Technische Universität Graz

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1 Tecnisce Universität Graz Boundary element metods for Diriclet boundary control problems Günter Of, Tan Pan Xuan, Olaf Steinbac Bericte aus dem Institut für Numerisce Matematik Berict 2009/2

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3 Tecnisce Universität Graz Boundary element metods for Diriclet boundary control problems Günter Of, Tan Pan Xuan, Olaf Steinbac Bericte aus dem Institut für Numerisce Matematik Berict 2009/2

4 Tecnisce Universität Graz Institut für Numerisce Matematik Steyrergasse 30 A 8010 Graz WWW: ttp:// c Alle Recte vorbealten. Nacdruck nur mit Genemigung des Autors.

5 Boundary element metods for Diriclet boundary control problems Günter Of, Tan Pan Xuan, Olaf Steinbac Institut für Numerisce Matematik, TU Graz, Steyrergasse 30, 8010 Graz, Austria Abstract For te solution of elliptic Diriclet boundary control problems, we propose and analyze two boundary element approaces. Te state equation, te adjoint equation, and te optimality condition are rewritten as systems of boundary integral equations involving te standard boundary integral operators of te Laplace equation and of te Bi Laplace equation. Wile te first approac is based on te use of te weakly singular Bi Laplace boundary integral equation, te additional use of te ypersingular Bi Laplace boundary integral equation results in a symmetric formulation, wic is also symmetric in te discrete case. We prove te unique solvability of bot boundary integral approaces and discuss related boundary element discretizations. In particular, we prove stability and related error estimates wic are confirmed by a numerical example. 1 Introduction Optimal control problems of elliptic or parabolic partial differential equations wit a Diriclet boundary control play an important role, for example, in te context of computational fluid mecanics, see, e.g., [1, 6, 10]. A difficulty in te andling of Diriclet control problems by finite element metods lies in te essential caracter of Diriclet boundary conditions. Wile Neumann or Robin type boundary conditions can be incorporated naturally in te weak formulation of te state equation, given Diriclet data on te boundary ave to be extended into te domain in a suitable way. For a discussion of several finite element approaces for Diriclet boundary control problems, see, e.g., [2, 4, 8, 11, 14, 15, 16]. In most cases, te Diriclet control is considered in L 2 (), but te energy space H 1/2 () seems to be more natural. In [18], a finite element approac was considered, were te energy norm was realized by using some stabilized ypersingular boundary integral operator. 1

6 Since te unknown function in Diriclet boundary control problems is to be found on te boundary = of te computational domain R n, n = 2, 3, te use of boundary integral equations seems to be a natural coice. But to our knowledge, tere are only a few results known on te use of boundary integral equations to solve optimal boundary control problems, see, e.g., [5, 23] for problems wit point observations. In tis paper, we consider te Poisson equation as a model problem, owever, tis approac can be applied to any elliptic partial differential equation, if a fundamental solution is known. In tis case, solutions of partial differential equations can be described by te means of surface and volume potentials. To find te complete Caucy data, boundary integral equations ave to be solved. For an overview on boundary integral equations, see, e.g., [12, 17] and te references given terein. Te numerical solution of boundary integral equations results in boundary element metods, see, e.g., [20, 22]. In tis paper, we formulate and analyze a boundary element approac to solve Diriclet boundary control problems were te control is considered in te energy space H 1/2 (). Te model problem is described in Section 2, were we also discuss te adjoint problem wic caracterizes te solution of te reduced minimization problem. In Section 3, we present te representation formulae to describe te solutions of bot te primal and adjoint Diriclet boundary value problems. To find te unknown normal derivatives of te state variable and of te adjoint variable, weakly singular boundary integral equations are formulated. Since te state enters te adjoint boundary value problem as a volume density, an additional volume integral as to be considered. By applying integration by parts, tis Newton potential can be reformulated by using boundary potentials of te Bi Laplace operator. Hence we recall some properties of boundary integral operators for te Bi Laplace operator in Section 4. In Section 5, we analyze a first boundary integral formulation to solve te Diriclet boundary control problem, and we discuss stability and error estimates of te related Galerkin boundary element metod. Since tis boundary element approximation leads to a non symmetric matrix representation of a self adjoint operator, we introduce and analyze a symmetric boundary element approac, wic includes a second, te so called ypersingular boundary integral equation, in te optimality condition in Section 6. Again we discuss te related stability and error analysis. Finally, we present a numerical example in Section 7. 2 Diriclet control problems As a model problem, we consider te Diriclet boundary control problem to minimize J(u, z) = 1 [u(x) u(x)] 2 dx z 2 A for (u, z) H1 () H 1/2 () (2.1) subject to te constraint u(x) = f(x) for x, u(x) = z(x) for x =, (2.2) 2

7 were u L 2 () is a given target, f L 2 () is a given volume density, R + is a fixed parameter, and R n, n = 2, 3, is a bounded Lipscitz domain wit boundary =. Moreover, A is an equivalent norm in H 1/2 () wic is induced by an elliptic, self adjoint, and bounded operator A : H 1/2 () H 1/2 (), i.e., γ A 1 w 2 H 1/2 () Aw, w, Aw H 1/2 () γ A 2 w H 1/2 () for all w H 1/2 (). For example, we may consider te stabilized ypersingular boundary integral operator A = D, see [19], were and Dz, w := Dz, w + z, 1 w, 1 for all z, w H 1/2 () (Dz)(x) = n x n y U (x, y)z(y)ds y for x, 1 log x y for n = 2, U 2π (x, y) = 1 1 for n = 3 4π x y (2.3) is te fundamental solution of te Laplace operator [22]. Note tat for τ H 1/2 () and w H 1/2 () τ, w = τ(x)w(x)ds x denotes te related duality pairing. Let u f H0 1 () be te weak solution of te omogeneous Diriclet boundary value problem u f (x) = f(x) for x, u f (x) = 0 for x. Te solution of te Diriclet boundary value problem (2.2) is ten given by u = u z + u f, were u z H 1 () is te unique solution of te Diriclet boundary value problem u z (x) = 0 for x, u z (x) = z(x) for x. (2.4) Note tat te solution of te Diriclet boundary value problem (2.4) defines a linear map u z = Sz wit S : H 1/2 () H 1 () L 2 (). Ten, by using u = Sz + u f, we consider te problem to find te minimizer z H 1/2 () of te reduced cost functional J(z) = 1 [(Sz)(x) + u f (x) u(x)] 2 dx Az, z. (2.5) Since te reduced cost functional J( ) is convex, te unconstrained minimizer z can be found from te optimality condition S Sz + S (u f u) + Az = 0, (2.6) 3

8 were S : L 2 () H 1/2 () is te adjoint operator of S : H 1/2 () L 2 (), i.e., S ψ, ϕ = ψ, Sϕ = ψ(x)(sϕ)(x)dx for all ϕ H 1/2 (), ψ L 2 (). Note tat te operator T : A + S S : H 1/2 () H 1/2 () is bounded and H 1/2 () elliptic, see, e.g., [18]. Hence, te operator equation (2.6), i.e., T z = ( A + S S)z = S (ū u f ) =: g (2.7) admits a unique solution z H 1/2 (). Inserting te primal variable u = Sz + u f, and introducing te adjoint variable τ = S (u u) H 1/2 (), we ave to solve te coupled problem τ + Az = 0, τ = S (u u), u = Sz + u f (2.8) instead of (2.7) and (2.6), respectively. Note tat for given z H 1/2 () and f L 2 () te application of u = Sz +u f corresponds to te solution of te Diriclet boundary value problem (2.2). Te application of te adjoint operator τ = S (u u) is caracterized by te Neumann datum τ(x) = n x p(x) for x, were p is te unique solution of te adjoint Diriclet boundary value problem p(x) = u(x) ū(x) for x, p(x) = 0 for x. (2.9) Hence we can rewrite te optimality condition τ + Az = 0 as n x p(x) = (Az)(x) for x. (2.10) Terefore, we ave to solve a coupled system, in particular of te state equation (2.2), of te adjoint boundary value problem (2.9), and of te optimality condition (2.10), to find te minimizer (u, z) H 1 () H 1/2 () of te cost functional (2.1) subject to te constraint (2.2). Since te unknown control z H 1/2 () is considered on te boundary =, te use of boundary integral equations to solve bot te primal boundary value problem (2.2) and te adjoint boundary value problem (2.9) seems to be a natural coice. 3 Laplace boundary integral equations 3.1 Primal boundary value problem Te solution of te Diriclet boundary value problem (2.2), u(x) = f(x) for x, u(x) = z(x) for x, 4

9 is given by te representation formula for x, see, e.g., [22], u( x) = U ( x, y) u(y)ds y U ( x, y)z(y)ds y + n y n y U ( x, y)f(y)dy, (3.1) were U (x, y) is te fundamental solution of te Laplace operator as given in (2.3). To find te related Neumann datum t = u n H 1/2 () for a given Diriclet datum z H 1/2 (), we consider te representation formula (3.1) for x x to obtain te boundary integral equation z(x) = u(x) = U (x, y)t(y)ds y z(x) U (x, y)z(y)ds y + U (x, y)f(y)dy n y for almost all x, wic can be written as (V t)(x) = ( 1 2 I + K)z(x) (N 0f)(x) for x. (3.2) Note tat (V t)(x) = U (x, y)t(y)ds y for x is te Laplace single layer potential V : H 1/2 () H 1/2 () satisfying V t H 1/2 () c V 2 t H 1/2 () for all t H 1/2 (), and (Kz)(x) = n y U (x, y)z(y)ds y for x is te Laplace double layer potential K : H 1/2 () H 1/2 () satisfying ( 1 2 I + K)z H 1/2 () c K 2 z H 1/2 () for all z H 1/2 (). Moreover, (N 0 f)(x) = U (x, y)f(y)dy for x is te related Newton potential. Note tat te single layer potential V is H 1/2 () elliptic, see, e.g., [22], were for n = 2 we assume te scaling condition diam < 1 to ensure tis: V t, t c V 1 t 2 H 1/2 () for all t H 1/2 (). Note tat in general we ave te mapping properties V : H 1/2+s () H 1/2+s (), K : H 1/2+s () H 1/2+s (), were s 1 2 in te case of a Lipscitz boundary, see, e.g., [3, 12, 17]. 5

10 3.2 Adjoint boundary value problem Te solution of te adjoint Diriclet boundary value problem (2.9), p(x) = u(x) ū(x) for x, p(x) = 0 for x, is given correspondingly by te representation formula for x, p( x) = U ( x, y) p(y)ds y + U ( x, y)[u(y) ū(y)]dy. (3.3) n y As in (3.2), we obtain a boundary integral equation (V q)(x) = (N 0 ū)(x) (N 0 u)(x) for x (3.4) to determine te unknown Neumann datum q = n p H 1/2 (). Remark 3.1 Wile te boundary integral equation (3.2) can be used to determine te unknown Neumann datum t H 1/2 () of te primal Diriclet boundary value problem (2.2), te unknown Neumann datum q H 1/2 () of te adjoint Diriclet boundary value problem (2.9) is given as te solution of te boundary integral equation (3.4). Ten, te control z H 1/2 () is determined by te optimality condition (2.10). However, since te solution u of te primal Diriclet boundary value problem (2.2) enters te volume potential N 0 u in te boundary integral equation (3.4), we also need to include te representation formula (3.1). Hence we ave to solve a coupled system of boundary and domain integral equations. Instead, we will now describe a system of only boundary integral equations to solve te adjoint boundary value problem (2.9). To end up wit a system of boundary integral equations only, instead of (3.3), we will introduce a modified representation formula for te adjoint state p as follows. First we note tat 1 V (x, y) = 8π x y 2( log x y 1 ) for n = 2, (3.5) 1 8π x y for n = 3 is a solution of te Poisson equation y V (x, y) = U (x, y) for x y, (3.6) i.e., V (x, y) is te fundamental solution of te Bi Laplacian. Hence we can rewrite te volume integral for u in (3.3), by using Green s second formula, as follows: U ( x, y)u(y)dy = [ y V ( x, y)]u(y)dy = = n y V ( x, y)u(y)ds y n y V ( x, y)z(y)ds y V ( x, y) u(y)ds y + V ( x, y)[ u(y)]dy n y V ( x, y)t(y)ds y V ( x, y)f(y)dy. 6

11 Terefore, we now obtain from (3.3) te modified representation formula p( x) = U ( x, y)q(y)ds y + V ( x, y)z(y)ds y V ( x, y)t(y)ds y n y U ( x, y)ū(y)dy V ( x, y)f(y)dy (3.7) for x, were te volume potentials involve given data only, and q = p is te unknown n Neumann datum wic is related to te adjoint Diriclet boundary value problem (2.9). Te representation formula (3.7) results, wen taking te limit x x, in te boundary integral equation 0 = p(x) = U (x, y)q(y)ds y + V (x, y)z(y)ds y V (x, y)t(y)ds y n y U (x, y)ū(y)dy V (x, y)f(y)dy for almost all x, wic can be written as (V q)(x) = (V 1 t)(x) (K 1 z)(x) + (N 0 ū)(x) + (M 0 f)(x) for x. (3.8) Note tat (V 1 t)(x) = V (x, y)t(y)ds y for x is te Bi Laplace single layer potential V 1 : H 3/2 () H 3/2 () satisfying, see, for example, [12, Teorem 5.7.3], and V 1 t H 3/2 () c V 1 2 t H 3/2 () for all t H 3/2 (), (3.9) (K 1 z)(x) = n y V (x, y)z(y)ds y for x is te Bi Laplace double layer potential K 1 : H 1/2 () H 3/2 () satisfying K 1 z H 3/2 () c K 1 2 z H 1/2 () for all z H 1/2 (). (3.10) In addition, we ave introduced a second Newton potential, wic is related to te fundamental solution of te Bi Laplace operator, (M 0 f)(x) = V (x, y)f(y)dy for x. 7

12 3.3 Optimality system Now we are in a position to reformulate te primal Diriclet boundary value problem (2.2), te adjoint Diriclet boundary value problem (2.9), and te optimality condition (2.10) as a system of boundary integral equations for x, V 1 V K 1 V 1I K 2 I A t q z = N 0 u + M 0 f N 0 f 0. (3.11) To investigate te unique solvability of (3.11), we first consider te associated Scur complement of (3.11). Since te Laplace single layer potential V is H 1/2 () elliptic and terefore invertible, we first obtain t = V 1 ( 1 2 I + K)z V 1 N 0 f (3.12) from te second equation in (3.11). Inserting tis into te first equation of (3.11) gives and terefore V q = V 1 V 1 ( 1 2 I + K)z K 1z + N 0 u + M 0 f V 1 V 1 N 0 f, q = V 1 V 1 V 1 ( 1 2 I + K)z V 1 K 1 z + V 1 N 0 u + V 1 M 0 f V 1 V 1 V 1 N 0 f. (3.13) Hence it remains to solve te Scur complement system T z = g, (3.14) were T := V 1 K 1 V 1 V 1 V 1 ( 1 I + K) + A (3.15) 2 is te boundary integral representation of te operator T as defined in (2.7), and g := V 1 N 0 u + V 1 M 0 f V 1 V 1 V 1 N 0 f (3.16) is te related rigt and side. To investigate te unique solvability of te Scur complement boundary integral equation (3.14), we first will recall some mapping properties of boundary integral operators wic are related to te Bi Laplace partial differential equation, see also [13]. 4 Bi Laplace boundary integral equations In tis section, we will consider a representation formula and related boundary integral equations for te Bi Laplace equation 2 u(x) = 0 for x, (4.1) 8

13 wic can be written as a system, w(x) = 0, u(x) = w(x) for x. (4.2) As for te Laplace equation we first find a representation formula for x, w( x) = U ( x, y) w(y)ds y U ( x, y)w(y)ds y, (4.3) n y n y wic results in te boundary integral equation w(x) = (V τ)(x) + 1 w(x) (Kw)(x) for x. (4.4) 2 Note tat w = u and τ = w = n w = n u n are te associated Caucy data on. Wen taking te normal derivative of te representation formula (4.3), we get a second, te so called ypersingular boundary integral equation were τ(x) = 1 2 τ(x) + (K τ)(x) + (Dw)(x) for x, (4.5) (K τ)(x) = n x U (x, y)τ(y)ds y for x is te adjoint Laplace double layer potential K : H 1/2 () H 1/2 (), and (Dw)(x) = U (x, y)w(y)ds y for x n x n y is te related ypersingular boundary integral operator D : H 1/2 () H 1/2 (). To obtain a representation formula for te solution u of te Bi Laplace equation (4.1), we first consider te related Green s first formula u(y) v(y)dy = n y u(y) v(y)ds y n y v(y)u(y)ds y + and in te sequel Green s second formula, u(y) v(y)ds y v(y)u(y)ds y + [ 2 v(y)]u(y)dy n y n y = n y v(y) u(y)ds y 9 n y u(y)v(y)ds y + [ 2 v(y)]u(y)dy, [ 2 u(y)]v(y)dy. (4.6)

14 Wen coosing v(y) = V ( x, y) for x, i.e., te fundamental solution (3.5) of te Bi Laplace operator, te solution of te Bi Laplace partial differential equation (4.1) is given by te representation formula for x by u( x) = n y u(y) y V ( x, y)ds y By using (3.6), tis can be written as u( x) = U ( x, y)t(y)ds y n y V ( x, y) u(y)ds y + n y y V ( x, y)u(y)ds y n y u(y)v ( x, y)ds y. n y U ( x, y)u(y)ds y (4.7) n y V ( x, y)w(y)ds y + Hence we obtain te boundary integral equation V ( x, y)τ(y)ds y. u(x) = (V t)(x) u(x) (Ku)(x) (K 1w)(x) + (V 1 τ)(x) (4.8) for almost all x. Moreover, wen taking te normal derivative of te representation formula (4.7), tis gives anoter boundary integral equation for x, were t(x) = 1 2 t(x) + (K t)(x) + (Du)(x) + (D 1 w)(x) + (K 1 τ)(x), (4.9) (K 1 τ)(x) = n x V (x, y)τ(y)ds y for x is te adjoint Bi Laplace double layer potential K 1 : H 3/2 () H 1/2 (), and (D 1 w)(x) = V (x, y)w(y)ds y for x n x n y is te Bi Laplace ypersingular boundary integral operator D 1 : H 1/2 () H 1/2 (). Te boundary integral equations (4.4), (4.5), (4.8), and (4.9) can now be written as a system, including te so called Calderon projection C, 1 u I K V K 2 1 V 1 u t 1 w = D 2 I + K D 1 K 1 t 1 I K V w. (4.10) 2 1 τ D I + 2 K τ 10

15 Lemma 4.1 Te Calderon projection C as defined in (4.10) is a projection, i.e., C 2 = C. Proof. Te proof follows as in te case of te Laplace equation [17, 22], for te Bi Laplace equation see also [13]. From te projection property as stated in Lemma 4.1 we obtain some well known relations of all boundary integral operators wic were introduced for bot te Laplace and te Bi Laplace equation. Lemma 4.2 For all boundary integral operators tere old te relations and KV = V K, DK = K D, V D = 1 4 I K2, DV = 1 4 I K 2 (4.11) K 1 V V K 1 = V 1 K KV 1, (4.12) K 1 D DK 1 = D 1 K K D 1, (4.13) V D 1 + V 1 D + KK 1 + K 1 K = 0, (4.14) DV 1 + D 1 V + K K 1 + K 1K = 0. (4.15) Proof. Te relations of (4.11) for te Laplace operator are well known, see, e.g., [22], for te Bi Laplace operator, see also [13]. To prove te ellipticity of te Scur complement boundary integral operator T as defined in (3.15), we need te following result: Lemma 4.3 For any t H 1/2 () tere olds te equality Ṽ t 2 L 2 () = K 1V t, t V 1 ( 1 2 I + K )t, t (4.16) were (Ṽ t)(x) = U (x, y)t(y)ds y for x. Proof. For x and t H 1/2 (), we define te Bi Laplace single layer potential u t (x) = (Ṽ1t)(x) = V (x, y)t(y)ds y wic is a solution of te Bi Laplace differential equation (4.1). Ten, te related Caucy data are given by u t (x) = (V 1 t)(x), n x u t (x) = (K 1t)(x) for x. 11

16 On te oter and, for x w t (x) = x u t (x) = x V (x, y)t(y)ds y = U (x, y)t(y)ds y = (Ṽ t)(x) is a solution of te Laplace equation. Hence, te related Caucy data are given by w t (x) = (V t)(x), n x w t (x) = 1 2 t(x) + (K t)(x) for x. Now, for u = v = u t, Green s first formula (4.6) reads [ u t (x)] 2 dx = u t (x) u t (x)ds x n x n x u t (x)u t (x)ds x, and terefore we conclude [w t (x)] 2 dx = = u t (x)w t (x)ds x w t (x)u t (x)ds x n x n x (K 1t)(x)(V t)(x)ds x [ 1 2 t(x) + (K t)(x)](v 1 t)(x)ds x = K 1 t, V t 1 2 t + K t, V 1 t = t, K 1 V t V 1 ( 1 2 I + K )t, t. Te assertion follows wit w t = Ṽ t. 5 Non symmetric boundary integral formulation Now we able to prove te unique solvability of te Scur complement boundary integral equation (3.14), were te operator T is defined by (3.15). Teorem 5.1 Te composed boundary integral operator T := A + V 1 K 1 V 1 V 1 V 1 ( 1 2 I + K) : H1/2 () H 1/2 () is self adjoint, bounded and H 1/2 () elliptic, i.e., T z, z c T 1 z 2 H 1/2 () for all z H 1/2 (). 12

17 Proof. Te mapping properties of T : H 1/2 () H 1/2 () follow from te boundedness of all used boundary integral operators [17, 20, 22]. In addition, we use te compact embedding of H 3/2 () in H 1/2 (). Next we will sow te self adjointness of T. For u, v H 1/2 () we ave T u, v = Au, v + V 1 K 1 u, v 1 2 V 1 V 1 V 1 u, v V 1 V 1 V 1 Ku, v = u, Av + u, K 1V 1 v 1 2 u, V 1 V 1 V 1 v u, K V 1 V 1 V 1 v = u, Av 1 2 u, V 1 V 1 V 1 v + u, [K 1 V 1 K V 1 V 1 V 1 ]v. Now, we conclude by using te relations (4.11) and (4.12) K 1V 1 K V 1 V 1 V 1 = K 1V 1 V 1 KV 1 V 1 = V 1 [V K 1 KV 1 ]V 1 = V 1 [K 1 V V 1 K ]V 1 = V 1 K 1 V 1 V 1 K V 1 = V 1 K 1 V 1 V 1 V 1 K. Hence we ave T u, v = u, Av 1 2 u, V 1 V 1 V 1 v + u, [V 1 K 1 V 1 V 1 V 1 K]v = u, [ A + V 1 K 1 V 1 V 1 V 1 ( 1 2 I + K)]v = u, T v, i.e., T is self adjoint. Moreover, for z H 1/2 () we ave, by using (4.11), t = V 1 z, and by Lemma 4.3, T z, z = Az, z + V 1 K 1 z, z V 1 V 1 V 1 ( 1 2 I + K)z, z = Az, z + K 1 V V 1 z, V 1 z V 1 ( 1 2 I + K )V 1 z, V 1 z = Az, z + K 1 V t, t V 1 ( 1 2 I + K )t, t = Az, z + Ṽ t 2 L 2 () z 2 A, i.e., te H 1/2 () ellipticity of T, since A defines an equivalent norm in H 1/2 (). Due to Teorem 5.1, we conclude te unique solvability of te Scur complement boundary integral equation (3.14) by applying te Lax Milgram lemma and terefore of te coupled system (3.11). 5.1 Galerkin boundary element discretization For te Galerkin discretization of (3.14) based on te boundary integral representation (3.15), let S 1 H() = span{ϕ i } M i=1 H 1/2 () 13

18 be some boundary element space of, e.g., piecewise linear and continuous basis functions ϕ i, wic are defined wit respect to a globally quasi uniform and sape regular boundary element mes of mes size H. Te Galerkin discretization of te Scur complement system (3.14) is to find z H SH 1 () suc tat T z H, v H = g, v H for all v H SH 1 (). (5.1) Wile te Galerkin variational formulation (5.1) admits a unique solution z H due to Cea s lemma satisfying te error estimate te composed boundary integral operator z z H H 1/2 () c inf z v H H (), (5.2) v H SH 1 1/2 () T := A + V 1 K 1 V 1 V 1 V 1 ( 1 2 I + K) does not allow a direct boundary element discretization in general. Instead, we may introduce an appropriate boundary element approximation T as follows. 5.2 Boundary element approximation of T For an arbitrary but fixed z H 1/2 (), te application of T z reads T z = Az + V 1 K 1 z V 1 V 1 V 1 ( 1 2 I + K)z = Az + q z, were q z, t z H 1/2 () are te unique solutions of te boundary integral equations For a Galerkin approximation of (5.3), let V q z = K 1 z V 1 t z, V t z = ( 1 I + K)z. (5.3) 2 S 0 () = span{ψ k} N k=1 H 1/2 () be anoter boundary element space of, e.g., piecewise constant basis functions ψ k, wic are defined wit respect to a second globally quasi uniform and sape regular boundary element mes of mes size. Now, t z, S 0 () is te unique solution of te Galerkin formulation V t z,, τ = ( 1 2 I + K)z, τ for all τ S 0 (). (5.4) Moreover, q z, S 0 () is te unique solution of te Galerkin formulation V q z,, τ = K 1 z V 1 t z,, τ for all τ S 0 (). (5.5) Hence we can define an approximation T of te operator T by T z := Az + q z,. (5.6) 14

19 Lemma 5.2 Te approximate operator T : H 1/2 () H 1/2 () as defined in (5.6) is bounded, i.e., T z H 1/2 () c e T 2 z H 1/2 () for all z H 1/2 (). Proof. From te Galerkin formulation (5.4) we first find, by coosing τ = t z, and by using te H 1/2 () ellipticity of te single layer potential, t z, H 1/2 () 1 c V 1 ( 1 2 I + K)z H 1/2 () ck 2 c V 1 z H 1/2 (). From (5.5), we now find for τ = q z,, by using H 3/2 () H 1/2 () and (3.9), (3.10), q z, H 1/2 () 1 c V 1 1 c V 1 1 c V 1 K 1 z V 1 t z, H 1/2 () K 1 z V 1 t z, H 3/2 () [ c K 1 2 z H 1/2 () + c V 1 2 t z, H 3/2 ()]. Te assertion now follows from H 1/2 () H 1/2 () and H 1/2 () H 3/2 (). Lemma 5.3 Let T : H 1/2 () H 1/2 () be given by (3.15), and let T be defined by (5.6). Ten tere olds te error estimate T z T z H 1/2 () cv 2 c V 1 inf q z τ H τ S 0 1/2 () () + cv1 2 c V 1 t z t z, H 3/2 (), (5.7) were q z, t z H 1/2 () are defined as in (5.3), and t z, S 0 () is te unique solution of te Galerkin variational problem (5.4). Proof. For an arbitrary cosen but fixed z H 1/2 () we ave, by definition, T z = Az + q z, q z = V 1 [K 1 z V 1 t z ], t z = V 1 ( 1 2 I + K)z. In particular, t z H 1/2 () is te unique solution of te variational problem V t z, τ = ( 1 2 I + K)z, τ for all τ H 1/2 (), and q z H 1/2 () is te unique solution of te variational problem V q z, τ = K 1 z, τ V 1 t z, τ for all τ H 1/2 (). By using definition (5.6), we also ave T z = Az + q z,, 15

20 were q z, is te unique solution of te Galerkin variational problem V q z,, τ = K 1 z, τ V 1 t z,, τ for all τ S 0 (), and t z, S 0 () is te unique solution of te variational problem V t z,, τ = ( 1 2 I + K)z, τ for all τ S 0 (). By applying Cea s lemma, we first conclude te error estimate t z t z, H 1/2 () cv 2 c V 1 inf t z τ H τ S 0 1/2 () (). Let us furter define q z, S 0 () as te unique solution of te variational problem V q z,, τ = K 1 z, τ V 1 t z, τ for all τ S 0 (). (5.8) Again, by using Cea s lemma we ave q z q z, H 1/2 () cv 2 c V 1 inf τ S 0 () q z τ H 1/2 (). By subtracting (5.5) from (5.8) we obtain te perturbed Galerkin ortogonality V (q z, q z, ), τ = V 1 (t z, t z ), τ for all τ S 0 (), from wic we furter conclude te estimate q z, q z, H 1/2 () 1 V c V 1 (t z t z, ) H 1/2 () 1 1 c V 1 V 1 (t z t z, ) H 3/2 () cv 2 1 c V 1 Hence we find, by applying te triangle inequality, t z t z, H 3/2 (). q z q z, H 1/2 () cv 2 c V 1 inf q z τ H τ S 0 1/2 () () + cv1 2 c V 1 t z t z, H 3/2 (), and te assertion follows from T z T z = q z q z,. By using te approximation property of te trial space S 0 () and te Aubin Nitsce trick, we conclude an error estimate from (5.7) wen assuming some regularity of q z and t z, respectively. Corollary 5.4 Assume q z, t z Hpw s () for some s [0, 1]. Ten tere olds te error estimate T z T z H 1/2 () c 1 s+1 2 qz H s pw () + c 2 s+3 2 tz H s pw (). (5.9) 16

21 5.3 Boundary element approximation of g As in (5.6), we may also define a boundary element approximation of te rigt and side g as defined in (3.16) g = V 1 N 0 ū + V 1 M 0 f V 1 V 1 V 1 N 0 f. In particular, g H 1/2 () is te unique solution of te variational problem V g, τ = N 0 ū + M 0 f V 1 V 1 N 0 f, τ = N 0 ū + M 0 f, τ V 1 t f, τ for all τ H 1/2 (), were t f = V 1 N 0 f H 1/2 () solves te variational problem V t f, τ = N 0 f, τ for all τ H 1/2 (). Hence we can define a boundary element approximation g S 0 () as te unique solution of te Galerkin variational problem V g, τ = N 0 ū + M 0 f, τ V 1 t f,, τ for all τ S 0 (), (5.10) were t f, S 0 () is te unique solution te Galerkin problem V t f,, τ = N 0 f, τ for all τ S 0 (). (5.11) Lemma 5.5 Let g be te rigt and side as defined by (3.16), and let g be te boundary element approximation as defined in (5.10). Ten tere olds te error estimate g g H 1/2 () cv 2 c V 1 inf g τ H τ S 0 1/2 () () + cv1 2 t f t f, H (). (5.12) 3/2 Proof. In addition to (5.10), let us consider te Galerkin formulation to find g S 0 () suc tat V g, τ = N 0 ū + M 0 f, τ V 1 t f, τ for all τ S 0 (). (5.13) Again, by using Cea s lemma, we obtain g g H 1/2 () cv 2 c V 1 c V 1 inf τ S 0 () g τ H 1/2 (). Subtracting (5.10) from (5.13) gives te perturbed Galerkin ortogonality V (g g ), τ = V 1 (t f t f, ), τ for all τ S 0 (). For τ = g g and by using te H 1/2 () ellipticity of te single layer potential V and te estimate (3.9), we furter obtain g g H 1/2 () 1 V c V 1 (t f t f, ) H 1/2 () cv 1 2 t 1 c V f t f, H (). 3/2 1 Te assertion finally follows from te triangle inequality. By using te approximation property of te trial space S 0 () and te Aubin Nitsce trick, we conclude an error estimate from (5.12) wen assuming some regularity of g and t f, respectively. 17

22 Corollary 5.6 Assume g, t f H s pw() for some s [0, 1]. Ten tere olds te error estimate g g H 1/2 () c 1 s+1 2 g H s pw () + c 2 s+3 2 tf H s pw (). (5.14) 5.4 Perturbed Galerkin variational problem Instead of te Galerkin variational problem (5.1), we now consider a perturbed Galerkin formulation to find z H SH 1 () suc tat T z H, v H = g, v H for all v H S 1 H(). (5.15) By combining te boundary element approximations (5.4) and (5.5) wit (5.10) and (5.11), it is sufficient to consider te Galerkin boundary element formulation of (3.11): Find (t, q, z H ) S 0() S0 () S1 H () suc tat V 1 t, w + V q, w + K 1 z H, w = N 0 ū + M 0 f, w, (5.16) V t, τ ( 1 2 I + K) z H, τ = N 0 f, τ, (5.17) q, v H + A z H, v H = 0 (5.18) is satisfied for all (w, τ, v H ) S 0() S0 () S1 H (). Te Galerkin formulation (5.16) (5.18) is equivalent to a system of linear equations were and V 1, V K 1, V ( 1 M 2 + K ) M A H t q z = V [l, k] = V ψ k, ψ l, K [l, i] = Kϕ i, ψ l, V 1, [l, k] = V 1 ψ k, ψ l, K 1, [l, i] = K 1 ϕ i, ψ l, A H [j, i] = Aϕ i, ϕ j, M [l, i] = ϕ i, ψ l, f 1,l = N 0 ū + M 0 f, ψ l, f 1 f 2 0 f 2,l = N 0 f, ψ l, (5.19) for k, l = 1,...,N and i, j = 1,...,M. Since te Laplace single layer potential V is H 1/2 () elliptic, te related Galerkin matrix V is positive definite and terefore invertible. Hence, we can resolve te second equation in (5.19) to obtain t = V 1 (1 2 M + K ) z + V 1 f. 2 Inserting tis into te first equation of (5.19) gives V q = V 1, V 1 (1 2 M + K ) z + V 1, V 1 f K 2 1, z + f 1 18

23 and terefore q = V 1 V 1,V 1 (1 2 M + K ) z + V 1 V 1,V 1 Hence it remains to solve te Scur complement system [ ] A H M V 1 V 1,V 1 were T,H = A H M V 1 (1 2 M + K ) + M V 1 K 1, V 1,V 1 f 2 V 1 K 1,z + V 1 f 1. [ ] z = M V 1 f 1 + V 1, V 1 f 2 (5.20) (1 2 M + K ) + M V 1 K 1, (5.21) defines a non symmetric Galerkin boundary element approximation of te self adjoint Scur complement boundary integral operator T. Teorem 5.7 Te approximate Scur complement T,H as defined in (5.21) is positive definite, i.e., ( T,H z, z) 1 2 ct 1 z H 2 H 1/2 () for all z R M z H S 1 H (), if c 0H is sufficiently small. Proof. For an arbitrary cosen but fixed z R M let z H SH 1 () be te associated boundary element function. Ten we ave ( T,H z, z) = T z H, z H = T z H, z H (T T )z H, z H c T 1 z H 2 H 1/2 () (T T )z H H 1/2 () z H H 1/2 (). Since z H S 1 H () is a continuous function, we ave z H H 1 (). Hence we find and t zh = V 1 ( 1 2 I + K)z H L 2 (), q zh = V 1 [K 1 z H V 1 t zh ] L 2 () according to (5.4) and (5.5). Terefore we can apply te error estimate (5.9) for s = 0 to obtain T z H T z H H 1/2 () c 1 1/2 q zh L2 () + c 2 3/2 t zh L2 () c 3 1/2 z H H 1 (). Now, by applying te inverse inequality for S 1 H (), z H H 1 () c I H 1/2 z H H 1/2 (), 19

24 we obtain T z H T z H H 1/2 () c 3 c I ( H ) 1/2 z H H 1/2 (). Hence we finally obtain ( T,H z, z) [ c T 1 c 3c I ( H ) 1/2 ] z H 2 H 1/2 () 1 2 ct 1 z H 2 H 1/2 (), if is satisfied. c 3 c I ( H ) 1/2 1 2 ct 1 Note tat Teorem 5.7 ensures te unique solvability of te linear system (5.20) and terefore of te perturbed variational problem (5.15). Since te approximate operator T is S 1 H () elliptic, an error estimate for te approximate solution z H of te perturbed Galerkin variational problem (5.15) follows from te Strang lemma, see, e.g., [22, Teorem 8.2, Teorem 8.3], wic reads for our problem as: Teorem 5.8 Let z be te unique solution of te operator equation (3.14). Let c 0 H be satisfied suc tat te approximate Scur complement T,H as defined in (5.21) is positive definite, and let z H SH 1 () be te unique solution of te perturbed Galerkin variational formulation (5.15). Ten tere olds te error estimate z z H H 1/2 () c 1 inf z v H H v H SH 1 1/2 () () + c 2 (T T )z H 1/2 () + c 3 g g H (). 1/2 (5.22) Corollary 5.9 Wen combining te error estimate (5.22) wit te approximation property of te ansatz space SH 1 (), and wit te error estimates (5.9) and (5.14), we finally obtain te error estimate z z H H 1/2 () c 1 H 3/2 z H 2 () + c 2 3/2 q z H 1 pw () + c 3 5/2 t z H 1 pw () +c 4 3/2 g H 1 pw () + c 5 5/2 t f H 1 pw () wen assuming z H 2 (), and q z, t z, g, t f H 1 pw (), respectively. For c 0H we can expect te convergence rate 1.5 wen measuring te error in te energy norm z z H H 1/2 () c(z, u, f) H 3/2. (5.23) Moreover, we are also able to derive an error estimate in L 2 (), i.e., wen applying te Aubin Nitsce trick [22]. z z H L2 () c(z, u, f) H 2. (5.24) 20

25 Remark 5.1 Te error estimates (5.23) and (5.24) provide optimal convergence rates wen approximating te control z by using piecewise linear basis functions. However, we ave to assume c 0 H to ensure te unique solvability of te perturbed Galerkin formulation (5.15), were te constant c 0 is in general unknown. Moreover, te matrix T,H as given in (5.21) defines a non symmetric approximation of te exact symmetric stiffness matrix T,H as used in (5.1). Hence we are interested in deriving a symmetric boundary element metod wic is stable witout any additional constraints in te coice of te boundary element trial spaces. 6 Symmetric boundary integral formulation Te boundary integral formulation of te primal boundary value problem (2.2) is given by (3.2), wile te adjoint boundary value problem (2.9) corresponds to te modified boundary integral equation (3.8). In wat follows, we will rewrite te optimality condition (2.10) by using a ypersingular boundary integral equation for te adjoint problem to obtain a symmetric boundary integral formulation for te coupled problem. Since te adjoint variable p, as defined in te representation formula (3.7), is a solution of te adjoint Diriclet boundary value problem (2.9), te normal derivative q(x) = lim n x ex p( x) = ex x n p(x) for x is well defined. Wen computing te normal derivative of te representation formula (3.7), tis gives a second boundary integral equation for x q(x) = ( 1 2 I + K )q(x) (D 1 z)(x) (K 1t)(x) (N 1 ū)(x) (M 1 f)(x), (6.1) were we introduce Newton potentials for x (N 1 ū)(x) = lim n x ex U ( x, y)ū(y)dy ex x for x and (M 1 f)(x) = lim n x ex ex x V ( x, y)f(y)dy for x in addition to te boundary integral operators used in (4.10). Combining te optimality condition (2.10) and te boundary integral equation (6.1) gives a boundary integral equation for x, (Az)(x) = ( 1 2 I + K )q(x) (D 1 z)(x) (K 1 t)(x) (N 1ū)(x) (M 1 f)(x). (6.2) 21

26 Now, to find te yet unknown triple (z, t, q) H 1/2 () H 1/2 () H 1/2 () we solve te system of boundary integral equations (3.2), (3.8), and (6.2) wic can be written as V 1 V K 1 t N 0 ū + M 0 f V 1I K q = N 2 0 f. (6.3) K 1 1I 2 K A + D 1 z N 1 ū M 1 f To investigate te unique solvability of (6.3), we consider te related Scur complement. As in (3.12) and (3.13) we obtain and t = V 1 ( 1 2 I + K)z V 1 N 0 f q = V 1 V 1 V 1 ( 1 2 I + K)z V 1 K 1 z + V 1 N 0 ū + V 1 M 0 f V 1 V 1 V 1 N 0 f. Hence it remains to solve te Scur complement system [ A + D 1 + K 1 V 1 ( 12 I + K) + (12 I + K )V 1 K 1 ( 12 I + K )V 1 V 1 V 1 ( 12 I + K) ] z = K 1 V 1 N 0 f N 1 ū M 1 f + ( 1 2 I + K )V 1 [ N 0 ū + M 0 f V 1 V 1 N 0 f ]. (6.4) Note tat (6.4) corresponds to a symmetric boundary integral formulation of te operator equation (2.7) representing te optimality condition. Teorem 6.1 Te composed boundary integral operator T = A+D 1 ( 1 2 I +K )V 1 V 1 V 1 ( 1 2 I +K)+K 1 V 1 ( 1 2 I +K)+(1 2 I +K )V 1 K 1 (6.5) is self adjoint, bounded, i.e., T : H 1/2 () H 1/2 (), and H 1/2 () elliptic. Proof. Wile te self adjointness of T in te symmetric representation (6.5) is obvious, te boundedness and ellipticity estimates follow as in te proof of Teorem 5.1. In particular, te Scur complement operators T in te symmetric representation (6.5) and in te non symmetric representation (3.15) coincide. Indeed, by using (4.11) and (4.12) we obtain T = A + D 1 ( 1 2 I + K )V 1 V 1 V 1 ( 1 2 I + K) + K 1V 1 ( 1 2 I + K) + (1 2 I + K )V 1 K 1 = A + D 1 + [K 1 ( 12 ] I + K )V 1 V 1 V 1 ( 1 2 I + K) + (1 2 I + K )V 1 K 1 = A + D 1 + V [V 1 K 1 KV 1 1 ] 2 V 1 V 1 ( 1 2 I + K) + (1 2 I + K )V 1 K 1 ] = A + D 1 + V 1 [K 1 V V 1 K 1 2 V 1 V 1 ( 1 2 I + K) + (1 2 I + K )V 1 K 1 = A + D 1 + V 1 K 1 ( 1 2 I + K) V 1 V 1 ( 1 2 I + K )V 1 ( 1 2 I + K) + (1 2 I + K )V 1 K 1. 22

27 Due to te representation of te Laplace Steklov Poincaré operator, see, e.g., [22], we furter conclude S = V 1 ( 1 2 I + K) = D + (1 2 I + K )V 1 ( 1 2 I + K), ( 1 2 I + K )V 1 ( 1 2 I + K) = V 1 ( 1 I + K) D. 2 Terefore, by using (4.11) and (4.14) we ave T = A + D 1 + V 1 K 1 ( 1 [ 2 I + K) V 1 V 1 V 1 ( 1 ] 2 I + K) D + V 1 ( 1 2 I + K)K 1 ] = A + V 1 [V D 1 + V 1 D + K 1 ( 1 2 I + K) V 1V 1 ( 1 2 I + K) + (1 2 I + K)K 1 ] = A + V [ KK 1 1 K 1 K + K 1 ( 1 2 I + K) V 1V 1 ( 1 2 I + K) + (1 2 I + K)K 1 [ = A + V 1 K 1 V 1 V 1 ( 1 ] 2 I + K), and we finally obtain te non symmetric representation (3.15). Terefore, te ellipticity of T follows as in Teorem 5.1. Due to te H 1/2 () ellipticity of te symmetric representation (6.5) of T, we can conclude te unique solvability of te Scur complement boundary integral equation (6.4), and terefore of te coupled system (6.3). 6.1 Galerkin boundary element discretization In wat follows, we will consider a boundary element discretization of te boundary integral equation system (6.3). Again, let S 0 () = span{ψ k} N k=1 H 1/2 (), S 1 () = span{ϕ i} M i=1 H1/2 () be some boundary element spaces of piecewise constant and piecewise linear basis functions ψ k and ϕ i, wic are defined wit respect to some admissible boundary element mes of mes size. Te Galerkin boundary element formulation of (6.3) ten reads to find (t, q, ẑ ) S 0() S0 () S1 () suc tat V 1 t, w + V q, w + K 1 ẑ, w = N 0 ū + M 0 f, w, (6.6) V t, τ ( 1 2 I + K)ẑ, τ = N 0 f, τ, (6.7) K 1 t, v ( 1 2 I + K )q, v + ( A + D 1 )ẑ, v = N 1 ū + M 1 f, v (6.8) 23

28 is satisfied for all (w, τ, v ) S 0() S0 () S1 (). Te Galerkin formulation (6.6) (6.8) is equivalent to a system of linear equations, V 1, V K 1, V ( 1M 2 + K ) K 1, ( 1 2 M + K ) A + D 1, t q ẑ = were we used, in addition to tose entries of te linear system (5.19), D 1, [j, i] = D 1 ϕ i, ϕ j, f 3,j = N 1 ū + M 1 f, ϕ j for i, j = 1,...,M. f 1 f 2 f 3, (6.9) To investigate te unique solvability of te linear system (6.9), we consider te invertibility of te related Scur complement. In particular, te second equation in (6.9) gives and we obtain from te first equation q = V 1 [V 1,V 1 t = V 1 (1 2 M + K )ẑ + V 1 f, 2 (1 2 M + K ) K 1, ]ẑ + V 1 f + V 1 1 V 1,V 1 f 2. Hence, by inserting tese results into te tird equation of (6.9), we finally end up wit te Scur complement system of te symmetric boundary integral formulation were te Scur complement is given by T, ẑ = f, (6.10) T, = A + D 1, + K1,V 1 (1 2 M + K ) + ( 1 2 M + K )V 1 K 1, (6.11) ( 1 2 M + K )V 1 V 1,V 1 (1 2 M + K ), and te rigt and side is f = f 3 K 1,V 1 Lemma 6.2 Te symmetric matrix f + (1 2 2 M + K )V 1 f + (1 1 2 M + K )V 1 V 1,V 1 f 2. T = T, A = D 1, + K1, V 1 (1 2 M + K ) + ( 1 2 M + K 1 )V K 1, ( 1 2 M + K is positive semi definite, i.e., all eigenvalues of T are non negative, ( T z, z) 0 for all z R M. 1 )V V 1,V 1 (1 2 M + K ) 24

29 Proof. We consider te generalized eigenvalue problem [ T z = µ S + ( 1 ] 2 M + K 1 )V (1 2 M + K ) z, (6.12) were te stabilized discrete Steklov Poincaré operator S = D + ( 1 2 M + K )V 1 (1 2 M + K ) is symmetric and positive definite. Since te eigenvalue problem (6.12) can be written as ( ( 1 M 2 + K 1 )V I )( ) ( V 1, K 1, V 1 (1M 2 + K ) D 1, I = µ ( ( 1 2 M + K K 1, ) z )V 1 I )( V S ) ( V 1 (1 2 M + K ) I ) z, it is sufficient to consider te generalized eigenvalue problem ( ) ( ) ( V1, K 1, w V w = µ D 1, z S z K 1, ), (6.13) were From (6.13), we ten conclude w = V 1 (1 2 M + K )z. and by taking te difference we obtain (K 1, z, w) (V 1, w, w) = µ(v w, w), (K 1, w, z) + (D 1,z, z) = µ( S z, z), (D 1, z, z) + (V 1, w, w) = µ[( S z, z) (V w, w)] = µ( D z, z). Since te Galerkin matrices D 1, and V 1, of te Bi Laplace boundary integral operators D 1 and V 1 are positive semi definite and positive definite, respectively, and since te stabilized Laplace ypersingular integral operator D is positive definite, we conclude Hence, we finally obtain ( T z, z) µ min [ for all z R M. ( S z, z) + (V 1 µ 0. (1 2 M + K )z, ( 1 ] 2 M + K )z) As a corollary of Lemma 6.2, we find te positive definiteness of te symmetric Scur complement matrix T, as defined in (6.11). 25 0

30 Corollary 6.3 Te approximate Scur complement T, as defined in (6.11) is positive definite, i.e., for all z R M z S 1 (). ( T, z, z) (A z, z) = Az, z γ A 1 z 2 H 1/2 () Hence, we can ensure te unique solvability of te Scur complement system (6.10) and terefore of te system (6.9) as well as of te Galerkin variational problem (6.6) (6.8). To find an error estimate for te approximate solution ẑ S 1 (), as for te non symmetric boundary element formulation, we will consider a perturbed variational problem of te operator equation (6.4) leading to te Scur complement system (6.10). 6.2 Symmetric boundary element approximation of T For an arbitrary but fixed given z H 1/2 () te application of T z reads, by using te symmetric representation (6.5), T z = Az + D 1 z + K 1 t z ( 1 2 I + K )q z, were q z, t z H 1/2 () are te unique solutions of te boundary integral equations (5.3). Hence, by using te unique solutions q z,, t z, S 0 () of te related Galerkin variational formulations (5.4) and (5.5) we can define te approximation T z := Az + D 1 z + K 1 t z, ( 1 2 I + K ) q z,. (6.14) Lemma 6.4 Te approximate operator T : H 1/2 () H 1/2 () as defined in (6.14) is bounded, i.e., T z H 1/2 () c b T 2 z H 1/2 () for all z H 1/2 (). Moreover, tere olds te error estimate T z T z H 1/2 () c 1 inf q z τ H τ S 0 1/2 () () + c 2 t z t z, H (). (6.15) 3/2 Proof. Te proof follows as for te boundary element approximation of te non symmetric formulation, see Lemma 5.2 and Lemma 5.3. By using te approximation property of te trial space S 0 () and te Aubin Nitsce trick, we ten conclude an error estimate from (6.15) wen assuming some regularity of q z and t z, respectively. Corollary 6.5 Assume q z, t z Hpw s () for some s [0, 1]. Ten tere olds te error estimate T z T z H 1/2 () c 1 s+1 2 qz H s pw () + c 2 s+3 2 tz H s pw (). (6.16) 26

31 6.3 Boundary element approximation of g As in te approximation (6.14), we can define a boundary element approximation of te related rigt and side, see (6.4), g = K 1 V 1 N 0 f N 1 u M 1 f + ( 1 2 I + K )V 1 [N 0 u + M 0 f V 1 V 1 N 0 f] = K 1 t f N 1 u M 1 f + ( 1 2 I + K )q f, were t f = V 1 N 0 f H 1/2 () is te unique solution of te variational problem V t f, τ = N 0 f, τ for all τ H 1/2 (), and q f = V 1 [N 0 u + M 0 f V 1 t f ] H 1/2 () is te unique solution of te variational problem V q f, τ = N 0 u + M 0 f V 1 t f, τ for all τ H 1/2 (). Hence we can define an approximation ĝ := K 1t f, N 1 u M 1 f + ( 1 2 I + K ) q f,, (6.17) were q f, S 0 () is te unique solution of te Galerkin variational problem V q f,, τ = N 0 u + M 0 f V 1 t f,, τ for all τ S 0 (), and t f, S 0 () is te unique solution of te Galerkin variational problem As in (5.14), we conclude te error estimate V t f,, τ = N 0 f, τ for all τ S 0 (). g ĝ H 1/2 () c 1 s+1 2 qf H s pw () + c 2 s+3 2 tf H s pw () (6.18) wen assuming q f, t f Hpw s () for some s [0, 1]. 6.4 Perturbed Galerkin variational problem We now consider a perturbed Galerkin formulation instead of te operator equation (6.4) to find z S 1 () suc tat T z, v = ĝ, v for all v S 1 (). (6.19) Note tat te Galerkin discretization of te perturbed variational problem (6.19) results in te linear system (6.10). Now we are in a position to formulate an error estimate for te approximate solution z by applying te Strang lemma. 27

32 Teorem 6.6 Let z H 1/2 () be te unique solution of te operator equation (6.4), and let z S 1 () be te unique solution of te perturbed Galerkin variational problem (6.19). Ten tere olds te error estimate z ẑ H 1/2 () c 1 inf z v H ()+c v S 1 1/2 2 (T T )z H ()+c 1/2 3 g ĝ H (). (6.20) 1/2 () Corollary 6.7 Wen combining te error estimate (6.20) wit te approximation property of te ansatz space S 1 (), and wit te error estimates (6.16) and (6.18), we finally obtain te error estimate z ẑ H 1/2 () c 1 3/2 z H 2 () + c 2 3/2 q z H 1 pw () + c 3 5/2 t z H 1 pw () +c 4 3/2 q f H 1 pw () + c 5 5/2 t f H 1 pw () wen assuming z H 2 (), and q z, t z, q f, t f Hpw 1 (), respectively. Hence we can expect te convergence order 1.5 wen measuring te error in te energy norm, z ẑ H 1/2 () c(z, u, f) 3/2. (6.21) Moreover, applying te Aubin Nitsce trick [22] we are also able to derive an error estimate in L 2 (), i.e., z ẑ L2 () c(z, u, f) 2. (6.22) 7 Numerical results We consider te Diriclet boundary control problem (2.1) and (2.2) for te domain = (0, 1 2 )2 R 2 were ( u(x) = ) [x 1 (1 2x 1 ) + x 2 (1 2x 2 )], f(x) = 8, = For te boundary element discretization, we introduce a uniform triangulation of = on several levels were te mes size is L = 2 (L+1). Since te minimizer of (2.1) is not known in tis case, we use te boundary element solution z of te 9t level as reference solution. Te boundary element discretization is done by using te trial space S 0 () of piecewise constant basis functions, and S 1 () of piecewise linear and continuous functions. In particular we use te same boundary element mes to approximate te control z by a piecewise linear approximation, and piecewise constant approximations for te fluxes t and q. Note tat we ave = H in tis case, and terefore we can not ensure te S 1 () ellipticity of te non symmetric boundary element approximation, see Teorem 5.7. However, te numerical example sows stability. In Table 1, we present te errors for te control z in te L 2 () norm and te estimated order of convergence (eoc). Tese results correspond to te error estimate (5.24) of te non symmetric boundary element approximation, and to te error estimate (6.22) of te 28

33 Non symmetric BEM (5.20) Symmetric BEM (6.10) FEM [18] L z L z 9 L2 () eoc ẑ L ẑ 9 L2 () eoc z FEM L z FEM 9 L2 () eoc Table 1: Comparison of BEM/FEM errors of te Diriclet control. symmetric boundary element approximation. In addition, we give te error of te related finite element solution, see [18]. Te results sow a quadratic order of convergence, wic confirm te teoretical estimates. Note tat in te finite element approac only a convergence order of 1.5 can be proved [18]. In Table 2, we present te errors for te flux t of te primal boundary value problem, again in te L 2 () norm. Since te computation of t corresponds to te solution of a Diriclet boundary value problem wit approximated Diriclet data, we can expect and observe a linear order of convergence wen using piecewise constant basis functions, see, e.g., [22]. Non symmetric BEM (5.20) Symmetric BEM (6.10) L t L t 9 L2 () eoc t L t 9 L2 () eoc Table 2: Comparison of non symmetric/symmetric BEM. Acknowledgement Tis work as been supported by te Austrian Science Fund (FWF) under te Grant SFB Matematical Optimization and Applications in Biomedical Sciences, Subproject Fast Finite Element and Boundary Element Metods for Optimality Systems. Te autors would like to tank K. Kunisc for elpful discussions. 29

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