Partial data inverse problems and simultaneous recovery of boundary and coefficients for

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Partial data inverse problems and simultaneous recovery of boundary and coefficients for

semilinear elliptic equations

Matti Lassas, Tony Liimatainen, Yi-Hsuan Lin, and Mikko Salo

Abstract. We study various partial data inverse boundary value problems for the semilinear elliptic equation ∆u + a(x, u) = 0 in a domain in Rⁿ by using the higher order linearization technique introduced in [LLLS19, FO20]. We show that the Dirichlet-to-Neumann map of the above equation determines the Taylor series of a(x, z) at z = 0 under general assumptions on a(x, z). The determination of the Taylor series can be done in parallel with the detection of an unknown cavity inside the domain or an unknown part of the boundary of the domain. The method relies on the solution of the linearized partial data Calder´on problem [FKSU09], and implies the solution of partial data problems for certain semilinear equations ∆u + a(x, u) = 0 also proved in [KU20].

The results that we prove are in contrast to the analogous inverse problems for the linear Schr¨odinger equation. There recovering an unknown cavity (or part of the boundary) and the potential simultaneously are long- standing open problems, and the solution to the Calder´on problem with partial data is known only in special cases when n ≥ 3.

Keywords. Calder´on problem, inverse obstacle problem, Schiffer’s problem, simultaneous recovery, partial data.

1. Introduction

In this paper, we extend the recent studies [LLLS19, FO20] to various partial data inverse problems for the semilinear elliptic equation

∆u + a(x, u) = 0 in Ω ⊂ Rⁿ,

for n ≥ 2. The proofs rely on higher order linearization. This method reduces inverse problems for semilinear elliptic equations to related problems for the Laplace equation, with artificial source terms produced by the nonlinear interaction, and then employs the exponential solutions introduced in [Cal80] to solve these problems. Hence, one can regard the nonlinearity as a tool to solve inverse problems for elliptic equations with certain nonlinearities.

As a matter of fact, many researchers have studied inverse problems for nonlinear elliptic equations. A classical method, introduced in [Isa93] in the parabolic case, is to show that the first linearization of the nonlinear DN map is actu- ally the DN map of the corresponding linearized equation, and then to adapt the theory of inverse problems for linear equations. For the semilinear equation

∆u + a(x, u) = 0, the problem of recovering the potential a(x, u) was studied in [IS94, IN95, Sun10, IY13a]. Further results are available for inverse problems for quasilinear elliptic equations [Sun96, SU97, KN02, LW07, MU], for the degener- ate elliptic p-Laplace equation [SZ12, BHKS18], and for the fractional semilinear Schr¨odinger equation [LL19]. Certain inverse problems for quasilinear elliptic equations on Riemannian manifolds were considered in [LLS19]. We refer to the surveys [Sun05, Uhl09] for more details on inverse problems for nonlinear elliptic equations.

Inverse problems for hyperbolic equations with various nonlinearities have also been studied. Many of the results mentioned above rely on a solution to a related inverse problem for a linear equation, which is in contrast to the study of inverse problems for nonlinear hyperbolic equations. In fact, it has been realized that the nonlinearity can be beneficial in solving inverse problems for nonlinear hyperbolic equations.

By regarding the nonlinearity as a tool, some unsolved inverse problems for hyperbolic linear equations have been solved for their nonlinear analogues. Kurylev- Lassas-Uhlmann [KLU18] studied the scalar wave equation with a quadratic nonlinearity. In [LUW18], the authors studied inverse problems for general semilinear wave equations on Lorentzian manifolds, and in [LUW17] they studied similar problems for the Einstein-Maxwell equations. We also refer readers to [CLOP19, dHUW18, KLOU14, WZ19] and references therein for further results on inverse problems of nonlinear hyperbolic equations.

In this work we employ the method introduced independently in [LLLS19] and [FO20] which uses nonlinearity as a tool that helps in solving inverse problems for certain nonlinear elliptic equations. The method is based on higher order linearizations of the DN map, and essentially amounts to using sources with several parameters and obtaining new linearized equations after differentiating with respect to these parameters. The works [LLLS19, FO20] considered inverse problems with boundary measurements on the whole boundary, also on manifolds of certain type.

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In this article we will consider similar problems in Euclidean domains with differ- ent types of assumptions. We consider situations where the data is given on the full boundary or just on a part of the boundary. We also consider cases when the domain includes an unknown cavity or an unknown part of the boundary. Espe- cially we will prove uniqueness in the partial data case. Moreover, just before this article was submitted to arXiv, the preprint [KU20] of Krupchyk and Uhlmann appeared on arXiv. The work [KU20] considers the partial data Calder´on problem for certain semilinear equations and proves Corollary 1.1 below.

In this work, we do not pursue optimal regularity assumptions for our inverse problems. Instead, we want to demonstrate how the nonlinearity helps us in un- derstanding related inverse problems.

Let us describe more precisely the semilinear equations studied in this article.

Let Ω ⊂ Rⁿ be a bounded domain with C^∞boundary ∂Ω, where n ≥ 2. Consider the following second order boundary value problem

(∆u + a(x, u) = 0 in Ω,

u = f on ∂Ω.

(1.1)

We will assume that the boundary data satisfies f ∈ C^s(∂Ω) and kf k_Cs(∂Ω)≤ δ, where s > 2 is not an integer and δ > 0 is a sufficiently small number. Here C^s(∂Ω) is the standard H¨older space on ∂Ω (the precise definition is given in Appendix A).

For the function a = a(x, z), we assume that a is C^∞ in Ω × R and satisfies one of the following conditions: Either a = a(x, z) satisfies

a(x, 0) = 0, and 0 is not a Dirichlet eigenvalue of ∆ + ∂_za(x, 0) in Ω, (1.2)

or a = a(x, z) satisfies

a(x, 0) = ∂za(x, 0) = 0.

(1.3)

Note that the condition (1.3) is stronger than (1.2). Nonlinearities satisfying (1.3) together with the condition ∂_z^ka(x, 0) 6= 0 for some k ≥ 2 are called genuinely nonlinear in [LUW18] in the context of inverse problems of nonlinear hyperbolic equations. The benefit of assuming (1.3) is that the linearized equation will be just the Laplace equation.

For nonlinearities satisfying (1.2), it follows from [LLLS19, Proposition 2.1]

(restated in Appendix A) that there are C, δ > 0 such that the boundary value problem (1.1) is well-posed for small boundary data f ∈ C^s(∂Ω), kf k_Cs(∂Ω)< δ, and there is a unique solution u of (1.1) satisfying kuk_Cs(Ω)< Cδ. The solution u is called the unique small solution. See Appendix A for a more detailed discussion.

We define the corresponding Dirichlet-to-Neumann map (DN map) Λa such that Λa : {f ∈ C^s(∂Ω) ; kf k_Cs(∂Ω)< δ} → C^s−1(∂Ω),

Λ_a(f ) = ∂_νu|_∂Ω, (1.4)

where ∂_ν is the normal derivative on the boundary ∂Ω.

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We have three main theorems. The first one considers a full data inverse problem and the latter two theorems consider cases where measurements are made only on subsets of a boundary.

The first theorem is a full data uniqueness result that follows from the method of [LLLS19, FO20] for Euclidean domains (in fact the theorem is contained in [FO20]

when n ≥ 3 also for H¨older continuous a(x, z), and in the case a(x, z) = q(x)z^m where q ∈ C^∞(Ω), m ∈ N and m ≥ 2, it is contained in [LLLS19, Theorem 1.2]).

The result considers the Calder´on problem for semilinear elliptic equations for a large class of nonlinear coefficients a(x, z), and shows that the Taylor series of a(x, z) at z = 0 can be recovered from the DN map. This theorem is not covered by earlier results on inverse problems for semilinear equations [IS94, IN95, Sun10, IY13a], which often assume a sign condition such as ∂ua(x, u) ≤ 0.

Theorem 1.1 (Global uniqueness). Let Ω ⊂ Rⁿ be a bounded domain with C^∞ boundary ∂Ω, where n ≥ 2. Let aj(x, z) be C^∞ functions in x, z satisfying (1.2) for j = 1, 2. Let Λa_j : C^s(∂Ω) → C^s−1(∂Ω), s > 2, be the (full data) DN maps of

∆u + aj(x, u) = 0 in Ω, for j = 1, 2, and assume that

Λa₁(f ) = Λa₂(f )

for any f ∈ C^s(∂Ω) with kf k_Cs(∂Ω)< δ, where δ > 0 is a sufficiently small number.

Then we have

∂_z^ka1(x, 0) = ∂_z^ka2(x, 0) in Ω, for k ≥ 1.

(1.5)

Even though the above result is mostly contained in [LLLS19, FO20], it will be helpful for the partial data results to give a proof of Theorem 1.1 as well as a reconstruction algorithm to recover the coefficients ∂_z^ka(x, 0) for all k ≥ 2 in Section 2.

Next, we introduce an inverse obstacle problem for semilinear elliptic equations.

Let Ω and D be a bounded open sets with C^∞ boundaries ∂Ω and ∂D such that D ⊂⊂ Ω. Assume that ∂Ω and Ω \ D are connected. Let a(x, z) ∈ C^∞((Ω \ D) × R) be a function satisfying (1.3) for x ∈ Ω \ D. Consider the following semilinear elliptic equation







∆u + a(x, u) = 0 in Ω \ D,

u = 0 on ∂D,

u = f on ∂Ω.

(1.6)

For s > 2 and s /∈ N, let f ∈ C^s(∂Ω) with kf k_Cs(∂Ω) < δ, where δ > 0 is a sufficiently small number. The condition (1.3) yields the well-posedness of (1.6) for small solutions by the results in Appendix A, and one can define the corresponding DN map Λ^D_a, with Neumann values measured only on ∂Ω, by

Λ^D_a : {f ∈ C^s(∂Ω) ; kf k_Cs(∂Ω)< δ} → C^s−1(∂Ω), Λ^D_a : f 7→ ∂_νu|_∂Ω. The inverse obstacle problem is to determine the unknown cavity D and the coefficient a from the DN map Λ^D_a. Our second main result is as follows.

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Theorem 1.2 (Simultaneous recovery: Unknown cavity and coefficients). Assume that Ω ⊂ Rⁿ, n ≥ 2, is a bounded domain with connected C^∞ boundary ∂Ω. Let D1, D2⊂⊂ Ω be nonempty open subsets with C^∞ boundaries such that Ω \ Dj are connected. For j = 1, 2, let

a_j(x, z) ∈ C^∞((Ω \ D_j) × R)

satisfy (1.3) and denote by Λ^Daj^j the DN maps of the following Dirichlet problems







∆uj+ aj(x, uj) = 0 in Ω \ Dj,

u_j = 0 on ∂D_j,

uj = f on ∂Ω

defined for any f ∈ C^s(∂Ω) with kf k_Cs(∂Ω)< δ, where δ > 0 is a sufficiently small number. Assume that

Λ^D_a₁¹(f ) = Λ^D_a₂²(f ), whenever kf kC^s(∂Ω)< δ.

Then

D := D₁= D₂ and ∂_z^ka₁(x, 0) = ∂_z^ka₂(x, 0) in Ω \ D for k ≥ 2.

The proof is based on higher order linearizations, and relies on the solution of the linearized Calder´on problem with partial data given in [FKSU09]. We remark that the analogous simultaneously recovering problem stays open when the lower order coefficient a(x, u) = q(x)u. More specifically, when the elliptic equation

∆u + a(x, u) = 0 becomes the (linear) Schr¨odinger equation ∆u + q(x)u = 0, one does not know how to determine the obstacle D ⊂⊂ Ω and q(x) by using the knowledge of the corresponding DN map.

The inverse problem of determining the obstacle D from the DN map Λ_D,a is usually regarded as the obstacle problem. The obstacle problem with a single measurement, i.e., determining the obstacle D by a single Cauchy data {u|_∂Ω, ∂_νu|_∂Ω} is a long-standing problem in inverse scattering theory. This type problem is also known as Schiffer’s problem, and the problem has been widely studied when the surrounding coefficients are known a priori. We refer the readers to [CK12, Isa06, LZ08] for introduction and discussion.

Many researchers have made significant progress in recent years on Schiffer’s problem for the case with general polyhedral obstacles. For the uniqueness and stability results, see [AR05, CY03, LZ06, LZ07, Ron03, Ron08]. Under the assumption that ∂D is nowhere analytic, Schiffer’s problem was solved in [HNS13].

However, Schiffer’s problem still remains open for the case with general obstacles (i.e., when obstacles have no geometrical assumptions). Furthermore, a nonlo- cal type Schiffer’s problem was solved by [CLL19]. We also want to point out that the simultaneous recovery of an obstacle and an unknown surrounding potential is also a long-standing problem in the literature. This problem is closely related to the partial data Calder´on problem [KSU07, IUY10]. Unique recovery results in the literature are based on knowing the embedded obstacle to recover

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the unknown potential [IUY10], knowing the surrounding potential to recover the unknown obstacle [KL13, KP98, LZZ15, LZ10, O’D06], or using multiple spectral data to recover both the obstacle and potential [LL17].

Based on the connection of simultaneous recovery problems and the partial data Calder´on problem, i.e., we do not know the DN map on the full boundary, we will next study a partial data problem for semilinear elliptic equations. In fact, we will consider the case where both the coefficients of the equation and a part of the boundary are unknown. In the study of partial data inverse problems for (linear) elliptic equations one usually assumes that the non-accessible part of the boundary is a priori known. This is not always a reasonable assumption in practical situations. For example, in medical imaging the body shape outside of the attached measurement device may not be precisely known.

Let Ω ⊂ Rⁿ be a bounded connected domain with C^∞ boundary ∂Ω. Let Γ ⊂ ∂Ω be nonempty open set (the known part of the boundary), and assume that we do not know ∂Ω \ Γ a priori. We consider the following semilinear elliptic equation







∆u + a(x, u) = 0 in Ω,

u = 0 on ∂Ω \ Γ,

u = f on Γ,

(1.7)

where a(x, u) is a smooth function fulfilling (1.3). For s > 2 and s /∈ N, let f ∈ C_c^s(Γ) with kf k_Cs(Γ)< δ, where δ > 0 is any sufficiently small number. Here

C_c^s(Γ) := {f ∈ C^s(Γ) : supp(f ) ⊂ Γ} .

Then by the well-posedness of (1.7) for small solutions (see Appendix A again), one can define the corresponding DN map Λ^Ω,Γ_a with

Λ^Ω,Γ_a : {f ∈ C_c^s(Γ) ; kf k_Cs(Γ)< δ} → C^s−1(Γ), f 7→ ∂νu|Γ.

The inverse problem is to determine unknown part of the boundary ∂Ω \ Γ and the coefficient a from the DN map Λ^Ω,Γ_a .

Theorem 1.3 (Simultaneous recovery: Unknown boundary and coefficients). Let Ωj ⊂ Rⁿ, n ≥ 2, be a bounded domain with C^∞ boundary ∂Ωj for j = 1, 2, and let Γ be a nonempty open subset of both ∂Ω₁ and ∂Ω₂. Let G be the connected component of Ω₁∩ Ω2 whose boundary contains Γ and suppose that G 6= ∅. Let aj(x, z) be smooth functions satisfying (1.3). Let Λ^Ωaj^j^,Γ be the DN maps of the following problems







∆uj+ aj(x, uj) = 0 in Ωj,

u_j= 0 on ∂Ω_j\ Γ,

uj= f on Γ,

for j = 1, 2. Assume that

Λ^Ω_a¹^,Γ

1 (f ) = Λ^Ω_a²^,Γ

2 (f )

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for any f ∈ C_c^s(Γ) with kf k_Cs(Γ)< δ, for a sufficiently small number δ > 0. Then we have

Ω1= Ω2:= Ω and ∂_z^ka1(x, 0) = ∂_z^ka2(x, 0) in Ω for k ≥ 2.

The proof again relies on higher order linearizations and on the solution of the linearized Calder´on problem with partial data [FKSU09]. By using Theorem 1.3, we immediately have the following result, which was first proved in the preprint [KU20] that appeared on arXiv just before this preprint was submitted.

Corollary 1.1 (Partial data). Let Ω ⊂ Rⁿ, n ≥ 2, be a bounded domain with C^∞ boundary ∂Ω, and let Γ ⊂ Ω be a nonempty open subset. Let aj(x, z) be smooth functions satisfying (1.3) and let Λ^Ω,Γ_a_j be the partial data DN map for the Dirichlet problem







∆uj+ aj(x, uj) = 0 in Ω,

u_j= 0 on ∂Ω \ Γ,

uj= f on Γ,

for j = 1, 2. Assume that

Λ^Ω,Γ_a₁ (f ) = Λ^Ω,Γ_a₂ (f ),

for any f ∈ C_c^s(Γ) with kf k_Cs(Γ)< δ, for a sufficiently small number δ > 0. Then

∂_z^ka₁(x, 0) = ∂_z^ka₂(x, 0) in Ω for k ≥ 2.

For the corresponding linear equation, i.e., a(x, u) = q(x)u, the partial data problem of determining q from the DN map Λ^Ω,Γ_q (f )|Γ for any f supported in Γ, where Γ is an arbitrary nonempty open subset of ∂Ω, was solved in [IUY10] for n = 2 and qj ∈ C^2,α. For n ≥ 3, the partial data problem stays open, but there are partial results [BU02, KSU07, Isa07, KS14a] when ∂Ω is assumed to be known.

We refer to the surveys [IY13b, KS14b] for further references.

Remark 1.2. If we assume that a_j(x, z) are real analytic in z for j = 1, 2, then one can completely recover the nonlinearity and show that a₁(x, z) = a₂(x, z) in Theorems 1.1, 1.2 and 1.3. In particular this applies to equations of the type

∆u + q(x)u^m= 0, where m ≥ 2 is an integer.

The paper is structured as follows. In Section 2 we prove Theorem 1.1. We also provide reconstruction algorithms for ∂_z^ka(x, z)|z=0 for all k ≥ 2. Theorem 1.2 and Theorem 1.3 will be proved in Section 3 and Section 4, respectively. Appendix A contains the proof of a topological lemma required in the arguments.

Acknowledgments. All authors were supported by the Finnish Centre of Ex- cellence in Inverse Modelling and Imaging (Academy of Finland grant 284715).

M.S. was also supported by the Academy of Finland (grant 309963) and by the European Research Council under Horizon 2020 (ERC CoG 770924). Y.-H. L. is supported by the Ministry of Science and Technology Taiwan, under the Columbus Program: MOST-109-2636-M-009-006.

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2. Proof of Theorem 1.1

We use higher order linearizations to prove Theorem 1.1. Before the proof we recall Calder´on’s exponential solutions ([Cal80]) to the equation ∆v = 0 in Rⁿ, and the complex geometrical optics solutions (CGOs) that solve ∆v + qv = 0 on a domain Ω in Rⁿ. These solutions will be used in the proof of Theorem 1.1. The exponential solutions of Calder´on are of the form

v1(x) := exp((η + iξ) · x), v2(x) := exp((−η + iξ) · x), (2.1)

where η and ξ are any vectors in Rⁿthat satisfy η ⊥ ξ and |η| = |ξ|. The functions v₁ and v₂ solve the Laplace equation

∆v₁= ∆v₂= 0 in Rⁿ.

The linear span of the products v1v2 = exp(2iξ · x), ξ ∈ Rⁿ, of Calder´on’s exponential solutions forms a dense set in L¹(Ω). In particular, if

Z

Ω

f v1v2dx = 0

holds for all Calder´on’s exponential solutions v1 and v2, then f = 0.

The complex geometrical optics solutions (CGOs) generalize Calder´on’s exponential solutions. For n ≥ 3, they are of the form (see e.g. [SU87])

(2.2) V1(x) = e^ρ¹^·x(1 + r1), V2(x) = e^ρ²^·x(1 + r2),

where ρ1 = η + i (ξ + ζ) ∈ Cⁿ and ρ2 = −η + i (ξ − ζ) ∈ Cⁿ. Here η, ξ, ζ ∈ Rⁿ satisfy

η · ξ = ξ · ζ = ζ · η = 0, and |η|²= |ξ|²+ |ζ|².

The idea is that ξ is fixed but |η|, |ζ| → ∞. If q ∈ L^∞, the CGO solutions V1 and V2 satisfy

(∆ + q)V1= (∆ + q)V2= 0 in Ω and kr_jk_L2(Ω)≤ _|ρ^C

j| for some constant C > 0 depending on q_j, for j = 1, 2. Thus the product V1V2 converges to e^2ix·ξ as |η|, |ζ| → ∞. For n = 2 one needs to use CGOs of the form e^{iτ θ(x)}a(x), where τ > 0 is large and the phase function θ is quadratic in x, instead of CGOs with linear phase functions of the form ρ · x described above. See [Buk08] for more details.

The products of pairs of CGOs form a complete set in L¹(Ω) by [SU87] for n ≥ 3 and in L²(Ω) by [Buk08, BTW19] for n = 2. Regarding this we record the following.

Proposition 2.1 (Density of products of CGO solutions). Let Ω ⊂ Rⁿ, n ≥ 2, be a bounded domain with a C^∞boundary ∂Ω and let q₁, q₂∈ C^∞(Ω). Let f ∈ L^∞(Ω).

Suppose that

Z

Ω

f V1V2dx = 0, (2.3)

for all V_j solving (−∆ + q_j)V_j= 0 in Ω. Then f ≡ 0 in Ω.

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Proof. The result for n ≥ 3 follows from [SU87]. The argument for n = 2 is based on the stationary phase method and requires some care, but following closely the argument in [BTW19, Section 5] (where q1− q2is replaced by our function f which is in L^p(Ω) for 1 ≤ p ≤ ∞) gives the required result.

We refer to the survey [Uhl09] for more details and references on CGOs.

Before the proof, we need to discuss a minor issue: the equation ∆u+a(x, u) = 0 involves real valued solutions (a is defined on Ω×R), whereas exponential solutions and CGOs are complex valued. However, in the proof we can just use the real and imaginary parts of these solutions (which are solutions themselves, since the coefficients are real valued) by virtue of the following simple lemma.

Lemma 2.2. Let f ∈ L^∞(Ω), v1, v2∈ L²(Ω), and v3, . . . , vm∈ L^∞(Ω) be complex valued functions where m ≥ 2. Then

Z

Ω

f v₁· · · vmdx =

2^m

X

j=1

Z

Ω

c_jf w₁^(j)· · · w_m^(j)dx

where cj ∈ {±1, ±i} and w^(j)₁ ∈ {Re(v1), Im(v1)}, · · · , w^(j)m ∈ {Re(vm), Im(vm)}

for 1 ≤ j ≤ 2^m.

Proof. The result follows by writing Z

Ω

f v1· · · vmdx = Z

Ω

f (Re(v1) + iIm(v1)) · · · (Re(vm) + iIm(vm)) dx and by multiplying out the right hand side.

Now, we can prove Theorem 1.1.

Proof of Theorem 1.1. We split the proof into two parts, where in the first part we assume that the linear terms of the operators ∆ + a_j(x, z) vanish: ∂_za_j(x, 0) ≡ 0, j = 1, 2. The proof in this case is based on Calder´on’s exponential solutions. In the second part we consider the case ∂_za_j(x, 0) 6= 0 and use CGOs instead of Calder´on’s exponential solutions.

Case 1. ∂_za_j(x, 0) ≡ 0.

The proof is by induction on the order of the order of differentiation k ∈ N. By assumption, we have that

∂_za₁(x, 0) = 0 = ∂_za₂(x, 0).

Let then N ∈ N and assume that

(2.4) ∂_z^ka1(x, 0) = ∂_z^ka2(x, 0) for all k = 1, 2, · · · , N.

The induction step is to show that (2.4) holds for k = N + 1.

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For ` = 1, . . . , N + 1, let _`be small positive real numbers, and let f_`∈ C^s(∂Ω) be functions on the boundary. Let us denote = (₁, ₂, . . . , _{N +1}) and let the function

uj:= uj(x; ), j = 1, 2, be the unique small solution of the Dirichlet problem







∆uj+ aj(x, uj) = 0 in Ω, uj =

N +1

X

`=1

`f` on ∂Ω.

(2.5)

The existence of the unique small solution (by redefining ` to be smaller if nec- essary) is given in Appendix A together with an explanation what unique small solutions mean. To prove the induction step, we will differentiate the equation (2.5) with respect to the ` parameters several times. The differentiation is justified by Proposition A.2.

We begin with the first order linearization as follows. Let us differentiate (2.5) with respect to _`, so that

(∆

∂

∂_`uj

+ ∂zaj(x, uj)

∂

∂_`uj

= 0 in Ω,

∂

∂`uj = f` on ∂Ω.

(2.6)

Evaluating (2.6) at = 0 shows that

∆v_j^(`)= 0 in Ω with v^(`)_j = f` on ∂Ω, where

v_j^(`)(x) = ∂

∂_`

₌₀u_j(x; ) .

Here we have used u_j(x; )|₌₀≡ 0 so that ∂_za_j(x, u_j)|₌₀≡ 0 in Ω. The functions v_j^(`)are harmonic functions defined in Ω with boundary data f`|∂Ω. By uniqueness of the Dirichlet problem for the Laplace operator we have that

v^(`):= v^(`)₁ = v^(`)₂ in Ω for ` = 1, 2, · · · , N + 1.

(2.7)

For illustrative purposes we show next how to prove that ∂_z²a1(x, 0) = ∂_z²a2(x, 0), which corresponds to the special case N = 1. The second order linearization is given by differentiating (2.6) with respect to k for arbitrary k 6= ` where k, ` ∈ {1, 2, · · · , N + 1}. Doing so yields

(∆

∂²

∂_k∂_`uj

+ ∂zaj(x, uj)

∂²

∂_k∂_`uj

+ ∂_z²a(x, uj)_∂u

j

∂_k

_∂u

j

∂_`

= 0 in Ω,

∂²

∂k∂`uj = 0 on ∂Ω.

(2.8)

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By evaluating (2.8) at = 0 we have that

(∆w^(k`)_j + ∂_z²aj(x, 0)v^(k)v^(`)= 0 in Ω,

w^(k`)_j = 0 on ∂Ω,

(2.9)

where we have denoted w_j^(k`)(x) = _∂^∂²

k∂_`uj(x; )

₌₀ and used uj(x; )|=0 ≡ 0 in Ω for j = 1, 2. By using the fact that Λ_a₁

PN +1

`=1 _`f_`

= Λ_a₂ PN +1

`=1 _`f_` , we have that

(2.10) ∂νu1|∂Ω= ∂νu2|∂Ω.

By applying ∂_k∂_`|=0 to the equation (2.10) above shows that

∂νw^(k`)₁

_∂Ω= ∂νw₂^(k`)

_∂Ω, for k, ` = 1, . . . , N + 1.

(We remind that this formal looking calculation is justified by [LLLS19, Proposi- tion 2.1].) Hence, by integrating the equation (2.9) over Ω and by using integration by parts we obtain the equation

0 = Z

∂Ω

∂νw₁^(k`)− ∂νw^(k`)₂ dS =

Z

Ω

∆

w₁^(k`)− w₂^(k`) dx

= Z

Ω

∂_z²a2(x, 0) − ∂_z²a1(x, 0) v^(k)v^(`)dx (2.11)

where v^(k)and v^(`)are defined in (2.7). (More generally, as in [LLLS19] we could as well have integrated against a third harmonic function v^(m).) Therefore, by choos- ing fk and f` as the boundary values of the real or imaginary parts of Calder´on’s exponential solutions v1and v2in (2.1) (note that the real and imaginary parts of v1 and v2 are also harmonic), and by using Lemma 2.2, we obtain that

Z

Ω

∂²_za2(x, 0) − ∂_z²a1(x, 0) v1v2dx = 0.

It follows that the Fourier transform of the difference ∂_z²a₁(x, 0) − ∂_z²a₂(x, 0) is zero. Thus ∂_z²a1(x, 0) = ∂_z²a2(x, 0). We define

∂_z²a(x, 0) := ∂_z²a₁(x, 0) = ∂_z²a₂(x, 0).

(2.12)

We also note that by using (2.12), the equation (2.9) shows that the function w₁^(k`)− w^(k`)₂ solves

∆

w^(k`)₁ − w^(k`)₂

= 0, with w^(k`)₁ − w^(k`)₂ = 0 on ∂Ω.

Thus we have that

w^(k`):= w₁^(k`)= w^(k`)₂ in Ω.

(2.13)

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We have now shown how to prove the special case of how to go from N = 1 to N = 2. Let us return to the general case N ∈ N. To prove the general case, we first show by induction within induction, call it subinduction, that

(2.14) ∂^ku₁(x; 0)

∂`1· · · ∂`_k

= ∂^ku₂(x; 0)

∂`1· · · ∂`_k

in Ω,

for all k = 1, . . . , N . The claim holds for k = 1 by (2.7). Let us then assume that (2.14) holds for all k ≤ K < N . The linearization of order K + 1 evaluated at = 0 reads

∆ ∂^K+1u_j(x, 0)

∂`₁· · · ∂`_K+1

+ R_K(u_j, a_j, 0) + ∂_z^K+1a_j(x, 0)

Π^K+1_k=1v^(`^k⁾

= 0 in Ω, (2.15)

where R_K(u_j, a_j, 0) is a polynomial of the functions ∂_z^ka_j(x, 0) and _∂^∂^k^u^j^(x;0)

`1···∂_`k for all k ≤ K. By the induction assumptions (2.4) and (2.14) these functions agree for j = 1, 2. Thus it follows that







∆

∂^K+1

`1···_`K+1u₁(x, 0) − ∂^K+1

`1···_`K+1u₂(x, 0)

= 0 in Ω

∂^K+1

`1···_`K+1u1(x, 0) − ∂^K+1

`1···_`K+1u2(x, 0) = 0 on ∂Ω.

(Above we have used the abbreviation ∂^K+1

`1···_`K+1uj(x, 0) = _∂^∂^K+1^u^j^(x,0)

`1···∂_`K+1 for j = 1, 2 and for K ∈ N, which will also be used later in the proof). Thus by the uniqueness of solutions to the Laplace equation we have that (2.14) holds for k = 1, . . . , K + 1, which concludes the induction step of the subinduction. Thus (2.14) holds for all k = 1, . . . , N .

Let us then continue with the main induction argument of the proof. The linearization of order N + 1 at = 0 yields the equation (2.15) with N in place of K. By the subinduction, we have that R_N(u₁, a₁, 0) = R_N(u₂, a₂, 0). By using this fact, it follows by subtracting the equations (2.15) with j = 1 and j = 2 from each other (with K = N ) that

Z

Ω

∂_z^{N +1}a1(x, 0) − ∂_z^{N +1}a2(x, 0)

Π^{N +1}_k=1v^(`^k⁾

dx = 0.

Here we used integration by parts and the assumption Λ_a₁ = Λ_a₂. We choose two of the functions v^(`^k⁾to be the real or imaginary parts of the exponential solutions (2.1), and the remaining N − 1 of them to be the constant function 1. Using Lemma 2.2 again, it follows that ∂_z^{N +1}a1(x, 0) = ∂_z^{N +1}a2(x, 0) in Ω as desired.

This concludes the main induction step.

Case 2. ∂_za_j(x, 0) 6≡ 0.

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The proof is similar to the Case 1, and therefore we keep exposition short. As said before, the main difference is that we use CGOs (2.2) instead of Calder´on’s exponential solutions (2.1). We consider `to be small numbers, ` = 1, 2, · · · , N +1, and = (1, · · · , N +1) and f` ∈ C^s(∂Ω), for all ` = 1, 2, · · · , N + 1. Let the function uj:= uj(x; ) be the unique small solution of

(∆uj+ aj(x, uj) = 0 in Ω, uj=PN +1

`=1 `f` on ∂Ω,

for j = 1, 2. We begin with the first order linearization as follows, which at = 0 yields:

((∆ + ∂zaj(x, 0)) v^(`)_j = 0 in Ω,

v^(`)_j = f` on ∂Ω,

(2.16)

where

v_j^(`)(x) = ∂

∂`

₌₀uj(x; ) .

The functions v_j^` are the solutions of the Schr¨odinger equation with potential

∂zaj(x, 0) in Ω with boundary data f`|∂Ω.

We show that ∂za1(x, 0) = ∂za2(x, 0) for x ∈ Ω. Since the DN maps Λa₁ and Λa₂ agree, we have by [LLLS19, Proposition 2.1] that the DN maps corresponding to the equation (2.16) are the same. Let us consider the function

g(x) := ∂za1(x, 0) − ∂za2(x, 0) ∈ L^∞(Ω),

then by Proposition 2.1, it is easy to see that g = 0 for n ≥ 3. On the other hand, by using the boundary determination (for example, see [GT11, Appendix]), then one has ∂_za₁(x, 0) = ∂_za₂(x, 0) for x ∈ ∂Ω for n = 2. Now, with this g = 0 on ∂Ω at hand, by using Proposition 2.1 again, it follows that g = 0 in Ω ⊂ R², which implies that

(2.17) ∂_za₁(x, 0) = ∂_za₂(x, 0).

Moreover, by using (2.17) and the uniqueness of solutions to the Dirichlet problem (2.16), we have that

v^(`):= v₁^(`)= v₂^(`) in Ω for ` = 1, 2, · · · , N + 1, (2.18)

and we simply denote

q(x) := ∂_za₁(x, 0) = ∂_za₂(x, 0), for x ∈ Ω.

Here we used the assumption (1.2), which says that operators ∆ + ∂_za_j(x, 0) are injective on H₀¹(Ω), , j = 1, 2.

Since ∂_za₁(x, 0) = ∂_za₂(x, 0), we have that the claim (1.5) of the theorem holds for k = 1. We proceed by induction on k. To do that, we assume that (1.5) holds

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for all k = 1, . . . , N . Again, we do the N = 1 case separately to explain how the induction works. The second order linearization yields the equations for j = 1, 2:

(∆w_j^(k`)+ q(x)w^(k`)_j + ∂_z²aj(x, 0)v^(k)v^(`) = 0 in Ω,

w^(k`)_j = 0 on ∂Ω,

(2.19)

where w^(k`)_j (x) = _∂^∂²

k∂`u_j(x; )

₌₀ and we used u_j(x; )|₌₀ ≡ 0 in Ω. Since Λa₁= Λa₂, we have (as in Case 1) that

∂_νw₁^(k`)

_∂Ω= ∂_νw₂^(k`)

_∂Ω, for k, ` ∈ 1, . . . , N.

Fix x0 ∈ Ω. We claim that there exists a solution v⁽⁰⁾ ∈ H^s(Ω), where s can be chosen arbitrarily large, of the Schr¨odinger equation

∆v⁽⁰⁾+ q(x)v⁽⁰⁾= 0 in Ω (2.20)

with

v⁽⁰⁾(x₀) 6= 0.

By the Runge approximation property (see e.g. [LLS19, Proposition A.2]), it is enough to construct such a solution in some small neighborhood U of x0. Since q is smooth, by a perturbation argument it is enough to construct a nonvanishing solution of ∆w+q(x₀)w = 0 near x₀. Writing q(x₀) = λ²for some complex number λ, it is enough to take w = e^iλx¹. This completes the construction of v⁽⁰⁾.

Now, multiplying (2.19) by v⁽⁰⁾ and integrating by parts yields that 0 =

Z

∂Ω

v⁽⁰⁾∂ν

w₁^(k`)− w^(k`)₂ dS

= Z

Ω

v⁽⁰⁾∆

w^(k`)₁ − w^(k`)₂ dx +

Z

Ω

∇v⁽⁰⁾· ∇

w^(k`)₁ − w^(k`)₂ dx

= Z

Ω

q(x)(w^(k`)₂ − w₁^(k`))v⁽⁰⁾dx + Z

Ω

∂_z²a2− ∂_z²a1 v^(k)v^(`)v⁽⁰⁾dx

− Z

Ω

w^(k`)₁ − w^(k`)₂

∆v⁽⁰⁾dx

= Z

Ω

∂_z²a2(x, 0) − ∂²_za1(x, 0) v^(k)v^(`)v⁽⁰⁾dx.

(2.21)

Here v^(k) and v^(`) are harmonic functions defined (2.18), which we now choose specifically to be real or imaginary parts of the CGOs (since q is real valued, the real and imaginary parts of CGOs are also solutions of ∆v + qv = 0). Then, by using Lemma 2.2 we can reduce to the case where v^(k) and v^(`) are the actual complex valued CGOs, and by applying the completeness of products of pairs of CGOs [Buk08, SU87] we obtain that

∂_z²a₂(x, 0)v⁽⁰⁾(x) = ∂²_za₁(x, 0)v⁽⁰⁾(x) for x ∈ Ω.

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In particular, when x = x₀, we have ∂_z²a₂(x₀, 0) = ∂_z²a₁(x₀, 0) since v⁽⁰⁾(x₀) 6= 0.

Since x₀∈ Ω was arbitrary, we have that

∂_z²a(x, 0) := ∂_z²a1(x, 0) = ∂_z²a2(x, 0) for x ∈ Ω.

(2.22)

This concludes the induction step in the special case of how to go from N = 1 to N = 2. We also have from (2.19) and (2.22) that w^(k`)₁ − w^(k`)₂ solves

(∆

w^(k`)₁ − w₂^(k`)

+ q(x)

w^(k`)₁ − w^(k`)₂

= 0 in Ω,

w₁^(k`)− w₂^(k`)= 0 on ∂Ω,

then the uniqueness of the solution to the Schr¨odinger equation yields that w^(k`):= w₁^(k`)= w^(k`)₂ in Ω.

Let us return to general case N ∈ N. As in the Case 1, we first prove by the subinduction that

∂^k_`₁_···`

ku1(x; 0) = ∂_`^k₁_···`

ku2(x; 0) in Ω,

for all k ≤ N . Then the linearization for j = 1, 2 of order N + 1 shows that

(∆ + q)

∂^{N +1}

`1···_`N+1uj(x, 0)

+ RN(uj, aj, 0) + ∂^{N +1}_z aj(x, 0)

Π^{N +1}_k=1v^(`^k⁾

= 0, (2.23)

for x ∈ Ω, and where R_N(u_j, a_j, 0) is a polynomial of the functions ∂_z^ka_j(x, 0) and

∂^k

`1···_`ku_j(x; 0) for k ≤ N . By the subinduction we have that R_N(u₁, a₁, 0) = RN(u2, a2, 0).

Finally, by multiplying (2.23) by v⁽⁰⁾ and repeating an integration by parts argument similar to that in (2.21) shows that we have the following integral identity

Z

Ω

∂_z^{N +1}a1(x, 0) − ∂_z^{N +1}a2(x, 0)

Π^{N +1}_k=1v^(`^k⁾

v⁽⁰⁾dx = 0.

With the help of Lemma 2.2 we can choose v^(`¹⁾and v^(`²⁾to be the CGOs as before, and we choose the remaining N − 1 solutions as v^(`³⁾= · · · = v^(`^{N +1}⁾= v⁽⁰⁾, where v⁽⁰⁾ is the solution in (2.20). We conclude that ∂_z^{N +1}a₁(x₀, 0) = ∂_z^{N +1}a₂(x₀, 0).

Since x₀ ∈ Ω was arbitrary, we obtain that ∂_z^{N +1}a₁(x, 0) = ∂_z^{N +1}a₂(x, 0) in Ω.

This concludes the proof.

Remark 2.3. In the proof of Theorem 1.1, we have used the Runge approximation property to construct solutions to the Schr¨odinger equation that are nonzero at a given point x0. An alternative method is to construct a nonvanishing solution of

∆v + q(x)v = 0. This can be done by considering a complex geometrical optics solution

v(x) = e^ρ·x(1 + r) in Ω, where ρ ∈ Cⁿ. Then r solves

e^−ρ·x(∆ + q)e^ρ·xr = −q in Ω,

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with the estimate (see [SU87, Theorem 1.1], the argument applies also in our case when n ≥ 2)

krk_Hs(Ω)≤ C

|ρ|kqk_Hs(Ω),

for s > n/2 and |ρ| large enough. Then by the Sobolev embedding we have that krkL^∞(Ω)≤ 1

2,

for |ρ| large enough. This implies that v(x) is nonvanishing in Ω, and the solution v⁽⁰⁾ in the proof of Theorem 1.1 could be replaced by v here.

In the case q(x) ≤ 0 in Ω (with the sign convention ∆ =Pn k=1∂_x²

k), another alternative is to apply the maximum principle to construct a positive solution to the Schr¨odinger equation in Ω.

Furthermore, when the coefficient a = a(x, z) of the operator ∆+a(x, ·) satisfies

∂za(x, 0) ≡ 0 one has the following reconstruction result.

Theorem 2.1 (Reconstruction). Let n ≥ 2, and let Ω ⊂ Rⁿ be a bounded domain with C^∞ boundary ∂Ω. Let Λa be the DN map of the equation

∆u + a(x, u) = 0 in Ω,

and assume that a(x, z) ∈ C^∞(Ω × R) satisfies (1.3). Then we can reconstruct

∂_z^ka(x, 0) from the knowledge of Λ_a, for all k ≥ 2.

Proof. For k = 2, the reconstruction formula can be easily obtained by reviewing the argument between the equations (2.9) and (2.11). Formally we have

∂d_z²a(·, 0)(−2ξ) = − Z

∂Ω

∂²

∂₁∂₂

1=₂=0

Λ_a(₁f₁+ ₂f₂) dS,

which reconstructs the coefficient ∂_z²a(x, 0). Here d∂_z²a denotes the Fourier trans- formation of a(x, z) in the x-variable, and f1 and f2 are the boundary values of the Calder´on’s exponential solutions (2.1). More precisely, we can take f1 and f2

to be the real or imaginary parts of the boundary values of the solutions (2.1), and we can then use a suitable combination as in Lemma 2.2 to recover d∂_z²a(·, 0)(−2ξ).

Moreover, by using (2.9), one can solve the boundary value problem (2.9) uniquely to construct the function w^(kl) given by (2.13).

The proof for general k is by recursion, but let us show separately how to reconstruct ∂_z³a(x, z) corresponding to k = 3. To reconstruct ∂_z³a(x, 0), we apply third order linearization for the equation







∆u + a(x, uj) = 0 in Ω, u =

N +1

X

`=1

`f` on ∂Ω,

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at = (₁, . . . , _{N +1}) = 0, where _` are small and f_`∈ C^s(∂Ω). For k = 3, we can take N to be 2. This shows that

∆w^(ik`)+ ∂²_za(x, 0)

w^(ik)v^(`)+ w^(i`)v^(k)+ w^(k`)v⁽ⁱ⁾ + ∂_z³a(x, 0)

v⁽ⁱ⁾v^(k)v^(`)

= 0 in Ω,

(2.24)

holds, where w^(ik`)(x) = _∂ ^∂³

i∂k∂`u(x; 0). An integration by parts formula now yields that

Z

∂Ω

∂³

∂i∂k∂`

i=k=`=0

Λa(ifi+ kfk+ `f`) dS +

Z

Ω

∂_z²a(x, 0)

w^(ik)v^(`)+ w^(i`)v^(k)+ w^(k`)v⁽ⁱ⁾ dx

= − Z

Ω

∂_z³a(x, 0)v⁽ⁱ⁾v^(k)v^(`)dx

Let v⁽ⁱ⁾and v^(k)be real or imaginary parts of Calder´on’s exponential solutions (2.1) and let v^(`) = 1. By using Lemma 2.2 and the fact that we have already reconstructed ∂_z²a(x, 0) and w^(k`), we see that we can reconstruct the Fourier transform of ∂_z³a(x, 0). Consequently, we know the all the coefficients of the equation (2.24) for w^(ik`), thus we may solve (2.24) to reconstruct also w^(ik`). (The boundary value for w^(ik`) is 0.)

To reconstruct ∂_z^kaj(x, 0) for any k ∈ N, one proceeds recursively. Let us assume that we have reconstructed ∂_z^ka(x, 0) and w^(`¹^···`^k⁾ for all k ≤ N . The linearization of order N + 1 then yields that (cf. (2.15))

(∆w^(`¹^···`^{N +1}⁾+ RN(u, a, 0) + ∂_z^{N +1}a(x, 0) Π^{N +1}_k=1v^(`^k⁾ = 0 in Ω,

w^(`¹^···`^{N +1}⁾= 0 on ∂Ω,

(2.25)

where w^(`¹^···`^{N +1}⁾(x) = ∂^{N +1}

`1···_`N+1u(x, 0) and where RN(u, a, 0) is a polynomial of the functions ∂_z^ka(x, 0) and ∂^k

`1···_`ku(x, 0) for k ≤ N . By the recursion assumption we have thus already recovered RN(u, a, 0). Finally, integrating the equation (2.25) over Ω shows that

Z

∂Ω

∂^{N +1}

∂`₁· · · ∂`_{N +1}

₌₀

Λa N +1

X

k=1

`_kf`_k

! dS +

Z

Ω

RN(u, a, 0) dx

= − Z

Ω

∂_z^La(x, 0)

Π^L_`=1v⁽ⁱ^`⁾ dx.

We choose v^(`¹⁾, v^(`²⁾ to be real or imaginary parts of exponential solutions (2.1) and v^(`³⁾ = · · · = v^(`^{N +1}⁾ = 1 in Ω. Using Lemma 2.2 again, this recovers

∂_z^{N +1}a(x, 0). To end the reconstruction argument, we insert the now reconstructed

∂_z^{N +1}a(x, 0) into (2.25) and solve the equation for w^(`¹^···`^{N +1}⁾ with zero Dirichlet boundary value.

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3. Simultaneous recovery of cavity and coefficients

We prove Theorem 1.2 by first recovering the cavity D from the first linearization of the equation

∆u(x) + a(x, u) = 0.

After that the function a = a(x, z) is recovered by higher order linearization.

Proof of Theorem 1.2. Let f =PN +1

`=1 _`f_`, where _`are sufficiently small numbers and let f_`∈ C^s(∂Ω) for all ` = 1, 2, · · · , N + 1. We denote = (₁, . . . , _{N +1}) and let u_j(x) = u_j(x; ) be the solution of







∆u_j+ a_j(x, u_j) = 0 in Ω \ D_j,

uj= 0 on ∂Dj,

u_j= f on ∂Ω

(3.1)

with f =PN +1

`=1 `f`, j = 1, 2.

Step 1. Recovering the cavity.

Let us differentiate (3.1) with respect to `, for ` = 1, · · · , N + 1. We obtain











∆

∂

∂_`uj

+ ∂zaj(x, uj)

∂

∂_`uj

= 0 in Ω \ Dj,

∂

∂`u_j= 0 on ∂D_j,

∂

∂_`u_j= f_` on ∂Ω,

(3.2)

for all ` = 1, 2, · · · , N + 1 and j = 1, 2. Note that by (1.3), the function u_j(x; 0) solves (3.1) with zero Dirichlet condition ∂Ω and ∂D_j. Thus we have u_j(x; 0) ≡ 0 in Ω \ Dj, for j = 1, 2. By letting = 0 and by denoting v^(`)_j (x) := _∂^∂

`

₌₀uj, the equation (3.2) becomes







∆v_j^(`)= 0 in Ω \ Dj, v^(`)_j = 0 on ∂Dj, v^(`)_j = f` on ∂Ω.

(3.3)

We show that D₁= D₂. This follows by a standard argument (see for instance [BV99, ABRV00]), but we include a proof for completeness. Let G be the connected component of Ω \ (D1∪ D2) whose boundary contains ∂Ω and letev^(`):= v₁^(`)−v₂^(`). Thenev^(`) solves

(∆ev^(`)= 0 in G, ev^(`)= ∂νve^(`)= 0 on ∂Ω since Λ^D_a¹

1(f ) = Λ^D_a²

2(f ) on ∂Ω for small f . By the unique continuation principle for harmonic functions, one has thatev^(`)= 0 in G. Thus

(3.4) v^(`)₁ = v^(`)₂ in G,

Partial data inverse problems and simultaneous recovery of boundary and coefficients for