Karl–Franzens Universitat GrazTechnische Universitat Graz

Medizinische Universitat Graz

SpezialForschungsBereich F32

Boundary element methods for

Dirichlet boundary control

problems

Gunther Of Thanh Phan Xuan

Olaf Steinbach

SFB-Report No. 2009–003 February 2009

A–8010 GRAZ, HEINRICHSTRASSE 36, AUSTRIA

Supported by theAustrian Science Fund (FWF)

SFB sponsors:

• Austrian Science Fund (FWF)

• University of Graz

• Graz University of Technology

• Medical University of Graz

• Government of Styria

• City of Graz

Boundary element methods for

Dirichlet boundary control problems

Gunther Of, Thanh Phan Xuan, Olaf Steinbach

Institut fur Numerische Mathematik, TU Graz,Steyrergasse 30, 8010 Graz, Austria

of,thanh.phanxuan,o.steinbach@tugraz.at

Abstract

For the solution of elliptic Dirichlet boundary control problems, we propose and

analyze two boundary element approaches. The state equation, the adjoint equation,

and the optimality condition are rewritten as systems of boundary integral equations

involving the standard boundary integral operators of the Laplace equation and of

the Bi–Laplace equation. While the first approach is based on the use of the weakly

singular Bi–Laplace boundary integral equation, the additional use of the hyper-

singular Bi–Laplace boundary integral equation results in a symmetric formulation,

which is also symmetric in the discrete case. We prove the unique solvability of both

boundary integral approaches and discuss related boundary element discretizations.

In particular, we prove stability and related error estimates which are confirmed by

a numerical example.

1 Introduction

Optimal control problems of elliptic or parabolic partial differential equations with a Dirich-let boundary control play an important role, for example, in the context of computationalfluid mechanics, see, e.g., [1, 6, 10]. A difficulty in the handling of Dirichlet control problemsby finite element methods lies in the essential character of Dirichlet boundary conditions.While Neumann or Robin type boundary conditions can be incorporated naturally in theweak formulation of the state equation, given Dirichlet data on the boundary have to beextended into the domain in a suitable way. For a discussion of several finite element ap-proaches for Dirichlet boundary control problems, see, e.g., [2, 4, 8, 11, 14, 15, 16]. In mostcases, the Dirichlet control is considered in L2(Γ), but the energy space H1/2(Γ) seems tobe more natural. In [18], a finite element approach was considered, where the energy normwas realized by using some stabilized hypersingular boundary integral operator.

1

Since the unknown function in Dirichlet boundary control problems is to be found onthe boundary Γ = ∂Ω of the computational domain Ω ⊂ R

n, n = 2, 3, the use of boundaryintegral equations seems to be a natural choice. But to our knowledge, there are only afew results known on the use of boundary integral equations to solve optimal boundarycontrol problems, see, e.g., [5, 23] for problems with point observations. In this paper, weconsider the Poisson equation as a model problem, however, this approach can be appliedto any elliptic partial differential equation, if a fundamental solution is known. In thiscase, solutions of partial differential equations can be described by the means of surfaceand volume potentials. To find the complete Cauchy data, boundary integral equationshave to be solved. For an overview on boundary integral equations, see, e.g., [12, 17] andthe references given therein. The numerical solution of boundary integral equations resultsin boundary element methods, see, e.g., [20, 22].

In this paper, we formulate and analyze a boundary element approach to solve Dirichletboundary control problems where the control is considered in the energy space H1/2(Γ).The model problem is described in Section 2, where we also discuss the adjoint problemwhich characterizes the solution of the reduced minimization problem. In Section 3, wepresent the representation formulae to describe the solutions of both the primal and ad-joint Dirichlet boundary value problems. To find the unknown normal derivatives of thestate variable and of the adjoint variable, weakly singular boundary integral equationsare formulated. Since the state enters the adjoint boundary value problem as a volumedensity, an additional volume integral has to be considered. By applying integration byparts, this Newton potential can be reformulated by using boundary potentials of the Bi–Laplace operator. Hence we recall some properties of boundary integral operators for theBi–Laplace operator in Section 4. In Section 5, we analyze a first boundary integral formu-lation to solve the Dirichlet boundary control problem, and we discuss stability and errorestimates of the related Galerkin boundary element method. Since this boundary elementapproximation leads to a non–symmetric matrix representation of a self–adjoint operator,we introduce and analyze a symmetric boundary element approach, which includes a sec-ond, the so–called hypersingular boundary integral equation, in the optimality conditionin Section 6. Again we discuss the related stability and error analysis. Finally, we presenta numerical example in Section 7.

2 Dirichlet control problems

As a model problem, we consider the Dirichlet boundary control problem to minimize

J(u, z) =1

2

∫

Ω

[u(x) − u(x)]2dx+1

2‖z‖2

A for (u, z) ∈ H1(Ω) ×H1/2(Γ) (2.1)

subject to the constraint

−∆u(x) = f(x) for x ∈ Ω, u(x) = z(x) for x ∈ Γ = ∂Ω, (2.2)

2

where u ∈ L2(Ω) is a given target, f ∈ L2(Ω) is a given volume density, ∈ R+ is afixed parameter, and Ω ⊂ R

n, n = 2, 3, is a bounded Lipschitz domain with boundaryΓ = ∂Ω. Moreover, ‖ · ‖A is an equivalent norm in H1/2(Γ) which is induced by an elliptic,self–adjoint, and bounded operator A : H1/2(Γ) → H−1/2(Γ), i.e.,

γA1 ‖w‖

2H1/2(Γ) ≤ 〈Aw,w〉Γ, ‖Aw‖H−1/2(Γ) ≤ γA

2 ‖w‖H1/2(Γ) for all w ∈ H1/2(Γ).

For example, we may consider the stabilized hypersingular boundary integral operatorA = D, see [19],

〈Dz, w〉Γ := 〈Dz,w〉Γ + 〈z, 1〉Γ〈w, 1〉Γ for all z, w ∈ H1/2(Γ)

where

(Dz)(x) = −∂

∂nx

∫

Γ

∂

∂nyU∗(x, y)z(y)dsy for x ∈ Γ,

and

U∗(x, y) =

−1

2πlog |x− y| for n = 2,

1

4π

1

|x− y|for n = 3

(2.3)

is the fundamental solution of the Laplace operator [22]. Note that for τ ∈ H−1/2(Γ) andw ∈ H1/2(Γ)

〈τ, w〉Γ =

∫

Γ

τ(x)w(x)dsx

denotes the related duality pairing.Let uf ∈ H1

0 (Ω) be the weak solution of the homogeneous Dirichlet boundary valueproblem

−∆uf (x) = f(x) for x ∈ Ω, uf(x) = 0 for x ∈ Γ.

The solution of the Dirichlet boundary value problem (2.2) is then given by u = uz + uf ,where uz ∈ H1(Ω) is the unique solution of the Dirichlet boundary value problem

−∆uz(x) = 0 for x ∈ Ω, uz(x) = z(x) for x ∈ Γ. (2.4)

Note that the solution of the Dirichlet boundary value problem (2.4) defines a linear mapuz = Sz with S : H1/2(Γ) → H1(Ω) ⊂ L2(Ω). Then, by using u = Sz + uf , we considerthe problem to find the minimizer z ∈ H1/2(Γ) of the reduced cost functional

J(z) =1

2

∫

Ω

[(Sz)(x) + uf(x) − u(x)]2dx+1

2 〈Az, z〉Γ. (2.5)

Since the reduced cost functional J(·) is convex, the unconstrained minimizer z can befound from the optimality condition

S∗Sz + S∗(uf − u) + Az = 0, (2.6)

3

where S∗ : L2(Ω) → H−1/2(Γ) is the adjoint operator of S : H1/2(Γ) → L2(Ω), i.e.,

〈S∗ψ, ϕ〉Γ = 〈ψ,Sϕ〉Ω =

∫

Ω

ψ(x)(Sϕ)(x)dx for all ϕ ∈ H1/2(Γ), ψ ∈ L2(Ω).

Note that the operator

T : A+ S∗S : H1/2(Γ) → H−1/2(Γ)

is bounded and H1/2(Γ)–elliptic, see, e.g., [18]. Hence, the operator equation (2.6), i.e.,

Tz = (A + S∗S)z = S∗(u− uf) =: g (2.7)

admits a unique solution z ∈ H1/2(Γ). Inserting the primal variable u = Sz + uf , andintroducing the adjoint variable τ = S∗(u − u) ∈ H−1/2(Γ), we have to solve the coupledproblem

τ + Az = 0, τ = S∗(u− u), u = Sz + uf (2.8)

instead of (2.7) and (2.6), respectively. Note that for given z ∈ H1/2(Γ) and f ∈ L2(Ω)the application of u = Sz+uf corresponds to the solution of the Dirichlet boundary valueproblem (2.2). The application of the adjoint operator τ = S∗(u − u) is characterized bythe Neumann datum

τ(x) = −∂

∂nxp(x) for x ∈ Γ,

where p is the unique solution of the adjoint Dirichlet boundary value problem

−∆p(x) = u(x) − u(x) for x ∈ Ω, p(x) = 0 for x ∈ Γ. (2.9)

Hence we can rewrite the optimality condition τ + Az = 0 as

∂

∂nx

p(x) = (Az)(x) for x ∈ Γ. (2.10)

Therefore, we have to solve a coupled system, in particular of the state equation (2.2),of the adjoint boundary value problem (2.9), and of the optimality condition (2.10), tofind the minimizer (u, z) ∈ H1(Ω) × H1/2(Γ) of the cost functional (2.1) subject to theconstraint (2.2). Since the unknown control z ∈ H1/2(Γ) is considered on the boundaryΓ = ∂Ω, the use of boundary integral equations to solve both the primal boundary valueproblem (2.2) and the adjoint boundary value problem (2.9) seems to be a natural choice.

3 Laplace boundary integral equations

3.1 Primal boundary value problem

The solution of the Dirichlet boundary value problem (2.2),

−∆u(x) = f(x) for x ∈ Ω, u(x) = z(x) for x ∈ Γ,

4

is given by the representation formula for x ∈ Ω, see, e.g., [22],

u(x) =

∫

Γ

U∗(x, y)∂

∂nyu(y)dsy −

∫

Γ

∂

∂nyU∗(x, y)z(y)dsy +

∫

Ω

U∗(x, y)f(y)dy, (3.1)

where U∗(x, y) is the fundamental solution of the Laplace operator as given in (2.3). To findthe related Neumann datum t = ∂

∂nu ∈ H−1/2(Γ) for a given Dirichlet datum z ∈ H1/2(Γ),

we consider the representation formula (3.1) for Ω ∋ x → x ∈ Γ to obtain the boundaryintegral equation

z(x) = u(x) =

∫

Γ

U∗(x, y)t(y)dsy +1

2z(x) −

∫

Γ

∂

∂ny

U∗(x, y)z(y)dsy +

∫

Ω

U∗(x, y)f(y)dy

for almost all x ∈ Γ, which can be written as

(V t)(x) = (1

2I +K)z(x) − (N0f)(x) for x ∈ Γ. (3.2)

Note that

(V t)(x) =

∫

Γ

U∗(x, y)t(y)dsy for x ∈ Γ

is the Laplace single layer potential V : H−1/2(Γ) → H1/2(Γ) satisfying

‖V t‖H1/2(Γ) ≤ cV2 ‖t‖H−1/2(Γ) for all t ∈ H−1/2(Γ),

and

(Kz)(x) =

∫

Γ

∂

∂nyU∗(x, y)z(y)dsy for x ∈ Γ

is the Laplace double layer potential K : H1/2(Γ) → H1/2(Γ) satisfying

‖(1

2I +K)z‖H1/2(Γ) ≤ cK2 ‖z‖H1/2(Γ) for all z ∈ H1/2(Γ).

Moreover,

(N0f)(x) =

∫

Ω

U∗(x, y)f(y)dy for x ∈ Γ

is the related Newton potential. Note that the single layer potential V is H−1/2(Γ)–elliptic,see, e.g., [22], where for n = 2 we assume the scaling condition diam Ω < 1 to ensure this:

〈V t, t〉Γ ≥ cV1 ‖t‖2H−1/2(Γ) for all t ∈ H−1/2(Γ).

Note that in general we have the mapping properties

V : H−1/2+s(Γ) → H1/2+s(Γ), K : H1/2+s(Γ) → H1/2+s(Γ),

where |s| ≤ 12

in the case of a Lipschitz boundary Γ, see, e.g., [3, 12, 17].

5

3.2 Adjoint boundary value problem

The solution of the adjoint Dirichlet boundary value problem (2.9),

−∆p(x) = u(x) − u(x) for x ∈ Ω, p(x) = 0 for x ∈ Γ,

is given correspondingly by the representation formula for x ∈ Ω,

p(x) =

∫

Γ

U∗(x, y)∂

∂nyp(y)dsy +

∫

Ω

U∗(x, y)[u(y)− u(y)]dy. (3.3)

As in (3.2), we obtain a boundary integral equation

(V q)(x) = (N0u)(x) − (N0u)(x) for x ∈ Γ (3.4)

to determine the unknown Neumann datum q = ∂∂np ∈ H−1/2(Γ).

Remark 3.1 While the boundary integral equation (3.2) can be used to determine the

unknown Neumann datum t ∈ H−1/2(Γ) of the primal Dirichlet boundary value problem

(2.2), the unknown Neumann datum q ∈ H−1/2(Γ) of the adjoint Dirichlet boundary value

problem (2.9) is given as the solution of the boundary integral equation (3.4). Then, the

control z ∈ H1/2(Γ) is determined by the optimality condition (2.10). However, since the

solution u of the primal Dirichlet boundary value problem (2.2) enters the volume potential

N0u in the boundary integral equation (3.4), we also need to include the representation

formula (3.1). Hence we have 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 the adjoint boundary value problem (2.9).

To end up with a system of boundary integral equations only, instead of (3.3), we willintroduce a modified representation formula for the adjoint state p as follows. First wenote that

V ∗(x, y) =

−1

8π|x− y|2

(log |x− y| − 1

)for n = 2,

1

8π|x− y| for n = 3

(3.5)

is a solution of the Poisson equation

∆yV∗(x, y) = U∗(x, y) for x 6= y, (3.6)

i.e., V ∗(x, y) is the fundamental solution of the Bi–Laplacian. Hence we can rewrite thevolume integral for u in (3.3), by using Green’s second formula, as follows:

∫

Ω

U∗(x, y)u(y)dy =

∫

Ω

[∆yV∗(x, y)]u(y)dy

=

∫

Γ

∂

∂nyV ∗(x, y)u(y)dsy −

∫

Γ

V ∗(x, y)∂

∂nyu(y)dsy +

∫

Ω

V ∗(x, y)[∆u(y)]dy

=

∫

Γ

∂

∂ny

V ∗(x, y)z(y)dsy −

∫

Γ

V ∗(x, y)t(y)dsy −

∫

Ω

V ∗(x, y)f(y)dy .

6

Therefore, we now obtain from (3.3) the modified representation formula

p(x) =

∫

Γ

U∗(x, y)q(y)dsy +

∫

Γ

∂

∂ny

V ∗(x, y)z(y)dsy −

∫

Γ

V ∗(x, y)t(y)dsy

−

∫

Ω

U∗(x, y)u(y)dy −

∫

Ω

V ∗(x, y)f(y)dy (3.7)

for x ∈ Ω, where the volume potentials involve given data only, and q = ∂∂np is the unknown

Neumann datum which is related to the adjoint Dirichlet boundary value problem (2.9).The representation formula (3.7) results, when taking the limit Ω ∋ x→ x ∈ Γ, in the

boundary integral equation

0 = p(x) =

∫

Γ

U∗(x, y)q(y)dsy +

∫

Γ

∂

∂nyV ∗(x, y)z(y)dsy −

∫

Γ

V ∗(x, y)t(y)dsy

−

∫

Ω

U∗(x, y)u(y)dy −

∫

Ω

V ∗(x, y)f(y)dy

for almost all x ∈ Γ, which can be written as

(V q)(x) = (V1t)(x) − (K1z)(x) + (N0u)(x) + (M0f)(x) for x ∈ Γ. (3.8)

Note that

(V1t)(x) =

∫

Γ

V ∗(x, y)t(y)dsy for x ∈ Γ

is the Bi–Laplace single layer potential V1 : H−3/2(Γ) → H3/2(Γ) satisfying, see, for exam-ple, [12, Theorem 5.7.3],

‖V1t‖H3/2(Γ) ≤ cV1

2 ‖t‖H−3/2(Γ) for all t ∈ H−3/2(Γ), (3.9)

and

(K1z)(x) =

∫

Γ

∂

∂nyV ∗(x, y)z(y)dsy for x ∈ Γ

is the Bi–Laplace double layer potential K1 : H−1/2(Γ) → H3/2(Γ) satisfying

‖K1z‖H3/2(Γ) ≤ cK1

2 ‖z‖H−1/2(Γ) for all z ∈ H−1/2(Γ). (3.10)

In addition, we have introduced a second Newton potential, which is related to the funda-mental solution of the Bi–Laplace operator,

(M0f)(x) =

∫

Ω

V ∗(x, y)f(y)dy for x ∈ Γ.

7

3.3 Optimality system

Now we are in a position to reformulate the primal Dirichlet boundary value problem (2.2),the adjoint Dirichlet boundary value problem (2.9), and the optimality condition (2.10) asa system of boundary integral equations for x ∈ Γ,

−V1 V K1

V −12I −K

−I A

t

q

z

=

N0u+M0f

−N0f

0

. (3.11)

To investigate the unique solvability of (3.11), we first consider the associated Schur com-plement of (3.11). Since the Laplace single layer potential V is H−1/2(Γ)–elliptic andtherefore invertible, we first obtain

t = V −1(1

2I +K)z − V −1N0f (3.12)

from the second equation in (3.11). Inserting this into the first equation of (3.11) gives

V q = V1V−1(

1

2I +K)z −K1z +N0u+M0f − V1V

−1N0f,

and therefore

q = V −1V1V−1(

1

2I +K)z − V −1K1z + V −1N0u+ V −1M0f − V −1V1V

−1N0f . (3.13)

Hence it remains to solve the Schur complement system

Tz = g, (3.14)

where

T := V −1K1 − V −1V1V−1(

1

2I +K) + A (3.15)

is the boundary integral representation of the operator T as defined in (2.7), and

g := V −1N0u+ V −1M0f − V −1V1V−1N0f (3.16)

is the related right hand side.To investigate the unique solvability of the Schur complement boundary integral equa-

tion (3.14), we first will recall some mapping properties of boundary integral operatorswhich are related to the Bi–Laplace partial differential equation, see also [13].

4 Bi–Laplace boundary integral equations

In this section, we will consider a representation formula and related boundary integralequations for the Bi–Laplace equation

∆2u(x) = 0 for x ∈ Ω, (4.1)

8

which can be written as a system,

∆w(x) = 0, ∆u(x) = w(x) for x ∈ Ω. (4.2)

As for the Laplace equation we first find a representation formula for x ∈ Ω,

w(x) =

∫

Γ

U∗(x, y)∂

∂ny

w(y)dsy −

∫

Γ

∂

∂ny

U∗(x, y)w(y)dsy, (4.3)

which results in the boundary integral equation

w(x) = (V τ)(x) +1

2w(x) − (Kw)(x) for x ∈ Γ. (4.4)

Note that

w = ∆u and τ =∂

∂nw = n · ∇w = n · ∇∆u

are the associated Cauchy data on Γ. When taking the normal derivative of the repre-sentation formula (4.3), we get a second, the so–called hypersingular boundary integralequation

τ(x) =1

2τ(x) + (K ′τ)(x) + (Dw)(x) for x ∈ Γ, (4.5)

where

(K ′τ)(x) =

∫

Γ

∂

∂nx

U∗(x, y)τ(y)dsy for x ∈ Γ

is the adjoint Laplace double layer potential K ′ : H−1/2(Γ) → H−1/2(Γ), and

(Dw)(x) = −∂

∂nx

∫

Γ

∂

∂ny

U∗(x, y)w(y)dsy for x ∈ Γ

is the related hypersingular boundary integral operator D : H1/2(Γ) → H−1/2(Γ).To obtain a representation formula for the solution u of the Bi–Laplace equation (4.1),

we first consider the related Green’s first formula∫

Ω

∆u(y)∆v(y)dy =

∫

Γ

∂

∂nyu(y)∆v(y)dsy −

∫

Γ

∂

∂ny∆v(y)u(y)dsy +

∫

Ω

[∆2v(y)]u(y)dy,

(4.6)and in the sequel Green’s second formula,

∫

Γ

∂

∂nyu(y)∆v(y)dsy −

∫

Γ

∂

∂ny∆v(y)u(y)dsy +

∫

Ω

[∆2v(y)]u(y)dy

=

∫

Γ

∂

∂ny

v(y)∆u(y)dsy −

∫

Γ

∂

∂ny

∆u(y)v(y)dsy +

∫

Ω

[∆2u(y)]v(y)dy .

9

When choosing v(y) = V ∗(x, y) for x ∈ Ω, i.e., the fundamental solution (3.5) of the Bi–Laplace operator, the solution of the Bi–Laplace partial differential equation (4.1) is givenby the representation formula for x ∈ Ω by

u(x) =

∫

Γ

∂

∂nyu(y)∆yV

∗(x, y)dsy −

∫

Γ

∂

∂ny∆yV

∗(x, y)u(y)dsy

−

∫

Γ

∂

∂nyV ∗(x, y)∆u(y)dsy +

∫

Γ

∂

∂ny∆u(y)V ∗(x, y)dsy .

By using (3.6), this can be written as

u(x) =

∫

Γ

U∗(x, y)t(y)dsy −

∫

Γ

∂

∂nyU∗(x, y)u(y)dsy (4.7)

−

∫

Γ

∂

∂ny

V ∗(x, y)w(y)dsy +

∫

Γ

V ∗(x, y)τ(y)dsy .

Hence we obtain the boundary integral equation

u(x) = (V t)(x) +1

2u(x) − (Ku)(x) − (K1w)(x) + (V1τ)(x) (4.8)

for almost all x ∈ Γ. Moreover, when taking the normal derivative of the representationformula (4.7), this gives another boundary integral equation for x ∈ Γ,

t(x) =1

2t(x) + (K ′t)(x) + (Du)(x) + (D1w)(x) + (K ′

1τ)(x), (4.9)

where

(K ′

1τ)(x) =

∫

Γ

∂

∂nxV ∗(x, y)τ(y)dsy for x ∈ Γ

is the adjoint Bi–Laplace double layer potential K ′1 : H−3/2(Γ) → H1/2(Γ), and

(D1w)(x) = −∂

∂nx

∫

Γ

∂

∂ny

V ∗(x, y)w(y)dsy for x ∈ Γ

is the Bi–Laplace hypersingular boundary integral operator D1 : H−1/2(Γ) → H1/2(Γ).The boundary integral equations (4.4), (4.5), (4.8), and (4.9) can now be written as a

system, including the so–called Calderon projection C,

u

t

w

τ

=

12I −K V −K1 V1

D 12I +K ′ D1 K ′

112I −K V

D 12I +K ′

u

t

w

τ

. (4.10)

10

Lemma 4.1 The Calderon projection C as defined in (4.10) is a projection, i.e., C2 = C.

Proof. The proof follows as in the case of the Laplace equation [17, 22], for the Bi–Laplaceequation see also [13].

From the projection property as stated in Lemma 4.1 we obtain some well–known relationsof all boundary integral operators which were introduced for both the Laplace and the Bi–Laplace equation.

Lemma 4.2 For all boundary integral operators there hold the relations

KV = V K ′, DK = K ′D, V D =1

4I −K2, DV =

1

4I −K ′2 (4.11)

and

K1V − V K ′

1 = V1K′ −KV1, (4.12)

K ′

1D −DK1 = D1K −K ′D1, (4.13)

V D1 + V1D +KK1 +K1K = 0, (4.14)

DV1 +D1V +K ′K ′

1 +K ′

1K′ = 0. (4.15)

Proof. The relations of (4.11) for the Laplace operator are well–known, see, e.g., [22],for the Bi–Laplace operator, see also [13].

To prove the ellipticity of the Schur complement boundary integral operator T as definedin (3.15), we need the following result:

Lemma 4.3 For any t ∈ H−1/2(Γ) there holds the equality

‖V t‖2L2(Ω) = 〈K1V t, t〉Γ − 〈V1(

1

2I +K ′)t, t〉Γ (4.16)

where

(V t)(x) =

∫

Γ

U∗(x, y)t(y)dsy for x ∈ Ω .

Proof. For x ∈ Ω and t ∈ H−1/2(Γ), we define the Bi–Laplace single layer potential

ut(x) = (V1t)(x) =

∫

Γ

V ∗(x, y)t(y)dsy

which is a solution of the Bi–Laplace differential equation (4.1). Then, the related Cauchydata are given by

ut(x) = (V1t)(x),∂

∂nxut(x) = (K ′

1t)(x) for x ∈ Γ.

11

On the other hand, for x ∈ Ω

wt(x) = ∆xut(x) = ∆x

∫

Γ

V ∗(x, y)t(y)dsy =

∫

Γ

U∗(x, y)t(y)dsy = (V t)(x)

is a solution of the Laplace equation. Hence, the related Cauchy data are given by

wt(x) = (V t)(x),∂

∂nx

wt(x) =1

2t(x) + (K ′t)(x) for x ∈ Γ.

Now, for u = v = ut, Green’s first formula (4.6) reads

∫

Ω

[∆ut(x)]2dx =

∫

Γ

∂

∂nxut(x)∆ut(x)dsx −

∫

Γ

∂

∂nx∆ut(x)ut(x)dsx,

and therefore we conclude∫

Ω

[wt(x)]2dx =

∫

Γ

∂

∂nx

ut(x)wt(x)dsx −

∫

Γ

∂

∂nx

wt(x)ut(x)dsx

=

∫

Γ

(K ′

1t)(x)(V t)(x)dsx −

∫

Γ

[1

2t(x) + (K ′t)(x)](V1t)(x)dsx

= 〈K ′

1t, V t〉Γ − 〈1

2t+K ′t, V1t〉Γ

= 〈t,K1V t〉Γ − 〈V1(1

2I +K ′)t, t〉Γ.

The assertion follows with wt = V t.

5 Non–symmetric boundary integral formulation

Now we able to prove the unique solvability of the Schur complement boundary integralequation (3.14), where the operator T is defined by (3.15).

Theorem 5.1 The composed boundary integral operator

T := A+ V −1K1 − V −1V1V−1(

1

2I +K) : H1/2(Γ) → H−1/2(Γ)

is self–adjoint, bounded and H1/2(Γ)–elliptic, i.e.,

〈Tz, z〉Γ ≥ cT1 ‖z‖2H1/2(Γ) for all z ∈ H1/2(Γ).

12

Proof. The mapping properties of T : H1/2(Γ) → H−1/2(Γ) follow from the boundednessof all used boundary integral operators [17, 20, 22]. In addition, we use the compactembedding of H3/2(Γ) in H1/2(Γ).

Next we will show the self–adjointness of T. For u, v ∈ H1/2(Γ) we have

〈Tu, v〉Γ = 〈Au, v〉Γ + 〈V −1K1u, v〉Γ −1

2〈V −1V1V

−1u, v〉Γ − 〈V −1V1V−1Ku, v〉Γ

= 〈u, Av〉Γ + 〈u,K ′

1V−1v〉Γ −

1

2〈u, V −1V1V

−1v〉Γ − 〈u,K ′V −1V1V−1v〉Γ

= 〈u, Av〉Γ −1

2〈u, V −1V1V

−1v〉Γ + 〈u, [K ′

1V−1 −K ′V −1V1V

−1]v〉Γ.

Now, we conclude by using the relations (4.11) and (4.12)

K ′

1V−1 −K ′V −1V1V

−1 = K ′

1V−1 − V −1KV1V

−1 = V −1[V K ′

1 −KV1]V−1

= V −1[K1V − V1K′]V −1 = V −1K1 − V −1V1K

′V −1

= V −1K1 − V −1V1V−1K .

Hence we have

〈Tu, v〉Γ = 〈u, Av〉Γ −1

2〈u, V −1V1V

−1v〉Γ + 〈u, [V −1K1 − V −1V1V−1K]v〉Γ

= 〈u, [A+ V −1K1 − V −1V1V−1(

1

2I +K)]v〉Γ = 〈u, Tv〉Γ,

i.e., T is self–adjoint.Moreover, for z ∈ H1/2(Γ) we have, by using (4.11), t = V −1z, and by Lemma 4.3,

〈Tz, z〉Γ = 〈Az, z〉Γ + 〈V −1K1z, z〉Γ − 〈V −1V1V−1(

1

2I +K)z, z〉Γ

= 〈Az, z〉Γ + 〈K1V V−1z, V −1z〉Γ − 〈V1(

1

2I +K ′)V −1z, V −1z〉Γ

= 〈Az, z〉Γ + 〈K1V t, t〉Γ − 〈V1(1

2I +K ′)t, t〉Γ

= 〈Az, z〉Γ + ‖V t‖2L2(Ω)

≥ ‖z‖2A,

i.e., the H1/2(Γ)–ellipticity of T, since ‖ · ‖A defines an equivalent norm in H1/2(Γ).

Due to Theorem 5.1, we conclude the unique solvability of the Schur complement boundaryintegral equation (3.14) by applying the Lax–Milgram lemma and therefore of the coupledsystem (3.11).

5.1 Galerkin boundary element discretization

For the Galerkin discretization of (3.14) based on the boundary integral representation(3.15), let

S1H(Γ) = spanϕi

Mi=1 ⊂ H1/2(Γ)

13

be some boundary element space of, e.g., piecewise linear and continuous basis functions ϕi,which are defined with respect to a globally quasi–uniform and shape regular boundaryelement mesh of mesh size H . The Galerkin discretization of the Schur complement sys-tem (3.14) is to find zH ∈ S1

H(Γ) such that

〈TzH , vH〉Γ = 〈g, vH〉Γ for all vH ∈ S1H(Γ). (5.1)

While the Galerkin variational formulation (5.1) admits a unique solution zH due to Cea’slemma satisfying the error estimate

‖z − zH‖H1/2(Γ) ≤ c infvH∈S1

H(Γ)‖z − vH‖H1/2(Γ), (5.2)

the composed boundary integral operator

T := A+ V −1K1 − V −1V1V−1(

1

2I +K)

does not allow a direct boundary element discretization in general. Instead, we may intro-duce an appropriate boundary element approximation T as follows.

5.2 Boundary element approximation of T

For an arbitrary but fixed z ∈ H1/2(Γ), the application of Tz reads

Tz = Az + V −1K1z − V −1V1V−1(

1

2I +K)z = Az + qz,

where qz, tz ∈ H−1/2(Γ) are the unique solutions of the boundary integral equations

V qz = K1z − V1tz, V tz = (1

2I +K)z. (5.3)

For a Galerkin approximation of (5.3), let

S0h(Γ) = spanψk

Nk=1 ⊂ H−1/2(Γ)

be another boundary element space of, e.g., piecewise constant basis functions ψk, whichare defined with respect to a second globally quasi–uniform and shape regular boundaryelement mesh of mesh size h. Now, tz,h ∈ S0

h(Γ) is the unique solution of the Galerkinformulation

〈V tz,h, τh〉Γ = 〈(1

2I +K)z, τh〉Γ for all τh ∈ S0

h(Γ). (5.4)

Moreover, qz,h ∈ S0h(Γ) is the unique solution of the Galerkin formulation

〈V qz,h, τh〉Γ = 〈K1z − V1tz,h, τh〉Γ for all τh ∈ S0h(Γ). (5.5)

Hence we can define an approximation T of the operator T by

Tz := Az + qz,h . (5.6)

14

Lemma 5.2 The approximate operator T : H1/2(Γ) → H−1/2(Γ) as defined in (5.6) is

bounded, i.e.,

‖Tz‖H−1/2(Γ) ≤ ceT

2 ‖z‖H1/2(Γ) for all z ∈ H1/2(Γ).

Proof. From the Galerkin formulation (5.4) we first find, by choosing τh = tz,h and byusing the H−1/2(Γ)–ellipticity of the single layer potential,

‖tz,h‖H−1/2(Γ) ≤1

cV1‖(

1

2I +K)z‖H1/2(Γ) ≤

cK2cV1

‖z‖H1/2(Γ).

From (5.5), we now find for τh = qz,h, by using H3/2(Γ) ⊂ H1/2(Γ) and (3.9), (3.10),

‖qz,h‖H−1/2(Γ) ≤1

cV1‖K1z − V1tz,h‖H1/2(Γ)

≤1

cV1‖K1z − V1tz,h‖H3/2(Γ)

≤1

cV1

[cK1

2 ‖z‖H−1/2(Γ) + cV1

2 ‖tz,h‖H−3/2(Γ)

].

The assertion now follows from H1/2(Γ) ⊂ H−1/2(Γ) and H−1/2(Γ) ⊂ H−3/2(Γ).

Lemma 5.3 Let T : H1/2(Γ) → H−1/2(Γ) be given by (3.15), and let T be defined by

(5.6). Then there holds the error estimate

‖Tz − Tz‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖qz − τh‖H−1/2(Γ) +

cV1

2

cV1‖tz − tz,h‖H−3/2(Γ), (5.7)

where qz, tz ∈ H−1/2(Γ) are defined as in (5.3), and tz,h ∈ S0h(Γ) is the unique solution of

the Galerkin variational problem (5.4).

Proof. For an arbitrary chosen but fixed z ∈ H1/2(Γ) we have, by definition,

Tz = Az + qz , qz = V −1[K1z − V1tz], tz = V −1(1

2I +K)z.

In particular, tz ∈ H−1/2(Γ) is the unique solution of the variational problem

〈V tz, τ〉Γ = 〈(1

2I +K)z, τ〉Γ for all τ ∈ H−1/2(Γ),

and qz ∈ H−1/2(Γ) is the unique solution of the variational problem

〈V qz, τ〉Γ = 〈K1z, τ〉Γ − 〈V1tz, τ〉Γ for all τ ∈ H−1/2(Γ).

By using definition (5.6), we also have

Tz = Az + qz,h,

15

where qz,h is the unique solution of the Galerkin variational problem

〈V qz,h, τh〉Γ = 〈K1z, τh〉Γ − 〈V1tz,h, τh〉Γ for all τh ∈ S0h(Γ),

and tz,h ∈ S0h(Γ) is the unique solution of the variational problem

〈V tz,h, τh〉Γ = 〈(1

2I +K)z, τh〉Γ for all τh ∈ S0

h(Γ).

By applying Cea’s lemma, we first conclude the error estimate

‖tz − tz,h‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖tz − τh‖H−1/2(Γ) .

Let us further define qz,h ∈ S0h(Γ) as the unique solution of the variational problem

〈V qz,h, τh〉Γ = 〈K1z, τh〉Γ − 〈V1tz, τh〉Γ for all τh ∈ S0h(Γ). (5.8)

Again, by using Cea’s lemma we have

‖qz − qz,h‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖qz − τh‖H−1/2(Γ).

By subtracting (5.5) from (5.8) we obtain the perturbed Galerkin orthogonality

〈V (qz,h − qz,h), τh〉Γ = 〈V1(tz,h − tz), τh〉Γ for all τh ∈ S0h(Γ),

from which we further conclude the estimate

‖qz,h − qz,h‖H−1/2(Γ) ≤1

cV1‖V1(tz − tz,h)‖H1/2(Γ)

≤1

cV1‖V1(tz − tz,h)‖H3/2(Γ) ≤

cV1

2

cV1‖tz − tz,h‖H−3/2(Γ).

Hence we find, by applying the triangle inequality,

‖qz − qz,h‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖qz − τh‖H−1/2(Γ) +

cV1

2

cV1‖tz − tz,h‖H−3/2(Γ),

and the assertion follows from Tz − Tz = qz − qz,h.

By using the approximation property of the trial space S0h(Γ) and the Aubin–Nitsche trick,

we conclude an error estimate from (5.7) when assuming some regularity of qz and tz,respectively.

Corollary 5.4 Assume qz, tz ∈ Hspw(Γ) for some s ∈ [0, 1]. Then there holds the error

estimate

‖Tz − Tz‖H−1/2(Γ) ≤ c1 hs+ 1

2 ‖qz‖Hspw(Γ) + c2 h

s+ 32 ‖tz‖Hs

pw(Γ). (5.9)

16

5.3 Boundary element approximation of g

As in (5.6), we may also define a boundary element approximation of the right hand side gas defined in (3.16)

g = V −1N0u+ V −1M0f − V −1V1V−1N0f.

In particular, g ∈ H−1/2(Γ) is the unique solution of the variational problem

〈V g, τ〉Γ = 〈N0u+M0f − V1V−1N0f, τ〉Γ = 〈N0u+M0f, τ〉Γ − 〈V1tf , τ〉Γ

for all τ ∈ H−1/2(Γ), where tf = V −1N0f ∈ H−1/2(Γ) solves the variational problem

〈V tf , τ〉Γ = 〈N0f, τ〉Γ for all τ ∈ H−1/2(Γ).

Hence we can define a boundary element approximation gh ∈ S0h(Γ) as the unique solution

of the Galerkin variational problem

〈V gh, τh〉Γ = 〈N0u+M0f, τh〉Γ − 〈V1tf,h, τh〉Γ for all τh ∈ S0h(Γ), (5.10)

where tf,h ∈ S0h(Γ) is the unique solution the Galerkin problem

〈V tf,h, τh〉Γ = 〈N0f, τh〉Γ for all τh ∈ S0h(Γ). (5.11)

Lemma 5.5 Let g be the right hand side as defined by (3.16), and let gh be the boundary

element approximation as defined in (5.10). Then there holds the error estimate

‖g − gh‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖g − τh‖H−1/2(Γ) +

cV1

2

cV1‖tf − tf,h‖H−3/2(Γ). (5.12)

Proof. In addition to (5.10), let us consider the Galerkin formulation to find gh ∈ S0h(Γ)

such that

〈V gh, τh〉Γ = 〈N0u+M0f, τh〉Γ − 〈V1tf , τh〉Γ for all τh ∈ S0h(Γ). (5.13)

Again, by using Cea’s lemma, we obtain

‖g − gh‖H−1/2(Γ) ≤cV2cV1

infτh∈S0

h(Γ)‖g − τh‖H−1/2(Γ).

Subtracting (5.10) from (5.13) gives the perturbed Galerkin orthogonality

〈V (gh − gh), τh〉Γ = −〈V1(tf − tf,h), τh〉Γ for all τh ∈ S0h(Γ).

For τh = gh − gh and by using the H−1/2(Γ)–ellipticity of the single layer potential V andthe estimate (3.9), we further obtain

‖gh − gh‖H−1/2(Γ) ≤1

cV1‖V1(tf − tf,h)‖H1/2(Γ) ≤

cV1

2

cV1‖tf − tf,h‖H−3/2(Γ).

The assertion finally follows from the triangle inequality.

By using the approximation property of the trial space S0h(Γ) and the Aubin–Nitsche trick,

we conclude an error estimate from (5.12) when assuming some regularity of g and tf ,respectively.

17

Corollary 5.6 Assume g, tf ∈ Hspw(Γ) for some s ∈ [0, 1]. Then there holds the error

estimate

‖g − gh‖H−1/2(Γ) ≤ c1 hs+ 1

2 ‖g‖Hspw(Γ) + c2 h

s+ 32 ‖tf‖Hs

pw(Γ). (5.14)

5.4 Perturbed Galerkin variational problem

Instead of the Galerkin variational problem (5.1), we now consider a perturbed Galerkinformulation to find zH ∈ S1

H(Γ) such that

〈TzH , vH〉Γ = 〈gh, vH〉Γ for all vH ∈ S1H(Γ). (5.15)

By combining the boundary element approximations (5.4) and (5.5) with (5.10) and (5.11),it is sufficient to consider the Galerkin boundary element formulation of (3.11): Find(th, qh, zH) ∈ S0

h(Γ) × S0h(Γ) × S1

H(Γ) such that

−〈V1th, wh〉Γ + 〈V qh, wh〉Γ + 〈K1zH , wh〉Γ = 〈N0u+M0f, wh〉Γ, (5.16)

〈V th, τh〉Γ − 〈(1

2I +K)zH , τh〉Γ = −〈N0f, τh〉Γ, (5.17)

−〈qh, vH〉Γ + 〈AzH , vH〉Γ = 0 (5.18)

is satisfied for all (wh, τh, vH) ∈ S0h(Γ)× S0

h(Γ)× S1H(Γ). The Galerkin formulation (5.16)–

(5.18) is equivalent to a system of linear equations

−V1,h Vh K1,h

Vh −(12Mh +Kh)

−M⊤h AH

t

q

z

=

f1

f2

0

, (5.19)

whereVh[ℓ, k] = 〈V ψk, ψℓ〉Γ, Kh[ℓ, i] = 〈Kϕi, ψℓ〉Γ,V1,h[ℓ, k] = 〈V1ψk, ψℓ〉Γ, K1,h[ℓ, i] = 〈K1ϕi, ψℓ〉Γ,AH [j, i] = 〈Aϕi, ϕj〉Γ, Mh[ℓ, i] = 〈ϕi, ψℓ〉Γ,

andf1,ℓ = 〈N0u+M0f, ψℓ〉Γ, f2,ℓ = −〈N0f, ψℓ〉Γ

for k, ℓ = 1, . . . , N and i, j = 1, . . . ,M .Since the Laplace single layer potential V is H−1/2(Γ)–elliptic, the related Galerkin

matrix Vh is positive definite and therefore invertible. Hence, we can resolve the secondequation in (5.19) to obtain

t = V −1h (

1

2Mh +Kh)z + V −1

h f2.

Inserting this into the first equation of (5.19) gives

Vhq = V1,hV−1h (

1

2Mh +Kh)z + V1,hV

−1h f

2−K1,hz + f

1

18

and therefore

q = V −1h V1,hV

−1h (

1

2Mh +Kh)z + V −1

h V1,hV−1h f

2− V −1

h K1,hz + V −1h f

1.

Hence it remains to solve the Schur complement system

[AH −M⊤

h V−1h V1,hV

−1h (

1

2Mh +Kh) +M⊤

h V−1h K1,h

]z = M⊤

h V−1h

[f

1+ V1,hV

−1h f

2

]

(5.20)where

T,H = AH −M⊤

h V−1h V1,hV

−1h (

1

2Mh +Kh) +M⊤

h V−1h K1,h (5.21)

defines a non–symmetric Galerkin boundary element approximation of the self–adjointSchur complement boundary integral operator T.

Theorem 5.7 The approximate Schur complement T,H as defined in (5.21) is positive

definite, i.e.,

(T,Hz, z) ≥1

2cT1 ‖zH‖

2H1/2(Γ)

for all z ∈ RM ↔ zH ∈ S1

H(Γ), if h ≤ c0H is sufficiently small.

Proof. For an arbitrary chosen but fixed z ∈ RM let zH ∈ S1

H(Γ) be the associatedboundary element function. Then we have

(T,Hz, z) = 〈TzH , zH〉Γ

= 〈TzH , zH〉Γ − 〈(T − T)zH , zH〉Γ

≥ cT1 ‖zH‖2H1/2(Γ) − ‖(T − T)zH‖H−1/2(Γ)‖zH‖H1/2(Γ).

Since zH ∈ S1H(Γ) is a continuous function, we have zH ∈ H1(Γ). Hence we find

tzH= V −1(

1

2I +K)zH ∈ L2(Γ),

andqzH

= V −1[K1zH − V1tzH] ∈ L2(Γ)

according to (5.4) and (5.5). Therefore we can apply the error estimate (5.9) for s = 0 toobtain

‖TzH − TzH‖H−1/2(Γ) ≤ c1 h1/2 ‖qzH

‖L2(Γ) + c2 h3/2 ‖tzH

‖L2(Γ) ≤ c3 h1/2 ‖zH‖H1(Γ).

Now, by applying the inverse inequality for S1H(Γ),

‖zH‖H1(Γ) ≤ cI H−1/2 ‖zH‖H1/2(Γ),

19

we obtain

‖TzH − TzH‖H−1/2(Γ) ≤ c3cI

(h

H

)1/2

‖zH‖H1/2(Γ).

Hence we finally obtain

(T,Hz, z) ≥

[cT1 − c3cI

(h

H

)1/2]‖zH‖

2H1/2(Γ) ≥

1

2cT1 ‖zH‖2

H1/2(Γ),

if

c3cI

(h

H

)1/2

≤1

2cT1

is satisfied.

Note that Theorem 5.7 ensures the unique solvability of the linear system (5.20) andtherefore of the perturbed variational problem (5.15). Since the approximate operator

T is S1H(Γ)–elliptic, an error estimate for the approximate solution zH of the perturbed

Galerkin variational problem (5.15) follows from the Strang lemma, see, e.g., [22, Theorem8.2, Theorem 8.3], which reads for our problem as:

Theorem 5.8 Let z be the unique solution of the operator equation (3.14). Let h ≤ c0H be

satisfied such that the approximate Schur complement T,H as defined in (5.21) is positive

definite, and let zH ∈ S1H(Γ) be the unique solution of the perturbed Galerkin variational

formulation (5.15). Then there holds the error estimate

‖z − zH‖H1/2(Γ) ≤ c1 infvH∈S1

H(Γ)‖z − vH‖H1/2(Γ) + c2‖(T − T)z‖H−1/2(Γ) + c3‖g − gh‖H−1/2(Γ).

(5.22)

Corollary 5.9 When combining the error estimate (5.22) with the approximation property

of the ansatz space S1H(Γ), and with the error estimates (5.9) and (5.14), we finally obtain

the error estimate

‖z − zH‖H1/2(Γ) ≤ c1H3/2 |z|H2(Γ) + c2 h

3/2 ‖qz‖H1pw

(Γ) + c3 h5/2 ‖tz‖H1

pw(Γ)

+c4 h3/2 ‖g‖H1

pw(Γ) + c5 h

5/2 ‖tf‖H1pw

(Γ)

when assuming z ∈ H2(Γ), and qz, tz, g, tf ∈ H1pw(Γ), respectively. For h ≤ c0H we can

expect the convergence rate 1.5 when measuring the error in the energy norm

‖z − zH‖H1/2(Γ) ≤ c(z, u, f)H3/2 . (5.23)

Moreover, we are also able to derive an error estimate in L2(Γ), i.e.,

‖z − zH‖L2(Γ) ≤ c(z, u, f)H2 . (5.24)

when applying the Aubin–Nitsche trick [22].

20

Remark 5.1 The error estimates (5.23) and (5.24) provide optimal convergence rates

when approximating the control z by using piecewise linear basis functions. However, we

have to assume h ≤ c0H to ensure the unique solvability of the perturbed Galerkin formu-

lation (5.15), where the constant c0 is in general unknown. Moreover, the matrix T,H as

given in (5.21) defines a non–symmetric approximation of the exact symmetric stiffness

matrix T,H as used in (5.1). Hence we are interested in deriving a symmetric bound-

ary element method which is stable without any additional constraints in the choice of the

boundary element trial spaces.

6 Symmetric boundary integral formulation

The boundary integral formulation of the primal boundary value problem (2.2) is givenby (3.2), while the adjoint boundary value problem (2.9) corresponds to the modifiedboundary integral equation (3.8). In what follows, we will rewrite the optimality condition(2.10) by using a hypersingular boundary integral equation for the adjoint problem toobtain a symmetric boundary integral formulation for the coupled problem.

Since the adjoint variable p, as defined in the representation formula (3.7), is a solutionof the adjoint Dirichlet boundary value problem (2.9), the normal derivative

q(x) = limΩ∋ex→x∈Γ

nx · ∇exp(x) =∂

∂np(x) for x ∈ Γ

is well defined. When computing the normal derivative of the representation formula (3.7),this gives a second boundary integral equation for x ∈ Γ

q(x) = (1

2I +K ′)q(x) − (D1z)(x) − (K ′

1t)(x) − (N1u)(x) − (M1f)(x), (6.1)

where we introduce Newton potentials for x ∈ Γ

(N1u)(x) = limΩ∋ex→x∈Γ

nx · ∇ex

∫

Ω

U∗(x, y)u(y)dy for x ∈ Γ

and

(M1f)(x) = limΩ∋ex→x∈Γ

nx · ∇ex

∫

Ω

V ∗(x, y)f(y)dy for x ∈ Γ

in addition to the boundary integral operators used in (4.10). Combining the optimal-ity condition (2.10) and the boundary integral equation (6.1) gives a boundary integralequation for x ∈ Γ,

(Az)(x) = (1

2I +K ′)q(x) − (D1z)(x) − (K ′

1t)(x) − (N1u)(x) − (M1f)(x). (6.2)

21

Now, to find the yet unknown triple (z, t, q) ∈ H1/2(Γ)×H−1/2(Γ)×H−1/2(Γ) we solve thesystem of boundary integral equations (3.2), (3.8), and (6.2) which can be written as

−V1 V K1

V −12I −K

K ′1 −1

2I −K ′ A+D1

t

q

z

=

N0u+M0f

−N0f

−N1u−M1f

. (6.3)

To investigate the unique solvability of (6.3), we consider the related Schur complement.As in (3.12) and (3.13) we obtain

t = V −1(1

2I +K)z − V −1N0f

and

q = V −1V1V−1(

1

2I +K)z − V −1K1z + V −1N0u+ V −1M0f − V −1V1V

−1N0f.

Hence it remains to solve the Schur complement system[A+D1 +K ′

1V−1(

1

2I +K) + (

1

2I +K ′)V −1K1 − (

1

2I +K ′)V −1V1V

−1(1

2I +K)

]z

= K ′

1V−1N0f −N1u−M1f + (

1

2I +K ′)V −1

[N0u+M0f − V1V

−1N0f]. (6.4)

Note that (6.4) corresponds to a symmetric boundary integral formulation of the operatorequation (2.7) representing the optimality condition.

Theorem 6.1 The composed boundary integral operator

T = A+D1 − (1

2I+K ′)V −1V1V

−1(1

2I+K)+K ′

1V−1(

1

2I+K)+(

1

2I+K ′)V −1K1 (6.5)

is self–adjoint, bounded, i.e., T : H1/2(Γ) → H−1/2(Γ), and H1/2(Γ)–elliptic.

Proof. While the self–adjointness of T in the symmetric representation (6.5) is obvious,the boundedness and ellipticity estimates follow as in the proof of Theorem 5.1. In par-ticular, the Schur complement operators T in the symmetric representation (6.5) and inthe non–symmetric representation (3.15) coincide. Indeed, by using (4.11) and (4.12) weobtain

T = A+D1 − (1

2I +K ′)V −1V1V

−1(1

2I +K) +K ′

1V−1(

1

2I +K) + (

1

2I +K ′)V −1K1

= A+D1 +

[K ′

1 − (1

2I +K ′)V −1V1

]V −1(

1

2I +K) + (

1

2I +K ′)V −1K1

= A+D1 + V −1

[V K ′

1 −KV1 −1

2V1

]V −1(

1

2I +K) + (

1

2I +K ′)V −1K1

= A+D1 + V −1

[K1V − V1K

′ −1

2V1

]V −1(

1

2I +K) + (

1

2I +K ′)V −1K1

= A+D1 + V −1K1(1

2I +K) − V −1V1(

1

2I +K ′)V −1(

1

2I +K) + (

1

2I +K ′)V −1K1.

22

Due to the representation of the Laplace Steklov–Poincare operator, see, e.g., [22],

S = V −1(1

2I +K) = D + (

1

2I +K ′)V −1(

1

2I +K),

we further conclude

(1

2I +K ′)V −1(

1

2I +K) = V −1(

1

2I +K) −D.

Therefore, by using (4.11) and (4.14) we have

T = A+D1 + V −1K1(1

2I +K) − V −1V1

[V −1(

1

2I +K) −D

]+ V −1(

1

2I +K)K1

= A+ V −1

[V D1 + V1D +K1(

1

2I +K) − V1V

−1(1

2I +K) + (

1

2I +K)K1

]

= A+ V −1

[−KK1 −K1K +K1(

1

2I +K) − V1V

−1(1

2I +K) + (

1

2I +K)K1

]

= A+ V −1

[K1 − V1V

−1(1

2I +K)

],

and we finally obtain the non–symmetric representation (3.15). Therefore, the ellipticityof T follows as in Theorem 5.1.

Due to the H1/2(Γ)–ellipticity of the symmetric representation (6.5) of T, we can concludethe unique solvability of the Schur complement boundary integral equation (6.4), andtherefore of the coupled system (6.3).

6.1 Galerkin boundary element discretization

In what follows, we will consider a boundary element discretization of the boundary integralequation system (6.3). Again, let

S0h(Γ) = spanψk

Nk=1 ⊂ H−1/2(Γ), S1

h(Γ) = spanϕiMi=1 ⊂ H1/2(Γ)

be some boundary element spaces of piecewise constant and piecewise linear basis functionsψk and ϕi, which are defined with respect to some admissible boundary element mesh ofmesh size h. The Galerkin boundary element formulation of (6.3) then reads to find(th, qh, zh) ∈ S0

h(Γ) × S0h(Γ) × S1

h(Γ) such that

−〈V1th, wh〉Γ + 〈V qh, wh〉Γ + 〈K1zh, wh〉Γ = 〈N0u+M0f, wh〉Γ, (6.6)

〈V th, τh〉Γ − 〈(1

2I +K)zh, τh〉Γ = −〈N0f, τh〉Γ, (6.7)

〈K ′

1th, vh〉Γ − 〈(1

2I +K ′)qh, vh〉Γ + 〈(A +D1)zh, vh〉Γ = −〈N1u+M1f, vh〉Γ (6.8)

23

is satisfied for all (wh, τh, vh) ∈ S0h(Γ)×S0

h(Γ)×S1h(Γ). The Galerkin formulation (6.6)–(6.8)

is equivalent to a system of linear equations,

−V1,h Vh K1,h

Vh −(12Mh +Kh)

K⊤1,h −(1

2M⊤

h +K⊤h ) Ah +D1,h

t

q

z

=

f1

f2

f3

, (6.9)

where we used, in addition to those entries of the linear system (5.19),

D1,h[j, i] = 〈D1ϕi, ϕj〉Γ, f3,j = −〈N1u+M1f, ϕj〉Γ for i, j = 1, . . . ,M.

To investigate the unique solvability of the linear system (6.9), we consider the invertibilityof the related Schur complement. In particular, the second equation in (6.9) gives

t = V −1h (

1

2Mh +Kh)z + V −1

h f2,

and we obtain from the first equation

q = V −1h [V1,hV

−1h (

1

2Mh +Kh) −K1,h]z + V −1

h f1+ V −1

h V1,hV−1h f

2.

Hence, by inserting these results into the third equation of (6.9), we finally end up withthe Schur complement system of the symmetric boundary integral formulation

T,hz = f, (6.10)

where the Schur complement is given by

T,h = Ah +D1,h +K⊤

1,hV−1h (

1

2Mh +Kh) + (

1

2M⊤

h +K⊤

h )V −1h K1,h (6.11)

−(1

2M⊤

h +K⊤

h )V −1h V1,hV

−1h (

1

2Mh +Kh),

and the right hand side is

f = f3−K⊤

1,hV−1h f

2+ (

1

2M⊤

h +K⊤

h )V −1h f

1+ (

1

2M⊤

h +K⊤

h )V −1h V1,hV

−1h f

2.

Lemma 6.2 The symmetric matrix

Th = T,h − Ah = D1,h +K⊤

1,hV−1h (

1

2Mh +Kh) + (

1

2M⊤

h +K⊤

h )V −1h K1,h

−(1

2M⊤

h +K⊤

h )V −1h V1,hV

−1h (

1

2Mh +Kh)

is positive semi–definite, i.e., all eigenvalues of Th are non–negative,

(Thz, z) ≥ 0 for all z ∈ RM .

24

Proof. We consider the generalized eigenvalue problem

Thz = µ

[Sh + (

1

2M⊤

h +K⊤

h )V −1h (

1

2Mh +Kh)

]z, (6.12)

where the stabilized discrete Steklov–Poincare operator

Sh = Dh + (1

2M⊤

h +K⊤

h )V −1h (

1

2Mh +Kh)

is symmetric and positive definite.Since the eigenvalue problem (6.12) can be written as

((1

2M⊤

h +K⊤h )V −1

h I)(

−V1,h K1,h

K⊤1,h D1,h

) (V −1

h (12Mh +Kh)I

)z

= µ(

(12M⊤

h +K⊤h )V −1

h I)(

Vh

Sh

) (V −1

h (12Mh +Kh)I

)z,

it is sufficient to consider the generalized eigenvalue problem(

−V1,h K1,h

K⊤1,h D1,h

) (w

z

)= µ

(Vhw

Shz

), (6.13)

where

w = V −1h (

1

2Mh +Kh)z .

From (6.13), we then conclude

(K1,hz, w) − (V1,hw,w) = µ(Vhw,w),

(K⊤

1,hw, z) + (D1,hz, z) = µ(Shz, z),

and by taking the difference we obtain

(D1,hz, z) + (V1,hw,w) = µ[(Shz, z) − (Vhw,w)] = µ(Dhz, z).

Since the Galerkin matrices D1,h and V1,h of the Bi–Laplace boundary integral operatorsD1

and V1 are positive semi–definite and positive definite, respectively, and since the stabilizedLaplace hypersingular integral operator Dh is positive definite, we conclude

µ ≥ 0 .

Hence, we finally obtain

(Thz, z) ≥ µmin

[(Shz, z) + (V −1

h (1

2Mh +Kh)z, (

1

2Mh +Kh)z)

]≥ 0

for all z ∈ RM .

As a corollary of Lemma 6.2, we find the positive definiteness of the symmetric Schurcomplement matrix T,h as defined in (6.11).

25

Corollary 6.3 The approximate Schur complement T,h as defined in (6.11) is positive

definite, i.e.,

(T,hz, z) ≥ (Ahz, z) = 〈Azh, zh〉Γ ≥ γA1 ‖zh‖

2H1/2(Γ)

for all z ∈ RM ↔ zh ∈ S1

h(Γ).

Hence, we can ensure the unique solvability of the Schur complement system (6.10) andtherefore of the system (6.9) as well as of the Galerkin variational problem (6.6)–(6.8). Tofind an error estimate for the approximate solution zh ∈ S1

h(Γ), as for the non–symmetricboundary element formulation, we will consider a perturbed variational problem of theoperator equation (6.4) leading to the Schur complement system (6.10).

6.2 Symmetric boundary element approximation of T

For an arbitrary but fixed given z ∈ H1/2(Γ) the application of Tz reads, by using thesymmetric representation (6.5),

Tz = Az +D1z +K ′

1tz − (1

2I +K ′)qz,

where qz, tz ∈ H−1/2(Γ) are the unique solutions of the boundary integral equations (5.3).Hence, by using the unique solutions qz,h, tz,h ∈ S0

h(Γ) of the related Galerkin variationalformulations (5.4) and (5.5) we can define the approximation

Tz := Az +D1z +K ′

1tz,h − (1

2I +K ′)qz,h. (6.14)

Lemma 6.4 The approximate operator T : H1/2(Γ) → H−1/2(Γ) as defined in (6.14) is

bounded, i.e.,

‖Tz‖H−1/2(Γ) ≤ cbT

2 ‖z‖H1/2(Γ) for all z ∈ H1/2(Γ).

Moreover, there holds the error estimate

‖Tz − Tz‖H−1/2(Γ) ≤ c1 infτh∈S0

h(Γ)‖qz − τh‖H−1/2(Γ) + c2 ‖tz − tz,h‖H−3/2(Γ). (6.15)

Proof. The proof follows as for the boundary element approximation of the non–symmetric formulation, see Lemma 5.2 and Lemma 5.3.

By using the approximation property of the trial space S0h(Γ) and the Aubin–Nitsche trick,

we then conclude an error estimate from (6.15) when assuming some regularity of qz andtz, respectively.

Corollary 6.5 Assume qz, tz ∈ Hspw(Γ) for some s ∈ [0, 1]. Then there holds the error

estimate

‖Tz − Tz‖H−1/2(Γ) ≤ c1 hs+ 1

2 ‖qz‖Hspw(Γ) + c2 h

s+ 32 ‖tz‖Hs

pw(Γ). (6.16)

26

6.3 Boundary element approximation of g

As in the approximation (6.14), we can define a boundary element approximation of therelated right hand side, see (6.4),

g = K ′

1V−1N0f −N1u−M1f + (

1

2I +K ′)V −1[N0u+M0f − V1V

−1N0f ]

= K ′

1tf −N1u−M1f + (1

2I +K ′)qf ,

where tf = V −1N0f ∈ H−1/2(Γ) is the unique solution of the variational problem

〈V tf , τ〉Γ = 〈N0f, τ〉Γ for all τ ∈ H−1/2(Γ),

and qf = V −1[N0u + M0f − V1tf ] ∈ H−1/2(Γ) is the unique solution of the variationalproblem

〈V qf , τ〉Γ = 〈N0u+M0f − V1tf , τ〉Γ for all τ ∈ H−1/2(Γ).

Hence we can define an approximation

g := K ′

1tf,h −N1u−M1f + (1

2I +K ′)qf,h, (6.17)

where qf,h ∈ S0h(Γ) is the unique solution of the Galerkin variational problem

〈V qf,h, τh〉Γ = 〈N0u+M0f − V1tf,h, τh〉Γ for all τh ∈ S0h(Γ),

and tf,h ∈ S0h(Γ) is the unique solution of the Galerkin variational problem

〈V tf,h, τh〉Γ = 〈N0f, τh〉Γ for all τh ∈ S0h(Γ).

As in (5.14), we conclude the error estimate

‖g − g‖H−1/2(Γ) ≤ c1 hs+ 1

2 ‖qf‖Hspw(Γ) + c2 h

s+ 32 ‖tf‖Hs

pw(Γ) (6.18)

when assuming qf , tf ∈ Hspw(Γ) for some s ∈ [0, 1].

6.4 Perturbed Galerkin variational problem

We now consider a perturbed Galerkin formulation instead of the operator equation (6.4)to find zh ∈ S1

h(Γ) such that

〈Tzh, vh〉Γ = 〈g, vh〉Γ for all vh ∈ S1h(Γ). (6.19)

Note that the Galerkin discretization of the perturbed variational problem (6.19) resultsin the linear system (6.10). Now we are in a position to formulate an error estimate forthe approximate solution zh by applying the Strang lemma.

27

Theorem 6.6 Let z ∈ H1/2(Γ) be the unique solution of the operator equation (6.4), and

let zh ∈ S1h(Γ) be the unique solution of the perturbed Galerkin variational problem (6.19).

Then there holds the error estimate

‖z−zh‖H1/2(Γ) ≤ c1 infvh∈S1

h(Γ)‖z−vh‖H1/2(Γ)+c2‖(T−T)z‖H−1/2(Γ)+c3‖g−g‖H−1/2(Γ). (6.20)

Corollary 6.7 When combining the error estimate (6.20) with the approximation property

of the ansatz space S1h(Γ), and with the error estimates (6.16) and (6.18), we finally obtain

the error estimate

‖z − zh‖H1/2(Γ) ≤ c1 h3/2 |z|H2(Γ) + c2 h

3/2 ‖qz‖H1pw

(Γ) + c3 h5/2 ‖tz‖H1

pw(Γ)

+c4 h3/2 ‖qf‖H1

pw(Γ) + c5 h

5/2 ‖tf‖H1pw

(Γ)

when assuming z ∈ H2(Γ), and qz, tz, qf , tf ∈ H1pw(Γ), respectively. Hence we can expect

the convergence order 1.5 when measuring the error in the energy norm,

‖z − zh‖H1/2(Γ) ≤ c(z, u, f) h3/2 . (6.21)

Moreover, applying the Aubin–Nitsche trick [22] we are also able to derive an error estimate

in L2(Γ), i.e.,

‖z − zh‖L2(Γ) ≤ c(z, u, f) h2 . (6.22)

7 Numerical results

We consider the Dirichlet boundary control problem (2.1) and (2.2) for the domain Ω =(0, 1

2)2 ⊂ R

2 where

u(x) = −

(4 +

1

)[x1(1 − 2x1) + x2(1 − 2x2)] , f(x) = −

8

, = 0.01 .

For the boundary element discretization, we introduce a uniform triangulation of Γ = ∂Ωon several levels where the mesh size is hL = 2−(L+1). Since the minimizer of (2.1) is notknown in this case, we use the boundary element solution zh of the 9th level as referencesolution. The boundary element discretization is done by using the trial space S0

h(Γ) ofpiecewise constant basis functions, and S1

h(Γ) of piecewise linear and continuous functions.In particular we use the same boundary element mesh to approximate the control z bya piecewise linear approximation, and piecewise constant approximations for the fluxest and q. Note that we have h = H in this case, and therefore we can not ensure theS1

h(Γ)–ellipticity of the non–symmetric boundary element approximation, see Theorem 5.7.However, the numerical example shows stability.In Table 1, we present the errors for the control z in the L2(Γ) norm and the estimatedorder of convergence (eoc). These results correspond to the error estimate (5.24) of thenon–symmetric boundary element approximation, and to the error estimate (6.22) of the

28

Non–symmetric BEM (5.20) Symmetric BEM (6.10) FEM [18]L ‖zhL

− zh9‖L2(Γ) eoc ‖zhL

− zh9‖L2(Γ) eoc ‖zFEM

hL− zFEM

h9‖L2(Γ) eoc

2 4.52 –1 2.25 –0 3.89 –13 1.28 –1 1.82 4.66 –1 2.27 1.07 –1 1.864 3.54 –2 1.85 8.84 –2 2.39 2.81 –2 1.935 9.03 –3 1.97 1.63 –2 2.44 7.28 –3 1.956 2.18 –3 2.05 3.02 –3 2.43 1.87 –3 1.967 5.16 –4 2.08 5.73 –4 2.40 4.69 –4 2.008 1.39 –4 1.89 1.24 –4 2.20 1.06 –4 2.15

Table 1: Comparison of BEM/FEM errors of the Dirichlet control.

symmetric boundary element approximation. In addition, we give the error of the re-lated finite element solution, see [18]. The results show a quadratic order of convergence,which confirm the theoretical estimates. Note that in the finite element approach only aconvergence order of 1.5 can be proved [18].

In Table 2, we present the errors for the flux t of the primal boundary value problem,again in the L2(Γ) norm. Since the computation of t corresponds to the solution of aDirichlet boundary value problem with approximated Dirichlet data, we can expect andobserve a linear order of convergence when using piecewise constant basis functions, see,e.g., [22].

Non–symmetric BEM (5.20) Symmetric BEM (6.10)L ‖thL

− th9‖L2(Γ) eoc ‖thL

− th9‖L2(Γ) eoc

2 33.51 39.873 24.03 0.48 28.67 0.484 14.80 0.70 16.22 0.825 8.44 0.81 8.88 0.876 4.59 0.88 4.73 0.917 2.40 0.93 2.45 0.958 1.17 1.04 1.19 1.05

Table 2: Comparison of non–symmetric/symmetric BEM.

Acknowledgement

This work has been supported by the Austrian Science Fund (FWF) under the GrantSFB Mathematical Optimization and Applications in Biomedical Sciences, Subproject FastFinite Element and Boundary Element Methods for Optimality Systems. The authorswould like to thank K. Kunisch for helpful discussions.

29

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