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arXiv:1002.1541v2 [math.NA] 22 Feb 2010
Shape derivatives of boundary integral operators in
electromagnetic scattering
M. Costabel1, F. Le Louër2
1IRMAR, University of Rennes 1
2POEMS, INRIA-ENSTA, Paris
Abstract
We develop the shape derivative analysis of solutions to the problem of scattering
of time-harmonic electromagnetic waves by a bounded penetrable obstacle. Since
boundary integral equations are a classical tool to solve electromagnetic scattering
problems, we study the shape differentiability properties of the standard electro-
magnetic boundary integral operators. To this end, we start with the Gâteaux
differentiability analysis with respect to deformations of the obstacle of boundary
integral operators with pseudo-homogeneous kernels acting between Sobolev spaces.
The boundary integral operators of electromagnetism are typically bounded on the
space of tangential vector fields of mixed regularity TH−1
decomposition, we can base their analysis on the study of scalar integral operators
in standard Sobolev spaces, but we then have to study the Gâteaux differentiability
of surface differential operators. We prove that the electromagnetic boundary in-
tegral operators are infinitely differentiable without loss of regularity and that the
solutions of the scattering problem are infinitely shape differentiable away from the
boundary of the obstacle, whereas their derivatives lose regularity on the boundary.
We also give a characterization of the first shape derivative as a solution of a new
electromagnetic scattering problem.
2(divΓ,Γ). Using Helmholtz
Keywords :
operators, shape derivatives, Helmholtz decomposition.
Maxwell’s equations, boundary integral operators, surface differential
Contents
1 The dielectric scattering problem4
2 Boundary integral operators and main properties5
3 Some remarks on shape derivatives10
4 Gâteaux differentiability of pseudo-homogeneous kernels13
5 Shape differentiability of the solution22
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Introduction
Consider the scattering of time-harmonic electromagnetic waves by a bounded obstacle
Ω in R3with a smooth and simply connected boundary Γ filled with an homogeneous
dielectric material. This problem is described by the system of Maxwell’s equations,
valid in the sense of distributions, with two transmission conditions on the boundary or
the obstacle guaranteeing the continuity of the tangential components of the electric and
magnetic fields across the interface. The transmission problem is completed by the Silver-
Müller radiation condition at infinity (see [23] and [24]). Boundary integral equations are
an efficient method to solve such problems for low and high frequencies. The dielectric
scattering problem is usually reduced to a system of two boundary integral equations for
two unknown tangential vector fields on the interface (see [6] and [24]). We refer to [9] and
[10] for methods developed by the authors to solve this problem using a single boundary
integral equation.
Optimal shape design with the modulus of the far field pattern of the dielectric scat-
tering problem as goal is of practical interest in some important fields of applied mathe-
matics, as for example telecommunication systems and radars. The utilization of shape
optimization methods requires the analysis of the dependency of the solution on the shape
of the dielectric scatterer. An explicit form of the shape derivatives is required in view
of their implementation in a shape optimization algorithms such as gradient methods or
Newton’s method.
In this paper, we present a complete analysis of the shape differentiability of the
solution of the dielectric scattering problem using an integral representation. Even if nu-
merous works exist on the calculus of shape variations [14, 25, 26, 31, 32], in the framework
of boundary integral equations the scientific literature is not extensive. However, one can
cite the papers [27], [29] and [28], where R. Potthast has considered the question, starting
with his PhD thesis [30], for the Helmholtz equation with Dirichlet or Neumann boundary
conditions and the perfect conductor problem, in spaces of continuous and Hölder con-
tinuous functions. Using the integral representation of the solution, one is lead to study
the Gâteaux differentiability of boundary integral operators and potential operators with
weakly and strongly singular kernels.
The natural space of distributions (energy space) which occurs in the electromagnetic
potential theory is TH−1
are in the Sobolev space H−1
main difficulties: On one hand, to be able to construct shape derivatives of the solution –
which is given in terms of products of boundary integral operators and their inverses – it is
imperative to prove that the derivatives are bounded operators between the same spaces
as the boundary integral operators themselves. On the other hand, the very definition of
shape differentiability of operators defined on TH−1
Our approach consists in using the Helmholtz decomposition of this Hilbert space. In this
way, we split the analysis in two main steps: First the Gâteaux differentiability analysis
of scalar boundary integral operators and potential operators with pseudo-homogeneous
kernels, and second the study of derivatives with respect to smooth deformations of the
obstacle of surface differential operators in the classical Sobolev spaces.
2(divΓ,Γ), the set of tangential vector fields whose components
2(Γ) and whose surface divergence is in H−1
2(Γ). We face two
2(divΓ,Γ) poses non-trivial problems.
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This work contains results from the thesis [21] where this analysis has been used to
develop a shape optimization algorithm of dielectric lenses in order to obtain a prescribed
radiation pattern.
The paper is organized as follows:
In section 1 we define the scattering problem of time-harmonic electromagnetic waves
at a dielectric interface and the appropriate spaces. In section 2 we recall some results
about trace mappings and boundary integral operators in electromagnetism, following the
notation of [6, 24]. We then give an integral representation of the solution following [9].
In section 3, we introduce the notion of shape derivative and its connection to Gâteaux
derivatives. We also recall elementary results about differentiability in Fréchet spaces.
The section 4 is dedicated to the Gâteaux differentiability analysis of a class of bound-
ary integral operators with respect to deformations of the boundary. We generalize the
results proved in [27, 29] for the standard acoustic boundary integral operators, to the
class of integral operators with pseudo-homogenous kernels. We also give higher order
Gâteaux derivatives of coefficient functions such as the Jacobian of the change of variables
associated with the deformation, or the components of the unit normal vector. These re-
sults are new and allow us to obtain explicit forms of the derivatives of the integral
operators.
The last section contains the main results of this paper: the shape differentiability
properties of the solution of the dielectric scattering problem. We begin by discussing the
difficulties of defining the shape dependency of operators defined on the shape-dependent
space TH−1
(see [11]) on the boundary of smooth domains. We then analyze the differentiability of
a family of surface differential operators. Again we prove their infinite Gâteaux differen-
tiability and give an explicit expression of their derivatives. These results are new and
important for the numerical implementation of the shape derivatives. Using the chain
rule, we deduce the infinite shape differentiability of the solution of the scattering prob-
lem away from the boundary and an expression of the shape derivatives. More precisely,
we prove that the boundary integral operators are infinitely Gâteaux differentiable with-
out loss of regularity, whereas previous results allowed such a loss [28], and we prove that
the shape derivatives of the potentials are smooth far from the boundary but they lose
regularity in the neighborhood of the boundary.
These new results generalize existing results: In the acoustic case, using the variational
formulation, a characterization of the first Gâteaux derivative was given by A. Kirsch in
[20] for the Dirichlet problem and then for a transmission problem by F. Hettlich in
[15, 16]. R. Potthast used the integral equation method to obtain a characterisation of
the first shape derivative of the solution of the perfect conductor scattering problem.
We end the paper by formulating a characterization of the first shape derivative as
the solution of a new electromagnetic scattering problem. We show that both by directly
deriving the boundary values and by using the integral representation of the solution, we
obtain the same characterization.
2(divΓ,Γ), and we present an altervative using the Helmholtz decomposition
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1The dielectric scattering problem
Let Ω denote a bounded domain in R3and let Ωcdenote the exterior domain R3\Ω. In
this paper, we will assume that the boundary Γ of Ω is a smooth and simply connected
closed surface, so that Ω is diffeomorphic to a ball. Let n denote the outer unit normal
vector on the boundary Γ.
In Ω (resp. Ωc) the electric permittivity ǫi(resp. ǫe) and the magnetic permeability
µi(resp. µe) are positive constants. The frequency ω is the same in Ω and in Ωc. The
interior wave number κiand the exterior wave number κeare complex constants of non
negative imaginary part.
Notation:
For a domain G ⊂ R3we denote by Hs(G) the usual L2-based Sobolev
space of order s ∈ R, and by Hs
loc(G) the space of functions whose restrictions to any
bounded subdomain B of G belong to Hs(B). Spaces of vector functions will be denoted
by boldface letters, thus
Hs(G) = (Hs(G))3.
If D is a differential operator, we write:
Hs(D,Ω)
Hs
loc(D,Ωc)
={u ∈ Hs(Ω) : Du ∈ Hs(Ω)}
{u ∈ Hs
=
loc(Ωc) : Du ∈ Hs
loc(Ωc)}
The space Hs(D,Ω) is endowed with the natural graph norm. When s = 0, this defines
in particular the Hilbert spaces H(curl,Ω) and H(curlcurl,Ω). We denote the L2scalar
product on Γ by ?·,·?Γ.
The time-harmonic Maxwell’s sytem can be reduced to second order equations for the
electric field only. The time-harmonic dielectric scattering problem is then formulated as
follows.
The dielectric scattering problem : Given an incident field Einc∈ Hloc(curl,R3)
that satisfies curlcurlEinc− κ2
Ei∈ H(curl,Ω) and Es∈ Hloc(curl,Ωc) satisfying the time-harmonic Maxwell equations
eEinc= 0 in a neighborhood of Ω, we seek two fields
curlcurlEi− κ2
curlcurlEs− κ2
iEi
eEs
= 0
in Ω,
in Ωc,
(1.1)
= 0
(1.2)
the two transmission conditions,
n × Ei= n × (Es+ Einc)
i(n × curlEi) = µ−1
on Γ
(1.3)
µ−1
en × curl(Es+ Einc)
on Γ
(1.4)
and the Silver-Müller radiation condition:
lim
|x|→+∞|x|
????curlEs(x) ×
x
|x|− iκeEs(x)
????= 0.
(1.5)
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The interior and exterior magnetic fields are then given by Hi=
1
iωµiEiand Hs=
1
iωµeEs.
It is well known that this problem admit a unique solution for any positive real values of
the exterior wave number [6, 21, 24].
An important quantity, which is of interest in many shape optimization problems, is
the far field pattern of the electric solution, defined on the unit sphere of R3, by
E∞(ˆ x) = lim
|x|→∞4π|x|Es(x)
eiκe|x|,
with
x
|x|= ˆ x.
2 Boundary integral operators and main properties
2.1Traces and tangential differential calculus
We use surface differential operators and traces. More details can be found in [8, 24].
For a vector function v ∈ (Ck(R3))qwith k,q ∈ N∗, we note [∇v] the matrix whose
the i-th column is the gradient of the i-th component of v and we set [Dv] =T[∇v]. The
tangential gradient of any scalar function u ∈ Ck(Γ) is defined by
∇Γu = ∇˜ u|Γ−?∇˜ u|Γ· n?n,
and the tangential vector curl by
(2.1)
curlΓu = ∇˜ u|Γ× n,
(2.2)
where ˜ u is an extension of u to the whole space R3. For a vector function u ∈ (Ck(Γ))3,
we note [∇Γu] the matrix whose the i-th column is the tangential gradient of the i-th
component of u and we set [DΓu] =T[∇Γu].
We define the surface divergence of any vectorial function u ∈ (Ck(Γ))3by
divΓu = div ˜ u|Γ−?[∇˜ u|Γ]n · n?,
and the surface scalar curl curlΓrurby
(2.3)
curlΓu = n · (curl ˜ u))
where ˜ u is an extension of u to the whole space R3. These definitions do not depend on
the extension.
Definition 2.1 For a vector function v ∈ (C∞(Ω))3and a scalar function v ∈ C∞(Ω)
we define the traces :
γv = v|Γ,
γDv := (n × v)|Γ(Dirichlet) and
γNκv := κ−1(n × curlv)|Γ(Neumann).
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We introduce the Hilbert spaces Hs(Γ) = γ
For s > 0, the traces
?
Hs+1
2(Ω)
?
, and THs(Γ) = γD
?
Hs+1
2(Ω)
?
γ : Hs+1
2(Ω) → Hs(Γ),
γD: Hs+1
2(Ω) → THs(Γ)
are then continuous. The dual of Hs(Γ) and THs(Γ) with respect to the L2(or L2) scalar
product is denoted by H−s(Γ) and TH−s(Γ), respectively.
The surface differential operators defined here above can be extended to the Sobolev
spaces: The tangential gradient and the tangential vector curl are linear and continuous
from Hs+1(Γ) to THs(Γ), the surface divergence and the surface scalar curl are linear
and continuous from THs+1(Γ) to Hs(Γ).
Definition 2.2 We define the Hilbert space
TH−1
2(divΓ,Γ) =
?
j ∈ TH−1
2(Γ),divΓj ∈ H−1
2(Γ)
?
endowed with the norm
|| · ||TH−1
2(divΓ,Γ)= || · ||TH−1
2(Γ)+ ||divΓ·||H−1
2(Γ).
Lemma 2.3 The operators γD and γN are linear and continuous from C∞(Ω,R3) to
TL2(Γ) and they can be extended to continuous linear operators from H(curl,Ω) and
H(curl,Ω) ∩ H(curlcurl,Ω), respectively, to TH−1
2(divΓ,Γ).
For u ∈ Hloc(curl,Ωc) and v ∈ Hloc(curlcurl,Ωc)) we define γc
same way and the same mapping properties hold true.
Recall that we assume that the boundary Γ is smooth and topologically trivial. For
a proof of the following result, we refer to [3, 8, 24].
Du and γc
Nv in the
Lemma 2.4 Let t ∈ R. The Laplace-Beltrami operator
∆Γ= divΓ∇Γ= −curlΓcurlΓ
(2.4)
is linear and continuous from Ht+2(Γ) to Ht(Γ).
It is an isomorphism from Ht+2(Γ)/R to the space Ht
∗(Γ) defined by
?
u ∈ Ht
∗(Γ)⇐⇒u ∈ Ht(Γ) and
Γ
u = 0.
This result is due to the surjectivity of the operators divΓand curlΓfrom THt+1(Γ) to
Ht
∗(Γ).
We note the following equalities:
curlΓ∇Γ= 0 and divΓcurlΓ= 0
(2.5)
divΓ(n × j) = −curlΓj and curlΓ(n × j) = divΓj
(2.6)
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2.2Pseudo-homogeneous kernels
In this paper we are concerned with boundary integral operators of the form :
?
where the integral is assumed to exist in the sense of a Cauchy principal value and
the kernel k is weakly singular, regular with respect to the variable y ∈ Γ and quasi-
homogeneous with respect to the variable z = x − y ∈ R3. We recall the regularity
properties of these operators on the Sobolev spaces Hs(Γ), s ∈ R available also for their
adjoints operators:
?
We use the class of weakly singular kernel introduced by Nedelec ([24] p. 176). More
details can be found in [13, 17, 19, 22, 35, 34].
Definition 2.5 The homogeneous kernel k(y,z) defined on Γ×?R3\{0}?is said of class
Definition 2.6 The kernel k ∈ C∞?Γ ×?R3\{0}??is pseudo-homogeneous of class −m
asymptotic expansion when z tends to 0:
KΓu(x) = vp.
Γ
k(y,x − y)u(y)dσ(y), x ∈ Γ
(2.7)
K∗
Γ(u)(x) = vp.
Γ
k(x,y − x)u(y)dσ(y), x ∈ Γ.
(2.8)
−m with m ≥ 0 if
sup
y∈Rdsup
|z|=1
?????
∂|α|
∂yα
∂|β|
∂zβk(y,z)
?????≤ Cα,β, for all multi-index α and β,
∂|β|
∂zβk(y,z) is homogeneous of degree − 2 with respect to the variable z
for all |β| = m and Dm
zk(y,z) is odd with respect to the variable z.
for an integer m such that m ? 0, if for all integer s the kernel k admit the following
k(y,z) = km(y,z) +
N−1
?
j=1
km+j(y,z) + km+N(y,z),
(2.9)
where for j = 0,1,...,N − 1 the function km+jis homogeneous of class −(m + j) and N
is chosen such that km+N is s times differentiables.
For the proof of the following theorem, we refer to [24].
Theorem 2.7 Let k be a pseudo-homogeneous kernel of class −m. The associated oper-
ator KΓgiven by (2.7) is linear and continuous from Hs(Γ) to Hs+m(Γ) for all s ∈ R.
We have similar results for the adjoint operators K∗
Γ.
The following theorem is established in [13].
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Theorem 2.8 Let k be a pseudo-homogeneous kernel of class −m. The potential operator
P defined by
?
is continuous from Hs−1
P(u)(x) =
Γ
k(y,x − y)u(y)dσ(y), x ∈ R3\Γ
(2.10)
2(Γ) to Hs+m(Ω) ∪ Hs+m
loc(Ωc) for all positive real number s.
2.3 The electromagnetic boundary integral operators
We use some well known results about electromagnetic potentials. Details can be found
in [3, 4, 5, 6, 24].
Let κ be a complex number such that Im(κ) ≥ 0 and let
G(κ,|x − y|) =
eiκ|x−y|
4π|x − y|
be the fundamental solution of the Helmholtz equation
∆u + κ2u = 0.
The single layer potential ψκis given by :
(ψκu)(x) =
?
Γ
G(κ,|x − y|)u(y)dσ(y)x ∈ R3\Γ,
and its trace by
Vκu(x) =
?
Γ
G(κ,|x − y|)u(y)dσ(y)x ∈ Γ.
The fundamental solution is pseudo-homogeneous of class −1 (see [18, 24]). As conse-
quence we have the following result :
Lemma 2.9 Let s ∈ R. The operators
ψκ
: Hs−1
2(Γ) → Hs+1
loc(R3)
Vκ
: Hs−1
2(Γ) → Hs+1
2(Γ)
are continuous.
We define the electric potential ΨEκgenerated by j ∈ TH−1
2(divΓ,Γ) by
ΨEκj := κψκj + κ−1∇ψκdivΓj
This can be written as ΨEκj := κ−1curlcurlψκj because of the Helmholtz equation and
the identity curlcurl = −∆ + ∇div (cf [3]).
We define the magnetic potential ΨMκgenerated by m ∈ TH−1
2(divΓ,Γ) by
ΨMκm := curlψκm.
We denote the identity operator by I.
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Lemma 2.10 The potentials ΨEκet ΨMκare continuous from TH−1
For j ∈ TH−1
2(divΓ,Γ) to Hloc(curl,R3).
2(divΓ,Γ) we have
(curlcurl−κ2I)ΨEκj = 0 and (curlcurl−κ2I)ΨMκm = 0 in R3\Γ
and ΨEκj and ΨMκm satisfy the Silver-Müller condition.
We define the electric and the magnetic far field operators for j ∈ TH−1
element ˆ x of the unit sphere S2of R3by
2(divΓ,Γ) and an
Ψ∞
Eκj(ˆ x) = κ ˆ x ×
??
??
Γ
e−iκˆ x·yj(y)dσ(y)
?
?
× ˆ x,
Ψ∞
Mκej(ˆ x) = iκ ˆ x ×
Γ
e−iκˆ x·yj(y)dσ(y).
(2.11)
These operators are bounded from TH−1
We can now define the main boundary integral operators:
2(divΓ,Γ) to T(C∞(S2))3.
Cκ= −1
2{γD+ γc
D}ΨEκ= −1
2{γN+ γc
N}ΨMκ,
Mκ= −1
2{γD+ γc
D}ΨMκ= −1
2{γN+ γc
N}ΨEκ.
These are bounded operators in TH−1
?
=?−κ n × Vκj + κ−1curlΓVκdivΓj?(x)
and
Mκj(x)
2(divΓ,Γ). We have
Cκj(x) = −κ
Γ
n(x) × (G(κ,|x − y|)j(y))dσ(y) + κ−1
?
Γ
curlx
Γ(G(κ,|x − y|)divΓj(y))dσ(y)
= −
?
Γ
n(x) × curlx(G(κ,|x − y|)j(y))dσ(y)
= (Dκj − Bκj)(x),
with
Bκj(x)=
?
?
Γ
∇xG(κ,|x − y|)(j(y) · n(x))dσ(y),
Dκj(x)=
Γ
(∇xG(κ,|x − y|) · n(x))j(y)dσ(y).
The kernel of Dκis pseudo-homogeneous of class −1 and the operator Mκhas the same
regularity as Dκon TH−1
2(divΓ,Γ), that is compact.
We describe briefly the boundary integral equation method developped by the autors
[9] to solve the dielectric scattering problem.
Boundary integral equation method : This is based on the Stratton-Chu formula,
the jump relations of the electromagnetic potentials and the Calderón projector’s formula
(see [6, 24]).
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We need a variant of the operator Cκdefined for j ∈ TH−1
2(divΓ,Γ) by :
C*
0j = n × V0j + curlΓV0divΓj.
The operator C∗
integral representation of the exterior electric field Es:
0is bounded in TH−1
2(divΓ,Γ). We use the following ansatz on the
Es= −ΨEκej − iη ΨMκeC*
0j in R3\¯Ω
(2.12)
η is a positive real number and j ∈ TH−1
we have the integral representation of the interior field
2(divΓ,Γ). Thanks to the transmission conditions
E1= −1
ρ(ΨEκi{γc
NeEinc+ Nej}) − (ΨMκi{γc
DEinc+ Lej}) in Ω
(2.13)
where ρ =κiµe
κeµi
and
Le= Cκe−iη
?1
2I − Mκe
?
?
C*
0,
Ne=
?1
2I − Mκe
+ iη CκeC*
0.
We apply the exterior Dirichlet trace to the righthandside (2.13). The density j then
solves the following boundary integral equation:
Sj = ρ
?
−1
2I + Mκi
?
Lej + CκiNej = −ρ
?
−1
2I + Mκi
?
γDEinc+ CκiγNκeEincsur Γ.
The operator S is linear, bounded and invertible on TH−1
If we are concerned with the far field pattern E∞of the solution, it suffices to re-
place the potential operators ΨEκeand ΨMκeby the far field operators Ψ∞
respectively.
The solution E(Ω) = (Ei(Ω),Es(Ω)) and the far field pattern E∞(Ω) consists of ap-
plications defined by integrals on the boundary Γ and if the incident field is a fixed data,
these quantities depend on the scatterrer Ω only.
2(divΓ,Γ).
Eκeand Ψ∞
Mκe
3 Some remarks on shape derivatives
We want to study the dependance of any functionals F with respect to the shape of the
dielectric scatterer Ω. The Ω-dependance is highly nonlinear. The standard differential
calculus tools need the framework of topological vector spaces which are locally convex at
least [33], framework we do not dispose in the case of shape functionals. An interesting
approach consists in representing the variations of the domain Ω by elements of a function
space. We consider variations generated by transformations of the form
x ?→ x + r(x)
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of any points x in the space R3, where r is a vectorial function defined (at least) in the
neiborhood of Ω. This transformation deforms the domain Ω in a domain Ωrof boundary
Γr. The functions r are assumed to be a small enough elements of a Fréchet space X in
order that (I + r) is an isomorphism from Γ to
Γr= (I + r)Γ = {xr= x + r(x);x ∈ Γ}.
Since we consider smooth surfaces, in the remaining of this paper, the space X will be
the Fréchet space C∞
k∈N
seminorms (|| · ||k)k∈Nwhere Ck
differentiable functions whose the derivatives are bounded and
b(R3,R3) =
?
Ck
b(R3,R3) undowed with the set of non decreasing
b(R3,R3) with k ∈ N is the space of k-times continuously
||r||k= sup
0≤p≤k
sup
x∈R3
???r(p)(x)
?3, d∞(0,r) < ǫ
???.
For ǫ small enough we set
B∞
ǫ =
?
r ∈
?
C∞(R3)
?
,
where d∞is the metric induced by the seminorms.
We introduce the application
r ∈ B∞
ǫ ?→ FΩ(r) = F(Ωr).
We define the shape derivative of the functional F trough the deformation Ω → Ωξas
the Gâteaux derivative of the application FΩin the direction ξ ∈ X. We write:
DF[Ω;ξ] =∂
∂t|t=0FΩ(tξ).
3.1Gâteaux differentiability: elementary results
Fréchet spaces are locally convex, metrisable and complete topological vector spaces on
which we can extend any elementary results available on Banach spaces. We recall some
of them. We refer to the Schwarz’s book [33] for more details.
Let X and Y be Fréchet spaces and let U be a subset of X.
Definition 3.1 (Gâteaux semi-derivatives) The application f
have Gâteaux semiderivative at r0∈ U in the direction ξ ∈ X if the following limit exists
and is finite
∂
∂rf[r0;ξ] = lim
t→0
t
: U → Y is said to
f(r0+ tξ) − f(r0)
=∂
∂t??t=0f(r0+ tξ).
Definition 3.2 (Gâteaux differentiability) The application f : U → Y is said to be
Gâteaux differentiable at r0∈ U if it has Gâteaux semiderivatives in all direction ξ ∈ X
and if the map
∂
∂rf[r0;ξ] ∈ Y
is linear and continuous.
ξ ∈ X ?→
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We say that f is continuously (or C1-) Gâteaux differentiable if it is Gâteaux differentiable
at all r0∈ U and the application
∂
∂rf : (r0;ξ) ∈ U × X ?→
∂
∂rf[r0;ξ] ∈ Y
is continuous.
Remark 3.3 Let us come to shape functionals. In calculus of shape variation, we usually
consider the Gâteaux derivative in r = 0 only. This is due to the result : If FΩis Gâteaux
differentiable on B∞
ǫ
then for all ξ ∈ X we have
∂
∂rFΩ[r0;ξ] = DF(Ωr0;ξ ◦ (I + r0)−1) =
∂
∂rFΩr0[0;ξ ◦ (I + r0)−1].
Definition 3.4 (higher order derivatives) Let m ∈ N. We say that f is (m+1)-times
continuously (or Cm+1-) Gâteaux differentiable if it is Cm-Gâteaux differentiable and
r ∈ U ?→
∂m
∂rmf[r;ξ1,...,ξm]
is continuously Gâteaux differentiable for all m-uple (ξ1,...,ξm) ∈ Xm. Then for all
r0∈ U the application
(ξ1,...,ξm+1) ∈ Xm+1?→
∂m+1
∂rm+1f[r0;ξ1,...,ξm+1] ∈ Y
is (m + 1)-linear, symetric and continuous. We say that f is C∞-Gâteaux differentiable
if it is Cm-Gâteaux differentiable for all m ∈ N.
We use the notation
∂m
∂rmf[r0,ξ] =∂m
∂tm??t=0f(r0+ tξ).
?
(3.1)
If it is Cm-Gâteaux differentiable we have
∂m
∂rmf[r0,ξ1,...,ξm] =
1
m!
m
?
p=1
(−1)m−p
1≤i1<···<ip≤m
∂m
∂rmf[r0;ξi1+ ... + ξip].
(3.2)
To determine higher order Gâteaux derivatives it is more easy to use this equality.
The chain and product rules and the Taylor expansion with integral remainder are still
available for Cm-Gâteaux differentiable maps ([33] p. 30). We use the following lemma
to study the Gâteaux differentiability of any applications mapping r on the inverse of an
element in a unitary topological algebra.
Lemma 3.5 Let X be a Fréchet space and Y be a unitary Fréchet algebra. Let U be an
open set of X. Assume that the application f : U → Y is Gâteaux differentiable at r0∈ U
and that f(r) is invertible in Y for all r ∈ U and that the application g : r ?→ f(r)−1
12
Page 13
is continuous at r0. Then g is Gâteaux differentiable at r0and its first derivative in the
direction ξ ∈ X is
∂
∂rf[r0,ξ] = −f(r0)−1◦∂
∂rf[r0,ξ] ◦ f(r0)−1.
(3.3)
Moreover if f is Cm-Gâteaux differentiable then g is too.
Proof.
Let ξ ∈ X and t > 0 small enough such that (r0+ tξ) ∈ U, on a:
g(r0+ tξ) − g(r0) =f(r0)−1◦ f(r0) ◦ f(r0+ tξ)−1− f(r0)−1◦ f(r0+ tξ) ◦ f(r0+ tξ)−1
=f(r0)−1◦ (f(r0) − f(r0+ tξ)) ◦ f(r0+ tξ)−1
=f(r0)−1◦ (f(r0) − f(r0+ tξ)) ◦ f(r0)−1
+f(r0)−1◦ (f(r0) − f(r0+ tξ)) ◦?f(r0+ tξ)−1− f(r0)−1?.
Since g is continuous in r0, we have lim
t→0
Gâteaux differentiable in r0we have
?f(r0+ tξ)−1− f(r0)−1?= 0 and since f is
lim
t→0
f(r0)−1◦ (f(r0) − f(r0+ tξ)) ◦ f(r0)−1
t
= −(f(r0))−1◦∂
∂rf[r0,ξ] ◦ (f(r0))−1.
As a consequence
lim
t→0
g(r0+ tξ) − g(r0)
t
= −(f(r0))−1◦∂
∂rf[r0,ξ] ◦ (f(r0))−1.
?
4Gâteaux differentiability of pseudo-homogeneous kernels
Let xrdenote an element of Γrand let nrbe the outer unit normal vector to Γr. When
r = 0 we write n0= n. We note again dσ the area element on Γr.
In this section we want to study the Gâteaux differentiability of the application map-
ping r ∈ B∞
ǫ
to the integral operator KΓrdefined for a function ur∈ Hs(Γr) by:
?
and of the application mapping r ∈ B∞
ǫ to the potential operator Prdefined for a function
ur∈ Hs(Γr) by:
?
where kr∈ C∞?Γr×?R3\{0}??is a pseudo-homogeneous kernel of class −m with m ∈ N.
KΓrur(xr) = vp.
Γr
kr(yr,xr− yr)ur(yr)dσ(yr), xr∈ Γr
(4.1)
Prur(x) =
Γr
kr(yr,x − yr)ur(yr)dσ(yr), x ∈ K,
(4.2)
13
Page 14
We want to differentiate applications of the form r ?→ FΩ(r) where the domain of
definition of FΩ(r) varies with r.How do we do?
[25, 27, 29]), is that instead of studying the application
A first idea, quite classical (see
r ∈ B∞
ǫ ?→ FΩ(r) ∈ Ck(Γr)
we consider the application
r ∈ B∞
ǫ ?→ FΩ(r) ◦ (I + r) ∈ Ck(Γ).
An example is r ?→ nr. This point of view can be extended to Sobolev spaces Hs(Γ),
s ∈ R. From now we use the transformation τrwhich maps a function urdefined on Γr
to the function ur◦(I+r) defined on Γ. For all r ∈ B∞
inverse. We have
ǫ, this transformation τradmit an
(τrur)(x) = ur(x + r(x)) and (τ−1
r u)(xr) = u(x).
Then, instead of studying the application
r ∈ B∞
ǫ ?→ KΓr∈ Lc
?Hs(Γr),Hs+m(Γr)?
?Hs(Γ),Hs+m(Γ)?.
we consider the conjugate application
r ∈ B∞
ǫ ?→ τrKΓrτ−1
r
∈ Lc
In the framework of boundary integral equations, this approach is sufficient to obtain the
shape differentability of any solution to scalar boundary value problems [27, 29].
Using the change of variable x ?→ xr= x + r(x), we have for u ∈ Hs(Γ):
?
where Jris the jacobian (the determinant of the Jacobian matrix) of the change of variable
mapping x ∈ Γ to x + r(x) ∈ Γr. The differentiablility analysis of these operators begins
with the jacobian one. We have
τrKrτ−1
r (u)(x) =
Γ
kr(y + r(y),x + r(x) − y − r(y))u(y)Jr(y)dσ(y), x ∈ Γ
Jr= JacΓ(I + r) = ||ωr|| with ωr= com(I + Dr|Γ)n0= det(I + Dr|Γ)T(I + Dr|Γ)−1n,
and the normal vector nris given by
nr= τ−1
r
?
ωr
?ωr?
?
.
The first derivative at r = 0 of these applications are well known [12, 25]. Here we present
one method to obtain higher order derivative.
Lemma 4.1 The application J mapping r ∈ B∞
Gâteaux differentiable and its first derivative at r0is defined for ξ ∈ C∞
ǫ
to the jacobian Jr∈ C∞(Γ,R) is C∞
b(R3,R3) by:
∂J
∂r[r0,ξ] = Jr0(τr0divΓr0τ−1
r0)ξ.
14
Page 15
Proof.
do the proof for hypersurfaces Γ of Rn, n ∈ N, n ≥ 2. We use local coordinate system.
Assume that Γ is parametrised by an atlas (Oi,φi)1≤i≤pthen Γrcan be parametrised by
the atlas (Oi,(I + r) ◦ φi)1≤i≤p. For any x ∈ Γ, let us note e1(x),e2(x),...,en−1(x) the
vector basis of the tangent plane to Γ at x. The vector basis of the tangent plane to Γr
at x + r(x) are given by
We just have to prove the C∞-Gâteaux differentiability of W : r ?→ wr. We
ei(r,x) = [(I + Dr)(x)]ei(x)
for i = 1,...,n − 1.
Thus, we have ωr(x) =
n−1
?
n−1
?
i=1
????
ei(r,x)
i=1
ei(x)
????
. Since the applications r ?→ ei(r,x), for i = 1,...,n−1
are C∞-Gâteaux differentiable, the application W is too. Now want to compute the
derivatives using the formula (3.2). Let ξ ∈ C∞
r0∈ B∞
ǫ
b(Rn,Rn) and t small enough. We have at
∂mW
∂rm[r0,ξ] =∂m
∂tm???t=0
n−1
?
i=1
(I + Dr0+ tDξ)ei(x)
????
n−1
?
i=1
ei(x)
????
.
To simplify this expression one have to note that
[Dξ(x)]ei(x) = [Dξ(x)][(I + Dr0)(x)]−1[(I + Dr0)(x)]ei(x)
= [Dξ(x)][D(I + r0)−1(x + r0(x))][(I + Dr0)(x)]ei(x)
= [(τr0Dτ−1
r0)ξ(x)]ei(r0,x) = [(τr0DΓr0τ−1
r0)ξ(x)]ei(r0,x).
NB: given a (n × n) matrix A we have
n−1
?
i=1
··· × ei−1× Aei× ei+1× ··· = (Trace(A)I −TA)
n−1
?
i=1
ei.
Thus we have with A = [τr0DΓr0τ−1
∂rm[r0,ξ]
r0ξ] and B0= I, B1(A) = Trace(A)I −TA
(#)
W(r0)
∂W
∂r[r0,ξ]
=Jr0(τr0nr0),
?(τr0divΓr0τ−1
[B1(A)ξ]W(r0),
=Jr0
r0)ξ · τr0nr0− [(τr0∇Γr0τ−1
r0)ξ]τr0nr0
?
=
∂mW
∂rm[r0,ξ]=[Bm(A)ξ]W(r0)
m ?
0 for all m ≥ n.
=
i=1
(−1)i+1(m − 1)!
(m − i)![B1(Ai)Bm−i(A)ξ]W(r0) for m = 1,...,n − 1,
∂mW
≡
15
Page 16
It follows that
∂J
∂r[r0,ξ] =
1
?W(r0)?
∂W
∂r[r0,ξ] · W(r0) =∂W
∂r[r0,ξ] · τr0nr0= Jr0(τr0divΓr0τ−1
r0)ξ.
?
Thanks to (#) we deduce easily the Gâteaux differentiability of r ?→ τrnr.
Lemma 4.2 The application N mapping r ∈ B∞
C∞Gâteaux-differentiable and its first derivative at r0is defined for ξ ∈ C∞
ǫ
to τrnr= nr◦ (I + r) ∈ C∞(Γ,R3) is
b(R3,R3) by:
∂N
∂r[r0,ξ] = −?τr0∇Γr0τ−1
r0ξ?N(r0).
Proof.
This results from the precedent proof and we have
∂N
∂r[r0,ξ]=
1
?W(r0)?
?∂W
−?τr0∇Γr0τ−1
∂W
∂r[r0,ξ] −
1
?W(r0)?3
?∂W
?∂W
∂r[r0,ξ] · W(r0)
??
?
W(r0)
=J−1
r0
∂r[r0,ξ] −
r0ξ?τr0nr0.
∂r[r0,ξ] · (τr0nr0)τr0nr0
=
?
To obtain higher order shape derivatives of these applications one can use the equalities
(#) and
As for exemple we have at r = 0 in the direction ξ ∈ C∞
(∗)
?τrnr?≡1,
∂mN · N
∂rm
[r0,ξ]≡0 for all m ≥ 1.
b(R3,R3):
∂J
∂r[0,ξ] = divΓξ and∂N
∂r[0,ξ] = −[∇Γξ]n,
∂2J
∂r2[0,ξ1,ξ2] = −Trace([∇Γξ2][∇Γξ1]) + divΓξ1· divΓξ2+ ([∇Γξ1]n · [∇Γξ2]n).
Notice that Trace([∇Γξ2][∇Γξ1]) = Trace([∇Γξ1][∇Γξ2]),
∂2N
∂r2[0,ξ1,ξ2] = [∇Γξ2][∇Γξ1]n + [∇Γξ1][∇Γξ2]n − ([∇Γξ1]n · [∇Γξ2]n)n.
For n ≥ 3 it needs too long calculations to simplify the expression of the derivatives and
we only obtain the quadratic expression. In the last section we give a second method to
obtain higher order derivatives using the Gâteaux derivatives of the surface differential
operators.
16
Page 17
Remark 4.3 We do not need more than the first derivative of the deformations ξ. As a
consequence for hypersurfaces of class Ck+1, it suffice to consider deformations of class
Ck+1to conserve the regularity of the jacobian and of the normal vector by differentiation.
The following theorem establish sufficient conditions for the Gâteaux differentiabil-
ity of the boundary integral operators described here above and that we obtain their
derivatives by deriving their kernels.
Theorem 4.4 Let k ∈ N. We set (Γ × Γ)∗= {(x,y) ∈ Γ × Γ; x ?= y}. Assume that
1) For all fixed (x,y) ∈ (Γ × Γ)∗the function
f :B∞
ǫ
r
→
?→
C
kr(y + r(y),x + r(x) − y − r(y))Jr(y)
is Ck+1-Gâteaux differentiable.
2) The functions (y,x − y) ?→ f(r0)(y,x − y) and
(y,x − y) ?→
∂α
∂rαf[r0,ξ1,...,ξα](y,x − y)
are pseudo-homogeneous of class −m for all r0∈ B∞
ξ1,...,ξk+1∈ C∞
Then the application
ǫ, for all α = 1,...,k +1 and for all
b(R3,R3).
B∞
ǫ
r
→
?→
Lc(Hs(Γ),Hs+m(Γ))
τrKΓrτ−1
r
is Ck-Gâteaux differentiable and
∂k
∂rk
?τrKΓrτ−1
We use the linearity of the integral and Taylor expansion with integral remain-
der. We do the proof k = 1 only. Let r0∈ B∞
that r0+ tξ ∈ B∞
ǫ. We have
r
?[r0,ξ1,...,ξk]u(x) =
?
Γ
∂k
∂rkf[r0,ξ1,...,ξk](y,x − y)u(y)dσ(y).
Proof.
ǫ, ξ ∈ C∞(Rn,Rn) and t small enough such
f(r0+tξ,x,y)−f(r0,y,x−y) = t∂f
∂r[r0,ξ](y,x−y)+t2
?1
0
(1−λ)∂2f
∂r2[r0+λtξ,ξ](y,x−y)dλ.
We have to verify that each terms in the equality here above is a kernel of an operator
mapping Hs(Γ) onto Hs+m(Γ). The two first terms in the left hand side are kernels of
class −m and by hypothesis∂2f
∂r2[0,ξ] is also a kernel of class −m. It remains to prove
that the operator with kernel
?1
(x,y) ?→
0
(1 − λ)∂2f
∂r2[r0+ λtξ,ξ](x,y)
17
Page 18
acts from Hs(Γ) to Hs+m(Γ). Since∂2f
∂r2[r0+λtξ,ξ] is pseudo-homogeneous of class −m
for all λ ∈ [0,1], it suffice to use Lebesgue’s theorem in order to invert the integration
with respect to the variable λ and the integration with respect to y on Γ.
????
=
0
Γ
????
≤C||u||Hs(Γ).
?
?1
Γ
??1
0
(1 − λ)∂2f
∂r2[r0+ λtξ,ξ](x,y)dλ
?
u(y)dσ(y)
????Hs+m(Γ)
dλ
????
supλ∈[0,1]
(1 − λ)
??
??
∂2f
∂r2[r0+ λtξ,ξ](x,y)u(y)dσ(y)
?
????Hs+m(Γ)
≤
Γ
∂2f
∂r2[r0+ λtξ,ξ](x,y)u(y)dσ(y)
?????Hs+m(Γ)
We then have
??
?
We pass to the limit in t = 0 and we obtain the first Gâteaux derivative. For higher order
∂k
∂rkf[r0,ξ1,...,ξk] instead of f. The linearity,
the symetry and the continuity of the first derivative is deduced from the kernel one. ?
1
t
Γ
f(r0+ tξ,x,y)u(y)dσ(y) −
?
Γ
??1
f(r0,x,y)u(y)dσ(y)
?
=
Γ
∂f
∂r[r0,ξ](x,y)u(y)dσ(y) + t
?
Γ
0
(1 − λ)∂2f
∂r2[r0+ λtξ,ξ](x,y)dλ
?
u(y)dσ(y).
derivative it suffice to write the proof with
Now we will consider some particular classes of pseudo-homogeneous kernels.
Corollary 4.5 Assume that the kernels krare of the form
kr(yr,xr− yr) = G(xr− yr)
where G is pseudo-homogeneous kernel which do not depend on r. Then the application
B∞
ǫ
r
→
?→
Lc(Hs(Γ),Hs+m(Γ))
τrKΓrτ−1
r
is C∞-Gâteaux differentiable and the kernel of the first derivative at r = 0 is defined for
ξ ∈ C∞
b(R3,R3) by
∂ {G(x + r(x) − y − r(y))}
∂r
[0,ξ] = (ξ(x) − ξ(y)) · ∇zG(x − y) + G(x − y)divΓξ(y).
Proof.
For all fixed (x,y) ∈ (Γ × Γ)∗, consider the application
f : U ?→ f(r,x,y) = G(x + r(x) − y − r(y))Jr(y) ∈ C.
We have to prove that r ?→ f(r) is C∞-Gâteaux differentiable and that each derivative
define a pseudo-homogeneous kernel of class −m.
18
Page 19
⊲Step 1:
First of all we prove that for (x,y) ∈ (Γ × Γ)∗fixed the application r ?→ f(r,x,y) is
infinitely Gâteaux differentiable on B∞
ǫ. By lemma 4.1 the application r ?→ Jr(y) is
infinitely Gâteaux differentiable on B∞
ǫ, the application r ?→ x + r(x) is also infinitely
Gâteaux differentiable on B∞
of applications infinitely Gâteaux differentiable, the application r ?→ f(r,x,y) is too and
using Leibniz formula we have :
ǫ and the kernel G is of class C∞on R3\{0}. Being composed
∂k
∂rkf[r0,ξ1,...,ξk](x,y) =
k ?
where S+
kdenote the non decreasing permutations of {1,...,k} and
α=0
?
σ∈S+
k
∂α
∂rα{G(x + r(x) − y − r(y))}[r0,ξσ(1),...,ξσ(α)]∂k−αJr(y)
∂rk−α
[r0,ξσ(α+1),...,ξσ(k)]
∂α
∂rα{G(xr− yr)}[r0;ξ1,...,ξα] = Dα
zG[x+r0(x)−y−r0(y);ξ1(x)−ξ1(y),...,ξα(x)−ξα(y)].
⊲Step 2:
We then prove that each derivative define a new pseudo-homogeneous kernel of class −m
that is to say that for all k ∈ N and for all k-uple (ξ1,...,ξk) the application
(x,y) ?→
∂k
∂rkf[r0,ξ1,...,ξk](x,y)
is pseudo-homogeneous of class −m. Since∂k−αJr
to prove that
∂rk−α[r0,ξ1,...,ξk−α] ∈ C∞(Γ,R) we have
(x,y) ?→
∂α
∂rα{G(x + r(x) − y − r(y))}[r0,ξ1,...,ξα]
defines a pseudo-homogeneous kernel of class −m. By definition, G(z) admit the following
asymptotic expansion when z tends to zero:
G(z) = Gm(z) +
N−1
?
j=1
Gm+j(z) + Gm+N(h,z)
where Gm+j is homogeneous of class −(m + j) for j = 0,...,N − 1 and Gm+N is of
arbitrary regularity. Using Taylor formula we obtain the following result :
Proposition 4.6 Let Gm(z) be an homogeneous kernel of class −m and any deformations
ξ = (ξ1,...,ξα) ∈ C∞
b(R3,R3). The function
(x,y) ?→ DαGm[x + r0(x) − y − r0(y);ξ1(x) − ξ1(y),...,ξα(x) − ξα(y)]
is pseudo-homogeneous of class −m.
19
Page 20
The application mapping (ξ1,...,ξα) ∈ C∞
b(R3,R3) to the integral operator of kernel
∂k{G(x + r(x) − y − r(y))}
∂rm
[r0,ξ1,...,ξk]
is clearly linear and continuous for all r0∈ B∞
ǫ.
?
Example 4.7 (Single layer kernel) We note Vr
Hs(Γr) by
Vr
κur(x) =
κthe integral operator defined for ur∈
?
Γr
G(κ,|xr− yr|)ur(yr)dσ(yr).
The application
Bδ
r
→
?→
Lc(Hs(Γ),Hs+1(Γ))
τrVr
rκτ−1
is C∞-Gâteaux differentiable and its first derivative at r = 0 in the direction ξ ∈ C∞
is
∂τrVr
r
∂r
where in R3we have
?(ξ(x) − ξ(y)) · (x − y)
Example 4.8 (Double layer kernel) We note Dr
Hs(Γr) by
Dr
κur(x) =
Γr
The application
Bδ
→
Lc(Hs(Γ),Hs+1(Γ))
r?→τrDr
is C∞Gâteaux-differentiable .
b(R3,R3)
κτ−1
[0,ξ]u(x) =
?
Γ
k′(y,x − y)u(y)dσ(y)
(4.3)
k′(x,y) =G(κ,|x − y|)
|x − y|
?
iκ −
1
|x − y|
?
+ divΓξ(y)
?
.
κthe integral operator defined for ur∈
?
nr(xr) · ∇zG(κ,|xr− yr|)ur(yr)dσ(yr).
κτ−1
r
Proof.
We have
nr(xr) · ∇zG(κ,|xr− yr|)ur(yr) = nr(xr) · (xr− yr)G(κ,|xr− yr|)
|xr− yr|
?
iκ −
1
|xr− yr|
?
.
We have to prove that
r ∈ B∞
ǫ ?→(τrnr)(x) · (x + r(x) − y − r(y))
|x + r(x) − y − r(y)|3
is C∞Gâteaux differentiable and that the derivatives are pseudo-homogeneous of class
−1. To do so we use local coordinates as Potthast did in [29] and prove that
∂k(τrnr)(x) · (x + r(x) − y − r(y))
∂rk
[r0,ξ1,...,ξk]
behaves as |x − y|2when x − y tends to zero.
?
20
Page 21
Each domain Ω is a countable union of compact subset of Ω: Ω =
?
p≥1
Kp. Instead of
studying the application
r ∈ B∞
ǫ ?→ FΩ(r) ∈ Lc
?Hs(Γr),Hs+m(Ωr)?
we consider the application
r ∈ B∞
ǫ ?→ FΩ(r)τ−1
r
∈ Lc
?Hs(Γ),Hs+m(Kp)?.
We use this approach for potential operators. We have for u ∈ Hs−1
?
Theorem 4.9 Let s ∈ R. Let G(z) be a pseudo-homogeneous kernel of class −(m + 1)
with m ∈ N. Assume that for all r ∈ B∞
ǫ, we have kr(yr,x − yr) = G(x − yr). Then the
application
B∞
ǫ
→
Lc
Hs−1
r ?→Prτ−1
r
2(Γ)
Prτ−1
r (u)(x) =
Γ
kr(y + r(y),x − y − r(y))u(y)Jr(y)dσ(y), x ∈ Kp.
?
2(Γ),C∞(Kp)
?
is infinitely Gâteaux differentiable and
∂kPrτ−1
∂rk
r
[r0,ξ1,...,ξk]u(x) =
?
Γ
∂k
∂rk{G(x − y − r(y))Jr(y)}[r0,ξ1,...,ξk]u(y)dσ(y).
Its first derivative at r = 0 in the direction ξ ∈ C∞
denoted by P1with the kernel
b(R3,R3) is the integral operator
∂
∂r{G(x − y − r(y))}[r0,ξ] = −ξ(y) · ∇zG(x − y) + G(x − y)divΓξ(y).
(4.4)
The operator P(1)can be extended in a linear and continuous integral operator from
Hs−1
loc(Ω).
2(Γ) to Hs+m(Ω) and Hs+m
Proof.
Since Ω is an increasing union of compact manifolds we can define a shape derivative on
the whole domain Ω. Let us look at the first derivative : the term G(x−y)divΓξ(y) has
the same regularity than G(x − y) when x − y tends to zero wheareas ξ(y) · ∇G(x − y)
loose one order of regularity. As a consequence the kernel must be of class −m + 1 in
order that its first derivative acts from Hs−1
The kernel and its higher order derivatives are of class C∞on Kp.
2(Γ) to Hs+m(Ω) and Hs+m
loc(Ωc).
?
Remark 4.10 We conclude that the boundary integral operators are smooth with respect
to the domain whereas the potential operators loose one order of regularity at each deriva-
tion. We point out that we do not need more than the first derivative of the deformations
ξ to compute the Gâteaux derivatives of these integral operators.
21
Page 22
Example 4.11 (Single layer potential) We denote by ψr
defined for ur∈ Hs(Γr) by
?
The application
B∞
ǫ
r
κthe single layer potential
ψr
κur(x) =
Γr
G(κ,|x − yr|)ur(yr)dσ(yr), x ∈ R3\Γr.
→
?→
Lc(Hs(Γ),C∞(Kp))
τrψr
rκτ−1
is infinitely Gâteaux differentiable. Its first derivative at r = 0 can be extended in a linear
and continuous operator from Hs−1
2(Γ) to Hs(Ω) ∪ Hs
loc(Ωc).
Since the potential operators are infinitely Gâteaux differentiable far from the bound-
ary we have the following result by inverting the derivation with respect to r and the
passage to the limit |x| → ∞.
Example 4.12 Let s ∈ R. We denote by ψ∞,r
single layer potential defined for ur∈ Hs(Γr) by
κ
the far field operator associated to the
ψ∞,r
κ
ur(ˆ x) =
?
Γr
e−iκˆ x·yrur(yr)dσ(yr), ˆ x ∈ S2.
The application
B∞
ǫ
r
→
?→
Lc(Hs(Γ),C∞(S2))
Ψ∞,r
κ
τ−1
r
is infinitely Gâteaux differentiable and its first deriavtive at r = 0 is defined for u ∈ Hs(Γ)
by:
∂Ψ∞,r
κ
τ−1
r
∂r
Γ
[0,ξ]u(ˆ x) =
??
e−iκˆ x·y(divΓξ(y) − iκˆ x · ξ(y))u(y)dσ(y)
?
.
5Shape differentiability of the solution
Let Eincbe an incident electric field which is a fixed data. The aim of this section is
to study the shape differentiation properties of the application E mapping the bounded
scatterer Ω to the solution E(Ω) =
scattering problem by the obstacle Ω lit by the incident field Eincestablished in section
1. To do so we use the integral representation of the solution.
We set Ei(r) = Ei(Ωr) and Es(r) = Es(Ωr) and we denote Ψr
Mr
the γc,r
Nκtrace mappings on Γr. We have :
?Ei(Ω),Es(Ω)?
∈ Hloc(curl,R3) to the dielectric
Eκ, Ψr
Mκ, C∗r
0, Cr
D, γr
κand
Nκ, γc,r
κthe potential operators and the boundary integral operators on Γrand γr
D
Etot(r) = Einc+ Es(r)
(5.1)
with
Es(r) =
?
−Ψr
Eκe− iηΨr
MκeC∗r
0
?
jr
dans Ωc
r= R3\Ωr
(5.2)
22
Page 23
where jrsolves the integral equation
Srjr= −ρ
?
−1
2I + Mr
κi
?
γr
DEinc− Cr
κiγr
NκeEinc,
and
Ei(r) = −1
ρΨr
Eκiγc,r
NκeEtot(r) − Ψr
Mκiγc,r
DEtot(r)
dans Ωr
(5.3)
Recall that the operator Sris composed of the operators Cr
these last ones are defined on the space TH−1
κe, Mr
κe, Cr
κiet Mr
κiand that
2(divΓr,Γr).
5.1 Variations of Helmholtz decomposition
We have to turn out many difficulties. On one hand, to be able to construct shape
derivatives of the solution it is necessary to prove that the derivatives are defined on the
same spaces than the boundary integral operators themselves, that is TH−1
we derive at r = 0). On the other hand, the very definition of the differentiability of
operators defined on TH−1
to derive applications defined on the variable space TH−1
A first idea is to insert the identity τ−1
r τr= IH−1
representation of the solution in order to consider integral operators on the fixed boundary
Γ only and to study the differentiability of the applications
2(divΓ,Γ) (if
2(divΓ,Γ) raises non-trivial questions. The first one is : How
2(divΓr,Γr)?
2(Γr)between each operator in the integral
r
r
r
r
?→
?→
?→
?→
τrCr
τrMr
Ψr
Ψr
κτ−1
κτ−1
Eκτ−1
Mκτ−1
r ,
r ,
r ,
r
but many difficulties persist as Potthast pointed out [28]. The electromagnetic boundary
integral operators are defined and bounded on tangential functions to Γr.The restriction of
the operator τrMr
r
to tangential densities to Γr, has the same regularity of the double
layer potential operator. If we differentiate τrMr
the same regularity than Mκand acting on TH−1
κτ−1
κτ−1
2(divΓ,Γ) since:
r , we will not obtain an operator with
τr(TH−1
2(divΓr,Γr)) ?= TH−1
2(divΓ,Γ).
The incident field Eincis analytic in the neighborhood of Γ thus γr
for all r ∈ B∞
ǫ. Set f(r) = τr
∂f(tξ)
∂t
have the same difficulties we the Neuman trace γr
As an alternative, the idea of R. Potthast was to introduce projectors on the tangent
planes of the surfaces Γ and Γr.Let us note π(r) the orthogonal projection of any
functions defined on Γr onto the tangent plane to Γ. This is a linear and continuous
DEinc∈ TH−1
2(divΓr,Γr)
?γr
DEinc?. For ξ ∈ C∞
b(R3,R3), the Gâteaux semiderivative
|t=0is not tangent to Γ anymore it follows that Mκ∂f(tξ)
∂t
|t=0is not defined. We
Nκand the other operators.
23
Page 24
operator from the continous vector function space on Γrto the the space of continuous
tangential function to Γ and for ur∈ (C(Γr))3we have
(π(r)ur)(x) = ur(x + r(x)) − (n(x) · ur(x + r(x)))n(x).
Proposition 5.1 The restriction of π(r) to the continuous and tangential functions to
Γradmit an inverse, denoted by π−1(r). The application π−1(r) is defined for a tangential
function u to Γ by
(π−1(r)u)(x + r(x)) = u(x) − n(x)nr(x + r(x)) · u(x)
nr(x + r(x)) · n(x).
And we have π−1(r)u ∈ THs(Γr) if and only if u ∈ THs(Γ).
In the framework of the space of tangential continuous functions it suffices to insert the
product π−1(r)π(r) = ITC0(Γr)in the integral representation of the solution to lead us to
study boundary integral operators defined on TC0(Γ) which do not depend on r anymore
but here we would obtain operators defined on
?
This space depends again on the variable r and do not correspond to TH−1
approach consist in using the Helmholtz decomposition of the spaces TH−1
r ∈ B∞
ǫ
and to introduce a new invertible operator Prdefined on TH−1
which is not a projection operator.
π(r)
TH−1
2(divΓr,Γr)
?
=
?
u ∈ TH−1
2(Γ),divΓr(π−1(r)u) ∈ H−1
2(Γr)
?
.
2(divΓ,Γ). Our
2(divΓr,Γr) for
2(divΓr,Γr) and
We have the following decomposition. We refer to [11] for the proof.
Theorem 5.2 The Hilbert space TH−1
sition:
TH−1
2(divΓ,Γ) admit the following Helmholtz decompo-
??
ǫ
the surfaces Γrare still regular and
2(divΓr,Γr) admit the same decomposition.
2(divΓ,Γ) = ∇Γ
?
H
3
2(Γ)/R
curlΓ
?
H
1
2(Γ)/R
?
.
(5.4)
Since the real ǫ is chosen such that for all r ∈ B∞
simply connected, then the spaces TH−1
Let jr∈ TH−1
pr∈ H
are in H
2(divΓr,Γr) and let ∇Γrpr+curlΓrqrits Helmholtz decomposition. Since
1
2(Γr), their change of variables from Γrto Γ, τr(pr) and τr(qr),
1
2(Γ) respectively. The following operator :
3
2(Γr) and qr∈ H
3
2(Γ) and H
Pr:
TH−1
jr= ∇Γrpr+ rotΓrqr
2(divΓr,Γr)−→
?→
TH−1
j = ∇Γτrpr+ rotΓτrqr
2(divΓ,Γ)
(5.5)
is well-defined.
The operator Prtransforms a tangential vector field jrto Γrin a tangential vector
field j to Γ. This operator is linear, continuous and admit an inverse P−1
r
given by :
P−1
r
:
TH−1
j = ∇Γp + rotΓq
2(divΓ,Γ)−→
?→
TH−1
jr= ∇Γrτ−1
2(divΓr,Γr)
r (p) + rotΓrτ−1
r (q).
(5.6)
24
Page 25
Obviously we have when r = 0 that Pr= P−1
r
= ITH−1
2(divΓ,Γ). We insert the identity
ITH−1
tion (Ei(r),Es(r)). Finaly we have to study the Gâteaux differentiality properties of the
following applications :
2(divΓr,Γr)= P−1
rPrbetween each operator in the integral representation of the solu-
B∞
ǫ
B∞
ǫ
B∞
ǫ
B∞
ǫ
→
→
→
→
Lc(TH−1
Lc(TH−1
Lc(THs(divΓ,Γ),TH−1
Lc(TH−1
2(divΓ,Γ),H(curl,Kp)):
:
:
:
r ?→ Ψr
r ?→ Ψr
r ?→ PrMr
r ?→ PrCr
EκP−1
MκP−1
κP−1
κP−1
r
2(divΓ,Γ),H(curl,Kp))
r
2(divΓ,Γ))
r
2(divΓ,Γ),TH−1
2(divΓ,Γ))
r
(5.7)
where Kpis a compact subset of R3\Γ.
Now let us look at the integral representation of these operators .
⊲ Integral representation of Ψr
The operator Ψr
by:
?
+ κ
Γr
EκP−1
r
.
EκPr−1is defined for j = ∇Γp + curlΓq ∈ TH−1
2(divΓ,Γ) and x ∈ Kp
Ψr
EκP−1
rj(x) = κ
Γr
G(κ,|x − yr|)?∇Γrτ−1
G(κ,|x − yr|)?rotΓrτ−1
− κ−1∇
Γr
r p?(yr)dσ(yr)
r q?(yr)dσ(yr)
r p?(yr)dσ(yr).
?
?
G(κ,|x − yr|)?∆Γrτ−1
⊲ Integral representation of Ψr
The operator Ψr
by:
MκP−1
r.
MκPr−1is defined for j = ∇Γp + curlΓq ∈ TH−1
2(divΓ,Γ) and x ∈ Kp
Ψr
MκP−1
rj(x) = curl
?
Γr
?
G(κ,|x − yr|)?∇Γrτ−1
G(κ,|x − yr|)?curlΓrτ−1
r p?(yr)dσ(yr)
r q?(yr)(yr)dσ(yr).+ curl
Γr
⊲ Integral representation of PrCr
Recall that for jr∈ TH−1
κP−1
r
.
2(divΓr,Γr), the operator Cr
?
−κ−1nr(xr) × ∇xr
κis defined by
Cr
κjr(xr) =−κnr(xr) ×
Γr
G(κ,|xr− yr|)jr(yr)dσ(yr)
?
Γr
Γr
G(κ,|xr− yr|)divΓrjr(yr)dσ(yr).
We want to write Cr
deduce that :
κjrof the form ∇ΓrPr+ curlΓrQr. Using the formula (2.5)-(2.6) we
divΓrCr
κjr= ∆ΓrPr et curlΓrCr
κjr= −∆ΓrQr.
25
Page 26
As a consequence we have for xr∈ Γr:
Pr(xr) =−κ ∆−1
ΓrdivΓr
?
nr(xr) ×
?
Γr
G(κ,|xr− yr|)jr(yr)dσ(yr)
?
(5.8)
and
Qr(xr) =−κ (−∆−1
Γr)curlΓr
?
nr(xr) ×
?
Γr
?
G(κ,|xr− yr|)jr(yr)dσ(yr)
?
−κ−1(−∆Γr)curlΓr(−curlΓr)
?
+κ−1
Γr
Γr
G(κ,|xr− yr|)divΓrjr(yr)dσ(yr),
=κ ∆−1
ΓrcurlΓr
?
is defined for j = ∇Γp + curlΓq ∈ TH−1
nr(xr) ×
?
Γr
G(κ,|xr− yr|)jr(yr)dσ(yr)
?
G(κ,|xr− yr|)divΓrjr(yr)dσ(yr).
The operator PrCr
κP−1
r
2(divΓ,Γ) by:
PrCr
κP−1
r
= ∇ΓP(r) + curlΓQ(r),
with
P(r)(x) =−κ
?τr∆−1
ΓrdivΓrτ−1
r
??
(τrnr)(x) × τr
??
G(κ,| · −yr|)(rotΓrτ−1
Γr
G(κ,| · −yr|)(∇Γrτ−1
r p)(yr)dσ(yr)
+
?
Γr
r q)(yr)dσ(yr)
?
(x)
?
and
Q(r)(x) =κ
?τr∆−1
ΓrcurlΓrτ−1
r
??
(τrnr)(x) × τr
??
G(κ,| · −yr|)(rotΓrτ−1
?
Γr
G(κ,| · −yr|)(∇Γrτ−1
r p)(yr)dσ(yr)
+
?
Γr
r q)(yr)dσ(yr)
?
(x)
?
+κ−1τr
??
Γr
G(κ,| · −yr|)(∆Γrτ−1
r p)(yr)dσ(yr)(x).
⊲ Integral representation of PrMr
Recall that for all jr∈ TH−1
κP−1
r
.
2(divΓr,Γr), the operator Mr
?
?
κis defined by
Mr
κjr(xr) =
Γr
((∇xrG(κ,|xr− yr|)) · nr(xr))jr(yr)dσ(yr)
−
Γr
∇xrG(κ,|xr− yr|)(nr(xr) · jr(yr))dσ(yr).
26
Page 27
Using the equalities (2.6) and the identity curlcurl = −∆ + ∇div, we have
?
=κ2nr(xr) ·
divΓrMr
κjr(xr) =
nr(xr) ·
Γr
curlcurlxr(G(κ,|xr− yr|)jr(yr))dσ(yr)
?
Γr
(G(κ,|xr− yr|)jr(yr))dσ(yr)
+
∂
∂nr
?
Γr
(G(κ,|xr− yr|)divΓrjr(yr))dσ(yr)
Proceeding by the same way than with the operator PrCr
PrMr
r
is defined for j = ∇Γp + curlΓq ∈ TH−1
κP−1
r, we obtain that the operator
2(divΓ,Γ) by:
κP−1
PrMr
κP−1
rj = ∇ΓP′(r) + rotΓQ′(r),
with
P′(r)(x) =
?τr∆−1
Γrτ−1
r
?τr
?
κ2
?
Γr
?
∂nr(·)G(κ,| · −yr|)(∆Γrτ−1
nr(·) ·?G(κ,| · −yr|)curlΓrτ−1
nr(·) ·?G(κ,| · −yr|)∇Γrτ−1
∂
r q(yr)?dσ(yr)
r p(yr)?dσ(yr)
r p)(yr)dσ(yr)
+κ2
Γr
+
?
Γr
?
(x),
and
Q′r(x) =
?τr∆−1
ΓrcurlΓrτ−1
r
?τr
((∇G(κ,| · −yr|) · nr(·))(∇Γrτ−1
??
Γr
(∇G(κ,| · −yr|) · nr(·))(rotΓrτ−1
r q)(yr)dσ(yr)
+
?
Γr
r p)(yr)dσ(yr)
−
?
∇G(κ,| · −yr|)?nr(·) · (∇Γrτ−1
Γr
∇G(κ,| · −yr|)?nr(·) · (rotΓrτ−1
r q)(yr)?dσ(yr)
r p)(yr)?dσ(yr)−
?
Γr
?
(x).
These operators are composed of boundary integral operators with weakly singular and
pseudo-homogeneous kernels of class -1 and of the surface differential operators defined
in section 2. By a change of variables in the integral, we then have to study the differen-
tiability properties of the applications
r
r
r
r
r
?→
?→
?→
?→
?→
τr∇Γrτ−1
τrcurlΓrτ−1
τrdivΓrτ−1
τrcurlΓrτ−1
τr∆Γrτ−1
r
r
r
r
r
27
Page 28
5.2Gâteaux differentiability of the surface differential operators
Lemma 5.3 The application
G :B∞
ǫ
r
→
?→
Lc(Hs+1(Γ),Hs(Γ))
τr∇Γrτ−1
r
is C∞-Gâteaux differentiable and its first derivative at r0is defined for ξ ∈ C∞
by
∂G
∂r[r0,ξ]u = −[G(r0)ξ]G(r0)u + (G(r0)u · [G(r0)ξ]N(r0))N(r0).
b(R3,R3)
Remark 5.4 Note that we can write∂N
tive of N and G can be expressed in function of N and G we obtain the Gâteaux derivative
of all order iteratively.
∂r[r0,ξ] = −[G(r0)ξ]N(r0). Since the first deriva-
Proof.
Gâteaux differentiability of G we have to prove the C∞- Gâteaux differentiability of the
application
?
In accordance to the definition (2.1) and the lemma 4.2, to prove the C∞-
f : r ∈ B∞
ǫ ?→u ?→ τr
?
∇?
τ−1
r u
?
|Γr
?
∈ Lc(Hs+1(Γ),Hs(Γ)).
Let x ∈ Γ, we have
?
τr
∇?
τ−1
r u
?
|Γr
(x) = ∇?? u ◦ (I + r)−1?
(I + Dr)−1
|Γr(x + r(x)) =
T(I + Dr)−1
|Γr(x + r(x)) ◦ ∇? u|Γ(x),
and
|Γr(x + r(x)) =?(I + Dr)|Γ(x)?−1.
?→ (I + Dr)|Γ∈ C∞(Γ) is continuous, and C∞-Gâteaux
∂
∂rg[0,ξ] = [Dξ]|Γand its higher order derivatives
ǫ ?→?x ?→ [g(r)(x)]−1?∈ C∞(Γ) is
∂r[r0,ξ] = −h(r0) ◦∂g
The application g : r ∈ B∞
ǫ
differentiable. Its first derivative is
vanish. One can show that the application h : r ∈ B∞
also C∞Gâteaux-differentiable and that we have at r0and in the direction ξ:
∂h
∂r[r0,ξ] ◦ h(r0) = −h(r0) ◦ [Dξ]|Γ◦ h(r0).
and
∂nh
∂rn[r0,ξ1,...,ξn] = (−1)n?
where Snis the permutation groupe of {1,...,n}. Finally we obtain the C∞− Gâteaux
differentiability of f and we have
σ∈Sn
(I + Dr0)−1◦ [τr0Dτ−1
r0ξσ(1)] ◦ ... ◦ [τr0Dτ−1
r0ξσ(n)]
∂f
∂r[r0,ξ]u = −[f(r0)ξ]f(r0)u.
28
Page 29
To obtain the expression of the first derivative of G we have to derive the following
expression:
?
= f(r0)u − (f(r0)u · N(r0))N(r0).
G(r)u = (τr∇Γrτ−1
r u) = τr∇
?
τ−1
r u
?
−
?
τrnr·
?
τr∇
?
?
τ−1
r u
???
τrnr
By lemma 4.2 and the chain and product rules we have
∂G
∂r[r0,ξ] =−[f(r0)ξ]f(r0)u + ([f(r0)ξ]f(r0)u · N(r0))N(r0)
+(f(r0)u · [G(r0)ξ]N(r0))N(r0) + (f(r0)u · N(r0))[G(r0)ξ]N(r0)
We had the first two terms in the right handside, it gives :
∂G
∂r[r0,ξ] =−[G(r0)ξ]f(r0)u + (f(r0)u · N(r0))[G(r0)ξ]N(r0)
+(f(r0)u · [G(r0)ξ]N(r0))N(r0)
=−[G(r0)ξ]G(r0)u + (f(r0)u · [G(r0)ξ]N(r0))N(r0).
To conclude it suffice to note that (f(r0)u · [G(r0)ξ]N(r0)) = (G(r0)u · [G(r0)ξ]N(r0)).
?
Lemma 5.5 The application
D :B∞
ǫ
r
→
?→
Lc(Hs+1(Γ,R3),Hs(Γ))
τrdivΓrτ−1
r
is C∞-Gâteaux differentiable and its first derivative at r0is defined for ξ ∈ C∞
by
∂D
∂r[r0,ξ]u = −Trace([G(r0)ξ][G(r0)u]) + ([G(r0)u]N(r0) · [G(r0)ξ]N(r0)).
b(R3,R3)
Proof. For u ∈ Hs+1(Γ,Rn) we have D(r)u = Trace(G(r)u). Then we use the differen-
tiation rules.
Remark 5.6 Since the first derivative of D is composed of G and N and the first deriva-
tive of J is composed of J and D, we can obtain an expression of higher order derivatives
of the jacobian iteratively.
?
Lemma 5.7 The application
R :B∞
ǫ
r
→
?→
Lc(Hs+1(Γ),Hs(Γ))
τrcurlΓrτ−1
r
is C∞-Gâteaux differentiable and its first derivative at r0is defined for ξ ∈ C∞
by
∂R
∂r[r0,ξ]u =T[G(r0)ξ]R(r0)u − D(r0)ξ · R(r0)u.
b(R3,R3)
29
Page 30
Proof.
lemmas 4.2 and 5.3 this application is C∞Gâteaux differentiable. We have in r0and in
the direction ξ ∈ C∞(Γ,R3)
Let u ∈ Hs+1(Γ). By definition, we have R(r0)u = G(r0)u × N(r0). By
∂R
∂r[r0,ξ]u = −T[G(r0)ξ]G(r0)u × N(r0) − G(r0)u × [G(r0)ξ]N(r0).
NB: recall that given a (3 × 3) matrix A and vectors b and c we have
Ab × c + b × Ac = Trace(A)(b × c) −TA(b × c).
We deduce the expression of the first derivatives with A = −[G(r0)ξ], b = G(r0)u et
c = N(r0).
?
Lemma 5.8 The application
R :B∞
ǫ
r
→
?→
Lc(Hs+1(Γ),Hs(Γ))
τrcurlΓrτ−1
r
is C∞-Gâteaux differentiable and its first derivative at r0is defined for ξ ∈ C∞
by
∂R
∂r[r0,ξ]u = −
i=1
b(R3,R3)
3
?
?G(r0)ξi· R(r0)ui
?− D(r0)ξ · R(r0)u
where u = (u1,u2,u3) and ξ = (ξ1,ξ2,ξ3).
Proof.
Let u ∈ Hs+1(Γ,R3). By definition of the surface rotational we have
R(r0)u = −Trace(R(r0)u).
We deduce the C∞differentiability of R and the first derivative in r0in the direction ξ is
?∂R
= − Trace
∂R
∂r[r0,ξ]u = − Trace
∂r[r0,ξ]u
?T[G(r0)ξ][R(r0)u]
(G(r0)ξi· R(r0)ui) − D(r0)ξ · R(r0)u.
?
?
− D(r0)ξ · Trace(−R(r0)u])
= −
3
?
i=1
?
Here again we can obtain higher order derivatives of these operators iteratively.
Remark 5.9 One can see that we do not need more than the first derivative of the defor-
mations ξ. Thus these results hold true for boundaries and deformations of class Ck+1,
k ∈ N∗with differential operators considered in Lc(Ck+1(Γ),Ck(Γ)).
30
Page 31
When ξ = n we obtain the commutators
∂
∂n(∇Γu) − ∇Γ
?∂
?∂
?∂
?∂
∂nu
?
?
?
?
= − RΓ∇Γu
∂
∂n(curlΓu) − curlΓ
∂
∂n(divΓu) − divΓ
∂
∂n(curlΓu) − curlΓ
∂nu= RΓcurlΓu − HΓcurlΓu
∂nu
= − Trace(RΓ[∇Γu])
∂nu
= − Trace(RΓ[curlΓu]) − HΓcurlΓu
(5.9)
where RΓ= [∇Γn] and HΓ= divΓn.
From the precedent results we have:
Lemma 5.10 The application
L :B∞
ǫ
r
→
?→
Lc(Hs+2(Γ),Hs(Γ))
τr∆Γrτ−1
r
is C∞-Gâteaux differentiable.
Proof.
It suffice to write:
τr∆Γrτ−1
r
= (τrdivΓrτ−1
r )(τr∇Γrτ−1
r ) = −(τrcurlΓrτ−1
r )(τrcurlΓrτ−1
r ).
The operators τr∆Γrτ−1
r
is composed of operators infinitely Gâteaux differentiable.
?
View the integral representations of the operators PrCr
to study the Gâteaux differentiability of the applications r ?→ τr∆−1
r ?→ τr∆−1
and continuous from Hs+1(Γr) to Hs
uous from THs+1(Γr) in Hs
To use the chain rules, it is important to construct derivatives in r = 0 between the
spaces Hs+1(Γ) and Hs
Hs
Laplace-Beltrami operator. As an alternative we use the :
κP−1
r
and PrMr
κP−1
r
we have
ΓrcurlΓrτ−1
r
and
ΓrcurlΓrτ−1
r . We have seen that for r ∈ B∞
∗(Γr), that the operator divΓris linear and contin-
∗(Γr) and that ∆−1
ǫ
the operator curlΓris linear
Γris defined from Hs
∗(Γr) in Hs+2(Γr)/R.
∗(Γ) for the scalar curl operator, between the spaces THs+1(Γ) and
∗(Γ) for the divergence operator and between the spaces Hs
∗(Γ) and Hs+2(Γ)/R for the
Proposition 5.11 Let u be a scalar function defined on Γr. Then ur∈ Hs
only if Jrτrur= Jrur◦ (I + r) ∈ Hs
∗(Γ).
∗(Γr) if and
As a consequence the applications r ?→ JrτrcurlΓrτ−1
well-defined from Hs+1(Γ) and THs+1(Γ) respectively to Hs
r
and r ?→ JrτrdivΓrπ−1(r) are
∗(Γ).
Lemma 5.12 The applications
B∞
ǫ
r
→
?→
Lc(Hs+1(Γ),Hs
JrτrcurlΓrτ−1
∗(Γ))
r
and
B∞
ǫ
r
→
?→
Lc(THs+1(Γ),Hs
JrτrdivΓrπ−1(r)
∗(Γ))
31
Page 32
are C∞-Gâteaux differentiable end their first derivatives at r = 0 defined for ξ ∈ C∞
by
∂JrτrcurlΓrτ−1
r
∂r
b(R3,R3)
[0,ξ]u = −
3
?
i=1
∇Γξi· curlΓui.
and
∂JrτrdivΓrπ−1(r)
∂r
[0,ξ]u =−Trace([∇Γξ]∇Γu) + divΓξ divΓu + ([∇Γu]n · [∇Γξ]n)
+(u · [∇Γξ]n)HΓ.
Proof.
Let u ∈ THs+1(Γ) We have
∂τrπ−1(r)
∂r
[0,ξ]u = (u · [∇Γξ]n)n.
Next we use the lemma 4.1, 5.5 and 5.8. For u ∈ Hs(Γ), it is clear that?3
L2scalar product. An other argument whithout using the explicit form of the derivatives
is : for all u ∈ Hs+1(Γ), we derive the application
?
and for u ∈ THs+1(Γ) we derive the application
?
i=1∇Γξi·curlΓui
is of vanishing mean value since the space ∇ΓHs(Γ) is orthogonal to curlΓHs(Γ) for the
r ?→
Γ
JrτrcurlΓrτ−1
r udσ ≡ 0.
r ?→
Γ
JrτrdivΓrπ−1(r)udσ ≡ 0.
?
Let us note that ur∈ Hs(Γr)/R if and only if τrur∈ Hs(Γ)/R.
Lemma 5.13 The application
B∞
ǫ
r
→
?→
Lc(Hs+2
∗
(Γ),Hs(Γ)/R)
τr∆−1
Γrτ−1
r J−1
r
is C∞-Gâteaux differentiable.
Proof. We have seen in section 2, that the Laplace-Beltrami operator is invertible from
Hs+2(Γr)/R to Hs
Hs
and that we have
∂ τr∆−1
r
∂r
∗(Γr). As a consequence Jrτr∆Γrτ−1
∗(Γ). By lemma 3.5 we deduce that r ?→ τr∆−1
r
is invertible from Hs+2(Γ)/R to
r J−1
r
is C∞-Gâteaux differentiable
Γrτ−1
Γrτ−1
r J−1
[0,ξ] = −∆−1
Γ◦
?∂ Jrτr∆Γrτ−1
r
∂r
[0,ξ]
?
◦ ∆−1
Γ.
?
Now we have all the tools to establish the differentiability properties of the electro-
magnetic boundary integral operators and then of the solution to the dielectric scattering
problem.
32
Page 33
5.3Shape derivatives of the solution to the dielectric problem
For more simplicity in the writing we use the following notations :
ΨEκ(r) = Ψr
EκP−1
r, ΨMκ(r) = Ψr
MκP−1
r, Cκ(r) = PrCr
κP−1
r, et Mκ(r) = PrMr
κP−1
r.
Theorem 5.14 The applications
B∞
ǫ
→
?→
?→
Lc(TH−1
ΨEκ(r)
ΨMκ(r)
2(divΓ,Γ),H(curl,Kp))
r
r
are infinitely Gâteaux differentiable. Moreover, their first derivative at r = 0 can be ex-
tended in linear an bounded operators from TH
2(divΓ,Γ) the potentials∂ΨEκ
∂r
equations
curlcurlu − κ2u = 0
1
2(divΓ,Γ) to H(curl,Ω) and Hloc(curl,Ωc)
[0,ξ]j et∂ΨMκ
∂r
and given j ∈ TH
1
[0,ξ]j satisfy the Maxwell’s
in Ω and Ωcand the Silver-Müller condition.
Proof.
that ΨEκ(r)j and ΨMκ(r)j can be written:
Let j ∈ TH−1
2(divΓ,Γ) and ∇Γp+curlΓq its Helmholtz decomposition. Recall
ΨEκ(r)j =κΨr
ΨEκ(r)j =curlψr
κτ−1
r (τrPr−1j) − κ−1∇Ψr
κτ−1
κτ−1
r (τr∆Γrτ−1
r p),
r (τrPr−1j).
By composition of differentiable applications, we deduce that r ?→ ΨEκ(r) and r ?→ ΨMκ(r)
are infinitely Gâteaux differentiable far from the boundary and that their first derivatives
are continuous from TH
1
2(divΓ,Γ) to L2(Ω) ∪ L2
loc(Ωc). Recall that we have,
curlΨEκ(r)j = κΨMκ(r)j and curlΨMκ(r)j = κΨEκ(r)j.
Far from the boundary we can invert the differentiation with respect to x and the deriva-
tion with respect to r and it gives:
curl∂ΨEκ
∂r
[0,ξ]j = κ∂ΨMκ
∂r
[0,ξ]j et curl∂ΨMκ
∂r
[0,ξ]j = κ∂Ψ
∂r[0,ξ]j.
It follows that∂ΨEκ
they satisfy the Maxwell equations and the Silver-Müller condition.
∂r
[0,ξ]j and∂ΨMκ
∂r
[0,ξ]j are in H(curl,Ω) ∪ Hloc(curl,Ωc) and that
?
We recall that the operator Cκ(r) admit the following representation :
Cκ(r) = PrCr
κPr−1j = ∇ΓP(r) + curlΓQ(r),
(5.10)
33
Page 34
where
P(r) = −κ (Jrτr∆Γrτ−1
r )−1(JrτrcurlΓrτ−1
r )?τrVr
κτ−1
r
???τr∇Γrτ−1
r p?+?τrcurlΓrτ−1
r q??
and Q(r) =
−κ (Jrτr∆Γrτ−1
+κ−1?τrVr
Remark 5.15 Let j ∈ THs(divΓ,Γ) and ∇Γ p + curlΓ q its helmholtz decomposition.
We want to derive:
r )−1(JrτrdivΓrπ−1(r))π(r)?τrVr
κτ−1
r
κτ−1
r
???τr∇Γrτ−1
r p?+?τrcurlΓrτ−1
r q??
??τr∆Γrτ−1
r p?.
PrCr
κP−1
rj
= PrCr
= Pr(∇ΓrPr+ curlΓrQr)
= ∇ΓP(r) + curlΓQ(r).
κ(∇Γrτ−1
r p + curlΓrτ−1
r q)
We have:
∂PrCr
κPr−1j
∂r
[0,ξ] = ∇Γ∂P
∂r[0,ξ] + rotΓ∂Q
κPr−1j is given by the derivatives of the
∂r[0,ξ].
The derivative with respect to r à r de PrCr
functions P(r) = τr(Pr) and of Q(r) = τr(Qr).
We also have
∂ π(r)f(r)
∂r
[0,ξ] = π(0)∂ f(r)
∂r
[0,ξ]
By composition of infinite differentiable applications we obtain the
Theorem 5.16 The application:
B∞
ǫ
r
→
Lc
PrCr
?
TH−1
2(divΓ,Γ),TH−1
2(divΓ,Γ)
?
?→
κP−1
r
is infinitely Gâteaux differentiable.
Recall that the operator PrMr
κPr−1admit the following representation :
PrMr
κPr−1j = ∇ΓP′(r) + curlΓQ′(r),
where
P′(r) =
?Jrτr∆Γrτ−1
r
?−1(κ2Jrτrnr· (τrVr
r
κτ−1
r )??τr∇Γrτ−1
r p)
r p?+?τrcurlΓrτ−1
r q??
+?Jrτr∆Γrτ−1
?−1(JrτrDr
κτ−1
r )(τr∆Γrτ−1
and Q′(r) =
?Jrτr∆Γrτ−1
r
?−1(JrτrcurlΓrτ−1
r )(τr(Br
κ− Dr
κ)τ−1
r )??τr∇Γrτ−1
r p?+?τrcurlΓrτ−1
r q??
34
Page 35
with
τrBr
kP−1
rj =τr
??
Γr
?
∇G(κ,| · −yr|)?nr(·) · (∇Γrτ−1
∇G(κ,| · −yr|)?nr(·) · (curlΓrτ−1
r p)(yr)?dσ(yr)
r q)(yr)?dσ(yr)
+
Γr
??
.
Theorem 5.17 The application:
B∞
ǫ
r
→
Lc
PrMκP−1
?
TH−1
2(divΓ,Γ),TH−1
2(divΓ,Γ)
?
?→
r
is infinitely Gâteaux differentiable and the Gâteaux derivatives have the same regularity
than Mκso that it is compact.
Proof.
infinite Gâteaux differentiability of the application
By composition of infinite differentiable applications it remains to prove the
Bδ
r
→
Lc
τrBr
?
TH−1
2(divΓ,Γ),H
1
2(Γ)
?
?→
κP−1
r.
The function (x,y−x) ?→ ∇G(κ,|x−y|) is pseudo-homogeneous of class 0. We then have
to prove that for any fixed (x,y) ∈ (Γ × Γ)∗and any function p ∈ H
derivatives of
r ?→ (τrnr)(x) ·?τr∇Γrτ−1
behave as |x − y|2when x − y tends to zero. To do so, either we write
(τrnr)(x) ·?τr∇Γrτ−1
or we use lemmas 4.2 and 2.1.
3
2(Γ) the Gâteaux
r p?(y)
r p?(y) = ((τrnr)(x) − (τrnr)(y)) ·?τr∇Γrτ−1
r p?(y)
?
Theorem 5.18 Assume that :
1) Einc∈ H1
2) the applications
loc(curl,R3) and
B∞
ǫ
→
TH−1
2(divΓ,Γ)
nr× Einc
?
r ?→
Pr
?
|Γr
?
r ?→
Pr
nr×?curlEinc?
|Γr
?
are Gâteaux differentiable at r = 0. Then the application mapping r onto the solution
E(r) = E(Ωr) ∈ H(curl,Ω) ∪ Hloc(curl,Ωc) to the scattering problem the obstacle Ωris
Gâteaux differentiable at r = 0.
35
Page 36
Proof.
Es:
By composition of differentiable applications. We write for the exterior field
∂Es
∂r[0,ξ] =
?
+ (−ΨEκe− iηΨMκeC∗
−∂ΨEκe
∂r
[0,ξ] − iη∂ΨMκe
∂r
[0,ξ]C0− iηΨMκe
?
0)S−1
?
?
∂C0
∂r[0,ξ]
?
j
0)S−1
−∂S
∂r[0,ξ]j
−ρ∂Mκi
∂r
?1
?
+ (−ΨEκe− iηΨMκeC∗
?
[0,ξ]γDEinc−∂Cκi
?∂Prγr
∂Prγr
∂r
∂r
[0,ξ]γNκeEinc
?
?
+ (−ΨEκe− iηΨMκeC∗
0)S−1
−ρ
2+ Mκi
DEinc
∂r
?
[0,ξ]
+ (−ΨEκe− iηΨMκeC∗
0)S−1
−Cκi
NκeEinc
[0,ξ].
The condition 1) guarantees that the solution j ∈ TH
the right handside are in Hloc(curl,Ωc) and the second condition guarantees that the last
two term is in Hloc(curl,Ωc). Of the same for the interior field we write:
1
2(divΓ,Γ) so that the first term in
∂Ei
∂r[0,ξ] = −1
ρ
∂ΨEκi
∂r
[0,ξ]γc
Nκe
?Es+ Einc?−∂ΨMκi
∂r
[0,ξ]γc
D
?Es+ Einc?
∂Prγr
D
−1
ρΨEκi
∂Prγr
Nκe
?Es(r) + Einc?
∂r
?Es+ Einc?and γc
[0,ξ] − ΨMκi
?Es(r) + Einc?
∂r
[0,ξ]
The condition 1) guarantees that γc
so that the first two terms are in H(curl,Ω) and the second condition guarantees that
the last two term is in H(curl,Ω).
Nκe
D
?Es+ Einc?are in TH
1
2(divΓ,Γ)
?
Theorem 5.19 The application mapping r to the far field pattern E∞(Ωr) ∈ TC∞(S2)
of the solution to the scattering problem the obstacle Ωris C∞-Gâteaux differentiable.
5.4Characterisation of the first derivative
The following theorem give a caracterisation of the first Gâteaux derivative of r ?→ E(r)
in r = 0.
Theorem 5.20 Under the hypothesis of theorem 5.18 the first derivative at r = 0 in the
direction ξ solve the following scattering problem :
curlcurl∂Ei
∂r[0,ξ] − κ2
i
∂Ei
∂r[0,ξ] = 0
∂Es
∂r[0,ξ] = 0
curlcurl∂Es
∂r[0,ξ] − κ2
e
(5.11)
36
Page 37
with the boundary conditions :
n ×∂Ei
∂r[0,ξ] − n ×∂Es
en × curl∂Es
∂r[0,ξ] = gD
µ−1
in × curl∂Ei
∂r[0,ξ] − µ−1
∂r[0,ξ] = gN,
(5.12)
where
gD= − (ξ · n)n ×
+ curlΓ(ξ · n) n ·?Ei− Es− Einc?,
∂
∂n
?Ei− Es− Einc?
and
gN= − (ξ · n)n ×
+ curlΓ(ξ · n) n ·?µ−1
and where∂Es
∂r[0,ξ] satisfies the Silver-Müller condition.
∂
∂n
?µ−1
i
curlEi− µ−1
e curlEs− µ−1
e curlEinc?
i
curlEi− µ−1
e curlEs− µ−1
e curlEinc?.
Proof.
their Gâteaux derivatives satisfy the Maxwell’ equations and the Silver-Müller condition.
It remains to compute the boundary conditions. We could use the integral representation
as Potthast did but it would need to write too long formula. For x ∈ Γ we derive in r = 0
the expression:
nr(x + r(x)) ×?Ei(r)(x + r(x)) − Es(r)(x + r(x)) − Einc(x + r(x))?= 0.
It gives in the direction ξ:
We have shown in the previous paragraph that the potential operators and
(5.13)
0 =
∂τrnr
∂r
[0,ξ](x) ×?Ei(x) − Es(x) − Einc(x)?
+n(x) ×
+n ×?ξ(x) · ∇?Ei− Es− Einc??.
[0,ξ](x) = −[∇Γξ]n and we use
?∂Ei
∂r[0,ξ](x) −∂Es
∂r[0,ξ](x)
?
We recall that∂τrnr
∂r
∇u = ∇Γu +
?∂u
∂n
?
n.
We obtain :
n(x) ×
?∂Ei
∂r[0,ξ](x) −∂Es
∂r[0,ξ](x)
?
=[∇Γξ]n ×?Ei(x) − Es(x) − Einc(x)?
∂
∂n
−n ×?ξ(x) · ∇Γ
−(ξ · n)n ×
?Ei(x) − Es(x) − Einc(x)??
?Ei(x) − Es(x) − Einc(x)?.
37
Page 38
Since the tangential component of Ei− Es− Eincvanish we have:
?ξ(x) · ∇Γ
and
[∇Γξ]n ×?Ei(x) − Es(x) − Einc(x)?= ([∇Γξ]n) × n?Ei(x) − Es(x) − Einc(x)?· n.
For regular surface we have ∇Γn =T∇Γn and
([∇Γξ]n) × n − n ×??T∇Γn?ξ?= curlΓ(ξ · n).
We deduce the first boundary conditions. The second boundary condition corresponds to
the same computation with the magnetic fields.
?Ei(x) − Es(x) − Einc(x)??=??T∇Γn?ξ??n ·?Ei(x) − Es(x) − Einc(x)??
?
Using the commutators (5.9) one can verify that the solution of this problem is in
Hloc(curl,Ω) since the trace
u ∈ H1(curl,Ω) ?→ n ×
∂
∂nu ∈ TH−1
2(divΓ,Γ)
is linear and continuous.
Conclusion
In this paper we have presented a complete shape differentiability analysis of the solu-
tion to the dielectric scattering problem using the boundary integral equation approach.
These results can be extended to many others electromagnetic boundary value problems.
Thanks to the numerous computations of Gâteaux derivatives we obtain two alternatives
to compute the first shape derivative of the solution : either we derive the integral repre-
sentation or we solve the new boundary value problem associated to the shape derivatives
with boundary integral equation method. Whereas this last alternative needs boundaries
of class C2at least since it appears any derivatives of the normal vector, many results in
this paper are still available for Lipschitz domains as for example the computations of the
Gâteaux derivatives of all the surface differential operators of order 1 with deformations
of class C1only and other functionals viewed in section 4. One can find in the litterature,
the theory of pseudo-differential operators on Lipschitz domain [34], it remains to find
the optimal regularity of the deformations in order that this integral operators are still
Gâteaux differentiable. According to the Helmholtz decomposition we have on Lipschitz
domain:
TH−1
2(divΓ,Γ) = ∇Γ(H(Γ))
?
curlΓ
?
H
1
2(Γ)/R
?
.
where
H(Γ) = {u ∈ H1(Γ)\R; ∆Γ∈ H−1
2
∗
(Γ)}.
If we want to extend the result to Lipschitz domain, we have to construct another invert-
ible operator between H(Γr) and H(Γ).
38
Page 39
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