• Nebyly nalezeny žádné výsledky

∂Ω

∂vκ

∂ny

(x,y)t(y) dsy for x∈∂Ω and σ is defined by (3.8).

Proof. The proof can be found in [9], following Lemma 2.32.

Remark 3.13. Similarly as in Remark 3.11, for Lipschitz domains we get simplified formulae γ0,int(Wκt)(x) =−1

2t(x) + (Kκt)(x) for x∈∂Ω, (3.11) γ0,ext(Wκt)(x) = 1

2t(x) + (Kκt)(x) for x∈∂Ω. (3.12) For the jump of the Dirichlet trace of the double layer potential Wκt on the boundary we have

0Wκt] :=γ0,extWκt−γ0,intWκt=t for all t∈H1/2(∂Ω).

3.4 Hypersingular Integral Operator

Finally, we consider the hypersingular integral operator defined as the negative Neumann trace of the double layer potential operator, i.e.,

Dκ:H1/2(∂Ω)→H−1/2(∂Ω), Dκ:=−γ1Wκ.

34 3 Boundary Integral Equations

Again, the hypersingular operator is linear and there exists a constantc∈R+ such that

∥Dκt∥H−1/2(∂Ω)≤c∥t∥H1/2(∂Ω) for allt∈H1/2(∂Ω).

The hypersingular operator cannot be represented in the same way as the preceding operators. For Dκtwith a smooth enough density functiont we have

(Dκt)(x) =−γ1(Wκt)(x) =− lim

Ω∋x→x∈∂Ω˜ ⟨n(x),∇x˜(Wκt)( ˜x)⟩.

Recall, that for the double layer potential we have the representation (Wκt)( ˜x) = Interchanging the limit with the computation of the normal derivative we obtain

Ω∋x→x∈∂Ω˜lim (Wκt)( ˜x)

By computing the normal derivative of the integrand we get (Dκt)(x) = 1 Unfortunately, for ε→0+ the integrand from (3.13) is not an integrable function. To ex-pressDκtexplicitly we need some regularization procedure. For the Galerkin discretization of the boundary integral equations we need to evaluate the bilinear form

⟨Dκt, s⟩∂Ω :=

∂Ω

(Dκt)(x)s(x) dsx

induced by the hypersingular operator. The following theorem gives us a representation using surface curl operators.

Theorem 3.14. For functions s, t∈H1/2(∂Ω) there holds the representation

with the surface curl operator

curl∂Ωt(y) :=n(y)× ∇˜t(y) for y∈∂Ω,

where ˜t is some locally defined extension of tinto the neighbourhood of ∂Ω.

Proof. For the proof and more details see [13], Theorem 3.4.2 and [15], Theorem 3.3.22 and Corrolary 3.3.24.

Theorem 3.15. The hypersingular operator Dκ:H1/2(∂Ω)→H−1/2(∂Ω) is coercive.

Proof. The proof is taken from [15], Lemma 3.9.8. The regularised hypersingular operator D0+I corresponding to the Laplace equation is H1/2(∂Ω)-elliptic. Due to the compact embedding H1/2(∂Ω) ↩→↩→ H−1/2(∂Ω) and the compactness of Dκ −D0, the operator C:=I+D0−Dκ:H1/2(∂Ω)→H−1/2(∂Ω) is compact. Thus, we have

⟨(Dκ+C)t, t⟩=⟨(D0+I)t, t⟩ ≥c∥t∥2

H1/2(∂Ω) for all t∈H1/2(∂Ω), which completes the proof.

To conclude, for the jump of the Neumann trace of the double layer potential Wκt on the boundary we have

1Wκt] :=γ1,extWκt−γ1,intWκt= 0 for all t∈H1/2(∂Ω). (3.14) 3.5 Boundary Integral Equations

Recall that the solution to an interior boundary value problem for the Helmholtz equation can be represented as

u=−Wκγ0,intu+Vκγ1,intu inΩ.

Applying the interior Dirichlet trace operator and using the definition ofVκ and the prop-erty (3.11) we get the boundary integral equation

γ0,intu=

1

2I−Kκ

γ0,intu+Vκγ1,intu on ∂Ω (3.15) with the identity operator I. Similarly, applying the Neumann trace operator and using (3.9) and the definition of the hypersingular integral operator we obtain

γ1,intu=Dκγ0,intu+

1

2I+Kκ

γ1,intu on ∂Ω. (3.16)

The boundary integral equations (3.15), (3.16) can be rewritten as

γ0,intu γ1,intu

=Cint

γ0,intu γ1,intu

36 3 Boundary Integral Equations

with the Calderon projection matrix Cint:=

1

2I−Kκ Vκ

Dκ 12I+Kκ

 .

For the solution in an unbounded domain we have the formula u=Wκγ0,extu−Vκγ1,extu inΩext.

Applying the Dirichlet and Neumann trace operators, respectively, and using properties (3.12), (3.10) and the definitions of the boundary integral operators Vκ and Dκ we get

γ0,extu=

1

2I+Kκ

γ0,extu−Vκγ1,extu on∂Ωext, (3.17) γ1,extu=−Dκγ0,extu+

1

2I−Kκ

γ1,extu on∂Ωext. (3.18) Again, the boundary integral equations (3.17), (3.18) can be rewritten as

γ0,extu γ1,extu

=Cext

γ0,extu γ1,extu

with the Calderon projection matrix Cext:=

1

2I+Kκ −Vκ

−Dκ 12I−Kκ

 .

Theorem 3.16. The Calderon operators Cint and Cext are projection operators, i.e., (Cint)2=Cint (Cext)2 =Cext.

Proof. The proof is analogous to the proof of Lemma 6.18 in [18].

Corollary 3.17. For the boundary integral operators we have the following identities VκDκ =

1

2I+Kκ

 1

2I−Kκ

 ,

DκVκ =

1

2I+Kκ

 1

2I−Kκ

 , DκKκ =KκDκ,

KκVκ =VκKκ.

Proof. The relations follow directly from the comparison of the matrices(Cint)2 and Cint.

The following theorems are standard results of functional analysis and will be used to prove solvability of boundary integral equations studied in the next sections. In the theorems we assume thatX denotes a Hilbert space.

Theorem 3.18 (Lax-Milgram Lemma). Let A: X → X denote a linear, bounded and X-elliptic operator. Then for any f ∈X there exists a unique elementu∈X satisfying

Au=f.

Moreover, there holds the estimate

∥u∥X ≤ 1 cA1 ∥f∥X with the ellipticity constant cA1.

Proof. For the proof of the Lax-Milgram Lemma see [18], proof following Theorem 3.4 or any standard functional analysis textbook.

Theorem 3.19 (Fredholm Alternative). Let K:X →X denote a compact operator. Ei-ther the homogeneous equation

(I −K)u= 0

has a nontrivial solution u∈X or the inhomogeneous problem (I−K)u=f

has a unique solutionu∈X for allf ∈X. In the latter case there exists a constantc∈R+

such that

∥u∥X ≤c∥f∥X for all f ∈X.

Proof. For the proof of the Fredholm Alternative see, e.g., [7], Theorem 2.2.9.

Theorem 3.20. Let A:X → X denote a linear, bounded and coercive operator and let A be injective, i.e., from Au = 0 it follows u = 0. Then for every f ∈ X there exists a unique solution to the equation

Au=f. (3.19)

Moreover, there exists a constant c∈R+ such that

∥u∥X ≤c∥f∥X for all f ∈X. (3.20) Proof. From coercivity ofAwe know that there exists a compact operatorC such that the linear operator D:= A+C:X → X is X-elliptic. From the Lax-Milgram Lemma 3.18 we get that there exists the inverse operator D−1: X → X. Therefore, we obtain that the equation (3.19) is equivalent to

Bu=D−1f

38 3 Boundary Integral Equations

with

B :=D−1A=D−1(D−C) =I−D−1C.

FromX-ellipticity ofDand boundedness ofAwe get thatB is bounded. Since we assume thatA is injective, it follows that the homogeneous equation

D−1Au= (I−D−1C)u= 0

only has the trivial solution and thus D−1C is injective. The Fredholm Alternative 3.19 with the compact operatorK :=D−1C then guarantees a unique solution to (3.19) satis-fying the estimate (3.20).

Theorem 3.21. Let X denote a Hilbert space. Then it holds (∀x, y∈X, x̸=y)(∃f ∈X) :⟨f, x⟩ ̸=⟨f, y⟩.

Proof. The theorem is a direct consequence of the well-known Hahn-Banach Theorem (see, e.g., [21], Theorem 1.B, Corollary 2 on pages 4–5).

In Section 2.3 we introduced representation formulae for the solution to the Helmholtz equation with smooth enough data. The following theorems generalize the previous results for a broader range of functions. For more details see [12], Theorems 6.10 and 7.12.

Theorem 3.22. Let Ω be a bounded Lipschitz domain and let u∈H1(Ω, ∆+κ2) satisfy the Helmholtz equation in the weak sense. Then we have the representation

u(x) = (Vκγ1,intu)(x)−(Wκγ0,intu)(x) for x∈Ω. (3.21) Theorem 3.23. Let Ω be a bounded Lipschitz domain and let us denote Ωext := R3\Ω.

Let u∈ Hloc1 (Ωext, ∆+κ2) satisfy the Sommerfeld radiation condition and the Helmholtz equation in the weak sense. Then we have the representation

u(x) =−(Vκγ1,extu)(x) + (Wκγ0,extu)(x) for x∈Ωext. (3.22) Before we proceed to individual boundary value problems, we provide theorems regard-ing existence and uniqueness of the weak solution to general boundary value problems for the Helmholtz equation (see [12] and [15]).

Theorem 3.24. The interior boundary value problem





∆u+κ2u= 0 inΩ, γ0,intu=gD on ΓD, γ1,intu=gN on ΓN

with a bounded Lipschitz domainΩ, non-overlapping setsΓD, ΓN satisfying ΓD∪ΓN=∂Ω and boundary conditions gD ∈ H1/2D), gN ∈ H−1/2N) has a unique weak solution

u∈H1(Ω, ∆+κ2)if and only ifκ2 does not coincide with an eigenvalueλof the eigenvalue problem for the Laplace equation





−∆uλ =λuλ in Ω, γ0,intuλ = 0 on ΓD, γ1,intuλ = 0 on ΓN.

(3.23)

Furthermore, if κ2 coincides with an eigenvalueλof (3.23), the solution exists if and only if

⟨γ0,intuλ, gNΓN =⟨γ1,intuλ, gDΓD for all corresponding eigenfunctions uλ.

Theorem 3.25. Let us denote Ωext := R3\Ω with a bounded Lipschitz domain Ω. The exterior boundary value problem













∆u+κ2u= 0 in Ωext, γ0,extu=gD on ΓD, γ1,extu=gN on ΓN,

∇u(x), x

∥x∥

−iκu(x)

=O

 1

∥x∥2

for ∥x∥ → ∞

with non-overlapping sets ΓD, ΓN satisfying ΓD∪ΓN=∂Ω and boundary conditions gD∈ H1/2D), gN∈H−1/2N) has a unique weak solution u∈Hloc1 (Ωext, ∆+κ2).

3.5.1 Interior Dirichlet Boundary Value Problem

Let us first consider the interior Dirichlet boundary value problem for the Helmholtz

equa-tion 

∆u+κ2u= 0 inΩ,

γ0,intu=gD on∂Ω (3.24)

with a bounded Lipschitz domainΩand the Dirichlet boundary conditiongD∈H1/2(∂Ω).

The solution to (3.24) is given by the representation formula (3.21) withγ0,intu=gD, i.e., u(x) = (Vκγ1,intu)(x)−(WκgD)(x) for x∈Ω (3.25) with unknown Neumann trace data γ1,intu ∈ H−1/2(∂Ω). Applying the Dirichlet trace operator γ0,int to (3.25) we obtain the Fredholm integral equation of the first kind

(Vκγ1,intu)(x) = 1

2gD(x) + (KκgD)(x) forx∈∂Ω. (3.26) Similarly, applying the Neumann trace operator γ1,int to (3.25) we obtain the Fredholm integral equation of the second kind

1

1,intu(x)−(Kκγ1,intu)(x) = (DκgD)(x) for x∈∂Ω. (3.27)

40 3 Boundary Integral Equations

According to Theorem 3.21, the boundary integral equations (3.26) and (3.27) are equivalent to the variational problems

⟨Vκγ1,intu, s⟩∂Ω =

1

2I+Kκ

 gD, s

∂Ω

for all s∈H−1/2(∂Ω) and

1

2I−Kκ

γ1,intu, t

∂Ω

=⟨DκgD, t⟩∂Ω for allt∈H1/2(∂Ω), respectively, where

⟨u, v⟩∂Ω :=

∂Ω

u(x)v(x) ds denotes the duality pairing betweenH1/2(∂Ω) and H−1/2(∂Ω).

The equations (3.26) and (3.27) are not uniquely solvable for all κ∈R+. To show this, let us consider the Dirichlet eigenvalue problem for the Laplace equation

 −∆uλ=λuλ inΩ,

γ0,intuλ= 0 on∂Ω, (3.28)

which can be understood as the Helmholtz boundary value problem with κ2 =λand the homogeneous Dirichlet boundary condition. Letλ∈R+ be an eigenvalue of (3.28) anduλ the corresponding eigensolution. From the boundary integral equations (3.26) and (3.27) we get

(Vκγ1,intuλ)(x) = 1

0,intuλ(x) + (Kκγ0,intuλ)(x) = 0 for x∈∂Ω, 1

1,intuλ(x)−(Kκγ1,intuλ)(x) = (Dκγ0,intuλ)(x) = 0 for x∈∂Ω

and hence, ifκ2 coincides with the eigenvalueλ, the operatorsVκ and(1/2I−Kκ)are not injective and thus not invertible.

On the other hand, for other values ofκboth operators are injective. The coercivity of the single layer potential operator then ensures a unique solution to the boundary integral equation (3.26) (see Theorem 3.20). The following theorem describes the relation between the weak solution and the solution given by the representation formula (see [12], Theorem 7.5).

Theorem 3.26. Ifu∈H1(Ω, ∆+κ2)is a solution to the interior Dirichlet problem (3.24), then the Neumann trace γ1,intu satisfies the boundary integral equation (3.26) and u has the representation (3.25).

Conversely, if γ1,intu satisfies the boundary integral equation (3.26), then the represen-tation formula (3.25)defines a solutionu∈H1(Ω, ∆+κ2)to the interior Dirichlet problem (3.24).

Another approach to computing the missing Neumann data is based on Theorems 3.6 and 3.7 suggesting that we may seek the solution in the form of a single layer potential

u(x) = (Vκs)(x) for x∈Ω (3.29) or a double layer potential

u(x) =−(Wκt)(x) for x∈Ω (3.30) with unknown density functionss, t. The density functions usually have no physical mean-ing and these methods are thus called indirect. The representations (3.29) and (3.30) give rise to the boundary integral equations

Vκs(x) =γ0,intu(x) for x∈∂Ω, (3.31) 1

2t(x)−(Kκt)(x) =γ0,intu(x) for x∈∂Ω and the corresponding variational problems

⟨Vκs, t⟩∂Ω =⟨γ0,intu, t⟩∂Ω for allt∈H−1/2(∂Ω),

1

2I−Kκ

 t, s

∂Ω

=⟨γ0,intu, s⟩∂Ω for alls∈H−1/2(∂Ω).

Note that the structure of (3.31) and (3.26) is similar, only the right-hand sides are differ-ent. Thus, for values of κ2 not coinciding with an eigenvalue of (3.28) we have a unique solution to (3.31).

Although the indirect approach can also be used for other types of boundary value problems, in the following sections we only consider the direct boundary element methods.

3.5.2 Interior Neumann Boundary Value Problem Solution to the interior Neumann boundary value problem

∆u+κ2u= 0 inΩ,

γ1,intu=gN on ∂Ω (3.32)

with a bounded Lipschitz domain Ω and the Neumann tracegN∈H−1/2(∂Ω) is given by the representation formula

u(x) = (VκgN)(x)−(Wκγ0,intu)(x) for x∈Ω (3.33) with unknown Dirichlet trace dataγ0,intu. Applying the Dirichlet trace operator to (3.33) we obtain the Fredholm integral equation of the second kind

1

0,intu(x) + (Kκγ0,intu)(x) = (VκgN)(x) forx∈∂Ω. (3.34)

42 3 Boundary Integral Equations

Similarly, applying the Neumann trace operator we get the Fredholm integral equation of the first kind

(Dκγ0,intu)(x) = 1

2gN(x)−(KκgN)(x) for x∈∂Ω. (3.35) Alternatively, we may consider the variational problems

1

2I+Kκ

γ0,intu, s

∂Ω

=⟨VκgN, s⟩∂Ω for alls∈H−1/2(∂Ω), and

⟨Dκγ0,intu, t⟩∂Ω =

1

2I−Kκ

 gN, t

∂Ω

for all t∈H1/2(∂Ω).

corresponding to (3.34) and (3.35), respectively.

To study the solvability of (3.34) and (3.35) let us first consider the Neumann eigenvalue problem for the Laplace equation

 −∆uµ=µuµ inΩ,

γ1,intuµ= 0 on ∂Ω. (3.36)

Let µ ∈ R+ be an eigenvalue of (3.36) with the corresponding eigensolution uµ. From (3.34) and (3.35) we obtain forκ2

1

0,intuµ(x) + (Kκγ0,intuµ)(x) = (Vκγ1,intuµ)(x) = 0 for x∈∂Ω, (Dκγ0,intuµ)(x) = 1

1,intuµ(x)−(Kκγ1,intuµ)(x) = 0 forx∈∂Ω,

implying that the functionγ0,intuµbelongs to the kernel of the boundary integral operators Dκ and (1/2I+Kκ) and thus these operators are not invertible.

For values ofκ2 not coinciding with eigenvalues of (3.36) both operators are injective.

Since the hypersingular operator Dκ is coercive, the boundary integral equation (3.35) is uniquely solvable.

Theorem 3.27. If u ∈ H1(Ω, ∆+κ2) is a solution to the interior Neumann problem (3.32), then the Dirichlet traceγ0,intu satisfies the boundary integral equation (3.35)andu has the representation (3.33).

Conversely, if γ0,intu satisfies the boundary integral equation (3.35), then the repre-sentation formula (3.33) defines a solution u ∈ H1(Ω, ∆+κ2) to the interior Neumann problem (3.32).

3.5.3 Interior Mixed Boundary Value Problem

In many engineering problems we have to deal with mixed boundary conditions, i.e., with problems of the type





∆u+κ2u= 0 inΩ, γ0,intu=gD on ΓD, γ1,intu=gN on ΓN

(3.37)

with a bounded Lipschitz domainΩ, non-overlapping setsΓD, ΓNsuch thatΓD∪ΓN=∂Ω, and functions gD ∈H1/2D) and gN ∈ H−1/2N). In the smooth case, the solution to (3.37) is given by the representation formula

u(x) =

ΓD

γ1,intu(y)vκ(x,y) dsy+

ΓN

gN(y)vκ(x,y) dsy

ΓD

gD(y) ∂vκ

∂ny

(x,y) dsy

ΓN

γ0,intu(y) ∂vκ

∂ny

(x,y) dsy for x∈Ω with unknown Dirichlet dataγ0,intu|ΓN and Neumann dataγ1,intu|ΓD. The missing Cauchy data can be obtained using any of the previously mentioned boundary integral equations (3.26), (3.27), (3.34) or (3.35). However, in this section we describe the so-called symmetric formulation using the single layer potential equation (3.26) and the hypersingular equation (3.35) simultaneously (see [17], Section 1.1.3).

From Sections 3.5.1 and 3.5.2 we have the relations (Vκγ1,intu)(x) = 1

0,intu(x) + (Kκγ0,intu)(x) for x∈ΓD⊂∂Ω, (Dκγ0,intu)(x) = 1

1,intu(x)−(Kκγ1,intu)(x) for x∈ΓN⊂∂Ω.

(3.38)

Let us define functions

H1/2N)∋t:=γ0,intu−g˜D, H−1/2D)∋s:=γ1,intu−g˜N

(3.39) with some suitable extensionsg˜D,˜gNdefined on∂Ω. Inserting the functionst, sinto (3.38) we obtain the system of equations

(Vκs)(x)−(Kκt)(x) = 1

2˜gD(x) + (Kκ˜gD)(x)−(Vκ˜gN)(x) for x∈ΓD, (Kκs)(x) + (Dκt)(x) = 1

2˜gN(x)−(Kκ˜gN)(x)−(Dκ˜gD)(x) for x∈ΓN

(3.40)

equivalent to the variational formulation

a(s, t, q, r) =F(q, r) for all q∈H−1/2D), r ∈H1/2N) (3.41) with unknown functions s∈H−1/2D),t∈H1/2N), the bilinear form

a(s, t, q, r) :=⟨Vκs, q⟩ΓD− ⟨Kκt, q⟩ΓD+⟨Kκs, r⟩ΓN+⟨Dκt, r⟩ΓN

44 3 Boundary Integral Equations

and the right-hand side F(q, r) := the functions s, t satisfy the system of boundary integral equations (3.40) and u has the representation

u(x) = (Vκ(s+ ˜gN))(x)−(Wκ(t+ ˜gD))(x) for x∈Ω. (3.42) Conversely, if s, t satisfy the system of boundary integral equations (3.40), then the representation formula (3.42) defines a solution u∈H1(Ω, ∆+κ2) to the interior mixed problem (3.37).

3.5.4 Exterior Dirichlet Boundary Value Problem

Let us denoteΩext:=R3\Ωwith a simply connected bounded Lipschitz domain Ω⊂R3 so that∂Ωext =∂Ω. Now we consider the exterior Dirichlet boundary value problem

with the Dirichlet boundary condition gD ∈ H1/2(∂Ω). The solution can be represented as

u(x) =−(Vκγ1,extu)(x) + (WκgD)(x) for x∈Ωext (3.44) with unknown Neumann trace dataγ1,extu. Applying the Dirichlet trace operator to (3.44) we obtain the Fredholm integral equation of the first kind

(Vκγ1,extu)(x) =−1

2gD(x) + (KκgD)(x) for x∈∂Ω. (3.45) Applying the Neumann trace gives rise to the Fredholm integral equation of the second

kind 1

1,extu(x) + (Kκγ1,extu)(x) =−(DκgD)(x) for x∈∂Ω. (3.46) The boundary integral equations (3.45) and (3.46) are equivalent to the variational problems

respectively.

The unique solvability of the boundary integral equation (3.45) and the related varia-tional problem (3.47) for κ2 not coinciding with an eigenvalue of (3.28) follows from the discussion in Section 3.5.1.

Theorem 3.29. If u ∈Hloc1 (Ωext, ∆+κ2) is a solution to the exterior Dirichlet problem (3.43), then the Neumann trace γ1,extu satisfies the boundary integral equation (3.45)and u has the representation (3.44).

Conversely, if γ1,extu satisfies the boundary integral equation (3.45), then the represen-tation formula (3.44) defines a solution u ∈ Hloc1 (Ωext, ∆+κ2) to the exterior Dirichlet problem (3.43).

Note that Theorems 3.25 and 3.29 ensure that for all κ∈R+ it holds

−1

2I+Kκ

gD∈ImVκ

even though the equation (3.45) is not uniquely solvable for some values of κ. To over-come this difficulty, several approaches have been proposed. The approach of Brakhage and Werner (see [5]) suggests seeking the solution in the form of some complex linear combination of a single and double layer potentials, i.e.,

u(x) = (Wκw)(x)−iη(Vκw)(x) for x∈Ωext

with an unknown density function w ∈ L2(∂Ω). In [4], Burton and Miller suggest an alternative to direct methods, using a complex linear combination of the boundary integral equations (3.17) and (3.18). To conclude, we also mention the CHIEF method proposed by Schenck (see [2]).

3.5.5 Exterior Neumann Boundary Value Problem

Now we consider the exterior Neumann boundary value problem









∆u+κ2u= 0 inΩext, γ1,extu=gN on ∂Ω,

∇u(x), x

∥x∥

−iκu(x)

=O

 1

∥x∥2

for ∥x∥ → ∞

(3.48)

with an unbounded domainΩext :=R3\Ωand the Neumann tracegN∈H−1/2(∂Ω). The solution is given by

u(x) =−(VκgN)(x) + (Wκγ0,extu)(x) for x∈Ωext (3.49) with unknown Dirichlet trace dataγ0,extu. Applying the Dirichlet trace operator to (3.49) we obtain the Fredholm integral equation of the second kind

−1

0,extu(x) + (Kκγ0,extu)(x) = (VκgN)(x) for x∈∂Ω. (3.50)

46 3 Boundary Integral Equations

Similarly, applying the Neumann trace operator we get the Fredholm integral equation of the first kind

(Dκγ0,extu)(x) =−1

2gN(x)−(KκgN)(x) for x∈∂Ω. (3.51) Instead of equations (3.50) and (3.51) we may consider the variational problems



−1

2I+Kκ

γ0,extu, s

∂Ω

=⟨VκgN, s⟩∂Ω for all s∈H−1/2(∂Ω), and

⟨Dκγ0,extu, t⟩∂Ω =



−1

2I −Kκ

 gN, t

∂Ω

for allt∈H1/2(∂Ω), (3.52) respectively.

The arguments for the unique solvability of the boundary integral equation (3.51) and the corresponding variational problem (3.52) for κ2 not coinciding with an eigenvalue of (3.36) are the same as in Section 3.5.2.

Theorem 3.30. If u∈Hloc1 (Ωext, ∆+κ2) is a solution to the exterior Neumann problem (3.48), then the Dirichlet trace γ0,extu satisfies the boundary integral equation (3.51) and u has the representation (3.49).

Conversely, if γ0,extu satisfies the boundary integral equation (3.51), then the represen-tation formula (3.49) defines a solution u ∈ Hloc1 (Ωext, ∆+κ2) to the interior Neumann problem (3.48).

Note that Theorems 3.25 and 3.30 ensure that for all κ∈R+ it holds

−1

2I−Kκ

gN∈ImDκ

and thus one of the methods mentioned at the end of the previous section can be used to compute a solution to (3.51) withκ2 coinciding with an eigenvalue of (3.36).

3.5.6 Exterior Mixed Boundary Value Problem

Finally, let us consider the exterior mixed boundary value problem













∆u+κ2u= 0 inΩext, γ0,extu=gD onΓD, γ1,extu=gN onΓN,

∇u(x), x

∥x∥

−iκu(x)

=O

 1

∥x∥2

for ∥x∥ → ∞

(3.53)

with an unbounded domain Ωext := R3 \Ω and boundary conditions gD ∈ H1/2D), gN∈H−1/2N). In the smooth case the solution to (3.53) is given by

u(x) =−

ΓD

γ1,extu(y)vκ(x,y) dsy

ΓN

gN(y)vκ(x,y) dsy +

ΓD

gD(y) ∂vκ

∂ny(x,y) dsy+

ΓN

γ0,extu(y) ∂vκ

∂ny(x,y) dsy for x∈Ωext with unknown Cauchy data γ0,extu|ΓN and γ1,extu|ΓD. Similarly as for the interior mixed problem we have the boundary integral equations

(Vκγ1,extu)(x) =−1

0,extu(x) + (Kκγ0,extu)(x) for x∈ΓD⊂∂Ω, (Dκγ0,extu)(x) =−1

1,extu(x)−(Kκγ1,extu)(x) for x∈ΓN⊂∂Ω.

(3.54)

Inserting the functions t, sdefined by (3.39) into (3.54) yields (Vκs)(x)−(Kκt)(x) =−1

2g˜D(x) + (KκD)(x)−(VκN)(x) for x∈ΓD, (Kκs)(x) + (Dκt)(x) =−1

2g˜N(x)−(KκN)(x)−(DκD)(x) for x∈ΓN

(3.55)

and the equivalent variational formulation

a(s, t, q, r) =F(q, r) for all q∈H−1/2D), r ∈H1/2N) (3.56) with unknown functions s∈H−1/2D),t∈H1/2N), the bilinear form

a(s, t, q, r) :=⟨Vκs, q⟩ΓD− ⟨Kκt, q⟩ΓD+⟨Kκs, r⟩ΓN+⟨Dκt, r⟩ΓN and the right-hand side

F(q, r) :=



−1

2I+Kκ

˜ gD, q

ΓD

− ⟨VκN, q⟩ΓD+



−1

2I−Kκ

˜ gN, r

ΓN

− ⟨DκD, r⟩ΓN. Theorem 3.31. If u∈Hloc1 (Ωext, ∆+κ2) is a solution to the mixed problem (3.53), then the functions s, t satisfy the system of boundary integral equations (3.55) and u has the representation

u(x) =−(Vκ(s+ ˜gN))(x) + (Wκ(t+ ˜gD))(x) for x∈Ωext. (3.57) Conversely, if s, t satisfy the system of boundary integral equations (3.55), then the representation formula (3.57) defines a solution u ∈ Hloc1 (Ωext, ∆+κ2) to the interior mixed problem (3.53).

4 Discretization and Numerical Realization

In this section we will describe the discretization of the boundary integral equations covered in the previous part. We will discuss the collocation method as well as the more complicated Galerkin scheme. The solution will be sought in approximating function spaces introduced below.

We restrict ourselves to triangular meshes approximating the corresponding boundary

∂Ω, i.e.,

∂Ω ≈

E

k=1

τk

and assume that neighbouring triangular elements either share a whole edge or a sin-gle vertex. Each element τk ⊂ R3 with nodes xk1,xk2,xk3 can be described via the parametrization

Rk(ξ) :=xk1+Rkξ for ξ∈ˆτ , (4.1) whereRk denotes the matrix

Rk :=

xk2−xk1 xk3 −xk1

=

xk12 −xk11 xk13 −xk11 xk22 −xk21 xk23 −xk21 xk32 −xk31 xk33 −xk31

∈R3×2 (4.2) and τˆ⊂R2 represents the reference triangle (see Figure 4.1a)

ˆ

τ :={ξ∈R2: 0< ξ1 <1,0< ξ2<1−ξ1}.

Remark 4.1. Note that a function f defined on τk can be identified with a function fˆ defined on τˆand vice versa, i.e.,

f(x) =f(Rk(ξ)) =: ˆf(ξ) for x∈τk,ξ ∈τ .ˆ

In the following text we use the symbol ∂Ω to denote the discretized boundary.

4.1 Piecewise Constant Basis Functions

For every elementτk we define the functionψk (see Figure 4.2a) as ψk(x) :=

1 for x∈τk, 0 otherwise.

Following Remark 4.1, the functionψk can alternatively be defined via the reference func-tion

ψ(ξ) :=ˆ

1 for ξ∈τ ,ˆ 0 otherwise.

50 4 Discretization and Numerical Realization

ξ1

ξ2

ˆ τ

1 1

(a) Reference triangle.

τk xk1

xk3

xk2 x1

x3

x2

(b) General triangle.

Figure 4.1: Triangular mesh elements.

τk ψk(x)

(a) Piecewise constant basis function.

xk ϕk(x)

(b) Piecewise affine basis function.

Figure 4.2: Basis functions.

Furthermore, we define the linear spaceTψ(∂Ω) as Tψ(∂Ω) := span{ψk}Ek=1,

whereE denotes the number of elements. Obviously,dimTψ(∂Ω) =Eand every complex-valued function gψ ∈Tψ(∂Ω)can be represented as

gψ =

E

k=1

gkψk

and thus it can be identified with a vectorg:= [g1, . . . , gE]T∈CE.

A smooth enough function gcan be approximated by a linear combination of piecewise constant functions in several ways. The simplest approach is to set

gk=g(xk)

withxk denoting the midpoint ofτk. Alternatively, we can set

However, since we deal with functions in Lebesgue and Sobolev spaces, where we do not distinguish between functions differing on a measure zero set, the above mentioned approxi-mations may prove impossible. Hence, we introduce the projectionPψ:L2(∂Ω)→Tψ(∂Ω) minimizing the error

∥g−Pψg∥L2(∂Ω), i.e.,

Pψg:= arg min

gψ∈Tψ(∂Ω)

∥g−gψL2(∂Ω).

To find the coefficients gk for a real-valued function g we introduce the function f:RE →Ras

For the partial derivatives off we get

∂f

Moreover, for the Hessian of f we have Hf[j, i] = ∂2f

∂gj∂gi = 2⟨ψi, ψj⟩,

which is the double Gram matrix corresponding to the basis ofTψ(∂Ω). Hence, the Hessian is positive definite and to find the minimum of f we set ∂g∂f

j = 0 to obtain the system of

52 4 Discretization and Numerical Realization

where∆k denotes the surface ofτk, we get for the piecewise constant approximation gk= 1

k

τk

g(x) dsx.

Let us now consider the case when g= Reg+ i Img is complex-valued. We thus search for coefficientsgk= Regk+ i Imgk. For theL2 error norm we get

and therefore, the problem is reduced to searching for appropriate coefficients of the two real-valued functions Reg andImg as was described in the previous paragraph.

The spaceTψ(∂Ω)will be used for the approximation of the Neumann data, i.e., of the normal derivatives on ∂Ω.

4.2 Continuous Piecewise Affine Basis Functions

LetN denote the number of all nodes of a given discretization. Then we define the family of functions {ϕk}Nk=1 continuous over the whole discretized boundary (see Figure 4.2b) as

ϕk(x) := piecewise affine otherwise.

Furthermore, we define the linear spaceTϕ(∂Ω)as Tϕ(∂Ω) := span{ϕk}Nk=1.

Obviously, dimTϕ(∂Ω) = N. A restriction of a function ϕk to an element τ ⊂ suppϕk can be identified with one of the functions ϕˆ1,ϕˆ2,ϕˆ3 defined on the reference triangle as

ˆ

ϕ1(ξ) := 1−ξ1−ξ2, ϕˆ2(ξ) :=ξ1, ϕˆ3(ξ) :=ξ2 for ξ∈τ .ˆ

k\τk xk1 xk2 xk3 local indices

1 1 2 3

global indices

2 2 4 3

3 4 5 3

... ... ... ...

τ1 τ3

τ2

x3

x4 x2

x1 x5

Figure 4.3: Connectivity table.

We use the connectivity table (see Figure 4.3) to determine the mapping between global and local indices of nodes building the triangular mesh. For example, the point x4 is the second vertex of the triangleτ2, i.e., x4=x22. Furthermore, forϕ4|τ2 we have

ϕ4|τ2(x) =ϕ4|τ2(R4(ξ)) = ˆϕ2(ξ) =ξ1 for x∈τ2,ξ∈ˆτ .

As in the case of piecewise constant basis functions we can identify a function Tϕ(∂Ω)∋gϕ=

N

k=1

gkϕk

with a vector [g1, . . . , gN]T ∈ CN. For the approximation of a smooth enough functiong we can either use an interpolation, i.e.,

gk=g(xk).

For a function g ∈L2(∂Ω) it is more appropriate to define the projection Pϕ: L2(∂Ω)→ Tϕ(∂Ω) as

Pϕg:= arg min

gϕ∈Tϕ(∂Ω)

∥g−gϕL2(∂Ω).

The coefficients for real-valued functions can be found in the same way as for the piecewise constant functions as the solution to the system of linear equations

N

k=1

gk⟨ϕk, ϕjL2(∂Ω)=⟨g, ϕjL2(∂Ω) for j∈ {1, . . . , N}.

In the case of complex-valued functions we can separately search for the real parts and the imaginary parts of the coefficients.

The space Tϕ(∂Ω) will be used for the approximation of the Dirichlet data, i.e., the values of the solution on∂Ω.

4.3 Discretized Boundary Integral Equations

In the following sections we describe the discretization of the boundary integral equations derived in Section 3.5. Although the collocation method will be mentioned, we will prefer

54 4 Discretization and Numerical Realization

the well studied Galerkin scheme based on the corresponding variational formulations. To

the well studied Galerkin scheme based on the corresponding variational formulations. To