A Partial Data Result for the Magnetic Schrodinger Inverse Problem
AA PARTIAL DATA RESULT FOR THE MAGNETICSCHR ¨ODINGER INVERSE PROBLEM
FRANCIS J. CHUNG
Abstract.
This article shows that knowledge of the Dirichlet-Neumann (DN)map on certain subsets of the boundary for functions supported roughly onthe rest of the boundary uniquely determines a magnetic Schr¨odinger operator.With some geometric conditions on the domain, either the subset on which theDN map is measured or the subset on which the input functions have supportmay be made arbitrarily small. This is a response to a question posed in[DKSU]. The method involves modifying the Carleman estimate in that paperby conjugation with certain pseudodifferential-like operators. Introduction
Let n ≥
2, and let Ω be a simply-connected bounded domain in R n +1 , withsmooth boundary. If W is a C vector field on R n +1 , and q is an L ∞ function on R n +1 , then let L W,q denote the magnetic Schr¨odinger operator L W,q = ( D + W ) + q, where D = − i ∇ . I will assume that q and W are such that zero is not an eigenvalueof L W,q on Ω. Then the Dirichlet problem L W,q u = 0 u | ∂ Ω = g has a unique solution u ∈ H (Ω) for each g ∈ H ( ∂ Ω). Therefore for g ∈ H ( ∂ Ω)we can define the Dirichlet-Neumann map Λ
W,q byΛ
W,q g = ( ∂ ν + iW · ν ) u | ∂ Ω , where ν is the outward unit normal, and u is the unique solution to the Dirichletproblem with boundary value g . This gives a well-defined map from H ( ∂ Ω) to H − ( ∂ Ω).The basic inverse problem associated to the magnetic Schr¨odinger operator L W,q is to recover dW and q from knowledge of Λ W,q . (Here W is identified with the1-form W dx + . . . + W n +1 dx n +1 .) Note that we cannot hope to recover W itselfsince the Dirichlet-Neumann map is invariant under the gauge transformation W (cid:55)→ W + ∇ Ψ whenever Ψ ∈ C (Ω) and Ψ | ∂ Ω = 0. However, identifying dW identifies W up to this gauge transformation.In [Su], Sun first showed that full knowledge of the Dirichlet-Neumann mapdetermines dW and q when W is small enough, in a certain sense. In [NSuU],Nakamura, Sun, and Uhlmann removed the smallness assumption, and showed Mathematics Subject Classification.
Primary 35R30.
Key words and phrases.
Dirichlet-Neumann map, Magnetic Schr¨odinger operator, Inverseproblems, Semiclassical analysis, Pseudodifferential operators. a r X i v : . [ m a t h . A P ] N ov CHUNG that full knowledge of the Dirichlet-Neumann map determines dW and q for C W and L ∞ q . Tolmasky [To] and Salo [Sa1] improved the regularity conditions on W to C / ε and Dini continuous, respectively. Salo also gave in [Sa2] a proof for W ∈ C ε involving a reconstruction method.In [DKSU], Dos Santos Ferreira, Kenig, Sj¨ostrand, and Uhlmann proved a partialdata result for this operator. Assume that x is not in the closure of the convexhull of Ω. Define the front and back of ∂ Ω (with respect to x ) by F Ω = { x ∈ ∂ Ω | ( x − x ) · ν ( x ) ≤ } B Ω = { x ∈ ∂ Ω | ( x − x ) · ν ( x ) ≥ } , where ν ( x ) is the outward unit normal at x . Then Theorem 1.1 in [DKSU] says(with different notation) the following. Theorem [DKSU].
Let W and W be C vector fields on Ω , and let q and q be L ∞ functions on Ω . Suppose U is a neighbourhood of F Ω such that Λ W ,q g | U = Λ W ,q g | U for all g ∈ H ( ∂ Ω) . Then q = q and dW = dW . However, in the case that W ≡
0, [KSU] had already given a better partial dataresult, which says that we only needΛ ,q g | U = Λ ,q g | U for g ∈ H ( ∂ Ω) with support in a neighbourhood of B Ω , to conclude that q = q .This paper will show that this sort of partial data result also holds for themagnetic Schr¨odinger inverse problem.Define Z Ω = { x ∈ ∂ Ω | ( x − x ) · ν ( x ) = 0 } . The main results of this work are the following two theorems.
Theorem 1.1.
Let W and W be C vector fields on Ω , and let q and q be L ∞ functions on Ω . Let U ⊂ ∂ Ω be a neighbourhood of F Ω , and let E ⊂ ∂ Ω be acompact subset of F Ω \ Z Ω . Suppose Λ W ,q g | U = Λ W ,q g | U for all g ∈ H ( ∂ Ω) with support contained in ∂ Ω \ E .Then q = q , and dW = dW . Theorem 1.2.
Let W and W be C vector fields on Ω , and let q and q be L ∞ functions on Ω . Let U ⊂ ∂ Ω be a neighbourhood of B Ω , and let E ⊂ ∂ Ω be acompact subset of B Ω \ Z Ω . Suppose Λ W ,q g | U = Λ W ,q g | U for all g ∈ H ( ∂ Ω) with support contained in ∂ Ω \ E .Then q = q , and dW = dW . The second theorem is essentially the first theorem after the conformal transfor-mation on Ω given by inversion in x .Roughly speaking, the first theorem says that if the Dirichlet-Neumann map isknown on a neighbourhood of the front for functions supported on a neighbour-hood of the back, then potentials can be determined. The second says that if the PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 3
Dirichlet-Neumann map is known on a neighbourhood of the back for functionssupported on a neighbourhood of the front, then the potentials can be determined.If the domain Ω is nice enough, then the front can be made arbitrarily small.For example, if Ω is strongly convex (convex, and the intersection of the boundarywith any tangent hyperplane to the boundary consists only of one point), then thefront can be contained in any non-empty open subset of the boundary. This givesus the following corollary.
Corollary 1.3.
Suppose Ω is a smooth bounded strongly convex domain in R n +1 .Let W and W be C vector fields on Ω , and let q and q be L ∞ functions on Ω . Then for any non-empty open subset U of the boundary, there exists compact E ⊂ U such that if Λ W ,q g | U = Λ W ,q g | U for all g ∈ H ( ∂ Ω) with support contained in ∂ Ω \ E , then q = q , and dW = dW .Alternatively, for any compact proper subset E of the boundary, there exists U ⊂ ∂ Ω with E ⊂ U such that if Λ W ,q g | U = Λ W ,q g | U for all g ∈ H ( ∂ Ω) with support contained in ∂ Ω \ E , then q = q , and dW = dW . The first part of the corollary says that in particular, the Dirichlet-Neumannmap can be measured on an arbitrarily small subset of the boundary. The secondpart of the corollary says that alternatively, the input functions may be restrictedto an arbitrarily small subset of the boundary.Theorem 1.2 can either be proved from Theorem 1.1 by the change of vari-ables mentioned above, or proved in the same manner as Theorem 1.1, makingthe changes indicated at the end of section 6. Therefore most of this paper willbe devoted to proving Theorem 1.1. From here on, unless otherwise noted, I willassume U , E and Ω are as in Theorem 1.1.The key to the proof of Theorem 1.1 is the construction of complex geometricaloptics solutions to the system L W,q u = 0 on Ω u | E = 0 . (1.1)In [DKSU], these are constructed using a Carleman estimate for L W,q and aHahn-Banach argument. However, the initial Carleman estimate proved in [DKSU]creates L solutions. These turn out to be good enough in the case W = 0, but notwhen the magnetic potential is present. Modifications to the Carleman estimate tocreate H solutions in that paper destroy information about the behaviour of thesolutions on the boundary, which explains the difference between the [KSU] and[DKSU] results. Proving theorems 1.1 and 1.2 will require more careful modifica-tion. The Carleman estimate proved here for L W,q can be described as follows.Let ϕ be a limiting Carleman weight on Ω; that is, a real-valued smooth functionwhich has nonvanishing gradient on Ω and satisfies (cid:104) ϕ (cid:48)(cid:48) ∇ ϕ, ∇ ϕ (cid:105) + (cid:104) ϕ (cid:48)(cid:48) ξ, ξ (cid:105) = 0 CHUNG whenever | ξ | = |∇ ϕ | and ∇ ϕ · ξ = 0. Define L ϕ,W,q = h e ϕh L W,q e − ϕh Here h is a semiclassical parameter; henceforth all Sobolev spaces and Fouriertransforms in this note are semiclassical, unless otherwise specified, with h beingthe semiclassical parameter. For the rest of this paper, I will fix ϕ to be thelogarithmic weight ϕ ( x ) = log | x − x | unless otherwise stated.I want to prove the following Carleman estimate. Theorem 1.4.
There exists a smooth domain Ω (cid:48) with Ω ⊂ Ω (cid:48) , and E ⊂ ∂ Ω (cid:48) , suchthat if w ∈ C ∞ (Ω) , h (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,W,q w (cid:107) H − (Ω (cid:48) ) . Theorem 1.4 will be proved over the next five sections. In section 7, I will usethis estimate to construct solutions to (1.1). Once these are constructed, the proofof Theorem 1 follows by more or less the identical argument as in [DKSU]. Thatargument is presented in section 8 for completeness.
Acknowledgements . This research was partially supported by a Doctoral Post-graduate Scholarship from the Natural Science and Engineering Research Councilof Canada. The author would also like to thank Carlos Kenig for his guidance,support, and patience.2.
An Initial Carleman Estimate
I want to begin by considering a special version of Theorem 1.4, where the set E coincides with a graph. Without loss of generality, I will assume x = 0. We canequip R n +1 with spherical coordinates ( r, θ ), with r ∈ [0 , ∞ ) and θ ∈ S n .Define L ϕ,ε,W,q = e ϕ ε L ϕ,W,q e − ϕ ε Proposition 2.1.
Suppose that f : S n → (0 , ∞ ) is a C ∞ function such that Ω lies entirely in the region A O = { ( r, θ ) | r ≥ f ( θ ) } ⊂ R n +1 , and E is a subset of thegraph r = f ( θ ) . (See the diagram below.) If w ∈ C ∞ (Ω) , then h √ e (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − ( A O ) . In addition to L ϕ,W,q , define L ϕ,ε,W,q = e ϕ ε L ϕ,W,q e − ϕ ε L ϕ = h e ϕh (cid:52) e − ϕh and L ϕ,ε = e ϕ ε L ϕ e − ϕ ε . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 5
We will need to do some work with a set which is slightly larger than Ω, but stillbounded. Let Ω ⊂ A O be a smooth bounded domain which contains Ω, such that E ⊂ ∂ Ω , as indicated in the diagram. Since 0 lies outside the closure of the convexhull of Ω, I can pick Ω so 0 stays outside of the closure of the convex hull of Ω .We have the following Carleman estimate from [DKSU]. Lemma 2.2. If w ∈ C ∞ (Ω ) , then h √ ε (cid:107) w (cid:107) H (Ω ) (cid:46) (cid:107)L ϕ,ε w (cid:107) L (Ω ) . A note on inequalities here: inequalities of the form F ( w, h ) (cid:46) G ( w, h ) meanthat there exists h > w , such that for h ≤ h , the inequality F ( w, h ) ≤ CG ( w, h ) holds for some positive constant C independent of w and h .In the case of Lemma 2.2, the constant implied in the (cid:46) sign is independent of ε as well.I can make a change of variables using the map ( r, θ ) (cid:55)→ ( rf ( θ ) , θ ). This is adiffeomorphism from A O to R n +1 \ B , where B is the open ball of radius 1 centredat the origin, with inverse ( r, θ ) (cid:55)→ ( rf ( θ ) , θ ). Let ˜Ω and ˜Ω be the images of Ωand Ω , respectively, under this map. This diffeomorphism maps E to a part of theunit sphere S n . Note that since 0 is outside of the closure of the convex hull of Ω ,it is also outside of the closure of the convex hull of ˜Ω . Lemma 2.3.
For w ∈ C ∞ ( ˜Ω ) , (2.1) h √ ε (cid:107) w (cid:107) H (˜Ω ) (cid:46) (cid:107) ˜ L ϕ,ε w (cid:107) L (˜Ω ) where ˜ L ϕ,ε = (cid:0) |∇ S n log f ( θ ) | S n (cid:1) h ∂ r − r ( α + ( ∇ S n log f ( θ )) · S n h ∇ S n ) h∂ r + 1 r ( α + h (cid:52) S n ) and α = 1 + hε log( rf ( θ )) . Here ∇ S n is the gradient operator on the unit sphere; | · | S n and · S n indicate the use of the Riemannian metric on S n , and (cid:52) S n is theLaplace-Beltrami operator on the unit sphere S n .Proof. Let w ∈ C ∞ (Ω ), and let˜ w ( r, θ ) = w ( rf ( θ ) , θ ) . CHUNG
Then ˜ w ∈ C ∞ ( ˜Ω ). Now by a change of variables, (cid:107) ˜ w (cid:107) L (˜Ω ) = (cid:90) S n (cid:90) ∞ | w ( rf ( θ ) , θ ) | r n dr dθ = (cid:90) S n (cid:90) ∞ | w ( r, θ ) | r n ( f ( θ )) − n − dr dθ (cid:39) (cid:90) S n (cid:90) ∞ | w ( r, θ ) | r n dr dθ. Therefore(2.2) (cid:107) ˜ w (cid:107) L (˜Ω ) (cid:39) (cid:107) w (cid:107) L (Ω ) . The constants implied in the (cid:39) sign depend on f ( θ ). In addition, (cid:107) ˜ w (cid:107) H (˜Ω ) = (cid:107) ˜ w (cid:107) L (˜Ω ) + (cid:107) h ∇ ˜ w (cid:107) L (˜Ω ) = (cid:107) ˜ w (cid:107) L (˜Ω ) + (cid:107) h∂ r ˜ w (cid:107) L (˜Ω ) + (cid:107) h ∇ t ˜ w (cid:107) L (˜Ω ) , where ∇ t is the orthogonal projection of ∇ onto the plane orthogonal to the radialdirection. Note that h∂ r ˜ w = hf ( θ ) (cid:103) ∂ r w, and h ∇ t ˜ w = h (cid:103) ∇ t w + h (cid:103) ∂ r w ∇ t f ( θ ) , so(2.3) (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:46) (cid:107) w (cid:107) H (Ω ) . Since w ( r, θ ) = ˜ w ( rf ( θ ) , θ ), the same argument shows that (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:38) (cid:107) w (cid:107) H (Ω ) ,and therefore that (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:39) (cid:107) w (cid:107) H (Ω ) . where the constants implied in the (cid:39) sign again depend on f .Now L ϕ,ε w ∈ L (Ω ), so using the reasoning in (2.2) gives us that (cid:94) L ϕ,ε w ∈ L ( ˜Ω ), and (cid:107)L ϕ,ε w (cid:107) L (Ω ) (cid:39) (cid:107) (cid:94) L ϕ,ε w (cid:107) L (˜Ω ) . Therefore, by Lemma 2.2, h √ ε (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:46) (cid:107) (cid:94) L ϕ,ε w (cid:107) L (˜Ω ) . Now a calculation shows that L ϕ,ε = h ∂ r − r − (cid:18) − hn + 2 hε log r (cid:19) h∂ r + r − (cid:18) h − hn + h ε ((log r ) − ε ) + h ε log r + (2 − hn ) hε log r + h (cid:52) S n (cid:19) . and then that (cid:94) L ϕ,ε w = f − ( θ ) ˜ L ϕ,ε ˜ w − hE ˜ w where E is a first order semiclassical differential operator with coefficients whichhave bounds independent of h and ε . Therefore h √ ε (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:46) (cid:107) f − ( θ ) ˜ L ϕ,ε ˜ w (cid:107) L (˜Ω ) + h (cid:107) ˜ w (cid:107) H (˜Ω ) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 7
For small enough ε , the last term on the right side can be absorbed into the leftside. Also, | f − | is bounded above, so h √ ε (cid:107) ˜ w (cid:107) H (˜Ω ) (cid:46) (cid:107) ˜ L ϕ,ε ˜ w (cid:107) L (˜Ω ) . for all w ∈ C ∞ (Ω ). Now any w ∈ C ∞ ( ˜Ω ) can be written as ˜ v for some v ∈ C ∞ (Ω ) just by taking v ( r, θ ) = w ( rf ( θ ) , θ ).This finishes the proof. (cid:3) For the next step, we need to fix coordinates on R n +1 \ B . Since ˜Ω lies entirelyon one side of a hyperplane through the origin, we can choose Cartesian coordinates x , . . . , x n +1 so that ˜Ω lies entirely in the intersection of R n +1 \ B with the halfspace x n +1 >
0. We have a map σ : R n +1 \ B ∩ { x n +1 > } → [1 , ∞ ) × (0 , π ) × . . . × (0 , π )by σ ( x , . . . , x n +1 ) = ( r, θ , . . . , θ n )where x = r cos θ x = r sin θ cos θ ... x n = r sin θ . . . sin θ n − cos θ n x n +1 = r sin θ . . . sin θ n . This fixes a set of spherical coordinates on R n +1 \ B ∩ { x n +1 > } . Note thatthis map σ is a diffeomorphism between the two open sets. Since ˜Ω is compactlycontained in R n +1 \ B ∩ { x n +1 > } this diffeomorphism has bounded derivativeson ˜Ω .We will make another change of variables to the Carleman estimate in Lemma2.3 to get a estimate for functions supported in σ ( ˜Ω ). For this we need somenotation for expressing differential operators on R n +1 \ B ∩ { x n +1 > } in sphericalcoordinates. On the portion of the unit sphere S n in R n +1 which lies in the region { x n +1 > } , we can express the Riemannian metric in these coordinates. Then themetric takes the form . . .
00 (sin θ ) . . . . . . θ . . . sin θ n − ) In these coordinates the metric has no dependence on θ n . Note that in the regionnear θ = . . . = θ n − = π , the metric is nearly the Euclidean metric. In particular,if (cid:52) S n is the Laplace-Beltrami operator on the sphere, then h (cid:52) S n , which hascoordinate expression h ∂ θ + h sin θ ∂ θ + . . . + h (sin θ . . . sin θ n − ) ∂ θ n + hE CHUNG on this domain, for some first order semiclassical differential operator E , will differfrom h (cid:52) θ = h ∂ θ + . . . + h ∂ θ n by a second order semiclassical differential operatorwith small coefficients.Functions in g ∈ C ∞ ( ˜Ω ) can be pushed forward to functions in C ∞ ( σ ( ˜Ω ))by taking g ( σ − ( r, θ , . . . , θ n )). It will be helpful to think of these pushed forwardfunctions as functions on R n +11+ = { ( r, θ ) | θ ∈ R n , r ≥ } . Now we can state thefollowing corollary. Corollary 2.4.
Let α = 1+ hε log( rf ( θ )) , β f be a vector field on R n +11+ which agreeswith the coordinate expression of ∇ S n log f ( θ ) on σ ( ˜Ω ) , γ f be a function on R n +11+ which agrees with the coordinate expression of |∇ S n log f ( θ ) | S n on σ ( ˜Ω ) , and L S n be a second order differential operator of R n +11+ which agrees with the coordinateexpression of the Laplacian on the sphere on σ ( ˜Ω ) .Let L ϕ,ε,σ = (cid:0) | γ f | (cid:1) h ∂ r − r ( α + β f · h ∇ θ ) h∂ r + 1 r ( α + h L S n ) . Then for all w ∈ C ∞ ( σ ( ˜Ω )) , (2.4) h √ ε (cid:107) w (cid:107) H ( σ (˜Ω )) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) L ( σ (˜Ω )) The proof follows from the same kind of argument made above.The expression for L ϕ,ε,σ is somewhat messy because of its dependence on θ .Therefore I want to first work with a version where the functions of θ that appearin L ϕ,ε,σ are nearly constant in some sense. Let e n denote the vector field on S n ∩ { x n +1 > } given in coordinates by (0 , . . . , , Proposition 2.5.
Fix K ≥ , and let µ > . Suppose that for all θ such that some ( r, θ ) ∈ Ω , the following conditions hold: | sin θ j − | ≤ µ where j = 1 , . . . , n − and |∇ S n log f − Ke n | S n ≤ µ. If µ is small enough, then for all w ∈ C ∞ σ ( ˜Ω) , h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) . Note that the hypotheses imply that on σ ( ˜Ω ), | β f − (0 , . . . , , K ) | ≤ C µ , | γ f − K | ≤ C µ , and if h L S n = a h ∂ θ + . . . + a n h ∂ θ n + b h ∂ θ + . . . + b n h ∂ θ n then | a j − | ≤ C µ for some constant C µ which goes to zero if µ goes to zero. C µ may depend on K ,but we are treating K as fixed, so this will be ok. PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 9
We may as well assume that β f , γ f , and the coefficients of L S n are extended tothe rest of R n +11+ in such a way that these conditions continue to hold.To prove this proposition, I want to divide w into small and large frequency parts,and prove the estimate for each part separately. Recall that R n +11+ = { ( r, θ ) | θ ∈ R n , r ≥ } . Let S ( R n +11+ ) be the restrictions to R n +11+ of Schwartz functions on R n +1 . Note that functions in C ∞ ( σ ( ˜Ω )) are in S ( R n +11+ ).Let r and r be such that | K | | K | < r < r ≤
12 + | K | | K | ) < , and let δ and δ be such that δ > δ >
0. Let ρ ∈ C ∞ ( R n ) be a cutoff functionsuch that ρ ( ξ ) = 0 if | ξ | > r or | ξ n | > δ , and ρ ( ξ ) = 1 if | ξ | ≤ r or | ξ n | ≤ δ .Let the hat ˆ indicate the (semiclassical) Fourier transform in the θ variablesonly. (In general, Fourier transforms here will be in the θ variables only unlessotherwise indicated.) For w ∈ C ∞ ( σ ( ˜Ω)), define w s and w (cid:96) by ˆ w s = ρ ( ξ ) ˆ w andˆ w (cid:96) = (1 − ρ ( ξ )) ˆ w , so w = w s + w (cid:96) . Lemma 2.6.
There exists µ > and choices of r , r , δ , and δ such that if thehypotheses of Proposition 2.5 hold, then h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( σ (˜Ω)) , for all w ∈ C ∞ ( σ ( ˜Ω)) , where w s is defined as above. Lemma 2.7.
There exists µ > such that if the hypotheses of Proposition 2.5hold, then h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( σ (˜Ω)) , for all w ∈ C ∞ ( σ ( ˜Ω)) , where w (cid:96) is defined as above. Taken together, these two lemmas imply Proposition 2.5. To see why, first weneed a lemma.
Lemma 2.8. If A is a pseudodifferential operator of order m ≥ on R n , it can beapplied to Schwartz functions on R n +1 , by taking Af ( x , . . . , x n +1 ) for each fixed x n +1 . Then as an operator defined for functions on R n +1 it extends to an operatorfrom H k + m ( R n +1 ) to H k ( R n +1 ) , (cid:107) Af (cid:107) H k ( R n +1 ) (cid:46) (cid:107) f (cid:107) H k + m ( R n +1 ) Proof.
I will give a proof for the case where k is a non-negative integer. Let( x, y ); x ∈ R n , y ∈ R denote the coordinates on R n +1 . Then if f is Schwartz, (cid:107) Af (cid:107) H k ( R n +1 ) = (cid:88) ≤| α | + j ≤ k (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) Now (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) = (cid:90) ∞−∞ (cid:90) R n | h | α | + j ∂ αx ∂ jy Af | dx dy. Here A and ∂ jy commute, so (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) = (cid:90) ∞−∞ (cid:90) R n | h | α | ∂ αx A ( h j ∂ jy f ) | dx dy (cid:46) (cid:90) ∞−∞ (cid:107) A ( h j ∂ jy f ) (cid:107) H | α | ( R n ) dy. Therefore, by the boundedness of A , (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) (cid:46) (cid:90) ∞−∞ (cid:107) ( h j ∂ jy f ) (cid:107) H | α | + m ( R n ) dy. Since | α | + m ≥ (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) (cid:46) (cid:107) ( h j ∂ jy f ) (cid:107) H | α | + m ( R n +1 ) . Therefore (cid:107) h | α | + j ∂ αx ∂ jy Af (cid:107) L ( R n +1 ) (cid:46) (cid:107) f (cid:107) H | α | + j + m ( R n +1 ) (cid:46) (cid:107) f (cid:107) H k + m ( R n +1 ) , so (cid:107) Af (cid:107) H k ( R n +1 ) (cid:46) (cid:107) f (cid:107) H k + m ( R n +1 ) . (cid:3) Note that this still holds if A depends on y , as long as A has bounds uniform in y . Proof of Proposition 2.5.
Adding together the estimates from Lemma 2.6 and Lemma2.7 gives h √ ε ( (cid:107) w s (cid:107) L ( R n +11+ ) + (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) ) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( σ (˜Ω)) Since w s + w (cid:96) = w , h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( σ (˜Ω)) . For small enough ε , we can absorb the last term into the left side to give h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) . Since (1 + | γ f | ) > K − C µ , for µ small enough, we have h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107) (1+ | γ f | ) − L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + (cid:107) (1+ | γ f | ) − L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) . Now w s = P w , where P is the semiclassical pseudodifferential operator of order 0on R n with symbol ρ ( ξ ). P commutes with ∂ r . Therefore there are some operators PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 11 E and E which for each fixed r ∈ [1 , ∞ ) are semiclassical pseudodifferentialoperators of order 1 and 0 respectively, on R n such that (cid:107) (1 + | γ f | ) − L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) = (cid:107) (1 + | γ f | ) − L ϕ,ε,σ P w (cid:107) H − ( R n +11+ ) (cid:46) (cid:107) P (1 + | γ f | ) − L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) E h∂ r + E w (cid:107) H − ( R n +11+ ) There is no hE − h ∂ r in the error term because the coefficient of h ∂ r in (1 + | γ f | ) − L ϕ,ε,σ is just 1.By the lemma above, E ∗ is bounded from H ( R n +11+ ) to L ( R n +11+ ), so by duality, E is bounded from L ( R n +11+ ) to H − ( R n +11+ ).In addition, E ∗ is bounded from H ( R n +11+ ) to H ( R n +11+ ). Moreover, E ∗ takesfunctions with trace 0 on the boundary of R n +11+ to other functions with trace 0 onthe boundary of R n +11+ , so by duality, E is bounded from H − ( R n +11+ ) to H − ( R n +11+ ).These bounds must be uniform in r for the range of r allowed on the support of w .Therefore (cid:107) (1+ | γ f | ) − L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) (cid:46) (cid:107) P (1+ | γ f | ) − L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) Now by the same lemma, P is bounded from H ( R n +11+ ) to H ( R n +11+ ). Moreover, if u has trace zero on the boundary of R n +11+ , then so does P u , so P is bounded from H ( R n +11+ ) to H ( R n +11+ ). Since ρ is real valued, P is also self adjoint, so by duality P is bounded from H − ( R n +11+ ) to H − ( R n +11+ ). Therefore (cid:107) (1+ | γ f | ) − L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) (cid:46) (cid:107) (1+ | γ f | ) − L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) . and thus (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) . Similarly, (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) . Therefore h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) . Again the last term can be absorbed into the left side for small enough ε , so h √ ε (cid:107) w (cid:107) L ( σ (˜Ω)) (cid:46) (cid:107)L ϕ,ε,σ w (cid:107) H − ( R n +11+ ) for each w ∈ C ∞ ( σ ( ˜Ω)) as desired. (cid:3) Therefore we need only deal with the proofs of Lemmas 2.6 and 2.7.3.
Small Frequency Operators
I want to begin by describing some operators for use in proving the small fre-quency case.Consider the function F : R n → C given by F ( ξ ) = 11 + | K | (cid:16) iKξ n + (cid:112) iKξ n − ( Kξ n ) + (1 + | K | ) | ξ | − | K | (cid:17) . where the square root is taken to mean the branch of the square root function withnonnegative imaginary part. If r and δ are chosen small enough, then this isnearly continuous on the support of ρ . To be more precise, F is smooth exceptwhere τ K ( ξ ) = 2 iKξ n − ( Kξ n ) + (1 + | K | ) | ξ | − | K | lies on the nonnegative real axis, where this branch of the square root has its branchcut. This occurs when ξ n = 0 and | ξ | ≥ | K | | K | , and gives a jump discontinuityof size 2 (cid:112) (1 + | K | ) | ξ | − | K | . However, on the support of ρ , | ξ | ≤ r , so for r close to | K | | K | the maximum possible size of the jump discontinuity is small.Therefore for any δ >
0, we can pick a smooth function F s ( ξ ) such that | F s ( ξ ) − F ( ξ ) | ≤ δ on the support of ρ , by choosing r small enough. Note that the derivatives of F s may depend on r , r , δ , and δ . Since the choice of these in turn depends on δ ,the derivatives of F s are bounded by a quantity that depends on δ .Now consider the necessary bounds on F s . On the support of ρ , the imaginarypart of τ K must lie in the interval [ − Kδ , Kδ ]. The real part of τ K is given by − ( Kξ n ) − | K | + (1 + | K | ) | ξ | . We have that | ξ | ≤ r on the support of ρ . We can choose r so close to K K that(1 + K ) r − K ≤ δ . Then the real part of τ K is bounded above by δ on the support of ρ . Therefore onthe support of ρ , (Re( τ K ) , Im( τ K )) ∈ ( −∞ , δ ] × [ − Kδ , Kδ ], and so by taking δ small enough, we can ensure that the real part of √ τ K has absolute value lessthan on the support of ρ .Therefore if δ is small enough, Re( F s ) , | F s | > K on the support of ρ . Wecan define F s outside the support of ρ so that Re( F s ) , | F s | >
12 11+ | K | for all ξ , andRe( F s ) , | F s | (cid:39) | ξ | for large | ξ | .Now for u ∈ S ( R n +11+ ), define J s u by (cid:100) J s u ( r, ξ ) = (cid:18) F s ( ξ ) r + h∂ r (cid:19) ˆ u ( r, ξ ) . This operator has adjoint J ∗ s given by (cid:100) J ∗ s u ( r, ξ ) = (cid:32) F s ( ξ ) r − h∂ r (cid:33) ˆ u ( r, ξ ) . These operators have (right) inverses defined by (cid:91) J − s u ( r, ξ ) = h − (cid:90) r ˆ u ( t, ξ ) (cid:18) tr (cid:19) Fs ( ξ ) h dt and (cid:92) J ∗− s u ( r, ξ ) = h − (cid:90) ∞ r ˆ u ( t, ξ ) (cid:16) rt (cid:17) Fs ( ξ ) h dt. Each of these is well defined for functions in S ( R n +11+ ). PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 13
Define the weighted Sobolev space H r ( R n +11+ ) by the norm (cid:107) u (cid:107) H r ( R n +11+ ) = (cid:13)(cid:13)(cid:13) ur (cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:107) h∂ r u (cid:107) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) hr ∇ θ u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) . Note that since σ ( ˜Ω ) lies in the set 1 ≤ r ≤ C σ (˜Ω ) for some C σ (˜Ω ) , H and H r norms are comparable for functions supported on σ ( ˜Ω ), with constantsof comparability depending only on C σ (˜Ω ) . This holds more generally for anyfunctions supported in 1 ≤ r ≤ C σ (˜Ω ) . Lemma 3.1. J s , J ∗ s , J − s , and J ∗− s extend as bounded maps J s , J ∗ s : H r ( R n +11+ ) → L ( R n +11+ ) and J − s , J ∗− s : L ( R n +11+ ) → H r ( R n +11+ ) . Moreover, the extensions of J ∗ s and J ∗− s are isomorphisms.Proof. Consider J s first. If u ∈ S ( R n +11+ ), then (cid:107) J s u (cid:107) L ( R n +11+ ) = h − n (cid:107) (cid:100) J s u (cid:107) L ( R n +11+ ) = h − n (cid:13)(cid:13)(cid:13)(cid:13) F s ( ξ ) r ˆ u + h∂ r ˆ u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) ≤ h − n (cid:13)(cid:13)(cid:13)(cid:13) F s ( ξ ) r ˆ u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + h − n (cid:107) h∂ r ˆ u (cid:107) L ( R n +11+ ) (cid:46) h − n (cid:13)(cid:13)(cid:13)(cid:13) | ξ | r ˆ u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:107) h∂ r u (cid:107) L ( R n +11+ ) (cid:46) (cid:13)(cid:13)(cid:13) ur (cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) hr ∇ θ u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:107) h∂ r u (cid:107) L ( R n +11+ ) = (cid:107) u (cid:107) H r ( R n +11+ ) By the density of S ( R n +11+ ) in H r ( R n +11+ ), J s extends to a bounded map J s : H r ( R n +11+ ) → L ( R n +11+ ). The proof for J ∗ s is similar.Now consider J − s . If u ∈ S ( R n +11+ ), then (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) r (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr = (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) h − (cid:90) r ˆ u ( t, ξ ) (cid:18) tr (cid:19) Fs ( ξ ) h dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r − dr ≤ (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) h − (cid:90) r ˆ u ( t, ξ ) (cid:18) tr (cid:19) Fs ( ξ ) h dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r − dr. By a change of variables, we get (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) r (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr = (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) h − (cid:90) ˆ u ( rt, ξ ) t Fs ( ξ ) h dt (cid:12)(cid:12)(cid:12)(cid:12) dr. Then Minkowski’s inequality gives us (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) r (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr ≤ h − (cid:32)(cid:90) (cid:18)(cid:90) ∞ | ˆ u ( rt, ξ ) t Fs ( ξ ) h | dr (cid:19) dt (cid:33) = h − (cid:32)(cid:90) (cid:18)(cid:90) ∞ | ˆ u ( rt, ξ ) | dr (cid:19) t Re Fs ( ξ ) h dt (cid:33) . Changing variables again, we get (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) r (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr ≤ h − (cid:32)(cid:90) (cid:18)(cid:90) ∞ | ˆ u ( r, ξ ) | dr (cid:19) t Re Fs ( ξ ) h t − dt (cid:33) = h − (cid:90) ∞ | ˆ u ( r, ξ ) | dr (cid:32) h Re F s ( ξ ) + h (cid:33) (cid:39) (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) ˆ u ( r, ξ )1 + | ξ | (cid:12)(cid:12)(cid:12)(cid:12) dr ≤ (cid:90) ∞ | ˆ u ( r, ξ ) | dr Therefore (cid:13)(cid:13)(cid:13)(cid:13) r J − s u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) = h − n (cid:13)(cid:13)(cid:13)(cid:13) r (cid:91) J − s u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) = h − n (cid:90) R n (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) r (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr dξ (cid:46) h − n (cid:90) R n (cid:90) ∞ | ˆ u | dr dξ = h − n (cid:107) ˆ u (cid:107) L ( R n +11+ ) = (cid:107) u (cid:107) L ( R n +11+ ) Similarly, (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12) ξr (cid:91) J − s u (cid:12)(cid:12)(cid:12)(cid:12) dr (cid:46) (cid:90) ∞ | ˆ u ( r, ξ ) | dr so (cid:13)(cid:13)(cid:13)(cid:13) hr ∇ θ J − s u (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) (cid:46) (cid:107) u (cid:107) L ( R n +11+ ) . Finally, h∂ r (cid:91) J − s u = − (cid:18) F s ( ξ ) r (cid:19) (cid:91) J − s u + ˆ u so (cid:90) ∞ (cid:12)(cid:12)(cid:12) h∂ r (cid:91) J − s u (cid:12)(cid:12)(cid:12) dr (cid:46) (cid:90) ∞ | ˆ u ( r, ξ ) | dr PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 15 and (cid:13)(cid:13) h∂ r J − s u (cid:13)(cid:13) L ( R n +11+ ) (cid:46) (cid:107) u (cid:107) L ( R n +11+ ) by the same logic.Putting all of this together gives (cid:107) J − s u (cid:107) H r ( R n +11+ ) (cid:46) (cid:107) u (cid:107) L ( R n +11+ ) for u ∈ S ( R n +11+ ). Therefore by the density of S ( R n +11+ ) in L ( R n +11+ ), J − s extendsto a bounded map J − s : L ( R n +11+ ) → H r ( R n +11+ ) . Again the proof for J ∗− s is similar.It remains to show that the extensions of J ∗ s and J ∗− s are isomorphisms. Notethat if u ∈ S ( R n +11+ ), then F ( J ∗ s J ∗− s u ) = (cid:32) F s ( ξ ) r − h∂ r (cid:33) (cid:92) J ∗− s u ( r, ξ )= (cid:32) F s ( ξ ) r − h∂ r (cid:33) h − (cid:90) ∞ r ˆ u ( t, ξ ) (cid:16) rt (cid:17) Fs ( ξ ) h dt = (cid:32) F s ( ξ ) r (cid:33) (cid:92) J ∗− s u − (cid:32) F s ( ξ ) r (cid:33) (cid:92) J ∗− s u + ˆ u ( r, ξ )= ˆ u ( r, ξ ) . (Here F stands for the semiclassical Fourier transform in the θ variables, just likethe hatˆ. I will use this notation when the hat becomes unwieldy.) Therefore(3.1) J ∗ s J ∗− s u = u for all u ∈ S ( R n +11+ ).On the other hand, integration by parts gives F ( J ∗− s J ∗ s u ) = h − (cid:90) ∞ r (cid:32) F s ( ξ ) t − h∂ t (cid:33) ˆ u ( t, ξ ) (cid:16) rt (cid:17) Fs ( ξ ) h dt = h − (cid:90) ∞ r (cid:32) F s ( ξ ) t (cid:33) ˆ u ( t, ξ ) (cid:16) rt (cid:17) Fs ( ξ ) h dt + h − (cid:90) ∞ r ˆ u ( t ) (cid:16) rt (cid:17) Fs ( ξ ) h (cid:32) − F s ( ξ ) t (cid:33) dt − ˆ u ( t, ξ ) (cid:16) rt (cid:17) Fs ( ξ ) h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ t = r = ˆ u ( r, ξ ) , so(3.2) J ∗− s J ∗ s u = u for all u ∈ S ( R n +11+ ). Now consider J ∗− s . By (3.1), (cid:107) J ∗− s u (cid:107) H r ( R n +11+ ) (cid:38) (cid:107) u (cid:107) L ( R n +11+ ) for all u ∈ S ( R n +11+ ). By the density of S ( R n +11+ ) in L ( R n +11+ ), this holds for all u ∈ L ( R n +11+ ). Therefore J ∗− s is 1-1 with closed range. Now (3.2) implies that S ( R n +11+ )is in the range of J ∗− s , so by the density of S ( R n +11+ ) in H r ( R n +11+ ), H r ( R n +11+ ) is inthe range of J ∗− s . Therefore J ∗− s : L ( R n +11+ ) → H r ( R n +11+ ) is an isomorphism.Similarly, (3.2) shows that (cid:107) J ∗ s u (cid:107) L ( R n +11+ ) (cid:38) (cid:107) u (cid:107) H r ( R n +11+ ) for all u ∈ S ( R n +11+ ), and hence for all u ∈ H r ( R n +11+ ). Therefore J ∗ s is 1-1 with closedrange. Then (3.1) implies that S ( R n +11+ ) is in the range of J ∗ s , and so L ( R n +11+ ) isin the range of J ∗ s . Therefore J ∗ s : H r ( R n +11+ ) → L ( R n +11+ ) is also an isomorphism.Note that J − s J s u (cid:54) = u in general, because integration by parts will pick up aboundary term at r = 1. Therefore the extensions of J s and J − s are not isomor-phisms. (cid:3) Let H r, ( R n +11+ ) denote the subspace of H r ( R n +11+ ) consisting of functions withtrace zero on the hyperplane r = 1, and let H − r ( R n +11+ ) denote the dual space to H r, ( R n +11+ ).The operator J s is closely related to a pseudodifferential operator. In particular,it has the following properties, which will be needed later. Lemma 3.2. i) Suppose w ∈ S ( R n +11+ ) . Then if Q is a second order semiclassicaldifferential operator with smooth bounded coefficients on R n +11+ , then (cid:107) ( J s Q − QJ s ) w (cid:107) H − r ( R n +11+ ) (cid:46) hC δ (cid:107) rw (cid:107) H ( R n +11+ ) ii) Suppose w ∈ S ( R n +11+ ) . Let χ ∈ S ( R n +11+ ) . Then (cid:107) J s χJ − s w (cid:107) L ( R n +11+ ) (cid:38) (cid:107) χw (cid:107) L ( R n +11+ ) − hC δ (cid:107) rw (cid:107) L ( R n +11+ ) The C δ factor is written explicitly to track the δ dependence. Proof. i) First, note that multiplication by 1 /r is a bounded operator from H r, ( R n +11+ )to H ( R n +11+ ). Therefore by duality, it is a bounded operator from H − ( R n +11+ ) to H − r ( R n +11+ ), and so (cid:107) ( J s Q − QJ s ) w (cid:107) H − r ( R n +11+ ) (cid:46) (cid:107) r ( J s Q − QJ s ) w (cid:107) H − ( R n +11+ ) Note that J s = h∂ r + r T , where T is a semiclassical pseudodifferential operatoron R n of order 1. Meanwhile, Q can be written as a combination of ∂ r derivativesand differential operators on R n : Q = Ah ∂ r + Bh∂ r + C where A, B, and C are (perhaps r-dependent) differential operators of orders 0 , , and 2 respectively on R n for each fixed r , with bounds uniform in r .If w ∈ S ( R n +11+ ), then Qw ∈ S ( R n +11+ ).Then (cid:107) r ( J s Q − QJ s ) w (cid:107) H − ( R n +11+ ) = (cid:107) r [ h∂ r + 1 r T, Ah ∂ r + Bh∂ r + C ] w (cid:107) H − ( R n +11+ ) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 17
Note that T commutes with ∂ r . Therefore we have (cid:107) r ( J s Q − QJ s ) w (cid:107) H − ( R n +11+ ) ≤ (cid:107) r [ h∂ r , Q ] w (cid:107) H − ( R n +11+ ) + (cid:107) [ T, A ] h ∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) hr [ T, A ] h∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) h r [ T, A ] w (cid:107) H − ( R n +11+ ) + (cid:107) [ T, B ] h∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) hr [ T, B ] w (cid:107) H − ( R n +11+ ) + (cid:107) [ T, C ] w (cid:107) H − ( R n +11+ ) Now r [ h∂ r , Q ] = hrE = hEr + h E (cid:48) , where E and E (cid:48) are second and first or-der semiclassical differential operators, by the product rule. Meanwhile, [ T, A ] = hE , [ T, B ] = hE , and [ T, C ] = hE , where E , E , and E are semiclassical pseu-dodifferential operators on R n of orders 0 , , and 2 respectively. Therefore (cid:107) r ( J s Q − QJ s ) w (cid:107) H − ( R n +11+ ) ≤ (cid:107) hErw (cid:107) H − ( R n +11+ ) + (cid:107) h E (cid:48) w (cid:107) H − ( R n +11+ ) + (cid:107) hE h ∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) h r E h∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) h r E w (cid:107) H − ( R n +11+ ) + (cid:107) hE h∂ r w (cid:107) H − ( R n +11+ ) + (cid:107) hr E w (cid:107) H − ( R n +11+ ) + (cid:107) hE w (cid:107) H − ( R n +11+ ) Now E is bounded from H ( R n +11+ ) to H − ( R n +11+ ), and E (cid:48) is bounded from L ( R n +11+ ) to H − ( R n +11+ ). In addition, by the lemma on the boundedness of pseudo-differential operators on R n in R n +1 , E ∗ is bounded from H ( R n +11+ ) to L ( R n +11+ ),so by duality, E is bounded from L ( R n +11+ ) to H − ( R n +11+ ). Meanwhile, if G isan invertible semiclassical pseudodifferential operator of order 1 on R n , then G isbounded from L ( R n +11+ ) to H − ( R n +11+ ), and G − E is bounded from H ( R n +11+ ) to L ( R n +11+ ), so E is bounded from H ( R n +11+ ) to H − ( R n +11+ ). Finally, E ∗ is boundedfrom H ( R n +11+ ) to H ( R n +11+ ) and maps functions with trace 0 on the boundary of R n +11+ to other functions with trace 0 on that boundary, so it is bounded from H ( R n +11+ ) to H ( R n +11+ ). Therefore by duality, E is bounded from H − ( R n +11+ ) to H − ( R n +11+ ). Also, note that r ≤ R n +11+ .Therefore (cid:107) ( J s Q − QJ s ) w (cid:107) H − r ( R n +11+ ) (cid:46) h (cid:107) rw (cid:107) H ( R n +11+ ) + h (cid:107) w (cid:107) L ( R n +11+ ) + hC δ (cid:107) h ∂ r w (cid:107) H − ( R n +11+ ) + h C δ (cid:107) h∂ r w (cid:107) H − ( R n +11+ ) + h C δ (cid:107) w (cid:107) H − ( R n +11+ ) + hC δ (cid:107) h∂ r w (cid:107) L ( R n +11+ ) + h C δ (cid:107) w (cid:107) L ( R n +11+ ) + hC δ (cid:107) w (cid:107) H ( R n +11+ ) (cid:46) hC δ (cid:107) rw (cid:107) H ( R n +11+ ) The C δ comes from the dependence of T upon δ . ii) Now (cid:107) J s χJ − s w (cid:107) L ( R n +11+ ) = (cid:107) ( h∂ r + Tr ) χJ − s w (cid:107) L ( R n +11+ ) ≥ (cid:107) χ ( h∂ r + Tr ) J − s w (cid:107) L ( R n +11+ ) −(cid:107) hEJ − s w (cid:107) L ( R n +11+ ) where for each fixed r , E is an order zero pseudodifferential operator on S n . There-fore (cid:107) J s χJ − s w (cid:107) L ( R n +11+ ) ≥ (cid:107) χJ s J − s w (cid:107) L ( R n +11+ ) − hC δ (cid:107) J − s w (cid:107) L ( R n +11+ ) ≥ (cid:107) χw (cid:107) L ( R n +11+ ) − hC δ (cid:107) rw (cid:107) L ( R n +11+ ) (cid:3) It would be nice if (cid:107) J s w (cid:107) H − r ( R n +11+ ) (cid:39) (cid:107) w (cid:107) L ( R n +11+ ) , but this is not quite true:since J s is not an isomorphism from H ,r ( R n +11+ ) to L ( R n +11+ ), there is no reason toexpect this to be true. Instead, we have the following result. Lemma 3.3.
Suppose u ∈ S ( R n +11+ ) . If g is defined by ˆ g ( r, ξ ) = 2Re F s ( ξ ) − hh (cid:90) ∞ ˆ u ( t, ξ ) r − Fs ( ξ ) h t − Fs ( ξ ) h dt, then (cid:107) J s u (cid:107) H − r ( R n +11+ ) (cid:39) (cid:107) u − g (cid:107) L ( R n +11+ ) Proof.
Suppose u ∈ S ( R n +11+ ). Define g as above. We have (cid:90) ∞ | ˆ g ( r, ξ ) | dr = (cid:12)(cid:12)(cid:12)(cid:12) F s − hh (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞ ˆ u ( t, ξ ) r − Fsh t − Fsh dt (cid:12)(cid:12)(cid:12)(cid:12) dr ≤ (cid:12)(cid:12)(cid:12)(cid:12) F s − hh (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) ∞ (cid:90) ∞ | ˆ u ( t, ξ ) | dt (cid:90) ∞ ( rt ) − Re Fsh dt dr = (cid:12)(cid:12)(cid:12)(cid:12) F s − hh (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) ∞ | ˆ u ( t, ξ ) | dt (cid:90) ∞ ( t ) − Re Fsh dt (cid:90) ∞ ( r ) − Re Fsh dr = (cid:12)(cid:12)(cid:12)(cid:12) F s − hh (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) ∞ | ˆ u ( t, ξ ) | dt (cid:18) h F s − h (cid:19) (cid:18) h F s − h (cid:19) = (cid:90) ∞ | ˆ u ( t, ξ ) | dt Therefore g makes sense as an element of L ( R n +11+ ), and (cid:107) g (cid:107) L ( R n +11+ ) ≤ (cid:107) u (cid:107) L ( R n +11+ ) .Note that (cid:100) J s g = (cid:18) F s ( ξ ) r + h∂ r (cid:19) ˆ g = 0 . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 19
Therefore (cid:107) J s u (cid:107) H − r ( R n +11+ ) = sup w ∈ H ,r ( R n +11+ ) ,w (cid:54) =0 | ( J s u, w ) |(cid:107) w (cid:107) H r ( R n +11+ ) = sup w ∈ H ,r ( R n +11+ ) ,w (cid:54) =0 | ( J s ( u − g ) , w ) |(cid:107) w (cid:107) H r ( R n +11+ ) = sup w ∈ H ,r ( R n +11+ ) ,w (cid:54) =0 | ( u − g, J ∗ s w ) |(cid:107) w (cid:107) H r ( R n +11+ ) . Since J ∗ s : H r ( R n +11+ ) → L ( R n +11+ ) is an isomorphism,(3.3) (cid:107) J s u (cid:107) H − r ( R n +11+ ) (cid:39) sup w ∈ H ,r ( R n +11+ ) ,J ∗ s w (cid:54) =0 | ( u − g, J ∗ s w ) |(cid:107) J ∗ s w (cid:107) L ( R n +11+ ) Now J ∗ s w ∈ L ( R n +11+ ), so (cid:107) J s u (cid:107) H − r ( R n +11+ ) (cid:46) (cid:107) u − g (cid:107) L ( R n +11+ ) . On the other hand, note that u − g = J ∗ s J ∗− s ( u − g ). J ∗− s ( u − g ) ∈ H r ( R n +11+ ),and (cid:92) J ∗− s g (1 , ξ ) = h − (cid:90) ∞ h − (2Re F s − h ) (cid:18)(cid:90) ∞ ˆ u ( t, ξ ) a − Fsh t − Fsh dt (cid:19) a − Fsh da = h − (cid:90) ∞ ˆ u ( t, ξ ) t − Fsh dt ( h − )(2Re F s − h ) (cid:90) ∞ a − Fsh da = h − (cid:90) ∞ ˆ u ( t, ξ ) t − Fsh dt = (cid:92) J ∗− s u (1 , ξ )Therefore J ∗− s ( u − g ) ∈ H r, ( R n +11+ ). If u − g = 0, then the lemma is true by (3.3).Otherwise, we can pick w = J ∗− s ( u − g ) in (3.3) to show that (cid:107) J s u (cid:107) H − r ( R n +11+ ) (cid:38) (cid:107) u − g (cid:107) L ( R n +11+ ) . This finishes the proof. (cid:3) The Small Frequency Case
Now we are ready to prove the small frequency case. Suppose χ ( r, θ ) ∈ C ∞ ( R n +11+ )is a cutoff function which is 1 on σ ( ˜Ω) and has support inside σ ( ˜Ω ).If w ∈ C ∞ ( σ ( ˜Ω)), then w s ∈ S ( R n +11+ ), supported away from r = 1. There-fore J − s w s ∈ S ( R n +11+ ), and is supported away from r = 1. Then χ J − s w s is in C ∞ ( σ ( ˜Ω )). Therefore by (2.4), h √ ε (cid:107) χ J − s w s (cid:107) H ( σ (˜Ω )) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( σ (˜Ω )) . Since χ J − s w s ∈ C ∞ ( σ ( ˜Ω )), the H and H r norms are comparable, so h √ ε (cid:107) χ J − s w s (cid:107) H r ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( R n +11+ ) . Using the boundedness properties from Lemma 3.1, h √ ε (cid:107) J s χ J − s w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( R n +11+ ) so applying the second part of Lemma 3.2 h √ ε (cid:107) χ w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( R n +11+ ) + C δ h ε (cid:107) rw s (cid:107) L ( R n +11+ ) . Now χ w s = χ P w . Since w is only supported on the region where χ is identically1, χ w s = P w + O ( h ∞ ) Ew = w s + O ( h ∞ ) Ew where E is an pseudodifferential operator of order 0 (actually a smoothing operator)on R n . Therefore h √ ε (cid:107) χ w s (cid:107) L ( R n +11+ ) (cid:38) h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) − O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) , and so h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( R n +11+ ) + C δ h ε (cid:107) rw s (cid:107) L ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . For small enough h , the second last term can be absorbed into the left side ( r isbounded on the support of w s ) to give h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ χ J − s w s (cid:107) L ( σ (˜Ω )) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . By the product rule, h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107) χ L ϕ,ε,σ J − s w s (cid:107) L ( σ (˜Ω )) + h (cid:107) J − s w s (cid:107) H ( σ (˜Ω )) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . On σ ( ˜Ω ), the H and H r norms are comparable, so h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ J − s w s (cid:107) L ( R n +11+ ) + h (cid:107) J − s w s (cid:107) H r ( R n +11+ )) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . Using the boundedness properties again, h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ J − s w s (cid:107) L ( R n +11+ ) + h (cid:107) w s (cid:107) L ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . The second last term can be absorbed into the left side to give(4.1) h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ J − s w s (cid:107) L ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . I want to combine this last inequality with Lemma 3.3 to get h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107) J s L ϕ,ε,σ J − s w s (cid:107) H − r ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . To do this I need to show that if u = L ϕ,ε,σ J − s w s , then the function g defined inLemma 3.3 satisfies a bound like (cid:107) g (cid:107) L ( R n +11+ ) ≤ (cid:107) u (cid:107) L ( R n +11+ ) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 21
This is not true in general, but happens in this case because of the particular formof u . Let v = J − s w s , and consider ˆ g ( r, ξ ).ˆ g = 2Re F s − hh (cid:90) ∞ (cid:92) L ϕ,ε,σ v ( t, ξ ) r − Fsh t − Fsh dt = 2Re F s − hh (cid:90) ∞ h ∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ F ( | γ f | h ∂ t v )( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ t h∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ t F (cid:18) hε log( tf ( θ )) h∂ t v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ t F ( β f · h ∇ θ h∂ t v ) ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ t F ((1 + h L S n ) v )( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ t F (cid:18)(cid:18) (2 hε log( tf ( θ )) + h ε log ( tf ( θ )) (cid:19) v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt.
Here F represents the same thing the hat ˆ does in the case where LaTeX’s widehat looks too strange to be used.Rewriting, we haveˆ g = 2Re F s − hh (cid:90) ∞ (1 + K ) h ∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ t (1 + iKξ n ) h∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ t (1 − | ξ | )ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ F (( | γ f | − K ) h ∂ t v )( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ F (cid:18) t hε log( tf ( θ )) h∂ t v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ F (cid:18) t ( β f − Ke n ) · h ∇ θ h∂ t v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ F (cid:18) t (cid:18) (2 hε log( tf ( θ )) + h ε log ( tf ( θ )) (cid:19) v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt. + 2Re F s − hh (cid:90) ∞ t F ( h ( L S n − (cid:52) θ ) v )( t, ξ ) r − Fsh t − Fsh dt Integrating by parts in the first term gives2Re F s − hh (cid:90) ∞ (1 + K ) h ∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt = 2Re F s − hh (cid:90) ∞ F s t (1 + K ) h∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt = 2Re F s − hh (cid:90) ∞ (cid:18) F s t (cid:19) (1 + K )ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ h F s t (1 + K )ˆ v ( t, ξ ) r − Fsh t − Fsh dt.
There are no boundary terms from the integration by parts, because w is supportedaway from r = 1, and hence w s and v are as well. Integrating by parts in the secondterm gives 2Re F s − hh (cid:90) ∞ t (1 + iKξ n ) h∂ t ˆ v ( t, ξ ) r − Fsh t − Fsh dt = 2Re F s − hh (cid:90) ∞ F sk t (1 + iKξ n )ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ ht (1 + iKξ n )ˆ v ( t, ξ ) r − Fsh t − Fsh dt Therefore we haveˆ g = 2Re F s − hh (cid:90) ∞ t ((1 + K ) F s − iKξ n ) F s + 1 − | ξ | )ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ h F s t (1 + K )ˆ v ( t, ξ ) r − Fsh t − Fsh dt. − F s − hh (cid:90) ∞ ht (1 + iKξ n )ˆ v ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ F (( | γ f | − K ) h ∂ t v )( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ F (cid:18) t hε log( tf ( θ )) h∂ t v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt − F s − hh (cid:90) ∞ F (cid:18) t ( β f − Ke n ) · h ∇ θ h∂ t v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt + 2Re F s − hh (cid:90) ∞ F (cid:18) t (cid:18) (2 hε log( tf ( θ )) + h ε log ( tf ( θ )) (cid:19) v (cid:19) ( t, ξ ) r − Fsh t − Fsh dt. + 2Re F s − hh (cid:90) ∞ t F ( h ( L S n − (cid:52) θ ) v )( t, ξ ) r − Fsh t − Fsh dt PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 23
Applying the same reasoning as in Lemma 3.3, (cid:107) g (cid:107) L ( R n +11+ ) ≤ h − n (cid:13)(cid:13)(cid:13)(cid:13) r ((1 + K ) F s − iKξ n ) F s + 1 − | ξ | )ˆ v ( r, ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + h − n (cid:13)(cid:13)(cid:13)(cid:13) h F s r (1 + K )ˆ v ( r, ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) hr (1 + Kh∂ θ n ) v ( r, θ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:107) ( | γ f | − K ) h ∂ r v (cid:107) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) r hε log( rf ( θ )) h∂ r v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) r ( β f − Ke n ) · h ∇ θ h∂ r v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:13)(cid:13)(cid:13)(cid:13) r (cid:18) hε log( rf ( θ )) + h ε log ( rf ( θ )) (cid:19) v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) + (cid:107) h r ( L S n − (cid:52) θ ) v (cid:107) L ( R n +11+ ) (cid:46) h − n (cid:107) ((1 + K ) F s − iKξ n ) F s + 1 − | ξ | )ˆ v ( r, ξ ) (cid:107) L ( R n +11+ ) + h − n (cid:107) hF s (1 + K )ˆ v ( r, ξ ) (cid:107) L ( R n +11+ ) + h (cid:107) v (cid:107) H ( R n +11+ ) + C µ (cid:107) h ∂ r v (cid:107) L ( R n +11+ ) + h ε (cid:107) h∂ r v (cid:107) L ( R n +11+ ) + C µ (cid:107) h ∇ θ h∂ r v (cid:107) L ( R n +11+ ) + h ε (cid:107) v (cid:107) L ( R n +11+ ) + C µ (cid:107) v (cid:107) H ( R n +11+ ) where C µ goes to zero as µ does.Now F s ( ξ ) is designed so that F s ( ξ ) is very nearly a solution to (1 + K ) X − iKξ n ) X + 1 − | ξ | = 0 when ˆ w s (cid:54) = 0 and hence when ˆ v (cid:54) = 0. More precisely, | (1 + K ) F s ( ξ ) − iKξ n ) F s ( ξ ) + 1 − | ξ | | (cid:46) δ ( | F s ( ξ ) | + | ξ n | ) (cid:46) δ | F s ( ξ ) | , so (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) h − n (cid:107) δ | F s ( ξ ) | ˆ v ( r, ξ ) (cid:107) L ( R n +11+ ) + h − n (cid:107) hF s (1 + K )ˆ v ( r, ξ ) (cid:107) L ( R n +11+ ) + C µ (cid:107) v (cid:107) H ( R n +11+ ) (cid:46) δ (cid:107) v (cid:107) H ( R n +11+ ) + h (cid:107) v (cid:107) H ( R n +11+ ) + C µ (cid:107) v (cid:107) H ( R n +11+ ) (cid:46) ( δ + C µ ) (cid:107) v (cid:107) H ( R n +11+ )4 CHUNG This gives an estimate for g in terms of v . However, we want the estimate to bein terms of u . We have u = L ϕ,ε,σ v , so (cid:107) u (cid:107) L ( R n +11+ ) = (cid:107)L ϕ,ε,σ v (cid:107) L ( R n +11+ ) ≥ (cid:13)(cid:13)(cid:13)(cid:13) ((1 + K ) h ∂ r − r (1 + Kh∂ θ n ) h∂ r + 1 r (1 + h (cid:52) θ ) v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) −(cid:107) ( | γ f | − K ) h ∂ r v (cid:107) L ( R n +11+ ) − (cid:13)(cid:13)(cid:13)(cid:13) r hε log( rf ( θ )) h∂ r v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) − (cid:13)(cid:13)(cid:13)(cid:13) r ( β f − Ke n ) · h ∇ θ h∂ r v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) − (cid:13)(cid:13)(cid:13)(cid:13) r ((2 hε log( rf ( θ )) + h ε log ( rf ( θ ))) v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) − (cid:13)(cid:13)(cid:13)(cid:13) h r ( L S n − (cid:52) θ ) v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) (cid:38) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) (1 + K ) h ∂ r − r (1 + Kh∂ θ n ) h∂ r + 1 r (1 + h (cid:52) θ (cid:19) v (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) − C µ (cid:107) v (cid:107) H ( R n +11+ ) Writing the last expression in terms of ˆ v , we get h − n (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) (1 + K ) h ∂ r − r (1 + iKξ n ) h∂ r + 1 r (1 − | ξ | ) (cid:19) ˆ v ( r, ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( R n +11+ ) − C µ (cid:107) v (cid:107) H ( R n +11+ ) Now ˆ v ( r, ξ ) = F ( J − s P w )( r, ξ ) is only non-zero for ξ such that | ξ | ≤ + | K | | K | <
1. The operator(1 + K ) h ∂ r − r (1 + iKξ n ) h∂ r + 1 r (1 − | ξ | )coincides, for r >
1, with a differential operator in r of the form(1 + K ) h ∂ r − ω (1 + iKξ n ) h∂ r + ω (1 − | ξ | )where ω is a smooth function that coincides with r for r >
1. This is second orderelliptic for each | ξ | such that ˆ v ( r, ξ ) is nonzero, and its symbol (in r ) is bounded PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 25 below, therefore (cid:90) ∞ | ((1 + K ) h ∂ r − r (1 + iKξ n ) h∂ r + 1 r (1 − | ξ | ))ˆ v ( r, ξ ) | dr (cid:39) (cid:90) ∞ | (1 − h ∂ r )ˆ v ( r, ξ ) | dr (cid:39) (cid:90) ∞ | (1 − h ∂ r + | ξ | )ˆ v ( r, ξ ) | dr (cid:39) (cid:90) ∞ |F ((1 − h (cid:52) ) v )( r, ξ ) | dr. Then h − n (cid:107) ((1 + K ) h ∂ r − r (1 + iKξ n ) h∂ r + 1 r (1 − | ξ | ))ˆ v ( r, ξ ) (cid:107) L ( R n +11+ ) (cid:39) (cid:107) v (cid:107) H ( R n +11+ ) and so (cid:107) u (cid:107) L ( R n +11+ ) (cid:38) (cid:107) v (cid:107) H ( R n +11+ ) − C µ (cid:107) v (cid:107) H ( R n +11+ ) (cid:38) (cid:107) v (cid:107) H ( R n +11+ ) for µ small enough.Plugging this into the inequality for g gives (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) ( δ + C µ ) (cid:107) u (cid:107) L ( R n +11+ ) . Taking µ and δ small enough means (cid:107) g (cid:107) L ( R n +11+ ) ≤ (cid:107) u (cid:107) L ( R n +11+ ) Combining this with (4.1) now gives h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107) J s L ϕ,ε,σ J − s w s (cid:107) H − r ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . Now using the first part of Lemma 3.2 gives h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ J s J − s w s (cid:107) H − r ( R n +11+ ) + C δ h (cid:107) rJ − s w s (cid:107) H ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − r ( R n +11+ ) + C δ h (cid:107) rJ − s w s (cid:107) H ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) L ϕ,ε,σ w s is supported in the r direction only for those r which can come from ˜Ω ,since w s is. Therefore the H − r and H − norms are comparable, and so h √ ε (cid:107) w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w s (cid:107) H − ( R n +11+ ) + C δ h (cid:107) rJ − s w s (cid:107) H ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ )6 CHUNG Meanwhile, (cid:92) J − s w s ( r, ξ ) = 1 h (cid:90) r ˆ w s ( t, ξ ) (cid:18) tr (cid:19) Fs ( ξ ) h dt, and ˆ w s ( t, ξ ) is supported only for 1 ≤ t ≤ C for some C depending on σ ( ˜Ω ).Therefore for r > C , | (cid:92) J − s w s ( r, ξ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) h (cid:90) C ˆ w s ( t, ξ ) (cid:18) t C (cid:19) Fsh dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) Re Fsh (cid:12)(cid:12)(cid:12)(cid:12) Cr (cid:12)(cid:12)(cid:12)(cid:12) Re Fsh , so | (cid:92) J − s w s ( r, ξ ) | (cid:46) (cid:90) C | ˆ w ( t, ξ ) | dt (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) Re Fsh (cid:12)(cid:12)(cid:12)(cid:12) Cr (cid:12)(cid:12)(cid:12)(cid:12) Re Fsh
Therefore (cid:107) rJ − s w s (cid:107) L ( R n +11+ ) (cid:46) (cid:107) rJ − s w s (cid:107) L (1 The Large Frequency Case Consider again the function F : R n → C given by F ( ξ ) = 11 + K (cid:16) iKξ n + (cid:112) iKξ n − ( Kξ n ) + (1 + K ) | ξ | − | K | (cid:17) , but this time take the branch of the square root which has nonnegative real part.Now F is smooth except where τ K ( ξ ) = 2 iKξ n − ( Kξ n ) + (1 + | K | ) | ξ | − | K | lies on the nonpositive real axis. This happens when ξ n = 0 and | ξ | ≤ | K | | K | . Therefore on the support of 1 − ρ ( ξ ), F is smooth. Moreover, since the real part ofthe square root is nonnegative, both | F | and the real part of F and are boundedbelow by K . Therefore we can pick a smooth function F (cid:96) such that F (cid:96) ( ξ ) = F ( ξ )on the support of 1 − ρ ( ξ ), and Re F (cid:96) , | F (cid:96) | > 12 11+ K . Note that for large | ξ | , wehave Re F ( ξ ) , | F ( ξ ) | (cid:39) | ξ | , F (cid:96) can satisfyRe F (cid:96) ( ξ ) , | F (cid:96) ( ξ ) | (cid:39) | ξ | In fact, if K K < r < r and 0 < δ < δ , we can arrange for F (cid:96) = F and F (cid:96) to be smooth for | ξ | ≥ r and ξ n ≥ δ . (cid:100) J (cid:96) u ( r, ξ ) = (cid:18) F (cid:96) ( ξ ) r + h∂ r (cid:19) ˆ u ( r, ξ ) . This operator has adjoint J ∗ s given by (cid:100) J ∗ (cid:96) u ( r, ξ ) = (cid:32) F (cid:96) ( ξ ) r − h∂ r (cid:33) ˆ u ( r, ξ ) . These operators have (right) inverses defined by (cid:91) J − (cid:96) u ( r, ξ ) = h − (cid:90) r ˆ u ( t, ξ ) (cid:18) tr (cid:19) F(cid:96) ( ξ ) h dt and (cid:92) J ∗− (cid:96) u ( r, ξ ) = h − (cid:90) ∞ r ˆ u ( t, ξ ) (cid:16) rt (cid:17) F(cid:96) ( ξ ) h dt. Each of these is well defined for functions in S ( R n +11+ ).We have the following lemmas. Lemma 5.1. J (cid:96) , J ∗ (cid:96) , J − (cid:96) , and J ∗− (cid:96) extend as bounded maps J (cid:96) , J ∗ (cid:96) : H r ( R n +11+ ) → L ( R n +11+ ) and J − (cid:96) , J ∗− (cid:96) : L ( R n +11+ ) → H r ( R n +11+ ) . Moreover, the extensions of J ∗ (cid:96) and J ∗− (cid:96) are isomorphisms. Lemma 5.2. i) Suppose w ∈ S ( R n +11+ ) . Then if Q is a second order semiclassicaldifferential operator with bounded coefficients, then (cid:107) ( J (cid:96) Q − QJ (cid:96) ) w (cid:107) H − r ( R n +11+ ) (cid:46) hC δ (cid:107) rw (cid:107) H ( R n +11+ ) ii) Suppose w ∈ S ( R n +11+ ) . Let χ ∈ S ( R n +11+ ) . Then (cid:107) J (cid:96) χJ − (cid:96) w (cid:107) L ( R n +11+ ) (cid:38) (cid:107) χw (cid:107) L ( R n +11+ ) − hC δ (cid:107) rw (cid:107) L ( R n +11+ )8 CHUNG Lemma 5.3. Suppose u ∈ S ( R n +11+ ) . If g is defined by ˆ g ( r, ξ ) = 2Re F (cid:96) ( ξ ) − hh (cid:90) ∞ ˆ u ( t, ξ ) r − F(cid:96) ( ξ ) h t − F(cid:96) ( ξ ) h dt, then (cid:107) J (cid:96) u (cid:107) H − r ( R n +11+ ) (cid:39) (cid:107) u − g (cid:107) L ( R n +11+ ) The proofs of these lemmas are identical to the proofs of the equivalent lemmasin the small frequency case.Now consider the Carleman estimate (2.4). By a similar argument as in thesmall frequency case, we get(5.1) h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ J − (cid:96) w (cid:96) (cid:107) L ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . Again I want to combine this last inequality with Lemma 5.3 to get h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107) J (cid:96) L ϕ,ε,σ J − (cid:96) w (cid:96) (cid:107) H − r ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . To do this I need to show that if u is of the form u = L ϕ,ε,σ J − (cid:96) w (cid:96) , then the function g defined in Lemma 5.3 satisfies a bound like (cid:107) g (cid:107) L ( R n +11+ ) ≤ (cid:107) u (cid:107) L ( R n +11+ ) + O ( h ) (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) . The approach used in the small frequency case does not work here, because L ϕ,ε,σ isnot at all elliptic from the point of view of w (cid:96) . However, now L ϕ,ε,σ can be factoredinto a composition of two operators, one of which has the desired properties.Let ζ ( ξ ) be a smooth cutoff function which is identically 1 on the set where | ξ | ≥ r or | ξ n | ≥ δ , and vanishes if | ξ | ≤ r or | ξ n | ≤ δ . Let G s = (1 − ζ ( ξ )) F (cid:96) ( ξ )and consider the symbols G ± = ζ ( ξ ) α + iβ f · ξ ± (cid:112) ( α + iβ f · ξ ) − (1 + ( γ f ) )( α − L S n ( θ, ξ ))1 + | γ f | + G s ( ξ )where L S n ( θ, ξ ) represents the symbol of the differential operator L S n . The squareroot represents the branch of the square root with nonnegative real part. Theargument of the square root lies on the nonpositive real axis only when β f · ξ = 0and L S n ≤ α | γ f | | γ f | . For µ small enough, this cannot happen on the support of ζ . Therefore G ± reallyare smooth, and hence they really are symbols of order 1 on R n . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 29 Now if T a is the operator associated to the symbol a ,( h∂ r − r T G + )(1 + | γ f | )( h∂ r − r T G − )= (1 + | γ f | ) h ∂ r − r (1 + | γ f | )( T G + + T G − ) h∂ r + 1 r (1 + | γ f | ) T G + G − + hE = (1 + | γ f | ) h ∂ r − r ( α + β f · h ∇ θ ) h∂ r T ζ + 1 r ( α + h L S n ) T ζ − r (1 + | γ f | ) T G s + 1 r (1 + | γ f | )( T G + G s + T G s G − + T G s ) + hE = (1 + | γ f | ) h ∂ r − r ( α + β f · h ∇ θ ) h∂ r T ζ + 1 r ( α + h L S n ) T ζ − r (1 + | γ f | ) T G s + 1 r (1 + | γ f | )( T G + T G s + T G − T G s + T G s T G s ) + hE where E is a operator (which changes from line to line as necessary) built of firstorder semiclassical pseudodifferential operators in R n and ∂ r derivatives which isbounded from H ( R n +11+ ) to L ( R n +11+ ).Now let v = J − (cid:96) w (cid:96) . Then( h∂ r − r T G + )(1 + | γ f | )( h∂ r − r T G − ) v = (1 + | γ f | ) h ∂ r v − r ( α + β f · h ∇ θ ) h∂ r T ζ v + 1 r ( α + h L S n ) T ζ v − r (1 + | γ f | ) T G s v + 1 r (1 + | γ f | )( T G + + T G − + T G s ) T G s v + hE v. Note that ˆ w (cid:96) ( r, ξ ) is only supported for ξ in the support of (1 − ρ ), and therefore v = J − (cid:96) w (cid:96) is supported only for ξ in the support of (1 − ρ ). Therefore T ζ v = v. since ζ ≡ − ρ . Similarly, T ζ v = v . In addition, T G s v = 0 , since G s is 0 on the support of 1 − ρ . Therefore( h∂ r − r T G + )(1 + | γ f | )( h∂ r − r T G − ) v = (1 + | γ f | ) h ∂ r v − r ( α + β f · h ∇ θ ) h∂ r v + 1 r ( α + h L S n ) v + hE v = L ϕ,ε,σ v + hE v where E is bounded from H ( R n +11+ ) to L ( R n +11+ ).Therefore L ϕ,ε,σ v = ( h∂ r − r T G + ) z + hE v for some function z , given by z = (1 + | γ f | )( h∂ r − r T G − ) v. Then ˆ g ( r, ξ ) = 2Re F (cid:96) − hh (cid:90) ∞ (cid:92) L ϕ,ε,σ v ( t, ξ ) r − F(cid:96)h t − F(cid:96)h dt = 2Re F (cid:96) − hh (cid:90) ∞ F (cid:18)(cid:18) h∂ t − t T G + (cid:19) z (cid:19) ( t, ξ ) r − F(cid:96)h t − F(cid:96)h dt + 2Re F (cid:96) − hh (cid:90) ∞ h (cid:100) E v ( t, ξ ) r − F(cid:96)h t − F(cid:96)h dt Integrating by parts givesˆ g ( r, ξ ) = 2Re F (cid:96) − hh (cid:90) ∞ t F (( T F (cid:96) − T G + ) z )( t, ξ ) r − F(cid:96)h t − F(cid:96)h dt + 2Re F (cid:96) − hh (cid:90) ∞ h (cid:100) E v ( t, ξ ) r − F(cid:96)h t − F(cid:96)h dt There are no boundary terms because z is supported away from r = 1. Thereforeby the reasoning used to prove Lemma 5.3, (cid:107) g (cid:107) L ( R n +11+ ) ≤ (cid:107) r ( T F (cid:96) − G + ) z (cid:107) L ( R n +11+ ) + h (cid:107) E v (cid:107) L ( R n +11+ ) We need an estimate for (cid:107) r ( T F (cid:96) − G + ) z (cid:107) L ( R n +11+ ) . Examine the symbol F (cid:96) − G + . F (cid:96) − G + = F (cid:96) ( ξ ) − ζ α + iβ f · ξ + (cid:112) ( α + iβ f · ξ ) − (1 + | γ f | )( α + L S n ( θ, ξ ))1 + | γ f | − (1 − ζ ) F (cid:96) ( ξ )= ζ (cid:32) F (cid:96) ( ξ ) − α + iβ f · ξ + (cid:112) ( α + iβ f · ξ ) − (1 + | γ f | )( α + L S n ( θ, ξ ))1 + | γ f | (cid:33) On the support of ζ , F (cid:96) ( ξ ) = 11 + K (1 + iKξ n + (cid:112) iKξ n − ( Kξ n ) + (1 + K ) | ξ | − | K | ) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 31 Therefore F (cid:96) − G + = ζ (cid:32) iKξ n + (cid:112) iKξ n − ( Kξ n ) − (1 + K ) | ξ | − | K | K − α + iβ f · ξ + (cid:112) ( α + iβ f · ξ ) − (1 + ( γ f ) )( α + L S n ( θ, ξ ))1 + | γ f | (cid:33) = ζ (cid:18) iKξ n K − α + iβ f · ξ | γ f | (cid:19) + ζ (cid:32) (cid:112) iKξ n − ( Kξ n ) − (1 + K ) | ξ | − | K | K − (cid:112) ( α + iβ f · ξ ) − (1 + ( γ f ) )( α + L S n ( θ, ξ ))1 + | γ f | (cid:33) Consider the first term.1 + iKξ n K − α + iβ f · ξ | γ f | = (1 + | γ f | )(1 + iKξ n ) − (1 + K )( α + iβ f · ξ )(1 + K )(1 + | γ f | )= ( | γ f | − K )(1 + iKξ n )(1 + K )(1 + | γ f | )+ ((1 + K )((1 − α ) + i ( β f − Ke n ) · ξ )(1 + K )(1 + | γ f | )The first order operators with symbols( | γ f | − K )(1 + iKξ n )(1 + K )(1 + | γ f | )and ((1 + K )((1 − α ) + i ( β f − Ke n ) · ξ )(1 + K )(1 + | γ f | )have bounds (cid:46) C µ , because they involve multiplication by a function of θ which isbounded by C K C µ .Similarly, consider the first order operator with symbol ζ (cid:32) (cid:112) iKξ n − ( Kξ n ) − (1 + K ) | ξ | − | K | K − (cid:112) ( α + iβ f · ξ ) − (1 + ( γ f ) )( α + L S n ( θ, ξ ))1 + | γ f | (cid:33) To fit everything horizontally on the page, denote τ K := 2 iKξ n − ( Kξ n ) − (1 + K ) | ξ | − | K | and τ f := ( α + iβ f · ξ ) − (1 + ( γ f ) )( α + L S n ( θ, ξ )) . Then √ τ K K − √ τ f | γ f | = (1 + | γ f | ) √ τ K − (1 + K ) √ τ f (1 + K )(1 + | γ f | )= (1 + | γ f | ) τ K − (1 + K ) τ f (1 + K )(1 + | γ f | )((1 + | γ f | ) √ τ K + (1 + K ) √ τ f )= (1 + K ) τ K − τ f (1 + | γ f | )((1 + | γ f | ) √ τ K + (1 + K ) √ τ f )+ ((1 + | γ f | ) − (1 + K ) ) τ K (1 + K )(1 + | γ f | )((1 + | γ f | ) √ τ K + (1 + K ) √ τ f ) . Expanding, τ K − τ f = 2 i ( Ke n − αβ f ) · ξ + (( β f · ξ ) − ( Ke n · ξ ) ) + ( | γ f | − K ) L ( θ, iξ )+( | γ f | − | K | ) + (1 + K )( | ξ | − L ( θ, ξ )) . Therefore the second term has operator bounds (cid:46) C µ , because each term involvesmultiplication by a function of θ which is bounded by C K C µ .Therefore (cid:107) r ( T F (cid:96) − G + ) z (cid:107) L ( R n +11+ ) ≤ δ (cid:107) z (cid:107) H ( R n +11+ ) for µ small enough. Then (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) (cid:107) r ( T F (cid:96) − G + ) z (cid:107) L ( R n +11+ ) + h (cid:107) E v (cid:107) L ( R n +11+ ) (cid:46) δ (cid:107) z (cid:107) H ( R n +11+ ) + h (cid:107) E v (cid:107) L ( R n +11+ ) (cid:46) δ (cid:107) z (cid:107) H ( R n +11+ ) + h (cid:107) v (cid:107) H ( R n +11+ ) Since L ϕ,ε,σ v = ( h∂ r − r T G + ) z + hE v, we have (cid:107)L ϕ,ε,σ v (cid:107) L ( R n +11+ ) ≥ (cid:107) ( h∂ r − r T G + ) z (cid:107) L ( R n +11+ ) − h (cid:107) E v (cid:107) L ( R n +11+ ) ≥ (cid:107) J ∗ (cid:96) z (cid:107) L ( R n +11+ ) − (cid:107) r T F (cid:96) − G + z (cid:107) L ( R n +11+ ) − h (cid:107) v (cid:107) H ( R n +11+ ) (cid:38) (cid:107) z (cid:107) H ( R n +11+ ) − δ (cid:107) z (cid:107) H ( R n +11+ ) − h (cid:107) v (cid:107) H ( R n +11+ ) (cid:38) (cid:107) z (cid:107) H ( R n +11+ ) − h (cid:107) v (cid:107) H ( R n +11+ ) PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 33 for δ small enough. Therefore (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) δ (cid:107)L ϕ,ε,σ v (cid:107) H ( R n +11+ ) + h (cid:107) v (cid:107) H ( R n +11+ ) (cid:46) δ (cid:107)L ϕ,ε,σ v (cid:107) H ( R n +11+ ) + h (cid:107) J − s (1 − P ) w (cid:107) H ( R n +11+ ) Using similar reasoning as for the small frequency case, h (cid:107) J − s (1 − P ) w (cid:107) H ( R n +11+ ) (cid:46) h (cid:107) J − s (1 − P ) w (cid:107) H r ( R n +11+ ) . Therefore (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) δ (cid:107)L ϕ,ε,σ v (cid:107) H ( R n +11+ ) + h (cid:107) J − s (1 − P ) w (cid:107) H r ( R n +11+ ) (cid:46) δ (cid:107)L ϕ,ε,σ v (cid:107) H ( R n +11+ ) + h (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) Then for δ small enough, (cid:107) g (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ v (cid:107) L ( R n +11+ ) + h (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) . Now using (5.1) and Lemma 5.3, h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107) J (cid:96) L ϕ,ε,σ χ J − (cid:96) w (cid:96) (cid:107) H − r ( R n +11+ ) + h (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . Absorbing the second last term into the left side gives h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107) J (cid:96) L ϕ,ε,σ χ J − (cid:96) w (cid:96) (cid:107) H − r ( R n +11+ ) + O ( h ∞ ) (cid:107) w (cid:107) L ( R n +11+ ) . We can finish the argument as in the small frequency case to get h √ ε (cid:107) w (cid:96) (cid:107) L ( R n +11+ ) (cid:46) (cid:107)L ϕ,ε,σ w (cid:96) (cid:107) H − ( R n +11+ ) + O ( h ) (cid:107) w (cid:107) L ( R n +11+ ) . This finishes the proof of Lemma 2.7, and thus of Proposition 2.5.6. Proof of Theorem 1.4 Now I can prove Proposition 2.1 essentially by gluing together estimates of theform in Proposition 2.5. First note that by a change of variables similar to the onesin Section 2, we can show that if for all θ ∈ S n such that some ( r, θ ) is in Ω , | sin( θ k ) − | ≤ µ where k = 1 , . . . , n − |∇ S n log f − Ke n | S n ≤ µ, where µ is small enough, then(6.1) h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε w (cid:107) H − ( A O ) for all w ∈ C ∞ (Ω).Now let Ω be as in Proposition 2.1. We can take an open cover U , . . . , U m of Ω such that on each Ω ∩ U j , there exists K j such that under some choice ofcoordinates, |∇ S n log f − K j e n | ≤ µ K j and | sin( θ k ) − | ≤ µ K j , where µ K j is thevalue of µ from Proposition 1 which works for K = K j . (Since |∇ S n log f | mustbe bounded above, µ K j must be bounded below, and therefore this is possible withonly finitely many U j .) Let ζ , . . . ζ m be a smooth partition of unity subordinate to the cover U , . . . U m .Now for w ∈ C ∞ (Ω), w = ζ w + . . . ζ m w =: w + . . . + w m , where each w j ∈ C ∞ (Ω ∩ U j ). Applying the result (6.1) to the domain Ω ∩ U j , h √ ε (cid:107) w j (cid:107) L (Ω ∩ U j ) (cid:46) (cid:107)L ϕ,ε w j (cid:107) H − ( A O ) for each j = 1 , . . . , m . Then (cid:88) j h √ ε (cid:107) w j (cid:107) L (Ω) (cid:46) (cid:88) j (cid:107)L ϕ,ε w j (cid:107) H − ( A O ) , so h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:88) j (cid:107)L ϕ,ε w j (cid:107) H − ( A O ) . Now by the product rule, (cid:107)L ϕ,ε w j (cid:107) H − ( A O ) = (cid:107)L ϕ,ε ζ j w (cid:107) H − ( A O ) ≤ (cid:107) ζ j L ϕ,ε w (cid:107) H − ( A O ) + Ch (cid:107) w (cid:107) A O ≤ (cid:107)L ϕ,ε w (cid:107) H − ( A O ) + Ch (cid:107) w (cid:107) A O . Therefore(6.2) h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε w (cid:107) H − ( A O ) . for ε small enough, for every w ∈ C ∞ (Ω).To treat the case where W and q are non-zero, note that L ϕ,ε,W,q = L ϕ,ε + h ( W · hD + hD · W ) + 2 ihW · ∇ (log r + h log r ε ) + h ( q + W ) . Therefore h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − ( A O ) + hC (cid:107) w (cid:107) L ( A O ) and the last term can be absorbed into the left side to give h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − ( A O ) . This completes the proof of Proposition 2.1.Finally, I can prove Theorem 1.4 by gluing together estimates of the form inProposition 2.1. If E is a compact subset of F Ω \ Z Ω , then define Ω (cid:48) to be a smoothdomain containing Ω, with ∂ Ω ∩ ∂ Ω (cid:48) = E .Then let U , . . . , U m be an open cover of Ω such that each ∂U j ∩ E coincideswith a graph of the form r = f j ( θ ). For each U j , Proposition 2.1 gives us h √ ε (cid:107) w (cid:107) L ( U j ) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − ( A j ) for w ∈ C ∞ ( U j ).Each A j is defined by the graph of a function r = f j ( θ ), and since ∂ Ω (cid:48) is smoothand coincides with ∂ Ω on E , and E is a compact subset of F Ω \ Z Ω , ∂ Ω (cid:48) mustbe locally a graph in a neighbourhood of E . Therefore we can assume that A j PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 35 coincides with Ω (cid:48) in a neighbourhood of each U j , in the sense that their charac-teristic functions are equal in that neighbourhood. Then there is a smooth cutofffunction χ j defined on A j ∩ Ω (cid:48) which is identically one on U j but vanishes out-side on the complements of A j and Ω (cid:48) . Multiplication by this function provides abounded map from H ( A j ) to H (Ω (cid:48) ) and vice versa, and therefore for w ∈ C ∞ ( U j ), (cid:107) w (cid:107) H − (Ω (cid:48) ) (cid:39) (cid:107) w (cid:107) H − ( A j ) . Therefore we have h √ ε (cid:107) w (cid:107) L ( U j ) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − (Ω (cid:48) ) for w ∈ C ∞ ( U j ).Gluing together these estimates in the matter used above gives h √ ε (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,ε,W,q w (cid:107) H − (Ω (cid:48) ) for w ∈ C ∞ (Ω).Finally, note that if w ∈ C ∞ (Ω), then e (log r )2 ε w ∈ C ∞ (Ω), so h √ ε (cid:107) e (log r )2 ε w (cid:107) L (Ω) (cid:46) (cid:107) e (log r )2 ε L ϕ,W,q w (cid:107) H − (Ω (cid:48) ) . On Ω, there exists some C Ω such that 1 ≤ e (log r )2 ε ≤ e C Ω ε , so h (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,W,q w (cid:107) H − (Ω (cid:48) ) as desired. This establishes Theorem 1.4. Remark If we want to prove Theorem 1.2 instead of Theorem 1.1, then wecould begin by supposing that f : S n → (0 , ∞ ) is a C ∞ function such that Ω liesentirely in the region A I = { ( r, θ ) | r ≤ f ( θ ) } ⊂ R n +1 , and E is a subset of the graph r = f ( θ ). Then by the change of variables ( r, θ ) (cid:55)→ ( r , θ ), Ω maps to a region ˆΩofthe form described in Proposition 2.1. Therefore by (6.2), h (cid:107) w (cid:107) L (ˆΩ) (cid:46) (cid:107)L ϕ,ε w (cid:107) H − ( ˆ A O ) . for w ∈ C ∞ ˆΩ, where ϕ = log r . Changing variables back gives the Carlemanestimate h (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L − log r,ε w (cid:107) H − ( A I ) for w ∈ C ∞ Ω. Therefore by the same kind of argument as above, we get h (cid:107) w (cid:107) L (Ω) (cid:46) (cid:107)L ϕ,W,q w (cid:107) H − (Ω (cid:48) ) where ϕ = − log r , and Ω (cid:48) is a domain containing Ω, with E ⊂ ∂ Ω (cid:48) ∩ ∂ Ω, whenever E is of the form described in Theorem 1.2. Using this Carleman estimate in theplace of Theorem 1.4 in the remainder of the argument proves Theorem 1.2 insteadof Theorem 1.1. 7. Complex Geometric Optics Solutions Theorem 1.4 can be used to construct solutions to equations of the system (1.1).The key is the following proposition. Proposition 7.1. For every v ∈ L (Ω) , there exists u ∈ H (Ω) such that L ∗ ϕ,W,q u = v on Ω u | E = 0 and (cid:107) u (cid:107) H (Ω) (cid:46) h (cid:107) v (cid:107) L (Ω) . Proof. The proof is based on a Hahn-Banach argument. Suppose v ∈ L (Ω). Thenfor all w ∈ C ∞ (Ω), | ( w | v ) Ω | (cid:46) h (cid:107) v (cid:107) L (Ω) h (cid:107) w (cid:107) L (Ω) . Therefore, by Theorem 1.4,(7.1) | ( w | v ) Ω | (cid:46) h (cid:107) v (cid:107) L (Ω) (cid:107)L ϕ,W,q w (cid:107) H − ( A O ) . Now consider the subspace {L ϕ,W,q w | w ∈ C ∞ (Ω) } ⊂ H − ( A O ) . By the estimate from Theorem 1.4, the map L ϕ,W,q w (cid:55)−→ ( w | v ) Ω is well-defined onthis subspace. It is a linear functional, and by (7.1), it is bounded by Ch (cid:107) v (cid:107) L (Ω) .Therefore by Hahn-Banach, there exists an extension of this functional to thewhole space H − ( A O ) with the same bound. This can be represented by an elementof the dual space H ( A O ), so there exists u ∈ H ( A O ) such that (cid:107) u (cid:107) H ( A O ) (cid:46) h (cid:107) v (cid:107) L (Ω) and ( w | v ) Ω = ( L ϕ,W,q w | u ) A O = ( L ϕ,W,q w | u ) Ω for all w ∈ C ∞ (Ω). Note that u ∈ H ( A ) implies that u | E = 0. Then( w | v ) Ω = ( w |L ∗ ϕ,W,q u ) Ω since w ∈ C ∞ (Ω), and thus ( w | v − L ∗ ϕ,W,q u ) Ω = 0for all v ∈ C ∞ (Ω). Therefore v = L ∗ ϕ,W,q u on Ω, and (cid:107) u (cid:107) H ( R n +1 ) (cid:46) h (cid:107) v (cid:107) L (Ω) as desired. (cid:3) Now I can construct the complex geometrical optics solutions. Proposition 7.2. There exists a solution of the problem L W,q u = 0 on Ω u | E = 0 PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 37 of the form u = e h ( ϕ + iψ ) ( a + r ) − e (cid:96)h b , where ϕ ( x, y ) = log r , ψ is a solution to theeikonal equation ∇ ϕ · ∇ ψ = 0 , |∇ ϕ | = |∇ ψ | ; a and b are C functions on Ω ; and Re (cid:96) ( x, y ) = ϕ ( x, y ) − k ( x, y ) where k ( x ) (cid:39) dist( x, E ) in a neighbourhood of E , and b has its support in that neigh-bourhood. Finally, r ∈ H (Ω) , with r | E = 0 , (cid:107) r (cid:107) H (Ω) = O ( h ) , and (cid:107) r (cid:107) L ( ∂ Ω) = O ( h ) . The proof is a combination of the proofs of the equivalent theorems in [DKSU]and [KSU]. Proof. Let ϕ ( r, θ ) = log r , and take ψ ( r, θ ) = d S n ( θ, ω ) for some fixed point ω ∈ S n .If ω (cid:54) = θ for all ( r, θ ) ∈ Ω, then ψ solves the eikonal equation ∇ ϕ · ∇ ψ = 0 , |∇ ϕ | = |∇ ψ | . Then h L W,q e h ( ϕ + iψ ) = e h ( ϕ + iψ ) ( h ( D + W ) · ( ∇ ψ − i ∇ ϕ )+ h ( ∇ ψ − i ∇ ϕ ) · ( D + W ) + h L W,q ) . Therefore if a is a C solution to( ∇ ψ − i ∇ ϕ ) · Da + ( ∇ ψ − i ∇ ϕ ) · W a + 12 i ( (cid:52) ψ − i (cid:52) ϕ ) a = 0 , then h L W,q e h ( ϕ + iψ ) a = e h ( ϕ + iψ ) h L W,q a = O ( h ) e h ( ϕ + iψ ) . We can look for an exponential solution a = e Φ , in which case the relevant equationbecomes ( ∇ ϕ + i ∇ ψ ) · ∇ Φ + i ( ∇ ϕ + i ∇ ψ ) · W + 12 (cid:52) ( ϕ + iψ ) = 0 . Now suppose x ∈ R n +1 , and write x = ( x ω , x (cid:48) ), where x ω is the component of x in the ω direction, and x (cid:48) are the remaining components. Then by considering z = x ω + i | x (cid:48) | as a complex variable, we get ϕ = Re log z and ψ = Im log z . Nowour equation is an inhomogeneous Cauchy-Riemann equation in the z variable, andcan be solved by the Cauchy formula. Then a is C , since W is. Note that thesolution is only unique up to addition of terms g a with(7.2) ( ∇ ϕ + i ∇ ψ ) · ∇ g a = 0 . Now I want to construct a (complex valued) function (cid:96) to be an approximatesolution to the equation ∇ (cid:96) · ∇ (cid:96) = 0 (cid:96) | E = ϕ + iψ. In order to avoid duplicating the solution ϕ + iψ , we can ask for ∂ ν (cid:96) | E = − ∂ ν ( ϕ + iψ ) | E . To construct an approximate solution, pick coordinates ( t, s ) near E such that t are the coordinates along E and s is perpendicular to E . Suppose (cid:96) takes the formof a power series (cid:96) ( t, s ) = ∞ (cid:88) j =0 a j ( t ) s j . Then ∇ (cid:96) = ( ∇ t (cid:96), ∂ s (cid:96) )= ( ∞ (cid:88) j =0 ∇ t a j ( t ) s j , ∞ (cid:88) j =0 a j ( t ) js j − )Expanding the equation ∇ (cid:96) · ∇ (cid:96) = 0 gives0 = (cid:88) j + k =0 ∇ t a j ∇ t a k + (cid:88) j + k =1 ∇ t a j ∇ t a k s + (cid:88) j + k =2 ∇ t a j ∇ t a k s + . . . + (cid:88) j + k =2 jka j a k + (cid:88) j + k =3 jka j a k s + (cid:88) j + k =4 jka j a k s + . . . Considering each power of s separately gives a sequence of equations(7.3) (cid:88) j + k = m ∇ t a j ∇ t a k + (cid:88) j + k = m +2 jka j a k = 0 . for each m = 0 , , , . . . . The boundary conditions determine a and a , so we cansolve this recursively. If m ≥ 1, and all a j are known for j ≤ m , the only part of(7.3) which contains an unknown looks like 2( m + 1) a a m +1 . Note that a = − ∂ ν ( ϕ + iψ )Since E coincides with a graph r = f ( θ ) for some smooth function f , and ϕ = log r ,there exists some ε > | a | > ε on E , and so we can divide by a to solvefor a m +1 .This gives a formal power series for (cid:96) , which may or may not converge outside s = 0. However, we can construct a C ∞ function in Ω whose Taylor series in s coincides with this formal power series at s = 0.Let χ : R → [0 , 1] be a smooth cutoff function which is identically one on [ − , ],and identically zero outside ( − , (cid:96) = ∞ (cid:88) j =0 a j ( t ) χ ( b j s ) s j = ∞ (cid:88) j =0 c j ( t, s ) s j where b j = max k ≤ j {(cid:107) a k ( t ) (cid:107) C k (Ω) , } . This defines a C ∞ function on Ω whose Taylor series at s = 0 coincides with theformal power series calculated earlier. Now examine the expression ∇ (cid:96) · ∇ (cid:96) . Thecoefficient of s m in this expansion is (cid:88) j + k = m ∇ t c j ∇ t c k + (cid:88) j + k = m ( ∂ s c j + ( j + 1) c j +1 )( ∂ s c k + ( k + 1) c k +1 ) . For s ≤ b p , all c j ( t, s ) = a j ( t ) for j ≤ p , so this is exactly zero by design when s ≤ b p and m < p . Therefore on the region where s ≤ b p , ∇ (cid:96) · ∇ (cid:96) is a smoothfunction which is O ( s p ) as s goes to zero. Since this can be done for any p , ∇ (cid:96) · ∇ (cid:96) = O (dist( x, E ) ∞ ) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 39 Moreover, ∂ ν Re (cid:96) | E = − ∂ ν ϕ | E < − ε , and Re (cid:96) | E = ϕ | E so in a neighbourhood of of E ,(7.4) Re (cid:96) ( x, y ) = ϕ ( x, y ) − k ( x, y )where k ( x ) (cid:39) dist( x, E ) in a neighbourhood of E .By a similar method, we can construct an approximate solution b for the problem ∇ (cid:96) · Db + ∇ (cid:96) · W b = 0 b | E = a | E . so ∇ (cid:96) · Db + ∇ (cid:96) · W b = O (dist( x, E ) ∞ ) b | E = a | E . Multiplying b by a smooth cutoff function does not change these properties, so wemay as well assume that b is only supported close to E for (7.4) to hold. Then − h L W,q ( e (cid:96)h b ) = e (cid:96)h ( O (dist( x, E ) ∞ ) + O ( h )) , so | h L W,q ( e (cid:96)h b ) | = e ϕh e − kh ( O (dist( x, E ) ∞ ) + O ( h )) . If dist( x, E ) ≤ h , for h small, this is e ϕh O ( h ), because of the O (dist( x, E ) ∞ )term. On the other hand, if dist( x, E ) ≥ h , this is still e ϕh O ( h ), because of e − kh .Now e h ( ϕ + iψ ) a − e (cid:96)h b = 0 on E , and e − ϕh h L W,q ( e h ( ϕ + iψ ) a + e (cid:96)h b ) = v where (cid:107) v (cid:107) L (Ω) = O ( h ). By Proposition 7.1, the problem L ∗ ϕ,W,q r = e − ϕh h L W,q e ϕh r = − v on Ω r | E = 0has an H solution r with (cid:107) r (cid:107) H (Ω) (cid:46) h (cid:107) v (cid:107) L (Ω) = O ( h )Set r = e − iψh r , and u = e h ( ϕ + iψ ) ( a + r ) − e (cid:96)h b . Then (cid:107) r (cid:107) H (Ω) = O ( h ) , so (cid:107) r (cid:107) L ( ∂ Ω) = O ( h ) by the trace theorem, and L W,q u = 0 on Ω u | E = 0 . This finishes the proof. (cid:3) If the boundary condition is not needed, then the result is as follows: Proposition 7.3. There exists a solution of the problem L W,q u = 0 on Ω of the form u = e h ( ϕ + iψ ) ( a + r ) , where ϕ ( x, y ) is any limiting Carleman weight, ψ is any solution to the eikonal equation, a is a C function on Ω , and r ∈ H (Ω) ,with (cid:107) r (cid:107) H (Ω) = O ( h ) , and (cid:107) r (cid:107) L ( ∂ Ω) = O ( h ) . This is essentially lemma 3.4 from [DKSU]. Note that we can always replace a by γa , where γ is a solution to( ∇ ϕ + i ∇ ψ ) · ∇ γ = 0 . on Ω. 8. Proof of Theorem 1.1 For convenience, (cid:107) · (cid:107) will denote the L norm in this section, unless otherwiseindicated. The tilde as used in this section has nothing to do with the notationfrom section 2.Using Proposition 7.2, we can construct ˜ u = e h ( ϕ + iψ ) ( a + r ) − e (cid:96)h b =: u + u r to be a solution to L W ,q ˜ u = 0 on Ω˜ u | E = 0 . Then − ϕ is also a Carleman weight, and if ϕ and ψ satisfy the eikonal equation,then so do − ϕ and ψ . Therefore using Proposition 7.3, we can construct u = e h ( − ϕ + iψ ) ( a + r ) to be a solution to L W ,q u = 0 . Let w be the unique solution to L W ,q w = 0 w | ∂ Ω = ˜ u | ∂ Ω . (This is where we use the assumption that L W ,q does not have a zero eigenvalue.)Note that in particular, w | E = ˜ u | E = 0, so by the hypothesis on the Dirichlet-Neumann map, ∂ ν ( w − ˜ u ) | U = 0 . Now L W ,q ( w − ˜ u ) = −L W ,q ˜ u = ( L W ,q − L W ,q )˜ u = ( W − W ) · D ˜ u + D · ( W − W )˜ u + ( W − W + q − q )˜ u . (8.1)On the other hand, the Green’s formula from [DKSU] gives us that (cid:90) Ω L W ,q ( w − ˜ u ) u dV = (cid:90) ∂ Ω ∂ ν (˜ u − w ) u dS = (cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS. (8.2) PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 41 Combining (8.1) with (8.2) gives (cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS = (cid:90) Ω ( W − W ) · ( D ˜ u u + ˜ u Du ) dV + (cid:90) Ω ( W − W + q − q )˜ u u dV. Expanding ˜ u as ˜ u = u + u r on the right side gives (cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS = (cid:90) Ω ( W − W ) · ( Du u + u Du ) dV + (cid:90) Ω ( W − W + q − q ) u u dV + (cid:90) Ω ( W − W ) · ( Du r u + u r Du ) dV + (cid:90) Ω ( W − W + q − q ) u r u dV (8.3)We can label the terms as follows: 1 (cid:13) = 2 (cid:13) + 3 (cid:13) + 4 (cid:13) + 5 (cid:13) . Consider the termson the right side first. 2 (cid:13) is bounded above by (cid:107) ( W − W ) e − ϕh Du (cid:107) Ω (cid:107) a + r (cid:107) Ω + (cid:107) ( W − W ) e ϕh Du (cid:107) Ω (cid:107) a + r (cid:107) Ω . Since W − W is bounded on Ω, (cid:107) a (cid:107) Ω and (cid:107) a (cid:107) Ω are O (1), and (cid:107) r (cid:107) Ω and (cid:107) r (cid:107) Ω are O ( h ), | (cid:13)| (cid:46) (cid:107) e − ϕh Du (cid:107) Ω + (cid:107) e ϕh Du (cid:107) Ω . Meanwhile, 3 (cid:13) is bounded above by | (cid:13)| ≤ (cid:107) ( W − W + q − q )( a + r ) (cid:107) Ω (cid:107) a + r (cid:107) Ω = O (1) . Similarly, | (cid:13)| (cid:46) (cid:107) e − ϕh Du r (cid:107) Ω + (cid:107) e ϕh Du (cid:107) Ω (cid:107) e − βyh (cid:107) Ω (cid:46) (cid:107) e − ϕh Du r (cid:107) Ω + h (cid:107) e ϕh Du (cid:107) Ω and 5 (cid:13) is bounded above by | (cid:13)| ≤ (cid:107) ( W − W + q − q ) e − βyh b (cid:107) Ω (cid:107) a + r (cid:107) Ω = O ( h ) . Now examine term 1 (cid:13) : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:107) ∂ ν (˜ u − w ) e − ϕh (cid:107) ∂ Ω \ U (cid:107) a + r (cid:107) ∂ Ω \ U . The factor (cid:107) a + r (cid:107) ∂ Ω \ U is O (1). Furthermore, on ∂ Ω \ U , ∂ ν ϕ ≥ ε , so (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:46) √ ε (cid:107) (cid:112) ∂ ν ϕe − ϕh ∂ ν (˜ u − w ) (cid:107) ∂ Ω \ U (cid:46) √ ε (cid:107) (cid:112) ∂ ν ϕe − ϕh ∂ ν (˜ u − w ) (cid:107) B Ω . Then using the original Carleman estimate from [DKSU], 1 (cid:13) is bounded above by C √ ε (cid:16) √ h (cid:107) e − ϕh L A ,q (˜ u − w ) (cid:107) Ω + (cid:107) (cid:112) − ∂ ν ϕe − ϕh ∂ ν (˜ u − w ) (cid:107) F Ω (cid:17) . The last term on the right side is zero, because ∂ ν (˜ u − w ) = 0 on U , and F Ω ⊂ U .Therefore the bound on 1 (cid:13) can be written as C √ h √ ε (cid:107) e − ϕh L W ,q (˜ u − w ) (cid:107) Ω = C √ h √ ε (cid:107) e − ϕh (( W − W ) · D ˜ u + D · ( W − W )˜ u + ( W − W + q − q )˜ u ) (cid:107) Ω ≤ C √ h √ ε (cid:107) e − ϕh (( W − W ) · Du + D · ( W − W ) u + ( W − W + q − q ) u ) (cid:107) Ω + C √ h √ ε (cid:107) e − ϕh (( W − W ) · Du r + D · ( W − W ) u r + ( W − W + q − q ) u r ) (cid:107) Ω ≤ C √ h √ ε (cid:16) (cid:107) e − ϕh Du (cid:107) Ω + (cid:107) e − ϕh u (cid:107) Ω + (cid:107) e − ϕh Du r (cid:107) Ω + (cid:107) e − ϕh u r (cid:107) Ω (cid:17) ≤ C √ h √ ε (cid:16) (cid:107) e − ϕh Du (cid:107) Ω + (cid:107) a + r (cid:107) Ω + (cid:107) e − ϕh Du r (cid:107) Ω + (cid:107) e − βyh b (cid:107) Ω (cid:17) ≤ C √ h √ ε (cid:16) (cid:107) e − ϕh Du (cid:107) Ω + O (1) + (cid:107) e − ϕh Du r (cid:107) Ω + O ( h ) (cid:17) where the constant C mutates as necessary to preserve the bound. Therefore inorder to bound the terms 1 (cid:13) , 2 (cid:13) , and 4 (cid:13) , we need to calculate (cid:107) e ϕh Du (cid:107) Ω , (cid:107) e − ϕh Du (cid:107) Ω , and (cid:107) e − ϕh Du r (cid:107) Ω . We have (cid:107) e ϕh Du (cid:107) Ω = (cid:107) e ϕh h D ( − ϕ + iψ ) e h ( − ϕ + iψ ) ( a + r ) + e iψh D ( a + r ) (cid:107) Ω (cid:46) h (cid:107) D ( − ϕ + iψ )( a + r ) (cid:107) Ω + (cid:107) D ( a + r ) (cid:107) Ω = O ( h − ) + O (1)= O ( h − )since (cid:107) r (cid:107) H (Ω) is O ( h ). Similarly, (cid:107) e − ϕh Du (cid:107) Ω = O ( h − ) . Finally, (cid:107) e − ϕh Du r (cid:107) Ω = (cid:107) e − ϕh h D(cid:96)e (cid:96)h b + e − ϕh e (cid:96)h Db (cid:107) Ω (cid:46) h (cid:107) e − kh bD(cid:96) (cid:107) Ω + (cid:107) e − kh Db (cid:107) Ω = O (1) + O ( h )= O (1) . PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 43 Putting all of this together gives1 (cid:13) = O ( h − )2 (cid:13) = O ( h − )3 (cid:13) = O (1)4 (cid:13) = O (1)5 (cid:13) = O ( h )Therefore multiplying (8.3) through by h and taking the limit as h goes to zerogives lim h → h (cid:90) Ω ( W − W ) · ( Du u + u Du ) dV = 0 . From here on the proof follows verbatim from [DKSU]; it is outlined here only forconvenience.We know from the proofs of Propositions 7.2 and 7.3 that a and a have theform a = γe Φ and a = e Φ , where γ solves( −∇ ϕ + i ∇ ψ ) · ∇ γ = 0on Ω. Therefore(8.4) (cid:90) Ω ( W − W ) · ( ∇ ϕ + i ∇ ψ ) e Φ +Φ gdV = 0 . where g solves ( ∇ ϕ + i ∇ ψ ) · ∇ g = 0 . on Ω.Meanwhile, Φ and Φ solve( ∇ ϕ − i ∇ ψ ) · ∇ Φ + i ( ∇ ϕ − i ∇ ψ ) · W + 12 (cid:52) ( ϕ − iψ ) = 0and ( ∇ ϕ + i ∇ ψ ) · ∇ Φ + i ( ∇ ϕ + i ∇ ψ ) · W + 12 (cid:52) ( ϕ + iψ ) = 0on Ω, respectively. Conjugating the first equation and adding it to the second gives(8.5) ( ∇ ϕ + i ∇ ψ ) · ( ∇ (Φ + Φ ) + i ( W − W )) + (cid:52) ( ϕ + iψ ) = 0 . Therefore ∇ · (( ∇ ϕ + i ∇ ψ ) e Φ +Φ ) = − i ( W − W ) · ( ∇ ϕ + i ∇ ψ ) e Φ +Φ . Combining this with (8.4) gives(8.6) (cid:90) Ω g ∇ · (( ∇ ϕ + i ∇ ψ ) e Φ +Φ ) dV = 0whenever ( ∇ ϕ + i ∇ ψ ) · ∇ g = 0.Choose cylindrical coordinates s, t, and η on R n +1 such that s = x ω , t = | x (cid:48) | > η = x (cid:48) | x (cid:48) | , and let z be the complex variable s + it . Then (8.6) becomes (cid:90) Ω gz ( ∂ z − n − z − z ) e Φ +Φ ( z − z ) n − dV = 0where ∂ z g = 0 on Ω. Since this is the only restriction on g , we can pick any g of the form g ( z ) g ( η )where ∂ z g ( z ) = 0on Ω. Therefore if Ω η represent the slices of constant η , then (cid:90) Ω η gz ( ∂ z − n − z − z ) e Φ +Φ ( z − z ) n − dz ∧ dz = 0for a.e. η , whenever g ( z ) is holomorphic on Ω η . Actually, since this holds for anyholomorphic g on Ω η , we can write that (cid:90) Ω η g ( ∂ z − n − z − z ) e Φ +Φ ( z − z ) n − dz ∧ dz = 0for a.e. η , whenever g ( z ) is holomorphic on Ω η .Now d ( g ( z ) e Φ +Φ ( z − z ) n − dz )= (cid:18) ∂ z − n − z − z (cid:19) e Φ +Φ g ( z )( z − z ) n − dz ∧ dz so by Stokes’ theorem,(8.7) (cid:90) ∂ Ω η g ( z ) e Φ +Φ ( z − z ) n − dz = 0 . Lemma 8.1. If (8.7) holds for every g which is holomorphic on Ω θ , then thereexists a nonvanishing holomorphic function F in Ω η , continuous on Ω η , such that F | ∂ Ω η = ( z − z ) n − e Φ +Φ | ∂ Ω η . Proof. Let f = ( z − z ) n − e Φ +Φ . Define F ( z ) = 12 πi (cid:90) ∂ Ω η f ( ζ ) ζ − z dζ for all z ∈ C \ ∂ Ω η . Then F is holomorphic on the interior and exterior of Ω η , and(8.8) lim z → z ,z ∈ Ω η F ( z ) − lim z → z ,z / ∈ Ω η F ( z ) = f ( z )for z ∈ ∂ Ω θ .Now for z / ∈ Ω η , ( ζ − z ) − is a holomorphic function of ζ on Ω η . Therefore (8.7)implies that F ( z ) = 0 for z / ∈ Ω η , and so (8.8) means thatlim z → z ,z ∈ Ω η F ( z ) = f ( z ) . Thus it remains only to prove that F does not vanish on Ω η . To see this, note firstthat since F is holomorphic on Ω η , the number of zeroes of F enclosed by ∂ Ω η isgiven by n ( F ( ∂ Ω η ) , n ( γ, z ) represents the winding number of γ around PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 45 z . Since F = f on the boundary, n ( F ( ∂ Ω η ) , 0) = n ( f ( ∂ Ω η ) , n ( f ( ∂ Ω η ) , 0) = (cid:90) f ( ∂ Ω η ) dzz = (cid:90) ∂ Ω η dff = (cid:90) ∂ Ω η ∂ z (Φ + Φ ) dz + ∂ z (Φ + Φ ) dz + (cid:90) ∂ Ω η ( n − z − z ) − dz − ( n − z − z − dz = (cid:90) ∂ Ω η d (Φ + Φ + ( n − 1) log( z − z )= 0by Stokes’ theorem. Therefore F is nonvanishing in Ω η . (cid:3) This argument works for a.e. η , so we can develop a function F ( z, η ) on Ω.Then dFF is closed on Ω, so it is exact, and therefore dFF = da for some function a satisfying F = e a . This defines a holomorphic logarithm of F .Therefore (cid:90) ∂ Ω η g ( z )(log( z − z ) n − + Φ + Φ ) dz = (cid:90) ∂ Ω η g ( z ) log F dz = 0 . Then by Stokes’ theorem, (cid:90) Ω η g ( z ) (cid:18) ∂ z (Φ + Φ ) − n − z − z (cid:19) dz ∧ dz = 0 . Using (8.5), we can rewrite this as(8.9) (cid:90) Ω η g ( z )( W − W ) · ( e s + ie t ) dz ∧ dz = 0 . Setting g = 1 and separating this into real and imaginary parts gives that (cid:90) Ω η ( W − W ) · e s dsdt = 0and (cid:90) Ω η ( W − W ) · e t dsdt = 0separately. This holds for a.e. η , where Ω η are the intersections of Ω with transla-tions of P = span { e s , e t } . Therefore(8.10) (cid:90) v + P Ω ( W − W ) · ξdm = 0for all ξ ∈ P , where dm represents the Lebesgue measure on v + P .Now we can vary our choice of coordinates to move the origin around in a smallneighbourhood, and vary ω slightly, without changing the fact that E coincideswith a graph of the form r = f ( θ ) for smooth f . Therefore (8.10) holds for eachplane P in a small neighbourhood of the original. Then Lemma 5.2 from [DKSU]gives us that dW = dW . It remains only to prove that q = q . Note that dW = dW implies that W = W + ∇ Ψ for some function Ψ. Therefore by (8.9) (cid:90) Ω η g ( z ) ∂ z Ψ( z, η ) dz ∧ dz = 0for a.e. η , whenever g is holomorphic in Ω η .By reasoning as in Lemma 8.1, there exists a holomorphic ˜Ψ on Ω η such that˜Ψ = Ψ on ∂ Ω η . Ψ is real, since W and W are, and so ˜Ψ is real on ∂ Ω η . Thereforethe imaginary part of ˜Ψ is harmonic and zero on ∂ Ω η , so it must be identically 0.Then the real part of ˜Ψ (and thus ˜Ψ itself) is constant. Therefore Ψ is constant oneach ∂ Ω η .Varying ω and the origin as before, we get that Ψ is constant on ∂ Ω, and sinceit is only defined up to a constant anyway, we can assume that Ψ = 0 on ∂ Ω.Therefore, up to a gauge transformation, Ψ = 0, so W = W . Going back to (8.3),we now have(8.11) (cid:90) ∂ Ω \ U ∂ ν (˜ u − w ) u dS = (cid:90) Ω ( q − q ) u u dx + (cid:90) Ω ( q − q ) u r u dV. The first and second terms on the right side are O (1) and O ( h ) as before. Theleft side is now bounded by √ h √ ε (cid:16) (cid:107) e − ϕh ( q − q ) u (cid:107) Ω + (cid:107) e − ϕh ( q − q ) u r (cid:107) Ω (cid:17) = √ h ( O (1) + O ( h )) = O ( h ) , so taking the limit of (8.11) as h goes to zero giveslim h → (cid:90) Ω ( q − q ) u u dV = 0 . Using the explicit forms of u and u gives (cid:90) Ω ( q − q ) e Φ +Φ gdV = 0whenever g solves ( ∇ ϕ + i ∇ ψ ) · ∇ g = 0on Ω.Choose coordinates s, t, and η as before. Again we can pick any g of the form g ( s, t ) g ( θ ) where ( ∂ s + i∂ t ) g ( s, t ) = 0 . on Ω. Therefore if Ω η represent the slices of constant η , then (cid:90) (cid:90) Ω η g ( s, t )( q − q ) e Φ +Φ dsdt = 0for a.e. θ . If z = s + it , then we can rewrite this as (cid:90) (cid:90) Ω θ g ( z )( q − q ) e Φ +Φ dz ∧ dz = 0for a.e. θ , whenever g ( z ) is holomorphic on Ω θ . Now the equations for Φ and Φ read ∂ z Φ = 0and ∂ z Φ = 0 , PARTIAL DATA RESULT FOR THE MAGNETIC SCHR ¨ODINGER INVERSE PROBLEM 47 so ∂ z (Φ + Φ ) = 0 , or in other words, Φ + Φ is holomorphic. Therefore e − (Φ +Φ ) is holomorphic,and so in particular (cid:90) (cid:90) Ω θ ( q − q ) dz ∧ dz = 0By varying ϕ as before, we get (cid:90) v + P Ω ( q − q ) dm = 0for all planes P through the origin containing a vector v in a neighbourhood of e y .Then reasoning as in [DKSU] gives q = q on Ω.9. References [DKSU] Dos Santos Ferreira, D., Kenig, C.E., Sj¨ostrand, J., and Uhlmann, G. De-termining a Magnetic Schr¨odinger Operator from Partial Cauchy Data. Comm. Math. Phys. Ann. of Math. Math. Ann. Ann. Acad. Scient. Fenn. Math. Dissertations 139 (2004).[Sa2] Salo, M. Semiclassical pseudodifferential calculus and the reconstruction ofa magnetic field. Comm. PDE 31, 1639-1666 (2006).[Su] Sun, Z. An inverse boundary value problem for the Schr¨odinger operatorwith vector potentials. Trans. Amer. Math. Soc. SIAM J. Math. Anal. Department of Mathematics, University of Chicago, Chicago, IL 60637 USA E-mail address ::