LAPLACE EQUATION ON PLANAR DOMAINS WITH
SMOOTH BOUNDARY AND INSIDE CRACKS AND
MODIFIED JUMP CONDITIONS ON CRACKS
DAGMAR MEDKOV ´AReceived 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 do-main 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.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
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
2. Formulation of the problem
Let M⊂Rm. We denote byC0(M) the set of all continuous functions on M. If k is a
positive integer, denote byCk(M) the set of all functions f such that there is continuous
Dαf onMfor each multi-indexαwith the length at mostk. If 0< β <1, denote byCβ(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 byCk+β(M) the set of all functions f ∈
Ck(M) such thatDαf ∈Cβ(M) for each multi-indexαwith the lengthk.
We say that a bounded open setH⊂R2 hasCα boundary∂H if there exist a finite
number of “local” coordinate systems (xk,yk) (k=1,. . .,m) and a finite number of
func-tionsϕkof classCα,k=1, 2,. . .,m, defined on (−δ,δ) (whereδ >0) such that
(1) (xk,yk)∈Hfor|xk|< δ,ϕk(xk)−δ < yk< ϕk(xk),
(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,
ϕk(xk)) 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.
3. Uniqueness
Denote by Ᏼk the k-dimensional Hausdorff measure normalized so that Ᏼk is the
Lebesgue measure inRk.
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 dᏴ 1+
Γu
n+·[∇u]
+−n+·[∇u]−+hu+dᏴ1
=
∂Ghu
2dᏴ 1+
∂H+
un+·[∇u]
+dᏴ1+
∂H−
un−·[∇u]−dᏴ1
=
∂Ghu
2dᏴ 1+
G|∇u|
2dᏴ 2.
(3.1)
Sinceh≥0, we have
∂Ghu
2dᏴ 1=0,
G|∇u|
2dᏴ
2=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,
fK=0 on∂H−\K. We say thatf ∈C0α(K) if fK∈Cα(∂H−). If 0< α <1, denote
f Cα
0(K)=sup
x∈K
f(x)+ sup
x,y∈K;x=y
f(x)−f(y)
|x−y|α ,
f C1+α
0 (K)=sup
x∈K
f(x)+∂ fK
∂τ
Cα 0(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∈C1
0(Γ). Ifh∈C00(∂G), then f ∈C00(∂G). Ifh≡0, then
∂Gf dᏴ1=0. (4.2)
Proof. SincegΓ=u+−u−in∂H+andu+,u−∈C1(∂H+), we deduce thatgΓ∈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−
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 writeGRf. (ForR=1, we get the usual single-layer potential.) Note thatRf differs by
constants. The functionRf is harmonic inR2\∂G.
Lemma 5.1. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,R >0. If
f ∈C0β(∂G), thenRf ∈C0(R2),Rf ∈C1(clH+),Rf ∈C1(clH−), and
sup
x∈G
Rf(x)+ sup
x∈G
∇Rf(x)≤M f
Cβ0(∂G), (5.2)
whereMdepends onG,R, andβ.
Proof. According to [3, Lemma 2.18], the functionRf is continuous inR2. According
to [10, page 227], we haveRf ∈C1(clH+),Rf ∈C1(clH−). ThusR: f →Rf is a
linear operator fromCβ0(∂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 classC1+γ, where 0< γ <1. Let 0< β < γ,R >0,
h∈Cβ0(∂G). IfVhf =hSRf for f ∈Cβ0(∂G) is denoted, thenVhis a compact linear operator
onC0β(∂G).
Proof. Lemma 5.1yields thatR represents a bounded linear operator fromCβ0(∂G) to
C1(clH
−). Since the identity operator is a compact operator fromC1(clH−) toCβ(∂H−), the operatorRis a compact linear operator fromC0β(∂G) toCβ(∂H−) by [13, Chapter
X, Section 2]. Sinceh∂G∈Cβ(∂H−), the operatorH:g→hgis a bounded linear operator inCβ(∂H−). SinceH
R is a compact linear operator fromCβ0(∂G) toCβ(∂H−) by [13,
Chapter X, Section 2], the operatorVhis a compact linear operator onCβ0(∂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)=RG1ϕ(P(x)) for x∈
[10, Lemma 4], we conclude that
|ϕ|G
1|ϕ|dᏴ<∞, (5.4)
and thus the real measure ϕᏴhas finite energy (see [7, Chapter 1, Section 4]). Since
Ᏼ1({x∈∂G;|ϕ(x) |>0})>0, [7, Theorem 1.16] yields
ϕG
1ϕ d Ᏼ>0. (5.5)
Now we use the fact that
∂Gϕ(x)SRϕ(x)dᏴ1(x)=R
2
∂Gϕ(x)S
G
1ϕ(x)d Ᏼ1(x). (5.6)
6. Double-layer potentials
Ifg∈C0
0(Γ), denote, forx∈R2\Γ,
Ᏸg(x)=(2π)−1
Γg(y)|x−y|
−2n+(y)·(y−x)dᏴ
1(y) (6.1)
as the double-layer potential with densityg.
IfV is a bounded open set withC1 boundary,g ∈C(∂V), andnV(y) denotes the
exterior unit normal ofVaty, define onR2\∂V
ᏰVg(x)=(2π)−1
∂Vg(y)|x−y|
−2nV(y)·(y−x)dᏴ
1(y) (6.2)
as the double-layer potential corresponding toVwith densityg.
Lemma 6.1. LetH,H+ have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,g∈
C1+0 β(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
C01+β(∂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 ∈Cβ(T) becauseg∈C1+β(∂H
+). Forx∈R2\Tand
{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−H+
1 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∈C
1(clH
+),ᏰH+g∈C
1(clH
−), andn+(x)·[∇Ᏸ
H+g]+(x)−n
+(x)·[∇Ᏸ
H+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 R
2\T. So, Ᏸg
T =ᏰH+g+ᏰH+g∈C
1(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 fromC01+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 classC1+γ, where 0< γ <1. Let 0< β < γ, f ∈Cβ0(∂G),h∈
Cβ0(∂G),h≥0,g∈C 1+β
0 (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)
F∈C0β(∂G) because f ∈C
β
0(∂G), h∈C
β
0(∂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 potentialRw, wherew∈C0β(∂G) andR >diamG.
Define forw∈C0β(∂G),x∈∂H,
KG∗w(x)=(2π)−1
∂Gw(y)|x−y|
−2n(x)·(y−x)dᏴ
1(y). (7.6)
According to [9, Theorem 2],
n(x)· ∇Rw(x)=w(x)2 +KG∗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 potentialRwis a harmonic function inG, we get using
Lemma 5.1thatRwis a solution of the problem (7.2) if and only if
Th,Rw=F, (7.9)
where
Th,Rw=w
2 +K ∗
Gw+hRw on∂G\Γ,
Th,Rw=w+hRw onΓ.
(7.10)
8. Solvability of the problem
Lemma 8.1. LetH,H+ have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,h∈
Cβ0(∂G),h≥0,R >diamG. ThenKG∗is a compact linear operator fromC β
0(∂G) toCβ(∂H)
andTh,Ris a bounded linear operator inC0β(∂G).
Proof. Fixα∈(β,γ). ThenKH∗is a linear operator fromCβ(∂H) toCα(∂H) by [11,
Theo-rem 14.IV]. SinceKH∗is a bounded linear operator fromCβ(∂H) toC0(∂H) by (7.7) and
Lemma 5.1, the operatorKH∗is a closed operator fromCβ(∂H) toCα(∂H). This gives that
the operatorKH∗ is a closed linear operator fromCβ0(∂G) toCα(∂H). The closed graph
theorem (see [13, Chapter II, Section 6, Theorem 1]) shows thatKH∗is a bounded linear operator fromCβ0(∂G) toCα(∂H). Since the identity operator is a compact operator from
Cα(∂H) toCβ(∂H), the operatorK∗
Gis a compact linear operator fromC β
0(∂G) toCβ(∂H)
by [13, Chapter X, Section 2]. SinceKG∗ is a bounded linear operator fromC0β(∂G) to
Cβ(∂H), the operatorT
0,R is a bounded linear operator inCβ0(∂G). SinceTh,R−T0,R is a
bounded linear operator inC0β(∂G) byLemma 5.2, the operatorTh,Ris a bounded linear
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 op-erator 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 classC1+γ, where 0< γ <1. Let 0< β < γ,h∈
Cβ0(∂G),h≥0,R >diamG. Ifλ∈C,λ=1/2,λ=1, thenTh,R−λIis a Fredholm operator
with index 0 in complC0β(∂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 complCβ0(∂G). SinceT0,R−Lis a compact operator inC0β(∂G) by
Lemma 8.1, the operatorT0,R−λIis a Fredholm operator with index 0 inCβ0(∂G) by [12,
Theorem 5.10]. SinceTh,R−T0,Ris a compact linear operator inC0β(∂G) byLemma 5.2,
the operatorTh,R−λIis a Fredholm operator with index 0 inCβ0(∂G) by [12, Theorem
5.10].
Lemma 8.5. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,R >
diamG,ϕ∈C0β(∂G). IfT0,2Rϕ=0, thenT0,Rϕ=0.
Proof. According toSection 7, the modified single-layer potentialRT0,Rϕis a solution of
the problem (2.2) withh≡0,g≡0, and f =T0,2Rϕ=0. According toTheorem 3.1, there
is a constantcsuch thatRT0,Rϕ=conG.
According toSection 7, the modified single-layer potentialRϕis a solution of the
problem (2.2) withh≡0,g≡0, and f =T0,Rϕ. Thus
∂GT0,Rϕ=0 (8.1)
byproposition 4.1. SinceRT0,Rϕ=cinGandRT0,Rϕis continuous inR2byLemma
5.1, we obtain
∂G
T0,RϕRTh,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 classC1+γ, where 0< γ <1. Let 0< β < γ,
h∈Cβ0(∂G),h≥0,h(x)>0 for somex∈∂G,R >diamG. ThenTh,R is continuosly
invert-ible inC0β(∂G). Denote byC
β
0,0(∂G) the space of all f ∈C
β
0(∂G) for which (4.2) holds. Then
T0,Ris continuously invertible inCβ0,0(∂G).
Proof. Letϕ∈Cβ0(∂G) be such thatRϕ=0 inG. ThenRϕ=0 in∂GbyLemma 5.1.
Ifϕ∈C0β(∂G) andTh,Rϕ=0, thenRϕis a solution of the problem (2.2) with f ≡0,
g≡0 bySection 7. SinceRϕ=0 inGbyTheorem 3.1, we haveϕ≡0. Since Th,R is a
Fredholm operator inCβ0(∂G) with index 0 by Lemma 8.4, we obtain Th,R(Cβ0(∂G))=
Cβ0(∂G). Thus, Th,R is continuously invertible inCβ0(∂G) by [13, Chapter II, Section 6,
Proposition 3] and [13, Chapter II, Section 6, Theorem 1].
Ifϕ∈Cβ0(∂G), thenRϕis a solution of the problem (2.2) withh≡0,g≡0, and f =
T0,Rϕ(seeSection 7). ThusT0,Rϕ∈C0,0β (∂G) byProposition 4.1. Hence,T0,R(C0β(∂G))⊂
Cβ0,0(∂G) andβ(T0,R)≥1. SinceT0,R is a Fredholm operator inCβ0(∂G) with index 0 by
Lemma 8.4, there is a nontrivialϕ1∈C0β(∂G) such thatT0,Rϕ1=0. According toSection 7
andTheorem 3.1, there is a constantc1such thatRϕ1=c1inG. Sincec1=0 yieldsϕ1=0
we deduce thatc1=0. Ifϕ∈C0β(∂G),T0,Rϕ=0, thenSection 7andTheorem 3.1imply
that there is a constantcsuch thatRϕ=cin G. SinceR(ϕ−(c/c1)ϕ1)=0 inG, we
obtain thatϕ−(c/c1)ϕ1≡0. This means thatα(T0,R)=1. SinceT0,R(Cβ0(∂G))⊂C
β
0,0(∂G),
β(T0,R)=α(T0,R)=1 shows thatT0,R(C0β(∂G))=C
β
0,0(∂G).
According toLemma 8.5, the operator T0,R is injective in T0,R(C0β(∂G))=C
β
0,0(∂G).
SinceCβ0,0(∂G) is aT0,R-invariant closed linear subspace of finite codimension inCβ0(∂G)
and T0,R is a Fredholm operator with index 0, the restriction of T0,R onto C0,0β (∂G) is
a Fredholm operator with index 0 by [8, Proposition 3.7.1.]. Since T0,R is injective in
Cβ0,0(∂G), it is surjective inC
β
0,0(∂G). The closed graph theorem (see [13, Chapter II,
Sec-tion 6, Theorem 1]) gives thatT0,Ris continuously invertible inCβ0,0(∂G).
Theorem 8.7. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,h∈
Cβ0(∂G),h≥0,h(x)>0 for somex∈∂G, f ∈C
β
0(∂G),g∈C 1+β
0 (Γ). Then there is a unique
solution of the problem (2.2).
Proof. FixR >diamG. According to Proposition 8.6, there isTh−,R1 in C β
0(∂G). LetF be
given by (7.3). Then
u=Ᏸg+RTh−,R1F (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 classC1+γ, where 0< γ <1. Let 0< β < γ,h≡0,
f ∈Cβ0(∂G),g∈C 1+β
0 (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
ontoCβ0,0(∂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
9. Solution of the problem
Lemma 9.1. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,R >
diamG,h∈Cβ0(∂G),h≥0, f ∈complC
β
0(∂G), f(x)=0 for somex∈∂G. Ifλis a complex
number such thatTh,Rf =λ f, thenλ≥0.
Proof. Take f1,f2∈Cβ0(∂G) such that f =f1+i f2. SinceR(f1−i f2)∈complC1(clH+),
R(f1−i f2)∈complC1(clH−),R(f1−i f2)∈complC0(R2) byLemma 5.1, we get
us-ing Green’s formula that
∂H+
R(f1−i f2) n+·
∇R(f1+i f2) +=
H+
∇Rf1 2
+∇Rf2 2
dᏴ2,
∂H−
Rf1−i f2 n−·∇Rf1+i f2 −=
H−
∇Rf12+∇Rf22
dᏴ2.
(9.1)
Summing,
∂G
Rf1−i f2 T0,Rf1+i f2
dᏴ1=
G
∇Rf1 2
+∇Rf2 2
dᏴ2. (9.2)
Using Fubini’s theorem,
λ
∂G
f1Rf1+ f2Rf2
dᏴ1=
∂G
R
f1−i f2 Th,R
f1+i f2
dᏴ1
=
G
∇Rf1 2
+∇Rf2 2
dᏴ1
+
∂Gh
Rf1
2
+Rf2 2
dᏴ1.
(9.3) Since 0< ∂G
f1Rf1+ f2Rf2
dᏴ1<∞ (9.4)
byLemma 5.3andh≥0, we get
λ=
G
∇Rf1 2
+∇Rf2 2
dᏴ1+
∂Gh
Rf1
2
+Rf2 2
dᏴ1
∂G
f1Rf1+f2Rf2
dᏴ1 ≥
0. (9.5)
Lemma 9.2. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,R >
diamG, f ∈complC0β(∂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 ∈C0β(∂G). SinceRλ−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). SinceT0,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
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
∇Rf2
dᏴ1=
∂V
Rfn· ∇RfdᏴ1
=
∂H
Rf
1 2I−K
∗
H
f dᏴ1+
∂Ωr(0) ∂Rf
∂n Rf dᏴ1
=(1−λ)
∂GfRf dᏴ1+
∂Ωr(0)
Rf
∂Rf
∂n dᏴ1.
(9.6)
Since (4.2) forcesRf(x)=o(1),∇Rf(x)=O(1/|x|) as|x| → ∞, we get forr→ ∞that
R2\clG
∇Rf2
dᏴ1=(1−λ)
∂GfRf dᏴ1. (9.7)
UsingLemma 5.3, we get
(1−λ)=
R2\clG∇Rf2dᏴ1
∂GfRf 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.
Lemma 9.4. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,R >
diamG,h∈C0β(∂G),h≥0. PutVhf =hRf for f ∈complC0β(∂G). Then
rVh
≤sup
x∈∂G
SRh(x). (9.9)
Proof. Letλ∈σ(Vh) in complC0β(∂G),λ=0. Since Vh is a compact linear operator in
complCβ0(∂G) byLemma 5.2, there is f ∈complC
β
0(∂G) such that f(z)=0 for somez∈
∂GandVhf =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|dᏴ1xsup∈∂GRh(x).
(9.10)
Since f ∈complC0β(∂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)
Theorem 9.5. LetH,H+have boundary of classC1+γ, where 0< γ <1. Let 0< β < γ,h∈
Cβ0(∂G),h≥0, f ∈C
β
0(∂G),g∈C 1+β
0 (Γ). Ifh≡0, suppose moreover that (4.2) holds. Put
α0≡1 + sup
x∈∂G
Rh(x). (9.12)
Fixα > α0/2,R >diamG. LetFbe given by (7.3) and put
ϕ=α−1∞
n=0
I−α−1T
h,R
n
F, u=Ᏸg+Rϕ,
ϕm=α−1 m
n=0
I−α−1T
h,R
n
F, um=Ᏸg+Rϕm.
(9.13)
Thenuis a solution of the problem (2.2). Moreover, there are constantsq∈(0, 1),C∈(1,∞) dependent onG,h,R, andαsuch that
α−1I−α−1Th,R
n
F
C0β(∂G)≤
Cqn f Cβ
0(∂G)+ g C1+β0 (Γ)
,
sup
x∈G
u(x)+ sup
x∈G
∇u(x)≤C f
Cβ0(∂G)+ g C 1+β 0 (Γ)
,
ϕ−ϕm
Cβ0(∂G)≤Cq
m f
C0β(∂G)+ g C 1+β 0 (Γ)
,
sup
x∈G
u−um+ sup
x∈G
∇u(x)− ∇um(x)≤Cqm f
Cβ0(∂G)+ g C1+β0 (Γ)
.
(9.14)
Proof. Ifλ∈σ(T0,R−(1/2)I),λ=0,λ=1/2, thenλis an eigenvalue ofT0,R−(1/2)Iby
Lemma 8.4. Since |λ| ≤1/2 for each eigenvalue ofT0,R−(1/2)I byLemma 9.2, we get
r(T0,R−(1/2)I)≤1/2 in complC0β(∂G). According toLemma 9.4, [13, Chapter VIII,
Sec-tion 2], and [13, Satz 45.1 and Lemma 9.4], we haver(Th,R−(1/2)I)≤r(Vh) +r(T0,R−
(1/2)I)≤α0−(1/2) in complCβ0(∂G).
Ifλ∈σ(Th,R−(1/2)I),λ=0,λ=1/2, thenλis an eigenvalue ofTh,RbyLemma 8.4.
Lemma 9.1forcesσ(Th,R−(1/2)I)⊂ −1/2,r(Th,R−(1/2)I) ⊂ −1/2,α0−1/2. Using
the spectral mapping theorem (see [13, Chapter VIII, Section 7]), we getσ(Th,R)⊂ 0,α0.
PutX=complCβ0,0(∂G) forh≡0 andX=complC
β
0(∂G) in the opposite case. Denote by
Th,Rthe restriction ofTh,RontoX. Letλ∈C\σ(Th,R). SinceXis aTh,R-invariant closed
linear subspace of finite codimension inC0β(∂G) (seeSection 7andProposition 4.1) and
Th,R−λIis a Fredholm operator with index 0, the operatorTh,R−λIis a Fredholm
op-erator with index 0 inXby [8, Proposition 3.7.1.] SinceTh,R−λIis injective, it is
sur-jective and the closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]) gives thatTh,R−λIis continuously invertible. Thusσ(Th,R)⊂σ(Th,R)⊂ 0,α0.Proposition 8.6
shows thatσ(Th,R)⊂(0,α0. Using the spectral mapping theorem (see [13, Chapter VIII,
Section 7]), we getσ(α−1T
h,R−I)⊂ {λ∈C;|λ|<1}. Thusr(α−1Th,R−I)<1. According
there are constantsq∈(0, 1),M1∈(1,∞) so that
α−1T
h,R−I
n
≤M1qn (9.15)
for arbitrary nonnegative integern.
F∈Cβ0(∂G) bySection 7. Since there is a solution of the problem (2.2) byTheorem 8.7
andTheorem 8.8, there is a solution of the problem (7.2) bySection 7. ThusF∈Xby Proposition 4.1. According to (9.15),
α−1I−α−1T
h,R
n
F
C0β(∂G)
≤M1qn F Cβ
0(∂G). (9.16) Hence,ϕis well defined and
ϕ Cβ 0(∂G)≤
∞
n=0
M1qn F Cβ
0(∂G)≤M1(1−q) −1 F
C0β(∂G), (9.17)
ϕ−ϕm
Cβ0(∂G)≤ ∞
n=m+1
M1qn F Cβ 0(∂G)≤
M1q
(1−q)q
m F
C0β(∂G). (9.18)
Easy calculation yields thatTh,Rϕ=F. So,uis a solution of the problem (2.2) bySection 7.
According toLemma 5.1, there is a constantM2such that
sup
x∈G
Rψ(x)+ sup
x∈G
∇Rψ(x)≤M2 ψ
Cβ0(∂G) (9.19)
for eachψ∈C0β(∂G). Sinceu−um=R(ϕ−ϕm), we get from (9.18) and (9.19) that
sup
x∈G
u−um+ sup
x∈G
∇u(x)− ∇um(x)≤M2M1q(1−q)−1qm F
Cβ0(∂G). (9.20)
According toLemma 6.1, there is a constantM3such that
sup
x∈G|Ᏸ
ψ(x)|+ sup
x∈G|∇Ᏸ
ψ(x)| ≤M3 ψ C1+β
0 (∂G) (9.21)
for eachψ∈C01+β(∂G). Using (9.17), (9.19), and (9.21), we get
sup
x∈G|
u|+ sup
x∈G|∇
u(x)| ≤M2M1(1−q)−1 F Cβ
0(∂G)+M3 g C 1+β
0 (∂G). (9.22) Denote
Tψ(x)=∂Ᏸψ(x)
∂n +h(x)Ᏸψ(x), x∈∂G\Γ, (9.23)
Tψ(x)=h(x)
1
2ψ(x) + 1 2π
Γ
n+(y)·(y−x)
|x−y|2 ψ(y)dᏴ1(y)
ThenTis a linear operator fromC1+0 β(∂G) toC
β
0(∂G) bySection 7. Since
[Ᏸψ]+(x)=
ψ(x) 2 + (2π)
−1
Γψ(y)|x−y|
−2n+(y)·(y−x)dᏴ
1(y) (9.25)
forx∈Γ(see [9, Theorem 1]), the operatorTis a bounded operator fromC01+β(∂G) to
C0(∂G) byLemma 6.1. Therefore,Tis a closed linear operator fromC1+β
0 (∂G) toC0β(∂G).
According to the closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]), there is a constantM3such that Tψ Cβ
0(∂G)≤M3 ψ C 1+β
0 (∂G)for eachψ∈C
1+β0(∂G). There-fore,
F Cβ
0(∂G)≤ f C β
0(∂G)+M3 g C 1+β
0 (∂G). (9.26) We get requested inequalities using (9.18), (9.20), (9.22), and (9.26).
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Dagmar Medkov´a: Department of Technical Mathematics, Faculty of Mechanical Engineering, Czech Technical University, Karlovo N´am. 13, 12135 Praha 2, Czech Republic