# Inverse Problem for the Schrödinger Equation in Dimension 3

Authors: Fagueye Ndiaye
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we consider the Schrödinger equation in the unit ball inℝ3. We study the inverse problem of identifying the potential q from the Dirichlet to Neumann map which associates to all possible functions f on the boundary ∂B and the measurements of the normal derivative of the solution of Schrödinger equation ∂u/∂ν on ∂B. Using spherical harmonics tools, we determine an explicit expression for the potential qx on the edge of the domain from an explicit formula for the Dirichlet to Neumann map in a unit ball in dimension 3. We theoretically and numerically present an example.


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## Body

## 1. Introduction

The electrical impedance tomography is an imaging method which consists in determining the electrical proprerty of a medium by making voltage and current measurements on its boundary. In the mathematical literature, this is also known as “Calderón’s problem” from Calderón’s pioneer contribution [1, 2]. Calderón was motivated by oil prospection.This pioneer contribution motivated many developments in inverse problems, particularly in the construction of “complex geometrical optics” solutions of partial differential equations to solve several inverse problems or solutions of the Faddeev type [3]. The problem that Calderón considered was whether one can determine the electrical conductivity of a medium by making voltage and current measurements at the boundary of the medium [2]. EIT is also used in medical imaging, including early diagnosis of breast cancer [4] and monitoring pulmonary function [5]. It is also used to detect leaks in buried pipes [6]. It is applied in the location of oil and mineral deposits in the Earth’s interior, creation of astrophysical images from telescope data, finding cracks and interfaces within materials, shape optimization, model identification in growth processes, and modeling in the life sciences among others.Studying an inverse problem requires good knowledge of the theory for the corresponding direct problem.The direct problem in EIT is to find the potential differences measured at the electrodes by knowing these following informations: the shape of the medium, the conductivity distribution of the medium, the position of the electrodes, and the intensity of currents injected at the electrodes. Therefore, solving the direct problem involves modeling the behavior of the tomography when taking EIT measurements. If the current is injected into the medium, then the tomography takes the EIT measurements, and the voltage is measured.To describe the mathematical problem, we consider a conductor filling a bounded isotropic domainΩ⊂ℝn with a smooth boundary (or Lipschitz boundaries), and the electrical conductivity at each point of Ω¯ is represented by a positive C2 function γ satisfying 1/M<γ<M and γC2Ω¯≤M [7].Under the assumption of no source or sink of current inΩ, when a voltage potential f∈H1/2∂Ω is applied on the boundary ∂Ω, the induced potential in Ω is u∈H1Ω; that is the weak solution of(1)∇.γ∇u=0,inΩ,u=f,on∂Ω.Here, we can define the Dirichlet to Neumann map (DN map) formally as(2)Λγf=γ∂u∂ν∂Ω,and Calderón’s problem would recover γ from Λγ.The study of the problem (1), for n≥2 and regular γ, is reduced to the corresponding problem for the Schrödinger equation (3) with a so called potential of conductivity type qx=Δγ1/2/γ1/2 [7]. That is why we are motivated to study the problem of(3)−Δ+qu=0,inΩ,u=f,on∂Ω.The inverse problem for the Schrödinger equation is analogous to Calderón’s problem. The inverse problem for the Schrödinger equation is to determine the potential function q from the measurements of ∂u/∂ν for all possible functions f on the boundary of Ω.That is, the knowledge of the mapΛq, also called Dirichlet-to-Neumann map for the Schrödinger equation, which to any f∈H1/2∂Ω associates the normal derivative ∂u/∂ν∈H−1/2∂Ω. This is why we are motived to determine an explicit formula of the potential knowing of an explicit formula of the Dirichlet-to-Neumann map Λqf, for all given f∈H1/2S2.In this paper, our aim is to determine the potentialq on the edge of the unit ball B from the Dirichlet to Neumann map Λqf, for any given f∈H1/2S2.As contribution, we give an explicit expression for the potential radialqx near the edge of the unit ball in dimension 3, from an explicit formula for the Dirichlet to Neumann map Λqf, corresponding to the Schrödinger operator −Δ+q.The paper is organized as follows: In Section2, we study the existence of a solution of the inverse problem for the Schrödinger equation and its uniqueness and the stability of the inverse potential problem. In Section 3, we establish a result giving the explicit expression of the potential qx on the boundary of the unit ball in dimension 3. In Section 4, we present an example of potential both theoretically and numerically. In Section 5, we present conclusions and perspectives.


## 2. Inverse Potential Problem

LetΩ be the unit ball B in ℝ3 defined by B=x∈ℝ3:|x|<1 and assumed that q∈L∞B. 0 is not a Dirichlet eigenvalue of −Δ+q in Ω.We focus to determineq∈L∞B from the knowledge of the function f∈H1/2∂B, and the explicit formula of the Dirichlet to Neumann map Λq for the Schrödinger equation is defined formally as(4)Λq:H1/2∂B⟶H−1/2∂B,f⇝Λqf=∂u∂ν|∂Ω,where ν is the outer unit normal vector on ∂B, and u the unique solution of(5)−Δ+qu=0,inx<1,u=f,onx=1.These choices guarantee the existence of a solution u of (5) by the Fourier method, and 0 is not an eigenvalue that ensures the uniqueness of the solution.The mapf⟶Λqf depends linearly on f. Λq encodes the measurements of ∂u/∂ν for all possible functions f on the boundary of Ω.The inverse potential problem we pose is to obtain information on the potentialqx from f to Λq.Let us introduce some results that will be useful in the study.


### 2.1. Preliminaries

In the following, the solutionu of (5) is written in polar coordinates r>0,θ∈S2. A scaled potential is defined by a piecewise constant function, which has points of discontinuity.For allℓ∈ℕ,pℓmr denotes the Bessel function of the first type jℓγmr or the Bessel function of the second type iℓγmr, and qℓmr denotes the modified Bessel function of the first type yℓγmr or the modified Bessel function of the second type −1ℓ+1kℓγmr (see equations (22) and (23)).Yℓkθ is the spherical harmonics defined on S2.Let us introduce this theorem for the explicit formula of the DN map in the unit ball inℝ3 given by Fagueye in [8].Theorem 1.
LetB be the unit ball in ℝ3 and the scaled potential(6)qr=∑m=1nγmχrm−1,rm,r=x,where n≥1and finite,γm,rm∈ℝ, with m=1,2,…,n and 0=r0<r1<···<rn−1<rn=1 such that the Dirichlet problem for −Δ+q is well posed.Then, there is an explicit formula for the Dirichlet-to-Neumann map defined as follows:(7)ΛqYℓkθ=Cknpℓ−1n1−knpℓn1qℓn1qℓ−1n1+knqℓ−1n1−ℓqℓn1qℓn1Yℓkθ,where ℓ=1,2,…, with kn=γn,C depending on n and ℓ.Proof.
Theorem 1: for details, see [8].Lemma 1.
In polar coordinatesr,θ with r>0 and θ∈S2, if fθ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ, then equation (5) admits a unique solution of the form(8)vr,θ=∑ℓ=0∞∑k=−ℓℓvℓkrYℓkθ,where vℓkr satisfies the problem of(9)r2vℓk″+2rvℓk′−ℓℓ+1vℓk−qrr2vℓk=0,r∈0,1,limr>0,r⟶0vℓkr<∞,vℓk1=fℓk.Proof Lemma 1.
We have assumed 0 is not an eigenvalue of−Δ+q, then the problem (5) admits a unique solution vr,θ with r>0 and θ∈S2 in the unit ball B.
Iffθ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ, then we can write v1,θ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ.
Using the separation of variablesvr,θ=vrgθ, the completeness property of the spherical harmonics implies that any well-behaved function g defined in S2, that is which is single valued, continuous, and finite, can be written as gθ=∑ℓ=0∞∑k=−ℓℓaℓkYℓkθ,θ∈S2 for some choice of coefficients aℓk.
Choosingaℓk=1, then we have gθ=∑ℓ=0∞∑k=−ℓℓYℓkθ,θ∈S2. Taking r=1 in vr,θ=vrgθ, we obtain v1,θ=∑ℓ=0∞∑k=−ℓℓvℓk1Yℓkθ, and then, vℓk1=fℓk in S2.
Fromv∈H1B0,1¯, we obtain vr,θ=∑ℓ=0∞∑k=−ℓℓvℓkrYℓkθ. And vr,θ satisfying (5) gives us(10)∑ℓ=0∞∑k=−ℓℓr2∂2vℓkr∂r2Yℓkθ+∑ℓ=0∞∑k=−ℓℓ2r∂vℓkr∂rYℓkθ+∑ℓ=0∞∑k=−ℓℓvℓkrΔSYℓkθ−∑ℓ=0∞∑k=−ℓℓqrr2vℓkrYℓkθ=0,where ΔS=1/sinα∂/∂αsinα∂/∂α+1/sin2αα2/∂ϕ2, and θ=α,ϕ∈S2.
Since−ΔSYℓk=ℓℓ+1Yℓk, we have(11)∑ℓ=0∞∑k=−ℓℓr2∂2vℓkr∂r2+2r∂vℓkr∂r−ℓℓ+1vℓkr−qrr2vℓkrYℓkθ=0.
We knowYℓkℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ is an orthonormal basis, then vℓkr verifies (9).Proposition 1.
LetΩ be a bounded Lipschitz domain in ℝn,q∈L∞Ω complex value potential, and f∈H1/2∂Ω. We assume that 0 is not an eigenvalue for the operator −Δ+q. Then, there exists a unique weak solution u∈H1Ω to the Dirichlet problem of(12)−Δ+qu=0,inΩ,u=f,on∂Ω.Moreover,(13)uH1Ω≤CfH1/2∂Ω,where C depends on n,Ω, and qL∞Ω.(For details of Proposition1, see [9]).Lemma 2.
LetΩ⊆ℝn be a bounded open set with a smooth boundary, where n≥3, and q1 and q2 be two functions in L∞Ω such that the Dirichlet problems for −Δ+q1 and −Δ+q2 in Ω are well posed. Then, for any f1,f2∈H1/2∂Ω, one has(14)Λq1−Λq2f1,f2=∫Ωq1−q2u1u2dx.where uj∈H2Ω is the solution of −Δ+qjuj=0 in Ω with boundary values uj|∂Ω=fj,j=1,2.(For details of lemma, see [7]).Theorem 2.
LetΩ⊆ℝn be a C1 boundary. ν denotes the outer unit normal vector on ∂Ω, and νj is one of its components.(1)
(Green formula) Ifu∈C1Ω¯ and v∈C2Ω¯, then(15)∫ΩuΔvdx=−∫Ω∇u.∇vdx+∫∂Ωu∂v∂νds,(2)
(Integration by parts) Ifu∈C1Ω¯ and v∈C1Ω¯, then(16)∫Ωu∂v∂xjdx=−∫Ω∂u∂xjvdx+∫∂Ωuvνjds,(3)
(Divergence theorem) IfW:Ω¯⟶ℝn is a C1 vector field, then(17)∫ΩdivWdx=∫∂ΩW.νds.Lemma 3.
LetΩ be included in ℝn. Let gx be a continuous function in Ω. If for all function φ∈C∞Ω with compact support in Ω, we have(18)∫Ωgxφxdx=0.Then, g is zero in Ω.(For proof of Lemma 3, see [10]).Let us suppose that there is x0∈Ω such that gx0≠0. We assume that gx0>0 (otherwise, we take −g). By continuity, there exists a small open neighborhood Γ⊂Ω of x0 such that gx>0 for all x∈Γ. Thus, for any positive and nonzero test function f, whose support is included in g, we have (19)∫Ωgxφxdx=∫Γgxφxdx=0,which is a contradiction with the hypothesis on g. Then, gx=0 for all x∈Ω. In order to reconstruct the potential q from Λq and f∈H1/2∂B, let us give results concerning its existence.


### 2.2. Existence of Solution of the Inverse Problem

In this section, the inverse potential problem of the Schrödinger equation (5) we pose is to determine the potential function qx from the knowledge of Λqf for all given f∈H1/2S2.LetCq be the subset of H1/2S2×H−1/2S2 such that(20)Cq=f,Λqf:f∈H1/2S2.Iff=Yℓ and the potential q a piecewise constant radial, then ΛqYℓ is defined by (7) in Theorem 1, and the subset Cq≠∅.Our aim is to determineq∈L∞B from the Cauchy data Cq.The question what we will want to answer is as follows:Is thereq∈L∞B nonzero and u∈H1B solution of problem 4 such that Λqf=∂u/∂ν|S2?According to [11, 12], there is q∈L∞B nonzero and u∈H1B solution of problem 4 such that Λqf=∂u/∂ν|S2.Now, we are interested first in the uniqueness result and the stability of the reconstruction ofq.


### 2.3. Uniqueness and Stability for the Inverse Potential Problem

Forn≥3 and the potential function q∈L∞B, the uniqueness result for the inverse potential problem, i.e., Λq uniquely determines q, was proved by Salo in [7], Sylvester and Uhlmann in [13], Uhlmann in [2], and Haberman and Tataru in [14].Uhlmann in [2] gives the following uniqueness result in dimension n≥3Theorem 3.
LetΩ be a bounded Lipschitz domain in ℝn,qi∈L∞Ω,i=1,2. We ssume that Cq1=Cq2, then q1=q2.Salo in [7] also gives the result as follows:Theorem 4.
LetB be the unit ball in ℝ3, and let q1 and q2 be two functions in L∞B such that the Dirichlet problems for −Δ+q1 and −Δ+q2 in B are well posed. If Λq1=Λq2, then q1=q2 in B.After ensuring the uniqueness, we will be interested in the stability of the reconstruction ofq; that is, we will want to answer the question as follows:Ifq1,q2 are two potentials such that Λq1 is close to Λq2, does this imply that q1 is close to q2?In [15], Beretta et al. studied the Lipschitz stability of an inverse boundary value problem for a Schrödinger type equation by assuming that the potential q is a piecewise constant complex function in dimension n≥2.In [16], Beretta and Francini proved the Lipschitz stability for the electrical impedance tomography problem for a complex piecewise conductivity γ satisfying λ−1≤ℜγ and γ≤λ for some λ≥1, in a bounded domain Ω⊂ℝn with a smooth boundary.In [17] Uhlmann has given a depth dependent logarithmic type stability estimate of an inverse boundary value problem for a Schrödinger type equation.The type of depth dependent stability estimate has been proved in [18] for the case of some electrical inclusions.In [8], Fagueye confirmed the Lipschitz stability in the following theorem for the reconstruction of the potential q near to the border S2 by giving an estimate constant.Theorem 5.
LetB be the unit ball in ℝ3, and the scaled potential qi,i=1,2 verifies(21)qir=∑m=1nγmiχrm−1,rm,i=1,2,r=x,where n≥1and finite,γm,rm∈ℝ, with m=1,2,…,n and 0=r0<r1<···<rn−1<rn=1, and kmi=γmi such that the Dirichlet problem for −Δ+qi is well posed. We assume that γn1×γn2>0, and there is a positive constant M such that(22)qiL∞B≤M.
Then, there is a constantC=Cn,M,ℓ such that(23)Λq1−Λq2H1/2⟶H−1/2≥Cγn1−γn2.Estimate (23) solves the stability problem near to the border.Taking qx=Δγ1/2/γ1/2 when γ∈C2Ω¯ in 1, this following theorem ensures the Lipschitz stability in the ball B [7].Theorem 6.
Kenig–Sjöstrand–Uhlmann assume thatΩ is convex and Γ is any open subset of ∂Ω. If Λγ1f|Γ=Λγ2f|Γ for all f∈H1/2∂Ω, and if γ1|∂Ω=γ2|∂Ω, then γ1=γ2 in Ω.In the next section, we will look at the reconstruction of the potential from the Cauchy dataCq.


## 2.1. Preliminaries

In the following, the solutionu of (5) is written in polar coordinates r>0,θ∈S2. A scaled potential is defined by a piecewise constant function, which has points of discontinuity.For allℓ∈ℕ,pℓmr denotes the Bessel function of the first type jℓγmr or the Bessel function of the second type iℓγmr, and qℓmr denotes the modified Bessel function of the first type yℓγmr or the modified Bessel function of the second type −1ℓ+1kℓγmr (see equations (22) and (23)).Yℓkθ is the spherical harmonics defined on S2.Let us introduce this theorem for the explicit formula of the DN map in the unit ball inℝ3 given by Fagueye in [8].Theorem 1.
LetB be the unit ball in ℝ3 and the scaled potential(6)qr=∑m=1nγmχrm−1,rm,r=x,where n≥1and finite,γm,rm∈ℝ, with m=1,2,…,n and 0=r0<r1<···<rn−1<rn=1 such that the Dirichlet problem for −Δ+q is well posed.Then, there is an explicit formula for the Dirichlet-to-Neumann map defined as follows:(7)ΛqYℓkθ=Cknpℓ−1n1−knpℓn1qℓn1qℓ−1n1+knqℓ−1n1−ℓqℓn1qℓn1Yℓkθ,where ℓ=1,2,…, with kn=γn,C depending on n and ℓ.Proof.
Theorem 1: for details, see [8].Lemma 1.
In polar coordinatesr,θ with r>0 and θ∈S2, if fθ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ, then equation (5) admits a unique solution of the form(8)vr,θ=∑ℓ=0∞∑k=−ℓℓvℓkrYℓkθ,where vℓkr satisfies the problem of(9)r2vℓk″+2rvℓk′−ℓℓ+1vℓk−qrr2vℓk=0,r∈0,1,limr>0,r⟶0vℓkr<∞,vℓk1=fℓk.Proof Lemma 1.
We have assumed 0 is not an eigenvalue of−Δ+q, then the problem (5) admits a unique solution vr,θ with r>0 and θ∈S2 in the unit ball B.
Iffθ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ, then we can write v1,θ=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ.
Using the separation of variablesvr,θ=vrgθ, the completeness property of the spherical harmonics implies that any well-behaved function g defined in S2, that is which is single valued, continuous, and finite, can be written as gθ=∑ℓ=0∞∑k=−ℓℓaℓkYℓkθ,θ∈S2 for some choice of coefficients aℓk.
Choosingaℓk=1, then we have gθ=∑ℓ=0∞∑k=−ℓℓYℓkθ,θ∈S2. Taking r=1 in vr,θ=vrgθ, we obtain v1,θ=∑ℓ=0∞∑k=−ℓℓvℓk1Yℓkθ, and then, vℓk1=fℓk in S2.
Fromv∈H1B0,1¯, we obtain vr,θ=∑ℓ=0∞∑k=−ℓℓvℓkrYℓkθ. And vr,θ satisfying (5) gives us(10)∑ℓ=0∞∑k=−ℓℓr2∂2vℓkr∂r2Yℓkθ+∑ℓ=0∞∑k=−ℓℓ2r∂vℓkr∂rYℓkθ+∑ℓ=0∞∑k=−ℓℓvℓkrΔSYℓkθ−∑ℓ=0∞∑k=−ℓℓqrr2vℓkrYℓkθ=0,where ΔS=1/sinα∂/∂αsinα∂/∂α+1/sin2αα2/∂ϕ2, and θ=α,ϕ∈S2.
Since−ΔSYℓk=ℓℓ+1Yℓk, we have(11)∑ℓ=0∞∑k=−ℓℓr2∂2vℓkr∂r2+2r∂vℓkr∂r−ℓℓ+1vℓkr−qrr2vℓkrYℓkθ=0.
We knowYℓkℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ is an orthonormal basis, then vℓkr verifies (9).Proposition 1.
LetΩ be a bounded Lipschitz domain in ℝn,q∈L∞Ω complex value potential, and f∈H1/2∂Ω. We assume that 0 is not an eigenvalue for the operator −Δ+q. Then, there exists a unique weak solution u∈H1Ω to the Dirichlet problem of(12)−Δ+qu=0,inΩ,u=f,on∂Ω.Moreover,(13)uH1Ω≤CfH1/2∂Ω,where C depends on n,Ω, and qL∞Ω.(For details of Proposition1, see [9]).Lemma 2.
LetΩ⊆ℝn be a bounded open set with a smooth boundary, where n≥3, and q1 and q2 be two functions in L∞Ω such that the Dirichlet problems for −Δ+q1 and −Δ+q2 in Ω are well posed. Then, for any f1,f2∈H1/2∂Ω, one has(14)Λq1−Λq2f1,f2=∫Ωq1−q2u1u2dx.where uj∈H2Ω is the solution of −Δ+qjuj=0 in Ω with boundary values uj|∂Ω=fj,j=1,2.(For details of lemma, see [7]).Theorem 2.
LetΩ⊆ℝn be a C1 boundary. ν denotes the outer unit normal vector on ∂Ω, and νj is one of its components.(1)
(Green formula) Ifu∈C1Ω¯ and v∈C2Ω¯, then(15)∫ΩuΔvdx=−∫Ω∇u.∇vdx+∫∂Ωu∂v∂νds,(2)
(Integration by parts) Ifu∈C1Ω¯ and v∈C1Ω¯, then(16)∫Ωu∂v∂xjdx=−∫Ω∂u∂xjvdx+∫∂Ωuvνjds,(3)
(Divergence theorem) IfW:Ω¯⟶ℝn is a C1 vector field, then(17)∫ΩdivWdx=∫∂ΩW.νds.Lemma 3.
LetΩ be included in ℝn. Let gx be a continuous function in Ω. If for all function φ∈C∞Ω with compact support in Ω, we have(18)∫Ωgxφxdx=0.Then, g is zero in Ω.(For proof of Lemma 3, see [10]).Let us suppose that there is x0∈Ω such that gx0≠0. We assume that gx0>0 (otherwise, we take −g). By continuity, there exists a small open neighborhood Γ⊂Ω of x0 such that gx>0 for all x∈Γ. Thus, for any positive and nonzero test function f, whose support is included in g, we have (19)∫Ωgxφxdx=∫Γgxφxdx=0,which is a contradiction with the hypothesis on g. Then, gx=0 for all x∈Ω. In order to reconstruct the potential q from Λq and f∈H1/2∂B, let us give results concerning its existence.


## 2.2. Existence of Solution of the Inverse Problem

In this section, the inverse potential problem of the Schrödinger equation (5) we pose is to determine the potential function qx from the knowledge of Λqf for all given f∈H1/2S2.LetCq be the subset of H1/2S2×H−1/2S2 such that(20)Cq=f,Λqf:f∈H1/2S2.Iff=Yℓ and the potential q a piecewise constant radial, then ΛqYℓ is defined by (7) in Theorem 1, and the subset Cq≠∅.Our aim is to determineq∈L∞B from the Cauchy data Cq.The question what we will want to answer is as follows:Is thereq∈L∞B nonzero and u∈H1B solution of problem 4 such that Λqf=∂u/∂ν|S2?According to [11, 12], there is q∈L∞B nonzero and u∈H1B solution of problem 4 such that Λqf=∂u/∂ν|S2.Now, we are interested first in the uniqueness result and the stability of the reconstruction ofq.


## 2.3. Uniqueness and Stability for the Inverse Potential Problem

Forn≥3 and the potential function q∈L∞B, the uniqueness result for the inverse potential problem, i.e., Λq uniquely determines q, was proved by Salo in [7], Sylvester and Uhlmann in [13], Uhlmann in [2], and Haberman and Tataru in [14].Uhlmann in [2] gives the following uniqueness result in dimension n≥3Theorem 3.
LetΩ be a bounded Lipschitz domain in ℝn,qi∈L∞Ω,i=1,2. We ssume that Cq1=Cq2, then q1=q2.Salo in [7] also gives the result as follows:Theorem 4.
LetB be the unit ball in ℝ3, and let q1 and q2 be two functions in L∞B such that the Dirichlet problems for −Δ+q1 and −Δ+q2 in B are well posed. If Λq1=Λq2, then q1=q2 in B.After ensuring the uniqueness, we will be interested in the stability of the reconstruction ofq; that is, we will want to answer the question as follows:Ifq1,q2 are two potentials such that Λq1 is close to Λq2, does this imply that q1 is close to q2?In [15], Beretta et al. studied the Lipschitz stability of an inverse boundary value problem for a Schrödinger type equation by assuming that the potential q is a piecewise constant complex function in dimension n≥2.In [16], Beretta and Francini proved the Lipschitz stability for the electrical impedance tomography problem for a complex piecewise conductivity γ satisfying λ−1≤ℜγ and γ≤λ for some λ≥1, in a bounded domain Ω⊂ℝn with a smooth boundary.In [17] Uhlmann has given a depth dependent logarithmic type stability estimate of an inverse boundary value problem for a Schrödinger type equation.The type of depth dependent stability estimate has been proved in [18] for the case of some electrical inclusions.In [8], Fagueye confirmed the Lipschitz stability in the following theorem for the reconstruction of the potential q near to the border S2 by giving an estimate constant.Theorem 5.
LetB be the unit ball in ℝ3, and the scaled potential qi,i=1,2 verifies(21)qir=∑m=1nγmiχrm−1,rm,i=1,2,r=x,where n≥1and finite,γm,rm∈ℝ, with m=1,2,…,n and 0=r0<r1<···<rn−1<rn=1, and kmi=γmi such that the Dirichlet problem for −Δ+qi is well posed. We assume that γn1×γn2>0, and there is a positive constant M such that(22)qiL∞B≤M.
Then, there is a constantC=Cn,M,ℓ such that(23)Λq1−Λq2H1/2⟶H−1/2≥Cγn1−γn2.Estimate (23) solves the stability problem near to the border.Taking qx=Δγ1/2/γ1/2 when γ∈C2Ω¯ in 1, this following theorem ensures the Lipschitz stability in the ball B [7].Theorem 6.
Kenig–Sjöstrand–Uhlmann assume thatΩ is convex and Γ is any open subset of ∂Ω. If Λγ1f|Γ=Λγ2f|Γ for all f∈H1/2∂Ω, and if γ1|∂Ω=γ2|∂Ω, then γ1=γ2 in Ω.In the next section, we will look at the reconstruction of the potential from the Cauchy dataCq.


## 3. Reconstruction of the Potential

First, we show that the Dirichlet to Neumann map depends continuously on the potentialq.Theorem 7.
The assumptions are in Theorem5.Then, there is a constantC′=C′n,M,ℓ>0 such that(24)Λq1−Λq2H1/2⟶H−1/2≤C′q1−q2.Inequality (24) gives the continuous dependence of the Dirichlet to Neumann map with respect to the potential q.Proof of Theorem 7.
using the results of Lemma2 with Ω=B, the unit ball, and f1=u1|S2,f2=u2|S2, we have(25)Λq1−Λq2f1,f2=∫Bq1−q2u1u2dx.
Using Cauchy–Schwarz theorem, we have(26)Λq1−Λq2f1,f2≤q1−q2L∞Bu1H1Bu2H1B.
Using the inequality (9) in the Proposition1 we have also(27)Λq1−Λq2f1,f2≤C′q1−q2L∞Bf1H1/2S2f2H1/2S2,where C′ depends on B and qiL∞B.
Takingf1=f2=Yℓ,ℓ=1,2,…., we obtain(28)Λq1−Λq2YℓH−1/2S2≤C′q1−q2L∞B.
We know that for allYℓk∈H1/2S2,ℓ≥1,(29)Λq1−Λq2H1/2⟶H−1/2=supYℓk∈H1/2S2,ℓ≥1Λq1−Λq2YℓkH−1/2S2.
Then, the relation (28) implies that(30)Λq1−Λq2H1/2⟶H−1/2≤C′q1−q2L∞B.
Let us introduce this result by giving the explicit formula of the radial potentialq for our inverse problem.Theorem 8.
Let the unit ballB be defined by B=x∈ℝ3:x<1; its surface S2=x∈ℝ3:x=1 and f,Λqf∈Cq with f=∑ℓ=0∞∑k=−ℓℓfℓkYℓkθ∈H1/2S2 and Λqf=∑ℓ=0∞∑k=−ℓℓcℓkfℓkYℓkθ∈H−1/2S2.We assume thatfℓk,ℓ=1,2,…,−ℓ≤k≤ℓ, are not all zero.If 0 is not a Dirichlet eigenvalue of−Δ+q in B, then there exist a unique nonzero solution u∈H1B of the problem(31)−Δ+qu=0,inx<1,u=f,onx=1,and a unique nonzero radial function q∈L∞B, which is uniquely determined at the edge from the data f,Λqf by(32)q1=Λqf2+∑ℓ=0∞∑k=−ℓℓfℓk2p˜ℓk2f2,where p˜ℓk is a constant for any ℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ.And in the bull,(33)qr=∂2u/∂r2+2/r∂u/∂rur−ℓℓ+1r2,for all0<r<1,θ∈S2andℓ=0,1,2,…Proof of Theorem 8.
Let us consider the problem (5). The existence and uniqueness of its solution u are ensured by the assumptions.
According to Lemma1, we have(34)ur,θ=∑ℓ=0∞∑k=−ℓℓuℓkrYℓkθ.
According to [11, 12], there is q∈L∞B nonzero and u∈H1B solution of problem (5) such that Λqf=∂u/∂ν|S2.
We know thatq∈L∞B with suppq⊂B0,1¯⊂ℝ3 and ur,θ∈H1B is the unique solution of 4 with u1,θ=fθ,∂ur,θ/∂r|S2=Λqfθ.
Then, using the weak reformulation of the equation−Δ+qu=0inx<1, we have(35)∫BuΔu=∫Bqu2.
According to the Divergence theorem, Green’s formula, and the orthonormal basisYℓkℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ, this implies that(36)∫S2u∂u∂nds−∫B∇u2=∫Bqu2,or(37)∫S2Λqff−∫B∇u2=∫Bqu2.
We know that the gradient of the multipole expansion of the scalar(38)u=∑ℓ=0∞∑k=−ℓℓuℓkrYℓkθ,can be expressed in terms of spherical vector harmonics as follows:(39)∇u=∑ℓ=0∞∑k=−ℓℓduℓkdrYℓk+uℓkrΨℓk,where Yℓk=e^rYℓk,Ψℓk=r∇Yℓke^θ,e^r, and e^θ are unit vectors such that e^r.e^θ=0 and ∇Yℓke^θ=pℓkθ,pℓk a function depending on θ.
Then, using Bessel Parseval’s identity, we can write(40)∫S2Λqff−∫B∑ℓ=0∞∑k=−ℓℓduℓkdr2+uℓkrrpℓk2=∫Bq∑ℓ=0∞∑k=−ℓℓuℓk2,which is equivalent to(41)∫S2Λqff−∫01∫S2∑ℓ=0∞∑k=−ℓℓduℓkdr2+uℓkpℓk2r2dθdr=∫01∫S2q∑ℓ=0∞∑k=−ℓℓuℓk2r2dθdr.
We note that the first term of left-hand side of this equality does not depend onr. By differentiating both members of the equality with respect to r, we obtain(42)∫S2∑ℓ=0∞∑k=−ℓℓduℓkdr2+uℓkpℓk2dθ=∫S2q∑ℓ=0∞∑k=−ℓℓuℓk2dθ.
By integrating overθ, since only pℓk depends on θ, we have(43)∑ℓ=0∞∑k=−ℓℓduℓkdr2+uℓkp˜ℓkθ2=q∑ℓ=0∞∑k=−ℓℓuℓk2,inS2,where(44)p˜ℓk2=1S2∫S2pℓkθ2dθ.
Then, sinceur,θ≠0 for all ur,θ∈H1B in the solution of (5), we have(45)qr=∑ℓ=0∞∑k=−ℓℓuℓk′r2+uℓkp˜ℓk2∑ℓ=0∞∑k=−ℓℓuℓk2r,inS2.
And, sincer=1 in S2, we have(46)q1=∑ℓ=0∞∑k=−ℓℓuℓk′12+uℓk1p˜ℓk2∑ℓ=0∞∑k=−ℓℓuℓk21,where p˜ℓk is a constant for any ℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ.
Using the Cauchy dataf,Λqf∈Cq and Bessel Parseval’s identity, since f≠0, we have(47)q1=Λqf2+∑ℓ=0∞∑k=−ℓℓfℓk2p˜ℓk2f2,where p˜ℓk is a constant for any ℓ=0,1,2,…;k=−ℓ,−ℓ+1,…,ℓ−1,ℓ.
This result shows thatq can be recovered from the Cauchy data Cq near the edge S2.
To recover it in the unit ball, we use always the Laplacian method in spherical coordinates as follows:(48)Δ=∂2∂r2+2r∂∂r+1r2ΔS,where(49)ΔS=1sinα∂∂αsinα∂∂α+1sin2α∂2∂ϕ2.
Since(50)u=∑ℓ=0∞∑k=−ℓℓuℓkrYℓkθ,and −ΔSYℓk=ℓℓ+1Yℓk, then(51)−Δ+qu=0is equivalent to(52)−∂2u∂r2−2r∂u∂r+ℓℓ+1r2u+qru=0.
As the solution must be finite at the origin, then we have the following condition:(53)limr>0,r⟶0ur<∞.
Sinceur,θ≠0 for all ur,θ∈H1B in the solution of (5), we can write(54)qr=∂2u/∂r2+2/r∂u/∂rur−ℓℓ+1r2,for all0<r<1andθ∈S2.
From (32) and (33), it is easy to verify that q∈L∞B.
In the below section, we will determineq both theoretically and numerically from a Cauchy data example Cq.


## 4. Numerical Simulations

In this section, we will numerically determine the radial potential withℓ=0,1,… in order to simplify the calculations. Since Theorem 3, the Λ map which associates a Dirichlet-to-Neumann map to all potentials q is injective.Let us considerPℓk as the solution to Laplace’s equation in spherical coordinates and introduce the spherical harmonics, Yℓk:(55)Yℓkα,ϕ=−1k2ℓ+1ℓ−k!4πℓ+k!Pℓkcosαeikϕ,forℓ=0,1,2,3,…,k=−ℓ,−ℓ+1,…,ℓ−1,ℓ.The phase factor−1k, introduced originally by Condon and Shortly, is convenient for applications in quantum mechanics. Let us introduce(56)Yℓ−kα,ϕ=−1kYℓkα,ϕ∗,where the ∗ means the complex conjugation.Let us list the spherical harmonics forℓ=0,1,... and non-negative values of k as(57)ℓ=0,Y00α,ϕ=14π,ℓ=1,Y10α,ϕ=34πcosα,Y11α,ϕ=−38πsinαeiϕ.Let us assume that the Dirichlet-to-Neumann map can be characterized by the firstn eigenvalues defined in theorem (7).Then, for allθ=α,ϕ∈S2with0≤α≤πand0≤ϕ≤2π, we define fθ=∑ℓ=01∑k=−ℓℓfℓkYℓkθ such that(58)fθ=14π+34πcosα−2i38πsinαsinϕ,Λqfθ=−2i38πsinαsinϕ+34πcosα.Let us introduce the first vector spherical harmonics introduced in equation (39) and defined by(59)Yℓk=e^rYℓk,Ψℓk=r∇Yℓke^θ,with pℓk depending on θ=α,ϕ,ande^θ is a combination of unit vectors e^α and e^ϕ. We have for(60)ℓ=0,Y00=14πe^r,Ψ00=0,ℓ=1,Y10=34πcosαe^r,Y11=−38πsinαeiϕe^r,Ψ10=−34πsinαe^α,Ψ11=−38πcosαe^α+ie^ϕeiϕ.According the explicit expression of the radial potentialqx near the boundary of the unit ball defined in (32), Theorem 1, and the infectivity of the map Λ, we have by calculations(61)q1=1π14+sin2ϕ,for each parameterϕ∈0,2π.Some values of the potentialq1 on the edge of B following the values of ϕ are presented in Table 1.Table 1
Some values of the potential near the edge.
ϕ0π/12π/6π/4π/3π/25π/64π/3q11/4π1/4π3−31/2π3/4π1/π5/4π1/2π1/4π5−3They are also represented in Figure1.Figure 1
Potential near the edge forl = 0, 1.The same approach allows to determine both theoretically and numerically the radial potential near the edge forℓ=0,1,2 or ℓ=0,1,2,3,….


## 5. Conclusion

We can conclude that the potentialqr can be recovered explicitly from the Cauchy data Cq near the edge of the unit ball in dimension 3 both theoretically and numerically. This paper can potentially provide accurate physical knowledge of the nature and properties of the potential q, near the edge of the unit ball, that is subject to a particle with a wave function u and that is a solution of the Schrödinger equation satisfying the boundary conditions. For perspectives, we will determine an explicit formula of the potential in any domain Ω with a Lipschitzian border in dimension 3 and in dimension 2. We will be also interested in the case where the domain Ω edge is not Lipschitzian [19].Another project is to see how to apply these results in the resolution of problems related to greenhouse gases and air pollution.



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*Source: 2935392-2022-02-15.xml*