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Weaver wins 15th as Halos blank A'sAug 7, 2012 - 5:21 AM Oakland, CA (Sports Network) - Jered Weaver won his ninth consecutive start by pitching a four-hit shutout in the Angels' 4-0 triumph over the Oakland Athletics.
Weaver (15-1), who became the first 15-game winner in the majors this year, fanned nine without allowing a walk. It was the right-hander's second shutout this year and sixth of his career.
"Obviously Weav set the tone and more," Angels manager Mike Scioscia said. "It's a great time for a game like that anytime from Weav, but to get it what we've been through this week with our bullpen, gives us a big lift."
Weaver, who threw 80 of his 117 pitches for strikes, is 10-0 over his last 12 starts. The 10 straight wins established a career-best. His only loss came May 13 at Texas. The Angels are 17-3 in Weaver's starts this season.
"He kept us off balance all night," A's manager Bob Melvin said.
The Angels scored twice in the second and Torii Hunter had a two-run single in the seventh inning in the opener of this three-game set between these AL West teams who are a half-game apart in the wild card standings.
Jarrod Parker (7-6) gave up nine hits and four runs over 6 2/3 innings in Oakland's third straight loss.
Kendrys Morales doubled to start the top of the second, and with two outs Erick Aybar lined an RBI single to center. Chris Iannetta singled and then Mike Trout, playing on the eve of his 21st birthday, tacked on a run-scoring base hit to right field.
Trout was visibly upset after being thrown out trying to stretch a single into a double in the fifth inning. Josh Reddick delivered a one-bounce throw, but Trout seemed to get his hand on the base before the tag was applied to his fingertips.
Hunter's ground single to center in the seventh plated Aybar and Iannetta.
Game Notes
Major League Baseball denied the Angels' protest of an 8-6, 10-inning loss to the White Sox Friday night. Scioscia argued that Paul Konerko wasn't in the baseline, as required while running to first base on a grounder with the bases loaded. Had Konerko been ruled out it would have been a double play...Weaver has 11 career complete games...Trout extended his club record of consecutive steals to 27 with two more swipes Monday. He hasn't been caught stealing since June 4 vs. Seattle and has 36 steals this season...Aybar, who was on the 15- day disabled list with a fractured big right toe, was activated before the game. The Angels optioned infielder Andrew Romine to Triple-A Salt Lake.
Aug 7 12:42 AM
- Torii Hunter single to center scored Erick Aybar and Chris Iannetta with two out. Runner on first and Albert Pujols due up.
Angels 4, Athletics 0 Top 7, 2 OutsAug 6 11:55 PM
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- Mike Trout stole second. Runners on second and third with two outs and Torii Hunter at the plate.
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- Erick Aybar single to center scored Kendrys Morales with two out. Runner on first and Chris Iannetta due up.
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Angels vs. AthleticsDec 20 6:03 AM
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Calderón problem. Mikko Salo
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1 Calderón problem Lecture notes, Spring 2008 Mikko Salo Department of Mathematics and Statistics University of Helsinki
3 Contents Chapter 1. Introduction 1 Chapter 2. Multiple Fourier series Fourier series in L Sobolev spaces 11 Chapter 3. Uniqueness Reduction to Schrödinger equation Complex geometrical optics Uniqueness proof 26 Chapter 4. Stability Schrödinger equation More facts on Sobolev spaces Conductivity equation 37 Chapter 5. Partial data Carleman estimates Uniqueness with partial data 50 Bibliography 55 v
5 CHAPTER 1 Introduction Electrical Impedance Tomography (EIT) is an imaging method with potential applications in medical imaging and nondestructive testing. The method is based on the following important inverse problem. Calderón problem: Is it possible to determine the electrical conductivity of a medium by making voltage and current measurements on its boundary? In this course we will prove a fundamental uniqueness result due to Sylvester and Uhlmann, which states that the conductivity is determined by the boundary measurements. We will also consider stable dependence of the conductivity on boundary measurements, and the case where measurements are only made on part of the boundary. In addition, we will discuss useful techniques in partial differential equations, Fourier analysis, and inverse problems. Let us begin by recalling the mathematical model of EIT, see [5] for details. The purpose is to determine the electrical conductivity γ(x) at each point x Ω, where Ω R n represents the body which is imaged (in practice n = 3). We assume that Ω is a bounded open subset of R n with C boundary, and that γ is a positive C 2 function in Ω. Under the assumption of no sources or sinks of current in Ω, a voltage potential f at the boundary Ω induces a voltage potential u in Ω, which solves the Dirichlet problem for the conductivity equation, (1.1) { γ u = 0 in Ω, u = f on Ω. Since γ is positive, there is a unique weak solution u H 1 (Ω) for any boundary value f H 1/2 ( Ω). One can define the Dirichlet to Neumann map (DN map) formally as Λ γ f = γ u. ν Ω 1
6 2 1. INTRODUCTION This is the current flowing through the boundary. More precisely, the DN map is defined weakly as (Λ γ f, g) Ω = γ u v dx, f, g H 1/2 ( Ω), Ω where u is the solution of (1.1), and v is any function in H 1 (Ω) with v Ω = g. The pairing on the boundary is integration with respect to the surface measure, (f, g) Ω = fg ds. With this definition Λ γ is a bounded linear map from H 1/2 ( Ω) into H 1/2 ( Ω). The Calderón problem (also called the inverse conductivity problem) is to determine the conductivity function γ from the knowledge of the map Λ γ. That is, if the measured current Λ γ f is known for all boundary voltages f H 1/2 ( Ω), one would like to determine the conductivity γ. There are several aspects of this inverse problem which are interesting both for mathematical theory and practical applications. Ω 1. Uniqueness. If Λ γ1 = Λ γ2, show that γ 1 = γ Reconstruction. Given the boundary measurements Λ γ, find a procedure to reconstruct the conductivity γ. 3. Stability. If Λ γ1 is close to Λ γ2, show that γ 1 and γ 2 are close (in a suitable sense). 4. Partial data. If Γ is a subset of Ω and if Λ γ1 f Γ = Λ γ2 f Γ for all boundary voltages f, show that γ 1 = γ 2. Starting from the work of Calderón in 1980, the inverse conductivity problem has been studied intensively. In the case where n 3 and all conductivities are in C 2 (Ω), the following positive results are an example of what can be proved. Theorem. (Sylvester-Uhlmann 1987) If Λ γ1 = Λ γ2, then γ 1 = γ 2 in Ω. Theorem. (Nachman 1988) There is a convergent algorithm for reconstructing γ from Λ γ. Theorem. (Alessandrini 1988) Let γ j H s (Ω) for s > n + 2, and 2 assume that γ j H s (Ω) M and 1/M γ j M (j = 1, 2). Then γ 1 γ 2 L (Ω) ω( Λ γ1 Λ γ2 H 1/2 ( Ω) H 1/2 ( Ω))
7 1. INTRODUCTION 3 where ω(t) = C log t σ for small t > 0, with C = C(Ω, M, n, s) > 0, σ = σ(n, s) (0, 1). Theorem. (Kenig-Sjöstrand-Uhlmann 2007) Assume that Ω is convex and Γ is any open subset of Ω. If Λ γ1 f Γ = Λ γ2 f Γ for all f H 1/2 ( Ω), and if γ 1 Ω = γ 2 Ω, then γ 1 = γ 2 in Ω. During this course we will discuss the methods involved in these results. The main tool will be the construction of special solutions, called complex geometrical optics solutions, to the conductivity equation and related equations. This will involve Fourier analysis, and we will begin the course with a discussion of n-dimensional Fourier series. References. This course a continuation of the class Impedanssitomografian perusteet (Principles of EIT) given by Petri Ola [5]. From [5] we will mainly use the solvability of the Dirichlet problem for elliptic equations, the definition of the DN map, and results which state that the boundary values of the conductivity can be recovered from the DN map. Chapter 2 on multiple Fourier series is classical, an excellent reference is Zygmund [8]. In Chapter 3 we prove the uniqueness result of Sylvester-Uhlmann [6]. The proof of the main estimate, Theorem 3.7, follows Hähner [4]. The stability question is taken up in Chapter 4, where the main results are due to Alessandrini [1]. The treatment also benefited from Feldman-Uhlmann [3]. In the final Chapter 5 we introduce Carleman estimates and prove the result in Bukhgeim-Uhlmann [2], which states that it is enough to measure currents on roughly half of the boundary to determine the conductivity.
9 CHAPTER 2 Multiple Fourier series 2.1. Fourier series in L 2 Joseph Fourier laid the foundations of the mathematical field now known as Fourier analysis in his 1822 treatise on heat flow. The basic question is to represent periodic functions as sums of elementary pieces. If the period of f is 2π and the elementary pieces are sine and cosine functions, then the desired representation would be f(x) = (a k cos(kx) + b k sin(kx)). k=0 Since e ikx = cos(kx)+i sin(kx), we may alternatively consider the series (2.1) f(x) = c k e ikx. k= We will need to represent functions of n variables as Fourier series. If f is a function in R n which is 2π-periodic in each variable, then a natural analog of (2.1) would be f(x) = k Z n c k e ik x. This is the form of Fourier series which we will study. Note that the terms on the right-hand side are 2π-periodic in each variable. There are many subtle issues related to various modes of convergence for the series above. However, we will mostly just need the case of convergence in L 2 norm for Fourier series of L 2 functions, and in this case no problems arise. Consider the cube Q = [ π, π] n, and define an inner product on L 2 (Q) by (f, g) = (2π) n fḡ dx, f, g L 2 (Q). Q With this inner product, L 2 (Q) is a separable infinite-dimensional Hilbert space. The space of functions which are locally square integrable and 5
10 6 2. MULTIPLE FOURIER SERIES 2π-periodic in each variable may be identified with L 2 (Q). Therefore, we will consider Fourier series of functions in L 2 (Q). Lemma 2.1. The set {e ik x } is an orthonormal subset of L 2 (Q). Proof. A direct computation: if k, l Z n then (e ik x, e il x ) = (2π) n e i(k l) x dx = (2π) n π = π { 1, k = l, 0, k l. π π Q e i(k 1 l 1 )x1 e i(kn ln)xn dx n dx 1 We recall a Hilbert space fact: if {e j } j=1 is an orthonormal subset of a separable Hilbert space H, then the following are equivalent: (1) {e j } j=1 is an orthonormal basis, in the sense that any f H may be written as the series f = (f, e j )e j with convergence in H, (2) for any f H one has f 2 = j=1 (f, e j ) 2, j=1 (3) if f H and (f, e j ) = 0 for all j, then f 0. If the condition (3) is satisfied, the orthonormal system {e j } is called complete. The main point is that {e ik x } k Z n is complete in L 2 (Q). Lemma 2.2. If f L 2 (Q) satisfies (f, e ik x ) = 0 for all k Z n, then f 0. The proof is given below. The main result on Fourier series of L 2 functions is now immediate. Below we denote by l 2 (Z n ) the space of complex sequences c = (c k ) k Z n with norm ( ) 1/2. c l 2 (Z n ) = k Z n c k 2
11 2.1. FOURIER SERIES IN L 2 7 Theorem 2.3. If f L 2 (Q), then one has the Fourier series f(x) = k Z n ˆf(k)e ik x with convergence in L 2 (Q), where the Fourier coefficients are given by ˆf(k) = (f, e ik x ) = (2π) n f(x)e ik x dx. One has the Plancherel formula f 2 L 2 (Q) = ˆf(k) 2. k Z n Q Conversely, if c = (c k ) l 2 (Z n ), then the series f(x) = k Z n c k e ik x converges in L 2 (Q) to a function f satisfying ˆf(k) = c k. Proof. The facts on the Fourier series of f L 2 (Q) follow directly from the discussion above, since {e ik x } k Z n is a complete orthonormal system in L 2 (Q). For the converse, if (c k ) l 2 (Z n ), then k Zn M k N c k e ik x 2 L 2 (Q) = k Zn M k N c k 2 by orthogonality. Since the right hand side can be made arbitrarily small by choosing M and N large, we see that f N = k Z n, k N c ke ik x is a Cauchy sequence in L 2 (Q), and converges to f L 2 (Q). One obtains ˆf(k) = (f, e ik x ) = c k again by orthogonality. It remains to prove Lemma 2.2. We begin with the most familiar case, n = 1. The partial sums of the Fourier series of a function f L 2 ([ π, π]), extended as a 2π-periodic function into R, are given by S m f(x) = m k= m ˆf(k)e ikx = m k= m ( 1 π ) f(y)e iky dy e ikx 2π π = π π D m (x y)f(y) dy
12 8 2. MULTIPLE FOURIER SERIES where D m (x) is the Dirichlet kernel D m (x) = 1 2π m k= m = 1 ei(2m+1)x 1 2π e imx e ix 1 e ikx = 1 2π e imx (1 + e ix e i2mx ) = 1 e i(m+ 1 2 )x e i(m+ 1 2 )x = 1 sin((m + 1)x) 2 2π e i 1 2 x e i 1 2 x 2π sin( 1x). 2 Thus, definining the convolution of two 2π-periodic locally integrable functions by f g(x) = π π f(x y)g(y) dy, we see that S m f(x) = (D m f)(x). The Dirichlet kernel acts in a similar way as an approximate identity, but is problematic because it takes both positive and negative values. Definition. A sequence (K N (x)) N=1 of 2π-periodic continuous functions on the real line is called an approximate identity if (1) K N 0 for all N, (2) π K π N(x) dx = 1 for all N, and (3) for all δ > 0 one has lim sup N δ x π K N (x) 0. Thus, an approximate identity (K N ) for large N resembles a Dirac delta at 0, extended in a 2π-periodic way. It is possible to approximate L p functions by convolving them against an approximate identity. Lemma 2.4. Let (K N ) be an approximate identity. If f L p ([ π, π]) where 1 p <, or if f is a continuous 2π-periodic function and p =, then K N f f L p ([ π,π]) 0 as N. Proof. Since K N has integral 1, we have (K N f)(x) f(x) = π π K N (y)[f(x y) f(x)] dy
13 2.1. FOURIER SERIES IN L 2 9 Let first f be continuous and p =. To estimate the L norm of K N f f, we fix ε > 0 and compute π (K N f)(x) f(x) K N (y) f(x y) f(x) dy π K N (y) f(x y) f(x) dy+ K N (y) f(x y) f(x) dy. y δ Here δ > 0 is chosen so that δ y π f(x y) f(x) < ε whenever x R and y δ. 2 This is possible because f is uniformly continuous. Further, we use the definition of an approximate identity and choose N 0 so that ε sup K N (x) <, for N N 0. δ x π 4π f L With these choices, we obtain (K N f)(x) f(x) ε 2 y δ K N (y) dy + 4π f L sup K N (x) < ε δ x π whenever N N 0. The result is proved in the case p =. Let now f L p ([ π, π]) and 1 p <. We need the integral form of Minkowski s inequality, ( F (x, y) dν(y) p ) 1/p ( 1/p dµ(x) F (x, y) dµ(x)) p dν(y), X Y which is valid on σ-finite measure spaces (X, µ) and (Y, ν), cf. the usual Minkowski inequality y F (, y) L p y F (, y) Lp. Using this, we obtain π K N f f L p ([ π,π]) K N (y) f( y) f L p ([ π,π]) dy π = K N (y) f( y) f L p dy + K N (y) f( y) f L p dy δ y π 4π f L p sup δ x π Y X y δ Now, for any ε > 0, there is δ > 0 such that K N (x) + sup f( y) f L p. y δ f( y) f L p ([ π,π]) < ε whenever y δ. 2 Thus the second term can be made arbitrarily small by choosing δ sufficiently small, and then the first term is also small if N is large. This shows the result.
14 10 2. MULTIPLE FOURIER SERIES Now, if the Dirichlet kernels (D m ) were an approximate identity, by Lemma 2.4 one would have S m f f in L 2 for any f L 2. This would in particular imply that any f L 2 which satisfies (f, e ikx ) = ˆf(k) = 0 for all k Z n, would be the zero function. However, D m is not an approximate identity because it takes negative values. One does get an approximate identity if a different summation method used: instead of the partial sums S m f consider the Cesàro sums σ N f(x) = 1 N + 1 N S m f(x). m=0 This can be written in convolution form as σ N f(x) = 1 N + 1 where F N is the Fejér kernel, F N (x) = = = = 1 2π(N + 1) N (D m f)(x) = (F N f)(x) m=0 N e i(m+ 1 2 )x e i(m+ 1 2 )x m=0 e i 1 2 x e i 1 2 x 1 e i 1 2 x ei(n+1)x 1 e i 1 e ix 2 x e i(n+1)x 1 1 e ix 1 2π(N + 1) e i 1 2 x e i 1 2 x 1 e i(n+1)x 1 + e i(n+1)x 1 2π(N + 1) (e i 1 2 x e i 1 2 x ) 2 1 sin 2 ( N+1x) 2 2π(N + 1) sin 2 ( 1x). 2 Clearly this is nonnegative, and in fact F N is an approximate identity (exercise). It follows from Lemma 2.4 that Cesàro sums of the Fourier series an L p function always converge in the L p norm if 1 p <. Lemma 2.5. If f L p ([ π, π]) where 1 p <, or if f is a continuous 2π-periodic function and p =, then σ N f f L p ([ π,π]) 0 as N. Proof of Lemma 2.2. We begin with the case n = 1. If f L 2 ([ π, π]) and (f, e ikx ) = 0 for all k Z, then S m f = 0 and also σ N f = 0 for all N. By Lemma 2.5 it follows that f 0.
15 2.2. SOBOLEV SPACES 11 Now let n 2, and assume that f L 2 (Q) and (f, e ik x ) = 0 for all k Z n. Since e ik x = e ik 1x1 e iknxn, we have π π h(x 1 ; k 2,..., k n )e ik 1x 1 dx 1 = 0 for all k 1 Z, where h(x 1 ; k 2,..., k n ) = f(x 1, x 2,..., x n )e i(k 2x k nx n) dx 2 dx n. [ π,π] n 1 Now h( ; k 2,..., k n ) is in L 2 ([ π, π]) by the Cauchy-Schwarz inequality. By the completeness of the system {e ik 1x 1 } in one dimension, we obtain that h( ; k 2,..., k n ) = 0 for all k 2,..., k n Z. Applying the same argument in the other variables shows that f 0. We have now proved the main facts on Fourier series of L 2 functions Sobolev spaces In this section we wish to consider Sobolev spaces of periodic functions. In fact, most of the results given here will not be used in their present form, but they provide good motivation for the developments in Chapter 3. Let T n = R n /2πZ n be the n-dimensional torus. Note that L 2 (Q) above may be identified with L 2 (T n ). However, C(Q) is different from C(T n ); in fact C(T n ) (resp. C (T n )) can be identified with the continuous (resp. C ) 2π-periodic functions in R n. First we need to define weak derivatives of periodic functions. This is similar to the nonperiodic case, except that here the test function space will be C (T n ). We use the notation D j = 1 i x j, D α = D α 1 1 D αn n. Definition. Let f L 2 (T n ). We say that D α f L 2 (T n ) if there is a function v L 2 (T n ) which satisfies fd α ϕ dx = ( 1) α vϕ dx T n T n for all ϕ C (T n ). In this case we define D α f = v. As in the nonperiodic case, weak derivatives are unique, and for smooth functions the definition coincides with the usual derivative.
16 12 2. MULTIPLE FOURIER SERIES This uses the fact that there are no boundary terms arising from integration by parts, because of periodicity. Definition. If m 0 is an integer, we denote by H m (T n ) the space of functions f L 2 (T n ) such that D α f L 2 (T n ) for all α N n satisfying α m. We equip H m (T n ) with the inner product (f, g) H m (T n ) = (D α f, D α g). α m Then H m (T n ) is a Hilbert space. If f and D α f are in L 2 (T n ), one may compute the Fourier coefficients of D α f to be (D α f)ˆ(k) = (2π) n T n D α f(x)e ik x dx = (2π) n ( 1) α T n f(x)d α (e ik x ) dx = k α ˆf(k) by the definition of weak derivative, since e ik x C (T n ). This motivates the following characterization of H m (T n ) in terms of Fourier coefficients. Lemma 2.6. Let f L 2 (T n ). Then f H m (T n ) if and only if k m ˆf l 2 (Z n ), where k = (1 + k k n ) 1/2. Proof. One has f H m (T n ) D α f L 2 (T n ) k α ˆf(k) l 2 (Z n ) for α m for α m (k 2 1,..., k 2 n) α ˆf(k) 2 l 1 (Z n ) for α m. If the last condition is satisfied, then k 2m ˆf(k) 2 = c β (k1, 2..., kn) 2 β ˆf(k) 2 l 1 (Z n ), β m consequently k m ˆf(k) l 2 (Z n ). Conversely, if k m ˆf(k) l 2 (Z n ), then k α ˆf(k) l 2 (Z n ) for α m because k j k. It is now easy to prove a version of the Sobolev embedding theorem. Theorem 2.7. If m > n/2 then H m (T n ) C(T n ).
17 2.2. SOBOLEV SPACES 13 Proof. Let f H m (T n ), so that k m ˆf l 2 (Z n ) and f(x) = k Z n ˆf(k)e ik x. Let M k = ˆf(k)e ik x = k m ( k m ˆf(k)). We have k Z n M k k m l 2 (Z n ) k m ˆf(k) l 2 (Z n ) <, by Lemma 2.6 and since m > n/2. Since the terms in the Fourier series of f are continuous functions, this Fourier series converges absolutely and uniformly into a continuous function in T n by the Weierstrass M- test. The final result in this section will be elliptic regularity in the periodic case. Consider a second order differential operator P (D) acting on 2π-periodic functions in R n, P (D) = a α D α, α 2 where a α are complex constants. The principal part of P (D) is P 2 (D) = a α D α. α =2 We say that P (D) is elliptic if P 2 (D) has real coefficients, and P 2 (k) > 0 whenever k Z n {0}. The following proof also indicates how Fourier series are used in the solution of partial differential equations. Theorem 2.8. Let P (D) be an elliptic second order differential operator with constant coefficients, and assume that u L 2 (T n ) solves the equation P (D)u = f in T n, for some f L 2 (T n ). Then u H 2 (T n ). Proof. Taking the Fourier coefficients on both sides of P (D)u = f gives (2.2) P (k)û(k) = ˆf(k), k Z n. We have P 2 (k) = k 2 P 2 (k/ k ) c k 2
18 14 2. MULTIPLE FOURIER SERIES for some c > 0, by ellipticity. Then for k 0, n P (k) = P 2 (k) + a j k j + a 0 P 2 (k) ( j=1 n a j ) k a 0 j=1 c k 2 C k. If C > 0 is sufficiently large, it follows that P (k) 1 2 c k 2, for k C. From (2.2) we obtain û(k) = ˆf(k) 2 P (k) c k ˆf(k), k C. 2 Since ˆf(k) l 2 (Z n ) this shows that k 2 û(k) l 2 (Z n ), which implies u H 2 (T n ) as required.
19 CHAPTER 3 Uniqueness In this chapter, we will discuss the proof of the following uniqueness result of Sylvester and Uhlmann. Theorem 3.1. Let Ω R n be a bounded open set with smooth boundary, where n 3, and let γ 1 and γ 2 be two positive functions in C 2 (Ω). If Λ γ1 = Λ γ2, then γ 1 = γ 2 in Ω. In fact, this theorem will be reduced to a uniqueness result for the Schrödinger equation (also due to Sylvester and Uhlmann). Theorem 3.2. Let Ω R n be a bounded open set with smooth boundary, where n 3, and let q 1 and q 2 be two functions in L (Ω) such that the Dirichlet problems for +q 1 and +q 2 in Ω are well posed. If Λ q1 = Λ q2, then q 1 = q 2 in Ω. The reduction of Theorem 3.1 to Theorem 3.2 is presented in Section 3.1 along with the relevant definitions. The proof of the uniqueness results relies on complex geometrical optics (CGO) solutions, which are constructed in Section 3.2 for the Schrödinger equation by Fourier analysis and perturbation arguments. Section 3.3 includes an integral identity relating boundary measurements to interior information about the coefficients, and this identity is used to prove Theorem 3.2 by taking an asymptotic limit with suitable CGO solutions Reduction to Schrödinger equation The first step in the proof is the reduction of the conductivity equation to a Schrödinger equation, ( + q)u = 0 in Ω, where q is a function in L (Ω). This equation turns out to be easier to handle, since the principal part is the Laplacian. 15
20 16 3. UNIQUENESS Lemma 3.3. If γ C 2 (Ω) and u H 1 (Ω) then (3.1) γ (γ 1/2 u) = γ 1/2 ( + q)u in the weak sense, where q = γ γ. Proof. This is a direct computation: if u C 2 (Ω) then j (γ j ( 1 γ u)) = j ( γ j u) + j ( j ( γ)u) = γ 2 j u j ( γ) j u + j ( γ) j u + 2 j ( γ)u. The claim follows by taking the sum over j. The case where u H 1 (Ω) can be proved by approximation 1. If q L (Ω), we consider the Dirichlet problem for the Schrödinger equation, { ( + q)u = 0 in Ω, (3.2) u = f on Ω, where f H 1/2 ( Ω). We say that u H 1 (Ω) is a weak solution of (3.2) if u Ω = f, and if for all ϕ H0(Ω) 1 one has ( u ϕ + quϕ) dx = 0. Ω 1 Let (u k ) C 2 (Ω) be a sequence with u k u in H 1 (Ω). We have γ (γ 1/2 u k ) = γ 1/2 ( + q)u k. If a C 1 (Ω) then u au is a continuous map on H 1 (Ω), since au H1 (Ω) = au L2 (Ω)+ ( a)u+a u L2 (Ω) 2( a L (Ω)+ a L (Ω)) u H1 (Ω). Using that γ is in C 2 (Ω) and that : H m (Ω) H m 1 (Ω) is a continuous map, we have u k u in H 1 (Ω) = γ 1/2 u k γ 1/2 u in H 1 (Ω) = (γ 1/2 u k ) (γ 1/2 u) in L 2 (Ω) = γ (γ 1/2 u k ) γ (γ 1/2 u) in L 2 (Ω) = γ (γ 1/2 u k ) γ (γ 1/2 u) in H 1 (Ω). Similarly, γ 1/2 ( + q)u k γ 1/2 ( + q)u in H 1 (Ω), which shows that (3.1) holds in the sense of H 1 (Ω).
21 3.1. REDUCTION TO SCHRÖDINGER EQUATION 17 The problem (3.2) is said to be well-posed if the following three conditions hold: 1. (existence) there is a weak solution u in H 1 (Ω) for any boundary value f in H 1/2 ( Ω), 2. (uniqueness) the solution u is unique, 3. (stability) the solution operator f u is continuous from H 1/2 ( Ω) H 1 (Ω), that is, u H 1 (Ω) C f H 1/2 ( Ω). If the problem is well-posed, we can define a DN map formally by Λ q : H 1/2 ( Ω) H 1/2 ( Ω), f u. ν Ω This is analogous to the conductivity equation, and also here the precise definition of the DN map is given by the weak formulation (Λ q f, g) Ω = ( u v + quv) dx, f, g H 1/2 ( Ω), Ω where u solves (3.2) and v is any function in H 1 (Ω) with v Ω = g. The next proof shows that this is a valid definition and Λ q is a bounded linear map. Lemma 3.4. If q L (Ω) is such that the problem (3.2) is wellposed, then Λ q is a bounded linear map from H 1/2 ( Ω) to H 1/2 ( Ω), and satisfies by (Λ q f, g) Ω = (f, Λ q g) Ω, f, g H 1/2 ( Ω). Proof. Fix f H 1/2 ( Ω), and define a map T : H 1/2 ( Ω) C T (g) = Ω ( u v + quv) dx, where u solves (3.2) and v is any function in H 1 (Ω) with v Ω = g. Since u is a solution, we have ( u ϕ + quϕ) dx Ω for any ϕ H0(Ω). 1 Therefore we may replace v by v + ϕ in the definition of T (g). If v, ṽ H 1 (Ω) and v Ω = ṽ Ω = g, then v ṽ H0(Ω), 1 so indeed the definition does not depend on the particular choice of v (as long as v H 1 (Ω) and v Ω = g).
22 18 3. UNIQUENESS Now, for g H 1/2 ( Ω), use the one-sided inverse to the trace operator to obtain v g H 1 (Ω) with v g H 1 (Ω) C g H ( Ω). Then by 1/2 Cauchy-Schwarz T (g) u v g + quv g dx C u H 1 (Ω) v g H 1 (Ω) Ω C f H 1/2 ( Ω) g H 1/2 ( Ω). Thus T : H 1/2 ( Ω) C is a continuous map, and there is an element Λ q f H 1/2 ( Ω) satisfying (Λ q f, g) = T (g). Consequently, the map Λ q : f Λ q f is a bounded linear map H 1/2 ( Ω) H 1/2 ( Ω). To show the last identity, let f, g H 1/2 (Ω), and let u, v be solutions of (3.2) with boundary values u Ω = f, v Ω = g. Then (Λ q f, g) Ω = ( u v + quv) dx = ( v u + qvu) dx = (Λ q g, f) Ω by the definition of Λ q. The problem (3.2) is not always well-posed, consider for instance the case q = λ where λ > 0 is an eigenvalue of the Laplacian in Ω (then there exists a nonzero u H 1 (Ω) with u = λu in Ω and u Ω = 0). However, for potentials q coming from conductivities, the Dirichlet problem is always well-posed, and there is a relation between the DN maps Λ γ and Λ q. Lemma 3.5. If γ C 2 (Ω) and q = γ/ γ, then the Dirichlet problem (3.2) is well-posed and Λ q f = γ 1/2 Λ γ (γ 1/2 f) + 1 γ γ 1 2 ν f Ω, Ω f H 1/2 (Ω). Proof. Fix f H 1/2 ( Ω). We need to show that there is u in H 1 (Ω) which solves (3.2). Motivated by Lemma 3.3, we take v be the solution of γ v = 0 in Ω with v Ω = γ 1/2 f. Then u = γ 1/2 v is a function in H 1 (Ω) which solves ( + q)u = 0 with the right boundary values. The same argument with f = 0 shows that the solution is unique. Further, since γ C 2 (Ω), we have the estimate u H 1 (Ω) C v H 1 (Ω) C γ 1/2 f H 1/2 ( Ω) C f H 1/2 ( Ω). If u solves (3.2), then v = γ 1/2 u solves the conductivity equation and Λ γ (γ 1/2 f) = γ v ν u = γ1/2 ν 1 γ γ 1/2 2 ν f.
23 3.1. REDUCTION TO SCHRÖDINGER EQUATION 19 Since Λ q f = u/ ν, we obtain the relation between Λ γ and Λ q. We may now show that uniqueness in the inverse conductivity problem can be deduced from the corresponding result for the Schrödinger equation. Proof that Theorem 3.2 implies Theorem 3.1. Let γ 1, γ 2 be positive functions in C 2 (Ω), and assume that Λ γ1 = Λ γ2. If q j = γ j / γ j, then q j L (Ω) and by Theorem 3.5 the Dirichlet problems for + q j are well posed. Also, the boundary reconstruction result ([5], Theorem 6.6.1) implies that γ 1 = γ 2 and γ 1 / ν = γ 2 / ν on Ω. Thus, for any f in H 1/2 ( Ω) one has Λ q1 f = γ 1/2 1 Λ γ1 (γ 1/2 1 f) + 1 γ 1 2 γ 1 1 ν f Ω = γ 1/2 2 Λ γ2 (γ 1/2 = Λ q2 f. 2 f) γ 1 Theorem 3.2 gives that q 1 = q 2 in Ω. Therefore, 2 γ 2 ν f Ω (3.3) γ 1 γ1 = γ 2 γ2 in Ω. We would like to conclude from (3.3) that γ 1 = γ 2. Let q = γ 1/2 1 γ 1 = γ 1/2 2 γ 2, and consider the equation ( + q)u = 0 in Ω. Both γ 1 and γ 2 solve this equation, and γ 1 Ω = γ 2 Ω by boundary determination. Since the Schrödinger equation with potential q coming from a conductivity is well-posed, we obtain γ 1 = γ 2 by uniqueness of solutions. Remark 3.6. We record for later use another argument showing that (3.3) implies γ 1 = γ 2. The equation (3.3) looks like a nonlinear PDE involving γ 1 and γ 2. To simplify the equation, we note that (log γ) = The equation becomes n j=1 j ( 1 γ j γ) = γ γ (log γ) 2. (log γ 1 log γ 2 ) + (log γ 1 ) 2 (log γ 2 ) 2 = 0.
24 20 3. UNIQUENESS Rewriting slightly, we get γ1 γ1 (log ) + a (log ) = 0, γ2 γ2 where a = log γ 1 γ 2. This is a linear equation for the C 2 function v = log γ 1 γ2. Further, noting the identity (e a v) = e a ( v + a v), and using the fact that γ 1 = γ 2 on Ω, we see that v solves the Dirichlet problem { ((γ1 γ 2 ) 1/2 v) = 0 in Ω, v = 0 on Ω. This problem is well-posed, and we get v 0 and γ 1 γ Complex geometrical optics In this section, we will construct complex geometrical optics (CGO) solutions to the Schrödinger equation ( + q)u = 0 in Ω. The potential q is assumed to be in L (Ω). Motivation. First let q = 0. We try as a solution to the equation u = 0 the complex exponential at frequency ζ C n, This satisfies u(x) = e iζ x. Du(x) = ζe iζ x, D 2 u(x) = (ζ ζ)e iζ x. Thus, if ζ C n satisfies ζ ζ = 0, then u(x) = e iζ x solves u = 0. Writing ζ = Re ζ + i Im ζ, we see that ζ ζ = 0 Re ζ = Im ζ, Re ζ Im ζ = 0. Now suppose q is nonzero. The function u = e iζ x is not an exact solution of ( + q)u = 0 anymore, but we can find solutions which resemble complex exponentials. These are the CGO solutions, which have the form (3.4) u(x) = e iζ x (1 + r(x)). Here r is a correction term which is needed to convert the approximate solution e iζ x to an exact solution.
25 3.2. COMPLEX GEOMETRICAL OPTICS 21 In fact, we are interested in solutions in the asymptotic limit as ζ. This follows the principle that it is usually not possible to obtain explicit formulas for solutions to general equations, but in suitable asymptotic limits explicit expressions for solutions may exist. We note that (3.4) is a solution of ( + q)u = 0 iff (3.5) e iζ x ( + q)e iζ x (1 + r) = 0. It will be convenient to conjugate the exponentials e iζ x into the Laplacian. By this we mean that We can rewrite (3.5) as e iζ x D j (e iζ x v) = (D j + ζ j )v, e iζ x D 2 (e iζ x v) = (D + ζ) 2 v = (D 2 + 2ζ D)v. (D 2 + 2ζ D + q)(1 + r) = 0. This implies the following equation for r: (3.6) (D 2 + 2ζ D + q)r = q. The solvability of (3.6) is the most important step in the construction of CGO solutions. We proceed in several steps Basic estimate. We first consider the free case in which there is no potential on the left hand side of (3.6). Theorem 3.7. There is a constant C 0 depending only on Ω and n, such that for any ζ C n satisfying ζ ζ = 0 and ζ 1, and for any f L 2 (Ω), the equation (3.7) (D 2 + 2ζ D)r = f in Ω has a solution r H 1 (Ω) satisfying r L 2 (Ω) C 0 ζ f L 2 (Ω), r L 2 (Ω) C 0 f L 2 (Ω). The idea of the proof is that (3.7) is a linear equation with constant coefficients, so one can try to solve it by the Fourier transform. Since (D j u)ˆ(ξ) = ξ j û(ξ), the Fourier transformed equation is (ξ 2 + 2ζ ξ)ˆr(ξ) = ˆf(ξ).
26 22 3. UNIQUENESS We would like to divide by ξ 2 + 2ζ ξ and use the inverse Fourier transform to get a solution r. However, the symbol ξ 2 + 2ζ ξ vanishes for some ξ R n, and the division cannot be done directly. It turns out that we can divide by the symbol if we use Fourier series in a large cube instead of the Fourier transform, and moreover take the Fourier coefficients in a shifted lattice instead of the usual integer coordinate lattice. Proof of Theorem 3.7. Write ζ = s(ω 1 +iω 2 ) where s = ζ / 2 and ω 1 and ω 2 are orthogonal unit vectors in R n. By rotating coordinates in a suitable way, we can assume that ω 1 = e 1 and ω 2 = e 2 (the first and second coordinate vectors). Thus we need to solve the equation (D 2 + 2s(D 1 + id 2 ))r = f. We assume for simplicity that Ω is contained in the cube Q = [ π, π] n. Extend f by zero outside Ω into Q, which gives a function in L 2 (Q) also denoted by f. We need to solve (3.8) (D 2 + 2s(D 1 + id 2 ))r = f in Q. Let w k (x) = e i(k+ 1 2 e2) x for k Z n. That is, we consider Fourier series in the lattice Z n + 1e 2 2. Writing (u, v) = (2π) n u v dx, u, v L 2 (Q), Q we see that (w k, w l ) = 0 if k l and (w k, w k ) = 1, so {w k } is an orthonormal set in L 2 (Q). It is also complete: if v L 2 (Q) and (v, w k ) = 0 for all k Z n then (ve 1 2 ix 2, e ik x ) = 0 for all k Z n, which implies v = 0. Hilbert space theory gives that f can be written as the series f = k Z n f k w k, where f k = (f, w k ) and f 2 L 2 (Q) = k Z n f k 2. Seeking also r in the form r = k Z n r k w k, and using that the equation (3.8) results in Dw k = (k e 2)w k, p k r k = f k, k Z n, where p k := (k e 2) 2 + 2s(k 1 + i(k )).
27 3.2. COMPLEX GEOMETRICAL OPTICS 23 Note that Im p k = 2s(k ) is never zero, which was the reason for 2 considering the shifted lattice. We define r k := 1 p k f k and r := k Z n r k w k. The last series converges in L 2 (Q) to a function r L 2 (Q) since r k 1 p k f k and then ( ) 1/2 r L 2 (Q) = r k 2 1 s k 1 2s(k ) f k 1 s f k, ( k f k 2 ) 1/2 = 1 s f L 2 (Q). This shows the desired estimate in L 2 (Q). It remains to show that Dr L 2 (Q) with correct bounds. We have Dr = k Z n (k e 2)r k w k. The derivative is justified since this is a convergent series in L 2 (Q): we claim (3.9) (k e 2)r k 4 f k, k Z n, which implies that Dr L 2 (Q) 4 f L 2 (Q). To show (3.9) we consider two cases: if k e 2 4s we have (k e 2)r k and if k e 2 4s then 4s 2s k 2 + 1/2 f k 4 f k, k e sk 1 k e 2 2 2s k e k e 2 2 which implies The statement is proved. (k e 2)r k k e k e 2 2 f k 1 2s f k.
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Doctor insights on:
Remedies For Burning Sensation In Calf Muscles Symptoms
Burning sensation in left calf muscle?
Do you have varicose: Veins in the leg, if you do, this could be superficial phlebitis, if you do not it could muscular origin, both you could treat with Motrin and moist heat. If did not go away check with your doctor. ...Read moreSee 1 more doctor answer
Muscle (Definition)
It is a body tissue that has the ability to contract. It shortens and generates force. It relaxes and returns to its original length. Muscles move joints, stabilize the body, move air and food through the organs, act as valves for bladder, bowel and other organs. They control movement of the eyes. They help us express ourselves by changing the shape of our ...Read more
My husband has been having a baker's cyst behind his knee. Today he began to feel a burning sensation radiating throughout his calf muscle. ?
Baker's cyst: Time to see an Orthopedic surgeon!Get a more detailed answer ›
What causes a burning sensation in the viens in my fore arms? Also the muscles in my calfs keep locking up.
Don't know: I have never heard of burning in forearm veins unless something has been injected into them. Assuming you are not doing that (and maybe especially if you are) then you should see a clinician to find the cause of this and the leg cramps. Probably need blood tests as well. Meanwhile, keep your arms above your heart level and try some heat on them as well as your leg muscles to see if it helps. ...Read more
I have a burning sensation in my leg when i stretch. Could this be a muscle pull?
Muscle or nerve: Muscle 'pulls' or strains can be irritated when you stretch or use the muscle. Tight muscles can also pull on the local nerves causing burning with stretch. Other times, nerve irritation can cause burning when the leg is pulled. A physiatrist or doctor of physical medicine and rehabilitation can help determine the cause and recommend treatment. ...Read more
Restless legs when i go to sleep like a burning sensation in thighs and calves have to put one leg up quite high to stretch the muscles to sleep?
You might try: a simple OTC remedy BENADRYL Capsules (Diphenhydramine) an ANTI-HISTAMINE...at night has been known to help with "Restless legs syndrome" Just a thought...hope it helps!) Dr Z ...Read more
Can I still have my anxiety symptoms (tremors, muscle twitches, burning sensation) when I'm not anxious at the moment. I have had anxiety for months.
Eval: please ask your parents to schedule a visit with a child/adolescent psychiatrist. if you don't want to do that then i'd recommend the counselor at your school. if you can't do that then find an adult that you trust and ask them to assist you in arranging an eval or getting you into counseling. ...Read more
Dull ache, sometimes mild burning sensation based on diet, in my lower left abdomen. On and off. Not severe. No other symptoms. Type of pulled muscle?
Lower and pain: if based on diet, doubt just a pulled muscle, could be bowel related. More info would be needed ...Read more
My bowel cancer missed symptoms since januaury had colonoscopy think they missed it loss in more weight notably muscle including burning sensation in?
Need treatments: Colonoscopy will show only colon tumors not bowel cancers, small bowel cancers are less common , difficult to diagnose , most of the time not diagnosed in early stages , from what you are telling , your problem appear to be nutritional need to take more calories , seek the advise of your treatment team, . ...Read more
I am experience warm & burning sensation (thigh) in my left leg. It does not pain at all. The above symptom may last for an hour or less in a day?
Thigh burning: Consider sensory nerve impingement--possible nerve compression or tight clothing. Called meuralgia paresthetica. ...Read more
Burning Sensation (Definition)
A burning sensation can be descriptive of nerve pain, some variations of migraine, and even central brain problems after a stroke. Have heard this said associated with phantom limb syndrome, reflex sympatheric dystrophy, and trigeminal neuralgia. Other patients might say ...Read more
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Delhi Polytechnic Result 2015
Candidates who are going to appear in the Delhi Polytechnic common entrance test which will be held in the month of May 2015, they can get all details about Delhi Polytechnic Result 2015 of CET Exam from this page. All the participating candidates can check and download their Delhi Polytechnic CET Exam Result from the official website by following the instructions mentioned below by the team of bharatrecruitment.com.
Steps to check Delhi Polytechnic Result 2015:
- Candidates can download their result from the official website www.cetdelhiexam.nic.in
- Check "important link" section and press on CET result 2015.
- Enter your roll no., Application Number, security pin in the text box and submit it.
- Your outcome of exam will be displayed on your system screen.
- Take a printout of your result for future use.
Procedure To Check Result Online
Conducting Body:
Department of Training & Technical Education are performing this CET for those who desire to engage in polytechnic course. Delhi Common Entrance Test (Delhi CET) provided admission in various engineering and non-engineering diploma courses in private and govt polytechnics colleges.
Mode To Check The Result:
Every year Technical Government of NCT of Delhi, Department of Training and Technical Education conducts the CET exam and many students who had filled CET exam application form may appear in the exam. After the exam, all appeared aspirants wait for result. Candidates can check their result through the online mode.
Purpose To Conduct The Exam:
The Delhi Polytechnic common entrance test is performed by the Department of Training & Technical Education in the month of May 2015 at various locations in Delhi State for admission in numerous polytechnic programs across various institutions in Delhi.
Top Polytechnic Colleges in Delhi
Selection Process:
Merit list will be prepared by the organization to admit the students in various polytechnic courses. Candidates who are appearing in entrance exam must check the merit ranking online and ensure your admission in 2015 sessions for Full-time Diploma programmes being offered at Institutes.
To check the result visit the Official Link
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We wish all the Best to all the students who are going to appear in the Delhi Polytechnic Entrance Test. Applicants are advised to keep in touch with our web portal to remain updated. You can also follow us on Facebook and Google plus.


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spirited away
This pup has cosplay potential! Keep at it little pup, and you might rise to the level of Dog Jareth.
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hello all, I have a problem with mapped drives on a few of my user computers. I have the same 3 drives that I can map normally on any machine. the mapping will go successfully but when I try to access the mapping they say access denied. if I go to the dos prompt and access the drive I have no issues, I can also access them via network neighborhood.
7 Replies
May 21, 2012 at 9:55 UTC
Is it the same users on the same computers?
May 21, 2012 at 9:56 UTC
Are there stored credentials that could be wrong or corrupt?
May 21, 2012 at 10:12 UTC
What has changed recently on those computers?
May 21, 2012 at 10:12 UTC
3 different computers, 3 different users.
my environment is a nt 4.0 domain. I am in the process of upgrading that, though this issue started before that. could there be a virus issue? One user this started when I added her account to a temporary laptop that did not have network connections previously, another used had it start last week out of the blue and it happened on my workstation several months ago but I have yet to try to very hard to fix my machine, too many other things going on. it's always the same three drives I am mapping that fail but other drives on the same servers work.....
May 21, 2012 at 10:18 UTC
I just checked on my machine with a different login, maps drives fine. If I go to another machine and log in as me I can map the drive fine. must be a local setting on the machine.... I have been trying to chase down a macro virus this past week as well possibly related??
May 22, 2012 at 3:25 UTC
It sounds like it could be a corrupt user file on the comp. Its very quick to determine so we might as well rule it out. On one of the machines rename the usres profile and to be on the safe side remove them from the registry also. local machine > software > microsoft > windows nt > current version > profile list.
May 22, 2012 at 7:11 UTC
I agree with Leif, sounds like a user profile issue.
Sounds like you tested your own account on different machines, what happens for those other users if you have them try on another machine?
Slightly related, We have workstations that use the Citrix Receiver client. At random times they will just no longer mount the printer into the session when connecting to a remote server. The only solution we have found so far is to delete the user's profile and start them over. Fortunately it doesn't happen all that often.
This discussion has been inactive for over a year.
You may get a better answer to your question by starting a new discussion.


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Home Improvement
Home Improvement Questions? Ask a Handyman for Answers
- What your going to need to do here is clean the filters if they are clogged up. If the filters are clogged, then the heat will not have any way of being taken out resulting in your coil and surrounding lines freezing up.
- Next you will want to make sure that your coil itself is clean. To access this, you will need to remove the side panel on the air-handler. Then coil sits above the filter and looks like the radiator of a car. If there is any residue, mold or plain old gunk built up on it, then you will need to call a service tech to come out and remove the coil, acid wash it and your unit will run fine.
- One last thing, if your unit has a leak, again you will have to have an AC tech come out. When he re-pressurizes your unit, he will inject a dye to see where the leak is so he can fix it. Any one of these conditions will cause this problem. Start with the most obvious first (the filter) and work your way from there.


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Making the megacity sustainable through IoT
32.6 trillion liters. That, roughly, is the amount of water annually lost through leaky pipes before it gets to homes, businesses and hospitals. Put another way, if you could detect those leaks and plug them, you could almost fill China's Three Gorges Dam to the brim every year with clean, treated water.* The "leaky pipe" problem highlights… Read more »


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Author Tags: Environment
As a Victoria resident, she described the journeys of 100 B.C. artists to Robson Bight and Tsitika Valley to promote conservation of the area in Beyond the Gate, an 'art-eco' project undertaken by the Victoria branch of the Western Canada Wilderness Committee in 1990. She is a graduate of UVic's creative writing program.
[BCBW 1992] "Environment"


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Hermes Hermasillo was a restaurateur most famous for founding the multi-trillon credit Arterial Blockage chain of unhealthy eating establishments in the early 22nd Century.[1]
At the age of 36 Hermes suffered a heart seizure and died, as did a number of his regulars. What remained of his fortune was quickly taxed away by EarthGov and the Earth Senate later passed legislation banning any such establishments opening on Earth again.
It wasn't until the 2240s that Julian Hermasillo, one of Hermes's great-grandchildren migrated to Cestus and opened what would be the only Arterial Blockage restaurant in the galaxy at the Fortune City mining colony.


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As a partner, albamy will help Maritz Travel Company design and deliver exceptional meeting experiences in Germany to its clients. Together, albamy and Maritz Travel Company will offer flexible, cost-effective and reliable global services to clients. Maritz Travel Company will rely on albamy to provide deep, regional industry knowledge of Germany to help develop global solutions for its clients.
For more information on the Maritz Global Meetings Network, please visit www.mtc-goglobal.com/GMN or contact us under firstname.lastname@example.com
albamy holds individual memberships with Meeting Professionals International. Being part of MPI is being part of the global meeting and event community. Apart from excellent networking opportunities, albamy employees also take part in the extensive educational opportunities offered to MPI members.
For more information on MPI, please visit www.mpiweb.org
albamy is official partner of Messe Berlin. This partnership allows albamy to provide government agencies and corporate clients with a one-stop-solution, when presenting their destination or products at trade shows or exhibitions held at Messe Berlin.
For more information on Messe Berlin, please visit www.messe-berlin.de
Michele, born in Italy grew up in four countries and learned to speak five languages. After spending 15 years in the beautiful city of Vienna, he decided to move to the new and hip hot-spot in Europe, Berlin.
Michele holds a degree in Economics, which together with his language skills has come in handy in his different career steps within the meetings industry. His professional highlight so far was working for one of the largest international event management companies, where he was part of the project management team organizing European medical conferences and corporate meetings.
With his 12 years of experience in the meeting industry and his hands-on mentality, he is an invaluable asset to the team and its clients.
Of Swedish nationality, Stephanie grew up in Stockholm and later Berlin. She found her passion for the hospitality and meeting industry early in life and kicked off this career with an international hospitality management school in Switzerland.
After a few stops within the hotel industry, she ended up in the meeting planning world. Here she spent 15 years in one of the largest international meeting management companies, combining her passion for meeting planning with her passion for travelling.
The highlight of these 15 years was to be part of the organizing team of the International AIDS Conference with around 30'000 delegates, which meant spending over a year in South Africa as well as one year as Conference Director in Bangkok with full responsibility for meeting planning.
Dan started his career as a professional equestrian rider, but soon found his passion for sales and the hospitality industry. He holds a degree in Business Administration and Management and with his excellent people skills he has held various sales director positions within the meeting and hospitality industry for the last 15 years.
From his experience of working for a large Swedish corporate venue sourcing company as well as 8 years with an international hotel chain, Dan contributes with great venue sourcing and contracting expertise.
He supports the two managing directors from his base in Stockholm, where he is also involved is some exciting new hospitality concepts for the meeting industry. Dan is passionate about connecting people and identifying new opportunities that lead to new exciting projects.
We are constantly looking for new team members and free lancers for our data base. We look forward to your application.
albamy GmbH is a meeting and event management company operating in Europe, Middle East and Africa. Our clients are large international corporations, outsourcing their venue search for their meetings to us, starting with searching the right venue to planning the event. For one of our accounts, we are looking for a Hotel & Venue Sourcing Coordinator (m/f) for the EMEA region, starting as soon as possible.
Please send your application (including CV, reference letters, salary expectations and possible starting date) to Mr. Michele Borzi at firstname.lastname@example.com. For questions, please send an email or call Michele Borzi directly at +49 174 1783512. We are looking forward to your application!
Gleimstr. 61
10437 Berlin
17 Wheelers Park
High Wycombe
Bucks, HP13 6GH
United Kingdom


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You may be an antique...
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'You may be an antique but you are in wonderful company'.
Greeting card designed and made by Designist, printed in Dublin.
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You can collect your items in the shop in georges street, dublin (map), or we can deliver it straight to you or your someone special anywhere in Ireland or the rest of Europe (delivery info here). If you want it sent further afield drop us a line and we can arrange that for you. If you are cutting it fine we also have an next day delivery option anywhere in Ireland.
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- sidearm
I. noun
Date: 1689
a weapon (as a sword, revolver, or bayonet) worn at the side or in the belt
II. adjective
Date: 1908
of, relating to, using, or being a throw (as in baseball) in which the arm is not raised above the shoulder and the ball is thrown with a sideways sweep of the arm between shoulder and hip <sidearm delivery> <a sidearm pass> • sidearm adverb
New Collegiate Dictionary. 2001.


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Description: The title says it all. This tenth edition of the best-sellingSociologyis a comprehensive book and multimedia package that offers readers aglobal perspectiveto help them better understand their own lives, provides a strong focus onsocial diversitythat allows them to see the impact of race, class, and gender, and focuses oncritical thinking. With technology integratedthroughout,this new edition features new research citations, more social theory applications, additional references to pop culture linking directly to sociological material.Covering all aspects of sociology, this book describes sociological investigation, culture, society, interaction in everyday life, groups and organizations, deviance, sexuality, social stratification, social class in the United States, global inequality, gender stratification, race and ethnicity, aging, the economy, politics and government, family, religion, education, health and medicine, population and urbanization, collective behavior and social movements, and social change.An excellent resource for professionals in the field of sociology, this book is also an excellent read for non-academic hobbyists and life-long learners.
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Apartment Antonija A1
Located 400 metres from ACI Marina Vodice in Vodice, this air-conditioned apartment features a balcony. Providing free private parking, the apartment is 2.2 km from Club Hacienda Vodice.
The unit equipped with a kitchen with an oven and fridge. A TV with satellite channels is available. Other facilities at Apartment Antonija A1 include a barbecue.
Split Airport is 49 km away.


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Jeremy Long with the main oil painting at his reception last night at the Glen House Inn. (photos by Suzanne Ehalt)
The nooks and crannies throughout the Village that often get overlooked. Today's blog showcases a peak at the Glen House Inn last night during Jeremy Long's reception. His paintings and drawings will be there until December 2, 2012.
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Cool kids
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This site will show you photos, our menu, our address, and much more.
You can order from our menu, or you can come to our buffet in our restsurant.
About Us
China Star is a chinese restaurant in Dunbar. It was opened in 1998. Most customers say it is the best chinese restaurant around. They love our food.