THE SOLUTION OF THE THIRD PROBLEM FOR THE LAPLACE EQUATION ON PLANAR DOMAINS WITH SMOOTH BOUNDARY AND INSIDE CRACKS AND MODIFIED JUMP CONDITIONS ON CRACKS

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LAPLACE EQUATION ON PLANAR DOMAINS WITH SMOOTH BOUNDARY AND INSIDE CRACKS AND MODIFIED JUMP CONDITIONS ON CRACKS

DAGMAR MEDKOV ´A

Received 23 November 2005; Revised 13 March 2006; Accepted 4 April 2006

This paper studies the third problem for the Laplace equation on a bounded planar domain with inside cracks. The third condition∂u/∂n+hu=f is given on the boundary of the domain. The skip of the functionu+−u₋=g and the modified skip of the normal derivatives (∂u/∂n)+−(∂u/∂n)₋+hu+= f are given on cracks. The solution is looked for in the form of the sum of a modified single-layer potential and a double-layer potential.

The solution of the corresponding integral equation is constructed.

1. Introduction

Krutitskii studied in [6] the boundary value problem for the Helmholtz equation out- side several cuts in the plane. Two boundary conditions were given on the cuts. One of them specified the jump of the unknown function. Another one of the type of the Robin condition contained the jump of the normal derivative of an unknown function and the one-side limit of this function on the cuts. He looked for a solution of the problem in the form of the sum of a single-layer potential and an angular potential. He reduced the problem to the solution of a uniquely solvable integral equation. So, he proved the unique solvability of the problem.

This paper studies the boundary value problem for the Laplace equation in a bounded planar domain with several cuts inside. The cuts in [6] are smooth open arcs. The cracks in this paper are arbitrary closed subsets of such sets. The Robin condition is given on the boundary of the domain. The same conditions as that in [6] are on the cuts. The case when the jumps of the solution and its normal derivatives are given on cracks is included. So, this problem is a generalization of the problem studied in [4,5]. I looked for a solution in a similar form like in [6] but instead of a single-layer potential, I used a modified single-layer potential. I reduced the problem to the solution of an integral equation and constructed the solution of this equation. So, I constructed a solution of the problem which is a new result even if there are no cuts inside the domain.

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 91983, Pages1–14

DOI10.1155/IJMMS/2006/91983

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2. Formulation of the problem

Let M⊂R^m. We denote byC⁰(M) the set of all continuous functions on M. If k is a positive integer, denote byC^k(M) the set of all functions f such that there is continuous D^αf onMfor each multi-indexαwith the length at mostk. If 0< β <1, denote byC^β(M) the set ofβ-H¨older functions onM, that is, the set of continuous functions f onMsuch that

sup

x∈M

f(x)+ sup

x,y∈M;x=y

f(x)−f(y)

|x−y|^β <∞. (2.1) Ifk is a positive integer and 0< β <1, denote byC^k+β(M) the set of all functions f ∈ C^k(M) such thatD^αf ∈C^β(M) for each multi-indexαwith the lengthk.

We say that a bounded open setH⊂R² hasC^α boundary∂H if there exist a finite number of “local” coordinate systems (xk,yk) (k=1,. . .,m) and a finite number of func- tionsϕkof classC^α,k=1, 2,. . .,m, defined on (−δ,δ) (whereδ >0) such that

(1) (x_k,y_k)∈Hfor|x_k|< δ,ϕ_k(x_k)−δ < y_k< ϕ_k(x_k),

(2) (xk,yk)∈/clH, the closure ofH, for|xk|< δ,ϕk(xk)< yk< ϕk(xk) +δ,

(3) for everyz∈∂H, there existk(k=1,. . .,m) andxk∈(−δ,δ) such thatz=(xk, ϕ_k(x_k)) in the corresponding coordinate system.

LetH,H+⊂R² be bounded open sets withC¹ boundaries and let clH+⊂H,H be connected. PutH₋=H\clH+. Denote byn(x) (n⁺(x),n⁻(x)) the unit exterior normal ofH(H+,H₋) atx, respectively. ForΓ, the closed subset of∂H+, putG₌H_\Γ. Ifuis a function inG,x∈∂G(x∈∂H+,x∈∂H₋), denote byu(x) (u+(x),u₋(x)) the limit ofu atxwith respect toG(H+,H₋), respectively. We will study the Robin problem for the Laplace equation on the cracked open setG.

Let h∈C⁰(∂G), h≥0, f ∈C⁰(∂G), g∈C⁰(Γ). We say that a function u in G is a solution of the problem

Δu=0 inG, ∂u

∂n

+hu=f on∂G\Γ, u+−u₋=g onΓ, ∂u

∂n⁺

+− ∂u

∂n⁺

−+hu+=f onΓ,

(2.2)

if

(1)uis harmonic inG;

(2) there is an extension ofuonto the function fromC¹(clH+);

(3) there is an extension ofuonto the function fromC¹(clH₋);

(4)n(x)· ∇u(x) +h(x)u(x)=f(x) in∂G\Γ;

(5)u+(x)−u₋(x)=g(x) inΓ;

(6)n⁺(x)·[∇u]+(x)−n⁺(x)·[∇u]₋(x) +h(x)u+(x)=f(x) inΓ.

Ifh=0, we will talk about the Neumann problem. In the opposite case, we will talk about the Robin problem.

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3. Uniqueness

Denote by Ᏼk the k-dimensional Hausdorﬀ measure normalized so that Ᏼk is the Lebesgue measure inR^k.

Theorem 3.1. Leth∈C⁰(∂G),h≥0, f ≡0,g≡0, and letube a solution of the problem (2.2). Thenuis constant inG. If there isx∈∂Gso thath(x)>0, thenu=0 inG.

Proof. u∈C⁰(clG) becauseg≡0. Sincen⁻= −n⁺on∂H+, we get, using Green’s formula forH+andH₋,

0=

∂H

u(n· ∇u) +hu² dᏴ1+

Γun⁺·[∇u]+−n⁺·[∇u]₋+hu+ dᏴ1

=

∂Ghu²dᏴ1+

∂H+

un⁺·[∇u]+ dᏴ1+

∂H−

un⁻·[∇u]₋dᏴ1

=

∂Ghu²dᏴ1+

G|∇u|²dᏴ2.

(3.1)

Sinceh≥0, we have

∂Ghu²dᏴ1=0,

G|∇u|²dᏴ2=0. (3.2) Since ∇u=0 in G,uis constant in each component of G. Since u∈C⁰(H) and H is connected, there is a constantcsuch thatu=conH. If there isx∈∂Gso thath(x)>

0, then 0=n⁺(x)·[∇u]+(x)−n⁺(x)·[∇u]₋(x) +h(x)u+(x)=h(x)c. Therefore,c=0.

4. Necessary conditions for the solvability

LetK be a closed subset of∂H₋. For a function f defined onK, denote fK= f onK, f_K=0 on∂H₋\K. We say thatf ∈C₀^α(K) if f_K∈C^α(∂H₋). If 0< α <1, denote

f C0^α(K)=sup

x∈K

f(x)+ sup

x,y∈K;x=y

f(x)−f(y)

|x−y|^α , f _C^1+α₀ _(K)=sup

x∈K

f(x)+∂ fK

∂τ

C0^α(K)

,

(4.1)

whereτ(x)=(−n2(x),n1(x)) is the unit tangential vector ofH₋atx.

Proposition 4.1. Leth∈C⁰(∂G),h≥0, f ∈C⁰(∂G),g∈C⁰(Γ), and letube a solution of the problem (2.2). Theng∈C¹₀(Γ). Ifh∈C₀⁰(∂G), then f ∈C⁰₀(∂G). Ifh≡0, then

∂Gf dᏴ1=0. (4.2)

Proof. Sinceg_Γ=u+−u₋in∂H+andu+,u₋∈C¹(∂H+), we deduce thatg_Γ∈C¹(∂H+). If h≡0, we get, using Green’s theorem and the fact thatn⁻= −n⁺on∂H+,

∂Gf dᏴ1=

∂H+

n⁺·[∇u]+dᏴ1+

∂H−

n⁻·[∇u]₋dᏴ1=0. (4.3)

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Suppose now thath∈C⁰₀(∂H+). Since f∂G=n⁺·[∇u]+−n⁺·[∇u]₋+hu+in∂H+and u+,n⁺, [∇u]+, [∇u]₋∈C⁰(∂H+), we get f∂G∈C⁰(∂H+).

5. Single-layer potentials FixR >0. For f ∈C⁰(∂G), define

᏿Rf(x)=(2π)⁻¹

∂Gf(y) log R

|x−y|

dᏴ1(y) (5.1)

as the modified single-layer potential with density f. In the case of several sets, we will write᏿^GRf. (ForR=1, we get the usual single-layer potential.) Note that᏿Rf diﬀers by constants. The function᏿Rf is harmonic inR²\∂G.

Lemma 5.1. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,R >0. If f ∈C₀^β(∂G), then᏿Rf ∈C⁰(R²),᏿Rf ∈C¹(clH+),᏿Rf ∈C¹(clH₋), and

sup

x∈G

᏿Rf(x)+ sup

x∈G

∇᏿Rf(x)≤M f _C^β

0(∂G), (5.2)

whereMdepends onG,R, andβ.

Proof. According to [3, Lemma 2.18], the function᏿Rf is continuous inR². According to [10, page 227], we have᏿Rf ∈C¹(clH+),᏿Rf ∈C¹(clH₋). Thus᏿R: f →᏿Rf is a linear operator fromC^β0(∂G) to C¹(clH₋). Easy calculation yields that this operator is closed. According to the closed graph theorem (see [13, Chapter II, Section 6, Theorem

1]), it is bounded, similarly forH+.

Lemma 5.2. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,R >0, h∈C^β0(∂G). IfVhf =hSRf for f ∈C^β0(∂G) is denoted, thenVhis a compact linear operator onC₀^β(∂G).

Proof. Lemma 5.1yields that᏿R represents a bounded linear operator fromC^β0(∂G) to C¹(clH₋). Since the identity operator is a compact operator fromC¹(clH₋) toC^β(∂H₋), the operator᏿Ris a compact linear operator fromC₀^β(∂G) toC^β(∂H₋) by [13, Chapter X, Section 2]. Sinceh∂G∈C^β(∂H₋), the operatorH:g→hgis a bounded linear operator inC^β(∂H₋). SinceH᏿R is a compact linear operator fromC^β₀(∂G) toC^β(∂H₋) by [13, Chapter X, Section 2], the operatorVhis a compact linear operator onC^β₀(∂G).

Lemma 5.3. Letϕ∈C⁰₀(∂G),R >diamG, the diameter ofG. If there isy∈∂Gsuch that ϕ(y)=0, then

0<

∂Gϕ(x)SRϕ(x)dᏴ1(x)<∞. (5.3) Proof. Fixx0in∂G. PutP(x)=(x−x0)/R,P₋1(x)=Rx+x0 forx∈R²,G=P(G). For x∈∂G, put ϕ(x) =ϕ(P₋1(x)). Then ϕ∈C⁰(∂G) and ᏿Rϕ(x)=R᏿^G1ϕ(P(x)) for x∈ G. Denote by Ᏼ the restriction of Ᏼ1 onto ∂G. Since ᏿^G1|ϕ| is continuous in R² by

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[10, Lemma 4], we conclude that

|ϕ|᏿^G1|ϕ|dᏴ<∞, (5.4) and thus the real measure ϕᏴ has finite energy (see [7, Chapter 1, Section 4]). Since Ᏼ1({x∈∂G;|ϕ(x) |>0})>0, [7, Theorem 1.16] yields

ϕ᏿^G1ϕ d Ᏼ>0. (5.5)

Now we use the fact that

∂Gϕ(x)SRϕ(x)dᏴ1(x)=R²

∂Gϕ(x)S ^G1ϕ(x)dᏴ 1(x). (5.6)

6. Double-layer potentials

Ifg∈C₀⁰(Γ), denote, forx∈R²\Γ, Ᏸg(x)=(2π)⁻¹

Γg(y)|x−y|⁻²n⁺(y)·(y−x)dᏴ1(y) (6.1) as the double-layer potential with densityg.

IfV is a bounded open set withC¹ boundary,g ∈C(∂V), andn^V(y) denotes the exterior unit normal ofVaty, define onR²\∂V

ᏰVg(x)=(2π)⁻¹

∂Vg(y)|x−y|⁻²n^V(y)·(y−x)dᏴ1(y) (6.2) as the double-layer potential corresponding toVwith densityg.

Lemma 6.1. LetH,H+ have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,g∈ C^1+β₀ (Γ). Then Ᏸg is a harmonic function in R²\Γ, Ᏸg ∈C¹(clH+), Ᏸg ∈C¹(clH₋), [Ᏸg]+(x)−[Ᏸg]₋(x)=g(x) onΓ,n⁺(x)·[∇Ᏸg]+(x)−n⁺(x)·[∇Ᏸg]−(x)=0 onΓ, and

sup

x∈G

Ᏸg(x)+ sup

x∈G

∇Ᏸg(x)≤M g _C^1+β

0 (∂G), (6.3)

whereMdepends onG,R, andβ.

Proof. Easy calculations yield that Ᏸg is a harmonic function in R²\Γ. Since Ᏸg= ᏰH+g=ᏰH−g, [9, Theorem 1] yields thatᏰg∈C⁰(clH+),Ᏸg∈C⁰(clH₋), and [Ᏸg]+(x)

−[Ᏸg]₋(x)=g(x) inΓ.

The boundary ofH+is formed by finitely many Jordan curves. Fix one of these curves T. Denote gT =g on T, gT =0 elsewhere. Let T be parametrized by the arc length s:T= {ϕ(s); s∈[a,b]}, andH+ is to the right when the parameters increases onT.

Put f(ϕ(s))= −[g(ϕ)](s). Then f ∈C^β(T) becauseg∈C^1+β(∂H+). Forx∈R²\Tand s∈[a,b], denote byv(x,ϕ(s)) the increment of the argument of y−xalong the curve

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{y=ϕ(t); t∈[a,s]}, and

V f(x)=(2π)⁻¹

Tv(x,y)f(y)dᏴ1(y) (6.4) is the angular potential corresponding to f. Define f =0 onR²\T. SinceV f is a con- jugate function to−᏿^H1⁺f (see [10, pages 226-227]), we have∂V f /∂x1= −∂᏿^H1⁺f /∂x2,

∂V f /∂x2=∂᏿^H1⁺f /∂x1. Lemma 5.1 gives ∇V f ∈C⁰(clH+), ∇V f ∈C⁰(clH₋). Using boundary properties of single-layer potentials (see [2, Theorem 2.2.13]), we can deduce thatn⁺(x)·[∇V f]+(x)−n⁺(x)·[∇V f]₋(x)=0 inT. Putg=g−g(ϕ(a)),g=g(ϕ(a)) onT,g,g=0 elsewhere. Since

gT

ϕ(s)−gT

ϕ(a)= _s

a

−f(t)dt= _b

s f(t)dt, (6.5)

we haveᏰH+g=V f by [10, page 226]. ThereforeᏰH+g∈C¹(clH+),ᏰH+g∈C¹(clH₋), andn⁺(x)·[∇ᏰH+g]+(x)−n⁺(x)·[∇ᏰH+g]₋(x)=0 inT. According to [9, page 137], the functionᏰgis constant in the interior of T and in the exterior ofT as well. Thus

∇ᏰH+g=0 in R²\T. So, ᏰgT =ᏰH+g+ᏰH+g∈C¹(clH+)∩C¹(clH₋) and n⁺(x)·

+(x)−n⁺(x)·[∇ᏰgT]₋(x)=0 on ∂H+. Summing over all T, we getᏰg ∈C¹(clH+), Ᏸg∈C¹(clH₋), andn⁺(x)·[∇Ᏸg]+(x)−n⁺(x)·[∇Ᏸg]−(x)=0 onΓ.

SinceᏰ:g→Ᏸg is a closed linear operator fromC₀^1+β(Γ) toC¹(clH+), it is bounded by the closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]), similarly for

H₋.

7. Reduction of the problem

LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ, f ∈C^β₀(∂G),h∈ C^β₀(∂G),h≥0,g∈C₀^1+β(Γ). We will look for a solutionuof the problem (2.2) in the form

u=Ᏸg+v. (7.1)

According toLemma 6.1and [9, Theorem 1], the functionuis a solution of the problem (2.2) if and only if the functionvis a solution of the problem

Δv=0 inG,

∂v

∂n+hv=F on∂G\Γ, v+−v₋=0 onΓ, ∂v

∂n⁺

+− ∂v

∂n⁺

−+hv+=F onΓ,

(7.2)

where

F=f −∂Ᏸg

∂n ⁻hᏰg on∂G\Γ, (7.3)

F(x)= f(x)−h(x) 1

2g(x) +

Γ

n⁺(y)·(y−x)

2π|x−y|² g(y)dᏴ1(y)

, x∈Γ. (7.4)

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F∈C₀^β(∂G) because f ∈C^β₀(∂G), h∈C^β₀(∂G), Ᏸg ∈C^∞(R²\Γ), Ᏸg ∈C¹(clH+) by Lemma 6.1, and

[Ᏸg]+(x)=1

2g(x) + 1 2π

Γ

n⁺(y)·(y−x)

|x−y|² g(y)dᏴ1(y) (7.5) forx∈Γby [1, Theorem].

We will look for a solution of the problem (7.2) in the form of a modified single-layer potential᏿Rw, wherew∈C₀^β(∂G) andR >diamG.

Define forw∈C0^β(∂G),x∈∂H, K_G^∗w(x)=(2π)⁻¹

∂Gw(y)|x−y|⁻²n(x)·(y−x)dᏴ1(y). (7.6) According to [9, Theorem 2],

n(x)· ∇᏿Rw(x)=w(x)

2 +K_G^∗w(x) forx∈∂G\Γ, (7.7) n⁺(x)·

∇᏿Rw ₊(x)−n⁺(x)·

∇᏿Rw ₋(x)=w(x) inΓ. (7.8) Since the modified single layer potential᏿Rwis a harmonic function inG, we get using Lemma 5.1that᏿Rwis a solution of the problem (7.2) if and only if

Th,Rw=F, (7.9)

where

T_h,Rw=w

2 +K_G^∗w+h᏿Rw on∂G\Γ,

Th,Rw=w+h᏿Rw onΓ. (7.10)

8. Solvability of the problem

Lemma 8.1. LetH,H+ have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,h∈ C^β₀(∂G),h≥0,R >diamG. ThenK_G^∗is a compact linear operator fromC₀^β(∂G) toC^β(∂H) andTh,Ris a bounded linear operator inC0^β(∂G).

Proof. Fixα∈(β,γ). ThenK_H^∗is a linear operator fromC^β(∂H) toC^α(∂H) by [11, Theo- rem 14.IV]. SinceK_H^∗is a bounded linear operator fromC^β(∂H) toC⁰(∂H) by (7.7) and Lemma 5.1, the operatorK_H^∗is a closed operator fromC^β(∂H) toC^α(∂H). This gives that the operatorK_H^∗ is a closed linear operator fromC^β₀(∂G) toC^α(∂H). The closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]) shows thatK_H^∗is a bounded linear operator fromC^β₀(∂G) toC^α(∂H). Since the identity operator is a compact operator from C^α(∂H) toC^β(∂H), the operatorK_G^∗is a compact linear operator fromC₀^β(∂G) toC^β(∂H) by [13, Chapter X, Section 2]. SinceK_G^∗ is a bounded linear operator fromC₀^β(∂G) to C^β(∂H), the operatorT0,R is a bounded linear operator inC^β0(∂G). SinceTh,R−T0,R is a bounded linear operator inC₀^β(∂G) byLemma 5.2, the operatorTh,Ris a bounded linear

operator inC0^β(∂G).

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Notation 8.2. LetXbe a real Banach space. Denote by complX= {x+iy; x,y∈X}the complexification ofXwith the norm x+iy = x + y . IfT is a bounded linear operator onX, we defineT(x+iy)=Tx+iT yas the bounded linear extension ofTonto complX.

Definition 8.3. The bounded linear operatorTon the Banach spaceXis called Fredholm ifα(T), the dimension of the kernel ofT, is finite, the rangeT(X) ofTis a closed subspace ofX, andβ(T), the codimension ofT(X), is finite. The numberi(T)=α(T)−β(T) is the index ofT.

Lemma 8.4. LetH,H+ have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,h∈ C^β0(∂G),h≥0,R >diamG. Ifλ∈C,λ=1/2,λ=1, thenTh,R−λIis a Fredholm operator with index 0 in complC₀^β(∂G). (HereIdenotes the identity operator.)

Proof. DenoteL f(x)= f(x)/2 for x∈∂H, L f(x)= f(x) for x∈Γ. Then L−λI is a Fredholm operator in complC^β₀(∂G). SinceT0,R−Lis a compact operator inC₀^β(∂G) by Lemma 8.1, the operatorT0,R−λIis a Fredholm operator with index 0 inC^β0(∂G) by [12, Theorem 5.10]. SinceTh,R−T0,Ris a compact linear operator inC₀^β(∂G) byLemma 5.2, the operatorTh,R−λIis a Fredholm operator with index 0 inC^β0(∂G) by [12, Theorem

5.10].

Lemma 8.5. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,R >

diamG,ϕ∈C₀^β(∂G). IfT_0,R² ϕ=0, thenT0,Rϕ=0.

Proof. According toSection 7, the modified single-layer potential᏿RT0,Rϕis a solution of the problem (2.2) withh≡0,g≡0, and f =T_0,R² ϕ=0. According toTheorem 3.1, there

is a constantcsuch that᏿RT_0,Rϕ=conG.

According toSection 7, the modified single-layer potential᏿Rϕis a solution of the problem (2.2) withh≡0,g≡0, and f =T0,Rϕ. Thus

∂GT0,Rϕ=0 (8.1)

byproposition 4.1. Since᏿RT_0,Rϕ₌cinGand᏿RT_0,Rϕis continuous inR²byLemma 5.1, we obtain

∂G

T0,Rϕ᏿RTh,RϕdᏴ1=c

∂GT0,Rϕ=0. (8.2)

According toLemma 5.3, we haveT0,Rϕ=0 a.e. in∂G.

Proposition 8.6. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ, h_∈C^β₀(∂G),h_≥0,h(x)>0 for somex_∈∂G,R >diamG. ThenT_h,R is continuosly invert- ible inC₀^β(∂G). Denote byC^β_0,0(∂G) the space of all f ∈C₀^β(∂G) for which (4.2) holds. Then T_0,Ris continuously invertible inC^β_0,0(∂G).

Proof. Letϕ∈C^β₀(∂G) be such that᏿Rϕ=0 inG. Then᏿Rϕ=0 in∂GbyLemma 5.1.

According toLemma 5.3, we haveϕ≡0.

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Ifϕ∈C₀^β(∂G) andT_h,Rϕ=0, then᏿Rϕis a solution of the problem (2.2) with f ≡0, g≡0 bySection 7. Since᏿Rϕ=0 inGbyTheorem 3.1, we haveϕ≡0. Since Th,R is a Fredholm operator inC^β0(∂G) with index 0 by Lemma 8.4, we obtain Th,R(C^β0(∂G))= C^β₀(∂G). Thus, T_h,R is continuously invertible inC^β₀(∂G) by [13, Chapter II, Section 6, Proposition 3] and [13, Chapter II, Section 6, Theorem 1].

Ifϕ∈C^β0(∂G), then᏿Rϕis a solution of the problem (2.2) withh≡0,g≡0, and f = T_0,Rϕ(seeSection 7). ThusT_0,Rϕ∈C_0,0^β (∂G) byProposition 4.1. Hence,T_0,R(C₀^β(∂G))⊂ C^β0,0(∂G) andβ(T0,R)≥1. SinceT0,R is a Fredholm operator inC^β0(∂G) with index 0 by Lemma 8.4, there is a nontrivialϕ1∈C0^β(∂G) such thatT0,Rϕ1=0. According toSection 7 andTheorem 3.1, there is a constantc1such that᏿Rϕ1=c1inG. Sincec1=0 yieldsϕ1=0 we deduce thatc1=0. Ifϕ∈C₀^β(∂G),T_0,Rϕ=0, thenSection 7andTheorem 3.1imply that there is a constantcsuch that᏿Rϕ=cin G. Since᏿R(ϕ−(c/c1)ϕ1)=0 inG, we obtain thatϕ−(c/c1)ϕ1≡0. This means thatα(T0,R)=1. SinceT0,R(C^β0(∂G))⊂C0,0^β (∂G), β(T0,R)=α(T0,R)=1 shows thatT0,R(C₀^β(∂G))=C^β_0,0(∂G).

According toLemma 8.5, the operator T_0,R is injective in T_0,R(C₀^β(∂G))=C_0,0^β (∂G).

SinceC^β0,0(∂G) is aT0,R-invariant closed linear subspace of finite codimension inC^β0(∂G) and T0,R is a Fredholm operator with index 0, the restriction of T0,R onto C0,0^β (∂G) is a Fredholm operator with index 0 by [8, Proposition 3.7.1.]. Since T_0,R is injective in C^β_0,0(∂G), it is surjective inC_0,0^β (∂G). The closed graph theorem (see [13, Chapter II, Sec- tion 6, Theorem 1]) gives thatT_0,Ris continuously invertible inC^β_0,0(∂G).

Theorem 8.7. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,h∈ C^β₀(∂G),h≥0,h(x)>0 for somex∈∂G, f ∈C₀^β(∂G),g∈C^1+β₀ (Γ). Then there is a unique solution of the problem (2.2).

Proof. FixR >diamG. According to Proposition 8.6, there isT_h,R⁻¹ in C^β0(∂G). LetF be given by (7.3). Then

u=Ᏸg+᏿RT_h,R⁻¹F (8.3)

is a solution of the problem (2.2) bySection 7. This solution is unique byTheorem 3.1.

Theorem 8.8. LetH,H+have boundary of classC^1+γ, where 0< γ <1. Let 0< β < γ,h≡0, f ∈C^β₀(∂G),g∈C^1+β₀ (Γ). Then there is a solution of the problem (2.2) if and only if (4.2) holds. This solution is unique up to an additive constant.

Proof. If there is a solution of the problem (2.2), the relation (4.2) holds byProposition 4.1. Suppose now that (4.2) holds. FixR >diamG. Denote byT the restriction ofT0,R

ontoC^β_0,0(∂G). Then there isT⁻¹byProposition 8.6. LetF be given by (7.4). Thenu= Ᏸg+᏿RT⁻¹Fis a solution of the problem (2.2) bySection 7. This solution is unique up

to an additive constant byTheorem 3.1.

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9. Solution of the problem

diamG,h∈C^β0(∂G),h≥0, f ∈complC^β0(∂G), f(x)=0 for somex∈∂G. Ifλis a complex number such thatT_h,Rf =λ f, thenλ≥0.

Proof. Take f1,f2∈C^β₀(∂G) such that f =f1+i f2. Since᏿R(f1−i f2)∈complC¹(clH+),

᏿R(f1−i f2)∈complC¹(clH₋),᏿R(f1−i f2)∈complC⁰(R²) byLemma 5.1, we get using Green’s formula that

∂H+

᏿R(f1−i f2) n⁺·

∇᏿R(f1+i f2) ₊=

H+

∇᏿Rf1²+∇᏿Rf2² dᏴ2,

∂H−

᏿R

f1−i f2 n⁻·

∇᏿R

f1+i f2 ₋=

H−

∇᏿Rf1²+∇᏿Rf2² dᏴ2.

(9.1) Summing,

∂G

᏿R

f1−i f2 T0,R f1+i f2

dᏴ1=

G

∇᏿Rf1²+∇᏿Rf2²

dᏴ2. (9.2) Using Fubini’s theorem,

λ

∂G

f1᏿Rf1+ f2᏿Rf2

dᏴ1=

∂G

᏿R

f1−i f2 Th,R

f1+i f2

dᏴ1

=

G

∇᏿Rf1²+∇᏿Rf2² dᏴ1

+

∂Gh᏿Rf1

2

+᏿Rf2

2 dᏴ1.

(9.3)

Since

0<

∂G

f1᏿Rf1+ f2᏿Rf2

dᏴ1<∞ (9.4)

byLemma 5.3andh≥0, we get λ=

G

∇᏿Rf1²+∇᏿Rf2²

dᏴ1+_∂Gh᏿Rf1

2

+᏿Rf2

2 dᏴ1

∂G

f1᏿Rf1+f2᏿Rf2

dᏴ1 ≥0. (9.5)

diamG, f ∈complC₀^β(∂G), f(x)=0 for somex∈∂G. Ifλis a complex number such that T0,Rf =λ f, then 0≤λ≤1.

Proof. We can suppose thatλ=0.Lemma 9.1yieldsλ >0 and we thus can suppose that f ∈C0^β(∂G). Since᏿Rλ⁻¹f is a solution of the problem (2.2) with h≡0, g ≡0 (see Section 7), Proposition 4.1gives that f fulfills (4.2). SinceT_0,Rf = f on Γ, we deduce from T0,Rf =λ f thatλ=1 or f =0 onΓ. We can restrict ourselves to the case when

f =0 onΓ.

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Fixr >0 such that clG⊂Ωr(0)≡ {y∈R²;|x|< r} and putV =Ωr(0)\clG. Then

᏿Rf ∈C¹(clV) byLemma 5.1. Using Green’s formula,

V

∇᏿Rf²dᏴ1=

∂V

᏿Rfn· ∇᏿RfdᏴ1

=

∂H

᏿Rf1 2I−K_H^∗

f dᏴ1+

∂Ωr(0)

∂᏿Rf

∂n ᏿Rf dᏴ1

=(1−λ)

∂Gf᏿Rf dᏴ1+

∂Ωr(0)᏿Rf∂᏿Rf

∂n dᏴ1.

(9.6)

Since (4.2) forces᏿Rf(x)=o(1),∇᏿Rf(x)=O(1/|x|) as|x| → ∞, we get forr→ ∞that

R²\clG

∇᏿Rf²dᏴ1=(1−λ)

∂Gf᏿Rf dᏴ1. (9.7) UsingLemma 5.3, we get

(1−λ)=

R²\clG∇᏿Rf²dᏴ1

∂Gf᏿Rf dᏴ1 ≥0. (9.8)

So,λ≤1.

Notation 9.3. LetXbe a complex Banach space and letTbe a bounded linear operator onX. Denote byσ(T) the spectrum ofTand byr(T)=sup{|λ|;λ∈σ(T)}the spectral radius ofT.

diamG,h∈C₀^β(∂G),h≥0. PutV_hf =h᏿Rf for f ∈complC₀^β(∂G). Then rVh

≤sup

x∈∂G

SRh(x). (9.9)

Proof. Letλ∈σ(V_h) in complC₀^β(∂G),λ=0. Since V_h is a compact linear operator in complC^β0(∂G) byLemma 5.2, there is f ∈complC0^β(∂G) such that f(z)=0 for somez∈

∂GandV_hf =λ f (see [13, Chapter X, Section 5, Theorem 2]). Using Fubini’s theorem,

∂G|λ f|dᏴ1=

∂G

h(x)

∂G(2π)⁻¹f(y) log R

|x−y|

dᏴ1(y)dᏴ1(x)

≤

∂Gh(x)(2π)⁻¹f(y)log R

|x−y|

dᏴ1(x)dᏴ1(y)

≤

∂G|f|dᏴ1sup

x∈∂G

᏿Rh(x).

(9.10)

Since f ∈complC₀^β(∂G) and f(z)=0, there isδ >0 such that|f(x)|>0 forx∈Ωδ(z)∩

∂GandᏴ1(Ωδ(z)∩∂G)>0. Dividing the inequality above by|f|, we obtain

|λ| ≤sup

x∈∂G

SRh(x). (9.11)