Representation Formulae

−^π₂

1

εe^iκε1−iκε ε ϕ

y+ε(cosϑcosψ,sinϑcosψ,sinψ)

ε²cosψdψdϑ

= 1 4π

 2π 0

 ^π

2

−^π₂

e^iκε(1−iκε)ϕ

y+ε(cosϑcosψ,sinϑcosψ,sinψ)

cosψdψdϑ.

The Lebesgue dominated convergence theorem allows us to interchange the limit and the integration. Moreover, becauseϕ∈C₀^∞(R³)and

ε→0lim+

e^iκε(1−iκε) = 1, we obtain

ε→0lim+

I2= 1 4π

 2π 0

 ^π

2

−^π

2

ϕ(y) cosψdψdϑ=ϕ(y) and finally

I =⟨∆˜v_κ+κ²v˜_κ, ϕ⟩= lim

ε→0₊I₁− lim

ε→0₊I₂ =−ϕ(y) =⟨−δ_y, ϕ⟩, which was to be proved.

2.3 Representation Formulae

The following two theorems form the basis of the boundary element method, providing formulae for calculating the solution to a boundary value problem in any point of the given domain. Firstly, we focus on interior problems, i.e., problems on bounded domains.

Theorem 2.6 (Representation Formula for Bounded Domains). LetΩ⊂R³ be a bounded C¹ domain, let v_κ denote the fundamental solution for the Helmholtz equation in R³ and let n denote the unit outward normal vector to ∂Ω. Then for u ∈ C²(Ω) we have the representation formula

u(x) =



∂Ω

∂u

∂n(y)vκ(x,y) dsy−



∂Ω

u(y)∂vκ

∂ny

(x,y) dsy

−



Ω

∆u(y) +κ²u(y)

v_κ(x,y) dy for x∈Ω.

In particular, if u satisfies the Helmholtz equation Proof. The proof is constructed in the same way as in the case of Theorem 2.5. Letx∈Ω be an arbitrary fixed point. Let us choose ε >0 such that

B_ε(x) :={y∈R³:∥x−y∥< ε} ⊂B_ε(x)⊂Ω

and let us denoteΩε:=Ω\Bε(x)(similar situation as in Figure 2.2 with points xand y swapped). From Theorem 2.2 and the symmetry of v_κ we have

∆yvκ(x,y) +κ²vκ(x,y) = 0 for ally∈Ωε.

Using the second Green’s identity with functions uand vκ on Ωε we obtain

 it holds that

v_κ(x,y) = 1

for ally∈∂B_ε(x). Using the parametrization as in the proof of Theorem 2.4 (but withx and yswapped) we have for the integralI1

I1= 1

24 2 Helmholtz Equation

and due to properties ofu

ε→0lim₊I₁ = 0.

For the integralI₂ we obtain I₂= 1 and using the Lebesgue dominated convergence theorem and qualities ofu we have

ε→0lim₊I₂=u(x).

Since the pointx∈Ω was arbitrary, we have proved Theorem 2.6.

When solving a problem of sound scattering or wave propagation described by the Helmholtz equation we are usually interested in the solution in an unbounded domain.

The following theorem gives us the representation formula for such domains (see [6]).

Theorem 2.7 (Representation Formula for Unbounded Domains). Let Ω ⊂ R³ be a bounded C¹ domain, let v_κ denote the fundamental solution for the Helmholtz equation in R³ and letn denote the unit outward normal vector to ∂Ω. Let us define Ω^ext :=R³\Ω.

Then for u∈C²(Ω^ext) satisfying

∆u+κ²u= 0 in Ω^ext and the Sommerfeld radiation condition

 we have the representation formula

u(x) =

Ωε

Figure 2.3: Illustration for the proof of Theorem 2.7.

Proof. LetB_ε(0) :={y∈R³:∥y∥< ε}, where εis taken such that Ω⊂B_ε(0). Further-more, let Ω_ε :=B_ε(0)\Ω (see Figure 2.3). We start with showing that



∂Bε(0)

|u(y)|²ds=O(1) for ε→ ∞.

From the radiation condition we deduce

ε→∞lim we get for the integrand from (2.9)



26 2 Helmholtz Equation Now we apply the first Green’s identity (1.7) in Ωε for functionsuand u¯ to obtain

 where we had to take into account the opposite direction ofnfory∈∂Ω (see Figure 2.3).

Because the left-hand side of (2.12) is real, we obtain Im Inserting (2.13) into (2.10) we get

ε→∞lim

Because both terms on the left-hand side are non-negative, they have to be bounded for ε→ ∞. Therefore, we have proved that



∂Bε(0)

|u(y)|²ds=O(1) for ε→ ∞. (2.14) From Theorem 2.3 we know that the radiation condition and thus also (2.14) is valid for the fundamental solution v_κ. Using the Hölder inequality we thus obtain

I1:=

and I₂:=



∂Bε(0)

v_κ(x,y)

∂u

∂n(y)−iκu(y)

 ds_y

≤



∂Bε(0)

|v_κ(x,y)|²ds_y

1/2



∂Bε(0)









∂u

∂n(y)−iκu(y)









2

ds_y

1/2

→0 for ε→ ∞.

(2.16) Finally, for the bounded domain Ωε we can apply the representation formula (2.7) to obtain

u(x) =



∂Ωε

∂u

∂n(y)vκ(x,y) dsy−



∂Ωε

u(y)∂v_κ

∂ny

(x,y) dsy

=−



∂Ω

∂u

∂n(y)vκ(x,y) dsy+



∂Ω

u(y)∂v_κ

∂ny

(x,y) dsy+I2−I1

for all x∈Ωε. Lettingε→ ∞, using (2.15) and (2.16), we eventually get u(x) =



∂Ω

u(y)∂v_κ

∂n_y(x,y) ds_y−



∂Ω

∂u

∂n(y)v_κ(x,y) ds_y for all x∈Ω^ext, which was to be proved.

Theorems 2.6 and 2.7 provide representation of the solution to the Helmholtz equation by means of boundary integrals. The functions

(V_κs)(x) :=



∂Ω

v_κ(x,y)s(y) ds_y and (W_κt)(x) :=



∂Ω

∂v_κ

∂n_y(x,y)t(y) ds_y are called potentials with density functionss, t. Although we have only discussed the case of smooth data, in the following section we will show that the potentials can also be defined for more general density functions and domains. The properties of the integral operators V_κ and W_κ will play a crucial role in obtaining the missing Cauchy data.

3 Boundary Integral Equations

At the beginning of this section we introduce some operator properties, which will be used to prove existence and uniqueness of solutions to boundary integral equations derived later in Section 3.5. In the following definitions we assumeX andY to be some Hilbert spaces.

Definition 3.1. A linear operator A: X → X^∗ is X-elliptic if there exists a constant c^A₁ ∈R+ such that the inequality

Re⟨Au, u⟩ ≥c^A₁∥u∥²_X holds for allu∈X.

Definition 3.2. A linear operatorA:X →Y is bounded if there exists a constantc^A₂ ∈R+

such that the inequality

∥Au∥_Y ≤c^A₂∥u∥_X holds for allu∈X.

Remark 3.3. Note that for linear operators boundedness is equivalent to continuity.

Definition 3.4. A linear operator A: X → X^∗ is coercive if there exists a compact operator C:X →X^∗ and a constant c^A₁ ∈R+ such that the Gårdings inequality

Re⟨(A+C)u, u⟩ ≥c^A₁∥u∥²_X holds for allu∈X.

In the previous section we showed that in the smooth case the solution to the Helmholtz equation on a bounded domain can be represented as

u(x) =



∂Ω

∂u

∂n(y)v_κ(x,y) ds_y−



∂Ω

u(y)∂v_κ

∂n_y(x,y) ds_y for x∈Ω. (3.1) For the solution on an unbounded domain we have similarly

u(x) =−



∂Ω

∂u

∂n(y)vκ(x,y) dsy+



∂Ω

u(y)∂vκ

∂ny

(x,y) dsy for x∈Ω^ext. (3.2) In both cases the solution u is determined by the Cauchy data, i.e., by the values of the solution on the boundary and its normal derivatives. However, these data are not given beforehand as the boundary value problem would be overdetermined. When considering a mixed problem, we are only given Dirichlet data on a part of the boundary and Neumann data on the remaining part. In this section we show how the Cauchy couple can be obtained from boundary integral equations derived from the representation formulae (3.1) and (3.2).

Afterwards, these formulae can be used to compute the solution in any point x ∈ Ω or x∈Ω^ext, respectively.

30 3 Boundary Integral Equations

Introducing the integral operators

Vκ:H^−1/2(∂Ω)→H_loc¹ (Ω), (Vκs)(x) :=



∂Ω

vκ(x,y)s(y) dsy, (3.3) W_κ:H^1/2(∂Ω)→H_loc¹ (Ω), (W_κt)(x) :=



∂Ω

∂v_κ

∂ny

(x,y)t(y) ds_y (3.4) with density functions s, t:∂Ω →Rallows us to rewrite the representation formulae (3.1) and (3.2) as

u=V_κγ^1,intu−W_κγ^0,intu inΩ, u=−V_κγ^1,extu+W_κγ^0,extu inΩ^ext.

Remark 3.5. The operatorsV_κandW_κare usually called the single layer potential operator and the double layer potential operator, respectively. However, there is a naming conflict.

In the following sections we also use these terms for composite operatorsγ⁰Vκ andγ⁰Wκ, whereγ⁰ denotes the interior or exterior Dirichlet trace operator.

Theorem 3.6. The operatorVκ:H^−1/2(∂Ω)→H_loc¹ (Ω)is linear and continuous. Hence, for bounded domains there exists a constant c∈R+ such that

∥V_κs∥_H1(Ω)≤c∥s∥_H−1/2(∂Ω) for all s∈H^−1/2(∂Ω).

Moreover, for all s∈H^−1/2(∂Ω) the function Vκs satisfies the Helmholtz equation in the weak sense (including the Sommerfeld radiation condition in the case of an unbounded domain).

Theorem 3.7. The operator Wκ:H^1/2(∂Ω)→H_loc¹ (Ω) is linear and continuous. Hence, for bounded domains there exists a constant c∈R+ such that

∥W_κt∥_H1(Ω)≤c∥t∥_H1/2(∂Ω) for all t∈H^1/2(∂Ω).

Moreover, for all t ∈ H^1/2(∂Ω) the function Wκt satisfies the Helmholtz equation in the weak sense (including the Sommerfeld radiation condition in the case of an unbounded domain).

The preceding theorems allow us to seek the solution to the Helmholtz equation in the formu=V_κsor u=W_κtwith unknown density functionss, t. This approach gives rise to indirect boundary element methods, which will be mentioned later.

Properties of the above given potential operators, which will be discussed in the follow-ing sections, can be found in [12] (see also [9] and [18]). These properties will be crucial for the derivation of boundary integral equations (both direct and indirect), which will be used for the computation of the missing Cauchy data.

3.1 Single Layer Potential Operator

Let us first consider the operator V_κ defined by (3.3). Recall, that for the Dirichlet trace operator γ⁰ we have

γ⁰:H_loc¹ (Ω)→H^1/2(∂Ω).

Combining this operator withV_κ we obtain the single layer potential operator Vκ:H^−1/2(∂Ω)→H^1/2(∂Ω), Vκ:=γ⁰Vκ.

From linearity and continuity ofγ⁰andV_κwe get that the single layer potential operator is linear and continuous, i.e., there exists a constant c∈R+ such that

∥V_κs∥_H1/2(∂Ω)≤c∥s∥_H−1/2(∂Ω) for alls∈H^−1/2(∂Ω).

Theorem 3.8. The single layer potential operator V_κ:H^−1/2(∂Ω) → H^1/2(∂Ω) is coer-cive.

Proof. From Theorem 6.22 in [18] we have that he operatorV₀corresponding to the Laplace equation, i.e., the Helmholtz equation with κ = 0, is H^−1/2(∂Ω)-elliptic. Moreover, the operator C:=V0−Vκ:H^−1/2(∂Ω)→H^1/2(∂Ω) is compact (see [18], Section 6.9). Thus, we have

⟨(V_κ+C)s, s⟩=⟨V₀s, s⟩ ≥c∥s∥²_H−1/2(∂Ω) for alls∈H^−1/2(∂Ω), which completes the proof.

Theorem 3.9. For s∈L^∞(∂Ω) there holds the representation (V_κs)(x) =



∂Ω

v_κ(x,y)s(y) ds_y for x∈∂Ω.

Proof. The proof is similar to the proof of Lemma 6.7 in [18].

Moreover, for the jump of the Dirichlet trace of the single layer potential Vκs on the boundary we have

[γ⁰Vκs] :=γ^0,extVκs−γ^0,intVκs= 0 for all s∈H^−1/2(∂Ω). (3.5) 3.2 Adjoint Double Layer Potential Operator

In Section 1.4 we introduced the Neumann trace operator γ¹:H_loc¹ (Ω, ∆+κ²)→H^−1/2(∂Ω).

32 3 Boundary Integral Equations

Because the function Vκs satisfies the Helmholtz equation in the weak sense, we actually have

V_κ:H^−1/2(∂Ω)→H_loc¹ (Ω, ∆+κ²)

and thus we can compose the Neumann trace operator with the single layer potential operator to obtain the linear continuous mapping

γ¹Vκ:H^−1/2(∂Ω)→H^−1/2(∂Ω).

Theorem 3.10. For s∈H^−1/2(∂Ω) there holds

γ^1,int(V_κs)(x) =σ(x)s(x) + (K_κ^∗s)(x) for x∈∂Ω, (3.6) γ^1,ext(Vκs)(x) = (σ(x)−1)s(x) + (K_κ^∗s)(x) for x∈∂Ω, (3.7) where K_κ^∗ denotes the adjoint double layer potential operator

(K_κ^∗s)(x) :=



∂Ω

∂v_κ

∂n_x(x,y)s(y) ds_y for x∈∂Ω and

σ(x) := lim

ε→0+



y∈Ω:∥x−y∥=ε

∂v_κ

∂n_y(x,y) ds_y. (3.8) Proof. The proof can be found in [9], following Lemma 2.29.

Remark 3.11. For the function σ we get σ(x) = lim

ε→0+

1 4π



y∈Ω:∥x−y∥=ε

⟨x−y,n(y)⟩e^{iκ∥x−y∥}

∥x−y∥²

 1

∥x−y∥ −iκ

 dsy. Fornwe can substitute

n(y) = x−y

∥x−y∥

to obtain

σ(x) = lim

ε→0+

1 4π



y∈Ω:∥x−y∥=ε

e^{iκ∥x−y∥}

∥x−y∥

 1

∥x−y∥ −iκ

 ds_y

= lim

ε→0+

1 4πe^iκε

1 ε² −iκ

ε

 

y∈Ω:∥x−y∥=ε

1 dsy.

Thus, the function σ depends on the interior angle of Ω in x ∈ ∂Ω. In particular, for domains with boundary smooth in x ∈ ∂Ω we get σ(x) = 1/2. For Lipschitz domains the relation σ(x) = 1/2is valid almost everywhere and we can simplify (3.6) and (3.7) to obtain

γ^1,int(Vκs)(x) = 1

2s(x) + (K_κ^∗s)(x) for x∈∂Ω, (3.9) γ^1,ext(Vκs)(x) =−1

2s(x) + (K_κ^∗s)(x) for x∈∂Ω. (3.10)

For the jump of the Neumann trace of the single layer potential Vκson the boundary we thus get

[γ¹Vκs] :=γ^1,extVκs−γ^1,intVκs=−s for all s∈H^−1/2(∂Ω).

3.3 Double Layer Potential Operator

Let us now consider the operatorWκ defined by (3.4). We define the corresponding bound-ary integral operator as the composition of γ⁰ andWκ to get the mapping

γ⁰Wκ:H^1/2(∂Ω)→H^1/2(∂Ω).

From the properties of the Dirichlet trace operator and the integral operator Wκ we get thatγ⁰W_κ is linear and continuous.

Theorem 3.12. For t∈H^1/2(∂Ω) there holds

γ^0,int(Wκt)(x) = (σ(x)−1)t(x) + (Kκt)(x) for x∈∂Ω, γ^0,ext(Wκt)(x) =σ(x)t(x) + (Kκt)(x) forx∈∂Ω, where Kκ denotes the double layer potential operator

(K_κt)(x) :=



∂Ω

∂v_κ

∂ny

(x,y)t(y) ds_y 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 γ^0,int(Wκt)(x) =−1

2t(x) + (Kκt)(x) for x∈∂Ω, (3.11) γ^0,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

[γ⁰W_κt] :=γ^0,extW_κt−γ^0,intW_κt=t for all t∈H^1/2(∂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.,

D_κ:H^1/2(∂Ω)→H^−1/2(∂Ω), D_κ:=−γ¹W_κ.

34 3 Boundary Integral Equations

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

∥D_κt∥_H−1/2(∂Ω)≤c∥t∥_H1/2(∂Ω) for allt∈H^1/2(∂Ω).

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

(D_κt)(x) =−γ¹(W_κt)(x) =− lim

Ω∋x→x∈∂Ω˜ ⟨n(x),∇_x˜(W_κt)( ˜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 (W_κt)( ˜x)

By computing the normal derivative of the integrand we get (D_κt)(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

⟨D_κt, s⟩_∂Ω :=



∂Ω

(D_κt)(x)s(x) ds_x

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

Theorem 3.14. For functions s, t∈H^1/2(∂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:H^1/2(∂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 H^1/2(∂Ω)-elliptic. Due to the compact embedding H^1/2(∂Ω) ↩→↩→ H^−1/2(∂Ω) and the compactness of Dκ −D0, the operator C:=I+D₀−D_κ:H^1/2(∂Ω)→H^−1/2(∂Ω) is compact. Thus, we have

⟨(D_κ+C)t, t⟩=⟨(D₀+I)t, t⟩ ≥c∥t∥²

H^1/2(∂Ω) for all t∈H^1/2(∂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

[γ¹W_κt] :=γ^1,extW_κt−γ^1,intW_κt= 0 for all t∈H^1/2(∂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=−W_κγ^0,intu+Vκγ^1,intu inΩ.

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

γ^0,intu=

1

2I−Kκ



γ^0,intu+Vκγ^1,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

γ^1,intu=D_κγ^0,intu+

1

2I+K_κ^∗



γ^1,intu on ∂Ω. (3.16)

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

γ^0,intu γ^1,intu



=C^int

γ^0,intu γ^1,intu



36 3 Boundary Integral Equations

with the Calderon projection matrix C^int:=

₁

2I−Kκ Vκ

D_κ ¹₂I+K_κ^∗

 .

For the solution in an unbounded domain we have the formula u=W_κγ^0,extu−V_κγ^1,extu inΩ^ext.

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

γ^0,extu=

1

2I+K_κ



γ^0,extu−V_κγ^1,extu on∂Ω^ext, (3.17) γ^1,extu=−D_κγ^0,extu+

1

2I−K_κ^∗



γ^1,extu on∂Ω^ext. (3.18) Again, the boundary integral equations (3.17), (3.18) can be rewritten as

γ^0,extu γ^1,extu



=C^ext

γ^0,extu γ^1,extu



with the Calderon projection matrix C^ext:=

₁

2I+K_κ −V_κ

−D_κ ¹₂I−K_κ^∗

 .

Theorem 3.16. The Calderon operators C^int and C^ext are projection operators, i.e., (C^int)²=C^int (C^ext)² =C^ext.

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 V_κD_κ =

1

2I+K_κ

 1

2I−K_κ

 ,

DκVκ =

1

2I+K_κ^∗

 1

2I−K_κ^∗

 , DκKκ =K_κ^∗Dκ,

KκVκ =VκK_κ^∗.

Proof. The relations follow directly from the comparison of the matrices(C^int)² and C^int.

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 c^A₁ ∥f∥_X^∗ with the ellipticity constant c^A₁.

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κγ^1,intu)(x)−(Wκγ^0,intu)(x) for x∈Ω. (3.21) Theorem 3.23. Let Ω be a bounded Lipschitz domain and let us denote Ω^ext := R³\Ω.

Let u∈ H_loc¹ (Ω^ext, ∆+κ²) satisfy the Sommerfeld radiation condition and the Helmholtz equation in the weak sense. Then we have the representation

u(x) =−(V_κγ^1,extu)(x) + (W_κγ^0,extu)(x) for x∈Ω^ext. (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, γ^0,intu=g_D on Γ_D, γ^1,intu=g_N on Γ_N

with a bounded Lipschitz domainΩ, non-overlapping setsΓD, ΓN satisfying ΓD∪ΓN=∂Ω and boundary conditions g_D ∈ H^1/2(Γ_D), g_N ∈ H^−1/2(Γ_N) 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







−∆u_λ =λu_λ in Ω, γ^0,intu_λ = 0 on ΓD, γ^1,intuλ = 0 on ΓN.

(3.23)

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

⟨γ^0,intuλ, gN⟩_ΓN =⟨γ^1,intuλ, gD⟩_ΓD for all corresponding eigenfunctions uλ.

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



















∆u+κ²u= 0 in Ω^ext, γ^0,extu=g_D on Γ_D, γ^1,extu=gN on ΓN,











∇u(x), x

∥x∥



−iκu(x)









=O

 1

∥x∥²



for ∥x∥ → ∞

with non-overlapping sets Γ_D, Γ_N satisfying Γ_D∪Γ_N=∂Ω and boundary conditions g_D∈ H^1/2(ΓD), gN∈H^−1/2(ΓN) has a unique weak solution u∈H_loc¹ (Ω^ext, ∆+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,

γ^0,intu=gD on∂Ω (3.24)

with a bounded Lipschitz domainΩand the Dirichlet boundary conditiongD∈H^1/2(∂Ω).

The solution to (3.24) is given by the representation formula (3.21) withγ^0,intu=g_D, i.e., u(x) = (Vκγ^1,intu)(x)−(WκgD)(x) for x∈Ω (3.25) with unknown Neumann trace data γ^1,intu ∈ H^−1/2(∂Ω). Applying the Dirichlet trace operator γ^0,int to (3.25) we obtain the Fredholm integral equation of the first kind

(Vκγ^1,intu)(x) = 1

2gD(x) + (KκgD)(x) forx∈∂Ω. (3.26) Similarly, applying the Neumann trace operator γ^1,int to (3.25) we obtain the Fredholm integral equation of the second kind

1

2γ^1,intu(x)−(K_κ^∗γ^1,intu)(x) = (D_κg_D)(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

⟨V_κγ^1,intu, s⟩_∂Ω =

1

2I+Kκ

 gD, s



∂Ω

for all s∈H^−1/2(∂Ω) and

1

2I−K_κ^∗



γ^1,intu, t



∂Ω

=⟨D_κgD, t⟩_∂Ω for allt∈H^1/2(∂Ω), respectively, where

⟨u, v⟩_∂Ω :=



∂Ω

u(x)v(x) ds denotes the duality pairing betweenH^1/2(∂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

 −∆u_λ=λuλ inΩ,

γ^0,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) andu_λ the corresponding eigensolution. From the boundary integral equations (3.26) and (3.27) we get

(V_κγ^1,intu_λ)(x) = 1

2γ^0,intu_λ(x) + (K_κγ^0,intu_λ)(x) = 0 for x∈∂Ω, 1

2γ^1,intu_λ(x)−(K_κ^∗γ^1,intu_λ)(x) = (D_κγ^0,intu_λ)(x) = 0 for x∈∂Ω

and hence, ifκ² coincides with the eigenvalueλ, the operatorsVκ and(1/2I−K_κ^∗)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 γ^1,intu satisfies the boundary integral equation (3.26) and u has the representation (3.25).

Conversely, if γ^1,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) =−(W_κt)(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

V_κs(x) =γ^0,intu(x) for x∈∂Ω, (3.31) 1

2t(x)−(K_κt)(x) =γ^0,intu(x) for x∈∂Ω and the corresponding variational problems

⟨V_κs, t⟩_∂Ω =⟨γ^0,intu, t⟩_∂Ω for allt∈H^−1/2(∂Ω),

1

2I−K_κ

 t, s



∂Ω

=⟨γ^0,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,

γ^1,intu=g_N on ∂Ω (3.32)

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

u(x) = (V_κg_N)(x)−(W_κγ^0,intu)(x) for x∈Ω (3.33) with unknown Dirichlet trace dataγ^0,intu. Applying the Dirichlet trace operator to (3.33) we obtain the Fredholm integral equation of the second kind

1

2γ^0,intu(x) + (K_κγ^0,intu)(x) = (V_κg_N)(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

(Dκγ^0,intu)(x) = 1

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

1

2I+Kκ



γ^0,intu, s



∂Ω

=⟨V_κgN, s⟩_∂Ω for alls∈H^−1/2(∂Ω), and

⟨D_κγ^0,intu, t⟩_∂Ω =

1

2I−K_κ^∗

 gN, t



∂Ω

for all t∈H^1/2(∂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

 −∆u_µ=µuµ inΩ,

γ^1,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=µ

1

2γ^0,intu_µ(x) + (K_κγ^0,intu_µ)(x) = (V_κγ^1,intu_µ)(x) = 0 for x∈∂Ω, (D_κγ^0,intu_µ)(x) = 1

2γ^1,intu_µ(x)−(K_κ^∗γ^1,intu_µ)(x) = 0 forx∈∂Ω,

implying that the functionγ^0,intu_µbelongs 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 D_κ 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 traceγ^0,intu satisfies the boundary integral equation (3.35)andu has the representation (3.33).

Conversely, if γ^0,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, γ^0,intu=gD on ΓD, γ^1,intu=gN on ΓN

(3.37)

with a bounded Lipschitz domainΩ, non-overlapping setsΓ_D, Γ_Nsuch thatΓ_D∪Γ_N=∂Ω, and functions gD ∈H^1/2(ΓD) and gN ∈ H^−1/2(ΓN). In the smooth case, the solution to (3.37) is given by the representation formula

with a bounded Lipschitz domainΩ, non-overlapping setsΓ_D, Γ_Nsuch thatΓ_D∪Γ_N=∂Ω, and functions gD ∈H^1/2(ΓD) and gN ∈ H^−1/2(ΓN). In the smooth case, the solution to (3.37) is given by the representation formula

In document 2011JanZapletal TheBoundaryElementMethodfortheHelmholtzEquationin3DMetodahraničníchprvkůproHelmholtzovurovnicive3D VŠB–TechnicalUniversityofOstravaFacultyofElectricalEngineeringandComputerScienceDepartmentofAppliedMathematics (Stránka 34-0)