Copyright c1999 by Academia Scientiarum Fennica ISSN??????? ISBN??????? Received 4 October Mathematics Subject Classication: 35J5,35R3,78A7,

Transcription

1 ANNALES ACADEMISCIENTIARUM FENNIC MATHEMATICA DISSERTATIONES??? ELECTRICAL IMPEDANCE TOMOGRAPHY AND FADDEEV GREEN'S FUNCTIONS SAMULI SILTANEN Dissertation for the Degree of Doctor of Technology to be presented with due permission for public examination and debate in Auditorium N at Helsinki University of Technology (Espoo, Finland) on the 6th of November, at 1 o'clock noon. HELSINKI 1999 SUOMALAINEN TIEDEAKATEMIA

2 Copyright c1999 by Academia Scientiarum Fennica ISSN??????? ISBN??????? Received 4 October Mathematics Subject Classication: 35J5,35R3,78A7,65R1,65R. YLIOPISTOPAINO HELSINKI 1999

3 Acknowledgements I am grateful to Erkki Somersalo for being the advisor of my work and for introducing me to the international inverse problems research community. Also, our combined boxing and mathematics sessions with him and Lassi Paivarinta were extremely inspiring. I thank Petri Ola and Margaret Cheney for carefully reading my manuscript and giving valuable suggestions I also thank Rainer Kress for being my opponent. I am grateful to Adrian Nachman for his interest to this work and the enlightening discussions about his results. This work led to wonderful collaboration with Jennifer Mueller and David Isaacson. Thanks to them together with Jon Newell and Margaret Cheney, my stay at the Rensselaer Polytechnic Institute was very pleasant. I appreciate the opportunity to be measured with the ACT 3 impedance imaging device (it didn't hurt). I thank Gunther Uhlmann for discussions about the inverse conductivity problem and John Sylvester for his advice in the numerical aspects of my work. I wish to thank my dear friend Matti Lassas for always having time for my questions, mathematical and others. I thank all the people in the Finnish inverse problems research society for creating a great social atmosphere. I thank Eero Saksman, Hans-Olav Tylli and Juha Kinnunen for their help in questions related to singular integrals and functional analysis. I am grateful to my wife and friends for supporting me during this work. Finally, I thank Olavi Nevanlinna for arranging an excellent working environment at the Helsinki University of Technology and the Finnish Ministry of Education, Finnish Academy of Science and Letters, the Emil Aaltonen Foundation, the Instrumentarium Science Foundation and the Finnish Cultural Foundation for nancial support. Tuusula, October 1999 Samuli Siltanen

4 Contents 1 Introduction Electrical impedance tomography Mathematics of EIT A brief history of the problem Applying A. Nachman's method to EIT A. Nachman's reconstruction method 11.1 Denitions From to t From t to Conditional stability Practical aspects Theoretical observations Faddeev Green's functions Denitions and basic properties Derivatives Representation formulas Estimates An asymptotic expansion Determining H k () The scattering transform Large jkj estimates Behaviour near k = Symmetries Computational methods Evaluating Faddeev Green's functions Computing g 1 for jxj > 1= Computing g 1 in the unit disc Numerical results An alternative algorithm for g Computing and t from Finding in a bounded domain Description of the algorithm Proof of convergence Generating test problems Conclusion 5

5 1. Introduction 1.1. Electrical impedance tomography Magnetic resonance imaging (MRI and fmri), X-ray tomography or computed tomography (CT), magnetic encephalography (MEG) and ultrasound imaging are examples of medical imaging methods that produce a picture of a patient's inner organs without intruding into the body. Images coming from dierent systems complement each other since each of these methods measures the distribution of some specic physical parameter, such as mass density (CT) or sound speed (ultrasound imaging). Some of the methods are more harmful to the patient than others: X-ray measurements expose her to dangerous radiation whereas ultrasound imaging has little or no health risks. Another dierence between the systems is that some of the machines are very expensive. For these reasons it is valuable to have as many dierent imaging methods available as possible. Electrical impedance tomography (EIT) is a medical imaging system that produces a picture of electric conductivity distribution inside the patient. For the EIT measurement an array of electrodes is attached for instance around the chest of the patient as shown in Figure 1. Measurements are done by feeding electric currents Figure 1: EIT measurements can be done with an array of electrodes around the chest. through the electrodes and measuring the corresponding voltages at the electrodes. This measurement is repeated with several dierent choices of current patterns. The

6 6 Samuli Siltanen resulting image is an approximate map of the electric conductivity in a horizontal slice on the level of heart and lungs. To get a feeling about why this should be possible, note that if the lungs are full of air, no electric current will pass through them. On the other hand, empty lungs do conduct electricity to a somewhat greater extent, so at least these two situations will give dierent measurements. Of course it is a lot more challenging task to reconstruct every detail inside the chest this requires precise measurements and nontrivial mathematics. Measuring around the chest can be used in monitoring the heart or to improve ECG measurements. For other medical purposes of EIT, such as screening for breast cancer, see [4]. Attractive features of EIT include the possibility of harmless monitoring and good temporal resolution. Moreover, EIT gives complementary information to other imaging modalities. There are working EIT machines, most notably the Adaptive Current Tomography (ACT) system built in Rensselaer Polytechnic Institute (RPI). It is capable of real-time imaging and has been tested on people. See [11] for more details. Impedance imaging is also used in a commercial breast cancer detector called T- Scan that was launched recently. EIT is not yet in wide clinical use because practically reliable and fast imaging is extremely dicult and needs still more research. This work is part of a project aiming to produce a new imaging algorithm for EIT starting from the latest results of applied mathematics. Hopefully this will help the ongoing eort to build accurate, cheap and portable EIT devices. 1.. Mathematics of EIT The mathematical task of interpreting EIT measurements is an inverse boundaryvalue problem called the inverse electric conductivity problem. It is a challenging question in the theory of partial dierential equations since it is nonlinear (Figure ) and ill-posed. Designing a practical algorithm for EIT is hard because nonlinearity makes the mathematical solution theoretically dicult and ill-posedness means that a small error in the practical measurements may cause a large error in the reconstructed image. Due to its challenging theoretical nature and medical applications, EIT is a good problem in applied mathematics. + = Figure : EIT is a nonlinear problem. The boundary measurements corresponding to the two leftmost objects do not add up to the measurement corresponding to the rightmost object.

7 Electrical Impedance Tomography and Faddeev Green's Functions 7 Let us state the inverse conductivity problem in mathematical terms: let R be a bounded, simply connected Lipschitz domain and :! R a bounded, measurable real function with a strictly positive lower bound, that is, <c (x) a.e. The forward electric conductivity problem is to determine, given any f H 1= (@), the H 1 solution u of the Dirichlet problem rru = in = f: (1) Note that the boundary value problem (1) is elliptic due to the lower bound of and its solution under various assumptions on is discussed in [17, ]. Dene now the Dirichlet-to-Neumann map of by : H 1= (@)! H ;1= (@) h f gi = ru rv where v is any H 1 () function with trace g on the boundary. The inverse electric conductivity problem is to nd out if (satisfying perhaps some additional regularity assumptions) is uniquely determined from the knowledge of and if so, to reconstruct. Physically f represents a voltage distribution on the boundary of a two-dimensional physical body with an unknown inhomogeneous conductivity distribution. Furthermore, u is the electric potential inside the body that solves the conductivity equation (1) and agrees with f on the boundary. Thus the Dirichlet-to-Neumann map can be interpreted as the voltage-to-current map taking the boundary value of the voltage to the resulting electric current through the boundary. In this way represents static electric measurements done on the boundary of and the inverse conductivity problem is to determine the conductivity that caused those measurements. Let us make some observations: The physical signicance of the assumption that the conductivity is bounded away from zero is that there cannot appear annulus-shaped areas of zero conductivity this would make it impossible to reconstruct anything inside the annulus. As mentioned above, in practical measurements one feeds currents and measures voltages and not the other way around. This is due to the fact that although the two measurements are mathematically equivalent, the latter procedure is unstable to measure

8 8 Samuli Siltanen Although physical bodies are three-dimensional, from the point of view of EIT the problem of producing the picture of a horizontal slice of a person's chest is approximately two-dimensional. This is why we consider only the twodimensional case in this work. For an analysis of avoiding this simplication by using approximate Dirichlet-to-Neumann maps, see [33]. In addition to EIT, there are other physical applications of the inverse conductivity problem, such as geological prospecting and non-destructive material testing. All further references to applications will, however, concentrate on EIT A brief history of the problem Research on the inverse conductivity problem has two traditions: a purely mathematical one providing uniqueness and reconstruction results and a practical approach having as its goal algorithms that perform well on physical data. The theoretically oriented line of research was started by Calderon [8], who studied electrical engineering before changing to mathematics. He stated the inverse conductivity problem for the rst time in the above form (in dimensions n ) and gave some preliminary results for conductivities close to a constant. Mathematically his work continued the tradition of Gel'fand, Levitan and Borg [16, 6]. Once Calderon asked his question, new results started to appear. Kohn and Vogelius [35] showed that for piecewise analytic conductivities the Dirichlet-to- Neumann map uniquely determines the conductivity in dimensions or more. Sylvester and Uhlmann [5] showed in dimensions n 3 that is C 1 then uniquely determines C 1 (). The rst reconstruction method was published by A. Nachman [4] this result is valid in dimensions strictly greater than two. The work of R. G. Novikov [3] contains similar results. Meanwhile the case n = remained unsolved. Why is the two-dimensional case more dicult than the higher-dimensional ones? A naive reason for this is that in n dimensions the Schwartz kernel of depends on (n ; 1) variables while is a function of n variables this means that in a sense there is no overdeterminacy in two dimensions and all the data has to be used for the reconstruction. In 1996 A. Nachman published a reconstruction procedure for the two-dimensional problem [43]. This method is especially interesting because it is constructive in nature. The smoothness assumption in [43] is W p () for some p > 1 which implies that is continuous. This result was soon sharpened for conductivities in W 1 () < < 1 by R. Brown and G. Uhlmann [7]. These conductivities are also continuous. A more detailed introduction to the mathematical development of the problem is given in [43, 51]. Also V. Isakov's article [3] and book [31] are good references

9 Electrical Impedance Tomography and Faddeev Green's Functions 9 they cover also results aiming at partial answers such as nding domains with zero or innite conductivity. See also [6, 7]. The practical branch of EIT research started with papers like [], although the inverse conductivity problem has been treated earlier in the engineering literature [38, 9]. Since the inverse conductivity problem is very complicated mathematically, we can expect the the practical EIT imaging task to be dicult. In addition to the nonlinearity and ill-posedness of the theoretical model there are further problems: the Dirichlet-to-Neumann map is not given instead its inverse operator has to be approximated from noisy and nite-precision current-to-voltage measurements. Furthermore, only a nite number of current patterns can be applied through a nite number of electrodes. This situation places restrictions on the resolution of the images [9] and requires the physical model of the electrode system to be rened [46]. This leads to the notion of distinguishability [8] describing the resolution it is possible to obtain in principle with a given system. With these restrictions known, many algorithms have been developed and tested on real data. They fall roughly in three categories: 1. Solve the linearized inverse problem. This approach works best for conductivities close to constant examples include back-projection [3] and one-step Newton schemes like NOSER [14].. Solve the nonlinear problem by nding iteratively a conductivity distribution that matches the physical measurements best in the least squares sense. These methods require regularization due to the ill-posedness of the inverse problem. For a list of references to methods of this type see [11, p.9] note also [1, 34]. See [55, 56] for reports on using anatomical a priori information in iterative EIT imaging. 3. Solve the problem directly. The methods of this category are called direct methods as opposed to iterative ones they aim as well to solve the full nonlinear problem. An existing example of this type is layer stripping [47, 49]. The numerical realization of Nachman's reconstruction method will fall into this category. Some of these ideas combined with nontrivial electrical engineering make it possible to build reasonably working EIT devices. This has been done in several laboratories around the world, e.g. RPI, Sheeld, Oxford and Kuopio, Finland, see [11, 39, 56]. We mention that the ACT system of the RPI group performs imaging in real time and has been tested in a clinical environment. For ideas for dynamical measurements using Kalman lters see [57]. The above treatment of the practical aspects of EIT is far from complete. For a more thorough review see [59, 1, 11].

10 1 Samuli Siltanen 1.4. Applying A. Nachman's method to EIT The two main lines of EIT research, the purely mathematical and the applied one, have not beneted much from each other. The methods used by the theoretical reconstructions tend to be hard or impossible to realize on a computer. On the other hand it has turned out to be dicult to prove convergence results for the existing practical algorithms in addition it is not easy to describe the relation between the actual conductivity and a reconstruction computed with a convergent but regularized algorithm. Nevertheless, Nachman's mathematical proof [43] for the two-dimensional case is throughout constructive and seems not to have steps that are impossible to implement practically. This work is part of a project to develop a practical reconstruction algorithm based on Nachman's method, the scope being to provide both theoretical insight and numerical algorithms for various steps of the reconstruction. In this way this thesis helps to build a bridge between theory and practice of the inverse conductivity problem. The larger project of realizing Nachman's whole reconstruction scheme is carried out by the author in collaboration with professor David Isaacson and PhD Jennifer Mueller from Rensselaer Polytechnic Institute, NY, USA. The results of the project will be published later. We present here briey the contents of this work. In Chapter we give an introduction to Nachman's proof [43]. A more detailed treatment can be found in [45]. Chapter 3 presents new results of practical importance related to Nachman's method: estimates, representation formulas and symmetry properties for Faddeev fundamental solutions g k and G k information about the size, smoothness and asymptotic behaviour of the functions (x k) (important intermediate objects in Nachman's proof) observations on howvarious properties of the conductivity manifest themselves in the non-physical scattering transform t(k), another function needed in the reconstruction new estimates on jt(k)j for both large and small jkj. These ideas are used in Chapter 4 for designing numerical algorithms useful for various steps of the reconstruction, namely computing numerically values of the Faddeev Green's functions g k and G k, computing the functions (x k) and t(k) for given test conductivities. One radially symmetric example is computed numerically. The reason for choosing a symmetric example is to avoid intensive computations and utilize the symmetry theoretically the algorithms work for nonsymmetric cases, too.

11 Electrical Impedance Tomography and Faddeev Green's Functions 11. A. Nachman's reconstruction method.1. Denitions In 1996, A. Nachman published a constructive proof of the inverse conductivity problem in two-dimensional space. The main theorem of his work is Theorem.1 [43, Theorem 1] Let be a bounded, simply connected Lipschitz domain in R. Let 1 and be in W p () for some p > 1, and have positive lower bounds. If 1 = then 1 = : The proof gives a constructive procedure for recovering from the knowledge of. We will make some simplifying assumptions. The rst part of the proof of Nachman's theorem is to nd on the boundary of and reduce the problem to the case 1 in a neighbourhood the reduction is done in such a way that we can to be smooth. We will thus assume 1 near the C 1 Furthermore, we prefer working with smooth functions instead of the Sobolev ones whenever possible and will therefore consider only smooth conductivities: C 1 (): From the point of view of applications there is not much dierence in considering C 1 conductivities instead of W p ones since in practice the conductivities (inner organs of people) are discontinuous and by Sobolev imbedding theorem the W p functions are continuous. The solution of the inverse problem is based on the use of an intermediate object called the non-physical scattering transform t, which is not directly measurable in experiments. See Figure 3 for a schematic illustration. We give the denition of t and explain the two main steps of the procedure, namely nding t from the boundary measurements and determining from the knowledge of t. To dene t we consider the Schrodinger equation associated with the inverse conductivity problem. If u is a solution of rru = in, dening ~u = 1= u yields (;+q)~u =in () with the potential q C 1 () given by q = 1= : (3) 1= Moreover, our assumption that be one near allows us to extend smoothly by one and q by zero to the whole plane. The exponentially behaving solutions of () introduced by Faddeev [15] are the key to the reconstruction:

12 1 Samuli Siltanen (x) forward ; ; ; ; t(k) melborp esrevni Figure 3: Overall strategy of the proof. Theorem. [43, Theorem 1.1] Let q L p (R ) 1 <p< be such that there exists a real-valued L 1 (R ) with q = ;1= 1=, (x) c> and r L p (R ). Then for any k C n there is a unique solution (x k) of with ( k) ; 1 L ~p \ L 1. Moreover, ; 1 W 1 ~p and for s 1 and k suciently large. (;+q)e ikx (x k) = (4) k( k) ; 1k W s ~p cjkj s;1 kqk L p (5) Here e ikx = exp(i(k 1 + ik )(x 1 + ix )) and ~p is dened by 1=~p = 1=p ; 1=. The Sobolev space W 1 ~p is dened by W 1 ~p (R )=ff L ~p (R ) i f L ~p (R ) i =1 g and the space W s ~p is the interpolation space between L ~p and W 1 ~p [4]. Note that the condition ; 1 W 1 ~p combined with Sobolev imbedding theorem means that ; 1 is continuous and tends to zero asymptotically when jkj! 1. Furthermore, since 1, we see that the solutions (x k) :=e ikx (x k) e ikx have somesortof exponential behaviour. Perhaps the most important content of Theorem. is the existence of the Faddeev solutions for all nonzero k since this was known only for large jkj from [5]. We dene the non-physical scattering transform for all k 6= by t(k) := e k (x)(x k)q(x)dx (6) R where e k (x) = exp(i(kx + kx)). Note that because is asymptotically near to one, t(k) resembles the Fourier transform of q(x)evaluated at the point(;k 1 k ) R, for large jkj. We will take a closer look at the construction of the functions in section 3. and prove that they are smooth in x.

13 Electrical Impedance Tomography and Faddeev Green's Functions 13.. From to t First we want to nd a relation between the Dirichlet-to-Neumann maps of the original conductivity problem and the Schrodinger problem described above. It turns out that the maps coincide. Note that zero is not a Dirichlet eigenvalue of the operator ;+q, that is, there are no nontrivial solutions ~u of () with vanishing trace. This is evident because for any such ~u the function ;1= ~u would be a nonvanishing solution with trace zero of the conductivity equation, which is a contradiction to the unique solvability of the Dirichlet problem (1). Thus we can dene the Dirichlet-to-Neumann map q : H 1= (@)! H ;1= (@) for equation () by hg q fi = rv r~u + qv~u (7) where ~u H 1 () is the weak solution of () with Dirichlet data = f and v is any H 1 () function with trace g. Furthermore, ;1= u @ since Thus q =, so our task is to nd the scattering transform out of q. Dene a single-layer operator S k for k C n by (S k )(x) G k (x ; y)(y)d(y) (8) where the function G k (x)isfaddeev's exponential fundamental solution for ;. We will discuss this function in detail in section Denote the Dirichlet-to-Neumann map of the zero potential (corresponding to homogeneous conductivity 1) in by. We have Theorem.3 [43, Theorem 5] Let be a bounded Lipschitz domain in R and let q L p () for some p > 1. Assume is not a Dirichlet eigenvalue of ; +q in and denote (x k) = e ikx (x k) with given by Theorem.. Then, for any k C n : (i) S k ( q ; ) is a compact operator: H 1= (@)! H 1= (@) (ii) The trace of the function ( k) satises the integral equation ( = e ikx ; S k ( q ; ) ( k) (9) (iii) I +S k ( q ; ) is invertible on H 1= (@) if and only if the Schrodinger problem (4) has a solution (here q is extended to be zero outside )

14 14 Samuli Siltanen (iv) If ( has been determined, then t(k) can be recovered from the formula t(k) e i kx ( q ; ) ( k)d: (1) Clearly the numerical realization of Theorem.3 is an important application for the Faddeev Green's function algorithm developed in section 3.1. Moreover, the possibility to compute t(k) numerically from known presented in section 4. is vital in testing any algorithm for Theorem From t to The non-physical scattering transform (6) contains enough information to recover. Namely, Theorem.1 of [43] shows that k (x k) = 4k t(k)e ;k(x)(x k) k 6= (11) where e k (x) = exp(i(kx+ kx)), holds (with respect to the variable k) intheweighted Sobolev space W 1 ~p ; = fhi; f W 1 ~p g > =~p where 1=~p =1=p ; 1= andhi =(1+jj ) 1=. By Theorem 4.1 of [43] the solution of (11) is given by (x k) =1; 1 () for all x R k C nfg. Taking formally k to zero in (4) yields t(k ) R (k ; k) k e ;x(k )(x k )dk1dk (1) (;+q)(x ) = (;+ 1= )(x ) = 1= which suggests that lim k! (x k) =(x ) = p (x) because of the uniqueness in Theorem. and the fact that ; 1inL ~p \ L 1. Section 3 of [43] is devoted to the proof that this indeed is the case. Thus we can recover the conductivity by taking k to zero in (1): 1= (x) =lim k! (x k) =1; 1 4 C t(k) jkj e ;x(k)(x k)dk: (13) Note that to evaluate the last expression, (x k) is needed for all values of k C n. An application of the numerical methods presented in Chapter 4 is to compute and consequently t for known conductivities. This way a computational realization of this step can be tested.

15 Electrical Impedance Tomography and Faddeev Green's Functions Conditional stability G. Alessandrini showed in [] an example demonstrating the ill-posedness of the inverse conductivity problem: let be the unit disc in R and dene two conductivities r L 1 () by 1 and s for jxj r r (x) = 1forr<jxj 1 where <s<1 and <r<1. By explicitly calculating the Dirichlet-to-Neumann map for both conductivities he showed that k r ; k H 1= (@)!H ;1= (@) j1 ; sjr! r! while k ; r k L 1 () = j1 ; sj is independent of r. It follows that there can be no continuous real function with f() = and an inequality oftype k ; r k L 1 () f(k r ; k H 1= (@)!H ;1= (@)) (14) ensuring that a small error in measured data leads to small L 1 error in the reconstruction. From the practical point of view the above seems devastating. However, it is possible to derive stability estimates of the form (14) if there is a priori knowledge about the conductivity. This is in general a reasonable assumption because in medical EIT there is a lot of knowledge about the type of conductivity changes inside the patient. There is a stability analysis available for the reconstruction method at hand: the work of Nachman and Liu [43, 4] shows that Theorem.4 [4, Theorem 1] Let 1 satisfy the a priori assumptions k j k W p () M for some xed 1 < p < and j (x) 1=M in j = 1. If we have j j =@ =, then The function! is such that k r ; k L 1 () C M!(k r ; k H 1= (@)!H ;1= (@)): (15)!(") jlog "j ; (16) for every <<=p and suciently small " the constant C M depends only on M p and. It is interesting to note that, roughly speaking, the inversion step from to t(k) has logarithmic stability and the step from t(k) to has linear stability.

16 16 Samuli Siltanen.5. Practical aspects Nachman's conductivities are continuous, and so are the ones considered by Brown and Uhlmann [7]. In practice the interesting conductivities (inner organs of a person) are discontinuous, and it not clear how numerical realizations of these reconstruction methods will perform with data corresponding to conductivities with jumps. In Nachman's approach the boundary data is the Dirichlet-to-Neumann map, which cannot be measured practically in a stable way. Its smoothing inverse operator R, the Neumann-to-Dirichlet map, is the one that has to be used in physical experiments. From this map, can be computed after the measurements since the two are inverse maps to each other. It is easier to recover details of located near the In fact, as pointed out in [18] and the references therein, the contribution of conductivity details to the boundary measurements gets exponentially smaller as the distance from the boundary increases. This is one reason why exponentially growing solutions are needed in the reconstruction. However, in EIT measurements it is for safety reasons impossible to use arbitrary large electric currents, so the boundary values of the exponentially growing solutions have to be approximated. This process is numerically unstable for large jkj. The computational cost of Nachman's reconstruction will be great. This is a consequence of the four-dimensional nature of the method: both the x- and k- variables range in a D grid. It may not be a possible choice for a real-time EIT application, but in any case it provides a reliable initial choice for a faster iterative approach. 3. Theoretical observations The purpose of this chapter is to derive theoretical results that help the realization of Nachman's reconstruction scheme. Our goal is to answer such practical questions as How to compute eciently and accurately point values of Faddeev Green's functions G k and g k? Finding the functions from equation involves solving a weakly singular integral equation with t(k) appearing in the kernel. How singular is the kernel at k =? How much error does the truncation of the integral introduce? In other words, what is the exact rate of decay of jt(k)j when jkj grows? How much can we deduce qualitative information about from symmetries of t(k)?

17 Electrical Impedance Tomography and Faddeev Green's Functions 17 Section 3.1 is an analysis of the Faddeev Green's functions g k and G k. There are more results than needed in this work we believe that they are valuable for the eld of inverse problems in D electrical prospection or nonlinear evolution equations of Novikov-Veselov type [58, 53]. In section 3. we derive both quantitative and qualitative facts concerning t(k) and (x k) Faddeev Green's functions The two fundamental solutions g k (x)andg k (x) originally introduced byfaddeev [15] are in the core of Nachman's reconstruction procedure. In this section we dene and analyse these functions and use the results later in both analytical and numerical considerations. In this work we deal with functions g k (x) and G k (x) dened on R. Similar results on their three-dimensional analogues can be found in [44] Denitions and basic properties Dene the function g k (x) for all x R and k C n by the formula g k (x) = 1 () R e ix ( d (17) +k) where x = x 1 1 +x in the exponent and in the denominator we have = 1 +i and complex multiplication. The function g k (x) is a fundamental solution for the operator L k = ; ; 4ik@ = ; ; 4ik 1 + ) that is, L k g k (x) =(x): (19) This can be seen by noting that the Fourier symbol of L k is cl k () =jj +k( 1 + i )=( +k) and formula (17) is essentially the inverse Fourier transform of 1=c Lk understood in the sense of tempered distributions: g k (x) =F ;1 (( +k)) ;1 : In Nachman's reconstruction scheme the function g k is a key object. It is used to nd the exponentially behaving solutions (x k) =e ikx (x k)mentioned in Chapter we will discuss this construction in sections 3. and 4.. Another important application for g k (x) is going from the boundary measurements to the nonphysical scattering transform t(k) see Theorem.3 for details. The function G k (x) needed for this is dened for all x R and k C n by G k (x) :=e ikx g k (x): ()

18 18 Samuli Siltanen Note that we used again the complex notation kx = (k 1 + ik )(x 1 + ix ) so that the modulus of the exponential multiplier is not necessarily one. Given this way, G k will be a fundamental solution for the negative Laplacian: ;G k (x) ikx g k (x) =;4@e (x) =e ikx (; ; k (x) =(x) (1) where we used (19) and the fact ikx = since e ikx is complex analytic in x = x 1 + ix. To compare the Faddeev fundamental solution with the regular one, let G the standard fundamental solution for ;: G (x) =; 1 log jxj and denote the dierence between the two by H k (x) :=G k (x) ; G (x) k C n : () Now H k is a harmonic distribution since H k = ; + =,andby Weyl's lemma we know that H k (x) is a smooth, harmonic function in the whole plane. This helps us understand the behaviour of G k and g k at the origin: both have a logarithmic singularity there. This follows from equation () and the identity g k (x) =e ;ikx H k (x) ; e;ikx log jxj (3) respectively. For convenience, we collect our observations about H k together as a lemma. Lemma 3.1 Let x R and k C n. Then the function H k (x) given in R n by () is smooth and harmonic at every point x R. In section we will compute H k (). Change of variables in the dening integral (17) shows that the Faddeev fundamental solution g k has several symmetries: Theorem 3. Let x R n and k C n. The Faddeev Green's function g k given by (17) has the properties where e k (x) = exp(i(kx + kx)). g k (x) = g 1 (kx) (4) g k (x) = g k(;x) (5) g k (x) = e ;k (x)g k (x) (6)

19 Electrical Impedance Tomography and Faddeev Green's Functions 19 This together with the computation H k (x) =e ikx g 1 (kx)+ 1 1 log jkjjxj; log jkj = H 1(kx) ; 1 log jkj shows that we can restrict our attention to the functions g 1 G 1 and H 1 : G k (x) = G 1 (kx) (7) H k (x) = H 1 (kx) ; 1 log jkj: (8) The functions G k (x) and H k (x) are real-valued for all k C n. This is seen by computing G k (x) =e ;i kx g k (x) =e ;i kx e k (x)g k (x) =e ikx g k (x) =G k (x): 3.1. Derivatives In this section we will derive explicit formulas for the rst-order partial derivatives of g k (x): Theorem 3.3 Let x R n k C n and dene the function g k (x) by (17). The rst order partial derivatives of g k (x) are given by where e ;k (x) =exp(;i(kx + k (x) = ; 1 4x ; e ;k(x) 4x ; ikg k(x) k 1 (x) 4ix ; e ;k(x) 4ix + kg k(x) (3) Before the proof we need a couple of lemmas. Lemma 3.4 We have F ;1 i = 1 x and F ;1 i = 1 x (31) where x = x 1 + ix and the inverse Fourier transforms are taken in the sense of tempered distributions. Proof. We will prove the rst of the identities in (31). The second one follows from this by taking complex conjugates. Denote h(x) :=F ;1 i = F ;1 i( 1 + i ) and note that the properties of the Fourier transform imply that h is a tempered distribution =. On the other hand, it is well-known [5, Chapter III] that f(x) :=1=x is a fundamental solution for the that =. (3)

20 Samuli Siltanen Our calculations show that h(x) =f(x)+u(x), where u = h ; f = ; =: (33) By elliptic regularity we know that u is dierentiable and so (33) implies that u is an entire function. On the Fourier transform side (33) takes the form i ( 1 + i )^u = (34) which implies that ^u is supported in the origin, so u(x) is a polynomial of x 1 and x. Lemma in [5] states that if u S(R ) is homogeneous of degree ;1, then the Fourier transform ^u is also homogeneous of degree ;1. This is the case for f(x) and cannot be true for the sum of f and a nonzero polynomial so u and the proof is complete. Q.E.D. Lemma 3.5 Let x R n and dene the function g 1 by (17). We 1 (x) = ; e;ix 1 4x 1 (x) = ; 1 4x ; ig 1(x) (36) g 1 (x) = ie;ix 1 Proof. We start with 1 (x). Compute c@g 1 = x : (37) i= ( +) = i ( +) and use Lemma 3.4 and the translation property of the Fourier transform to 1 (x) = i F 1 )=;e;i( )x ;1 ( F ;1 + 4 i = 1 ;e;ix 4x (38) which is (35). The Laplacian of g 1 is now easily computed with aid of (38) and the fact that x ;1 is antiholomorphic at x 6= : g 1 (x) 1 (x) =;4@ e;i(x+x) 4x and identity (37)is proved. = ; e;ix = ie;ix1 x

21 Electrical Impedance Tomography and Faddeev Green's Functions 1 To 1 (x) atx 6=,take complex conjugate of equation (19) and apply equation (6) of Theorem 3. to get =;g 1 (x)+4i@g 1 (x) = ieix 1 x ; 4eix 1 g 1 (x)+4ie ix 1 (x): Now we can solve 1 and the claim follows. Q.E.D. Proof. (Theorem 3.3) We start with the special case k = 1. Then by Lemma 1 (x) 1 (x) =; 1 1 4x ; e;ix1 4x ; ig 1 (x) 1 (x) = 1 4ix ; e;ix1 4ix + g 1(x): Now the claim follows from g k (x) =g 1 (kx) and the chain rule. Q.E.D Representation formulas We use residue calculus to nd formulas similar to [5, formula (3.1)]. Theorem 3.6 Let x =(x 1 x ) R n with x 1 >. Then g 1 (x) is given by g 1 (x) = e;ix 1 1 Re e ix e ;t(x 1+ix ) 1 dt : (39) t ; i Formula (39) combined with a Laplace transform identity gives another representation of g k : Corollary 3.7 Let x R n x 1 >. Then g 1 (x) is given by g 1 (x) = e;ix 1 1 x 1 s cos s + x sin s ds: (4) s + x Proof. (Theorem 3.6) Write formally () g 1 (x) = 1 1 ;1 = e ;ix 1 ;1 1 ;1 To evaluate the inner integral note that f(z) = e i(x 1 1 +x ) ( 1 +1) +( + i) d 1d 1 e ix e ix 1t t +( + i) dt d : ;1 e ix 1z z +( + i) = e ix1z (z ; z )(z + z ) where z = ;1+i, is a meromorphic complex function. Dene for all R> the path ;(R) =[;R R] [fre i' : ' g with positive orientation. Observe that

22 Samuli Siltanen x 1 > implies the exponential decay of f(z) in the domain Im (z) >. Compute for > and large R i Res z=z f(z) = ;(R) Taking the limit R!1we see that 1 ;1 f(z)dz = R ;R e ix 1t dt + O(1=R): t +( + i) e ix 1t eix1(;1+i) dt =i t +( + i) (;+i ) = e;ix1;x1 + i In a similar fashion we obtain Thus for x 1 > 1 ;1 () g 1 (x) = e ;ix 1 = e ;ix 1 e ix 1t dt = ;eix1+x1 t +( + i) + i ; e ix 1 = e ;ix 1 Re e ix +ix 1 +x 1 + i ;1 1 e ;ix ;x 1 e ix 1 1 d + < : d + e ;ix 1 ; i e ;ix ;x 1 d ; i > : 1 e ix ;ix 1 ;x 1 1 d + i e ix ;x 1 d + i and the claim follows. Q.E.D. Proof. (Corollary 3.7) Notice that the Laplace transform of the constant function 1is1=z for Re (z) >, i.e. Using this we can compute z = e ;zw dw Re (z) > : e ;xt 1 1 t ; i dt = e iw e ;t(x 1+ix +w) dtdw = 1 e iw x 1 + ix + w dw: Substitution to formula (39) and a change of coordinates s = x 1 + w yield 1 e ix 1 e i(x 1+w) 1 g 1 (x) =Re ix +(x 1 + w) dw e is =Re ix + s ds = x 1 = 1 x 1 Re [(s ; ix )e is ] s + x ds = e;ix 1 1 x 1 s cos s + x sin s ds: s + x Q.E.D.

23 Electrical Impedance Tomography and Faddeev Green's Functions Estimates We start with a simple estimate for jg k (x)j which is a straightforward consequence of Theorem 3.6. Theorem 3.8 Let x R n and k C n. Then jg k (x)j 1 jkjjxj : (41) Combining this result with Theorem 3.3 yields immediately Corollary 3.9 Let k C n. For x R satisfying jxj 1 and an arbitrary multi-index we have j@ C( k) g k (x)j : jxj (4) The integral formula (17) is formal because the integrand belongs neither to L 1 (R ) nor L (R ). Using the Riesz-Thorin interpolation theorem, however, the inverse Fourier transform can be understood as an interpolated operator from L to L for 1 < <, << 1 and 1= +1= = 1. Since the integrand in (17) belongs to L, we see that g k L : Theorem 3.1 Let k C n and <<1. The Faddeev Green's function given by (17) satises the estimate where 1= +1= =1. kg k k L jkj;= ; = ; (43) We now turn to the proofs. Proof. (Theorem 3.8) Because g k (x) =g 1 (kx) forany k C n, it is enough to prove jg 1 (x)j 1 jxj x R n : (44) Let x 1 >. From formula (39) we see at once that jg 1 (x)j 1 1 e ;x 1t jt ; ij dt 1 1 e ;x 1t dt = 1 x 1 : To prove a similar estimate in terms of jx j we divide the consideration to the cases x > and x <. Let x >. The complex function f(z) = e ;ix z;x 1 z (z ; i) ;1 is analytic for z C n i and decays exponentially in the quadrant Re(z) > Im(z) <. Dene for any R>thepath ;(R) =[ ;ir] [fre i' : ;= ' g[[r ]:

24 4 Samuli Siltanen Then we have 1 This yields as before jg 1 (x)j 1 e ;ix t;x 1 t t ; i 1 dt = 1 e ;x s+ix 1 s ds: s +1 e ;x s ds = 1 x x > : (45) Let x <. In this case f(z) decays exponentially in the quadrant Re(z) > Im(z) >. Dene for any R>thepath Then we have 1 ;(R) =[ R] [fre i' : ' =4g[e i=4 [R ]: e ;ix t;x 1 t 1 dt =(1+i) t ; i and so, since js + is ; ij 1= p for all s>, jg 1 (x)j 1 1 p e s(x ;x 1 ) js + is ; ij ds e ;is(x +x 1 )+s(x ;x 1 ) 1 s + is ; i e s(x ;x 1) ds Now for points x satisfying x 1 > andjx jx 1 we have jg 1 (x)j 1 p x 1 jxj 1 jxj ds 1 p jx j : since jxj= p x 1 in this domain. For points satisfying x 1 > and jx j > x 1 we have by a similar argument jg 1 (x)j 1 p jx j 1 jxj : We have thus shown that the desired estimate holds for all (x 1 x ) satisfying x 1 >. By continuityofg 1 outside the origin and the particular case g 1 (;x 1 x )=g 1 (x 1 x ) of formula (5) we see that the estimate (44) is valid at all x 6=. Q.E.D. Proof. (Theorem 3.1) Note that kg k k L = kg jkj k L. Furthermore, by (4) we have R kg jkj k L = jg 1 (jkjx)j dx = jkj ; jg 1 (x)j dx = jkj ; kg 1 k L R and thus it is enough to prove the claim for g 1. By the Riesz-Thorin interpolation theorem we have acontinuous operator F ;1 : L! L kf ;1 k L!L 1=(): (46)

25 Electrical Impedance Tomography and Faddeev Green's Functions 5 Since <<1 we have 1 < <. We will nd a bound for the L norm of ^g 1 as follows: d k^g 1 k L =( + ) : (47) jj j +j jj>4 To estimate the rst integral in (47) note that in the domain of integration we have j + j (jj;) (jj;jj=) = jj=: Thus jj>4 d jj j +j For the second integral in (47) compute jj4 1 1 d jj j +j jj4 1 4 jj4 j+j>1 r 1; dr < d d + jj j+j1 j +j ; 1 : (48) 4 1 ( r 1; dr + r 1; dr) =( 4; ; + 1 ; ) < 1 ; : (49) Combining (48) and (49) with (46) yields the claim. Q.E.D An asymptotic expansion Integrating by parts in formula (39) yields for g 1 an asymptotic expansion with error estimate. This is especially suitable for numerical purposes indeed, we will take Theorem 3.11, below, as the starting point for the numerical algorithm for g k (x) developed in section Theorem 3.11 Let x = x 1 + ix R n x 1 >. For any integer N we have " # g 1 (x) = e;ix 1 NX Re ;e ix j! 1 + E N (x) (5) (ix) j+1 j= where the remainder E N (x) is given by E N (x) = e;ix 1 Re (N + 1)!e ix1 (;x) N +1 1 e ;tx (t ; i) dt N + (51) and satises the estimate je N (x)j (N + 1)!(N +1)= jxj N + : (5) Since g 1 (;x 1 x )=g 1 (x 1 x ) similar formulas hold for x 1 <.

26 6 Samuli Siltanen Proof. We claim that for any N 1 e ;xt t ; i dt = ; N X j= j! (N +1)! 1 + j+1 (ix) (;x) N +1 e ;xt dt: (53) (t ; i) N + We give an induction proof keeping carefully track of the signs. First we consider the case N =: 1 e ;xt t ; i dt = e ;xt 1 1 ;e ;xt ; (;x)(t ; i) (;x)(t ; i) dt = ; 1 ix ; 1 1 e ;xt x (t ; i) dt: Now we make the induction assumption: let the claim hold for all integers smaller or equal to M. Then the claim for M + 1 follows from ; MX j= 1 e ;xt X M t ; i dt = ; j= (;x) M +1 j! (M + 1)! + j+1 (ix) M +1 X ; j= j! (M + 1)! 1 + j+1 (ix) (;x) M +1 e ;xt (;x)(t ; i) M + 1 ; j! (M + )! 1 + j+1 (ix) (;x) M + e ;xt dt = (t ; i) M + (M + 1)! 1 ;(M +)e ;xt dt = (;x) M +1 (;x)(t ; i) M +3 e ;xt dt: (t ; i) M +3 We have thus established formulas (5) and (51). The error estimate follows similarly to the proof of Theorem 3.8. The choices of integration paths in the complex plane are the same, but this time we need the estimate js + is ; ij N + (1= p ) N + for all s>. Q.E.D Determining H k () Lemma 3.1 Let k C n. Then the value of the function H k (x) at the origin is where : is Euler's constant. H k () = ; ; 1 log jkj (54) Proof. Lemma 3.4 in [43] states that for any ", < "<1, there is a constant C " such that for all < jkj 1= there is an inequality jg k (x) + (log jkj + )= ; G (x)j C " jkj " hxi " (55) where hxi = (1 + jxj ) 1=. Fix now any x R n and let k C n be arbitrary. Then formula (4) gives H 1 (kx) =e ikx g k (x)+ 1 log jkjjxj =

27 Electrical Impedance Tomography and Faddeev Green's Functions 7 = e ikx (g k (x) + (log jkj + )= log jxj)+ (1 ; eikx )logjkjjxj; e ikx : (56) Choose now any sequence fk j g 1 j= of nonzero complex numbers converging to zero. Note that e ikx ; 1=O(jkxj) for kx! and that lim r!+ r log r =,so lim (1 ; j!1 eik jx )logjk j jjxj =: Then estimate (55) together with (56) imply H 1 () = lim j!1 H 1 (k j x)=; and the claim is proved for H 1. The general result follows from (8) combined with the continuity of H k (x) with respect to x. Q.E.D. Recall from Chapter the denition 3.. The scattering transform t(k) := e k (x)(x k)q(x)dx k C n (57) R of the non-physical scattering transform. Here the functions are the unique solutions of (;+q)e ikx (x k) = xr, satisfying ( k) ; 1 W 1 ~p (see Theorem.). The function t(k) isthekey intermediate object of the inversion procedure. We want to understand how the non-physical scattering transform represents the corresponding conductivity. It turns out that t(k) resembles closely the Fourier transform of q = ;1= 1= evaluated at (;k 1 k ) R. Roughly speaking, we can expect this since the exponential in (57) can be written in the form exp(;i(;k 1 k ) x) and the functions ( k)areclose to 1 asymptotically. That t(k) actually diers from the Fourier transform of q is evident from the fact that the map q 7! t(k) is nonlinear and 6 1ifq6. The results of this section are aimed to make these speculations more precise and prove such properties of t(k) that are important from the point of view of numerical work. The proof of Theorem 4 in [43] implies that t(k) iscontinuous outside the origin. We need additional information about its behaviour for both large and small jkj for solving equation (1) numerically. In Section 3..1 we present quantitative information about the dierence between t(k) and ^q which leads to an estimate of type jt(k)j Cjkj ;m for large jkj. This is a consequence of the smoothness of. In section 3.. we sharpen results of [43] to get the estimate jt(k)j Cjkj 1+" for any <"<1andjkj small. Thus when solving (1) numerically we don't have to deal with singularities at the origin also writing formula (13) in polar coordinates yields an integrand vanishing at jkj =. In Section 3..3 we see that symmetries of manifest themselves in the shape of t(k) almost as in the Fourier transform of q:

28 8 Samuli Siltanen if is rotationally symmetric, so is t(k) if is preserved in a reection, roughly the same is true for t(k). Two further properties of Fourier transform carry over to t(k): for any > the non-physical scattering transform corresponding to the scaled conductivity (x) ist(k=) translating results in multiplication of t(k) by an exponential function of modulus one. For a more precise statement see Theorem These results are valuable in numerical work for several reasons: given we have a priori knowledge about properties of t(k) and vice versa. This is important for both the forward and the inverse problem. On the other hand, it is possible to produce several examples of pairs ( t) out of one initial example this is useful for testing inversion algorithms Large jkj estimates Theorem 3.13 Let R be a bounded, simply connected C 1 domain. Let C m+ (R ) m 1 have a positive lower bound and satisfy ; 1 C m+ (). Dene q = ;1= 1= and let t(k) be given by (57). Then there is such a constant C = C(m jj kqk C m ) that jt(k)j Cjkj ;m for large jkj: (58) The rest of this section is devoted to the proof of Theorem Let us recall the construction of the functions appearing in (57). k C n and 1 <p<we want to nd a solution of For any (; ; + q)(x k) = x R (59) satisfying ( k) ; 1 W 1 ~p, where 1=~p = 1=p ; 1=. This is done in the spirit of classical scattering theory: we think of = i + s as a sum of an \incoming wave" i 1 and a \scattered wave" satisfying s W 1 ~p, an analogue of the Sommerfeld radiation condition. We will nd as the solution of the Lippmann-Schwinger type equation ; 1= s = ;g k (q): (6) Applying ; ; to (6) and recalling (19) shows that a solution of (6) solves (59), too. To solve (6), note that the operator I + g k (q ):W 1 ~p! W 1 ~p

29 Electrical Impedance Tomography and Faddeev Green's Functions 9 is Fredholm of index zero and injective [43, Theorem 1.1 and Lemma 1.5], thus invertible for every k C n. Furthermore, by Lemma 1.3 of [43] the function g k q W 1 ~p and we can dene ( k)=1; [I + g k (q )] ;1 (g k q): (61) Applying now I + g k (q ) to both sides of (61) yields (6). Since our potential is innitely smooth, we are able to prove stronger estimates with a simpler proof. For this we need a bound for the norm of the operator I + g k (q ) acting on the Sobolev spaces W m ~p for m 1. Denition 3.14 Let D R be a domain, m an integer and 1 r<1: The Sobolev space W m r (D) consists of distributions having all partial derivatives of order less or equal to m in L r (D). We X will use the norm kuk W m r (D) = k@ uk L r (D) : (6) jjm The Banach space L(W m r (D)) is normed with the operator norm induced by (6). Theorem 3.15 Let R be a bounded, simply connected C 1 domain. Let C 1 (R ) have a positive lower bound and satisfy ; 1 C 1 (). Dene q = ;1= 1= and Faddeev Green's function g k by (17), and let 1 <p<. Fix a k C n and let (x k) be the unique solution of (; ; + q)(x k) = x R ; 1 W 1 ~p : Then ; 1 W m ~p for all m 1. Furthermore, satises k( k) ; 1k W m ~p jkj ;=~p kqk W m 1 for large jkj: (63) Proof. We start by checking the invertibility of the operator I + g k (q ):W m ~p! W m ~p : (64) By [43, Theorem 1.1] we know that this holds for m = 1, and in particular we have injectivity W m ~p! W 1 ~p. The inequalities of Young and Holder imply for any multi-indices with jj + j j = j m kg k ((@ q)(@ f))k L ~p kg k k L ~pk(@ q)(@ f)k L 1 kg k k L ~pk@ qk L ~pk@ fk L ~p kg k k L ~pkqk W m ~pkfk W m ~p (65) so Theorem 3.1 shows that the restriction of I +g k (q )tow m ~p has W m ~p as target space. Thus it is enough to prove that g k (q ) is compact and invertibility of (64) follows by a Fredholm argument. Compactness is evident from the decomposition W m ~p ;! C m;1 q ;! W m;1 p g k ;! W m ~p

30 3 Samuli Siltanen where we used the Sobolev imbedding theorem and Lemma 1.3 of [43]. To prove (63) we estimate rst the operator norm of [I + g k (q )] ;1. Applying Theorem 3.1 to (65) gives kg k (q )k W m ~p!w m ~p Ckg kk L ~pkqk W m ~p Cjkj ;=~p kqk W m ~p: Thus k[i + g k (q )] ;1 k W m ~p!w m ~p for large jkj: (66) An application of X Young's inequality and Theorem 3.1 once more gives X kg k qk W m ~p = kg k (@ q)k L ~p kg k k L ~p k@ qk L 1 = Cjkj ;=~p kqk W m 1: (67) jjm jjm Substituting (66) and (67) to (61) yields the claim. Q.E.D. Proof. (Theorem 3.13) Let m 1 be arbitrary. It is enough to derive for large jkj = jk 1 + ik j an estimate of type Integration by parts gives k 1 jk m j t(k)j C = C (m jj kqk C m ) j =1 : (68) R e k (x) (x)dx = i R e 1 (x)dx C 1 (R ) and for k the same with aminus sign. Write q = q( ; 1) + q to get jkj m t(k)j ;m q(x) m jdx + ;m j m (((x k) ; 1)q(x))jdx: (69) j We can bound the rst term by jj sup m q(x) x. An application of m j 3.15 shows that for large enough jkj the integral in the second term of (69) is bounded by mx ((x k) ; 1)k L = m; q(x)k L ~p j Ck( k) ; 1k W m ~pkqk W m ~p Ckqk W m 1kqk W m ~p: (7) Estimating the absolute values of derivatives of q by supremums and xing a p satisfying 1 <p< in (7) yields the claim. Q.E.D. A digression. We remark that it is possible to follow a path nearer to Nachman's original proof. Namely, we can formally write using the resolvent identity g k f =(; ; ;1 f = ; 1 4 (@ + ;1 f = ; 1 4ik (I ; (@ ;1 f: Suppose we had an explicit estimate of type k@ ;1 fk L ~p A p kfk L p (71)