)·J(f~), (J(f~) denoting the Jacobian matrix of f~). Recall that (gij)=diag(h2,1,1) (see [7]) with
(3.3)h(s,u2,u3):=1-k(s)(cos⁡(θ(s))u2+sin⁡(θ(s))u3).
Note that Assumption 3.2 implies that 0<1-a∥k∥∞≤1-h(s,u2,u3)≤1+a∥k∥∞ for all s∈ℝ and u=(u2,u3)∈ω. Moreover, setting
(3.4)g:=h2,
we can replace the Dirichlet Laplacian operator -ΔDΩ acting on L2(Ω) by the Laplace-Beltrami operator Kg acting on L2(Ω0,hdsdu) relative to the metric tensor (gij). We can rewrite Kg into a Schrödinger-type operator acting on L2(Ω0,dsdu). Indeed, using the unitary transformation
(3.5)Wg:L2(Ω0,hdsdu)⟶L2(Ω0,dsdu)ψ↦g1/4ψ,
setting
(3.6)Hk:=WgKgWg-1,
we get
(3.7)Hk=-∂s(h-2∂s)-∂u22-∂u32+Vk,
where ∂s denotes the derivative relative to s and ∂ui denotes the derivative relative to ui and with
(3.8)Vk:=-k24h2+∂s2h2h3-5(∂sh)24h4.
We assume also throughout all this section that the following hypotheses hold:Assumption 3.3.
One has the following.
(i)
k
′
∈
L
∞
(
ℝ
)
,
k
′′
∈
L
∞
(
ℝ
)
(ii)
θ
∈
C
2
(
ℝ
)
,
θ
′
=
τ
∈
L
∞
(
ℝ
)
,
θ
′′
∈
L
∞
(
ℝ
).Remarks 2.
Note that as for the 2-dimensional case, such operatorHk (defined by (3.3)−(3.8)) admits bound states and that the minimum eigenvalue λ1 is simple and associated with a positive eigenfunction ϕ1 (see [7, 9]). Still note that (λ,ϕ) is an eigenpair of the operator Hk acting on L2(Ω0,dsdu) means that (λ,Wg-1ϕ) is an eigenpair of -ΔDΩ acting on L2(Ω) (with Wg defined by (3.5)). Finally, note that by [10, Theorem 7.1] any eigenfunction of Hk is continuous and by [11, Remark 25 page 182] any eigenfunction of Hk belongs to H2(Ω0).
As for the 2-dimensional case, first we prove that the data of one eigenpair determines uniquely the curvature.Theorem 3.4.
LetΩ be the curved guide in ℝ3 defined as above. Let k be the first curvature function of Ω. Assume that Assumptions 3.1 to 3.3 are satisfied. Let Hk be the operator defined by (3.3)−(3.8) and (λ,ϕ) be an eigenpair of Hk.
Thenk2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4λ for all s when ϕ(s,0,0)≠0.Then, One has the following.Assumption 3.5.
One has the following.(i)
k∈C5(ℝ),k(i)∈L∞(ℝ) for all i=0,…,5,(ii)
θ∈C5(ℝ),θ(i)∈L∞(ℝ) for all i=1,…,5,where k(i) (resp., θ(i)) denotes the ith derivative of k (resp. of θ), we obtain the following result.Theorem 3.6.
LetΩ be the curved guide in ℝ3 defined as above. Let k be the first curvature function of Ω. Assume that Assumptions 3.1 to 3.5 are satisfied. Let Hk be the operator defined by (3.3)−(3.8). Let f∈H3(Ω0)∩C(Ω0) and let ϕ∈H01(Ω0) be a weak solution of Hkϕ=f in Ω0.
Thenϕ is a classical solution and k2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4(f(s,0,0)/ϕ(s,0,0)) for all s when ϕ(s,0,0)≠0.Remarks 3.
Recall that inℝ3, k is a nonnegative function and that the condition imposed on ϕ (ϕ(s,0,0)≠0) in Theorems 3.4 and 3.6 is satisfied by the positive eigenfunction ϕ1.
As for the two-dimensional case, we can restrain the hypotheses upon the regularity of the functionsk and θ.
For a guide with a known torsion, we obtain the following result.Theorem 3.7.
LetΩ be the curved guide in ℝ3 defined as above. Let k be the first curvature function of Ω and let τ be the second curvature function (i.e., the torsion) of Ω. Denote by θ a primitive of τ and suppose that 0≤θ(s)≤π/2 for all s∈ℝ. Assume that Assumptions 3.1 to 3.3 are satisfied. Let Hk be the operator defined by (3.3)−(3.8). Let f∈L2(Ω0) be a non null function and let ϕ∈H01(Ω0) be a weak solution of Hkϕ=f in Ω0. Assume that there exists a positive constant M such that |f(s,u)|≤M|ϕ(s,u)| a.e. in Ω0.
Then the data(f,ϕ) determines uniquely the first curvature function k if the torsion τ is given.


## 4. Proofs of Theorems3.4, 3.6, and 3.7

### 4.1. Proof of Theorem3.4

Recall thatϕ is an eigenfunction of Hk. Since ϕ is continuous, Hkϕ=λϕ and ϕ∈H2(Ω0) then Hkϕ is continuous. Therefore, for u=(u2,u3)=(0,0), we get: -Δϕ(s,0,0)-(k2(s)/4)ϕ(s,0,0)=λϕ(s,0,0) and equivalently, k2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4λ if ϕ(s,0,0)≠0.


### 4.2. Proof of Theorem3.6

We follow the proof of Theorem1.5. We have Hkϕ=f with ϕ∈H01(Ω0). So
(4.1)∫Ω0‍[h-2(∂sϕ)(∂sψ)+(∂u2ϕ)(∂u2ψ)+(∂u3ϕ)(∂u3ψ)]=∫Ω0‍[f-Vkϕ]ψ,∀ψ∈H01(Ω0),
with h defined by (3.3) and Vk defined by (3.8).From Assumptions3.2 and 3.3, since k,k′,k′′,θ′,θ′′ are bounded, we deduce that Vk∈L∞(Ω0). Therefore, f-Vkϕ∈L2(Ω0). Moreover, we have also h-2∈C1(Ω0¯) and Dα(h-2)∈L∞(Ω0) for any α,|α|≤1. Thus, using Lemma 2.1 for (4.1), we obtain that ϕ∈H2(Ω0).By the same way, we get thatf-Vkϕ∈H1(Ω0),h-2∈C2(Ω0¯) and Dα(h-2)∈L∞(Ω0) for any α,|α|≤2 (since k∈C3(ℝ),θ∈C3(ℝ) and all of their derivatives are bounded). Using Lemma 2.1, we obtain that ϕ∈H3(Ω0).We apply again Lemma2.1 to get that ϕ∈H4(Ω0) (since f-Vγϕ∈H2(Ω0),cγ∈C3(Ω0¯),Dαcγ∈L∞(Ω0) for all α,|α|≤3, from the hypotheses γ∈C4(ℝ) and γ(k)∈L∞(ℝ) for k=0,…,4).Finally, using Assumption3.5 and Lemma 2.1, we obtain that ϕ∈H5(Ω0). Due to the regularity of Ω0 (see [11, Note page 169]), we have ϕ∈H5(ℝ3) and Δϕ∈H3(ℝ3). Since ∇(Δϕ)∈(H2(ℝ3))3 and H2(ℝ3)⊂L∞(ℝ3), we can deduce that Δϕ is continuous (see [11, Remark  8 page 154]).Thus, we conclude as in Theorem1.5 and for u=(u2,u3)=(0,0), we get -Δϕ(s,0,0)-(k2(s)/4)ϕ(s,0,0)=f(s,0,0) and equivalently, k2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4(f(s,0,0)/ϕ(s,0,0)) if ϕ(s,0,0)≠0.


### 4.3. Proof of Theorem3.7

We prove here that(f,ϕ,θ) determines uniquely k.Assume thatΩ1 and Ω2 are two guides in ℝ3. We denote by k1 and k2 the first curvatures functions associated with Ω1 and Ω2 and we denote by θ a primitive of τ the common torsion of Ω1 and Ω2. We suppose that k1,k2 and θ satisfy Assumptions 3.2 and 3.3 and that 0≤θ(s)≤π/2 for all s∈ℝ. Assume that Hk1ϕ=f=Hk2ϕ.Thenϕ satisfies
(4.2)-∂s((h1-2(s,u2,u3)-h2-2(s,u2,u3))∂sϕ(s,u2,u3))+(Vk1(s,u2,u3)-Vk2(s,u2,u3))ϕ(s,u2,u3)=0,
where h1 (associated with k1) is defined by (3.3), Vk1 is defined by (3.8), h2 (associated with k2) is defined by (3.3), and Vk2 is defined by (3.8).Assume thatk1≢k2.Step 1.
First, we consider the case where (for example)k1(s)<k2(s) for all s∈ℝ. Recall that each ki is a nonnegative function.
Letϵ>0 and denote by Jϵ:=]-ϵ,0[×]-ϵ,0[,Oϵ:=ℝ×Jϵ with ϵ small enough to have Jϵ⊂ω (recall that Ω0=ℝ×ω).
Multiplying (4.2) by ϕ and integrating over Oϵ, we get
(4.3)∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs+∫Oϵ‍(Vk1-Vk2)ϕ2=0.
Sinceϵ≪1,Vki≃-ki2(s)/4 for i=1,2, and so Vk1(s,u2,u3)-Vk2(s,u2,u3)>0 in Oϵ.
Moreover, note that(4.4)h1-2(s,u2,u3)-h2-2(s,u2,u3)=α(s,u2,u3)(k1(s)-k2(s))(h1(s,u2,u3)+h2(s,u2,u3))h12(s,u2,u3)h22(s,u2,u3),
with α(s,u2,u3):=cos(θ(s))u2+sin(θ(s))u3.
Since(u2,u3)∈Jϵ and 0≤θ(s)≤π/2 for all s∈ℝ, we have α(s,u2,u3)<0. Therefore, by (4.4), we deduce that h1-2-h2-2>0 in Oϵ.
Thus,∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫Oϵ‍(Vk1-Vk2)ϕ2≥0.
Note also that(4.5)∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs=0.
Therefore, from (4.3) and (4.5) we get
(4.6)∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫Oϵ‍(Vk1-Vk2)ϕ2=0,
with h1-2-h2-2>0 in Oϵ and Vk1-Vk2>0 in Oϵ.
From (4.6) we can deduce that ϕ=0 in Oϵ. Using a unique continuation theorem (see [12, Theorem XIII.63 page 240]), from Hk1ϕ=f, noting that -Δ(Wg-1ϕ)=Wg-1f=g-1/4f, by |f|≤M|ϕ| a.e. in Ω0, we can deduce that ϕ=0 in Ω0. So we get a contradiction since f is assumed to be a non null function.Step 2.
From Step1, we obtain that there exists at least one point s0∈ℝ such that k1(s0)=k2(s0). Since k1≢k2, we can choose a∈ℝ and b∈ℝ∪{+∞} such that (for example) k1(a)=k2(a),k1(s)<k2(s) for all s∈]a,b[ and k1(b)=k2(b) if b∈ℝ. We proceed as in Step 1, considering in this case Oϵ:=]a,b[×Jϵ. From k1(a)=k2(a), we get that h1-2(a,u2,u3)=h2-2(a,u2,u3). Therefore, we obtain ∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs=0. So (4.3) becomes (4.6) with h1-2-h2-2>0 in Oϵ and Vk1-Vk2>0 in Oϵ. So ϕ=0 in Oϵ and as in Step 1, by a unique continuation theorem, we obtain that ϕ=0 in Ω0. Therefore, we get a contradiction.


## 4.1. Proof of Theorem3.4

Recall thatϕ is an eigenfunction of Hk. Since ϕ is continuous, Hkϕ=λϕ and ϕ∈H2(Ω0) then Hkϕ is continuous. Therefore, for u=(u2,u3)=(0,0), we get: -Δϕ(s,0,0)-(k2(s)/4)ϕ(s,0,0)=λϕ(s,0,0) and equivalently, k2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4λ if ϕ(s,0,0)≠0.


## 4.2. Proof of Theorem3.6

We follow the proof of Theorem1.5. We have Hkϕ=f with ϕ∈H01(Ω0). So
(4.1)∫Ω0‍[h-2(∂sϕ)(∂sψ)+(∂u2ϕ)(∂u2ψ)+(∂u3ϕ)(∂u3ψ)]=∫Ω0‍[f-Vkϕ]ψ,∀ψ∈H01(Ω0),
with h defined by (3.3) and Vk defined by (3.8).From Assumptions3.2 and 3.3, since k,k′,k′′,θ′,θ′′ are bounded, we deduce that Vk∈L∞(Ω0). Therefore, f-Vkϕ∈L2(Ω0). Moreover, we have also h-2∈C1(Ω0¯) and Dα(h-2)∈L∞(Ω0) for any α,|α|≤1. Thus, using Lemma 2.1 for (4.1), we obtain that ϕ∈H2(Ω0).By the same way, we get thatf-Vkϕ∈H1(Ω0),h-2∈C2(Ω0¯) and Dα(h-2)∈L∞(Ω0) for any α,|α|≤2 (since k∈C3(ℝ),θ∈C3(ℝ) and all of their derivatives are bounded). Using Lemma 2.1, we obtain that ϕ∈H3(Ω0).We apply again Lemma2.1 to get that ϕ∈H4(Ω0) (since f-Vγϕ∈H2(Ω0),cγ∈C3(Ω0¯),Dαcγ∈L∞(Ω0) for all α,|α|≤3, from the hypotheses γ∈C4(ℝ) and γ(k)∈L∞(ℝ) for k=0,…,4).Finally, using Assumption3.5 and Lemma 2.1, we obtain that ϕ∈H5(Ω0). Due to the regularity of Ω0 (see [11, Note page 169]), we have ϕ∈H5(ℝ3) and Δϕ∈H3(ℝ3). Since ∇(Δϕ)∈(H2(ℝ3))3 and H2(ℝ3)⊂L∞(ℝ3), we can deduce that Δϕ is continuous (see [11, Remark  8 page 154]).Thus, we conclude as in Theorem1.5 and for u=(u2,u3)=(0,0), we get -Δϕ(s,0,0)-(k2(s)/4)ϕ(s,0,0)=f(s,0,0) and equivalently, k2(s)=-4(Δϕ(s,0,0)/ϕ(s,0,0))-4(f(s,0,0)/ϕ(s,0,0)) if ϕ(s,0,0)≠0.


## 4.3. Proof of Theorem3.7

We prove here that(f,ϕ,θ) determines uniquely k.Assume thatΩ1 and Ω2 are two guides in ℝ3. We denote by k1 and k2 the first curvatures functions associated with Ω1 and Ω2 and we denote by θ a primitive of τ the common torsion of Ω1 and Ω2. We suppose that k1,k2 and θ satisfy Assumptions 3.2 and 3.3 and that 0≤θ(s)≤π/2 for all s∈ℝ. Assume that Hk1ϕ=f=Hk2ϕ.Thenϕ satisfies
(4.2)-∂s((h1-2(s,u2,u3)-h2-2(s,u2,u3))∂sϕ(s,u2,u3))+(Vk1(s,u2,u3)-Vk2(s,u2,u3))ϕ(s,u2,u3)=0,
where h1 (associated with k1) is defined by (3.3), Vk1 is defined by (3.8), h2 (associated with k2) is defined by (3.3), and Vk2 is defined by (3.8).Assume thatk1≢k2.Step 1.
First, we consider the case where (for example)k1(s)<k2(s) for all s∈ℝ. Recall that each ki is a nonnegative function.
Letϵ>0 and denote by Jϵ:=]-ϵ,0[×]-ϵ,0[,Oϵ:=ℝ×Jϵ with ϵ small enough to have Jϵ⊂ω (recall that Ω0=ℝ×ω).
Multiplying (4.2) by ϕ and integrating over Oϵ, we get
(4.3)∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs+∫Oϵ‍(Vk1-Vk2)ϕ2=0.
Sinceϵ≪1,Vki≃-ki2(s)/4 for i=1,2, and so Vk1(s,u2,u3)-Vk2(s,u2,u3)>0 in Oϵ.
Moreover, note that(4.4)h1-2(s,u2,u3)-h2-2(s,u2,u3)=α(s,u2,u3)(k1(s)-k2(s))(h1(s,u2,u3)+h2(s,u2,u3))h12(s,u2,u3)h22(s,u2,u3),
with α(s,u2,u3):=cos(θ(s))u2+sin(θ(s))u3.
Since(u2,u3)∈Jϵ and 0≤θ(s)≤π/2 for all s∈ℝ, we have α(s,u2,u3)<0. Therefore, by (4.4), we deduce that h1-2-h2-2>0 in Oϵ.
Thus,∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫Oϵ‍(Vk1-Vk2)ϕ2≥0.
Note also that(4.5)∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs=0.
Therefore, from (4.3) and (4.5) we get
(4.6)∫Oϵ‍(h1-2-h2-2)(∂sϕ)2+∫Oϵ‍(Vk1-Vk2)ϕ2=0,
with h1-2-h2-2>0 in Oϵ and Vk1-Vk2>0 in Oϵ.
From (4.6) we can deduce that ϕ=0 in Oϵ. Using a unique continuation theorem (see [12, Theorem XIII.63 page 240]), from Hk1ϕ=f, noting that -Δ(Wg-1ϕ)=Wg-1f=g-1/4f, by |f|≤M|ϕ| a.e. in Ω0, we can deduce that ϕ=0 in Ω0. So we get a contradiction since f is assumed to be a non null function.Step 2.
From Step1, we obtain that there exists at least one point s0∈ℝ such that k1(s0)=k2(s0). Since k1≢k2, we can choose a∈ℝ and b∈ℝ∪{+∞} such that (for example) k1(a)=k2(a),k1(s)<k2(s) for all s∈]a,b[ and k1(b)=k2(b) if b∈ℝ. We proceed as in Step 1, considering in this case Oϵ:=]a,b[×Jϵ. From k1(a)=k2(a), we get that h1-2(a,u2,u3)=h2-2(a,u2,u3). Therefore, we obtain ∫∂Oϵ‍(h1-2-h2-2)(∂sϕ)ϕνs=0. So (4.3) becomes (4.6) with h1-2-h2-2>0 in Oϵ and Vk1-Vk2>0 in Oϵ. So ϕ=0 in Oϵ and as in Step 1, by a unique continuation theorem, we obtain that ϕ=0 in Ω0. Therefore, we get a contradiction.



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*Source: 651390-2012-08-01.xml*