An inverse problem for the minimal surface equation

This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ An inverse problem for the minimal surface equation © 2022 The Author(s). Published by Elsevier Ltd. Published version Nurminen, Janne Nurminen, J. (2023). An inverse problem for the minimal surface equation. Nonlinear Analysis : Theory, Methods and Applications, 227, Article 113163. https://doi.org/10.1016/j.na.2022.113163 2023 Nonlinear Analysis 227 (2023) 113163 Contents lists available at ScienceDirect Nonlinear Analysis www.elsevier.com/locate/na An inverse problem for the minimal surface equation Janne Nurminen Department of Mathematics and Statistics, University of Jyväskylä, Finland a r t i c l e i n f o Article history: Received 25 March 2022 Accepted 10 October 2022 Communicated by Francesco Maggi Keywords: Inverse problem Higher order linearization Quasilinear elliptic equation Minimal surface equation abstract We use the method of higher order linearization to study an inverse boundary value problem for the minimal surface equation on a Riemannian manifold (Rn, g), where the metric gis conformally Euclidean. In particular we show that with the knowledge of Dirichlet-to-Neumann map associated to the minimal surface equation, one can determine the Taylor series of the conformal factor c(x)at xn= 0 up to a multiplicative constant. We show this both in the full data case and in some partial data cases. ©2022 The Author(s). Published by Elsevier Ltd. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). 1. Introduction This article focuses on an inverse problem for the minimal surface equation (MSE), which is a quasilinear elliptic PDE. In particular we consider MSE on a manifold (Rn, g), n ≥3, where gij(x) = c(x)δij, 1≤i, j ≤n, with c∈C∞(Rn), c(x)>0 for all x∈Rn, that is the metric is conformally Euclidean. The aim is to use the method of higher order linearization to recover information about the conformal factor c from boundary measurements. This method, which uses the nonlinearity of the partial differential equation as a tool, was first introduced in [18] in the case of a nonlinear wave equation and was further developed in [7,22] for nonlinear elliptic equations. The novelty of this work is that we use higher order linearization in the case of MSE. For a sufficiently smooth function u:Ω⊂Rn−1→R,Ωa bounded domain with C∞boundary, consider Graphu:= {(x′, u(x′)) : x′∈Ω} ⊂ Rn. If c≡1 we would call Graphua minimal surface if and only if the function u solves the Euclidean MSE div ⎛ ⎝ ∇u √1 + |∇u|2⎞ ⎠= 0,in Ω. E-mail address: janne.s.n[email protected]. https://doi.org/10.1016/j.na.2022.113163 0362-546X/©2022 The Author(s). Published by Elsevier Ltd. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). J. Nurminen Nonlinear Analysis 227 (2023) 113163 Define then a function F:Rn2→R, F(x′, u, p, P) := − n−1 ∑ i=1 Pii −n−1 2c(x′, u)(n−1 ∑ i=1 pi∂xic(x′, u)−∂xnc(x′, u))(1.1) +1 1 + |p|2 n−1 ∑ i,j=1 Pijpipj, where p= (p1, . . . , pn−1)∈Rn−1, P = (Pij) is an (n−1) ×(n−1) matrix and x′∈Rn−1. With the conformally Euclidean metric, MSE takes the form F(x′, u, ∇u, ∇2u) = −∆u−n−1 2c(∇x′c· ∇u−∂xnc) + ∇uT∇2u∇u 1 + |∇u|2(1.2) =−divgn−1(∇u (1 + |∇u|2)1/2)+(n−1)∂xnc 2c(1 + |∇u|2)1/2 = 0. for x′∈Ω. Here divgn−1(a) = ∑n−1 i=1 (∂xiai+∑n−1 j=1 ajΓi ij)is the Riemannian divergence with respect to the first n−1 variables and Γi ij is the Christoffel symbol corresponding to the metric g. The derivation of this equation is done in Section 3. In this work we consider a boundary value problem {F(x′, u, ∇u, ∇2u) = 0 in Ω u=f, on ∂Ω, and prove that it is well-posed (Section 2) for a certain class of small boundary values f. To be more precise, we show that there is δ > 0 such that whenever f∈Cs(∂Ω), s > 3, s /∈N, with ∥f∥Cs(∂Ω)≤δ, there exists a unique small solution u∈Cs(¯ Ω) with sufficiently small norm. Let Uδ:= {h∈Cs(∂Ω) : ∥h∥Cs(∂Ω)< δ}. Thus the Dirichlet-to-Neumann (DN) map can now be defined for these small solutions as Λc:Uδ→Cs−1(∂Ω), f ↦→ ∂νuf⏐⏐∂Ω.(1.3) Here Cs=Ck,α,k∈Z, 0 < α < 1, is the standard H¨older space (see for example [6, Section 5.1]) and ∂νuf is the Euclidean boundary normal derivative. One can think of the normal derivative on the boundary as tension on the boundary caused by the minimal surface. From the knowledge of the DN map, can we recover information about the metric g? It is worth noting that there is a small gauge invariance for Eq. (1.2) and thus for the DN map. That is, if you instead of cput λc,λ= 0, into (1.2), the equation stays the same. Thus also the DN maps Λcand Λλc are the same. We also consider partial data cases, that is, if we have knowledge of the DN map in an open subset Γof the boundary ∂Ω. In this case the partial DN map is defined for f∈Uδ, spt(f)⊂Γ, as ΛΓ c:Uδ→Cs−1(∂Ω), f ↦→ ∂νuf⏐⏐Γ.(1.4) Can we recover information about the metric gif we have knowledge of this partial DN map? These are our inverse problems for the MSE and our main result gives the following answers. Before stating it, we denote by Fjthe function Fwith creplaced by cj. Theorem 1.1. Let Ω⊂Rn−1,n≥3, be a bounded domain with C∞boundary, (Rn, g1),(Rn, g2)be two Riemannian manifolds with (gj)ik(x) = cj(x)δik, where cj∈C∞(Rn),cj(x)>0for j= 1,2and for all x∈Rn. Assume that ∂xncj(x′,0) = ∂2 xncj(x′,0) = 0 for x′∈Ω. We have four cases: 2 J. Nurminen Nonlinear Analysis 227 (2023) 113163 (1) Let n > 3and Λcjbe the DN maps associated to {Fj(x′, u, ∇u, ∇2u)=0,in Ω u=f, on ∂Ω,(1.5) j= 1,2, and assume that Λc1(f) = Λc2(f) for all f∈Uδ:= {h∈Cs(∂Ω) : ∥h∥Cs(∂Ω)< δ}, where δ > 0is sufficiently small. (2) Assume either that (a) n= 3,Γ⊂∂Ωbe open and Γ=∅or (b) n > 3,Ω⊂ {xn−1>0},Γ⊂∂Ωbe open, Γ=∅and that ∂Ω\Γ⊂ {xn−1= 0}or (c) n > 3,Ωis a strict subset of some ball B⊂Rn−1,Γ⊂∂Ωbe open, Γ=∅and that ∂Ω\Γ⊂∂B. In addition assume that ΛΓ c1(f) = ΛΓ c2(f) for all f∈Uδ,spt(f)⊂Γ, where δ > 0is sufficiently small and ΛΓ cjare the partial DN maps associated to (1.5) for j= 1,2. Then in the cases (1) and (2) we have for λ= 0 ∂m xnc1(x′,0) = λ∂m xnc2(x′,0),in Ω, m ≥0. The assumption ∂xncj(x′,0) = 0 is needed in order for u≡0 to be a solution to (1.5), and this is used to prove the well-posedness. The condition ∂2 xncj(x′,0) = 0 is assumed in order for the method to work and it is not known if it could be removed. As an immediate corollary of Theorem 1.1 we get the following. Corollary 1.2. Assume the conditions in Theorem 1.1 and assume additionally that cjare real analytic with respect to xn. Then for λ= 0 we have c1(x) = λc2(x), x ∈Ω×R. In Section 5we give a full proof of Theorem 1.1 and as mentioned, it will use higher order linearization together with complex geometric optics (CGO) solutions. In the proof we first linearize (1.5) at u≡0 and the DN map at f= 0. We see that the linearization of (1.5) correspond to a conductivity equation where the conductivity is cj(x′,0). The first linearization of the DN map maps a boundary value fto ∂νv|∂Ωwhere v is a solution to the conductivity equation. We will show that cj(x′,0) can be recovered up to a multiplicative constant with the knowledge of this (partial) DN map with the help of boundary determination for a first order perturbation of the Laplacian from [2] ([10] for n= 3 and [12,29] for n > 3). In the full data case the higher order linearizations lead to an integral equality ∫Ω(∂m+4 xnc1(x′,0) −λ∂m+4 xnc2(x′,0)) m+3 ∏ N=1 vlNdx′= 0. where vlNare solutions to the first linearization. For the partial data cases, we need a special solution v(0) which is positive in Ωand vanishes on ∂Ω\Γ. With the help of this function we get the integral identity ∫Ω(∂m+4 xnc1(x′,0) −λ∂m+4 xnc2(x′,0))v(0) m+3 ∏ N=1 vlNdx′= 0. 3 J. Nurminen Nonlinear Analysis 227 (2023) 113163 Again vlNare solutions to the first linearization. In both cases, choosing two of vlNto be real or imaginary parts of CGO solutions and the rest equal to 1 we get that ∂m+4 xnc1(x′,0) = λ∂m+4 xnc2(x′,0) (for n= 3 [3], for n > 3 [29]). This method has received a lot of attention in various situations lately. Linearization has already been used in a parabolic case in [11] where the author shows that the first linearization of the nonlinear DN map is the DN map of a linear equation. Thus one can use the theory of inverse problems for linear equations. Also nonlinear elliptic cases have been studied, for example in [13,28]. As mentioned above, the method of higher order linearization was first used in [18] for a nonlinear wave equation. After that there were two simultaneously published articles [7,22] in which higher order linearization was introduced to nonlinear elliptic equations of the type ∆u+a(x, u) = 0. The important thing in this method was that it used the nonlinearity as a tool. In [16,23] the method was further developed for the case ∆u+a(x, u) = 0 in inverse problems with partial data. See also [24,27] for more results on the special case of a power type nonlinearity. After these, there have been several articles using this method for different nonlinear elliptic equations. Different cases of nonlinear conductivity equations have had a treatment in [4,14]. This method has also been used in the case of a nonlinear magnetic Schr¨odinger equation [21] and in inverse transport and diffusion problems [20]. See also [17] for a semilinear elliptic equation with gradient nonlinearities and [19] for the case of fractional semilinear elliptic equations. There are also works in inverse problems that have considered the minimal surface equation. The Euclidean case has had a treatment in [25] where the authors consider a quasilinear conductivity depending on a function uand its gradient. Also while writing this article we have learned that C˘at˘alin I. Cˆarstea, Matti Lassas, Tony Liimatainen and Lauri Oksanen are working on an inverse problem involving minimal surface equation on a Riemannian manifold in their upcoming preprint [5]. They simultaneously and independently prove a result similar to Theorem 1.1. In their work it is shown that from the knowledge of the DN map of the minimal surface equation it is possible to determine a 2-dimensional Riemannian manifold (Σ, g). We agreed with them to publish our preprints at the same time on the same preprint server. This article is organized as follows. In Section 2we prove well-posedness for a general nonlinear boundary value problem and we describe the first and second order linearizations for the general case. Section 3is dedicated to the derivation of the minimal surface equation on a manifold with conformally Euclidean metric. Section 4consists of describing the setting for Theorem 1.1 and then calculating the first and second order linearizations in this setting. Finally, we will use higher order linearization to prove Theorem 1.1 in Section 5. 2. Well-posedness and linearizations In this section, we consider general equations F(x, u, ∇u, ∇2u) = 0 and in later sections apply these methods. Let Ω⊂Rn, n ≥2 be a bounded domain with C∞boundary and let F:¯ Ω×R×Rn×Rn2→R, be a C∞function. Consider next the boundary value problem {F(x, u, ∇u, ∇2u)=0,in Ω u=f, on ∂Ω,(2.1) where f∈Cs(∂Ω) and ∇u, ∇2udenote the gradient and Hessian of u, respectively. In addition let F(x, 0,0,0) = 0 which guarantees that u≡0 is a solution to (2.1) with f= 0. Next we prove well-posedness for (2.1) using the implicit function theorem on Banach spaces [26, Theorem 10.6 and Remark 10.5]. In what follows, we denote for m×nmatrices A= (aij), B = (bij) the matrix product A:B= m ∑ i=1 n ∑ j=1 aijbij 4 J. Nurminen Nonlinear Analysis 227 (2023) 113163 and ∇PFis the matrix with elements ∂Pij F. Also a linear differential operator Lu =A(x) : ∇2u+b(x)· ∇u+c(x)uis strictly elliptic [8] in Ωif for some constants λ, Λ>0 we have λ|ξ|2≤ξTAξ ≤Λ|ξ|2, x ∈¯ Ω, for all ξ∈Rn\ {0}. Here Ais a symmetric n×nmatrix. Proposition 2.1. Let F:Ω×R×Rn×Rn2→Rbe a C∞mapping with F(x, 0,0,0) = 0. Furthermore assume that the map v↦→ L(v) := ∂uF(x, 0,0,0)v+∇F(x, 0,0,0) · ∇v+∇PF(x, 0,0,0) : ∇2v is injective on H1 0(Ω)and that the operator Lis strictly elliptic. Let s > 3, s /∈N. Then there exists C, δ > 0 such that for any f∈Uδ:= {h∈Cs(∂Ω) : ∥h∥Cs(∂Ω)< δ} the boundary value problem {F(x, u, ∇u, ∇2u)=0,in Ω u=f, on ∂Ω, has a unique small solution u=ufwhich satisfies ∥u∥Cs(¯ Ω)≤C∥f∥Cs(∂Ω). Moreover the following mappings are C∞maps S:Uδ→Cs(¯ Ω), f ↦→ uf, Λ:Uδ→Cs−1(∂Ω), f ↦→ ∂νuf|∂Ω. Proof. Let X=Cs(∂Ω), Y =Cs(¯ Ω), Z =Cs−2(¯ Ω)×Cs(∂Ω) and T:X×Y→Z, T (f, u) = (F(x, u, ∇u, ∇2u), u|∂Ω−f) Since u|∂Ω, f ∈Cs(∂Ω), u∈Cs(¯ Ω) and F∈C∞, the map Treally has this mapping property. Next we show that the map u↦→ F(x, u, ∇u, ∇2u) is a C∞map Cs(¯ Ω)→Cs−2(¯ Ω). This is done by using a Taylor expansion. Write λ= (z, p, P)∈R×Rn×Rn2and expand F(x, ·) at µ∈R×Rn×Rn2: F(x, λ +µ) = ∑ |α|≤k Dα µF(x, λ) α!µα+∑ |β|=k+1 Rβ(λ+µ)µβ, where Rβ(λ+µ) = |β| β!∫1 0 (1 −t)|β|−1Dβ µF(x, λ +tµ)dt. Now let u∈Cs(¯ Ω) be fixed, λ= (u, ∇u, ∇2u) and let µ= (h, ∇h, ∇2h), h ∈Cs(¯ Ω) be such that ∥µ∥C1,α(¯ Ω)≤1. It is enough to show that the map u↦→ Dα µF(x, u, ∇u, ∇2u) is continuous for all αand Rβ(λ+µ) = o(µk) in Cs(¯ Ω). Firstly, since the composition of a C∞function Fwith a Cs−2function is again a Cs−2function [9, Theorem A.8], we have the continuity. The space Cs(¯ Ω) is an algebra under pointwise multiplication 5 J. Nurminen Nonlinear Analysis 227 (2023) 113163 [9, Theorem A.7], and thus ∥Rβ(λ+µ)µβ∥Cs(¯ Ω)≤C(∥Rβ(λ+µ)∥C(¯ Ω)∥µβ∥Cs(¯ Ω)+∥Rβ(λ+µ)∥Cs(¯ Ω)∥µβ∥C(¯ Ω)) ≤C∥Rβ(λ+µ)∥Cs(¯ Ω)∥µ∥|β| Cs(¯ Ω) ≤C∥µ∥|β| Cs(¯ Ω) |β| β!∫1 0 (1 −t)|β|−1∥Dβ µF(x, λ +tµ)∥Cs(¯ Ω)dt ≤CF,u∥µ∥|β| Cs(¯ Ω) |β| β!∫1 0 (1 −t)|β|−1dt where ∥Dβ µF(x, λ +tµ)∥Cs(¯ Ω)is uniformly bounded in t∈(0,1) and the bounding constant may depend on uand F. This is due to Fbeing a C∞function and that u∈Cs(¯ Ω). Now the remainder satisfies ⏐⏐⏐⏐⏐⏐∑ |β|=k+1 Rβ(λ+µ)µβ⏐⏐⏐⏐⏐⏐Cs(¯ Ω)≤C∥(h, ∇h, ∇2h)∥k+1 Cs(¯ Ω) and hence the map u↦→ F(x, u, ∇u, ∇2u) is a C∞map Cs(¯ Ω)→R. By the assumption F(x, 0,0,0) = 0 we have T(0,0) = 0. Also DuT(0,0) is linear and DuT(0,0)v= (∂uF(x, 0,0,0)v+∇pF(x, 0,0,0) · ∇v+∇PF(x, 0,0,0) : ∇2v, v|∂Ω). The mapping v↦→ L(v) is injective and v≡0 is a solution to {∂uF(x, 0,0,0)v+∇pF(x, 0,0,0) · ∇v+∇PF(x, 0,0,0) : ∇2v=H, in Ω v=g, on ∂Ω,(2.2) when H=g= 0. Using Fredholm alternative [8, Theorem 6.15] the boundary value problem (2.2) has a unique solution for all Hand g. Thus DuT(0,0) is surjective. Then by the implicit function theorem there exist δ > 0 and Uδ:= B(0, δ)⊂X=Cs(∂Ω) and a C∞map S:Uδ→Y=Cs(¯ Ω) such that T(f, S(f)) = 0. Also, for small enough f∈Uδ(not necessarily the same δ) and uf∈Cs(¯ Ω), S(f) = ufis the only solution of T(f, uf) = 0. Moreover, since Sis Lipschitz continuous and S(0) = 0, for u=S(f) we have ∥u∥Cs(¯ Ω)≤C∥f∥Cs(∂Ω). Also the mapping Λis a well defined C∞map between Uδand Cs−1(∂Ω) since taking a normal derivative is a linear map from Cs(Ω) to Cs−1(∂Ω). □ In order to use the method of higher order linearization, we calculate formally the first and second order linearizations of (2.1) and the corresponding DN map. This formal looking calculation can be justified as in [22]. Let us begin by assuming that for {Fj(x, u, ∇u, ∇2u)=0,in Ω u=f, on ∂Ω,(2.3) j= 1,2, we have ΛF1(f) = ΛF2(f) for all f∈Cs(∂Ω) with ∥f∥Cs(∂Ω)≤δ, for δ > 0 sufficiently small. In order to find the linearizations, let ε1, . . . , εkbe sufficiently small numbers and f1, . . . , fk∈Cs(∂Ω). Let uj(x, ε1, . . . , εk) be the unique small solution to {Fj(x, uj,∇uj,∇2uj)=0,in Ω uj=∑k m=1 εmfm,on ∂Ω,(2.4) 6 J. Nurminen Nonlinear Analysis 227 (2023) 113163 for j= 1,2. Differentiate this with respect to εl,l∈ {1, . . . , k}, and evaluate at ε1=· · · =εk= 0 to get {∂uFj(x, 0,0,0)vl j+∇pFj(x, 0,0,0) · ∇vl j+∇PFj(x, 0,0,0) : ∇2vl j= 0,in Ω vl j=fl,on ∂Ω,(2.5) where vl j:= ∂εluj(x, ε1, . . . , εk)⏐⏐ε1=···=εk=0. The boundary value problem (2.5) has a unique solution if we assume that the map v↦→ L(v) = ∂uFj(x, 0,0,0)v+∇pFj(x, 0,0,0) · ∇v+∇PFj(x, 0,0,0) : ∇2v is injective on H1 0(Ω) and assume strict ellipticity of the operator L. At this point, we would like to see what exactly is the first linearization and see if some information can be recovered about the coefficients ∂uFj(x, 0,0,0), ∇pFj(x, 0,0,0), ∇PFj(x, 0,0,0) from the knowledge of the DN maps corresponding to (2.3) for j= 1,2. What actually can be recovered depends on the equation at hand. Let us next differentiate (2.4) first with respect to εland then with respect to εa, a =l: Ij:= ∂2 εaεlFj(x, uj,∇uj,∇2uj) =∂εa(∂uFj(x, uj,∇uj,∇2uj)∂εluj) +∂εa(n ∑ i=1 ∂piFj(x, uj,∇uj,∇2uj)∂xi∂εluj) +∂εa⎛ ⎝ n ∑ j,k=1 ∂Pjk Fj(x, uj,∇uj,∇2uj)∂2 xjxk∂εluj⎞ ⎠ := Ij,1+Ij,2+Ij,3. Then we expand these one by one: Ij,1=∂uFj∂2 εaεluj+∂2 uFj∂εauj∂εluj+ n ∑ i=1 ∂pi∂uFj∂xi∂εauj∂εluj + n ∑ j,k=1 ∂Pjk ∂uFj∂2 xjxk∂εauj∂εluj, Ij,2= n ∑ i=1(∂piFj∂xi∂2 εaεluj+∂u∂piFj∂εauj∂xi∂εluj + n ∑ r=1 ∂pr∂piFj∂xr∂εauj∂xi∂εluj+ n ∑ j,k=1 ∂Pjk ∂piFj∂2 xjxk∂εauj∂xi∂εluj), Ij,3= n ∑ j,k=1(∂Pjk Fj∂2 xjxk∂2 εaεluj+∂u∂Pjk F ∂εauj∂2 xjxk∂εluj + n ∑ i=1 ∂pi∂Pjk Fj∂xi∂εauj∂2 xjxk∂εluj+ n ∑ r,t=1 ∂Prt ∂Pjk Fj∂2 xrxt∂εauj∂2 xjxk∂εluj). Evaluate Ijat ε1=· · · =εk= 0 and denote w(al) j= (∂2 εaεluj)(x, ε1, . . . , εk)|ε1=···=εk=0 to have Ij=∂uFj(x, 0,0,0)w(al) j+∂2 uFj(x, 0,0,0)vlva(2.6) +((∇p(∂uFj))(x, 0,0,0) · ∇va+(∇P(∂uFj))(x, 0,0,0) : ∇2va)vl +∇pFj(x, 0,0,0) · ∇w(al) j+(∇p(∂uFj))(x, 0,0,0) · ∇vlva 7 J. Nurminen Nonlinear Analysis 227 (2023) 113163 + n ∑ i=1 ((∇p(∂piFj))(x, 0,0,0) · ∇va+ (∇P(∂piFj)(x, 0,0,0) : ∇2va))∂xivl +∇PFj(x, 0,0,0) : ∇2w(al) j+(∇P(∂uFj))(x, 0,0,0) : ∇2vlva + n ∑ j,k=1 ((∇p(∂Pjk Fj))· ∇va+(∇P(∂Pjk Fj))(x, 0,0,0) : ∇2va)∂2 xixjvl Thus w(al) jsatisfies the boundary value problem {Ij= 0,in Ω w(al) j= 0,on ∂Ω.(2.7) Next we would like to integrate I1−I2against a solution to the adjoint of ∂uFj(x, 0,0,0)vl j+∇pFj(x, 0,0,0) · ∇vl j+∇PFj(x, 0,0,0) : ∇2vl j= 0 and use the assumption that the DN maps associated to (2.3) coincide for j= 1,2 together with a completeness result to recover information about the coefficients of I1and I2. Again the information that can be recovered depends on the equation and below this method is applied in the case of the minimal surface equation. What we would do next is to use an induction argument to show that from higher order linearizations it is possible to recover more information. This too will be specified below. 3. Mimimal surface equation on a Riemannian manifold In this section we derive Eq. (1.2). Let (M, g), M=Rn,n≥3, be a Riemannian manifold with the metric gij(x′, xn) = c(x′, xn)δij ,(3.1) where (x′, xn)∈Rn−1×R, c ∈C∞(Rn), c(x)>0 for all x∈Rn. These assumptions are valid for the rest of the article, unless otherwise stated. Let u:Ω⊂Rn−1→R,u∈C2(¯ Ω), and consider the graph of the function u Graphu={(x′, u(x′)): x′∈Ω} ⊂ M. This graph is a minimal surface if and only if its mean curvature His equal to zero at all points on the graph. By defining f:Ω×R→R, f(x′, xn) = xn−u(x′), the graph of uis the surface Σ:= {(x′, xn)∈Ω×R:f(x′, xn)=0}. The mean curvature of Σat x∈Σis the sum of principal curvatures. We omit the normalizing factor 1 n−1when calculating the mean curvature. In order to calculate the principal curvatures, we introduce the Riemannian gradient and Hessian of a function f:M→R: ∇gf=gij∂xif∂xj,∇2 gf=(∂2 xixjf−Γm ij∂xmf)n i,j=1 , 8 J. Nurminen Nonlinear Analysis 227 (2023) 113163 Case (1): Assume now that Λc1(f) = Λc2(f) (5.1) for all f∈Cs(∂Ω) sufficiently small. Now we have the corresponding F1, F2and the first linearization of Λcjis (4.4), with c(x′,0) replaced by cj(x′,0), which corresponds to the conductivity equation ⎧ ⎨ ⎩ div (cj(x′,0)n−1 2∇vl j)= 0,in Ω vl j=fl,on ∂Ω, (5.2) for j= 1,2. Using boundary determination from [2] for the case of Laplacian with a convection term we get for x′∈∂Ω ∇x′c1(x′,0) c1(x′,0) =∇x′c2(x′,0) c2(x′,0) ⇐⇒ ∇x′(ln(c1(x′,0)))=∇x′(ln(c2(x′,0))). From this we get that ∇x′(ln(c1(x′,0)) −ln(c2(x′,0))) = ∇x′(ln c1(x′,0) c2(x′,0) )= 0 which then implies that c1(x′,0) = λc2(x′,0) for x′∈∂Ωand λ= 0. (We cannot use boundary determination for the conductivity equation (e.g. [15]) because the DN maps are different: here f↦→ ∂νvf⏐⏐∂Ωinstead of f↦→ c1(x′,0)∂νvf⏐⏐∂Ω.) It is known that the knowledge of this linearized DN map combined with c1(x′,0)|∂Ω=λc2(x′,0)|∂Ωgives us c1(x′,0) = λc2(x′,0) in Ω[29]. By the gauge invariance of (4.3) (replacing c2(x′,0) = λ−1c1(x′,0)) we have that vl jsolves the equation ⎧ ⎨ ⎩ div (c1(x′,0)n−1 2∇vl j)= 0,in Ω vl j=fl,on ∂Ω. Since solutions to this are unique, we define vl:= vl 1=vl 2. For recovering the higher order derivatives of cj(x′,0) we can use the second linearizations (from (4.6)) ⎧ ⎨ ⎩ div (cj(x′,0)n−1 2∇w(al) j)−n−1 2cj(x′,0)n−1 2−1∂3 xncj(x′,0)vlva= 0,in Ω w(al) j= 0,on ∂Ω.(5.3) corresponding to j= 1,2. Notice that if we replace c2(x′,0) by λ−1c1(x′,0) in (4.5), except in the third order derivative, we get that wal 2solves ⎧ ⎨ ⎩ div (c1(x′,0)n−1 2∇w(al) 2)−λn−1 2c1(x′,0)n−1 2−1∂3 xnc2(x′,0)vlva= 0,in Ω w(al) 2= 0,on ∂Ω. (5.4) Subtract now (5.3) for j= 1 from (5.4), integrate against v≡1 (solution to the first linearization) over Ω ∫Ω div(c1(x′,0)n−1 2∇w(al) 1−c1(x′,0)n−1 2∇w(al) 2) −(n−1 2c1(x′,0)n−1 2−1∂3 xnc1(x′,0) −λn−1 2c1(x′,0)n−1 2−1∂3 xnc2(x′,0))vlvadx′ = 0 15 J. Nurminen Nonlinear Analysis 227 (2023) 113163 and use integration by parts to have 0 = ∫∂Ω c1(x′,0)n−1 2(∇w(al) 1·ν− ∇w(al) 2·ν)dS =∫Ω div(c1(x′,0)n−1 2∇w(al) 1−c2(x′,0)n−1 2∇w(al) 2)dx′ =∫Ω n−1 2c1(x′,0)n−1 2−1(∂3 xnc1(x′,0) −λ∂3 xnc2(x′,0))vlvadx′. This is true since by (5.1) ∂νu1⏐⏐∂Ω=∂νu2⏐⏐∂Ω and applying ∂εa∂εl⏐⏐ε=0 to this implies ∂νw(al) 1⏐⏐∂Ω=∂νw(al) 2⏐⏐∂Ω,for a, l ∈ {1, . . . , k}. Thus ∫Ω c1(x′,0)n−1 2−1(∂3 xnc1(x′,0) −λ∂3 xnc2(x′,0))vlvadx′= 0 (5.5) for any va, vlsolving the conductivity equation (5.2). A solution to (5.2) is equivalently a solution to ⎧ ⎪ ⎨ ⎪ ⎩ (−∆x′+∆x′cj(x′,0)α/2 cj(x′,0)α/2)gl= 0,in Ω gl=cj(x′,0)α/2fl,on ∂Ω, where α=n−1 2,gl=cj(x′,0)α/2vland ∆x′denotes the Laplacian with respect to the first two variables. Hence by using the fact that a product of a pair of solutions (Proposition 5.1) is dense in L1(Ω), we get ∂3 xnc1(x′,0) = λ∂3 xnc2(x′,0), x′∈Ω. Also (5.3),(5.4) together with the previous equality and c1(x′,0) = λc2(x′,0), gives the following boundary value problem ⎧ ⎨ ⎩ div (c1(x′,0)n−1 2∇(w(al) 1−w(al) 2))= 0,in Ω w(al) 1−w(al) 2= 0,on ∂Ω. This has a unique solution and thus w(al) 1=w(al) 2. Next we use induction to show ∂k xnc1(x′,0) = λ∂k xnc2(x′,0) for all k∈N. By the above this already holds for k= 0,1,2,3. Our assumption now is ∂k xnc1(x′,0) = λ∂k xnc2(x′,0), x′∈Ω,for all k= 0,1,2, . . . , N ∈N, N > 3. Let us do a subinduction to prove ∂k l1...lku1(x′,0) = ∂k l1...lku2(x′,0), x′∈Ω, for all k= 1, . . . , N, where ∂k l1...lkuj(x′,0) = ∂kuj(x′,0) ∂εl1...∂εlk . Above we have shown this for k= 1,2. Assume that it holds for k≤K < N. Then the linearization of order K+ 1 is, when evaluated at ε1=· · · =εK+1 = 0, div (cj(x′,0)n−1 2∇(∂K+1 l1...lK+1 uj(x′,0)))+RK(uj, cj(x′,0),0) (5.6) +Ccj(x′,0)n−1 2−1∂K+2 xncj(x′,0) (K+1 ∏ k=1 v(lk))= 0, 16 J. Nurminen Nonlinear Analysis 227 (2023) 113163 x′∈Ω, where C= 0. Actually C=n−1 2, since it comes from the uderivatives of Fand is the constant n−1 2 appearing in front of the second term of F. Also, here Ris a polynomial of ∂k xncj(x′,0), ∂k l1...lku1(x′,0) and the components of ∇x′(∂k xncj(x′,0)). Now an integration by parts argument similar to the case of the second linearization and together with Proposition 5.1 (choosing v3=· · · =vK+1 = 1) gives ∂K+2 xnc1(x′,0) = λ∂K+2 xnc2(x′,0). Subtracting Eqs. (5.6) (similarly as for Eqs. (5.3) and (5.4)) for j= 1,2 we get ⎧ ⎨ ⎩ div (c1(x′,0)n−1 2∇(∂K+1 l1...lK+1 u1(x′,0) −∂K+1 l1...lK+1 u2(x′,0)))= 0,in Ω ∂K+1 l1...lK+1 u1(x′,0) −∂K+1 l1...lK+1 u2(x′,0) = 0,on ∂Ω. This is true, since by induction assumptions for all x′∈Ωwe have ∇x′(∂k xnc1(x′,0) −∂k xnc2(x′,0))= 0 and the other terms agree for j= 1,2, k≤K. Again, by the uniqueness of solutions, ∂K+1 l1...lK+1 u1(x′,0) = ∂K+1 l1...lK+1 u2(x′,0), x′∈Ω, which ends the subinduction. Returning to the original induction, the linearization of order N+ 1 at ε1=· · · =εN+1 = 0 is div (cj(x′,0)n−1 2∇(∂N+1 l1...lN+1 uj(x′,0))) +RN+1(uj, cj(x′,0),0) + Ccj(x′,0)n−1 2−1∂N+2 xncj(x′,0) (N+1 ∏ k=1 v(lk))= 0, x′∈Ω. By the subinduction, the terms RN(uj, cj(x′,0),0) agree for j= 1,2. Thus by subtracting, using integration by parts and that ∂ν∂N+1 l1...lN+1 u1(x′,0)|∂Ω=∂ν∂N+1 l1...lN+1 u2(x′,0)|∂Ωwe get ∫Ω cj(x′,0)n−1 2−1(∂N+2 xnc1(x′,0) −λ∂N+2 xnc2(x′,0)) N+1 ∏ k=1 vlkdx′= 0. Choosing all but two of the functions vlkto be equal to 1, we have by the completeness of such solutions (Proposition 5.1) that ∂N+2 xnc1(x′,0) = λ∂N+2 xnc2(x′,0), x′∈Ω, which ends the proof for case (1). Case (2): Now we assume that the partial DN maps coincide for f∈Uδ, spt(f)⊂Γ. Then, as in the previous case, from the first linearization we get c1(x′,0) = λc2(x′,0) in Ω(using first the boundary determination from [2], then [10] for n= 3 and [12] for n > 3). Now define vl:= vl 1=vl 2again by uniqueness of solutions. Moving to the second order linearizations and recovering higher order derivatives produces some extra work since we only have partial data. From the assumption that the DN maps coincide we get ∂νwal 1⏐⏐Γ=∂νwal 2⏐⏐Γ.(5.7) If we would now integrate the difference of (5.3), for j= 1, and (5.4) against v≡1 and integrate by parts, some terms would not cancel out. Let us instead introduce the function v(0) which is a solution to ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ div (c1(x′,0)n−1 2∇v(0))= 0,in Ω v(0) = 0,on ∂Ω\Γ v(0) =g, on Γ, (5.8) 17 J. Nurminen Nonlinear Analysis 227 (2023) 113163 where g∈C∞ c(Γ) such that g≥0 and g≡ 0. Then by the maximum principle v(0) >0 in Ω. Now we integrate against this and use integration by parts to have ∫Ω n−1 2c1(x′,0)n−1 2−1(∂3 xnc1(x′,0) −λ∂3 xnc2(x′,0))v(0)vlvadx′ =∫Ω div (c1(x′,0)n−1 2∇(w(al) 1−w(al) 2))v(0) dx′ =∫Ω (wal 1−wal 2) div (c1(x′,0)n−1 2∇v(0))dx′ +∫Γ c1(x′,0)n−1 2(∂ν(wal 1−wal 2)v(0) −(wal 1−wal 2)∂νv(0))dS +∫∂Ω\Γ c1(x′,0)n−1 2(∂ν(wal 1−wal 2)v(0) −(wal 1−wal 2)∂νv(0))dS = 0. In the last inequality we used Eq. (5.7), the fact that v(0) solves (5.8) and that wal j= 0 on ∂Ωfor j= 1,2. Then using Proposition 5.1 and the positivity of v(0) we can conclude ∂3 xnc1(x′,0) = λ∂3 xnc2(x′,0), x′∈Ω. As in the previous case we use induction to show ∂k xnc1(x′,0) = λ∂k xnc2(x′,0) for all k∈N. By the above this already holds for k= 0,1,2,3. Our assumption now is ∂k xnc1(x′,0) = λ∂k xnc2(x′,0), x′∈Ω,for all k= 0,1,2, . . . , N ∈N, N > 3. By a subinduction we can show that ∂k l1...lku1(x′,0) = ∂k l1...lku2(x′,0), x′∈Ω, for all k= 1, . . . , N, where ∂k l1...lkuj(x′,0) = ∂kuj(x′,0) ∂εl1...∂εlk . This goes in the same way as in the previous case except the integration by parts argument needs to be done as shown in this case. Returning to the original induction, the rest of the proof is again the same as in case (2). We only need to modify the integration by parts argument using again the function v(0) and Proposition 5.1 which finishes the proof. □ Acknowledgements The author was supported by the Finnish Centre of Excellence in Inverse Modelling and Imaging (Academy of Finland grant 284715). The author would like to thank Mikko Salo for helpful discussions on the minimal surface equation and everything related to inverse problems. 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