Journal of Mathematical Analysis and Applications

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1 J. Math. Anal. Appl. 393 (2012) Contents lists available at SciVerse ScienceDirect Journal of Mathematical Analysis and Applications journal homepage: On Riesz minimal energy problems H. Harbrecht c, W.L. Wendland a,, N. Zorii b a Institut für Angewandte Analysis und Numerische Simulation, Universität Stuttgart, Pfaffenwaldring 57, Stuttgart, Germany b Institute of Mathematics, National Academy of Sciences of Ukraine, Tereshchenkivska 3, 01601, Kyiv-4, Ukraine c Mathematisches Institut, Universität Basel, Rheinsprung 21, CH-4051 Basel, Switzerland a r t i c l e i n f o a b s t r a c t Article history: Received 21 December 2010 Available online 17 April 2012 Submitted by Andrea Cianchi Dedicated to Professor Bogdan Bojarski on the occasion of his 80th birthday Keywords: Minimal Riesz energy problem External field Pseudodifferential operator Penalty method Simple layer boundary integral operator Boundary element approximation In R n, n 2, we study the constructive and numerical solution of minimizing the energy relative to the Riesz kernel x y α n, where 1 < α < n, for the Gauss variational problem, considered for finitely many compact, mutually disjoint, boundaryless (n 1)- dimensional C -manifolds l, l L, each l being charged with Borel measures with the sign α l := ±1 prescribed. We show that the Gauss variational problem over an affine cone of Borel measures can alternatively be formulated as a minimum problem over an affine cone of surface distributions belonging to the Sobolev Slobodetski space H ε/2 ( ), where ε := α 1 and := Ɣ l L l. This allows the application of simple layer boundary integral operators on and, hence, a penalty approximation. A corresponding numerical method is based on the Galerkin Bubnov discretization with piecewise constant boundary elements. Wavelet matrix compression is applied to sparsify the system matrix. To the discretized problem, a gradient-projection method is applied. Numerical results are presented to illustrate the approach Elsevier Inc. All rights reserved. 1. Introduction Carl Friedrich Gauss investigated in [1] the variational problem of minimizing the Newtonian energy evaluated in the presence of an external field, nowadays called the Gauss functional, over nonnegative charges ϕ ds on the boundary surface of a given domain. For this problem, later on the sign condition was given up in connection with boundary integral equation methods where distributional boundary charges had been introduced for solving boundary value problems. (For the history, see Costabel s article [2].) A different generalization of the original Gauss variational problem, maintaining the sign restriction but employing Borel or Radon measures µ as charges and replacing the Newtonian kernel by a much more general one (e.g., by the Riesz or Green kernel) has independently grown into an eminent branch of modern potential theory (see, e.g., [3] and the extensive works [4 7]; for two dimensions, see [8]). In this paper, we consider the Gauss variational problem with respect to the Riesz kernel x y α n, 1 < α < n, on := l L l, where l, l L, are finitely many compact, connected, mutually disjoint, boundaryless (n 1)-dimensional orientable manifolds, immersed into R n, n 2, which are assumed to be at least Lipschitz, and is loaded by charges µ = l L α lµ l, where α l is a function of l taking the value +1 or 1 and µ l is a nonnegative Borel measure supported by l. We first show that, if each l is a C -manifold, then to every Borel measure ν on with finite Riesz energy there corresponds a unique distribution σ on belonging to the Sobolev Slobodetski space H ε/2 ( ), where ε := α 1, such that the linear functional defined by the measure ν and the one defined by the L 2 ( )-duality with σ coincide on C ( ). Moreover, the Gauss problem over Borel measures is equivalent to the problem of minimizing the Gauss functional over the Corresponding author. addresses: helmut.harbrecht@unibas.ch (H. Harbrecht), wendland@mathematik.uni-stuttgart.de (W.L. Wendland), natalia.zorii@gmail.com (N. Zorii) X/$ see front matter 2012 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 398 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) corresponding affine cone in H ε/2 ( ), and then the Gauss functional can be expressed in terms of a simple layer boundary integral operator on. This allows us to approximate the Gauss problem by employing a penalty formulation. The latter corresponds to a nonlinear variational problem on the convex cone of all ϕ = l L α lϕ l where ϕ l H ε/2 ( l ) and ϕ l 0. For the penalty method, we present error estimates depending on the penalizing parameter. For the penalized variational problem, the use of piecewise constant boundary elements on triangular or quadrangular meshes on and a corresponding Galerkin Bubnov discretization results in a convex, finite-dimensional minimization problem with a symmetric, positive definite system matrix. The latter can be solved by the gradient-projection method applied to the discrete system (see [9, Chapter I]). The convergence of the gradient-projection method depends on the penalty parameter as well as on the boundary element s mesh-width and becomes very slow for large penalty and small mesh-width. In applications, the numerical solution of the Gauss variational problem is of great interest if for practical reasons in electrical engineering on some of the l only nonnegative while on the others only nonpositive charges are allowed (see capacitors in [10]). It also has applications in approximation theory and the development of efficient numerical integration and various other fields as mentioned in [11]. In case n = 3, we present numerical experiments for the Gauss problem and determine the equilibrium state by an appropriate boundary element approximation. Since the gradient-projection method turns out to converge very slowly, we apply the wavelet matrix compression [12,13] to realize fast matrix-vector multiplications. In particular, as already observed in [14], the charges converge faster for smaller penalty whereas the total charge values are better approximated for larger penalty. Therefore we are using a cascadic approach. Remark 1. The analytic part of this article extends the results from [14], obtained for the Newtonian kernel x y 2 n and a Lipschitz surface in R 3 (n = 3, α = 2), to the case of the Riesz kernels x y α n with α (1, n) arbitrary; however, the Borel measures are required to be supported now on a C -manifold. To compare these two cases, observe that the Newtonian potential of a Borel measure ν on with finite energy satisfies Laplace s partial differential equation in R n \ while its Dirichlet integral provides an equivalent representation of the Newtonian energy of ν (see [15, Theorem 1.20]), which enables us (cf. [14]) to use the results for boundary value problems in Lipschitz domains. In the case of the Riesz kernels (with α 2), however, the energy of a Borel measure on can be characterized only by means of pseudodifferential operators operating within the manifold, which usually requires to be C (see, e.g., [16]). Namely, in order to approximate Borel measures on with finite Riesz energy by absolutely continuous surface charges supported by, we exploit the representation of pseudodifferential operators on in terms of local pseudohomogeneous kernel expansions in one chart (which goes back to Seeley [17]), and then we use the approximability of measures in R n 1, i.e., in corresponding local coordinates, by absolutely continuous measures on R n 1 as constructed by Cartan (see Lemma 1.2 and Corollary 2 in [15, Chapter 1]). 2. Gauss variational problem We consider the problem of minimizing the energy relative to the Riesz kernel x y α n, 1 < α < n, for signed Borel measures on a given (n 1)-dimensional (in general, non-connected) manifold in R n, n 2, in the presence of an external field. The corresponding admissible measures (or charges) are associated with a (generalized) condenser, which is treated here as an ordered collection A = (A i ) of finitely many mutually disjoint plates A i, i I, and each A i is the finite union of compact, nonintersecting, boundaryless, connected Lipschitz (n 1)-dimensional orientable manifolds l, l L i, immersed into R n. That is, = A i, where A i = l L i l. Each plate A i, i I, is treated with the sign α i prescribed, where α i takes the value +1 for i I + and 1 for i I. Here, I = I + I, I + I =, and I is allowed to be empty. Changing notation if necessary, we assume the index sets L i, i I, to be mutually disjoint. Write L := L i, L + := + L i, L := L i and define α l := +1 for l L + and α l := 1 for l L. In the sequel, we also need the notation + := l L + l and := l L l. To formulate the problem, let M = M( ) be the σ -algebra of Borel measures ν on, equipped with the weak topology, i.e., the topology of pointwise convergence on the class C( ) of all real-valued continuous functions on (see, e.g., [18]). For ν 1, ν 2 M( ), the mutual Riesz energy is given by the formula I α (ν 1, ν 2 ) := x y α n d(ν 1 ν 2 )(x, y), provided the integral on the right is well defined (as a finite number or ± ). For ν 1 = ν 2, we get the Riesz energy I α (ν 2 ) := I α (ν 2, ν 2 ) of ν 2. Let E α = E α ( ) consist of all ν M( ) with < I α (ν) <. Since the Riesz kernel is strictly positive definite (see, e.g., [15]), the bilinear form I α (ν 1, ν 2 ) defines on E α a scalar product and, hence, E α is a pre-hilbert space with the norm ν Eα := ν Eα ( ) := I α (ν). The topology on E α ( ) defined by the norm Eα ( ) will be called strong.

3 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Given a Borel set B, let M(B) consist of all ν M( ) concentrated in B, and let M + (B) be the convex cone of all nonnegative ν M(B). We also write E + α (B) := M+ (B) E α. The condenser A = (A i ) is supposed to be loaded by charges µ = α i µ i, where µ i E + α (A i). The set of all those µ will be denoted by E α (A); it is a convex cone in E α. Further, let g be a given continuous, positive function on and let a = (a i ) be a given vector with a i > 0, i I. Then the set of admissible charges for the Gauss problem is defined by E α (A, a, g) := µ E α (A) : A i g dµ i = a i for all i I Note that E α (A, a, g) is an affine, convex cone in the pre-hilbert space E α. In addition, let f denote a given continuous function on, characterizing an exterior source of energy. Then G f (µ) := I α (µ) 2 f dµ defines the value of the Gauss functional at µ E α (A). The Gauss problem now reads as follows: Problem 1. Find µ 0 that minimizes G f (µ) in E α (A, a, g), i.e., µ 0 E α (A, a, g) with G f (µ 0 ) =. inf G f (µ) =: G f (A, a, g). (1) µ E α (A,a,g) A minimizer µ 0 is unique (if exists). This follows from the strict positive definiteness of the Riesz kernel and the convexity of the class of admissible measures; see [5, Lemma 6]. But what about the existence of µ 0? Assume for a moment that at least one of the A i is noncompact. Then it is not clear at all whether the equilibrium state in the Gauss variational problem can be attained. Moreover, it has been shown by the third author that, in this case, a minimizing measure µ 0 in general does not exist; necessary and sufficient conditions for µ 0 to exist were given in [4,6,7]. However, in the case under consideration, where all the A i are assumed to be compact, the Gauss variational problem has a (unique) solution µ 0. Indeed, this follows from the weak compactness of E α (A, a, g) when combined with the fact that the Gauss functional G f is weakly lower semicontinuous on E α (A); cf. [3]. Under additional restrictions on, g, and f, in Section 5 we shall find an equivalent formulation of the Gauss variational problem (1) (for the Riesz kernel x y α n, where 1 < α < n), based on distributions concentrated on with densities from the Sobolev Slobodetski space H ε/2 ( ), where ε := α 1. These distributions define bounded linear functionals on H ε/2 ( ), whereas Borel measures µ M( ) define bounded linear functionals on C( ); however, C( ) H ε/2 ( ) C( ) (for more details, see the section below). 3. Riesz potentials in R n and on manifolds For any s > 0 and 0 < R <, write H s (B R ) := ϕ H s (R n ) : suppϕ B R, where BR := {x R n : x < R} (see, e.g., [19, (4.1.17)]). Here, H s (R n ) is the Sobolev space of order s in R n (see, e.g., [20]). For the Riesz potentials of order α (1, n) in R n, n 2, we have the following Lemma 1. The operator V α, given by the formula V α ϕ(x) := x y α n ϕ(y) dy, where ϕ C R n 0 (Rn ) and x R n, is a strongly elliptic classical pseudodifferential operator of order α. Moreover, for any given R (0, ) there exist positive constants c 0 and c 1 depending on R only such that c 0 ϕ 2 H α/2 (B R ) V α ϕ, ϕ L 2 (B R ) c 1 ϕ 2 H α/2 (B R ) for all ϕ H α/2 (B R ). (2) Proof. Observe that the Schwartz kernel of the integral operator V α is homogeneous of degree α n < 0 and, by Seeley [17], the homogeneous symbol of V α is given by a α (x, ξ) = 2 α π α n/2 2 ξ α, ξ R n, n α 2 where ( ) denotes the Gamma function. Since for ξ = 1, a α (x, ξ) is a positive constant, V α is strongly elliptic and, as a pseudodifferential operator on the (bounded) domain B R, it is a continuous mapping from H α/2 (B R ) into H α/2 (B R ). This yields the inequality on the right in (2) with a constant c 1 depending on R only. The one on the left follows with the Fourier transform and Parseval s equality (see [19, Section 7.1.1]); actually, c 0 does not depend on R.

4 400 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Recall that l, l L, are compact, connected, mutually disjoint, boundaryless, Lipschitz (n 1)-dimensional orientable manifolds, immersed into R n, and = l L l. Given α (1, n), write ε := ε(α) := α 1. Then 0 < ε < n 1. Let Ω R n be the domain (bounded or unbounded) with the boundary R nω = and let H ε/2 ( ) be the space of traces of elements from the Sobolev space H α/2 (Ω) on (see [20,21]). Let C ( ) be the trace space of C 0 (Rn ) on, and define for ϕ C ( ) ϕ H ε/2 ( ) := inf ϕ H α/2 (Ω), where ϕ C 0 (Rn ) and ϕ = ϕ. (3) Since is Lipschitz, C ( ) is dense in the trace space H ε/2 ( ), its closure with respect to the norm given by (3) (see [20]). Moreover, the surface measure ds on is well defined and generates on C ( ) the L 2 -scalar product, (ϕ, ψ) := (ϕ, ψ) L2 ( ) := ϕψ ds, where ϕ, ψ C ( ). (4) In fact, H ε/2 ( ) is a Hilbert space equipped with the scalar product (ϕ(x) ϕ(y)) (ψ(x) ψ(y)) ((ϕ, ψ)) H ε/2 ( ) := (ϕ, ψ) + ds(x) ds(y) x y n 1+ε and the norms given by (3) and by ((ϕ, ϕ)) H ε/2 ( ) are equivalent (see [20, Theorem 7.48]). The L 2 -scalar product (4) continuously extends to the duality between H ε/2 ( ) and its dual space H ε/2 ( ), which is equipped with the norm ϕ H ε/2 ( ) := sup (ϕ, ψ), where ψ H ε/2 ( ) and ψ H ε/2 ( ) 1. We denote that extension by the same symbol (, ) = (, ) L2 ( ). Note that the function space C ( ) is also dense in each of the spaces L 2 ( ) and H ε/2 ( ). We shall show below that, under additional restrictions on g, f, and, the solution to the Gauss problem (1) can be obtained with the help of the simple layer potential V α ψ(x) := x y α n ψ(y) ds(y), where x R n. Write V := γ 0 V α, where γ 0 is the Gagliardo trace operator onto (see [22,21,23]). Theorem 2. Let Ω C k 1,1 (R n ), where k = 1 if α < 3 and 2k α if α 3. Then the operator V is a linear, continuous, invertible mapping V : H ε/2 ( ) H ε/2 ( ). Moreover, it is H ε/2 ( )-elliptic; i.e., there exist positive constants c c and c V depending on only such that c V ψ 2 H ε/2 ( ) ψ 2 V c c ψ 2 H ε/2 ( ) for all ψ H ε/2 ( ), (5) where ψ 2 V := (ψ, Vψ) L 2 ( ). Proof. Choose R (0, ) such that B R, and let γ 0 be the adjoint to the Gagliardo trace operator γ 0, given by (γ 0 ψ, Φ) L 2 (R n ) = (ψ, γ 0 Φ) L2 ( ), where Φ C 0 (Rn ) and ψ H ε/2 ( ). Then, because of α/2 > 1/2, respectively ε/2 > 0, the trace theorem [20,23] can be applied and we obtain that for any ψ H ε/2 ( ), γ 0 ψ H ε (R n ) = H α 2 (R n ) and supp(γ 0 ψ), hence, γ 0 ψ H α/2 (B R ); moreover, there exist positive constants c and c such that c ψ H ε/2 ( ) γ 0 ψ H α/2 (B R ) c ψ H ε/2 ( ). Because of Lemma 1, we have V α γ 0 ψ Hα/2 (B R ). Therefore the trace of V α γ 0 ψ on exists and (2) implies (Vψ, ψ) L2 ( ) = (γ 0 V α γ 0 ψ, ψ) L 2 ( ) = (V α γ 0 ψ, γ 0 ψ) L 2 (R n ) c 0 γ 0 ψ 2 H α/2 (B R ) c 0c 2 ψ 2 H ε/2 ( )

5 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) and also (Vψ, ψ) L2 ( ) V α γ 0 ψ 2 H α/2 (B R ) V α c 2 ψ 2 H ε/2 ( ), which is (5). The invertibility of V then follows with the Lax Milgram lemma. This completes the proof. Remark 2. For n = 2 or n = 3, Theorem 2 is valid if Ω is just Lipschitz. See [22, Theorem 3.6] and [23, pp ]. 4. Relations between E α ( ) and H ε/2 ( ) The main purpose of this section is to characterize the Borel measures on with finite Riesz energy, i.e. Σ E α ( ) where 1 < α < n, via distributions in H ε/2 ( ). We have succeeded in this under the additional restriction that all the l, l L, are C -manifolds, which from now on is always assumed to be satisfied. The characterization obtained is given by the following principal result. Theorem 3. Let Σ M( ) have finite Riesz energy x y α n dσ(x) dσ(y) = Σ 2 E α ( ) <. Then there exists a unique element σ H ε/2 ( ), where ε = α 1, such that Σ(ϕ) = ϕ dσ = (ϕ, σ ) L2 ( ) for all ϕ C ( ). (6) Moreover, Σ 2 E α ( ) = σ 2 V σ 2 H ε/2 ( ) (7) where denotes equivalence. Note that for the Newtonian kernel, i.e., for α = 2 and n 3, the corresponding result can be obtained for Lipschitz manifolds (see [14], where this was done for n = 3). The proof of Theorem 3 is based on the following result, which is of independent interest. Theorem 4. For every Σ E α ( ) there exists a sequence of absolutely continuous measures Σ k ϕ k C( ) (i.e., dσ k (x) = ϕ k (x) ds(x)) such that {Σ k } k N converges to Σ weakly and strongly, i.e. Σ k (ϕ) Σ(ϕ) for all ϕ C( ) and lim k Σ k Σ 2 E α ( ) = 0. E α ( ) with densities To provide proofs of Theorems 3 and 4, to be given in Sections 4.3 and 4.4, respectively, we need the following preliminaries Preliminaries For each of the orientable C -manifolds l, l L, immersed into R n, we associate a family of finite-dimensional atlases A l. Each atlas A l is a family of local charts (O lr,, X lr ), where r ranges through a finite index set R l. The open sets O lr l define an open covering of l, while X lr is a C -diffeomorphism of O lr onto R n 1. Let {β lr } r Rl be a C -partition of unity of l subordinate to the atlas A l. In addition to the partition of unity, let {γ lr } r Rl be a second system of functions γ lr C 0 (O lr) with the property Then γ lr (x) = 1 for all x supp β lr. γ lr (x)β lr (x) = β lr (x) and β lr (x)γ lr (x) = β lr (x) for all x l. (8) With respect to the atlas A l, let X lr denote the corresponding pushforwards and X lr the pullbacks. Then X lr β lr C (U 0 lr). Without loss of generality, the local parametric representations can always be chosen in such a way that at one point x 0 lr O lr where β lr (x 0 ) = lr 1 we have X lr(x 0 lr ) = 0 and, moreover, at this point the tangent bundle x x = X 1 lr (x ) x, where x := X lr (x), x =0 x =0

6 402 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) forms a positively oriented system of n 1 mutually orthogonal unit vectors. This implies that the Riemannian tensor of l in the local coordinates at the point x 0 lr is the unity matrix. Hence, for the surface measure ds l (x) = J lr (x ) dx, where x, we have J lr (0) = 1. Given an arbitrary finite-dimensional atlas A l on l, write d l := min r R l diam. Then one can choose δ 0 > 0 so that for any given 0 < δ < δ 0 there exists a finite-dimensional atlas A δ l satisfying all the properties formulated above and such that d l = δ. Hence, we have a whole family of atlases A δ l, δ (0, δ 0), and from now on only these atlases will be under consideration. For the sake of brevity, we shall omit the upper index δ in the notation. Note that the Jacobians J lr depend on the geometric properties of the C -manifold l only, and J lr together with their derivatives are uniformly continuous relative to δ (0, δ 0 ). Corresponding to the partition of unity, the pseudodifferential operator V on can be decomposed into a finite sum of localized operators as follows: V = β lr Vγ lr + Z 1 = γ lr Vβ lr + Z 2, (9) l L r R l l L r R l where Z 1, Z 2 are smoothing operators of order, whose Schwartz kernels are C -functions. This follows from [19, p. 418] in view of the facts that x y α n C ( p k ) for all p k (where p, k L) and supp (1 γ lr ) supp (β lr ) = provided x, y l. Note that each of the operators β lr Vγ lr and γ lr Vβ lr is a pseudodifferential operator of order ε on. Then, with the pullback and pushforward mappings the pseudodifferential operator V takes the following form: V = β lr X lr V lrx lr γ lr + Z 1 = γ lr X lr V lrx lr β lr + Z 2, l L r R l l L r R l where V lr ϕ(x ) = X 1 lr (x ) X 1 lr (y ) α n ϕ(y )J lr (y ) dy, ϕ C 0 (), (10) are the localized pseudodifferential operators, defined by the operator V in the parametric domains Auxiliary results To prove Theorems 3 and 4, we also need the following auxiliary results, the first one being well known (see [24, Proposition 24.30]). Lemma 5. Let Σ be a Borel measure on with supp Σ O lr. Then the pushforward X lr Σ defines a Borel measure Σ lr in R n 1 with supp Σ lr and, furthermore, dσ(x) = J lr (x ) dσ lr (x ), dσ lr (x ) = J 1 lr (x) dσ(x). (11) Conversely, if Σ lr is a Borel measure in R n 1 with supp Σ lr, then the pullback X lr Σ lr defines a Borel measure Σ on with supp Σ O lr and (11) holds. Lemma 6. A Borel measure Σ on can be identified with an element of the Sobolev Slobodetski space H n 2 q ( ) for every q > 1 2. Proof. The lemma is a simple consequence of Sobolev s embedding theorem (see, e.g., [20]). Indeed, since Σ defines a bounded linear functional on the space C( ), we have for every ψ C( ) ψ dσ Σ BV ψ C( ) c Σ ψ n H 2 +q ( ) with an arbitrary q > 1 2. Hence Σ belongs to the dual space H n 2 q ( ) of H n 2 +q ( ). Lemma 7. Let Σ M + ( ) have finite Riesz energy. Then for all r R l and l L, β lr Σ 2 E α ( ) = x y α n β lr (x)β lr (y) dσ(x) dσ(y) <.

7 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Proof. Using relations (8) and (9) and Lemma 6 for q = 0, we get β lr Σ 2 E α ( ) = γ lr (x) x y α n β lr (y) dσ(y)β lr (x) dσ(x) γ kt (x) x y α n β pm (y) dσ(y)β kt (x) dσ(x) k L t R k p L m R p = x y α n dσ(x) dσ(y) (Z 2 Σ) dσ. Hence, β lr Σ 2 E α ( ) Σ 2 E α ( ) + c Σ H n/2 ( ) Z 2 Σ H n/2 ( ) Σ 2 E α ( ) + c Σ 2 H n/2 ( ) < due to the fact that Z 2 is a pseudodifferential operator of order. For all l L, r R l and δ (0, δ 0 ), where δ 0 has been defined in Section 4.1, we define β lr (x ) := (X lr β lr )(x ) = β lr X 1 lr (x ), k lr (x, x y ) := X 1 lr (x ) X 1 lr (y ) α n, klr1 (x, x y ) := k lr (x, x y )J lr (x )J lr (y ) x y α n. (12) Theorem 8. Let Σ belong to E α + ( ). Then for any l L, r R l, and δ (0, δ 0 ) we have β lr Σ 2 E α ( ) = β lr Σ 2 E α ( ) + where and β lr Σ 2 E α ( ) := klr1 (x, x y )β lr (x ) dσ lr (x )β lr (y ) dσ lr (y ), (13) x y α n β lr (x ) dσ lr (x )β lr (y ) dσ lr (y ) klr1 (x, x y )β lr (x ) dσ lr (x )β lr (y ) dσ lr (y ) C δ β lr Σ 2 E α ( ), (14) the constant C depending on the geometry of only, but neither on δ nor on Σ. Proof. Recall that V lr in (10) is a classical pseudodifferential operator of order ε on. Hence, it admits the Schwartz kernel k lr (x, x y ), which because of 0 > ε > n + 1 has the homogeneous asymptotic expansion k lr (x, z ) = z α n k lr0 (x, Θ ) + J z α n+j k lrj (x, Θ ) + z α n+1 K lrj (x, z ), (15) j=1 where z := x y, Θ := z / z and k lrj, 0 j J, are C -functions such that k lrj (x, z ) C 1, the constant C 1 depending on the geometry of l only. Having chosen J so that J > n α + 2, we find the remainder K lrj (x, z ) to be continuously differentiable with respect to x and z (see [19, Section 7.11] and [17, p. 209]). Define z α n+1ˆklr (x, z ) := k lr (x, z ) z α n k lr0 (x, Θ ). (16) Then ˆklr (x, z ) C 2 for all x, y, where C 2 does not depend on δ, and z ˆklr (x, z ) is continuous with respect to x and y. Note that, due to the particularly chosen local representation of X lr at x lr0 O lr (see Section 4.1), for x = x lr0 we have x = X lr (x lr0 ) = 0 and k lr0 (0, z ) = z α n. Moreover, since k lr0 (x, Θ ) is a C -function, one can write where k lr0 (x, z ) = z α n + z α n klr0 (x, Θ ), (17) klr0 (x, Θ ) = k lr0 (x, Θ ) k lr0 (0, Θ ).

8 404 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Then klr0 C ( S n 2 ), klr0 (0, Θ ) = 0, and the function klr0 (x, Θ ) is measurable and bounded by C 3 δ for x, y, the constant C 3 being independent of δ. (Here S n 2 denotes the unit sphere in R n 1.) Combining relations (15) (17), on account of definition (12) we obtain klr1 (x, z ) = z α n k lr1 (x, z ), where the function k lr1 (x, z ) := z ˆklr (x, z ) + klr0 x, z z J lr (x )J lr (y ) + J lr (x ) 1 J lr (y ) 1 + J lr (x ) 1 + J lr (y ) 1 is measurable and its modulus is bounded by C δ, with C not depending on δ. In conclusion, in consequence of Lemma 7 one finds > β lr Σ 2 E α ( ) = = + β lr (x )k lr (x, x y )J lr (x ) dσ(x )β lr (y )J lr (y ) dσ(y ) x y α n β lr (x ) dσ(x )β lr (y ) dσ(y ) U lr x y α n k lr1 (x, x y )β lr (x ) dσ(x )β lr (y ) dσ(y ). Since β lr Σ is a nonnegative measure with support in, this yields estimate (14) as proposed. From now on, we fix δ = δ so that C δ 1/2, where C is the constant in Theorem 8. Substituting (14) with such a δ into (13), we arrive at the following assertion. Corollary 9. For this δ and for any Σ E α + ( ) we have the estimate 1 2 β lr Σ 2 E α ( ) β lrσ 2 E α ( ) 3 2 β lr Σ 2 E α ( ). (18) Now we are in a position to prove Theorem Proof of Theorem 4 Without loss of generality, we can assume Σ E α ( ) to be nonnegative, i.e. Σ E α + ( ). Then we obtain from Lemma 7 and Corollary 9 that β lr Σ 2 E α ( ) < for all r R l, l L and for the δ, chosen above. Now, according to Lemma 1.2 from [15] (see also Corollary 2 there, attached to the lemma), applied in R n 1, one can find absolutely continuous measures Σ lrk E α + (), k N, concentrated in, with densities ϕ lrk C 0 ( ), such that β lr Σ Σ lrk 0 for all k N and lim β lr Σ Σ lrk 2 E k α ( ) = 0. Then (18) implies lim β lrσ X lrσ lrk Eα ( ) = 0. k Since Σ l = r R l β lr Σ, the last relation and the triangle inequality yield lim k Σ l X lrσ lrk lim β lr Σ X lrσ lrk Eα ( ) = 0. k r R l r R l E α ( ) Hence, if we define Σ k := ϕ lrk X lr (x)j 1 lr (x) ds(x), k N, l L r R l then Σ k Σ strongly. Since for the Riesz kernel the strong convergence of nonnegative measures implies the weak convergence to the same limit (see, e.g., Lemma 1.2 in [15]), these Σ k, k N, are as desired.

9 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Proof of Theorem 3 Let ϕ C ( ). Since V is a pseudodifferential operator of order ε = 1 α on and coercive, the equation ϕ(x) = Vψ(x) = x y α n ψ(y) ds(y) is uniquely solvable due to the Lax Milgram lemma, and for its solution we have ψ H ε/2 ( ). Moreover, since ϕ C ( ), we actually obtain ψ C ( ). Therefore, Σ(ϕ) = ϕ dσ = x y α n ψ(y) ds(y) dσ(x) = I α (Ψ, Σ), where Ψ, Ψ (B) := ψ(y) ds(y) for Borel sets B, B is a Borel measure on because ψ C ( ) L 1 ( ). Moreover, and hence I α (Ψ, Ψ ) = (ψ, Vψ) L2 ( ) c c ψ 2 H ε/2 ( ) c2 0 ϕ 2 H ε/2 ( ) < Σ(ϕ) = I α (Σ, Ψ ) Σ Eα ( ) Ψ Eα ( ) c Σ c 0 ϕ H ε/2 ( ) for all ϕ C ( ), where c Σ and c 0 are independent of ϕ. Since C ( ) is dense in H ε/2 ( ), the measure Σ defines a bounded linear functional on H ε/2 ( ), and there exists a uniquely determined element σ H ε/2 ( ) such that (6) holds. In order to establish (7), we approximate Σ in both the strong and weak topologies of the space E α ( ) by a sequence of absolutely continuous measures Σ k with densities ϕ k C( ), which is possible due to Theorem 4, and use the fact that C( ) is a dense subspace of H ε/2 ( ). Then {ϕ k } defines a Cauchy sequence also in the Hilbert space H ε/2 ( ) because of (5), and so {ϕ k } converges to an element σ H ε/2 ( ). In consequence of (6), the weak convergence of Σ k to Σ implies that, for any ϕ C ( ), (σ, ϕ) L2 ( ) = Σ(ϕ) = lim k Σ k(ϕ) = lim k (ϕ k, ϕ) L2 ( ) = (σ, ϕ) L2 ( ), because ϕ k σ in H ε/2 ( ). Hence, σ = σ. Now, from the strong convergence of Σ k to Σ we get Σ Eα ( ) = lim k Σ k Eα ( ) = lim k ϕ k V = σ V, for ϕ k σ in H ε/2 ( ). This proves (7) and completes the proof of Theorem Variational formulation in a trace space From now on, for the given functions g and f we require that f, g C( ) H ε/2 ( ). Define V f (ϕ) := ϕ 2 V 2(f, ϕ) L 2 ( ), where ϕ H ε/2 ( ). The following theorem shows that the Gauss problem (1) (for the Riesz kernel x y α n of order α (1, n)) can alternatively be formulated as the problem of minimizing the functional V f over the affine cone K(A, a, g) in H ε/2 ( ), where K(A, a, g) := ϕ = α l ϕ l, where ϕ l H ε/2 ( l ), ϕ l 0 and gϕ l ds = a i for all i I. l L l Such a problem will be referred to as the dual Gauss problem. Theorem 10. Under the stated assumptions on g, f, and, to the (unique) solution µ 0 E α (A, a, g) of the Gauss problem (1) there corresponds a unique element ϕ 0 K(A, a, g) H ε/2 ( ) with the properties and µ 0 (ϕ) = (ϕ 0, ϕ) L2 ( ) for all ϕ C ( ) V f (ϕ 0 ) = G f (µ 0 ) = G f (A, a, g). This ϕ 0 is the minimizer of the functional V f over K(A, a, g), i.e., V f (ϕ 0 ) = min V f (ϕ) =: V f (A, a, g). (20) ϕ K(A,a,g) l L i (19)

10 406 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Proof. By Theorem 3, to any Borel measure µ = α l L lµ l E α (A, a, g) there corresponds a unique element ϕ µ = α l L lϕµ l H ε/2 ( ) with ϕµ l H ε/2 ( l ), satisfying (6) and (7). Since µ l 0 for all l L, relation (6) yields 0 (ϕµ l, ϕ) = µl (ϕ) for all nonnegative ϕ C ( ); hence, ϕµ l 0 for all l L. Moreover, since g belongs to C( ) H ε/2 ( ) while C ( ) is dense in both the spaces C( ) and H ε/2 ( ), we also conclude from (6) that (ϕµ l, g) L 2 ( l ) = µ l (g) = a i, i I. l L i l L i Therefore, ϕ µ K(A, a, g). Furthermore, applying (7) and the same arguments as just above, but now with f instead of g, one also gets V f (ϕ µ ) = ϕ µ 2 V 2(ϕ µ, f ) L2 ( ) = µ 2 E α 2µ(f ) = G f (µ). (21) If now µ 0 E α (A, a, g) is the (unique) solution of the Gauss problem (1), then for the corresponding element ϕ 0 := ϕ µ0 K(A, a, g) we obtain (19) in consequence of (21), which is a part of the desired conclusion. The proof will be complete once we establish (20). To this end, observe that one can construct a sequence ϕ k C ( ) K(A, a, g) converging to ϕ 0 in H ε/2 ( ). Hence, by (19), V f (ϕ k ) V f (ϕ 0 ) = G f (A, a, g). Moreover, ϕds E α (A, a, g) for all ϕ C ( ) K(A, a, g) and so, by (21), G f (A, a, g) which implies with k inf V ϕ K(A,a,g) C f (ϕ) V f (ϕ k ) for all k N, ( ) inf V ϕ K(A,a,g) C f (ϕ) = V f (ϕ 0 ) = G f (A, a, g). ( ) Repeated application of the fact that C ( ) is a dense subspace of H ε/2 ( ) yields (20) as required. 6. Penalty approximation In order to find a solution of the dual Gauss problem (20) with a suitable algorithm, we replace the affine cone K(A, a, g) by a cone K(A) with vertex at 0 by employing Lagrange multipliers for the side conditions gϕ l ds = a i, i I. (22) l L i l Then we arrive at the following Problem 2. Find ϕ ϱ K(A) := ϕ = l L α l ϕ l, where ϕ l H ε/2 ( l ) and ϕ l 0, which is the minimizer of V f,ϱ (ϕ) := V f (ϕ) + ϱ 2 2 gϕ l ds a i (23) l l L i over K(A). Namely, V f,ϱ (ϕ ϱ ) = min ϕ K(A) V f,ϱ(ϕ) =: V f,ϱ (A, a, g). (24) Here ϱ > 0 is a penalty parameter to be appropriately chosen later on. Lemma 11. For the extremal values V f (A, a, g) and V f,ϱ (A, a, g) and the solutions ϕ 0 and ϕ ϱ of the problems (20) and (24), respectively, the a priori estimates V f,ϱ (A, a, g) V f (A, a, g), ϕ 0 H ε/2 ( ) C and ϕ ϱ H ε/2 ( ) C

11 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) hold true, where C := 1 c D f H c ε/2 ( ) + V Here c D is the duality constant in c 2 D f 2 H ε/2 ( ) + 1/2 c V V f (A, a, g). (25) (χ, ψ) L2 ( ) c D χ H ε/2 ( ) ψ H ε/2 ( ) for χ H ε/2 ( ), ψ H ε/2 ( ). The proof of this lemma as well as that of the following theorem will be presented in the Appendix. Theorem 12. For the given ϕ ϱ = α l L lϕϱ l and every i I we denote δ i := ϕϱ l g ds a i, a 1 i := a 1 i ϕϱ l g ds l Li l l L i l and define δ 2 := δ 2 i, ϕ ϱ := l L α l ϕ l ϱ, where ϕ l ϱ := a iϕ l ϱ for all l L i and i I. Then ϕ ϱ K(A, a, g) and δ i δ 2 2Cc D f H ε/2 ( ) + V f (A, a, g) 1/2 ϱ 1 2 =: C1 ϱ 1/2. (26) Moreover, if ϱ is sufficiently large, then there exist constants C 2, C 3, and C 4 not depending on δ and ϱ such that ϕ ϱ ϕ 0 H ε/2 ( ) C 2δ 1/2 C 3 ϱ 1/4, (27) V f,ϱ (A, a, g) V f (A, a, g) V f,ϱ (A, a, g) + C 4 ϱ 1/4. (28) Theorem 12 shows that the solution ϕ 0 of the dual Gauss problem (20) can be approximated by the solutions of Problem 2 by choosing ϱ large enough. The advantage lies in the fact that the cone K(A) is a cone with vertex 0 and not an affine cone as in (20). For solving this problem we use the gradient-projection method, described in Section Approximation of the Gauss problem by the use of piecewise constant charges Since Problem 2 is quadratic and of the type (3.8) in [9, Chapter I], one could compute the minimizer by using the gradientprojection method: ϕ k+1 := P K ϕ k ηv f,ϱ ϕ k, k = 0, 1,..., where P K is the H ε/2 ( )-orthogonal projection onto K(A) (see [9, Chapter I, (3.13)]) and V f,ϱ is the Frechet derivative of the functional V f,ϱ. But a numerical realization of P K is as difficult as solving the original Gauss problem itself. Therefore we apply a gradient-projection iteration to the discretized formulation instead. For solving Problem 2 numerically, we use a quasiregular family of meshes T h on the manifold (see [25, Chapter 10]), where h denotes the maximum mesh width of the elements in T h. On the mesh, we introduce piecewise constant functions, forming the space S 0( ) h L 2( ) H ε/2 ( ), as the trial as well as the test space. Correspondingly, we define K h (A) := ϕ h = α l ϕ l h S0( ) h with supp ϕl h l and ϕ l h 0. (29) l L The finite-dimensional approximation to Problem 2 then reads as follows. Problem 3. Find the minimizer ϕ ϱh K h (A) of the quadratic functional V f,ϱ (ϕ h ) = i,j I l L i m L j where ϕ h ranges over K h (A). l α i α j (Vϕ, h ϕm) h 2 l L i α i (f, ϕ l h ) + ϱ 2 l L i (g, ϕ l h ) a i 2, (30) If M denotes the number of elements in T h, which here is also the dimension of S 0 h ( ), then minimizing the functional (30) on K h (A) defines a quadratic programming problem with M linear constraints, given by (29). Since the quadratic functional V f,ϱ (ϕ) is H ε/2 ( )-elliptic, Theorem 5.2 in [9] implies the following statement on convergence.

12 408 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Lemma 13. For any ϱ 0 fixed, the solutions ϕ ϱh of Problem 3 converge in H ε/2 ( ) as h 0 to ϕ ϱ, the solution of Problem 2. Because M, in general, is rather large, standard optimization methods for finding the minimizer of Problem 3 require long computing times (see [11], where just a few hundred degrees of freedom were used). Instead we apply the gradientprojection method in R M. To this end, insert into (30) the representation of ϕ h in terms of the basis of S 0 h ( ) given by the characteristic functions χ m of τ m T h, m = 1,..., M, where the coefficients x m are forming the vector x R M. Then one gets a discrete problem in the following form: Problem 4. Minimize x A 0 x 2f x + ϱ (g i x a i ) 2, 2 when x ranges over R M + := y R M : y j 0 for all j = 1,..., M. Here A 0 is the Galerkin M M matrix of V on, whose elements are given by α i α j (Vχ m, χ p ), where m L i, p L j, i, j I, while g i x and f x are the representations of l L i (g, ϕ l) h and l L i α i (f, ϕ l h ), respectively, in terms of the vector x. Note that the components of the vector g i that correspond to l L i are zero. Problem 4 is a problem in the Euclidian space R M with a positive definite symmetric matrix and, hence, it is of the type (3.8) in [9]. Therefore, we now may use the gradient-projection method in R M (see [9, Chapter I, (3.12)]): x k+1 := x k η A 0 x k + ϱ g i g i x k F, 2 (31) x k+1 := P R M + x k+1 for all k = 0, 1,..., where F := f + ϱ a i g i, 2 the projection P R M is defined by + yj if y (P R M y) j := j 0, + 0 if y j < 0, and η is appropriately chosen within the interval given by A = A 0 + ϱ g i g i R M R M, 2 0, 2γ h / A 2RM,RM. Here A is a full, symmetric, positive definite matrix R M,R M is the spectral norm, and γ h is a positive constant, not depending on x, such that x Ax γ h x 2 R M for all x R M. Observe that, obviously, P R M + R M,R M = 1. Due to the error analysis for Galerkin Bubnov methods applied to equations with strongly elliptic pseudodifferential operators, it follows from [26, Lemma 2.12] that the γ h can be chosen so that γ h = c 0 h α 1 > 0 (32) with a constant c 0 not depending on h. Therefore, x k in (31) converges in R M as the series k=0 (1 η)k. The matrix times vector multiplications in (31) can be executed extremely fast since we apply wavelet matrix compression to sparsify the system matrix A, cf. [12,13]. Nevertheless, because of (32), the convergence turns out to be rather slow. Future research will be spend on more elaborated and faster optimization methods. 8. Numerical results As in [11] we consider n = 3 and a single torus as +, i.e., I + = L + = {1} and =. We do not apply an external field, i.e. f = 0, and choose the total charge on the torus equal to 1, where g = 1. We compute the charge distribution for several settings of α, namely α = 2.9/2.5/2/1.1. The charge distribution is approximated by piecewise constant ansatz functions on a mesh which stems from dyadic subdivision of 9 four-sided curved coarse grid patches. To approximate efficiently the system matrix A, we apply a wavelet

13 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Fig. 1. Charge distribution in case α = 2.9. Fig. 2. Charge distribution in case α = 2.5. Fig. 3. Charge distribution in case α = 2.0. boundary element method [12,13]. Since the charges converge faster for smaller penalty (see [14]), we use a cascadic approach by increasing successively the penalty parameter ϱ during the iteration. We compute the approximate minimizer ϕ J = M J m=1 x J,mχ J,m on refinement level J = 5 which yields about M J piecewise constant boundary elements. The approximate solutions are shown in Figs. 1 4.

14 410 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Fig. 4. Charge distribution in case α = 1.1. Via coarsening we can restrict the minimizer ϕ J to the coarse grids which yields coarse grid functions ϕ j = M j m=1 x j,mχ j,m for j < 5. To illustrate the charge distribution in addition to the colors we also put a small ball at the vertices of those underlying elements τ j,k, k = 1,..., M j, where the charge satisfies x j,k j J max x J,m : m = 1,..., M J. By this procedure we arrive at the charge distributions seen in the figures below. As a comparison with the illustrations in [11] confirms, our method yields basically the same distributions as the ones obtained in [11]. Moreover, for 1 < α 2, the charges completely cover the toroidal surface whereas for 2 < α < 3, the support of ϕ 0 is a strict subset of in accordance with the results by Brauchart et al. in [27]. Acknowledgments The first author acknowledges the support of the DFG Cluster of Excellence Simulation Technologies. The third author acknowledges the support of the University of Stuttgart for her visits in Stuttgart. Appendix Proof of Lemma 11. Since ϕ 0 K(A, a, g) K(A), from (22) and (23) with ϕ = ϕ 0 we get V f,ϱ (A, a, g) = V f,ϱ (ϕ ϱ ) V f,ϱ (ϕ 0 ) = V f (ϕ 0 ) = V f (A, a, g), (33) the first of the proposed estimates. Observing that V f,ϱ (ϕ) V f (ϕ) (where ϕ K(A) is given), with the help of (5) we then obtain c V ϕ ϱ 2 H ε/2 ( ) ϕ ϱ 2 V = V f (ϕ ϱ ) + 2(f, ϕ ϱ ) L2 ( ) V f,ϱ (ϕ ϱ ) + 2(f, ϕ ϱ ) L2 ( ) V f (A, a, g) + 2c D f H ε/2 ( ) ϕ ϱ H ε/2 ( ), which yields the estimate ϕ ϱ H ε/2 ( ) C with C defined by (25). Since c V ϕ 0 2 H ε/2 ( ) ϕ 0 2 V = V f (ϕ 0 ) + 2(f, ϕ 0 ) L2 ( ) V f (A, a, g) + 2c D f H ε/2 ( ) ϕ 0 H ε/2 ( ), the estimate ϕ 0 H ε/2 ( ) C holds as well. Proof of Theorem 12. Since one gets ϱ V f,ϱ (ϕ ϱ ) = V f (ϕ ϱ ) + ϱ 2 δ2 = ϕ ϱ 2 V 2(f, ϕ ϱ) + ϱ 2 δ2, 2 δ2 V f,ϱ (ϕ ϱ ) + 2(f, ϕ ϱ ), which together with (33) and the a priori estimate for ϕ ϱ yields the desired relation (26). Since we consider increasing penalty ϱ, we now require ϱ > 4C 2 1 c c L 2, where a c V a 2 min := min a i. min (34)

15 H. Harbrecht et al. / J. Math. Anal. Appl. 393 (2012) Note that the modified solution ϕ ϱ satisfies the side conditions (22), and so ϕ ϱ K(A, a, g). Having denoted ψ := ϕ ϱ ϕ 0, we then get ϕ 0 + tψ K(A, a, g) for all t [0, 1]. The quadratic polynomial in t, defined as F(t) := V f,ϱ (ϕ 0 + tψ) V f,ϱ (ϕ 0 ) = V f (ϕ 0 + tψ) V f (ϕ 0 ), satisfies F(0) = 0 and F(t) 0 for all t [0, 1]. Hence, the right derivative at 0 is nonnegative, i.e., df/dt + (0) 0. Moreover, the second derivative is given by d 2 F/dt 2 = 2 ψ 2 V and it is constant in t. Then ψ 2 V = 1 2 F (0) F(0) + df dt + (0) F (0) = F(1) = V f (ϕ ϱ ) V f (ϕ 0 ) = V f,ϱ (ϕ ϱ ) ϱ 2 δ2 V f (A, a, g) + V f (ϕ ϱ ) V f (ϕ ϱ ). Due to (33), this implies the estimate ψ 2 V V f (ϕ ϱ ) V f (ϕ ϱ ) = ϕ ϱ 2 ϕ V ϱ 2 V 2(f,ϕ ϱ ϕ ϱ ) ϕ ϱ V + ϕ ϱ V + 2c D c 1/2 V f H ε/2 ( ) ϕ ϱ ϕ ϱ V. (35) On account of (34), from (a i 1) l Li l ϕ l ϱ g ds = a i l L i l ϕ l ϱ g ds = δ i we obtain Since a i 1 δ i a i + δ i 2 δ i a i. ϕ ϱ ϕ ϱ = α l (ϕ ϱ l ϕl ϱ ) = (a i 1) l L i l Li α l ϕ l ϱ, this together with (5), (26), and the inequality ϕϱ l H ε/2 ( l ) ϕ ϱ H ε/2 ( ) yields ϕ ϱ ϕ ϱ V = (a i 1) α l ϕϱ l 2δ ϕϱ l a V 2δc1/2 c ϕϱ l l Li min a H ε/2 ( l ) min and, by (26) and (34), 2δc1/2 c a min V 1/2 l L L 2δcc L ϕ ϱ H ε/2 ( ) c 1/2 V a min ϕ ϱ V ϕ ϱ V + ϕ ϱ ϕ ϱ V C + 2δCc1/2 c L c 1/2 V 2C. a min Returning to (35), from the last two relations we conclude that where ϕ ϱ V 2δc1/2 c l L L c 1/2 V C a min ϕ ϱ ϕ 0 2 = V ψ 2 V C 5 ϕ ϱ ϕ ϱ V C 2 2 δ, (36) C 5 := 3C + 2c D c 1/2 V f H ε/2 ( ) and C 2 := 2 2CC 5a 1 min c1/2 c L c 1/2 V. Now, substituting (26) into (36) gives (27) as desired. Finally, from what has already been proved we get V f (A, a, g) = V f (ϕ 0 ) V f (ϕ ϱ ) V f,ϱ (ϕ ϱ ) + V f (ϕ ϱ ) V f (ϕ ϱ ) V f,ϱ (A, a, g) + C 5 ϕ ϱ ϕ ϱ V. When combined with (27), this yields (28) and completes the proof of the theorem.

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