The linearized Calderón problem for polyharmonic operators

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/ The linearized Calderón problem for polyharmonic operators © 2023 The Author(s). Published by Elsevier Published version Sahoo, Suman Kumar; Salo, Mikko Sahoo, S. K., & Salo, M. (2023). The linearized Calderón problem for polyharmonic operators. Journal of Differential Equations, 360, 407-451. https://doi.org/10.1016/j.jde.2023.03.017 2023 Available online at www.sciencedirect.com ScienceDirect Journal of Differential Equations 360 (2023) 407–451 www.elsevier.com/locate/jde The linearized Calderón problem for polyharmonic operators Suman Kumar Sahoo ∗, Mikko Salo Department of Mathematics and Statistics, University of Jyväskylä, Finland Received 23 August 2022; accepted 6 March 2023 Abstract In this article we consider a linearized Calderón problem for polyharmonic operators of order 2m (m ≥2) in the spirit of Calderón’s original work [7]. We give a uniqueness result for determining coefficients of order ≤2m −1up to gauge, based on inverting momentum ray transforms. ©2023 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons .org /licenses /by /4 .0/). MSC: primary 35R30, 31B20, 31B30, 35J40 Keywords: Calderón problem; Perturbed polyharmonic operator; Anisotropic perturbation; Tensor tomography; Momentum ray transform 1. Introduction Let n ≥3 and let  ⊂Rnbe a bounded domain with smooth boundary ∂. For m ≥2, we consider the following polyharmonic operator Lwith lower order anisotropic perturbations up to order 2m −1: L(x, D) =(−)m+Q(x, D), (1.1) where *Corresponding author. E-mail addresses: [email protected] (S.K. Sahoo), [email protected] (M. Salo). https://doi.org/10.1016/j.jde.2023.03.017 0022-0396/©2023 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons .org /licenses /by /4 .0/). S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Q(x, D) = 2m−1  l=0 al i1···il(x) Di1···il(1.2) is a differential operator of order 2m −1 with 1 ≤i1, ···, il≤nand alis a smooth symmetric tensor field of order lin . Einstein summation convention is assumed for repeated indices throughout the article. The boundary measurements corresponding to the equation L(x, D)u =0in may be encoded in terms of the Cauchy data set (see e.g. [27]for the general case) CL={(u|∂,∂ νu|∂,···,∂2m−1 νu|∂):u∈H2m(), Lu=0}. The inverse problem of interest is to determine some information on the coefficients of the operator L, up to suitable gauge transformations, from the knowledge of the Cauchy data set L. For various special choices of boundary conditions, one could define a Dirichlet-to-Neumann type operator for solutions of Lu =0in . This requires that the boundary conditions lead to an elliptic boundary value problem (Lopatinskii-Shapiro condition) and that 0is not an eigenvalue for this problem. For example, one could consider solutions with the clamped boundary conditions u|∂ =f0,∂ νu|∂ =f1, ... , ∂m−1 νu|∂ =fm−1 and consider the boundary map C L:(f0,...,f m−1)→ (∂m νu|∂,...,∂2m−1 νu|∂). Alternatively, one could consider Navier boundary conditions u|∂ =f0,(−)u|∂ =f1,...,(−)m−1u|∂ =fm−1 and consider the boundary map N L:(f0,...,f m−1)→ (∂νu|∂,∂ ν(−)u|∂,...,∂ ν(−)m−1u|∂). If 0is not an eigenvalue of the corresponding elliptic boundary value problem, then knowing the Cauchy data set is equivalent to knowing the boundary map. One can view the Cauchy data set as a generalization of such boundary maps that is independent of the choice of (elliptic) boundary conditions. The most classical case is m =1, so that Lis a second order elliptic operator. The prototypical inverse problem for such operators is the inverse conductivity problem posed by Calderón [7]. The original problem was stated for the equation div(γ ∇u) =0, but if the conductivity function γis C2and positive one can reduce matters to the Schrödinger equation ( +q)u =0for some potential q. One can further consider equations of the form u +A(x) ·∇u +qu =0for some vector field Aand potential q. This includes the magnetic Schrödinger equation and equations with convection terms, which are clearly of the form (1.1). Various results are known for related inverse problems for determining the vector field A(x), sometimes up to a gauge of the form 408 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 A → A +∇φ, and the potential qfrom boundary measurements. We refer the reader to the survey [34,35]for more information and references on the second order case. In this paper we consider a Calderón type problem for polyharmonic operators of order 2m with a lower order perturbation up to order 2m −1. There are many previous results on inverse problems for polyharmonic operators, including [18,19,14,15,2–4]. In all of these results one only determines lower order coefficients up to order ≤mfrom boundary measurements. The reason for this restriction is that the method of complex geometrical optics solutions used in these inverse problems requires certain L2Carleman estimates, which can become highly delicate for higher order operators. In the previous works the required Carleman estimate for the polyharmonic operator has been obtained by using a well known Carleman estimate for second order equations several times in a row. Since there is a loss of half derivative in the original Carleman estimate (it involves a limiting Carleman weight), iterating it many times leads to a loss of several derivatives and thus restricts the method to lower order terms of degree ≤m. In this article we will consider a general higher order elliptic operator given by (1.1) and recover several lower order coefficients up to order 2m −1. However, because of the restriction mentioned above, we are not able to consider the full nonlinear inverse problem. Rather, we will only consider the linearization of this inverse problem. In the linearized problem it is sufficient to use solutions of the “free” equation (−)mv=0in the recovery of the coefficients, and this avoids the need to use Carleman estimates. However, even in the linearized problem the recovery of higher order terms becomes intricate. We will follow the ideas of [5] where it was observed that momentum ray transforms (MRT) appear naturally in solving inverse problem for polyharmonic operators. The inversion of various MRT is crucial for recovering the coefficients. In particular, to handle the coefficient of order 2m −1, we study the kernel of a partial MRT. To do so, we demonstrate a new trace free Helmholtz type decomposition result for symmetric tensor fields. The rest of the article is organised as follows. In Section 2we derive the linearized inverse problem and state our main results. Section 3is devoted to the unique recovery of tensor fields up to order 2m −2. Then Section 4deals with the recovery of coefficients up to order 2m −1 under certain assumptions. Section 5deals with gauge transformations and recovery of coefficients up to natural obstructions; see Theorem 2.3. Finally, in Section 6we describe the kernel of MRT which is the key tool for proving our main results. The required Helmholtz type decomposition result for tensor fields is then proved in Section 7. In Appendix Awe review some known results and give the construction of special solutions (known as CGO solutions) of (−)mu =0. The Navier to Neumann map is then linearized in Appendix Bby computing the Fréchet derivative. 2. Linearization and main results We now derive the formal linearization of the polyharmonic inverse problem. The forward operator is formally given by the map L → CL. Since CLis a set and not necessarily an element in a Banach space, we cannot directly compute the Fréchet derivative of the forward operator. We instead assume that CL=CL0for all ∈(−a,a), a > 0,(2.1) where L0=(−)m, L=(−)m+Q(x, , D) and Q(x, , D) = 2m−1  l=0 al i1···il(x, ) Di1···il. We assume that aldepend smoothly on xand and that al(x, 0) =0for all 0 ≤l≤2m −1, i.e. we 409 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 are computing the linearization at the case of zero lower order coefficients. Using that CL= CL0, from [27, Lemma 2.8] we have the integral identity (L−L0)u, vL2() =0,(2.2) for any u, v∈H2m() satisfying Lu =0 and L∗ 0v=0in . This can be rewritten as 2m−1  l=0 al i1···il(x, ) Di1···ilu(x, ), v(x)=0 for all ∈(−a,a). Differentiating this with respect to we obtain 2m−1  l=0 ∂(al i1···il(x, )) Di1···ilu(x, ), v(x)+2m−1  l=0 al i1···il(x, ) Di1···il∂u(x, ), v(x)=0. Writing w=u|=0and setting =0, we have 2m−1  l=0 ∂al i1···il(x, 0)Di1···ilw(x),v(x)=0.(2.3) Setting =0in the equation Lu =0 and using al(x, 0) =0for all 0 ≤l≤2m −1, we see that wsolves (−)mw=0. It follows that the identity (2.3) holds for any wand vsolving (−)mw=(−)mv=0in . We can now formulate (with slightly different notation) the main uniqueness question for the linearized inverse problem considered in this article. Question 2.1. If al i1···ilfor 0 ≤l≤2m −1are smooth tensor fields in and if   2m−1  l=0 al i1···il(Di1···ilu)v dx =0 for all u, v∈H2m() satisfying (−)mu =(−)mv=0, is it true that the coefficients al i1···il vanish possibly up to suitable gauge transformations? We mention that if the Cauchy data set is the graph of suitable Dirichlet-to-Neumann type map, one can make the above formal derivation rigorous and the linearized problem is still given by Question 2.1. See Appendix Bwhere this is done for the Navier boundary conditions. We now show that it is not in general possible to recover all the coefficients in Question 2.1 due to the presence of a gauge. One possible gauge is obtained by replacing Lby e−φLeφfor a suitable function φ. Note that if φ∈C2m() and ∂j νφ|∂ =0for0≤j≤2m−1,(2.4) 410 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 then uand eφuwill have the same Cauchy data up to order 2m −1. Thus for such functions φ one has CL=Ce−φLeφ. It is easy to compute the highest order terms for the conjugated operator: e−φLeφu=e−φ((−)m+a2m−1 i1···i2m−1Di1···i2m−1+...)eφu =(e−φ(−)eφ)mu+e−φ(a2m−1 i1···i2m−1Di1···i2m−1+...)(eφu) =(−1)m(φ +|∇φ|2+2∇φ·∇+)mu+e−φ(a2m−1 i1···i2m−1Di1···i2m−1+...)(eφu) =(−)mu+(a2m−1 i1···i2m−1Di1···i2m−1+(−1)m2m∇φ·∇()m−1)u +.... Above ... denotes lower order terms. Thus there is always a gauge invariance, where one can add terms of the form (−1)m2m∇φ·∇()m−1to the term of order 2m −1. There will be corresponding changes in the lower order terms as well. However, if we only consider operators Lfor which a2m−1=0, then this type of gauge does not arise: if the term (−1)m2m∇φ·∇()m−1vanishes identically, then necessarily ∇φ=0 and hence φ=0using the condition φ|∂ =0. Our first main result states that for operators with a2m−1=0, one can recover all the lower order terms completely in the linearized inverse problem. Theorem 2.1. Suppose that   2m−2  l=0 al i1···ilDi1···iluv=0whenever mu=mv=0.(2.5) Then al=0in for 0 ≤l≤2m −2. As an immediate corollary of Theorem 2.1 one obtains the following density result on certain spaces of symmetric tensor fields. For density results involving harmonic functions and tensor fields see e.g. [8]. One may also expect other related density results as a consequence of the method of proof of Theorem 2.1. See Remark 3.3. Corollary 2.2. For each even integer M≥0consider the set AM:=span v⊕0≤|α|≤MDαu:M+2 2u=M+2 2v=0in . Then AMis dense in SMC∞()=⊕ M p=0SpC∞(), where SpC∞()stands for the space of smooth symmetric ptensor fields in and ⊕is the direct sum. 411 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Note that for m =1, Theorem 2.1 reduces to the statement that if f∈C∞() satisfies   fuv=0 whenever u =v =0in, then fvanishes identically in . This is just the linearized Calderón problem for the Schrödinger equation (linearized at the zero potential), which has been studied in various settings including partial data [11,32] and Riemannian manifolds [16,17]. Results for such linearized problems have recently become important in the context of inverse problems for nonlinear PDEs, see e.g. [24,25]. Now we state our second main result. It also includes terms of order 2m −1 and gives a complete answer (modulo an assumption for a2m−1on ∂) for the linearized Calderón problem for polyharmonic operators when m =2, 3, showing that one can determine the coefficients uniquely up to the gauge transform L →e−φLeφ. In particular, there are no other gauge invariances in this problem. We also obtain a partial answer when m ≥4. Theorem 2.3. Let Lbe as in (1.1)–(1.2)and suppose ∂r νa2m−1=0on ∂ for all 0 ≤r≤2m −1. Assume that   2m−1  l=0 al i1···ilDi1···iluv=0whenever mu=mv=0. 1. For m =2or m =3, one has L=e−φ(−)meφ for some φsatisfying (2.4). In particular, a2m−1=im−1 δ(∇φ) where iδis the symmetrization with Kronecker delta tensor; see (3.1)for the definition. 2. For m ≥4, if we additionally assume that a2m−1=im−1 δA1for some vector field A1∈ C∞(), then L =e−φ(−)meφfor some φsatisfying (2.4). In particular, A1=∇φ. Remark 2.4. In Theorem 2.3, the boundary assumption on the coefficient a2m−1can be removed through a boundary determination result. In the next section we present the proof of Theorem 2.1. Then, in Section 4, we prove few results (mainly Proposition 4.3), which we then employ in Section 5to demonstrate Theorem 2.3. 412 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 3. Proof of Theorem 2.1 In this section we demonstrate the proof of Theorem 2.1. The proof will be based on the trace free decomposition on Sm, the space of symmetric mtensor fields over . To this end, we define two operators iδ:Sm→Sm+2and jδ:Sm→Sm−2as follows: (iδf) i1···im+2:=σ(fi1···im⊗δim+1im+2)(3.1) (jδf) i1···im−2:= n  k=1 fi1···im−2kk,(3.2) where σdenotes the symmetrization of a tensor field. Here δij is the Kronecker delta tensor which is equal to 1for i=jand 0 otherwise. We also define jδf=0for f∈S0or f∈S1. Note that the operators iδand jδare dual to each other with respect to the L2inner product. Based on this observation we write the trace free decomposition of tensor fields from [9,28]as follows: for l≥2 one has al= [l 2]  kl=0 ikl δbl,kl(3.3) where bl,klis a symmetric tensor field of order l−2klwith jδbl,kl=0. Any f∈Smsatisfies jδf=0is known as trace free tensor field. For l=0, 1any tensor field alis trace free. Note that (3.3)is an orthogonal decomposition on the space of square integrable symmetric tensor fields over , denoted by L2(Sm). This implies that al=0if and only if bl,kl=0for all klwith 0 ≤kl≤[l 2]; see [30, Equation 6.4.2]. Thus in the remainder of this section we show that under the conditions in Theorem 2.1, if alhas the decomposition (3.3)for 0 ≤l≤2m −2, then bl,kl=0for all klwith 0 ≤kl≤[l 2]. This completes the proof of Theorem 2.1. Proof of Theorem 2.1.We insert (3.3)in the integral identity (2.5) and obtain 0= 2m−2  l=0 [l 2]  kl=0 (ikl δbl,kl)i1···ilDi1···iluv= 2m−2  l=0 [l 2]  kl=0 bl,kl i1···il−2kl Di1···il−2kl(−)kluv (3.4) whenever mu =mv=0. We now recall suitable complex geometrical optics (CGO) solutions to the polyharmonic equation given in (A.5), u(x;h) =e1 h(e1+iη2)·xa0(x) +ha1(x) +···+hm−1am−1(x) +r(x;h)=e1 h(e1+iη2)·x˜ A, v(x;h) =e−1 h(e1+iη2)·xb0(x) +hb1(x) +···+hm−1bm−1(x) +r(x;h)=e−1 h(e1+iη2)·x˜ B, where the error terms r(x; h) and r(x; h) satisfy the estimate r(x, h)H2m scl ≤chmand r(x, h)H2m scl ≤chm. Moreover, we also have that ajand bjare smooth functions in for all jwith 0 ≤j≤m −1. Substituting the expressions for uand vgiven above in the identity (3.4) and using (A.2) and (A.3)we obtain 413 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 0= 2m−2  l=0 [l 2]  kl=0 bl,kl i1···il−2kl Di1···il−2kle1 h(e1+iη2)·x(−1 hT−)kl˜ Ae−1 h(e1+iη2)·x˜ B = 2m−2  l=0 [l 2]  kl=0 bl,kl i1···il−2kl (Di1+1 h(e1+iη2)i1)···(Dil−2kl+1 h(e1+iη2)il−2kl) ×(−1 hT−)kl˜ A˜ B = 2m−2  l=0 [l 2]  kl=0 l−2kl  j=0 l−2kl jbl,kl i1···il−2kl hl−2kl−j(e1+iη2)i1···(e1+iη2)il−2kl−j ×Dil−2kl−j+1···Dil−2kl((−1 hT−)kl˜ A) ˜ B. (3.5) Here η2can be any unit vector with e1·η2=0. From now on we write η=η2. We recover the coefficients one by one multiplying (3.5) with suitable powers of h, by letting h →0 and by inverting momentum ray transforms (MRT, see Lemma 3.1). This will be done in several steps. Step 1. We first show that b2m−2,0=0. We start by multiplying (3.5) by h2m−2. Next utilizing the estimates given in Lemma A.4, r(x, h)H2m scl ≤chm, ˜r(x, h)H2m scl ≤chmand letting h →0, we see that only  Rn b2m−2,0 i1···i2m−2(e1+ iη)i1···(e1+iη)i2m−2a0b0term survives and other terms will be 0. This implies  Rn b2m−2,0 i1···i2m−2(e1+iη)i1···(e1+iη)i2m−2a0b0=0.(3.6) Adapting ideas from [21]we write, b2m−2,0 i1···i2m−2(e1+iη)i1···(e1+iη)i2m−2 = 2m−2  p=0 (i)p2m−2 pb2m−2,0 i1···ip1···1ηi1···ηip = 2m−2  p=0 b2m−2,0,p i1···ipηi1···ηip,2≤i1,···,i p≤nfor each 0 ≤p≤2m−2,(3.7) where b2m−2,0,p i1···ip:= (i)p2m−2 pb2m−2,0 i1···ip1···1for each pwith 0 ≤p≤2m −2. Recall that a0and b0solve the transport equations Tma0=Tmb0=0. We utilize (A.6) and choose the following specific solutions a0and b0, where y2denotes the direction of ηand y = (y3, ..., yn)are orthogonal directions: 414 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Remark 3.2. This remark suggests that there is a another way of avoiding the gauge appearing in the Theorem 2.3. Instead of assuming a2m−1=0, one can assume that al=iδbl−2for m ≤l≤ 2m −1in Theorem 2.1 and conclude in a similar way that aj=0for all jwith 0 ≤j≤2m −1. Remark 3.3. Corollary 2.2 gave a density result for even integers. For odd integers K≥0, Theorem 2.1 implies the following (non-optimal) result: the set AK=spanv⊕|α|≤KDαu:K+2 2+1u=K+2 2+1v=0in is dense in the space SKC∞()=⊕ K p=0SpC∞(). One could also ask if the linear span of the set Ap1p2:= Dαv⊗Dβu:p1+1u=p2+1v=0in, for all multi-indices α,βsuch that |α|=p1,|β|=p2!, is dense in the space smooth symmetric tensor fields defined in of order p1+p2. Here ⊗ denotes the symmetric tensor product. In particular when p1=p2=1, this is asking if the linear span of tensor products of gradients of any two biharmonic functions is dense in the space of smooth symmetric two tensor fields. However, a similar density result is not true if one replace biharmonic functions by harmonic functions. The latter density question arises in the linearized version of the boundary rigidity problem; see [30, Chapter 1] and [28, Chapter 11]. This question also appears while linearizing certain anisotropic elliptic PDE from the Cauchy data set; see [8, Section 6]. 4. Recovery of coefficients under an additional assumption In this section we present the proof of Proposition 4.3 which will be the main ingredient to prove Theorem 2.3 in the following section. To proceed further, we define two differential operators on C∞(Sm), the space of smooth symmetric tensor fields of order m. Definition 4.1. [30, Chapter 2] The symmetric covariant derivative d :C∞(Sm) →C∞(Sm+1) is given by (df) i1···im+1=σ(i1···im+1)∂fi1···im ∂xim+1 ,1≤i1···,i m+1≤n. Definition 4.2. [30, Chapter 2] The divergence operator δ:C∞(Sm+1) →C∞(Sm)is given by (δf )i1···im= n  k=1 ∂fi1···imk ∂xk ,1≤i1···,i m≤n. The operators −d and δare dual to each other with respect to L2inner product. We also define dk:C∞(Sm) →C∞(Sm+k)by taking composition of d with itself ktimes. Similarly we define δk:C∞(Sm) →C∞(Sm−k). In particular, 421 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 δmf= n  i1,···,im=1 ∂mfi1···im ∂xi1···∂xim ,for any f∈C∞(Sm), dmφ=∂mφ ∂xi1···∂xim ,for any φ∈C∞(). Our next result is as follows. Proposition 4.3. Let m ≥1. Suppose   2m−1  l=0 al i1···ilDi1···iluv=0whenever mu=mv=0.(4.1) 1. Then we have a2m−1=0in if we additionally assume δ2m−1a2m−1=0and jδa2m−1=0.(4.2) As a result, this implies aj=0in for 0 ≤j≤2m −1by Theorem 2.1. 2. Moreover, we also have a2m−1=d2m−1φ+iδa2m−1,1where φfulfils ∂l νφ|∂ =0for 0≤l≤2m−2, (4.3) where φis a scalar function and a2m−1,1is a symmetric tensor of order 2m −3. Proof. Step 1. Extend a2m−1by zero to Rn. We first prove that (4.1) implies  Rn a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0,whenever Tma0=Tmb0=0.(4.4) This step can be proved by choosing uand vto be CGO solutions as in (A.5) and then multiplying (4.1)by h2m−1and letting h →0. Step 2. Here we show that (4.4) implies that certain MRTs of a2m−1,odd := m−1  p=0 im−1−p δa2m−1,2p+1(0,y 2,y)and a2m−1,even := m−1  p=0 im−p δa2m−1,2p(0,y 2,y) become 0. Recall that ˆadenotes the Fourier transform of ain x1, and note that a2m−1,odd and a2m−1,even are tensor fields of order 2m −1 and 2m −2 respectively. Also i δand j δare same as (3.1) and (3.2) respectively. However in this case the indices vary from 2to n. 422 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Similar to (3.7)) we next write a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1 = 2m−1  p=0 a2m−1,p i1···ipηi1···ηip,2≤i1,···,i p≤nfor each 0 ≤p≤2m−1, where a2m−1,p i1···ip=(i)p2m−1 pa2m−1 i1···ip1···1=cp(i)pa2m−1 i1···ip1···1. This together with (4.4) implies  Rn 2m−1  p=0 a2m−1,p i1···ipηi1···ηipa0b0=0. Next we choose a0=ym−1 2g(y)e−iλ(y1+iy2), b0=ym−1 2. Then utilizing the arbitrariness of g and taking partial Fourier transform in the first variable, we obtain  R eλy2 2m−1  p=0a2m−1,p i1···ip(λ, y2,y)η i1···ηipy2m−2 2dy2=0.(4.5) Next, we specify λ =0in above, after that we replace ηby −ηand again put λ =0in above. As a result, we obtain following two equations. 0=⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ R 2m−1  p=0a2m−1,p i1···ip(0,y 2,y)η i1···ηipy2m−2 2dy2 R 2m−1  p=0 (−1)pa2m−1,p i1···ip(0,y 2,y)η i1···ηipy2m−2 2dy2. We add and subtract above two relations. This entails  R 2m−1  p=0 (1±(−1)p)a2m−1,p i1···ip(0,y 2,y)η i1···ηipy2m−2 2dy2=0. This allows us to separate even and odd modes of a2m−1. After a re-indexing this entails 0=⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ R m−1  p=0a2m−1,2p+1 i1···i2p+1(0,y 2,y)η i1···ηipy2m−2 2dy2 R m−1  p=0a2m−1,2p i1···i2p(0,y 2,y)η i1···ηipy2m−2 2dy2. Utilizing Lemma 6.5, the next identities follow from the definition of a2m−1,odd and a2m−1,even. 423 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451  Ra2m−1,odd i1···i2m−1(0,y 2,y)η i1···ηi2m−1y2m−2 2dy2=0,(4.6)  Ra2m−1,even i1···i2m−2(0,y 2,y)η i1···ηi2m−2y2m−2 2dy2=0.(4.7) Step 3. We now assume the additional conditions (4.2). Then we claim that a2m−1(0,x)=0. Observe that, combining Lemma 6.2 and (4.6)we have Ra2m−1,odd i1···i2m−1(0, y2, y)ηi1···ηi2m−1yp 2dy2 =0for all 0 ≤p≤2m −2. In other words, vanishing the highest order moment is enough. This along with [30, Theorem 2.17.2] entails that there is a distribution φ(0, x)satisfying a2m−1,odd (0, x) =d2m−1 xφ(0, x). This implies a2m−1,2m−1=− m−2  p=0 im−1−p δa2m−1,2p+1+d2m−1 xφ. (4.8) From [30, Theorem 2.17.2] we also have φ(0, x) =0 outside of some ball in Rn−1. Using the fact that a2m−1,odd (0, x) ∈L2(Rn−1)and d2m−1 xφ(0, x) =(∂j1...j2m−1φ(0, x))2≤j1,...,j2m−1≤n, it follows that φ∈H2m−1(Rn−1). Here Hk(Rn−1)denotes the L2based Sobolev space of order k; see e.g. [12, Chapter 5] for precise definition. We next consider the L2 inner product of a2m−1,2m−1in the following way: a2m−1,2m−1(0,x),a2m−1,2m−1(0,x) = n  i1,···i2m−1=2 Rn−1a2m−1,2m−1 i1···i2m−1(0,x)a2m−1,2m−1 i1···i2m−1(0,x)dx. This together with (4.8) entails a2m−1,2m−1,a2m−1,2m−1 =a2m−1,2m−1,− m−2  p=0 im−1−p δa2m−1,2p+1+d2m−1 xφ =a2m−1,2m−1,− m−2  p=0 im−1−p δa2m−1,2p+1+&a2m−1,2m−1,d2m−1 xφ' =− m−2  p=0&a2m−1,2m−1,im−1−p δa2m−1,2p+1'+&a2m−1,2m−1,d2m−1 xφ'. (4.9) Now we take the Fourier transform of jδa2m−1=0in the x1variable and evaluate at the origin to obtain jδa2m−1(0, x) =0. This implies 424 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 j δa2m−1,2m−1= n  k=2a2m−1,2m−1 i1···i2m−3kk = n  k=2 (i)2m−1a2m−1 i1···i2m−3kk =−(i)2m−1a2m−1 i1···i2m−311 =−2m−1 2m−3−1 (i)2m−1−(2m−3)a2m−1,2m−3=1 c2m−3a2m−1,2m−3. Proceeding in this way we obtain jm−1−p δa2m−1,2m−1=1 cpa2m−1,2p+1for 0 ≤p≤m−2.(4.10) Taking Fourier transform of the relation δ2m−1a2m−1=0 with respect to x1variable gives 2m−1  k=02m−1 kλkδ2m−1−k xa2m−1(λ, x) =0. This for λ =0 entails δ2m−1 xa2m−1(0,x)=0=⇒ δ2m−1 xa2m−1,2m−1(0,x)=0. Let φN∈C∞ c(Rn−1)such that φN→φin H2m−1, then from above as N→∞we see that 0=(−1)2m−1δ2m−1 xa2m−1,2m−1,φ N=a2m−1,2m−1,d2m−1 xφN→a2m−1,2m−1,d2m−1 xφ. This implies a2m−1,2m−1, d2m−1 xφ =0. The combination of this along with (4.9) and (4.10) implies a2m−1,2m−1,a2m−1,2m−1=− 1 cp m−2  p=0&a2m−1,2p+1,a2m−1,2p+1'. To put it another way, it gives a2m−1,2m−1,a2m−1,2m−1+ m−2  p=0 1 cp&a2m−1,2p+1,a2m−1,2p+1'=0. This in addition to the fact that cp>0for 0 ≤p≤m −2 implies a2m−1,2p+1(0,x)=0for0≤p≤m−1. By Lemma 3.1 and jδa2m−1=0 from (4.7)we obtain a2m−1,even(0, x) =0. This together with jδa2m−1=0 implies m−1  p=0 1 cp(a2m−1,2p,a2m−1,2p)=0. We have that cp>0for all pwith 0 ≤p≤m −1, this entails a2m−1,2p(0, x) =0for all 0 ≤p≤m −1. The combination of a2m−1,2p(0, x) =a2m−1,2p+1(0, x) =0for all pwith 0 ≤p≤m −1 implies a2m−1(0, x) =0. Step 4. We still assume the additional conditions in (4.2), and show that a2m−1=0. Since a2m−1is compactly supported in the x1variable, this implies a2m−1(λ, x)is analytic in λby Paley-Wiener theorem. Therefore it is enough to show that 425 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 dl dλla2m−1(0,x)=0 for all non-negative integers l. From previous step we have a2m−1(0, x) =0. We now argue by induction and assume that dl dλla2m−1(0,x)=0 for all integers lwith 0 <l≤M. (4.11) For M+1, differentiating (4.5)M+1 times with respect to λgives M+1  k=0 RM+1 kyM+1−k 2eλy2dk dλk⎛ ⎝ 2m−1  p=0a2m−1,p i1···ip(λ, y2,y)⎞ ⎠ηi1···ηipy2m−2 2dy2=0. We now specify λ =0in preceding equation and then utilize (4.11)to conclude  R 2m−1  p=0dM+1 dλM+1a2m−1,p i1···ip(0,y 2,y)ηi1···ηipy2m−2 2dy2=0.(4.12) Differentiating the relation 2m−1  k=02m−1 kλkδ2m−1−k xa2m−1(λ, x) =0, M+1 times more with respect to λgives M+1  r=0 2m−1  k=r2m−1 kk! (k −r)!λk−rdM+1−r dλM+1−rδ2m−1−k xa2m−1(λ, x)=0.(4.13) From (4.11)we obtain dM+1−r dλM+1−rδ2m−1−k xa2m−1(0,x)=0for1≤r≤M+1 and 1 ≤k≤2m−1. In the last relation above we have used the fact that, d dλcommutes with ∂x. Setting λ =0in (4.13), then utilizing above findings we obtain dM+1 dλM+1δ2m−1 xa2m−1(0, x) =0. Since jδa2m−1=0, this implies jδdM+1 dλM+1a2m−1(0,x)=0. Using the last two relations and repeating the similar analysis as before for the integral (4.12)we obtain dM+1 dλM+1a2m−1(0, x) =0. This completes the induction step and implies dl dλla2m−1(0,x)=0 for all non-negative integers l. Combining this with Paley-Wiener theorem we obtain a2m−1(x) =0 in . In combining all of the above steps with Theorem 2.1, we conclude that aj(x) =0 for all 0 ≤j≤2m −1. To complete the proof of Proposition 4.3 we will use a decomposition result proved in Lemma 7.1. Step 5. In this step we prove (4.3). 426 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 We first substitute uand vfrom (A.5)in the integral identity (4.1) and obtain   2m−1  l=0 al i1···ilDi1···ile (e1+iη)·x h˜ Ae−(e1+iη)·x h˜ B=0. Multiplying above by h2m−1and letting h →0we arrive at a2m−1 i1···i2m−1(e1+iη)i1···(e1+ iη)i2m−1a0b0=0, where a0and b0solve the transport equation Tma0=Tmb0=0. Thanks to Lemma 7.1 we can decompose a2m−1in the following way: a2m−1=˜a2m−1+iδb2m−3+d2m−1φwith ∂l νφ|∂ =0forl=0,1,··· ,2m−2, and ˜a2m−1meets the following conditions δ2m−1˜a2m−1=jδ˜a2m−1=0in . This is known as trace free Helmholtz type decomposition of symmetric tensor fields proved in Section 7. Plugging this decomposition of a2m−1into a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0 and utilizing e1+iη, e1+iη =0we obtain   (˜a2m−1+d2m−1φ)i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0.(4.14) Since φfulfils ∂l νφ|∂ =0for 0 ≤l≤2m −2, this along with integration by parts entails (−1)2m−1  (d2m−1φ)i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0 =  φ 2m−1  k=02m−1 kT2m−1−ka0Tkb0. We next make use of transport equations Tma0=0 and Tmb0=0to deduce   φ 2m−1  k=02m−1 kT2m−1−ka0Tkb0=0. This implies (d2m−1φ)i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0. Combining this with (4.14)we arrive at ˜a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0. We next extend ˜a2m−1to be 0 outside of and obtain  Rn ˜a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0,where a0and b0solve Tm(·)=0. Since ˜a2m−1satisfies the conditions in (4.2), utilizing previous steps we obtain ˜a2m−1=0. This implies a2m−1=iδb2m−3+d2m−1φ, where φsatisfies ∂l νφ|∂ =0for 0 ≤l≤2m −2. This completes the proof.  427 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 5. Proof of Theorem 2.3 The proof of Theorem 2.3 will be presented in this section. However, before we get to the proof of Theorem 2.3, let us have a look at the gauge transformation for the case of m =2. 5.1. Gauge transformation To this end we denote La3,a2,a1,a0=(−)2+a3 ij k Dij k +a2 ij Dij +a1 iDi+a0, L0=(−)2. Note that Cauchy data of vand eφvup to order 3are the same if φ∈C4() satisfies ∂j νφ|∂ =0 for 0 ≤j≤3. Now we compute e−φL0eφv=e−φ(−)2eφv =(e−φ(−)eφ)2v =(φ +|∇φ|2+2∇φ·∇+)(φ +|∇φ|2+2∇φ·∇+)v =()2v+(∇φ·∇v) +(f (φ)v) +2∇φ·∇()v +4∇φ·∇(∇φ·∇v) +2∇φ·∇(f (φ)v) +f2(φ)v +2f(φ)∇φ·∇v+f(φ)v =()2v+4∇φ·∇v *+, - third order term +4∇2φ·∇2+4∇φ⊗∇φ·∇2+2f(φ) +2∇φ ·∇v+4∇φ·∇2φ·∇v+f(φ)∇φ·∇v+2∇f(φ)·∇v +4f(φ)∇φ·∇v+v(f (φ)) +2v∇φ·∇f(φ)+vf2(φ), where f(φ) =φ +|∇φ|2. The above calculation shows that if φis as above, then Ca3,a2,a1,a0=C0, where a3=4iδ(∇φ), and a2, a1, a0can be read off from the computation above. Such calculations can also be done for m ≥3, but it will be somewhat cumbersome. 5.2. Summary of results Here we summarize few important results we have gotten so far from previous sections since they will help us and the reader when we apply them in this section. Lemma 5.1. Let aM∈C∞() be a symmetric Mtensor field where M≥0is an integer. 428 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 1. Let M=2m −1and assume that  Rn a2m−1 i1···i2m−1(e1+iη)i1···(e1+iη)i2m−1a0b0=0for all unit vectors η⊥e1 where Tma0=Tmb0=0. Then we have a2m−1=d2m−1φ+iδa2m−1,1, where a2m−1,1is a smooth symmetric tensor field of order 2m −3, and φ∈C∞() satisfies ∂l νφ|∂ =0for 0 ≤l≤2m −2. 2. Let M=2mand assume that  Rn a2m i1···i2m(e1+iη)i1···(e1+iη)i2mTa 0b0=0for all unit vectors η⊥e1 where Tm+1a0=Tm+1b0=0. Then we have a2m=d2mφ1+iδa2m,1, where a2m,1is a smooth symmetric tensor field of order 2m −2, and φ1∈C∞() satisfies ∂l νφ1|∂ =0for 0 ≤l≤2m −1. Proof. For first part we refer Step 2-5 in the proof of Proposition 4.3. One can employ similar line of arguments for second part as well.  Lemma 5.2. Let albe the same as above such that l≤m. Suppose there holds m  j=1  aj i1···ij(e1+iη)i1···(e1+iη)ijTm−ja0b0=0for all unit vectors η⊥e1 where Tma0=Tmb0=0. Then for some tensor aj,1of order j−2we have aj=djφj+iδaj,1where φjmatches the boundary condition ∂lφj|∂ =0for 0≤l≤j−1. Note that, a1,1=0. Proof. The proof relies on the careful choice of solution of transport equations a0and b0along with second condition of Lemma 6.7 and Lemma 5.1. The proof of Theorem 2.1 which uses the first hypothesis of Lemma 6.7, involves a similar argument; see in Step 2from equation (3.13)-(3.16).  5.3. Proof of Theorem 2.3 Proof. The proof of Theorem 2.3 will be divided into several steps. Note that, for simplicity, in the proof we may use same symbols such as φ0, φ1, to express tensor fields of different order. Step 1. The case of m =2. 429 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 By the assumption of Theorem 2.3 we have   3  l=0 al i1···ilDi1···iluv=0 whenever 2u=2v=0. We insert CGO solutions of uand vgiven in (A.5) and derive   3  l=0 al i1···ilDi1···ile (e1+iη)·x h˜ Ae−(e1+iη)·x h˜ B=0.(5.1) To continue we then multiply (5.1) by h3and let h →0to obtain   a3 i1···i3(e1+iη)i1···(e1+iη)i3a0b0=0. Here a0and b0solve T2a0=T2b0=0. This together with Proposition 4.3 entails a3=d3φ0+iδa3,1where φ0fulfils ∂l νφ0|∂ =0for0≤l≤2 (5.2) where a3,1is a smooth vector field in . Recall that, am,k denotes a symmetric tensor field of order m −2kfor every non-negative integers mand kfor which m −2k≥0. After that, we see that the coefficient of h3is indeed zero by plugging (5.2)into (5.1). However, it can be seen from the discussion at the beginning of this section that we have not yet obtained the exact gauge, which is iδ(∇φ), for some scalar φ. This differs from inverse problems involving second order elliptic partial differential equations, where the gauge can be obtained in a single step by multiplying a particular integral equation by hand then letting h →0. The form of a3presented in (5.2) can eliminate the coefficient of h3, but not the one corresponding to h2. As a result, multiplying (5.1) by h2and setting h →0 yields 3  a3 i1···i3(e1+iη)i1(e1+iη)i2∂xi3a0b0+  a2 i1i2(e1+iη)i1(e1+iη)i2a0b0=0. This together with (5.2) implies 0=3  ∂3 i1i2i3φ0(e1+iη)i1(e1+iη)i2∂xi3a0b0+  a2 i1i2(e1+iη)i1(e1+iη)i2a0b0 +  a3,1 i(e1+iη)iTa 0b0.(5.3) Combining integration by parts, boundary conditions of φ0and T2a0=T2b0=0 from above we derive 430 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Lemma 6.1. [5, Theorem 4.5] Let F∈Sm(E)and m ≥0. Then JkF=0holds for all integers kwith 0≤k≤m=⇒ F=0. Lemma 6.2. [5, Lemma 4.8] Let F∈Sm(E), then the operator JkFsatisfies the following relation ξ, ∂ ∂xpJkF=(−1)pk pp!Jk−pFif p≤k 0if p>k. After taking restrictions on Rn×Sn−1this gives ξ, ∂ ∂xpIkF=(−1)pk pp!Ik−pFif p≤k 0if p>k. This lemma entails that highest order MRT (J mF) uniquely determines all the lower order MRT (J kF) 0 ≤k≤m. Analogous result holds if one replaces JkFby IkF. Consequently, one can obtain next results. Lemma 6.3. Suppose F∈Sm(E)and m ≥0. Then JmF=0 =⇒ F=0. Lemma 6.4. [5, Theorem 4.18] Let m ≥2and F= m  l=0 f(l) ∈Sm(E). Then ImF=JmF|Rn×Sn−1=0 if and only if f2m 2=−iδ⎛ ⎝m 2  l=1 im 2−l δf(2l−2)⎞ ⎠and f2m−1 2+1=−iδ⎛ ⎜ ⎜ ⎝m−1 2  l=1 im 2−l δf(2l−1)⎞ ⎟ ⎟ ⎠. Moreover, ImF=0if and only if F=0for m =0, 1. Lemma 6.5. The following equality Ik(ip δf(l))(x, ξ) =Ikf(l)(x, ξ) holds for any f(l) ∈Sland for all integers pwith p≥1. For p=0this holds trivially because i0 δis the identity operator. Remark 6.6. We know that any symmetric m-tensor field in Rnhas m+n−1 mdistinct components. Suppose Fis the same as in (6.1). Then Fhas a total of m  l=0l+n−1 l=m+n mdistinct components. As a consequence, recovering Fin Rnis equivalent to recovering a symmetric 437 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 m-tensor field in Rn+1. However to recover a symmetric m-tensor field in Rn+1, MRT information is required on the tangent bundle of unit sphere which is a 2(n +1) −2 =2ndimensional data set. As a result, the kernel of JmFis trivial (see Lemma 6.3), whereas the kernel of ImF is nontrivial when m ≥2(see Lemma 6.4) because ImFis specified on the unit sphere bundle. Next we prove the following result, which was used to segregate MRT of different order tensor fields in the previous sections. Lemma 6.7. Let Fm∈Sm(E)and m ≥0. 1. Suppose cmIk+mFm+···+c1Ik+1F1+c0IkF0=0 holds for certain non-zero constants cjwhere 0 ≤j≤mand for all 0 ≤k≤m. Then IpFp=0for p=0,1,···,m. (6.3) 2. Additionally, if we assume dmIk+m−1Fm+···+d1IkF1=0 holds for certain non-zero constants djwhere 0 ≤j≤m −1and for all 0 ≤k≤m −1. Then IpFp+1=0for p=0,1,···,m−1.(6.4) Proof. We only give the proof of (6.3) and that of (6.4) follows similarly. We have that cmIk+mFm+···+c1Ik+1F1+c0IkF0=0. Applying ξ, ∂xkto the above equation and then using Lemma 6.2 we obtain cmk+m kImFm+cm−1k+m−1 kIm−1Fm−1+··· +c1k+1 kI1F1+c0k kI0F0=0, for all kwith 0 ≤k≤m. The above relation can be written as the matrix equation AX =0 where 438 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 X=(ImFm,···,I0F0)t, A= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ cmcm−1··· c1c0 cmm+1 1cm−1m 1··· c12 1c0 cmm+2 2cm−1m+1 2··· c13 2c0 . . .. . .··· . . .. . . cm2m mcm−12m−1 m··· c1m+1 mc0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . We now complete the proof by showing detA = 0. This follows from Lemma 6.8. This implies X=0, i.e. IpFp=0forp=0,1,··· ,m.  Lemma 6.8. detA =(−1)m(m+1) 2c0···cm. Proof. We have detA=c0···cmdetAm+1 where Am+1= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 11··· 11 m+1 1 m 1··· 2 11 m+2 2 m+1 2··· 3 21 . . .. . .··· . . . 2m m 2m−1 m··· m+1 m1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . We now subtract the (m −k)-th column from the (m −k−1)-th column for 1 ≤k≤m −2to obtain Am+1−→ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 10··· 0 m+1 1−1··· −1 m+2 2−m+1 1··· −2 1 . . .. . .··· . . . 2m m−2m−1 m−1··· −m m−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . This implies detAm+1=(−1)mdetAm, where Amis obtained from Am+1by considering first mrows and mcolumns respectively. Proceeding in this way after finally many steps we obtain detAm+1=(−1)m+(m−1)+···+2+1. This implies detA =(−1)m(m+1) 2c0···cm. 7. A new decomposition of symmetric tensor fields We now prove a suitable trace free Helmholtz type decomposition of symmetric tensor fields, which we used in the previous section to deal with partial data MRT. 439 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Lemma 7.1. Let fbe a smooth symmetric mtensor field in . Then the following decomposition holds: f=˜ f+iδv+dmφ, with jδ˜ f=δm˜ f=0 and ∂l νφ|∂ =0, l=0, 1, ···, m −1, where ˜ f∈Sm, v∈Sm−2and φ∈S0are smooth symmetric tensor fields in . This is a generalization of the decomposition shown in [9]. In [26], they proved certain decomposition of symmetric tensors. Remark 7.2. For m =1, this is the well known Helmholtz (or solenoidal) decomposition. Proof. We closely follow the arguments used in [9]. We first assume that fcan be written in the given form as f=˜ f+iδv+dmφ, (7.1) and we derive an equation for φ. Applying jδto (7.1) and using jδ˜ f=0we get jδf=jδiδv+jδdmφ. Since jδiδis invertible by [9, Lemma 2.3], we obtain v=(jδiδ)−1(jδf−jδdmφ). (7.2) This together with (7.1) implies f=˜ f+iδ((jδiδ)−1(jδf−jδdmφ)) +dmφ. (7.3) Denote q=iδ(jδiδ)−1jδand p=(Id −q). From [9, equation (2.15)] we have that pis the projection to the trace free component of a symmetric tensor and that p(iδv) =0for any v. Thus from (7.3)we obtain pf =˜ f+pdmφ. Applying δmto the above identity and using δm˜ f=0we obtain δmpdmφ=δmpf. Hence, φsolves the following boundary value problem. (−1)mδmpdmφ=(−1)mδmpf in  ∂l νφ=0on∂ for l=0,1,···,m−1.(7.4) 440 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 We next show that δmpdmacting on scalar functions in is a strongly elliptic operator in the sense of [33, Formula 11.79, Section 5.11]. Then by [33, Exercise 3, Section 5.11] the map H2m() ∩Hm 0() −→ L2() φ−→ δmpdmφ (7.5) will be Fredholm operator of index zero. To prove the ellipticity of δmpdmwe follow [9, Section 5] and introduce the following. For x∈Rn, we define the symmetric multiplication operators ix:Sm→Sm+1by (ixf) i1i2...im+1=σ(i1,...,i m,i m+1)(xim+1fi1i2...im). We also define the dual of the operator ix, the contraction operator jx:Sm→Sm−1by (jxf) i1i2...im−1=fi1i2...imxim. Similarly, ix⊗kand jx⊗kare defined by taking the composition of ixand jxwith itself k times, respectively. The principal symbol of δmpdmat (x, ξ) ∈ ×(Rn\{0})is given by (−1)mjξ⊗mpiξ⊗m, where (i)mjξ⊗mand (i)miξ⊗mare the principal symbols of δmand dmrespectively. Thus the principal symbol of (−1)mδmpdmis jξ⊗mpiξ⊗m, which is real valued and non-negative since for any φ jξ⊗mpiξ⊗mφ,φ=piξ⊗mφ,iξ⊗mφ=piξ⊗mφ,piξ⊗mφ. Also by Lemma 7.3 we obtain jξ⊗mpiξ⊗m= 0for ξ= 0. Thus we have shown the ellipticity of the boundary value problem (7.4). By elliptic regularity any element in the kernel of (7.4) will be smooth. Now (−1)mδmpdmφ=0in and ∂l νφ|∂ =0for 0 ≤l≤m −1gives (−1)mδmpdmφ,φ=pdmφ,pdmφ=0 using integration by parts. This implies pdmφ=0. From [9, Lemma 3.4, Equation 3.8] we have that dpf =pdf+m n+2m−2iδδpf for f∈Sm. Replacing fby dm−1φentails dpdm−1φ=m−1 n+2m−4iδδpdm−1φ. Denote u =pdm−1φand v=m−1 n+2m−4δpdm−1φ. Note that, trace of u =0. This gives du=iδv. Hence uis a trace free conformal Killing tensor field. Now using the boundary conditions ∂l νφ|∂ =0for 0 ≤l≤m −1we conclude u|∂ =0. By [9, Theorem 1.3] we have u =0. Repeating this process finitely many times we will obtain pdφ=dφ=0. Since φ|∂ =0, this 441 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 implies that φ=0in . Thus the boundary value problem has (7.4) has zero kernel. Since it has Fredholm index zero, this says (7.4)also has zero co-kernel. This shows that the mapping (7.5) is an isomorphism. Finally, given f∈Sm, the fact that (7.5)is an isomorphism together with ellipticity shows that there exists a unique φ∈C∞() solving (7.4). We then define vby (7.2), which implies that iδv=qf −qdmφ. We also define ˜ f=f−iδv−dmφ. It follows that ˜ f=f−(qf −qdmφ) −dmφ=pf −pdmφ. Then jδ˜ f=0 and δm˜ f=δmpf −δmpdmφ=0. This proves the required decomposition.  Lemma 7.3. For any ξ= 0, if piξ⊗mφ=0, then φ=0. Proof. For m =1we have piξφ=iξφ. Since ξ= 0 and (iξφ)k=ξkφ, we obtain φ=0. We proceed by induction and assume that ξ= 0,pi ξ⊗mφ=0=⇒ φ=0. Suppose that piξ⊗m+1φ=0. Then by [9, Lemma 5.3] we have iξpiξ⊗mφ−2 m+1iδ(jδiδ)−1jξpiξ⊗mφ=0. Denote f=piξ⊗mφ. Taking the inner product of the above equation with iξfgives iξf,iξf− 2 m+1iδ(jδiδ)−1jξf,iξf=0 =⇒  jξiξf,f − 2 m+1(jδiδ)−1jξf,jδiξf=0.(7.6) From [30, Lemma 3.3.3] we have jξiξf=|ξ|2 m+1f+m m+1iξjξf. Combining jδf=0 with the formula jδiξ=2 m+1jξ+m−1 m+1iξjδon Smgiven in [9, Equation 5.6], we get jδiξf=2 m+1jξf. From [9, Equation 5.13], which can be used since jδf=0, we obtain 442 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 (jδiδ)−1jξf=m(m +1) 2(n +2m−2)jξf. The combination of last two displayed equations and (7.6)give |ξ|2|f|2+m1−2 n+2m−2|jξf|2=0. Since 1 −2 n+2m−2≥0for n ≥2 and m ≥1, this implies f=piξ⊗mφ=0. Hence by induction we have φ=0.  Data availability The authors are unable or have chosen not to specify which data has been used. Acknowledgments Both authors were partly supported by the Academy of Finland (Centre of Excellence in Inverse Modelling and Imaging, grant 284715) and by the European Research Council under Horizon 2020 (ERC CoG 770924). Appendix A. Construction of special solutions In this section, we describe the construction of special solutions known as complex geometric optic (CGO) solutions of polyharmonic operators that we have already utilized in the earlier sections. A.1. Carleman estimate Let  ⊂Rn, n ≥3 be a bounded domain with smooth boundary. In this section, we construct Complex Geometric Optics (CGO) solutions for (1.1) following a Carleman estimate approach from [6,10]. We first introduce semiclassical Sobolev spaces. For h >0be a small parameter, we define the semiclassical Sobolev space Hs scl(Rn), s∈Ras the space Hs(Rn)endowed with the semiclassical norm u2 Hs scl(Rn)=hDsu2 L2(Rn),ξ=(1+|ξ|2)1 2. For open sets  ⊂Rnand for non-negative integers m, the semiclassical Sobolev space Hm scl() is the space Hm() endowed with the following semiclassical norm u2 Hm scl() = |α|≤m (hD)αu2 L2(). These two norms are equivalent when  =Rnand for every integers m ≥0. Let  be an open subset containing in its interior and let ϕ∈C∞( ) with ∇ϕ= 0in . We consider the semi-classical conjugated Laplacian P0,ϕ =eϕ h(−h2)e −ϕ h, where 0 <h 1. The 443 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 semi-classical principal symbol of this operator P0,ϕ is given by p0,ϕ(x, ξ) =|ξ|2−|∇ xϕ|2+ 2iξ·∇ xϕ. Definition A.1 ([22]). We say that ϕ∈C∞( ) is a limiting Carleman weight for P0,ϕ in if ∇ϕ= 0in and Re(p0,ϕ), Im(p0,ϕ )satisfies Re(p0,ϕ), Im(p0,ϕ)!(x, ξ) =0 whenever p0,ϕ(x, ξ) =0for(x, ξ) ∈ ×(Rn\{0}), where {·, ·} denotes the Poisson bracket. Examples of such ϕare linear weights ϕ(x) =α·x, where 0 = α∈Rnor logarithmic weights ϕ(x) =log|x−x0|with x0/∈ . As mentioned already, we consider the limiting Carleman weight in this paper to be ϕ(x) =x1. Proposition A.2 (Interior Carleman estimate). Let ϕ(x) be a limiting Carleman weight for the conjugated semiclassical Laplacian. Then there exists a constant C=C,Aj,q such that for 0 <h 1, we have hmuL2() ≤Ch2meϕ h(−)me−ϕ huH−2m scl ,for all u∈C∞ 0(). This follows by iterating a Carleman estimate for the semiclassical Laplacian with a gain of two derivatives proved in [31]. We omit the proof here. We refer the reader to [19,18,15]. A.2. Construction of CGO solutions Next we use Proposition A.2 to construct CGO solutions for the equation (−)mu =0. To this end, we state an existence result whose proof is standard; see [10,18]for instance. Proposition A.3. Let ϕbe as defined in Proposition A.2. Then for any v∈L2() and small enough h >0one has u ∈H2m() such that e−ϕ h(−)meϕ hu=vin , with uH2m scl () ≤ChmvL2(). We now use Proposition A.3 to construct a solution for (−)mu =0of the form u(x;h) =eϕ+iψ ha0(x) +ha1(x) +···+hm−1am−1(x) +r(x;h),(A.1) =eϕ+iψ h(A(x;h) +r(x;h)), where A(x;h) = m−1  j=0 hjaj(x). Here {aj(x)}and r(x; h) will be determined later. We choose ψ(x) ∈C∞() in such a way that p0,ϕ(x, ∇ψ) =0. This implies |∇ϕ| =|∇ψ|and ∇ϕ·∇ψ=0in . We calculate the term (−)meϕ+iψ hA(x; h) in . Due to the choices of ϕand ψwe have ∇(ϕ +iψ) ·∇(ϕ +iψ) =0 in and thus we obtain 444 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 e−ϕ+iψ h(−)meϕ+iψ hA(x;h) =−1 hT−m A(x;h) (A.2) where T=2∇x(ϕ +iψ)·∇ x+x(ϕ +iψ). (A.3) We now make the coefficient of h−m+jto be 0for 0 ≤j≤m −1 and obtain the following system of transport equations. Tma0(x) =0in, Tmaj(x) =−j k=1Mkaj−k(x) in , and 1 ≤j≤m−1,(A.4) where Mj’s are certain differential operator of order m +j(1 ≤j≤m −1)and can be computed from (A.2). It is well known that equations in (A.4)have smooth solutions; see for instance [10]. We provide an explicit form for the smooth solution a0solving (A.4)for our inverse problem, which will be effective in getting the generalized MRT of the coefficients. Using a0(x), ··· , am−1(x) ∈ C∞() satisfying (A.4), we see that e−ϕ+iψ h(−)meϕ+iψ ha(x;h) ≃O(1). Now if u(x; h) as in (A.1)is a solution of (−)mu(x; h) =0in , we see that 0=e−ϕ+iψ h(−)mu=e−ϕ+iψ h(−)meϕ+iψ h(A(x;h) +r(x;h)). This implies e−ϕ+iψ h(−)meϕ+iψ hr(x;h) =F(x;h), for some F(x;h) ∈L2(), for all h>0small. By our choices, aj(x) annihilates all the terms of order h−m+jin e−ϕ+iψ hL(x; D)eϕ+iψ ha(x; h) in for j=0, ..., m −1. Thus we get F(x, h)L2() ≤C, where C>0is uniform in hfor h 1. Using Proposition A.3 we have the existence of r(x; h) ∈H2m() solving e−ϕ+iψ h(−)meϕ+iψ hr(x;h) =F(x;h), with the estimate r(x;h)H2m scl () ≤Chm,for h>0 small enough. Similarly, we can construction CGO solution of the adjoint equation (e−ϕ+iψ h(−)meϕ+iψ h)∗u = eϕ+iψ h(−)me−ϕ+iψ hu =0. We now sum up the above calculation in the next lemma. 445 S.K. Sahoo and M. Salo Journal of Differential Equations 360 (2023) 407–451 Lemma A.4. Let h >0small enough and ϕ, ψ∈C∞() satisfy p0,ϕ(x, ∇ψ) =0. There are suitable choices of a0(x), ..., am−1(x), b0(x), ..., bm−1(x) ∈C∞() and r(x; h), r(x; h) ∈ H2m() such that u(x;h) =eϕ+iψ ha0(x) +ha1(x) +···+hm−1am−1(x) +r(x;h)=eϕ+iψ h˜ A, v(x;h) =e−ϕ+iψ hb0(x) +hb1(x) +···+hm−1bm−1(x) +r(x;h)=e−ϕ+iψ h˜ B solving (−)mu(x) =(−)mv(x) =0in for h >0small enough, with the estimates r(x; h)H2m scl (), r(x; h)H2m scl () ≤Chm. Moreover, a0(x) and b0(x) solve the transport equations Tma0=Tmb0=0. A.3. Solutions with linear phase Now we choose a suitable coordinate system and express the solutions in that system. We choose ϕ(x) =x·e1=x1, where e1=(1, 0, 0 ··· , 0). Choose an orthonormal frame {η1= e1, η2, ··· , ηn}, where {η2, ··· , ηn}are unit vectors on the hyperplane perpendicular to e1. Then we choose ψ(x) =x·η2. We denote y=(y1, y2, ··· , yn)is the coordinate with respect to new basis and y=(y2, ···, yn), y =(y3, ··· , yn). With this choice of ϕand ψthe solutions in Lemma A.4 take the form u(x;h) =e1 h(e1+iη2)·xa0(x) +ha1(x) +···+hm−1am−1(x) +r(x;h)=e1 h(e1+iη2)·x˜ A, v(x;h) =e−1 h(e1+iη2)·xb0(x) +hb1(x) +···+hm−1bm−1(x) +r(x;h)=e−1 h(e1+iη2)·x˜ B (A.5) The transport equation Tma0=0 becomes Tma0=2m(∂y1+i∂y2)ma0=0. Denote the complex variable z=y1+iy2. Then the above transport equation reduces to ∂m ¯za0=0. A complex valued function satisfying ∂m ¯za0=0is known as poly-analytic function; see [1]for more details. The general solution of ∂m ¯za0=0is given in the following lemma. Lemma A.5. The general solution of ∂m ¯za0=0is given by a0= m−1  k=0 (z −¯z)kfk(z), where fkis a holomorphic function for all 0 ≤k≤m −1. We do not give the proof of this lemma as it follows from standard induction argument; see [5, Lemma 2.6]. This immediately gives the following particular solution of Tma0=Tmb0=0 having the form 446