Mixed Boundary Value Problems for the Helmholtz Equation in a

Mixed Boundary Value Problems for the Helmholtz Equation in a Quadrant L. P. Castro, F.-O. Speck and F. S. Teixeira Dedicated to the memory of Ernst L¨ uneburg

Abstract. The main objective is the study of a class of boundary value problems in weak formulation where two boundary conditions are given on the halflines bordering the first quadrant that contain impedance data and oblique derivatives. The associated operators are reduced by matricial coupling relations to certain boundary pseudodifferential operators which can be analyzed in detail. Results are: Fredholm criteria, explicit construction of generalized inverses in Bessel potential spaces, eventually after normalization, and regularity results. Particular interest is devoted to the impedance problem and to the oblique derivative problem, as well. Mathematics Subject Classification (2000). Primary 35J25; Secondary 30E25, 47G30, 45E10, 47A53, 47A20. Keywords. Boundary value problem, Helmholtz equation, half-line potential, pseudodifferential operator, Fredholm property, normalization, diffraction problem.

1. Introduction A class of quite basic model problems from diffraction theory gave rise to the present studies, see [16] for the physical background, history and early references. Let Qj , j = 1, 2, 3, 4, denote the four open quadrants in R2 bordered by the coordinate semi-axes Γ1 = {x = (x1 , x2 ) ∈ R2 : x1 ≥ 0 , x2 = 0}, etc., that are numbered counter-clockwise. For open sets Ω ⊂ Rn and s ∈ R, let H s (Ω) and HΩs denote the common Bessel potential spaces of H s = H s (Rn ) elements restricted This work was supported in part by Centro de Matem´ atica e Aplica¸c˜ oes of Instituto Superior T´ ecnico and Unidade de Investiga¸c˜ ao Matem´ atica e Aplica¸c˜ oes of Universidade de Aveiro, through the Portuguese Science Foundation (FCT–Funda¸c˜ ao para a Ciˆ encia e a Tecnologia), co-financed by the European Community.

2

Castro, Speck and Teixeira

to Ω and supported on Ω, respectively. Our central aim is solving the following boundary value problem (BVP) for the Helmholtz equation (HE) in Q1 . Problem P1 (B1 , B2 ). Determine (all weak solutions) u ∈ H 1 (Q1 ) (explicitly and in closed analytical form) such that 2 ∂2 ∂ 2 2 Au(x) = (∆ + k )u(x) = + + k u(x) = 0 in Q1 ∂x21 ∂x22 ∂u ∂u +γ (x) = g1 (x) on Γ1 (1.1) B1 u(x) = αu + β ∂x2 ∂x1 ∂u ∂u + γ′ (x) = g2 (x) on Γ2 . B2 u(x) = α′ u + β ′ ∂x1 ∂x2 Herein the following data are given: a complex wave number k with ℑm k > 0, constant coefficients α, β, γ, α′ , β ′ , γ ′ as fixed parameters and arbitrary gj ∈ H −1/2 (Γj ), justified later in Proposition 2.1, see Remark 2.2 (b). Note that β and β ′ are the coefficients of the normal derivatives, whilst γ and γ ′ are those of the tangential derivatives. In case of a Dirichlet condition, i.e., β = γ = 0, we assume g1 ∈ H 1/2 (Γ1 ). The impedance problem plays a key role; for convenience it will be denoted by P1 (ℑ1 , ℑ2 ) with boundary conditions (BC): ∂u(x) + ip1 u(x) = g1 (x) ∂x2 ∂u(x) ℑ2 u(x) = + ip2 u(x) = g2 (x) ∂x1

ℑ1 u(x) =

on

Γ1

on

Γ2

(1.2)

where the imaginary part of pj turns out to be important: (i) ℑm pj > 0: physically most reasonable due to positive finite conductance in electromagnetic theory for instance; (ii) p1 = 0 or/and p2 = 0: Neumann condition(s) allow a much simpler solution [19], [6]; (iii) if both ℑm pj are negative the potential approach has to be modified in a cumbersome way (in contrast to the mixed case which can be solved like (i) or (ii)). (iv) if pj ∈ R \ {0} for j = 1, 2, the problem needs another kind of normalization that is not carried out in this paper as considered to be less important. The treatment of example (1.2) carries a great part of the hitherto unknown structure of mixed BVPs (1.1). For brevity and symmetry reasons we shall write the two BCs of (1.1) in the following form: + + αu+ 0 + βu1 + γuτ = g1

on Γ1

′ + ′ + α ′ u+ 0 + β u 1 + γ u τ = g2

on

Γ2

(1.3)

+ where u+ 0 = T0,Γj u denotes the trace of u, u1 = T1,Γj u the trace of the normal + derivative and uτ the trace of the tangential derivative, respectively, on the positive bank of Γj .

Mixed Boundary Value Problems for the Helmholtz Equation

3

All considerations will be carried out for BVPs of normal type1 where the so-called pre-symbol σ1 of B1 satisfies σ1 (ξ) = α − βt(ξ) + γϑ(ξ) 6= 0 , ξ∈R ±1 σ1±1 (ξ) = O |ξ| as |ξ| → ∞ ,

(1.4)

and the pre-symbol σ2 of B2 is supposed to satisfy the same condition with dashed 1/2 coefficients. Herein t(ξ) = (ξ 2 − k 2 ) denotes the common branch in C (with vertical cuts from k via ∞ to −k and t(ξ) ∼ ξ at +∞) and ϑ(ξ) = −iξ, ξ ∈ R. In brief we write t−1 σj ∈ G L∞ = G L∞ (R) – the group of invertible L∞ functions – or say that σj is 1-regular. This assumption can be proved to be necessary for the Fredholm property in the spaces under consideration (for first order BCs) and it holds in physically relevant cases. Furthermore we shall study the question of low regularity simultaneously with P1 (B1 , B2 ), i.e., looking for u ∈ H 1+ǫ (Q1 ) and assuming gj ∈ H −1/2+ǫ (Γj ) where ǫ ∈ [0, 1[, provided order Bj = 1. This will be helpful particularly when the problem is not Fredholm, namely for normalization in the data space. High regularity, due to ǫ ≥ 1, is planned to be studied in a forthcoming paper by a modified reasoning. In those considerations we speak about the parameter-dependent problem P1ǫ (B1 , B2 ). The explicit analytical solution of P1 (B1 , B2 ) is known for the case (j) where both conditions are Dirichlet or Neumann type [19] and for the case (jj) where only one condition is Dirichlet, Neumann, or tangential (β = 0) type and the other one may be arbitrary in the sense of (1.1) and (1.4) [6]. In particular the impedance problem P1 (ℑ1 , ℑ2 ) and the oblique derivative problem (α = α′ = 0) were still open. Only well-posedness of P1 (ℑ1 , ℑ2 ) has been proved in [23] for the case (1.2) (i). The present results are complete in the sense to solve the normal type problems P1 (B1 , B2 ) for arbitrary complex coefficients in (1.3) and any data in (1.4) by closed analytical formulas as it was carried out for Sommerfeld problems [24], [25], [26]. The present article represents an extension of the paper [6] where we started working with coupling relations between (i) the operator associated to P1ǫ (B1 , B2 ): T

Lǫ = (B1 , B2 ) : H 1+ǫ (Q1 ) → H −1/2+ǫ (Γ1 ) × H −1/2+ǫ (Γ2 ) , 1+ǫ

1

1+ǫ

(1.5)

that acts from a subspace H (Q1 ) = H (Q1 ) ∩ H (Q1 ) of weak solutions of the HE into the corresponding data space, and (ii) certain boundary pseudodifferential operators (generalizing those which result from the single/double layer potentials). That approach was based on a particular representation formula (in terms of Dirichlet/Neumann data) which is now replaced by a new general ansatz in terms of so-called half-line potentials. It enables us to reduce Lǫ by explicit operator matrix identities to (pure) convolution type operators with symmetry (CTOS, for short; see Section 3 and 4). These can be analyzed with respect to their Fredholm characteristics and explicitly inverted (in the sense of generalized inverses), 1 In

view of some literature it could be called “piecewise elliptic” [3], [6].

4


eventually after normalization according to a recent paper of the authors [7]. The impedance problem is treated firstly in Section 4 because of technical reasons and its importance as well. The general case (1.1) is outlined in Section 5, and oblique derivative problems with real coefficients are analyzed in Section 6 as a prototype where all the principal features and difficulties appear.

2. Basic Notation and Previous Results e s (Ω) denotes2 the subspace of Let us recall some known results [28], [11], [6], [7]. H s H (Ω) functionals f that are extensible by zero in H s , i.e., ℓ0 f ∈ S ′ (Rn ) belongs to H s , whilst HΩs stands for the H s distributions supported on Ω, s ∈ R. Let r± denote the restriction operator to R± , in corresponding Bessel potential spaces. We know that for Ω = R± e s (Ω) = H s (Ω) r± HΩs = H

iff

|s| < 1/2

(2.1)

n

and the same holds for (special) Lipschitz domains Ω ⊂ R [27], e.g., for Ω = Q1 . Furthermore (still for Ω = R± ) e s (Ω) ⊂ r± HΩs = H 6=

H s (Ω)

if |s| = 1/2

(2.2)

dense

where the embedding is continuous. The same, except density, holds for s > 1/2. The case of s < −1/2 is less important here, but note that the δ-functional belongs s to H± = HRs ± if and only if s < −1/2. Therefore, the zero extension operator s ℓ0 : H s (R+ ) → H+

(2.3)

is bounded and invertible by restriction r+ ℓ0 = IH s (R+ ) ,

ℓ0 r+ = IH+s

iff

|s| < 1/2 .

(2.4)

As a matter of fact, even and odd extension have wider ranges, precisely 1 3 e s s,e ℓ : H (R+ ) → H for s ∈ − , 2 2 (2.5) 3 1 ℓo : H s (R+ ) → H s,o for s ∈ − , 2 2 where they are invertible by r+ [7, §2]. Here we are using the notation H s,e = {ϕ ∈ H s : ϕ = Jϕ} ,

H s,o = {ϕ ∈ H s : ϕ = −Jϕ}

(2.6)

with Jϕ(x) = ϕ(−x) for ϕ ∈ H s , s ≥ 0, and Jϕ(ψ) = ϕ(Jψ) for test functions ψ in the case of s < 0, respectively. 2 The

notation of the “tilde spaces” is not uniform in the existing literature. Sometimes it is used s , see e.g. [8], [28]. The present notation is compatible with the early sources for what we call HΩ that can be found in [12].


5

Note that the even and odd extension operators can be also used for extending functionals from Q1 to the upper half-plane Q12 = int Q1 ∪ Q2 or to the right halfplane Q14 ⊂ R2 , respectively, in a similar way. The following result tells us that the mixed Dirichlet/Neumann problem P1 (D, N ) is well-posed and explicitly solved in a very simple way. Proposition 2.1 ([19], [6]). There is a toplinear isomorphism KDN,Q1 : X = H 1/2 (Γ1 ) × H −1/2 (Γ2 ) → H 1 (Q1 ) T

u = KDN,Q1 (f, g) = KD,Q12 ℓe f + KN,Q14 ℓo g e c KD,Q12 ℓe f (x) = Fξ7−1 →x1 exp [−t(ξ)x2 ]ℓ f (ξ) ,

−1 o c KN,Q14 ℓo g(x) = −Fξ7−1 →x2 exp [−t(ξ)x1 ] t (ξ) ℓ g(ξ) ,

(2.7) x ∈ Q12 x ∈ Q14

that satisfies

T

(T0,Γ1 , T1,Γ2 ) KDN,Q1 = IX T

(2.8)

KDN,Q1 (T0,Γ1 , T1,Γ2 ) = IH 1 (Q1 ) . Herein we used the common Fourier transformation F in H s (R) and vector T transposition (·, ·) for the operator format. Remarks 2.2. (a) The representation formula (2.7) can be employed as a potential ansatz of H 1 (Q1 ) functions to solve other BVPs, see the next results. (b) It justifies the choice of data spaces in the formulation of P1 (B1 , B2 ). (c) Formulas (2.8) can be interpreted in the way that the operator L = L0 associated with the BVP, see (1.5), is (two-sided bounded) invertible by the Poisson operator (2.7) – a very rare but not unique situation as we shall see in the next two sections. (d) Low regularity is evident: The result holds for solutions u ∈ H 1+ǫ (Q1 ) and data (f, g) ∈ X ǫ = H 1/2+ǫ (Γ1 ) × H −1/2+ǫ (Γ2 ), if ǫ ∈ [0, 1[, and formally even for ǫ ∈ ] − 1, 1[. Classically speaking: If we put now the ansatz (2.7) into the general BCs of P1 (B1 , B2 ), we obtain a two-by-two system of boundary pseudodifferential equations with a particular structure described by the form of the boundary pseudodifferential operator (BΨDO) T given in the next theorem. Theorem 2.3 ([6]). The operator L = L0 in (1.5) that is associated with the BVP P1 (B1 , B2 ) can be factorized as T

L = T (T0,Γ1 , T1,Γ2 )

T = LKDN,Q1 : H 1/2 (Γ1 ) × H −1/2 (Γ2 ) → H −1/2 (Γ1 ) × H −1/2 (Γ2 ) T 1 K1 r+ Aφ1 ℓe C0 Aψ1 ℓo = = K2 T 2 C0 Aψ2 ℓe r+ Aφ2 ℓo

(2.9)

6


where we meet convolution operators Aφ = F −1 φ·F of order 1 and 0, respectively, with Fourier symbols φ1 = α − βt + γϑ , ψ2 = α ′ − γ ′ t , (recalling that t(ξ) = (ξ 2 − k 2 )

ψ1 = βθ

φ2 = −α′ t−1 + β ′ + γ ′ θ

(2.10)

1/2

and ϑ(ξ) = −iξ, for ξ ∈ R), θ = −ϑ/t, and Z −1 C0 f (x) = (2π) exp[−t(ξ)x]fb(ξ) dξ , x > 0. (2.11) R

Remarks 2.4. (a) Due to the last two results, L and T are toplinear equivalent (i.e., they coincide up to composition with boundedly invertible operators) [1], [14, Chapter IV- §1]. (b) The operators C0 and Kj were named around the corner operators in [19], since they related data from Γ1 and Γ2 , e.g., C0 ℓe = T0,Γ2 KD,Q12 ℓe : H 1/2 (Γ1 ) → H 1/2 (Γ2 ) K1 = βT1,Γ2 KN,Q12 ℓo : H −1/2 (Γ1 ) → H −1/2 (Γ2 ) .

(2.12)

(c) The so-called convolution type operators with symmetry Tj were treated in [7] (scalar case), and [5] (matrix case). They have similar properties as Wiener-Hopf operators in Bessel potential spaces [22]. Some of these properties are outlined in the appendix. (d) Therefore, all those BVPs could be analyzed in detail for which one of the operators Kj is zero, i.e., precisely if one of the boundary operators (BOs) is Dirichlet, Neumann or tangential type, see (2.10) with β = γ = 0, α′ = γ ′ = 0 or β = 0, respectively, and think of a possible exchange of the roles of x1 and x2 . The results are summarized as follows. Theorem 2.5 ([6]). Let P1 (B1 , B2 ) be of normal type, see (1.1) and (1.4), and the operator T defined in (2.9) be triangular due to Remark 2.4 (d). The following cases occur: (i) Tj are both invertible and so is T . (ii) Tj are both one-sided invertible and Fredholm, thus T is Fredholm. (iii) At least one of the operators Tj is not normally solvable. Then there exists an ǫ0 > 0 such that the operators Tjǫ , T ǫ , that result from (1.5), satisfy (ii) for ǫ ∈ ]0, ǫ0 [. In all cases, a generalized inverse of T or T ǫ , respectively, can be explicitly represented in terms of factorizations of φj and algebraic composition formulas. Moreover, in the last case, T : X → Y can be normalized by choosing a dense subspace ≺

Y ⊂ Y with a different topology such that the continuous extension of a generalized ≺ − inverse (T ǫ ) of T ǫ in L (Y , X) represents a generalized inverse of the normalized ≺

≺

operator T : X →Y .


7

There are further direct results about the computation of defect numbers and the description of low regularity in all cases where T is triangular. The cases that are not covered by the preceding result are BVPs where B1 is such that • α 6= 0, β 6= 0, γ = 0 – impedance condition • α = 0, β 6= 0, γ 6= 0 – oblique derivative condition • α 6= 0, β 6= 0, γ 6= 0 – “general” BC and B2 (with dashed coefficients) belongs also to one of these cases. They will be analyzed in what follows.

3. The Half-line Potential Approach We propose an ansatz for the weak solutions of the HE, which is more general than the DN-representation in (2.7), in order to solve the BVPs P1 (B1 , B2 ) that could not be treated before, see [6], Theorem 5.7. Definition 3.1. Let mj ∈ N0 , ψj : R → C be measurable functions such that ψj is mj -regular, i.e., t−mj ψj ∈ G L∞ and let ℓj : H 1/2−mj (R+ ) → H 1/2−mj (R) be continuous extension operators for j = 1, 2. Then n o −1 d u(x) = Fξ7−1 →x1 exp[−t(ξ)x2 ]ψ1 (ξ)ℓ1 f1 (ξ) (3.1) n o −1 d exp[−t(ξ)x ]ψ (ξ) ℓ f (ξ) +Fξ7−1 1 2 2 2 →x2

with fj ∈ H 1/2−mj (R+ ) and x = (x1 , x2 ) ∈ Q1 is said to be a half-line potential (HLP) in Q1 with density (f1 , f2 ). We call it strict for H 1 (Q1 ) if (3.1) defines a bijective mapping, written in the form K

= K1 + K2 = K :

ψ1 ,ψ2

X = H 1/2−m1 (R+ ) × H 1/2−m2 (R+ ) → H 1 (Q1 )

(3.2)

specifying ℓj when necessary. Keeping in mind low regularity properties: K ǫ : X ǫ = H 1/2−m1 +ǫ (R+ ) × H 1/2−m2 +ǫ (R+ ) → H 1+ǫ (Q1 ), we speak about a strict HLP for H 1+ǫ (Q1 ) in the corresponding case. Remarks 3.2. (a) Under the assumptions of Definition 3.1, K is a bounded linear operator in the setting of (3.2), as well as K ǫ for ǫ ∈ ] − 1, 1[ if ℓj = ℓe , mj = 0 or ℓj = ℓo , mj = 1, which follows by elementary estimation. (b) If K is strict, the operator L associated to the BVP (1.1) is toplinear equivalent to a BΨDO T = LK that will be analyzed and optimized by the choice of ψ1 and ψ2 later on. (c) As we shall see, in general it is not evident to recognize the strictness of K and it turns out to be convenient to admit non-strict potentials. A similar observation was recently used by S. E. Mikhailov in the method of reduction by parametrices [20].

8


(d) We are mainly interested in the case mj = 1 and ℓj = ℓo due to (1.1), but also make use of m1 = 0 (for incorporating Dirichlet data) and mj ≥ 2 in the context of regularity properties and higher order BOs in future work. (e) One can prove that t−mj ψj ∈ G L∞ is necessary for the strictness of K (similarly to the theory of singular integral operators of normal type [21], [2]). T

Proposition 3.3. Let L = (B1 , B2 ) be given by (1.1) and K by (3.1) and (3.2). Then the composed operator T = LK has the form: ! r+ Aφ11 ℓ1 C0 Aφ12 ℓ2 T = :X→Y (3.3) C0 Aφ21 ℓ1 r+ Aφ22 ℓ2 2

where Y = H −1/2 (R+ ) identifying Γj with R+ and φ11 = σ1 ψ1−1 = (α − βt + γϑ)ψ1−1 ,

φ12 = σ1∗ ψ2−1 = (α + βϑ − γt)ψ2−1

φ21 = σ2∗ ψ1−1 = (α′ + β ′ ϑ − γ ′ t)ψ1−1 , φ22 = σ2 ψ2−1 = (α′ − β ′ t + γ ′ ϑ)ψ2−1 . Proof. It is a straightforward computation substituting (3.1) into the BCs.

L=

B1 B2

−−−−→ Y տ ր = K1 + K2 T = H 1 (Q1 )

K

(3.4)

X

T1 K2

K1 T2

Figure 1. Operator composition T = LK . In general it is unknown how to analyze the analytical properties (like invertibility, Fredholmness, etc.) of (3.3) unless T is upper/lower triangular [6], [7]. However (3.3) provides a lot of information about P1 (B1 , B2 ) due to the great variety of possible choices of ψj , as we shall see. Figure 1 illustrates the operator composition in use. Lemma 3.4. For any φ ∈ L∞ the operator given by K (s) = C0 Aφ ℓo : H s (R+ ) → H s (R+ ) Z −1 o f (ξ) dξ , x ∈ R K (s) f (x1 ) = (2π) exp[−t(ξ)x1 ]φ(ξ)ℓd 1 +

(3.5)

R

is well-defined and bounded if s ∈ ] − 3/2, 1/2[. In this case, K (s) = 0 if and only if φ is an even function. Replacing ℓo by ℓe , we have boundedness of K (s) for s ∈ ] − 1/2, 3/2[ being zero if and only if φ is odd.


9

Proof. K (s) is bounded as a composition of bounded operators ℓo : H s (R+ ) → H s [6], Aφ and C0 . If φ is even, then g = Aφ ℓo f is odd and C0 g = 0, since the integrand of (3.5) is odd for any x1 > 0. If φ is not even (a.e.), there exists an interval Iǫ = ]ξ0 − ǫ, ξ0 + ǫ[⊂ R+ where Z Z φ(ξ) dξ − φ(ξ) dξ 6= 0 . Iǫ

−Iǫ

(s) o f (ξ) = exp[t(ξ)] sgn(ξ) χ Choosing x1 = 1 and ℓd f Iǫ ∪−Iǫ (ξ) we obtain that K does not vanish in a neighborhood of 1. χΓ denotes the characteristic function of a set Γ ⊂ R or Rn . The choice of ℓo f is admissible since F ℓo : H s (R+ ) → L2,o (R, ts ) is surjective (where the exponent notation “o” in the last space refers to the odd functions in the corresponding weighted L2 space). The second statement is proved by analogy.

Example 3.5. We can interpret the formulas of Proposition 2.1 in a way that the DN ansatz (2.7) represents the simplest possible HLP that is strict for H 1 (Q1 ) due to (2.8) and, moreover, “reproduces the data”, i.e., T = I in (3.3) and Figure 1, choosing ℓ1 = ℓe , ℓ2 = ℓo , σ1 = 1 = ψ1 , σ2 = −t = ψ2 . Looking at (3.3) and (3.5) it seems promising to search for more HLPs with the two nice properties (2.8) or, at least, one of them. Definition 3.6. The HLP defined by (3.1) and (3.2) is said to be reproducing if there are BOs B1 , B2 such that T = I in (3.3). For given BOs (1.1) we say that the BVP P1 (B1 , B2 ) admits a reproducing HLP ansatz if T = I for ψj = σj , i.e., when B1 u = f1 , B2 u = f2 in (3.1). Note that this definition can be modified if the orders of Bj are different from 1, see Remark 3.2 (d). However, for the operators given by (1.1), we introduce the companion operators due to B1 and B2 : + + B1∗ u = αu+ 0 + γu1 + βuτ

B2∗ u =

α ′ u+ 0

+

γ ′ u+ 1

+

β ′ u+ τ

on

Γ2

on Γ1 ,

(3.6)

respectively, exchanging the coefficients of the normal and tangential derivatives on the other branch of the boundary. Theorem 3.7. Consider a BVP P1 (B1 , B2 ) given by (1.1). Then the following conditions are equivalent: (i) The BVP admits a reproducing HLP ansatz (3.1); in this case, ψj coincides with the pre-symbol σj of Bj . (ii) The BVP is of normal type and B2 equals (up to a constant factor) the companion operator B1∗ due to B1 ; thus they are companion to each other and, therefore, compositions of the corresponding trace operator with the same

10

Castro, Speck and Teixeira spacial differential operator: ∂u ∂u +γ B1 u = T0,Γ1 αu + β ∂x2 ∂x1 ∂u ∂u B2 u = T0,Γ2 αu + β +γ · const . ∂x2 ∂x1

(3.7)

(iii) The pre-symbols of Bj have the form σ1 = α − βt + γϑ and are 1-regular, i.e., t

−1

σ2 = (α − γt + βϑ) · const

(3.8)

∞

σj ∈ G L , cf. (1.4).

Proof. (i) implies relations (3.3), (3.4) with T = I, these yield that σ1 = ψ1 , σ2 = ψ2 are 1-regular due to the boundedness of the corresponding operators; further σ1∗ ψ2−1 and σ2∗ ψ1−1 have to be even functions, which leads to (iii) and, vice versa, (iii) implies (i). (ii) is a reformulation by definition. So we arrived at the conjecture that the DN ansatz (2.7) is not the only one with properties (2.8) – being reproducing and strict (for H 1 (Q1 ) and in modification of the present definitions, since the order of the Dirichlet BO is zero). Apparently there is at least a considerable class of reproducing HLPs K ψ,ψ∗ that can be used to solve further problems. An important and non-trivial question is whether K ψ,ψ∗ is strict and, if not, how to overcome the difficulty that not all u ∈ H 1 (Q1 ) are represented by such potentials. Let us start with some natural conclusions. Corollary 3.8. Let ψ be the pre-symbol of a BO B on Γ1 (see (1.4)) such that t−1 ψ ∈ G L∞ . Then K ψ,ψ∗ is strict for H 1 (Q1 ) if and only if P1 (B, B∗ ) is well-posed. T

Proof. Consider, by analogy of (2.8b) the composed operator K ψ,ψ∗ (B, B∗ ) which equals the unity operator in H 1 (Q1 ) if and only if it is bounded and surjective, i.e., P1 (B, B∗ ) is well-posed (note also that t−1 ψ ∈ G L∞ does not imply t−1 ψ∗ ∈ G L∞ , cf. (3.8)). Corollary 3.9. Let P1 (B1 , B2 ) be of normal type with BOs of first order. Then K = K σ1 ,σ1∗ is a reproducing HLP (and K σ2∗ ,σ2 as well), provided σ1∗ (or σ2∗ , respectively) is 1-regular. Furthermore, the BΨDO corresponding to (3.3) has the form I 0 T 1 K1 or T = T = , (3.9) 0 I K2 T 2

respectively, where

T1 = r+ Aσ1 σ−1 ℓo ,

K1 = C0 Aσ1∗ σ−1 ℓo

T2 = r+ Aσ2 σ−1 ℓo ,

K2 = C0 Aσ2∗ σ−1 ℓo .

2∗

1∗

Proof. By inspection.

2

(3.10)

1


11

Definition 3.10 (see also [1]). Two linear operators W1 and W2 (acting between Banach spaces) are said to be toplinear equivalent after extension if there exist additional Banach spaces Z1 and Z2 , and boundedly invertible linear operators E and F such that W2 0 W1 0 F. (3.11) =E 0 IZ2 0 IZ1 Corollary 3.11. If P1 (B1 , B2 ) is of normal type and K K σ2∗ ,σ2 with analogue results), then

σ1 ,σ1∗

is strict (respectively

T

(i) L = (B1 , B2 ) and T2 are toplinear equivalent after extension, (ii) they have isomorphic kernels and cokernels, (iii) they are simultaneously either (1) invertible or (2) only one-sided invertible and Fredholm or (3) not normally solvable (but easily normalized by help of the parameter ǫ, see the appendix). Proof. All is based on the operator matrix identity I 0 0 I T2 T T = (B1 , B2 ) K = = K2 T 2 I K2 0

0 I

0 I

I 0

(3.12)

with obvious consequences and on the discussion of scalar CTOS in [6] and [7] outlined in the appendix.

Corollary 3.12. If P1 (B1 , B1∗ ) is of normal type and K = K for H 1 (Q1 ), then

·

H 1 (Q1 ) = im K + ker (B1 , B1∗ )

T

σ1 ,σ1∗

is not strict (3.13)

is a toplinear decomposition of the weak solutions space. Proof. Since K is left invertible by K IH 1 (Q1 )

−

T

= (B1 , B1∗ ) we can decompose · =K K − + I −K K − . (3.14)

T

Remark 3.13. Although a projector onto ker (B1 , B1∗ ) is known as to be the second part in the right-hand side of (3.14) for any P1 (B1 , B2 ) of normal type, it is not yet clear how to determine its rank or a basis, in general (see examples in Section 4 and further results in Section 5). T

The principle remaining question is: How to “invert” an operator L= (B1 , B2 ) that satisfies the relation (3.12) where K is explicitly left invertible and Fredholm and T2 is a CTOS that has one of the properties mentioned in Corollary 3.11 (iii). We prepare the concrete answer by some auxiliary formulas as follows. Lemma 3.14. Let K ∈ L (X, Z), L ∈ L (Z, Y ) be bounded linear operators in Banach spaces and T = LK .

(3.15)

12


(i) If T − is a right regularizer of T , i.e., T T − = I − F1 where F1 is compact (or even a finite rank operator or projector), then L− = K T − is a right regularizer of L. (ii) If T − is a left regularizer of T and K is Fredholm, then L− = K T − is a left regularizer of L. (iii) If K is Fredholm, then L is normally solvable if and only if T is normally solvable. Proof. Proposition (i) is evident, by inspection: L(K T − ) = (LK )T − = T T − = I − F1 . −

−

(ii) Let T T = I − F2 , K K = I − F3 , K K −

−

(3.16)

= I − F4 . Then

−

S = K K T LK = (I − F3 )(I − F2 ) = I − F5 K SK

−

(3.17)

= I − F6 = (I − F4 )K T − L(I − F4 ) ,

i.e., K T − L = I − F7

(3.18)

where Fj are compact (or finite rank operators, respectively). (iii) Since K is Fredholm and K K − K = K , we have a direct (algebraic and topological) sum Z = Z1 ⊕ Z0 = im K ⊕ F4 Z −

where F4 = I − K K has finite characteristic. (3.15) implies that im T ⊂ im L. A direct algebraic decomposition of im L is given by

·

·

im L = LZ1 + LZ0 = im T + span{y1 , . . . , yn }

(3.19)

where {yj } is a basis of LZ0 . Thus im L and im T are simultaneously closed or not. Corollary 3.15. (i) If T has the form of (3.12), we have corresponding conclusions T between properties of L = (B1 , B2 ) and T2 . (ii) If K is Fredholm, the operators L, T and T2 are simultaneously Fredholm or not. (iii) The explicit form of F7 in (3.18) is: F7 F6

= F6 − F4 K T − L − K T − LF4 + F4 K T − LF4 = F4 + K F5 K − = F4 + K (F2 + F3 − F3 F2 )K = I − K K −K T −T K −

(3.20) −

which simplifies to the following form, if K is bijective, i.e., in the concrete setting (3.2), it is reproducing (F3 = 0) and strict for H 1 (Q1 ) (F4 = 0): F7

= K F2 K

−

=KK

−

− K T −T K

In the last case, F7 is a projector if F2 is a projector.

−

.

(3.21)


13

For fixed complex parameters p, with ℑm p > 0, from now on we will use the function ζp defined by ζp (ξ) = (ξ − p)/(ξ + p), ξ ∈ R (which has winding number 1 as in the case p = i). Knowing the structure of T2 very well, we obtain the following result. Theorem 3.16. Let P1 (B1 , B2 ) be of normal type, L defined by (1.5) (case ǫ = 0), 2 and K σ1 ,σ1∗ : H −1/2 (R+ ) → H 1 (Q1 ) be reproducing and strict. Then L is Fredholm if and only if Z −1 −1 ℜe (2πi) d log σ2 σ1∗ − 1/4 ∈ / Z. (3.22) R

In this case, L is left/right invertible by I 0 − − L =KT =K −T2− K2 T2−   r+ Aσe−1 ℓo r+ Aζ −κ σ−1 ℓo , − i − −1 o −1 o −1 o T2 = r+ Aσe ℓ r+ Aζ κ ℓ r+ Aσ− ℓ = i  r+ A −1 −κ ℓo r+ A −1 ℓo , σe ζ σ −

i

κ≥0

(3.23)

κ≤0

where (see the appendix):

−1 σ = σ2 σ1∗ = σ− ζiκ σe ,

κ∈Z

(3.24)

and Aσ = Aσ− Aζiκ Aσe : H −1/2 → H 2η → H −1/2

(3.25)

is an asymmetric factorization through H 2η , 2η ∈ ]−3/2, 1/2[. The integer number κ and the intermediate space order 2η are given by the splitting Z −1 −1 d log σ2 σ1∗ ℜe (2πi) =κ+η . (3.26) R

Proof. Corollary 3.11 implies that L and T2 are toplinear equivalent after extension by (3.12) and therefore both are Fredholm or not, respectively. The Fredholm criterion for T2 is known [7], see the appendix, as well as a representation of a generalized inverse which is a left/right inverse for κ ≥ 0 or κ ≤ 0, respectively. The formula (3.23) for L− then yields LL− L = L provided T2 T2− T2 = T2 which applies in both cases of left or right invertibility. Corollary 3.17. Under the previous assumptions, the Fredholmness of L implies that Lǫ is also Fredholm with the same characteristics α(Lǫ ) = dim ker Lǫ = α(L) β(Lǫ ) = codim im Lǫ = β(L) as long as η + ǫ ∈ ] − 3/4, 1/4[.

(3.27)

14


Corollary 3.18. If, under the same assumptions, L is not Fredholm, i.e., (3.22) is violated, then L is not normally solvable. However β(L) = codim im L < ∞

(3.28)

and, admitting η = −3/4, Lǫ is Fredholm for ǫ ∈ ]0, 1[. The operator L can be image normalized and the normalized operator has a left inverse defined by a continuous − extension of (Lǫ ) , ǫ > 0 [22], [7].

4. Explicit Solution of the Impedance Problem We come back to the BVP (1.1) with particular BO (1.2) that have pre-symbols of the form σ1 (ξ) = ip1 − t(ξ) σ2 (ξ) = ip2 − t(ξ) ,

(4.1)

ξ∈R,

such that pj 6= 0 and t−1 σj ∈ G L∞ , j = 1, 2, i.e., the BVP is of normal type. It was recognized as an open canonical problem of particular interest in mathematical physics [15], [16], [6]. The DN ansatz (2.7) yielded a fully equipped (non-triangular) BΨDO (2.9). With the new HLP ansatz we shall succeed to present a complete and most adequate analytical solution that reflects the various aspects pointed out before. Therefore it can be seen as an important reference problem within the class P1 (B1 , B2 ). A crucial point is that the companion BOs Bj∗ are tangential, i.e., σj∗ (ξ) = ipj + ϑ(ξ) = ipj − iξ ,

ξ∈R,

j = 1, 2

(4.2)

which fact simplifies the computations tremendously. Corollary 3.9 tells us that the HLPs K σ1 ,σ1∗ and K σ2∗ ,σ2 are reproducing. Due to Corollary 3.8 they are strict for H 1 (Q1 ) if and only if P1 (B1 , B1∗ ) and P1 (B2∗ , B2 ), respectively, are well-posed. This is equivalent to the fact that ℑm p1 > 0 and ℑm p2 > 0, respectively, see Corollary 5.6 in [6]. So, let us first consider the case where one of these conditions is satisfied, say: 4.1. Case ℑm p2 > 0 According to the last observation the operator L = (ℑ1 , ℑ2 ) 2 H −1/2 (R+ ) is toplinear equivalent to T 1 K1 T = 0 I

T

: H 1 (Q1 ) →

(4.3)

where T1 and K1 are given by (3.10): T1 = r+ Aσ ℓo ,

K1 = C0 Aσ1∗ σ−1 ℓo , 2

t(ξ) − ip1 −1 (ξ) = −i σ(ξ) = σ1 (ξ)σ2∗ . ξ − p2

(4.4)


15

One can see that the sign of ℑm p1 does not matter so far. The form of the pre-symbol σ yields already a so-called asymmetric factorization through an intermediate space (AFIS) of Aσ [6], [7] (cf. also the appendix), namely: Aσ = Aσ−1 Aσ1 : H −1/2+ǫ → H −3/2+ǫ → H −1/2+ǫ 2∗

(4.5)

in the sense of bounded operator composition where ǫ ∈ ]0, 1[, σ1 is an even function −1 and σ2∗ is “minus type”, i.e., holomorphically extensible into the lower complex half-plane. The corresponding factor (invariance) properties imply that (T1ǫ )

−1

= r+ Aσ−1 ℓo r+ Aσ2∗ ℓo : H −1/2+ǫ (R+ ) → H −1/2+ǫ (R+ ) 1

(4.6)

represents the bounded inverse of the operator T1ǫ that is restricted to spaces of low regularity. The CTOS theory [7] implies that ǫ ∈ ]0, 1[ is also necessary for the invertibility of T1ǫ . So we obtain the first result: Proposition 4.1. Let pj ∈ C, t−1 σj ∈ G L∞ , ℑm p1 > 0 or ℑm p2 > 0, and ǫ ≥ 0. Then P1ǫ (ℑ1 , ℑ2 ) is well-posed if and only if ǫ ∈ ]0, 1[. Remark 4.2. The condition t−1 σj ∈ G L∞ is guaranteed in diffraction theory, e.g., if pj and k are taken from the first quadrant of the complex plane [17]. Next let us study the explicit solution formulas, i.e., solve the equation Lu = LK f = T f = g

(4.7)

where T is given by (4.3). We obtain under the assumptions of Proposition 4.1 (dropping the ǫ-dependence for short) −1 g1 T1 −T1−1 K1 u = K f = K T −1 g = K σ2∗ ,σ2 g2 0 I −1 T1 (g1 − K1 g2 ) = K σ2∗ ,σ2 g2 (4.8) −1 −1 (ξ)Fx1 7→ξ ℓo r+ Aσ−1 ℓo r+ Aσ2∗ ℓo ge1 (x1 ) u(x1 , x2 ) = Fξ7→x1 exp[−t(ξ)x2 ]σ2∗ 1

−1 o +Fξ7−1 →x2 exp[−t(ξ)x1 ]σ2 (ξ)Fx2 7→ξ ℓ g2 (x2 )

where ge1 = g1 − K1 g2 = g1 − C0 Aσ1∗ σ−1 ℓo g2 . Further we can omit the first term 2 ℓo r+ , since Aσ−1 transforms odd functions into odd functions, and replace ℓo ge1 by 1

an arbitrary extension of ge1 ∈ H −1/2+ǫ (R+ ) into H −1/2+ǫ (R), commonly denoted by ℓge1 [11], since Aσ2∗ is minus type. Thus we arrive at the explicit solution formula (which can also be verified straightforwardly as to present a solution of P1 (ℑ1 , ℑ2 ) in H 1+ǫ (Q1 )):

16


Proposition 4.3. Under the assumptions of Proposition 4.1 the unique solution of P1ǫ (ℑ1 , ℑ2 ) for ǫ ∈ ]0, 1[ and arbitrary gj ∈ H −1/2+ǫ (R+ ) , j = 1, 2, is given by −1 −1 u(x1 , x2 ) = Fξ7−1 →x1 exp[−t(ξ)x2 ]σ2∗ (ξ)σ1 (ξ) Fx1 7→ξ ℓo r+ Aσ2∗ ℓ g1 − C0 Aσ1∗ σ−1 ℓo g2 (x1 )

(4.9)

2

−1 o +Fξ7−1 →x2 exp[−t(ξ)x1 ]σ2 (ξ)Fx2 7→ξ ℓ g2 (x2 ) ,

(x1 , x2 ) ∈ Q1 .

Remark 4.4. This formula can be used to discuss smoothness and singular behavior of u near the boundary and near the edge, analogously to considerations in [18]. Regarding the problem in the original H 1 setting (ǫ = 0), it is known from the Neumann problem P1 (N, N ) [6] that a compatibility condition is necessary for the solution, namely e −1/2 (R+ ) g1 + g2 ∈ H

(4.10) 1/2

considering the principal parts of the BOs and using the fact that H (R+ ) ⊂ e −1/2 (R+ ). H Condition (4.10) can be re-discovered here for P1 (ℑ1 , ℑ2 ), in a little complicated but systematic way, by so-called minimal normalization of T or T1 in their image spaces [22], [7], see the final result in Theorem 4.13 at the end of this section. 4.2. Case ℑm p2 < 0 3 We know already that K

σ2∗ ,σ2

is reproducing, i.e., left invertible: T

(ℑ2∗ , ℑ2 ) K

σ2∗ ,σ2

= IH −1/2 (R+ )2 .

(4.11) 2

The formula remains obviously true in Y ǫ = H −1/2+ǫ (R+ ) for ǫ ∈ ] − 1, 1[ as a composition of bounded operators continuously extended or restricted, respectively, from Y 0 , see (2.5). Fortunately we know more about the auxiliary BO T

Lǫa = (ℑ2∗ , ℑ2 ) : H 1+ǫ → Y ǫ ,

|ǫ| < 1

(4.12)

from [6], because ℑ2∗ is tangential: Lemma 4.5. Let P1 (ℑ2∗ , ℑ2 ) be of normal type, σ2 = ip2 − t be the pre-symbol of ℑ2 , ℑm p2 < 0, and ǫ ∈ ] − 1, 1[. Then Lǫa is Fredholm with characteristics α (Lǫa ) = 1 ,

β (Lǫa ) = 0 .

(4.13)

Proof. Clearly, Lǫa is surjective from the observations before. The (strict) DNansatz yields a toplinear equivalent BΨDO ǫ Ta1 0 ǫ ǫ 1,−t : Xaǫ = H 1/2+ǫ (R+ )×H −1/2+ǫ (R+ ) → Y ǫ (4.14) Ta = La K = ǫ ǫ Ta2 Ka2 where

ǫ Ta1 = r+ Aσ2∗ ℓe , 3 and

ǫ Ta2 = r+ A−σ2 t−1 ℓo ,

ℑm p1 6= 0 with main interest in ℑm p1 < 0.

ǫ Ka2 = C0 Aσ2 ℓe

(4.15)


17

ǫ cf. [6], Corollary 5.6. By inspection we see that Ta2 is invertible by r+ A−σ−1 t ℓo 2 ∞ ǫ since the Fourier symbol is even and belongs to G L . The other CTOS Ta1 has a “plus type” pre-symbol that must be factorized (see the appendix):

σ2∗ (ξ) = −iξ + ip2 = −i(ξ − p2 ) = −i(ξ + p2 )

ξ − p2 −1 (ξ) (4.16) = σ− (ξ)ζ−p 2 ξ + p2

where wind ζ−p2 = 1 since ℑm (−p2 ) > 0. The corresponding AFIS Aσ2∗ = Aσ− Aζ −1 : H 1/2+ǫ → H 1/2+ǫ → H −1/2+ǫ −p2

(4.17)

yields a right inverse −

ǫ (Ta1 ) = r+ Aζ−p2 ℓe r+ Aσ−1 ℓ

(4.18)

−

ǫ and a projector onto the kernel of Ta1 −

ǫ ǫ Π = I − (Ta1 ) Ta1

(4.19)

−1 ǫ = 1. which has characteristic α (Ta1 ) = −wind ζ−p 2

Corollary 4.6. The kernel of

ǫ Ta1

ǫ ker Ta1

is explicitly given by

= ker r+ Aζ −1 ℓe = span{f0 } −p2

−1 fb0 (ξ) = (ξ − p2 )

(4.20)

f0 (x1 ) = −i exp[−ip2 x1 ] χR+ (x1 ) so that ker Taǫ ϕ01 = f0 ,

ϕ01 = µ :µ∈C ϕ02

ǫ −1 ǫ ϕ02 = −(Ta2 ) Ka2 ϕ01 = r+ Aσ−1 t ℓo C0 Aσ2 ℓe f0 2

and the kernel of Lǫa consists of the multiples of the function u0 = u01 + u02 = K

1,−t

(ϕ01 , ϕ02 ) .

(4.21)

The first part has the form u01 (x1 , x2 )

e d = Fξ7−1 →x1 exp[−t(ξ)x2 ]ℓ f0 (ξ) Z −1 = (2π) exp[−iξx1 − t(ξ)x2 ] R

1 dξ , ξ 2 − p22

xj > 0 .

The next result is a consequence of (4.11), (4.12) and Corollary 3.12 admitting restricted or extended spaces due to ǫ ∈]0, 1[ or ǫ ∈] − 1, 0[, respectively. Corollary 4.7. Under the assumptions of Lemma 4.5, K ǫ = (K

σ2∗ ,σ2 ǫ

2

) : H −1/2+ǫ (R+ ) → H 1+ǫ (Q1 )

(4.22)

is Fredholm with characteristics α(K ǫ ) = 0 ,

β(K ǫ ) = 1

(4.23)

18


and span{u0 } represents a complement of im K ǫ . Next, let us look at the composed operator T

T = LK = (ℑ1 , ℑ2 ) K

σ2∗ ,σ2

=

T1 K2

K1 T2

(4.24)

bearing in mind the ǫ-dependence, T ǫ : X ǫ → Y ǫ , as in (4.12). I.e., we consider the BΨDO of the impedance problem P1 (ℑ1 , ℑ2 ) resulting from the (ℑ2∗ , ℑ2 ) data ansatz. Proposition 3.3 gives us the form of the involved operators that appear in (4.24): T1 = r+ Aσ1 σ−1 ℓo ,

T2 = I

K1 = C0 Aσ1∗ σ−1 ℓo ,

K2 = 0 .

2∗

2

(4.25)

Lemma 4.8. Let P1 (ℑ1 , ℑ2 ) be of normal type (i.e., σj = ipj − t are 1-regular, j = 1, 2), let ℑm p2 < 0 (such that σ2∗ = ip2 + ϑ is 1-regular) and ǫ ∈ ] − 1, 1[. Then T1ǫ is left invertible by −

(T1ǫ ) = r+ Aσe−1 ℓo r+ Aζ −1 ℓo r+ Aσ−1 ℓ = r+ Aσe−1 ℓo r+ Aσ−1 ℓo −p2

−

+

−1

(4.26)

−1

σ− (ξ) = −i(ξ + p2 ) , σe (ξ) = t(ξ) − ip1 = −σ1 (ξ), σ+ (ξ) = −i(ξ − p2 ) and β(T1ǫ ) = 1. −1 Proof. The pre-symbol σ = σ1 σ2∗ of T1 admits the factorization t − ip1 ip1 − t −1 ξ + p2 = −i = −i(ξ + p2 ) (t − ip1 ) σ(ξ) = ip2 − iξ ξ − p2 ξ − p2

Aσ = Aσ− Aζ−p2 Aσe : H −1/2+ǫ → H −3/2+ǫ → H −3/2+ǫ → H −1/2+ǫ .

(4.27)

which is an AFIS with κ = 1 (see the appendix), so we can apply formulas (A.11) and (A.12). Moreover, the second term ℓo r+ in (4.26) can be omitted, since Aζ −1 is −p2 minus type, which yields the second formula including a plus type symbol σ+ . Remark 4.9 (The HLP paradox). We have “reduced” the impedance problem (with ℑm p2 < 0) to a boundary pseudodifferential equation T u = g where T is not surjective but only left invertible (with β(T ) = β(T1ǫ ) = 1). Could it be that the problem is well-posed anyway? The answer is yes, since the reduction of L to T1 in (4.24) is not an equivalence (after extension) relation. We have seen already in Proposition 4.1 that this happens if ℑm p1 ·ℑm p2 < 0 and suppose now that the non-strict HLPs will be useful in general. Theorem 4.10. Let P1 (ℑ1 , ℑ2 ) be of normal type, ℑm pj 6= 0, j = 1, 2, and ǫ ∈ [0, 1[. Then P1ǫ (ℑ1 , ℑ2 ) is well-posed if and only if ǫ 6= 0. Proof. First we realize that the problem is not well-posed if ǫ = 0, because the compatibility condition (4.11) is also necessary for the impedance problem. This can be seen from the fact that the sum g1 +g2 of the given impedance data belongs e −1/2 (R+ ) up to a function in H 1/2 (R+ ), which is a subspace of the first one. to H


19

It remains to prove the result for both ℑm pj < 0 and ǫ ∈ ]0, 1[. We have, in the notation of Corollary 4.7, that T ǫ = Lǫ K

ǫ

where T ǫ and K ǫ are left invertible with one-dimensional defect spaces. This implies that Lǫ is Fredholm with ind Lǫ = α(Lǫ ) − β(Lǫ ) = 0 β(Lǫ ) ≤ β(T ǫ ) = 1. I.e., α(Lǫ ) = β(Lǫ ) = 0 or 1. We have to disprove that the latter can happen. To this end consider (dropping ǫ) L = L(K K

−

+ (I − K K − )) = T K

−

+ LP1

T

−

(4.28) −

where K = (ℑ2∗ , ℑ2 ) is a left inverse of K and P1 = I −K K is the projector along im K onto ker K − , that consists of the multiples of u0 given by (4.21). The first term on the right hand side of (4.28) is an operator in full rank factorization (known from matrix theory), i.e., T is left invertible, K − right invertible. Moreover α(T K − ) = α(K − ) = 1 , β(T K − ) = β(T ) = 1 ,

ker(T K − ) = ker(K − ) im (T K − ) = im (T )

Now, there are two possibilities: either (i) LP1 “fills the gap”, i.e., it maps u0 into Y \ im T and L is bijective, or (ii) L maps u0 into im T (including 0) and α(L) = β(L) = 1 according to (4.28). It remains to disprove that the latter can happen. Thus we like to show that, for u0 given by (4.21), Lu0 ∈ / im T

(4.29)

(I − T T − )Lu0 6= 0

(4.30)

or, equivalently (where T

−

is the left inverse of T ), for which it suffices to prove that g0 = (I − T1 T1− )ℑ1 u01 6= 0 .

(4.31)

where u01 represents the first part of u0 , see Corollary 4.6. Written in full we have g0 = r+ Aσ− ζ−p2 (I − ℓo r+ )Aζ −1

−1 −p2 σ−

e

ℓf+

f+ (x1 ) = ℑ1 u01 (x1 ) = r+ Aσ1 ℓ ϕ01 (x1 ) =

r+ Fξ7−1 →x1

ip1 − t(ξ) 6= 0 . ξ 2 − p22

(4.32)

−1 −1 σ (ξ) = i(ξ − p2 ) The extension ℓf+ ∈ H −1/2+ǫ does not matter for g0 , since ζ−p 2 − is minus type, i.e., annulated by r+ later on in the operator composition. −1/2+ǫ Choosing the zero extension ψ+ = ℓ0 f+ ∈ H+ (recall that ǫ ∈ ]0, 1[), we arrive at

g0 = r+ Aσ− ζ−p2 (I − ℓo r+ )Aζ −1

−1 −p2 σ−

ψ+ 6= 0

(4.33)

20


since (I − ℓo r+ )Aζ −1

−1 −p2 σ−

ψ+ is supported in R− and σ− ζ−p2 is plus type.

Remark 4.11. We like to point out that the operator associated to the impedance problem P1 (ℑ1 , ℑ2 ) where both ℑm pj < 0, j = 1, 2, cannot be reduced by a strict, reproducing HLP to a triangular BΨDO of the form (3.3). The proof will be given later in Section 5 (Example 5.7 (a)). Let us determine the explicit solution formulas in the last case, starting with non-vanishing ǫ ∈]0, 1[. They are just a modification of (4.9) by incorporation of a one-dimensional operator K0 . Theorem 4.12. Let P1 (ℑ1 , ℑ2 ) be of normal type, ℑm p1 6= 0, ℑm p2 < 0, and ǫ ∈ ]0, 1[. The unique solution uǫ of P1ǫ (ℑ1 , ℑ2 ) for arbitrary gj ∈ H −1/2+ǫ (R+ ) is given by (dropping the parameter ǫ in various terms) uǫ = K

σ2∗ ,σ2 T

(f1 , f2 ) + µ u0 −

−

f = (f1 , f2 ) = T g − µ T Lu0 where

(4.34) (4.35)

− T 1 K1 T1 −T1− K1 − T = , T = (4.36) 0 I 0 I as given in Lemma 4.8, u0 is taken from Corollary 4.6 and µ ∈ C such that4 µ I − T T − Lu0 = I − T T − g . (4.37)

Proof. We know from Theorem 4.10 and Corollary 4.7 that the unique solution can be uniquely represented in the form (4.34) where fj ∈ H −1/2+ǫ (R+ ) and u0 ∈ ker(ℑ2∗ , ℑ2 )T . Applying L, we obtain LK

σ2∗ ,σ2

(f1 , f2 ) + µLu0 = g .

(4.38)

−

Now (4.35) results from the application of T to (4.38) and (4.37) follows from the rank 1 projection I − T T − along im T onto ker T − . 4.3. Explicit Solution of P1 (ℑ1 , ℑ2 ) The impedance problem in the genuine setting (ǫ = 0) is not Fredholm, but can be normalized by changing the data space “slightly”, i.e., choosing a suitable dense subspace of Y with a different topology, as described in the appendix. Theorem 4.13. Let P1 (ℑ1 , ℑ2 ) be of normal type and ℑm pj 6= 0. Then the associ2 ated operator L : H 1 (Q1 ) → Y = H −1/2 (R+ ) is not normally solvable. However α(L) = 0 ,

β(L) = dim Y / im L = 0 .

(4.39)

In the case ℑm p2 > 0, the related operators T = LK and T1 , see (4.7) etc., have the same properties. The minimal image normalized operator (which is unique up to equivalent norms in the new image space) of T1 , given by (4.25), reads ≺

4 The

−1/2 e −1/2 (R+ ) (R+ ) → H T 1 = Rst T1 : H

(4.40)

parameter µ can be expressed in a complicated way as the value of a linear functional acting on g.


21

and is boundedly invertible by ≺

e −1/2 (R+ ) → H −1/2 (R+ ) :H

(4.41)

e −1/2+ǫ (R+ ) → H −1/2+ǫ (R+ ) , = r+ Aσ−1 ℓo r+ Aσ2∗ ℓo : H

ǫ ∈ ]0, 1[ . (4.42)

−1

T1−1 = Ext (T1ǫ )

defined as continuous extension of −1

(T1ǫ )

1

≺

≺

Explicit representations of inverses T −1 , L−1 of the corresponding normalized operators result from the relations (4.7), (4.8), and the explicit solution of P1 (ℑ1 , ℑ2 ) from (4.9). In the case ℑm p2 < 0, all runs the same up to β(T ) = β(T1 ) = 1 and the modified solution formulas (4.34)–(4.37). Proof. The characteristics of the operators L, T , T1 are known from the foregoing. −1 The normalization of T1 = r+ Aσ ℓo , where σ = σ1 σ2∗ and ℑm p2 > 0, results from the AFIS of Aǫσ = Aǫσ−1 Aǫσ1 : H −1/2+ǫ → H −3/2+ǫ → H −1/2+ǫ

(4.43)

2∗

for ǫ ∈ ]0, 1[, such that r+ ℓo is well-defined and we have common invariance properties in these spaces: (Aǫσ ) ℓo r+

±1 o

±1

ℓ r+ = ℓo r+ (Aǫσ ) ℓo r+ ±1 ±1 o Aǫσ2∗ = ℓo r+ Aǫσ2∗ ℓ r+ .

(4.44)

In the case ǫ = 0, the formal inverse (4.42) is not defined on the full space H −1/2 (R+ ) due to the unboundedness of ℓo , see (2.5), but as a bounded operator ≺ e −1/2 (R+ ) into H −1/2 (R+ ), invertible by T1 . The rest of the proof is evident. from H e 1/2 (R+ ) for the density f0 of u0 Note that (4.35) yields the condition f0 ∈ H given in Corollary 4.6 in the case ǫ = 0 such that ℓo f0 ∈ H 1/2 (R) and K0 f0 ∈ H 1 (Q1 ). ≺

Remark 4.14. The compatibility condition g ∈ Y is equivalent to (4.10), which is known to be necessary from the Neumann problem (considering principal parts) and sufficient by verifying that u ∈ H 1 (Q1 ) in (4.34) for ǫ = 0 under the condition (4.10). Remark 4.15. It is clear that the present method holds completely for problems P1 (B1 , ℑ2 ) (see (1.1), (1.2)) of normal type with ℑm p2 6= 0. The resulting operator T1 has the form (4.25) where σ1 is not even anymore and needs to be factorized (see [6] and the appendix). However, the results can be carried out by analogy taking into account an additional case: T1 may be Fredholm and right invertible.

22


5. General Boundary Conditions including Oblique Derivatives Now we tackle the problem P1 (B1 , B2 ) where all principal coefficients β, γ, β ′ and γ ′ do not disappear, still considering normal type problems where σ1 , σ2 are 1–regular, see (1.4). The idea of using a reproducing HLP with ψ1 = σ1 = ψ2∗ or ψ2 = σ2 = ψ1∗ leads to the difficulty of analyzing P1 (B1 , B1∗ ) or P1 (B2∗ , B2 ) where no boundary operator is tangential neither of impedance type that are not included in the classes treated before [6]. I.e., there is no direct way to determine (at least) the codimension of the image of that kind of potential which we needed for generalizing the previous results. Thus we look for a new type of HLP K that makes T triangular in Figure 1 of Proposition 3.3 and can be analyzed in what its mapping properties are concerned. This will be carried out in detail for a prototype ansatz, referring to analogy in the other cases. As before, we prefer a strict HLP, however must accept Fredholm K which now might be left or right invertible (having a low dimensional kernel) or just Fredholm (with α(K ) · β(K ) 6= 0 which we like to avoid). The strategy for the choice of the HLP is based upon the following lemma: Lemma 5.1. Formula (3.1) represents a strict HLP for H 1 (Q1 ) (or H 1+ǫ (Q1 )) if and only if for one (and thus for any) well-posed problem P1 (B1 , B2 ) the operator T in (3.3) is boundedly invertible. Proof. It is clear from Figure 1.

So we can test if an ansatz is strict and modified methods hold for one-sided invertible K . On the other hand it is clear how to obtain a triangular BΨDO T : Lemma 5.2. Let P1 (B1 , B2 ) be given by (1.1), where β, γ, β ′ , γ ′ ∈ C \ {0}, and K be a HLP as defined in (3.1). Then the BΨDO T = LK (see (3.3)) is lower/upper triangular if and only if ψ1 = σ2∗ or ψ2 = σ1∗ , respectively (up to constant factors). Proof. This is an application of Lemma 3.4 noting that the pre-symbols σ1 , σ2 , σ1∗ , σ2∗ are all not even nor odd. Remark 5.3. Plenty of possible choices result from the two preceding lemmas, but many lead to the same or analogous conclusions. Choosing the second variant in Lemma 5.2, the point is to find a well-posed problem P1 (G1 , G2 ) and a symbol ψ1 , such that T Ttest = (G1 , G2 ) K ψ1 ,σ1∗ (5.1) is upper/lower triangular. This is easily achieved by I K1 T 1,σ1∗ Ttest 1 = (D, N ) K = 0 r+ A−tσ−1 ℓo 1∗ I K 2 T −t,σ1∗ Ttest 2 = (N, D) K = 0 r+ Aσ−1 ℓo 1∗

(5.2)


23

where the form of Kj does not matter. Also tangential or impedance operators are possible instead of D or N , respectively. Another suitable choice is, e.g., 0 r+ A−σ1 t−1 ℓo T Ttest 3 = (B1 , D) K −t,σ1∗ = (5.3) 0 r+ Aσ−1 ℓo 1∗

provided P1 (B1 , D) is well-posed. One finds several variants. However, let us focus (only) the first two, which are not equivalent in general – fortunately. Let us analyze and compare them: Lemma 5.4. Let σ1∗ = α + βϑ − γt be 1–regular, i.e., −1 ϑ α −1 ∈ G L∞ . φ1 = −tσ1∗ = γ − β − t t

(5.4)

Then the main terms in the matrices of (5.2), denoted by T1 and T2 , satisfy T1 = r+ A−tσ−1 ℓo = r+ At1/2 ℓo r+ Aφ10 ℓo r+ At−1/2 ℓo : H −1/2 (R+ ) → H −1/2 (R+ ) 1∗ − (5.5) −1/4 −1/2 −1 −1 1/2 , = ζk −tσ1∗ −tσ1∗ t φ10 = t−

i.e., the operator T1 is toplinear equivalent to the (lifted) CTOS T10 = r+ Aφ10 ℓo : L2 (R+ ) → L2 (R+ ). Further T2 = r+ Aσ−1 ℓo = r+ At−1/2 ℓo r+ Aφ20 ℓo r+ At−1/2 ℓo : H −1/2 (R+ ) → H 1/2 (R+ ) 1∗ − (5.6) 1/4 1/2 1/2 −1 1/2 −1 = −ζk −tσ1∗ φ20 = t− σ1∗ t = −ζk φ10

and T20 = r+ Aφ20 ℓo : L2 (R+ ) → L2 (R+ ) is toplinear equivalent to T2 .

Proof. The formulas are verified straightforwardly due to the fact that the terms ℓo r+ in the factorizations of Tj may be omitted. Proposition 5.5. Let σ1∗ = α + βϑ − γt be 1–regular and use the notation as before in (5.2)–(5.6). Then: (i) K 1,σ1∗ is strict if and only if Z 1 1 −1 −1 ℜe (2πi) d log tσ1∗ ∈ − , ; (5.7) 2 2 R (ii) K

−t,σ1∗

is strict if and only if Z −1 −1 ∈ ]−1, 0[ . ℜe (2πi) d log tσ1∗

(5.8)

R

Proof. (i) Lemma 5.4 yields that K 1,σ1∗ is toplinear equivalent after extension to T10 . The appendix tells us that the invertibility of S = T10 is equivalent to the fact that its pre-symbol σ = φ10 satisfies (A.3) with κ = 0 and η ∈ ]−3/4, 1/4[, i.e., (5.7) holds. (ii) Analogously, K −t,σ1∗ is toplinear equivalent after extension 1/2 to T20 with a different pre-symbol φ20 = −ζk φ10 given by (5.6), which leads to (5.8).

24


Corollary 5.6. For any first order BO B1 of normal type we can find a strict HLP K ψ1 ,σ1∗ (at least) if Z 1 −1 −1 d log tσ1∗ ∈ −1, (5.9) ℜe (2πi) 2 R

putting ψ1 = 1 or ψ1 = −t in the corresponding case.

Examples 5.7. (a) Proof of the statement in Remark 4.11: Recall the impedance problem with both ℑm pj < 0. If there was a strict HLP that reduces L to an upper triangular T , it necessarily has the form K ψ1 ,σ1∗ . Now the DN test yields that ! o o −1 ℓ −1 ℓ r A C A + 0 T ψ σ 1 1 Ttest = (D, N ) K ψ1 ,σ1∗ = 0 r+ A−tσ−1 ℓo 1∗

o

is invertible, in particular that r+ A−tσ−1 ℓ is invertible which is not the case as 1∗

−1 we obtain we can see from (5.7): For ℑm k > 0, ℑm p1 < 0, σ = −tσ1∗ 1/2 1/2 1/2 ξ−k ξ+k −(ξ 2 − k 2 ) = −i σ(ξ) = ip1 − iξ ξ − p1 ξ − p1 1/2 Z Z Z ξ−k 1 −1 −1 −1 ℜe (2πi) d logσ = (2π) d arg σ = (2π) d arg = . ξ − p 2 1 R R R

(b) The graphs of the pre-symbols σ1∗ of oblique derivative operators B1∗ = β

∂ ∂ +γ ∂x1 ∂x2

(5.10)

(i.e., α = 0) were analyzed in [22] so that the conditions (5.7)–(5.9) can be easily checked. It is not difficult to see that the factorization indices κ (cf. (3.25), (3.26)) −1 of φ10 or φ20 satisfy |κ| ≤ 2 since, for tσ1∗ , we have from the Figure 2 Z −1 −1 (5.11) d log tσ1∗ |κ + η| = ℜe (2πi) < 1. R

t−1 σ1∗ (−∞) t−1 σ1∗ (+∞) κ+

η

Figure 2. The oriented graph of t−1 σ1∗ .


25

Another possibility to enlarge the interval (5.9) where one of the standard HLPs is strict is to think about the “low regularity parameter” ǫ introduced in (1.5). Corollary 5.8. Let B1 be a first order BO of normal type, then we can find an ǫ ∈] − 1, 1[ and a HLP (K

ψ1 ,σ1∗ ǫ

) : H ∓1/2+ǫ (R+ ) × H −1/2+ǫ (R+ ) → H 1+ǫ (Q1 )

that is bijective if κ + η = ℜe (2πi)

−1

Z

R

d log

−1 tσ1∗

3 ∈ − ,1 2

(5.12)

(5.13)

where the ∓ sign in (5.12) corresponds with ψ1 = −t and ǫ ∈]−1, 1[ or with ψ1 = 1 and ǫ ∈] − 3/2, 1/2[, respectively. Proof. Starting with Lǫ from (1.5) we come to the toplinear equivalent after ex−1/4+ǫ/2 tension operators (5.5) or (5.6) where φ10 carries a factor ζk and φ20 carries 1/4+ǫ/2 a factor ζk instead of those with ǫ = 0. This gives the wider range of (5.13) by analogy to the preceding considerations. ∂ ∂ + γ ∂x (α = 0) of Remark 5.9. A pure oblique derivative operator B1 = β ∂x 2 1 normal type satisfies (5.13) as observed in (5.11).

Due to the overlapping intervals (5.7), (5.8) it is not necessary to work with potential operators which are not normally solvable. It remains an open question whether it is always possible to find a strict HLP of the form (5.1) by the present ideas for any BVP P1 (B1 , B2 ) of normal type. However the described methods cover already a wide subclass of physically interesting cases. Let us see the consequences of a strict ansatz again starting with ǫ = 0 for simplicity. Theorem 5.10. Let P1 (B1 , B2 ) be of normal type, order Bj = 1, K = K 1,σ1∗ : T H 1/2 (R+ ) × H −1/2 (R+ ) → H 1 (Q1 ) be strict, L = (B1 , B2 ) : H 1 (Q1 ) → 2 H −1/2 (R+ ) and r+ Aσ1 ℓe 0 T1 0 T = LK = = (5.14) C0 Aσ2∗ ℓe r+ Aσ2 σ−1 ℓo K T2 1∗ 2

: H 1/2 (R+ ) × H −1/2 (R+ ) → H −1/2 (R+ ) . Then the following assertions are equivalent: (i) L is normally solvable, (ii) T is normally solvable, (iii) T1 and T2 are normally solvable. In this case, all these operators are Fredholm and T1 and T2 have one-sided inverses given by the appendix, especially by (A.11) and its analogue for ℓe . Then, T is also Fredholm.

26


Proof. Let L be normally solvable. Hence the image of T1 is closed due to the triangular form of T and thus Fredholm, see the appendix. This implies that I 0 T ∼ + F1 (5.15) K T2

are toplinear equivalent operators, where F1 is a finite rank operator. Thus im T2 must be closed and T2 is also Fredholm. Moreover scalar Fredholm CTOS are onesided invertible. Finally, if T1 and T2 are Fredholm and one-sided invertible, then T is Fredholm and L as well. Remark 5.11. If a bounded linear operator in Banach spaces has the form T1 0 T = (5.16) K T2 and T1 and T2 are (only) one-sided invertible, then T is not necessarily normally solvable, see the example of Wiener-Hopf operators with oscillating pre-symbols arising from convolution type operators on finite intervals [4]. In the present context, T1 and T2 are also Fredholm and so is T . Theorem 5.12. In the situation of Theorem 5.10 let T1 be left invertible or T2 be right invertible, i.e., there are generalized inverses Tj− such that T1− T1 = I ,

T2 T2− T2 = T2

(5.17)

T2 T2− = I .

(5.18)

(5.19)

or T1 T1− T1 = T1 , Then, for U =

−T2− KT1− , T− =

T1− U

0 T2−

represents a generalized inverse of T . In particular T − is the inverse of T if T1 and T2 are both invertible. Proof. A direct computation yields T1 T1− T1 − TT T = − KT1 T1 + T2 U T1 + T2 T2− K

0 T2 T2− T2

(5.20)

and insertion of U gives T in both situations. The special case of invertible Tj is now evident. Proposition 5.13. In the situation of Theorem 5.10 let T1 T1− = I and T2− T2 = I. Then (5.19) represents a two-sided regularizer of the Fredholm operator T . Proof. With T1− T1 = I − F1 , T2 T2− = I − F2 we have I 0 I − TT = = KT1− − T2 T2− KT1− I − F2 F2 KT1−

0 I − F2

(5.21)

Mixed Boundary Value Problems for the Helmholtz Equation where F2 is a finite rank operator, and similarly I + F1 0 T −T = . T2− KF1 I

27

(5.22)

It is obvious how these results can be employed to determine the explicit solution of P1 (B1 , B2 ) in general, under the present assumptions, and for its variants (5.2), (5.3) by analogy. In order to avoid the situation of Proposition 5.13 where none of the two operators is two-sided invertible, it may help to exchange the roles of σ1 and σ2 in (5.14) or to search for a different ansatz. There is a certain stability with respect to the parameter ǫ which we describe only in the principal case: Corollary 5.14. Under the assumptions of Theorem 5.12 there exists an interval ]ǫ1 , ǫ2 [, ǫ1 < 0 < ǫ2 , such that Lǫ , T ǫ , Tjǫ are Fredholm for ǫ ∈]ǫ1 , ǫ2 [ with Fredholm characteristics independent of ǫ. The operators L, T , Tj and their generalized inverses given before map the corresponding subspaces into each other (low regularity) for ǫ ∈]0, ǫ2 [ and have continuous extensions for ǫ ∈]ǫ1 , 0[5 . It remains to say something about the case where we cannot find a strict HLP. As a prototype, let us consider again K 1,σ1∗ of (5.2) when it is Fredholm and not invertible, i.e., only one-sided invertible: Lemma 5.15. The operator Ttest1 in (5.2) is left or right invertible if and only if −1 tσ1∗ ∈ G L∞ and Z −1 −1 ℜe (2πi) d log tσ1∗ =κ+η (5.23) R

where η ∈] − 1/2, 1/2[ and κ ∈ N0 or −κ ∈ N0 , respectively. In this case, − I K1 I −K1 T1− − Ttest1 = = 0 T1 0 T1−

(5.24)

represents a left/right inverse of Ttest1 if T1− is a left/right inverse of T1 = r+ A−tσ−1 ℓo, and we can choose explicitly 1∗

o o −1 o T1− = r+ A−1 ℓ r+ A−1 e ℓ r+ C − ℓ

(5.25)

where A−tσ−1 = A− CAe is an AFIS (see explicitly analytical formulas in the 1∗ appendix), C = Aζiκ and we may drop the second/first term ℓo r+ in this formula. Proof. The criteria (5.23) is known from the appendix, see (5.7), the formulas (5.24) and (5.25) are verified straightforwardly. 5 The largest interval can be determined as to be the intersection of the two intervals resulting −1 from the factorization of the lifted pre-symbols of σ1 and σ2 σ1∗ due to the appendix taking into account the possible changes of the parameter s in (A.10) (cf. Section 6 for the class of oblique derivative problems).

28


Theorem 5.16. Let P1 (B1 , B2 ) be of normal type, β, γ, β ′ , γ ′ ∈ C \ {0}, Ttest1 = T (D, N ) K 1,σ1∗ be left invertible (cf. (5.2) and Lemma 5.15). As before put 2

T

L = (B1 , B2 ) : H 1 (Q1 ) → H −1/2 (R+ ) = Y K = K 1,σ1∗ : X = H 1/2 (R+ ) × H −1/2 (R+ ) → H 1 (Q1 ) T = LK : X → Y

(5.26)

as given in (5.14). Then (i) either both T and L are not normally solvable or both are Fredholm; (ii) if T − is a left/right regularizer of T , then L− = K T − is a left/right regularizer of L (and all are two-sided regularizers); (iii) in this case α(T ) ≤ α(L) ≤ α(T ) + β(K ) β(T ) − β(K ) ≤ β(L) ≤ β(T ) β(K ) = α(K − ) if K − K = I .

(5.27)

Particularly, if T is right invertible then L is also right invertible, and, if T is left invertible then α(L) ≤ β(K ); (iv) in the other case ( im T and im L are not closed), the formulas (5.27) hold with β(T ), β(L) exchanged by β(T ) = dim Y / im T and β(L) = dim Y / im L, respectively. Proof. (i) is a consequence of Theorem 5.10 noting that K is Fredholm if Ttest1 is left invertible according to Lemma 5.15. (ii) follows from Lemma 3.14. (iii) is generally true for Fredholm operators which satisfy the relations T = LK , K − K = I and the additional assumptions, respectively. (iv) can be seen by analogy to Corollary 3.18. Let us outline the explicit solution of P1 (B1 , B2 ) under the assumptions of Theorem 5.16 in the Fredholm case (characterized by Lemma 5.15). One can try to determine a generalized inverse L− of L (till now we have only the regularizer L− = K T − but not LL− L = L). However it seems most convenient for practical purposes to proceed as follows. Our HLP K yields a unique decomposition of any u ∈ H 1 (Q1 ) by u = K f + u0 = K K − u + (I − K K − )u where K

−

(5.28)

is the left inverse of K given by K

−

− = Ttest1 (D, N )

T

(5.29)

− and Ttest1 is the left inverse of Ttest1 obtained in (5.24) by AFIS. I.e., we search for f = K − u ∈ im K − = H 1 (Q1 ) and u0 ∈ ker K − such that

Lu = T f + Lu0 = g .

(5.30)


29

Note that ker K − is finite dimensional and explicitly determined by the analysis of (5.29). Now there are two cases: (i) either im T ∩ L ker K

−

= {0}

(5.31)

or (ii) there exist linear independent h1 , . . . , hm in this intersection (explicitly calculable by Lemma 5.15 etc., by determining the form of u0 ∈ ker K − and solving T T − Lu0 = Lu0 ). In the simpler case (i) of (5.31), a necessary and sufficient solvability condition for the equation (5.30) as well as the form of the solution, if existing, can be determined by means of linear algebra as follows. Since im L = im T ⊕ L ker K

−

(5.32)

equation (5.30) decomposes by the ansatz (5.28) into the system of two independent equations LK f = T f = g1 ∈ im T Lu0 = g0 = g − g1 ∈ L ker K

−

.

(5.33)

Evidently, the resulting solvability condition reads g − T T − g ∈ L ker K

−

,

(5.34)

−

because T f = g1 is solvable for g1 = T T g ∈ im T , and can be verified or disproved analytically. If it holds, the general solution of Lu = g is obviously given by u = K T − g + v + u0 (5.35)

where T − g = T − T T − g = T − g1 = f is a particular solution of T f = g1 , since T − is a reflexive generalized inverse due to factorization theory, v = (I − T − T )v is an arbitrary element in the kernel of T , and u0 is obtained from the finite system Lu0 = g0 (see Section 6 for a special case). In the second case (ii) where (5.31) differs from the zero space, we can follow the same strategy noting that now the kernel of L may contain additional terms due to the appearance of h1 , . . . , hm in (5.31) that can be determined by algebraic means computing the intersection of span{h1 , . . . , hm } and L ker K − . However, the solvability condition (5.34) remains the same and (5.35) may be replaced by X u = K T − g + v + u0 + cj uj (5.36)

where uj ∈ ker K − such that Luj ∈ im T and u0 belongs to a complement of span{uj } in ker K − . The exact numbers α(L) and β(L) in general are still unknown but can be determined in concrete cases, cf. Section 6. So we have proved: Corollary 5.17. Under the assumptions of Theorem 5.16, in the Fredholm case, P1 (B1 , B2 ) is solvable if and only if (5.34) is satisfied. Then the general solution is given by (5.35) where u0 ∈ ker K − , which is a linear combination of H 1 (Q1 ) elements determined from (5.29) with coefficients provided by Lu0 = g0 .

30


The final case, where the HLP is not left but right invertible, is similar and even simpler. We refine ourselves to resuming some central results. Corollary 5.18. Under the assumptions of Theorem 5.16, up to Ttest1 being right instead of left invertible, i.e., K K − = I instead of K − K = I, we obtain the results (i) and (ii) as before. (iii) is replaced by α(T ) − α(K ) ≤ α(L) ≤ α(T ) β(L)

= β(T ) .

(5.37)

Particularly, if T is left invertible, then L is left invertible, T − T = I implies that L− = K T − is a left inverse of L. If T is right invertible, then L− = K T − is a right regularizer (but K T − is not necessarily a right inverse of L). (iv) holds by analogy, referring to (5.28). Note that we have here L = T K − beside of T = LK and therefore easier conclusions from im L = im T , particularly the solvability condition T T −g = g

(5.38)

for the solution of Lu = g. Furthermore any u ∈ H 1 (Q1 ) can be represented by the HLP K , however not uniquely in general: u=Kf f = f1 + f0 = K − K f + (I − K − K )f where f1 ∈ im K

−

(5.39)

is unique and f0 ∈ ker K arbitrary. So we get:

Corollary 5.19. Under the assumptions of Corollary 5.18, in the Fredholm case, P1 (B1 , B2 ) is solvable if and only if (5.29) is satisfied where T − is known from an AFIS. Then the general solution is given by u = K f1 f1 = T − g + f2

(5.40)

where f2 ∈ ker T (and this representation is only unique up to elements f0 ∈ ker K ∩ ker T ).

6. Oblique Derivative Problems with Real Coefficients As an example we treat the class of BVPs (1.1) where α = α′ = 0 and β, γ, β ′ , γ ′ ∈ R, as illustrated in Figure 3, which is of particular interest from the physical point of view. T Again we abbreviate L = (B1 , B2 ) , σ1 = −βt + γϑ , ′

′

σ2 = −β t + γ ϑ ,

σ1∗ = −γt + βϑ σ2∗ = −γ ′ t + β ′ ϑ

(6.1)

PSfrag

Mixed Boundary Value Problems for the Helmholtz Equation Γ2

31

(β ′ , γ ′ )–direction Q1 ∂u ∂u β ′ ∂x + γ ′ ∂x = g2 1 2

(γ, β)–direction

∂u ∂u + γ ∂x = g1 β ∂x 2 1

Γ1 Figure 3. Q1 with the boundary conditions on Γ1 and Γ2 , and the oblique directions. and see that P1 (B1 , B2 ) is of normal type, i.e., t−1 σj ∈ G L∞ ,

j = 1, . . . , 2∗

(6.2)

which is clear from a glance at their graphs, see Figure 4. The cases where γ = 0 or γ ′ = 0 (Neumann condition) are known already from [6] and are excluded only for technical reasons (the companion symbols are not 1–regular then, and we needed a different ansatz). The complex winding numbers defined by Z −1 ωj = (2πi) d log t−1 σj (6.3) R

shall turn out to be responsible for the behavior of P1 (B1 , B2 ).

β + iγ at +∞ arctan(γ/β)

β at 0 gr(−t−1 σ1 ) β − iγ at −∞

Figure 4. The oriented graph of −t−1 σ1 . We know already that, if σ2 = σ1∗ , the potential u=K

σ1 ,σ2

g

(6.4) −1/2

2

solves the problem P1 (B1 , B2 ) for any given data g ∈ H (R) – even in the case of complex coefficients (provided the problem is normal type), since K σ1 ,σ2 is reproducing in that case (see Definition 3.6 and Theorem 3.7). However, the question of uniqueness needs more attention, already in this special situation.

32


Lemma 6.1. Let β, γ ∈ R \ {0}, σ1 and σ1∗ be given by (6.1). If βγ > 0 then Z −1 ω1 = (2πi) d log t−1 σ1 ∈ ]0, 1/2[ , R

ω1∗ = 1/2 − ω1 ∈ ]0, 1/2[ .

(6.5)

If βγ < 0, then ω1 = −1/2 − ω1∗ ∈] − 1/2, 0[

(6.6)

(and analogue results hold for γ = 2, 2∗ and dashed coefficients). I.e., all ωj ∈ R, ωj 6= 0, |ωj | < 1/2, j = 1, . . . , 2∗. Proof. In the first case βγ > 0, Figure 4 shows that Z 1 d arg(t−1 σ1 ) ∈ R ω1 = 2π R Z 1 d arg(−t−1 σ1 ) = 2π R 1 γ 1 = 2 arctan ∈ 0, 2π β 2 1 β 1 π = − ω1∗ − arctan = π 2 γ 2

(6.7)

by changing the roles of β and γ. The rest is evident.

Theorem 6.2. The oblique derivative problem with non-vanishing real coefficients is well-posed in H 1 (Q1 ) if and only if θ = θ(β, γ, β ′ , γ ′ ) =

γ γ′ + ′ > 0. β β

(6.8) 2

Otherwise the associate operator L : H 1 (Q1 ) → H −1/2 (R+ ) is right invertible with α(L) = dim ker L = 1 (θ < 0) or not normally solvable with a dense image and injective (θ = 0), respectively. Proof. First we prove that the HLP K 1,σ1∗ is strict in any case. For this reason consider Ttest1 of (5.2) and confirm that r+ A−tσ−1 ℓo : H −1/2 (R+ ) → H −1/2 (R+ ) 1∗

(6.9)

is bijective. This is guaranteed by the appendix, (A.1)–(A.3) and (A.12) with s = −1/2, and Lemma 6.1, since Z 1 1 3 1 1 1 −1 −1 = − , , (6.10) d log −tσ1∗ (2πi) ∈ − + , + 4 4 4 4 2 2 R i.e., in this case κ = 0, τ = 0, ω = η in (A.3), independent of β, γ ∈ R \ {0}.


33

2

T

Hence the invertibility of L = (B1 , B2 ) : H 1 (Q1 ) → H −1/2 (R+ ) is equivalent to the invertibility of the operator r+ Aσ1 ℓe 0 T11 0 1,σ1∗ T = LK = = (6.11) C0 Aσ2∗ ℓe r+ Aσ2 σ−1 ℓo K T22 1∗ 2

: H 1/2 (R+ ) × H −1/2 (R+ ) → H −1/2 (R+ )

according to Theorem 5.10 and we have to study the main diagonal elements Tjj (the notation is chosen to avoid confusion with formerly used Tj ). By analogy to Lemma 5.4 (lifting) we have the operator toplinear equivalence T11 = r+ Aσ1 ℓe ∼ r+ Aσ10 ℓe : L2 (R+ ) → L2 (R+ ) −1/2

−1/4

σ1 t−1/2 = ζk t−1 σ1 ( 1 1 if βγ > 0 −4, 4 1 = − + ω1 ∈ 3 1 4 −4, −4 if βγ < 0

(6.12)

T22 = r+ Aσ2 σ−1 ℓo : H −1/2 (R+ ) → H −1/2 (R+ )

(6.13)

σ10 = t− ω10

according to Lemma 6.1 and Figure 4 (and ω10 = −1/4 in the exceptional case γ = 0 of the Neumann condition). Furthermore the operator 1∗

−1 has an (unlifted) pre-symbol σ22 = σ2 σ1∗ with complex winding number Z −1 −1 ω22 = (2πi) d log σ2 σ1∗ ZR Z −1 −1 = (2πi) d log t−1 σ2 − (2πi) d log t−1 σ1∗ R R ( ω1 + ω2 − 12 if βγ > 0 = ω2 − ω1∗ = ω1 + ω2 + 21 if βγ < 0

(6.14)

due to Lemma 6.1. Now let us study the three cases corresponding to the sign of θ: Case θ > 0. We can assume γ/β > 0 (otherwise exchange the roles of B1 and B2 ). Thus we have 1 3 1 1 ⊂ − , , (6.15) ω10 ∈ − , 4 4 4 4

i.e., it belongs to the parameter range where T11 is invertible (see the appendix, case ℓe ). Furthermore (6.8) yields γ γ′ 1 arctan + arctan ′ > 0 π β β ω1 + ω2 > 0 (6.16) 1 1 1 ω22 = ω1 + ω2 − ∈ − , 2 2 2

34


according to Figure 4, (6.14) and Lemma 6.1, i.e., ω22 belongs exactly to the parameter range where T22 is invertible (for s = −1/2). Case θ < 0. We can assume γ/β < 0 (otherwise exchange the roles of B1 and B2 ). (6.12) implies 3 1 ω10 ∈ − , − , (6.17) 4 4 i.e., ω10 = −1 + η where η ∈]1/4, 3/4[⊂] − 1/4, 3/4[ and T11 is right invertible with α(T11 ) = dim ker T11 = 1. It remains to show that T22 is invertible anyway for T in (6.11) having the same characteristics as T11 , cf. Theorem 5.12. Analogously to (6.16) we conclude ω1 + ω2 < 0 ω22

1 ∈ = ω1 + ω2 + 2

1 1 − , 2 2

(6.18)

with the help of the last line of (6.14) and Lemma 6.1, i.e., it belongs also to the parameter range where T22 is invertible (for s = −1/2). Case θ = 0. Let γ/β > 0 and γ ′ /β ′ = −γ/β < 0. Again T11 is invertible (see the case θ > 0), ω1 + ω2 = 0 due to (6.16) and ω22 = −1/2, see (6.14). Thus we −1 have the case ω22 = η22 = −1/2 with κ = τ = 0 in (A.3) for σ = σ2 σ1∗ and the conclusions of the statement. Remark 6.3. There is a nice geometrical interpretation of the well-posedness condition (6.8). Letting β > 0 and β ′ > 0 without loss of generality (otherwise multiply the BC by −1), we can say: Either the vectors v = (γ, β), v ′ = (β ′ , γ ′ ) of direction of the oblique derivatives are both pointing into the interior of Q1 (γ > 0 and γ ′ > 0), or one is not, but the other one points into Q1 and is “more flat”, i.e., nearer to the axis direction, i.e., the tangential component is relatively longer. Remark 6.4. It is possible to include the case γ ′ = 0 (or γ = 0 by symmetry) in the statement of Theorem 6.2, i.e., in this case one of the BCs is Neumann type. Obviously, γ ′ = 0 does not effect the strictness of K 1,σ1∗ nor the other conclusions of the proof of Theorem 6.2 (ω2 = 0). However this result can also be obtained from [6]. Further, the case of parallel directions (σ1 = σ2∗ ) mentioned in the context of (6.4) really splits into the two cases θ > 0 and θ < 0 corresponding to a well-posed problem and a solvable problem with one-dimensional kernel, respectively. Let us think about the explicit solution to the oblique derivative problem in closed analytical form, namely by the help of an AFIS of Aσ2 σ−1 and of At−1 σ1 as 1∗ well. Fortunately our HLP K 1,σ1∗ is strict for any choice of β, γ ∈ R \ {0}6 , see (6.11), but there are three cases of different nature corresponding with the sign of θ. However, the results are extremely transparent: We have just to “invert” the two scalar operators Tjj of (6.11) in a sense. 6 It

is strict also for β = 0 and γ 6= 0 which is a joke.


35

Lemma 6.5. Let β, γ, β ′ , γ ′ ∈ R \ {0}, the pre-symbols σ1 , . . . , σ2∗ be given by (6.1) and T22 = r+ Aσ22 ℓo : H −1/2 (R+ ) → H −1/2 (R+ ) be defined by (6.13) with −1 σ22 = σ2 σ1∗ . Then T22 is boundedly invertible if and only if θ=

γ′ γ + ′ 6= 0 β β

(6.19)

and the inverse reads −1 T22 = r+ Aσ−1 ℓo r+ Aσ−1 ℓo 22e

22−

(6.20)

where Aσ22 = Aσ22− Aσ22e is an AFIS according to the appendix with κ = 0. If θ = 0, T22 is not normally solvable (and will be considered later in propositions 6.12– 6.14). Proof. We collect and assemble the formulas. In the case θ > 0 we know from (6.16) that the complex winding number ω22 of σ22 is real and belongs to the interval ] − 1/2, 1/2[. The appendix yields for σ = σ22 and s = −1/2 that κ = 0, τ = 0, η = ω22 and 3 1 1 s + 2η = − + 2ω22 ∈ − , (6.21) 2 2 2 such that (A.10) is satisfied. Factorizing σ22 = σ22− σ22e

(6.22)

as σ in (A.4)–(A.8) we obtain an AFIS of Aσ22 and the inverse (6.20). The case θ < 0 leads to ω22 ∈] − 1/2, 1/2[ as well, see (6.18), and we receive the same formulas as before. The case θ = 0 corresponds with ω22 = −1/2 and s + 2η = −1/2 + 2ω22 = −3/2 discussed subsequent to (A.10). Lemma 6.6. Let β, γ ∈ R \ {0}, i.e., σ1 and σ1∗ as given by (6.1) are 1–regular. Then T11 = r+ Aσ1 ℓe : H 1/2 (R+ ) → H −1/2 (R+ ) is right invertible by − T11 = r+ Aσ−1 ℓe r+ Aζ −κ ℓe r+ Aσ−1 ℓe i

11e

where κ=

(

11−

0 if βγ > 0 −1 if βγ < 0

(6.23)

(6.24)

and the factors σ11e , σ11− are given in the proof. Moreover the formula (6.23) simplifies to −1 T11 = r+ Aσ−1 ℓe r+ Aσ−1 ℓe 11e

11−

(6.25)

if (and only if ) βγ > 0, representing the inverse of T11 . Proof. In both cases, due to the sign of βγ, we obtain an AFIS (case ℓe ) of the (lifted) operator Aσ10 = Aσ10− CAσ10e

(6.26)

36


from the appendix formulas (A.4)–(A.8) putting σ = σ10 = σ10− ζiκ σ10e , because the complex winding number of σ10 satisfies ω10 = κ + η

(6.27)

with κ given by (6.24) and η ∈] − 1/4, 3/4[ according to (6.12) and the splitting in (6.17), respectively. Due to the lifting in (6.12), T11 and T10 = r+ Aσ10 ℓe are related by T11 = r+ At1/2 ℓe T10 r+ At1/2 ℓe . −

(6.28)

Hence, the right inverse − = r+ Aσ−1 ℓe r+ Aζ −κ ℓe r+ Aσ−1 ℓe T10 i

10e

10−

(6.29)

of T10 yields (6.23) with 1/2

σ11e = t1/2 σ10e ,

σ11− = t− σ10−

and (6.25) is an obvious consequence of κ = 0.

(6.30)

Theorem 6.7. Let β, γ, β ′ , γ ′ ∈ R \ {0}, θ = β/γ + β ′ /γ ′ , T be defined by (6.11) and − T11 0 − (6.31) T = −1 − −1 T22 KT11 −T22 − −1 where T11 and T22 are given by (6.23) and (6.20), respectively.

(i) If θ > 0 let βγ > 0 (without loss of generality). The unique solution of the 2 oblique derivative problem (in H 1 (Q1 ) for arbitrary g ∈ H −1/2 (R+ ) ) is given by u=K

1,σ1∗

f

−

f = T g.

(6.32)

− −1 Here T11 and T22 are the inverses of T11 and T22 due formula (6.23) or (6.25) and (6.20) inserted into (6.31). (ii) If θ < 0 let βγ < 0 (without loss of generality). The oblique derivative problem 2 is solvable (in H 1 (Q1 ) for arbitrary g ∈ H −1/2 (R+ ) ) and the solutions are given by

u=K

1,σ1∗

f

−

f = T g + µf0 , µ ∈ C f01 f01 = f0 = −1 f02 −T22 Kf01 −1 f01 = r+ F −1 σ10e t−5/2

where σ10e is the factor found in (6.26).

(6.33)


37

Proof. In the first case, the problem is uniquely solvable according to Theorem 6.2 and the solution is given by (6.32) due to the first part of its proof until (6.11). Both Tjj are invertible with inverses given by Lemma 6.5 and Lemma 6.6. In the second case, u is also represented by the strict HLP K 1,σ1∗ and the density f is obtained from T f = g where T is right invertible with a onedimensional kernel, spanf0 say, where T11 0 f01 T f0 = =0 (6.34) K T22 f02 leads to the representation in terms of f01 since T22 is invertible by (6.20). The equation T11 f01 = 0 is equivalent to each of the following conditions resulting from (6.26)–(6.28): T10 r+ At1/2 ℓe f01 = 0 r+ Aζ −1 ℓe r+ Aσ10e t1/2 ℓe f01 = 0 i

(6.35)

r+ Aσ10e t1/2 ℓe f01 = µ r+ F −1 t−2 which implies the representation of f01 in (6.33).

Remark 6.8. All factors mentioned before can be directly expressed in terms of explicit analytical formulas of the appendix. Particularly, in the present case, −1/2

σ10 = t−

−3/4 −1/4 t+ σ 1

σ1 t−1/2 = t−

= σ10− ζi−1 σ10e

(6.36)

due to κ = −1, where (put σ = σ10 in (A.4)–(A.8)) −1 σ10e = ζi σ10− σ10

n o 2η e−1 σ10− = t2η − ψ10− = t− exp P− log ψ10 ψ10

(6.37)

−1 ψ10 = σ10 (+∞) ζi−ω10 σ10

and ω10 = η − 1 with η ∈] − 1/4, 3/4[. If κ = 0 (case θ > 0), we have ω10 = η and the factors ζi−1 in (6.36) and ζi in (6.37) have to be omitted. −1 In what concerns the representation of T22 (continuing Lemma 6.5) it is even simpler to factor the pre-symbol

σ22 = σ22− σ22e according to s = −1/2, κ = 0, η = ω22 , cf. (6.21), where n o 2η−1/2 −1 exp P− log ψ22 ψe22 σ22− = ts+2η ψ22− = t− − −1 ψ22 = σ22 (+∞) ζi−ω22 σ22

(6.38)

(6.39)

with η − 1/4 ∈] − 3/4, 1/4[. Before thinking about the critical case (θ = 0), it is useful to discuss low ǫ ǫ regularity of the solutions (for θ 6= 0). Again we start considering T11 and T22

38


separately, as restrictions (cf. (1.5)) or continuous extensions, respectively, writing ǫ T11 = Rst T11 : H 1/2+ǫ (R+ ) → H −1/2+ǫ (R+ ) ,

ǫ ∈]0, 1[

ǫ T11 = Ext T11 : H 1/2+ǫ (R+ ) → H −1/2+ǫ (R+ ) ,

ǫ ∈] − 1, 0[

(6.40)

ǫ and T22 analogously with domain H −1/2+ǫ due to the even/odd extension operator in (6.12) and (6.13), respectively. It is possible to determine exactly the parameter ranges of ǫ in terms of the numbers ω1 and ω2 rather than regarding the existence result of Corollary 5.14. However, we carry it out only for the case where θ > 0, referring to analogy for θ < 0.

Proposition 6.9. Under the assumptions of Theorem 6.7, (i), i.e., θ > 0, βγ > 0, ω1 > 0 and ω1 + ω2 > 0, the following operators are invertible ǫ T11 : H 1/2+ǫ (R+ ) → H −1/2+ǫ (R+ )

iff

ǫ ∈] − ω1 , 1 − ω1 [

ǫ T22 : H −1/2+ǫ (R+ ) → H −1/2+ǫ (R+ )

iff

ǫ ∈] − (ω1 + ω2 ), 1 − (ω1 + ω2 )[ (6.41)

T ǫ : H 1/2+ǫ (R+ ) × H −1/2+ǫ (R+ ) → H −1/2+ǫ (R+ )

2

if

ǫ ∈] − ω1 , 1 − (ω1 + ω2 )[

where ω1 + ω2 < 1. Proof. From Lemma 6.6 and formulas (6.12), (6.15) we have that T11 is invertible if and only if 1 1 1 (6.42) ω10 = ω1 − ∈ − , 4 4 4

belongs to ] − 1/4, 3/4[. Therefore, the restricted/extended operator is invertible if and only if 1 1 1 3 and ǫ > − − ω1 − (6.43) ǫ < − ω1 − 4 4 4 4

which proves the first result. Similarly, T22 is invertible if and only if (6.16) is satisfied which gives the second result. Thus, if both conditions are fulfilled then T ǫ is also invertible due to its form (6.11) and (6.31). Lemma 6.1 yields ω1 + ω2 < 1.

Remark 6.10. It is also sufficient to assume ǫ ∈] − ω2 , 1 − (ω1 + ω2 )[ for T ǫ to be invertible instead of the last condition in Proposition 6.9, due to the symmetry argument. Corollary 6.11 (low regularity of the solution). Under the assumption of Proposition 6.9, the weak solution u of the oblique derivative problem belongs to H 1+ǫ (Q1 ) 2 provided g ∈ H −1/2+ǫ (R+ ) and ǫ < 1 − (ω1 + ω2 ) where ωj ∈]0, 1/2[, see (6.7). Now let us consider the critical case where γ′ γ θ= + ′ =0 β β

(6.44)


39

which corresponds with directions of the two oblique derivatives that are orthogo2 nal to each other. We know from Theorem 6.2 that L : H 1 (Q1 ) → H −1/2 (R+ ) is not normally solvable, but injective with a dense image. Thus L can be normalized by restriction of its image space ≺

≺

1 L= Rst L : H (Q1 ) →Y = im L

(6.45)

with the norm induced by im L. This idea implies a “natural compatibility condition” for the data g = (g1 , g2 ) to make the oblique derivative problem well-posed, the analogue of (4.10) known from the Neumann problem. We like to derive this necessary and sufficient condition. First we normalize L by a simultaneous positive ǫ–shift in the orders of all spaces: Proposition 6.12. Let β, γ, β ′ , γ ′ ∈ R \ {0}, βγ > 0, θ = 0 and ǫ ∈]0, 1[. Then T

2

Lǫ = Rst (B1 , B2 ) : H 1+ǫ (Q1 ) → H −1/2+ǫ (R+ )

(6.46)

is boundedly invertible. Proof. First we recall that the restricted/extended operator T

(D, N ) : H 1+ǫ (Q1 ) → H 1/2+ǫ (R+ ) × H −1/2+ǫ (R+ )

(6.47)

is invertible for |ǫ| < 1, see Proposition 2.1 and (2.5). Therefore, Remark 5.3 implies that the HLP ǫ K 1,σ1∗ : H 1/2+ǫ (R+ ) × H −1/2+ǫ (R+ ) → H 1+ǫ (Q1 ) (6.48)

ǫ is toplinear equivalent to the modified operator Ttest1 in (5.2). We know from (6.9)– (6.10) that (6.48) is strict for ǫ = 0 since −ω1∗ ∈] − 1/2, 1/2[. Replacing s = −1/2 by s = −1/2 + ǫ we find that this HLP is strict if and only if ω1 − 1/2 + ǫ/2 = −ω1∗ + ǫ/2 ∈] − 1/2, 1/2[, i.e., if and only if ǫ ∈] − ω1 , 1 − ω1 [ , (6.49) 2 an ǫ–interval of length 2, indeed, as in (6.47). ǫ ǫ Modifying (6.12) for T11 we find that ω10 = −1/4 + ǫ/2 + ω1 ∈] − 1/4 + ǫ ǫ ǫ/2, 1/4 + ǫ/2[ and that T11 is invertible if and only if ω10 ∈] − 1/4, 3/4[, see the e ℓ variant of (A.3) for s = 0. This is also equivalent to (6.49). ǫ ǫ Generalizing T22 in (6.13) for ǫ > 0 we get invertibility for ω22 = ω2 − ω1∗ + ǫ/2 ∈] − 1/2, 1/2[ where ω2 − ω1∗ = ω1 + ω2 − 1/2 = −1/2, since βγ > 0 and θ = 0, i.e., simply ǫ/2 ∈]0, 1[. The intersection with (6.49) yields that Lǫ ∼ T ǫ is boundedly invertible for ǫ ∈ ]0, 2(1 − ω1 )[ where ω1 ∈]0, 1/2[ due to Lemma 6.1. ǫ Remark 6.13. The ǫ–intervals where Tjj , j = 1, 2, are bijective, have both length 2. In the present situation, their intersection has length more than 1. However, that result depends on the ansatz, and the determination of the maximal ǫ–interval where Lǫ is invertible remains open.

40


Now let us determine the “natural compatibility conditions” mentioned in the context of (6.45). Proposition 6.14. Under the assumptions of Proposition 6.12 the image normalization of L (for ǫ = 0) reads ≺

≺

T

1 L= Rst (B1 , B2 ) : H (Q1 ) →Y

(6.50)

where ≺

Y⊂H

−1/2

2

(R+ )

(6.51)

is a dense subspace characterized by the compatibility condition ≺

−1 e −1/2 (R+ ) g2 − KT11 g1 ∈ H

(6.52)

−1 − is given by T11 in Lemma 6.6 with κ = 0 (regarding for (g1 , g2 ) ∈ Y . Herein, T11 ≺

ǫ ∈ [0, 1[). L is boundedly invertible in that space setting and the oblique derivative ≺

problem is well-posed for data given in Y . Proof. Considering the family of boundedly invertible operators 2 T11 0 ǫ T = Rst : H 1/2+ǫ (R+ ) × H −1/2+ǫ (R+ ) → H −1/2+ǫ (R+ ) (6.53) K T22

ǫ ǫ for ǫ ∈]0, 1[ being is invertible for ǫ ∈ [0, 1[ and T22 for ǫ ∈]0, 1[, we know that T11 −1/2 injective for ǫ = 0 with dense image in H (R+ ) and given by the appendix: ≺

−1/2

o Y 22 = im T22 = r+ Aλ−s−iτ −1/2 ℓ r+ H+ −

−1/2

= r+ H+

e −1/2 (R+ ) (6.54) =H

since s = −1/2, τ = 0, i.e., −s − iτ − 1/2 = 0. Thus, the necessary and sufficient condition (6.52) for the solvability of T f = g results from (6.54) and T11 0 g1 f =g= K T22 g2 Kf1 + T22 f2 = g2

(6.55)

≺

−1 T22 f2 = g2 − KT11 g1 ∈Y 22 .

The consequences for L are evident.

Appendix A. Here is a brief outline of the constructive (generalized) inversion of a scalar CTOS under special assumptions that are most convenient for the present purpose [7]. A generalization to matrix CTOS can be found in [5] and more constructive methods are planned to be published in near future combining results from [9], [10] and [13]. The central case is S = r+ Aσ ℓo : H s (R+ ) → H s (R+ )

(A.1)


41

(or ℓe instead of ℓo , respectively) where ¨ , σ ∈ G C ν (R)

ν ∈]0, 1[

s ∈] − 3/2, 1/2[

(or ] − 1/2, 3/2[ respectively)

(A.2)

which yields that S is a composition of bounded linear operators and Aσ : H s → H s is bijective. Defining Z −1 ω = (2πi) d logσ = κ + η + iτ (A.3) R

where κ ∈ Z, η ∈ [−3/4 − s/2, 1/4 − s/2[ (or η ∈ [−1/4 − s/2, 3/4 − s/2[, respectively), and τ ∈ R, one can write σ = σ(+∞) ζiω ψ

(A.4)

˙ ψ(∞) = 1 and7 with ψ ∈ G C ν (R), ζi (ξ) =

ξ−i λ− (ξ) = , λ+ (ξ) ξ+i

ξ ∈ R.

e = ψ(−ξ) for ξ ∈ R and factorize We put ψ(ξ) ψ = ψ− ψe

(A.5)

(A.6)

˙ and a minus type function ψ− ∈ G C ν (R) ˙ by into an even function ψe ∈ G C ν (R) o n (A.7) ψ− = exp P− log ψ ψe−1

where P− = (I − SR )/2 is the Hilbert projector. Now define 2(η+iτ )

σ − = λ− σe =

ψ−

−1 σ ζi−κ σ−

(A.8)

where σe turns out to be even [7] and consider the factorization Aσ = Aσ− CAσe : H s → H s+2η → H s+2η → H s with C =

F −1 ζiκ

· F and 3 1 s + 2η ∈ − , 2 2

1 3 (or s + 2η ∈ − , , respectively). 2 2

(A.9)

(A.10)

(A.9) was named an asymmetric factorization through an intermediate space (AFIS) in [7], [6] and has the following properties: If s + 2η 6= −3/2, i.e., s + 2η ∈] − 3/2, 1/2[ (or s + 2η ∈] − 1/2, 3/2[), the composed operator (replace ℓo by ℓe in the alternative case): o −1 o s s S − = r+ A−1 ℓ r+ A−1 σe ℓ r+ C σ− ℓ : H (R+ ) → H (R+ ) 7 Instead

of λ± (ξ) one can use t± (ξ) = ξ ± k, ξ ∈ R, since ℑm k > 0.

(A.11)

42


represents a composition of bounded linear operators and a one-sided (left/right) inverse of S (if κ ≥ 0 or κ ≤ 0, respectively, and the second/first term ℓo r+ in (A.11) can be omitted). In this case S is Fredholm with α(S) = dim ker S = max{0, −κ} β(S) = dim H s / im S = max{0, κ}

(A.12)

and projectors I − S − S onto the kernel and SS − onto the image of S are obtained in the standard way, as an advantage of considering all operators as to be bounded in suitable (Sobolev) spaces. κ is referred to as the factorization index and directly given by 3 s 3 s = sup z ∈ Z : z ≤ ℜe ω + + (A.13) κ = Int ℜe ω + + 4 2 4 2

(replace 3/4 by 1/4 in the ℓe case). If s + 2η = −3/2 (or −1/2, respectively), the operator S is not normally solvable. It can be easily normalized by a simultaneous change of s in the two spaces of (A.1), e.g., replacing s by s + ǫ, ǫ 6= 0, such that s + ǫ ∈] − 3/2, 1/2[ (or s + ǫ ∈] − 1/2, 3/2[, respectively) and applying the previous result with the same formulas (A.12) if ǫ > 0. However, S can also be normalized by only changing the image space, namely replacing it by a dense subspace [7, Theorem 5.4] ≺

−1/2

H s(R+ ) = r+ Aλ−s−iτ −1/2 H+ −

−1/2

with the norm induced by H+

(A.14)

. Then, a one sided inverse of the normalized

≺

operator S is obtained by continuous extension of the one-sided inverse of the previous operator in H s+ǫ (R+ ) where ǫ > 0, in brief ≺− ≺ = Ext S (s+ǫ) : H s (R+ ) → H s (R+ ) . (A.15) S

A dual method for domain normalization is described in [22] in the context of Wiener-Hopf operators which is not carried out here for CTOS. Both are called minimal (image/domain) normalization of S and unique up to norm equivalence. There are various consequences, e.g., for operators of the form (A.1) acting ¨ which between different spaces, S : H r (R+ ) → H s (R+ ) where ts−r σ ∈ G C ν (R) are toplinear equivalent to so-called lifted operators on L2 (R+ ) (i.e., (A.1) with 1/2 1/2 s = 0). Choosing λ(ξ) = (ξ 2 + 1) or, alternatively, λ(ξ) = t(ξ) = (ξ 2 − k 2 ) , we have S = r+ Aλ−s ℓS0 r+ Aλr ℓo −

S0 = r+ Aσ0 ℓo : L2 (R+ ) → L2 (R+ ) σ0 = λs− σλ−r

(A.16)

provided r ∈] − 3/2, 1/2[ (or r ∈] − 1/2, 3/2[ if we replace ℓo by ℓe ), such that the extension operator is left invertible by r+ .


43

References [1] H. Bart and V. E. Tsekanovskii, Matricial coupling and equivalence after extension, Oper. Theory Adv. Appl. 59 (1992), 143–160. [2] A. B¨ ottcher and B. Silbermann, Analysis of Toeplitz Operators, Springer-Verlag, Berlin, 1990. [3] L. P. Castro, R. Duduchava and F.-O. Speck, Localization and minimal normalization of some basic mixed boundary value problems. In: Factorization, Singular Operators and Related Problems (S. Samko, A. Lebre and A. F. dos Santos, eds.), pp. 73–100. Kluwer Acad. Publ., Dordrecht, 2003. [4] L. P. Castro and F.-O. Speck, Relations between convolution type operators on intervals and on the half-line, Integral Equations Operator Theory 37 (2000), 169–207. [5] L. P. Castro and F.-O. Speck, Inversion of matrix convolution type operators with symmetry, Port. Math. (N.S.) 62 (2005), 193–216. [6] L. P. Castro, F.-O. Speck and F. S. Teixeira, On a class of wedge diffraction problems posted by Erhard Meister, Oper. Theory Adv. Appl. 147 (2004), 211–238. [7] L. P. Castro, F.-O. Speck and F. S. Teixeira, A direct approach to convolution type operators with symmetry, Math. Nachr. 269-270 (2004), 73–85. [8] R. Duduchava, Integral Equations in Convolution with Discontinuous Presymbols, Singular Integral Equations with Fixed Singularities, and their Applications to Some Problems of Mechanics, Teubner Verlagsgesellschaft, Leipzig, 1979. [9] T. Ehrhardt, Invertibility theory for Toeplitz plus Hankel operators and singular integral operators with flip, J. Funct. Anal. 208 (2004), 64–106. [10] T. Ehrhardt and F.-O. Speck, Transformation techniques towards the factorization of non-rational 2×2 matrix functions, Linear Algebra Appl. 353 (2002), 53–90. ` [11] G. I. Eskin, Boundary Value Problems for Elliptic Pseudodifferential Equations, Translations of Mathematical Monographs 52, AMS, Providence, R.I., 1981. [12] P. Grisvard, Elliptic Problems in Nonsmooth Domains, Monographs and Studies in Mathematics 24, Pitman (Advanced Publishing Program), Boston, 1985. [13] I. Gohberg, M. A. Kaashoek and I. M. Spitkovsky, An overview of matrix factorization theory and operator applications, Oper. Theory Adv. Appl. 141 (2003), 1–102. [14] S. Lang, Real and Functional Analysis, 3rd ed., Grad. Texts in Math. 142, SpringerVerlag, New York, 1993. [15] E. L¨ uneburg and R. A. Hurd, On the diffraction problem of a half-plane with different face impedances, Can. J. Phys. 62 (1984), 853–860. [16] E. Meister, Some solved and unsolved canonical problems of diffraction theory, Lecture Notes in Math. 1285 (1987), 320–336. [17] E. Meister and F.-O. Speck, Diffraction problems with impedance conditions, Appl. Anal. 22 (1986), 193–211. [18] E. Meister and F.-O. Speck, Modern Wiener-Hopf methods in diffraction theory, Pitman Res. Notes Math. Ser. 216 (1989), 130–171. [19] E. Meister, F. Penzel, F.-O. Speck and F. S. Teixeira, Some interior and exterior boundary value problems for the Helmholtz equation in a quadrant, Proc. Roy. Soc. Edinburgh Sect. A 123 (1993), 275–294.

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[20] S. E. Mikhailov, Analysis of united boundary-domain integro-differential and integral equations for a mixed BVP with variable coefficient, Preprint CMS/MAT/2004/11, Glasgow Caledonian University (2004), 26 p. [21] S. G. Mikhlin and S. Pr¨ ossdorf, Singular Integral Operators, Springer-Verlag, Berlin, 1986. [22] A. Moura Santos, F.-O. Speck and F. S. Teixeira, Minimal normalization of WienerHopf operators in spaces of Bessel potentials, J. Math. Anal. Appl. 225 (1998), 501– 531. [23] F. Penzel and F. S. Teixeira, The Helmholtz equation in a quadrant with Robin’s conditions, Math. Methods Appl. Sci. 22 (1999), 201–216. [24] A. F. dos Santos, A. B. Lebre and F. S. Teixeira, The diffraction problem for a half-plane with different face impedances revisited, J. Math. Anal. Appl. 140 (1989), 485–509. [25] F.-O. Speck, Sommerfeld diffraction problems with first and second kind boundary conditions, SIAM J. Math. Anal. 20 (1989), 396–407. [26] F.-O. Speck, R. A. Hurd and E. Meister, Sommerfeld diffraction problems with third kind boundary conditions, SIAM J. Math. Anal. 20 (1989), 589–607. [27] E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, Princeton, 1970. [28] H. Triebel, Interpolation Theory, Function Spaces, Differential Operators, NorthHolland, Amsterdam, 1978. L. P. Castro Department of Mathematics University of Aveiro 3810–193 Aveiro Portugal e-mail: [email protected] F.-O. Speck Department of Mathematics I.S.T., Technical University of Lisbon 1049–001 Lisbon Portugal e-mail: [email protected] F. S. Teixeira Department of Mathematics I.S.T., Technical University of Lisbon 1049–001 Lisbon Portugal e-mail: [email protected]

Mixed Boundary Value Problems for the Helmholtz Equation in a

Mixed Boundary Value Problems for the Helmholtz Equation in a

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