# On the shape derivative of polygonal inclusions in the conductivity problem

Martin Hanke Institut für Mathematik, Johannes Gutenberg-Universität Mainz,
55099 Mainz, Germany (hanke@math.uni-mainz.de).

###### Abstract

We consider the conductivity problem for a homogeneous body with an inclusion
of a different, but known, conductivity. Our interest concerns the associated
shape derivative, i.e., the derivative of the corresponding electrostatic
potential with respect to the shape of the inclusion. For a smooth inclusion
it is known that the shape derivative is the solution of a specific
inhomogeneous transmission problem. We show that this characterization of the
shape derivative is also valid when the inclusion is a polygonal domain, but
due to singularities at the vertices of the polygon, the shape derivative
fails to belong to $H^{1}$ in this case.

###### keywords:

Impedance tomography, Lipschitz domain, transmission problem, shape
optimization, corner singularities

###### AMS:

35B65, 49N60, 49Q10, 65N21

Last modified.

##  1 Introduction

The inverse conductivity problem aims at information about the spatial
electric conductivity distribution within a body, which is only accessible to
electrostatic or quasi-electrostatic measurements at its boundary. For
example, in electric impedance tomography a series of probing currents is
applied through electrodes attached to the surface of the body, and the
resulting boundary potentials are measured (or the other way round). As shown
by Astala and Päivärinta [2] the full set of these current/voltage pairs
completely determines an isotropic conductivity distribution in two space
dimensions. See the monographs by Mueller and Siltanen [26] and Kirsch [20]
for an exposition of electric impedance tomography. Except for the D-bar [21]
and the layer stripping methods [30] most of the pertinent inversion
algorithms are iterative in nature: The spatial conductivity in question is
represented by an element from the nonnegative cone of a reasonable infinite
dimensional vector space of functions, and in each iteration the gradient of
some loss function is used to update the current approximation of the
conductivity distribution; cf., e.g., [7, 10, 27] for some early references.
This is the method of choice when the material is very inhomogeneous.

When the object is homogeneous, except for an inclusion of some other known
material say, then one can hope that only one or a few pairs of Cauchy data
may suffice to determine the location and the shape of the inclusion.
Sophisticated noniterative reconstruction methods like pole fitting methods
[3, 19, 13] or the computation of the convex source support [15] have been
developed for this setting, but iterative methods can also be an option,
provided they exploit gradient information with respect to the shape of the
inclusion (or its parameterization).

Such derivatives have been termed _shape derivative_ or _domain derivative_ ,
and have originally been developed in the optimization community, cf., e.g.,
[29, 16]. The idea is to analyze the impact of a pointwise perturbation of the
boundary of the inclusion in the direction of a vector field that is attached
to it. Then a gradient descent or Newton type method can be used to shift and
deform an approximate inclusion so as to match the given data as good as
possible.

Shape derivatives for inverse conductivity problems have been determined in
the late 90’s of the previous century. However, up to recently their rigorous
foundation has been limited to inclusions with a certain smoothness, e.g.,
$C^{1,1}$-domains. For this smooth case it has been shown by Hettlich and
Rundell [17] that the shape derivative of an inclusion with prescribed
conductivity can be characterized via the solution of a transmission problem
with distributed sources sitting on the boundary of the inclusion: The
solution of this differential equation provides the perturbation of the
electric potential at each individual point within and on the boundary of the
body under consideration. More details and references follow in the subsequent
section.

In the optimization community shape derivatives for less regular domains,
e.g., Lipschitz domains, have been studied, e.g., by Delfour and Zolésio [9],
Lamboley, Novruzi, and Pierre [24], and Laurain [23]; these works focus on the
derivative of certain integrated shape functionals. In the application that we
are interested in, such a functional can be, for example, the inner product of
the given data with a certain test function on the boundary of the two-
dimensional body. For this particular case Beretta, Francini, and Vessella [5]
have determined in a _tour de force_ the impact of a perturbation of a
conducting polygonal inclusion. In a subsequent paper, Beretta, Micheletti,
Perotto, and Santacesaria [6] gave a somewhat more elegant derivation of the
same result using the established shape derivative theory. Still, this
functional only provides the shape derivative of the given data in a weak
(i.e., variational) form, which lacks the interpretation of [17] via an
underlying transmission problem. Further, the impact of the perturbation of
the polygon on the electric potential in the interior of the body (i.e., the
state variable) remained unresolved.

In this paper we show that the same transmission problem as in the smooth case
describes the sensitivity of the electric potential at each point within the
body; its trace on the outer boundary provides an explicit definition of the
shape derivative of the electrostatic measurements. Due to singularities at
the vertices of the polygon, the solution of this transmission problem is less
regular than in the smooth case. This presents some difficulties in proving
the existence of a solution, which can be overcome by using a technique which
goes back to Kondratiev [22] (see also Grisvard [12]), and which has also been
employed in [23].

Our results are relevant for investigating the stability of numerical
algorithms for solving the inverse problem. As we will show in a companion
paper [14], the outcome of the present work implies that the determination of
a polygon from two pairs of current/voltage boundary measurements – which is
possible according to Seo [28] – is actually well-posed, i.e., the inverse
operator is locally Lipschitz continuous.

The outline of this paper is as follows. In Section 2 we set up the problem
under consideration and review known results about the associated shape
derivative. After that we focus on polygonal inclusions only. In Section 3 we
recapitulate the properties of the electric potential for such inclusions.
Then, in Section 4 we derive some estimates for the shape derivative of the
electrostatic boundary measurements determined in [5, 6], and provide
preliminary results about the shape derivative of the electric potential
within the body. In Section 5 we turn to the transmission problem formulated
in [17], and show that it has a unique solution when the inclusion is a
polygon, and we determine its regularity. That this solution coincides with
the shape derivative of the electric potential is the topic of Section 6. In
the final section we briefly discuss the case of an insulating or a perfectly
conducting polygonal inclusion; all our results essentially extend to these
two degenerate cases.

##  2 The conductivity problem and associated shape derivatives

Let $\mathchar 266\relax\subset{\mathord{\mathbb{R}}}^{2}$ be a simply
connected Lipschitz domain, and ${\mathscr{D}}$ be another Lipschitz domain
with simply connected closure $\overline{\mathscr{D}}\subset\mathchar
266\relax$. The outer normal vector on $\partial{\mathscr{D}}$ and
$\partial\mathchar 266\relax$, respectively, will be denoted by $\nu$. We call
${\mathscr{D}}$ the inclusion, and we assume that the conductivity in
$\mathchar 266\relax$ is given by

$\sigma(x)\,=\,\begin{cases}k\,,&\text{for $x\in{\mathscr{D}}$}\,,\\\
1\,,&\text{for $x\in\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\end{cases}$ (1)

where $k$ is taken to be greater than zero and different from one; in Section
7, we will briefly treat the degenerate cases $k=0$ and $k=+\infty$,
respectively.

Given a fixed (nontrivial) driving boundary current

$f\,\in\,L^{2}_{\diamond}(\partial\mathchar 266\relax)\,=\,\bigl{\\{}\,f\in
L^{2}(\partial\mathchar 266\relax)\,:\,\int_{\partial\mathchar
266\relax}f\,\mathrm{d}s=0\,\bigr{\\}}\,,$

the induced quasistatic electric potential

$u\,\in\,H^{1}_{\diamond}(\mathchar 266\relax)\,=\,\bigl{\\{}\,u\in
H^{1}(\mathchar 266\relax)\,:\,\int_{\partial\mathchar
266\relax}u\,\mathrm{d}s=0\,\bigr{\\}}$

is the unique solution of the conductivity equation

$\nabla\cdot(\sigma\nabla u)\,=\,0\quad\text{in $\mathchar
266\relax$}\,,\qquad\frac{\partial}{\partial\nu}u\,=\,f\quad\text{on
$\partial\mathchar 266\relax$}\,,$ (2)

with vanishing mean on $\partial\mathchar 266\relax$. The restrictions
$u^{-}=u|_{\mathscr{D}}$ and $u^{+}=u|_{\mathchar
266\relax\setminus\overline{\mathscr{D}}}$ are both harmonic (smooth)
functions, which satisfy the transmission conditions

$\bigl{[}u\bigr{]}_{\partial{\mathscr{D}}}=u^{+}|_{\partial{\mathscr{D}}}-u^{-}|_{\partial{\mathscr{D}}}=0\qquad\text{and}\qquad\bigl{[}D_{\nu}u\bigr{]}_{\partial{\mathscr{D}}}=\frac{\partial}{\partial\nu}u^{+}\,-\,k\,\frac{\partial}{\partial\nu}u^{-}=0$
(3)

on $\partial{\mathscr{D}}$. Alternatively, $u$ can be characterized as the
unique solution in $H^{1}_{\diamond}(\mathchar 266\relax)$ of the variational
problem

$\int_{\mathchar 266\relax}\sigma\,\nabla u\cdot\nabla
w\,\mathrm{d}x\,=\,\int_{\partial\mathchar
266\relax}fw\,\mathrm{d}s\qquad\text{for all $w\in H^{1}(\mathchar
266\relax)$}\,.$ (4)

Assuming that the (normalized) background conductivity and its anomalous value
$k>0$ in the inclusion are fixed (and known), we are interested in the
sensitivity of the boundary data

$u|_{\partial\mathchar 266\relax}\,=:\,\mathchar
259\relax_{f}({\mathscr{D}})\,\in\,L^{2}_{\diamond}(\partial\mathchar
266\relax)$

of the electrostatic potential with respect to perturbations of
${\mathscr{D}}$. To be specific, assume that a vector field
$h:\partial{\mathscr{D}}\to{\mathord{\mathbb{R}}}^{2}$ is prescribed on the
Lipschitz boundary of ${\mathscr{D}}$, say, and define the perturbation

$\mathchar
256\relax_{h}\,=\,\bigl{\\{}x+h(x)\,:\,x\in\partial{\mathscr{D}}\\}$

of $\partial{\mathscr{D}}$ in the direction of $h$. When $h$ belongs to
$W^{1,\infty}(\partial{\mathscr{D}})$ then $\mathchar 256\relax_{h}$ is a
Jordan curve in $\mathchar 266\relax$ for $h$ sufficiently small, and we
denote its interior domain by ${\mathscr{D}}_{h}$. Let $u_{h}$ be the solution
of the corresponding problem (2) with ${\mathscr{D}}$ replaced by
${\mathscr{D}}_{h}$; then the so-called shape derivative $u_{h}^{\prime}$ of
the electric potential $u=u({\mathscr{D}})$ in direction $h$ is given by

$u_{h}^{\prime}\,=\,\lim_{t\to 0}\frac{u_{th}-u}{t}\qquad\text{in $\mathchar
266\relax$}\,,$ (5)

provided this limit exists. Likewise,

$\partial\mathchar 259\relax_{f}({\mathscr{D}})h\,=\,\lim_{t\to
0}\frac{1}{t}\bigl{(}\mathchar 259\relax_{f}({\mathscr{D}}_{th})-\mathchar
259\relax_{f}({\mathscr{D}})\bigr{)}$

is the shape derivative of $\mathchar 259\relax_{f}({\mathscr{D}})$ in
direction $h$, if the latter limit exists.

Under the assumption that the boundary of ${\mathscr{D}}$ is smooth, Hettlich
and Rundell [17] (see also Afraites, Dambrine, and Kateb [1]) established the
existence of $u_{h}^{\prime}$ and showed that it is the unique solution of the
inhomogeneous transmission problem

$\begin{array}[]{c}\Delta u_{h}^{\prime}\,=\,0\quad\text{in $\mathchar
266\relax\setminus\partial{\mathscr{D}}$}\,,\qquad\dfrac{\partial}{\partial\nu}u_{h}^{\prime}\,=\,0\quad\text{on
$\partial\mathchar 266\relax$}\,,\qquad{\displaystyle\int_{\partial\mathchar
266\relax}u_{h}^{\prime}\,\mathrm{d}s\,=\,0\,,}\\\\[12.91663pt]
\bigl{[}u_{h}^{\prime}\bigr{]}_{\partial{\mathscr{D}}}\,=\,(1-k)(h\cdot\nu)\dfrac{\partial}{\partial\nu}u^{-}\,,\quad\bigl{[}D_{\nu}u_{h}^{\prime}\bigr{]}_{\partial{\mathscr{D}}}\,=\,(1-k)\,\dfrac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\dfrac{\partial}{\partial\tau}u\Bigr{)}\,.\end{array}$
(6)

Due to the $H^{2}$-regularity of the forward problem at either side of the
smooth boundary of ${\mathscr{D}}$, compare, e.g., McLean [25, Theorem 4.20],
the inhomogeneous data for this problem belong to the appropriate Sobelev
spaces $H^{\pm 1/2}(\partial{\mathscr{D}})$, and this garantees existence of a
unique weak solution of (6), whose restriction to $\mathchar
266\relax\setminus\overline{\mathscr{D}}$ belongs to $H^{1}$, so that its
trace $u^{\prime}|_{\partial\mathchar 266\relax}\in
L^{2}_{\diamond}(\partial\mathchar 266\relax)$ is well-defined. If
${\mathscr{D}}$ is merely a Lipschitz domain, then this chain of arguments is
no longer valid, and it is not immediately clear, whether the transmission
problem (6) admits a solution, and if so, in which space. And even if there is
a unique solution of (6), one may still wonder whether this solution is the
shape derivative (5) of $u$.

As a step towards less regular inclusions Beretta, Francini, and Vessella [5]
studied the conductivity equation (2) with a polygonal anomaly and a
perturbation field $h$, whose vector components are linear splines over
$\partial{\mathscr{D}}$ with their nodes attached to the vertices of
${\mathscr{D}}$; i.e., both components of $h$ are affine linear on each of the
edges of ${\mathscr{D}}$ and continuous at the vertices. With such a field the
perturbed domains ${\mathscr{D}}_{h}$ also are polygons – at least for $h$
sufficiently small. It is proved in [5] that the associated shape derivative
$\partial\mathchar 259\relax_{f}({\mathscr{D}})$ of $\mathchar 259\relax_{f}$
at ${\mathscr{D}}$ exists in a Fréchet sense; more precisely they showed that

$\|\mathchar 259\relax_{f}({\mathscr{D}}_{h})-\mathchar
259\relax_{f}({\mathscr{D}})-\partial\mathchar
259\relax_{f}({\mathscr{D}})h\|_{L^{2}(\partial\mathchar
266\relax)}\,\leq\,C\hskip 0.43057pt\|h\|^{1+\delta}$ (7)

for $h$ sufficiently small, where the constants $C>0$ and $\delta\in(0,1)$ are
independent of the particular choice of $h$. (Note that in (7) it is
irrelevant which norm of $h$ is used, because the admissible spline functions
from [5] constitute a finite dimensional vector space.) Finally, the authors
of [5] came up with a weak (variational) definition of this shape derivative,
namely***Beware of the sign error in [5, Theorem 4.6].

$\langle\,\partial\mathchar
259\relax_{f}({\mathscr{D}})h,g\,\rangle_{L^{2}(\mathchar
266\relax)}\,=\,(1-k)\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\nabla
u^{-}\cdot(M\nabla v_{g}^{-})\,\mathrm{d}s$ (8)

for all $g\in L^{2}_{\diamond}(\partial\mathchar 266\relax)$, where

$v_{g}\,=\,\begin{cases}v_{g}^{-}&\text{in ${\mathscr{D}}$}\,,\\\
v_{g}^{+}&\text{in $\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\end{cases}$

is the corresponding solution in $H^{1}_{\diamond}(\mathchar 266\relax)$ of
the conductivity equation

$\nabla\cdot(\sigma\nabla v_{g})\,=\,0\quad\text{in $\mathchar
266\relax$}\,,\qquad\frac{\partial}{\partial\nu}v_{g}\,=\,g\quad\text{on
$\partial\mathchar 266\relax$}\,,$ (9)

with driving current $g$, and $M$ is the symmetric $2\times 2$-matrix with
eigenvalues $1$ and $k$ and corresponding eigenvectors in tangential and
normal directions, respectively; see (19) below for a justification that the
expression (8) is well-defined.

We mention that the analysis in [5] avoids the use of the transmission problem
(6) and the general theory of shape derivatives. An alternative derivation of
(8) on the grounds of this latter theory was subsequently handed in by
Beretta, Micheletti, Perotto, and Santacesaria [6], who determined the so-
called material derivative ${\dot{u}}$ of $u$ in $\mathchar 266\relax$, cf.
(22) below, but the existence of the shape derivative $u^{\prime}$ of $u$
remained unsettled. In Section 4 we will also exploit the material derivative
to show that the estimate (7) for the Taylor remainder of the shape derivative
of the boundary data can be improved to the order $\|h\|^{2}$.

In Sections 5 and 6 we analyze the transmission problem (6) for polygonal
inclusions as treated in [5, 6], and we prove that it has a unique solution
$u_{h}^{\prime}$. In contrast to the smooth case its restrictions to
${\mathscr{D}}$ and to $\mathchar 266\relax\setminus\overline{\mathscr{D}}$
merely belong to some Sobolev space $H^{\gamma}$, where $\gamma\in(1/2,1)$
depends on the interior angles of the vertices of ${\mathscr{D}}$. We further
show that $u_{h}^{\prime}$ has a well-defined trace which satisfies the same
variational problem (8) as $\partial\mathchar 259\relax_{f}({\mathscr{D}})h$;
from this we then conclude that the solution of the transmission problem is
the shape derivative of $u$ in $\mathchar 266\relax$, i.e., that the main
result by Hettlich and Rundell carries over to polygonal inclusions.

##  3 The conductivity problem with polygonal inclusions

We recapitulate some basic facts about the two-dimensional conductivity
problem (2) with a polygonal inclusion ${\mathscr{D}}$ with $n$ vertices,
$n\geq 3$, and a simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$.

We start with some notation: We denote the (relatively open) edges of
${\mathscr{D}}$ by $\mathchar 256\relax_{i}$ and its vertices by $x_{i}$,
$i=1,\dots,n$, with the convention that $x_{i}$ connects $\mathchar
256\relax_{i}$ and $\mathchar 256\relax_{i+1}$, where we identify $\mathchar
256\relax_{n+1}$ with $\mathchar 256\relax_{1}$; likewise we identify
$x_{n+1}$ with $x_{1}$, when necessary. On $\mathchar 256\relax_{i}$ the
tangent vector $\tau$ is pointing in the direction of $x_{i}$. The interior
angle of ${\mathscr{D}}$ at $x_{i}$ is denoted by
$\alpha_{i}\in(0,2\pi)\setminus\\{\pi\\}$, and we use a local coordinate
system

$x\,=\,x_{i}\,+\,\bigl{(}r\cos(\theta_{i}+\theta),r\sin(\theta_{i}+\theta)\bigr{)}\,,\qquad
0<r<r_{i}\,,\ 0\leq\theta<2\pi\,,$ (10)

near $x=x_{i}$, where $\theta_{i}$ is such that $\theta=0$ corresponds to
points on $\mathchar 256\relax_{i+1}$, $\theta=\alpha_{i}$ for $x\in\mathchar
256\relax_{i}$, and the range $0<\theta<\alpha_{i}$ corresponds to points
$x\in{\mathscr{D}}$, while the interval $\alpha_{i}<\theta<2\pi$ corresponds
to points $x\in\mathchar 266\relax\setminus\overline{\mathscr{D}}$. We will
also make use of cut-off functions $\chi_{i}=\chi_{i}(x)\in
C_{0}^{\infty}({\mathord{\mathbb{R}}}^{2})$, which only depend on the distance
$|x-x_{i}|$, are one near $x=x_{i}$ and zero for $|x-x_{i}|\geq r_{i}$.
Without loss of generality we assume that the radius $r_{i}$ is so small that
the supports of any two cut-off functions $\chi_{i}$ and $\chi_{j}$ with
$i\neq j$ have no points in common, and that the intersection of
$\partial{\mathscr{D}}$ with the support of $\chi_{i}$ is a subset of
$\mathchar 256\relax_{i}\cup\\{x_{i}\\}\cup\mathchar 256\relax_{i+1}$.

Since the two components $u^{\pm}$ of the solution $u$ of (2) are harmonic and
satisfy the transmission conditions (3) they can both be continued as harmonic
functions across each of the edges $\mathchar 256\relax_{i}$ of
${\mathscr{D}}$ by appropriate reflections. This proves that all their
derivatives extend continuously to every edge from either side. In particular
this implies that $u^{+}|_{\mathchar 256\relax_{i}}=u^{-}|_{\mathchar
256\relax_{i}}$ is a well-defined $C^{\infty}$ function; we therefore drop the
superscripts $\pm$ when dealing with this trace.

We quote from Bellout, Friedman, and Isakov [4] that near a fixed vertex
$x_{i}$ of ${\mathscr{D}}$, $i\in\\{1,\dots,n\\}$, the potential $u$ satisfies

$u(x)\,=\,u(x_{i})\,+\,\sum_{j=1}^{\infty}\beta_{ij}\,y_{ij}(\theta)\,r^{\gamma_{ij}}\,,$
(11)

when using the local coordinate system (10) for

$x\in{\cal B}_{r_{i}}(x_{i})\,=\,\bigl{\\{}x\,:\,|x-x_{i}|<r_{i}\bigr{\\}}\,.$

The functions $y_{ij}\in H^{1}(0,2\pi)$ in (11) are the eigenfunctions of the
eigenvalue problem

$-y^{\prime\prime}=\gamma^{2}y\qquad\text{in
$(0,2\pi)\setminus\\{\alpha_{i}\\}$}\,,$ (12a) subject to the transmission
conditions $\displaystyle\phantom{k^{\prime}\,}y(0+)=y(2\pi-)\,,$
$\displaystyle\phantom{k^{\prime}\,}y(\alpha_{i}-)=y(\alpha_{i}+)\,,$ (12b)
$\displaystyle k\,y^{\prime}(0+)=y^{\prime}(2\pi-)\,,$ $\displaystyle
k\,y^{\prime}(\alpha_{i}-)=y^{\prime}(\alpha_{i}+)\,,$

$\gamma_{ij}^{2}$ are the corresponding eigenvalues in increasing order, and
$\gamma_{ij}$ its square roots. The latter are the nonnegative solutions of
the nonlinear equation

$\bigl{|}\sin\gamma_{ij}(\alpha_{i}-\pi)\bigr{|}\,=\,\lambda\,\bigl{|}\sin\gamma_{ij}\pi\bigr{|}\,,\qquad\lambda\,=\,\Bigl{|}\frac{k+1}{k-1}\Bigr{|}\,.$
(13)

There holds

$\gamma_{i0}\,=\,0\,,\quad\frac{1}{2}\,<\,\gamma_{i1}\,<\,1\,,\quad\text{and}\quad\gamma_{i2}\,>\,1\,,$
(14)

and the eigenfunctions have the form

$y_{ij}(\theta)\,=\,\begin{cases}A_{ij}^{-}\cos\gamma_{ij}\theta\,+\,B_{ij}^{-}\sin\gamma_{ij}\theta\,,&\quad\phantom{\alpha_{i}}0<\theta<\alpha_{i}\,,\\\\[4.30554pt]
A_{ij}^{+}\cos\gamma_{ij}\theta\,+\,B_{ij}^{+}\sin\gamma_{ij}\theta\,,&\quad\phantom{0}\alpha_{i}<\theta<2\pi\,,\end{cases}$

with appropriate values of $A_{ij}^{\pm}$ and $B_{ij}^{\pm}$. In particular,
$y_{i0}$ is constant in the entire interval $(0,2\pi)$. We normalize these
eigenfunctions in such a way that they define an orthonormal basis with
respect to the inner product

$\langle\,y,\widetilde{y}\,\rangle\,=\,\int_{0}^{\alpha_{i}}k\,y(\theta)\widetilde{y}(\theta)\,\mathrm{d}\theta\,+\,\int_{\alpha_{i}}^{2\pi}y(\theta)\widetilde{y}(\theta)\,\mathrm{d}\theta\,.$

From the transmission conditions (12b) it follows that

$\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols c}1&0&-\cos
2\pi\gamma&-\sin 2\pi\gamma\\\ 0&k&\phantom{-}\sin 2\pi\gamma&-\cos
2\pi\gamma\\\
\phantom{-k}\cos\gamma\alpha&\phantom{k}\sin\gamma\alpha&-\cos\gamma\alpha&-\sin\gamma\alpha\\\
-k\sin\gamma\alpha&k\cos\gamma\alpha&\phantom{-}\sin\gamma\alpha&-\cos\gamma\alpha\end{array}\right]\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}A^{-}\\\ B^{-}\\\ A^{+}\\\
B^{+}\end{array}\right]\,=\,\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}0\\\ 0\\\ 0\\\ 0\end{array}\right],$ (15)

where we have omitted all subscripts $i$ and $j$ for the ease of readability;
in order that this system has a nontrivial solution the matrix
$Y\in{\mathord{\mathbb{R}}}^{4\times 4}$ in (15) must be singular. In fact, we
state the following result, which is implicit in Seo [28].

######  Lemma 1.

Let $Y=Y(\gamma,\alpha)$ be given by the $4\times 4$-matrix in (15) with
$\alpha\in(0,2\pi)\setminus\\{\pi\\}$ and $\gamma>0$. Then $Y$ is singular, if
and only if $\gamma$ is a solution of

$\bigl{|}\sin(\gamma(\alpha-\pi))\bigr{|}\,=\,\lambda\,\bigl{|}\sin\gamma\pi\bigr{|}$
(16)

with $\lambda$ as in (13). If $\alpha$ and $\gamma$ satisfy (16) then
$\operatorname{rank}(Y)=3$, unless
$\gamma\in{\mathord{\mathbb{N}}}\setminus\\{1\\}$ and $\alpha=l\pi/\gamma$ for
some $l\in\\{1,\dots,2\gamma-1\\}\setminus\\{\gamma\\}$, in which case
$\operatorname{rank}(Y)=2$ and the corresponding eigenvalue problem (12) has a
two dimensional eigenspace.

###### Proof.

Since we need this result in a slightly different context below (see (40)) we
provide a sketch of its proof. Introducing

$D\,=\,\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols c}1&0\\\
0&k\end{array}\right]\qquad\text{and}\qquad
R(\omega)\,=\,\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}\phantom{-}\cos\omega&\sin\omega\\\
-\sin\omega&\cos\omega\end{array}\right]$

we can write

$Y\,=\,\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}D&-R(2\pi\gamma)\\\ DR(\gamma\alpha)&-R(\gamma\alpha)\end{array}\right],$

and since its $(1,1)$-block $D$ is nonsingular, $Y$ is singular, if and only
if the Schur complement

$\displaystyle S$
$\displaystyle\,=\,DR(\gamma\alpha)D^{-1}R(2\pi\gamma)-R(\gamma\alpha)$
$\displaystyle\,=\,DR(\gamma\pi)\Bigl{(}R(\gamma(\alpha-\pi))D^{-1}R(\gamma\pi)\,-\,R(-\gamma\pi)D^{-1}R(\gamma(\alpha-\pi)\Bigr{)}R(\gamma\pi)$

of $D$ in $Y(\gamma,\alpha)$ is singular. Computing the products in the inner
paranthesis and taking the determinant of their difference it is readily seen
that $S$ is singular, if and only if $\gamma$ is a solution of (16). The Schur
complement $S$ vanishes completely, if and only if
$\gamma\in{\mathord{\mathbb{N}}}$ and $\gamma\alpha$ is an integer multiple of
$\pi$, in which case $y(\theta)=\cos\gamma\theta$ is one associated
eigenfunction of (12), and $y(\theta)=B^{\pm}\sin\gamma\theta$ with
$B^{+}=kB^{-}$ is a second one. ∎

As mentioned in [4] the series (11) can be differentiated termwise to any
order with respect to $\theta$ and $r$ (with one-sided derivatives at
$\theta=\alpha_{i},0,2\pi$, respectively). We thus compute

$\displaystyle\frac{\partial}{\partial\nu}u^{-}(x)$
$\displaystyle\,=\,-\frac{1}{r}\frac{\partial}{\partial\theta}u^{-}(x)\,=\,-\sum_{j=1}^{\infty}\beta_{ij}\gamma_{ij}B_{ij}^{-}r^{\gamma_{ij}-1}\,,\qquad
x\in\mathchar 256\relax_{i+1}\,,$ (17a) and
$\displaystyle\frac{\partial}{\partial\nu}u^{-}(x)$
$\displaystyle\,=\,\frac{1}{r}\frac{\partial}{\partial\theta}u^{-}(x)\,=\,\sum_{j=1}^{\infty}\beta_{ij}\gamma_{ij}(B_{ij}^{-}c_{ij}-A_{ij}^{-}s_{ij})r^{\gamma_{ij}-1}\,,\qquad
x\in\mathchar 256\relax_{i}\,,$ (17b)

for $x$ close to $x_{i}$, where we have set

$c_{ij}=\cos\gamma_{ij}\alpha_{i}\qquad\text{and}\qquad
s_{ij}=\sin\gamma_{ij}\alpha_{i}\,.$

Likewise, we have

$\displaystyle\frac{\partial}{\partial\tau}u(x)$
$\displaystyle\,=\,\frac{\partial}{\partial
r}u(x)\,=\,\sum_{j=1}^{\infty}\beta_{ij}\gamma_{ij}A_{ij}^{-}r^{\gamma_{ij}-1}\,,\qquad
x\in\mathchar 256\relax_{i+1}\,,$ (18a) and
$\displaystyle\frac{\partial}{\partial\tau}u(x)$
$\displaystyle\,=\,-\frac{\partial}{\partial
r}u(x)\,=\,-\sum_{j=1}^{\infty}\beta_{ij}\gamma_{ij}(A_{ij}^{-}c_{ij}+B_{ij}^{-}s_{ij})r^{\gamma_{ij}-1}\,,\qquad
x\in\mathchar 256\relax_{i}\,,$ (18b)

near $x=x_{i}$.

From (14), (17), (18), and (3), and from the fact that the gradient of $u$ has
continuous extensions from either side to every edge of
$\partial{\mathscr{D}}$ it follows that

$\frac{\partial}{\partial\nu}u^{\pm}\Big{|}_{\partial{\mathscr{D}}}\,,\
\frac{\partial}{\partial\tau}u\Big{|}_{\partial{\mathscr{D}}}\,\in\,L^{2}(\partial{\mathscr{D}})\,,$
(19)

and, of course, the same is true for every solution $v_{g}$ of (9); compare
also Escauriaza, Fabes, and Verchota [11]. Accordingly, the right-hand side of
(8) is well-defined.

##  4 The shape derivative of $u=u({\mathscr{D}})$ for a polygonal inclusion
in $\mathchar 266\relax$

In the sequel we investigate the existence of the shape derivative of the
solution $u=u({\mathscr{D}})$ of the conductivity equation (2) at a polygonal
inclusion ${\mathscr{D}}$. We also derive bounds for the corresponding shape
derivative $\partial\mathchar 259\relax_{f}({\mathscr{D}})$ and its associated
Taylor remainder, the latter improving upon the estimate (7) provided in [5].

We start by making the assumption that a vector field $h:\mathchar
266\relax\to{\mathord{\mathbb{R}}}^{2}$ with

$h\,\in\,{W_{c}^{1,\infty}(\mathchar 266\relax)}\,=\,\bigl{\\{}\,h\in
W^{1,\infty}(\mathchar 266\relax)\,:\,\mathop{\operator@font supp}\nolimits
h\subset\mathchar 266\relax\,\bigr{\\}}$

is given, and define

$\mathchar 264\relax_{h}(x)\,=\,x\,+\,h(x)\,,\qquad x\in\mathchar
266\relax\,.$ (20)

Note that $\mathchar 264\relax_{h}:\mathchar 266\relax\to\mathchar 266\relax$
is bijective, if, for example,

$\|h\|_{W^{1,\infty}(\mathchar 266\relax)}\,=\,\sup_{x\in\mathchar
266\relax}\,\bigl{(}\|h(x)\|_{2}\,+\,\|h^{\prime}(x)\|_{2}\bigr{)}\,\leq\,1/2\,,$
(21)

by virtue of Banach’s fixed point theorem.

We now follow the standard argumentation from [29], which has been worked out
in [6] for the present application. Let $h$ satisfy (21), define
${\mathscr{D}}_{h}=\mathchar 264\relax_{h}({\mathscr{D}})$, and denote by
$u_{h}\in H^{1}_{\diamond}(\mathchar 266\relax)$ the solution of the forward
problem

$\nabla\cdot(\sigma_{h}\nabla u_{h})\,=\,0\quad\text{in $\mathchar
266\relax$}\,,\qquad\frac{\partial}{\partial\nu}u_{h}\,=\,f\quad\text{on
$\partial\mathchar 266\relax$}\,,$

where

$\sigma_{h}(x)\,=\,\begin{cases}k\,,&\text{for $x\in{\mathscr{D}}_{h}$}\,,\\\
1\,,&\text{for $x\in\mathchar
266\relax\setminus\overline{\mathscr{D}}_{h}$}\,.\end{cases}$

Introducing

${\widetilde{u}}_{h}\,=\,u_{h}\circ\mathchar 264\relax_{h}\,:\,\mathchar
266\relax\to{\mathord{\mathbb{R}}}\,,$

it has been shown in [6] that the _material derivative_

${\dot{u}}_{h}\,=\,\lim_{t\to
0}\frac{{\widetilde{u}}_{th}-u}{t}\,\in\,H^{1}_{\diamond}(\mathchar
266\relax)$ (22)

is well-defined, the convergence being in $H^{1}(\mathchar 266\relax)$, and
that ${\dot{u}}_{h}$ satisfies the variational problem

$\int_{\mathchar 266\relax}\sigma\,\nabla{\dot{u}}_{h}\cdot\nabla
w\,\mathrm{d}x\,=\,-\int_{\mathchar 266\relax}\sigma\,\nabla u\cdot({\cal
A}_{h}\nabla w)\,\mathrm{d}x\qquad\text{for every $w\in H^{1}(\mathchar
266\relax)$}\,,$ (23)

where $\sigma$ is as in (1) and

${\cal A}_{h}\,=\,(\nabla\cdot h)I\,-\,h^{\prime}\,-\,{h^{\prime}}^{T}$

is an $L^{\infty}$ function of $2\times 2$ matrices with

$\|{\cal A}_{h}\|_{L^{\infty}(\mathchar 266\relax)}\,=\,\sup_{x\in\mathchar
266\relax}\|{\cal A}_{h}(x)\|_{2}\,\leq\,4\,\|h\|_{W^{1,\infty}(\mathchar
266\relax)}\,.$ (24)

It has further been proved in [6] that ${\dot{u}}_{h}|_{\partial\mathchar
266\relax}$ coincides with the shape derivative of $\mathchar
259\relax_{f}({\mathscr{D}})$ in the direction of
$h|_{\partial{\mathscr{D}}}$, i.e.,

$\langle\,{\dot{u}}_{h},g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,\langle\,\partial\mathchar
259\relax_{f}({\mathscr{D}}_{t})h|_{\partial{\mathscr{D}}},g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}$ (25)

is given by (8) for every $g\in L^{2}_{\diamond}(\partial\mathchar
266\relax)$.

This approach via the material derivative allows for a stronger result than
(7).

######  Theorem 2.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$. Then the material
derivative in ${\cal L}({W_{c}^{1,\infty}(\mathchar
266\relax)},H_{\diamond}^{1}(\mathchar 266\relax))$ is a Fréchet derivative.
Moreover, there are constants $c,C>0$, which only depend on $\mathchar
266\relax$, ${\mathscr{D}}$, $f$, and $k$, such that

$\|\mathchar 259\relax_{f}({\mathscr{D}}_{h})-\mathchar
259\relax_{f}({\mathscr{D}})-\partial\mathchar
259\relax_{f}({\mathscr{D}})h\|_{L^{2}(\partial\mathchar
266\relax)}\,\leq\,C\,\|h\|_{W^{1,\infty}(\partial{\mathscr{D}})}^{2}$ (26)

for every $h\in W^{1,\infty}(\partial{\mathscr{D}})$ with
$\|h\|_{W^{1,\infty}(\partial{\mathscr{D}})}\leq c$.

###### Proof.

Let $h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ satisfy (21). Then the
Jacobian $\mathchar 264\relax_{h}^{\prime}$ is invertible, and it follows from
[6, Eq. (2.11)] that the auxiliary function $w_{h}={\widetilde{u}}_{h}-u$
solves the variational problem

$\int_{\mathchar 266\relax}\sigma\,\nabla w_{h}\cdot(A_{h}\nabla
w)\,\mathrm{d}x\,=\,\int_{\mathchar 266\relax}\sigma\,\nabla
u\cdot\bigl{(}(I-A_{h})\nabla w\bigr{)}\,\mathrm{d}x$ (27)

for every $w\in H^{1}(\mathchar 266\relax)$, where

$A_{h}\,=\,\mathchar 264\relax_{h}^{\prime}{}^{-1}\mathchar
264\relax_{h}^{\prime}{}^{-T}\det\mathchar 264\relax_{h}^{\prime}\,.$ (28)

From (27) we deduce that

$\displaystyle\int_{\mathchar 266\relax}\sigma\,\nabla w_{h}\cdot\nabla
w\,\mathrm{d}x$ $\displaystyle\,=\,\int_{\mathchar
266\relax}\sigma\,\nabla({\widetilde{u}}_{h}-u)\cdot\bigl{(}(I-A_{h})\nabla
w\bigr{)}\,\mathrm{d}x\,+\,\int_{\mathchar 266\relax}\sigma\,\nabla
w_{h}\cdot(A_{h}\nabla w)\,\mathrm{d}x$ $\displaystyle\,=\,\int_{\mathchar
266\relax}\sigma\,\nabla{\widetilde{u}}_{h}\cdot\bigl{(}(I-A_{h})\nabla
w\bigr{)}\,\mathrm{d}x\,,$

and hence, (23) implies that the Taylor remainder

$e_{h}\,=\,{\widetilde{u}}_{h}-u-{\dot{u}}_{h}\,=\,w_{h}-{\dot{u}}_{h}$

of the material derivative satisfies

$\displaystyle\int_{\mathchar 266\relax}\sigma\,\nabla e_{h}\cdot\nabla
w\,\mathrm{d}x\,=\,\int_{\mathchar
266\relax}\sigma\,\nabla{\widetilde{u}}_{h}\cdot\bigl{(}(I-A_{h})\nabla
w\bigr{)}\,\mathrm{d}x\,+\,\int_{\mathchar 266\relax}\sigma\,\nabla
u\cdot({\cal A}_{h}\nabla w)\,\mathrm{d}x$ (29)
$\displaystyle\qquad\,=\,\int_{\mathchar 266\relax}\sigma\,\nabla
w_{h}\cdot\bigl{(}(I-A_{h})\nabla w\bigr{)}\,\mathrm{d}x\,+\,\int_{\mathchar
266\relax}\sigma\,\nabla u\cdot\bigl{(}(I-A_{h}+{\cal A}_{h})\nabla
w\bigr{)}\,\mathrm{d}x$

for every $w\in H^{1}(\mathchar 266\relax)$.

From (20) and (28) we have

$A_{h}\,=\,(I+h^{\prime})^{-1}(I+{h^{\prime}}^{T})^{-1}(1+\nabla\cdot h+\det
h^{\prime})\,,$

and hence, using (21), it follows that there exists some constant
$C^{\prime}$, which we take to be greater than one, such that

$\|A_{h}-I\|_{L^{\infty}(\mathchar
266\relax)}\,\leq\,C^{\prime}\,\|h\|_{W^{1,\infty}(\mathchar
266\relax)}\quad\text{and}\quad\|A_{h}-I-{\cal A}_{h}\|_{L^{\infty}(\mathchar
266\relax)}\,\leq\,C^{\prime}\,\|h\|_{W^{1,\infty}(\mathchar
266\relax)}^{2}\,,$

independent of the particular choice of $h$. Accordingly, the right-hand side
of (29) can be estimated by

$C^{\prime}\bigl{(}\|\sqrt{\sigma}\,\nabla w_{h}\|_{L^{2}(\mathchar
266\relax)}\|h\|_{W^{1,\infty}(\mathchar
266\relax)}\,+\,\|\sqrt{\sigma}\,\nabla u\|_{L^{2}(\mathchar
266\relax)}\|h\|_{W^{1,\infty}(\mathchar
266\relax)}^{2}\bigr{)}\,\|\sqrt{\sigma}\,\nabla w\|_{L^{2}(\mathchar
266\relax)}\,,$

and using $w=e_{h}$ in (29) we thus obtain the estimate

$\displaystyle\|\sqrt{\sigma}\,\nabla e_{h}\|_{L^{2}(\mathchar
266\relax)}\,\leq\,$ (30) $\displaystyle\qquad
C^{\prime}\bigl{(}\|\sqrt{\sigma}\,\nabla w_{h}\|_{L^{2}(\mathchar
266\relax)}\|h\|_{W^{1,\infty}(\mathchar
266\relax)}\,+\,\|\sqrt{\sigma}\,\nabla u\|_{L^{2}(\mathchar
266\relax)}\|h\|_{W^{1,\infty}(\mathchar 266\relax)}^{2}\bigr{)}$

for the Taylor remainder.

Similarly, we conclude from (27) and (28) that

$\frac{1}{2}\,\|\sqrt{\sigma}\,\nabla w_{h}\|_{L^{2}(\mathchar
266\relax)}\,\leq\,C^{\prime}\,\|\sqrt{\sigma}\,\nabla u\|_{L^{2}(\mathchar
266\relax)}\|h\|_{W^{1,\infty}(\mathchar 266\relax)}$

for every $h$ satisfying

$\|h\|_{W^{1,\infty}(\mathchar 266\relax)}\,\leq\,\frac{1}{2C^{\prime}}\,,$
(31)

whereas the weak form (4) of (2) implies that

$\displaystyle\|\sqrt{\sigma}\,\nabla u\|_{L^{2}(\mathchar 266\relax)}^{2}$
$\displaystyle\,\leq\,\|f\|_{L^{2}(\partial\mathchar
266\relax)}\|u\|_{L^{2}(\partial\mathchar 266\relax)}$ (32)
$\displaystyle\,\leq\,\frac{c_{\mathchar
266\relax}}{\min\\{1,\sqrt{k}\\}}\,\|f\|_{L^{2}(\partial\mathchar
266\relax)}\|\sqrt{\sigma}\,\nabla u\|_{L^{2}(\mathchar 266\relax)}\,,$

where $c_{\mathchar 266\relax}$ depends on the norm of the trace operator from
$H^{1}(\mathchar 266\relax)$ to $L^{2}(\partial\mathchar 266\relax)$ and the
Poincaré constant for $H^{1}_{\diamond}(\mathchar 266\relax)$. With these two
estimates and another use of the Poincaré inequality it follows from (30) that

$\|e_{h}\|_{H^{1}(\mathchar
266\relax)}\,\leq\,C^{\prime\prime}\|h\|_{W^{1,\infty}(\mathchar
266\relax)}^{2}$ (33)

with a constant $C^{\prime\prime}$ which is independent of $h$, as long as $h$
satisfies (31). This shows that the material derivative in ${\cal
L}({W_{c}^{1,\infty}(\mathchar 266\relax)},H_{\diamond}^{1}(\mathchar
266\relax))$ is a Fréchet derivative.

Next, let $h\in W^{1,\infty}(\partial{\mathscr{D}})$ be given. Then we
construct an extension $h_{e}\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ of
this given function in the following way. We first choose a polygonal domain
${\mathscr{D}}_{1}$ with vertices $x_{i}^{(1)}$ in the vicinity of $x_{i}$,
$i=1,\dots,n$, respectively, such that
$\overline{\mathscr{D}}_{1}\subset{\mathscr{D}}$ and each (open) quadrangle
${\mathscr{Q}}_{i}$ with vertices $x_{i},x_{i+1},x_{i}^{(1)}$, and
$x_{i+1}^{(1)}$ is convex; using the associated convex representation

$x\,=\,c(1-d)\hskip 0.43057ptx_{i}\,+\,cd\,x_{i+1}\,+\,(1-c)(1-d)\hskip
0.43057ptx_{i}^{(1)}\,+\,(1-c)d\,x_{i+1}^{(1)}$

with $0<c,d<1$ of a general point $x\in{\mathscr{Q}}_{i}$ we can extend $h$ to
${\mathscr{Q}}_{i}$ via

$h_{e}(x)\,=\,c\,h\bigl{(}(1-d)\hskip
0.43057ptx_{i}\,+\,d\,x_{i+1}\bigr{)}\,,\qquad x\in{\mathscr{Q}}_{i}\,,$

and in a second step, extend $h_{e}$ continuously by zero to
$\overline{\mathscr{D}}_{1}$. In the same manner we can construct a polygonal
domain ${\mathscr{D}}_{2}$ with
$\overline{\mathscr{D}}\subset{\mathscr{D}}_{2}$ and
$\overline{\mathscr{D}}_{2}\subset\mathchar 266\relax$ and a corresponding
extension of $h_{e}$ to ${\mathscr{D}}_{2}$, and finally extend $h_{e}$ by
zero to the rest of $\mathchar 266\relax$. Note that this construction makes
sure that there exists a constant ${C_{\mathscr{D}}}\geq 1$, independent of
$h$, such that the extended vector field satisfies

$\|h_{e}\|_{W^{1,\infty}(\mathchar
266\relax)}\,\leq\,{C_{\mathscr{D}}}\,\|h\|_{W^{1,\infty}(\partial{\mathscr{D}})}\,.$
(34)

In particular, $h_{e}$ satisfies (31) if

$\|h\|_{W^{1,\infty}(\partial{\mathscr{D}})}\,\leq\,c\,:=\,\frac{1}{2C^{\prime}{C_{\mathscr{D}}}}\,.$

Using this extended vector field the trace of the associated material
derivative ${\dot{u}}_{h_{e}}$ coincides with $\partial\mathchar
259\relax_{f}({\mathscr{D}}_{t})h$, and hence, the left hand side of (26) is
the norm of the trace on $\partial\mathchar 266\relax$ of

$e_{h_{e}}\,=\,{\widetilde{u}}_{h_{e}}\,-\,u\,-\,{\dot{u}}_{h_{e}}\,.$

Accordingly, the desired inequality (26) is a consequence of (33), (34), and
the trace theorem. ∎

We also have the following estimate that is of independent interest.

######  Theorem 3.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$. Then there is a constant
$C>0$, depending only on $\mathchar 266\relax$, $f$, and $k$, such that

$\|\partial\mathchar 259\relax_{f}({\mathscr{D}}^{\prime})\|_{{\cal
L}(W^{1,\infty}(\partial{\mathscr{D}}^{\prime}),L^{2}(\partial\mathchar
266\relax))}\,\leq\,C$

for all polygons ${\mathscr{D}}^{\prime}$ with the same number of vertices,
which are sufficiently close to those of ${\mathscr{D}}$.

###### Proof.

For the given polygon ${\mathscr{D}}$ we choose ${\mathscr{D}}_{1}$ and
${\mathscr{D}}_{2}$ as in the proof of Theorem 2. Then every polygon
${\mathscr{D}}^{\prime}$ which is a sufficiently small perturbation of
${\mathscr{D}}$ as specified in the statement of this theorem satisfies
$\partial{\mathscr{D}}^{\prime}\subset{\mathscr{D}}_{2}\setminus\overline{\mathscr{D}}_{1}$,
and every vector field $h\in W^{1,\infty}(\partial{\mathscr{D}}^{\prime})$ can
be extended to a vector field $h_{e}\in{W_{c}^{1,\infty}(\mathchar
266\relax)}$ supported in $\overline{\mathscr{D}}_{2}$ as described in the
previous proof. Moreover, the corresponding extension satisfies

$\|h_{e}\|_{W^{1,\infty}(\mathchar
266\relax)}\,\leq\,{C_{{\mathscr{D}}^{\prime}}}\|h\|_{W^{1,\infty}(\partial{\mathscr{D}}^{\prime})}\,\leq\,2{C_{\mathscr{D}}}\,\|h\|_{W^{1,\infty}(\partial{\mathscr{D}}^{\prime})}\,,$
(35)

provided ${\mathscr{D}}^{\prime}$ is sufficiently close to ${\mathscr{D}}$. In
particular, if we stipulate that
$\|h\|_{W^{1,\infty}(\partial{\mathscr{D}}^{\prime})}=1/(4{C_{\mathscr{D}}})$,
then $h_{e}$ satisfies (21).

It therefore follows from (25) and (23) with $w={\dot{u}}_{h_{e}}$ that

$\displaystyle\|\partial\mathchar
259\relax_{f}({\mathscr{D}}^{\prime})h\|_{L^{2}(\partial\mathchar 266\relax)}$
$\displaystyle\,\leq\,\frac{c_{\mathchar
266\relax}}{\min\\{1,\sqrt{k}\\}}\,\|\sqrt{\sigma}\,\nabla{\dot{u}}_{h_{e}}\|_{L^{2}(\mathchar
266\relax)}$ $\displaystyle\,\leq\,\frac{c_{\mathchar
266\relax}}{\min\\{1,\sqrt{k}\\}}\,\|\sqrt{\sigma}\,\nabla
u\|_{L^{2}(\mathchar 266\relax)}\|{\cal A}_{h_{e}}\|_{L^{\infty}(\mathchar
266\relax)}\,,$

where $c_{\mathchar 266\relax}$ is the same constant as in (32). From (32),
(24), and (35) we therefore deduce that

$\displaystyle\|\partial\mathchar
259\relax_{f}({\mathscr{D}}^{\prime})h\|_{L^{2}(\partial\mathchar 266\relax)}$
$\displaystyle\,\leq\,\frac{4c_{\mathchar
266\relax}^{2}}{\min\\{1,k\\}}\,\|f\|_{L^{2}(\mathchar
266\relax)}\,\|h_{e}\|_{W^{1,\infty}(\mathchar 266\relax)}$
$\displaystyle\,\leq\,\frac{8c_{\mathchar
266\relax}^{2}\,{C_{\mathscr{D}}}}{\min\\{1,k\\}}\,\|f\|_{L^{2}(\mathchar
266\relax)}\,\|h\|_{W^{1,\infty}(\partial{\mathscr{D}}^{\prime})}\,.$

This proves the assertion. ∎

######  Remark 4.

In the same way one can show that the constants $c$ and $C$ of Theorem 2 can
be chosen in such a way that the same Taylor remainder estimate (26) is valid
for all polygons ${\mathscr{D}}^{\prime}$ sufficiently close to
${\mathscr{D}}$.  $\diamond$

Now we turn to the existence of the shape derivative $u^{\prime}_{h}$ of $u$,
formally defined in (5).

######  Theorem 5.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$. Then the solution
$u=u({\mathscr{D}})$ of (2) has a Fréchet shape derivative $\partial
u({\mathscr{D}})\in{\cal L}({W_{c}^{1,\infty}(\mathchar
266\relax)},L^{2}(\mathchar 266\relax))$. The derivative
$u^{\prime}_{h}=\partial u({\mathscr{D}})h$ of $u$ in the direction of
$h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ is given by

$u_{h}^{\prime}\,=\,{\dot{u}}_{h}\,-\,h\cdot\nabla u\,.$ (36)

This function is harmonic in $\mathchar
266\relax\setminus\partial{\mathscr{D}}$, and it belongs to $H^{1}(\mathchar
266\relax\setminus\mathop{\operator@font supp}\nolimits h)$. The Neumann
boundary values of $u_{h}^{\prime}$ on $\partial\mathchar 266\relax$ vanish,
and the trace $u_{h}^{\prime}|_{\partial\mathchar 266\relax}\in
L^{2}_{\diamond}(\mathchar 266\relax)$ is the shape derivative
$\partial\mathchar 259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}}$ of
the given measurements on the boundary.

###### Proof.

The statement on the Fréchet derivative of $u=u({\mathscr{D}})$ and its
representation (36) follows with the aid of, e.g., [16, Lemme 5.3.3] from the
fact that the material derivative is a Fréchet derivative.

Let $h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ be fixed, and consider any
domain $\mathchar 266\relax^{\prime}$ with $\overline{\mathchar
266\relax}^{\prime}\subset{\mathscr{D}}$, and let $w\in
C_{0}^{\infty}(\mathchar 266\relax^{\prime})$. Then $\mathchar
266\relax^{\prime}\subset{\mathscr{D}}_{th}$ for every $0<t<t^{\prime}$ with
$t^{\prime}$ sufficiently small, and hence, both $u_{th}$ and $u$ are harmonic
in $\mathchar 266\relax^{\prime}$ for $0<t<t^{\prime}$. Accordingly,

$\int_{\mathchar 266\relax^{\prime}}\frac{u_{th}-u}{t}\,\Delta
w\,\mathrm{d}x\,=\,0\qquad\text{for $0<t<t^{\prime}$}\,,$

and in the limit $t\to 0$ this yields

$\int_{\mathchar 266\relax^{\prime}}u^{\prime}_{h}\Delta
w\,\mathrm{d}x\,=\,0\qquad\text{for every $w\in C_{0}^{\infty}(\mathchar
266\relax^{\prime})$}\,.$

It thus follows from Weil’s lemma that $u_{h}^{\prime}$ is harmonic in
$\mathchar 266\relax^{\prime}$, and hence in ${\mathscr{D}}$. The same
argument applies for any domain $\mathchar 266\relax^{\prime}$ with
$\overline{\mathchar 266\relax}^{\prime}\subset\mathchar
266\relax\setminus\overline{\mathscr{D}}$, showing that $u_{h}^{\prime}$ is
also harmonic in $\mathchar 266\relax\setminus\overline{\mathscr{D}}$.

By virtue of (36) $u_{h}^{\prime}$ coincides with ${\dot{u}}_{h}$ on
$\mathchar 266\relax\setminus\mathop{\operator@font supp}\nolimits h$.
Accordingly, ${\dot{u}}_{h}$ is a harmonic function in $\mathchar
266\relax\setminus\mathop{\operator@font supp}\nolimits h$, $u_{h}^{\prime}$
belongs to $H^{1}(\mathchar 266\relax\setminus\mathop{\operator@font
supp}\nolimits h)$, and there holds $u_{h}^{\prime}|_{\partial\mathchar
266\relax}={\dot{u}}_{h}|_{\partial\mathchar 266\relax}=\partial\mathchar
259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}}$, and

$\frac{\partial}{\partial\nu}u_{h}^{\prime}\,=\,\frac{\partial}{\partial\nu}{\dot{u}}_{h}\,\in\,H^{-1/2}(\partial\mathchar
266\relax)\,.$

This Neumann derivative can be determined from the variational definition (23)
of ${\dot{u}}_{h}$: Choosing an arbitrary test function $w\in
C^{\infty}({\mathord{\mathbb{R}}}^{2})$, which vanishes on
${\mathscr{D}}\cup\mathop{\operator@font supp}\nolimits h$, it follows from
(23) that

$0\,=\,\int_{\mathchar
266\relax\setminus\overline{\mathscr{D}}}\nabla{\dot{u}}_{h}\cdot\nabla
w\,\mathrm{d}x\,=\,\int_{\partial\mathchar
266\relax}w\,\frac{\partial}{\partial\nu}{\dot{u}}_{h}\,\mathrm{d}s\,.$

Since the traces on $\partial\mathchar 266\relax$ of all these admissible test
functions are dense in $L^{2}(\partial\mathchar 266\relax)$, the Neumann
derivative on $\partial\mathchar 266\relax$ of ${\dot{u}}_{h}$, and hence of
$u_{h}^{\prime}$, must vanish. ∎

We will see below (in Theorem 7) that $u_{h}^{\prime}$ actually only depends
on $h|_{\partial{\mathscr{D}}}$.

##  5 The transmission problem for polygonal inclusions

We now investigate the transmission problem (6) for a polygonal inclusion.

######  Theorem 6.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$, and let $h\in
W^{1,\infty}(\partial{\mathscr{D}})$. Then the transmission problem

$\begin{array}[]{c}\Delta w\,=\,0\quad\text{in $\mathchar
266\relax\setminus\partial{\mathscr{D}}$}\,,\qquad\dfrac{\partial}{\partial\nu}w\,=\,0\quad\text{on
$\partial\mathchar 266\relax$}\,,\qquad{\displaystyle\int_{\partial\mathchar
266\relax}w\,\mathrm{d}s\,=\,0\,,}\\\\[12.91663pt]
\bigl{[}w\bigr{]}_{\partial{\mathscr{D}}}\,=\,(1-k)(h\cdot\nu)\dfrac{\partial}{\partial\nu}u^{-}\,,\quad\bigl{[}D_{\nu}w\bigr{]}_{\partial{\mathscr{D}}}\,=\,(1-k)\,\dfrac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\dfrac{\partial}{\partial\tau}u\Bigr{)}\,,\end{array}$
(37)

has a unique solution $w$, which belongs to $H^{\gamma}(\mathchar
266\relax\setminus\partial{\mathscr{D}})$ for some $\gamma\in(1/2,1)$ and also
to $H^{1}\bigl{(}\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal
B}_{r_{i}}(x_{i})}\bigr{)}$.

###### Proof.

Uniqueness is obvious: If $w_{1}$ and $w_{2}$ are two solutions of (37), then
$v=w_{1}-w_{2}$ solves the homogeneous Neumann problem (9) with $g=0$. This
proves that $v=0$.

To establish existence, let

$h_{i}^{-}\,=\,\lim_{x\to x_{i},\,x\in\mathchar
256\relax_{i}}(h\cdot\nu)(x)\,,\qquad h_{i}^{+}\,=\,\lim_{x\to
x_{i},\,x\in\mathchar 256\relax_{i+1}}(h\cdot\nu)(x)\,,$ (38)

for every $i=1,\dots,n$; take note that $h_{i}^{-}\neq h_{i}^{+}$ in general,
and that

$h(x)\,=\,\begin{cases}h_{i}^{-}\,+\,O(|x-x_{i}|)\,,&x\in\mathchar
256\relax_{i}\,,\\\ h_{i}^{+}\,+\,O(|x-x_{i}|)\,,&x\in\mathchar
256\relax_{i+1}\,,\end{cases}$ (39)

for $x$ near $x_{i}$. Furthermore, for $i=1,\dots,n$ let
$A_{i}^{\prime}{}^{\pm}$ and $B_{i}^{\prime}{}^{\pm}$ be the entries of the
solution of the linear system

$Y(\gamma_{i1}-1,\alpha_{i})\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}A_{i}^{\prime}{}^{-}\\\ B_{i}^{\prime}{}^{-}\\\ A_{i}^{\prime}{}^{+}\\\
B_{i}^{\prime}{}^{+}\end{array}\right]\,=\,(1-k)\gamma_{i1}\left[\hskip-5.0pt\begin{array}[]{*\c@MaxMatrixCols
c}h_{i}^{+}B_{i1}^{-}\\\ h_{i}^{+}A_{i1}^{-}\\\
\phantom{-}h_{i}^{-}(A_{i1}^{-}s_{i1}-B_{i1}^{-}c_{i1})\\\
-h_{i}^{-}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\end{array}\right]$ (40)

with $Y\in{\mathord{\mathbb{R}}}^{4\times 4}$ defined in (15). Since
$\gamma^{\prime}=\gamma_{i1}-1\in(-1/2,0)$ according to (14),
$\gamma^{\prime}$ cannot be a solution of (13), and hence the inhomogeneous
linear system (40) has a unique solution by virtue of Lemma 1. The way this
system is set up, the function

$w_{i}(x)\,=\,\begin{cases}\beta_{i1}\chi_{i}(x){\widetilde{y}}_{i}(\theta)r^{\gamma_{i1}-1}\,,&x\in{\cal
B}_{r_{i}}(x_{i})\,,\\\ 0\,,&x\in\mathchar 266\relax\setminus\overline{{\cal
B}_{r_{i}}(x_{i})}\,,\end{cases}$ (41)

defined in terms of the local coordinate system (10), with

${\widetilde{y}}_{i}(\theta)\,=\,\begin{cases}A_{i}^{\prime}{}^{-}\cos(\gamma_{ij}-1)\theta\,+\,B_{i}^{\prime}{}^{-}\sin(\gamma_{ij}-1)\theta\,,&\quad\phantom{\alpha_{i}}0<\theta<\alpha_{i}\,,\\\\[4.30554pt]
A_{i}^{\prime}{}^{+}\cos(\gamma_{ij}-1)\theta\,+\,B_{i}^{\prime}{}^{+}\sin(\gamma_{ij}-1)\theta\,,&\quad\phantom{0}\alpha_{i}<\theta<2\pi\,,\end{cases}$
(42)

is such that $\bigl{[}w_{i}\bigr{]}_{\partial{\mathscr{D}}}$ and
$\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}$ coincide near $x=x_{i}$
with the leading order terms of the prescribed transmission data in (37),
compare (17) and (18). In fact, from (39) and (14) follows that

$\displaystyle\varphi$
$\displaystyle\,=\,(1-k)\,(h\cdot\nu)\frac{\partial}{\partial\nu}u^{-}\,-\,\sum_{i=1}^{n}\bigl{[}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\in\,H^{1/2}(\partial{\mathscr{D}})$
(43a) and $\displaystyle\psi$
$\displaystyle\,=\,(1-k)\,\dfrac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\dfrac{\partial}{\partial\tau}u\Bigr{)}\,-\,\sum_{i=1}^{n}\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\in\,H^{-1/2}(\partial{\mathscr{D}})\,.$
(43b) Further, since ${\widetilde{y}}_{i}r^{\gamma_{i1}-1}$ is harmonic in
$\mathop{\operator@font supp}\nolimits\chi_{i}\setminus\partial{\mathscr{D}}$,
$\displaystyle F_{i}$ $\displaystyle\,=\,\sigma\Delta
w_{i}\,=\,\sigma\beta_{i1}\bigl{(}2\nabla\chi_{i}\cdot\nabla({\widetilde{y}}_{i}r^{\gamma_{i1}-1})\,+\,{\widetilde{y}}_{i}r^{\gamma_{i1}-1}\Delta\chi_{i}\bigr{)}\,,$
(43c)

defined in $\mathchar 266\relax\setminus\partial{\mathscr{D}}$, is smooth in
both subdomains and supported in a small annulus around $x_{i}$. In
particular, $F_{i}\in L^{2}(\mathchar 266\relax)$, and hence the transmission
problem

$\begin{array}[]{c}-\nabla\cdot(\sigma\nabla
w_{0})\,=\,{\displaystyle\sum_{i=1}^{n}F_{i}}\qquad\text{in $\mathchar
266\relax\setminus\partial{\mathscr{D}}$}\,,\\\\[12.91663pt]
\dfrac{\partial}{\partial\nu}w_{0}=0\quad\text{on $\partial\mathchar
266\relax$}\,,\qquad{\displaystyle\int_{\partial\mathchar
266\relax}w_{0}\,\mathrm{d}s\,=\,0}\,,\\\\[17.22217pt]
\bigl{[}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\,=\,\varphi\,,\qquad\bigl{[}D_{\nu}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\,=\,\psi\,,\end{array}$
(44)

has a unique solution $w_{0}\in H^{1}(\mathchar
266\relax\setminus\partial{\mathscr{D}})$, if and only if the integrability
condition

$\sum_{i=1}^{n}\int_{\mathchar
266\relax\setminus\partial{\mathscr{D}}}F_{i}\,\mathrm{d}x\,-\,\int_{\partial{\mathscr{D}}}\psi\,\mathrm{d}s\,=\,0$
(45)

is satisfied: compare Costabel and Stephan [8].

In order to check (45) one has to be careful, because although the two
integrals in (45) are well-defined, the two individual terms of $\psi$ in
(43b) fail to be integrable in a classical sense. To take this into account we
choose $\delta>0$ so small that $\chi_{i}$ is equal to one in ${\cal
B}_{\delta}(x_{i})$ for every $i=1,\dots,n$, in which case

$\frac{\partial}{\partial\nu}w_{i}(x)\,=\,\beta_{i1}(\gamma_{i1}-1)\,{\widetilde{y}}_{i}(\theta)\,\delta^{\gamma_{i1}-2}\qquad\text{on
$\partial{\cal B}_{\delta}(x_{i})$}\,,$

where $\nu$ denotes the exterior normal vector on $\partial{\cal
B}_{\delta}(x_{i})$. Let ${\cal B}_{\delta}$ be the union of these disks
${\cal B}_{\delta}(x_{i})$, $i=1,\dots,n$. Then Green’s formula, applied in
${\mathscr{D}}\setminus\overline{\cal B}_{\delta}$ and in $\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup{\cal B}_{\delta}}$, yields

$\displaystyle\int_{\mathchar
266\relax\setminus\partial{\mathscr{D}}}F_{i}\,\mathrm{d}x\,=\,\int_{\mathchar
266\relax\setminus\overline{{\cal
B}_{\delta}\cup\partial{\mathscr{D}}}}F_{i}\,\mathrm{d}x\,=\,\int_{\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup{\cal B}_{\delta}}}\Delta
w_{i}\,\mathrm{d}x\,+\,\int_{{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}k\,\Delta w_{i}\,\mathrm{d}x$
$\displaystyle\qquad\,=\,\int_{\partial\mathchar
266\relax}\frac{\partial}{\partial\nu}w_{i}\,\mathrm{d}s\,-\,\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma\frac{\partial}{\partial\nu}w_{i}\,\mathrm{d}s$
$\displaystyle\qquad\,=\,-\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,-\,\beta_{i1}(\gamma_{i1}-1)\,\delta^{\gamma_{i1}-1}\Bigl{(}\int_{0}^{\alpha_{i}}k\,{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\,+\,\int_{\alpha_{i}}^{2\pi}{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\Bigr{)}$

for every $i=1,\dots,n$. From (43b) we further have

$\displaystyle\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\psi\,\mathrm{d}s$
$\displaystyle\,=\,(1-k)\sum_{i=1}^{n}\int_{\mathchar
256\relax_{i}\setminus\overline{\cal
B}_{\delta}}\frac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\frac{\partial}{\partial\tau}u\Bigr{)}\,\mathrm{d}s\,-\,\sum_{i=1}^{n}\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s$
$\displaystyle\,=\,(1-k)\sum_{i=1}^{n}\Bigl{[}(h\cdot\nu)\frac{\partial}{\partial\tau}u\Bigr{]}_{i}\\!-\,\sum_{i=1}^{n}\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\bigl{[}D_{\nu}w_{i}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,,$

where

$\Bigl{[}(h\cdot\nu)\frac{\partial}{\partial\tau}u\Bigr{]}_{i}\,=\,-\beta_{i1}\gamma_{i1}\bigl{(}h_{i}^{-}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\,+\,h_{i}^{+}A_{i1}^{-}\bigr{)}\delta^{\gamma_{i1}-1}\,+\,o(1)$
(46)

as $\delta\to 0$ is a short-hand notation for the difference of the values of
$(h\cdot\nu)\frac{\partial}{\partial\tau}u$ at $\mathchar
256\relax_{i}\cap{\cal B}_{\delta}(x_{i})$ and $\mathchar
256\relax_{i+1}\cap{\cal B}_{\delta}(x_{i})$, which amounts to the right-hand
side of (46) according to (18) and (39). Since (42) and the second and fourth
equations of (40) give

$\displaystyle(\gamma_{i1}-1)\Bigl{(}\int_{0}^{\alpha_{i}}k\,{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\,+\,\int_{\alpha_{i}}^{2\pi}{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\Bigr{)}$
(47)
$\displaystyle\qquad\qquad\,=\,(1-k)\gamma_{i1}\bigl{(}h_{i}^{+}A_{i1}^{-}\,+\,h_{i}^{-}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\bigr{)}\,,$

it follows that

$\displaystyle\sum_{i=1}^{n}\int_{\mathchar
266\relax\setminus\partial{\mathscr{D}}}F_{i}\,\mathrm{d}x\,-\,\int_{\partial{\mathscr{D}}\setminus\overline{\cal
B}_{\delta}}\psi\,\mathrm{d}s$ $\displaystyle\
\,=\,(k-1)\sum_{i=1}^{n}\Bigl{[}(h\cdot\nu)\frac{\partial}{\partial\tau}u\Bigr{]}_{i}\,-\,\sum_{i=1}^{n}\beta_{i1}(\gamma_{i1}-1)\,\delta^{\gamma_{i1}-1}\Bigl{(}\int_{0}^{\alpha_{i}}k\,{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\,+\,\int_{\alpha_{i}}^{2\pi}{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\Bigr{)}$

converges to zero as $\delta\to 0$.

We thus have shown that (44) has a solution $w_{0}\in H^{1}(\mathchar
266\relax\setminus\partial{\mathscr{D}})$, and hence,

$w\,=\,w_{0}\,+\,\sum_{i=1}^{n}w_{i}$ (48)

solves (37). Since $w_{i}$ belongs to $H^{\gamma}(\mathchar
266\relax\setminus\partial{\mathscr{D}})$ for every

$\gamma\,<\,\min\\{\gamma_{i1}\,:\,i=1,\dots,n\\}\,,$

cf., e.g., [12, Theorem 1.2.18], this is also true for $w$. Finally, since the
cut-off functions $\chi_{i}$, $i=1,\dots,n$, are supported in ${\cal
B}_{r_{i}}(x_{i})$, the solution $w$ coincides with $w_{0}$ in $\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal B}_{r_{i}}(x_{i})}$,
and hence $w$ belongs to $H^{1}\bigl{(}\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal
B}_{r_{i}}(x_{i})}\bigr{)}$. ∎

##  6 The shape derivative of $u$ solves the transmission problem

In Section 4 we have demonstrated that $u=u({\mathscr{D}})$ admits a well-
defined shape derivative $u_{h}^{\prime}=\partial u({\mathscr{D}})h$ in the
direction of any given vector field $h\in{W_{c}^{1,\infty}(\mathchar
266\relax)}$. In the sequel it will be shown that $u_{h}^{\prime}$ coincides
with the solution $w$ of the transmission problem (37), if the inhomogeneous
transmission data in (37) are defined in terms of the trace of $h$ on
$\partial{\mathscr{D}}$.

######  Theorem 7.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$. Then the shape derivative
of $u=u({\mathscr{D}})$ in direction $h\in{W_{c}^{1,\infty}(\mathchar
266\relax)}$ is the solution of the transmission problem (37).

###### Proof.

Let $w$ denote the solution of the transmission problem (37), where the vector
field which enters the inhomogeneous transmission data in (37) is the trace of
$h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ on $\partial{\mathscr{D}}$. In
the first and major step of this proof we show that the boundary values of $w$
on $\partial\mathchar 266\relax$ coincide with
$u_{h}^{\prime}|_{\partial\mathchar 266\relax}=\partial\mathchar
259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}}$, compare Theorem 5.

Since $w$ belongs to $H^{1}\bigl{(}\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal
B}_{r_{i}}(x_{i})}\bigr{)}$ by virtue of Theorem 6, its trace on
$\partial\mathchar 266\relax$ is well-defined in $H^{1/2}(\partial\mathchar
266\relax)\subset L^{2}(\partial\mathchar 266\relax)$, and it has vanishing
mean according to (37). In view of (8) we need to show that

$\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(1-k)\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\nabla
u^{-}\cdot(M\nabla v_{g}^{-})\,\mathrm{d}s$ (49)

for every $g\in L^{2}_{\diamond}(\partial\mathchar 266\relax)$, where $v_{g}$
is the solution of (9) and $M\in{\mathord{\mathbb{R}}}^{2\times 2}$ is the
symmetric $2\times 2$-matrix with eigenvalues $1$ and $k$ and eigenvectors
$\tau$ and $\nu$, respectively.

As in the proof of Theorem 6 let $\delta>0$ be so small that $\chi_{i}(x)=1$
for every $x\in{\cal B}_{\delta}(x_{i})$ and every $i=1,\dots,n$, set ${\cal
B}_{\delta}=\cup_{i}{\cal B}_{\delta}(x_{i})$, and denote by $\nu$ the
exterior normal vector on $\partial{\cal B}_{\delta}$. Further, let
$w^{-}=w|_{\mathscr{D}}$ and $w^{+}=w|_{\mathchar
266\relax\setminus\overline{\mathscr{D}}}$. Since the Neumann derivative of
$w$ vanishes on $\partial\mathchar 266\relax$ by virtue of (37), Green’s
formula in $\mathchar 266\relax\setminus\overline{{\mathscr{D}}\cup{\cal
B}_{\delta}}$ and in ${\mathscr{D}}\setminus\overline{\cal B}_{\delta}$ yields

$\displaystyle\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,\int_{\partial\mathchar
266\relax}w\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,-\,\int_{\partial\mathchar
266\relax}v_{g}\,\frac{\partial}{\partial\nu}w\,\mathrm{d}s$
$\displaystyle\qquad\,=\,-\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}v_{g}\bigl{[}D_{\nu}w\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,+\,\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}w{}^{+}\,\frac{\partial}{\partial\nu}v_{g}^{+}\,\mathrm{d}s\,-\,\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}kw{}^{-}\,\frac{\partial}{\partial\nu}v_{g}^{-}\,\mathrm{d}s$
$\displaystyle\qquad\phantom{\,=\,}\,-\,\int_{\partial{\cal B}_{\delta}}\sigma
v_{g}\,\frac{\partial}{\partial\nu}w\,\mathrm{d}s\,+\,\int_{\partial{\cal
B}_{\delta}}\sigma w\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,.$

Using

$w^{-}|_{\partial{\mathscr{D}}}\,=\,w^{+}|_{\partial{\mathscr{D}}}\,-\,\bigl{[}w\bigr{]}_{\partial{\mathscr{D}}}$

and the transmission conditions (37) as well as the fact that
$\bigl{[}D_{\nu}v_{g}\bigr{]}_{\partial{\mathscr{D}}}=0$ we can transform this
equation further into

$\displaystyle\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(k-1)\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}v_{g}\,\frac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\dfrac{\partial}{\partial\tau}u\Bigr{)}\,\mathrm{d}s$
$\displaystyle\quad\,\,+\,k(1-k)\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}(h\cdot\nu)\dfrac{\partial}{\partial\nu}u^{-}\,\frac{\partial}{\partial\nu}v_{g}^{-}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma
v_{g}\,\frac{\partial}{\partial\nu}w\,\mathrm{d}s\,+\,\int_{\partial{\cal
B}_{\delta}}\sigma w\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,.$

The first term on the right-hand side consists of $n$ integrals over the
individual edges of ${\mathscr{D}}$, and partial integration on each of these
edges gives

$\displaystyle\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(k-1)\sum_{i=1}^{n}\Bigl{[}(h\cdot\nu)\hskip
0.43057ptv_{g}\frac{\partial}{\partial\tau}u\Bigr{]}_{i}+\,(1-k)\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}\\!(h\cdot\nu)\frac{\partial}{\partial\tau}u\,\frac{\partial}{\partial\tau}v_{g}\,\mathrm{d}s$
$\displaystyle\quad\ \ \,+\,k(1-k)\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}\\!(h\cdot\nu)\dfrac{\partial}{\partial\nu}u^{-}\,\frac{\partial}{\partial\nu}v_{g}^{-}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\\!\sigma
v_{g}\,\frac{\partial}{\partial\nu}w\,\mathrm{d}s\,+\,\int_{\partial{\cal
B}_{\delta}}\\!\sigma w\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,,$

where we utilize the notation introduced in (46). Note that this identity can
be rewritten as

$\displaystyle\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(1-k)\int_{\partial{\mathscr{D}}\setminus{\cal
B}_{\delta}}(h\cdot\nu)\,\nabla u^{-}\cdot(M\nabla v_{g}^{-})\,\mathrm{d}s$
(50) $\displaystyle\ \,+\,(k-1)\sum_{i=1}^{n}\Bigl{[}(h\cdot\nu)\hskip
0.43057ptv_{g}\frac{\partial}{\partial\tau}u\Bigr{]}_{i}\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma
v_{g}\,\frac{\partial}{\partial\nu}w\,\mathrm{d}s\,+\,\int_{\partial{\cal
B}_{\delta}}\sigma w\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s$

with $M$ as in (8), and the first integral converges to the right-hand side of
(49) as $\delta\to 0$ by virtue of (19). It therefore remains to show that the
sum of terms in the second line of (50) converges to zero as $\delta\to 0$.

We now investigate these three summands individually. For the first one we
utilize (11), (18), and (38) to obtain

$\displaystyle\sum_{i=1}^{n}\Bigl{[}(h\cdot\nu)\hskip
0.43057ptv_{g}\frac{\partial}{\partial\tau}u\Bigr{]}_{i}$ $\displaystyle\
=\,\sum_{i=1}^{n}\bigl{(}h_{i}^{-}+O(\delta)\bigr{)}\bigl{(}v_{g}(x_{i})+O(\delta^{\gamma_{i1}})\bigr{)}\bigl{(}-\beta_{i1}\gamma_{i1}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\delta^{\gamma_{i1}-1}+O(\delta^{\gamma_{i2}-1})\bigr{)}$
$\displaystyle\
\phantom{=\,}\,-\,\sum_{i=1}^{n}\bigl{(}h_{i}^{+}+O(\delta)\bigr{)}\bigl{(}v_{g}(x_{i})+O(\delta^{\gamma_{i1}})\bigr{)}\bigl{(}\beta_{i1}\gamma_{i1}A_{i1}^{-}\delta^{\gamma_{i1}-1}+O(\delta^{\gamma_{i2}-1})\bigr{)}$
$\displaystyle\
=\,\displaystyle{-\sum_{i=1}^{n}\beta_{i1}\gamma_{i1}v_{g}(x_{i})\bigl{(}h_{i}^{-}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\,+\,h_{i}^{+}A_{i1}^{-}\bigr{)}\delta^{\gamma_{i1}-1}}\,+\,o(1)$
(51a) as $\delta\to 0$, where the estimate of the remainder follows from (14).

Concerning the two integrals over the boundary of ${\cal B}_{\delta}$ in (50)
we decompose $w$ as in (48) and investigate the corresponding parts
separately. Consider $w_{0}$ first, i.e., the solution of the transmission
problem (44), and let $w_{0}^{-}=w_{0}|_{\mathscr{D}}$ and
$w_{0}^{+}=w_{0}|_{\mathchar 266\relax\setminus\overline{\mathscr{D}}}$. Since
the source terms $F_{i}$ of (44) vanish in ${\cal B}_{\delta}$, compare (43c),
$w_{0}$ and $v_{g}$ are both harmonic in ${\cal
B}_{\delta}\setminus\partial{\mathscr{D}}$. Since both functions also belong
to $H^{1}({\cal B}_{\delta}\setminus\partial{\mathscr{D}})$ we can apply
Green’s formula in ${\cal B}_{\delta}\cap{\mathscr{D}}$ and in ${\cal
B}_{\delta}\setminus\overline{\mathscr{D}}$ to obtain

$\displaystyle\int_{\partial{\cal B}_{\delta}}\sigma
w_{0}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma v_{g}\,\frac{\partial}{\partial\nu}w_{0}\,\mathrm{d}s$
$\displaystyle\qquad\,=\,-\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}v_{g}\bigl{[}D_{\nu}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,-\,\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}k\,w_{0}^{-}\,\frac{\partial}{\partial\nu}v_{g}^{-}\,\mathrm{d}s\,+\,\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}w_{0}^{+}\,\frac{\partial}{\partial\nu}v_{g}^{+}\,\mathrm{d}s\,.$

Using (44) and the fact that

$w_{0}^{+}|_{\partial{\mathscr{D}}}\,=\,w_{0}^{-}|_{\partial{\mathscr{D}}}\,+\,\bigl{[}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\,,$

we thus arrive at

$\displaystyle\int_{\partial{\cal B}_{\delta}}\sigma
w_{0}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma v_{g}\,\frac{\partial}{\partial\nu}w_{0}\,\mathrm{d}s$
$\displaystyle\quad\,=\,-\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}v_{g}\bigl{[}D_{\nu}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,+\,\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}w_{0}^{-}\bigl{[}D_{\nu}v_{g}\bigr{]}_{\partial{\mathscr{D}}}\,\mathrm{d}s\,+\,\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}\bigl{[}w_{0}\bigr{]}_{\partial{\mathscr{D}}}\frac{\partial}{\partial\nu}v_{g}^{+}\,\mathrm{d}s$
$\displaystyle\quad\,=\,-\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}v_{g}\psi\,\mathrm{d}s\,+\,\int_{\partial{\mathscr{D}}\cap{\cal
B}_{\delta}}\varphi\,\frac{\partial}{\partial\nu}v_{g}^{+}\,\mathrm{d}s\,.$

The two products $v_{g}\psi$ and
$\varphi\frac{\partial}{\partial\nu}v_{g}^{+}$ are integrable over
$\partial{\mathscr{D}}$ because $\psi\in H^{-1/2}(\partial{\mathscr{D}})$ and
$\varphi\in H^{1/2}(\partial{\mathscr{D}})$, cf. (43), and from this we
conclude that

$\int_{\partial{\cal B}_{\delta}}\sigma
w_{0}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,-\,\int_{\partial{\cal
B}_{\delta}}\sigma
v_{g}\,\frac{\partial}{\partial\nu}w_{0}\,\mathrm{d}s\,=\,o(1)\,,\qquad\delta\to
0\,.$ (51b)

Next, consider

$\int_{\partial{\cal B}_{\delta}}\sigma
w_{i}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,=\,\int_{\partial{\cal
B}_{\delta}(x_{i})}\sigma
w_{i}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s$

for any fixed $i\in\\{1,\dots,n\\}$. It follows from (41) and (11) that

$|w_{i}(x)|\,=\,O(\delta^{\gamma_{i1}-1})$

and

$\left|\frac{\partial}{\partial\nu}v_{g}(x)\right|\,=\,\left|\frac{\partial}{\partial
r}v_{g}(x)\right|\,=\,O(\delta^{\gamma_{i1}-1})$

on $\partial{\cal B}_{\delta}(x_{i})$. Inserting these estimates into the
integral in question shows that

$\int_{\partial{\cal B}_{\delta}}\sigma
w_{i}\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s\,=\,O(\delta^{2\gamma_{i1}-1})\,=\,o(1)\,,\qquad\delta\to
0\,,$ (51c)

by virtue of (14).

Finally, since $\delta$ is so small that $\chi_{i}=1$ in ${\cal
B}_{\delta}(x_{i})$ the Neumann boundary derivative of $w_{i}$ satisfies

$\frac{\partial}{\partial\nu}w_{i}\,=\,\begin{cases}\beta_{i1}(\gamma_{i1}-1){\widetilde{y}}_{i}(\theta)\,\delta^{\gamma_{i1}-2}\,,&\text{on
$\partial{\cal B}_{\delta}(x_{i})$}\,,\\\\[4.30554pt] 0\,,&\text{on
$\partial{\cal B}_{\delta}(x_{j})$, $j\neq i$}\,,\end{cases}$

while

$v_{g}\,=\,v_{g}(x_{i})\,+\,O(\delta^{\gamma_{i1}})\qquad\text{on
$\partial{\cal B}_{\delta}(x_{i})$}\,,$

and hence,

$\displaystyle\int_{\partial{\cal B}_{\delta}}\\!\\!\sigma
v_{g}\,\frac{\partial}{\partial\nu}w_{i}\,\mathrm{d}s\,=\,\beta_{i1}(\gamma_{i1}-1)v_{g}(x_{i})\left(\int_{0}^{\alpha_{i}}k\,{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta+\int_{\alpha_{i}}^{2\pi}{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\right)\delta^{\gamma_{i1}-1}$
(51d) $\displaystyle\qquad\qquad\qquad\qquad\
\,+\,O(\delta^{2\gamma_{i1}-1})\,.$

Assembling the pieces of (51) and inserting them into (50) we thus obtain

$\displaystyle\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(1-k)\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\nabla
u^{-}\cdot(M\nabla v_{g}^{-})\,\mathrm{d}s$
$\displaystyle\qquad\,-\,(k-1)\sum_{i=1}^{n}\beta_{i1}\gamma_{i1}v_{g}(x_{i})\bigl{(}h_{i}^{-}(A_{i1}^{-}c_{i1}+B_{i1}^{-}s_{i1})\,+\,h_{i}^{+}A_{i1}^{-}\bigr{)}\delta^{\gamma_{i1}-1}$
$\displaystyle\qquad\,-\,\sum_{i=1}^{n}\beta_{i1}(\gamma_{i1}-1)v_{g}(x_{i})\left(\int_{0}^{\alpha_{i}}k\,{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\,+\,\int_{\alpha_{i}}^{2\pi}{\widetilde{y}}_{i}(\theta)\,\mathrm{d}\theta\right)\delta^{\gamma_{i1}-1}\,+\,o(1)\,.$

The two integrals over ${\widetilde{y}}_{i}$ in the bottom line have already
been evaluated in the course of the proof of Theorem 6, compare (47). Their
value is such that all terms of order $\delta^{\gamma_{i1}-1}$ cancel, proving
that

$\langle\,w,g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,(1-k)\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\nabla
u^{-}\cdot(M\nabla v_{g}^{-})\,\mathrm{d}s\,+\,o(1)\,,\qquad\delta\to 0\,.$

Since the left-hand side of this equation is independent of $\delta$, the
desired identity (49) must hold true.

As shown in Theorem 5 the shape derivative $u_{h}^{\prime}$ of $u$ is harmonic
in ${\mathscr{D}}$ and in $\mathchar 266\relax\setminus\overline{\mathscr{D}}$
and has homogeneous Neumann boundary values. Accordingly, $u_{h}^{\prime}$ and
$w$ share the same Cauchy data on $\partial\mathchar 266\relax$. Since they
are both harmonic in $\mathchar 266\relax\setminus\overline{\mathscr{D}}$ they
coincide up to the boundary of ${\mathscr{D}}$ by virtue of Holmgren’s
theorem. In particular, their restrictions to $\mathchar
266\relax\setminus\overline{\mathscr{D}}$ have the same trace on
$\partial{\mathscr{D}}$. Since the material derivative ${\dot{u}}_{h}$ belongs
to $H^{1}(\mathchar 266\relax)$ and the traces of $u^{\pm}$ (and their
tangential derivatives) coincide on $\partial{\mathscr{D}}$ it follows from
(36) and (3) that

$\bigl{[}u_{h}^{\prime}\bigr{]}_{\partial{\mathscr{D}}}\,=\,-h\cdot\bigl{[}\nabla
u\bigr{]}_{\partial{\mathscr{D}}}\,=\,-(h\cdot\nu)\Bigl{(}\frac{\partial}{\partial\nu}u^{+}\,-\,\frac{\partial}{\partial\nu}u^{-}\Bigr{)}\,=\,(1-k)(h\cdot\nu)\frac{\partial}{\partial\nu}u^{-}\,.$

According to (37) this shows that $w$ and $u_{h}^{\prime}$ experience the same
jump across $\partial{\mathscr{D}}$, and therefore the traces on
$\partial{\mathscr{D}}$ of $w|_{\mathscr{D}}$ and
$u_{h}^{\prime}|_{\mathscr{D}}$ also coincide. Both being harmonic functions
we thus conclude that $w=u_{h}^{\prime}$ in ${\mathscr{D}}$, and hence, in all
of $\mathchar 266\relax$. ∎

Note that it follows from Theorem 7 that the shape derivative $\partial
u({\mathscr{D}})h$ of $u$ in direction $h\in{W_{c}^{1,\infty}(\mathchar
266\relax)}$ only depends on $h|_{\partial{\mathscr{D}}}$.

##  7 The case of an insulating or a perfectly conducting polygonal inclusion

So far we have assumed that the conductivity of the inclusion is a known
positive value $k>0$. Now we turn to the limiting cases which are formally
described by $k=0$ (the insulating case) and $k=+\infty$ (the perfectly
conducting case).

###  7.1 The insulating case

The appropriate formulation of the conductivity equation (2) when
${\mathscr{D}}$ is an insulating inclusion, is in terms of the boundary value
problem

$\begin{array}[]{c}\Delta u\,=\,0\quad\text{in $\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\qquad{\displaystyle\int_{\partial\mathchar
266\relax}u\,\mathrm{d}s\,=\,0}\,,\\\\[8.61108pt]
\dfrac{\partial}{\partial\nu}u\,=\,0\quad\text{on
$\partial{\mathscr{D}}$}\,,\qquad\dfrac{\partial}{\partial\nu}u\,=\,f\quad\text{on
$\partial\mathchar 266\relax$}\,.\end{array}$ (52)

When ${\mathscr{D}}$ is smooth the existence of an associated shape derivative
of $u=u({\mathscr{D}})$ is implicit in the result of [29, Section 3.2]: this
shape derivative in direction $h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$ is
given by the solution $u_{h}^{\prime}\in H^{1}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ of the Neumann boundary value
problem

$\begin{array}[]{c}\Delta u_{h}^{\prime}\,=\,0\quad\text{in $\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\qquad{\displaystyle\int_{\partial\mathchar
266\relax}u_{h}^{\prime}\,\mathrm{d}s\,=\,0}\,,\\\\[8.61108pt]
\dfrac{\partial}{\partial\nu}u_{h}^{\prime}\,=\,\dfrac{\partial}{\partial\tau}\Bigl{(}(h\cdot\nu)\dfrac{\partial}{\partial\tau}u\Bigr{)}\quad\text{on
$\partial{\mathscr{D}}$}\,,\qquad\dfrac{\partial}{\partial\nu}u_{h}^{\prime}\,=\,0\quad\text{on
$\partial\mathchar 266\relax$}\,.\end{array}$ (53)

To the best of our knowledge the existence of a corresponding shape derivative
for a polygonal inclusion ${\mathscr{D}}$ has not yet been investigated.
However, the analysis in [6] can be adapted in a straightforward manner to
derive the respective material derivative ${\dot{u}}_{h}\in H^{1}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ in the direction of $h$ for the
insulating case and the corresponding variational definition of
$\partial\mathchar
259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}}={\dot{u}}_{h}|_{\partial\mathchar
266\relax}$, i.e.,

$\langle\,\partial\mathchar
259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}},g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\frac{\partial}{\partial\tau}u\,\frac{\partial}{\partial\tau}v_{g}\,\mathrm{d}s$

for every $g\in L^{2}_{\diamond}(\partial\mathchar 266\relax)$, where $v_{g}$
is the corresponding solution of (52) with $f$ replaced by $g$. This is the
natural analog of (8) for $k=0$.

For the proof of the corresponding version of Theorem 6 analogous properties
of the solution of the forward problem (52), as recollected in Section 3 for
the case $k>0$, are needed. The corresponding results can be found, e.g., in
[12]: In the case $k=0$ the eigenvalues $\gamma_{ij}^{2}$ and eigenfunctions
$y_{ij}$ which enter into the expansion of $u$ near any of the vertices, are
known explicitly, cf. [12, p. 50], namely

$\gamma_{ij}\,=\,j\pi/(2\pi-\alpha_{i})\,,\qquad
j\in{\mathord{\mathbb{N}}}_{0}\,,$ (54)

and, for $j\in{\mathord{\mathbb{N}}}$,

$y_{ij}(\theta)\,=\,\Bigl{(}\frac{2}{2\pi-\alpha_{i}}\Bigr{)}^{1/2}\cos\gamma_{ij}(\theta-\alpha_{i})\,,\qquad\alpha_{i}<\theta<2\pi\,.$

Take note that the nonnegative roots $\gamma_{ij}$ of these eigenvalues still
satisfy (14) when $0<\alpha_{i}<\pi$, whereas they are all greater than one,
if $\pi<\alpha_{i}<2\pi$. Using these properties of the solution of the
forward problem the proof of Theorem 6 carries over to the insulating case
with straightforward modifications. Since at least one of the interior angles
of ${\mathscr{D}}$ is smaller than $\pi$, the corresponding singular function
$w_{i}$ only belongs to $H^{\gamma}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ for some $\gamma\in(1/2,1)$
according to [12, Theorem 1.2.18], and hence the overall smoothness of the
solution $u_{h}^{\prime}$ of (53) is of similar type than in the case of a
polygonal inclusion with positive conductivity.

If the inhomogeneous Neumann boundary data on $\partial{\mathscr{D}}$ in (53)
are defined by some vector field $h\in{W_{c}^{1,\infty}(\mathchar
266\relax)}$, a little extra care is necessary to show that the corresponding
solution of (53) is the shape derivative of $u$ in direction $h$, because the
function $(u_{th}-u)/t$ is not defined in all of $\mathchar 266\relax$; in
fact, the two functions $u_{th}$ and $u$ live on different domains. To
overcome this problem one can consider an arbitrary domain $\mathchar
266\relax^{\prime}\subset\mathchar 266\relax\setminus\overline{\mathscr{D}}$
which satisfies $\partial\mathchar
266\relax^{\prime}\cap\partial{\mathscr{D}}=\emptyset$, in which case

$\frac{u_{th}-u}{t}\,:\,\mathchar 266\relax^{\prime}\to{\mathord{\mathbb{R}}}$

is well-defined for $t$ sufficiently close to zero. The analog of Theorems 5,
6, and 7 then reads as follows.

######  Theorem 8.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$ and
$h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$. Then the boundary value problem
(53) has a unique solution $u^{\prime}\in H^{\gamma}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ for some $\gamma\in(1/2,1)$. This
solution is the shape derivative of $u=u({\mathscr{D}})$ in the direction of
$h$, i.e.,

$\frac{u_{th}-u}{t}\,\to\,u_{h}^{\prime}\,=\,{\dot{u}}_{h}\,-\,h\cdot\nabla
u\,,\qquad t\to 0\,,$

in $L^{2}(\mathchar 266\relax^{\prime})$ for every domain $\mathchar
266\relax^{\prime}\subset\mathchar 266\relax$ with $\partial\mathchar
266\relax^{\prime}\cap\partial{\mathscr{D}}=\emptyset$. This shape derivative
also belongs to $H^{1}\bigl{(}\mathchar
266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal
B}_{r_{i}}(x_{i})}\bigr{)}$, and its trace on $\partial\mathchar 266\relax$ is
the shape derivative of $\mathchar 259\relax_{f}({\mathscr{D}})$ in direction
$h|_{\partial{\mathscr{D}}}$.

The other results of Section 4 apply verbatim to the insulating case; the
corresponding modifications of the proofs are straightforward.

###  7.2 The perfectly conducting case

When $k$ becomes infinitely large, on the other hand, the transmission
condition (3) forces $u^{-}$ to approximate a constant but unknown value; this
is in agreement with our intuition that in a perfect conductor potential
differences equilibrate immediately. Up to an additive constant the
corresponding potential $u\in H^{1}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ is therefore given by the solution
of the boundary value problem

$\Delta u\,=\,0\quad\text{in $\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\qquad u\,=\,0\quad\text{on
$\partial{\mathscr{D}}$}\,,\qquad\frac{\partial}{\partial\nu}u\,=\,f\quad\text{on
$\partial\mathchar 266\relax$}\,.$ (55)

Note that $u$ can be extended by zero to a function in $H^{1}(\mathchar
266\relax)$. Ito, Kunisch, and Li [18] pointed out that the solution of the
boundary value problem

$\begin{array}[]{c}\Delta u_{h}^{\prime}\,=\,0\quad\text{in $\mathchar
266\relax\setminus\overline{\mathscr{D}}$}\,,\\\\[8.61108pt]
u_{h}^{\prime}\,=\,-(h\cdot\nu)\dfrac{\partial}{\partial\nu}u\quad\text{on
$\partial{\mathscr{D}}$}\,,\qquad\dfrac{\partial}{\partial\nu}u_{h}^{\prime}\,=\,0\quad\text{on
$\partial\mathchar 266\relax$}\,.\end{array}$ (56)

is the shape derivative of $u=u({\mathscr{D}})$ of (55) in the direction of
$h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$, provided that ${\mathscr{D}}$
has a smooth boundary.

When ${\mathscr{D}}$ is a polygon the solution of the forward problem (55)
admits a similar expansion near the vertices of ${\mathscr{D}}$ as in the
insulating case; again, see [12] for details. The eigenvalues
$\gamma_{ij}^{2}$ which are relevant for this expansion are the same as in the
insulating case, cf. (54), this time the eigenfunctions being the
corresponding sine functions

$y_{ij}(\theta)\,=\,\Bigl{(}\frac{2}{2\pi-\alpha_{i}}\Bigr{)}^{1/2}\sin\gamma_{ij}(\theta-\alpha_{i})\,,\qquad\alpha_{i}<\theta<2\pi\,,$

where the index $j$ runs through the natural numbers only; all eigenvalues are
strictly positive.

The analysis in [6] also extends to the perfectly conducting case, showing
that the shape derivative $\partial\mathchar 259\relax_{f}({\mathscr{D}})$ of
a polygonal perfect conductor ${\mathscr{D}}$ exists, and that it satisfies

$\langle\,\partial\mathchar
259\relax_{f}({\mathscr{D}})h|_{\partial{\mathscr{D}}},g\,\rangle_{L^{2}(\partial\mathchar
266\relax)}\,=\,-\int_{\partial{\mathscr{D}}}(h\cdot\nu)\,\frac{\partial}{\partial\nu}u\,\frac{\partial}{\partial\nu}v_{g}\,\mathrm{d}s$

for every $g\in L^{2}_{\diamond}(\partial\mathchar 266\relax)$, where $v_{g}$
solves the boundary value problem (55) with $f$ replaced by $g$. Note that the
projection of $\partial\mathchar 259\relax_{f}({\mathscr{D}})h$ onto
$L^{2}_{\diamond}(\partial\mathchar 266\relax)$ is independent of the chosen
grounding in the definition (55) of $u$ and $v_{g}$, respectively.

Further, the analysis from Sections 4 to 6 can be modified in a
straightforward way to achieve the following result.

######  Theorem 9.

Let ${\mathscr{D}}$ be a polygon with simply connected closure
$\overline{\mathscr{D}}\subset\mathchar 266\relax$ and
$h\in{W_{c}^{1,\infty}(\mathchar 266\relax)}$. Then the shape derivative of
the solution $u=u({\mathscr{D}})$ of (55) in direction $h$ is given by the
unique solution $u_{h}^{\prime}$ of the boundary value problem (56). This
solution belongs to $H^{\gamma}(\mathchar
266\relax\setminus\overline{\mathscr{D}})$ for some $\gamma\in(1/2,1)$ and to
$H^{1}\bigl{(}\mathchar 266\relax\setminus\overline{{\mathscr{D}}\cup_{i}{\cal
B}_{r_{i}}(x_{i})}\bigr{)}$. Its trace on $\partial\mathchar 266\relax$ is the
shape derivative of $\mathchar 259\relax_{f}({\mathscr{D}})$ in direction
$h|_{\partial{\mathscr{D}}}$.

Like in the insulating case the other results of Section 4 are valid in the
perfectly conducting case, too.

## Acknowledgement

The author is grateful to Elena Beretta for pointing out references [6, 23] as
a possible starting point to determine the shape derivative $\partial\mathchar
259\relax_{f}$ for insulating or perfectly conducting polygonal inclusions.

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