Inverse problems for the perturbed polyharmonic operator with coefficients in Sobolev spaces with non-positive order
aa r X i v : . [ m a t h . A P ] M a r INVERSE PROBLEMS FOR THE PERTURBEDPOLYHARMONIC OPERATOR WITH COEFFICIENTS INSOBOLEV SPACES WITH NON-POSITIVE ORDER
YERNAT M. ASSYLBEKOV
Abstract.
We show that the knowledge of the Dirichlet-to-Neumann mapon the boundary of a bounded open set in R n , n ≥
3, for the perturbedpolyharmonic operator ( − ∆) m + A · D + q , m ≥
2, with n > m , A ∈ W − m − , nm and q ∈ W − m + δ, nm , with 0 < δ < /
2, determines the potentials A and q in the set uniquely. The proof is based on a Carleman estimate with linearweights and with a gain of two derivatives and on the property of products offunctions in Sobolev spaces. Introduction
Let Ω ⊂ R n , n ≥
3, be a bounded open set with C ∞ boundary. Consider thepolyharmonic operator ( − ∆) m , where m ≥ − ∆) m ispositive and self-adjoint on L (Ω) with domain H m (Ω) ∩ H m (Ω), where H m (Ω) = { u ∈ H m (Ω) : γu = 0 } . This operator can be obtained as the Friedrichs extension starting from the spaceof test functions. This fact can be found, for example, in [10]. Here and in whatfollows, γ is the Dirichlet trace operator γ : H m (Ω) → m − Y j =0 H m − j − / ( ∂ Ω) , γu = ( u | ∂ Ω , ∂ ν u | ∂ Ω , . . . , ∂ m − ν u | ∂ Ω ) , where ν is the unit outer normal to the boundary ∂ Ω, and H s (Ω) and H s ( ∂ Ω) arethe standard L -based Sobolev spaces on Ω and its boundary ∂ Ω for s ∈ R .Throughout the paper we shall assume that n > m .Let A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ), where E ′ (Ω) = { u ∈ D ′ ( R n ) : supp( u ) ⊆ Ω } and W s,p ( R n ) is the standard L p -basedSobolev space on R n , s ∈ R and 1 < p < ∞ , which is defined by the Besselpotential operator. Thus W s,p ( R n ) is the space of all distributions u on R n suchthat J − s u ∈ L p ( R n ), where J s is the operator defined as u ((1 + | · | ) − s/ b u ( · )) ∨ . In the case s ≥ W s,p ( R n ) coincides with the space of all functions whoseall derivatives of order less or equal to s is in L p ( R n ). The reader is referred to [33]for properties of these spaces. Before stating the problem, we consider the bilinear forms B A and b q on H m (Ω)which are defined by B A ( u, v ) = h A, ˜ v D ˜ u i R n , b q ( u, v ) = h q, ˜ u ˜ v i R n , u, v ∈ H m (Ω) , where h· , ·i R n is the distributional duality on R n , and ˜ u, ˜ v ∈ H m ( R n ) are extensionsof u and v , respectively. In Appendix A, we show that these definitions are well-defined (i.e. independent of the choice of extensions ˜ u, ˜ v ). Using a property onmultiplication of functions in Sobolev spaces, we show that the forms B A and b q are bounded on H m (Ω); see Proposition A.2.Consider the operator D A which is formally A · D , where D j = − i∂ x j , and theoperator m q of multiplication by q . To be precise, for u ∈ H m (Ω), D A ( u ) and m q ( u ) are defined as h D A ( u ) , ψ i Ω = B A ( u, ψ ) and h m q ( u ) , ψ i Ω = b q ( u, ψ ) , ψ ∈ C ∞ (Ω) . Then the operators D A and m q are bounded H m (Ω) → H − m (Ω) (see Corol-lary A.3), and hence, standard arguments show that the operator L A,q = ( − ∆) m + D A + m q : H m (Ω) → H − m (Ω) = ( H m (Ω)) ′ , is Fredholm operator with zero index; see Appendix B.For f = ( f , . . . , f m − ) ∈ Q m − j =0 H m − j − / ( ∂ Ω), consider the Dirichlet problem L A,q u = 0 in Ω ,γu = f on ∂ Ω . (1.1)If 0 is not in the spectrum of L A,q , the Dirichlet problem (1.1) has a unique solution u ∈ H m (Ω). We define the Dirichlet-to-Neumann map N A,q as follows hN A,q f, h i ∂ Ω := X | α | = m m ! α ! ( D α u, D α v h ) L (Ω) + B A ( u, v h ) + b q ( u, v h ) , (1.2)where h = ( h , . . . , h m − ) ∈ Q m − j =0 H m − j − / ( ∂ Ω), and v h ∈ H m (Ω) is an extensionof h , that is γv h = h . It is shown in Appendix B that N A,q is a well-defined (i.e.independent of the choice of v h ) bounded operator N A,q : m − Y j =0 H m − j − / ( ∂ Ω) → m − Y j =0 H m − j − / ( ∂ Ω) ′ = m − Y j =0 H − m + j +1 / ( ∂ Ω) . The inverse boundary problem for the perturbed polyharmonic operator L A,q is todetermine A and q in Ω from the knowledge of the Dirichlet-to-Neumann map N A,q .When m = 1 the operator L A,q is the first order perturbation of the Laplacianand N A,q u is formally given by N A,q f = ( ∂ ν u + i ( A · ν ) u ) | ∂ Ω , where u is an H m (Ω)solution to the equation L A,q u = 0. It was shown in [29] that in this case there is anobstruction to uniqueness in this problem given by the following gauge equivalenceof the set of the Cauchy data: if ψ ∈ W , ∞ in a neighbourhood of Ω and ψ | ∂ Ω = 0,then C A,q = C A + ∇ ψ,q ; see also [19, Lemma 3.1]. Hence, given C A,q , we may only
NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 3 hope to recover the magnetic field dA and electric potential q . Here and in whatfollows the magnetic field dA is defined by dA = X ≤ j,k, ≤ n (cid:0) ∂ x j A k − ∂ x k A j (cid:1) dx j ∧ dx k . Due to the lack smoothness of A , this definition is in the sense of distributions.Starting with the paper of Sun [29], inverse boundary value problems for the mag-netic Schr¨odinger operators have been extensively studied. It was shown in [29] thatthe hope mentioned above is justified provided that A ∈ W , ∞ , q ∈ L ∞ and dA satisfies a smallness condition. The smallness condition was removed in [22] for C ∞ magnetic and electric potentials, and also for compactly supported C magneticpotentials and essentially bounded electric potentials. The regularity assumptionon magnetic potentials were subsequently weakened to C in [32], and then to Dinicontinuous in [27]. All these results were obtained under the assumption that zerois not a Dirichlet eigenvalue for the magnetic Schr¨odinger operator in Ω. Thereare two best result by now. One is due to Krupchyk and Uhlmann [19], wherethey prove uniqueness under the assumption that magnetic and electric potentialsare of class L ∞ . Another is due to Haberman [12] in dimension n = 3, where theuniqueness is shown for the case when q ∈ W − , and A ∈ W s, for some s > m = 1 can be eliminated by considering operators of higher order.More precisely, they show that for m ≥ C A,q determines A and q uniquely provided that A ∈ W , ∞ (Ω , C n ) ∩ E ′ (Ω , C n ) and q ∈ L ∞ (Ω). Theyalso show that the uniqueness result holds without the assumption A = 0 on ∂ Ωbut for C ∞ magnetic and electric potentials. This is also true for A ∈ W , ∞ (Ω , C n )and q ∈ L ∞ (Ω) when the boundary of the domain Ω is connected.The purpose of this paper is to relax the regularity assumption on A from W , ∞ to W − m − , nm class and q from L ∞ to W − m + δ, nm , 0 < δ < /
2, for the perturbedpolyharmonic operator L A,q with m ≥
2. Therefore, throughout the paper weassume that m ≥
2. Our main result is as follows.
Theorem 1.1.
Let Ω ⊂ R n , n ≥ , be a bounded open set with C ∞ bound-ary, and let m ≥ be an integer such that n > m . Suppose that A , A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω) and q , q ∈ W − m + δ, nm ( R n , C ) ∩E ′ (Ω) , for some < δ < / , are such that is not in the spectrums of L A ,q and L A ,q . If N A ,q = N A ,q ,then A = A and q = q . The assumption n > m is related to the dual space of W s, nm and the estimate onproducts of functions in different Sobolev spaces. It seems to the author that thetechniques of the present paper can be adopted to the case n ≤ m by changingregularity assumptions. We hope to consider this problem in future work.The key ingredient in the proof of Theorem 1.1 is a construction of complex geo-metric optics solutions for the operator A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and YERNAT M. ASSYLBEKOV q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ). For this, we use the method of Carleman estimateswhich is based on the corresponding Carleman estimate for the Laplacian, with again of two derivatives, due to Salo and Tzou [28]. Another important tool in ourproof is the property of products of functions in Sobolev spaces [26]. This was usedin the paper of Brown and Torres [2]. The idea of constructing such solutions forthe Schr¨odinger operator goes back to the fundamental paper due to Sylvester andUhlmann [30]. Such solutions were first introduced in [6] in the setting of quantuminverse scattering problem.The inverse boundary value problem of the recovery of a zeroth order perturbation ofthe biharmonic operator, that is when m = 2, has been studied by Isakov [15], whereuniqueness result was obtained, similarly to the case of the Schr¨odinger operator.In [14], the uniqueness result was extended to q ∈ L n/ (Ω), n > q ∈ L n/ m , n > m by Krupchyk and Uhlmann [18]. In the case m = 1, that isfor zeroth order perturbation of the Schr¨odinger operator, global uniqueness resultwas established by Lavine and Nachman [20] for q ∈ L n/ (Ω), following an earlierresult of Chanillo [5] for q ∈ L n/ ε (Ω), ε > q ∈ L ∞ (Ω).Higher ordered polyharmonic operators occur in the areas of physics and geometrysuch as the study of the Kirchhoff plate equation in the theory of elasticity, and thestudy of the Paneitz-Branson operator in conformal geometry; for more details seemonograph [8].We would like to remark that the problem considered in this paper can be con-sidered as generalization of the Calder´on’s inverse conductivity problem [3], knownalso as electrical impedance tomography, for which the reduction of regularity havebeen studied extensively. In the fundamental paper by Sylvester and Uhlmann [30]it was shown that C conductivities can be uniquely determined from boundarymeasurements. The regularity assumptions were weakened to C / ε conductivi-ties by Brown [1], and corresponding result for C / conductivities was obtainedby P¨aiv¨arinta, Panchenko and Uhlmann [24]. Uniqueness result for C ε conormalconductivities was shown by Greenleaf, Lassas and Uhlmann [9]. There is a recentwork by Haberman and Tataru [13] which gives a uniqueness result for Calder´onsproblem with C conductivities and with Lipschitz continuous conductivities, whichare close to the identity in a suitable sense. Very recent work of Caro and Rogers[4] shows that Lipschitz conductivities can be determined from the Dirichlet-to-Neumann map. Finally, Haberman [11] gives uniquess results for conductivitieswith unbounded gradient. In particular, uniqueness for conductivities in W ,n (Ω)with n = 3 , L A,q with A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ). Thenthe proof of Theorem 1.1 is given in Section 3. In Appendix A, we study mapping NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 5 properties of the operators D A and m q . Finally, Appendix B is devoted to the well-posedness of the Dirichlet problem for L A,q with A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω , C n )and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ).2. Carleman estimates and Complex geometric optics solutions
In this section we construct the complex geometric optics solutions for the equation L A,q u = 0 in Ω with A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩E ′ (Ω , C ), m ≥
2. When constructing such solutions, we shall first derive Carlemanestimates for the operator L A,q .We start by recalling the Carleman estimate for the semiclassical Laplace operator − h ∆ with a gain of two derivatives, established in [28]. Let e Ω be an open set in R n such that Ω ⊂⊂ e Ω and let ϕ ∈ C ∞ ( e Ω , R ). Consider the conjugated operator P ϕ = e ϕ/h ( − h ∆) e − ϕ/h , and its semiclassical principal symbol p ϕ ( x, ξ ) = ξ + 2 i ∇ ϕ · ξ − |∇ ϕ | , x ∈ e Ω , ξ ∈ R n . Following [16], we say that ϕ is a limiting Carleman weight for − h ∆ in e Ω, if ∇ ϕ = 0in e Ω and the Poisson bracket of Re p ϕ and Im p ϕ satisfies { Re p ϕ , Im p ϕ } ( x, ξ ) = 0 when p ϕ ( x, ξ ) = 0 , ( x, ξ ) ∈ e Ω × R n . In this paper we shall consider only the linear Carleman weights ϕ ( x ) = α · x , α ∈ R n , | α | = 1.In what follows we consider the semiclassical norm on the standard Sobolev space H s ( R n ), s ∈ R , k u k H s scl ( R n ) = kh hD i s u k L ( R n ) , h ξ i = (1 + | ξ | ) / . Our starting point is the following Carleman estimate for the semiclassical Laplaceoperator − h ∆ with a gain of two derivatives, which is due to Salo and Tzou [28]. Proposition 2.1.
Let ϕ be a limiting Carleman weight for − h ∆ in e Ω , and let ϕ ε = ϕ + h ε ϕ . Then for < h ≪ ε ≪ and s ∈ R , we have h √ ε k u k H s +2scl ( R n ) ≤ C k e ϕ ε /h ( − h ∆) e − ϕ ε /h u k H s scl ( R n ) , C > , for all u ∈ C ∞ (Ω) . Next, we state theorem on products of functions in Sobolev spaces. This result iswell-known, see Theorem 2 in [26, Subsection 4.4.4].
Proposition 2.2.
Let < p, q < ∞ and < s ≤ s < n min(1 /p, /q ) . Then W s ,p ( R n ) · W s ,q ( R n ) ֒ → W s ,r ( R n ) where /r = 1 /p + 1 /q − s /n . Now we shall derive Carleman estimate for the perturbed operator L A,q with A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ). To that endwe shall iterate ( m times) inequality in Proposition 2.1 and use it with s = − m , YERNAT M. ASSYLBEKOV and with fixed ε > h . We have thefollowing result. Proposition 2.3.
Let ϕ be a limiting Carleman weight for − h ∆ in e Ω , and supposethat A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ) . Thenfor < h ≪ , we have k u k H m scl ( R n ) . h m k e ϕ/h ( h m L A,q ) e − ϕ/h u k H − m scl ( R n ) , (2.1) for all u ∈ C ∞ (Ω) .Proof. Iterating the Carleman estimate in Proposition 2.1 m times, m ≥
2, we getthe following Carleman estimate for the polyharmonic operator, h m ε m/ k u k H s +2 m scl ( R n ) ≤ C k e ϕ ε /h ( − h ∆) m e − ϕ ε /h u k H s scl ( R n ) , for all u ∈ C ∞ (Ω), s ∈ R and 0 < h ≪ ε ≪
1. We shall use this estimate with s = − m , and with fixed ε > h : h m ε m/ k u k H m scl ( R n ) ≤ C k e ϕ ε /h ( − h ∆) m e − ϕ ε /h u k H − m scl ( R n ) , (2.2)for all u ∈ C ∞ (Ω) and 0 < h ≪ ε ≪ H − m ( R n ) k v k H − m scl ( R n ) = sup = ψ ∈ C ∞ ( R n ) |h v, ψ i R n |k ψ k H m scl ( R n ) , (2.3)where h· , ·i R n is the distribution duality on R n .Let ϕ ε = ϕ + h ε ϕ be the convexified weight with ε > < h ≪ ε ≪ u ∈ C ∞ (Ω). Then for all 0 = ψ ∈ C ∞ ( R n ), by duality of the spaces W − m , nm ( R n , C ) and W m , n n − m ( R n , C ) and by Proposition 2.2, we have |h e ϕ ε /h h m m q ( e − ϕ ε /h u ) , ψ i R n | ≤ Ch m k q k W − m , nm ( R n ) k uψ k W m , n n − m ( R n ) ≤ Ch m k q k W − m , nm ( R n ) k u k H m ( R n ) k ψ k H m ( R n ) ≤ Ch m k q k W − m , nm ( R n ) k u k H m ( R n ) k ψ k H m ( R n ) ≤ Ch m k q k W − m , nm ( R n ) k u k H m scl ( R n ) k ψ k H m scl ( R n ) . Therefore, by (2.3), we obtain k e ϕ ε /h h m m q ( e − ϕ ε /h u ) k H − m scl ( R n ) ≤ Ch m k q k W − m , nm ( R n ) k u k H m scl ( R n ) . (2.4)For all 0 = ψ ∈ C ∞ (Ω), we can show |h e ϕ ε /h h m D A ( e − ϕ ε /h u ) , ψ i R n | = |h h m A, e ϕ ε /h ψD ( e − ϕ ε /h u ) i R n |≤ |h h m − A, e ϕ ε /h ψ [ − u (1 + hϕ/ε ) Dϕ + hDu ] i R n |≤ Ch m − k A k W − m − , nm ( R n ) k − u (1 + hϕ/ε ) Dϕψ + hDuψ k W m − , n n − m ( R n ) . NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 7
In the last step we used duality between W − m − , nm ( R n , C n ) and W m − , n n − m ( R n , C n ).In the case m ≥
3, we use Proposition 2.2 with p = q = 2, s = m − and s = m (this is exactly the moment where we need the stronger assumption m < n ratherthan m < n ), to get k − u (1 + hϕ/ε ) Dϕψ + hDuψ k W m − , n n − m ( R n ) ≤ C k − u (1 + hϕ/ε ) Dϕ + hDu k H m − ( R n ) k ψ k H m ( R n ) ≤ Ch m − k u k H m ( R n ) k ψ k H m ( R n ) ≤ Ch m − k u k H m scl ( R n ) k ψ k H m scl ( R n ) , for some constant C > ϕ . When m = 2, we use H¨older’s in-equality and Sobolev embedding H ( R n ) ⊂ L nn − ( R n ) (see [31, Chapter 13, Propo-sition 6.4]), and obtain k − u (1 + hϕ/ε ) Dϕψ + hDuψ k L n n − ( R n ) ≤ C k − u (1 + hϕ/ε ) Dϕ + hDu k L nn ( R n ) k ψ k L nn − ( R n ) ≤ C k − u (1 + hϕ/ε ) Dϕ + hDu k L ( R n ) k ψ k H ( R n ) ≤ Ch k u k H ( R n ) k ψ k H ( R n ) ≤ Ch k u k H ( R n ) k ψ k H ( R n ) , for some constant C > ϕ . Therefore, for m ≥
2, we get |h e ϕ ε /h h m D A ( e − ϕ ε /h u ) , ψ i R n | ≤ Ch m k A k W − m − , nm ( R n ) k u k H m scl ( R n ) k ψ k H m scl ( R n ) . Hence, by (2.3), we obtain k e ϕ ε /h h m D A ( e − ϕ ε /h u ) k H − m scl ( R n ) ≤ Ch m k A k W − m − , nm ( R n ) k u k H m scl ( R n ) . Combining these estimates with (2.2) and (2.4) we get that for small enough h > k u k H m scl ( R n ) . h m k e ϕ ε /h ( h m L A,q ) e − ϕ ε /h u k H − m scl ( R n ) . Using that e − ϕ ε /h u = e − ϕ/h e − ϕ / ε u we obtain (2.1). (cid:3) Let ϕ ∈ C ∞ ( e Ω , R ) be a limiting Carleman weight for − h ∆. Set L ϕ := e ϕ/h ( h m L A,q ) e − ϕ/h . Then by Proposition A.4 we have hL ϕ u, v i Ω = h u, L ∗ ϕ v i Ω , u, v ∈ C ∞ (Ω) , YERNAT M. ASSYLBEKOV where L ∗ ϕ = e − ϕ/h ( h L A,q + D · A ) e ϕ/h is the formal adjoint of L ϕ , and h· , ·i Ω is thedistribution duality on Ω. We have that L ∗ ϕ : C ∞ (Ω) → H − m ( R n ) ∩ E ′ (Ω)is bounded. Therefore, the estimate (2.1) holds for L ∗ ϕ , since − ϕ is a limitingCarleman weight as well.To construct the complex geometric optics solutions for the operator L A,q , we needto convert the Carleman estimate (2.1) for L ∗ ϕ into the following solvability result.The proof is essentially well-known, and we include it here for the convenience ofthe reader. In what follows, we shall write k u k H m scl (Ω) = X | α |≤ m k ( h∂ ) α u k L (Ω) , k v k H − m scl (Ω) = sup = φ ∈ C ∞ (Ω) |h v, φ i Ω |k φ k H m scl (Ω) = sup = f ∈ H m (Ω) |h v, f i Ω |k f k H m scl (Ω) . Proposition 2.4.
Let A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩E ′ (Ω , C ) , and let ϕ be a limiting Carleman weight for the semiclassical Laplacianon ˜Ω . If h > is small enough, then for any v ∈ H − m (Ω) , there is a solution u ∈ H m (Ω) of the equation e ϕ/h ( h m L A,q ) e − ϕ/h u = v in Ω , which satisfies k u k H m scl (Ω) . h m k v k H − m scl (Ω) . Proof.
Let v ∈ H − m (Ω) and let us consider the following complex linear functional, L : L ∗ ϕ C ∞ (Ω) → C , L ∗ ϕ,α w
7→ h w, v i Ω . By the Carleman estimate (2.1) for L ∗ ϕ , the map L is well-defined. Let w ∈ C ∞ (Ω).Then we have | L ( L ∗ ϕ w ) | = |h w, v i Ω | ≤ k w k H m scl ( R n ) k v k H − m scl (Ω) ≤ Ch m k v k H − m scl (Ω) kL ∗ ϕ w k H − m scl ( R n ) . By the Hahn-Banach theorem, we may extend L to a linear continuous functional˜ L on H − m ( R n ), without increasing its norm. By the Riesz representation theorem,there exists u ∈ H m ( R n ) such that for all ψ ∈ H − m ( R n ),˜ L ( ψ ) = h ψ, u i R n , and k u k H m scl ( R n ) ≤ Ch m k v k H − m scl (Ω) . Let us now show that L ϕ u = v in Ω. To that end, let w ∈ C ∞ (Ω). Then hL ϕ u, w i Ω = h u, L ∗ ϕ w i R n = ˜ L ( L ∗ ϕ w ) = h w, v i Ω = h v, w i Ω . The proof is complete. (cid:3)
NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 9
Our next goal is to construct the complex geometric optics solutions for the equation L A,q u = 0 in Ω with A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩E ′ (Ω , C ) using the solvability result Proposition 2.4. Complex geometric opticssolutions are the solutions of the following form, u ( x, ζ ; h ) = e ix · ζh ( a ( x, ζ ) + h m/ r ( x, ζ ; h )) , (2.5)where ζ ∈ C n such that ζ · ζ = 0, | ζ | ∼ a ∈ C ∞ (Ω) is an amplitude, r is acorrection term, and h > h m L A,q by e ix · ζ/h . We have e − ix · ζh h m L A,q e ix · ζh = ( − h ∆ − iζ · h ∇ ) m + h m D A + h m − m A · ζ + h m m q . (2.6)We shall consider ζ depending slightly on h , i.e. ζ = ζ + ζ with ζ being inde-pendent of h and ζ = O ( h ) as h →
0. We also assume that | Re ζ | = | Im ζ | = 1.Then we can write (2.6) as follows e − ix · ζh h m L A,q e ix · ζh = ( − h ∆ − iζ · h ∇ − iζ · h ∇ ) m + h m D A + h m − m A · ( ζ + ζ ) + h m m q . Then (2.5) is a solution to L A,q u = 0 if and only if e − ix · ζh h m L A,q ( e ix · ζh h m/ r ) = − e − ix · ζh h m L A,q ( e ix · ζh a )= − m X k =0 m ! k !( m − k )! ( − h ∆ − iζ · h ∇ ) m − k ( − iζ · h ∇ ) k a − h m D A a − h m − m A · ( ζ + ζ ) ( a ) − h m m q ( a ) in Ω . (2.7)If a ∈ C ∞ (Ω) satisfies ( ζ · ∇ ) k a = 0 in Ωfor some k ≥ ζ = O ( h ), one can show that thelowest order of h on the right-hand side of (2.7) is k − m − k +1) = 2 m − k +1.In order to get k e − ix · ζh h m L A,q ( e ix · ζh a ) k H − m scl (Ω) ≤ O ( h m + m/ ) , we should choose k satisfying 2 m − k + 1 ≥ m + m/ k ≤ ( m + 2) / m ≥
2, we should choose a ∈ C ∞ (Ω), satisfying the following transportequation, ( ζ · ∇ ) a = 0 in Ω . (2.8)The choice of such a is clearly possible. Having chosen the amplitude a in this way,we obtain the following equation for r , e − ix · ζh h m L A,q ( e ix · ζh h m/ r ) = − e − ix · ζh h m L A,q ( e ix · ζh a )= − ( − h ∆ − iζ · h ∇ ) m a − m ( − h ∆ − iζ · h ∇ ) m − ( − iζ · h ∇ ) a − h m D A a − h m − m A · ( ζ + ζ ) ( a ) − h m m q ( a ) =: g in Ω . (2.9) Notice that g belongs to H − m (Ω) and we would like to estimate k g k H − m scl (Ω) . Tothat end, we let ψ ∈ C ∞ (Ω) such that ψ = 0. Then using the fact that ζ = O ( h ),we get by the Cauchy-Schwarz inequality, | (cid:0) ( − h ∆ − iζ · h ∇ ) m a, ψ (cid:1) L (Ω) | + | (cid:0) ( − h ∆ − iζ · h ∇ ) m − ( − iζ · h ∇ ) a, ψ (cid:1) L (Ω) |≤ O ( h m − ) k ψ k L (Ω) ≤ O ( h m − ) k ψ k H m scl (Ω) . (2.10)In the case m ≥
3, we use Proposition 2.2 with p = q = 2, s = m − and s = m (assuming m < n ), to get |h h m − m A · ( ζ + ζ ) ( a ) , ψ i Ω | ≤ Ch m − k A k W − m − , nm ( R n ) k aψ k W m − , n n − m ( R n ) ≤ O ( h m − ) k ψ k H m − (Ω) ≤ O ( h m ) k ψ k H m (Ω) ≤ O ( h m ) k ψ k H m scl (Ω) . When m = 2, we use H¨older’s inequality, and obtain |h h m A · ( ζ + ζ ) ( a ) , ψ i Ω | ≤ Ch k A k L n ( R n ) k aψ k L n n − ( R n ) ≤ O ( h ) k a k L nn − ( R n ) k ψ k L nn ( R n ) ≤ O ( h ) k ψ k L (Ω) ≤ O ( h ) k ψ k H (Ω) . Therefore, for m ≥
2, we get |h h m − m A · ( ζ + ζ ) ( a ) , ψ i Ω | ≤ O ( h m ) k ψ k H m scl (Ω) . (2.11)Similarly, in the case m ≥
3, we use Proposition 2.2 with p = q = 2, s = m − and s = m (assuming m < n ), to get |h h m D A ( a ) , ψ i Ω | ≤ Ch m k A k W − m − , nm ( R n ) k ψDa k W m − , n n − m ( R n ) ≤ O ( h m ) k ψ k H m (Ω) ≤ O ( h m ) k ψ k H m (Ω) ≤ O ( h m ) k ψ k H m scl (Ω) , When m = 2, we again use H¨older’s inequality and Sobolev embeddings H ( R n ) ⊂ L nn − ( R n ) (see [31, Chapter 13, Proposition 6.4]), and obtain |h h D A ( a ) , ψ i Ω | ≤ h k A k L n ( R n ) k ψDa k L n n − ( R n ) ≤ O ( h ) k Da k L nn ( R n ) k ψ k L nn − ( R n ) ≤ O ( h ) k ψ k H (Ω) ≤ O ( h ) k ψ k H (Ω) . Therefore, for m ≥
2, we get |h h m − D A ( a ) , ψ i Ω | ≤ O ( h m ) k ψ k H m scl (Ω) . (2.12) NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 11
Finally, using Proposition 2.2, we show that |h h m m q ( a ) , ψ i Ω | ≤ h m k q k W − m , nm ( R n ) k aψ k W m , n n − m ( R n ) ≤ O ( h m ) k ψ k H m ( R n ) ≤ O ( h m ) k ψ k H m scl (Ω) . Thus, combining this together with the estimates (2.10) , (2.11), (2.12) in (2.9),and using (2.3) and m ≥
2, we can conclude that k g k H − m scl (Ω) ≤ O ( h m ) ≤ O ( h m + m/ ) . Thanks to this and Proposition 2.4, for h > r ∈ H m (Ω) of (2.9) such that k h m/ r k H m scl (Ω) . h m k e − ix · ζh h m L A,q ( e ix · ζh a ) k H − m scl (Ω) = 1 h m k g k H − m scl (Ω) . h m/ . Therefore, k r k H m scl (Ω) = O (1). The discussion of this section can be summarized inthe following proposition. Proposition 2.5.
Let Ω ⊂ R n , n ≥ , be a bounded open set with C ∞ boundary,and let m ≥ be an integer such that n > m . Suppose that A ∈ W − m − , nm ( R n , C n ) ∩E ′ (Ω) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω) , and let ζ ∈ C n be such that ζ · ζ = 0 , ζ = ζ + ζ with ζ being independent of h > , | Re ζ | = | Im ζ | = 1 , and ζ = O ( h ) as h → . Then for all h > small enough, there exists a solution u ( x, ζ ; h ) ∈ H m (Ω) to the equation L A,q u = 0 in Ω , of the form u ( x, ζ ; h ) = e ix · ζ/h ( a ( x, ζ ) + h m/ r ( x, ζ ; h )) , where the function a ( · , ζ ) ∈ C ∞ (Ω) satisfies (2.8) and the remainder term r is suchthat k r k H m scl (Ω) = O (1) as h → . Proof of Theorem 1.1
The first ingredient in the proof of Theorem 1.1 is a standard reduction to a largerdomain; see [30]. For the proof we follow [19, Proposition 3.2] and [27, Lemma 4.2].
Proposition 3.1.
Let Ω , Ω ′ ⊂ R n be two bounded opens sets such that Ω ⊂⊂ Ω ′ and ∂ Ω being C ∞ . Let A , A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω) and q , q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω) . If N A ,q = N A ,q , then N ′ A ,q = N ′ A ,q , where N ′ A j ,q j denotes the set of the Dirichlet-to-Neumann map for L A j ,q j in Ω ′ , j = 1 , .Proof. Let f ′ ∈ Q mj =0 H m − j − / ( ∂ Ω ′ ) and let u ′ ∈ H m (Ω ′ ) be a unique solutionto L A ,q u ′ = 0 in Ω ′ with γu ′ = f ′ on ∂ Ω ′ , where γ ′ denotes the Dirichlet traceon ∂ Ω ′ . Let u = u ′ | Ω ∈ H m (Ω) and f = γu . Since N A ,q = N A ,q , we canguarantee the existence of u ∈ H m (Ω) satisfying L A ,q u = 0 and γu = f . Inparticular ϕ := u − u ∈ H m (Ω) ⊂ H m (Ω ′ ). We define u ′ = u ′ + ϕ ∈ H m (Ω ′ ) , (3.1)so that we have u ′ = u in Ω. It follows that γ ′ u ′ = γ ′ u ′ = f ′ on ∂ Ω ′ . Next, we show that u ′ satisfies L A ,q u ′ = 0 in Ω ′ . For this, let ψ ∈ C ∞ (Ω ′ ). Thenwe have hL A ,q u ′ , ψ i Ω ′ = X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) + h D A ( u ′ ) , ψ i Ω ′ + h m q ( u ′ ) , ψ i Ω ′ . Since A = 0 and q = 0 outside of Ω, by (3.1), with ϕ ∈ H m (Ω), we can rewritethe above equality as hL A ,q u ′ , ψ i Ω ′ = X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) + X | α | = m m ! α ! ( D α ϕ, D α ( ψ | Ω )) L (Ω) + B A ( u ′ , ψ | Ω ) + b q ( u ′ , ψ | Ω )= X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) − X | α | = m m ! α ! ( D α u , D α ( ψ | Ω )) L (Ω) + X | α | = m m ! α ! ( D α u , D α ( ψ | Ω )) L (Ω) + B A ( u , ψ | Ω )+ b q ( u , ψ | Ω ) . Note that hN A ,q f, γ ( ψ | Ω ) i ∂ Ω = X | α | = m m ! α ! ( D α u , D α ( ψ | Ω )) L (Ω) + B A ( u , ψ | Ω ) + b q ( u , ψ | Ω ) . Therefore, we have hL A ,q u ′ , ψ i Ω ′ = X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) − X | α | = m m ! α ! ( D α u , D α ( ψ | Ω )) L (Ω) + hN A ,q f, ψ | ∂ Ω i ∂ Ω . Since N A ,q = N A ,q and since hN A ,q f, γ ( ψ | Ω ) i ∂ Ω = X | α | = m m ! α ! ( D α u , D α ( ψ | Ω )) L (Ω) + B A ( u , ψ | Ω ) + b q ( u , ψ | Ω ) , we come to hL A ,q u ′ , ψ i Ω ′ = X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) + B A ( u , ψ | Ω ) + b q ( u , ψ | Ω ) . Using that A = 0 and q = 0 outside Ω, we obtain hL A ,q u ′ , ψ i Ω ′ = X | α | = m m ! α ! ( D α u ′ , D α ψ ) L (Ω ′ ) + h D A ( u ′ ) , ψ i Ω ′ + h m q ( u ′ ) , ψ i Ω ′ = hL A ,q u ′ , ψ i Ω ′ = 0 . This shows that L A ,q u ′ = 0 in Ω ′ . NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 13
Using the analogous arguments one can show that N ′ A ,q f ′ = N ′ A ,q f ′ on ∂ Ω ′ ,which finishes the proof. (cid:3) The second ingredient is the derivation of the following integral identity based onthe assumption that N A ,q = N A ,q . Proposition 3.2.
Let Ω ⊂ R n , n ≥ , be a bounded open set with C ∞ boundary.Assume that A , A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω) and q , q ∈ W − m , nm ( R n , C ) ∩E ′ (Ω) . If N A ,q = N A ,q , then the following integral identity holds B A − A ( u , u ) + b q − q ( u , u )= ( D A − A ( u ) , u ) Ω + ( m q − q ( u ) , u ) Ω (3.2) for any u , u ∈ H m (Ω) satisfying L A ,q u = 0 in Ω and L ∗ A ,q u = 0 in Ω ,respectively. Recall that L ∗ A,q = L A,q + D · A is the formal adjoint of L A,q . Proof.
Since u ∈ H m (Ω) satisfies L − A ,q − D · A u = 0, the following0 = hL − A ,q − D · A u , ψ i Ω = X | α | = m m ! α ! ( D α u , D α ψ ) L (Ω) − h D A ( u ) , ψ i Ω + h m q − D · A ( u ) , ψ i Ω , (3.3)holds for every ψ ∈ C ∞ (Ω). Density and continuity imply that (3.3) holds also forall ψ ∈ H m (Ω).The hypothesis that N A ,q = N A ,q implies the existence of v ∈ H m (Ω) satisfying L A ,q v = 0 in Ω such γu = γv and N A ,q γu = N A ,q γv . Then ψ = u − v ∈ H m (Ω). Hence, applying (3.3) we obtain0 = X | α | = m m ! α ! ( D α u , D α ( u − v )) L (Ω) − h D A ( u ) , ( u − v ) i Ω + h m q − D · A ( u ) , ( u − v ) i Ω . (3.4)The equality hN A ,q γ , γu i ∂ Ω = hN A ,q γv , γu i ∂ Ω together with the definition(1.2) gives0 = X | α | = m m ! α ! ( D α ( u − v ) , D α u ) L (Ω) − ( B A ( u , u ) − B A ( v , u ))+ b q ( u , u ) − b q ( v , u ) . Combining this with (3.4) and using Proposition A.4, we derive the integral identity(3.2) as desired. (cid:3)
Let A , A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω) and q , q ∈ W − m + δ, nm ( R n , C ) ∩ E ′ (Ω),0 < δ < /
2, as in the statement of Theorem 1.1. In order to show that A = A ,we will need to use the Poincar´e lemma for currents [25]. We need this reductionto apply the Poincar´e lemma for currents , which requires the domain to be simply connected. Therefore, we reduce the problem to a larger simply connected domain.In particular, to a ball.Let B be an open ball in R n such that Ω ⊂⊂ B . According to Proposition 3.1, weknow that N BA ,q = N BA ,q , where N BA j ,q j denotes the Dirichlet-to-Neumann mapfor L A j ,q j in B , j = 1 ,
2. Now, by Proposition 3.2, the following integral identityholds B BA − A ( u , u ) + b Bq − q ( u , u ) = 0 (3.5)for any u , u ∈ H m ( B ) satisfying L A ,q u = 0 in B and L ∗ A ,q u = 0 in B ,respectively. Here and in what follows, by B BA − A and b Bq − q we denote the bilinearforms corresponding to A − A and q − q , respectively, defined (by means of (A.1))in the ball B .The main idea of the proof of Theorem 1.1 is to use the integral identity (3.5)with u , u ∈ H m ( B ) being complex geometric optics solutions for the equations L A ,q u = 0 in B and L ∗ A ,q u = 0 in B , respectively. In order to construct thesesolutions, consider ξ, µ , µ ∈ R n such that | µ | = | µ | = 1 and µ · µ = ξ · µ = ξ · µ = 0. For h >
0, we set ζ = hξ r − h | ξ | µ + iµ , ζ = − hξ r − h | ξ | µ − iµ . So we have ζ j · ζ j = 0, j = 1 ,
2, and ζ − ζ = hξ .By Proposition 2.5, for all h > u ( · , ζ ; h ) and u ( · , ζ ; h ) in H m ( B ) to the equations L A ,q u = 0 in B and L ∗ A ,q u = 0 in B, respectively, of the form u ( x, ζ ; h ) = e ix · ζ /h ( a ( x, µ + iµ ) + h m/ r ( x, ζ ; h )) u ( x, ζ ; h ) = e ix · ζ /h ( a ( x, µ − iµ ) + h m/ r ( x, ζ ; h )) , (3.6)where the amplitudes a ( x, µ + iµ ) , a ( x, µ − iµ ) ∈ C ∞ ( B ) satisfy the transportequations (( µ + iµ ) · ∇ ) a ( x, µ + iµ ) = 0 , in B, (3.7)and (( µ − iµ ) · ∇ ) a ( x, µ − iµ ) = 0 , in B, (3.8)and the remainder terms r ( · , ζ ; h ) and r ( · , ζ ; h ) satisfy k r j k H m scl ( B ) = O (1) , j = 1 , . (3.9)We substitude u and u given by (3.6) into (3.2), and get0 = 1 h b Bζ · ( A − A ) (cid:16) a + h m/ r , e ix · ξ ( a + h m/ r ) (cid:17) + B BA − A (cid:16) a + h m/ r , e ix · ξ ( a + h m/ r ) (cid:17) + b Bq − q (cid:16) a + h m/ r , e ix · ξ ( a + h m/ r ) (cid:17) . (3.10) NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 15
Multiplying this by h and letting h → +0, we obtain that b B ( µ + iµ ) · ( A − A ) (cid:0) a , e ix · ξ a (cid:1) = 0 . (3.11)Here we have used (3.9), Proposition A.2 and the fact that a , a ∈ C ∞ ( B ) toconclude that | B BA − A (cid:16) a + h m/ r , e ix · ξ ( a + h m/ r ) (cid:17) |≤ C k A − A k W − m − , nm ( R n ) k a + h m/ r k H m ( B ) k a + h m/ r k H m ( B ) ≤ C ( k a k H m ( B ) + k r k H m ( B ) )( k a k H m ( B ) + k r k H m ( B ) ) ≤ O (1) . and | b Bq − q (cid:16) a + h m/ r , e ix · ξ ( a + h m/ r ) (cid:17) |≤ C k q − q k W − m , nm ( R n ) k a + h m/ r k H m ( B ) k a + h m/ r k H m ( B ) ≤ C ( k a k H m ( B ) + h m/ k r k H m ( B ) )( k a k H m ( B ) + h m/ k r k H m ( B ) ) ≤ O (1) . Substituting a = a = 1 in (3.11), we obtain (cid:10) ( µ + iµ ) · ( A − A ) , e ix · ξ (cid:11) B = 0 . (3.12)This implies that( µ + iµ ) · ( b A ( ξ ) − b A ( ξ )) = 0 , for all µ, ξ ∈ R n , µ · ξ = 0 , (3.13)where b A j stands for the Fourier transform of A j , j = 1 ,
2. It follows from (3.13)that ∂ x j ( A ,k − A ,k ) − ∂ x k ( A ,j − A ,j ) = 0 in Ω , ≤ j, k ≤ n, (3.14)in the sense of distributions. Indeed, for each ξ = ( ξ , . . . , ξ n ) and for j = k ,1 ≤ j, k ≤ n , consider the vector µ = µ ( ξ, j, k ) such that µ j = − ξ k , µ k = ξ j andall other components are equal to zero. Therefore, µ satisfy µ · ξ = 0. Hence, using(3.13) we obtain ξ j · ( b A ,k ( ξ ) − b A ,k ( ξ )) − ξ k · ( b A ,j ( ξ ) − b A ,j ( ξ )) = 0 , which proves (3.14) in the sense of distributions.Our goal is to show that A = A . Considering A − A as a 1-current and usingthe Poincar´e lemma for currents, we conclude that there is ψ ∈ D ′ ( R n ) such that dψ = A − A ∈ W − m − , nm ( R n , C n ) ∩E ′ ( B, C n ); see [25]. Note that ψ is a constant,say c ∈ C , outside B since A − A = 0 in R n \ B (and also near ∂B ). Considering ψ − c instead of ψ , we may and shall assume that ψ ∈ E ′ ( B, C ).To show that A = A , consider (3.11) with a ( · , µ − iµ ) = 1 and a ( · , µ + iµ )satisfying (( µ + iµ ) · ∇ ) a ( x, µ + iµ ) = 1 in B. (3.15) The latter choice is possible thanks to (3.7), (3.8) and the assumption that m ≥ ∂ -equation and one can solve it bysetting a ( x, µ + iµ ) = 12 π Z R χ ( x − y µ − y µ ) y + iy dy dy , where χ ∈ C ∞ ( R n ) is such that χ ≡ B ; see [27, Lemma 4.6].From (3.11), we have b B ( µ + iµ ) ·∇ ψ ( a , e ix · ξ ) = 0 . Using the fact that µ · ξ = µ · ξ = 0, we obtain0 = − b B ( µ + iµ ) ·∇ ψ ( a , e ix · ξ ) = − (cid:10) ( µ + iµ ) · ∇ ψ, e ix · ξ a (cid:11) B = (cid:10) ψ, e ix · ξ ( µ + iµ ) · ∇ a (cid:11) B = h ψ, e ix · ξ i B . This gives ˆ ψ = 0, and hence we have ψ = 0 in B , which completes the proof of A = A .To show that q = q , we substitude A = A and a = a = 1 into the identity(3.10) and obtain b Bq − q (cid:16) h m/ r , (1 + h m/ r ) e ix · ξ (cid:17) = 0 . Letting h → + , we get b q ( ξ ) − b q ( ξ ) = 0 for all ξ ∈ R n . Let us justify this laststatement. We will only consider the term b Bq − q ( h m/ r , e ix · ξ ). The justificationfor the other two terms follows similarly. Since q , q ∈ W − m + δ, nm ( R n , C ) ∩ E ′ (Ω)for some 0 < δ < /
2, we use Proposition 2.2 with p = q = 2, s = m − δ , s = m and r = 2 n/ (2 n − m ) to get | b Bq − q ( h m/ r , e ix · ξ ) | ≤ Ch m/ k q − q k W − m δ, nm ( R n ) k r e ix · ξ k W m − δ, n n − m ( R n ) ≤ Ch m/ k q − q k W − m δ, nm ( R n ) k e ix · ξ k H m ( B ) k r k H m − δ ( B ) ≤ Ch δ k q − q k W − m δ, nm ( R n ) k e ix · ξ k H m ( B ) k h m − δ r k H m − δ ( B ) ≤ O ( h δ ) k r k H m − δ scl ( B ) ≤ O ( h δ ) . This implies that q = q in B completing the proof of Theorem 1.1. Appendix A. Mapping properties of D A and m q Let Ω ⊂ R n , n ≥
3, be a bounded open set with C ∞ boundary, and m ≥ n > m . Let A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ), where W s,p ( R n ) is the standard L p -based Sobolevspace on R n , s ∈ R and 1 < p < ∞ . The reader is referred to [33] for properties ofthese spaces.We start with considering the bilinear forms B R n A and b R n q on H m ( R n ) which aredefined by B R n A ( u, v ) = h A, vDu i R n , b R n q ( u, v ) = h q, uv i R n , u, v ∈ H m ( R n ) . NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 17
The following result shows that the forms B R n A and b R n q are bounded on H m ( R n ).The proof is based on a property on multiplication of functions in Sobolev spacesgiven in Proposition 2.2. Proposition A.1.
The bilinear forms B R n A and b R n q on H m ( R n ) are bounded andsatisfy | b R n q ( u, v ) | ≤ C k q k W − m , nm ( R n ) k u k H m ( R n ) k v k H m ( R n ) and | B R n A ( u, v ) | ≤ C k A k W − m − , nm ( R n ) k u k H m ( R n ) k v k H m ( R n ) for all u, v ∈ H m ( R n ) .Proof. First, we give the proof for the form b R n q . Using the duality between W − m , nm ( R n )and W m , n n − m ( R n ), and using Proposition 2.2, we conclude that for all u, v ∈ H m ( R n ) | b R n q ( u, v ) | = |h q, uv i R n | ≤ k q k W − m , nm ( R n ) k uv k W m , n n − m ( R n ) ≤ C k q k W − m , nm ( R n ) k u k H m ( R n ) k v k H m ( R n ) . Now, we give the proof for the form B R n A . In the case m ≥
3, using the dualitybetween W − m − , nm ( R n ) and W m − , n n − m ( R n ), and using Proposition 2.2 with p = q = 2, s = m − and s = m (assuming m < n ), we conclude that | B R n A ( u, v ) | = |h A, vDu i R n | ≤ k A k W − m − , nm ( R n ) k vDu k W m − , n n − m ( R n ) ≤ C k A k W − m − , nm ( R n ) k Du k H m − ( R n ) k v k H m ( R n ) ≤ C k A k W − m − , nm ( R n ) k u k H m ( R n ) k v k H m ( R n ) . When m = 2, we use H¨older’s inequality and Sobolev embedding H ( R n ) ⊂ L nn − ( R n ) (see [31, Chapter 13, Proposition 6.4]), and obtain | B R n A ( u, v ) | = |h A, vDu i R n | ≤ k A k L n ( R n ) k vDu k L n n − ( R n ) ≤ k A k L n ( R n ) k Du k L nn ( R n ) k v k L nn − ( R n ) ≤ C k A k L n ( R n ) k Du k L ( R n ) k v k H ( R n ) ≤ C k A k L n ( R n ) k u k H ( R n ) k v k H ( R n ) . Therefore, | B R n A ( u, v ) | ≤ C k A k W − m − , nm ( R n ) k u k H m ( R n ) k ψ k H m ( R n ) for all u, v ∈ H m ( R n ). (cid:3) The bilinear forms B A and b q on H m (Ω), which were defined in the introduction,can be rewritten as B A ( u, v ) = B R n A (˜ u, ˜ v ) , b q ( u, v ) = b R n q (˜ u, ˜ v ) , u, v ∈ H m (Ω) , (A.1)where ˜ u, ˜ v ∈ H m ( R n ) are extensions of u and v , respectively. First, we show thatthese definitions are well-defined, i.e. independent of the choice of extensions ˜ u, ˜ v . Indeed, let u , u ∈ H m ( R n ) be such that u = u = u in Ω, and let v , v ∈ H m ( R n )be such that v = v = v in Ω. Then we need to show that B R n A ( u − u , , v − v ) = 0 and b R n q ( u − u , v − v ) = 0 . (A.2)Since A and q are supported in Ω, for any φ, ψ ∈ S ( R n ) with φ = ψ = 0 in Ω, wehave B R n A ( φ, ψ ) = h A, ψDφ i R n = 0 and b R n q ( φ, ψ ) = h q, ψφ i R n = 0 . Since S ( R n ) is dense in H m ( R n ) and B R n A and b R n q are continuous bilinear forms(by Proposition A.1), we get (A.2).The next result shows that the bilinear forms B A and b q are bounded on H m (Ω).This is a consequence of Proposition A.1. Proposition A.2.
The bilinear forms B A and b q on H m (Ω) are bounded and satisfy | b q ( u, v ) | ≤ C k q k W − m , nm ( R n ) k u k H m (Ω) k v k H m (Ω) and | B A ( u, v ) | ≤ C k A k W − m − , nm ( R n ) k u k H m (Ω) k v k H m (Ω) for all u, v ∈ H m (Ω) .Proof. Let u, v ∈ H m (Ω) and let e Ω be a bounded open neigbhborhood of Ω. Thenthere is a bounded linear map E : H m (Ω) → H m ( e Ω) such that E | Ω = Id; see [7,Theorem 6.44]. Then according to estimates proven in Proposition A.1, we obtain | b q ( u, v ) | = | b R n q ( E ( u )) , E ( v )) |≤ C k q k W − m , n ( R n ) k E ( u ) k H m ( R n ) k E ( v ) k H m ( R n ) ≤ C k q k W − m , n ( R n ) k u k H m (Ω) k v k H m (Ω) and | B A ( u, v ) | = | B R n A ( E ( u ) , E ( v )) |≤ C k A k W − m − , nm ( R n ) k E ( u ) k H m ( R n ) k E ( v ) k H m ( R n ) ≤ C k A k W − m − , nm ( R n ) k u k H m (Ω) k v k H m (Ω) . These estimates finish the proof. (cid:3)
Now, for u ∈ H m (Ω), we define D A ( u ) and m q ( u ), for any v ∈ H m (Ω) by h D A ( u ) , v i Ω = B A ( u, v ) and h m q ( u ) , v i Ω = b q ( u, v ) , The following result, which is an immediate corollary of Proposition A.2, impliesthat D A , m q are bounded operators from H m (Ω) to H − m (Ω). The norm on H − m (Ω) is the usual dual norm given by k v k H − m (Ω) = sup = φ ∈ H m (Ω) |h v, φ i Ω |k φ k H m (Ω) . NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 19
Corollary A.3.
The operators B A and b q are bounded from H m (Ω) to H − m (Ω) and satisfy k m q ( u ) k H − m (Ω) ≤ C k q k W − m , nm ( R n ) k u k H m (Ω) and k D A ( u ) k H − m (Ω) ≤ C k A k W − m − , nm ( R n ) k u k H m (Ω) for all u ∈ H m (Ω) . Finally, we record and give the proof of the following useful identities.
Proposition A.4.
For any u, v ∈ H m (Ω) , the forms B A and b q satisfy the followingidentities B A ( u, v ) = − B A ( v, u ) − b D · A ( u, v ) and b q ( u, v ) = b q ( v, u ) . Proof.
According to the definitions (A.1) and density of S ( R n ) in H m ( R n ), it issufficient to prove for the case u, v ∈ S ( R n ). This follows by straightforward com-putations b R n q ( u, v ) = h q, uv i R n = h m q ( v ) , u i R n , and using product rule B R n A ( u, v ) = h A, vDu i R n = −h A, uDv i R n + h A, D ( uv ) i R n = − B R n A ( v, u ) − h D · A, uv i R n = − B R n A ( v, u ) − b D · A ( u, v ) . The proof is thus finished. (cid:3)
Appendix B. Well-posedness and Dirichlet-to-Neumann map
Let Ω ⊂ R n , n ≥
3, be a bounded open set with C ∞ boundary, and let A ∈ W − m − , nm ( R n , C n ) ∩ E ′ (Ω , C n ) and q ∈ W − m , nm ( R n , C ) ∩ E ′ (Ω , C ), where n > m .For f = ( f , . . . , f m − ) ∈ Q m − j =0 H m − j − / ( ∂ Ω), consider the Dirichlet problem L A,q u = 0 in Ω ,γu = f on ∂ Ω . (B.1)Here, by γ we denote the Dirichlet trace operator, given by γ : H m (Ω) → m − Y j =0 H m − j − / ( ∂ Ω) , γu = ( u | ∂ Ω , ∂ ν u | ∂ Ω , . . . , ∂ m − ν u | ∂ Ω ) , which is bounded and surjective, see [10, Theorem 9.5, page 226].First aim of this appendix is to use the standard variational arguments to show thewell-posedness of the problem (B.1). First, consider the following inhomogeneousproblem L A,q u = F in Ω ,γu = 0 on ∂ Ω , (B.2)with u ∈ H − m (Ω). To define a sesquilinear form a , associated to the problem (B.2), for u, v ∈ C ∞ (Ω),we integrate by parts and get hL A,q u, v i Ω = X | α | = m m ! α ! ( D α u, D α v ) L (Ω) + h D A ( u ) , v i Ω + h m q ( u ) , v i Ω := a ( u, v ) . Therefore, a is defined on H m (Ω) by a ( u, v ) := X | α | = m m ! α ! ( D α u, D α v ) L (Ω) + B A ( u, v ) + b q ( u, v ) , u, v ∈ H m (Ω) . Note that this is not a unique way to define a sesquilinear form associated to theproblem (B.2).Now, we show that the sesquilinear form a can be extended to a bounded form on H m (Ω). Using duality and Proposition A.1, for u, v ∈ H m (Ω), we obtain | a ( u, v ) | ≤ X | α | = m m ! α ! k D α u k L (Ω) k D α v k L (Ω) + (cid:18) k A k W − m − , nm ( R n ) + k q k W − m , nm ( R n ) (cid:19) k u k H m (Ω) k v k H m (Ω) ≤ C k u k H m (Ω) k v k H m (Ω) . (B.3)Thus, the sesquilinear form a is a bounded form on H m (Ω).Applying Poincar´e’s inequality, we have k u k H m (Ω) ≤ C X | α | = m k D α u k L (Ω) , u ∈ H m (Ω) . (B.4)Write q = q ♯ + ( q − q ♯ ) with q ♯ ∈ L ∞ (Ω , C ) and k q − q ♯ k W − m , nm ( R n ) small enough,and write A = A ♯ + ( A − A ♯ ) with A ♯ ∈ L ∞ (Ω , C n ) and k A − A ♯ k W − m − , nm ( R n ) small enough. Using (B.4) and Proposition A.2, for ε >
0, we obtain thatRe a ( u, u ) ≥ X | α | = m m ! α ! k D α u k L (Ω) − | B A ( u, u ) | − | b q ( u, u ) |≥ C X | α | = m k D α u k L (Ω) − | B A ♯ ( u, u ) | − | b q ♯ ( u, u ) |− | B A − A ♯ ( u, u ) | − | b q − q ♯ ( u, u ) |≥ C k u k H m (Ω) − k A ♯ k L ∞ (Ω) k Du k L (Ω) k u k L (Ω) − k q ♯ k L ∞ (Ω) k u k L (Ω) − C ′ k A − A ♯ k W − m − , nm ( R n ) k u k H m (Ω) − C ′ k q − q ♯ k W − m , nm ( R n ) k u k H m (Ω) ≥ C k u k H m (Ω) − k A ♯ k L ∞ (Ω) ε k Du k L (Ω) − k A ♯ k L ∞ (Ω) ε k u k L (Ω) − k q ♯ k L ∞ (Ω) k u k L (Ω) − C ′ k A − A ♯ k W − m − , nm ( R n ) k u k H m (Ω) − C ′ k q − q ♯ k W − m , nm ( R n ) k u k H m (Ω) , C, C ′ > , u ∈ H m (Ω) . NVERSE PROBLEMS FOR POLYHARMONIC OPERATOR 21
Taking ε > a ( u, u ) ≥ C k u k H m (Ω) − C k u k L (Ω) , C, C > , u ∈ H m (Ω) . Therefore, the form a is coercive on H m (Ω). As the inclusion map H m (Ω) ֒ → L (Ω)is compact, the operator L A,q = ( − ∆) m + D A + m q : H m (Ω) → H − m (Ω) = ( H m (Ω)) ′ , is Fredholm operator with zero index; see [21, Theorem 2.34].Positivity of the operator L A,q + C : H m (Ω) → H − m (Ω) and an application of theLax-Milgram lemma implies that L A,q + C has a bounded inverse. By compactSobolev embedding H m (Ω) ֒ → H − m (Ω) and the Fredholm theorem, the equation(B.2) has a unique solution u ∈ H m (Ω) for any F ∈ H − m (Ω) if one is outside acountable set of eigenvalues.Now, consider the Dirichlet problem L A,q u = 0 in Ω ,γu = f on ∂ Ω , (B.5)with f = ( f , . . . , f m − ) ∈ Q m − j =0 H m − j − / ( ∂ Ω). We assume that 0 is not inthe spectrum of L A,q : H m (Ω) → H − m (Ω). By [10, Theorem 9.5, page 226],there is w ∈ H m (Ω) such that γw = f . According to Corollary A.3, we have L A,q w ∈ H − m (Ω). Therefore, u = v + w , with v ∈ H m (Ω) being the uniquesolution of the equation L A,q v = −L A,q w ∈ H − m (Ω), is the unique solution of theDirichlet problem (B.5).Under the assumption that 0 is not in the spectrum of L A,q , the Dirichlet-to-Neumann map N A,q is defined as follows: let f, h ∈ Q m − j =0 H m − j − / ( ∂ Ω). Thenwe set hN A,q f, h i ∂ Ω := X | α | = m m ! α ! ( D α u, D α v h ) L (Ω) + B A ( u, v h ) + b q ( u, v h ) , (B.6)where u is the unique solution of the Dirichlet problem (B.5) and v h ∈ H m (Ω) is anextension of h , that is γv h = h . In this appendix we show that N A,q is a well-defined(i.e. independent of the choice of v h ) bounded operator N A,q : m − Y j =0 H m − j − / ( ∂ Ω) → m − Y j =0 H m − j − / ( ∂ Ω) ′ = m − Y j =0 H − m + j +1 / ( ∂ Ω) . Let us first show that the definition (B.6) of N A,q f is independent of the choice of anextension v h of h . For this, let v h, , v h, ∈ H m (Ω) be such that γv h, = γv h, = h .Note that v h, − v h, ∈ H m (Ω). Then we have to show that X | α | = m m ! α ! ( D α u, D α ( v h, − v h, ) h ) L (Ω) + h D A ( u ) , ( v h, − v h, ) i Ω + h m q ( u ) , ( v h, − v h, ) i Ω = 0 . (B.7) For any w ∈ C ∞ (Ω) and for u ∈ H m (Ω) solution of the Dirichlet problem (B.5), wehave0 = hL A,q u, ¯ w i = X | α | = m m ! α ! ( D α u, D α w ) L (Ω) + h D A ( u ) , w i Ω + h m q ( u ) , w i Ω . Density of C ∞ (Ω) in H m (Ω) and continuity of the form on H m (Ω) give (B.7).Now we show that N A,q f is a well-defined element of Q m − j =0 H − m + j +1 / ( ∂ Ω). From(B.3), it follows that |hN
A,q u, h i ∂ Ω | ≤ C k u k H m (Ω) k v h k H m (Ω) ≤ C k γu k Q m − j =0 H m − j − / ( ∂ Ω) k h k Q m − j =0 H m − j − / ( ∂ Ω) , where k h k Q m − j =0 H m − j − / ( ∂ Ω) = ( k h k H m − / ( ∂ Ω) + · · · + k h m − k H / ( ∂ Ω) ) / is the product norm on the space Q m − j =0 H m − j − / ( ∂ Ω). Here we have used the factthat the extension operator Q m − j =0 H m − j − / ( ∂ Ω) ∋ h v h ∈ H m (Ω) is bounded,again see [10, Theorem 9.5, page 226]. Hence, we have that N A,q f belongs to (cid:0) Q m − j =0 H m − j − / ( ∂ Ω) (cid:1) ′ = Q m − j =0 H − m + j +1 / ( ∂ Ω). The proof given above alsoshows that N A,q : m − Y j =0 H m − j − / ( ∂ Ω) → m − Y j =0 H m − j − / ( ∂ Ω) ′ = m − Y j =0 H − m + j +1 / ( ∂ Ω)is bounded.
Acknowledgements
The author thanks his advisor Professor Gunther Uhlmann for all his encourage-ment and support. Many thanks to Professor Katya Krupchyk whose unvaluablecomments substantially improved the main result of the paper. The author grate-fully acknowledges Professor Winfried Sickel for sharing his copy of the monograph[26]. Finally, the author is very grateful to Karthik Iyer for pointing out some mis-takes in the previous version of the paper and for suggesting the ways to fix them.Specifically, he pointed out that D A becomes a bounded operator only for m < n (see the proof of Proposition A.1 for details) and that we need to assume little moreregularity of the zeroth order perturbation in order to get uniqueness (for details seethe last paragraph of Section 3). The work of the author was partially supportedby the National Science Foundation. References [1] R. Brown,
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