Spectral resolution of the Neumann-Poincar\'{e} operator on intersecting disks and analysis of plasmon resonance

0.34 MB·English

by  Hyeonbae Kang

Spectral resolution of the Neumann-Poincar´e operator on
intersecting disks and analysis of plasmon resonance
∗
Hyeonbae Kang† Mikyoung Lim‡ Sanghyeon Yu‡
5
1
January 14, 2015
0
2

n
a
J Abstract

3
1 The purpose of this paper is to investigate the spectral nature of the Neumann-Poincar´e
operator on the intersecting disks, which is a domain with the Lipschitz boundary. The

] complete spectral resolution of the operator is derived, which shows in particular that it
P
admits only the absolutely continuous spectrum, no singularly continuous spectrum and
A
no pure point spectrum. We then quantitatively analyze using the spectral resolution the
.
h plasmon resonance at the absolutely continuous spectrum.
t
a
m AMS subject classifications. 35J47 (primary), 35P05 (secondary)
Key words. Neumann-Poincar´e operator, Lipschitz domain, spectrum, spectral resolution, plasmon resonance,
[

bipolar coordinates
1
v
2 1 Introduction
5
9
2 The Neumann-Poincar´e (NP) operator is a boundary integral operator naturally arising when
0
solving classical Dirichlet or Neumann boundary value problems using layer potentials. This
.
1 operator is not self-adjoint with respect to the usual L2 inner product unless the domain is a
0
disk or a ball. However, it is recently found that by introducing a new inner product the NP
5
1 operator can besymmetrized and admits a spectral resolution. If the boundaryof the domain is
:
v smooth, thenthecorrespondingNP operator iscompact, andhencehas only thepointspectrum
i accumulating to 0. But, if the boundary of the domain is Lipschitz, then the NP operator is
X
not compact (it is a singular integral operator), and it is quite interesting to investigate the
r
a nature of the spectrum of the NP operator, in particular, whether it admits both point and
continuous spectrum. In this paper we derive a complete spectral resolution of the NP operator
on the intersecting disks which is a Lipschitz domain, from which we are able to show that the
NP operator on the intersecting disks has only absolutely continuous spectrum. As far as we
are aware of, this is the first example of a domain with the Lipschitz boundary for which the
complete spectrum of the NP operator is derived.
The study of the NP operator goes back to Poincar´e as the name of the operator alludes.
It has been a central object in the theory of singular integral operators developed during the
last century (see [13]). Recently there is growing interest in the spectral property of the NP
operator in relation to plasmonics. The plasmonic structure consists of an inclusion of the
∗This work is supported by the Korean Ministry of Education, Sciences and Technology through NRFgrants
Nos. 2010-0017532 (to H.K) and 2012003224 (to S.Y) and by the Korean Ministry of Science, ICT and Future
Planning through NRFgrant No. NRF-2013R1A1A3012931 (toM.L).
†Department of Mathematics, InhaUniversity,Incheon 402-751, S. Korea ([e-mail]).
‡DepartmentofMathematicalSciences,KoreaAdvancedInstituteofScienceandTechnology,Daejeon305-701,
Korea ([e-mail], [e-mail]).
1
negative dielectric constant (with dissipation) embedded in the matrix of the positive dielectric
constant. TheeigenvalueoftheNPoperatorisrelatedtotheplasmonresonance(seeforexample
[8]). The negativity of the dielectric constant of the inclusion makes it necessary to look into
the spectrum of the corresponding NP operator. If the dielectric constant is positive, which
corresponds to the ellipticity of the problem, only the resolvent of the NP operator matters. In
this paperwe derive quantitatively precise estimates of theplasmon resonance at thecontinuous
spectrum of the NP operator on intersecting disks in the quasi-static limit. Particularly we
obtain precise rates of resonance as the dissipation parameter tends to zero. As far as we are
aware of, this is thefirstsuch a result. We mention that the plasmon resonance(with a non-zero
frequency) at the continuous spectrum on the intersecting disks was also studied in [15]. But
there quantitative estimate of resonance is missing.
As mentioned before the NP operator is not self-adjoint with respect to the usual L2 inner
product, unless the domain is a disc or a ball (see [16]). However, it is recently found in [14] (see
also [9]) that NP operator can be symmetrized using Plemelj’s symmetrization principle. There
are a few domains with smooth boundaries on which the spectrum of the NP operator is known;
a complete spectrum is known on disks, ellipses, and balls, and a few eigenvalues on ellipsoids
(see [9] and references therein). The spectrum of the NP operator on two disjoint disks has
been computed recently, and applied to analysis of high concentration of the stress in between
closely located disks [5, 6, 17]. The spectral property of the NP operator is barely known. But,
the bounds on the essential spectrum of the NP operator on the curvilinear polygonal domains
have been derived in [19]. The intersecting disks is among domains considered in the paper,
and interestingly the result of this paper reveals that the bounds there are optimal for the
intersecting disks. Recently, the spectral theory of the NP operator finds new applications. In
[1] it was efficiently used for analysis of the cloaking by anomalous localized resonance on coated
plasmonic structure. More recently, uniformity (with respect to the elliptic constants) of elliptic
estimates and boundary perturbation formula has been proved using the spectral property of
the NP operator [10]. It is also applied to analysis of plasmon resonance on smooth domains [4].
The finding of this paper shows that the NP operator on intersecting disks does not admits
−1/2
a point spectrum on H (∂Ω). We suspect that this is rather exceptional and domains with
0
corners, especially 4 or more corners like squares, admit point spectrums. This is an interesting
question to investigate. In this regards, it is also interesting to see whether all the non-smooth
Lipschitz domains admit non-empty continuous spectrum. It is worth mentioning that the NP
operator on H−1/2(∂Ω) space on any domain Ω (with possibly multiple connected components)
with the Lipschitz boundary has 1/2 as an eigenvalue and its multiplicity is the same as the
number of connected boundary components [2].
This paper is organized as follows. In the next section we introduce the NP operator and
the bipolar coordinate system which is an essential ingredient of this paper. In section 3 we
explicitly compute the single layer potential and the NP operator on the intersecting disks in
terms of the bipolar coordinates. In section 4 we derive a resolution of the identity and the
spectral resolution of the NP operator. The last two sections are to derive estimates for the
plasmon resonance.
2 Preliminaries
Let us fix some notation first. Let Ω be a bounded domain in R2 with the Lipschitz boundary.
We denoteby H−1/2(∂Ω) thedualspaceof H1/2(∂Ω) wherethe latter denotes the usualSobolev
space. We use the notation , for either the L2 inner product or the duality pairing of H−1/2
h· ·i
and H1/2 on ∂Ω. We denote by the Sobolev norm on Hs(∂Ω) for s = 1/2 or 1/2. Let
s
k·k −
H−1/2(∂Ω) be the space of ψ H−1/2(∂Ω) satisfying ψ,1 = 0.
0 ∈ h i
2
2.1 The NP operator and symmetrization
For x = (x ,x ) = 0, let Γ(x) be the fundamental solution to the Laplacian, i.e.,
1 2
6
1
Γ(x) = ln x . (2.1)
2π | |
A classical way of solving the Neumann boundary value problem on Ω is to use the single layer
potential [ϕ] which is defined by
∂Ω
S
[ϕ](x) := Γ(x y)ϕ(y)dσ(y) , x R2, (2.2)
∂Ω
S − ∈
Z∂Ω
where ϕ L2(∂Ω) or H−1/2(∂Ω). It is well known (see, for example, [3, 7]) that [ϕ] satisfies
∂Ω
∈ S
the jump relation
∂ 1
[ϕ] (x) = I + ∗ [ϕ](x), a.e. x ∂Ω , (2.3)
∂νS∂Ω ± ±2 K∂Ω ∈
(cid:18) (cid:19)
(cid:12)
where the operator ∗ is defi(cid:12)ned by
K∂Ω
1 (x y) ν
∗ [ϕ](x) = − · xϕ(y)dσ(y) . (2.4)
K∂Ω 2π x y 2
Z∂Ω | − |
Here indicates thelimits (to∂Ω)fromoutsideandinsideofΩ,respectively, andν denotesthe
x
±
outward unitnormalvector to∂Ω atx ∂Ω. Theoperator ∗ iscalled theNeumann-Poincar´e
∈ K∂Ω
(NP) operator associated with the domain Ω. Sometimes it is called the adjoint NP operator to
distinguish it from the NP operator ( is the L2 adjoint of ∗ ).
K∂Ω K∂Ω K∂Ω
Even though ∗ is not self-adjoint on the usual L2-space, it can be symmetrized by intro-
K∂Ω
ducing a new inner product. Recently it is proved in [14] that ∗ can be symmetrized using
K∂Ω
Plemelj’s symmetrization principle (also known as Caldero´n’s identity)
∗ = . (2.5)
S∂ΩK∂Ω K∂ΩS∂Ω
In fact, if we define
ϕ,ψ := ϕ, [ψ] (2.6)
H∗ ∂Ω
h i −h S i
for ϕ,ψ H−1/2(∂Ω), then it is an inner product on H−1/2(∂Ω) and ∗ is self-adjoint with
∈ 0 0 K∂Ω
respect to this inner product.
Let H∗ = H∗(∂Ω) be the space H0−1/2(∂Ω) equipped with the inner product h , iH∗ and
denote the norm associated with , H∗ by H∗. It is proved in [10] that H∗ is equivalent
h i k · k k · k
to H−1/2 norm, i.e., there are constants C and C such that
1 2
C1 ϕ −1/2 ϕ H∗ C2 ϕ −1/2. (2.7)
k k ≤ k k ≤ k k
−1/2
for all ϕ H (∂Ω).
∈ 0
Since ∗ is self-adjoint on ∗, its spectrum σ( ∗ ) is real and consists of point and con-
K∂Ω H K∂Ω
tinuous spectra. Moreover, by the spectral resolution theorem there is a family of projection
operators on ∗ (called a resolution of the identity) such that
t
E H
1/2
∗ = td . (2.8)
K∂Ω Et
Z−1/2
See [21]. We emphasize that the spectrum of ∗ on ∗ lies in ( 1,1) (see [12]).
K∂Ω H −2 2
3
2.2 Intersecting disks and bipolar coordinates
Let B = B (c ) and B = B (c ) be two disks of the same radii a centered at c and c ,
1 a 1 2 a 2 1 2
respectively, which are intersecting, namely, c c < 2a. The domain Ω we consider in this
1 2
| − |
paper is defined to be
Ω := B B . (2.9)
1 2
∪
(We may consider B B instead.) Note that ∂Ω is Lipschitz continuous with corners at the
1 2
∩
intersecting points. We assume that c is on the x -axis and c = c so that the intersecting
1 2 1 2
−
points lie on the x -axis and they are symmetric with respect to the origin. Let 2θ (θ > 0)
1 0 0
be the angle between two circles at the intersecting point (see Fig. 2.1). One can easily see that
the intersecting points are given by asinθ .
0
±
Ξ=0
G Θ=-Π(cid:144)4
-
5
a Ξ=-0.8 Θ=-3Π(cid:144)4 Θ=-Π(cid:144)2 Ξ=0.8
Ξ=-1.6 Ξ=1.6
0
2Θ
0
Θ=Π
Θ=0 Θ=0
Θ=3Π(cid:144)4 Θ=Π(cid:144)2
a
-5
Θ=Π(cid:144)4
W G
+
-5 0 5
Figure 2.1: The left figure is the intersecting disks Ω. Level coordinate curves of ζ (when
α= √7) are illustrated in z-plane in the right figure.
For a given positive number α the bilinear transformation
z+α
λ(z) = (2.10)
z α
−
is a conformal map from the Riemann sphere onto itself, and maps the interval [ α,α] onto the
−
negative real axis. So,
ζ = ξ+iθ := Logλ(z) (2.11)
is well defined for z C [ α,α], where Log is the logarithm with the principal branch. The
∈ \ −
coordinate ζ or (ξ,θ) is the bipolar coordinates we will use. The level coordinate curves for ζ
are illustrated in Fig. 2.1. Here and afterwards, we take
α = asinθ , (2.12)
0
so that (α,0) and ( α,0) are two intersecting points of the circles. Note that Logλ maps
−
C [ α,α] onto λ = ξ+iθ, < ξ < , π < θ < π . If we let Γ and Γ be the lower and
+ −
\ − { −∞ ∞ − }
upper halves of ∂Ω, respectively, one can see that Logλ maps Γ onto θ = θ , and Γ onto
+ 0 −
{ }
θ = θ , and outside of Ω onto λ = ξ+iθ, < ξ < , θ < θ <θ (see Fig. 2.2). Note
0 0 0
{ − } { −∞ ∞ − }
that we denote the lower half by Γ because it is mapped to θ = +θ .
+ 0
{ }
Let Φ be the inverse of Logλ, i.e.,
eζ +1
z = x +ix = Φ(ζ):= α . (2.13)
1 2 eζ 1
−
4
Θ
x2
Θ=-Θ0 Θ=Π
Θ=Θ
0
Θ=0
x1 Ξ
Θ=±Π Θ=-Θ
0
Θ=Θ
0 Θ=-Π
Figure 2.2: The exterior of the intersecting disks (the gray region in the left figure) is the image
of the infinite strip (the gray region in the right figure) via the conformal map z = x +ix =
1 2
Φ(ξ+iθ).
We also regard Φ as a mapping from R2 into R2 by identifying x = (x ,x ) R2 with z =
1 2
∈
x +ix C. Here we recall a few facts to be used later. We define the scale factor h by
1 2
∈
1 coshξ cosθ
h(ξ,θ):= = − . (2.14)
Φ′(ξ+iθ) α
| |
SinceΦ is a conformal transformation, h is the Jacobian of thecoordinate change. In particular,
one can see that the outward normal derivative on ∂Ω is transformed in the following way:
∂u ∂(u Φ)
= h(ξ,θ ) ◦ on Γ , (2.15)
0 ±
∂ν ∓ ∂θ
and the line element dσ on ∂Ω is transformed by
1
dσ = dξ. (2.16)
h(ξ,θ )
0
Define
Logλ(z) =: Ψ (z)+iΨ (z). (2.17)
1 2
Then we have
Φ(Ψ (x),θ ) = x if x Γ ,
1 0 +
∈ (2.18)
(Φ(Ψ1(x), θ0) = x if x Γ−.
− ∈
3 Some explicit computations
3.1 Single layer potential on R2
The purpose of this section is to derive an explicit formula for the single layer potential on ∂Ω
using the bipolar coordinates. For ϕ ∗ let
∈ H
u(x) := [ϕ](x), x R2.
∂Ω
S ∈
5
Then u is the solution to the following transmission problem:
∆u= 0 in Ω (R2 Ω),
∪ \
u+ = u− on ∂Ω,
| |
∂u ∂u (3.1)
 = ϕ on ∂Ω,
∂ν +− ∂ν −
u(x(cid:12)) = O(x(cid:12)−1) as x .
(cid:12) (cid:12)
 (cid:12) |(cid:12)| | |→ ∞


It is worth mentioning thatthe last condition follows since ϕ,1 = 0.
h i
Let us define
u˜(ξ,θ) := (u Φ)(ξ,θ),
◦
where Φ is the conformal mapping defined by (2.13). Since Φ is conformal, u˜ satisfies
∆u˜(ξ,θ)= 0 for (ξ,θ) R ( π, θ ) ( θ ,θ ) (θ ,π) , (3.2)
0 0 0 0
∈ × − − ∪ − ∪
where ∆ =∂2/∂ξ2 +∂2/∂θ2. The continuity o(cid:0)f the potential in (3.1) is equiv(cid:1)alent to
u˜ = u˜ on θ = θ . (3.3)
+ − 0
| | { ± }
In view of (2.15), the continuity of the flux is equivalent to
∂u˜ ∂u˜ 1
= − ϕ˜ on θ = θ (3.4)
+ 0
∂θ +− ∂θ − h(,θ0) { }
(cid:12) (cid:12) ·
(cid:12) (cid:12)
and (cid:12) (cid:12)
∂u˜ ∂u˜ 1
= ϕ˜ on θ = θ , (3.5)
− 0
∂θ +− ∂θ − h(,θ0) { − }
(cid:12) (cid:12) ·
where (cid:12) (cid:12)
(cid:12) (cid:12)
ϕ˜ (ξ)= (ϕ Φ)(ξ,θ ) and ϕ˜ (ξ) = (ϕ Φ)(ξ, θ ).
+ 0 − 0
◦ ◦ −
Since (x ,x ) if and only if (ξ,θ) 0, it follows from the last condition in (3.1) that
1 2
| | → ∞ | | →
u˜(ξ,θ) 0 as (ξ,θ) (0,0), or simply
→ →
u˜(0,0) = 0. (3.6)
We now impose boundary conditions on θ = π . Since lines θ = π correspond to the line
{ ± } { ± }
segment ( α,α) in (x ,x )-plane, we have
1 2
−
∂u˜ ∂u˜
u˜(ξ,π) = u˜(ξ, π), (ξ,π) = (ξ, π) for all ξ R, (3.7)
− ∂θ ∂θ − ∈
which implies that u˜(ξ,θ) is periodic in θ with the period 2π.
It is straightforward to find the solution u˜ to (3.2) satisfying (3.3)-(3.7). Denote by the
F
Fourier transformation in ξ-variable, namely,
1 ∞
[f](s)= f(ξ)e−isξdξ.
F √2π
Z−∞
If we apply to the both sides of the equation (3.2) and denote
F
v(s,θ)= [u˜(,θ)](s), s R,
F · ∈
then we have
d2v
s2v = 0.
dθ2 −
6
We solve this ordinary differential equation with conditions (3.3)-(3.7). In doing so, it is conve-
nient to consider separately the cases when ϕ is odd and even with respect to x -axis.
1
Suppose that ϕ is odd with respect to x -axis, namely
1
ϕ(x ,x ) = ϕ(x , x ).
1 2 1 2
− −
Then, one can see easily that
ϕ˜ = ϕ˜ , (3.8)
+ −
−
and the solution is given by
a(s)sinhs(θ π), θ < θ < π,
0
−
 sinhs(θ π)
v(s,θ)= a(s) sinhs0θ− sinhsθ, −θ0 ≤ θ ≤ θ0, (3.9)
 0
a(s)sinhs(θ+π), π < θ < θ ,
0
− −


where 

sinhsθ ϕ˜
0 +
a(s) = (s).
ssinhsπF h(,θ )
0
h · i
If ϕ is even with respect to x -axis, then we have
1
ϕ˜ = ϕ˜ , (3.10)
+ −
and the solution is given by
b(s)coshs(θ π), θ < θ < π,
0
−
 coshs(θ π)
v(ξ,θ) = (const.)+b(s) coshs0θ− coshsθ, −θ0 ≤ θ ≤ θ0, (3.11)
 0
b(s)coshs(θ+π), π < θ < θ ,
0
− −


where 

coshsθ ϕ˜
0 +
b(s) = (s).
−ssinhsπF h(,θ )
0
h · i
In general case, we decompose ϕ into the odd and even parts, namely,
ϕ = ϕo+ϕe (3.12)
where
1 1
ϕo(x ,x ):= (ϕ(x ,x ) ϕ(x , x )) and ϕe(x ,x ) := (ϕ(x ,x )+ϕ(x , x )),
1 2 1 2 1 2 1 2 1 2 1 2
2 − − 2 −
and then obtain the solution by superposingsolutions corresponding to the odd and even parts,
respectively. By taking the inverse Fourier transform we obtain u˜(ξ,θ), namely,
u˜(ξ,θ)= u(Φ(ξ,θ)) = −1[v(,θ)](ξ). (3.13)
F ·
We emphasize that
ϕedσ = 0. (3.14)
ZΓ+
In fact, since ϕ ∗, it holds that
∈ H
ϕodσ+ ϕedσ = 0.
ZΓ ZΓ
Since ϕo is odd with respect to the x -axis, we have ϕodσ = 0, and hence ϕedσ =0. Since
1 Γ Γ
ϕe is even, we have (3.14).
R R
7
3.2 Single layer potential on ∂Ω
In particular, we obtain from (3.13) the following formula [ϕ] on ∂Ω:
∂Ω
S
[ϕ](Φ(ξ, θ )) = −1[ Ao+Ae](ξ), (3.15)
∂Ω 0
S ± F ±
where
sinhsθ sinhs(π θ ) ϕ˜o
Ao(s):= − 0 − 0 + (s),
ssinhsπ F h(,θ )
0
coshsθ coshs(π θ ) h ·ϕ˜e i
Ae(s):= − 0 − 0 + (s).
ssinhsπ F h(,θ )
0
h · i
It is worth reminding here that
ϕ˜o(ξ) = ϕo(Φ(ξ,θ )),
+ 0
and likewise for ϕ˜e (ξ).
+
Above formula suggest to introduce an operator as follows: for ϕ ∗
C ∈ H
[ϕ](ξ)
[ϕ](ξ) := C+
C −[ϕ](ξ)
(cid:20)C (cid:21)
where
ϕ(Φ(ξ, θ ))
0
[ϕ](ξ) := ± .
±
C h(ξ,θ )
0
Note that −1 is given by
C
g h(Ψ (x),θ )g (Ψ (x)), x Γ ,
−1 1 (x) = 1 0 1 1 ∈ + (3.16)
C (cid:20)g2(cid:21) (h(Ψ1(x),θ0)g2(Ψ1(x)), x ∈ Γ−.
We also define a multiplication operator by
h
M
[f]:= h(ξ,θ )f(ξ).
h 0
M
Let
sinhsθ sinhs(π θ )
0 0
p (s) := − , (3.17)
1
ssinhsπ
coshsθ coshs(π θ )
0 0
p (s) := − . (3.18)
2
ssinhsπ
Observe that p is a smooth function on R1 and p is smooth except at 0 where the following
1 2
holds:
lim s 2p (s) = 1. (3.19)
2
s→±0| |
Both of them are positive and satisfy
1
lim s p (s)= , j = 1,2. (3.20)
j
s→±∞| | 2
Then, since ϕo (ξ) = ϕo (ξ) and ϕe (ξ) = ϕe (ξ), we see from (3.15) that
+ − − + −
[ϕ](s) = p (s) [ϕo](s) p (s) [ϕe](s),
h + ∂Ω 1 + 2 +
FM C S − FC − FC
[ϕ](s) = p (s) [ϕo](s) p (s) [ϕe](s).
h − ∂Ω 1 − 2 −
FM C S − FC − FC
8
Let
f(s) = f1(s) = FC+[ϕo](s) . (3.21)
(cid:20)f2(s)(cid:21) "FC+[ϕe](s)#
Then we have
f +f = [ϕ] and f +f = [ϕ],
1 2 + 1 2 −
FC − FC
in other words,
f
ϕ= 2 −1 −1Λ−1 1 , (3.22)
C F f2
(cid:20) (cid:21)
where
1 1
Λ= − . (3.23)
1 1
(cid:20) (cid:21)
So, we obtain
f p f p f
2 (Λ )−1 1 = − 1 1− 2 2 .
FMhCS∂Ω FC f2 p1f1 p2f2
(cid:20) (cid:21) (cid:20) − (cid:21)
Note that
p f p f
− 1 1− 2 2 = 2Λ−1P(s)f(s)
p1f1 p2f2 −
(cid:20) − (cid:21)
where
p 0
P = 1 .
0 p
2
(cid:20) (cid:21)
If we define the transformed single layer potential S by
S := Λ (Λ )−1, (3.24)
h ∂Ω
FM CS FC
it follows that
S[f](s)= P(s)f(s). (3.25)
−
3.3 Inner product
Observe that if we make changes of variables x =Φ(ξ, θ ), then we see from (2.16) that
0
±
ϕ(x)ψ(x)dσ(x) = [ϕ](ξ) [ψ](ξ)dξ+ [ϕ](ξ) [ψ](ξ)dξ.
+ h + − h −
Z∂Ω ZRC M C ZRC M C
It then follows from Parseval’s identity that
ϕ(x)ψ(x)dσ(x) = [ϕ](s) [ψ](s)ds+ [ϕ](s) [ψ](s)ds
+ h + − h −
Z∂Ω ZRFC FM C ZRFC FM C
= [ϕ](s) [ψ](s)ds.
h
RFC ·FM C
Z
Since Λ−1 = 1ΛT, we have
2
1
ϕ(x)ψ(x)dσ(x) = Λ [ϕ](s) Λ [ψ](s)ds.
h
Z∂Ω 2 ZR FC · FM C
Let
f = (f ,f )T := Λ [ϕ] and g = (g ,g )T := Λ [ψ]. (3.26)
1 2 1 2
FC FC
9
Then, we have from (3.25)
ϕ,ψ H∗ = ϕ, ∂Ω[ψ]
h i −h S i
1
= Λ [ϕ](s) Λ [ψ](s)ds
h ∂Ω
−2 R FC · FM CS
Z
1
= f(s) S[g](s)ds
−2 R ·
Z
1
= f(s) P(s)g(s)ds.
2 R ·
Z
Let
1 1
f,g := f(s) P(s)g(s) ds = p (s)f (s)g (s)+p (s)f (s)g (s) ds, (3.27)
W∗ 1 1 1 2 2 2
h i 2 R · 2 R
Z Z h i
and let W∗ be the norm induced by this inner product. Let W∗ be the space of f such that
k·k
f W∗ < . In fact, W∗ = L2(R,p1) L2(R,p2), where L2(R,pj) is the L2 space weighted by
k k ∞ ×
p . It is worth mentioning that because of (3.20) L2(R,p ) is the Fourier transform of the space
j 1
H−1/2(R). On the other hand, because of (3.19), if g L2(R,p ), g(0) = 0 if g is continuous at
2
∈
0. So, L2(R,p ) is the Fourier transform of H−1/2(R). This is natural in view of (3.14).
2 0
Define
U := Λ . (3.28)
FC
We have shown that U is a unitary transformation from ∗ onto W∗, i.e.,
H
ϕ,ψ H∗ = U[ϕ],U[ψ] W∗. (3.29)
h i h i
3.4 Neumann-Poincar´e operator
We now compute the NP operator and show that
∗ [ϕ](Φ(ξ, θ )) = h(ξ,θ ) −1[ Bo Be](ξ), (3.30)
K∂Ω ± 0 0 F ± −
where
ϕ˜o
Bo(s):= η(s) + (s),
F h(,θ )
0
h ϕ·˜e i
Be(s):= η(s) + (s).
F h(,θ )
0
h · i
Here, η is defined by
1sinhs(π 2θ )
0
η(s) := − . (3.31)
2 sinhsπ
We only show (3.30) for θ = θ . Let u = [ϕ] and u˜(ξ,θ)= [ϕ](Φ(ξ,θ)) as before. We
0 ∂Ω ∂Ω
S S
see from (2.3) and (2.15) that
1 ∂u ∂u
∗ [ϕ] Φ(ξ,θ ) = + Φ(ξ,θ )
K∂Ω 0 2 ∂ν + ∂ν − 0
(cid:18) (cid:12) (cid:12) (cid:19)
(cid:0) (cid:1) h(ξ,θ(cid:12) ) ∂u˜(cid:12) (cid:0) ∂u˜(cid:1)
0
= (cid:12) (cid:12) (ξ,θ )+ (ξ,θ ) . (3.32)
0 0
− 2 ∂θ + ∂θ −
(cid:18) (cid:12) (cid:12) (cid:19)
(cid:12) (cid:12)
Recall that the region R2 Ω outside the intersecting(cid:12)disks correspo(cid:12)nds to the region θ < θ .
0
\ {| | }
Sothesubscript+in ∂ indicates thelimitfromtheregion θ < θ . Similarly, thesubscript
∂θ|± {| | 0}
indicates the limit from the region θ < θ < π .
0
− { | | }
10