# Integral and Variational Formulations for the Helmholtz Equation Inverse Source Problem

Authors: Marcelo L. Rainha; Nilson C. Roberty
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

The purpose of this paper is to explore the Hilbert space functional structure of the Helmholtz equation inverse source problem. An integral equation for the sources reconstruction based on the composition of the trace and Green's function operators is introduced and compared with the reciprocity source reconstruction methodologies. An equivalence theorem comparing the integral inverse source equation with the variational weak reciprocity gap functional equation is then demonstrated. Some examples on applications to the unitary disk are presented.


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## Body

## 1. Introduction

The inverse source problem for the Helmholtz Dirichlet equation is a basic tool for the investigation of transient source problems [1–4]. In order to investigate this class of problems, let Ω⊂ℝN be a bounded domain with smooth boundary Γ. Let κ be a complex number, g∈H1/2(Γ), and f∈L2(Ω). The direct problem with the Helmholtz operator: to find a regular field u that satisfy the system
(1.1)-Δu-κ2u=finΩ,γ0u=gon⁡Γ,
is well posed and has a unique solution u∈H1(Ω) when κ2 is not an eigenvalue of the Laplacian. In this paper we consider γ0:H1(Ω)→H1/2(Γ) the trace operator and simplify the notation by calling the boundary data g of problem (1.1) γ0u. The trace theorem [5] assures the existence of a function gνx∈H-1/2(Γ) which is the normal trace, that is,
(1.2)γ1u=∂νxu.Whenκ is a real positive or an imaginary number we have, respectively, the proper or the modified Helmholtz equation. When κ=0, we obtain the Laplace equation. For the complete setting of complex values, we consider the problem as the Helmholtz equation direct problem.The inverse source problem consist, in knowing the Cauchy data in the boundaryΓ, that is, the Dirichlet to Neumann map in at least one Dirichlet datum g, to recover the source f. It may be formally posed as follows:
(1.3)given a Cauchy data set{(g,gνx)}⊂H1/2(Γ)×H-1/2(Γ),with compatibility condition∂g∂νx=gνxon⁡Γ,for the equation model in the system(1.1),to find(u,f)∈H1(Ω)×L2(Ω).The two problems, direct and inverse, can also be formulated with only one system of equations: to find(u,f)∈H1(Ω)×L2(Ω) such that
(1.4)-Δu-k2u=finΩ,γ0u=gon⁡Γ,γ1u=gνxon⁡Γ.Since in the inverse problem the Cauchy data are known, we may associate these data with a fourth-order Dirichlet problem with the Bilaplacian operator(1.5)Δ2u-κ4u=hinΩ,γ0u=gon⁡Γ,γ1u=gνxon⁡Γ,
where h∈H-2(Ω) is an arbitrary given function. This problem is well posed and has a unique solution u∈H2(Ω) when κ4 is not an eigenvalue of the Bilaplacian. This motivates the following naive existence resultRemark 1.1.
Suppose that a Cauchy data pair(g,gνx)∈H3/2(Γ)×H1/2(Γ) is more regular than normal case and κ4∉Σ4 is not an eigenvalue of the Bilaplacian, then there exists a solution to inverse source problem (1.3).Proof.
Since data is regular, we may consider the fourth-order direct Dirichlet problem (1.5) with the given Cauchy data and some h∈H-2(Ω). The inverse source solution will be f=-Δu+κ2u∈L2(Ω).To obtain the existence of global solution for problems (1.1), and (1.5), please see [6]. For more information about linear integral equations, see [7]. For the inverse source problems, see [8]. For functional analysis, please see [9].In Section2 we develop a Hilbert space functional framework to the problem based on special L2(Ω) decomposition. The analysis of the Dirichlet to Newman map and of the Source to Neumann maps is done in Sections 2.1 and 2.2, respectively. It is based only on the analysis of the direct problem structure. The analysis of the Adjoint Source to Neumann map is done in Section 2.3. In Section 3 we use the Green function operator to put together the results found in Section 2. There in Section 3.1 we present an integral equation for the inverse problem based on the relative Dirichlet to Newman map. The reciprocity gap functional is introduced in Section 4, where an equivalence theorem between this and the integral formulation is proved. In Section 5 some particular results for the unitary disk in ℝ2 are presented. Finally, we conclude the paper in Section 6.


## 2. The Dirichlet and the Source to Neumann Map

Forl∈ℝ, the space Hl(Ω) is the Sobolev class of the functions of the spatial variable x. For more information, see [5]. Let us consider for future use the following sets of eigenvalues:
(2.1)Σ2∶={λ∈C:-Δu=λuinH2(Ω)∩H01(Ω)},Σ4∶={λ∈C:Δ2u=λuinH4(Ω)∩H02(Ω)}.Definition 2.1.
One says that a function is metaharmonic when it is in the set(2.2)H-Δ-κ2(Ω)={v∈L2(Ω);-Δv-κ2v=0},
where κ2∉Σ2 and κ4∉Σ4.


### 2.1. The Dirichlet to Neumann Map

Definition 2.2.
Consider problem (1.1) with zero source, that is, f=0 and g∈H1/2(Γ).This problem has a solution w0∈H1(Ω). One defines the Dirichlet to Neumann map for the Helmholtz equation as the operator
(2.3)Λ0:H1/2(Γ)⟶H-1/2(Γ),Λ0[g](x)=∂w0∂νx(x),x∈Γ.
By the trace theorem [5], it is well defined, linear, and continuous.Remark 2.3.
We alternatively can define the Dirichlet to Neumann map (2.2) for the problem (1.1) as an operator with a nonzero source problem Λf:H1/2(Γ)→H-1/2(Γ) defined by the graph (g,Λfg)∶=(u,∂νu)∣∂Ω.Definition 2.4.
One has(2.4)M∶={(g,gνx)∈H1/2(Γ)×H-1/2(Γ);gνx-Λ0g∈H1/2(Γ)},
where Λ0 is a Dirichlet to Neumann map.Theorem 2.5.
Let(g,gνx)∈H1/2(Γ)×H-1/2(Γ). A function  f∈H-Δ+κ2(Ω)is a solution to the inverse source problem (1.3) if and only if  (g,gνx)∈ℳProof.
Suppose(g,gνx)∈ℳ. Consider the following fourth-order problem:
(2.5)(Δ2-k4)w=0inΩ,γ0w=0on⁡Γ,γ1w=gνx-Λ0gon⁡Γ.
Let us use this problem to define a source f0=(-Δ-k2)w, where w∈H2(Ω) is solution of problem (2.5). Note that f0∈H-Δ+κ2(Ω). Let (u,f)∈H1(Ω)×L2(Ω) be a solution of the inverse source problem (1.4). Then u=w0+w, where w0 is the solution of the homogeneous source problem in the definition of the Dirichlet to Neumann map (2.2) and f=f0. The sufficiency is proved.
To proof necessity, suppose that there exists a(u,f)∈H1(Ω)×H-Δ+κ2(Ω) which is solution of the inverse problem (1.4).
Consider the second-order problem with homogeneous boundary Dirichlet data(2.6)(-Δ-k2)w0=f,x∈Ω,γ0w0=0,x∈Γ,
with a unique solution w0∈H2(Ω)∩H01(Ω). By the trace theorem, we have γ1w0∈H1/2(Γ). Note that gνx=γ1w0+γ1w0, where γ1w0=Λ0g. So, γ1w0=gνx-Λ0g.Remark 2.6.
We have proved the existence and uniqueness of solution to (1.4) in H1(Ω)×H-Δ+κ2(Ω). However, this does not mean that when we do the search in a larger space H1(Ω)×L2(Ω), we will continue having uniqueness. In fact, we will prove in the next proposition that f-f0⊥v∈H-Δ-κ2(Ω) and consequently
(2.7)f∈f0+[H-Δ-κ2(Ω)]⊥,
where f∈L2(Ω) and f0∈H-Δ+κ2(Ω). This leads us to conclude that the solution of (1.4) is actually a class of functions, where we can “reconstruct” or “observe” only part of the solution in H-Δ+κ2(Ω).Proposition 2.7.
Ifκ2∉Σ2 and  κ4∉Σ4, then
(2.8)L2(Ω)=[H-Δ-κ2(Ω)]⊥⊕H-Δ+κ2(Ω).Proof.
Consequence of Lemmas2.8 and 2.9.From now on, we are supposing always that κ2∉Σ2 and κ4∉Σ4.Lemma 2.8.
One hasL2(Ω)=H-Δ+κ2(Ω)⊕(-Δ-κ2)[H02(Ω)].Proof.
For an arbitrarily givenf∈L2(Ω), the problem (1.1) with zero Dirichlet datum g=0 is well posed and has an w0∈H2(Ω) with normal derivative trace Λf=∂νw0∈H1/2(Γ). Also, since κ4∉Σ4, the fourth-order problem (1.5) with Cauchy datum (0,Λf)∈H3/2(Γ)×H1/2(Γ) and zero source has solution v and is well posed in H2(Ω). These two H2(Ω) solutions may be used to define a function
(2.9)w:=w0-v∈H02(Ω),
and since by problem (1.1)
(2.10)f=(-Δ-κ2)w0=(-Δ-κ2)v+(-Δ-κ2)w,
we obtain that arbitrary L2(Ω) function f is the sum of a function in H-Δ-κ2(Ω) and a function in (-Δ-κ2)[H02(Ω)]. If we show that
(2.11)H-Δ+κ2(Ω)∩(-Δ-κ2)[H02(Ω)]={0},
we prove that L2(Ω)=H-Δ+κ2⊕(-Δ-κ2)[H02(Ω)]. For this, take a u in the intersection H-Δ-κ2∩(-Δ+κ2)[H02(Ω)]. Then
(2.12)(-Δ-κ2)u=0,u=(-Δ+κ2)v,for somev∈H02(Ω)
which means that v is a solution of completely homogeneous fourth-order problem (1.5), with no source and zero Cauchy datum. Since κ4∉Σ4, the unique solution is trivial v=0, and u=(-Δ+κ2)v=0. We have proved that L2(Ω)=H-Δ-κ2(Ω)⊕(-Δ+κ2)[H02(Ω)].Lemma 2.9.
One has(-Δ-κ2)[H02(Ω)]=[H-Δ-κ2(Ω)]⊥.Proof.
Using the second Green theorem withf∈H-Δ-κ2(Ω) and v=(-Δ-κ2)w for some w∈H02(Ω), we obtain
(2.13)∫Ωfvdx=∫Ωf(-Δ-κ2)wdx=∫Ωf(-Δw)dx-κ2∫Ωfwdx=∫Ω(-Δf)wdx-κ2∫Ωfwdx=0,
and since the scalar product with arbitrary h∈(-Δ-κ2)[H02(Ω)] is zero, the inclusion (-Δ-κ2)[H02(Ω)]⊂H-Δ-κ2⊥(Ω) follows.
To prove the other inclusion, letf∈L2(Ω), suppose f is orthogonal to v for all v∈H-Δ-κ2(Ω), and take a function w∈H2(Ω) such that w is solution of (1.1) with source f and Dirichlet datum g=0.
By using the second Green formulas forv∈H-Δ-κ2(Ω), we have
(2.14)0=∫Ωfvdx=∫Ω(-Δ-κ2)wvdx=∫Ωw(-Δ-κ2)vdx=∫Γv∂νxwdσx.
Note that, since the test functions are dense on H-Δ-κ2(Ω), we may consider a variational formulation using the dual system (D(Ω),D(Ω)*). Because v∈L2(Ω), by trace theorem v|∂Ω∈H-1/2(Γ) and
(2.15)∫Γv∂νxwdσx=〈v,∂νxw〉(D(Ω),[D(Ω)]*)=0⟹∂νxw=0on⁡H1/2(Γ).
So, if an arbitrary L2(Ω) function is orthogonal to all functions in H-Δ-κ2(Ω), then it is in (-Δ-κ2)[H02(Ω)] and the reverse inclusion follows.


### 2.2. The Source to Neumann Map

Let us consider the simultaneous solution of the direct and inverse source problem (1.4) and search for a solution (u,f)∈H1(Ω)×L2(Ω). It follows from Theorem 2.5 that if we restrict the source search to the subspace H-Δ+κ2(Ω)⊂L2(Ω), we will find a unique solution.Definition 2.10.
Consider problem (1.1) with zero Dirichlet data, that is, g=0 and f∈H-Δ+κ2(Ω)⊂L2(Ω) and solution w0f∈H2(Ω)∩H01(Ω). One defines the Source to Neumann map for the Helmholtz equation as the operator
(2.16)Λ0:H-Δ+κ2(Ω)⟶H1/2(Γ),Λ0[f](x)=∂w0f∂νx(x)=γ1[w0f](x),x∈Γ.
By the trace theorem [5], it is well defined, linear, and continuous.Theorem 2.11.
Λ0:H-Δ+κ2(Ω)→H1/2(Γ),  is an isomorphism.Remark 2.12.
If we consider problem (1.1) with Dirichlet data g∈H3/2(Γ) and f∈H-Δ+κ2(Ω)⊂L2(Ω) and solution wgf∈H2(Ω)∩H01(Ω), we can define the Source to Neumann map for the Helmholtz equation as the operator
(2.17)Λg:H-Δ+κ2(Ω)⟶H-1/2(Γ),Λg[f](x)=∂wgf∂νx(x)=γ1[wgf](x),x∈Γ.
Note that this more general situation will produce results such as(2.18)Λg[0](x)=Λ0[g](x)=γ1[g]=∂wg0∂νx(x).
As consequence,(i)
a functional such as a Source-Dirichlet to Neumann map may be defined(2.19)Λ[⋅,⋅]:H-Δ+κ2(Ω)×H1/2(Γ)⟶H-1/2(Γ),[f,g]⟼Λ[f,g]=γ1[wgf],(ii)
and restrieted to be a functional such as a Dirichlet to Neumann map(2.20)Λ[0,⋅]:{0}×H1/2(Γ)⟶H-1/2(Γ),g⟼Λ[0,g]≡Λ0[g]=γ1[wg0],(iii)
or to be a functional such as a Source to Neumann map(2.21)Λ[⋅,0]:H-Δ+κ2(Ω)×{0}⟶H-1/2(Γ),f⟶Λ[0,f]≡Λ0[f]=γ1[w0f]. It is important to note that in this more general definition it is not possible to prove that Λ[·,·] is an isomorphism, since the fact that the trace γ0[wgf]=0 is used in the proof.Lemma 2.13.
H-Δ+κ2(Ω) is a Hilbert space with L2(Ω) norm.Proof.
Let us consider the canonical projection(2.22)π2:L2(Ω)=H-Δ+κ2(Ω)⊕H-Δ-κ2⊥(Ω)⟶H-Δ-κ2⊥(Ω)[f,f⊥]⟼f⊥.
Note that π2 is continuous, π2-1[0] is closed. Since
(2.23)π2-1[0]=[f,0]
for all f∈H-Δ+κ2(Ω), it follows that {0}×HΔ+κ2(Ω) is closed. Consequently H-Δ+κ2(Ω) is closed subspace of the Banach space L2(Ω). So, it is Banach and, consequently, is a Hilbert space with the scalar product induced by L2(Ω).Proof of Theorem2.11.
(i)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is continuous.
Note thatΛ0=γ1∘i is a composition of the normal trace γ1:H2(Ω)→H1/2(Γ) and the canonical embedding i:H-Δ+κ2(Ω)→H2(Ω). The normal trace is continuous by trace theorem. The canonical embedding is also continuous since H-Δ+κ2(Ω) is closed by Lemma 2.13. So, Λ0 is continuous.(ii)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is one to one.
Take some arbitraryf∈Ker(Λ0). Then Λ0[f]=γ1[w0]=0 is the normal derivative of the problem (1.1) with g=0. By hypotheses, f∈H-Δ+κ2(Ω) and, consequently, (-Δ+κ2)f=(-Δ+κ2)(-Δ-κ2)w0f=0. The fourth-order problem
(2.24)(Δ2-k4)w0f=0inΩ,γ0w0f=0on⁡Γ,γ1w0f=0on⁡Γ,
is well posed and has a unique w0f=0. Then f=(-Δ-κ2)w0f=0. Since f is arbitrary, Ker(Λ0)={0} and the injectivity is proved.(iii)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is onto.
Consider an arbitrarygνx∈H1/2(Γ) and κ≠0, where gνx does not necessarily satisfy the compatibility condition. Let wgνx∈H-Δ+κ2(Ω) be a solution of the well-posed Neumann data problem,
(2.25)(-Δ+κ2)wgνx=0on⁡Ω,γ1wgνx=gνxinΓ.
Note that with f=wgνx we obtain Λ0[f]=γ1[wgνx]=gνx. So, Λ0 is surjective.
It remains to prove the following.(iv)Λ0-1:H1/2(Γ)→H-Δ+κ2(Ω) is continuous.
In fact, this is a consequence of the Banach open map theorem, since.(2.26)Λ0:H-Δ+κ2(Ω)⟶H1/2(Γ)
is a linear continuous bijective application between Banach spaces.


### 2.3. The Adjoint Source to Neumann Operator

Definition 2.14.
Consider again the problem (1.1) with zero Dirichlet data, that is, g=0 and f∈L2(Ω)and  solution w0f∈H2(Ω)∩H01(Ω).One defines the extension of the Source to Neumann map for the Helmholtz equation as the operator
(2.27)Λ0¯:L2(Ω)⟶H1/2(Γ),Λ0¯[f](x)=∂w0f∂νx(x)=γ1[w0f](x),x∈Γ.Remark 2.15.
By the trace theorem [5], it is well defined, linear, and continuous. As an extension of Λ0, surjectivity is preserved. Also Λ0¯=Λ0∘π1.Corollary 2.16.
The quotient ofL2(Ω) by  H-Δ-κ2⊥(Ω) is a copy of  H1/2(Γ).Proof.
Consider the following chain of embeddings:(2.28)
where c0 is a canonic embedding and c1 is an isomorphism by the Banach isomorphism theorem. Since Λ0 also is an isomorphism by Theorem 2.11, the corollary is proved, that is, L2(Ω)/H-Δ+κ2⊥(Ω)≅H1/2(Γ).Corollary 2.17.
Ker(Λ0¯)=H-Δ-κ2⊥(Ω) is a closed subspace of  L2(Ω).Remark 2.18.
SinceΛ0¯∈ℒ(L2(Ω);H1/2(Γ)) is bounded, then its adjoint operator Λ0¯*∈ℒ(H-1/2(Γ);L2(Ω)) is well defined and continuous and one to one.Corollary 2.19.
Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω)  is an isomorphism.Proof.
(i)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is well defined.
We know that(2.29)H-Δ-k2⊥(Ω)=Ker⁡(Λ0¯)=Rg(Λ0¯*),[⊥Rg(Λ0¯*)]⊥=H-Δ-k2(Ω),
where Rg and Ker denote the operator range and kernel, respectively. It is well known that if X is a Banach space and N, that is, a subspace of its dual X*, then [⊥N]⊥=N¯w* is the weak star closure of N in X*. Applying this result to our case, we have H-Δ-κ2(Ω)=Rg(Λ0¯*)¯w* in L2(Ω). Particularly, Rg(Λ0¯*)⊂H-Δ-κ2(Ω) assures that Λ0¯* is well defined.(ii)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is one to one.
Note that,Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is onto and H-Δ-κ2(Ω) is a closed subspace of L2(Ω),(iii)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is onto.
Note that since,Rg(Λ0¯) is closed in L2(Ω),
(2.30)Ker⁡(Λ0¯)=H-Δ-κ2⊥(Ω)⟹H-Δ-κ2(Ω)=Ker⁡(Λ0¯)⊥⊂Rg(Λ0¯*)=Rg(Λ0¯*)¯w*=H-Δ-κ2(Ω).
Since Λ0¯* is linear, continuous, and bijective from H-1/2(Γ) to H-Δ-κ2(Ω), by the open mapping Banach theorem, Λ0¯* is an isomorphism.Proposition 2.20.
Ifκ1≠κ2,  then H-Δ-κ12(Ω)∩H-Δ-κ22(Ω)={0}.Proof.
Suppose thatv∈H-Δ-κ12(Ω)∩H-Δ-κ22(Ω). Then
(2.31)(-Δ-κ12)v=0,ifv∈H-Δ-κ12(Ω),(-Δ-κ22)v=0,ifv∈H-Δ-κ22(Ω),
and, consequently, v=0.Remark 2.21.
If we substituteκ by iκ in problem (1.1) and use the same argument already used in the precedent proofs, then(i)
L2(Ω)=H-Δ-κ2(Ω)⊕(-Δ+κ2)[H02(Ω)],(ii)
H-Δ+κ2⊥(Ω)=(-Δ+κ2)[H02(Ω)],(iii)
H-Δ+κ2(Ω) is a closed subspace of L2(Ω),(iv)
L2(Ω)=H-Δ-κ2(Ω)⊕H-Δ+κ2⊥(Ω),(v)
ifκ1≠κ2, H-Δ-κ12(Ω)∩H-Δ-κ22(Ω)={0}.


## 2.1. The Dirichlet to Neumann Map

Definition 2.2.
Consider problem (1.1) with zero source, that is, f=0 and g∈H1/2(Γ).This problem has a solution w0∈H1(Ω). One defines the Dirichlet to Neumann map for the Helmholtz equation as the operator
(2.3)Λ0:H1/2(Γ)⟶H-1/2(Γ),Λ0[g](x)=∂w0∂νx(x),x∈Γ.
By the trace theorem [5], it is well defined, linear, and continuous.Remark 2.3.
We alternatively can define the Dirichlet to Neumann map (2.2) for the problem (1.1) as an operator with a nonzero source problem Λf:H1/2(Γ)→H-1/2(Γ) defined by the graph (g,Λfg)∶=(u,∂νu)∣∂Ω.Definition 2.4.
One has(2.4)M∶={(g,gνx)∈H1/2(Γ)×H-1/2(Γ);gνx-Λ0g∈H1/2(Γ)},
where Λ0 is a Dirichlet to Neumann map.Theorem 2.5.
Let(g,gνx)∈H1/2(Γ)×H-1/2(Γ). A function  f∈H-Δ+κ2(Ω)is a solution to the inverse source problem (1.3) if and only if  (g,gνx)∈ℳProof.
Suppose(g,gνx)∈ℳ. Consider the following fourth-order problem:
(2.5)(Δ2-k4)w=0inΩ,γ0w=0on⁡Γ,γ1w=gνx-Λ0gon⁡Γ.
Let us use this problem to define a source f0=(-Δ-k2)w, where w∈H2(Ω) is solution of problem (2.5). Note that f0∈H-Δ+κ2(Ω). Let (u,f)∈H1(Ω)×L2(Ω) be a solution of the inverse source problem (1.4). Then u=w0+w, where w0 is the solution of the homogeneous source problem in the definition of the Dirichlet to Neumann map (2.2) and f=f0. The sufficiency is proved.
To proof necessity, suppose that there exists a(u,f)∈H1(Ω)×H-Δ+κ2(Ω) which is solution of the inverse problem (1.4).
Consider the second-order problem with homogeneous boundary Dirichlet data(2.6)(-Δ-k2)w0=f,x∈Ω,γ0w0=0,x∈Γ,
with a unique solution w0∈H2(Ω)∩H01(Ω). By the trace theorem, we have γ1w0∈H1/2(Γ). Note that gνx=γ1w0+γ1w0, where γ1w0=Λ0g. So, γ1w0=gνx-Λ0g.Remark 2.6.
We have proved the existence and uniqueness of solution to (1.4) in H1(Ω)×H-Δ+κ2(Ω). However, this does not mean that when we do the search in a larger space H1(Ω)×L2(Ω), we will continue having uniqueness. In fact, we will prove in the next proposition that f-f0⊥v∈H-Δ-κ2(Ω) and consequently
(2.7)f∈f0+[H-Δ-κ2(Ω)]⊥,
where f∈L2(Ω) and f0∈H-Δ+κ2(Ω). This leads us to conclude that the solution of (1.4) is actually a class of functions, where we can “reconstruct” or “observe” only part of the solution in H-Δ+κ2(Ω).Proposition 2.7.
Ifκ2∉Σ2 and  κ4∉Σ4, then
(2.8)L2(Ω)=[H-Δ-κ2(Ω)]⊥⊕H-Δ+κ2(Ω).Proof.
Consequence of Lemmas2.8 and 2.9.From now on, we are supposing always that κ2∉Σ2 and κ4∉Σ4.Lemma 2.8.
One hasL2(Ω)=H-Δ+κ2(Ω)⊕(-Δ-κ2)[H02(Ω)].Proof.
For an arbitrarily givenf∈L2(Ω), the problem (1.1) with zero Dirichlet datum g=0 is well posed and has an w0∈H2(Ω) with normal derivative trace Λf=∂νw0∈H1/2(Γ). Also, since κ4∉Σ4, the fourth-order problem (1.5) with Cauchy datum (0,Λf)∈H3/2(Γ)×H1/2(Γ) and zero source has solution v and is well posed in H2(Ω). These two H2(Ω) solutions may be used to define a function
(2.9)w:=w0-v∈H02(Ω),
and since by problem (1.1)
(2.10)f=(-Δ-κ2)w0=(-Δ-κ2)v+(-Δ-κ2)w,
we obtain that arbitrary L2(Ω) function f is the sum of a function in H-Δ-κ2(Ω) and a function in (-Δ-κ2)[H02(Ω)]. If we show that
(2.11)H-Δ+κ2(Ω)∩(-Δ-κ2)[H02(Ω)]={0},
we prove that L2(Ω)=H-Δ+κ2⊕(-Δ-κ2)[H02(Ω)]. For this, take a u in the intersection H-Δ-κ2∩(-Δ+κ2)[H02(Ω)]. Then
(2.12)(-Δ-κ2)u=0,u=(-Δ+κ2)v,for somev∈H02(Ω)
which means that v is a solution of completely homogeneous fourth-order problem (1.5), with no source and zero Cauchy datum. Since κ4∉Σ4, the unique solution is trivial v=0, and u=(-Δ+κ2)v=0. We have proved that L2(Ω)=H-Δ-κ2(Ω)⊕(-Δ+κ2)[H02(Ω)].Lemma 2.9.
One has(-Δ-κ2)[H02(Ω)]=[H-Δ-κ2(Ω)]⊥.Proof.
Using the second Green theorem withf∈H-Δ-κ2(Ω) and v=(-Δ-κ2)w for some w∈H02(Ω), we obtain
(2.13)∫Ωfvdx=∫Ωf(-Δ-κ2)wdx=∫Ωf(-Δw)dx-κ2∫Ωfwdx=∫Ω(-Δf)wdx-κ2∫Ωfwdx=0,
and since the scalar product with arbitrary h∈(-Δ-κ2)[H02(Ω)] is zero, the inclusion (-Δ-κ2)[H02(Ω)]⊂H-Δ-κ2⊥(Ω) follows.
To prove the other inclusion, letf∈L2(Ω), suppose f is orthogonal to v for all v∈H-Δ-κ2(Ω), and take a function w∈H2(Ω) such that w is solution of (1.1) with source f and Dirichlet datum g=0.
By using the second Green formulas forv∈H-Δ-κ2(Ω), we have
(2.14)0=∫Ωfvdx=∫Ω(-Δ-κ2)wvdx=∫Ωw(-Δ-κ2)vdx=∫Γv∂νxwdσx.
Note that, since the test functions are dense on H-Δ-κ2(Ω), we may consider a variational formulation using the dual system (D(Ω),D(Ω)*). Because v∈L2(Ω), by trace theorem v|∂Ω∈H-1/2(Γ) and
(2.15)∫Γv∂νxwdσx=〈v,∂νxw〉(D(Ω),[D(Ω)]*)=0⟹∂νxw=0on⁡H1/2(Γ).
So, if an arbitrary L2(Ω) function is orthogonal to all functions in H-Δ-κ2(Ω), then it is in (-Δ-κ2)[H02(Ω)] and the reverse inclusion follows.


## 2.2. The Source to Neumann Map

Let us consider the simultaneous solution of the direct and inverse source problem (1.4) and search for a solution (u,f)∈H1(Ω)×L2(Ω). It follows from Theorem 2.5 that if we restrict the source search to the subspace H-Δ+κ2(Ω)⊂L2(Ω), we will find a unique solution.Definition 2.10.
Consider problem (1.1) with zero Dirichlet data, that is, g=0 and f∈H-Δ+κ2(Ω)⊂L2(Ω) and solution w0f∈H2(Ω)∩H01(Ω). One defines the Source to Neumann map for the Helmholtz equation as the operator
(2.16)Λ0:H-Δ+κ2(Ω)⟶H1/2(Γ),Λ0[f](x)=∂w0f∂νx(x)=γ1[w0f](x),x∈Γ.
By the trace theorem [5], it is well defined, linear, and continuous.Theorem 2.11.
Λ0:H-Δ+κ2(Ω)→H1/2(Γ),  is an isomorphism.Remark 2.12.
If we consider problem (1.1) with Dirichlet data g∈H3/2(Γ) and f∈H-Δ+κ2(Ω)⊂L2(Ω) and solution wgf∈H2(Ω)∩H01(Ω), we can define the Source to Neumann map for the Helmholtz equation as the operator
(2.17)Λg:H-Δ+κ2(Ω)⟶H-1/2(Γ),Λg[f](x)=∂wgf∂νx(x)=γ1[wgf](x),x∈Γ.
Note that this more general situation will produce results such as(2.18)Λg[0](x)=Λ0[g](x)=γ1[g]=∂wg0∂νx(x).
As consequence,(i)
a functional such as a Source-Dirichlet to Neumann map may be defined(2.19)Λ[⋅,⋅]:H-Δ+κ2(Ω)×H1/2(Γ)⟶H-1/2(Γ),[f,g]⟼Λ[f,g]=γ1[wgf],(ii)
and restrieted to be a functional such as a Dirichlet to Neumann map(2.20)Λ[0,⋅]:{0}×H1/2(Γ)⟶H-1/2(Γ),g⟼Λ[0,g]≡Λ0[g]=γ1[wg0],(iii)
or to be a functional such as a Source to Neumann map(2.21)Λ[⋅,0]:H-Δ+κ2(Ω)×{0}⟶H-1/2(Γ),f⟶Λ[0,f]≡Λ0[f]=γ1[w0f]. It is important to note that in this more general definition it is not possible to prove that Λ[·,·] is an isomorphism, since the fact that the trace γ0[wgf]=0 is used in the proof.Lemma 2.13.
H-Δ+κ2(Ω) is a Hilbert space with L2(Ω) norm.Proof.
Let us consider the canonical projection(2.22)π2:L2(Ω)=H-Δ+κ2(Ω)⊕H-Δ-κ2⊥(Ω)⟶H-Δ-κ2⊥(Ω)[f,f⊥]⟼f⊥.
Note that π2 is continuous, π2-1[0] is closed. Since
(2.23)π2-1[0]=[f,0]
for all f∈H-Δ+κ2(Ω), it follows that {0}×HΔ+κ2(Ω) is closed. Consequently H-Δ+κ2(Ω) is closed subspace of the Banach space L2(Ω). So, it is Banach and, consequently, is a Hilbert space with the scalar product induced by L2(Ω).Proof of Theorem2.11.
(i)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is continuous.
Note thatΛ0=γ1∘i is a composition of the normal trace γ1:H2(Ω)→H1/2(Γ) and the canonical embedding i:H-Δ+κ2(Ω)→H2(Ω). The normal trace is continuous by trace theorem. The canonical embedding is also continuous since H-Δ+κ2(Ω) is closed by Lemma 2.13. So, Λ0 is continuous.(ii)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is one to one.
Take some arbitraryf∈Ker(Λ0). Then Λ0[f]=γ1[w0]=0 is the normal derivative of the problem (1.1) with g=0. By hypotheses, f∈H-Δ+κ2(Ω) and, consequently, (-Δ+κ2)f=(-Δ+κ2)(-Δ-κ2)w0f=0. The fourth-order problem
(2.24)(Δ2-k4)w0f=0inΩ,γ0w0f=0on⁡Γ,γ1w0f=0on⁡Γ,
is well posed and has a unique w0f=0. Then f=(-Δ-κ2)w0f=0. Since f is arbitrary, Ker(Λ0)={0} and the injectivity is proved.(iii)Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is onto.
Consider an arbitrarygνx∈H1/2(Γ) and κ≠0, where gνx does not necessarily satisfy the compatibility condition. Let wgνx∈H-Δ+κ2(Ω) be a solution of the well-posed Neumann data problem,
(2.25)(-Δ+κ2)wgνx=0on⁡Ω,γ1wgνx=gνxinΓ.
Note that with f=wgνx we obtain Λ0[f]=γ1[wgνx]=gνx. So, Λ0 is surjective.
It remains to prove the following.(iv)Λ0-1:H1/2(Γ)→H-Δ+κ2(Ω) is continuous.
In fact, this is a consequence of the Banach open map theorem, since.(2.26)Λ0:H-Δ+κ2(Ω)⟶H1/2(Γ)
is a linear continuous bijective application between Banach spaces.


## 2.3. The Adjoint Source to Neumann Operator

Definition 2.14.
Consider again the problem (1.1) with zero Dirichlet data, that is, g=0 and f∈L2(Ω)and  solution w0f∈H2(Ω)∩H01(Ω).One defines the extension of the Source to Neumann map for the Helmholtz equation as the operator
(2.27)Λ0¯:L2(Ω)⟶H1/2(Γ),Λ0¯[f](x)=∂w0f∂νx(x)=γ1[w0f](x),x∈Γ.Remark 2.15.
By the trace theorem [5], it is well defined, linear, and continuous. As an extension of Λ0, surjectivity is preserved. Also Λ0¯=Λ0∘π1.Corollary 2.16.
The quotient ofL2(Ω) by  H-Δ-κ2⊥(Ω) is a copy of  H1/2(Γ).Proof.
Consider the following chain of embeddings:(2.28)
where c0 is a canonic embedding and c1 is an isomorphism by the Banach isomorphism theorem. Since Λ0 also is an isomorphism by Theorem 2.11, the corollary is proved, that is, L2(Ω)/H-Δ+κ2⊥(Ω)≅H1/2(Γ).Corollary 2.17.
Ker(Λ0¯)=H-Δ-κ2⊥(Ω) is a closed subspace of  L2(Ω).Remark 2.18.
SinceΛ0¯∈ℒ(L2(Ω);H1/2(Γ)) is bounded, then its adjoint operator Λ0¯*∈ℒ(H-1/2(Γ);L2(Ω)) is well defined and continuous and one to one.Corollary 2.19.
Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω)  is an isomorphism.Proof.
(i)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is well defined.
We know that(2.29)H-Δ-k2⊥(Ω)=Ker⁡(Λ0¯)=Rg(Λ0¯*),[⊥Rg(Λ0¯*)]⊥=H-Δ-k2(Ω),
where Rg and Ker denote the operator range and kernel, respectively. It is well known that if X is a Banach space and N, that is, a subspace of its dual X*, then [⊥N]⊥=N¯w* is the weak star closure of N in X*. Applying this result to our case, we have H-Δ-κ2(Ω)=Rg(Λ0¯*)¯w* in L2(Ω). Particularly, Rg(Λ0¯*)⊂H-Δ-κ2(Ω) assures that Λ0¯* is well defined.(ii)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is one to one.
Note that,Λ0:H-Δ+κ2(Ω)→H1/2(Γ) is onto and H-Δ-κ2(Ω) is a closed subspace of L2(Ω),(iii)Λ0¯*:H-1/2(Γ)→H-Δ-κ2(Ω) is onto.
Note that since,Rg(Λ0¯) is closed in L2(Ω),
(2.30)Ker⁡(Λ0¯)=H-Δ-κ2⊥(Ω)⟹H-Δ-κ2(Ω)=Ker⁡(Λ0¯)⊥⊂Rg(Λ0¯*)=Rg(Λ0¯*)¯w*=H-Δ-κ2(Ω).
Since Λ0¯* is linear, continuous, and bijective from H-1/2(Γ) to H-Δ-κ2(Ω), by the open mapping Banach theorem, Λ0¯* is an isomorphism.Proposition 2.20.
Ifκ1≠κ2,  then H-Δ-κ12(Ω)∩H-Δ-κ22(Ω)={0}.Proof.
Suppose thatv∈H-Δ-κ12(Ω)∩H-Δ-κ22(Ω). Then
(2.31)(-Δ-κ12)v=0,ifv∈H-Δ-κ12(Ω),(-Δ-κ22)v=0,ifv∈H-Δ-κ22(Ω),
and, consequently, v=0.Remark 2.21.
If we substituteκ by iκ in problem (1.1) and use the same argument already used in the precedent proofs, then(i)
L2(Ω)=H-Δ-κ2(Ω)⊕(-Δ+κ2)[H02(Ω)],(ii)
H-Δ+κ2⊥(Ω)=(-Δ+κ2)[H02(Ω)],(iii)
H-Δ+κ2(Ω) is a closed subspace of L2(Ω),(iv)
L2(Ω)=H-Δ-κ2(Ω)⊕H-Δ+κ2⊥(Ω),(v)
ifκ1≠κ2, H-Δ-κ12(Ω)∩H-Δ-κ22(Ω)={0}.


## 3. Integral Representation

Definition 3.1.
The Dirichlet Green functionG(x,ζ) for the problem (1.1) is its solution with source δ(x-ζ),x,ζ∈Ω,and homogeneous Dirichlet data, that is, G(x,ζ)=0 for x on Γ.Definition 3.2.
LetG(x,ζ) be the Green function for problem (1.1) with homogeneous Dirichlet boundary conditions. Then, the solution
(3.1)S:L2(Ω)×H1/2(Ω)⟶H1(Ω),(f,g)⟼u=S[f,g],S[f,g](x)=∫Ωf(ζ)G(x,ζ)dζ+∫Γg(ζ)∂G(x,ζ)∂νζdσζforx∈Ω.Remark 3.3.
By using problem (1.1) linearity, we formally decompose the solution in two additive parts
(3.2)u=S[f,g]∶=S[f,0]+S[0,g]:L2(Ω)×H1/2(Γ)⟶H1(Ω),
where S[f,0]:L2(Ω)×{0}→H1(Ω) is the homogeneous Dirichlet source auxiliary problem solution and S[0,g]:{0}×H1/2(Γ)→H1(Ω) is the zero source auxiliary Dirichlet problem solution. For simplicity, for a fixed f∈L2(Ω) or g∈H1/2(Γ), we will denote
(3.3)Sf=S[f,⋅]:H1/2(Γ)⟶H1(Ω),Sg=S[⋅,g]:L2(Ω)⟶H1(Ω).
By taking the normal trace of the solution (3.1), we obtain
(3.4)Λ[f,g](x)=γ1∘S[f,g]=∫Ωf(ζ)∂G(x,ζ)∂νxdζ+∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσζforx∈Γ
which is an explicit representation to the Dirichlet to Newman map with arbitrary f and g.
By using the same notation adopted for the additive decomposition of the solution map, fixedf∈L2(Ω) or g∈H1/2(Γ), we will denote
(3.5)Λf=Λ[f,⋅]:H1/2(Γ)⟶H1(Ω),Λg=Λ[⋅,g]:L2(Ω)⟶H1(Ω).Remark 3.4.
Note that(3.6)S0[f](x)=S[f,0](x)=∫Ωf(ζ)G(x,ζ)dζ,x∈Ω,S0[g](x)=S[0,g](x)=∫Γg(ζ)∂G(x,ζ)∂νζdσ(ζ),x∈Ω.
With this decomposition, we obtain the following explicit representation to operators in Section2:(1)
(3.7)Λ[f,g]=γ1∘S[f,g](x)=∂S[f,g]∂νx(x)=∫Ωf(ζ)∂G(x,ζ)∂νxdζ+∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ),
with x∈Γ;(2)
(3.8)Λ0[g](x)=Λ[0,g](x)=γ1∘S[0,g](x)=∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ),x∈Γ,
is an explicit representation to the Dirichlet to Neumann map;(3)
(3.9)Λ0[f]=(x)Λ[f,0](x)=γ1∘S[f,0](x)=∫Ωf(ζ)∂G(x,ζ)∂νxdζ,x∈Γ,
is an explicit representation to the Source to Neumann map.


### 3.1. The Inverse Source Integral Equation

Lemma 3.5.
Letuj, j=1,2, be two solutions of problem (1.1) with the same source f and different Dirichlet data gj, j=1,2, respectively. Then(i)
Λf[g1]-Λ0[g1]=Λf[g2]-Λ0[g2]onΓ, that is, the relative Dirichlet to Newman operator Λf-Λ0:H1/2(Γ)→H1/2(Γ) is constant operator whose functional value is independent of the Dirichlet datum g and depends only on the source function f;(ii)
∫Ωf(ζ)(∂G(x,ζ)/∂νx)dζ=Λf[gj]-Λ0[gj] for all solutions of (1.1) with arbitrary Dirichlet data but the same source, that is, the integral is the function given by the relative Dirichlet to Newman map.Proof.
The equality∫Ωf(ζ)(∂G(x,ζ)/∂νx)dζ=Λf[g1]-Λ0[g1]=Λf[g2]-Λ0[g2] in both (i) and (ii) is a trivial consequence of (3.4). Note that in this case the unique information available for source reconstruction is given by only one measurement, say that Neumann boundary measurement
(3.10)∂νxu=Λf[g],
corresponding to some specific Dirichlet datum g, which without loss of generality can be assumed zero. Note also that problem (1.1) with Dirichlet datum g=0 and source f∈L2(Ω) has solution u∈H2(Ω). The normal trace of this regular solution is in H1/2(Γ). So we have proved that the range of Λf-Λ0 is in H1/2(Γ). The domain of Λf-Λ0 is H1/2(Ω) since this is the set of nonzero Dirichlet data that gives the same function in the range.Definition 3.6 (strong integral equation problem).
Since in the inverse source problem the exact Cauchy data pair(g,gνx) is given, the relative Dirichlet to Newman map value for the source problem (1.1)
(3.11)Λ=(Λf-Λ0)[g]=gνx(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσζforx∈Γ,
is known and Lemma 3.5 suggests the following integral equation formulation for the source reconstruction problem: to find f∈L2(Ω) such that
(3.12)F[f]=Λon⁡Γ,
where F∶=Λ0¯=Λ[·,0]:L2(Ω)→H1/2(Γ),
(3.13)F[f](x)∶=∫Ωf(ζ)∂G(x,ζ)∂νxdζ,
for x∈Γ.Remark 3.7.
Note that we introduce hereF as a simplified notation to the extended Source to Neumann map Λ0¯. This notation is more usual.The following corollary resumes all that has been discussed.Corollary 3.8.
Supposed thatκ2∉Σ2 and  κ4∉Σ4. Then, (i)
for a Cauchy datum(v,∂νv)∣Γ=(0,Λ)∈H3/2(Γ)×H1/2(Γ), there exists a unique function f∈H-Δ+κ2(Ω) solution of the inverse source problem (1.3) for the Helmholtz equation (1.1),(ii)
the associated mapping(3.14)F-1:H1/2(Γ)⟶H-Δ+κ2(Ω)
defines a linear homeomorphism between these spaces,(iii)
and is a right inverse of the mappingF:L2(Ω)→H1/2(Ω) defined by the strong inverse source equation (3.12),(iv)
the projectionQ:L2(Ω)→L2(Ω)∖H-Δ+κ2(Ω) is well defined and constant in the level set
(3.15)Ch={f∈L2:(-Δ+κ2)f=h},(v)
if the sourcef is known to be in the class Ch, then a single boundary measurement (0,Λ) is sufficient to identify f,(vi)
for a Cauchy datum(v,∂νv)∣Γ=(0,Λ)∈H3/2(Γ)×H1/2(Γ), there are many functions f¯=f+h, where f∈H-Δ+κ2(Ω) is an observed consequence of (ii) and h∈(-Δ+κ2)[H02(Ω)]=L2(Ω)∖H-Δ+κ2(Ω) is an arbitrary nonobserved function.Proof.
The items are trivial consequences of the results already proved.Remark 3.9.
Givenh∈Ch, the unique solution referred to in Section 3.1 is the unique solution of the fourth-order direct problem [1]:
(3.16)(Δ2-κ4)wf=hinΩ,wf=0on⁡Γ,∂νwf=0on⁡Γ.Remark 3.10.
The adopted Hilbert space framework for solution of the problem may be understood as an a priori information about the criteria for selecting the observable and the nonobservable part of the source. Other Sobolev spaces that induced partitions of the pivot spaceLp(Ω),1≤p≤∞ (in this work p=2) will modify this observability relation.Remark 3.11 (relation between star-shaped and metaharmonic functions).
Let us define the set(3.17)UType={χω:ω⊂ΩisoftypeType}.
The Usquares⊂Ustar  shape⊂Ucharacteristic. Note that Usquares is dense in L2(Ω). If, for all ω⊂Ω, there exists a family of metaharmonic functions in H-Δ+κ2(Ω) that approach χω, then H-Δ+κ2(Ω) is dense in L2(Ω).Remark 3.12.
The most important classes of sources that may be reconstructed uniquely from boundary data occur whenf is metaharmonic or when f≡fχω, where χω is the characteristic function of an open star-shaped set ω⊂Ω with C2 being boundary and f a C2(Ω¯) function. We will discuss these classes when establishing uniqueness.Remark 3.13 (the adjoint integral equation).
This equation may be used for the source reconstruction independent of solution of the direct problem (1.1). By substituting the explicit integral definition of F in the duality definition of adjoint
(3.18)〈F[f],ψ〉H1/2×H-1/2=〈f,F*[ψ]〉L2×L2,
we obtain that F*:H-1/2(Γ)→L2(Ω) is explicitly given by
(3.19)h(ζ)=F*[ψ](ζ)=∫Γψ(x)∂G(x,ζ)∂νxdσx.Remark 3.14.
From these formulas, we can deduce that, for a fixedf∈L2(Ω), the operator
(3.20)Λgf(x)-Λg0(x)=F[f](x),x∈Γ,
for any g∈H2(Γ), we will call this operator Extended Dirichlet to Neumann map.Remark 3.15.
Once we know the integral formulation toF, we can determine the integral formulation to F*. In fact,
(3.21)〈F[f],ψ〉H1/2×H-1/2=〈f,F*[ψ]〉L2×L2,
and from
(3.22)F[f](x)=∫Ω(ζ)∂G(x,ζ)∂νxdζ,x∈Γ,
it follows that
(3.23)F*[ψ](ζ)=∫Ωψ(x)∂G(x,ζ)∂νxdσ(x),x∈Γ.Remark 3.16.
Consider the following direct problem(3.24)-Δw+k2w=0inΩ,γ0w=ψon⁡Γ,
with ψ∈H-1/2(Γ). This problem has a unique solution w∈L2(Ω). Let
(3.25)w(ζ)=S1,ψ(ζ)=∫Ωψ(x)∂G(x,ζ)∂νxdσ(x)=F*[ψ](ζ),
where G is the associated Green function. This happens for each ψ∈H-1/2(Γ) since F*[ψ](ζ)=S1,ψ(ζ). From this we deduce that the integral S1,ψ inherits all good properties from F*.


## 3.1. The Inverse Source Integral Equation

Lemma 3.5.
Letuj, j=1,2, be two solutions of problem (1.1) with the same source f and different Dirichlet data gj, j=1,2, respectively. Then(i)
Λf[g1]-Λ0[g1]=Λf[g2]-Λ0[g2]onΓ, that is, the relative Dirichlet to Newman operator Λf-Λ0:H1/2(Γ)→H1/2(Γ) is constant operator whose functional value is independent of the Dirichlet datum g and depends only on the source function f;(ii)
∫Ωf(ζ)(∂G(x,ζ)/∂νx)dζ=Λf[gj]-Λ0[gj] for all solutions of (1.1) with arbitrary Dirichlet data but the same source, that is, the integral is the function given by the relative Dirichlet to Newman map.Proof.
The equality∫Ωf(ζ)(∂G(x,ζ)/∂νx)dζ=Λf[g1]-Λ0[g1]=Λf[g2]-Λ0[g2] in both (i) and (ii) is a trivial consequence of (3.4). Note that in this case the unique information available for source reconstruction is given by only one measurement, say that Neumann boundary measurement
(3.10)∂νxu=Λf[g],
corresponding to some specific Dirichlet datum g, which without loss of generality can be assumed zero. Note also that problem (1.1) with Dirichlet datum g=0 and source f∈L2(Ω) has solution u∈H2(Ω). The normal trace of this regular solution is in H1/2(Γ). So we have proved that the range of Λf-Λ0 is in H1/2(Γ). The domain of Λf-Λ0 is H1/2(Ω) since this is the set of nonzero Dirichlet data that gives the same function in the range.Definition 3.6 (strong integral equation problem).
Since in the inverse source problem the exact Cauchy data pair(g,gνx) is given, the relative Dirichlet to Newman map value for the source problem (1.1)
(3.11)Λ=(Λf-Λ0)[g]=gνx(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσζforx∈Γ,
is known and Lemma 3.5 suggests the following integral equation formulation for the source reconstruction problem: to find f∈L2(Ω) such that
(3.12)F[f]=Λon⁡Γ,
where F∶=Λ0¯=Λ[·,0]:L2(Ω)→H1/2(Γ),
(3.13)F[f](x)∶=∫Ωf(ζ)∂G(x,ζ)∂νxdζ,
for x∈Γ.Remark 3.7.
Note that we introduce hereF as a simplified notation to the extended Source to Neumann map Λ0¯. This notation is more usual.The following corollary resumes all that has been discussed.Corollary 3.8.
Supposed thatκ2∉Σ2 and  κ4∉Σ4. Then, (i)
for a Cauchy datum(v,∂νv)∣Γ=(0,Λ)∈H3/2(Γ)×H1/2(Γ), there exists a unique function f∈H-Δ+κ2(Ω) solution of the inverse source problem (1.3) for the Helmholtz equation (1.1),(ii)
the associated mapping(3.14)F-1:H1/2(Γ)⟶H-Δ+κ2(Ω)
defines a linear homeomorphism between these spaces,(iii)
and is a right inverse of the mappingF:L2(Ω)→H1/2(Ω) defined by the strong inverse source equation (3.12),(iv)
the projectionQ:L2(Ω)→L2(Ω)∖H-Δ+κ2(Ω) is well defined and constant in the level set
(3.15)Ch={f∈L2:(-Δ+κ2)f=h},(v)
if the sourcef is known to be in the class Ch, then a single boundary measurement (0,Λ) is sufficient to identify f,(vi)
for a Cauchy datum(v,∂νv)∣Γ=(0,Λ)∈H3/2(Γ)×H1/2(Γ), there are many functions f¯=f+h, where f∈H-Δ+κ2(Ω) is an observed consequence of (ii) and h∈(-Δ+κ2)[H02(Ω)]=L2(Ω)∖H-Δ+κ2(Ω) is an arbitrary nonobserved function.Proof.
The items are trivial consequences of the results already proved.Remark 3.9.
Givenh∈Ch, the unique solution referred to in Section 3.1 is the unique solution of the fourth-order direct problem [1]:
(3.16)(Δ2-κ4)wf=hinΩ,wf=0on⁡Γ,∂νwf=0on⁡Γ.Remark 3.10.
The adopted Hilbert space framework for solution of the problem may be understood as an a priori information about the criteria for selecting the observable and the nonobservable part of the source. Other Sobolev spaces that induced partitions of the pivot spaceLp(Ω),1≤p≤∞ (in this work p=2) will modify this observability relation.Remark 3.11 (relation between star-shaped and metaharmonic functions).
Let us define the set(3.17)UType={χω:ω⊂ΩisoftypeType}.
The Usquares⊂Ustar  shape⊂Ucharacteristic. Note that Usquares is dense in L2(Ω). If, for all ω⊂Ω, there exists a family of metaharmonic functions in H-Δ+κ2(Ω) that approach χω, then H-Δ+κ2(Ω) is dense in L2(Ω).Remark 3.12.
The most important classes of sources that may be reconstructed uniquely from boundary data occur whenf is metaharmonic or when f≡fχω, where χω is the characteristic function of an open star-shaped set ω⊂Ω with C2 being boundary and f a C2(Ω¯) function. We will discuss these classes when establishing uniqueness.Remark 3.13 (the adjoint integral equation).
This equation may be used for the source reconstruction independent of solution of the direct problem (1.1). By substituting the explicit integral definition of F in the duality definition of adjoint
(3.18)〈F[f],ψ〉H1/2×H-1/2=〈f,F*[ψ]〉L2×L2,
we obtain that F*:H-1/2(Γ)→L2(Ω) is explicitly given by
(3.19)h(ζ)=F*[ψ](ζ)=∫Γψ(x)∂G(x,ζ)∂νxdσx.Remark 3.14.
From these formulas, we can deduce that, for a fixedf∈L2(Ω), the operator
(3.20)Λgf(x)-Λg0(x)=F[f](x),x∈Γ,
for any g∈H2(Γ), we will call this operator Extended Dirichlet to Neumann map.Remark 3.15.
Once we know the integral formulation toF, we can determine the integral formulation to F*. In fact,
(3.21)〈F[f],ψ〉H1/2×H-1/2=〈f,F*[ψ]〉L2×L2,
and from
(3.22)F[f](x)=∫Ω(ζ)∂G(x,ζ)∂νxdζ,x∈Γ,
it follows that
(3.23)F*[ψ](ζ)=∫Ωψ(x)∂G(x,ζ)∂νxdσ(x),x∈Γ.Remark 3.16.
Consider the following direct problem(3.24)-Δw+k2w=0inΩ,γ0w=ψon⁡Γ,
with ψ∈H-1/2(Γ). This problem has a unique solution w∈L2(Ω). Let
(3.25)w(ζ)=S1,ψ(ζ)=∫Ωψ(x)∂G(x,ζ)∂νxdσ(x)=F*[ψ](ζ),
where G is the associated Green function. This happens for each ψ∈H-1/2(Γ) since F*[ψ](ζ)=S1,ψ(ζ). From this we deduce that the integral S1,ψ inherits all good properties from F*.


## 4. Integral and Variational Solutions: The Equivalence Theorem

### 4.1. Integral Formulation

With the integral formulas (3.8), (3.2), Definition (2.2), and supposing that the compatibility condition has been verified, we obtain that
(4.1)F[f](x)=Λgf(x)-Λ0[g](x)
is the integral equation
(4.2)∫Ωf(ζ)∂G(x,ζ)∂νxdζ=gνx(x)-∫Γg(ζ)∂G(x,ζ)∂η(x)∂η(ζ)dσ(ζ).


### 4.2. Variational Formulation

Definition 4.1 (reciprocity gap functional problem).
One may use the second Green theorem(4.3)∫Ω(uΔv-vΔu)dx=∫Γ(u∂v∂νx-v∂u∂νx)dσx,
valid for all u,v∈H2(Ω) with u a solution of problem (1.1) to formulate the reciprocity gap functional inverse problem: to find f∈L2(Ω)(4.4)∫Ωfvdx=∫Γ(gνxv-g∂v∂νx)dσx,
for all v∈H-Δ-κ2(Ω).
Note that the Lax-Milgram theorem assures the existence of a solution in this case.


### 4.3. The Equivalence Theorem

Theorem 4.2.
Let one consider the two inverse source problems related with problem (1.1) with relative Dirichlet to Newman map  Λ∈L2(Ω): (i)
integral equation problem given by (3.12);(ii)
reciprocity gap functional problem given by (4.4). Then (i)⇒(ii). Suppose additionally that the relative Dirichlet to Newman data Λ∈H1/2(Γ). Then (ii)⇒(i).Proof.
Let us consider the inverse source problem: to find(u,f)∈H1(Ω)×L2(Ω) such that
(4.5)-Δu-k2u=finΩ,γ0u=gon⁡Γ,γ1u=gνxon⁡Γ,
where g∈H1/2(Γ), gνx∈H-1/2(Γ) with compatibility condition gν=∂g/∂ν.
(i)⇒ (ii).
We start the demonstration by supposing that (i) is true; that is, there existsf∈L2(Ω) such that
(4.6)∫Ωf(ζ)∂G(x,ζ)∂νxdζ=gνx-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ),
where G is the Green function associated with the Helmholtz operator in Ω.
Letv∈HΔ-κ2(Ω) be extended to the boundary of Ω. We then have the following integral representation for v:
(4.7)v(ζ)=∫Γv(x)∂G(x,ζ)∂νx(ζ)dσ(x),ζ∈Γ.
By taking the normal trace
(4.8)∂v∂νζ(ζ)=∫Γv(x)∂2G(x,ζ)∂νx∂νζdσ(x),ζ∈Γ.
We now multiply (4.2) by v and integrate on Γ to obtain
(4.9)∫Γv(x)[∫Ωf(ζ)∂G∂νx(x,ζ)dζ]dσ(x)=∫Γv(x)[φ(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x)=∫Γv(x)φ(x)dσ(x)-∫Γv(x)[∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x).
Now applying the Fubini theorem, we obtain(4.10)∫Ωf(ζ)[∫Γv(x)∂G(x,ζ)∂νxdσ(x)]︷v(x)dζ=∫Γv(x)φ(x)︷∂g(x)/∂νxdσ(x)-∫Γg(ζ)[∫Γv(x)∂2G(x,ζ)∂νx∂η(ζ)dσ(x)]︷∂v(ζ)/∂νζdσ(ζ),
which implies
(4.11)∫Ωf(ζ)v(ζ)dζ=∫Γ(v(ζ)∂g(ζ)∂η(ζ)-g(ζ)∂v(ζ)∂η(ζ))dσ(ζ).
Since v∈H-Δ+κ2(Ω) is arbitrary, we obtain the weak formulation. Note that this is almost expected, since the integral formulation is stronger than the variational, and, as usual, strong ⇒ weak.
(ii)⇒ (i).
Let us now suppose that for all test functionsv∈H-Δ+κ2(Ω) we have the weak reciprocity integral equation (4.4)
(4.12)∫Ωf(x)v(x)dx=∫Γgν(x)v(x)dσ(x)-∫Γg(x)∂v(x)∂νdσ(x).
Consider the substitution of the Green function integral representation
(4.13)v(ζ)=∫Γv(x)∂G(x,ζ)∂νxdσ(x),
whose normal derivative is
(4.14)∂v(ζ)∂νζ=∫Γv(x)∂2G(x,ζ)∂νζ∂νxdσ(x),
in (4.4) and apply Fubini’s theorem to obtain
(4.15)∫Γv(x)∫Ωf(ζ)∂G(x,ζ)∂νxdζdσ(x)=∫Γv(x)[∫Γgν(ζ)∂G(x,ζ)∂νxdσ(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x).
Note that since(i)
(4.16)∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)=Λ0[g],
is the Dirichlet to Neumann map,(ii)
and(4.17)∫Γgν(ζ)∂G(x,ζ)∂νxdσ(x)=gν(ζ),
we obtain
(4.18)∫Γv(x)[∫Ωf(ζ)∂G(x,ζ)νxdζ-gν+Λ0[g]]dσ(x)=0,
for all v∈H1/2(Γ)≅H-Δ+κ2(Ω). By property of the integral, we obtain (3.12).


## 4.1. Integral Formulation

With the integral formulas (3.8), (3.2), Definition (2.2), and supposing that the compatibility condition has been verified, we obtain that
(4.1)F[f](x)=Λgf(x)-Λ0[g](x)
is the integral equation
(4.2)∫Ωf(ζ)∂G(x,ζ)∂νxdζ=gνx(x)-∫Γg(ζ)∂G(x,ζ)∂η(x)∂η(ζ)dσ(ζ).


## 4.2. Variational Formulation

Definition 4.1 (reciprocity gap functional problem).
One may use the second Green theorem(4.3)∫Ω(uΔv-vΔu)dx=∫Γ(u∂v∂νx-v∂u∂νx)dσx,
valid for all u,v∈H2(Ω) with u a solution of problem (1.1) to formulate the reciprocity gap functional inverse problem: to find f∈L2(Ω)(4.4)∫Ωfvdx=∫Γ(gνxv-g∂v∂νx)dσx,
for all v∈H-Δ-κ2(Ω).
Note that the Lax-Milgram theorem assures the existence of a solution in this case.


## 4.3. The Equivalence Theorem

Theorem 4.2.
Let one consider the two inverse source problems related with problem (1.1) with relative Dirichlet to Newman map  Λ∈L2(Ω): (i)
integral equation problem given by (3.12);(ii)
reciprocity gap functional problem given by (4.4). Then (i)⇒(ii). Suppose additionally that the relative Dirichlet to Newman data Λ∈H1/2(Γ). Then (ii)⇒(i).Proof.
Let us consider the inverse source problem: to find(u,f)∈H1(Ω)×L2(Ω) such that
(4.5)-Δu-k2u=finΩ,γ0u=gon⁡Γ,γ1u=gνxon⁡Γ,
where g∈H1/2(Γ), gνx∈H-1/2(Γ) with compatibility condition gν=∂g/∂ν.
(i)⇒ (ii).
We start the demonstration by supposing that (i) is true; that is, there existsf∈L2(Ω) such that
(4.6)∫Ωf(ζ)∂G(x,ζ)∂νxdζ=gνx-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ),
where G is the Green function associated with the Helmholtz operator in Ω.
Letv∈HΔ-κ2(Ω) be extended to the boundary of Ω. We then have the following integral representation for v:
(4.7)v(ζ)=∫Γv(x)∂G(x,ζ)∂νx(ζ)dσ(x),ζ∈Γ.
By taking the normal trace
(4.8)∂v∂νζ(ζ)=∫Γv(x)∂2G(x,ζ)∂νx∂νζdσ(x),ζ∈Γ.
We now multiply (4.2) by v and integrate on Γ to obtain
(4.9)∫Γv(x)[∫Ωf(ζ)∂G∂νx(x,ζ)dζ]dσ(x)=∫Γv(x)[φ(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x)=∫Γv(x)φ(x)dσ(x)-∫Γv(x)[∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x).
Now applying the Fubini theorem, we obtain(4.10)∫Ωf(ζ)[∫Γv(x)∂G(x,ζ)∂νxdσ(x)]︷v(x)dζ=∫Γv(x)φ(x)︷∂g(x)/∂νxdσ(x)-∫Γg(ζ)[∫Γv(x)∂2G(x,ζ)∂νx∂η(ζ)dσ(x)]︷∂v(ζ)/∂νζdσ(ζ),
which implies
(4.11)∫Ωf(ζ)v(ζ)dζ=∫Γ(v(ζ)∂g(ζ)∂η(ζ)-g(ζ)∂v(ζ)∂η(ζ))dσ(ζ).
Since v∈H-Δ+κ2(Ω) is arbitrary, we obtain the weak formulation. Note that this is almost expected, since the integral formulation is stronger than the variational, and, as usual, strong ⇒ weak.
(ii)⇒ (i).
Let us now suppose that for all test functionsv∈H-Δ+κ2(Ω) we have the weak reciprocity integral equation (4.4)
(4.12)∫Ωf(x)v(x)dx=∫Γgν(x)v(x)dσ(x)-∫Γg(x)∂v(x)∂νdσ(x).
Consider the substitution of the Green function integral representation
(4.13)v(ζ)=∫Γv(x)∂G(x,ζ)∂νxdσ(x),
whose normal derivative is
(4.14)∂v(ζ)∂νζ=∫Γv(x)∂2G(x,ζ)∂νζ∂νxdσ(x),
in (4.4) and apply Fubini’s theorem to obtain
(4.15)∫Γv(x)∫Ωf(ζ)∂G(x,ζ)∂νxdζdσ(x)=∫Γv(x)[∫Γgν(ζ)∂G(x,ζ)∂νxdσ(x)-∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)]dσ(x).
Note that since(i)
(4.16)∫Γg(ζ)∂2G(x,ζ)∂νx∂νζdσ(ζ)=Λ0[g],
is the Dirichlet to Neumann map,(ii)
and(4.17)∫Γgν(ζ)∂G(x,ζ)∂νxdσ(x)=gν(ζ),
we obtain
(4.18)∫Γv(x)[∫Ωf(ζ)∂G(x,ζ)νxdζ-gν+Λ0[g]]dσ(x)=0,
for all v∈H1/2(Γ)≅H-Δ+κ2(Ω). By property of the integral, we obtain (3.12).


## 5. Examples on the Unitary Disk

### 5.1. The Green Function for the Helmholtz Equation Dirichlet Problem in Circular Domains

In this section we will consider the Green function determination when the domainΩ⊂ℝ2 in problem (1.1) is circular with respect to the polar coordinate system.A Green function to problem (1.1) is a solution of(5.1)-ΔG(ζ,x)-κ2G(ζ,x)=δ(ζ-x),ζ∈Ω,G(ζ,x)=0,ζ∈∂Ω,
where x∈Ω is the localization of the delta Dirac source. We may use the linearity of the problem for decomposing the solution in two additive parts
(5.2)G(ζ,x)=F(ζ,x)+12(GF(ζ,x)+GF(x,ζ)).
Here F is the fundamental solution for the free space Helmholtz equation
(5.3)-ΔF(ζ,x)-κ2F(ζ,x)=δ(ζ-x),
and GF is a homogeneous source regular solution of the Helmholtz equation
(5.4)-ΔGF(ζ,x)-κ2GF(ζ,x)=0,ζ∈Ω,G(ζ,x)=-F(ζ,x),ζ∈∂Ω,
and x∈Ω¯. In polar coordinates with r=|ζ-x|(5.5)-1r∂∂r(r∂F(r)∂r)-κ2F(r)=δ(r)forr≥0,
has at r=0 singular behavior solution
(5.6)F(κr)=-14Y0(κr),
which is a Bessel function of second kind.Remark 5.1.
Whenκ is small, this solution has a singularity that has the same behavior of the logarithmic function, that is,
(5.7)Y0(κr)≈-2πlog⁡r,lim⁡κ→0+(Δ+κ2)=Δ⟹lim⁡κ→0+14Y0(κr)=-12log⁡r.Remark 5.2 (addition theorem).
Letx=(ρcos(θ),ρsin(θ)) and ζ=(σcos(β),σsin(β)). Then
(5.8)r=σ2+ρ2-2σρcos⁡(β-θ),(5.9)Y0(κr)=∑n=-∞+∞Yn(κσ)Jn(κρ)cos⁡(n(β-θ)),0=∑n=-∞+∞Yn(κσ)Jn(κρ)sin⁡(n(β-θ)),Y0(κr)=∑n=-∞+∞Yn(κσ)Jn(κρ)ein(β-θ).
The nonhomogeneous Dirichlet boundary condition in (5.4) is
(5.10)G(ζ,x)=14Y0(κ1+ρ2-2ρcos⁡(β-θ)),ζ∈∂Ω.
The regular solution of (5.4) is
(5.11)GF(σ,β;ρ,θ)=∑n=-∞+∞cn(ρ,θ)Jn(κσ)einβ,
where the coefficients c¯n=-c-n are determined by the Dirichlet boundary condition:
(5.12)GF(1,β;ρ,θ)=14Y0(κ1+ρ2-2ρcos⁡(β-θ))=14∑n=-∞+∞Yn(κ)Jn(κρ)ein(β-θ)=14∑n=-∞+∞Yn(κρ)Jn(κ)ein(β-θ)=∑n=-∞+∞cn(ρ,θ)Jn(κ)einβ.
Since the solutions {einβ,n=-∞,+∞} are linearly independent, we obtain
(5.13)Jn(κ)cn=14Yn(κ)Jn(κρ)e-inθ=14Yn(κρ)Jn(κ)e-inθ,
and when κ is not a root of the Bessel function Jn,
(5.14)cn(ρ,θ)=14Yn(κ)Jn(κρ)Jn(κ)e-inθ=14Yn(κρ)Jn(κ)Jn(κ)e-inθ,(5.15)GF(σ,β;ρ,θ)=14∑n=-∞+∞Yn(κ)Jn(κρ)Jn(κ)Jn(κσ)ein(β-θ)=14∑n=-∞+∞Yn(κρ)Jn(κ)Jn(κ)Jn(κσ)ein(β-θ),(5.16)GF(ρ,θ;σ,β)=14∑n=-∞+∞Yn(κ)Jn(κσ)Jn(κ)Jn(κρ)ein(β-θ)=14∑n=-∞+∞Yn(κσ)Jn(κ)Jn(κ)Jn(κρ)ein(β-θ),
by noting that by addition theorem
(5.17)F(ζ,x)=-14∑n=-∞+∞Yn(κσ)Jn(κρ)ein(β-θ)=-14∑n=-∞+∞Yn(κρ)Jn(κσ)ein(β-θ).
By substituting (5.6) and (5.15) and using the addition theorem (5.9) in (5.4), we finally obtain the Green function for the circular domain Helmholtz equation problem (5.1)
(5.18)G(σ,β;ρ,θ)=-18∑n=-∞+∞Jn(κ)Yn(κσ)-Yn(κ)Jn(κσ)Jn(κ)Jn(κρ)ein(β-θ)-18∑n=-∞+∞Jn(κ)Yn(κρ)-Yn(κ)Jn(κρ)Jn(κ)Jn(κσ)ein(θ-β),
if we define r>=maxσ,ρ and r<=minσ,ρ, which can also be rewritten as
(5.19)G(σ,β;ρ,θ)=-14∑n=-∞+∞Jn(κ)Yn(κr>)-Yn(κ)Jn(κr>)Jn(κ)Jn(κr<)ein(β-θ).


### 5.2. The Integral Equation Kernel for the Unitary Disk

The kernel of the integral equation is obtain by taking the normal derivative trace of the Green function (5.18). By using the Wronskian identity for the Bessel functions
(5.20)Jn(z)Yn′(z)-Jn′(z)Yn(z)=2πz,
we obtain
(5.21)Pκ(ρ,β-θ)=∂G(x,ζ)∂νx=∂G(σ,β;ρ,θ)∂σ∣σ=1=12π∑n=-∞+∞Jn(κρ)Jn(κ)ein(β-θ),
which is the modified Poisson kernel for Dirichlet Helmholtz equation problem in the disk. Note that, when κ→0, it tends to the classical Poisson kernel for the disk.Note that, whenκ≡iκ is substitute in (1.1), it becomes a modified Helmholtz equation for which extensions of the Maximum Modulo Principle and the Strong Maximum Principle for metaharmonics functions are applicable. In this case the Poisson Kernel may be rewrit with modified Bessel function of first kind
(5.22)In(r)=e-in(π/2)Jn(ir),(5.23)Piκ(ρ,β-θ)=12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ).Remark 5.3.
Substituting the kernel (5.23) in (3.12), we obtain the strong inverse source equation (3.12) for unitary disk
(5.24)∫02π∫01(12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ))f(ρ,θ)ρdρdθ=Λ(β),β∈[0,2π).Remark 5.4.
Another important class of sources to be reconstructed is the characteristic source with star-shaped boundary with parametrization given by(x,y)=(r(θ)cos(θ),r(θ)sin(θ)), with r∈H1(0,2π). The strong inverse star-shaped characteristic source equation (3.12) for unitary disk is
(5.25)∫02π∫0r(θ)(12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ))ρdρdθ=Λ(β),β∈[0,2π).Remark 5.5.
Note that the kernel with finite sum(5.26)PiκN(ρ,β-θ)=12π∑n=-N+NI|n|(κρ)I|n|(κ)ein(β-θ),
converges with respect to (ρ,θ) pointwise to Piκ(ρ,β-θ). Since
(5.27)|Piκ(ρ,β-θ)|=12π∑n=-N+NI|n|(κρ)I|n|(κ)=12π∑n=-N+Nρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)≤12π∑n=-N+Nρ|n|,
it is bounded by the function ρdρ integrable 2/(1-ρ), in such a way that the limit and the integral can be transposed.Remark 5.6.
The transposed equation(5.28)12π∑n=-∞+∞(∫02π∫0r(θ)I|n|(κρ)I|n|(κ)e-inθρdρdθ)einβ=Λ(β),β∈[0,2π),
can be Fourier transformed giving
(5.29)∫02π∫0r(θ)I|n|(κρ)I|n|(κ)e-inθρdρdθ=Λ̂(n),n∈Z.


### 5.3. The Variational Formulation for the Unitary Disk

TakeΩ=D as the unitary disk. The weak equation (4.4) can also be specialized for the characteristic star-shaped source χω=χr inside the unitary disk Modified Helmholtz problem
(5.30)∫Dv(ρ,θ)χω(ρ,θ)ρdρdθ=∫02π∫0r(θ)v(ρ,θ)dρdθ=∫02π(gνx(θ)v(1,θ)-g(θ)∂v∂νx(θ))dθ,
for all v∈H-Δ+κ2(D). As we have proved in Lemma 3.5, without loss of generality we can consider the data from the homogeneous Dirichlet problem g=0.  Note that the Modified Poisson Dirichlet kernel
(5.31)Piκ(ρ,β-θ)=12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ),
is a function inside H-Δ+κ2(D) and can be substituted in this equation giving
(5.32)∫02π∫0r(θ)12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ)ρdρdθ=∫02πgνx(θ)12π∑n=-∞+∞ein(β-θ)dθ.Proposition 5.7.
Ifgνx∈Hl(0,2π) for  l≥1/2, then
(5.33)∫02πgνx(θ)12π∑n=-∞+∞ein(β-θ)dθ=gνx(β).Proof.
This is consequence from the fact thatDN=∑n=-N+Nein(β-θ) is the Dirichlet kernel of order N. When it acts on a continuous function, the fast oscillations of DN far from θ=β do not contribute to N-truncated Fourier series
(5.34)gνxN(β)=12π∫02πgνx(θ)DN(θ-β)dθ.
As N grows, the dominant contribution to this integral comes from an arbitrarily small neighborhood of θ-β. This behavior gives the Dirac delta distribution character to the series
(5.35)δ(θ-β)=lim⁡N⟶∞DN(θ-β),
and the integral is well defined only when gνx is a continuous function. By Sobolev embedding, the minimal admissible index is 1/2.Remark 5.8.
Note that this result is in agreement with the sufficient condition for (ii)⇒ (i) in the equivalence theorem Section 4.3.


### 5.4. The Integral Operator for Star-Shaped Characteristic Source

Let us consider the nonlinear mapping(5.36)F[r]=F[χr],
where F is the function in the formulation of problem given by (3.12).  Note that the strong integral equation (5.25) can be formally stated as the nonlinear problem: to find r(θ) such that
(5.37)F[r]=Λ.
We can investigate the operator ℱ:H1(0,2π)→Hl(0,2π) with respect to a possible set of values l∈ℝ for which the regularity of the source boundary r influences the regularity in the range of the functional.Proposition 5.9.
One considers two possible cases for the star-shaped source inside the unitary disk in which the source boundary can do or do not touch the disk border(i)
ifr(θ)≤1, then ℱ[r]∈Hl(0,2π) with l<1/2;(ii)
whenr(θ)<1,ℱ[r]∈C∞(0,2π).Proof.
We will estimate the functional norm inHl(0,2π) by using its Fourier transform (5.29). Note that
(5.38)I|n|(κρ)I|n|(κ)=ρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)≤ρ|n|,∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ≤r(θ)|n|+2|n|+2.
Since, for r≤1, ∫0r(θ)(I|n|(κρ)/I|n|(κ))ρdρ≤1/(|n|+2) and (1+n2)l/(|n|+2)2≤2l|n|2l-2, we obtain that
(5.39)‖F[r]‖Hl(0,2π)=12π∑-∞∞(1+n2)l|∫02π(∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ)e-inθdθ|2≤∑-∞∞(1+n2)l|∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ|2≤∑-∞∞(1+n2)l(|n|+2)2≤∑-∞∞2l|n|2l-2,
converges when 2l-2<-1, which proves (i). If 2l-2=-1, that is, l=1/2, the upper bound series results in the divergent harmonic series.  Note that the result (i) in the proposition gives only a sufficient condition, since
(5.40)‖F[r]‖Hl(0,2π)=12π∑-∞∞(1+n2)l|∫02π(∫0r(θ)ρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)ρdρ)e-inθdθ|2=∑-∞∞(1+n2)l×|∫02πr(θ)|n|+2|n|+21+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!r(θ)2m)/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)e-inθdθ|2,
if r(θ)=1 for all θ∈[0,2π), then
(5.41)‖F[r]‖Hl(0,2π)=1∀l∈R.
For (ii) we note that r∈H1(0,2π)⊂C(0,2π) and that supθr(θ)=∥r∥∞<1.  In this case, for each m∈ℕ, we have
(5.42)lim⁡n→∞(|n|m|F[r]̂(n)|)=lim⁡n→∞(|n|m∣‖r‖∞|n|+2|n|+2)=0,
which proves that ℱ[r]∈C∞(0,2π).Proposition 5.10.
Suppose that the source star-shaped boundaries do not touch the unitary circumference. Then(5.43)F:N[r]⊂H1(0,2π)⟶Hl(0,2π),
is locally Lipschitz continuous for every l∈ℝ; that is, for all r, there exists a neighborhood 𝒩(r) of r in H1(0,2π) such that ℱ is Lipschitz.Proof.
Letr1 and r2 be two star-shaped boundaries in the same neighborhood. Then
(5.44)‖F[r1]-F[r2]‖Hl(0,2π)=(12π∑-∞∞(1+n2)l|∫02π(F[r1]-F[r2])e-inθdθ|2)1/2,
where
(5.45)F[r1]-F[r2]=(r1(θ)|n|+2-r2(θ)|n|+2)/(|n|+2)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)+∑m=1∞(1/(|n|+2m+2))((κ2/4)m|n|!(r1(θ)|n|+2+2m-r2(θ)|n|+2+2m)/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!),
can be factored in (r1(θ)-r2(θ))(5.46)F[r1]-F[r2]=(r1(θ)-r2(θ))×f|n|(θ,r1(θ),r2(θ)),
where the factor function
(5.47)f|n|(θ,r1(θ),r2(θ))∶=∑t=0|n|+1r1(θ)|n|+1-tr2(θ)t(|n|+2)(1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!))+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!/m!(|n|+m)!)∑t=0|n|+1+2mr1(θ)|n|+1+2m-tr2(θ)t(|n|+2)(1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!))≥0
is nonnegative. So
(5.48)‖F[r1]-F[r2]‖Hl(0,2π)=(12π∑-∞∞(1+n2)l|∫02π(r1(θ)-r2(θ))×f|n|(θ,r1(θ),r2(θ))e-inθdθ|2)1/2‖F[r1]-F[r2]‖Hl(0,2π)≤(12π∑-∞∞(1+n2)l∫02π|r1(θ)-r2(θ)|2f|n|(θ,r1(θ),r2(θ))2dθ)1/2,
or
(5.49)‖F[r1]-F[r2]‖Hl(0,2π)≤C0‖r1(θ)-r2(θ)‖C(0,2π),
where
(5.50)C0≥(∑-∞∞(1+n2)l∫02π(f|n|(θ,r1(θ),r2(θ)))2dθ)1/2
is the Lipschitz constant. By denoting r0=maxθ∈(0,2π){r1(θ),r2(θ)}, we have the following estimate:
(5.51)f|n|(θ,r0,r0)=r0|n|+11+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!/m!(|n|+m)!)r02m1+∑m=1∞(κ2/4)m|n|!/m!(|n|+m)!≤r0|n|+1
which gives
(5.52)C0=(∑-∞∞(1+n2)lr02(|n|+1))1/2<∞∀l∈R.
Note that we have used the boundedness of the embedding of H1(0,2π) in C(0,2π) to find a neighborhood 𝒩[r1] of r1<1∈H1(0,2π) such that sup{∥r2∥∞:r2∈𝒩[r1]}<1. This embedding says that there also exists a neighborhood in H1(0,2π) such that the nonlinear mapping ℱ[r] is Lipschitz continuous, as enunciated in the proposition.


## 5.1. The Green Function for the Helmholtz Equation Dirichlet Problem in Circular Domains

In this section we will consider the Green function determination when the domainΩ⊂ℝ2 in problem (1.1) is circular with respect to the polar coordinate system.A Green function to problem (1.1) is a solution of(5.1)-ΔG(ζ,x)-κ2G(ζ,x)=δ(ζ-x),ζ∈Ω,G(ζ,x)=0,ζ∈∂Ω,
where x∈Ω is the localization of the delta Dirac source. We may use the linearity of the problem for decomposing the solution in two additive parts
(5.2)G(ζ,x)=F(ζ,x)+12(GF(ζ,x)+GF(x,ζ)).
Here F is the fundamental solution for the free space Helmholtz equation
(5.3)-ΔF(ζ,x)-κ2F(ζ,x)=δ(ζ-x),
and GF is a homogeneous source regular solution of the Helmholtz equation
(5.4)-ΔGF(ζ,x)-κ2GF(ζ,x)=0,ζ∈Ω,G(ζ,x)=-F(ζ,x),ζ∈∂Ω,
and x∈Ω¯. In polar coordinates with r=|ζ-x|(5.5)-1r∂∂r(r∂F(r)∂r)-κ2F(r)=δ(r)forr≥0,
has at r=0 singular behavior solution
(5.6)F(κr)=-14Y0(κr),
which is a Bessel function of second kind.Remark 5.1.
Whenκ is small, this solution has a singularity that has the same behavior of the logarithmic function, that is,
(5.7)Y0(κr)≈-2πlog⁡r,lim⁡κ→0+(Δ+κ2)=Δ⟹lim⁡κ→0+14Y0(κr)=-12log⁡r.Remark 5.2 (addition theorem).
Letx=(ρcos(θ),ρsin(θ)) and ζ=(σcos(β),σsin(β)). Then
(5.8)r=σ2+ρ2-2σρcos⁡(β-θ),(5.9)Y0(κr)=∑n=-∞+∞Yn(κσ)Jn(κρ)cos⁡(n(β-θ)),0=∑n=-∞+∞Yn(κσ)Jn(κρ)sin⁡(n(β-θ)),Y0(κr)=∑n=-∞+∞Yn(κσ)Jn(κρ)ein(β-θ).
The nonhomogeneous Dirichlet boundary condition in (5.4) is
(5.10)G(ζ,x)=14Y0(κ1+ρ2-2ρcos⁡(β-θ)),ζ∈∂Ω.
The regular solution of (5.4) is
(5.11)GF(σ,β;ρ,θ)=∑n=-∞+∞cn(ρ,θ)Jn(κσ)einβ,
where the coefficients c¯n=-c-n are determined by the Dirichlet boundary condition:
(5.12)GF(1,β;ρ,θ)=14Y0(κ1+ρ2-2ρcos⁡(β-θ))=14∑n=-∞+∞Yn(κ)Jn(κρ)ein(β-θ)=14∑n=-∞+∞Yn(κρ)Jn(κ)ein(β-θ)=∑n=-∞+∞cn(ρ,θ)Jn(κ)einβ.
Since the solutions {einβ,n=-∞,+∞} are linearly independent, we obtain
(5.13)Jn(κ)cn=14Yn(κ)Jn(κρ)e-inθ=14Yn(κρ)Jn(κ)e-inθ,
and when κ is not a root of the Bessel function Jn,
(5.14)cn(ρ,θ)=14Yn(κ)Jn(κρ)Jn(κ)e-inθ=14Yn(κρ)Jn(κ)Jn(κ)e-inθ,(5.15)GF(σ,β;ρ,θ)=14∑n=-∞+∞Yn(κ)Jn(κρ)Jn(κ)Jn(κσ)ein(β-θ)=14∑n=-∞+∞Yn(κρ)Jn(κ)Jn(κ)Jn(κσ)ein(β-θ),(5.16)GF(ρ,θ;σ,β)=14∑n=-∞+∞Yn(κ)Jn(κσ)Jn(κ)Jn(κρ)ein(β-θ)=14∑n=-∞+∞Yn(κσ)Jn(κ)Jn(κ)Jn(κρ)ein(β-θ),
by noting that by addition theorem
(5.17)F(ζ,x)=-14∑n=-∞+∞Yn(κσ)Jn(κρ)ein(β-θ)=-14∑n=-∞+∞Yn(κρ)Jn(κσ)ein(β-θ).
By substituting (5.6) and (5.15) and using the addition theorem (5.9) in (5.4), we finally obtain the Green function for the circular domain Helmholtz equation problem (5.1)
(5.18)G(σ,β;ρ,θ)=-18∑n=-∞+∞Jn(κ)Yn(κσ)-Yn(κ)Jn(κσ)Jn(κ)Jn(κρ)ein(β-θ)-18∑n=-∞+∞Jn(κ)Yn(κρ)-Yn(κ)Jn(κρ)Jn(κ)Jn(κσ)ein(θ-β),
if we define r>=maxσ,ρ and r<=minσ,ρ, which can also be rewritten as
(5.19)G(σ,β;ρ,θ)=-14∑n=-∞+∞Jn(κ)Yn(κr>)-Yn(κ)Jn(κr>)Jn(κ)Jn(κr<)ein(β-θ).


## 5.2. The Integral Equation Kernel for the Unitary Disk

The kernel of the integral equation is obtain by taking the normal derivative trace of the Green function (5.18). By using the Wronskian identity for the Bessel functions
(5.20)Jn(z)Yn′(z)-Jn′(z)Yn(z)=2πz,
we obtain
(5.21)Pκ(ρ,β-θ)=∂G(x,ζ)∂νx=∂G(σ,β;ρ,θ)∂σ∣σ=1=12π∑n=-∞+∞Jn(κρ)Jn(κ)ein(β-θ),
which is the modified Poisson kernel for Dirichlet Helmholtz equation problem in the disk. Note that, when κ→0, it tends to the classical Poisson kernel for the disk.Note that, whenκ≡iκ is substitute in (1.1), it becomes a modified Helmholtz equation for which extensions of the Maximum Modulo Principle and the Strong Maximum Principle for metaharmonics functions are applicable. In this case the Poisson Kernel may be rewrit with modified Bessel function of first kind
(5.22)In(r)=e-in(π/2)Jn(ir),(5.23)Piκ(ρ,β-θ)=12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ).Remark 5.3.
Substituting the kernel (5.23) in (3.12), we obtain the strong inverse source equation (3.12) for unitary disk
(5.24)∫02π∫01(12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ))f(ρ,θ)ρdρdθ=Λ(β),β∈[0,2π).Remark 5.4.
Another important class of sources to be reconstructed is the characteristic source with star-shaped boundary with parametrization given by(x,y)=(r(θ)cos(θ),r(θ)sin(θ)), with r∈H1(0,2π). The strong inverse star-shaped characteristic source equation (3.12) for unitary disk is
(5.25)∫02π∫0r(θ)(12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ))ρdρdθ=Λ(β),β∈[0,2π).Remark 5.5.
Note that the kernel with finite sum(5.26)PiκN(ρ,β-θ)=12π∑n=-N+NI|n|(κρ)I|n|(κ)ein(β-θ),
converges with respect to (ρ,θ) pointwise to Piκ(ρ,β-θ). Since
(5.27)|Piκ(ρ,β-θ)|=12π∑n=-N+NI|n|(κρ)I|n|(κ)=12π∑n=-N+Nρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)≤12π∑n=-N+Nρ|n|,
it is bounded by the function ρdρ integrable 2/(1-ρ), in such a way that the limit and the integral can be transposed.Remark 5.6.
The transposed equation(5.28)12π∑n=-∞+∞(∫02π∫0r(θ)I|n|(κρ)I|n|(κ)e-inθρdρdθ)einβ=Λ(β),β∈[0,2π),
can be Fourier transformed giving
(5.29)∫02π∫0r(θ)I|n|(κρ)I|n|(κ)e-inθρdρdθ=Λ̂(n),n∈Z.


## 5.3. The Variational Formulation for the Unitary Disk

TakeΩ=D as the unitary disk. The weak equation (4.4) can also be specialized for the characteristic star-shaped source χω=χr inside the unitary disk Modified Helmholtz problem
(5.30)∫Dv(ρ,θ)χω(ρ,θ)ρdρdθ=∫02π∫0r(θ)v(ρ,θ)dρdθ=∫02π(gνx(θ)v(1,θ)-g(θ)∂v∂νx(θ))dθ,
for all v∈H-Δ+κ2(D). As we have proved in Lemma 3.5, without loss of generality we can consider the data from the homogeneous Dirichlet problem g=0.  Note that the Modified Poisson Dirichlet kernel
(5.31)Piκ(ρ,β-θ)=12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ),
is a function inside H-Δ+κ2(D) and can be substituted in this equation giving
(5.32)∫02π∫0r(θ)12π∑n=-∞+∞I|n|(κρ)I|n|(κ)ein(β-θ)ρdρdθ=∫02πgνx(θ)12π∑n=-∞+∞ein(β-θ)dθ.Proposition 5.7.
Ifgνx∈Hl(0,2π) for  l≥1/2, then
(5.33)∫02πgνx(θ)12π∑n=-∞+∞ein(β-θ)dθ=gνx(β).Proof.
This is consequence from the fact thatDN=∑n=-N+Nein(β-θ) is the Dirichlet kernel of order N. When it acts on a continuous function, the fast oscillations of DN far from θ=β do not contribute to N-truncated Fourier series
(5.34)gνxN(β)=12π∫02πgνx(θ)DN(θ-β)dθ.
As N grows, the dominant contribution to this integral comes from an arbitrarily small neighborhood of θ-β. This behavior gives the Dirac delta distribution character to the series
(5.35)δ(θ-β)=lim⁡N⟶∞DN(θ-β),
and the integral is well defined only when gνx is a continuous function. By Sobolev embedding, the minimal admissible index is 1/2.Remark 5.8.
Note that this result is in agreement with the sufficient condition for (ii)⇒ (i) in the equivalence theorem Section 4.3.


## 5.4. The Integral Operator for Star-Shaped Characteristic Source

Let us consider the nonlinear mapping(5.36)F[r]=F[χr],
where F is the function in the formulation of problem given by (3.12).  Note that the strong integral equation (5.25) can be formally stated as the nonlinear problem: to find r(θ) such that
(5.37)F[r]=Λ.
We can investigate the operator ℱ:H1(0,2π)→Hl(0,2π) with respect to a possible set of values l∈ℝ for which the regularity of the source boundary r influences the regularity in the range of the functional.Proposition 5.9.
One considers two possible cases for the star-shaped source inside the unitary disk in which the source boundary can do or do not touch the disk border(i)
ifr(θ)≤1, then ℱ[r]∈Hl(0,2π) with l<1/2;(ii)
whenr(θ)<1,ℱ[r]∈C∞(0,2π).Proof.
We will estimate the functional norm inHl(0,2π) by using its Fourier transform (5.29). Note that
(5.38)I|n|(κρ)I|n|(κ)=ρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)≤ρ|n|,∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ≤r(θ)|n|+2|n|+2.
Since, for r≤1, ∫0r(θ)(I|n|(κρ)/I|n|(κ))ρdρ≤1/(|n|+2) and (1+n2)l/(|n|+2)2≤2l|n|2l-2, we obtain that
(5.39)‖F[r]‖Hl(0,2π)=12π∑-∞∞(1+n2)l|∫02π(∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ)e-inθdθ|2≤∑-∞∞(1+n2)l|∫0r(θ)I|n|(κρ)I|n|(κ)ρdρ|2≤∑-∞∞(1+n2)l(|n|+2)2≤∑-∞∞2l|n|2l-2,
converges when 2l-2<-1, which proves (i). If 2l-2=-1, that is, l=1/2, the upper bound series results in the divergent harmonic series.  Note that the result (i) in the proposition gives only a sufficient condition, since
(5.40)‖F[r]‖Hl(0,2π)=12π∑-∞∞(1+n2)l|∫02π(∫0r(θ)ρ|n|1+∑m=1∞((κ2/4)m|n|!ρ2m/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)ρdρ)e-inθdθ|2=∑-∞∞(1+n2)l×|∫02πr(θ)|n|+2|n|+21+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!r(θ)2m)/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)e-inθdθ|2,
if r(θ)=1 for all θ∈[0,2π), then
(5.41)‖F[r]‖Hl(0,2π)=1∀l∈R.
For (ii) we note that r∈H1(0,2π)⊂C(0,2π) and that supθr(θ)=∥r∥∞<1.  In this case, for each m∈ℕ, we have
(5.42)lim⁡n→∞(|n|m|F[r]̂(n)|)=lim⁡n→∞(|n|m∣‖r‖∞|n|+2|n|+2)=0,
which proves that ℱ[r]∈C∞(0,2π).Proposition 5.10.
Suppose that the source star-shaped boundaries do not touch the unitary circumference. Then(5.43)F:N[r]⊂H1(0,2π)⟶Hl(0,2π),
is locally Lipschitz continuous for every l∈ℝ; that is, for all r, there exists a neighborhood 𝒩(r) of r in H1(0,2π) such that ℱ is Lipschitz.Proof.
Letr1 and r2 be two star-shaped boundaries in the same neighborhood. Then
(5.44)‖F[r1]-F[r2]‖Hl(0,2π)=(12π∑-∞∞(1+n2)l|∫02π(F[r1]-F[r2])e-inθdθ|2)1/2,
where
(5.45)F[r1]-F[r2]=(r1(θ)|n|+2-r2(θ)|n|+2)/(|n|+2)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!)+∑m=1∞(1/(|n|+2m+2))((κ2/4)m|n|!(r1(θ)|n|+2+2m-r2(θ)|n|+2+2m)/m!(|n|+m)!)1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!),
can be factored in (r1(θ)-r2(θ))(5.46)F[r1]-F[r2]=(r1(θ)-r2(θ))×f|n|(θ,r1(θ),r2(θ)),
where the factor function
(5.47)f|n|(θ,r1(θ),r2(θ))∶=∑t=0|n|+1r1(θ)|n|+1-tr2(θ)t(|n|+2)(1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!))+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!/m!(|n|+m)!)∑t=0|n|+1+2mr1(θ)|n|+1+2m-tr2(θ)t(|n|+2)(1+∑m=1∞((κ2/4)m|n|!/m!(|n|+m)!))≥0
is nonnegative. So
(5.48)‖F[r1]-F[r2]‖Hl(0,2π)=(12π∑-∞∞(1+n2)l|∫02π(r1(θ)-r2(θ))×f|n|(θ,r1(θ),r2(θ))e-inθdθ|2)1/2‖F[r1]-F[r2]‖Hl(0,2π)≤(12π∑-∞∞(1+n2)l∫02π|r1(θ)-r2(θ)|2f|n|(θ,r1(θ),r2(θ))2dθ)1/2,
or
(5.49)‖F[r1]-F[r2]‖Hl(0,2π)≤C0‖r1(θ)-r2(θ)‖C(0,2π),
where
(5.50)C0≥(∑-∞∞(1+n2)l∫02π(f|n|(θ,r1(θ),r2(θ)))2dθ)1/2
is the Lipschitz constant. By denoting r0=maxθ∈(0,2π){r1(θ),r2(θ)}, we have the following estimate:
(5.51)f|n|(θ,r0,r0)=r0|n|+11+∑m=1∞((|n|+2)/(|n|+2m+2))((κ2/4)m|n|!/m!(|n|+m)!)r02m1+∑m=1∞(κ2/4)m|n|!/m!(|n|+m)!≤r0|n|+1
which gives
(5.52)C0=(∑-∞∞(1+n2)lr02(|n|+1))1/2<∞∀l∈R.
Note that we have used the boundedness of the embedding of H1(0,2π) in C(0,2π) to find a neighborhood 𝒩[r1] of r1<1∈H1(0,2π) such that sup{∥r2∥∞:r2∈𝒩[r1]}<1. This embedding says that there also exists a neighborhood in H1(0,2π) such that the nonlinear mapping ℱ[r] is Lipschitz continuous, as enunciated in the proposition.


## 6. Conclusions

The central question investigated in this paper is nonuniqueness of the inverse source problem, which is related with nonobservability of the source by using only boundary data. The Hilbert space framework constrains the class of functions that can be reconstructed and may be considered a kind of a priory information about the source. For more generic Banach spaces and other optimal formulations, different sources may be obtained. The demonstrated equivalence Theorem4.2 can be used to investigate questions such as stability and regularization. Further numerical studies for the unitary disk based on the equations presented in Section 5 remain as future work.



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*Source: 808913-2012-02-22.xml*